Solutions for one class of nonlinear fourth-order partial differential equations
aa r X i v : . [ m a t h . C A ] O c t Solutions for one class of nonlinearfourth-order partial differential equations
S. Suksern a a Department of Mathematics, Faculty of Science, Naresuan University,Phitsanulok, 65000, Thailand
Abstract
Some solutions for one class of nonlinear fourth-order partial differential equations u tt = (cid:0) κu + γu (cid:1) xx + νuu xxxx + µu xxtt + αu x u xxx + βu xx where α, β, γ, µ, ν and κ are arbitrary constants are presented in the paper. Thisequation may be thought of as a fourth-order analogue of a generalization of theCamassa-Holm equation, about which there has been considerable recent interest.Furthermore, this equation is a Boussinesq-type equation which arises as a modelof vibrations of harmonic mass-spring chain. The idea of travelling wave solutionsand linearization criteria for fourth-order ordinary differential equations by pointtransformations are applied to this problem. Key words:
Linearization problem, point transformation, nonlinear ordinarydifferential equation, travelling wave solution
Almost all important governing equations in physics take the form of nonlineardifferential equations, and, in general, are very difficult to solve explicitly.While solving problems related to nonlinear ordinary differential equations itis often expedient to simplify equations by a suitable change of variables.Many methods of solving differential equations use a change of variables thattransforms a given differential equation into another equation with knownproperties. Since the class of linear equations is considered to be the simplestclass of equations, there is the problem of transforming a given differential
Email address: [email protected] (S. Suksern).
Preprint submitted to Symmetry, Integrability and Geometry: Methods and Applications7 November 2018 quation into a linear equation. This problem, which is called a linearizationproblem. The reduction of an ordinary differential equation to a linear ordi-nary differential equation besides simplification allows constructing an exactsolution of the original equation.One of the most interested nonlinear problems but also difficultly in solvingis the problem of nonlinear fourth-order partial differential equations [1] u tt = (cid:16) κu + γu (cid:17) xx + νuu xxxx + µu xxtt + αu x u xxx + βu xx (1)where α, β, γ, µ, ν and κ are arbitrary constants. The main difficulty insolving this problem comes from the terms of nonlinear partial differentialequations and the large number of order. Because of this difficulty, there areonly a few attempts to solve this problem.In 2008, Suksern, Meleshko and Ibragimov [2,3] found the explicit form ofthe criteria for linearization of fourth-order ordinary differential equations bypoint transformations. Moreover, the procedure for the construction of thelinearizing transformation are presented. By the virtue of [2,3] to bring aboutthe idea for solving the problem of nonlinear fourth-order partial differentialequations (1).The way of solving the problem is organized as follows. Firstly, reducing thenonlinear fourth-order partial differential equations to the nonlinear fourth-order ordinary differential equations by substituting the form of travellingwave solutions. Secondly, reducing the nonlinear fourth-order ordinary differ-ential equations to the linear fourth-order ordinary differential equations byapplying the criteria for linearization in [2,3]. Finally, finding the exact solu-tions of linear equations and then substituting back to the exact solutions ofthe original problem. The important tools for this research is the linearization criteria for fourth-order ordinary differential equations by point transformations. From [2,3] wehave the following theorems.
Theorem 1
Any fourth-order ordinary differential equation y (4) = f ( x, y, y ′ , y ′′ , y ′′′ ) , an be reduced by a point transformation t = ϕ ( x, y ) , u = ψ ( x, y ) , (2) to the linear equation u (4) + α ( t ) u ′ + β ( t ) u = 0 , (3) where t and u are the independent and dependent variables, respectively, if itbelongs to the class of equations y (4) +( A y ′ + A ) y ′′′ + B y ′′ + ( C y ′ + C y ′ + C ) y ′′ + D y ′ + D y ′ + D y ′ + D y ′ + D = 0 , (4) or y (4) + y ′ + r ( − y ′′ + F y ′ + F y ′ + F ) y ′′′ + y ′ + r ) [15 y ′′ + ( H y ′ + H y ′ + H ) y ′′ +( J y ′ + J y ′ + J y ′ + J y ′ + J ) y ′′ + K y ′ + K y ′ + K y ′ + K y ′ + K y ′ + K y ′ + K y ′ + K ] = 0 , (5) where A j = A j ( x, y ) , B j = B j ( x, y ) , C j = C j ( x, y ) , D j = D j ( x, y ) , r = r ( x, y ) , F j = F j ( x, y ) , H j = H j ( x, y ) , J j = J j ( x, y ) and K j = K j ( x, y ) arearbitrary functions of x, y . Since this research deals with the first class, let us emphasize to the first classfor other theorems that we need to use.
Theorem 2
Equation (4) is linearizable if and only if its coefficients obey the ollowing conditions: A y − A x = 0 , (6)4 B − A = 0 , (7)12 A y + 3 A − C = 0 , (8)12 A x + 3 A A − C = 0 , (9)32 C y + 12 A x A − C x + 3 A A − A C = 0 , (10)4 C y + A C − D = 0 , (11)4 C y + A C − D = 0 , (12)16 C x − A x A − A A + 4 A C + 8 A C − D = 0 , (13)192 D x + 36 A x A A − A x C − C x A − D y + 9 A A − A C − A A C + 48 A D + 32 C C = 0 , (14)384 D xy − (cid:20) A A − C ) A + 16(2 A D + C C ) − A C − D ) A ) A − C D − C D + C D )+ (3 A D − C ) A ) − D y A + 384 D y A + 1536 D yy − A A − C ) C x + 12((3 A A − C ) A − A C − D )) A x (cid:21) = 0 . (15) Theorem 3
Provided that the conditions (6)-(15) are satisfied, the linearizingtransformation (2) is defined by a fourth-order ordinary differential equationfor the function ϕ ( x ) , namely by the Riccati equation dχdx − χ = 8 C − A − A x , (16) for χ = ϕ xx ϕ x , (17) and by the following integrable system of partial differential equations for thefunction ψ ( x, y ) 4 ψ yy = ψ y A , ψ xy = ψ y ( A + 6 χ ) , (18)4 nd ψ xxxx = 9600 ψ xxx χ + 160 ψ xx ( − A x − A − χ + 8 C )+ 40 ψ x (12 A x A + 72 A x χ − C x + 3 A + 18 A χ − A C + 120 χ − χC + 24 D − ψ (144 A x + 72 A x A − A x C − C xx − C x A − D y + 640 D x − x + 9 A − A C + 160 A D + 30 A Ω − A D + 300 χ Ω + 144 C ) + 1600 ψ y D , (19) where χ is given by equation (17) and Ω is the following expression Ω = A − A C + 8 D − C x + 6 A x A + 4 A xx . (20) Finally, the coefficients α and β of the resulting linear equation (3) are α = Ω8 ϕ x , (21) β = (1600 ϕ x ) − ( − A x − A x A + 352 A x C + 160 C xx + 80 C x A + 1600 D y − D x + 80Ω x − A + 88 A C − A D − A Ω+ 400 A D − χ Ω − C ) . (22) Let us consider the nonlinear fourth-order partial differential equation (Clark-son and Priestley, 1999) u tt = (cid:16) κu + γu (cid:17) xx + νuu xxxx + µu xxtt + αu x u xxx + βu xx , (23)where α, β, γ, µ, ν and κ are arbitrary constants.Of particular interest among solutions of equation (23) are traveling wavesolutions: u ( x, t ) = H ( x − Dt ) , where D is a constant phase velocity and the argument x − Dt is a phase ofthe wave. Substituting the representation of a solution into equation (23), onefinds( νH + µD ) H (4) + αH ′ H ′′′ + βH ′′ + (2 γH + κ − D ) H ′′ + 2 γH ′ = 0 . (24)This is an equation of the form (4) with coefficients A = ανH + µD , A = 0 , B = βνH + µD , = C = 0 , C = 2 γH + κ − D νH + µD ,D = D = 0 , D = 2 γνH + µD , D = D = 0 . Substituting these coefficients into the linearization conditions (6)-(15), oneobtains the following results. ν = 0If ν = 0, then equation (24) is linearizable if and only if α = β = γ = 0 . This means, with theses conditions the original equation (24) becomes thelinear equation ( µD ) H (4) + ( κ − D ) H ′′ = 0 . The solution of this equation is H ( x − Dt ) = C sin s κ − D µD ( x − Dt ) + C cos s κ − D µD ( x − Dt ) , where C and C are arbitrary constants. Hence, u ( x, t ) = C sin s κ − D µD ( x − Dt ) + C cos s κ − D µD ( x − Dt ) . ν = 0 Case 2.1 : γ = 0 • If ν = 0, γ = 0 and β = 0, then equation (24) is linearizable if and only if α = 0 , κ = D . Next, to find the linear form of equation (24) and it’s solutions.
Note that : For applying the theorems to our problem, here x = x − Dt and y ( x ) = H ( x − Dt ).From (16) one has 2 dχdx − χ = 0 . χ = 0 . Then invoking (17), we let ϕ = x. Now the equations (18)-(19) are written as ψ yy = 0 , ψ xy = 0 , and yield ψ y = K , K = const. Hence, ψ = K y + K ( x ) . Since one can use any particular solution, we set K = 1 , K ( x ) = 0 andtake ψ = y. Noting that (20) yields Ω = 0, one can readily verify that the function ψ = y solves equation (19) as well. Hence, one obtains the following transforma-tions ˜ t = x, ˜ u = y. (25)Since Ω = 0, equations (21) and (22) give˜ α = 0 , ˜ β = 0 . Hence, the equation (24) is mapped by the transformations (25) to the linearequation ˜ u (4) = 0 . The solution of this linear equation is˜ u = C + C ˜ t + C ˜ t + C ˜ t , where C , C , C and C are arbitrary constants. That is H ( x − Dt ) = C + C ( x − Dt ) + C ( x − Dt ) + C ( x − Dt ) . Hence, u ( x, t ) = C + C ( x − Dt ) + C ( x − Dt ) + C ( x − Dt ) . • If ν = 0, γ = 0 and β = 3 ν , then equation (24) is linearizable if and only if α = 4 ν, κ = D . Next, to find the linear form of equation (24) and it’s solutions. Considering(16) one has 2 dχdx − χ = 0 . χ = 0 . Then invoking (17), we let ϕ = x. Now the equations (18)-(19) are written as ψ yy = νψ y ( νy + D µ ) , ψ xy = 0 , and yield ψ y = K ( x ) y + D µν ! . Since ψ xy = 0, that means K ( x ) = K = const . One arrives at ψ = K y D µν y ! + K ( x ) . Since one can use any particular solution, we set K = 1 , K ( x ) = 0 andtake ψ = y D µν y. Observing that (20) yields Ω = 0, one can readily verify that the function ψ = y + D µν y solves equation (19) as well. Hence, one obtains the followingtransformations ˜ t = x, ˜ u = y D µν y. (26)Since Ω = 0, equations (21) and (22) give˜ α = 0 , ˜ β = 0 . Hence, the equation (24) is mapped by the transformations (26) to the linearequation ˜ u (4) = 0 . The solution of this linear equation is˜ u = C + C ˜ t + C ˜ t + C ˜ t , where C , C , C and C are arbitrary constants. So that we obtain the implicitsolution in the form H D µν H = C + C ( x − Dt ) + C ( x − Dt ) + C ( x − Dt ) . Hence, u D µν u = C + C ( x − Dt ) + C ( x − Dt ) + C ( x − Dt ) . ase 2.2 : γ = 0Equation (24) is linearizable if and only if α = 4 ν, β = 3 ν, κ = (2 γµ + ν ) D ν . Because of (16) one obtains dχdx − χ = 2 γ ν . Solving this equation, one gets χ = 2 r γ ν tan (cid:18)r γ ν ( x + C ) (cid:19) , where C is an arbitrary constant. Since one can use any particular solution,we set C = 0, so that χ = 2 r γ ν tan (cid:18)r γ ν x (cid:19) . Then invoking (17), ϕ xx ϕ x = 2 r γ ν tan (cid:18)r γ ν x (cid:19) . Thus, ϕ x = K sec (cid:18)r γ ν x (cid:19) , where K is an arbitrary constant. By solving this equation, one obtains ϕ = s νγ K tan (cid:18)r γ ν x (cid:19) + K , where K is an arbitrary constant. One can choose K = q γ ν and K = 0.One arrives at ϕ = tan (cid:18)r γ ν x (cid:19) . Now the equations (18)-(19) are written as ψ xy = 3 r γ ν tan (cid:18)r γ ν x (cid:19) ψ y , (27)and ψ yy = νψ y D µ + νy . (28)Equation (27) and (28) give ψ = sec (cid:18)r γ ν x (cid:19) " y D µν y . ψ solves equation (19) as well. Hence,one obtains the following transformations˜ t = tan (cid:18)r γ ν x (cid:19) , ˜ u = sec (cid:18)r γ ν x (cid:19) " y D µν y . (29)Equations (21) and (22) give˜ α = 0 , ˜ β = − r γ ν x. Hence, the equation (24) is mapped by the transformations (29) to the linearequation ˜ u (4) − r γ ν x ˜ u = 0 . (30)Since this is an unsolved linear equation, so that the result of this case weobtained only at the linear form of nonlinear equation (24). In the present work, we found some following solutions for nonlinear fourth-order partial differential equations (23). • If ν = α = β = γ = 0, then the solution of (23) is u ( x, t ) = C sin s κ − D µD ( x − Dt ) + C cos s κ − D µD ( x − Dt ) . • If ν = 0, α = β = γ = 0 and κ = D , then the solution of (23) is u ( x, t ) = C + C ( x − Dt ) + C ( x − Dt ) + C ( x − Dt ) . • If ν = 0, α = 4 ν , β = 3 ν, γ = 0 and κ = D , then the solution of (23) is u ( x, t )2 + D µν u ( x, t ) = C + C ( x − Dt ) + C ( x − Dt ) + C ( x − Dt ) . • If γ = 0 , α = 4 ν, β = 3 ν and κ = (2 γµ + ν ) D ν , then the linear form of (23) is(30).An interesting aspect of the results in this paper is that the class of exactsolutions of the original nonlinear problems, which can not find by the classicalmethods. 10 Acknowledgements
This work was financially supported by Faculty of Science, Naresuan Univer-sity. The author wishes to express thanks to Prof.Dr. Sergey V. Meleshko,Suranaree University of Technology for his guidance during the work.
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