# An algebraic proof for the Umemura polynomials
for the third Painlevé equation

Peter A. Clarkson††School of Mathematics, Statistics and Actuarial Science,
University of Kent, Canterbury, CT2 7NF, UK. Email<EMAIL_ADDRESS>Chun-Kong Law††Department of Applied Mathematics, National Sun Yat-sen
University, Kaohsiung, Taiwan 804, R.O.C. Email<EMAIL_ADDRESS>Chia-
Hua Lin††Department of Applied Mathematics, National Sun Yat-sen University,
Kaohsiung, Taiwan 804, R.O.C.

###### Abstract

We are concerned with the Umemura polynomials associated with the third
Painlevé equation. We extend Taneda’s method, which was developed for the
Yablonskii–Vorob’ev polynomials associated with the second Painlevé equation,
to give an algebraic proof that the rational functions generated by the
nonlinear recurrence relation satisfied by Umemura polynomials are indeed
polynomials.

††AMS Subject Classification (2010) : 34A55, 34B24, 47A75.††Keywords: Umemura
polynomials, third Painlevé equation, recurrence relation

##  1 Introduction

The third Painlevé equation (PIII)

$\frac{d^{2}{w}}{{d{z}}^{2}}=\frac{1}{w}\left(\frac{d{w}}{{d{z}}}\right)^{2}-\frac{1}{z}\frac{d{w}}{{d{z}}}+\dfrac{\alpha
w^{2}+\beta}{z}+\gamma w^{3}+\dfrac{\delta}{w},$ (1.1)

where ${}^{\prime}=d/dz$ and $\alpha$, $\beta$, $\gamma$ and $\delta$ are
arbitrary parameters. We are concerned with the generic case when
$\gamma\delta\neq 0$, so we set $\gamma=1$ and $\delta=-1$, without loss of
generality (by rescaling $w$ and $z$ if necessary), and so consider

$\frac{d^{2}{w}}{{d{z}}^{2}}=\frac{1}{w}\left(\frac{d{w}}{{d{z}}}\right)^{2}-\frac{1}{z}\frac{d{w}}{{d{z}}}+\dfrac{\alpha
w^{2}+\beta}{z}+w^{3}-\dfrac{1}{w}.$ (1.2)

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

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

where $F$ is rational in $dw/dz$ and $w$ and analytic in $z$. The Painlevé
equations can be thought of as nonlinear analogues of the classical special
functions. Indeed Iwasaki, Kimura, Shimomura and Yoshida [35] characterize the
six Painlevé equations as “the most important nonlinear ordinary differential
equations” and state that “many specialists believe that during the twenty-
first century the Painlevé functions will become new members of the community
of special functions”. Subsequently this has happened as the Painlevé
equations are a chapter in the NIST Digital Library of Mathematical Functions
[60, §32].

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
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. [15, 23, 33] 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.

Vorob’ev [73] and Yablonskii [75] expressed the rational solutions of PII

$\frac{d^{2}{w}}{{d{z}}^{2}}=2w^{3}+zw+\alpha,$ (1.4)

with $\alpha$ an arbitrary constant, in terms of special polynomials, now
known as the Yablonskii–Vorob’ev polynomials, which are defined through the
recurrence relation (a second-order, bilinear differential-difference
equation)

$Q_{n+1}{Q_{n-1}}=zQ_{n}^{2}-4\left[Q_{n}\frac{d^{2}{Q_{n}}}{{d{z}}^{2}}-\left(\frac{d{Q_{n}}}{{d{z}}}\right)^{2}\right],$
(1.5)

with $Q_{0}(z)=1$ and $Q_{1}(z)=z$. It is clear from the recurrence relation
(1.5) that the $Q_{n}$ are rational functions, though it is not obvious that
they are polynomials since one is dividing by $Q_{n-1}$ at every iteration. In
fact it is somewhat remarkable that the $Q_{n}$ are polynomials. Taneda [68],
see also [26], used an algebraic method to prove that the functions $Q_{n}$
defined by (1.5) are indeed polynomials. Clarkson and Mansfield [20]
investigated the locations of the roots of the Yablonskii–Vorob’ev polynomials
in the complex plane and showed that these roots have a very regular,
approximately triangular structure; the term “approximate” is used since the
patterns are not exact triangles as the roots lie on arcs rather than straight
lines. An earlier study of the distribution of the roots of the
Yablonskii–Vorob’ev polynomials is given in [41]; see also [35, p. 255, 339].
Recently Bertola and Bothner [3] and Buckingham and Miller [8, 9] have studied
the Yablonskii–Vorob’ev polynomials $Q_{n}(z)$ in the limit as $n\to\infty$
and shown that the roots lie in a “triangular region” with elliptic sides
which meet with interior angle $\tfrac{2}{5}\pi$, suggesting a limit to a
solution of PI. Indeed Buckingham & Miller [8, 9] show that in the limit as
$n\to\infty$, the rational solution of PII tends to the tritronquée solution
of PI.

Okamoto [57] obtained special polynomials, analogous to the
Yablonskii–Vorob’ev polynomials, which are associated with some of the
rational solutions of PIV. Noumi and Yamada [54] generalized Okamoto’s results
and expressed all rational solutions of PIV in terms of special polynomials,
now known as the generalized Hermite polynomials and generalized Okamoto
polynomials. The structure of the roots of these polynomials is studied in
[12], where it is shown that the roots of the generalized Hermite polynomials
have an approximate rectangular structure and the roots of the generalized
Okamoto polynomials have a combination of approximate rectangular and
triangular structures. As for PII, “approximate” is used since the patterns
are not exact triangles and rectangles as the roots lie on arcs rather than
straight lines.

Umemura [71] derived special polynomials with certain rational and algebraic
solutions of PIII, PV and PVI which have similar properties to the
Yablonskii–Vorob’ev polynomials and the Okamoto polynomials. Recently there
have been further studies of the special polynomials associated with rational
solutions of PII [3, 8, 9, 26, 37, 39, 42, 68], rational and algebraic
solutions of PIII [1, 11, 38, 51, 52, 59, 56, 72], rational solutions of PIV
[14, 22, 26, 54, 40, 55], rational and algebraic solutions of PV [13, 49, 53,
58, 74], algebraic solutions of PVI [43, 48, 67, 69]; a comprehensive review
is given in [16]. Several of these papers are concerned with the combinatorial
structure and determinant representation of the polynomials, often related to
the Hamiltonian structure and affine Weyl symmetries of the Painlevé
equations. Additionally the coefficients of these special polynomials have
some interesting, indeed somewhat mysterious, combinatorial properties [70,
71].

Further these special polynomials associated with rational solutions of the
Painlevé equations arise in several applications:

  * •

the Yablonskii–Vorob’ev polynomials arise in the transition behaviour for the
semi-classical sine-Gordon equation [7] and in moving boundary problems [65,
66];

  * •

the Umemura polynomials associated with rational solutions of PIII and PV
arise as multivortex solutions of the complex sine-Gordon equation [2, 4, 5,
61];

  * •

the generalized Hermite polynomials associated with rational solutions of PIV
arise as multiple integrals in random matrix theory [6, 24], in the
description of vortex dynamics with quadrupole background flow [18], and as
coefficients of recurrence relations for semi-classical orthogonal polynomials
[10, 19];

  * •

the generalized Okamoto polynomials associated with rational solutions of PIV
generate previously unknown rational-oscillatory solutions of the de-focusing
nonlinear Schrödinger equation [17];

  * •

these special polynomials associated with rational solutions of Painlevé
equations are examples of exceptional orthogonal polynomials [27, 28, 29, 30,
31, 32, 45, 46, 47], for which there is much current interest.

We emphasize that the fact that the nonlinear recurrence relation (1.5)
generates polynomials also follows from the $\tau$-function theory associated
with the theory of Painlevé equations. The $\tau$-functions are in general
entire functions. It can be shown that for PII with $\alpha=m$, the associated
$\tau$-function is

$\tau_{m}(z)=Q_{m}(z)\exp\left(-\frac{z^{3}}{24}\right),$

Consequently the rational function $Q_{m}(z)$ has to be a polynomial. Taneda
[68] and Fukutani, Okamoto and Umemura [26] independently gave a direct
algebraic proof, which is one of the first studies of nonlinear recurrence
relations for polynomials. In particular, Taneda [68] defined a Hirota-like
operator

$\ell(f)=f\frac{d^{2}{f}}{{d{z}}^{2}}-\left(\frac{d{f}}{{d{z}}}\right)^{2},$

and showed that if $f(z)$ is a polynomial in $z$, and $g=zf^{2}-4\,\ell(f)$,
then $f$ divides $2z\,g^{2}-4\,\ell(g)$. Hence if $f(z)=Q_{m-1}(z)$, then
$g(z)=Q_{m}(z)Q_{m-2}(z)$ and

$2z\,g^{2}-4\,\ell(g)=Q_{m}^{2}Q_{m-3}Q_{m-1}+Q_{m-2}^{2}(zQ_{m}^{2}-4\,\ell(Q_{m})),$

so that $Q_{m}$ divides $zQ_{m}^{2}-4\,\ell(Q_{m})$, implying that $Q_{m+1}$
is a polynomial. This is based on the assumption that each $Q_{m}$ has simple
zeros (implying that $Q_{m}$ and $Q_{m-1}$ have no common zeros), which in
turn can be proved using another identity derived from PII,

$\frac{d{Q_{m+1}}}{{d{z}}}Q_{m-1}-Q_{m+1}\frac{d{Q_{m-1}}}{{d{z}}}=(2m+1)Q_{m}^{2},$
(1.6)

which is proved in [26, 68] (see also [42]).

Now for PIII (1.2), the recurrence relation becomes

$S_{n+1}S_{n-1}=-z\left[S_{n}\frac{d^{2}{S_{n}}}{{d{z}}^{2}}-\left(\frac{d{S_{n}}}{{d{z}}}\right)^{2}\right]-S_{n}\frac{d{S_{n}}}{{d{z}}}+(z+\mu)S_{n}^{2},$
(1.7)

where $\mu$ is a complex parameter. Here there is one more term $\displaystyle
S_{n}\frac{d{S_{n}}}{{d{z}}}$, and $z$ in the main term implies that the root
$z=0$ of $S_{n}$, if exists, will accumulate. We continue to employ Taneda’s
trick, by defining another Hirota-like operator

$\mathcal{L}_{z}(f)=f\frac{d^{2}{f}}{{d{z}}^{2}}-\left(\frac{d{f}}{{d{z}}}\right)^{2}-\frac{f}{z}\frac{d{f}}{{d{z}}}.$

Also we need another identity. We find that it is suitable to use the fourth
order differential equation in $S_{n}$ (4.6) found in [11]. This higher order
equation in fact comes from the $\sigma$-equation equivalent to PIII

$\left(z\frac{d^{2}{\sigma}}{{d{z}}^{2}}-\frac{d{\sigma}}{{d{z}}}\right)^{2}+4\left(\frac{d{\sigma}}{{d{z}}}\right)^{2}\left(z\frac{d{\sigma}}{{d{z}}}-2\sigma\right)+4z\lambda_{1}\,\frac{d{\sigma}}{{d{z}}}-z^{2}\left(z\frac{d{\sigma}}{{d{z}}}-2\sigma+2\lambda_{0}\right)=0,$
(1.8)

with $\lambda_{0}=\frac{1}{8}(\alpha^{2}+(\beta-2)^{2})$ and
$\lambda_{1}=-\frac{1}{4}\alpha(\beta-2)$; or equivalently

$z^{2}\frac{d^{3}{\sigma}}{{d{z}}^{3}}-z\frac{d^{2}{\sigma}}{{d{z}}^{2}}+6z\left(\frac{d{\sigma}}{{d{z}}}\right)^{2}-8\sigma\frac{d{\sigma}}{{d{z}}}+\frac{d{\sigma}}{{d{z}}}-\tfrac{1}{2}z^{3}-2z\lambda_{1}=0,$
(1.9)

which is obtained by differentiating (1.8). The higher order equation

$\displaystyle z^{2}$
$\displaystyle\left[S_{n}\frac{d^{4}{S_{n}}}{{d{z}}^{4}}-4\frac{d{S_{n}}}{{d{z}}}\frac{d^{3}{S_{n}}}{{d{z}}^{3}}+3\left(\frac{d^{2}{S_{n}}}{{d{z}}^{2}}\right)^{2}\right]+2z\left(S_{n}\frac{d^{3}{S_{n}}}{{d{z}}^{3}}-\frac{d{S_{n}}}{{d{z}}}\frac{d^{2}{S_{n}}}{{d{z}}^{2}}\right)$
$\displaystyle\quad-4z(z+\mu)\left[S_{n}\frac{d^{2}{S_{n}}}{{d{z}}^{2}}-\left(\frac{d{S_{n}}}{{d{z}}}\right)^{2}\right]-2S_{n}\frac{d^{2}{S_{n}}}{{d{z}}^{2}}+4\mu
S_{n}\frac{d{S_{n}}}{{d{z}}}=2n(n+1)S_{n}^{2},$

which is derived from (1.9), is also instrumental in the analysis of the case
when $z=0$ is a root of $S_{n}$ (see §4 below).

In §2 we describe rational solutions of equation (1.2). In §§3 and 4 we extend
Taneda’s algebraic proof for equation (1.5) to equation (2.3) and in §5 we
discuss our results.

##  2 Rational solutions of PIII

The location of rational solutions of equation (1.2), which is PIII with
$\gamma=1$ and $\delta=-1$, are given in the following theorem.

######  Theorem 2.1.

Equation (1.2) has a rational solution if and only if $\alpha\pm\beta=4n$,
$n\in\mathbb{Z}$.

###### Proof.

See Gromak, Laine and Shimomura [33, p. 174]; also [51, 52]. ∎

Umemura [71] derived special polynomials associated with rational solutions of
PIII (1.2), which are defined in Theorem 2.2, and states that these
polynomials are the analogues of the Yablonskii–Vorob’ev polynomials
associated with rational solutions of PII [73, 75] and the Okamoto polynomials
associated with rational solutions of PIV [57].

######  Theorem 2.2.

Suppose that $T_{n}(z;\mu)$ satisfies the recurrence relation

$zT_{n+1}T_{n-1}=-z\left[T_{n}\frac{d^{2}{T_{n}}}{{d{z}}^{2}}-\left(\frac{d{T_{n}}}{{d{z}}}\right)^{2}\right]-T_{n}\frac{d{T_{n}}}{{d{z}}}+(z+\mu)T_{n}^{2},$
(2.1)

with $T_{-1}(z;\mu)=1$ and $T_{0}(z;\mu)=1$. Then

$w_{n}(z;\mu)\equiv
w(z;\alpha_{n},\beta_{n})=\frac{T_{n}(z;\mu-1)\,T_{n-1}(z;\mu)}{T_{n}(z;\mu)\,T_{n-1}(z;\mu-1)}\equiv
1+\frac{d{}}{{d{z}}}\ln\frac{T_{n}(z;\mu-1)}{z^{n}\,T_{n}(z;\mu)},$ (2.2)

satisfies PIII (1.2), with $\alpha_{n}=2n+2\mu-1$ and $\beta_{n}=2n-2\mu+1$.

###### Proof.

See Umemura [71]; also [11, 38]. ∎

The “polynomials” $T_{n}(z;\mu)$ are rather unsatisfactory since they are
actually polynomials in $\xi=1/z$ rather than polynomials in $z$, which would
be more natural. However it is straightforward to determine a recurrence
relation which generates functions $S_{n}(z;\mu)$ which are polynomials in
$z$. These are given in the following theorem which generalizes the work of
Kajiwara and Masuda [38].

######  Theorem 2.3.

Suppose that $S_{n}(z;\mu)$ satisfies the recurrence relation

$S_{n+1}S_{n-1}=-z\left[S_{n}\frac{d^{2}{S_{n}}}{{d{z}}^{2}}-\left(\frac{d{S_{n}}}{{d{z}}}\right)^{2}\right]-S_{n}\frac{d{S_{n}}}{{d{z}}}+(z+\mu)S_{n}^{2},$
(2.3)

with $S_{-1}(z;\mu)=S_{0}(z;\mu)=1$. Then

$\begin{split}w_{n}=w(z;\alpha_{n},\beta_{n})&=\frac{S_{n}(z;\mu-1)\,S_{n-1}(z;\mu)}{S_{n}(z;\mu)\,S_{n-1}(z;\mu-1)}\equiv
1+\frac{d{}}{{d{z}}}\ln\frac{S_{n-1}(z;\mu-1)}{S_{n}(z;\mu)},\end{split}$
(2.4)

satisfies PIII (1.2) with $\alpha_{n}=2n+2\mu-1$ and $\beta_{n}=2n-2\mu+1$.

###### Proof.

See Clarkson [11] and Kajiwara [36]. ∎

######  Remarks 2.4.

  1.   2. 1.

The polynomials $S_{n}(z;\mu)$ and $T_{n}(z;\mu)$, defined by (2.3) and (2.1),
respectively, are related through $S_{n}(z;\mu)=z^{n(n+1)/2}T_{n}(z;\mu)$. The
polynomials $S_{n}(z;\mu)$ also have the symmetry property
$S_{n}(z;\mu)=S_{n}(-z;-\mu)$.

  3. 2.

The rational solutions of PIII (1.2) lie on the lines
$\alpha+\epsilon\beta=4n$, with $\epsilon=\pm 1$, in the $\alpha$-$\beta$
plane. Further, letting $w(z)=u(\zeta)/\sqrt{\zeta}$, with
$\zeta=\tfrac{1}{4}z^{2}$, in PIII (1.2) yields

$\frac{d^{2}{u}}{{d{\zeta}}^{2}}=\frac{1}{u}\\!\left(\frac{d{u}}{{d{\zeta}}}\right)^{2}-\frac{1}{\zeta}\frac{d{u}}{{d{\zeta}}}+\frac{\alpha
u^{2}}{2\zeta^{2}}+\frac{\beta}{2\zeta}+\frac{u^{3}}{\zeta^{2}}-\frac{1}{u},$
(2.5)

which is known as $\mbox{P}_{\rm III^{\prime}}$ (cf. Okamoto [59]) and is
frequently used to determine properties of solutions of PIII. However (2.5)
has algebraic solutions rather than rational solutions [51, 52].

  4. 3.

For any $n\in\mathbb{N}\cup\\{0\\}$, if $\alpha_{n}=2n+2\mu-1$ and
$\beta_{n}=2n-2\mu+1$, with $\mu\in\mathbb{C}$, then
$\alpha_{n}+\beta_{n}=4n$.

Kajiwara and Masuda [38] derived representations of rational solutions for
PIII (1.2) in the form of determinants, which are described in the following
theorem.

######  Theorem 2.5.

Let $p_{k}(z;\mu)$ be the polynomial defined by

$\sum_{j=0}^{\infty}p_{j}(z;\mu)\lambda^{j}=(1+\lambda)^{\mu}\exp\left(z\lambda\right),$
(2.6)

with $p_{j}(z;\mu)=0$ for $j<0$, and $\tau_{n}(z)$, for $n\geq 1$, be the
$n\times n$ determinant

$\tau_{n}(z;\mu)=\mathcal{W}\left(p_{1}(z;\mu),p_{3}(z;\mu),\ldots,p_{2n-1}(z;\mu)\right),$
(2.7)

where $\mathcal{W}(\phi_{1},\phi_{2},\ldots,\phi_{n})$ is the Wronskian. Then

$\begin{split}w_{n}=w(z;\alpha_{n},\beta_{n},1,-1)&=1+\frac{d{}}{{d{z}}}\ln\frac{\tau_{n-1}(z;\mu-1)}{\tau_{n}(z;\mu)},\end{split}$
(2.8)

for $n\geq 1$, satisfies PIII (1.2) with $\alpha_{n}=2n+2\mu-1$ and
$\beta_{n}=2n-2\mu+1$.

###### Proof.

See Kajiwara and Masuda [38]. ∎

######  Remarks 2.6.

  1.   2. 1.

We note that $p_{k}(z;\mu)=L_{k}^{(\mu-k)}(-z)$, where $L_{k}^{(m)}(\zeta)$ is
the associated Laguerre polynomial, cf. [60, §18].

  3. 2.

The relationship between the polynomial $S_{n}(z,\mu)$ and the Wronskian
$\tau_{n}(z;\mu)$ is

$S_{n}(z,\mu)=c_{n}\tau_{n}(z;\mu),\qquad c_{n}=\prod_{j=1}^{n}(2j+1)^{n-j}.$

  4. 3.

In the special case when $\mu=1$, then

$S_{n}(z,1)=z^{n(n-1)/2}\theta_{n}(z),$

where $\theta_{n}(z)$ is the Bessel polynomial (sometimes known as the reverse
Bessel polynomial, cf. [34]) given by

$\theta_{n}(z)=\sqrt{\frac{2}{\pi}}\,z^{n+1/2}e^{z}K_{n+1/2}(z)\equiv\frac{n!}{(-2)^{n}}L_{n}^{(-2n-1)}(2x),$

with $K_{\nu}(z)$ the modified Bessel function, which recently arose in the
description of point vortex equilibria [62].

The recurrence relation (2.3) is nonlinear, so in general, there is no
guarantee that the rational function $S_{n+1}$ thus derived is indeed a
polynomial (since one is dividing by $S_{n-1}$), as was the case for the
recurrence relation (1.5). However, the Painlevé theory guarantees that this
is the case through an analysis of the $\tau$-function. A few of these Umemura
polynomials $S_{n}(z;\mu)$, with $\mu$ an arbitrary complex parameter, are
given in Table 2.1.

$\displaystyle S_{1}(z;\mu)$ $\displaystyle=z+\mu,$ $\displaystyle
S_{2}(z;\mu)$ $\displaystyle=\xi^{3}-\mu,$ $\displaystyle S_{3}(z;\mu)$
$\displaystyle=\xi^{6}-5\mu\xi^{3}+9\mu\xi-5\mu^{2},$ $\displaystyle
S_{4}(z;\mu)$
$\displaystyle=\xi^{10}-15\mu\xi^{7}+63\mu\xi^{5}-225\mu\xi^{3}+315\mu^{2}\xi^{2}-175\mu^{3}\xi+36\mu^{2},$
$\displaystyle S_{5}(z;\mu)$
$\displaystyle=\xi^{15}-35\mu\xi^{12}+252\mu\xi^{10}+175{\mu}^{2}\xi^{9}-2025\mu\xi^{8}+945{\mu}^{2}\xi^{7}$
$\displaystyle\qquad\;-1225\mu(\mu^{2}-9)\xi^{6}-26082{\mu}^{2}\xi^{5}+33075{\mu}^{3}\xi^{4}-350{\mu}^{2}(35{\mu}^{2}+36)\xi^{3}$
$\displaystyle\qquad\;+11340{\mu}^{3}\xi^{2}-225{\mu}^{2}(49\mu^{2}-36)\xi+7{\mu}^{3}(875{\mu}^{2}-828).$
Table 2.1:  The first few Umemura polynomials $S_{n}(z;\mu)$, with
$\xi=z+\mu$.

Plots of the roots of the polynomials $S_{n}(z;\mu)$ for various $\mu$ are
given in [11]. Initially for $\mu$ sufficiently large and negative, the
$\tfrac{1}{2}n(n+1)$ roots form an approximate triangle with $n$ roots on each
side. Then as $\mu$ increases, the roots in turn coalesce and eventually for
$\mu$ sufficiently large and positive they form another approximate triangle,
similar to the original triangle, though with its orientation reversed. It is
straightforward to determine when the roots of $S_{n}(z;\mu)$ coalesce using
discriminants of polynomials. Suppose that

$f(z)=z^{m}+a_{m-1}z^{m-1}+\ldots+a_{1}z+a_{0},$

is a monic polynomial of degree $m$ with roots
$\alpha_{1},\alpha_{2},\ldots,\alpha_{m}$, so
$f(z)=\prod_{j=1}^{m}(z-\alpha_{j})$. Then the discriminant of $f(z)$ is

$\mbox{\rm Dis}(f)=\prod_{1\leq j<k\leq m}(\alpha_{j}-\alpha_{k})^{2}.$

Hence the polynomial $f$ has a multiple root when $\mbox{\rm Dis}(f)=0$.

$\displaystyle\mbox{\rm Dis}(S_{2}(z;\mu))$ $\displaystyle=-3^{3}\mu^{2},$
$\displaystyle\mbox{\rm Dis}(S_{3}(z;\mu))$
$\displaystyle=3^{12}5^{5}\mu^{6}(\mu^{2}-1)^{2},$ $\displaystyle\mbox{\rm
Dis}(S_{4}(z;\mu))$
$\displaystyle=3^{27}5^{20}7^{7}\mu^{14}(\mu^{2}-1)^{6}(\mu^{2}-4)^{2},$
$\displaystyle\mbox{\rm Dis}(S_{5}(z;\mu))$
$\displaystyle=3^{66}5^{45}7^{28}\mu^{26}(\mu^{2}-1)^{14}(\mu^{2}-4)^{6}(\mu^{2}-9)^{2},$
$\displaystyle\mbox{\rm Dis}(S_{6}(z;\mu))$
$\displaystyle=-3^{147}5^{80}7^{63}11^{11}\mu^{44}(\mu^{2}-1)^{26}(\mu^{2}-4)^{14}(\mu^{2}-9)^{6}(\mu^{2}-16)^{2}.$
Table 2.2:  The discriminants of the Umemura polynomials $S_{n}(z;\mu)$.

The discriminants of the first few Umemura polynomials $S_{n}(z;\mu)$ are
given in Table 2.2. Thus $S_{2}(z;\mu)$ has multiple roots when $\mu=0$,
$S_{3}(z;\mu)$ has multiple roots when $\mu=0,\pm 1$, $S_{4}(z;\mu)$ has
multiple roots when $\mu=0,\pm 1,\pm 2$, $S_{5}(z;\mu)$ has multiple roots
when $\mu=0,\pm 1,\pm 2,\pm 3$, and $S_{6}(z;\mu)$ has multiple roots when
$\mu=0,\pm 1,\pm 2,\pm 3,\pm 4$. Further the multiple roots occur at $z=0$.
This leads to the following theorem.

######  Theorem 2.7.

The discriminant of the polynomial $S_{n}(z;\mu)$ is given by

$|\mbox{\rm
Dis}(S_{n})|=\prod_{j=1}^{n}(2j+1)^{(2j+1)(n-j)^{2}}\prod_{k=-n}^{n}(\mu-k)^{c_{n-|k|}},$

where
$c_{n}=\frac{1}{6}n^{3}+\tfrac{1}{4}n^{2}-\tfrac{1}{6}n-\tfrac{1}{8}[1-(-1)^{n}]$
and $\mbox{\rm Dis}(S_{n})<0$ if and only $n=2\ \mbox{mod}\ 4$. Further the
polynomial $S_{n}(z;\mu)$ has multiple roots at $z=0$ when $\mu=0,\pm 1,\pm
2,\ldots,\pm(n-2)$.

###### Proof.

See Amdeberhan [1].∎

Using the Hamiltonian formalism for PIII, it is shown in [11] that the
polynomials $S_{n}(z;\mu)$ satisfy an fourth order bilinear equation and a
sixth order, hexa-linear (homogeneous of degree six) difference equation.

##  3 Application of Taneda’s method

In this section, we use the algebraic method due to Taneda [68] to prove that
the rational functions $S_{n}$ satisfying (2.3) are indeed polynomials,
assuming all the functions $S_{n}$’s have simple zeros.

We define an operator $\mathcal{L}_{z}$ as follows

$\mathcal{L}_{z}(f)=f\frac{d^{2}{f}}{{d{z}}^{2}}-\left(\frac{d{f}}{{d{z}}}\right)^{2}+\frac{f}{z}\frac{d{f}}{{d{z}}}.$
(3.1)

######  Lemma 3.1.

Let $f(z)$ and $g(z)$ be arbitrary polynomials. Then

  1. (a)

$\displaystyle\mathcal{L}_{z}(kf)=k^{2}\,\mathcal{L}_{z}(f)$, with $k$ a
constant;

  2. (b)

$\displaystyle\mathcal{L}_{z}(f\,g)=f^{2}\mathcal{L}_{z}(g)+g^{2}\mathcal{L}_{z}(f)$;

  3. (c)

If $\displaystyle h=-z\mathcal{L}_{z}(f)+k(z+\mu)f^{2}$, with $k$ and $\mu$
constants, then

$f\,|\,z\mathcal{L}_{z}(h)-2k(z+\mu)h^{2},$ (3.2)

where the symbol $|$ means that the right hand side is divisible by the left
hand side.

###### Proof.

  1.   2. (a)

This follows directly from the definition.

  3. (b)

We observe that

$\displaystyle\mathcal{L}_{z}(fg)$
$\displaystyle=fg\frac{d^{2}{}}{{d{z}}^{2}}(fg)-\left[\frac{d{}}{{d{z}}}(fg)\right]^{2}+\dfrac{fg}{z}\frac{d{}}{{d{z}}}(fg)$
$\displaystyle=f^{2}\left[g\frac{d^{2}{g}}{{d{z}}^{2}}-\left(\frac{d{g}}{{d{z}}}\right)^{2}+\frac{g}{z}\frac{d{g}}{{d{z}}}\right]+g^{2}\left[f\frac{d^{2}{f}}{{d{z}}^{2}}-\left(\frac{d{f}}{{d{z}}}\right)^{2}+\frac{f}{z}\frac{d{f}}{{d{z}}}\right]$
$\displaystyle=f^{2}\mathcal{L}_{z}(g)+g^{2}\mathcal{L}_{z}(f),$

so the result is valid.

  4. (c)

Finally, by definition

$\displaystyle h$
$\displaystyle=-z\left[f\frac{d^{2}{f}}{{d{z}}^{2}}-\left(\frac{d{f}}{{d{z}}}\right)^{2}+\dfrac{f}{z}\frac{d{f}}{{d{z}}}\right]+k(z+\mu)f^{2}$
$\displaystyle=z\left(\frac{d{f}}{{d{z}}}\right)^{2}+f\times(a\ polynomial)$
$\displaystyle\frac{d{h}}{{d{z}}}$
$\displaystyle=-2f\frac{d^{2}{f}}{{d{z}}^{2}}-z\left(f\frac{d^{3}{f}}{{d{z}}^{3}}-\frac{d{f}}{{d{z}}}\frac{d^{2}{f}}{{d{z}}^{2}}\right)+kf^{2}+2k(z+\mu)f\frac{d{f}}{{d{z}}}$
$\displaystyle=z\frac{d{f}}{{d{z}}}\frac{d^{2}{f}}{{d{z}}^{2}}+f\times(a\
polynomial)$ $\displaystyle\frac{d^{2}{h}}{{d{z}}^{2}}$
$\displaystyle=-\frac{d{f}}{{d{z}}}\frac{d^{2}{f}}{{d{z}}^{2}}-3f\frac{d^{3}{f}}{{d{z}}^{3}}-z\left[f\frac{d^{4}{f}}{{d{z}}^{4}}-\left(\frac{d^{2}{f}}{{d{z}}^{2}}\right)^{2}\right]+4kf\frac{d{f}}{{d{z}}}+2k(z+\mu)\left[f\frac{d^{2}{f}}{{d{z}}^{2}}+\left(\frac{d{f}}{{d{z}}}\right)^{2}\right]$
$\displaystyle=-\frac{d{f}}{{d{z}}}\frac{d^{2}{f}}{{d{z}}^{2}}+z\left(\frac{d^{2}{f}}{{d{z}}^{2}}\right)^{2}+2k(z+\mu)\left(\frac{d{f}}{{d{z}}}\right)^{2}+f\times(a\
polynomial).$

Then we can see

$\displaystyle\mathcal{L}_{z}(h)$
$\displaystyle=h\frac{d^{2}{h}}{{d{z}}^{2}}-\left(\frac{d{h}}{{d{z}}}\right)^{2}+\dfrac{h}{z}\frac{d{h}}{{d{z}}}$
$\displaystyle=2z(z+\mu)\left(\frac{d{f}}{{d{z}}}\right)^{4}-f\frac{d{f}}{{d{z}}}\left(\frac{d^{2}{f}}{{d{z}}^{2}}\right)^{2}+\dfrac{2k(z+\mu)f}{z}\left(\frac{d{f}}{{d{z}}}\right)^{3}+f\times(a\
polynomial)$
$\displaystyle=2z(z+\mu)\left(\frac{d{f}}{{d{z}}}\right)^{4}+\dfrac{2k\mu
f}{z}\left(\frac{d{f}}{{d{z}}}\right)^{3}+f\times(a\ polynomial).$

Since $z\mathcal{L}_{z}(h)-2k(z+\mu)h^{2}=f\times(a\ polynomial)$, then

$f\,|\,z\mathcal{L}_{z}(h)-2k(z+\mu)h^{2},$ (3.3)

as required.

∎

######  Theorem 3.2.

Suppose $\\{S_{n}\\}$ is a sequence of rational functions with simple nonzero
roots, satisfying (2.3). For all $N\in\mathbb{N}\cup\\{0\\}$, if $z=0$ is not
a root of any $S_{n}(z)$ for $0\leq n\leq N$, then:

  1. (a)

$S_{N+1}(z)$ is a polynomial of degree $\tfrac{1}{2}(N+1)(N+2)$;

  2. (b)

$S_{N+1}(z)$ and $S_{N}(z)$ do not have a common root.

###### Proof.

We shall prove these using induction. We have $S_{-1}(z)=S_{0}(z)=1$, then
$S_{1}(z)=z+\mu$ and $S_{2}(z)=(z+\mu)^{3}-\mu$. Clearly, (a) and (b) hold for
$n=0,1,2$, when $\mu\neq 0,\pm 1$. We assume that (a) and (b) hold for
$n=N-2,N-1,N$, with $N\geq 2$. Then we will prove that the statements also
hold for $n=N+1$.

We let $f$ be $S_{N-1}$. Then $n=N-1$ and $h=S_{N}S_{N-2}$ in Lemma 3.1. Then
(3.3) becomes

$S_{N-1}\,|\,z\mathcal{L}_{z}(S_{N}S_{N-2})-2k(z+\mu)(S_{N}S_{N-2})^{2}.$
(3.4)

Hence

$\displaystyle\left[\mathcal{L}_{z}(S_{N}S_{N-2})-\dfrac{2(z+\mu)}{z}(S_{N}S_{N-2})^{2}\right](-z)$
$\displaystyle\qquad=-z\left[S_{N-2}^{2}\mathcal{L}_{z}(S_{N})+S_{N}^{2}\mathcal{L}_{z}(S_{N-2})\right]+2(z+\mu)(S_{N}S_{N-2})^{2}$
$\displaystyle\qquad=S_{N-2}^{2}[-z\mathcal{L}_{z}(S_{N})+(z+\mu)S^{2}_{N}]+S_{N}^{2}[-z\mathcal{L}_{z}(S_{N-2})+(z+\mu)S^{2}_{N-2}]$
$\displaystyle\qquad=S_{N-2}^{2}[-z\mathcal{L}_{z}(S_{N})+(z+\mu)S^{2}_{N}]+S_{N}^{2}S_{N-1}S_{N-3}.$

Then by (3.3) and (b) with $n=N-1$, we have

$S_{N-1}\,|\,-z\mathcal{L}_{z}(S_{N})+(z+\mu)S^{2}_{N}=-z\left[S_{N}\frac{d^{2}{S_{N}}}{{d{z}}^{2}}-\left(\frac{d{S_{N}}}{{d{z}}}\right)^{2}\right]-S_{N}\frac{d{S_{N}}}{{d{z}}}+(z+\mu)S_{N}^{2}$

So, according to (2.3), $S_{N+1}$ is a polynomial by induction.

To prove (b), we use a proof by contradiction. If $S_{N}$ and $S_{N-1}$ have
the same root $z_{0}\neq 0$, then by (2.3), $z_{0}$ is also a root of

$S_{N}\frac{d^{2}{S_{n}}}{{d{z}}^{2}}-\left(\frac{d{S_{N}}}{{d{z}}}\right)^{2},$

and hence $\displaystyle\frac{d{S_{N}}}{{d{z}}}$. This implies $z_{0}$ is at
least a double root of $S_{N}$, which contradicts our assumption about
$S_{N}$. ∎

##  4 Further analysis on $S_{n}(0,\mu)$

######  Theorem 4.1.

Let $\phi_{n}=S_{n}(0,\mu)$, and

$\phi_{n}^{\prime}:=\frac{\partial S_{n}}{\partial
z}(0,\mu),\qquad\phi_{n}^{\prime\prime}:=\frac{\partial^{2}S_{n}}{\partial
z^{2}}(0,\mu),$

etc. Then for all $n\geq 3$,

$\displaystyle\phi_{n+1}$
$\displaystyle=\frac{\phi_{n}\phi_{n-1}}{\phi_{n-2}}\,\left(2\mu^{2}-2n^{2}+2n-1-\frac{\phi_{n}\phi_{n-3}}{\phi_{n-1}\phi_{n-2}}\right);$
(4.1) $\displaystyle\phi^{\prime}_{n+1}$
$\displaystyle=-\frac{\phi_{n}\phi_{n+2}}{\phi_{n+1}}+\mu\phi_{n+1}.$ (4.2)

###### Proof.

Differentiating (2.3) with respect to $z$ gives

$\displaystyle\frac{d{S_{n+1}}}{{d{z}}}=\frac{1}{S_{n-1}}$
$\displaystyle\left\\{S_{n}^{2}+2(z+\mu)S_{n}\frac{d{S_{n}}}{{d{z}}}-2S_{n}\frac{d^{2}{S_{n}}}{{d{z}}^{2}}\right.$
$\displaystyle\qquad\left.+\
z\left(\frac{d{S_{n}}}{{d{z}}}\frac{d^{2}{S_{n}}}{{d{z}}^{2}}-S_{n}\frac{d^{3}{S_{n}}}{{d{z}}^{3}}\right)-S_{n+1}\frac{d{S_{n-1}}}{{d{z}}}\right\\}.$
(4.3)

Substitute $z=0$ into (2.3) and (4.3). We obtain

$\phi_{n+1}=\frac{\phi_{n}}{\phi_{n-1}}\left(\mu\phi_{n}-\phi_{n}^{\prime}\right),$
(4.4)

and

$\phi_{n+1}^{\prime}=\frac{\phi_{n}}{\phi_{n-1}}\left(\phi_{n}+2\mu\phi_{n}^{\prime}-2\phi_{n}^{\prime\prime}-\frac{\phi_{n-1}^{\prime}\phi_{n+1}}{\phi_{n}}\right).$
(4.5)

Now (4.4) implies that (4.2) is valid. Furthermore, in [11, p. 9519],

$\displaystyle z^{2}$
$\displaystyle\left[S_{n}\frac{d^{4}{S_{n}}}{{d{z}}^{4}}-4\frac{d{S_{n}}}{{d{z}}}\frac{d^{3}{S_{n}}}{{d{z}}^{3}}+3\left(\frac{d^{2}{S_{n}}}{{d{z}}^{2}}\right)^{2}\right]+2z\left(S_{n}\frac{d^{3}{S_{n}}}{{d{z}}^{3}}-\frac{d{S_{n}}}{{d{z}}}\frac{d^{2}{S_{n}}}{{d{z}}^{2}}\right)$
$\displaystyle\quad-4z(z+\mu)\left[S_{n}\frac{d^{2}{S_{n}}}{{d{z}}^{2}}-\left(\frac{d{S_{n}}}{{d{z}}}\right)^{2}\right]-2S_{n}\frac{d^{2}{S_{n}}}{{d{z}}^{2}}+4\mu
S_{n}\frac{d{S_{n}}}{{d{z}}}=2n(n+1)S_{n}^{2}.$ (4.6)

This implies, as $\phi_{n}$ is not identically zero, that

$2\mu\phi_{n}^{\prime}-\phi_{n}^{\prime\prime}=n(n+1)\phi_{n}.$ (4.7)

Hence by (4.2),

$\phi_{n}^{\prime\prime}=2\mu\phi_{n}^{\prime}-n(n+1)\phi_{n}=[2\mu^{2}-n(n+1)]\phi_{n}-\frac{2\mu\phi_{n-1}\phi_{n+1}}{\phi_{n}}.$

Now substitute this equation and (4.2) into (4.5) to obtain, after
simplification,

$-\frac{\phi_{n}\phi_{n+2}}{\phi_{n+1}}=\frac{\phi_{n}^{2}}{\phi_{n-1}}\,(2n^{2}+2n+3-2\mu^{2})+\frac{\phi_{n+1}\phi_{n}\phi_{n-2}}{\phi_{n-1}^{2}}.$

Therefore we have

$\phi_{n+2}=\frac{\phi_{n}\phi_{n+1}}{\phi_{n-1}}\,\left(2\mu^{2}-2n^{2}-2n-3-\frac{\phi_{n+1}\phi_{n-2}}{\phi_{n}\phi_{n-1}}\right).$

Thus (4.1) is also valid. ∎

######  Corollary 4.2.

  1.   2. (a)

For all $n\in\mathbb{N}$,

$\phi_{n}(\mu)=\mu^{\gamma_{0}^{n}}(\mu^{2}-1)^{\gamma_{1}^{n}}(\mu^{2}-4)^{\gamma_{2}^{n}}\ldots\left[\mu^{2}-(n-1)^{2}\right]^{\gamma_{n-1}^{n}},$

where for $0\leq j<k$,

$\displaystyle\gamma_{2j}^{n}$
$\displaystyle=\lceil\frac{n}{2}\rceil-j=k-j\quad\mbox{if }n=2k\mbox{ or
}2k-1;$ $\displaystyle\gamma_{2j+1}^{n}$
$\displaystyle=\lfloor\frac{n}{2}\rfloor-j=k-j\quad\mbox{if }n=2k\mbox{ or
}2k+1.$

  3. (b)

$\displaystyle\phi_{n}^{\prime}(\mu)=\phi_{n-1}(\mu)g(\mu)$, where $g(\mu)$ is
a polynomial of degree $n$ or $n-2$ according as $n$ is even or odd.

###### Proof.

If $n=2k$ is even, then by induction hypothesis,

$\begin{split}\phi_{2k}&=\mu^{k}(\mu^{2}-1)^{k}(\mu^{2}-4)^{k-1}(\mu^{2}-9)^{k-1}\ldots\left[\mu^{2}-(2k-2)^{2}\right]\left[\mu^{2}-(2k-1)^{2}\right].\\\
&=\mu^{k}(\mu^{2}-1)^{k}\prod_{j=1}^{k-1}\left[\mu^{2}-(2j)^{2}\right]^{k-j}\left[\mu^{2}-(2j+1)^{2}\right]^{k-j}.\end{split}$

while $\phi_{n-1}$ and $\phi_{n-2}$ have similar expressions. Thus

$\begin{split}\frac{\phi_{n}\phi_{n-1}}{\phi_{n-2}}&=\mu^{k+1}(\mu^{2}-1)^{k}(\mu^{2}-4)^{k}(\mu^{2}-9)^{k-1}\ldots\\\
&\qquad\qquad\qquad\ldots\left[\mu^{2}-(2k-2)^{2}\right]^{2}\left[\mu^{2}-(2k-1)^{2}\right].\end{split}$

Hence it follows from Theorem 4.1 that

$\begin{split}\phi_{n+2}&=\mu^{k+1}(\mu^{2}-1)^{k}(\mu^{2}-4)^{k}(\mu^{2}-9)^{k-1}\ldots\\\
&\qquad\qquad\qquad\ldots\left[\mu^{2}-(2k-2)^{2}\right]^{2}\left[\mu^{2}-(2k-1)^{2}\right]\left[\mu^{2}-(2k)^{2}\right].\end{split}$

The situation is similar for odd $n$.

We also consider odd $n$’s and even $n$’s in part (b). When $n=2k$, we note
that

$\phi_{2k}=\phi_{2k-1}\prod_{j=1}^{k}\left[\mu^{2}-(2j-1)^{2}\right].$

Hence by Theorem 4.1,

$\phi_{2k}^{\prime}=\mu\phi_{2k-1}\left\\{\prod_{j=1}^{k}\left[\mu^{2}-(2j-1)^{2}\right]-\prod_{j=1}^{k-1}\left[\mu^{2}-(2j)^{2}\right]\right\\}.$

When $n=2k+1$, then

$\phi_{2k+1}=\phi_{2k}\prod_{j=0}^{k}\left[\mu^{2}-(2j)^{2}\right],$

and

$\phi_{2k+1}^{\prime}=\phi_{2k}\left\\{\prod_{j=0}^{k}\left[\mu^{2}-(2j)^{2}\right]-\prod_{j=0}^{k}\left[\mu^{2}-(2j+1)^{2}\right]\right\\}.$

∎

######  Remark 4.3.

Part (a) of Theorem 4.2 above means that $z=0$ is a root of $S_{n}$ if and
only if $\mu=0,\pm 1,\pm 2,\ldots,\pm(n-1)$. In particular, the first few
$\phi_{n}(\mu)$’s are

$\displaystyle\phi_{1}(\mu)$ $\displaystyle=\mu,$ $\displaystyle\phi_{2}(\mu)$
$\displaystyle=\mu(\mu^{2}-1),$ $\displaystyle\phi_{3}(\mu)$
$\displaystyle=\mu^{2}(\mu^{2}-1)(\mu^{2}-4),$ $\displaystyle\phi_{4}(\mu)$
$\displaystyle=\mu^{2}(\mu^{2}-1)^{2}(\mu^{2}-4)(\mu^{2}-9),$
$\displaystyle\phi_{5}(\mu)$
$\displaystyle=\mu^{3}(\mu^{2}-1)^{2}(\mu^{2}-4)^{2}(\mu^{2}-9)(\mu^{2}-16).$

######  Theorem 4.4.

Fix $n\in\mathbb{N}$. If $\mu=\pm n$, then for the recurrence relation (2.3)
and starting polynomials $S_{-1}=S_{0}=1$, we have:

  1. (a)

all the zeros of rational functions $S_{k}$ are simple, when $k=0,1,\ldots,n$;

  2. (b)

each $S_{k}$ is a monic polynomial in $z$ of degree $\tfrac{1}{2}k(k+1)$, for
$k=0,1,\ldots,n$.

###### Proof.

We shall make use of the identity (4.6) again. Suppose $z_{0}\neq 0$ is a root
of $S_{n}$. Then from (4.6),

$3\left[\frac{d^{2}{S_{n}}}{{d{z}}^{2}}(z_{0})\right]^{2}=\frac{d{S_{n}}}{{d{z}}}(z_{0})\left[4\frac{d^{3}{S_{n}}}{{d{z}}^{3}}(z_{0})+2\frac{d^{2}{S_{n}}}{{d{z}}^{2}}(z_{0})+4(z_{0}+\mu)\frac{d{S_{n}}}{{d{z}}}(z_{0})\right].$

Hence if $z_{0}$ is a root of $\displaystyle\frac{d{S_{n}}}{{d{z}}}$, then it
also has to be a root of $\displaystyle\frac{d^{2}{S_{n}}}{{d{z}}^{2}}$. That
is, if $z_{0}$ is not a simple root of $S_{n}$, then its order $\geq 3$. Let
$k$ be the order of the root $z_{0}$. Analyzing on the identity (4.6), the
term

$S_{n}\frac{d^{3}{S_{n}}}{{d{z}}^{3}}-4\frac{d{S_{n}}}{{d{z}}}\frac{d^{3}{S_{n}}}{{d{z}}^{3}}+3\left(\frac{d^{2}{S_{n}}}{{d{z}}^{2}}\right)^{2},$

has the zero $z_{0}$ with order $2k-4$, while the other terms has order
greater than or equal to $2k+3$. Therefore let $S_{n}(z)=(z-z_{0})^{k}g(z)$,
where $g(z)$ is a polynomial and $\displaystyle\frac{d{g}}{{d{z}}}(z_{0})\neq
0$. The there exists a polynomial $h(z)$ such that

$S_{n}\frac{d^{4}{S_{n}}}{{d{z}}^{4}}-4\frac{d{S_{n}}}{{d{z}}}\frac{d^{3}{S_{n}}}{{d{z}}^{3}}+3\left(\frac{d^{2}{S_{n}}}{{d{z}}^{2}}\right)^{2}=(z-z_{0})^{2k-4}\left[(z-z_{0})h(z)+6k(k-1)g^{2}(z)\right].$

But it must have $z_{0}$ as a root, which gives a contradiction. We see that
every $z_{0}$ is at most a simple root. ∎

######  Theorem 4.5.

Let $\mu\in\mathbb{Z}\setminus\\!\\{0\\}$. Suppose that
$S_{n}(z)=z^{\sigma}g(z)$, where $g(z)=\sum_{j=0}^{k}a_{j}z^{j}$, with
$a_{0}\neq 0$. Then

  1. (a)

$a_{1}=\mu a_{0}$; and

  2. (b)

if $\sigma=\tfrac{1}{2}(n-|\mu|)(n-|\mu|+1)$, then

$a_{2}=\tfrac{1}{2}\left(\mu^{2}-\frac{|\mu|}{2m+1}\right)a_{0}.$

###### Proof.

We use the auxilliary identity (4.6) for the proof. First

$\frac{d{S_{n}}}{{d{z}}}=z^{\sigma}\frac{d{g}}{{d{z}}}+\sigma z^{\sigma-1}g,$
(4.8a) and
$\displaystyle\frac{d^{2}{S_{n}}}{{d{z}}^{2}}=z^{\sigma}\frac{d^{2}{g}}{{d{z}}^{2}}$
$\displaystyle+2\sigma
z^{\sigma-1}\frac{d{g}}{{d{z}}}+\sigma(\sigma-1)z^{\sigma-2}g,$ (4.8b)
$\displaystyle\frac{d^{3}{S_{n}}}{{d{z}}^{3}}=z^{\sigma}\frac{d^{3}{g}}{{d{z}}^{3}}$
$\displaystyle+3\sigma
z^{\sigma-1}\frac{d^{2}{g}}{{d{z}}^{2}}+3\sigma(\sigma-1)z^{\sigma-2}\frac{d{g}}{{d{z}}}+\sigma(\sigma-1)(\sigma-2)z^{\sigma-3}g,$
(4.8c)
$\displaystyle\frac{d^{4}{S_{n}}}{{d{z}}^{4}}=z^{\sigma}\frac{d^{4}{g}}{{d{z}}^{4}}$
$\displaystyle+4\sigma
z^{\sigma-1}\frac{d^{3}{g}}{{d{z}}^{3}}+6\sigma(\sigma-1)z^{\sigma-2}\frac{d^{2}{g}}{{d{z}}^{2}}+4\sigma(\sigma-1)(\sigma-2)z^{\sigma-3}\frac{d{g}}{{d{z}}}$
$\displaystyle+\sigma(\sigma-1)(\sigma-2)(\sigma-3)z^{\sigma-4}g.$ (4.8d)

Express (4.6) as

$\displaystyle 2n(n+1)S_{n}^{2}$
$\displaystyle+4z^{2}\left[S_{n}\frac{d^{2}{S_{n}}}{{d{z}}^{2}}-\left(\frac{d{S_{n}}}{{d{z}}}\right)^{2}\right]$
$\displaystyle=z^{2}\left[S_{n}\frac{d^{4}{S_{n}}}{{d{z}}^{4}}-4\frac{d{S_{n}}}{{d{z}}}\frac{d^{3}{S_{n}}}{{d{z}}^{3}}+3\left(\frac{d^{2}{S_{n}}}{{d{z}}^{2}}\right)^{2}\right]+2z\left(S_{n}\frac{d^{3}{S_{n}}}{{d{z}}^{3}}-\frac{d{S_{n}}}{{d{z}}}\frac{d^{2}{S_{n}}}{{d{z}}^{2}}\right)$
$\displaystyle\qquad\qquad-4\mu
z\left[S_{n}\frac{d^{2}{S_{n}}}{{d{z}}^{2}}-\left(\frac{d{S_{n}}}{{d{z}}}\right)^{2}\right]-2S_{n}\frac{d^{2}{S_{n}}}{{d{z}}^{2}}+4\mu
S_{n}\frac{d{S_{n}}}{{d{z}}}.$ (4.9)

Then we substitute (4.8) into (4.9) and obtain, after simplification,

$\displaystyle[2n(n+1)-4\sigma]z^{2\sigma}g^{2}+\ldots$
$\displaystyle\qquad=-8\sigma z^{2\sigma-1}g\frac{d{g}}{{d{z}}}+8\mu\sigma
z^{2\sigma-1}g^{2}-(8\sigma+2)z^{2\sigma}g\frac{d^{2}{g}}{{d{z}}^{2}}+8\sigma
z^{2\sigma}\left(\frac{d{g}}{{d{z}}}\right)^{2}+4\mu
z^{2\sigma}g\frac{d{g}}{{d{z}}}+\ldots$

Comparing coefficients of $\displaystyle z^{2\sigma-1}$ in the resulting
polynomials, we obtain

$8\sigma a_{0}a_{1}-8\mu\sigma a_{0}^{2}=0.$

This implies part (a). Next we compare coefficients of $z^{2\sigma}$ to get

$[2n(n+1)-4\sigma]a_{0}^{2}=-8\sigma(a_{1}^{2}+2a_{0}a_{2})+(16\sigma+4)\sigma
a_{0}a_{1}-(16\sigma+4)a_{0}a_{2}+8\sigma a_{1}^{2}.$

Since $a_{1}=\mu a_{0}$, we have

$(32\sigma+4)a_{0}a_{2}=[\mu^{2}(16\sigma+4)-2n(n+1)+4\sigma]a_{0}^{2},$

so that

$a_{2}=\frac{\mu^{2}(16\sigma+4)-2n(n+1)+4\sigma}{32\sigma+4}\,a_{0}.$

Now let $m=n-|\mu|$ and $\displaystyle\sigma=\tfrac{1}{2}m(m+1)$,

$2n(n+1)=2[\mu^{2}+(2m+1)|\mu|+m(m+1)],$

while

$8\sigma+1=(2m+1)^{2}.$

Therefore part (b) is valid. ∎

######  Theorem 4.6.

Let $\mu\in\mathbb{Z}$. Then for all $n>|\mu|$:

  1. (a)

$z=0$ is a root of order $\displaystyle\tfrac{1}{2}(n-|\mu|)(n-|\mu|+1)$ for
$S_{n}$;

  2. (b)

$S_{n}$ is a monic polynomial of degree $\tfrac{1}{2}n(n+1)$;

  3. (c)

all other roots of $S_{n}$ are simple.

###### Proof.

Part (c) follows from the proof of Theorem 4.4. For parts (a) and (b), the
case when $\mu=0$ is simple. For then, $S_{n}(z)=z^{\sigma_{n}}$. In general,
let $\mu\in\mathbb{Z}\setminus\\{0\\}$. Let $k=|\mu|$, by Corollary 4.2, $z=0$
is a root of $S_{k+1}$. Let $S_{k+1}=zg_{1,\mu}(z)$. By (4.3) and (4.7),

$\frac{d{S_{k+1}}}{{d{z}}}(0;k)=\frac{\phi_{k}(k)\left[\phi_{k}(k)+2\mu\phi_{k}^{\prime}(k)-\phi_{k}^{\prime\prime}(k)\right]}{\phi_{k-1}(k)}=\frac{\phi_{k}^{2}(k)}{\phi_{k-1}(k)}(k^{2}+k+1)\neq
0.$

Thus $z$ is a simple root of $S_{k+1}$. So $S_{k+1}=zg_{1,\mu}(z)$ where
$g_{1,\mu}=\sum a_{j}^{1,\mu}x^{j}$ is a degree $\sigma_{k+1}-1$ polynomial
with nonzero roots.

Let $n=k+1$. By (2.3),

$S_{k}S_{k+2}=z^{3}\left[\left(\frac{d{g_{1,\mu}}}{{d{z}}}\right)^{2}-g_{1,\mu}\frac{d^{2}{g_{1,\mu}}}{{d{z}}^{2}}+g_{1,\mu}^{2}\right]+z^{2}\left(\mu
g_{1,\mu}^{2}-g_{1,\mu}\frac{d{g_{1,\mu}}}{{d{z}}}\right).$

For the right hand side, the coefficient of $z^{3}$ is given by

$a_{0}^{1,\mu}\left(a_{0}^{1,\mu}+2\mu
a_{1}^{1,\mu}-4a_{2}^{1,\mu}\right)=\left(a_{0}^{1,\mu}\right)^{2}(1+\tfrac{2}{3}\mu)\neq
0,$

by Lemma 4.5. Hence $\displaystyle S_{k+2}=z^{3}g_{2,\mu}(z)$ where
$g_{2,\mu}$ had nonzero constant term. Use induction on $m=n-|\mu|>1$. Let

$S_{n-1}=z^{\sigma_{m-1}}g_{m-1,\mu},\qquad S_{n}=z^{\sigma_{m}}g_{m,\mu},$

where $g_{m-1,\mu}$ and $g_{m,\mu}$ are polynomials of degrees $\sigma_{m-1}$
and $\sigma_{m}$ respectively with nonzero roots. Then by (2.3),

$z^{\sigma_{m-1}}g_{m-1,\mu}S_{m-1}=z^{2\sigma_{m}}\left(\mu
g_{m,\mu}^{2}-g_{m,\mu}\frac{d{g_{m,\mu}}}{{d{z}}}\right)+z^{2\sigma_{m}+1}\left[\left(\frac{d{g_{m,\mu}}}{{d{z}}}\right)^{2}-g_{m,\mu}\frac{d^{2}{g_{m,\mu}}}{{d{z}}^{2}}+g_{m,\mu}^{2}\right].$

Let $a_{j}^{m,\mu}$ be the coefficients of $g_{m,\mu}$. By Lemma 4.5,
$\displaystyle a_{1}^{m,\mu}=\mu a_{0}^{m,\mu}$. So we may write
$S_{n+1}=z^{\sigma_{m+1}}g_{m+1,\mu}$, where

$\displaystyle a_{0}^{m+1,\mu}a_{0}^{m-1,\mu}$ $\displaystyle=2\mu
a_{0}^{m,\mu}a_{1}^{m,\mu}-4a_{0}^{m\mu}a_{2}^{m,\mu}+\left(a_{0}^{m,\mu}\right)^{2}$
$\displaystyle=a_{0}^{m,\mu}\left(2\mu
a_{1}^{m,\mu}+a_{0}^{m,\mu}-4a_{2}^{m,\mu}\right)$
$\displaystyle=\left(a_{0}^{m,\mu}\right)^{2}\left(1+\frac{2|\mu|}{2m+1}\right).$

So $a_{0}^{m+1,\mu}$ is nonzero, and the rational function $g_{m+1,\mu}$ does
not have $z=0$ as a root.

Next we want to show that $g_{m+1,\mu}$ is a polynomial. From the proof of
Theorem 4.4 and Theorem 3.2(b), we know that all nonzero roots of $S_{n}$ and
$S_{n-1}$ are simple and not common. Furthermore, we still have

$S_{n-1}\,|\,S_{n-2}^{2}\left[-z\mathcal{L}_{z}(S_{n})+(z+\mu)S_{n}^{2}\right],$

where

$\Delta:=-z\mathcal{L}_{z}(S_{n})+(z+\mu)S_{n}^{2}=z\left[\left(\frac{d{S_{n}}}{{d{z}}}\right)^{2}-S_{n}\frac{d^{2}{S_{n}}}{{d{z}}^{2}}\right]-S_{n}\frac{d{S_{n}}}{{d{z}}}+(z+\mu)S_{n}^{2}.$

We conclude that $g_{m-1,\mu}$ divides $\Delta$, which means that
$g_{m+1,\mu}$ is also a polynomial. Consequently parts (a) and (b) follow by
induction. ∎

##  5 Conclusions

We have given a direct algebraic proof that the nonlinear recurrence relation
(2.3) generates polynomials $S_{n}$, rather than rational functions without
direct resort to the $\tau$-function theory of Painlevé equations. However we
critically needed a higher order equation derived from the corresponding
$\sigma$-equation, which seems to be inevitable in the nonlinear scenario. We
believe that the method can be developed to apply to the fifth Painlevé
equation (PV) as well, though we shall not pursue this further here.

## Acknowledgements

PAC thanks the Department of Applied Mathematics, National Sun Yat-sen
University, Kaohsiung, Taiwan and the Department of Mathematics, National
Taiwan University, Taipei, Taiwan, for their hospitality during his visit
where some of this work was done. CKL is partially supported by Ministry of
Science and Technology, Taiwan, under contract number MOST104-2115-M-110-008.

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