Numerical Solutions of Variable-Coefficient Fractional-in-Space KdV Equation with the Caputo Fractional Derivative

Abstract

In this paper, numerical solutions of the variable-coefficient Korteweg-De Vries (vcKdV) equation with space described by the Caputo fractional derivative operator is developed. The propagation and interaction of vcKdV equation in different cases, such as breather soliton and periodic suppression soliton, are numerically simulated. Especially, the Fourier spectral method is used to solve the fractional-in-space vcKdV equation with breather soliton. From numerical simulations and compared with other methods, it can be easily seen that our method has low computational complexity and higher precision.

1. Introduction

The KdV equation is one of the most notable integrable equations and has found numerous applications in many fields of science, such as plasma physics, nonlinear optics, telecommunications, fluid mechanics, condensed matter physics and dust plasma. The KdV equation was derived by Korteweg and de Vires in 1895 [1]. There are many analytical methods to obtain the analytical solutions of the KdV equation, including Hirota method, Darboux transformation and so on [2,3]. The symmetry method is used to solve many fractional differential equations, such as the seventh-order generalized KdV equation [4], the generalized KdV-Burgers-Kuramoto equation [5] and the time fractional generalized fifth-order KdV equation [6]. Qin [7] used the Hirota method to obtain the N-soliton solutions of the coupled KdV-mKdV system based on Bell polynomials. Chen [8] used the test function method combined with the bilinear to obtain the lump solutions to the generalized variable coefficient Burgers equation.

There are also many good numerical methods to solve the KdV equation. Yan [9] used a local discontinuous Galerkin method to solve the KdV equation. Jackaman [10] advanced the design of the conservative finite element discretizations for the vectorial modified KdV equation. Energy-conserving Hamiltonian boundary value methods were used to solve the KdV equation by Brugnano [11]. However, the numerical method that can be applied to solve the space fractional vcKdV equation with breather soliton and periodic suppression soliton is seldom researched. Therefore, a Fourier spectral method is developed in this paper which has low computational complexity and higher precision. We consider the KdV equation with the following form:

$$v_t+{\alpha }_1{v}^2{D}_x^{{\beta }_1}v+\epsilon v{D}_x^{{\beta }_2}v+{\alpha }_2{v}_{xxx}+{\alpha }_3{v}_x=0,$$

(1)

equipped with the following initial and boundary conditions,

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}\hfill v(x,0)=v_0,a\le x\le b,\hfill \\ v(a,t)=v(b,t)=0,\hfill \end{array}\right.\hfill \end{array}\end{array}$$

(2)

where $D_x^{{\beta }_1}v$, $D_x^{{\beta }_2}vs.$ denote Caputo fractional derivative operator which has the following form,

$$\begin{array}{c}\hfill D_x^{\beta }v(x)=\left\{\begin{array}{c}\hfill \frac{1}{\mathsf{\Gamma }(1-{\beta }_1)}{\int }_0^x{(x-\tau )}^{-{\beta }_1}\frac{\partial v(\tau )}{\partial \tau }d\tau ,0<{\beta }_1<1,\hfill \\ \\ \frac{\partial v(x)}{\partial x},{\beta }_1=1,\hfill \end{array}\right.\end{array}$$

(3)

and $\epsilon$, ${\alpha }_1$, ${\alpha }_2$, ${\alpha }_3$ are positive parameters.

When ${\alpha }_1={\alpha }_3={\alpha }_4={\beta }_1=0$, ${\alpha }_2=\mu$ and ${\beta }_2=1$, Equation (1) becomes the KdV equation,

$$v_t+\epsilon v{v}_x+\mu v_{xxx}=0.$$

(4)

When $\epsilon =\frac{3}{2}$, ${\alpha }_1=0$, ${\alpha }_2=\frac{1}{6}$, ${\alpha }_3=-{\omega }_0$, ${\beta }_1=0$ and ${\beta }_2=1$, Equation (1) is reduced to the gKdV equation [12],

$$\begin{array}{c}\hfill \begin{array}{c}\hfill v_t-{\omega }_0{v}_x+\frac{3}{2}v{v}_x+\frac{1}{6}{v}_{xxx}=0.\hfill \end{array}\end{array}$$

(5)

When $\epsilon ={\alpha }_3={\alpha }_4=0$ and ${\beta }_1=1$, Equation (1) becomes the modified KdV(mKdV) equation,

$$v_t+{\alpha }_1{v}^2{v}_x+{\alpha }_2{v}_{xxx}=0,$$

(6)

which has important applications in studying nonlinear optics [13,14], and quantum mechanics [15].

When $\epsilon =\epsilon (t)$, ${\beta }_2=1$, ${\alpha }_1={\alpha }_1(t)$, ${\beta }_1=1$, ${\alpha }_2={\alpha }_2(t)$, ${\alpha }_3={\alpha }_3(t)$ and ${\alpha }_1(t){\alpha }_2(t)\ne 0$, Equation (1) becomes the generalized variable-coefficient KdV-modified KdV equation [16],

$$v_t+\epsilon (t)v{v}_x+{\alpha }_1(t)v^2{v}_x+{\alpha }_2(t)v_{xxx}+{\alpha }_3(t)v_x=0.$$

(7)

Consulting materials in the literature, we discover that the numerical solutions of the space fractional vcKdV equation are rarely researched. Therefore, some numerical solutions of the vcKdV equation are given in this manuscript. From the numerical simulations and compared with other methods, it can be easily seen that our method has low computational complexity and higher precision.

This paper is organized as follows: In Section 2, we introduce the numerical method. Many numerical experiments are provided in Section 3. Section 4 concludes the paper.

2. Numerical Approximation Method

We use the Fourier spectral method [17,18,19] for spatial discretization. In order to standardize the space period $[a,b]$ to $[0,2\pi ]$, we use the transformations $x\to \frac{2\pi (x-a)}{L}$ and $L=b-a$. v is converted into a Fourier space with respect to x. We use Fast Fourier transform [20] to complete this operation. We use Fourier transform for Equation (1) in spatial domain,

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}\hfill \frac{\partial \widehat{v}}{\partial t}=-\epsilon \mathcal{F}[v{\mathcal{F}}^{-1}({(ik)}^{{\beta }_2}\widehat{v})]-{\alpha }_1\mathcal{F}[v^2{\mathcal{F}}^{-1}({(ik)}^{{\beta }_1}\widehat{v})]-{\alpha }_2{(ik)}^3\widehat{v}-{\alpha }_3(ik\widehat{v}),\hfill \\ \\ \widehat{v}(k,0)={\widehat{v}}_0,\hfill \end{array}\right.\hfill \end{array}\end{array}$$

(8)

Let $x_j=j\Delta x=\frac{2\pi Lj}{N}$, $N>0$ and N is an integer, $L=b-a$, $j=0,1,\cdots \cdots ,N-1$. We use the discrete Fourier transform for v,

$$\widehat{v}(k,t)=\mathcal{F}(v)=\frac{1}{N}\sum _{j=1}^{N-1}v(x_j,t)e^{-ik{x}_j},-\frac{N}{2}\le k\le \frac{N}{2}-1,$$

(9)

and the inverse formula is

$$v(x_j,t)={\mathcal{F}}^{-1}(\widehat{v})=\sum _{k=-\frac{N}{2}}^{\frac{N}{2}-1}\widehat{v}(k,t)e^{ik{x}_j},0\le j\le N-1.$$

(10)

Next, the fourth-order Runge–Kutta method is used to solve the ordinary differential Equation (8),

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{k}_1=r(t_n,{\widehat{v}}_n),\hfill \\ k_2=r(t_n+\frac{\tau }{2},{\widehat{v}}_n+\frac{\tau k_1}{2}),\hfill \\ k_3=r(t_n+\frac{\tau }{2},{\widehat{v}}_n+\frac{\tau k_2}{2}),\hfill \\ k_4=r(t_n+\tau ,{\widehat{v}}_n+\tau k_3),\hfill \\ {\widehat{v}}_{n+1}={\widehat{v}}_n+\frac{\tau }{6}(k_1+2k_2+2k_3+k_4),\hfill \end{array}\right.\hfill \end{array}\end{array}$$

(11)

where $\tau$ is step-size and $r(t,\widehat{v})=-\epsilon \mathcal{F}[v{\mathcal{F}}^{-1}({(ik)}^{{\beta }_2}\widehat{v})]-{\alpha }_1\mathcal{F}[v^2{\mathcal{F}}^{-1}({(ik)}^{{\beta }_1}\widehat{v})]-{\alpha }_2{(ik)}^3\widehat{v}-{\alpha }_3(ik\widehat{v})$.

We mark

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}V={({\widehat{v}}_0(k,t),{\widehat{v}}_1(k,t),\cdots ,{\widehat{v}}_{N-1}(k,t))}^T,\hfill \\ R(t,V)={(r_0(t,{\widehat{v}}_0(t)),r_1(t,{\widehat{v}}_0(t)),\cdots ,r_{N-1}(t,{\widehat{v}}_{N-1}(t)))}^T,\hfill \\ V_0(k)={({\widehat{v}}_{00}(k),{\widehat{v}}_{01}(k),\cdots ,{\widehat{v}}_{0(N-1)}(k))}^T,\hfill \end{array}\right.\hfill \end{array}\end{array}$$

(12)

where $n=1,\cdots ,\frac{T}{\tau }$.

Equation (8) can be reduced to

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}\hfill \frac{\partial V}{\partial t}=R(t,V),\hfill \\ \hfill V(k,0)=V_0.\hfill \end{array}\right.\hfill \end{array}\end{array}$$

(13)

Then, we can obtain the solving formula [21]

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{k}_{j1}=r_j(t_n,{\widehat{v}}_{0,n},{\widehat{v}}_{1,n},\cdots ,{\widehat{v}}_{N-1,n}),\hfill \\ k_{j2}=r_j(t_n+\frac{h}{2},{\widehat{v}}_{0,n}+\frac{h{k}_{01}}{2},\cdots ,{\widehat{v}}_{N-1,n}+\frac{h{k}_{(N-1)1}}{2}),\hfill \\ k_{j3}=r_j(t_n+\frac{h}{2},{\widehat{v}}_{0,n}+\frac{h{k}_{02}}{2},\cdots ,{\widehat{v}}_{N-1,n}+\frac{h{k}_{(N-1)2}}{2}),\hfill \\ k_{j4}=r_j(t_n+h,{\widehat{v}}_{0,n}+h{k}_{03},\cdots ,{\widehat{v}}_{N-1,n}+h{k}_{(N-1)3}),\hfill \\ {\widehat{v}}_{j,n+1}={\widehat{v}}_{j,n}+\frac{h}{6}(k_{j1}+2k_{j2}+2k_{j3}+k_{j4}),\hfill \end{array}\right.\hfill \end{array}\end{array}$$

(14)

where $j=0,1,\cdots ,N-1$.

Definition 1.

A class of the single-step method for solving an ordinary differential equation in the form of:

$$\begin{array}{c}\hfill V_{n+1}=V_n+h\varphi (V_n,t_n,h),\end{array}$$

(15)

where incremental function ϕ is determined by $R(t,V)$, that is a function of $V_n$, $t_n$, τ.

Theorem 1.

If $\varphi (V,t,\tau )$ satisfies the Lipschitz condition in V, then the numerical method that is given by Equation (15) is stable.

Proof. 

We refer the reader to [22,23,24] for the details of the proof. □

Lemma 1.

Let $e_n=V_n-V(t_n)$ and D is constant [22]. If $\parallel e_{n+1}\parallel \le (1+\tau L)\parallel e_n\parallel +D$, then

$$\begin{array}{c}\hfill \parallel e_n\parallel \le \frac{D{(1+\tau L)}^n}{\tau L}+{(1+\tau L)}^n\parallel e_0\parallel \le \frac{D}{\tau L}-(e^{LT}-1)+e^{LT}\parallel e_0\parallel .\end{array}$$

(16)

Theorem 2.

If $\varphi (V,t,\tau )$ is a continuous function in V, t, τ, which satisfies the Lipschitz condition on $0\le \tau \le {\tau }_0$, $0\le t\le T$, $-\infty \le v_i\le \infty$, then Equation (15) is convergent [22].

Theorem 3.

Let $V(t_n)$ be the analytical solution of problem (13) and $V_n$ is the numerical solution. If $\varphi (V,t,\tau )$ satisfies the Lipschitz condition on $0\le \tau \le {\tau }_0$, $0\le t\le T$ and Equation (15) is the fourth-order method, then the error estimate has the following form

$$\begin{array}{c}\hfill \left|e_n\right|\le e^{LT}(\left|e_0\right|+cT{\tau }^4+L\tau M_0),\end{array}$$

(17)

where $M_0=max(\left|e_0\right|,\left|e_1\right|,\cdots ,\left|e_{n-1}\right|)$ and L is lipschitz constant.

Proof. 

If Equation (15) is the fourth-order one-step method, then $V(t)$ satisfies

$$\begin{array}{c}\hfill V(t_{n+1})=V(t_n)+\tau \varphi (t_n,V(t_n),\tau )+\mathit{O}({\tau }^5).\end{array}$$

(18)

By denoting $e_n=V(t_n)-V_n$, we have

$$\begin{array}{c}\hfill \left|e_{n+1}\right|-\left|e_n\right|\le \left|e_{n+1}-e_n\right|=\left|\tau \left(\varphi (t_n,V(t_n),\tau )-\varphi (t_n,V_n,\tau )\right)+\mathit{o}({\tau }^5)\right|\le \tau L\left|e_n\right|+\mathit{O}({\tau }^5).\end{array}$$

(19)

Summing over n, we get

$$\begin{array}{c}\hfill \left|e_n\right|-\left|e_0\right|\le \tau L\sum _{k=0}^{n-1}\left|e_k\right|+nc{\tau }^5,\end{array}$$

(20)

$$\begin{array}{c}\hfill \left|e_n\right|\le \tau L\sum _{k=0}^{n-1}\left|e_k\right|+cT{\tau }^4+\left|e_0\right|.\end{array}$$

(21)

Using the Gronwall inequality, we have

$$\begin{array}{c}\hfill \left|e_n\right|\le e^{LT}(\left|e_0\right|+cT{\tau }^4+L\tau M_0).\end{array}$$

(22)

□

Finally, we find the numerical solution using the inverse discrete Fourier transform [25].

3. Simulation Results

Numerical solutions of the space fractional vcKdV equation are obtained in this chapter. We use the error norms, $L_2$, $L_{\infty }$ and $\mathrm{GRE}$ (global relative error) to test the accuracy of the method: $L_2=\sqrt{\frac{1}{N}{\sum }_{j=1}^N{[v(x_j,t)-v^{*}(x_j,t)]}^2}$, $L_{\infty }=\underset{1\le j\le N}{\max }\mid v(x_j,t)-v^{*}(x_j,t)\mid$, $\mathrm{GRE}=\frac{{\sum }_{j=1}^N\mid v(x_j,t)-v^{*}(x_j,t)\mid }{{\sum }_{j=1}^N\mid v^{*}(x_j,t)\mid }$, where $v(x_j,t)$ and $v^{*}(x_j,t)$ are the numerical solution and analytical solution. The order of convergence in space is computed by

$$\begin{array}{c}\hfill order={\displaystyle \frac{\log \frac{\parallel v_N(x,t)-v^{*}(x,t)\parallel }{\parallel v_{\frac{N}{2}}(x,t)-v^{*}(x,t)\parallel }}{\log 2}}.\end{array}$$

(23)

Example 1.

Let $\epsilon =1$, ${\beta }_2=1$ and ${\alpha }_1={\alpha }_3={\alpha }_4={\beta }_1=0$ in Equation (1). The initial value is as follows,

$$\begin{array}{c}\hfill v(x,0)=3c(sec{h}^2((kx-x_0)),\end{array}$$

(24)

with

$$\begin{array}{c}\hfill k=\frac{1}{2}\sqrt{\frac{c}{{\alpha }_2}},\omega =ck,\end{array}$$

(25)

where c, $x_0$ are constant parameters.

Referring to the numerical experiment in [26,27,28,29,30], the analytical solution is as follows:

$$\begin{array}{c}\hfill v^{*}(x,t)=3c(sec{h}^2((kx-\omega t-x_0)),-2\le x\le 2.\end{array}$$

(26)

In this simulation, we set ${\alpha }_2=4.84\times {10}^{-4}$, $c=0.5$, $x_0=6$, $\tau =0.01$ and $N=512$. By present method, the numerical results are given in

Table 1. Spatial numerical errors $L_{\infty }$, $L_2$ and their corresponding convergence rates at $t=0.001$ for Example 1.

N	${\mathit{L}}_{\infty }$	$\mathit{O}\mathit{r}\mathit{d}\mathit{e}\mathit{r}$	${\mathit{L}}_2$	$\mathit{O}\mathit{r}\mathit{d}\mathit{e}\mathit{r}$
32	$7.3429\times {10}^{-4}$	−	$4.1640\times {10}^{-4}$	−
64	$6.9269\times {10}^{-4}$	$N^{-0.0842}$	$2.4502\times {10}^{-4}$	$N^{-0.7651}$
128	$2.6108\times {10}^{-5}$	$N^{-4.7297}$	$8.3098\times {10}^{-6}$	$N^{-4.8819}$
256	$1.1068\times {10}^{-9}$	$N^{-14.5257}$	$3.8198\times {10}^{-10}$	$N^{-14.4090}$
512	$1.0904\times {10}^{-10}$	$N^{-3.3436}$	$3.2646\times {10}^{-11}$	$N^{-3.5486}$

.

Table 2. Comparison of absolute errors at $t=0.005$.

x	Absolute Error
	HBI [30]	ANS [29]	Hybrid [26]	B–Spline [27]	Present Method
$0.1$	$9.7700\times {10}^{-6}$	$2.0000\times {10}^{-8}$	$1.5900\times {10}^{-6}$	$1.1900\times {10}^{-6}$	$8.0000\times {10}^{-8}$
$0.5$	$6.7880\times {10}^{-3}$	$5.2021\times {10}^{-4}$	$4.8032\times {10}^{-4}$	$6.5700\times {10}^{-5}$	$1.0000\times {10}^{-8}$
$0.6$	$5.8627\times {10}^{-3}$	$4.9210\times {10}^{-5}$	$1.1316\times {10}^{-4}$	$1.8600\times {10}^{-5}$	$1.9000\times {10}^{-7}$
$0.7$	$5.9116\times {10}^{-4}$	$3.6400\times {10}^{-6}$	$1.3030\times {10}^{-5}$	$1.6500\times {10}^{-6}$	$1.0000\times {10}^{-8}$
$0.8$	$4.9840\times {10}^{-5}$	$6.4000\times {10}^{-7}$	$1.4100\times {10}^{-6}$	$1.3000\times {10}^{-7}$	$1.0000\times {10}^{-8}$
$0.9$	$4.1400\times {10}^{-6}$	$5.0000\times {10}^{-8}$	$1.2000\times {10}^{-7}$	$1.0000\times {10}^{-8}$	$1.0000\times {10}^{-8}$
$1.0$	$3.5000\times {10}^{-7}$	$1.0000\times {10}^{-8}$	$1.0000\times {10}^{-8}$	$0.0000$	$1.0000\times {10}^{-8}$

and

Table 3. Comparison of absolute errors at $t=0.01$.

x	Absolute Error
	HBI [30]	ANS [29]	Hybrid [26]	B–Spline [27]	Present Method
$0.2$	$2.3064\times {10}^{-4}$	$1.7400\times {10}^{-6}$	$5.0200\times {10}^{-6}$	$7.2000\times {10}^{-7}$	$8.0000\times {10}^{-8}$
$0.5$	$1.3048\times {10}^{-2}$	$9.8242\times {10}^{-4}$	$8.9174\times {10}^{-4}$	$1.1370\times {10}^{-4}$	$6.0000\times {10}^{-8}$
$0.6$	$1.1900\times {10}^{-2}$	$5.5860\times {10}^{-5}$	$2.3296\times {10}^{-4}$	$3.5900\times {10}^{-5}$	$1.9000\times {10}^{-7}$
$0.7$	$1.2044\times {10}^{-3}$	$7.4500\times {10}^{-6}$	$2.5860\times {10}^{-5}$	$4.0100\times {10}^{-6}$	$5.0000\times {10}^{-8}$
$0.8$	$1.0158\times {10}^{-4}$	$1.3900\times {10}^{-6}$	$2.9200\times {10}^{-6}$	$3.4000\times {10}^{-7}$	$4.0000\times {10}^{-8}$
$0.9$	$8.4300\times {10}^{-6}$	$1.2000\times {10}^{-7}$	$2.5000\times {10}^{-7}$	$3.0000\times {10}^{-8}$	$4.0000\times {10}^{-8}$
$1.0$	$7.0000\times {10}^{-7}$	$1.0000\times {10}^{-8}$	$3.0000\times {10}^{-8}$	$0.0000$	$4.0000\times {10}^{-8}$

show the absolute error by our numerical method, Hybrid method [26], B-spline method [27], ANS method [29], HBI method [30] at $t=0.005,0.01$. Then, we shall investigate the space fractional KdV equation.

Table 4. Comparison of $L_{\infty }$ of $v(x,t)$ at ${\alpha }_1=1$, ${\alpha }_2=1$, $k_1=0.5$, $k_2=1$ for Example 1.

t	${\mathit{\beta }}_1=0.9999$	${\mathit{\beta }}_1=0.99999$	${\mathit{\beta }}_1=0.999999$	${\mathit{\beta }}_1=0.9999999$	${\mathit{\beta }}_1=1$
$1.0$	$3.7351\times {10}^{-3}$	$3.7415\times {10}^{-4}$	$3.6993\times {10}^{-5}$	$4.0937\times {10}^{-6}$	$2.0209\times {10}^{-6}$
$1.5$	$6.9479\times {10}^{-3}$	$6.9721\times {10}^{-4}$	$7.0101\times {10}^{-5}$	$7.4340\times {10}^{-6}$	$1.2061\times {10}^{-6}$
$2.0$	$1.1371\times {10}^{-2}$	$1.1409\times {10}^{-3}$	$1.1363\times {10}^{-4}$	$1.1542\times {10}^{-5}$	$8.8914\times {10}^{-7}$
$2.5$	$1.6674\times {10}^{-2}$	$1.6785\times {10}^{-3}$	$1.6737\times {10}^{-4}$	$1.6364\times {10}^{-5}$	$8.7256\times {10}^{-7}$
$3.0$	$2.2625\times {10}^{-2}$	$2.2826\times {10}^{-3}$	$2.2962\times {10}^{-4}$	$2.4065\times {10}^{-5}$	$1.1692\times {10}^{-6}$

shows comparison of $L_{\infty }$ at different ${\beta }_2$. For the space fractional KdV equation, we take (26) as reference solution because the analytical solution can not be obtained. Figure 1 and Figure 2 show the logarithm of absolute errors by present method, Hybrid numerical method [26], B-spline method [27], ANS method [29], HBI method [30] at the the selected notes for $t=0.005,0.01$, which shows that our numerical method has higher accuracy than other methods. Comparisons are made between numerical solutions and analytical solutions at $t=0,1,2$ in Figure 3. Figure 4 presents absolute error at $x=2$. Numerical solutions at $t=1,2$ and ${\beta }_2=0.9,0.99,1$ are plotted in Figure 5 and Figure 6. Numerical solutions at different ${\beta }_2$ are presented in Figure 7.

From Example 1, we know that our numerical method has higher accuracy than other methods for the one-soliton solution. Next, we will study the generalized vc KdV-mKdV equation and the influence of ${\beta }_1$, ${\beta }_2$ on the numerical solution of this equation.

Example 2.

Let $\epsilon (t)=-6$, ${\alpha }_1(t)=0.06$, ${\beta }_1=2$, ${\beta }_2=1$, ${\alpha }_2(t)=0.01$, ${\alpha }_3(t)=-15\cos \frac{\pi t}{2}$ in Equation (1) and $\tau =0.01$, $N=512$. The initial value is as follows,

$$\begin{array}{c}\hfill v(x,0)=-\frac{202}{\mathsf{\Psi }}\exp (x+\frac{1}{100}).\end{array}$$

(27)

Wang [31] obtained one periodic depression soliton solution of the generalized vcKdV-mKdV equation,

$$\begin{array}{c}\hfill v^{*}(x,t)=-\frac{202}{\mathsf{\Psi }}\exp (\frac{t}{100}+\frac{30}{\pi }\sin \frac{\pi t}{2}+x+\frac{1}{100}),-20\le x\le 20,\end{array}$$

(28)

with

$$\begin{array}{c}\hfill \mathsf{\Psi }=20200\exp (\frac{t}{100}+\frac{30}{\pi }\sin \frac{\pi t}{2}+x+\frac{1}{100})+10001\exp (\frac{60}{\pi }\sin \frac{\pi t}{2}+2x+\frac{1}{50})+10201\exp (\frac{t}{50}).\end{array}$$

(29)

Table 5. Error norms $L_{\infty }$ and $\mathrm{GRE}$ at different times for Example 2.

	${\mathit{L}}_{\infty }$		$\mathbf{GRE}$
	Present Method	Ref. [32]	Present Method	Ref. [32]
$t=1.0$	$8.2189\times {10}^{-7}$	$5.1096\times {10}^{-6}$	$2.9499\times {10}^{-4}$	$7.4281\times {10}^{-4}$
$t=2.0$	$9.0010\times {10}^{-7}$	$7.3044\times {10}^{-6}$	$4.1717\times {10}^{-4}$	$1.2207\times {10}^{-3}$
$t=3.0$	$1.0152\times {10}^{-6}$	$1.1358\times {10}^{-5}$	$5.2455\times {10}^{-4}$	$1.8416\times {10}^{-3}$
$t=4.0$	$9.2295\times {10}^{-7}$	$1.4313\times {10}^{-5}$	$6.8842\times {10}^{-4}$	$2.4443\times {10}^{-3}$

shows $L_{\infty }$ and $GRE$ at different times.

Table 6. Numerical results of $v(x,t)$ obtained by the present method at $t=1$ for Example 2.

x	Analytical Solution	Numerical Solution	Absolute Error
$-13.75$	$-0.0003$	$-0.0003$	$1.9412\times {10}^{-9}$
$-12.5$	$-0.0009$	$-0.0009$	$6.7763\times {10}^{-9}$
$-11.25$	$-0.0026$	$-0.0026$	$1.3403\times {10}^{-8}$
$-10$	$-0.0047$	$-0.0047$	$1.9307\times {10}^{-9}$
$-8.75$	$-0.0043$	$-0.0043$	$3.8465\times {10}^{-8}$
$-7.5$	$-0.0020$	$-0.0020$	$1.6845\times {10}^{-8}$
$-6.25$	$-0.0007$	$-0.0007$	$2.2970\times {10}^{-8}$
$-5$	$-0.0002$	$-0.0002$	$3.6866\times {10}^{-9}$

shows numerical results. Figure 8, Figure 9 and Figure 10 represent numerical solution. Absolute errors at $t=3.4,3.5,3.6$ are shown in Figure 11, Figure 12, Figure 13 and Figure 14 present absolute errors at $x=-10,-5,10$. From Figure 15 which shows the numerical solutions at different ${\beta }_1$ and ${\beta }_2$, we can find that the change of ${\beta }_1$ and ${\beta }_2$ has minimal effect on the shape of periodic depression soliton.

From Example 2, we find that our numerical method has higher accuracy and low computational complexity than other methods for one periodic depression soliton solution of the generalized vc KdV-mKdV equation. Next, we will investigate the breather-type solution of the mKdV equation.

Example 3.

Let ${\alpha }_3={\alpha }_4=\epsilon =0$ and ${\beta }_1=1$ in Equation (1).

Case I The initial value is as follows,

$$\begin{array}{c}\hfill v(x,-4)=\pm 2\sqrt{\frac{6{\alpha }_2}{{\alpha }_1}}\frac{\partial }{\partial x}\arctan \left(\frac{k_2}{k_1}\frac{\sin (k_1{\alpha }_2^{-\frac{1}{3}}x-{\omega }_1(-4))}{\cosh ({\omega }_2(-4)-k_2{\alpha }_2^{-\frac{1}{3}}x)}\right),-20\le x\le 20,\end{array}$$

(30)

with

$$\begin{array}{c}\hfill {\omega }_1=3k_2^2{k}_1-k_1^3,{\omega }_2=k_2^3-3k_1^2{k}_2,\end{array}$$

(31)

where $k_1$, $k_2$ are constant parameters.

The breather-type solution is as follows [33]:

$$\begin{array}{c}\hfill v^{*}(x,t)=\pm 2\sqrt{\frac{6{\alpha }_2}{{\alpha }_1}}\frac{\partial }{\partial x}\arctan \left(\frac{k_2}{k_1}\frac{\sin (k_1{\alpha }_2^{-\frac{1}{3}}x-{\omega }_{1}t)}{\cosh ({\omega }_{2}t-k_2{\alpha }_2^{-\frac{1}{3}}x)}\right).\end{array}$$

(32)

In this simulation, we set ${\alpha }_1=1$, ${\alpha }_2=1$, $k_1=0.5$, $k_2=1$, $\tau =0.01$ and $N=512$.

Table 7. Numerical results of $v(x,t)$ at $t=1$,${\alpha }_1=1$, ${\alpha }_2=1$, $k_1=0.5$, $k_2=1$ for Case I.

x	Analytical Solution	Numerical Solution	Absolute Error
$-10.0$	$0.0003$	$0.0003$	$4.7709\times {10}^{-9}$
$-7.5$	$0.0094$	$0.0094$	$1.7568\times {10}^{-8}$
$-5.0$	$0.0306$	$0.0306$	$1.7562\times {10}^{-8}$
$-2.5$	$-1.1357$	$-1.1357$	$4.1767\times {10}^{-9}$
0	$-0.2941$	$-0.2941$	$2.4809\times {10}^{-8}$
$2.5$	$1.2591$	$1.2591$	$6.0335\times {10}^{-8}$
$5.0$	$-0.1163$	$-0.1163$	$7.8856\times {10}^{-8}$
$7.5$	$-0.0147$	$-0.0147$	$5.0883\times {10}^{-8}$

gives the numerical results and Figure 16, Figure 17 and Figure 18 show the numerical solution. Figure 19 shows the absolute errors at $t=1$. The absolute error at $x=5$ is plotted in Figure 20.

Table 8. Error norms, $L_2$, $L_{\infty }$ and $\mathrm{GRE}$ at ${\alpha }_1=1$, ${\alpha }_2=1$, $k_1=0.5$, $k_2=1$ for Example 3.

	$\mathit{t}=0$	$\mathit{t}=1.0$	$\mathit{t}=2.0$	$\mathit{t}=3.0$	$\mathit{t}=4.0$
$L_2$	$4.3364\times {10}^{-7}$	$4.1878\times {10}^{-8}$	$3.6983\times {10}^{-7}$	$7.3778\times {10}^{-7}$	$7.7441\times {10}^{-7}$
$L_{\infty }$	$5.0048\times {10}^{-7}$	$9.0930\times {10}^{-8}$	$4.5768\times {10}^{-7}$	$8.1363\times {10}^{-7}$	$8.4852\times {10}^{-7}$
$GRE$	$1.5727\times {10}^{-6}$	$1.1964\times {10}^{-7}$	$1.3157\times {10}^{-6}$	$2.6396\times {10}^{-6}$	$2.7113\times {10}^{-6}$

gives $L_2$, $L_{\infty }$ and $\mathrm{GRE}$ at different times. Then, we shall investigate the space fractional modified KdV equation. We take (32) as the analytical solution for the space fractional modified KdV equation. Figure 21 and Figure 22 show the numerical solutions at ${\beta }_1=0.6,0.99,1$ and $t=0,1$.

Table 9. Comparison of $L_{\infty }$ of $v(x,t)$ at ${\alpha }_1=1$, ${\alpha }_2=1$, $k_1=0.5$, $k_2=1$ for Example 3.

t	${\mathit{\beta }}_1=0.999999$	${\mathit{\beta }}_1=0.9999999$	${\mathit{\beta }}_1=0.99999999$	${\mathit{\beta }}_1=0.999999999$	${\mathit{\beta }}_1=1$
$-2$	$1.7658\times {10}^{-3}$	$1.7716\times {10}^{-4}$	$1.8260\times {10}^{-5}$	$2.3703\times {10}^{-6}$	$1.2617\times {10}^{-6}$
$-1$	$1.8815\times {10}^{-3}$	$1.8875\times {10}^{-4}$	$1.9565\times {10}^{-5}$	$2.6702\times {10}^{-6}$	$9.2254\times {10}^{-7}$
0	$5.4293\times {10}^{-3}$	$5.4241\times {10}^{-4}$	$5.3860\times {10}^{-5}$	$5.7137\times {10}^{-6}$	$5.0048\times {10}^{-7}$
1	$6.9070\times {10}^{-3}$	$6.9092\times {10}^{-4}$	$6.9067\times {10}^{-5}$	$6.8848\times {10}^{-6}$	$9.0930\times {10}^{-8}$
2	$9.6478\times {10}^{-3}$	$9.6405\times {10}^{-4}$	$9.6092\times {10}^{-5}$	$9.9524\times {10}^{-6}$	$5.5768\times {10}^{-6}$

gives $L_{\infty }$ at different ${\beta }_1$. Figure 23 shows numerical solutions at different ${\beta }_1$.

Case II The initial value is as follows,

$$\begin{array}{c}\hfill v(x,-2)=\pm 2\sqrt{\frac{6{\alpha }_2}{{\alpha }_1}}\frac{\partial }{\partial x}\arctan \left(\frac{k_2}{k_1}\frac{\sin (k_1{\alpha }_2^{-\frac{1}{3}}x-{\omega }_1(-2))}{\cosh ({\omega }_2(-2)-k_2{\alpha }_2^{-\frac{1}{3}}x)}\right),-10\le x\le 10.\end{array}$$

(33)

In this simulation, we set ${\alpha }_1=1$, ${\alpha }_2=1$, $k_1=1$, $k_2=-2$, $\tau =0.01$ and $N=256$.

Table 10. Numerical results of $v(x,t)$ at ${\alpha }_1=1$, ${\alpha }_2=1$, $k_1=1$, $k_2=-2$, $t=0$ for Case II.

x	Analytical Solution	Numerical Solution	Absolute Error
$-5.625$	$-0.0005$	$-0.0005$	$9.8195\times {10}^{-7}$
$-5.0$	$-0.0020$	$-0.0020$	$1.5196\times {10}^{-6}$
$-3.75$	$-0.0035$	$-0.0035$	$1.0557\times {10}^{-6}$
$-1.875$	$1.0078$	$1.0078$	$1.2865\times {10}^{-6}$
$-1.25$	$2.2706$	$2.2706$	$1.9922\times {10}^{-6}$
$0$	$-9.7980$	$-9.7980$	$1.6831\times {10}^{-5}$
$1.25$	$2.2706$	$2.2706$	$2.1611\times {10}^{-6}$
$1.875$	$1.0078$	$1.0078$	$1.6041\times {10}^{-6}$
$3.75$	$-0.0035$	$-0.0035$	$1.5075\times {10}^{-6}$

shows numerical results.

Table 11. Error norms, $L_2$, $L_{\infty }$ and $\mathrm{GRE}$ at ${\alpha }_1=1$, ${\alpha }_2=1$, $k_1=1$, $k_2=-2$ for Example 3.

	$\mathit{t}=-2$	$\mathit{t}=-1$	$\mathit{t}=0$	$\mathit{t}=1$	$\mathit{t}=2$
$L_2$	$4.0774\times {10}^{-16}$	$8.5229\times {10}^{-7}$	$2.7727\times {10}^{-6}$	$1.6213\times {10}^{-6}$	$1.5889\times {10}^{-6}$
$L_{\infty }$	$1.7764\times {10}^{-15}$	$2.1440\times {10}^{-6}$	$1.6831\times {10}^{-5}$	$3.5398\times {10}^{-6}$	$2.5029\times {10}^{-6}$
$GRE$	$5.2480\times {10}^{-16}$	$1.3009\times {10}^{-6}$	$3.2589\times {10}^{-6}$	$2.7586\times {10}^{-6}$	$2.7296\times {10}^{-6}$

gives $L_2$, $L_{\infty }$ and $\mathrm{GRE}$ at different times. Absolute error at $t=-1$ is plotted in Figure 24. Figure 25 and Figure 26 show absolute errors at $x=-5,5$. Figure 27 shows numerical solutions at ${\beta }_1=0.9,0.99,1$ and $t=0$.

Table 12. Comparison of $L_{\infty }$ of $v(x,t)$ at ${\alpha }_1=1$, ${\alpha }_2=1$, $k_1=1$, $k_2=-2$ for Example 3.

t	${\mathit{\beta }}_1=0.9999999$	${\mathit{\beta }}_1=0.99999999$	${\mathit{\beta }}_1=0.99999999$	${\mathit{\beta }}_1=0.9999999999$	${\mathit{\beta }}_1=1$
$-1$	$3.8303\times {10}^{-3}$	$3.8396\times {10}^{-4}$	$3.9336\times {10}^{-5}$	$4.8743\times {10}^{-6}$	$2.1440\times {10}^{-6}$
0	$1.6335\times {10}^{-2}$	$1.6269\times {10}^{-3}$	$1.7066\times {10}^{-4}$	$2.7020\times {10}^{-5}$	$1.6831\times {10}^{-5}$
1	$3.7859\times {10}^{-2}$	$3.7830\times {10}^{-3}$	$3.7557\times {10}^{-4}$	$3.8415\times {10}^{-5}$	$3.5398\times {10}^{-6}$
2	$5.7068\times {10}^{-2}$	$5.7102\times {10}^{-3}$	$5.6925\times {10}^{-4}$	$5.6869\times {10}^{-5}$	$2.5029\times {10}^{-6}$

presents $L_{\infty }$ at different ${\beta }_1$. Numerical solution at different ${\beta }_1$ are plotted in Figure 28. Figure 29, Figure 30 and Figure 31 present the numerical solution.

Through Example 3, we can obtain that if ${\beta }_1$ tends to 1, the numerical solution of the spatial fractional mKdV equation tends to the analytical solution of Equation (6); in addition, the numerical solution of the spatial fractional mKdV equation is very sensitive to a change in ${\beta }_1$.

4. Conclusions

In this manuscript, we study the influence of ${\beta }_1$, ${\beta }_2$ on the numerical solutions of the spatial fractional vcKdV equation. Comparisons are made between the present method and others methods; it can be easily seen that our method has low computational complexity and higher precision. Through Examples 1–3, we know that if ${\beta }_1,{\beta }_2$ tends to 1, the numerical solution of the spatial fractional vcKdV equation tends to the analytical solution of the original equation. Through Example 3, the solutions of the space fractional KdV equation are very sensitive to a change in ${\beta }_1$, ${\beta }_2$. From Example 2, we can find that a change in ${\beta }_1$ and ${\beta }_2$ has a minimal effect on the shape of a periodic depression soliton. These results are consistent with the numerical simulation of other scholars [24,27,32].

All computations are performed by the MatlabR2017b software.