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.

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)