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Separation of variables for nonlinear equations of hyperbolic and Korteweg–de Vries type
Vol. 68 (2011)
REPORTS ON MATHEMATICAL PHYSICS
No. 1
SEPARATION OF VARIABLES FOR NONLINEAR EQUATIONS OF HYPERBOLIC AND KORTEWEG–DE VRIES TYPE A NATOLIY BARANNYK Institute of Mathematics, Pomeranian University, Arciszewskiego Str., 76-200 Slupsk, Poland
TATJANA BARANNYK Poltava National Pedagogical University, 2 Ostrogradskogo Str., 36000 Poltava, Ukraine
and I VAN Y URYK National University of Food Technologies, 68 Volodymyrska Str., 01033 Kyiv, Ukraine (e-mail: [email protected]) (Received November 19, 2010 — Revised May 12, 2011) We propose substitutions that have been used for constructing wide classes of exact solutions with the generalized separation of variables for nonlinear equations of the hyperbolic and Korteweg–de Vries type (KdV-type). These solutions cannot be obtained by means of S. Lie method or by the method of conditional symmetries. Keywords: nonlinear hyperbolic equations; Korteweg–de Vries equations; generalized separation of variables; Lie method.
1.
Introduction One of the effective methods for constructing exact solutions of linear PDE is separation of variables. For equations with two independent variables x, t and a dependent variable u, this method is based on finding exact solutions in the form of product of functions of different arguments u = a(x)b(t).
(1)
A method of constructing exact solutions of nonlinear PDE generalizing the classic variable separation method is described in [1]. The solutions are sought in the form of a finite sum of k terms, u(x, t) =
k
fi (t)ai (x),
(2)
i=1
where fi (t), ai (x) are smooth functions that have to be determined. Exact solutions in the form of generalized separation of variables (2) that contain more than two [97]
98
A. BARANNYK, T. BARANNYK and I. YURYK
terms are given in [2–5]. The results given in [3–5] are based on finding finitedimensional subspaces that are invariant under a corresponding nonlinear differential operator. The assumption (1) can be considered as an ansatz which reduces the equation under study to an ordinary differential equation with an unknown function a = a(x) (or with an unknown function b = b(t)). Using the idea of reduction, in the present paper we propose the following generalization of (1) for nonlinear equations, u=
m
ωi (t)ai (x) + f (x, t),
m ≥ 1.
(3)
i=1
The postulate (3) contains an unknown function f (x, t), m unknown functions ai (x) and m unknown functions ωi (t). They are found from the condition that (3) reduces the equation under consideration to a system of m ordinary differential equations with unknown functions ωi (t). If m = 1, this system is reduced to an ordinary differential equation with one unknown function ω1 (t). In many cases formula (3) makes easy to find exact solutions in the form of generalized separation of variables (2). If in a given differential equation we set x = x(u, t), then we get a differential equation with the independent variables u, t and the dependent variable x. To construct solutions of such equation, we can use the substitution x=
m
ωi (t)ai (u) + f (u, t),
(4)
i=1
which is obtained from (3) as a result of replacement u → x, x → u, t → t. In this paper we use substitutions of type (3) and (4) to solve exact by the nonlinear hyperbolic-type equation 2 ∂ 2u ∂ 2u ∂u = au 2 + b +c (5) 2 ∂t ∂x ∂x and the KdV-type equation
k ∂ 3u ∂u ∂u + 3 = 0, + F (u) ∂t ∂x ∂x
(6)
where k is a real parameter. Eq. (5) was the subject of investigation in [4, 5], while Eq. (6) was investigated in [6] using the conditional symmetry method. The solutions of Eqs. (5) and (6) given in the present work cannot be obtained by means of the group analysis method. 2.
Separation of variables for Eq. (5) To construct exact solutions of Eq. (5), we assume that u = ω(t) + f (x, t),
(7)
SEPARATION OF VARIABLES FOR NONLINEAR EQUATIONS
which is a particular case of the following one,
99
(3). This substitution is in turn a generalization of u(x, t) = ϕ(x) + ψ(t)
that is often used to find exact solutions of nonlinear PDE. Substituting (7) into (5), we get (8) ω + ftt = a (ω + f )fxx + bfx2 + c, which has to be an ordinary differential equation with unknown function ω = ω(t). It follows that the coefficient afxx at ω in Eq. (8) can be written in the form of afxx = 2aμ2 (t), and then f = μ2 (t)x 2 + μ1 (t)x + μ˜ 0 (t), where μ˜ 0 (t), μ1 (t), μ2 (t) are some functions of t. From (7) we obtain u = μ2 (t)x 2 + μ1 (t)x + μ0 (t),
(9)
μ0 (t) = μ˜ 0 (t) + ω(t). This was found in [4] in the study of the structure of invariant subspaces for the operator which corresponds to the right-hand side of Eq. (5). Let us consider a particular case of Eq. (5), 2 ∂ 2u ∂u ∂ 2u = au 2 + b . (10) 2 ∂t ∂x ∂x To construct exact solutions of Eq. (10) we set u = ω(t)d(x) + f (x, t),
(11)
which is a particular case of (3). This assumption (11) reduces (10) to the equation ω d − ω(afxx d + 2bfx d + af d ) − ω2 (add + bd 2 ) + ftt − affxx − bfx2 = 0. (12) Eq. (12) has to be an ordinary differential equation with an unknown function ω = ω(t). It follows that add + bd 2 = βd, β ∈ R, (13) adfxx + 2bd fx + ad f = γ˜ (t)d.
(14)
As a result, the problem of finding exact solutions for Eq. (10) in the form of (11) is reduced to integration of the system of two ordinary differential equations, one of which is linear. Eq. (13) has the particular solution, α a = b, β = 0, α = 1. (15) d = xα, 1−α Setting d = x α into Eq. (14), we get x 2 fxx + 2(1 − α)xfx + α(α − 1)f = γ (t)x 2 , 1−α γ˜ (t). b Let us consider three cases.
where γ (t) =
(16)
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A. BARANNYK, T. BARANNYK and I. YURYK
a) Case α = 2. From (15), it follows that a = −2b. The general solution of Eq. (16) for α = 2 is the function 2 ˜ f = γ (t)x 2 ln |x| + β(t)x + δ(t)x,
and therefore, according to (11), we have u = γ (t)x 2 ln |x| + β(t)x 2 + δ(t)x,
(17)
˜ + ω(t). Substituting (17) into (10), we find the equality δ(t) = 0 where β(t) = β(t) and a system of equations determining the functions β(t) and γ (t): γ = −2bγ 2 ,
β = −2bβγ + bγ 2 .
(18)
The system (18) has the following particular solution: 9 3 β = c1 t 3 + c2 t −2 − t −2 ln |t|. γ = − t −2 , b 5b As a result, we find an exact solution of Eq. (10) for a = −2b: 3 −2 2 9 −2 3 −2 u = − t x ln |x| + c1 t + c2 t − t ln |t| x 2 , b 5b where c1 , c1 are arbitrary constants. 3 b) Case α = 3. From (15) it follows that a = − b. The general solution of 2 (16) for α = 3 is the function 3 ˜ , f = −γ (t)x 2 ln |x| + β(t)x 2 + δ(t)x and therefore according to (11) we have u = −γ (t)x 2 ln |x| + β(t)x 2 + δ(t)x 3 ,
(19)
˜ + ω(t). Substituting (19) into Eq. (10) we find that γ (t) = 0, and where δ(t) = δ(t) then (19) has the form u = β(t)x 2 + δ(t)x 3 . This is a particular case of thea more general expression u = μ3 (t)x 3 + μ2 (t)x 2 + μ1 (t)x + μ0 (t), which is given in [5]. c) Case α = 1, 2, 3. The general solution of (16) is the function γ (t) x 2 + μ˜ 2 (t)x α + μ0 (t)x −1+α , f = (α − 2)(α − 3) and therefore according to (11) we have u = μ1 (t)x 2 + μ2 (t)x α + μ0 (t)x −1+α , where the unknown functions μ0 (t), μ1 (t) =
γ (t) , (α − 2)(α − 3)
(20)
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SEPARATION OF VARIABLES FOR NONLINEAR EQUATIONS
and μ2 (t) = μ˜ 2 (t) + ω(t) have to be determined. Substituting (20) into (10), we find that μ0 (t) = 0, and a system of equations determining the functions μ1 (t) and μ2 (t): μ1 = (2a + 4b)μ21 , a2b 2a 2 + 6ab μ2 = − + μ1 μ2 . (a + b)2 a+b
(21) (22)
A particular solution of (21) is the function 3 μ1 = (23) t −2 . a + 2b Substituting (23) into (22) and using (15), we obtain a linear equation for the function μ2 (t), t 2 μ2 = (9α − 3α 2 )μ2 . (24) Eq. (24) has the following solutions: μ2 = t 1/2 [c1 cos(σ ln t) + c2 sin(σ ln t)], if σ 2 = 3α 2 − 9α −
1 > 0, 4
1 − 3α 2 + 9α > 0, 4 √ 9 ± 2 21 1/2 , μ2 = t [c1 + c2 ln t], if α = 6 where c1 , c2 are arbitrary constants. So, Eq. (10) has the following exact solutions: 3 t −2 x 2 + t 1/2 x α [c1 cos(σ ln t) + c2 sin(σ ln t)], u= a + 2b 1 if σ 2 = 3α 2 − 9α − > 0; 4 3 u= t −2 x 2 + t 1/2 x α [c1 t σ + c2 t −σ ], a + 2b 1 if σ 2 = − 3α 2 + 9α > 0; 4 3 u= t −2 x 2 + t 1/2 x α [c1 + c2 ln t], a + 2b √ 9 ± 2 21 if α = , where c1 , c2 are arbitrary constants. 6 Note that u = μ1 x 2 + μ2 x 4 is a particular case of the more general substitution μ2 = t 1/2 [c1 t σ + c2 t −σ ], if σ 2 =
u = μ4 (t)x 4 + μ3 (t)x 3 + μ2 (t)x 2 + μ1 (t)x + μ0 (t), which was found in [5].
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The results obtained for (10) can be generalized to equation 2 ∂ 2u ∂ 2u ∂u = au 2 + b + φ(t)u. 2 ∂t ∂x ∂x
(25)
We have the following cases. a) Case a = −2b. The substitution u = γ (t)x 2 ln |x| + β(t)x 2 , reduces Eq. (25) to the system γ = −2bγ 2 + φγ ,
β = −2bβγ + bγ 2 + φβ.
b) Case a = − 32 b. The substitution u = β(t)x 2 + δ(t)x 3 , reduces Eq. (25) to the system δ = φδ, β = bβ 2 + φβ. 3 c) Case a = −b; −2b; − b. The substitution 2 u = μ1 (t)x 2 + μ2 (t)x α , a , reduces Eq. (25) to the system where α = a+b μ1 = (2a + 4b)μ21 + φμ1 , a2b 2a 2 + 6ab μ1 μ2 + φμ2 . μ2 = − + (a + b)2 a+b 3.
Separation of variables for Eq. (6) Let us find functions F (u) for which Eq. (6) admits the substitution in the form x = ω1 (t)d(u) + ω2 (t).
This is a particular case of ω1 (t), ω2 (t) and d(u) from of two ordinary differential ω2 = ω2 (t). Let us put (26) −
(26)
the general formula (4). We determine the functions the condition that (26) reduces Eq. (6) to a system equations with the unknown functions ω1 = ω1 (t), into (6),
ω1 d ω2 1 1 F (u) 1 d 1 3(d )2 − + − + = 0. ω1 d ω1 d ω1k (d )k ω13 (d )4 ω13 (d )5
(27)
ω1 ω2 d 1 , at − , − in Eq. (27) are linearly independent. Let k = 3 d d ω1 ω1 1 in Eq. (27). We lay down the conditions on the coefficients at the functions k , ω1 The functions
103
SEPARATION OF VARIABLES FOR NONLINEAR EQUATIONS
1 d so that they can be represented as a linear combination of the functions , 3 d ω1 1 on the field of real numbers. So we have d − or
d (d )2 d 1 + 3 = λ + μ , 4 5 (d ) (d ) d d
λ, μ ∈ R,
−d d + 3(d )2 = λd(d )4 + μ(d )4 .
(28)
Taking into account (28), we find from (27) that F (u) = ω1 ω1k−1 − λω1k−3 d(d )k−1 + ω2 ω1k−1 − μω1k−3 (d )k−1 , and then
ω1 ω1k−1 − λω1k−3 = λ1 , ω2 ω1k−1 − μω1k−3 = λ2 ,
where λ1 , λ2 are constants. So, in the case k = 3, Eq. (6) admits (26) if the function F (u) has the form F (u) = λ1 d(d )k−1 + λ2 (d )k−1 ,
(29)
where d = d(u) is an arbitrary solution of Eq. (28). Let us consider the particular solutions of Eq. (28): (a) (b) (c) (d)
d = ln u, if λ = 0, μ = 1; d = u1/2 , if λ = μ = 0; d = arcsin u, if λ = 0, μ = −1; d = Arsh u, if λ = 0, μ = 1.
In the case (a) we get from (29) that F (u) = (λ1 ln u + λ2 )u1−k ,
(30)
and the solutions of Eq. (6) are k(kλ1 )−3/k (k−3)/k λ2 u = exp − , + ct −1/k + (kλ1 t)−1/k x − t k−2 λ1 λ2 −3/2 −1/2 −1/2 −1/2 u = exp −(2λ1 ) t ln t + ct + (2λ1 t) x− , λ1
k = 2, k = 2,
where c is an arbitrary constant. The first of these functions is also a solution of (6) in the case k = 3, which follows from the consideration of this case below. Similarly, we determine that in the case (b), F (u) = λ1 u(2−k)/2 + λ2 u(1−k)/2 , in the case (c)
λ1 = 21−k λ1 ,
F (u) = (λ1 arcsin u + λ2 )(1 − u2 )(1−k)/2 ,
λ2 = 21−k λ2 ,
(31) (32)
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A. BARANNYK, T. BARANNYK and I. YURYK
and in the case (d) F (u) = (λ1 Arsh u + λ2 )(1 + u2 )(1−k)/2 .
(33)
The functions (30)–(33) and the solutions of Eq. (6) corresponding to them have been obtained by the method of conditional symmetry in [6]. Let k = 3 in Eq. (6). From Eq. (27) we find F (u) = ω1 ω12 d(d )2 + ω2 ω12 (d )2 +
d 3(d )2 − , d (d )2
and then as a consequence of linear independence of the functions d(d )2 and (d )2 we have (34) ω1 ω12 = λ1 , ω2 ω12 = λ2 , where λ1 , λ2 are constants. So, in the case k = 3 Eq. (6) admits substitution (26) if F (u) has the form F (u) = λ1 d(d )2 + λ2 (d )2 +
d 3(d )2 − , d (d )2
(35)
where d = d(u) is an arbitrary smooth function. For example, if d = exp u then from (35) we get F (u) = λ1 exp(3u) + λ2 exp(2u) − 2, and then because (26) and (34) the equation
3 ∂u ∂u ∂ 3u + 3 =0 + [λ1 exp(3u) + λ2 exp(2u) − 2] ∂t ∂x ∂x
has an exact solution
x + c1 λ2 , u = ln − (3λ1 t + c2 )1/3 λ1
where c1 , c2 are arbitrary constants. 4.
Conclusion All exact solutions of Eqs. (5) represented in Section 2 of this paper are solutions on invariant subspaces in the meaning of [5]. The substitution (3) and the one obtained from it as a result of rearrangement of the variables u, x, t can be also used for finding exact solutions of other nonlinear partial differential equations. As a rule, the substitution (3) can be applied to equations with the polynomial nonlinearity. If it is impossible to find a solution of the form (3), then, at first, we look for transformations that reduce equation under consideration to an equation with polynomial nonlinearity, in particular with quadratic nonlinearity, and then we search solutions of the latter equation in the form (3). Many nonlinear equations that by means of an appropriate transformation can be reduced to equations with quadratic nonlinearity have been described in [3, 4, 7].
SEPARATION OF VARIABLES FOR NONLINEAR EQUATIONS
105
REFERENCES [1] V. A. Galaktionov, S. A. Posashkov and S. R. Svirshchevskii: Generalized separation of variables for differential equations with polynomial nonlinearities, Differ. Equat. 31 (1995), 233. [2] S. I. Pokhozhaev: On a problem of L. V. Ovsyannikov, Zh. Prikl. Mekh. i Tekhn. Fiz. 2 (1989), 5 (in Russian). [3] V. A. Galaktionov and S. A. Posashkov: New exact solutions of parabolic equations with quadratic nonlinearities, U.S.S.R. Comput. Math. and Math. Phys. 29 (1989), 112. [4] V. A. Galaktionov: Invariant subspace and new explicit solutions to evolution equations with quadratic nonlinearities, Proc. Roy. Soc. Edinburgh 125A no. 2 (1995), 225. [5] V. A. Galaktionov and S. R. Svirshchevskii: Exact solutions and invariant subspaces of nonlinear partial differential equations in mechanics and physics, Chapman & Hall/CRC, Boca Raton, London, New York 2007. [6] W. I. Fushchich, N. I. Serov and T. K. Ahmerov: On the conditional symmetry of the generalized KdV equation, Rep. Ukr. Acad. Sci. A 12 (1991), 15. [7] V. A. Galaktionov and S. A. Posashkov: Exact solutions and invariant spaces for nonlinear equations of gradient diffusion, Comput. Math. Math. Phys. 34 (1994), 313.
REPORTS ON MATHEMATICAL PHYSICS
No. 1
SEPARATION OF VARIABLES FOR NONLINEAR EQUATIONS OF HYPERBOLIC AND KORTEWEG–DE VRIES TYPE A NATOLIY BARANNYK Institute of Mathematics, Pomeranian University, Arciszewskiego Str., 76-200 Slupsk, Poland
TATJANA BARANNYK Poltava National Pedagogical University, 2 Ostrogradskogo Str., 36000 Poltava, Ukraine
and I VAN Y URYK National University of Food Technologies, 68 Volodymyrska Str., 01033 Kyiv, Ukraine (e-mail: [email protected]) (Received November 19, 2010 — Revised May 12, 2011) We propose substitutions that have been used for constructing wide classes of exact solutions with the generalized separation of variables for nonlinear equations of the hyperbolic and Korteweg–de Vries type (KdV-type). These solutions cannot be obtained by means of S. Lie method or by the method of conditional symmetries. Keywords: nonlinear hyperbolic equations; Korteweg–de Vries equations; generalized separation of variables; Lie method.
1.
Introduction One of the effective methods for constructing exact solutions of linear PDE is separation of variables. For equations with two independent variables x, t and a dependent variable u, this method is based on finding exact solutions in the form of product of functions of different arguments u = a(x)b(t).
(1)
A method of constructing exact solutions of nonlinear PDE generalizing the classic variable separation method is described in [1]. The solutions are sought in the form of a finite sum of k terms, u(x, t) =
k
fi (t)ai (x),
(2)
i=1
where fi (t), ai (x) are smooth functions that have to be determined. Exact solutions in the form of generalized separation of variables (2) that contain more than two [97]
98
A. BARANNYK, T. BARANNYK and I. YURYK
terms are given in [2–5]. The results given in [3–5] are based on finding finitedimensional subspaces that are invariant under a corresponding nonlinear differential operator. The assumption (1) can be considered as an ansatz which reduces the equation under study to an ordinary differential equation with an unknown function a = a(x) (or with an unknown function b = b(t)). Using the idea of reduction, in the present paper we propose the following generalization of (1) for nonlinear equations, u=
m
ωi (t)ai (x) + f (x, t),
m ≥ 1.
(3)
i=1
The postulate (3) contains an unknown function f (x, t), m unknown functions ai (x) and m unknown functions ωi (t). They are found from the condition that (3) reduces the equation under consideration to a system of m ordinary differential equations with unknown functions ωi (t). If m = 1, this system is reduced to an ordinary differential equation with one unknown function ω1 (t). In many cases formula (3) makes easy to find exact solutions in the form of generalized separation of variables (2). If in a given differential equation we set x = x(u, t), then we get a differential equation with the independent variables u, t and the dependent variable x. To construct solutions of such equation, we can use the substitution x=
m
ωi (t)ai (u) + f (u, t),
(4)
i=1
which is obtained from (3) as a result of replacement u → x, x → u, t → t. In this paper we use substitutions of type (3) and (4) to solve exact by the nonlinear hyperbolic-type equation 2 ∂ 2u ∂ 2u ∂u = au 2 + b +c (5) 2 ∂t ∂x ∂x and the KdV-type equation
k ∂ 3u ∂u ∂u + 3 = 0, + F (u) ∂t ∂x ∂x
(6)
where k is a real parameter. Eq. (5) was the subject of investigation in [4, 5], while Eq. (6) was investigated in [6] using the conditional symmetry method. The solutions of Eqs. (5) and (6) given in the present work cannot be obtained by means of the group analysis method. 2.
Separation of variables for Eq. (5) To construct exact solutions of Eq. (5), we assume that u = ω(t) + f (x, t),
(7)
SEPARATION OF VARIABLES FOR NONLINEAR EQUATIONS
which is a particular case of the following one,
99
(3). This substitution is in turn a generalization of u(x, t) = ϕ(x) + ψ(t)
that is often used to find exact solutions of nonlinear PDE. Substituting (7) into (5), we get (8) ω + ftt = a (ω + f )fxx + bfx2 + c, which has to be an ordinary differential equation with unknown function ω = ω(t). It follows that the coefficient afxx at ω in Eq. (8) can be written in the form of afxx = 2aμ2 (t), and then f = μ2 (t)x 2 + μ1 (t)x + μ˜ 0 (t), where μ˜ 0 (t), μ1 (t), μ2 (t) are some functions of t. From (7) we obtain u = μ2 (t)x 2 + μ1 (t)x + μ0 (t),
(9)
μ0 (t) = μ˜ 0 (t) + ω(t). This was found in [4] in the study of the structure of invariant subspaces for the operator which corresponds to the right-hand side of Eq. (5). Let us consider a particular case of Eq. (5), 2 ∂ 2u ∂u ∂ 2u = au 2 + b . (10) 2 ∂t ∂x ∂x To construct exact solutions of Eq. (10) we set u = ω(t)d(x) + f (x, t),
(11)
which is a particular case of (3). This assumption (11) reduces (10) to the equation ω d − ω(afxx d + 2bfx d + af d ) − ω2 (add + bd 2 ) + ftt − affxx − bfx2 = 0. (12) Eq. (12) has to be an ordinary differential equation with an unknown function ω = ω(t). It follows that add + bd 2 = βd, β ∈ R, (13) adfxx + 2bd fx + ad f = γ˜ (t)d.
(14)
As a result, the problem of finding exact solutions for Eq. (10) in the form of (11) is reduced to integration of the system of two ordinary differential equations, one of which is linear. Eq. (13) has the particular solution, α a = b, β = 0, α = 1. (15) d = xα, 1−α Setting d = x α into Eq. (14), we get x 2 fxx + 2(1 − α)xfx + α(α − 1)f = γ (t)x 2 , 1−α γ˜ (t). b Let us consider three cases.
where γ (t) =
(16)
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A. BARANNYK, T. BARANNYK and I. YURYK
a) Case α = 2. From (15), it follows that a = −2b. The general solution of Eq. (16) for α = 2 is the function 2 ˜ f = γ (t)x 2 ln |x| + β(t)x + δ(t)x,
and therefore, according to (11), we have u = γ (t)x 2 ln |x| + β(t)x 2 + δ(t)x,
(17)
˜ + ω(t). Substituting (17) into (10), we find the equality δ(t) = 0 where β(t) = β(t) and a system of equations determining the functions β(t) and γ (t): γ = −2bγ 2 ,
β = −2bβγ + bγ 2 .
(18)
The system (18) has the following particular solution: 9 3 β = c1 t 3 + c2 t −2 − t −2 ln |t|. γ = − t −2 , b 5b As a result, we find an exact solution of Eq. (10) for a = −2b: 3 −2 2 9 −2 3 −2 u = − t x ln |x| + c1 t + c2 t − t ln |t| x 2 , b 5b where c1 , c1 are arbitrary constants. 3 b) Case α = 3. From (15) it follows that a = − b. The general solution of 2 (16) for α = 3 is the function 3 ˜ , f = −γ (t)x 2 ln |x| + β(t)x 2 + δ(t)x and therefore according to (11) we have u = −γ (t)x 2 ln |x| + β(t)x 2 + δ(t)x 3 ,
(19)
˜ + ω(t). Substituting (19) into Eq. (10) we find that γ (t) = 0, and where δ(t) = δ(t) then (19) has the form u = β(t)x 2 + δ(t)x 3 . This is a particular case of thea more general expression u = μ3 (t)x 3 + μ2 (t)x 2 + μ1 (t)x + μ0 (t), which is given in [5]. c) Case α = 1, 2, 3. The general solution of (16) is the function γ (t) x 2 + μ˜ 2 (t)x α + μ0 (t)x −1+α , f = (α − 2)(α − 3) and therefore according to (11) we have u = μ1 (t)x 2 + μ2 (t)x α + μ0 (t)x −1+α , where the unknown functions μ0 (t), μ1 (t) =
γ (t) , (α − 2)(α − 3)
(20)
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SEPARATION OF VARIABLES FOR NONLINEAR EQUATIONS
and μ2 (t) = μ˜ 2 (t) + ω(t) have to be determined. Substituting (20) into (10), we find that μ0 (t) = 0, and a system of equations determining the functions μ1 (t) and μ2 (t): μ1 = (2a + 4b)μ21 , a2b 2a 2 + 6ab μ2 = − + μ1 μ2 . (a + b)2 a+b
(21) (22)
A particular solution of (21) is the function 3 μ1 = (23) t −2 . a + 2b Substituting (23) into (22) and using (15), we obtain a linear equation for the function μ2 (t), t 2 μ2 = (9α − 3α 2 )μ2 . (24) Eq. (24) has the following solutions: μ2 = t 1/2 [c1 cos(σ ln t) + c2 sin(σ ln t)], if σ 2 = 3α 2 − 9α −
1 > 0, 4
1 − 3α 2 + 9α > 0, 4 √ 9 ± 2 21 1/2 , μ2 = t [c1 + c2 ln t], if α = 6 where c1 , c2 are arbitrary constants. So, Eq. (10) has the following exact solutions: 3 t −2 x 2 + t 1/2 x α [c1 cos(σ ln t) + c2 sin(σ ln t)], u= a + 2b 1 if σ 2 = 3α 2 − 9α − > 0; 4 3 u= t −2 x 2 + t 1/2 x α [c1 t σ + c2 t −σ ], a + 2b 1 if σ 2 = − 3α 2 + 9α > 0; 4 3 u= t −2 x 2 + t 1/2 x α [c1 + c2 ln t], a + 2b √ 9 ± 2 21 if α = , where c1 , c2 are arbitrary constants. 6 Note that u = μ1 x 2 + μ2 x 4 is a particular case of the more general substitution μ2 = t 1/2 [c1 t σ + c2 t −σ ], if σ 2 =
u = μ4 (t)x 4 + μ3 (t)x 3 + μ2 (t)x 2 + μ1 (t)x + μ0 (t), which was found in [5].
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The results obtained for (10) can be generalized to equation 2 ∂ 2u ∂ 2u ∂u = au 2 + b + φ(t)u. 2 ∂t ∂x ∂x
(25)
We have the following cases. a) Case a = −2b. The substitution u = γ (t)x 2 ln |x| + β(t)x 2 , reduces Eq. (25) to the system γ = −2bγ 2 + φγ ,
β = −2bβγ + bγ 2 + φβ.
b) Case a = − 32 b. The substitution u = β(t)x 2 + δ(t)x 3 , reduces Eq. (25) to the system δ = φδ, β = bβ 2 + φβ. 3 c) Case a = −b; −2b; − b. The substitution 2 u = μ1 (t)x 2 + μ2 (t)x α , a , reduces Eq. (25) to the system where α = a+b μ1 = (2a + 4b)μ21 + φμ1 , a2b 2a 2 + 6ab μ1 μ2 + φμ2 . μ2 = − + (a + b)2 a+b 3.
Separation of variables for Eq. (6) Let us find functions F (u) for which Eq. (6) admits the substitution in the form x = ω1 (t)d(u) + ω2 (t).
This is a particular case of ω1 (t), ω2 (t) and d(u) from of two ordinary differential ω2 = ω2 (t). Let us put (26) −
(26)
the general formula (4). We determine the functions the condition that (26) reduces Eq. (6) to a system equations with the unknown functions ω1 = ω1 (t), into (6),
ω1 d ω2 1 1 F (u) 1 d 1 3(d )2 − + − + = 0. ω1 d ω1 d ω1k (d )k ω13 (d )4 ω13 (d )5
(27)
ω1 ω2 d 1 , at − , − in Eq. (27) are linearly independent. Let k = 3 d d ω1 ω1 1 in Eq. (27). We lay down the conditions on the coefficients at the functions k , ω1 The functions
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1 d so that they can be represented as a linear combination of the functions , 3 d ω1 1 on the field of real numbers. So we have d − or
d (d )2 d 1 + 3 = λ + μ , 4 5 (d ) (d ) d d
λ, μ ∈ R,
−d d + 3(d )2 = λd(d )4 + μ(d )4 .
(28)
Taking into account (28), we find from (27) that F (u) = ω1 ω1k−1 − λω1k−3 d(d )k−1 + ω2 ω1k−1 − μω1k−3 (d )k−1 , and then
ω1 ω1k−1 − λω1k−3 = λ1 , ω2 ω1k−1 − μω1k−3 = λ2 ,
where λ1 , λ2 are constants. So, in the case k = 3, Eq. (6) admits (26) if the function F (u) has the form F (u) = λ1 d(d )k−1 + λ2 (d )k−1 ,
(29)
where d = d(u) is an arbitrary solution of Eq. (28). Let us consider the particular solutions of Eq. (28): (a) (b) (c) (d)
d = ln u, if λ = 0, μ = 1; d = u1/2 , if λ = μ = 0; d = arcsin u, if λ = 0, μ = −1; d = Arsh u, if λ = 0, μ = 1.
In the case (a) we get from (29) that F (u) = (λ1 ln u + λ2 )u1−k ,
(30)
and the solutions of Eq. (6) are k(kλ1 )−3/k (k−3)/k λ2 u = exp − , + ct −1/k + (kλ1 t)−1/k x − t k−2 λ1 λ2 −3/2 −1/2 −1/2 −1/2 u = exp −(2λ1 ) t ln t + ct + (2λ1 t) x− , λ1
k = 2, k = 2,
where c is an arbitrary constant. The first of these functions is also a solution of (6) in the case k = 3, which follows from the consideration of this case below. Similarly, we determine that in the case (b), F (u) = λ1 u(2−k)/2 + λ2 u(1−k)/2 , in the case (c)
λ1 = 21−k λ1 ,
F (u) = (λ1 arcsin u + λ2 )(1 − u2 )(1−k)/2 ,
λ2 = 21−k λ2 ,
(31) (32)
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and in the case (d) F (u) = (λ1 Arsh u + λ2 )(1 + u2 )(1−k)/2 .
(33)
The functions (30)–(33) and the solutions of Eq. (6) corresponding to them have been obtained by the method of conditional symmetry in [6]. Let k = 3 in Eq. (6). From Eq. (27) we find F (u) = ω1 ω12 d(d )2 + ω2 ω12 (d )2 +
d 3(d )2 − , d (d )2
and then as a consequence of linear independence of the functions d(d )2 and (d )2 we have (34) ω1 ω12 = λ1 , ω2 ω12 = λ2 , where λ1 , λ2 are constants. So, in the case k = 3 Eq. (6) admits substitution (26) if F (u) has the form F (u) = λ1 d(d )2 + λ2 (d )2 +
d 3(d )2 − , d (d )2
(35)
where d = d(u) is an arbitrary smooth function. For example, if d = exp u then from (35) we get F (u) = λ1 exp(3u) + λ2 exp(2u) − 2, and then because (26) and (34) the equation
3 ∂u ∂u ∂ 3u + 3 =0 + [λ1 exp(3u) + λ2 exp(2u) − 2] ∂t ∂x ∂x
has an exact solution
x + c1 λ2 , u = ln − (3λ1 t + c2 )1/3 λ1
where c1 , c2 are arbitrary constants. 4.
Conclusion All exact solutions of Eqs. (5) represented in Section 2 of this paper are solutions on invariant subspaces in the meaning of [5]. The substitution (3) and the one obtained from it as a result of rearrangement of the variables u, x, t can be also used for finding exact solutions of other nonlinear partial differential equations. As a rule, the substitution (3) can be applied to equations with the polynomial nonlinearity. If it is impossible to find a solution of the form (3), then, at first, we look for transformations that reduce equation under consideration to an equation with polynomial nonlinearity, in particular with quadratic nonlinearity, and then we search solutions of the latter equation in the form (3). Many nonlinear equations that by means of an appropriate transformation can be reduced to equations with quadratic nonlinearity have been described in [3, 4, 7].
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