Page 53 - Mathematical Techniques of Fractional Order Systems

Nonlinear Fractional Order Boundary-Value Problems Chapter | 2 43

at p 5 0 and p 5 1 respectively. Taylor expansion of ϕ(t, E, p) can be written
in the form
N
X
m
ϕðt; E; pÞ 5 ϕðt; E; 0Þ 1 u m ðt; EÞp ; ð2:21Þ
m51
m
where u m (t, E) 5 D ϕ(t, E, p). From the Eq. (2.20), the power series (2.21)
becomes
N
X
m
ϕðt; EÞ 5 u 0 ðt; EÞ 1 u m ðt; EÞp ; ð2:22Þ
m51
and they are solutions of the nonlinear problems (2.1) with boundary condi-
@u
tions ℬ u; E; 5 0, but the functions u m (t,E) are still unknowns. These
0
@n
unknowns functions u m (t, E) are governed by the mth-order deformation
equation. Take mth-order homotopy derivative (2.7) for both sides of the
general zero-order deformation Eq. (2.18), we obtain
α
m
m
D fð1 2 pÞL½ϕðt; E; pÞ 2 u 0 ðt; Eފg 5 HðtÞhD ½pfD ϕðt; E; pÞ 1 fðt; ϕðt; E; pÞÞgŠ;
t
ð2:23Þ
since L is a linear operator independent of p and by using the properties of
the mth-order homotopy derivative (2.7), then the left side of Eq. (2.23)
becomes
m m
D fð1 2 pÞL½ϕðt; E; pÞ 2 u 0 ðt; Eފg 5 L½D ϕðt; E; pÞ
m m m
2 D ðpϕðt; E; pÞÞ 2 u 0 ðt; EÞD ðpފ 5 L½u m ðt; EÞ 2 u m21 ðt; EÞ 2 u 0 ðt; EÞD ðpފ 5

L½u m ðt; Eފ; m # 1
5 L½u m ðt; EÞ 2 χ u m21 ðt; Eފ;
L½u m ðt; EÞ 2 u m21 ðt; Eފ; m , 1 m
ð2:24Þ
and the mth-order deformation Eq. (2.23) becomes in the following form:
α m
L½u m ðt; EÞ 2 χ u m21 ðt; Eފ 5 HðtÞhfD u m21 ðt; EÞ 1 D ½pfðt; ϕðt; E; pÞފg ð2:25Þ
m t
subject to the boundary conditions

@ m @ϕðt; E; pÞ
ℬ ϕðt; E; pÞ; E; p50 5 0: ð2:26Þ
0
@p m @n
It should be emphasized, the mth-order deformation Eq. (2.25) is just the
linear ordinary differential equation with boundary condition (2.26) and can
be easily solved by using some software programs such as Matlap,
Mathematica, etc. Starting by u 0 (t,E) which satisfies the boundary conditions
@u
ℬ u; E; 5 0 and from Eqs. (2.25) and (2.26), the unknowns functions
0
@n
u m (t, E) for m : 1(1)M are obtained. The mth-order approximate solution of
the problems (2.1) and with boundary conditions ℬ u; E; @u 5 0 is given by

0
@n