# On properties of a deformed Freud weight

Mengkun<EMAIL_ADDRESS><EMAIL_ADDRESS>Department of
Mathematics, University of Macau, Avenida da Universidade, Taipa, Macau, China
Yang<EMAIL_ADDRESS>Department of Mathematics, University of
Macau, Avenida da Universidade, Taipa, Macau, China

###### Abstract

We study the recurrence coefficients of the monic polynomials $P_{n}(z)$
orthogonal with respect to the deformed (also called semi-classical) Freud
weight

$w_{\alpha}(x;s,N)=|x|^{\alpha}{\rm
e}^{-N\left[x^{2}+s\left(x^{4}-x^{2}\right)\right]},~{}~{}x\in\mathbb{R},$

with parameters $\alpha>-1,~{}N>0,~{}s\in[0,1]$. We show that the recurrence
coefficients $\beta_{n}(s)$ satisfy the first discrete Painlevé equation
(denoted by d${\rm P_{I}}$), a differential-difference equation and a second
order nolinear ordinary differential equation (ODE) in $s$. Here $n$ is the
order of the Hankel matrix generated by $w_{\alpha}(x;s,N)$. We describe the
asymptotic behavior of the recurrence coefficients in three situations, (i)
$s\rightarrow 0$, $n,N$ finite, (ii) $n\rightarrow\infty$, $N$ finite, (iii)
$n,N\rightarrow\infty$, such that the radio $r:=\frac{n}{N}$ is bounded away
from $0$ and closed to $1$. We also investigate the existence and uniqueness
for the positive solutions of the d${\rm P_{I}}$.

Further more, we derive, using the ladder approach, a second order linear ODE
satisfied by the polynomials $P_{n}(z)$. It is found as $n\rightarrow\infty$,
the linear ODE turns to be a biconfluent Heun equation. This paper concludes
with the study of the Hankel determinant, $D_{n}(s)$, associated with
$w_{\alpha}(x;s,N)$ when $n$ tends to infinity.

Key words: Deformed Freud Weight; Unitary Random Matrices; Hankel Determinant;
Discrete and Continuous Painlevé Equation; Integrable Systems

##  1 Introduction

The orthogonal polynomials, recurrence coefficients, eigenvalues and the
relevant properties associated with the exponential weights ${\rm
e}^{-|x|^{m}}$, $m\in\mathbb{N}$ as well as various generalized forms of this
weight function on $\mathbb{R}$, have been investigated by many authors, Refs.
[3, 5, 9, 12, 13, 17, 19, 27, 29, 34$-$36].

Let $P_{n}(z)$ be the monic polynomials of degree $n$ orthogonal with respect
to the deformed Freud weight

$w(x)=w_{\alpha}(x;s,N)=|x|^{\alpha}{\rm
e}^{-N\left[x^{2}+s\left(x^{4}-x^{2}\right)\right]},~{}x\in\mathbb{R},~{}\alpha>-1,~{}N>0,~{}s\in[0,1],$
(1.1)

where the deformation $s\left(x^{4}-x^{2}\right)$, given by $0\leq s\leq 1$,
interpolates between the generalized Gaussian (Hermite) weight
($|x|^{\alpha}{\rm e}^{-Nx^{2}},~{}x\in\mathbb{R},~{}\alpha>-1,~{}N>0$) when
$s=0$ and the Freud weight ($|x|^{\alpha}{\rm
e}^{-Nx^{4}},~{}x\in\mathbb{R},~{}\alpha>-1,~{}N>0$) when $s=1$.

From the orthogonality condition, given by

$\int_{\mathbb{R}}P_{j}(x)P_{k}(x)w_{\alpha}(x;s,N)dx=h_{j}(s;\alpha,N)\delta_{jk},~{}~{}h_{j}(s;\alpha,N)>0,~{}~{}j,k\in\\{0,1,2,\ldots\\},$
(1.2)

where $\delta_{jk}$ denotes the Kronecker delta, and $h_{j}$ is the square of
$L^{2}$ norm of the monic polynomial $P_{j}(x)$, there follows the three term
recurrence relation,

$zP_{n}(z;s,\alpha,N)=P_{n+1}(z;s,\alpha,N)+\beta_{n}(s;\alpha,N)P_{n-1}(z;s,\alpha,N),~{}~{}n\geq
0,$ (1.3)

subject to the initial conditions

$P_{0}(z):=1~{}~{}~{}{\rm and}~{}~{}~{}\beta_{0}P_{-1}(z):=0.$

Multiplying both sides of Eq. (1.3) by $P_{n-1}(z)w_{\alpha}(z;s,N)$ and
integrating this with respect to $z$ on $\mathbb{R}$, which, due to the
orthogonality condition (1.2), gives us

$\beta_{n}(s)=\frac{1}{h_{n-1}(s)}\int_{\mathbb{R}}zP_{n}(z)P_{n-1}(z)w_{\alpha}(z;s,N)dx=\frac{h_{n}(s)}{h_{n-1}(s)}>0.$

It should be pointed out that $P_{n}(z)$ contains only the terms $z^{n-j}$,
$j\leq n$ and even, since our weight function $w_{\alpha}(x;s,N)$ is even on
$\mathbb{R}$. This implies that

$P_{n}(-z)=(-1)^{n}P_{n}(z)~{}~{}~{}{\rm and}~{}~{}~{}P_{n}(0)P_{n-1}(0)=0.$

Then we note the monic polynomials $P_{n}(z)$, associated with
$w_{\alpha}(x;s,N)$, can be normalized as [11],

$P_{2j}(z)=z^{2j}+\textbf{p}(2j;s)z^{2j-2}+\cdots+P_{2j}(0),$

and

$\begin{split}P_{2j+1}(z)&=z^{2j+1}+\textbf{p}(2j+1;s)z^{2j-1}+\cdots+{\rm
const.}z\\\ &=z\left(z^{2j}+\textbf{p}(2j+1;s)z^{2j-2}+\cdots+{\rm
const.}\right).\end{split}$

In unitary random matrix theory with the Hermitian ensemble, the joint
probability density function of the $n$ eigenvalues
$\left\\{x_{j}\right\\}_{j=1}^{n}$ is given in [31] by

$p\left(x_{1},\ldots,x_{n}\right)\prod_{j=1}^{n}dx_{j}=\frac{1}{D_{n}(s)}\frac{1}{n!}\prod_{1\leq
j<k\leq n}\left(x_{j}-x_{k}\right)^{2}\prod_{\ell=1}^{n}w(x_{\ell})dx_{\ell},$

where $D_{n}(s)$ denotes the normalization constant (also called partition
function), which reads

$D_{n}(s)=\frac{1}{n!}\int_{\mathbb{R}^{n}}\prod_{1\leq j<k\leq
n}\left(x_{k}-x_{j}\right)^{2}\prod_{\ell=1}^{n}w(x_{\ell})dx_{\ell},$

so that

$\int_{\mathbb{R}^{n}}p\left(x_{1},\ldots,x_{n}\right)\prod_{j=1}^{n}dx_{j}=1.$

For the problem at hand, the weight function $w(x)$ is given by

$w(x)=|x|^{\alpha}{\rm
e}^{-N\left[x^{2}+s\left(x^{4}-x^{2}\right)\right]},~{}x\in\mathbb{R},~{}\alpha>-1,~{}N>0,~{}s\in[0,1],$

and the moments are defined by

$\mu_{j}(s;\alpha,N)=\int_{\mathbb{R}}x^{j}|x|^{\alpha}{\rm
e}^{-N\left[x^{2}+s\left(x^{4}-x^{2}\right)\right]}dx,~{}~{}j\in\\{0,1,2,\ldots\\}.$

$D_{n}(s)$ admits two more alternative representations [41],

$D_{n}(s):=\det\left(\mu_{j+k}\right)_{j,k=0}^{n-1}=\prod_{j=0}^{n-1}h_{j},$
(1.4)

where $\det\left(\mu_{j+k}\right)_{j,k=0}^{n-1}$ is the determinant of the
Hankel matrix (or moment matrix).

For the deformed Freud weight (1.1), $\mu_{0}(s;\alpha,N)$ can be evaluated in
term of the parabolic cylinder (Hermite-Weber) function $D_{\nu}(\zeta)$. By
definition

$\begin{split}\mu_{0}(s;\alpha,N):&=\int_{-\infty}^{\infty}|x|^{\alpha}{\rm
e}^{-N\left[x^{2}+s\left(x^{4}-x^{2}\right)\right]}dx\\\
&=2\int_{0}^{\infty}x^{\alpha}{\rm
e}^{-N\left[x^{2}+s\left(x^{4}-x^{2}\right)\right]}dx\\\
&=\int_{0}^{\infty}y^{\frac{\alpha-1}{2}}{\rm
e}^{-N\left[y+s\left(y^{2}-y\right)\right]}dy\\\
&=(2Ns)^{-\frac{\alpha+1}{4}}\Gamma\left(\frac{\alpha+1}{2}\right)\exp\left[\frac{N(1-s)^{2}}{8s}\right]D_{-\frac{\alpha+1}{2}}\left[\frac{N(1-s)}{\sqrt{2Ns}}\right],\end{split}$

since the parabolic cylinder function $D_{\nu}(z)$ has the integral
representation [37, §12.5 (i)]

$D_{\nu}(z)=\frac{\exp\left(-\frac{z^{2}}{4}\right)}{\Gamma(-\nu)}\int_{0}^{\infty}t^{-\nu-1}\exp\left(-\frac{t^{2}}{2}-tz\right)dt,~{}~{}~{}\text{$\Re\nu<0$}.$

We note that the even moments are

$\mu_{2n}(s;\alpha,N)=\int_{-\infty}^{\infty}x^{2n}|x|^{\alpha}{\rm
e}^{-N\left[x^{2}+s\left(x^{4}-x^{2}\right)\right]}dx=\mu_{0}\left(s;\alpha+2n,N\right),~{}n\in\mathbb{N},$

whilst the odd ones are

$\mu_{2n+1}(s;\alpha,N)=\int_{-\infty}^{\infty}x^{2n+1}|x|^{\alpha}{\rm
e}^{-N\left[x^{2}+s\left(x^{4}-x^{2}\right)\right]}dx=0,~{}n\in\mathbb{N}.$

It should be pointed out our recurrence coefficients, moments, $L^{2}$ norm of
orthogonal polynomials, Hankel determinant depend on $s$, also parameters $N$
and $\alpha$, but we may not display them unless we have to.

######  Remark 1.1.

The dependence of the orthogonal polynomials $P_{n}(x;s,\alpha,N)$ on $s$,
$\alpha$ and $N$ can be seen from its determinant representation in terms of
the moments, or alternatively, from the Heine formula [41, Eq. 2.2.10]

$\begin{split}P_{n}(z;s,\alpha,N)&=\frac{1}{D_{n}(s;\alpha,N)}\frac{1}{n!}\int_{\mathbb{R}^{n}}\prod_{1\leq
j<k\leq
n}\left(x_{j}-x_{k}\right)^{2}\prod_{\ell=1}^{n}\left(z-x_{\ell}\right)w_{\alpha}(x_{\ell};s,N)dx_{\ell}\\\
&=\frac{\det\left(\int_{\mathbb{R}}x^{j+k}(z-x)w_{\alpha}(x;s,N)dx\right)_{j,k=0}^{n-1}}{\det\left(\int_{\mathbb{R}}x^{j+k}w_{\alpha}(x;s,N)dx\right)_{j,k=0}^{n-1}}.\end{split}$

The remainder of this paper is organized as follows. In Sect. 2 we prove that
the recurrence coefficients satisfy the first discrete Painlevé equation
(${\rm dP_{I}}$), a differential-difference equation and a second order ODE.
Sects. 3-5 discuss the asymptotic behavior of the recurrence coefficients in
three cases, i.e., $s\rightarrow 0$, $n,N$ fixed; $n\rightarrow\infty$, $N$
fixed; $n,N\rightarrow\infty$ but the quantity $n/N$ fixed. In Sect. 6, we
obtain a second order ODE satisfied by the deformed Freud orthogonal
polynomials applying the ladder operator approach, as well as a biconfluent
Heun equation when $n$ tends to infinity. Sect. 7 investigates the existence
and uniqueness for the solutions of the ${\rm dP_{I}}$. Finally, in Sect. 8,
with the aid of the compatibility conditions for the ladder operators, we find
the asymptotic expansion of the logarithm of the Hankel determinant associated
with our weight.

##  2 Recurrence coefficients of the deformed Freud polynomials

It is known [43, 17, 29] that the recurrence coefficients satisfy the
following second order nonlinear difference equation,

######  Proposition 2.1.

The recurrence coefficients $\beta_{n}(s)$ satisfy the first discrete Painlevé
equation, ${\rm dP_{I}}$,

$\beta_{n+1}(s)+\beta_{n}(s)+\beta_{n-1}(s)=\frac{z_{n}+\gamma(-1)^{n}}{\beta_{n}(s)}+\delta,$
(2.1)

with

$z_{n}=\frac{2n+\alpha}{8Ns},~{}~{}\gamma=-\frac{\alpha}{8Ns}~{}~{}{\rm
and}~{}~{}\delta=\frac{s-1}{2s}.$

###### Proof.

Writing

$w(x)=|x|^{\alpha}w_{0}(x),$

where

$w_{0}(x):={\rm e}^{-v_{0}(x)}~{}~{}~{}{\rm
with}~{}~{}~{}v_{0}(x):=Nsx^{4}+N(1-s)x^{2}.$

With the aid of the three term recurrence relation (1.3), repeatedly, we have

$\begin{split}\int_{-\infty}^{\infty}\left[P_{n}(x)P_{n-1}(x)|x|^{\alpha}\right]^{\prime}w_{0}(x)dx=&4Ns\int_{-\infty}^{\infty}x^{3}P_{n}(x)P_{n-1}(x)w(x)dx\\\
&+2N(1-s)\int_{-\infty}^{\infty}xP_{n}(x)P_{n-1}(x)w(x)dx\\\
=&4Ns\sqrt{\beta_{n}}\left(\beta_{n+1}+\beta_{n}+\beta_{n-1}\right)+2N(1-s)\sqrt{\beta_{n}},\end{split}$

Combining this result and using the identity [43, Eq. (13)],

$\int_{-\infty}^{\infty}\left[P_{n}(x)P_{n-1}(x)|x|^{\alpha}\right]^{\prime}\widetilde{w}(x)dx=\frac{n+\alpha\Delta_{n}}{\sqrt{\beta_{n}}},~{}~{}\Delta_{n}=\frac{1-(-1)^{n}}{2},$

which holds whenever $\widetilde{w}(x)$ is a symmetric weight on the real
line, then we get

$\frac{n+\alpha\Delta_{n}}{4Ns}=\beta_{n}\left[\beta_{n+1}+\beta_{n}+\beta_{n-1}+\frac{1-s}{2s}\right].$
(2.2)

∎

######  Remark 2.1.

Multiplying Eq. (2.2) by $s$, we see that
$\beta_{n}(0;\alpha,N)=\frac{n+\alpha\Delta_{n}}{2N}$.

######  Remark 2.2.

The nonlinear discrete equation (2.1) can be found from [17, Eq. (23), P. 5],
written by Freud, see also [1, §2]. Joshi and Lustri [25] studied the large
$n$ behavior of ${\rm dP_{I}}$, see also [13].

######  Proposition 2.2.

The recurrence coefficients $\beta_{n}(s)$ satisfy the differential-difference
equation,

$\beta_{n}^{\prime}(s)=\frac{\beta_{n}(s)}{2s}\left[N(s+1)\left(\beta_{n+1}(s)-\beta_{n-1}(s)\right)-1\right].$
(2.3)

###### Proof.

Differentiating $h_{n}(s)$ with respect to $s$,

$\begin{split}\frac{dh_{n}(s)}{ds}&=\frac{d}{ds}\int_{-\infty}^{\infty}P_{n}^{2}(x;s)w(x;s)dx\\\
&=N\left[\int_{-\infty}^{\infty}\left(x^{2}-x^{4}\right)P_{n}^{2}(x;s)w(x;s)dx\right]\\\
\
&=Nh_{n}(s)\left[\beta_{n+1}\left(1-\beta_{n+2}-\beta_{n+1}-\beta_{n}\right)+\beta_{n}\left(1-\beta_{n+1}-\beta_{n}-\beta_{n-1}\right)\right]\\\
&=Nh_{n}(s)\left[\frac{1+s}{2s}\left(\beta_{n+1}+\beta_{n}\right)-\frac{2n+1}{4Ns}-\frac{\alpha}{4Ns}\right].\end{split}$
(2.4)

Substituting this into

$\begin{split}\frac{d\beta_{n}(s)}{ds}=\beta_{n}(s)\left[\frac{h^{\prime}_{n}(s)}{h_{n}(s)}-\frac{h^{\prime}_{n-1}(s)}{h_{n-1}(s)}\right],\end{split}$

the result of (2.3) will be obtained. ∎

Based on the formulas (2.2) and (2.3), with $n$ and $n-1$, eliminating the
term $\beta_{n+1}$ and $\beta_{n-1}$, then we can get a second order ODE for
the coefficients $\beta_{n}(s)$.

######  Lemma 2.1.

The recurrence coefficients $\beta_{n}(s)$ satisfy the equation

$\begin{split}\beta_{n}^{\prime\prime}(s)&=\frac{\beta_{n}^{\prime
2}(s)}{2\beta_{n}(s)}-\frac{2+s}{s(1+s)}\beta_{n}^{\prime}(s)+\frac{3N^{2}(1+s)^{2}}{8s^{2}}\beta_{n}^{3}(s)+\frac{N^{2}(1+s)^{2}(1-s)}{4s^{3}}\beta_{n}^{2}(s)\\\
&+\bigg{[}\frac{N^{2}(1-s)^{2}(1+s)^{3}-4s^{2}(3-s)}{32s^{4}(1+s)}+\frac{N(1+s)^{2}}{16s^{3}}p_{n}\bigg{]}\beta_{n}(s)-\frac{(1+s)^{2}q_{n}}{128s^{4}\beta_{n}(s)},\end{split}$
(2.5)

where the parameters $p_{n}$ and $q_{n}$ are given by

$p_{n}=\left\\{\begin{array}[]{l@{\hspace{1cm}}l}n+2\alpha\hfil\hskip
28.45274pt&\text{$n$ {\rm even}}\\\ n-\alpha\hfil\hskip 28.45274pt&\text{$n$
{\rm odd}}\\\ \end{array}\right.$ and
$q_{n}=\left\\{\begin{array}[]{l@{\hspace{1cm}}l}n^{2}\hfil\hskip
28.45274pt&\text{$n$ {\rm even}}\\\ (n+\alpha)^{2}\hfil\hskip
28.45274pt&\text{$n$ {\rm odd}}~{}.\\\ \end{array}\right.$

###### Proof.

From Eq. (2.2), we have

$\beta_{n-1}=\frac{n+\alpha\Delta_{n}}{4Ns\beta_{n}}-\beta_{n+1}-\beta_{n}-\frac{1-s}{2s},$
(2.6)

and

$\beta_{n-2}=\frac{n-1+\alpha\Delta_{n-1}}{4Ns\beta_{n-1}}-\beta_{n}-\beta_{n-1}-\frac{1-s}{2s}.$
(2.7)

From Eq. (2.3), we obtain

$\beta_{n+1}=\frac{2s}{N(1+s)}\frac{\beta^{\prime}_{n}}{\beta_{n}}+\frac{1}{N(1+s)}+\beta_{n-1}.$
(2.8)

Substituting Eq. (2.8) into Eq. (2.6), we get

$\beta_{n-1}=-\frac{s}{N(1+s)}\frac{\beta^{\prime}_{n}}{\beta_{n}}+\frac{n+\alpha\Delta_{n}}{8Ns\beta_{n}}-\frac{\beta_{n}}{2}-\frac{1}{2N(1+s)}-\frac{1-s}{4s}.$
(2.9)

Based on the formulas (2.8) and (2.7), we find

$\begin{split}\beta_{n}=&\frac{2s}{N(s+1)}\frac{\beta^{\prime}_{n-1}}{\beta_{n-1}}+\frac{1}{N(1+s)}+\beta_{n-2}\\\
=&\frac{2s}{N(1+s)}\frac{\beta^{\prime}_{n-1}}{\beta_{n-1}}+\frac{1}{N(1+s)}+\frac{n-1+\alpha_{n-1}}{4Ns\beta_{n-1}}-\beta_{n}-\beta_{n-1}-\frac{1-s}{2s}.\end{split}$
(2.10)

Substituting the expression of $\beta_{n-1}$, given by Eq. (2.9), with its
derivative into Eq. (2.10), one will find the result (2.5) after some
simplification. ∎

##  3 Asymptotics for the recurrence coefficient $\beta_{n}(s)$ as
$s\rightarrow 0$

######  Theorem 3.1.

Assuming $s\rightarrow 0$, the recurrence coefficient $\beta_{n}(s;\alpha,N)$
satisfying the second order ODE (2.5), has the asymptotic expansion

$\begin{split}\beta_{n}=&\frac{n}{2}\bigg{\\{}\frac{1}{N}+\frac{N-3n-2\alpha}{N^{2}}s+\frac{18n^{2}-9nN+N^{2}+22n\alpha+6\left(-N\alpha+\alpha^{2}+1\right)}{N^{3}}s^{2}\\\
&+\frac{1}{N^{4}}\Big{[}N^{3}+nN\left(40n^{2}+\frac{15}{2}n-18N+\frac{85}{2}\right)+30N-135n^{3}-162n-\big{(}12N^{2}\\\
&-30nN+236n^{2}-80n^{2}N+100\big{)}\alpha+\big{(}30N-126n\big{)}\alpha^{2}-20\alpha^{3}\Big{]}s^{3}\\\
&+\mathcal{O}\big{(}s^{4}\big{)}\bigg{\\}},~{}~{}~{}~{}n~{}{\rm
even},\end{split}$

whilst

$\begin{split}\beta_{n}=&\frac{n}{2}\bigg{\\{}\frac{1}{N}+\frac{N-3n-\alpha}{N^{2}}s+\frac{18n^{2}-9nN+N^{2}+14n\alpha-3N\alpha+2\alpha^{2}+6}{N^{3}}s^{2}\\\
&+\frac{1}{N^{4}}\bigg{[}N^{3}+nN\left(40n^{2}+\frac{15}{2}n-18N+\frac{85}{2}\right)+30N-135n^{3}-162n\\\
&-\big{(}6N^{2}-40n^{2}N+15nN-\frac{85}{2}N+169n^{2}+62\big{)}\alpha+\left(\frac{15}{2}N-40nN-59n\right)\alpha^{2}\\\
&-5\alpha^{3}(1+8N)\bigg{]}s^{3}+\mathcal{O}\big{(}s^{4}\big{)}\bigg{\\}},~{}~{}~{}~{}n~{}{\rm
odd}.\end{split}$

###### Proof.

We rewrite the Eq. (2.5) as follows:

$\begin{split}0&=\frac{1}{s^{4}}\left[\frac{N^{2}(1-s)^{2}(1+s)^{2}\beta_{n}}{32}-\frac{(1+s)^{2}q_{n}}{128\beta_{n}}\right]+\frac{1}{s^{3}}\left[\frac{N^{2}(1+s)^{2}(1-s)}{4}\beta_{n}^{2}+\frac{N(1+s)^{2}p_{n}}{16}\beta_{n}\right]\\\
&+\frac{1}{s^{2}}\left[\frac{3N^{2}(1+s)^{2}}{8}\beta_{n}^{3}-\frac{3-s}{8(1+s)}\beta_{n}\right]-\frac{1}{s}\frac{(2+s)\beta^{\prime}_{n}}{1+s}+\frac{\beta_{n}^{\prime
2}}{2\beta_{n}}-\beta_{n}^{\prime\prime}.\end{split}$ (3.1)

Considering the coefficient of the term $s^{-4}$ then one finds that

$\beta_{n}\simeq\frac{\sqrt{q_{n}}}{2N(1-s)}=\frac{\sqrt{q_{n}}}{2N}\left(1+s+s^{2}+s^{3}+\cdots\right).$

Then we suppose $\beta_{n}$ has the following expansion

$\beta_{n}=\frac{\sqrt{q_{n}}}{2N}\left(\sum_{k=0}^{\infty}c_{k}s^{k}\right),~{}~{}s\rightarrow
0.$

Substituting this into the formula (2.5) with directing calculating, we get
the asymptotic expansion of $\beta_{n}$, reads

$\begin{split}\beta_{n}(s)=&\frac{\sqrt{q_{n}}}{2N}\bigg{[}1+\left(\frac{N-p_{n}-2\sqrt{q_{n}}}{N}\right)s+\frac{1}{2N^{2}}\big{(}2N^{2}-6Np_{n}+3p_{n}^{2}-12N\sqrt{q_{n}}\\\
&+16p_{n}\sqrt{q_{n}}+17q_{n}\big{)}s^{2}+\frac{1}{2N^{3}}\Big{(}64N+2N^{3}-64p_{n}-12N^{2}p_{n}+15Np_{n}^{2}-5p_{n}^{3}\\\
&-128\sqrt{q_{n}}-24N^{2}\sqrt{q_{n}}+80Np_{n}\sqrt{q_{n}}-48p_{n}^{2}\sqrt{q_{n}}+85Nq_{n}-125p_{n}q_{n}\\\
&-92q_{n}^{\frac{3}{2}}\Big{)}s^{3}+\mathcal{O}\left(s^{4}\right)\bigg{]},~{}~{}s\rightarrow
0.\\\ \end{split}$

which completes the proof. ∎

######  Remark 3.1.

For $s=0,~{}\alpha=0,~{}N=1$, which are the Hermite polynomials orthogonal
with the weight function ${\rm e}^{-x^{2}}$, one finds
$\beta_{n}=\frac{n}{2}$, see [17, 43].

Figure 1: Plots of the recurrence coefficients $\beta_{n}(s;\alpha,N)$, with
$\alpha=12$, $N=6$, $s$ small.

##  4 Asymptotics for the recurrence coefficient $\beta_{n}(s;\alpha,N)$ as
$n\rightarrow\infty$ with $N$ fixed.

The asymptotic expansion of $\beta_{n}(s;\alpha,N)$, satisfied by Eq. (2.1),
for the special case when $\alpha=0$ and $s=N=1$ was studied by Lew and
Quarles [26], see also [35, 36]. To analyze the asymptotic properties of
$\beta_{n}(s;\alpha,N)$ as $n\rightarrow\infty$. We first give a brief
description of Coulomb fluid, see e.g. [10, 8, 7]. The energy of a system of
$n$ logarithmically repelling particles on the line confined by an external
potential $v(x)$ reads

$E\left(x_{1},x_{2},\ldots,x_{n}\right)=-2\sum_{1\leq j<k\leq
n}\log\left|x_{j}-x_{k}\right|+\sum_{j=1}^{n}v\left(x_{j}\right),$

The collection particles, for large enough $n$, can be approximated as a
continuous fluid with a certain density $\sigma(x)$ supported on a single
interval $(a,b)\in\mathbb{R}$, see [16]. This density, $\sigma(x)$,
corresponding to the equilibrium density of the fluid, is obtained by the
constrained minimization of the free-energy function, $F[\sigma]$,

$F[\sigma]:=\int_{a}^{b}\sigma(x)v(x)dx-\int_{a}^{b}\int_{a}^{b}\sigma(x)\log|x-y|\sigma(y)dxdy.$

subject to

$\int_{a}^{b}\sigma(x)dx=n,~{}~{}~{}\sigma(x)>0.$

Upon minimization [42], the equilibrium density $\sigma(x)$ is found to
satisfy the integral equation,

$L:=v(x)-2\int_{a}^{b}\log|x-y|\sigma(y)dy,~{}~{}~{}~{}x\in[a,b],$ (4.1)

where $L$ is the Lagrange multiplier. The derivative of this equation over $x$
gives rise to a singular integral equation

$v^{\prime}(x)-2{\rm
p.v.}\int_{a}^{b}\frac{\sigma(y)}{x-y}dy=0,~{}~{}~{}x\in[a,b].$

where ${\rm p.v.}$ denotes the Cauchy principal value. Based on the theory of
singular integral equations [33], we find

$\sigma(x)=\frac{1}{2\pi^{2}}\sqrt{\frac{b-x}{x-a}}{\rm
p.v.}\int_{a}^{b}\frac{v^{\prime}(y)}{y-x}\sqrt{\frac{y-a}{b-y}}dy.$ (4.2)

Thus the normalization, $\int_{a}^{b}\sigma(x)dx=n$, becomes

$\frac{1}{2\pi}\int_{a}^{b}\sqrt{\frac{y-a}{b-y}}v^{\prime}(y)dy=n.$

with a supplementary condition, see [7, 8],

$\int_{a}^{b}\frac{v^{\prime}(x)}{\sqrt{(b-x)(x-a)}}=0.$

Consequently, the normalization condition becomes

$\int_{a}^{b}\frac{xv^{\prime}(x)}{\sqrt{(b-x)(x-a)}}=2\pi n.$ (4.3)

For the weight at hand, the equilibrium density $\sigma(x)$ supported on
$(-b,b)$, and

$v(x)=-\log w_{\alpha}(x;s)=-\alpha\log|x|+Nsx^{4}+N(1-s)x^{2},$ (4.4)

we find,

######  Lemma 4.1.

For sufficiently large $n$, with the parameters $N$ and $\alpha$ finite, we
have

$b^{2}\simeq\frac{2\sqrt{3}}{3}N^{-\frac{1}{2}}s^{-\frac{1}{2}}n^{\frac{1}{2}}.$
(4.5)

###### Proof.

It follows from Eq. (4.4) that

$v^{\prime}(x)=-{\rm p.v.}\frac{\alpha}{x}+4Nsx^{3}+2N(1-s)x.$

Putting $a=-b$, we find from Eq. (4.3) that

$\begin{split}2\pi n=&\lim_{\varepsilon\rightarrow
0^{+}}\left(\int_{-b}^{-\varepsilon}+\int_{\varepsilon}^{b}\right)\frac{-\alpha+4Nsx^{4}+2N(1-s)x^{2}}{\sqrt{b^{2}-x^{2}}}dx\\\
=&\lim_{\varepsilon\rightarrow
0^{+}}\int_{\varepsilon^{2}}^{b^{2}}\frac{-\alpha+4Nsy^{2}+2N(1-s)y}{\sqrt{\left(b^{2}-y\right)y}}dy\\\
=&-\alpha\pi+\frac{3\pi}{2}Nsb^{4}+\pi N(1-s)b^{2}.\end{split}$ (4.6)

Since $b^{2}>0$, we have

$\begin{split}b^{2}=&\frac{s-1+\sqrt{(1-s)^{2}+6s\left(\frac{2n+\alpha}{N}\right)}}{3s}\\\
\simeq&\frac{2\sqrt{3}}{3}N^{-\frac{1}{2}}s^{-\frac{1}{2}}n^{\frac{1}{2}},~{}~{}n\rightarrow\infty.\end{split}$
(4.7)

∎

######  Lemma 4.2.

A upper bound for the recurrence coefficient $\beta_{n}(s;\alpha,N)$ with
respect to the weight $w_{\alpha}(x;s,N)$ is given by

$\beta_{n}<\frac{n^{\frac{1}{2}}}{2\sqrt{Ns}}+\frac{s-1}{4s}+\frac{\alpha\Delta_{n}+\frac{N(1-s)^{2}}{4s}}{4\sqrt{Ns}}n^{-\frac{1}{2}}+\sum_{j=2}^{\infty}\frac{b_{j}}{\sqrt{Ns}}\frac{\left[\alpha\Delta_{n}+\frac{N(1-s)^{2}}{4s}\right]^{j}}{n^{j-\frac{1}{2}}},$
(4.8)

where

$b_{j}=\frac{(-1)^{j-1}(2j-3)!}{2^{2j-1}j!(j-2)!}.$

###### Proof.

From Eq. (2.1), we obtain

$\beta_{n}^{2}+\frac{1-s}{2s}\beta_{n}+\beta_{n}\left(\beta_{n+1}+\beta_{n-1}\right)=\frac{n+\alpha\Delta_{n}}{4Ns},$

it follows that

$\beta_{n}^{2}+\frac{1-s}{2s}\beta_{n}<\frac{n+\alpha\Delta_{n}}{4Ns},$

due to $\beta_{n}=\frac{h_{n}}{h_{n-1}}>0$, where

$h_{n}(s;\alpha,N)=\int_{-\infty}^{\infty}P_{n}^{2}(x)w_{\alpha}(x;s,N)dx.$

So

$\begin{split}0<\beta_{n}&<-\frac{1-s}{4s}+\sqrt{\left(\frac{1-s}{4s}\right)^{2}+\frac{n+\alpha\Delta_{n}}{4Ns}}\\\
&=\frac{s-1}{4s}+\frac{n^{\frac{1}{2}}}{2\sqrt{Ns}}\left[1+\frac{\alpha\Delta_{n}+\frac{N(1-s)^{2}}{4s}}{n}\right]^{\frac{1}{2}}\\\
&=\frac{n^{\frac{1}{2}}}{2\sqrt{Ns}}+\frac{s-1}{4s}+\frac{\alpha\Delta_{n}+\frac{N(1-s)^{2}}{4s}}{4\sqrt{Ns}}n^{-\frac{1}{2}}+\sum_{j=2}^{\infty}\frac{b_{j}}{\sqrt{Ns}}\frac{\left[\alpha\Delta_{n}+\frac{N(1-s)^{2}}{4s}\right]^{j}}{n^{j-\frac{1}{2}}},\end{split}$

where

$b_{j}=\frac{(-1)^{j-1}(2j-3)!!}{2^{j+1}j!}=\frac{(-1)^{j-1}(2j-3)!}{2^{2j-1}j!(j-2)!},$

which completes the proof. ∎

In the following, we provide the asymptotic expansion of
$\beta_{n}(s;\alpha,N)$ in Eq. (1.3) as $n\rightarrow\infty$ with $N$ fixed,
for $s\in[0,1],\alpha>-1$.

######  Theorem 4.1.

Let $n\rightarrow\infty$ with $N$ fixed, the recurrence coefficient
$\beta_{n}(s;\alpha,N)$ associated with monic deformed Freud polynomials
satisfying the nonlinear discrete equation (2.1), i.e.

$\beta_{n}\left(\beta_{n+1}+\beta_{n}+\beta_{n-1}+\frac{1-s}{2s}\right)=\frac{n+\alpha\Delta_{n}}{4Ns},~{}~{}~{}\Delta_{n}=\frac{1-(-1)^{n}}{2},$
(4.9)

has the asymptotic expansion

$\begin{split}\beta_{n}=&\frac{1}{2\sqrt{3s}}\sqrt{\frac{n}{N}}-\frac{1-s}{12s}+\frac{(1-s)^{2}}{48\sqrt{3}s^{\frac{3}{2}}}\sqrt{\frac{N}{n}}+\left[\frac{1}{48\sqrt{3Ns}}-\frac{N^{\frac{3}{2}}(1-s)^{4}}{2304\sqrt{3}s^{\frac{5}{2}}}\right]n^{-\frac{3}{2}}\\\
&-\frac{1-s}{288s}n^{-2}+\frac{N^{\frac{5}{2}}(1-s)^{6}-144\sqrt{N}s^{2}(1-s)^{2}}{55296\sqrt{3}s^{3}}n^{-\frac{5}{2}}+\frac{N(1-s)^{3}}{1728s^{2}}n^{-3}\\\
&+\mathcal{O}\left(n^{-\frac{7}{2}}\right),~{}~{}~{}\text{$n$ {\rm
even}},\end{split}$ (4.10)

whilst

$\begin{split}\beta_{n}=&\frac{1}{2\sqrt{3s}}\sqrt{\frac{n}{N}}-\frac{1-s}{12s}+\left[\frac{\sqrt{N}(1-s)^{2}}{48\sqrt{3}s^{\frac{3}{2}}}+\frac{\alpha}{4\sqrt{3Ns}}\right]\frac{1}{\sqrt{n}}+\bigg{\\{}-\frac{\left[N(1-s)^{2}+12s\alpha\right]^{2}}{2304\sqrt{3N}s^{\frac{5}{2}}}\\\
&+\frac{1}{48\sqrt{3Ns}}\bigg{\\}}n^{-\frac{3}{2}}-\frac{1-s}{288s}n^{-2}+\Psi(s)n^{-\frac{5}{2}}+\frac{N(1-s)^{3}+12s(1-s)\alpha}{1728s^{2}}n^{-3}\\\
&+\mathcal{O}\left(n^{-\frac{7}{2}}\right),~{}~{}~{}\text{$n$ {\rm
odd}},\end{split}$ (4.11)

with

$\Psi(s)=\frac{N^{\frac{5}{2}}(1-s)^{6}+36N^{\frac{3}{2}}(1-s)^{4}s\alpha+144\sqrt{N}(1-s)^{2}s^{2}\left(3\alpha^{2}-1\right)+1728s^{3}\alpha\left(\alpha^{2}-1\right)}{55296\sqrt{3}s^{\frac{7}{2}}}.$

###### Proof.

In fact, Lemma 4.1 gives the first term in the asymptotic expansion of
$\beta_{n}(s;\alpha,N)$, since

$\beta_{n}\simeq\left(\frac{b-a}{4}\right)^{2}\simeq\frac{1}{2\sqrt{3Ns}}n^{\frac{1}{2}},~{}a=-b.$

Hence it seems reasonable to assume $\beta_{n}(s;\alpha,N)$ has the expansion
of the form

$\beta_{n}=\frac{n^{\frac{1}{2}}}{2\sqrt{3Ns}}\left(1+\sum_{k=1}^{\infty}d_{k}n^{-\frac{k}{2}}\right),~{}~{}n\rightarrow\infty.$
(4.12)

Replacing $n$ by $n\pm 1$ in Eq. (4.12), we obtain

$\begin{split}\beta_{n\pm 1}=&\frac{(n\pm
1)^{\frac{1}{2}}}{2\sqrt{3Ns}}\left[1+\sum_{k=1}^{\infty}d_{k}(n\pm
1)^{-\frac{k}{2}}\right]\\\
=&\frac{n^{\frac{1}{2}}\left(1\pm\frac{1}{n}\right)^{\frac{1}{2}}}{2\sqrt{3Ns}}\left[1+\sum_{k=1}^{\infty}d_{k}n^{-\frac{k}{2}}\left(1\pm\frac{1}{n}\right)^{-\frac{k}{2}}\right]\\\
=&\frac{n^{\frac{1}{2}}}{2\sqrt{3Ns}}\bigg{[}1+\frac{d_{1}}{n^{\frac{1}{2}}}+\frac{2d_{2}\pm
1}{2n}+\frac{d_{3}}{n^{\frac{3}{2}}}+\frac{8d_{4}\mp
4d_{2}-1}{8n^{2}}+\frac{d_{5}\mp d_{3}}{n^{\frac{5}{2}}}\\\ &+\frac{16d_{6}\mp
24d_{4}+6d_{2}\pm 1}{16n^{3}}+\frac{d_{7}\mp
2d_{5}+d_{3}}{n^{\frac{7}{2}}}+\mathcal{O}\left(n^{-4}\right)\bigg{]},\end{split}$
(4.13)

by doing an asymptotic expansion in the above equation. Substituting Eqs.
(4.12) and (4.13) into Eq. (4.9), followed by comparing the corresponding
coefficients on both sides, we find

$\begin{split}&d_{1}=-\frac{\sqrt{N}(1-s)}{2\sqrt{3s}},~{}~{}d_{2}=\frac{N(1-s)^{2}}{24s}+\frac{\alpha\Delta_{n}}{2},~{}~{}d_{3}=0,\\\
&d_{4}=\frac{1}{24}-\frac{\left[N(1-s)^{2}+12s\alpha\Delta_{n}\right]^{2}}{1152s^{2}},~{}~{}d_{5}=-\frac{\sqrt{N}(1-s)}{48\sqrt{3s}},\\\
&d_{6}=\frac{\left[N(1-s)^{2}+12s\alpha\Delta_{n}\right]\left[N(1-s)^{2}-12s+12s\alpha\Delta_{n}\right]\left[N(1-s)^{2}+12s+12s\alpha\Delta_{n}\right]}{27648s^{3}},\\\
&d_{7}=\frac{\sqrt{N}(1-s)\left[N(1-s)^{2}+12s\alpha\Delta_{n}\right]}{288\sqrt{3}s^{\frac{3}{2}}}.\end{split}$
(4.14)

Hence, we obtain the asymptotic expansions (4.10) and (4.11), followed by some
computation. ∎

######  Corollary 4.1.

Assume that $\beta_{n}(s;\alpha,N)$ satisfying Eq. (4.9). Then for
$s\in[0,1],~{}\alpha>-1,~{}N>0$:

(i) the sequence
$\left\\{\frac{\beta_{n}(s;\alpha,N)}{\sqrt{n}}\right\\}_{n=1}^{\infty}$ is
bounded;

(ii)
$\lim_{n\rightarrow\infty}\frac{\beta_{n}(s;\alpha,N)}{\sqrt{n}}=\frac{1}{2\sqrt{3Ns}}$.

######  Remark 4.1.

Putting $s=1,~{}\alpha=0,~{}N=1$, the classical result, obtained by Lew and
Quarles [26] for the Freud weight ${\rm e}^{-x^{4}}$, is recovered, i.e.

$\lim_{n\rightarrow\infty}\frac{\beta_{n}(1;0,1)}{\sqrt{n}}=\frac{1}{2\sqrt{3}}.$

######  Corollary 4.2.

For $s\in[0,1],~{}\alpha>-1,~{}N>0$, the recurrence coefficients
$\beta_{n}(s;\alpha,N)$ in Eq. (4.9) satisfy

$\begin{split}\frac{\beta_{n+1}(s;\alpha,N)}{\beta_{n}(s;\alpha,N)}&=1+\mathcal{O}(n^{-1}),~{}~{}~{}~{}\text{$n\rightarrow\infty$},\\\
\frac{\beta_{n}(s;\alpha,N)}{a_{\mu}^{2}}&=\frac{1}{4}+\mathcal{O}(n^{-1}),~{}~{}~{}~{}\text{$n\rightarrow\infty$}.\end{split}$
(4.15)

where $a_{\mu}$ is sometimes called the Mhaskar-Rakhmanov-Saff number [40,
34], the unique positive root of the equation

$\mu=\frac{2}{\pi}\int_{0}^{1}\frac{a_{\mu}yQ^{\prime}(a_{\mu}y)}{\sqrt{1-y^{2}}}dy,$

with $Q(x)=Nsx^{4}+N(1-s)x^{2}$.

###### Proof.

See [14, Thm. 2.1]. ∎

##  5 Asymptotics for the recurrence coefficient $\beta_{n}(s;\alpha,N)$ as
$n\rightarrow\infty,N\rightarrow\infty$ with $\frac{n}{N}$ fixed.

In Section 2 we have derived that the recurrence coefficient
$\beta_{n}(s;\alpha,N)$ satisfies the second order ODE (2.5). Let $0<r_{0}\leq
r:=\frac{n}{N}\leq 1$, i.e. the quantity $\frac{n}{N}$ is bounded away from
$0$ and close to $1$. Substituting $n=Nr$ into Eq. (2.5) gives,

$\begin{split}&\bigg{[}\frac{3(1+s)^{2}}{8s^{2}}\beta_{n}^{3}(s)+\frac{(1+s)^{2}(1-s)}{4s^{3}}\beta_{n}^{2}(s)+\frac{(1+s)^{2}(1-2s+2rs+s^{2})}{32s^{4}}\beta_{n}(s)\\\
&-\frac{(1+s)^{2}r^{2}}{128s^{4}\beta_{n}(s)}\bigg{]}N^{2}+\left[\frac{\alpha(1+s)^{2}}{8s^{3}}\beta_{n}(s)\right]N-\beta_{n}^{\prime\prime}(s)+\frac{\beta_{n}^{\prime
2}(s)}{2\beta_{n}(s)}-\frac{2+s}{s(1+s)}\beta_{n}^{\prime}(s)\\\
&-\frac{3-s}{8s^{2}(1+s)}\beta_{n}(s)=0,~{}~{}~{}\text{$n$ is
even},\end{split}$ (5.1)

and

$\begin{split}&\bigg{[}\frac{3(1+s)^{2}}{8s^{2}}\beta_{n}^{3}(s)+\frac{(1+s)^{2}(1-s)}{4s^{3}}\beta_{n}^{2}(s)+\frac{(1+s)^{2}(1-2s+2rs+s^{2})}{32s^{4}}\beta_{n}(s)\\\
&-\frac{(1+s)^{2}r^{2}}{128s^{4}\beta_{n}(s)}\bigg{]}N^{2}-\left[\frac{\alpha(1+s)^{2}}{16s^{3}}\beta_{n}(s)+\frac{r\alpha(1+s)^{2}}{64s^{4}\beta_{n}(s)}\right]N-\beta_{n}^{\prime\prime}(s)+\frac{\beta_{n}^{\prime
2}(s)}{2\beta_{n}(s)}\\\
&-\frac{2+s}{s(1+s)}\beta_{n}^{\prime}(s)-\frac{3-s}{8s^{2}(1+s)}\beta_{n}(s)-\frac{\alpha^{2}(1+s)^{2}}{128s^{4}\beta_{n}(s)}=0,~{}~{}~{}\text{$n$
is odd},\end{split}$ (5.2)

which imply that

$\begin{split}&\frac{3(1+s)^{2}}{8s^{2}}\beta_{n}^{3}(s)+\frac{(1+s)^{2}(1-s)}{4s^{3}}\beta_{n}^{2}(s)+\frac{(1+s)^{2}(1-2s+2rs+s^{2})}{32s^{4}}\beta_{n}(s)\\\
&-\frac{(1+s)^{2}r^{2}}{128s^{4}\beta_{n}(s)}=0,~{}~{}~{}\text{$N\rightarrow\infty$}.\end{split}$
(5.3)

Solving this equation, we get the solution

$\begin{split}\beta_{n}(s)=\frac{s-1+\sqrt{1-2s+12rs+s^{2}}}{12s},\end{split}$
(5.4)

which is consistent with the result (4.7) since

$\begin{split}\beta_{n}(s)\simeq\frac{b^{2}}{4},~{}N\rightarrow\infty.\end{split}$
(5.5)

Therefore, we assume that $\beta_{n}(s)$ has the asymptotic expansion

$\begin{split}\beta_{n}(s)=a_{0}(s)+\sum_{k=1}^{\infty}\frac{a_{k}(s)}{N^{k}},~{}N\rightarrow\infty.\end{split}$
(5.6)

where

$\begin{split}a_{0}(s):=\frac{s-1+\sqrt{1-2s+12rs+s^{2}}}{12s}.\end{split}$

Let $g(s):=\sqrt{1-2s+12rs+s^{2}}$ and $f(s):=1-2s-4rs+s^{2}$. Then
substituting the asymptotic expansion Eq. (5.6) into Eqs. (5.1) and (5.2),
respectively, we obtain

$a_{1}(s)=\begin{cases}\frac{\left[s-1+g(s)\right]\alpha}{4(s-1)g(s)-2g^{2}(s)},~{}~{}~{}~{}~{}~{}~{}~{}n~{}{\rm
even},\\\ \\\ \frac{\alpha(1-s)}{2f(s)}-\frac{4\alpha sr}{f(s)g(s)},\ \
~{}~{}~{}~{}~{}n~{}{\rm odd}.\end{cases}$

and, for $n$ is even

$\begin{split}a_{2}(s)=\frac{s(s-1)}{2g^{4}(s)}+\frac{s\left(4-3\alpha^{2}\right)}{8g^{3}(s)}+\frac{3\alpha^{2}s(s-1)}{2f^{2}(s)}+\frac{15\alpha^{2}s+3\alpha^{2}s^{2}(12r+5s-10)}{8f(s)g(s)},\end{split}$

else, if $n$ is odd,

$\begin{split}a_{2}(s)=\frac{s(s-1)}{2g^{4}(s)}+\frac{s(4-3\alpha^{2})}{8g^{3}(s)}+\frac{\alpha^{2}s(1-s)}{2f^{2}(s)}-\frac{3\alpha^{2}s}{8f(s)g(s)}.\end{split}$

######  Theorem 5.1.

As $n,N\rightarrow\infty$, with the quantity $\frac{n}{N}$ is bounded away
from $0$ and close to $1$, the recurrence coefficients have the asymptotic
expansion

$\begin{split}\beta_{n}(s)=&\frac{s-1+g(s)}{12s}+\frac{\left[s-1+g(s)\right]\alpha}{4(s-1)g(s)-2g^{2}(s)}\frac{1}{N}+\bigg{[}\frac{s(s-1)}{2g^{4}(s)}+\frac{s\left(4-3\alpha^{2}\right)}{8g^{3}(s)}\\\
&+\frac{3\alpha^{2}s(s-1)}{2f^{2}(s)}+\frac{15\alpha^{2}s+3\alpha^{2}s^{2}(12r+5s-10)}{8f(s)g(s)}\bigg{]}\frac{1}{N^{2}}+\mathcal{O}\left(N^{-3}\right),~{}\text{$n$
{\rm even}},\end{split}$

whilst

$\begin{split}\beta_{n}(s)=&\frac{s-1+g(s)}{12s}+\left[\frac{\alpha(1-s)}{2f(s)}-\frac{4\alpha
sr}{f(s)g(s)}\right]\frac{1}{N}+\bigg{[}\frac{s(s-1)}{2g^{4}(s)}+\frac{s(4-3\alpha^{2})}{8g^{3}(s)}\\\
&+\frac{\alpha^{2}s(1-s)}{2f^{2}(s)}-\frac{3\alpha^{2}s}{8f(s)g(s)}\bigg{]}\frac{1}{N^{2}}+\mathcal{O}\left(N^{-3}\right),~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\text{$n$
{\rm odd}},\end{split}$

with

$\begin{split}g(s)=\sqrt{1-2s+12rs+s^{2}}~{}~{}~{}{\rm
and}~{}~{}~{}f(s)=1-2s-4rs+s^{2}.\end{split}$

######  Remark 5.1.

Higher order corrections can be calculated systematically in Mathematica, but
will be quite cumbersome. Hence we only give the expansion terms up to order
$N^{-2}$.

##  6 The deformed Freud polynomials

Based on an extension of the ladder operators technique developed in [6], Chen
and Feigin [5] studied the weight
$\widetilde{w}(x)|x-t|^{K},x,t,K\in\mathbb{R}$, for any smooth reference
weight $\widetilde{w}(x)$. They showed that when $\widetilde{w}(x)$ is the
Gaussian (Hermite) weight (${\rm e}^{-x^{2}},x\in\mathbb{R}$), the recurrence
coefficients satisfy a particular two-parameter Painlevé IV equation. Filipuk
et al [19] found that the recurrence coefficients for the Freud weight
$|x|^{2\alpha+1}{\rm e}^{-x^{4}+tx^{2}},x,t\in\mathbb{R},\alpha>-1$ are
related to the solutions of the Painlevé IV and the first discrete Painlevé
equation. In [12, 13], Clarkson et al gave a systematic study on Freud weight
$|x|^{2\alpha+1}{\rm e}^{-x^{4}+tx^{2}}$, and some generalized work for [5].
In the following theorem, we give the differential-difference equation (i.e.,
lowing operator) satisfied by the deformed Freud orthogonal polynomials
associated with the weight $w_{\alpha}(x;s,N)$ given in Eq. (1.1).

######  Lemma 6.1.

The monic orthogonal polynomials $P_{n}(z;s,\alpha,N)$ with respect to the
deformed Freud weight $w_{\alpha}(x;s,N)$ on $\mathbb{R}$ satisfy the
differential-differentce recurrence relation

$P^{\prime}_{n}(z)=\beta_{n}(s)A_{n}(z)P_{n-1}(z)-B_{n}(z)P_{n}(z),$ (6.1)

where

$\begin{split}&A_{n}(z):=\frac{1}{h_{n}}\int_{-\infty}^{\infty}\frac{v_{0}^{\prime}(z)-v_{0}^{\prime}(y)}{z-y}P^{2}_{n}(y)w(y)dy,\\\
&B_{n}(z):=\frac{1}{h_{n-1}}\int_{-\infty}^{\infty}\frac{v_{0}^{\prime}(z)-v_{0}^{\prime}(y)}{z-y}P_{n}(y)P_{n-1}(y)w(y)dy+\frac{\alpha\left[1-(-1)^{n}\right]}{2z},\end{split}$

with

$v_{0}(x)=Nsx^{4}+N(1-s)x^{2},~{}x\in\mathbb{R},~{}s\in[0,1],~{}N>0.$ (6.2)

###### Proof.

Since the derivative of $P_{n}(z)$ is a polynomial of degree $n-1$ in $z$,
hence $P^{\prime}_{n}(z)$ can be given by

$P^{\prime}_{n}(z)=\sum_{j=0}^{n-1}c_{n,j}P_{j}(z).$ (6.3)

Using the orthogonality relations, and integrating by parts, we have

$c_{n,j}=\frac{1}{h_{j}}\int_{-\infty}^{\infty}P^{\prime}_{n}(y)P_{j}(y)w(y)dy=\frac{1}{h_{j}}\int_{-\infty}^{\infty}P_{n}(y)P_{j}(y)\left[v_{0}^{\prime}(y)-\frac{\alpha}{y}\right]w(y)dy,$
(6.4)

where

$v_{0}(y):=Nsy^{4}+N(1-s)y^{2}.$ (6.5)

Substituting Eq. (6.4) into Eq. (6.3) and summating over $j$ applying the
Christoffel-Darboux formula [23, Theorem 2.2.2],

$\sum_{j=0}^{n-1}\frac{P_{j}(z)P_{j}(y)}{h_{j}}=\frac{P_{n}(z)P_{n-1}(y)-P_{n}(y)P_{n-1}(z)}{(z-y)h_{n-1}},$

we have

$\begin{split}P^{\prime}_{n}(z)=&\int_{-\infty}^{\infty}P_{n}(y)\sum_{j=0}^{n-1}\frac{P_{j}(z)P_{j}(y)}{h_{j}}\left[v_{0}^{\prime}(y)-\frac{\alpha}{y}\right]w(y)dy\\\
=&\int_{-\infty}^{\infty}P_{n}(y)\sum_{j=0}^{n-1}\frac{P_{j}(z)P_{j}(y)}{h_{j}}\left[v_{0}^{\prime}(y)-v_{0}^{\prime}(z)\right]w(y)dy\\\
&-\alpha\int_{-\infty}^{\infty}\frac{P_{n}(y)}{y}\sum_{j=0}^{n-1}\frac{P_{j}(z)P_{j}(y)}{h_{j}}w(y)dy\\\
=&-\frac{1}{h_{n-1}}\int_{-\infty}^{\infty}\left[P_{n}(z)P_{n}(y)P_{n-1}(y)-P^{2}_{n}(y)P_{n-1}(z)\right]\frac{v_{0}^{\prime}(z)-v_{0}^{\prime}(y)}{z-y}w(y)dy\\\
&-\frac{\alpha}{z}\int_{-\infty}^{\infty}\frac{[(z-y)+y]P_{n}(y)}{y}\left[\sum_{j=0}^{n-1}\frac{P_{j}(z)P_{j}(y)}{h_{j}}\right]w(y)dy\\\
=&\frac{P_{n-1}(z)}{h_{n-1}}\int_{-\infty}^{\infty}P^{2}_{n}(y)\frac{v_{0}^{\prime}(z)-v_{0}^{\prime}(y)}{z-y}w(y)dy-\frac{\alpha}{z}\frac{P_{n}(z)}{h_{n-1}}\int_{-\infty}^{\infty}\frac{P_{n}(y)P_{n-1}(y)}{y}w(y)dy\\\
&-\frac{P_{n}(z)}{h_{n-1}}\int_{-\infty}^{\infty}P_{n}(y)P_{n-1}(y)\frac{v_{0}^{\prime}(z)-v^{\prime}_{0}(y)}{z-y}w(y)dy.\end{split}$
(6.6)

Furthermore, by an inductive argument based on the three term recurrence
relation Eq. (1.3) with the initial conditions

$P_{0}(z):=1,~{}~{}~{}{\rm and}~{}~{}~{}\beta_{0}P_{-1}(z)=0,$ (6.7)

we obtain

$\begin{split}\frac{1}{h_{n-1}}\int_{-\infty}^{\infty}\frac{P_{n}(y)P_{n-1}(y)}{y}w(y)dy&=\frac{1}{h_{n-1}}\int_{-\infty}^{\infty}\frac{\left[yP_{n-1}(y)-\beta_{n-1}P_{n-2}(y)\right]P_{n-1}(y)}{y}w(y)dy\\\
&=\begin{cases}1,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}n=1,\\\
1-\frac{1}{h_{n-2}}\int_{-\infty}^{\infty}\frac{P_{n-1}(y)P_{n-2}(y)}{y}w(y)dy,~{}~{}n\geq
2,\ \ \end{cases}\\\
&=\begin{cases}0,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}n=2,\\\
\frac{1}{h_{n-3}}\int_{-\infty}^{\infty}\frac{P_{n-2}(y)P_{n-3}(y)}{y}w(y)dy,~{}~{}~{}~{}~{}~{}~{}~{}n\geq
3,\ \ \end{cases}\\\ &\ldots\\\
&=\begin{cases}0,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}n~{}{\rm
even},\\\
1,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}n~{}{\rm
odd}.\ \ \end{cases}\\\ \end{split}$ (6.8)

Therefore, (bearing in mind $\beta_{n}=h_{n}/h_{n-1}$) the result is derived
by Eqs. (6.6) and (6.8) immediately. ∎

######  Lemma 6.2.

$A_{n}(z)$ and $B_{n}(z)$ defined by Lemma 6.1 satisfy the relation:

$A_{n}(z)=\frac{v_{0}^{\prime}(z)}{z}+\frac{B_{n}(z)+B_{n+1}(z)}{z}-\frac{\alpha}{z^{2}}.$
(6.9)

###### Proof.

Be the definition of $A_{n}(z)$, we rewrite it as

$\begin{split}A_{n}(z)=&\frac{1}{zh_{n}}\left\\{\int_{-\infty}^{\infty}\frac{v_{0}^{\prime}(z)-v_{0}^{\prime}(y)}{z-y}yP_{n}^{2}(y)w(y)dy+\int_{-\infty}^{\infty}\left[v_{0}^{\prime}(z)-v_{0}^{\prime}(y)\right]P_{n}^{2}(y)w(y)dy\right\\}\\\
=&\frac{1}{zh_{n}}\bigg{\\{}\int_{-\infty}^{\infty}\frac{v_{0}^{\prime}(z)-v_{0}^{\prime}(y)}{z-y}\left[P_{n+1}(y)+\beta_{n}P_{n-1}(y)\right]P_{n}(y)w(y)dy+v_{0}^{\prime}(z)h_{n}\\\
&-\int_{-\infty}^{\infty}P_{n}^{2}(y)\left[\frac{\alpha}{y}w(y)-w^{\prime}(y)\right]dy\bigg{\\}}\\\
=&\frac{v_{0}^{\prime}(z)}{z}+\frac{1}{z}\left\\{B_{n+1}(z)-\frac{\alpha}{2z}\left[1-(-1)^{n+1}\right]+B_{n}(z)-\frac{\alpha}{2z}\left[1-(-1)^{n}\right]\right\\}\\\
=&\frac{v_{0}^{\prime}(z)}{z}+\frac{B_{n}(z)+B_{n+1}(z)}{z}-\frac{\alpha}{z^{2}},\end{split}$

which completes the proof. ∎

######  Theorem 6.1.

The monic orthogonal polynomials $P_{n}(z;s,\alpha,N)$ with respect to
deformed Freud weight (1.1) satisfy the linear second order ODE:

$P^{\prime\prime}_{n}(z)+R_{n}(z)P^{\prime}_{n}(z)+Q_{n}(z)P_{n}(z)=0,$ (6.10)

where

$R_{n}(z)=-4Nsz^{3}-2N(1-s)z+\frac{\alpha}{z}-\frac{2z}{z^{2}+\frac{1-s}{2s}+\beta_{n}+\beta_{n+1}},$
(6.11)
$\begin{split}Q_{n}(z)=&4Ns\beta_{n}\left[1+\alpha(-1)^{n}\right]+16N^{2}s^{2}\beta_{n}\left[\frac{1-s}{2s}+\beta_{n}+\beta_{n-1}\right]\left[\frac{1-s}{2s}+\beta_{n}+\beta_{n+1}\right]\\\
&-\alpha\left[1-(-1)^{n}\right]\left[N(1-s)+\frac{1}{2z^{2}}\right]-\frac{8Ns\beta_{n}z^{2}+\alpha\left[1-(-1)^{n}\right]}{z^{2}+\frac{1-s}{2s}+\beta_{n}+\beta_{n+1}}+4nNsz^{2}.\\\
\end{split}$ (6.12)

###### Proof.

From the differential-difference equation (6.1), we have the following type of
rasing operator,

$\begin{split}P_{n-1}^{\prime}(z)=&-B_{n-1}(z)P_{n-1}(z)+A_{n-1}(z)\beta_{n-1}(s)P_{n-2}(z)\\\
=&-B_{n-1}(z)P_{n-1}(z)+zA_{n-1}(z)P_{n-1}(z)-A_{n-1}(z)P_{n}(z)\\\
\end{split}$

Using the relation between $A_{n}(z)$ and $B_{n}(z)$, see Lemma 6.2, we get

$\begin{split}P_{n-1}^{\prime}(z)=&-B_{n-1}(z)P_{n-1}(z)+z\left[\frac{v_{0}^{\prime}(z)}{z}+\frac{B_{n-1}(z)+B_{n}(z)}{z}-\frac{\alpha}{z^{2}}\right]P_{n-1}(z)-A_{n-1}(z)P_{n}(z)\\\
=&\left[v_{0}^{\prime}(z)+B_{n}(z)-\frac{\alpha}{z}\right]P_{n-1}(z)-A_{n-1}(z)P_{n}(z).\\\
\end{split}$ (6.13)

Differentiating both sides of Eq. (6.1) with respect to $z$ we obtain

$\begin{split}P_{n}^{\prime\prime}(z)=&-B_{n}^{\prime}(z)P_{n}(z)-B_{n}(z)P_{n}^{\prime}(z)+\beta_{n}(s)A_{n}^{\prime}(z)P_{n-1}(z)+\beta_{n}(s)A_{n}(z)P_{n-1}^{\prime}(z).\\\
\end{split}$

Substituting Eq. (6.13) into above, we obtain

$\begin{split}P_{n}^{\prime\prime}(z)=&-B_{n}^{\prime}(z)P_{n}(z)-B_{n}(z)P_{n}^{\prime}(z)-\beta_{n}(s)A_{n}(z)A_{n-1}(z)P_{n}(z)\\\
&+\left\\{\beta_{n}(s)A_{n}^{\prime}(z)+\beta_{n}(s)A_{n}(z)\left[v_{0}^{\prime}(z)+B_{n}(z)-\frac{\alpha}{z}\right]\right\\}P_{n-1}(z).\\\
\end{split}$ (6.14)

Eliminating $P_{n-1}(z)$ from Eqs. (6.1) and (6.14), we find

$\begin{split}P_{n}^{\prime\prime}(z)+R_{n}(z)P_{n}^{\prime}(z)+Q_{n}(z)P_{n}(z)=0,\end{split}$
(6.15)

where

$\begin{split}R_{n}(z):=&-\frac{A_{n}^{\prime}(z)}{A_{n}(z)}-v_{0}^{\prime}(z)+\frac{\alpha}{z},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
\end{split}$ (6.16)
$\begin{split}Q_{n}(z):=&B_{n}^{\prime}(z)-\frac{A_{n}^{\prime}(z)B_{n}(z)}{A_{n}(z)}-\left[v_{0}^{\prime}(z)+B_{n}(z)-\frac{\alpha}{z}\right]B_{n}(z)+\beta_{n}A_{n}(z)A_{n-1}(z).\\\
\end{split}$ (6.17)

Furthermore,

$\begin{split}A_{n}(z)=&\frac{4Ns}{h_{n}}\int_{-\infty}^{\infty}\left(z^{2}+zy+y^{2}\right)P_{n}^{2}(y)w(y)dy+\frac{2N(1-s)}{h_{n}}\int_{-\infty}^{\infty}P_{n}^{2}(y)w(y)dy\\\
=&4Nsz^{2}+\frac{4Ns}{h_{n}}\int_{-\infty}^{\infty}\left[P_{n+1}(y)+\beta_{n}P_{n-1}(y)\right]\left[P_{n+1}(y)+\beta_{n}P_{n-1}(y)\right]w(y)dy\\\
&+\frac{4Nsz}{h_{n}}\int_{-\infty}^{\infty}\left[P_{n+1}(y)+\beta_{n}P_{n-1}(y)\right]P_{n}(y)w(y)dy+2N(1-s)\\\
=&4Nsz^{2}+4Ns\left(\beta_{n+1}+\beta_{n}\right)+2N(1-s),\end{split}$ (6.18)

since

$\frac{v_{0}^{\prime}(z)-v_{0}^{\prime}(y)}{z-y}=4Ns\left(z^{2}+zy+y^{2}\right)+2N(1-s).$

Similarly, we get

$B_{n}(z)=4Ns\beta_{n}z+\frac{\alpha\left[1-(-1)^{n}\right]}{2z}.$ (6.19)

Substituting Eqs. (6.5) and (6.18) into Eq. (6.16), we obtain

$R_{n}(z)=-4Nsz^{3}-2N(1-s)z+\frac{\alpha}{z}-\frac{2z}{z^{2}+\frac{1-s}{2s}+\beta_{n}+\beta_{n+1}}.$

Substituting Eqs. (6.5), (6.18) and (6.19) into Eq. (6.17), we get

$\begin{split}Q_{n}(z)=&4Ns\beta_{n}\left[1+\alpha(-1)^{n}\right]+16N^{2}s^{2}\beta_{n}\left[\frac{1-s}{2s}+\beta_{n}+\beta_{n-1}\right]\left[\frac{1-s}{2s}+\beta_{n}+\beta_{n+1}\right]\\\
&-\alpha\left[1-(-1)^{n}\right]\left[N(1-s)+\frac{1}{2z^{2}}\right]-\frac{8Ns\beta_{n}z^{2}+\alpha\left[1-(-1)^{n}\right]}{z^{2}+\frac{1-s}{2s}+\beta_{n}+\beta_{n+1}}\\\
&+Nsz^{2}\big{\\{}16Ns\beta_{n}\left(\beta_{n+1}+\beta_{n}+\beta_{n-1}\right)-2\alpha\left[1-(-1)^{n}\right]+8N(1-s)\beta_{n}\big{\\}}.\end{split}$

The proof will be completed by using Eq. (4.9) to instead the last term of the
above formula. ∎

Since from Theorem 4.1 we have

$\beta_{n}(s)=\frac{1}{2\sqrt{3Ns}}n^{\frac{1}{2}}+\mathcal{O}(1),~{}~{}~{}n\rightarrow\infty,$

it follows from Eq.(6.11) and Eq.(6.12), as $n\rightarrow\infty$, that

$\begin{split}R_{n}(z)&=-4Nsz^{3}-2N(1-s)z+\frac{\alpha}{z}+\mathcal{O}\left(n^{-\frac{1}{2}}\right),\\\
Q_{n}(z)&=\left(\frac{4N^{\frac{1}{3}}s^{\frac{1}{3}}n}{3}\right)^{\frac{3}{2}}+\mathcal{O}(n).\end{split}$

Now we reconsider the Eq. (6.10), with denoting $P_{n}(z)$ by
$\widetilde{P}_{n}(z)$ as $n\rightarrow\infty$,

$\widetilde{P}_{n}^{\prime\prime}(z)-\left[4Nsz^{3}+2N(1-s)z-\frac{\alpha}{z}\right]\widetilde{P}_{n}^{\prime}(z)+\left[\frac{4(Ns)^{\frac{1}{3}}n}{3}\right]^{\frac{3}{2}}\widetilde{P}_{n}(z)=0.$
(6.20)

Let

$\kappa:=(Ns)^{\frac{1}{3}}n,$ (6.21)

where $Ns\rightarrow 0^{+}$, $n\rightarrow\infty$, such that $\kappa$ fixed.
We find the Eq. (6.20) is equivalent to the biconfluent Heun equation (BHE)
cf. [38, §31.12, 31.12.3]

$u^{\prime\prime}(x)+\left(\frac{\gamma}{x}+\delta+x\right)u^{\prime}(x)+\left(\eta-\frac{\rho}{x}\right)u(x)=0$
(6.22)

through the transformation

$\widetilde{P}_{n}(z;s,\alpha)=u(x;\gamma,\delta,\eta,\rho),~{}~{}~{}x=\sqrt{2}z^{2},$

with parameters

$\gamma=-\frac{1}{2}-\frac{\alpha}{2},~{}~{}\delta=\frac{\sqrt{2}N(1-s)}{2},~{}~{}\eta=0,~{}~{}{\rm
and}~{}~{}\rho=-\frac{\sqrt{6}}{9}\kappa^{\frac{3}{2}}.$ (6.23)

######  Remark 6.1.

The BHE is widely encountered in contemporary mathematics and physics
research. For example, many Schrödinger equations can be solved applying the
BHE, and in atomic and nuclear physics, the BHE frequently appears in studying
the motion of quantum particles in $1-$, $2-$ or $3-$dimensional confinement
potentials, see [39]. The solutions of the BHE attract many authors, see e.g.
[24, 4, 18]. Based on [24], we show that Eq. (6.22), with parameters given in
Eq. (6.23), has the solutions in terms of the Hermite functions, reads (${\rm
i}=\sqrt{-1}$)

$u(x)=k_{1}\sum_{j=0}^{\infty}c_{j}H_{\frac{2j-\alpha-1}{2}}\left({\rm
i}\left(\frac{x}{\sqrt{2}}+\frac{N(1-s)}{2}\right)\right)+k_{2}\sum_{j=0}^{\infty}c_{j}H_{\frac{2j-\alpha-1}{2}}\left(-{\rm
i}\left(\frac{x}{\sqrt{2}}+\frac{N(1-s)}{2}\right)\right)$ (6.24)

where $k_{1}$ and $k_{2}$ are constants, whilst the coefficients $c_{n}$
satisfied a three term recurrence relation:

$L_{j}c_{j}+M_{j-1}c_{j-1}+T_{j-2}c_{j-2}=0,$

with initial conditions $c_{-2}=c_{-1}=0$, and here

$\begin{split}&L_{j}=-{\rm i}\frac{j(2j-\alpha-1)}{\sqrt{2}},\\\
&M_{j}=\mp\left[\frac{9N(1-s)(2j-\alpha-1)-4\sqrt{3}\kappa^{\frac{3}{2}}}{18\sqrt{2}}\right],\\\
&T_{j}=-{\rm i}\frac{2j-\alpha-1}{2\sqrt{2}},\end{split}$

where the signs $\mp$ in $M_{j}$ corresponding to $\pm{\rm i}$ in Eq. (6.24).
See details in [24].

######  Remark 6.2.

Note that if we make the transformation
$\widetilde{P}_{n}(z)=\widehat{P}_{n}(y)$ in Eq. (6.20), with
$y=\left(\frac{4\kappa}{3}\right)^{\frac{3}{4}}z$, then in the limit as
$n\rightarrow\infty$ ($Ns\rightarrow 0$, with $\kappa$ fixed), we obtain

$\widehat{P}_{n}^{\prime\prime}(y)+\frac{\alpha}{y}\widehat{P}_{n}^{\prime}(y)+\widehat{P}_{n}(y)=0,$

which has the solution

$\widehat{P}_{n}(y)=y^{\frac{1-\alpha}{2}}\left[c_{1}J_{\frac{1-\alpha}{2}}(y)+c_{2}Y_{\frac{1-\alpha}{2}}(y)\right],$

where $J_{n}(x)$, $Y_{n}(x)$ are corresponding with the Bessel function of the
first and second kind, respectively. Hence the polynomials $P_{n}(z)$ might
have Mehler-Heine type asymptotic formulae.

##  7 Existence uniqueness of positive solutions

In the paper [1], Alsulami et al. discussed the existence and uniqueness of a
positive solution for the nonhomogeneous nonlinear second order difference
equations of the form,

$\ell_{n}=x_{n}\left(\sigma_{n,1}x_{n+1}+\sigma_{n,0}x_{n}+\sigma_{n,-1}\beta_{n-1}\right)+\kappa_{n}x_{n}$
(7.1)

with the initial conditions $x_{0}\in\mathbb{R},~{}x_{1}\geq 0$, whilst
$\kappa_{n}\in\mathbb{R}$, $\sigma_{n,j}>0,~{}j\in\\{0,\pm 1\\}$, or
$\left\\{\sigma_{n,0}>0,\sigma_{n,-1}\geq 0,\sigma_{n,1}\geq 0\right\\}$.
Following the results presented in [1], the solution of Eq. (4.9) is obtained.

######  Theorem 7.1.

For $\alpha>-1$ and $\beta_{0}=0$, there exists a unique positive $\beta_{1}$,
restricted by

$\beta_{1}(s;\alpha,N)=\frac{\mu_{2}(s;\alpha,N)}{\mu_{0}(s;\alpha,N)}=\frac{\mu_{0}(s;\alpha+2,N)}{\mu_{0}(s;\alpha,N)}=\frac{\alpha+1}{2\sqrt{2Ns}}\frac{D_{-\frac{\alpha+3}{2}}\left[\frac{N(1-s)}{\sqrt{2Ns}}\right]}{D_{-\frac{\alpha+1}{2}}\left[\frac{N(1-s)}{\sqrt{2Ns}}\right]}>0,$
(7.2)

such that the sequence $\left\\{\beta_{n}(s;\alpha,N)\right\\}$ satisfied by

$\beta_{n}\left(\beta_{n+1}+\beta_{n}+\beta_{n-1}+\frac{1-s}{2s}\right)=\frac{n+\alpha\Delta_{n}}{4Ns},~{}~{}~{}\Delta_{n}=\frac{1-(-1)^{n}}{2},$
(7.3)

is positive and the solution increases.

###### Proof.

Obviously, Eq. (7.3) is a special case of Eq. (7.1) with

$\sigma_{n,j}=1,~{}j\in\\{0,\pm
1\\},~{}~{}\kappa_{n}=\frac{1-s}{2s},~{}~{}\ell_{n}=\frac{n+\alpha\Delta_{n}}{4Ns},$

where $\Delta_{n}=\frac{1-(-1)^{n}}{2}$. The existence of
$\beta_{1}(s;\alpha,N)>0$ such that the sequence
$\left\\{\beta_{n}(s;\alpha,N)\right\\}$ is positive follows directly from [1,
Thm. 4.1] and the uniqueness of the solution of Eq. (7.3) follows immediately
from [1, Thm. 5.2]. ∎

Figure 2: Plots of $\beta_{1}(s;\alpha,1)$, $s\in[0,1]$ with various $\alpha$.
Figure 3: Plots of $\beta_{1}(s;1,N)$, $s\in[0,1]$ with various $N$.

Figure 4: Plots of the recurrence coefficients $\beta_{n}(s;\alpha,1)$, in
given initial condition (7.2) with various choices of $(s;\alpha)$, and $N=1$.

Figure 5: Plots of the recurrence coefficients
$\beta_{n}\left(\frac{1}{2};3,N\right)$, in given initial condition (7.2) with
various choices of $N$, and $s=\frac{1}{2},\alpha=3$.

Figure 6: Plots of the recurrence coefficients
$\beta_{n}\left(\frac{1}{2};3,1\right)$, with initial conditions
$\widetilde{\beta}_{0}\left(\frac{1}{2};3,1\right)=0,\widetilde{\beta_{1}}\left(\frac{1}{2};3,1\right)=\beta_{1}\left(\frac{1}{2};3,1\right)\pm\varepsilon,\varepsilon\in\left\\{10^{-1},10^{-3},10^{-5}\right\\}$.

Figure 7: Plots of the recurrence coefficients
$\beta_{n}\left(\frac{1}{2};3,1\right)$, with initial conditions
$\widetilde{\beta}_{0}\left(\frac{1}{2};3,1\right)=\pm
10^{-10},\widetilde{\beta_{1}}\left(\frac{1}{2};3,1\right)=\beta_{1}\left(\frac{1}{2};3,1\right)$.

######  Remark 7.1.

The figures are plotted by Mathematica taking 500 digits precision.

##  8 The Hankel determinant

The well-known supplementary conditions ($S_{1}$), ($S_{2}$) and the “sum
rule” ($S^{\prime}_{2}$) for the ladder operators, satisfied by the functions
$A_{n}(z)$ and $B_{n}(z)$, continue to hold by a slight modification. These
are,

$~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}B_{n+1}(z)+B_{n}(z)=zA_{n}(z)-v_{0}^{\prime}(z)+\frac{\alpha}{z},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(S_{1})$
$~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}1+z\left[B_{n+1}(z)-B_{n}(z)\right]=\beta_{n+1}(s)A_{n+1}(z)-\beta_{n}(s)A_{n-1}(z),~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(S_{2})$

and

$~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}B_{n}^{2}(z)+\left[v_{0}^{\prime}(z)-\frac{\alpha}{z}\right]B_{n}(z)+\sum_{j=0}^{n-1}A_{j}(z)=\beta_{n}(s)A_{n}(z)A_{n-1}(z).~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(S^{\prime}_{2})$

where $v_{0}(z)=Nsz^{4}+N(1-s)z^{2},~{}z\in\mathbb{R},~{}s\in[0,1]$.
$A_{n}(z)$ and $B_{n}(z)$ have been given in Eq. (6.18) and Eq. (6.19), which
read

$A_{n}(z)=4Nsz^{2}+4Ns\left(\beta_{n+1}+\beta_{n}\right)+2N(1-s),$ (8.1)

and

$B_{n}(z)=4Ns\beta_{n}z+\frac{\alpha}{z}\Delta_{n},~{}{\rm
with}~{}\Delta_{n}=\frac{\left[1-(-1)^{n}\right]}{2}.$ (8.2)

The three identities (S1), (S2) and (S${}^{\prime}_{2}$) were also stated in
Magnus [28]. Based on this and the formula (1.4), we first display the
relation between $D_{n}(s;\alpha,N)$ and $\beta_{n}(s;\alpha,N)$ in the next
lemma.

######  Lemma 8.1.

The logarithmic derivative of the Hankel determinant can be expressed as

$\begin{split}\frac{d}{ds}\log
D_{n}(s)=&\frac{N\left(s^{2}-1\right)(n+\alpha\Delta_{n})-2sn(n+\alpha)}{8s^{2}}+\frac{(1+s)\left(n+\alpha\Delta_{n}\right)^{2}}{32s^{2}\beta_{n}}\\\
&+\frac{N^{2}\left(1-s^{2}\right)^{2}+2Ns(1+s)^{2}\left(n+2\alpha-3\alpha\Delta_{n}\right)-4s^{2}}{8s^{2}(1+s)}\beta_{n}\\\
&+\frac{N^{2}(1-s^{2})}{2s}\beta_{n}^{2}+\frac{1}{2}N^{2}(1+s)\beta_{n}^{3}-\frac{2s}{1+s}\left(\frac{s{\beta_{n}^{\prime}}^{2}}{\beta_{n}}+\beta_{n}^{\prime}\right).\end{split}$
(8.3)

###### Proof.

It follows from Eq. (1.4) that,

$\log D_{n}(s)=\sum_{j=0}^{n-1}\log h_{j}(s),$ (8.4)

we find by taking the derivative of Eq. (8.4) with respect to $s$,

$\begin{split}&\frac{D_{n}^{\prime}(s)}{D_{n}(s)}=\sum_{j=0}^{n-1}\frac{h^{\prime}_{j}(s)}{h_{j}(s)}=\frac{N(1+s)}{2s}\sum_{j=0}^{n-1}\left(\beta_{j+1}+\beta_{j}\right)-\frac{n^{2}}{4s}-\frac{n\alpha}{4s}.\\\
\end{split}$

Here we have used the identity (2.4), which gives

$\frac{h^{\prime}_{n}(s)}{h_{n}(s)}=\frac{N(1+s)}{2s}\left(\beta_{n+1}+\beta_{n}\right)-\frac{2n+1}{4s}-\frac{\alpha}{4s}.$

Substituting Eqs. (8.1), (8.2) and $v_{0}(x)=Nsx^{4}+N(1-s)x^{2}$ into
($S^{\prime}_{2}$), we obtain

$\begin{split}&\left[4nNs+16N^{2}s^{2}\beta_{n}^{2}+4Ns\alpha\Delta_{n}+8N^{2}s(1-s)\beta_{n}\right]z^{2}+4Ns\alpha\beta_{n}(2\Delta_{n}-1)+2N(1-s)\alpha\Delta_{n}\\\
&+2nN(1-s)+4Ns\sum_{j=0}^{n-1}\left(\beta_{j+1}+\beta_{j}\right)=16N^{2}s^{2}\beta_{n}\left[\beta_{n+1}+2\beta_{n}+\beta_{n-1}+\frac{1-s}{s}\right]z^{2}\\\
&+16N^{2}s^{2}\beta_{n}\left[\beta_{n+1}+\beta_{n}+\frac{1-s}{2s}\right]\left[\beta_{n}+\beta_{n-1}+\frac{1-s}{2s}\right].\end{split}$

Equating coefficients on $z$ gives

$\begin{split}\beta_{n}\left(\beta_{n+1}+\beta_{n}+\beta_{n-1}+\frac{1-s}{2s}\right)=\frac{n+\alpha\Delta_{n}}{4Ns},\end{split}$
(8.5)
$\begin{split}\sum_{j=0}^{n-1}\left(\beta_{j+1}+\beta_{j}\right)=&4Ns\beta_{n}\left[\beta_{n+1}+\beta_{n}+\frac{1-s}{2s}\right]\left[\beta_{n}+\beta_{n-1}+\frac{1-s}{2s}\right]\\\
&+\alpha\beta_{n}(1-2\Delta_{n})-\frac{1-s}{2s}\left(n+\alpha\Delta_{n}\right).\end{split}$
(8.6)

Hence we have

$\begin{split}\frac{d}{ds}\log
D_{n}(s)=&2N^{2}(1+s)\beta_{n}\left(\beta_{n+1}+\beta_{n}+\frac{1-s}{2s}\right)\left(\beta_{n}+\beta_{n-1}+\frac{1-s}{2s}\right)\\\
&+\frac{N(1+s)}{2s}\alpha\beta_{n}(1-2\Delta_{n})-\frac{N\left(1-s^{2}\right)}{4s^{2}}(n+\alpha\Delta_{n})-\frac{n(n+\alpha)}{4s}.\end{split}$
(8.7)

From the Eq. (8.5), we find

$\begin{split}\beta_{n}\left(\beta_{n+1}+\beta_{n}+\frac{1-s}{2s}\right)=\frac{n+\alpha\Delta_{n}}{4Ns}-\beta_{n}\beta_{n-1}.\end{split}$

Substituting this into Eq. (8.7) and eliminating $\beta_{n-1}$ by Eq. (2.9),
the result is obtained. ∎

Let $N\rightarrow\infty,~{}n\rightarrow\infty$ but the term $n/N$ tends to a
fixed number $r$, which is bounded away from $0$ and close to $1$. If we
substitute the asymptotic expansion of $\beta_{n}(s)$, given by Theorem 5.1,
into the right hand of Eq. (8.3), we find

$\begin{split}\frac{D_{n}^{\prime}(s)}{D_{n}(s)}=A(s)N^{2}+B(s)N+C(s)+\frac{D(s)}{N}+\mathcal{O}\left(N^{-2}\right),~{}N\rightarrow\infty,\end{split}$
(8.8)

where the constants $A(s),B(s),C(s)$ and $D(s)$ can be found in Appendix B.

Therefore integrating from $s=0$ to $s=1$, and substituting the constants
given in Appendix B, we have

######  Lemma 8.2.

For any $\alpha>-1$, as $n\rightarrow\infty$ with $n=rN$, $r$ fixed, we have

$\begin{split}&\log\frac{D_{n}(1)}{D_{n}(0)}=\int_{0}^{1}\frac{d}{ds}\log
D_{n}(s)ds\\\ &=\frac{\left(3-2\log 3-2\log
r\right)}{8}n^{2}+\frac{\alpha\left(-1+\log 3+\log
r\right)}{4}n+\frac{\left(3\alpha^{2}-1\right)\log 2}{12}-\frac{\alpha^{2}\log
3}{4}\\\
&~{}~{}~{}+\frac{\alpha\left(\alpha^{2}+3\right)}{48}\frac{1}{n}+\mathcal{O}\left(n^{-2}\right),~{}~{}~{}\text{$n$
{\rm even}},\\\ \end{split}$ (8.9)

and

$\begin{split}\log\frac{D_{n}(1)}{D_{n}(0)}=&\frac{\left(3-2\log 3-2\log
r\right)}{8}n^{2}-\frac{\alpha\left(-1+\log 3+\log
r\right)}{4}n+\frac{\left(3\alpha^{2}-1\right)\log 2}{12}\\\
&-\frac{\alpha^{2}\log
3}{4}+\frac{\alpha\left(\alpha^{2}-3\right)}{48}\frac{1}{n}+\mathcal{O}\left(n^{-2}\right),~{}~{}~{}\text{$n$
{\rm odd}}.\\\ \end{split}$ (8.10)

###### Proof.

We calculate the integrals in Eq. (8.8) with the aid of Mathematica,

$\begin{split}&\log\frac{D_{n}(1)}{D_{n}(0)}=\int_{0}^{1}\frac{d}{ds}\log
D_{n}(s)ds\\\
&=N^{2}\int_{0}^{1}A(s)ds+N\int_{0}^{1}B(s)ds+\int_{0}^{1}C(s)ds+\frac{1}{N}\int_{0}^{1}D(s)ds+\mathcal{O}\left(N^{-2}\right)\\\
&=\frac{r^{2}\left(3-2\log 3-2\log
r\right)}{8}N^{2}+\frac{r\alpha\left(-1+\log 3+\log
r\right)}{4}N+\frac{\left(3\alpha^{2}-1\right)\log 2}{12}-\frac{\alpha^{2}\log
3}{4}\\\
&~{}~{}~{}+\frac{\alpha\left(\alpha^{2}+3\right)}{48r}\frac{1}{N}+\mathcal{O}\left(N^{-2}\right)\\\
&=\frac{\left(3-2\log 3-2\log r\right)}{8}n^{2}+\frac{\alpha\left(-1+\log
3+\log r\right)}{4}n+\frac{\left(3\alpha^{2}-1\right)\log
2}{12}-\frac{\alpha^{2}\log 3}{4}\\\
&~{}~{}~{}+\frac{\alpha\left(\alpha^{2}+3\right)}{48}\frac{1}{n}+\mathcal{O}\left(n^{-2}\right),~{}~{}~{}\text{$n$
even},\\\ \end{split}$

Similarly, we get the result for the case $n$ is odd. ∎

Mehta and Normand [32] gives

$\begin{split}D_{n}(0)&=\frac{1}{n!}\int_{\mathbb{R}^{n}}\prod_{1\leq j<k\leq
n}\left(x_{k}-x_{j}\right)^{2}\prod_{\ell=1}\left|x_{\ell}\right|^{\alpha}{\rm
e}^{-Nx_{\ell}^{2}}dx_{\ell}\\\
&=\frac{(2\pi)^{\frac{n}{2}}}{(2N)^{\frac{n^{2}}{2}}N^{\frac{\alpha
n}{2}}\Gamma(n+1)}\prod_{j=1}^{n}\frac{\Gamma\left(\frac{\alpha+1}{2}+\left\lfloor\frac{j}{2}\right\rfloor\right)}{\Gamma\left(\frac{1}{2}+\left\lfloor\frac{j}{2}\right\rfloor\right)}j!,\\\
\end{split}$

where $\lfloor n\rfloor$ denotes the integer part of $n$.

Alternatively, when $n$ is even, Han and Chen [20], see also [15], gives the
expression of $D_{n}(0)$ as

$\begin{split}D_{n}(0)=\frac{(2\pi)^{\frac{n}{2}}}{(2N)^{\frac{n^{2}}{2}}N^{\frac{\alpha
n}{2}}}\frac{G\left(\frac{3}{2}\right)G\left(\frac{1}{2}\right)}{G\left(\frac{\alpha+3}{2}\right)G\left(\frac{\alpha+1}{2}\right)}\frac{G(n+1)G\left(\frac{\alpha+n+3}{2}\right)G\left(\frac{\alpha+n+1}{2}\right)}{G\left(\frac{n+3}{2}\right)G\left(\frac{n+1}{2}\right)},\end{split}$
(8.11)

where $G$ is the Barnes $G-$function, that satisfies the functional relation
$G(z+1)=\Gamma(z)G(z)$, with $G(0)=1$, see Voros [44]. It is well-known that
the Barnes G-function has the asymptotic expansion:

$\begin{split}\log
G(z+1)\simeq&\frac{z^{2}}{4}+z\log\Gamma(z+1)-\left[\frac{z(z+1)}{2}+\frac{1}{12}\right]\log
z-\log A\\\
&+\sum_{k=1}^{\infty}\frac{\mathbb{B}_{2k+2}}{2k(2k+1)(2k+2)z^{2k}},\end{split}$
(8.12)

as $z\rightarrow\infty$ with $|\arg z|<\pi$. Here $\mathbb{B}_{n}$ are
Bernoulli numbers and

$\begin{split}A=\exp\left(\frac{1}{12}-\zeta^{\prime}(-1)\right),\end{split}$

is the Glaisher-Kinkelin constant, $A=1.2824271291\ldots$, where $\zeta(x)$ is
the Riemann zeta function, see for example [38, E.q. 5.17.5]. Then
substituting Eq. (8.12) into Eq. (8.11), we obtain,

$\begin{split}\log D_{n}(0)=&\frac{2\log r-2\log
2-3}{4}n^{2}+\left[\log(2\pi)-\frac{\alpha\left(1+\log 2-\log
r\right)}{2}\right]n\\\ &+\frac{3\alpha^{2}-1}{12}\log n-\log
A+\frac{1+6\alpha\log(2\pi)-3\alpha^{2}\log 2}{12}\\\
&+\log\frac{G\left(\frac{3}{2}\right)G\left(\frac{1}{2}\right)}{G\left(\frac{\alpha+3}{2}\right)G\left(\frac{\alpha+1}{2}\right)}+\frac{\alpha^{3}+\alpha}{12n}+\mathcal{O}\left(n^{-2}\right),~{}\text{$n$
even}.\end{split}$ (8.13)

A similar calculation gives

$\begin{split}D_{n}(0)=\frac{(2\pi)^{\frac{n}{2}}}{(2N)^{\frac{n^{2}}{2}}N^{\frac{\alpha
n}{2}}}\frac{G\left(\frac{3}{2}\right)G\left(\frac{1}{2}\right)}{G\left(\frac{\alpha+3}{2}\right)G\left(\frac{\alpha+1}{2}\right)}\frac{G(n+1)\left[G\left(\frac{\alpha+n+2}{2}\right)\right]^{2}}{\left[G\left(\frac{n+2}{2}\right)\right]^{2}},~{}~{}~{}\text{$n$
odd}.\end{split}$

which has the asymptotic expansion

$\begin{split}\log D_{n}(0)=&\frac{2\log r-2\log
2-3}{4}n^{2}+\left[\log(2\pi)-\frac{\alpha\left(1+\log 2-\log
r\right)}{2}\right]n\\\ &+\frac{3\alpha^{2}-1}{12}\log n-\log
A+\frac{1+6\alpha\log(2\pi)-3\alpha^{2}\log 2}{12}\\\
&+\log\frac{G\left(\frac{3}{2}\right)G\left(\frac{1}{2}\right)}{G\left(\frac{\alpha+3}{2}\right)G\left(\frac{\alpha+1}{2}\right)}+\frac{\alpha^{3}-2\alpha}{12n}+\mathcal{O}\left(n^{-2}\right),~{}\text{$n$
odd}.\end{split}$ (8.14)

Consequently, we get the asymptotics expansion for the Hankel determinant
$D_{n}(1)$,

######  Theorem 8.1.

As $r=\frac{n}{N}$ tends to $1$, we have

$\begin{split}\log D_{n}(1)=&\frac{2\log r-\log
144-3}{8}n^{2}+\frac{1}{4}\left[3\alpha\log
r+4\log(2\pi)-\alpha\left(3+\log\frac{4}{3}\right)\right]n\\\
&+\frac{3\alpha^{2}-1}{12}\log n+\frac{1}{12}\left[1-\log
2+6\alpha\log(2\pi)-3\alpha^{2}\log 3\right]-\log A\\\
&+\log\frac{G\left(\frac{3}{2}\right)G\left(\frac{1}{2}\right)}{G\left(\frac{\alpha+3}{2}\right)G\left(\frac{\alpha+1}{2}\right)}+\frac{\alpha\left(5\alpha^{2}+7\right)}{48}+\mathcal{O}\left(n^{-2}\right),~{}~{}~{}~{}~{}~{}\text{$n$
{\rm even}},\end{split}$ (8.15)

whilst

$\begin{split}\log D_{n}(1)=&\frac{2\log r-\log
144-3}{8}n^{2}+\frac{1}{4}\left[\alpha\log r+4\log(2\pi)-\alpha-\alpha\log
12\right]n\\\ &+\frac{3\alpha^{2}-1}{12}\log n+\frac{1}{12}\left[1-\log
2+6\alpha\log(2\pi)-3\alpha^{2}\log 3\right]-\log A\\\
&+\log\frac{G\left(\frac{3}{2}\right)G\left(\frac{1}{2}\right)}{G\left(\frac{\alpha+3}{2}\right)G\left(\frac{\alpha+1}{2}\right)}+\frac{\alpha\left(5\alpha^{2}-11\right)}{48}+\mathcal{O}\left(n^{-2}\right),~{}~{}~{}~{}~{}~{}\text{$n$
{\rm odd}}.\end{split}$ (8.16)

###### Proof.

The result is obtained by direct computation based on Eqs.(8.13) and (8.14)
and Lemma 8.2. ∎

##  9 Acknowledgements

The financial support of the Macau Science and Technology Development Fund
under grant number FDCT 130/2014/A3 and FDCT 023/2017/A1 are gratefully
acknowledged. We would also like to thank the University of Macau for generous
support: MYRG 2014-00011 FST, MYRG 2014-00004 FST.

##  10 Appendix A

$\displaystyle{\rm
dP_{I}}~{}~{}~{}~{}~{}~{}~{}x_{n+1}+x_{n}+x_{n-1}=\frac{z_{n}+\gamma(-1)^{n}}{x_{n}}+\delta.$

##  11 Appendix B

Here we list the coefficients of the Eq. (8.8).

For $A(s)$, letting $\widetilde{g}(s):=s-1+g(s)$, where
$g(s)=\sqrt{1-2s+12rs+s^{2}}$, then we have

$\begin{split}A(s)=&\frac{3r^{2}(1+s)}{8s\widetilde{g}(s)}-\frac{r\left(1+2rs-s^{2}\right)}{8s^{2}}+\frac{(1+s)\left(1-2s+2rs+s^{2}\right)\widetilde{g}(s)}{96s^{3}}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
&+\frac{\left(1-s^{2}\right)\widetilde{g}^{2}(s)}{288s^{3}}+\frac{(1+s)\widetilde{g}^{3}(s)}{3456s^{3}},~{}~{}~{}\text{$n$
even or odd}.\end{split}$

For $B(s)$, defining

$\begin{split}B_{0}(s):=&\frac{\alpha}{6s^{2}g(s)\widetilde{g}^{2}(s)}\Big{\\{}1-g(s)+s(6r-1)(2g(s)-3)+s^{2}\left[2+\left(6r-18r^{2}\right)(g(s)-4)\right]\\\
&+s^{3}(6r-2)(g(s)-1)+s^{4}(g(s)+12r-3)+s^{5}\Big{\\}},\end{split}$

then we obtain $B(s)=-B_{0}(s)$ if $n$ is even and $B(s)=B_{0}(s)$ if $n$ is
odd.

For $C(s)$, we first define

$\begin{split}c_{0}(s):=&\frac{1+s}{6sg\widetilde{g}^{2}\left[1+2g-2s(1-6r+g)+s^{2}\right]},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\end{split}$
$\begin{split}c_{1}(s):=&-1+6s-3rs-15s^{2}+12rs^{2}+72r^{2}s^{2}+20s^{3}-18rs^{3}-144r^{2}s^{3}-432r^{3}s^{3}-15s^{4}\\\
&+12rs^{4}+72r^{2}s^{4}+6s^{5}-3rs^{5}-s^{6},\end{split}$
$\begin{split}c_{2}(s):=&g\left(1-5s-3rs+10s^{2}+9rs^{2}-36r^{2}s^{2}-10s^{3}-9rs^{3}+36r^{2}s^{3}+5s^{4}+3rs^{4}-s^{5}\right),\end{split}$
$\begin{split}c_{3}(s):=&g\big{(}-3+15s-27rs-30s^{2}+81rs^{2}+108r^{2}s^{2}+30s^{3}-81rs^{3}-108r^{2}s^{3}-15s^{4}\\\
&+27rs^{4}+3s^{5}\big{)}\alpha^{2},\end{split}$
$\begin{split}c_{4}(s):=&\big{(}3-18s+45rs+45s^{2}-180rs^{2}-60s^{3}+270rs^{3}-1296r^{3}s^{3}+45s^{4}-180rs^{4}-18s^{5}\\\
&+45rs^{5}+3s^{6}\big{)}\alpha^{2},\end{split}$

then we find that
$C(s)=c_{0}(s)\left[c_{1}(s)+c_{2}(s)+c_{3}(s)+c_{4}(s)\right]$, either $n$ is
even or odd.

For $D(s)$, we get $D(s):=d_{0}(s)\left[d_{1}(s)+d_{2}(s)\right]$ when $n$ is
even, with

$\begin{split}d_{0}(s):=&-\frac{(1+s)\alpha}{g\widetilde{g}^{2}\left[1+2g-2s(1-6r+g)+s^{2}\right]^{3}},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\end{split}$
$\begin{split}d_{1}(s):=&g\big{(}-3+15s-15rs-30s^{2}+45rs^{2}-36r^{2}s^{2}+30s^{3}-45rs^{3}+36r^{2}s^{3}-15s^{4}\\\
&+15rs^{4}+3s^{5}-\alpha^{2}+5s\alpha^{2}-15rs\alpha^{2}-10s^{2}\alpha^{2}+45rs^{2}\alpha^{2}+36r^{2}s^{2}\alpha^{2}+10s^{3}\alpha^{2}\\\
&-45rs^{3}\alpha^{2}-36r^{2}s^{3}\alpha^{2}-5s^{4}\alpha^{2}+15rs^{4}\alpha^{2}+s^{5}\alpha^{2}\big{)},\end{split}$
$\begin{split}d_{2}(s):=&3-18s+33rs+45s^{2}-132rs^{2}+72r^{2}s^{2}-60s^{3}+198rs^{3}-144r^{2}s^{3}-2160r^{3}s^{3}\\\
&+45s^{4}-132rs^{4}+72r^{2}s^{4}-18s^{5}+33rs^{5}+3s^{6}+\alpha^{2}-6s\alpha^{2}+21rs\alpha^{2}+15s^{2}\alpha^{2}\\\
&-84rs^{2}\alpha^{2}+36r^{2}s^{2}\alpha^{2}-20s^{3}\alpha^{2}+126rs^{3}\alpha^{2}-72r^{2}s^{3}\alpha^{2}-864r^{3}s^{3}\alpha^{2}+15s^{4}\alpha^{2}\\\
&-84rs^{4}\alpha^{2}+36r^{2}s^{4}\alpha^{2}-6s^{5}\alpha^{2}+21rs^{5}\alpha^{2}+s^{6}\alpha^{2}.\end{split}$

else if $n$ is odd, we have
$D(s):=\widehat{d}_{0}(s)\left[\widehat{d}_{1}(s)+\widehat{d}_{2}(s)\right]$,
where

$\begin{split}\widehat{d}_{0}(s):=\frac{\alpha}{4f^{2}g^{5}},~{}{\rm
with}~{}f=1-2s-4rs+s^{2},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\end{split}$
$\begin{split}\widehat{d}_{1}(s):=&g\big{(}1-4s+56rs+5s^{2}-112rs^{2}+272r^{2}s^{2}-5s^{4}+112rs^{4}-272r^{2}s^{4}+4s^{5}\\\
&-56rs^{5}-s^{6}-\alpha^{2}+4s\alpha^{2}-24rs\alpha^{2}-5s^{2}\alpha^{2}+48rs^{2}\alpha^{2}-144r^{2}s^{2}\alpha^{2}+5s^{4}\alpha^{2}\\\
&-48rs^{4}\alpha^{2}+144r^{2}s^{4}\alpha^{2}-4s^{5}\alpha^{2}+24rs^{5}\alpha^{2}+s^{6}\alpha^{2}\big{)},\end{split}$
$\begin{split}\widehat{d}_{2}(s):=&-2+10s-64rs-18s^{2}+192rs^{2}-544r^{2}s^{2}+10s^{3}-128rs^{3}+544r^{2}s^{3}-768r^{3}s^{3}\\\
&+10s^{4}-128rs^{4}+544r^{2}s^{4}-768r^{3}s^{4}-18s^{5}+192rs^{5}-544r^{2}s^{5}+10s^{6}-64rs^{6}\\\
&-2s^{7}+\alpha^{2}-5s\alpha^{2}+32rs\alpha^{2}+9s^{2}\alpha^{2}-96rs^{2}\alpha^{2}+272r^{2}s^{2}\alpha^{2}-5s^{3}\alpha^{2}+64rs^{3}\alpha^{2}\\\
&-272r^{2}s^{3}\alpha^{2}+384r^{3}s^{3}\alpha^{2}-5s^{4}\alpha^{2}+64rs^{4}\alpha^{2}-272r^{2}s^{4}\alpha^{2}+384r^{3}s^{4}\alpha^{2}+9s^{5}\alpha^{2}\\\
&-96rs^{5}\alpha^{2}+272r^{2}s^{5}\alpha^{2}-5s^{6}\alpha^{2}+32rs^{6}\alpha^{2}+s^{7}\alpha^{2}.\end{split}$

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# Galerkin method for linear Integral-Algebraic Equations of index $1$

B. Shiri  Faculty of mathematical science, University of Tabriz, Tabriz - Iran
Faculty of mathematical science, University of Tabriz, Tabriz - Iran
<EMAIL_ADDRESS>
###### Abstract.

In this paper, we study direct and indirect Galerkin method for solving linear
Integral-Algebraic Equations of index $1.$ Convergence of indirect method is
also analyzed.

###### Key words and phrases:

Galerkin method, Differential Index, Integral-Algebraic Equations.

##  1\. Introduction

Recently numerical solution for Integral-Algebraic Equation (IAE) has been
studied by many researchers [1]. In physical models with constrained time or
space variables, Algebraic Equations including Integral or Differential
Equations arise. Therefore, the investigation of mixed types of equations is
important. This can be seen by observing the number of papers that are about
Differential-Algebraic Equations, Integro Differential Equations, and so on.
One of these systems which mixes Volterra Integral Equations of first and
second type is the system of Integral- Algebraic Equations. However, we don t
find any application for this systems except [6, 7, 8, 9], but it seems that
they be very important in the future. In this paper we use the standard
Galerkin method, for the system of Integral Algebraic Equations of index 1. We
implement the Galerkin method by considering the structure of IAE of index 1
in two ways, direct and indirect and we analyze indirect method here. We
Consider the system

(1.1) $A(t)y(t)+\int_{0}^{t}k(t,s)y(s)ds=f(t),\ \ \ \ t\in I:=[0,T],$

where $A\in\mathbf{C}(I,\mathbb{R}^{r\times r}),$
$f\in\mathbf{C}(I,\mathbb{R}^{r})$ and
$k\in\mathbf{C}(\mathbb{D},\mathbb{R}^{r\times r})$ with
$\mathbb{D}:=\\{(t,s):0\leq s\leq t\leq T\\}.$ If $A(t)$ is nonsingular for
all $t\in I$ then multiplying (1.1) by $A.^{-1}$ changes it to a system of the
second kind Volterra Integral Equations that it’s theoretical and numerical
analysis have been almost completely investigated [3, 4]. Otherwise, if $A(t)$
is a singular matrix with constant rank for all $t\in I$, then, the system
(1.1) is an IAE. The classification of this system done by introducing index
notation. The differential index of differential-algebraic equation can be
extended to this system. The following extension is done by Gear in [5].

######  Definition 1.1.

We say the system (1.1) has index m if, m is the minimum possible number of
differentiating (1.1) to obtain a system of second kind Volterra Integral
equation.

For example the following system

(1.2) $x(t)=f_{1}(t)+\int_{0}^{t}k^{11}(t,s)x(s)+k^{12}(t,s)y(s)ds,$ (1.3)
$0=f_{2}(t)+\int_{0}^{t}k^{21}(t,s)x(s)+k^{22}(t,s)y(s)ds.$

with the functions $f_{i}$, $k^{ij}$, $i,j=1,2$ are sufficiently smooth and
$f_{2}(0)=0$, $|k^{22}(t,t)|\geq k_{0}>0$ for all $t\in[0,T];$ (1.2) and (1.3)
is the system of Integral-Algebraic Equations of index 1. For this system the
Piecewise Polynomial Collocation method is investigated by Kauthen in [2]. In
this paper we investigate the Galerkin method for this system.

##  2\. preliminary

The Galerkin method is a projection type method [3, 4] that we apply to IAE of
index 1. Let $\mathcal{K}:\mathbf{X}\rightarrow\mathbf{X}$ be an operator,
where $\mathbf{X}$ is a Hilbert space with orthonormal base
$\\{\phi_{i}\\}_{i=1}^{\infty}$ and $\mathbf{X}_{n}$ be an $n$-dimensional
subspace of X generated by $\\{\phi_{i}\\}_{i=1}^{n}.$ Then the system (1.2)
and (1.3) changes to

(2.1) $\mathcal{K}x=f.$

For using the Galerkin method’s we define the projection
$P_{n}:\mathbf{X}\rightarrow\mathbf{X}_{n}$ by $P_{n}x=\sum_{i=1}^{n}\langle
x,\phi_{i}\rangle\phi_{i}.$ Then it is clear that $\|P_{n}x-x\|\rightarrow 0$
as $n\rightarrow\infty$. We search $\mathbf{x}_{n}\in\mathbf{X}_{n}$,
$\mathbf{x}_{n}=\overline{x}\overline{\phi}$ with
$\overline{x}=(x_{1},\ldots,x_{n})$ and
$\overline{\phi}=(\phi_{1},\ldots,\phi_{n})^{T},$ such that

(2.2) $P_{n}\mathcal{K}\mathbf{x}_{n}=P_{n}f.$

The corresponding operator of the system (1.1) is defined by

(2.3)
$\mathcal{K}([x,y])=\left(x-\int_{0}^{t}k^{11}(t,s)x(s)+k^{12}(t,s)y(s)ds,\
-\int_{0}^{t}k^{21}(t,s)x(s)+k^{22}(t,s)y(s)ds\right)^{T},$

and the assumption that $f=(f_{1},f_{2})^{T}$,
$\mathbf{X}\subseteq(\mathcal{L}^{2}[0,T])$ and the functions $f_{i}$,
$k^{ij}$, $i,j=1,2$ are sufficiently smooth and $f_{2}(0)=0$,
$|k^{22}(t,t)|\geq k_{0}>0$ for all $t\in[0,T],$ guarantee us that
$\mathcal{K}x=f$ has a continuous solution as it follows. We first
differentiate equation (1.3) w. r. to t and we have

$\begin{split}0=&f^{{}^{\prime}}_{2}(t)+k^{21}(t,t)x(t)+k^{22}(t,t)y(t)\\\
&+\int_{0}^{t}\frac{\partial k^{21}(t,s)}{\partial t}x(s)+\frac{\partial
k^{22}(t,s)}{\partial t}y(s)ds.\end{split}$

Because $|k^{22}(t,t)|\geq k_{0}>0$ we obtain

(2.4)
$\begin{split}y(t)=&-\frac{f^{{}^{\prime}}_{2}(t)}{k^{22}(t,t)}-\frac{k^{21}(t,t)}{k^{22}(t,t)}x(t)\\\
&-\int_{0}^{t}\frac{1}{k^{22}(t,t)}\frac{\partial k^{21}(t,s)}{\partial
t}x(s)-\frac{1}{k^{22}(t,t)}\frac{\partial k^{22}(t,s)}{\partial
t}y(s)ds,\end{split}$

as a Volterra integral equations of the second kind for $y(t)$. Now it is easy
to show that this system has a unique solution (see [4]).

##  3\. Implementation and Convergence

We can implement the Galerkin method for two systems (1.2),(1.3) and
(1.2),(2.4). For the former the Galerkin method is called direct and for the
further, is called indirect method. For the second system, the convergence and
order of convergence can be obtained by similar arguments as discussed [6] and
[3]. In [3] the order of convergence has been explained for Fredholm integral
equation. Since, by using suitable kernel, each Volterra integral equation can
be changed to a Fredholm integral equation, Galerkin method for both are the
same. Hence the convergence order of the approximate solution is equal to the
convergence order of the best approximation in space $X_{n}$. This is
confirmed by numerical examples. Let $\\{V_{i}\\}_{i=1}^{\infty}$ be Legendre
polynomials as orthonormal base for $\mathcal{L}^{2}[-1,1]$. For
$\mathcal{L}^{2}[0,T]$ we use shifted Legendre polynomials
$\\{V_{T_{i}}\\}_{i=1}^{\infty}$,
$V_{T_{i}}(s)=\sqrt{\frac{2}{T}}V_{i}(\frac{2*s-T}{T})$, $s\in[0,T]$ and for
$(\mathcal{L}^{2}[0,T])^{2},$ we use
$\\{(V_{T_{i}},0),(0,V_{T_{i}})\\}_{i=1}^{\infty}$ with
$\langle(x,y),(z,t)\rangle=\langle x,z\rangle_{2}+\langle y,t\rangle_{2}$ and
$\langle x,y\rangle_{2}=\int_{0}^{T}x\overline{y}dt$. Also we suppose
$P_{n}([x,y])=\left(\sum_{i=1}^{n}x_{i}V_{T_{i}},\sum_{i=1}^{n}y_{i}V_{T_{i}}\right).$
For equations (1.2), (1.3) and (2.4) we have.

(3.1)
$\begin{split}\sum_{i=1}^{n}x_{i}V_{T_{i}}(t)=&f_{1}(t)+\sum_{i=1}^{n}x_{i}\int_{0}^{t}k^{11}(t,s)V_{T_{i}}(s)ds\\\
&+\sum_{i=1}^{n}y_{i}\int_{0}^{t}k^{12}(t,s)V_{T_{i}}(s)ds,\end{split}$ (3.2)
$0=f_{2}(t)+\sum_{i=1}^{n}x_{i}\int_{0}^{t}k^{21}(t,s)V_{T_{i}}(s)ds+\sum_{i=1}^{n}y_{i}\int_{0}^{t}k^{22}(t,s)V_{T_{i}}(s)ds,$
(3.3)
$\begin{split}\sum_{i=1}^{n}y_{i}V_{T_{i}}(t)=&-\frac{f^{{}^{\prime}}_{2}(t)}{k^{22}(t,t)}-\frac{k^{21}(t,t)}{k^{22}(t,t)}\sum_{i=1}^{n}x_{i}V_{T_{i}}(t)\\\
&-\sum_{i=1}^{n}x_{i}\int_{0}^{t}\frac{1}{k^{22}(t,t)}\frac{\partial
k^{21}(t,s)}{\partial t}V_{T_{i}}(s)ds\\\
&-\sum_{i=1}^{n}y_{i}\int_{0}^{t}\frac{1}{k^{22}(t,t)}\frac{\partial
k^{22}(t,s)}{\partial t}V_{T_{i}}(s)ds.\end{split}$

Now we can use (3.1),(3.2) for the first system and (3.1),(3.3) for the second
system introduced above. It is important for analysis to note that formula
(3.3) is derivative of (3.2). Now if we multiply equations (3.1)-(3.3) to
$V_{T_{j}}(t)$ and integrate on the $[0,T]$, we have obtained

(3.4)
$\begin{split}x_{j}=&\int_{0}^{T}f_{1}(t)V_{T_{j}}(t)dt+\sum_{i=1}^{n}x_{i}\int_{0}^{T}\int_{0}^{t}k^{11}(t,s)V_{T_{i}}(s)V_{T_{j}}(t)dsdt\\\
&+\sum_{i=1}^{n}y_{i}\int_{0}^{T}\int_{0}^{t}k^{12}(t,s)V_{T_{i}}(s)V_{T_{j}}(t)dsdt,\end{split}$
(3.5)
$\begin{split}0=&\int_{0}^{T}f_{2}(t)V_{T_{j}}(t)dt+\sum_{i=1}^{n}x_{i}\int_{0}^{T}\int_{0}^{t}k^{21}(t,s)V_{T_{i}}(s)V_{T_{j}}(t)dsdt\\\
&+\sum_{i=1}^{n}y_{i}\int_{0}^{T}\int_{0}^{t}k^{22}(t,s)V_{T_{i}}(s)V_{T_{j}}(t)dsdt,\end{split}$
(3.6)
$\begin{split}y_{j}=&-\int_{0}^{T}\frac{f^{{}^{\prime}}_{2}(t)}{k^{22}(t,t)}dt-\sum_{i=1}^{n}x_{i}\int_{0}^{T}\int_{0}^{t}\frac{1}{k^{22}(t,t)}\frac{\partial
k^{21}(t,s)}{\partial t}V_{T_{i}}(s)V_{T_{j}}(t)dsdt\\\
&-\sum_{i=1}^{n}x_{i}\int_{0}^{T}\frac{k^{21}(t,t)}{k^{22}(t,t)}V_{T_{i}}(t)V_{T_{j}}(t)dt\\\
&-\sum_{i=1}^{n}y_{i}\int_{0}^{T}\int_{0}^{t}\frac{1}{k^{22}(t,t)}\frac{\partial
k^{22}(t,s)}{\partial t}V_{T_{i}}(s)V_{T_{j}}(t)dsdt.\end{split}$

The direct and indirect Galerkin method for system (1.2),(1.3) are solving the
linear systems (3.4),(3.5) and (3.4),(3.6) respectively.

######  Lemma 3.1.

[4] Let $V$ and $W$ be normed spaces, with $W$ complete. Let
$K\in\mathcal{L}(V,W)$, let $\\{K_{n}\\}$ be a sequence of compact operators
in $\mathcal{L}(V,W),$ and assume $K_{n}\rightarrow K$ in $\mathcal{L}(V,W).$
Then K is compact.

######  Lemma 3.2.

[4] Let $V$ be a Banach space, and let $\\{P_{n}\\}$ be a family of bounded
projections on $V$ with

$P_{n}u\rightarrow u\ \ \ \mbox{as}\ \ \ n\rightarrow\infty,u\in V$

If $K:V\rightarrow V$ is compact, then

$\|K-P_{n}K\|\rightarrow 0\ \ \ \mbox{as}\ \ \ \ n\rightarrow\infty.$

######  Theorem 3.3.

[4] Assume $K:V\rightarrow V$ is bounded, with V a Banach space; and assume
$\lambda-K:V\rightarrow V$ is one to one and surjective. Further assume

$\|K-P_{n}K\|\rightarrow 0\ \mbox{as}\ n\rightarrow\infty.$

Then for all sufficiently large $n$, say, $n\geq N,$ the operator
$(\lambda-K)^{-1}$ exists as a bounded operator from $V$ to $V.$ Moreover, it
is uniformly bounded:

$\sup_{n\geq N}\|(\lambda-P_{n}K)^{-1}\|\leq\infty.$

For the solutions $u_{n}$ (n sufficiently large) of
$P_{n}(\lambda-K)u_{n}=P_{n}F$ and the solutions $u$ of $(\lambda-K)u=F,$ we
have

$u-u_{n}=\lambda(\lambda-P_{n}K)^{-1}(u-P_{n}u)$

and the two-sided error estimate

$\frac{|\lambda|}{\|\lambda-
P_{n}K\|}\|u-P_{n}u\|\leq\|u-u_{n}\|\leq|\lambda|\left\|(\lambda-
P_{n}K)^{-1}\right\|\|u-P_{n}u\|.$

We have following Theorem for indirect method.

######  Theorem 3.4.

Let for the system (1.2) and (1.3) the following conditions are satisfied:

  1. (1)

$f_{i}\in C^{1}(0,T),\ K^{i,j}[t,s]\in C^{1}(\mathbb{D}),$ for all
$i,j=\\{1,2\\},$

  2. (2)

$k^{22}(t,t)\geq k_{0}>0,$

where $\mathbb{D}:=\\{(t,s):0\leq s\leq t\leq T\\}.$ Then the convergence
order of the approximate solution is equal to the convergence order of the
best approximation in the space $X_{n}=\\{V_{T_{i}}\\}_{i=1}^{\infty}.$

###### Proof.

The space $\mathcal{L}^{2}[0,T]$ with $\mathcal{L}^{2}$ Norm is Banach space,
hence the Cartesian product on this space, i.e.,
$V=(\mathcal{L}^{2}[0,T])^{2},$ with introduced norm generate also Banach
space. Now we change the system (1.2) and (2.4) to the following system

(3.7) $(I-K)x=F,\ \ F=[F_{1},F_{2}]^{T},\ \ x=[x_{1},x_{2}]^{T},$

with

$F_{1}(t)=f_{1}(t),\ \ \ \ \ \
F_{2}(t)=-\frac{f^{{}^{\prime}}_{2}(t)}{k^{22}(t,t)}-\frac{k^{21}(t,t)}{k^{22}(t,t)}f_{1}(t),$
$K^{11}(t,s)=\left\\{\begin{array}[]{lc}k^{11}(t,s),&s\leq t,\\\
0,&s>t,\end{array}\right.\ \
K^{12}(t,s)=\left\\{\begin{array}[]{lc}k^{12}(t,s),&s\leq t,\\\
0,&s>t,\end{array}\right.\ \ $
$K^{21}(t,s)=\left\\{\begin{array}[]{lc}-\frac{k^{21}(t,t)}{k^{22}(t,t)}k^{11}(t,s)-\frac{1}{k^{22}(t,t)}\frac{\partial
k^{21}(t,s)}{\partial t},&s\leq t,\\\ 0,&s>t,\end{array}\right.$
$K^{22}(t,s)=\left\\{\begin{array}[]{lc}-\frac{k^{21}(t,t)}{k^{22}(t,t)}k^{12}(t,s)-\frac{1}{k^{22}(t,t)}\frac{\partial
k^{22}(t,s)}{\partial t},&s\leq t,\\\ 0.&s>t,\end{array}\right.$

and with the integral operator

$Kx=\int_{0}^{T}\left[\begin{array}[]{cc}K^{11}(t,s)&K^{12}(t,s)\\\
K^{21}(t,s)&K^{22}(t,s)\\\
\end{array}\right]\left[\begin{array}[]{c}x_{1}(s)\\\ x_{2}(s)\\\
\end{array}\right]ds.$

By introducing

$K_{n}x:=\int_{0}^{T}\left[\begin{array}[]{cc}K^{11}_{n}(t,s)&K^{12}_{n}(t,s)\\\
K^{21}_{n}(t,s)&K^{22}_{n}(t,s)\\\
\end{array}\right]\left[\begin{array}[]{c}x_{1}(s)\\\ x_{2}(s)\\\
\end{array}\right]ds,$

with continuous kernels

$K^{11}_{n}(t,s)=\left\\{\begin{array}[]{lc}k^{11}(t,s),&s\leq t,\\\
-n(k^{11}(t,t))((t+\frac{1}{n})-s),&s\in[t,t+\frac{1}{n}],\\\
0,&s>t+\frac{1}{n},\end{array}\right.$
$K^{12}_{n}(t,s)=\left\\{\begin{array}[]{lc}k^{12}_{n}(t,s),&s\leq t\\\
-n(k^{12}(t,t))((t+\frac{1}{n})-s),&s\in[t,t+\frac{1}{n}],\\\
0,&s>t+\frac{1}{n},\end{array}\right.\ \ $
$K^{21}_{n}(t,s)=\left\\{\begin{array}[]{lc}-\frac{k^{21}(t,t)}{k^{22}(t,t)}k^{11}(t,s)-\frac{1}{k^{22}(t,t)}\frac{\partial
k^{21}(t,s)}{\partial t},&s\leq t,\\\
-n(\frac{k^{21}(t,t)}{k^{22}(t,t)}k^{11}(t,t)+\frac{1}{k^{22}(t,t)}\frac{\partial
k^{21}(t,t)}{\partial t})((t+\frac{1}{n})-s),&s\in[t,t+\frac{1}{n}],\\\
0,&s>t+\frac{1}{n},\end{array}\right.$
$K^{22}_{n}(t,s)=\left\\{\begin{array}[]{lc}-\frac{k^{21}(t,t)}{k^{22}(t,t)}k^{12}(t,s)-\frac{1}{k^{22}(t,t)}\frac{\partial
k^{22}(t,s)}{\partial t},&s\leq t,\\\
-n(\frac{k^{21}(t,t)}{k^{22}(t,t)}k^{12}(t,t)+\frac{1}{k^{22}(t,t)}\frac{\partial
k^{22}(t,t)}{\partial t})((t+\frac{1}{n})-s),&s\in[t,t+\frac{1}{n}],\\\
0,&s>t+\frac{1}{n}.\end{array}\right.$

which is compact and using lemma 3.1, imply K is compact.

Applying the Galerkin method to the Fredholm system (3.7) leads to the linear
system equivalent with the system (3.4),(3.6). So it is enough to show that
for this system the convergence order of the approximate solution by the
Galerkin method is equal to the convergence order of the best approximation.
Since, the system (3.7) also is a volterra system, using the same line of [4],
$(I-K)$ is invertible so is surjective. The Lemma 3.2 provide the remainder
condition. Finally all conditions of the theorem 3.3 are satisfied and this
completes the proof. ∎

##  4\. Numerical results

.

Table 1.  Numerical result obtained from (3.4),(3.5)

$\begin{array}[]{|c|c|c|c|c|c|}\hline\cr n&2&4&6&8&10\\\
\hline\cr\|x_{n}-x\|&4.0e-002&3.0e-004&7.4e-007&9.4e-010&7.6e-013\\\
\hline\cr\|y_{n}-y\|&1.6e-001&2.2e-003&8.1e-006&1.4e-008&1.4e-011\\\
\hline\cr\end{array}$

.

Table 2. Numerical result obtained from (3.4),(3.6)

$\begin{array}[]{|c|c|c|c|c|c|}\hline\cr n&2&4&6&8&10\\\
\hline\cr\|x_{n}-x\|&3.8e-002&2.6e-004&6.7e-007&8.8e-010&6.8e-013\\\
\hline\cr\|y_{n}-y\|&7.3e-002&5.1e-004&1.3e-006&1.7e-009&1.3e-012\\\
\hline\cr\end{array}$

.

Table 3. Logarithm of the error by best approximation using Legendre
polynomial)

$\begin{array}[]{|c|c|c|c|c|c|}\hline\cr n&2&4&6&8&10\\\
\hline\cr\|P_{n}\sin(t)-\sin(t)\|&5.1e-002&3.3e-004&8.1e-007&1.0e-009&8.0e-013\\\
\hline\cr\|P_{n}\cos(t)-\cos(t)\|&7.8e-002&5.4e-004&1.4e-006&1.7e-009&1.3e-012\\\
\hline\cr\end{array}$

Figure 1. plots of convergence order by direct Galerkin method

Figure 2. plots of convergence order by indirect Galerkin method. We note that
the slope of this figures is same as the slope of figure 3 as claimed

Figure 3. plots of convergence order of best the approximation

In this section we illustrate the methods by numerical examples.
Example 1.
We consider (2.3) with $k^{11}=s+t$, $k^{12}=s^{2}+t^{2}$, $k^{21}=s-t^{2}$,
$k^{31}=s+t+1$, $f_{1}=-t-2sin(t)t^{2}+2sin(t)$ and
$f_{2}=t^{2}-2sin(t)+cos(t)t-cos(t)t^{2}+1-cos(t)-2sin(t)t$ where the exact
solutions are $x=sin(t)$ and $y=cos(t)$. Numerical results obtained by two
methods are in table 1 and 2 which show rapid convergence. In these tables
$\|f\|=\max_{T\in[0,T]}|f(t)|$. In the following figures, we also illustrate
$\log_{10}(\|x_{n}-x\|)$ and $\log_{10}(\|y_{n}-y\|)$ . For the second system,
we expect that this figure to be similar to the figure of
$\log_{10}(\|P_{n}x-x\|)$ and $\log_{10}(\|P_{n}y-y\|)$. This numerical
experiment shows that the second system is more accurate than the first. But
we need more computations in comparison with the first.

######  Remark 4.1.

In equations (3.4),(3.5) and (3.6), we find two type of integrals that should
be computed, single and double integrals. The integral of type
$\int_{0}^{T}f(t)dt$ can be computed by Gaussian Integration Methods (see
[11]) of the form

$\int_{-1}^{1}f(t)dt\simeq\sum_{i=1}^{n}w_{i}f(x_{i})$

and by changing of variable we have

(4.1) $\int_{0}^{T}f(t)dt\simeq\sum_{i=1}^{n}w_{i}f(Tx_{i}/2+T/2)$

where $x_{1},...,x_{n}$ are the roots of $V_{T_{(}n+1)},$ and
$w_{1},...,w_{n}$ are computed by the system expressed in (3.6.13) of [11]. An
approximate values of are computed as follows
$\int_{0}^{T}\int_{0}^{t}f(t,s)dsdt$ in this calculation is as it follows

(4.2)
$\int_{0}^{T}\int_{0}^{t}f(t,s)dsdt=\sum_{i=1}^{n}\sum_{j=1}^{n}w_{i}w_{j}f(\frac{Tx_{i}}{2}+\frac{T}{2},\frac{(\frac{Tx_{i}}{2}+\frac{T}{2})x_{i}}{2}+\frac{\frac{Tx_{i}}{2}+\frac{T}{2}}{2})$

Also we notice that the term $n$ is not fixed for different $P_{m}$, indeed in
our calculation $n=m$ that means for improving accuracy of the solution we
improve the accurate approximate solutions of integrals.

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