Travelling wave solutions of (2 1)-dimensional generalised

Preparing to load PDF file. please wait...

0 of 0
100%
Travelling wave solutions of (2 1)-dimensional generalised

Transcript Of Travelling wave solutions of (2 1)-dimensional generalised

Pramana – J. Phys. (2018) 90:34 https://doi.org/10.1007/s12043-018-1522-4

© Indian Academy of Sciences

Travelling wave solutions of (2+1)-dimensional generalised time-fractional Hirota equation

YOUWEI ZHANG
School of Mathematics and Statistics, Hexi University, Beihuan Road, 846, Zhangye 734000, China E-mail: [email protected]

MS received 24 July 2017; revised 28 September 2017; accepted 10 October 2017; published online 9 February 2018
Abstract. In this article, we have developed new exact analytical solutions of a nonlinear evolution equation that appear in mathematical physics, a (2 + 1)-dimensional generalised time-fractional Hirota equation, which describes the wave propagation in an erbium-doped nonlinear fibre with higher-order dispersion. By virtue of the tanh-expansion and complete discrimination system by means of fractional complex transform, travelling wave solutions are derived. Wave interaction for the wave propagation strength and angle of field quantity under the long wave limit are analysed: Bell-shape solitons are found and it is found that the complex transform coefficient in the system affects the direction of the wave propagation, patterns of the soliton interaction, distance and direction.
Keywords. Time-fractional Hirota equation; fractional complex transform; complete discrimination system; tanhexpansion; travelling wave.
PACS Nos 02.30.Jr; 05.45.Yv; 04.20.Jb

1. Introduction
We consider the solution of the (2 + 1)-dimensional generalised time-fractional Hirota equation
i ∂tαu + uxy + i uxxx + uv − i |u|2ux = 0, (1)
3vx + (|u|2)y = 0,
where 0 < α ≤ 1, u = u(x, y, t), v = v(x, y, t) are two field variables, (x, y) ∈ ( ⊆ R2) are spatial coordinates in propagation direction of the fields and t ∈ T (= [0, t0] (t0 > 0)) are temporal coordinates, which occur in different contexts in mathematical physics, ∂tα is a modified Riemann–Liouville fractional derivative. The subscripts denote the partial differentiation of the functions u, v with respect to x, y and t.
Hirota equation [1], a typical model of mathematical physics, which encompasses the well-known Schrödinger equation and modified KdV equation, especially contains the nonlinear derivative Schrödinger–KdV equation [2–14]. In plasma physics, Hirota equation shows interaction of large-amplitude lower-hybrid waves with finite-frequency density perturbations. It takes into account higher-order dispersion and timedelay corrections to the cubic nonlinearity, describing

wave propagation in ocean and in optical fibres. Furthermore, it accurately describes evolution of light in a two-mode fibre with defocussing Kerr effect and some high-order effects. The equation can be used for studying travelling wave and condensed mathematical physics and other scientific applications [15–22].
Fractional calculus has attracted much attention in recent years due to their numerous applications: the fluid dynamic traffic model with fractional derivatives can eliminate the deficiency arising from the assumption of continuum traffic flow [23], the nonlinear oscillation of earthquake can be modelled with fractional derivatives [24]. Other applications in finance, physics and engineering can be seen in [25–29] and the references therein. Finding a new mathematical algorithm to construct exact solution of nonlinear fractional-order evolution equations is important and may have significant impact on future research. Many effective methods have been proposed for fractional calculus: Using the homotopy analysis, Bakkyaraj and Sahadevan [30] considered two coupled time-fractional Hirota equations and constructed their periodic wave solution and approximate solitary wave solution. Cui et al [31] presented exact solutions of generalised Hirota–Satsuma coupled Korteweg–de Vries equations by an improved extended

34 Page 2 of 12

Pramana – J. Phys. (2018) 90:34

tan-function, combines fractional complex transform calculus and various definitions of fractional integration

and extended tanh-function. Based on a relatively analytical technique [32], Ganji et al solved nonlinear partial differential equations of fractional order, then

and differentiation [40–42]. Now we state the definition and some important properties for the modified Riemann–Liouville derivative of order α as follows.

pointed out that this method can be used as an alternative to obtain analytic and approximate solutions of differ- DEFINITION 1

ent types of fractional differential equations which when applied in engineering mathematics, the corresponding solutions of the integer order equations are found to follow as special cases of fractional-order equations (also see [33]). In [34], the fractional differential transformation which can be applied to many complicated

A real multivariable function f (x, t), t > 0 is said to be
in the space Cγ , γ ∈ R with respect to t if there exists a real number p (>γ ), such that f (x, t) = t p f1(x, t), where f1(x, t) ∈ C( × T ). Obviously, Cγ ⊂ Cδ if δ ≤ γ.

nonlinear partial differential equations and does not

require linearisation, discretisation, restrictive assump- DEFINITION 2 tions or perturbation is implemented to the solution

of time-fractional generalised Hirota–Satsuma coupled KdV with a number of initial and boundary values has been proved. Mirzazadeh [35] developed He’s semiinverse variational principle method and ansatz method to construct exact solutions of fractional Klein–Gordon equation and generalised Hirota–Satsuma coupled KdV system. Eslami et al [36] obtained the exact solutions of nonlinear fractional generalised reaction Duffing

Assume f (x, t) ∈ Cγ ( × T ) (γ ≥ −1). We use the following equality for the integral with respect to (dt)α

Itα f (x, t) =

1

t
(t − τ )α−1 f (x, τ )dτ

(α) 0

=1

t
f (x, τ )(dτ )α, 0 < α < 1.

(α + 1) 0

model and nonlinear fractional diffusion reaction equation with quadratic and cubic nonlinearity by first DEFINITION 3

integral approach [37–39]. The above results have basically proved that the fractional calculus theory is nonconservative among most of the physical phenomena.

Jumarie’s derivative for multivariate function f (x, t) ∈ Cγ ( × T ) (γ ≥ −1) is defined as

⎧ ⎪⎪⎪⎪⎪⎨ ∂tα f (x, t) = ⎪⎪ ⎪⎪⎪⎩

1



t
(t − τ )−α−1( f (x, τ ) −

(−α) ∂t 0

1



t
(t − τ )−α( f (x, τ ) −

(1 − α) ∂t 0

∂ α−n

(n)

∂tα−n f (x, t) , n ≤ α < n + 1,

f (x, 0))dτ, f (x, 0))dτ, n ≥ 1.

α < 0, 0 < α < 1,

This paper is organised as follows. Section 2 states some background material from fractional calculus. Section 3 presents the algorithm of tanh-expansion and complete discrimination system methods for determining the solution of (2 + 1)-dimensional generalised time-fractional Hirota equation. The implementation of the proposed methods for establishing the exact solutions of eq. (1) are proposed in §4 and 5. The description of solutions for the obtained graphs are presented in §6. In §7, brief conclusions of the study are given.
2. Preliminaries
Fractional calculus is defined as the generalisation of classical calculus and it can be extended to complex set. Many mathematicians developed fractional

The fundamental mathematical operations and results of Jumarie’s derivative are given as follows:

∂αc = 0, ∂αtβ = (1 + β) tβ−α,

t

t

(1 + β − α)

∂tα(c f (x, t)) = c∂tα f (x, t), (dt)α = tα,

where c is a constant, 0 < α, β < 1. Leibnitz’ formula of the fractional Riemann–Liouville differential takes the form

α

∞ α α−n

n

∂t f (x, t)g(x, t) = n ∂t f (x, t)∂t g(x, t),

n=0

where α (−1)n−1α (n − α) n = (1 − α) (n + 1) .

Pramana – J. Phys. (2018) 90:34

Page 3 of 12 34

The generalisation of the well-known chain rule for the composite function is given as

m

m k k1

r ∂m gk−r dk f (g)

∂t f (g(x, t)) =

r k! (−g) ∂tm dgk .

k=0 r =0

where ak (k = 0, 1, . . . , n) are some undetermined constants, the integer n can be determined by balanc-
ing the highest-order derivative term and nonlinear term appearing in eq. (4), Z = tanh(ξ ) is a new independent variable. Then we can find the derivative with respect to ξ as follows:

3. Analytical methods

3.1 Tanh-expansion

The tanh-expansion [43,44] provides an effective procedure for explicit and travelling wave solutions of a wide and general class of differential systems representing real physical problems. Moreover, the foregoing restrictions and limitations of approximate techniques are overcome so that it provides us with a possibility to analyse strongly nonlinear evolution equations. Therefore, we extend this method to solve the (2 + 1)-dimensional generalised time-fractional Hirota equation, and the basic features of the tanh-expansion are outlined as follows.
Assume that a nonlinear time-fractional partial differential equation in three independent variables x, y ∈ and t ∈ T is given as

F(u, ∂tαu, ux , u y, uxx , uxy, u yy, uxxx , . . .) = 0,

0 < α ≤ 1.

(2)

Using the fractional complex transform [45–48], we have

u(x, y, t) = U = U (ξ ),

ξ = λx + μy − ν tα,

(3)

(α + 1)

where λ, μ, ν are undetermined constants. By using the chain rule formula for composite function, we have

∂tαu = σt Uξ ∂tαξ,

where σt is the fractal index. Without loss of generality we can take σt = κ, where κ is a constant. Then eq. (2) is reduced to a nonlinear ordinary differential equation
for U

F(U, −νU , λU , μU , λ2U , λμU , μ2U ,

λ3U , . . .)

= 0,

(4)

where = d/dξ . Assume that the solution of eq. (4) can be written by
a polynomial in Z given as

n

U = ak Z k,

(5)

k=0

dU = (1 − Z 2) dU ,



dZ

d2U = (1 − Z 2)2 d2U − 2Z (1 − Z 2) dU ,

dξ 2

dZ2

dZ

d3U = (1 − Z 2)3 d3U − 6Z (1 − Z 2)2 d2U

dξ 3

dZ3

dZ2

+ 2Z (1 − Z 2)(3Z 2 − 1) dU , dZ

··· .

(6)

By replacing eq. (5) in eq. (4) and using eq. (6) fol-
lowed by bringing together all the like terms with the same degree of Z k (k = 0, 1, 2, . . .), eq. (4) is changed into a polynomial in Z k (k = 0, 1, 2, . . .). Equating
every coefficient of this polynomial to zero yields a set of algebraic equations for ak (k = 0, 1, . . . , n) and λ, μ, ν.
By solving the algebraic equations obtained and substituting these constants ak (k = 0, 1, . . . , n), λ, μ, ν in eq. (5), we can obtain the explicit solution of eq. (2)
instantly.

3.2 Complete discrimination system

Using complete discrimination system [49–51] for the polynomial, classifications of all single travelling wave solutions to nonlinear differential equations, such as mKdV equation, sine-Gordon equation, Fujimoto– Watanabe equation, coupled Harry–Dym equation, coupled KdV equation, etc. are obtained. In this section, we have modified the complete discrimination system to show how to construct the hierarchy of travelling wave solutions of the (2 + 1)-dimensional generalised time-fractional Hirota equation, and presented the classification of explicit forms for the two lower-order solutions. Each one is a regular (non-singular) wave solution with a single maximum that can describe a rogue wave in this model, and numerical simulations reveal the appearance of these solutions in a chaotic field generated from a perturbed continuous wave solution.
Assume that the nonlinear time-fractional partial differential equations (2) can be reduced to a nonlinear constant coefficient ordinary differential equation with variable ξ under the fractional complex transform (3)

(U )2 = a1U 2 + a2U + a3,

(7)

34 Page 4 of 12

Pramana – J. Phys. (2018) 90:34

where ak (k = 1, 2, 3) are constants, and eq. (7) can be written as

dU

ξ − ξ0 = ±

.

(8)

a1U 2 + a2U + a3

Set G(U ) = a1U 2 + a2U + a3, = a22 − 4a1a3, then the solution of (8) has the following cases:
Case 1. = 0. There exist double real roots of the equation G(U ) = 0,

a2 2

G(U ) = a1 U +

,

2a1

then the solution of eq. (7) is U (ξ ) = ± exp(±√a1(ξ − ξ0)) − a2 ,
2a1

(a1 > 0).

Case 2. > 0. There exist two different real roots of the equation G(U ) = 0,

a2 2

G(U ) = a1 U +

−,

2a1

4a1

then the solutions of eq. (7) are

U1(ξ ) = ± 1 exp(±√a1(ξ − ξ0)) 2

+ exp(∓√a1(ξ − ξ0)) ∓ a2 , (a1 > 0),

8a12

2a1

1√



U2(ξ ) = − 2a1 (± sin(± −a1(ξ − ξ0)) + a2),

(a1 < 0).

Case 3. < 0. There exist no real roots of the equation G(U ) = 0, and the solution of eq. (7) can be given as

U (ξ ) = ± 1 exp(±√a1(ξ − ξ0)) 2

+ exp(∓√a1(ξ − ξ0)) ∓ a2 ,

8a12

2a1

(a1 > 0).

For (2 + 1)-dimensional generalised time-fractional Hirota equation (1), we provide a fractional complex transform

u = exp(iη)U (ξ ), v = V (ξ ),

(9)

where η = λ1x + μ1 y − ξ = λ2x + μ2 y −

ν1 tα, (α + 1)
ν2 tα. (α + 1)

4. Implementation of tanh-expansion for the exact solutions of (2 + 1)-dimensional generalised time-fractional Hirota equation

In this section, we use the tanh-expansion to determine the new exact solutions for (2 + 1)-dimensional generalised time-fractional Hirota equation. By applying the transform (9), eq. (1) can be reduced to the following form:

λ32U − λ2U 2U + (−ν2 + λ1μ2 + λ2μ1 − 3λ21λ2)U

= 0,

(10)

(λ2μ2 − 3λ1λ22)U + (ν1 − λ1μ1 + λ31)U + U V

+ λ1U 3 = 0,

(11)

V = − μ2 U 2 + c,

(12)

3λ2

where c is an integral constant. Integrating eq. (10), and substituting (12) into (11), we have

λ32U − λ32 U 3 + (−ν2 + λ1μ2 + λ2μ1

− 3λ21λ2)U = d,

(13)

(λ2μ2 − 3λ1λ2)U + λ1 − μ2 U 3 + (ν1 − λ1μ1

2

3λ2

+ λ31 + c)U = 0.

(14)

For the compatibility of eqs (13) and (14), let

d = 0,

c = μ2 − 3λ1λ2 −ν2 + λ1μ2 + λ2μ1 − 3λ2λ2

λ22

1

− ν1 + λ1μ1 − λ31.

(15)

Under condition (15), eqs (13) and (14) are consistent. Further, eq. (13) yields

U 2 = 1 U 4 − −ν2 + λ1μ2 + λ2μ1 − 3λ21λ2 U 2

6λ22

λ32

+ c1,

(16)

where c1 is a constant. For solving eq. (16), we set

U = a0 +

n k=1

ak

Z

k

.

By

balancing

the

highest-order

derivative term and nonlinear term in eq. (16), the value

of n can be determined, which is n = 1 in this problem,

U = a0 + a1 Z .

(17)

In view of (6), we have

U = 1 − Z 2 a1.

(18)

Substituting (17) and (18) into eq. (16), then collecting all the like terms with the same degree of Z k
(k = 0, 1, 2, . . .), we can obtain a system of algebraic

Pramana – J. Phys. (2018) 90:34

equations for ak (k = 0, 1, 2, . . .), λm = λm, μm = μm, νm = νm (m = 1, 2) as follows:

Z 0 : 1 a4 + ν2 − λ1μ2 − λ2μ1 + 3λ21λ2 a2

6λ22 0

λ32

0

− a12 + c1 = 0,

Z 1: 32λ2 a03a1 + 2 ν2 − λ1μ2 −λλ32μ1 + 3λ21λ2 a0a1

2

2

= 0,

Z 2: 1 a2a2 + ν2 − λ1μ2 − λ2μ1 + 3λ21λ2 + 2a2

λ22 0 1

λ32

1

= 0,

Z 3: 2 a0a3 = 0, 3λ22 1
Z 4: 1 a4 − a2 = 0. 6λ22 1 1

Solving these algebraic equations, we have the sets of
coefficients for the solutions of eq. (17) under the compatibility condition 2λ2(μ2−3λ1λ2)−ν1+λ1μ1−λ31 = 1 as given below:

λm = λm, μm = μm, νm = νm (m = 1, 2),

a0 = 0, a1 = ±1,

and so we have the solutions to eq. (1) as

u(x, y, t) = ± exp i λ1x + μ1 y − ν1 tα (α + 1)
× tanh λ2x + μ2 y − ν2 tα , (α + 1)
v(x, y, t) = − μ2 tanh2 λ2x + μ2 y 3λ2
− ν2 tα + 1. (α + 1)

5. Implementation of complete discrimination system for the exact solutions of (2 + 1)-dimensional generalised time-fractional Hirota equation

In this section, we use the complete discrimination system to determine new travelling wave solutions for (2 + 1)-dimensional generalised time-fractional Hirota equation. By applying the transform (9), eq. (1) can be reduced to the following form:

λ32U − λ2U 2U + (−ν2 + λ1μ2 + λ2μ1

− 3λ21λ2)U = 0,

(19)

(λ2μ2 − 3λ1λ22)U + (ν1 − λ1μ1 + λ31)U + U V

+ λ1U 3 = 0,

(20)

Page 5 of 12 34

V = − μ2 U 2 + c,

(21)

3λ2

where c is an integral constant. Integrating eq. (19), and substituting (21) into (20), we have

λ32U − λ32 U 3 + (−ν2 + λ1μ2 + λ2μ1 − 3λ21λ2)U

= d,

(22)

(λ2μ2 − 3λ1λ22)U + + λ31 + c)U = 0.

λ1 − μ2 3λ2

U 3 + (ν1 − λ1μ1 (23)

For the compatibility of eqs (22) and (23), let

d = 0,

c = μ2 − 3λ1λ2 (−ν2 + λ1μ2 + λ2μ1 − 3λ2λ2)

λ22

1

−ν1 + λ1μ1 − λ31.

(24)

Under condition (24), eqs (22) and (23) are consistent, then eq. (22) yields

U 2 = 1 U 4 − −ν2 + λ1μ2 + λ2μ1 − 3λ21λ2 U 2

6λ22

λ32

+ c1,

(25)

where c1 is a constant. For solving eq. (25), we set

U = 4 6λ22W (ζ ),

ζ = 1 ξ, 4 6λ22

then eq. (25) becomes

Wζ2 = W 4 + b1W 2 + b2,

(26)

where

b1 = − −ν2 + λ1μ2 +λ3λ2μ1 − 3λ21λ2
2

6λ22,

b2 = c1.

Set ϕ = W 2, substituting into (26), it yields

ϕζ2 = 4ϕ(ϕ2 + b1ϕ + b2).

(27)

Integrating eq. (27), we have

ζ − ζ0 = ± 1 √ dϕ , (28) 2 ϕ H (ϕ)

where H (ϕ) = ϕ2 + b1ϕ + b2, ζ0 is an integral constant. Set ˜ = b12 − 4b2 is a discrimination for the secondorder polynomial of H (ϕ). Thus, the solutions of eq. (28) have the following four cases:

Case 1. ˜ = 0. As ϕ > 0, then

1



ζ − ζ0 = ± 2

√ϕ(ϕ

+

b1

. )

(29)

2

34 Page 6 of 12

When b1 < 0, eq. (29) reads as

1 ζ − ζ0 = ±
2

√√

2 ln √2ϕ − √−b1 .

−b1

2ϕ + −b1

Corresponding solutions of eq. (27) are

ϕ(ζ ) = − b1 tanh2 2
ϕ(ζ ) = − b1 coth2 2

− b1 (ζ − ζ0) , 2
− b1 (ζ − ζ0) . 2

Thus, the exact solutions to eq. (1) are

u (x, y, t) = ± 3(−ν2 + λ1μ2 + λ2μ1 − 3λ21λ2)

1

λ2

× exp i λ1x + μ1 y − ν1 tα (α + 1)

× tanh −ν2 + λ1μ2 + λ2μ1 − 3λ21λ2 2λ32
× λ2x + μ2 y − ν2 tα − ξ0 , (α + 1)
v1(x, y, t) = − μ2(−ν2 + λ1μ2λ+2 λ2μ1 − 3λ21λ2)
2
× tanh2 −ν2 + λ1μ2 + λ2μ1 −3λ21λ2 2λ32
× λ2x + μ2 y − ν2 tα − ξ0 (α + 1)
+ μ2 − 3λ1λ2 (−ν2 + λ1μ2 + λ2μ1 λ22
− 3λ21λ2) − ν1 + λ1μ1 − λ31,

u (x, y, t) = ± 3(−ν2 + λ1μ2 + λ2μ1 − 3λ21λ2)

2

λ2

× exp i λ1x + μ1 y − ν1 tα (α + 1)

× coth −ν2 + λ1μ2 + λ2μ1 −3λ21λ2 2λ32
× λ2x + μ2 y − ν2 tα − ξ0 , (α + 1)
v2(x, y, t) = − μ2(−ν2 + λ1μ2λ+2 λ2μ1 − 3λ21λ2)
2
× coth2 −ν2 + λ1μ2 + λ2μ1 −3λ21λ2 2λ32
× λ2x + μ2 y − ν2 tα − ξ0 (α + 1)

Pramana – J. Phys. (2018) 90:34
+ μ2 − 3λ1λ2 (−ν2 + λ1μ2 + λ2μ1 λ22
− 3λ21λ2) − ν1 + λ1μ1 − λ31. When b1 > 0, eq. (29) becomes

ζ − ζ0 = ∓ 2 arctan 2ϕ .

b1

b1

Corresponding solution of eq. (27) is

ϕ(ζ ) = b1 tan2 2

b1 (ζ − ζ0) . 2

Thus, the exact solutions to eq. (1) are

u (x, y, t) = ± 3(−ν2 + λ1μ2 + λ2μ1 − 3λ21λ2)

3

λ2

× exp i λ1x + μ1 y − ν1 tα (α + 1)

× tan ν2 − λ1μ2 − λ2μ1 + 3λ21λ2 2λ32
× λ2x + μ2 y − ν2 tα − ξ0 , (α + 1)
v3(x, y, t) = μ2(−ν2 + λ1μ2λ+2 λ2μ1 − 3λ21λ2)
2
× tan2 ν2 − λ1μ2 − λ2μ1 + 3λ21λ2 2λ32
× λ2x + μ2 y − ν2 tα − ξ0 (α + 1)
+ μ2 − 3λ1λ2 (−ν2 + λ1μ2 + λ2μ1 λ22
− 3λ21λ2) − ν1 + λ1μ1 − λ31.

When b1 = 0, eq. (29) becomes

1 ζ − ζ0 = ∓ √ϕ .

Corresponding solution of eq. (27) is

1 ϕ(ζ ) = (ζ − ζ0)2 . Thus, the exact solutions to eq. (1) are

u4(x, y, t) = ± 6λ22 exp i λ1x + μ1 y − (αν+1 1) tα 1 × λ2x + μ2 y − ν2 tα − ξ0 ,
(α+1)

Pramana – J. Phys. (2018) 90:34

v4(x, y, t)

=−

2λ2μ2

+ μ2 − 3λ1λ2

λ2x + μ2 y − (αν+2 1) t α − ξ0 2

λ22

× (−ν2 + λ1μ2 + λ2μ1 − 3λ21λ2)

− ν1 + λ1μ1 − λ31.

Case 2. ˜ > 0 and b2 = 0. Since ϕ > −b1, then eq. (28) becomes

ζ − ζ0 = ± 1 √dϕ . (30) 2 ϕ ϕ + b1

When b1 > 0, eq. (30) becomes

ζ − ζ0 = ± 1 2





2 ln √2(ϕ + b1) − √b1 .

b1

2(ϕ + b1) + b1

Corresponding solutions of eq. (27) are

ϕ(ζ ) = b1 tanh2 2

b1 (ζ − ζ0) − b1, 2

ϕ(ζ ) = b1 coth2 2

b1 (ζ − ζ0) − b1. 2

Thus, the exact solutions to eq. (1) are

u5(x, y, t)

= ± 6(−ν2 + λ1μ2 + λ2μ1 − 3λ21λ2) λ2
× exp i λ1x + μ1 y − ν1 tα (α + 1)

× 1 tanh2 2

ν2 − λ1μ2 − λ2μ1 + 3λ21λ2 2λ32

× λ2x + μ2 y − ν2 tα − ξ0 (α + 1)

1/2
−1 ,

v5(x, y, t)

= 2μ2(−ν2 + λ1μ2 + λ2μ1 − 3λ21λ2) λ22

× 1 tanh2 2

ν2 − λ1μ2 − λ2μ1 + 3λ21λ2 2λ32

× λ2x + μ2 y − ν2 tα − ξ0 − 1 (α + 1)

+ μ2 − 3λ1λ2 (−ν2 + λ1μ2 + λ2μ1 − 3λ2λ2)

λ22

1

− ν1 + λ1μ1 − λ31,

u6(x, y, t)

= ± 6(−ν2 + λ1μ2 + λ2μ1 − 3λ21λ2) λ2

Page 7 of 12 34

× exp i λ1x + μ1 y − ν1 tα (α + 1)

× 1 coth2 2

ν2 − λ1μ2 − λ2μ1 + 3λ21λ2 2λ32

× λ2x + μ2 y − ν2 tα − ξ0 (α + 1)

1/2
−1 ,

v6(x, y, t)

= 2μ2(−ν2 + λ1μ2 + λ2μ1 − 3λ21λ2) λ22

× 1 coth2 2

ν2 − λ1μ2 − λ2μ1 + 3λ21λ2 2λ32

× λ2x + μ2 y − ν2 tα − ξ0 − 1 (α + 1)

+ μ2 − 3λ1λ2 (−ν2 + λ1μ2 + λ2μ1 − 3λ2λ2)

λ22

1

− ν1 + λ1μ1 − λ31.

When b1 < 0, eq. (30) becomes

ζ − ζ0 = ∓

2 arctan

2(ϕ + b1) .

−b1

−b1

Corresponding solution of eq. (27) is

ϕ(ζ ) = − b1 tan2 2

− b1 (ζ − ζ0) − b1. 2

Thus, the exact solutions to eq. (1) are

u7(x, y, t)

= ± 6(−ν2 + λ1μ2 + λ2μ1 − 3λ21λ2) λ2
× exp i λ1x + μ1 y − ν1 tα (α + 1)

× 1 tan2 2

−ν2 + λ1μ2 + λ2μ1 − 3λ21λ2 2λ32

× λ2x + μ2 y − ν2 tα − ξ0 (α + 1)

1/2
+1 ,

v7(x, y, t)

= − 2μ2(−ν2 + λ1μ2 + λ2μ1 − 3λ21λ2) λ22

× 1 tan2 2

−ν2 + λ1μ2 + λ2μ1 − 3λ21λ2 2λ32

× λ2x + μ2 y − ν2 tα − ξ0 + 1 (α + 1)

34 Page 8 of 12

+ μ2 − 3λ1λ2 (−ν2 + λ1μ2 + λ2μ1 − 3λ2λ2)

λ22

1

− ν1 + λ1μ1 − λ31.

Case 3. ˜ > 0 and b2 = 0. Let ω1 < ω2 < ω3, and between ω1, ω2, ω3, one equals to zero, others are the root of H (ϕ) = 0. Since ω1 < W < ω2, set ϕ = ω1 + (ω2 − ω1) sin2 φ, by eq. (28),

1 ζ − ζ0 = ± √

dφ , (31)

ω3 − ω1

1 − m2 sin2 φ

where Jacobi’ elliptic function modulus [52]

m = ω2 − ω1 . ω3 − ω1
Taking into account the elliptic function and eq. (31), then we obtain the solution of eq. (27) as

ϕ(ζ ) = ω1 + (ω2 − ω1)

× sn2

√ ± ω3 − ω1(ζ − ζ0), m .

(32)

If W > ω3, then we set

−ω2 sin2 φ + ω3

ϕ=

cos2 φ ,

substituting into eq. (28), there has the solution of eq.

(27)

−ω2sn2(±√ω3 − ω1(ζ − ζ0), m) + ω3

ϕ(ζ ) =

cn2(±√ω3 − ω1(ζ − ζ0), m) .

Thus, the exact solutions to eq. (1) are

u8(x, y, t)

= ± 4 6λ22 exp i λ1x + μ1 y − (αν+1 1) tα

×

ω1 + (ω2 − ω1)sn2

√ ± ω3 − ω1

1

4 6λ22

× λ2x + μ2 y − (αν+2 1) tα − ξ0 , m 1/2,

v8(x, y, t)

= − μ2 3λ2
× sn2

6λ22 ω1 + (ω2 − ω1) ± √ω3 − ω1 1 λ2x + μ2 y
4 6λ22

− ν2 tα − ξ0 , m , (α + 1)

Pramana – J. Phys. (2018) 90:34

u9(x, y, t) = ± 4 6λ22 exp i
× − ω2sn2

λ1x + μ1 y − ν1 tα (α + 1)

± √ω3 − ω1 1 4 6λ22

λ2x + μ2 y

− ν2 tα − ξ0 , m + ω3 (α + 1)

cn2

√ ± ω3 − ω1

1

λ2x + μ2 y

4 6λ22

− (αν+2 1) tα − ξ0 , m 1/2,

v9(x, y, t) = − μ2 6λ2 3λ2 2

×

− ω2sn2

√ ± ω3 − ω1

1

4 6λ22

λ2x + μ2 y

− ν2 tα − ξ0 , m + ω3 (α + 1)

cn2

√ ± ω3 − ω1

1

λ2x + μ2 y

4 6λ22

− ν2 tα − ξ0 , m . (α + 1)

Case 4. ˜ < 0. As ϕ > 0, let

ϕ = b2 tan2 φ .

(33)

2

Substituting (33) into eq. (28) yields

1 ζ − ζ0 = ± √
2 4 b2

dφ , 1 − m2 sin2 φ

where the modulus

m = 1 1 − √b1 .

2

2 b2

√ In view of cn 2 4 b2(ζ − ζ0), m have
√ cos φ = 2 √b2 − 1.
ϕ + b2

= cos φ and (32), we

Then, we obtain the solution of eq. (27) as

ϕ(ζ ) =



1 − cn(2 4 b2(ζ − ζ0), m)

b2



.

1 + cn(2 4 b2(ζ − ζ0), m)

Pramana – J. Phys. (2018) 90:34

Thus, the exact solutions to eq. (1) are

u10(x, y, t) = ± 4 6b2λ22 exp i λ1x + μ1 y −

ν1 t α (α + 1)

1−cn

2 4 6bλ22

λ2x +μ2 y −

ν2 (α+1

)

t

α



ξ

0

,m

×2 ,

1+cn

2 4 6bλ22

λ2x +μ2 y −

ν2 (α+1

)

t

α



ξ

0

,m

2

v10(x, y, t) = − μ2 6b2λ2

3λ2

2

1−cn

24

b2
2

λ2x +μ2 y −

6λ2

×

1+cn

24

b2
2

λ2x + μ2 y −

6λ2

(αν+2 1) t α − ξ0 , m (αν+2 1) t α − ξ0 , m

+ μ2 − 3λ1λ2 (−ν2 + λ1μ2 + λ2μ1 − 3λ2λ2)

λ22

1

− ν1 + λ1μ1 − λ31.

|u(x,y=0,t)|

|u(x,y=0,t)|

Page 9 of 12 34

1

0.8

0.6

0.4

0.2

0 2

1.5

4

1

2

t

0.5

0

-2

x

0 -4

(α=0.98)

1

0.8

0.6

0.4

0.2

0 2

1.5

4

1

2

t

0.5

0

-2

x

0 -4

(α=0.90)

6. The description for solutions of time-fractional Hirota equation

The aim of this section is to explore the effect of fractional-order derivative on the structure and propagation of the resulting travelling waves obtained from fractional Hirota equation [53]. In §4 and 5, our derivations concerning the solution of the time-fractional Hirota equation are carried out by taking into account the values of some meaningful values to fractional order, namely, α = 0.98, 0.90, and so on. Due to the presence of complex variable solutions for u, which may be useful for describing certain nonlinear physical phenomena in fluid, we apply the strength and angle to deal with the properties of these travelling wave solutions, such as

|u10(x, y, t)| = 4 6b2λ22

1−cn

2 4 6bλ22

λ2x +μ2 y −

ν2 (α+1

)

t

α



ξ

0

,m

×2 ,

1+cn

2 4 6bλ22

λ2x +μ2 y −

ν2 (α+1

)

t

α



ξ

0

,m

2

Arg(u10(x, y, t)) = arctan λ1x + μ1 y − (k = 0, ±1, ±2, . . .).

ν1 t α (α + 1)

+ 2kπ,

v(x,y=0,t)

0.95 0.9
0.85 0.8
0.75 0.7
0.65

2

1.5

4

1

2

t

0.5

0

-2

x

0 -4

(α=0.98)

v(x,y=0,t)

0.95 0.9
0.85 0.8
0.75 0.7
0.65

2

1.5

4

1

2

t

0.5

0

-2

x

0 -4

(α=0.90)

Figure 1. The tanh-expansion surfaces of the travelling wave solutions for |u(x, y, t)| and v(x, y, t) for different values of α.

Three-dimensional representation of the solutions |u|, v to the time-fractional Hirota equation for different

34 Page 10 of 12

Pramana – J. Phys. (2018) 90:34

arg(u(x,y=0,t))

1.5

1

0.5

0

-0.5

-1

-1.5 2

1.5

10

1

t

0.5

-5

0 -10

5 0
x

(α=0.98)

|u10(x,y=0,t)|

2

1.5

1

0.5

0 2

1.5

4

1

2

t

0.5

0

-2

x

0 -4

(α=0.98)

arg(u(x,y=0,t))

1.5

1

0.5

0

-0.5

-1

-1.5 2

1.5

10

1

t

0.5

-5

0 -10

5 0
x

(α=0.90)

Figure 2. The tanh-expansion angle surfaces of the travelling wave solution for u(x, y, t) for different values of α.

values of α are presented in figure 1. It shows that the solution |u| is still a single soliton wave solution for different values of α, the balancing scenario between nonlinearity and dispersion are still valid via
tanh-expansion. Figure 2 presents the change of ampli-
tude and width of the soliton due to the variation of α. Three-dimensional graphs depict the behaviour of the solution u corresponding to different values of α. This behaviour indicates that the order α can be used to modify the shape (strength and angle) of the travelling
wave without changing the nonlinearity and dispersion
effects in some medium. Figure 3 shows the expression
between the amplitude of the soliton and the fractional
order at different time values by using the complete
discrimination system method. Under the long wave limit, i.e. m → 1, the strength |u10| of field quantity u10 is reduced to a single domain soliton structure. Certainly, we can consider the strength of field quantity to the wave limit m → 0, but we omitted it. Figure 4 also presents the change of amplitude and width of the soliton due to the variation of α, and the three-dimensional graphs depict the behaviour of the
solution u10 corresponding to different values of order α. These figures show that at the same time, the change

|u10(x,y=0,t)|

2

1.5

1

0.5

0 2

1.5

4

1

2

t

0.5

0

-2

x

0 -4

(α=0.90)

v10(x,y=0,t)

2

0

-2

-4

-6 2

1.5

4

1

2

t

0.5

0

-2

x

0 -4

(α=0.98)

2

0

v10(x,y=0,t)

-2

-4

-6 2

1.5

4

1

2

t

0.5

0

-2

x

0 -4

(α=0.90)

Figure 3. The complete discrimination system surfaces of
the travelling wave solutions for |u10(x, y, t)| and v10(x, y, t) for different values of α.

of fractional α increases the amplitude of the travelling wave.
SolutionsSolutionWave SolutionsPhysNonlinear