The New G-Double-Laplace Transforms and One-Dimensional Coupled Sine-Gordon Equations

Abstract

This paper establishes a novel technique, which is called the G-double-Laplace transform. This technique is an extension of the generalized Laplace transform. We study its properties with examples and various theorems related to the G-double-Laplace transform that have been addressed and proven. Finally, we apply the G-double-Laplace transform decomposition method to solve the nonlinear sine-Gordon and coupled sine-Gordon equations. This method is a combination of the G-double-Laplace transform and decomposition method. In addition, some examples are examined to establish the accuracy and effectiveness of this technique.

1. Introduction

The nonlinear partial differential equation models in many areas of mathematical physics perform an essential role in theoretical sciences. The sine-Gordon equation is one of the widely important nonlinear equations that plays a vital role in physical science and engineering. This equation has many applications such as relativistic field theory, the Josephson junction, and mechanical transmission lines. The nonlinear sine-Gordon equation (SGE) is a class of hyperbolic partial differential equations, which is usually utilized to explain the physical phenomena in several areas of engineering and science, for instance, nonlinear problems in physics and the diffusion of fluxions; see [1]. The sine-Gordon equation was first introduced in the course of study of various problems of differential geometry [2]. The approximate solution of the one-dimensional sine-Gordon equation was obtained by utilizing the combined Laplace transform method and VIM [3]. The author in [4] studied the two-dimensional SGE on Cartesian grids by improving an energy-conserving local discontinuous Galerkin method. The numerical solutions of the one-dimensional coupled nonlinear sine-Gordon equations are obtained by using the geometric meshless method [5]. Many techniques have been applied to obtain the approximate analytical solutions of sine-Gordon equations, such as the reduced differential transform method [6] and the modified Adomian decomposition method applied to obtain approximate Jacobi elliptic function solutions of the (N + 1)-dimensional sine-Gordon equation [7], and the authors in [8] used an analytical method called the Natural Decomposition Method (NDM) for solving the nonlinear sine-Gordon equation. The nonlinear coupled sine-Gordon equations were first presented in [9]. The author in [10] studied the numerical solution of the coupled sine-Gordon equation by applying the modified decomposition method. Recently, many authors have utilized diverse methods to obtain the solution of nonlinear coupled sine-Gordon equations; for example, the authors in [11] introduced rational exponential techniques to find the exact solutions of a coupled sine-Gordon equation; the authors in [12] discussed the solution of sine-Gordon and coupled sine-Gordon equations by using the homotopy perturbation method. The author [13] employed the double-Sumudu transform iterative method for solving the one-dimensional coupled nonlinear sine-Gordon equation. The authors [14] applied the triple-Laplace transform method to obtain the solution of the two-dimensional coupled nonlinear sine-Gordon equation. In recent years, various mathematicians have applied the double-Laplace transform to solve different types of partial differential equations (see [15,16,17]). The generalized Laplace transform was introduced first in [18]; further properties of this transform are given in [19] and were subsequently utilized to solve certain nonlinear dynamical models with non-integer order in [20]. This work aims to introduce a new integral transform called the G-double-Laplace transform. This new integral transform is derived from the generalized Laplace transform. It is proposed with the decomposition method to facilitate the procedure of solving sine-Gordon and coupled sine-Gordon equations and to help researchers solve several problems related to physics and engineering in the future.

Here, we recall the definition of the generalized Laplace transform as follows.

Definition 1

([18]). Let $f\left(t\right)$ be an integrable function, for all $t\ge 0$. The generalized integral transform $G_{\alpha }$ of the function $f\left(t\right)$ is given by

$$F\left(s\right)=G_{\alpha }\left(f\right)=s^{\alpha }\underset{0}{\overset{\infty }{\int }}f\left(t\right)e^{-\frac{t}{s}}d\nu ,$$

(1)

for $s\in \mathbb{C}$ and $\alpha \in Z$.

2. Main Results

2.1. The Idea of G-Double-Laplace Transform

Here, we introduce some necessary definitions and theorems of the existence of G-double-Laplace transform, which are useful in this paper.

Definition 2.

Let $f\left(x,t\right)$ be an integrable function on $[0,\infty )\times [0,\infty )$. The G-double-Laplace transform of

$f\left(x,t\right)$ is given by

$$G_x{G}_t\left[f\left(x,t\right)\right]=F\left(p,s\right)=p^{\alpha }{s}^{\alpha }\underset{0}{\stackrel{\infty }{\int }}\underset{0}{\stackrel{\infty }{\int }}{e}^{-(\frac{x}{p}+\frac{t}{s})}f(x,t)dxdt,$$

(2)

where $\alpha \in \mathbb{Z}$, $p,s\in \mathbb{C}$, the symbol $G_x{G}_t$ indicates the transform of x and t, respectively, and the function $F\left(p,s\right)$ is denoted as the G-double Laplace transform of $f\left(x,t\right).$

A prominent advantage of the G-double-Laplace transform is that it is easy to generate the following transformations:

• If we put $\alpha =0$, $s=\frac{1}{s}$, and $p=\frac{1}{p}$, we obtain the double-Laplace transform:

  $$L_x{L}_t\left(f\left(x,t\right)\right)=F\left(p,s\right)=\underset{0}{\stackrel{\infty }{\int }}\underset{0}{\stackrel{\infty }{\int }}f\left(x,t\right)e^{-\left(px+st\right)}dtdx,$$

  (3)
• If we put $\alpha =0,$ $q=\frac{1}{p}$ and replace s with $\varpi$, we obtain the Laplace–Yang transform:

  $$L_{x}Y\left(f\left(x,t\right)\right)=F\left(p,\varpi \right)=\underset{0}{\stackrel{\infty }{\int }}\underset{0}{\stackrel{\infty }{\int }}f\left(x,t\right)e^{-\left(qx+\frac{t}{\varpi }\right)}dtdx,$$

  (4)
• With $\alpha =-1$ and replacing $p,s$ with $u,$ v, respectively, we obtain the double-Sumudu transform:

$$S_x{S}_t\left(f\left(x,t\right)\right)=F\left(u,v\right)=\frac{1}{uv}\underset{0}{\stackrel{\infty }{\int }}\underset{0}{\stackrel{\infty }{\int }}f\left(x,t\right)e^{-\left(\frac{x}{u}+\frac{t}{v}\right)}dtdx.$$

(5)

The inverse G-double-Laplace transform (IDGLT) is

$$G_p^{-1}{G}_s^{-1}\left(F\left(p,s\right)\right)=f\left(x,t\right)=\frac{1}{{\left(2\pi i\right)}^2}{\int }_{\tau -i\infty }^{\tau +i\infty }{\int }_{\varsigma -i\infty }^{\varsigma +i\infty }{e}^{\frac{1}{\rho }x+\frac{1}{s}t}F\left(p,s\right)dsd\rho ,$$

where $G_p^{-1}{G}_s^{-1}$ indicates the IDGLT.

The existence conditions of the G-double-Laplace transform of the function $f\left(x,t\right)$ are defined as follows. Assume that $f\left(x,t\right)$ is piecewise continuous on $[0,\infty )\times [0,\infty )$ and has an exponential order at infinity with

$$\left|f\left(x,t\right)\right|\le K{e}^{m_{1}x+m_{2}t},$$

(6)

for $x>X$, $t>T$, where $K\ge 0,$ and $m_1,m_2,X,T$ are constants.

We write

$$f\left(x,t\right)=O\left(e^{m_{1}x+m_{2}t}\right)\mathrm{as}x\to \infty ,t\to \infty$$

or similarly,

$$\underset{\begin{array}{c}x\to \infty \\ t\to \infty \end{array}}{lim}{e}^{-\frac{1}{\mu }x-\frac{1}{\lambda }t}\left|f\left(x,t\right)\right|=K\underset{\begin{array}{c}x\to \infty \\ t\to \infty \end{array}}{lim}{e}^{-\left(\frac{1}{\mu }-m_1\right)x-\left(\frac{1}{\lambda }-m_2\right)t}=0,$$

(7)

whenever $\frac{1}{\mu }>m_1$ and $\frac{1}{\lambda }>m_2$. Such a function $f\left(x,t\right)$ is simply called an exponential order as $x\to \infty$, $t\to \infty$, and clearly, it does not grow faster than $K{e}^{m_{1}x+m_{2}t}$ as $x\to \infty$, $t\to \infty$.

Theorem 1.

If the function $f\left(x,t\right)$ is a continuous function defined on $(0,X)$ and $(0,T)$ with exponential order $e^{m_{1}x+m_{2}t}$, then the G-double-Laplace transform of $f\left(x,t\right)$ exists for all $\mathrm{Re}\left(\frac{1}{p}\right)>$$\frac{1}{\mu }$ and $\mathrm{Re}\left(\frac{1}{s}\right)>$$\frac{1}{\lambda }$.

Proof. 

By placing Equation (2) into Equation (6), we obtain

$$\begin{array}{ccc}\hfill \left|F\left(p,s\right)\right|& =& \left|p^{\alpha }{s}^{\alpha }\underset{0}{\stackrel{\infty }{\int }}\underset{0}{\stackrel{\infty }{\int }}{e}^{-(\frac{x}{p}+\frac{t}{s})}f(x,t)dxdt\right|\le K\left|e^{\left(m_1-\frac{1}{p}\right)x+\left(m_2-\frac{1}{s}\right)t}\right|\hfill \\ & =& \frac{R{p}^{\alpha +1}{s}^{\alpha +1}}{\left(1-m_{1}p\right)\left(1-m_{2}s\right)}.\hfill \end{array}$$

Utilizing the condition $\mathrm{Re}\left(\frac{1}{p}\right)>$$\frac{1}{\mu }$ and $\mathrm{Re}\left(\frac{1}{s}\right)>$$\frac{1}{\lambda }$, we obtain

$$\underset{\begin{array}{c}p\to \infty \\ s\to \infty \end{array}}{lim}\left|F\left(p,s\right)\right|=0\mathrm{or}\underset{\begin{array}{c}p\to \infty \\ s\to \infty \end{array}}{lim}F\left(p,s\right)=0,$$

□

2.2. Some Properties of G-Double-Laplace Transform

In this part, several properties of the G-double-Laplace transform are explained and proven, such as the linearity property, the change of scale property, and the convolution theorem.

Linearity property: Consider $f(x,t)$ and $g(x,t)$ to be functions of the variables x and t, then

$$G_x{G}_t\left[af\left(x,t\right)+bg(x,t)\right]=aF\left(p,s\right)+bG\left(p,s\right).$$

Theorem 2.

Let $f(x,t)$ be a periodic function of periods a and b, where $f(x+a,t+b)=f(x,t)$ for all x and t, and if $F\left(p,s\right)$ exists, then

$$F\left(p,s\right)=\frac{p^{\alpha }{s}^{\alpha }{\int }_0^a{\int }_0^b{e}^{-\frac{x}{p}-\frac{t}{s}}f(x,t)dtdx}{\left(1-e^{-\frac{a}{p}-\frac{b}{s}}\right)}.$$

Proof. 

Using Definition 1, we obtain

$$G_x{G}_t\left[f\left(x,t\right)\right]=p^{\alpha }{s}^{\alpha }{\int }_0^a{\int }_0^b{e}^{-\frac{x}{p}-\frac{t}{s}}f(x,t)dtdx+p^{\alpha }{s}^{\alpha }{\int }_a^{\infty }{\int }_b^{\infty }{e}^{-\frac{x}{p}-\frac{t}{s}}f(x,t)dtdx.$$

Putting $x=u+a$, $t=v+b$ in the last double-integral in the above equation, we obtain

$$\begin{array}{ccc}\hfill F\left(p,s\right)& =& p^{\alpha }{s}^{\alpha }{\int }_0^a{\int }_0^b{e}^{-\frac{x}{p}-\frac{t}{s}}f(x,t)dtdx+p^{\alpha }{s}^{\alpha }{\int }_0^{\infty }{\int }_0^{\infty }{e}^{-\frac{u}{p}-\frac{v}{s}}f(u+a,v+b)dvdu\hfill \\ & =& p^{\alpha }{s}^{\alpha }{\int }_0^a{\int }_0^b{e}^{-\frac{x}{p}-\frac{t}{s}}f(x,t)dtdx+p^{\alpha }{s}^{\alpha }{e}^{-\frac{a}{p}-\frac{b}{s}}{\int }_0^{\infty }{\int }_0^{\infty }{e}^{-\frac{u}{p}-\frac{v}{s}}f(u,v)dvdu\hfill \\ \hfill F\left(p,s\right)& =& p^{\alpha }{s}^{\alpha }{\int }_0^a{\int }_0^b{e}^{-\frac{x}{p}-\frac{t}{s}}f(x,t)dtdx+e^{-\frac{a}{p}-\frac{b}{s}}F\left(p,s\right),\hfill \end{array}$$

therefore

$$F\left(p,s\right)=\frac{p^{\alpha }{s}^{\alpha }{\int }_0^a{\int }_0^b{e}^{-\frac{x}{p}-\frac{t}{s}}f(x,t)dtdx}{\left(1-e^{-\frac{a}{p}-\frac{b}{s}}\right)}.$$

□

Theorem 3.

If $G_x{G}_t\left[f\left(x,t\right)\right]=F\left(p,s\right),$ then

$$G_x{G}_t\left[f(x-\rho ,t-\sigma )H\left(x-\rho ,t-\sigma \right)\right]=e^{-\frac{\rho }{p}-\frac{\sigma }{s}}F\left(p,s\right)$$

where $H(x,t)$ is the Heaviside unit step function.

Proof. 

Using Equation (38), we obtain

$$\begin{array}{ccc}& & G_x{G}_t\left[f(x-\rho ,t-\sigma )H\left(x-\rho ,t-\sigma \right)\right]\hfill \\ & =& p^{\alpha }{s}^{\alpha }\underset{0}{\stackrel{\infty }{\int }}\underset{0}{\stackrel{\infty }{\int }}{e}^{-(\frac{x}{p}+\frac{t}{s})}f(x-\rho ,t-\sigma )H\left(x-\rho ,t-\sigma \right)dxdt\hfill \\ & =& p^{\alpha }{s}^{\alpha }\underset{\rho }{\stackrel{\infty }{\int }}\underset{\sigma }{\stackrel{\infty }{\int }}{e}^{-(\frac{x}{p}+\frac{t}{s})}f(x-\rho ,t-\sigma )dxdt\hfill \end{array}$$

where $H(x,t)$ is the Heaviside unit step function defined by $H\left(x-\rho ,t-\sigma \right)=1,$ at $x>\rho$ and $t>\rho$, and $H\left(x-\rho ,t-\sigma \right)=0$ when $x<\rho$ and $t<\sigma$. Putting $\zeta =x-\rho$ and $\eta =t-\sigma ,$ we obtain

$$\begin{array}{ccc}\hfill G_x{G}_t\left[f(x-\rho ,t-\sigma )H\left(x-\rho ,t-\sigma \right)\right]& =& p^{\alpha }{s}^{\alpha }{e}^{-\frac{\rho }{p}-\frac{\sigma }{s}}\underset{0}{\stackrel{\infty }{\int }}\underset{0}{\stackrel{\infty }{\int }}{e}^{-(\frac{x}{p}+\frac{t}{s})}f(\zeta ,\eta )d\zeta d\eta \hfill \\ & =& e^{-\frac{\rho }{p}-\frac{\sigma }{s}}F\left(p,s\right)\hfill \end{array}$$

□

Theorem 4.

(Convolution theorem) Let $G_x{G}_t\left[f\left(x,t\right)\right]$ and $G_x{G}_t\left[y\left(x,t\right)\right]$ exist and $G_x{G}_t\left[f\left(x,t\right)\right]=$ $F\left(p,s\right)$, $G_x{G}_t\left[y\left(x,t\right)\right]=$ $Y\left(p,s\right)$, then

$$p^{\alpha }{s}^{\alpha }{G}_x{G}_t\left[f\left(x,t\right)\ast \ast y\left(x,t\right)\right]=F\left(p,s\right)Y\left(p,s\right),$$

(8)

where

$$f\left(x,t\right)\ast \ast y\left(x,t\right)=\underset{0}{\overset{x}{\int }}\underset{0}{\overset{t}{\int }}f(x-\zeta ,t-\eta )y(\zeta ,\eta )d\zeta d\eta ,$$

and the symbol $\ast \ast$ indicates the double-convolution with respect to x and t.

Proof. 

Using the definition of the G-double-Laplace transform, we obtain

$$\begin{array}{ccc}{G}_x{G}_t\left[f\left(x,t\right)\ast \ast y\left(x,t\right)\right]\hfill & =\hfill & p^{\alpha }{s}^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{e}^{-\frac{x}{p}-\frac{t}{s}}f\left(x,t\right)\ast \ast y\left(x,t\right)dxdt\hfill \\ \hfill & =\hfill & p^{\alpha }{s}^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{e}^{-\frac{x}{p}-\frac{t}{s}}\left(\underset{0}{\overset{x}{\int }}\underset{0}{\overset{t}{\int }}f(x-\zeta ,t-\eta )y(\zeta ,\eta )d\zeta d\eta \right)dxdt.\hfill \end{array}$$

(9)

Set $\rho =x-\zeta$ and $\sigma =t-\eta$, and apply the upper bound of integrals to $x\to \infty$ and $t\to \infty$; Equation (42) can be written as

$$\begin{array}{ccc}\hfill & \hfill & G_x{G}_t\left[f\left(x,t\right)\ast \ast y\left(x,t\right)\right]\hfill \\ \hfill & =\hfill & p^{\alpha }{s}^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{e}^{-\frac{\zeta }{p}-\frac{\eta }{s}}f(x-\rho ,t-\sigma )d\zeta d\eta \underset{-\zeta }{\overset{\infty }{\int }}\underset{-\eta }{\overset{\infty }{\int }}{e}^{-\frac{\rho }{p}-\frac{\sigma }{s}}y(\rho ,\sigma )d\rho d\sigma ;\hfill \end{array}$$

(10)

the functions $f\left(x,t\right)$ and $g\left(x,t\right)$ equal zero at x < $0,$ t < 0; therefore, it follows with respect to the lower limit of integrations that

$$\begin{array}{ccc}& & G_x{G}_t\left[f\left(x,t\right)\ast \ast y\left(x,t\right)\right]\hfill \\ & =& \frac{1}{p^{\alpha }{s}^{\alpha }}\left(p^{\alpha }{s}^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{e}^{-\zeta \mu -\frac{\eta }{s}}f(\zeta ,\eta )d\zeta d\eta \right)\left(p^{\alpha }{s}^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{e}^{-p\rho -\frac{\sigma }{s}}y(\rho ,\sigma )d\rho d\sigma \right),\hfill \end{array}$$

and it is easy to see that

$$p^{\alpha }{s}^{\alpha }{G}_x{G}_t\left[f\left(x,t\right)\ast \ast g\left(x,t\right)\right]=F\left(p,s\right)Y\left(p,s\right).$$

□

Example 1

([21]). The G-double-Laplace transform of $f(x,t)={\left(xt\right)}^n$ is determined by

$$G_x{G}_t\left[{\left(xt\right)}^n\right]={\left(n!\right)}^2{s}^{\alpha +n+1}{p}^{\alpha +n+1},$$

where n is a non-negative integer. If $\mu (>-1)$ and $\nu (>-1)$$\in \mathbb{R}$, then

$$G_x{G}_t\left[x^{\mu }{t}^{\nu }\right]=\Gamma \left(\mu +1\right)\Gamma \left(\nu +1\right)p^{\alpha +\mu +1}{s}^{\alpha +\nu +1},$$

can be derived from the definition of the G-double-Laplace transform, and we have

$$\begin{array}{ccc}\hfill G_x{G}_t\left[x^{\mu }{t}^{\nu }\right]& =& p^{\alpha }{s}^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{x}^{\mu }{t}^{\nu }{e}^{-\frac{x}{p}-\frac{t}{s}}dtdx\hfill \\ & =& \left(p^{\alpha }\underset{0}{\overset{\infty }{\int }}{x}^{\mu }{e}^{-\frac{x}{p}}dx\right)\left(s^{\alpha }\underset{0}{\overset{\infty }{\int }}{t}^{\nu }{e}^{-\frac{t}{s}}dt\right).\hfill \end{array}$$

(11)

Now. by substituting $k=\frac{x}{p},$ $x=kp,$ $r=\frac{t}{s},$ $t=rs$ into Equation (11), we obtain

$$\begin{array}{ccc}\hfill G_x{G}_t\left[x^{\mu }{t}^{\nu }\right]& =& \left(p^{\alpha +\mu +1}\underset{0}{\overset{\infty }{\int }}{k}^{\mu }{e}^{-k}dk\right)\left(s^{\alpha +\nu +1}\underset{0}{\overset{\infty }{\int }}{r}^{\nu }{e}^{-r}dr\right)\hfill \\ & =& \Gamma \left(\mu +1\right)\Gamma \left(\nu +1\right)p^{\alpha +\mu +1}{s}^{\alpha +\nu +1},\hfill \end{array}$$

where the Gamma functions of $\mu +1$ and $\Gamma \left(\nu +1\right)$ are defined by the convergent integrals:

$$\begin{array}{ccc}\hfill \Gamma \left(\mu +1\right)& =& \underset{0}{\overset{\infty }{\int }}{k}^{\mu }{e}^{-k}dk,\mu >0\hfill \\ \hfill \Gamma \left(\nu +1\right)& =& \underset{0}{\overset{\infty }{\int }}{r}^{\nu }{e}^{-r}dk,\nu >0\hfill \end{array}$$

Example 2.

The G-double-Laplace transform for the following function:

$$f(x,t)=H\left(x\right)\otimes H\left(t\right)lnxlnt,$$

is given by

$$G_x{G}_t\left[H\left(x\right)\otimes H\left(t\right)lnxlnt\right]=s^{\alpha }{p}^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}lnxlnt{e}^{-\frac{x}{p}-\frac{\nu }{s}}dtdx,$$

where $H(x,t)=H\left(x\right)\otimes H\left(t\right)$ is the dimensional Heaviside function and ⊗ is a tensor product; see [22].

Let $\zeta =\frac{1}{p}x$ and $\eta =\frac{1}{s}t;$ the integral becomes

$$\begin{array}{ccc}\hfill G_x{G}_t\left[H\left(x\right)\otimes H\left(t\right)lnxlnt\right]& =& p^{\alpha +1}{s}^{\alpha +1}{\int }_0^{\infty }{e}^{-\eta }ln\left(s\eta \right)\left({\int }_0^{\infty }{e}^{-\zeta }ln\left(p\zeta \right)d\zeta \right)d\eta \hfill \\ & =& p^{\alpha +1}{s}^{\alpha +1}\left(-\gamma +lnp\right)\left(-\gamma +lns\right),\hfill \end{array}$$

(12)

where $\gamma ={\int }_0^{\infty }{e}^{-\eta }ln\left(\eta \right)d\eta =$ 0.5772 … is Euler’s constant.

Theorem 5.

If the G-double-Laplace transform of the function $f(x,t)$ is given by $G_x{G}_t\left[f(x,t)\right]=F\left(p,s\right)$, then $\left(GDLT\right)$ of $\frac{{\partial }^{n}f(x,t)}{\partial x^n}$ and $\frac{{\partial }^{n}f\left(x,t\right)}{\partial t^n}$ are given:

$$G_x{G}_t\left[\frac{{\partial }^{n}f(x,t)}{\partial x^n}\right]=\frac{F\left(p,s\right)}{p^n}-p^{\alpha }\sum _{k=1}^n\frac{1}{p^{n-k}}{G}_t\left[\frac{{\partial }^{k-1}f\left(0,t\right)}{\partial x^{k-1}}\right],$$

(13)

and

$$G_x{G}_t\left[\frac{{\partial }^{n}f(x,t)}{\partial t^n}\right]=\frac{F\left(p,s\right)}{s^n}-s^{\alpha }\sum _{k=1}^n\frac{1}{s^{n-k}}{G}_x\left[\frac{{\partial }^{k-1}f\left(x,0\right)}{\partial t^{k-1}}\right].$$

(14)

Proof. 

Now, by substituting $n=1$ in Equation (13),

$$G_x{G}_t\left[\frac{\partial f(x,t)}{\partial x}\right]=s^{\alpha }\underset{0}{\overset{\infty }{\int }}{e}^{-\frac{t}{s}}\left[p^{\alpha }\underset{0}{\overset{\infty }{\int }}{e}^{-\frac{x}{p}}\frac{\partial f\left(x,t\right)}{\partial x}dx\right]dt,$$

(15)

first, we calculate the integral inside the bracket for Equation (15), and we obtain

$$\begin{array}{ccc}\hfill p^{\alpha }\underset{0}{\overset{\infty }{\int }}{e}^{-\frac{x}{p}}\frac{\partial f\left(x,t\right)}{\partial x}dx& =& p^{\alpha }{\left[e^{-\frac{x}{p}}f\left(x,t\right)\right]}_0^{\infty }+\frac{p^{\alpha }}{p}\underset{0}{\overset{\infty }{\int }}{e}^{-\frac{x}{p}}f\left(x,t\right)dx\hfill \\ & =& \frac{1}{p}F\left(p,t\right)-p^{\alpha }f\left(0,t\right),\hfill \end{array}$$

(16)

then, by substituting Equation (16) into Equation (15), we have

$$\begin{array}{ccc}\hfill G_x{G}_t\left[\frac{\partial f(x,t)}{\partial x}\right]& =& s^{\alpha }\underset{0}{\overset{\infty }{\int }}{e}^{-\frac{t}{s}}\left[\frac{1}{p}{F}_{\alpha }\left(p,t\right)-p^{\alpha }f\left(0,t\right)\right]dt\hfill \\ & =& s^{\alpha }\underset{0}{\overset{\infty }{\int }}{e}^{-\frac{t}{s}}\frac{1}{p}{F}_{\alpha }\left(p,t\right)dt-s^{\alpha }\underset{0}{\overset{\infty }{\int }}{e}^{-\frac{t}{s}}{p}^{\alpha }f\left(0,t\right)dt;\hfill \end{array}$$

we conclude that

$$G_x{G}_t\left[\frac{\partial f(x,t)}{\partial x}\right]=\frac{1}{p}F\left(p,s\right)-p^{\alpha }{G}_t\left[f\left(0,t\right)\right].$$

(17)

Let us consider the case of $n=2$. In a similar manner,

$$G_x{G}_t\left[\frac{{\partial }^{2}f(x,t)}{\partial x^2}\right]=\frac{1}{p^2}{F}_{2\alpha }\left(p,s\right)-p^{\alpha -1}{G}_t\left[f\left(0,t\right)\right]-p^{\alpha }{G}_t\left[\frac{\partial f\left(0,t\right)}{\partial x}\right].$$

Let us suppose that $n=m$ is valid for some m. Thus,

$$\begin{array}{ccc}{G}_x{G}_t\left[\frac{{\partial }^{m}f\left(x,t\right)}{\partial x^m}\right]\hfill & =\hfill & \frac{1}{p^m}{F}_{2\alpha }\left(p,s\right)-\frac{p^{\alpha }}{p^{m-1}}{G}_t\left[f\left(0,t\right)\right]-\frac{p^{\alpha }}{p^{m-2}}{G}_t\left[\frac{\partial f\left(0,t\right)}{\partial x}\right]\hfill \\ \hfill & \hfill & -\cdots -p^{\alpha -1}{G}_t\left[\frac{{\partial }^{m-2}f\left(0,t\right)}{\partial {\mu }^{m-2}}\right]-p^{\alpha }{G}_t\left[\frac{{\partial }^{m-1}f\left(0,t\right)}{\partial {\mu }^{m-1}}\right],\hfill \end{array}$$

(18)

holds for $\frac{{\partial }^0}{\partial }=1$; see [19]; now we show that

$$\begin{array}{ccc}{G}_x{G}_t\left[\frac{{\partial }^{m+1}f\left(x,t\right)}{\partial x^{m+1}}\right]\hfill & =\hfill & \frac{1}{p^{m+1}}F\left(p,s\right)-\frac{p^{\alpha }}{p^m}{G}_t\left[f\left(0,\nu \right)\right]-\frac{p^{\alpha }}{p^{m-1}}{G}_t\left[\frac{\partial f\left(0,\nu \right)}{\partial \mu }\right]\hfill \\ \hfill & \hfill & -\cdots -p^{\alpha -2}{G}_t\left[\frac{{\partial }^{m-1}f\left(0,\nu \right)}{\partial {\mu }^{m-1}}\right]-p^{\alpha -1}{G}_t\left[\frac{{\partial }^{m}f\left(0,\nu \right)}{\partial {\mu }^m}\right].\hfill \end{array}$$

(19)

By the notion of $n=1$, we have

$$\begin{array}{ccc}\hfill G_x{G}_t\left[\frac{{\partial }^{m+1}f\left(x,t\right)}{\partial x^{m+1}}\right]& =& \frac{1}{p}\left[\begin{array}{c}\frac{1}{p^m}F\left(p,s\right)-\frac{p^{\alpha }}{p^{m-1}}{G}_t\left[f\left(0,t\right)\right]-\frac{p^{\alpha }}{p^{m-2}}{G}_t\left[\frac{\partial f\left(0,t\right)}{\partial x}\right]\\ -\cdots -p^{\alpha -1}{G}_t\left[\frac{{\partial }^{m-2}f\left(0,t\right)}{\partial x^{m-2}}\right]-p^{\alpha }{G}_t\left[\frac{{\partial }^{m-1}f\left(0,t\right)}{\partial x^{m-1}}\right]\end{array}\right]\hfill \\ & & -p^{\alpha }{G}_t\left[\frac{{\partial }^{m}f\left(0,t\right)}{\partial x^m}\right],\hfill \end{array}$$

From the above equation, the formula inside the bracket can be written in the following form:

$$\begin{array}{ccc}\hfill G_x{G}_t\left[\frac{{\partial }^{m}f\left(x,t\right)}{\partial x^m}\right]& =& \frac{1}{p^m}F\left(p,s\right)-\frac{p^{\alpha }}{p^{m-1}}{G}_t\left[f\left(0,t\right)\right]-\frac{p^{\alpha }}{p^{m-2}}{G}_t\left[\frac{\partial f\left(0,t\right)}{\partial x}\right]\hfill \\ & & \cdots -p^{\alpha }{G}_t\left[\frac{{\partial }^{m-1}f\left(0,t\right)}{\partial x^{m-1}}\right];\hfill \end{array}$$

therefore,

$$G_x{G}_t\left[\frac{{\partial }^{m+1}f\left(x,t\right)}{\partial x^{m+1}}\right]=\frac{1}{p}{G}_x{G}_t\left[\frac{{\partial }^{m}f\left(x,t\right)}{\partial x^m}\right]-p^{\alpha }{G}_x{G}_t\left[\frac{{\partial }^{m}f\left(0,t\right)}{\partial x^m}\right];$$

hence, Equation (13) can be written as follows:

$$G_x{G}_t\left[\frac{{\partial }^{n}f(x,t)}{\partial x^n}\right]=\frac{F\left(p,s\right)}{p^n}-p^{\alpha }\sum _{k=1}^n\frac{1}{p^{n-k}}{G}_t\left[\frac{{\partial }^{k-1}f\left(0,t\right)}{\partial x^{k-1}}\right].$$

Similarly, we can prove Equation (14). □

In the following example, we apply the G-double-Laplace transform to solve the one-dimensional telegraphic equation.

Example 3.

Consider the homogeneous telegraphic equation denoted by

$${\psi }_{xx}-{\psi }_{tt}-{\psi }_t-\psi =0,$$

(20)

with boundary conditions:

$$\psi \left(0,t\right)=e^{-t},{\psi }_x\left(0,t\right)=e^{-t},$$

(21)

and initial conditions:

$$\psi \left(x,0\right)=e^x,{\psi }_t\left(x,0\right)=-e^x.$$

(22)

Using the G-double-Laplace transform on Equation (20), we have

$$\begin{array}{ccc}& & \frac{1}{p^2}\mathsf{\Psi }\left(p,s\right)-p^{\alpha -1}{G}_t\left[\psi \left(0,t\right)\right]-p^{\alpha }{G}_t\left[{\psi }_x\left(0,t\right)\right]\hfill \\ & & -\frac{1}{s^2}\mathsf{\Psi }\left(p,s\right)-s^{\alpha -1}{G}_x\left[\psi \left(x,0\right)\right]-s^{\alpha }{G}_x\left[{\psi }_t\left(x,0\right)\right]\hfill \\ & & -\frac{1}{s}\mathsf{\Psi }\left(p,s\right)-s^{\alpha }{G}_x\left[\psi \left(x,0\right)\right]-\mathsf{\Psi }\left(p,s\right)\hfill \\ & =& 0.\hfill \end{array}$$

(23)

Taking the G-Laplace transform for Equations (21) and (22), we obtain

$$\begin{array}{ccc}\hfill G_t\left[\psi \left(0,t\right)\right]& =& \frac{s^{\alpha +1}}{\left(1+s\right)},G_t\left[{\psi }_x\left(0,t\right)\right]=\frac{s^{\alpha +1}}{\left(1+s\right)}\hfill \\ \hfill G_x\left[\psi \left(x,0\right)\right]& =& \frac{p^{\alpha +1}}{\left(1-p\right)},G_x\left[{\psi }_t\left(x,0\right)\right]=-\frac{p^{\alpha +1}}{\left(1-p\right)},\hfill \end{array}$$

(24)

substituting Equation (24) into Equation (23), we obtain

$$\begin{array}{ccc}& & \frac{1}{p^2}\mathsf{\Psi }\left(p,s\right)-p^{\alpha -1}\frac{s^{\alpha +1}}{\left(1+s\right)}-p^{\alpha }\frac{s^{\alpha +1}}{\left(1+s\right)}\hfill \\ & & -\left[\frac{1}{s^2}\mathsf{\Psi }\left(p,s\right)-s^{\alpha -1}\frac{p^{\alpha +1}}{\left(1-p\right)}+s^{\alpha }\frac{p^{\alpha +1}}{\left(1-p\right)}\right]\hfill \\ & & -\left[\frac{1}{s}\mathsf{\Psi }\left(p,s\right)-s^{\alpha }\frac{p^{\alpha +1}}{\left(1-p\right)}\right]-\mathsf{\Psi }\left(p,s\right)\hfill \\ & =& 0;\hfill \end{array}$$

(25)

through arranging Equation (25), we obtain

$$\begin{array}{ccc}& & \left[\frac{1}{p^2}-\frac{1}{s^2}-\frac{1}{s}-1\right]\mathsf{\Psi }\left(p,s\right)\hfill \\ & =& p^{\alpha -1}\frac{s^{\alpha +1}}{\left(1+s\right)}+p^{\alpha }\frac{s^{\alpha +1}}{\left(1+s\right)}-s^{\alpha -1}\frac{p^{\alpha +1}}{\left(1-p\right)};\hfill \end{array}$$

(26)

by simplifying Equation (26), we have

$$\mathsf{\Psi }\left(p,s\right)=\frac{s^{\alpha +1}{p}^{\alpha +1}}{\left(1+s\right)\left(1-p\right)}.$$

(27)

Taking the inverse G-double-Laplace transform for Equation (27), we obtain the solution of Equation (20) as follows:

$$\psi \left(x,t\right)=e^{x-t}.$$

The solution is in agreement with the result obtained by [16].

3. G-Double-Laplace Transform Decomposition Method and One-Dimensional Coupled Sine-Gordon Equation

This section discusses the nonlinear initial-boundary-value problem (IBVP) of the one-dimensional sine-Gordon and coupled sine-Gordon equations by applying the G-double-Laplace transform decomposition methods. Here, we offer two problems.

Problem 1: In this part, the essential notion of the G-double-Laplace decomposition method is demonstrated in the following nonlinear sine-Gordon equation:

$${\psi }_{tt}-{\psi }_{xx}=\sin \left(\psi \left(x,t\right)\right)$$

(28)

subject to the initial condition:

$$\psi \left(x,0\right)=f\left(x\right),{\psi }_t\left(x,0\right)=g\left(x\right).$$

(29)

By using the G-double-Laplace transform for Equation (28) and the G-Laplace transform for Equation (29), we obtain

$$\begin{array}{ccc}\hfill \mathsf{\Psi }\left(p,s\right)& =& s^{\alpha +1}\mathsf{\Psi }\left(p,0\right)+s^{\alpha +2}{\mathsf{\Psi }}_t\left(p,0\right)\hfill \\ & & +s^2{G}_x{G}_t\left[{\psi }_{xx}+\sin \left(\psi \left(x,t\right)\right)\right],\hfill \end{array}$$

(30)

and

$$\mathsf{\Psi }\left(p,0\right)=F\left(p\right)\mathrm{and}{\mathsf{\Psi }}_t\left(p,0\right)=G\left(p\right),$$

(31)

respectively. Then, substituting Equation (31) into Equation (30), we obtain

$$\begin{array}{ccc}\hfill \mathsf{\Psi }\left(p,s\right)& =& s^{\alpha +1}F\left(p\right)+s^{\alpha +2}G\left(p\right)\hfill \\ & & +s^2{G}_x{G}_t\left[{\psi }_{xx}+\sin \left(\psi \left(x,t\right)\right)\right].\hfill \end{array}$$

(32)

Applying the inverse G-double-Laplace transform of Equation (32), we achieve

$$\psi \left(x,t\right)=f\left(x\right)+tg\left(x\right)+G_p^{-1}{G}_s^{-1}\left[s^2{G}_x{G}_t\left[{\psi }_{xx}+\sin \left(\psi \right)\right]\right],$$

(33)

where the symbol $G_p^{-1}{G}_s^{-1}$ indicates the G-double-Laplace transform, and we assume here the inverse exists. Now, we suppose a series solution is given by

$$\psi \left(x,t\right)={\displaystyle \sum }_{m=0}^{\infty }{\psi }_m\left(x,t\right),$$

(34)

By substituting Equation (34) into Equation (33), we obtain

$${\displaystyle \sum }_{m=0}^{\infty }{\psi }_m=f\left(x\right)+tg\left(x\right)+G_p^{-1}{G}_s^{-1}\left[s^2{G}_x{G}_t\left[{\displaystyle \sum }_{m=0}^{\infty }{\psi }_{mxx}+{\displaystyle \sum }_{m=0}^{\infty }{A}_m\right]\right],$$

(35)

where $A_m$ is a nonlinear term given by

$$\sin \left(\psi \left(x,t\right)\right)={\displaystyle \sum }_{m=0}^{\infty }{A}_m,$$

(36)

where the first few terms of the Adomian polynomials $A_m$ are denoted by

$$\begin{array}{ccc}\hfill A_0& =& \sin \left({\psi }_0\left(x,t\right)\right),A_1={\psi }_1\left(x,t\right)\cos \left({\psi }_0\left(x,t\right)\right)\hfill \\ \hfill A_2& =& {\psi }_2\left(x,t\right)\cos \left({\psi }_0\left(x,t\right)\right)-\frac{1}{2!}{\psi }_1^2\left(x,t\right)\sin \left({\psi }_0\left(x,t\right)\right)\hfill \\ \hfill A_3& =& {\psi }_2\left(x,t\right)\cos \left({\psi }_0\left(x,t\right)\right)-{\psi }_1\left(x,t\right){\psi }_2\left(x,t\right)\sin \left({\psi }_0\left(x,t\right)\right)\hfill \\ & & -\frac{1}{3!}{\psi }_1^3\left(x,t\right)\cos \left({\psi }_0\left(x,t\right)\right).\hfill \end{array}$$

(37)

For more details, see [23]. Then, by matching both sides of Equation (35) above, we can easily create the recursive relation as follows:

$${\psi }_0\left(x,t\right)=f\left(x\right)+tg\left(x\right),$$

and the rest of the components are given by

$${\psi }_{m+1}=G_p^{-1}{G}_s^{-1}\left[s^2{G}_x{G}_t\left[{\psi }_{mxx}+A_m\right]\right],$$

for all $m\ge 0.$

Example 4.

Consider the following nonlinear sine-Gordon equation of the form:

$${\psi }_{tt}-{\psi }_{xx}=\sin \left(\psi \left(x,t\right)\right),$$

(38)

with the initial conditions:

$$\psi \left(x,0\right)=\frac{\pi }{2},{\psi }_t\left(x,0\right)=1.$$

(39)

Using the G-double-Laplace transform of Equation (38) with the initial conditions, we obtain

$$\begin{array}{ccc}\hfill \mathsf{\Psi }\left(p,s\right)& =& s^{\alpha +1}\mathsf{\Psi }\left(p,0\right)+s^{\alpha +2}{\mathsf{\Psi }}_t\left(p,0\right)\hfill \\ & & +s^2{G}_x{G}_t\left[{\psi }_{xx}+\sin \left(\psi \left(x,t\right)\right)\right],\hfill \end{array}$$

(40)

where $\mathsf{\Psi }\left(p,0\right)$ and ${\mathsf{\Psi }}_t\left(p,0\right)$ are given by

$$\mathsf{\Psi }\left(p,0\right)=\frac{\pi p^{\alpha +1}}{2}and{\mathsf{\Psi }}_t\left(p,0\right)=p^{\alpha +1};$$

(41)

substituting Equation (41) into Equation (40), we obtain

$$\mathsf{\Psi }\left(p,s\right)=\frac{\pi p^{\alpha +1}{s}^{\alpha +1}}{2}+p^{\alpha +1}{s}^{\alpha +2}+s^2{G}_x{G}_t\left[{\psi }_{xx}+\sin \left(\psi \left(x,t\right)\right)\right].$$

(42)

Applying the inverse G-double-Laplace transform of Equation (42), we achieve

$$\psi \left(x,t\right)=\frac{\pi }{2}+t+G_p^{-1}{G}_s^{-1}\left[s^2{G}_x{G}_t\left[{\psi }_{xx}+\sin \left(\psi \right)\right]\right].$$

(43)

The solution of Equation (38) is given by infinite series as follows:

$$\psi \left(x,t\right)={\displaystyle \sum }_{m=0}^{\infty }{\psi }_m\left(x,t\right),$$

(44)

By substituting Equation (44) into Equation (43), we obtain

$${\displaystyle \sum }_{m=0}^{\infty }{\psi }_m=\frac{\pi }{2}+t+G_p^{-1}{G}_s^{-1}\left[s^2{G}_x{G}_t\left[{\displaystyle \sum }_{m=0}^{\infty }{\psi }_{mxx}+{\displaystyle \sum }_{m=0}^{\infty }{A}_m\right]\right],$$

(45)

where $A_m$ is the Adomian polynomial that is given by Equations (36) and (37). Then, by matching both sides of Equation (45), we can easily create the iterative relation as follows:

$$\begin{array}{ccc}\hfill {\psi }_0& =& \frac{\pi }{2}+t\hfill \\ \hfill {\psi }_{m+1}& =& G_p^{-1}{G}_s^{-1}\left[s^2{G}_x{G}_t\left[{\psi }_{mxx}+A_m\right]\right],\hfill \end{array}$$

(46)

where $m\ge 0;$ at $m=0,$ we have

$$\begin{array}{ccc}\hfill {\psi }_1& =& G_p^{-1}{G}_s^{-1}\left[s^2{G}_x{G}_t\left[{\psi }_{0xx}+A_0\right]\right]\hfill \\ & =& G_p^{-1}{G}_s^{-1}\left[s^2{G}_x{G}_t\left[\sin \left(\frac{\pi }{2}+t\right)\right]\right]\hfill \\ & =& G_p^{-1}{G}_s^{-1}\left[s^2{G}_x{G}_t\left[\cos \left(t\right)\right]\right]\hfill \\ & =& G_p^{-1}{G}_s^{-1}\left[s^2{G}_x{G}_t\left[1-\frac{t^2}{2!}+\frac{t^4}{4!}-\dots \right]\right]\hfill \\ & =& G_p^{-1}{G}_s^{-1}\left[s^{\alpha +3}-s^{\alpha +5}+s^{\alpha +7}-\dots \right]\hfill \\ & =& \frac{t^2}{2!}-\frac{t^4}{4!}+\frac{t^6}{6!}-\dots \hfill \\ \hfill {\psi }_1& =& 1-\cos t;\hfill \end{array}$$

at $m=1$,

$$\begin{array}{ccc}\hfill {\psi }_2& =& G_p^{-1}{G}_s^{-1}\left[s^2{G}_x{G}_t\left[{\psi }_{1xx}+A_1\right]\right]\hfill \\ & =& G_p^{-1}{G}_s^{-1}\left[s^2{G}_x{G}_t\left[\left(1-\cos t\right)\cos \left(\frac{\pi }{2}+t\right)\right]\right]\hfill \\ & =& G_p^{-1}{G}_s^{-1}\left[s^2{G}_x{G}_t\left[\sin t\cos t-\sin t\right]\right]\hfill \\ & =& G_p^{-1}{G}_s^{-1}\left[s^2{G}_x{G}_t\left[\frac{1}{2}\sin 2t-\sin t\right]\right]\hfill \\ & =& G_p^{-1}{G}_s^{-1}\left[s^2{G}_x{G}_t\left[\begin{array}{c}\frac{1}{2}\left(2t-\frac{{\left(2t\right)}^3}{3!}+\frac{{\left(2t\right)}^5}{5!}-\dots \right)\\ -\left(t-\frac{{\left(t\right)}^3}{3!}+\frac{{\left(t\right)}^5}{5!}-\dots \right)\end{array}\right]\right]\hfill \\ & =& G_p^{-1}{G}_s^{-1}\left[\begin{array}{c}\frac{1}{8}\left(8s^{\alpha +4}-32s^{\alpha +6}+128s^{\alpha +8}-\dots \right)-\\ -\left(s^{\alpha +4}-3s^{\alpha +6}+s^{\alpha +8}-\dots \right)\end{array}\right]\hfill \\ & =& -\frac{3}{4}t-\frac{1}{8}\left(2t-\frac{{\left(2t\right)}^3}{3!}+\frac{{\left(2t\right)}^5}{5!}-\dots \right)\hfill \\ & & +\left(t-\frac{{\left(t\right)}^3}{3!}+\frac{{\left(t\right)}^5}{5!}-\dots \right)\hfill \\ & =& \sin t-\frac{\sin 2t}{8}-\frac{3}{4}t,\hfill \end{array}$$

and so on. Then, the approximation solution of Equations (38) and (39) is given by

$$\psi ={\psi }_0+{\psi }_1+{\psi }_2+\dots .$$

Therefore,

$$\psi =\frac{\pi }{2}+t+1-\cos t+\sin t-\frac{\sin 2t}{8}-\frac{3}{4}t+\dots ;$$

hence,

$$\psi =\frac{\pi }{2}+t+\frac{t^2}{2!}-\frac{t^4}{4!}+\dots .$$

The approximate solution is in agreement with the result obtained by [8,23].

Problem 2: Consider the following general form of the one- dimensional coupled sine-Gordon equation:

$$\begin{array}{ccc}\hfill {\psi }_{tt}-{\psi }_{xx}& =& -\sin \left(\psi -\varphi \right)+f\left(x,t\right)\hfill \\ \hfill {\varphi }_{tt}-{\varphi }_{xx}& =& \sin \left(\psi -\varphi \right)+g\left(x,t\right),\hfill \end{array}$$

(47)

with initial conditions:

$$\begin{array}{ccc}\hfill \psi \left(x,0\right)& =& f_1\left(x\right),{\psi }_t\left(x,0\right)=f_2\left(x\right)\hfill \\ \hfill \varphi \left(x,0\right)& =& g_1\left(x\right),{\varphi }_t\left(x,0\right)=g_2\left(x\right),\hfill \end{array}$$

(48)

and boundary conditions:

$$\begin{array}{ccc}\hfill \psi \left(0,t\right)& =& f_3\left(t\right),{\psi }_x\left(0,t\right)=f_4\left(t\right)\hfill \\ \hfill \varphi \left(0,t\right)& =& g_3\left(t\right),{\varphi }_x\left(0,t\right)=g_4\left(t\right).\hfill \end{array}$$

(49)

Taking the G-double-Laplace transform on both sides of the system of Equation (47) and the single G-Laplace transform for the initial, boundary conditions and simplifying the output, we obtain

$$\begin{array}{ccc}\hfill \left(\frac{1}{s^2}-\frac{1}{p^2}\right)\mathsf{\Psi }\left(p,s\right)& =& \frac{p^2{s}^2}{p^2-s^2}\left(s^{\alpha -1}{F}_1\left(p\right)+s^{\alpha }{F}_2\left(p\right)-p^{\alpha -1}{F}_3\left(s\right)\right)\hfill \\ & & -p^{\alpha }{F}_4\left(s\right)+F\left(p,s\right)+G_x{G}_t\left[{\psi }_{xx}\right]\hfill \\ & & -G_x{G}_t\left[\sin \left(\psi -\varphi \right)\right],\hfill \end{array}$$

(50)

and

$$\begin{array}{ccc}\hfill \left(\frac{1}{s^2}-\frac{1}{p^2}\right)\mathsf{\Phi }\left(p,s\right)& =& \frac{p^2{s}^2}{p^2-s^2}\left(s^{\alpha -1}{G}_1\left(p\right)+s^{\alpha }{G}_2\left(p\right)-p^{\alpha -1}{G}_3\left(s\right)\right)\hfill \\ & & -p^{\alpha }{G}_4\left(s\right)+G\left(p,s\right)+G_x{G}_t\left[{\varphi }_{xx}\right]\hfill \\ & & +G_x{G}_t\left[\sin \left(\psi -\varphi \right)\right];\hfill \end{array}$$

(51)

by arranging Equations (50) and (51), we obtain

$$\begin{array}{ccc}\hfill \mathsf{\Psi }\left(p,s\right)& =& \frac{p^2{s}^2}{p^2-s^2}\left(s^{\alpha -1}{F}_1\left(p\right)+s^{\alpha }{F}_2\left(p\right)-p^{\alpha -1}{F}_3\left(s\right)\right)\hfill \\ & & -\frac{p^2{s}^2}{p^2-s^2}\left(p^{\alpha }{F}_4\left(s\right)-F\left(p,s\right)\right)\hfill \\ & & +\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[{\psi }_{xx}-\sin \left(\psi -\varphi \right)\right],\hfill \end{array}$$

(52)

and

$$\begin{array}{ccc}\hfill \mathsf{\Phi }\left(p,s\right)& =& \frac{p^2{s}^2}{p^2-s^2}\left(s^{\alpha -1}{G}_1\left(p\right)+s^{\alpha }{G}_2\left(p\right)-p^{\alpha -1}{G}_3\left(s\right)\right)\hfill \\ & & -\frac{p^2{s}^2}{p^2-s^2}\left(p^{\alpha }{G}_4\left(s\right)+G\left(p,s\right)\right)\hfill \\ & & +\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[{\varphi }_{xx}+\sin \left(\psi -\varphi \right)\right].\hfill \end{array}$$

(53)

Using the inverse G-double-Laplace transform in Equations (52) and (53),

$$\begin{array}{ccc}\hfill \psi \left(x,t\right)& =& G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}\left(s^{\alpha -1}{F}_1\left(p\right)+s^{\alpha }{F}_2\left(p\right)-p^{\alpha -1}{F}_3\left(s\right)\right)\right]\hfill \\ & & -G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}\left(p^{\alpha }{F}_4\left(s\right)-F\left(p,s\right)\right)\right]\hfill \\ & & +G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[{\psi }_{xx}-\sin \left(\psi -\varphi \right)\right]\right],\hfill \end{array}$$

(54)

and

$$\begin{array}{ccc}\hfill \varphi \left(x,t\right)& =& G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}\left(s^{\alpha -1}{G}_1\left(p\right)+s^{\alpha }{G}_2\left(p\right)-p^{\alpha -1}{G}_3\left(s\right)\right)\right]\hfill \\ & & -G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}\left(p^{\alpha }{G}_4\left(s\right)+G\left(p,s\right)\right)\right]\hfill \\ & & +G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[{\varphi }_{xx}+\sin \left(\psi -\varphi \right)\right]\right],\hfill \end{array}$$

(55)

The solution of Equation (47) is given by an infinite series as follows:

$$\psi \left(x,t\right)={\displaystyle \sum }_{m=0}^{\infty }{\psi }_m\left(x,t\right),\varphi \left(x,t\right)={\displaystyle \sum }_{m=0}^{\infty }{\varphi }_m\left(x,t\right);$$

(56)

by substituting Equation (56) into Equations (54) and (55), we obtain

$$\begin{array}{ccc}{\displaystyle {\displaystyle \sum }_{m=0}^{\infty }}{\psi }_m\hfill & =\hfill & G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}\left(s^{\alpha -1}{F}_1\left(p\right)+s^{\alpha }{F}_2\left(p\right)-p^{\alpha -1}{F}_3\left(s\right)\right)\right]\hfill \\ \hfill & \hfill & -G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}\left(p^{\alpha }{F}_4\left(s\right)-F\left(p,s\right)\right)\right]\hfill \\ \hfill & \hfill & +G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[{\displaystyle {\displaystyle \sum }_{m=0}^{\infty }}{\psi }_{mxx}-\sin {\displaystyle {\displaystyle \sum }_{m=0}^{\infty }}\left({\psi }_m-{\varphi }_m\right)\right]\right],\hfill \end{array}$$

(57)

and

$$\begin{array}{ccc}{\displaystyle {\displaystyle \sum }_{m=0}^{\infty }}{\varphi }_m\hfill & =\hfill & G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}\left(s^{\alpha -1}{G}_1\left(p\right)+s^{\alpha }{G}_2\left(p\right)-p^{\alpha -1}{G}_3\left(s\right)\right)\right]\hfill \\ \hfill & \hfill & -G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}\left(p^{\alpha }{G}_4\left(s\right)+G\left(p,s\right)\right)\right]\hfill \\ \hfill & \hfill & +G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[{\displaystyle {\displaystyle \sum }_{m=0}^{\infty }}{\varphi }_{mxx}+\sin {\displaystyle {\displaystyle \sum }_{m=0}^{\infty }}\left({\psi }_m-{\varphi }_m\right)\right]\right];\hfill \end{array}$$

the iterative relation of Equations (57) and (58) is defined by

$$\begin{array}{ccc}\hfill {\psi }_0& =& G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}\left(s^{\alpha -1}{F}_1\left(p\right)+s^{\alpha }{F}_2\left(p\right)-p^{\alpha -1}{F}_3\left(s\right)\right)\right]\hfill \\ & & -G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}\left(p^{\alpha }{F}_4\left(s\right)-F\left(p,s\right)\right)\right],\hfill \end{array}$$

(58)

$$\begin{array}{ccc}\hfill {\varphi }_0& =& G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}\left(s^{\alpha -1}{G}_1\left(p\right)+s^{\alpha }{G}_2\left(p\right)-p^{\alpha -1}{G}_3\left(s\right)\right)\right]\hfill \\ & & -G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}\left(p^{\alpha }{G}_4\left(s\right)+G\left(p,s\right)\right)\right],\hfill \end{array}$$

(59)

and

$$\begin{array}{ccc}\hfill {\psi }_{m+1}& =& G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[{\psi }_{mxx}-\sin \left({\psi }_m-{\varphi }_m\right)\right]\right]\hfill \\ \hfill {\varphi }_{m+1}& =& G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[{\varphi }_{mxx}+\sin \left({\psi }_m-{\varphi }_m\right)\right]\right],\hfill \end{array}$$

(60)

where $m\ge 0;$ we suppose that the inverse of Equations (58)–(60) exist.

In the following examples, the one-dimensional coupled sine-Gordon equations are studied by applying the G-double-Laplace transform decomposition method to show the effectiveness of this method.

Example 5.

Consider the following IBVP for a one-dimensional coupled sine-Gordon equation:

$$\begin{array}{ccc}\hfill {\psi }_{tt}-{\psi }_{xx}& =& -\sin \left(\psi -\varphi \right)+\sin \left(t\sin \left(x\right)-e^{-t}\sin \left(x\right)\right)\hfill \\ & & +t\sin \left(x\right)\hfill \\ \hfill {\varphi }_{tt}-{\varphi }_{xx}& =& \sin \left(\psi -\varphi \right)-\sin \left(t\sin \left(x\right)-e^{-t}\sin \left(x\right)\right)\hfill \\ & & +2e^{-t}\sin \left(x\right),\hfill \end{array}$$

(61)

with IC:

$$\begin{array}{ccc}\hfill \psi \left(x,0\right)& =& 0,{\psi }_t\left(x,0\right)=\sin \left(x\right)\hfill \\ \hfill \varphi \left(x,0\right)& =& \sin \left(x\right),{\varphi }_t\left(x,0\right)=-\sin \left(x\right),\hfill \end{array}$$

(62)

and BC:

$$\begin{array}{ccc}\hfill \psi \left(0,t\right)& =& 0,{\psi }_x\left(0,t\right)=t\hfill \\ \hfill \varphi \left(0,t\right)& =& 0,{\varphi }_x\left(0,t\right)=e^{-t}.\hfill \end{array}$$

(63)

Applying the G-double-Laplace transform of Equation (61) and the single G-Laplace transform for initial, boundary conditions and simplifying the output, we acquire

$$\begin{array}{ccc}\hfill \mathsf{\Psi }\left(p,s\right)& =& \frac{p^{\alpha +4}{s}^{\alpha +2}}{\left(p^2-s^2\right)\left(1+p^2\right)}-\frac{p^{\alpha +2}{s}^{\alpha +4}}{\left(p^2-s^2\right)}+\frac{p^{\alpha +4}{s}^{\alpha +4}}{\left(p^2-s^2\right)\left(1+p^2\right)}\hfill \\ & & +\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[{\psi }_{xx}-\sin \left(\psi -\varphi \right)+\sin \left(t\sin \left(x\right)-e^{-t}\sin \left(x\right)\right)\right]\hfill \\ & & ,\hfill \end{array}$$

(64)

and

$$\begin{array}{ccc}\mathsf{\Phi }\left(p,s\right)\hfill & =\hfill & \frac{p^{\alpha +4}{s}^{\alpha +1}\left(1+s\right)-p^{\alpha +4}{s}^{\alpha +2}\left(1+s\right)}{\left(p^2-s^2\right)\left(1+p^2\right)\left(1+s\right)}\hfill \\ \hfill & \hfill & \frac{-p^{\alpha +2}{s}^{\alpha +3}\left(1+p^2\right)+2p^{\alpha +4}{s}^{\alpha +3}}{\left(p^2-s^2\right)\left(1+p^2\right)\left(1+s\right)}\hfill \\ \hfill & \hfill & +\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[{\varphi }_{xx}+\sin \left(\psi -\varphi \right)-\sin \left(t\sin \left(x\right)-e^{-t}\sin \left(x\right)\right)\right].\hfill \end{array}$$

(65)

Arranging Equations (64) and (65), one can obtain

$$\mathsf{\Psi }\left(p,s\right)=\frac{p^{\alpha +2}{s}^{\alpha +2}}{\left(1+p^2\right)}+\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[\Delta \right],$$

(66)

and

$$\mathsf{\Phi }\left(p,s\right)=\frac{p^{\alpha +2}{s}^{\alpha +1}}{\left(1+p^2\right)\left(1+s\right)}+\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[\Lambda \right],$$

(67)

where

$$\begin{array}{ccc}\hfill \Delta & =& {\psi }_{xx}-\sin \left(\psi -\varphi \right)+\sin \left(t\sin \left(x\right)-e^{-t}\sin \left(x\right)\right)\hfill \\ \hfill \Lambda & =& {\varphi }_{xx}+\sin \left(\psi -\varphi \right)-\sin \left(t\sin \left(x\right)-e^{-t}\sin \left(x\right)\right).\hfill \end{array}$$

Using the inverse G-double-Laplace transform in Equations (52) and (53),

$$\psi \left(x,t\right)=t\sin x+G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[\Delta \right]\right],$$

(68)

and

$$\varphi \left(x,t\right)=e^{-t}\sin x+G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[\Lambda \right]\right];$$

(69)

by substituting Equation (56) into Equations (68) and (69), we obtain

$${\displaystyle \sum }_{m=0}^{\infty }{\psi }_m=t\sin x+G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[{\displaystyle \sum }_{m=0}^{\infty }{\Delta }_m\right]\right],$$

(70)

and

$${\displaystyle \sum }_{m=0}^{\infty }{\varphi }_m=e^{-t}\sin x+G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[{\displaystyle \sum }_{m=0}^{\infty }{\Lambda }_m\right]\right];$$

(71)

we obtain the following approximations:

$${\psi }_0=t\sin x,{\varphi }_0=e^{-t}\sin x,$$

and

$$\begin{array}{ccc}\hfill {\psi }_{m+1}& =& G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[{\Delta }_m\right]\right]\hfill \\ \hfill {\varphi }_{m+1}& =& G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[{\Lambda }_m\right]\right],\hfill \end{array}$$

where $m\ge 0;$ at $m=0$

$$\begin{array}{ccc}\hfill {\psi }_1& =& G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[{\psi }_{0xx}-\sin \left({\psi }_0-{\varphi }_0\right)+\sin \left(t\sin \left(x\right)-e^{-t}\sin \left(x\right)\right)\right]\right]\hfill \\ & =& G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[0\right]\right]=0\hfill \\ \hfill {\varphi }_1& =& G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[{\varphi }_{0xx}+\sin \left({\psi }_0-{\varphi }_0\right)-\sin \left(t\sin \left(x\right)-e^{-t}\sin \left(x\right)\right)\right]\right]\hfill \\ & =& G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[0\right]\right]=0,\hfill \end{array}$$

and in a similar way, we obtain ${\psi }_2={\psi }_3=\cdots ={\psi }_m=0$ and ${\varphi }_2={\varphi }_3=\cdots ={\varphi }_m=0$, which means that method’s convergence is very fast. Therefore, the series solution is given by

$$\begin{array}{ccc}\hfill \psi \left(x,t\right)& =& {\psi }_0+{\psi }_1+{\psi }_2+\dots \hfill \\ \hfill \varphi \left(x,t\right)& =& {\varphi }_0+{\varphi }_1+{\varphi }_2+\dots ;\hfill \end{array}$$

hence,

$$\begin{array}{ccc}\hfill \psi \left(x,t\right)& =& t\sin x\hfill \\ \hfill \varphi \left(x,t\right)& =& e^{-t}\sin x.\hfill \end{array}$$

The approximate solution is in agreement with the result obtained by [13].

Example 6.

Consider the following IBVP for the one-dimensional coupled sine-Gordon equation on the domain $\mathsf{\Omega }=\left[0,\pi \right],$ $t>0:$

$$\begin{array}{ccc}\hfill {\psi }_{tt}-{\psi }_{xx}& =& -\sin \left(\psi -\varphi \right)+\sin \left(\cos \left(x+2t\right)-\sin \left(x\right)sinh\left(t\right)\right)\hfill \\ & & -3\cos \left(x+2t\right)\hfill \\ \hfill {\varphi }_{tt}-{\varphi }_{xx}& =& \sin \left(\psi -\varphi \right)-\sin \left(\cos \left(x+2t\right)-\sin \left(x\right)sinh\left(t\right)\right)\hfill \\ & & +2\sin \left(x\right)sinh\left(t\right),\hfill \end{array}$$

(72)

with IC:

$$\begin{array}{ccc}\hfill \psi \left(x,0\right)& =& \cos x,{\psi }_t\left(x,0\right)=-2\sin \left(x\right)\hfill \\ \hfill \varphi \left(x,0\right)& =& 0,{\varphi }_t\left(x,0\right)=\sin \left(x\right)\hfill \end{array}$$

(73)

and BC:

$$\begin{array}{ccc}\hfill \psi \left(0,t\right)& =& \cos 2t,{\psi }_x\left(0,t\right)=-\sin 2t\hfill \\ \hfill \varphi \left(0,t\right)& =& 0,{\varphi }_x\left(0,t\right)=sinht.\hfill \end{array}$$

(74)

By employing the proposed method for Equation (72), we obtain

$$\mathsf{\Psi }\left(p,s\right)=\frac{p^{\alpha +1}{s}^{\alpha +1}\left(1-2ps\right)}{\left(1+p^2\right)\left(1+4s^2\right)}+\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[\mathsf{\Pi }\right],$$

(75)

and

$$\mathsf{\Phi }\left(p,s\right)=\frac{p^{\alpha +2}{s}^{\alpha +2}}{\left(1+p^2\right)\left(1-s^2\right)}+\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[\mathsf{\Theta }\right],$$

(76)

where

$$\begin{array}{ccc}\hfill \mathsf{\Pi }& =& {\psi }_{xx}-\sin \left(\psi -\varphi \right)+\sin \left(\cos \left(x+2t\right)-\sin \left(x\right)sinh\left(t\right)\right)\hfill \\ \hfill \mathsf{\Theta }& =& {\varphi }_{xx}+\sin \left(\psi -\varphi \right)+\sin \left(\cos \left(x+2t\right)-\sin \left(x\right)sinh\left(t\right)\right),\hfill \end{array}$$

Taking the inverse G-double-Laplace transform into Equations (75) and (76),

$$\psi \left(x,t\right)=\cos \left(x+2t\right)+G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[\mathsf{\Pi }\right]\right],$$

(77)

and

$$\varphi \left(x,t\right)=\sin xsinht+G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[\mathsf{\Theta }\right]\right];$$

(78)

by substituting Equation (56) into Equations (77) and (78), one can obtain

$${\displaystyle \sum }_{m=0}^{\infty }{\psi }_m=\sin \left(x+2t\right)+G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[{\displaystyle \sum }_{m=0}^{\infty }{\mathsf{\Pi }}_m\right]\right],$$

(79)

and

$${\displaystyle \sum }_{m=0}^{\infty }{\varphi }_m=\sin xsinht+G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[{\displaystyle \sum }_{m=0}^{\infty }{\mathsf{\Theta }}_m\right]\right],$$

(80)

By comparing both sides of Equations (79) and (80), we have

$${\psi }_0=\sin \left(x+2t\right),{\varphi }_0=\sin xsinht,$$

and in general, the iteration relation is given by

$$\begin{array}{ccc}\hfill {\psi }_{m+1}& =& G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[\begin{array}{c}{\psi }_{mxx}-\sin \left({\psi }_m-{\varphi }_m\right)\\ +\sin \left(\cos \left(x+2t\right)-\sin \left(x\right)sinh\left(t\right)\right)\end{array}\right]\right]\hfill \\ \hfill {\varphi }_{m+1}& =& G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[\begin{array}{c}{\varphi }_{mxx}+\sin \left({\psi }_m-{\varphi }_m\right)\\ -\sin \left(\cos \left(x+2t\right)-\sin \left(x\right)sinh\left(t\right)\right)\end{array}\right]\right],\hfill \end{array}$$

where $m\ge 0;$ at $m=0$,

$$\begin{array}{ccc}\hfill {\psi }_1& =& G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[\begin{array}{c}{\psi }_{xx}-\sin \left({\psi }_0-{\varphi }_0\right)\\ +\sin \left(\cos \left(x+2t\right)-\sin \left(x\right)sinh\left(t\right)\right)\end{array}\right]\right]\hfill \\ & =& G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[0\right]\right]=0,\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill {\varphi }_1& =& G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[\begin{array}{c}{\varphi }_{xx}+\sin \left({\psi }_0-{\varphi }_0\right)\\ -\sin \left(\cos \left(x+2t\right)-\sin \left(x\right)sinh\left(t\right)\right)\end{array}\right]\right]\hfill \\ & =& G_p^{-1}{G}_s^{-1}\left[\frac{p^2{s}^2}{p^2-s^2}{G}_x{G}_t\left[0\right]\right]=0;\hfill \end{array}$$

in the same manner, we obtain ${\psi }_2={\psi }_3=\cdots ={\psi }_m=0$ and ${\varphi }_2={\varphi }_3=\cdots ={\varphi }_m=0$.

Hence, the series solution is given by

$$\begin{array}{ccc}\hfill \psi \left(x,t\right)& =& {\psi }_0+{\psi }_1+{\psi }_2+\dots \hfill \\ \hfill \varphi \left(x,t\right)& =& {\varphi }_0+{\varphi }_1+{\varphi }_2+\cdots ;\hfill \end{array}$$

hence.

$$\begin{array}{ccc}\hfill \psi \left(x,t\right)& =& \cos \left(x+2t\right)\hfill \\ \hfill \varphi \left(x,t\right)& =& \sin xsinht.\hfill \end{array}$$

The approximate solution is in agreement with the result obtained by [13].

4. Conclusions

In the course of this work, we introduced a new technique called the G-double-Laplace transform, which accurately describes its various definitions, theorems, existence conditions, partial derivatives, and the double-convolution theorems. Taking advantage of these new insights, we successfully obtained the exact solution of the one-dimensional coupled nonlinear sine-Gordon equation. To check the applicability of our technique, we provided some demonstrative examples. From the results we obtained, we strongly advise applying this method in the future so it can be extended to solve different nonlinear fractional partial differential equations related to physics and engineering problems.