Double Elzaki Transform Decomposition Method for Solving Third Order Korteweg-De-Vries Equations

Abstract

In this study, we used Double Elzaki Transform (DET) coupled with Adomian polynomial to produce a new method to solve Third Order Korteweg-De Vries Equations (KdV) equations. We will provide the necessary explanation for this method with addition some examples to demonstrate the effectiveness of this method.

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Hassan, M. and Elzaki, T. (2021) Double Elzaki Transform Decomposition Method for Solving Third Order Korteweg-De-Vries Equations. Journal of Applied Mathematics and Physics, 9, 21-30. doi: 10.4236/jamp.2021.91003.

1. Introduction

In 1985, Two Dutchmen, D. J. Korteweg and G. de Vrie derived a nonlinear partial differential equation, well known by the Korteweg-de Vries (KdV) equation, to model the height of the surface of shallow water in the presence of solitary wave’s. The KdV equation also describes the propagation of plasma waves in a dispersive medium.

Third order Korteweg-de-Vries (KdV) equation of the form

u t + a u u x + b u x x x = 0 (1)

with the initial conditions:

u ( x , 0 ) = f ( x ) (2)

where a and b are constants.

So many methods and approaches have been made to find the approximate analytic solutions and numerical solutions of KdV equations, such as Adomian Decomposition Method (ADM) [1], Variation Iteration Method (VIM) [1], Homotopy Perturbation Method (HPM) [1], Homotopy Perturbation Method using Elzaki Transform [2], Homotopy Perturbation Method using Laplace Transform [3], Adomian Decomposition Method using Elzaki Transform [4], Numerical solutions to a linear KdV equation on unbounded domain [5], The numerical solutions of KdV equation using radial basis functions [6], Numerical solution of separated solitary waves for KdV equation through finite element technique [7].

In this paper, we study a new method to solutions of KdV equations namely Double Elzaki Transform Decomposition Method (DETDM), this method is its capability of combining easy integral transform Double ELzaki Transform (DET) [8] and an effective method for solving non-linear partial differential equations, namely Adomian Decomposition Method [1].

Several examples are given as follows to illustrate this method to explain its effectiveness.

2. Basic Definitions of Double Elzaki Transform

Definition: Let f ( x , t ) , t , x R + be a function which can be expressed as a convergent infinite series, then its Double Elzaki Transform given by:

E 2 [ f ( x , t ) , u , v ] = T ( u , v ) = u v 0 0 f ( x , t ) e ( x u + t v ) d x d t , x , t > 0. (3)

where u , v are complex values.

To obtain double Elzaki transform of partial derivatives we use integration by parts [8], and then we have:

E 2 [ f x ] = 1 u T ( u , v ) u T ( 0 , v )

E 2 [ 2 f x 2 ] = 1 u 2 T ( u , v ) T ( 0 , v ) u x T ( 0 , v )

E 2 [ f t ] = 1 v T ( u , v ) v T ( u , 0 ) (4)

E 2 [ 2 f t 2 ] = 1 v 2 T ( u , v ) T ( u , 0 ) v t T ( u , 0 )

E 2 [ 2 f x t ] = 1 u v T ( u , v ) v u T ( u , 0 ) u v T ( 0 , v ) + u v T ( 0 , 0 )

Proof:

E 2 [ f x ] = u v 0 0 x f ( x , t ) e ( x u + t v ) d x d t = v 0 e t v [ u 0 e x u x f ( x , t ) d x ] d t

The inner integral gives 1 u T ( u , t ) u f ( 0 , t )

E 2 [ f x ] = v u 0 e t v T ( u , t ) d t u v 0 e t v f ( 0 , t ) d t

E 2 [ f x ] = 1 u T ( u , v ) u T ( 0 , v )

Also E 2 [ f t ] = 1 v T ( u , v ) v T ( u , 0 )

E 2 [ 2 f ( x , t ) x 2 ] = u v 0 0 2 f ( x , t ) x 2 e ( x u + t v ) d x d t = v 0 e t v [ u 0 2 f ( x , t ) x 2 e x u d x ] d t

The inner integral: u 0 2 f ( x , t ) x 2 e x u d x = T ( u , t ) u 2 f ( 0 , t ) u f ( 0 , t ) x .

By taking Elzaki transform with respect to t for above integral we get:

E 2 [ 2 f ( x , t ) x 2 ] = 1 u 2 T ( u , v ) T ( 0 , v ) u x T ( 0 , v )

Similarly:

E 2 [ 2 f ( x , t ) t 2 ] = 1 v 2 T ( u , v ) T ( u , 0 ) v t T ( u , 0 )

3. Theorems of Convergence of Double Elzaki Transform:

Theorem 3.1. Let the function f ( x , t ) is continuous in the x t plane, if the integral converges at u = u 0 , v = v 0 then the integral, u v 0 0 f ( x , t ) e ( x u + t v ) d x d t is convergence for u < u 0 , v < v 0 .

For the proof we will use the following theorems.

Theorem 3.2. Suppose that: v 0 f ( x , t ) e t v d t , converges at v = v 0 , then the integral converges for v < v 0

Proof

Let α ( x , t ) = v 0 0 t f ( x , s ) e s v 0 d s , 0 < t < (5)

Clearly α ( x , 0 ) = 0 and lim t α ( x , t ) exist.

By fundamental theorem of calculus we have:

α t ( x , t ) = v 0 f ( x , t ) e t v 0 (6)

If we choose 1 and R 1 such that ( 0 < 1 < R 1 ) and using Equation (6) we get:

v 1 R 1 f ( x , t ) e t v d t = v 1 R 1 1 v 0 e t v 0 α t ( x , t ) e t v d t = v v 0 1 R 1 α t ( x , t ) e ( v 0 v v v 0 ) t d t (7)

Integrating the last integral by parts to gives:

v v 0 1 R 1 α t ( x , t ) e ( v 0 v v v 0 ) t d t = v v 0 [ ( α ( x , t ) e ( v 0 v v v 0 ) t ) 1 R 1 1 R 1 α ( x , t ) e ( v 0 v v v 0 ) t ( ( v 0 v v v 0 ) ) d t ] = v v 0 [ α ( x , R 1 ) e ( v 0 v v v 0 ) R 1 α ( x , 1 ) e ( v 0 v v v 0 ) 1 + ( v 0 v v v 0 ) 1 R 1 α ( x , t ) e ( v 0 v v v 0 ) t d t ] (8)

Now let 1 0 , R 1 , if v < v 0 , then we have

v 0 f ( x , t ) e t v d t = ( v 0 v v 0 2 ) 0 α ( x , t ) e ( v 0 v v v 0 ) t d t (9)

Now if the integral on the right converges then the theorem is proved.

By using limit test for convergence we get:

lim t t 2 α ( x , t ) e ( v 0 v v v 0 ) t = lim t ( t 2 e ( v 0 v v v 0 ) t ) lim t ( α ( x , t ) )

The first limit equal zero at t if v < v 0 and the second limit exist, then

lim t t 2 α ( x , t ) e ( v 0 v v v 0 ) t = 0 , finite.

Then the integral v 0 f ( x , t ) e t v d t is converges at v < v 0 .

Theorem 3.3. Suppose that: u 0 f ( x , t ) e x u d x , converges at u = u 0 , then the integral converges for u < u 0

Proof

Prove, of this theorem is same as the method in theorem (3.2).

Now the proof of the theorem (3.1) is as follows

u v 0 0 f ( x , t ) e ( x u + t v ) d x d t = u 0 e x u [ v 0 e t v f ( x , t ) d t ] d x (10)

By using theorem (3.2) and theorem (3.3) we see the integral in RHS of Equation (10) is converges for u < u 0 , v < v 0 , hence the integral u v 0 0 f ( x , t ) e ( x u + t v ) d x d t converges for u < u 0 , v < v 0 [9] [10].

4. Double Elzaki Transform Decomposition Method (DETDM)

The main focus of this study is to solve the Third Order Korteweg-De Vries Equations (KdV) equations. Firstly we show how to use Double Elzaki Transform Decomposition Method (DETDM) to solve the general nonlinear partial differential equations [11].

Consider a general partial differential equation with the initial condition of the following form:

L u ( x , t ) + R u ( x , t ) + N u ( x , t ) = g ( x , t ) , (11)

u ( x , 0 ) = h ( x ) , u t ( x , 0 ) = f ( x ) . (12)

where, L is the second order linear differential operator L = 2 t 2 , R is the linear differential operator of less order then L , N represents the general nonlinear differential operator and g ( x , t ) is the source term.

Taking the double Elzaki Transform on both sides of Equation (11) and single Elzaki Transform of Equation (12), we get:

E 2 ( L u ( x , t ) ) + E 2 ( R u ( x , t ) ) + E 2 ( N u ( x , t ) ) = E 2 ( g ( x , t ) ) , (13)

E ( u ( x , 0 ) ) = E ( h ( x ) ) = T ( u , 0 ) and E ( u t ( x , 0 ) ) = E ( f ( x ) ) = t T ( u , 0 ) . (14)

To substitute Equation (14) in (13), after using Equation (4), we get:

E 2 ( u ( x , t ) ) = v 2 E 2 ( g ( x , t ) ) + v 2 E ( h ( x ) ) + v 3 E ( f ( x ) ) v 2 E 2 ( R u ( x , t ) ) v 2 E 2 ( N u ( x , t ) ) . (15)

Now, with the application of the inverse Double Elzaki Transform on both side of Equation (15) we get:

u ( x , t ) = G ( x , t ) E 2 1 [ v 2 E 2 [ R u ( x , t ) + N u ( x , t ) ] ] . (16)

where G ( x , t ) represents the terms arising from the source term and the prescribed initial conditions.

After that we represent solution as an infinite series given below,

u ( x , t ) = n = 0 u n ( x , t ) , (17)

and the nonlinear term can be written as follow,

N u ( x , t ) = n = 0 A n ( u ) , (18)

where, A n ( u ) are Adomian polynomial and it can be calculated by formula given below:

A n = 1 n ! d n d λ n [ N ( i = 0 λ i u i ) ] λ = 0 , n = 0 , 1 , 2 , 3 , (19)

To substitute (17) and (18) in (16), we get:

n = 0 u n ( x , t ) = G ( x , t ) E 2 1 [ v 2 E 2 ( R n = 0 u n ( x , t ) + n = 0 A n ) ] . (20)

Then from Equation (20) we get:

u 0 ( x , t ) = G ( x , t ) , u 1 ( x , t ) = E 2 1 [ v 2 E 2 [ R u 0 ( x , t ) + A 0 ] ] , u 2 ( x , t ) = E 2 1 [ v 2 E 2 [ R u 1 ( x , t ) + A 1 ] ] . (21)

In general, the recursive relation is given by:

u n ( x , t ) = E 2 1 [ v 2 E 2 [ R u n 1 ( x , t ) + A n 1 ] ] , n 1. (22)

Finally, we approximate the solution u ( x , t ) by the series:

u ( x , t ) = lim N n = 0 u n ( x , t ) . (23)

5. Applications

Now we are demonstrated the effectiveness of this method, by solving the following third Order Korteweg-De Vries Equations (KdV) equations.

Example 1: Consider the following KdV equations

u t + 6 u u x + u x x x = 0 , (24)

with initial condition:

u ( x , 0 ) = x . (25)

Take the double Elzaki transform to both sides of Equation (24), we get:

T ( u , v ) v v T ( u , 0 ) = E 2 ( 6 u u x + u x x x ) , (26)

Take single Elzaki transform to initial condition we get:

E ( u ( x , 0 ) ) = T ( u , 0 ) = E ( x ) = u 3 , (27)

Substitute Equation (27) in Equation (26), we obtain:

T ( u , v ) = v 2 u 3 v E 2 ( 6 u u x + u x x x ) . (28)

Take the inverse double Elzaki transform to both sides of Equation (28), we obtain:

u ( x , t ) = x E 2 1 [ v E 2 ( 6 u u x + u x x x ) ] . (29)

From the Adomian decomposition method, rewrite Equation (29) as follows,

n = 0 u n ( x , t ) = x E 2 1 [ v E 2 ( 6 n = 0 A n ( u ) + n = 0 ( u n ) x x x ) ] . (30)

where, A n ( u ) areAdomian polynomials that represent the nonlinear terms.

The first few components of A n ( u ) are given by:

A 0 ( u ) = u 0 ( u 0 ) x , A 1 ( u ) = ( u 0 ) x u 1 + u 0 ( u 1 ) x , A 2 ( u ) = ( u 0 ) x u 2 + ( u 1 ) x u 1 + ( u 2 ) x u 0 , A 3 ( u ) = ( u 0 ) x u 3 + ( u 1 ) x u 2 + ( u 2 ) x u 1 + ( u 3 ) x u 0 , (31)

By comparing both sides of Equation (22), we get:

u 0 ( x , t ) = x , (32)

u n + 1 ( x , t ) = E 2 1 [ v E 2 [ 6 A n ( u ) + ( u n ) x x x ] ] , n 0. (33)

Then:

u 1 ( x , t ) = E 2 1 [ v E 2 [ 6 A 0 ( u ) + ( u 0 ) x x x ] ] = E 2 1 [ v E 2 ( 6 x ) ] = E 2 1 [ 6 v 3 u 3 ] = 6 x t , (34)

u 2 ( x , t ) = E 2 1 [ v E 2 [ 6 A 1 ( u ) + ( u 1 ) x x x ] ] = E 2 1 [ v E 2 ( 72 x t ) ] = E 2 1 [ 72 v 4 u 3 ] = 36 x t 2 , (35)

By similar way we get:

u 3 ( x , t ) = 216 x t 3 . (36)

And so on, then the first four terms of the decomposition series for Equation (24) are given by:

u ( x , t ) = x 6 x t + 36 x t 2 216 x t 3 + , (37)

This can be written as:

u ( x , t ) = x [ 1 6 t + ( 6 t ) 2 ( 6 t ) 3 + ] , (38)

The solution in a closed form is given by:

u ( x , t ) = x 1 + 6 t , | t | < 1. (39)

Example 2: Consider the following KdV equations

u t 6 u u x + u x x x = 0 , (40)

with initial condition:

u ( x , 0 ) = 1 6 ( x 1 ) . (41)

Take the double Elzaki transform to both sides of equation (40), we get:

T ( u , v ) v v T ( u , 0 ) = E 2 ( 6 u u x u x x x ) , (42)

Take single Elzaki transform to initial condition we get:

E ( u ( x , 0 ) ) = T ( u , 0 ) = E ( 1 6 ( x 1 ) ) = 1 6 ( u 3 u 2 ) , (43)

Substitute Equation (42) in Equation (41), we obtain:

T ( u , v ) = 1 6 ( v 2 u 3 v 2 u 2 ) + v E 2 ( 6 u u x u x x x ) . (44)

Take the inverse double Elzaki transform to both sides of Equation (44), we obtain:

u ( x , t ) = 1 6 ( x 1 ) + E 2 1 [ v E 2 ( 6 u u x u x x x ) ] . (45)

From the Adomian decomposition method, rewrite Equation (45) as follows,

n = 0 u n ( x , t ) = 1 6 ( x 1 ) + E 2 1 [ v E 2 ( 6 n = 0 A n ( u ) n = 0 ( u n ) x x x ) ] . (46)

where, A n ( u ) are the Adomian polynomials that represent the nonlinear terms.

The first few components of A n ( u ) are given by:

A 0 ( u ) = u 0 ( u 0 ) x , A 1 ( u ) = ( u 0 ) x u 1 + u 0 ( u 1 ) x , A 2 ( u ) = ( u 0 ) x u 2 + ( u 1 ) x u 1 + ( u 2 ) x u 0 , A 3 ( u ) = ( u 0 ) x u 3 + ( u 1 ) x u 2 + ( u 2 ) x u 1 + ( u 3 ) x u 0 , (47)

By comparing both sides of Equation (46), we get:

u 0 ( x , t ) = 1 6 ( x 1 ) , (48)

u n + 1 ( x , t ) = E 2 1 [ v E 2 [ 6 A n ( u ) ( u n ) x x x ] ] , n 0. (49)

Then:

u 1 ( x , t ) = E 2 1 [ v E 2 [ 6 A 0 ( u ) ( u 0 ) x x x ] ] = E 2 1 [ v E 2 ( 6 1 36 ( x 1 ) ) ] = E 2 1 [ 1 6 ( v 3 u 3 v 3 u 2 ) ] = 1 6 ( x 1 ) t , (50)

u 2 ( x , t ) = E 2 1 [ v E 2 [ 6 A 1 ( u ) ( u 1 ) x x x ] ] = E 2 1 [ v E 2 ( 6 1 6 ( x 1 ) 1 3 t ) ] = E 2 1 [ v E 2 ( 1 3 x t 1 3 t ) ] = E 2 1 [ 1 3 u 3 v 4 1 3 u 2 v 4 ] = 1 6 ( x 1 ) t 2 , (51)

By similar way we get:

u 3 ( x , t ) = 1 6 ( x 1 ) t 3 . (52)

And so on. Then the first four terms of the decomposition series for Equation (40), is given by:

u ( x , t ) = 1 6 ( x 1 ) ( 1 + t + t 2 + t 3 + ) , (53)

The solution in a closed form is given by:

u ( x , t ) = 1 6 ( x 1 1 t ) , | t | < 1. (54)

6. Conclusion

This method is very effective for solving non-linear partial differential equations in general, and as a special case, the Third Order Korteweg-De Vries (KdV) equations. It can be applied to higher order Korteweg-De Vries Equations.

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Authors’ Contributions

The authors read and agreed the final manuscript.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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