C      ALGORITHM 810, COLLECTED ALGORITHMS FROM ACM.
C      THIS WORK PUBLISHED IN TRANSACTIONS ON MATHEMATICAL SOFTWARE,
C      VOL. 27,NO. 2,      June, 2001, P.  143--192.
#! /bin/sh
# This is a shell archive, meaning:
# 1. Remove everything above the #! /bin/sh line.
# 2. Save the resulting text in a file.
# 3. Execute the file with /bin/sh (not csh) to create the files:
#	Doc/
#	Doc/README.first
#	Doc/autoinput.txt
#	Doc/help.tex
#	Doc/intro.tex
#	Doc/readme.txt
#	Doc/xamples.tex
#	Fortran/
#	Fortran/Sp/
#	Fortran/Sp/Drivers/
#	Fortran/Sp/Drivers/data2
#	Fortran/Sp/Drivers/driver1.f
#	Fortran/Sp/Drivers/driver2.f
#	Fortran/Sp/Drivers/driver3.f
#	Fortran/Sp/Drivers/makepqw.f
#	Fortran/Sp/Drivers/res1
#	Fortran/Sp/Drivers/res2
#	Fortran/Sp/Drivers/res3
#	Fortran/Sp/Drivers/xamples.f
#	Fortran/Sp/Src/
#	Fortran/Sp/Src/src.f
# This archive created: Tue Nov 13 09:51:12 2001
export PATH; PATH=/bin:$PATH
if test ! -d 'Doc'
then
	mkdir 'Doc'
fi
cd 'Doc'
if test -f 'README.first'
then
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else
cat << "SHAR_EOF" > 'README.first'
In order to make this code compatible with the other CALGO codes
there has been a renaming of filenames. This hasn't gone through to
the documentation; the mappings are

coupdr -> driver1
drive  -> driver2
sepdr  -> driver3
sleign2.f -> src.f

data2 needs to be renamed auto.in before running driver2 in
automatic mode.
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		SLEIGN2: THE AUTOINPUT.TXT FILE

01 March 2001: P.B. BAILEY, W.N. EVERITT AND A. ZETTL

     The experienced user of SLEIGN2 who would like to avoid the
"question & answer" format in favor of a more direct (and faster) system
can do so by simply creating a very brief text file containing the
parameters of his particular problem. The file must be called
   
                   auto.in

and must reside in the same directory as the executable SLEIGN2 file
(possibly xamples.x or bloggs.x) .  Once such a file has been created,
the user simply types

        xamples.x  <ENTER>

as usual (or bloggs.x <ENTER>), whereupon the prompt 

   WOULD YOU LIKE AN OVERVIEW OF HELP ? (Y/N) (h?)

appears -- as usual.  But instead of replying to the question
with y, or n, or h, one simply types in the response

           a  <ENTER>

The "a" here stands for "automatic" operation of SLEIGN2;
at this point the computation of the requested eigenvalue(s) 
proceeds without further action by the user.
     The construction of the file "auto.in" consists of simply
defining a number of "KEYWORDS", each on a separate line, which
together constitute a complete set of input parameters defining
the eigenvalue problem to be solved. (The differential equation
coefficients have, presumably, been already defined in either
xamples.f, or bloggs.f, or.. .)   
          -------------------------------------------
     The KEYWORDS, all of which end in a colon and must be
followed by at least one space, are:

   a:  -- The left end-point of the interval (a,b); 
          Value is any real number;
          Default value is -infinity.
   b:  -- The right end-point of the interval (a,b);
          Value is any real number;
          Default value is +infinity.
   classa:  -- The end-point classification of a;
               Value is one of { r, wr, lcno, lco, lp, d }.
   classb:  -- The end-point classification of b:
               Value is one of { r, wr, lcno, lco, lp, d }.
   bca: -- boundary condition for the end-point a;
           Value is either d (for Dirichlet),
                           n (for Neuman),
                        or two real numbers A1, A2 . 
   bcb: -- boundary condition for the end-point b;
           Value is either d (for Dirichlet),
                           n (for Neuman),
                        or two real numbers B1, B2 . 
   bcc:  -- coupled boundary condition;
            Value is either p (for periodic),
                            s (for semi-periodic),
                        or five real numbers
                           alpha, k11, k12, k21, k22 .
   numeig:  -- Index (or range of indices) of desired eigenvalue;
               Value is an integer N1, or pair of integers N1,N2 .
   param:  -- Parameter(s) appropriate for the problem;
              Value is one or two real numbers, param1, param2.
   np:  --  Problem number;
            (Appropriate only if one of the differential equations
              in xamples.f is being used.);
            Value is an integer from 1 to 32.
   output:  -- Name of output file;
               Value is a character string, the name of the output
                 file where the results of the computation are to
                 be written;
               Default value is "auto.out" .
   end:  -- Last line of file "auto.in" ; no value set.
            
   again:  --  Terminates the input for one eigenvalue problem and
                 begins the input for another ; no value set.

Although the KEYWORDS used to define any one problem can be defined
in any order, there are a few rules to be observed:  Namely,

  Only those KEYWORDS whose values are necessary need be mentioned.
  Any KEYWORD definition remains in effect until redefined;
  To erase a previous definition of a KEYWORD, redefine it to 
     have the value "null".  
          -----------------------------------------------
     
     Perhaps a simple example would help to make clear what such
a file would look like.  One such could be:

output: bessel.rep
np: 2
param: 0.75
a: 0.0
b: 1.0
classa: lcno
classb: r
bca: 1.0,0.0
bcb: d
numeig: 2,5
end:

     This file (which must be called "auto.in", of course), would
be suitable for running Bessel's equation in xamples.f.   
Evidently the problem selected is #2 of xamples.f (Bessel's equation),
with the parameter nu = 0.75, on the interval (0.0,1.0) .  The
end-point a is asserted to be LCNO; end-point b is R; the boundary
condition at a is defined by A1 = 1.0 & A2 = 0.0; boundary condition
at b is Dirichlet; and the eigenvalues lambda(2), lambda(3), lambda(4),
lambda(5) are to be computed.
 
     If the user wishes to run several problems, one after another, 
things can become a little complicated, but can be done.  Here is
a more complicated example of a file auto.in :
 
output: Legendre 
np: 1
a: -1.0
classa: lcno
bca: 1.0,0.0
b: 1.0
classb: lcno
bcb: 1.0,0.0
numeig: 0,5
 
again:
np: 6
output: Sears-Titchmarsh
a: 1.0
classa: r
classb: lco
bcb: 1.0,0.0
numeig: -1,2 
bca: d
b: null
 
again:
numeig: 1,5
np: 12 
a: 0.0 
output: Mathieu.rep
classa: r
classb: r
b: pi
bcc: p 
param: 5.0
 
again:
np: 16
output: Jacobi.rep
a: -1.0 
b:  1.0
param: 0.5,-1.2
classa: lp
classb: lcno
bcb:  1.0,0.0
numeig: 0,2
 
again:
param: 1.2,-0.5
classa: wr
classb: lp
bca: d
bcb: null
 
end:
      ---------------------------------------------------
 
     The user must be  aware, however, that this "automatic" system
for running SLEIGN2 has no built in safety features.  
Whereas the "question & answer" format can and does catch many kinds
of simple errors on the part of the user, there are no safety
devices of any kind in place when SLEIGN2 is run in automatic mode.
If the program runs, and if an output file has been written, then
the user has at least a record of the parameters of the problem which
were used for the run.  However if the auto.in file is constructed with
any errors in it, almost anything can happen. 
   
Department of Mathematical Sciences, Northern Illinois University,
DeKalb, IL 60115-2888, USA.


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else
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%% This document created by Scientific Word (R) Version 3.5

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\newtheorem{acknowledgement}{Acknowledgement}
\newtheorem{algorithm}{Algorithm}[section]
\newtheorem{axiom}{Axiom}[section]
\newtheorem{case}{Case}[section]
\newtheorem{claim}{Claim}[section]
\newtheorem{conclusion}{Conclusion}[section]
\newtheorem{condition}{Condition}[section]
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\newtheorem{corollary}{Corollary}[section]
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\begin{document}
\title[HELP file]{Sleign2: the HELP file}
\author{P.B. Bailey}
\author{W.N. Everitt}
\author{A. Zettl}
\address{Department of Mathematical Sciences, Northern Illinois University, DeKalb, IL
60115-2888, USA}
\date{01 March 2001 (File: help.tex)}
\maketitle


This copy of the HELP data has been written in AMS-LaTeX in view of the
considerable amount of mathematical formulae involved in the text. The user
should note that when HELP is accessed in MAKEPQW and/or DRIVE the
corresponding data is given in Fortran notation. The best use of this HELP
data can be made by printing out a copy of this AMS-LaTeX file to have
available when the code files are in use.

HELP may be called at any point where the program halts and displays (h?), by
pressing ``h
%TCIMACRO{\TEXTsymbol{<}}%
%BeginExpansion
$<$%
%EndExpansion
ENTER%
%TCIMACRO{\TEXTsymbol{>}}%
%BeginExpansion
$>$%
%EndExpansion
''. To RETURN from HELP, press ``r
%TCIMACRO{\TEXTsymbol{<}}%
%BeginExpansion
$<$%
%EndExpansion
ENTER%
%TCIMACRO{\TEXTsymbol{>}}%
%BeginExpansion
$>$%
%EndExpansion
''. To QUIT at any program halt, press ``q
%TCIMACRO{\TEXTsymbol{<}}%
%BeginExpansion
$<$%
%EndExpansion
ENTER%
%TCIMACRO{\TEXTsymbol{>}}%
%BeginExpansion
$>$%
%EndExpansion
''.

This AMS-LaTeX file is supplied as a separate text file within the SLEIGN2
package; it can be accessed on-line in both the MAKEPQW (if used) and DRIVE files.

HELP contains information to aid the user in entering data: on the coefficient
functions $p,q,w;$ on the self-adjoint, separated and coupled, regular and
singular, boundary conditions; on the limit-circle boundary condition
functions $u,v$ at the endpoint $a$ and $U,V$ at the endpoint $b$ of the
interval $(a,b)$; on the endpoint classifications of the differential
equation; on DEFAULT entry; on eigenvalue indexes; on IFLAG information; and
on the general use of the program SLEIGN2.

\medskip

The 17 sections of HELP are:

H1: Overview of HELP. H2: File name entry. H3: The differential equation. H4:
endpoint classification. H5: DEFAULT entry. H6: Self-adjoint limit-circle
boundary conditions. H7: General self-adjoint boundary conditions. H8:
Recording the results. H9: Type and choice of interval. H10: Entry of
endpoints. H11: endpoint values of $p,q,w$. H12: Initial value problems. H13:
Indexing of eigenvalues. H14: Entry of eigenvalue index, initial guess, and
tolerance. H15: IFLAG information. H16: Plotting. H17: Indexing of eigenvalues.

HELP can be accessed at each point in MAKEPQW and DRIVE where the user is
asked for input, by pressing ``h
%TCIMACRO{\TEXTsymbol{<}}%
%BeginExpansion
$<$%
%EndExpansion
ENTER%
%TCIMACRO{\TEXTsymbol{>}}%
%BeginExpansion
$>$%
%EndExpansion
''; this places the user at the appropriate HELP section. Once in HELP, the
user can scroll the further HELP sections by repeatedly pressing ``h
%TCIMACRO{\TEXTsymbol{<}}%
%BeginExpansion
$<$%
%EndExpansion
ENTER%
%TCIMACRO{\TEXTsymbol{>}}%
%BeginExpansion
$>$%
%EndExpansion
'', or jump to a specific HELP section Hn (n =1,2,...17) by typing ``hn
%TCIMACRO{\TEXTsymbol{<}}%
%BeginExpansion
$<$%
%EndExpansion
ENTER%
%TCIMACRO{\TEXTsymbol{>}}%
%BeginExpansion
$>$%
%EndExpansion
''; to return to the place in the program from which HELP is called, press
``r
%TCIMACRO{\TEXTsymbol{<}}%
%BeginExpansion
$<$%
%EndExpansion
ENTER%
%TCIMACRO{\TEXTsymbol{>}}%
%BeginExpansion
$>$%
%EndExpansion
''.

\medskip

H2: File name entry.

MAKEPQW is used to create a Fortran file containing the coefficients $p,q,w$
defining the differential equation, and the boundary condition functions $u,v$
and $U,V$ if required. The file must be given a NEW filename which is
acceptable to your Fortran compiler. For example, it might be called bessel.f
or bessel.for depending upon your compiler.

The same naming considerations apply if the Fortran file is prepared other
than with the use of MAKEPQW.

\medskip

H3: The differential equation.

The prompt ``Input $p$ (or $q$ or $w$) ='' requests you to type in a Fortran
expression defining the function $p$, which is one of the three coefficient
functions defining the Sturm-Liouville differential equation%
\begin{equation}
-(py^{\prime})^{\prime}+qy=\lambda wy \tag{*}%
\end{equation}
to be considered on some interval $(a,b)$ of the real line. The actual
interval used in a particular problem can be chosen later, and may be either
the whole interval $(a,b)$ where the coefficient functions $p,q,w$ are
defined, or on any sub-interval $(a^{\prime},b^{\prime})$ of $(a,b)$;
$a=\infty$ and/or $b=+\infty$ are allowable choices for the endpoints.

The coefficient functions $p,q,w$ of the differential equation may be chosen
arbitrarily but must satisfy the following conditions:

(1) $p,q,w$ are real-valued throughout $(a,b)$.

(2) $p,q,w$ are piece-wise continuous and defined throughout the interior of
the interval $(a,b)$.

(3) $p$ and $w$ are strictly positive in $(a,b)$.

\noindent For better error analysis in the numerical procedures, condition (2)
above is often replaced with

(2$^{\prime}$) $p,q,w$ are four times continuously differentiable on $(a,b)$.

\noindent The behaviour of $p,q,w$ near the endpoints $a$ and $b$ is critical
to the classification of the differential equation (see H4 and H11).

\medskip

H4: endpoint classification.

The correct classification of the endpoints $a$ and $b$ is essential to the
working of the SLEIGN2 program. To classify the endpoints, it is convenient to
choose a point $c$ in $(a,b)$; \textit{i.e.} $a<c<b$. \quad\quad$\quad
\;\;$Subject to the general conditions on the coefficient functions $p,q,w$
(see H3):

(1) a is REGULAR (say R) if $-\infty<a,$ $p,q,w$ are piece-wise continuous on
$[a,c]$, and $p(a)>0,w(a)>0.$

(2) $a$ is WEAKLY REGULAR (say WR) if $-\infty<a,$ $a$ is not R and%
\[
\int_{a}^{c}\left\{  p^{-1}+\left|  q\right|  +w\right\}  <+\infty.
\]

If endpoint $a$ is neither R nor WR, then $a$ is SINGULAR; that is, either
$-\infty=a,$ or $-\infty<a$ and%
\[
\int_{a}^{c}\left\{  p^{-1}+\left|  q\right|  +w\right\}  =+\infty.
\]

(3) The SINGULAR endpoint $a$ is LIMIT-CIRCLE NON-OSCILLATORY (say LCNO) if
for some real value of the spectral parameter $\lambda$ ALL real-valued
solutions $y$ of the differential equation%
\[
-(py^{\prime})^{\prime}+qy=\lambda wy\;\text{on}\;(a,c]
\]

satisfy the conditions:%
\[
\int_{a}^{c}w\left|  y\right|  ^{2}<+\infty
\]

\noindent and $y$ has at most a finite number of zeros in $(a,c].$

(4) The SINGULAR endpoint a is LIMIT-CIRCLE OSCILLATORY (say LCO) if for some
real $\lambda$ ALL real-valued solutions of the differential equation (*)
satisfy the conditions:%
\[
\int_{a}^{c}w\left|  y\right|  ^{2}<+\infty
\]

\noindent and $y$ has an infinite number of zeros in $(a,c].$

(5) The SINGULAR endpoint a is LIMIT POINT (say LP) if for some real $\lambda$
at least one solution of the differential equation $(\ast)$ satisfies the
condition:%
\[
\int_{a}^{c}w\left|  y\right|  ^{2}=+\infty.
\]

There is a similar classification of the endpoint $b$ into one of the five
distinct cases R, WR, LCNO, LCO, LP.

Although the classification of singular endpoints invokes a real value of the
parameter $\lambda$, this classification is invariant in $\lambda$; all real
choices of $\lambda$ lead to the same classification.

In determining the classification of singular endpoints for the differential
equation $(\ast)$, it is often convenient to start with the choice $\lambda=0$
in attempting to find solutions (particularly when $q=0$ on $(a,b)$); however,
see example 7 below.

See H6 on the use of maximal domain functions to determine the classification
at singular endpoints.

EXAMPLES:

1. $-y^{\prime\prime}=\lambda y$ is R at both endpoints of $(a,b)$ when $a$
and $b$ are finite.

2. $-y^{\prime\prime}=\lambda y$ on $(-\infty,+\infty)$ is LP at both endpoints.

3. $-\left(  x^{1/2}y^{\prime}(x)\right)  ^{\prime}=\lambda x^{-1/2}y(x)$ for
all$\;x\in(0,+\infty)$ is WR at $0$ and LP at $+\infty$ (take $\lambda=0$ in
$(\ast)$). See the file xamples.f, \#10 (Weakly Regular).

4. $-((1-x^{2})y^{\prime}(x))^{\prime}=\lambda y(x)\;$for all$\;x\in(-1,+1)$
is LCNO at both ends (take $\lambda=0$ in $(\ast)$). See xamples.f, \#1 (Legendre).

5. $-y^{\prime\prime}(x)+Cx^{-2}y(x)=\lambda y(x)\;$for all$\;x\in(0,+\infty)$
is LP at $+\infty;$ however at $0$ the equation illustrates a number of
different classifications(take $\lambda=0$ in $(\ast)$) as follows:

LP $C\geq3/4$ ; LCNO for $-1/4\leq C<3/4$ (but $C\neq0$); LCO for $C<-1/4.$

6. $-(xy^{\prime}(x))^{\prime}-x^{-1}y(x)=\lambda y(x)\;$for all$\;x\in
(0,\infty)$ is LCO at $0$ and LP $+\infty$ (take $\lambda=0$ in $(\ast)$ with
solutions $\cos(\ln(x))$ and $\sin(\ln(x))$. See xamples.f, \#7 (BEZ).

7. $-(xy^{\prime}(x))^{\prime}-xy(x)=\lambda x^{-1}y(x)\;$for all$\;x\in
(0,+\infty)$ is LP at $0$ and LCO at $+\infty$ (take $\lambda=-1/4$ in
$(\ast)$ with solutions $x^{-1/2}\cos(x)$ and $x^{-1/2}\sin(x)$). See
xamples.f, \#6 (Sears-Titchmarsh).

8. $-y^{\prime\prime}(x)+x\sin(x)y(x)=\lambda y(x)$ on $(0,+\infty)$ is R at
$0$ and LP at $+\infty.$ See xamples.f \#30 (Littlewood-McLeod).

\medskip

H5: DEFAULT entry.

The complete range of problems for which SLEIGN2 is applicable can only be
reached by appropriate entries under endpoint classification and boundary
conditions. However, there is a DEFAULT application which requires no detailed
entry of endpoint classification or boundary conditions, subject to:

1) The DEFAULT application CANNOT be used at a LCO endpoint.

2) If an endpoint $a$ is R, then the Dirichlet boundary condition $y(a)=0$ is
automatically used.

3) If an endpoint $a$ is WR, then the following boundary condition is
automatically applied:

(i) if $p(a)=0$, and both $q$ and $w$ are bounded near $a$, then the Dirichlet
boundary condition $y(a)=0$ is used

(ii) if $p(a)>0$, and $q$ and/or $w$ are not bounded near $a$, then the
Neumann boundary condition $y^{\prime}(a)=0$ is used.

If $p(a)=0$, and $q$ and/or $w$ are not bounded near $a$, then no reliable
information, in general, can be given on the DEFAULT boundary condition.

4) If an endpoint is LCNO, then in most cases the principal or Friedrichs
boundary condition is applied (see H6).

5) If an endpoint is LP, then the normal LP procedure is applied (see H7(1.)).

If you choose the DEFAULT condition, then no entry is required for the $u,v$
and $U,V$ boundary condition functions.

\medskip

H6: Limit-circle (LC) boundary conditions.

At an endpoint $a$, the LC type separated boundary condition is of the form
(similar remarks throughout apply to the endpoint $b$ with $U,V$ being
boundary condition functions at $b$)%
\begin{equation}
A1[y,u](a)+A2[y,v](a)=0 \tag{**}%
\end{equation}

\noindent where $y$ is a solution of the differential equation%
\begin{equation}
-(py^{\prime})^{\prime}+qy=\lambda wy\;\text{on}\;(a,b) \tag{*}%
\end{equation}

Here $A1,A2$ are real numbers, not both zero; $u$ and $v$ are boundary
condition functions at $a$; and for real-valued $y$ and $u$ the form
$[y,u](\cdot)$ is defined by%
\[
\lbrack y,u](x)=y(x)(pu^{\prime})(x)-(py^{\prime})(x)u(x)\;\text{for
all}\;x\in(a,b).
\]

If neither endpoint is LP then there are also self-adjoint coupled boundary
conditions. These have a canonical form given by%
\[
Y(b)=\exp(i\alpha)\mathbf{K}Y(a)
\]

\noindent where

(i) $\mathbf{K}$ is a real $2\times2$ matrix with $\det(\mathbf{K})=1$

(ii) the parameter $\alpha$ is restricted to $\alpha\in(-\pi,\pi]$

(iii) $Y$ is the solution column vector $Y(a)=[y(a),(py^{\prime})(a)]^{T}$ at
a regular R endpoint $a$, and $Y$ is the ``singular solution vector''
$Y(a)=[[y,u](a),[y,v](a)]^{T}$ at a singular LC endpoint $a$. Similarly at the
right endpoint $b$ with $U,V.$

The object of this section is to provide help in choosing appropriate
functions $u$ and $v$ in $(\ast\ast)$ (or in choosing $U,V$) given the
differential equation $(\ast)$. Full details of the boundary conditions for
$(\ast)$ are discussed in H7; here it is sufficient to say that the
limit-circle type boundary condition $(\ast\ast)$ must be applied at any
endpoint in the LCNO, LCO classification, but can also be used in the R, WR
classification subject to the appropriate choice of $u$ and $v,$ and $U$ and $V.$

Let $(\ast)$ be R, WR, LCNO, or LCO at endpoint $a$ and choose $c$ in
$(a,b).$. Then \textit{either} $u$ and $v$ are a pair of linearly independent
real solutions of $(\ast)$ on $(a,c]$ for any chosen real value of $\lambda$,
\textit{or }$u$ and $v$ are a pair of real-valued maximal domain functions
defined on $(a,c]$ satisfying $[u,v](a)\neq0.$

The maximal domain $D(a,b)$ is defined by%
\[%
\begin{array}
[c]{lll}%
D(a,b)= & \{f:(a,b)\rightarrow\mathbb{C}: & (i)\;f\text{ and }pf^{\prime}\in
AC_{\text{loc}}(a,b)\\
&  & (ii)\;f\text{ and }w^{-1}(-(pf^{\prime})^{\prime}+qf)\in L^{2}%
((a,b);w)\}.
\end{array}
\]
The domains $D(a,c]$ and $D[c,b)$ are the restrictions of the functions
$D(a,b)$ to the sub-intervals. It is known that for all $f,g\in D(a,c]$ the
limit%
\[
\lbrack f,g](a)=\lim_{x\rightarrow a}[f,g](x)
\]

\noindent exists and is finite. If $(\ast)$ is LCNO or LCO at $a$, then all
solutions of (*) belong to $D(a,c]$ for all values of $\lambda.$ The boundary
condition $(\ast\ast)$ is essential in the LCNO and LCO cases but can also be
used with advantage in some R and WR cases.

In the R, WR, and LCNO cases, but not in the LCO case, the boundary condition
functions can always be chosen so that%
\[
\lim_{x\rightarrow a}\frac{u(x)}{v(x)}=0
\]
and it is recommended that this normalisation be effected, but this is not
essential; this normalization has been entered in the examples given below. In
this case, the boundary condition $[y,u](a)=0$ (\textit{i.e. }$A1=1,A2=0$ in
$(\ast\ast)$) is called the principal or Friedrichs boundary condition at $a$.

In the case when endpoints $a$ and $b$ are, independently, in the R, WR, LCNO,
or LCO classification, it may be that symmetry or other reasons permit one set
of boundary condition functions to be used at both end-points (see xamples.f,
\#1 (Legendre)). In other cases, different pairs must be chosen for each
endpoint (see xamples.f: \#16 (Jacobi), \#18 (Dunsch), and \#19 (Donsch)).

Note that a solution pair $u,v$ is always a maximal domain pair, but not
necessarily vice versa. EXAMPLES:

1. $-y^{\prime\prime}=\lambda y$ on $[0,\pi]$ is R at $0$ and R at $\pi.$ At
$0$, with $\lambda=0$, a solution pair is
\[
u(x)=x,v(x)=1\;\text{for all}\;x\in\lbrack0,\pi].
\]
At $\pi,$with $\lambda=1$, a solution pair is%
\[
U(x)=\sin(x),V(x)=\cos(x)\;\text{for all}\;x\in\lbrack0,\pi].
\]

2. $-(x^{1/2}y^{\prime}(x))^{\prime}=\lambda x^{-1/2}y(x)$ on $(0,1]$ is WR at
$0$ and R at $1.$ (The general solutions of this equation are $u(x)=\cos
(2x^{1/2}\sqrt{\lambda})$ and $v(x)=\sin(2x^{1/2}\sqrt{\lambda})$.)

At $0$, $\lambda=0$, a solution pair is%
\[
u(x)=2x^{1/2},v(x)=1.
\]
At $1$, with $\lambda=\pi^{2}/4$, a solution pair is%
\[
U(x)=\sin(\pi x^{1/2}),V(x)=\cos(\pi x^{1/2}).
\]
Also at 1, with $\lambda=0$, a solution pair is%
\[
U(x)=2(1-x^{1/2}),V(x)=1.
\]

See also xamples.f, \#10 (Weakly Regular).

3. $-((1-x^{2})y^{\prime}(x))^{\prime}=\lambda y(x)$ on $(-1,+1)$ is LCNO at
both endpoints.

At both $\pm1$, $\lambda=0$, a solution pair is%
\[
u(x)=1,v(x)=\ln((1+x)/(1-x))/2
\]

At $+1$, a maximal domain pair is $U(x)=1,V(x)=\ln(1-x)$. At $-1$, a maximal
domain pair is $u(x)=1,v(x)=\ln(1+x).$

See also xamples.f, \#1 (Legendre).

4. $-y^{\prime\prime}(x)-(4x^{2})^{-1}y(x)=\lambda y(x)$ on $(0,+\infty)$ is
LCNO at $0$ and LP at $+\infty$.

At $0$, a maximal domain pair is%
\[
u(x)=x^{1/2},v(x)=x^{1/2}\ln(x).
\]

See also xamples.f, \#2 (Bessel).

5. $-y^{\prime\prime}(x)-5(4x^{2})^{-1}y(x)=\lambda y(x)$ on $(0,+\infty)$ LCO
at $0$ and LP at $+\infty.$

At $0$, $\lambda=0$, a solution pair is%
\[
u(x)=x^{1/2}\cos(\ln(x)),v(x)=\sin(\ln(x)).
\]

See also xamples.f, \#20 (Krall).

6. $-y^{\prime\prime}(x)=\lambda y(x)$ on $(0,+\infty)$ is LCNO at 0 and LP at +infinity.

At $0$, a maximal domain pair is%
\[
u(x)=x,v(x)=1-x\ln(x).
\]

See also xamples.f, \#4(Boyd).

7. $-(x^{-1}y^{\prime}(x))^{\prime}+(kx^{-2}+k^{2}x^{-1})y(x)=\lambda y(x)$ on
$(0,1]$ with $k$ real and $k\neq0$, is LCNO at 0 and R at 1.

At $0$, a maximal domain pair is%
\[
u(x)=x^{2},v(x)=x-k^{-1}%
\]

See also xamples.f, \#8 (Laplace Tidal Wave).

\medskip

H7: General self-adjoint boundary conditions.

Boundary conditions for Sturm-Liouville boundary value problems%
\begin{equation}
-(py^{\prime})^{\prime}+qy=\lambda wy\;\text{on}\;(a,b) \tag{*}%
\end{equation}

\noindent are \textit{either }SEPARATED, with at most one condition at
endpoint $a$ and at most one condition at endpoint $b$,

\noindent\textit{or }COUPLED, when both $a$ and $b$ are, independently, in one
of the end-point classifications R, WR, LCNO, LCO, in which case two
independent boundary conditions are required which link the solution values
near $a$ to those near $b$. The SLEIGN2 program allows for all self-adjoint
boundary conditions; separated self-adjoint conditions and all cases of
coupled self-adjoint conditions.

Separated Conditions: the boundary conditions to be selected depend upon the
classification of the differential equation at the endpoint, say, $a$:

1. If the endpoint a is LP, then no boundary condition is required or allowed.

2. If the endpoint $a$ is R or WR, then a separated self-adjoint boundary
condition is of the form%
\[
A1y(a)+A2(py^{\prime})(a)=0
\]

\noindent where $A1,A2$ are real constants the user must choose, not both zero.

3. If the endpoint a is LCNO or LCO, then a separated boundary condition is of
the form%
\[
A1[y,u](a)+A2[y,v](a)=0
\]

\noindent where $A1,A2$ are real constants the user must choose, not both
zero; here $u,v$ are the pair of boundary condition functions the user has
previously selected when the input Fortran file was being prepared with makepqw.f.

4. If the endpoint a is LCNO and the boundary condition pair $u,v$ has been
chosen so that%
\[
\lim_{x\rightarrow a}\frac{u(x)}{v(x)}=0
\]

\noindent(which is always possible), then $A1=1,A2=0$ (\textit{i.e.
}$[y,u](a)=0$) gives the principal (Friedrichs) boundary condition at $a.$

5. If $a$ is R or WR and boundary condition functions $u,v$ have been entered
in the Fortran input file, then 3 and 4 above apply to entering separated
boundary conditions at such an endpoint; the boundary conditions in this form
are equivalent to the point-wise conditions in 2 (subject to care in choosing
$A1,A2$). This singular form of a regular boundary condition may be
particularly effective in the WR case if the boundary condition form in 2
leads to numerical difficulties. Conditions 2,3,4,5 apply similarly at
endpoint $b$ (with $U,V$ as the boundary condition functions at $b$).

6. If $a$ is R, WR, LCNO, or LCO and $b$ is LP, then only a separated
condition at $a$ is required and allowed (or instead at $b$ if $a$ and $b$ are interchanged).

7. If both endpoints $a$ and $b$ are LP, then no boundary conditions are
required or allowed.

The indexing of eigenvalues for boundary value problems with separated
conditions is discussed in H13.

Coupled Conditions:

8. Coupled regular self-adjoint boundary conditions on $(a,b)$ apply only when
both endpoints $a$ and $b$ are R or WR.

\medskip

H8: Recording the results.

If you choose to have a record kept of the results, then the following
information is stored in a file with the name you select:

1. The file name.

2. The header line prompted for (up to 32 characters of your choice).

3. The interval $(a,b)$ selected by the user.

For SEPARATED boundary conditions:

4. The endpoint classification.

5. A summary of coefficient information at WR, LCNO, LCO endpoints.

6. The boundary condition constants $(A1,A2),(B1,B2)$ if entered.

7. (NUMEIG,EIG,TOL) or (NUMEIG1,NUMEIG2,TOL), as entered.

For COUPLED boundary conditions:

8. The boundary condition parameter $\alpha$ and the coupling matrix
$\mathbf{K}$, see H6.

For ALL self-adjoint boundary conditions:

9. The computed eigenvalue, EIG, and its estimated accuracy, TOL.

10. IFLAG reported (see H15).

\medskip

H9: Type and choice of interval.

You may enter any interval $(a,b)$ for which the coefficients $p,q,w$ are well
defined by your Fortran statements in the input file, provided that $(a,b)$
contains no interior singularities.

\medskip

H10: Entry of endpoints.

Endpoints a and b should generally be entered as real numbers to an
appropriate number of decimal places.

\medskip

H11: endpoint values of $p,q,w.$

The program SLEIGN2 needs to know whether the coefficient functions $p,q,w$ as
defined by the Fortran expressions entered in the input file, can be evaluated
numerically without running into difficulty. If, for example, either $q$ or
$w$ is unbounded at $a$, or $p(a)=0$, then SLEIGN2 needs to know this
information so that $a$ is not chosen for functional evaluation.

\medskip

H12: Initial value problems.

The initial value problem facility for Sturm-Liouville problems%
\begin{equation}
-(py^{\prime})^{\prime}+qy=\lambda wy\;\text{on}\;(a,b) \tag{*}%
\end{equation}

\noindent allows for the computation of a solution of $(\ast)$ with a
user-chosen value $\lambda$ and any one of the following initial conditions:

1. From endpoint $a$ of any classification except LP towards endpoint $b$ of
any classification

2. From endpoint $b$ of any classification except LP back towards endpoint $a$
of any classification

3. From endpoints $a$ and $b$ of any classifications except LP towards an
interior point of $(a,b)$ selected by the program.

Initial values at $a$ are of the form $y(a)=\alpha1,(py^{\prime})(a)=\alpha2$
when $a$ is R or WR; and $[y,u](a)=\alpha1,[y,v](a)=\alpha2$ when $a$ is LCNO
or LCO.

Initial values at $b$ are of the form $y(b)=\beta1,(py^{\prime})(b)=\beta2$
when $b$ is R or WR; and $[y,u](b)=\beta1,[y,v](b)=\beta2$ when $b$ is LCNO or LCO.

In $(\ast)$, $\lambda$ is a user-chosen real number; while in the above
initial values, $(\alpha1,\alpha2)$ and $(\beta1,\beta2)$ are user-chosen
pairs of real numbers not both zero.

In the initial value case 3 above when the interval $(a,b)$ is finite, the
interior point selected by the program is generally near the midpoint of
$(a,b)$; when $(a,b)$ is infinite, no general rule can be given. Also if,
given $(\alpha1,\alpha2)$ and $(\beta1,\beta2)$, the $\lambda$ chosen is an
eigenvalue of the associated boundary value problem, the computed solution may
not be the corresponding eigenfunction -- the signs of the computed solutions
on either side of the interior point may be opposite.

The output for a solution of an initial value problem is in the form of stored
numerical data which can be plotted on the screen (see H16), or printed out in
graphical form if graphics software is available.

\medskip

H13: Indexing of eigenvalues.

The indexing of eigenvalues is an automatic facility in SLEIGN2. The following
general results hold for the separated boundary condition problem (see H7):

1. If neither endpoint $a$ or $b$ is LP or LCO, then the spectrum of the
eigenvalue problem is discrete (eigenvalues only), simple (eigenvalues all of
multiplicity 1), and bounded below with a single cluster point at $+\infty.$
The eigenvalues are indexed as $\{\lambda_{n}:n=0,1,2,...\}$, where
$\lambda_{n}<\lambda_{n+1}$ $(n=0,1,2,..)$, $\lim_{n\rightarrow+\infty}%
\lambda_{n}=+\infty$; and if $\{\psi_{n}:n=0,1,2,...\}$ are the corresponding
eigenfunctions, then $\psi_{n}$ has exactly $n$ zeros in the open interval $(a,b).$

2. If neither endpoint $a$ or $b$ is LP but at least one endpoint is LCO, then
the spectrum is discrete and simple as for 1, but with cluster points at both
$\pm\infty.$ The eigenvalues are indexed as $\{\lambda_{n}:n=0,\pm
1,\pm2,...\}$, where $\lambda_{n}<\lambda_{n+1}$ $($for $n<n+1$ and
$n=0,\pm1,\pm2,..),$ with $\lambda_{0}$ the smallest non-negative eigenvalue;
$\lim_{n\rightarrow-\infty}\lambda_{n}=-\infty$ and $\lim_{n\rightarrow
+\infty}\lambda_{n}=+\infty$; and if $\{\psi_{n}:n=0,\pm1,\pm2,...\}$ are the
corresponding eigenfunctions, then every $\psi_{n}$ has infinitely many zeros
in $(a,b).$

3. If one or both endpoints is LP, then there can be one or more intervals of
continuous spectrum for the boundary value problem in addition to some
(necessarily simple) eigenvalues. For these essentially more difficult
problems, SLEIGN2 can be used as an investigative tool to give qualitative and
possibly quantitative information on the spectrum.

For example, if a problem has continuous spectrum starting at a real number
$L$, then there may be no eigenvalues below $L$, any finite number of
eigenvalues below $L$, or an infinite (but countable) number of eigenvalues
below $L.$ SLEIGN2 can be used to compute $L$ (see the paper BEWZ on the
SLEIGN2 home page for an algorithm to compute $L$), and to determine the
number of these eigenvalues and compute them. In this respect, see xamples.f:
\#13 (Hydrogen Atom), \#17 (Morse Oscillator), \#21 (Fourier), and \#27
(J\"{o}rgens) as examples of success; and \#2 (Mathieu), \#14 (Marletta), and
\#28 (Behnke-Goerisch) as examples of failure.

The problem need not have a continuous spectrum, in which case if its discrete
spectrum is bounded below, then the eigenvalues are indexed and the
eigenfunctions have zero counts as in 1. If, on the other hand, the discrete
spectrum is unbounded below, then all the eigenfunctions have infinitely many
zeros in the open interval $(a,b)$. SLEIGN2 can, in principle, compute these
eigenvalues if neither endpoint is LP although this is a computationally
difficult problem. Note however, that SLEIGN2 has no algorithm when the
spectrum is discrete, unbounded above and below and one endpoint is LP, as in
xamples.f \#30.

In respect to the five classes of endpoints, the following identified examples
from xamples.f illustrate the spectral property of these boundary value problems:

1. Neither endpoint is LP or LCO:

\#1 (Legendre), \#2 (Bessel) with $-1/4<C<3/4,$ \#4 (Boyd), \#5 (Latzko).

2. Neither endpoint is LP, but at least one is LCO:

\#6 (Sears-Titchmarsh), \#7 (BEZ), \#19 (Donsch)

3. At least one endpoint is LP:

\#13 (Hydrogen Atom), \#14 (Marletta), \#20 (Krall), \#21 (Fourier) on [0,infinity).

\medskip

H14: Entry of eigenvalue index, initial guess, and tolerance.

For all self-adjoint boundary condition problems (see H7), SLEIGN2 calls for
input information options to compute \textit{either}

1. a single eigenvalue, \textit{or}

2. a series of eigenvalues.

\noindent In each case indexing of eigenvalues is called for (see H13).

1 above asks for data triples NUMEIG, EIG, TOL separated by commas. Here
NUMEIG is the integer index of the desired eigenvalue; NUMEIG can be negative
only when the problem is LCO at one or both endpoints. EIG allows for the
entry of an initial guess for the requested eigenvalue (if an especially good
one is available), or can be set to 0 in which case an initial guess is
generated by SLEIGN2 itself. TOL is the desired accuracy of the computed
eigenvalue. It is an absolute accuracy if the magnitude of the eigenvalue is 1
or less, and is a relative accuracy otherwise. Typical values for TOL might be
0.001 for moderate accuracy and 0.0000001 for high accuracy in single
precision. If TOL is set to 0, the maximum achievable accuracy is requested.

If the input data list is truncated with a ``\thinspace/\thinspace'' after
NUMEIG or EIG, then the remaining elements default to 0.

2 above asks for data triples NUMEIG1, NUMEIG2, TOL separated by commas. Here
NUMEIG1 and NUMEIG2 are the first and last integer indices of the sequence of
desired eigenvalues, with NUMEIG1
%TCIMACRO{\TEXTsymbol{<}}%
%BeginExpansion
$<$%
%EndExpansion
NUMEIG2; they can be negative only when the problem is LCO at one or both
endpoints. TOL is the desired accuracy of the computed eigenvalues. It is an
absolute accuracy if the magnitude of an eigenvalue is 1 or less, and is a
relative accuracy otherwise. Typical values for TOL might be 0.001 for
moderate accuracy and 0.0000001 for high accuracy in single precision. If TOL
is set to 0, the maximum achievable accuracy is requested.

If the input data list is truncated with a ``\thinspace/\thinspace'' after
NUMEIG2, then TOL defaults to 0.

For COUPLED self-adjoint boundary condition problems (see H7 and H17), SLEIGN2
also reports which eigenvalues are double. Double eigenvalues can occur only
for coupled boundary conditions with the parameter $\alpha=0$ or $\pi.$

\medskip

H15: IFLAG information.

All results are reported by SLEIGN2 with a flag identification. There are four
values of IFLAG:

1. The computed eigenvalue has an estimated accuracy within the

tolerance requested.

2. The computed eigenvalue does not have an estimated accuracy within the
tolerance requested, but is the best the program could obtain.

3. There seems to be no eigenvalue of index equal to NUMEIG.

4. The program has been unable to compute the requested eigenvalue.

\medskip

H16: Plotting.

After computing a single eigenvalue (see H14(1.)), but not a sequence of
eigenvalues (see H14(2.)), the eigenfunction can be plotted for separated
conditions and for coupled ones with $\alpha=0,\pi.$ If this data is desired,
respond $y$ when asked and SLEIGN2 will then compute eigenfunction data and
store them for subsequent use.

The user can ask that the eigenfunction data be in the form of either points
$(x,y)$ for $x\in(a,b)$, or points $(t,y)$ for $t$ in the standardized
interval $(-1,+1)$ mapped onto from $(a,b)$; the $t$-choice can be especially
helpful when the original interval is infinite. Additionally, the user can ask
for a plot of the so-called Pr\"{u}fer angle, in the $x$- or $t$-variables.

In both forms, once the choice has been made of the function to be plotted, a
crude plot is displayed on the monitor screen and the user is asked whether or
not to save the computed plot points in a file.

\medskip

H17: Indexing of eigenvalues for coupled self-adjoint problems.

The indexing of eigenvalues is an automatic facility in SLEIGN2. The following
general result holds for coupled boundary condition problems (see H7):

The spectrum of the eigenvalue problem is discrete (eigenvalues only). In
general the spectrum is not simple, but no eigenvalue exceeds multiplicity
$2$. The eigenvalues are indexed as $\{\lambda_{n}:n=0,1,2,...\}$ where
$\lambda_{n}<\lambda_{n+1}$ for $n=0,1,2,...,$and $\lim_{n\rightarrow+\infty
}\lambda_{n}=+\infty$ if neither endpoint is LCO. If one or both endpoints are
LCO the eigenvalues cluster at both $-\infty$ and at $+\infty,$ and all
eigenfunctions have infinitely many zeros.

If neither endpoint is LCO and $\alpha=0,\pi$, then the $n$-th eigenfunction
has $n-1,n,n+1$ zeros in the half open interval $[a,b)$ (also in $(a,b]$). All
three possibilities occur. Recall that in the case of double eigenvalues,
although the $n$-th eigenvalue is well defined there is an ambiguity about
which solution is declared as the $n$-th eigenfunction. If $\alpha\neq0,\pi$,
then the eigenfunction is non-real and has no zero in $(a,b)$, but each of the
real and imaginary parts of the $n$-th eigenfunction have the same zero
properties as mentioned above when $\alpha=0,\pi.$

The following identified examples from xamples.f are of special interest:

\#11 (Plum) on $[0,\pi],$ \#21 (Fourier) on $[0,\pi],$ \#25 (Meissner) on $[-1/2,+1/2].$
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\begin{document}
\title[Introduction to SLEIGN2]{Introduction to SLEIGN2}
\author{P.B. Bailey}
\author{W.N. Everitt}
\author{A. Zettl}
\address{Department of Mathematical Sciences, Northern Illinois University, DeKalb, IL
60115-2888, USA}
\date{01 March 2001 (File: intro.tex).}
\maketitle


The main purpose of this code is to compute eigenvalues and eigenfunctions of
regular and singular self-adjoint Sturm-Liouville problems (SLP) and to
approximate the continuous spectrum in the singular case. For a general
description of the analytical and numerical properties of the SLEIGN2 code see
\cite{BEZ} and \cite{BEZ3}.

These problems consist of a second order linear differential equation
\[
-(py^{\prime})^{\prime}+qy=\lambda wy\;\text{on}\;(a,b)
\]
together with boundary conditions (BC). The nature of the BC depends on the
regular or singular classification of the end points a and b. For both cases
the BC fall into two major classes: separated and coupled. The former are two
separate conditions, one at each endpoint; the latter are two coupled
conditions linking the values of the solution near the two endpoints, e.g.
periodic and semi-periodic boundary conditions. SLEIGN2 seems to be the only
general purpose code available for arbitrary self-adjoint BC, separated or
coupled, and for both regular and singular problems.

A number $\lambda$ for which there is a nontrivial solution satisfying the BC
is called an eigenvalue and such a solution is a (corresponding)
eigenfunction. If one or both endpoints are LP (see below or section 4 of HELP
for a definition) there may be points $\lambda$ in the spectrum in addition to
eigenvalues i.e. there may be continuous spectrum.

In the theory of SLP the coefficients $1/p$ and $q$ and the weight function
$w$ are assumed to be real-valued and locally Lebesgue integrable on $(a,b).$

To meet the needs of numerical computing techniques we make the stronger assumptions:

(i) The interval $(a,b)$ of R may be bounded or unbounded

(ii) $p,q$ and $w$ are real-valued functions on $(a,b)$

(iii) $p,q$ and $w$ are piecewise continuous on $(a,b)$

(iv) $p$ and $w$ are strictly positive on $(a,b)$.

For better error analysis in the numerical procedures, condition (iii) above
is replaced with

(iii)$^{\prime}$ $p,q$ and $w$ are four times continuously differentiable on
$(a,b)$.

To study a SLP using operator theory one associates a self-adjoint operator in
the weighted Hilbert space of square-integrable functions, with respect to the
weight $w$, on $(a,b)$, with each SLP in such a way that the spectrum of the
problem is the spectrum of the operator. In the case of a regular problem the
spectrum consists entirely of eigenvalues and these are bounded below (when
$p>0$ and $w>0$). This is still so for the case when each endpoint is either
regular (R) or singular limit-circle nonoscillatory (LCNO). In case one
endpoint is limit-circle oscillatory (LCO) and the other is not limit-point
(LP) then there are still only eigenvalues in the spectrum but these are not
bounded below. (The spectrum is never bounded above.)

If one or both endpoints is LP the spectrum may be extremely complicated.
There may be no eigenvalues, finitely many, or infinitely many. Some may be
embedded in the continuous spectrum. For $p=1,w=1,q(x)=\sin(x)$ on
$(-\infty,+\infty)$ there are no eigenvalues and the continuous spectrum
consists of the union of an infinite number of disjoint compact intervals.
(SLEIGN2 can be used to approximate this spectrum - see example 12 in the file
xamples.tex and the references quoted there.)

See the HELP file help.tex for a definition of the terms regular (R),
limit-circle (LC), limit-circle non-oscillatory (LCNO), limit-circle
oscillatory (LCO), limit-point (LP).

SLP problems are classified into various classes based on the classification
of the endpoints and on whether the boundary conditions are separated (S) or
coupled (C). We have the following categories:

1. R/R, S

2. R/R, C

3. R/LCNO or LCNO/R, S

4. R/LCNO or LCNO/R, C

5. R/LCO or LCO/R , S

6. R/LCO or LCO/R , C

7. LCNO/LCO or LCO/LCNO or LCO/LCO, S

8. LCNO/LCO or LCO/LCNO or LCO/LCO, C

9. LP/R or LP/LCNO or LP/LCO or R/LP or LCNO/LP or LCO/LP

10. LP/LP

For 9 there is only a separated condition at the non-LP endpoint and for 10
there are no boundary conditions at either end.

There are only two other major general purpose codes for computing eigenvalues
and eigenfunctions of Sturm-Liouville problems : SLEDGE (Fulton and Pruess)
and the earlier code SLEIGN (Bailey \textit{et al).} (There was another code,
written by Pryce and Marletta, available from the NAG library but the current
NAG library no longer includes such a code.)

SLEDGE uses a method based on piecewise constant approximations of the
coefficients of the differential equation; SLEIGN and SLEIGN2 both use the
Pr\"{u}fer transformation.

Both SLEIGN and SLEDGE are designed for separated regular boundary conditions
and both have a mechanism to automatically handle endpoints which are either
regular or singular but non-oscillatory. In the latter case if an endpoint is
LCNO the Friedrichs condition is usually, but not always, the one chosen by
the code. SLEDGE can also determine the LP/LC classification but only in a
restricted number of cases (the method requires that the coefficients $p,q$
and $w$ are analytic on $(a,b)$ and depends on the Frobenius method for series
solutions at regular singular endpoints); SLEIGN and SLEIGN2 do not have such
a facility but, for the sake of generality, rely on the user input of this information.

SLEIGN2 is the only general purpose code in existence which can, in principle,
handle arbitrary self-adjoint, separated or coupled, regular or singular,
boundary conditions, see the HELP file help.tex and \cite{BEZ3}. Problems with
coupled boundary conditions, see\cite{BEZ2}, at singular endpoints are
difficult to handle numerically, especially if one or both of the endpoints is LCO.

In addition to the above mentioned capabilities to compute eigenvalues and
eigenfunctions SLEIGN2 also computes the solution of an initial value problem
with the users choice of $\lambda$ and either a regular or singular initial condition.

When combined with the algorithm established in \cite{BEWZ}, SLEIGN2 can be
used to approximate the continuous spectrum.

An important feature of the SLEIGN2 program is its user friendly interface
contained in makepqw.f and drive.f. Users who want to bypass this interface
and use their own driver may wish to look at a sample driver; two such drivers
are provided, see (7) and (8) below.

The whole package consists of the following files:

1. A brief ``readme'' file readme.txt with basic information on how to run the code.

2. This introduction file intro.tex.

3. makepqw.f - This is an interactive Fortran file to input the coefficient
functions $p,q,w$ and, if necessary, the functions $u,v$ which define the
singular boundary conditions

4. drive.f - This is an interactive Fortran file containing the driver, a HELP
file help.tex, and a ``user friendly'' interface.

5. sleign2.f - The main code for the computation of eigenvalues and eigenfunctions.

6. xamples.f - A Fortran file with 32 examples ready to run. These examples
were chosen to illustrate various features of the code.

7. sepdr.f - A sample driver for separated regular or singular boundary conditions.

8. coupdr.f - A sample driver for coupled regular or singular boundary conditions.

9. xamples.tex - A LaTeX file containing information about the 32 examples.

10. help.tex - A LaTeX file with information about endpoint classifications,
boundary conditions etc. It is a separate text file and it can also be
accessed from both makepqw.f and drive.f.

11. autoinput.txt- A file describing an ``automatic'' method for using SLEIGN2
which avoids the user friendly ``question and answer'' format; this is
recommended for experienced users only.

When an eigenfunction has been computed it is stored. If it is real-valued, it
can be examined:

(i) by printing out the numerical data

(ii) by using the discrete graph plotter in the program

(iii) by using a local graph plotter.

Full details are given in HELP; see the file help.tex.

All six of the Fortran files in the SLEIGN2 package are in single precision.

\begin{thebibliography}{9}                                                                                                %

\bibitem {BEZ}P.B. Bailey, W.N. Everitt and A. Zettl, \emph{Computing
eigenvalues of singular Sturm-Liouville problems}, Results in Mathematics,
\textbf{20 }(1991), 391-423.

\bibitem {BEZ2}P.B. Bailey, W.N. Everitt and A. Zettl, \emph{Regular and
singular Sturm-Liouville problems with coupled boundary conditions}, Proc.
Royal Soc. Edinburgh (A) \textbf{126} (1996), 505-514.

\bibitem {BEZ3}P.B. Bailey, W.N. Everitt and A. Zettl, \emph{The SLEIGN2
Sturm-Liouville code.} (To appear in ACM Trans. Math. Software).

\bibitem {BEWZ}P.B. Bailey, W.N. Everitt, J. Weidmann and A. Zettl,
\emph{Regular approximation of singular Sturm-Liouville problems}. Results in
Mathematics \textbf{23} (1993), 3-22.
\end{thebibliography}
\end{document}
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                   SLEIGN2: THE README.TXT FILE
  
   01 March 2001; P.B. BAILEY, W.N. EVERITT AND A. ZETTL
  
  PLEASE read carefully this readme.txt file and the intro.tex file before 
 using the SLEIGN2 package for the first time.
  
 	There are eleven files in the SLEIGN2 package as follows:
 
 	Two ASCII files:
 	autoinput.txt  readme.txt
 
 	Six FORTRAN files:
 	coupdr.f  drive.f  makepqw.f  sepdr.f  sleign2.f  xamples.f
 
 	Three AMS-LaTeX files (these files can be compiled in the UNIX latex
 compiler, and then printed out in hard copy):
 	help.tex  intro.tex  xamples.tex   
  
   To run one of the examples in the xamples.f file in an UNIX environment 
 with a Fortran77 (or Fortran77) compiler, enter the following command:
  
 f77 xamples.f  drive.f  sleign2.f  -o  xamples.x
 (Replace f77 by f90 if you want to use the Fortran90 compiler.)
  
   This will create the executable file xamples.x and object files drive.o
 and sleign2.o. Now run xamples.x whenever you want to work an example from 
 the list of examples in the xamples.f file. The hard printed copy of the
 file xamples.tex provides detailed information on each one of the 32
 chosen examples in the file xamples.x.
  
   To run your own problem proceed as follows:
  Step 1: Enter the command 
        f77  makepqw.f  -o  makepqw.x
  Step 2: Run
        makepqw.x  
   (This interactive program will ask you for a file name - this must end 
 in .f - for example, if your chosen problem has the name bloggs then enter 
 bloggs.f). This file, after makepqw.x has been run, will contain the 
 subroutines for p,q,w and, if necessary, the functions u, v and U, V which 
 are used to define singular limit-circle boundary conditions at one or both 
 endpoints.
  Step 3: Enter the command
        f77  bloggs.f  drive.f  sleign2.f  -o  bloggs.x
   (drive.f and sleign2.f can be replaced with drive.o and sleign2.o 
 if these .o files are available to speed up the compilation; these -o 
 files are created by the first compilation.)
  Step 4: Run
        bloggs.x  
   (The user is asked to provide information for the code to run
 bloggs.x, for example: boundary conditions, eigenvalue indexes, 
 numerical tolerances, name for report file if desired. See the autoinput.txt 
 file for instructions on how to automate the input of the required 
 information).
  
  The file sepdr.f is a sample driver for separated regular and singular
 boundary conditions; coupdr.f is a sample driver program for coupled regular
 and singular boundary conditions. Experienced users who want to bypass the 
 extensive user friendly interface provided in drive.f and makepqw.f, and
 use their own driver may wish to look at these two sample drivers.

  Note that the above procedures may have to be modified for non UNIX 
 environments, e.g. DOS or APPLE.
 
  Note also that in running xamples.x or bloggs.x the user has access to
 the interactive help device; at any point where the program halts  
 type h <ENTER> for access; to return to the point in the program where
 help was accessed, type r <ENTER>. Additional information on the help 
 device can be found in the intro.tex file. The whole of the help data
 can be printed out separately from the file help.tex and the user is
 advised to have a copy available for consultation when working with the
 code files for the first time.
 
  See the autoinput.txt file for instructions on how to bypass these 
 program halts in xamples.x and bloggs.x
  
  All six of the FORTRAN .f files are supplied in single precision; 
 to convert these files to double precision replace the string `  REAL' by 
 the string `  DOUBLE PRECISION' throughout. (Note the two spaces in front of 
 REAL and in front of DOUBLE PRECISION - these spaces are important.)  
 In UNIX this can be effected within the vi editor as follows:
  
 :1,$ s/  REAL/  DOUBLE PRECISION
  
  It is recommended that the user try the program in single precision and 
 switch to double precision as required.
  
  Additional information on the SLEIGN2 package can be found in the intro.tex
 and help.tex files. See also a hard printed copy of the file bailey.tex,
 available in the Zettl/Sleign2 directory detailed below, which contains
 an account of the analytical and numerical properties of the SLEIGN2
 code, together with an extensive list of references.
  
  All of the eleven files in the SLEIGN2 package can be obtained by anonymous
 ftp from the computer
        ftp.math.niu.edu
 and the directory
        /pub/papers/Zettl/Sleign2
  
  A sample session using the traditional UNIX or DOS ftp client is as 
 follows:
  
  Step 1. At the UNIX or DOS prompt type
              fop  ftp.math.niu.edu
  
  Step 2. In response to the request for user id type
              fop
  
  Step 3. In response to the request for username type your e-mail address;
             for example:
             w.n.everitt@bham.ac.uk
  
  Step 4. At the ftp> prompt enter:
             cd  pub/papers/Zettl/Sleign2
             get  readme.txt
             quit
  
  This will transfer the readme.txt file to your current directory and close 
 the connection.
  
  At the ftp> prompt you can also enter 'ls' to see the listing of other 
 files available in the Sleign2 directory.
  
  Users who prefer World Wide Web software such as Netscape 
 or lynx can specify the URL
            ftp://ftp.math.niu.edu/pub/papers/Zettl/Sleign2
 to access the Sleign2 directory; this software will then show the list 
 of files available in it.
 
  Replacing the directory "Sleign2" with the directory "Pub papers" and
 repeating the above procedures will make available a number of recent papers
 which are related to the SLEIGN2 package. For example, the paper "BEWZ" 
 (Bailey, Everitt, Weidmann and Zettl) contains some results which can be 
 combined with SLEIGN2 to approximate the continuous (essiential) spectrum 
 of singular limit-point Sturm-Liouville problems.
  
  All eleven files of the SLEIGN2 package and a number of recent publications
 related to it, can also be accessed through the web page:
           http://www.math.niu.edu/~zettl/SL2/
  
  All suggestions, comments and criticisms are welcome; please send all 
 comments to Tony Zettl at
          zettl@math.niu.edu
  
     Paul Bailey, Norrie Everitt and Tony Zettl 
 (with the assistance of Burt Garbow.)
  
  Acknowledgement. The authors are grateful to their colleagues Eric Behr, 
 Qingkai Kong and Hongyou Wu for help and advice at a number of stages 
 in the development of this program. Some of the theoretical underpinnings 
 of the algorithm for coupled boundary conditions were obtained jointly 
 with Michael Eastham, Qingkai Kong and Hongyou Wu.

  A special thanks to Eric Behr for his help throughout the development of
 the code, for setting up the public access through the Internet and the
 world wide web, and for informed advice.

 Department of Mathematical Sciences, Northern Illinois University 
 DeKalb, IL 60115-2888, USA



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\begin{document}
\title[SLEIGN2]{SLEIGN2\\Commentary on the individual examples in xamples.f}
\author{P.B. Bailey}
\address{P.B. Bailey, c/o Department of Mathematical Sciences, Northern Illinois
University, DeKalb, IL 60155-2888, USA}
\email{70621.3674@compuserve.com}
\author{W.N. Everitt}
\address{W.N. Everitt, School of Mathematics and Statistics, University of Birmingham,
Edgbaston, Birmingham B15 2TT, England, UK}
\email{w.n.everitt@bham.ac.uk}
\author{A. Zettl}
\address{A. Zettl, Department of Mathematical Sciences, Northern Illinois University,
DeKalb, IL 60155-2888, USA}
\email{zettl@math.niu.edu}
\date{01 March 2001 (File: xamples.tex)\quad\AmS  -\LaTeX  {}; prepared in
\textsl{Scientific Word\/}{}}
\maketitle


\section{Introduction}

The examples in this commentary have been chosen to illustrate the
capabilities and limitations of the program SLEIGN2. Many of the examples have
been chosen from special cases of the well known and well studied ``special
functions'' of mathematical analysis. All possible cases of endpoint
classification are represented; all types of self-adjoint boundary conditions
are included, \textit{i.e.} regular or singular, and separated or coupled. In
the limit-circle case examples are given for which the endpoints may be
oscillatory or non-oscillatory.

For a general account of both the analytical and numerical properties of the
SLEIGN2 code see the paper by Bailey, Everitt and Zettl \cite{BEZ3}.

For all 32 examples in this commentary the following data have been entered:

(i) the Sturm-Liouville differential equation and associated interval on the
real line $\mathbb{R}$

(ii) the range of any parameters in the differential equation; this serves to
remind the reader that numerical values for these parameters have to be
entered in some of the examples given in xamples.x, for any such example to run

(iii) the endpoint classification of the differential equation, in the
relevant Hilbert function space $L^{2}((a,b);w)$

(iv) the boundary condition functions $u,v$ required for any LCNO or LCO endpoint

(v) comments on any particular features of the numbered example.

The data in items (i) to (iv) above can also be found in the file xamples.f,
but this search requires scrolling through the file as the data items, for any
particular example, are located in different sections.

For some of these examples it is possible to give explicit information on the
spectrum of associated boundary value problems; this can take the form of
providing explicit formulas for eigenvalues against which the program
calculated results can be compared.

In all cases of limit-circle endpoints, boundary condition functions $u$ and
$v$ have been entered as part of the example data. In the case of limit-circle
non-oscillatory endpoints we use the convention that the boundary condition
function $u$ determines the principal or Friedrichs boundary condition.

On selecting a numbered example in the file xamples.x, the differential
equation is displayed in Fortran, and details of the endpoint classification
given. If information on the form of the boundary condition functions $u$ and
$v$ is required then the user should scroll separately through the file
xamples.f to the appropriate numbered part of the $u,v$ section or refer to
the information in this commentary.

Some regular and weakly regular problems can be more successfully run using
the limit-circle non-oscillatory (LCNO) algorithm; details are given below for
some of the examples.

It should be noted that for limit-circle oscillatory problems it is sometimes
difficult to compute numerically more than a few of the eigenvalues. This is
due, at least in part, to the rapid growth of the eigenvalues in both the
positive and the negative directions; but particularly in the negative direction.

The Laguerre problem, Example 22, has a discrete spectrum and for one
particular boundary condition the eigenvalues are known explicitly, leading to
the classical Laguerre orthogonal polynomials. In this case numerical values
to confirm the details of the spectrum can also be obtained by use of the
Liouville transformation; this leads to the Laguerre/Liouville Example 23, for
which the program is successful over a wide range of boundary conditions.

The Liouville transformation has also been applied to the Jacobi equation,
Example 16, to yield the Jacobi/Liouville Example 24.

The Liouville transformation is sometimes useful in other cases to put a
Sturm-Liouville differential equation into a form more suitable for numerical
computation; see in particular the Bessel Example 2.

\textbf{Parameters.} The reader is reminded that many of the examples involve
the choice of one or more parameters; the range of these parameters is given
when the numbered differential equation is displayed in xamples.x; if a choice
of parameter is made outside of the stated range the program may abort.

\section{Remarks on the individual examples.}

\begin{enumerate}
\item \textbf{Classical Legendre equation} (see \cite[Chapter IV]{T})
\[
-\left(  \left(  1-x^{2}\right)  y^{\prime}(x)\right)  ^{\prime}+\tfrac{1}%
{4}y(x)=\lambda y(x)\;\text{for all}\;x\in(-1,+1).
\]

Endpoint classification in $L^{2}(-1,+1)$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$-1$ & LCNO\\
$+1$ & LCNO\\\hline
\end{tabular}
\]
$\quad$

For both endpoints the boundary condition functions $u,v$ are given by (note
that $u$ and $v$ are solutions of the Legendre equation for $\lambda=1/4$)%
\[
u(x)=1\quad\quad v(x)=\frac{1}{2}\ln\left(  \frac{1+x}{1-x}\right)
\;\text{for all}\;x\in(-1,+1).
\]

\begin{itemize}
\item[(i)] The Legendre polynomials are obtained by taking the principal
(Friedrichs) boundary condition at both endpoints $\pm1:$ enter
$A1=1,A2=0,\;B1=1,B2=0;$ \textit{i.e.} take the boundary condition function
$u$ at $\pm1$; eigenvalues: $\lambda_{n}=(n+1/2)^{2}\ ;$ $n=0,1,2,\cdots;$
eigenfunctions: Legendre polynomials $P_{n}(x)$.

\item[(ii)] Enter $A1=0,\;A2=1,\;B1=0,\;B2=1,$ \textit{i.e.} use the boundary
condition function $v$ at $\pm1$; eigenvalues: $\mu_{n};$ $n=0,1,2,\cdots$ but
no explicit formula is available; eigenfunctions are logarithmically unbounded
at $\pm1$.

\item[(iii)] Observe that $\mu_{n}<\lambda_{n}<\mu_{n+1};$ $n=0,1,2\cdots$.
\end{itemize}

\item \textbf{The Bessel equation }(see \cite[Chapter IV]{T})
\[
-y^{\prime\prime}(x)+\left(  \nu^{2}-1/4\right)  x^{-2}y(x)=\lambda
y(x)\;\text{for all}\;x\in(0,+\infty)
\]
with the parameter $\nu\in\lbrack0,+\infty).$ This is the Liouville form of
the classical Bessel equation.

Endpoint classification in $L^{2}(0,+\infty)$:%
\[%
\begin{tabular}
[c]{ccc}\hline
Endpoint & Parameter $\nu$ & Classification\\\hline
$0$ & For $\nu=1/2$ & R\\
$0$ & For all $\nu\in\lbrack0,1)$ but $\nu\neq1/2$ & LCNO\\
$0$ & For all $\nu\in\lbrack1,\infty)$ & LP\\\hline
$+\infty$ & For all $\nu\in\lbrack0,\infty)$ & LP\\\hline
\end{tabular}
\]

For endpoint $0$ and $\nu\in(0,1)$ but $\nu\neq1/2,$ the LCNO boundary
condition functions $u,v$ are determined by, for all$\;x\in(0,+\infty),$
\[%
\begin{tabular}
[c]{ccc}\hline
Parameter & $u$ & $v$\\\hline
$\nu\in(0,1)$ but $\nu\neq1/2$ & $x^{\nu+1/2}$ & $x^{-\nu+1/2}$\\
$\nu=0$ & $x^{1/2}$ & $x^{1/2}\ln(x)$\\\hline
\end{tabular}
\
\]

(a) Problems on $(0,1]$ with $y(1)=0$:

\noindent For $0\leq\nu<1,\nu\neq\frac{1}{2}:$ the Friedrichs case:
$A1=1,A2=0$ yields the classical Fourier-Bessel series; here $\lambda
_{n}=j_{\nu,n}^{2}$ where $\{j_{\nu,n}:n=0,1,2,\ldots\}$ are the zeros
(positive) of the Bessel function $J_{\nu}(\cdot).$

\noindent For $\nu\geq1;$ LP at $0$ so that there is a unique boundary value
problem with $\lambda_{n}=j_{\nu,n}^{2}$ as before.

(b) Problems on $[1,\infty)$ all have continuous spectrum on $[0,\infty)$:

\noindent For Dirichlet and Neumann boundary conditions there are no eigenvalues.

\noindent For $A1=A2=1$ at $1$ there is one isolated negative eigenvalue.

(c) Problems on $(0,\infty)$ all have continuous spectrum on $[0,\infty)$:

\noindent For $\nu\geq1$ there are no eigenvalues.

\noindent For $0\leq\nu<1$ the Friedrichs case is given by $A1=1,A2=0;$ there
are no eigenvalues.

\noindent For $\nu=0.45$ and $A1=10,A2=-1$ there is one isolated eigenvalue
near to the value $-175.57.$\noindent

\item \textbf{The Halvorsen equation}%
\[
-y^{\prime\prime}(x)=\lambda x^{-4}\exp(-2/x)y(x)\;\text{for all }%
x\in(0,+\infty)
\]

The endpoint classification in the weighted space $L^{2}((0,+\infty;x^{-4}%
\exp(-2/x))$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$0$ & WR\\
$+\infty$ & LCNO
\end{tabular}
\]

For the endpoints $0$ and $+\infty$ in the WR and LCNO classification the
boundary condition functions $u,v$ are determined by%
\[%
\begin{tabular}
[c]{ccc}\hline
Endpoint & $u$ & $v$\\\hline
$0$ & $x$ & $1$\\
$+\infty$ & $1$ & $x$\\\hline
\end{tabular}
\]

Since this equation is WR at $0$ and LCNO at $+\infty$ the spectrum is
discrete and bounded below for all boundary conditions. However, this example
illustrates that even a R or WR endpoint can cause difficulties for
computation. The program fails on R at $0$; is successful for WR at $0$; is
successful for LCNO at $0.$

At $0$, the principal boundary condition entry is $A1=1,\ A2=0$; at $\infty$
with $u(x)=1,\ v(x)=x$ the principal boundary condition entry is also
$A1=1,\;A2=0,$ but note the interchange of the definitions of $u$ and $v$ at
these two endpoints.

\item \textbf{The Boyd equation}%
\[
-y^{\prime\prime}(x)-x^{-1}y(x)=\lambda y(x)\;\text{for all}\;x\in
(-\infty,0)\cup(0,+\infty).
\]

Endpoint classification in $L^{2}(-\infty,0)\cup$ $L^{2}(0,+\infty)$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$-\infty$ & LP\\
$0-$ & LCNO\\
$0+$ & LCNO\\
$+\infty$ & LP\\\hline
\end{tabular}
\]

For both endpoints $0-$ and $0+$%
\[
u(x)=x\quad\quad v(x)=x\ln(\left|  x\right|  )\;\text{for all}\;x\in
(-\infty,0)\cup(0,+\infty).
\]

This equation arises in a model studying eddies in the atmosphere; see
\cite{B}. There is no explicit formula for the eigenvalues of any particular
boundary condition; eigenfunctions can be given in terms of Whittaker
functions; see \cite[Example 3]{BEZ}.

\item \textbf{The regularized Boyd equation}%
\[
-(p(x)y^{\prime}(x))^{\prime}+q(x)y(x)=\lambda w(x)y(x)\;\text{for all}%
\;x\in(-\infty,0)\cup(0,+\infty)
\]
where%
\[
p(x)=r(x)^{2}\quad q(x)=-r(x)^{2}\left(  \ln(\left|  x\right|  \right)
^{2}\quad w(x)=r(x)^{2}%
\]
with%
\[
r(x)=\exp\left(  -(x\ln(\left|  x\right|  )-x)\right)  \;\text{for all}%
\;x\in(-\infty,0)\cup(0,+\infty).
\]

Endpoint classification in $L^{2}(-\infty,0)\cup$ $L^{2}(0,+\infty)$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$-\infty$ & LP\\
$0-$ & WR\\
$0+$ & WR\\
$+\infty$ & LP\\\hline
\end{tabular}
\]

This is a WR form of Example 4; the singularity at zero has been regularized
using quasi-derivatives. There is a close relationship between the examples 4
and 5; in particular they have the same eigenvalues - see \cite{AEZ}. For a
general discussion of regularization using non-principal solutions see
\cite{NZ}. For numerical results see \cite[Example 3]{BEZ}.

\item \textbf{The Sears-Titchmarsh equation}%
\[
-(xy^{\prime}(x))^{\prime}-xy(x)=\lambda x^{-1}y(x)\;\text{for all}%
\;x\in(0,+\infty).
\]

Endpoint classification in $L^{2}(0,\infty)$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$0$ & LP\\
$+\infty$ & LCO\\\hline
\end{tabular}
\]

For the endpoint $+\infty$%
\[
u(x)=x^{-1/2}\left(  \cos(x)+\sin(x)\right)  \quad v(x)=x^{-1/2}\left(
\cos(x)-\sin(x)\right)  \;\text{for all}\;x\in(0,+\infty).
\]

This differential equation has one LP and one LCO endpoint. For details of
boundary value problems on $[1,\infty)$ see \cite[Example 4]{BEZ}. The
equation was studied originally in \cite[Chapter IV]{T}; but see \cite{ST}.

For problems on $[1,\infty)$ the spectrum is simple and discrete but unbounded
both above and below.

Numerical results are given in \cite[Example 4]{BEZ}.

\item \textbf{The BEZ equation}%
\[
-(xy^{\prime}(x))^{\prime}-x^{-1}y(x)=\lambda y(x)\;\text{for all}%
\;x\in(-\infty,0)\cup(0,+\infty).
\]

Endpoint classification in $L^{2}(-\infty,0)\cup$ $L^{2}(0,+\infty)$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$-\infty$ & LP\\
$0-$ & LCO\\
$0+$ & LCO\\
$+\infty$ & LP\\\hline
\end{tabular}
\]

For both endpoints $0-$ and $0+$:%
\[
u(x)=\cos\left(  \ln(\left|  x\right|  )\right)  \quad\quad v(x)=\sin\left(
\ln(\left|  x\right|  )\right)  \;\text{for all}\;x\in(-\infty,0)\cup
(0,+\infty).
\]

This example is similar to the differential equation of Example 6. On the
interval $(0,1]$ there is a singularity at $0$ in LCO; the equation is R at 1.

For numerical results see \cite[Example 5]{BEZ}.

\item \textbf{The Laplace tidal wave equation}%
\[
-(x^{-1}y^{\prime}(x))^{\prime}+\left(  kx^{-2}+k^{2}x^{-1}\right)
y(x)=\lambda y(x)\;\text{for all}\;x\in(0,+\infty)
\]
where the parameter $k\in(-\infty,0)\cup(0,+\infty)$

Endpoint classification in $L^{2}(0,\infty)$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$0$ & LCNO\\
$+\infty$ & LP\\\hline
\end{tabular}
\]

For the endpoint $0$:%
\[
u(x)=x^{2}\quad\quad v(x)=x-k^{-1}\;\text{for all}\;(0,+\infty).
\]

This equation is a particular case of the more general equation with this
name; for details and references see \cite{H}.

There are no representations for solutions of this differential equation in
terms of the well-known special functions. Thus to determine boundary
conditions at the LCNO endpoint $0$ use has to be made of maximal domain
functions; see the $u,\ v$ functions given above. Numerical results for some
boundary value problems and certain values of the parameter $k,$ are given in
\cite[Example 8]{BEZ}.

\item \textbf{The Latzko equation}%
\[
-((1-x^{7})y^{\prime}(x))^{\prime}=\lambda x^{7}y(x)\;\text{for all}%
\;x\in(0,1].
\]

Endpoint classification in $L^{2}(0,1]$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$0$ & WR\\
$1$ & LCNO\\\hline
\end{tabular}
\]

For the endpoint $1$:%
\[
u(x)=1\quad\quad v(x)=-\ln(1-x)\;\text{for all}\;(0,1).
\]

This differential equation has a long and celebrated history; see \cite[Pages
43 to 45]{F}. There is a LCNO singularity at the endpoint $1$ which requires
the use of maximal domain functions; see the $u,\ v$ functions given above.
The endpoint $0$ is WR due to the fact that $w(0)=0$.

This example is similar in some respects to the Legendre equation of Example 1 above.

For numerical results see \cite[Example 7]{BEZ}.

\item \textbf{A weakly regular equation}%
\[
-(x^{1/2}y^{\prime}(x))^{\prime}=\lambda x^{-1/2}y(x)\;\text{for all}%
\;x\in(0,+\infty).
\]

Endpoint classification in $L^{2}(0,1]$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$0$ & WR\\
$+\infty$ & LP\\\hline
\end{tabular}
\]

This is a devised example to illustrate the computational difficulties of
weakly regular problems.

The differential equation gives $p(0)=0$ and $w(0)=\infty$ but nevertheless
$0$ is a regular endpoint in the Lebesgue integral sense; however $0$ has to
be classified as weakly regular in the computational sense.

The Liouville normal form of this equation is the Fourier equation, see
Example 21 below; thus numerical results for this WR problem can be checked
against numerical results from (i) a R problem, (ii) the roots of
trigonometrical equations, and (iii) a LCNO problem (see below).

There are explicit solutions of this equation given by%
\[
\cos(2x^{1/2}\surd\lambda)\ \ ;\ \sin(2x^{1/2}\surd\lambda)/\surd\lambda.
\]

If $0$ is treated as a LCNO endpoint then $u,\ v$ boundary condition functions
are%
\[
u(x)=2x^{1/2}\quad\quad\ v(x)=1.
\]

The regular Dirichlet condition$\;y(0)=0$ is equivalent to the singular
condition $[y,u](0)=0$. Similarly the regular Neumann condition $(py^{\prime
})(0)=0$ is equivalent to the singular condition $[y,\ v](0)=0$.

The following indicated boundary value problems have the given explicit
formulae for the eigenvalues:
\begin{align*}
y(0)  &  =0\text{{ or }}[y,\ u](0)=0,\text{{ and }}y(1)=0\text{ gives}\\
\;\;\lambda_{n}  &  =((n+1)\pi)^{2}/4\ (n=0,1,...)
\end{align*}%
\begin{align*}
(py^{\prime})(0)  &  =0\text{{ or }}[y,v](0)=0,\text{{ and }}(py^{\prime
})(1)=0\text{ gives}\\
\;\;\lambda_{n}  &  =\left(  (n+\tfrac{1}{2})\pi\right)  ^{2}/4\ (n=0,1,...).
\end{align*}

\item \textbf{The Plum equation}%
\[
-(y^{\prime}(x))^{\prime}+100\cos^{2}(x)y(x)=\lambda y(x)\;\text{for
all}\;x\in(-\infty,+\infty).
\]

Endpoint classification in $L^{2}(-\infty,+\infty)$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$-\infty$ & LP\\
$+\infty$ & LP\\\hline
\end{tabular}
\]

Plum \cite{P} computed the first seven eigenvalues for periodic eigenvalues on
the interval $[0,\pi],$\textit{i.e.}%
\[
y(0)=y(\pi)\quad\quad\quad y^{\prime}(0)=y^{\prime}(\pi),
\]
using a numerical homotopy method together with interval arithmetic, and
obtained rigorous bounds for these seven computed eigenvalues. In double
precision the SLEIGN2 computed eigenvalues are in good agreement with these
guaranteed bounds.

\item \textbf{The Mathieu equation}%
\[
-y^{\prime\prime}(x)+2k\cos(2x)y(x)=\lambda y(x)\;\text{for all}\;x\in
(-\infty,+\infty)
\]
where that parameter $k\in(-\infty,0)\cup(0,+\infty).$

Endpoint classification in $L^{2}(-\infty,+\infty)$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$-\infty$ & LP\\
$+\infty$ & LP\\\hline
\end{tabular}
\]

The classical Mathieu equation has a celebrated history and voluminous
literature. There are no eigenvalues for this problem on $(-\infty,+\infty)$.
There may be one negative eigenvalue of the problem on $[0,\infty)$ depending
on the boundary condition at the endpoint $0$. The continuous (essential)
spectrum is the same for the whole line or half-line problems and consists of
an infinite number of disjoint closed intervals. The endpoints of these - and
thus the spectrum of the problem - can be characterized in terms of periodic
and semi-periodic eigenvalues of Sturm-Liouville problems on the compact
interval $[0,2\pi]$; these can be computed with SLEIGN2.

The above remarks also apply to the general Sturm-Liouville equation with
periodic coefficients of the same period; the so-called Hill's equation.

Of special interest is the starting point of the continuous spectrum - this is
also the oscillation number of the equation. For the Mathieu equation
($p=1,q=\cos(x),w=1$) on both the whole line and the half line it is
approximately -0.378; this result may be obtained by computing the first
eigenvalue $\lambda_{0}$ of the periodic problem on the interval $[0,2\pi]$.

\item \textbf{The hydrogen atom equation}

It is convenient to take this equation in two forms:%
\begin{equation}
-y^{\prime\prime}(x)+(kx^{-1}+hx^{-2})y(x)=\lambda y(x)\;\text{for all}%
\;x\in(0,+\infty) \tag{1}%
\end{equation}
where the two independent parameters $h\in\lbrack-1/4,+\infty)$ and
$k\in\mathbb{R},$ and%
\begin{equation}
-y^{\prime\prime}(x)+(kx^{-1}+hx^{-2}+1)y(x)=\lambda y(x)\;\text{for
all}\;x\in(0,+\infty) \tag{2}%
\end{equation}
where the two independent parameters $h\in(-\infty,-1/4)$ and $k\in\mathbb{R}.$

Note that form (2) is introduced as a device to aid the numerical computations
in the difficult LCO case; it forces the boundary value problem to have a
non-negative eigenvalue.

Endpoint classification, for both forms (1) and (2), in $L^{2}(0,+\infty)$:%
\[%
\begin{tabular}
[c]{cccc}\hline
Endpoint & Form & Parameters & Classification\\\hline
$0$ & 1 & $h=k=0$ & R\\
$0$ & 1 & $h=0,k\in\mathbb{R}\,\backslash\,\{0\}$ & LCNO\\
$0$ & 1 & $-1/4\leq h<3/4,h\neq0,k\in\mathbb{R}$ & LCNO\\
$0$ & 1 & $h\geq3/4,k\in\mathbb{R}$ & LP\\
$0$ & 2 & $h<-1/4,k\in\mathbb{R}$ & LCO\\\hline
$+\infty$ & 1 and 2 & $h,k\in\mathbb{R}$ & LP\\\hline
\end{tabular}
\]

This is the two parameter version of the classical one-dimensional equation
for quantum modelling of the hydrogen atom; see \cite[Section 10]{J}.

For form (1) and all $h,k$ there are no positive eigenvalues; form (2) is best
considered in the single LCO case when some eigenvalues are positive; in form
(1) there is a continuous spectrum on $[0,\infty)$; in form (2) there is a
continuous spectrum on $[1,\infty).$

If $k=0$ the equation reduces to Bessel, see Example 2 above with $h=\nu
^{2}-1/4$.

\noindent\textbf{Results for form (1)}

In all cases below $\rho$ is defined by
\[
\rho:=(h+1/4)^{1/2}\text{ for }h\geq-1/4.
\]

\begin{enumerate}
\item For $h\geq3/4$ and $k\geq0$ no boundary conditions are required; there
is at most one negative eigenvalue and $\lambda=0$ may be an eigenvalue; for
$h\geq3/4$ and $k<0$ there are infinitely many negative eigenvalues given by
\[
\lambda_{n}=\frac{-k^{2}}{(2n+2\rho+1)^{2}},\ \rho=(h+1/4)^{1/2}%
>0,\ n=0,1,2,3,\ldots
\]
and $\lambda=0$ is not an eigenvalue.

\item For $h=0$ and $k\in\mathbb{R}\,\backslash\,\{0\}$ a boundary condition
is required at $0$ for which
\[
u(x)=x\quad\quad\ v(x)=1+k\,x\ln(x).
\]
For some computed eigenvalues see \cite{BEZ} and \cite[Section 10]{J}.

\item For $-1/4<h<3/4,$ \textit{i.e.} $0<\rho<1,$ and $h\neq0,$ \textit{i.e.}
$\rho\not =1/2$, then a boundary condition is required at $0$ for which, for
all $x\in(0,+\infty),$%
\[
u(x)=x^{\frac{1}{2}+\rho}\quad\quad v(x)=x^{\frac{1}{2}-\rho}+\frac{k}%
{1-2\rho}x^{\frac{3}{2}-\rho};
\]

the following results hold for the non-Friedrichs boundary condition
$[y,v](0)=0$, \textit{i.e.} $A1=0,A2=1$:

\begin{itemize}
\item[(i)] $k>0,\ 0<\rho<1/2$ there are no negative eigenvalues

\item[(ii)] $k>0,\ 1/2<\rho<1$ there is exactly one negative eigenvalue given
by
\[
\lambda_{0}=\frac{-k^{2}}{(2\rho-1)^{2}}%
\]

\item[(iii)] if $k<0,\ 0<\rho<1/2$ there are infinitely many negative
eigenvalues given by
\[
\lambda_{n}=\frac{-k^{2}}{(2n-2\rho+1)^{2}},\ n=0,1,2,3,\ldots
\]

\item[(iv)] if $k<0,\ 1/2<\rho<1$ there are infinitely many negative
eigenvalues given by
\[
\lambda_{n}=\frac{-k^{2}}{(2n-2\rho+3)^{2}}\,,\ n=0,1,2,3,\ldots
\]

\item[(v)] for $k=0$ and $(A1)(A2)<0$ there is exactly one negative eigenvalue
given by:
\[
\lambda_{0}=\frac{4\left(  A1\right)  \Gamma(1+\rho)}{\left(  A2\right)
\Gamma(1-\rho)^{1/\rho}}.
\]
\end{itemize}

\item For $h=-1/4,\ k\in R$ , the LCNO classification at $0$ prevails and a
boundary condition is required for which, for all $x\in(0,+\infty),$%
\[
u(x)=x^{1/2}+kx^{3/2}\quad\quad v(x)=2x^{1/2}+\left(  x^{1/2}+kx^{3/2}\right)
\ln(x).
\]
\noindent For $k=0$ and $(A1)(A2)<0$ there is exactly one negative eigenvalue
given by:
\[
\lambda_{0}=-c\exp(2(A1)/A2),\quad c=4\exp(4-2\gamma)
\]
where $\gamma$ is Euler's constant: $\gamma=0.5772156649\ldots$.

\noindent\noindent\textbf{Results for form (2)}

\item For $h<-1/4,\ k\in R$, the equation is LCO at $0$ (recall that we added
$1$ to the coefficient $q(\cdot)$ for this case, thus moving the start of the
continuous spectrum from $0$ to $1$) for which, defining%
\[
\sigma:=(-h-1/4)^{1/2},
\]
then, for all $x\in(0,+\infty),$%
\begin{align*}
u(x)  &  =x^{1/2}\left[  (1-(4h)^{-1}kx)\cos(\sigma\ln(x))+k\sigma
x\sin(\sigma\ln(x))/2\right] \\
v(x)  &  =x^{1/2}\left[  (1-(4h)^{-1}kx)\sin(\sigma\ln(x))+k\sigma
x\cos(\sigma\ln(x))/2\right]  ;
\end{align*}

\begin{itemize}
\item[(i)] when $k=0$ this equation reduces to the Krall equation Example 20
below (but note that the notation is different).

\item[(ii)] When $k\not =0$ explicit formulas for the eigenvalues are not
available; however we report here on the qualitative properties of the
spectrum for any boundary condition at $0$:

$(\alpha)$ for all $k\in R$ there are infinitely many negative eigenvalues
tending exponentially to $-\infty$

$(\beta)$ for $k>0$ there are only a finite number of eigenvalues in any
bounded interval, in particular they do not accumulate at $1$

$(\gamma)$ for $k\leq0$ the eigenvalues accumulate also at $1$.

$(\delta)$ for $k=0$ and $(A1)(A2)<0$ there is exactly one negative eigenvalue
given by:
\[
\lambda_{0}=\frac{4\left(  A1\right)  \Gamma(1+\rho)}{\left(  A2\right)
\Gamma(1-\rho)^{1/\rho}}.
\]
\end{itemize}

Most of these results are due to J\"{o}rgens, see \cite[Section 10]{J}; a few
new results were established by the authors.
\end{enumerate}

\item \textbf{The Marletta equation}%
\[
-y^{\prime\prime}(x)+\frac{3(x-31)}{4(x+1)(x+4)^{2}}y(x)=\lambda
y(x)\;\text{for all}\;x\in\lbrack0,+\infty).
\]

Endpoint classification in $L^{2}(0,+\infty)$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$0$ & R\\
$+\infty$ & LP\\\hline
\end{tabular}
\]
Since $q(x)\rightarrow0$ as $x\rightarrow\infty$ the continuous spectrum
consists of $[0,\infty)$ and every negative number is an eigenvalue for some
boundary condition at $0.$

For the boundary condition $A1=5,A2=8$ there is a negative eigenvalue
$\lambda_{0}$ near $-1.185.$ However the equation with $\lambda=0$ has a
solution
\[
y(x)=\frac{1-x^{2}}{(1+x/4)^{5/2}}\text{ for all }x\in\lbrack0,\infty)
\]
that satisfies this boundary condition which is NOT in $L^{2}(0,\infty)$ but
is ``nearly'' in this space. This solution deceives SLEIGN and SLEIGN2 in
single precision, and SLEDGE in double precision, into reporting $\lambda=0$
as a second eigenvalue; in double precision SLEIGN and SLEIGN2 correctly
report that $\lambda_{0}$ is the only eigenvalue, and SLEIGN2 reports the
start of the continuous spectrum at $0.$

Additional details of this example are to be found in the Marletta
certification report on SLEIGN (not SLEIGN2) \cite{M}.

\item \textbf{The harmonic oscillator equation}%
\[
-y^{\prime\prime}(x)+x^{2}y(x)=\lambda y(x)\;\text{for all}\;x\in
(-\infty,+\infty).
\]

Endpoint classification in $L^{2}(-\infty,+\infty)$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$-\infty$ & LP\\
$+\infty$ & LP\\\hline
\end{tabular}
\
\]

This is another classical equation; it is also the Liouville normal form of
the differential equation for the Hermite orthogonal polynomials. On the whole
real line the boundary value problem requires no boundary conditions at the
endpoints of $\pm\infty$. Thus there is a unique self-adjoint extension with
discrete spectrum given by :
\[
\{\lambda_{n}=2n+1;\;n=0,1,2,...\}.
\]
For a classical treatment see \cite[Chapter IV, Section 2]{T}.

\item \textbf{The Jacobi equation}%
\[
-\left(  (1-x)^{\alpha+1}(1+x)^{\beta+1}y^{\prime}(x)\right)  ^{\prime
}=\lambda(1-x)^{\alpha}(1+x)^{\beta}y(x)\;\text{for all}\;x\in(-1,+1)
\]
where the parameters $\alpha,\beta\in(-\infty,+\infty).$

Endpoint classification in the weighted space $L^{2}((-1,+1);(1-x)^{\alpha
}(1+x)^{\beta}))$:%
\[%
\begin{tabular}
[c]{ccc}\hline
Endpoint & Parameter & Classification\\\hline
$-1$ & $\beta\leq-1$ & LP\\
$-1$ & $-1<\beta<0$ & WR\\
$-1$ & $0\leq\beta<1$ & LCNO\\
$-1$ & $1\leq\beta$ & LP\\\hline
\end{tabular}
\quad\quad%
\begin{tabular}
[c]{ccc}\hline
Endpoint & Parameter & Classification\\\hline
$+1$ & $\alpha\leq-1$ & LP\\
$+1$ & $-1<\alpha<0$ & WR\\
$+1$ & $0\leq\alpha<1$ & LCNO\\
$+1$ & $1\leq\alpha$ & LP\\\hline
\end{tabular}
\]

For the endpoint $-1$ and for the WR and LCNO cases the boundary condition
functions $u,v$ are determined by%
\[%
\begin{tabular}
[c]{ccc}\hline
Parameter & $u$ & $v$\\\hline
$-1<\beta<0$ & $(1+x)^{-\beta}$ & $1$\\
$\beta=0$ & $1$ & $\ln\left(  \dfrac{1+x}{1-x}\right)  $\\
$0<\beta<1$ & $1$ & $(1+x)^{-\beta}$\\\hline
\end{tabular}
.
\]

For the endpoint $+1$ and for the WR and LCNO cases the boundary condition
functions $u,v$ are determined by%
\[%
\begin{tabular}
[c]{ccc}\hline
Parameter & $u$ & $v$\\\hline
$-1<\alpha<0$ & $(1-x)^{-\alpha}$ & $1$\\
$\alpha=0$ & $1$ & $\ln\left(  \dfrac{1+x}{1-x}\right)  $\\
$0<\alpha<1$ & $1$ & $(1-x)^{-\alpha}$\\\hline
\end{tabular}
.
\]

To obtain the classical Jacobi orthogonal polynomials it is necessary to take
$-1<\alpha,\,\beta$; then note the required boundary conditions:

Endpoint $-1$:%
\[%
\begin{tabular}
[c]{cc}\hline
Parameter & Boundary condition\\\hline
$-1<\beta<0$ & $(py^{\prime})(-1)=0\;$or$\;[y,v](-1)=0$\\
$0\leq\beta<1$ & $[y,u](-1)=0$\\\hline
\end{tabular}
\]

Endpoint $+1$:%
\[%
\begin{tabular}
[c]{cc}\hline
Parameter & Boundary condition\\\hline
$-1<\alpha<0$ & $(py^{\prime})(+1)=0\;$or$\;[y,v](+1)=0$\\
$0\leq\alpha<1$ & $[y,u](+1)=0$\\\hline
\end{tabular}
\]
\newline For the classical Jacobi orthogonal polynomials the eigenvalues are
given by:
\[
\lambda_{n}=n(n+\alpha+\beta+1)\;\text{for}\;n=0,1,2,\ldots
\]
and this explicit formula can be used to give an independent check on the
accuracy of the results from the SLEIGN2 code.

It is interesting to note that the required boundary condition for these
Jacobi polynomials is the Friedrichs condition in the LCNO case but not in the
WR case.

\item \textbf{The rotation Morse oscillator equation}%
\[
-y^{\prime\prime}(x)+(2x^{-2}-2000(2e(x)-e(x)^{2}))y(x)=\lambda
y(x)\;\text{for all}\;x\in(0,+\infty)
\]
where%
\[
e(x)=\exp(-1.7(x-1.3))\;\text{for all}\;x\in(0,+\infty).
\]

Endpoint classification in the space $L^{2}(0,+\infty)$%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$0$ & LP\\
$+\infty$ & LP\\\hline
\end{tabular}
\
\]

This classical problem has continuous spectrum on $[0,\infty)$ and exactly 26
negative eigenvalues. Enter NUMEIG1 = 0, NUMEIG2 = 28 and observe the 26
eigenvalues and the start of the continuous spectrum at 0.

\item \textbf{The Dunsch equation}%
\[
-\left(  (1-x^{2})y^{\prime}(x)\right)  ^{\prime}+\left(  \frac{2\alpha^{2}%
}{(1+x)}+\frac{2\beta^{2}}{(1-x)}\right)  y(x)=\lambda y(x)\;\text{for
all}\;x\in(-1,+1)
\]
where the independent parameters $\alpha,\beta\in\lbrack0,+\infty).$

Boundary value problems for this differential equation are discussed in
\cite[Chapter VIII, Pages 1515-20]{DS}.

Endpoint classification in the space $L^{2}(-1,+1)$ for $-1$:%
\[%
\begin{tabular}
[c]{cc}\hline
Parameter & Classification\\\hline
$0\leq\alpha<1/2$ & LCNO\\
$1/2\leq\alpha$ & LP\\\hline
\end{tabular}
\]

Endpoint classification in the space $L^{2}(-1,+1)$ for $+1$:%
\[%
\begin{tabular}
[c]{cc}\hline
Parameter & Classification\\\hline
$0\leq\beta<1/2$ & LCNO\\
$1/2\leq\beta$ & LP\\\hline
\end{tabular}
\]

For the LCNO cases the boundary condition functions $u,v$ are given by%
\[%
\begin{tabular}
[c]{cccc}\hline
Endpoint & Parameter & $u$ & $v$\\\hline
$-1$ & $\alpha=0$ & $1.0$ & $\dfrac{1}{2}\ln\left(  \dfrac{1+x}{1-x}\right)
$\\
$-1$ & $0<\alpha<1/2$ & $(1+x)^{\alpha}$ & $(1+x)^{-\alpha}$\\
$+1$ & $\beta=0$ & $1.0$ & $\dfrac{1}{2}\ln\left(  \dfrac{1+x}{1-x}\right)
$\\
$+1$ & $0<\beta<1/2$ & $(1-x)^{\beta}$ & $(1-x)^{-\beta}$\\\hline
\end{tabular}
\]

Note that these $u$ and $v$ are not solutions of the differential equation but
maximal domain functions. In \cite[Page 1519]{DS} it is stated that the
boundary value problem determined by the boundary conditions
\[
\lbrack y,u](-1)=0=[y,u](1)
\]
has eigenvalues given by the explicit formula
\[
\lambda_{n}=(n+\alpha+\beta+1)(n+\alpha+\beta)\;\text{for}\;n=0,1,2,\ldots
\]

\item \textbf{The Donsch equation}%
\[
-\left(  (1-x^{2})y^{\prime}(x)\right)  ^{\prime}+\left(  \frac{-2\gamma^{2}%
}{(1+x)}+\frac{2\beta^{2}}{(1-x)}\right)  y(x)=\lambda y(x)\;\text{for
all}\;x\in(-1,+1)
\]
where the independent parameters $\gamma,\beta\in\lbrack0,+\infty).$

Endpoint classification in the space $L^{2}(-1,+1)$ for $-1$:%
\[%
\begin{tabular}
[c]{cc}\hline
Parameter & Classification\\\hline
$\gamma=0$ & LCNO\\
$0<\gamma$ & LCO\\\hline
\end{tabular}
\]

Endpoint classification in the space $L^{2}(-1,+1)$ for $+1$:%
\[%
\begin{tabular}
[c]{cc}\hline
Parameter & Classification\\\hline
$0\leq\beta<1/2$ & LCNO\\
$1/2\leq\beta$ & LP\\\hline
\end{tabular}
\]

For these LCNO/LCO cases the boundary condition functions $u,v$ are given by%
\[%
\begin{tabular}
[c]{cccc}\hline
Endpoint & Parameter & $u$ & $v$\\\hline
$-1$ & $\gamma=0$ & $1$ & $\dfrac{1}{2}\ln\left(  \dfrac{1+x}{1-x}\right)  $\\
$-1$ & $0<\gamma$ & $\cos(\gamma\ln(1+x))$ & $\sin(\gamma\ln(1+x))$\\
$+1$ & $\beta=0$ & $1$ & $\dfrac{1}{2}\ln\left(  \dfrac{1+x}{1-x}\right)  $\\
$+1$ & $0<\beta<1/2$ & $(1-x)^{\beta}$ & $(1-x)^{-\beta}$\\\hline
\end{tabular}
\
\]

This is a modification of Example 18 above which illustrates an LCNO/LCO mix
obtained by replacing $\alpha$ with $i\gamma$; this changes the singularity at
$-1$ from LCNO to LCO.

Again these $u$ and $v$ are not solutions of the differential equation but
maximal domain functions.

\item \textbf{The Krall equation}%
\[
-y^{\prime\prime}(x)+(1-(k^{2}+1/4)x^{-2})y(x)=\lambda y(x)\;\text{for
all}\;x\in(0,+\infty)
\]
where the parameter $k\in(0,+\infty).$

Endpoint classification in the space $L^{2}(0,+\infty)$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$0$ & LCO\\
$+\infty$ & LP\\\hline
\end{tabular}
\]

This example should be seen as a special case of the Bessel Example 2 above;
solutions can be obtained in terms of the modified Bessel functions.

To help with the computations for this example the spectrum is translated by a
term $+1$; this simple device is used for numerical convenience.

For problems with separated boundary conditions at endpoints $0$ and $\infty$
there is a continuous spectrum on $[1,\infty)$ with a discrete (and simple)
spectrum on $(-\infty,1)$. This discrete spectrum has cluster points at both
$-\infty$ and $1$.

For the LCO endpoint at $0$ the boundary condition functions are given by%
\[
u(x)=x^{1/2}\cos(k\ln(x))\quad\quad v(x)=x^{1/2}\sin(k\ln(x)).
\]

For the boundary value problem with boundary condition $[y,u](0)=0$ the
eigenvalues are given explicitly by:

(i) suppose $\Gamma(1+i)=\alpha+i\beta$ and $\mu>0$ satisfies $\tan\left(
\ln(\frac{1}{2}\mu)\right)  =-\alpha/\beta$

(ii) $\theta=\operatorname{Im}(\log(\Gamma(1+i)))$

(iii) $\ln(\frac{1}{2}\mu)=\tfrac{1}{2}\pi+\theta+s\pi\;$for$\;\;s=0,\pm1,\pm2,...$

(iv) $\mu_{s}^{2}=\left(  2\exp(\theta+\frac{1}{2}\pi)\right)  ^{2}\exp
(2s\pi)\;\,s=0,\pm1,\pm2,...$

\noindent then the eigenvalues are $\lambda_{n}=-\mu_{-(n+1)}^{2}%
+1\ (n=0,\pm1,\pm2,...)$.

SLEIGN2 can compute only six of these eigenvalues in a normal UNIX server,
even in double precision, $\lambda_{-3}$ to $\lambda_{2}$; other eigenvalues
are, numerically, too close to 1 or too close to $-\infty$. Here we list these
SLEIGN2 computed eigenvalues in double precision in a normal UNIX server and
compare them with the same eigenvalues computed from the transcendental
equation; for the problem on $(0,\infty)$ with $k=1$ and $A1=1.0,\ A2=0.0$.
\[%
\begin{array}
[c]{cccc}%
\text{NUMEIG} & \text{eig from SLEIGN2} & \text{eig from trans. equ.} &
\text{iflag}\\
-3 & -276,562.5 & -14,519,130 & 4\\
-2 & -27,114.48 & -27,114.67 & 2\\
-1 & -49.62697 & -49.63318 & 2\\
0 & 0.9054452 & 0.9054454 & 1\\
1 & 0.9998234 & 0.9998234 & 1\\
2 & 0.9999997 & 0.9999997 & 3
\end{array}
.
\]

\item \textbf{The Fourier equation}%
\[
-y^{\prime\prime}(x)=\lambda y(x)\;\text{for all}\;x\in(-\infty,+\infty)
\]

Endpoint classification in $L^{2}(-\infty,+\infty)$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$-\infty$ & LP\\
$+\infty$ & LP\\\hline
\end{tabular}
\]

This is a simple constant coefficient equation whose eigenvalues, for any
self-adjoint boundary condition, can be characterized in terms of a
transcendental equation involving only trigonometric functions.

\item \textbf{The Laguerre equation}%
\[
-(x^{\alpha+1}\exp(-x)y^{\prime}(x))^{\prime}=\lambda x^{\alpha}%
\exp(-x)y(x)\;\text{for all}\;x\in(0,+\infty)
\]
where the parameter $\alpha\in(-\infty,+\infty).$

Endpoint classification in the weighted space $L^{2}((0,+\infty);x^{\alpha
}\exp(-x))$:%
\[%
\begin{tabular}
[c]{ccc}\hline
Endpoint & Parameter & Classification\\\hline
$0$ & $\alpha\leq-1$ & LP\\
$0$ & $-1<\alpha<0$ & WR\\
$0$ & $0\leq\alpha<1$ & LCNO\\
$0$ & $1\leq\alpha$ & LP\\
$+\infty$ & $\alpha\in(-\infty,+\infty)$ & LP\\\hline
\end{tabular}
\]

For these WR/LCNO cases the boundary condition functions $u,v$ are given by:%
\[%
\begin{tabular}
[c]{cccc}\hline
Endpoint & Parameter & $u$ & $v$\\\hline
$0$ & $-1<\alpha<0$ & $x^{-\alpha}$ & $1$\\
$0$ & $\alpha=0$ & $1$ & $\ln(x)$\\
$0$ & $0<\alpha<1$ & $1$ & $x^{-\alpha}$\\\hline
\end{tabular}
\]

This is the classical form of the differential equation which for parameter
$\alpha>-1$ produces the classical Laguerre polynomials as eigenfunctions; for
the boundary condition $[y,1](0)=0$ at $0$, when required, the eigenvalues are
then (remarkably!) independent of $\alpha$ and given by $\lambda
_{n}=n\;(n=0,1,2,...)$; see \cite[Chapter 22, Section 22.6]{AB}.

SLEIGN2 does not compute eigenvalues well with this differential equation on
$(0,\infty)$, with the code in a UNIX server; this appears to be due to
numerical problems resulting from the exponentially small coefficients;
however, see Example 23 below.

\item \textbf{The Laguerre/Liouville equation}%
\[
-y^{\prime\prime}(x)+\left(  \frac{\alpha^{2}-1/4}{x^{2}}-\frac{\alpha+1}%
{2}+\frac{x^{2}}{16}\right)  y(x)=\lambda y(x)\;\text{for all}\;x\in
(0,+\infty)
\]
where the parameter $\alpha\in(-\infty,+\infty).$

Endpoint classification in the space $L^{2}(0,+\infty)$:%
\[%
\begin{tabular}
[c]{ccc}\hline
Endpoint & Parameter & Classification\\\hline
$0$ & $\alpha\leq-1$ & LP\\
$0$ & $-1<\alpha<1,$ but $\alpha^{2}\neq1/4$ & LCNO\\
$0$ & $\alpha^{2}=1/4$ & R\\
$0$ & $1\leq\alpha$ & LP\\
$+\infty$ & $\alpha\in(-\infty,+\infty)$ & LP\\\hline
\end{tabular}
\]

For these WR/LCNO cases the boundary condition functions $u,v$ are given by:%
\[%
\begin{tabular}
[c]{cccc}\hline
Endpoint & Parameter & $u$ & $v$\\\hline
$0$ & $-1<\alpha<0$ but $\alpha\neq-1/2$ & $x^{\frac{1}{2}-\alpha}$ &
$x^{\frac{1}{2}+\alpha}$\\
$0$ & $\alpha=-1/2$ & $x$ & $1$\\
$0$ & $\alpha=0$ & $x^{1/2}$ & $x^{1/2}\ln(x)$\\
$0$ & $0<\alpha<1$ but $\alpha\neq1/2$ & $x^{\frac{1}{2}+\alpha}$ &
$x^{\frac{1}{2}-\alpha}$\\
$0$ & $\alpha=1/2$ & $x$ & $1$\\\hline
\end{tabular}
\]

This is the Liouville normal form of the Laguerre equation; the two forms are
unitarily equivalent so that the spectrum and the eigenfunctions of equivalent
boundary value problems are identical. This Liouville form is more suitable
for eigenvalue computations in contrast to the previous example.

The Laguerre polynomials are produced as eigenfunctions only when $\alpha>-1$.
For $\alpha\geq1$ the LP condition holds at $0$. For $0\leq\alpha<1$ the
appropriate boundary condition is the Friedrichs condition: $[y,u](0)=0;$ for
$-1<\alpha<0$ use the non-Friedrichs condition: $[y,\ v](0)=0$. In all these
cases $\lambda_{n}=n$ for $n=0,1,2,...$.

\item \textbf{The Jacobi/Liouville equation}%
\[
-y^{\prime\prime}(x)+q(x)y(x)=\lambda y(x)\;\text{for all}\;x\in(-\pi
/2,+\pi/2)
\]
where the coefficient $q$ is given by, for all$\;x\in(-\pi/2,+\pi/2),$%
\[
q(x)=\frac{\beta^{2}-1/4}{4\tan^{2}((x+\pi)/2)}+\frac{\alpha^{2}-1/4}%
{4\tan^{2}((x-\pi)/2)}-\frac{4\alpha\beta+4\beta+4\alpha+3}{8}.
\]
Here the parameters $\alpha,\beta\in(-\infty+,\infty).$

Endpoint classification in the space $L^{2}(-\pi/2,+\pi/2)$:%
\[%
\begin{tabular}
[c]{ccc}\hline
Endpoint & Parameter & Classification\\\hline
$-\pi/2$ & $\beta\leq-1$ & LP\\
$-\pi/2$ & $-1<\beta<1$ but $\beta^{2}\neq1/4$ & LCNO\\
$-\pi/2$ & $\beta^{2}=1/4$ & R\\
$-\pi/2$ & $1\leq\beta$ & LP\\\hline
\end{tabular}
\]%
\[%
\begin{tabular}
[c]{ccc}\hline
Endpoint & Parameter & Classification\\\hline
$+\pi/2$ & $\alpha\leq-1$ & LP\\
$+\pi/2$ & $-1<\alpha<1$ but $\alpha^{2}\neq1/4$ & LCNO\\
$+\pi/2$ & $\alpha^{2}=1/4$ & R\\
$+\pi/2$ & $1\leq\alpha$ & LP\\\hline
\end{tabular}
\]

For the endpoint $-\pi/2$ and for LCNO cases the boundary condition functions
$u,v$ are determined by, here $b(x)=2\tan^{-1}(1)+x$ for all $x\in(-\pi
/2,+\pi/2),$%
\[%
\begin{tabular}
[c]{ccc}\hline
Parameter & $u$ & $v$\\\hline
$-1<\beta<0$ & $b(x)^{\frac{1}{2}-\beta}$ & $b(x)^{\frac{1}{2}+\beta}$\\
$\beta=0$ & $\sqrt{b(x)}$ & $\sqrt{b(x)}\ln(b(x))$\\
$0<\beta<1$ & $b(x)^{\frac{1}{2}+\beta}$ & $b(x)^{\frac{1}{2}-\beta}$\\\hline
\end{tabular}
\]

For the endpoint $+\pi/2$ and for LCNO cases the boundary condition functions
$u,v$ are determined by, here $a(x)=2\tan^{-1}(1)-x$ for all $x\in(-\pi
/2,+\pi/2),$%
\[%
\begin{tabular}
[c]{ccc}\hline
Parameter & $u$ & $v$\\\hline
$-1<\alpha<0$ & $a(x)^{\frac{1}{2}-\alpha}$ & $a(x)^{\frac{1}{2}+\alpha}$\\
$\alpha=0$ & $\sqrt{a(x)}$ & $\sqrt{a(x)}\ln(a(x))$\\
$0<\alpha<1$ & $a(x)^{\frac{1}{2}+\alpha}$ & $a(x)^{\frac{1}{2}-\alpha}%
$\\\hline
\end{tabular}
\]

This is the Liouville normal form of the Jacobi equation of Example 16.

The classical Jacobi orthogonal polynomials are produced only when both
$\alpha,\beta>-1.$ For $\alpha,\beta>+1$ the LP condition holds and no
boundary condition is required to give the polynomials. If $-1<\alpha,\beta<1$
then the LCNO condition holds and boundary conditions are required to produce
the Jacobi polynomials; these conditions are as follows:

Endpoint $-\pi/2$%
\[%
\begin{tabular}
[c]{cc}\hline
Parameter & Boundary condition\\\hline
$-1<\beta<0$ & $[y,v](-\pi/2)=0$\\
$0\leq\beta<1$ & $[y,u](-\pi/2)=0$\\\hline
\end{tabular}
\]

Endpoint $+\pi/2$%
\[%
\begin{tabular}
[c]{cc}\hline
Parameter & Boundary condition\\\hline
$-1<\alpha<0$ & $[y,v](+\pi/2)=0$\\
$0\leq\alpha<1$ & $[y,u](+\pi/2)=0$\\\hline
\end{tabular}
\]

Recall from Example 16 for the classical orthogonal Jacobi polynomials the
eigenvalues are given explicitly by:%
\[
\lambda_{n}=n(n+\alpha+\beta+1)\;\text{for}\;n=0,1,2,\ldots
\]

\item \textbf{The Meissner equation}%
\[
-y^{\prime\prime}(x)=\lambda w(x)y(x)\;\text{for all}\;x\in(-\infty,+\infty)
\]
where the weight coefficient $w$ is defined by%
\begin{align*}
w(x)  &  =1\;\text{for all}\;x\in(-\infty,0]\\
&  =9\;\text{for all}\;x\in(0,+\infty).
\end{align*}

Endpoint classification in the space $L^{2}(-\infty,+\infty)$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$-\infty$ & LP\\
$+\infty$ & LP\\\hline
\end{tabular}
\]

This equation arose in a model of a one dimensional crystal. For this constant
coefficient equation with a weight function which has a jump discontinuity the
eigenvalues can be characterized as roots of a transcendental equation
involving only trigonometrical and inverse trigonometrical functions. There
are infinitely many simple eigenvalues and infinitely many double ones for the
periodic case; they are given by:

\textbf{Periodic boundary conditions on} $(-1/2,+1/2)$\textbf{, }\textit{i.e.}%
\[
y(-1/2)=y(+1/2)\quad\quad y^{\prime}(-1/2)=y^{\prime}(+1/2).
\]

We have $\lambda_{0}=0$ and for $n=0,1,2,\ldots$
\[
\lambda_{4n+1}=(2m\pi+\alpha)^{2};\ \ \lambda_{4n+2}=(2(n+1)\pi-\alpha))^{2};
\]%
\[
\ \lambda_{4n+3}\ =\ \ \lambda_{4n+4}=(2(n+1)\pi))^{2}.
\]
where $\alpha\,=\,\cos^{-1}(-7/8)$

\textbf{Semi-periodic boundary conditions on }$($\textbf{ }$-1/2,+1/2)$%
\textbf{, }\textit{i.e.}%
\[
y(-1/2)=-y(+1/2)\quad\quad y^{\prime}(-1/2)=-y^{\prime}(+1/2).
\]

With $\beta=\cos^{-1}((1+\sqrt{(}33))/16)$ and $\gamma=\cos^{-1}((1-\sqrt
{(}33))/16)$ these are all simple and given by, for $n=0,1,2,\ldots$
\[
\lambda_{4n}\,=\,(2n\pi\,+\,\beta)^{2};\ \lambda_{4n+1}\,=\,(2n\pi
\,+\,\gamma)^{2};\ \
\]%
\[
\lambda_{4n+2}\,=\,(2(n+1)\pi\,-\,\gamma)^{2};\text{ }\lambda_{4n+3}%
\,=\,(2(n+1)\pi\,-\,\beta)^{2}.
\]
See \cite{E} and \cite{Hoc}.

\item \textbf{The Lohner equation}%
\[
-y^{\prime\prime}(x)-1000xy(x)=\lambda y(x)\;\text{for all}\;x\in
(-\infty,+\infty)
\]

Endpoint classification in the space $L^{2}(-\infty,+\infty)$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$-\infty$ & LP\\
$+\infty$ & LP\\\hline
\end{tabular}
\]

In \cite{L} Lohner computed the Dirichlet eigenvalues of index (in SLEIGN2
notation) 0, 9, 49 and 99 using interval arithmetic and obtained rigorous
bounds. In double precision SLEIGN2 computed eigenvalues are in good agreement
with these guaranteed bounds.

\item \textbf{The J\"{o}rgens equation}%
\[
-y^{\prime\prime}(x)+(\exp(2x)/4-k\exp(x))y(x)=\lambda y(x)\;\text{for
all}\;x\in(-\infty,+\infty)
\]
where the parameter $k\in(-\infty,+\infty).$

Endpoint classification in the space $L^{2}(-\infty,+\infty),$ for all
$k\in(-\infty,+\infty)$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$-\infty$ & LP\\
$+\infty$ & LP\\\hline
\end{tabular}
\]

This is a remarkable example from J\"{o}rgens and SLEIGN2 obtains excellent
results. Details of this problem are given in \cite[Part II, Section 10]{J}.
For all $k\in(-\infty,+\infty)$ the boundary value problem on the interval
$(-\infty,+\infty)$ has a continuous spectrum on $[0,+\infty)$; for $k\leq1/2$
there are no eigenvalues; for $h=0,1,2,3,\ldots$ and then $k$ chosen by
$h<k-1/2\leq h+1,$ there are exactly $h+1$ eigenvalues and these are all below
the continuous spectrum; these eigenvalues are given explicitly by
\[
\lambda_{n}=-(k-1/2-n)^{2},\quad n=0,1,2,3,\ldots,h.
\]

\item \textbf{The Behnke-Goerisch equation}%
\[
-y^{\prime\prime}(x)+k\cos^{2}(x)y(x)=\lambda y(x)\;\text{for all}%
\;x\in(-\infty,+\infty)
\]
where the parameter $k\in(-\infty,+\infty),$

Endpoint classification in the space $L^{2}(-\infty,+\infty),$ for all
$k\in(-\infty,+\infty)$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$-\infty$ & LP\\
$+\infty$ & LP\\\hline
\end{tabular}
\]

This is a form of the Mathieu equation. In \cite{BG} these authors computed a
number of Neumann eigenvalues of this problem using interval arithmetic with
rigorous bounds. In double precision SLEIGN2 computed eigenvalues are in good
agreement with these guaranteed bounds.

\item \textbf{The Whittaker equation}%
\[
-y^{\prime\prime}(x)+\left(  \frac{1}{4}+\frac{k^{2}-1}{x^{2}}\right)
y(x)=\lambda\frac{1}{x}y(x)\;\text{for all}\;x\in(0,+\infty)
\]
where the parameter $k\in\lbrack1,+\infty).$

Endpoint classification in the space $L^{2}(0,+\infty),$ for all $k\in
\lbrack1,+\infty)$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$0$ & LP\\
$+\infty$ & LP\\\hline
\end{tabular}
\]

This equation is studied in \cite[Part II, Section 10]{J}. There it is shown
that the LP case holds at $+\infty$ and also at $0$ for $k\geq1$. The spectrum
is discrete and is given explicitly by:
\[
\lambda_{n}=n+(k+1)/2,\quad n=0,1,2,3,\ldots.
\]

\item \textbf{The Littlewood-McLeod equation}%
\[
-y^{\prime\prime}(x)+x\sin(x)y(x)=\lambda y(x)\;\text{for all}\;x\in
\lbrack0,+\infty).
\]

Endpoint classification in the space $L^{2}(0,+\infty)$:%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$0$ & R\\
$+\infty$ & LP\\\hline
\end{tabular}
\]

The spectral analysis of this differential equation is considered in
\cite{JEL} and \cite{JBM}; the equation is R at $0$ and LP at $\pm\infty.$ All
self-adjoint operators in $L^{2}[0,\infty)$ have a simple, discrete spectrum
$\{\lambda_{n}:n=0,\pm1,\pm2,\ldots\}$ that is unbounded both above and below,
\textit{i.e.}%
\[
\lim_{n\rightarrow-\infty}\lambda_{n}=-\infty\quad\quad\quad\lim
_{n\rightarrow+\infty}\lambda_{n}=+\infty.
\]
Every eigenfunction has infinitely many zeros in $(0,\infty).$

SLEIGN2, and other codes, fail to compute the eigenvalues for this type of LP
oscillatory problem. However there is qualitative information to be obtained
by considering regular problems on $[0,X]$ with, say, Dirichlet boundary
conditions $y(0)=Y(X)=0.$

\item \textbf{The Morse equation}%
\[
-y^{\prime\prime}(x)+(9\exp(-2x)-18\exp(-x))y(x)=\lambda y(x)\;\text{for
all}\;x\in(-\infty,+\infty)
\]

Endpoint classification in the space $L^{2}(-\infty,+\infty)$%
\[%
\begin{tabular}
[c]{cc}\hline
Endpoint & Classification\\\hline
$-\infty$ & LP\\
$+\infty$ & LP\\\hline
\end{tabular}
\]

This differential equation is studied in \cite[Example 6]{PBB1}; the spectrum
has exactly three negative, simple eigenvalues, and a continuous spectrum on
$[0,\infty);$ the eigenvalues are given explicitly by%
\[
\lambda_{n}=-(n-2.5)^{2}\;\text{for}\;n=0,1,2.
\]

\item \textbf{The Heun equation}%
\[
-(py^{\prime})^{\prime}+qy=\lambda wy\;\text{on}\;(0,1)
\]
where the coefficients $p,q,w$ are given explicitly by, for all $x\in(0,1),$%
\[%
\begin{array}
[c]{l}%
p(x)=x^{c}(1-x)^{d}(x+s)^{e}\\
q(x)=abx^{c}(1-x)^{d-1}(x+s)^{e-1}\\
w(x)=x^{c-1}(1-x)^{d-1}(x+s)^{e-1}.
\end{array}
\]
The parameters $a,b,c,d,e$ and $s$ are all real numbers and satisfy the
following two conditions%
\[
(i)\;s>0\;\text{and}\;c\geq1,d\geq1,a\geq b
\]
and%
\[
(ii)\;a+b+1-c-d-e=0.
\]
From these conditions it follows that%
\[
a\geq1,b\geq1,e\geq1\;\text{and}\;a+b-d\geq1.
\]

The differential equation above is a special case of the general Heun equation%
\[
\frac{d^{2}w(z)}{dz^{2}}+\left(  \frac{\gamma}{z}+\frac{\delta}{z-1}%
+\frac{\varepsilon}{z-a}\right)  \frac{dw(z)}{dz}+\frac{\alpha\beta
z-q}{z(z-1)(z-a)}w(z)=0
\]
with the general parameters $\alpha,\beta,\gamma,\delta,\varepsilon$ replaced
by the real numbers $a,b,c,d,e,$ $a$ replaced by $-s,$ and $q$ replaced by the
spectral parameter $\lambda.$ For general information concerning the Heun
equation see the compendium \cite{AR1}; for the special form of the Heun
equation considered here, and for the connection with confluence of
singularities and applications, see the recent paper \cite{LS1}.

We note that the coefficients of the Sturm-Liouville differential equation
above satisfy the conditions

\begin{itemize}
\item[$(i)$] $q,w\in C[0,1]$ and $w(x)>0$ for all $x\in(0,1)$

\item[$(ii)$] $p^{-1}\in L_{\text{loc}}^{1}(0,1),p(x)>0$ for all $x\in(0,1)$

\item[$(iii)$] $p^{-1}\notin L^{1}(0,1/2]$ and $p^{-1}\notin L^{1}[1/2,1).$
\end{itemize}

Thus both endpoints $0$ and $1$ are singular for the differential equation.
Analysis shows that the endpoint classification for this equation is%
\[%
\begin{tabular}
[c]{ccc}\hline
Endpoint & Parameter & Classification\\\hline
$0$ & $c\in\lbrack1,2)$ & LCNO\\
$0$ & $c\in\lbrack2,+\infty)$ & LP\\
$1$ & $d\in\lbrack1,2)$ & LCNO\\
$1$ & $d\in\lbrack2,+\infty)$ & LP\\\hline
\end{tabular}
\ \
\]

For the endpoint $0$ and for LCNO cases the boundary condition functions $u,v$
are determined by:%
\[%
\begin{tabular}
[c]{ccc}\hline
Parameter & $u$ & $v$\\\hline
$c=1$ & $1$ & $\ln(x)$\\
$1<c<2$ & $1$ & $x^{1-c}$\\\hline
\end{tabular}
\]

For the endpoint $1$ and for LCNO cases the boundary condition functions $u,v$
are determined by:%
\[%
\begin{tabular}
[c]{ccc}\hline
Parameter & $u$ & $v$\\\hline
$d=1$ & $1$ & $\ln(1-x)$\\
$1<d<2$ & $1$ & $(1-x)^{1-d}$\\\hline
\end{tabular}
\]

Further it may be shown that the spectrum of any self-adjoint problem on
$(0,1),$ with the parameters $a,b,c,d,e$ and $s$ satisfying the above
conditions, and considered in the space $L^{2}((0,1);w)$ with either separated
or coupled boundary conditions, is bounded below and discrete. For the
analytic properties, and proofs of the spectral properties of this Heun
differential equation, see the paper \cite{BEHZ}.

In xamples.f, under Example 32, the user can enter a choice for the parameters
$a,b,c,d,e$ and $s,$ subject to the conditions above being satisfied, and
obtain numerical results for the eigenvalues.
\end{enumerate}

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Equations, Y. Alavi and P. Hsieh editors, World Scientific, (1994), 393-406.
\end{thebibliography}
\end{document}
SHAR_EOF
fi # end of overwriting check
cd ..
if test ! -d 'Fortran'
then
	mkdir 'Fortran'
fi
cd 'Fortran'
if test ! -d 'Sp'
then
	mkdir 'Sp'
fi
cd 'Sp'
if test ! -d 'Drivers'
then
	mkdir 'Drivers'
fi
cd 'Drivers'
if test -f 'data2'
then
	echo shar: will not over-write existing file "'data2'"
else
cat << "SHAR_EOF" > 'data2'
(output: again: )
(np: param: a: b: classa: classb: bca: bcb: bcc: numeig: end:)
(bca: d n a1,a2)
(bcc:  p s alfa k11,k12,k21,k22)
 
output: results.txt
np: 1 
a: -1.0 
classa:  lcno 
b: 1.0 
classb: lcno 
bca: 1,0  
bcb: 1,0
numeig: 0,2 

again:
a: 0.0
classa: r
bca: d
bcb: 0,1
 
again:

 
np: 2 
param: 0.75
a: 0.0 
classa:  lcno 
b: 1.0 
classb: r 
bca: 1,0  
bcb: d 
numeig: 0,2 

again:
bca: 0,1
 
again:
b: null
bcb: null
classb: lp

again:

np: 3 
param: null
a: 0.0 
classa:  wr 
b: null
classb: lcno 
bca: d  
bcb: 1,0 
numeig: 0,2 

again:

 
np: 4 
a: 0.0 
b: 1.0
classa:  lcno 
classb: r 
bca: 1,0  
bcb: d 
numeig: 0,4 

again:
bca: 0,1
 
again:

 
np: 5 
a: 0.0 
b: 1.0
classa:  wr 
classb: r 
bca: d  
bcb: d 
numeig: 0,4 

again:
bca: 1,1
 
again:
np: 6 
param: null
a: 1.0 
classa:  r 
b: null
classb: lco 
bca: d  
bcb: 1,0 
numeig: -2,2 

again:
bcb: 0,1
 
again:

 
np: 7 
a: 0.0 
b: 1.0
classa: lco 
classb: r 
bca: 1,0  
bcb: d 
numeig: -2,3 

again:
bca: 0,1
numeig: -1,5
 
again:
bca: null
bcb: null
bcc: p
numeig: -1,2

again:
 
bcc: null
param: 1.0
np: 8 
a: 0.0 
b: 1.0
classa:  lcno 
classb: r 
bca: 1,0  
bcb: d 
numeig: 0,1 

again:
bca: 0,1
numeig: 0,2 

again:
param: -0.5
bca: 1,0
numeig: 0,1
 
again:
bca: 0,1
numeig: 0,2

again:

 
param: null
np: 9 
a: 0.0 
b: 1.0
classa: wr
classb: lcno 
bca: d  
bcb: 1,0 
numeig: 0,4 

again:
bcb: 0,1
numeig: 0,4 

again:

 
np: 10 
a: 0.0 
b: 1.0
classa: wr
classb: r 
bca: d  
bcb: d 
numeig: 0,2 

 
again:
np: 11 
a: 0.0 
bca: null
b: pi 
bcb: null
classa: r
classb: r 
bcc: p 
numeig: 0,2 

again:

 
param: 5.0
np: 12 
a: 0.0 
b: pi 
classa: r
classb: r 
bcc: p 
numeig: 0,6 

again:

 
param: -1.0,2.0 
np: 13 
a: 0.0 
classa: lp 
b: null
classb: lp 
bcc: null
numeig: 0,2 

again:

 
param: null
np: 14 
a: 0.0 
classa: r
bca: 5.0,8.0 
classb: lp 
numeig: 0,2 

again:

 
np: 15 
a: null
classa: lp 
bca: null
classb: lp 
numeig: 0,2 

 
again:

np: 16 
param: -0.5,-1.2
a: -1.0
b: 1.0
classa: lp 
classb: wr 
bcb: d
numeig: 0,2 

again:
param: 0.5,-1.2
classb: lcno
bcb: 1,0

again:
param: 1.2,-1.2
classb: lp
bcb: null

again:
param: .5,-.5
classa: wr
classb: lcno
bca: d
bcb: 1,0
 
again:
param: 1.2,-0.5
classb: lp
bcb: null
 
again:
param: 1.2,0.5
classa: lcno
bca: 1,0
 
again:

 
param: null
np: 17 
a: 0.0
classa: lp 
b: null
classb: lp 
bca: null
bcb: null
numeig: 0,3 

again:

 
np: 18 
param: 0.2,0.6
a: -1.0
b: 1.0 
classa: lcno 
classb: lp 
bca: 1,0
numeig: 0,2 

again:

 
np: 19 
param: 0.5,0.2
a: -1.0
b: 1.0 
classa: lco 
classb: lcno 
bca: 1,0
bcb: 1,0
numeig: -1,2 

again:
param: 0.5,0.6
classb: lp
bcb: null
numeig: 0,3

again:

np: 20 
param: 1.0
a: 0.0
classa: lco 
b: null
classb: lp 
bca: 1,0
bcb: null
numeig: -2,1 

 
again:
param: null
np: 21 
a: -pi 
b: pi 
classa: r 
classb: r 
bca: d
bcb: d 
numeig: 0,4
 
again:
bca: null
bcb: null
bcc: p
numeig: 0,4 

again:
bcc: 0.7853982,2.0000000,1.0000000,1.0000000,1.0000000

again:
 
np: 22 
param: 0.0
a: 0.0
classa: lcno 
b: null
classb: lp 
bca: 1,0 
bcb: null
bcc: null
numeig: 0,2
 
again:
bca: 0,1
numeig: 0,2 

again:
param: -1.2
classa: lp
bca: null
 
again:
param: -0.6
classa: wr
bca: d
 
again:
param: 0.5
classa: lcno
bca: 1,0
 
again:
bca: 0,1

again:
param: 1.2
classa: lp
bca: null
 
again:

 
np: 23 
param: 0.0 
a: 0.0
classa: lcno 
b: null
classb: lp 
bca: 1,0 
bcb: null
numeig: 1,2
 
again:
param: -1.2
classa: lp
bca: null
numeig: 0,2 

again:
param: 0.6 
classa: lcno 
bca: 1,0
 
again:
param: 0.5
classa: r
bca: d
 
again:
param: 1.2
classa: lp
bca: null

again:

 
np: 24 
param: -0.6,-1.2
a: -pi/2
b:  pi/2 
classa: lp 
bca: null
classb: lcno 
bcb: 1,0 
numeig: 0,2
 
again:
param: 0.5,-1.2
classb: r 
bcb: d
numeig: 0,2 

again:
param: 1.2,-1.2 
classb: lp 
bcb: null

again:
param: 0.5,-0.6
classa: lcno 
classb: r
bca: 1,0
bcb: d 
 
again:
param: 1.2,-0.6
classb: lp
bca: 1,0
bcb: null

again:
param: 1.2,0.5
classa: r
bca: d

again:

 
param: null
np: 25 
a: -0.5 
b:  0.5 
classa: r 
classb: r 
bca: null
bcb: null
bcc: p
numeig: 0,6

 
again:
param: null
np: 26 
a: 0.0 
b:  1.0 
classa: r 
classb: r 
bca: d
bcb: d
bcc: null
numeig: 0,4

again:
 
np: 27
param: 2.0
a: null
b: null
classa: lp
classb: lp
numeig: 0,2

again:

np: 28 
param: 2.0
a: 0.0
b: pi
classa: r 
classb: r 
bca: n
bcb: n
numeig: 0,6

again:

 
np: 29 
param: 1.0
a: 0.0
classa: lp 
b: null
classb: lp 
bca: null
bcb: null
numeig: 0,2

again:

param: null
np: 30 
a: 0.0
classa: r
b: 10.0 
classb: r 
bca: d 
bcb: d 
numeig: 0,9
 
again:

b: 20.0

again:
b: 30.0

again:
b: 40.0

again:
param: null
np: 31
a: null
b: null
classa: lp
classb: lp
bca: null
bcb: null
numeig: 0,4

again:
np: 32
param: 1.0,4.0,2.0,1.5,4.5
a: 0.0
classa: lcno
bca: 1.0,0.0
b: 1.0
classb: lp
numeig: 0,2

again:
numeig: 18

end:
 


SHAR_EOF
fi # end of overwriting check
if test -f 'driver1.f'
then
	echo shar: will not over-write existing file "'driver1.f'"
else
cat << "SHAR_EOF" > 'driver1.f'
C  PROGRAM COUPDR

C     **********
C     MARCH 1, 2001; P.B. BAILEY, W.N. EVERITT AND A. ZETTL
C     VERSION 1.2
C     **********

      PROGRAM COUPDR

C     This program is for the purpose of indicating how SLCOUP
C     can be called for the purpose of obtaining eigenvalues
C     of Sturm-Liouville problems with regular and singular
C     coupled boundary conditions.
C
C     The call is of the form:
C
C     CALL SLCOUP(A, B, INTAB, P0ATA, QFATA, P0ATB, QFATB,
C    1    A1, A2, B1, B2, NUMEIG, EIG, TOL, IFLAG,
C    2    CPFUN, NCA, NCB, ALFA, K11, K12, K21, K22)
C
C     DECLARE THE NEEDED VARIABLES:

C
C     DECLARE THE NEEDED EXTERNALS:

C
C     SET THE PARAMETERS DEFINING THE INTERVAL OF
C       DEFINITION OF THE DIFFERENTIAL EQUATION:
C       (BESSEL'S EQUATION WITH NU = 0.75)
C
C     .. Scalars in Common ..
      REAL A1S,A2S,AA,ASAV,B1S,B2S,BB,BSAV,DTHDAA,DTHDBB,
     +                 EIGSAV,EPSMIN,FA,FB,GQA,GQB,
     +                 GWA,GWB,HPI,LPQA,
     +                 LPQB,LPWA,LPWB,P0ATAS,P0ATBS,PI,QFATAS,QFATBS,
     +                 TMID,TSAVEL,TSAVER,TWOPI,Z
      INTEGER IND,INTSAV,ISAVE,MDTHZ,MFS,MLS,MMWD,T21
      LOGICAL ADDD,PR
C     ..
C     .. Arrays in Common ..
      REAL TEE(100),TT(7,2),YS(200),YY(7,3,2),ZEE(100)
      INTEGER JAY(100),MMW(100),NT(2)
C     ..
C     .. Local Scalars ..
      REAL A,A1,A2,ALFA,B,B1,B2,EIG,K11,K12,K21,K22,P0ATA,
     +                 P0ATB,QFATA,QFATB,TOL
      INTEGER I,ICPFUN,IFLAG,INTAB,NCA,NCB,NP,NUMEIG
C     ..
C     .. Local Arrays ..
      REAL CPFUN(100),X(100)
C     ..
C     .. External Subroutines ..
      EXTERNAL SLCOUP
C     ..
C     .. Common blocks ..
C     COMMON /ALBE/LPWA,LPQA,LPWB,LPQB
      COMMON /ALBE/LPWA,LPQA,FA,GWA,GQA,LPWB,LPQB,FB,GWB,GQB
      COMMON /BCDATA/A1S,A2S,P0ATAS,QFATAS,B1S,B2S,P0ATBS,QFATBS
      COMMON /DATADT/ASAV,BSAV,INTSAV
      COMMON /DATAF/EIGSAV,IND
      COMMON /LP/MFS,MLS
      COMMON /PASS/YS,MMW,MMWD
      COMMON /PIE/PI,TWOPI,HPI
      COMMON /PRIN/PR,T21
      COMMON /RNDOFF/EPSMIN
      COMMON /TDATA/AA,TMID,BB,DTHDAA,DTHDBB,MDTHZ,ADDD
      COMMON /TEEZ/TEE
      COMMON /TEMP/TT,YY,NT
      COMMON /TSAVE/TSAVEL,TSAVER,ISAVE
      COMMON /Z1/Z
      COMMON /ZEEZ/JAY,ZEE
C     ..
      T21 = 21
      OPEN (T21,FILE='test.out')
      PR = .true.
      A = 0.0D0
      B = 1.0D0

C  ----------------------------------------------------------------C
C     SET THE PARAMETERS DEFINING THE COUPLED BOUNDARY CONDITIONS:
C  ----------------------------------------------------------------C

      A1 = 1.0D0
      A2 = 0.0D0
      B1 = 1.0D0
      B2 = 0.0D0

C     SET THE REMAINING PARAMETERS NEEDED BY SLCOUP:

      P0ATA = -1.0D0
      QFATA = -1.0D0
      P0ATB = -1.0D0
      QFATB = 1.0D0
      INTAB = 1
      NUMEIG = 3
      EIG = 0.0D0
      TOL = 1.D-5
      NCA = 3
      NCB = 1
      ICPFUN = 0

C  ----------------------------------------------------------------C
C     SET THE COUPLED BOUNDARY CONDITION PARAMETERS:
C     (THESE ARE THE "PERIODIC" CONDITIONS.)
C  ----------------------------------------------------------------C
      ALFA = 0.0D0
      K11 = 1.0D0
      K12 = 0.0D0
      K21 = 0.0D0
      K22 = 1.0D0

C ---------------------------------------------------------C
C  If eigenfunction values are wanted, then:

      NP = 42
      DO 10 I = 2,NP - 1
          X(I-1) = A + (I-1)* (B-A)/ (NP-1)
   10 CONTINUE
      ICPFUN = NP - 2

      DO 15 I = 1,ICPFUN
          CPFUN(I) = X(I)
   15 CONTINUE
C  ----------------------------------------------------------------C
C     CALL SLCOUP:
C  ----------------------------------------------------------------C

      CALL SLCOUP(A,B,INTAB,P0ATA,QFATA,P0ATB,QFATB,A1,A2,B1,B2,NUMEIG,
     +            EIG,TOL,IFLAG,ICPFUN,CPFUN,NCA,NCB,ALFA,K11,K12,K21,
     +            K22)

C     WRITE OUT THE RETURNED EIGENVALUE, IT'S ESTIMATED ACCURACY,
C     AND THE IFLAG STATUS:

      WRITE (*,FMT=*) ' NUMEIG, EIG, TOL, IFLAG = ',NUMEIG,EIG,TOL,IFLAG

C
      IF (ICPFUN.GT.0) THEN
          WRITE (*,FMT=*) ' EIGENFUNCTION '
          DO 30 I = 1,ICPFUN
              WRITE (*,FMT=*) X(I),CPFUN(I)
   30     CONTINUE
      END IF
C
      CLOSE (T21)
      STOP
      END
C
      REAL FUNCTION P(X)

C     .. Scalar Arguments ..
      REAL X
C     ..
      P = 1.0D0
      RETURN
      END
C
      REAL FUNCTION Q(X)

C     .. Scalar Arguments ..
      REAL X
C     ..
C     .. Local Scalars ..
      REAL NU
C     ..
      NU = 0.75D0
      Q = (NU*NU-0.25D0)/X**2
      RETURN
      END
C
      REAL FUNCTION W(X)

C     .. Scalar Arguments ..
      REAL X
C     ..
      W = 1.0D0
      RETURN
      END
C
      SUBROUTINE UV(X,U,PUP,V,PVP,HU,HV)

C     .. Scalar Arguments ..
      REAL HU,HV,PUP,PVP,U,V,X
C     ..
C     .. Local Scalars ..
      REAL NU
C     ..
      NU = 0.75D0
      U = X** (NU+0.5D0)
      PUP = (NU+0.5D0)*X** (NU-0.5D0)
      V = X** (-NU+0.5D0)
      PVP = (-NU+0.5D0)*X** (-NU-0.5D0)
      HU = 0.0D0
      HV = 0.0D0
      RETURN
      END
SHAR_EOF
fi # end of overwriting check
if test -f 'driver2.f'
then
	echo shar: will not over-write existing file "'driver2.f'"
else
cat << "SHAR_EOF" > 'driver2.f'
C  PROGRAM DRIVE

C     **********
C     MARCH 1, 2001; P.B. BAILEY, W.N. EVERITT AND A. ZETTL
C     VERSION 1.2
C     **********

      PROGRAM DRIVE
C
C     THIS PROGRAM IS AN "INTERACTIVE" DRIVER FOR THE USE OF
C     SLEIGN2. IT CAN BE COMPILED WITH SLEIGN2.F AND XAMPLES.F,
C     FOR EXAMPLE, WHICH MAKES IT EASY TO RUN PROBLEMS WITH ANY
C     ONE OF 32 DIFFERENT DIFFERENTIAL EQUATIONS.  OR, IN PLACE
C     OF XAMPLES.F, ANY OTHER STURM-LIOUVILLE DIFFERENTIAL
C     EQUATION CAN BE EMPLOYED.  THE EASIEST WAY TO CREATE A FILE
C     CONTAINING AN ARBITRARY DIFFERENTIAL EQUATION, IN PLACE OF
C     THE FILE XAMPLES.F, IS BY USING THE INTERACTIVE PROGRAM
C     MAKEPQW.F.
C          IN EITHER CASE, AS SOON AS THE "EXECUTABLE" IS ACTIVATED,
C     THIS PROGRAM DISPLAYS PROMPTS ON THE SCREEN WHICH INVITE THE
C     USER TO SUPPLY, VIA THE KEYBOARD, THE DATA WHICH DEFINE THE
C     PARTICULAR EIGENVALUE PROBLEM WANTED.  DATA SUCH AS
C        THE INTERVAL (a,b),
C        WHETHER THE ENDPOINTS ARE REGULAR, LIMIT CIRCLE, LIMIT
C          POINT, ETC,
C        WHAT KIND OF BOUNDARY CONDITIONS ARE WANTED AT EACH END,
C        WHICH EIGENVALUES ARE WANTED,
C        AND WHETHER OR NOT A PLOT OF AN EIGENFUNCTION IS WANTED.
C        ETC.
C
C     THERE IS ALSO ANOTHER WAY OF USING THIS DRIVER, WHICH AVOIDS
C     THE "QUESTION & ANSWER" FORMAT, IF A USER WOULD PREFER.  IT
C     REQUIRES PUTTING THE PROBLEM DATA (SUCH AS a, b, Regular,
C     Limit Circle, Limit Point, Boundary Conditions, etc.) IN A
C     VERY BRIEF TEXT FILE, CALLED auto.in .  FOR MORE DETAILS
C     ABOUT THIS METHOD, SEE THE COMMENTS AT THE BEGINNING OF THE
C     SUBROUTINE AUTO() WHICH IS IN THE SAME FILE AS THE ONE
C     CONTAINING THIS DRIVER.  THIS "AUTOMATIC" MODE IS, OF COURSE,
C     MUCH FASTER, BUT PROBABLY SHOULD BE USED ONLY BY SOMEONE WHO
C     IS ALREADY FAMILIAR WITH USING THE MORE USUAL "QUESTION &
C     ANSWER" MODE.
C
C     .. Scalars in Common ..
      REAL A,A1,A1S,A2,A2S,AA,ALF,ALFA1,ALFA2,ALPH0,
     +     ASAV,B,B1,B1S,B2,
     +     B2S,BB,BETA0,BETA1,BETA2,BSAV,DTHDAA,DTHDBB,EIG,EIGSAV,
     +     EPSMIN,FA,FB,GAMM0,GQA,GQB,GWA,GWB,H0,HPI,K0,K11,K12,K21,K22,
     +     L0,LPQA,LPQB,LPWA,LPWB,NU0,P0ATA,P0ATAS,P0ATB,P0ATBS,P10,P20,
     +     P30,P40,P50,P60,PI,QFATA,QFATAS,QFATB,QFATBS,SLF9,TMID,TOL,
     +     TOLL,TSAVEL,TSAVER,TWOPI,Z
      INTEGER ILAST,IND,INTAB,INTSAV,ISAVE,ISLFN,JFLAG,MDTHZ,MFS,MLS,
     +        MMWD,NCA,NCB,NEIG1,NEIG2,NEND,NF,NIVP,NLAST,NUMB0,NUMEIG,
     +        NV,T21,T22,T23,T24,T25
      LOGICAL ADDD,LCA,LCB,PEIGF,PR,REGA,REGB,RITE
      CHARACTER*9 INFM,INFP
      CHARACTER*19 FILLA,FILLB,FILLC
      CHARACTER*32 CH1,CH2,CH3,CH4,CH5,CH6
C     ..
C     .. Arrays in Common ..
      REAL EES(50),SLFUN(9),TEE(100),TT(7,2),TTS(50),
     +     YS(200),YY(7,3,2),
     +     ZEE(100)
      INTEGER IIS(50),JAY(100),MMW(100),NT(2)
      CHARACTER*39 BLNK(2),STAR(2),STR(2)
      CHARACTER*55 XC(8)
C     ..
C     .. Local Scalars ..
      REAL ONE
      INTEGER I,IFLAG,NEIGS,NTMP,NUMB,RESP
      LOGICAL EIGV,PERIOD,SKIPB
      CHARACTER*32 TAPE23
C     ..
C     .. Local Arrays ..
      REAL PLOTF(1000,6),XT(1000,2)
C     ..
C     .. External Subroutines ..
      EXTERNAL AUTO,DRAW,DSPLAY,EXAMP,PERIO,QPLOT,SLEIGN,STAGE
C     ..
C     .. Intrinsic Functions ..
      INTRINSIC ATAN,MIN
C     ..
C     .. Common blocks ..
      COMMON /ALBE/LPWA,LPQA,FA,GWA,GQA,LPWB,LPQB,FB,GWB,GQB
      COMMON /BCDATA/A1S,A2S,P0ATAS,QFATAS,B1S,B2S,P0ATBS,QFATBS
      COMMON /BCONDS/A1,A2,B1,B2,ALFA1,ALFA2,BETA1,BETA2,NUMEIG,EIG,TOL,
     +       TOLL,NEIG1,NEIG2,ALF,K11,K12,K21,K22
      COMMON /DATADT/ASAV,BSAV,INTSAV
      COMMON /DATAF/EIGSAV,IND
      COMMON /EIGS/EES,TTS,IIS,ILAST,NLAST,JFLAG,SLF9,NIVP,NEND,PEIGF,
     +       SLFUN,RITE,ISLFN,NF,NV
      COMMON /FLAG/NUMB0
      COMMON /LP/MFS,MLS
      COMMON /PAR/NU0,H0,K0,L0,ALPH0,BETA0,GAMM0,P10,P20,P30,P40,P50,P60
      COMMON /PARAM/INTAB,A,NCA,P0ATA,QFATA,B,NCB,P0ATB,QFATB,REGA,LCA,
     +       REGB,LCB
      COMMON /PASS/YS,MMW,MMWD
      COMMON /PIE/PI,TWOPI,HPI
      COMMON /PRIN/PR,T21
      COMMON /RNDOFF/EPSMIN
      COMMON /SHAR/INFM,INFP,CH1,CH2,CH3,CH4,CH5,CH6,STAR,BLNK,STR,
     +       FILLA,FILLB,FILLC,XC
      COMMON /TAPES/T22,T23,T24,T25
      COMMON /TDATA/AA,TMID,BB,DTHDAA,DTHDBB,MDTHZ,ADDD
      COMMON /TEEZ/TEE
      COMMON /TEMP/TT,YY,NT
      COMMON /TSAVE/TSAVEL,TSAVER,ISAVE
      COMMON /Z1/Z
      COMMON /ZEEZ/JAY,ZEE
C     ..
      ONE = 1.0D0
      HPI = 2.0D0*ATAN(ONE)
      PI = 2.0D0*HPI
      TWOPI = 2.0D0*PI
C
      T21 = 21
      T22 = 22
      T23 = 23
      T24 = 24
      T25 = 25
C
C         DEFINITIONS OF SOME STRINGS.
C      (TO BE USED IN SUBROUTINE DSPLAY.)
C
      INFP = '+INFINITY'
      INFM = '-INFINITY'
      CH1 = 'REGULAR                       * '
      CH2 = 'WEAKLY REGULAR                * '
      CH3 = 'LIMIT CIRCLE, NON-OSCILLATORY * '
      CH4 = 'LIMIT CIRCLE, OSCILLATORY     * '
      CH5 = 'LIMIT POINT                   * '
      CH6 = 'UNSPEC.(NOT LCO), DEFAULT B.C.* '
      STAR(1) = ' **************************************'
      STAR(2) = '***************************************'
      BLNK(1) = ' *                                     '
      BLNK(2) = '                                      *'
      FILLA = '*******************'
      FILLB = '                  *'
      FILLC = '------------------ '
C
      XC(1) = ' * (1) THE SOLUTION Y                               * '
      XC(2) = ' * (2) THE QUASI-DERIVATIVE p*Y''                    *'
      XC(3) = ' * (3) THE BOUNDARY CONDITION FUNCTION Y OR [Y,U]   * '
      XC(4) = ' * (4) THE BOUNDARY CONDITION FUNCTION p*Y'' OR [Y,V]*'
      XC(5) = ' * (5) THE PRUFER ANGLE, THETA                      * '
      XC(6) = ' * (6) THE PRUFER MODULUS, RHO                      * '
      XC(7) = ' * (1) x IN THE INTERVAL (a,b)                      * '
      XC(8) = ' * (2) t IN THE INTERVAL (-1,1)                     * '
C
C
      OPEN (T21,FILE='test.out')
C
C     INTRODUCTION :-
      CALL DSPLAY(1,RESP)
      IF (RESP.EQ.0) CALL AUTO

      CALL EXAMP

C     IS MORE INFORMATION REQUIRED? :-
      CALL DSPLAY(2,RESP)

C     SHOULD THE RESULTS BE RECORDED IN A FILE? :=
      CALL DSPLAY(3,RESP)
C     IF SO, GET THE NAME OF THE FILE :-
      IF (RESP.EQ.1) CALL DSPLAY(4,RESP)

  100 CONTINUE
      SKIPB = .FALSE.
      P0ATA = -1.0D0
      QFATA = 1.0D0
      P0ATB = -1.0D0
      QFATB = 1.0D0
C     WHAT KIND OF INTERVAL IS THE PROBLEM ON? :-
      CALL DSPLAY(5,RESP)

  120 CONTINUE
C     GET ENDPOINT A, IF FINITE :-
      IF (INTAB.EQ.1 .OR. INTAB.EQ.2) CALL DSPLAY(6,RESP)

C     CLASSIFICATION OF A :-
      CALL DSPLAY(7,RESP)

C       (IS P0 AT A, OR QF AT A? :- )
      IF (.NOT.REGA .AND. INTAB.LE.2) CALL DSPLAY(8,RESP)

      IF (SKIPB) GO TO 150
  140 CONTINUE
C     (GET ENDPOINT B, IF FINITE :- )
      IF (INTAB.EQ.1 .OR. INTAB.EQ.3) CALL DSPLAY(9,RESP)

C     (CLASSIFICATION OF B :- )
      CALL DSPLAY(10,RESP)

C       (IS P0 AT B, OR QF AT B :- )
      IF (.NOT.REGB .AND. (INTAB.EQ.1.OR.INTAB.EQ.3)) CALL DSPLAY(11,
     +    RESP)

  150 CONTINUE
C     BRIEF SUMMARY OF PROBLEM PARAMETERS :-
C     (IS THIS CORRECT SO FAR?  :- )
      CALL DSPLAY(12,RESP)

C     (IF THIS IS NOT THE RIGHT PROBLEM, DO IT OVER :- )
      IF (RESP.NE.1) THEN
          CALL DSPLAY(13,RESP)
          IF (RESP.EQ.1) THEN
              SKIPB = .TRUE.
              GO TO 120
          ELSE IF (RESP.EQ.2) THEN
              GO TO 140
          ELSE IF (RESP.EQ.3) THEN
              GO TO 100
          END IF
      END IF

C ----------------------------------------------------C
C
C     AT THIS POINT THE DIFFERENTIAL EQUATION AND THE INTERVAL OF
C       INTEREST HAVE BEEN DEFINED AND CHARACTERIZED.
C

C     SHOULD WE COMPUTE AN EIVENVALUE? OR THE SOLUTION
C       TO AN INITIAL VALUE PROBLEM? :-
      CALL DSPLAY(14,RESP)
      EIGV = (RESP.EQ.1)
      IF (EIGV) THEN
          NIVP = 0
          NEND = 0
          PERIOD = .FALSE.
      END IF

  200 CONTINUE
      IF (EIGV .AND. NCA.LE.4 .AND. NCB.LE.4) THEN
C        (ARE THE BC'S SEPARATED?, OR COUPLED?)
          CALL DSPLAY(15,RESP)
          IF (RESP.EQ.1) THEN
C           (THIS MEANS THE BC'S ARE COUPLED)
              PERIOD = .TRUE.
          ELSE
              PERIOD = .FALSE.
          END IF
      END IF
C

  300 CONTINUE
C       SET BOUNDARY CONDITIONS, OR INITIAL CONDITIONS
      IF (EIGV) THEN
          IF (PERIOD) THEN
C           (SET COUPLED BC'S :- )
              CALL DSPLAY(18,RESP)

          ELSE
C           (SET SEPARATED BC'S AT A :- )
              CALL DSPLAY(16,RESP)

C           (SET SEPARATED BC'S AT B :- )
              CALL DSPLAY(17,RESP)
          END IF
      ELSE
C        (SET INITIAL CONDITIONS FOR IVP :- )
          PERIOD = .FALSE.
          CALL DSPLAY(19,RESP)
      END IF
C ----------------------------------------------------C
C
C     PRESUMABLY WE NOW HAVE ASSEMBLED ALL THE DATA
C     NEEDED FOR DEFINING
C          THE EIGENVALUE PROBLEM,
C                     OR
C          THE INITIAL VALUE PROBLEM,
C     WHICHEVER IS WANTED.

  400 CONTINUE
C          COMPUTE A SINGLE EIGENVALUE?, OR SERIES? :-
      NEIGS = 0
      IF (EIGV .AND. .NOT.PERIOD) THEN
          CALL DSPLAY(20,RESP)
          IF (RESP.EQ.1) THEN
              CALL DSPLAY(21,RESP)
              NEIGS = 1
          ELSE
              CALL DSPLAY(23,RESP)
              NEIGS = NEIG2 - NEIG1 + 1
          END IF
      ELSE IF (EIGV) THEN
          CALL DSPLAY(24,RESP)
          IF (RESP.EQ.1) THEN
              CALL DSPLAY(25,RESP)
              NEIGS = 1
          ELSE
              CALL DSPLAY(27,RESP)
              NEIGS = NEIG2 - NEIG1 + 1
              EIG = 0.0D0
          END IF
      END IF
C
      IF (EIGV) THEN
C        (COMPUTE REQUESTED EIGENVALUES)

          IF (NEIGS.EQ.1) THEN
              NEIG1 = NUMEIG
              NEIG2 = NUMEIG
              NLAST = NUMEIG
          END IF
          ILAST = NEIG2 - NEIG1 + 1

          IF (.NOT.PERIOD) THEN
              DO 410 I = 1,ILAST
                  NUMB = NEIG1 + I - 1
                  TOL = TOLL
                  NTMP = NUMB
                  IF (NEIGS.GT.1) EIG = 0.0D0
                  PEIGF = .FALSE.

                  CALL SLEIGN(A,B,INTAB,P0ATA,QFATA,P0ATB,QFATB,A1,A2,
     +                        B1,B2,NTMP,EIG,TOL,IFLAG,0,SLFUN,NCA,NCB)

                  IF (IFLAG.EQ.0 .OR. IFLAG.GE.15) THEN
                      IF (IFLAG.EQ.0) THEN
                          WRITE (*,FMT=*) ' IMPROPER INPUT PARAMETERS '
                          IF (RITE) THEN
                              WRITE (T22,FMT=*)
     +                          ' IMPROPER INPUT PARAMETERS '
                              WRITE (T22,FMT=*)
     +                        '----------------------------------------'
                          END IF
                      ELSE IF (IFLAG.EQ.15) THEN
                          WRITE (*,FMT=*)
     +                      ' WE CANNOT HANDLE THIS KIND OF ENDPOINT '
                          IF (RITE) THEN
                              WRITE (T22,FMT=*)
     +                        ' WE CANNOT HANDLE THIS KIND OF ENDPOINT '
                              WRITE (T22,FMT=*)
     +                        '----------------------------------------'
                          END IF
                      ELSE IF (IFLAG.EQ.16) THEN
                          WRITE (*,FMT=*) ' COULD NOT GET STARTED '
                          IF (RITE) THEN
                              WRITE (T22,FMT=*)
     +                          ' COULD NOT GET STARTED '
                              WRITE (T22,FMT=*)
     +                        '----------------------------------------'
                          END IF
                      ELSE IF (IFLAG.EQ.17) THEN
                          WRITE (*,FMT=*) ' FAILED TO GET A BRACKET '
                          IF (RITE) THEN
                              WRITE (T22,FMT=*)
     +                          ' FAILED TO GET A BRACKET '
                              WRITE (T22,FMT=*)
     +                        '----------------------------------------'
                          END IF
                      ELSE IF (IFLAG.EQ.18) THEN
                          WRITE (*,FMT=*) ' ESTIMATOR FAILED '
                          IF (RITE) THEN
                              WRITE (T22,FMT=*) ' ESTIMATOR FAILED '
                              WRITE (T22,FMT=*)
     +                        '----------------------------------------'
                          END IF
                      END IF
                      GO TO 700
                  END IF

                  IFLAG = MIN(IFLAG,4)
                  JFLAG = 0

C   JFLAG = 1 OR 2 MEANS ONE OF THE END-POINTS IS LP .
C   JFLAG = 1 MEANS THERE IS NO CONTINUOUS SPECTRUM .
C         = 2 MEANS THERE IS A CONTINUOUS SPECTRUM .

                  SLF9 = SLFUN(9)
                  IF (SLF9.GT.-10000.0D0) JFLAG = 1
                  IF (SLF9.LT.10000.0D0 .AND. JFLAG.EQ.1) JFLAG = 2

                  EES(I) = EIG
                  TTS(I) = TOL
                  IIS(I) = IFLAG

                  IF (IFLAG.EQ.3) THEN
C  (THIS MEANS THAT THERE IS NO EIGENVALUE )
C  (  WITH THIS INDEX                      )
                      ILAST = I
                      NLAST = NTMP
                      GO TO 420
                  END IF
  410         CONTINUE
  420         CONTINUE
              CALL DSPLAY(28,RESP)

          ELSE

              DO 430 I = 1,ILAST
                  NUMEIG = NEIG1 + I - 1
                  TOL = TOLL
                  EIG = 0.0D0
                  CALL PERIO(A,B,INTAB,P0ATA,QFATA,P0ATB,QFATB,A1,A2,B1,
     +                       B2,NUMEIG,EIG,TOL,IFLAG,SLFUN,NCA,NCB,ALF,
     +                       K11,K12,K21,K22)
                  EES(I) = EIG
                  TTS(I) = TOL
                  IIS(I) = IFLAG
C      (IFLAG = 2 MEANS THAT THE EIGENVALUE IS A DOUBLE)
  430         CONTINUE
              CALL DSPLAY(29,RESP)

          END IF
      END IF
C                                                              C
C -------------------------------------------------------------C
  500 CONTINUE
C
C  WE SHOULD NOW HAVE THE EIGENVALUES COMPUTED, OR HAVE        C
C  THE INITIAL VALUE PROBLEM DEFINED, AS THE CASE MAY BE.      C
C  IF ONLY ONE EIGENVALUE HAS BEEN COMPUTED, THE CORRESPONDING C
C  EIGENFUNCTION CAN NOW BE PLOTTED.  OR IF AN INITIAL VALUE   C
C  PROBLEM HAS BEEN DEFINED, IT MAY BE PLOTTED NOW.            C
C
      IF (.NOT.EIGV) THEN
          CALL DSPLAY(32,RESP)
          CALL STAGE(A,B,INTAB,P0ATA,QFATA,P0ATB,QFATB,A1,A2,B1,B2,EIG,
     +               IFLAG,SLFUN,NCA,NCB,NIVP,NEND)
      ELSE IF (NEIGS.EQ.1 .AND. (IFLAG.EQ.1.OR.IFLAG.EQ.2)) THEN
          IF (.NOT.PERIOD) THEN
              CALL DSPLAY(22,RESP)
          ELSE
              CALL DSPLAY(26,RESP)
          END IF
      ELSE IF (IFLAG.EQ.0) THEN
          PEIGF = .FALSE.
      END IF

      IF (PEIGF) THEN
          CALL DRAW(A1,A2,B1,B2,NUMEIG,EIG,SLFUN,NIVP,NEND,EIGV,NCA,NCB,
     +              ISLFN,XT,PLOTF,K11,K12,K21,K22,PERIOD)
  600     CONTINUE
C         PLOT WHICH FUNCTION ?
          CALL DSPLAY(33,RESP)
          IF (RESP.EQ.1) CALL QPLOT(ISLFN,XT,NV,PLOTF,NF)
C         SAVE THE PLOT FILE ?
          CALL DSPLAY(34,RESP)
          IF (RESP.EQ.1) THEN
              WRITE (*,FMT=*) ' SPECIFY NAME OF FILE FOR PLOTTING '
              READ (*,FMT='(A)') TAPE23
              OPEN (T23,FILE=TAPE23,STATUS='NEW')
              DO 610 I = 1,ISLFN
                  WRITE (T23,FMT=*) XT(9+I,NV),PLOTF(9+I,NF)
  610         CONTINUE
              CLOSE (T23)
              WRITE (*,FMT=*) ' THE PLOT FILE HAS BEEN WRITTEN TO ',
     +          TAPE23
              IF (RITE) WRITE (T22,FMT=*)
     +            ' THE PLOT FILE HAS BEEN WRITTEN TO ',TAPE23
          END IF
C         PLOT ANOTHER FUNCTION ?
          CALL DSPLAY(35,RESP)
          IF (RESP.EQ.1) GO TO 600
      END IF

  700 CONTINUE
C  ARE WE FINISHED?, OR DO WE HAVE MORE PROBLEMS TO DO?         C
C
      IF (EIGV) THEN
          CALL DSPLAY(30,RESP)
          IF (RESP.EQ.1) GO TO 400
          IF (RESP.EQ.2) GO TO 200
          IF (RESP.EQ.3) GO TO 100
          EIGV = .FALSE.
          IF (RESP.EQ.4) GO TO 300
      ELSE
          CALL DSPLAY(31,RESP)
          IF (RESP.EQ.1) GO TO 500
          IF (RESP.EQ.2) GO TO 300
          IF (RESP.EQ.3) GO TO 100
          EIGV = .TRUE.
          IF (RESP.EQ.4) GO TO 200
      END IF
C
      CLOSE (T22)
      CLOSE (T21)
      STOP
      END
C
      SUBROUTINE DSPLAY(DIS,RESP)
C     .. Scalars in Common ..
      REAL A,A1,A2,ALF,ALFA1,ALFA2,B,B1,B2,BETA1,BETA2,
     +     EIG,HPI,K11,K12,
     +     K21,K22,P0ATA,P0ATB,PI,QFATA,QFATB,SLF9,TOL,TOLL,TWOPI
      INTEGER ILAST,INTAB,ISLFN,JFLAG,NCA,NCB,NEIG1,NEIG2,NEND,NF,NIVP,
     +        NLAST,NUMEIG,NV,T22,T23,T24,T25
      LOGICAL LCA,LCB,PEIGF,REGA,REGB,RITE
      CHARACTER*9 INFM,INFP
      CHARACTER*19 FILLA,FILLB,FILLC
      CHARACTER*32 CH1,CH2,CH3,CH4,CH5,CH6
C     ..
C     .. Local Scalars ..
      REAL CC,DETK,R1,R2,RHO,THA,THB,TMP
      INTEGER I,I1,I2,IFLAG,NANS
      LOGICAL DUBBLE,YEH
      CHARACTER ANSCH,HQ,YN
      CHARACTER*32 CHA,CHANS,CHB,FMAT,TAPE22
      CHARACTER*70 CHTXT
C     ..
C     .. Local Arrays ..
      INTEGER ICOL(2)
      CHARACTER*2 COL(32)
C     ..
C     .. External Subroutines ..
      EXTERNAL HELP,LST,LSTDIR
C     ..
C     .. External Functions ..
      CHARACTER*32 FMT2
      EXTERNAL FMT2
C     ..
C     .. Intrinsic Functions ..
      INTRINSIC ABS,ATAN2,COS,SIN,SQRT
C     ..

C
C     .. Scalar Arguments ..
      INTEGER DIS,RESP
C     ..
C     .. Arrays in Common ..
      REAL EES(50),SLFUN(9),TTS(50)
      INTEGER IIS(50)
      CHARACTER*39 BLNK(2),STAR(2),STR(2)
      CHARACTER*55 XC(8)
C     ..
C     .. Common blocks ..
      COMMON /BCONDS/A1,A2,B1,B2,ALFA1,ALFA2,BETA1,BETA2,NUMEIG,EIG,TOL,
     +       TOLL,NEIG1,NEIG2,ALF,K11,K12,K21,K22
      COMMON /EIGS/EES,TTS,IIS,ILAST,NLAST,JFLAG,SLF9,NIVP,NEND,PEIGF,
     +       SLFUN,RITE,ISLFN,NF,NV
      COMMON /PARAM/INTAB,A,NCA,P0ATA,QFATA,B,NCB,P0ATB,QFATB,REGA,LCA,
     +       REGB,LCB
      COMMON /PIE/PI,TWOPI,HPI
      COMMON /SHAR/INFM,INFP,CH1,CH2,CH3,CH4,CH5,CH6,STAR,BLNK,STR,
     +       FILLA,FILLB,FILLC,XC
      COMMON /TAPES/T22,T23,T24,T25
C     ..
C     .. Data statements ..
      DATA COL/'01','02','03','04','05','06','07','08','09','10','11',
     +     '12','13','14','15','16','17','18','19','20','21','22','23',
     +     '24','25','26','27','28','29','30','31','32'/
C     ..
      RESP = 1
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,
     +       23,24,25,26,27,28,29,30,31,32,33,34,35) DIS
C
    1 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' This program solves the boundary value problem '
      WRITE (*,FMT=*) '   defined by the differential equation '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) '   -(py'')'' + q*y = lambda*w*y  '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' together with appropriate boundary conditions. '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
     +  ' HELP may be called at any point where the program   '
      WRITE (*,FMT=*)
     +  '    halts and displays (h?) by pressing "h <ENTER>". '
      WRITE (*,FMT=*)
     +  '    To RETURN from HELP, press "r <ENTER>".          '
      WRITE (*,FMT=*)
     +  '    To QUIT at any program halt, press "q <ENTER>".  '
      WRITE (*,FMT=*)
     +  ' WOULD YOU LIKE AN OVERVIEW OF HELP ? (Y/N) (h?) '
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H' .OR. HQ.EQ.'y' .OR.
     +    HQ.EQ.'Y') CALL HELP(1)
      WRITE (*,FMT=*)
      YEH = HQ .EQ. 'y' .OR. HQ .EQ. 'Y' .OR. HQ .EQ. 'n' .OR.
     +      HQ .EQ. 'N' .OR. HQ .EQ. 'a' .OR. HQ .EQ. 'A'
      IF (.NOT.YEH) GO TO 1
      IF (HQ.EQ.'a' .OR. HQ.EQ.'A') RESP = 0
      RETURN
C
    2 CONTINUE
      WRITE (*,FMT=*) ' DO YOU REQUIRE INFORMATION ON THE RANGE OF '
      WRITE (*,FMT=*) '    BOUNDARY CONDITIONS AVAILABLE ? (Y/N) (h?) '
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H' .OR. HQ.EQ.'y' .OR.
     +    HQ.EQ.'Y') CALL HELP(7)
      YEH = HQ .EQ. 'y' .OR. HQ .EQ. 'Y' .OR. HQ .EQ. 'n' .OR.
     +      HQ .EQ. 'N' .OR. HQ .EQ. 'h' .OR. HQ .EQ. 'H'
      IF (.NOT.YEH) GO TO 2
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') CALL HELP(7)
      RETURN
C
    3 CONTINUE
  105 CONTINUE
      WRITE (*,FMT=*) ' DO YOU WANT A RECORD KEPT OF THE PROBLEMS '
      WRITE (*,FMT=*) '    AND RESULTS ? (Y/N) (h?) '
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      YEH = HQ .EQ. 'y' .OR. HQ .EQ. 'Y' .OR. HQ .EQ. 'n' .OR.
     +      HQ .EQ. 'N' .OR. HQ .EQ. 'h' .OR. HQ .EQ. 'H'
      IF (.NOT.YEH) GO TO 105
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(8)
          GO TO 105
      END IF
      RITE = HQ .EQ. 'y' .OR. HQ .EQ. 'Y' .OR. HQ .EQ. 'yes' .OR.
     +       HQ .EQ. 'YES'
      IF (.NOT.RITE) RESP = 0
      RETURN
C
    4 CONTINUE
  205 CONTINUE
      WRITE (*,FMT=*) '  SPECIFY NAME OF THE OUTPUT RECORD FILE: (h?) '
      READ (*,FMT=9010) CHANS
      IF (CHANS.EQ.'q' .OR. CHANS.EQ.'Q') THEN
          STOP
      ELSE IF (CHANS.EQ.'h' .OR. CHANS.EQ.'H') THEN
          CALL HELP(8)
          GO TO 205
      ELSE
          TAPE22 = CHANS
          WRITE (*,FMT=*) '  YOU MAY ENTER SOME HEADER LINE FOR THE '
          WRITE (*,FMT=*) '  OUTPUT RECORD FILE (<=70 CHARACTERS)   '
          WRITE (*,FMT=*) '  IF YOU WISH; OTHERWISE JUST HIT "RETURN".'
          WRITE (*,FMT=*)
          READ (*,FMT=9030) CHTXT
      END IF

C
      OPEN (T22,FILE=TAPE22,STATUS='NEW')
C
      WRITE (T22,FMT=*) '                         ',TAPE22
      WRITE (T22,FMT=*)
      WRITE (T22,FMT=*) CHTXT
      WRITE (T22,FMT=*)
C
      RETURN
C
    5 CONTINUE
  405 CONTINUE
      WRITE (*,FMT=*)
     +  ' ************************************************** '
      WRITE (*,FMT=*)
     +  ' * INDICATE THE KIND OF PROBLEM INTERVAL (a,b):   * '
      WRITE (*,FMT=*)
     +  ' *                                                * '
      WRITE (*,FMT=*)
     +  ' *  (CHECK THAT THE COEFFICIENTS p,q,w ARE WELL   * '
      WRITE (*,FMT=*)
     +  ' *  DEFINED THROUGHOUT THE INTERVAL open(a,b).)   * '
      WRITE (*,FMT=*)
     +  ' *                                                * '
      WRITE (*,FMT=*)
     +  ' *    (1) FINITE, (a,b)                           * '
      WRITE (*,FMT=*)
     +  ' *                                                * '
      WRITE (*,FMT=*)
     +  ' *    (2) SEMI-INFINITE, (a,+INFINITY)            * '
      WRITE (*,FMT=*)
     +  ' *                                                * '
      WRITE (*,FMT=*)
     +  ' *    (3) SEMI-INFINITE, (-INFINITY,b)            * '
      WRITE (*,FMT=*)
     +  ' *                                                * '
      WRITE (*,FMT=*)
     +  ' *    (4) DOUBLY INFINITE, (-INFINITY,+INFINITY)  * '
      WRITE (*,FMT=*)
     +  ' *                                                * '
      WRITE (*,FMT=*)
     +  ' * ENTER THE NUMBER OF YOUR CHOICE: (h?)          * '
      WRITE (*,FMT=*)
     +  ' ************************************************** '
      WRITE (*,FMT=*)
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(9)
          GO TO 405
      END IF
      YEH = HQ .EQ. '1' .OR. HQ .EQ. '2' .OR. HQ .EQ. '3' .OR.
     +      HQ .EQ. '4'
      IF (.NOT.YEH) GO TO 405
      READ (CHANS,FMT='(I32)') INTAB
      IF (RITE) THEN
          IF (INTAB.EQ.1) WRITE (T22,FMT=*) ' The interval is (a,b) .'
          IF (INTAB.EQ.2) WRITE (T22,FMT=*) ' The interval is (a,+inf).'
          IF (INTAB.EQ.3) WRITE (T22,FMT=*
     +        ) ' The interval is (-inf,b). '
          IF (INTAB.EQ.4) WRITE (T22,FMT=*
     +        ) ' The interval is (-inf,+inf).'
      END IF
      RESP = INTAB
      RETURN
C
    6 CONTINUE
   60 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' ********************************************* '
      WRITE (*,FMT=*) ' * INPUT a: (h?)                             * '
      WRITE (*,FMT=*) ' ********************************************* '
      WRITE (*,FMT=*) ' a = '
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(10)
          GO TO 60
      END IF
      CALL LST(CHANS,A)
      IF (RITE) WRITE (T22,FMT=*) ' a = ',A
      IF (INTAB.EQ.2) B = A + 1.0D0

      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      RETURN
C
    7 CONTINUE
   70 CONTINUE
      STR(1) = ' * IS THIS PROBLEM:                    '
      STR(2) = '                                      *'
      WRITE (*,FMT=9110) STAR
      WRITE (*,FMT=9110) STR
      WRITE (*,FMT=9110) BLNK
      STR(1) = ' *   (1) REGULAR AT a ?                '
      STR(2) = '                                      *'
      WRITE (*,FMT=9110) STR
      STR(1) = ' *       (I.E., THE FUNCTIONS p, q, & w'
      STR(2) = ' ARE BOUNDED CONTINUOUS NEAR a;       *'
      WRITE (*,FMT=9110) STR
      STR(1) = ' *        p & w ARE POSITIVE AT a.)    '
      STR(2) = '                                      *'
      WRITE (*,FMT=9110) STR
      WRITE (*,FMT=9110) BLNK
      STR(1) = ' *   (2) WEAKLY REGULAR AT a ?         '
      STR(2) = '                                      *'
      WRITE (*,FMT=9110) STR
      STR(1) = ' *       (I.E., THE FUNCTIONS 1/p, q, &'
      STR(2) = ' w ALL ARE FINITELY INTEGRABLE ON     *'
      WRITE (*,FMT=9110) STR
      STR(1) = ' *        SOME INTERVAL [a,a+e] FOR e >'
      STR(2) = ' 0; p & w ARE POSITIVE NEAR a.)       *'
      WRITE (*,FMT=9110) STR
      WRITE (*,FMT=9110) BLNK
      STR(1) = ' *   (3) LIMIT CIRCLE, NON-OSCILLATORY '
      STR(2) = 'AT a ?                                *'
      WRITE (*,FMT=9110) STR
      WRITE (*,FMT=9110) BLNK
      STR(1) = ' *   (4) LIMIT CIRCLE, OSCILLATORY AT a'
      STR(2) = ' ?                                    *'
      WRITE (*,FMT=9110) STR
      WRITE (*,FMT=9110) BLNK
      STR(1) = ' *   (5) LIMIT POINT AT a ?            '
      STR(2) = '                                      *'
      WRITE (*,FMT=9110) STR
      WRITE (*,FMT=9110) BLNK
      STR(1) = ' *   (6) NOT SPECIFIED (BUT NOT LIMIT C'
      STR(2) = 'IRCLE OSCILLATORY) WITH DEFAULT       *'
      WRITE (*,FMT=9110) STR
      STR(1) = ' *       BOUNDARY CONDITION AT a ?     '
      STR(2) = '                                      *'
      WRITE (*,FMT=9110) STR
      WRITE (*,FMT=9110) BLNK
      STR(1) = ' * ENTER THE NUMBER OF YOUR CHOICE: (h)'
      STR(2) = '?                                     *'
      WRITE (*,FMT=9110) STR
      WRITE (*,FMT=9110) STAR
      WRITE (*,FMT=*)
C
C     SPECIFY TYPE OF BOUNDARY CONDITION AT a.
C
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(4)
          GO TO 70
      END IF
      YEH = HQ .EQ. '1' .OR. HQ .EQ. '2' .OR. HQ .EQ. '3' .OR.
     +      HQ .EQ. '4' .OR. HQ .EQ. '5' .OR. HQ .EQ. '6'
      IF (.NOT.YEH) GO TO 70
      READ (CHANS,FMT='(I32)') NANS
C
C     SET CHARACTER STRING CHA ACCORDING TO BOUNDARY CONDITION AT a.
C
      IF (NANS.EQ.1) THEN
          REGA = .TRUE.
          NCA = 1
          P0ATA = -1.0D0
          QFATA = 1.0D0
          CHA = CH1
          IF (RITE) WRITE (T22,FMT=*) ' Endpoint a is Regular. '
      ELSE IF (NANS.EQ.2) THEN
          NCA = 2
          CHA = CH2
          IF (RITE) WRITE (T22,FMT=*) ' Endpoint a is Weakly Regular. '
      ELSE IF (NANS.EQ.3) THEN
          LCA = .TRUE.
          CHA = CH3
          NCA = 3
          IF (RITE) WRITE (T22,FMT=*) ' Endpoint a is Limit Circle,',
     +        ' Non-Oscillatory. '
      ELSE IF (NANS.EQ.4) THEN
          LCA = .TRUE.
          CHA = CH4
          NCA = 4
          IF (RITE) WRITE (T22,FMT=*) ' Endpoint a is Limit Circle,',
     +        ' Oscillatory. '
      ELSE IF (NANS.EQ.5) THEN
          CHA = CH5
          NCA = 5
          IF (RITE) WRITE (T22,FMT=*) ' Endpoint a is Limit Point. '
      ELSE
          CHA = CH6
          NCA = 6
          IF (RITE) WRITE (T22,FMT=*)
     +        ' Endpoint a is Singular, unspecified. '
      END IF
      RESP = NANS
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      RETURN
C
    8 CONTINUE
   80 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' ********************************************* '
      WRITE (*,FMT=*) ' * IS p = 0. AT a ? (Y/N) (h?)               * '
      WRITE (*,FMT=*) ' ********************************************* '
      WRITE (*,FMT=*)
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(11)
          GO TO 80
      END IF
      YEH = HQ .EQ. 'y' .OR. HQ .EQ. 'Y' .OR. HQ .EQ. 'n' .OR.
     +      HQ .EQ. 'N'
      IF (.NOT.YEH) GO TO 80
      YEH = HQ .EQ. 'y' .OR. HQ .EQ. 'Y'
      P0ATA = -1.0D0
      IF (YEH) P0ATA = 1.0D0
      IF (RITE) THEN
          IF (YEH) THEN
              WRITE (T22,FMT=*) ' p is zero at a. '
          ELSE
              WRITE (T22,FMT=*) ' p is not zero at a. '
          END IF
      END IF

   90 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' ********************************************* '
      WRITE (*,FMT=*) ' * IS EITHER OF THE COEFFICIENT FUNCTIONS    * '
      WRITE (*,FMT=*) ' *                                           * '
      WRITE (*,FMT=*) ' *  q OR w UNBOUNDED NEAR a ? (Y/N) (h?)     * '
      WRITE (*,FMT=*) ' ********************************************* '
      WRITE (*,FMT=*)
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(11)
          GO TO 90
      END IF
      YEH = HQ .EQ. 'y' .OR. HQ .EQ. 'Y' .OR. HQ .EQ. 'n' .OR.
     +      HQ .EQ. 'N'
      IF (.NOT.YEH) GO TO 90
      YEH = HQ .EQ. 'y' .OR. HQ .EQ. 'Y'
      QFATA = 1.0D0
      IF (YEH) QFATA = -1.0D0
      IF (RITE) THEN
          IF (YEH) THEN
              WRITE (T22,FMT=*) ' either q or w is unbounded near a. '
          ELSE
              WRITE (T22,FMT=*) ' both q and w are bounded near a.  '
          END IF
      END IF
      IF (.NOT.YEH) RESP = 0
      RETURN
C
    9 CONTINUE
  110 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' ********************************************* '
      WRITE (*,FMT=*) ' * INPUT b: (h?)                             * '
      WRITE (*,FMT=*) ' ********************************************* '
      WRITE (*,FMT=*) ' b = '
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(10)
          GO TO 110
      END IF
      CALL LST(CHANS,B)
      IF (RITE) WRITE (T22,FMT=*) ' b = ',B
      IF (INTAB.EQ.3) A = B - 1.0D0
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      RETURN
C
   10 CONTINUE
  120 CONTINUE
      STR(1) = ' * IS THIS PROBLEM:                    '
      STR(2) = '                                      *'
      WRITE (*,FMT=9110) STAR
      WRITE (*,FMT=9110) STR
      WRITE (*,FMT=9110) BLNK
      STR(1) = ' *   (1) REGULAR AT b ?                '
      STR(2) = '                                      *'
      WRITE (*,FMT=9110) STR
      STR(1) = ' *       (I.E., THE FUNCTIONS p, q, & w'
      STR(2) = ' ARE BOUNDED CONTINUOUS NEAR b;       *'
      WRITE (*,FMT=9110) STR
      STR(1) = ' *        p & w ARE POSITIVE AT b.)    '
      STR(2) = '                                      *'
      WRITE (*,FMT=9110) STR
      WRITE (*,FMT=9110) BLNK
      STR(1) = ' *   (2) WEAKLY REGULAR AT b ?         '
      STR(2) = '                                      *'
      WRITE (*,FMT=9110) STR
      STR(1) = ' *       (I.E., THE FUNCTIONS 1/p, q, &'
      STR(2) = ' w ALL ARE FINITELY INTEGRABLE ON     *'
      WRITE (*,FMT=9110) STR
      STR(1) = ' *        SOME INTERVAL [b-e,b] FOR e >'
      STR(2) = ' 0; p & w ARE POSITIVE NEAR b.)       *'
      WRITE (*,FMT=9110) STR
      WRITE (*,FMT=9110) BLNK
      STR(1) = ' *   (3) LIMIT CIRCLE, NON-OSCILLATORY '
      STR(2) = 'AT b ?                                *'
      WRITE (*,FMT=9110) STR
      WRITE (*,FMT=9110) BLNK
      STR(1) = ' *   (4) LIMIT CIRCLE, OSCILLATORY AT b'
      STR(2) = ' ?                                    *'
      WRITE (*,FMT=9110) STR
      WRITE (*,FMT=9110) BLNK
      STR(1) = ' *   (5) LIMIT POINT AT b ?            '
      STR(2) = '                                      *'
      WRITE (*,FMT=9110) STR
      WRITE (*,FMT=9110) BLNK
      STR(1) = ' *   (6) NOT SPECIFIED (BUT NOT LIMIT C'
      STR(2) = 'IRCLE OSCILLATORY) WITH DEFAULT       *'
      WRITE (*,FMT=9110) STR
      STR(1) = ' *       BOUNDARY CONDITION AT b ?     '
      STR(2) = '                                      *'
      WRITE (*,FMT=9110) STR
      WRITE (*,FMT=9110) BLNK
      STR(1) = ' * ENTER THE NUMBER OF YOUR CHOICE: (h?'
      STR(2) = ')                                     *'
      WRITE (*,FMT=9110) STR
      WRITE (*,FMT=9110) STAR
      WRITE (*,FMT=*)
C
C     SPECIFY TYPE OF BOUNDARY CONDITION AT b.
C
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(4)
          GO TO 120
      END IF
      YEH = HQ .EQ. '1' .OR. HQ .EQ. '2' .OR. HQ .EQ. '3' .OR.
     +      HQ .EQ. '4' .OR. HQ .EQ. '5' .OR. HQ .EQ. '6'
      IF (.NOT.YEH) GO TO 120
      READ (CHANS,FMT='(I32)') NANS
C
C     SET CHARACTER STRING CHB ACCORDING TO BOUNDARY CONDITION AT b.
C
      WRITE (*,FMT=*)
      IF (NANS.EQ.1) THEN
          REGB = .TRUE.
          CHB = CH1
          NCB = 1
          P0ATB = -1.0D0
          QFATB = 1.0D0
          IF (RITE) WRITE (T22,FMT=*) ' Endpoint b is Regular. '
      ELSE IF (NANS.EQ.2) THEN
          CHB = CH2
          NCB = 2
          IF (RITE) WRITE (T22,FMT=*) ' Endpoint b is Weakly Regular. '
      ELSE IF (NANS.EQ.3) THEN
          LCB = .true.
          CHB = CH3
          NCB = 3
          IF (RITE) WRITE (T22,FMT=*) ' Endpoint b is Limit Circle,',
     +        ' Non-Oscillatory. '
      ELSE IF (NANS.EQ.4) THEN
          LCB = .true.
          CHB = CH4
          NCB = 4
          IF (RITE) WRITE (T22,FMT=*) ' Endpoint b is Limit Circle,',
     +        ' Oscillatory. '
      ELSE IF (NANS.EQ.5) THEN
          CHB = CH5
          NCB = 5
          IF (RITE) WRITE (T22,FMT=*) ' Endpoint b is Limit Point. '
      ELSE
          CHB = CH6
          NCB = 6
          IF (RITE) WRITE (T22,FMT=*)
     +        ' Endpoint b is Singular, unspecified. '
      END IF
      RESP = NANS
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      RETURN
C
   11 CONTINUE
  130 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' ********************************************* '
      WRITE (*,FMT=*) ' * IS p = 0. AT b ? (Y/N) (h?)               * '
      WRITE (*,FMT=*) ' ********************************************* '
      WRITE (*,FMT=*)
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(11)
          GO TO 130
      END IF
      READ (CHANS,FMT=9020) YN
      YEH = YN .EQ. 'y' .OR. YN .EQ. 'Y'
      IF (.NOT. (YEH.OR.YN.EQ.'n'.OR.YN.EQ.'N')) GO TO 130
      P0ATB = -1.0D0
      IF (YEH) P0ATB = 1.0D0
      IF (RITE) THEN
          IF (YEH) THEN
              WRITE (T22,FMT=*) ' p is zero at b. '
          ELSE
              WRITE (T22,FMT=*) ' p is not zero at b. '
          END IF
      END IF

  140 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' ********************************************* '
      WRITE (*,FMT=*) ' * IS EITHER OF THE COEFFICIENT FUNCTIONS    * '
      WRITE (*,FMT=*) ' *                                           * '
      WRITE (*,FMT=*) ' *  q OR w UNBOUNDED NEAR b ? (Y/N) (h?)     * '
      WRITE (*,FMT=*) ' ********************************************* '
      WRITE (*,FMT=*)
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(11)
          GO TO 140
      END IF
      YEH = HQ .EQ. 'y' .OR. HQ .EQ. 'Y' .OR. HQ .EQ. 'n' .OR.
     +      HQ .EQ. 'N'
      IF (.NOT.YEH) GO TO 140
      YEH = HQ .EQ. 'y' .OR. HQ .EQ. 'Y'
      QFATB = 1.0D0
      IF (YEH) QFATB = -1.0D0
      IF (RITE) THEN
          IF (YEH) THEN
              WRITE (T22,FMT=*) ' either q or w is unbounded near b. '
          ELSE
              WRITE (T22,FMT=*) ' both q and w are bounded near b. '
          END IF
      END IF
      IF (.NOT.YEH) RESP = 0
      RETURN
C
   12 CONTINUE
  150 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
     +  ' ************************************************ '
      WRITE (*,FMT=*)
     +  ' * THIS PROBLEM IS ON THE INTERVAL              * '
      WRITE (*,FMT=*)
     +  ' *                                              * '
      IF (INTAB.EQ.1) THEN
          WRITE (*,FMT=9070) A,B
          WRITE (*,FMT=*) ' *                            ',FILLB
          WRITE (*,FMT=*) ' * ENDPOINT a IS  ',CHA
          WRITE (*,FMT=*) ' *                            ',FILLB
          IF (P0ATA.GT.0.0D0) THEN
              WRITE (*,FMT=*) ' * p IS ZERO AT a             ',FILLB
              WRITE (*,FMT=*) ' *                            ',FILLB
          END IF
          IF (QFATA.LT.0.0D0) THEN
              WRITE (*,FMT=*) ' * q OR w IS NOT BOUNDED AT a ',FILLB
              WRITE (*,FMT=*) ' *                            ',FILLB
          END IF
          WRITE (*,FMT=*) ' * ENDPOINT b IS  ',CHB
          WRITE (*,FMT=*) ' *                            ',FILLB
          IF (P0ATB.GT.0.0D0) THEN
              WRITE (*,FMT=*) ' * p IS ZERO AT b             ',FILLB
              WRITE (*,FMT=*) ' *                            ',FILLB
          END IF
          IF (QFATB.LT.0.0D0) THEN
              WRITE (*,FMT=*) ' * q OR w IS NOT BOUNDED AT b ',FILLB
              WRITE (*,FMT=*) ' *                            ',FILLB
          END IF
      ELSE IF (INTAB.EQ.2) THEN
          WRITE (*,FMT=9080) A,INFP
          WRITE (*,FMT=*) ' *                            ',FILLB
          WRITE (*,FMT=*) ' * ENDPOINT a IS  ',CHA
          WRITE (*,FMT=*) ' *                            ',FILLB
          IF (P0ATA.GT.0.0D0) THEN
              WRITE (*,FMT=*) ' * p IS ZERO AT a             ',FILLB
              WRITE (*,FMT=*) ' *                            ',FILLB
          END IF
          IF (QFATA.LT.0.0D0) THEN
              WRITE (*,FMT=*) ' * q OR w IS NOT BOUNDED AT a ',FILLB
              WRITE (*,FMT=*) ' *                            ',FILLB
          END IF
          WRITE (*,FMT=*) ' * ENDPT +INF IS  ',CHB
          WRITE (*,FMT=*) ' *                            ',FILLB
      ELSE IF (INTAB.EQ.3) THEN
          WRITE (*,FMT=9090) INFM,B
          WRITE (*,FMT=*) ' *                            ',FILLB
          WRITE (*,FMT=*) ' * ENDPT -INF IS  ',CHA
          WRITE (*,FMT=*) ' *                            ',FILLB
          WRITE (*,FMT=*) ' * ENDPOINT b IS  ',CHB
          WRITE (*,FMT=*) ' *                            ',FILLB
          IF (P0ATB.GT.0.0D0) THEN
              WRITE (*,FMT=*) ' * p IS ZERO AT b             ',FILLB
              WRITE (*,FMT=*) ' *                            ',FILLB
          END IF
          IF (QFATB.LT.0.0D0) THEN
              WRITE (*,FMT=*) ' * q OR w IS NOT BOUNDED AT b ',FILLB
              WRITE (*,FMT=*) ' *                            ',FILLB
          END IF
      ELSE
          WRITE (*,FMT=9100) INFM,INFP
          WRITE (*,FMT=*) ' *                            ',FILLB
          WRITE (*,FMT=*) ' * ENDPT -INF IS  ',CHA
          WRITE (*,FMT=*) ' *                            ',FILLB
          WRITE (*,FMT=*) ' * ENDPT +INF IS  ',CHB
          WRITE (*,FMT=*) ' *                            ',FILLB
      END IF
      WRITE (*,FMT=*)
     +  ' * IS THIS THE PROBLEM YOU WANT ? (Y/N) (h?)    * '
      WRITE (*,FMT=*)
     +  ' ************************************************ '
      WRITE (*,FMT=*)
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(3)
          GO TO 150
      END IF
      YEH = HQ .EQ. 'y' .OR. HQ .EQ. 'Y' .OR. HQ .EQ. 'n' .OR.
     +      HQ .EQ. 'N'
      IF (.NOT.YEH) GO TO 150
      IF (HQ.NE.'y' .AND. HQ.NE.'Y') RESP = 0
      RETURN
C
   13 CONTINUE
  160 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' ********************************************* '
      WRITE (*,FMT=*) ' * DO YOU WANT TO RE-DO                      * '
      WRITE (*,FMT=*) ' *                                           * '
      WRITE (*,FMT=*) ' *    (1)  ENDPOINT a                        * '
      WRITE (*,FMT=*) ' *                                           * '
      WRITE (*,FMT=*) ' *    (2)  ENDPOINT b                        * '
      WRITE (*,FMT=*) ' *                                           * '
      WRITE (*,FMT=*) ' *    (3)  BOTH ENDPOINTS a AND b ?          * '
      WRITE (*,FMT=*) ' *                                           * '
      WRITE (*,FMT=*) ' * ENTER THE NUMBER OF YOUR CHOICE: (h?)     * '
      WRITE (*,FMT=*) ' ********************************************* '
      WRITE (*,FMT=*)
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(3)
          GO TO 160
      END IF
      YEH = HQ .EQ. '1' .OR. HQ .EQ. '2' .OR. HQ .EQ. '3'
      IF (.NOT.YEH) GO TO 160
      READ (CHANS,FMT='(I32)') NANS
      RESP = NANS
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      RETURN
C
   14 CONTINUE
  170 CONTINUE
      IF (RITE) WRITE (T22,FMT=*)
     +    '----------------------------------------'
      IF (RITE) WRITE (T22,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
     +  ' *************************************************'
      WRITE (*,FMT=*)
     +  ' * DO YOU WANT TO COMPUTE                        *'
      WRITE (*,FMT=*)
     +  ' *                                               *'
      WRITE (*,FMT=*)
     +  ' *   (1) AN EIGENVALUE, OR SERIES OF EIGENVALUES *'
      WRITE (*,FMT=*)
     +  ' *                                               *'
      WRITE (*,FMT=*)
     +  ' *   (2) SOLUTION TO AN INITIAL VALUE PROBLEM ?  *'
      WRITE (*,FMT=*)
     +  ' *                                               *'
      WRITE (*,FMT=*)
     +  ' * ENTER THE NUMBER OF YOUR CHOICE: (h?)         *'
      WRITE (*,FMT=*)
     +  ' *************************************************'
      WRITE (*,FMT=*)
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(12)
          GO TO 170
      END IF
      YEH = HQ .EQ. '1' .OR. HQ .EQ. '2'
      IF (.NOT.YEH) GO TO 170
      READ (CHANS,FMT='(I32)') NANS
      RESP = NANS
      RETURN
C
   15 CONTINUE
  200 CONTINUE
      WRITE (*,FMT=*) ' ******************************************** '
      WRITE (*,FMT=*) ' * IS THE BOUNDARY CONDITION PERIODIC ?     * '
      WRITE (*,FMT=*) ' *   OR COUPLED ?                           * '
      WRITE (*,FMT=*) ' *                                          * '
      WRITE (*,FMT=*) ' *   (I.E., y(b) = c*y(a)                   * '
      WRITE (*,FMT=*) ' *    &   p(b)*y''(b) = (1/c)*p(a)*y''(a)     *'
      WRITE (*,FMT=*) ' *                                          * '
      WRITE (*,FMT=*) ' *   OR SOME OTHER COUPLED CONDITION)       * '
      WRITE (*,FMT=*) ' *                                          * '
      WRITE (*,FMT=*) ' * ANSWER (Y/N): (h?)                       * '
      WRITE (*,FMT=*) ' ******************************************** '
      WRITE (*,FMT=*)
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(7)
          GO TO 200
      END IF
      YEH = HQ .EQ. 'y' .OR. HQ .EQ. 'Y' .OR. HQ .EQ. 'n' .OR.
     +      HQ .EQ. 'N'
      IF (.NOT.YEH) GO TO 200
      YEH = HQ .EQ. 'y' .OR. HQ .EQ. 'Y'
      IF (.NOT.YEH) RESP = 0
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      RETURN
C
   16 CONTINUE
      IF (NCA.LE.2 .OR. CHA.EQ.CH6) THEN
  190     CONTINUE
          WRITE (*,FMT=*) ' ****************************************** '
          WRITE (*,FMT=*) ' * IS THE BOUNDARY CONDITION AT a         * '
          WRITE (*,FMT=*) ' *                                        * '
          WRITE (*,FMT=*) ' *    (1) THE DIRICHLET CONDITION         * '
          WRITE (*,FMT=*) ' *        (I.E., y(a) = 0.0)              * '
          WRITE (*,FMT=*) ' *                                        * '
          WRITE (*,FMT=*) ' *    (2) THE NEUMANN CONDITION           * '
          WRITE (*,FMT=*) ' *        (I.E., y''(a) = 0.0)             *'
          WRITE (*,FMT=*) ' *                                        * '
          WRITE (*,FMT=*) ' *    (3) A MORE GENERAL LINEAR           * '
          WRITE (*,FMT=*) ' *        BOUNDARY CONDITION              * '
          WRITE (*,FMT=*) ' *        A1*y(a) + A2*(py'')(a) = 0 ?   *'
          WRITE (*,FMT=*) ' *                                        * '
          WRITE (*,FMT=*) ' * ENTER THE NUMBER OF YOUR CHOICE: (h?)  * '
          WRITE (*,FMT=*) ' ****************************************** '
          WRITE (*,FMT=*)
          READ (*,FMT=9010) CHANS
          READ (CHANS,FMT=9020) HQ
          IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
          IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
              CALL HELP(7)
              GO TO 190
          END IF
          YEH = HQ .EQ. '1' .OR. HQ .EQ. '2' .OR. HQ .EQ. '3'
          IF (.NOT.YEH) GO TO 190
          READ (CHANS,FMT='(I32)') NANS
          IF (NANS.EQ.1) THEN
              A1 = 1.0D0
              A2 = 0.0D0
              IF (RITE) WRITE (T22,FMT=*) ' Dirichlet B.C. at a. '
          ELSE IF (NANS.EQ.2) THEN
              A1 = 0.0D0
              A2 = 1.0D0
              IF (RITE) WRITE (T22,FMT=*) ' Neumann B.C. at a. '
          ELSE
  207         CONTINUE
              WRITE (*,FMT=*)
     +          ' *************************************** '
              WRITE (*,FMT=*)
     +          ' * CHOOSE A1,A2: (h?)                  * '
              WRITE (*,FMT=*)
     +          ' *************************************** '
              WRITE (*,FMT=*)
              WRITE (*,FMT=*) ' A1,A2 = '
              READ (*,FMT=9010) CHANS
              READ (CHANS,FMT=9020) HQ
              IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
              IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
                  CALL HELP(7)
                  GO TO 207
              END IF
              CALL LSTDIR(CHANS,I,ICOL)
              I1 = ICOL(1) - 1
              FMAT = FMT2(I1)
              READ (CHANS,FMT=FMAT) A1,A2
              IF (RITE) WRITE (T22,FMT=*) ' A1,A2 = ',A1,A2
          END IF
      ELSE IF (LCA) THEN
  210     CONTINUE
          IF (RITE) THEN
              WRITE (T22,FMT=*) ' The B.C. at a is '
              WRITE (T22,FMT=*) ' A1*[y,u](a) + A2*[y,v](a) = 0. '
          END IF
          WRITE (*,FMT=*) ' ****************************************** '
          WRITE (*,FMT=*) ' * THE BOUNDARY CONDITION AT a IS         * '
          WRITE (*,FMT=*) ' *                                        * '
          WRITE (*,FMT=*) ' *    A1*[y,u](a) + A2*[y,v](a) = 0,      * '
          WRITE (*,FMT=*) ' *                                        * '
          WRITE (*,FMT=*) ' *  WHERE THE CONSTANTS A1 AND A2         * '
          WRITE (*,FMT=*) ' *    MAY BE CHOSEN ARBITRARILY.          * '
          WRITE (*,FMT=*) ' *                                        * '
          WRITE (*,FMT=*) ' * CHOOSE A1,A2: (h?)                     * '
          WRITE (*,FMT=*) ' ****************************************** '
          WRITE (*,FMT=*)
          WRITE (*,FMT=*) ' A1,A2 = '
          READ (*,FMT=9010) CHANS
          READ (CHANS,FMT=9020) HQ
          IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
          IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
              CALL HELP(7)
              GO TO 210
          END IF
          CALL LSTDIR(CHANS,I,ICOL)
          I1 = ICOL(1) - 1
          FMAT = FMT2(I1)
          READ (CHANS,FMT=FMAT) A1,A2
          IF (RITE) WRITE (T22,FMT=*) ' A1,A2 = ',A1,A2
      END IF
      RETURN
C
   17 CONTINUE
      IF (NCB.LE.2 .OR. CHB.EQ.CH6) THEN
  220     CONTINUE
          WRITE (*,FMT=*) ' ****************************************** '
          WRITE (*,FMT=*) ' * IS THE BOUNDARY CONDITION AT b         * '
          WRITE (*,FMT=*) ' *                                        * '
          WRITE (*,FMT=*) ' *    (1) THE DIRICHLET CONDITION         * '
          WRITE (*,FMT=*) ' *        (I.E., y(b) = 0.0)              * '
          WRITE (*,FMT=*) ' *                                        * '
          WRITE (*,FMT=*) ' *    (2) THE NEUMANN CONDITION           * '
          WRITE (*,FMT=*) ' *        (I.E., y''(b) = 0.0)             *'
          WRITE (*,FMT=*) ' *                                        * '
          WRITE (*,FMT=*) ' *    (3) A MORE GENERAL LINEAR           * '
          WRITE (*,FMT=*) ' *        BOUNDARY CONDITION              * '
          WRITE (*,FMT=*) ' *        B1*y(b) + B2*(py'')(b) = 0 ?   *'
          WRITE (*,FMT=*) ' *                                        * '
          WRITE (*,FMT=*) ' * ENTER THE NUMBER OF YOUR CHOICE: (h?)  * '
          WRITE (*,FMT=*) ' ****************************************** '
          WRITE (*,FMT=*)
          READ (*,FMT=9010) CHANS
          READ (CHANS,FMT=9020) HQ
          IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
          IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
              CALL HELP(7)
              GO TO 220
          END IF
          YEH = HQ .EQ. '1' .OR. HQ .EQ. '2' .OR. HQ .EQ. '3'
          IF (.NOT.YEH) GO TO 220
          READ (CHANS,FMT='(I32)') NANS
          IF (NANS.EQ.1) THEN
              B1 = 1.0D0
              B2 = 0.0D0
              IF (RITE) WRITE (T22,FMT=*) ' Dirichlet B.C. at b. '
          ELSE IF (NANS.EQ.2) THEN
              B1 = 0.0D0
              B2 = 1.0D0
              IF (RITE) WRITE (T22,FMT=*) ' Neumann B.C. at b. '
          ELSE
  230         CONTINUE
              WRITE (*,FMT=*)
     +          ' *************************************** '
              WRITE (*,FMT=*)
     +          ' * CHOOSE B1,B2: (h?)                  * '
              WRITE (*,FMT=*)
     +          ' *************************************** '
              WRITE (*,FMT=*)
              WRITE (*,FMT=*) ' B1,B2 = '
              READ (*,FMT=9010) CHANS
              READ (CHANS,FMT=9020) HQ
              IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
              IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
                  CALL HELP(7)
                  GO TO 230
              END IF
              CALL LSTDIR(CHANS,I,ICOL)
              I1 = ICOL(1) - 1
              FMAT = FMT2(I1)
              READ (CHANS,FMT=FMAT) B1,B2
              IF (RITE) WRITE (T22,FMT=*) ' B1,B2 = ',B1,B2
          END IF
      ELSE IF (LCB) THEN
  240     CONTINUE
          IF (RITE) THEN
              WRITE (T22,FMT=*) ' The B.C. at b is '
              WRITE (T22,FMT=*) ' B1*[y,u](b) + B2*[y,v](b) = 0. '
          END IF
          WRITE (*,FMT=*) ' ****************************************** '
          WRITE (*,FMT=*) ' * THE BOUNDARY CONDITION AT b IS         * '
          WRITE (*,FMT=*) ' *                                        * '
          WRITE (*,FMT=*) ' *    B1*[y,u](b) + B2*[y,v](b) = 0,      * '
          WRITE (*,FMT=*) ' *                                        * '
          WRITE (*,FMT=*) ' *  WHERE THE CONSTANTS B1 AND B2         * '
          WRITE (*,FMT=*) ' *    MAY BE CHOSEN ARBITRARILY.          * '
          WRITE (*,FMT=*) ' *                                        * '
          WRITE (*,FMT=*) ' * CHOOSE B1,B2: (h?)                     * '
          WRITE (*,FMT=*) ' ****************************************** '
          WRITE (*,FMT=*)
          WRITE (*,FMT=*) ' B1,B2 = '
          READ (*,FMT=9010) CHANS
          READ (CHANS,FMT=9020) HQ
          IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
          IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
              CALL HELP(7)
              GO TO 240
          END IF
          CALL LSTDIR(CHANS,I,ICOL)
          I1 = ICOL(1) - 1
          FMAT = FMT2(I1)
          READ (CHANS,FMT=FMAT) B1,B2
          IF (RITE) WRITE (T22,FMT=*) ' B1,B2 = ',B1,B2
      END IF
      RETURN
C
   18 CONTINUE
  310 CONTINUE
      WRITE (*,FMT=*) ' ******************************************** '
      WRITE (*,FMT=*) ' * IS THIS PROBLEM:                         * '
      WRITE (*,FMT=*) ' *                                          * '
      WRITE (*,FMT=*) ' *   (1) PERIODIC ?                         * '
      WRITE (*,FMT=*) ' *         (I.E., y(b) = y(a)               * '
      WRITE (*,FMT=*) ' *          &   p(b)*y''(b) = p(a)*y''(a) )   *'
      WRITE (*,FMT=*) ' *                                          * '
      WRITE (*,FMT=*) ' *   (2) SEMI-PERIODIC ?                    * '
      WRITE (*,FMT=*) ' *         (I.E., y(b) = -y(a)              * '
      WRITE (*,FMT=*) ' *          &   p(b)*y''(b) = -p(a)*y''(a) )  *'
      WRITE (*,FMT=*) ' *                                          * '
      WRITE (*,FMT=*) ' *   (3) GENERAL PERIODIC TYPE ?            * '
      WRITE (*,FMT=*) ' *         (I.E., y(b) = c*y(a)             * '
      WRITE (*,FMT=*) ' *          &   p(b)*y''(b) = p(a)*y''(a)/c   *'
      WRITE (*,FMT=*) ' *            for some number c .NE. 0. )   * '
      WRITE (*,FMT=*) ' *                                          * '
      WRITE (*,FMT=*) ' *   (4) MORE GENERAL COUPLED TYPE ?        * '
      WRITE (*,FMT=*) ' *                                          * '
      WRITE (*,FMT=*) ' * ENTER THE NUMBER OF YOUR CHOICE: (h)?    * '
      WRITE (*,FMT=*) ' ******************************************** '
      WRITE (*,FMT=*)
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(7)
          GO TO 310
      END IF
      YEH = HQ .EQ. '1' .OR. HQ .EQ. '2' .OR. HQ .EQ. '3' .OR.
     +      HQ .EQ. '4'
      IF (.NOT.YEH) GO TO 310
      READ (CHANS,FMT='(I32)') NANS
      IF (NANS.EQ.1) THEN
          CC = 1.0D0
          ALF = 0.0D0
          K11 = 1.0D0
          K12 = 0.0D0
          K21 = 0.0D0
          K22 = 1.0D0
          IF (RITE) WRITE (T22,FMT=*) ' The B.C. is Periodic. '
      ELSE IF (NANS.EQ.2) THEN
          CC = -1.0D0
          ALF = 0.0D0
          K11 = -1.0D0
          K12 = 0.0D0
          K21 = 0.0D0
          K22 = -1.0D0
          IF (RITE) WRITE (T22,FMT=*) ' The B.C. is Semi-Periodic. '
      ELSE IF (NANS.EQ.3) THEN
  320     CONTINUE
          WRITE (*,FMT=*) ' ****************************************** '
          WRITE (*,FMT=*) ' * INPUT c: (h?)                          * '
          WRITE (*,FMT=*) ' ****************************************** '
          WRITE (*,FMT=*) ' c = '
          READ (*,FMT=9010) CHANS
          READ (CHANS,FMT=9020) HQ
          IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
          IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
              CALL HELP(7)
              GO TO 320
          END IF
          READ (CHANS,FMT='(F32.0)') CC
          ALF = 0.0D0
          K11 = CC
          K12 = 0.0D0
          K21 = 0.0D0
          K22 = 1.0D0/CC
          IF (RITE) WRITE (T22,FMT=*)
     +        ' The B.C. is General Periodic type. '
          IF (RITE) WRITE (T22,FMT=*) ' Parameter c = ',CC
      ELSE
  322     CONTINUE
          WRITE (*,FMT=*)
     +      ' ************************************************** '
          WRITE (*,FMT=*)
     +      ' * FOR THIS PROBLEM, THE GENERAL COUPLED          * '
          WRITE (*,FMT=*)
     +      ' *   BOUNDARY CONDITIONS ARE:                     * '
          WRITE (*,FMT=*)
     +      ' *                                                * '
          IF (NCA.LE.2 .AND. NCB.LE.2) THEN
              WRITE (*,FMT=*)
     +          ' * y( b)  =  EX*{k11*y(a) + k12*py''(a)}           *'
              WRITE (*,FMT=*)
     +          ' * py''(b) = EX*{k21*y(a) + k22*py''(a)}            *'
          ELSE IF (NCA.GE.3 .AND. NCB.LE.2) THEN
              WRITE (*,FMT=*)
     +          ' * y(b) = EX*{k11*n*[y,u](a)+k12[y,v](a)}/d       *'
              WRITE (*,FMT=*)
     +          ' * py''(b) = EX*{k21*n*[y,u](a) + k22*[y,v](a)}/d  *'
              WRITE (*,FMT=*)
     +          ' *                                                * '
              WRITE (*,FMT=*)
     +          ' * WHERE d = sqrt(abs([u,v](a)))                  * '
              WRITE (*,FMT=*)
     +          ' *       n = +1 or -1                             * '
          ELSE IF (NCA.LE.2 .AND. NCB.GE.3) THEN
              WRITE (*,FMT=*)
     +          ' * [y,U](b)*N/D = EX*{k11*y(a) + k12*py''(a)}      *'
              WRITE (*,FMT=*)
     +          ' * [y,V](b)/D = EX*{k21*y(a) + k22*py''(a)}        *'
              WRITE (*,FMT=*)
     +          ' *                                                * '
              WRITE (*,FMT=*)
     +          ' * WHERE D = sqrt(abs([U,V](b)))                  * '
              WRITE (*,FMT=*)
     +          ' *       N = +1 OR -1                             * '
          ELSE
              WRITE (*,FMT=*)
     +          ' * [y,U](b)*N/D=EX*{k11*n*[y,u](a)+k12*[y,v](a)}/d*'
              WRITE (*,FMT=*)
     +          ' * [y,V](b)/D=EX*{k21*n*[y,u](a)+k22*[y,v](a)}/d  *'
              WRITE (*,FMT=*)
     +          ' *                                                * '
              WRITE (*,FMT=*)
     +          ' *       D = sqrt(abs([U,V](b)))                  * '
              WRITE (*,FMT=*)
     +          ' *       N = +1 OR -1                             * '
              WRITE (*,FMT=*)
     +          ' *       D = SQRT(ABS([U,V](A)))                  * '
              WRITE (*,FMT=*)
     +          ' *       N = +1 OR -1                             * '
          END IF
          WRITE (*,FMT=*)
     +      ' *                                                * '
          WRITE (*,FMT=*)
     +      ' * WHERE EX = EXP(I*ALFA)                         * '
          WRITE (*,FMT=*)
     +      ' * AND WHERE     ALFA                             * '
          WRITE (*,FMT=*)
     +      ' * AND THE       K11,K12                          * '
          WRITE (*,FMT=*)
     +      ' *               K21,K22                          * '
          WRITE (*,FMT=*)
     +      ' *                                                * '
          WRITE (*,FMT=*)
     +      ' * NEED TO BE CHOSEN. IT IS NECESSARY THAT        * '
          WRITE (*,FMT=*)
     +      ' *                                                * '
          WRITE (*,FMT=*)
     +      ' *      0.0 .LE. ALFA .LT. PI                     * '
          WRITE (*,FMT=*)
     +      ' *                                                * '
          WRITE (*,FMT=*)
     +      ' * INPUT ALFA : (H?)                              * '
          WRITE (*,FMT=*)
     +      ' ************************************************** '
          WRITE (*,FMT=*) ' ALFA = '
          READ (*,FMT=9010) CHANS
          READ (CHANS,FMT=9020) HQ
          IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
          IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
              CALL HELP(7)
              GO TO 322
          END IF
          READ (CHANS,FMT='(1F32.0)') ALF
          WRITE (*,FMT=*)
     +      ' ********************************************* '
          WRITE (*,FMT=*)
     +      ' * FOR SELF ADJOINTNESS IT IS NECESSARY      * '
          WRITE (*,FMT=*)
     +      ' * THAT                                      * '
          WRITE (*,FMT=*)
     +      ' *                                           * '
          WRITE (*,FMT=*)
     +      ' *      K11*K22 - K12*K21 =  1               * '
          WRITE (*,FMT=*)
     +      ' *                                           * '
          WRITE (*,FMT=*)
     +      ' * INPUT K11,K12 :                           * '
          WRITE (*,FMT=*)
     +      ' ********************************************* '
          WRITE (*,FMT=*) ' K11,K12 = '
          READ (*,FMT=9010) CHANS
          READ (CHANS,FMT=9020) HQ
          IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
          IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
              CALL HELP(7)
              GO TO 322
          END IF
          CALL LSTDIR(CHANS,I,ICOL)
          I1 = ICOL(1) - 1
          FMAT = FMT2(I1)
          READ (CHANS,FMT=FMAT) K11,K12
          WRITE (*,FMT=*)
     +      ' ********************************************* '
          WRITE (*,FMT=*)
     +      ' * INPUT K21,K22 :                           * '
          WRITE (*,FMT=*)
     +      ' ********************************************* '
          WRITE (*,FMT=*) ' K21,K22 = '
          READ (*,FMT=9010) CHANS
          READ (CHANS,FMT=9020) HQ
          IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
          IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
              CALL HELP(7)
              GO TO 322
          END IF
          CALL LSTDIR(CHANS,I,ICOL)
          I1 = ICOL(1) - 1
          FMAT = FMT2(I1)
          READ (CHANS,FMT=FMAT) K21,K22
          WRITE (*,FMT=*)
          WRITE (*,FMT=*)
          WRITE (*,FMT=*)
          WRITE (*,FMT=*) ' ALFA = ',ALF
          WRITE (*,FMT=*) ' K11,K12 = ',K11,K12
          WRITE (*,FMT=*) ' K21,K22 = ',K21,K22
      END IF
      DETK = K11*K22 - K21*K12
      TMP = ABS(DETK-1.0D0)
      IF (TMP.GT.0.01D0) THEN
          WRITE (*,FMT=*) ' WARNING: K11*K22-K12*K21 IS NOT = 1 '
      END IF
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      IF (RITE) THEN
          WRITE (T22,FMT=*)
          WRITE (T22,FMT=*) '   ALFA = ',ALF
          WRITE (T22,FMT=*)
          WRITE (T22,FMT=*) ' K11,K12 = ',K11,K12
          WRITE (T22,FMT=*) ' K21,K22 = ',K21,K22
          WRITE (T22,FMT=*) '   DET(K) = ',DETK
          IF (TMP.GT.0.01D0) THEN
              WRITE (T22,FMT=*) ' WARNING: K11*K22-K12*K21 IS NOT = 1 '
          END IF
      END IF
      IF (RITE) WRITE (T22,FMT=*) ' -------------------------------'//
     +    FILLC
      RETURN
C
   19 CONTINUE
  350 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
     +  ' *********************************************** '
      WRITE (*,FMT=*)
     +  ' * DO YOU WANT TO COMPUTE THE SOLUTION TO:     * '
      WRITE (*,FMT=*)
     +  ' *                                             * '
      WRITE (*,FMT=*)
     +  ' *    (1) AN INITIAL VALUE PROBLEM FROM ONE    * '
      WRITE (*,FMT=*)
     +  ' *        END OF THE INTERVAL TO THE OTHER     * '
      WRITE (*,FMT=*)
     +  ' *                                             * '
      WRITE (*,FMT=*)
     +  ' *    (2) INITIAL VALUE PROBLEMS FROM BOTH     * '
      WRITE (*,FMT=*)
     +  ' *        ENDS TO A MIDPOINT ?                 * '
      WRITE (*,FMT=*)
     +  ' *                                             * '
      WRITE (*,FMT=*)
     +  ' * ENTER THE NUMBER OF YOUR CHOICE: (h?)       * '
      WRITE (*,FMT=*)
     +  ' *********************************************** '
      WRITE (*,FMT=*)
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(12)
          GO TO 350
      END IF
      YEH = HQ .EQ. '1' .OR. HQ .EQ. '2'
      IF (.NOT.YEH) GO TO 350
      READ (CHANS,FMT='(I32)') NIVP
      IF (NIVP.EQ.1) THEN
          WRITE (*,FMT=*)
          WRITE (*,FMT=*)
          WRITE (*,FMT=*)
          WRITE (*,FMT=*)
  360     CONTINUE
          WRITE (*,FMT=*)
          WRITE (*,FMT=*)
     +      ' ********************************************* '
          WRITE (*,FMT=*)
     +      ' * WHICH IS THE INITIAL POINT: a OR b ? (h?) * '
          WRITE (*,FMT=*)
     +      ' ********************************************* '
          WRITE (*,FMT=*)
          WRITE (*,FMT=*) ' INITIAL POINT IS '
          READ (*,FMT=9010) CHANS
          READ (CHANS,FMT=9020) HQ
          IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
          IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
              CALL HELP(12)
              GO TO 360
          END IF
          READ (CHANS,FMT=9020) ANSCH
          IF (ANSCH.EQ.'a' .OR. ANSCH.EQ.'A') THEN
              NEND = 1
              IF (RITE) WRITE (T22,FMT=*) ' The Initial Point for this',
     +            ' Initial Value Problem is a. '
              IF (NCA.LE.4 .OR. NCA.EQ.6) THEN
                  WRITE (*,FMT=*)
                  WRITE (*,FMT=*)
                  WRITE (*,FMT=*)
                  WRITE (*,FMT=*)
  370             CONTINUE
                  WRITE (*,FMT=*)
                  WRITE (*,FMT=*)
     +              ' ************************************ '
                  WRITE (*,FMT=*)
     +              ' * THE INITIAL CONDITIONS AT a ARE  * '
                  WRITE (*,FMT=*)
     +              ' *                                  * '
                  IF (NCA.LE.2 .OR. NCA.EQ.6) THEN
                      WRITE (*,FMT=*)
     +                  ' *    y(a)=alfa1, py''(a)=alfa2      *'
                  ELSE
                      WRITE (*,FMT=*)
     +                  ' *  [y,u](a)=alfa1, [y,v](a)=alfa2  * '
                  END IF
                  WRITE (*,FMT=*)
     +              ' *                                  * '
                  WRITE (*,FMT=*)
     +              ' * WHERE THE CONSTANTS alfa1, alfa2 * '
                  WRITE (*,FMT=*)
     +              ' *    MAY BE CHOSEN ARBITRARILY.    * '
                  WRITE (*,FMT=*)
     +              ' *                                  * '
                  WRITE (*,FMT=*)
     +              ' * CHOOSE alfa1,alfa2: (h?)         * '
                  WRITE (*,FMT=*)
     +              ' ************************************ '
                  WRITE (*,FMT=*)
                  WRITE (*,FMT=*) ' alfa1,alfa2 = '
                  READ (*,FMT=9010) CHANS
                  READ (CHANS,FMT=9020) HQ
                  IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
                  IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
                      CALL HELP(12)
                      GO TO 370
                  END IF
                  CALL LSTDIR(CHANS,I,ICOL)
                  I1 = ICOL(1) - 1
                  FMAT = FMT2(I1)
                  READ (CHANS,FMT=FMAT) ALFA1,ALFA2
                  IF (RITE) WRITE (T22,FMT=*) ' alfa1,alfa2 = ',ALFA1,
     +                ALFA2
                  A1 = ALFA2
                  A2 = -ALFA1
              END IF
          ELSE IF (ANSCH.EQ.'b' .OR. ANSCH.EQ.'B') THEN
              NEND = 2
              IF (RITE) WRITE (T22,FMT=*) ' The Initial Point for this',
     +            ' Initial Value Problem is b. '
              IF (NCB.LE.4 .OR. NCB.EQ.6) THEN
                  WRITE (*,FMT=*)
                  WRITE (*,FMT=*)
                  WRITE (*,FMT=*)
                  WRITE (*,FMT=*)
  380             CONTINUE
                  WRITE (*,FMT=*)
                  WRITE (*,FMT=*)
     +              ' ************************************ '
                  WRITE (*,FMT=*)
     +              ' * THE INITIAL CONDITIONS AT b ARE  * '
                  WRITE (*,FMT=*)
     +              ' *                                  * '
                  IF (NCB.LE.2 .OR. NCB.EQ.6) THEN
                      WRITE (*,FMT=*)
     +                  ' *    y(b)=beta1, py''(b)=beta2      *'
                  ELSE
                      WRITE (*,FMT=*)
     +                  ' *  [y,u](b)=beta1, [y,v](b)=beta2  * '
                  END IF
                  WRITE (*,FMT=*)
     +              ' *                                  * '
                  WRITE (*,FMT=*)
     +              ' * WHERE THE CONSTANTS beta1, beta2 * '
                  WRITE (*,FMT=*)
     +              ' *    MAY BE CHOSEN ARBITRARILY.    * '
                  WRITE (*,FMT=*)
     +              ' *                                  * '
                  WRITE (*,FMT=*)
     +              ' * CHOOSE beta1,beta2: (h?)         * '
                  WRITE (*,FMT=*)
     +              ' ************************************ '
                  WRITE (*,FMT=*)
                  WRITE (*,FMT=*) ' beta1,beta2 = '
                  READ (*,FMT=9010) CHANS
                  READ (CHANS,FMT=9020) HQ
                  IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
                  IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
                      CALL HELP(12)
                      GO TO 380
                  END IF
                  CALL LSTDIR(CHANS,I,ICOL)
                  I1 = ICOL(1) - 1
                  FMAT = FMT2(I1)
                  READ (CHANS,FMT=FMAT) BETA1,BETA2
                  IF (RITE) WRITE (T22,FMT=*) ' beta1,beta2 = ',BETA1,
     +                BETA2
                  B1 = BETA2
                  B2 = -BETA1
              END IF
          END IF
      ELSE IF (NIVP.EQ.2) THEN
          NEND = 3
          IF (NCA.LE.4 .OR. NCA.EQ.6) THEN
              WRITE (*,FMT=*)
              WRITE (*,FMT=*)
              WRITE (*,FMT=*)
              WRITE (*,FMT=*)
  390         CONTINUE
              WRITE (*,FMT=*)
              WRITE (*,FMT=*) ' ************************************ '
              WRITE (*,FMT=*) ' * THE INITIAL CONDITIONS AT a ARE  * '
              WRITE (*,FMT=*) ' *                                  * '
              IF (NCA.LE.2 .OR. NCA.EQ.6) THEN
                  WRITE (*,FMT=*)
     +              ' *    y(a)=alfa1, py''(a)=alfa2      *'
              ELSE
                  WRITE (*,FMT=*)
     +              ' *  [y,u](a)=alfa1, [y,v](a)=alfa2  * '
              END IF
              WRITE (*,FMT=*) ' *                                  * '
              WRITE (*,FMT=*) ' * WHERE THE CONSTANTS alfa1, alfa2 * '
              WRITE (*,FMT=*) ' *    MAY BE CHOSEN ARBITRARILY.    * '
              WRITE (*,FMT=*) ' *                                  * '
              WRITE (*,FMT=*) ' * CHOOSE alfa1,alfa2: (h?)         * '
              WRITE (*,FMT=*) ' ************************************ '
              WRITE (*,FMT=*)
              WRITE (*,FMT=*) ' alfa1,alfa2 = '
              READ (*,FMT=9010) CHANS
              READ (CHANS,FMT=9020) HQ
              IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
              IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
                  CALL HELP(12)
                  GO TO 390
              END IF
              CALL LSTDIR(CHANS,I,ICOL)
              I1 = ICOL(1) - 1
              FMAT = FMT2(I1)
              READ (CHANS,FMT=FMAT) ALFA1,ALFA2
              A1 = ALFA2
              A2 = -ALFA1
          END IF
          IF (NCB.LE.4 .OR. NCB.EQ.6) THEN
              WRITE (*,FMT=*)
              WRITE (*,FMT=*)
              WRITE (*,FMT=*)
              WRITE (*,FMT=*)
  400         CONTINUE
              WRITE (*,FMT=*)
              WRITE (*,FMT=*) ' ************************************ '
              WRITE (*,FMT=*) ' * THE INITIAL CONDITIONS AT b ARE  * '
              WRITE (*,FMT=*) ' *                                  * '
              IF (NCB.LE.2 .OR. NCB.EQ.6) THEN
                  WRITE (*,FMT=*)
     +              ' *    y(b)=beta1, py''(b)=beta2      *'
              ELSE
                  WRITE (*,FMT=*)
     +              ' *  [y,u](b)=beta1, [y,v](b)=beta2  * '
              END IF
              WRITE (*,FMT=*) ' *                                  * '
              WRITE (*,FMT=*) ' * WHERE THE CONSTANTS beta1, beta2 * '
              WRITE (*,FMT=*) ' *    MAY BE CHOSEN ARBITRARILY.    * '
              WRITE (*,FMT=*) ' *                                  * '
              WRITE (*,FMT=*) ' * CHOOSE beta1,beta2: (h?)         * '
              WRITE (*,FMT=*) ' ************************************ '
              WRITE (*,FMT=*)
              WRITE (*,FMT=*) ' beta1,beta2 = '
              READ (*,FMT=9010) CHANS
              READ (CHANS,FMT=9020) HQ
              IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
              IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
                  CALL HELP(12)
                  GO TO 400
              END IF
              CALL LSTDIR(CHANS,I,ICOL)
              I1 = ICOL(1) - 1
              FMAT = FMT2(I1)
              READ (CHANS,FMT=FMAT) BETA1,BETA2
              B1 = BETA2
              B2 = -BETA1
          END IF
      END IF
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      RETURN
C
   20 CONTINUE
  250 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' ********************************************* '
      WRITE (*,FMT=*) ' * DO YOU WANT TO COMPUTE                    * '
      WRITE (*,FMT=*) ' *                                           * '
      WRITE (*,FMT=*) ' *    (1) A SINGLE EIGENVALUE                * '
      WRITE (*,FMT=*) ' *                                           * '
      WRITE (*,FMT=*) ' *    (2) A SERIES OF EIGENVALUES ?          * '
      WRITE (*,FMT=*) ' *                                           * '
      WRITE (*,FMT=*) ' * ENTER THE NUMBER OF YOUR CHOICE: (h?)     * '
      WRITE (*,FMT=*) ' ********************************************* '
      WRITE (*,FMT=*)
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(13)
          GO TO 250
      END IF
      YEH = HQ .EQ. '1' .OR. HQ .EQ. '2'
      IF (.NOT.YEH) GO TO 250
      READ (CHANS,FMT='(I32)') NANS
      RESP = NANS
      RETURN
   21 CONTINUE
  260 CONTINUE
      WRITE (*,FMT=*) ' ****************************************** '
      WRITE (*,FMT=*) ' * INPUT NUMEIG, EIG, TOL: (h?)           * '
      WRITE (*,FMT=*) ' ****************************************** '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' NUMEIG,EIG,TOL = '
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(14)
          GO TO 260
      END IF
      EIG = 0.0D0
      TOL = 0.0D0
      CALL LSTDIR(CHANS,I,ICOL)
      I1 = ICOL(1) - 1
      IF (I.EQ.3) THEN
          I2 = ICOL(2) - ICOL(1) - 1
          FMAT = '(I'//COL(I1)//',1X,F'//COL(I2)//'.0,1X,F'//
     +           COL(30-I1-I2)//'.0)'
          READ (CHANS,FMT=FMAT) NUMEIG,EIG,TOL
      ELSE IF (I.EQ.2) THEN
          FMAT = '(I'//COL(I1)//',1X,F'//COL(31-I1)//'.0)'
          READ (CHANS,FMT=FMAT) NUMEIG,EIG
      ELSE
          FMAT = '(I'//COL(I1)//')'
          READ (CHANS,FMT=FMAT) NUMEIG
      END IF
      IF (RITE) WRITE (T22,FMT=*) ' NUMEIG,EIG,TOL = ',NUMEIG,EIG,TOL
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      RETURN
C
   22 CONTINUE
  270 CONTINUE
      WRITE (*,FMT=*) ' *******************************'//FILLA
      WRITE (*,FMT=*) ' *                              '//FILLB
      WRITE (*,FMT=*) ' * PLOT EIGENFUNCTION ? (Y/N) (h?)'//
     +  '                *'
      WRITE (*,FMT=*) ' *******************************'//FILLA
      WRITE (*,FMT=*)
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(16)
          GO TO 270
      END IF
      YEH = HQ .EQ. 'y' .OR. HQ .EQ. 'Y' .OR. HQ .EQ. 'n' .OR.
     +      HQ .EQ. 'N'
      IF (.NOT.YEH) GO TO 270
      PEIGF = (HQ.EQ.'y' .OR. HQ.EQ.'Y')
      IF (.NOT.PEIGF) RESP = 0
      RETURN
C
   23 CONTINUE
  280 CONTINUE
      WRITE (*,FMT=*) ' ****************************************** '
      WRITE (*,FMT=*) ' * INPUT numeig1, numeig2, TOL (h?)       * '
      WRITE (*,FMT=*) ' ****************************************** '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' numeig1,numeig2,TOL = '
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(14)
          GO TO 280
      END IF
      TOLL = 0.0D0
c           TOLL = 1.E-5
      CALL LSTDIR(CHANS,I,ICOL)
      I1 = ICOL(1) - 1
      IF (I.EQ.3) THEN
          I2 = ICOL(2) - ICOL(1) - 1
          FMAT = '(I'//COL(I1)//',1X,I'//COL(I2)//',1X,F'//
     +           COL(30-I1-I2)//'.0)'
          READ (CHANS,FMT=FMAT) NEIG1,NEIG2,TOLL
      ELSE
          FMAT = '(I'//COL(I1)//',1X,I'//COL(31-I1)//')'
          READ (CHANS,FMT=FMAT) NEIG1,NEIG2
      END IF
      IF (NEIG1.NE.NEIG2) PEIGF = .FALSE.
      IF (RITE) WRITE (T22,FMT=*) ' numeig1,numeig2,TOL = ',NEIG1,NEIG2,
     +    TOLL
      RETURN
C
   24 CONTINUE
  253 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' ********************************************* '
      WRITE (*,FMT=*) ' * DO YOU WANT TO COMPUTE                    * '
      WRITE (*,FMT=*) ' *                                           * '
      WRITE (*,FMT=*) ' *    (1) A SINGLE EIGENVALUE                * '
      WRITE (*,FMT=*) ' *                                           * '
      WRITE (*,FMT=*) ' *    (2) A SERIES OF EIGENVALUES ?          * '
      WRITE (*,FMT=*) ' *                                           * '
      WRITE (*,FMT=*) ' * ENTER THE NUMBER OF YOUR CHOICE: (h?)     * '
      WRITE (*,FMT=*) ' ********************************************* '
      WRITE (*,FMT=*)
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(13)
          GO TO 253
      END IF
      YEH = HQ .EQ. '1' .OR. HQ .EQ. '2'
      IF (.NOT.YEH) GO TO 253
      READ (CHANS,FMT='(I32)') NANS
      RESP = NANS
      RETURN
C
   25 CONTINUE
  330 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' ********************************************* '
      WRITE (*,FMT=*) ' * INPUT NUMEIG,TOL: (h?)                    * '
      WRITE (*,FMT=*) ' ********************************************* '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' NUMEIG,TOL = '
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(17)
          GO TO 330
      END IF
      TOL = 0.0D0
      CALL LSTDIR(CHANS,I,ICOL)
      I1 = ICOL(1) - 1
      IF (I.EQ.2) THEN
          FMAT = '(I'//COL(I1)//',1X,F'//COL(31-I1)//'.0)'
          READ (CHANS,FMT=FMAT) NUMEIG,TOL
      ELSE
          FMAT = '(I'//COL(I1)//')'
          READ (CHANS,FMT=FMAT) NUMEIG
      END IF
      IF (RITE) WRITE (T22,FMT=*) ' NUMEIG,TOL = ',NUMEIG,TOL
      IF (NUMEIG.LT.0 .AND. (NCA.NE.4.AND.NCB.NE.4)) THEN
          WRITE (*,FMT=*) ' NUMEIG MUST BE .GE. 0 '
          GO TO 330
      END IF
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      RETURN
C
   26 CONTINUE
  340 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
     +  ' *************************************************'
      WRITE (*,FMT=*)
     +  ' * PLOT EIGENFUNCTION ? (Y/N) (h?)               *'
      WRITE (*,FMT=*)
     +  ' *************************************************'
      WRITE (*,FMT=*)
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(16)
          GO TO 340
      END IF
      YEH = HQ .EQ. 'y' .OR. HQ .EQ. 'Y' .OR. HQ .EQ. 'n' .OR.
     +      HQ .EQ. 'N'
      IF (.NOT.YEH) GO TO 340
      PEIGF = HQ .EQ. 'y' .OR. HQ .EQ. 'Y'
      IF (.NOT.PEIGF) RESP = 0
      IF (PEIGF) THEN
          THA = ATAN2(A2,-A1)
          IF (THA.LT.0.0D0) THA = THA + PI
          THB = ATAN2(B2,-B1)
          IF (THB.LE.0.0D0) THB = THB + PI
          A1 = COS(THA)
          A2 = -SIN(THA)
          B1 = COS(THB)
          B2 = -SIN(THB)
          R1 = K11*SIN(THA) + K12*COS(THA)
          R2 = K21*SIN(THA) + K22*COS(THA)
          RHO = SQRT(R1**2+R2**2)
          B1 = RHO*B1
          B2 = RHO*B2
          NIVP = 2
          NEND = 3
          SLFUN(1) = 0.0D0
          SLFUN(2) = -1.0D0
          SLFUN(3) = THA
          SLFUN(5) = 1.0D0
          SLFUN(6) = THB + NUMEIG*PI
          SLFUN(8) = 0.00001D0
          SLFUN(9) = 1.0D0
      END IF
      RETURN
C
   27 CONTINUE
  283 CONTINUE
      WRITE (*,FMT=*) ' ****************************************** '
      WRITE (*,FMT=*) ' * INPUT numeig1, numeig2, TOL (h?)       * '
      WRITE (*,FMT=*) ' ****************************************** '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' numeig1,numeig2,TOL = '
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(14)
          GO TO 283
      END IF
      CALL LSTDIR(CHANS,I,ICOL)
      I1 = ICOL(1) - 1
      IF (I.EQ.3) THEN
          I2 = ICOL(2) - ICOL(1) - 1
          FMAT = '(I'//COL(I1)//',1X,I'//COL(I2)//',1X,F'//
     +           COL(30-I1-I2)//'.0)'
          READ (CHANS,FMT=FMAT) NEIG1,NEIG2,TOLL
      ELSE
          FMAT = '(I'//COL(I1)//',1X,I'//COL(31-I1)//')'
          READ (CHANS,FMT=FMAT) NEIG1,NEIG2
      END IF
      IF (NEIG1.NE.NEIG2) PEIGF = .FALSE.
      IF (RITE) WRITE (T22,FMT=*) ' numeig1,numeig2,TOL = ',NEIG1,NEIG2,
     +    TOLL
      RETURN
C
   28 CONTINUE
      WRITE (*,FMT=*)
     +  ' **************************************************'
      DO 285 I = 1,ILAST
          NUMEIG = NEIG1 + (I-1)
          EIG = EES(I)
          TOL = TTS(I)
          IFLAG = IIS(I)
          IF (IFLAG.LE.2) THEN
              WRITE (*,FMT=9055) NUMEIG,EIG
              WRITE (*,FMT=9050) TOL,IFLAG

              IF (RITE) THEN
                  WRITE (T22,FMT=9055) NUMEIG,EIG
                  WRITE (T22,FMT=9050) TOL,IFLAG
              END IF

          ELSE IF (IFLAG.EQ.4) THEN
              WRITE (*,FMT=9045) NUMEIG
              IF (RITE) WRITE (T22,FMT=9045) NUMEIG
          ELSE
              WRITE (*,FMT=9040) NUMEIG,IFLAG
              IF (RITE) WRITE (T22,FMT=9040) NUMEIG,IFLAG
              IF (IFLAG.EQ.3) THEN
                IF(NLAST.GE.0) THEN
                  WRITE (*,FMT=9230) NLAST
                  IF (RITE) WRITE (T22,FMT=9230) NLAST
                ELSE IF(.NOT.(NCA.EQ.4 .OR. NCB.EQ.4)) THEN
                  WRITE (*,FMT=9235) 
                  IF (RITE) WRITE (T22,FMT=9235)
                ELSE
                  WRITE (*,FMT=9240)
                  IF (RITE) WRITE (T22,FMT=9240)
                ENDIF
              END IF
          END IF

  285 CONTINUE
      WRITE (*,FMT=*)
     +  ' **************************************************'

      WRITE (*,FMT=*)
      IF (RITE) WRITE (T22,FMT=*)
      IF (JFLAG.EQ.1) WRITE (*,FMT=9260)
      IF (JFLAG.EQ.1 .AND. RITE) WRITE (T22,FMT=9260)
      IF (JFLAG.EQ.2) WRITE (*,FMT=9250) SLF9
      IF (JFLAG.EQ.2 .AND. RITE) WRITE (T22,FMT=9250) SLF9
      IF (RITE) WRITE (T22,FMT=*) ' *******************************'//
     +    FILLA
      WRITE (*,FMT=*)
      RETURN
C
   29 CONTINUE
      WRITE (*,FMT=*)
     +  ' **************************************************'
      DO 303 I = 1,ILAST
          NUMEIG = NEIG1 + I - 1
          EIG = EES(I)
          TOL = TTS(I)
          IFLAG = IIS(I)
          DUBBLE = .FALSE.
          IF (IFLAG.EQ.2) THEN
              DUBBLE = .TRUE.
              IFLAG = 1
          END IF
          WRITE (*,FMT=9055) NUMEIG,EIG
          WRITE (*,FMT=9050) TOL,IFLAG
          IF (DUBBLE) THEN
              WRITE (*,FMT=*)
     +          ' * This eigenvalue appears to be double.          *'
          END IF
          IF (RITE) THEN
              WRITE (T22,FMT=9055) NUMEIG,EIG
              WRITE (T22,FMT=9050) TOL,IFLAG
              IF (DUBBLE) WRITE (T22,FMT=*
     +            ) ' * This eigenvalue appears to be double. * '
          END IF
          WRITE (*,FMT=*)
  303 CONTINUE
      WRITE (*,FMT=*)
     +  ' **************************************************'
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
C
      IF (RITE) WRITE (T22,FMT=*) ' *******************************'//
     +    FILLA
      RETURN
C
   30 CONTINUE
      WRITE (*,FMT=*) ' Press any key to continue. '
      READ (*,FMT=9010) CHANS
      WRITE (*,FMT=*)
     +  ' *********************************************** '
      WRITE (*,FMT=*)
     +  ' * WHAT WOULD YOU LIKE TO DO NOW ?             * '
      WRITE (*,FMT=*)
     +  ' *                                             * '
      WRITE (*,FMT=*)
     +  ' *    (1)  SAME EIGENVALUE PROBLEM, DIFFERENT  * '
      WRITE (*,FMT=*)
     +  ' *           NUMEIG, EIG, OR TOL               * '
      WRITE (*,FMT=*)
     +  ' *                                             * '
      WRITE (*,FMT=*)
     +  ' *    (2)  SAME EIGENVALUE PROBLEM, SAME (a,b) * '
      WRITE (*,FMT=*)
     +  ' *           AND p,q,w,u,v BUT DIFFERENT       * '
      WRITE (*,FMT=*)
     +  ' *           BOUNDARY CONDITIONS A1,A2,B1,B2   * '
      WRITE (*,FMT=*)
     +  ' *                                             * '
      WRITE (*,FMT=*)
     +  ' *    (3)  INTERVAL CHANGE, PROBLEM RESTART    * '
      WRITE (*,FMT=*)
     +  ' *                                             * '
      WRITE (*,FMT=*)
     +  ' *    (4)  AN INITIAL VALUE PROBLEM            * '
      WRITE (*,FMT=*)
     +  ' *                                             * '
      WRITE (*,FMT=*)
     +  ' *    (5)  QUIT                                * '
      WRITE (*,FMT=*)
     +  ' *                                             * '
      WRITE (*,FMT=*)
     +  ' * ENTER THE NUMBER OF YOUR CHOICE: (h?)       * '
      WRITE (*,FMT=*)
     +  ' *********************************************** '
      WRITE (*,FMT=*)
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(3)
          GO TO 30
      END IF
      YEH = HQ .EQ. '1' .OR. HQ .EQ. '2' .OR. HQ .EQ. '3' .OR.
     +      HQ .EQ. '4' .OR. HQ .EQ. '5'
      IF (.NOT.YEH) GO TO 30
      READ (CHANS,FMT='(I32)') NANS
      RESP = NANS
      RETURN
C
   31 CONTINUE
      WRITE (*,FMT=*) ' Press any key to continue. '
      READ (*,FMT=9010) CHANS
      WRITE (*,FMT=*)
     +  ' *********************************************** '
      WRITE (*,FMT=*)
     +  ' *    (1)  SAME INITIAL VALUE PROBLEM,         * '
      WRITE (*,FMT=*)
     +  ' *           DIFFERENT LAMBDA                  * '
      WRITE (*,FMT=*)
     +  ' *                                             * '
      WRITE (*,FMT=*)
     +  ' *    (2)  NEW INITIAL VALUE PROBLEM           * '
      WRITE (*,FMT=*)
     +  ' *                                             * '
      WRITE (*,FMT=*)
     +  ' *    (3)  INTERVAL CHANGE, PROBLEM RESTART    * '
      WRITE (*,FMT=*)
     +  ' *                                             * '
      WRITE (*,FMT=*)
     +  ' *    (4)  AN EIGENVALUE PROBLEM               * '
      WRITE (*,FMT=*)
     +  ' *                                             * '
      WRITE (*,FMT=*)
     +  ' *    (5)  QUIT                                * '
      WRITE (*,FMT=*)
     +  ' *                                             * '
      WRITE (*,FMT=*)
     +  ' * ENTER THE NUMBER OF YOUR CHOICE: (h?)       * '
      WRITE (*,FMT=*)
     +  ' *********************************************** '
      WRITE (*,FMT=*)
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(3)
          GO TO 31
      END IF
      YEH = HQ .EQ. '1' .OR. HQ .EQ. '2' .OR. HQ .EQ. '3' .OR.
     +      HQ .EQ. '4' .OR. HQ .EQ. '5'
      IF (.NOT.YEH) GO TO 31
      READ (CHANS,FMT='(I32)') NANS
      RESP = NANS
      RETURN
C
   32 CONTINUE
  410 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
     +  ' ************************************************ '
      WRITE (*,FMT=*)
     +  ' * WHAT VALUE SHOULD BE USED FOR THE            * '
      WRITE (*,FMT=*)
     +  ' *   EIGENPARAMETER, EIG ?                      * '
      WRITE (*,FMT=*)
     +  ' *                                              * '
      WRITE (*,FMT=*)
     +  ' * INPUT EIG = (h?)                             * '
      WRITE (*,FMT=*)
     +  ' ************************************************ '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' EIG = '
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(12)
          GO TO 410
      END IF
      READ (CHANS,FMT='(F32.0)') EIG
      PEIGF = .TRUE.
      RETURN
C
   33 CONTINUE
  450 CONTINUE
      WRITE (*,FMT=*)
     +  ' ****************************************************'
      WRITE (*,FMT=*)
     +  ' * WHICH FUNCTION DO YOU WANT TO PLOT ?             *'
      WRITE (*,FMT=*)
     +  ' *                                                  *'
      WRITE (*,FMT=*) XC(1)
      WRITE (*,FMT=*) XC(2)
      WRITE (*,FMT=*) XC(3)
      WRITE (*,FMT=*) XC(4)
      WRITE (*,FMT=*) XC(5)
      WRITE (*,FMT=*) XC(6)
      WRITE (*,FMT=*)
     +  ' *                                                  *'
      WRITE (*,FMT=*)
     +  ' * ENTER THE NUMBER OF YOUR CHOICE: (h?)            *'
      WRITE (*,FMT=*)
     +  ' ****************************************************'
      WRITE (*,FMT=*)
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') RESP = 0
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(16)
          GO TO 450
      END IF
      YEH = HQ .EQ. '1' .OR. HQ .EQ. '2' .OR. HQ .EQ. '3' .OR.
     +      HQ .EQ. '4' .OR. HQ .EQ. '5' .OR. HQ .EQ. '6'
      IF (.NOT.YEH) GO TO 450
      READ (CHANS,FMT='(I32)') NF
  470 CONTINUE
      WRITE (*,FMT=*)
     +  ' ****************************************************'
      WRITE (*,FMT=*)
     +  ' * WHICH DO YOU WANT AS THE INDEPENDENT VARIABLE ?  *'
      WRITE (*,FMT=*)
     +  ' *                                                  *'
      WRITE (*,FMT=*) XC(7)
      WRITE (*,FMT=*) XC(8)
      WRITE (*,FMT=*)
     +  ' *                                                  *'
      WRITE (*,FMT=*)
     +  ' * ENTER THE NUMBER OF YOUR CHOICE: (h?)            *'
      WRITE (*,FMT=*)
     +  ' ****************************************************'
      WRITE (*,FMT=*)
C
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') RESP = 0
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(16)
          GO TO 470
      END IF
      YEH = HQ .EQ. '1' .OR. HQ .EQ. '2'
      IF (.NOT.YEH) GO TO 470
      READ (CHANS,FMT='(I32)') NV
      RETURN
C
   34 CONTINUE
  480 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' DO YOU WANT TO SAVE THE PLOT FILE ? (Y/N) (h?)'
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(16)
          GO TO 480
      END IF
      YEH = HQ .EQ. 'y' .OR. HQ .EQ. 'Y' .OR. HQ .EQ. 'n' .OR.
     +      HQ .EQ. 'N'
      IF (.NOT.YEH) GO TO 480
      RESP = 0
      IF (HQ.EQ.'y' .OR. HQ.EQ.'Y') RESP = 1
      RETURN
C
   35 CONTINUE
  490 CONTINUE
      WRITE (*,FMT=*) ' PLOT ANOTHER FUNCTION ? (Y/N) (h?)'
      READ (*,FMT=9010) CHANS
      READ (CHANS,FMT=9020) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') STOP
      IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
          CALL HELP(16)
          GO TO 490
      END IF
      YEH = HQ .EQ. 'y' .OR. HQ .EQ. 'Y' .OR. HQ .EQ. 'n' .OR.
     +      HQ .EQ. 'N'
      IF (.NOT.YEH) GO TO 490
      RESP = 0
      IF (HQ.EQ.'y' .OR. HQ.EQ.'Y') RESP = 1
      RETURN
C
 9010 FORMAT (A32)
 9020 FORMAT (A1)
 9030 FORMAT (A70)
 9040 FORMAT (1X,'   NUMEIG = ',I5,'   IFLAG = ',I3)
 9045 FORMAT (1X,'   NUMEIG = ',I5,' : COMPUTATION FAILED ')
 9050 FORMAT (1X,' * TOL = ',D14.5,2X,' IFLAG = ',I3,'             *')
 9055 FORMAT (1X,' * NUMEIG = ',I5,' EIG = ',D18.9,'        *')
 9070 FORMAT (1X,1X,'*   (',F12.7,',',F12.7,')','                *')
 9080 FORMAT (1X,1X,'*   (',F12.7,',',A12,')','                *')
 9090 FORMAT (1X,1X,'*   (',A12,',',F12.7,')','                *')
 9100 FORMAT (1X,1X,'*   (',A12,',',A12,')','                *')
 9110 FORMAT (1X,2A39)
 9230 FORMAT (1X,' * THERE SEEMS TO BE NO EIGENVALUE OF INDEX       *',
     +       /,1X,' * GREATER THAN',I5,'                              *'
     +       )
 9235 FORMAT (1X,' * THERE SEEM TO BE NO EIGENVALUES               *')
 9240 FORMAT (1X,' * THERE SEEMS TO BE NO EIGENVALUE OF THIS INDEX *')
 9250 FORMAT (1X,'* THERE SEEMS TO BE CONTINUOUS SPECTRUM BEGINNING*',/,
     +       1X,'* AT ABOUT',1P,D8.1,'                               *')
 9260 FORMAT (1X,'* THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *')
      END
C
      SUBROUTINE STAGE(A,B,INTAB,P0ATA,QFATA,P0ATB,QFATB,A1,A2,B1,B2,
     +                 EIG,IFLAG,SLFUN,NCA,NCB,NIVP,NEND)

C
C     THE FOLLOWING CALL SETS THE STAGE IN sleign -- I.E., SAMPLES
C     THE COEFFICIENTS AND SETS THE INITIAL INTERVAL.
C
C     .. Scalar Arguments ..
      REAL A,A1,A2,B,B1,B2,EIG,P0ATA,P0ATB,QFATA,QFATB
      INTEGER IFLAG,INTAB,NCA,NCB,NEND,NIVP
C     ..
C     .. Array Arguments ..
      REAL SLFUN(9)
C     ..
C     .. Scalars in Common ..
      REAL AA,BB,DTHDAA,DTHDBB,HPI,PI,TMID,TWOPI
      INTEGER MDTHZ
      LOGICAL ADDD
C     ..
C     .. Local Scalars ..
      REAL ALFA,BETA,TOL
      INTEGER NUMEIG
C     ..
C     .. External Subroutines ..
      EXTERNAL SLEIGN
C     ..
C     .. Common blocks ..
      COMMON /PIE/PI,TWOPI,HPI
      COMMON /TDATA/AA,TMID,BB,DTHDAA,DTHDBB,MDTHZ,ADDD
C     ..
      NUMEIG = 0
      TOL = .001D0
      CALL SLEIGN(A,B,INTAB,P0ATA,QFATA,P0ATB,QFATB,A1,A2,B1,B2,NUMEIG,
     +            EIG,TOL,IFLAG,-1,SLFUN,NCA,NCB)
      IF (NIVP.EQ.1 .AND. NEND.EQ.1) TMID = BB
      IF (NIVP.EQ.1 .AND. NEND.EQ.2) TMID = AA
C
C     ACTUALLY, WE MAY HAVE TO AVOID AA OR BB IN SOME CASES,
C     WHICH WILL BE TAKEN CARE OF LATER.
C
      ALFA = 0.0D0
      IF ((NIVP.EQ.1.AND.NEND.EQ.1.AND.NCA.LE.2) .OR.
     +    NIVP.EQ.2) ALFA = SLFUN(3)
      BETA = PI
      IF ((NIVP.EQ.1.AND.NEND.EQ.2.AND.NCB.LE.2) .OR.
     +    NIVP.EQ.2) BETA = SLFUN(6)
      SLFUN(3) = ALFA
      SLFUN(6) = BETA
      SLFUN(4) = 0.0D0
      SLFUN(7) = 0.0D0
      SLFUN(8) = .01D0
      RETURN
      END
C
      SUBROUTINE LSTDIR(CHANS,I,ICOL)
C     .. Local Scalars ..
      INTEGER J
C     ..
C     .. Scalar Arguments ..
      INTEGER I
      CHARACTER*32 CHANS
C     ..
C     .. Array Arguments ..
      INTEGER ICOL(2)
C     ..
      I = 1
      DO 10 J = 1,32
          IF (CHANS(J:J).EQ.',' .OR. CHANS(J:J).EQ.'/') THEN
              ICOL(I) = J
              IF (CHANS(J:J).EQ.'/') THEN
                  CHANS(J:J) = ' '
                  RETURN
              END IF
              I = I + 1
          END IF
   10 CONTINUE
      RETURN
      END
      CHARACTER*32 FUNCTION FMT2(I1)
C     .. Local Arrays ..
      CHARACTER*2 COL(32)
C     ..
C
C     .. Scalar Arguments ..
      INTEGER I1
C     ..
C     .. Data statements ..
      DATA COL/'01','02','03','04','05','06','07','08','09','10','11',
     +     '12','13','14','15','16','17','18','19','20','21','22','23',
     +     '24','25','26','27','28','29','30','31','32'/
C     ..
      FMT2 = '(F'//COL(I1)//'.0,1X,F'//COL(31-I1)//'.0)'
      RETURN
      END
C
      SUBROUTINE CONVT(CHIN,CHINN)
C   THIS PROGRAM CONVERTS AN 8 CHARACTER STRING WITHOUT A ':'
C     TO A BLANK STRING.  BUT IF IT HAS A ':', THEN IT KEEPS THE
C     STRING UP TO AND INCLUDING THE ':' .
C   THE INPUT IS STRING CHIN, AND THE OUTPUT IS CHINN.

C     .. Scalar Arguments ..
      CHARACTER*8 CHIN,CHINN
C     ..
C     .. Local Scalars ..
      INTEGER I,J,K
C     ..
      CHINN = ' '
      J = 0
      K = 0
      DO 10 I = 1,8
          IF (CHIN(I:I).EQ.' ' .AND. J.EQ.0) K = I
          IF (CHIN(I:I).EQ.':') J = I
   10 CONTINUE
      IF (J.GE.1 .AND. J.LE.8) THEN
          DO 20 I = 1,J - K
              CHINN(I:I) = CHIN(I+K:I+K)
   20     CONTINUE
      ELSE
          CHINN = ' '
      END IF
      RETURN
      END
C
      SUBROUTINE AUTO
C  THIS SUBROUTINE IS, IN EFFECT, A VERY SPECIAL KIND OF DRIVER
C    FOR SLEIGN2. IT ENABLES ONE TO BYPASS THE QUESTION
C    AND ANSWER FORMAT IN DRIVE (FOR EXPERIENCED USERS ONLY!).
C    INSTEAD OF ENTERING THE NEEDED INPUT FOR DRIVE FROM THE
C    KEYBOARD, A USER CAN INSTEAD CREATE A VERY BRIEF FILE,
C    CALLED auto.in, IN THE SAME DIRECTORY, WHICH CONTAINS ALL
C    THE DATA THAT WOULD OTHERWISE BE ENTERED FROM THE
C    KEYBOARD.
C
C    ONCE SUCH A FILE HAS BEEN CREATED, THE USER BEGINS AS USUAL,
C    TYPING THE NAME OF THE EXECUTABLE, AS IN

C       XAMPLES.X  <ENTER>

C    FOR EXAMPLE.  THIS WOULD CAUSE THE PROMPT

C        WOULD YOU LIKE AN OVERVIEW OF HELP?(Y/N)(h?)

C    TO APPEAR, AS USUAL.  BUT INSTEAD OF REPLYING TO THE
C    QUESTION WITH y, OR n, OR h, ONE SIMPLY TYPES IN THE
C    RESPONSE
C       a  <ENTER>
C    (The "a" here stands for "automatic" operation of SLEIGN2.);
C    and at this point the computation of the requested
C    eigenvalues(s) proceeds without further action by the user,
C    taking the needed data from the auto.in file instead.
C         The construction of the file "auto.in" consists of simply
C    defining a number of "KEYWORDS", each on a separate line, which
C    together constitute a complete set of input parameters defining
C    the eigenvalue problem to be solved. (The Differential Equation
C    coefficients have, presumably, been already defined in either
C    XAMPLES.F, or BLOGGS.F, or.. .)  The order in which the
C    needed keywords are defined is of no importance.
C         -------------------------------------------
C    The KEYWORDS, all of which end in a colon and must be
C    followed by at least one space, are:

C    a:  -- The left endpoint of the interval (a,b);
C         Value is any real number;
C         Default value is -infinity.
C    b:  -- The right endpoint of the interval (a,b);
C         Value is any real number;
C         Default value is +infinity.
C    classa:  -- The endpoint classification of a;
C              Value is one of { r, wr, lcno, lco, lp, d }.
C    classb:  -- The endpoint classification of b:
C              Value is one of { r, wr, lcno, lco, lp, d }.
C    bca: -- Boundary Condition for the endpoint a;
C          Value is either d (for Dirichlet),
C                          n (for Neuman),
C                       or two real numbers A1, A2 .
C    bcb: -- Boundary Condition for the endpoint b;
C          Value is either d (for Dirichlet),
C                          n (for Neuman),
C                       or two real numbers B1, B2 .
C    bcc:  -- Coupled Boundary Condition;
C           Value is either p (for Periodic),
C                           s (for Semi-Periodic),
C                       or five real numbers
C                          alpha, k11, k12, k21, k22 .
C    numeig:  -- Index (or range of Indices) of desired Eigenvalue;
C              Value is an integer N1, or pair of integers N1,N2 .
C    param:  -- Parameter(s) appropriate for the problem;
C             Value is one or two real numbers, param1, param2 .
C    np:  --  Problem Number;
C           (Appropriate only if one of the Differential Equations
C             in XAMPLES.F is being used.);
C           Value is an integer from 1 to 32 .
C    output:  -- Name of output file;
C              Value is a character string, the name of the output
C                file where the results of the computation are to
C                be written;
C              Default value is "auto.out" .
C    end:  -- Last line of file "auto.in" ; no value set.

C    again:  --  Terminates the input for one eigenvalue problem and
C                begins the input for another ; no value set.

C Although the KEYWORDS used to define any one problem can be defined
C in any order, there are a few rules to be observed:  Namely,
C Only those KEYWORDS whose values are necessary need be mentioned.
C Any KEYWORD definition remains in effect until redefined;
C To erase a previous definition of a KEYWORD, redefine it to
C have the value "null". (For instance, if one problem has
C the endpoint a defined as in "a: 0.0" , and a following
C problem needs to have a undefined (so that a = -Infinity),
C then it would be necessary to set "a: null" .)

C  An example of such a file is exhibited in the box below.

C   ________________________
C  |                        |
C  |output: Bessel.rep      |
C  |np: 2                   |
C  |param: 0.75             |
C  |a: 0.0                  |
C  |b: 1.0                  |
C  |classa: lcno            |
C  |classb: r               |
C  |bca: 1.0,0.0            |
C  |bcb: d                  |
C  |numeig: 2,5             |
C  |end:                    |
C  |________________________|

C     This file (which must be called "auto.in", of course), would
C be suitable for running Bessel's equation in XAMPLES.F .
C Evidently the problem selected is #2 of XAMPLES.F (Bessel's
C equation), with the parameter nu = 0.75, on the interval (0.0,1.0).
C The endpoint a is asserted to be LCNO; endpoint b is R; the
C Boundary Condition at a is defined by A1 = 1.0 & A2 = 0.0;
C Boundary Condition at b is Dirichlet; and the eigenvalues
C lambda(2), lambda(3), lambda(4), lambda(5) are to be computed.

C  --------------------------------------------------------------- C
C     .. Scalars in Common ..
      REAL A,ALPHA,B,BETA,GAMMA,H,KK,L,NU,P1,P2,P3,P4,P5,P6
      INTEGER INTAB,NUMBER,T21,T22,T23,T24,T25
      LOGICAL PR
C     ..
C     .. Local Scalars ..
      REAL A1,A2,ALF,AS,B1,B2,BS,DET,EIG,
     +     K11,K12,K21,K22,ONE,P0ATA,
     +     P0ATB,PARAM1,PARAM2,QFATA,QFATB,TOL,ZER
      INTEGER I,I1,IFLAG,II,INTABS,J,JFLAG,K,LAST,LASTI,M,NCA,NCB,NEIG1,
     +        NEIG2,NP,NUMEIG
      CHARACTER*8 CH8,CHH8
      CHARACTER*32 FMAT
      CHARACTER*62 BCC,BLANK,CH1,CHANS,CHEND,CHIN,CHT,CLASSA,CLASSB,
     +             TAPE25
C     ..
C     .. Local Arrays ..
      REAL PP(5),SLFUN(9)
      INTEGER ICOL(2),VAL(60)
      CHARACTER*2 COL(62)
      CHARACTER*62 CH(60)
C     ..
C     .. External Functions ..
      CHARACTER*32 FMT2
      EXTERNAL FMT2
C     ..
C     .. External Subroutines ..
      EXTERNAL CHAR,CONVT,LST,LSTDIR,PERIO,PQ,SLEIGN,STR2R
C     ..
C     .. Intrinsic Functions ..
      INTRINSIC MIN
C     ..
C     .. Common blocks ..
      COMMON /DATADT/A,B,INTAB
      COMMON /FLAG/NUMBER
      COMMON /PAR/NU,H,KK,L,ALPHA,BETA,GAMMA,P1,P2,P3,P4,P5,P6
      COMMON /PRIN/PR,T21
      COMMON /TAPES/T22,T23,T24,T25
C     ..
C     .. Data statements ..
c

      DATA COL/'01','02','03','04','05','06','07','08','09','10','11',
     +     '12','13','14','15','16','17','18','19','20','21','22','23',
     +     '24','25','26','27','28','29','30','31','32','33','34','35',
     +     '36','37','38','39','40','41','42','43','44','45','46','47',
     +     '48','49','50','51','52','53','54','55','56','57','58','59',
     +     '60','61','62'/
C     ..
c
      OPEN (T21,FILE='test.out')
      OPEN (T24,FILE='auto.in')
c  Set the name of the Output File to 'auto.out' as default.
      TAPE25 = 'auto.out'
c
      ONE = 1.0D0
      ZER = 0.0D0
      ALF = 0.0D0
      CHEND = 'end:'
      BLANK = '                                '
C
C  READ THE auto.in FILE (ASSUMED LESS THAN 50 LINES
C       FOR ANY ONE PROBLEM):

      READ (T24,FMT=9010,END=75) (CH(I),I=1,50)
   75 CONTINUE
C
C  THE VALUES OF THE FUNCTION VAL(*) BELOW INDICATE WHETHER
C    OR NOT CERTAIN KEYWORDS ARE PRESENT IN THE FILE auto.in.
C  WHEN I = 1, VAL(I) REFERS TO THE PROBLEM # IN XAMPLES.F;
C           2, REFERS TO THE POSSIBLE PARAMETERS ;
C              (THERE MAY BE UP TO 2 INPUT PARAMETERS IN
C               THE PARTICULAR DIFFERENTIAL EQUATION.)
C           3, REFERS TO THE ENDPOINT a;
C           4, REFERS TO THE ENDPOINT b;
C           5, REFERS TO THE CLASSIFICATION OF ENDPOINT a;
C           6, REFERS TO THE CLASSIFICATION OF ENDPOINT b;
C           7, REFERS TO A SEPARATED BOUNDARY CONDITION AT a;
C           8, REFERS TO A SEPARATED BOUNDARY CONDITION AT b;
C           9, REFERS TO COUPLED BOUNDARY CONDITIONS;
C          10, REFERS TO THE INDEX(ES) OF THE EIGENVALUES WANTED;
C          11, REFERS TO THE OUTPUT FILE FOR THE RESULTS.
C          12, REFERS TO WHETHER OR NOT THERE IS ANOTHER
C                PROBLEM TO BE RUN.

      DO 15 I = 1,50
          VAL(I) = 0
   15 CONTINUE
      LAST = 0
C
  300 CONTINUE
      A1 = ONE
      A2 = ZER
      B1 = ONE
      B2 = ZER
      DO 60 I = 1,50
          IF (CH(I).NE.BLANK) THEN
              READ (CH(I),FMT=9010) CH1
              READ (CH1,FMT='(A8)') CH8
          ELSE
              CH1 = ' '
              CH8 = ' '
          END IF
          CALL CONVT(CH8,CHH8)
          IF (CHH8.EQ.'np:') THEN
              CALL CHAR(CH(I),K,M,CHANS)
              READ (CHANS,FMT='(I7)') NP
              VAL(1) = 1
          ELSE IF (CHH8.EQ.'param:') THEN
              CALL CHAR(CH(I),K,M,CHANS)
              IF (CHANS.EQ.'null') THEN
                  VAL(2) = 0
              ELSE IF (M.EQ.0) THEN
                  READ (CHANS,FMT='(F32.0)') PARAM1
                  VAL(2) = 1
              ELSE IF (M.EQ.1) THEN
                  CALL LSTDIR(CHANS,II,ICOL)
                  I1 = ICOL(1) - 1
                  FMAT = FMT2(I1)
                  READ (CHANS,FMT=FMAT) PARAM1,PARAM2
                  VAL(2) = 2
              ELSE IF (M.EQ.4) THEN
                  CHIN = CHANS
                  IF (CHIN.NE.' ') THEN
                      DO 400 J = 1,5
                          IF (CHIN.NE.' ') CALL STR2R(CHIN,PP(J))
  400                 CONTINUE
                  END IF
                  P1 = PP(1)
                  P2 = PP(2)
                  P3 = PP(3)
                  P4 = PP(4)
                  P5 = PP(5)
                  P6 = P2 + P3 + 1.0D0 - P4 - P5
C  THESE NUMBERS ARE THE PARAMETERS s,a,b,c,d,e in the Heun eqn.
C    WITH e = a+b+1-c-d
                  VAL(2) = 5
              END IF
          ELSE IF (CHH8.EQ.'a:') THEN
              CALL CHAR(CH(I),K,M,CHANS)
              IF (CHANS.EQ.'null') THEN
                  VAL(3) = 0
              ELSE
                  CALL LST(CHANS,A)
                  VAL(3) = 1
              END IF
          ELSE IF (CHH8.EQ.'b:') THEN
              CALL CHAR(CH(I),K,M,CHANS)
              IF (CHANS.EQ.'null') THEN
                  VAL(4) = 0
              ELSE
                  CALL LST(CHANS,B)
                  VAL(4) = 1
              END IF
          ELSE IF (CHH8.EQ.'classa:') THEN
              CALL CHAR(CH(I),K,M,CHANS)
              READ (CHANS,FMT='(A32)') CLASSA
              VAL(5) = 1
          ELSE IF (CHH8.EQ.'classb:') THEN
              CALL CHAR(CH(I),K,M,CHANS)
              READ (CHANS,FMT='(A32)') CLASSB
              VAL(6) = 1
          ELSE IF (CHH8.EQ.'bca:') THEN
              CALL CHAR(CH(I),K,M,CHANS)
              VAL(7) = 2
              IF (CHANS.EQ.'null') THEN
                  VAL(7) = 0
              ELSE IF (M.EQ.0) THEN
                  IF (CHANS.EQ.'d') THEN
                      A1 = ONE
                      A2 = ZER
                  ELSE IF (CHANS.EQ.'n') THEN
                      A1 = ZER
                      A2 = ONE
                  END IF
              ELSE
                  CALL LSTDIR(CHANS,II,ICOL)
                  I1 = ICOL(1) - 1
                  FMAT = FMT2(I1)
                  READ (CHANS,FMT=FMAT) A1,A2
              END IF
          ELSE IF (CHH8.EQ.'bcb:') THEN
              CALL CHAR(CH(I),K,M,CHANS)
              VAL(8) = 2
              IF (CHANS.EQ.'null') THEN
                  VAL(8) = 0
              ELSE IF (M.EQ.0) THEN
                  IF (CHANS.EQ.'d') THEN
                      B1 = ONE
                      B2 = ZER
                  ELSE IF (CHANS.EQ.'n') THEN
                      B1 = ZER
                      B2 = ONE
                  END IF
              ELSE
                  CALL LSTDIR(CHANS,II,ICOL)
                  I1 = ICOL(1) - 1
                  FMAT = FMT2(I1)
                  READ (CHANS,FMT=FMAT) B1,B2
              END IF
          ELSE IF (CHH8.EQ.'bcc:') THEN
              CALL CHAR(CH(I),K,M,CHANS)
              IF (CHANS.EQ.'null') THEN
                  VAL(9) = 0
              ELSE IF (M.EQ.0) THEN
                  VAL(9) = 1
                  READ (CHANS,FMT='(A32)') BCC
                  IF (BCC.EQ.'p') THEN
                      K11 = ONE
                      K12 = ZER
                      K21 = ZER
                      K22 = ONE
                  ELSE IF (BCC.EQ.'s') THEN
                      K11 = -ONE
                      K12 = ZER
                      K21 = ZER
                      K22 = -ONE
                  END IF
              ELSE
                  CHIN = CHANS
                  DO 405 J = 1,5
                     CALL STR2R(CHIN,PP(J))
 405              CONTINUE
                  ALF = PP(1)
                  K11 = PP(2)
                  K12 = PP(3)
                  K21 = PP(4)
                  K22 = PP(5)
                  VAL(9) = 5
              END IF
          ELSE IF (CHH8.EQ.'numeig:') THEN
              CALL CHAR(CH(I),K,M,CHANS)
              IF (M.EQ.0) THEN
                  READ (CHANS,FMT='(I32)') NUMEIG
                  NEIG1 = NUMEIG
                  NEIG2 = NEIG1
                  VAL(10) = 1
              ELSE
                  CALL LSTDIR(CHANS,II,ICOL)
                  I1 = ICOL(1) - 1
                  FMAT = '(I'//COL(I1)//',1X,I'//COL(61-I1)//')'
                  READ (CHANS,FMT=FMAT) NEIG1,NEIG2
                  VAL(10) = 2
              END IF
          ELSE IF (CHH8.EQ.'output:') THEN
              CLOSE (T25)
              CALL CHAR(CH(I),K,M,CHANS)
              READ (CHANS,FMT='(A32)') TAPE25
              OPEN (T25,FILE=TAPE25)
              VAL(11) = 1
          ELSE IF (CHH8.EQ.'again:') THEN
              LASTI = I
              VAL(12) = VAL(12) + 1
              GO TO 70
          ELSE IF (CHH8.EQ.CHEND) THEN
              GO TO 70
          END IF
   60 CONTINUE
   70 CONTINUE
c
C  WHEN AN ENDPOINT IS LP, NO BOUNDARY CONDITION CAN BE GIVEN;
C    AND COUPLED BOUNDARY CONDITIONS ARE NOT PERMISSIBLE,
C    SO VAL(9) = 0.
      IF (CLASSA.EQ.'lp') THEN
          VAL(7) = 0
          VAL(9) = 0
          NCA = 5
      END IF
      IF (CLASSB.EQ.'lp') THEN
          VAL(8) = 0
          VAL(9) = 0
          NCB = 5
      END IF
      IF ((CLASSA.EQ.'n') .OR. (CLASSA.EQ.'d')) THEN
          VAL(9) = 0
      END IF
      IF ((CLASSB.EQ.'n') .OR. (CLASSB.EQ.'d')) THEN
          VAL(9) = 0
      END IF
c
C  IF ENDPOINT a = -INF, THEN a WAS OMITTED, SO VAL(3) = 0)
C  IF ENDPOINT b = +INF, THEN b WAS OMITTED, SO VAL(4) = 0)
      IF (VAL(3).NE.0 .AND. VAL(4).NE.0) THEN
          INTAB = 1
      ELSE IF (VAL(3).EQ.0 .AND. VAL(4).NE.0) THEN
          INTAB = 3
      ELSE IF (VAL(3).NE.0 .AND. VAL(4).EQ.0) THEN
          INTAB = 2
      ELSE
          INTAB = 4
      END IF
      P0ATA = -1.0D0
      QFATA = 1.0D0
      P0ATB = -1.0D0
      QFATB = 1.0D0
      NCA = 1
      NCB = 1
      NUMBER = NP

C  N.B. FOR classa, or classb, THE VALUE 'd' MEANS DEFAULT.
C    (NOT DIRICHLET)

      IF (VAL(5).NE.0) THEN
          IF (CLASSA.EQ.'r') THEN
              NCA = 1
          ELSE IF (CLASSA.EQ.'wr') THEN
              NCA = 2
          ELSE IF (CLASSA.EQ.'lcno') THEN
              NCA = 3
          ELSE IF (CLASSA.EQ.'lco') THEN
              NCA = 4
          ELSE IF (CLASSA.EQ.'lp') THEN
              NCA = 5
          ELSE IF (CLASSA.EQ.'d') THEN
              NCA = 6
          END IF
      END IF
      IF (VAL(6).NE.0) THEN
          IF (CLASSB.EQ.'r') THEN
              NCB = 1
          ELSE IF (CLASSB.EQ.'wr') THEN
              NCB = 2
          ELSE IF (CLASSB.EQ.'lcno') THEN
              NCB = 3
          ELSE IF (CLASSB.EQ.'lco') THEN
              NCB = 4
          ELSE IF (CLASSB.EQ.'lp') THEN
              NCB = 5
          ELSE IF (CLASSB.EQ.'d') THEN
              NCB = 6
          END IF
      END IF
C
C   WRITE THE RESULTS TO THE OUTPUT FILE:
C
      IF (VAL(11).EQ.0) OPEN (T25,FILE=TAPE25)
C
      IF (VAL(1).NE.0) THEN
          WRITE (T25,FMT=*) ' np = ',NP
      END IF
      IF (VAL(2).EQ.1) THEN
          WRITE (T25,FMT=*) ' param = ',PARAM1
          NU = PARAM1
          KK = PARAM1
          ALPHA = PARAM1
          GAMMA = PARAM1
      ELSE IF (VAL(2).EQ.2) THEN
          WRITE (T25,FMT=*) ' param = ',PARAM1,PARAM2
          NU = PARAM1
          KK = PARAM1
          ALPHA = PARAM1
          GAMMA = PARAM1
          H = PARAM2
          BETA = PARAM2
      ELSE IF (VAL(2).EQ.5) THEN
          WRITE (T25,FMT=*) ' param = ',P1,P2,P3
          WRITE (T25,FMT=*) '         ',P4,P5,P6
      END IF
      IF (VAL(3).NE.0) THEN
          WRITE (T25,FMT=*) ' a = ',A
          IF (CLASSA.NE.'r') THEN
              CALL PQ(-ONE,P0ATA,QFATA)
              WRITE (T25,FMT=*) '   P0ATA,QFATA = ',P0ATA,QFATA
          END IF
      ELSE
          WRITE (T25,FMT=*) ' A = ',' -INF'
      END IF
      IF (VAL(5).NE.0) WRITE (T25,FMT=*) '   CLASSA = ',CLASSA
      IF (VAL(7).EQ.2) WRITE (T25,FMT=*) '   A1,A2 = ',A1,A2
      IF (VAL(4).NE.0) THEN
          WRITE (T25,FMT=*) ' b = ',B
          IF (CLASSB.NE.'r') THEN
              CALL PQ(ONE,P0ATB,QFATB)
              WRITE (T25,FMT=*) '   P0ATB,QFATB = ',P0ATB,QFATB
          END IF
      ELSE
          WRITE (T25,FMT=*) ' B = ','+INF'
      END IF
      IF (VAL(6).NE.0) WRITE (T25,FMT=*) '   CLASSB = ',CLASSB
      IF (VAL(8).EQ.2) WRITE (T25,FMT=*) '   B1,B2 = ',B1,B2
      IF (VAL(9).NE.0) THEN
          IF (VAL(9).EQ.5) WRITE (T25,FMT=*) ' BCC = G '
          WRITE (T25,FMT=*) '    ALFA = ',ALF
          WRITE (T25,FMT=*) ' K11,K12 = ',K11,K12
          WRITE (T25,FMT=*) ' K21,K22 = ',K21,K22
          DET = K11*K22 - K12*K21
          WRITE (T25,FMT=*) ' DET = ',DET
      END IF
      IF (VAL(10).EQ.1) THEN
          WRITE (T25,FMT=*) ' NUMEIG = ',NEIG1
      ELSE IF (VAL(10).EQ.2) THEN
          WRITE (T25,FMT=*) ' NUMEIG1,NUMEIG2 = ',NEIG1,NEIG2
      END IF
C
      WRITE (T25,FMT=*)

C  HERE, VAL(9) = 0 MEANS THE BOUNDARY CONDITIONS ARE SEPARATED.

      IF (VAL(9).EQ.0) THEN
          DO 10 I = NEIG1,NEIG2
              NUMEIG = I
              EIG = 0.0D0
              TOL = ZER
              IFLAG = 1
              AS = A
              BS = B
              INTABS = INTAB
c             CALL SLEIGN(A,B,INTAB,P0ATA,QFATA,P0ATB,QFATB,A1,A2,B1,B2,
              CALL SLEIGN(AS,BS,INTABS,P0ATA,QFATA,P0ATB,QFATB,A1,A2,B1,
     +                    B2,NUMEIG,EIG,TOL,IFLAG,0,SLFUN,NCA,NCB)
c    +                    NUMEIG,EIG,TOL,IFLAG,0,SLFUN,NCA,NCB)
              IF (IFLAG.EQ.15) THEN
                  WRITE (T25,FMT=*)
     +              ' WE CANNOT HANDLE THIS KIND OF ENDPOINT. '
                  WRITE (*,FMT=*)
                  GO TO 12
              END IF
              IFLAG = MIN(IFLAG,4)
              JFLAG = 0
              IF (SLFUN(9).GT.-10000.0D0) JFLAG = 1
              IF (SLFUN(9).LT.10000.0D0 .AND. JFLAG.EQ.1) JFLAG = 2
              IF (IFLAG.EQ.4) WRITE (*,FMT=9045) NUMEIG
              IF (IFLAG.EQ.4) WRITE (T25,FMT=9045) NUMEIG
              IF (IFLAG.LE.3) WRITE (*,FMT=*) ' IFLAG = ',IFLAG
              IF (IFLAG.LE.3) WRITE (T25,FMT=*) ' IFLAG = ',IFLAG
              IF (IFLAG.EQ.3) THEN
                IF (NUMEIG.GE.0) THEN
                  WRITE (T25,FMT=9230) NUMEIG
                ELSE IF( .NOT.(NCA.EQ.4 .OR. NCB.EQ.4)) THEN
                  WRITE (T25,FMT=9235)
                ELSE
                  WRITE (T25,FMT=9240)
                ENDIF
              ELSE IF (IFLAG.LE.2) THEN
                  WRITE (*,FMT=*) ' NUMEIG,EIG,TOL = ',NUMEIG,EIG,TOL
                  WRITE (T25,FMT=*) ' NUMEIG,EIG,TOL = ',NUMEIG,EIG,TOL
              END IF
   10     CONTINUE
   12     CONTINUE
          WRITE (*,FMT=*)
          WRITE (T25,FMT=*)
          IF (JFLAG.EQ.1) WRITE (*,FMT=9260)
          IF (JFLAG.EQ.2) WRITE (*,FMT=9250) SLFUN(9)
          IF (JFLAG.EQ.1) WRITE (T25,FMT=9260)
          IF (JFLAG.EQ.2) WRITE (T25,FMT=9250) SLFUN(9)
      ELSE
          DO 20 I = NEIG1,NEIG2
              NUMEIG = I
              EIG = 0.0D0
              TOL = ZER
              AS = A
              BS = B
              INTABS = INTAB
              IFLAG = 1
              CALL PERIO(AS,BS,INTABS,P0ATA,QFATA,P0ATB,QFATB,A1,A2,B1,
     +                   B2,NUMEIG,EIG,TOL,IFLAG,SLFUN,NCA,NCB,ALF,K11,
     +                   K12,K21,K22)
              IFLAG = MIN(IFLAG,4)
              IF (IFLAG.EQ.0) IFLAG = 4
              IF (IFLAG.EQ.4) WRITE (*,FMT=9045) NUMEIG
              IF (IFLAG.EQ.4) WRITE (T25,FMT=9045) NUMEIG
              IF (IFLAG.LE.2) THEN
                  WRITE (*,FMT=*) ' IFLAG = ',IFLAG
                  WRITE (T25,FMT=*) ' IFLAG = ',IFLAG
                  WRITE (*,FMT=*) ' NUMEIG,EIG,TOL = ',NUMEIG,EIG,TOL
                  WRITE (T25,FMT=*) ' NUMEIG,EIG,TOL = ',NUMEIG,EIG,TOL
              END IF
              IF (IFLAG.EQ.2) WRITE (T25,FMT=*
     +            ) ' THIS EIGENVALUE APPEARS TO BE DOUBLE. '
   20     CONTINUE
      END IF
      WRITE (T25,FMT=*) '______________________________________________'
C
C  HERE, VAL(12) .NE. 0 MEANS THAT ANOTHER PROBLEM IS TO BE RUN.
C  SO ANY CHANGES IN THE ENDPOINTS, OR BOUNDARY CONDITIONS, OR
C  ETC., MUST BE READ IN.  ANY PARAMETERS THAT ARE NOT CHANGED
C  ARE KEPT AS BEFORE.

      IF (VAL(12).NE.0) THEN
          REWIND T24
          DO 80 I = 1,50
              CH(I) = BLANK
   80     CONTINUE
          LAST = LAST + LASTI
          READ (T24,FMT=9010) (CHT,I=1,LAST)
          READ (T24,FMT=9010,END=85) (CH(I),I=1,50)
   85     CONTINUE
          VAL(12) = 0
          GO TO 300
      END IF
C
      CLOSE (T21)
      CLOSE (T24)
      WRITE (T25,FMT=*) 'end ',TAPE25
      CLOSE (T25)
C
      STOP
 9010 FORMAT (A62)
 9045 FORMAT (1X,'   NUMEIG = ',I5,' : COMPUTATION FAILED ')
 9230 FORMAT (1X,' * THERE SEEMS TO BE NO EIGENVALUE OF INDEX       *',
     +       /,1X,' * GREATER THAN',I5,'                              *'
     +       )
 9235 FORMAT (1X,' * THERE SEEM TO BE NO EIGENVALUES                *')
 9240 FORMAT (1X,' * THERE SEEMS TO BE NO EIGENVALUE OF THIS INDEX  *')
 9250 FORMAT (1X,' * THERE SEEMS TO BE CONTINUOUS SPECTRUM BEGINNING*',
     +       /,1X,' * AT ABOUT',1P,D8.1,
     +       '                               *')
 9260 FORMAT (1X,' * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *')
      END
c
      SUBROUTINE STR2R(CHIN,P)
C  THIS PROGRAM EXPECTS A CHARACTER STRING, CHIN, CONTAINING
C    DIGITS AND, POSSIBLY, COMMAS.  IT CONVERTS THE FIRST SUBSTRING
C    TO BE THE CORRESPONDING REAL NUMBER P.
C    THUS THE STRING
C        4.0,2.0,1.5,2.5
C    WOULD CAUSE P TO BE SET EQUAL TO THE REAL NUMBER 4.0,
C    AND CHIN ON OUTPUT WOULD BE THE STRING
C        2.0,1.5,2.5
C     .. Scalar Arguments ..
      REAL P
      CHARACTER*62 CHIN
C     ..
C     .. Local Scalars ..
      INTEGER I,J,K
      CHARACTER CH1
      CHARACTER*62 CHINN,CHOUT
C     ..
      CH1 = ','
      CHINN = '                               '
      CHOUT = '                               '
      DO 10 I = 1,60
          IF ((CHIN(I:I).EQ.CH1) .OR. (I.GT.1.AND.CHIN(I:I).EQ.' '))
     +        THEN
              J = I - 1
              GO TO 20
          END IF
   10 CONTINUE
   20 CONTINUE
      DO 30 I = 1,J
          CHINN(I:I) = CHIN(I:I)
   30 CONTINUE
      READ (CHINN,FMT=9020) P
      DO 40 I = J + 2,62
          K = I - J - 1
          CHOUT(K:K) = CHIN(I:I)
   40 CONTINUE
      CHIN = CHOUT
 9020 FORMAT (5F12.7)
      RETURN
      END
c
      SUBROUTINE CHAR(CHIN,K,M,CHOUT)
C
C  THIS PROGRAM IS INTENDED TO READ A LINE LIKE
C     '  abcd:  0.532,12.34,-57.000,0.7693 '
C  AND RETURN WITH
C    K = 9     (the number of characters up to the first non-blank
C                one after the : )
C    M = 3     (the number of commas after the : )
C  AND WITH
C    CHOUT = '0.532,12.34,-57.000,0.7693'
C  THUS READING THE LINE
C    'param:  1.23,-0.43,22.7,0.0037,-11.21'
C  THE RESULT WOULD BE 
C     K = 8
C     M = 4
C     CHOUT = 1.23,-0.43,22.7,0.0037,-11.21
C
C     .. Scalar Arguments ..
      INTEGER K,M
      CHARACTER*62 CHIN,CHOUT
C     ..
C     .. Local Scalars ..
      INTEGER I,J,L
      CHARACTER*16 CH2
C     ..
C     .. Local Arrays ..
      CHARACTER CH(62)
C     ..
      M = 0
      DO 10 I = 1,8
          READ (CHIN,FMT=100) (CH(J),J=1,I)
          IF (CH(I).EQ.':') GO TO 20
          M = I
   10 CONTINUE
   20 CONTINUE
      M = 0
      J = 0
      K = 0
      DO 30 I = 1,12
          READ (CHIN,FMT=100) (CH(L),L=1,I)
          IF (J.NE.0 .AND. K.EQ.0 .AND. CH(I).NE.' ') K = I
          IF (M.NE.0 .AND. K.EQ.0 .AND. CH(I).EQ.' ') J = I
          IF (CH(I).EQ.':') M = I
   30 CONTINUE
      M = 0
      DO 45 I = 1,62
          READ (CHIN,FMT=100) (CH(L),L=1,I)
          IF (CH(I).EQ.',') M = M + 1
   45 CONTINUE
      K = J
      IF (K.EQ.1) THEN
          READ (CHIN,FMT=101) CH2,CHOUT
      ELSE IF (K.EQ.2) THEN
          READ (CHIN,FMT=102) CH2,CHOUT
      ELSE IF (K.EQ.3) THEN
          READ (CHIN,FMT=103) CH2,CHOUT
      ELSE IF (K.EQ.4) THEN
          READ (CHIN,FMT=104) CH2,CHOUT
      ELSE IF (K.EQ.5) THEN
          READ (CHIN,FMT=105) CH2,CHOUT
      ELSE IF (K.EQ.6) THEN
          READ (CHIN,FMT=106) CH2,CHOUT
      ELSE IF (K.EQ.7) THEN
          READ (CHIN,FMT=107) CH2,CHOUT
      ELSE IF (K.EQ.8) THEN
          READ (CHIN,FMT=108) CH2,CHOUT
      ELSE IF (K.EQ.9) THEN
          READ (CHIN,FMT=109) CH2,CHOUT
      ELSE IF (K.EQ.10) THEN
          READ (CHIN,FMT=110) CH2,CHOUT
      ELSE IF (K.EQ.11) THEN
          READ (CHIN,FMT=111) CH2,CHOUT
      ELSE IF (K.EQ.12) THEN
          READ (CHIN,FMT=112) CH2,CHOUT
      END IF
      RETURN
  100 FORMAT (62A1)
  101 FORMAT (A1,A61)
  102 FORMAT (A2,A60)
  103 FORMAT (A3,A59)
  104 FORMAT (A4,A58)
  105 FORMAT (A5,A57)
  106 FORMAT (A6,A56)
  107 FORMAT (A7,A55)
  108 FORMAT (A8,A54)
  109 FORMAT (A9,A53)
  110 FORMAT (A10,A52)
  111 FORMAT (A11,A51)
  112 FORMAT (A12,A50)
      END
C
      SUBROUTINE LST(CHANS,A)
C  THIS PROGRAM CONVERTS A CHARACTER STRING CHANS TO A REAL
C    NUMBER A.  IT JUST ALLOWS THE NUMBERS PI, PI/2, PI/4,
C    2PI TO BE READ IN AS CHARACTERS INSTEAD OF HAVING TO
C    ENTER THEM AS DECIMAL DIGITS.
C
C     .. Scalar Arguments ..
      REAL A
      CHARACTER*32 CHANS
C     ..
C     .. Local Scalars ..
      REAL ONE,PI,PI4
      INTEGER I
C     ..
C     .. Local Arrays ..
      REAL Y(48)
      CHARACTER*8 X(48)
C     ..
C     .. Intrinsic Functions ..
      INTRINSIC ATAN
C     ..
      ONE = 1.0D0
      PI4 = ATAN(ONE)
      PI = 4.0D0*PI4
c
      X(1) = '-PI'
      Y(1) = -PI
      X(2) = '-2.0*PI'
      Y(2) = -2.0D0*PI
      X(3) = '-2.*PI'
      Y(3) = -2.0D0*PI
      X(4) = '-.5*PI'
      Y(4) = -0.5D0*PI
      X(5) = '-0.5*PI'
      Y(5) = -0.5D0*PI
      X(6) = '-.25*PI'
      Y(6) = -0.25D0*PI
      X(7) = '-0.25*PI'
      Y(7) = -0.25D0*PI
      X(8) = '-PI/2'
      Y(8) = -0.5D0*PI
      X(9) = '-PI/4'
      Y(9) = -0.25D0*PI
c
      X(10) = '-pi'
      Y(10) = -PI
      X(11) = '-2.0*pi'
      Y(11) = -2.0D0*PI
      X(12) = '-2.*pi'
      Y(12) = -2.0D0*PI
      X(13) = '-.5*pi'
      Y(13) = -0.5D0*PI
      X(14) = '-0.5*pi'
      Y(14) = -0.5D0*PI
      X(15) = '-.25*pi'
      Y(15) = -0.25D0*PI
      X(16) = '-0.25*pi'
      Y(16) = -0.25D0*PI
      X(17) = '-pi/2'
      Y(17) = -0.5D0*PI
      X(18) = '-pi/4'
      Y(18) = -0.25D0*PI
c
      X(19) = 'PI'
      Y(19) = PI
      X(20) = '2.0*PI'
      Y(20) = 2.0D0*PI
      X(21) = '2.*PI'
      Y(21) = 2.0D0*PI
      X(22) = '.5*PI'
      Y(22) = 0.5D0*PI
      X(23) = '.25*PI'
      Y(23) = 0.25D0*PI
      X(24) = 'PI/2'
      Y(24) = 0.5D0*PI
      X(25) = 'PI/4'
      Y(25) = 0.25D0*PI
c
      X(26) = 'pi'
      Y(26) = PI
      X(27) = '2.0*pi'
      Y(27) = 2.0D0*PI
      X(28) = '2.*pi'
      Y(28) = 2.0D0*PI
      X(29) = '.5*pi'
      Y(29) = 0.5D0*PI
      X(30) = '.25*pi'
      Y(30) = 0.25D0*PI
      X(31) = 'pi/2'
      Y(31) = 0.5D0*PI
      X(32) = 'pi/4'
      Y(32) = 0.25D0*PI
c
      X(33) = '0.5*PI'
      Y(33) = 0.5D0*PI
      X(34) = '0.5*pi'
      Y(34) = 0.5D0*PI
      X(35) = '-0.5*PI'
      Y(35) = -0.5D0*PI
      X(36) = '-0.5*pi'
      Y(36) = -0.5D0*PI
      X(37) = 'PI/2.0'
      Y(37) = PI/2.0D0
      X(38) = 'pi/2.0'
      Y(38) = PI/2.0D0
      X(39) = '-PI/2.0'
      Y(39) = -PI/2.0D0
      X(40) = '-pi/2.0'
      Y(40) = -PI/2.0D0
      X(41) = 'PI/4.0'
      Y(41) = PI/4.0D0
      X(42) = 'pi/4.0'
      Y(42) = PI/4.0D0
      X(43) = '-PI/4.0'
      Y(43) = -PI/4.0D0
      X(44) = '-pi/4.0'
      Y(44) = -PI/4.0D0
      X(45) = '-0.25*PI'
      Y(45) = -0.25D0*PI
      X(46) = '-0.25*pi'
      Y(46) = -0.25D0*PI
      X(47) = '0.25*PI'
      Y(47) = 0.25D0*PI
      X(48) = '0.25*pi'
      Y(48) = 0.25D0*PI
C
      DO 10 I = 1,48
          IF (CHANS.EQ.X(I)) THEN
              A = Y(I)
              RETURN
          END IF
   10 CONTINUE
      READ (CHANS,FMT='(F32.0)') A
C
      RETURN
      END
C
      SUBROUTINE HELP(NH)
C
C     .. Scalar Arguments ..
      INTEGER NH
C     ..
C     .. Local Scalars ..
      INTEGER I,N
      CHARACTER ANS
C     ..
C     .. Local Arrays ..
      CHARACTER*36 X(23),Y(23)
C     ..
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) NH
c
    1 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) 'H1:  Overview of HELP.'
      X(1) = '  This ASCII text file is supplied a'
      Y(1) = 's a separate file with the sleign2  '
      X(2) = 'package; it can be accessed on-line '
      Y(2) = 'in both MAKEPQW (if used) and DRIVE.'
      X(3) = '  HELP contains information to aid t'
      Y(3) = 'he user in entering data on the co- '
      X(4) = 'efficient functions p,q,w; on the se'
      Y(4) = 'lf-adjoint, separated and coupled,  '
      X(5) = 'regular and singular, boundary condi'
      Y(5) = 'tions; on the limit circle boundary '
      X(6) = 'condition functions u,v at a and U,V'
      Y(6) = ' at b; on the end-point classifica- '
      X(7) = 'tions of the differential equation; '
      Y(7) = 'on DEFAULT entry; on eigenvalue in- '
      X(8) = 'dexes; on IFLAG information; and on '
      Y(8) = 'the general use of the program      '
      X(9) = 'sleign2.                            '
      Y(9) = ' '
      X(10) = '  The 17 sections of HELP are:      '
      Y(10) = ' '
      X(11) = ' '
      Y(11) = ' '
      X(12) = '    H1: Overview of HELP.           '
      Y(12) = ' '
      X(13) = '    H2: File name entry.            '
      Y(13) = ' '
      X(14) = '    H3: The differential equation.  '
      Y(14) = ' '
      X(15) = '    H4: End-point classification.   '
      Y(15) = ' '
      X(16) = '    H5: DEFAULT entry.              '
      Y(16) = ' '
      X(17) = '    H6: Self-adjoint limit-circle bo'
      Y(17) = 'undary conditions.                  '
      DO 101 I = 1,17
          WRITE (*,FMT=*) X(I),Y(I)
  101 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
      X(1) = '    H7: General self-adjoint boundar'
      Y(1) = 'y conditions. '
      X(2) = '    H8: Recording the results.      '
      Y(2) = ' '
      X(3) = '    H9: Type and choice of interval.'
      Y(3) = ' '
      X(4) = '   H10: Entry of end-points.        '
      Y(4) = ' '
      X(5) = '   H11: End-point values of p,q,w.  '
      Y(5) = ' '
      X(6) = '   H12: Initial value problems.     '
      Y(6) = ' '
      X(7) = '   H13: Indexing of eigenvalues for '
      Y(7) = 'separated boundary conditions.      '
      X(8) = '   H14: Entry of eigenvalue index, i'
      Y(8) = 'nitial guess, and tolerance.     '
      X(9) = '   H15: IFLAG information.          '
      Y(9) = ' '
      X(10) = '   H16: Plotting.                  '
      Y(10) = ' '
      X(11) = '   H17: Indexing of eigenvalues for'
      Y(11) = 'coupled boundary conditions.       '
      X(12) = ' '
      Y(12) = ' '
      X(13) = '  HELP can be accessed at each point'
      Y(13) = ' in MAKEPQW and DRIVE where the user'
      X(14) = 'is asked for input, by pressing "h <'
      Y(14) = 'ENTER>"; this places the user at the'
      X(15) = 'appropriate HELP section.  Once in H'
      Y(15) = 'ELP, the user can scroll the further'
      X(16) = 'HELP sections by repeatedly pressing'
      Y(16) = ' "h <ENTER>", or jump to a specific '
      X(17) = 'HELP section Hn (n=1,2,...17) by typ'
      Y(17) = 'ing "Hn <ENTER>"; to return to the  '
      X(18) = 'place in the program from which HELP'
      Y(18) = ' is called, press "r <ENTER>".      '
      DO 102 I = 1,18
          WRITE (*,FMT=*) X(I),Y(I)
  102 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
    2 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) 'H2:  File name entry.'
      X(1) = '  MAKEPQW is used to create a FORTRA'
      Y(1) = 'N file containing the coefficients  '
      X(2) = 'p(x),q(x),w(x), defining the differe'
      Y(2) = 'ntial equation, and the boundary    '
      X(3) = 'condition functions u(x),v(x) and U('
      Y(3) = 'x),V(x) if required.  The file must '
      X(4) = 'be given a NEW filename which is acc'
      Y(4) = 'eptable to your FORTRAN compiler.   '
      X(5) = 'For example, it might be called bess'
      Y(5) = 'el.f or bessel.for, depending upon  '
      X(6) = 'your compiler.                      '
      Y(6) = ' '
      X(7) = '  The same naming considerations app'
      Y(7) = 'ly if the FORTRAN file is prepared  '
      X(8) = 'other than with the use of MAKEPQW. '
      Y(8) = ' '
      DO 201 I = 1,8
          WRITE (*,FMT=*) X(I),Y(I)
  201 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
    3 CONTINUE
      WRITE (*,FMT=*) 'H3:  The differential equation.'
      X(1) = '  The prompt "Input p (or q or w) ="'
      Y(1) = ' requests you to type in a FORTRAN  '
      X(2) = 'expression defining the function p(x'
      Y(2) = '), which is one of the three coeffi-'
      X(3) = 'cient functions defining the Sturm-L'
      Y(3) = 'iouville differential equation      '
      X(4) = ' '
      Y(4) = ' '
      X(5) = '                   -(p*y'')'' + q*y = '
      Y(5) = ' lambda*w*y              (*)        '
      X(6) = ' '
      Y(6) = ' '
      X(7) = 'to be considered on some interval (a'
      Y(7) = ',b) of the real line.  The actual   '
      X(8) = 'interval used in a particular proble'
      Y(8) = 'm can be chosen later, and may be   '
      X(9) = 'either the whole interval (a,b) wher'
      Y(9) = 'e the coefficient functions p,q,w,  '
      X(10) = 'etc. are defined or any sub-interval'
      Y(10) = ' (a'',b'') of (a,b); a = -infinity  '
      X(11) = 'and/or b = +infinity are allowable c'
      Y(11) = 'hoices for the end-points.          '
      X(12) = '  The coefficient functions p,q,w of'
      Y(12) = ' the differential equation may be   '
      X(13) = 'chosen arbitrarily but must satisfy '
      Y(13) = 'the following conditions:           '
      X(14) = '  1. p,q,w are real-valued throughou'
      Y(14) = 't (a,b).                            '
      X(15) = '  2. p,q,w are piece-wise continuous'
      Y(15) = ' and defined throughout the         '
      X(16) = '     interior of the interval (a,b).'
      Y(16) = '                                    '
      X(17) = '  3. p and w are strictly positive i'
      Y(17) = 'n (a,b).                            '
      X(18) = '     For better error analysis in th'
      Y(18) = 'e numerical procedures, condition 2.'
      X(19) = '     above is often replaced with   '
      Y(19) = ' '
      X(20) = '  2''. p,q,w are four times continuou'
      Y(20) = 'sly differentiable on (a,b).        '
      X(21) = '  The behavior of p,q,w near the end'
      Y(21) = '-points a and b is critical to the  '
      X(22) = 'classification of the differential e'
      Y(22) = 'quation (see H4 and H11).           '
      DO 301 I = 1,22
          WRITE (*,FMT=*) X(I),Y(I)
  301 CONTINUE
      WRITE (*,FMT=*)
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
    4 CONTINUE
      WRITE (*,FMT=*) 'H4:  End-point classification.'
      X(1) = '  The correct classification of the '
      Y(1) = 'end-points a and b is essential to  '
      X(2) = 'the working of the sleign2 program. '
      Y(2) = ' To classify the end-points, it is  '
      X(3) = 'convenient to choose a point c in (a'
      Y(3) = ',b); i.e.,  a < c < b.  Subject to  '
      X(4) = 'the general conditions on the coeffi'
      Y(4) = 'cient functions p,q,w (see H3):     '
      X(5) = '  1. a is REGULAR (say R) if -infini'
      Y(5) = 'ty < a, p,q,w are piece-wise        '
      X(6) = '     continuous on [a,c], and p(a) >'
      Y(6) = ' 0 and w(a) > 0.                    '
      X(7) = '  2. a is WEAKLY REGULAR (say WR) if'
      Y(7) = ' -infinity < a, a is not R, and     '
      X(8) = '                     |c             '
      Y(8) = ' '
      X(9) = '            integral | {1/p+|q|+w} <'
      Y(9) = ' +infinity.                         '
      X(10) = '                     |a             '
      Y(10) = ' '
      X(11) = '     If end-point a is neither R nor'
      Y(11) = ' WR, then a is SINGULAR; that is,   '
      X(12) = '     either -infinity = a, or -infin'
      Y(12) = 'ity < a and                         '
      X(13) = '                     |c             '
      Y(13) = ' '
      X(14) = '            integral | {1/p+|q|+w} ='
      Y(14) = ' +infinity.                         '
      X(15) = '                     |a             '
      Y(15) = ' '
      X(16) = '  3. The SINGULAR end-point a is LIM'
      Y(16) = 'IT-CIRCLE NON-OSCILLATORY (say      '
      X(17) = '     LCNO) if for some real lambda A'
      Y(17) = 'LL real-valued solutions y of the   '
      X(18) = '     differential equation          '
      Y(18) = ' '
      X(19) = ' '
      Y(19) = ' '
      X(20) = '                   -(p*y'')'' + q*y = '
      Y(20) = 'lambda*w*y on (a,c]     (*)         '
      X(21) = ' '
      Y(21) = ' '
      X(22) = '     satisfy the conditions:        '
      Y(22) = ' '
      DO 401 I = 1,22
          WRITE (*,FMT=*) X(I),Y(I)
  401 CONTINUE
      WRITE (*,FMT=*)
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
      X(1) = '                     |c             '
      Y(1) = ' '
      X(2) = '            integral | { w*y*y } < +'
      Y(2) = 'infinity, and                       '
      X(3) = '                     |a             '
      Y(3) = ' '
      X(4) = '     y has at most a finite number o'
      Y(4) = 'f zeros in (a,c].                   '
      X(5) = '  4. The SINGULAR end-point a is LIM'
      Y(5) = 'IT-CIRCLE OSCILLATORY (say LCO) if  '
      X(6) = '     for some real lambda ALL real-v'
      Y(6) = ' alued solutions of the differential'
      X(7) = '     equation (*) satisfy the condit'
      Y(7) = 'ions:                               '
      X(8) = '                     |c             '
      Y(8) = ' '
      X(9) = '            integral | { w*y*y } < +'
      Y(9) = 'infinity, and                       '
      X(10) = '                     |a             '
      Y(10) = ' '
      X(11) = '     y has an infinite number of zer'
      Y(11) = 'os in (a,c].                        '
      X(12) = '  5. The SINGULAR end-point a is LIM'
      Y(12) = 'IT POINT (say LP) if for some real  '
      X(13) = '     lambda at least one solution of'
      Y(13) = ' the differential equation (*) sat- '
      X(14) = '     isfies the condition:          '
      Y(14) = ' '
      X(15) = '                     |c             '
      Y(15) = ' '
      X(16) = '            integral | {w*y*y} = +in'
      Y(16) = 'finity.                             '
      X(17) = '                     |a             '
      Y(17) = ' '
      X(18) = '  There is a similar classification '
      Y(18) = 'of the end-point b into one of the  '
      X(19) = 'five distinct cases R, WR, LCNO, LCO'
      Y(19) = ', LP.                               '
      DO 402 I = 1,19
          WRITE (*,FMT=*) X(I),Y(I)
  402 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
      X(1) = '  Although the classification of sin'
      Y(1) = 'gular end-points invokes a real     '
      X(2) = 'value of the parameter lambda, this '
      Y(2) = 'classification is invariant in      '
      X(3) = 'lambda; all real choices of lambda l'
      Y(3) = 'ead to the same classification.     '
      X(4) = '  In determining the classification '
      Y(4) = 'of singular end-points for the      '
      X(5) = 'differential equation (*), it is oft'
      Y(5) = 'en convenient to start with the     '
      X(6) = 'choice lambda = 0 in attempting to f'
      Y(6) = 'ind solutions (particularly when    '
      X(7) = 'q = 0 on (a,b)); however, see exampl'
      Y(7) = 'e 7 below.                          '
      X(8) = '  See H6 on the use of maximal domai'
      Y(8) = 'n functions to determine the        '
      X(9) = 'classification at singular end-point'
      Y(9) = 's.                                  '
      DO 403 I = 1,9
          WRITE (*,FMT=*) X(I),Y(I)
  403 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
      WRITE (*,FMT=*) ' EXAMPLES: '
      X(1) = '  1. -y'''' = lambda*y is R at both en'
      Y(1) = 'd-points of (a,b) when a and b are  '
      X(2) = '     finite.                        '
      Y(2) = ' '
      X(3) = '  2. -y'''' = lambda*y on (-infinity,i'
      Y(3) = 'nfinity) is LP at both end-points.  '
      X(4) = '  3. -(sqrt(x)*y''(x))'' = lambda*(1./'
      Y(4) = 'sqrt(x))*y(x) on (0,infinity) is    '
      X(5) = '     WR at 0 and LP at +infinity (ta'
      Y(5) = 'ke lambda = 0 in (*)).  See         '
      X(6) = '     examples.f, #10 (Weakly Regular'
      Y(6) = ').                                  '
      X(7) = '  4. -((1-x*x)*y''(x))'' = lambda*y(x)'
      Y(7) = ' on (-1,1) is LCNO at both ends     '
      X(8) = '     (take lambda = 0 in (*)).  See '
      Y(8) = 'xamples.f, #1 (Legendre).           '
      X(9) = '  5. -y''''(x) + C*(1/(x*x))*y(x) = la'
      Y(9) = 'mbda*y(x) on (0,infinity) is LP at  '
      X(10) = '     infinity and at 0 is (take lamb'
      Y(10) = 'da = 0 in (*)):                     '
      X(11) = '       LP for C .ge. 3/4 ;          '
      Y(11) = ' '
      X(12) = '       LCNO for -1/4 .le. C .lt. 3/4'
      Y(12) = ' (but C .ne. 0);                    '
      X(13) = '       LCO for C .lt. -1/4.         '
      Y(13) = ' '
      X(14) = '  6. -(x*y''(x))'' - (1/x)*y(x) = lamb'
      Y(14) = 'da*y(x) on (0,infinity) is LCO at 0 '
      X(15) = '     and LP at +infinity (take lambd'
      Y(15) = 'a = 0 in (*) with solutions         '
      X(16) = '     cos(ln(x)) and sin(ln(x))).  Se'
      Y(16) = 'e xamples.f, #7 (BEZ).              '
      X(17) = '  7. -(x*y''(x))'' - x*y(x) = lambda*('
      Y(17) = '1/x)*y(x) on (0,infinity) is LP at 0'
      X(18) = '     and LCO at infinity (take lambd'
      Y(18) = 'a = -1/4 in (*) with solutions      '
      X(19) = '     cos(x)/sqrt(x) and sin(x)/sqrt('
      Y(19) = 'x)).  See xamples.f,                '
      X(20) = '     #6 (Sears-Titchmarsh).         '
      Y(20) = ' '
      X(21) = '  8. -y''''(x) + x*sin(x)*y(x) = lambd'
      Y(21) = 'a*y(x) on (0,infinity) is R at 0 and'
      X(22) = '     LP at infinity.  See examples.f'
      Y(22) = ' #30 (Littlewood-McLeod).           '
      DO 404 I = 1,22
          WRITE (*,FMT=*) X(I),Y(I)
  404 CONTINUE
      WRITE (*,FMT=*) ' '
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
    5 CONTINUE
      WRITE (*,FMT=*) 'H5:  DEFAULT entry.'
      X(1) = '  The complete range of problems for'
      Y(1) = ' which sleign2 is applicable can    '
      X(2) = 'only be reached by appropriate entri'
      Y(2) = 'es under end-point classification   '
      X(3) = 'and boundary conditions.  However, t'
      Y(3) = 'here is a DEFAULT application which '
      X(4) = 'requires no detailed entry of end-po'
      Y(4) = 'int classification or boundary      '
      X(5) = 'conditions, subject to:             '
      Y(5) = ' '
      X(6) = '  1. The DEFAULT application CANNOT '
      Y(6) = 'be used at a LCO end-point.         '
      X(7) = '  2. If an end-point a is R, then th'
      Y(7) = 'e Dirichlet boundary condition      '
      X(8) = '     y(a) = 0 is automatically used.'
      Y(8) = ' '
      X(9) = '  3. If an end-point a is WR, then t'
      Y(9) = 'he following boundary condition     '
      X(10) = '     is automatically applied:      '
      Y(10) = ' '
      X(11) = '       if p(a) = 0, and both q(a),w('
      Y(11) = 'a) are bounded, then the Neumann    '
      X(12) = '       boundary condition (py'')(a) '
      Y(12) = '= 0 is used, or                     '
      X(13) = '       if p(a) > 0, and q(a) and/or '
      Y(13) = 'w(a)) are not bounded, then the     '
      X(14) = '       Dirichlet boundary condition '
      Y(14) = 'y(a) = 0 is used.                   '
      X(15) = '       If p(a) = 0, and q(a) and/or '
      Y(15) = 'w(a) are not bounded, then no simple'
      X(16) = '       information, in general, can '
      Y(16) = ' be given on the DEFAULT boundary   '
      X(17) = '       condition.                   '
      Y(17) = ' '
      X(18) = '  4. If an end-point is LCNO, then i'
      Y(18) = 'n most cases the principal or       '
      X(19) = '     Friedrichs boundary condition i'
      Y(19) = 's applied (see H6).                 '
      X(20) = '  5. If an end-point is LP, then the'
      Y(20) = ' normal LP procedure is applied     '
      X(21) = '     (see H7(1.)).                  '
      Y(21) = ' '
      X(22) = 'If the DEFAULT condition is chosen, '
      Y(22) = 'then no entry is required for the   '
      X(23) = 'u,v and U,V boundary condition funct'
      Y(23) = 'ions. '
      DO 501 I = 1,23
          WRITE (*,FMT=*) X(I),Y(I)
  501 CONTINUE
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
    6 CONTINUE
      WRITE (*,FMT=*) 'H6:  Limit-circle boundary conditions.'
      X(1) = '  At an end-point a, the limit-circl'
      Y(1) = 'e type separated boundary condition '
      X(2) = 'is of the form (similar remarks thro'
      Y(2) = 'ughout apply to the end-point b with'
      X(3) = ' U,V being boundary condition functi'
      Y(3) = 'ons at b)                           '
      X(4) = ' '
      Y(4) = ' '
      X(5) = '       A1*[y,u](a) + A2*[y,v](a) = 0'
      Y(5) = ',                          (**)     '
      X(6) = ' '
      Y(6) = ' '
      X(7) = 'where y is a solution of the differe'
      Y(7) = 'ntial equation                      '
      X(8) = ' '
      Y(8) = ' '
      X(9) = '     -(p*y'')'' + q*y = lambda*w*y  on'
      Y(9) = ' (a,b).                     (*)     '
      X(10) = ' '
      Y(10) = ' '
      X(11) = 'Here A1, A2 are real numbers; u and '
      Y(11) = 'v are boundary condition functions  '
      X(12) = 'at a; and for real-valued y and u th'
      Y(12) = 'e form [y,u] is defined by          '
      X(13) = '                                    '
      Y(13) = '                                    '
      X(14) = '   [y,u](x) = y(x)*(pu'')(x) - u(x)*'
      Y(14) = '(py'')(x)   for x in (a,b).         '
      X(15) = '                                    '
      Y(15) = '                                    '
      X(16) = '  If neither end-point is LP then th'
      Y(16) = 'ere are also self-adjoint coupled   '
      X(17) = 'boundary conditions.  These have a c'
      Y(17) = 'anonical form given by              '
      X(18) = '                                    '
      Y(18) = '                                    '
      X(19) = '    Y(b) = exp(i*alpha)*K*Y(a),     '
      Y(19) = ' '
      X(20) = '                                    '
      Y(20) = '                                    '
      X(21) = 'where K is a real 2 by 2 matrix with'
      Y(21) = ' determinant 1, alpha is restricted '
      X(22) = 'to the interval (-pi,pi], and Y is  '
      Y(22) = '  '
      DO 551 I = 1,22
          WRITE (*,FMT=*) X(I),Y(I)
  551 CONTINUE
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
      X(1) = ' (i) the solution vector Y = transpo'
      Y(1) = 'se [y(a), (py'')(a)] at a regular   '
      X(2) = '     end-point a, or                '
      Y(2) = ' '
      X(3) = ' (ii) the "singular solution vector"'
      Y(3) = ' Y(a) = transpose [[y,u](a),        '
      X(4) = '      [y,v](a)] at a singular LC end'
      Y(4) = ' -point a.  Similarly at the right  '
      X(5) = '      end-point b with U,V.         '
      Y(5) = ' '
      X(6) = '  The object of this section is to p'
      Y(6) = 'rovide help in choosing appropriate '
      X(7) = 'functions u and v in (**) (or U,V), '
      Y(7) = 'given the differential equation (*).'
      X(8) = 'Full details of the boundary conditi'
      Y(8) = 'ons for (*) are discussed in H7;    '
      X(9) = 'here it is sufficient to say that th'
      Y(9) = 'e limit-circle type boundary condi- '
      X(10) = 'tion (**) can be applied at any end-'
      Y(10) = 'point in the LCNO, LCO classifica-  '
      X(11) = 'tion, but also in the R, WR classifi'
      Y(11) = 'cation subject to the appropriate   '
      X(12) = 'choice of u,v or U,V.               '
      Y(12) = ' '
      X(13) = '  Let (*) be R, WR, LCNO, or LCO at '
      Y(13) = 'end-point a and choose c in (a,b).  '
      X(14) = 'Then either                         '
      Y(14) = ' '
      X(15) = '  u and v are a pair of linearly ind'
      Y(15) = 'ependent solutions of (*) on (a,c]  '
      X(16) = '  for any chosen real values of lamb'
      Y(16) = 'da, or                              '
      X(17) = '  u and v are a pair of real-valued '
      Y(17) = ' maximal domain functions defined on'
      X(18) = '  (a,c] satisfying [u,v](a) .ne. 0. '
      Y(18) = ' The maximal domain D(a,c] is de-   '
      X(19) = '  defined by                        '
      Y(19) = ' '
      X(20) = ' '
      Y(20) = ' '
      X(21) = '       D(a,c] = {f:(a,c]->R:: f,pf'' '
      Y(21) = 'in AC(a,c];                         '
      X(22) = '                   f, ((-pf'')''+qf)/w'
      Y(22) = ' in L2((a,c;w)}  .                  '
      X(23) = ' '
      Y(23) = ' '
      DO 601 I = 1,23
          WRITE (*,FMT=*) X(I),Y(I)
  601 CONTINUE
      READ (*,FMT=999) ANS,N
      WRITE (*,FMT=*)
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
      X(1) = '  It is known that for all f,g in D('
      Y(1) = 'a,c] the limit                      '
      X(2) = ' '
      Y(2) = ' '
      X(3) = '       [f,g](a) = lim[f,g](x) as x->'
      Y(3) = 'a                                   '
      X(4) = ' '
      Y(4) = ' '
      X(5) = 'exists and is finite.  If (*) is LCN'
      Y(5) = 'O or LCO at a, then all solutions of'
      X(6) = 'of (*) belong to D(a,c] for all valu'
      Y(6) = 'es of lambda.                       '
      X(7) = '  The boundary condition (**) is ess'
      Y(7) = 'ential in the LCNO and LCO cases but'
      X(8) = 'can also be used with advantage in s'
      Y(8) = 'ome R and WR cases.  In the R, WR, '
      X(9) = 'and LCNO cases, but not in the LCO c'
      Y(9) = 'ase, the boundary condition         '
      X(10) = 'functions can always be chosen so th'
      Y(10) = 'at                                  '
      X(11) = '        lim u(x)/v(x) = 0 as x->a,  '
      Y(11) = ' '
      X(12) = 'and it is recommended that this norm'
      Y(12) = 'alisation be effected, but this is  '
      X(13) = 'not essential; this normalisation ha'
      Y(13) = 's been entered in the examples given'
      X(14) = 'below.  In this case, the boundary c'
      Y(14) = 'ondition [y,u](a) = 0 (i.e., A1 = 1,'
      X(15) = 'A2 = 0 in (**) is called the princip'
      Y(15) = 'al or Friedrichs boundary condition '
      X(16) = 'at a.                              '
      Y(16) = ' '
      X(17) = '  In the case when end-points a and '
      Y(17) = 'b are, independently, in R, WR,     '
      X(18) = 'LCNO, or LCO classification, it may '
      Y(18) = 'be that symmetry or other reasons   '
      X(19) = 'permit one set of boundary condition'
      Y(19) = ' functions to be used at both end-  '
      X(20) = 'points (see xamples.f, #1 (Legendre)'
      Y(20) = ').  In other cases, different pairs '
      X(21) = 'must be chosen for each end-point (s'
      Y(21) = 'ee xamples.f: #16 (Jacobi),         '
      X(22) = '#18 (Dunsch), and #19 (Donsch)).    '
      Y(22) = ' '
      DO 602 I = 1,22
          WRITE (*,FMT=*) X(I),Y(I)
  602 CONTINUE
      WRITE (*,FMT=*)
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
      X(1) = ' '
      Y(1) = ' '
      X(2) = '  Note that a solution pair u,v is a'
      Y(2) = 'lways a maximal domain pair, but not'
      X(3) = 'necessarily vice versa.             '
      Y(3) = ' '
      X(4) = ' '
      Y(4) = ' '
      X(5) = 'EXAMPLES:                           '
      Y(5) = ' '
      X(6) = '1. -y''''(x) = lambda*y(x) on [0,pi] i'
      Y(6) = 's R at 0 and R at pi.               '
      X(7) = '   At 0, with lambda = 0, a solution'
      Y(7) = ' pair is u(x) = x, v(x) = 1.        '
      X(8) = '   At pi, with lambda = 1, a solutio'
      Y(8) = 'n pair is                           '
      X(9) = '     u(x) = sin(x), v(x) = cos(x).  '
      Y(9) = ' '
      X(10) = '2. -(sqrt(x)*y''(x))'' = lambda*y(x)/s'
      Y(10) = 'qrt(x) on (0,1] is                  '
      X(11) = '     WR at 0 and R at 1.            '
      Y(11) = ' '
      X(12) = '   (The general solutions of this eq'
      Y(12) = 'uation are                          '
      X(13) = '     u(x) = cos(2*sqrt(x*lambda)), v'
      Y(13) = '(x) = sin(2*sqrt(x*lambda)).)       '
      X(14) = '   At 0, with lambda = 0, a solution'
      Y(14) = ' pair is                            '
      X(15) = '     u(x) = 2*sqrt(x), v(x) = 1.    '
      Y(15) = ' '
      X(16) = '   At 1, with lambda = pi*pi/4, a so'
      Y(16) = 'lution pair is                      '
      X(17) = '     u(x) = sin(pi*sqrt(x)), v(x) = '
      Y(17) = 'cos(pi*sqrt(x)).                    '
      X(18) = '   At 1, with lambda = 0, a solution'
      Y(18) = ' pair is                            '
      X(19) = '     u(x) = 2*(1-sqrt(x)), v(x) = 1.'
      Y(19) = ' '
      X(20) = '   See also xamples.f, #10 (Weakly R'
      Y(20) = 'egular).                            '
      X(21) = '3. -((1-x*x)*y''(x))'' = lambda*y(x) o'
      Y(21) = 'n (-1,1) is LCNO at both ends.      '
      X(22) = '   At +-1, with lambda = 0, a soluti'
      Y(22) = 'on pair is                          '
      X(23) = '     u(x) = 1, v(x) = 0.5*log((1+x)/'
      Y(23) = '(1-x)).                             '
      DO 603 I = 1,23
          WRITE (*,FMT=*) X(I),Y(I)
  603 CONTINUE
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
      X(1) = '   At 1, a maximal domain pair is u('
      Y(1) = 'x) = 1, v(x) = log(1-x)             '
      X(2) = '   At -1, a maximal domain pair is u'
      Y(2) = '(x) = 1, v(x) = log(1+x).           '
      X(3) = '   See also xamples.f, #1 (Legendre)'
      Y(3) = '.                                   '
      X(4) = '4. -y''''(x) - (1/(4x*x))*y(x) = lambd'
      Y(4) = 'a*y(x) on (0,infinity) is           '
      X(5) = '     LCNO at 0 and LP at +infinity. '
      Y(5) = ' '
      X(6) = '   At 0, a maximal domain pair is   '
      Y(6) = ' '
      X(7) = '     u(x) = sqrt(x), v(x) = sqrt(x)*'
      Y(7) = 'log(x).                             '
      X(8) = '   See also xamples.f, #2 (Bessel). '
      Y(8) = ' '
      X(9) = '5. -y''''(x) - 5*(1/(4*x*x))*y(x) = la'
      Y(9) = 'mbda*y(x) on (0,infinity) is        '
      X(10) = '     LCO at 0 and LP at +infinity.  '
      Y(10) = ' '
      X(11) = '   At 0, with lambda = 0, a solution'
      Y(11) = ' pair is                            '
      X(12) = '     u(x) = sqrt(x)*cos(log(x)), v(x'
      Y(12) = ') = sqrt(x)*sin(log(x))             '
      X(13) = '   See also xamples.f, #20 (Krall). '
      Y(13) = ' '
      X(14) = '6. -y''''(x) - (1/x)*y(x) = lambda*y(x'
      Y(14) = ') on (0,infinity) is               '
      X(15) = '     LCNO at 0 and LP at +infinity.'
      Y(15) = ' '
      X(16) = '   At 0, a maximal domain pair is   '
      Y(16) = ' '
      X(17) = '     u(x) = x, v(x) = 1 -x*log(x).  '
      Y(17) = ' '
      X(18) = '   See also xamples.f, #4(Boyd).    '
      Y(18) = ' '
      X(19) = '7. -((1/x)*y''(x))'' + (k/(x*x) + k*k/'
      Y(19) = 'x)*y(x) = lambda*y(x) on (0,1],     '
      X(20) = '     with k real and .ne. 0, is LCNO'
      Y(20) = ' at 0 and R at 1.                   '
      X(21) = '   At 0, a maximal domain pair is   '
      Y(21) = ' '
      X(22) = '     u(x) = x*x, v(x) = x - 1/k.    '
      Y(22) = ' '
      X(23) = '   See also xamples.f, #8 (Laplace T'
      Y(23) = 'idal Wave).                         '
      DO 604 I = 1,23
          WRITE (*,FMT=*) X(I),Y(I)
  604 CONTINUE
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
    7 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) 'H7:  General self-adjoint boundary conditions.'
      X(1) = '  Boundary conditions for Sturm-Liou'
      Y(1) = 'ville boundary value problems       '
      X(2) = ' '
      Y(2) = ' '
      X(3) = '                   -(p*y'')'' + q*y = '
      Y(3) = 'lambda*w*y              (*)         '
      X(4) = ' '
      Y(4) = ' '
      X(5) = 'on an interval (a,b) are either     '
      Y(5) = ' '
      X(6) = '  SEPARATED, with at most one condit'
      Y(6) = 'ion at end-point a and at most one  '
      X(7) = '    condition at end-point b, or    '
      Y(7) = ' '
      X(8) = '  COUPLED, when both a and b are, in'
      Y(8) = 'dependently, in one of the end-point'
      X(9) = '    classifications R, WR, LCNO, LCO'
      Y(9) = ', in which case two independent     '
      X(10) = '    boundary conditions are required'
      Y(10) = ' which link the solution values near'
      X(11) = '    a to those near b.              '
      Y(11) = ' '
      X(12) = '  The sleign2 program allows for all'
      Y(12) = ' self-adjoint boundary conditions:  '
      X(13) = 'separated self-adjoint conditions an'
      Y(13) = 'd all cases of coupled self-adjoint '
      X(14) = 'conditions.                         '
      Y(14) = ' '
      DO 701 I = 1,14
          WRITE (*,FMT=*) X(I),Y(I)
  701 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
      WRITE (*,FMT=*) '    Separated Conditions:      '
      WRITE (*,FMT=*) '    ---------------------      '
      X(1) = '  The boundary conditions to be sele'
      Y(1) = 'cted depend upon the classification '
      X(2) = 'of the differential equation at the '
      Y(2) = 'end-point, say, a:                  '
      X(3) = '  1. If the end-point a is LP, then '
      Y(3) = 'no boundary condition is required or'
      X(4) = '     allowed.                       '
      Y(4) = ' '
      X(5) = '  2. If the end-point a is R or WR, '
      Y(5) = 'then a separated boundary condition '
      X(6) = '     is of the form                 '
      Y(6) = ' '
      X(7) = '         A1*y(a) + A2*(py'')(a) = 0,'
      Y(7) = ' '
      X(8) = '     where A1, A2 are real constants'
      Y(8) = ' the user must choose, not both zero.  '
      X(9) = '  3. If the end-point a is LCNO or'
      Y(9) = ' LCO, then a separated boundary     '
      X(10) = '  condition is of the form          '
      Y(10) = ' '
      X(11) = '         A1*[y,u](a) + A2*[y,v](a) ='
      Y(11) = ' 0,                                 '
      X(12) = '  where A1, A2 are real constants th'
      Y(12) = 'e user must choose, not both zero;  '
      X(13) = '  here, u,v are the pair of boundary'
      Y(13) = ' condition functions you have       '
      X(14) = '  previously selected when the input'
      Y(14) = ' FORTRAN file was being prepared    '
      X(15) = '  with makepqw.f.                   '
      Y(15) = ' '
      X(16) = '  4. If the end-point a is LCNO an'
      Y(16) = 'd the boundary condition pair       '
      X(17) = '  u,v has been chosen so that       '
      Y(17) = ' '
      X(18) = '         lim u(x)/v(x) = 0  as x->a '
      Y(18) = ' '
      X(19) = '  (which is always possible), then A'
      Y(19) = '1 = 1, A2 = 0 (i.e., [y,u](a) = 0)  '
      X(20) = '  gives the principal (Friedrichs) b'
      Y(20) = 'oundary condition at a.             '
      DO 702 I = 1,20
          WRITE (*,FMT=*) X(I),Y(I)
  702 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
      X(1) = '  5. If a is R or WR and boundary '
      Y(1) = 'condition functions u,v have been   '
      X(2) = '  entered in the FORTRAN input file,'
      Y(2) = ' then (3.,4.) above apply to        '
      X(3) = '  entering separated boundary condit'
      Y(3) = 'ions at such an end-point; the      '
      X(4) = '  boundary conditions in this form a'
      Y(4) = 're equivalent to the point-wise     '
      X(5) = '  conditions in (2.) (subject to car'
      Y(5) = 'e in choosing A1, A2).  This        '
      X(6) = '  singular form of a regular boundar'
      Y(6) = 'y condition may be particularly     '
      X(7) = '  effective in the WR case if the bo'
      Y(7) = 'undary condition form in (2.) leads '
      X(8) = '  to numerical difficulties.        '
      Y(8) = ' '
      X(9) = '  Conditions (2.,3.,4.,5.) apply sim'
      Y(9) = 'ilarly at end-point b (with U,V as  '
      X(10) = 'the boundary condition functions at '
      Y(10) = 'b.                                  '
      X(11) = '    6. If a is R, WR, LCNO, or LCO a'
      Y(11) = 'nd b is LP, then only a separated   '
      X(12) = '  condition at a is required and all'
      Y(12) = 'owed (or instead at b if a and b    '
      X(13) = '  are interchanged).                '
      Y(13) = ' '
      X(14) = '    7. If both end-points a and b ar'
      Y(14) = 'e LP, then no boundary conditions   '
      X(15) = '  are required or allowed.          '
      Y(15) = ' '
      X(16) = '  The indexing of eigenvalues for bo'
      Y(16) = 'undary value problems with separated'
      X(17) = '  conditions is discussed in H13.   '
      Y(17) = ' '
      X(18) = '    Coupled Conditions:             '
      Y(18) = ' '
      X(19) = '    -------------------             '
      Y(19) = ' '
      X(20) = '    8. Coupled regular self-adjoint '
      Y(20) = 'boundary conditions on (a,b) apply  '
      X(21) = 'only when both end-points a and b ar'
      Y(21) = 'e R or WR.                          '
      X(22) = ' '
      Y(22) = ' '
      DO 704 I = 1,22
          WRITE (*,FMT=*) X(I),Y(I)
  704 CONTINUE
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
    8 CONTINUE
      WRITE (*,FMT=*) 'H8:  Recording the results.'
      X(1) = '  If you choose to have a record kep'
      Y(1) = 't of the results, then the following'
      X(2) = 'information is stored in a file with'
      Y(2) = ' the name you select:               '
      X(3) = ' '
      Y(3) = ' '
      X(4) = '  1. The file name.                 '
      Y(4) = ' '
      X(5) = '  2. The header line prompted for (u'
      Y(5) = 'p to 32 characters of your choice). '
      X(6) = '  3. The interval (a,b) which was us'
      Y(6) = 'ed.                                 '
      X(7) = ' '
      Y(7) = ' '
      X(8) = '  For SEPARATED boundary conditions:'
      Y(8) = ' '
      X(9) = '  4. The end-point classification.  '
      Y(9) = ' '
      X(10) = '  5. A summary of coefficient inform'
      Y(10) = 'ation at WR, LCNO, LCO end-points.  '
      X(11) = '  6. The boundary condition constant'
      Y(11) = 's (A1,A2), (B1,B2) if entered.      '
      X(12) = '  7. (NUMEIG,EIG,TOL) or (NUMEIG1,NU'
      Y(12) = 'MEIG2,TOL), as entered.             '
      X(13) = ' '
      Y(13) = ' '
      X(14) = '  For COUPLED boundary conditions:  '
      Y(14) = ' '
      X(15) = '  8. The boundary condition paramete'
      Y(15) = 'r alpha and the coupling matrix K;  '
      X(16) = '     see H6.                        '
      Y(16) = ' '
      X(17) = '  For ALL self-adjoint boundary cond'
      Y(17) = 'itions: '
      X(18) = '  9. The computed eigenvalue, EIG, a'
      Y(18) = 'nd its estimated accuracy, TOL.     '
      X(19) = ' 10. IFLAG reported (see H15).      '
      Y(19) = ' '
      DO 801 I = 1,19
          WRITE (*,FMT=*) X(I),Y(I)
  801 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' '
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
    9 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) 'H9:  Type and choice of interval.'
      X(1) = '  You may enter any interval (a,b) f'
      Y(1) = 'or which the coefficients p,q,w are '
      X(2) = 'well-defined by your FORTRAN stateme'
      Y(2) = 'nts in the input file, provided that'
      X(3) = '(a,b) contains no interior singulari'
      Y(3) = 'ties.                               '
      DO 901 I = 1,3
          WRITE (*,FMT=*) X(I),Y(I)
  901 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' '
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
   10 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) 'H10:  Entry of end-points.'
      X(1) = '  End-points a and b should generall'
      Y(1) = 'y be entered as real numbers to an  '
      X(2) = 'appropriate number of decimal places'
      Y(2) = '. '
      DO 1001 I = 1,2
          WRITE (*,FMT=*) X(I),Y(I)
 1001 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' '
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
   11 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) 'H11:  End-point values of p,q,w.'
      X(1) = '  The program sleign2 needs to know '
      Y(1) = 'whether the coefficient functions   '
      X(2) = 'p(x),q(x),w(x) defined by the FORTRA'
      Y(2) = 'N expressions entered in the input  '
      X(3) = 'file can be evaluated numerically wi'
      Y(3) = 'thout running into difficulty.  If, '
      X(4) = 'for example, either q or w is unboun'
      Y(4) = 'ded at a, or p(a) is 0, then sleign2'
      X(5) = 'needs to know this so that a is not '
      Y(5) = 'chosen for functional evaluation.   '
      DO 1101 I = 1,5
          WRITE (*,FMT=*) X(I),Y(I)
 1101 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' '
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
   12 CONTINUE
      WRITE (*,FMT=*) 'H12:  Initial value problems.'
      X(1) = '  The initial value problem facility'
      Y(1) = ' for Sturm-Liouville problems       '
      X(2) = ' '
      Y(2) = ' '
      X(3) = '                   -(p*y'')'' + q*y = '
      Y(3) = ' lambda*w*y              (*)        '
      X(4) = ' '
      Y(4) = ' '
      X(5) = 'allows for the computation of a solu'
      Y(5) = 'tion of (*) with a user-chosen      '
      X(6) = 'value lambda and any one of the foll'
      Y(6) = 'owing initial conditions:           '
      X(7) = '  1. From end-point a of any classif'
      Y(7) = 'ication except LP towards           '
      X(8) = 'end-point b of any classification,  '
      Y(8) = ' '
      X(9) = '  2. From end-point b of any classif'
      Y(9) = 'ication except LP back towards      '
      X(10) = 'end-point a of any classification,  '
      Y(10) = ' '
      X(11) = '  3. From end-points a and b of any '
      Y(11) = 'classifications except LP towards an'
      X(12) = 'interior point of (a,b) selected by '
      Y(12) = 'the program.                        '
      X(13) = ' '
      Y(13) = ' '
      X(14) = '  Initial values at a are of the for'
      Y(14) = 'm y(a) = alpha1, (p*y'')(a) =alpha2,'
      X(15) = 'when a is R or WR; and [y,u](a) = al'
      Y(15) = 'pha1, [y,v](a) = alpha2, when a is  '
      X(16) = 'LCNO or LCO.                        '
      Y(16) = ' '
      X(17) = '  Initial values at b are of the for'
      Y(17) = 'm y(b) = beta1, (p*y'')(b) = beta2, '
      X(18) = 'when b is R or WR; and [y,u](b) = be'
      Y(18) = 'ta1, [y,v](b) = beta2, when b is    '
      X(19) = 'LCNO or LCO.                        '
      Y(19) = ' '
      X(20) = '  In (*), lambda is a user-chosen re'
      Y(20) = 'al number; while in the above ini-  '
      X(21) = 'tial values, (alpha1,alpha2) and (be'
      Y(21) = 'ta1,beta2) are user-chosen pairs of '
      X(22) = 'real numbers not both zero.         '
      Y(22) = ' '
      DO 1201 I = 1,22
          WRITE (*,FMT=*) X(I),Y(I)
 1201 CONTINUE
      WRITE (*,FMT=*)
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      X(1) = '  In the initial value case (3.) abo'
      Y(1) = 've when the interval (a,b) is       '
      X(2) = 'finite, the interior point selected '
      Y(2) = 'by the program is generally near the'
      X(3) = 'midpoint of (a,b); when (a,b) is inf'
      Y(3) = 'inite, no general rule can be given.'
      X(4) = 'Also if, given (alpha1,alpha2) and ('
      Y(4) = 'beta1,beta2), the lambda chosen is  '
      X(5) = 'an eigenvalue of the associated boun'
      Y(5) = 'dary value problem, the computed    '
      X(6) = 'solution may not be the correspondin'
      Y(6) = 'g eigenfunction -- the signs of the '
      X(7) = 'computed solutions on either side of'
      Y(7) = ' the interior point may be opposite.'
      X(8) = '  The output for a solution of an in'
      Y(8) = 'itial value problem is in the form  '
      X(9) = 'of stored numerical data which can b'
      Y(9) = 'e plotted on the screen (see H16),  '
      X(10) = 'or printed out in graphical form if '
      Y(10) = 'graphics software is available.     '
      DO 1202 I = 1,10
          WRITE (*,FMT=*) X(I),Y(I)
 1202 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' '
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
   13 CONTINUE
      WRITE (*,FMT=*) 'H13:  Indexing of eigenvalues.'
      X(1) = '  The indexing of eigenvalues is an '
      Y(1) = 'automatic facility in sleign2.  The '
      X(2) = 'following general results hold for t'
      Y(2) = 'he separated boundary condition     '
      X(3) = 'problem (see H7):                   '
      Y(3) = ' '
      X(4) = '  1. If neither end-point a or b is '
      Y(4) = 'LP or LCO, then the spectrum of the '
      X(5) = 'eigenvalue problem is discrete (eige'
      Y(5) = 'nvalues only), simple (eigenvalues  '
      X(6) = 'all of multiplicity 1), and bounded '
      Y(6) = 'below with a single cluster point at'
      X(7) = '+infinity.  The eigenvalues are inde'
      Y(7) = 'xed as {lambda(n): n=0,1,2,...},    '
      X(8) = 'where lambda(n) < lambda(n+1) (n=0,1'
      Y(8) = ',2,...), lim lambda(n) -> +infinity;'
      X(9) = 'and if {psi(n): n=0,1,2,...} are the'
      Y(9) = ' corresponding eigenfunctions, then '
      X(10) = 'psi(n) has exactly n zeros in the op'
      Y(10) = 'en interval (a,b).                  '
      X(11) = '  2. If neither end-point a or b is '
      Y(11) = 'LP but at least one end-point is    '
      X(12) = 'LCO, then the spectrum is discrete a'
      Y(12) = 'nd simple as for (1.), but with     '
      X(13) = 'cluster points at both +infinity and'
      Y(13) = ' -infinity.  The eigenvalues are in-'
      X(14) = 'dexed as {lambda(n): n=0,1,-1,2,-2,.'
      Y(14) = '..}, where                          '
      X(15) = 'lambda(n) < lambda(n+1) (n=...-2,-1,'
      Y(15) = '0,1,2,...) with lambda(0) the small-'
      X(16) = 'est non-negative eigenvalue and lim '
      Y(16) = 'lambda(n) -> +infinity or -> -infi- '
      X(17) = 'nity with n; and if {psi(n): n=0,1,-'
      Y(17) = '1,2,-2,...} are the corresponding   '
      X(18) = 'eigenfunctions, then every psi(n) ha'
      Y(18) = 's infinitely many zeros in (a,b).   '
      DO 1301 I = 1,18
          WRITE (*,FMT=*) X(I),Y(I)
 1301 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
      X(1) = '  3. If one or both end-points is LP'
      Y(1) = ', then there can be one or more in- '
      X(2) = 'tervals of continuous spectrum for t'
      Y(2) = 'he boundary value problem in addi-  '
      X(3) = 'tion to some (necessarily simple) ei'
      Y(3) = 'genvalues.  For these essentially   '
      X(4) = 'more difficult problems, sleign2 can'
      Y(4) = ' be used as an investigative tool to'
      X(5) = 'give qualitative and possibly quanti'
      Y(5) = 'tative information on the spectrum. '
      X(6) = '     For example, if a problem has a'
      Y(6) = ' continuous spectrum starting at L, '
      X(7) = 'then there may be no eigenvalue belo'
      Y(7) = 'w L, any finite number of eigenval- '
      X(8) = 'ues below L, or an infinite (but cou'
      Y(8) = 'ntable) number of eigenvalues below '
      X(9) = 'L. sleign2 can be used to compute L '
      Y(9) = '(see the paper bewz on the sleign2  '
      X(10) = 'home page for an algorithm to comput'
      Y(10) = 'e L), and determine the number of   '
      X(11) = 'these eigenvalues and compute them. '
      Y(11) = ' In this respect, see xamples.f: #13'
      X(12) = '(Hydrogen Atom), #17 (Morse Oscillat'
      Y(12) = 'or), #21 (Fourier), #27 (Joergens)  '
      X(13) = 'as examples of success; and #12 (Mat'
      Y(13) = 'hieu), #14 (Marletta), and #28      '
      X(14) = '(Behnke-Goerisch) as examples of fai'
      Y(14) = 'lure.                               '
      X(15) = '     The problem need not have a con'
      Y(15) = 'tinuous spectrum, in which case if  '
      X(16) = 'its discrete spectrum is bounded bel'
      Y(16) = 'ow, then the eigenvalues are indexed'
      X(17) = 'and the eigenfunctions have zero cou'
      Y(17) = 'nts as in (1.).  If, on the other   '
      X(18) = 'hand, the discrete spectrum is unbou'
      Y(18) = 'nded below, then all the eigenfunc- '
      X(19) = 'tions have infinitely many zeros in '
      Y(19) = 'the open interval (a,b).  sleign2   '
      X(20) = 'can, in principle, compute these eig'
      Y(20) = 'envalues if neither end-point is LP,'
      X(21) = 'although this is a computationally d'
      Y(21) = 'ifficult problem.  Note, however,   '
      X(22) = 'that sleign2 has no algorithm when t'
      Y(22) = 'he spectrum is discrete, unbounded  '
      X(23) = 'above and below, and one end-point i'
      Y(23) = 's LP, as in xamples.f #30.          '
      DO 1302 I = 1,23
          WRITE (*,FMT=*) X(I),Y(I)
 1302 CONTINUE
      WRITE (*,FMT=*)
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
      X(1) = '  In respect to the three different '
      Y(1) = 'types of indexing discussed above,  '
      X(2) = 'the following identified examples fr'
      Y(2) = 'om xamples.f illustrate the spectral'
      X(3) = 'property of these boundary problems:'
      Y(3) = '                                    '
      X(4) = '  1. Neither end-point is LP or LCO.'
      Y(4) = ' '
      X(5) = '       #1 (Legendre)                '
      Y(5) = ' '
      X(6) = '       #2 (Bessel) with -1/4 < c < 3'
      Y(6) = '/4                                  '
      X(7) = '       #4 (Boyd)                    '
      Y(7) = ' '
      X(8) = '       #5 (Latzko)                  '
      Y(8) = ' '
      X(9) = '  2. Neither end-point is LP, but at'
      Y(9) = ' least one is LCO.                  '
      X(10) = '       #6 (Sears-Titchmarsh)        '
      Y(10) = ' '
      X(11) = '       #7 (BEZ)                     '
      Y(11) = ' '
      X(12) = '      #19 (Donsch)                  '
      Y(12) = ' '
      X(13) = '  3. At least one end-point is LP.  '
      Y(13) = ' '
      X(14) = '      #13 (Hydrogen Atom)           '
      Y(14) = ' '
      X(15) = '      #14 (Marletta)                '
      Y(15) = ' '
      X(16) = '      #20 (Krall)                   '
      Y(16) = ' '
      X(17) = '      #21 (Fourier) on [0,infinity) '
      Y(17) = ' '
      DO 1303 I = 1,17
          WRITE (*,FMT=*) X(I),Y(I)
 1303 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' '
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
   14 CONTINUE
      WRITE (*,FMT=*) 'H14:  Entry of eigenvalue index, initial guess,'
     +  //' and tolerance.'
      X(1) = '  For all self-adjoint boundary cond'
      Y(1) = 'ition problems (see H7), sleign2    '
      X(2) = 'calls for input information options '
      Y(2) = 'to compute either                   '
      X(3) = '  1. a single eigenvalue, or        '
      Y(3) = ' '
      X(4) = '  2. a series of eigenvalues.       '
      Y(4) = ' '
      X(5) = 'In each case indexing of eigenvalues'
      Y(5) = ' is called for (see H13).           '
      X(6) = '  (1.) above asks for data triples N'
      Y(6) = 'UMEIG, EIG, TOL separated by commas.'
      X(7) = 'Here NUMEIG is the integer index of '
      Y(7) = 'the desired eigenvalue; NUMEIG can  '
      X(8) = 'be negative only when the problem is'
      Y(8) = ' LCO at one or both end-points.     '
      X(9) = 'EIG allows for the entry of an initi'
      Y(9) = 'al guess for the requested          '
      X(10) = 'eigenvalue (if an especially good on'
      Y(10) = 'e is available), or can be set to 0 '
      X(11) = 'in which case an initial guess is ge'
      Y(11) = 'nerated by sleign2 itself.          '
      X(12) = 'TOL is the desired accuracy of the c'
      Y(12) = 'omputed eigenvalue.  It is an       '
      X(13) = 'absolute accuracy if the magnitude o'
      Y(13) = 'f the eigenvalue is 1 or less, and  '
      X(14) = 'is a relative accuracy otherwise.  T'
      Y(14) = 'ypical values might be .001 for     '
      X(15) = 'moderate accuracy and .0000001 for h'
      Y(15) = 'igh accuracy in single precision.   '
      X(16) = 'If TOL is set to 0, the maximum achi'
      Y(16) = 'evable accuracy is requested.       '
      X(17) = '  If the input data list is truncate'
      Y(17) = 'd with a "/" after NUMEIG or EIG,   '
      X(18) = 'then the remaining elements default '
      Y(18) = 'to 0.                               '
      DO 1401 I = 1,18
          WRITE (*,FMT=*) X(I),Y(I)
 1401 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
      X(1) = '  (2.) above asks for data triples N'
      Y(1) = 'UMEIG1, neig2, TOL separated by   '
      X(2) = 'commas.  Here numeig1 and numeig2 ar'
      Y(2) = 'e the first and last integer indices'
      X(3) = 'of the sequence of desired eigenvalu'
      Y(3) = 'es, numeig1 < numeig2; they can be  '
      X(4) = 'negative only when the problem is LC'
      Y(4) = 'O at one or both end-points.        '
      X(5) = 'TOL is the desired accuracy of the c'
      Y(5) = 'omputed eigenvalues.  It is an      '
      X(6) = 'absolute accuracy if the magnitude o'
      Y(6) = 'f an eigenvalue is 1 or less, and   '
      X(7) = 'is a relative accuracy otherwise.  T'
      Y(7) = 'ypical values might be .001 for     '
      X(8) = 'moderate accuracy and .0000001 for h'
      Y(8) = 'igh accuracy in single precision.   '
      X(9) = 'If TOL is set to 0, the maximum achi'
      Y(9) = 'evable accuracy is requested.       '
      X(10) = '  If the input data list is truncate'
      Y(10) = 'd with a "/" after neig2, then TOL'
      X(11) = 'defaults to 0.                      '
      Y(11) = ' '
      X(12) = '  For COUPLED self-adjoint boundary '
      Y(12) = 'condition problems (see H7 and H17),'
      X(13) = 'sleign2 also reports which eigenvalu'
      Y(13) = 'es are double.  Double eigenvalues  '
      X(14) = 'can occur only for coupled boundary '
      Y(14) = 'conditions with alpha = 0 or pi.    '
      DO 1402 I = 1,14
          WRITE (*,FMT=*) X(I),Y(I)
 1402 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' '
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
   15 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) 'H15:  IFLAG information.'
      X(1) = '  All results are reported by sleign2'
      Y(1) = '2 with a flag identification.  There'
      X(2) = 'are four values of IFLAG:           '
      Y(2) = ' '
      X(3) = ' '
      Y(3) = ' '
      X(4) = '  1 - The computed eigenvalue has an'
      Y(4) = ' estimated accuracy within the      '
      X(5) = '      tolerance requested.          '
      Y(5) = ' '
      X(6) = ' '
      Y(6) = ' '
      X(7) = '  2 - The computed eigenvalue does n'
      Y(7) = 'ot have an estimated accuracy within'
      X(8) = '      the tolerance requested, but i'
      Y(8) = 's the best the program could obtain.'
      X(9) = ' '
      Y(9) = ' '
      X(10) = '  3 - There seems to be no eigenvalu'
      Y(10) = 'e of index equal to NUMEIG.         '
      X(11) = ' '
      Y(11) = ' '
      X(12) = '  4 - The program has been unable to'
      Y(12) = ' compute the requested eigenvalue.  '
      DO 1501 I = 1,12
          WRITE (*,FMT=*) X(I),Y(I)
 1501 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' '
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
   16 CONTINUE
      WRITE (*,FMT=*) 'H16:  Plotting.'
      X(1) = '  After computing a single eigenvalu'
      Y(1) = 'e (see H14(1.)), but not a sequence '
      X(2) = 'of eigenvalues (see H14(2.)), the ei'
      Y(2) = 'genfunction can be plotted for sepa-'
      X(3) = 'rated conditions and for coupled one'
      Y(3) = 's with alpha = 0 or pi.  When alpha '
      X(4) = 'is not zero or pi the eigenfunctions'
      Y(4) = ' are not real-valued and so no plot '
      X(5) = 'is provided.  If this is desired, re'
      Y(5) = 'spond "y" when asked so that sleign2'
      X(6) = 'will compute some eigenfunction data'
      Y(6) = ' and store them.                    '
      X(7) = '  One can ask that the eigenfunction'
      Y(7) = ' data be in the form of either      '
      X(8) = 'points (x,y) for x in (a,b), or poin'
      Y(8) = 'ts (t,y) for t in the standardized  '
      X(9) = 'interval (-1,1) mapped onto from (a,'
      Y(9) = 'b); the t- choice can be especially '
      X(10) = 'helpful when the original interval i'
      Y(10) = 's infinite.  Additionally, one can  '
      X(11) = 'ask for a plot of the so-called Prue'
      Y(11) = 'fer angle, in x- or t- variables.   '
      X(12) = '  In both forms, once the choice has'
      Y(12) = ' been made of the function to be    '
      X(13) = 'plotted, a crude plot is displayed o'
      Y(13) = 'n the monitor screen and you are    '
      X(14) = 'asked whether you wish to save the c'
      Y(14) = 'omputed plot points in a file.      '
      DO 1601 I = 1,14
          WRITE (*,FMT=*) X(I),Y(I)
 1601 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' '
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
   17 CONTINUE
      WRITE (*,FMT=*) 'H17:  Indexing of eigenvalues for'//
     +  ' coupled self-adjoint problems.'
      X(1) = '  The indexing of eigenvalues is an '
      Y(1) = 'automatic facility in sleign2.  The '
      X(2) = 'following general result holds for c'
      Y(2) = 'oupled boundary condition problems  '
      X(3) = '(see H7):                           '
      Y(3) = ' '
      X(4) = '  The spectrum of the eigenvalue pro'
      Y(4) = 'blem is discrete (eigenvalues only).'
      X(5) = 'In general the spectrum is not simpl'
      Y(5) = 'e, but no eigenvalue exceeds multi- '
      X(6) = 'plicity 2.The eigenvalues are indexe'
      Y(6) = 'd as {lambda(n): n=0,1,2,...}, where'
      X(7) = 'lambda(n) .le. lambda(n+1) (n=0,1,2,'
      Y(7) = '...), lim lambda(n) -> +infinity if '
      X(8) = 'neither end-point is LCO.  If one or'
      Y(8) = ' both end-points are LCO the eigen- '
      X(9) = 'values cluster at both +infinity and'
      Y(9) = ' at -infinity, and all eigenfunc-   '
      X(10) = 'tions have infinitely many zeros.   '
      Y(10) = '                                    '
      X(11) = '  If neither end-point is LCO and al'
      Y(11) = 'pha = 0 or pi, then the n-th eigen- '
      X(12) = 'function has n-1, n, or n+1 zeros in'
      Y(12) = ' the half-open interval [a,b) (also '
      X(13) = 'in (a,b]).  All three possibilities '
      Y(13) = 'occur.  Recall that in the case of  '
      X(14) = 'double eigenvalues, although the n-t'
      Y(14) = 'h eigenvalue is well-defined, there '
      X(15) = 'is an ambiguity about which solution'
      Y(15) = ' is declared the n-th eigenfunction.'
      X(16) = '  If alpha is not 0 or pi, then the '
      Y(16) = 'eigenfunction is non-real and has no'
      X(17) = 'zero in (a,b); but each of the real '
      Y(17) = 'and imaginary parts of the n-th ei- '
      X(18) = 'genfunction have the same zero prope'
      Y(18) = 'rties as mentioned above when alpha '
      X(19) = '= 0 or pi.                          '
      Y(19) = ' '
      DO 1651 I = 1,19
          WRITE (*,FMT=*) X(I),Y(I)
 1651 CONTINUE
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      X(1) = '  The following identified examples '
      Y(1) = 'from xamples.f are of special inter-'
      X(2) = 'est:                                '
      Y(2) = ' '
      X(3) = '    #11 (Plum) on [0,pi]           '
      Y(3) = ' '
      X(4) = '    #21 (Fourier) on [0,pi]        '
      Y(4) = ' '
      X(5) = '    #25 (Meissner) on [-0.5,0.5]   '
      Y(5) = ' '
      DO 1701 I = 1,5
          WRITE (*,FMT=*) X(I),Y(I)
 1701 CONTINUE
      WRITE (*,FMT=*) ' '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
      READ (*,FMT=999) ANS,N
      IF (ANS.EQ.'r' .OR. ANS.EQ.'R') RETURN
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17) N
      GO TO 1
  999 FORMAT (A1,I2)
      END
C
      SUBROUTINE PQ(END,P0,QF)
C  THIS PROGRAM IS INTENDED TO DETERMINE NUMERICALLY WHETHER OR
C    NOT THE FUNCTION P(X) IS ZERO AT A NON-REGULAR ENDPOINT,
C    AND WHETHER OR NOT THE FUNCTION Q(X)/W(X) IS INFINITE THERE.
C    IF P IS ZERO, P0 IS SET TO +1.0; ELSE TO -1.0.
C    IF Q AND/OR W IS FINITE, QF IS SET TO +1.0; ELSE TO -1.0.
C     .. Scalar Arguments ..
      REAL END,P0,QF
C     ..
C     .. Local Scalars ..
      REAL DT,HALF,ONE,PV,QV,QVP,QVSAV,T,TMP,TSAV,X,ZER
      INTEGER I
C     ..
C     .. External Functions ..
      REAL P,Q,W
      EXTERNAL P,Q,W
C     ..
C     .. External Subroutines ..
      EXTERNAL DXDT
C     ..
C     .. Intrinsic Functions ..
      INTRINSIC ABS
C     ..
      ONE = 1.0D0
      HALF = 0.5D0
      ZER = 0.0D0
c
      P0 = -1.0D0
      QF = 1.0D0
      DT = 0.1D0
      PV = 1.0D0
      IF (END.LT.0.0D0) DT = -DT
      DO 10 I = 1,25
          DT = HALF*DT
          T = END - DT
          IF (T.EQ.-1.0D0) THEN
              IF (PV.LT.0.0001D0) THEN
                  P0 = 1.0D0
                  GO TO 15
              ELSE
                  GO TO 15
              END IF
          END IF
          CALL DXDT(T,TMP,X)
          PV = ABS(P(X))
          IF (PV.GT. (1.D+5)) THEN
              GO TO 15
          ELSE IF (PV.LT. (.0001D0) .AND. I.GT.15) THEN
              P0 = 1.0D0
              GO TO 15
          END IF
   10 CONTINUE
   15 CONTINUE
c
      DT = 0.1D0
      IF (END.LT.0.0D0) DT = -DT
      QVP = ONE
      QVSAV = ZER
      TSAV = END
      QV = 1.0D0
      DO 20 I = 1,17
          DT = HALF*DT
          T = END - DT
          IF (T.EQ.1.0D0 .AND. (QV.GT. (1.D+5))) THEN
              QF = -1.0D0
              GO TO 25
          END IF
          CALL DXDT(T,TMP,X)
          QV = ABS(Q(X)) + ABS(W(X))
          QVP = ABS((QV-QVSAV)/ (T-TSAV))
          QVSAV = QV
          IF (QVP.GT. (1.D+5)) THEN
              QF = -1.0D0
              GO TO 25
          END IF
   20 CONTINUE
   25 CONTINUE
c
      RETURN
      END
SHAR_EOF
fi # end of overwriting check
if test -f 'driver3.f'
then
	echo shar: will not over-write existing file "'driver3.f'"
else
cat << "SHAR_EOF" > 'driver3.f'
C  PROGRAM SEPDR

C     **********
C     MARCH 1, 2001; P.B. BAILEY, W.N. EVERITT AND A. ZETTL
C     VERSION 1.2
C     **********

      PROGRAM SEPDR

C     This program is for the purpose of indicating how SLEIGN2
C     can be called for the purpose of obtaining eigenvalues
C     of Sturm-Liouville problems with regular and singular
C     separated boundary conditions.
C
C     The call is of the form:
C
C     CALL SLEIGN(A, B, INTAB, P0ATA, QFATA, P0ATB, QFATB,
C    1    A1, A2, B1, B2, NUMEIG, EIG, TOL, IFLAG,
C    2    ISLFUN, SLFUN, NCA, NCB)
C
C     SET THE PARAMETERS DEFINING THE INTERVAL OF
C       DEFINITION OF THE DIFFERENTIAL EQUATION:
C       (BESSEL'S EQUATION WITH NU = 0.75)
C
C     .. Scalars in Common ..
      INTEGER T21
      LOGICAL PR
C     ..
C     .. Local Scalars ..
      REAL A,A1,A2,B,B1,B2,EIG,P0ATA,P0ATB,QFATA,QFATB,TOL
      INTEGER I,IFLAG,INTAB,ISLFUN,NCA,NCB,NP,NUMEIG
C     ..
C     .. Local Arrays ..
      REAL SLFUN(100),X(100)
C     ..
C     .. External Subroutines ..
      EXTERNAL SLEIGN
C     ..
C     .. Common blocks ..
      COMMON /PRIN/PR,T21
C     ..
      T21 = 21
      OPEN (T21,FILE='test.out')
      PR = .TRUE.
C
      A = 0.0D0
      B = 1.0D0

C ------------------------------------------------------------------
C     SET THE PARAMETERS DEFINING THE SEPARATED BOUNDARY CONDITIONS:
C ------------------------------------------------------------------

      A1 = 1.0D0
      A2 = 0.0D0
      B1 = 1.0D0
      B2 = 0.0D0

C     SET THE REMAINING PARAMETERS NEEDED BY SLEIGN2:

      P0ATA = -1.0D0
      QFATA = -1.0D0
      P0ATB = -1.0D0
      QFATB = 1.0D0
      INTAB = 1
      NUMEIG = 2
      EIG = 0.0D0
      TOL = 1.D-5
      ISLFUN = 0
      NCA = 3
      NCB = 1

C ---------------------------------------------------------
C  If eigenfunction values are wanted, then:

      NP = 22
      DO 10 I = 2,NP - 1
          X(I-1) = A + (I-1)* (B-A)/ (NP-1)
   10 CONTINUE
      ISLFUN = NP - 2

      DO 15 I = 1,ISLFUN
          SLFUN(9+I) = X(I)
   15 CONTINUE
C ---------------------------------------------------------

C     NOW CALL SLEIGN2:

      CALL SLEIGN(A,B,INTAB,P0ATA,QFATA,P0ATB,QFATB,A1,A2,B1,B2,NUMEIG,
     +            EIG,TOL,IFLAG,ISLFUN,SLFUN,NCA,NCB)

C     WRITE OUT THE RETURNED EIGENVALUE, IT'S ESTIMATED ACCURACY,
C     AND THE IFLAG STATUS:

      WRITE (*,FMT=*) ' NUMEIG, EIG, TOL, IFLAG = ',NUMEIG,EIG,TOL,IFLAG

C ---------------------------------------------------------
      IF (ISLFUN.GT.0) THEN
C     IN THIS CASE, THE EIGENFUNCTION WAS COMPUTED :
C
          WRITE (*,FMT=*) ' EIGENFUNCTION = '
          DO 25 I = 1,ISLFUN
              WRITE (*,FMT=*) X(I),SLFUN(9+I)
   25     CONTINUE
      END IF

C ---------------------------------------------------------
C
      CLOSE (T21)
      STOP
      END
C
      REAL FUNCTION P(X)

C     .. Scalar Arguments ..
      REAL X
C     ..
      P = 1.0D0
      RETURN
      END
C
      REAL FUNCTION Q(X)

C     .. Scalar Arguments ..
      REAL X
C     ..
C     .. Local Scalars ..
      REAL NU
C     ..
      NU = 0.75D0
      Q = (NU*NU-0.25D0)/X**2
      RETURN
      END
C
      REAL FUNCTION W(X)

C     .. Scalar Arguments ..
      REAL X
C     ..
      W = 1.0D0
      RETURN
      END
C
      SUBROUTINE UV(X,U,PUP,V,PVP,HU,HV)

C     .. Scalar Arguments ..
      REAL HU,HV,PUP,PVP,U,V,X
C     ..
C     .. Local Scalars ..
      REAL NU
C     ..
      NU = 0.75D0
      U = X** (NU+0.5D0)
      PUP = (NU+0.5D0)*X** (NU-0.5D0)
      V = X** (-NU+0.5D0)
      PVP = (-NU+0.5D0)*X** (-NU-0.5D0)
      HU = 0.0D0
      HV = 0.0D0
      RETURN
      END
SHAR_EOF
fi # end of overwriting check
if test -f 'makepqw.f'
then
	echo shar: will not over-write existing file "'makepqw.f'"
else
cat << "SHAR_EOF" > 'makepqw.f'
C  PROGRAM MAKEPQW 
 
C     **********
C     MARCH 1, 2001; P.B. BAILEY, W.N. EVERITT AND A. ZETTL
C     VERSION 1.2
C     **********

      PROGRAM MAKEPQW
C
C     THIS PROGRAM GENERATES THE FORTRAN COEFFICIENT FUNCTIONS
C     P(X), Q(X), W(X), AND THE SUBROUTINE UV WHICH DEFINES THE
C     BOUNDARY CONDITION FUNCTIONS u(X), v(X) AND/OR U(X), V(X)
C     FOR SLEIGN2.
C
C     THE DIFFERENTIAL EQUATION IS OF THE FORM
C
C           -(p*y')' + q*y = lambda*w*y
C
C     .. Local Scalars ..
      CHARACTER*1 HQ
      CHARACTER*16 CHANS,TAPE1
      CHARACTER*62 STR
      REAL C
C     ..
C     .. External Subroutines ..
      EXTERNAL LC
C     ..
      WRITE(*,*)
      WRITE(*,*)
      WRITE(*,*)
      WRITE(*,*)
      WRITE(*,*)
      WRITE(*,*)
      WRITE(*,*)
      WRITE(*,*)
      WRITE(*,*)
      WRITE(*,*)
      WRITE(*,*)
      WRITE(*,*) ' HELP may be called at any point where the program   '
      WRITE(*,*) '    halts and displays (h?) by pressing "h <ENTER>". '
      WRITE(*,*) '    To RETURN from HELP, press "r <ENTER>".          '
      WRITE(*,*) '    To QUIT at any program halt, press "q <ENTER>".  '
      WRITE(*,*) ' WOULD YOU LIKE AN OVERVIEW OF HELP ? (Y/N) (h?)     '
      WRITE(*,*)
      WRITE(*,*)
      WRITE(*,*)
      WRITE(*,*)
      WRITE(*,*)
      WRITE(*,*)
      WRITE(*,*)
      WRITE(*,*)
      READ(*,1) HQ
      IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') THEN
            STOP
         ELSE IF (HQ.EQ.'y' .OR. HQ.EQ.'Y' .OR.
     1            HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
            CALL HELP(1)
            END IF
C
  100 CONTINUE
         WRITE(*,*) ' SPECIFY OUTPUT FILE NAME (h?)  '
         READ(*,16) CHANS
         IF (CHANS.EQ.'q' .OR. CHANS.EQ.'Q') THEN
            STOP
         ELSE IF (CHANS.EQ.'h' .OR. CHANS.EQ.'H') THEN
            CALL HELP(2)
            GO TO 100
         ELSE
            TAPE1 = CHANS
            END IF
         OPEN(1,FILE=TAPE1,STATUS='NEW')
         WRITE(1,'(A)') 'C'
         WRITE(1,'(A)') 'C               ' // TAPE1
C
      WRITE(*,*) ' THE DIFFERENTIAL EQUATION IS OF THE FORM: '
      WRITE(*,*) '        -(p*y'')'' + q*y = lambda*w*y      '
      WRITE(*,*)
C
  200 CONTINUE
         WRITE(*,*) ' INPUT (h?) p = '
         READ(*,62) STR
         HQ = STR
         IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') THEN
            STOP
         ELSE IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
            CALL HELP(3)
            GO TO 200
         ELSE
            WRITE(1,'(A)') 'C'
            WRITE(1,'(A)') '      FUNCTION P(X)'
            WRITE(1,'(A)') '      REAL P,X'
            WRITE(1,'(A)') '      P = ' // STR
            WRITE(1,'(A)') '      RETURN'
            WRITE(1,'(A)') '      END'
         END IF
C
  300 CONTINUE
         WRITE(*,*) ' INPUT (h?) q = '
         READ(*,62) STR
         HQ = STR
         IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') THEN
            STOP
         ELSE IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
            CALL HELP(3)
            GO TO 300
         ELSE
            WRITE(1,'(A)') 'C'
            WRITE(1,'(A)') '      FUNCTION Q(X)'
            WRITE(1,'(A)') '      REAL Q,X'
            WRITE(1,'(A)') '      Q = ' // STR
            WRITE(1,'(A)') '      RETURN'
            WRITE(1,'(A)') '      END'
            END IF
C
  400 CONTINUE
         WRITE(*,*) ' INPUT (h?) w = '
         READ(*,62) STR
         HQ = STR
         IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') THEN
            STOP
         ELSE IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
            CALL HELP(3)
            GO TO 400
         ELSE
            WRITE(1,'(A)') 'C'
            WRITE(1,'(A)') '      FUNCTION W(X)'
            WRITE(1,'(A)') '      REAL W,X'
            WRITE(1,'(A)') '      W = ' // STR
            WRITE(1,'(A)') '      RETURN'
            WRITE(1,'(A)') '      END'
            END IF
C
  500 CONTINUE
         WRITE(*,*) ' DO YOU REQUIRE INFORMATION ON END-POINT '
         WRITE(*,*) '    CLASSIFICATION ? (Y/N) (h?)  '
         READ(*,1) HQ
         IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') THEN
            STOP
         ELSE IF (HQ.EQ.'y' .OR. HQ.EQ.'Y' .OR.
     1            HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
            CALL HELP(4)
            GO TO 500
         ELSE IF (HQ.EQ.'n' .OR. HQ.EQ.'N') THEN
            GO TO 800
         ELSE
            END IF
C
  600 CONTINUE
         WRITE(*,*) ' DO YOU REQUIRE INFORMATION ON DEFAULT CLASSIFI-'
         WRITE(*,*) '    CATION AND BOUNDARY CONDITIONS ? (Y/N) (h?)  '
         READ(*,1) HQ
         IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') THEN
            STOP
         ELSE IF (HQ.EQ.'y' .OR. HQ.EQ.'Y' .OR.
     1            HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
            CALL HELP(5)
            GO TO 600
         ELSE
            END IF
C
  700 CONTINUE
         WRITE(*,*) ' DO YOU REQUIRE INFORMATION ON LIMIT CIRCLE '
         WRITE(*,*) '    BOUNDARY CONDITIONS ? (Y/N) (h?)  '
         READ(*,1) HQ
         IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') THEN
            STOP
         ELSE IF (HQ.EQ.'y' .OR. HQ.EQ.'Y' .OR.
     1            HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
            CALL HELP(6)
            GO TO 700
         ELSE
            END IF
C
  800 CONTINUE
         WRITE(*,*) ' DO YOU WANT TO USE A LIMIT CIRCLE  '
         WRITE(*,*) '    BOUNDARY CONDITION ? (Y/N) (h?)  '
         READ(*,1) HQ
         IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') THEN
            STOP
         ELSE IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
            CALL HELP(6)
            GO TO 800
         ELSE IF (HQ.EQ.'y' .OR. HQ.EQ.'Y') THEN
            WRITE(1,'(A)') 'C'
            WRITE(1,'(A)') '      SUBROUTINE UV(X,U,PUP,V,PVP,HU,HV)'
            WRITE(1,'(A)') '      REAL X,U,PUP,V,PVP,HU,HV'
  900       CONTINUE
               WRITE(*,*) ' DO YOU WANT TO USE TWO DIFFERENT PAIRS OF '
               WRITE(*,*) '    FUNCTIONS U(X),V(X) ? (Y/N) (h?) '
               READ(*,1) HQ
               IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') THEN
                  STOP
               ELSE IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
                  CALL HELP(6)
                  GO TO 900
               ELSE IF (HQ.EQ.'y' .OR. HQ.EQ.'Y') THEN
 1000             CONTINUE
                     WRITE(*,*) ' ASSUMING THAT ONE PAIR OF FUNCTIONS  '
                     WRITE(*,*) '    U(X),V(X) IS FOR a < X < c, AND   '
                     WRITE(*,*) '    THE OTHER PAIR IS FOR c <= X < b, '
                     WRITE(*,*) '    WHAT IS THE VALUE OF c ? (h?)     '
                     WRITE(*,*)
                     WRITE(*,*) ' c = '
                     READ(*,16) CHANS
                     HQ = CHANS
                     IF (HQ.EQ.'q' .OR. HQ.EQ.'Q') THEN
                        STOP
                     ELSE IF (HQ.EQ.'h' .OR. HQ.EQ.'H') THEN
                        CALL HELP(6)
                        GO TO 1000
                     ELSE
                        READ(CHANS,'(F16.0)') C
                        END IF
                  WRITE(*,*)
                  WRITE(*,*) ' FOR a < X < c :'
                  WRITE(*,*)
                  WRITE(1,'(A,1PE12.5,A)') '      IF (X.LT.',C,') THEN'
                  CALL LC(9,1)
                  WRITE(*,*)
                  WRITE(*,*) ' FOR c <= X < b :'
                  WRITE(*,*)
                  WRITE(1,'(A)') '      ELSE'
                  CALL LC(9,2)
                  WRITE(1,'(A)') '      END IF'
               ELSE
                  WRITE(*,*)
                  CALL LC(6,1)
                  END IF
         ELSE
            WRITE(1,'(A)') 'C'
            WRITE(1,'(A)') '      SUBROUTINE UV'
            END IF
      WRITE(1,'(A)') '      RETURN'
      WRITE(1,'(A)') '      END'
C
      WRITE(1,'(A)') 'C'
      WRITE(1,'(A)') '      SUBROUTINE EXAMP'
      WRITE(1,'(A)') '      RETURN'
      WRITE(1,'(A)') '      END'
      CLOSE(1)
      STOP
    1 FORMAT(A1)
   16 FORMAT(A16)
   62 FORMAT(A62)
      END
C
      SUBROUTINE LC(INDENT,N)
      INTEGER INDENT,N
C     .. Local Scalars ..
      CHARACTER*57 STR
C     ..
      IF (INDENT.EQ.6) THEN
         WRITE(*,*) ' INPUT u = '
         READ(*,57) STR
         WRITE(1,'(A)') '      u = ' // STR
         WRITE(*,*) ' INPUT v = '
         READ(*,57) STR
         WRITE(1,'(A)') '      v = ' // STR
         WRITE(*,*) ' INPUT pu'' = '
         READ(*,57) STR
         WRITE(1,'(A)') '      pup = ' // STR
         WRITE(*,*) ' INPUT pv'' = '
         READ(*,57) STR
         WRITE(1,'(A)') '      pvp = ' // STR
         WRITE(*,*) ' INPUT -(pu'')'' + q*u = '
         READ(*,57) STR
         WRITE(1,'(A)') '      hu = ' // STR
         WRITE(*,*) ' INPUT -(pv'')'' + q*v = '
         READ(*,57) STR
         WRITE(1,'(A)') '      hv = ' // STR
      ELSE IF(N.EQ.1) THEN
         WRITE(*,*) ' INPUT u = '
         READ(*,57) STR
         WRITE(1,'(A)') '         u = ' // STR
         WRITE(*,*) ' INPUT v = '
         READ(*,57) STR
         WRITE(1,'(A)') '         v = ' // STR
         WRITE(*,*) ' INPUT pu'' = '
         READ(*,57) STR
         WRITE(1,'(A)') '         pup = ' // STR
         WRITE(*,*) ' INPUT pv'' = '
         READ(*,57) STR
         WRITE(1,'(A)') '         pvp = ' // STR
         WRITE(*,*) ' INPUT -(pu'')'' + q*u = '
         READ(*,57) STR
         WRITE(1,'(A)') '         hu = ' // STR
         WRITE(*,*) ' INPUT -(pv'')'' + q*v = '
         READ(*,57) STR
         WRITE(1,'(A)') '         hv = ' // STR
      ELSE
         WRITE(*,*) ' INPUT U = '
         READ(*,57) STR
         WRITE(1,'(A)') '         U = ' // STR
         WRITE(*,*) ' INPUT V = '
         READ(*,57) STR
         WRITE(1,'(A)') '         V = ' // STR
         WRITE(*,*) ' INPUT PU'' = '
         READ(*,57) STR
         WRITE(1,'(A)') '         PUP = ' // STR
         WRITE(*,*) ' INPUT PV'' = '
         READ(*,57) STR
         WRITE(1,'(A)') '         PVP = ' // STR
         WRITE(*,*) ' INPUT -(PU'')'' + Q*U = '
         READ(*,57) STR
         WRITE(1,'(A)') '         HU = ' // STR
         WRITE(*,*) ' INPUT -(PV'')'' + Q*V = '
         READ(*,57) STR
         WRITE(1,'(A)') '         HV = ' // STR
         END IF
      RETURN
   57 FORMAT(A57)
      END
c
      subroutine help(nh)
      integer i,n,nh
      character*36 x(23),y(23)
      character*1 ans
c
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),NH
c
    1 CONTINUE
      write(*,*)
      write(*,*) 'H1:  Overview of HELP.'
      x(1)='  This ASCII text file is supplied a'
      y(1)='s a separate file with the SLEIGN2  '
      x(2)='package; it can be accessed on-line '
      y(2)='in both MAKEPQW (if used) and DRIVE.'
      x(3)='  HELP contains information to aid t'
      y(3)='he user in entering data on the co- '
      x(4)='efficient functions p,q,w; on the se'
      y(4)='lf-adjoint, separated and coupled,  '
      x(5)='regular and singular, boundary condi'
      y(5)='tions; on the limit circle boundary '
      x(6)='condition functions u,v at a and U,V'
      y(6)=' at b; on the end-point classifica- '
      x(7)='tions of the differential equation; '
      y(7)='on DEFAULT entry; on eigenvalue in- '
      x(8)='dexes; on IFLAG information; and on '
      y(8)='the general use of the program      '
      x(9)='SLEIGN2.                            '
      y(9)=' '
      x(10)='  The 17 sections of HELP are:      '
      y(10)=' '
      x(11)=' '
      y(11)=' '
      x(12)='    H1: Overview of HELP.           '
      y(12)=' '
      x(13)='    H2: File name entry.            '
      y(13)=' '
      x(14)='    H3: The differential equation.  '
      y(14)=' '
      x(15)='    H4: End-point classification.   '
      y(15)=' '
      x(16)='    H5: DEFAULT entry.              '
      y(16)=' '
      x(17)='    H6: Self-adjoint limit-circle bo' 
      y(17)='undary conditions.                  '
      do 101 i = 1,17
        write(*,*) x(i),y(i)
  101   continue
      write(*,*)
      write(*,*)
      write(*,*)
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
      x(1)='    H7: General self-adjoint boundar'
      y(1)='y conditions. '
      x(2)='    H8: Recording the results.      '
      y(2)=' '
      x(3)='    H9: Type and choice of interval.'
      y(3)=' '
      x(4)='   H10: Entry of end-points.        '
      y(4)=' '
      x(5)='   H11: End-point values of p,q,w.  '
      y(5)=' '
      x(6)='   H12: Initial value problems.     '
      y(6)=' '
      x(7)='   H13: Indexing of eigenvalues for '
      y(7)='separated boundary conditions.      '
      x(8)='   H14: Entry of eigenvalue index, i'
      y(8)='nitial guess, and tolerance.     '
      x(9)='   H15: IFLAG information.          '
      y(9)=' '
      x(10)='   H16: Plotting.                  '
      y(10)=' '
      x(11)='   H17: Indexing of eigenvalues for'
      y(11)='coupled boundary conditions.       '
      x(12)=' '
      y(12)=' '
      x(13)='  HELP can be accessed at each point'
      y(13)=' in MAKEPQW and DRIVE where the user'
      x(14)='is asked for input, by pressing "h <'
      y(14)='ENTER>"; this places the user at the'
      x(15)='appropriate HELP section.  Once in H'
      y(15)='ELP, the user can scroll the further'
      x(16)='HELP sections by repeatedly pressing'
      y(16)=' "h <ENTER>", or jump to a specific '
      x(17)='HELP section Hn (n=1,2,...17) by typ'
      y(17)='ing "Hn <ENTER>"; to return to the  '
      x(18)='place in the program from which HELP'
      y(18)=' is called, press "r <ENTER>".      '
      do 102 i = 1,18
        write(*,*) x(i),y(i)
  102   continue
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*) 
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
    2 CONTINUE
      write(*,*)
      write(*,*) 'H2:  File name entry.'
      x(1)='  MAKEPQW is used to create a FORTRA'
      y(1)='N file containing the coefficients  '
      x(2)='p(x),q(x),w(x), defining the differe'
      y(2)='ntial equation, and the boundary    '
      x(3)='condition functions u(x),v(x) and U('
      y(3)='x),V(x) if required.  The file must '
      x(4)='be given a NEW filename which is acc'
      y(4)='eptable to your FORTRAN compiler.   '
      x(5)='For example, it might be called bess'
      y(5)='el.f or bessel.for, depending upon  '
      x(6)='your compiler.                      '
      y(6)=' '
      x(7)='  The same naming considerations app'
      y(7)='ly if the FORTRAN file is prepared  '
      x(8)='other than with the use of MAKEPQW. ' 
      y(8)=' '
      do 201 i = 1,8
        write(*,*) x(i),y(i)
  201   continue
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
    3 CONTINUE
      write(*,*) 'H3:  The differential equation.'
      x(1)='  The prompt "Input p (or q or w) ="'
      y(1)=' requests you to type in a FORTRAN  '
      x(2)='expression defining the function p(x'
      y(2)='), which is one of the three coeffi-'
      x(3)='cient functions defining the Sturm-L'
      y(3)='iouville differential equation      '
      x(4)=' '
      y(4)=' '
      x(5)='                   -(p*y'')'' + q*y = '
      y(5)=' lambda*w*y              (*)        '
      x(6)=' '
      y(6)=' '
      x(7)='to be considered on some interval (a'
      y(7)=',b) of the real line.  The actual   '
      x(8)='interval used in a particular proble'
      y(8)='m can be chosen later, and may be   '
      x(9)='either the whole interval (a,b) wher'
      y(9)='e the coefficient functions p,q,w,  '
      x(10)='etc. are defined or any sub-interval'
      y(10)=' (a'',b'') of (a,b); a = -infinity  '
      x(11)='and/or b = +infinity are allowable c'
      y(11)='hoices for the end-points.          '
      x(12)='  The coefficient functions p,q,w of'
      y(12)=' the differential equation may be   '
      x(13)='chosen arbitrarily but must satisfy '
      y(13)='the following conditions:           '
      x(14)='  1. p,q,w are real-valued throughou'
      y(14)='t (a,b).                            '
      x(15)='  2. p,q,w are piece-wise continuous'
      y(15)=' and defined throughout the         '
      x(16)='     interior of the interval (a,b).'
      y(16)='                                    '
      x(17)='  3. p and w are strictly positive i'
      y(17)='n (a,b).                            '
      x(18)='     For better error analysis in th'
      y(18)='e numerical procedures, condition 2.'
      x(19)='     above is often replaced with   '
      y(19)=' '
      x(20)='  2''. p,q,w are four times continuou'
      y(20)='sly differentiable on (a,b).        '
      x(21)='  The behavior of p,q,w near the end'
      y(21)='-points a and b is critical to the  '
      x(22)='classification of the differential e'
      y(22)='quation (see H4 and H11).           '
      do 301 i = 1,22
        write(*,*) x(i),y(i)
  301   continue
      write(*,*)
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
    4 CONTINUE
      write(*,*) 'H4:  End-point classification.'
      x(1)='  The correct classification of the '
      y(1)='end-points a and b is essential to  '
      x(2)='the working of the SLEIGN2 program. '
      y(2)=' To classify the end-points, it is  '
      x(3)='convenient to choose a point c in (a'
      y(3)=',b); i.e.,  a < c < b.  Subject to  '
      x(4)='the general conditions on the coeffi'
      y(4)='cient functions p,q,w (see H3):     '
      x(5)='  1. a is REGULAR (say R) if -infini'
      y(5)='ty < a, p,q,w are piece-wise        '
      x(6)='     continuous on [a,c], and p(a) >'
      y(6)=' 0 and w(a) > 0.                    '
      x(7)='  2. a is WEAKLY REGULAR (say WR) if'
      y(7)=' -infinity < a, a is not R, and     '
      x(8)='                     |c             '
      y(8)=' '
      x(9)='            integral | {1/p+|q|+w} <'
      y(9)=' +infinity.                         '
      x(10)='                     |a             '
      y(10)=' '
      x(11)='     If end-point a is neither R nor'
      y(11)=' WR, then a is SINGULAR; that is,   '
      x(12)='     either -infinity = a, or -infin'
      y(12)='ity < a and                         '
      x(13)='                     |c             '
      y(13)=' '
      x(14)='            integral | {1/p+|q|+w} ='
      y(14)=' +infinity.                         '
      x(15)='                     |a             '
      y(15)=' '
      x(16)='  3. The SINGULAR end-point a is LIM'
      y(16)='IT-CIRCLE NON-OSCILLATORY (say      '
      x(17)='     LCNO) if for some real lambda A'
      y(17)='LL real-valued solutions y of the   '
      x(18)='     differential equation          '
      y(18)=' '
      x(19)=' '
      y(19)=' '
      x(20)='                   -(p*y'')'' + q*y = '
      y(20)='lambda*w*y on (a,c]     (*)         '
      x(21)=' '
      y(21)=' '
      x(22)='     satisfy the conditions:        '
      y(22)=' '
      do 401 i = 1,22
        write(*,*) x(i),y(i)
  401   continue
      write(*,*)
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
      x(1)='                     |c             '
      y(1)=' '
      x(2)='            integral | { w*y*y } < +'
      y(2)='infinity, and                       '
      x(3)='                     |a             '
      y(3)=' '
      x(4)='     y has at most a finite number o'
      y(4)='f zeros in (a,c].                   '
      x(5)='  4. The SINGULAR end-point a is LIM'
      y(5)='IT-CIRCLE OSCILLATORY (say LCO) if  '
      x(6)='     for some real lambda ALL real-v'
      y(6)=' alued solutions of the differential'
      x(7)='     equation (*) satisfy the condit'
      y(7)='ions:                               '
      x(8)='                     |c             '
      y(8)=' '
      x(9)='            integral | { w*y*y } < +'
      y(9)='infinity, and                       '
      x(10)='                     |a             '
      y(10)=' '
      x(11)='     y has an infinite number of zer'
      y(11)='os in (a,c].                        '
      x(12)='  5. The SINGULAR end-point a is LIM'
      y(12)='IT POINT (say LP) if for some real  '
      x(13)='     lambda at least one solution of'
      y(13)=' the differential equation (*) sat- '
      x(14)='     isfies the condition:          '
      y(14)=' '
      x(15)='                     |c             '
      y(15)=' '
      x(16)='            integral | {w*y*y} = +in'
      y(16)='finity.                             '
      x(17)='                     |a             '
      y(17)=' '
      x(18)='  There is a similar classification '
      y(18)='of the end-point b into one of the  '
      x(19)='five distinct cases R, WR, LCNO, LCO'
      y(19)=', LP.                               '
      do 402 i = 1,19
        write(*,*) x(i),y(i)
  402   continue
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
      x(1)='  Although the classification of sin'
      y(1)='gular end-points invokes a real     '
      x(2)='value of the parameter lambda, this '
      y(2)='classification is invariant in      '
      x(3)='lambda; all real choices of lambda l'
      y(3)='ead to the same classification.     '
      x(4)='  In determining the classification '
      y(4)='of singular end-points for the      '
      x(5)='differential equation (*), it is oft'
      y(5)='en convenient to start with the     '
      x(6)='choice lambda = 0 in attempting to f'
      y(6)='ind solutions (particularly when    '
      x(7)='q = 0 on (a,b)); however, see exampl'
      y(7)='e 7 below.                          '
      x(8)='  See H6 on the use of maximal domai'
      y(8)='n functions to determine the        '
      x(9)='classification at singular end-point'
      y(9)='s.                                  '
      do 403 i = 1,9
        write(*,*) x(i),y(i)
  403   continue
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
      write(*,*) ' EXAMPLES: '
      x(1)='  1. -y'''' = lambda*y is R at both en'
      y(1)='d-points of (a,b) when a and b are  '
      x(2)='     finite.                        '
      y(2)=' '
      x(3)='  2. -y'''' = lambda*y on (-infinity,i'
      y(3)='nfinity) is LP at both end-points.  '
      x(4)='  3. -(sqrt(x)*y''(x))'' = lambda*(1./'
      y(4)='sqrt(x))*y(x) on (0,infinity) is    '
      x(5)='     WR at 0 and LP at +infinity (ta'
      y(5)='ke lambda = 0 in (*)).  See         '
      x(6)='     examples.f, #10 (Weakly Regular'
      y(6)=').                                  '
      x(7)='  4. -((1-x*x)*y''(x))'' = lambda*y(x)'
      y(7)=' on (-1,1) is LCNO at both ends     '
      x(8)='     (take lambda = 0 in (*)).  See '
      y(8)='xamples.f, #1 (Legendre).           '
      x(9)='  5. -y''''(x) + C*(1/(x*x))*y(x) = la'
      y(9)='mbda*y(x) on (0,infinity) is LP at  '
      x(10)='     infinity and at 0 is (take lamb'
      y(10)='da = 0 in (*)):                     '
      x(11)='       LP for C .ge. 3/4 ;          '
      y(11)=' '
      x(12)='       LCNO for -1/4 .le. C .lt. 3/4'
      y(12)=' (but C .ne. 0);                    '
      x(13)='       LCO for C .lt. -1/4.         '
      y(13)=' '
      x(14)='  6. -(x*y''(x))'' - (1/x)*y(x) = lamb'
      y(14)='da*y(x) on (0,infinity) is LCO at 0 '
      x(15)='     and LP at +infinity (take lambd'
      y(15)='a = 0 in (*) with solutions         '
      x(16)='     cos(ln(x)) and sin(ln(x))).  Se'
      y(16)='e xamples.f, #7 (BEZ).              '
      x(17)='  7. -(x*y''(x))'' - x*y(x) = lambda*('
      y(17)='1/x)*y(x) on (0,infinity) is LP at 0'
      x(18)='     and LCO at infinity (take lambd'
      y(18)='a = -1/4 in (*) with solutions      '
      x(19)='     cos(x)/sqrt(x) and sin(x)/sqrt('
      y(19)='x)).  See xamples.f,                '
      x(20)='     #6 (Sears-Titchmarsh).         '
      y(20)=' '
      x(21)='  8. -y''''(x) + x*sin(x)*y(x) = lambd'
      y(21)='a*y(x) on (0,infinity) is R at 0 and'
      x(22)='     LP at infinity.  See examples.f'
      y(22)=' #30 (Littlewood-McLeod).           '
      do 404 i = 1,22
        write(*,*) x(i),y(i)
  404   continue
      write(*,*) ' '
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
    5 CONTINUE
      write(*,*) 'H5:  DEFAULT entry.'
      x(1)='  The complete range of problems for'
      y(1)=' which SLEIGN2 is applicable can    '
      x(2)='only be reached by appropriate entri'
      y(2)='es under end-point classification   '
      x(3)='and boundary conditions.  However, t'
      y(3)='here is a DEFAULT application which '
      x(4)='requires no detailed entry of end-po'
      y(4)='int classification or boundary      '
      x(5)='conditions, subject to:             '
      y(5)=' '
      x(6)='  1. The DEFAULT application CANNOT '
      y(6)='be used at a LCO end-point.         '
      x(7)='  2. If an end-point a is R, then th'
      y(7)='e Dirichlet boundary condition      '
      x(8)='     y(a) = 0 is automatically used.'
      y(8)=' '
      x(9)='  3. If an end-point a is WR, then t'
      y(9)='he following boundary condition     '
      x(10)='     is automatically applied:      '
      y(10)=' '
      x(11)='       if p(a) = 0, and both q(a),w('
      y(11)='a) are bounded, then the Neumann    '
      x(12)='       boundary condition (py'')(a) '
      y(12)='= 0 is used, or                     '
      x(13)='       if p(a) > 0, and q(a) and/or '
      y(13)='w(a)) are not bounded, then the     '
      x(14)='       Dirichlet boundary condition '
      y(14)='y(a) = 0 is used.                   '
      x(15)='       If p(a) = 0, and q(a) and/or '
      y(15)='w(a) are not bounded, then no simple'
      x(16)='       information, in general, can '
      y(16)=' be given on the DEFAULT boundary   '
      x(17)='       condition.                   '
      y(17)=' '
      x(18)='  4. If an end-point is LCNO, then i'
      y(18)='n most cases the principal or       '
      x(19)='     Friedrichs boundary condition i'
      y(19)='s applied (see H6).                 '
      x(20)='  5. If an end-point is LP, then the'
      y(20)=' normal LP procedure is applied     '
      x(21)='     (see H7(1.)).                  '
      y(21)=' '
      x(22)='If the DEFAULT condition is chosen, '
      y(22)='then no entry is required for the   '
      x(23)='u,v and U,V boundary condition funct'
      y(23)='ions. '
      do 501 i = 1,23
        write(*,*) x(i),y(i)
  501   continue
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
    6 CONTINUE
      write(*,*) 'H6:  Limit-circle boundary conditions.'
      x(1)='  At an end-point a, the limit-circl'
      y(1)='e type separated boundary condition '
      x(2)='is of the form (similar remarks thro'
      y(2)='ughout apply to the end-point b with'
      x(3)=' U,V being boundary condition functi'
      y(3)='ons at b)                           '
      x(4)=' '
      y(4)=' '
      x(5)='       A1*[y,u](a) + A2*[y,v](a) = 0'
      y(5)=',                          (**)     '
      x(6)=' '
      y(6)=' '
      x(7)='where y is a solution of the differe'
      y(7)='ntial equation                      '
      x(8)=' '
      y(8)=' '
      x(9)='     -(p*y'')'' + q*y = lambda*w*y  on'
      y(9)=' (a,b).                     (*)     '
      x(10)=' '
      y(10)=' '
      x(11)='Here A1, A2 are real numbers; u and '
      y(11)='v are boundary condition functions  '
      x(12)='at a; and for real-valued y and u th'
      y(12)='e form [y,u] is defined by          '
      x(13)='                                    '
      y(13)='                                    '
      x(14)='   [y,u](x) = y(x)*(pu'')(x) - u(x)*'
      y(14)='(py'')(x)   for x in (a,b).         '
      x(15)='                                    '
      y(15)='                                    '
      x(16)='  If neither end-point is LP then th'
      y(16)='ere are also self-adjoint coupled   '
      x(17)='boundary conditions.  These have a c'
      y(17)='anonical form given by              '
      x(18)='                                    '
      y(18)='                                    '
      x(19)='    Y(b) = exp(i*alpha)*K*Y(a),     '
      y(19)=' '
      x(20)='                                    '
      y(20)='                                    '
      x(21)='where K is a real 2 by 2 matrix with'
      y(21)=' determinant 1, alpha is restricted '
      x(22)='to the interval (-pi,pi], and Y is  '
      y(22)='  '
      do 551 i = 1,22
        write(*,*) x(i),y(i)
  551   continue
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
      x(1)=' (i) the solution vector Y = transpo'
      y(1)='se [y(a), (py'')(a)] at a regular   '
      x(2)='     end-point a, or                ' 
      y(2)=' '
      x(3)=' (ii) the "singular solution vector"'
      y(3)=' Y(a) = transpose [[y,u](a),        '
      x(4)='      [y,v](a)] at a singular LC end'
      y(4)=' -point a.  Similarly at the right  '
      x(5)='      end-point b with U,V.         '
      y(5)=' '
      x(6)='  The object of this section is to p'
      y(6)='rovide help in choosing appropriate '
      x(7)='functions u and v in (**) (or U,V), '
      y(7)='given the differential equation (*).'
      x(8)='Full details of the boundary conditi'
      y(8)='ons for (*) are discussed in H7;    '
      x(9)='here it is sufficient to say that th'
      y(9)='e limit-circle type boundary condi- '
      x(10)='tion (**) can be applied at any end-'
      y(10)='point in the LCNO, LCO classifica-  '
      x(11)='tion, but also in the R, WR classifi'
      y(11)='cation subject to the appropriate   '
      x(12)='choice of u,v or U,V.               '
      y(12)=' '
      x(13)='  Let (*) be R, WR, LCNO, or LCO at '
      y(13)='end-point a and choose c in (a,b).  '
      x(14)='Then either                         '
      y(14)=' '
      x(15)='  u and v are a pair of linearly ind'
      y(15)='ependent solutions of (*) on (a,c]  '
      x(16)='  for any chosen real values of lamb'
      y(16)='da, or                              '
      x(17)='  u and v are a pair of real-valued '
      y(17)=' maximal domain functions defined on'
      x(18)='  (a,c] satisfying [u,v](a) .ne. 0. '
      y(18)=' The maximal domain D(a,c] is de-   '
      x(19)='  defined by                        '
      y(19)=' '
      x(20)=' '
      y(20)=' '
      x(21)='       D(a,c] = {f:(a,c]->R:: f,pf'' '
      y(21)='in AC(a,c];                         '
      x(22)='                   f, ((-pf'')''+qf)/w'
      y(22)=' in L2((a,c;w)}  .                  '
      x(23)=' '
      y(23)=' '
      do 601 i = 1,23
        write(*,*) x(i),y(i)
  601   continue
      read(*,999) ans,n
      write(*,*)
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
      x(1)='  It is known that for all f,g in D('
      y(1)='a,c] the limit                      '
      x(2)=' '
      y(2)=' '
      x(3)='       [f,g](a) = lim[f,g](x) as x->'
      y(3)='a                                   '
      x(4)=' '
      y(4)=' '
      x(5)='exists and is finite.  If (*) is LCN'
      y(5)='O or LCO at a, then all solutions of'
      x(6)='of (*) belong to D(a,c] for all valu'
      y(6)='es of lambda.                       '
      x(7)='  The boundary condition (**) is ess'
      y(7)='ential in the LCNO and LCO cases but'
      x(8)='can also be used with advantage in s'
      y(8)='ome R and WR cases.  In the R, WR, '
      x(9)='and LCNO cases, but not in the LCO c'
      y(9)='ase, the boundary condition         '
      x(10)='functions can always be chosen so th'
      y(10)='at                                  '
      x(11)='        lim u(x)/v(x) = 0 as x->a,  '
      y(11)=' '
      x(12)='and it is recommended that this norm'
      y(12)='alisation be effected, but this is  '
      x(13)='not essential; this normalisation ha'
      y(13)='s been entered in the examples given'
      x(14)='below.  In this case, the boundary c'
      y(14)='ondition [y,u](a) = 0 (i.e., A1 = 1,'
      x(15)='A2 = 0 in (**) is called the princip'
      y(15)='al or Friedrichs boundary condition '
      x(16)='at a.                              '
      y(16)=' '
      x(17)='  In the case when end-points a and '
      y(17)='b are, independently, in R, WR,     '
      x(18)='LCNO, or LCO classification, it may '
      y(18)='be that symmetry or other reasons   '
      x(19)='permit one set of boundary condition'
      y(19)=' functions to be used at both end-  '
      x(20)='points (see xamples.f, #1 (Legendre)'
      y(20)=').  In other cases, different pairs '
      x(21)='must be chosen for each end-point (s'
      y(21)='ee xamples.f: #16 (Jacobi),         '
      x(22)='#18 (Dunsch), and #19 (Donsch)).    '
      y(22)=' '
      do 602 i = 1,22
        write(*,*) x(i),y(i)
  602   continue
      write(*,*)
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
      x(1)=' '
      y(1)=' '
      x(2)='  Note that a solution pair u,v is a'
      y(2)='lways a maximal domain pair, but not'
      x(3)='necessarily vice versa.             '
      y(3)=' '
      x(4)=' '
      y(4)=' '
      x(5)='EXAMPLES:                           '
      y(5)=' '
      x(6)='1. -y''''(x) = lambda*y(x) on [0,pi] i'
      y(6)='s R at 0 and R at pi.               '
      x(7)='   At 0, with lambda = 0, a solution'
      y(7)=' pair is u(x) = x, v(x) = 1.        '
      x(8)='   At pi, with lambda = 1, a solutio'
      y(8)='n pair is                           '
      x(9)='     u(x) = sin(x), v(x) = cos(x).  '
      y(9) =' '
      x(10)='2. -(sqrt(x)*y''(x))'' = lambda*y(x)/s'
      y(10)='qrt(x) on (0,1] is                  '
      x(11)='     WR at 0 and R at 1.            '
      y(11)=' '
      x(12)='   (The general solutions of this eq'
      y(12)='uation are                          '
      x(13)='     u(x) = cos(2*sqrt(x*lambda)), v'
      y(13)='(x) = sin(2*sqrt(x*lambda)).)       '
      x(14)='   At 0, with lambda = 0, a solution'
      y(14)=' pair is                            '
      x(15)='     u(x) = 2*sqrt(x), v(x) = 1.    '
      y(15)=' '
      x(16)='   At 1, with lambda = pi*pi/4, a so'
      y(16)='lution pair is                      '
      x(17)='     u(x) = sin(pi*sqrt(x)), v(x) = '
      y(17)='cos(pi*sqrt(x)).                    '
      x(18)='   At 1, with lambda = 0, a solution'
      y(18)=' pair is                            '
      x(19)='     u(x) = 2*(1-sqrt(x)), v(x) = 1.'
      y(19)=' '
      x(20)='   See also xamples.f, #10 (Weakly R'
      y(20)='egular).                            '
      x(21)='3. -((1-x*x)*y''(x))'' = lambda*y(x) o'
      y(21)='n (-1,1) is LCNO at both ends.      '
      x(22)='   At +-1, with lambda = 0, a soluti'
      y(22)='on pair is                          '
      x(23)='     u(x) = 1, v(x) = 0.5*log((1+x)/'
      y(23)='(1-x)).                             '
      do 603 i = 1,23
        write(*,*) x(i),y(i)
  603   continue
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
      x(1)='   At 1, a maximal domain pair is u('
      y(1)='x) = 1, v(x) = log(1-x)             '
      x(2)='   At -1, a maximal domain pair is u'
      y(2)='(x) = 1, v(x) = log(1+x).           '
      x(3)='   See also xamples.f, #1 (Legendre)'
      y(3)='.                                   '
      x(4)='4. -y''''(x) - (1/(4x*x))*y(x) = lambd'
      y(4)='a*y(x) on (0,infinity) is           '
      x(5)='     LCNO at 0 and LP at +infinity. '
      y(5)=' '
      x(6)='   At 0, a maximal domain pair is   '
      y(6)=' '
      x(7)='     u(x) = sqrt(x), v(x) = sqrt(x)*'
      y(7)='log(x).                             '
      x(8)='   See also xamples.f, #2 (Bessel). '
      y(8)=' '
      x(9)='5. -y''''(x) - 5*(1/(4*x*x))*y(x) = la'
      y(9)='mbda*y(x) on (0,infinity) is        '
      x(10)='     LCO at 0 and LP at +infinity.  '
      y(10)=' '
      x(11)='   At 0, with lambda = 0, a solution'
      y(11)=' pair is                            '
      x(12)='     u(x) = sqrt(x)*cos(log(x)), v(x'
      y(12)=') = sqrt(x)*sin(log(x))             '
      x(13)='   See also xamples.f, #20 (Krall). '
      y(13)=' '
      x(14)='6. -y''''(x) - (1/x)*y(x) = lambda*y(x'
      y(14)=') on (0,infinity) is               '
      x(15)='     LCNO at 0 and LP at +infinity.'
      y(15)=' '
      x(16)='   At 0, a maximal domain pair is   '
      y(16)=' '
      x(17)='     u(x) = x, v(x) = 1 -x*log(x).  '
      y(17)=' '
      x(18)='   See also xamples.f, #4(Boyd).    '
      y(18)=' '
      x(19)='7. -((1/x)*y''(x))'' + (k/(x*x) + k*k/'
      y(19)='x)*y(x) = lambda*y(x) on (0,1],     '
      x(20)='     with k real and .ne. 0, is LCNO'
      y(20)=' at 0 and R at 1.                   '
      x(21)='   At 0, a maximal domain pair is   '
      y(21)=' '
      x(22)='     u(x) = x*x, v(x) = x - 1/k.    '
      y(22)=' '
      x(23)='   See also xamples.f, #8 (Laplace T'
      y(23)='idal Wave).                         '
      do 604 i = 1,23
        write(*,*) x(i),y(i)
  604   continue
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
    7 CONTINUE
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*) 'H7:  General self-adjoint boundary conditions.'
      x(1)='  Boundary conditions for Sturm-Liou'
      y(1)='ville boundary value problems       '
      x(2)=' '
      y(2)=' '
      x(3)='                   -(p*y'')'' + q*y = '
      y(3)='lambda*w*y              (*)         '
      x(4)=' '
      y(4)=' '
      x(5)='on an interval (a,b) are either     '
      y(5)=' '                                    
      x(6)='  SEPARATED, with at most one condit'
      y(6)='ion at end-point a and at most one  '
      x(7)='    condition at end-point b, or    '
      y(7)=' '
      x(8)='  COUPLED, when both a and b are, in'
      y(8)='dependently, in one of the end-point'
      x(9)='    classifications R, WR, LCNO, LCO'
      y(9)=', in which case two independent     '
      x(10)='    boundary conditions are required'
      y(10)=' which link the solution values near'
      x(11)='    a to those near b.              '
      y(11)=' '
      x(12)='  The SLEIGN2 program allows for all'
      y(12)=' self-adjoint boundary conditions:  '
      x(13)='separated self-adjoint conditions an'
      y(13)='d all cases of coupled self-adjoint '
      x(14)='conditions.                         '
      y(14)=' '
      do 701 i = 1,14
        write(*,*) x(i),y(i)
  701   continue
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
      write(*,*) '    Separated Conditions:      '
      write(*,*) '    ---------------------      '
      x(1)='  The boundary conditions to be sele'
      y(1)='cted depend upon the classification '
      x(2)='of the differential equation at the '
      y(2)='end-point, say, a:                  '
      x(3)='  1. If the end-point a is LP, then '
      y(3)='no boundary condition is required or'
      x(4)='     allowed.                       '
      y(4)=' '
      x(5)='  2. If the end-point a is R or WR, '
      y(5)='then a separated boundary condition '
      x(6)='     is of the form                 '
      y(6)=' '
      x(7)='         A1*y(a) + A2*(py'')(a) = 0,'
      y(7)=' '
      x(8)='     where A1, A2 are real constants'
      y(8)=' the user must choose, not both zero.  '
      x(9)='  3. If the end-point a is LCNO or'
      y(9)=' LCO, then a separated boundary     '
      x(10)='  condition is of the form          '
      y(10)=' '
      x(11)='         A1*[y,u](a) + A2*[y,v](a) ='
      y(11)=' 0,                                 '
      x(12)='  where A1, A2 are real constants th'
      y(12)='e user must choose, not both zero;  '
      x(13)='  here, u,v are the pair of boundary'
      y(13)=' condition functions you have       '
      x(14)='  previously selected when the input'
      y(14)=' FORTRAN file was being prepared    '
      x(15)='  with makepqw.f.                   '
      y(15)=' ' 
      x(16)='  4. If the end-point a is LCNO an'
      y(16)='d the boundary condition pair       '
      x(17)='  u,v has been chosen so that       '
      y(17)=' '
      x(18)='         lim u(x)/v(x) = 0  as x->a '
      y(18)=' '
      x(19)='  (which is always possible), then A'
      y(19)='1 = 1, A2 = 0 (i.e., [y,u](a) = 0)  '
      x(20)='  gives the principal (Friedrichs) b'
      y(20)='oundary condition at a.             '
      do 702 i = 1,20
        write(*,*) x(i),y(i)
  702   continue
      write(*,*)
      write(*,*)
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
      x(1)='  5. If a is R or WR and boundary '
      y(1)='condition functions u,v have been   '
      x(2)='  entered in the FORTRAN input file,'
      y(2)=' then (3.,4.) above apply to        '
      x(3)='  entering separated boundary condit'
      y(3)='ions at such an end-point; the      '
      x(4)='  boundary conditions in this form a'
      y(4)='re equivalent to the point-wise     '
      x(5)='  conditions in (2.) (subject to car'
      y(5)='e in choosing A1, A2).  This        '
      x(6)='  singular form of a regular boundar'
      y(6)='y condition may be particularly     '
      x(7)='  effective in the WR case if the bo'
      y(7)='undary condition form in (2.) leads '
      x(8)='  to numerical difficulties.        '
      y(8)=' '
      x(9)='  Conditions (2.,3.,4.,5.) apply sim'
      y(9)='ilarly at end-point b (with U,V as  '
      x(10)='the boundary condition functions at '
      y(10)='b.                                  '
      x(11)='    6. If a is R, WR, LCNO, or LCO a'
      y(11)='nd b is LP, then only a separated   '
      x(12)='  condition at a is required and all'
      y(12)='owed (or instead at b if a and b    '
      x(13)='  are interchanged).                '
      y(13)=' '
      x(14)='    7. If both end-points a and b ar'
      y(14)='e LP, then no boundary conditions   '
      x(15)='  are required or allowed.          '
      y(15)=' '
      x(16)='  The indexing of eigenvalues for bo'
      y(16)='undary value problems with separated'
      x(17)='  conditions is discussed in H13.   '
      y(17)=' '
      x(18)='    Coupled Conditions:             '
      y(18)=' '
      x(19)='    -------------------             '
      y(19)=' '
      x(20)='    8. Coupled regular self-adjoint '
      y(20)='boundary conditions on (a,b) apply  '
      x(21)='only when both end-points a and b ar'
      y(21)='e R or WR.                          '
      x(22)=' '
      y(22)=' '
      do 704 i = 1,22
        write(*,*) x(i),y(i)
  704   continue
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
    8 CONTINUE
      write(*,*) 'H8:  Recording the results.'
      x(1)='  If you choose to have a record kep'
      y(1)='t of the results, then the following'
      x(2)='information is stored in a file with'
      y(2)=' the name you select:               '
      x(3)=' '
      y(3)=' '
      x(4)='  1. The file name.                 '
      y(4)=' '
      x(5)='  2. The header line prompted for (u'
      y(5)='p to 32 characters of your choice). '
      x(6)='  3. The interval (a,b) which was us'
      y(6)='ed.                                 '
      x(7)=' '
      y(7)=' '
      x(8)='  For SEPARATED boundary conditions:'
      y(8)=' '
      x(9)='  4. The end-point classification.  '
      y(9)=' '
      x(10)='  5. A summary of coefficient inform'
      y(10)='ation at WR, LCNO, LCO end-points.  '
      x(11)='  6. The boundary condition constant'
      y(11)='s (A1,A2), (B1,B2) if entered.      '
      x(12)='  7. (NUMEIG,EIG,TOL) or (NUMEIG1,NU'
      y(12)='MEIG2,TOL), as entered.             '
      x(13)=' '
      y(13)=' '
      x(14)='  For COUPLED boundary conditions:  '
      y(14)=' '
      x(15)='  8. The boundary condition paramete'
      y(15)='r alpha and the coupling matrix K;  '
      x(16)='     see H6.                        '
      y(16)=' '
      x(17)='  For ALL self-adjoint boundary cond'
      y(17)='itions: '
      x(18)='  9. The computed eigenvalue, EIG, a'
      y(18)='nd its estimated accuracy, TOL.     '
      x(19)=' 10. IFLAG reported (see H15).      '
      y(19)=' '
      do 801 i = 1,19
        write(*,*) x(i),y(i)
  801   continue
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*) ' '
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
    9 CONTINUE
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*) 'H9:  Type and choice of interval.'
      x(1)='  You may enter any interval (a,b) f'
      y(1)='or which the coefficients p,q,w are '
      x(2)='well-defined by your FORTRAN stateme'
      y(2)='nts in the input file, provided that'
      x(3)='(a,b) contains no interior singulari'
      y(3)='ties.                               '
      do 901 i = 1,3
        write(*,*) x(i),y(i)
  901   continue
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*) ' '
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
   10 CONTINUE
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*) 'H10:  Entry of end-points.'
      x(1)='  End-points a and b should generall'
      y(1)='y be entered as real numbers to an  '
      x(2)='appropriate number of decimal places'
      y(2)='. '
      do 1001 i = 1,2
        write(*,*) x(i),y(i)
 1001   continue
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*) ' '
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
   11 CONTINUE
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*) 'H11:  End-point values of p,q,w.'
      x(1)='  The program SLEIGN2 needs to know '
      y(1)='whether the coefficient functions   '
      x(2)='p(x),q(x),w(x) defined by the FORTRA'
      y(2)='N expressions entered in the input  '
      x(3)='file can be evaluated numerically wi'
      y(3)='thout running into difficulty.  If, '
      x(4)='for example, either q or w is unboun'
      y(4)='ded at a, or p(a) is 0, then SLEIGN2'
      x(5)='needs to know this so that a is not '
      y(5)='chosen for functional evaluation.   '
      do 1101 i = 1,5
        write(*,*) x(i),y(i)
 1101   continue
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*) ' '
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
   12 CONTINUE
      write(*,*) 'H12:  Initial value problems.'
      x(1)='  The initial value problem facility'
      y(1)=' for Sturm-Liouville problems       '
      x(2)=' '
      y(2)=' '
      x(3)='                   -(p*y'')'' + q*y = '
      y(3)=' lambda*w*y              (*)        '
      x(4)=' '
      y(4)=' '
      x(5)='allows for the computation of a solu'
      y(5)='tion of (*) with a user-chosen      '
      x(6)='value lambda and any one of the foll'
      y(6)='owing initial conditions:           '
      x(7)='  1. From end-point a of any classif'
      y(7)='ication except LP towards           '
      x(8)='end-point b of any classification,  '
      y(8)=' '
      x(9)='  2. From end-point b of any classif'
      y(9)='ication except LP back towards      '
      x(10)='end-point a of any classification,  '
      y(10)=' '
      x(11)='  3. From end-points a and b of any '
      y(11)='classifications except LP towards an'
      x(12)='interior point of (a,b) selected by '
      y(12)='the program.                        '
      x(13)=' '
      y(13)=' '
      x(14)='  Initial values at a are of the for'
      y(14)='m y(a) = alpha1, (p*y'')(a) =alpha2,'
      x(15)='when a is R or WR; and [y,u](a) = al'
      y(15)='pha1, [y,v](a) = alpha2, when a is  '
      x(16)='LCNO or LCO.                        '
      y(16)=' '
      x(17)='  Initial values at b are of the for'
      y(17)='m y(b) = beta1, (p*y'')(b) = beta2, '
      x(18)='when b is R or WR; and [y,u](b) = be'
      y(18)='ta1, [y,v](b) = beta2, when b is    '
      x(19)='LCNO or LCO.                        '
      y(19)=' '
      x(20)='  In (*), lambda is a user-chosen re'
      y(20)='al number; while in the above ini-  '
      x(21)='tial values, (alpha1,alpha2) and (be'
      y(21)='ta1,beta2) are user-chosen pairs of '
      x(22)='real numbers not both zero.         '
      y(22)=' '
      do 1201 i = 1,22
        write(*,*) x(i),y(i)
 1201   continue
      write(*,*)
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      x(1)='  In the initial value case (3.) abo'
      y(1)='ve when the interval (a,b) is       '
      x(2)='finite, the interior point selected '
      y(2)='by the program is generally near the'
      x(3)='midpoint of (a,b); when (a,b) is inf'
      y(3)='inite, no general rule can be given.'
      x(4)='Also if, given (alpha1,alpha2) and ('
      y(4)='beta1,beta2), the lambda chosen is  '
      x(5)='an eigenvalue of the associated boun'
      y(5)='dary value problem, the computed    '
      x(6)='solution may not be the correspondin'
      y(6)='g eigenfunction -- the signs of the '
      x(7)='computed solutions on either side of'
      y(7)=' the interior point may be opposite.'
      x(8)='  The output for a solution of an in'
      y(8)='itial value problem is in the form  '
      x(9)='of stored numerical data which can b'
      y(9)='e plotted on the screen (see H16),  '
      x(10)='or printed out in graphical form if '
      y(10)='graphics software is available.     '
      do 1202 i = 1,10
        write(*,*) x(i),y(i)
 1202   continue
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*) ' '
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
   13 continue
      write(*,*) 'H13:  Indexing of eigenvalues.'
      x(1)='  The indexing of eigenvalues is an '
      y(1)='automatic facility in SLEIGN2.  The '
      x(2)='following general results hold for t'
      y(2)='he separated boundary condition     '
      x(3)='problem (see H7):                   '
      y(3)=' '
      x(4)='  1. If neither end-point a or b is '
      y(4)='LP or LCO, then the spectrum of the '
      x(5)='eigenvalue problem is discrete (eige'
      y(5)='nvalues only), simple (eigenvalues  '
      x(6)='all of multiplicity 1), and bounded '
      y(6)='below with a single cluster point at'
      x(7)='+infinity.  The eigenvalues are inde'
      y(7)='xed as {lambda(n): n=0,1,2,...},    '
      x(8)='where lambda(n) < lambda(n+1) (n=0,1'
      y(8)=',2,...), lim lambda(n) -> +infinity;'
      x(9)='and if {psi(n): n=0,1,2,...} are the'
      y(9)=' corresponding eigenfunctions, then '
      x(10)='psi(n) has exactly n zeros in the op'
      y(10)='en interval (a,b).                  '
      x(11)='  2. If neither end-point a or b is '
      y(11)='LP but at least one end-point is    '
      x(12)='LCO, then the spectrum is discrete a'
      y(12)='nd simple as for (1.), but with     '
      x(13)='cluster points at both +infinity and'
      y(13)=' -infinity.  The eigenvalues are in-'
      x(14)='dexed as {lambda(n): n=0,1,-1,2,-2,.'
      y(14)='..}, where                          '
      x(15)='lambda(n) < lambda(n+1) (n=...-2,-1,'
      y(15)='0,1,2,...) with lambda(0) the small-'
      x(16)='est non-negative eigenvalue and lim '
      y(16)='lambda(n) -> +infinity or -> -infi- '
      x(17)='nity with n; and if {psi(n): n=0,1,-'
      y(17)='1,2,-2,...} are the corresponding   '
      x(18)='eigenfunctions, then every psi(n) ha'
      y(18)='s infinitely many zeros in (a,b).   '
      do 1301 i = 1,18
        write(*,*) x(i),y(i)
 1301   continue
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
      x(1)='  3. If one or both end-points is LP'
      y(1)=', then there can be one or more in- '
      x(2)='tervals of continuous spectrum for t'
      y(2)='he boundary value problem in addi-  '
      x(3)='tion to some (necessarily simple) ei'
      y(3)='genvalues.  For these essentially   '
      x(4)='more difficult problems, SLEIGN2 can'
      y(4)=' be used as an investigative tool to' 
      x(5)='give qualitative and possibly quanti'
      y(5)='tative information on the spectrum. '
      x(6)='     For example, if a problem has a'
      y(6)=' continuous spectrum starting at L, '
      x(7)='then there may be no eigenvalue belo'
      y(7)='w L, any finite number of eigenval- '
      x(8)='ues below L, or an infinite (but cou'
      y(8)='ntable) number of eigenvalues below '
      x(9)='L. SLEIGN2 can be used to compute L '
      y(9)='(see the paper bewz on the sleign2  '
      x(10)='home page for an algorithm to comput'
      y(10)='e L), and determine the number of   '
      x(11)='these eigenvalues and compute them. '
      y(11)=' In this respect, see xamples.f: #13'
      x(12)='(Hydrogen Atom), #17 (Morse Oscillat'
      y(12)='or), #21 (Fourier), #27 (Joergens)  '
      x(13)='as examples of success; and #12 (Mat'
      y(13)='hieu), #14 (Marletta), and #28      '
      x(14)='(Behnke-Goerisch) as examples of fai'
      y(14)='lure.                               '
      x(15)='     The problem need not have a con'
      y(15)='tinuous spectrum, in which case if  '
      x(16)='its discrete spectrum is bounded bel'
      y(16)='ow, then the eigenvalues are indexed'
      x(17)='and the eigenfunctions have zero cou'
      y(17)='nts as in (1.).  If, on the other   '
      x(18)='hand, the discrete spectrum is unbou'
      y(18)='nded below, then all the eigenfunc- '
      x(19)='tions have infinitely many zeros in '
      y(19)='the open interval (a,b).  SLEIGN2   '   
      x(20)='can, in principle, compute these eig'
      y(20)='envalues if neither end-point is LP,'
      x(21)='although this is a computationally d'
      y(21)='ifficult problem.  Note, however,   '
      x(22)='that SLEIGN2 has no algorithm when t'
      y(22)='he spectrum is discrete, unbounded  '
      x(23)='above and below, and one end-point i'
      y(23)='s LP, as in xamples.f #30.          '
      do 1302 i = 1,23
        write(*,*) x(i),y(i)
 1302   continue
      write(*,*)
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
      x(1)='  In respect to the three different '
      y(1)='types of indexing discussed above,  '
      x(2)='the following identified examples fr'
      y(2)='om xamples.f illustrate the spectral'
      x(3)='property of these boundary problems:'
      y(3)='                                    '
      x(4)='  1. Neither end-point is LP or LCO.'
      y(4)=' '
      x(5)='       #1 (Legendre)                '
      y(5)=' '
      x(6)='       #2 (Bessel) with -1/4 < c < 3'
      y(6)='/4                                  '
      x(7)='       #4 (Boyd)                    '
      y(7)=' '
      x(8)='       #5 (Latzko)                  '
      y(8)=' '
      x(9)='  2. Neither end-point is LP, but at'
      y(9)=' least one is LCO.                  '
      x(10)='       #6 (Sears-Titchmarsh)        '
      y(10)=' '
      x(11)='       #7 (BEZ)                     '
      y(11)=' '
      x(12)='      #19 (Donsch)                  '
      y(12)=' '
      x(13)='  3. At least one end-point is LP.  '
      y(13)=' '
      x(14)='      #13 (Hydrogen Atom)           '
      y(14)=' '
      x(15)='      #14 (Marletta)                '
      y(15)=' '
      x(16)='      #20 (Krall)                   '
      y(16)=' '
      x(17)='      #21 (Fourier) on [0,infinity) '
      y(17)=' '
      do 1303 i = 1,17
        write(*,*) x(i),y(i)
 1303   continue
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*) ' '
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
   14 CONTINUE
      write(*,*) 'H14:  Entry of eigenvalue index, initial guess,'//
     1           ' and tolerance.'
      x(1)='  For all self-adjoint boundary cond'
      y(1)='ition problems (see H7), SLEIGN2    '
      x(2)='calls for input information options '
      y(2)='to compute either                   '
      x(3)='  1. a single eigenvalue, or        '
      y(3)=' '
      x(4)='  2. a series of eigenvalues.       '
      y(4)=' '
      x(5)='In each case indexing of eigenvalues'
      y(5)=' is called for (see H13).           '
      x(6)='  (1.) above asks for data triples N'
      y(6)='UMEIG, EIG, TOL separated by commas.'
      x(7)='Here NUMEIG is the integer index of '
      y(7)='the desired eigenvalue; NUMEIG can  '
      x(8)='be negative only when the problem is'
      y(8)=' LCO at one or both end-points.     '
      x(9)='EIG allows for the entry of an initi'
      y(9)='al guess for the requested          '
      x(10)='eigenvalue (if an especially good on'
      y(10)='e is available), or can be set to 0 '
      x(11)='in which case an initial guess is ge'
      y(11)='nerated by SLEIGN2 itself.          '
      x(12)='TOL is the desired accuracy of the c'
      y(12)='omputed eigenvalue.  It is an       '
      x(13)='absolute accuracy if the magnitude o'
      y(13)='f the eigenvalue is 1 or less, and  '
      x(14)='is a relative accuracy otherwise.  T'
      y(14)='ypical values might be .001 for     '
      x(15)='moderate accuracy and .0000001 for h'
      y(15)='igh accuracy in single precision.   '
      x(16)='If TOL is set to 0, the maximum achi'
      y(16)='evable accuracy is requested.       '
      x(17)='  If the input data list is truncate'
      y(17)='d with a "/" after NUMEIG or EIG,   '
      x(18)='then the remaining elements default '
      y(18)='to 0.                               '
      do 1401 i = 1,18
        write(*,*) x(i),y(i)
 1401   continue
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
      x(1)='  (2.) above asks for data triples N'
      y(1)='UMEIG1, NUMEIG2, TOL separated by   '
      x(2)='commas.  Here NUMEIG1 and NUMEIG2 ar'
      y(2)='e the first and last integer indices'
      x(3)='of the sequence of desired eigenvalu'
      y(3)='es, NUMEIG1 < NUMEIG2; they can be  '
      x(4)='negative only when the problem is LC'
      y(4)='O at one or both end-points.        '
      x(5)='TOL is the desired accuracy of the c'
      y(5)='omputed eigenvalues.  It is an      '
      x(6)='absolute accuracy if the magnitude o'
      y(6)='f an eigenvalue is 1 or less, and   '
      x(7)='is a relative accuracy otherwise.  T'
      y(7)='ypical values might be .001 for     '
      x(8)='moderate accuracy and .0000001 for h'
      y(8)='igh accuracy in single precision.   '
      x(9)='If TOL is set to 0, the maximum achi'
      y(9)='evable accuracy is requested.       '
      x(10)='  If the input data list is truncate'
      y(10)='d with a "/" after NUMEIG2, then TOL'
      x(11)='defaults to 0.                      '
      y(11)=' '
      x(12)='  For COUPLED self-adjoint boundary '
      y(12)='condition problems (see H7 and H17),'
      x(13)='SLEIGN2 also reports which eigenvalu'
      y(13)='es are double.  Double eigenvalues  '
      x(14)='can occur only for coupled boundary '
      y(14)='conditions with alpha = 0 or pi.    ' 
      do 1402 i = 1,14
        write(*,*) x(i),y(i)
 1402   continue
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*) ' '
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
   15 CONTINUE
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*) 'H15:  IFLAG information.'
      x(1)='  All results are reported by SLEIGN'
      y(1)='2 with a flag identification.  There'
      x(2)='are four values of IFLAG:           '
      y(2)=' '
      x(3)=' '
      y(3)=' '
      x(4)='  1 - The computed eigenvalue has an'
      y(4)=' estimated accuracy within the      '
      x(5)='      tolerance requested.          '
      y(5)=' '
      x(6)=' '
      y(6)=' '
      x(7)='  2 - The computed eigenvalue does n'
      y(7)='ot have an estimated accuracy within'
      x(8)='      the tolerance requested, but i'
      y(8)='s the best the program could obtain.'
      x(9)=' '
      y(9)=' '
      x(10)='  3 - There seems to be no eigenvalu'
      y(10)='e of index equal to NUMEIG.         '
      x(11)=' '
      y(11)=' '
      x(12)='  4 - The program has been unable to'
      y(12)=' compute the requested eigenvalue.  '
      do 1501 i = 1,12
        write(*,*) x(i),y(i)
 1501   continue
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*) ' '
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
   16 CONTINUE
      write(*,*) 'H16:  Plotting.'
      x(1)='  After computing a single eigenvalu'
      y(1)='e (see H14(1.)), but not a sequence '
      x(2)='of eigenvalues (see H14(2.)), the ei'
      y(2)='genfunction can be plotted for sepa-'
      x(3)='rated conditions and for coupled one'
      y(3)='s with alpha = 0 or pi.  When alpha '
      x(4)='is not zero or pi the eigenfunctions'
      y(4)=' are not real-valued and so no plot '
      x(5)='is provided.  If this is desired, re'
      y(5)='spond "y" when asked so that SLEIGN2'
      x(6)='will compute some eigenfunction data'
      y(6)=' and store them.                    '
      x(7)='  One can ask that the eigenfunction'
      y(7)=' data be in the form of either      '
      x(8)='points (x,y) for x in (a,b), or poin'
      y(8)='ts (t,y) for t in the standardized  '
      x(9)='interval (-1,1) mapped onto from (a,'
      y(9)='b); the t- choice can be especially '
      x(10)='helpful when the original interval i'
      y(10)='s infinite.  Additionally, one can  '
      x(11)='ask for a plot of the so-called Prue'
      y(11)='fer angle, in x- or t- variables.   '
      x(12)='  In both forms, once the choice has'
      y(12)=' been made of the function to be    '
      x(13)='plotted, a crude plot is displayed o'
      y(13)='n the monitor screen and you are    '
      x(14)='asked whether you wish to save the c'
      y(14)='omputed plot points in a file.      '
      do 1601 i = 1,14
        write(*,*) x(i),y(i)
 1601   continue
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*) ' '
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
   17 CONTINUE
      write(*,*) 'H17:  Indexing of eigenvalues for'//
     1           ' coupled self-adjoint problems.'
      x(1)='  The indexing of eigenvalues is an '
      y(1)='automatic facility in SLEIGN2.  The '
      x(2)='following general result holds for c'
      y(2)='oupled boundary condition problems  '
      x(3)='(see H7):                           '
      y(3)=' '
      x(4)='  The spectrum of the eigenvalue pro'
      y(4)='blem is discrete (eigenvalues only).'
      x(5)='In general the spectrum is not simpl'
      y(5)='e, but no eigenvalue exceeds multi- '
      x(6)='plicity 2.The eigenvalues are indexe'
      y(6)='d as {lambda(n): n=0,1,2,...}, where'
      x(7)='lambda(n) .le. lambda(n+1) (n=0,1,2,'
      y(7)='...), lim lambda(n) -> +infinity if '
      x(8)='neither end-point is LCO.  If one or'
      y(8)=' both end-points are LCO the eigen- '
      x(9)='values cluster at both +infinity and'
      y(9)=' at -infinity, and all eigenfunc-   '
      x(10)='tions have infinitely many zeros.   '
      y(10)='                                    '
      x(11)='  If neither end-point is LCO and al'
      y(11)='pha = 0 or pi, then the n-th eigen- '
      x(12)='function has n-1, n, or n+1 zeros in'
      y(12)=' the half-open interval [a,b) (also '
      x(13)='in (a,b]).  All three possibilities '
      y(13)='occur.  Recall that in the case of  '
      x(14)='double eigenvalues, although the n-t'
      y(14)='h eigenvalue is well-defined, there '
      x(15)='is an ambiguity about which solution'
      y(15)=' is declared the n-th eigenfunction.'
      x(16)='  If alpha is not 0 or pi, then the '
      y(16)='eigenfunction is non-real and has no'
      x(17)='zero in (a,b); but each of the real '
      y(17)='and imaginary parts of the n-th ei- '
      x(18)='genfunction have the same zero prope'
      y(18)='rties as mentioned above when alpha '
      x(19)='= 0 or pi.                          '
      y(19)=' '
      do 1651 i = 1,19
        write(*,*) x(i),y(i)
 1651   continue    
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      x(1)='  The following identified examples '
      y(1)='from xamples.f are of special inter-'
      x(2)='est:                                '
      y(2)=' '
      x(3)='    #11 (Plum) on [0,pi]           '
      y(3)=' '
      x(4)='    #21 (Fourier) on [0,pi]        '
      y(4)=' '
      x(5)='    #25 (Meissner) on [-0.5,0.5]   '
      y(5)=' '
      do 1701 i = 1,5
        write(*,*) x(i),y(i)
 1701   continue
      write(*,*) ' '
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      write(*,*)
      read(*,999) ans,n
      if (ans.eq.'r' .or. ans.eq.'R') return
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),N
      go to 1
  999 FORMAT(A1,I2)
      end
SHAR_EOF
fi # end of overwriting check
if test -f 'res1'
then
	echo shar: will not over-write existing file "'res1'"
else
cat << "SHAR_EOF" > 'res1'
 
  ******************** 
  EIGENVALUE            3
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
  NUMBER OF ITERATIONS WAS            1
 -----------------------------------------------
 
  ******************** 
  EIGENVALUE            3
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
  NUMBER OF ITERATIONS WAS            2
 -----------------------------------------------
 
  ******************** 
  EIGENVALUE            3
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
  NUMBER OF ITERATIONS WAS            3
 -----------------------------------------------
 
  ******************** 
  EIGENVALUE            3
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
  BRACKET 
  DO SECANT METHOD 
  NUMBER OF ITERATIONS WAS            4
 -----------------------------------------------
 
  ******************** 
  EIGENVALUE            3
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
  NEWTON'S METHOD 
  NUMBER OF ITERATIONS WAS            5
 -----------------------------------------------
 
  ******************** 
  EIGENVALUE            3
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
  NEWTON'S METHOD 
  NUMBER OF ITERATIONS WAS            6
 -----------------------------------------------
 
  ******************** 
  EIGENVALUE            3
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
  NEWTON'S METHOD 
  NUMBER OF ITERATIONS WAS            7
 -----------------------------------------------
 
  ******************** 
  EIGENVALUE            3
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
 
  ******************** 
  EIGENVALUE            3
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
  NUMBER OF ITERATIONS WAS            1
 -----------------------------------------------
 
  ******************** 
  EIGENVALUE            3
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
  BRACKET 
  DO SECANT METHOD 
  NUMBER OF ITERATIONS WAS            2
 -----------------------------------------------
 
  ******************** 
  EIGENVALUE            3
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
  NRAY =            2
 
  ******************** 
  EIGENVALUE            3
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
  NUMBER OF ITERATIONS WAS            1
 -----------------------------------------------
 
  ******************** 
  EIGENVALUE            3
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
  NUMBER OF ITERATIONS WAS            2
 -----------------------------------------------
 
  ******************** 
  EIGENVALUE            3
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
  BRACKET 
  DO SECANT METHOD 
  NUMBER OF ITERATIONS WAS            3
 -----------------------------------------------
 
  ******************** 
  EIGENVALUE            3
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
  NEWTON'S METHOD 
  NUMBER OF ITERATIONS WAS            4
 -----------------------------------------------
 
  ******************** 
  EIGENVALUE            3
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
  NUMBER OF ITERATIONS WAS            5
 -----------------------------------------
  tol =    1.000000E-02
  NUMEIG, EIG, TOL, IFLAG =            3   129.1077       1.000000E-02
           1
  EIGENFUNCTION 
   2.439024E-02  0.1843239    
   4.878049E-02  0.3875616    
   7.317073E-02  0.5925087    
   9.756097E-02  0.7730039    
  0.1219512      0.9096436    
  0.1463415      0.9882429    
  0.1707317       1.000000    
  0.1951219      0.9418139    
  0.2195122      0.8163515    
  0.2439024      0.6317391    
  0.2682927      0.4008913    
  0.2926829      0.1404982    
  0.3170732     -0.1302928    
  0.3414634     -0.3913654    
  0.3658537     -0.6231784    
  0.3902439     -0.8082872    
  0.4146341     -0.9326829    
  0.4390244     -0.9868982    
  0.4634146     -0.9667394    
  0.4878049     -0.8736328    
  0.5121951     -0.7145320    
  0.5365854     -0.5014054    
  0.5609756     -0.2503364    
  0.5853658       1.968684E-02
  0.6097561      0.2882183    
  0.6341463      0.5348929    
  0.6585366      0.7409933    
  0.6829268      0.8908585    
  0.7073171      0.9730875    
  0.7317073      0.9814129    
  0.7560976      0.9151979    
  0.7804878      0.7794352    
  0.8048781      0.5844262    
  0.8292683      0.3449862    
  0.8536586       7.931010E-02
  0.8780488     -0.1923980    
  0.9024390     -0.4494697    
  0.9268293     -0.6723418    
  0.9512195     -0.8440480    
  0.9756098     -0.9515125    
SHAR_EOF
fi # end of overwriting check
if test -f 'res2'
then
	echo shar: will not over-write existing file "'res2'"
else
cat << "SHAR_EOF" > 'res2'
  np =            1
  a =   -1.000000    
    P0ATA,QFATA =    1.000000       1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    1.000000       0.000000E+00
  b =    1.000000    
    P0ATB,QFATB =    1.000000       1.000000    
    CLASSB = lcno                                                          
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  0.2500000       0.000000E+00
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   2.250013       1.623200E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   6.250016       1.237014E-07
 
 ______________________________________________
  np =            1
  a =    0.000000E+00
    CLASSA = r                                                             
    A1,A2 =    1.000000       0.000000E+00
  b =    1.000000    
    P0ATB,QFATB =    1.000000       1.000000    
    CLASSB = lcno                                                          
    B1,B2 =    0.000000E+00   1.000000    
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  0.2499998       1.023036E-06
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   9.137038       1.466703E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   26.07486       1.291642E-08
 
 ______________________________________________
  np =            2
  param =   0.7500000    
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    1.000000       0.000000E+00
  b =    1.000000    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   12.18714       2.341864E-08
  IFLAG =            2
  NUMEIG,EIG,TOL =            1   44.25756       4.609519E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   96.07160       3.908298E-08
 
 ______________________________________________
  np =            2
  param =   0.7500000    
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    0.000000E+00   1.000000    
  b =    1.000000    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            2
  NUMEIG,EIG,TOL =            0   1.120459       1.548081E-04
  IFLAG =            2
  NUMEIG,EIG,TOL =            1   18.35443       3.618022E-06
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   55.36758       2.305809E-06
 
 ______________________________________________
  np =            2
  param =   0.7500000    
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    1.000000       0.000000E+00
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            3
  * THERE SEEM TO BE NO EIGENVALUES                *
  IFLAG =            3
  * THERE SEEM TO BE NO EIGENVALUES                *
  IFLAG =            3
  * THERE SEEM TO BE NO EIGENVALUES                *
 
  * THERE SEEMS TO BE CONTINUOUS SPECTRUM BEGINNING*
  * AT ABOUT 4.7D-11                               *
 ______________________________________________
  np =            3
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000       1.000000    
    CLASSA = wr                                                            
    A1,A2 =    1.000000       0.000000E+00
  B = +INF
    CLASSB = lcno                                                          
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   5.783186       4.544096E-09
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   30.47117       1.332183E-07
  IFLAG =            2
  NUMEIG,EIG,TOL =            2   74.88702       2.554860E-07
 
 ______________________________________________
  np =            4
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    1.000000       0.000000E+00
  b =    1.000000    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           4
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   7.373990       3.599457E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   36.33601       1.710510E-10
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   85.29259       1.761736E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            3   154.0986       1.039524E-11
  IFLAG =            2
  NUMEIG,EIG,TOL =            4   242.7056       2.162225E-07
 
 ______________________________________________
  np =            4
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    0.000000E+00   1.000000    
  b =    1.000000    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           4
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  0.9843206       9.071464E-07
  IFLAG =            2
  NUMEIG,EIG,TOL =            1   21.96709       3.192975E-07
  IFLAG =            2
  NUMEIG,EIG,TOL =            2   62.12804       4.399757E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            3   121.8128       3.678471E-09
  IFLAG =            2
  NUMEIG,EIG,TOL =            4   201.1194       8.829408E-07
 
 ______________________________________________
  np =            5
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = wr                                                            
    A1,A2 =    1.000000       0.000000E+00
  b =    1.000000    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           4
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   7.374030       3.380070E-09
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   36.33622       1.018603E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   85.29340       1.945537E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            3   154.0990       2.410782E-08
  IFLAG =            2
  NUMEIG,EIG,TOL =            4   242.7072       1.852725E-06
 
 ______________________________________________
  np =            5
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = wr                                                            
    A1,A2 =    1.000000       1.000000    
  b =    1.000000    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           4
 
  IFLAG =            2
  NUMEIG,EIG,TOL =            0  0.9845399       5.873397E-06
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   21.97102       1.285538E-06
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   62.12859       2.196669E-09
  IFLAG =            1
  NUMEIG,EIG,TOL =            3   121.8127       9.634303E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            4   201.1203       3.318861E-08
 
 ______________________________________________
  np =            6
  a =    1.000000    
    CLASSA = r                                                             
    A1,A2 =    1.000000       0.000000E+00
  B = +INF
    CLASSB = lco                                                           
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =           -2           2
 
  IFLAG =            2
  NUMEIG,EIG,TOL =           -2  -16.00223       3.667952E-05
  IFLAG =            2
  NUMEIG,EIG,TOL =           -1  -3.802630       1.203780E-05
  IFLAG =            2
  NUMEIG,EIG,TOL =            0   4.435460       4.053261E-05
  IFLAG =            2
  NUMEIG,EIG,TOL =            1   14.82855       2.022616E-04
  IFLAG =            2
  NUMEIG,EIG,TOL =            2   27.72313       2.671632E-05
 
 ______________________________________________
  np =            6
  a =    1.000000    
    CLASSA = r                                                             
    A1,A2 =    1.000000       0.000000E+00
  B = +INF
    CLASSB = lco                                                           
    B1,B2 =    0.000000E+00   1.000000    
  NUMEIG1,NUMEIG2 =           -2           2
 
  IFLAG =            2
  NUMEIG,EIG,TOL =           -2  -25.02998       5.644441E-05
  IFLAG =            2
  NUMEIG,EIG,TOL =           -1  -8.987889       1.573901E-06
  IFLAG =            2
  NUMEIG,EIG,TOL =            0  0.2879478       3.573658E-06
  IFLAG =            2
  NUMEIG,EIG,TOL =            1   9.277247       3.890050E-05
  IFLAG =            2
  NUMEIG,EIG,TOL =            2   20.99024       4.662531E-05
 
 ______________________________________________
  np =            7
  a =    0.000000E+00
    P0ATA,QFATA =    1.000000      -1.000000    
    CLASSA = lco                                                           
    A1,A2 =    1.000000       0.000000E+00
  b =    1.000000    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =           -2           3
 
  IFLAG =            2
  NUMEIG,EIG,TOL =           -2  -127.0453       5.035825E-03
  IFLAG =            1
  NUMEIG,EIG,TOL =           -1  -5.426962       2.222912E-04
  IFLAG =            2
  NUMEIG,EIG,TOL =            0   4.397848       1.330141E-04
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   18.12068       2.112323E-04
  IFLAG =            2
  NUMEIG,EIG,TOL =            2   37.92352       2.845487E-04
  IFLAG =            1
  NUMEIG,EIG,TOL =            3   63.47567       3.562436E-04
 
 ______________________________________________
  np =            7
  a =    0.000000E+00
    P0ATA,QFATA =    1.000000      -1.000000    
    CLASSA = lco                                                           
    A1,A2 =    0.000000E+00   1.000000    
  b =    1.000000    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =           -1           5
 
  IFLAG =            1
  NUMEIG,EIG,TOL =           -1  -26.36498       2.824862E-04
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   2.995483E-06   8.032315E-11
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   10.44947       4.732026E-05
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   27.29205       6.729619E-05
  IFLAG =            1
  NUMEIG,EIG,TOL =            3   50.02732       8.700747E-05
  IFLAG =            2
  NUMEIG,EIG,TOL =            4   78.38776       1.063048E-04
  IFLAG =            1
  NUMEIG,EIG,TOL =            5   112.2355       1.232737E-04
 
 ______________________________________________
  np =            7
  a =    0.000000E+00
    P0ATA,QFATA =    1.000000      -1.000000    
    CLASSA = lco                                                           
  b =    1.000000    
    CLASSB = r                                                             
     ALFA =    0.000000E+00
  K11,K12 =    1.000000       0.000000E+00
  K21,K22 =    0.000000E+00   1.000000    
  DET =    1.000000    
  NUMEIG1,NUMEIG2 =           -1           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =           -1  -1.752737       1.000000E-02
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   14.57461       9.999999E-04
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   21.87494       9.999999E-04
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   55.57359       9.999999E-04
 ______________________________________________
  np =            8
  param =    1.000000    
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    1.000000       0.000000E+00
  b =    1.000000    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           1
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   30.39583       4.436753E-09
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   102.4421       1.380663E-07
 
 ______________________________________________
  np =            8
  param =    1.000000    
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    0.000000E+00   1.000000    
  b =    1.000000    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            2
  NUMEIG,EIG,TOL =            0  -1.514330       2.616173E-04
  IFLAG =            2
  NUMEIG,EIG,TOL =            1   42.21439       4.138282E-06
  IFLAG =            2
  NUMEIG,EIG,TOL =            2   127.6121       3.991765E-07
 
 ______________________________________________
  np =            8
  param =  -0.5000000    
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    1.000000       0.000000E+00
  b =    1.000000    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           1
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   24.46245       2.042590E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   94.09742       2.174501E-07
 
 ______________________________________________
  np =            8
  param =  -0.5000000    
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    0.000000E+00   1.000000    
  b =    1.000000    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            2
  NUMEIG,EIG,TOL =            0   4.912322       1.612672E-05
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   45.86612       5.685998E-09
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   131.0852       1.559603E-07
 
 ______________________________________________
  np =            9
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000       1.000000    
    CLASSA = wr                                                            
    A1,A2 =    1.000000       0.000000E+00
  b =    1.000000    
    P0ATB,QFATB =    1.000000       1.000000    
    CLASSB = lcno                                                          
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           4
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   8.727497       4.274168E-08
  IFLAG =            2
  NUMEIG,EIG,TOL =            1   152.4240       2.811454E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   435.0685       1.277678E-07
  IFLAG =            2
  NUMEIG,EIG,TOL =            3   855.5906       9.064525E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            4   1413.968       2.679770E-07
 
 ______________________________________________
  np =            9
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000       1.000000    
    CLASSA = wr                                                            
    A1,A2 =    1.000000       0.000000E+00
  b =    1.000000    
    P0ATB,QFATB =    1.000000       1.000000    
    CLASSB = lcno                                                          
    B1,B2 =    0.000000E+00   1.000000    
  NUMEIG1,NUMEIG2 =            0           4
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  -11.44399       4.113226E-07
  IFLAG =            2
  NUMEIG,EIG,TOL =            1   110.5386       6.395486E-07
  IFLAG =            2
  NUMEIG,EIG,TOL =            2   376.0186       4.195959E-07
  IFLAG =            2
  NUMEIG,EIG,TOL =            3   780.8220       6.603556E-07
  IFLAG =            2
  NUMEIG,EIG,TOL =            4   1323.937       1.283997E-06
 
 ______________________________________________
  np =           10
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = wr                                                            
    A1,A2 =    1.000000       0.000000E+00
  b =    1.000000    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            2
  NUMEIG,EIG,TOL =            0   2.467553       6.290385E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   9.870468       2.631066E-07
  IFLAG =            2
  NUMEIG,EIG,TOL =            2   22.20890       1.097374E-06
 
 ______________________________________________
  np =           11
  a =    0.000000E+00
    CLASSA = r                                                             
  b =    3.141593    
    CLASSB = r                                                             
     ALFA =    0.000000E+00
  K11,K12 =    1.000000       0.000000E+00
  K21,K22 =    0.000000E+00   1.000000    
  DET =    1.000000    
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   9.743221       1.000000E-02
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   28.68514       9.999999E-05
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   46.47783       1.000000E-02
 ______________________________________________
  np =           12
  param =    5.000000    
  a =    0.000000E+00
    CLASSA = r                                                             
  b =    3.141593    
    CLASSB = r                                                             
     ALFA =    0.000000E+00
  K11,K12 =    1.000000       0.000000E+00
  K21,K22 =    0.000000E+00   1.000000    
  DET =    1.000000    
  NUMEIG1,NUMEIG2 =            0           6
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  -5.800044       1.000000E-02
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   2.099460       9.999999E-04
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   7.449109       1.000000E-02
  IFLAG =            1
  NUMEIG,EIG,TOL =            3   16.64823       9.999999E-05
  IFLAG =            1
  NUMEIG,EIG,TOL =            4   17.09656       9.999999E-05
  IFLAG =            2
  NUMEIG,EIG,TOL =            5   36.36221       9.999999E-05
  THIS EIGENVALUE APPEARS TO BE DOUBLE. 
  IFLAG =            2
  NUMEIG,EIG,TOL =            6   36.35990       9.999999E-05
  THIS EIGENVALUE APPEARS TO BE DOUBLE. 
 ______________________________________________
  np =           13
  param =   -1.000000       2.000000    
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lp                                                            
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  -6.250001E-02   8.235754E-09
  IFLAG =            1
  NUMEIG,EIG,TOL =            1  -2.777767E-02   1.189267E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            2  -1.562498E-02   8.267769E-09
 
  * THERE SEEMS TO BE CONTINUOUS SPECTRUM BEGINNING*
  * AT ABOUT-1.2D-05                               *
 ______________________________________________
  np =           14
  a =    0.000000E+00
    CLASSA = r                                                             
    A1,A2 =    5.000000       8.000000    
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            2
  NUMEIG,EIG,TOL =            0  -1.185215       4.098948E-07
  IFLAG =            2
  NUMEIG,EIG,TOL =            1  -1.476557E-08   3.537459E-08
  IFLAG =            3
  * THERE SEEMS TO BE NO EIGENVALUE OF INDEX       *
  * GREATER THAN    1                              *
 
  * THERE SEEMS TO BE CONTINUOUS SPECTRUM BEGINNING*
  * AT ABOUT 1.1D-10                               *
 ______________________________________________
  np =           15
  A =  -INF
    CLASSA = lp                                                            
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  0.9999987       6.143419E-09
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   3.000000       2.887227E-09
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   5.000000       8.709401E-08
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           16
  param =  -0.5000000      -1.200000    
  a =   -1.000000    
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lp                                                            
  b =    1.000000    
    P0ATB,QFATB =    1.000000      -1.000000    
    CLASSB = wr                                                            
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   1.700078       2.990736E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   5.400301       8.815109E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   11.10050       1.304223E-07
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           16
  param =   0.5000000      -1.200000    
  a =   -1.000000    
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lp                                                            
  b =    1.000000    
    P0ATB,QFATB =    1.000000       1.000000    
    CLASSB = lcno                                                          
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   1.799999       5.350955E-09
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   5.499990       9.523299E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   11.19997       1.594841E-07
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           16
  param =    1.200000      -1.200000    
  a =   -1.000000    
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lp                                                            
  b =    1.000000    
    P0ATB,QFATB =    1.000000       1.000000    
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   2.640180       4.102007E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   7.040290       1.490146E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   13.44038       3.013832E-07
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           16
  param =   0.5000000     -0.5000000    
  a =   -1.000000    
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = wr                                                            
    A1,A2 =    1.000000       0.000000E+00
  b =    1.000000    
    P0ATB,QFATB =    1.000000       1.000000    
    CLASSB = lcno                                                          
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  0.7499469       4.251627E-07
  IFLAG =            2
  NUMEIG,EIG,TOL =            1   3.750189       3.303244E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   8.750423       6.691114E-10
 
 ______________________________________________
  np =           16
  param =    1.200000     -0.5000000    
  a =   -1.000000    
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = wr                                                            
    A1,A2 =    1.000000       0.000000E+00
  b =    1.000000    
    P0ATB,QFATB =    1.000000       1.000000    
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   1.099910       3.079692E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   4.800034       5.082312E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   10.50054       1.541088E-07
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           16
  param =    1.200000      0.5000000    
  a =   -1.000000    
    P0ATA,QFATA =    1.000000       1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    1.000000       0.000000E+00
  b =    1.000000    
    P0ATB,QFATB =    1.000000       1.000000    
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  -5.398488E-08   1.390756E-16
  IFLAG =            2
  NUMEIG,EIG,TOL =            1   3.700003       4.963928E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   9.400015       3.097163E-08
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           17
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lp                                                            
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           3
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  -1923.530       3.816839E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            1  -1777.291       2.584533E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            2  -1636.832       5.421725E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            3  -1502.155       1.277586E-07
 
  * THERE SEEMS TO BE CONTINUOUS SPECTRUM BEGINNING*
  * AT ABOUT 3.0D-10                               *
 ______________________________________________
  np =           18
  param =   0.2000000      0.6000000    
  a =   -1.000000    
    P0ATA,QFATA =    1.000000      -1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    1.000000       0.000000E+00
  b =    1.000000    
    P0ATB,QFATB =    1.000000      -1.000000    
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   1.440026       7.188100E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   5.040562       4.372558E-09
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   10.64384       1.023286E-07
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           19
  param =   0.5000000      0.2000000    
  a =   -1.000000    
    P0ATA,QFATA =    1.000000      -1.000000    
    CLASSA = lco                                                           
    A1,A2 =    1.000000       0.000000E+00
  b =    1.000000    
    P0ATB,QFATB =    1.000000      -1.000000    
    CLASSB = lcno                                                          
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =           -1           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =           -1  -8.299860E-02   6.689788E-06
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   6.000289E-02   4.372909E-06
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   3.281972       1.093854E-06
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   8.923641       1.553110E-06
 
 ______________________________________________
  np =           19
  param =   0.5000000      0.6000000    
  a =   -1.000000    
    P0ATA,QFATA =    1.000000      -1.000000    
    CLASSA = lco                                                           
    A1,A2 =    1.000000       0.000000E+00
  b =    1.000000    
    P0ATB,QFATB =    1.000000      -1.000000    
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           3
 
  IFLAG =            2
  NUMEIG,EIG,TOL =            0  0.9507382       7.565559E-06
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   5.211924       5.710576E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   11.79338       2.395671E-06
  IFLAG =            1
  NUMEIG,EIG,TOL =            3   20.61071       3.544208E-06
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           20
  param =    1.000000    
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lco                                                           
    A1,A2 =    1.000000       0.000000E+00
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =           -2           1
 
  IFLAG =            2
  NUMEIG,EIG,TOL =           -2  -27122.01       5.806144E-05
  IFLAG =            1
  NUMEIG,EIG,TOL =           -1  -49.63317       7.235560E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  0.9054454       2.386840E-07
  IFLAG =            2
  NUMEIG,EIG,TOL =            1  0.9999983       8.324409E-05
 
  * THERE SEEMS TO BE CONTINUOUS SPECTRUM BEGINNING*
  * AT ABOUT 1.0D+00                               *
 ______________________________________________
  np =           21
  a =   -3.141593    
    CLASSA = r                                                             
    A1,A2 =    1.000000       0.000000E+00
  b =    3.141593    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           4
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  0.2500001       1.144485E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   1.000000       2.448057E-09
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   2.250000       6.745866E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            3   4.000002       2.448005E-09
  IFLAG =            1
  NUMEIG,EIG,TOL =            4   6.250000       0.000000E+00
 
 ______________________________________________
  np =           21
  a =   -3.141593    
    CLASSA = r                                                             
  b =    3.141593    
    CLASSB = r                                                             
     ALFA =    0.000000E+00
  K11,K12 =    1.000000       0.000000E+00
  K21,K22 =    0.000000E+00   1.000000    
  DET =    1.000000    
  NUMEIG1,NUMEIG2 =            0           4
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   2.320417E-08   1.000000E-02
  IFLAG =            2
  NUMEIG,EIG,TOL =            1  0.9999427       9.999999E-05
  THIS EIGENVALUE APPEARS TO BE DOUBLE. 
  IFLAG =            2
  NUMEIG,EIG,TOL =            2   1.000098       9.999999E-05
  THIS EIGENVALUE APPEARS TO BE DOUBLE. 
  IFLAG =            2
  NUMEIG,EIG,TOL =            3   3.999958       9.999999E-05
  THIS EIGENVALUE APPEARS TO BE DOUBLE. 
  IFLAG =            2
  NUMEIG,EIG,TOL =            4   4.000043       9.999999E-05
  THIS EIGENVALUE APPEARS TO BE DOUBLE. 
 ______________________________________________
  np =           21
  a =   -3.141593    
    CLASSA = r                                                             
  b =    3.141593    
    CLASSB = r                                                             
  BCC = G 
     ALFA =   0.7853982    
  K11,K12 =    2.000000       1.000000    
  K21,K22 =    1.000000       1.000000    
  DET =    1.000000    
  NUMEIG1,NUMEIG2 =            0           4
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  -6.854099       1.000000E-02
  IFLAG =            1
  NUMEIG,EIG,TOL =            1 -0.1724424       9.999999E-05
  IFLAG =            1
  NUMEIG,EIG,TOL =            2  0.6465324       1.000000E-02
  IFLAG =            1
  NUMEIG,EIG,TOL =            3   1.418499       1.000000E-02
  IFLAG =            1
  NUMEIG,EIG,TOL =            4   3.565771       1.000000E-02
 ______________________________________________
  np =           22
  param =    0.000000E+00
  a =    0.000000E+00
    P0ATA,QFATA =    1.000000       1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    1.000000       0.000000E+00
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   6.124560E-08   2.575899E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   1.000186       1.568899E-04
  IFLAG =            2
  NUMEIG,EIG,TOL =            2   2.010715       2.899273E-03
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           22
  param =    0.000000E+00
  a =    0.000000E+00
    P0ATA,QFATA =    1.000000       1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    0.000000E+00   1.000000    
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0 -0.7238111       2.172801E-07
  IFLAG =            2
  NUMEIG,EIG,TOL =            1  0.6250747       2.286314E-05
  IFLAG =            2
  NUMEIG,EIG,TOL =            2   1.679729       1.398322E-03
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           22
  param =   -1.200000    
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lp                                                            
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            2
  NUMEIG,EIG,TOL =            0   1.202147       1.578277E-04
  IFLAG =            2
  NUMEIG,EIG,TOL =            1   2.212156       5.083272E-03
  IFLAG =            2
  NUMEIG,EIG,TOL =            2   3.340242       2.599581E-02
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           22
  param =  -0.6000000    
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = wr                                                            
    A1,A2 =    1.000000       0.000000E+00
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  0.6000011       4.396054E-06
  IFLAG =            2
  NUMEIG,EIG,TOL =            1   1.600965       3.857370E-04
  IFLAG =            2
  NUMEIG,EIG,TOL =            2   2.625299       4.002534E-03
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           22
  param =   0.5000000    
  a =    0.000000E+00
    P0ATA,QFATA =    1.000000       1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    1.000000       0.000000E+00
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            2
  NUMEIG,EIG,TOL =            0   1.780892E-07   1.820222E-07
  IFLAG =            2
  NUMEIG,EIG,TOL =            1   1.000051       2.895401E-05
  IFLAG =            2
  NUMEIG,EIG,TOL =            2   2.003115       4.195821E-04
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           22
  param =   0.5000000    
  a =    0.000000E+00
    P0ATA,QFATA =    1.000000       1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    0.000000E+00   1.000000    
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0 -0.5000198       6.353588E-06
  IFLAG =            1
  NUMEIG,EIG,TOL =            1  0.4999023       3.181945E-06
  IFLAG =            2
  NUMEIG,EIG,TOL =            2   1.500451       1.243966E-04
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           22
  param =    1.200000    
  a =    0.000000E+00
    P0ATA,QFATA =    1.000000       1.000000    
    CLASSA = lp                                                            
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   8.458892E-08   4.269161E-08
  IFLAG =            2
  NUMEIG,EIG,TOL =            1   1.000191       5.363456E-05
  IFLAG =            2
  NUMEIG,EIG,TOL =            2   2.007833       8.566619E-04
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           23
  param =    0.000000E+00
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    1.000000       0.000000E+00
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            1           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   1.000001       2.562651E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   1.999999       2.370820E-07
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           23
  param =   -1.200000    
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lp                                                            
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   1.200000       7.809699E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   2.200002       1.111408E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   3.200001       2.459040E-08
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           23
  param =   0.6000000    
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    1.000000       0.000000E+00
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  -8.116149E-08   1.256651E-09
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   1.000001       6.992908E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   2.000000       1.059875E-07
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           23
  param =   0.5000000    
  a =    0.000000E+00
    CLASSA = r                                                             
    A1,A2 =    1.000000       0.000000E+00
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  -4.362323E-08   2.176583E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   1.000001       7.033938E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   2.000000       6.459348E-08
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           23
  param =    1.200000    
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lp                                                            
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            2
  NUMEIG,EIG,TOL =            0   6.865588E-08   9.872200E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   1.000001       7.061217E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   2.000002       2.261668E-07
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           24
  param =  -0.6000000      -1.200000    
  a =   -1.570796    
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lp                                                            
  b =    1.570796    
    P0ATB,QFATB =   -1.000000      -1.000000    
    CLASSB = lcno                                                          
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   1.800051       2.420128E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   5.600335       3.084510E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   11.40111       1.382167E-07
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           24
  param =   0.5000000      -1.200000    
  a =   -1.570796    
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lp                                                            
  b =    1.570796    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   1.800046       7.490858E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   5.500312       3.105890E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   11.20105       9.322546E-08
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           24
  param =    1.200000      -1.200000    
  a =   -1.570796    
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lp                                                            
  b =    1.570796    
    P0ATB,QFATB =   -1.000000      -1.000000    
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   2.640177       8.986748E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   7.040959       3.714351E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   13.44289       2.551236E-08
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           24
  param =   0.5000000     -0.6000000    
  a =   -1.570796    
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    1.000000       0.000000E+00
  b =    1.570796    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  0.9000002       8.022207E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   4.000000       1.603900E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   9.100000       9.002807E-08
 
 ______________________________________________
  np =           24
  param =    1.200000     -0.6000000    
  a =   -1.570796    
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    1.000000       0.000000E+00
  b =    1.570796    
    P0ATB,QFATB =   -1.000000      -1.000000    
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   1.320051       1.021804E-07
  IFLAG =            2
  NUMEIG,EIG,TOL =            1   5.120334       1.505382E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   10.92110       7.599379E-08
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           24
  param =    1.200000      0.5000000    
  a =   -1.570796    
    CLASSA = r                                                             
    A1,A2 =    1.000000       0.000000E+00
  b =    1.570796    
    P0ATB,QFATB =   -1.000000      -1.000000    
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   4.536696E-05   1.335018E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   3.700312       4.947385E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   9.401054       1.054291E-08
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           25
  a =  -0.5000000    
    CLASSA = r                                                             
  b =   0.5000000    
    CLASSB = r                                                             
     ALFA =   0.7853982    
  K11,K12 =    1.000000       0.000000E+00
  K21,K22 =    0.000000E+00   1.000000    
  DET =    1.000000    
  NUMEIG1,NUMEIG2 =            0           6
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  0.1223064       1.000000E-02
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   6.363539       9.999999E-04
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   14.14196       1.000000E-02
  IFLAG =            1
  NUMEIG,EIG,TOL =            3   35.20601       9.999999E-04
  IFLAG =            1
  NUMEIG,EIG,TOL =            4   43.99551       9.999999E-04
  IFLAG =            1
  NUMEIG,EIG,TOL =            5   77.54198       1.000000E-02
  IFLAG =            1
  NUMEIG,EIG,TOL =            6   100.8771       1.000000E-02
 ______________________________________________
  np =           26
  a =    0.000000E+00
    CLASSA = r                                                             
    A1,A2 =    1.000000       0.000000E+00
  b =    1.000000    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           4
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  -766.1892       1.912871E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            1  -591.2048       4.530190E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            2  -447.9438       7.309772E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            3  -321.3208       2.424300E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            4  -205.2622       9.640413E-08
 
 ______________________________________________
  np =           27
  param =    2.000000    
  A =  -INF
    CLASSA = lp                                                            
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  -2.250000       8.233463E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            1 -0.2500021       2.631279E-07
  IFLAG =            3
  * THERE SEEMS TO BE NO EIGENVALUE OF INDEX       *
  * GREATER THAN    1                              *
 
  * THERE SEEMS TO BE CONTINUOUS SPECTRUM BEGINNING*
  * AT ABOUT 0.0D+00                               *
 ______________________________________________
  np =           28
  param =    2.000000    
  a =    0.000000E+00
    CLASSA = r                                                             
    A1,A2 =    0.000000E+00   1.000000    
  b =    3.141593    
    CLASSB = r                                                             
    B1,B2 =    0.000000E+00   1.000000    
  NUMEIG1,NUMEIG2 =            0           6
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  0.8782346       2.335631E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   2.466767       8.934072E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   5.100900       7.405345E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            3   10.01761       7.667311E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            4   17.00836       7.130479E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            5   26.00521       3.833846E-09
  IFLAG =            2
  NUMEIG,EIG,TOL =            6   37.00360       1.648778E-07
 
 ______________________________________________
  np =           29
  param =    1.000000    
  a =    0.000000E+00
    P0ATA,QFATA =   -1.000000      -1.000000    
    CLASSA = lp                                                            
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  0.9999998       1.103903E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   2.000000       2.793699E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   3.000001       8.900191E-08
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           30
  a =    0.000000E+00
    CLASSA = r                                                             
    A1,A2 =    1.000000       0.000000E+00
  b =    10.00000    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           9
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  -3.282521       6.474966E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            1 -0.4411472       1.335193E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   1.695664       2.232077E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            3   2.576408       1.189793E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            4   4.218426       3.216729E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            5   4.704751       4.592186E-09
  IFLAG =            1
  NUMEIG,EIG,TOL =            6   6.049498       6.147590E-09
  IFLAG =            1
  NUMEIG,EIG,TOL =            7   7.899601       6.038611E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            8   9.502640       1.569990E-09
  IFLAG =            1
  NUMEIG,EIG,TOL =            9   11.08074       1.424450E-07
 
 ______________________________________________
  np =           30
  a =    0.000000E+00
    CLASSA = r                                                             
    A1,A2 =    1.000000       0.000000E+00
  b =    20.00000    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           9
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  -14.42148       4.630085E-08
  IFLAG =            2
  NUMEIG,EIG,TOL =            1  -8.788376       5.441314E-03
  IFLAG =            1
  NUMEIG,EIG,TOL =            2  -8.739977       5.730926E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            3  -4.288173       1.229574E-08
  IFLAG =            2
  NUMEIG,EIG,TOL =            4  -3.444830       1.178764E-02
  IFLAG =            2
  NUMEIG,EIG,TOL =            5  -3.282521       4.012191E-05
  IFLAG =            2
  NUMEIG,EIG,TOL =            6 -0.4411488       4.359679E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            7 -0.1536089       2.419883E-06
  IFLAG =            2
  NUMEIG,EIG,TOL =            8   1.579559       1.385482E-05
  IFLAG =            1
  NUMEIG,EIG,TOL =            9   1.695661       1.092819E-08
 
 ______________________________________________
  np =           30
  a =    0.000000E+00
    CLASSA = r                                                             
    A1,A2 =    1.000000       0.000000E+00
  b =    30.00000    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           9
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  -20.45948       1.222829E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            1  -20.20757       2.214055E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            2  -14.42148       2.849872E-09
  IFLAG =            2
  NUMEIG,EIG,TOL =            3  -13.59181       6.664543E-02
  IFLAG =            1
  NUMEIG,EIG,TOL =            4  -8.788380       9.831155E-07
  IFLAG =            2
  NUMEIG,EIG,TOL =            5  -8.739976       5.591357E-03
  IFLAG =            2
  NUMEIG,EIG,TOL =            6  -7.283415       1.000359    
  IFLAG =            2
  NUMEIG,EIG,TOL =            7  -7.255684      0.3026575    
  IFLAG =            2
  NUMEIG,EIG,TOL =            8  -4.288174       1.982404E-03
  IFLAG =            2
  NUMEIG,EIG,TOL =            9  -3.444951       6.913358E-07
 
 ______________________________________________
  np =           30
  a =    0.000000E+00
    CLASSA = r                                                             
    A1,A2 =    1.000000       0.000000E+00
  b =    40.00000    
    CLASSB = r                                                             
    B1,B2 =    1.000000       0.000000E+00
  NUMEIG1,NUMEIG2 =            0           9
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  -31.95168       1.585298E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            1  -26.05754       5.790772E-08
  IFLAG =            2
  NUMEIG,EIG,TOL =            2  -23.70292      0.1020873    
  IFLAG =            1
  NUMEIG,EIG,TOL =            3  -20.20758       8.424398E-08
  IFLAG =            2
  NUMEIG,EIG,TOL =            4  -18.58225      0.1109363    
  IFLAG =            2
  NUMEIG,EIG,TOL =            5  -15.72533      0.1145801    
  IFLAG =            1
  NUMEIG,EIG,TOL =            6  -14.42148       3.767827E-07
  IFLAG =            2
  NUMEIG,EIG,TOL =            7  -13.59178       7.223222E-02
  IFLAG =            2
  NUMEIG,EIG,TOL =            8  -11.38139      0.2459849    
  IFLAG =            1
  NUMEIG,EIG,TOL =            9  -8.788383       2.692789E-09
 
 ______________________________________________
  np =           31
  A =  -INF
    CLASSA = lp                                                            
  B = +INF
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           4
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0  -6.250001       3.659573E-08
  IFLAG =            1
  NUMEIG,EIG,TOL =            1  -2.249999       9.381459E-08
  IFLAG =            2
  NUMEIG,EIG,TOL =            2 -0.2500000       2.891981E-07
  IFLAG =            3
  * THERE SEEMS TO BE NO EIGENVALUE OF INDEX       *
  * GREATER THAN    2                              *
  IFLAG =            3
  * THERE SEEMS TO BE NO EIGENVALUE OF INDEX       *
  * GREATER THAN    2                              *
 
  * THERE SEEMS TO BE CONTINUOUS SPECTRUM BEGINNING*
  * AT ABOUT 0.0D+00                               *
 ______________________________________________
  np =           32
  param =    1.000000       4.000000       2.000000    
             1.500000       4.500000       1.000000    
  a =    0.000000E+00
    P0ATA,QFATA =    1.000000       1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    1.000000       0.000000E+00
  b =    1.000000    
    P0ATB,QFATB =    1.000000       1.000000    
    CLASSB = lp                                                            
  NUMEIG1,NUMEIG2 =            0           2
 
  IFLAG =            1
  NUMEIG,EIG,TOL =            0   1.801087       2.738954E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            1   10.27988       6.118663E-07
  IFLAG =            1
  NUMEIG,EIG,TOL =            2   21.70573       2.382815E-07
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
  np =           32
  param =    1.000000       4.000000       2.000000    
             1.500000       4.500000       1.000000    
  a =    0.000000E+00
    P0ATA,QFATA =    1.000000       1.000000    
    CLASSA = lcno                                                          
    A1,A2 =    1.000000       0.000000E+00
  b =    1.000000    
    P0ATB,QFATB =    1.000000       1.000000    
    CLASSB = lp                                                            
  NUMEIG =           18
 
  IFLAG =            2
  NUMEIG,EIG,TOL =           18   600.9661       3.787571E-06
 
  * THERE SEEMS TO BE NO CONTINUOUS SPECTRUM       *
 ______________________________________________
 end results.txt                                                   
SHAR_EOF
fi # end of overwriting check
if test -f 'res3'
then
	echo shar: will not over-write existing file "'res3'"
else
cat << "SHAR_EOF" > 'res3'
 
  ******************** 
  EIGENVALUE            2
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
  NUMBER OF ITERATIONS WAS            1
 -----------------------------------------------
 
  ******************** 
  EIGENVALUE            2
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
  BRACKET 
  DO SECANT METHOD 
  NUMBER OF ITERATIONS WAS            2
 -----------------------------------------------
 
  ******************** 
  EIGENVALUE            2
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
  BRACKET 
  DO SECANT METHOD 
  NUMBER OF ITERATIONS WAS            3
 -----------------------------------------------
 
  ******************** 
  EIGENVALUE            2
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
  NEWTON'S METHOD 
  NUMBER OF ITERATIONS WAS            4
 -----------------------------------------------
 
  ******************** 
  EIGENVALUE            2
  SET TMID AND BOUNDARY CONDITIONS 
  OBTAIN DTHETA 
  SET BRACKET 
  NUMBER OF ITERATIONS WAS            5
 -----------------------------------------
  NUMEIG, EIG, TOL, IFLAG =            2   96.07165       4.492589E-07
           1
  EIGENFUNCTION = 
   4.761905E-02  -1.416879    
   9.523810E-02  -1.359300    
  0.1428571      -1.228830    
  0.1904762      -1.032400    
  0.2380952     -0.7804999    
  0.2857143     -0.4866511    
  0.3333333     -0.1666390    
  0.3809524      0.1623511    
  0.4285714      0.4825928    
  0.4761905      0.7768494    
  0.5238096       1.029226    
  0.5714286       1.226115    
  0.6190476       1.356875    
  0.6666667       1.414441    
  0.7142857       1.395683    
  0.7619048       1.301583    
  0.8095238       1.137223    
  0.8571429      0.9114306    
  0.9047619      0.6364385    
  0.9523810      0.3270384    
SHAR_EOF
fi # end of overwriting check
if test -f 'xamples.f'
then
	echo shar: will not over-write existing file "'xamples.f'"
else
cat << "SHAR_EOF" > 'xamples.f'
C  SUBROUTINE EXAMP

C     **********
C     MARCH 1, 2001; P.B. BAILEY, W.N. EVERITT AND A. ZETTL
C     VERSION 1.2
C     **********

      SUBROUTINE EXAMP
C
C
C     THIS SUBROUTINE CONTAINS A SELECTION OF COEFFICIENT FUNCTIONS
C     p,q,w (AND POSSIBLY SUITABLE FUNCTIONS u,v WITH WHICH TO 
C     DEFINE BOUNDARY CONDITIONS AT LIMIT CIRCLE ENDPOINTS) 
C     WHICH DEFINE SOME INTERESTING STURM-LIOUVILLE BOUNDARY
C     VALUE PROBLEMS. IT CAN BE CALLED BY THE MAIN PROGRAM, DRIVE.
C
C     .. Scalars in Common ..
      REAL A,ALPHA,B,BETA,C,D,E,GAMMA,H,K,L,NU,S
      INTEGER NUMBER
C     ..
C     .. Local Scalars ..
      REAL TMP
      INTEGER MUMBER
      CHARACTER ANS
C     ..
C     .. Common blocks ..
      COMMON /FLAG/NUMBER
      COMMON /PAR/NU,H,K,L,ALPHA,BETA,GAMMA,S,A,B,C,D,E
C     ..
      WRITE (*,FMT=*)
     +  ' Here is a collection of 32 differential equations '
      WRITE (*,FMT=*) ' which can be used with SLEIGN2.  By typing an '
      WRITE (*,FMT=*)
     +  ' integer from 1 to 32, one of these differential '
      WRITE (*,FMT=*)
     +  ' equations is selected, whereupon its coefficient '
      WRITE (*,FMT=*) ' functions p,q,w will be displayed along with a '
      WRITE (*,FMT=*) ' brief description of its singular points.  The'
      WRITE (*,FMT=*) ' endpoints a, b of the interval over which the '
      WRITE (*,FMT=*)
     +  ' differential equation is integrated are specified '
      WRITE (*,FMT=*) ' later; any interval which does not contain '
      WRITE (*,FMT=*) ' singular points in its interior is acceptable. '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' DO YOU WISH TO CONTINUE ? (Y/N) '
      READ (*,FMT=*) ANS
      IF (.NOT. (ANS.EQ.'y'.OR.ANS.EQ.'Y')) STOP
   35 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) '  1 IS THE LEGENDRE EQUATION '
      WRITE (*,FMT=*) '  2 IS THE BESSEL EQUATION '
      WRITE (*,FMT=*) '  3 IS THE HALVORSEN EQUATION '
      WRITE (*,FMT=*) '  4 IS THE BOYD EQUATION '
      WRITE (*,FMT=*) '  5 IS THE REGULARIZED BOYD EQUATION '
      WRITE (*,FMT=*) '  6 IS THE SEARS-TITCHMARSH EQUATION '
      WRITE (*,FMT=*) '  7 IS THE BEZ EQUATION '
      WRITE (*,FMT=*) '  8 IS THE LAPLACE TIDAL WAVE EQUATION '
      WRITE (*,FMT=*) '  9 IS THE LATZKO EQUATION '
      WRITE (*,FMT=*) ' 10 IS A WEAKLY REGULAR EQUATION '
      WRITE (*,FMT=*) ' 11 IS THE PLUM EQUATION '
      WRITE (*,FMT=*) ' 12 IS THE MATHIEU PERIODIC EQUATION '
      WRITE (*,FMT=*) ' 13 IS THE HYDROGEN ATOM EQUATION '
      WRITE (*,FMT=*) ' 14 IS THE MARLETTA EQUATION '
      WRITE (*,FMT=*) ' 15 IS THE HARMONIC OSCILLATOR EQUATION '
      WRITE (*,FMT=*) ' 16 IS THE JACOBI EQUATION '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' Press any key to continue. '
      READ (*,FMT=9010) ANS
 9010 FORMAT (A1)
      WRITE (*,FMT=*) ' 17 IS THE ROTATION MORSE OSCILLATOR EQUATION '
      WRITE (*,FMT=*) ' 18 IS THE DUNSCH EQUATION '
      WRITE (*,FMT=*) ' 19 IS THE DONSCH EQUATION '
      WRITE (*,FMT=*) ' 20 IS THE KRALL EQUATION '
      WRITE (*,FMT=*) ' 21 IS THE FOURIER EQUATION '
      WRITE (*,FMT=*) ' 22 IS THE LAGUERRE EQUATION '
      WRITE (*,FMT=*) ' 23 IS THE LAGUERRE/LIOUVILLE FORM EQUATION '
      WRITE (*,FMT=*) ' 24 IS THE JACOBI/LIOUVILLE FORM EQUATION '
      WRITE (*,FMT=*) ' 25 IS THE MEISSNER EQUATION '
      WRITE (*,FMT=*) ' 26 IS THE LOHNER EQUATION '
      WRITE (*,FMT=*) ' 27 IS THE JOERGENS EQUATION '
      WRITE (*,FMT=*) ' 28 IS THE BEHNKE-GOERISCH EQUATION '
      WRITE (*,FMT=*) ' 29 IS THE WHITTAKER EQUATION '
      WRITE (*,FMT=*) ' 30 IS THE LITTLEWOOD-MCLEOD EQUATION '
      WRITE (*,FMT=*) ' 31 IS THE MORSE EQUATION '
      WRITE (*,FMT=*) ' 32 IS THE HEUN EQUATION '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' ENTER THE NUMBER OF YOUR CHOICE: '
      READ (*,FMT=*) NUMBER
      IF (NUMBER.LT.1 .OR. NUMBER.GT.32) GO TO 35
      MUMBER = NUMBER
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,
     +       23,24,25,26,27,28,29,30,31,32) NUMBER
C
    1 CONTINUE
      WRITE (*,FMT=*) '    -(p*y'')'' + q*y = lambda*w*y on (-1,1) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1 - x*x, q = 1/4, w = 1 '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' LIMIT CIRCLE, NON-OSCILLATORY AT -1. '
      WRITE (*,FMT=*) ' LIMIT CIRCLE, NON-OSCILLATORY AT +1. '
      GO TO 40
C
    2 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (0,+infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1, q = (nu*nu-0.25)/x*x, w = 1 '
      WRITE (*,FMT=*) '            (nu a parameter) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' THE ENDPOINT X = 0: '
      WRITE (*,FMT=*) '    LIMIT CIRCLE, NON-OSCILLATORY '
      WRITE (*,FMT=*) '       FOR -1.LT.nu.LT.1 BUT nu*nu.NE.0.25. '
      WRITE (*,FMT=*) '    REGULAR FOR nu*nu = 0.25. '
      WRITE (*,FMT=*) '    LIMIT POINT FOR nu*nu.GE.1.0. '
      WRITE (*,FMT=*) ' THE ENDPOINT X = +INFINITY: '
      WRITE (*,FMT=*) '    LIMIT POINT FOR ALL nu. '
      MUMBER = 101
      GO TO 40
C
    3 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (0,+infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1, q = 0, w = exp(-2/x)/x**4 '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' WEAKLY REGULAR AT 0. '
      WRITE (*,FMT=*) ' LIMIT CIRCLE, NON-OSCILLATORY AT +INFINITY. '
      GO TO 40
C
    4 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (-infinity,0) '
      WRITE (*,FMT=*)
     +  '                            AND on (0,+infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1, q = -1/x, w = 1 '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' LIMIT POINT AT -INFINITY. '
      WRITE (*,FMT=*) ' LIMIT CIRCLE, NON-OSCILLATORY AT 0+ AND 0-. '
      WRITE (*,FMT=*) ' LIMIT POINT AT +INFINITY. '
      GO TO 40
C
    5 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (-infinity,0) '
      WRITE (*,FMT=*)
     +  '                            AND on (0,+infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = r*r, q = -r*r*(ln|x|)**2, w = r*r '
      WRITE (*,FMT=*) '      where r = exp(-(x*ln(|x|)-x)) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' LIMIT POINT AT -INFINITY. '
      WRITE (*,FMT=*) ' WEAKLY REGULAR AT 0+ AND 0-. '
      WRITE (*,FMT=*) ' LIMIT POINT AT +INFINITY. '
      GO TO 40
C
    6 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (0,+infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = x, q = -x, w = 1/x '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' LIMIT POINT AT 0. '
      WRITE (*,FMT=*) ' LIMIT CIRCLE, OSCILLATORY AT +INFINITY. '
      GO TO 40
C
    7 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (-infinity,0) '
      WRITE (*,FMT=*)
     +  '                            AND on (0,+infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = x, q = -1/x, w = 1 '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' LIMIT POINT AT -INFINITY. '
      WRITE (*,FMT=*) ' LIMIT CIRCLE, OSCILLATORY AT 0+ AND 0-. '
      WRITE (*,FMT=*) ' LIMIT POINT AT +INFINITY. '
      GO TO 40
C
    8 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (0,+infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1/x, q = (k/x**2) + (k**2/x), w = 1 '
      WRITE (*,FMT=*) '         (k a non-zero parameter) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' LIMIT CIRCLE, NON-OSCILLATORY AT 0. '
      WRITE (*,FMT=*) ' LIMIT POINT AT +INFINITY. '
      MUMBER = 102
      GO TO 40
C
    9 CONTINUE
      WRITE (*,FMT=*) '    -(p*y'')'' + q*y = lambda*w*y on (0,1) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1 - x**7, q = 0, w = x**7 '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' WEAKLY REGULAR AT 0. '
      WRITE (*,FMT=*) ' LIMIT CIRCLE, NON-OSCILLATORY AT +1. '
      GO TO 40
C
   10 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (0,+infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = sqrt(x), q = 0, w = 1/sqrt(x) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' WEAKLY REGULAR AT 0. '
      WRITE (*,FMT=*) ' LIMIT POINT AT +INFINITY. '
      GO TO 40
C
   11 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (-infinity, '
      WRITE (*,FMT=*) '                                    +infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1, q = 100*cos(x)**2, w = 1 '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' LIMIT POINT AT -INFINITY. '
      WRITE (*,FMT=*) ' LIMIT POINT AT +INFINITY. '
      GO TO 40
C
   12 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (-infinity, '
      WRITE (*,FMT=*) '                                    +infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1, q = 2*k*cos(2x), w = 1 '
      WRITE (*,FMT=*) '       (k a non-zero parameter) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' LIMIT POINT AT -INFINITY. '
      WRITE (*,FMT=*) ' LIMIT POINT AT +INFINITY. '
      MUMBER = 103
      GO TO 40
C
   13 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (0,+infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1, q = k/x + h/x**2, w = 1 '
      WRITE (*,FMT=*) '        q = k/x + h/x**2 + 1  if h.lt.-0.25  '
      WRITE (*,FMT=*) '        (h,k parameters) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' THE ENDPOINT X = 0: '
      WRITE (*,FMT=*) '    REGULAR for h = k = 0. '
      WRITE (*,FMT=*) '    LIMIT-CIRCLE, NON-OSCILLATORY '
      WRITE (*,FMT=*) '       FOR h = 0 AND ALL k.NE.0. '
      WRITE (*,FMT=*) '    LIMIT-CIRCLE, NON-OSCILLATORY '
      WRITE (*,FMT=*)
     +  '       FOR -0.25.LE.h.LT.0.75 BUT h.NE.0, AND ALL k. '
      WRITE (*,FMT=*) '    LIMIT-CIRCLE, OSCILLATORY '
      WRITE (*,FMT=*) '       FOR h.LT.-0.25 AND ALL k. '
      WRITE (*,FMT=*) '      (Here, 1 has been added to the usual q so '
      WRITE (*,FMT=*) '       at least some eigenvalues are positive.) '
      WRITE (*,FMT=*) '    LIMIT POINT FOR h.GE.0.75 AND ALL k. '
      WRITE (*,FMT=*) ' THE ENDPOINT X = +INFINITY: '
      WRITE (*,FMT=*) '    LIMIT POINT FOR ALL h,k. '
      MUMBER = 104
      GO TO 40
C
   14 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (0,+infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
     +  ' p = 1, q = 3.0*(X-31.0)/(4.0*(X+1.0)*(4.0+X)**2), '
      WRITE (*,FMT=*) ' w = 1 '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' REGULAR AT 0. '
      WRITE (*,FMT=*) ' LIMIT POINT AT +INFINITY. '
      GO TO 40
C
   15 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (-infinity, '
      WRITE (*,FMT=*) '                                    +infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1, q = x*x, w = 1 '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' LIMIT POINT AT -INFINITY. '
      WRITE (*,FMT=*) ' LIMIT POINT AT +INFINITY. '
      GO TO 40
C
   16 CONTINUE
      WRITE (*,FMT=*) '    -(p*y'')'' + q*y = lambda*w*y on (-1,1) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = (1-x)**(alpha+1)*(1+x)**(beta+1), '
      WRITE (*,FMT=*) ' q = 0, w = (1-x)**alpha*(1+x)**beta '
      WRITE (*,FMT=*) '            (alpha, beta parameters) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' THE ENDPOINT X = -1.0: '
      WRITE (*,FMT=*) '    LIMIT POINT FOR beta.LE.-1. '
      WRITE (*,FMT=*) '    WEAKLY REGULAR FOR -1.LT.beta.LT.0. '
      WRITE (*,FMT=*) '    LIMIT CIRCLE, NON-OSCILLATORY '
      WRITE (*,FMT=*) '       FOR 0.LE.beta.LT.1. '
      WRITE (*,FMT=*) '    LIMIT POINT FOR beta.GE.1. '
      WRITE (*,FMT=*) ' THE ENDPOINT X = +1.0: '
      WRITE (*,FMT=*) '    LIMIT POINT FOR alpha.LE.-1. '
      WRITE (*,FMT=*) '    WEAKLY REGULAR FOR -1.LT.alpha.LT.0. '
      WRITE (*,FMT=*) '    LIMIT CIRCLE, NON-OSCILLATORY '
      WRITE (*,FMT=*) '       FOR 0.LE.alpha.LT.1. '
      WRITE (*,FMT=*) '    LIMIT POINT FOR alpha.GE.1. '
      MUMBER = 105
      GO TO 40
C
   17 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (0,+infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1, q = 2/x**2 - 2000(2e-e*e), w = 1 '
      WRITE (*,FMT=*) '       where e = exp(-1.7(x-1.3)) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' LIMIT POINT AT 0. '
      WRITE (*,FMT=*) ' LIMIT POINT AT +INFINITY. '
      GO TO 40
C
   18 CONTINUE
      WRITE (*,FMT=*) '    -(p*y'')'' + q*y = lambda*w*y on (-1,1) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
     +  ' p = 1 - x*x, q = 2*alpha**2/(1+x) + 2*beta**2/(1-x),'
      WRITE (*,FMT=*)
     +  ' w = 1        (alpha, beta non-negative parameters) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' THE ENDPOINT X = -1.0: '
      WRITE (*,FMT=*) '    LIMIT CIRCLE, NON-OSCILLATORY '
      WRITE (*,FMT=*) '       FOR 0.LE.alpha.LT.0.5. '
      WRITE (*,FMT=*) '    LIMIT POINT FOR alpha.GE.0.5. '
      WRITE (*,FMT=*) ' THE ENDPOINT X = +1.0: '
      WRITE (*,FMT=*) '    LIMIT CIRCLE, NON-OSCILLATORY '
      WRITE (*,FMT=*) '       FOR 0.LE.beta.LT.0.5. '
      WRITE (*,FMT=*) '    LIMIT POINT FOR beta.GE.0.5. '
      MUMBER = 106
      GO TO 40
C
   19 CONTINUE
      WRITE (*,FMT=*) '    -(p*y'')'' + q*y = lambda*w*y on (-1,1) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
     +  ' p = 1 - x*x, q = -2*gamma**2/(1+x) +2*beta**2/(1-x),'
      WRITE (*,FMT=*) ' w = 1            (gamma, beta parameters) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' THE ENDPOINT X = -1.0: '
      WRITE (*,FMT=*)
     +  '    LIMIT CIRCLE, NON-OSCILLATORY FOR gamma = 0. '
      WRITE (*,FMT=*) '    LIMIT CIRCLE, OSCILLATORY FOR gamma.GT.0. '
      WRITE (*,FMT=*) ' THE ENDPOINT X = +1.0: '
      WRITE (*,FMT=*) '    LIMIT CIRCLE, NON-OSCILLATORY '
      WRITE (*,FMT=*) '       FOR 0.LE.beta.LT.0.5. '
      WRITE (*,FMT=*) '    LIMIT POINT FOR beta.GE.0.5. '
      MUMBER = 107
      GO TO 40
C
   20 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (0,+infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1, q = 1 - (k**2+0.25)/x**2, w = 1 '
      WRITE (*,FMT=*) '        (k a positive parameter) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' LIMIT CIRCLE, OSCILLATORY AT 0. '
      WRITE (*,FMT=*) ' LIMIT POINT AT +INFINITY. '
      MUMBER = 108
      GO TO 40
C
   21 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (-infinity, '
      WRITE (*,FMT=*) '                                    +infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1, q = 0, w = 1 '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' LIMIT POINT AT -INFINITY. '
      WRITE (*,FMT=*) ' LIMIT POINT AT +INFINITY. '
      GO TO 40
C
   22 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (0,+infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = x**(alpha+1)*exp(-x), w = x**alpha*exp(-x) '
      WRITE (*,FMT=*) ' q = 0        (alpha a parameter) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' THE ENDPOINT X = 0.: '
      WRITE (*,FMT=*) '    LIMIT POINT FOR alpha.LE.-1. '
      WRITE (*,FMT=*) '    WEAKLY REGULAR FOR -1.LT.alpha.LT.0. '
      WRITE (*,FMT=*) '    LIMIT CIRCLE, NON-OSCILLATORY '
      WRITE (*,FMT=*) '       FOR 0.LE.alpha.LT.1. '
      WRITE (*,FMT=*) '    LIMIT POINT FOR alpha.GE.1. '
      WRITE (*,FMT=*) ' THE ENDPOINT X = +INFINITY: '
      WRITE (*,FMT=*) '    LIMIT POINT FOR ALL alpha. '
      MUMBER = 109
      GO TO 40
C
   23 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (0,+infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1, w = 1 '
      WRITE (*,FMT=*)
     +  ' q = (alpha**2-0.25)/x**2 - (alpha+1)/2 + x**2/16'
      WRITE (*,FMT=*) '              (alpha a parameter) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' THE ENDPOINT X = 0.: '
      WRITE (*,FMT=*) '    LIMIT POINT FOR alpha.LE.-1. '
      WRITE (*,FMT=*) '    LIMIT CIRCLE, NON-OSCILLATORY '
      WRITE (*,FMT=*)
     +  '       FOR -1.LT.alpha.LT.1 BUT alpha**2.NE.0.25. '
      WRITE (*,FMT=*) '    REGULAR FOR alpha**2 = 0.25. '
      WRITE (*,FMT=*) '    LIMIT POINT FOR alpha.GE.1. '
      WRITE (*,FMT=*) ' THE ENDPOINT X = +INFINITY: '
      WRITE (*,FMT=*) '    LIMIT POINT FOR ALL alpha. '
      MUMBER = 110
      GO TO 40
C
   24 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (-pi/2,pi/2) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1, w = 1 '
      WRITE (*,FMT=*) ' q = (beta**2-0.25)/(4*tan((x+pi/2)/2)**2)+ '
      WRITE (*,FMT=*) '     (alpha**2-0.25)/(4*tan((x-pi/2)/2)**2)- '
      WRITE (*,FMT=*) '     (4*alpha*beta+4*alpha+4*beta+3)/8 '
      WRITE (*,FMT=*) '            (alpha, beta parameters) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' THE ENDPOINT X = -pi/2: '
      WRITE (*,FMT=*) '    LIMIT POINT FOR beta.LE.-1. '
      WRITE (*,FMT=*) '    LIMIT CIRCLE, NON-OSCILLATORY '
      WRITE (*,FMT=*) '       FOR -1.LT.beta.LT.1 BUT beta**2.NE.0.25. '
      WRITE (*,FMT=*) '    REGULAR FOR beta**2 = 0.25. '
      WRITE (*,FMT=*) '    LIMIT POINT FOR beta.GE.1. '
      WRITE (*,FMT=*) ' THE ENDPOINT X = +pi/2: '
      WRITE (*,FMT=*) '    LIMIT POINT FOR alpha.LE.-1. '
      WRITE (*,FMT=*) '    LIMIT CIRCLE, NON-OSCILLATORY '
      WRITE (*,FMT=*)
     +  '       FOR -1.LT.alpha.LT.1 BUT alpha**2.NE.0.25. '
      WRITE (*,FMT=*) '    REGULAR FOR alpha**2 = 0.25. '
      WRITE (*,FMT=*) '    LIMIT POINT FOR alpha.GE.1. '
      MUMBER = 111
      GO TO 40
C
   25 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (-infinity, '
      WRITE (*,FMT=*) '                                    +infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1, q = 0 '
      WRITE (*,FMT=*) ' w = 1 when x.le.0. '
      WRITE (*,FMT=*) '   = 9 when x.gt.0. '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' LIMIT POINT AT -INFINITY. '
      WRITE (*,FMT=*) ' LIMIT POINT AT +INFINITY. '
      GO TO 40
C
   26 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (-infinity, '
      WRITE (*,FMT=*) '                                    +infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1, q = -1000*x, w = 1 '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' LIMIT POINT AT -INFINITY. '
      WRITE (*,FMT=*) ' LIMIT POINT AT +INFINITY. '
      GO TO 40
C
   27 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (-infinity, '
      WRITE (*,FMT=*) '                                    +infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1, q = 0.25*exp(2*x) - k*exp(x), w = 1 '
      WRITE (*,FMT=*) '                (k a parameter) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' LIMIT POINT AT -INFINITY. '
      WRITE (*,FMT=*) ' LIMIT POINT AT +INFINITY. '
      MUMBER = 112
      GO TO 40
C
   28 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (-infinity, '
      WRITE (*,FMT=*) '                                    +infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1, q = k*cos(x)**2, w = 1 '
      WRITE (*,FMT=*) '          (k a parameter) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' LIMIT POINT AT -INFINITY. '
      WRITE (*,FMT=*) ' LIMIT POINT AT +INFINITY. '
      MUMBER = 113
      GO TO 40
C
   29 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (0,+infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1, q = 0.25 + (k**2-1)/(4*x**2)), w = 1/x '
      WRITE (*,FMT=*) '       (k a positive parameter .GE. 1) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' LIMIT POINT AT 0. '
      WRITE (*,FMT=*) ' LIMIT POINT AT +INFINITY. '
      MUMBER = 114
      GO TO 40
C
   30 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (0,+infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1, q = x*sin(x), w = 1 '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' REGULAR AT 0. '
      WRITE (*,FMT=*) ' LIMIT POINT AT +INFINITY. '
      GO TO 40
C
   31 CONTINUE
      WRITE (*,FMT=*)
     +  '    -(p*y'')'' + q*y = lambda*w*y on (-infinity,+infinity) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = 1, q = 9*exp(-2*x) - 18*exp(-x), w = 1 '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' LIMIT POINT AT -INFINITY. '
      WRITE (*,FMT=*) ' LIMIT POINT AT +INFINITY. '
      GO TO 40
C
   32 CONTINUE
      WRITE (*,FMT=*) '    -(p*y'')'' + q*y = lambda*w*y on (0,1) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' p = x**c*(1-x)**d*(x+s)**e '
      WRITE (*,FMT=*) ' q = a*b*x**c*(1-x)**(d-1)*(x+s)**(e-1) '
      WRITE (*,FMT=*) ' w = x**(c-1)*(1-x)**(d-1)*(x+s)**(e-1) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' where the parameters s,a,b,c,d,e satisfy '
      WRITE (*,FMT=*) ' the conditions '
      WRITE (*,FMT=*) '   s > 0, a.ge.b.ge.1, c.ge.1, d.ge.1, '
      WRITE (*,FMT=*) '   a+b+1-c-d-e = 0. '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' THE ENDPOINT 0: '
      WRITE (*,FMT=*) '   LIMIT CIRCLE, NON-OSCILLATORY '
      WRITE (*,FMT=*) '     FOR 1.LE.c.LT.2; '
      WRITE (*,FMT=*) '   LIMIT POINT FOR 2.LE.c. '
      WRITE (*,FMT=*) ' THE ENDPOINT 1: '
      WRITE (*,FMT=*) '   LIMIT CIRCLE, NON-OSCILLATORY '
      WRITE (*,FMT=*) '     FOR 1.LE.d.LT.2; '
      WRITE (*,FMT=*) '   LIMIT POINT FOR 2.LE.d. '
      MUMBER = 115
      GO TO 40
C
   40 CONTINUE
      WRITE (*,FMT=*)
      WRITE (*,FMT=*)
     +  ' IS THIS THE CORRECT DIFFERENTIAL EQUATION ? (Y/N) '
      READ (*,FMT=*) ANS
      IF (.NOT. (ANS.EQ.'y'.OR.ANS.EQ.'Y')) GO TO 35
      IF (MUMBER.EQ.NUMBER) RETURN
C
C     Now enter any parameters needed for these D.E.'s, or defaults.
C
      MUMBER = MUMBER - 100
      GO TO (41,42,43,44,45,46,47,48,49,50,51,52,53,54,55) MUMBER
C
   41 CONTINUE
      NU = 1.0D0
      WRITE (*,FMT=*) ' Choose real parameter nu, nu = '
      READ (*,FMT=*) NU
      RETURN
C
   42 CONTINUE
      K = 1.0D0
      WRITE (*,FMT=*) ' Choose real parameter k.ne.0., k = '
      READ (*,FMT=*) K
      RETURN
C
   43 CONTINUE
      K = 1.0D0
      WRITE (*,FMT=*) ' Choose real parameter k = '
      READ (*,FMT=*) K
      RETURN
C
   44 CONTINUE
      H = 1.0D0
      K = 1.0D0
      WRITE (*,FMT=*) ' Choose real parameters k,h = '
      READ (*,FMT=*) K,H
      RETURN
C
   45 CONTINUE
      BETA = 0.1D0
      ALPHA = 0.1D0
      WRITE (*,FMT=*) ' Choose real parameters alpha,beta = '
      READ (*,FMT=*) ALPHA,BETA
      RETURN
C
   46 CONTINUE
      ALPHA = 0.1D0
      BETA = 0.1D0
      WRITE (*,FMT=*) ' Choose real parameters alpha,beta = '
      READ (*,FMT=*) ALPHA,BETA
      RETURN
C
   47 CONTINUE
      GAMMA = 0.1D0
      BETA = 0.1D0
      WRITE (*,FMT=*) ' Choose real parameters gamma,beta = '
      READ (*,FMT=*) GAMMA,BETA
      RETURN
C
   48 CONTINUE
      K = 1.0D0
      WRITE (*,FMT=*) ' Choose real parameter k.gt.0., k = '
      READ (*,FMT=*) K
      RETURN
C
   49 CONTINUE
      ALPHA = 0.1D0
      WRITE (*,FMT=*) ' Choose real parameter alpha = '
      READ (*,FMT=*) ALPHA
      RETURN
C
   50 CONTINUE
      ALPHA = 0.1D0
      WRITE (*,FMT=*) ' Choose real parameter alpha = '
      READ (*,FMT=*) ALPHA
      RETURN
C
   51 CONTINUE
      BETA = 0.1D0
      ALPHA = 0.1D0
      WRITE (*,FMT=*) ' Choose real parameters alpha,beta = '
      READ (*,FMT=*) ALPHA,BETA
      RETURN
C
   52 CONTINUE
      K = 0.1D0
      WRITE (*,FMT=*) ' Choose real parameter k = '
      READ (*,FMT=*) K
      RETURN
C
   53 CONTINUE
      K = 0.1D0
      WRITE (*,FMT=*) ' Choose real parameter k = '
      READ (*,FMT=*) K
      RETURN
C
   54 CONTINUE
      K = 0.1D0
      WRITE (*,FMT=*) ' Choose real parameter k = '
      READ (*,FMT=*) K
      RETURN
C
   55 CONTINUE
      WRITE (*,FMT=*) ' Choose real parameter s > 0 '
      READ (*,FMT=*) S
      WRITE (*,FMT=*) ' Before entering the numerical values of the '
      WRITE (*,FMT=*) '   parameters a,b,c,d,e the user is advised to '
      WRITE (*,FMT=*) '   make a preliminary choice of these numbers '
      WRITE (*,FMT=*) '   that are consistent with the conditions '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) '    a.ge.b, b.ge.1 '
      WRITE (*,FMT=*) '    1.le.d .le. (a+b-1) '
      WRITE (*,FMT=*) '    1.le.c .le. (a+b-d) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' (Parameter e will be set equal to a+b+1-c-d.) '
      WRITE (*,FMT=*)
      WRITE (*,FMT=*) ' Choose parameters a, b  = '
      READ (*,FMT=*) A,B
      TMP = A + B - 1.0D0
      WRITE (*,FMT=*) ' Choose parameter d, 1.LE.d .LE. ',TMP
      READ (*,FMT=*) D
      TMP = A + B - D
      WRITE (*,FMT=*) ' Choose parameter c, 1.LE.c .LE. ',TMP
      READ (*,FMT=*) C
      E = A + B + 1.0D0 - C - D
      RETURN
      END
C
      REAL FUNCTION P(X)

C     .. Scalar Arguments ..
      REAL X
C     ..
C     .. Scalars in Common ..
      REAL A,ALPHA,B,BETA,C,D,E,GAMMA,H,K,L,NU,S
      INTEGER NUMBER
C     ..
C     .. Intrinsic Functions ..
      INTRINSIC ABS,EXP,LOG,SQRT
C     ..
C     .. Common blocks ..
      COMMON /FLAG/NUMBER
      COMMON /PAR/NU,H,K,L,ALPHA,BETA,GAMMA,S,A,B,C,D,E
C     ..
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,
     +       23,24,25,26,27,28,29,30,31,32) NUMBER
C
    1 CONTINUE
      P = 1.0D0 - X*X
      RETURN
C
    2 CONTINUE
      P = 1.0D0
      RETURN
C
    3 CONTINUE
      P = 1.0D0
      RETURN
C
    4 CONTINUE
      P = 1.0D0
      RETURN
C
    5 CONTINUE
      P = EXP(-2.0D0* (X*LOG(ABS(X))-X))
      RETURN
C
    6 CONTINUE
      P = X
      RETURN
C
    7 CONTINUE
      P = X
      RETURN
C
    8 CONTINUE
      P = 1.0D0/X
      RETURN
C
    9 CONTINUE
      P = 1.0D0 - X**7
      RETURN
C
   10 CONTINUE
      P = SQRT(X)
      RETURN
C
   11 CONTINUE
      P = 1.0D0
      RETURN
C
   12 CONTINUE
      P = 1.0D0
      RETURN
C
   13 CONTINUE
      P = 1.0D0
      RETURN
C
   14 CONTINUE
      P = 1.0D0
      RETURN
C
   15 CONTINUE
      P = 1.0D0
      RETURN
C
   16 CONTINUE
      IF (ABS(X).EQ.1.0D0) THEN
          P = 0.0D0
      ELSE IF (ALPHA.NE.-1.0D0 .AND. BETA.NE.-1.0D0) THEN
          P = (1.0D0-X)** (ALPHA+1.0D0)* (1.0D0+X)** (BETA+1.0D0)
      ELSE IF (ALPHA.NE.-1.0D0 .AND. BETA.EQ.-1.0D0) THEN
          P = (1.0D0-X)** (ALPHA+1.0D0)
      ELSE IF (ALPHA.EQ.-1.0D0 .AND. BETA.NE.-1.0D0) THEN
          P = (1.0D0+X)** (BETA+1.0D0)
      ELSE
          P = 1.0D0
      END IF
      RETURN
C
   17 CONTINUE
      P = 1.0D0
      RETURN
C
   18 CONTINUE
      P = 1.0D0 - X*X
      RETURN
C
   19 CONTINUE
      P = 1.0D0 - X*X
      RETURN
C
   20 CONTINUE
      P = 1.0D0
      RETURN
C
   21 CONTINUE
      P = 1.0D0
      RETURN
C
   22 CONTINUE
      P = EXP(-X)
      IF (ALPHA.NE.-1.0D0) P = P*X** (ALPHA+1.0D0)
      RETURN
C
   23 CONTINUE
      P = 1.0D0
      RETURN
C
   24 CONTINUE
      P = 1.0D0
      RETURN
C
   25 CONTINUE
      P = 1.0D0
      RETURN
C
   26 CONTINUE
      P = 1.0D0
      RETURN
C
   27 CONTINUE
      P = 1.0D0
      RETURN
C
   28 CONTINUE
      P = 1.0D0
      RETURN
C
   29 CONTINUE
      P = 1.0D0
      RETURN
C
   30 CONTINUE
      P = 1.0D0
      RETURN
C
   31 CONTINUE
      P = 1.0D0
      RETURN
C
   32 CONTINUE
      P = X**C* (1.0D0-X)**D* (X+S)**E
      RETURN
      END
C
      REAL FUNCTION Q(X)
c
C     .. Scalar Arguments ..
      REAL X
C     ..
C     .. Scalars in Common ..
      REAL A,ALPHA,B,BETA,C,D,E,GAMMA,H,K,L,NU,S
      INTEGER NUMBER
C     ..
C     .. Local Scalars ..
      REAL EE,HPI
C     ..
C     .. Intrinsic Functions ..
      INTRINSIC ABS,ATAN,COS,EXP,LOG,SIN,TAN
C     ..
C     .. Common blocks ..
      COMMON /FLAG/NUMBER
      COMMON /PAR/NU,H,K,L,ALPHA,BETA,GAMMA,S,A,B,C,D,E
C     ..
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,
     +       23,24,25,26,27,28,29,30,31,32) NUMBER
C
    1 CONTINUE
      Q = 0.25D0
      RETURN
C
    2 CONTINUE
      Q = 0.0D0
      IF (NU.NE.-0.5D0 .AND. NU.NE.0.5D0) Q = (NU*NU-0.25D0)/X**2
      RETURN
C
    3 CONTINUE
      Q = 0.0D0
      RETURN
C
    4 CONTINUE
      Q = -1.0D0/X
      RETURN
C
    5 CONTINUE
      Q = -EXP(-2.0D0* (X*LOG(ABS(X))-X))*LOG(ABS(X))**2
      RETURN
C
    6 CONTINUE
      Q = -X
      RETURN
C
    7 CONTINUE
      Q = -1.0D0/X
      RETURN
C
    8 CONTINUE
      Q = K/X**2 + K**2/X
      RETURN
C
    9 CONTINUE
      Q = 0.0D0
      RETURN
C
   10 CONTINUE
      Q = 0.0D0
      RETURN
C
   11 CONTINUE
      Q = 100.0D0*COS(X)**2
      RETURN
C
   12 CONTINUE
      Q = 2.0D0*K*COS(2.0D0*X)
      RETURN
C
   13 CONTINUE
      Q = K/X + H/ (X*X)
      IF (H.LT.-0.25D0) Q = Q + 1.0D0
      RETURN
C
   14 CONTINUE
      Q = 3.0D0* (X-31.0D0)/ (4.0D0* (X+1.0D0)* (X+4.0D0)**2)
      RETURN
C
   15 CONTINUE
      Q = X*X
      RETURN
C
   16 CONTINUE
      Q = 0.0D0
      RETURN
C
   17 CONTINUE
      L = 1.0D0
      EE = EXP(-1.7D0* (X-1.3D0))
      Q = L* (L+1.0D0)/X**2 - 2000.0D0*EE* (2.0D0-EE)
      RETURN
C
   18 CONTINUE
      IF (ALPHA.NE.0.0D0 .AND. BETA.NE.0.0D0) THEN
          Q = 2.0D0*ALPHA**2/ (1.0D0+X) + 2.0D0*BETA**2/ (1.0D0-X)
      ELSE IF (ALPHA.EQ.0.0D0 .AND. BETA.NE.0.0D0) THEN
          Q = 2.0D0*BETA**2/ (1.0D0-X)
      ELSE IF (ALPHA.NE.0.0D0 .AND. BETA.EQ.0.0D0) THEN
          Q = 2.0D0*ALPHA**2/ (1.0D0+X)
      ELSE
          Q = 0.0D0
      END IF
      RETURN
C
   19 CONTINUE
      Q = -2.0D0*GAMMA**2/ (1.0D0+X) + 2.0D0*BETA**2/ (1.0D0-X)
      RETURN
C
   20 CONTINUE
      Q = 1.0D0 - (K**2+0.25D0)/X**2
      RETURN
C
   21 CONTINUE
      Q = 0.0D0
      RETURN
C
   22 CONTINUE
      Q = 0.0D0
      RETURN
C
   23 CONTINUE
      Q = - (ALPHA+1.0D0)/2.0D0 + X**2/16.0D0
      IF (ALPHA.NE.0.5D0 .AND. ALPHA.NE.-0.5D0) Q = Q +
     +    (ALPHA**2-0.25D0)/X**2
      RETURN
C
   24 CONTINUE
      HPI = 2.0D0*ATAN(1.0D0)
      IF (BETA*BETA.NE.0.25D0 .AND. ALPHA*ALPHA.NE.0.25D0) THEN
          Q = (BETA**2-0.25D0)/ (4.0D0*TAN((X+HPI)/2.0D0)**2) +
     +        (ALPHA**2-0.25D0)/ (4.0D0*TAN((X-HPI)/2.0D0)**2) -
     +        (ALPHA*BETA+ALPHA+BETA+0.75D0)/2.0D0
      ELSE IF (BETA*BETA.EQ.0.25D0 .AND. ALPHA*ALPHA.NE.0.25D0) THEN
          Q = (ALPHA**2-0.25D0)/ (4.0D0*TAN((X-HPI)/2.0D0)**2) -
     +        (ALPHA*BETA+ALPHA+BETA+0.75D0)/2.0D0
      ELSE IF (BETA*BETA.NE.0.25D0 .AND. ALPHA*ALPHA.EQ.0.25D0) THEN
          Q = (BETA**2-0.25D0)/ (4.0D0*TAN((X+HPI)/2.0D0)**2) -
     +        (ALPHA*BETA+ALPHA+BETA+0.75D0)/2.0D0
      ELSE
          Q = - (ALPHA*BETA+ALPHA+BETA+0.75D0)/2.0D0
      END IF
      RETURN
C
   25 CONTINUE
      Q = 0.0D0
      RETURN
C
   26 CONTINUE
      Q = -1000.0D0*X
      RETURN
C
   27 CONTINUE
      EE = EXP(X)
      Q = EE* (0.25D0*EE-K)
      RETURN
C
   28 CONTINUE
      Q = K*COS(X)**2
      RETURN
C
   29 CONTINUE
      Q = 0.25D0 + (K**2-1.0D0)/ (4.0D0*X**2)
      RETURN
C
   30 CONTINUE
      Q = X*SIN(X)
      RETURN
C
   31 CONTINUE
      EE = EXP(-X)
      Q = 9.0D0*EE*EE - 18.0D0*EE
      RETURN
C
   32 CONTINUE
      Q = A*B*X**C* (1.0D0-X)** (D-1.0D0)* (X+S)** (E-1.0D0)
      RETURN
      END
C
      REAL FUNCTION W(X)

C     .. Scalar Arguments ..
      REAL X
C     ..
C     .. Scalars in Common ..
      REAL A,ALPHA,B,BETA,C,D,E,GAMMA,H,K,L,NU,S
      INTEGER NUMBER
C     ..
C     .. Intrinsic Functions ..
      INTRINSIC ABS,EXP,LOG,SQRT
C     ..
C     .. Common blocks ..
      COMMON /FLAG/NUMBER
      COMMON /PAR/NU,H,K,L,ALPHA,BETA,GAMMA,S,A,B,C,D,E
C     ..
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,
     +       23,24,25,26,27,28,29,30,31,32) NUMBER
C
    1 CONTINUE
      W = 1.0D0
      RETURN
C
    2 CONTINUE
      W = 1.0D0
      RETURN
C
    3 CONTINUE
      W = 0.0D0
      IF (X.NE.0.0D0) W = EXP(-2.0D0/X)/X**4
      RETURN
C
    4 CONTINUE
      W = 1.0D0
      RETURN
C
    5 CONTINUE
      W = EXP(-2.0D0* (X*LOG(ABS(X))-X))
      RETURN
C
    6 CONTINUE
      W = 1.0D0/X
      RETURN
C
    7 CONTINUE
      W = 1.0D0
      RETURN
C
    8 CONTINUE
      W = 1.0D0
      RETURN
C
    9 CONTINUE
      W = X**7
      RETURN
C
   10 CONTINUE
      W = 1.0D0/SQRT(X)
      RETURN
C
   11 CONTINUE
      W = 1.0D0
      RETURN
C
   12 CONTINUE
      W = 1.0D0
      RETURN
C
   13 CONTINUE
      W = 1.0D0
      RETURN
C
   14 CONTINUE
      W = 1.0D0
      RETURN
C
   15 CONTINUE
      W = 1.0D0
      RETURN
C
   16 CONTINUE
      IF (ALPHA.NE.0.0D0 .AND. BETA.NE.0.0D0) THEN
          W = (1.0D0-X)**ALPHA* (1.0D0+X)**BETA
      ELSE IF (ALPHA.NE.0.0D0 .AND. BETA.EQ.0.0D0) THEN
          W = (1.0D0-X)**ALPHA
      ELSE IF (ALPHA.EQ.0.0D0 .AND. BETA.NE.0.0D0) THEN
          W = (1.0D0+X)**BETA
      ELSE
          W = 1.0D0
      END IF
      RETURN
C
   17 CONTINUE
      W = 1.0D0
      RETURN
C
   18 CONTINUE
      W = 1.0D0
      RETURN
C
   19 CONTINUE
      W = 1.0D0
      RETURN
C
   20 CONTINUE
      W = 1.0D0
      RETURN
C
   21 CONTINUE
      W = 1.0D0
      RETURN
C
   22 CONTINUE
      W = EXP(-X)
      IF (ALPHA.NE.0.0D0) W = W* (X**ALPHA)
      RETURN
C
   23 CONTINUE
      W = 1.0D0
      RETURN
C
   24 CONTINUE
      W = 1.0D0
      RETURN
C
   25 CONTINUE
      W = 9.0D0
      IF (X.LE.0.0D0) W = 1.0D0
      RETURN
C
   26 CONTINUE
      W = 1.0D0
      RETURN
C
   27 CONTINUE
      W = 1.0D0
      RETURN
C
   28 CONTINUE
      W = 1.0D0
      RETURN
C
   29 CONTINUE
      W = 1.0D0/X
      RETURN
C
   30 CONTINUE
      W = 1.0D0
      RETURN
C
   31 CONTINUE
      W = 1.0D0
      RETURN
C
   32 CONTINUE
      W = X** (C-1.0D0)* (1.0D0-X)** (D-1.0D0)* (X+S)** (E-1.0D0)
      RETURN
      END
C
      SUBROUTINE UV(X,U,PUP,V,PVP,HU,HV)
C
C     HERE, HU MEANS -(pu')' + qu.
C
C     .. Scalar Arguments ..
      REAL HU,HV,PUP,PVP,U,V,X
C     ..
C     .. Scalars in Common ..
      REAL A,ALPHA,B,BETA,C,D,E,GAMMA,H,K,L,NU,S
      INTEGER NUMBER
C     ..
C     .. Local Scalars ..
      REAL CC,EE,HPI,L2,SQ,SS,TX
C     ..
C     .. External Functions ..
      REAL Q,W
      EXTERNAL Q,W
C     ..
C     .. Intrinsic Functions ..
      INTRINSIC ABS,ATAN,COS,EXP,LOG,SIN,SQRT
C     ..
C     .. Common blocks ..
      COMMON /FLAG/NUMBER
      COMMON /PAR/NU,H,K,L,ALPHA,BETA,GAMMA,S,A,B,C,D,E
C     ..
      GO TO (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,
     +       23,24,25,26,27,28,29,30,31,32) NUMBER
C
    1 CONTINUE
      U = 1.0D0
      PUP = 0.0D0
      V = 0.5D0*LOG((1.0D0+X)/ (1.0D0-X))
      PVP = 1.0D0
      HU = 0.25D0*U
      HV = 0.25D0*V
      RETURN
C
    2 CONTINUE
      IF (NU.NE.-0.5D0 .AND. NU.NE.0.5D0 .AND. NU.NE.0.0D0) THEN
          U = X** (NU+0.5D0)
          PUP = (NU+0.5D0)*X** (NU-0.5D0)
          V = X** (-NU+0.5D0)
          PVP = (-NU+0.5D0)*X** (-NU-0.5D0)
      ELSE IF (NU.EQ.-0.5D0) THEN
          U = X
          PUP = 1.0D0
          V = 1.0D0
          PVP = 0.0D0
      ELSE IF (NU.EQ.0.5D0) THEN
          U = X
          PUP = 1.0D0
          V = -1.0D0
          PVP = 0.0D0
      ELSE IF (NU.EQ.0.0D0) THEN
          U = SQRT(X)
          PUP = 0.5D0/U
          V = U*LOG(X)
          PVP = (0.5D0*LOG(X)+1.0D0)/U
      END IF
      HU = 0.0D0
      HV = 0.0D0
      RETURN
C
    3 CONTINUE
      U = 1.0D0
      V = X
      PUP = 0.0D0
      PVP = 1.0D0
      HU = 0.0D0
      HV = 0.0D0
      RETURN
C
    4 CONTINUE
      TX = LOG(ABS(X))
      U = X
      PUP = 1.0D0
      V = 1.0D0 - X*TX
      PVP = -1.0D0 - TX
      HU = -1.0D0
      HV = TX
      RETURN
C
    5 CONTINUE
      RETURN
C
    6 CONTINUE
      U = (COS(X)+SIN(X))/SQRT(X)
      V = (COS(X)-SIN(X))/SQRT(X)
      PUP = -0.5D0*U + X*V
      PVP = -0.5D0*V - X*U
      HU = -0.25D0*U/X
      HV = -0.25D0*V/X
      RETURN
C
    7 CONTINUE
      TX = LOG(ABS(X))
      U = COS(TX)
      V = SIN(TX)
      PUP = -V
      PVP = U
      HU = 0.0D0
      HV = 0.0D0
      RETURN
C
    8 CONTINUE
      V = X - 1.0D0/K
      U = X*X
      PVP = 1.0D0/X
      PUP = 2.0D0
      HU = K + X*K**2
      HV = K**2
      RETURN
C
    9 CONTINUE
      U = 1.0D0
      V = -LOG(1.0D0-X)
      PUP = 0.0D0
      PVP = (((((X+1.0D0)*X+1.0D0)*X+1.0D0)*X+1.0D0)*X+1.0D0)*X + 1.0D0
      HU = 0.0D0
      HV = - (((((6.0D0*X+5.0D0)*X+4.0D0)*X+3.0D0)*X+2.0D0)*X+1.0D0)
      RETURN
C
   10 CONTINUE
      U = 2.0D0*SQRT(X)
      V = 1.0D0
      PUP = 1.0D0
      PVP = 0.0D0
      HU = 0.0D0
      HV = 0.0D0
      RETURN
C
   11 CONTINUE
      RETURN
C
   12 CONTINUE
      RETURN
C
   13 CONTINUE
      IF (H.GT.-0.25D0) THEN
          L = SQRT(H+0.25D0)
          IF (H.EQ.0.0D0) THEN
              U = X
              V = 1.0D0 + K*X*LOG(X)
              PUP = 1.0D0
              PVP = K* (1.0D0+LOG(X))
              HU = K
              HV = K*K*LOG(X)
          ELSE
              U = X** (0.5D0+L)
              V = X** (0.5D0-L) + (K/ (1.0D0-2.0D0*L))*X** (1.5D0-L)
              PUP = (0.5D0+L)*X** (L-0.5D0)
              PVP = (0.5D0-L)*X** (-L-0.5D0) +
     +              K* (1.5D0-L)/ (1.0D0-2.0D0*L)*X** (0.5D0-L)
              HU = K*X** (L-0.5D0)
              HV = K**2/ (1.0D0-2.0D0*L)*X** (0.5D0-L)
          END IF
      ELSE IF (H.LT.-0.25D0) THEN
          L2 = - (H+0.25D0)
          L = SQRT(L2)
          CC = COS(L*LOG(X))
          SS = SIN(L*LOG(X))
          SQ = SQRT(X)
          U = SQ* ((1.D0-0.25D0*K*X/H)*CC+0.5D0*K*L*X*SS)
          V = SQ* ((1.D0-0.25D0*K*X/H)*SS+0.5D0*K*L*X*CC)
          PUP = (0.5D0*CC-L*SS)/SQ - 0.5D0*K*SQ* ((0.5D0-H)*CC+L*SS)/H
          PVP = (0.5D0*SS+L*CC)/SQ + 0.5D0*K*SQ* ((H-0.5D0)*SS+L*CC)/H
          HU = 0.5D0*K*K*SQ* ((H+0.5D0)*CC+L*SS)/ (H*H) + U
          HV = 0.5D0*K*K*SQ* ((H+0.5D0)*SS-L*CC)/ (H*H) + V
      ELSE IF (H.EQ.-0.25D0) THEN
          SQ = SQRT(X)
          U = SQ + K*X*SQ
          V = 2.0D0*SQ + (SQ+K*X*SQ)*LOG(X)
          PUP = 0.5D0* (1.0D0/SQ+3.D0*K*SQ)
          PVP = 2.0D0/SQ + K*SQ + 0.5D0* (1.0D0/SQ+3.0D0*K*SQ)*LOG(X)
          HU = K*K*SQ
          HV = K*K*SQ*LOG(X)
      END IF
      RETURN
C
   14 CONTINUE
      RETURN
C
   15 CONTINUE
      RETURN
C
   16 CONTINUE
      IF (X.LT.0.0D0) THEN
          IF (BETA.GT.-1.0D0 .AND. BETA.LT.0.0D0) THEN
              U = (1.0D0+X)** (-BETA)
              V = 1.0D0
              IF (ALPHA.NE.-1.0D0) THEN
                  PUP = -BETA* (1.0D0-X)** (ALPHA+1.0D0)
                  HU = -BETA* (ALPHA+1.0D0)* (1.0D0-X)**ALPHA
              ELSE
                  PUP = -BETA
                  HU = 0.0D0
              END IF
              PVP = 0.0D0
              HV = 0.0D0
          ELSE IF (BETA.EQ.0.0D0) THEN
              U = 1.0D0
              V = LOG((1.0D0+X)/ (1.0D0-X))
              PUP = 0.0D0
              PVP = 2.0D0* (1.0D0-X)**ALPHA
              HU = 0.0D0
              HV = 2.0D0*ALPHA* (1.0D0-X)** (ALPHA-1.0D0)
          ELSE IF (BETA.GT.0.0D0 .AND. BETA.LT.1.0D0) THEN
              U = 1.0D0
              V = (1.0D0+X)** (-BETA)
              PUP = 0.0D0
              IF (ALPHA.NE.-1.0D0) THEN
                  PVP = -BETA* (1.0D0-X)** (ALPHA+1.0D0)
                  HV = -BETA* (ALPHA+1.0D0)* (1.0D0-X)**ALPHA
              ELSE
                  PVP = -BETA
                  HV = 0.0D0
              END IF
              HU = 0.0D0
          END IF
      ELSE IF (X.GE.0.0D0) THEN
          IF (ALPHA.GT.-1.0D0 .AND. ALPHA.LT.0.0D0) THEN
              U = (1.0D0-X)** (-ALPHA)
              V = 1.0D0
              IF (BETA.NE.-1.0D0) THEN
                  PUP = ALPHA* (1.0D0+X)** (BETA+1.0D0)
                  HU = -ALPHA* (BETA+1.0D0)* (1.0D0+X)**BETA
              ELSE
                  PUP = ALPHA
                  HU = 0.0D0
              END IF
              PVP = 0.0D0
              HV = 0.0D0
          ELSE IF (ALPHA.EQ.0.0D0) THEN
              U = 1.0D0
              V = LOG((1.0D0+X)/ (1.0D0-X))
              PUP = 0.0D0
              PVP = 2.0D0* (1.0D0+X)**BETA
              HU = 0.0D0
              HV = -2.0D0*BETA* (1.0D0+X)** (BETA-1.0D0)
          ELSE IF (ALPHA.GT.0.0D0 .AND. ALPHA.LT.1.0D0) THEN
              U = 1.0D0
              V = (1.0D0-X)** (-ALPHA)
              PUP = 0.0D0
              HU = 0.0D0
              IF (BETA.NE.-1.0D0) THEN
                  PVP = ALPHA* (1.0D0+X)** (BETA+1.0D0)
                  HV = -ALPHA* (BETA+1.0D0)* (1.0D0+X)**BETA
              ELSE
                  PVP = ALPHA
                  HV = 0.0D0
              END IF
          END IF
      END IF
      RETURN
C
   17 CONTINUE
      RETURN
C
   18 CONTINUE
      IF (X.LT.0.0D0) THEN
          IF (ALPHA.EQ.0.0D0) THEN
              U = 1.0D0
              V = 0.5D0*LOG((1.0D0+X)/ (1.0D0-X))
              PUP = 0.0D0
              PVP = 1.0D0
              HU = Q(X)
              HV = Q(X)*V
          ELSE IF (ALPHA.GT.0.0D0 .AND. ALPHA.LT.0.5D0) THEN
              U = (1.0D0+X)**ALPHA
              V = (1.0D0+X)** (-ALPHA)
              PUP = ALPHA* (1.0D0-X)*U
              PVP = -ALPHA* (1.0D0-X)*V
              HU = ALPHA* (ALPHA+1.0D0)*U + 2.0D0*BETA**2*U/ (1.0D0-X)
              HV = ALPHA* (ALPHA-1.0D0)*V + 2.0D0*BETA**2*V/ (1.0D0-X)
          ELSE
          END IF
      ELSE IF (X.GE.0.0D0) THEN
          IF (BETA.EQ.0.0D0) THEN
              U = 1.0D0
              V = 0.5D0*LOG((1.0D0+X)/ (1.0D0-X))
              PUP = 0.0D0
              PVP = 1.0D0
              HU = Q(X)
              HV = Q(X)*V
          ELSE IF (BETA.GT.0.0D0 .AND. BETA.LT.0.5D0) THEN
              U = (1.0D0-X)**BETA
              V = (1.0D0-X)** (-BETA)
              PUP = -BETA* (1.0D0+X)*U
              PVP = BETA* (1.0D0+X)*V
              HU = BETA* (BETA+1.0D0)*U + 2.0D0*ALPHA**2*U/ (1.0D0+X)
              HV = BETA* (BETA-1.0D0)*V + 2.0D0*ALPHA**2*V/ (1.0D0+X)
          ELSE
          END IF
      END IF
      RETURN
C
   19 CONTINUE
      IF (X.LT.0.0D0) THEN
          IF (GAMMA.EQ.0.0D0) THEN
              U = 1.0D0
              V = 0.5D0*LOG((1.0D0+X)/ (1.0D0-X))
              PUP = 0.0D0
              PVP = 1.0D0
              HU = Q(X)*U
              HV = Q(X)*V
          ELSE
              U = COS(GAMMA*LOG(1.0D0+X))
              V = SIN(GAMMA*LOG(1.0D0+X))
              PUP = -GAMMA* (1.0D0-X)*V
              PVP = GAMMA* (1.0D0-X)*U
              HU = -GAMMA**2*U - GAMMA*V + 2.0D0*BETA**2*U/ (1.0D0-X)
              HV = -GAMMA**2*V + GAMMA*U + 2.0D0*BETA**2*V/ (1.0D0-X)
          END IF
      ELSE IF (X.GE.0.0D0) THEN
          IF (BETA.EQ.0.0D0) THEN
              U = 1.0D0
              V = 0.5D0*LOG((1.0D0+X)/ (1.0D0-X))
              PUP = 0.0D0
              PVP = 1.0D0
              HU = Q(X)*U
              HV = Q(X)*V
          ELSE IF (BETA.GT.0.0D0 .AND. BETA.LT.0.5D0) THEN
              U = (1.0D0-X)**BETA
              V = (1.0D0-X)** (-BETA)
              PUP = -BETA* (1.0D0+X)*U
              PVP = BETA* (1.0D0+X)*V
              HU = BETA* (BETA+1.0D0)*U - 2.0D0*GAMMA**2*U/ (1.0D0+X)
              HV = BETA* (BETA-1.0D0)*V - 2.0D0*GAMMA**2*V/ (1.0D0+X)
          ELSE
          END IF
      END IF
      RETURN
C
   20 CONTINUE
      U = SQRT(X)*COS(K*LOG(X))
      V = SQRT(X)*SIN(K*LOG(X))
      PUP = 0.5D0*U/X - K*V/X
      PVP = 0.5D0*V/X + K*U/X
      HU = U
      HV = V
      RETURN
C
   21 CONTINUE
      RETURN
C
   22 CONTINUE
      EE = EXP(-X)
      IF (ALPHA.GT.-1.0D0 .AND. ALPHA.LT.0.0D0) THEN
          U = X** (-ALPHA)
          V = 1.0D0
          PUP = -ALPHA*EE
          PVP = 0.0D0
          HU = -ALPHA*EE
          HV = 0.0D0
      ELSE IF (ALPHA.EQ.0.0D0) THEN
          U = 1.0D0
          V = LOG(X)
          PUP = 0.0D0
          PVP = EE
          HU = 0.0D0
          HV = EE
      ELSE IF (ALPHA.GT.0.0D0 .AND. ALPHA.LT.1.0D0) THEN
          U = 1.0D0
          V = X** (-ALPHA)
          PUP = 0.0D0
          PVP = -ALPHA*EE
          HU = 0.0D0
          HV = -ALPHA*EE
      ELSE
      END IF
      RETURN
C
   23 CONTINUE
      IF (ALPHA.GT.-1.0D0 .AND. ALPHA.LT.0.0D0) THEN
          IF (ALPHA.NE.-0.5D0) THEN
              U = X** (0.5D0-ALPHA)
              V = X** (0.5D0+ALPHA)
              PUP = (0.5D0-ALPHA)*X** (-0.5D0-ALPHA)
              PVP = (0.5D0+ALPHA)*X** (-0.5D0+ALPHA)
          ELSE
              U = X
              V = 1.0D0
              PUP = 1.0D0
              PVP = 0.0D0
          END IF
          TX = X**2/16.0D0 - (ALPHA+1.0D0)/2.0D0
          HU = TX*U
          HV = TX*V
      ELSE IF (ALPHA.EQ.0.0D0) THEN
          U = SQRT(X)
          V = U*LOG(X)
          PUP = 0.5D0/U
          PVP = (1.0D0+0.5D0*LOG(X))/U
          HU = (X**2/16.0D0-0.5D0)*U
          HV = (X**2/16.0D0-0.5D0)*V
      ELSE IF (ALPHA.GT.0.0D0 .AND. ALPHA.LT.1.0D0) THEN
          IF (ALPHA.NE.0.5D0) THEN
              U = X** (0.5D0+ALPHA)
              V = X** (0.5D0-ALPHA)
              PUP = (0.5D0+ALPHA)*X** (-0.5D0+ALPHA)
              PVP = (0.5D0-ALPHA)*X** (-0.5D0-ALPHA)
          ELSE
              U = X
              V = 1.0D0
              PUP = 1.0D0
              PVP = 0.0D0
          END IF
          TX = X**2/16.0D0 - (ALPHA+1.0D0)/2.0D0
          HU = TX*U
          HV = TX*V
      ELSE
      END IF
      RETURN
C
   24 CONTINUE
      HPI = 2.0D0*ATAN(1.0D0)
      IF (X.GE.0.0D0) THEN
          IF (ALPHA.GT.-1.0D0 .AND. ALPHA.LT.0.0D0) THEN
              U = (HPI-X)** (0.5D0-ALPHA)
              V = (HPI-X)** (0.5D0+ALPHA)
              PUP = - (0.5D0-ALPHA)* (HPI-X)** (-0.5D0-ALPHA)
              PVP = - (0.5D0+ALPHA)* (HPI-X)** (-0.5D0+ALPHA)
              HU = (0.25D0-ALPHA**2)* (HPI-X)** (-1.5D0-ALPHA) + Q(X)*U
              HV = (0.25D0-ALPHA**2)* (HPI-X)** (-1.5D0+ALPHA) + Q(X)*V
          ELSE IF (ALPHA.EQ.0.0D0) THEN
              U = SQRT(HPI-X)
              V = U*LOG(HPI-X)
              PUP = -0.5D0/U
              PVP = - (1.0D0+0.5D0*LOG(HPI-X))/U
              HU = 0.25D0/ ((HPI-X)*U) + Q(X)*U
              HV = 0.25D0*LOG(HPI-X)/ ((HPI-X)*U) + Q(X)*V
          ELSE IF (ALPHA.GT.0.0D0 .AND. ALPHA.LT.1.0D0) THEN
              U = (HPI-X)** (0.5D0+ALPHA)
              V = (HPI-X)** (0.5D0-ALPHA)
              PUP = - (0.5D0+ALPHA)* (HPI-X)** (-0.5D0+ALPHA)
              PVP = - (0.5D0-ALPHA)* (HPI-X)** (-0.5D0-ALPHA)
              HU = (0.25D0-ALPHA**2)* (HPI-X)** (-1.5D0+ALPHA) + Q(X)*U
              HV = (0.25D0-ALPHA**2)* (HPI-X)** (-1.5D0-ALPHA) + Q(X)*V
          END IF
      ELSE
          IF (BETA.GT.-1.0D0 .AND. BETA.LT.0.0D0) THEN
              U = (HPI+X)** (0.5D0-BETA)
              V = (HPI+X)** (0.5D0+BETA)
              PUP = (0.5D0-BETA)* (HPI+X)** (-0.5D0-BETA)
              PVP = (0.5D0+BETA)* (HPI+X)** (-0.5D0+BETA)
              HU = (0.25D0-BETA**2)* (HPI+X)** (-1.5D0-BETA) + Q(X)*U
              HV = (0.25D0-BETA**2)* (HPI+X)** (-1.5D0+BETA) + Q(X)*V
          ELSE IF (BETA.EQ.0.0D0) THEN
              U = SQRT(HPI+X)
              V = U*LOG(HPI+X)
              PUP = 0.5D0/U
              PVP = (1.0D0+0.5D0*LOG(HPI+X))/U
              HU = 0.25D0/ ((HPI+X)*U) + Q(X)*U
              HV = 0.25D0*LOG(HPI+X)/ ((HPI+X)*U) + Q(X)*V
          ELSE IF (BETA.GT.0.0D0 .AND. BETA.LT.1.0D0) THEN
              U = (HPI+X)** (0.5D0+BETA)
              V = (HPI+X)** (0.5D0-BETA)
              PUP = (0.5D0+BETA)* (HPI+X)** (-0.5D0+BETA)
              PVP = (0.5D0-BETA)* (HPI+X)** (-0.5D0-BETA)
              HU = (0.25D0-BETA**2)* (HPI+X)** (-1.5D0+BETA) + Q(X)*U
              HV = (0.25D0-BETA**2)* (HPI+X)** (-1.5D0-BETA) + Q(X)*V
          END IF
      END IF
      RETURN
C
   25 CONTINUE
      RETURN
C
   26 CONTINUE
      RETURN
C
   27 CONTINUE
      RETURN
C
   28 CONTINUE
      RETURN
C
   29 CONTINUE
      RETURN
C
   30 CONTINUE
      RETURN
C
   31 CONTINUE
      RETURN
C
   32 CONTINUE
      U = 1.0D0
      PUP = 0.0D0
      HU = Q(X)
      IF (X.LE.0.5D0) THEN
          IF (C.EQ.1.0D0) THEN
              V = LOG(X)
              PVP = (1.0D0-X)* (X+S)*W(X)
              HV = D* (X+S)*W(X) - E* (1.0D0-X)*W(X) + V*Q(X)
          ELSE
              V = X** (1.0D0-C)
              PVP = (1.0D0-C)* (1.0D0-X)**D* (X+S)**E
              HV = (1.0D0-C)* (D* (1.0D0-X)** (D-1.0D0)* (X+S)**E-
     +             E* (1.0D0-X)**D* (X+S)** (E-1.0D0)) + V*Q(X)
          END IF
      ELSE
          IF (D.EQ.1.0D0) THEN
              V = LOG(1.0D0-X)
              PVP = -X* (X+S)*W(X)
              HV = C* (X+S)*W(X) + E*X*W(X) + V*Q(X)
          ELSE
              V = (1.0D0-X)** (1.0D0-D)
              PVP = - (1.0D0-D)*X**C* (X+S)**E
              HV = (1.0D0-D)* (C*X** (C-1.0D0)* (X+S)**E+
     +             E*X**C* (X+S)** (E-1.0D0)) + V*Q(X)
          END IF
      END IF
      RETURN
      END
SHAR_EOF
fi # end of overwriting check
cd ..
if test ! -d 'Src'
then
	mkdir 'Src'
fi
cd 'Src'
if test -f 'src.f'
then
	echo shar: will not over-write existing file "'src.f'"
else
cat << "SHAR_EOF" > 'src.f'
c-------------------------------------------------------------------
C  SUBROUTINE SLEIGN

C     **********
C     MARCH 1, 2001; P.B. BAILEY, W.N. EVERITT AND A. ZETTL
C     VERSION 1.2
C     **********

      SUBROUTINE SLEIGN(A,B,INTAB,P0ATA,QFATA,P0ATB,QFATB,A1,A2,B1,B2,
     +                  NUMEIG,EIG,TOL,IFLAG,ISLFUN,SLFUN,NCA,NCB)

C     **********
C
C     This subroutine is designed for the calculation of a specified
C     eigenvalue, EIG, of a Sturm-Liouville problem for the equation
C
C       -(p(x)*y'(x))' + q(x)*y(x) = eig*w(x)*y(x)  on (a,b)
C
C     with user-supplied coefficient functions p, q, and w,
C     and with separated boundary conditions.  (For coupled
C     boundary conditions, see the companion subroutine SLCOUP.)
C          The problem may be either nonsingular or singular.  In
C     the nonsingular case, boundary conditions are of the form
C
C        A1*y(a) + A2*p(a)*y'(a) = 0
C        B1*y(b) + B2*p(b)*y'(b) = 0,
C
C     but are of the form
C
C        A1*[y,u](a) + A2*[y,v](a) = 0
C        B1*[y,U](b) + B2*[y,V](a) = 0,
C
C     when the endpoints are singular, of type Limit Circle.
C     In either case the boundary conditions are prescribed by
C     specifying the numbers A1, A2, B1, B2.  In the singular
C     case the user must also supply the "boundary condition
C     functions" u, v near a and/or U, V near b, whichever is
C     singular, of type Limit Circle.
C     The index of the desired eigenvalue is specified in NUMEIG
C     and its requested accuracy in TOL.  Initial data for the
C     associated eigenfunction are also computed along with values
C     at selected points, if desired, in array SLFUN.
C
C     In addition to the coefficient functions p, q, and w, the user
C     must supply subroutine UV to describe the boundary condition
C     when the problem is limit circle.  UV can be a dummy subroutine
C     if the problem is not limit circle.
C
C     The SUBROUTINE statement is
C
C     SUBROUTINE SLEIGN(A,B,INTAB,P0ATA,QFATA,P0ATB,QFATB,A1,A2,B1,B2,
C    1    NUMEIG,EIG,TOL,IFLAG,ISLFUN,SLFUN,NCA,NCB)
C
C     where
C
C       A and B are input variables defining the interval.  If the
C         interval is finite, A must be less than B.  (See INTAB below.)
C
C       INTAB is an integer input variable specifying the nature of the
C         interval.  It can have four values.
C
C         INTAB = 1 - A and B are finite.
C         INTAB = 2 - A is finite and B is infinite (+).
C         INTAB = 3 - A is infinite (-) and B is finite.
C         INTAB = 4 - A is infinite (-) and B is infinite (+).
C
C         If either A or B is infinite, it is classified singular and
C         its value is ignored.
C
C       P0ATA, QFATA, P0ATB, and QFATB are input variables set to
C         1.0 or -1.0 as the following properties of p, q, and w at
C         the interval endpoints are true or false, respectively.
C
C         P0ATA -  p(a) is zero.              (If true, A is singular.)
C         QFATA -  q(a) and w(a) are finite.  (If false, A is singular.)
C         P0ATB -  p(b) is zero.              (If true, B is singular.)
C         QFATB -  q(b) and w(b) are finite.  (If false, B is singular.)
C
C       A1 and A2 are input variables set to prescribe the boundary
C         condition at A.
C
C       B1 and B2 are input variables set to prescribe the boundary
C         condition at B.
C
C       NUMEIG is an integer variable.  On input, it should be set to
C         the index of the desired eigenvalue (increasing sequence where
C         index 0 corresponds to the lowest eigenvalue -- if the
C         eigenvalues are bounded below -- or to the smallest nonegative
C         eigenvalue otherwise).  On output, it is unchanged unless the
C         problem (apparently) lacks eigenvalue NUMEIG, in which case it
C         is reset to the index of the largest eigenvalue that seems to
C         exist.
C
C       EIG is a variable set on input to 0.0 or to an initial guess of
C         the eigenvalue.  If EIG is set to 0.0, SLEIGN2 will generate
C         the initial guess.  On output, EIG holds the calculated
C         eigenvalue if IFLAG (see below) signals success.
C
C       TOL is a variable set on input to the desired accuracy of the
C         eigenvalue.  On output, TOL is reset to the accuracy estimated
C         to have been achieved if IFLAG (see below) signals success.
C         This accuracy estimate is absolute if EIG is less than one
C         in magnitude, and relative otherwise.  In addition, prefixing
C         TOL with a negative sign, removed after interrogation, serves
C         as a flag to request trace output from the calculation.
C
C       IFLAG is an integer output variable set as follows:
C
C         IFLAG = 0 -  improper input parameters.
C         IFLAG = 1  - successful problem solution, within tolerance.
C         IFLAG = 2  - best problem result, not within tolerance.
C         IFLAG = 3  - NUMEIG exceeds actual highest eigenvalue index.
C         IFLAG = 4  - RAY and EIG fail to agree after 5 tries.
C         IFLAG = 6  - in SECANT-METHOD, ABS(DE) .LT. EPSMIN .
C         IFLAG = 7  - iterations are stuck in a loop.
C         IFLAG = 8  - number of iterations has reached the set limit.
C         IFLAG = 9  - residual truncation error dominates.
C         IFLAG = 10 - integrator tolerance cannot be reduced.
C         IFLAG = 11 - no more improvement.
C         IFLAG = 13 - AA cannot be moved in any further.
C         IFLAG = 14 - BB cannot be moved in any further.
c         iflag = 15 - Bad behavior of coefficients at an endpoint.
C         IFLAG = 16 - Could not get started.
C         IFLAG = 17 - Failed to get a bracket.
C         IFLAG = 18 - Estimator failed.
C         IFLAG = 51 - integration failure after 1st call to INTEG.
C         IFLAG = 52 - integration failure after 2nd call to INTEG.
C         IFLAG = 53 - integration failure after 3rd call to INTEG.
C         IFLAG = 54 - integration failure after 4th call to INTEG.
C
C       ISLFUN is an integer input variable set to the number of
C         selected eigenfunction values desired.  If no values are
C         desired, set ISLFUN to zero.
c         (If ISLFUN is set to -1, the result will be that SLEIGN2
c           will return to the calling program directly after sampling
c           the coefficients p,q,w .  This device is used only once,
c           in SUBROUTINE PERIO.)
C
C       SLFUN is an array of length at least 9.  On output, the first 9
C         locations contain the integration interval and initial data
C         that completely determine the eigenfunction.
C
C         SLFUN(1) - point where two pieces of eigenfunction Y match.
C         SLFUN(2) - left endpoint XAA of the (truncated) interval.
C         SLFUN(3) - value of THETA at XAA.  (Y = RHO*sin(THETA))
C         SLFUN(4) - value of F at XAA.  (RHO = exp(F))
C         SLFUN(5) - right endpoint XBB of the (truncated) interval.
C         SLFUN(6) - value of THETA at XBB.
C         SLFUN(7) - value of F at XBB.
C         SLFUN(8) - final value of integration accuracy parameter EPS.
C         SLFUN(9) - the constant Z in the polar form transformation.
C
C         F(XAA) and F(XBB) are chosen so that the eigenfunction is
C         continuous in the interval (XAA,XBB) and has weighted (by W)
C         L2-norm of 1.0 on the interval.  If ISLFUN is positive, then
C         on input the further ISLFUN locations of SLFUN specify the
C         points, in ascending order, where the eigenfunction values
C         are desired and on output contain the values themselves.
C
c       nca & ncb are integers which indicate the nature of the
c         endpoints a & b, respectively.  Namely:
c       nca = 1 : Endpoint a is REGULAR .
C             2 :     "         WEAKLY REGULAR .
C             3 :     "         LIMIT CIRCLE, NON-OSC .
C             4 :     "         LIMIT CIRCLE, OSC .
C             5 :     "         LIMIT POINT, REGULAR, AT FINITE PT.
C             6 :     "         LIMIT POINT, DEFAULT .
C             7 :     "         LIMIT POINT, AT INFINITY OR IRREG.
C             8 :     "         LIMIT POINT, BAD BEHAVIOR AT ENDPT.
C        ncb = 1 : Endpoint b is REGULAR .
C                  etc. as for nca .
C
C     **********
C  INPUT QUANTITIES: A,B,INTAB,P0ATA,QFATA,P0ATB,QFATB,A1,A2,B1,B2,
C                   NUMEIG,EIG,TOL,IFLAG,ISLFUN,SLFUN,NCA,NCB,PR
C  OUTPUT QUANTITIES: NUMEIG,EIG,TOL,IFLAG,SLFUN,
C                   MFS,MLS,PI,TWOPI,HPI,EPSMIN,Z,JAY,ZEE,
C                   AA,TMID,BB,DTHDAA,DTHDBB,MDTHZ,ADDD
C
C     .. Scalar Arguments ..
      REAL A,A1,A2,B,B1,B2,EIG,P0ATA,P0ATB,QFATA,QFATB,TOL
      INTEGER IFLAG,INTAB,ISLFUN,NCA,NCB,NUMEIG
C     ..
C     .. Array Arguments ..
      REAL SLFUN(9)
C     ..
C     .. Scalars in Common ..
      REAL A1S,A2S,AA,ASAV,B1S,B2S,BB,BSAV,DTHDAA,DTHDBB,
     +                 EIGSAV,EPSMIN,FA,FB,GQA,GQB,GWA,GWB,HPI,LPQA,
     +                 LPQB,LPWA,LPWB,P0ATAS,P0ATBS,PI,QFATAS,QFATBS,
     +                 TMID,TSAVEL,TSAVER,TWOPI,Z
      INTEGER IND,INTSAV,ISAVE,MDTHZ,MFS,MLS,MMWD,T21
      LOGICAL ADDD,PR
C     ..
C     .. Arrays in Common ..
      REAL TEE(100),TT(7,2),YS(200),YY(7,3,2),ZEE(100)
      INTEGER JAY(100),MMW(100),NT(2)
C     ..
C     .. Local Scalars ..
      REAL AA1,AAA,AAF,AAL,AAS,ALFA,ATHETA,BALLPK,BB1,BBB,
     +                 BBF,BBL,BBS,BESTAA,BESTBB,BETA,BSTEIG,BSTEPS,
     +                 BSTEST,BSTMID,CHLIM,CHNG,CONVC,DE,DEDW,DIST,
     +                 DTHDA,DTHDB,DTHDE,DTHDEA,DTHDEB,DTHETA,DTHOLD,
     +                 DTHOLY,EEE,EIGLO,EIGLT,EIGPI,EIGRT,EIGUP,EL,
     +                 ELIMUP,EMAX,EMIN,EOLD,EPS,EPSL,EPSM,ER1,ER1M,
     +                 ESTERR,FLO,FLOUP,FMAX,FUP,GUESS,OLDEST,OLDRAY,
     +                 OLRAYS,ONE,PIN,PT2,PT3,RAY,RLX,SAVAA,SAVBB,
     +                 SAVERR,SUM,T1,T2,T3,TAU,TAU0,TAUM,TMID1,TMP,U,V,
     +                 WL,ZZ
      INTEGER I,IMAX,IMID,IMIN,J,JFLAG,JJL,JJR,K,LOOP2,LOOP3,MF,ML,NEIG,
     +        NEIGST,NITER,NRAY,NTMP
      LOGICAL AOK,BOK,BRACKT,BRS,BRSS,CHEPS,CONVRG,ENDA,ENDB,EXIT,
     +        FIRSTT,GESS0,IOSC,LCOA,LCOB,LIMUP,LOGIC,NEWTF,NEWTON,
     +        OLNEWT,ONEDIG,THEGT0,THELT0,TRUNKA,TRUNKB
C     ..
C     .. Local Arrays ..
      REAL DELT(100),DS(100),EIGEST(200,4),PS(100),PSS(100),
     +                 QS(100),WS(100),XS(100)
C     ..
C     .. External Functions ..
      REAL EPSLON
      EXTERNAL EPSLON
C     ..
C     .. External Subroutines ..
      EXTERNAL AABB,EFDATA,EIGFCN,ESTIM,OBTAIN,PRELIM,RESET,SAMPLE,SAVE
C     ..
C     .. Intrinsic Functions ..
      INTRINSIC ABS,ATAN,INT,MAX,MIN,SIGN
C     ..
C     .. Common blocks ..
c     COMMON /ALBE/LPWA,LPQA,LPWB,LPQB
      COMMON /ALBE/LPWA,LPQA,FA,GWA,GQA,LPWB,LPQB,FB,GWB,GQB
      COMMON /BCDATA/A1S,A2S,P0ATAS,QFATAS,B1S,B2S,P0ATBS,QFATBS
      COMMON /DATADT/ASAV,BSAV,INTSAV
      COMMON /DATAF/EIGSAV,IND
      COMMON /LP/MFS,MLS
      COMMON /PASS/YS,MMW,MMWD
      COMMON /PIE/PI,TWOPI,HPI
      COMMON /PRIN/PR,T21
      COMMON /RNDOFF/EPSMIN
      COMMON /TDATA/AA,TMID,BB,DTHDAA,DTHDBB,MDTHZ,ADDD
      COMMON /TEEZ/TEE
      COMMON /TEMP/TT,YY,NT
      COMMON /TSAVE/TSAVEL,TSAVER,ISAVE
      COMMON /Z1/Z
      COMMON /ZEEZ/JAY,ZEE
C     ..
C     To produce printout, set PR = .TRUE.
      PR = .true.
C
      IFLAG = 1
      ONE = 1.0D0
      EPSMIN = EPSLON(ONE)
        if(epsmin .le. 1.0d-12) epsmin = 1.0d-12
      PI = 4.0D0*ATAN(ONE)
      TWOPI = 2.0D0*PI
      HPI = 0.5D0*PI
      Z = 1.0D0
      NEIG = NUMEIG - 1
C
      LOGIC = 1 .LE. INTAB .AND. INTAB .LE. 4 .AND.
     +        P0ATA*QFATA*P0ATB*QFATB .NE. 0.0D0
      IF (INTAB.EQ.1) LOGIC = LOGIC .AND. A .LT. B
      IF (.NOT.LOGIC) THEN
          IFLAG = 0
          GO TO 150
      END IF
C
C     (JFLAG = 0 INDICATES FAILURE OF THE ESTIMATOR)
      JFLAG = 1
C
      CALL SAVE(A,B,INTAB,P0ATA,QFATA,P0ATB,QFATB,A1,A2,B1,B2)
C
      DO 5 I = 1,100
          JAY(I) = 0
          ZEE(I) = 1.0D0
    5 CONTINUE
C
      CALL SAMPLE(NCA,NCB,MF,ML,XS,PS,QS,WS,DS,DELT,EMIN,EMAX,IMIN,IMAX,
     +            AAA,BBB)

C       (MAKE MF,ML AVALABLE THROUGH COMMON/LP/ )
      MFS = MF
      MLS = ML
C
      EIGPI = NUMEIG*PI
      PIN = EIGPI + PI
      TAU = ABS(TOL)
      TAU0 = 0.001D0
      TAUM = MAX(TAU,EPSMIN)
      LIMUP = .FALSE.
      ELIMUP = EMAX
      GUESS = EIG
      GESS0 = ABS(EIG) .LE. (1D-7)
      LCOA = NCA .EQ. 4
      LCOB = NCB .EQ. 4
      IOSC = LCOA .OR. LCOB

      IF (GESS0) THEN
          CALL ESTIM(IOSC,PIN,MF,ML,PS,QS,WS,PSS,DS,DELT,TAU0,JJL,JJR,
     +               SUM,LIMUP,ELIMUP,EMAX,IMAX,IMIN,EL,WL,DEDW,BALLPK,
     +               EEE,JFLAG)
      END IF

      IF (JFLAG.EQ.0) THEN
          IF (PR) WRITE (T21,FMT=*) ' ESTIMATOR FAILED '
          IFLAG = 18
          RETURN
      END IF

      CALL PRELIM(MF,ML,EEE,BALLPK,XS,QS,WS,DS,DELT,PS,PSS,TAU0,JJL,JJR,
     +            LIMUP,ELIMUP,EL,WL,DEDW,GUESS,EIG,AAA,AA,BBB,BB,NCA,
     +            NCB,EIGPI,ALFA,BETA,DTHDEA,DTHDEB,IMID,TMID)

      IF (JAY(3).EQ.0 .AND. ZEE(2).NE.1.0D0) THEN
          IF (PR) WRITE (T21,FMT=*) ' COULD NOT GET STARTED. '
          IFLAG = 16
          RETURN
      END IF
C
      IF (NCA.EQ.8 .OR. NCB.EQ.8) THEN
C       (WE CAN'T HANDLE THIS KIND OF LIMIT POINT)
          IFLAG = 15
          RETURN
      END IF
C
      IF (ISLFUN.EQ.-1) THEN
          SLFUN(1) = TMID
          SLFUN(2) = AA
          SLFUN(3) = ALFA
          SLFUN(5) = BB
          SLFUN(6) = BETA + EIGPI
          SLFUN(9) = Z
          ADDD = .FALSE.
          MDTHZ = 0
          IFLAG = 1
          RETURN
      END IF
C
C     SET LOGICAL VARIABLES:
      AOK = INTAB .LT. 3.D0 .AND. P0ATA .LT. 0.0D0 .AND.
     +      QFATA .GT. 0.0D0
      BOK = (INTAB.EQ.1 .OR. INTAB.EQ.3) .AND. P0ATB .LT. 0.0D0 .AND.
     +      QFATB .GT. 0.0D0
      LCOA = NCA .EQ. 4
      LCOB = NCB .EQ. 4
      TRUNKA = LCOA .OR. (NCA.EQ.6)
      TRUNKB = LCOB .OR. (NCB.EQ.6)
C
C     END PRELIMINARY WORK, BEGIN MAIN TASK OF COMPUTING EIG.
C
C     LOGICAL VARIABLES HAVE THE FOLLOWING MEANINGS IF TRUE.
C        AOK    - ENDPOINT A IS NOT SINGULAR.
C        BOK    - ENDPOINT B IS NOT SINGULAR.
C        BRACKT - EIG HAS BEEN BRACKETED.
C        CONVRG - CONVERGENCE TEST FOR EIG HAS BEEN SUCCESSFULLY PASSED.
C        NEWTON - NEWTON ITERATION MAY BE EMPLOYED.
C        THELT0 - LOWER BOUND FOR EIG HAS BEEN FOUND.
C        THEGT0 - UPPER BOUND FOR EIG HAS BEEN FOUND.
C        LIMIT  - UPPER BOUND EXISTS WITH BOUNDARY CONDITIONS SATISFIED.
C        ONEDIG - MOST SIGNIFICANT DIGIT CAN BE EXPECTED TO BE CORRECT.
C
C     INITIALIZE SOME OF THE CONTROL VARIABLES
      EXIT = .FALSE.
      FIRSTT = .TRUE.
      LOOP2 = 0
      LOOP3 = 0
      PT2 = 0.0D0
      PT3 = 0.0D0
      ENDA = .FALSE.
      ENDB = .FALSE.
      EPSM = EPSMIN
      CHEPS = .FALSE.
      NEWTF = .FALSE.
      BRS = .FALSE.
      BRSS = .FALSE.
      SAVERR = 1.0D+9
      ATHETA = 1.0D+9
      BSTEST = 1.0D+9
      OLDEST = 1.0D+9
      OLDRAY = 1.0D+9
      NRAY = 1
      AAL = AA
      BBL = BB
      AAS = AA
      BBS = BB
      AAF = AAA
      BBF = BBB
      NEIGST = 0
      EIG = EEE
      BSTEIG = EIG
      EPS = 0.0001D0
      EPSL = EPS
      BESTAA = AA
      BESTBB = BB
      BSTMID = TMID
      BSTEPS = EPS
C
  110 CONTINUE
C       (INITIAL-IZE)
      BRACKT = .FALSE.
      CONVRG = .FALSE.
      THELT0 = .FALSE.
      THEGT0 = .FALSE.
      NEWTON = .FALSE.
      EIGLO = EMIN - 1.0D0
      FLO = -5.0D0
      FUP = 5.0D0
      EIGLT = 0.0D0
      EIGRT = 0.0D0
      EIGUP = EMAX + 1.0D0
      IF (LIMUP) EIGUP = MIN(EMAX,ELIMUP)
      DTHOLD = 1.0D0
C
      IF (PR) WRITE (T21,FMT=*)
      IF (PR) WRITE (T21,FMT=*)
     +    '---------------------------------------------'
      IF (PR) WRITE (T21,FMT=*) ' INITIAL GUESS FOR EIG = ',EIG
      IF (PR) WRITE (T21,FMT=*) ' AA,BB = ',AA,BB

C           (UNTIL(CONVRG .OR. EXIT)
      DO 120 NITER = 1,40
          IF (PR) WRITE (*,FMT=*)
          IF (PR) WRITE (*,FMT=*) ' ******************** '
          IF (PR) WRITE (*,FMT=*) ' EIGENVALUE ',NUMEIG
C
          CALL RESET(AA,BB,ALFA,BETA,EIG,GUESS,MF,ML,QS,WS,IMID,TMID,
     +               LIMUP,ELIMUP,DTHDEA,DTHDEB,THELT0,EIGLO,EIGUP,
     +               BRACKT,NCA,NCB,IFLAG)
          IF (IFLAG.EQ.11) THEN
c             EXIT = .TRUE.
c             GO TO 130
              GO TO 333
          END IF
C
          AA1 = AA
          BB1 = BB
          TMID1 = TMID
          IFLAG = 1
          CALL OBTAIN(NCA,ALFA,DTHDEA,NCB,BETA,DTHDEB,AA1,BB1,EIG,EIGPI,
     +                TMID1,EPS,DTHETA,DTHDE,ONEDIG,DTHDA,DTHDB,ER1,
     +                ER1M,IFLAG)
          IF (IFLAG.EQ.2) THEN
              AAS = AA
              BBS = BB
          END IF
          ATHETA = ABS(DTHETA)
          IF (51.LE.IFLAG .AND. IFLAG.LE.54) THEN
              EXIT = .TRUE.
              GO TO 130
          ELSE
              FIRSTT = .FALSE.
          END IF
C-----------------------------------------------------------
          CHEPS = .FALSE.
          CONVRG = .FALSE.
          OLNEWT = NEWTON
          NEWTON = ABS(DTHETA) .LT. 0.06D0 .AND. BRACKT
          IF (NEWTON) ONEDIG = ONEDIG .OR.
     +                         ABS(DTHETA+ER1) .LT. 0.5D0*DTHOLD
          IF (.NOT.ONEDIG .AND. BRS) THEN
              EXIT = .TRUE.
              IF (PR) WRITE (T21,FMT=*) ' NOT ONEDIG '
              GO TO 130
          END IF
C
          IF (PR) WRITE (*,FMT=*) ' SET BRACK