Classical-Quantum Correspondence and Functional Relations for Painlevé Equations

Abstract

In light of the quantum Painlevé–Calogero correspondence, we investigate the inverse problem. We imply that this type of the correspondence (classical-quantum correspondence) holds true, and we find out what kind of potentials arise from the compatibility conditions of the related linear problems. The latter conditions are written as functional equations for the potentials depending on a choice of a single function—the left-upper element of the Lax connection. The conditions of the correspondence impose restrictions on this function. In particular, it satisfies the heat equation. It is shown that all natural choices of this function (rational, hyperbolic, and elliptic) reproduce exactly the Painlevé list of equations. In this sense, the classical-quantum correspondence can be regarded as an alternative definition of the Painlevé equations.

1 Introduction

The Painlevé equations (PI–PVI) discovered by P. Painlevé, R. Fuchs and B. Gambier [13, 15, 43] have been studied extensively during the last century [7, 20]. Their applications include self-similar reductions of nonlinear integrable partial differential equations [11], correlation functions of integrable models [3, 23], quantum gravity and string theory [5], topological field theories [8], 2D polymers [59], random matrices [12, 51] and stochastic growth processes [30], conformal field theories and KZ equations [14, 24], the AGT conjecture [10, 37] and spectral duality [38–40], to mention only a few applications and references.

As is known from classical works [13, 16, 44], the Painlevé equations describe the monodromy preserving deformations of a system of linear differential equations with rational coefficients. The monodromy approach was developed by H. Flaschka and A. Newell and by M. Jimbo, T. Miwa, K. Ueno [11, 21, 22, 25], see also [19]. At present, different types of linear problems are known (scalar [13, 16], 2$\times$2-matrix [21] (see also [32, 60]), 3$\times$3-matrix [9, 26], 8$\times$8-matrix [41], and higher-dimensional Lax pairs [35]).

We deal with the linear problems depending on a spectral parameter [7–43]:

$$\begin{aligned} \left\{ \begin{array}{l} \partial _{x}\varvec{\Psi } = \mathbf{U} (x,t)\varvec{\Psi },\\ \partial _{\,t}\varvec{\Psi } =\mathbf{V} (x,t)\varvec{\Psi }, \end{array}\right. \, \quad \quad \varvec{\Psi } =\left( \begin{array}{l} \psi _1 \\ \psi _{2} \end{array}\right) , \end{aligned}$$

(1.1)

where $\mathbf{U},\mathbf{V}\in {sl}_2$ explicitly depend on the spectral parameter $x$ and on the deformation parameter $t$ (time variable) and contain an unknown function $u(t)$ to be constrained by the condition that the two equations have a family of common solutions.¹ In fact, the latter is equivalent to the compatibility of the linear problems expressed as the zero curvature equation (integrability condition):

$$\begin{aligned} \partial _x \mathbf{V} -\partial _t \mathbf{U} +[\mathbf{V}, \mathbf{U}]=0. \end{aligned}$$

(1.2)

Set

$$\begin{aligned} \mathbf{U}=\left( \begin{array}{cc} a &{} b\\ c &{} d \end{array}\right) , \quad \quad \mathbf{V}=\left( \begin{array}{cc} A &{} B\\ C &{} D \end{array}\right) . \end{aligned}$$

The matrices $\mathbf{U},\mathbf{V}$ are traceless, i.e., $a+d=0$, $A+D=0$. Then the zero curvature equation gives:

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle {a_t -A_x +b C - c B =0},\\ \displaystyle {b_t -B_x +2aB -2bA=0},\\ \displaystyle {c_t -C_x +2cA -2a C=0.} \end{array} \right. \end{aligned}$$

(1.3)

In [57, 58], by applying the diagonal gauge transformation $\Omega =\hbox {diag}(\omega ,\omega ^{-1})$, we chose the matrices $\mathbf{U}$, $\mathbf{V}$ such that

$$\begin{aligned} b_x =2B. \end{aligned}$$

(1.4)

Then the linear system (1.1) for the vector function $\varvec{\Psi }=(\psi _1 , \psi _2)^\mathsf{t}$ can be reduced to two scalar equations for $\psi :=\psi _1$:

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle { \left( \frac{1}{2}\, \partial _{x}^{2} -\frac{1}{2}\, (\partial _x \log b) \, \partial _x \, + W (x,t)\right) \psi =0},\\ \displaystyle { \partial _t \psi =\left( \frac{1}{2}\, \partial _x^2 + U(x,t)\right) \psi ,} \end{array} \right. \end{aligned}$$

(1.5)

where

$$\begin{aligned} W=U(x,t)-\frac{1}{2}\, \partial _t \log b + \frac{1}{4}\, \partial _{x}^2\log b+ \frac{1}{4}(\partial _x \log b)^2 \end{aligned}$$

and

$$\begin{aligned} U(x,t)=\frac{1}{2}\,(ad-bc -a_x) +A \, =\, \frac{1}{2}\, \det \mathbf{U}-\frac{a_x}{2}+A. \end{aligned}$$

(1.6)

The second equation in (1.5) has the form of a nonstationary Schrödinger equation in imaginary time with the potential $U(x,t)$. It describes the isomonodromic deformations of the first one, and their compatibility implies the Painlevé equation (in the Calogero form) for the function $u=u(t)$:

$$\begin{aligned} {\ddot{u}}=-\partial _u {\tilde{V}}(u,t), \end{aligned}$$

(1.7)

generated by the Hamiltonian $H({\dot{u}},u,t)=\frac{1}{2}{\dot{u}}^2+{\tilde{V}}(u,t)$. The function $u=u(t)$ is defined as a (simple) zero of the function $b(x)$:

$$\begin{aligned} b(u)=0. \end{aligned}$$

The second important condition we are going to use together with (1.4) is

$$\begin{aligned} U(x,t)=U(x,{\dot{u}}(t),u(t),t)=V(x,t)-H({\dot{u}},u,t), \end{aligned}$$

(1.8)

where $H({\dot{u}},u,t)$ is the classical Hamiltonian. The $x$-dependent part of the potential $V(x,t)$ does not contain the dependent variable $u$. Therefore, the second equation in (1.5) acquires the form

$$\begin{aligned} \partial _t \Psi (x,t) =\left( \frac{1}{2}\, \partial _x^2 +V(x,t)\right) \Psi (x,t), \end{aligned}$$

(1.9)

with

$$\begin{aligned} \Psi (x,t)=e^{\int ^t H (\dot{u} , u,t')dt'}\psi (x,t). \end{aligned}$$

(1.10)

Notice that condition (1.4) can be easily satisfied by choosing a suitable gauge. However, together with (1.8), it becomes a nontrivial condition and leads to the quantum Painlevé–Calogero correspondence (see below), which relates the potentials of the classical problem $\tilde{V}$ with $V$ in the quantum one. It appears that the potentials differ only by “quantum corrections” of the coupling constants. Therefore, (1.9) is the quantization of (1.7) with the unit Planck constant.

In [57, 58], it was shown that there exists a choice of gauge and variables $(x,t)$ such that the nonstationary Schrödinger equation becomes a quantized Painlevé equation. Thus, the linear problem (1.1) leads to both classical and quantum Painlevé equations. The classical one is written in the variable $u(t)$ and follows from the zero-curvature equation (1.1) valid for all $x$. The quantum one is written in terms of the spectral parameter $x$ for a component of the common solution $\psi _1$ of the linear problems. We have called this construction the quantum Painlevé–Calogero correspondence. It is a quantum version of the classical correspondence introduced by A. Levin and M. Olshanetsky [31] and developed by K. Takasaki [49]. It should be mentioned that a phenomenon similar to the quantum Painlevé–Calogero correspondence was first observed by B. Suleimanov [46, 47] in terms of rational linear problems. See also S. Slavyanov’s papers [45].

Let us note that the phenomenon of the classical-quantum correspondence is also known in the theory of integrable systems in some other contexts. There are interrelations between classical and quantum problems of a simingly different type [1, 54, 55], where Bethe vectors of integrable quantum spin chains are related to some data of classical integrable many-body systems. A similarity between quantum transfer matrices and classical $\tau$-functions was pointed out in [27, 29, 56].

The aim of this paper is to address the inverse problem. We start with the system of scalar equations (1.5) and assume that the quantum Painlevé–Calogero correspondence takes place, i.e., equations (1.7)–(1.8) hold true. (In this paper, we refer to it as classical-quantum correspondence, since it is not clear initially which equations satisfy the conditions). Then we derive and solve functional equations² for the potential $V$ searching through possible choices of the function $b$. In other words, we assume that the classical-quantum correspondence holds true and find out what kind of potentials arise from the compatibility conditions.

We prove the following:

Theorem 1.1

Let the compatibility condition for the system (1.5) with

$$\begin{aligned} U(x,{\dot{u}}(t),u(t),t)=V(x,t)-H({\dot{u}},u,t) \end{aligned}$$

and

$$\begin{aligned} H({\dot{u}},u,t)=\frac{1}{2}{\dot{u}}^2+{\tilde{V}}(u,t) \end{aligned}$$

be equivalent to

$$\begin{aligned} {\ddot{u}}=-\partial _u {\tilde{V}}(u,t), \end{aligned}$$

where $u$ is defined as a simple zero of the function $b(x,t)$: $b(x,t)\left. \right| _{x=u}=0$. Then there are two possibilities:

1.

$$\begin{aligned} b(x,u,t)=b(x-u,t). \end{aligned}$$

(1.11)

The function $b(z,t)$ satisfies the heat equation

$$\begin{aligned} 2\partial _t b(z,t)=\partial _z^2b(z,t); \end{aligned}$$

the quantum potential coincides with the classical one,

$$\begin{aligned} \tilde{V}(u,t)=V(u,t), \end{aligned}$$

and satisfies the functional equation

(1.12)

where $f(x,t)=b_x(x,t)/b(x,t)$.

2.

$$\begin{aligned} b(x,u,t)=b(x-u,t)b(x+u,t). \end{aligned}$$

(1.13)

The function $b(z,t)$ satisfies the heat equation

$$\begin{aligned} 2\partial _t b(z,t)=\partial _z^2b(z,t); \end{aligned}$$

the classical and quantum potentials are related by

$$\begin{aligned} \left. \tilde{V}(u,t)=V(u,t)+\frac{1}{2}\, \partial _x^2 \log b(x,t)\right| _{x=2u}, \end{aligned}$$

and $V(x,t)$ satisfies the functional equation

(1.14)

where $f(x,t)=b_x(x,t)/b(x,t)$.

The proof of the theorem is based on Propositions 3.1, 3.2, and 3.3. For possibilities different from (1.11) and (1.13), see the remark after (3.33) and Appendix A.

Solving equations (1.12) and (1.14), we get the following results: for the rational (in $x$) function $b$, we obtain PI and PII from (1.12) and PIV from (1.14); for the hyperbolic, we obtain PIII from (1.12) and PV from (1.14). The most general equation PVI arises for the $\theta$-functional ansatz for $b$ from (1.14), while the equation from (1.12) is shown to have only trivial solutions in this case.

Finally, it is shown that all natural choices of the function $b$ (rational, hyperbolic, and elliptic) reproduce exactly the Painlevé list of equations. In this sense, the classical-quantum correspondence can be viewed as an alternative definition for the Painlevé equations.

The paper is organized as follows. In the next section, we recall the quantum Painlevé–Calogero correspondence. In Section 3, we derive the functional equations from (1.12) and (1.14) and then solve these equations in Sections 4-6. In the appendices, we give the definitions and identities for necessary elliptic functions, discuss some special cases of the $b$-function, and list the $\mathbf{U}{-}\mathbf{V}$ pairs for PI–PV which are acceptable for the quantum Painlevé–Calogero correspondence.

2 Quantum Painlevé–Calogero Correspondence

The quantum Painlevé–Calogero correspondence states that for Painlevé equations, the nonstationary Baxter equation at $\hbar =1$ represents a classical linear problem. Let us start from example.

2.1 Example of Painlevé V

The PV equation is conventionally written as:

$$\begin{aligned} \partial ^2_{T} y=\left( \frac{1}{2y}+\frac{1}{y\!-\! 1}\right) (\partial _{T} y)^2 -\frac{\partial _{T} y}{T} +\frac{y(y\!-\! 1)^2}{T^2} \left( \alpha +\frac{\beta }{y^2}+\frac{\gamma T}{(y\!-\! 1)^2} +\frac{\delta T^2 (y+1)}{(y\!-\! 1)^3}\right) , \end{aligned}$$

where $\alpha$, $\beta , \gamma , \delta$ are parameters.³ Making change of variable

$$\begin{aligned} y=\coth ^2 u \end{aligned}$$

(2.1)

together with

$$\begin{aligned} T =e^{2t}, \end{aligned}$$

(2.2)

PV acquires the form

$$\begin{aligned} \ddot{u} =-\frac{2\alpha \cosh u}{\sinh ^3 u}- \frac{2\beta \sinh u}{\cosh ^3 u} -\gamma e^{2t}\sinh (2u) -\frac{1}{2}\, \delta e^{4t}\sinh (4u). \end{aligned}$$

(2.3)

The later equation is Hamiltonian with $H_\mathrm{V}(p, x)=\frac{p^2}{2}+V_\mathrm{V}(u,t)$, where

$$\begin{aligned} V_\mathrm{V}(u,t)=-\, \frac{\alpha }{\sinh ^2 u} -\, \frac{\beta }{\cosh ^2 u}+\frac{\gamma e^{2t}}{2} \cosh (2u)+\frac{\delta e^{4t}}{8}\cosh (4u). \end{aligned}$$

(2.4)

The zero curvature representation is known from [21]. It is rational in spectral parameter $X$. As was shown in [57], the change

$$\begin{aligned} X=\cosh ^2 x \end{aligned}$$

with (2.1) and (2.2) and some special gauge transformation brings the Jimbo–Miwa $\mathbf{U}{-}\mathbf{V}$ pair to the one given in (C.1)–(C.4). Then the first component of the linear problem (1.1) $\psi$ satisfies the nonstationary Schrödinger equation

$$\begin{aligned} \partial _t \psi =\left( H_\mathrm{V}^{(\alpha -\frac{1}{8}, \, \beta +\frac{1}{8}, \, \gamma ,\, \frac{1}{2} )}(\partial _x , x)- H_\mathrm{V}^{(\alpha , \beta , \gamma , \frac{1}{2} )}(\dot{u} , u )\right) \psi , \end{aligned}$$

and, therefore,

$$\begin{aligned} \partial _t \Psi = H_\mathrm{V}^{(\alpha -\frac{1}{8}, \, \beta +\frac{1}{8}, \, \gamma ,\, \frac{1}{2} )}(\partial _x , x)\Psi =\left( \frac{1}{2}\, \partial _x^2 + V_\mathrm{V}^{(\alpha -\frac{1}{8}, \, \beta +\frac{1}{8}, \, \gamma ,\, \frac{1}{2} )}(x,t)\right) \Psi \end{aligned}$$

for $\Psi (x,t)=e^{\int ^t H_\mathrm{V}^{(\alpha , \, \beta , \, \gamma ,\, \frac{1}{2} )} (\dot{u} , u)dt'}\psi (x,t)$; i.e., the linear problem admits the form of the quantized equation (in spectral parameter). Notice that the parameters $\alpha$, $\beta$ are shifted by $\pm \frac{1}{8}$ in the quantum Hamiltonian.

2.2 Summary

The following theorem summarizes the results of [57, 58] for all Painlevé equations, see also the table of changes of variables below.

Theorem

[57, 58] For any of the six equations from the Painlevé list written in the Calogero form as classical nonautonomous Hamiltonian systems with time-dependent Hamiltonians $H(p,u,t)$, there exists a pair of compatible linear problems

$$\begin{aligned} \left\{ \begin{array}{l} \partial _{x}\varvec{\Psi } =\mathbf{U} (x,t,u,\dot{u},\{c_k\})\varvec{\Psi },\\ \partial _{t}\varvec{\Psi } =\mathbf{V} (x,t,u,\dot{u},\{c_k\})\varvec{\Psi }, \end{array}\right. \, \quad \quad \varvec{\Psi } =\left( \begin{array}{l} \psi _1 \\ \psi _{2} \end{array}\right) , \end{aligned}$$

where $\mathbf{U}$ and $\mathbf{V}$ are $sl_2$-valued functions, $x$ is a spectral parameter, $t$ is the time variable and $\{c_k\}=\{\alpha , \beta , \gamma , \delta \}$ is the set of parameters involved in the Painlevé equation, such that:

• 1) The zero curvature condition

  $$\begin{aligned} \partial _t\mathbf{U}-\partial _x \mathbf{V}+[\mathbf{U},\mathbf{V}]=0 \end{aligned}$$

  is equivalent to the Painlevé equation for the variable $u$ defined as any (simple) zero of the right-upper element of the matrix $\mathbf{U}(x,t)$ in the spectral parameter: $\mathbf{U}_{12}(u, t)=0$;

• 2) The function $\Psi =e^{\int ^t H(\dot{u}, u,t')dt'}\psi _1$, where $\psi _1$ is the first component of $\varvec{\Psi }$, satisfies the nonstationary Schrödinger equation in imaginary time

  $$\begin{aligned} \partial _t\Psi =\left( \frac{1}{2}\, \partial _x^2 +\tilde{V} (x,t) \right) \Psi \end{aligned}$$

  with the potential

  $$\begin{aligned}&\tilde{V}(x,t)=V(x,t, \{\tilde{c}_k\}),\\&V(x,t, \{\tilde{c}_k\})-\left( \frac{1}{2}{\dot{u}}^2+V(u,t, \{c_k\})\right) = \frac{1}{2}\Bigl [\det (\mathbf{U})-\partial _x \mathbf{U}_{11}+2\mathbf{V}_{11}\Bigr ], \end{aligned}$$

  which coincides with the classical potential $V(u,t)=V(u,t, \{c_k\})$ up to possible shifts of the parameters $\{c_k\}$:

  $$\begin{aligned} \begin{array}{l} (\tilde{\alpha }, \tilde{\beta }) =(\alpha , \beta +\frac{1}{2}) \quad \text{ for } \text{ PIV },\\ (\tilde{\alpha }, \tilde{\beta }, \tilde{\gamma }, \tilde{\delta })= \left( \alpha -\frac{1}{8}, \beta +\frac{1}{8}, \gamma , \delta \right) \quad \text{ for } \text{ PV },\\ (\tilde{\alpha }, \tilde{\beta }, \tilde{\gamma }, \tilde{\delta })= \left( \alpha -\frac{1}{8}, \beta +\frac{1}{8}, \gamma -\frac{1}{8}, \delta +\frac{1}{8}\right) \quad \text{ for } \text{ PVI }. \end{array} \end{aligned}$$

The list of changes of variables is summarized in the following table:

Equation	$y(u,t)$	$T(t)$	$X(x,t)$	$\mathbf{U}_{12}(x,t)$
PI	$u$	$t$	$x$	$x-u$
PII	$u$	$t$	$x$	$x-u$
PIV	$u^2$	$t$	$x^2$	$x^2 -u^2$
PIII	$e^{2u}$	$e^t$	$e^{2x}$	$2e^{t/2}\sinh (x-u)$
PV	$\coth ^2 u$	$e^{2t}$	$\cosh ^2 x$	$2e^{t} \sinh (x\! -\! u)\sinh (x\! +\! u)$
PVI	$\frac{\wp (u)-\wp (\omega _1)}{\wp (\omega _2)-\wp (\omega _1)}$	$\frac{\wp (\omega _3)-\wp (\omega _1)}{\wp (\omega _2)-\wp (\omega _1)}$	$\frac{\wp (x)-\wp (\omega _1)}{\wp (\omega _2)-\wp (\omega _1)}$	$\vartheta _1(x-u)\vartheta _1(x+u) h(u,t)$

Function $h(u,t)$ for the PVI case can be found in [58]. Notice that the changes of variables given above can be derived in a general form from (1.4) and the requirement that the potential (1.6) could be presented as a sum of two parts depending on $x,t$ and $u,t$ separately. This calculation was made in [58] for the most general Painlevé VI equation. The appropriate $\mathbf{U}{-}\mathbf{V}$ pairs for PI–PV are given in Appendix C.

3 The Scalar Linear Problems and Functional Equations

It was shown in [57, 58] that each of the six equations from the Painlevé list, hereinafter referred to as PI–PVI, written in the so-called Calogero form, can be obtained as integrability conditions for two Schrödinger-like equations

$$\begin{aligned} \left\{ \begin{array}{l}\displaystyle { \left( \frac{1}{2}\, \partial _{x}^{2} -\frac{b_x}{2b}\, \partial _x \, + W (x,t)\right) \Psi =0},\\ \displaystyle { \partial _t \Psi =\left( \frac{1}{2}\, \partial _x^2 + V(x,t)\right) \Psi ,} \end{array} \right. \end{aligned}$$

(3.1)

stationary and nonstationary. The time-dependent potentials $W$ and $V$ are related by

$$\begin{aligned} W(x,t)={U}-\frac{2{\dot{b}}-b_{xx}}{4b}=V(x,t)-H- \frac{2{\dot{b}}-b_{xx}}{4b}, \end{aligned}$$

(3.2)

where $H$ does not depend on $x$ and $b$ is some function of the spectral parameter $x$ and time $t$ to be chosen in such a way that the two linear problems are compatible for some $V(x,t)$. Suppose it has a (simple) zero at the point $x=u=u(t)$: $b(u,t)=0$, and let $V(x,t)$ be a function that depends on $x,t$ in an explicit way only (i.e., $V(x,t)$ does not contain $u$). Also let $H$ be a function of $u$ and $\dot{u}$.

Remark

Note that function $b$ may depend on $t$ in two ways — explicit and implicit. The latter means the time dependence through the unknown functions of $t$ (dependent variables). Writing $\partial _t b$, we mean the derivative with respect to the explicit dependence only. For example, $\partial _t(z-u)=0$. The lower index $t$ means the same ($\partial _t b(z,u(t),t)=b_t$), while the dot is the full time derivative: ${\dot{b}}(z,u(t),t)={\dot{u}}\partial _u b+b_t$. The same notation is used for other functions depending on $t$ and $u(t)$ apart from $\Psi$ in the linear problem (where the partial derivative symbols $\partial _x$, $\partial _t$ are traditionally used, but, in fact, the operator $\partial _t$ acts as the full time derivative).

Combining equations (3.1), one can write another pair of linear problems whose compatibility implies the Painlevé equations:

$$\begin{aligned} \left\{ \begin{array}{l}\displaystyle { \left( \frac{1}{2}\, \partial _{x}^{2} -\frac{b_x}{2b}\, \partial _x \, + W (x,t)\right) \Psi =0},\\ \displaystyle { \partial _t \Psi =\left( \frac{b_x}{2b}\, \partial _x +\frac{2{\dot{b}}-b_{xx}}{4b}+H \right) \Psi ,} \end{array} \right. \end{aligned}$$

(3.3)

(The first equation is the same, while the other one is a first-order equation.) Passing to the function $\tilde{\Psi }=\Psi /\sqrt{b}$, we can write these linear problems in the Fuchs–Garnier form:

$$\begin{aligned} \left\{ \begin{array}{l}\displaystyle { \left( \frac{1}{2}\, \partial _{x}^{2} + S(x,t)\right) \tilde{\Psi }=0},\\ \displaystyle { \partial _t \tilde{\Psi }= \left( \frac{1}{2}f \partial _x - \frac{1}{4}\, f_{x} \right) \tilde{\Psi }}, \end{array} \quad \quad f= \partial _x \log b, \quad f _{x} \equiv \partial _x f , \right. \end{aligned}$$

(3.4)

where we have introduced the function $S =S(x,t)$ by the formula

$$\begin{aligned} S={U}-\frac{{\dot{b}}}{2b}+\frac{b_{xx}}{2b}-\frac{3}{8} \left( \frac{b_x}{b}\right) ^2=V-H -\frac{{\dot{b}}}{2b}+\frac{b_{xx}}{2b}-\frac{3}{8} \left( \frac{b_x}{b}\right) ^2. \end{aligned}$$

(3.5)

Their integrability is equivalent to the condition

$$\begin{aligned} \Bigl [ \frac{1}{2}\, \partial _{x}^2 +S, \, \partial _t -\frac{1}{2}f \partial _x +\frac{1}{4}\, f _{x}\Bigr ]\tilde{\Psi }=0, \end{aligned}$$

which implies

$$\begin{aligned} {\dot{S}} = S f_{x}+\frac{1}{2}f S_x + \frac{1}{8}\, f_{xxx} \end{aligned}$$

(3.6)

or

(3.7)

This equation is our main interest in this paper. In the next sections, we determine the potential $V$ making one or another ansatz for $b$.

Notice that the equation (3.7) can be obtained from the compatibility of initial matrix linear problem (1.3) with ${U}$ defined by (1.6). One can express all elements of $\mathbf{U}$ and $\mathbf{V}$ in terms of three functions $a=\mathbf{U}_{11}$, $b=\mathbf{U}_{12}$, and ${U}$:

$$\begin{aligned}&\mathbf{U}:\ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{c} \mathbf{U}_{11}=a,\ \ \mathbf{U}_{12}=b,\\ \ \\ \mathbf{U}_{21}=- \frac{1}{2b^2}\left( 2a^2b+2a_xb-2ab_x+4{U}b-2{\dot{b}}+b_{xx} \right) , \end{array}\\&\mathbf{V}:\ \ \ \ \ \ \ \ \begin{array}{c} \mathbf{V}_{11}=\frac{1}{4b}\left( 2ab_x+2{\dot{b}}-b_{xx} \right) ,\ \ \mathbf{V}_{12}=\frac{1}{2}b_x,\\ \ \\ \mathbf{V}_{21}=-\frac{1}{4b^2}\left( 4{\dot{a}}b-2{\dot{b}}_x+b_{xxx}-2ab_{xx}+2a^2b_x+4b_x{U} \right) . \end{array} \end{aligned}$$

The function $a$ cancels out from compatibility condition (1.3).

Recall that the dynamical variable $u$ is defined as a zero of the function $b(x,t)=b(x,u(t),t)$: $b(u,u,t)=0$. Suppose $b$ is an analytical function near $x=u$; then in the vicinity of $x=u$,

$$\begin{aligned} b=b_1(u,t)(x-u)+b_2(u,t)(x-u)^2+b_3(u,t)(x-u)^3+\cdots . \end{aligned}$$

(3.8)

Consider equation (3.7) at $x=u$:

$$\begin{aligned} \left( -2b_{x}^2{U}-\frac{1}{2}b_{x}b_{x x x}+b_{x}{\dot{b}}_{x}-{\dot{b}}^2+\frac{1}{4}b_{x x}^2\right) \left. \right| _{x=u}=0, \end{aligned}$$

(3.9)

where we used that ${U}\!=\!V(x,t)\!-\!H({\dot{u}},u,t)$, and, therefore, it is a regular function at $x=u$ and $\left[ b{U}\right] (x=u)=0$. From the expansion (3.8), we get

$$\begin{aligned} b_x\left. \right| _{x=u}=b_1,\ \ b_{xxx}\left. \right| _{x=u}=6b_3,\ \ {\dot{b}}\left. \right| _{x=u}=-{\dot{u}}b_1,\ \ {\dot{b}}_x\left. \right| _{x=u}={\dot{u}}(b_1'-2b_2)+\partial _t b_1. \end{aligned}$$

Plugging this into (3.9), we obtain:

$$\begin{aligned} {U}\left. \right| _{x=u}=-\frac{1}{2}v^2+\frac{1}{2b_1^2}\left[ \left( b_2-\frac{1}{2}b_1'\right) ^2-3b_1b_3+b_1\partial _t b_1 +\frac{1}{4}b_2^2 \right] , \end{aligned}$$

where

$$\begin{aligned} v={\dot{u}}+\frac{b_2}{b_1}-\frac{b_1'}{2b_1}. \end{aligned}$$

(3.10)

The latter expression is the “momentum.” Notice that this local evaluation at $x=u$ fixes the dependence $H({\dot{u}})$, since $V(x,t)$ is independent of ${\dot{u}}$. We consider some nontrivial cases ($v\ne \dot{u}$) in Appendix A.

Let us find out what kind of restriction on the behavior of $b=b(x,u(t),t)$ arises from the classical-quantum correspondence. First, recall that the quantum Hamiltonian which we use in (1.5), (3.1) has the form $\hat{H}=\frac{1}{2}\partial _x^2+V(x,t)$. Therefore, the classical one is $H(p_x,x,t)=\frac{1}{2}p_x^2+V(x,t)$. The classical-quantum correspondence implies that the classical equations for $u(t)$ arising from the compatibility condition (1.2) (or (3.6) or (3.7)) are generated by $H({\dot{u}}, u,t)$, which differs from $H(p_x,x,t)$ by only possible “quantum corrections” of the potential. Thus, the classical Hamiltonian should have the “Calogero form,” i.e., $H({\dot{u}}, u,t)=\frac{1}{2}{\dot{u}}^2+{\tilde{V}}(u,t)$. At the moment, we do not assume any relations between $V(x,t)$ and ${\tilde{V}}(u,t)$. However, the Calogero form of the Hamiltonian provides some special properties of $b(x,u(t),t)$.

Proposition 3.1

Let the compatibility condition (3.7) describe nonautonomous dynamics

$$\begin{aligned} {\dot{u}}= & {} v,\nonumber \\ {\dot{v}}= & {} {\ddot{u}}=-\partial _u {\tilde{V}}(u,t) \end{aligned}$$

(3.11)

generated by the Hamiltonian

$$\begin{aligned} H({\dot{u}}, u,t)=\frac{1}{2}{v}^2+{\tilde{V}}(u,t). \end{aligned}$$

(3.12)

Then $b(x,u(t),t)$ factorizes into the product

$$\begin{aligned} b(x,u(t),t)=b_1(x-u(t),t)\, b_2(x+u(t),t), \end{aligned}$$

(3.13)

and each of the factors satisfies the heat equation:

$$\begin{aligned} 2\partial _t b_{1,2}(z,t)=\partial _z^2 b_{1,2}(z,t). \end{aligned}$$

Proof

Substituting (3.11) and (3.12) into (3.7), we get an equation where the left-hand side. is quadratic in $v=\dot{u}$ . Since $v$ is an independent variable, all the coefficients in front of $v^k$ ($k=2,1,0$) vanish. The coefficient in front of $v^2$ gives (it comes from terms $2b b_{xx}U$, $-2b_x^2 U$, $-\dot{b}^2$ and $b\ddot{b}$ in (3.7))

$$\begin{aligned} b_x^2-b_u^2+bb_{uu}-bb_{xx}=0 \end{aligned}$$

or

$$\begin{aligned} \left( \partial _x^2-\partial _u^2\right) \log b=\left( \partial _x-\partial _u\right) \left( \partial _x+\partial _u\right) \log b=0, \end{aligned}$$

which is equivalent to (3.13). The coefficient in front of $v$ gives

$$\begin{aligned} b_xb_{xu}-bb_{xxu}+2bb_{tu}-2b_ub_t=0 \end{aligned}$$

or

$$\begin{aligned} \left( \frac{b_{xu}}{b}\right) _x=2\left( \frac{b_t}{b}\right) _u. \end{aligned}$$

(3.14)

Plugging (3.13) into (3.14), we obtain:

$$\begin{aligned} 2\left( \frac{{b_2}_t}{b_2}\right) '-2\left( \frac{{b_1}_t}{b_1}\right) '=\left( \frac{b_2''}{b_2}\right) '- \left( \frac{b_1''}{b_1}\right) '. \end{aligned}$$

The variables $x-u$ and $x+u$ are independent. Therefore,

$$\begin{aligned} 2\left( \frac{{b_k}_t}{b_k}\right) '=\left( \frac{b_k''}{b_k}\right) ',\ \ k=1,2. \end{aligned}$$

Then

$$\begin{aligned} 2{b_k}_t={b_k''}+c(t)b_k,\ \ k=1,2, \end{aligned}$$

where $c(t)$ is the integration constant. The term with $c(t)$ can be removed by the substitution $b\rightarrow b e^{\int _t c(t)}$.$\square$

The coefficient in front of $v^0$ gives rise to equations for $V(x,t)$ and ${\tilde{V}}(u(t),t)$. We study these equations in the next sections.

3.1 One Simple Zero

Let us first consider the case when $b$ has only a simple zero at $u(t)$. The reason for this behavior of $b(z,t)$ is partly explained in Section 6.2.

Proposition 3.2

Let $b(z,t)$ satisfy the heat equation

$$\begin{aligned} 2\partial _t b(z,t)=\partial _z^2 b(z,t), \end{aligned}$$

(3.15)

and let $u$ be a simple zero of the function $b$: $b(x-u,t)\left. \right| _{x=u}=0$. Then integrability condition (3.7) implies that

$$\begin{aligned} H= & {} \frac{1}{2}\dot{u}^2+V(u),\end{aligned}$$

(3.16)

$$\begin{aligned} \ddot{u}= & {} -V'(u), \end{aligned}$$

(3.17)

and

(3.18)

where $f(x)=f(x,t)=b_x(x,t)/b(x,t)$ (for brevity we do not indicate the $t$-dependence of $f$ explicitly). In particular, if $f(x)=\frac{1}{x}+c_1x+c_3x^3+\cdots$, then

$$\begin{aligned} V'_t= & {} \frac{1}{12}V'''+2c_1V',\end{aligned}$$

(3.19)

$$\begin{aligned} \frac{1}{120}V^{(5)}= & {} \frac{1}{2}c_1V'''+24c_3V', \end{aligned}$$

(3.20)

where $V_t(u)=\partial _t V(u,t)$.

Proof

Direct substitution of $b=b(x-u(t),t)$ into (3.6) together with (3.15) yields

$$\begin{aligned} V_t(x)-\dot{H}-\frac{1}{2}f\left( V'(x)-\ddot{u}\right) - f_x\left( V(x)+\frac{1}{2}\dot{u}^2-H\right) =0. \end{aligned}$$

Locally, $f=\frac{b_x}{b}\sim \frac{1}{z-u}$. Therefore, the cancellation of the second-order pole leads to (3.16). At this stage, we have

$$\begin{aligned} V_t(x)-V_t(u)-\dot{u}(\ddot{u}+V'(u))-\frac{1}{2}f \left( V'(x)-\ddot{u}\right) - f_x\left( V(x)-V(u)\right) =0. \end{aligned}$$

From the last two terms, it is easy to see that the cancellation of the first-order term gives (3.17). Substituting (3.17) into the above equation, we get (3.18). The differential equations (3.19), (3.20) follow from the local expansion of (3.18) near $x=u$. To be exact, (3.20) follows from (3.19) and $V_t'''=\frac{3}{40}V^{(5)}+\frac{5}{2}c_1V'''+24c_3V'$. $\square$

In this proof, only the heat equation was used. In what follows, we need some more properties that follow from the heat equation.

Lemma 3.1

Let $b$ satisfy the heat equation (3.15) and $f=\partial _x \log b$. Then

$$\begin{aligned} \partial _t f= & {} \frac{1}{2}\partial _x\left( f^2+ f_x\right) =f_x f+ \frac{1}{2}f_{xx},\end{aligned}$$

(3.21)

$$\begin{aligned} \partial _t f_x= & {} f_{xx}f+ f^2+\frac{1}{2}f_{xxx}. \end{aligned}$$

(3.22)

Suppose also that $b$ is an odd function of $x$ and has a simple zero at $x=0$. Then

$$\begin{aligned} \frac{1}{2}f_x(x-w)f_x(x+w)= & {} (f_x(x-w)+f_x(x+w))f_{xx}(2w)\nonumber \\&+\left( f(x+w)-f(x-w) \right) f_{xx}(2w)- \partial _tf_{x}(2w).\nonumber \\ \end{aligned}$$

(3.23)

Proof

The proof of (3.21) and (3.22) is direct. Identity (3.23) is proved via consideration of the local expansion and comparing of the poles taking into account (3.22). $\square$

3.2 Two Simple Zeros

Suppose $b$ has two simple poles. Let us derive an analogue of (3.16)–(3.18) for this case.

Proposition 3.3

Let $b=b_1(z,t)b_2(z,t)$, and let each factor satisfy the heat equation

$$\begin{aligned} 2\partial _t b_{1,2}(z,t)=\partial _z^2 b_{1,2}(z,t). \end{aligned}$$

(3.24)

Suppose that $b_{1,2}$ has a simple zero $u_{1,2}$: $b_{1,2}(x-u_{1,2},t)\left. \right| _{x=u_{1,2}}=0$. Then equation (3.6) has the following solution:

$$\begin{aligned}&u_1=-u_2,\ \ V(u)=V(-u),\ \ b_1=b(x-u(t),t),\ \ b_2=b(x+u(t),t),\end{aligned}$$

(3.25)

$$\begin{aligned}&H=\frac{1}{2}\dot{u}^2+V(u)+\frac{1}{2}f_x(2u),\end{aligned}$$

(3.26)

$$\begin{aligned}&\ddot{u}=-V'(u)-f_{xx}(2u), \end{aligned}$$

(3.27)

where $b(x,t)$ is an odd function of $x$, $f=\partial _x \log b$, and the potential satisfies

(3.28)

In particular, if $f=\frac{1}{x}+c_1x+c_3x^3+\cdots$, then

$$\begin{aligned} V'_t= & {} \frac{1}{12}V'''+\frac{1}{2}\, f (2x)V''+ \left( 2c_1+f_x(2x)\right) V',\end{aligned}$$

(3.29)

$$\begin{aligned} V'''_t= & {} \frac{3}{40}V^{(5)}+\frac{1}{2}\, f(2x)V^{(4)}+ \frac{5}{2}\left( c_1+f_x(2x)\right) V'''\nonumber \\&+ \frac{9}{2}f_{xx}(2x)V''+\left( 24c_3+3f_{xxx}(2x)\right) V'. \end{aligned}$$

(3.30)

Proof

The direct substitution leads to

$$\begin{aligned}&V_t(x)-\dot{H} -\frac{1}{2}\frac{b_{1,x}}{b_1}\left( V'(x) -\ddot{u}_1\right) -\frac{1}{2}\frac{b_{2,x}}{b_2}\left( V'(x)-\ddot{u}_2\right) \nonumber \\&\quad -\left( \frac{b_{1,x}}{b_1}\right) _x\left( V(x)+\frac{1}{2}\dot{u}_1^2 -H+\frac{1}{2}(\dot{u}_1+\dot{u}_2)\frac{b_{2,x}}{b_2}\right) \nonumber \\&\quad -\left( \frac{b_{2,x}}{b_2}\right) _x\left( V(x)+\frac{1}{2}\dot{u}_2^2 -H+\frac{1}{2}(\dot{u}_1+\dot{u}_2)\frac{b_{1,x}}{b_1}\right) \nonumber \\&\quad -\frac{1}{2}\left( \frac{b_{1,x}}{b_1}\right) _x\left( \frac{b_{2,x}}{b_2} \right) _x=0. \end{aligned}$$

(3.31)

From cancellation of the second-order poles, we get

$$\begin{aligned} H= & {} \frac{1}{2}\dot{u}_1^2+V(u_1)+\frac{1}{2}(\dot{u}_1+\dot{u}_2)\frac{b_{2,x}}{b_2}(u_1) +\frac{1}{2}\left( \frac{b_{2,x}}{b_2}\right) _x(u_1),\nonumber \\ H= & {} \frac{1}{2}\dot{u}_2^2+V(u_2)+\frac{1}{2}(\dot{u}_1+\dot{u}_2)\frac{b_{1,x}}{b_1}(u_2) +\frac{1}{2}\left( \frac{b_{1,x}}{b_1}\right) _x(u_2). \end{aligned}$$

(3.32)

Comparing these two expressions, one can see that (3.25) and (3.26) indeed satisfy (3.6). Then vanishing of the first-order poles at $\pm u$ gives (3.27). Substituting (3.27) into (3.31), we get

$$\begin{aligned}&V_t(x)-\left. \partial _t\left( V(z)+\frac{1}{2}f(2z)\right) \right| _{z=u(t)} \nonumber \\&\quad -\frac{1}{2}\frac{b_{1,x}}{b_1}\left( V'(x)+V'(u)+f_{xx}(2u)\right) -\frac{1}{2}\frac{b_{2,x}}{b_2}\left( V'(x)-V'(u)-f_{xx}(2u)\right) \nonumber \\&\quad +\left( \left( \frac{b_{1,x}}{b_1}\right) _x+ \left( \frac{b_{2,x}}{b_2}\right) _x\right) \left( V(u)-V(x)+\frac{1}{2}f_{xx}(2u)\right) \nonumber \\&\quad -\frac{1}{2}\left( \frac{b_{1,x}}{b_1}\right) _x \left( \frac{b_{2,x}}{b_2}\right) _x=0. \end{aligned}$$

(3.33)

All terms that do not contain $V$ cancel because of (3.23), and we get (3.28). Differential equations (3.29), (3.30) follow from the local expansion of (3.28) near $x=u$. $\square$

Remark

To investigate the case more general than (3.25), one should solve the equation emerging from equality of right-hand sides of (3.32) (see Appendix A).

Notice also that the right-hand side of (3.29) and (3.30) are full derivatives:

$$\begin{aligned} V'_t= & {} \partial _x\left( \frac{1}{12}V''+2c_1V+ \frac{1}{2} f(2x)V'\right) ,\\ V'''_t= & {} \partial _x\left( \frac{3}{40}V^{(4)}+\frac{5}{2}c_1V''+24c_3V + \frac{3}{2}f_{xx}(2x)V'+\frac{3}{2} f_{x}(2x)V''+ \frac{1}{2}f(2x)V'''\right) . \end{aligned}$$

In particular, this leads to the following equation:

$$\begin{aligned} V^{(4)}-60c_1V''+60f_{xx}(2x)V'+60f_{x}(2x)V''+24c_3V=\hbox {const}(t). \end{aligned}$$

4 Rational Solutions

4.1 The Simplest Case: $b=x-u(t)$

The simplest possibility is to set

$$\begin{aligned} b=x-u(t). \end{aligned}$$

(4.1)

We will see that already this case is meaningful and leads to PI and PII equations.

In this case, integrability condition (3.18) turns into

$$\begin{aligned} \Bigl (V_t(x)-V_t(u)\Bigr )-\frac{1}{2(x-u)} \Bigl (V'(x)+V'(u)\Bigr )+\frac{1}{(x-u)^2}\Bigl (V(x)-V(u)\Bigr )=0 \end{aligned}$$

(4.2)

or

$$\begin{aligned} 2(x-u)^2 \Bigl (V_t(x)-V_t(u)\Bigr )-(x-u)\Bigl (V'(x)+V'(u)\Bigr )+ 2\Bigl (V(x)-V(u)\Bigr )=0. \end{aligned}$$

(4.3)

It should be an identity for all $x,u$ which enter here as independent variables on equal footing. The way to proceed is to take the third derivative of (4.3) with respect to $x$. The result is

$$\begin{aligned}&2u^2 V_{t}'''(x) +u \left( V^\mathrm{IV}(x)-4x V_{t}'''(x) -12V_{t}''(x)\right) \\&\quad \displaystyle {+\,\,\,\, 12 V_{t}'(x)+12x V_{t}''(x)+2x^2 V_{t}'''(x)-V'''(x)-x V^\mathrm{IV}(x) =0}. \end{aligned}$$

The equality holds identically if the coefficients in front of $u^2$, $u$, and the free term in $u$ vanish. This implies the conditions

$$\begin{aligned} \left\{ \begin{array}{l} V_{t}'''(x)=0, \\ \\ 12V_{t}'(x) =V'''(x). \end{array}\right. \end{aligned}$$

From the first equation, it follows that $V_{t}(x)$ is a polynomial in $x$ of second degree at most, while from the second one, it then follows that $V(x)$ is a polynomial in $x$ of fourth degree at most. There are three possibilities:

• 1) $V_{t}'(x)\equiv 0$; then $V(x)$ is a quadratic polynomial $V(x)=a_2 x^2 +a_1 x +a_0$ with $\dot{a}_2 =\dot{a}_1 =0$. Plugging it into equation (4.3), we see that the equation holds identically for any constants $a_2, a_1$, with the irrelevant free term $a_0$ being an arbitrary function of $t$. This is the potential for the harmonic oscillator.

• 2) $V_{t}''(x)\equiv 0$; then $V(x)$ is a 3-d degree polynomial $V(x)=a_3 x^3 +a_2 x^2 +a_1 x +a_0$ with $\dot{a}_3 =\dot{a}_2 =0$. By rescaling and shift of the variable $x$, we can set $a_3=1$, $a_2=0$. The free term, $a_0$, is irrelevant since it cancels in equation (4.3). Plugging the potential in the form $V(x)=x^3 +a_1 x$ into equation (4.3), we get $(x-u)^2 (2\dot{a}_1 -1)=0$. Therefore, $a_1 =t/2$, and

  $$\begin{aligned} V(x)=x^3 +\frac{tx}{2}. \end{aligned}$$

  This is, up to a common factor, the potential for the PI equation.

• 3) $V_{t}''(x)\ne 0$; then $V(x)$ is a 4-th degree polynomial $V(x)=a_4 x^4 +a_3 x^3 +a_2 x^2 +a_1 x +a_0$ with $\dot{a}_4 =\dot{a}_3 =0$. Again, we can set $a_4=1$, $a_3=0$, and $a_0 =0$. Plugging the potential in the form $V(x)=x^4 +a_2 x^2 +a_1 x$ into equation (4.3), we get $(x^2 -u^2)(\dot{a}_2 -1) +\dot{a}_1 =0$. Therefore, $a_2 =t$, $a_1 =-2\alpha$, where $\alpha$ is an arbitrary constant. Up to a common factor, we obtain the potential

  $$\begin{aligned} V(x)=x^4 +tx^2 -2\alpha x \end{aligned}$$

  for the PII equation with the parameter $\alpha$.

4.2 The Case $b=(x-u_1(t))(x-u_2(t))$

Let us make the similar calculations for $b=(x-u_1 (t))(x-u_2(t))$. Instead of

$$\begin{aligned} V_t(x)-H_t -\frac{V'(x)-\ddot{u}}{2(x-u)} +\frac{V(x)-H +\dot{u}^2/2}{(x-u)^2}=0, \end{aligned}$$

we get, after cancellation of third- and fourth-order poles:

$$\begin{aligned}&V_t(x)-H_t -\frac{1}{2(x-u_1)}\Big (V'(x)-\ddot{u}_1-\frac{2}{(u_1-u_2)^3}\Big ) \nonumber \\&\quad -\frac{1}{2(x-u_2)}\Big (V'(x)-\ddot{u}_2 -\frac{2}{(u_2-u_1)^3}\Big )\nonumber \\&\quad +\frac{1}{(x-u_1)^2}\Big ( V(x)-H +\frac{1}{2}\frac{\dot{u}_1+\dot{u}_2}{u_1-u_2}+\frac{1}{2}\dot{u}_1^2-\frac{1}{2(u_1-u_2)^{2}}\Big )\nonumber \\&\quad +\frac{1}{(x-u_2)^2}\Big ( V(x)-H +\frac{1}{2}\frac{\dot{u}_1+\dot{u}_2}{u_2-u_1}+\frac{1}{2}\dot{u}_2^2-\frac{1}{2(u_1-u_2)^{2}}\Big )=0. \end{aligned}$$

(4.4)

Cancellation of the second-order poles at $x=u_{1,2}$ yields

$$\begin{aligned} H=\frac{1}{2}\dot{u}_1^2+\frac{1}{2}\frac{\dot{u}_1+\dot{u}_2}{u_1-u_2}+V(u_1)-\frac{1}{2(u_1-u_2)^{2}} \end{aligned}$$

and

$$\begin{aligned} H=\frac{1}{2}\dot{u}_2^2 +\frac{1}{2}\frac{\dot{u}_1+\dot{u}_2}{u_2-u_1}+V(u_2)-\frac{1}{2(u_1-u_2)^{2}}. \end{aligned}$$

By equating the two “kinetic” terms, we get the following two possibilities:

$$\begin{aligned} \mathbf{\hbox {1)}}\ \ \dot{u}_1+\dot{u}_2=0 , \quad \mathbf{\hbox {2)}}\ \ \partial _t(u_1-u_2)^2=-4. \end{aligned}$$

In the first case, $u_1+u_2=\hbox {const}$, and one can shift $x$ in the initial problem to set $u_1=-u_2\equiv u$. Therefore, the two possibilities are rewritten as

$$\begin{aligned} \mathbf{\hbox {1)}}\ \ \left\{ \begin{array}{l}u_1=-u_2\equiv u,\\ \\ V(u)=V(-u), \end{array}\right. \quad \mathbf{\hbox {2)}} \ \ \left\{ \begin{array}{l}u_1=u_2+\sqrt{c-4t}, \\ \\ V(u)=V(u-\sqrt{c-4t}), \end{array}\right. \end{aligned}$$

(4.5)

where $c$ is some constant. The second case is given in Appendix A. Here we consider the first one. In this case, (4.4) leads to integrability condition (3.28):

$$\begin{aligned}&\displaystyle { V_t(x)-V_t(u)-\frac{1}{2(x-u)}\left( V'(x)+V'(u) -2\, \frac{V(x)-V(u)}{x-u}\right) }\\&\quad \displaystyle { -\,\,\, \frac{1}{2(x+u)}\left( V'(x)-V'(u) -2\, \frac{V(x)-V(u)}{x+u} \right) =0}, \end{aligned}$$

or, equivalently,

$$\begin{aligned}&\displaystyle {2(x^2-u^2)^2(V_t(x)-V_t(u))- (x+u)(x^2-u^2)\left( V'(x)+V'(u)\right) }\nonumber \\&\quad \displaystyle {-\,\, (x-u)(x^2-u^2)\left( V'(x)-V'(u)\right) +4(x^2+u^2)(V(x)-V(u))\, =\, 0}. \end{aligned}$$

(4.6)

Since the maximal degree of $x$ in (4.6) is $4$, the differential operator $\partial _{X}^{5}$ applied to this equation kills all terms containing $V(u)$, and we are left with

$$\begin{aligned} \partial _{X}^{5}\Bigl [(x^2-u^2)^2 V_t (x)-x(x^2 -u^2) V'(x)+2(x^2 +u^2)V(x)\Bigr ]=0. \end{aligned}$$

Equating the coefficients in front of $u^4$, $u^2$, and $u^0$ to zero, we get the following conditions:

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle { \partial _{x}^{5} V_t (x)=0}, \\ \displaystyle { \partial _{x}^{5}\Bigl [-2x^2 V_t (x)+x V'(x) +2V(x) \Bigr ]=0},\\ \displaystyle { \partial _{x}^{5}\Bigl [ x^4 V_t (x)-x^3 V'(x) +2x^2 V(x) \Bigr ]=0}. \end{array}\right. \end{aligned}$$

They mean that the expressions in the square brackets are polynomials in $x$ of at most fourth degree:

$$\begin{aligned} V_t (x)= & {} P_4(x),\\ -2x^2 V_t (x)+x V'(x) +2V(x)= & {} Q_4(x), \\ x^4 V_t (x)-x^3 V'(x) +2x^2 V(x)= & {} R_4(x). \end{aligned}$$

Combining these conditions, we find that $x^2 V(x)$ must be a polynomial of at most $8$-th degree such that its highest and lowest coefficients do not depend on $t$. We also recall that it must contain only even powers of $x$. So we can write

$$\begin{aligned} V(x)=\mu x^6 +a_4x^4 +a_2x^2 +a_0 +\frac{\nu }{x^2}, \quad \quad \dot{\mu }=\dot{\nu }=0. \end{aligned}$$

Plugging this potential back into equation (4.6), we obtain

$$\begin{aligned} (x^4-u^4)(x^2 -u^2)(\dot{a}_4-4\mu )+(x^2-u^2)^2(\dot{a}_2-2a_4)=0. \end{aligned}$$

The solution is $a_4 =4\mu t +\alpha _4$, $a_2 = 4\mu t^2 +2\alpha _4 t+\alpha _2$, with integration constants $\alpha _4$, $\alpha _2$, and $a_0$ is arbitrary. There are three cases:

• 1) $\mu \ne 0$ (the case of general position); then one can set it equal to $1$ by rescaling and set $\alpha _4=0$ by a shift of the $t$-variable. Then the potential acquires the form

  $$\begin{aligned} V(x,t)=x^6 +4tx^4 +(4t^2 +\alpha _2)x^2 +a_0(t) + \frac{\nu }{x^2}. \end{aligned}$$

  (4.7)

  This is the potential for the PIV equation.

• 2) $\mu = 0$ but $\alpha _4 \ne 0$; then one can set $\alpha _4$ equal to $1$ by rescaling and set $\alpha _2=0$ by a shift of the $t$-variable. The potential is

  $$\begin{aligned} V(x,t)=x^4 +2tx^2 +a_0(t) + \frac{\nu }{x^2}. \end{aligned}$$

  It generates the equation

  $$\begin{aligned} \ddot{u} =-4u^3 -4tu +\frac{2\nu }{u^3}. \end{aligned}$$

  The change of the dependent variable $u\rightarrow y$ such that $u^2+ y^2+\frac{1}{2}\dot{y}+t=0$ (a version of a similar change in [17], section14.331]) brings the equation to the form $\ddot{y} =8y^3 +8ty +\sqrt{-32\nu }-2$, which is equivalent to the PII equation.

• 3) $\mu =\alpha _4 = 0$; then

  $$\begin{aligned} V(x,t)=\alpha _2 x^2+\frac{\nu }{x^2} \, +a_0(t). \end{aligned}$$

  This gives the exactly solvable rational 2-particle Calogero model in the harmonic potential. The $x$-independent term $a_0(t)$ is irrelevant.

5 Hyperbolic Solutions

5.1 The Case $b=e^{t/2} \sinh (x-u(t))$

Let us consider the case when $b$ is a trigonometric (hyperbolic, to be exact) function with one simple zero in the strip of periodicity:

$$\begin{aligned} b=e^{t/2} \sinh (x-u(t)). \end{aligned}$$

We will see that it leads to the PIII equation. Since $b$ satisfies the heat equation (3.15), Proposition 3.2 can be applied. The integrability condition (3.18) with $\frac{b_x}{b}=\coth (x)$ becomes

$$\begin{aligned} 2\sinh ^2 (x-u) \Bigl (V_t(x)-V_t(u)\Bigr )-\sinh (x-u)\cosh (x-u) \Bigl (V'(x)+V'(u)\Bigr )\nonumber \\ +2\Bigl (V(x)-V(u)\Bigr )=0.\nonumber \\ \end{aligned}$$

(5.1)

Let us make the change of variables $V \rightarrow \mathcal{V}$, $x \rightarrow X$, $u \rightarrow U$ such that

$$\begin{aligned} V(x)\equiv e^{-4x}\mathcal{V}(e^{2x}), \quad X=e^{2x}, \quad U=e^{2u}; \end{aligned}$$

then equation (5.1) is rewritten as

$$\begin{aligned}&(X-U)^2 \left( U^2 \mathcal{V}_t (X)-X^2 \mathcal{V}_t (U)\right) -UX(X^2 -U^2)\left( U\mathcal{V}'(X)+ X\mathcal{V}'(U)\right) \nonumber \\&\quad +\, 2(X^2 -U^2)\left( U^2 \mathcal{V}(X)+X^2 \mathcal{V}(U)\right) +4UX \left( U^2 \mathcal{V}(X)-X^2 \mathcal{V}(U)\right) \, =\, 0. \end{aligned}$$

(5.2)

Since the maximal degree of $X$ here equals $4$, the differential operator $\partial _{X}^{5}$ applied to this equation kills all terms containing $\mathcal{V}(U)$, and we are left with

$$\begin{aligned} \partial _{X}^{5}\Bigl [(X-U)^2 \mathcal{V}_t (X)-X(X^2 -U^2) \mathcal{V}'(X)+2(X^2 -U^2+2UX)\mathcal{V}(X)\Bigr ]=0. \end{aligned}$$

Equating the coefficients in front of $U^2$, $U^1$, and $U^0$ to zero, we get the following conditions:

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle { \partial _{X}^{5}\Bigl [\mathcal{V}_t (X)+X\mathcal{V}'(X)-2\mathcal{V}(X)\Bigr ]=0}, \\ \displaystyle { \partial _{X}^{5}\Bigl [-X \mathcal{V}_t (X)+2X \mathcal{V}(X) \Bigr ]=0}, \\ \displaystyle { \partial _{X}^{5}\Bigl [ X^2 \mathcal{V}_t (X)-X^3 \mathcal{V}'(X) +2X^2\mathcal{V}(X) \Bigr ]=0}. \end{array}\right. \end{aligned}$$

They mean that the expressions in the square brackets are polynomials in $X$ of at most fourth degree:

$$\begin{aligned}&\displaystyle { \mathcal{V}_t (X)+X\mathcal{V}'(X)-2\mathcal{V}(X)=P_4(X)}, \nonumber \\&\displaystyle { -X \mathcal{V}_t (X)+2X \mathcal{V}(X) =Q_4(X)}, \nonumber \\&\displaystyle { X^2 \mathcal{V}_t (X)-X^3 \mathcal{V}'(X) +2X^2\mathcal{V}(X) = R_4(X)}. \end{aligned}$$

(5.3)

Combining these conditions, we obtain that $X^2 \mathcal{V}'(X)$ and $X^2 \mathcal{V}_t(X)$ are polynomials of at most $5$-th and $6$-th degrees, respectively. It is easy to see that the former polynomial must be divisible by $X^2$. Indeed, let it be $X^2 \mathcal{V}'(X)=X^2 P_3(X) +p_1 X +p_0$ with some nonzero $p_{0,1}$. Then the first equation in (5.3) implies $p_0=0$ (otherwise the left-hand side contains a nonpolynomial term $\propto X^{-1}$) and the second equation multiplied by $X$ implies $p_1 =0$ (otherwise the left-hand side contains a nonpolynomial term $\propto X^{2}\log X$). Therefore, we conclude that $\mathcal{V}'(X)$ is a polynomial of at most third degree, and, thus, $\mathcal{V}(X)$ itself is a polynomial of at most fourth degree:

$$\begin{aligned} \mathcal{V}(X)=a_4 X^4 +a_3 X^3 +a_2 X^2 +a_1 X +a_0. \end{aligned}$$

Let us plug it in equation (5.2). After simple transformations, we obtain the relation

$$\begin{aligned}&(X-U)(X^2-U^2)(\dot{a}_4 -2a_4)+(X-U)^2 (\dot{a}_3 -a_3) \nonumber \\&\quad \displaystyle { -\,\,\, \frac{(X-U)^2}{UX}\, (\dot{a}_1 -a_1)- \frac{(X-U)(X^2-U^2)}{U^2X^2}\, (\dot{a}_0 -2a_0)\, =\, 0}. \end{aligned}$$

It must be satisfied identically for all $X,U$. This implies $\dot{a}_4 =2a_4$, $\dot{a}_3 =a_3$, $\dot{a}_1 =a_1$, $\dot{a}_0 =2a_0$, and no condition for $a_2$. Therefore, the potential $V(x,t)$ is fixed to be

$$\begin{aligned} V(x,t)=\alpha _1 e^{2t +4x} +\alpha _2 e^{2t -4x} +\alpha _3 e^{t +2x} +\alpha _4 e^{t -2x} +a(t), \end{aligned}$$

where $\alpha _i$ are arbitrary constants. This is precisely the potential for the PIII equation.

5.2 The Case $b=e^{t}\sinh (x-u(t))\sinh (x+u(t))$

In this case, $b=\left( e^{t/2}\sinh (x-u)\right) \left( e^{t/2}\sinh (x+u)\right)$. Each of the multiples satisfies the heat equation (3.24). Therefore, Proposition 3.3 can be applied. Then equation (3.28) assumes the form

$$\begin{aligned}&V_t(x)-V_t(u) -\frac{1}{2}\coth (x-u)\Big ( V'(x)+V'(u)\Big ) -\frac{1}{2}\coth (x+u)\Big ( V'(x)-V'(u) \Big ) \\&\quad +\Big (V(x)-V(u)\Big )\Big (\frac{1}{\sinh ^2(x-u)} +\frac{1}{\sinh ^2(x+u)}\Big )=0. \end{aligned}$$

Multiplying by $32\sinh ^2(x-u)\sinh ^2(x+u)$ and making change of variables $X=\cosh ^2(x)$, $y=\coth ^2(u)$, we get

$$\begin{aligned}&(Xy-X-y)^2(V_t(X)-V_t(y)) -2X(X-1)(y-1)(Xy-X-y)V'(X)\\&\quad +2y(y-1)(Xy-X-y)V'(y)+2(y-1)(Xy+X-y)(V(X)-V(y))=0. \end{aligned}$$

Now one can apply the calculation method similar to the previous cases. That is to take the third derivative with respect to $X$ and analyze the differential equations (the later equations appear as the coefficients behind different powers of $y$). This analysis gives the potential of the Painlevé V equation after some tedious evaluations. Instead of proceeding in this manner, let us simplify the problem by assuming that the solution is a sum of terms of the form $V(x)=e^{kt}v(X)$. Making this substitution, one gets:

$$\begin{aligned}&k(Xy-X-y)^2(v(X)-v(y)) -2X(X-1)(y-1)(Xy-X-y)v'(X)\\&\quad +2y(y-1)(Xy-X-y)v'(y)+2(y-1)(Xy+X-y)(v(X)-v(y))=0. \end{aligned}$$

We will see that nontrivial solutions exist for $k=0,2,4$. The way to proceed is to take the third derivative of the expression with respect to $X$. The equality holds identically if the coefficients in front of $y^2$, $y$ and the free term in $y$ vanish. This implies the following conditions:

$$\begin{aligned} \left\{ \begin{array}{l} X(X-1)v'''(X)+3(2X-1)v''(X)+3(2-k)v'(X)=0,\\ \ \\ X(X-1)v^{(4)}(X)+4(2X-1)v'''(X)+3(4-k)v''(X)=0,\\ \ \\ kv'''(X)=0. \end{array} \right. \end{aligned}$$

(5.4)

Consider the last equation. If $k=0$, one gets

$$\begin{aligned} v(X)=\frac{c_1}{X}+\frac{c_2}{X-1}+c_3=\frac{\tilde{c}_1}{\sinh ^2 x}+\frac{\tilde{c}_2}{\cosh ^2 x}+\tilde{c}_3, \end{aligned}$$

else $v'''(X)=0$. The later case leads to $(k-4)v''(X)=0$ (from the second equation in (5.4)). Then, $k=4$ or $v''(X)=0$. In the latter case, one gets $(k-2)v'(X)=0$ (from the first equation in (2.4)). In this way, one can easily recover the potential of the Painlevé V equation (2.4):

$$\begin{aligned} V(x,t)=-\frac{2(\xi +\sigma )^2}{\sinh ^2 x} +\frac{2\zeta ^2}{\cosh ^2 x}+ \frac{e^{2t}}{2}(2\sigma -1)\cosh (2x )- \frac{e^{4t}}{16}\cosh (4x ). \end{aligned}$$

6 Elliptic Solutions

6.1 The Case $b=\vartheta _1(x-u(t),2\pi i t)$

Consider an elliptic curve with moduli $\tau =2\pi i t$,

$$\begin{aligned} \Sigma _\tau :\ \ {{\mathbb {C}}}/{{\mathbb {Z}}}+{{\mathbb {Z}}}\tau , \end{aligned}$$

and let $b=\vartheta _1(x-u(t),2\pi i t)$. Definitions and properties of elliptic functions are given in Appendix B. Then from (3.18), we have

$$\begin{aligned} V_t(x)-V_t(u)-\frac{1}{2}E_1(x-u)(V'(u)+V'(x)) +E_2(x-u)\left( V(x)-V(u)\right) =0. \end{aligned}$$

We will show that this equation has only trivial solutions $V(x,t)=f(t)$. For this purpose, consider the same equation at $x+\tau$ and subtract it from the initial one. Then, using the behavior of $E_1(z)$ (B.1) and $E_2(z)$ (B.2) on the torus lattice, we get

$$\begin{aligned}&V_t(x+\tau )-V_t(x)-\frac{1}{2}E_1(x-u)(V'(x+\tau )-V'(x))\nonumber \\&\quad +E_2(x-u)(V(x+\tau )-V(x)) +\pi i (V'(u)+V'(x+\tau ))=0. \end{aligned}$$

(6.1)

Let us now differentiate the obtained equality with respect to $x$:

$$\begin{aligned}&V'_t(x+\tau )-V'_t(x)-\frac{1}{2}E_1(x-u)(V''(x+\tau )-V''(x))+\pi i V''(x+\tau )\nonumber \\&\quad +\frac{3}{2}E_2(x-u)(V'(x+\tau )-V'(x)) +E'_2(x-u)(V(x+\tau )-V(x))=0.\qquad \quad \end{aligned}$$

(6.2)

Similarly, let us shift the argument $u\rightarrow u+\tau$ in the equation (6.2) and subtract it from (6.2) itself (keeping in mind that $E_2'$ is the double-periodic function). This gives

$$\begin{aligned} V''(x+\tau )-V''(x)=0 \end{aligned}$$

or

$$\begin{aligned} V(x+\tau )-V(x)=a(\tau )x+b(\tau ). \end{aligned}$$

Plugging this back into (6.1), one can easily get that $a(\tau )=b(\tau )=0$ by analyzing coefficients behind the poles at $x-u$ of the second and the first orders. Therefore, the potential should be a double-periodic function. If it is, then (6.1) reduces to

$$\begin{aligned} \partial _t(V(x+\tau )-V(x)) +\pi i (V'(u)-V'(x))=0, \end{aligned}$$

since $\partial _t(V(x+\tau )-V(x))=V_t(x+\tau )-V_t(x)+2\pi i V'(x+\tau )$. The latter equation should hold for all $x$ and $u$. Then the only solution is

$$\begin{aligned} V(x,t)=f(t). \end{aligned}$$

6.2 The Case $b=\vartheta _1(x-u(t),2\pi i t)\vartheta _1(x+u(t),2\pi i t)$

Equation (3.28) in this case has the form

$$\begin{aligned}&V_t(x)-V_t(u) -\frac{1}{2}E_1(x\!-\!u)(V'(u)+V'(x))-\frac{1}{2}E_1(x\!+\!u)(-V'(u)+V'(x)) \nonumber \\&\quad +(E_2(x\!-\!u)+E_2(x\!+\!u))(V(x)-V(u))=0. \end{aligned}$$

(6.3)

Let us make a change of variables:

$$\begin{aligned} X(x,t)=\frac{\wp (x)-e_1}{e_2-e_1},\ \ Q(u,t)=\frac{\wp (u)-e_1}{e_2-e_1},\ \ T(t)=\frac{e_3-e_1}{e_2-e_1}. \end{aligned}$$

Then

$$\begin{aligned}&E_1(x+u)+E_1(x-u)=2E_1(x)+\frac{\wp '(x)}{\wp (x)-\wp (u)} =2E_1(x)+\frac{X_x}{X-Q},\\&E_1(x+u)-E_1(x-u)=2E_1(u)+\frac{\wp '(u)}{\wp (u)-\wp (x)} =2E_1(x)-\frac{Q_u}{X-Q},\\&E_2(x+u)+E_2(x-u)=2E_2(u)+\frac{Q_{uu}}{X-Q}+\frac{Q_u^2}{(X-Q)^2}. \end{aligned}$$

Therefore, equation (6.3) is written as

$$\begin{aligned}&(V_T(X)-V_T(Q))T_t+V_X(X)X_t-V_Q(Q)Q_t\\&\quad -\frac{1}{2}V_X(X)X_x\left( 2E_1(x)+\frac{X_x}{X-Q}\right) +\frac{1}{2}V_Q(Q)Q_u\left( 2E_1(u)-\frac{Q_u}{X-Q}\right) \\&\quad +\left( 2\left[ 2\eta _1+e_1+(e_2-e_1)Q\right] + \frac{Q_{uu}}{X-Q}+\frac{Q_u^2}{(X-Q)^2}\right) (V(x)-V(u))=0. \end{aligned}$$

It follows from (B.8) that

$$\begin{aligned} X_t-X_xE_1(x)= & {} X_x(E_1(x+\omega _3)\!-\!E_1(x)\!-\!E_1(\omega _3))=\frac{1}{2}X_x\frac{\wp '(x)}{\wp (x)-\wp (\omega _3)}\\ {}= & {} \frac{1}{2}\frac{X_x^2}{X-T}. \end{aligned}$$

Therefore,

$$\begin{aligned}&(V_T(X)-V_T(Q))T_t+V_X(X)\frac{1}{2}\frac{X_x^2}{X-T} -V_Q(Q)\frac{1}{2}\frac{Q_u^2}{Q-T}\nonumber \\&\quad -\frac{1}{2}V_X(X)\frac{X_x^2}{X-Q} -\frac{1}{2}V_Q(Q)\frac{Q_u^2}{X-Q} \nonumber \\&\quad +\left( 2\left[ 2\eta _1+e_1+(e_2-e_1)Q\right] + \frac{Q_{uu}}{X-Q}+\frac{Q_u^2}{(X-Q)^2}\right) (V(x)-V(u))=0. \end{aligned}$$

(6.4)

Now let us proceed as in the previous examples. First, multiply (6.4) by $(X-Q)^2$. Secondly, take the third derivative with respect to $X$. This excludes $V(Q)$. Thirdly, substitute $Q_u^2=4(e_2-e_1)Q(Q-1)(Q-T)$ and $Q_{uu}=2(e_2-e_1)(3Q^2-2Q(T+1)+T)$. Then, the coefficients in front of $Q^2$, $Q^1$, and $Q^0$ should vanish independently:

$$\begin{aligned} \left\{ \begin{array}{l} F+2V(X)(2\eta _1+e_1+X(e_2-e_1))=P_2(X),\\ {} -2XF+\frac{1}{2}V_X(X)X_x^2\\ {} +2V(X)(X^2(e_2-e_1)+4X(e_1-\eta _1)+T(e_2-e_1))=Q_2(X), \\ {} X^2 F-\frac{1}{2}V_X(X)X_x^2X+2V(X)((2\eta _1+e_1)X^2+(e_2-e_1)XT)=R_2(X), \end{array}\right. \end{aligned}$$

(6.5)

where $P_2(X)$, $Q_2(X)$, $R_2(X)$ are the second-order polynomials in $X$ with time-dependent coefficients and $F=V_T(X)T_t+V_X(X)\frac{1}{2}\frac{X_x^2}{X-T}$.

Excluding $F$ from the two upper equations in (6.5), we obtain the following equality:

$$\begin{aligned}&V_X(X)X(X\!-\!1)(X\!-\!T)+V(X)\Big (X(X\!-\!1)+X(X\!-\!T)+(X\!-\!1)(X\!-\!T)\Big )\\&\quad =\frac{1}{e_2-e_1}\Big (\frac{1}{2}Q_2(X)+XP_2(X)\Big ). \end{aligned}$$

General solution of the latter equation has a form:

$$\begin{aligned} V(X)\!=\!\frac{1}{X(X\!-\!1)(X\!-\!T)}\int ^XdZ\frac{1}{e_2\!-\!e_1}\Big (\frac{1}{2}Q_2(Z)\!+\!ZP_2(Z)\Big )\!=\! \frac{H_4(X)}{X(X\!-\!1)(X\!-\!T)}, \end{aligned}$$

where $H_4(X)$ are the forth-order polynomials in $X$ with time-dependent coefficients. Therefore, $V(X)$ can be presented as

$$\begin{aligned} V(X)=a(T)X+\frac{b(T)}{X}+\frac{c(T)}{X-1}+\frac{d(T)}{X-T}+h(T). \end{aligned}$$

(6.6)

The last term $h(T)$ is not fixed by (6.3); i.e., $h(T)$ is arbitrary.

Plugging (6.6) into (6.4) and multiplying the result by $(X-Q)X(X-1)(X-T)Q(Q-1)(Q-T)$, we get a polynomial function in $X$ and $Q$. The coefficients in front of $Q^kX^j$ provides differential equations. It can be verified that all of them are equivalent to the following system:

$$\begin{aligned} \left\{ \begin{array}{l} a_T(T) T(T-1)(e_2-e_1)+a(T)(e_3+2\eta _1)=0,\\ b_T(T) T(T-1)(e_2-e_1)+b(T)(e_2+2\eta _1)=0,\\ c_T(T) T(T-1)(e_2-e_1)+c(T)(e_1+2\eta _1)=0,\\ d_T(T) T(T-1)(e_2-e_1)+d(T)(-2e_3+2\eta _1)=0. \end{array}\right. \end{aligned}$$

Its solutions (see (B.6)–(B.7)) are

$$\begin{aligned} \left\{ \begin{array}{l} a(T)=\alpha (e_2-e_1), \ \ \alpha =\hbox {const},\\ b(T)=\beta (e_2-e_1)T, \ \ \beta =\hbox {const},\\ c(T)=\gamma (e_2-e_1)(T-1), \ \ \gamma =\hbox {const},\\ d(T)=\delta (e_2-e_1)T(T-1), \ \ \delta =\hbox {const}. \end{array}\right. \end{aligned}$$

(6.7)

Then, in view of (B.5), we have

$$\begin{aligned} V(x)=\alpha \wp (x)+\beta \wp (x+\omega _1)+\gamma \wp (x+\omega _2) +\delta \wp (x+\omega _3)+h(t). \end{aligned}$$

(6.8)

This is the potential of the Painlevé VI equation in the elliptic form [18, 36, 53, 60] (see also [32, 50] and [9]). We remark that the nonstationary Lamé equation in connection with the PVI equation (and with the 8-vertex model) was discussed in [4]. Recently, the nonstationary Lamé equation has appeared [10, 37–40] in the context of the AGT conjecture. The results of [38–40] allow one in principle to construct higher Painlevé equations⁴ in terms of 2x2 linear problems related to spin chains via spectral duality transformation. We are going to study this possibility in our future publications.

Appendices

Appendix A: Special Cases

1.1 $b=(x-u(t))e^{g(t)x}$ and $b=(x-u(t))e^{g(t)x^2}$

Let $b=(x-u(t))e^{g(t)x}$. The calculation similar to the one leading to (4.2) gives in this case

$$\begin{aligned} V_t(x)-V_t(u)- \frac{V'(x)+V'(u)}{2(x-u)}+ \frac{V(x)\! -\! V(u)}{(x-u)^2}-\frac{\ddot{g}}{2}\, (x-u) -\frac{g}{2}\left( V'(x)\! -\! V'(u)\right) =0 \end{aligned}$$

(A.1)

and

$$\begin{aligned} H=\frac{1}{2}\left( {\dot{u}}+\frac{g}{2}\right) ^2+V(u,t)-\frac{1}{2}u{\dot{g}}+\frac{1}{8}g^2, \end{aligned}$$

(A.2)

with equation of motion

$$\begin{aligned} {\ddot{u}}=-V'(u). \end{aligned}$$

It is easy to see that equation (A.1) becomes equivalent to (4.2) for the potential $\tilde{V}(\tilde{x})$ after the change of variables

$$\begin{aligned} x \rightarrow \tilde{x} = x-\frac{1}{2}\, G(t) , \quad \quad V(x)\rightarrow \tilde{V}(x)=V\Bigl (x-\frac{1}{2}\, G(t)\Bigr ) - \frac{\dot{g}}{2}\, x, \end{aligned}$$

where $\dot{G}=g$. Notice also that the dependence $H({\dot{u}})$ in (A.2) can be obtained from (3.10) via the local expansion (3.8). The later gives $b_1=e^{ug}$ and $b_2=g\, e^{ug}$. Then $v={\dot{u}}+\frac{g}{2}$.

Consider now the case $b=(x-u(t))e^{g(t)x^2}$. Let us perform the calculation similar to the one leading to (4.2) again. In this case, we have:

$$\begin{aligned}&V_t(x)-V_t(u)-\frac{V'(x)+V'(u)}{2(x-u)}+ \frac{V(x)\! -\! V(u)}{(x-u)^2}-2{g}\left( V(x)\! -\! V(u)\right) \nonumber \\&\quad -g\Big (x V'(x)+u V'(u)\Big ) +(x^2-u^2)\left[ 3g{\dot{g}}-\frac{1}{2}\,{\ddot{g}}-2g^3 \right] =0 \end{aligned}$$

(A.3)

and

$$\begin{aligned} H=\frac{1}{2}\left( {\dot{u}}+{g}u\right) ^2+V(u,t)+\frac{1}{2}(g^2-{\dot{g}})u^2+\frac{3}{2}g, \end{aligned}$$

(A.4)

with equation of motion

$$\begin{aligned} {\ddot{u}}=-V'(u). \end{aligned}$$

As in the previous example, it can be shown that equation (A.3) becomes equivalent to (4.2) for the potential $\tilde{V}(\tilde{x})$ after the following change of variables:

$$\begin{aligned} x \rightarrow \tilde{x}= & {} \alpha x=x\,e^{\int _t g(t)},\ \alpha =e^{\int _t g(t)},\\ V(x)\rightarrow \tilde{V}(x)= & {} \alpha ^2\Bigl ( V(\alpha x)- x^2\left( g^2-\int _t \frac{{\ddot{g}} -2g{\dot{g}}}{2\alpha ^2} \right) \Bigr )\\= & {} e^{2\int _t g(t)}\left( V(x\,e^{\int _t g(t)})- x^2\left( g^2-\int _t\left[ e^{-2\int _t g(t)}\left( \frac{1}{2}{\ddot{g}} -g{\dot{g}}\right) \right] \right) \right) . \end{aligned}$$

Notice also that the dependence $H({\dot{u}})$ in (A.4) can be obtained from (3.10) via the local expansion (3.8). The latter gives $b_1=e^{g u^2}$ and $b_2=2gu\, e^{g u^2}$. Then $v={\dot{u}}+gu$.

1.2 $b=(x-u_1(t))(x-u_2(t))(x-u_3(t))$

When $b=(x-u_1)(x-u_2)(x-u_3)$, the coefficients behind the second-order pole $\frac{1}{(x-u_1)^2}$ in (3.6) have the following form:

$$\begin{aligned}&V(x,t)-H+\frac{1}{2}{\dot{u}}_1^2+\frac{1}{2}\frac{{\dot{u}}_1+{\dot{u}}_2}{{ u}_1-{ u}_2}+\frac{1}{2}\frac{{\dot{u}}_1+{\dot{u}}_3}{{u}_1-{u}_3} -\frac{1}{2}\frac{1}{({u}_1-{ u}_2)^2}-\frac{1}{2}\frac{1}{({ u}_1-{ u}_3)^2}\\&\quad +\frac{1}{2}\frac{1}{({u}_1-{u}_2)({u}_1-{u}_3)}, \end{aligned}$$

and two other coefficients can be obtained by the cyclic permutations. All three coefficients cannot vanish simultaneously. Therefore, some other anzats for $W$ (3.2) should be used in this case. This reflects the fact that (3.1)–(3.2) imply the one degree of freedom case.

1.3 $b=(x-u_1(t))^{\gamma }$ and $b=(x-u_1(t))^{\gamma _1}(x-u_2(t))^{\gamma _2}$

Let us study the case $b=(x-u_1(t))^{\gamma }$, where $\gamma \in {\mathbb {C}}^*$ (the case $\gamma =0$ is trivial). Notice that under change $b\rightarrow b^\gamma$ the functions $f$ (3.4) and $S$ (3.5) transform as follows:

$$\begin{aligned} f= & {} \frac{b_x}{b}\longrightarrow \gamma \frac{b_x}{b},\\&S\longrightarrow V-H-\frac{1}{2}\gamma \frac{b_t}{b}+\frac{1}{2}\gamma \frac{b_{xx}}{b}+ \frac{1}{2}\left( \frac{1}{4}\gamma ^2-\gamma \right) \left( \frac{b_x}{b}\right) ^2. \end{aligned}$$

For the case under consideration, we have $f={\gamma }\frac{1}{x-u}$ and

$$\begin{aligned} S=V-H+\frac{\gamma }{2}{\dot{u}}\frac{1}{x-u}+\frac{1}{2}\left( \frac{1}{4}\gamma ^2-\gamma \right) \frac{1}{(x-u)^2}. \end{aligned}$$

Substituting it into (3.6), we obtain the following condition for cancellation of the fourth- and the third-order poles:

$$\begin{aligned}&(x-u)^{-4}:\ \ 0=\frac{1}{4}\gamma (\gamma ^2-4\gamma +3),\\&(x-u)^{-3}:\ \ {\dot{u}}\left( \frac{1}{4}\gamma ^2-\gamma \right) =-\frac{3}{4}\gamma ^2 {\dot{u}}. \end{aligned}$$

The first equation gives $\gamma =\{0,1,3\}$, while the second one $\gamma =\{0,1\}$. Therefore, the nontrivial solution is

$$\begin{aligned} \gamma =1. \end{aligned}$$

Similarly, the case $b=(x-u_1(t))^{\gamma _1}(x-u_2(t))^{\gamma _2}$ leads to the following conditions:

$$\begin{aligned} \begin{array}{l} (x-u_1)^{-4}: \frac{1}{4}\gamma _1(\gamma _1^2-4\gamma _1+3)=0,\\ (x-u_1)^{-3}:\frac{1}{2}\frac{\gamma _1(\gamma _1-1)}{u_1-u_2}\left( 2{\dot{u}}_1(u_1-u_2)+\gamma _2\right) , \\ \ \\ (x-u_2)^{-4}: \frac{1}{4}\gamma _2(\gamma _2^2-4\gamma _2+3)=0,\\ (x-u_2)^{-3}: \frac{1}{2}\frac{\gamma _2(\gamma _2-1)}{u_2-u_1}\left( 2{\dot{u}}_2(u_2-u_1)+\gamma _1\right) , \end{array} \end{aligned}$$

which give

$$\begin{aligned} \gamma _1=\gamma _2=1. \end{aligned}$$

1.4 $b=\exp \Big (\left( z/u(t)\right) ^\gamma \Big )$

First, it can be shown that $\gamma =0,1,2,3$...

Consider $\gamma =1$. Substituting $b(z,u(t),t)=\exp (z/u(t))$ into (3.6), we get

$$\begin{aligned} \left( -\frac{{\dot{u}}^2}{u^3}+\frac{1}{2}\frac{\ddot{u}}{u^2} \right) x+V_t-H_t-\frac{1}{2}\frac{{\dot{u}}}{u^3}-\frac{1}{2u}{V'}_x=0. \end{aligned}$$

(A.5)

Applying $\partial _x^2$ gives

$$\begin{aligned} V''_t-\frac{1}{2u}V'''=0. \end{aligned}$$

Notice that the function ${U}(z,{\dot{u}},{ u},t)$ satisfies the same equation even if we do not impose the condition ${U}=V(x,t)-H({\dot{u}},{ u},t)$. Under assumption ${U}=V(x,t)-H({\dot{u}},{ u},t)$, we have

$$\begin{aligned} V'''=V''_t=0. \end{aligned}$$

This leads to

$$\begin{aligned} V(x,t)=\frac{\alpha }{2}x^2+b(t)x+c(t),\ \ \alpha =\hbox {const}. \end{aligned}$$

Plugging it back into (A.5), we obtain the following two equations (as coefficients behind $x^1$ and $x^0$):

$$\begin{aligned} \left\{ \begin{array}{l} {\ddot{u}}=2\frac{{\dot{u}}^2}{u}-2{\dot{b}}u^2+\alpha u,\\ H_t=\frac{1}{2}\frac{{\dot{u}}^2}{u^3}+{\dot{c}}-\frac{1}{2u}b. \end{array}\right. \end{aligned}$$

1.5 Case 2 in (4.5)

Here it may be useful to use variable $u=u_1-\frac{1}{2}\sqrt{c-4t}$ (then $\dot{u}=\dot{u}_1+\frac{1}{\sqrt{c-4t}}$). Then

$$\begin{aligned} H=\frac{1}{2} \left( \dot{u}_1+\frac{1}{\sqrt{c-4t}}\right) ^2+V(u_1)=\frac{1}{2}\dot{u}^2+V\left( u+\frac{1}{2}\sqrt{c-4t}\right) , \end{aligned}$$

(A.6)

and, therefore,

$$\begin{aligned}&V_t(x)-H_t -\frac{1}{2(x-u_1)}\Big (V'(x)-\ddot{u}_1-2(c-4t)^{-\frac{3}{2}}-2\frac{V(x)-V(u_1)}{x-u_1}\Big ) \\&\quad -\frac{1}{2(x-u_2)}\Big (V'(x)-\ddot{u}_2+2(c-4t)^{-\frac{3}{2}}-2\frac{V(x)-V(u_2)}{x-u_2}\Big )=0 \end{aligned}$$

Cancellation of the first-order poles at $x=u_{1,2}$ yields $\ddot{u}_1=-V'(u_1)-2(c-4t)^{-\frac{3}{2}}$. On this equation, $H_t=V_t(u_1)-V'(u_1)\frac{1}{\sqrt{c-4t}}$. Thus we arrive at

$$\begin{aligned}&V_t(x)-V_t(u_1)+V'(u_1)\frac{1}{\sqrt{c-4t}}-\frac{1}{2(x-u_1)} \Big (V'(x)+V'(u_1)-2\frac{V(x)-V(u_1)}{x-u_1}\Big )\nonumber \\&\quad -\frac{1}{2(x-u_2)}\Big (V'(x)+V'(u_1)-2\frac{V(x)-V(u_2)}{x-u_2}\Big )=0. \end{aligned}$$

(A.7)

By analogy with (4.7), we get

$$\begin{aligned} \left\{ \begin{array}{l} V_{t}^\mathrm{V}(x)=0, \\ -20V_{t}^\mathrm{IV}(x)+V^\mathrm{VI}(x)=0, \\ -13V^\mathrm{V}(x)+120V_{t}^\mathrm{III}(x)=0,\\ V^\mathrm{IV}(x)-6V_{t}^\mathrm{II}(x)=0,\\ 6\partial _z V_t(x)-V^\mathrm{III}(x)+\Big (-\frac{16}{13}t+\frac{4}{13}\Big )V_{t}^\mathrm{III}(x)=0. \end{array}\right. \end{aligned}$$

(A.8)

From the two upper equations, it follows that $V(x)$ is the 6-th degree polynomial. Plugging it into (A.8) drops the degree to 4 (similar to the Painlevé I, II cases). However, after substituting it back into (A.7), we get only the trivial solution

$$\begin{aligned} V(x,t)=f(t). \end{aligned}$$

Appendix B: Elliptic Functions

Here we give a short version of the Appendix in [58].

1.1 Theta-functions

The Jacobi’s theta-functions $\vartheta _a (z)= \vartheta _a (z|\tau )$, $a=0,1,2,3$, are defined by the formulas

$$\begin{aligned} \vartheta _1(z)= & {} -\displaystyle {\sum _{k\in {{\mathbb Z}}}} \exp \left( \pi i \tau \left( k+\frac{1}{2}\right) ^2 +2\pi i \left( z+\frac{1}{2}\right) \left( k+\frac{1}{2}\right) \right) , \\ \vartheta _2(z)= & {} \displaystyle {\sum _{k\in {{\mathbb Z}}}} \exp \left( \pi i \tau \left( k+\frac{1}{2}\right) ^2 +2\pi i z\left( k+\frac{1}{2}\right) \right) , \\ \vartheta _3(z)= & {} \displaystyle {\sum _{k\in {{\mathbb Z}}}} \exp \left( \pi i \tau k^2 +2\pi i z k \right) , \\ \vartheta _0(z)= & {} \displaystyle {\sum _{k\in {{\mathbb Z}}}} \exp \left( \pi i \tau k^2 +2\pi i \left( z+\frac{1}{2}\right) k\right) , \end{aligned}$$

where $\tau$ is a complex parameter (the modular parameter) such that $\mathrm{Im}\, \tau >0$. Set

$$\begin{aligned} \omega _0 =0, \quad \omega _1 =\frac{1}{2}, \quad \omega _2=\frac{1+\tau }{2}, \quad \omega _3 =\frac{\tau }{2}; \end{aligned}$$

then the function $\vartheta _a(z)$ has simple zeros at the points of the lattice $\omega _{a-1}+{\mathbb {Z}}+{\mathbb {Z}}\tau$ (here $\omega _a \equiv \omega _{a+4}$).

1.2 Weierstrass $\wp$-function

The Weierstrass $\wp$-function is defined as

$$\begin{aligned} \wp (z)= -\partial _z^2 \log \vartheta _1 (z)-2\eta , \end{aligned}$$

where

$$\begin{aligned} \eta = -\, \frac{1}{6}\, \frac{\vartheta _{1}^{'''}(0)}{\vartheta _1' (0)}= -\, \frac{2\pi i}{3}\, \partial _{\tau } \log \theta _1'(0|\tau ). \end{aligned}$$

Its derivative is given by

$$\begin{aligned} \wp '(z)=-\, \frac{2\, (\vartheta _1'(0))^3}{\vartheta _2(0)\vartheta _3(0) \vartheta _0(0)}\, \frac{\vartheta _2(z)\vartheta _3(z) \vartheta _0(z)}{\vartheta _{1}^{3}(z)}. \end{aligned}$$

The values at the half-periods

$$\begin{aligned} e_1 =\wp (\omega _1), \quad e_2 =\wp (\omega _2), \quad e_3 =\wp (\omega _3) \end{aligned}$$

have special properties. For example, $e_1 +e_2 +e_3=0$. The differences $e_j-e_k$ can be represented in two different ways:

$$\begin{aligned}&\displaystyle {e_1 -e_2 =\pi ^2 \vartheta _0^4 (0)\, =\, 4\pi i \, \partial _{\tau } \log \frac{\vartheta _3 (0)}{\vartheta _2 (0)}},\\&\displaystyle {e_1 -e_3 =\pi ^2 \vartheta _3^4 (0)\, =\, 4\pi i \, \partial _{\tau } \log \frac{\vartheta _0 (0)}{\vartheta _2 (0)}},\\&\displaystyle {e_2 -e_3 =\pi ^2 \vartheta _2^4 (0)\, =\, 4\pi i \, \partial _{\tau } \log \frac{\vartheta _0 (0)}{\vartheta _3 (0)}}. \end{aligned}$$

The second representation is a consequence of the heat equation (B.3) (see below):

$$\begin{aligned} e_k = 4\pi i \, \partial _{\tau }\Bigl ( \frac{1}{3}\, \log \vartheta _1'(0) -\log \vartheta _{k+1}(0)\Bigr ) \end{aligned}$$

or

$$\begin{aligned} \pi i \, \partial _{\tau }\log (e_j -e_k)=-e_l -2\eta , \end{aligned}$$

where $\{jkl\}$—any cyclic permutation of $\{123\}$. The $\wp$-function satisfies the differential equation

$$\begin{aligned} (\wp '(z))^2 =4 (\wp (z)-e_1)(\wp (z)-e_2)(\wp (z)-e_3). \end{aligned}$$

We also mention the formulae

$$\begin{aligned} \wp (z)-e_k =\frac{(\vartheta _1'(0))^2}{\vartheta _{k+1}^2(0)}\, \frac{\vartheta _{k+1}^2(z)}{\vartheta _1^2(z)}. \end{aligned}$$

1.3 Eisenstein functions and $\Phi$-function

By definition,

$$\begin{aligned} E_1(z)=\partial _z\log \vartheta _1(z), \quad \quad E_2(z)=-\partial _zE_1(z)= -\partial _z^2\log \vartheta _1(z)=\wp (z)+2\eta . \end{aligned}$$

Behavior on the lattice:

$$\begin{aligned} E_1(z+1)= & {} E_1(z),\ \ E_1(z+\tau )=E_1(z)-2\pi i,\end{aligned}$$

(B.1)

$$\begin{aligned} E_2(z+1)= & {} E_2(z),\ \ E_2(z+\tau )=E_2(z). \end{aligned}$$

(B.2)

The local expansion near $z=0$:

$$\begin{aligned} E_1(z)=\frac{1}{z}-2\eta z + \cdots ,\quad \quad ~~E_2(z) =\frac{1}{z^2}+2\eta +\cdots \end{aligned}$$

Values at half-periods:

$$\begin{aligned} E_1(\omega _j)=-2\pi i\partial _\tau \omega _j, \end{aligned}$$

and, therefore,

$$\begin{aligned} E_1(\omega _j)+E_1(\omega _k)=E_1(\omega _j+\omega _k) \end{aligned}$$

holds true for any different $j,k =1,2,3$.

Another useful function is

$$\begin{aligned} \Phi (u,z)= \frac{\vartheta _1(u+z)\vartheta _1'(0)}{\vartheta _1(u)\vartheta _1(z)}. \end{aligned}$$

It has the following properties:

$$\begin{aligned} \Phi (u,z)= & {} \Phi (z,u),\\ \Phi (-u,-z)= & {} -\Phi (u,z),\\ \Phi (u,z)\Phi (-u,z)= & {} \wp (z)-\wp (u), \\ \Phi (u,z)\Phi (w,z)= & {} \Phi (u+w,z)(E_1(z)+E_1(u)+E_1(w)-E_1(z+u+w)),\\ \Phi (u,z)= & {} \frac{1}{z}+E_1(u)+\frac{z}{2}(E_1^2(u)-\wp (u))+ O(z^2),\\ \partial _z \Phi (u,z)= & {} \Phi (u,z)(E_1(u+z)-E_1(z)). \end{aligned}$$

Behavior on the lattice:

$$\begin{aligned} \Phi (u,z+1)=\Phi (u,z),~~~\Phi (u,z+\tau )=e^{-2\pi i u}\Phi (u,z). \end{aligned}$$

Is is also convenient to introduce

$$\begin{aligned} \varphi _j(z)=e^{2\pi i z\partial _\tau \omega _j}\Phi (z,\omega _j), \quad j=1,2,3, \end{aligned}$$

with properties:

$$\begin{aligned}&\varphi _j^2(z)=\wp (z)-e_j,\ \ \ \varphi _j^2(z)-\varphi _k^2(z)=e_k-e_j, \\&\varphi _j(z)\varphi _k(z)=\varphi _l(z)(E_1(z)+E_1(\omega _l)-E_1(z+\omega _l)),\\&\partial _z\varphi _j(z)=\varphi _j (z)\Bigl [ E_1(z+\omega _j) -E_1(\omega _j)-E_1(z)\Bigr ]=-\varphi _k(z)\varphi _l(z), \end{aligned}$$

where $j,k,l$ is any cyclic permutation of $1,2,3$.

1.4 Heat equation and related formulae

All the theta-functions satisfy the “heat equation”

$$\begin{aligned} 4\pi i \partial _{\tau }\vartheta _a(z|\tau )= \partial _{z}^{2}\vartheta _a(z|\tau ) \end{aligned}$$

(B.3)

or

$$\begin{aligned} 2\partial _{t}\vartheta _a(z )=\partial _{z}^{2}\vartheta _a(z)\ \ \ \displaystyle {t=\frac{\tau }{2\pi i}}. \end{aligned}$$

One can also introduce the “heat coefficient” $\displaystyle {\kappa =\frac{1}{2\pi i}}$ and rewrite the heat equation in the form $\displaystyle {\partial _{\tau }\vartheta _a(z|\tau )= \frac{\kappa }{2}\, \partial _{z}^{2}\vartheta _a(z|\tau )}$. All formulas for derivatives of elliptic functions with respect to the modular parameter are based on the heat equation.

The $\tau$-derivatives are given by the following:

Proposition B.1

The identities

$$\begin{aligned} \partial _\tau \Phi (z,u)=\kappa \partial _z\partial _u\Phi (z,u), \end{aligned}$$

(B.4)

$$\begin{aligned} \partial _\tau E_1(z)= & {} \frac{\kappa }{2}\, \partial _z(E_1^2(z)-\wp (z)),\\ \partial _\tau E_2(z)= & {} \kappa E_1(z)E_2'(z)-\kappa E_2^2(z)+\frac{\kappa }{2}\, \wp ''(z), \end{aligned}$$

with the “heat coefficient” $\displaystyle {\kappa =\frac{1}{2\pi i}}$, hold true.⁵

The proof can be found in [58].

Introduce now

$$\begin{aligned} X(x,t)=\frac{\wp (x)-e_1}{e_2 -e_1}, \quad \quad T(t)=\frac{e_3 -e_1}{e_2 -e_1}=\left( \frac{\vartheta _3(0|\tau )}{\vartheta _0 (0|\tau )}\right) ^4. \end{aligned}$$

Then we have

$$\begin{aligned} X=\frac{\wp (x)-e_1}{e_2 -e_1}, \quad \quad X\! - \! 1 =\frac{\wp (x)-e_2}{e_2 -e_1}, \quad \quad X\! - \! T =\frac{\wp (x)-e_3}{e_2 -e_1}, \end{aligned}$$

and, therefore,

$$\begin{aligned} \left( \frac{\partial X}{\partial x}\right) ^2= & {} 4(e_2 -e_1)\, X(X-1)(X-T),\\ \frac{\partial ^2 X}{\partial x^2}= & {} 2(e_2 -e_1)\, X(X-1)(X-T)\left( \frac{1}{X}+ \frac{1}{X-1}+\frac{1}{X-T}\right) . \end{aligned}$$

Let us give some more relations:

$$\begin{aligned} \displaystyle {\frac{(e_2 -e_1)T}{X}}= & {} \wp (x+\omega _1) -e_1,\nonumber \\ \displaystyle {-\, \frac{(e_2 -e_1)(T-1)}{X-1}}= & {} \wp (x+\omega _2) -e_2,\nonumber \\ \displaystyle {\frac{(e_2 -e_1)T(T-1)}{X-T}}= & {} \wp (x+\omega _3) -e_3,\end{aligned}$$

(B.5)

$$\begin{aligned} \frac{\partial T}{\partial t}= & {} 2(e_2 -e_1) T(T-1), \end{aligned}$$

(B.6)

$$\begin{aligned} \partial _T(e_2-e_1)= & {} \partial _t(e_2-e_1)\frac{1}{T_t}=-\frac{e_3+2\eta _1}{T(T-1)}. \end{aligned}$$

(B.7)

The following identity holds true⁶:

$$\begin{aligned} \frac{\partial X}{\partial t}=\frac{\partial X}{\partial x}\, \, \frac{\vartheta _0'(x)}{\vartheta _0(x)} \end{aligned}$$

or

$$\begin{aligned} \partial _\tau X =\kappa \partial _z X\left( E_1(z+\omega _3)-E_1(\omega _3)\right) =\kappa \, \partial _z X\, \partial _z \log \theta _0(z). \end{aligned}$$

(B.8)

Appendix C: $\mathbf{U}{-}\mathbf{V}$ pairs for PI–PV

Here we list the $\mathbf{U}{-}\mathbf{V}$ pairs for PI–PV satisfying zero curvature equation (1.2) and admitting the quantum Painlevé–Calogero correspondence. The PVI case is too complicated. In principle, it is gauge equivalent to different types of known elliptic $2\times 2$ $\mathbf{U}{-}\mathbf{V}$ pairs (see [32, 60]) which are in their turn related by Hecke transformations [33, 34].

Painlevé I

$$\begin{aligned} 4\ddot{u}= & {} 6u^2 +t, \\ H_\mathrm{I}(p, u)= & {} \frac{p^2}{2} -\frac{u^3}{2} -\frac{tu}{4}, \\ \mathbf{U}(x,t)= & {} \left( \begin{array}{cc} \dot{u} &{} x-u\\ &{}\\ x^2 \! +\! xu \! +\! u^2 \! +\! \frac{1}{2}\,t &{} -\dot{u} \end{array}\right) ,\quad \quad \mathbf{V} (x,t) =\left( \begin{array}{cc} 0&{} \frac{1}{2}\\ &{}\\ \frac{1}{2}\, x +u &{} 0 \end{array}\right) . \end{aligned}$$

Painlevé II

$$\begin{aligned} \ddot{u}= & {} 2u^3 +tu-\alpha , \\ H_\mathrm{II}(p, u)= & {} \displaystyle {\frac{p^2}{2} -\frac{1}{2} \left( u^2+\frac{t}{2}\right) ^2 +\alpha u},\\ \mathbf{U}= & {} \left( \begin{array}{cc} x^2+\dot{u} -u^2 &{} x-u\\ (x+u)(2u^2 \! -\! 2\dot{u} \! +\! t)\! -\! 2\alpha \! - \! 1 &{} -x^2\! -\! \dot{u} \! +\! u^2 \end{array}\right) ,\\ \mathbf{V}= & {} \left( \begin{array}{cc} \frac{x+u}{2}&{} \frac{1}{2}\\ u^2 \! - \! \dot{u} \! +\! \frac{t}{2} &{} -\, \frac{x+u}{2}\end{array}\right) . \end{aligned}$$

Painlevé III

$$\begin{aligned}&2\ddot{u} = e^t(\alpha e^{2u} +\beta e^{-2u}) + e^{2t}(\gamma e^{4u} +\delta e^{-4u}),\\&H_\mathrm{III}(p, u)= \displaystyle {\frac{p^2}{2} -\nu ^2 e^t \cosh (2u-2\varrho ) -\mu ^2 e^{2t}\cosh (4u)},\\&\mathbf{U}_{11} = \dot{u}e^{2u-2x}+\theta \Big (1-e^{2u-2x} \Big )+\frac{1}{2}\Big (e^{2x+t}-e^{2u-2x}-e^{4u+t-2x}+1\Big ),\\&\mathbf{U}_{12}= e^{\frac{t}{2}}\Big (e^{-u+x}-e^{u-x}\Big ),\\&\mathbf{U}_{21}=\dot{u}^2 e^{u-\frac{t}{2}-3x}\Big (e^{2x}+e^{2u}\Big ) -\dot{u}e^{u-\frac{t}{2}-3x}\Big (e^{2x}+e^{2u+t+2x}+(1+2\theta )e^{2u} +e^{4u+t}\Big )\\&\quad + \theta ^2\left( -e^{u-\frac{t}{2}-x}+e^{3u-\frac{t}{2}-3x} \right) +\theta \left( e^{3u-\frac{t}{2}-3x}+e^{5u+\frac{t}{2}-3x}\right) +4\lambda e^{-u+\frac{t}{2}-x}\\&\quad - 4\chi \left( e^{-3u+\frac{3t}{2}-x}+e^{-u+\frac{3t}{2}-3x} \right) \\&\quad + \frac{1}{4}\Big ( e^{u-\frac{t}{2}-x} +2e^{3u+\frac{t}{2}-x}+e^{5u+\frac{3t}{2}-x}+e^{3u-\frac{t}{2}-3x} +2e^{5u+\frac{t}{2}-3x}+e^{7u+\frac{3t}{2}-3x}\Big ).\\&\mathbf{V}_{11}=-\frac{1}{2}\dot{u}\Big (e^{2u\!-\!2x}\!+\!1\Big ) +\frac{\theta }{2}\Big (1\!+\!e^{2u\!-\!2x}\Big ) \!+\!\frac{1}{4}\Big ( e^{2x\!+\!t}\!+\!e^{2u\!-\!2x}\!+\!e^{4u\!+\!t\!-\!2x}\!+\!1\!+\!2e^{2u\!+\!t}\Big ),\\&\mathbf{V}_{12}= \frac{1}{2}e^{\frac{t}{2}}\Big (e^{-u+x}+e^{u-x}\Big ),\\&\mathbf{V}_{21}=\frac{1}{2}\dot{u}^2 e^{u-\frac{t}{2}-3x}\Big (e^{2x}\!-\!e^{2u}\Big ) -\frac{1}{2}\dot{u}e^{u-\frac{t}{2}-3x}\Big (e^{2x}\!+\!e^{2u+t+2x}\!-\!(1+2\theta )e^{2u} \!-\!e^{4u+t}\Big ) \\&\quad - \frac{\theta ^2}{2}\left( e^{u-\frac{t}{2}-x}+e^{3u-\frac{t}{2}-3x} \right) -\frac{\theta }{2}\left( e^{3u-\frac{t}{2}-3x}+e^{5u+\frac{t}{2}-3x}\right) +2\lambda e^{-u+\frac{t}{2}-x}\\&\quad - 2\chi \left( e^{-3u+\frac{3t}{2}-x}-e^{-u+\frac{3t}{2}-3x} \right) \\&\quad + \frac{1}{8}\Big ( e^{u-\frac{t}{2}-x} +2e^{3u+\frac{t}{2}-x}+e^{5u+\frac{3t}{2}-x}-e^{3u-\frac{t}{2}-3x}-2e^{5u+\frac{t}{2}-3x}-e^{7u+\frac{3t}{2}-3x}\Big ). \end{aligned}$$

Notice that an interesting equation holds:

$$\begin{aligned} \partial _x\Big (\mathbf{U}_{21}e^{2x} \Big )=2 \Big (\mathbf{V}_{21}e^{2x} \Big ) \end{aligned}$$

(in this case $X=e^{2x}$). Therefore, some relation exists between $\mathbf{U}_{21}$ and $\mathbf{V}_{21}$ elements just as for (12)-elements. For example, for PII we have $\partial _x \mathbf{U}_{21}=2 \mathbf{V}_{21}$.

Truncated Painlevé III [2]: $\ddot{u} =2\nu ^2 e^t\sinh (2u )$

$$\begin{aligned} \mathbf{U}(x,t)= & {} \left( \begin{array}{cc} \dot{u} &{}2\nu e^{t/2}\sinh (x-u )\\ 2\nu e^{t/2}\sinh (x+u ) &{} -\dot{u} \end{array}\right) ,\\ \mathbf{V }(x,t)= & {} \left( \begin{array}{cc} 0&{} \nu e^{t/2}\cosh (x-u )\\ \nu e^{t/2} \cosh (x+u )&{} 0 \end{array}\right) . \end{aligned}$$

Painlevé IV

$$\begin{aligned} \ddot{u}= & {} \frac{3}{4}\, u^5 +2tu^3 + (t^2 -\alpha ) u +\frac{\beta }{2u^3},\\ H_\mathrm{IV}^{(\alpha , \beta )}(p, u)= & {} \displaystyle {\frac{p^2}{2} -\frac{u^6}{8} -\frac{tu^4}{2} -\frac{1}{2}\left( t^2 -\alpha \right) u^2 +\frac{\beta }{4u^2}.}\\ \mathbf{U}= & {} \left( \begin{array}{cc}\displaystyle { \frac{x^3}{2}\! +\! tx \! +\! \frac{Q+\frac{1}{2}}{x}}&{}x^2 -u^2\\ \displaystyle {\frac{Q^2+\frac{\beta }{2}}{u^2x^2}\! -\! Q\! -\! \alpha \! -\! 1} &{} \,\,\,\,\displaystyle { -\frac{x^3}{2}\! -\! tx \! -\! \frac{Q+\frac{1}{2}}{x}} \end{array}\right) ,\\ \mathbf{V}= & {} \left( \begin{array}{cc} \displaystyle { \frac{x^2+u^2}{2} + t}&{}\,\, x\\ \displaystyle { -\, \frac{Q+\alpha +1}{x}}&{}\,\,\,\,\,\,\,\, \displaystyle { -\frac{x^2 +u^2}{2} - t} \end{array}\right) , \end{aligned}$$

where

$$\begin{aligned} Q=u\dot{u} -\frac{u^4}{2}-tu^2. \end{aligned}$$

Painlevé V

$$\begin{aligned} \ddot{u}= & {} -\frac{2\alpha \cosh u}{\sinh ^3 u}- \frac{2\beta \sinh u}{\cosh ^3 u} -\gamma e^{2t}\sinh (2u) -\frac{1}{2}\, \delta e^{4t}\sinh (4u),\nonumber \\ H_\mathrm{V}(p, u)= & {} {\frac{p^2}{2}-\, \frac{\alpha }{\sinh ^2 x} -\, \frac{\beta }{\cosh ^2 x}+\frac{\gamma e^{2t}}{2} \cosh (2x)+\frac{\delta e^{4t}}{8}\cosh (4x)},\nonumber \\ \mathbf{U}_{11}= & {} \dot{u}\frac{\sinh (2u)}{\sinh (2x)}-\frac{2\sigma }{\sinh (2x)}\Big ( \cosh (2x)-\cosh (2u)\Big )\nonumber \\&+\frac{e^{2t}}{4\sinh (2x)}\Big ( \cosh (4x)-\cosh (4u)\Big )+\coth (2x).\end{aligned}$$

(C.1)

$$\begin{aligned} \mathbf{U}_{12}= & {} e^t\Big ( \cosh (2x)-\cosh (2u)\Big ).\end{aligned}$$

(C.2)

$$\begin{aligned} \mathbf{U}_{21}= & {} \dot{u}^2\frac{e^{-t}}{\sinh ^2(2x)}\Big ( \cosh (2u)+\cosh (2x)\Big )\nonumber \\&+\,\dot{u}\frac{\sinh (2u)}{\sinh ^2(2x)}\Big ( 4\sigma e^{-t}-e^t \Big [\cosh (2u)+\cosh (2x)\Big ] \Big )\nonumber \\&+\,8\sigma ^2e^{-t}\frac{\coth ^2(u)}{\sinh ^2(2x)}\Big ( \sinh ^2(u)-\cosh ^2(x)\Big )-2\sigma e^t\frac{\sinh ^2(2u)}{\sinh ^2(2x)}\nonumber \\&-2e^{-t}\frac{\xi ^2+2\xi \sigma }{\sinh ^2(u)\sinh ^2(x)} +2e^{-t}\frac{\zeta ^2}{\cosh ^2(u)\cosh ^2(x)}\nonumber \\&+\, \frac{e^{3t}\sinh ^2(2u)}{4\sinh ^2(2x)}\Big ( \cosh (2u)+\cosh (2x)\Big ).\end{aligned}$$

(C.3)

$$\begin{aligned} \mathbf{V}_{11}= & {} \frac{1}{2}e^{2t}\Big ( \cosh (2x)+\cosh (2u)\Big )-2\sigma +\frac{1}{2},\nonumber \\ \mathbf{V}_{12}= & {} e^t\sinh (2x),\nonumber \\ \mathbf{V}_{21}= & {} \frac{e^{-t}}{\sinh (2x)}\Big ( \Big (\dot{u}^2\!-\!\frac{1}{2}\dot{u}e^{2t}\sinh (2u)\Big )^2\! +\!\frac{4\zeta ^2}{\cosh ^2(u)}\nonumber \\&-\,4\frac{\xi ^2\!+\!2\xi \sigma }{\sinh ^2(u)}\! -\!4\sigma ^2\coth ^2(u)\Big ). \end{aligned}$$

(C.4)