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The L aplace Decomposition Method [1] [2] [3] is applied to a system of nonlinear partial differential equations to demonstrate potential applicability to such systems.

Differential equations theory is an important Mathematical branch which is used to describe practical problems in physics, chemistry, and biology and so on [

In this study, I have used LDM to solve a system of nonlinear partial differential equations for two different initial conditions. Later, we compared the obtained results with exact solutions and solution obtained by Method of Differential Quadrature [

Consider a system of nonlinear partial differential equations on our interest of region given by:

u t = u u x + v u y (1)

v t = u v x + v v y (2)

with initial condition

u ( x , y , 0 ) = f ( x , y ) (3)

v ( x , y ,0 ) = g ( x , y ) (4)

here, we have consider the general form of boundary conditions. Taking Laplace transform of Equations (2.1) and (2.2) with respect to t, we get

L t [ u t ] = L t [ u u x + v u y ]

L t [ v t ] = L t [ u v x + v v y ]

s u ( x , y , s ) − u ( x , y ,0 ) = L t [ u u x + v u y ]

s v ( x , y , s ) − v ( x , y ,0 ) = L t [ u v x + v v y ]

s u ( x , y , s ) = 1 s f ( x , y ) + 1 s L t [ u u x + v u y ]

s v ( x , y , s ) = 1 s g ( x , y ) + 1 s L t [ u v x + v v y ]

Taking inverse Laplace transform of above system with respect to “t”, we get

u ( x , y , t ) = f ( x , y ) + L t − 1 [ 1 s L t [ u u x + v u y ] ] (5)

v ( x , y , t ) = g ( x , y ) + L t − 1 [ 1 s L t [ u v x + v v y ] ] (6)

Let us suppose that,

u ( x , y , t ) = ∑ n = 0 ∞ u n ( x , y , t ) , v ( x , y , t ) = ∑ n = 0 ∞ v n ( x , y , t ) (7)

be the solution of given system of Equations (2.1), (2.2) in series form. Also we can decompose the nonlinear terms appeared in given system by using adomian polynomials, namely

u u x = ∑ n = 0 ∞ A n , v u y = ∑ n = 0 ∞ B n , u v x = ∑ n = 0 ∞ C n , v v y = ∑ n = 0 ∞ D n (8)

where A n , B n , C n and D n are adomian polynomials [

∑ n = 0 ∞ u n ( x , y , t ) = f ( x , y ) + L t − 1 [ 1 s L t [ ∑ n = 0 ∞ A n + ∑ n = 0 ∞ B n ] ]

∑ n = 0 ∞ v n ( x , y , t ) = g ( x , y ) + L t − 1 [ 1 s L t [ ∑ n = 0 ∞ C n + ∑ n = 0 ∞ D n ] ]

Comparing the both sides of above system of equations, we get the following recursive relations

u 0 ( x , y , t ) = f ( x , y ) , u n + 1 ( x , y , t ) = L t − 1 [ 1 s L t [ ∑ n = 0 ∞ A n + ∑ n = 0 ∞ B n ] ] , n ≥ 0. (9)

v 0 ( x , y , t ) = g ( x , y ) , v n + 1 ( x , y , t ) = L t − 1 [ 1 s L t [ ∑ n = 0 ∞ C n + ∑ n = 0 ∞ D n ] ] , n ≥ 0. (10)

Note that the solution of (2.1), (2.2) can exhibit a shock phenomenon for finite t; we select f(x, y) and g(x, y) such that the shock occurs for a value of t far from our region of interest. Let

f ( x , y ) = g ( x , y ) = x + y (11)

Therefore from the recursive relation (2.9) and (2.10), we get

u 0 ( x , y , t ) = v 0 ( x , y , t ) = x + y

then u 1 ( x , y , t ) , v 1 ( x , y , t ) can be calculate as

u 1 ( x , y , t ) = L t − 1 [ 1 s L t [ A 0 + B 0 ] ] = L t − 1 [ 1 s L t [ u 0 u 0 x + v 0 u 0 y ] ] = L t − 1 [ 1 s L t [ ( x + y ) + ( x + y ) ] ] = 2 t ( x + y )

Similarly,

v 1 ( x , y , t ) = L t − 1 [ 1 s L t [ C 0 + D 0 ] ] = L t − 1 [ 1 s L t [ u 0 v 0 x + v 0 v 0 y ] ] = 2 ( x + y ) t

Also, u 2 ( x , y , t ) and v 2 ( x , y , t ) are calculated as

u 2 ( x , y , t ) = L t − 1 [ 1 s L t [ A 1 + B 1 ] ] = L t − 1 [ 1 s L t [ ( u 0 u 1 x + u 1 u 0 x ) + ( v 0 u 1 y + v 1 u 0 y ) ] ] = L t − 1 [ 1 s L t [ ( 2 t ( x + y ) + 2 t ( x + y ) ) + ( 2 t ( x + y ) + 2 t ( x + y ) ) ] ] = L t − 1 [ 1 s L t [ 8 t ( x + y ) ] ] = 2 t 2 ( x + y )

Similarly,

v 2 ( x , y , t ) = 4 t 2 ( x + y )

Substitute all the values of u 0 , u 1 , u 2 , ⋯ and v 0 , v 1 , v 2 , ⋯ in the Equation (2.7), we get

u ( x , y , t ) = ( x + y ) + 2 t ( x + y ) + 4 t 2 ( x + y ) + ⋯

v ( x , y , t ) = ( x + y ) + 2 t ( x + y ) + 4 t 2 ( x + y ) + ⋯

This implies,

u ( x , y , t ) = ( x + y ) [ 1 + 2 t + 4 t 2 + ⋯ ]

v ( x , y , t ) = ( x + y ) [ 1 + 2 t + 4 t 2 + ⋯ ]

u ( x , y , t ) = x + y 1 − 2 t

v ( x , y , t ) = x + y 1 − 2 t

This is an exact solution of the given system of nonlinear partial differential Equations (2.1) and (2.2). We have verified this through the substitution, which is identical to the solution obtained by R. E. Bellman using the method of differential quadrature [

f ( x , y ) = x 2 , g ( x , y ) = y (12)

From the recursive relation (2.9), (2.10) and above initial conditions, we get

u 0 ( x , y , t ) = x 2 , v 0 ( x , y , t ) = y

u 1 ( x , y , t ) = L t − 1 [ 1 s L t [ A 0 + B 0 ] ] = 2 x 3 t

Similarly,

v 1 ( x , y , t ) = L t − 1 [ 1 s L t [ C 0 + D 0 ] ] = y t

Also, u 2 ( x , y , t ) and v 2 ( x , y , t ) are calculated as

u 2 ( x , y , t ) = L t − 1 [ 1 s L t [ A 1 + B 1 ] ] = 5 x 4 t

Similarly,

v 2 ( x , y , t ) = y t 2 , u 3 ( x , y , t ) = 14 x 5 t 3

and so on. Substitute all the values of u 0 , u 1 , u 2 , ⋯ and v 0 , v 1 , v 2 , ⋯ in Equation (2.7), we get

u ( x , y , t ) = x 2 ( 1 + 2 t x + 5 t 2 x 2 + 14 x 2 t 2 + ⋯ )

v ( x , y , t ) = y ( 1 + t + t 2 + ⋯ ) = y 1 − t

(The shock occurs at t = 1 4 x ). This is an approximate solution of given system of equations.

From the examples above, we can clearly say that we can calculate u ( x , y , t ) and v ( x , y , t ) when explicitly solutions exist for given initial functions. More importantly, the methodology [

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

Handibag, S.S. (2019) Laplace Decomposition Method for the System of Non Linear PDEs. Open Access Library Journal, 6: e5954. https://doi.org/10.4236/oalib.1105954