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3.3 Space-time fractional foam drainage equation
Let us consider the following space-time fractional foam drainage equation
,
0
,
1
,
0
,
2
2
1
2
2
2
2
t
x
V
x
V
V
t
V
V
t
V
(41)
It has been studied by many authors [15, 34]. This equation have appeared as a simple model for
describing the flow of liquid through channels ( Plateau borders [35] ) and nodes ( intersection
of four channels) between the bubbles, driven by gravity and capillarity [36].
Introducing the fractional complex transformation
,
)
1
(
)
1
(
),
(
)
,
(
ct
kx
V
t
x
V
(42)
where
k
and
c
are constants in eq. (41), Eq. (41) can be transformed as
,
0
2
2
1
2
2
2
2
V
k
V
kV
V
V
k
V
c
(43)
According to the balancing principle, the solution of (43) can be written as
)
(
1
0
)
(
e
A
A
V
(44)
where
1
0
,
A
A
are constants to be determined later and
)
(
satisfies the nonlinear ODE (7).
By substituting eq. (44) into eq. (43) and using (7) commonly, the left-hand side of eq. (43)
becomes a polynomial in
)
(
e
. Collecting the coefficients of this polynomial to zero yields a
system of algebraic equations in terms of
c
r
q
p
k
A
A
,
,
,
,
,
,
1
0
. The algebraic equations are
overlooked for convenience. Solving the resulting algebraic equations with aid of symbolic
computation, such as Maple, one get
2
3
3
1
0
4
1
,
,
,
2
r
k
pq
k
c
k
k
kp
A
r
k
A
(45)
where
k,
q
p
,
and
r
are arbitrary constants.
By combining the equations (9), (42), (44) and (45), the space-time fractional foam drainage
equation (41) has the following traveling wave solutions:
For Type 1:
,
0
4
,
0
,
))
(
4
5
.
0
tanh(
4
2
2
)
,
(
2
0
2
2
1
q
r
q
r
q
r
q
r
q
k
r
k
t
x
V
(46)
,
0
4
,
0
,
))
(
4
5
.
0
tan(
4
2
2
)
,
(
2
0
2
2
2
q
r
q
r
r
q
r
q
q
k
r
k
t
x
V
(47)
,
0
4
,
0
,
1
))
(
exp(
2
)
,
(
2
0
3
q
r
q
r
r
k
r
k
t
x
V
(48)
,
0
4
,
0
,
0
,
4
))
(
2
)
(
2
)
,
(
2
0
0
2
4
q
r
r
q
r
r
k
r
k
t
x
V
(49)
where,
t
r
k
q
k
kx
)
1
(
4
4
)
1
(
2
3
3
.
For Type 2:
,
0
,
0
,
)
1
(
)
1
(
tan
)
,
(
0
3
5
q
p
t
pq
k
kx
pq
pq
k
t
x
V
(50)
,
0
,
0
,
)
1
(
)
1
(
cot
)
,
(
0
3
6
q
p
t
pq
k
kx
pq
pq
k
t
x
V
(51)
0
,
0
,
)
1
(
)
1
(
tanh
)
,
(
0
3
7
q
p
t
pq
k
kx
pq
pq
k
t
x
V
,
(52)
,
0
,
0
,
)
1
(
)
1
(
coth
)
,
(
0
3
8
q
p
t
pq
k
kx
pq
pq
k
t
x
V
(53)
For Type 3:
,
0
,
0
,
)
1
(
4
4
)
1
(
)
,
(
1
0
2
3
3
9
r
q
t
r
k
q
k
kx
k
t
x
V
(54)
3.4 Time fractional fifth order Sawada-Kotera (SK) equation
Finally, to illustrate more applicability of this proposed method, the following time fractional SK
equation is considered:
.
1
0
,
0
5
5
5
5
5
3
3
2
2
2
x
U
x
U
U
x
U
x
U
x
U
U
t
U
(55)
The time fractional equation (55) appeared as a model equation in many physical instances that
propagates in opposite directions. To obtain the analytical solutions of Eq. (55), one can
introduce the following fractional complex transformation:
],
)
1
(
[
),
(
)
,
(
t
c
x
k
w
t
x
U
(56)
where
k
and
c
are constants to be evaluated later. According to the transformation (55), one can
be obtained to the following nonlinear ODE:
.
0
)
(
5
)
(
3
5
5
5
5
2
2
3
3
d
w
d
k
d
w
d
w
d
d
k
w
d
d
k
d
dw
kc
(57)
Integrating the Eq. (57) once and setting the constant of integration to zero for simplicity, one get
.
0
5
3
5
4
4
4
2
2
2
3
d
w
d
k
d
w
d
w
k
w
cw
(58)
According to the balancing principle between
4
4
d
w
d
and
3
w
that are involved in eq. (58), one
obtain
.
2
N
Therefore the solution of (58) according to the proposed method can be written as
)
(
2
2
)
(
1
0
)
(
e
A
e
A
A
w
(59)
where
1
0
,
A
A
and
2
A
are constants to be determined later. Substituting Eq. (59) into Eq. (58) and
collecting the degree of the polynomial in
)
(
e
yields a system of algebraic nonlinear equations
which are omitted for simplicity. Solving the resulting algebraic equations, one obtains
Set 1:
2
2
2
2
1
2
0
2
2
4
2
4
6
,
6
,
6
),
16
8
(
p
k
A
pr
k
A
pq
k
A
q
p
r
pqr
k
c
(60)
Set2:
,
6
,
6
,
,
52
2
11
)
10
2
5
(
2
2
2
2
1
2
0
2
2
4
4
4
2
4
4
2
4
p
k
A
pr
k
A
k
A
q
p
k
r
k
qp
r
k
pq
k
r
k
c
(61)
where
20
840
1680
105
)
15
60
(
2
2
2
4
2
pqr
q
p
r
r
pq
,
q
p
,
and
r
are arbitrary constants.
Therefore, the time fractional SK equation according to the
Set 1
has the following traveling
wave solutions:
For Type 1:
,
0
,
0
4
,
))
(
4
5
.
0
tanh(
4
6
))
(
4
5
.
0
tanh(
4
6
6
)
,
(
2
2
0
2
2
2
0
2
2
2
2
1
1
q
q
r
r
q
r
q
r
q
k
r
q
r
q
r
q
r
k
q
k
t
x
U
(62)
,
0
,
0
4
,
))
(
)
4
(
5
.
0
tan(
)
4
(
2
6
))
(
)
4
(
5
.
0
tan(
)
4
(
6
6
)
,
(
2
2
0
2
2
2
0
2
2
2
2
1
2
q
r
r
q
r
q
r
q
k
r
q
r
q
r
q
r
k
q
k
t
x
U
(63)
0
)
4
(
,
0
,
0
,
1
6
1
6
)
,
(
2
2
))
(
(
2
))
(
(
2
2
1
0
0
3
q
r
r
q
e
r
k
e
r
k
t
x
U
r
r
(64)
,
0
4
,
0
,
0
,
4
))
(
2
)
(
6
4
))
(
2
)
(
6
6
)
,
(
2
2
0
0
2
2
0
0
3
2
2
1
4
q
r
r
q
r
r
k
r
r
k
q
k
t
x
U
(65)
where,
.
)
1
(
)
16
8
(
2
4
2
4
t
q
r
qr
k
x
k
For Type 2:
,
0
,
0
)),
(
(
tan
6
6
)
,
(
0
2
2
2
1
5
q
p
pq
pq
k
pq
k
t
x
U
(66)
,
0
,
0
)),
(
(
cot
6
6
)
,
(
0
2
2
2
1
6
q
p
pq
pq
k
pq
k
t
x
U
(67)
,
0
,
0
)),
(
(
tanh
6
6
)
,
(
0
2
2
2
1
7
q
p
pq
pq
k
pq
k
t
x
U
(68)
,
0
,
0
)),
(
(
coth
6
6
)
,
(
0
2
2
2
1
8
q
p
pq
pq
k
pq
k
t
x
U
(69)
where,
.
)
1
(
)
16
(
2
2
4
t
q
p
k
x
k
The other obtained solutions of Eq. (55) according to
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