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ANALYTICAL SOLUTIONS FOR THE FRACTIONAL
FISHER’S EQUATION
H. KHEIRI1∗
, A. MOJAVER2
, AND S. SHAHI3
Abstract. In this paper, we consider the inhomogeneous time-fractional nonlinear Fisher equation with three known boundary conditions. We first apply a modified Homotopy perturbation method for translating the proposed problem to a set of linear problems. Then we use the separation variables method to solve obtained problems. In examples, we illustrate that by right choice of source term in the modified Homotopy perturbation method, it is possible to get an exact solution.
1.
Introduction
Fractional partial differential equations (FPDEs) have recently aroused
considerable interest in mathematics and its applications. Scientists
used them to model many physical, biological and chemical processes
[16, 18, 19]. Besides, they have applications in sampling, hold
algo-rithms, and signal processing. There are various analytical method for
solving nonlinear FPDEs including the Adomian decomposition [2, 4, 9],
Homotopy perturbation method [5, 6, 13], Variation iteration method
[14, 15], and Homotopy analysis method [10, 11]. Finding exact solutions
for FPDEs are often too complicated and required so much calculations.
Nowadays, biological models have been the focus of many mathematical
scientists. Fisher’s equation
(1.1)
u
t=
u
xx+
u
(1
−
u
)
2010Mathematics Subject Classification. 34B24, 34B27.
Key words and phrases. Fractional Fisher’s equation, Mittag-Leffler, Method of separating variables.
Received: 5 November 2014, Accepted: 7 December 2014.
∗Corresponding author.
was first proposed by Fisher as a model for the propagation of a mutant
gene [10]. In this model,
u
(
x, t
) is the population density and
u
(
u
−
1)
denotes the logistic form.
In this paper, we consider the inhomogeneous time-fractional Fisher’s
equation which is examined in [22, 21]:
(1.2)
D
αtu
(
x, t
) =
u
xx(
x, t
)+
u
(
x, t
)(1
−
u
(
x, t
))+
f
(
x, t
)
,
0
< x < L,
0
< α
≤
1
,
where the fractional derivative in (1.2) is the Caputo derivative.
We introduce a scheme to solve (1.2) which is a combination of
mod-ified homotopy perturbation [5], an especial case of homotopy analysis
method, Laplace transform, and separation of variables. This kind of
modification of homotopy perturbation method has the capability of
transforming nonlinear terms into linear ones while homotopy analysis
method and homotopy perturbation method don’t possess this
charac-teristic.
2.
Preliminaries
In this section, we give some necessary definitions and lemmas about
fractional calculus. For some details, you can refer to [18, 20].
Definition 2.1
([3])
.
A function
f
:
R
→
R
+is said to be in the space
C
ν, with
ν
∈
R
, if it can be written as
f
(
x
) =
x
pf
1(
x
) with
p > ν
,
f
1(
x
)
∈
C
[0
,
∞
) and it is said to be in the space
C
νmif
f
(m)∈
C
νfor
m
∈
N
∪
{
0
}
.
Definition 2.2
([12])
.
The Riemann-Liouville fractional integral of
f
∈
C
νwith order
α >
0 and
ν
≥ −
1 is defined as:
J
αf
(
t
) =
1
Γ(
α
)
∫
t0
(
t
−
τ
)
α−1f
(
τ
)
dτ,
α >
0
,
t >
0
,
(2.1)
J
0f
(
t
) =
f
(
t
)
.
Definition 2.3
([20])
.
The Riemann-Liouville fractional derivative of
f
∈
C
m−1
with order
α >
0 and
m
∈
N
∪
{
0
}
, is defined as:
D
αt
f
(
t
) =
dm
dtm
J
m−α
f
(
t
)
,
m
−
1
< α
≤
m,
m
∈
N
.
(2.2)
Definition 2.4
([20])
.
The Caputo fractional derivative of
f
∈
C
m−1
with order
α >
0 and
m
∈
N
∪
{
0
}
, is defined as:
(2.3)
CD
tαf
(
t
) =
{
J
m−αf
(m)(
t
)
,
dmf(t) dtm
,
Definition 2.5
([20])
.
A two-parameter Mittag-Leffler function is
de-fined by the following series
E
α,β(
t
) =
∞
∑
k=0
t
kΓ(
αk
+
β
)
.
(2.4)
Definition 2.6
([12])
.
A multivariate Mittag-Leffler function is defined
as
E
(a1,a2,···,an),b(
z
1, z
2,
· · ·
, z
n)
(2.5)
=
∞
∑
k=0
∑
l1+l2+···+ln=k
k
!
l
1!
×
l
2!
× · · · ×
l
n!
n
∏
i=1
z
lii
Γ
(
b
+
n
∑
i=1
a
il
i)
,
where
b >
0,
l
1, l
2, . . . , l
n≥
0,
|
z
i|
<
∞
, i
= 1
,
2
, . . . , n.
Definition 2.7.
Let us define the Laplace-transform (LT) operator
ϕ
on a function
u
(
x, t
)
,
(
t
≥
0) by
ϕ
{
u
(
x, t
);
t
7→
s
}
=
∫
∞0
e
−stu
(
x, t
)
dt,
(2.6)
and denote it by
ϕ
{
u
(
x, t
);
t
7→
s
}
=
L
(
u
(
x, t
)), where
s
is the LT
parameter. For our purpose here, we shall take
s
to be real and positive.
As a consequence, the LT of Mittag-Leffler function takes the following
form
L
(
E
α,β(
t
)) =
∫
∞0
e
−stE
α,β
(
t
)
dt
(2.7)
=
∞
∑
k=0
1
s
k+1Γ(
αk
+
β
)
.
Lemma 2.8
(see [12])
.
Let
µ > µ
1> µ
2> . . . > µ
n≥
0
, m
i−
1
<
µ
i≤
m
i, m
i∈
N
0=
N
∪
{
0
}
,
d
i∈
R
, i
= 1
,
2
, . . . , n.
Consider the initial
value problem
(
D
µy
)(
x
)
−
n∑
i=1
λ
i(
D
µiy
)(
x
) =
g
(
x
)
,
y
(k)(0) =
c
k
∈
R
,
k
= 0
,
1
, . . . , m
−
1
, m
−
1
< µ
≤
m,
C
m−1
.
This has solution
y
(
x
) =
y
g(
x
) +
m−1∑
k=0
c
ku
k(
x
)
,
x
≥
0
,
where
y
g(
x
) =
∫
x0
t
µ−1E
(·),µ(
t
)
g
(
x
−
t
)
dt
and
u
k(
x
) =
x
kk
!
+
n∑
i=lk+1
d
ix
k+µ−µi
E
(·),k+1+µ−µi(
x
)
,
k
= 0
,
1
, . . . , m
−
1
,
fulfills the initial conditions
u
(kl)(0) =
δ
kl,
k, l
= 0
,
1
, . . . , m
−
1
.
The function
E
(·),σ(
x
) =
E
(µ−µ1,...,µ−µn),σ(
d
1x
µ−µ1
, . . . , d
nx
µ−µn)
,
is a particular case of the multivariate Mittag-Leffler function (see
[12]
)
and the natural numbers
l
k, k
= 0
,
1
, . . . , m
−
1
,
are determined from the
condition
{
m
lk≥
k
+ 1
,
m
lk+1≤
k.
In the case
m
i≤
k, i
= 1
,
2
, . . . , n,
we set
l
k:= 0
,
and if
m
i≥
k
+ 1
, i
=
1
,
2
, . . . , n,
then
l
k:=
n.
3.
Modified homotopy perturbation method (MHPM)
The homotopy perturbation method is one of the most effective
meth-ods for solving nonlinear problems. Several modifications of this method
is presented. In this paper, we used a modified one for solving the
frac-tional Fisher’s equation as follows:
D
αtu
(
x, t
) =
u
xx(
x, t
) +
u
(
x, t
) +
ph
(
u
(
x, t
))
(3.1)
+
f
1(
x, t
) +
pf
2(
x, t
)
,
where
f
1(
x, t
) +
f
2(
x, t
) =
f
(
x, t
),
f
(
x, t
) is the source term of Eq. (1.2),
and
p
is an embedding parameter that varies from zero to one. For more
details see [5]. By choosing proper functions
f
1and
f
2, we can improve
the success of our method.
converting the nonlinear problems to linear ones, we can also use a
special case of modified homotopy analysis method.
4.
Inhomogeneous fractional Fisher’s equation
4.1.
Dirichlet boundary condition.
In this subsection, we determine
the solution of the fractional nonlinear Fisher’s equation
(4.1)
D
tαu
(
x, t
) =
u
xx(
x, t
) +
u
(
x, t
)
−
u
2(
x, t
) +
f
(
x, t
)
,
with the initial and Dirichlet boundary conditions
u
(
x,
0) =
φ
(
x
)
,
u
(0
, t
) =
µ
1(
t
)
,
0
≤
x
≤
L,
u
(
L, t
) =
µ
2(
t
)
,
t
≥
0
.
(4.2)
In order to solve the problem with inhomogeneous boundary conditions,
first transform it into a problem with homogeneous boundary conditions.
For this purpose let
u
(
x, t
) =
W
(
x, t
) +
V
(
x, t
)
,
where
W
(
x, t
) is a new unknown function and
(4.3)
V
(
x, t
) =
µ
2(
t
)
−
µ
1(
t
)
L
x
+
µ
1(
t
)
,
satisfies the boundary conditions as
(4.4)
V
(0
, t
) =
µ
1(
t
)
,
V
(
L, t
) =
µ
2(
t
)
.
Furthermore, the function
W
(
x, t
) satisfies in problem with
homoge-neous boundary conditions as follows:
(4.5)
D
αtW
(
x, t
) =
W
xx(
x, t
) +
W
(
x, t
) +
h
(
W
+
V
) + ˜
f
(
x, t
)
,
W
(
x,
0) =
g
(
x
)
,
W
(0
, t
) = 0
,
0
≤
x
≤
L,
W
(
L, t
) = 0
,
where
˜
f
(
x, t
) =
f
(
x, t
) +
x
L
[
D
α
t
µ
1(
t
)
−
D
tαµ
2(
t
)]
−
D
tαµ
1(
t
)
(4.6)
+
µ
2(
t
)
−
µ
1(
t
)
L
x
+
µ
1(
t
)
,
and
g
(
x
) =
φ
(
x
)
−
x
L
[
µ
2(0)
−
µ
1(0)]
−
µ
1(0)
.
(4.7)
For solving (4.5) we use MHPM
By assuming
W
(
x, t
) =
∞
∑
i=0
W
ip
i, and substituting it in (4.8), we obtain
p
0:
D
αt
W
0(
x, t
) =
∂ 2W0(x,t)
∂x2
+
W
0(
x, t
) + ˜
f
1(
x, t
)
,
W
0(
x,
0) =
g
(
x
)
W
0(0
,
t) = 0
,
0
≤
x
≤
L,
W
0(L
,
t) = 0
,
(4.9)
p
1:
D
tαW
1(
x, t
) =
∂ 2W1(x,t)
∂x2
+
W
1(
x, t
) + ˜
f
2(
x, t
) +
A
0,
W
1(
x,
0) = 0
,
W
1(0
, t
) = 0
,
0
≤
x
≤
L,
W
1(
L, t
) = 0
,
(4.10)
..
.
p
k:
D
2αt
W
k(
x, t
) =
∂ 2Wk(x,t)
∂x2
+
W
k(
x, t
) +
A
k−1,
W
k(
x,
0) = 0
,
W
k(0
, t
) = 0
,
0
≤
x
≤
L,
W
k(
L, t
) = 0
,
(4.11)
..
.
where ˜
f
1(
x, t
) + ˜
f
2(
x, t
) = ˜
f
(
x, t
) and ˜
f
1(
x, t
) must be satisfied in
ini-tial and boundary conditions (4.9) [5] and
A
k, k
= 0
,
1
, . . .
are Adomian
polynomials [1] and are obtained
A
k=
d
kdp
kh
(
∞∑
i=0
W
ip
i+
V
)
p=0
,
k
= 0
,
1
, . . . .
(4.12)
We solve the corresponding homogeneous equation in (4.9) by the method
of separation of variables. By assuming
W
0(
x, t
) =
X
0(
x
)
T
0(
t
) and
sub-stituting it in (4.9), we obtain an ordinary linear differential equation
for
X
0(
x
) as
(4.13)
X
′′0
(
x
) +
λ
2X
0(
x
) = 0
,
X
0(0) =
X
0(
L
) = 0
,
and a fractional ordinary linear differential equation for
T
0(
t
) as follows:
(4.14)
D
αtT
0−
T
0+
λ
2T
0= 0
.
The Sturm-Liouville problem given by (4.13) has eigenvalues
(4.15)
λ
n=
n
2π
2L
2,
n
= 1
,
2
, . . . ,
and corresponding eigenfunctions are
(4.16)
(
X
0)
n(
x
) = sin
(
nπx
L
)
Now we seek a solution of the inhomogeneous problem in (4.9) of the
form
(4.17)
W
0(
x, t
) =
∞
∑
n=1
(
B
0)
n(
t
) sin
(
nπx
L
)
.
We assumed that the series can be differentiated term by term. In order
to determine (
B
0)
n(
t
), we expanded ˜
f
1(
x, t
) as a Fourier series by the
eigenfunctions sin(
nπxL) as follows:
(4.18)
f
˜
1(
x, t
) =
∞
∑
n=1
( ˜
f
1)
n(
t
) sin
(
nπx
L
)
,
then
(4.19)
( ˜
f
1)
n(
t
) =
2
L
∫
L0
˜
f
1(
x, t
) sin
(
nπx
L
)
dx.
Substituting (4.17) and (4.18) into (4.9) yields
∞
∑
n=1
D
α(
B
0)
n(
t
) sin
(
nπx
L
)
+
(
n
2π
2L
2−
1
)
∞∑
n=1
D
α(
B
0)
n(
t
) sin
(
nπx
L
)
(4.20)
=
∞
∑
n=1
( ˜
f
1)
n(
t
) sin
(
nπx
L
)
.
By orthogonality properties of sin(
nπxL
), we get
D
tα(
B
0)
n(
t
) +
(
n
2π
2L
−
1
)
(
B
0)
n(
t
) = ( ˜
f
1)
n(
t
)
.
(4.21)
Since
W
0(
x, t
) satisfies the initial conditions in (4.9), we have
(4.22)
∞
∑
n=1
(
B
0)
n(0) sin
(
nπx
L
)
=
g
(
x
)
,
which yields
(4.23)
(
B
0)
n(0) =
2
L
∫
X0
g
1(
x
) sin
(
nπx
L
)
dx.
According to Lemma 2.8, the fractional initial value problem with
µ
=
α, µ
1= 0
, m
1= 0
, λ
1= 1
−
n 2π2L
, m
= 1
,
has the solution
(
B
0)
n(
t
) =
∫
t0
τ
αE
(α,α)(
λ
1τ
α)( ˜
f
1)
n(
t
−
τ
)
dτ
(4.24)
+ (
B
0)
n(0)
[
1 +
λ
1E
(α,α+1)(
λ
1t
α)
]
.
Hence we get the solution of the initial boundary value problem (4.9) in
the form
(4.25)
W
0(
x, t
) =
∞
∑
n=1
(
B
0)
n(
t
) sin
(
nπx
L
)
.
In a similar way, we can get
W
k, k
= 1
, . . .
from (4.10) and (4.11). Since
in calculating
W
k+1the value of
A
kis known from pervious stages, then
all of problems in (4.9)-(4.11) are linear and hence solving them is
sim-pler than main problems. Note that the success of these methods relies
mainly on the proper choice of the functions ˜
f
1and ˜
f
2. Furthermore,
this proper choice of ˜
f
1and ˜
f
2may provide the solution only in one
iteration of MHPM.
4.2.
Neumann boundary condition.
Now, we obtain the solution of
the inhomogeneous fractional Fisher’s equation (1.2) with the initial and
Neumann boundary conditions as follows:
u
(
x,
0) =
φ
(
x
)
,
0
≤
x
≤
L,
(4.26)
u
x(0
, t
) =
µ
1(
t
)
,
u
x(
L, t
) =
µ
2(
t
)
,
t
≥
0
,
in which
φ
(
x
)
, µ
1(
t
)
, µ
2(
t
) are as defined in subsection 4.1.
For solving the problem with inhomogeneous boundary conditions,
as before, we transform it into a problem with homogeneous boundary
conditions. Thus we suppose that
u
(
x, t
) = ˜
W
(
x, t
) + ˜
V
(
x, t
)
,
where ˜
W
(
x, t
) is an unknown function and
(4.27)
V
˜
(
x, t
) =
µ
2(
t
)
−
µ
1(
t
)
2
L
x
2
+
µ
1(
t
)
x,
which satisfies the following boundary conditions:
(4.28)
V
˜
x(0
, t
) =
µ
1(
t
)
,
V
˜
x(
L, t
) =
µ
2(
t
)
.
(4.29)
D
αt
W
˜
(
x, t
) =
∂2W˜(x,t)
∂x2
+ ˜
W
(
x, t
) +
h
( ˜
W
(
x, t
) + ˜
V
(
x, t
)) = ˜
f
(
x, t
)
,
˜
W
(
x,
0) =
g
(
x
)
,
˜
W
x(0
, t
) = 0
,
0
≤
x
≤
L,
˜
W
x(
L, t
) = 0
,
t
≥
0
,
in which ˜
f
(
x, t
) and
g
(
x
) are as the form as below:
˜
f
(
x, t
) =
f
(
x, t
) +
x
2
2
L
(
D
α
t
µ
1(
t
)
−
D
αtµ
2(
t
))
−
D
αtµ
1(
t
) +
µ
2(
t
)
−
µ
1(
t
)
L
(4.30)
−
µ
2(
t
)
−
µ
1(
t
)
2
L
x
2
+
µ
1(
t
)
x,
g
(
x
) =
φ
(
x
)
−
x
22
L
[
µ
2(0)
−
µ
1(0)]
−
µ
1(0)
x.
Now, for solving (4.29) by MHPM, we have
D
αtW
˜
(
x, t
) = ˜
W
xx(
x, t
) + ˜
W
(
x, t
) +
ph
( ˜
W
(
x, t
) + ˜
V
(
x, t
))
(4.31)
+ ˜
f
1(
x, t
) +
p
f
˜
2(
x, t
)
.
If we assume ˜
W
(
x, t
) =
∞
∑
i=0
˜
W
ip
i, and substitute it in (4.29), we obtain
p
0:
D
αt
W
f
0(
x, t
) = (
W
f
0)
xx(
x, t
) +
W
f
0(
x, t
) +
f
e
1(
x, t
)
,
˜
W
0(
x,
0) =
g
(
x
)
,
( ˜
W
0)
x(0
, t
) = 0
,
0
≤
x
≤
L,
( ˜
W
0)
x(
L, t
) = 0
,
(4.32)
p
1:
D
αt
W
f
1(
x, t
) = (
W
f
1xx)(
x, t
) +
W
f
1(
x, t
) +
f
e
2(
x, t
) +
A
0˜
W
1(
x,
0) = 0
,
( ˜
W
1)
x(0
, t
) = 0
,
0
≤
x
≤
L,
( ˜
W
1)
x(
L, t
) = 0
,
(4.33)
..
.
p
k:
D
αt
W
f
k(
x, t
) = (
W
f
k)
xx(
x, t
) +
W
f
k(
x, t
) +
A
k−1,
˜
W
k(
x,
0) = 0
,
( ˜
W
k)
x(0
, t
) = 0
,
( ˜
W
k)
x(
L, t
) = 0
,
(4.34)
and so on, in which ˜
f
1(
x, t
) + ˜
f
2(
x, t
) = ˜
f
(
x, t
) and
A
k, k
= 0
,
1
, . . .
are
Adomian polynomials defined in (4.12).
For solving the corresponding homogeneous equation in (4.32) by the
method of separation of variables, we assume that ˜
W
0(
x, t
) =
X
0(
x
)
T
0(
t
)
FDE for
T
0(
t
) as
X
′′0
(
x
) +
λX
(
x
) = 0
,
X
(0) =
X
(
L
) = 0
,
D
tαT
0(
t
) + (
λ
−
1)
T
0(
t
) = 0
.
(4.35)
The Sturm-Liouville problem, which is given by (4.35), has eigenvalues
and corresponding eigenfunctions as:
λ
n=
n
2π
2L
2,
n
= 1
,
2
, . . . ,
(
X
0)
n(
x
) = cos
(
nπx
L
)
,
n
= 1
,
2
, . . . .
(4.36)
Now we are going to seek a solution of the inhomogeneous problem in
(4.29) which takes the form
(4.37)
W
˜
0(
x, t
) =
∞
∑
n=1
(
B
0)
n(
t
) cos
(
nπ
L
x
)
.
For determining (
B
0)
n(
t
), by expanding ˜
f
1(
x, t
) as a Fourier series by
the eigenfunctions cos(
nπLx
) we have:
(4.38)
f
˜
1(
x, t
) =
∞
∑
n=1
( ˜
f
1)
n(
t
) cos
(
nπ
L
x
)
,
in which the Fourier coefficients are as the following form:
(4.39)
( ˜
f
1)
n(
t
) =
2
L
∫
L0
˜
f
1(
x, t
) cos
(
nπ
L
x
)
dx.
Then substituting (4.37), (4.38) into (4.29) implies
D
αt(
B
0)
n(
t
) +
(
−
1 +
n
2
π
2L
2)
(
B
0)
n(
t
) = ( ˜
f
1)
n(
t
)
.
(4.40)
Since ˜
W
(
x, t
) fulfills the initial conditions in (4.29), we have
∞
∑
n=1
(
B
0)
n(0) cos
(
nπ
L
x
)
=
g
(
x
)
,
(4.41)
which yields
(
B
0)
n(0) =
2
L
∫
L0
g
(
x
) cos
(
nπ
L
x
)
dx.
Therefore, Lemma 2.8 implies that the fractional initial value problem
has the solution as follows:
u(x, t) = ∞
∑
n=1
(B0)n(t) cos
(nπx
L
)
(4.43)
=µ1(t)x+µ2(t)−µ1(t) 2L x
2
+ ∞
∑
n=1
cos(nπ
L x
) [∫ t
0
τα−1E(α,α)
((
1−n
2π2 L2
)
τα(f1˜)
n(t−τ)
)
dτ
]
.
4.3. Robin boundary condition. In this subsection, we try to solve (4.1) with the initial and Robin boundary conditions as
u(x,0) =φ(x),
u(0, t) +α1ux(0, t) =µ1(t), u(L, t) +β1ux(L, t) =µ2(t),
0≤x≤L, t≥0, t≥0,
where α1, β1 are nonzero constants. To solve this problem, we translate the inhomogeneous bondary conditions to the homogenouse ones. So, suppose that
u(x, t) = ¯W(x, t) + ¯V(x, t),
where ¯W(x, t) is a new unknown function and
(4.44) V¯(x, t) = µ1(t)−µ2(t)
α1−β1−Lx−
(L+β1)µ1(t)−α1µ2(t)
α1−β1−L .
Therefore, we will have
{ ¯
V(0, t) +α1Vx¯ (0, t) =µ1(t),
¯
V(L, t) +β1V¯(L, t) =µ2(t).
The function ¯W(x, t) is the solution of nonlinear problem with homogeneous boundary conditions:
(4.45)
Dα
tW¯(x, t) = ¯Wxx(x, t) + ¯W(x, t) +h( ¯W + ¯V) + ˜f(x, t),
¯
W(x,0) =g(x),
¯
W(0, t) +α1Wx¯ (0, t) = 0,
¯
W(L, t) +β1W¯(L, t) = 0,
0≤x≤L,
where
˜
f(x, t) =f(x, t)−Dα tV¯(x, t).
(4.46)
For solving (4.45), again we use MHPM. By assuming
¯
W(x, t) = ∞
∑
i=0
¯
and substituting it in (4.45), we obtain
p0:
Dα
tW0¯ (x, t) = ∂2¯
W0(x,t)
∂x2 + ¯W0(x, t) + ˜f1(x, t), ¯
W0(x,0) =g(x),
¯
W0(0, t) +α1( ¯W0)x(0, t) =µ1(t),
¯
W0(L, t) +β1( ¯W0)x(L, t) =µ2(t),
0≤x≤L,
t≥0,
(4.47)
p1:
Dα
tW1¯ (x, t) = ∂2¯
W1(x,t)
∂x2 + ¯W1(x, t) + ˜f2(x, t) +A0, ¯
W1(x,0) = 0,
¯
W1(0, t) +α1( ¯W1)x(0, t) =µ1(t),
¯
W1(L, t) +β1( ¯W1)x(L, t) =µ2(t),
0≤x≤L,
t≥0,
(4.48)
.. .
pk :
D2α
t Wk¯ (x, t) = ∂2¯
Wk(x,t)
∂x2 + ¯Wk(x, t) +Ak−1, ¯
Wk(x,0) = 0,
¯
Wk(0, t) +α1( ¯Wk)x(0, t) =µ1(t),
¯
Wk(L, t) +β1( ¯Wk)x(L, t) =µ2(t),
0≤x≤L,
t≥0,
(4.49)
.. .
where ˜f1(x, t) + ˜f2(x, t) = ˜f(x, t) and Ak, k = 0, 1, . . . are Adomian polyno-mials and are obtained from (4.12). We solve the corresponding homogeneous equation in (4.47) by the method of separation of variables. By assuming
¯
W0(x, t) =X0(x)T0(t) and substituting it in (4.9), we obtain an ordinary lin-ear differential equation forX0(x):
X′′
0(x) +λ2X0(x) = 0, X0(0) +α1X0′(0) = 0, X0(L) +β1X0′(L) = 0.
(4.50)
and a fractional ordinary linear differential equation for T0(t) as (4.14). The Sturm-Liouville problem given by (4.50) has eigenvaluesλ2
n and corresponding
eigenfunctions are
(4.51) (X0)n(x) =−α1λncos(λnx) + sin(λnx), n= 1,2, . . . .
Now we seek a solution for the nonhomogeneous problem in (4.47) of the form
(4.52) W0¯ (x, t) = ∞
∑
n=1
(B0)n(t)(X0)n(x).
Like previous section, in order to determine (B0)n(t), we expand ˜f1(x, t) as a
Fourier series by the eigenfunctions (X0)n(x) as follows
(4.53) f1˜(x, t) = ∞
∑
n=1
( ˜f1)n(t)(X0)n(x).
We know that
(4.54) ( ˜f1)n(t) =
2
L
∫ L
0
˜
By substituting (4.52) and (4.53) into (4.47) we have ∞
∑
n=1
Dα(B0)n(t)(X0)n(x) + (λ2n+ 1)
∞
∑
n=1
(B0)n(t)(X0)n(x)
= ∞
∑
n=1
( ˜f1)n(t)(X0)n(x).
(4.55)
We know that eigenfunctions of Stumr-Lioville equations are orthogonal, so
Dα
t(B0)n(t) + (λ2n−1)(B0)n(t) = ( ˜f1)n(t).
(4.56)
Since ¯W0(x, t) satisfies the initial conditions in (4.47), we have
(4.57)
∞
∑
n=1
(B0)n(0)(X0)n(x) =g(x),
which yields
(4.58) (B0)n(0) =
2
L
∫ L
0
g(x)(X0)n(x)dx.
According to Lemma 2.8, the fractional initial value problem with
µ=α, µ1= 0, m1= 0, λ1=−(1 +λ2
n), m= 1,has the solution
(B0)n(t) =
∫ t
0
τα−1E(α,α)(λ1τα)( ˜f1)
n(t−τ)dτ
(4.59)
+ (B0)n(0)
[
1 +λ1tαE(α,α+1)(λ1tα)].
Hence we get the solution of the initial boundary value problem (4.47) in the form
(4.60) W0¯ (x, t) = ∞
∑
n=1
(B0)n(t) (−α1λncos(λnx) + sin(λnx)).
By applying the same way we can get ¯Wk(x, t).
5. Examples
In this section, we consider three examples with different initial and bound-ary conditions and source term. We show that the solution obtained above agree with those established in these examples.
Example 5.1. Consider the fractional nonlinear Fisher’s equation (4.1) with the initial and Dirichlet boundary conditions
u(x,0) = 2,
u(0,t) = 2,
0≤x≤1,
u(1, t) =t2+ 2, t≥0,
(5.1)
where
f(x, t) = sin(3πx)
(
t6sin(3πx) + 2t5x+ (9π2+ 3)t3+ Γ(4) Γ(4−α)t
3−α
)
+x
(
t4x+ 3t2− Γ(3) Γ(3−α)t
2−α
)
In order to solve this problem, we first transform it into a problems homoge-neous boundary conditions as
u(x, t) =W(x, t) +V(x, t) (5.2)
=W(x, t) +t2x+ 2,
(5.3)
Dα
tW(x, t) = ∂2
W(x,t)
∂x2 +W(x, t) + (W +V)2+ ˜f(x, t),
W(x,0) =, W(0, t) = 0,
0≤x≤1, W(1, t) = 0,
where
˜
f(x, t) = sin(3πx)
(
t6sin(3πx) + 2t5x+ (9π2+ 3)t3+ Γ(4) Γ(4−α)t
3−α
)
+t4x2+ 4t2+ 4.
˜
f(x, t) = sin(3πx)
(
t6sin(3πx) + 2t5x+ (9π2+ 3)t3+ Γ(4) Γ(4−α)t
3−α
)
+t4x2+ 4t2+ 4.
For solving (5.3) we apply MHPM
(5.4) DαtW(x, t) =
∂2W(x, t)
∂x2 +W(x, t)−p(W+V)
2+ ˜f1(x, t) +pf2˜(x, t).
By assuming W(x, t) = ∞
∑
i=0
Wipi, and substituting it in (5.4), we obtain
p0:
Dα
tW0(x, t) =W0(x, t) + ∂2
W0(x,t)
∂x2 + ˜f1(x, t),
W0(x,0) = 2, W0(0, t) = 0,
0≤x≤1, W0(L, t) = 0,
(5.5)
p1:
Dα
tW1(x, t) =W1(x, t) +∂
2
W1(x,t)
∂x2 + ˜f2(x, t) +A0,
W1(x,0) = 0, W1(0, t) = 0,
0≤x≤1, W1(L, t) = 0,
(5.6)
.. .
pk:
Dα
tWk(x, t) =Wk(x, t) + ∂2
Wk(x,t)
∂x2 +Ak−1,
Wk(x,0) = 0, Wk(0, t) = 0,
0≤x≤1, Wk(L, t) = 0,
(5.7)
.. .
where ˜f1(x, t) + ˜f2(x, t) = ˜f(x, t) and
(5.8) f1˜(x, t) = sin(3πx)
(
Γ(4) Γ(4−α)t
3−α
+t3(9π2−1)
)
and
With similar calculation, in subsection 4.1, we obtain a Sturm-Liouville prob-lem and an ordinary linear differential equation respect toxandtrespectively. The eigenvalues and eigenfunctions of the Sturm-Liouville problem are
λn=n2π2, (X0)n(x) = sin(nπx), n= 1,2, . . . .
(5.10)
Furthermore, we have
(f1)n(t) =
2 1
∫ 1 0
˜
f1(x, t) sin(nπx)dx=
H(t), n= 3
0, n̸= 3
with
(5.11) H(t) = Γ(4) Γ(4−α)t
3−α+t3(9π2−1).
So
(B0)n(t) =
∫t 0Eα,α
(
(1−n2π2)τα)H(t−τ)dτ, n= 3
0, n̸= 3.
(5.12)
To evaluate (B0)3(t), we take laplace transform from both side of (5.12): L[(B0)3(t)] =L
[∫ t
0
τα−1E(α,α)((1−9π2)τα)H(t−τ)dτ
]
(5.13)
=L[τα−1E(α,α)((1−9π2)τα)]L[H(t)]
=L
[∞
∑
k=0
1−9π2
Γ(αk+α)t
α(k+1)−1
]
L[H(t)]
=
(∞
∑
k=0
(1−9π2)k sα(k+ 1)
)
L[H(t)]
=
(
1
sα
∞
∑
k=0
(1−9π
2 sα )
k
)
L[H(t)]
= 1
sα
1 1−1−9π2
sα
(
6
s4−α +
9π2−1 s4
)
= 1
sα−(1−9π2)
sα+ 9π2−1 s4
= 6
s4.
From (5.12) and (5.13), we get
(B0)n(t) =
{
t3, n= 3,
0, n̸= 3.
Therefore, the solution for (5.18) is ˜
Again by arguments in section 4.1, we have
(W)i(x, t) = 0, i= 1, 2, . . . .
Then the exact solution for the fractional Fisher’s equation given in Exam-ple 5.1 is
u(x, t) =t3sin(3πx) +t2x+ 2.
Example 5.2. Once again, we consider the fractional Fisher’s equation
(5.14) Dα
tu(x, t) =uxx(x, t) +u(x, t) (1−u(x, t)) +f(x, t),
with the initial and Neumann boundary conditions as
u(x,0) = cos(5πx),
(5.15)
ux(0, t) =t3, ux(1, t) = 2t4+t3, t≥0,
and
f(x, t) = cos(5πx)[ Γ(γ+ 1) Γ(γ+ 1−α)t
γ−α+ 25π2(tγ+ 1)−(tγ+ 1)
+ (tγ+1)2cos(5πx) + 2t3(tγ+1)x+ 2t4(tγ+1)x2]
+ Γ(5) Γ(5−α)t
4−αx2+ Γ(4)
Γ(4−α)t
3−αx−2t4−x2t4−t3x
+t6x2+t8x4+ 2t7x3.
Now, if we assume that
u(x, t) = ˜W(x, t) + ˜V(x, t) = ˜W(x, t) +t4x2+t3x,
we get (5.16)
Dα
tW˜(x, t) = ∂2˜
W(x,t)
∂x2 + ˜W(x, t)−( ˜W(x, t) +t4x2+t3x)2+ ˜f(x, t), ˜
W(x,0) = cos(5πx),
˜
Wx(0, t) = 0, Wx˜ (1, t) = 0, t≥0,
in which
˜
f(x, t) = cos(5πx)
[
Γ(γ+ 1) Γ(γ+ 1−α)t
γ−α
+ (25π2−1)(tγ+ 1)
]
+[(tγ+ 1) cos(5πx) +t4x2+t3x]2.
To solve (5.16) we use MHPM
(5.17) DtαW˜(x, t) =
∂2W˜(x, t)
Therefore if we assume ˜W(x, t) = ∞
∑
i=0
˜
Wipi, and substitute it in (5.17), we
derive
p0:
Dα
tW0˜ (x, t) = ∂2˜
W0(x,t)
∂x2 + ˜W0(x, t) + ˜f1(x, t), ˜
W0(x,0) = cos(3πx),
( ˜W0)x(0, t) = 0,
0≤x≤L,
( ˜W0)x(L, t) = 0,
(5.18)
p1:
Dα
tW1˜ (x, t) = ∂2˜
W1(x,t)
∂x2 + ˜W1(x, t) + ˜f2(x, t) +A0, ˜
W1(x,0) = 0,
( ˜W1)x(0, t) = 0,
0≤x≤L,
( ˜W1)x(L, t) = 0,
(5.19)
.. .
pk:
Dα
tWk˜ (x, t) = ∂2˜
Wk(x,t)
∂x2 + ˜Wk(x, t) +Ak−1, ˜
Wk(x,0) = 0,
( ˜Wk)x(0, t) = 0,
0≤x≤L,
( ˜Wk)x(L, t) = 0,
(5.20)
in which we have ˜f1(x, t) + ˜f2(x, t) = ˜f(x, t) and ˜f1(x, t) must be satisfied in initial and boundary conditions (5.18) [8]. We choose
˜
f1(x, t) = cos(5πx)
[
Γ(γ+ 1) Γ(γ+ 1−α)t
γ−α
+ (25π2−1)(tγ+ 1)
]
,
˜
f2(x, t) =[(tγ+ 1) cos(5πx) +t4x2+t3x]2.
With a similar manner as Example 5.1, we first apply separation method for the corresponding homogeneous equation in (5.18). We obtain the eigenvalue and eigenfunctions of the Sturm-Liouville problem as
λn=n2π2, (X0)n(x) = cos(nπx), n= 1,2, . . . .
(5.21)
By supposing that
(5.22) W0˜ (x, t) = ∞
∑
n=1
(B0)n(t) cos(nπx),
and substituting in (5.18) we derive
Dα
t(B0)n(t) +
[
(nπ)2−1](B0)
n(t) = ( ˜f1)n(t),
(5.23)
so, same as Example 5.1, we have
H(t) = Γ(γ+ 1) Γ(γ+ 1−α)t
γ−α
+ (25π2−1)(tγ+ 1).
Since ˜W0(x, t) satisfies the initial conditions in (5.18), we have
∞
∑
n=1
(B0)n(0) cos(nπx) = cos(5πx),
which gives
(B0)n(0) =
2 1
∫ 1 0
cos(5πx) cos(nπx)dx
(5.25)
=
{
1,
0,
n= 5, n̸= 5.
where
(f1)n(t) =
2 1
∫ 1 0
˜
f1(x, t) cos(nπx)dx
=
{
H(t),
0,
n= 5, n̸= 5.
Furthermore Lemma 2.8 implies that
(B0)n(t) = t
∫
0
τα−1Eα,α((1−25π2)τα)( ˜f1)
n(t−τ)dτ
(5.26)
+ cos(5πx)((B0)n)0(t)
=
t
∫
0
τα−1Eα,α((1−25π2)τα)
{
H(t−τ),
0,
n= 5, n̸= 5, dτ
+ 1 + (1−25π2)tαEα,α+1((1−25π2)tα).
Now, if we take the Laplace transform from both side of (5.26) we obtain
L[(B0)n(t)] = 0, n̸= 5
and
L[(B0)5(t)] =L
t
∫
0
τα−1Eα,α((1−25π2)τα)H(t−τ)dτ
(5.27)
+L[1 + (1−25π2)tαEα,α+1((1−25π2)tα)]
= 1
sα−1 + 25π2 ×L[H(t)] +
1
s
+ (1−25π2)L
[∞
∑
k=0
tα(1−25π2) k
tαk
Γ(αk+α+ 1)
]
= 1
sα−1 + 25π2 ×
(
Γ(γ+ 1)
sγ+1−α+ (25π2−1)( 1 sγ+1 +1s)
)
+1
s+
1−25π2 s(sα−(1−25π2))
= 1
s+
Γ(γ+ 1)
Hence from (5.26) and (5.27), we obtain
(B0)5(t) =
{
tγ+ 1, n= 5,
0, n̸= 5.
Thus, the solution for (5.18) with above Neumann boundary conditions takes the form as
˜
W0(x, t) = (tγ+ 1) cos(5πx).
Hence like as Example 5.1, and by some computational algebra we derive
( ˜W)i(x, t)≡0, i= 1,2, . . . .
Then the analytical solution for the fractional Fisher’s equation with given conditions is as follows:
u(x, t) = (tγ+ 1) cos(5πx) +t4x2+t3x.
Example 5.3. One more time we consider the fractional Fisher’s equation as follows
(5.28) Dαtu(x, t) =uxx(x, t) +u(x, t) (1−u(x, t)) +f(x, t),
with the initial and Robin boundary conditions as
u(x,0) = 0,
u(0, t)−π1ux(0, t) =t3− 1 πt
5, u(1, t)−π1ux(1, t) = (1−π1)t5+t3,
0≤x≤1, t≥0, t≥0,
(5.29)
and
f(x, t) = (cos(πx) + sin(πx))
[
Γ(5) Γ(5−α)t
4−α+π2t4−t4+ 2t9x+ 2t7
]
(5.30)
+ Γ(6) Γ(6−α)t
5−α
x+ Γ(4) Γ(4−α)t
3−α
−t5x−t3
+t8(cos(πx) + sinπx)2
+t10x2+t6+ 2t8x.
Next, by assuming
u(x, t) = ¯W(x, t) + ¯V(x, t)
= ¯W(x, t) +t5x+t3,
we get (5.31)
Dα
tW¯(x, t) = ∂2¯
W(x,t)
∂x2 + ¯W(x, t)−( ˜W(x, t) +t5x+t3)2+ ˜f(x, t), ¯
W(x,0) = 0,
¯
W(0, t)−π1Wx¯ (0, t) = 0,
¯
W(1, t)−π1Wx¯ (1, t) = 0,
0≤x≤1, t≥0, t≥0,
in which
˜
f(x, t) = (cos(πx) + sin(πx))
[
Γ(5) Γ(5−α)t
4−α+π2t4−t4+ 2t9x+ 2t7
]
To solve (5.31) we use MHPM as
(5.32) Dα
tW¯(x, t) =
∂2W¯(x, t)
∂x2 + ¯W(x, t) +ph( ¯W+ ¯V) + ˜f1(x, t) +pf2˜(x, t).
Therefore if we assume ˜W(x, t) = ∞
∑
i=0
˜
Wipi, and substitute it in (5.32), we
obtain
p0:
Dα
tW0¯ (x, t) = ∂2¯
W0(x,t)
∂x2 + ¯W0(x, t) + ˜f1(x, t), ¯
W0(x,0) = 0,
¯
W0(0, t)−1
π( ¯W0)x(0, t) = 0,
¯
W0(1, t)−1
π( ¯W0)x(1, t) = 0,
0≤x≤L, t≥0, t≥0.
(5.33)
p1:
Dα
tW1¯ (x, t) = ∂2¯
W1(x,t)
∂x2 + ¯W1(x, t) + ˜f2(x, t) +A0, ¯
W1(x,0) = 0,
¯
W1(0, t)−1
π( ¯W1)x(0, t) = 0,
¯
W1(1, t)−1
π( ¯W1)x(1, t) = 0,
0≤x≤L, t≥0, t≥0.
(5.34)
.. .
pk:
Dα
tWk¯ (x, t) = ∂2¯
Wk(x,t)
∂x2 + ¯Wk(x, t) +Ak−1, ¯
Wk(x,0) = 0,
¯
Wk(0, t)−1π( ¯Wk)x(0, t) = 0,
¯
Wk(1, t)−1π( ¯Wk)x(1, t) = 0,
0≤x≤L, t≥0, t≥0,
(5.35)
in which we have ˜f1(x, t) + ˜f2(x, t) = ˜f(x, t) and ˜f1(x, t) must be satisfied in initial and boundary conditions (5.33) [8]. Here,
˜
f1(x, t) = (cos(πx) + sin(πx))
[
Γ(5) Γ(5−α)t
4−α
+π2t4−t4
]
.
With a similar manner as Examples 5.1 and 5.2, we use separation method for the corresponding homogeneous equation in (5.33). We obtain the eigen-value and eigenfunction of the Sturm-Liouville problem as follows
λn =n2π2, (X0)n(x) = cos(nπx) + sin(nπx), n= 1,2, . . . .
(5.36)
By assuming that
(5.37) W0¯ (x, t) = ∞
∑
n=1
(B0)n(t) (cos(nπx) + sin(nπx)),
and substituting in (5.33) we derive
Dtα(B0)n(t) +
[
(nπ)2−1](B0)n(t) = ( ˜f1)n(t),
(5.38)
so, like as Examples 5.1 and 5.2, we have
H(t) = Γ(5) Γ(5−α)t
4−α
Since ¯W0(x, t) satisfies the initial conditions in (5.33), we have ∞
∑
n=1
(B0)n(0)(cos(nπx) + sin(nπx)) = cos(πx) + sin(nπx),
(5.39)
which gives
(B0)n(0) =
2 1
∫ 1 0
(cos(πx) + sin(πx)) (cos(nπx) + sin(nπx))dx
(5.40)
=
{
1,
0,
n= 1, n̸= 1,
where
( ˜f1)n(t) =
2 1
∫ 1 0
˜
f1(x, t) (cos(nπx) + sin(nπx))dx
=
{
H(t),
0,
n= 1, n= 1.
Furthermore Lemma 2.8 implies that
(B0)n(t) = t
∫
0
τα−1Eα,α((1−π2)τα)( ˜f1)
n(t−τ)dτ
(5.41)
=
t
∫
0
τα−1Eα,α((1−π2)τα)
{
H(t−τ),
0,
n= 1, n= 1, dτ.
Now, if we take the Laplace transform from both side of (5.41) we obtain
L[(B0)n(t)] = 0, n̸= 1
and
L[(B0)1(t)] =
24
sα−1 +π2×
sα+π2−1 s5
(5.42)
= 24
s5.
Thus from (5.41) and (5.42), we obtain
(B0)1(t) =
{
t4, n= 1,
0, n̸= 1.
Hence, the solution for (5.33) with above Robin boundary conditions takes the form as
¯
W0(x, t) =t4(cos(πx) + sin(πx)).
Therefore, we derive
( ¯W2)i(x, t)≡0, i= 1,2, . . . .
Then the analytical solution for the fractional Fisher’s equation with given conditions is as follows:
6. Conclusion
In this article, we obtained analytical solutions for the time-fractional Fisher’s nonlinear differential equation. We showed that by choosing proper functions
˜
f1and ˜f2, the solution can be obtained only in one iteration of MHPM. Finally, we illustrated the effectiveness of this method by some examples.
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1
Faculty of Mathematical Sciences, University of Tabriz, Tabriz, Iran.
E-mail address: h-kheiri@tabrizu.ac.ir
2
Faculty of Mathematical Sciences, University of Tabriz, Tabriz, Iran.
E-mail address: aida mojaver1987@yahoo.com
3
Faculty of Mathematical Sciences, University of Tabriz, Tabriz, Iran.
