On the discretization of Darboux Integrable Systems

TL;DR

This paper explores the discretization process of Darboux integrable systems using their integrals, leading to new semi-discrete examples and advancing understanding of their discrete analogs.

Contribution

It introduces a novel discretization approach based on integrals, resulting in new semi-discrete Darboux integrable systems.

Findings

New semi-discrete Darboux integrable systems identified

Discretization method based on x- and y-integrals developed

Enhanced understanding of discrete analogs of integrable systems

Abstract

We study the discretization of Darboux integrable systems. The discretization is done using $x$-, $y$-integrals of the considered continuous systems. New examples of semi-discrete Darboux integrable systems are obtained.

Equations

$$p_{xy}={\cal A}(p,p_{x},p_{y})\qquad\big{(}p^{i}_{xy}={\cal A}^{i}(p^{1}\dots p^{N},p^{1}_{x}\dots p^{N}_{x},p^{1}_{y}\dots p^{N}_{y})\quad i=1,\dots N\big{)},$$

$$D_{x}F(x,y,p,p_y,p_{yy},...)=0\text{mbox}onallthesolutionsofthesystem\text{eqref}ge{n}_{c}on{t}_{s}ystem.$$

$$q_{x1}={\cal B}(q,q_{x},q_{1}),\qquad\big{(}q^{i}_{x1}={\cal B}^{i}(q^{1}\dots q^{N},q^{1}_{x}\dots q^{N}_{x},q^{1}_{1}\dots q_{1}^{N})\quad i=1,\dots N\big{)},$$

$$DI(x,n,q,q_x,q_{xx},...)=I(x,n,q,q_x,q_{xx},...)\text{mbox}onallthesolutionsofthesystem\text{eqref}ge{n}_{s}em-disc{r}_{s}ystem.$$

$${\mu }_{xy}^i=e^{\sum a_{ij}{\mu }^j},i,j=1,2,\dots ,N.$$

$$\displaystyle{\left\{\begin{array}[]{l}u_{xy}=\frac{u_{x}u_{y}}{u+v+c}+\left(\frac{1}{u+v+c}+\frac{1}{u+v-c}\right)u_{x}v_{y}\\ v_{xy}=\frac{v_{x}v_{y}}{u+v-c}+\left(\frac{1}{u+v+c}+\frac{1}{u+v-c}\right)u_{x}v_{y}\,,\\ \end{array}\right.}$$

$$I_1=2v-\frac{v_x(u+v+c)}{u_x}+2c\ln \frac{u_x}{u+v+c}$$

$$I_2=\frac{u_{xx}}{u_x}-\frac{2u_x+v_x}{u+v+c}.$$

$$\left\{\begin{array}[]{l}u_{1x}=f(x,n,u,v,u_{1},v_{1},u_{x},v_{x})\\ v_{1x}=g(x,n,u,v,u_{1},v_{1},u_{x},v_{x})\end{array}\right.$$

$$\left\{\begin{array}[]{l}\displaystyle{u_{1x}=\frac{(u_{1}+v_{1}+c_{1})u_{x}}{u+v+c}}\\ \\ \displaystyle{v_{1x}=\frac{2(v_{1}-v)u_{x}}{u+v+c}+\frac{2(c_{1}-c)u_{x}}{u+v+c}\ln\frac{u_{x}}{u+v+c}+v_{x}}\,.\end{array}\right.$$

$$F_1=\frac{(c-c_1)(v_2-v)-(c-c_2)(v_1-v)}{(c-c_2)(v_3-v)-(c-c_3)(v_2-v)}$$

$$F_2=\frac{(c_1-c_2)u+(c_2-c)u_1+(c-c_1)u_2}{\sqrt{(c_1-c_2)v+(c_2-c)v_1+(c-c_1)v_2}}-\sqrt{(c_1-c_2)v+(c_2-c)v_1+(c-c_1)v_2}.$$

$$\left\{\begin{array}[]{l}\displaystyle{u_{1x}=\frac{(u_{1}+v_{1}+c)u_{x}}{u+v+c}}\\ \\ \displaystyle{v_{1x}=\frac{2(v_{1}-v)u_{x}}{u+v+c}+v_{x}}\end{array}\right.$$

$$\left\{\begin{array}[]{l}\displaystyle{u_{1x}=\frac{(u_{1}+v_{1}+c)Bu_{x}}{u+v+c}}\\ \\ \displaystyle{v_{1x}=\frac{2B(v_{1}-v+c\ln B)}{u+v+c}u_{x}+Bv_{x}}\,,\end{array}\right.$$

$$K_1=\frac{v_1-vB+B(1-B)u+c\ln B}{(B-1{)}^2}+c\ln (B-1)-c\ln B$$

$$K_2=\frac{u_1+cB-c-c\ln B}{B-1}+\frac{B^{2}v-B{v}_1-cB\ln B}{(B-1{)}^2}+c\ln (B-1)-c\ln B,$$

$$B(u,v,v_1)=\sum _{n=0}^{\infty }{a}_n(v_1-v{)}^n,$$

$$B(u,v,v_1)=1+\frac{1}{u+v-c}(v_1-v)+\sum _{n=2}^{\infty }{a}_n(v_1-v{)}^n.$$

$$\begin{array}[]{l}\mu_{xy}=e^{2\mu-\nu},\\ \nu_{xy}=e^{-c\mu-2\nu},\end{array}$$

$$\left\{\begin{array}[]{l}u_{1x}=\tilde{f}(u,v,u_{1},v_{1},u_{x},v_{x})\\ v_{1x}=\tilde{g}(u,v,u_{1},v_{1},u_{x},v_{x}),\end{array}\right.$$

$$I_1=u_{xx}+v_{xx}-u_x^2+u_x{v}_x-v_x^2$$

$$I_1^{\ast }=u_{xxx}+u_x(v_{xx}-2u_{xx})+u_x^2{v}_x-u_x{v}_x^2$$

$$\left\{\begin{array}[]{lll}u_{1x}&=&u_{x}+Ae^{u_{1}+u-v_{1}}\\ v_{1x}&=&v_{x}+Be^{-u+v+v_{1}}\,,\\ \end{array}\right.$$

$$\left\{\begin{array}[]{lll}u_{1x}&=&u_{x}+Ae^{u_{1}+u-v}\\ v_{1x}&=&v_{x}+Be^{-u_{1}+v+v_{1}}\,,\\ \end{array}\right.$$

$$I_2=2u_{xx}+v_{xx}-2u_x^2+2u_x{v}_x-v_x^2$$

$$\begin{gathered} I_2^{\ast }=u_{xxxx}+u_x(v_{xxx}-2u_{xxx})+u_{xx}(4u_x{v}_x-2u_x^2-v_x^2) \\ +u_{xx}(v_{xx}-u_{xx})+v_{xx}{u}_x(u_x-2v_x)+u_x^4+u_x^2{v}_x^2-2u_x^3{v}_x \end{gathered}$$

$$\left\{\begin{array}[]{lll}u_{1x}&=&u_{x}+Ae^{u+u_{1}-v_{1}}\\ v_{1x}&=&v_{x}+Be^{-2u+v+v_{1}}\,,\end{array}\right.$$

$$I_3=u_{xx}+\frac{1}{3}{v}_{xx}-u_x^2+u_x{v}_x-\frac{1}{3}{v}_x^2$$

$$\begin{gathered} I_3^{\ast }=u_{(6)}-2u_{(5)}{u}_x+v_{(5)}{u}_x+u_{(4)}(32(u_x{)}^2-30u_x{v}_x+11(v_x{)}^2-40u_{xx}-11v_{xx}) \\ +v_{(4)}(14(u_x{)}^2-15u_x{v}_x+(13/3)(v_x{)}^2-10u_{xx}-(13/3)v_{xx})+19(u_{(3)}{)}^2+(13/6)(v_{(3)}{)}^2+16u_{(3)}{v}_{(3)} \\ +u_{(3)}(-36u_{xx}{u}_x+18u_{xx}{v}_x+80v_{xx}{u}_x-45v_{xx}{v}_x)+v_{(3)}(-52u_{xx}{u}_x+33u_{xx}{v}_x-5v_{xx}{u}_x) \\ +u_{(3)}(-64(u_x{)}^3+102(u_x{)}^2{v}_x-62u_x(v_x{)}^2+13(v_x{)}^3)+v_{(3)}(32(u_x{)}^3-58(u_x{)}^2{v}_x \\ +38u_x(v_x{)}^2-(26/3)(v_x{)}^3)+66(u_{xx}{)}^3+(26/3)(v_{xx}{)}^3-35(u_{xx}{)}^2(v_{xx})-5u_{xx}(v_{xx}{)}^2 \\ +(u_{xx}{)}^2(30(u_x{)}^2-18u_x{v}_x-(11/2)(v_x{)}^2)+u_{xx}{v}_{xx}(-34(u_x{)}^2+32u_x{v}_x-2(v_x{)}^2)-2(v_{xx}{)}^2{u}_x{v}_x \\ +u_{xx}(6(u_x{)}^4-24(u_x{)}^3{v}_x+25(u_x{)}^2(v_x{)}^2-9u_x(v_x{)}^3+(v_x{)}^4)+v_{xx}(-(u_x{)}^4+8(u_x{)}^3{v}_x-8(u_x{)}^2(v_x{)}^2 \\ +2u_x(v_x{)}^3)+(-2(u_x{)}^6+6(u_x{)}^5{v}_x-(13/2)(u_x{)}^4(v_x{)}^2+3(u_x{)}^3(v_x{)}^3-(1/2)(u_x{)}^2(v_x{)}^4) \end{gathered}$$

$$\left\{\begin{array}[]{lll}u_{1x}&=&u_{x}+Ae^{u+u_{1}-v_{1}}\\ v_{1x}&=&v_{x}+Be^{-3u+v+v_{1}}\,,\end{array}\right.$$

Full text

On the discretization of Darboux Integrable Systems

Kostyantyn Zheltukhin111e-mail: zheltukh@metu.edu.tr

Department of Mathematics, Middle East Technical University, Ankara, Turkey

Natalya Zheltukhina222e-mail: natalya@fen.bilkent.edu.tr

Department of Mathematics, Faculty of Science, Bilkent University, Ankara, Turkey

Abstract We study the discretization of Darboux integrable systems. The discretization is done using $x$-, $y$-integrals of the considered continuous systems. New examples of semi-discrete Darboux integrable systems are obtained.

Keywords: semi-discrete system, Darboux integrability, $x$-integral, $n$-integral.

1 Introduction

The classification problem of Darboux integrable equations has attracted a considerable interest in the recent time, see the survey paper [1] and references there in. There are many classification results in the continuous case. The case of semi-discrete and discrete equations is not that well studied. Discrete models play a big role in many areas of physics and discretization of existing integrable continuous models is an important problem. There is a currently discussed conjecture saying that for each continuous Darboux integrable system it is possible to find a semi-discrete Darboux integrable system that admits the same set of $x$-integrals. To better understand properties of semi-discrete and discrete Darboux integrable systems it is important to have enough examples of such systems. We can test the conjecture and obtain new semi-discrete Darboux integrable systems, corresponding to given continuous ones, following an approach proposed by Habibullin et al., see [2]. In this case we take a Darboux integrable continuous equation and look for a semi-discrete equation admitting the same integrals. The method was successfully applied to many Darboux integrable continuous equations, see [2]-[4]. In almost all considered cases such semi-discrete equations exist and they are Darboux integrable.

In the present paper we apply this method of discretization to Darboux integrable systems to obtain new Darboux integrable semi-discrete systems. Let us give necessary definitions and formulate the main results of our work.

Consider a hyperbolic continuous system

[TABLE]

where $p^{i}(x,y)$, $i=1,\dots N$, are functions of continuous variables $x,y\in\mathbb{R}$. We say that a function $F(x,y,p,p_{y},p_{yy},\dots)$ is an $x$-integral of the system (1) if

[TABLE]

The operator $D_{x}$ represents the total derivative with respect to $x$. The $y$-integral of the system (1) is defined in a similar way. The system (1) is called Darboux integrable if it admits $N$ functionally independent non-trivial $x$-integrals and $N$ functionally independent non-trivial $y$-integrals.

Consider a hyperbolic semi-discrete system

[TABLE]

where $q^{i}(x,n)$, $i=1,\dots N$, are functions of a continuous variable $x\in\mathbb{R}$ and a discrete variable $n\in\mathbb{N}$. Note that we use notation $q_{1}(x,n)=Dq(x,n)=q(x,n+1)$ and $q_{k}(x,n)=D^{k}q(x,n)=q(x,n+k)$, where $D$ is the shift operator. To state the Darboux integrability of a semi-discrete system we need to define $x$- and $n$-integrals for such systems, see [5]. An $x$-integral is defined in the same way as in continuous case and a function $I(x,n,q,q_{x},q_{xx},\dots)$ is an $n$-integral of system (2) if

[TABLE]

The system (2) is called Darboux integrable if it admits $N$ functionally independent non-trivial $x$-integrals and $N$ functionally independent non-trivial $n$-integrals.

To find new Darboux integrable semi-discrete systems we applied the discretization method proposed in [2] to one of the continuous systems derived by Zhiber, Kostrigina in [6] and continuous systems derived by Shabat, Yamilov in [7]. In [6] the authors considered the classification problem for Darboux integrable continuous systems that admit the $x$- and $y$-integrals of the first and second order. In [7] the authors considered the exponential type system

[TABLE]

It was shown that such a system is Darboux integrable if and only if the matrix $A=(a_{ij})$ is a Cartan matrix of a semi-simple Lie algebra. Such systems are closely related to the classical Toda field theories, see [8]-[10] and references there in. In this case we obtain the Darboux integrable semi-discrete systems that were already described in [11].

First we consider the following system (see [6])

[TABLE]

where $c$ is an arbitrary constant. This system is Darboux integrable and admits the following $y$-integrals

[TABLE]

and

[TABLE]

The $x$- integrals have the same form in $u,\,v,\,u_{y},\,v_{y},\dots$ variables.

Now we look for semi-discrete systems admitting these functions as $n$-integrals. The obtained results are given in Theorems 1 and 2 below.

Theorem 1

The system

[TABLE]

possessing $n$-integrals (4) and (5), where $c$ is a function of $n$ satisfying $c(n)\neq c(n+1)$ for all $n\in\mathbb{Z}$, has the form

[TABLE]

Moreover, the system above also possesses $x$-integrals

[TABLE]

and

[TABLE]

Hence, semi-discrete system (7) is Darboux integrable.

Theorem 2

The system (6) possessing $n$-integrals (4) and (5), where $c$ is a constant, is either

[TABLE]

with $x$-integrals $\displaystyle{F_{1}=\frac{v_{1}-v}{v_{2}-v_{1}}}$ and $\displaystyle{F_{2}=\frac{u_{2}-u+v-v_{2}}{\sqrt{v_{1}-v}}}$, or

[TABLE]

where $B$ is defined by equality $H(K_{1},K_{2})=0$ with

[TABLE]

and

[TABLE]

and $H$ being any smooth function.

Remark 1

We considered some special cases of the system (11) and get Darboux integrable systems.

(I) System (11) with $B=\displaystyle{\frac{u-v+(-1)^{n}\sqrt{(u-v)^{2}+4uv_{1}}}{2u}}$ is Darboux Integrable. (The expression for $B$ is found from $K_{1}=0$, with $c=0$.)

(II) System (11) with $B=\displaystyle{\frac{v_{1}-u_{1}+(-1)^{n}\sqrt{(v_{1}-u_{1})^{2}+4u_{1}v}}{2v}}$ is Darboux Integrable. (The expression for $B$ is found from $K_{2}=0$, with $c=0$.)

Remark 2

Expansion of the function $B(u,v,v_{1})$, given implicitly by $(B-1)^{2}K_{1}=0$, into a series of the form

[TABLE]

where coefficients $a_{n}$ depend on variables $u$ and $v$, yields $a_{0}=1$ and $a_{1}=\dfrac{1}{u+v-c}$. So $B$ can be written as

[TABLE]

By letting $u_{1}=u+\epsilon u_{y}$ and $v=v+\epsilon v_{y}$ and taking $\epsilon\to 0$ one can see that the system (11) has a continuum limit (3).

Let us discuss the exponential type systems. We consider the discretization of such systems corresponding to $2\times 2$ matrices, namely,

[TABLE]

where $c=1,2,3$. The obtained results are given in Theorem 3 below. The discretization of such systems was also considered in [11], where the form of the corresponding semi-discrete system was directly postulated and then the Darboux integrability proved. In our approach we do not make any specific assumptions about the form of the corresponding semi-discrete system. Note that the integrals corresponding to Darboux integrable exponential systems are given in the statement of Theorem 3.

Theorem 3

(1) The system

[TABLE]

possessing $n$-integrals

[TABLE]

and

[TABLE]

has the form

[TABLE]

or

[TABLE]

where $A$ and $B$ are arbitrary constants.

(2) The system (15) possessing $n$-integrals

[TABLE]

and

[TABLE]

has the form

[TABLE]

where $A$ and $B$ are arbitrary constants.

(3) The system (15) possessing $n$-integrals

[TABLE]

and

[TABLE]

has the form

[TABLE]

where $A$ and $B$ are arbitrary constants.

Remark 3

We note that while considering systems with integrals (20) and (21) we also obtain two degenerate systems

[TABLE]

and

[TABLE]

where $A$, $B$ and $c$ are arbitrary constants,which are equivalent to a Darboux integrable equation.

Remark 4

By letting $u=\mu^{1}$, $u_{1}=\mu^{1}+\epsilon\mu^{1}_{y}$, $v=\mu^{2}$, $v_{1}=\mu^{2}+\epsilon\mu^{2}_{y}$ and $A=\epsilon$, $B=\epsilon$ in equations (18), (22), (25) and taking $\epsilon\to 0$ one can see that the considered systems have corresponding continuum limit given by (14).

2 Proof of Theorems 1.1 and 1.2

Let us find a semi-discrete system (6) possessing $n$-integrals (4) and (5), where $c$ is an arbitrary constant, possibly dependent on $n$. Let $Dc=c_{1}$. It follows from $DI_{2}=I_{2}$ that

[TABLE]

that is

[TABLE]

Compare the coefficients by $v_{xx}$ and $u_{xx}$, we get $f_{v_{x}}=0$ and $\dfrac{f_{u_{x}}}{f}=\dfrac{1}{u_{x}}$. Hence

[TABLE]

It follows from $DI_{1}=I_{1}$ that

[TABLE]

Using (29) we obtain

[TABLE]

and find $g$ as

[TABLE]

Substituting the expressions (29) and (31) into equality (28) and comparing coefficients by $u_{x}$, $v_{x}$, $u_{x}\ln\dfrac{u_{x}}{u+v+c}$ and free term we get the following equalities

[TABLE]

[TABLE]

[TABLE]

[TABLE]

We have two possibilities: $c_{1}\neq c$ and $c_{1}=c$.

2.1 $c$ depends on $n$

First we consider the case $c_{1}\neq c$, that is $c$ depends on $n$ and satisfies $c(n)\neq c(n+1)$ for all $n$. Then equations (33)-(35) are transformed into

[TABLE]

[TABLE]

[TABLE]

Equations (36) and (38) imply that

[TABLE]

Substituting the above $A$ into (37) we get that $M$ satisfies

[TABLE]

Differentiating equation (40) with respect to $v$ and $v_{1}$ we get that $M_{u}=0$ and $M_{u_{1}}=0$ respectively. Thus, equation (40) implies that $M=1$. So in the case $c_{1}\neq c$ we arrive to the system of equations (7). We note that the system (7) is Darboux integrable. It admits two $n$-integrals (4) and (5) and two $x$-integrals (8) and (9). The $x$-integrals can be found by considering the characteristic $x$-ring for system (7).

2.2 $c$ does not depend on $n$

Now we consider the case $c=c_{1}$, that is $c$ is a constant independent of $n$. Then we have equations (34) and (35). Introducing new variable $B=\dfrac{(u+v+c)}{(u_{1}+v_{1}+c)}A$ we can rewrite the equations as

[TABLE]

[TABLE]

The set of solutions of the above system is not empty, for example it admits a solution $B=1$. Setting $B=1$ we arrive to the system of equations (10). We note that the system (10) is Darboux integrable. It admits two $n$-integrals (4) and (5) and two $x$-integrals

[TABLE]

The $x$-integrals are calculated by considering the characteristic $x$-ring for system (10).

Now let us consider case when $B\neq 1$ identically. For function $W=W(u,v,u_{1},v_{1},B)$ equations (41) and (42) become

[TABLE]

[TABLE]

After the change of variables $\tilde{v}=v+c$, $\tilde{v_{1}}=v_{1}+c-(v+c)B$, $\tilde{u}=u$, $\tilde{u_{1}}=u_{1}$, $\tilde{B}=B$ equations (44) and (43) become $W_{\tilde{v}}=0$ and

[TABLE]

We differentiate the last equality with respect to $\tilde{v}$, use $W_{\tilde{v}}=0$, and find that $W$ satisfies the following equations

[TABLE]

[TABLE]

After doing another change of variables $u_{1}^{*}=\tilde{u_{1}}-\tilde{B}^{2}\tilde{u}$, $v_{1}^{*}=\tilde{v_{1}}+\tilde{B}(1-\tilde{B})\tilde{u}$, $u^{*}=\tilde{u}$, $B^{*}=\tilde{B}$, we obtain that $W_{u^{*}}=0$ and

[TABLE]

The first integrals of the last equation are

[TABLE]

and

[TABLE]

They can be rewritten in the original variables as

[TABLE]

and

[TABLE]

Therefore, system (6) becomes (11) due to (29) and (31).

2.3 Proof of Remark 1.1

Function $B$ is any function satisfying the equality $H(K_{1},K_{2})=0$, where $H$ is any smooth function.

(I) By taking function $H$ as $H(K_{1},K_{2})=K_{1}$ we obtain one possible function $B$. It satisfies the equality $-uB^{2}+(u-v)B+v_{1}=0$ and can be taken as $B=\displaystyle{\frac{u-v+(-1)^{n}\sqrt{(u-v)^{2}+4uv_{1}}}{2u}}$

(II) By taking function $H$ as $H(K_{1},K_{2})=K_{2}$ we obtain another possible function $B$. It satisfies the equality $vB^{2}+(u_{1}-v_{1})B-u_{1}=0$ and can be taken as $B=\displaystyle{\frac{v_{1}-u_{1}+(-1)^{n}\sqrt{(v_{1}-u_{1})^{2}+4u_{1}v}}{2v}}$.

In both cases ((I) and (II)) let us consider the corresponding $x$-rings. Denote by $X=D_{x}$, $Y_{1}=\displaystyle{\frac{\partial}{\partial u_{x}}}$, $Y_{2}=\displaystyle{\frac{\partial}{\partial v_{x}}}$, $E_{1}=\displaystyle{\frac{u+v}{B}[Y_{1},X]}$, $E_{2}=\displaystyle{\frac{1}{B}[Y_{2},X]}$, $E_{3}=[E_{1},E_{2}]$. Note that $X=u_{x}E_{1}+v_{x}E_{2}$. We have,

[TABLE]

where

[TABLE]

in case (I) and

[TABLE]

in case (II).

3 Proof of Theorem 1.3

3.1 Case (1)

Let us find a system

[TABLE]

possessing $n$-integrals (16) and (17). The equality $DI=I$ implies

[TABLE]

or the same

[TABLE]

We consider the coefficients by $u_{xx}$ and $u_{xx}$ in (47) to get

[TABLE]

The equality $DI_{1}^{*}=I_{1}^{*}$ implies

[TABLE]

Since $DI_{1}^{*}=u_{1xxx}+\dots=\tilde{f}_{u_{x}}u_{xxx}+\dots$, where the remaining terms do not depend on $u_{xxx}$, the equality (50) implies

[TABLE]

Note that $J=D_{x}I_{1}-I_{1}^{*}=v_{xxx}+v_{x}(u_{xx}-2v_{xx})+v_{x}^{2}u_{x}-u_{x}^{2}v_{x}$ is an $n$-integral as well. Since $DJ=J$ and $DJ=v_{1xxx}+\dots=\tilde{g}_{v_{x}}v_{xxx}+\dots$, where the remaining terms do not depend on $v_{xxx}$, then

[TABLE]

It follows from equalities (48), (49), (51) and (52) that $\tilde{f}_{v_{x}}=0$ and $\tilde{g}_{u_{x}}=0$. Therefore the system (45) and equality (47) become

[TABLE]

and

[TABLE]

By considering coefficients by $u_{x}$, $v_{x}$ and $u_{x}^{0}v_{x}^{0}$ in the last equality, we get

[TABLE]

Now let us rewrite inequality (50) for the system (53)

[TABLE]

By comparing the coefficients by $u_{xx}$ and $v_{xx}$ in the last equality, we get

[TABLE]

It follows from equality $DJ=J$ that

[TABLE]

By comparing the coefficients by $u_{xx}$ and $v_{xx}$ in the last equality, we get

[TABLE]

Note that the equalities (55) and (56) follow from equalities (59) and (61). Let us use equalities (59) and (61) to rewrite equality (58)

[TABLE]

We note that the consideration of the coefficients by $u_{xx}$, $v_{xx}$, $u_{x}^{2}$, $v_{x}^{2}$, $u_{x}v_{x}$ in the above equality give us equations that follow immediately from (59) and (61). Considering coefficient by $u_{x}$ we get

[TABLE]

Using equations (59) and (61) we get

[TABLE]

or using equation (57) ,

[TABLE]

Considering coefficient by $v_{x}$ we get

[TABLE]

Using equations (59) and (61) we get

[TABLE]

It follows from equations (64) and (66) that $f_{x}=0$ and $f(f_{u_{1}}-f)=0$. Thus either $f=0$ or

[TABLE]

Now we consider the coefficient by $u_{x}^{0}v_{x}^{0}$ in (58) we get

[TABLE]

First assume that $f\neq 0$ then using (67) we can rewrite the above equality as

[TABLE]

Also we can rewrite equality (60), using equations (59), (61) and (57) then considering coefficients by $u_{x}$ and $v_{x}$ we obtain

[TABLE]

From above equalities and (61) it follows that $g_{x}=0$, $g_{v_{1}}=g$ and $g_{v}=g$ (we assume that $g\neq 0$). We have

[TABLE]

Using (70), the equality (69) takes form $g_{u_{1}}f_{v_{1}}(-g+f)=0$. This equality implies that under assumptions that $f\neq 0$ and $g\neq 0$ we have three possibilities: (I) $g_{u_{1}}=0$, (II) $f_{v_{1}}=0$ and (III) $g=f$. Let us consider these possibilities.

Case (I) From $g_{u_{1}}=0$, using (70), we get that $g_{u}=-g$, $g_{v_{1}}=g$, $g_{v}=g$. Thus $g=Be^{-u+v+v_{1}}$, where $B$ is a constant. We also get that $f_{u}=f$, $f_{u_{1}}=f$, $f_{v}=0$ and $f_{v_{1}}=-f$. Thus $f=Ae^{u_{1}+u-v_{1}}$, where $A$ is a constant. So the system (53) takes form (18).

Case (II) From $f_{v_{1}}=0$, using (70), we get that $f_{u}=f$, $f_{u_{1}}=f$, $f_{v}=-f$. Thus $f=Ae^{u_{1}+u-v}$, where $A$ is a constant. We also get that $g_{u}=0$, $g_{u_{1}}=-g$, $g_{v}=g$ and $g_{v_{1}}=g$. Thus $g=Be^{-u_{1}+v_{1}+v}$, where $B$ is a constant. So the system (53) takes form (19).

Case (III) From $g=f$, using (70), we get that $f=0$ and $g=0$. So the system (53) takes form

[TABLE]

3.2 Case (2)

Let us find system (15) possessing $n$-integrals (20) and (21). We compare the coefficients in $DI_{2}=I_{2}$ by $u_{xx}$ and $v_{xx}$ and get

[TABLE]

We also compare the coefficients in $DI_{2}^{*}=I_{2}^{*}$ and

$D(D_{x}^{2}I_{2}-2I^{*}_{2})=(D_{x}^{2}I_{2}-2I^{*}_{2})$ by $u_{xxxx}$ and $v_{xxxx}$ respectively and get $\tilde{f}_{u_{x}}=1$ and $\tilde{g}_{v_{x}}=1$. It follows from (71) that $\tilde{f}_{v_{x}}=0$ and $\tilde{g}_{u_{x}}=0$. Therefore, our system (15) becomes

[TABLE]

We write equality $DI_{2}=I_{2}$ and get

[TABLE]

By comparing the coefficients by $u_{x}$, $v_{x}$ and $u^{0}_{x}v^{0}_{x}$ in the last equality we obtain the system of equations

[TABLE]

That suggests the following change of variables

[TABLE]

to be made. In new variables the system (15) becomes

[TABLE]

The comparison of coefficients in $DI_{2}=I_{2}$ by $P_{x}$, $S_{x}$ and $P^{0}_{x}S^{0}_{x}$ gives

[TABLE]

The coefficients in $DI_{2}^{*}=I^{*}_{2}$ by $S_{xxx}$ and $P_{xxx}$ are compared and we obtain the following equalities

[TABLE]

It follows from (73) and (74) that $G_{S}=2G$, $G_{P}=-2G$, $F_{S}=-F$ and $F_{P}=2F$. Therefore, system (72) can be written as

[TABLE]

We compare the coefficient in $DI^{*}_{2}=I^{*}_{2}$ by $S_{xx}$ and get

[TABLE]

that is $A=A_{Q}$. Hence, $A(Q,T)=e^{Q}\tilde{A}(T)$. Now we compare the coefficient in $DI_{2}=I_{2}$ by $P^{0}_{x}S^{0}_{x}$ and get

[TABLE]

Since functions $\tilde{A}(T)$ and $B(Q,T)$ do not depend on variable $P$, then it follows from (75) that $B=B_{T}$, that is $B=\tilde{B}(Q)e^{T}$. Now (75) becomes

[TABLE]

Note that the right side of the last equality depends on $Q$ only, while the left side depends on $T$ only. Hence, $-2\frac{\tilde{A}+\tilde{A}_{T}}{\tilde{A}}=c$ and $\frac{\tilde{B}_{Q}}{\tilde{B}}=c$, where $c$ is some constant. One can see that $\tilde{A}=c_{1}e^{-(2c+1)T}$ and $\tilde{B}=c_{2}e^{cQ}$ and therefore system (72) becomes

[TABLE]

where $c$, $c_{1}$ and $c_{2}$ are some constants. Equality $DI_{2}-I_{2}=0$ becomes $-3cc_{1}c_{2}e^{s+(c+1)Q-2cT}=0$, which implies that either $c=0$, or $c_{1}=0$, or $c_{2}=0$. Note that the $DI_{2}^{*}=I_{2}^{*}$ is also satisfied if either $c=0$ or $c_{1}=0$ or $c_{2}=0$. So we have three cases:

when $c=0$ the system (15) becomes (22) with $c_{1}=A$ and $c_{2}=B$.

when $c_{1}=0$ the system (15) becomes (26) with $c_{2}=B$.

when $c_{2}=0$ the system (15) becomes (27) with $c_{1}=A$.

3.3 Case (3)

Let us find system (15) possessing $n$-integrals (23) and (24). We compare the coefficients in $DI_{3}=I_{3}$ by $u_{xx}$ and $v_{xx}$ and get

[TABLE]

We also compare the coefficients in $DI_{3}^{*}=I_{3}^{*}$ and $D(D_{x}^{4}I_{3}-I^{*}_{3})=(D_{x}^{4}I_{3}-I^{*}_{3})$ by $u_{(6)}$ and $v_{(6)}$ respectively and get $\tilde{f}_{u_{x}}=1$ and $\tilde{g}_{v_{x}}=1$. It follows from (76) that $\tilde{f}_{v_{x}}=0$ and $\tilde{g}_{u_{x}}=0$. Therefore, our system (15) becomes

[TABLE]

By comparing the coefficients by $u_{x}$, $v_{x}$ and $u^{0}_{x}v^{0}_{x}$ in $DI_{3}=I_{3}$ we obtain the system of equations

[TABLE]

That suggests the following change of variables

[TABLE]

to be made. In new variables the system (15) becomes

[TABLE]

The comparison of coefficients in $DI_{3}=I_{3}$ by $P_{x}$, $S_{x}$ and $P^{0}_{x}S^{0}_{x}$ gives

[TABLE]

The comparison of coefficients in $DI_{3}^{*}=I^{*}_{3}$ by $S_{(5)}$ and $P_{(5)}$ gives

[TABLE]

Using equations (78) and (79) we get $G_{S}=2G$, $G_{P}=-3G$, $F_{S}=-F$, and $F_{P}=2F$. Therefore, system (77) can be written as

[TABLE]

where $A$ and $B$ are some functions depending on $Q$ and $T$ only. We compare the coefficients in $DI_{3}-I_{3}=0$ by $S_{x}^{0}P_{x}^{0}$ and the coefficients in $DI_{3}^{*}-I_{3}^{*}=0$ by $P_{(4)}$, $S_{(4)}$ and $P_{(3)}P_{x}$ respectively and get

[TABLE]

where

[TABLE]

and

[TABLE]

We solve the linear system of equations (80) with respect to $A_{T}$, $A_{Q}$, $B_{T}$ and $B_{Q}$ and get the following system of differential equations $A_{T}=-A$, $A_{Q}=A$, $B_{T}=B$ and $B_{Q}=0$. Thus the system (77) is written as

[TABLE]

where $c_{1}$ and $c_{2}$ are arbitrary constants. It is equivalent to system (25) with $A=c_{1}$ and $B=c_{2}$.