15.6. METHOD OF FUNCTIONAL SEPARATION OF VARIABLES 703 Let us dwell on the case (15.6.3.22b). According to (15.6.3.23), h = A 1 (z + C 1 ) 2 + A 2 z + C 1 , (15.6.3.25) where A 1 =–C 3 /3 and A 2 =–C 2 are any numbers. Substituting (15.6.3.22b) and (15.6.3.25) into (15.6.3.24) yields w = B 1 ln |z + C 1 | + B 2 , F = A 1 B 1 (z + C 1 )+ A 2 B 1 (z + C 1 ) 2 . Eliminating z, we arrive at the explicit form of the right-hand side of equation (15.6.3.16): F(w)=A 1 B 1 e u + A 2 B 1 e –2u ,whereu = w – B 2 B 1 . (15.6.3.26) For simplicity, we set C 1 = 0, B 1 = 1,andB 2 = 0 and denote A 1 = a and A 2 = b. Thus, we have w(z)=ln|z|, F(w)=ae w + be –2w , g(z)=–1/z, h(z)=az 2 + b/z. (15.6.3.27) It remains to determine ψ(t)andϕ(x). We substitute (15.6.3.27) into the functional differential equa- tion (15.6.3.18). Taking into account (15.6.3.17), we find [ψ tt ψ –(ψ t ) 2 – aψ 3 – b]–[ϕ xx ϕ –(ϕ x ) 2 + aϕ 3 ]+(ψ tt – 3aψ 2 )ϕ – ψ(ϕ xx + 3aϕ 2 )=0. (15.6.3.28) Differentiating (15.6.3.28) with respect to t and x yields the separable equation* (ψ ttt – 6aψψ t )ϕ x –(ϕ xxx + 6aϕϕ x )ψ t = 0, whose solution is determined by the ordinary differential equations ψ ttt – 6aψψ t = Aψ t , ϕ xxx + 6aϕϕ x = Aϕ x , where A is the separation constant. Each equation can be integrated twice, thus resulting in (ψ t ) 2 = 2aψ 3 + Aψ 2 + C 1 ψ + C 2 , (ϕ x ) 2 =–2aϕ 3 + Aϕ 2 + C 3 ϕ + C 4 , (15.6.3.29) where C 1 , , C 4 are arbitrary constants. Eliminating the derivatives from (15.6.3.28) using (15.6.3.29), we find that the arbitrary constants are related by C 3 =–C 1 and C 4 = C 2 + b. So, the functions ψ(t)andϕ(x)are determined by the first-order nonlinear autonomous equations (ψ t ) 2 = 2aψ 3 + Aψ 2 + C 1 ψ + C 2 , (ϕ x ) 2 =–2aϕ 3 + Aϕ 2 – C 1 ϕ + C 2 + b. The solutions of these equations are expressed in terms of elliptic functions. For the other cases in (15.6.3.22), the analysis is performed in a similar way. Table 15.6 presents the final results for the cases (15.6.3.22a)–(15.6.3.22e). Case 2. Integrating the third and fourth equations in (15.6.3.21) yields ψ = √ Bt+ D 1 , ϕ = √ B – Ct+ D 2 if A = 0; ψ = 1 4A (At + D 1 ) 2 – B A , ϕ =– 1 4A (Ax + D 2 ) 2 + B – C A if A ≠ 0; (15.6.3.30) where D 1 and D 2 are arbitrary constants. In both cases, the function F(w) in equation (15.6.3.16) is arbitrary. The first row in (15.6.3.30) corresponds to the traveling-wave solution w = w(kx + λt). The second row leads to a solution of the form w = w(x 2 – t 2 ). * To solve equation (15.6.3.28), one can use the solution of functional equation (15.5.4.4) [see (15.5.4.5)]. 704 NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS TABLE 15.6 Nonlinear Klein–Gordon equations ∂ tt w – ∂ xx w = F(w) admitting functional separable solutions of the form w = w(z), z = ϕ(x)+ψ(t). Notation: A, C 1 ,andC 2 are arbitrary constants; σ = 1 for z > 0 and σ =–1 for z < 0 No. Right-hand side F(w) Solution w(z) Equations for ψ(t)andϕ(x) 1 aw lnw + bw e z (ψ t ) 2 = C 1 e –2ψ + aψ – 1 2 a + b + A, (ϕ x ) 2 = C 2 e –2ϕ – aϕ + 1 2 a + A 2 ae w + be –2w ln |z| (ψ t ) 2 = 2aψ 3 + Aψ 2 + C 1 ψ + C 2 , (ϕ x ) 2 =–2aϕ 3 + Aϕ 2 – C 1 ϕ + C 2 + b 3 a sin w + b sin w ln tan w 4 + 2 sin w 4 4 arctan e z (ψ t ) 2 = C 1 e 2ψ + C 2 e –2ψ + bψ + a + A, (ϕ x ) 2 =–C 2 e 2ϕ – C 1 e –2ϕ – bϕ + A 4 a sinh w + b sinh w ln tanh w 4 + 2 sinh w 2 2 ln coth z 2 (ψ t ) 2 = C 1 e 2ψ + C 2 e –2ψ – σbψ + a + A, (ϕ x ) 2 = C 2 e 2ϕ + C 1 e –2ϕ + σbϕ + A 5 a sinh w + 2b sinh w arctan e w/2 +cosh w 2 2 ln tan z 2 (ψ t ) 2 = C 1 sin 2ψ + C 2 cos 2ψ + σbψ +a + A, (ϕ x ) 2 =–C 1 sin 2ϕ + C 2 cos 2ϕ – σbϕ + A 15.6.4. Splitting Method. Solutions of Some Nonlinear Functional Equations and Their Applications 15.6.4-1. Three-argument functional equations of special form. Splitting method. The substitution of the expression w = F(z), z = ϕ(x)+ψ(t)(15.6.4.1) into a nonlinear partial differential equation sometimes leads to functional differential equations of the form f(t)+Φ 1 (x)Ψ 1 (z)+···+ Φ k (x)Ψ k (z)=0,(15.6.4.2) where Φ j (x)andΨ j (z) are functionals dependent on the variables x and z, respectively, Φ j (x) ≡ Φ j x, ϕ, ϕ x , , ϕ (n) x , Ψ j (z) ≡ Ψ j F , F z , , F (n) z . (15.6.4.3) It is reasonable to solve equation (15.6.4.2) by the splitting method. At the first stage, we treat (15.6.4.2) as a purely functional equation, thus disregarding (15.6.4.3). Differenti- ating (15.6.4.2) with respect to x yields the standard bilinear functional differential equation in two independent variables x and z: Φ 1 (x)Ψ 1 (z)+···+Φ k (x)Ψ k (z)+Φ 1 (x)ϕ (x)Ψ 1 (z)+···+ϕ (x)Φ k (x)Ψ k (z)=0,(15.6.4.4) which can be solved using the results of Subsections 15.5.3–15.5.5. Then, substituting the solutions Φ m (x)andΨ m (z) into (15.6.4.2) and taking into account the second relation in (15.6.4.1), we find the function f(t). Further, substituting the functionals (15.6.4.3) into the solutions of the functional equation (15.6.4.2), we obtain determining systems of ordinary differential equations for F (z), ϕ(x), and ψ(t). 15.6. METHOD OF FUNCTIONAL SEPARATION OF VARIABLES 705 Below, we discuss several types of three-argument functional equations of the form (15.6.4.2) that arise most frequently in the functional separation of variables in nonlinear equations of mathematical physics. The results are used for constructing exact solutions for some classes of nonlinear heat and wave equations. 15.6.4-2. Functional equation f(t)+g(x)=Q(z), with z = ϕ(x)+ψ(t). Here, one of the two functions f(t)andψ(t) is prescribed and the other is assumed unknown, also one of the functions g(x)andψ(x) is prescribed and the other is unknown, and the function Q(z) is assumed unknown.* Differentiating the equation with respect to x and t yields Q zz = 0. Consequently, the solution is given by f(t)=Aψ(x)+B, g(x)=Aϕ(x)–B + C, Q(z)=Az + C,(15.6.4.5) where A, B,andC are arbitrary constants. 15.6.4-3. Functional equation f(t)+g(x)+h(x)Q(z)+R(z)=0, with z = ϕ(x)+ψ(t). Differentiating the equation with respect to x yields the two-argument equation g x + h x Q + hϕ x Q z + ϕ x R z = 0.(15.6.4.6) Such equations were discussed in Subsections 15.5.3 and 15.5.4. Hence, the following relations hold [see formulas (15.5.4.4) and (15.5.4.5)]: g x = A 1 hϕ x + A 2 ϕ x , h x = A 3 hϕ x + A 4 ϕ x , Q z =–A 1 – A 3 Q, R z =–A 2 – A 4 Q, (15.6.4.7) where A 1 , , A 4 are arbitrary constants. By integrating system (15.6.4.7) and substituting the resulting solutions into the original functional equation, one obtains the results given below. Case 1.IfA 3 = 0 in (15.6.4.7), the corresponding solution of the functional equation is given by f =– 1 2 A 1 A 4 ψ 2 +(A 1 B 1 + A 2 + A 4 B 3 )ψ – B 2 – B 1 B 3 – B 4 , g = 1 2 A 1 A 4 ϕ 2 +(A 1 B 1 + A 2 )ϕ + B 2 , h = A 4 ϕ + B 1 , Q =–A 1 z + B 3 , R = 1 2 A 1 A 4 z 2 –(A 2 + A 4 B 3 )z + B 4 , (15.6.4.8) where A k and B k are arbitrary constants and ϕ = ϕ(x)andψ = ψ(t) are arbitrary functions. * In similar equations with a composite argument, it is assumed that ϕ(x) const and ψ(t) const. 706 NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS Case 2.IfA 3 ≠ 0 in (15.6.4.7), the corresponding solution of the functional equation is f =–B 1 B 3 e –A 3 ψ + A 2 – A 1 A 4 A 3 ψ – B 2 – B 4 – A 1 A 4 A 2 3 , g = A 1 B 1 A 3 e A 3 ϕ + A 2 – A 1 A 4 A 3 ϕ + B 2 , h = B 1 e A 3 ϕ – A 4 A 3 , Q = B 3 e –A 3 z – A 1 A 3 , R = A 4 B 3 A 3 e –A 3 z + A 1 A 4 A 3 – A 2 z + B 4 , (15.6.4.9) where A k and B k are arbitrary constants and ϕ = ϕ(x)andψ = ψ(t) are arbitrary functions. Case 3. In addition, the functional equation has two degenerate solutions: f = A 1 ψ + B 1 , g = A 1 ϕ + B 2 , h = A 2 , R =–A 1 z – A 2 Q – B 1 – B 2 , (15.6.4.10a) where ϕ = ϕ(x), ψ = ψ(t), and Q = Q(z) are arbitrary functions; A 1 , A 2 , B 1 ,andB 2 are arbitrary constants; and f = A 1 ψ + B 1 , g = A 1 ϕ + A 2 h + B 2 , Q =–A 2 , R =–A 1 z – B 1 – B 2 , (15.6.4.10b) where ϕ = ϕ(x), ψ = ψ(t), and h = h(x) are arbitrary functions; and A 1 , A 2 , B 1 ,andB 2 are arbitrary constants. The degenerate solutions (15.6.4.10a) and (15.6.4.10b) can be obtained directly from the original equation or its consequence (15.6.4.6) using formulas (15.5.4.6). Example 1. Consider the nonstationary heat equation with a nonlinear source ∂w ∂t = ∂ 2 w ∂x 2 + F(w). (15.6.4.11) We look for exact solutions of the form w = w(z), z = ϕ(x)+ψ(t). (15.6.4.12) Substituting (15.6.4.12) into (15.6.4.11) and dividing by w z yields the functional differential equation ψ t = ϕ xx +(ϕ x ) 2 w zz w z + F(w(z)) w z . Let us solve it by the splitting method. To this end, we represent this equation as the functional equation f(t)+g(x)+h(x)Q(z)+R(z)=0,where f(t)=–ψ t , g(x)=ϕ xx , h(x)=(ϕ x ) 2 , Q(z)=w zz /w z , R(z)=f(w(z))/w z . (15.6.4.13) On substituting the expressions of g and h of (15.6.4.13) into (15.6.4.8)–(15.6.4.10), we arrive at overde- termined systems of equations for ϕ = ϕ(x). Case 1. The system ϕ xx = 1 2 A 1 A 4 ϕ 2 +(A 1 B 1 + A 2 )ϕ + B 2 , (ϕ x ) 2 = A 4 ϕ + B 1 following from (15.6.4.8) and corresponding to A 3 = 0 in (15.6.4.7) is consistent in the cases ϕ = C 1 x + C 2 for A 2 =–A 1 C 2 1 , A 4 = B 2 = 0, B 1 = C 2 1 , ϕ = 1 4 A 4 x 2 + C 1 x + C 2 for A 1 = A 2 = 0, B 1 = C 2 1 – A 4 C 2 , B 2 = 1 2 A 4 , (15.6.4.14) where C 1 and C 2 are arbitrary constants. 15.6. METHOD OF FUNCTIONAL SEPARATION OF VARIABLES 707 The first solution in (15.6.4.14) with A 1 ≠ 0 leads to a right-hand side of equation (15.6.4.11) containing the inverse of the error function [the form of the right-hand side is identified from the last two relations in (15.6.4.8) and (15.6.4.13)]. The second solution in (15.6.4.14) corresponds to the right-hand side of F(w)=k 1 w ln w+k 2 w in (15.6.4.11). In both cases, the first relation in (15.6.4.8) is, taking into account that f =–ψ t ,afirst-order linear solution with constant coefficients, whose solution is an exponential plus a constant. Case 2. The system ϕ xx = A 1 B 1 A 3 e A 3 ϕ + A 2 – A 1 A 4 A 3 ϕ + B 2 , (ϕ x ) 2 = B 1 e A 3 ϕ – A 4 A 3 , following from (15.6.4.9) and corresponding to A 3 ≠ 0 in (15.6.4.7), is consistent in the following cases: ϕ = √ –A 4 /A 3 x + C 1 for A 2 = A 1 A 4 /A 3 , B 1 = B 2 = 0, ϕ =– 2 A 3 ln |x| + C 1 for A 1 = 1 2 A 2 3 , A 2 = A 4 = B 2 = 0, B 1 = 4A –2 3 e –A 3 C 1 , ϕ =– 2 A 3 ln cos 1 2 √ A 3 A 4 x + C 1 + C 2 for A 1 = 1 2 A 2 3 , A 2 = 1 2 A 3 A 4 , B 2 = 0, A 3 A 4 > 0, ϕ =– 2 A 3 ln sinh 1 2 √ –A 3 A 4 x + C 1 + C 2 for A 1 = 1 2 A 2 3 , A 2 = 1 2 A 3 A 4 , B 2 = 0, A 3 A 4 < 0, ϕ =– 2 A 3 ln cosh 1 2 √ –A 3 A 4 x + C 1 + C 2 for A 1 = 1 2 A 2 3 , A 2 = 1 2 A 3 A 4 , B 2 = 0, A 3 A 4 < 0, where C 1 and C 2 are arbitrary constants. The right-hand sides of equation (15.6.4.11) corresponding to these solutions are represented in parametric form. Case 3. Traveling-wave solutions of the nonlinear heat equation (15.6.4.11) and solutions of the linear equa- tion (15.6.4.11) with F w = const correspond to the degenerate solutions of the functional equation (15.6.4.10). 15.6.4-4. Functional equation f(t)+g(x)Q(z)+h(x)R(z)=0, with z = ϕ(x)+ψ(t). Differentiating with respect to x yields the two-argument functional differential equation g x Q + gϕ x Q z + h x R + hϕ x R z = 0,(15.6.4.15) which coincides with equation (15.5.4.4), up to notation. Nondegenerate case. Equation (15.6.4.15) can be solved using formulas (15.5.4.5). In this way, we arrive at the system of ordinary differential equations g x =(A 1 g + A 2 h)ϕ x , h x =(A 3 g + A 4 h)ϕ x , Q z =–A 1 Q – A 3 R, R z =–A 2 Q – A 4 R, (15.6.4.16) where A 1 , , A 4 are arbitrary constants. The solution of equation (15.6.4.16) is given by g(x)=A 2 B 1 e k 1 ϕ + A 2 B 2 e k 2 ϕ , h(x)=(k 1 – A 1 )B 1 e k 1 ϕ +(k 2 – A 1 )B 2 e k 2 ϕ , Q(z)=A 3 B 3 e –k 1 z + A 3 B 4 e –k 2 z , R(z)=(k 1 – A 1 )B 3 e –k 1 z +(k 2 – A 1 )B 4 e –k 2 z , (15.6.4.17) where B 1 , , B 4 are arbitrary constants and k 1 and k 2 are roots of the quadratic equation (k – A 1 )(k – A 4 )–A 2 A 3 = 0.(15.6.4.18) 708 NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS In the degenerate case k 1 = k 2 ,thetermse k 2 ϕ and e –k 2 z in (15.6.4.17) must be replaced by ϕe k 1 ϕ and ze –k 1 z , respectively. In the case of purely imaginary or complex roots, one should extract the real (or imaginary) part of the roots in solution (15.6.4.17). On substituting (15.6.4.17) into the original functional equation, one obtains conditions that must be met by the free coefficients and identifies the function f (t), specifically, B 2 = B 4 = 0 =⇒ f (t)=[A 2 A 3 +(k 1 – A 1 ) 2 ]B 1 B 3 e –k 1 ψ , B 1 = B 3 = 0 =⇒ f (t)=[A 2 A 3 +(k 2 – A 1 ) 2 ]B 2 B 4 e –k 2 ψ , A 1 = 0 =⇒ f (t)=(A 2 A 3 + k 2 1 )B 1 B 3 e –k 1 ψ +(A 2 A 3 + k 2 2 )B 2 B 4 e –k 2 ψ . (15.6.4.19) Solution (15.6.4.17), (15.6.4.19) involves arbitrary functions ϕ = ϕ(x)andψ = ψ(t). Degenerate case. In addition, the functional equation has two degenerate solutions, f = B 1 B 2 e A 1 ψ , g = A 2 B 1 e –A 1 ϕ , h = B 1 e –A 1 ϕ , R =–B 2 e A 1 z – A 2 Q, where ϕ = ϕ(x), ψ = ψ(t), and Q = Q(z) are arbitrary functions; A 1 , A 2 , B 1 ,andB 2 are arbitrary constants; and f = B 1 B 2 e A 1 ψ , h =–B 1 e –A 1 ϕ – A 2 g, Q = A 2 B 2 e A 1 z , R = B 2 e A 1 z , where ϕ = ϕ(x), ψ = ψ(t), and g = g(x) are arbitrary functions; and A 1 , A 2 , B 1 ,andB 2 are arbitrary constants. The degenerate solutions can be obtained immediately from the original equation or its consequence (15.6.4.15) using formulas (15.5.4.6). Example 2. For the nonlinear heat equation (15.6.3.1), searching for exact solutions in the form w = w(z), with z = ϕ(x)+ψ(t), leads to the functional equation (15.6.3.3), which coincides with the equation f(t)+g(x)Q(z)+h(x)R(z)=0 if f(t)=–ψ t , g(x)=ϕ xx , h(x)=(ϕ x ) 2 , Q(z)=F(w), R(z)= [F(w)w z ] z w z , w = w(z). 15.7. Direct Method of Symmetry Reductions of Nonlinear Equations 15.7.1. Clarkson–Kruskal Direct Method 15.7.1-1. Simplified scheme. Examples of constructing exact solutions. The basic idea of the simplified scheme is as follows: for an equation with the unknown function w = w(x, t), an exact solution is sought in the form w = f(t)u(z)+g(x, t), z = ϕ(t)x + ψ(t). (15.7.1.1) The functions f(t), g(x, t), ϕ(t), and ψ(t) are found in the subsequent analysis and are chosen in such a way that, ultimately, the function u(z) would satisfy a single ordinary differential equation. Below we consider some cases in which it is possible to construct exact solutions of nonlinear equations of the form (15.7.1.1). 15.7. DIRECT METHOD OF SYMMETRY REDUCTIONS OF NONLINEAR EQUATIONS 709 Example 1. Consider the generalized Burgers–Korteweg–de Vries equation ∂w ∂t = a ∂ n w ∂x n + bw ∂w ∂x .(15.7.1.2) We seek its exact solution in the form (15.7.1.1). Inserting (15.7.1.1) into (15.7.1.2), we obtain afϕ n u (n) z + bf 2 ϕuu z + f(bgϕ – ϕ t x – ψ t )u z +(bfg x – f t )u + ag (n) x + bgg x – g t = 0.(15.7.1.3) Equating the functional coefficients of u (n) z and uu z in (15.7.1.3), we get f = ϕ n–1 .(15.7.1.4) Further, equating the coefficient of u z to zero, we obtain g = 1 bϕ (ϕ t x + ψ t ). (15.7.1.5) Inserting the expressions (15.7.1.4) and (15.7.1.5) into (15.7.1.3), we arrive at the relation ϕ 2n–1 (au (n) z + buu z )+(2 – n)ϕ n–2 ϕ t u + 1 bϕ 2 (2ϕ 2 t – ϕϕ tt )x + 2ϕ t ψ t – ϕψ tt = 0. Dividing each term by ϕ 2n–1 and then eliminating x with the help of the relation x =(z – ψ)/ϕ, we obtain au (n) z + buu z +(2 – n)ϕ –n–1 ϕ t u + 1 b ϕ –2n–2 (2ϕ 2 t – ϕϕ tt )z + 1 b ϕ –2n–2 (ϕψϕ tt – ϕ 2 ψ tt + 2ϕϕ t ψ t – 2ψϕ 2 t )=0. (15.7.1.6) Let us require that the functional coefficient of u and the last term be constant, ϕ –n–1 ϕ t =–A, ϕ –2n–2 (ϕψϕ tt – ϕ 2 ψ tt + 2ϕϕ t ψ t – 2ψϕ 2 t )=B, where A and B are arbitrary. As a result, we arrive at the following system of ordinary differential equations for ϕ and ψ: ϕ t =–Aϕ n+1 , ψ tt + 2Aϕ n ψ t + A 2 (1 – n)ϕ 2n ψ =–Bϕ 2n . (15.7.1.7) Using (15.7.1.6) and (15.7.1.7), we obtain an equation for u(z), au (n) z + buu z + A(n – 2)u + A 2 b (1 – n)z + B b = 0.(15.7.1.8) For A ≠ 0, the general solution of equations (15.7.1.7) has the form ϕ(t)=(Ant + C 1 ) – 1 n , ψ(t)=C 2 (Ant + C 1 ) n–1 n + C 3 (Ant + C 1 ) – 1 n + B A 2 (n – 1) , (15.7.1.9) where C 1 , C 2 ,andC 3 are arbitrary constants. Formulas (15.7.1.1), (15.7.1.4), (15.7.1.5), and (15.7.1.9), together with equation (15.7.1.8), describe an exact solution of the generalized Burgers–Korteweg–de Vries equation (15.7.1.2). Example 2. Consider the Boussinesq equation ∂ 2 w ∂t 2 + ∂ ∂x w ∂w ∂x + a ∂ 4 w ∂x 4 = 0. (15.7.1.10) Just as in Example 1, we seek its solutions in the form (15.7.1.1), where the functions f (t), g(x, t), ϕ(t), and ψ(t) are found in the subsequent analysis. Substituting (15.7.1.1) into (15.7.1.10) yields afϕ 4 u + f 2 ϕ 2 uu + f(z 2 t + gϕ 2 )u + f 2 ϕ 2 (u ) 2 +(fz tt + 2fg x ϕ + 2f t z t )u +(fg xx + f tt )u + g tt + gg xx + g 2 x + ag (4) x = 0. (15.7.1.11) Equating the functional coefficients of u and uu ,weget f = ϕ 2 . (15.7.1.12) . solutions of the form w = w(z), z = ϕ(x)+ψ(t). Notation: A, C 1 ,andC 2 are arbitrary constants; σ = 1 for z > 0 and σ =–1 for z < 0 No. Right-hand side F(w) Solution w(z) Equations for ψ(t )and (x) 1 aw. the solutions of the functional equation (15.6.4.2), we obtain determining systems of ordinary differential equations for F (z), ϕ(x), and ψ(t). 15.6. METHOD OF FUNCTIONAL SEPARATION OF VARIABLES. used for constructing exact solutions for some classes of nonlinear heat and wave equations. 15.6.4-2. Functional equation f(t)+g(x)=Q(z), with z = ϕ(x)+ψ(t). Here, one of the two functions f(t )and (t)