In this paper, we get many new analytical solutions of the space-time nonlinear fractional modified KDV-Zakharov Kuznetsov (mKDV-ZK) equation by means of a new approach namely method of undetermined coefficients based on a fractional complex transform. These solutions have physics meanings in natural sciences. This method can be used to other nonlinear fractional differential equations.
Nonlinear fractional differential equations (NFDEs) are universal models of the classical differential equations of integer order. In recent years, the fractional order derivative and integral is becoming a hot spot of international research; it can more accurately describe the nonlinear phenomena in physics. Such as chemical kinematics, chemical physics and geochemistry, communication, phy- sics, biology, engineering, mathematics, diffusion processes in porous media, in vibrations in a nonlinear string, power-law non-locality, and power-law long- term memory can use NFDEs as models to express these problem [
The present article aims to find out the modified KDV-Zakharov Kuznetsov [
Fractional calculus is a generalization of classical calculus. There are a lot of approaches developed over years to generalize the concept of fractional order derivative, such as, Riemann-Liouville, Grünwald-Letnikow, Caputo [
In the section, the some properties and definitions of the modified Riemann- Liouville derivative that will be applied in the sequel of the work were given.
The following is the modified Riemann-Liouville derivative defined by Jumarie [
D t α f ( t ) = { 1 Γ ( − α ) ∫ 0 t ( t − ξ ) − α − 1 [ f ( ξ ) − f ( 0 ) ] d ξ , α < 0 , 1 Γ ( 1 − α ) d d t ∫ 0 t ( t − ξ ) − α [ f ( ξ ) − f ( 0 ) ] d ξ , 0 < α < 1 , ( f ( n ) ( t ) ) ( α − n ) , n ≤ α ≤ n + 1 , n ≥ 1. (1)
Remark1. f : R → R , t → f ( t ) denote a continuous but not necessarily differentiable function.
The probability calculus, fractional Laplace problems, and fractional varia- tional calculus successfully applied Jumarie’s modified Riemann-Liouville derivative. To summarize a few useful formulae by Jumarie’s modified Riemann- Liouville derivative in [
D t α t γ = Γ ( γ + 1 ) Γ ( γ + 1 − α ) t γ − α , γ > 0 , (2)
D t α ( c f ( t ) ) = c D t α f ( t ) , c = constant , (3)
D t α f [ g ( t ) ] = f ′ g [ g ( t ) ] D t α g ( t ) , (4)
D t α f [ g ( t ) ] = D g α f [ g ( t ) ] ( g ′ ) α , (5)
D t α [ a f ( t ) + b g ( t ) ] = a D t α f ( t ) + b D t α g ( t ) . (6)
Remark 2. J. H. He et al. in [
D t α f [ g ( t ) ] = σ ′ t f ′ g [ g ( t ) ] D t α g ( t ) , (7)
where σ ′ t is called the sigma indexes (see [
D t α f [ g ( t ) ] = σ ′ t D g α f [ g ( t ) ] ( g ′ ) α . (8)
In the section, we introduce the generally steps of method of undetermined coefficients
Step 1: We set a nonlinear fractional order partial differential equation as follows
P ( u , D t α u , D x β u , D t α D t α u , D t α D x β u , D x β D x β u , ⋯ ) = 0 , 0 < α , β < 1 (9)
where u is an unknown function about x , t two independent variables, D t α u , D x α u modified Riemann-Liouville derivative of u , and P is a polyno- mial of u and its partial fractional derivatives, in which includes the highest order derivatives and the nonlinear terms.
Step 2: By using the traveling wave transformation
u ( x , t ) = U ( ξ ) , ξ = k x β Γ ( β + 1 ) − c t α Γ ( α + 1 ) , (10)
where k and c are non zero arbitrary constants. And by using the chain rule
D t α u = σ ′ t d U d ξ D t α ξ , D x α u = σ ′ x d U d ξ D x α ξ , (11)
where σ ′ t and σ ′ x are called the sigma index. The sigma index usually is determined by gamma function [
Substituting (10) along with (2) and (11) into (9), we can rewrite Equation (9) in the following nonlinear ordinary differential equation
Q ( U , U ′ , U ″ , U ‴ , ⋯ ) = 0 , (12)
where the prime denotes the derivative with respect to ξ . For the convenience of calculation, we should obtain a new equation by integrating Equation (12) term by term one or more times.
Step 3: By the following form [
U ( ξ ) = A sech m ξ , (13)
where A is nonzero constant, m is obtained by balancing the highest order term and nonlinear term of Equation (9) or Equation (12).
Step 4: Substituting the constant A and m into Equation (14), we can obtain the solution of the fractional order Equation (9).
In this current sub-section, we apply method of undetermined coefficients to solve the (3 + 1) dimensional space-time fractional mKDV-ZK equation of the form,
D t α u + d u 2 u x + e u x x x + f u x y y + g u x z z = 0 , t > 0 , 0 < α < 1 , (14)
where d , e , f and g are nonzero constants, α is a parameter describing the order of the fractional space-time-derivative. When f = 0 , g = 0 , d , e ≠ 0 , Equation (14) is called the fractional modified KDV equation
D t α u + d u 2 u x + e u x x x = 0 , t > 0 , 0 < α < 1 , (15)
when α = 1 , Equation (14) is called the modified KDV-ZK equation
u t + d u 2 u x + e u x x x + f u x y y + g u x z z = 0 , t > 0. (16)
The modified KDV-ZK equation is applied in many physical areas. Existence of the solutions for this equation has been considered in several papers, see references in [
Therefore, we use the following transformations,
u ( x , y , z , t ) = U ( ξ ) , ξ = k x + p y + q z − λ t α Γ ( 1 + α ) , (17)
Where k , p , q , λ are nonzero constants.
Substituting Equation (17) with Equation (2) and Equation (11) into Equation (14), we have
− λ U ′ + k d U 2 U ′ + k 3 e U ‴ + k f p 2 U ‴ + k g q 2 U ‴ = 0 , (18)
where “ U ′ ” = d U d ξ . By once integrating and setting the constants of integration
to zero, we obtain
− λ U + k d 3 U 3 + k ( e k 2 + f p 2 + g q 2 ) U ″ = 0. (19)
To get the non-topological soliton solution of Equation (19), we can make the assumption,
U ( ξ ) = A sech m ξ , (20)
where
ξ = k x + p y + q z − λ t α Γ ( 1 + α ) , (21)
where k , p , q , λ are nonzero constants coefficients. The m is unknown at this point and will be determined later. From the Equation (20)-(21), we obtain
d U ( ξ ) d ξ = − A m sech m + 1 ξ sinh ξ , (22)
and
d 2 U ( ξ ) d ξ 2 = − A m ( m + 1 ) sech m ξ ( − sech ξ tanh ξ ) sinh ξ + ( − A m sech m + 1 ξ cosh ξ ) = A m ( m + 1 ) sech m + 2 ξ sinh 2 ξ − A m sech m ξ = A m ( m + 1 ) sech m + 2 ξ ( 1 sec h 2 ξ − 1 ) − A m sech m ξ = A m 2 sech m ξ − A m ( m + 1 ) sech m + 2 ξ , (23)
and
U 3 ( ξ ) = A 3 sech 3 m ξ . (24)
Thus, substituting the ansatz (23)-(27) into Equation (21), yields to
− λ A sech m ξ + k d 3 A 3 sech 3 m ξ + k ( e k 2 + f p 2 + g q 2 ) ( A m 2 sech m ξ − A m ( m + 1 ) sech m + 2 ξ ) = 0. (25)
Now, from Equation (25), equating the exponents m + 2 and 3 m leads to
m + 2 = 3 m , (26)
so that
m = 1. (27)
From Equation (25), setting the coefficients of sech m + 2 ξ and sech 3 m ξ terms to zero, we obtain
k d 3 A 3 − A k ( e k 2 + f p 2 + g q 2 ) m ( m + 1 ) = 0 , (28)
by using Equation (27) and after some calculations, we have
A = ± 6 ( e k 2 + f p 2 + g q 2 ) d . (29)
We find, from setting the coefficients of sech m ξ terms in Equation (25) to zero
− λ A + A m 2 k ( e k 2 + f p 2 + g q 2 ) = 0 , (30)
also we get
λ = k ( e k 2 + f p 2 + g q 2 ) . (31)
From Equation (29), it is important to note that
d ( e k 2 + f p 2 + g q 2 ) > 0. (32)
Thus finally, the 1-soliton solution of Equation (14) is given by:
u 1 ( x , y , z , t ) = 6 ( e k 2 + f p 2 + g q 2 ) d sech ( k x + p y + q z − d ( e k 2 + f p 2 + g q 2 ) t α Γ ( 1 + α ) ) , (33)
u 2 ( x , y , z , t ) = − 6 ( e k 2 + f p 2 + g q 2 ) d sech ( k x + p y + q z − d ( e k 2 + f p 2 + g q 2 ) t α Γ ( 1 + α ) ) . (34)
In order to start off with the solution hypothesis, we use the solitary wave ansatz of the form
U ( ξ ) = A tanh m ξ , (35)
and
ξ = k x + p y + q z − λ t α Γ ( 1 + α ) , (36)
where k , p , q , λ are the free parameters. Also the m is unknown at this point and will be determined later.
From Equations (35)-(36), we obtain
d U ( ξ ) d ξ = A m ( tanh m − 1 ξ − tanh m + 1 ξ ) (37)
and
d 2 U ( ξ ) d ξ 2 = A m { ( m − 1 ) tanh m − 2 ξ − 2 m tanh m ξ + ( m + 1 ) tanh m + 2 ξ } , (38)
and
U 3 ( ξ ) = A 3 tanh 3 m ξ . (39)
Substituting Equations (35)-(39) into Equation (19), gives
− λ A tanh m ξ + k d 3 A 3 tanh 3 m ξ + A m k ( e k 2 + f p 2 + g q 2 ) { ( m − 1 ) tanh m − 2 ξ − 2 m tanh m ξ + ( m + 1 ) tanh m + 2 ξ } = 0. (40)
Now, from Equation (40), equating the exponents of tanh 3 m ξ and tanh m + 2 ξ gives,
3 m = m + 2 , (41)
which yields
m = 1. (42)
Setting the coefficients of tanh 3 m ξ and tanh m + 2 ξ terms in Equation (40) to zero, we have
k d 3 A 3 + A k m ( m + 1 ) ( e k 2 + f p 2 + g q 2 ) = 0 , (43)
then, we get
A = ± − 6 ( e k 2 + f p 2 + g q 2 ) d . (44)
Again, from Equation (40) setting the coefficients of tanh m ξ terms to zero,
− λ A − 2 m 2 A k ( e k 2 + f p 2 + g q 2 ) = 0 , (45)
and from Equation (45) we have
λ = − 2 k ( e k 2 + f p 2 + g q 2 ) . (46)
Equation (46) prompts the constraint
d ( e k 2 + f p 2 + g q 2 ) < 0. (47)
Thus finally, the dark soliton solution for the (3 + 1) dimensional space-time fractional mKDV-ZK equation is given by:
u 3 ( x , y , z , t ) = − 6 ( e k 2 + f p 2 + g q 2 ) d tanh ( k x + p y + q z + 2 k ( e k 2 + f p 2 + g q 2 ) Γ ( 1 + α ) ) , (48)
u 4 ( x , y , z , t ) = − − 6 ( e k 2 + f p 2 + g q 2 ) d tanh ( k x + p y + q z + 2 k ( e k 2 + f p 2 + g q 2 ) Γ ( 1 + α ) ) . (49)
In this article, we have got the new solutions for the (3 + 1) dimensional space- time fractional mKDV-ZK equation by using the method of undetermined coefficients. Up to now, we could not find that these solutions were reported in other papers. In order to solve many systems of nonlinear fractional partial differential equations in mathematical and physical sciences, such as, the space-time fractional mBBM equation, the time fractional mKDV equation, the nonlinear fractional Zoomeron equation and so on, we can use the method of undetermined coefficients recommended herein would be general to a certain extent.
This work is in part supported by the Natural Science Foundation of China (Grant Nos. 11271008, 61072147).
Jin, Q.Y., Xia, T.C. and Wang, J.B. (2017) The Exact Solution of the Space-Time Fractional Modified Kdv- Zakharov-Kuznetsov Equation. Journal of Applied Mathematics and Physics, 5, 844- 852. https://doi.org/10.4236/jamp.2017.54074