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 uxy=f(u,ux,uy) and the Klein-Gordon equations vxy=F(v) connected with one another by differential substitutions v=φ(u,ux,uy) such that φuxφuy≠0 over the ring of complex-valued variables.

Gespeichert in:

Bibliographische Detailangaben
Datum:	2012
Hauptverfasser:	Kuznetsova, M.N., Pekcan, A., Zhiber, A.V.
Format:	Artikel
Sprache:	English
Veröffentlicht:	Інститут математики НАН України 2012
Schriftenreihe:	Symmetry, Integrability and Geometry: Methods and Applications
Online Zugang:	http://dspace.nbuv.gov.ua/handle/123456789/148676
Tags:	Tag hinzufügen Keine Tags, Fügen Sie den ersten Tag hinzu!
Назва журналу:	Digital Library of Periodicals of National Academy of Sciences of Ukraine
Zitieren:	The Klein-Gordon Equation and Differential Substitutions of the Form v=φ(u,ux,uy) / M.N. Kuznetsova, A. Pekcan, A.V. Zhiber // Symmetry, Integrability and Geometry: Methods and Applications. — 2012. — Т. 8. — Бібліогр.: 20 назв. — англ.

Institution

Digital Library of Periodicals of National Academy of Sciences of Ukraine

id	irk-123456789-148676
record_format	dspace
spelling	irk-123456789-1486762019-02-19T01:27:58Z The Klein-Gordon Equation and Differential Substitutions of the Form v=φ(u,ux,uy) Kuznetsova, M.N. Pekcan, A. Zhiber, A.V. We present the complete classification of equations of the form uxy=f(u,ux,uy) and the Klein-Gordon equations vxy=F(v) connected with one another by differential substitutions v=φ(u,ux,uy) such that φuxφuy≠0 over the ring of complex-valued variables. 2012 Article The Klein-Gordon Equation and Differential Substitutions of the Form v=φ(u,ux,uy) / M.N. Kuznetsova, A. Pekcan, A.V. Zhiber // Symmetry, Integrability and Geometry: Methods and Applications. — 2012. — Т. 8. — Бібліогр.: 20 назв. — англ. 1815-0659 2010 Mathematics Subject Classification: 35L70 DOI: http://dx.doi.org/10.3842/SIGMA.2012.090 http://dspace.nbuv.gov.ua/handle/123456789/148676 en Symmetry, Integrability and Geometry: Methods and Applications Інститут математики НАН України
institution	Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection	DSpace DC
language	English
description	We present the complete classification of equations of the form uxy=f(u,ux,uy) and the Klein-Gordon equations vxy=F(v) connected with one another by differential substitutions v=φ(u,ux,uy) such that φuxφuy≠0 over the ring of complex-valued variables.
format	Article
author	Kuznetsova, M.N. Pekcan, A. Zhiber, A.V.
spellingShingle	Kuznetsova, M.N. Pekcan, A. Zhiber, A.V. The Klein-Gordon Equation and Differential Substitutions of the Form v=φ(u,ux,uy) Symmetry, Integrability and Geometry: Methods and Applications
author_facet	Kuznetsova, M.N. Pekcan, A. Zhiber, A.V.
author_sort	Kuznetsova, M.N.
title	The Klein-Gordon Equation and Differential Substitutions of the Form v=φ(u,ux,uy)
title_short	The Klein-Gordon Equation and Differential Substitutions of the Form v=φ(u,ux,uy)
title_full	The Klein-Gordon Equation and Differential Substitutions of the Form v=φ(u,ux,uy)
title_fullStr	The Klein-Gordon Equation and Differential Substitutions of the Form v=φ(u,ux,uy)
title_full_unstemmed	The Klein-Gordon Equation and Differential Substitutions of the Form v=φ(u,ux,uy)
title_sort	klein-gordon equation and differential substitutions of the form v=φ(u,ux,uy)
publisher	Інститут математики НАН України
publishDate	2012
url	http://dspace.nbuv.gov.ua/handle/123456789/148676
citation_txt	The Klein-Gordon Equation and Differential Substitutions of the Form v=φ(u,ux,uy) / M.N. Kuznetsova, A. Pekcan, A.V. Zhiber // Symmetry, Integrability and Geometry: Methods and Applications. — 2012. — Т. 8. — Бібліогр.: 20 назв. — англ.
series	Symmetry, Integrability and Geometry: Methods and Applications
work_keys_str_mv	AT kuznetsovamn thekleingordonequationanddifferentialsubstitutionsoftheformvphuuxuy AT pekcana thekleingordonequationanddifferentialsubstitutionsoftheformvphuuxuy AT zhiberav thekleingordonequationanddifferentialsubstitutionsoftheformvphuuxuy AT kuznetsovamn kleingordonequationanddifferentialsubstitutionsoftheformvphuuxuy AT pekcana kleingordonequationanddifferentialsubstitutionsoftheformvphuuxuy AT zhiberav kleingordonequationanddifferentialsubstitutionsoftheformvphuuxuy
first_indexed	2025-07-12T19:56:39Z
last_indexed	2025-07-12T19:56:39Z
_version_	1837472376849170432
fulltext	Symmetry, Integrability and Geometry: Methods and Applications SIGMA 8 (2012), 090, 37 pages The Klein–Gordon Equation and Differential Substitutions of the Form v = ϕ(u, ux, uy) ? Mariya N. KUZNETSOVA †, Aslı PEKCAN ‡ and Anatoliy V. ZHIBER § † Ufa State Aviation Technical University, 12 K. Marx Str., Ufa, Russia E-mail: kuznetsova@matem.anrb.ru ‡ Department of Mathematics, Istanbul University, Istanbul, Turkey E-mail: pekcan@istanbul.edu.tr § Ufa Institute of Mathematics, Russian Academy of Science, 112 Chernyshevskii Str., Ufa, Russia E-mail: zhiber@mail.ru Received April 25, 2012, in final form November 14, 2012; Published online November 26, 2012 http://dx.doi.org/10.3842/SIGMA.2012.090 Abstract. We present the complete classification of equations of the form uxy = f(u, ux, uy) and the Klein–Gordon equations vxy = F (v) connected with one another by differential sub- stitutions v = ϕ(u, ux, uy) such that ϕuxϕuy 6= 0 over the ring of complex-valued variables. Key words: Klein–Gordon equation; differential substitution 2010 Mathematics Subject Classification: 35L70 1 Introduction In this paper, we study the classification problem of equations of the form uxy = f(u, ux, uy) (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 fer- romagnetic 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 vxy = 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. ?This paper is a contribution to the Special Issue “Symmetries of Differential Equations: Frames, Invariants and Applications”. The full collection is available at http://www.emis.de/journals/SIGMA/SDE2012.html mailto:kuznetsova@matem.anrb.ru mailto:pekcan@istanbul.edu.tr mailto:zhiber@mail.ru http://dx.doi.org/10.3842/SIGMA.2012.090 http://www.emis.de/journals/SIGMA/SDE2012.html 2 M.N. Kuznetsova, A. Pekcan and A.V. Zhiber Before going further, let us give some definitions. Let u be a solution of equation (1.1). All the mixed derivatives of u ux, uy, uxx, uyy, . . . (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 and D̄, respectively. Definition 1. The symmetry of equation (1.1) of order (n,m) is the function g = g(u, u1, . . . , un, ū1, . . . , ūm), gun 6= 0, gūm 6= 0, satisfying the equation (DD̄ − fu1D − fū1D̄ − fu)g = 0. Here ui = ∂iu ∂xi and ūi = ∂iu ∂yi , 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 gun and gūm satisfy the so-called characteristic equations D̄(gun) = 0 and D(gūm) = 0, respectively. It actually can be shown that gun depends only on the variables u, u1, . . . , un, while gūm is a function of the variables u, ū1, . . . , ūm. Definition 2. The function ω(u, u1, u2, . . . , un), ωun 6= 0, is called an x-integral of order n of equation (1.1) if D̄(ω) = 0. Similarly, the y-integral of order m is the function ω̄(u, ū1, ū2, . . . , ūm), ω̄ūm 6= 0 which satisfies D(ω̄) = 0. Another important notion is the sequence of the Laplace invariants of equation (1.1). Definition 3. The main generalized Laplace invariants of equation (1.1) are the functions H0 and H1 given by the formulae H1 = −D ( ∂f ∂u1 ) + ∂f ∂u1 ∂f ∂ū1 + ∂f ∂u , H0 = −D̄ ( ∂f ∂ū1 ) + ∂f ∂u1 ∂f ∂ū1 + ∂f ∂u . Other Laplace invariants can be found recurring in the relation DD̄(lnHi) = −Hi+1 −Hi−1 + 2Hi, i ∈ Z. Sokolov and Zhiber [19] showed that the functions H1 and H0 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 (Hr = Hs ≡ 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 Vxy + a(x, y)Vx + b(x, y)Vy + 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. The Klein–Gordon Equation and Differential Substitutions 3 Definition 4. The relation v = ϕ ( u, ∂u ∂x , . . . , ∂nu ∂xn , ∂u ∂y , . . . , ∂mu ∂ym ) (1.4) is called a differential substitution from equation (1.1) to the equation vxy = g(v, vx, vy) (1.5) if function (1.4) satisfies equation (1.5) for every solution u(x, y) of equation (1.1). 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 equa- tions. 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, ux) were described. All these equations are contained in the following list: uxy = uF ′ ( F−1(ux) ) , vxy = F (v), v = F−1(ux); uxy = sinu √ 1− u2 x, vxy = sin v, v = u+ arcsinux; uxy = expu √ 1 + u2 x, vxy = exp v, v = u+ ln ( ux + √ 1 + u2 x ) ; uxy = √ 2uy s′(ux) , vxy = F (v), v = s(ux), where the functions s and f satisfy s′(ux)F (s(ux)) = 1; uxy = c− uyϕu(u, ux) ϕux(u, ux) , vxy = 0, v = ϕ(u, ux); uxy = ux ( ψ(u, uy)− uyα′(u) ) , vxy = exp v, v = α(u) + lnux, where ψu + ψψuy − α′uyψuy = expα; uxy = ux ( ψ(u, uy)− uyα′(u) ) , vxy = 0, v = α(u) + lnux, where ψu + ψψuy − α′uyψuy = 0; uxy = u, vxy = v, v = c1u+ c2ux; uxy = δ(uy), vxy = 1, v = c1u+ c2ux, δ(c1 + c2δ ′) = 1, 4 M.N. Kuznetsova, A. Pekcan and A.V. Zhiber 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, c1 and c2 are constants satisfying (c1, c2) 6= (0, 0), and the function ψ satisfies (ψu, ψuy) 6= (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, vy) are given in the following list: vxy = F (v), uxy = F ′ ( F−1(ux) ) u, u = vy; vxy = 1, uxy = ψ′′ ( ψ−1(u) ) uy ψ′ ( ψ−1(u) ) , u = ψ(vy); vxy = 0, uxy = 0, u = cv + µ(vy); vxy = 0, uxy = −ux expu, u = ln vy − ln v; vxy = v, uxy = u, u = c1v + c2vy; vxy = 1, uxy = 1, u = v + vy, 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, c1 and c2 are constants satisfying (c1, c2) 6= (0, 0). Based on the above lists, Bäcklund transformations have been constructed for some pairs of equations. For instance, the equations uxy = F ′ ( F−1(ux) ) u, vxy = F (v) (1.6) are connected by the Bäcklund transformation v = F−1(ux), u = vy. Kuznetsova [8] showed that linearizations of equation (1.6) are related by Laplace transforma- tions of the first order. For example, we give the equations uxy = ( λ− βnbn−1(ux) ) u, vxy = λv − βvn, n > 0, where λ and β are arbitrary constants, and the function b satisfies the equation λb(ux) − βbn(ux) = ux. The Bäcklund transformation is given by u = vy, v = b(ux). Note that the equation vxy = λv− βvn 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 v = ϕ(u, ux, uy), ϕuxϕuy 6= 0, (1.7) 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, ux, uy) (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 u = ψ(v, vy, vx), ψvyψvx 6= 0, (1.8) over the ring of complex-valued variables. The Klein–Gordon Equation and Differential Substitutions 5 2 Equations transformed into Klein–Gordon equations In this section, we give all possible cases when equation (1.1) is transformed into equation (1.2) by a differential substitution of form (1.7). The main result of this paper is the following theorem. Theorem 1. Suppose that equation (1.1) is transformed into the Klein–Gordon equation (1.2) by differential substitution (1.7). Then equations (1.1), (1.2), and substitution (1.7) take one of the following forms: uxy = √ u2 x + a √ u2 y + b, vxy = 1 2 ( exp v − ab exp(−v) ) , v = ln [( ux + √ u2 x + a )( uy + √ u2 y + b )] ; (2.1) uxy = √ uxuy, vxy = 1 4v, v = √ ux + √ uy; (2.2) uxy = √ ux, vxy = 1 2 , v = √ ux + uy; (2.3) uxy = 1, vxy = 0, v = ux + uy; (2.4) uxy = 1 γ′(uy) , vxy = 1, v = ux + γ(uy) + u, (2.5) where the function γ satisfies 1− γ′′ γ′2 = γ′; uxy = 0, vxy = 0, v = β(ux) + γ(uy) + c3u; (2.6) uxy = µ(u)uxuy, vxy = 0, v = c1 lnux + c2 lnuy + α(u), (2.7) where µ′(c1 + c2) + µ2(c1 + c2) + α′′ + α′µ = 0; uxy = µ(u)uxuy, vxy = exp v, v = ln(uxuy) + α(u), (2.8) where 2µ′ + 2µ2 + α′′ + α′µ = expα; uxy = u, vxy = v, v = c1uy + c2ux + c3u; (2.9) uxy = µ(u)(uy + c)ux, vxy = exp v, v = ln(uy + c) + lnux + α(u), (2.10) where 2µ′ + 2µ2 + α′′ + α′µ = expα, 2µ2 + µ′ + α′µ = expα; uxy = µ(u)(uy + c)ux, vxy = 0, v = c2 ln(uy+ c)+ c1 lnux+ α(u), (2.11) where (µ′ + µ2)(c1 + c2) + α′′ + α′µ = 0, c1µ ′ + µ2(c1 + c2) + α′µ = 0; uxy = µ(u)ux, vxy = 0, v = uy − lnux + α(u), (2.12) where α′′ + µ′ = 0, µ2 − µ′ + α′µ = 0; uxy = µ(u)ux γ′(uy) , vxy = 0, v = lnux + γ(uy) + α(u), (2.13) where c3 + γ′′ γ′2 + c4γ ′uy = 0, α′′ + µ′ + c4µ 2 = 0, and c3µ 2 + µ′ + µ2 + α′µ = 0; uxy = ux (au+ b)γ′(uy) , vxy = exp v, v = lnux + γ(uy)− 2 ln(au+ b), (2.14) 6 M.N. Kuznetsova, A. Pekcan and A.V. Zhiber where c3 + γ′′ γ′2 + c4γ ′uy = −γ′ exp γ, c3 + 1− 3a = 0, and c4 + 2a2 − a = 0; uxy = − 1 uβ′(ux)γ′(uy) , vxy = 0, v = β(ux) + γ(uy), (2.15) where β′′ β′2 = uxβ ′ + c1, γ′′ γ′2 = uyγ ′ − c1; uxy = µ(u) β′(ux)γ′(uy) , vxy = exp v, v = β(ux) + γ(uy) + α(u), (2.16) where ux + 1 β′(ux) = exp(β), uy + 1 γ′(uy) = exp γ, α′′ = expα, and µ = (expα)/α′; uxy = µ(u) β′(ux)γ′(uy) , vxy = exp v, v = β(ux) + γ(uy) + α(u), (2.17) where 2ux + 1 β′(ux) = expβ, 2uy + 1 γ′(uy) = exp γ, α′µ− 2µ2 = expα, and α′2 = 8 expα; uxy = s(u) √ 1− u2 x √ 1− u2 y, vxy = c sin v, v = arcsinux + arcsinuy + p(u), (2.18) where s′′ − 2s3 + λs = 0, p′2 = 2s′ − 2s2 + λ; uxy = s(u)b(ux)b̄(uy), vxy = c1 exp v + c2 exp(−2v), v = −1 2 ln(ux − b(ux))− 1 2 ln(uy − b̄(uy)) + p(u), (2.19) where (ux − b(ux))(b(ux) + 2ux)2 = 1, (uy − b̄(uy))(b̄(uy) + 2uy) 2 = 1, s′′ − 2ss′ − 4s3 = 0, and p′2 − 2sp′ − 3s′ − 2s2 = 0; uxy = ν(u)− qu(u, uy) quy(u, uy) ux, vxy = c3 exp v, v = lnux + q(u, uy), (2.20) where ν − qu quy ( ν − ν − qu q2 uy quyuy − 2 quuy quy ) + ν ′ quy − quu quy + ν ′uy = c3 exp q, quuy 6= 0, up to the point transformations u→ θ(u), v → κ(v), x→ ξx, and y → ηy, and the substitution u+ ξx+ ηy → u, where ξ and η are arbitrary constants. Here c3 and c4 are arbitrary constants, a and b are constants satisfying (a, b) 6= (0, 0), and c, c1, and c2 are nonzero constants; in cases (2.13) and (2.14) the function γ satisfies the condition ( γ′′/γ′2 )′ 6= 0; in cases (2.15)– (2.17) the functions β and γ satisfy the conditions ( β′′/β′2 )′ 6= 0 and ( γ′′/γ′2 )′ 6= 0 accordingly, the function µ satisfies µ′ 6= 0, and µ 6= 0 in all cases. Now, let us analyze some of the above equations in detail. Consider (2.1) with ab 6= 0. Using the point transformations √ ax→ x, √ by → y, and v − ln(ab)1/2 → v, we obtain uxy = √ u2 x + 1 √ u2 y + 1. (2.21) Equation (2.21) is transformed into the sine-Gordon equation vxy = 1 2 ( exp v − exp(−v) ) The Klein–Gordon Equation and Differential Substitutions 7 by the differential substitution v = ln [( ux + √ u2 x + 1 )( uy + √ u2 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− u2 x − iux ) = −i arcsinux we can also convert the above equations into uxy = √ 1− u2 x1− u2 y √ 1− u2 y, vxy = − sin v, v = arcsinux + arcsinuy. Now, assume that a = 0. Under the transformations v − ln 2→ v, √ by → y, and v − ln √ b→ v equations (2.1) take the form uxy = ux √ u2 y + 1, vxy = exp v, v = lnux + ln ( uy + √ u2 y + 1 ) . (2.22) Applying the transformation iy → y to the above equations we arrive at uxy = ux √ 1− u2 y, vxy = −i exp v, v = −i arcsinuy + lnux. As shown in [11], equation (2.221) has symmetries of the third order. In [11] the x- and y- integrals and the general solution of equation (2.221) were presented. Note that the equation (2.21) is the Goursat equation. Its symmetries of the third order can be found, for instance, in [11]. The equation (2.31) has symmetries of the third order [11]. The x- and y-integrals of this equation are given by ω = uxx√ ux , ω̄ = uyyy. Consider cases (2.7) and (2.8). The equation uxy = µ(u)uxuy possesses the x- and y-integrals of the first order, ω = lnux − σ(u), ω̄ = lnuy − σ(u). Here σ′ = µ. The equation uxy = µ(u)(uy + c)ux in cases (2.10) and (2.11) possess the y-integral of the first order ω̄ = ln(uy + c)− σ(u), where σ′ = µ. The x-integral in case (2.10) is ω = uxxx ux − 3 2 u2 xx u2 x − 1 2 ( µ2(u) + 2µ(u)α′(u) + α′2(u) ) u2 x, and in case (2.11) we get the x-integral ω = c2µ(u)ux + c1 uxx ux + α′(u)ux. The equation (2.141) possesses the y-integral of the first order and the x-integral of the third order ω̄ = γ(uy)− 1 a ln(au+ b), ω = uxxx ux − 3 2 u2 xx u2 x + u2 x(2a− 1) 2(au+ b)2 . Now, we consider the equation which appears in (2.16) and (2.17). The equation (2.161) is transformed into the equation presented in [19] by a point transformation and has the integrals of the second order ω = β′(ux)uxx − µ′(u) µ(u)β′(ux) , ω̄ = γ′(uy)uyy − µ′(u) µ(u)γ′(uy) . 8 M.N. Kuznetsova, A. Pekcan and A.V. Zhiber On the other hand, equation (2.171) can be transformed into the equation given in [19] uxy = 1 u B(ux)B̄(uy). (2.23) Here B(ux)B′(ux) + B(ux) − 2ux = 0, B̄(uy)B̄ ′(uy) + B̄(uy) − 2uy = 0. The integrals of equation (2.23) are [19] ω = uxxx B + 2(B − ux) B3 u2 xx + 2(2ux +B) uB + B(ux +B) u2 , ω̄ = uyyy B̄ + 2(B̄ − uy) B̄3 u2 yy + 2(2uy + B̄) uB̄ + B̄(uy + B̄) u2 . The equation (2.201) possesses the y-integral of the first order ω̄ = q(u, uy) − σ(u). Here σ′ = ν. If c3 6= 0 then we obtain the x-integral of the third order ω = uxxx ux − 3 2 u2 xx u2 x + ν ′(u)u2 x − 1 2 ν2(u)u2 x. If c3 = 0 then we have the x-integral of the second order ω = uxx ux + ν(u)ux. 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 uxy = uxuy, v = ln(2uxuy), vxy = exp v. The equation uxy = uxuy has the x-integral ω(x) = exp(−u)ux. Integrating this equation with respect to x and redenoting ∫ ω(x)dx by ω(x) we obtain exp(−u) = ω(x) + ω̄(y). Hence u = − ln ( ω(x) + ω̄(y) ) . Substituting the function u into the equation v = ln(2uxuy) we get the general solution of the Liouville equation vxy = exp v as v(x, y) = ln ( 2ω′(x)ω̄′(y)( ω(x) + ω̄(y) )2 ) . 3 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 ϕuf + ux ( ϕuuuy + ϕuuxf + ϕuuyuyy ) + uxx ( ϕuxuuy + ϕuxuxf + ϕuxuyuyy ) The Klein–Gordon Equation and Differential Substitutions 9 + ϕux ( fuux + fuxuxx + fuyf ) + ϕuy ( fuuy + fuxf + fuyuyy ) + f ( ϕuyuuy + ϕuyuxf + ϕuyuyuyy ) = F (ϕ). (3.1) Since the function F (ϕ) depends only on u, ux, and uy, the coefficients at uxx, uyy, and uxxuyy are equal to zero, i.e. ϕuxuy = 0, ϕuuxuy + ϕuxuxf + ϕuxfux = 0, ϕuuyux + ϕuyfuy + fϕuyuy = 0. Integration of these equations leads to ϕ = p(u, ux) + q(u, uy), (3.2) ϕuuy + ϕuxf = A(u, uy), (3.3) ϕuux + ϕuyf = B(u, ux). (3.4) The remaining terms in (3.1) give f ( ϕu+ uxϕuux + ϕuxfuy + ϕuyfux + uyϕuuy ) + ϕuuuxuy+ ( uxϕux + uyϕuy ) fu = F (ϕ). (3.5) 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( A− uyϕu ) ϕuy = ( B − uxϕu ) ϕux . (3.6) Applying the operator ∂2 ∂ux∂uy to equation (3.6) we arrive at the equation( uyϕuy ) uy ϕuux = ( uxϕux ) ux ϕuuy . (3.7) Relation (3.7) is satisfied if one of the following conditions hold: ϕuux = 0, ϕuuy = 0, (3.8) ϕuux = 0, ( uxϕux ) ux = 0, (3.9) (uyϕuy)uy = 0, ϕuuy = 0, (3.10) (uyϕuy)uy = 0, (uxϕux)ux = 0, (3.11) (uyϕuy)uy ϕuuy = (uxϕux)ux ϕuux = λ(u), λ(u) 6= 0. (3.12) First, let us analyze equation (3.12). By substituting the function ϕ given by (3.2) into equation (3.12) we get (uyquy)uy = λ(u)quuy , (uxpux)ux = λ(u)puux . (3.13) Now we integrate the first equation of (3.13) with respect to uy and the second one with respect to ux. This gives uyquy = λ(u)qu + C(u), uxpux = λ(u)pu + E(u). The general solutions of these equations are q = Φ1(uyκ(u)) + ε(u), p = Φ2(uxκ(u)) + µ(u), where κ(u) = λ(u)κ′(u), λ(u)ε′(u) +C(u) = 0, λ(u)µ′(u) +E(u) = 0. Therefore, the function ϕ defined by (3.2) takes the form ϕ = Φ(u) + Φ1(uyκ(u)) + Φ2(uxκ(u)). 10 M.N. Kuznetsova, A. Pekcan and A.V. Zhiber Here Φ(u) = ε(u) + µ(u). Furthermore, if we use the point transformation ∫ κ(u)du→ u in the above formula, we obtain ϕ = α(u) + β(ux) + γ(uy). (3.14) 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 puux = 0, (uxpux)ux = 0, which gives p = α(u) + c lnux. Here c is an arbitrary constant. Hence, function (3.2) takes the form ϕ = α(u)+c lnux+q(u, uy). Replacing α(u) + q(u, uy) by q(u, uy) in this equation we get ϕ = c lnux + q(u, uy). (3.15) Recall that ϕuxϕuy 6= 0. This property implies c 6= 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 (uyquy)uy = 0, (uxpux)ux = 0. By integrating these equations we get the functions q and p, q = µ(u) lnuy + ε(u), p = κ(u) lnux + δ(u). Consequently, the function ϕ defined by formula (3.2) takes the form ϕ = α(u) + κ(u) lnux + µ(u) lnuy. (3.16) Thus, to solve the original classification problem it is sufficient to consider three cases: (3.14), (3.15), and (3.16). 3.1 Case ϕ = α(u) + β(ux) + γ(uy) When we substitute (3.14) into equation (3.6), we obtain( A(u, uy)− uyα′(u) ) γ′(uy) = ( B(u, ux)− uxα′(u) ) β′(ux). Since ux and uy are regarded as independent variables, the above equation is equivalent to the system( A(u, uy)− uyα′(u) ) γ′(uy) = µ(u), ( B(u, ux)− uxα′(u) ) β′(ux) = µ(u). From this system we find the functions A and B as A = µ γ′ + uyα ′, B = µ β′ + uxα ′. By substituting A and B into equations (3.3) and (3.4) we determine f as follows f = µ(u) β′(ux)γ′(uy) . (3.17) The Klein–Gordon Equation and Differential Substitutions 11 Using (3.17) we transform equation (3.5) into α′µ β′γ′ − µ2 ( γ′′ γ′2 + β′′ β′2 ) 1 β′γ′ + α′′uxuy + µ′ ( ux γ′ + uy β′ ) = F (α+ β + γ). (3.18) Applying the operators ∂ ∂ux and ∂ ∂uy to equation (3.18) we obtain −α′µ β′′ β′2γ′ − µ2 ( − γ ′′ γ′3 β′′ β′2 + 1 γ′ ( β′′ β′3 )′) + α′′uy + µ′ ( 1 γ′ − uy β′′ β′2 ) = F ′(α+ β + γ)β′, −α′µ γ′′ β′γ′2 − µ2 (( γ′′ γ′3 )′ 1 β′ − β′′ β′3 γ′′ γ′2 ) + α′′ux + µ′ ( −ux γ′′ γ′2 + 1 β′ ) = F ′(α+ β + γ)γ′. By eliminating F ′ from these equations we get −α′µ β ′′ β′2 − µ2 ( β′′ β′3 )′ + α′′uyγ ′ − µ′uyγ′ β′′ β′2 = −α′µ γ ′′ γ′2 − µ2 ( γ′′ γ′3 )′ + α′′uxβ ′ − µ′uxβ′ γ′′ γ′2 . (3.19) Under the action of the operator ∂2 ∂ux∂uy , equation (3.19) takes the form µ′ ( (uxβ ′)′ ( γ′′ γ′2 )′ − (uyγ ′)′ ( β′′ β′2 )′) = 0. It can be easily seen that the above equation is true if one of the following conditions is met: µ′(u) = 0, (3.20) (uxβ ′)′ = 0, (uyγ ′)′ = 0, (3.21) (uxβ ′)′ = 0, ( β′′ β′2 )′ = 0, (3.22)( γ′′ γ′2 )′ = 0, (uyγ ′)′ = 0, (3.23)( γ′′ γ′2 )′ = 0, ( β′′ β′2 )′ = 0, (3.24) (uxβ ′)′( β′′ β′2 )′ = (uyγ ′)′( γ′′ γ′2 )′ 6= 0. (3.25) It should be noted that µ′ 6= 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. Lemma 1. By condition (3.20), equations (1.1), (1.2), and substitution (1.7) take one of the following forms: uxy = 0, vxy = exp v, v = α(u) + ln(uxuy), (3.26) where the function α satisfies α′′ = expα; uxy = uxuy, vxy = exp v, v = α(u) + ln(uxuy), (3.27) 12 M.N. Kuznetsova, A. Pekcan and A.V. Zhiber where α′′ + α′ + 2 = expα; uxy = −uxuy, vxy = 0, v = expu+ (a1 + b1)u+ a1 lnux + b1 lnuy; (3.28) uxy = c √ u2 x + a2 √ u2 y + b2, vxy = c2 2 ( exp v − a2b2 exp(−v) ) , v = ln [( ux + √ u2 x + a2 )( uy + √ u2 y + b2 )] ; (3.29) uxy = c √ 1− u2 x √ 1− u2 y, vxy = −c2 sin v, v = arcsinux + arcsinuy; (3.30) uxy = c √ uxuy, vxy = c2v 4 , v = √ ux + √ uy; (3.31) uxy = c √ ux, vxy = c2 2 , v = √ ux + uy; (3.32) uxy = c, vxy = 0, v = ux + uy; (3.33) uxy = cuy √ 1− u2 x, vxy = −ic2 exp v, v = −i arcsinux + lnuy; (3.34) uxy = a1 γ′(uy) , vxy = b1, v = ux + γ(uy) + u, (3.35) where a1 − a2 1 γ′′ γ′2 = b1γ ′; uxy = a(ux + c7)(uy + c9), vxy = 0, v = a1 ln(ux + c7) + b1 ln(uy + c9) + u, (3.36) where aa1 + ab1 + 1 = 0; uxy = 0, vxy = 0, v = β(ux) + γ(uy) + u, (3.37) up to the point transformations u → θ(u), v → κ(v), x → ξx, and y → ηy and the substitution u + ξx + ηy → u, where ξ and η are arbitrary constants. Here α′′, α′, and 1 are linearly independent functions, c, c1, c2, c7, c9, a1 6= 0, b1 6= 0, a 6= 0, b2, and a2 are arbitrary constants. Proof. If condition (3.20) holds then µ(u) = c, where c is an arbitrary constant. Rewri- ting (3.19) we obtain cα′(u) β′′(ux) β′2(ux) + c2 ( β′′(ux) β′3(ux) )′ + α′′(u)uxβ ′(ux) = cα′(u) γ′′(uy) γ′2(uy) + c2 ( γ′′(uy) γ′3(uy) )′ + α′′(u)uyγ ′(uy). Since we regard the variables ux, uy as independent, this equation is equivalent to the equations cα′(u) β′′(ux) β′2(ux) + c2 ( β′′(ux) β′3(ux) )′ + α′′(u)uxβ ′(ux) = σ(u), cα′(u) γ′′(uy) γ′2(uy) + c2 ( γ′′(uy) γ′3(uy) )′ + α′′(u)uyγ ′(uy) = σ(u). By the same fact that the variables ux, uy are considered as independent we define the func- tion σ as σ(u) = A1α ′(u) + B1α ′′(u) + C1. According to this we rewrite the above equations as α′ ( c β′′ β′2 −A1 ) + α′′ ( uxβ ′ −B1 ) = C1 − c2 ( β′′ β′3 )′ , The Klein–Gordon Equation and Differential Substitutions 13 α′ ( c γ′′ γ′2 −A1 ) + α′′ ( uyγ ′ −B1 ) = C1 − c2 ( γ′′ γ′3 )′ . (3.38) Here A1, B1, and C1 are constants. Let us assume that 1, α′, and α′′ are linearly independent functions. Clearly, equations (3.38) imply c β′′ β′2 = A1, uxβ ′ = B1, C1 − c2 ( β′′ β′3 )′ = 0, c γ′′ γ′2 = A1, uyγ ′ = B1, C1 − c2 ( γ′′ γ′3 )′ = 0. From these equations we get β′ = B1 ux , γ′ = B1 uy , − c B1 = A1, C1 + c2 B2 1 = 0. Using the above equations we transform equation (3.18) into the equation uxuy ( cα′ B2 1 + 2c2 B3 1 + α′′ ) = F (α+ β + γ). (3.39) Since 1, α′, and α′′ are linearly independent functions, the left-hand side of equation (3.39) does not vanish. Then F 6= 0. By differentiating (3.39) with respect to ux and using β′ = B1/ux we get the equation 1 = F ′(z)B1/F (z), where z = α+ β + γ. Its general solution is given by F (z) = C1 exp(z/B1). (3.40) Substituting function (3.40) into equation (3.39) and using β = B1 lnux +C2, γ = B1 lnuy +C3 we obtain cα′ B2 1 + 2c2 B3 1 + α′′ = C1 exp ( α B1 + C2 B1 + C3 B1 ) . Thus, equations (1.1), (1.2), and (1.7) have the following forms uxy = cuxuy B2 1 , vxy = C1 exp(v/B1), v = α(u) +B1 ln(uxuy) + C2 + C3, where c α′ B2 1 + 2c2 1 B3 1 + α′′ = C1 exp ( α+ C2 + C3 B1 ) . We redenote (α + C2 + C3)/B1 by α. Under the point transformation v → B1v the above equations take the forms uxy = cuxuy B2 1 , vxy = C1 B1 exp v, v = α(u) + ln(uxuy), where c α′ B2 1 + 2c2 1 B4 1 + α′′ = C1 B1 expα. 14 M.N. Kuznetsova, A. Pekcan and A.V. Zhiber The multiplier C1/B1 can be eliminated by the shift v → v + ln(B1/C1). Finally, redenoting α− ln(B1/C1) by α and c/B2 1 by c we get uxy = cuxuy, vxy = exp v, v = α(u) + ln(uxuy), where α′′ + cα′ + 2c2 = expα. 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 c2 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 C1α ′′ + C2α ′ + C3 = 0, (C1, C2, C3) 6= (0, 0, 0). If C1 = 0 then C2 6= 0 and we get α′ = c. Otherwise, α′′ = c1α ′ + c2. Case α′ = c is a subcase of α′′ = c1α ′ + c2. This equation has two families of solutions α = c3u 2 + c4u+ c5, α = 1 c1 exp(c1u) + c6u+ c7. The constants c5, c7 can be eliminated by β + c5 → β, β + c7 → β in equation (3.14). So there are two possibilities α = c2u 2 + c3u (3.41) and α = ( exp c1u ) /c1 + c4u, which takes the form α = exp(c1u) + c4u, c1 6= 0 (3.42) under the shifts u→ u+ (ln c1)/c1 and α→ α+ c4(ln c1)/c1. Now, let us concentrate on case (3.42), taking into account the fact that µ(u) = c. Equa- tion (3.18) can be rewritten as c(c1 exp(c1u) + c4) β′γ′ − c2 ( γ′′ γ′2 + β′′ β′2 ) 1 β′γ′ + c2 1 exp(c1u)uxuy = F (α+ β + γ). (3.43) Applying ∂ ∂u to equation (3.43) we obtain cc2 1 exp(c1u) β′γ′ + c3 1 exp(c1u)uxuy = F ′(α+ β + γ)(c1 exp(c1u) + c4). Therefore, cc2 1 β′γ′ + c3 1uxuy = (c1 + c4 exp(−c1u))F ′(α+ β + γ). Next, by applying the differentiation ∂ ∂u to both sides of this equation, we get −c1c4 exp(−c1u)F ′(α+ β + γ) + (c1 + c4 exp(−c1u))(c1 exp(c1u) + c4)F ′′(α+ β + γ) = 0. It is not difficult to see that the above equation implies (c1 exp(c1u) + c4)2F ′′(α+ β + γ) = c1c4F ′(α+ β + γ). The Klein–Gordon Equation and Differential Substitutions 15 Consequently, we have two possibilities F ′(α+ β + γ) = 0, (3.44) F ′′(α+ β + γ) F ′(α+ β + γ) = c1c4 (c1 exp(c1u) + c4)2 . (3.45) Equation (3.44) yields F = c5, where c5 is an arbitrary constant. In this case by using (3.43) we obtain cc1 β′γ′ + c2 1uxuy = 0, cc4 β′γ′ − c2 ( γ′′ γ′2 + β′′ β′2 ) 1 β′γ′ = c5. According to the fact that ux and uy are considered as independent variables we have β′(ux) = cc1 c6ux , γ′(uy) = − c6 c2 1uy , cc4 − c2 ( c2 1 c6 − c6 c1c ) = c5β ′(ux)γ′(uy). Moreover, since β′γ′ 6= 0 we get c5 = 0, hence F ≡ 0. Consequently, equations (1.1), (1.2), and (1.7) take the following forms uxy = −c1uxuy, vxy = 0, v = exp c1u+ c4u+ cc1 c6 lnux − c6 c2 1 lnuy + c7, where c4 − cc2 1 c6 + c6 c1 = 0, c1 6= 0. Using the point transformations u→ u/c1, v → v−cc1 ln(c1)/c6+c6 ln(c1)/c2 1+c7, and redenoting cc1/c6 by a1, −c6/c 2 1 by b1 we get equation (3.28). Now, suppose that (3.45) is true. Applying ∂ ∂ux to both sides of equation (3.45) we get( F ′′ F ′ )′ β′ = 0. Recall that β′ 6= 0, therefore F ′′/F ′ = 0. This equation has two families of solutions. Namely, F (z) = c6 exp c5z + c7, c5c6 6= 0, which turns into F (z) = exp c5z + c7, c5 6= 0 (3.46) by the shift z → z − (ln c6) /c5, and F (z) = c6z + c7, c6 6= 0. (3.47) Now consider equation (3.46). In this case, equation (3.43) takes the form c(c1 exp(c1u) + c4) β′γ′ − c2 ( γ′′ γ′2 + β′′ β′2 ) 1 β′γ′ + c2 1 exp(c1u)uxuy = exp ( c5(exp c1u+ c4u) ) exp(c5β) exp(c5γ). This equation is not satisfied because c5c1 6= 0. Let us focus on equation (3.47). Equation (3.43) can be written as c(c1 exp(c1u) + c4) β′γ′ − c2 ( γ′′ γ′2 + β′′ β′2 ) 1 β′γ′ + c2 1 exp(c1u)uxuy 16 M.N. Kuznetsova, A. Pekcan and A.V. Zhiber = c6(exp(c1u) + c4u+ β + γ) + c7. Applying the operator ∂ ∂u to the above equation gives cc2 1 exp(c1u) β′γ′ + c3 1uxuy exp(c1u) = c6(c1 exp c1u+ c4). Collecting the coefficients at exp(c1u) and rewriting the remaining terms we obtain cc2 1 β′γ′ + c3 1uxuy = c6c1, c6c4 = 0. Since ux and uy are considered as independent, the first equation is true if and only if c6 = 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 c β′γ′ (2c2u+ c3)− c2 ( γ′′ γ′2 + β′′ β′2 ) 1 β′γ′ + 2c2uxuy = F (c2u 2 + c3u+ β + γ). Differentiating this equation with respect to u and denoting c2u 2 + c3u+ β + γ by z we obtain 2c2 c β′γ′ = F ′(z)(2c2u+ c3). (3.48) Now we should analyze equation (3.48). First, we suppose that c2 = c3 = 0. The function α described by equation (3.41) vanishes. Equations (3.38) can be written as c1ux − c2 β ′′ β′3 = a1, c1uy − c2 γ ′′ γ′3 = b1. Here a1, b1 are arbitrary constants. The above equations imply β′(ux) = √ −c2 1√ c1u2 x − 2a1ux + 2a2 , γ′(uy) = √ −c2 1√ c1u2 y − 2b1uy + 2b2 . 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 (3.48) if c2 6= 0, c3 = 0, and c2c3 6= 0. Taking the logarithm of both sides of equation (3.48) leads to ln ( 2c2 c β′γ′ ) = lnF ′(z) + ln(2c2u+ c3). To eliminate β′(ux) and γ′(uy) we differentiate this equation with respect to u, 0 = F ′′ F ′ (2c2u+ c3) + 2c2 2c2u+ c3 . (3.49) Applying ∂ ∂ux to both sides of equation (3.49) we get (F ′′/F ′)′ = 0, which means that F ′′/F = c4. By virtue of this, equation (3.49) is written as c4(2c2u+ c3)2 + 2c2 = 0. Hence c2 = 0. This contradicts c2 6= 0. The Klein–Gordon Equation and Differential Substitutions 17 It remains to discuss the case if c2 = 0, c3 6= 0. It is clear that we have F (z) = c4 from equation (3.48). Here c4 is an arbitrary constant. Rewriting (3.18) with α = c3u, µ = c we get c3c− c2 ( γ′′ γ′2 + β′′ β′2 ) = c4β ′γ′. (3.50) The equation −c2 ( β′′ β′2 )′ = c4β ′′γ′, arises when we apply ∂ ∂ux to both the sides of equation (3.50). Suppose that β′′ = 0. Determining the function β as β(ux) = c5ux + c6, we transform equation (3.50) into an ordinary differential equation c3c− c2 γ ′′ γ′2 = c4c5γ ′. Thus, we find equations of forms (1.1), (1.2), and (1.7), uxy = c c5γ′(uy) , vxy = c4, v = c5ux + γ(uy) + c3u, where c3c − c2γ′′/γ′2 = c4c5γ ′, c5 6= 0. We use the transformations x/c5 → x, v/c3 → v. Then we redenote c4c5 by c2, γ/c3 by γ. To obtain (3.35) we apply the transformation c3x→ x once again. Finally, we redenote c/c2 3 by a1, c2/c 2 3 by b1. Let us assume that β′′ 6= 0. This assumption enables us to rewrite equation (3.50) in the form −c2 1 β′′(ux) ( β′′(ux) β′2(ux) )′ = c4γ ′(uy). Since ux, uy are regarded as independent variables, the above equation is equivalent to the system −c2 1 β′′ ( β′′ β′2 )′ = c5, c4γ ′ = c5. (3.51) If c4 = 0 then c5 = 0, which yields c = 0 or β′′/β′2 = −c6 6= 0. The last equation implies β(ux) = 1 c6 ln(c6ux + c7). Substituting this function into equation (3.50) and using c4 = 0 we can define the function γ as γ(uy) = 1 c8 ln(c8uy + c9), and the following equations result in uxy = c(c6ux + c7)(c8uy + c9), vxy = 0, v = 1 c6 ln(c6ux + c7) + 1 c8 ln(c8uy + c9) + c3u, where cc6 +cc8 +c3 = 0, c3 6= 0. We use the transformations v/c3 → v, x/c6 → x, and y/c8 → y. Replacing 1/(c3c6) by a1, 1/(c3c8) by b1, and cc6c8 by a, we get (3.36). If c = 0 then c5 = c4 = 0, and we obtain (3.37). 18 M.N. Kuznetsova, A. Pekcan and A.V. Zhiber Let us turn back to the system (3.51). Given the assumption c4 6= 0, this enables us to find the function γ, γ(uy) = c5 c4 uy + c6. We also have an ordinary differential equation defining the function β, −c2 β ′′ β′2 = c5β ′ + c7. Rewriting equation (3.50) by using these equations we get c7 + cc3 = 0 and, therefore, uxy = cc4 c5β′(ux) , v = c5 c4 uy + β(ux) + c3u, vxy = c4, where −c2β′′/β′2 = c5β ′ + c7, c7 + cc3 = 0, and c4c5 6= 0. Clearly, this case coincides with equation (3.35) up to the permutation of x and y. � Lemma 2. Assume that (3.21) is satisfied and µ′(u) 6= 0. Then equations (1.1), (1.2), and (1.7) take one of the following forms: uxy = µ(u)uxuy, vxy = 0, v = c1 lnux + c2 lnuy + α(u), (3.52) where the functions µ and α satisfy µ′(c1 + c2) + µ2(c1 + c2) + α′′ + α′µ = 0, µ′ 6= 0; uxy = µ(u)uxuy, vxy = exp v, v = ln(uxuy) + α(u), (3.53) where µ and α satisfy 2µ′ + 2µ2 + α′′ + α′µ = expα, up to the point transformations u→ θ(u), v → κ(v), x → ξx, and y → ηy, where ξ and η are arbitrary constants. Here c1 and c2 are nonzero constants. Proof. Condition (3.21) allows us to determine the functions β and γ as β(ux) = c1 lnux, γ(uy) = c2 lnuy. Using these equations (3.18) can be written in the form µ′(u)uxuy ( 1 c2 + 1 c1 ) + µ2(u)uxuy c1c2 ( 1 c2 + 1 c1 ) + α′′(u)uxuy + α′(u)µ(u) uxuy c1c2 = F ( c1 lnux + c2 lnuy + α(u) ) . (3.54) If we apply the operator ∂ ∂ux to both sides of equation (3.54), we obtain µ′(u)uy ( 1 c2 + 1 c1 ) + µ2(u)uy c1c2 ( 1 c2 + 1 c1 ) + α′′(u)uy + α′(u)µ(u) uy c1c2 = F ′c1 ux . Comparing the above equation with equation (3.54) we notice that F = c1F ′. Similarly, diffe- rentiating equation (3.54) with respect to uy we deduce that F = c2F ′. These equations yield F ′ = 0 or c2 = c1. If F ′ = 0, equation (3.54) takes the form uxuy ( µ′(u) ( 1 c2 + 1 c1 ) + µ2 c1c2 ( 1 c2 + 1 c1 ) + α′′ + α′µ c1c2 ) = c. The Klein–Gordon Equation and Differential Substitutions 19 Since u, ux, and uy 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 uxy = µ(u)uxuy c1c2 , vxy = 0, v = c1 lnux + c2 lnuy + α(u), where µ′ ( 1 c2 + 1 c1 ) + µ2 c1c2 ( 1 c2 + 1 c1 ) + α′′ + α′µ c1c2 = 0. Finally, replacing µ/c1c2 by µ we get equation (3.52). If we replace c2 with c1, we determine F = c3 exp(v/c1). Equation (3.54) turns into 2µ′uxuy c1 + 2µ2uxuy c3 1 + α′′uxuy + α′µuxuy c2 1 = c3uxuy exp(α(u)/c1). Thus, the following equations appear uxy = 1 c2 1 µ(u)uxuy, vxy = c3 exp(v/c1), v = c1 lnuxuy + α(u), where 2µ′ c1 + 2µ2 c3 1 + α′′ + α′µ c2 1 = c3 exp(α/c1). First, we redenote µ/c2 1 by µ and α/c1 by α. Second, use the transformation v → c1v and then the shift v → v− ln c. Finally, replace α+ ln c by α, c3/c by c1, and obtain the equations (3.53). � Lemma 3. Assume that condition (3.24) is satisfied but (3.20) and (3.21) are not. Then equations (1.1), (1.2), and (1.7) take one of the following forms: uxy = u, vxy = v, v = c1uy + c2ux + c3u; (3.55) uxy = µ(u)(uy + b)ux, vxy = exp v, v = ln(uy + b) + lnux + α(u), (3.56) where the functions µ and α satisfy 2µ′ + 2µ2 + α′′ + α′µ = expα, 2µ2 + µ′ + α′µ = expα; uxy = µ(u)(uy + b)ux, vxy = 0, v = c2 ln(uy + b) + c1 lnux + α(u), (3.57) where µ and α satisfy (µ′ + µ2)(c1 + c2) + α′′ + α′µ = 0, c1µ ′ + µ2(c1 + c2) + α′µ = 0; uxy = µ(u)ux, vxy = 0, v = uy − lnux + α(u), (3.58) where µ and α satisfy α′′+µ′ = 0, µ2−µ′+α′µ = 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 c3 is an arbitrary constant, c1, c2, and b are nonzero constants. Proof. Condition (3.24) implies the following three possibilities for functions β and γ γ(uy) = c1uy + c2, β(ux) = c3ux + c4, (3.59) γ(uy) = − 1 c1 ln(a1uy + b1), β(ux) = − 1 c2 ln(a2ux + b2), (3.60) γ(uy) = c1uy + c2, β(ux) = − 1 c3 ln(aux + b). (3.61) 20 M.N. Kuznetsova, A. Pekcan and A.V. Zhiber According to (3.59), equation (3.18) can be written as µ′(u)uy c3 + µ′(u)ux c1 + α′′(u)uxuy + α′(u)µ(u) c1c3 = F (c1uy + c3ux + α(u)). (3.62) Applying the operators ∂ ∂ux and ∂ ∂uy to both sides of (3.62) gives µ′ c1 + α′′uy = F ′c3, µ′ c3 + α′′ux = F ′c1. Eliminating F ′ from the above equations we obtain α′′(c1uy − c3ux) = 0. Clearly, we have α′′ = 0, hence α = c2u + c4. Furthermore, by using any of the above equations we obtain F ′ = µ′/c1c3. Consequently, F (z) = c5 c1c3 z + c7, z = c1uy + c3ux + α(u). The equation µ′′uy c3 + µ′′ux c1 + c2µ ′ c1c3 = F ′c2 arises after the differentiation of equation (3.62) with respect to u. Substituting F ′ = µ′/c1c3 into this equation yields µ(u) = c5u+ c6. Therefore, the equation (3.62) is equivalent to c2c6 c1c3 = c5c4 c1c3 + c7. Thus, we find that equations (1.1), (1.2), and the substitution (1.7) have the forms uxy = c5u+ c6 c1c3 , vxy = c5 c1c3 v + c7, v = c1uy + c3ux + c2u+ c4. Using the transformations u + c6/c5 → cu, v + c1c3c7/c5 → cv and replacing c5/c1 by c3 we get (3.55). Let us discuss the case when the functions γ and β are of form (3.60). It turns out that equation (3.18) takes the form −c2µ ′uy a2ux + b2 a2 − µ2 c1c 2 2 a1a2 (a2ux + b2)(a1uy + b1)− µ2 c 2 1c2 a1a2 (a1uy + b1)(a2ux + b2) − c1µ ′ux a1uy + b1 a1 + α′′uxuy + α′µ c1c2 a1a2 (a2ux + b2)(a1uy + b1) = F ( − 1 c1 ln(a1uy + b1)− 1 c2 ln(a2ux + b2) + α(u) ) . (3.63) Applying the operator ∂ ∂ux to both sides of equation (3.63) leads to −c2µ ′uy − µ2 c1c 2 2 a1 (a1uy + b1)− c1µ ′a1uy + b1 a1 − µ2 c 2 1c2 a1 (a1uy + b1) + α′′uy + α′µ c1c2 a1 (a1uy + b1) = F ′ ( − 1 c2 ) a2 a2ux + b2 . The last equation and equation (3.63) imply F ′ ( − 1 c2 ) − F = −c1µ ′a1uy + b1 a1 b2 a2 + α′′uy b2 a2 . The Klein–Gordon Equation and Differential Substitutions 21 Similarly, differentiating equation (3.63) with respect to uy we obtain F ′ ( − 1 c1 ) − F = −c2µ ′a2ux + b2 a2 b1 a1 + α′′ux b1 a1 . To eliminate ux and uy we apply the operators ∂ ∂ux and ∂ ∂uy to the two above equations, respec- tively. We get F ′′ ( − 1 c2 − F ′ ) = 0, F ′′ ( − 1 c1 − F ′ ) = 0, therefore F ′′(c2 − c1) = 0. Assuming that c1 = c2 = c we define F as follows F (z) = −1 c exp(−cz + c7) + c8. Substituting the above function F into equation (3.63) we get −µ′uyc ( ux + b2 a2 ) − 2µ2 c3 a1a2 (a2ux + b2)(a1uy + b1)− µ′uxc ( uy + b1 a1 ) + α′′uxuy + α′µ c2 a1a2 (a2ux + b2)(a1uy + b1) = −1 c (a2ux + b2)(a1uy + b1) exp(−cα+ c7) + c8. Since u, ux, and uy are considered as independent variables, the above equation is equivalent to the following system − 2cµ′ − 2µ2c3 + α′′ + α′µc2 = −a1a2 c exp(−cα+ c7), (3.64a) − 2µ2 c 3 a1 b1 − µ′c b1 a1 + α′µc2 b1 a1 = −1 c a2b1 exp(−cα+ c7), (3.64b) − µ′c b2 a2 − 2µ2c3 b2 a2 + α′µc2 b2 a2 = −1 c a1b2 exp(−cα+ c7), (3.64c) − 2µ2c3 b1b2 a1a2 + α′µc2 b1b2 a1a2 = −1 c b1b2 exp(−cα+ c7) + c8. (3.64d) Note that (b1, b2) 6= (0, 0). Otherwise, condition (3.21) is true, which contradicts the assumption of the lemma. If b2 = 0, b1 6= 0 then c8 = 0 and uxy = µ(u)c2 a1 (a1uy + b1)ux, vxy = −1 c exp(−cv + c7), v = −1 c ln(a1uy + b1)− 1 c ln(a2ux) + α(u), (3.65) where the functions µ and α satisfy the following equations −2cµ′ − 2µ2c3 + α′′ + α′µc2 = −a1a2 c exp(−cα+ c7), −2µ2c3 − µ′c+ α′µc2 = −a1a2 c exp(−cα+ c7). Applying the transformation −cv + c7 → v and redenoting −cα+ c7 + ln(a1a2) by α, µc2 by µ and b1/a1 by b, we transform (3.1) into (3.56). It is not hard to prove that system (3.64) has no solutions if b1b2 6= 0. 22 M.N. Kuznetsova, A. Pekcan and A.V. Zhiber 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 −c2µ ′uy ( ux + b2 a2 ) − µ2c1c 2 2 ( ux + b2 a2 )( uy + b1 a1 ) − c1µ ′ux ( uy + b1 a1 ) + α′′uxuy − µ2c2 1c2 ( ux + b2 a2 )( uy + b1 a1 ) + α′µc1c2 ( ux + b2 a2 )( uy + b1 a1 ) = c ( − 1 c1 ln(a1uy + b1)− 1 c2 ln(a2ux + b2) + α(u) ) + p. It is clear that the coefficients at ln(a1uy + b1) and ln(a2ux + b2) are equal to zero, i.e. c = 0. Since u, ux, and uy are regarded as independent variables, the above equation is equivalent to the system −c2µ ′ − µ2c1c 2 2 − c1µ ′ − µ2c2 1c2 + α′′ + α′µc1c2 = 0, −µ2c1c 2 2 b1 a1 − c1µ ′ b1 a1 − µ2c2 1c2 b1 a1 + α′µc1c2 b1 a1 = 0, −c2µ ′ b2 a2 − µ2c1c 2 2 b2 a1 − µ2c2 1c2 b2 a1 + α′µc1c2 b2 a2 = 0, −µ2c1c 2 2 − µ2c2 1c2 + α′µc1c2 b1b2 a1a2 = p. Note that (b1, b2) 6= (0, 0). Otherwise, condition (3.21) is satisfied, which contradicts the as- sumption of the lemma. If b2 = 0, b1 6= 0 then p = 0 and uxy = µ(u) c1c2 a1 (a1uy + b1)ux, vxy = 0, v = − 1 c1 ln(a1uy + b1)− 1 c2 ln(a2ux) + α(u), where the functions µ and α satisfy the equations µ′(c1 + c2) + µ2c1c2(c1 + c2)− α′′ − α′µc1c2 = 0, c1µ ′ + µ2c1c2(c1 + c2)− α′µc1c2 = 0. We replace c1c2µ by µ, −c1c2α+c2 ln a1 +c1 ln a2 by α. Using the transformation v → −v/(c1c2) and redenoting b1/a1 by b we transform the above equations into (3.57). If b1b2 6= 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), −c2 a µ′uy(aux + b) + c2 2 ac1 (aux + b) + 1 c1 µ′ux + α′′uxuy + α′µ(aux + b) ( − c2 c1a ) = F ( c1uy − 1 c2 ln(aux + b) + α(u) ) . (3.66) Applying the operators ∂ ∂ux and ∂ ∂uy to both sides of equation (3.66) we obtain −c2µ ′uy + c2 2 c1 µ2 + 1 c1 µ′ + α′′uy − c2 c1 α′µ = F ′ ( − 1 c2 ) a aux + b , (3.67) −c2 a µ′(aux + b) + α′′ux = F ′c1. (3.68) The Klein–Gordon Equation and Differential Substitutions 23 If F ′ = 0 then we obviously get F = c3 and −c2µ ′ + α′′ = 0, c22µ 2 + µ′ − c2α ′µ = 0, µ′b = 0. We analyze equation (3.66) based on these equations and find that c3 = 0. It allows us to determine equations (1.1), (1.2), and (1.7) as follows uxy = −c2 c1 µ(u)ux, vxy = 0, v = c1uy − 1 c2 ln(aux) + α(u), where the functions µ and α satisfy α′′ = c2µ ′, c2 2µ 2 + µ′ − c2α ′µ = 0. Point transformations enable us to represent the above equations in form (3.58). Assuming that F ′ 6= 0 we can eliminate F ′ from equations (3.67) and (3.68) c2 2 ( aux + b a ) µ′uy − c3 2 c1 ( aux + b a ) µ2 − c2 c1 ( aux + b a ) µ′ − c2 ( aux + b a ) α′′uy + c2 2 c1 α′µ ( aux + b a ) = − c2 c1a (aux + b)µ′ + α′′ c1 ux. Recall that variables u, ux, and uy are considered as independent. Hence, the above equation is equivalent to the system c2 2µ ′ − c2α ′′ = 0, (3.69a) −c 3 2 c1 µ2 − c2 c1 µ′ + c2 2 c1 α′µ = −c2 c1 µ′ + α′′ c1 , (3.69b) c2 2b a µ′ − c2 b a α′′ = 0, (3.69c) −c 3 2 c1 b a µ2 + c2 2 c1 α′µ b a = 0. (3.69d) If b = 0, we transform equation (3.66) into −c2µ ′uxuy + c2 2 c1 µ2ux + 1 c1 µ′ux + α′′uxuy − c2 c1 α′µux = F ( c1uy − 1 c2 ln(aux) + α(u) ) . Differentiating this equation with respect to ux we obtain c2 2 c1 µ2 − c2µ ′uy + 1 c1 µ′ + α′′uy − c2 c1 α′µ = − 1 c2 F ′ 1 ux . One can notice that these two equations imply F + F ′/c2 = 0 or F (z) = c3 exp(−c2z). Conse- quently, we get −c2µ ′uxuy + c2 2 c1 µ2ux + 1 c1 µ′ux + α′′uxuy − c2 c1 α′µux = c3 exp(−c2c1uy)aux exp(α). This equation is not realized because of the given assumptions c3 6= 0 and a 6= 0. Now, it remains only to consider the case when b 6= 0. System (3.69) takes the form c2µ ′ − α′′ = 0, −c3 2µ 2 + c2 2α ′µ = α′′, −c2µ 2 + α′µ = 0. These equations imply that µ′ = 0, which contradicts the given assumptions of the lemma. � 24 M.N. Kuznetsova, A. Pekcan and A.V. Zhiber Lemma 4. Suppose that condition (3.22) holds but (3.20), (3.21), and (3.24) do not. Then equations (1.1), (1.2), and (1.7) take one of the following forms: uxy = µ(u)ux γ′(uy) , vxy = 0, v = lnux + γ(uy) + α(u), (3.70) where c3 + γ′′ γ′2 + c4γ ′uy = 0, α′′ + µ′ + c4µ 2 = 0, and c3µ 2 + µ′ + µ2 + α′µ = 0; uxy = ux (au+ b)γ′(uy) , vxy = exp v, v = lnux + γ(uy)− 2 ln(au+ b) + ln(−c5), (3.71) where c3 + γ′′ γ′2 + c4γ ′uy = c5γ ′ exp γ, c3 + 1 − 3a = 0, and c4 + 2a2 − a = 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 c3, c4 are arbitrary constants, c5 6= 0, and (a, b) 6= (0, 0). Proof. According to (3.22), the function β is of the form β = c1 lnux + c2. Without loss of generality, we may set β = c1 lnux. Substituting β into equation (3.18) we obtain α′µux c1γ′ − µ2ux c1γ′ ( γ′′ γ′2 − 1 c1 ) + α′′uxuy + µ′ ( ux γ′ + uxuy c1 ) = F (α+ β + γ). (3.72) Applying the operator ∂ ∂ux to both sides of (3.72) leads to α′µ c1γ′ − µ2 c1γ′ ( γ′′ γ′2 − 1 c1 ) + α′′uy + µ′ ( 1 γ′ + uy c1 ) = F ′ ( c1 ux ) . (3.73) From equations (3.72) and (3.73) it follows that F = F ′c1/ux, hence F (z) = c2 exp(z/c1). By substituting F into equation (3.72) we get ux ( α′µ c1γ′ − µ2 c1γ′ ( γ′′ γ′2 − 1 c1 ) + α′′uy + µ′ ( 1 γ′ + uy c1 )) = c2ux exp(γ/c1) exp(α/c1). This equation can be written in the form µ′c1 + α′µ+ µ2 c1 − µ2 γ ′′ γ′2 + (α′′c1 + µ′)γ′uy = c2c1γ ′ exp(γ/c1) exp(α/c1). Having the fixed value of u we can determine γ as a solution of the ordinary differential equation c3 + γ′′ γ′2 + c4γ ′uy = c1c5γ ′ exp(γ/c1). Moreover, based on this equation we get α′µ+ µ2 c1 + c1µ ′ + c3µ 2 + γuy ( c1α ′′ + µ′ + c4µ 2 ) − c1γ ′ exp(γ/c1)(c5µ 2 + c2 exp(α/c1)) = 0. Note that if uy = κ exp(γ/c1) then γ = c1 ln(uy/κ) and (γ′uy) ′ = 0. Since the last equation con- tradicts the assumption of the lemma, we obtain that uy and exp(γ/c1) are linearly independent and that is why c1α ′′ + µ′ + c4µ 2 = 0, c5µ 2 + c2 exp(α/c1) = 0, c3µ 2 + c1µ ′ + µ2 c1 + α′µ = 0. The Klein–Gordon Equation and Differential Substitutions 25 In order to find equations (1.1), (1.2), and (1.7) we first set c5 = 0, hence c2 = 0 and uxy = µ(u) β′(ux)γ′(uy) , vxy = 0, v = β(ux) + γ(uy) + α(u), where the functions β and γ are solutions of the ordinary differential equations β′ = c1 ux , c3 + γ′′ γ′2 + c4γ ′uy = 0, and the functions µ and α satisfy the equations c1α ′′ + µ′ + c4µ 2 = 0, c3µ 2 + c1µ ′ + µ2 c1 + α′µ = 0. We use the transformation v → c1v. Next, we redenote α/c1 by α, γ/c1 by γ, and µ/c2 1 by µ. Finally, after replacing c4c 2 1 by c4 and c1c3 by c3, (3.70) is obtained. If c5 6= 0 then we get uxy = µ(u) β′(ux)γ′(uy) , vxy = c2 exp(v/c1), v = β(ux) + γ(uy) + α(u), where the functions β and γ are the solutions of the ordinary differential equations β′ = c1 ux , c3 + γ′′ γ′2 + c4γ ′uy = c1c5γ ′ exp(γ/c1), and the functions α and µ are given by the equations α = 2c1 ln(−2c1)− 2c1 ln ( −2 3 √ −c2 c5 c3c1 + 1 c1 u+ c6 ) , µ = √ −c2 c5  −2c1 −2 3 √ − c2 c5 ( c3c1+1 c1 ) u+ c6  , 2 9 ( c3c1 + 1 c1 )2 − 1 3 ( c3c1 + 1 c2 1 ) + c4 = 0. After point transformations we get (3.71). � Lemma 5. Suppose that condition (3.25) holds but (3.20)–(3.24) do not. Then equations (1.1), (1.2), and (1.7) take one of the following forms: uxy = − 1 uβ′(ux)γ′(uy) , vxy = 0, v = β(ux) + γ(uy), (3.74) where β′′ β′2 = uxβ ′ + c1, γ′′ γ′2 = uyγ ′ − c1; uxy = µ(u) β′(ux)γ′(uy) , vxy = exp v, v = β(ux) + γ(uy) + α(u), (3.75) where ux + 1 β′(ux) = exp(β), uy + 1 γ′(uy) = exp γ, α′′ = expα, and µ = (expα)/α′; uxy = µ(u) β′(ux)γ′(uy) , vxy = exp v, v = β(ux) + γ(uy) + α(u), (3.76) 26 M.N. Kuznetsova, A. Pekcan and A.V. Zhiber where −cux + 1 β′(ux) = expβ, −cuy + 1 γ′(uy) = exp γ, α′µ+ 2µ2(c+ 1) = expα, α′2 = 2c2 expα, c = −1 2 ,−2; uxy = µ(u) β′(ux)γ′(uy) , vxy = exp v + exp(−v), v = β(ux) + γ(uy) + α(u), (3.77) where A1 expβ + B1 exp(−β) = ux, A2 exp γ + B2 exp(−γ) = uy, α′′ = 1 4 ( exp(−α) B1B2 + expα A1A2 ) , µ = α′′ α′ ; uxy = µ(u) β′(ux)γ′(uy) , vxy = exp v + exp(−2v), v = β(ux) + γ(uy) + α(u), (3.78) where A1 expβ+B1 exp(−2β) = ux, A2 exp γ+B2 exp(−2γ) = uy, α′2 = 2 9 ( 4 expα A1A2 − 1 2 exp(−2α) B1B2 ) , −2µ2 + α′µ − 1 9 ( expα A1A2 + exp(−2α) B1B2 ) = 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 A1, A2, B1, and B2 are nonzero constants. Proof. Considering that ux and uy are independent variables, equation (3.25) yields (uxβ ′)′( β′′ β′2 )′ = c, (uyγ ′)′( γ′′ γ′2 )′ = c, c 6= 0. Integrating these equations we obtain β′′ β′2 = cuxβ ′ + c1, γ′′ γ′2 = cuyγ ′ + c2. (3.79) According to (3.79), equation (3.18) is rewritten in the form 1 β′ ( α′µ γ′ − µ2 γ′ ( c1 + cuyγ ′ + c2 ) + µ′uy ) + ux ( −cµ 2 γ′ + α′′uy + µ′ γ′ ) = F (α+ β + γ).(3.80) Having fixed values of u and uy we can define that F (β + c3) = c4ux + c5/β ′. Without loss of generality, we redenote β + c3 by β, therefore F (β) = c4ux + c5 β′ . (3.81) Applying the operator ∂ ∂ux to both sides of equation (3.81) and using (3.79) we obtain F ′(β) = −cc5ux + c4 − c1c5 β′ . We differentiate this equation with respect to ux, F ′′(β) = −c(c4 − c1c5)ux − cc5 + c1(c4 − c1c5) β′ . The above three equations allow us to establish that the function F satisfies the ordinary differential equation F ′′ = c7F ′ + c8F. (3.82) The Klein–Gordon Equation and Differential Substitutions 27 Equation (3.82) possesses two families of solutions F (v) = A1 exp(σ1v) +B1 exp(σ2v), σ1 6= σ2, and F (v) = (A2 +B2v) exp(σv). Setting definite values of the constants Ai, Bi, where i = 1, 2, we obtain that the function F can take only one of the following forms F (v) = 0, (3.83) F (v) = 1, (3.84) F (v) = v, (3.85) F (v) = v exp v, (3.86) F (v) = exp v, (3.87) F (v) = exp v + 1, (3.88) F (v) = exp v + exp(σv). (3.89) From equation (3.80) by setting different values of u and uy we obtain a set of equations αiux + βi β′(ux) = F (β(ux) + γi) . (3.90) Here αi, βi, and γi are constants, i = 1, 2, . . . , n. Thus, we will focus on (3.90). Let us assume that (αi, βi) are linearly dependent vectors. This means that a set of numbers µi satisfying (αi, βi) = µi(α1, β1), µ1 = 1, exists. Using this equation we rewrite (3.90) as µi ( α1ux + β1 β′(ux) ) = F (β + γi). (3.91) Now, we will deal with equations (3.83)–(3.89). We begin with (3.83). In this case we have µi ( α1ux + β1 β′(ux) ) = 0 (3.92) from the equation (3.91). Suppose that α1 = β1 = 0. In equation (3.80), we find µ′ − cµ2 + α′′uyγ ′ = 0, α′µ− µ2(c1 + c2 + cuyγ ′) + µ′uyγ ′ = 0. (3.93) If α′′ = 0 then α = εu+ δ, hence from (3.93) we have µ(u) = − 1 cu+ κ , ε cu+ κ + c1 + c2 (cu+ κ)2 = 0. Clearly, the last equation requires ε = 0 and c2 = −c1. Thus, we determine equations (1.1), (1.2), and (1.7) as follows uxy = µ(u) β′(ux)γ′(uy) , vxy = 0, v = β(ux) + γ(uy) + α(u), 28 M.N. Kuznetsova, A. Pekcan and A.V. Zhiber where µ(u) = − 1 cu+ κ , α(u) = δ, β′′ β′2 = cuxβ ′ + c1, γ′′ γ′2 = cuyγ ′ − c1. We replace β by aβ, γ by aγ. Take the constant a so that a2c→ 1. Using the transformations u+ κ/c→ u, v − δ → av and redenoting ac1 → c1 obtain equation (3.74). Now, assume that α′′ 6= 0. The equation uyγ ′(uy) = cµ2 − µ′ α′′ arises from (3.93). Since u and uy are regarded as independent variables, the last equation leads to uyγ ′(uy) = κ, where κ is a constant. This contradicts the assumption of the lemma. Consider the case where α1β1 6= 0. We have the equation β′(ux) = −β1/(α1ux) 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 µi ( α1ux + β1 β′(ux) ) = 1. This equation must be true for every i = 1, 2, . . . . This requirement implies that µi = 1, αi = α1, and βi = β1 for every i. Taking this into account we define β′ as follows: β′(ux) = β1 1− α1ux . (3.94) 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 α1ux + β1 β′(ux) = β(ux) + γ1, µi ( α1ux + β1 β′(ux) ) = β(ux) + γi. Consequently, β(ux)(µi − 1) + γ1µi − γi = 0. It is clear that µi = 1, γi = γ1. Hence, αi = α1, βi = β1 for every i. So we have β′ = β1 β(ux)− α1ux + γ1 . 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 α1ux + β1 β′(ux) = (β + γ1) exp(β + γ1), µi ( α1ux + β1 β′(ux) ) = (β + γi) exp(β + γi). Comparing these equations we conclude that (β(exp γi − µi exp γ1) + γi exp γi − µiγ1 exp γ1) expβ = 0. Recall that β depends on the variable ux, while the remaining terms of the above equations are constants. Hence, we have exp γi − µi exp γ1 = 0, γi exp γi − µiγ1 exp γ1 = 0. The Klein–Gordon Equation and Differential Substitutions 29 From these equations we obtain γi exp γi − γ1 exp γi = 0, hence γi = γ1 for all i. By (3.90) we determine that α(u) + γ(uy) = γ1, where γ1 is an arbitrary constant. This equation contradicts γuy 6= 0. Let the function F be defined by (3.87). From (3.90) we obtain α1ux + β1 β′(ux) = exp(β + γ1). (3.95) Note that β1 6= 0, otherwise (β′ux)′ = 0. Redenoting β + γ1 by β we rewrite equation (3.95) in the form α1ux + β1 β′(ux) = expβ. (3.96) From equations (3.79) and (3.96) we find that c = −α1/β1, c1 = −1 − c. Now, we rewrite equation (3.80) based on equation (3.96) 1 β′ expβ ( α′µ γ′ − µ2 γ′ ( c1 + cuyγ ′ + c2 ) + µ′uy ) + ux ( −cµ 2 γ′ + α′′uy + µ′ γ′ ) − α1 β1 ux ( α′µ γ′ − µ2 γ′ ( c1 + cuyγ ′ + c2 ) + µ′uy ) = exp(α+ γ) expβ. Since (β′ux)′ 6= 0, expβ and ux are linearly independent, the above equation is equivalent to the system α′µ γ′ − µ2 γ′ ( c1 + cuyγ ′ + c2 ) + µ′uy = exp(α+ γ)β1, −α1 β1 ( α′µ γ′ − µ2 γ′ ( c1 + cuyγ ′ + c2 ) + µ′uy ) + ( −cµ 2 γ′ + α′′uy + µ′ γ′ ) = 0. Hence, we get uxy = µ(u) β′(ux)γ′(uy) , vxy = exp v, v = β(ux) + γ(uy) + α(u), (3.97) where α1ux + β1 β′ = expβ, β′′ β′2 = cuxβ ′ + c1, c1 = −1− c, cβ1 = −α1, γ′′ γ′2 = cuyγ ′ + c2, α′µ γ′ − µ2 γ′ (cuyγ ′ + c1 + c2) + µ′uy = exp(α+ γ)β, −α1 exp(α+ γ) + α′′uy + µ′ − cµ2 γ′ = 0. Now, consider case (3.88). Equations (3.90) and (3.91) can be rewritten in the forms α1ux + β1 β′(ux) = exp(β + γ1) + 1, µi ( α1ux + β1 β′(ux) ) = exp(β + γi) + 1. It is not hard to show that expβ (µi exp γ1 − exp γi) + µi − 1 = 0. The dependence of β only on the variable ux implies that µi = 1 and γi = γ1 for every i. This gives α(u) + γ(uy) = γ1, where γ1 is a constant, which contradicts the assumption γuy 6= 0. 30 M.N. Kuznetsova, A. Pekcan and A.V. Zhiber It remains to consider the case when F is given by (3.89) to complete the analysis in the case when (αi, βi) are linearly dependent vectors. Using (3.89) we transform equations (3.90) and (3.91) into α1ux + β1 β′(ux) = exp(β + γ1) + exp(σ(β + γ1)), µi ( α1ux + β1 β′(ux) ) = exp(β + γi) + exp(σ(β + γi)). Consequently, we get expβ (µi exp γ1 − exp γi) + exp(σβ) (µi exp(σγ1)− exp(σγi)) = 0. Recall that σ 6= 1. Collecting coefficients at expβ and exp(σβ) yields µi exp γ1 = exp γi, µi exp(σγ1) = exp(σγi). The above equations provide µi exp(σγ1)(µσ−1 i − 1) = 0, hence µi = 1. It follows that γi = γ1 for every i. By (3.90) we find that α(u) + γ(uy) = γ1. This equation contradicts γuy 6= 0. Now, we must deal with the case when αi, βi, i = 1, 2, satisfying α1β2 − β1α2 6= 0 exist. Setting definite values of u, uy in (3.80) we obtain the system α1ux + β1 β′(ux) = F (β(ux) + γ1), α2ux + β2 β′(ux) = F (β(ux) + γ2). Because of the given assumption (uxβ ′)ux 6= 0 we get κ1F (β + γ1)− κ2F (β + γ2) = ux, κ3F (β + γ1)− κ4F (β + γ2) = 1 β′ . (3.98) We use κ1 = β2 α1β2− α2β1 , κ2 = β1 α1β2− α2β1 , κ3 = α2 β1α2− β2α1 , κ4 = α1 β1α2− β2α1 . 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 ux = 0. Thus, this case is not realized. Next, based on (3.84) we obtain that ux is a constant. So it is also not possible. If (3.85) is true then system (3.98) can be written as follows κ1(β + γ1)− κ2(β + γ2) = ux, κ3(β + γ1)− κ4(β + γ2) = 1 β′ . It is not hard to verify that β′(κ1 − κ2) = 1, β(κ3 − κ4) + γ1κ3 − γ2κ4 = κ1 − κ2. Note that we used the properties κ1 − κ2 6= 0, κ1 − κ2 6= 0, which result from α1β2 − β1α2 6= 0. Further, since κ3 − κ4 6= 0, β is a constant. This contradicts βux 6= 0. Let us discuss the case when the function F is defined by (3.86). Rewriting (3.98) we get κ1(β + γ1) exp(β + γ1)− κ2(β + γ2) exp(β + γ2) = ux, κ3(β + γ1) exp(β + γ1)− κ4(β + γ2) exp(β + γ2) = 1 β′ . The Klein–Gordon Equation and Differential Substitutions 31 Setting A = κ1 exp γ1 − κ2 exp γ2 and B = κ1γ1 exp γ1 − κ2γ2 exp γ2 we obtain ux = Aβ expβ +B expβ. (3.99) It is not difficult to determine that equations (3.98), (3.99) lead to (A+B) ( α′µ γ′ − µ2 γ′ ( c1+cuyγ ′+c2 ) + µ′uy ) +B ( α′′uy + µ′−cµ2 γ′ ) = (α+γ) exp(α+ γ), A ( α′µ γ′ − µ2 γ′ ( c1 + cuyγ ′ + c2 ) + µ′uy + α′′uy + µ′ − cµ2 γ′ ) = exp(α+ γ). Rewriting (3.79) by using (3.99) we find that c = 1, c1 = −2. Thus, we obtain the equations uxy = µ(u) β′(ux)γ′(uy) , vxy = v exp v, v = α(u) + β(ux) + γ(uy), (3.100) herewith (A+B) ( α′µ γ′ − µ2 γ′ (uyγ ′ − 2 + c2) + µ′uy ) +B ( α′′uy + µ′ − µ2 γ′ ) = (α+ γ) exp(α+ γ), A ( α′µ γ′ − µ2 γ′ (uyγ ′ − 2 + c2) + µ′uy + α′′uy + µ′ − µ2 γ′ ) = exp(α+ γ), ux = Aβ expβ +B expβ, β′′ β′2 = uxβ ′ − 2, γ′′ γ′2 = uyγ ′ + c2. Note that case (3.87) yields the equations (3.97). Next, assume that the function F is defined by (3.88). Hence, we write (3.98) as κ1 (exp(β + γ1) + 1)− κ2 (exp(β + γ2) + 1) = ux, κ3 (exp(β + γ1) + 1)− κ4 (exp(β + γ2) + 1) = 1 β′ . Eliminating β′ from the last equation we get expβ(κ3 exp γ1 − κ4 exp γ2 − κ1 exp γ1 + κ2 exp γ2) + κ3 − κ4 = 0. It is easy to show from this equation that β is a constant. This contradicts βux 6= 0. Assuming that (3.89) holds, we can write (3.98) as expβ(κ1 exp γ1 − κ2 exp γ2) + exp(σβ)(κ1 exp(σγ1)− κ2 exp(σγ2)) = ux, expβ(κ3 exp γ1 − κ4 exp γ2) + exp(σβ)(κ3 exp(σγ1)− κ4 exp(σγ2)) = 1 β′ . (3.101) And further, from (3.79) based on (3.101) we obtain (1 + c+ c1)(κ1 exp γ1 − κ2 exp γ2) expβ + ( σ2 + c+ c1σ )( κ1 exp(σγ1)− κ2 exp(σγ2) ) expσβ = 0. (3.102) From (3.80) using (3.101) again we get (κ1 exp γ1 − κ2 exp γ2) ( α′µ γ′ − µ2 γ′ (cuyγ ′ + c1 + c2) + µ′uy + µ′ − cµ2 γ′ + α′′uy ) = exp(α+ γ), (3.103) 32 M.N. Kuznetsova, A. Pekcan and A.V. Zhiber (κ1 expσγ1 − κ2 expσγ2) ( σ ( α′µ γ′ − µ2 γ′ (cuyγ ′ + c1 + c2) + µ′uy ) + µ′ − cµ2 γ′ + α′′uy ) = expσ(α+ γ). (3.104) Note that if κ1 exp(σγ1) − κ2 exp(σγ2) = 0 then equations (3.103) and (3.104) imply that expσ(α + γ) = 0. Consequently, the equalities 1 + c + c1 = 0 and σ2 + c1σ + c = 0 arise from equation (3.102). The solution of the last equation is found as σ = c, where c = −1− c1. Thus, denoting A = κ1 exp γ1 − κ2 exp γ2, B = κ1 exp(σγ1)− κ2 exp(σγ2) we obtain uxy = µ(u) β′(ux)γ′(uy) , vxy = exp v + exp(σv), v = α(u) + β(ux) + γ(uy), (3.105) where A expβ +B exp(σβ) = ux, β′′ β′2 = σuxβ ′ − 1− σ, γ′′ γ′2 = σuyγ ′ + c2, A ( α′µ γ′ − µ2 γ′ ( σuyγ ′ + c2 − 1 ) + µ′uy + µ′ γ′ + α′′uy ) = exp(α+ γ), B ( σ ( α′µ γ′ − µ2 γ′ ( σuyγ ′ + c2 − σ ) + µ′uy ) + µ′ γ′ + α′′uy ) = expσ(α+ γ). Let us discuss the results obtained. We should analyze the equations and conditions for the parameters found in cases (3.83)–(3.89) and use the fact that functions (3.14) and (3.17) are invariant under the permutation of β(ux) and γ(uy). In case (3.87) we obtained (3.97). By interchanging β(ux) and γ(uy) we get α2uy + β2 γ′ = exp γ, γ′′ γ′2 = cuyγ ′ + c2, c2 = −1− c, cβ2 = −α2, α′µ β′ − µ2 β′ (cuxβ ′ + c1 + c2) + µ′ux = exp(α+ β)β2, −α2 exp(α+ β) + α′′ux + µ′ − cµ2 β′ = 0. (3.106) We substitute γ satisfying the conditions for the parameters listed for equation (3.106) in- to (3.97). At the same time we substitute β satisfying the conditions for the parameters listed for equation (3.97) into (3.106). As a result, we obtain the system 1 β2 (exp γ − α2uy) ( α′µ+ 2µ2(1 + c) ) + µ′uy = exp(α+ γ)β1, −α1 exp(α+ γ) + α′′uy + ( µ′ − cµ2 ) 1 β2 (exp γ − α2uy) = 0, 1 β1 (expβ − α1ux) ( α′µ+ 2µ2(1 + c) ) + µ′ux = exp(α+ β)β2, −α2 exp(α+ β) + α′′ux + ( µ′ − cµ2 ) 1 β1 (expβ − α1ux) = 0. Since exp γ and uy, expβ and ux are independent, equations (1.1), (1.2), and (1.7) take the following forms: uxy = µ(u) β′(ux)γ′(uy) , vxy = exp v, v = α(u) + β(ux) + γ(uy), The Klein–Gordon Equation and Differential Substitutions 33 where α and β are solutions of the ordinary differential equations α1ux + β1 β′ = expβ, α2uy + β2 γ′(uy) = exp γ, −α1 β1 = −α2 β2 = c, and the functions µ and α satisfy α′µ+ 2µ2(c+ 1) = β1β2 expα, cβ1β2 expα+ µ′ − cµ2 = 0, α′′ + c(µ′ − cµ2) = 0. Analyzing the last system we obtain cases (3.75), (3.76). It is easy to verify that case (3.86) is not possible. Based on (3.89) we get (3.105). Interchanging β(ux) and γ(uy) implies A2 exp γ +B2 expσγ = uy, γ′′ γ′2 = σuyγ ′ + c2, c2 = −1− c, A2 β′ ( α′µ− µ2(c1 − 1) + µ′ ) +A2ux ( −σµ2 + µ′ + α′′ ) = exp(α+ β), B2 β′ ( σ(α′µ− µ2(c2 − σ)) + µ′ ) + uxB2 ( σ(−σµ2 + µ′) + α′′ ) = expσ(α+ β). (3.107) Similarly, we substitute β satisfying the conditions for the parameters listed for equation (3.105) into (3.107) and obtain (A expβ +Bσ expσβ)A2(α′µ+ µ2(2 + σ) + µ′) + (A expβ +B expσβ)A2(µ′ − σµ2 + α′′) = exp(α+ β), (A expβ +Bσ exp(σβ))B2 ( σ(α′µ+ µ2(1 + 2σ) + µ′) ) + (A expβ +B expσβ)B2 ( σ(µ′ − σµ2) + α′′ ) = expσ(α+ β). Taking into account the fact that expβ, expσβ are independent, we get AA2 ( α′µ+ µ2(2 + σ) + 2µ′ − σµ2 + α′′ ) = expα, σα′µ+ σ(σ + 1)µ2 + (σ + 1)µ′ + α′′ = 0, BB2 ( σ2α′µ+ 2σ3µ2 + 2σµ′ + α′′ ) = exp(σα). Solving the above system we obtain cases (3.77) and (3.78). � 3.2 Case ϕ = c lnux + q(u, uy) We have the following statement in this case. Lemma 6. Suppose that (3.15) is satisfied. Then equations (1.1), (1.2), and (1.7) take the following forms uxy = µ(u)− qu(u, uy) quy(u, uy) ux, vxy = c2 exp v, v = lnux + q(u, uy), (3.108) where µ− qu quy ( µ− µ− qu q2 uy quyuy − 2 quuy quy ) + µ′ quy − quu quy + µ′uy = c2 exp q, quuy 6= 0, up to the point transformations u → θ(u), v → κ(v), x → ξx, and y → ηy, where ξ and η are arbitrary constants. 34 M.N. Kuznetsova, A. Pekcan and A.V. Zhiber Proof. Substituting function (3.15) into equation (3.6) we obtain A(u, uy)quy(u, uy)− qu(u, uy)quy(u, uy)uy + cqu(u, uy) = B(u, ux) c ux . Recall that ux, uy are considered as independent variables. Hence, the above equation is equiv- alent to the system Aquy − ququyuy + cqu = µ(u), Bc ux = µ(u). From these equations we find the functions A and B, B = µux c , A = µ+ ququyuy − cqu quy . By using these equations in each of equations (3.3), (3.4) we determine the function f of equa- tion (1.1) as f = µ− cqu cquy ux. Substituting the functions (3.15) and f into (3.7) we have ux ( µ− cqu cquy ( µ c − µ− cqu q2 uy quyuy − 2c quuy quy ) + µ′ quy − quu quy + µ′uy c ) = F (c lnux + q). It is not difficult to prove by differentiating this equation with respect to ux that cF ′ = F . Consequently, F (z) = c2 exp(z/c). Here c2 is an arbitrary constant. Thus, equations (1.1), (1.2), and (1.7) are of the forms uxy = µ(u)− cqu(u, uy) cquy(u, uy) ux, vxy = c2 exp(v/c), v = c lnux + q(u, uy), where µ− cqu cquy ( µ c − µ− cqu q2 uy quyuy − 2c quuy quy ) + µ′ quy − cquu quy + µ′uy c = c2 exp(q/c). Finally, the transformations v → cv, q → cq, µ→ c2µ, and c2/c→ c2 transform these equations into (3.108). � 3.3 Case ϕ = α(u) + κ(u) lnux + µ(u) lnuy By substituting (3.16) into (3.6) we obtain ( A(u, uy)− (κ′(u) lnux + µ′(u) lnuy + α′(u))uy ) µ(u) uy = ( B(u, uy)− (κ′(u) lnux + µ′(u) lnuy + α′(u))ux ) κ(u) ux , which can be written as B(u, ux)κ(u) ux + ( κ′(u) lnux + α′(u) ) (µ(u)− κ(u)) The Klein–Gordon Equation and Differential Substitutions 35 = A(u, uy)µ(u) uy − µ′(u) lnuy (µ(u)− κ(u)) . Since ux and uy are regarded as independent variables, the above equation is equivalent to the system B(u, ux)κ(u) ux + ( κ′(u) lnux + α′(u) ) (µ(u)− κ(u)) = λ(u), A(u, uy)µ(u) uy − µ′(u) lnuy (µ(u)− κ(u)) = λ(u). The formulae B = (λ− (µ− κ)(κ′ lnux + α′))ux κ , A = (λ+ µ′(µ− κ) lnuy)uy µ thereby immediately follow. Substituting A and B into equations (3.3) and (3.4) we find f , f = λ− κµ′ lnuy − µκ′ lnux − µα′ κµ uxuy. (3.109) We apply the operator ∂ ∂ux to both sides of equation (3.5) and use the equations obtained. So we get F ′κ = F , while applying ∂ ∂uy implies F ′µ = F . This requires µ(u) = κ(u) = c. Thus ϕ takes the form ϕ = α(u)+c ln(uxuy), and case (3.16) is reduced to case (3.14) considered earlier. Theorem 1 follows from Lemmas 1–6. 4 Differential substitutions of the form u = ψ(v, vx, vy) In this section we consider the problem which is, in a sense, inverse to the original problem. The aim is to describe equations of form (1.2) which are transformed into equations of form (1.1) by differential substitutions (1.8). Theorem 2. Suppose that equation (1.2) is transformed into equation (1.1) by differential sub- stitution (1.8). Then equations (1.2), (1.1) and substitution (1.8) take one of the following forms: vxy = v, uxy = u, u = c1ux + c2uy + c3u; vxy = 0, uxy = 0, u = β(vx) + γ(vy) + c3v; vxy = 0, uxy = exp(u)uy, u = ln ( − p′(v)vx µ(vy) + p(v) ) , where p′(v) = exp(cv); vxy = 1, uxy = c1(ux − c2), u = exp(c1vx) + c2vy; vxy = exp v, uxy = uux, u = vy + µ(vx) exp v, where 2µ′ = µ2; vxy = 0, uxy = expu, u = ln(vxvy) + δ(v), where δ′′(v) = exp δ(v); vxy = 1, uxy = c1ux + c2uy − c1c2u, u = exp(c1vx) + exp(c2vy) up to the point transformations u → θ(u), v → κ(v), x → ξx, and y → ηy and the substitution u+ξx+ηy → u, where ξ and η are arbitrary constants. Here c is an arbitrary constant, c1 and c2 are nonzero constants. 36 M.N. Kuznetsova, A. Pekcan and A.V. Zhiber Note that symmetries, x- and y-integrals, and the general solutions of the equations uxy = uux and uxy = exp(u)uy 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 ψvF + ψvxF ′vx + ψvyF ′vy + vx ( ψvvvy + ψvvxF + ψvvyvyy ) + vxx ( ψvxvvy + ψvxvxF + ψvxvyvyy ) + ( ψvyvvy + ψvyvxF + ψvyvyvyy ) F = f ( ψ,ψvvx + ψvxvxx + ψvyF,ψvvy + ψvxF + ψvyvyy ) . (4.1) Denote the arguments of the function f by a, b, and c. Recall that we have ψvxψvy 6= 0. The equality f ′′bb = f ′′cc = 0 thereby immediately follows from equation (4.1). Hence, equation (1.1) takes the form uxy = α(u) + β(u)ux + γ(u)uy + ε(u)uxuy. After the point transformation u→ A(u) with A′′− εA′2 = 0 the above equation takes the form uxy = f = α(u) + β(u)ux + γ(u)uy. Next, taking into account the last equality which defines the function f we can rewrite equa- tion (4.1) as follows ψvF + ψvxF ′vx + ψvyF ′vy + vx ( ψvvvy + ψvvxF + ψvvyvyy ) + vxx ( ψvxvvy + ψvxvxF + ψvxvyvyy ) + ( ψvyvvy + ψvyvxF + ψvyvyvyy ) F = α(ψ) + β(ψ) ( ψvvx + ψvxvxx + ψvyF ) + γ(ψ) ( ψvvy + ψvxF + ψvyvyy ) . Since vxx and vyy are independent variables, this equation is equivalent to the system ψvxvy = 0, ψvxvvy + ψvxvxF = β(ψ)ψvx , ψvyvvx + Fψvyvy = γ(ψ)ψvy , ψvF + ψvxF ′vx + ψvyF ′vy + ψvvvxvy + vxψvvxF + vyψvvyF + F 2ψvyvx = α(ψ) + β(ψ) ( ψvvx + ψvxvxx + ψvyF ) + γ(ψ) ( ψvvy + ψvxF + ψvyvyy ) . Consequently, we have ψ = A(v, vx) +B(v, vy), Avvxvy +AvxvxF = β(A+B)Avx , Bvvyvx +BvyvyF = γ(A+B)Bvy , (Av +Bv)F +AvxF ′vx +BvyF ′vy + (Avv +Bvv)vxvy + vxAvvxF + vyBvvyF = α(A+B) + β(A+B) ( vx(Av +Bv) + FBvy ) + γ(A+B) ( vy(Av +Bv) +AvxF ) . By using the above equations we prove Theorem 2. � Acknowledgements This work is partially supported by the Russian Foundation for Basic Research (RFBR) (Grants 11-01-97005-Povolj’ie-a, 12-01-31208 mol-a). The Klein–Gordon Equation and Differential Substitutions 37 References [1] Anderson I.M., Kamran N., The variational bicomplex for hyperbolic second-order scalar partial differential equations in the plane, Duke Math. J 87 (1997), 265–319. [2] Bäcklund A.V., Einiges über Curven und Flächen Transformationen, Lund Universitëts Arsskrift 10 (1874), 1–12. [3] Bianchi L., Ricerche sulle superficie elicoidali e sulle superficie a curvatura costante, Ann. Scuola Norm. Sup. Pisa Cl. Sci. 2 (1879), 285–341. [4] Darboux G., Leçons sur la théorie générale des surfaces et les applications géométriques du calcul infinitésimal. II, Gauthier-Villars, Paris, 1889. [5] Drinfel’d V.G., Svinolupov S.I., Sokolov V.V., Classification of fifth-order evolution equations having an infinite series of conservation laws, Dokl. Akad. Nauk Ukrain. SSR Ser. A (1985), no. 10, 8–10. [6] Goursat E., Leçon sur l’intégration des équations aux dérivées partielles du second ordre á deux variables indépendantes, I, II, Hermann, Paris, 1896. [7] Khabirov S.V., Infinite-parameter families of solutions of nonlinear differential equations, Sb. Math. 77 (1994), 303–311. [8] Kuznetsova M.N., Laplace transformation and nonlinear hyperbolic equations, Ufa Math. J. 1 (2009), no. 3, 87–96. [9] Kuznetsova M.N., On nonlinear hyperbolic equations related with the Klein–Gordon equation by differential substitutions, Ufa Math. J. 4 (2012), no. 3, 86–103. [10] Liouville J., Sur l’equation aux différences partielles ∂2 log λ/∂u∂v ± λ/(aa2) = 0, J. Math. Pures Appl. 18 (1853), 71–72. [11] Meshkov A.G., Sokolov V.V., Hyperbolic equations with third-order symmetries, Theoret. Math. Phys. 166 (2011), 43–57. [12] Sokolov V.V., On the symmetries of evolution equations, Russian Math. Surveys 43 (1988), no. 5, 165–204. [13] Soliman A.A., Abdo H.A., New exact solutions of nonlinear variants of the RLN, the PHI-four and Boussinesq equations based on modified extended direct algebraic method, Int. J. Nonlinear Sci. 7 (2009), 274–282, arXiv:1207.5127. [14] Startsev S.Ya., Hyperbolic equations admitting differential substitutions, Theoret. Math. Phys. 127 (2001), 460–470. [15] Startsev S.Ya., Laplace invariants of hyperbolic equations linearizable by a differential substitution, Theoret. Math. Phys. 120 (1999), 1009–1018. [16] Svinolupov S.I., Second-order evolution equations with symmetries, Russian Math. Surveys 40 (1985), no. 5, 241–242. [17] Tzitzéica G., Sur une nouvelle classe de surfaces, C. R. Acad. Sci. 144 (1907), 1257–1259. [18] Zhiber A.V., Shabat A.B., Klein–Gordon equations with a nontrivial group, Soviet Phys. Dokl. 24 (1979), 607–609. [19] Zhiber A.V., Sokolov V.V., Exactly integrable hyperbolic equations of Liouville type, Russian Math. Surveys 56 (2001), no. 1, 61–101. [20] Zhiber A.V., Sokolov V.V., Startsev S.Ya., Darboux integrable nonlinear hyperbolic equations, Dokl. Math. 52 (1995), 128–130. http://dx.doi.org/10.1215/S0012-7094-97-08711-1 http://dx.doi.org/10.1070/SM1994v077n02ABEH003442 http://dx.doi.org/10.1007/s11232-011-0004-3 http://dx.doi.org/10.1070/RM1988v043n05ABEH001927 http://arxiv.org/abs/1207.5127 http://dx.doi.org/10.1023/A:1010359808044 http://dx.doi.org/10.1007/BF02557408 http://dx.doi.org/10.1007/BF02557408 http://dx.doi.org/10.1070/RM1985v040n05ABEH003693 http://dx.doi.org/10.1070/rm2001v056n01ABEH000357 1 Introduction 2 Equations transformed into Klein-Gordon equations 3 Proof of the main theorem 3.1 Case = (u) + (ux) + (uy) 3.2 Case = c lnux + q(u, uy) 3.3 Case = (u) + (u) lnux + (u) lnuy 4 Differential substitutions of the form u=(v,vx,vy) References

The Klein-Gordon Equation and Differential Substitutions of the Form v=φ(u,ux,uy)

Institution

Ähnliche Einträge