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S. S. Nourazar et al.

Subject to initial condition:
e

u ( x, 0 ) =



x
4

x

e4 + e



(3.11)

x
4

To solve Equation (3.10) we construct a homotopy in the following form:

 ∂υ ∂u 
H (υ , p ) =(1 − p )  − 0  +
∂t 
 ∂t

 ∂υ ∂ 2υ

∂υ
p  − 2 −υ
− υ (1 − υ )(υ − 1) 
∂x
 ∂t ∂x


(3.12)

The solution of Equation (3.10) can be written as a power series in p as:

υ =υ0 + pυ1 + p 2υ2 +

(3.13)

Substituting Equation (3.13) and Equation (3.11) in to Equation (3.12) and equating the term with identical
powers of p, leads to:

∂υ
p : 0
∂t
0

∂u0
=
υ 0 ( x, 0 )
,
∂t

∂υ ∂u
p1 : 1 + 0 =
∂t
∂t
∂υ
∂ 2υ
p 2 : 2 = 21
∂t
∂x
∂υ
∂ 2υ
p 3 : 3 = 22
∂t
∂x

e



x
4

x
4


x
4

,

e +e
∂ 2υ0
∂υ0
+ υ0
+ υ0 (1 − υ0 )(υ0 − 1) ,
υ1 ( x, 0 ) = 0,
(3.14)
∂x
∂x 2
∂υ
∂υ
+ υ0 1 + υ1 0 + 4υ0υ1 − υ1 − 3υ1υ02 ,
υ2 ( x, 0 ) =0,
∂x
∂x
∂υ
∂υ
∂υ
+ υ0 2 + υ2 0 + υ1 1 + 4υ0υ2 − υ2 − 3υ2υ02 + 2υ12 − 3υ0υ12 , υ3 ( x, 0 ) =
0.
∂x
∂x
∂x

Using the Maple package to solve recursive sequences, Equation (3.14), we obtain the followings:

e



x
4

3
1
t,
x 2
4 x
− 
e +e
 e 4 + e 4 


  x 2  − x 2

x
− 
 4x
4
4


4
+

e
e
4




 e − e 
  



9 
9    

 t 3.
2
υ 2 ( x, t ) =
t , υ3 ( x, t ) = −
3
4
x
x
x
32  x
128
− 
− 
 4
 e 4 + e 4 
 e + e 4 





υ 0 ( x, t ) =

x
4



x
4

, υ1 ( x, t ) = −

(3.15)

By setting p = 1 in Equation (3.13), the solution of Equation (3.10) can be obtained as υ = υ0 + υ1 + υ2 + υ3 +
Therefore the solution of Equation (3.10) is written as:



x

e 4
υ ( x, t ) =
x
e +e
4



x
4



3
1
x 2
4 x
− 
4
4
 e + e 



  x 2  − x 2

x
− 
 4x
4
4


4
e
e
4
+





 e − e 
  



9 
9    


2
t+
t −
t3 +
x 3
x
x 4
32  x
128
− 
− 
 4
 e 4 + e 4 
 e + e 4 





x 3t
x 3t
+
− −

1 1 e4 8 − e 4 8

The Taylor series expansion for  −
x 3t
x 3t
 2 2 e 4 + 8 + e− 4 − 8


289


 is written as:




(3.16)