# Linear systems, Hankel products and the sinh-Gordon equation

Gordon Blower<EMAIL_ADDRESS>Department of Mathematics and
Statistics, Lancaster University, Lancaster LA1 4YF, United Kingdom  and  Ian
Doust<EMAIL_ADDRESS>School of Mathematics and Statistics, University of
New South Wales, Sydney, NSW 2052, Australia

(Date: 13 October 2022)

###### Abstract.

Let $(-A,B,C)$ be a linear system in continuous time $t>0$ with input and
output space ${\mathbb{C}}^{2}$ and state space $H$. The scattering functions
$\phi_{(x)}(t)=Ce^{-(t+2x)A}B$ determines a Hankel integral operator
$\Gamma_{\phi_{(x)}}$; if $\Gamma_{\phi_{(x)}}$ is trace class, then the
Fredholm determinant $\tau(x)=\det(I+\Gamma_{\phi_{(x)}})$ determines the tau
function of $(-A,B,C)$. The paper establishes properties of algebras including
$R_{x}=\int_{x}^{\infty}e^{-tA}BCe^{-tA}dt$ on $H$. Thus the paper obtains
solutions of the sinh-Gordon PDE. The tau function for sinh-Gordon satisfies a
particular Painléve $\mathrm{III}^{\prime}$ nonlinear ODE and describes a
random matrix model, with asymptotic distribution found by the Coulomb fluid
method to be the solution of an electrostatic variational problem on an
interval.

Key words: Sinh-Gordon equation, tau function, linear systems, Howland
operators

AMS Classification: 47B35, 47A48, 34M55

##  1\. Introduction

This paper is concerned with the Fredholm determinants of operators that are
introduced via linear systems and with their applications to the sinh-Gordon
equations. The study has application in Tracy and Widom’s approach [41] to
random matrix theory.

We begin by fixing some notation concerning linear systems. Let $H$ be a
separable complex Hilbert space with orthonormal basis
$(e_{j})_{j=0}^{\infty}$, and let ${\mathcal{L}}(H)$ denote the algebra of
bounded linear operators on $H$. We shall denote the adjoint of
$B\in\mathcal{L}(H)$ by $B^{\dagger}$. Let $H_{0}$ be a separable complex
Hilbert space which serves as the input and output space; let
$B:H_{0}\rightarrow H$ and $C:H\rightarrow H_{0}$ be bounded linear operators.
On the state space $H$, let $(T_{t})_{t>0}$ be a strongly continuous and
bounded semigroup with infinitesimal generator $-A$, which is densely defined
on domain ${\mathcal{D}}(A)$, where ${\mathcal{D}}(A)$ is itself a Hilbert
space for the graph norm $\|f\|^{2}_{{\mathcal{D}}(A)}=\|f\|^{2}+\|Af\|^{2}$.

The continuous time linear system $(-A,B,C)$ is

$\displaystyle{\frac{dX}{dt}}$ $\displaystyle=-AX+BU$ $\displaystyle Y$
$\displaystyle=CX,$ $\displaystyle X(0)$ $\displaystyle=0.$ (1.1)

The scattering function of $(-A,B,C)$ is $\phi(t)=Ce^{-tA}B$, which is a
bounded and weakly continuous function
$\phi:(0,\infty)\rightarrow{\mathcal{L}}(H_{0})$.

Suppose that $\phi\in L^{2}((0,\infty);{\mathcal{L}}(H_{0}))$. Then the Hankel
integral operator with scattering function $\phi$ is the operator

$\Gamma_{\phi}f(x)=\int_{0}^{\infty}\phi(x+y)f(y)\,dy\qquad(f\in
L^{2}((0,\infty);H_{0})).$ (1.2)

Such Hankel operators do not themselves form an algebra, although they have an
algebraic structure which has been exploited in [35] and section 3.5 of [28].
An important fact is that every bounded self-adjoint Hankel integral operator
on $L^{2}(0,\infty)$ can be realised as the Hankel operator associated with a
linear system $(-A,B,C)$ in continuous time with state space $H$, and such
that $\phi(t)=Ce^{-tA}B$ where $(e^{-tA})$ is a strongly continuous semigroup
on $H$ (see [34] and [32]).

Recall that $K\in\mathcal{L}(H)$ is Hilbert–Schmidt if
$\left\|K\right\|_{{\mathcal{L}}^{2}}^{2}=\sum_{j=0}^{\infty}\left\|Ke_{j}\right\|^{2}$
is finite. We shall denote the set of all Hilbert–Schmidt operators on $H$ by
${\mathcal{L}}^{2}={\mathcal{L}}^{2}(H)$. This space contains the ideal of
trace class operators
${\mathcal{L}}^{1}=\\{K_{1}K_{2}\,:\,K_{1},K_{2}\in{\mathcal{L}}^{2}\\}$.
There are now several criteria for ensuring the boundedness of a Hankel
operator; see, for example [34] and [36]. We shall repeatedly use the basic
result that if $t\|\phi(t)\|^{2}_{{\mathcal{L}}^{2}(H_{0},H_{0})}$ is
integrable, then $\Gamma_{\phi}$ determines a Hilbert–Schmidt operator
$L^{2}((0,\infty);H_{0})\rightarrow L^{2}((0,\infty);H_{0})$.

######  Definition 1.1.

Let $(-A,B,C)$ be a linear system as in (1) with scattering function $\phi$,
and suppose that $\Gamma_{\phi}$ is trace class. Let
$\phi_{(x)}(t)=\phi(t+2x)$. Then the tau function of $(-A,B,C)$ is defined to
be $\tau(x)=\det(I+\Gamma_{\phi_{(x)}})$, for $x>0$.

This tau function is analogous to the tau function introduced by Jimbo, Miwa
and Ueno [25] to describe the isomonodromy of rational differential equations,
and generalizes the classical theta function. Mumford [33] constructed
solutions of the cubic nonlinear Schrödinger, $KdV$, $mKdV$ and sine-Gordon
equations in terms of classical theta functions on abelian varieties. Ercolani
and McKean [14] extended the analysis of $KdV$ to infinite-dimensional abelian
varieties in which case the scattering function $\phi$ is of rapid decay.

In [6], we reinterpreted the results of [35] and [28] in terms of an algebra
of operators on $H$ with a special associative product, and showed how
nonlinear differential equations such as $KdV$ emerge from algebraic
identities in this product. In the present paper, we continue this analysis by
addressing the sinh-Gordon equation in section 5. The simplest Darboux
transformation of a linear systems is $(-A,B,C)\mapsto(-A,B,-C)$ which takes
$R\mapsto-R$. In section 5, we introduce a $2\times 2$ block matrix system
which enables use to introduce a tau function for the sinh-Gordon equation in
the form $\det(I+\Gamma_{\phi_{(x)}})(I-\Gamma_{\phi_{(x)}})^{-1}$.

In [13], the authors consider the algebra of integrable operators of the form
$\lambda I+K$ where $K$ is an integral operator on $L^{2}(L)$ for a curve $L$
in ${\mathbb{C}}$ with kernel

$k_{K}(z,w)={\frac{\sum_{j=1}^{N}f_{j}(z)g_{j}(w)}{z-w}}$

where $\sum_{j=1}^{N}f_{j}(z)g_{j}(z)=0$. They achieve several results using
Riemann–Hilbert theory and make applications to some integrable operators in
random matrix theory. Integrable operators of this form include the
Christoffel–Darboux kernel from orthogonal polynomials when $L$ is a real
subinterval. In [41], the authors systematically consider these kernels and
their scaling limits for classical orthogonal polynomials and obtain
integrable operators that are fundamental to random matrix theory. In this
application, the Fredholm determinant $\det(I+\lambda K)$ is of primary
interest. In particular, they show that the Airy and Bessel kernels can be
expressed as products of Hankel integral operators on $L^{2}(0,\infty)$.

In section 3 we give sufficient conditions $T\in{\mathcal{A}}$ to be expressed
as a product of Hankel integral operators.

##  2\. Differential rings and Darboux addition

Let $u\in C^{\infty}({\mathbb{R}};{\mathbb{C}})$ be the potential in
Schrödinger’s equation. Gelfand and Dikii [18] considered the algebra
${\mathcal{A}}={\mathbb{C}}[u,u^{\prime},\dots]$ of differential polynomials
generated by the potential and its derivatives. In this section we consider a
generalization of this relating to linear systems. In [32] there is an
existence theory covering the self-adjoint case.

We suppose that $\phi$ is a scalar scattering function that can be realized
from a linear system $(-A,B,C)$ from (1) as $\phi(t)=Ce^{-tA}B$, where
$A:H\rightarrow H$, $B:{{\mathbb{C}}}\rightarrow H$ and
$C:H\rightarrow{{\mathbb{C}}}$ are bounded. Suppose further that
$\Xi_{x},\Theta_{x}:L^{2}((0,\infty);H_{0}^{2})\rightarrow
L^{2}((0,\infty);{\mathbb{C}})$ are Hilbert–Schmidt operators, where

$\displaystyle\Xi_{x}f$ $\displaystyle=\int_{x}^{\infty}e^{-tA}Bf(t)\,dt$
(2.1) $\displaystyle\Theta_{x}f$
$\displaystyle=\int_{x}^{\infty}e^{-tA^{\dagger}}C^{\dagger}f(t)\,dt.$ (2.2)

Then

$R_{x}=\Xi_{x}\Theta^{\dagger}_{x}=\int_{x}^{\infty}e^{-tA}BCe^{-tA}\,dt$
(2.3)

is trace class, and satisfies the Lyapunov equation

${\frac{d}{dx}}R_{x}=-AR_{x}-R_{x}A.$ (2.4)

######  Definition 2.1.

The tau function of $(-A,B,C)$ is $\tau(x)=\det(I+R_{x})$.

With $F_{x}=(I+R_{x})^{-1}$, (2.4) gives

$(d/dx)F_{x}=AF_{x}+F_{x}A-2F_{x}AF_{x}=e^{-xA}BCe^{-xA}.$

We introduce the vector space of linear operators

${\mathcal{A}}_{\Sigma}={\hbox{span}}_{{\mathbb{C}}}\\{A^{n_{1}},A^{n_{2}}F_{x}A^{n_{2}}\dots
F_{x}A^{n_{r}};n_{j}\in{{\mathbb{N}}}\\},$ (2.5)

and introduce the associative product

$P\ast Q=P(AF_{x}+F_{x}A-2F_{x}AF_{x})Q\qquad(P,Q\in{\mathcal{A}}_{\Sigma})$
(2.6)

and the derivation

$\partial
P=A(I-2F_{x})P+{\frac{dP}{dx}}+P(I-2F_{x})A\qquad(P\in{\mathcal{A}}_{\Sigma}).$
(2.7)

and the bracket operation

$\bigl{\lfloor}P\bigr{\rfloor}=Ce^{-xA}F_{x}PF_{x}e^{-xA}B,\qquad(P\in{\mathcal{A}}_{\Sigma})$
(2.8)

as in [6] Definition 4.3.

######  Lemma 2.2.

  1. (i)

$({\mathcal{A}}_{\Sigma},\ast,\partial)$ is a differential ring, and the
bracket operation
$\lfloor\,\cdot\rfloor:({\mathcal{A}}_{\Sigma},\ast,\partial)\rightarrow(C^{\infty}((0,\infty);{\mathbb{C}}),\cdot,d/dx)$
gives a homomorphism of differential rings.

  2. (ii)

Let the potential of the linear system $(-A,B,C)$ be

$u=-2\,\frac{d^{2}}{dx^{2}}\log\det(I+R_{x}).$

Then $u=-4\lfloor A\rfloor$.

###### Proof.

See [6] Lemma 4.2, and a related approach is developed in [12]. ∎

######  Definition 2.3.

[19]

  1. (i)

For a potential $u\in C^{\infty}({\mathbb{R}};{\mathbb{C}})$ the stationary
$KdV$ hierarchy for a sequence $(f_{\ell})$ is the recurrence relation
$f_{0}=1,$

${\frac{\partial f_{\ell}}{\partial
x}}=-{\frac{1}{4}}{\frac{\partial^{3}f_{\ell-1}}{\partial
x^{3}}}+u{\frac{\partial f_{\ell-1}}{\partial x}}+{\frac{1}{2}}{\frac{\partial
u}{\partial x}}f_{\ell-1}\qquad(\ell=1,2,3,\dots).$ (2.9)

  2. (ii)

The stationary hierarchy for the sine-Gordon and modified $KdV$ equation is
similar, with $u$ replaced by

$w_{+}=-{\frac{1}{4}}\Bigl{(}\Bigl{(}{\frac{\partial u}{\partial
x}}\Bigr{)}^{2}+2i{\frac{\partial^{2}u}{\partial x^{2}}}\Bigr{)}.$ (2.10)

The solutions sequence for these hierarchies are differential polynomials in
$u$ by [19] Remark 2.2, and by Lemma 2.2 there exists
$X_{j}\in{\mathcal{A}}_{\sigma}$ such that $f_{\ell}=\lfloor X_{\ell}\rfloor$.
For the stationary $KdV$ hierarchy, we showed in [6] that when $f_{0}=(1/2)u$,
the terms $f_{\ell}=(-1)^{\ell}2\lfloor A^{2\ell-1}\rfloor$ give a solution.
In the physical models discussed in [23], $KdV$ has a partition function which
is the square of a tau function; whereas $mKdV$ has a partition function that
is the product of tau functions; for sinh-Gordon, one considers quotients of
tau functions.

Suppose that $u\in C^{\infty}({\mathbb{R}};{\mathbb{R}})$ is bounded and let
$\lambda_{0}$ be the bottom of the spectrum of $L=-d^{2}/dx^{2}+u$. For
$\lambda<\min\\{\lambda_{0},0\\}$ and signs $\\{\pm\\}$, we consider
solutions:

  1. ($+$)

$h_{+}(x;\lambda)$ of $Lh_{+}=\lambda h_{+}$ such that $h_{+}\geq 0$, with

$\int_{0}^{\infty}h_{+}(x)^{2}\,dx<\infty,\quad{\hbox{and}}\quad\int_{-\infty}^{0}h_{+}(x)^{2}\,dx=\infty;$
(2.11)

  2. ($-$)

$h_{-}(x;\lambda)$ of $Lh_{-}=\lambda h_{-}$ such that $h_{-}\geq 0$, with

$\int_{-\infty}^{0}h_{-}(x)^{2}\,dx<\infty\quad{\hbox{and}}\quad\int_{0}^{\infty}h_{-}(x)^{2}\,dx=\infty.$
(2.12)

Let $G(x,y;\lambda)$ be the Greens integral kernel that implements $(\lambda
I-L)^{-1}$ on $L^{2}({\mathbb{R}};{\mathbb{C}}),$ so the diagonal of $G$
satisfies

$G(x,x;\lambda)={\frac{h_{+}(x,\lambda)h_{-}(x\,\lambda)}{{\hbox{Wr}}\,(h_{+}(x,\lambda),h_{-}(x,\lambda))}}.$
(2.13)

In the case where $u$ is a real continuous and $2\pi$-periodic potential, the
differential equation $-h^{\prime\prime}+uh=\lambda h$ is known as Hill’s
equation, and is associated with a hyperelliptic spectral curve $\Sigma$ with
points $\\{(\lambda,\pm):\lambda\in{\mathbb{C}}\\}$ giving a two-sheeted cover
of ${\mathbb{C}}$ as in [31]. The notion of Darboux addition refers to the
operation on potentials that corresponds to the addition rule for pole
divisors of Baker-Akhiezer functions on $\Sigma$, as discussed in [30]. In our
case, we have a scattering potential $u$, and we consider the corresponding
addition rule on the potentials, without seeking to interpret directly the
notion of the spectral curve. Nevertheless, there is a simple operation on the
linear system $(-A,B,C)$ that gives rise to Darboux addition on the associated
potential from Lemma 2.2 (ii).

######  Definition 2.4.

(Darboux addition) For $\sigma\in\\{+,-\\}$, we define
$e(x,\lambda,\sigma)=h_{\sigma}(x,\lambda)$. Associated with
$(\lambda,\sigma)$ there is a change in the potential

$T^{(\lambda,\sigma)}:u\mapsto u-2{\frac{d^{2}}{dx^{2}}}\log
e(x,\lambda,\sigma).$

corresponding to the change of linear systems

$T^{(\zeta,\sigma)}(-A,B,C)=(-A,(\zeta I+\sigma A)(\zeta I-\sigma
A)^{-1}B,C).$

Let ${\hbox{Wr}}\,(f,g)=fg^{\prime}-f^{\prime}g$ be the Wronskian determinant
of $f$ and $g$. Then for $(\lambda_{1},\sigma_{1})$ and
$(\lambda_{2},\sigma_{2})$, the effect of successive transformations is

$T^{(\lambda_{2},\sigma_{2})}T^{(\lambda_{1},\sigma_{1})}:u\mapsto
u-2{\frac{d^{2}}{dx^{2}}}\log{\hbox{Wr}}\,(e(x,\lambda_{2},\sigma_{2}),e(x,\lambda_{1},\sigma_{1})).$
(2.14)

See [6] Corollary 3.6 and [29]. In particular, with $p_{1}=(\lambda,-)$
followed by $p_{2}=(\lambda+\delta\lambda,+)$, we have an infinitesimal
addition $u\mapsto u-\delta\lambda{\bf X}(u)+o(\delta\lambda)$. We proceed to
compute the generating function for infinitesimal addition.

######  Proposition 2.5.

Suppose that the linear system $(-A,B,C)$ has potential $u=-4\lfloor
A\rfloor$. Then the infinitesimal addition satisfies

${\bf X}(u)=-2{\frac{d}{dx}}G(x,x;\lambda)$ (2.15)

or equivalently

${\bf X}(u)={\frac{-2}{\sqrt{-\lambda}}}\bigl{\lfloor}A(I-2F)A(\lambda
I+A^{2})^{-1}+A(\lambda
I+A^{2})^{-1}(I-2F)A\bigr{\rfloor}\qquad(\lambda<\lambda_{0}).$ (2.16)

###### Proof.

We begin by recalling McKean’s calculation. By the composition formula (2.14),
the infinitesimal addition satisfies

$\displaystyle u-\delta\lambda{\bf X}(u)+o(1)$
$\displaystyle=u-2{\frac{d^{2}}{dx^{2}}}\log{\hbox{Wr}}\,(h_{+}(x,\lambda+\delta\lambda),h_{-}(x,\lambda))$
(2.17)

where from the differential equation

${\frac{d}{dx}}{\hbox{Wr}}\,(h_{+}(x,\lambda+\delta\lambda),h_{-}(x,\lambda))=\delta\lambda
h_{+}(x,\lambda+\delta\lambda)h_{-}(x,\lambda).$ (2.18)

Combining (2.17), (2.18) and (2.13), and taking $\delta\lambda\rightarrow 0$,
one can express the infinitesimal transformation in terms of the diagonal of
the Greens function as in (2.15). In [6] Theorem 5.4, we computed this
diagonal Greens function, and we deduce

${\bf
X}(u)={\frac{-2}{\sqrt{-\lambda}}}{\frac{d}{dx}}\Bigl{(}{\frac{1}{2}}-{\frac{\lfloor
A\rfloor}{\lambda}}+{\frac{\lfloor A^{3}\rfloor}{\lambda^{2}}}-{\frac{\lfloor
A^{5}\rfloor}{\lambda^{3}}}+\dots\Bigr{)}\qquad(\lambda\rightarrow-\infty)$

where the summands of this series are

${\frac{d}{dx}}\lfloor A^{2m+1}\rfloor=\lfloor
A(I-2F)A^{2m+1}+A^{2m+1}(I-2F)A\rfloor$

as in (5.4) of [6]. Hence by summing the resulting geometric series, we have

${\bf X}(u)={\frac{-2}{\sqrt{-\lambda}}}\bigl{\lfloor}A(I-2F)A(\lambda
I+A^{2})^{-1}+A(\lambda I+A^{2})^{-1}(I-2F)A\bigr{\rfloor}.$

∎

##  3\. Products of Hankel operators

In what follows we shall require some of the concepts of abstract differential
calculus as developed in [11]. Let $S$ be an associative and unital complex
algebra. There is a natural multiplication map $\mu:S\otimes S\to S$,
$\mu\bigl{(}\sum a_{j}\otimes b_{j}\bigr{)}=\sum a_{j}b_{j}$. We shall denote
the nullspace of this map as $\Omega^{1}S$ and so we have an exact sequence

$0\longrightarrow\Omega^{1}S\longrightarrow S\otimes S\longrightarrow
S\longrightarrow 0;$ (3.1)

In this section, we shall mainly be interested in the case where
$S={\mathbb{C}}(x)$, the space of rational functions. In these cases, the
tensor product $S\otimes S$ may be regarded as consisting of functions of the
form $\sum f_{j}(x)g_{j}(y)$ acting on a product space. There are several
natural derivation maps that act on these spaces.

######  Definition 3.1.

For $S={\mathbb{C}}(x)$ define

  1. (1)

$d:S\rightarrow\Omega^{1}S$ $:f\mapsto df$ or $f(x)\mapsto f(x)-f(y)$, the
noncommutative differential;

  2. (2)

$\delta:S\otimes S\rightarrow S\otimes S:$ $f(x,y)\mapsto(x-y)f(x,y)$, the
inner derivation on $S\otimes S$ as an $S$-bimodule given by $x\in S$;

  3. (3)

$D:\Omega^{1}S\to S\otimes S:$ $fdg\mapsto f(x){\frac{g(x)-g(y)}{x-y}}$,
dividing by $x-y$; in particular, $D\circ d:s\rightarrow S\otimes S:$
$g(x)\mapsto{\frac{g(x)-g(y)}{x-y}}$ is the divided difference;

  4. (4)

$\partial:S\rightarrow S:$ $f(x)\mapsto f^{\prime}(x)$, the usual derivative.

See (6.13) for a significant example.

Products of Hankel operators arise from rational differential equations,
involving poles. As is common in control theory, we let ${\mathcal{S}}$ be the
unital complex algebra of stable rational functions, namely the space of
$f(s)/g(s)$ where $f(s),g(s)\in{{\mathbb{C}}}[s]$ satisfy
$\mathop{\mathrm{degree}}f(s)\leq\mathop{\mathrm{degree}}g(s)$, and all the
zeroes of $g(s)$ satisfy $\mathop{{\rm Re}}s<0$. We allow poles at $\infty$ by
adjoining $s$ to form ${\mathcal{S}}[s]$. Now let
$p_{1},\dots,p_{r}\in{\mathbb{C}}$ be distinct points satisfying $\mathop{{\rm
Re}}p_{j}>0$, and for $P=\\{p_{1},\dots,p_{r},\infty\\}$, introduce the
algebra

$\mathcal{S}_{P}=\mathcal{S}\left[s,\frac{s+1}{s-p_{1}},\dots,\frac{s+1}{s-p_{r}}\right]$
(3.2)

by formally adjoining the inverses of $(s-p_{j})/(s+1)\in\mathcal{S}$. The
possible poles of elements of $\mathcal{S}_{p}$ are in $P$ or the open left
half plane. For even $m=2n$, let
$\Omega_{\infty},\Omega_{1},\dots,\Omega_{r}\in M_{m\times m}(\mathcal{S})$
and define $\Omega(s)\in M_{n\times n}(\mathcal{S}_{p})$ by

$\Omega(s)=\Omega_{\infty}(s)s+\sum_{j=1}^{r}\frac{\Omega_{j}(s)}{s-p_{j}}.$
(3.3)

The residues are
$\Omega_{\infty}(\infty),\Omega_{1}(p_{1}),\dots,\Omega_{r}(p_{r})$. We assume
further that $\Omega(s)^{\top}=\Omega(s)$, where $\Omega^{\top}$ denotes the
transpose. Let

$J=\begin{bmatrix}0&-I_{n}\\\ I_{n}&0\end{bmatrix}.$ (3.4)

We also let
$\langle\cdot,\cdot\rangle:{\mathbb{C}}^{m}\times{\mathbb{C}}^{m}\rightarrow{\mathbb{C}}$
$\langle(z_{j})_{j=1}^{m},(w_{j})_{j=1}^{m}\rangle=\sum_{j=1}^{m}z_{j}w_{j}$
be the standard bilinear pairing, and $\|\cdot\|$ the operator norm on
$M_{m\times m}({{\mathbb{C}}})$. Given a solution of the differential equation

$J{\frac{d}{dx}}\Psi=\Omega\Psi,$ (3.5)

for $\Psi\in L^{2}((0,\infty);{\mathbb{C}})$ we can introduce a kernel
$k(x,y)=\langle J\Psi(x),\Psi(y)\rangle/(x-y)$ which operates on
$L^{2}((0,\infty);{\mathbb{C}})$.

######  Proposition 3.2.

Let $\Omega$ be as in (3.3) and suppose that
$\int_{0}^{\infty}t\left\|\Psi(t)\right\|^{2}\,dt$ is finite. Then there exist
Hankel integral operators
$\Gamma_{\psi_{j}},\Gamma_{\phi_{j}}\in{\mathcal{L}}^{2}(L^{2}((0,\infty);M_{m\times
m}({{\mathbb{C}}})))$ such that
$K=\sum_{j=1}^{N}\Gamma_{\psi_{j}}\Gamma_{\phi_{j}}$ determines a trace class
operator $K$ on $L^{2}((0,\infty);{{\mathbb{C}}}^{m})$ with kernel

$k(z,w)=\int_{0}^{\infty}\sum_{j=1}^{N}\psi_{j}(z+u)\phi_{j}(w+u)\,du\qquad(z,w>0).$
(3.6)

###### Proof.

Consider $\partial_{\Delta}:R\otimes R\rightarrow R\otimes R:$
$\partial_{\Delta}f(x,y)={\frac{\partial}{\partial
x}}f(x,y)+{\frac{\partial}{\partial y}}f(x,y)$, so
$\partial_{\Delta}=\partial\otimes Id+Id\otimes\partial$. We have

$\displaystyle\delta_{\Delta}K(z,w)$
$\displaystyle=\delta_{\Delta}D\bigl{(}\Psi(z)^{\top}J\Psi(w)-\Psi(w)^{\top}J\Psi(z)\bigr{)}$
$\displaystyle=\Psi(z)^{\top}\frac{-\Omega(z)+\Omega(w)}{z-w}\Psi(w)-\Psi(w)^{\top}\frac{\Omega(z)-\Omega(w)}{z-w}\Psi(z),$
(3.7)

which is a finite-rank $M_{m\times m}({{\mathbb{C}}})$-valued kernel. Here the
entries of the difference quotient $(\Omega(z)-\Omega(w))/(z-w)$ may be
expressed as a sum of products of rational functions in $z$ and rational
functions in $w$, which are bounded for $z,w>0$ by the partial fraction
decomposition of $\Omega(s)$ in (3.3). Note that

${\frac{(z-\alpha)^{-j}-(w-\alpha)^{-j}}{z-w}}=-\begin{bmatrix}{\frac{1}{z-\alpha}}&\dots&{\frac{1}{(z-\alpha)^{j}}}\end{bmatrix}\begin{bmatrix}0&\dots&1\\\
\vdots&\ddots&\vdots\\\
1&\dots&0\end{bmatrix}\begin{bmatrix}{\frac{1}{w-\alpha}}\\\ \vdots\\\
{\frac{1}{(w-\alpha)^{j}}}\end{bmatrix}$

in which the matrix $R$ is constant on cross diagonals, and is real symmetric
and satisfies $R^{2}=I_{r\times r}$, with ${\hbox{trace}}\,R=0$ for even $r$
and $1$ for odd $r$, hence $R$ is similar via a real orthogonal matrix to a
real diagonal matrix with signature $0$ or $1$, respectively. From the right-
hand side of (3.6) we extract coefficients depending on $w$ or $z$ such that

$\displaystyle\delta_{\Delta}K(z,w)$
$\displaystyle=-\sum_{j=1}^{N}\psi_{j}(z)\phi_{j}(w)$
$\displaystyle=\delta_{\Delta}\int_{0}^{\infty}\sum_{j=1}^{N}\psi_{j}(z+u)\phi_{j}(w+u)\,du$
(3.8)

Also, $k(z,w)\rightarrow 0$ as $z\rightarrow\infty$ or $w\rightarrow\infty$.
Likewise the right-hand side of (1.2) converges to $0$ as $z\rightarrow\infty$
or $w\rightarrow\infty$, so we have the identity (3.6). Recall that if
$\int_{0}^{\infty}t\left\|\phi(t)\right\|^{2}\,dt$ is finite then
$\Gamma_{\phi}$ is Hilbert–Schmidt. The representation
$K=\sum_{j=1}^{N}\Gamma_{\psi_{j}}\Gamma_{\phi_{j}}$ then shows that $K$ is a
trace-class operator. ∎

######  Theorem 3.3.

Let $k$ be a trace-class kernel as in Proposition 3.2, and let
$k_{x}(u,v)=k(x+u,x+v)$. Then for all $\lambda\in{\mathbb{C}}$, there exists a
matrix function $T(x,y)\in M_{N\times N}({\mathbb{C}})$ for $0\leq x\leq y$
that satisfies

${\frac{\partial^{2}T}{\partial x^{2}}}(x,y)-{\frac{\partial^{2}T}{\partial
y^{2}}}(x,y)=-2\Bigl{(}{\frac{d}{dx}}T(x,x)\Bigr{)}T(x,y)\qquad(0<x<y)$ (3.9)

and

${\frac{d}{dx}}\log\det(I+\lambda
K_{x})=\mathop{\mathrm{trace}}T(x,x)\qquad(x>0).$ (3.10)

###### Proof.

Using Proposition 3.2, write $K$ in the form
$\sum_{j=1}^{N}\Gamma_{\psi_{j}}\Gamma_{\phi_{j}}$ with $\psi_{j},\phi_{j}\in
L^{2}((0,\infty);{\mathbb{R}})$, or in matrix form

$\begin{bmatrix}K&0\\\
0&0\end{bmatrix}=\begin{bmatrix}\Gamma_{\psi_{1}}&\dots&\Gamma_{\psi_{N}}\\\
0&\dots&0\end{bmatrix}\begin{bmatrix}\Gamma_{\phi_{1}}&0\\\ \vdots&\vdots\\\
\Gamma_{\phi_{N}}&0\end{bmatrix}.$ (3.11)

Consider $N\times N$ matrix functions

$\Phi_{1}=\begin{bmatrix}\psi_{1}&\dots&\psi_{N}\\\
{}&0&{}\end{bmatrix},\Phi_{2}=\begin{bmatrix}\phi_{1}&{}\\\ \vdots&0\\\
\phi_{N}&{}\end{bmatrix}.$

First we factorize $K$ as a product of matrix Hankel operators. By Proposition
3.2 $K$ is a complex-linear combination of Hankel products
$\Gamma_{\psi}\Gamma_{\phi}$, where $\psi,\phi\in
L^{2}((0,\infty);{\mathbb{R}})$. Given measurable
$\phi:(0,\infty)\rightarrow{\mathbb{R}}$ such that
$\int_{0}^{\infty}t|\phi(t)|^{2}dt$ is finite, the Hankel operator with
scattering function $\phi$ is self-adjoint and Hilbert–Schmidt, hence can be
realised by a linear system $(-A,B,C)$ in continuous time with input and
output space ${\mathbb{C}}$; this follows from [32]. We can apply this to the
real and imaginary parts of the entries of $\Phi_{1}$ and $\Phi_{2}$, and
introduce for $j=1,2$ the matrices

$(-A_{j},B_{j},C_{j})=\Biggl{(}\begin{bmatrix}-A_{j1}&\dots&0\\\
\vdots&\ddots&\vdots\\\
0&\dots&-A_{jN}\end{bmatrix},\begin{bmatrix}B_{j1}&\dots&0\\\
\vdots&\ddots&\vdots\\\
0&\dots&B_{jN}\end{bmatrix},\begin{bmatrix}C_{j1}&\dots&C_{jN}\\\
\vdots&\dots&\vdots\\\ 0&\dots&0\end{bmatrix}\Biggr{)}.$

We can suppose that there are linear systems $(-A_{1},B_{1},C_{1})$ and
$(-A_{2},B_{2},C_{2})$ with state space $H$ and input and output spaces
${\mathbb{C}}^{N}$ such that $\Phi_{1}(t)=C_{1}e^{-tA_{1}}B_{1}$ and
$\Phi_{2}(t)=C_{2}e^{-tA_{2}}B_{2}$. Then

$\det(I+\lambda
K)=\det(I+\lambda\Gamma_{\Phi_{1}}\Gamma_{\Phi_{2}})=\det\begin{bmatrix}I&\lambda\Gamma_{\Phi_{1}}\\\
-\Gamma_{\Phi_{2}}&I\end{bmatrix}.$ (3.12)

Let $\bigl{(}-{\hat{A}},{\hat{B}},{\hat{C}}\bigr{)}$ be the linear system

$\bigl{(}-{\hat{A}},{\hat{B}},{\hat{C}}\bigr{)}=\Biggl{(}\begin{bmatrix}-A_{1}&0\\\
0&-A_{2}\end{bmatrix},\begin{bmatrix}0&B_{1}\\\
B_{2}&0\end{bmatrix},\begin{bmatrix}\lambda C_{1}&0\\\
0&-C_{2}\end{bmatrix}\Biggr{)}$

which has scattering function

$\hat{\Phi}(t)={\hat{C}}e^{-t{\hat{A}}}{\hat{B}}=\begin{bmatrix}0&\lambda
C_{1}e^{-tA_{1}}B_{1}\\\
-C_{2}e^{-tA_{2}}B_{2}&0\end{bmatrix}=\begin{bmatrix}0&\lambda\Phi_{1}(t)\\\
-\Phi_{2}(t)&0\end{bmatrix}.$ (3.13)

The operator function

${\hat{R}}_{x}=\int_{x}^{\infty}e^{-t{\hat{A}}}{\hat{B}}{\hat{C}}e^{-t{\hat{A}}}\,dt=\int_{x}^{\infty}\begin{bmatrix}0&-e^{-tA_{1}}B_{1}C_{2}e^{-tA_{2}}\\\
\lambda e^{-tA_{2}}B_{2}C_{1}e^{-tA_{1}}&0\end{bmatrix}\,dt$ (3.14)

satisfies the Lyapunov equation
$\frac{d{\hat{R}}_{x}}{dx}=-{\hat{R}}_{x}{\hat{A}}-{\hat{A}}{\hat{R}}_{x}$ and
we clearly have

$-{\frac{d{\hat{R}}_{x}}{dx}}=e^{-x{\hat{A}}}{\hat{B}}{\hat{C}}e^{-x{\hat{A}}}\qquad(x>0).$
(3.15)

Its determinant satisfies

$\det(I+R_{0})=\det(I+\Gamma_{\hat{\Phi}})=\det(I+\lambda K).$ (3.16)

Let

${\hat{T}}(x,y)=-{\hat{C}}e^{-x{\hat{A}}}(I+{\hat{R}}_{x})^{-1}e^{-y{\hat{A}}}{\hat{B}}$

To find this determinant, we consider the Gelfand–Levitan equation

${\hat{\Phi}}(x+y)+{\hat{T}}(x,y)+\int_{x}^{\infty}{\hat{T}}(x,z){\hat{\Phi}}(z+y)\,dz=0\qquad(0<x<y)$
(3.17)

which has solution

${\hat{T}}(x,y)=-{\hat{C}}e^{-x{\hat{A}}}(I+{\hat{R}}_{x})^{-1}e^{-y{\hat{A}}}{\hat{B}}$

as one verifies by substituting the given formula into the equation. Also, the
Lyapunov equation gives

$\displaystyle\mathop{\mathrm{trace}}\bigl{(}{\hat{T}}(x,x)\bigr{)}$
$\displaystyle=\mathop{\mathrm{trace}}\Bigl{(}(I+{\hat{R}}_{x})^{-1}{\frac{d{\hat{R}}_{x}}{dx}}\Bigr{)}$
$\displaystyle={\frac{d}{dx}}\log\det(I+{\hat{R}}_{x}).$

Also, $\hat{T}$ is a solution of

${\frac{\partial^{2}{\hat{T}}}{\partial
x^{2}}}(x,y)-{\frac{\partial^{2}{\hat{T}}}{\partial
y^{2}}}(x,y)=-2\Bigl{(}{\frac{d}{dx}}{\hat{T}}(x,x)\Bigr{)}{\hat{T}}(x,y)\qquad(0<x<y).$

∎

Theorem 3.3 applies in particular to the sinh-Gordon equation, as we discuss
in section 5.

######  Example 3.4.

For the weight $x^{\alpha}e^{-x}$ on $(0,\infty)$, Laguerre’s polynomials

$L^{(\alpha)}_{n}(x)=x^{-\alpha}e^{x}{\frac{d^{n}}{dx^{n}}}\bigl{(}x^{n+\alpha}e^{-x}\bigr{)}\qquad(\alpha\geq
0,n=0,1,2,\dots)$

are orthononal. Let
$u_{\alpha}(x)=e^{-x/2}x^{(\alpha+1)/2}L_{n}^{(\alpha)}(x)$ and
$w_{\alpha}(x)=u_{\alpha}^{\prime}(x);$ then these give a solution of
Laguerre’s differential equation

$\begin{bmatrix}0&-1\\\
1&0\end{bmatrix}{\frac{d}{dx}}\begin{bmatrix}u_{\alpha}\\\
w_{\alpha}\end{bmatrix}=\begin{bmatrix}{\frac{2n+\alpha+1}{2x}}+{\frac{1-\alpha^{2}}{4x^{2}}}-{\frac{1}{4}}&0\\\
0&1\end{bmatrix}\begin{bmatrix}u_{\alpha}\\\ w_{\alpha}\end{bmatrix},$

so one can apply Proposition 3.2. In particular, for $\alpha=1$, we obtain a
Hankel product

${\frac{u_{1}(x)u_{1}^{\prime}(y)-u_{1}^{\prime}(x)u_{1}(y)}{x-y}}=\int_{0}^{\infty}{\frac{u_{1}(x+t)u_{1}(y+t)}{(x+t)(y+t)}}dt.$

The Laguerre system is associated with orthogonal polynomials for the weight
$x^{\alpha}e^{-x}$ on $(0,\infty)$, which is the limit of the semi-classical
weights $x^{\alpha}(1-x/n)^{n}$ from [26]. In section 6, we consider the
singular weight $w_{\alpha,s}(s)=x^{\alpha}e^{-x-s/x}$, which has $\log
w_{\alpha,s}(x)$ not integrable over $(0,1)$, hence lies beyond the scope of
Szegö’s theory on asymptotic formulas for orthogonal polynomials.

##  4\. A linear system for Darboux addition

The pair of linear systems $(-A,B,C)$ giving $I+R_{x}$ and $(-A,B,-C)$ giving
$I-R_{x}$ are related by a Darboux transformation, as discussed in Theorem 3.4
of [6]. As there, we introduce the matricial system

$\bigl{(}-{\hat{A}},{\hat{B}},{\hat{C}}\bigr{)}=\Biggl{(}\begin{bmatrix}-A&0\\\
0&-A\end{bmatrix},\begin{bmatrix}0&B\\\ B&0\end{bmatrix},\begin{bmatrix}C&0\\\
0&C\end{bmatrix}\Biggr{)}$ (4.1)

which has scattering function

${\hat{\Phi}}(t)={\hat{C}}e^{-t{\hat{A}}}{\hat{B}}=\begin{bmatrix}0&Ce^{-tA}\\\
Ce^{-tA}B&0\end{bmatrix}=\begin{bmatrix}0&\phi(t)\\\ \phi(t)&0\end{bmatrix}.$

Let

$\displaystyle{\hat{R}}_{x}$
$\displaystyle=\int_{x}^{\infty}e^{-t{\hat{A}}}{\hat{B}}{\hat{C}}e^{-t{\hat{A}}}\,dt$
$\displaystyle=\int_{x}^{\infty}\begin{bmatrix}0&e^{-tA}BCe^{-tA}\\\
e^{-tA}BCe^{-tA}&0\end{bmatrix}\,dt=\begin{bmatrix}0&R_{x}\\\
R_{x}&0\end{bmatrix}.$ (4.2)

so that when $I\pm R_{x}$ are invertible, we have

$\hat{F}_{x}=(I+\hat{R}_{x})^{-1}=\begin{bmatrix}(I-R_{x}^{2})^{-1}&-R(I-R_{x}^{2})^{-1}\\\
-R(I-R_{x}^{2})^{-1}&(I-R_{x}^{2})^{-1}\end{bmatrix}.$

We can then define

$\displaystyle{\hat{T}}(x,y)$
$\displaystyle=-{\hat{C}}e^{-t{\hat{A}}}(I+{\hat{R}}_{x})^{-1}e^{-y{\hat{A}}}{\hat{B}}=\begin{bmatrix}W(x,y)&V(x,y)\\\
V(x,y)&W(x,y)\end{bmatrix}$ (4.3)

where

$\displaystyle W(x,y)$ $\displaystyle=Ce^{-xA}(I-R^{2}_{x})^{-1}R_{x}e^{-yA}B$
(4.4) $\displaystyle V(x,y)$
$\displaystyle=-Ce^{-xA}(I-R_{x}^{2})^{-1}e^{-yA}B.$ (4.5)

######  Proposition 4.1.

  1. (i)

The functions $V(x,x)$ and

$S(x)=\log\det(I+R_{x})-\log\det(I-R_{x})$

satisfy

${\frac{d}{dx}}S(x)=2V(x,x).$ (4.6)

  2. (ii)

The Fredholm determinant satisfies

$\det(I-\Gamma_{\phi_{x}}^{2})=\exp\Bigl{(}-4\int_{x}^{\infty}(t-x)V(t,t)^{2}\,dt\Bigr{)},$
(4.7)

  3. (iii)

and with $q(x)=-4\lfloor A\rfloor$, the function

$U(x)=\exp\Bigl{(}2\int_{x}^{\infty}V(t,t)\,dt\Bigr{)}=\exp(-S(x))$ (4.8)

satisfies

${\frac{d^{2}U}{dx^{2}}}=q(x)U(x).$ (4.9)

###### Proof.

(i) From Lyapunov’s equation, we have

$\displaystyle{\frac{d}{dx}}S(x)$
$\displaystyle={\frac{d}{dx}}\Bigl{(}\log\det(I+R_{x})-\log\det(I-R_{x})\Bigr{)}$
$\displaystyle={\hbox{trace}}\Bigl{(}(I+R_{x})^{-1}R^{\prime}_{x}+(I-R_{x})^{-1}R_{x}^{\prime}\Bigr{)}$
$\displaystyle=-2{\hbox{trace}}\Bigl{(}(I-R_{x}^{2})^{-1}e^{-xA}BCe^{-xA}\Bigr{)}$
$\displaystyle=-2Ce^{-xA}(I-R_{x}^{2})^{-1}e^{-xA}B$ $\displaystyle=2V(x,x).$
(4.10)

(ii) Then

$-V(x,x)=Ce^{-xA}(I-R_{x}^{2})^{-1}e^{-xA}B=\bigl{\lfloor}(I-R_{x})^{-1}(I+R_{x})\bigr{\rfloor}$
(4.11)

so by Lemma 2.5

$\displaystyle-{\frac{d}{dx}}V(x,x)$ $\displaystyle={\frac{\partial}{\partial
x}}\bigl{\lfloor}(I-R_{x})^{-1}(I+R_{x})\bigr{\rfloor}$
$\displaystyle=\bigl{\lfloor}-2A-2(I-R_{x})^{-1}(AR_{x}+R_{x}A)(I-R_{x})^{-1}\bigr{\rfloor}$
$\displaystyle=-2\bigl{\lfloor}A\bigr{\rfloor}-2Ce^{-xA}(I-R_{x}^{2})^{-1}e^{-xA}BCe^{-xA}e^{-xA}(I-R_{x}^{2})^{-1}e^{-xA}B$
$\displaystyle=-2\bigl{\lfloor}A\bigr{\rfloor}-2V(x,x)^{2}.$ (4.12)

and likewise

$2W(x,x)=Ce^{-xA}2R_{x}(I-R_{x}^{2})^{-1}e^{-xA}B=\bigl{\lfloor}2R_{x}(I-R_{x})^{-1}(I+R_{x})\bigr{\rfloor}$
(4.13)

so

$\displaystyle 2{\frac{d}{dx}}W(x,x)$ $\displaystyle={\frac{\partial}{\partial
x}}\bigl{\lfloor}2R_{x}(I-R_{x})^{-1}(I+R_{x})\bigr{\rfloor}$
$\displaystyle=\bigl{\lfloor}-4(I-R_{x})^{-1}(AR_{x}+R_{x}A)(I-R_{x})^{-1}\bigr{\rfloor}$
$\displaystyle=-4Ce^{-xA}(I-R_{x}^{2})^{-1}e^{-xA}BCe^{-xA}(I-R_{x}^{2})^{-1}e^{-xA}B$
$\displaystyle=-4V(x,x)^{2}$ (4.14)

We have

$\displaystyle-4V(x,x)^{2}$ $\displaystyle=2{\frac{d}{dx}}W(x,x)$
$\displaystyle={\hbox{trace}}\,\hat{T}(x,z)$
$\displaystyle=-{\hbox{trace}}\,\hat{C}e^{-x\hat{A}}(I+\hat{R}_{x})^{-1}e^{-x\hat{A}}\hat{B}$
$\displaystyle=-{\frac{d}{dx}}{\hbox{trace}}(I+\hat{R}_{x})^{-1}e^{-x\hat{A}}\hat{B}\hat{C}e^{-x\hat{A}}$
$\displaystyle={\frac{d^{2}}{dx^{2}}}{\hbox{trace}}\log(I+\hat{R}_{x})$
$\displaystyle={\frac{d^{2}}{dx^{2}}}\log\det(I+\hat{R}_{x})$
$\displaystyle={\frac{d^{2}}{dx^{2}}}\log\det(I-R_{x}^{2}).$ (4.15)

(iii) Also

$U(x)=\exp 2\int_{x}^{\infty}V(t,t)dt$

satisfies $U^{\prime}(x)/U(x)=-2V(x,x)$, so as in Riccati’s equation

$\displaystyle{\frac{U^{\prime\prime}(x)}{U(x)}}-\Bigl{(}{\frac{U^{\prime}(x)}{U(x)}}\Bigr{)}^{2}$
$\displaystyle=-2{\frac{d}{dx}}V(x,x)$ $\displaystyle=-4\lfloor
A\rfloor-4V(x,x)^{2}$ $\displaystyle=-4\lfloor
A\rfloor-\Bigl{(}{\frac{U^{\prime}(x)}{U(x)}}\Bigr{)}^{2}$ (4.16)

so $U^{\prime\prime}(x)=-4\lfloor A\rfloor U$. ∎

######  Example 4.2.

With $\phi(x)={\hbox{Ai}}(x/2)$, we have

${\frac{d}{dx}}{\mathop{\mathrm{trace}}}\Gamma_{\phi_{x}}={\frac{d}{dx}}\int_{0}^{\infty}{\hbox{Ai}}(x+t)dt=-{\hbox{Ai}}(x),$

so as $x\rightarrow\infty$ we have

$\displaystyle 2V(x,x)$
$\displaystyle={\frac{d}{dx}}{\mathop{\mathrm{trace}}}\Bigl{(}\log(I+\Gamma_{\phi_{x}})-\log(I-\Gamma_{\phi_{x}})\Bigr{)}$
$\displaystyle\approx{\frac{d}{dx}}2{\mathop{\mathrm{trace}}}\Gamma_{\phi_{x}}$
$\displaystyle=-2{\hbox{Ai}}(x).$ (4.17)

This example is considered in detail by Hastings and McLeod [22].

##  5\. Solutions of the sinh-Gordon equation

Howland [24] observed that Hankel matrices are analogous to one-dimensional
Schrödinger operators with the role of the Laplacian played by Carleman’s
operator

$\Gamma f(x)=\int_{0}^{\infty}{\frac{f(y)dy}{x+y}}\qquad(f\in
L^{2}(0,\infty)).$ (5.1)

Power [36] gives several unitarily equivalent forms of this operator. To
obtain such operators in terms of linear systems, we recall Proposition 2.1
from [3]. Let $H=L^{2}((0,\infty);{\mathbb{C}})$ which has dense linear
subspace ${\mathcal{D}}(A)=\\{f\in L^{2}((0,\infty);{{\mathbb{C}}}):yf(y)\in
L^{2}((0,\infty);{\mathbb{C}})$. Suppose that $h\in
L^{\infty}((0,\infty);{\mathbb{R}})$ and $h(y)/\sqrt{y}\in L^{2}(0,\infty)$.
Then we let

$\displaystyle A:{\mathcal{D}}(A)\rightarrow H:\quad f(y)\mapsto yf(y);$
$\displaystyle B_{t}:{\mathbb{C}}\rightarrow H:\quad\beta\mapsto
h(y)\exp\Bigl{(}-{\frac{t}{y}}\Bigr{)}\beta$ $\displaystyle
C_{t}:H\rightarrow{\mathbb{C}}:\quad
f\mapsto\int_{0}^{\infty}\exp\Bigl{(}-{\frac{t}{y}}\Bigr{)}h(y)f(y)\,dy$ (5.2)

where the input and output operators depend upon the real parameter $t>0$.
Then we introduce the scattering function

$\phi(x;t)=C_{t}e^{-xA}B_{t}=\int_{0}^{\infty}\exp\Bigl{(}-xy-{\frac{2t}{y}}\Bigr{)}h(y)^{2}\,dy\qquad(t>0)$
(5.3)

which satisfies the linear PDE

${\frac{\partial^{2}\phi}{\partial x\partial t}}=2\phi,$ (5.4)

which may be interpreted as a linear counterpart of the sinh-Gordon equation.
We also introduce

$R_{(x;t)}=\int_{x}^{\infty}e^{-uA}B_{t}C_{t}e^{-uA}\,du.$

In the following result, the pair $(x,t)$ may be regarded as light-cone co-
ordinates, rather than space and time.

######  Theorem 5.1.

For the linear system (5), let

$S(x;t)=\log\det(I+R_{(x;t)})-\log\det(I-R_{(x;t)}).$ (5.5)

Then $S$ gives a solution of the sinh-Gordon equation

${\frac{\partial^{2}S}{\partial x\,\partial t}}=2\sinh 2S.$ (5.6)

###### Proof.

As an integral operator on $L^{2}(0,\infty)$, $R_{(x;t)}$ has a kernel

${\frac{h(y)h(z)}{y+z}}\exp\Bigl{(}-x(y+z)-t\Bigl{(}{\frac{1}{y}}+{\frac{1}{z}}\Bigr{)}\Bigr{)}\qquad(y,z>0)$
(5.7)

which has the form of a Howland operator. This $R_{(x;t)}$ evidently defines a
Hilbert–Schmidt linear operator on $L^{2}(0,\infty)$. Suppose for simplicity
that $h$ is real-valued. Then $R_{(x;t)}$ is the Schur product of the Carleman
operator $\Gamma$ from (5.1) with kernel $1/(y+z)$ and the tensor product of
$h(y)e^{-xy-t/y}$ with itself. Power [36] showed that $\Gamma$ has spectrum
$[0,\pi]$, hence $R_{(x;t)}$ is self-adjoint and positive with trace

$\mathop{\mathrm{trace}}\,R_{(x;t)}=\int_{0}^{\infty}{\frac{h(y)^{2}}{2y}}e^{{}_{-}2xy-2t/y}\,dy\leq{\frac{1}{2}}\int_{0}^{\infty}\frac{h(y)^{2}}{y}\,dy.$

So $R_{(x;t)}$ is trace class and there exists $x_{0}>0$ such that for all
$x>x_{0}$, we have $\|R_{(x;t)}\|<1$; hence $I\pm R_{(x;t)}$ are invertible.
(Similar results hold for complex $h$ by polarization.)

Note that due to the special form of the linear system

$\displaystyle-{\frac{\partial R_{(x;t)}}{\partial t}}$
$\displaystyle=A^{-1}R_{(x;t)}+R_{(x;t)}A^{-1}$
$\displaystyle=A^{-1}(AR_{(x;t)}+R_{(x;t)}A)A^{-1}$
$\displaystyle=-A^{-1}{\frac{\partial R_{(x;t)}}{\partial x}}A^{-1}$
$\displaystyle=A^{-1}e^{-xA}B_{t}C_{t}e^{-xA}A^{-1}$ (5.8)

gives an operator of rank one so

$\displaystyle{\frac{\partial S}{\partial t}}$
$\displaystyle={\mathop{\mathrm{trace}}}\Bigl{(}(I+R_{(x;t)})^{-1}{\frac{\partial
R_{(x;t)}}{\partial t}}+(I-R_{(x;t)})^{-1}{\frac{\partial R_{(x;t)}}{\partial
t}}\Bigr{)}$
$\displaystyle=C_{t}e^{-xA}A^{-1}(I-R_{(x;t)}^{2})^{-1}A^{-1}e^{-xA}B_{t}.$

We write $2S(x;t)=\int_{x}^{\infty}\psi(s;t)ds$, where $\psi(x;t)=-4V(x,x)$ as
in (4). By (4), we have

$2{\frac{\partial\psi}{\partial x}}=-16\lfloor A\rfloor-\psi^{2}$ (5.9)

so by integrating with respect to $x$ we deduce

$-2\psi(x;t)=8{\frac{\partial}{\partial
x}}\log\det(I+R_{(x;t)})-\int_{x}^{\infty}\psi(y;t)^{2}\,dy,$

which implies when we differentiate with respect to $t$ and multiply by
$\psi$, that

$-2\psi{\frac{\partial\psi}{\partial t}}=8\psi{\frac{\partial^{2}}{\partial
x\partial
t}}\log\det(I+R_{(x;t)})-\psi(x;t)\int_{x}^{\infty}{\frac{\partial}{\partial
t}}\psi(y;t)^{2}\,dy.$ (5.10)

We also have

$\displaystyle{\frac{\partial\psi}{\partial t}}$ $\displaystyle=2\lfloor
A^{-1}(I+R_{(x;t)})\rfloor\lfloor(I+R_{(x;t)})A^{-1}\rfloor$
$\displaystyle\qquad+2\lfloor(I-R_{(x;t)})^{-1}A^{-1}(I+R_{(x;t)})\rfloor\lfloor(I+R_{(x;t)})A^{-1}(I-R_{(x;t)})^{-1}\rfloor$
$\displaystyle\qquad-4\lfloor(I-R_{(x;t)})^{-1}A^{-1}(I+R_{(x;t)})\rfloor-4\lfloor(I+R_{(x;t)})A^{-1}(I-R_{(x;t)})^{-1}\rfloor,$
(5.11)

and from (5),

$\displaystyle{\frac{\partial\lfloor A\rfloor}{\partial t}}$
$\displaystyle=\lfloor
A^{-1}(I+R_{(x;t)})\rfloor\lfloor(I+R_{(x;t)})A^{-1}(I+R_{(x;t)})^{-1}A\rfloor$
$\displaystyle\qquad+\lfloor
A(I+R_{(x;t)})^{-1}A^{-1}(I+R_{(x;t)})\rfloor\lfloor(I+R_{(x;t)})A^{-1}\rfloor$
$\displaystyle\qquad-\lfloor(I+R_{(x;t)})A^{-1}(I+R_{(x;t)})^{-1}A\rfloor-\lfloor
A(I+R_{(x;t)})^{-1}A^{-1}(I+R_{(x;t)})\rfloor.$ (5.12)

Also, from (5.9) we have

${\frac{\partial^{2}\psi}{\partial t\partial x}}=-8{\frac{\partial\lfloor
A\rfloor}{\partial t}}-\psi{\frac{\partial\psi}{\partial t}}.$ (5.13)

We need to combine the equations (5.10) and (5.13) and thereby obtain the
sinh-Gordon equation. The expression (5) involves factors $I-R$; whereas (5)
involves only $I+R$; we reconcile these by multiplying by
$\psi=4\lfloor(I+R)/(I-R)\rfloor$, which cancels out the factors of $I-R$.
From (5), we have

$\displaystyle{\frac{\partial^{2}}{\partial x\partial
t}}\log\det(I+R_{(x;t)})$ $\displaystyle=-\lfloor
A^{-1}(I+R_{(x;t)})\rfloor\lfloor(I+R_{(x;t)})A^{-1}\rfloor+\lfloor
A^{-1}(I+R_{(x;t)})\rfloor$ $\displaystyle+\lfloor(I+R_{(x;t)})A^{-1}\rfloor;$
(5.14)

hence from the $\ast$ product (2.6) we obtain

$\displaystyle\psi$ $\displaystyle{\frac{\partial^{2}}{\partial x\partial
x}}\log\det(I+R_{(x;t)})$
$\displaystyle=-4\bigl{\lfloor}(I+R_{(x;t)})A^{-1}\bigr{\rfloor}\bigl{\lfloor}(I+R_{(x;t)})(I-R_{(x;t)})^{-1}\bigr{\rfloor}\bigl{\lfloor}A^{-1}(I+R_{(x;t)})\rfloor$
$\displaystyle\qquad+4\bigl{\lfloor}(I+R_{(x;t)})A^{-1}\bigr{\rfloor}\bigl{\lfloor}(I+R_{(x;t)})(I-R_{(x;t)})^{-1}\bigr{\rfloor}$
$\displaystyle\qquad\qquad+4\bigl{\lfloor}(I+R_{(x;t)})(I-R_{(x;t)})^{-1}\bigr{\rfloor}\bigl{\lfloor}A^{-1}(I+R_{(x;t)})A^{-1}\bigr{\rfloor}$
$\displaystyle=-4\bigl{\lfloor}(I+R_{(x;t)})(I-R_{(x;t)})^{-1}-(I+R_{(x;t)})A^{-1}(I+R_{(x;t)})^{-1}A\bigr{\rfloor}\bigl{\lfloor}A^{-1}(I+R_{(x;t)})\bigr{\rfloor}$
$\displaystyle\qquad+\psi-4\bigl{\lfloor}(I+R_{(x;t)})A^{-1}(I+R_{(x;t)})^{-1}A\bigr{\rfloor}+\psi-4\bigl{\lfloor}A(I+R_{(x;t)})^{-1}A^{-1}(I+R_{(x;t)})\bigr{\rfloor}$
(5.15)

hence we obtain

$\displaystyle\psi$ $\displaystyle{\frac{\partial^{2}}{\partial x\partial
x}}\log\det(I+R_{(x;t)})$
$\displaystyle=4\bigl{\lfloor}(I+R_{(x;t)})A^{-1}(I+R_{(x;t)})^{-1}A\bigr{\rfloor}\bigl{\lfloor}A^{-1}(I+R_{(x;t)})\bigr{\rfloor}$
$\displaystyle\quad-4\bigl{\lfloor}(I+R_{(x;t)})(I-R_{(x;t)})^{-1}\bigr{\rfloor}\bigl{\lfloor}A^{-1}(I+R_{(x;t)})\bigr{\rfloor}$
$\displaystyle\qquad+\psi-4\bigl{\lfloor}(I+R_{(x;t)})A^{-1}(I+R_{(x;t)})^{-1}A\bigr{\rfloor}+\psi-4\bigl{\lfloor}A(I+R_{(x;t)})^{-1}A^{-1}(I+R_{(x;t)})\bigr{\rfloor}$
$\displaystyle=4\bigl{\lfloor}(I+R_{(x;t)})A^{-1}(I+R_{(x;t)})^{-1}A\bigr{\rfloor}\bigl{\lfloor}A^{-1}(I+R_{(x;t)})\bigr{\rfloor}$
$\displaystyle\qquad+4\bigl{\lfloor}A(I+R_{(x;t)})^{-1}A^{-1}(I+R_{(x;t)})\bigr{\rfloor}-\psi$
$\displaystyle\qquad+\psi-4\bigl{\lfloor}(I+R_{(x;t)})A^{-1}(I+R_{(x;t)})^{-1}A\bigr{\rfloor}+\psi-4\bigl{\lfloor}A(I+R_{(x;t)})^{-1}A^{-1}(I+R_{(x;t)})\bigr{\rfloor}$
$\displaystyle=\psi+4\bigl{\lfloor}(I+R_{(x;t)})A^{-1}(I+R_{(x;t)})^{-1}A\bigr{\rfloor}\bigl{\lfloor}A^{-1}(I+R_{(x;t)})\bigr{\rfloor}$
$\displaystyle\qquad-4\bigl{\lfloor}(I+R_{(x;t)})A^{-1}(I+R_{(x;t)})^{-1}A\bigr{\rfloor}$
$\displaystyle=\psi+2{\frac{\partial}{\partial
t}\bigl{\lfloor}A\bigr{\rfloor}}.$ (5.16)

Hence (5.10) gives

$-2\psi{\frac{\partial\psi}{\partial t}}=16{\frac{\partial}{\partial
t}}\bigl{\lfloor}A\bigr{\rfloor}+8\psi-\psi(x;t)\int_{x}^{\infty}{\frac{\partial\psi^{2}}{\partial
t}}\,dy$ (5.17)

so (5.13) becomes

$2{\frac{\partial^{2}\psi}{\partial x\partial
t}}=8\psi-2\psi(x;t)\int_{x}^{\infty}\psi(y;t){\frac{\partial\psi(y;t)}{\partial
t}}\,dy,$

so

$-{\frac{\partial\psi}{\partial
t}}+\int_{x}^{\infty}\psi(y;t)\int_{y}^{\infty}\psi(v;t){\frac{\partial\psi(v;t)}{\partial
t}}\,dvdy=4\int_{x}^{\infty}\psi(y;t)\,dy,$ (5.18)

so we need to compare the integral in (5.18) with the $\sinh 2S$ term in
(5.6).

We introduce the integration operator $F_{\psi}:L^{2}(0,\infty)\rightarrow
L^{2}(0,\infty)$, which depends upon $t>0$ by

$F_{\psi}f(x;t)=\int_{x}^{\infty}\psi(s;t)f(s)\,ds.$

Choosing $f=1$, we obtain by induction the formula

$F_{\psi}^{n}\circ
1(x;t)={\frac{1}{n!}}\Bigl{(}\int_{x}^{\infty}\psi(s;t)ds\Bigr{)}^{n}={\frac{1}{n!}}(F_{\psi}\circ
1)^{n}(x;t).$

Now we express the right-hand side of (5.6) as a composition of operators

$\displaystyle\sinh 2S$
$\displaystyle=\sum_{n=0}^{\infty}{\frac{2^{2n+1}S^{2n+1}}{(2n+1)!}}=\sum_{n=0}^{\infty}{\frac{(F_{\psi}\circ
1)^{2n+1}}{(2n+1)!}}$ $\displaystyle=\sum_{n=0}^{\infty}F_{\psi}^{2n+1}\circ
1(x;t)=(I-F_{\psi}^{2})^{-1}F_{\psi}\circ 1(x;t).$

Hence the equation (5.18) gives

$-{\frac{\partial\psi}{\partial t}}=4(I-F_{\psi}^{2})^{-1}F_{\psi}\circ
1(x;t)$

which gives

$2{\frac{\partial^{2}S}{\partial t\partial x}}=4\sinh 2S.$

which is the sinh-Gordon equation. ∎

######  Corollary 5.2.

Let $\phi_{(x;t)}(y)=\phi(y+2x;t)$, with $\phi$ as in Theorem 5.1 for (i) and
(ii).

  1. (i)

Then

$S(x;t)=\log\det(I+\Gamma_{\phi_{(x;t)}})-\log\det(I-\Gamma_{\phi_{(x;t)}})$
(5.19)

satisfies the sinh-Gordon equation (5.6).

  2. (ii)

The function $\varphi(z;t)=\log\det(I-\Gamma_{\phi_{(z;t)}}^{2})$ is
holomorphic on $\\{z:\mathop{{\rm Re}}z>x_{0}\\}$ for some $x_{0}>0$ and
$\varphi(z;t)\rightarrow 0$ as $\mathop{{\rm Re}}z\rightarrow\infty$.

  3. (iii)

Now let

$\phi(x;s/2)=\int_{0}^{\infty}e^{-xy-s/y}dy\qquad(\mathop{{\rm Re}}x>0,s>0).$

Then the conclusion of (ii) holds for the associated Hankel operator
$\Gamma_{\phi_{(z;t)}}$.

###### Proof.

(i) We have

$\det(I+\lambda\Gamma_{\phi_{(x;t)}})=\det(I+\lambda\Theta_{x}^{\dagger}\Xi_{x})=\det(I+\lambda\Xi_{x}\Theta_{x}^{\dagger})=\det(I+\lambda
R_{(x;t)})\qquad(\lambda\in{\mathbb{C}}),$

so the formulas (5.5) and (5.19) for $S(x;t)$ are equivalent, and the result
holds by Theorem 5.1. See also [42].

(ii) By Cauchy-Schwarz inequality

$|\phi_{(z;t)}(x)|^{2}\leq\|h\|^{2}_{\infty}\int_{0}^{\infty}e^{-xy-4t/y}|h(y)|^{2}dy\int_{0}^{\infty}e^{-(x+4\mathop{{\rm
Re}}z)y}dy;$

hence the Hilbert–Schmidt norm satisfies

$\displaystyle\|\Gamma_{\phi_{(z;t)}}\|^{2}_{{\mathcal{L}}^{2}}$
$\displaystyle=\int_{0}^{\infty}x|\phi_{(z;t)}(x)|^{2}dx$
$\displaystyle\leq\|h\|^{2}_{\infty}\int_{0}^{\infty}\Bigl{\\{}\int_{0}^{\infty}{\frac{xy}{x+4\mathop{{\rm
Re}}z}}e^{-xy}dx\Bigr{\\}}{\frac{e^{-4t/y}}{y}}|h(y)|^{2}dy$ (5.20)

where the inner integral is bounded and converges to zero as $\mathop{{\rm
Re}}z\rightarrow\infty$, so
$\|\Gamma_{\phi_{(z;t)}}\|^{2}_{{\mathcal{L}}^{2}}\rightarrow 0$ by the
dominated convergence theorem, hence the result.

(iii) The conclusion of Corollary 5.2(ii) holds, but the hypotheses are not
quite satisfied, so we provide a special argument. Let

$K_{1}(z)=\int_{0}^{\infty}\exp(-z\cosh u)\cosh u\,du\qquad(\mathop{{\rm
Re}}z>0)$ (5.21)

be the modified Bessel function of the third kind of order $1$, also known as
MacDonald’s function of the second kind, which is holomorphic on
$\\{z\>\mathop{{\rm Re}}z>0\\}$ with $|K_{1}(z)|\leq K_{1}(\mathop{{\rm
Re}}z)$. This also satisfies

${\frac{e^{-t}}{t}}\leq
K_{1}(t)\leq\sqrt{{\frac{\pi}{2t}}}e^{-t}+{\frac{e^{-t}}{t}}\qquad(t>0)$

by simple estimates following from $\cosh u=1+2\sinh^{2}(u/2)$. Then for
$\alpha=0$, we have

$\displaystyle\phi(x;s/2)$
$\displaystyle=\sqrt{{\frac{4s}{x}}}K_{1}(2\sqrt{sx})$ (5.22)

so $\int_{0}^{\infty}x|\phi(x+2x_{0};s/2)|^{2}dx$ is finite for all
$x_{0},s>0$ and we can deduce
$\|\Gamma_{\phi_{(x_{0})}}\|_{{\mathcal{L}}^{2}}\rightarrow 0$ as
$x_{0}\rightarrow\infty$, as in Corollary 5.2(ii).

∎

##  6\. Hankel determinants

The Wishart ensemble is a standard model in random matrix theory [16, page 91]
which produces the Laguerre ensemble of random eigenvalues on $(0,\infty)$.
The Laguerre ensemble is in turn used as a model in the theory of wireless
transmission, and in an integral model of quantum field theory at finite
temperature [8]. The $KP$, $KdV$ and sinh-Gordon differential equation can be
interpreted as aspects of a single hierarchy.

One can produce particular solutions to such differential equations in terms
of the Painlevé transcendental functions. Chen and Its [8] considered the
singularly perturbed Laguerre weight $y^{\alpha}e^{-y-s/y}$ for $y>0$ and
$\alpha>0$. Let $h(y)=y^{\alpha/2}$, and observe that

$\phi(x;s/2)=\int_{0}^{\infty}y^{\alpha}e^{-xy-s/y}\,dy$ (6.1)

is a moment generating function for moments of this weight, so we have

$(-1)^{j+k}\Bigl{(}{\frac{\partial^{j+k}}{\partial
x^{j+k}}}\Bigr{)}_{x=1}\phi(x;s/2)=\int_{0}^{\infty}y^{\alpha+j+k}e^{-y-s/y}\,dy.$

Then the corresponding Hankel determinants are given by

$D_{n}(s)=\det\Bigl{[}\int_{0}^{\infty}y^{\alpha+j+k}e^{-y-s/y}\,dy\Bigr{]}_{j,k=0}^{n-1}\qquad(s>0;n=1,2,\dots),$
(6.2)

which turns out to be the isomonodromic tau function for a particular sequence
of solutions of Painlevé $\text{III}^{\prime}$ differential equation. This
terminology refers to the work of Okamoto.

######  Example 6.1.

Let $s=0$ and $\alpha>0$, let $G(z)$ be Barnes $G$-function such that $G(0)=1$
and $G(z+1)=\Gamma(z)G(z)$, where $\Gamma(z)$ is Euler’s gamma function. Then
by (1.19) of [8]

$D_{n}(0)={\frac{G(n+1)G(n+\alpha+1)}{G(\alpha+1)}}.$

Let $\alpha=0$. By Andréief’s identity [15], we have

$\det\bigl{[}(-1)^{j+k-2}K_{1}^{(j+k-2)}(t)\bigr{]}_{j,k=1}^{n}\\\
={\frac{1}{n!}}\int_{0}^{\infty}\dots\int_{0}^{\infty}\prod_{1\leq j<k\leq
n}(\cosh u_{j}-\cosh u_{k})^{2}\prod_{j=1}^{n}e^{-t\cosh u_{j}}\cosh
u_{j}\,du_{j}$ (6.3)

with $K_{1}$ as in (5.21). With the change of variable $x_{j}=2/(1+\cosh
u_{j})$, this may be written as

${\frac{1}{n!}}\int_{0}^{\infty}\dots\int_{0}^{\infty}\exp\Bigl{(}-t\Bigl{(}{\frac{2}{x_{j}}}-1\Bigr{)}\Bigr{)}\prod_{j=1}^{n}{\frac{x_{j}^{-2n}(2-x_{j})}{\sqrt{1-x_{j}}}}\prod_{1\leq
j<k\leq n}(x_{j}-x_{k})^{2}\prod_{j=1}^{n}dx_{j}.$

This is associated with the generalized unitary ensemble for the scalar
potential

$u_{n}(x)=t\Bigl{(}{\frac{2}{x}}-1\Bigr{)}+2n\log x+(1/2)\log(1-x)-\log(2-x),$
(6.4)

with

$u_{n}^{\prime}(x)={\frac{-2t}{x^{2}}}+{\frac{2n}{x}}-{\frac{1/2}{1-x}}+{\frac{1}{2-x}},$
(6.5)

and

$u_{n}^{\prime\prime}(x)={\frac{4t}{x^{3}}}+{\frac{-2n}{x^{2}}}-{\frac{1/2}{(1-x)^{2}}}+{\frac{1}{(2-x)^{2}}},$
(6.6)

Dyson introduced a technique for funding the asymptotics of such determinants,
which has been refined by Chen [10, page 4603] and others into the Coulomb
fluid method. In this, the point distribution
$n^{-1}\sum_{j=1}^{n}\delta_{x_{j}}$ of the $x_{j}$ is approximated in the
weak topology on probability measures as $n\rightarrow\infty$ by a continuous
distribution $\rho$ which is found by potential theory. We write

$\mathop{\mathcal{H}}\rho(x)=\mathop{\mathrm{p.v.}}\int_{-\infty}^{\infty}{\frac{\rho(y)}{x-y}}{\frac{dy}{\pi}}$

for the Hilbert transform of $\rho$.

Fix $\xi\in(0,1)$ and write $t$ for $2n\xi$. The zeros of $u_{n}^{\prime}(x)$
may be approximated by the zeros of $h(x)=(2n-1/2)(x-1)(x-2)(x-t/n)$ by
Rouché’s theorem, so we locate a real zero of $u_{n}^{\prime}(x)$ near to
$2\xi$. Note that

${\frac{u_{n}^{\prime\prime}(2\xi)}{n}}={\frac{1}{2\xi^{2}}}-{\frac{1}{2n(1-2\xi)^{2}}}+{\frac{1}{n(2-\xi)^{2}}}$

is positive for all $0<\xi<1/2$ and all sufficiently large $n$, so $u_{n}$ is
convex near to this minimizer. For a continuous
$V:[0,1]\rightarrow{\mathbb{R}}$, we introduce the energy functional

$E_{V}(\rho)=\Bigl{\\{}\int_{0}^{1}V(x)\rho(x)\,dx+\int_{0}^{1}\int_{0}^{1}\log{\frac{1}{|x-y|}}\rho(x)\rho(y)\,dxdy\Bigr{\\}},$
(6.7)

where $V$ indicates an electric field and the $\log|x-y|$ term involves
electrostatic interaction between points $x,y\in[0,1]$, where the charge
distribution is $\rho$. The equilibrium distribution is defined to be the
minimizer of this energy functional; see [38].

We consider the approximate scaled potential that is given by the first two
summands in (6.4), namely

$u_{0}(x)=2\xi\Bigl{(}{\frac{2}{x}}-1\Bigr{)}+2\log x,$ (6.8)

with

$u_{0}^{\prime}(x)={\frac{-2\xi}{x^{2}}}+{\frac{2}{x}};\quad
u_{0}^{\prime\prime}(x)={\frac{8\xi}{x^{3}}}-{\frac{2}{x^{2}}}\qquad(0<x<1)$

which has a local minimum at $x=2\xi$.

######  Proposition 6.2.

  1. (i)

The equilibrium distribution for $u_{0}$ is $\sigma_{0}$, where

$\sigma_{0}(x)={\frac{\sqrt{(b-x)(x-a)}}{\pi\sqrt{ab}}}\Bigl{(}{\frac{\xi(a+b)}{abx}}+{\frac{2\xi}{x^{2}}}-{\frac{1}{x}}\Bigr{)}\qquad(a<x<b),$

is supported on $(a,b)$, where

$a,b={\frac{4\pi\xi}{(2\pi-1)^{2}}}\bigl{(}2\pi\pm\sqrt{4\pi-1}\bigr{)}.$

  2. (ii)

For $1/4<\xi<1/2$, the free logarithmic Sobolev inequality

$E_{u_{0}}(p)-E_{u_{0}}(\sigma_{0})\leq{\frac{2}{8\xi-2}}\int_{a}^{b}(2\pi\mathop{\mathcal{H}}p(x)-u_{0}^{\prime}(x))^{2}p(x)\,dx.$
(6.9)

holds for all probability density functions $p\in L^{3}(a,b)$.

  3. (iii)

Let $\rho_{n}$ be the minimizer of $E_{u_{n}/n}(\rho)$ over all continuous
$\rho$ such that $\rho\geq 0$ and $\int_{0}^{1}\rho(x)dx=1$. Suppose that
$\rho_{n}$ has a continuous density which is supported in a single interval.
Then $\rho_{n}$ converges weakly as $n\rightarrow\infty$ to $\sigma_{0}$.

###### Proof.

(i) By standard results [38] Theorem 1.3 and p. 215, there exists a unique
continuous probability density function on $[0,1]$ that attains the minimum of
$E_{V}(\rho)$ for continuous $V$. We aim to solve the singular integral
problem

$u_{0}^{\prime}(x)=\mathop{\mathrm{p.v.}}\int_{a}^{b}{\frac{2\sigma_{0}(y)}{x-y}}\,dy=2\pi\mathop{\mathcal{H}}\sigma_{0}(x)$

for the equilibrium density $\sigma_{0}$ subject to the constraint
$\int_{a}^{b}\sigma_{0}(y)\,dy=1$. Since $u_{0}$ is convex, the solution is
continuous and positive on a single interval $(a,b)$, where $0<a<b<1$, and we
have

$\displaystyle\sigma_{0}(x)$
$\displaystyle={\frac{\sqrt{(b-x)(x-a)}}{2\pi^{2}}}\int_{a}^{b}{\frac{u_{0}^{\prime}(x)-u_{0}^{\prime}(y)}{x-y}}{\frac{dy}{\sqrt{(b-y)(y-a)}}}$
$\displaystyle={\frac{\sqrt{(b-x)(x-a)}}{2\pi^{2}}}\int_{a}^{b}\Bigl{(}{\frac{4\xi(x+y)}{x^{2}y^{2}}}-{\frac{2}{xy}}\Bigr{)}{\frac{dy}{\sqrt{(b-y)(y-a)}}}$
$\displaystyle={\frac{\sqrt{(b-x)(x-a)}}{\pi\sqrt{ab}}}\Bigl{(}{\frac{\xi(a+b)}{abx}}+{\frac{2\xi}{x^{2}}}-{\frac{1}{x}}\Bigr{)}\qquad(a<x<b),$
(6.10)

where the final step follows from the substitution $y=a+(b-a)\sin^{2}\phi$ and
some elementary integrals. The initial factor has the form of a semicircular
distribution, which is modulated by a rational function with poles at $x=0$
outside of the support interval $(a,b)$. The endpoints of this interval are
subject to the constraints

$0=\int_{a}^{b}{\frac{u_{0}^{\prime}(x)\,dx}{\sqrt{(b-x)(x-a)}}}=-{\frac{2\pi\xi(a+b)}{ab\sqrt{ab}}}+{\frac{2\pi}{\sqrt{ab}}},$

and

$1=\int_{a}^{b}{\frac{xu_{0}^{\prime}(x)\,dx}{\sqrt{(b-x)(x-a)}}}=2\pi-{\frac{4\pi\xi}{\sqrt{ab}}},$

so that

$a,b={\frac{2\pi^{2}\xi}{(\pi-1/2)^{2}}}\mp{\frac{2\pi\xi}{\pi-1/2}}\sqrt{{\frac{\pi^{2}}{(\pi-1/2)^{2}}}-1}.$
(6.11)

(ii) We have $u_{0}^{\prime\prime}(x)=(8x\xi-2x)/x^{3}>8\xi-2>0$. The result
follows from Theorem 3.1 of [1].

(iii) Returning to the original potential, we have $u_{n}=nu_{0}+f$ where the
correction term $f(x)=(1/2)\log(1-x)-\log(2-x)$. There is a corresponding
correction to the equilibrium density, $\rho_{n}=\sigma_{0}+\tilde{\rho}/n$
where

$\displaystyle{\tilde{\rho}}(x)$
$\displaystyle={\frac{1}{2\pi^{2}\sqrt{(b-x)(x-a)}}}\mathop{\mathrm{p.v.}}\int_{a}^{b}{\frac{\sqrt{(b-y)(y-a)}}{y-x}}f^{\prime}(y)\,dy$
$\displaystyle={\frac{1}{2\pi^{2}\sqrt{(b-x)(x-a)}}}\
\mathop{\mathrm{p.v.}}\int_{a}^{b}{\frac{\sqrt{b-y)(y-a)}}{y-x}}\Bigl{(}{\frac{-1/2}{1-y}}+{\frac{1}{2-y}}\Bigr{)}\,dy$
$\displaystyle={\frac{1}{4\pi\sqrt{(b-x)(x-a)}}}\Bigl{(}{\frac{\sqrt{(1-a)(1-b)}}{1-x}}+1+{\frac{2\sqrt{(2-a)(2-b)}}{x-2}}\Bigr{)}$

where all the roots are positive square roots and we have used the evaluation
(258) in the Appendix to [10]. This has the form of an arcsine distribution,
modulated by a rational function with poles outside $(a,b)$. ∎

Biane [1] interprets the tangent space to $L^{2}(\sigma_{0})$ as a space of
functions with finite Dirichlet norm, which has a natural interpretation in
the present context. Let $\psi:(a,b)\rightarrow{\mathbb{R}}$ be a
differentiable function such that

$\sigma(\psi)^{2}={\frac{1}{2\pi^{2}}}\int_{a}^{b}{\frac{\psi(x)}{\sqrt{(b-x)(x-a)}}}\
\mathop{\mathrm{p.v.}}\int_{a}^{b}{\frac{\sqrt{(b-y)(y-x)}}{x-y}}\psi^{\prime}(y)\,dydx,$
(6.12)

is finite. Then by results of [9], the asymptotic distribution of a linear
statistic $\sum_{j=1}^{N}\psi(x_{j})$ is a Gaussian with mean
$N\int_{a}^{b}\psi(x)\sigma_{0}(x)\,dx$ and variance $\sigma(\psi)^{2}$ where
$\sigma(\psi)$ depends upon $[a,b]$ but not upon the equilibrium distribution
$\sigma_{0}$ itself. In terms of the Chebyshev polynomials, we introduce
$\psi((a+b)/2(+(b-a)t/2)=\sum_{k}a_{k}T_{k}(t)$ where

${\frac{a_{k}}{2}}=\int_{-1}^{1}\psi((a+b)/2+(b-a)t/2)T_{k}(t){\frac{dt}{\pi\sqrt{1-t^{2}}}}$

so the variance is $\sigma(\psi)^{2}=\sum_{k=1}^{\infty}ka_{k}^{2}/4.$ Then we
introduce $\phi(\theta)=\psi((a+b)/2+(b-a)\cos\theta/2)$ has an expansion

$\phi(\theta)=(1/2)a_{0}+\sum_{k=1}^{\infty}a_{k}\cos k\theta,$

which gives a self-adjoint and Hilbert–Schmidt Hankel matrix
$\Gamma_{\phi}=[a_{j+k}]_{j,k=0}^{\infty}$.

######  Remark 6.3.

In their study of the Laguerre unitary ensemble, Forrester and Witte [17]
obtain solutions of $\mathrm{PIII}^{\prime}$ from similar formulas involving
the modified Bessel function of the first kind $I_{\nu}$, although their
Corollary 4.5 has a Toeplitz rather than a Hankel determinant.

######  Example 6.4.

For $s,x,\alpha>0$ in (6.1), consider

$v(z)=-\alpha\log z+\sqrt{sx}\Bigl{(}z+{\frac{1}{z}}\Bigr{)}\qquad(z>0).$

By a simple scaling, this is the scalar potential that one needs to study the
determinant (6.2) with weight $y^{\alpha}e^{-y-s/y}$ on $(0,\infty)$. Then $v$
is convex with

${\frac{v^{\prime}(z)-v^{\prime}(y)}{z-y}}={\frac{\alpha}{z}}+\sqrt{sx}\Bigl{(}{\frac{z+y}{z^{2}y^{2}}}\Bigr{)}$
(6.13)

a sum of products of rational functions of $z$ and $y$, and there exist
$b>a>0$, such that the integral equation

$v(z)=2\int_{a}^{b}\log|z-y|\rho(y)dy+C\qquad(z\in(a,b))$ (6.14)

for some $C\in{\mathbb{R}}$ with the normalization $\rho(y)\geq 0$ and

$\int_{a}^{b}\rho(y)\,dy=1,$

has solution

$\rho(z)={\frac{\sqrt{(b-z)(z-a)}}{2\pi\sqrt{ab}}}\Bigl{[}{\frac{\alpha}{z}}+{\frac{\sqrt{sx}}{z^{2}}}+{\frac{\sqrt{sx}}{2z}}\Bigl{(}{\frac{1}{a}}+{\frac{1}{b}}\Bigr{)}\Bigr{]}\qquad(a<z<b)$

by a similar argument to (6). We need $a>0$ to ensure that the resulting
$\rho$ is integrable. The relevant integrals arise from (248) in the appendix
to [10]. Let $(P_{j,N})_{j=0}^{\infty}$ be the monic polynomials that are
orthogonal for the weight $e^{-Nv(z)}$, which gives rise to the integral
equation (6.14) multiplied through by $N$, with the normalizations preserved.
Then using the results of [9], we observe that $N\int_{a}^{b}\rho(x)\,dx$
gives the number of zeros of $P_{N,N}(z)$ and have an asymptotic formula

$\log P_{N,N}(z)\sim
N\int_{a}^{b}\log(z-y)\rho(y)dy\qquad(N\rightarrow\infty)$

for $z\in{\mathbb{C}}\setminus{\mathbb{R}}.$ This type of double scaling is
standard in random matrix theory and addresses the singularity of the weight;
see section 4 of [9] for more details.

Acknowledgements. Part of this document was written by GB during a visit to
the University of Macau, supported by a visiting scholarship. The results were
announced at IWOTA Chapman USA, and were partially supported by EPSRC Grant
EP/T007524/1 IWOTA Lancaster 2020. The authors acknowledge helpful
correspondence with S.J.A. Malham leading to the proof of Theorem 5.1.

Rights Retention Statement. This research was produced in whole or part by
UNSW Sydney researchers and is subject to the UNSW Intellectual property
policy. For the purposes of Open Access, the authors have applied a Creative
Commons Attribution CC-BY licence to any Author Accepted Manuscript(AAM)
version arising from this submission. Declarations of interest:none

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