Separation of Variables and Construction of Exact Solutions of Nonlinear Wave Equations

Transcription

1 Proceedings of Institute of Mathematics of NAS of Uraine 000, Vol 30, Part 1, 73 8 Separation of Variables and Construction of Eact Solutions of Nonlinear Wave Equations AF BARANNYK and II YURYK Institute of Math, Pedagogical University, barciszewsiego Str, S lups, Poland Urainian State University of Food Technologies, Kyiv, Uraine An approach to construction of eact solutions of nonlinear equations on the basis of separated variables is proposed 1 Introduction To construct the eact solutions of nonlinear equations in mathematical physics the following ansatz is commonly used u) =f)ϕω)+g), 1) where f), g), ω = ω, u) are certain functions, and functions ϕω) are undetermined If the eplicit form of variables ω = ω, u) and functions f), g) is determined on the basis of subalgebra of invariance algebra of this equation, then ansatz 1) is called as a symmetry or Lie one Not all ansatzes are symmetry ones In [1 4] a definition of conditional invariance of this differential equation was introduced If the eplicit form of new variables ω = ω, u) and functions f), g) are determined on the basis of conditional symmetry operators then ansatz 1) is called an arbitrary invariant or non-lie one By means of arbitrary invariant ansatzes new classes types) of eact solutions of many nonlinear equations in mathematical physics were constructed Let us note an effective algorithm for finding of arbitrary symmetry operators is not found yet In this paper an approach to the construction of eact solutions of nonlinear equations is proposed It is based on the method of separated variables and has a great advantage in view of its simplicity and possibility to be unchanged for construction of eact solutions for manydimensional equations We will consider this approach using the Boussinesq equation Eact solutions of the Boussinesq equation u 0 = λ u) + λu u Let us consider the Boussinesq equation u 0 = λ u) + λu u, ) where λ is an arbitrary constant, u = u 0, 1,, n ), u 0 = u,and 0 ) u ) u u) = + +, u = u 1 n + + u 1 n

2 74 AF Baranny and II Yury Certain partial solutions of Eq) for two variables 0, have been obtained in [5, 6], and for many variables in [1, 7] Now let us consider the one-dimensional Boussinesq equation ) u u 0 = λ + λu u 1 3) 1 1 We see for a solution of Eq3) in the form u = a 0 )b 1 ), where functions a 0 )and b 1 ) are not constants Substituting into Eq3) we have λa bb + λa b a b =0 4) It follows from 4) that the functions a, a are linearly dependent Consequently a = αa for a real number α and Eq3) has the form λbb + λb )a αa b =0 We find from this equation λbb + λb αb =0 Notice that the substitution a = αa suggests α = 1 Thus we will consider the equation λbb + λb b =0 5) The general solution of Eq5) has the form bdb = ± c + b 3 3λ 1 + c 1 ), 6) where c, c 1 are arbitrary constants If, for eample, c = 0, then b = 1 6λ 1 + c 1 ), and we obtain the solution of 3) u = 1 + c 1 ) 6λ 0 + c ), which is transformed into u = 1 6λ 0 7) The solution 7) is a partial case for u = 1 + f 0, 1 ) 6λ 0 Substituting into Eq3), we find f 0 = 1f 1 1 f 11 1 f + λf1 + λff 11 8) The solution of Eq8) can be found in the form f = a 0 )b 1 ) and we have a b = a 0 3 1b 1 6 b 1 ) 3 b + a λb + λbb ) Let a = α a 0, where α is a real number Hence, a = c α To determine the function b 1)we find the system of equations: 1 b +4 1 b + +6α)b =0, b + bb =0

3 Separation of Variables and Construction of Eact Solutions 75 Thus, the Boussinesq equation possesses the following solution u = c 5/8 0 1/ 1 1 6λ 0 If the function f in 8) depends on 0 only, then we obtain f 0 = fthus,eq3)hasa solution u = 1 + c 1/3 6λ 0 Now let us consider Eq) for the case n > 1 We shall loo for solution of ) in the form u = a 0 )b 1,, ), where the functions a 0 )andb 1,, )arenotconstant Substituting this epression into ) we find λa [ b) + b b ] a b =0 9) It follows from 9) that functions a, a are linearly dependent, thus a = αa and Eq9) has a form λb b + λ b) ) a λa b =0 It can be obtained from this equation that λb b + λ b) αb =0 10) The function b = ϕω), ω = 1 + +, b n satisfies Eq10) iff 4λωϕϕ +λϕϕ +4λωϕ αϕ =0 11) If α =λ +) then a particular solution of Eq11) is the function ϕ = ω Since the equation a = αa possesses the solution a = 1 α 0, then Eq) has a solution of the form u = λ +) 0 1) The solution 1) is a particular case of u = f 0,, ) λ +) 0 Substituting this epression into Eq) we obtain f 0 = 1f 1 f + λ f1 + + f λ +) 0 λ +) + f f ) n 0 +λ ) + f f f nn ) f λ +) 0 +) 0 Let the function f be independent of 1,,, then f 0 = λ f f n ) + λ f λ +) 0 Thus, ) f +1, f nn ) 13) +) 0 f f +1, f nn )=0, f 0 = λ f f n ) +) 0 f 14)

4 76 AF Baranny and II Yury The solution of 14) can be found in the form f = µ µ n n + ν, where µ +1,,µ n, ν are functions dependent on 0 only Substituting this epression into the second equation of 14) we have Thus, µ µ n n + ν = λ µ ) µ n µ µ n n + ν) +) 0 µ +1 = µ +1, 0 +) 0 µ n = µ n, 0 +) 0 ν = λ µ ) µ n ν 0 +) 0 The general solution of 15) has the following form: µ +1 = c ,, µ n = c n ν = λ +) + 0, c c + n) c + 0, where c, c +1,,c n are arbitrary constants Thus, we obtain the multiparameter set of solutions of Eq) 15) u = λ +) 0 λ +) + +c c n n + c) c c n) Moreover, if =1,n = 3 then solution 16) taes the form 16) u = 1 6λ 0 +c + c 3 3 ) 1/ λ If =,n = 3 then solution 16) has a form c + c 3) 1/3 u = c 3 3 1/ 0 +λc 8λ 3 0 If the function f in 13) does not depend on 1,,, then we have f 0 = f +) 0 Thus, f = c +

5 Separation of Variables and Construction of Eact Solutions 77 And the Boussinesq equation ) has also the following solution u = λ +) 0 + c + If, for eample, = then we have u = c 1/ 8λ 0 In the case of =3wehave u = λ 0 + c 3/5 3 Eact solutions of the Boussinesq equation u 00 + u) + u u + u) =0 Let us consider the Boussinesq equation u 00 + uu 11 + u 1 + u 1111 =0, 17) where u = u), = 0, 1 ), u 1 = u 1, u 11 = u, u 1111 = 4 u It is invariant with respect to the algebra with operators [8] P 0 =, P 1 =, D = u u Operators P 0, P 1 and D give rise to the one-parameter symmetry group of equations: G 0 : 0, 1,u) 0 + ε, 1,u), G 1 : 0, 1,u) 0, 1 + ε, u), G : 0, 1,u) e ε 0,e ε 1,e ε u ) 18) Eq17) is also invariant under the discrete transformations 0, 1,u) 0, 1,u), 0, 1,u) 0, 1,u), 0, 1,u) 0, 1,u) 19) One-parameter subgroups 18) and discrete transformations 19) give rise to the group G of Eq17) Therefore, the most general solution obtained from u = f 0, 1 ) by means of the transformations of the group G has the form u = α f α 0 + β 0,α 1 + β 1 ), where α, β 0, β 1 are arbitrary real numbers The derivation of eact solutions of Eq17) is discussed in [1 4] A new method of invariant reduction of the Boussinesq equation is proposed in [] Eact solutions of Eq17) on the basis of the conditional symmetry concept are obtained in [3 4]

6 78 AF Baranny and II Yury 31 We see a solution of Eq17) in the form u = a 0 )+b 1 ), where the functions a 0 )and b 1 ) are not constant Substituting this epression into Eq17) we have a + ab + bb + b + b ) =0 0) Since b is independent of 0, it is clear from Eq0) that a = α+βa for real α and β Therefore, we obtain from 0) that aβ + b )+α + bb + b + b ) = 0, ie b + β =0, bb + b + b + α =0 1) If β = 0 then the system of Eqs1) possesses a solution b = γ 1 + δ, where γ = α The function b 1 ) can be transformed into 1 by means of a transformation from the group G Then α = 4 anda = 0 + γ δ 1, where γ 1, δ 1 are real numbers Since a can be rewritten as a = 0 γ 1 /) + δ 1 γ1 /4, this solution can be transformed with the help of the group G to become u = 1 0) ) Let us construct another type of solutions to Eq17) with partial solution ) to the Boussinesq equation A partial solution of Eq17) can be found in the form u = 1 0) + f0, 1 ) 3) Ansatz 3) reduces Eq17) to the form f 00 + ff 11 + f 1 + f ) f11 +4f 1 =0 4) Ansatz ϕ = ϕω), ω = reduces Eq4) to the ordinary differential equation ϕϕ + ϕ + ϕ +ωϕ +6ϕ =0 5) A partial solution of Eq5) we find in the form ϕ = tω s, s 1 Substituting it into 5) we obtain s =, t = 1 Thus, the function u = 1 0) ) 6) is a solution of Eq17) 3 Now, we loo for a solution of Eq17) in the form u = a 0 )b 1 ), where the functions a 0 )andb 1 ) are not constant Substituting this epression into 17) we obtain a b + a bb + b ) + ab =0 7) In complete analogy with Subsection 31 we see that a = αa +βa Substituting a into Eq7) and taing into account the functions a and a are linearly independent we obtain the following system to determine the function b 1 ) b + βb =0, bb + b + αb =0 8) It may be easily seen from these equations that β = 0 and α 0 We can always set α =6 by multiplying the function a by the number α/6 and the function b by 6/α Since β =0,we see from the first of equations 8) that b is polynomial in 1 of degree not higher than three Plugging b in the form of the general polynomial of degree three into the second of equations 8), we see that in fact b = 1 Hence, Eq17) possesses the solution u = 1P 0 ), 9)

7 Separation of Variables and Construction of Eact Solutions 79 u = 1 0, 30) where P 0 ) is the Weierstrass function with invariants g = 0 and g 3 = c 1 A new class of solutions of Eq17) can be constructed using its partial solution 9) We loo for these new solutions in the form u = 1P 0 )+f 0, 1 ) 31) Ansatz 31) reduces Eq17) to f00 + ff 11 + f 1 + f 1111 ) P 1 f f 1 +f ) =0 3) If the function f is independent of 1, then we have f 00 =Pf This is the Lamé equation and its solutions are well-nown [11] Thus, the function u = 1P 0 )+Λ 0 ), Λ =PΛ 33) is a solution of the Boussinesq equation If the function f in 3) does not depend on 0, then we have a system of equations to determine the function f 1f f 1 +f =0, ff 11 + f 1 + f 1111 =0 34) The first equation of this system is linear and its complementary function is well-nown [11] Hence, f = 1 1, and Eq17) possesses a solution u = 1P 0 ) ) We obtain simultaneously that the function u = ) is a solution of the Boussinesq equation too Then we find a solution of Eq3) which is dependent on 0 and 1 It can be found in the form f = a 0 )b 1 )+c 0 ) where functions a 0 )andc 0 ) are linearly independent Substituting into Eq17) we obtain c + a b + a bb + b ) + acb + ab + ap 1 b 4 1 b b ) Pc =0 37) Without going into details let us suppose from the outset that b = 0 Then b = α 1 + β and consequently f = αa 0 ) 1 +βa 0 )+c 0 )) It means that setting α =1,β = 0 in Eq37) we arrive at Thus, c + αa 1 + a + ap ) Pc =0 a 6Pa =0, c = a +Pc 38) The equation a 6Pa = 0 is the Lamé equation with a solution a = P 0 ) Hence the complementary function of the Lamé equation can be written as a = γ 1 P 0 )+γ Λ 0 ), where P 0 )andλ 0 ) are linearly independent The corresponding solution of Eq17) has the form u = P 0 ) 1 γ 1 /) + γ 1 Λ 0 )+ c 0 )+γ 1/4P 0 ) )

8 80 AF Baranny and II Yury Under transformations from the group G it reduces to u = 1P 0 )+γ 1 Λ 0 )+d 0 ), 39) where the function d 0 ) is a solution of the following equation d = γ 1Λ +Pd In a similar manner from 30) a new class of the Boussinesq equation solutions can be constructed u = , 40) u = c c c 0 + c ) The solution of Eq17) is in the form u = a 0 )b 1 )+c 0 ), where functions a 0 )andc 0 ) are linearly independent By substituting in Eq17) we obtain a b + c + a b + bb ) + acb + ab =0 If c = αa, a = 0, then a α + b + bb ) + acb + ab =0 It follows from this equation that b + bb + a =0, b =0 The solution of this system up to transformations from the group G is a function b = 1 if α = 1 Then with the requirement that c = αa, a = 0, it is possible to obtain a = 0, c = γ 0 + δ Thus the function u = γ 0 + δ is the Boussinesq equation solution with arbitrary real numbers γ, δ 33 We go now to the construction of eact solutions of the Boussinesq equation for the case n>1 The generalization of Eq17) for arbitrary number of variables 0, 1,, n is the equation [10] u 00 + u) + u u + δu) =0, 4) where u u) = 1 ) ) u + +, u = u n u n The solution of 4) can be found in the form u = a 0 )b 1,, ), n Substituting this epression into 4) we have a b + a [ b b + b) ] + a b) =0 Hence c = αa + βa and as a result we obtain the following system to determine the function b 1,, ) b)+βb =0, b b + b) + αb =0

9 Separation of Variables and Construction of Eact Solutions 81 If b = 0, then α 0 and it may be considered that α = 6 The system has a solution b = ) for these values b and α Therefore, the Boussinesq equation solutions are functions u = ) P0 ), 43) u = ) 44) Let us construct another solution of Eq4) from 43) We will loo for it in the form u = ) P0 )+f 0, 1,, ) 45) Ansatz 45) reduces Eq4) to f 00 + f) + f f + f) [ ] 3 P ) f +41 f f )+f =0 If f does not depend on variables 1,, in Eq46) then f 00 = function u = ) 6 +f and, therefore, the ) P0 )+Λ 0 ), Λ = 6 PΛ 47) + is a solution of Eq4) If the function f depends on variables 0, 1,, in 46) then the solution of Eq4) can be obtained in the following form u = ) P0 )+α 1 Λ 0 )+c 0 ), 48) where P 11 =6P,Λ =4+)PΛ, c = α Λ +Pc Similarly we find a solution of Eq4) from 44): u = ) 0 + c c where c 1, c, c 3, c 4 are arbitrary real numbers; =1,,n A new type of solutions of Eq4) can be constructed using c , u = 1P 0 )+f 0,, 3 ) 49) Substituting anzats 49) into 4) we have f 00 + f) + f f) + f) 1P f) Pf =0 Since the function f does not depend on 1, then f = 0 and we obtain the following system of equations to determine the function f: f 00 + f + f 3 Pf =0, f + f 33 =0 50) We will see now a solution of Eqs50) in the form f = a 0 ) + b 0 ) 3 + c 0 ) Substitution of f into the first equation of 50) gives a + b 3 + c + a + b Pa + b 3 + c) =0

10 8 AF Baranny and II Yury It follows from this equation that a =Pa, b 11 =Pb, c = a b +Pc 51) Solving Eq51) we find the eplicit form of functions a 0 ), b 0 ), c 0 ) and the solution of Eq4) too If we use the ansatz u = f 0,, 3 ), we construct by analogy with the above the following solution of Eq4): u = c c 4 1 ) 0 + c c 4 1 ) 0 3 c 1 + c c 1c + c 3 c c + c And using the ansatz u = ) P0 )+f 0, 3 ) another solution of Eq4) can be obtained u = ) P0 )+Λ 0 ) 3 + c 0 ), where Λ =PΛ, c = Λ +Pc Maing use of u = ) 0 + f 0, 3 ) we find a solution of Eq4) in the form u = 1 References 1 + ) 0 + c c 1 ) c c 1c c [1] Fushchych WI, Shtelen VM and Serov NI, Symmetry Analysis and Eact Solutions ofequations of Nonlinear Mathematical Physics, Dordrecht, Kluwer, 1993 [] Fushchych WI and Niitin AG, Symmetries ofmawell s Equations, Dordrecht, Reidel, 1987 [3] Fushchych WI and Tsyfra IM, On a reduction and solutions of nonlinear wave equations with broen symmetry, J Phys A: Math Gen, 1987, V0, L45 L48 [4] Levi D and Winternitz P, Nonclassical symmetry reductions: eample ofthe Boussinesq equation, J Phys A: Math Gen, 1989, V, [5] Nishitani T and Tajiri M, On similarity solutions ofthe Boussinesq equation, Phys Lett A, 198, V89, N 4, [6] Soolov YuF, On some particular solutions ofthe Boussinesq equation, Ur Math J, 1956, V8, N 1, in Russian) [7] Baranni LF and Lahno HO, Symmetry reduction ofthe Bussinesq equation to ordinary differential equations, Reports on Math Phys, 1996, V38, N 1, 1 9 [8] Olver PJ and Rosenau Ph, The construction ofspecial solutions to partial differential equations, Phys Lett A, 1986, V114, N 3, [9] Clarson P and Krusal M, New similarity reductions ofthe Boussinesq equation, J Math Phys, 1989, V30, N 10, [10] Fushchych WI and Serov NI, The conditional invariance and eact solutions ofthe Boussinesq equation, in Symmetries and Solutions ofequations ofmathematical Physics, Kyiv, 1989, in Russian) [11] Kame E, Handboo ofthe Ordinary Differential Equations, Moscow, Naua, 1976 in Russian)