Soliton solutions of conformable time-fractional perturbed Radhakrishnan-Kundu-Lakshmanan equation

1.   Introduction

Due to the wide application in the fields of physics, communication and engineering, the study of soliton solutions of the Radhakrishnan-Kundu-Lakshmanan (RKL) equation has attracted much attention^{[1,2,3,4,5,6,7]}. Especially in nonlinear optical fibers, the RKL equation usually describes the propagation of optical pulses, which is represented by the higher-order nonlinear Schrödinger equation. In recent years, many powerful mathematical methods have been proposed to derive soliton solutions ^{[8,9,10,11,12,13,14,15,16,17,18,19,20,21]} for the RKL equation, such as the first integral method^[22], the generalized exponential rational function method ^[23], the Laplace-Adomian decomposition method ^[24], the dynamical system method^[25], the Painlevé analysis^[26], the auxiliary equation method and extended simple equation method^[27], the modified simple equation and exp$(-\varphi(q))$ method ^[28].

In the paper, we consider the fractional perturbed Radhakrishnan-Kundu-Lakshmanan (FPRKL) equation^{[29,30,31,32]}:

where $\phi$ is the complex-valued wave function.

Definition 1.1. Let $\psi:[0, \infty)\rightarrow \bf{R}$. Then, the conformable fractional derivative of $\psi$ of order $\alpha$ is defined as

for all $t > 0$ and $\alpha\in(0, 1]$. Further, some properties of conformable fractional derivative is given

(i) $D_{t}^{\alpha}(t^{\delta}) = \mu t^{\delta-\alpha}$, $\forall\delta\in \bf{R}$.

(ii) $D_{t}^{\alpha}(\psi(t)+\varphi(t)) = D_{t}^{\alpha}\psi(t)+D_{t}^{\alpha}\varphi(t)$.

(iii) $D_{t}^{\alpha}(\psi\circ \varphi)(t) = t^{1-\alpha}\varphi(t)^{\alpha-1}\varphi'(t)D_{t}^{\alpha}(\psi(t))|_{t = \varphi(t)}$.

This article is arranged as follows. In Section 2, we employ three different methods to solve the FPRKL equation. In Section 3, we draw three-dimensional graph of Eq (1.1). In Section 4, we give a brief conclusion.

2.   Soliton solutions of Eq (1.1)

Making the complex transformation

Substituting Eq (2.1) into Eq (1.1), separating into real and imaginary parts yields

Integrating Eq (2.3) once, we have

Since the function $U$ satisfies both Eq (2.3) and Eq (2.4), the following constraint condition is obtained

So, $k$ and $c$ in Eq (2.5) can be obtained

2.1.   Extended $(G'/G)$-expansion method

Balancing $u^{3}$ and $u^{''}$ in Eq (2.4), we have $N = 1$. So, the solution form of Eq (2.4) is

Here $G = G(\xi)$ satisfies the following nonlinear ordinary differential equation

where $A$, $B$, $C$ are real parameters, Eq (2.8) satisfies the following equation

Then we can obtain a nonlinear algebraic equations.

$(\frac{G'}{G})^{3}$: $6\mu^{2}-\gamma(A-1)^{2}a_{1}-(3\lambda+2\sigma)a_{1}^{3} = 0$.

$(\frac{G'}{G})^{2}$: $9\mu^{2}-\gamma B(A-1)^{2}a_{1}-3(3\lambda+2\sigma)a_{1}^{2}a_{0} = 0$.

$(\frac{G'}{G})^{1}$: $3\mu^{2}-\gamma [2C(A-1)+B^{2}]a_{1}-3(\nu+2ak+\delta+3k^{2}-\gamma)a_{1}-3(3\lambda+2\sigma)(a_{0}^{2}a_{1}+a_{1}^{2}a_{-1}) = 0$.

$(\frac{G'}{G})^{0}$: $3\mu^{2}-\gamma [BCa_{1}+B(A-1)a_{-1}]-3(\nu+2ak+\delta+3k^{2}-\gamma)a_{0}-(3\lambda+2\sigma)(6a_{0}^{2}a_{1}a_{-1}+a_{0}^{3}) = 0$.

$(\frac{G'}{G})^{-1}$: $3\mu^{2}-\gamma [2C(A-1)+B^{2}]a_{-1}-3(\nu+2ak+\delta+3k^{2}-\gamma)a_{-1}-3(3\lambda+2\sigma)(a_{0}^{2}a_{-1}+a_{-1}^{2}a_{1}) = 0$.

$(\frac{G'}{G})^{-2}$: $9\mu^{2}-\gamma B(A-1)a_{-1}-3(3\lambda+2\sigma)a_{-1}^{2}a_{0} = 0$.

$(\frac{G'}{G})^{-3}$: $6\mu^{2}-\gamma C^{2}a_{-1}-(3\lambda+2\sigma)a_{-1}^{3} = 0$.

Next, we get the following the results:

Case 1.1. $a_{1} = \pm\sqrt{\frac{6\mu^{2}-\gamma(A-1)^{2}}{3\lambda+2\sigma}}$, $a_{0} = \pm\sqrt{\frac{\mu^{2}-\gamma[B^{2}+2C(A-1)]-(\nu+2ak+\delta+3k^{2}-\gamma)}{3\lambda+2\sigma}}$, $a_{-1} = 0$,

$\gamma = -\frac{\mu^{2}-\gamma[B^{2}+2C(A-1)]+2(\nu+2ak+\delta+3k^{2}-\gamma)}{6\mu^{2}(A-1)C}$.

Case 1.2. $a_{1} = 0$, $a_{0} = \pm\sqrt{\frac{\mu^{2}-\gamma[B^{2}+2C(A-1)]-(\nu+2ak+\delta+3k^{2}-\gamma)}{3\lambda+2\sigma}}$, $a_{-1} = \pm\sqrt{\frac{6\mu^{2}-\gamma C^{2}}{3\lambda+2\sigma}}$,

$\gamma = -\frac{\mu^{2}-\gamma[B^{2}+2C(A-1)]+2(\nu+2ak+\delta+3k^{2}-\gamma)}{6\mu^{2}(A-1)C}$.

Family 1. When $A\neq1$, $\Delta_{1} = B^{2}+4C-4AC > 0$, we obtain

where $C_{1}$ and $C_{2}$ are arbitrary constants.

Especially, if $C_{1}\neq0$, and $C_{2} = 0$ in Eq (2.7), we have

Family 2. When $A\neq1$, and $\Delta_{2} = B^{2}+4C-4AC < 0$, we obtain

Family 3. When $A\neq1$, and $B^{2}+4C-4AC = 0$, we obtain the rational function solution of Eq (1.1) as

2.2.   First integral method

Now, Eq (2.4) is equivalent to the following of two dimensional system

where $Q(X, Y) = \sum_{i = 0}^{m}a_{i}(X)Y^{i}$,

Then, based on the division theorem, there exists a polynomial $g(X)+h(X)Y$ in $C[X, Y]$ such that

Case 2.1. If $m = 1$.

Substituting Eq (2.21) into Eq (2.22) and calculating the coefficients of $Y^{i}(\xi) = (i = 0, 1, 2)$ both sides of Eq (2.22), we have

Since $a_{i}(X)(i = 0, 1)$ are polynomials, then we obtain when $deg(g(X)) = 1$

Substituting$a _{0}(X), a _{1}(X), g(X)$ into Eq (2.23) and setting all the coefficients of powers $X$ to be zero, then, we obtain

Using the conditions Eq (2.28) in Eq (2.21), we obtain

Combining Eq (2.29) with Eq (2.20), we obtain the exact solution for FPRKL equation which can be written as

Type 1. If $\frac{c+3ak+\delta-3k^{2}\gamma}{\gamma} < 0, (3\lambda+2\sigma)\gamma < 0$, we get

Type 2. If $\frac{c+3ak+\delta-3k^{2}\gamma}{\gamma} > 0, (3\lambda+2\sigma)\gamma > 0$, we get

Type 3. If $\frac{\nu+3ak+\delta-3k^{2}\gamma}{\gamma} = 0, (3\lambda+2\sigma)\gamma < 0$, we get

Case 2.2. If $m = 2$.

Comparing the coefficients of $Y^{i}(\xi) = (0, 1, 2, 3)$ on both sides of Eq (2.22), we obtain

Balancing the degrees of $g(X)$ and $a_{1}(X)$, we get $deg(g(X)) = 1$, $deg(a_{1}(X)) = 2$, then

where $A_{1}, A_{0}, B_{0}$ are all constants to be determined.

Now, Eq (2.37) becomes

where $d$ is the constant of integration.

Substituting $a_{0}(X), g(X), a_{1}(X)$ into Eq (2.36) and setting all the coefficients of powers $X$ to be zero, we obtain

From Eq (2.43) into Eq (2.21), we obtain

This shows that the two cases $m = 1$ and $m = 2$ give the same solutions.

2.3.   Complete discrimination system for the polynomial method

Multiplying $u^{'}$ on both sides of Eq (2.4), and again integrating it on $\xi$, we can get

where $a_{4} = -\frac{3\lambda+2\sigma}{6\mu^{2}\gamma}$, $a_{2} = -\frac{\nu+2ak+\delta-3k^{2}\gamma}{\mu^{2}\gamma}$, and $a_{0}$ is the constant.

Making the transformation $\psi = \pm \sqrt{-(\frac{2(3\lambda+2\sigma)}{3\mu^{2}\gamma})^{-\frac{1}{3}}u}$, $\xi_{1} = -(\frac{2(3\lambda+2\sigma)}{3\mu^{2}\gamma})^{\frac{1}{3}}\xi$, Eq (2.45) becomes

where $p_{1} = -\frac{4(\nu+2ak+\delta-3k^{2}\gamma)}{\mu^{2}\gamma}(-\frac{2(3\lambda+2\sigma)}{3\mu^{2}\gamma})^{-\frac{2}{3}}$, $p_{0} = -4a_{0}(\frac{2(3\lambda+2\sigma)}{3\mu^{2}\gamma})^{-\frac{1}{3}}$.

Integrating Eq (2.46), we have

where $\xi_{0}$ is the integration constant.

Suppose that $\Delta = p_{1}^{2}-4p_{0}$ and $G(\psi) = \psi^{2}+p_{1}\psi+p_{0}$, there are four cases for the solutions of Eq (2.4).

Case 3.1. $\Delta = 0$.

When $\psi > 0$, we have

If $p_{1} < 0$, the corresponding solutions are

If $p_{1} > 0$, the corresponding solutions are

If $p_{1} = 0$, we get the corresponding solutions

Case 3.2. $\Delta > 0$ and $p_{0} = 0$. As for $\psi > -p_{1}$, we have

If $p_{1} > 0$, the solutions are given as follows

If $p_{1} < 0$, the solutions are given as follows

Case 3.3. $\Delta > 0$, $p_{0}\neq0$.

Suppose that $\lambda_{1} < \lambda_{2} < \lambda_{3}$, $\lambda_{1}$, $\lambda_{2}$ and $\lambda_{3}$ are two roots of $G(\psi) = 0$. Here we make the transformation $\psi = \lambda_{1}+(\lambda_{2}-\lambda_{1})\sin^{2}\varphi$, it is clear that

where $m^{2}_{1} = \frac{\lambda_{2}-\lambda_{1}}{\lambda_{3}-\lambda_{1}}$. We get the corresponding solutions

If $\psi > \lambda_{3}$, we take the following transformation $\psi = \frac{-\lambda_{2}\sin^{2}\varphi+\lambda_{3}}{\cos ^{2}\varphi}$, the corresponding solutions are

Case 3.4. $\Delta < 0$, taking the transformation $\psi = \sqrt{p_{0}}\tan^{2}\frac{\varphi}{2}$ it is clear that

where $m^{2}_{2} = \frac{1}{2}(1-\frac{p_{1}}{2\sqrt{p_{0}}})$. we get the corresponding solutions

3.   Physical explanations

In this section, the numerical simulations of some remarkable solutions for the FPRKL equation are presented. By the $(G'/G)$-expansion method, we obtained the solution $\phi_{1_{1}}(x, t)$ and $\phi_{1_{2}}(x, t)$ shown in Figure 1. The graphical solutions $\phi_{10}(x, t)$ and $\phi_{11}(x, t)$ are shown in Figure 2. Moreover, the graphical solutions $\phi_{16}(x, t)$ and $\phi_{17}(x, t)$ are shown in Figure 3.

4.   Conclusions

In this article, we have investigated the exact solutions to the FPRKL equation by three different methods. Many exact solutions have been obtained. In the paper, we get all the traveling wave solutions, which have not been seen in other literature. These solutions might be further useful and effective to study more about the various forms of solitary waves in physics. We have noticed that the proposed complete discrimination system for the polynomial method gives much more new and general exact solutions than the other two suggested methods. In future work, we will consider the bifurcation, phase diagrams and exact solutions of the FPRKL equation.

Acknowledgments

This work was supported by Science Research Fund of Education Department of Sichuan Province of China under grant No.18ZB0537 and Scientific Research Funds of Sichuan Vocational and Technical College under grant No.2022YZB009.

Conflict of interest

The author declare no conflict of interest.