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Abstract

In this work, the general algebraic method is proposed for finding some solutions of a conformable fractional coupled nonlinear Schrödinger equation with variable coefficients arising in inhomogeneous fibers with two orthogonal polarization states. With the aid of symbolic computations, many types of new soliton pulse and periodic pulse solutions including complex doubly periodic solutions, solitary wave solutions, and trigonometric function solutions are obtained. Some 3D and 2D numerical simulations about these solutions are portraited, which show the novelty and visibility of the dynamical structure and propagation behavior of the corresponding model. Moreover, we found the W-shaped soliton pulse and the periodic wave pulse can be effectively controlled by changing the relative parameters of the frequency coefficients and orders, which will help us to have a better understanding about the internal structure of this model.

Keywords

conformable derivative fractional coupled nonlinear Schrödinger equations W-shaped soliton pulse general algebraic method exact solutions

Introduction

Since Newton and Leibniz founded calculus, the development of natural science has entered a new era, many natural phenomena can be described by partial differential equation models, but in recent years, people realized that more and more problems of engineering science and mathematical physics need to be descripted by fractional differential equations(FDEs), such as chaotic oscillations,¹ engineering and physics ,² two-scale thermal science ,³ optical fibers, ⁴ and so on.^5–6 For better understanding and explaining the physical meaning of these models, searching for analytic solutions including approximate solutions and exact solutions of these FDEs plays an important and significant role in these areas. In the last two decades, many powerful methods have been proposed by scholars to handle this subject, such as homotopy analysis method(HAM),⁷ homotopy perturbation method(HPM),⁸ Adomian’s decomposition method(ADM),⁹ and variational iteration method(VIM)¹⁰ to find the approximate solutions of FDEs. Bäcklund transformation method,¹¹ Darboux transformation,¹² and Hirota bilinear method¹³ can be used to find the N-soliton solutions. Improved F-expansion method,¹⁴ projective Riccati equations method,¹⁵ Jacobi elliptic function expansion method,¹⁶ $G^{'} / G$ -expansion method,¹⁷ exp-function method^18–19 etc.,²⁰ can be used to find doubly periodic solutions, solitary wave solutions, and trigonometric function solutions of these models.

Up to now, scholars have built many types of definitions about the fractional derivative, the most classic definitions are

Riemann–Liouville fractional derivative²¹

D_{t}^{α} y (t) = {\begin{cases} \frac{1}{Γ (n - α)} \frac{d^{n}}{d t^{n}} \int_{0}^{t} {(t - ξ)}^{n - α - 1} y (ξ) d ξ, n - 1 < α < n, n \in N, \\ \frac{d^{(n)} y (t)}{d t^{n}}, α = n \in N . \end{cases}

Caputo fractional derivative²²

D_{t}^{α} y (t) = {\begin{cases} \frac{1}{Γ (n - α)} \int_{0}^{t} {(t - ξ)}^{n - α - 1} y^{(n)} (ξ) d ξ, n - 1 < α < n, n \in N, \\ \frac{d^{(n)} y (t)}{d t^{n}}, α = n \in N . \end{cases}

Jumaries’s fractional derivative²³

D_{t}^{α} y (t) = {\begin{cases} \frac{1}{Γ (1 - α)} \frac{d}{d t} \int_{0}^{t} {(t - ξ)}^{- α} [y (ξ) - y (0)] d ξ, 0 < α < 1, \\ {(y^{(n)} (t))}^{(α - n)}, n \leq α < n + 1, n \in N^{+} . \end{cases}

He’s fractional derivative²⁴

D_{t}^{α} y (t) = {\begin{cases} \frac{1}{Γ (n - α)} \frac{d^{n}}{d t^{n}} \int_{t_{0}}^{t} {(ξ - t)}^{n - α - 1} [y_{0} (ξ) - y (ξ)] d ξ, n - 1 < α < n, n \in N, \\ \frac{d^{(n)} y (t)}{d t^{n}}, α = n \in N . \end{cases}

Furthermore, Atangana fractional derivative,²⁵ M-fractional derivative,²⁶ Atangana–Baleanu derivative, ²⁷ and the conformable fractional derivative which will be utilized in this article,^28,29 etc ³⁰ are built recently, the geometrical explanation on fractional calculus has been discussed by He in Ref. [31].

We consider the conformable fractional coupled nonlinear Schrödinger equation in inhomogeneous fibers (CFCNLS) in the form^32–34

{\begin{cases} i D_{t}^{α} u + f (t) D_{x}^{2 β} u + 2 g (t) ({| u |}^{2} + {| v |}^{2}) u = 0, t > 0,0 < α, β \leq 1, \\ i D_{t}^{α} v + f (t) D_{x}^{2 β} v + 2 g (t) ({| u |}^{2} + {| v |}^{2}) v = 0 . \end{cases}

(1)

Here,

u = u (x, t), D_{x}^{2 β} u = D_{x}^{β} (D_{x}^{β} u),

v = v (x, t), D_{x}^{2 β} v = D_{x}^{β} (D_{x}^{β} v), i = \sqrt{- 1}

u

and

v

are two complex-valued functions represent the slowly varying amplitudes of two orthogonally polarized optical pulses, the coeffificients

f (t)

and

g (t)

are the group velocity dispersion and the Kerr nonlinearity coefficients. When

α = β = 1

, two class of solitary wave solutions of equation (1) are studied by using the modified sine-Gordon equation method in Ref. [32], the interactions of anti-dark soliton solutions are discussed through the bilinear Hirota operators in Ref. [33] and the asymmetric M-shaped and W-shaped soliton pulse are generated in Ref. [34]. Let us review some basic definitions and properties of the fractional calculus theory which will be used further in this paper.^28–29

Definition for a function $y (t) : [0, \infty) \to R$ .We defined the conformable fractional derivative operator and integral operator of order $α$ as²⁸

D_{t}^{α} y (t) = \lim_{ε \to 0} \frac{y (t + ε t^{1 - α}) - y (t)}{ε}, t > 0, α \in (0,1] .

I_{t}^{α} y (t) = \int_{0}^{t} τ^{α - 1} y (τ) d τ, t > 0, α \in (0,1] .

Also, we have the following important properties^28,29:

\begin{array}{l} (1) D_{t}^{α} y (t) = t^{1 - α} y^{'} (t) . \\ (2) D_{t}^{α} (a y (t) + b z (t)) = a D_{t}^{α} y (t) + b D_{t}^{α} z (t), \forall a, b \in R . \\ (3) D_{t}^{α} (y (t) z (t)) = y (t) D_{t}^{α} z (t) + z (t) D_{t}^{α} y (t) . \\ (4) D_{t}^{α} (y (t) / z (t)) = [z (t) D_{t}^{α} y (t) - y (t) D_{t}^{α} z (t)] / z^{2} (t) . \\ (5) D_{t}^{α} (y \circ z) (t) = t^{1 - α} y' (z (t)) z' (t) . \end{array}

The paper is organized as follows: In Section 2, we introduce the general algebraic method based on the sub-ode method. In Section 3, some new exact solutions of the CFCNLS are found and discussed by utilizing the proposed method. Finally, the conclusion is presented in Section 4.

The general algebraic method (GAM)

A brief description of the GAM is presented as follows:

Step 1. Consider the following nonlinear fractional partial differential equation

E (u, u_{t}^{α}, u_{x}^{β}, u_{x}^{2 β}, u u_{x}^{β}, \dots) = 0 .

(2)

Step 2. Using the fractional complex transformation^35–37

u (x, t) = u (ξ), ξ = \frac{k_{1}}{β} x^{β} + \frac{c_{1}}{α} t^{α} .

(3)

Here,

k_{1}, c_{1}

are constants or functions of

x, t

and equation. (2) can be converted to an ordinary differential equation(ODE) as follows:

O (u, u_{ξ}, u_{ξ ξ}, u u_{ξ}, \dots) = 0 .

(4)

Step 3. Assume that equation (4) has the following solution

u (ξ) = \sum_{i = 0}^{n} A_{i} φ^{i} (ξ) + \sum_{i = 1}^{n} A_{- i} φ^{- i} (ξ) .

(5)

Here,

n

is a balance number,

A_{i}

A_{- i}

and variable function

ξ = ξ (x, t)

are determined later. And

φ (ξ)

is a solution of the following auxiliary ODE:

φ^{' 2} (ξ) = \sum_{i = 0}^{4} a_{i} φ^{i} (ξ) .

(6)

Step 4. We assume that equation (6) have the following solutions:

φ (ξ) = b_{0} + b_{1} e + b_{2} f + b_{3} g + b_{4} h + b_{- 1} e^{- 1} + b_{- 2} f^{- 1} + b_{- 3} g^{- 1} + b_{- 4} h^{- 1} .

(7)

Here,

b_{i} (i = 0, \pm 1, \dots, \pm 4)

are constants of to be determined later, and the four functions

$e = e (ξ), f = f (ξ), g = g (ξ), h = h (ξ)$ are expressed as the follows³⁸:

e = \frac{1}{Z}, f = \frac{F}{Z}, g = \frac{F^{2}}{Z}, h = \frac{F^{'}}{Z}, Z = p + q F + r F^{2} + l F^{'} .

(8)

Here,

F'^{2} = A + B F^{2} + C F^{4} + 2 D F + 2 E F^{3} .

Step 5

After finding the new solutions of equation (6) by equations (7) and (8), let us substitute Equations (6) and (5) in equation (4), setting the coefficients of $φ^{i} (ξ)$ and $φ^{j} (ξ) \sqrt{\sum_{i = 0}^{4} a_{i} φ^{i} (ξ)}$ $(i, j = 0, \pm 1, \pm 2, \dots)$ to zero yield a set of algebraic equations(AEs) for $A_{i}$ , $A_{- i}, a_{0}, a_{1}, a_{2}, a_{3}, a_{4},$ $k_{1}$ and $c_{1}$ . After solving the AEs and substituting each of the solutions $φ (ξ)$ along with (3) in equation (2), we can get the solutions of equation (2).

Remark 1. In equation (8), the functions $F, e, f, g, h$ satisfy the restricted relation (6) (7) and (8a–8d) mentioned in i³⁸, if we select $p = r = l = 0, q = 1, h = \frac{F^{'}}{F}$ , this method turns to the $\frac{G^{'}}{G}$ method^39–44, if we select $p = q = l = 0, r = 1, h = \frac{F^{'}}{F^{2}}$ , it turns to the $\frac{G^{'}}{G^{2}}$ method⁴⁵, if we select $p = r = 0, q = 1, h = \frac{F^{'}}{F + l F^{'}}$ , it turns to the $\frac{G^{'}}{G + μ G^{'}}$ method,⁴⁶ and if we select $a_{0} = α^{2} {(L n A)}^{2}, a_{1} = 2 α β {(L n A)}^{2}, a_{2} = (2 α σ + β^{2}) {(L n A)}^{2}, a_{3} = 2 β σ {(L n A)}^{2},$ $a_{4} = σ^{2} {(L n A)}^{2},$ it turns to the direct algebraic method.⁴⁷

In the following, we will use this method to solve the CFCNLS.

Exact solutions to the CFCNLS

Exact solutions

If we let $D_{t}^{α} u = \frac{\partial^{α} u}{\partial t^{α}}, D_{x}^{2 β} u = \frac{\partial^{2 β} u}{\partial x^{2 β}}$ , equation (1) can be rewritten as follows

{\begin{cases} i u_{t}^{α} + f (t) u_{x}^{2 β} + 2 g (t) ({| u |}^{2} + {| v |}^{2}) u = 0, t > 0,0 < α, β \leq 1, \\ i v_{t}^{α} + f (t) v_{x}^{2 β} + 2 g (t) ({| u |}^{2} + {| v |}^{2}) v = 0 . \end{cases}

(9)

We can give the following complex variable transformation^48–50:

u = ϕ (ξ) e^{i η}, v = ψ (ξ) e^{i η},

(10)

ξ = \frac{k_{1}}{β} x^{β} + \int c_{1} t^{α - 1} d t, η = \frac{k_{2}}{β} x^{β} + \int c_{2} t^{α - 1} d t .

(11)

Here, constants

k_{1}, k_{2}, a n d c_{1} = c_{1} (t), c_{2} = c_{2} (t)

are to be determined later.

Substituting equations (10) and (11) in equation (9), separate the real part and the imaginary part, we obtain

{\begin{cases} k_{1}^{2} f ϕ_{ξ ξ} - (c_{2} + k_{2}^{2} f) ϕ + 2 g (ϕ^{3} + ψ^{2} ϕ) = 0, \begin{array}{c} (12.1) \end{array} \\ k_{1}^{2} f ψ_{ξ ξ} - (c_{2} + k_{2}^{2} f) ψ + 2 g (ψ^{3} + ϕ^{2} ψ) = 0, \begin{array}{c} (12.2) \end{array} \\ (c_{1} + 2 k_{1} k_{2} f) ϕ_{ξ} = 0, \begin{array}{c} (12.3) \end{array} \\ (c_{1} + 2 k_{1} k_{2} f) ψ_{ξ} = 0 . \begin{array}{c} (12.4) \end{array} \end{cases}

(12)

Where

ϕ_{ξ ξ} = \frac{d^{2} ϕ (ξ)}{d ξ^{2}}, ϕ_{ξ} = \frac{d ϕ (ξ)}{d ξ}, ψ_{ξ ξ} = \frac{d^{2} ψ (ξ)}{d ξ^{2}}, ψ_{ξ} = \frac{d ψ (ξ)}{d ξ}, f = f (t), g = g (t)

We assume that equation (12) have the following solutions:

\begin{array}{l} ϕ = ϕ (ξ) = A_{0} + A_{1} φ (ξ) + A_{- 1} φ^{- 1} (ξ) = A_{0} + A_{1} φ + A_{- 1} φ^{- 1}, \\ ψ = ψ (ξ) = B_{0} + B_{1} φ (ξ) + B_{- 1} φ^{- 1} (ξ) = B_{0} + B_{1} φ + B_{- 1} φ^{- 1} . \end{array}

(13)

Here,

φ'^{2} = \sum_{i = 0}^{4} a_{i} φ^{i}, φ = φ (ξ)

and

φ^{'} = \frac{d φ (ξ)}{d ξ}

Substituting both equations (6) and (13) in equation (12) and setting the coefficients of $φ^{i}$ to zero yields an AEs with respect to the unknowns $A_{0}, A_{1}, A_{- 1}, B_{0}, B_{1}, B_{- 1}, a_{0}, a_{1}, a_{2}, a_{3}, a_{4}, k_{1}, k_{2}, c_{1}, f, g$ , and $c_{2}$ as follows:

\begin{array}{l} φ^{0} : 2 g A_{0}^{3} + 12 g A_{- 1} A_{0} A_{1} + 4 g A_{1} B_{- 1} B_{0} + 2 g A_{0} B_{0}^{2} + 4 g A_{0} B_{- 1} B_{1} + 4 g A_{- 1} B_{0} B_{1} - A_{0} c_{2} \\ + \frac{1}{2} f a_{3} A_{- 1} k_{1}^{2} + \frac{1}{2} f a_{1} A_{1} k_{1}^{2} - f A_{0} k_{2}^{2} = 0, \\ φ^{1} : 6 g A_{0}^{2} A_{1} + 6 g A_{- 1} A_{1}^{2} + 2 g A_{1} B_{0}^{2} + 4 g A_{1} B_{- 1} B_{1} + 4 g A_{0} B_{0} B_{1} + 2 g A_{- 1} B_{1}^{2} - A_{1} c_{2} \\ + f a_{2} A_{1} k_{1}^{2} - f A_{1} k_{2}^{2} = 0, \\ φ^{2} : 6 g A_{0} A_{1}^{2} + 4 g A_{1} B_{0} B_{1} + 2 g A_{0} B_{1}^{2} + \frac{3}{2} f a_{3} A_{1} k_{1}^{2} = 0, \\ φ^{3} : 2 g A_{1}^{3} + 2 g A_{1} B_{1}^{2} + 2 f a_{4} A_{1} k_{1}^{2} = 0, \\ φ^{- 1} : 6 g A_{- 1} A_{0}^{2} + 6 g A_{- 1}^{2} A_{1} + 2 g A_{1} B_{- 1}^{2} + 4 g A_{0} B_{- 1} B_{0} + 2 g A_{- 1} B_{0}^{2} + 4 g A_{- 1} B_{- 1} B_{1} - A_{- 1} c_{2} \\ + f a_{2} A_{- 1} k_{1}^{2} - f A_{- 1} k_{2}^{2} = 0, \\ φ^{- 2} : 6 g A_{- 1}^{2} A_{0} + 2 g A_{0} B_{- 1}^{2} + 4 g A_{- 1} B_{- 1} B_{0} + \frac{3}{2} f a_{1} A_{- 1} k_{1}^{2} = 0, \\ φ^{- 3} : 2 g A_{- 1}^{3} + 2 g A_{- 1} B_{- 1}^{2} + 2 f a_{0} A_{- 1} k_{1}^{2} = 0, \end{array}

\begin{array}{l} φ^{0} : 4 g A_{0} A_{1} B_{- 1} + 2 g A_{0}^{2} B_{0} + 4 g A_{- 1} A_{1} B_{0} + 2 g B_{0}^{3} + 4 g A_{- 1} A_{0} B_{1} + 12 g B_{- 1} B_{0} B_{1} - B_{0} c_{2} \\ + \frac{1}{2} f a_{3} B_{- 1} k_{1}^{2} + \frac{1}{2} f a_{1} B_{1} k_{1}^{2} - f B_{0} k_{2}^{2} = 0, \\ φ^{1} : 2 g A_{1}^{2} B_{- 1} + 4 g A_{0} A_{1} B_{0} + 2 g A_{0}^{2} B_{1} + 4 g A_{- 1} A_{1} B_{1} + 6 g B_{0}^{2} B_{1} + 6 g B_{- 1} B_{1}^{2} - B_{1} c_{2} \\ + f a_{2} B_{1} k_{1}^{2} - f B_{1} k_{2}^{2} = 0, \\ φ^{2} : 2 g A_{1}^{2} B_{0} + 4 g A_{0} A_{1} B_{1} + 6 g B_{0} B_{1}^{2} + \frac{3}{2} f a_{3} B_{1} k_{1}^{2} = 0, \\ φ^{3} : 2 g A_{1}^{2} B_{1} + 2 g B_{1}^{3} + 2 f a_{4} B_{1} k_{1}^{2} = 0, \\ φ^{- 1} : 2 g A_{0}^{2} B_{- 1} + 4 g A_{- 1} A_{1} B_{- 1} + 4 g A_{- 1} A_{0} B_{0} + 6 g B_{- 1} B_{0}^{2} + 2 g A_{- 1}^{2} B_{1} + 6 g B_{- 1}^{2} B_{1} - B_{- 1} c_{2} \\ + f a_{2} B_{- 1} k_{1}^{2} - f B_{- 1} k_{2}^{2} = 0, \\ φ^{- 2} : 4 g A_{- 1} A_{0} B_{- 1} + 2 g A_{- 1}^{2} B_{0} + 6 g B_{- 1}^{2} B_{0} + \frac{3}{2} f a_{1} B_{- 1} k_{1}^{2} = 0, \\ φ^{- 3} : 2 g A_{- 1}^{2} B_{- 1} + 2 g B_{- 1}^{3} + 2 f a_{0} B_{- 1} k_{1}^{2} = 0 . \end{array}

After solving the above AEs along with equations (10) and (11), we could determine the following solutions:

Family 1 $ϕ (ξ) = ψ (ξ) = C$ , $c_{2} = 4 C^{2} g (t) - k_{2}^{2} f (t)$ .

We find the following trivial solution of equation (9)

u_{0} = v_{0} = C e^{i {\frac{k_{2}}{β} x^{β} + \int [4 C^{2} g (t) - k_{2}^{2} f (t)] t^{α - 1} d t}} .

Family 2 $c_{1} + 2 k_{1} k_{2} f = 0$ .

In this situation, we produce the following solutions of the above AEs:

Case 1

\begin{array}{l} a_{1} = a_{3} = 0, A_{0} = B_{0} = 0, f = λ g, B_{1} = \pm \sqrt{- A_{1}^{2} - λ a_{4} k_{1}^{2}}, k_{2} = \pm k_{1} \sqrt{\frac{a_{2}}{2}}, \\ c_{1} = \mp k_{1}^{2} λ \sqrt{2 a_{2}} g, c_{2} = \frac{a_{2} k_{1}^{2} λ g}{2} \end{array}

Case 2

\begin{array}{l} a_{1} = a_{3} = 0, A_{0} = B_{0} = 0, f = λ g, c_{2} = (a_{2} k_{1}^{2} - 6 k_{1}^{2} \sqrt{a_{0} a_{4}} - k_{2}^{2}) λ g \\ A_{1} = \overset{\pm}{\pm} k_{1} \sqrt{\frac{a_{4} λ}{2}}, A_{- 1} = \overset{\pm}{\pm} k_{1} \sqrt{\frac{a_{0} λ}{2}}, B_{1} = \pm k_{1} \sqrt{- \frac{3 a_{4} λ}{2}}, B_{- 1} = \pm k_{1} \sqrt{- \frac{3 a_{0} λ}{2}} . \end{array}

Thus, the following solutions of equation (9) are deduced:

Type 1

{\begin{cases} u_{1} = A_{1} φ [\frac{k_{1}}{β} x^{β} \mp k_{1}^{2} λ \sqrt{2 a_{2}} \int g (t) t^{α - 1} d t] e^{i [\pm \frac{k_{1}}{β} \sqrt{\frac{a_{2}}{2}} x^{β} + \frac{a_{2} k_{1}^{2} λ}{2} \int g (t) t^{α - 1} d t]}, \\ v_{1} = \pm \sqrt{- A_{1}^{2} - λ a_{4} k_{1}^{2}} φ [\frac{k_{1}}{β} x^{β} \mp k_{1}^{2} λ \sqrt{2 a_{2}} \int g (t) t^{α - 1} d t] e^{i [\pm \frac{k_{1}}{β} \sqrt{\frac{a_{2}}{2}} x^{β} + \frac{a_{2} k_{1}^{2} λ}{2} \int g (t) t^{α - 1} d t]} . \end{cases}

Type 2

{\begin{cases} u_{2} = [\pm k_{1} \sqrt{\frac{a_{4} λ}{2}} φ (ξ_{2}) \pm k_{1} \sqrt{\frac{a_{0} λ}{2}} φ^{- 1} (ξ_{2})] e^{i [\frac{k_{2}}{β} x^{β} + (a_{2} k_{1}^{2} - 6 k_{1}^{2} \sqrt{a_{0} a_{4}} - k_{2}^{2}) λ \int g (t) t^{α - 1} d t]}, \\ v_{2} = [\pm k_{1} \sqrt{- \frac{3 a_{4} λ}{2}} φ (ξ_{2}) \pm k_{1} \sqrt{- \frac{3 a_{0} λ}{2}} φ^{- 1} (ξ_{2})] e^{i [\frac{k_{2}}{β} x^{β} + (a_{2} k_{1}^{2} - 6 k_{1}^{2} \sqrt{a_{0} a_{4}} - k_{2}^{2}) λ \int g (t) t^{α - 1} d t]}, \\ ξ_{2} = \frac{k_{1}}{β} x^{β} - 2 k_{1} k_{2} λ \int g (t) t^{α - 1} d t . \end{cases}

Here, $φ_{i}$ is an arbitrary solution of the auxiliary equation $φ_{i}'^{2} = a_{0} + a_{2} φ_{i}^{2} + a_{4} φ_{i}^{4}$ in $u_{1}, u_{2}, v_{1}, v_{2}$ , the coefficients $a_{0}, a_{2}, a_{4}$ are arbitrary constants, we can find some new solutions of equation (6) by using equations (7) and (8), and many types of $φ_{i}$ have been found in a large number of papers such as Refs. [51, 52], and if we select different sets of coefficients, then the corresponding solutions of Equation (9) can be found. Let us take some typical cases for example.

Set 1 For $a_{0} = 1 - m^{2}, a_{2} = 2 m^{2} - 1, a_{4} = - m^{2},$ $φ_{1.1} = c n ξ$ , we have

{\begin{cases} u_{1.1} = A_{1} c n [\frac{k_{1}}{β} x^{β} \mp k_{1}^{2} λ \sqrt{4 m^{2} - 2} \int g (t) t^{α - 1} d t] e^{i [\pm \frac{k_{1}}{β} \sqrt{\frac{2 m^{2} - 1}{2}} x^{β} + \frac{(2 m^{2} - 1) k_{1}^{2} λ}{2} \int g (t) t^{α - 1} d t]}, \\ v_{1.1} = \pm \sqrt{λ m^{2} k_{1}^{2} - A_{1}^{2}} c n [\frac{k_{1}}{β} x^{β} \mp k_{1}^{2} λ \sqrt{4 m^{2} - 2} \int g (t) t^{α - 1} d t] e^{i [\pm \frac{k_{1}}{β} \sqrt{\frac{2 m^{2} - 1}{2}} x^{β} + \frac{(2 m^{2} - 1) k_{1}^{2} λ}{2} \int g (t) t^{α - 1} d t]} . \end{cases}

Obviously, $u_{1.1}$ is degenerated to the famous bell-shape-like solitary wave solutions when $α = β = 1, m = 1$ .

{\begin{cases} u_{1 . 1_{m = 1}} = A_{1} \sec h [k_{1} x \mp k_{1}^{2} λ \sqrt{2} \int g (t) d t] e^{i [\pm k_{1} \sqrt{\frac{1}{2}} x + \frac{k_{1}^{2} λ}{2} \int g (t) d t]}, \\ v_{1 . 1_{m = 1}} = \pm \sqrt{λ k_{1}^{2} - A_{1}^{2}} \sec h [k_{1} x \mp k_{1}^{2} λ \sqrt{2} \int g (t) d t] e^{i [\pm k_{1} \sqrt{\frac{1}{2}} x + \frac{k_{1}^{2} λ}{2} \int g (t) d t]} . \end{cases}

Set 2 For $a_{0} = m^{2} - 1, a_{2} = 2 - m^{2}, a_{4} = - 1,$ $φ_{1.2} = d n ξ$ , we have

{\begin{cases} u_{1.2} = A_{1} d n [\frac{k_{1}}{β} x^{β} \mp k_{1}^{2} λ \sqrt{2 (2 - m^{2})} \int g (t) t^{α - 1} d t] e^{i [\pm \frac{k_{1}}{β} \sqrt{\frac{2 - m^{2}}{2}} x^{β} + \frac{(2 - m^{2}) k_{1}^{2} λ}{2} \int g (t) t^{α - 1} d t]}, \\ v_{1.2} = \pm \sqrt{λ k_{1}^{2} - A_{1}^{2}} d n [\frac{k_{1}}{β} x^{β} \mp k_{1}^{2} λ \sqrt{2 (2 - m^{2})} \int g (t) t^{α - 1} d t] e^{i [\pm \frac{k_{1}}{β} \sqrt{\frac{2 - m^{2}}{2}} x^{β} + \frac{(2 - m^{2}) k_{1}^{2} λ}{2} \int g (t) t^{α - 1} d t]} . \end{cases}

Set 3 For $a_{0} = \frac{1 - m^{2}}{4}, a_{2} = \frac{1 + m^{2}}{2}, a_{4} = \frac{1 - m^{2}}{4},$ $φ_{1.3} = n c ξ \pm s c ξ$ , we have

{\begin{cases} u_{1.3} = A_{1} (n c ξ_{1.3} + s c ξ_{1.3}) e^{i [\pm \frac{k_{1}}{β} \sqrt{1 + m^{2}} x^{β} + \frac{(1 + m^{2}) k_{1}^{2} λ}{4} \int g (t) t^{α - 1} d t]}, \\ v_{1.3} = \pm \sqrt{- A_{1}^{2} - \frac{λ (1 - m^{2})}{4} k_{1}^{2}} (n c ξ_{1.3} + s c ξ_{1.3}) e^{i [\pm \frac{k_{1}}{β} \sqrt{1 + m^{2}} x^{β} + \frac{(1 + m^{2}) k_{1}^{2} λ}{4} \int g (t) t^{α - 1} d t]} \\ ξ_{1.3} = \frac{k_{1}}{β} x^{β} \mp k_{1}^{2} λ \sqrt{1 + m^{2}} \int g (t) t^{α - 1} d t . \end{cases}

Selecting $α = β = 1, A_{1} = 1, k_{1} = 1, λ = - 8, m = 0$ in the above solution, thus

u_{1.3, m = 0} = v_{1.3, m = 0} = [\frac{1}{\cos [x \pm 8 \int g (t) d t]} + \tan [x \pm 8 \int g (t) d t] e^{i [\pm x - 2 \int g (t) d t]}

Set 4 For $a_{0} = \frac{1}{4}, a_{2} = \frac{1}{2} - m^{2}, a_{4} = \frac{1}{4},$ $φ_{1.4} = \frac{s n ξ}{1 \pm c n ξ}$ , we have

{\begin{cases} u_{1.4} = \frac{A_{1} s n ξ_{1.4}}{1 \pm c n ξ_{1.4}} e^{i [\pm \frac{k_{1}}{β} \sqrt{\frac{1 - 2 m^{2}}{4}} x^{β} + \frac{(1 - 2 m^{2}) k_{1}^{2} λ}{4} \int g (t) t^{α - 1} d t]}, \\ v_{1.4} = \frac{\pm \sqrt{- A_{1}^{2} - \frac{1}{4} λ k_{1}^{2}} s n ξ_{1.4}}{1 \pm c n ξ_{1.4}} e^{i [\pm \frac{k_{1}}{β} \sqrt{\frac{1 - 2 m^{2}}{4}} x^{β} + \frac{(1 - 2 m^{2}) k_{1}^{2} λ}{4} \int g (t) t^{α - 1} d t],} \\ ξ_{1.4} = \frac{k_{1}}{β} x^{β} \mp k_{1}^{2} λ \sqrt{1 - 2 m^{2}} \int g (t) t^{α - 1} d t . \end{cases}

Set 5 For $a_{0} = 1, a_{2} = - 1 - 6 m - m^{2}, a_{4} = 4 m {(1 + m)}^{2},$ $φ_{1.5} = \frac{s n ξ}{1 + m s n^{2} ξ}$ , we have

{\begin{cases} u_{1.5} = \frac{A_{1} s n ξ_{1.3}}{1 + m s n^{2} ξ_{1.3}} e^{i η_{1.5}}, \\ v_{1.5} = \frac{\pm i \sqrt{A_{1}^{2} + λ 4 m {(1 + m)}^{2} k_{1}^{2}} s n ξ_{1.3}}{1 + m s n^{2} ξ_{1.3}} e^{i η_{1.5}}, \\ ξ_{1.5} = \frac{k_{1}}{β} x^{β} \mp k_{1}^{2} λ i \sqrt{2 (1 + 6 m + m^{2})} \int g (t) t^{α - 1} d t, \\ η_{1.5} = \pm i \frac{k_{1}}{β} \sqrt{\frac{1 + 6 m + m^{2}}{2}} x^{β} - \frac{(1 + 6 m + m^{2}) k_{1}^{2} λ}{2} \int g (t) t^{α - 1} d t . \end{cases}

Set 6 For $a_{0} = 1, a_{2} = - m^{2} - 1, a_{4} = m^{2}, φ_{2.1} = s n ξ$ , we have

{\begin{cases} u_{2.1} = (\pm k_{1} m \sqrt{\frac{λ}{2}} s n ξ_{2.1} \pm k_{1} \sqrt{\frac{λ}{2}} n s ξ_{2.1}) e^{i [\frac{k_{2}}{β} x^{β} + ((- m^{2} - 1) k_{1}^{2} - 6 k_{1}^{2} m - k_{2}^{2}) λ \int g (t) t^{α - 1} d t]}, \\ v_{2.1} = (\pm k_{1} m i \sqrt{\frac{3 λ}{2}} s n ξ_{2.1} \pm k_{1} i \sqrt{\frac{3 λ}{2}} s n ξ_{2.1}) e^{i [\frac{k_{2}}{β} x^{β} + ((- m^{2} - 1) k_{1}^{2} - 6 k_{1}^{2} m - k_{2}^{2}) λ \int g (t) t^{α - 1} d t]}, \\ ξ_{2.1} = \frac{k_{1}}{β} x^{β} - 2 k_{1} k_{2} λ \int g (t) t^{α - 1} d t . \end{cases}

Remark 2

If we let

$A_{1} = B_{1}, \sqrt{λ k_{1}^{2} - A_{1}^{2}} = E_{1}, k_{1} = μ, \sqrt{2} k_{1} = 2 α, f = λ g = β (t), u_{1.1, m = 1}, v_{1.1, m = 1}$ turn to the solution (14) mentioned in Ref. [32] which made a clerical error, then

{\begin{cases} u_{[32]} = B_{1} \sec h [μ (x \mp 2 α \int β (t) d t)] e^{i [\pm α x + α^{2} \int β (t) d t]}, \\ v_{[32]} = E_{1} \sec h [μ (x \mp 2 α \int β (t) d t)] e^{i [\pm α x + α^{2} \int β (t) d t]} . \end{cases}

Case 3

\begin{array}{l} a_{1} = \frac{a_{2} a_{3}}{2 a_{4}} - \frac{a_{3}^{3}}{8 a_{4}^{2}}, A_{0} = \pm B_{0} \sqrt{\frac{a_{3}^{3}}{16 a_{1} a_{4}^{2}} - 1}, A_{1} = \pm B_{0} \sqrt{\frac{a_{3}}{a_{1}} - \frac{16 a_{4}^{2}}{a_{3}^{2}}}, B_{1} = \frac{4 a_{4}}{a_{3}} B_{0}, f = λ g, \\ c_{1} = \mp 2 B_{0} k_{2} \sqrt{\frac{- λ a_{3}}{a_{1} a_{4}}} g, c_{2} = (\frac{2 B_{0}^{2} a_{3}^{3}}{16 a_{1} a_{4}^{2}} - 2 B_{0}^{2} - λ k_{2}^{2}) g, k_{1} = \pm B_{0} \sqrt{\frac{- a_{3}}{λ a_{1} a_{4}}} . \end{array}

We find the following solutions of equation (9):

Type 3

{\begin{cases} u_{3} = [\pm B_{0} \sqrt{\frac{a_{3}^{3}}{16 a_{1} a_{4}^{2}} - 1} \pm B_{0} \sqrt{\frac{a_{3}}{a_{1}} - \frac{16 a_{4}^{2}}{a_{3}^{2}}} φ (ξ_{3})] e^{i [\frac{k_{2}}{β} x^{β} + (\frac{2 B_{0}^{2} a_{3}^{3}}{16 a_{1} a_{4}^{2}} - 2 B_{0}^{2} - λ k_{2}^{2}) \int g (t) t^{α - 1} d t]}, \\ v_{3} = [B_{0} + \frac{4 a_{4}}{a_{3}} B_{0} φ (ξ_{3})] e^{i [\frac{k_{2}}{β} x^{β} + (\frac{2 B_{0}^{2} a_{3}^{3}}{16 a_{1} a_{4}^{2}} - 2 B_{0}^{2} - λ k_{2}^{2}) \int g (t) t^{α - 1} d t],} \\ ξ_{3} = \pm \frac{B_{0}}{β} \sqrt{\frac{- a_{3}}{λ a_{1} a_{4}}} x^{β} \mp 2 B_{0} k_{2} \sqrt{\frac{- λ a_{3}}{a_{1} a_{4}}} \int g (t) t^{α - 1} d t . \end{cases}

Set 7 For $a_{0} = \frac{1}{4} (1 - m^{2}), a_{1} = m^{2} - 1, a_{2} = 2 - m^{2}, a_{3} = - 2, a_{4} = 1, φ_{3.1} = \frac{c n ξ}{1 \pm s n ξ + c n ξ}$ , thus

{\begin{cases} u_{3.1} = \pm B_{0} \sqrt{\frac{2 m^{2} - 1}{2 - 2 m^{2}}} (1 - \frac{2 c n ξ_{3.1}}{1 \pm s n ξ_{3.1} + c n ξ_{3.1}}) e^{i [\frac{k_{2}}{β} x^{β} - (B_{0}^{2} (2 - \frac{1}{1 - m^{2}}) + k_{2}^{2} λ)) \int g (t) t^{α - 1} d t]}, \\ v_{3.1} = (B_{0} - \frac{2 B_{0} s n ξ_{3.1}}{1 - s n ξ_{3.1} + c n ξ_{3.1}}) e^{i [\frac{k_{2}}{β} x^{β} -) B_{0}^{2}) 2 - \frac{1}{1 - m^{2}}) + k_{2}^{2} λ)) \int g) t) t^{α - 1} d t]}, \\ ξ_{3.1} = \pm \frac{B_{0}}{β} \sqrt{\frac{2}{λ (m^{2} - 1)}} x^{β} - 2 B_{0} k_{2} \sqrt{\frac{2 λ}{m^{2} - 1}} \int g (t) t^{α - 1} d t . \end{cases}

Selecting $α = β = 1, k_{2} = 1, B_{0} = 1, λ = - 8, m = \frac{\sqrt{3}}{2}, g_{1} (t) = 1, g_{2} (t) = \cos t$ , we get

u_{3.1, m = \sqrt{3} / 2} = [1 - \frac{2 c n [\pm x - 16 \int g (t) d t]}{1 \pm s n [\pm x - 16 \int g (t) d t] + c n [\pm x - 16 \int g (t) d t]}] e^{i [x + 10 \int g (t) d t]} .

Set 8 For $a_{0} = \frac{1}{4}, a_{1} = 1, a_{2} = 2 - m^{2}, a_{3} = 2 - 2 m^{2}, a_{4} = 1 - m^{2}, φ_{3.2} = \frac{s n ξ}{1 - s n ξ + c n ξ}$ , we have

{\begin{cases} u_{3.2} = \pm i \sqrt{2 + 2 m^{2}} B_{0} (\frac{1}{2} + \frac{s n ξ_{3.2}}{1 - s n ξ_{3.2} + c n ξ_{3.2}}) e^{i [\frac{k_{2}}{β} x^{β} -) B_{0}^{2}) 1 + m^{2}) + k_{2}^{2} λ)) \int g (t) t^{α - 1} d t]}, \\ v_{3.2} =) B_{0} + \frac{2 B_{0} s n ξ_{3.2}}{1 - s n ξ_{3.2} + c n ξ_{3.2}}) e^{i [\frac{k_{2}}{β} x^{β} -) B_{0}^{2}) 1 + m^{2}) + k_{2}^{2} λ)) \int g) t) t^{α - 1} d t]}, \\ ξ_{3.2} = \pm \frac{B_{0}}{β} \sqrt{\frac{2}{- λ}} x^{β} - 2 B_{0} k_{2} \sqrt{- 2 λ} \int g (t) t^{α - 1} d t . \end{cases}

Set 9 For $a_{0} = 1, a_{1} = - 4, a_{2} = 8 - 4 m^{2}, a_{3} = 8 m^{2} - 8, a_{4} = 4 - 4 m^{2}, φ_{3.3} = \frac{c n ξ}{c n ξ \pm s n ξ d n ξ}$ , we have

{\begin{cases} u_{3.3} = i B_{0} \sqrt{2 + 2 m^{2}} (\frac{1}{2} - \frac{c n ξ_{3.3}}{c n ξ_{3.3} \pm s n ξ_{3.3} d n ξ_{3.3}}) e^{i [\frac{k_{2}}{β} x^{β} - (B_{0}^{2} (1 + m^{2}) + k_{2}^{2} λ) \int g (t) t^{α - 1} d t]}, \\ v_{3.3} = (B_{0} - 2 B_{0} \frac{c n ξ_{3.3}}{c n ξ_{3.3} \pm s n ξ_{3.3} d n ξ_{3.3}}) e^{i [\frac{k_{2}}{β} x^{β} - (B_{0}^{2} (1 + m^{2}) + k_{2}^{2} λ) \int g (t) t^{α - 1} d t]}, \\ ξ_{3.3} = - \frac{B_{0}}{β \sqrt{- 2 λ}} x^{β} + \sqrt{- 2 λ} B_{0} k_{2} \int g (t) t^{α - 1} d t . \end{cases}

Set 10 For $a_{0} = 1, a_{1} = - 4 m, a_{2} = 8 m^{2} - 4, a_{3} = 8 m - 8 m^{3}, a_{4} = 4 m^{4} - 4 m^{2}, φ_{3.4} = \frac{d n ξ}{d n ξ \pm m^{2} s n ξ c n ξ}$ , we have

{\begin{cases} u_{3.4} = i B_{0} \sqrt{2 m^{2} + 2} (\frac{1}{2 m} - \frac{d n ξ_{3.4}}{d n ξ_{3.4} \pm m^{2} s n ξ_{3.4} c n ξ_{3.4}}) e^{i [\frac{k_{2}}{β} x^{β} - (B_{0}^{2} (\frac{1}{m^{2}} + 1) + k_{2}^{2} λ) \int g (t) t^{α - 1} d t]}, \\ v_{3.4} = (B_{0} - \frac{2 B_{0} m d n ξ_{3.4}}{d n ξ_{3.4} \pm m^{2} s n ξ_{3.4} c n ξ_{3.4}}) e^{i [\frac{k_{2}}{β} x^{β} - (B_{0}^{2} (\frac{1}{m^{2}} + 1) + k_{2}^{2} λ) \int g (t) t^{α - 1} d t]} \\ ξ_{3.4} = - \frac{B_{0}}{β m \sqrt{- 2 λ}} x^{β} + \frac{B_{0} k_{2} \sqrt{- 2 λ}}{m} \int g (t) t^{α - 1} d t . \end{cases}

Selecting $α = β = 1, k_{2} = 1, B_{0} = 1, λ = - 2, m = 0.5$ , we get

u_{3.4, m = 0.5} = [1 - \frac{d n (- x + 4 \int g (t) d t)}{d n (- x + 4 \int g (t) d t) + 0.25 s n (- x + 4 \int g (t) d t) c n (- x + 4 \int g (t) d t)}] e^{i [x - 3 \int g (t) d t]} .

Set 11 For $a_{0} = 1, a_{1} = - 4 \sqrt{1 - m^{2}}, a_{2} = 8 - 4 m^{2}, a_{3} = - 8 \sqrt{1 - m^{2}}, a_{4} = 4 - 4 m^{2},$ $φ_{3.5} = \frac{s n ξ}{\sqrt{1 - m^{2}} s n ξ + c n ξ d n ξ}$ , we have

{\begin{cases} u_{3.5} = B_{0} \sqrt{\frac{4 m^{2} - 2}{1 - m^{2}}} (\frac{1}{2} - \frac{\sqrt{1 - m^{2}} s n ξ_{3.5}}{\sqrt{1 - m^{2}} s n ξ_{3.5} + c n ξ_{3.5} d n ξ_{3.5}}) e^{i [\frac{k_{2}}{β} x^{β} - (\frac{(1 - 2 m^{2}) B_{0}^{2}}{1 - m^{2}} + k_{2}^{2} λ) \int g (t) t^{α - 1} d t]}, \\ v_{3.5} = (B_{0} - \frac{2 B_{0} \sqrt{1 - m^{2}} s n ξ_{3.5}}{\sqrt{1 - m^{2}} s n ξ_{3.5} + c n ξ_{3.5} d n ξ_{3.5}}) e^{i [\frac{k_{2}}{β} x^{β} - (\frac{(1 - 2 m^{2}) B_{0}^{2}}{1 - m^{2}} + k_{2}^{2} λ) \int g (t) t^{α - 1} d t]} \\ ξ_{3.5} = \frac{B_{0}}{β \sqrt{- 2 λ (1 - m^{2})}} x^{β} + \frac{B_{0} k_{2} \sqrt{- 2 λ}}{\sqrt{1 - m^{2}}} \int g (t) t^{α - 1} d t . \end{cases}

Remark 3. All of these solutions have been checked by mathematical software, and they are noted for the first time to our knowledge.

We simulated some of these solutions in Figure (1), (2), (3), (4), (5), (6), and (7).

Figure 1.

The modulus of $u_{1.1}$ at $α = β = 1, A_{1} = 1, k_{1} = 1, λ = \frac{8}{3}, m = \frac{\sqrt{3}}{2}, g (t) = 1$ and $t = 0.1$ .

Figure 2.

The modulus of $u_{1.1}$ at $α = \frac{1}{3}, β = \frac{1}{2}, g (t) = t, A_{1} = k_{1} = 1, λ = \frac{8}{3}, m = \frac{\sqrt{3}}{2}$ and $x = 1$ .

Figure 3.

The modulus of $u_{1.1}$ at $α = β = 1, A_{1} = 1, k_{1} = 1, λ = \frac{8}{3}, m = 1, g (t) = 1$ and $t = 0.1$ .

Figure 4.

The real part of $u_{1.3}$ at $α = β = 1, A_{1} = 1, k_{1} = 1, λ = - 8, m = 0, g (t) = 1$ and $t = 0.1$ .

Figure 5.

The modulus of $u_{3.1}$ at $β = 1, k_{2} = 1, B_{0} = 1, λ = - 8, m = \frac{\sqrt{3}}{2}, g (t) = 1$ and $t = 0.1$ .

Figure 6.

The imaginary part of $u_{3.1}$ at $β = 1, k_{2} = 1, B_{0} = 1, λ = - 8, m = \frac{\sqrt{3}}{2}, g (t) = \cos t$ and $t = 0.1$ .

Figure 7.

The modulus of $v_{3.4}$ at $α = β = 1, k_{2} = 1, B_{0} = 1, λ = - 2, m = 0.5$ and $t = 0.1$ .

Results and discussion

After utilizing the GAM, we get many types of similar solutions of equation (1), and some structure of these solutions are simulated in Figure (1), (2), (3), (4), (5), (6), and (7). The visualization can help us to better understand the dynamic behavior and propagation property of these soliton pulse and periodic pulse solutions, and the doubly periodic pulse solution of equation (1) is shown in Figure 1, which can be degenerated to a bell-soliton solution when the modulus $m = 1$ is exhibited in Figure 3. We find that the Jacobi-like elliptic function solution $u_{1.3}$ is degenerated to an unbounded trigonometric-like function solutions when the modulus $m \to 0$ in Figure 4. Solution $u_{3.1}$ in Figures 5 and 6 shows that the evolution of wave signal obeys a stable and periodic state when the frequency $k_{2} = 1$ , while the amplitude of solution $v_{3.4}$ possesses periodic W-shaped feature are shown in Figure 7. Some changes of the modulus for $u_{1.1}$ and $v_{3.4}$ are simulated in Figure 8 at different selection of the order $α, β$ and the kerr nonlinear coefficient $g (t)$ . We select $g (t) = t, A_{1} = k_{1} = 1, λ = \frac{8}{3}, m = \frac{\sqrt{3}}{2}, t = 1$ for $u_{1.1}$ in Figure (8a) and $= β = 1, k_{2} = 1,$ $B_{0} = 1, λ = - 2, m = 0.5$ , $t = 0.1, g_{1} (t) = 1, g_{2} (t) = t$ for $v_{3.4}$ in Figure (8b), which show that the wave shape varies dramatically for different $α, β$ , but keeps stability for the different $g (t)$ .

Figure 8.

The change of modulus for $u_{1.1}$ and $v_{3.4}$ .

Conclusion

In conclusion, many types of new exact soliton pulse and periodic pulse solutions for the CFCNLS have been found after utilizing the GAM. Some propagation behavior of these solutions are discussed and simulated, and the graphs of which show that these doubly periodic wave solutions, solitary wave solutions, and single periodic solutions are propagated through different pattern which were controlled by the parameters of the two components. The efficient and significant method can be used by many other nonlinear models such as vmKdV equation, Ginzburg-Landau equation, and Burgers-BBM equation.

Footnotes

Author contribution

Baojian Hong: Completed all of the study, carried out the results and drafted the paper.

Declaration of conflicting interests

The author(s) declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.

Funding

The author(s) disclosed receipt of the following financial support for the research, authorship, and/or publication of this article: This work is supported by the practical innovation training program projects for the university students of Jiangsu Province (Grant No. 202211276054Y), Natural science research projects of Institutions in Jiangsu Province (Grant No. 18KJB110013) and Nanjing Institute of Technology (Grant No. ZKJ201513, YZKC2019086). The author wish to express his sincere appreciation to the editors and the anonymous referees for their valuable comments and suggestions.

Data availability

The author affirmed availability of data and material in deriving the solutions mentioned in this manuscript.

ORCID iD

Baojian Hong

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Exact solutions for the conformable fractional coupled nonlinear Schrödinger equations with variable coefficients

Abstract

Keywords

Introduction

The general algebraic method (GAM)

Exact solutions to the CFCNLS

Exact solutions

Results and discussion

Conclusion

Footnotes

Author contribution

Declaration of conflicting interests

Funding

Data availability

ORCID iD

References