Consider the following nonlinear PDE:
F(u,∂αu∂tα,∂βu∂xβ,∂2αu∂t2α,∂2βu∂x2β,.....)=0, 0<α,β≤1
(3.1)
Where F is a polynomial in u(x,t) and its partial derivatives, in which the highest order derivatives and the nonlinear terms are involved. In the following, we give the main steps of this method:
Step 1: We use the conformable space-time wave transformation:
u(x,t)=u(ξ), ξ=xββ−c1tαα,
(3.2)
Where
c1
is a constant, to reduce Equation (3.1) to the following ODE:
P(u,u',u'',.....)=0,
(3.3)
Where
P
is a polynomial in
u(ξ)
and its total derivatives, such that
'=ddξ
.
Step 2: We assume that Equation (3.3) has the formal solution:
u(ξ)=a0+N∑i=1fi−1(ξ)[aif(ξ)+big(ξ)],
(3.4)
Where
a0
,
ai
,
bi
(i=1,.....,N)
are constants to be determined later, such that
aN≠0
or
bN≠0,
while
f(ξ)
and
g(ξ)
satisfy the auxiliary ODEs:
f'(ξ)=f(ξ)g(ξ),
(3.5)
g'(ξ)=q+g2(ξ)+rf−2(ξ),
(3.6)
g2(ξ)=−[q+r2f−2(ξ)+cf2(ξ)]
(3.7)
Where
q
,
r
and
c
are constants.
Step 3: We determine the positive integer
N
in (3.4) by using the homogeneous balance between the highest order derivatives and the nonlinear terms in Equation (3.3). More precisely, we define the degree of
u(ξ)
as
D[u(ξ)]=N
, which gives rise to the degree of other expressions as follows:
D[up1(ξ)(dq1u(ξ)dξq1)s1]=NP1+s1(q1+N).
(3.8)
From (3.8) we can get the value of
N
in (3.4). In some nonlinear equations, the balance number
N
is not a positive integer. In this case, we make the following transformations:
(a) When
N=q1p1,
where
q1p1
is a fraction in the lowest terms, we let
u(ξ)=[v(ξ)]q1p1,
(3.9)
When
N
is a negative number, we let
u(ξ)=[v(ξ)]N,
(3.10)
And substitute (3.9) or (3.10) into Equation (3.3) to get a new equation in terms of the function
v(ξ)
with a positive integer balance number.
Step 4: We substitute (3.4) along with (3.5)-(3.7) into Equation (3.3) and collect all terms of the same order of
fi(ξ)
gj(ξ)
(i,j=0,1,.....)
and set them to zero, yield a set of algebraic equations which can be solved by using the Maple or Mathematical to find
a0
,
ai
,
bi
,
c1
,
q
,
r
,
c
.
Step 5: It is well-known40 that (3.5), (3.6) have the following Jacobi elliptic function solutions:
- If q=(1+m2), r=−2m2, c=−1, then
f1(ξ)=1sn(ξ,m) , g1(ξ)=−cn(ξ,m)dn(ξ,m)sn(ξ,m).
- If q=(1−2m2), r=2m2, c=(m2−1), then
f2(ξ)=1cn(ξ,m) , g2(ξ)=sn(ξ,m)dn(ξ,m)cn(ξ,m).
- If q=(−2+m2), r=2, c=(1−m2),
then
f3(ξ)=1dn(ξ,m), g3(ξ)=m2sn(ξ,m)cn(ξ,m)dn(ξ,m).
- If q=(1+m2), r=−2, c=−m2, then
f4(ξ)=sn(ξ,m) , g4(ξ)=cn(ξ,m)dn(ξ,m)sn(ξ,m) .
- If q=(1−2m2), r=(−2+2m2),c=m2,
then
f5(ξ)=cn(ξ,m), g5(ξ)=−sn(ξ,m)dn(ξ,m)cn(ξ,m).
- If q=(−2+m2), r=(2−2m2), c=1,
then
f6(ξ)=dn(ξ,m), g6(ξ)=−m2sn(ξ,m)cn(ξ,m)dn(ξ,m).
- If q=(−2+m2), r=(−2+2m2), c=−1,
then
f7(ξ)=cn(ξ,m)sn(ξ,m), g7(ξ)=−dn(ξ,m)sn(ξ,m)cn(ξ,m).
- If q=(1−2m2), r=(2m2−2m4), c=−1, then
f8(ξ)=dn(ξ,m)sn(ξ,m), g8(ξ)=−cn(ξ,m)sn(ξ,m)dn(ξ,m).
- If q=(−2+m2), r=−2, c=(−1+m2),
then
f9(ξ)=sn(ξ,m)cn(ξ,m), g9(ξ)=dn(ξ,m)sn(ξ,m)cn(ξ,m).
- If q=(1+m2), r=−2m2, c=−1,then
f10(ξ)=dn(ξ,m)cn(ξ,m), g10(ξ)=(1−m2)sn(ξ,m)cn(ξ,m)dn(ξ,m).
- If q=(1−2m2), r=−2, c=(m2−m4),then
f11(ξ)=sn(ξ,m)dn(ξ,m), g11(ξ)=cn(ξ,m)sn(ξ,m)dn(ξ,m).
- If q=12(−1+2m2), r=−12, c=−14,then
f12(ξ)=cn(ξ,m)±1sn(ξ,m) , g12(ξ)=∓ds(ξ,m).
- If q=−12(m2+1), r=12(1−m2), c=14(1−m2), then
f13(ξ)=dn(ξ,m)msn(ξ,m)±1, g13(ξ)=∓mcd(ξ,m).
- If q=−12(1+m2)
,r=12(m2−1)2
, c=14,
then
f14(ξ)=mcn(ξ,m)+dn(ξ,m) , g14(ξ)=−msn(ξ,m).
- If q=−12(1+m2), r=−12(m2−1)2,c=−14, then
f15(ξ)=cn(ξ,m)±dn(ξ,m)sn(ξ,m) , g15(ξ)=∓ns(ξ,m).
- If q=(m2−6m+1), r=−2,c=4m (m−1)2,then
f16(ξ)=sn(ξ,m)msn2(ξ,m)−1 , g16(ξ)=−cs(ξ,m)dn(ξ,m)[msn2(ξ,m)+1msn2(ξ,m)−1].
- If q=(m2+6m+1), r=−2, c=−4m,(m+1)2,
then
f17(ξ)=sn(ξ,m)msn2(ξ,m)+1 , g17(ξ)=−cs(ξ,m)dn(ξ,m)[msn2(ξ,m)−1msn2(ξ,m)+1].
- If q=−12(1+m2), r=12(m2−1),c=14(m2−1),then
f18(ξ)=cn(ξ,m)sn(ξ,m)±1 , g18(ξ)=∓dc(ξ,m).
- If q=12(2−m2), r=−m42,c=−14,then
f19(ξ)=dn(ξ,m)±1sn(ξ,m) , g19(ξ)=∓cs(ξ,m).
- If q=12(2m2−1), r=−12, c=−14,then
f20(ξ)=sn(ξ,m)1±cn(ξ,m) , g20(ξ)=±ds(ξ,m).
- If q=−12(1+m2), r=−12, c=−14(m2−1)2,
then
f21(ξ)=sn(ξ,m)cn(ξ,m)±dn(ξ,m) , g21(ξ)=±ns(ξ,m).
Where
sn(ξ,m),
cn(ξ,m)
and
dn(ξ,m)
are Jacobi elliptic sine function, Jacobi elliptic cosine function, Jacobi elliptic function of the third kind respectively, and
m
denotes the modulus of Jacobi elliptic functions, where
0≤m≤1
. It is well-known54−56 that the Jacobi-elliptic functions satisfy the following relations:
cn2(ξ,m)=1−sn2(ξ,m),dn2(ξ,m)=1−m2sn2(ξ,m),
sn'(ξ,m)=cn(ξ,m)dn(ξ,m),cn'(ξ,m)=−sn(ξ,m)dn(ξ,m),
dn'(ξ,m)=−m2sn(ξ,m)cn(ξ,m).
ns(ξ,m)=1sn(ξ,m),nc(ξ,m)=1cn(ξ,m),nd(ξ,m)=1dn(ξ,m),
sc(ξ,m)=sn(ξ,m)cn(ξ,m),sd(ξ,m)=sn(ξ,m)dn(ξ,m),cs(ξ,m)=cn(ξ,m)sn(ξ,m),
cd(ξ,m)=cn(ξ,m)dn(ξ,m),ds(ξ,m)=dn(ξ,m)sn(ξ,m),dc(ξ,m)=dn(ξ,m)cn(ξ,m),
The Jacobi elliptic functions degenerate into hyperbolic functions when
m→1
as follows:
sn(ξ,1)→tanh(ξ),cn(ξ,1)→sech(ξ),dn(ξ,1)→sech(ξ),ns(ξ,1)→coth(ξ),dc(ξ,1)→1,
ds(ξ,1)→cosech(ξ),sc(ξ,1)→sinh(ξ),sd(ξ,1)→sinh(ξ),cs(ξ,1)→cosech(ξ),
And into trigonometric functions when
m→0
as follows:
sn(ξ,0)→sin(ξ),cn(ξ,0)→cos(ξ),dn(ξ,0)→1,ns(ξ,0)→cosec(ξ),cs(ξ,0)→cot(ξ),
ds(ξ,0)→cosec(ξ),sc(ξ,0)→tan(ξ),sd(ξ,0)→sin(ξ),dc(ξ,0)→sec(ξ)
Step 6: We substitute the values
a0
,
and the solutions (1)−(21) given in step 5 into (3.4), to get the exact solutions of Equation (3.1).
In this section, we apply the method of section 3, to solve Equation (1.1).To this aim, we first use the conformable space-time wave transformation:
(4.1)
Where
is a non zero constant and
, to reduce Equation (1.1) into the following nonlinear ordinary differential equation (ODE):
(4.2)
Integrating Equation (4.2) twice with respect to
and vanshing the constants of integration, we get
(4.3)
Balancing
with
, we get
. According to (3.9) we use the transformation:
(4.4)
Where
is a new function of
, to reduce Equation (4.3) into the new ODE:
(4.5)
Balancing
with
in Equation (4.5), we get
. According to the form (3.4), Equation (4.5) has the formal solution:
(4.6)
Where
,
,
are constants to be determined, such that
or
. Substituting (4.6) along with (3.5)−(3.7) into Equation (4.5) and collecting all terms of the same order of
,
and setting them to zero, we have the following algebraic equations:
According to Step5 of Section 3, we have the following results:
Result 1: If we substitute
into the algebraic equations (4.7) and use the Maple, we have
(4.8)
or
(4.9)
Substituting (4.8) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(4.10)
Where
. If
, then we have the periodic solution:
(4.11)
while if
, then we have the solitary wave solution:
(4.12)
Substituting (4.9) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(4.13)
where
. If
, then we have the singular soliton solution:
(4.14)
Result 2: If we substitute
into the algebraic equations (4.7) and use the Maple, we have
(4.15)
Substituting (4.15) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(4.16)
where
If
, then we have the periodic solution:
(4.17)
Result 3: If we substitute
into the algebraic equations (4.7) and use the Maple, we have
(4.18)
Substituting (4.18) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(4.19)
where
If
, then we have the dark soliton solution:
(4.20)
Result 4: If we substitute
into the algebraic equations (4.7) and use the Maple, we have
(4.21)
Substituting (4.21) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(4.22)
where
If
, then we have the dark soliton solution:
(4.23)
Result 5: If we substitute
into the algebraic equations (4.7) and use the Maple, we have
(4.24)
Substituting (4.24) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(4.25)
where
or
If
, then we have the dark soliton solution:
(4.26)
Result 6: If we substitute
into the algebraic equations (4.7) and use the Maple, we have
(4.27)
Substituting (4.27) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(4.28)
where
If
, then we have the bright soliton solution:
(4.29)
Result 7: If we substitute
into the algebraic equations (4.7) and use the Maple, we have
(4.30)
Substituting (4.30) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(4.31)
where
If
, then we have the same solitary wave solution (4.14).
Result 8. If we substitute
into the algebraic equations (4.7) and use the Maple, we have
(4.32)
Substituting (4.32) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(4.33)
where
If
, then we have the periodic solution:
(4.34)
Result 9: If we substitute
into the algebraic equations (4.7) and use the Maple, we have
(4.35)
Substituting (4.35) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(4.36)
where
If
, then we have the singular soliton solution:
(4.37)
while if
, then we have the periodic solution:
(4.38)
Result 10: If we substitute
into the algebraic equations (4.7) and use the Maple, we have
(4.39)
Substituting (4.39) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(4.40)
where
If
, then we have the periodic solution:
(4.41)
Result 11: If we substitute
into the algebraic equations (4.7) and use the Maple, we have
(4.42)
Substituting (4.42) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(4.43)
If
, then we have the same periodic solution (4.11), while if
, then we have the same singular soliton solution (4.37).
Result 12: If we substitute
into the algebraic equations (4.7) and use the Maple, we have:
(4.44)
Substituting (4.44) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(3.45)
where
If
, then we have the same periodic solution (4.34).
Result 13: If we substitute
into the algebraic equations (4.7) and use the Maple, we have
(4.46)
Substituting (4.46) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(4.47)
where
Result 14: If we substitute
into the algebraic equations (4.7) and use the Maple, we have
(4.48)
Substituting (4.48) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(4.49)
where
If
, then we have the same dark soliton solution (4.26).
Result 15: If we substitute
into the algebraic equations (4.7) and use the Maple, we have
(4.50)
Substituting (4.50) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(4.51)
where
If
then we have the singular soliton solution:
(4.52)
while if
then we have the periodic solutions.
(4.53)
Result 16: If we substitute
into the algebraic equations (4.7) and use the Maple, we have
(4.54)
Substituting (4.54) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(4.55)
wher
If
, then we have the same periodic solution (4.11), while if
then we have the solitary wave solutions:
(4.56)
Result 17: If we substitute
into the algebraic equations (4.7) and use the Maple, we have
(4.57)
Substituting (4.57) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(4.58)
where
If
, then we have the solitary wave solution:
(4.59)
Result 18: If we substitute
into the algebraic equations (4.7) and use the Maple, we have
(4.60)
Substituting (4.60) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(4.61)
where
If
, then we have the same periodic solutions (4.17) and (4.41) respectively.
Result 19: If we substitute
into the algebraic equations (4.7) and use the Maple, we have
(4.62)
Substituting (4.62) into (4.4), (4.6),we have the Jacobi elliptic solutions:
(4.63)
where
If
, then we have the solitary wave solutions:
(4.64)
Result 20: If we substitute
into the algebraic equations (4.7) and use the Maple, we have
(4.65)
Substituting (4.65) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(4.66)
where
If
, then we have the solitary wave solution:
(4.67)
while if
, then we have the periodic solution:
(4.68)
Result 21: If we substitute
into the algebraic equations (4.7) and use the Maple, we have
(4.69)
Substituting (4.69) into (4.4), (4.6), we have the Jacobi elliptic solutions:
(4.70)
where
If
, then we have the singular soliton solution:
(4.71)
while if
, then we have the periodic solution:
(4.72)