# Classical Solutions of the Degenerate
Fifth Painlevé Equation

Peter A. Clarkson
School of Mathematics, Statistics and Actuarial Science,
University of Kent, Canterbury, CT2 7FS, UK
Email<EMAIL_ADDRESS>
###### Abstract

In this paper classical solutions of the degenerate fifth Painlevé equation
are classified, which include hierarchies of algebraic solutions and solutions
expressible in terms of Bessel functions. Solutions of the degenerate fifth
Painlevé equation are known to be expressible in terms of solutions of the
third Painlevé equation. The classification and description of the classical
solutions of the degenerate fifth Painlevé equation is done using the
Hamiltonian associated with third Painlevé equation. Two applications of these
classical solutions are discussed, deriving exact solutions of the complex
sine-Gordon equation and of the coefficients in the three-term recurrence
relation associated with generalised Charlier polynomials.

##  1 Introduction

In this paper we are concerned with solutions of the equation

$\frac{{\rm d}^{2}{w}}{{{\rm
d}{z}}^{2}}=\left(\frac{1}{2w}+\frac{1}{w-1}\right)\left(\frac{{\rm
d}{w}}{{{\rm d}{z}}}\right)^{\\!2}-\frac{1}{z}\frac{{\rm d}{w}}{{{\rm
d}{z}}}+\frac{(w-1)^{2}(\alpha w^{2}+\beta)}{z^{2}w}+\frac{\gamma w}{z},$
(1.1)

with $\alpha$, $\beta$ and $\gamma$ constants. Equation (1.1) is the special
case of the fifth Painlevé equation (PV)

$\frac{{\rm d}^{2}{w}}{{{\rm
d}{z}}^{2}}=\left(\frac{1}{2w}+\frac{1}{w-1}\right)\left(\frac{{\rm
d}{w}}{{{\rm d}{z}}}\right)^{\\!2}-\frac{1}{z}\frac{{\rm d}{w}}{{{\rm
d}{z}}}+\frac{(w-1)^{2}(\alpha w^{2}+\beta)}{z^{2}w}+\frac{\gamma
w}{z}+\frac{\delta w(w+1)}{w-1}.$ (1.2)

with $\alpha$, $\beta$, $\gamma$ and $\delta$ constants, when $\delta=0$ and
is known as the degenerate fifth Painlevé equation (deg-PV), cf. [40].

The six Painlevé equations (PI–PVI), were discovered by Painlevé, Gambier and
their colleagues whilst studying second order ordinary differential equations
of the form

$\frac{{\rm d}^{2}{w}}{{{\rm d}{z}}^{2}}=F\left(z,w,\frac{{\rm d}{w}}{{{\rm
d}{z}}}\right),$ (1.3)

where $F$ is rational in ${\rm d}w/{\rm d}z$ and $w$ and analytic in $z$. The
Painlevé functions can be thought of as nonlinear analogues of the classical
special functions. The general solutions of the Painlevé equations are
transcendental in the sense that they cannot be expressed in terms of known
elementary functions and so require the introduction of a new transcendental
function to describe their solution. However, it is well known that PII–PVI
possess rational solutions, algebraic solutions and solutions expressed in
terms of the classical special functions — Airy, Bessel, parabolic cylinder,
Kummer and hypergeometric functions, respectively — for special values of the
parameters, see, e.g. [12, 22] and the references therein. These hierarchies
are usually generated from “seed solutions” using the associated Bäcklund
transformations and frequently can be expressed in the form of determinants.
These solutions of the Painlevé equations are often called “classical
solutions”, cf. [51, 52].

It is well known that solutions of deg-PV (1.1) are related to solutions of
the third Painlevé equation

$\frac{{\rm d}^{2}{q}}{{{\rm d}{x}}^{2}}=\frac{1}{q}\left(\frac{{\rm
d}{q}}{{{\rm d}{x}}}\right)^{\\!2}-\frac{1}{x}\frac{{\rm d}{q}}{{{\rm
d}{x}}}+\frac{Aq^{2}+B}{x}+Cq^{3}+\frac{D}{q},$ (1.4)

with $A$, $B$, $C$ and $D$ constants, a result originally due to Gromak [21];
see also [22, §34]. The relationship between solutions of deg-PV and the third
Painlevé equation is given in Lemma LABEL:lem21 below. The objective of this
paper is to give a classification and description of the classical solutions
of deg-PV (1.1) using the associated Hamiltonian formalism, rather than
through solutions of the third Painlevé equation (1.4).

In §2, the relationship between deg-PV (1.1) and the third Painlevé equation
(1.4) is discussed using the associated Hamitonian. In §3, classical solutions
of the third Painlevé equation (1.4) are reviewed, the rational solutions in
§3.1 and the Bessel function solutions in §3.2. In §4, Bäcklund
transformations of deg-PV (1.1) are given, which can be used to derive a
hierarchy of solutions from a “seed solution”. In §5, classical solutions of
deg-PV (1.1) are classified, the algebraic solutions in §5.1 and the Bessel
function solutions in §5.2. In §6, two applications of classical solutions of
deg-PV (1.1) are given to derive exact solutions of the complex sine-Gordon
equation, which is equivalent to the Pohlmeyer-Lund-Regge model, and to derive
explicit representations of the coefficients in the three-term recurrence
relation satisfied by generalised Charlier polynomials, which are discrete
orthogonal polynomials.

##  2 The relationship between deg-PV and PIII

In the generic case when $CD\not=0$ in the third Painlevé equation (1.4), we
set $C=1$ and $D=-1$, without loss of generality (by rescaling the variables
if necessary), and so consider the equation

$\frac{{\rm d}^{2}{q}}{{{\rm d}{x}}^{2}}=\frac{1}{q}\left(\frac{{\rm
d}{q}}{{{\rm d}{x}}}\right)^{\\!2}-\frac{1}{x}\frac{{\rm d}{q}}{{{\rm
d}{x}}}+\frac{Aq^{2}+B}{x}+q^{3}-\frac{1}{q}.$ (2.1)

In the sequel, we shall refer to this equation as PIII since it is the generic
case.

Consider the Hamiltonian associated with PIII (2.1) given by

${\mbox{$\mathcal{H}_{\rm
III}$}}(q,p,x;a,b,\varepsilon)=q^{2}p^{2}-xq^{2}p-(2a+2b+1)qp+\varepsilon
xp+2bxq,$ (2.2)

with $a$ and $b$ parameters and $\varepsilon=\pm 1$, see [28, 44]. Then $p(x)$
and $q(x)$ satisfy the Hamiltonian system

$\displaystyle x\frac{{\rm d}{q}}{{{\rm d}{x}}}$
$\displaystyle=\frac{\partial{\mbox{$\mathcal{H}_{\rm
III}$}}}{{\partial{p}}}\phantom{-}=2q^{2}p-xq^{2}-(2a+2b+1)q+\varepsilon x,$
(2.3a) $\displaystyle x\frac{{\rm d}{p}}{{{\rm d}{x}}}$
$\displaystyle=-\frac{\partial{\mbox{$\mathcal{H}_{\rm
III}$}}}{{\partial{q}}}=-2qp^{2}+2xqp+(2a+2b+1)p-2bx.$ (2.3b)

Solving (2.3a) for $p(x)$ gives

$p(x)={\frac{1}{2q^{2}}}\left\\{x\frac{{\rm d}{q}}{{{\rm
d}{x}}}+xq^{2}+(2a+2b+1)q-\varepsilon x\right\\},$

and then substituting this in (2.3b) gives

$\frac{{\rm d}^{2}{q}}{{{\rm d}{x}}^{2}}=\frac{1}{q}\left(\frac{{\rm
d}{q}}{{{\rm d}{x}}}\right)^{\\!2}-\frac{1}{x}\frac{{\rm d}{q}}{{{\rm
d}{x}}}+\frac{2(a-b)q^{2}}{x}+\frac{2\varepsilon(a+b+1)}{x}+q^{3}-\frac{1}{q}.$
(2.4)

which is PIII (2.1), with parameters

$A=2(a-b),\qquad B=2\varepsilon(a+b+1).$ (2.5)

Solving (2.3b) for $q(x)$ gives

$q(x)=\frac{1}{2p(x-p)}\left\\{x\frac{{\rm d}{p}}{{{\rm
d}{x}}}-(2a+2b+1)p+2bx\right\\},$

and then substituting this in (2.3a) gives

$\frac{{\rm d}^{2}{p}}{{{\rm
d}{x}}^{2}}=\frac{1}{2}\left(\frac{1}{p}+\frac{1}{p-x}\right)\left(\frac{{\rm
d}{p}}{{{\rm d}{x}}}\right)^{2}-{\frac{p}{x(p-x)}}\frac{{\rm d}{p}}{{{\rm
d}{x}}}+2\varepsilon
p-\frac{2b^{2}}{p}-\frac{4a^{2}-1}{2(p-x)}+\frac{1-4(a^{2}-b^{2})-4\varepsilon
p^{2}}{2x}.$ (2.6)

Then making the transformation

$p(x)=\frac{2\sqrt{z}\,w(z)}{w(z)-1},\qquad x=2\sqrt{z},$ (2.7)

in (2.6) gives

$\frac{{\rm d}^{2}{w}}{{{\rm
d}{z}}^{2}}=\left(\frac{1}{2w}+\frac{1}{w-1}\right)\left(\frac{{\rm
d}{w}}{{{\rm d}{z}}}\right)^{2}-\frac{1}{z}\,\frac{{\rm d}{w}}{{{\rm
d}{z}}}+\frac{(w-1)^{2}(a^{2}w^{2}-b^{2})}{2z^{2}w}+\frac{\varepsilon w}{z},$
(2.8)

which is deg-PV (1.1) with parameters

$\alpha=\tfrac{1}{2}a^{2},\qquad\beta=-\tfrac{1}{2}b^{2},\qquad\gamma=\varepsilon.$
(2.9)

Hence we have the following result; see also [22, Theorem 34.2].

######  Lemma 2.1.

If $q(x)$ is a solution of (2.4) then

$w(z)=\frac{xq^{\prime}(x)+xq^{2}(x)+(2a+2b+1)q(x)-\varepsilon
x}{xq^{\prime}(x)-xq^{2}(x)+(2a+2b+1)q(x)-\varepsilon x},\qquad
z=\tfrac{1}{2}x^{2},$ (2.10)

with ${}^{\prime}\equiv{\rm d}/{\rm d}x$, is a solution of (2.8), provided
that

$x\frac{{\rm d}{q}}{{{\rm d}{x}}}-xq^{2}+(2a+2b+1)q-\varepsilon x\not=0.$

Conversely, if $w(z)$ is a solution of (2.8), then

$q(x)=\frac{1}{2\sqrt{z}\,w}\left\\{z\frac{{\rm d}{w}}{{{\rm
d}{z}}}+(w-1)(aw+b)\right\\},\qquad x=\sqrt{2z},$ (2.11)

is a solution of (2.4).

###### Proof.

Solving (2.3a) for $p(x)$, substituting in (2.7) and solving for $w(z)$ gives
(2.10). Also solving (2.3b) for $q(x)$ and substituting (2.7) into the
resulting expression gives (2.11). ∎

An alternative method of deriving solutions of (2.8) involves the second-
order, second-degree equation satisfied associated with the Hamiltonian system
(2.3), due to Jimbo and Miwa [28] and Okamoto [44], which is often called the
“$\sigma$-equation”.

######  Theorem 2.2.

If ${\mbox{$\mathcal{H}_{\rm III}$}}(q,p,x;a,b,\varepsilon)$ is given by
(2.2), then

$\sigma(x;a,b,\varepsilon)=\mbox{$\mathcal{H}_{\rm
III}$}(q,p,x;a,b,\varepsilon)+qp-\tfrac{1}{2}\varepsilon x^{2}+(a+b)^{2},$
(2.12)

where $q(x)$ and $p(x)$ satisfy the system (2.3), satisfies the second-order,
second-degree equation $(\mbox{\rm S${}_{\rm III}$})$

$\displaystyle\left(x\frac{{\rm d}^{2}{\sigma}}{{{\rm d}{x}}^{2}}-\frac{{\rm
d}{\sigma}}{{{\rm d}{x}}}\right)^{\\!2}+2\left\\{\left(\frac{{\rm
d}{\sigma}}{{{\rm d}{x}}}\right)^{\\!2}-x^{2}\right\\}\left(x\frac{{\rm
d}{\sigma}}{{{\rm d}{x}}}-2\sigma\right)$
$\displaystyle-8\varepsilon(a^{2}-b^{2})x\frac{{\rm d}{\sigma}}{{{\rm
d}{x}}}=8(a^{2}+b^{2})x^{2}.$ (2.13)

Conversely, if $\sigma(x;a,b,\varepsilon)$ satisfies (2.13) then the solution
of the Hamiltonian system (2.3) is given by

$\displaystyle q(x)$ $\displaystyle=\frac{\displaystyle\varepsilon
x\sigma^{\prime\prime}(x)-\varepsilon(2a+2b+1)\sigma^{\prime}(x)-2(a-b)x}{\displaystyle
x^{2}-\left[\sigma^{\prime}(x)\right]^{2}},$ (2.14a) $\displaystyle p(x)$
$\displaystyle=\tfrac{1}{2}\varepsilon\sigma^{\prime}(x)+\tfrac{1}{2}x.$
(2.14b)

###### Proof.

See Jimbo and Miwa [28] and Okamoto [44].∎

Consequently solutions of deg-PV (2.8) can be expressed in terms of solutions
of SIII (2.13).

######  Corollary 2.3.

If $\sigma(x;a,b,\varepsilon)$ is a solution of SIII (2.13), then

$w(z;a,b,\varepsilon)=\frac{\sigma^{\prime}(x;a,b,\varepsilon)+\varepsilon
x}{\sigma^{\prime}(x;a,b,\varepsilon)-\varepsilon x},\qquad
z=\tfrac{1}{2}x^{2},$ (2.15)

is a solution of (2.8).

###### Proof.

This immediately follows from (2.7) and Theorem 2.2.∎

######  Remark 2.4.

From Lemma LABEL:lem21 and Corollary 2.3, it’s clear that it’s simpler to
derive solutions of deg-PV (2.8) from equation (LABEL:PV0-SIII) rather than
equation (2.10). Further as shown in §3, classical solutions of SIII involve
one determinant, whereas classical solutions of PIII involve two determinants.

##  3 Classical solutions of PIII and SIII

###  3.1 Rational solutions of PIII and SIII

Rational solutions of PIII (2.1) are classified in the following theorem.

######  Theorem 3.1.

Equation (2.1) has a rational solution if and only if

$\varepsilon_{1}A+\varepsilon_{2}B=4n,$ (3.1)

with $n\in\mathbb{Z}$ and $\varepsilon_{1}^{2}=1$, $\varepsilon_{2}^{2}=1$,
independently.

###### Proof.

For details see Lukashevich [31]; see also [37, 38].∎

Umemura [53]111The original manuscript was written by Umemura in 1996 for the
proceedings of the conference “Theory of nonlinear special functions: the
Painlevé transcendents” in Montreal, which were not published; for further
details see [45]. derived special polynomials associated with rational
solutions of PIII (2.1), which we now define; see also [10, 29, 30].

######  Definition 3.2.

The Umemura polynomial $S_{n}(x;\mu)$ is given by the recursion relation

$S_{n+1}S_{n-1}=-x\left\\{S_{n}\frac{{\rm d}^{2}{S_{n}}}{{{\rm
d}{x}}^{2}}-\left(\frac{{\rm d}{S_{n}}}{{{\rm
d}{x}}}\right)^{\\!2}\right\\}-S_{n}\frac{{\rm d}{S_{n}}}{{{\rm
d}{x}}}+(x+\mu)S_{n}^{2},$ (3.2)

where $S_{-1}(x;\mu)=S_{0}(x;\mu)=1$, with $\mu$ an arbitrary parameter.

######  Remarks 3.3.

  1.   2. (i)

The Umemura polynomial $S_{n}(x;\mu)$ has the Wronskian representation

$S_{n}(x;\mu)=c_{n}\operatorname{Wr}\left(\varphi_{1},\varphi_{3},\ldots,\varphi_{2n-1}\right),\qquad
c_{n}=\prod_{k=0}^{n}(2k+1)^{n-k},$ (3.3)

where $\operatorname{Wr}(\varphi_{1},\varphi_{2},\ldots,\varphi_{n})$ is the
Wronskian defined by

$\operatorname{Wr}(\varphi_{1},\varphi_{2},\ldots,\varphi_{n})=\left|\begin{matrix}\varphi_{1}&\varphi_{2}&\ldots&\varphi_{n}\\\
\varphi_{1}^{\prime}&\varphi_{2}^{\prime}&\ldots&\varphi_{n}^{\prime}\\\
\vdots&\vdots&\ddots&\vdots\\\
\varphi_{1}^{(n-1)}&\varphi_{2}^{(n-1)}&\ldots&\varphi_{n}^{(n-1)}\end{matrix}\right|,\qquad\varphi_{j}^{(k)}=\frac{{\rm
d}^{k}{\varphi_{j}}}{{{\rm d}{x}}^{k}},$

and

$\varphi_{m}(x;\mu)=L_{2m-1}^{(\mu-2m+1)}(-x),$

with $L_{k}^{(\alpha)}(x)$ the Laguerre polynomial, for details see Kajiwara
and Masuda [30]; see also [10, 29].

  3. (ii)

Rational solutions of PIII (2.1) are expressed in terms of Umemura
polynomials. For example,

$w_{n}(z;\mu)=1+\frac{{\rm d}{}}{{{\rm
d}{z}}}\ln\left\\{\frac{S_{n-1}(z;\mu-1)}{S_{n}(z;\mu)}\right\\},$ (3.4)

satisfies (2.1) for the parameters

$A=2n+2\mu-1,\qquad B=2n-2\mu+1,$

for details see [10, 29, 30].

To describe rational solutions of deg-PV (1.1), it is more convenient to use
rational solutions of SIII (2.13), which involve one Umemura polynomial and
are discussed in the following theorem, whereas rational solutions of PIII
(2.1) which involve two Umemura polynomials, as shown in (3.4).

######  Theorem 3.4.

The rational function solution of SIII (2.13) is given by

$\sigma_{n}(x;\mu,\varepsilon)=2x\frac{{\rm d}{}}{{{\rm d}{x}}}\left\\{\ln
S_{n}(x;\mu)\right\\}-\tfrac{1}{2}x^{2}-2\mu x-\tfrac{1}{4},\qquad n\geq 0,$
(3.5a) with $S_{n}(x;\mu)$ the Umemura polynomial, for the parameters
$a=n+\tfrac{1}{2},\qquad b=\mu,\qquad\varepsilon=1.$ (3.5b)

###### Proof.

See Clarkson [10]. ∎

###  3.2 Special function solutions of PIII and SIII

Special function solutions of PIII (2.1), which are expressed in terms of
Bessel functions and are classified in the following Theorem.

######  Theorem 3.5.

Equation (2.1) has solutions expressible in terms of the Riccati equation

$x\frac{{\rm d}{q}}{{{\rm
d}{x}}}=\varepsilon_{1}xq^{2}+(A\varepsilon_{1}-1)q+\varepsilon_{2}x,$ (3.6)

if and only if

$\varepsilon_{1}A+\varepsilon_{2}B=4n+2,$ (3.7)

with $n\in\mathbb{Z}$ and $\varepsilon_{1}^{2}=1$, $\varepsilon_{2}^{2}=1$,
independently. Further, the Riccati equation (3.6) has the solution

$q(x)=-{\varepsilon_{1}}\,\frac{{\rm d}{}}{{{\rm d}{x}}}\ln\psi_{\nu}(x),$
(3.8)

where $\psi_{\nu}(x)$ satisfies

$x\frac{{\rm d}^{2}{\psi_{\nu}}}{{{\rm
d}{x}}^{2}}+(1-2\varepsilon_{1}\nu)\frac{{\rm d}{\psi_{\nu}}}{{{\rm
d}{x}}}+\varepsilon_{1}\varepsilon_{2}x\psi_{\nu}=0,$ (3.9)

which has solution

$\psi_{\nu}(x)=\begin{cases}x^{\nu}\left\\{C_{1}J_{\nu}(x)+C_{2}Y_{\nu}(x)\right\\},&\mbox{\rm
if}\quad\varepsilon_{1}=1,\phantom{-}\enskip\varepsilon_{2}=1,\\\
x^{-\nu}\left\\{C_{1}J_{\nu}(x)+C_{2}Y_{\nu}(x)\right\\},&\mbox{\rm
if}\quad\varepsilon_{1}=-1,\enskip\varepsilon_{2}=-1,\\\
x^{\nu}\left\\{C_{1}I_{\nu}(x)+C_{2}K_{\nu}(x)\right\\},&\mbox{\rm
if}\quad\varepsilon_{1}=1,\phantom{-}\enskip\varepsilon_{2}=-1,\\\
x^{-\nu}\left\\{C_{1}I_{\nu}(x)+C_{2}K_{\nu}(x)\right\\},\quad&\mbox{\rm
if}\quad\varepsilon_{1}=-1,\enskip\varepsilon_{2}=1,\end{cases}$ (3.10)

with $C_{1}$, $C_{2}$ arbitrary constants, and $J_{\nu}(x)$, $Y_{\nu}(x)$,
$I_{\nu}(x)$, $K_{\nu}(x)$ Bessel functions.

###### Proof.

For details see Okamoto [44]; see also [12, 22, 35, 37, 38].∎

Determinantal representations of special function solutions of PIII (2.1) were
given by Okamoto [44]; see also [19, 36]. As for rational solutions, to
describe special function solutions of deg-PV (1.1), it is more convenient to
use special function solutions of SIII (2.13), which are discussed in the
following theorem.

######  Theorem 3.6.

Suppose $\tau_{n}(x;\mu,\varepsilon)$ is the determinant given by

$\tau_{n}(x;\mu,\varepsilon)=\det\left[\left(x\frac{{\rm d}{}}{{{\rm
d}{x}}}\right)^{j+k}\varphi_{\mu}(x;\varepsilon)\right]_{j,k=0}^{n-1},$
(3.11a) where
$\varphi_{\mu}(x;\varepsilon)=\begin{cases}c_{1}J_{\mu}(x)+c_{2}Y_{\mu}(x),&\quad\text{if}\quad\varepsilon=1,\\\
c_{1}I_{\mu}(x)+c_{2}K_{\mu}(x),&\quad\text{if}\quad\varepsilon=-1,\end{cases}$
(3.11b)

with $c_{1}$, $c_{2}$ arbitrary constants, and $J_{\mu}(z)$, $Y_{\mu}(z)$,
$I_{\mu}(z)$, $K_{\mu}(z)$ Bessel functions.

The Bessel function solution of SIII (2.13) is given by

$\sigma_{n}(x;\mu,\varepsilon)=2x\frac{{\rm d}{}}{{{\rm
d}{x}}}\left\\{\ln\tau_{n}(x;\mu,\varepsilon)\right\\}+\tfrac{1}{2}\varepsilon
x^{2}+\mu^{2}-n^{2}+2n,$ (3.12a) for the parameters $a=n,\qquad b=\mu.$
(3.12b)

######  Lemma 3.7.

The determinant $\tau_{n}(x;\mu,\varepsilon)$ given by (3.11) satisfies the
equation

$x^{2}\left\\{\tau_{n}\frac{{\rm d}^{2}{\tau_{n}}}{{{\rm
d}{x}}^{2}}-\left(\frac{{\rm d}{\tau_{n}}}{{{\rm
d}{x}}}\right)^{2}\right\\}+x\tau_{n}\frac{{\rm d}{\tau_{n}}}{{{\rm
d}{x}}}=\tau_{n+1}\tau_{n-1},$ (3.13)

or equivalently the Toda equation

$\left(x\frac{{\rm d}{}}{{{\rm
d}{x}}}\right)^{\\!2}\ln\tau_{n}=\frac{\tau_{n+1}\tau_{n-1}}{\tau_{n}^{2}}.$
(3.14)

###### Proof.

See Okamoto [44, Theorem 2]. ∎

##  4 Bäcklund transformations

A Bäcklund transformation relates the solution of a Painlevé equation either
to another solution of the same equation with different values of the
parameters, or to another Painlevé equation. All Painlevé equations, except
the first Painlevé equation, have Bäcklund transformations. Hierarchies of
classical solutions of the Painlevé equations can be obtained by applying
Bäcklund transformations to a “seed solution”.

Let $w_{j}(z_{j}z;\alpha_{j},\beta_{j},\gamma_{j})$, $j=0,1,2$, be solutions
of deg-PV (1.1) with

$\begin{array}[]{ll}z_{1}=-z_{0},&({\alpha_{1}},{\beta_{1}},{\gamma_{1}})=(\alpha_{0},\beta_{0},-\gamma_{0}),\\\\[2.5pt]
z_{2}=z_{0},&({\alpha_{2}},{\beta_{2}},{\gamma_{2}})=(-\beta_{0},-\alpha_{0},-\gamma_{0}).\end{array}$

Then deg-PV (1.1) has the symmetries

$\displaystyle\mathcal{S}_{1}:\qquad$
$\displaystyle{w_{1}}({z_{1}})={w_{0}(-z_{0})},$ (4.1)
$\displaystyle\mathcal{S}_{2}:\qquad$
$\displaystyle{w_{2}}({z_{2}})={1}/{w_{0}(z_{0})}.$ (4.2)

######  Theorem 4.1.

Suppose that $W_{0}=w(z;\alpha,\beta,\gamma)$ satisfies deg-PV (1.1) with
parameters

$\alpha=\tfrac{1}{2}a^{2},\qquad\beta=-\tfrac{1}{2}b^{2},\qquad\gamma=c.$

Let $W_{j}=w(z;\alpha_{j},\beta_{j},\gamma_{j})$, $j=1,2,3,4$, be solutions of
(1.1) with parameters

$\begin{array}[]{l@{\qquad}l@{\qquad}l}\alpha_{1}=\tfrac{1}{2}(a+1)^{2},&\beta_{1}=-\tfrac{1}{2}b^{2},&\gamma_{1}=c,\\\\[2.5pt]
\alpha_{2}=\tfrac{1}{2}(a-1)^{2},&\beta_{2}=-\tfrac{1}{2}b^{2},&\gamma_{2}=c,\\\\[2.5pt]
\alpha_{3}=\tfrac{1}{2}a^{2},&\beta_{3}=-\tfrac{1}{2}(b+1)^{2},&\gamma_{3}=c,\\\\[2.5pt]
\alpha_{4}=\tfrac{1}{2}a^{2},&\beta_{4}=-\tfrac{1}{2}(b-1)^{2},&\gamma_{4}=c,\end{array}$

respectively. Then these solutions can be obtained from $W_{0}$ as follows

$\displaystyle\mathcal{W}_{1}:\quad W_{1}$
$\displaystyle=\frac{\displaystyle\left\\{zW_{0}^{\prime}+(W_{0}-1)(aW_{0}-b)\right\\}\left\\{zW_{0}^{\prime}+(W_{0}-1)(aW_{0}+b)\right\\}}{\displaystyle
z^{2}(W_{0}^{\prime})^{2}+2azW_{0}(W_{0}-1)W_{0}^{\prime}+2cz^{2}W_{0}(W_{0}-1)+(W_{0}-1)^{2}(a^{2}W_{0}^{2}-b^{2})},$
(4.3a) $\displaystyle\mathcal{W}_{2}:\quad W_{2}$
$\displaystyle=\frac{\displaystyle\left\\{zW_{0}^{\prime}-(W_{0}-1)(aW_{0}-b)\right\\}\left\\{zW_{0}^{\prime}-(W_{0}-1)(aW_{0}+b)\right\\}}{\displaystyle
z^{2}(W_{0}^{\prime})^{2}-2azW_{0}(W_{0}-1)W_{0}^{\prime}+2cz^{2}W_{0}(W_{0}-1)+(W_{0}-1)^{2}(a^{2}W_{0}^{2}-b^{2})},$
(4.3b) $\displaystyle\mathcal{W}_{3}:\quad W_{3}$
$\displaystyle=\frac{\displaystyle
z^{2}(W_{0}^{\prime})^{2}+2bz(W_{0}-1)W_{0}^{\prime}+2cz^{2}W_{0}^{2}(W_{0}-1)-(W_{0}-1)^{2}(a^{2}W_{0}^{2}-b^{2})}{\displaystyle\left\\{zW_{0}^{\prime}-(W_{0}-1)(aW_{0}-b)\right\\}\left\\{zW_{0}^{\prime}+(W_{0}-1)(aW_{0}+b)\right\\}},$
(4.3c) $\displaystyle\mathcal{W}_{4}:\quad W_{4}$
$\displaystyle=\frac{\displaystyle
z^{2}(W_{0}^{\prime})^{2}-2bz(W_{0}-1)W_{0}^{\prime}+2cz^{2}W_{0}^{2}(W_{0}-1)-(W_{0}-1)^{2}(a^{2}W_{0}^{2}-b^{2})}{\displaystyle\left\\{zW_{0}^{\prime}-(W_{0}-1)(aW_{0}-b)\right\\}\left\\{zW_{0}^{\prime}+(W_{0}-1)(aW_{0}+b)\right\\}}.$
(4.3d)

###### Proof.

See Adler [3]; also Filipuk and Van Assche [18]. ∎

##  5 Classical solutions of deg-PV

To discuss classical solutions of deg-PV (1.1), it is convenient to make the
transformation

$w(z)=u(x),\qquad z=\tfrac{1}{2}x^{2},$ (5.1)

in (1.1), which gives

$\frac{{\rm d}^{2}{u}}{{{\rm
d}{x}}^{2}}=\left(\frac{1}{2u}+\frac{1}{u-1}\right)\left(\frac{{\rm
d}{u}}{{{\rm d}{x}}}\right)^{\\!2}-\frac{1}{x}\frac{{\rm d}{u}}{{{\rm
d}{x}}}+\frac{4(u-1)^{2}(\alpha u^{2}+\beta)}{x^{2}u}+{2\gamma u}.$ (5.2)

We could fix the parameter $\gamma$ in (5.2), by rescaling $x$ if necessary,
but it is more convenient not to do so. Instead classical solutions will be
classified for $\gamma=\pm 1$. From Corollary 2.3 and (5.1), we have that if
$\sigma(x;a,b,\varepsilon)$ is a solution of SIII (2.13), then

$u(x;a,b,\varepsilon)=\frac{\sigma^{\prime}(x;a,b,\varepsilon)+\varepsilon
x}{\sigma^{\prime}(x;a,b,\varepsilon)-\varepsilon x},$ (5.3)

is a solution of (5.2) with $\gamma=\varepsilon$. As remarked above, it is
simpler to derive classical solutions of deg-PV (1.1) from SIII rather than
PIII, compare equations (LABEL:PV0-SIII) and (2.10).

######  Theorem 5.1.

Supppose that $u_{0}=u(x;\alpha,\beta,\gamma)$ satisfies (5.2) with parameters

$\alpha=\tfrac{1}{2}a^{2},\qquad\beta=-\tfrac{1}{2}b^{2},\qquad\gamma=c.$

Let $u_{j}=u(x;\alpha_{j},\beta_{j},\gamma_{j})$, $j=1.2.3,4$ be solutions of
(5.2) with parameters

$\begin{array}[]{l@{\qquad}l@{\qquad}l}\alpha_{1}=\tfrac{1}{2}(a+1)^{2},&\beta_{1}=-\tfrac{1}{2}b^{2},&\gamma_{1}=c,\\\\[2.5pt]
\alpha_{2}=\tfrac{1}{2}(a-1)^{2},&\beta_{2}=-\tfrac{1}{2}b^{2},&\gamma_{2}=c,\\\\[2.5pt]
\alpha_{3}=\tfrac{1}{2}a^{2},&\beta_{3}=-\tfrac{1}{2}(b+1)^{2},&\gamma_{3}=c,\\\\[2.5pt]
\alpha_{4}=\tfrac{1}{2}a^{2},&\beta_{4}=-\tfrac{1}{2}(b-1)^{2},&\gamma_{4}=c,\end{array}$

respectively. Then these can be obtained from $u_{0}$ as follows

$\displaystyle\mathcal{U}_{1}:\quad u_{1}$
$\displaystyle=\frac{\displaystyle\left\\{xu_{0}^{\prime}+2(u_{0}-1)(au_{0}-b)\right\\}\left\\{xu_{0}^{\prime}+2(u_{0}-1)(au_{0}+b)\right\\}}{\displaystyle
x^{2}(u_{0}^{\prime})^{2}+4axu_{0}(u_{0}-1)u_{0}^{\prime}+4cu_{0}(u_{0}-1)x^{2}+4(u_{0}-1)^{2}(a^{2}u_{0}^{2}-b^{2})},$
(5.4a) $\displaystyle\mathcal{U}_{2}:\quad u_{2}$
$\displaystyle=\frac{\displaystyle\left\\{xu_{0}^{\prime}-2(u_{0}-1)(au_{0}-b)\right\\}\left\\{xu_{0}^{\prime}-2(u_{0}-1)(au_{0}+b)\right\\}}{\displaystyle
x^{2}(u_{0}^{\prime})^{2}-4axu_{0}(u_{0}-1)u_{0}^{\prime}+4cu_{0}(u_{0}-1)x^{2}+4(u_{0}-1)^{2}(a^{2}u_{0}^{2}-b^{2})},$
(5.4b) $\displaystyle\mathcal{U}_{3}:\quad u_{3}$
$\displaystyle=\frac{\displaystyle
x^{2}(u_{0}^{\prime})^{2}+4bx(u_{0}-1)u_{0}^{\prime}+4cx^{2}u_{0}^{2}(u_{0}-1)-4(u_{0}-1)^{2}(a^{2}u_{0}^{2}-b^{2})}{\displaystyle\left\\{xu_{0}^{\prime}-2(u_{0}-1)(au_{0}-b)\right\\}\left\\{xu_{0}^{\prime}+2(u_{0}-1)(au_{0}+b)\right\\}},$
(5.4c) $\displaystyle\mathcal{U}_{4}:\quad u_{4}$
$\displaystyle=\frac{\displaystyle
x^{2}(u_{0}^{\prime})^{2}-4bx(u_{0}-1)u_{0}^{\prime}+4cx^{2}u_{0}^{2}(u_{0}-1)-4(u_{0}-1)^{2}(a^{2}u_{0}^{2}-b^{2})}{\displaystyle\left\\{xu_{0}^{\prime}-2(u_{0}-1)(au_{0}-b)\right\\}\left\\{xu_{0}^{\prime}+2(u_{0}-1)(au_{0}+b)\right\\}}.$
(5.4d)

satisfy (1.1) with parameters

$\begin{array}[]{l@{\qquad}l@{\qquad}l}\alpha_{1}=\tfrac{1}{2}(a+1)^{2},&\beta_{1}=-\tfrac{1}{2}b^{2},&\gamma_{1}=c,\\\\[2.5pt]
\alpha_{2}=\tfrac{1}{2}(a-1)^{2},&\beta_{2}=-\tfrac{1}{2}b^{2},&\gamma_{2}=c,\\\\[2.5pt]
\alpha_{3}=\tfrac{1}{2}a^{2},&\beta_{3}=-\tfrac{1}{2}(b+1)^{2},&\gamma_{3}=c,\\\\[2.5pt]
\alpha_{4}=\tfrac{1}{2}a^{2},&\beta_{4}=-\tfrac{1}{2}(b-1)^{2},&\gamma_{4}=c,\end{array}$

respectively.

###### Proof.

This is easily proved by applying (5.1) to the Bäcklund transformations in
Theorem 4.1. ∎

###  5.1 Algebraic solutions

Since deg-PV (1.1) and equation (5.2) are related by the transformation (5.1)
then algebraic solutions of deg-PV (1.1), which are rational functions of
$\sqrt{z}$, are equivalent to rational solutions of (5.2), which are rational
functions of $x$. Therefore we discuss rational solutions of (5.2), which are
classified in the following Theorem.

######  Theorem 5.2.

Necessary and sufficient conditions for the existence of rational solutions of
(5.2) are either

$(\alpha,\beta,\gamma)=\left(\tfrac{1}{2}(n+\tfrac{1}{2}),-\tfrac{1}{2}\mu^{2},1\right),$
(5.5)

or

$(\alpha,\beta,\gamma)=\left(\tfrac{1}{2}\mu^{2},-\tfrac{1}{2}(n+\tfrac{1}{2}),-1\right),$
(5.6)

where $n\in\mathbb{Z}$ and $\mu$ is an arbitrary constant.

###### Proof.

For details see Gromak, Laine and Shimomura [22, §38]; see also [37, 38]. ∎

We remark that the solutions of (5.2) satisfying (5.5) are related to those
satisfying (5.6) through the analog of the symmetry (4.2). Consequently we
shall be concerned only with rational solutions of (5.2) for the parameters
given by (5.5).

######  Theorem 5.3.

The rational solution of (5.2) for the parameters (5.5) is given by

$u_{n}(x;\mu)=1-\frac{xS_{n}^{2}(x;\mu)}{S_{n+1}(x;\mu)S_{n-1}(x;\mu)},\qquad
n\geq 0,$ (5.7)

where $S_{n}(x;\mu)$ is the Umemura polynomial (3.3).

###### Proof.

Substituting the rational solution of SIII (2.13) given by (3.5) into (5.3)
and then using the reccurence relation (3.2) gives the result. ∎

######  Remark 5.4.

The Umemura polynomial $S_{n}(x;\mu)$ satisfies the difference equation

$S_{n+1}(x;\mu)S_{n-1}(x;\mu)=xS_{n}^{2}(x;\mu)+\mu
S_{n}(x;\mu+1)\,S_{n}(x;\mu-1).$ (5.8)

Hence from (5.7) there are two alternative representations of the rational
solution

$\begin{split}u_{n}(x;\mu)&=\frac{\mu S_{n}(x;\mu+1)\,S_{n}(x;\mu-1)}{\mu
S_{n}(x;\mu+1)\,S_{n}(x;\mu-1)+xS_{n}^{2}(x;\mu)},\qquad
u_{n}(x;\mu)=\frac{\mu
S_{n}(x;\mu+1)\,S_{n}(x;\mu-1)}{S_{n+1}(x;\mu)S_{n-1}(x;\mu)}.\end{split}$

###  5.2 Bessel function solutions

######  Theorem 5.5.

Necessary and sufficient conditions for the existence of Bessel function
solutions of (5.2) are either

$(\alpha,\beta,\gamma)=\left(\tfrac{1}{2}n^{2},-\tfrac{1}{2}\mu^{2},\varepsilon\right),$
(5.9)

or

$(\alpha,\beta,\gamma)=\left(\tfrac{1}{2}\mu^{2},-\tfrac{1}{2}n^{2},-\varepsilon\right),$
(5.10)

with $\varepsilon=\pm 1$, and where $n\in\mathbb{Z}^{+}$ and $\mu$ is an
arbitrary constant.

###### Proof.

From (2.5) and (2.9), the parameters in PIII (2.1) and deg-PV (5.2) are given
by

$(A,B)=\big{(}2(a-b),2\varepsilon(a+b+1)\big{)},\qquad(\alpha,\beta,\gamma)=(\tfrac{1}{2}a^{2},-\tfrac{1}{2}b^{2},\varepsilon),$

respectively, for parameters $a$, $b$ and $\varepsilon$. The result then
follows from Theorem 3.5. ∎

######  Theorem 5.6.

The Bessel function solution of (5.2) for the parameters

$(\alpha,\beta,\gamma)=\left(\tfrac{1}{2}n^{2},-\tfrac{1}{2}\mu^{2},\varepsilon\right),$

is given by

$u_{n}(x;\mu,\varepsilon)=1+\frac{\varepsilon
x^{2}\tau_{n}^{2}(x;\mu,\varepsilon)}{\tau_{n+1}(x;\mu,\varepsilon)\,\tau_{n-1}(x;\mu,\varepsilon)},\qquad
n\geq 1,$ (5.11)

where

$\tau_{n}(x;\mu,\varepsilon)=\det\left[\left(x\frac{{\rm d}{}}{{{\rm
d}{x}}}\right)^{\\!j+k}\varphi_{\mu}(x;\varepsilon)\right]_{j,k=0}^{n-1},$
(5.12)

and $\tau_{0}(x;\mu,\varepsilon)=1$, with

$\varphi_{\mu}(x;\varepsilon)=\begin{cases}c_{1}J_{\mu}(x)+c_{2}Y_{\mu}(x),&\quad\text{if}\quad\varepsilon=1,\\\
c_{1}I_{\mu}(x)+c_{2}K_{\mu}(x),&\quad\text{if}\quad\varepsilon=-1,\end{cases}$
(5.13)

$c_{1}$ and $c_{2}$ arbitrary constants, and $J_{\mu}(x)$, $Y_{\mu}(x)$,
$I_{\mu}(x)$ and $K_{\mu}(x)$ Bessel functions.

###### Proof.

Substituting the Bessel function solution of SIII (2.13) given by (3.12) into
(5.3) and then using (3.13) gives the result. ∎

######  Corollary 5.7.

The Bessel function solution of (5.2) for the parameters

$(\alpha,\beta,\gamma)=\left(\tfrac{1}{2}n^{2},-\tfrac{1}{2}\mu^{2},2\varepsilon\right),$

is given by

$w_{n}(z;\mu,\varepsilon)=1+\frac{\varepsilon
z\mathcal{T}_{n}^{2}(z;\mu,\varepsilon)}{\mathcal{T}_{n+1}(z;\mu,\varepsilon)\,\mathcal{T}_{n-1}(z;\mu,\varepsilon)},\qquad
n\geq 1,$ (5.14)

where

$\mathcal{T}_{n}(z;\mu,\varepsilon)=\det\left[\left(z\frac{{\rm d}{}}{{{\rm
d}{z}}}\right)^{\\!j+k}\psi_{\mu}(z;\varepsilon)\right]_{j,k=0}^{n-1},$ (5.15)

and $\mathcal{T}_{0}(z;\mu,\varepsilon)=1$, with

$\varphi_{\mu}(z;\varepsilon)=\begin{cases}c_{1}J_{\mu}(2\sqrt{z})+c_{2}Y_{\mu}(2\sqrt{z}),&\quad\text{if}\quad\varepsilon=1,\\\
c_{1}I_{\mu}(2\sqrt{z})+c_{2}K_{\mu}(2\sqrt{z}),&\quad\text{if}\quad\varepsilon=-1,\end{cases}$
(5.16)

$c_{1}$ and $c_{2}$ arbitrary constants, and $J_{\mu}(x)$, $Y_{\mu}(x)$,
$I_{\mu}(x)$ and $K_{\mu}(x)$ Bessel functions.

In the next Lemma, it is shown that the first solution
$u_{1}(x;\mu,\varepsilon)$, the “seed solution”, satisfies a first-order,
second-degree equation.

######  Lemma 5.8.

The solution of (5.2) for the parameters

$(\alpha,\beta,\gamma)=\left(\tfrac{1}{2},-\tfrac{1}{2}\mu^{2},\varepsilon\right),$

is

$u_{1}(x;\mu,\varepsilon)=\frac{\varphi_{\mu+1}(x;\varepsilon)\left[x\varphi_{\mu+1}(x;\varepsilon)-2\varepsilon\mu\varphi_{\mu}(x;\varepsilon)\right]}{x\varphi_{\mu+1}^{2}(x;\varepsilon)-2\varepsilon\mu\varphi_{\mu+1}(x;\varepsilon)\varphi_{\mu}(x;\varepsilon)+\varepsilon
x\varphi_{\mu}^{2}(x;\varepsilon)},$ (5.17)

where

$\varphi_{\mu}(x;\varepsilon)=\begin{cases}c_{1}J_{\mu}(x)+c_{2}Y_{\mu}(x),&\quad\text{if}\quad\varepsilon=1,\\\
c_{1}I_{\mu}(x)+c_{2}K_{\mu}(x),&\quad\text{if}\quad\varepsilon=-1,\end{cases}$

with $c_{1}$ and $c_{2}$ constants, satisfies the first-order, second-degree
equation

$x^{2}\left(\frac{{\rm d}{u}}{{{\rm d}{x}}}\right)^{\\!2}-4xu(u-1)\frac{{\rm
d}{u}}{{{\rm d}{x}}}+4\varepsilon x^{2}u(u-1)+4(u-1)^{2}(u^{2}-\mu^{2})=0.$
(5.18)

###### Proof.

Define

$\Phi_{\mu}(x;\varepsilon)=\frac{\varphi_{\mu+1}(x;\varepsilon)}{\varphi_{\mu}(x;\varepsilon)},$

then from (5.17)

$u_{1}(x;\mu,\varepsilon)=1-\frac{x}{\varepsilon
x\Phi_{\mu}^{2}-2\mu\Phi_{\mu}+x},$ (5.19)

and $\Phi_{\mu}(x;\varepsilon)$ satisfies the Riccati equation

$x\frac{{\rm d}{\Phi_{\mu}}}{{{\rm d}{x}}}=\varepsilon
x\Phi_{\mu}^{2}-(2\mu+1)\Phi_{\mu}+x.$ (5.20)

Next we assume that $u_{1}(x;\mu,\varepsilon)$ satisfies a first-order,
second-degree equation of the form

$x^{2}\left(\frac{{\rm d}{u}}{{{\rm
d}{x}}}\right)^{\\!2}+x\left[f_{2}(x,\mu,\varepsilon)u^{2}+f_{1}(x,\mu,\varepsilon)u+f_{0}(x,\mu,\varepsilon)\right]\frac{{\rm
d}{u}}{{{\rm d}{x}}}+\sum_{j=0}^{4}g_{j}(x,\mu,\varepsilon)u^{j}=0,$ (5.21)

where $\left\\{f_{j}(x,\mu,\varepsilon)\right\\}_{j=0}^{2}$ and
$\left\\{g_{j}(x,\mu,\varepsilon)\right\\}_{j=0}^{4}$ are to be determined.
Then substituting (5.19) into (5.21), using the fact that
$\Phi_{\mu}(x;\varepsilon)$ satisfies (5.20) and equating coefficients of
powers of $\Phi_{\mu}$ yields

$\displaystyle f_{2}=-4,$ $\displaystyle f_{1}=4,$ $\displaystyle f_{0}=0,$
$\displaystyle g_{4}=4,$ $\displaystyle g_{3}=-8,$ $\displaystyle
g_{2}=4\varepsilon x^{2}-4\mu^{2}+4,$ $\displaystyle g_{1}=-4\varepsilon
x^{2}+8\mu^{2},$ $\displaystyle g_{0}=-4\mu^{2}.$

Hence we obtain equation (5.18), as required. ∎

This demonstrates that special function solutions of (5.2), and hence also
deg-PV (1.1), are different from special function solutions of PII–PVI where
the “seed solution” satisfies a Riccati equation, a first-order, first-degree
equation.

######  Remark 5.9.

Gromak, Laine and Shimomura [22, equation (38.7)] give, without proof, a
first-order, second-degree equation associated with Bessel function solutions
of deg-PV (1.1); see also Filipuk and Van Assche [18, §2.3].

##  6 Applications

###  6.1 Complex sine-Gordon equation

Consider the two-dimensional complex sine-Gordon equation

$\nabla^{2}\psi+\frac{(\nabla\psi)^{2}\overline{\psi}}{1-|\psi|^{2}}+\psi\left(1-|\psi|^{2}\right)=0,$
(6.1)

where $\nabla\psi=(\psi_{x},\psi_{y})$. Making the transformation

$\psi(x,y)=\cos(\varphi(x,y))\exp\\{{\rm
i}\eta(x,y)\\},\qquad\overline{\psi}(x,y)=\cos(\varphi(x,y))\exp\\{-{\rm
i}\eta(x,y)\\},$

in the complex sine-Gordon equation (6.1) yields

$\displaystyle\nabla^{2}\varphi+\frac{\cos\varphi}{\sin^{3}\varphi}(\nabla\eta)^{2}-\tfrac{1}{2}\sin(2\varphi)=0,$
$\displaystyle\sin(2\varphi)\,\nabla^{2}\eta=4(\varphi_{x}\eta_{x}+\varphi_{y}\eta_{y}),$

which is the Pohlmeyer-Lund-Regge model [32, 33, 48].

The complex sine-Gordon equation (6.1) has a separable solution in polar
coordinates given by $\psi(r,\theta)=R_{n}(r)\,{\rm e}^{{\rm i}n\theta}$,
where $R_{n}(r)$ satisfies

$\frac{{\rm d}^{2}{R_{n}}}{{{\rm d}{r}}^{2}}+\frac{1}{r}\,\frac{{\rm
d}{R_{n}}}{{{\rm d}{r}}}+\frac{R_{n}}{1-R_{n}^{2}}\left\\{\left(\frac{{\rm
d}{R_{n}}}{{{\rm
d}{r}}}\right)^{\\!2}-\frac{n^{2}}{r^{2}}\right\\}+R_{n}\big{(}1-R_{n}^{2}\big{)}=0,$
(6.2)

We remark that this equation also arises in extended quantum systems [5, 6,
7], in relativity [20] and in coefficients in the three-term recurrence
relation for orthogonal polynomials with respect to the weight $w(\theta)={\rm
e}^{t\cos\theta}$ on the unit circle, see [54, equation (3.13)]. The
orthogonal polynomials for this weight on the unit circle are related to
unitary random matrices [47].

Equation (6.2) can be shown to possess the Painlevé property, though it is not
in the list of 50 equations given in [25, Chapter 14]. Equation (6.2) can be
transformed to the fifth Painlevé equation (1.2) in two different ways.

  1. (i)

If $R_{n}(r)$ satisfies (6.2) then making the transformation

$R_{n}(r)=\frac{1+u_{n}(z)}{1-u_{n}(z)},\qquad r=\tfrac{1}{2}z,$ (6.3)

yields

$\frac{{\rm d}^{2}{u_{n}}}{{{\rm
d}{z}}^{2}}=\left(\frac{1}{2u_{n}}+\frac{1}{u_{n}-1}\right)\left(\frac{{\rm
d}{u_{n}}}{{{\rm d}{z}}}\right)^{\\!2}-\frac{1}{z}\,\frac{{\rm
d}{u_{n}}}{{{\rm
d}{z}}}+\frac{n^{2}(u_{n}-1)^{2}(u_{n}^{2}-1)}{8z^{2}u_{n}}-\frac{u_{n}(u_{n}+1)}{2(u_{n}-1)},$
(6.4)

which is PV (1.2) with $\alpha=\tfrac{1}{8}n^{2}$, $\beta=-\tfrac{1}{8}n^{2}$,
$\gamma=0$ and $\delta=-\tfrac{1}{2}$.

  2. (ii)

If $R_{n}(r)$ satisfies (6.2) then making the transformation

$R_{n}(r)=\frac{1}{\sqrt{1-v_{n}(x)}},\qquad r=\sqrt{x},$ (6.5)

yields

$\frac{{\rm d}^{2}{v_{n}}}{{{\rm
d}{x}}^{2}}=\left(\frac{1}{2v_{n}}+\frac{1}{v_{n}-1}\right)\left(\frac{{\rm
d}{v_{n}}}{{{\rm d}{x}}}\right)^{\\!2}-\frac{1}{x}\,\frac{{\rm
d}{v_{n}}}{{{\rm
d}{x}}}-\frac{n^{2}(v_{n}-1)^{2}}{2x^{2}v_{n}}+\frac{v_{n}}{2x},$ (6.6)

which is deg-PV (1.1) with $\alpha=0$, $\beta=-\tfrac{1}{2}n^{2}$ and
$\gamma=\tfrac{1}{2}$ so is equivalent to PIII (2.1), as mentioned above.

This shows that solutions of equations (6.4) and (6.6) are related by

$v_{n}(x)=\frac{4u_{n}(z)}{1+u^{2}_{n}(z)},\qquad x=\tfrac{1}{4}z^{2}.$

The function $R_{n}(r)$ satisfies the ordinary differential equation (6.2),
the differential-difference equations

$\displaystyle\frac{{\rm d}{R_{n}}}{{{\rm
d}{r}}}+\frac{n}{r}R_{n}-\big{(}1-R_{n}^{2}\big{)}R_{n-1}=0,$ (6.7a)
$\displaystyle\frac{{\rm d}{R_{n-1}}}{{{\rm
d}{r}}}-\frac{n-1}{r}R_{n-1}+\big{(}1-R_{n-1}^{2}\big{)}R_{n}=0,$ (6.7b)

since solving (6.7a) for $R_{n-1}(r)$ and substituting in (6.7b) yields
equation (6.2). Also eliminating the derivatives in (6.7), after letting $n\to
n+1$ in (6.7b), yields the difference equation

$R_{n+1}+R_{n-1}={\frac{2n}{r}\,\frac{R_{n}}{1-R_{n}^{2}}},$ (6.8)

which is known as the discrete Painlevé II equation [39, 47].

If $n=1$ then equations (6.7) have the solution

$R_{0}(r)=1,\qquad
R_{1}(r)=\frac{C_{1}I_{1}(r)-C_{2}K_{1}(r)}{C_{1}I_{0}(r)+C_{2}K_{0}(r)},$

where $I_{0}(r)$, $K_{0}(r)$, $I_{1}(r)$ and $K_{1}(r)$ are the imaginary
Bessel functions and $C_{1}$ and $C_{2}$ are arbitrary constants. For
solutions which are bounded at $r=0$ then necesssarily $C_{2}=0$ and so

$R_{0}(r)=1,\qquad R_{1}(r)=\frac{I_{1}(r)}{I_{0}(r)}.$ (6.9)

Hence one can use the difference equation (6.8) to determine $R_{n}(r)$, for
$n\geq 2$, which yields

$\displaystyle R_{2}(r)$
$\displaystyle=-\frac{rR_{1}^{2}(r)+2R_{1}(r)-r}{r\left[R_{1}^{2}(r)-1\right]},$
$\displaystyle R_{3}(r)$
$\displaystyle=\frac{R_{1}^{3}(r)-rR_{1}^{2}(r)-2R_{1}(r)+r}{R_{1}(r)\left[rR_{1}^{2}(r)+R_{1}(r)-r\right]},$
$\displaystyle R_{4}(r)$
$\displaystyle=\frac{r(r^{2}+5)R_{1}^{4}(r)+4R_{1}^{3}(r)-2r(r^{2}+3)R_{1}^{2}(r)+r^{3}}{r\left[(r^{2}-1)R_{1}^{4}(r)+4rR_{1}^{3}(r)-2(r^{2}+2)R_{1}^{2}(r)-4rR_{1}(r)+r^{2}\right]}.$

These results suggest that (6.2) should be solvable in terms of PIII (2.1),
which is illustrated in the following theorem.

######  Theorem 6.1.

If $R_{n}(r)$ satisfies (6.2) then $\displaystyle
w_{n}(r)={R_{n+1}(r)}/{R_{n}(r)}$ satisfies

$\frac{{\rm d}^{2}{w_{n}}}{{{\rm d}{r}}^{2}}=\frac{1}{w_{n}}\left(\frac{{\rm
d}{w_{n}}}{{{\rm d}{r}}}\right)^{\\!2}-\frac{1}{r}\,\frac{{\rm
d}{w_{n}}}{{{\rm
d}{r}}}-\frac{2n}{r}w_{n}^{2}+\frac{2(n+1)}{r}+w_{n}^{3}-\frac{1}{w_{n}},$
(6.10)

which is PIII (2.1) with parameters $\alpha=-2n$ and $\beta=2(n+1)$.

###### Proof.

See Hisakado [23] and Tracy & Widom [50]; see also [54, §3.1]. ∎

We note that since the parameters in (6.10) satisfy $-\alpha+\beta=4n+2$, with
$n\in\mathbb{Z}^{+}$, then the equation has solutions expressible in terms of
the modified Bessel functions $I_{0}(r)$ and $I_{1}(r)$ (as well as $K_{0}(r)$
and $K_{1}(r)$, but these are not needed here).

######  Theorem 6.2.

Let $\tau_{n}(r;\nu)$ be the $n\times n$ determinant

$\tau_{n}(r;\nu)=\det\left[\left(r\frac{{\rm d}{}}{{{\rm
d}{r}}}\right)^{j+k}I_{\nu}(r)\right]_{j,k=0}^{n-1},$ (6.11)

with $I_{\nu}(r)$ the modified Bessel function, then

$w_{n}(r;\nu)=\frac{\tau_{n+1}(r;\nu+1)\,\tau_{n}(r;\nu)}{\tau_{n+1}(r;\nu)\,\tau_{n}(r;\nu+1)}\equiv\frac{{\rm
d}{}}{{{\rm
d}{z}}}\left\\{\ln\frac{\tau_{n+1}(z;\nu)}{\tau_{n}(z;\nu+1)}\right\\}-\frac{n+\nu}{z},\qquad
n\geq 0,$ (6.12)

satisfies PIII (2.1) with $\alpha=2(\nu-n)$ and $\beta=2(\nu+n+1)$.

###### Proof.

See, for example, [19, 36]. ∎

######  Theorem 6.3.

Equation (6.2) has the solution

$R_{n}(r)=\frac{\tau_{n}(r;1)}{\tau_{n}(r;0)},$ (6.13)

where $\tau_{n}(r;\nu)$ is the determinant given by (6.11).

###### Proof.

The proof is straightforward using induction. From (6.9) we have

$R_{1}(r)=\frac{I_{1}(r)}{I_{0}(r)}=\frac{\tau_{1}(r;1)}{\tau_{1}(r;0)},$

so (6.13) is true if $n=1$. Assuming (6.13) holds then from Theorems 6.1 and
6.2

$R_{n+1}(r)=w_{n}(r;0)R_{n}(r)=\frac{\tau_{n+1}(r;1)\,\tau_{n}(r;0)}{\tau_{n+1}(r;0)\,\tau_{n}(r;1)}\times\frac{\tau_{n}(r;1)}{\tau_{n}(r;0)}=\frac{\tau_{n+1}(r;1)}{\tau_{n+1}(r;0)},$

as required, and so the result follows by induction. ∎

######  Corollary 6.4.

Equations (6.4) and (6.6) have the Bessel function solutions

$u_{n}(z)=\frac{\tau_{n}(\tfrac{1}{2}z;1)+\tau_{n}(\tfrac{1}{2}z;0)}{\tau_{n}(\tfrac{1}{2}z;1)-\tau_{n}(\tfrac{1}{2}z;0)},\qquad
v_{n}(x)=1-\frac{\tau_{n}^{2}(\sqrt{x};0)}{\tau_{n}^{2}(\sqrt{x};1)},$

respectively, with $\tau_{n}(r;\nu)$ the determinant given by (6.11).

######  Lemma 6.5.

The formal asymptotic behaviour of the vortex solution $R_{n}(r)$ is given by

$\displaystyle R_{n}(r)$
$\displaystyle=\frac{r^{n}}{2^{n}\,n!}\left\\{1-\frac{r^{2}}{4(n+1)}+\mathcal{O}\left(r^{4}\right)\right\\},$
$\displaystyle\mbox{as}\quad r\to 0,$ (6.14) $\displaystyle R_{n}(r)$
$\displaystyle=1-\frac{n}{2r}-\frac{n^{2}}{8r^{2}}-\frac{n(n^{2}+1)}{16r^{3}}+\mathcal{O}(r^{-4}),$
$\displaystyle\mbox{as}\quad r\to\infty.$ (6.15)

###### Proof.

These are determined from (6.8) and (6.9). ∎

###  6.2 Generalised Charlier polynomials

The Charlier polynomials $C_{n}(k;z)$ are a family of orthogonal polynomials
introduced in 1905 by Charlier [8] given by

$C_{n}(k;z)={}_{2}F_{0}\left(-n,-k;;-{1/z}\right)=(-1)^{n}n!L_{n}^{(-1-k)}\left(-{1/z}\right),\quad
z>0,$ (6.16)

where ${}_{2}F_{0}(a,b;;z)$ is the hypergeometric function and
$L_{n}^{(\alpha)}(z)$ is the associated Laguerre polynomial, see, for example,
[46, §18.19]. The Charlier polynomials are orthogonal on the lattice
$\mathbb{N}$ with respect to the Poisson distribution

$\omega(k)=\frac{z^{k}}{k!},\qquad z>0,$ (6.17)

and satisfy the orthogonality condition

$\sum_{k=0}^{\infty}C_{m}(k;z)C_{n}(k;z)\frac{z^{k}}{k!}=\frac{n!\,{\rm
e}^{z}}{z^{n}}\delta_{m,n}.$

Smet and Van Assche [49] generalized the Charlier weight (6.17) with one
additional parameter through the weight function

$\omega(k;\nu)=\frac{\Gamma(\nu+1)\,z^{k}}{\Gamma(\nu+k+1)\,\Gamma(k+1)},\qquad
z>0,$

with $\nu$ a parameter such that $\nu>-1$. This gives the discrete weight

$\omega(k;\nu)=\frac{z^{k}}{(\nu+1)_{k}\,k!},\qquad z>0,$ (6.18)

where $(\nu+1)_{k}=\Gamma(\nu+1+k)/\Gamma(\nu+1)$ is the Pochhammer symbol, on
the lattice $\mathbb{N}$. Discrete orthogonal polynomials are characterized by
the discrete Pearson equation

$\Delta\big{[}\sigma(k)\omega(k)\big{]}=\tau(k)\omega(k),$ (6.19)

where $\Delta$ is the forward difference operator

$\Delta f(k)=f(k+1)-f(k).$

The weight (6.18) satisfies the discrete Pearson equation (6.19) with

$\sigma(k)=k(k+\nu),\qquad\tau(k)=-k^{2}-\nu k+z,$

and so the generalised Charlier polynomials are semi-classical orthogonal
polynomials since $\tau(k)$ is a polynomial with deg$(\tau)>1$. The special
case $\nu=0$ was first considered by Hounkonnou, Hounga and Ronveaux [24] and
later studied by Van Assche and Foupouagnigni [55].

For the generalised Charlier weight (6.18), the orthonormal polynomials
$p_{n}(k;z)$ satisfy the orthogonality condition

$\sum_{k=0}^{\infty}p_{m}(k;z)p_{n}(k;z)\frac{z^{k}}{(\nu+1)_{k}\,k!}=\delta_{m,n},$

and the three-term recurrence relation

$kp_{n}(k;z)=a_{n+1}(z)p_{n+1}(k;z)+b_{n}(z)p_{n}(k;z)+a_{n}(z)p_{n-1}(k;z),$
(6.20)

with $p_{-1}(k;z)=0$ and $p_{0}(k;z)=1$. Our interest is in the coefficients
$a_{n}(z)$ and $b_{n}(z)$ in the recurrence relation (6.20).

Smet and Van Assche [49, Theorem 2.1] proved the following theorem for the
recurrence relation coefficients associated with the generalised Charlier
weight (6.18).

######  Theorem 6.6.

The recurrence relation coefficients $a_{n}(z)$ and $b_{n}(z)$ for orthonormal
polynomials associated with the generalised Charlier weight (6.18) on the
lattice $\mathbb{N}$ satisfy the discrete system

$\begin{array}[]{l}(a_{n+1}^{2}-z)(a_{n}^{2}-z)=z(b_{n}-n)(b_{n}-n+\nu),\\\\[5.0pt]
\displaystyle b_{n}+b_{n-1}-n+\nu+1={nz/a_{n}^{2}},\end{array}$ (6.21)

with initial conditions

$a_{0}^{2}=0,\qquad
b_{0}=\frac{\sqrt{z}\,I_{\nu+1}(2\sqrt{z})}{I_{\nu}(2\sqrt{z})}=z\frac{{\rm
d}{}}{{{\rm d}{z}}}\big{\\{}\ln I_{\nu}(2\sqrt{z})\big{\\}}-\frac{\nu}{2},$
(6.22)

with $I_{\nu}(k)$ the modified Bessel function.

######  Remark 6.7.

The discrete system such as (6.21) for recurrence relation coefficients is
sometimes known as the Laguerre-Freud equations, cf. [4, 24, 34].

The recurrence relation coefficients $a_{n}(z)$ and $b_{n}(z)$ also satisfy
the Toda lattice, cf. [54, Theorem 3.8]

$\displaystyle z\frac{{\rm d}{}}{{{\rm
d}{z}}}a_{n}^{2}=a_{n}^{2}(b_{n}-b_{n-1}),$ (6.23a) $\displaystyle z\frac{{\rm
d}{}}{{{\rm d}{z}}}b_{n}=a_{n+1}^{2}-a_{n}^{2}.$ (6.23b)

Letting $a_{n}^{2}(z)=x_{n}(z)$ and $b_{n}(z)=y_{n}(z)$ in (6.21) and (6.23)
yields

$\displaystyle(x_{n+1}-z)(x_{n}-z)=t(y_{n}-n)(y_{n}-n+\nu),$ $\displaystyle
z\frac{{\rm d}{x_{n}}}{{{\rm d}{t}}}=x_{n}(y_{n}-y_{n-1}),$ $\displaystyle
y_{n}+y_{n-1}-n+\nu+1=\frac{nz}{x_{n}},$ $\displaystyle z\frac{{\rm
d}{y_{n}}}{{{\rm d}{z}}}=x_{n+1}-x_{n}.$

Eliminating $x_{n+1}$ and $y_{n-1}$ in these equations yields the differential
system

$\displaystyle z\frac{{\rm d}{x_{n}}}{{{\rm d}{z}}}$
$\displaystyle=x_{n}(2y_{n}+\nu-n+1)-nz,$ (6.24a) $\displaystyle z\frac{{\rm
d}{y_{n}}}{{{\rm d}{z}}}$
$\displaystyle=-x_{n}+z+\frac{(y_{n}-n)(y_{n}-n+\nu)z}{x_{n}-z}.$ (6.24b)

Solving (6.24a) for $y_{n}$ gives

$y_{n}=\frac{z}{2x_{n}}\frac{{\rm d}{x_{n}}}{{{\rm
d}{z}}}+\frac{nz}{2x_{n}}+\frac{n-\nu-1}{2},$

and substituting this into (6.24b) yields

$\displaystyle\frac{{\rm d}^{2}{x_{n}}}{{{\rm
d}{z}}^{2}}=\frac{1}{2}\left(\frac{1}{x_{n}}+\frac{1}{x_{n}-z}\right)-\frac{x_{n}}{z(x_{n}-z)}\frac{{\rm
d}{x_{n}}}{{{\rm d}{z}}}$
$\displaystyle-\frac{2x_{n}^{2}}{z^{2}}+\frac{4x_{n}+n^{2}-\nu^{2}+1}{2z}-\frac{n^{2}}{2x_{n}}+\frac{1-\nu^{2}}{2(x_{n}-z)}.$
(6.25)

Making the transformation

$x_{n}(z)=\frac{z}{1-w_{n}(z)}.$ (6.26)

in (6.25) yields

$\displaystyle\frac{{\rm d}^{2}{w_{n}}}{{{\rm d}{z}}^{2}}$
$\displaystyle=\left(\frac{1}{2w_{n}}+\frac{1}{w_{n}-1}\right)\\!\left(\frac{{\rm
d}{w_{n}}}{{{\rm d}{z}}}\right)^{\\!2}-\frac{1}{z}\frac{{\rm d}{w_{n}}}{{{\rm
d}{z}}}+\frac{(w_{n}-1)^{2}(n^{2}w_{n}^{2}-\nu^{2})}{2w_{n}z^{2}}-\frac{2w_{n}}{z},$
(6.27)

which is deg-PV (1.1) with parameters $\alpha=\tfrac{1}{2}n^{2}$,
$\beta=-\tfrac{1}{2}{\nu}^{2}$ and $\gamma=-2$.

Solving (6.24b) for $x_{n}$ gives

$x_{n}=-\tfrac{1}{2}z\frac{{\rm d}{y_{n}}}{{{\rm d}{z}}}+z+\tfrac{1}{2}X_{n},$
(6.28)

where

$X_{n}^{2}=z^{2}\left(\frac{{\rm d}{y_{n}}}{{{\rm
d}{z}}}\right)^{\\!\\!2}+4z(y_{n}-n)(y_{n}-n+\nu).$ (6.29)

From (6.29) we get

$\displaystyle\frac{{\rm d}{X_{n}}}{{{\rm d}{z}}}$
$\displaystyle=\frac{z^{2}}{X_{n}}\frac{{\rm d}^{2}{y_{n}}}{{{\rm
d}{z}}^{2}}\frac{{\rm d}{y_{n}}}{{{\rm d}{z}}}+\frac{z}{X_{n}}\left(\frac{{\rm
d}{y_{n}}}{{{\rm
d}{z}}}\right)^{\\!\\!2}+\frac{2z(2y_{n}-2n+\nu)}{X_{n}}\frac{{\rm
d}{y_{n}}}{{{\rm d}{z}}}+\frac{2(y_{n}-n)(y_{n}-n+\nu)}{X_{n}}.$ (6.30)

Substituting (6.28) into (6.24a), then using (6.30), solving for $X_{n}$, and
substituting into (6.29) yields the second-order, second-degree equation

$\left(2z\frac{{\rm d}^{2}{y_{n}}}{{{\rm d}{z}}^{2}}+\frac{{\rm
d}{y_{n}}}{{{\rm
d}{z}}}+8y_{n}-8n+4\nu\right)^{\\!\\!2}=\frac{(4y_{n}-2n+2\nu+1)^{2}}{z}\left\\{z\left(\frac{{\rm
d}{y_{n}}}{{{\rm d}{z}}}\right)^{\\!\\!2}+4(y_{n}-n)(y_{n}-n+\nu)\right\\}.$
(6.31)

Making the transformation

$y_{n}(z)=\tfrac{1}{2}v_{n}(x)+\tfrac{1}{2}n-\tfrac{1}{2}{\nu}-\tfrac{1}{4},\qquad
x=2\sqrt{z},$

in (6.31) yields

$\left(\frac{{\rm d}^{2}{v_{n}}}{{{\rm
d}{x}}^{2}}+4v_{n}-4n-2\right)^{\\!\\!2}=\frac{4v_{n}^{2}}{x^{2}}\left\\{\left(\frac{{\rm
d}{v_{n}}}{{{\rm
d}{x}}}\right)^{\\!\\!2}+4v_{n}^{2}-4(2n+1)v_{n}+(2n+1)^{2}-4\nu^{2}\right\\}.$
(6.32)

Equation (A.5) in [15] is

$\left(\frac{{\rm d}^{2}{v}}{{{\rm
d}{x}}^{2}}-av-b\right)^{\\!\\!2}=\frac{4v^{2}}{x^{2}}\left\\{\left(\frac{{\rm
d}{v}}{{{\rm d}{x}}}\right)^{\\!\\!2}-av^{2}-2bv-c\right\\},$ (6.33)

with $a$, $b$ and $c$ parameters, an equation derived by Chazy [9], and is the
primed version of equation SD-III in [16]. Hence equation (6.32) is the
special case of equation (6.33) with

$a=-4,\qquad b=4n+2,\qquad c=4\nu^{2}-(2n+1)^{2}.$

Cosgrove [15] showed that equation (6.33) is solvable in terms of solutions of
PIII (2.1). Consequently, the solution of (6.32) is given by

$v_{n}(x)=\frac{x}{2q}\left(\frac{{\rm d}{q}}{{{\rm d}{x}}}+q^{2}+1\right),$

where $q(x)$ satisfies PIII (2.1) for the parameters $A=2\nu-2n-2$ and
$B=2\nu+2n$.

######  Theorem 6.8.

The recurrence relation coefficients $a_{n}(z)$ and $b_{n}(z)$ are given by

$\displaystyle a_{n}^{2}(z)$
$\displaystyle=x_{n}(z)=\frac{\mathcal{T}_{n+1}(z;\nu)\mathcal{T}_{n-1}(z;\nu)}{\mathcal{T}_{n}^{2}(z;\nu)},$
(6.34a) $\displaystyle b_{n}(z)$ $\displaystyle=y_{n}(z)=z\frac{{\rm
d}{}}{{{\rm
d}{z}}}\left\\{\ln\frac{\mathcal{T}_{n+1}(z;\nu)}{\mathcal{T}_{n}(z;\nu)}\right\\}-\frac{\nu}{2},$
(6.34b)

where

$\mathcal{T}_{n}(z;\nu)=\det\left[\left(z\frac{{\rm d}{}}{{{\rm
d}{z}}}\right)^{\\!j+k}I_{\nu}\big{(}2\sqrt{z}\big{)}\right]_{j,k=0}^{n-1},$

with $\mathcal{T}_{0}(z;\nu)=1$, and $I_{\nu}(x)$ is the modified Bessel
function.

###### Proof.

The expression (6.34a) for $a_{n}^{2}(z)$ follows immediately by substituting
(5.14) in (6.26). To prove the result (6.34b) for $b_{n}(z)$ we use induction
and the fact that from equation (6.23b), $a_{n}^{2}(z)=x_{n}(z)$ and
$b_{n}(z)=y_{n}(z)$ are related by

$z\frac{{\rm d}{x_{n}}}{{{\rm d}{t}}}=x_{n}(y_{n}-y_{n-1}),$

and initially

$y_{0}(z)=z\frac{{\rm d}{}}{{{\rm
d}{z}}}\big{\\{}\ln\mathcal{T}_{1}(z;\nu)\big{)}\\}-\frac{\nu}{2}.$

Hence

$\begin{split}y_{1}(z)&=z\frac{{\rm d}{}}{{{\rm d}{z}}}\big{\\{}\ln
x_{1}(z)\big{\\}}+y_{0}(z)\\\ &=z\frac{{\rm d}{}}{{{\rm
d}{z}}}\left\\{\ln\frac{\mathcal{T}_{2}(z;\nu)\mathcal{T}_{0}(z;\nu)}{\mathcal{T}_{1}^{2}(z;\nu)}\right\\}+z\frac{{\rm
d}{}}{{{\rm d}{z}}}\left\\{\ln\mathcal{T}_{1}(z;\nu)\right\\}-\frac{\nu}{2}\\\
&=z\frac{{\rm d}{}}{{{\rm
d}{z}}}\left\\{\ln\frac{\mathcal{T}_{2}(z;\nu)}{\mathcal{T}_{1}(z;\nu)}\right\\}-\frac{\nu}{2},\end{split}$

since $\mathcal{T}_{0}(z;\nu)=1$, so (6.34b) is true for $n=1$. Now suppose
that (6.34b) is true, then

$\begin{split}y_{n+1}(z)&=z\frac{{\rm d}{}}{{{\rm d}{z}}}\big{\\{}\ln
x_{n}(z)\big{\\}}+y_{n}(z)\\\ &=z\frac{{\rm d}{}}{{{\rm
d}{z}}}\left\\{\ln\frac{\mathcal{T}_{n+2}(z;\nu)\mathcal{T}_{n}(z;\nu)}{\mathcal{T}_{n+1}^{2}(z;\nu)}\right\\}+z\frac{{\rm
d}{}}{{{\rm
d}{z}}}\left\\{\ln\frac{\mathcal{T}_{n+1}(z;\nu)}{\mathcal{T}_{n}(z;\nu)}\right\\}-\frac{\nu}{2}\\\
&=z\frac{{\rm d}{}}{{{\rm
d}{z}}}\left\\{\ln\frac{\mathcal{T}_{n+2}(z;\nu)}{\mathcal{T}_{n+1}(z;\nu)}\right\\}-\frac{\nu}{2},\end{split}$

as required, and so the result follows by induction. We remark that equation
(6.23a) is identically satisfied by $a_{n}^{2}(z)$ and $b_{n}(z)$ given by
(6.34). ∎

In a recent paper, Fernández-Irisarri and Mañas [17, §2] discuss the
generalised Charlier weight (6.18), in particular properties of the
coefficients in the recurrence relation. The relationship between the
notations in [17] and those here are $x_{n}(z)=\gamma_{n}(\eta)$ and
$y_{n}(z)=\beta_{n}(\eta)$ with $z=\eta$. Fernández-Irisarri and Mañas [17]
relate $x_{n}(z)$ and $y_{n}(z)$ to Okamoto’s Hamiltonian for P${}_{\rm
III^{\prime}}$ [44] and derive two ordinary differential equations for
$x_{n}(z)$.

  1. (i)

Equation (45) in [17, Theorem 4] is the third-order equation

$\delta_{z}\left(\frac{x_{n}}{z}\left\\{\delta_{z}^{2}(\ln
x_{n})+2x_{n}\right\\}+\frac{n^{2}z}{x_{n}}\right)=2x_{n},\qquad\delta_{z}(f)=z\frac{{\rm
d}{f}}{{{\rm d}{z}}},$

i.e.

$\frac{{\rm d}^{3}{x_{n}}}{{{\rm
d}{z}}^{3}}=\frac{1}{zx_{n}^{2}}\left(z\frac{{\rm d}{x_{n}}}{{{\rm
d}{z}}}-x_{n}\right)\left\\{2x_{n}\frac{{\rm d}^{2}{x_{n}}}{{{\rm
d}{z}}^{2}}-\left(\frac{{\rm d}{x_{n}}}{{{\rm
d}{z}}}\right)^{2}+n^{2}\right\\}-\frac{4x_{n}}{z^{2}}\frac{{\rm
d}{x_{n}}}{{{\rm d}{z}}}+\frac{2x_{n}(x_{n}+z)}{z^{3}},$ (6.35)

and the authors state that this equation “should have the Painlevé property”.
Equation (6.35) can be integrated to give equation (6.25), with $\nu^{2}$ as
the constant of integration. Since equation (6.25) is equivalent to deg-PV
(5.2) then equation (6.35) does have the Painlevé property.

  2. (ii)

Equation (60) in [17, Theorem 5] is the second-order equation

$\begin{split}\left(1-\frac{x_{n}}{z}\right)&\left\\{\delta_{z}\left(\frac{\delta_{z}(x_{n})+nz}{x_{n}}\right)+2x_{n}\right\\}+2\\{x_{n}-z+(n-b)n\\}\\\
&=-\tfrac{1}{2}\left(\frac{\delta_{z}(x_{n})+nz}{x_{n}}\right)^{2}+(n+1)\left(\frac{\delta_{z}(x_{n})+nz}{x_{n}}\right)+(n-b-1)(3n-b+1),\end{split}$

which is equation (6.25) with

$\nu^{2}=2(b-n)^{2}+n^{2}-2n-1.$

##  7 Discussion

In this paper the classical solutions of deg-PV (5.2) have been classified.
Ohyama and Okumura [41, Theorem 2.1] give a list of classical solutions of PI
to PV and state that “deg-P5 with $\alpha=\tfrac{1}{2}a^{2}$,
$\beta=-\tfrac{1}{8}$, $\gamma=-2$ has the algebraic solution
$w(z)=1+2\sqrt{z}/a$”222As noted in [2], there is typo in [41] who say
$\beta=-8$ rather than $\beta=-\tfrac{1}{8}$. and “deg-P5 with $\beta=0$ has
the Riccati type solutions”. The results in this paper show that there are
more classical solutions of deg-PV (1.1). The algebraic solution is equivalent
to the “seed solution” obtained by setting $n=0$ in (5.7), i.e.

$u_{0}(x;\mu)=\frac{\mu}{x+\mu},$

and there is a more general hierarchy of “Riccati type solutions” which are
described in Theorem 5.6.

All solutions of PII–PVI that are expressible in terms of special functions
satisfy a first-order equation of the form

$\left(\frac{{\rm d}{u}}{{{\rm
d}{x}}}\right)^{\\!n}=\sum_{j=0}^{n-1}F_{j}(u,x)\left(\frac{{\rm d}{u}}{{{\rm
d}{x}}}\right)^{\\!j},$ (7.1)

where $F_{j}(u,x)$ is polynomial in $u$ with coefficients that are rational
functions of $x$. It can be shown that the Bessel function solutions of PIII
(2.1) satisfy a first-order equation of the form (7.1) for $n$ odd, whereas
the Bessel function solutions of deg-PV (5.2) satisfy a first-order equation
of the form (7.1) for $n$ even.

The relationship between PIII (2.1) and deg-PV (1.1) is similar to that
between the second Painlevé equation (PII)

$\frac{{\rm d}^{2}{q}}{{{\rm d}{x}}^{2}}=2q^{3}+xq,$ (7.2)

with $\alpha$ a parameter, and Painlevé XXXIV equation ($\mbox{\rm
P}_{\\!34}$)

$\frac{{\rm d}^{2}{p}}{{{\rm d}{x}}^{2}}=\frac{1}{2p}\left(\frac{{\rm
d}{p}}{{{\rm
d}{x}}}\right)^{\\!2}+2p^{2}-xp-\frac{(\alpha+\tfrac{1}{2})^{2}}{2p},$ (7.3)

which is equivalent to equation XXXIV of Chapter 14 in [25], in that both
pairs of equations arise from a Hamiltonian. The Hamiltonian associated with
PII (7.2) and $\mbox{\rm P}_{\\!34}$ (7.3) is

$\mbox{$\mathcal{H}_{\rm
II}$}(q,p,z;\alpha)=\tfrac{1}{2}p^{2}-(q^{2}+\tfrac{1}{2}z)p-(\alpha+\tfrac{1}{2})q$
(7.4)

and so

$\frac{{\rm d}{q}}{{{\rm d}{z}}}=p-q^{2}-\tfrac{1}{2}z,\qquad\frac{{\rm
d}{p}}{{{\rm d}{z}}}=2qp+\alpha+\tfrac{1}{2},$ (7.5)

see [28, 42]. It is known that PII (7.2) and $\mbox{\rm P}_{\\!34}$ (7.3) have
special function solutions in terms of Airy functions, cf. [14]. It can be
shown that the Airy function solutions of PII (7.2) satisfy first-order
equation of the form (7.1) for $n$ odd, whereas the Airy function solutions of
$\mbox{\rm P}_{\\!34}$ (7.3) satisfy a first-order equation of the form (7.1)
for $n$ even. Further the function $\sigma(z;\alpha)=\mbox{$\mathcal{H}_{\rm
II}$}(q,p,z;\alpha)$ given by (7.4), with $q$ and $p$ satisfying (7.5),
satisfies the second-order, second degree equation (SII)

$\left(\frac{{\rm d}^{2}{\sigma}}{{{\rm
d}{z}}^{2}}\right)^{\\!2}+4\left(\frac{{\rm d}{\sigma}}{{{\rm
d}{z}}}\right)^{\\!3}+2\frac{{\rm d}{\sigma}}{{{\rm d}{z}}}\left(z\frac{{\rm
d}{\sigma}}{{{\rm
d}{z}}}-\sigma\right)=\tfrac{1}{4}(\alpha+\tfrac{1}{2})^{2},$ (7.6)

see [28, 42]. Conversely, if $\sigma(z;\alpha)$ is a solution of (7.6), then

$q(z;\alpha)=\frac{4\sigma^{\prime\prime}(z;\alpha)+2\alpha+1}{8\sigma^{\prime}(z;\alpha)},\qquad
p(z;\alpha)=-2\sigma^{\prime}(z;\alpha),$ (7.7)

with ${}^{\prime}\equiv{\rm d}/{\rm d}z$, are solutions of (7.2) and (7.3),
respectively. Consequently it is simpler to express classical solutions of
$\mbox{\rm P}_{\\!34}$ (7.3) in terms of classical solutions of SII (7.6),
which involve one determinant, rather than solutions of PII (7.2), which
involve two determinants.

## Acknowledgements

I thank Clare Dunning and Steffen Krusch for helpful comments and illuminating
discussions. I also thank the anonymous reviewers whose comments were
invaluable in improving the manuscript.

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# Sampling The Lowest Eigenfunction to Recover the Potential in a
One–Dimensional Schrödinger Equation

Rob Rahm $\dagger$ College Station, TX<EMAIL_ADDRESS>
###### Abstract.

We consider the BVP $-y^{\prime\prime}+qy=\lambda y$ with $y(0)=y(1)=0$. The
inverse spectral problems asks one to recover $q$ from spectral information.
In this paper, we present a very simple method to recover a potential by
sampling one eigenfunction. The spectral asymptotics imply that for larger
modes, more and more information is lost due to imprecise measurements (i.e.
relative errors increases) and so it is advantageous to use data from lower
modes. Our method also allows us to recover "any" potential from one boundary
condition.

###### Key words and phrases:

Schrödinger Equation, Eigenvalue, Inverse Problem

###### 2010 Mathematics Subject Classification:

65L09, 65L10, 65L15

$\dagger$<EMAIL_ADDRESS>
##  1\. Introduction

The goal of this paper is to present a method for recovering the potential in
a one–dimensional Schrödinger equation by sampling only the lowest mode
eigenfunction. In particular, if $\\{\lambda_{k}\\}_{k=1}^{\infty}$ and
$\\{y(x,\lambda_{k};q)\\}_{k=1}^{\infty}$ are the eigenvalues (in increasing
order) and eigenfunctions for

$\displaystyle-y^{\prime\prime}+qy=\lambda y,\hskip 72.26999pty(0)=y(1)=0,$
(1.1)

then we want to recover $q$ by sampling $y(x,\lambda_{1};q)$ at a relatively
small number of points.

This equation (the Schrödinger equation) is ubiquitous both as a model problem
and as a sort of canonical form of any second order linear ODE. Typically, one
wants to determine the potential $q$ from some (finite) spectral data. For
example, in the context of strings, $q$ is related to the density of the
string via the Liouville transform and one would like to recover the density
by measuring properties of the resonances and corresponding eigenmodes.

The equation $-y^{\prime\prime}+qy=\lambda y$ is a perturbation of the
equation $-y^{\prime\prime}=\lambda y$ and as $k\to\infty$ the eigenfunctions
and eigenvalues all converge (somewhat rapidly) to the spectral data in the
$q\equiv 0$ case. Thus, in a sense, there is "more" usable information in the
lower modes than in the higher modes. So it is desirable to use only the lower
modes to recover $q$ (this is discussed more below).

Our method is based on derivative formulas from [PosTur1987]. In particular,
for a fixed $\lambda$ and $x$ we consider the map:

$\displaystyle q\mapsto y_{2}(x,\lambda,q).$

(Throughout, $y_{1}(x,\lambda,q)$ and $y_{2}(x,\lambda,q)$ will denote the
solutions to $-y^{\prime\prime}+qy=\lambda q$ with initial conditions
$y_{1}(0,\lambda,q)=1$ and $y_{1}^{\prime}(0,\lambda,q)=0$ and
$y_{2}(0,\lambda,q)=0$ and $y_{2}^{\prime}(0,\lambda,q)=1$.) The derivative of
this function is a linear map from $L^{2}$ to $\mathbb{C}$. By the Riesz
Representation Theorem, this can be given by an integral. In [PosTur1987], it
was shown that the derivative is:

$\displaystyle\frac{\partial y_{2}(x,\lambda,q)}{\partial
q}(v)=\int_{t=0}^{1}K(t,x,\lambda,q)v(t)dt,$

where:

$\displaystyle
K(t,x,\lambda,q)=y_{2}(t,\lambda,q)\left(y_{1}(t,\lambda,q)y_{2}(x,\lambda,q)-y_{1}(x,\lambda,q)y_{2}(t,\lambda,q)\right)1\\!\\!1_{[0,x]}(t).$

To apply this, we will consider a restricted map from a finite dimension
subspace of $L^{2}$ (i.e. we will consider $q$ to be a finite linear
combination of basis functions). Each sampling of an eigenfunction corresponds
to one equation. So, if we have $n$ sample points, we should be able to solve
the equation for $q$ in a $n$–dimensional subspace.

The paper is organized in the following way. In Section 2 we discuss some
previous work in this area (there has been a lot). In Section 3 we give our
inverse method and illustrate it with some examples. Finally, in Section 4 we
give the code listings used here.

##  2\. Previous Work and Current Work

The inverse problem that we discuss here is old and well-researched. We can’t
discuss all previous work, but we discuss some of the results that we find to
be most relevant.

In 1946, Borg proves in [Borg1946] that if two spectra are given
(corresponding to two different boundary conditions) then, in principle, one
can recover $q$. Levinson proved in [Lev1949] (with a different and much
shorter proof) that one can also recover the boundary condition. In 1951,
Gelfand and Levitan ([GelLev1951]) show that the potential can be recovered
from a spectrum and some information on the eigenfunctions (in particular the
norming constants).

In practice, to recover the potential, we are only given a finite set of
spectral data. So precisely speaking there isn’t a uniqueness theorem. So any
algorithm will necessarily only produce a potential that approximates the
target potential. In [Hald1978], Hald gives a method to recover an even
potential from a spectrum and boundary conditions. A result similar to ours by
Hald and McLaughlin in [HaldMcl1989] gives a way to determine a potential from
nodal data of the eigenfunctions. Another method that is similar to ours is in
[LowPilRun1992] and, for example, [RunSac1992].

As discussed above, the spectral data for the Schrödinger operator satisfy the
following asymptotics: (see [PosTur1987]):

$\displaystyle\lambda_{k}(q)=(k\pi)^{2}+\int q(x)dx+\alpha_{k}(q),\hskip
36.135pty_{2}(x,\lambda;q)=\frac{\sin\sqrt{\lambda}x}{\sqrt{\lambda}}+O(\frac{1}{\lambda}),$

where $\\{\alpha_{k}\\}$ is an $\ell^{2}$ sequence. In other words, spectral
data is "centered" around (and converges to) the corresponding data for the
constant potential equal to the average value of $q$. For example, if we
consider just well–behaved, mean–value zero potentials, then

$\displaystyle\lambda_{10}(q)=(10\pi)^{2}+\alpha_{k}(q)\approx
987+\alpha_{k}(q).$

Typically, we can take $\alpha_{k}$ to be bounded in absolute value by
$k^{-1}$. Therefore, in the tenth eigenvalue, the information that
distinguishes one potential from another is contained in an interval of radius
$\frac{1}{10}$ centered at $987$. In other words, for the tenth eigenvalue to
be usable, a relative precision of (at least) $.01$ percent is needed.

So, it is better to use lower–mode data. Such techniques were also considered
in, for example, [MclRun1987] where the lowest eigenvalue corresponding to
different boundary conditions is used and in [Run2015] where (in essence)
known masses are added to the potential and the lowest eigenvalue is measured.
The current paper uses sampling data from the lowest mode eigenfunction –
though as is shown below, we can also use higher modes.

Finally, we mention that in addition to the sample data, we need to know both
the eigenvalue and the value of $y^{\prime}(0)$. So in a sense, we can
approximate the projection of $q$ onto the first $n$ basis functions by
collecting $n+2$ pieces of data.

Acknowledgment. I’d like to thank Bill Rundell for helpful discussions and
information regarding the topic of this paper.

##  3\. The Method

In this section, we describe our method and provide a discussion of its
efficacy - mentioning both its advantages and disadvantages. Psychologically,
We concentrate our attention to the setting of Dirichlet–Dirichlet boundary
conditions since this corresponds to (for example) the setting of a clamped
string. However, the methods we use only really require a Dirichlet condition
at the left endpoint. After a discussion of the method, we present some graphs
of our algorithm’s reconstruction of some potentials.

###  3.1. Description of the Method.

Fix $\lambda$ and define an operator $T:L^{2}([0,1])\to C^{2}([0,1])$ by:

$\displaystyle Tq(x)=y_{2}(x,\lambda,q)$

where $y_{2}(x,\lambda,q)$ is the solution to $-y^{\prime\prime}+py=\lambda y$
with $y(0)=0$ and $y^{\prime}(0)=1$. Let $y_{1}$ be the solution with $y(0)=1$
and $y^{\prime}(0)=0$ (this is the notation used in [PosTur1987]). It was
shown in [PosTur1987] that the derivative of this operator (with respect to
the potential) is:

$\displaystyle
T_{p}^{\prime}(v)=\int_{t=0}^{x}y_{2}(t)\left(y_{1}(t)y_{2}(x)-y_{1}(x)y_{2}(t)\right)v(t)dt=:\int_{t=0}^{x}K_{p}(t,x;\lambda)v(t)dt.$

If $q$ is near $p$, linearization then gives:

$\displaystyle y_{2}(x,\lambda,q)\simeq
y_{2}(x,\lambda,p)+\int_{t=0}^{x}K_{p}(t,x;\lambda)(q(t)-p(t))dt.$ (3.1)

In practice, we will use a quasi–Newton method. In this case, the kernel
$K_{p}$ is replaced by the kernel $K_{0}$ which is
$K(x,t)=s_{\lambda}(t)s_{\lambda}(x-t)$ where
$s_{\lambda}(x):=\frac{\sin\sqrt{\lambda}x}{\sqrt{\lambda}}$. This gives the
quasi–Newton iteration:

$\displaystyle q_{0}=0,\hskip 7.22743ptq_{k+1}=q_{k}+\delta,\hskip
5.05942pt\textnormal{ where }\hskip
5.05942pty_{2}(x,\lambda,q)=y(x,\lambda,q_{k})+\int_{t=0}^{x}s_{\lambda}(t)s_{\lambda}(x-t)(t,x;\lambda)\delta(t)dt.$

To solve this equation for $\delta$, we assume that $q$ is a linear
combination of some basis functions $\\{\varphi_{l}\\}$ (e.g.
$\varphi_{l}(x)=\cos(2(l-1)\pi x)$ if we assume a priori that $q$ is even.)
Then $\delta$ is a sum of those same basis functions:
$\delta(x)=\sum_{l=1}^{n}d_{l}\varphi_{l}(x)$. Since there are $n$ unknowns,
we need to convert our spectral data into $n$ equations.

For example, assume that we know $y_{2}(x,\lambda,q)$ evaluated at $n$ points
- say $x_{1},\ldots,x_{n}$. The quasi–Newton method gives the system:

$\displaystyle\begin{pmatrix}y_{2}(x_{1},\lambda,q)\\\ \vdots\\\
y_{2}(x_{n},\lambda,q)\end{pmatrix}=\begin{pmatrix}y_{2}(x_{1},\lambda,q_{k})\\\
\vdots\\\
y_{2}(x_{1},\lambda,q_{k})\end{pmatrix}+\begin{pmatrix}\int_{t=0}^{x_{1}}s_{\lambda}(t)s_{\lambda}(x_{1}-t)\delta(t)dt\\\
\vdots\\\
\int_{t=0}^{x_{n}}s_{\lambda}(t)s_{\lambda}(x_{n}-t)\delta(t)dt\end{pmatrix}.$
(3.2)

This can be written as a matrix–vector equation. If $J$ is the matrix whose
$j,l$ entry is:

$\displaystyle\int_{t=0}^{x_{j}}s_{\lambda}(t)s_{\lambda}(1-t)\varphi_{l}(t)dt,$

and $\vec{y_{2}}(1,\lambda,q_{k})$ is the first term, we can write this as:

$\displaystyle\vec{0}=\vec{y_{2}}(1,\lambda,q_{k})+J\vec{d},$

where $\delta(t)=\sum_{l=1}^{n}d_{l}\varphi_{l}(t)$.

###  3.2. Discussion of Recovery.

In this subsection, we explain why our algorithm is able to recover the
potential.

Using the derivative formula and the mean value theorem, we can estimate:

$\displaystyle\left|y_{2}(x,\lambda,q)-y_{2}(x,\lambda,p)\right|\leq\int_{0}^{x}\left|K_{p}(x,t)(q(t)-p(t)\right|dt+O(\left\|q-p\right\|^{2}).$

In many cases, the kernel is bounded by $1$. If
$q(x)=\sum_{l=0}^{\infty}a_{l}\cos(l\pi k)$ and
$q_{N}(x)=\sum_{l=0}^{N-1}a_{l}\cos(l\pi x)$ then standard estimates show that
$\left\|q-q_{N}\right\|\leq\frac{1}{N}$.

In addition, by the injectivity of the Jacobian in the quasi–Newton iteration
scheme, if $q_{1}$ and $q_{2}$ are linear combinations of the first $N$ basis
functions, and if their values at $N$ sample points are close, then their
difference $q_{1}-q_{2}$ must be small (how small depends on precise
quantitative properties of the Jacobian - and this depends on the choice of
sample points which is discussed below.)

Putting these two facts together shows that the potential the algorithm
recovers is close to projection of the target potential on the first $N$ basis
functions.

Recall that the Jacobian is the matrix whose $j,l$ entry is:

$\displaystyle\int_{t=0}^{x_{j}}s_{\lambda}(t)s_{\lambda}(x_{j}-t)\varphi_{l}(t)dt,\hskip
18.06749pt\textnormal{ where }\hskip
18.06749pts_{\lambda}(t)=\frac{\sin\sqrt{\lambda}t}{\sqrt{\lambda}}.$

It is of course relevant to ask if this matrix is well–behaved. For example,
is it injective and what is its condition number? These data depend on
$\lambda$ and the sample points. Standard computations (i.e. explicitly
computing singular values) shows that for evenly spaces sample points at
$\sqrt{\lambda}=k\pi$ then this matrix is invective. Since $\lambda_{k}$ is
close to $(k\pi)^{2}$ this indicates (via continuity of the determinant
function), that for most potentials the Jacobian is injective. We have never
found a potential where the Jacobian isn’t invective and, in practice we use
the Moore–Penrose pseudoinverse so that we can delete smaller singular values.

In practice, if we let $C(x_{1},\ldots,x_{n})$ be the condition number of this
Jacobian with $\sqrt{\lambda}=k\pi$, then we can choose the sample points by
minimizing this condition number. This is illustrated in the examples below.

A limitation of this method is that we must know a good basis with which to
approximate $q$ a priori. In our method, once we have sampled the
eigenfunction, we can then run the algorithm using different basis functions
for the $\varphi_{l}(x)$. With out knowing $q$, how can we determine which
basis gives us a better approximation?

A related question is this: suppose we have two orthonormal bases
$\\{\varphi_{l}\\}$ and $\\{\psi_{l}\\}$ and we know the inner products
$\\{\left\langle q,\varphi_{l}\right\rangle\\}_{l=1}^{N}$ and $\\{\left\langle
q,\psi_{l}\right\rangle\\}_{l=1}^{N}$ (i.e. we know the projection of $q$ onto
the first $N$ basis functions.) We can determine which projection is a better
$L^{2}$ approximation in the following way. Since

$\displaystyle\left\|q\right\|_{L^{2}}=\sum_{l=1}^{\infty}(\left\langle
q,\varphi_{l}\right\rangle)^{2}=\sum_{l=1}^{\infty}(\left\langle
q,\psi_{l}\right\rangle)^{2}$

the better approximation is going to be the one that captures more of the
norm, that is whichever projection has the larger $L^{2}$ norm. Since the
algorithm produces an approximation to the projection of $q$ onto the basis
$\\{\varphi_{l}\\}$, if we have two (or more) approximations to $q$, we just
select the one that has the higher $L^{2}$ norm. This selection criteria is
illustrated in the final example.

###  3.3. Examples Using Sampling.

In this section, we give examples of our method. In addition to sampling the
lowest mode eigenfunction, we also sample higher modes to illustrate the
method further. We also use different bases.

For each potential and each sampled eigenfunction, we will plot the potential
and its reconstruction on the plot on the left, and the plot on the right is
the eigenfunction we sampled. The asterisks in the plot of the eigenfunction
indicate the location of the sample points.

######  Example 1.

The first example deals with $q(x)=1-\exp(-20(x-\frac{1}{2})^{2})$. We also
use the even cosine basis $\cos 2(k-1)\pi x$ for $k=1,2,\ldots$.

For the first reconstruction, we use the first eigenfunction and three sample
points chosen by the optimization procedure explained above.

For the next reconstruction, we use the second eigenfunction and three sample
points chosen by the optimization procedure explained above.

For the next reconstruction, we use the first eigenfunction and three
equally–spaced sample points.

######  Example 2.

For the next example, we use the function
$q(x)=1-\left|x-.25\right|1\\!\\!1_{[0,.5]}(x)-\left|x-.75\right|1\\!\\!1_{[.5,1]}(x)$.
We also use the even cosine basis $\cos 2(k-1)\pi x$ for $k=1,2,\ldots$.

In the first reconstruction, we use three optimally spaced sample points and
the first eigenfunction.

In the next reconstruction, we use three equally spaced sample points and the
first eigenfunction.

######  Example 3.

######  Example 4.

In the next example, we work with the potential:

$\displaystyle q(x)=p(x-.5),\hskip 18.06749pt\textnormal{ where }\hskip
18.06749ptp(x)=(2x)^{6}-3(2x)^{4}+(2x)^{2}-1.$

We use the even Legendre polynomial basis. In the first case, we use three
optimally spaced sample points and the first eigenfunction:

In the second case we use four optimally spaced points and the first
eigenfunction. Of course, this means we are using degree four Legendre
polynomials and so the recovery is perfect.

In the examples below we use equally spaced sample points:

######  Example 5.

Next we test our methods on the potential

$\displaystyle q(t)=1+t+.3\cos(2\pi t)-.1\sin(2\pi t)+\cos(4\pi
t)+.56\sin(4\pi t),$

and we use the Legendre basis again with six sample points.

######  Example 6.

Finally, we use the potential $q(t)=t^{6}+t^{5}-t$. Once again we use the
Legendre basis. Unsurprisingly, the reconstruction is perfect.

######  Example 7.

In the next example, we use the potential $q(x)=1+\frac{1}{2}\sin(4\pi x)$. We
show a reconstruction using the Legendre basis and one using the basis
$\\{\cos 2(l-1)\pi x,\sin 2l\pi x\\}_{l=1}^{2}$. Clearly the second basis
should be the better (and should recover $q$ exactly). But suppose we don’t
know this.

The reconstruction with the Legendre basis is:

The reconstruction with the trigonometric basis is:

The $L^{2}$ norm of the potential recovered using the Legendre basis is
$1.7306$ and the one using the trigonometric basis is $2$. Therefore, our
selection criteria indicates that the trigonometric basis gives the better
approximation (which is true in this case, of course).

######  Example 8.

In the final exam, we begin with a Legendre basis and a trigonometric basis.
Then we pick the four best basis functions (based on the criteria selection
mentioned above) and get a better approximation (in this case, it is exact).
The target potential is $q(x)=1+(t-.5)^{2}+\frac{1}{2}\sin(4\pi x)$. In the
Legendre basis, the approximation is:

And in the trigonometric basis, the approximation is:

By looking at the coefficients, we can pick the four largest which correspond
to the constant function from both bases, the quadratic Legendre polynomial,
and the $\sin(4\pi x)$ function from the trigonometric basis. Using these four
basis functions the reconstruction is:

##  4\. Appendix

In this section is the code for the inverse problem. Below, the function "jac"
computes the matrix in the iteration. As above, the function ivp is just an
initial value problem solve.

⬇

%

% Given N Dirichlet eigenvalues for the problem -y” + qy = l y

% X is the sample points and L is the data. We need to have

% a Y vector as well; this is the LHS.

% B is the basis functions.

%

% X must be a row vector and contains *only* the sample points

% L must be a row vector and contains the evals

% Y must be a column vector and contains the rhs of the solution

function [coefs, iters] = inverse_full(X, L, Y, B, mode, num_coefs)

%% estimate qbar

qbar_est = L(mode) - (mode.*pi).^2;

L = L - qbar_est;

mat = jac(X,L,B);

[U, S, V] = svd(mat);

d = diag(S);

CUT_OFF = 1e-6;

d(abs(d)<CUT_OFF) = 0;

S = diag(d);

mat = U*S*V’;

mati = pinv(mat);

tolerance = 1e-4;

iters = 0;

delta = ones(num_coefs, 1);

coefs = zeros(num_coefs, 1);

while ((iters < 10)) && (max(abs(delta)) > tolerance)

iters = iters + 1;

rhs = get_rhs(coefs, X, L, Y, B, num_coefs);

delta = mati*rhs;

coefs = coefs + delta;

end

coefs(1) = coefs(1) + qbar_est;

end

%

% Computes the RHS for the iteration

%

function rhs = get_rhs(coefs, X, L, Y, B, num_coefs)

tol = [1e-10 1e-10 1e-11];

N = num_coefs;

rhs = zeros(N, 1);

for row = 1:N

q = @(t) pot(t, coefs, B);

[~, u] = ivp(q, L(row), [0 X(row)], tol);

rhs(row) = u(end,1) - Y(row);

end

end

Below is the listing for the "jac" function:

⬇

%{

This will give the matrix in the iteration for the

Newton’s method in the inverse problem.

X is the sample points

L is the eigenvalues

B is the basis

%}

function jac = jac(X, L, B)

N = size(X, 2);

jac = zeros(N, N);

for row = 1:N

for col = 1:N

xpt = X(row);

jac(row, col) = integral(@(t)jf(t, xpt, B{col}, L(row)), 0, xpt);

end

end

end

%

% This will give the function for the entry in the jacobian that is to

% be integrated.

%

function val = jf(t, xpt, b, l)

val = sin(sqrt(l).*t) .* sin(sqrt(l).*(t-xpt)) .* b(t);

val = val ./ l;

end

## References