Handbook of mathematics for engineers and scienteists part 113 pot

752 NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS where it has been taken into account that ξ x 0. If follows from the first equation of (15.11.2.5) that w 0 =–2νξ x . (15.11.2.6) Substituting this into the second and third equations of (15.11.2.5), after some rearrangements we obtain ξ t + w 1 ξ x – νξ xx = 0, (ξ t + w 1 ξ x – νξ xx ) x = 0. (15.11.2.7) It is obvious that if the first equation of (15.11.2.7) is valid, then the second is satisfied identically. The last equation in (15.11.2.5) is the Burgers equation. Hence, formula (15.11.2.4) in view of (15.11.2.6) can be rewritten as w =–2ν ∂ ∂x ln ξ + w 1 , (15.11.2.8) where the functions w and w 1 satisfy the Burgers equation and the function ξ is described by the first equation of (15.11.2.7). Given a solution w 1 of the Burgers equation, formula (15.11.2.8) allows obtaining other solutions of it by solving the first equation in (15.11.2.7), which is linear in ξ. Taking into account that w 1 = 0 is a particular solution of the Burgers equation, let us substitute it into (15.11.2.7) and (15.11.2.8). This results in the Cole–Hopf transformation w =–2ν ξ x ξ . This transformation allows constructing solutions of the nonlinear Burgers equation (15.11.2.1) via solutions of the linear heat equation ξ t = νξ xx . Example 2. Consider the Korteweg–de Vries equation ∂w ∂t + w ∂w ∂x + ∂ 3 w ∂x 3 = 0. (15.11.2.9) First step. Let us substitute the leading term of the expansion (15.11.1.1) into equation (15.11.2.9) and then multiply the resulting relation by ξ p+3 (the product ξ p+3 w xxx gives a zeroth order quantity) to obtain w  0 ξ 3 + pw 0 x  0 ξ 2 – pw 2 0 ξ 2–p – p(p + 1)(p + 2)w 0 = 0, where ξ = x – x 0 , x 0 = x 0 (t), and w 0 = w 0 (t). From the balance of the highest-order terms (only the last two terms are taken into account) it follows that p = 2, w 0 =–12 (m = 0). Since p is a positive integer, the first necessary condition of the Painlev ´ e test is satisfied. Second step. To fine the Fuchs indices (resonances), we substitute the binomial w =–12ξ –2 + w m ξ m–2 into the leading terms of equation (15.11.2.9), where the second and the third term are taken into account. Isolating the term proportional to w m ,wehave (m + 1)(m – 4)(m – 6)w m ξ m–5 + ···= 0. Equating (m + 1)(m – 4)(m – 6) to zero gives the Fuchs indices m 1 = 4, m 2 = 6. Since they are both positive integers, the second necessary condition of the Painlev ´ e test is satisfied. Third step. We substitute the expansion (15.11.1.1), while considering, according to the second step, the terms up to number m = 6 inclusive, w =–12ξ –2 + w 1 ξ –1 + w 2 + w 3 ξ + w 4 ξ 2 + w 5 ξ 3 + w 6 ξ 4 + ···, ξ = x – x 0 (t)(15.11.2.10) 15.11. PAINLEV ´ E TEST FOR NONLINEAR EQUATIONS OF MAT HE MAT I C A L PHYSICS 753 into equation (15.11.2.9). Then we collect the terms of equal powers in ξ and equate the coefficients of the different powers of ξ to zero to arrive at a system of equations for the w m : ξ –4 : 2w 1 = 0, ξ –3 : 24w 2 – 24x  0 – w 2 1 = 0, ξ –2 : 12w 3 + w 1 x  0 – w 1 w 2 = 0, ξ –1 : 0×w 4 + w  1 = 0, 1:–6w 5 + w  2 – w 3 x  0 + w 1 w 4 + w 2 w 3 = 0, ξ: 0×w 6 + w  3 – 2w 4 x  0 + w 2 3 + 2w 1 w 5 + 2w 2 w 4 = 0. (15.11.2.11) Simple computations show that the equations with resonances corresponding to the powers ξ –1 and ξ are satisfied identically. Hence, the Korteweg–de Vries equation (15.11.2.9) passes the Painlev ´ etest. The solution of equation (15.11.2.11) results in the following expansion coefficients in (15.11.2.10): w 1 = 0, w 2 = x  0 (t), w 3 = 0, w 4 = w 4 (t), w 5 = 1 6 x  0 (t), w 6 = w 6 (t), where x 0 (t), w 4 (t), and w 6 (t) are arbitrary functions. Truncated series expansion and the B ¨ acklund transformation. For further analysis, let us use a truncated expansion of the general form (15.11.1.6) with p = 2: w = w 0 ξ 2 + w 1 ξ + w 2 .(15.11.2.12) Substituting (15.11.2.12) into (15.11.2.9) and equating the functional coefficients of the different powers of ξ to zero, in the same way as in Example 2, we arrive at the B ¨ acklund transformation w = 12(ln ξ) xx + w 2 , ξ t ξ x + w 2 ξ 2 x + 4ξ x ξ xxx – 3ξ 2 xx = 0, ξ xt + w 2 ξ xx + ξ xxx = 0, (w 2 ) t + w 2 (w 2 ) x +(w 2 ) xxx = 0. (15.11.2.13) It relates the solutions w and w 2 of the Korteweg–de Vries equations. Eliminating w 2 from the second and third equations in (15.11.2.13), one can derive an equation for ξ, which can further be reduced, via suitable transformations, to a system of linear equations. 15.11.3. Construction of Solutions of Nonlinear Equations That Fail the Painlev ´ e Test, Using Truncated Expansions In some cases truncated expansions of the form (15.11.1.6) may be effective in finding exact solutions to nonlinear equations of mathematical physics that fail the Painlev ´ etest. In such cases, the expansion parameter p must be a positive integer; it is determined in the same way as at the first step of performing the Painlev ´ e test. We illustrate this by a specific example below. Example. Consider the nonlinear diffusion equation with a cubic source w t = w xx – 2w 3 . (15.11.3.1) First step. Let us substitute the leading term of the expansion (15.11.1.1) into equation (15.11.3.1) and then multiply the resulting relation by ξ p+2 to obtain w  0 ξ 2 + pw 0 x  0 ξ = p(p + 1)w 0 – 2w 3 0 ξ 2–2p , where ξ = x – x 0 , x 0 = x 0 (t), and w 0 = w 0 (t). From the balance of the highest-order terms (both terms on the right-hand side are taken into account) it follows that p = 1, w 0 = 1 (m = 0). (15.11.3.2) Since p is a positive integer, the first necessary condition of the Painlev ´ e test is satisfied. 754 NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS Second step. The equation is invariant under the substitution of –w for w. Hence, it suffices to consider only the positive value of w 0 in (15.11.3.2). Therefore, in order to find resonances, we substitute the binomial w = ξ –1 + w m ξ m–1 in the leading terms w xx and bw 3 of equation (15.11.3.1). Collecting the terms proportional to w m ,weget (m + 1)(m – 4)w m ξ m–3 + ···= 0. Equating (m + 1)(m – 4) to zero gives the Fuchs index m 1 = 4. Since it is integer and positive, the second necessary condition of the Painlev ´ e test is satisfied. Third step. We substitute the expansion (15.11.1.1) into equation (15.11.3.1); according to the second step, the terms up to number m = 4 inclusive must be taken into account. It can be shown that the consistency condition (15.11.1.5) is not satisfied, and therefore the equation in question fails the Painlev ´ etest. Using a truncated expansion for finding exact solutions. For further analysis, we use a truncated expansion of the general form (15.11.1.6) with p = 1, which from the first step. This results in formula (15.11.2.4). Substituting (15.11.2.4) into the diffusion equation (15.11.3.1) and collecting the terms of equal powers in ξ, we obtain ξ –3  2w 0 ξ 2 x – 2w 3 0  + ξ –2  w 0 ξ t – 2(w 0 ) x ξ x – w 0 ξ xx – 6w 2 0 w 1  + ξ –1  –(w 0 ) t +(w 0 ) xx – 6w 0 w 2 1  –(w 1 ) t +(w 1 ) xx – 2w 3 1 = 0. Equating the coefficients of like powers of ξ to zero, we arrive at the system of equations w 0 (ξ 2 x – w 2 0 )=0, w 0 ξ t – 2(w 0 ) x ξ x – w 0 ξ xx – 6w 2 0 w 1 = 0, –(w 0 ) t +(w 0 ) xx – 6w 0 w 2 1 = 0, (w 1 ) t –(w 1 ) xx + 2w 3 1 = 0. (15.11.3.3) From the first equation in (15.11.3.3) we have w 0 = ξ x . (15.11.3.4) The other solution differs in sign only and gives rise to the same result, and therefore is not considered. Substituting (15.11.3.4) into the second and third equations of (15.11.3.3) and canceling by nonzero factors, we obtain ξ t – 3ξ xx – 6 w 1 ξ x = 0, –ξ xt + ξ xxx – 6w 2 1 ξ x = 0. (15.11.3.5) The latter equation in (15.11.3.3), which coincides with the original equation (15.11.3.1), is satisfied if w 1 = 0. (15.11.3.6) On inserting (15.11.3.4) and (15.11.3.6) in (15.11.2.4), we get the following representation for solutions: w = ξ x ξ , (15.11.3.7) where the function ξ is determined by an overdetermined linear system of equations resulting from the substi- tution of (15.11.3.6) into (15.11.3.5): ξ t – 3ξ xx = 0, –ξ xt + ξ xxx = 0. (15.11.3.8) Differentiate the first equation with respect to x and then eliminate the mixed derivative w xt using the second equation to obtain ξ xxx = 0. It follows that ξ = ϕ 2 (t)x 2 + ϕ 1 (t)x + ϕ 0 (t). (15.11.3.9) In order to determine the functions ϕ k (t), let us substitute (15.11.3.9) into equations (15.11.3.8) to obtain ϕ  2 x 2 + ϕ  1 x + ϕ  0 – 6ϕ 2 = 0, –ϕ  2 x – ϕ  1 = 0. Equating the functional coefficients of the different powers of x to zero and integrating the resulting equations, we get ϕ 2 = C 2 , ϕ 1 = C 1 , ϕ 0 = 6C 2 t + C 0 ,(15.11.3.10) where C 0 , C 1 ,andC 2 are arbitrary constants. Substituting (15.11.3.9) into (15.11.3.7) and taking into account (15.11.3.10), we find an exact solution of equation (15.11.3.1) in the form w = 2C 2 x + C 1 C 2 x 2 + C 1 x + 6C 2 t + C 0 . 15.12. METHODS OF THE INVERSE SCATTERING PROBLEM (SOLITON THEORY) 755 15.12. Methods of the Inverse Scattering Problem (Soliton Theory) Preliminary remarks. The methods of the inverse scattering problem rely on “implicit” linearization of equations. Main idea: Instead of the original nonlinear equation in the unknown w, one considers an auxiliary overdetermined linear system of equation for a (vector) function ϕ, with the coefficients of this system generally dependent on w and the derivatives of w with respect to the independent variables. The linear system for ϕ is chosen so that the compatibility condition for its equations gives rise to the original nonlinear equation for w. 15.12.1. Method Based on Using Lax Pairs 15.12.1-1. Method description. Consistency condition. Lax pairs. We will be studying a nonlinear evolution equation of the form ∂w ∂t =F(w), (15.12.1.1) where the right-hand side F(w) depends on w and its derivatives with respect to x. Consider two auxiliary linear differential equations, one corresponding to an eigenvalue problem and involving derivatives with respect to the space variable x only, Lϕ = λϕ,(15.12.1.2) and the other describing the evolution of the eigenfunction in time, ∂ϕ ∂t =–Mϕ.(15.12.1.3) The coefficients of the linear differential operators L and M in equations (15.12.1.2) and (15.12.1.3) depend on w and its derivatives with respect to x. Since system (15.12.1.2)–(15.12.1.3) is overdetermined (there are two equations for ϕ), the operators L and M cannot be arbitrary—they must satisfy a compatibility condition. In order to find this condition, let us first differentiate (15.12.1.2) with respect to t. Assuming that the eigenvalues λ are independent of time t,wehave L t ϕ +Lϕ t = λϕ t . Replacing ϕ t here by the right-hand side (15.12.1.3), we get L t ϕ –LMϕ =–λMϕ. Taking into account the relations λMϕ =M(λϕ)andλϕ =Lϕ, we arrive at the compatibility condition L t ϕ =LMϕ –MLϕ, which can be rewritten in the form of an operator equation: ∂L ∂t =LM–ML. (15.12.1.4) The linear operators L and M [or the linear equations (15.12.1.2) and (15.12.1.3)] are said to form a Lax pair for the nonlinear equation (15.12.1.1) if the compatibility condition (15.12.1.4) coincides with equation (15.12.1.1). The right-hand side of equation (15.12.1.4) represents the commutator of the operators L and M, which is denoted by [L, M] = LM–ML for short. Thus, if a suitable Lax pair is found, the analysis of the nonlinear equation (15.12.1.1) can be reduced to that of two simpler, linear equations, (15.12.1.2) and (15.12.1.3). 756 NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS Remark. The operator M in equations (15.12.1.3) and (15.12.1.3) is defi nedtoanadditivefunctionof time; it can be changed according to the rule M =⇒ M+p(t), where p(t) is an arbitrary function. This function is found in solving a Cauchy problem for equation (15.12.1.1); see Paragraph 15.12.3-2. 15.12.1-2. Examples of Lax pairs for nonlinear equations of mathematical physics. Example 1. Let us show that a Lax pair for the Korteweg–de Vries equation ∂w ∂t + ∂ 3 w ∂x 3 – 6w ∂w ∂x = 0 (15.12.1.5) is formed by the operators L=w – ∂ 2 ∂x 2 ,M=4 ∂ 3 ∂x 3 – 6w ∂ ∂x – 3 ∂w ∂x + p(t), (15.12.1.6) which generate the linear equations ϕ xx +(λ – w)ϕ = 0, ϕ t + 4ϕ xxx – 6wϕ x – 3w x ϕ + p(t)ϕ = 0. (15.12.1.7) Here, p(t) is an arbitrary function. It is not difficult to verify that the following formulas hold: LM(ϕ)=–4ϕ xxxxx + 10wϕ xxx +[15w x – p(t)]ϕ xx +(12w xx – 6w 2 )ϕ x +[3w xxx – 3ww x + wp(t)]ϕ, ML(ϕ)=–4ϕ xxxxx + 10wϕ xxx +[15w x – p(t)]ϕ xx +(12w xx – 6w 2 )ϕ x +[4w xxx – 9ww x + wp(t)]ϕ, where ϕ(x, t) is an arbitrary function. It follows that LM(ϕ)–ML(ϕ)=(–w xxx + 6ww x )ϕ. (15.12.1.8) From (15.12.1.6) and (15.12.1.8) we obtain L t = w t ,LM–ML=–w xxx + 6ww x . On inserting these expressions into (15.12.1.4), we arrive at the Korteweg–de Vries equation (15.12.1.5). Remark. A procedure for solving the Cauchy problem for equation (15.12.1.5) is outlined in Subsec- tion 15.12.3. The linear equations (15.12.1.2) and (15.12.1.3) for the auxiliary function ϕ,which form a Lax pair, can be vector; in this case, the linear operators L and M are represented by matrices. In other words, the individual equations (15.12.1.2) and (15.12.1.3) may be replaced by appropriate systems of linear equations. Example 2. The sinh-Gordon equation w xt = 4a sinh w can be represented as a vector Lax pair Lϕ = λϕ, ϕ t =–Mϕ, where ϕ =  ϕ 1 ϕ 2  ,L=  0 ∂ x + 1 2 w x ∂ x – 1 2 w x 0  ,M= a λ  0 e w e –w 0  . The determination of a Lax pair for a given nonlinear equation is a very complex problem that is basically solvable for isolated equations only. Therefore, the “implicit” linearization of equations is usually realized using a simpler method, described in Subsection 15.12.2. 15.12. METHODS OF THE INVERSE SCATTERING PROBLEM (SOLITON THEORY) 757 15.12.2. Method Based on a Compatibility Condition for Systems of Linear Equations 15.12.2-1. General scheme. Compatibility condition. Systems of two equations. Consider two systems of linear equations ϕ x =Aϕ,(15.12.2.1) ϕ t =Bϕ,(15.12.2.2) where ϕ is an n-dimensional vector and A and B are n×n matrices. The right-hand sides of systems (15.12.2.1) and (15.12.2.2) cannot be arbitrary—they must satisfy a compatibility condition. To fi nd this condition, let us differentiate systems (15.12.2.1) and (15.12.2.2) with respect to t and x, respectively, and eliminate the mixed derivative ϕ xt from the resulting equations. Then replacing the derivatives ϕ x and ϕ t by the right-hand sides of (15.12.2.1) and (15.12.2.2), we obtain A t –B x +[A,B]=0,(15.12.2.3) where [A, B] = AB – BA. It turns out that, given a matrix A, there is a simple deductive procedure for finding B as a result of which the compatibility condition (15.12.2.3) becomes a nonlinear evolution equation. Let us dwell on the special case where the vector function ϕ two components, ϕ =  ϕ 1 ϕ 2  . We choose a linear system of equations (15.12.2.1) in the form (ϕ 1 ) x =–iλϕ 1 + fϕ 2 , (ϕ 2 ) x = gϕ 1 + iλϕ 2 , (15.12.2.4) where λ is the spectral parameter, f and g are some (generally complex valued) functions of two variables x and t,andi 2 =–1. As system (15.12.2.2) we take the most general linear system involving the derivatives with respect to t: (ϕ 1 ) t = Aϕ 1 + Bϕ 2 , (ϕ 2 ) t = Cϕ 1 + Dϕ 2 , (15.12.2.5) where A, B, C,andD are some functions (dependent on the variables x, t and the parame- ter λ) to be determined in the subsequent analysis. Differentiating equations (15.12.2.4) with respect to t and equations (15.12.2.5) with respect to x and assuming that (ϕ 1,2 ) xt =(ϕ 1,2 ) tx , we obtain compatibility conditions in the form A x = Cf – Bg, B x + 2iλB = f t –(A – D)f, C x – 2iλC = g t +(A – D)g, –D x = Cf – Bg. (15.12.2.6) For simplicity, we set D =–A. In this case, the first and last equations in (15.12.2.6) coincide, so that (15.12.2.6) turns into a system of three determining equations: A x = Cf – Bg, B x + 2iλB = f t – 2Af, C x – 2iλC = g t + 2Ag. (15.12.2.7) The functions A, B,andC must be expressed in terms of f and g. The general solution of system (15.12.2.7) for arbitrary functions f and g cannot be found. So let us look for particular solutions in the form finite expansions in positive and negative powers of the parameter λ. 758 NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS 15.12.2-2. Solution of the determining equations in the form of polynomials in λ. The simplest polynomial representations of the unknown functions that give rise to nontrivial results are quadratic in the spectral parameter λ: A = A 2 λ 2 + A 1 λ + A 0 , B = B 2 λ 2 + B 1 λ + B 0 , C = C 2 λ 2 + C 1 λ + C 0 . (15.12.2.8) Let us substitute (15.12.2.8) into (15.12.2.7) and collect the terms with equal powers in λ to obtain λ 2 (A 2x – C 2 f + B 2 g)+λ(A 1x – C 1 f + B 1 g)+A 0x – C 0 f + B 0 g = 0,(15.12.2.9) 2iλ 3 B 2 + λ 2 (B 2x + 2iB 1 + 2A 2 f) + λ(B 1x + 2iB 0 + 2A 1 f)+B 0x + 2A 0 f – f t = 0,(15.12.2.10) –2iλ 3 C 2 + λ 2 (C 2x – 2iC 1 – 2A 2 g) + λ(C 1x – 2iC 0 – 2A 1 g)+C 0x – 2A 0 g – g t = 0.(15.12.2.11) Let us equate the coefficients of like powers of λ to zero starting from the highest power. Setting the coefficients of λ 3 to zero gives B 2 = C 2 = 0.(15.12.2.12) Equating the coefficients of λ 2 to zero and taking into account (15.12.2.12), we find that A 2 = a = const, B 1 = iaf, C 1 = iag.(15.12.2.13) Setting the coefficient of λ in (15.12.2.9) to zero and then replacing B 1 and C 1 in accordance with (15.12.2.13), we have A 1x = 0, whence A 1 = a 1 = const. For simplicity, we dwell on the special case a 1 = 0 (arbitrary a 1 gives rise to more general results), so that A 1 = 0.(15.12.2.14) By equating to zero the coefficients of λ in the equations obtained from (15.12.2.10) and (15.12.2.11) and taking into account (15.12.2.13) and (15.12.2.14), we get B 0 =– 1 2 af x , C 0 = 1 2 ag x .(15.12.2.15) Setting the coefficient of λ 0 in (15.12.2.9) to zero and then integrating, we find that A 0 = 1 2 afg + a 0 ,wherea 0 is an arbitrary constant. As before, we set a 0 = 0 for simplicity, which results in A 0 = 1 2 afg.(15.12.2.16) Then equations (15.12.2.10) and (15.12.2.11) in view of (15.12.2.15) and (15.12.2.16) become f t =– 1 2 af xx + af 2 g, g t = 1 2 ag xx – afg 2 . (15.12.2.17) Substituting (15.12.2.8) and (15.12.2.12)–(15.12.2.17) into (15.12.2.5) yields (ϕ 1 ) t = a(λ 2 + 1 2 fg)ϕ 1 + a(iλf – 1 2 f x )ϕ 2 , (ϕ 2 ) t = a(iλg + 1 2 g x )ϕ 1 – a(λ 2 + 1 2 fg)ϕ 2 . (15.12.2.18) . A, B,andC must be expressed in terms of f and g. The general solution of system (15.12.2.7) for arbitrary functions f and g cannot be found. So let us look for particular solutions in the form. x 0 (t), w 4 (t), and w 6 (t) are arbitrary functions. Truncated series expansion and the B ¨ acklund transformation. For further analysis, let us use a truncated expansion of the general form (15.11.1.6). satisfied, and therefore the equation in question fails the Painlev ´ etest. Using a truncated expansion for finding exact solutions. For further analysis, we use a truncated expansion of the general form