THE SOLUTION OF COUPLED MODIFIED KDV SYSTEM BY THE ...

126 TWMS JOUR. PURE APPL. MATH., V.3, N.1, 2012
Example 3.1. For the first example, we present the new coupled MKdV equations with analytical
solution to show the efficiency of HAM, which has been described in the section 2. We consider
the system (1) with initial conditions as follows [7, 9]
u(x, 0) = 1 + 1 tanh x,
2
v(x, 0) = 1 2 − 1 tanh x,
4
z(x, 0) = 2 − tanh x.
It can be verified that the exact solutions of this example are
u(x, t) = 1 + 1 (
2 tanh x − 11 )
2 t ,
v(x, t) = 1 2 − 1 (
4 tanh x − 11 )
2 t ,
(
z(x, t) = 2 − tanh x − 11 )
2 t .
We start with the following zeroth-components
u 0 (x, t) = 1 + 1 tanh x, (26)
2
v 0 (x, t) = 1 2 − 1 tanh x,
4
(27)
z 0 (x, t) = 2 − tanh x. (28)
According to section 2, we can define the operators N 1 , N 2 and N 3 as
N 1 [φ] = φ t − 1 2 φ xxx + 3φ 2 φ x − 3 2 (ϕψ) xx − 3(φϕψ) x , (29)
N 2 [ϕ] = ϕ t + ϕ xxx + 3(ϕφ x ) x + 3ϕ 2 ψ x − 6φϕφ x − 3φ 2 ϕ x , (30)
N 3 [ψ] = ψ t + ψ xxx + 3(ψφ x ) x + 3ψ 2 ϕ x − 6φψφ x − 3φ 2 ψ x , (31)
Now, we can obtain the zeroth-
where φ = φ(x, t; p), ϕ = ϕ(x, t; p) and ψ = ψ(x, t; p).
deformation equation (15)-(17) as
(
R 1m (⃗u m−1 ) = (u m−1 ) t − 1 2 (u m−1) xxx + 3
− 3 2
( m−1
)
∑
v i z m−1−i −
i=0
xx
R 2m (⃗v m−1 ) = (v m−1 ) t + (v m−1 ) xxx +3
+3
m−1
∑
i=0
(
∑
v i
m−1−i
k=0
v k (z m−1−i−k ) x
)
−6
−3
m−1
∑
i=0
( m−1 ∑
i=0
∑
u i
m−1−i
k=0
∑
u i
m−1−i
k=0
( m−1
)
∑
v i (u m−1−i ) x
i=0
m−1
∑
i=0
m−1
∑
i=0
(
(
∑
u i
m−1−i
k=0
∑
u i
m−1−i
k=0
u k (u m−1−i−k ) x
)
−
(v k z m−1−i−k )
x
+
)
v k (u m−1−i−k ) x
)
−
u k (v m−1−i−k ) x
)
,
x
,
(32)
(33)