Symmetry, Integrability and Geometry: Methods and Applications The Klein–Gordon Equation and Differential Substitutions of the Form v = ϕ(u, ux, uy) ⋆

We present the complete classification of equations of the form $u_{xy}=f(u,u_x,u_y)$ and the Klein-Gordon equations $v_{xy}=F(v)$ connected with one another by differential substitutions $v=\varphi(u,u_x,u_y)$ such that $\varphi_{u_x}\varphi_{u_y}\neq 0$ over the ring of complex-valued variables.


Introduction
In this paper, we study the classification problem of equations of the form u xy = f (u, u x , u y ) (1.1) over the ring of complex-valued variables. Such equations have applications in many fields of mathematics and physics. Liouville [10], Bäcklund [2], Darboux [4] and other authors [3,17] studying the surfaces of constant negative curvature discovered the first examples of integrable nonlinear hyperbolic equations. In the 1970s, one of the fundamental methods of mathematical physics, the inverse scattering method, was introduced. After that, since hyperbolic equations have many applications in physics (continuum mechanics, quantum field theory, theory of ferromagnetic materials etc.), many important studies were published. Existence of higher symmetries is a hallmark of integrability of an equation. Drinfel'd, Sokolov and Svinolupov [5,16] showed that symmetries can be effectively used for classification of evolution equations. Zhiber and Shabat [18] obtained the complete list of the Klein-Gordon equations v xy = F (v) (1.2) with higher symmetries. However, the symmetry method for the classification of equations of form (1.1) faces particular difficulties. Therefore, here we use differential substitutions to solve the classification problem.
Before going further, let us give some definitions. Let u be a solution of equation (1.1). All the mixed derivatives of u u x , u y , u xx , u yy , . . . (1.3) will be expressed through equation (1.1) with differential consequences of this equation. Here u and variables (1.3) will be regarded as independent.
We begin with an important notion of (infinitesimal) symmetry of equation (1.1). Denote the operators of total derivatives with respect to x and y by D andD, respectively. Definition 1. The symmetry of equation (1.1) of order (n, m) is the function g = g(u, u 1 , . . . , u n , u 1 , . . . ,ū m ), g un = 0, gū m = 0, satisfying the equation Here u i = ∂ i u ∂x i andū i = ∂ i u ∂y i , i ∈ N. If n ≤ 1 and m ≤ 1 then the function g is called a classical symmetry, otherwise we have a higher symmetry.
Assume that g is a symmetry of equation (1.1). It is easy to check that the derivatives g un and gū m satisfy the so-called characteristic equationsD(g un ) = 0 and D(gū m ) = 0, respectively. It actually can be shown that g un depends only on the variables u, u 1 , . . . , u n , while gū m is a function of the variables u,ū 1 , . . . ,ū m . Other Laplace invariants can be found recurring in the relation Sokolov and Zhiber [19] showed that the functions H 1 and H 0 are invariants of equation (1.1) under the point transformations u → ζ(x, y, u). Generalized Laplace invariants play a significant role in the investigation of integrability of equations. Namely, Anderson and Kamran [1], Zhiber, Sokolov and Startsev [20] proved that an equation has nontrivial x-and y-integrals if and only if the Laplace sequence of invariants terminates on both sides (H r = H s ≡ 0 for some values r and s), which is indeed a definition of the (Darboux) integrability of an equation. Equations satisfying the last condition are called Liouville type equations. Using this definition for linear equations V xy + a(x, y)V x + b(x, y)V y + c(x, y)V = 0, one can obtain equations with the finite Laplace sequence studied in detail by Goursat [6]. It should be noted that symmetries of Liouville type equations have two arbitrary functions, while the equations integrable by the inverse scattering method (for instance, the sine-Gordon equation) have a countable set of symmetries.
The main notion of the paper is the notion of differential substitutions. Before proceeding, let us briefly mention some works related to differential substitutions. Sokolov [12] showed that substitutions can be used in the study of integrability of nonlinear differential equations. There exist various different definitions of exact integrable hyperbolic equations. Sokolov and Zhiber [19] presented one of the most comprehensive reviews of such equations. As mentioned before, existence of higher symmetries is a hallmark of integrability of an equation. Meshkov and Sokolov [11] presented the complete list of one-field hyperbolic equations with generalized integrable x-and y-symmetries of the third order. One can find many examples of nonlinear equations and differential substitutions in [11,19]. Startsev [14,15] described properties of generalized Laplace invariants of nonlinear equations with differential substitutions. Bäcklund transformations and, in particular cases, differential substitutions were studied by Khabirov [7]. Kuznetsova [8] described coupled equations for which linearizations are related by Laplace transformations of the first and the second orders. A Bäcklund transformation was constructed for such pairs.
Although we know a considerable amount of nonlinear equations which are connected with one another by differential substitutions, the problem of classifying differential substitutions and Bäcklund transformations was solved only for evolution equations.
Recently, Zhiber and Kuznetsova [9] have applied differential substitutions to classify equations. Namely, all equations of form (1.1) are transformed into equations of form (1.2) by differential substitutions of the special form v = ϕ(u, u x ) were described. All these equations are contained in the following list: where the functions s and f satisfy s (u x )F (s(u x )) = 1; where ψ u + ψψ uy − α u y ψ uy = exp α; where ψ u + ψψ uy − α u y ψ uy = 0; up to the point transformations u → θ(u), v → κ(v), x → ξx, and y → ηy, where ξ and η are arbitrary constants. Here c is an arbitrary constant, c 1 and c 2 are constants satisfying (c 1 , c 2 ) = (0, 0), and the function ψ satisfies (ψ u , ψ uy ) = (0, 0). Furthermore, all equations of form (1.2) that can be transformed into equations of form (1.1) by differential substitutions of the form u = ψ(v, v y ) are given in the following list: x → ξx, and y → ηy, where ξ and η are arbitrary constants. Here c is an arbitrary constant, c 1 and c 2 are constants satisfying (c 1 , c 2 ) = (0, 0). Based on the above lists, Bäcklund transformations have been constructed for some pairs of equations. For instance, the equations Kuznetsova [8] showed that linearizations of equation (1.6) are related by Laplace transformations of the first order. For example, we give the equations where λ and β are arbitrary constants, and the function b satisfies the equation λb(u x ) − βb n (u x ) = u x . The Bäcklund transformation is given by Note that the equation v xy = λv − βv n is a version of the PHI-four equation [13]. The PHI-four equation and the corresponding Bäcklund transformation are obtained for n = 3. The purpose of this paper is to describe all equations of form (1.1) that are transformed into equations of form (1.2) by differential substitutions over the ring of complex-valued variables. It should be noted that most of the differential substitutions which connect the well-known integrable equations (1.1) have the form v = ϕ(u, u x , u y ) (see [11,19]). Therefore, we are interested just in this form of substitutions. This paper is organized as follows. Section 2 presents the complete list of equations (1.1) that are transformed into the Klein-Gordon equations by differential substitutions of form (1.7). In Section 3, the main theorem of the paper is proven. Section 4 is devoted to the problem which is, in a sense, inverse to the original problem. Namely, equations (1.2) are transformed into equations (1.1) by differential substitutions of the form where the function γ satisfies 1 − γ γ 2 = γ ;
Now, let us analyze some of the above equations in detail. Consider (2.1) with ab = 0. Using the point transformations √ ax → x, √ by → y, and v − ln(ab) 1/2 → v, we obtain u xy = u 2 x + 1 u 2 y + 1.
by the differential substitution v = ln u x + u 2 x + 1 u y + u 2 y + 1 .
Equation (2.21) is a S-integrable and possesses symmetries of the third order (see [11]). Note that applying the point transformations v → iv, ix → x, iy → y, and using the formula ln 1 − u 2 x − iu x = −i arcsin u x we can also convert the above equations into Applying the transformation iy → y to the above equations we arrive at As shown in [11], equation (2.22 1 ) has symmetries of the third order. In [11] the x-and yintegrals and the general solution of equation (2.22 1 ) were presented. Note that the equation (2.2 1 ) is the Goursat equation. Its symmetries of the third order can be found, for instance, in [11].
The equation (2.3 1 ) has symmetries of the third order [11]. The x-and y-integrals of this equation are given by Consider cases (2.7) and (2.8). The equation u xy = µ(u)u x u y possesses the x-and y-integrals of the first order, ω = ln u x − σ(u),ω = ln u y − σ(u). Here σ = µ.
The equation u xy = µ(u)(u y + c)u x in cases (2.10) and (2.11) possess the y-integral of the first orderω = ln(u y + c) − σ(u), where σ = µ. The x-integral in case (2.10) is and in case (2.11) we get the x-integral The equation (2.14 1 ) possesses the y-integral of the first order and the x-integral of the third orderω Now, we consider the equation which appears in (2.16) and (2.17). The equation (2.16 1 ) is transformed into the equation presented in [19] by a point transformation and has the integrals of the second order .
On the other hand, equation (2.17 1 ) can be transformed into the equation given in [19] u xy = 1 u B(u x )B(u y ). (2.23) The equation (2.20 1 ) possesses the y-integral of the first orderω = q(u, u y ) − σ(u). Here σ = ν. If c 3 = 0 then we obtain the x-integral of the third order If c 3 = 0 then we have the x-integral of the second order Note that equations in (2.18) and (2.19) are well-known equations, which are integrable by the inverse scattering method (see [19]).
All of the previously mentioned equations possessing x-and y-integrals are contained in the list of Liouville type equations given in [19]. Now we will show how to obtain a solution of an equation from a solution of another one by applying differential substitutions. As an example, we consider case (2.8) with specifying µ(u) = 1, α(u) = ln 2. So we have The equation u xy = u x u y has the x-integral ω(x) = exp(−u)u x . Integrating this equation with respect to x and redenoting ω(x)dx by ω(x) we obtain exp(−u) = ω(x) +ω(y).
Substituting the function u into the equation v = ln(2u x u y ) we get the general solution of the

Proof of the main theorem
In this section we prove Theorem 1. In order to do that we determine the functions f , F , and ϕ in (1.1), (1.2) and (1.7). By substituting function (1.7) into equation (1.2) and using equation (1.1) we get Since the function F (ϕ) depends only on u, u x , and u y , the coefficients at u xx , u yy , and u xx u yy are equal to zero, i.e.
Integration of these equations leads to The remaining terms in (3.1) give Hence, the original classification problem is reduced to the analysis of equations (3.2)-(3.5).
Eliminating the function f from equations (3.

3) and (3.4) we obtain the relation
Applying the operator ∂ 2 ∂ux∂uy to equation (3.6) we arrive at the equation Relation (3.7) is satisfied if one of the following conditions hold: First, let us analyze equation (3.12). By substituting the function ϕ given by (3.2) into equation (3.12) we get Now we integrate the first equation of (3.13) with respect to u y and the second one with respect to u x . This gives The general solutions of these equations are where κ(u) = λ(u)κ (u), λ(u) (u) + C(u) = 0, λ(u)µ (u) + E(u) = 0. Therefore, the function ϕ defined by (3.2) takes the form Here Φ(u) = (u) + µ(u). Furthermore, if we use the point transformation κ(u)du → u in the above formula, we obtain Clearly, function (3.2) satisfying (3.8) also takes form (3.14). Assume that condition (3.9) holds. In this case, the substitution of the functions ϕ defined by (3.2) into (3.9) yields Here c is an arbitrary constant. Hence, function (3.2) takes the form ϕ = α(u)+c ln u x +q(u, u y ).
Replacing α(u) + q(u, u y ) by q(u, u y ) in this equation we get Recall that ϕ ux ϕ uy = 0. This property implies c = 0. Clearly, case (3.10) coincides with (3.9) up to the permutation of x and y. It remains to consider the case when ϕ satisfies (3.11). Based on (3.2), we rewrite (3.11) as By integrating these equations we get the functions q and p, Consequently, the function ϕ defined by formula (3.2) takes the form Thus, to solve the original classification problem it is sufficient to consider three cases: (3.14), (3.15), and (3.16).

Case
When we substitute (3.14) into equation (3.6), we obtain Since u x and u y are regarded as independent variables, the above equation is equivalent to the system From this system we find the functions A and B as By substituting A and B into equations (3.3) and (3.4) we determine f as follows .
Applying the operators ∂ ∂ux and ∂ ∂uy to equation (3.18) we obtain By eliminating F from these equations we get Under the action of the operator ∂ 2 ∂ux∂uy , equation (3.19) takes the form It can be easily seen that the above equation is true if one of the following conditions is met: It should be noted that µ = 0 in cases (3.21)-(3.25). To analyze cases (3.20)-(3.25) in a unified manner we begin by giving the following lemma.
Proof . If condition (3.20) holds then µ(u) = c, where c is an arbitrary constant. Rewriting (3.19) we obtain Since we regard the variables u x , u y as independent, this equation is equivalent to the equations By the same fact that the variables u x , u y are considered as independent we define the function σ as σ(u) = A 1 α (u) + B 1 α (u) + C 1 . According to this we rewrite the above equations as Here A 1 , B 1 , and C 1 are constants. Let us assume that 1, α , and α are linearly independent functions. Clearly, equations (3.38) imply From these equations we get Using the above equations we transform equation (3.18) into the equation Its general solution is given by Substituting function (3.40) into equation (3.39) and using β = B 1 ln u x + C 2 , γ = B 1 ln u y + C 3 we obtain Thus, equations (1.1), (1.2), and (1.7) have the following forms We redenote (α + C 2 + C 3 )/B 1 by α. Under the point transformation v → B 1 v the above equations take the forms The multiplier C 1 /B 1 can be eliminated by the shift v → v + ln(B 1 /C 1 ). Finally, redenoting α − ln(B 1 /C 1 ) by α and c/B 2 1 by c we get where α + cα + 2c 2 = exp α. If c = 0 then these equations take the form (3.26). Otherwise, applying the point transformation u → u/c and redenoting α by α + ln c 2 we can reduce the above equations to form (3.27). Let us assume that 1, α , and α are linearly dependent functions. It means that This equation has two families of solutions The constants c 5 , c 7 can be eliminated by β + c 5 → β, β + c 7 → β in equation (3.14). So there are two possibilities which takes the form under the shifts u → u + (ln c 1 )/c 1 and α → α + c 4 (ln c 1 )/c 1 . Now, let us concentrate on case (3.42), taking into account the fact that µ(u) = c. Equation (3.18) can be rewritten as Applying ∂ ∂u to equation (3.43) we obtain Therefore, Next, by applying the differentiation ∂ ∂u to both sides of this equation, we get It is not difficult to see that the above equation implies Consequently, we have two possibilities (3.45) where c 5 is an arbitrary constant. In this case by using (3.43) we obtain According to the fact that u x and u y are considered as independent variables we have Moreover, since β γ = 0 we get c 5 = 0, hence F ≡ 0. Consequently, equations (1.1), (1.2), and (1.7) take the following forms Applying the operator ∂ ∂u to the above equation gives Collecting the coefficients at exp(c 1 u) and rewriting the remaining terms we obtain Since u x and u y are considered as independent, the first equation is true if and only if c 6 = 0. In this case, it is clear that we obtain the equations (3.28).
Assume that the function α satisfies equation (3.41). Using (3.41) and µ(u) = c we transform equation (3.18) into Differentiating this equation with respect to u and denoting c 2 u 2 + c 3 u + β + γ by z we obtain Now we should analyze equation (3.48). First, we suppose that c 2 = c 3 = 0. The function α described by equation (3.41) vanishes. Equations (3.38) can be written as Here a 1 , b 1 are arbitrary constants. The above equations imply Integrating these equations we obtain distinct formulae which determine the functions β and γ. Uniting these formulae in pairs we arrive at (3.29)-(3.34). Furthermore, we must consider equation ( To eliminate β (u x ) and γ (u y ) we differentiate this equation with respect to u,  .7), Then we redenote c 4 c 5 by c 2 , γ/c 3 by γ. To obtain (3.35) we apply the transformation c 3 x → x once again. Finally, we redenote c/c 2 3 by a 1 , c 2 /c 2 3 by b 1 . Let us assume that β = 0. This assumption enables us to rewrite equation (3.50) in the form Since u x , u y are regarded as independent variables, the above equation is equivalent to the system −c 2 1 β β β 2 = c 5 , c 4 γ = c 5 . Let us turn back to the system (3.51). Given the assumption c 4 = 0, this enables us to find the function γ, γ(u y ) = c 5 c 4 u y + c 6 .
Proof . Condition (3.21) allows us to determine the functions β and γ as β(u x ) = c 1 ln u x , γ(u y ) = c 2 ln u y .
Using these equations (3.18) can be written in the form (3.54) If we apply the operator ∂ ∂ux to both sides of equation (3.54), we obtain Comparing the above equation with equation (3.54) we notice that F = c 1 F . Similarly, differentiating equation (3.54) with respect to u y we deduce that F = c 2 F . These equations yield F = 0 or c 2 = c 1 . If F = 0, equation (3.54) takes the form Since u, u x , and u y are regarded as independent variables and the functions µ and α are functions depending on u, we conclude that c = 0. Consequently, we obtain the equations Finally, replacing µ/c 1 c 2 by µ we get equation (3.52). If we replace c 2 with c 1 , we determine F = c 3 exp(v/c 1 ). Equation (3.54) turns into Thus, the following equations appear First, we redenote µ/c 2 1 by µ and α/c 1 by α. Second, use the transformation v → c 1 v and then the shift v → v − ln c. Finally, replace α + ln c by α, c 3 /c by c 1 , and obtain the equations (3.53).

The last equation and equation (3.63) imply
Similarly, differentiating equation (3.63) with respect to u y we obtain To eliminate u x and u y we apply the operators ∂ ∂ux and ∂ ∂uy to the two above equations, respectively. We get Assuming that c 1 = c 2 = c we define F as follows Substituting the above function F into equation (3.63) we get Since u, u x , and u y are considered as independent variables, the above equation is equivalent to the following system Note that (b 1 , b 2 ) = (0, 0). Otherwise, condition (3.21) is true, which contradicts the assumption of the lemma. If b 2 = 0, b 1 = 0 then c 8 = 0 and where the functions µ and α satisfy the following equations Applying the transformation −cv + c 7 → v and redenoting −cα + c 7 + ln(a 1 a 2 ) by α, µc 2 by µ and b 1 /a 1 by b, we transform (3.1) into (3.56). It is not hard to prove that system (3.64) has no solutions if b 1 b 2 = 0.
Let us suppose that F = 0, hence F (z) = cz + p, where c and p are arbitrary constants. In this case equation (3.63) is represented as It is clear that the coefficients at ln(a 1 u y + b 1 ) and ln(a 2 u x + b 2 ) are equal to zero, i.e. c = 0. Since u, u x , and u y are regarded as independent variables, the above equation is equivalent to the system Note that (b 1 , b 2 ) = (0, 0). Otherwise, condition (3.21) is satisfied, which contradicts the assumption of the lemma. If b 2 = 0, b 1 = 0 then p = 0 and where the functions µ and α satisfy the equations We replace c 1 c 2 µ by µ, −c 1 c 2 α+c 2 ln a 1 +c 1 ln a 2 by α. Using the transformation v → −v/(c 1 c 2 ) and redenoting b 1 /a 1 by b we transform the above equations into (3.57). If b 1 b 2 = 0 then the last system has no solutions. Let us suppose that the functions γ and β are given by (3.61). We rewrite equation (3.18) using (3.61), If F = 0 then we obviously get F = c 3 and −c 2 µ + α = 0, We analyze equation (3.66) based on these equations and find that c 3 = 0. It allows us to determine equations (1.1), (1.2), and (1.7) as follows where the functions µ and α satisfy Point transformations enable us to represent the above equations in form (3.58).
Assuming that F = 0 we can eliminate F from equations (3.67) and (3.68) Recall that variables u, u x , and u y are considered as independent. Hence, the above equation is equivalent to the system Differentiating this equation with respect to u x we obtain One can notice that these two equations imply F + F /c 2 = 0 or F (z) = c 3 exp(−c 2 z). Consequently, we get This equation is not realized because of the given assumptions c 3 = 0 and a = 0. Now, it remains only to consider the case when b = 0. System (3.69) takes the form These equations imply that µ = 0, which contradicts the given assumptions of the lemma.

This equation can be written in the form
Having the fixed value of u we can determine γ as a solution of the ordinary differential equation Moreover, based on this equation we get Note that if u y = κ exp(γ/c 1 ) then γ = c 1 ln(u y /κ) and (γ u y ) = 0. Since the last equation contradicts the assumption of the lemma, we obtain that u y and exp(γ/c 1 ) are linearly independent and that is why In order to find equations (1.1), (1.2), and (1.7) we first set c 5 = 0, hence c 2 = 0 and where the functions β and γ are solutions of the ordinary differential equations and the functions µ and α satisfy the equations We use the transformation v → c 1 v. Next, we redenote α/c 1 by α, γ/c 1 by γ, and µ/c 2 1 by µ.
Having fixed values of u and u y we can define that F (β + c 3 ) = c 4 u x + c 5 /β . Without loss of generality, we redenote β + c 3 by β, therefore Applying the operator ∂ ∂ux to both sides of equation (3.81) and using (3.79) we obtain We differentiate this equation with respect to u x , The above three equations allow us to establish that the function F satisfies the ordinary differential equation Setting definite values of the constants A i , B i , where i = 1, 2, we obtain that the function F can take only one of the following forms Here α i , β i , and γ i are constants, i = 1, 2, . . . , n. Thus, we will focus on (3.90). Let us assume that (α i , β i ) are linearly dependent vectors. This means that a set of numbers µ i satisfying exists. Using this equation we rewrite (3.90) as Now, we will deal with equations (3.83)-(3.89). We begin with (3.83). In this case we have from the equation (3.91). Suppose that α 1 = β 1 = 0. In equation (3.80), we find µ − cµ 2 + α u y γ = 0, α µ − µ 2 (c 1 + c 2 + cu y γ ) + µ u y γ = 0.
We replace β by aβ, γ by aγ. Take the constant a so that a 2 c → 1. Using the transformations u + κ/c → u, v − δ → av and redenoting ac 1 → c 1 obtain equation (3.74). Now, assume that α = 0. The equation u y γ (u y ) = cµ 2 − µ α arises from (3.93). Since u and u y are regarded as independent variables, the last equation leads to u y γ (u y ) = κ, where κ is a constant. This contradicts the assumption of the lemma. Consider the case where α 1 β 1 = 0. We have the equation β (u x ) = −β 1 /(α 1 u x ) which results from (3.92), and it contradicts the assumptions of the lemma.
Let us discuss the case where F is determined by (3.84). Rewriting (3.91) we have This equation must be true for every i = 1, 2, . . . . This requirement implies that µ i = 1, α i = α 1 , and β i = β 1 for every i. Taking this into account we define β as follows: Rewriting (3.79) by using (3.94) we see that this case is not realized. Now, we assume that F is described by (3.85). Equations (3.90), (3.91) are presented in the forms Consequently, It is clear that µ i = 1, γ i = γ 1 . Hence, α i = α 1 , β i = β 1 for every i. So we have Trying to simplify (3.80) by using this equation gives a contradiction to the assumption of the lemma. Concentrate on the case when F satisfies (3.86). We can rewrite equations (3.90), (3.91) as Comparing these equations we conclude that Recall that β depends on the variable u x , while the remaining terms of the above equations are constants. Hence, we have From these equations we obtain γ i exp γ i − γ 1 exp γ i = 0, hence γ i = γ 1 for all i. By (3.90) we determine that α(u) + γ(u y ) = γ 1 , where γ 1 is an arbitrary constant. This equation contradicts γ uy = 0. Let the function F be defined by (3.87). From (3.90) we obtain Note that β 1 = 0, otherwise (β u x ) = 0. Redenoting β + γ 1 by β we rewrite equation (3.95) in the form Since (β u x ) = 0, exp β and u x are linearly independent, the above equation is equivalent to the system Hence, we get where Now, consider case (3.88). Equations (3.90) and (3.91) can be rewritten in the forms It is not hard to show that The dependence of β only on the variable u x implies that µ i = 1 and γ i = γ 1 for every i. This gives α(u) + γ(u y ) = γ 1 , where γ 1 is a constant, which contradicts the assumption γ uy = 0.
It remains to consider the case when F is given by (3.89) to complete the analysis in the case when (α i , β i ) are linearly dependent vectors. Using (3.89) we transform equations (3.90) and (3.91) into = exp(β + γ 1 ) + exp(σ(β + γ 1 )), Consequently, we get Recall that σ = 1. Collecting coefficients at exp β and exp(σβ) yields The above equations provide µ i exp(σγ 1 )(µ σ−1 i − 1) = 0, hence µ i = 1. It follows that γ i = γ 1 for every i. By (3.90) we find that α(u) + γ(u y ) = γ 1 . This equation contradicts γ uy = 0. Now, we must deal with the case when α i , β i , i = 1, 2, satisfying α 1 β 2 − β 1 α 2 = 0 exist. Setting definite values of u, u y in (3.80) we obtain the system Because of the given assumption (u x β ) ux = 0 we get We use Let us analyze equation (3.98) taking into account conditions (3.83)-(3.89). Consider the case when F is given by (3.83). It is not hard to show that equation (3.98) implies u x = 0. Thus, this case is not realized. Next, based on (3.84) we obtain that u x is a constant. So it is also not possible.
Solving the above system we obtain cases (3.77) and (3.78).
3.2 Case ϕ = c ln u x + q(u, u y ) We have the following statement in this case.
Proof . Substituting function (3.15) into equation (3.6) we obtain A(u, u y )q uy (u, u y ) − q u (u, u y )q uy (u, u y )u y + cq u (u, u y ) = B(u, u x ) c u x .
Recall that u x , u y are considered as independent variables. Hence, the above equation is equivalent to the system Aq uy − q u q uy u y + cq u = µ(u), Bc u x = µ(u).
From these equations we find the functions A and B, B = µu x c , A = µ + q u q uy u y − cq u q uy .
By using these equations in each of equations ( q uyuy − 2c q uuy q uy + µ q uy − c q uu q uy + µ u y c = c 2 exp(q/c).
3.3 Case ϕ = α(u) + κ(u) ln u x + µ(u) ln u y By substituting (3.16) into (3.6) we obtain A(u, u y ) − (κ (u) ln u x + µ (u) ln u y + α (u))u y µ(u) u y = B(u, u y ) − (κ (u) ln u x + µ (u) ln u y + α (u))u x κ(u) u x , which can be written as Note that symmetries, x-and y-integrals, and the general solutions of the equations u xy = uu x and u xy = exp(u)u y were given in [11]. The transformation connecting the Liouville equation to the wave equation is well known (see [19]).
Here we just give the outline of the proof.
Scheme of the proof. Substituting the function ψ given by (1.8) into equation (1.1) and using (1.2) we obtain After the point transformation u → A(u) with A − A 2 = 0 the above equation takes the form u xy = f = α(u) + β(u)u x + γ(u)u y .