The Solution of Coupled Burgers’ Equation by G-Laplace Transform

Abstract

The coupled Burgers’ equation is a fundamental partial differential equation with applications in various scientific fields. Finding accurate solutions to this equation is crucial for understanding physical phenomena and mathematical models. While different methods have been explored, this work highlights the importance of the G-Laplace transform. The G-transform is effective in solving a wide range of non-constant coefficient differential equations, setting it apart from the Laplace, Sumudu, and Elzaki transforms. Consequently, it stands as a powerful tool for addressing differential equations characterized by variable coefficients. By applying this transformative approach, the study provides reliable and exact solutions for both homogeneous and non-homogeneous coupled Burgers’ equations. This innovative technique offers a valuable tool for gaining deeper insights into this equation’s behavior and significance in diverse disciplines.

1. Introduction

The Burgers’ equation is a fundamental partial differential equation and convection–diffusion equation occurring in various areas of applied mathematics, such as fluid mechanics, nonlinear acoustics, gas dynamics, and traffic flow. The solution to this equation is quite important for mathematical models and physical phenomena. Several scientists have suggested analytical solutions to the one-dimensional coupled Burgers’ equation. Many analytical methods have been produced to obtain the solution to Burgers’ equation, see [1,2,3]. In their work [4], the authors employed the space–time Sinc–collocation method to address the fourth-order nonlocal heat model arising in viscoelasticity. Nonlinear terms were handled using the MacCormack method, while the Riemann–Liouville (R–L) fractional integral term was managed through the second-order convolution quadrature formula, as outlined in [5]. Furthermore, they discussed the modified Burgers’ model with nonlocal dynamic properties and utilized an implicit robust difference method with graded meshes, as elaborated in [6]. Additionally, in their study [7], the authors established the theory that the L1 scheme is effective in solving time-fractional partial differential equations with non-smooth data. In recent years, significant effort has been devoted to applying the Laplace decomposition method (LDM) and its modifications to study physical model equations [8]. The exact solution of the Burgers’ equation has been given in [9], employing the Adomian Decomposition method, and the authors of [10,11] offered a modified, expanded tanh-function method to receive its exact solution. The researchers in [12] suggested the homotopy perturbation method to achieve the exact solution of the nonlinear Burgers’ equation. Ref. [13] introduced a groundbreaking approach that combined the Laplace transform and new homotopy perturbation method (NHPM) for obtaining closed-form solutions of coupled viscous Burgers’ equations. This study holds significant importance as it sheds light on the understanding of polydispersity and its connection to gravity effects [14]. The study of coupled Burgers’ equations is paramount because it explains the precipitation of polydispersity commentary down to the effect of gravity [14]. The authors of [15,16] discussed the solution of the time-fractional two-mode coupled Burgers’ equation. G-transform was first proposed in [17] and later applied to solve certain nonlinear dynamical models with a non-integer order in [18].

The main goal of this work is to apply the G-Laplace transform in order to secure exact solutions with high reliability for homogeneous and nonhomogeneous coupled Burgers’ equations.

2. Some Basic Idea of the ${\mathit{G}}_{\mathbf{\alpha }}$-Laplace Transforms

In this work, we transact with $G_{\alpha }$ Transform and Laplace Transform to support us with solving some partial differential equations.

Definition 1.

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

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

(1)

for, $s\in \mathbb{C}$ and $\alpha \in Z$ (see [19]).

As an illustration, if we set $\alpha =0$ and $p=\frac{1}{s}$ in Equation (1), we can derive the Laplace transform as follows:

$$G_0\left(f\right)=\underset{0}{\overset{\infty }{\int }}f\left(\nu \right)e^{-\frac{v}{s}}d\nu .$$

This demonstrates that the $G_{\alpha }$-transform serves as a generalized version of both the Laplace and other transforms. It encompasses a broader and more fundamental range than existing transforms. For a more in-depth understanding, we recommend referring to [19].

Definition 2.

The Laplace transform of the function $f\left(\mu \right)$ is determined by

$$L_{\mu }\left[f\left(\mu \right)\right]=F\left(p\right)=\underset{0}{\overset{\infty }{\int }}f\left(\mu \right)e^{-p\mu }d\mu ,$$

where, $p\in \mathbb{C}$.

Definition 3.

The $G_{\alpha }$-Laplace Transform of the function $f(\mu ,\nu )$ is a well-behaved and integrable function defined for all $\mu ,\nu \ge 0$. It is represented by $L_{\mu }{G}_{\alpha }\left[f(\mu ,\nu )\right]=F_{\alpha }(p,s),$ where $F_{\alpha }(p,s)$ is given by the following expression:

$$F_{\alpha }(p,s)=s^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}f(\mu ,\nu )e^{-p\mu -\frac{\nu }{s}}d\nu d\mu .$$

(2)

Definition 4.

The inverse $G_{\alpha }$-Laplace Transform $F_{\alpha }\left(p,s\right)=f\left(\mu ,\nu \right)$ is defined as follows:

$$f(\mu ,\nu )=\frac{1}{2\pi i}\underset{T\to \infty }{lim}{\int }_{\gamma -iT}^{\gamma +iT}{e}^{p\mu }dp=\frac{1}{2\pi i}\underset{0}{\overset{\infty }{\int }}{s}^{\alpha }{e}^{\frac{\nu }{s}}{F}_{\alpha }\left(p,s\right)ds.$$

Example 1.

$L_{\mu }{G}_{\alpha }$ Transform of the $f(\mu ,\nu )=e^{i\left(a\mu +b\nu \right)}$ is given by

$$\begin{array}{ccc}\hfill L_{\mu }{G}_{\alpha }\left[e^{i\left(a\mu +b\nu \right)}\right]& =& \frac{s^{\alpha +1}}{\left(p-ai\right)\left(1-bsi\right)}\hfill \\ & =& \frac{s^{\alpha +1}\left(p+ai\right)\left(1+bsi\right)}{\left(p-ai\right)\left(1-bsi\right)\left(p+ai\right)\left(1+bsi\right)}\hfill \\ & =& \frac{s^{\alpha +1}\left(p-abs\right)+s^{\alpha +1}\left(a+pbs\right)i}{\left(p^2+a^2\right)\left(1+b^2{s}^2\right)}.\hfill \end{array}$$

where $L_{\mu }{G}_{\alpha }$ indicates the $G_{\alpha }$-Laplace Transform, and consequently,

$$L_{\mu }{G}_{\alpha }\left[cos(a\mu +b\nu )\right]=\frac{s^{\alpha +1}\left(p-abs\right)}{\left(p^2+a^2\right)\left(1+b^2{s}^2\right)},$$

and

$$L_{\mu }{G}_{\alpha }\left[sin(a\mu +b\nu )\right]=\frac{s^{\alpha +1}\left(a+pbs\right)}{\left(p^2+a^2\right)\left(1+b^2{s}^2\right)}.$$

Example 2.

The $L_{\mu }{G}_{\alpha }$ Transform of $f(\mu ,\nu )={\left(\mu \nu \right)}^n$ is determined by

$$L_{\mu }{G}_{\alpha }\left[{\left(\mu \nu \right)}^n\right]=\frac{{\left(n!\right)}^2{s}^{\alpha +n+1}}{p^{n+1}},$$

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

$$L_{\mu }{G}_{\alpha }\left[{\mu }^{\theta }{\nu }^{\beta }\right]=\frac{\mathsf{\Gamma }\left(\theta +1\right)\mathsf{\Gamma }\left(\beta +1\right)s^{\alpha +\beta +1}}{p^{\theta +1}}$$

can be derived from the definition of $G_{\alpha }$-Laplace Transform, so we have

$$\begin{array}{ccc}\hfill L_{\mu }{G}_{\alpha }\left[{\mu }^{\theta }{\nu }^{\beta }\right]& =& s^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{\mu }^{\theta }{\nu }^{\beta }{e}^{-p\mu -\frac{\nu }{s}}d\nu d\mu \hfill \\ & =& \left(\underset{0}{\overset{\infty }{\int }}{\mu }^{\theta }{e}^{-p\mu }d\mu \right)\left(s^{\alpha }\underset{0}{\overset{\infty }{\int }}{\nu }^{\beta }{e}^{-\frac{\nu }{s}}d\nu \right),\hfill \end{array}$$

(3)

and by using the definition of Laplace transform for the integral inside the first bracket of Equation (3), we obtain

$$L_{\mu }{G}_{\alpha }\left[{\mu }^{\theta }{\nu }^{\beta }\right]=\frac{\mathsf{\Gamma }\left(\theta +1\right)}{p^{\theta +1}}\left(s^{\alpha }\underset{0}{\overset{\infty }{\int }}{\nu }^{\beta }{e}^{-\frac{\nu }{s}}d\nu \right).$$

(4)

In Equation (4), put $r=\frac{\nu }{s},$$\nu =rs$ to obtain

$$\begin{array}{ccc}\hfill L_{\mu }{G}_{\alpha }\left[{\mu }^{\theta }{\nu }^{\beta }\right]& =& \frac{\mathsf{\Gamma }\left(\theta +1\right)}{p^{\theta +1}}\left(s^{\alpha +\beta +1}\underset{0}{\overset{\infty }{\int }}{r}^{\beta }{e}^{-r}dr\right)\hfill \\ & =& \frac{\mathsf{\Gamma }\left(\theta +1\right)\mathsf{\Gamma }\left(\beta +1\right)s^{\alpha +\beta +1}}{p^{\theta +1}},\hfill \end{array}$$

where, Gamma functions of $\beta +1$ are defined by the convergent integral:

$$\mathsf{\Gamma }\left(\beta +1\right)={\int }_0^{\infty }{e}^{-r}{r}^{\beta }dr,\beta >0.$$

Example 3.

$G_{\alpha }$-Laplace Transform for the the following function

$$f(\mu ,\nu )=H\left(\mu \right)\otimes H\left(\nu \right)ln\mu ln\nu$$

is given by

$$L_{\mu }{G}_{\alpha }\left[H\left(\mu \right)\otimes H\left(\nu \right)ln\mu ln\nu \right]=s^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}ln\mu ln\nu e^{-p\mu -\frac{\nu }{s}}d\nu d\mu ,$$

where the $H(\mu ,\nu )=H\left(\nu \right)\otimes H\left(\mu \right)$ is the dimensional Heaviside function and ⊗ is a tensor product (see [20]).

Let $\zeta =p\mu$ and $\eta =\frac{1}{s}\nu$, then the integral becomes

$$\begin{array}{ccc}\hfill L_{\mu }{G}_{\alpha }\left[H\left(\mu \right)\otimes H\left(\nu \right)ln\mu ln\nu \right]& =& \frac{s^{\alpha +1}}{p}{\int }_0^{\infty }{e}^{-\eta }ln\left(s\eta \right)\left({\int }_0^{\infty }{e}^{-\zeta }ln\left(\frac{1}{p}\zeta \right)d\zeta \right)d\eta \hfill \\ & =& \frac{s^{\alpha +1}}{p}\left(\gamma +lnp\right)\left(-\gamma +lns\right),\hfill \end{array}$$

(5)

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

Existence Condition for the $G_{\alpha }$-Laplace Transform

In the following theorem, we establish the conditions for the existence of the $G_{\alpha }$-Laplace Transform of $f(\mu ,\nu )$. Let $f(\mu ,\nu )$ be considered of exponential order a$(>0)$ and b$(>0)$ on $0\le \mu ,\nu <\infty$ if there exists a non-negative constant M, such that for all $\mu >X$ and $\nu >Y$, the inequality

$$\left|f(\mu ,\nu )\right|\le M{e}^{a\mu +b\nu }$$

(6)

holds. In this case, we can express $f(\mu ,\nu )$ as

$$f(\mu ,\nu )=O\left(e^{a\mu +b\nu }\right),$$

as $\mu \to \infty$ and $\nu \to \infty$. Equivalently, we have

$$\begin{array}{ccc}& & \underset{\begin{array}{c}\mu \to \infty \\ \nu \to \infty \end{array}}{lim}{e}^{-p\mu -\frac{\nu }{s}}\left|f(\mu ,\nu )\right|\hfill \\ & =& M\underset{\begin{array}{c}\mu \to \infty \\ \nu \to \infty \end{array}}{lim}{e}^{-(p-a)\mu -\left(\frac{1}{s}-b\right)\nu }=0,\hfill \end{array}$$

(7)

for $p>a$ and $s>\frac{1}{b}$. The function $f(\mu ,\nu )$ is simply referred to as having an exponential order as $\mu \to \infty$ and $\nu \to \infty$. Clearly, it does not grow faster than $M{e}^{a\mu +b\nu }$ as $\mu \to \infty$ and $\nu \to \infty$.

Theorem 1.

If $f(\mu ,\nu )$ is a continuous function in every bounded interval $,(0,\mu )$ and $(0,\nu )$ and of exponential order $e^{a\mu +b\nu }$, then the $L_{\mu }{G}_{\alpha }$ transform of $f(\mu ,\nu )$ exists for all p and s, provided that $p>a$ and $\frac{1}{s}>b$.

Proof. 

By using the definition of $G_{\alpha }$-Laplace Transform and Equations (6) and (7), one can obtain

$$\begin{array}{ccc}\hfill \left|F_{\alpha }\left(p,s\right)\right|& =& \left|s^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{e}^{-p\mu -\frac{\nu }{s}}f(\mu ,\nu )d\mu d\nu \right|\hfill \\ & & \le M{s}^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{e}^{-\left(p-a\right)\mu -\left(\frac{1}{s}-b\right)\nu }d\mu d\nu \hfill \\ & & =\frac{M{s}^{\alpha +1}}{\left(p-a\right)\left(1-bs\right)},\hfill \end{array}$$

for $p>a$ and $\frac{1}{s}>b$. □

Lemma 1.

If $f\left(\mu ,\nu \right)$ is a piecewise continuous function on $[0,\infty )\times$$[0,\infty )$ and has an exponential order at infinity with $M{e}^{a\mu +b\nu }$ for μ$\ge \mu$ and $\nu \ge \nu ,$ where μ and ν are constant, then for any real number ρ$\ge 0$ and σ$\ge 0$, we have

$$L_{\mu }{G}_{\alpha }\left[f\left(\mu -\rho ,\nu -\sigma \right)H\left(\mu -\rho ,\nu -\sigma \right)\right]=e^{-p\rho -\frac{\sigma }{s}}{F}_{\alpha }\left(p,s\right),$$

where $L_{\mu }{G}_{\alpha }\left(f\left(\mu ,\nu \right)\right)=F_{\alpha }\left(p,s\right)$ and $H\left(\mu -\rho ,\nu -\sigma \right)$ is the Heaviside function, defined by

$$H\left(\mu -\rho ,\nu -\sigma \right){=\{}_{0\mathrm{if}\mu <\rho ,\nu <\sigma }^{1\mathrm{if}\mu >\rho ,\nu >\sigma }.$$

Proof. 

By using the definition of $L_{\mu }{G}_{\alpha }$ Transform

$$\begin{array}{ccc}\hfill L_{\mu }{G}_{\alpha }\left[f\left(\mu -\rho ,\nu -\sigma \right)H\left(\mu -\rho ,\nu -\sigma \right)\right]& =& s^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{e}^{-p\mu -\frac{\nu }{s}}f\left(\mu -\rho ,\nu -\sigma \right)H\left(\mu -\rho ,\nu -\sigma \right)d\mu d\nu \hfill \\ & =& s^{\alpha }\underset{\rho }{\overset{\infty }{\int }}\underset{\sigma }{\overset{\infty }{\int }}{e}^{-p\mu -\frac{\nu }{s}}f\left(\mu -\rho ,\nu -\sigma \right)d\mu d\nu ,\hfill \end{array}$$

(8)

let $z=\mu -\rho$ and $r=\nu -\sigma ,$ Equation (8) becomes

$$\begin{array}{ccc}\hfill L_{\mu }{G}_{\alpha }\left[f\left(\mu -\rho ,\nu -\sigma \right)H\left(\mu -\rho ,\nu -\sigma \right)\right]& =& e^{-p\rho -\frac{\sigma }{s}}\left[s^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{e}^{-pz-\frac{r}{s}}f\left(z,r\right)dzdr\right]\hfill \\ & =& e^{-p\rho -\frac{\sigma }{s}}{F}_{\alpha }\left(p,s\right).\hfill \end{array}$$

□

Theorem 2.

Suppose $f(\mu ,\nu )$ is a periodic function with periods λ and μ. For this periodicity condition to hold, we require that:

$$f(\mu +\lambda ,\nu +\mu )=f(\mu ,\nu ),\mathit{for}\mathit{all}\mu ,\nu .$$

Then $G_{\alpha }$-Laplace Transform of $f(\mu ,\nu )$ is given by

$$F_{\alpha }\left(p,s\right)=\left(1-e^{-p\lambda -\frac{\mu }{\nu }}\right)s^{\alpha }\underset{0}{\overset{\lambda }{\int }}\underset{0}{\overset{\mu }{\int }}{e}^{-p\mu -\frac{\nu }{s}}f\left(\mu ,\nu \right)d\nu d\mu .$$

(9)

Proof. 

Utilizing the definition of the $G_{\alpha }$-Laplace Transform given by

$$L_{\mu }{G}_{\alpha }\left[f\left(\mu ,\nu \right)\right]=F_{\alpha }\left(p,s\right)=s^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}f\left(\mu ,\nu \right)e^{-p\mu -\frac{\nu }{s}}d\nu d\mu ,$$

(10)

we can apply the property of improper integrals to Equation (10), resulting in

$$L_{\mu }{G}_{\alpha }\left[f\left(\mu ,\nu \right)\right]=F_{\alpha }\left(p,s\right)=s^{\alpha }\left[\underset{0}{\overset{\lambda }{\int }}\underset{0}{\overset{\mu }{\int }}f\left(\mu ,\nu \right)e^{-p\mu -\frac{\nu }{s}}d\nu d\mu +\underset{\lambda }{\overset{\infty }{\int }}\underset{\mu }{\overset{\infty }{\int }}f\left(\mu ,\nu \right)e^{-p\mu -\frac{\nu }{s}}d\nu d\mu \right].$$

(11)

By setting $\mu =\lambda +\gamma$ and $\nu =\mu +\delta$ in the second part of the integral in Equation (11), we have

$$F_{\alpha }\left(p,s\right)=s^{\alpha }\left[\underset{0}{\overset{\lambda }{\int }}\underset{0}{\overset{\mu }{\int }}f\left(\mu ,\nu \right)e^{-p\mu -\frac{\nu }{s}}d\nu d\mu +\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}f\left(\mu ,\nu \right)e^{-\left(p\lambda +p\gamma \right)-\frac{\left(\mu +\delta \right)}{s}}d\delta d\gamma \right].$$

(12)

Equation (12) can then be rewritten as follows:

$$F_{\alpha }\left(p,s\right)=s^{\alpha }\underset{0}{\overset{\lambda }{\int }}\underset{0}{\overset{\mu }{\int }}f\left(\mu ,\nu \right)e^{-p\mu -\frac{\nu }{s}}d\nu d\mu +e^{-p\lambda -\frac{\mu }{s}}{s}^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}f\left(\mu ,\nu \right)e^{-p\gamma -\frac{\delta }{s}}d\delta d\gamma .$$

(13)

By the second integral in Equation (13), given the definition of $G_{\alpha }$-Laplace Transform, we obtain

$$F_{\alpha }\left(p,s\right)=s^{\alpha }\underset{0}{\overset{\lambda }{\int }}\underset{0}{\overset{\mu }{\int }}f\left(\mu ,\nu \right)e^{-p\mu -\frac{\nu }{s}}d\nu d\mu +e^{-p\lambda -\frac{\mu }{s}}{F}_{\alpha }\left(p,s\right),$$

and hence,

$$F_{\alpha }\left(p,s\right)=\left(1-e^{-p\lambda -\frac{\mu }{\nu }}\right)s^{\alpha }\underset{0}{\overset{\lambda }{\int }}\underset{0}{\overset{\mu }{\int }}{e}^{-p\mu -\frac{\nu }{s}}f\left(\mu ,\nu \right)d\nu d\mu .$$

□

Theorem 3 (Convolution Theorem).

Let $L_{\mu }{G}_{\alpha }\left[\varphi \left(\mu ,\nu \right)\right]$ and $L_{\mu }{G}_{\alpha }\left[\phi \left(\mu ,\nu \right)\right]$ exist and $L_{\mu }{G}_{\alpha }\left[\varphi \left(\mu ,\nu \right)\right]=$${\varphi }_{\alpha }\left(p,s\right)$, $L_{\mu }{G}_{\alpha }\left[\phi \left(\mu ,\nu \right)\right]=$${\phi }_{\alpha }\left(p,s\right)$, then

$$s^{\alpha }{L}_{\mu }{G}_{\alpha }\left[\varphi \left(\mu ,\nu \right)\ast \ast \phi \left(\mu ,\nu \right)\right]={\varphi }_{\alpha }\left(p,s\right){\phi }_{\alpha }\left(p,s\right),$$

where

$$\varphi \left(\mu ,\nu \right)\ast \ast \phi \left(\mu ,\nu \right)=\underset{0}{\overset{\mu }{\int }}\underset{0}{\overset{\nu }{\int }}\varphi (\mu -\zeta ,\nu -\eta )\phi (\zeta ,\eta )d\zeta d\eta ,$$

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

Proof. 

On using the definition of $G_{\alpha }$-Laplace Transform, we obtain

$$\begin{array}{ccc}\hfill L_{\mu }{G}_{\alpha }\left[\varphi \left(\mu ,\nu \right)\ast \ast \phi \left(\mu ,\nu \right)\right]& =& s^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{e}^{-p\mu -\frac{\nu }{s}}\varphi \left(\mu ,\nu \right)\ast \ast \phi \left(\mu ,\nu \right)d\mu d\nu \hfill \\ & =& s^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{e}^{-p\mu -\frac{\nu }{s}}\left(\underset{0}{\overset{\mu }{\int }}\underset{0}{\overset{\nu }{\int }}\varphi (\mu -\zeta ,\nu -\eta )\phi (\zeta ,\eta )d\zeta d\eta \right)d\mu d\nu .\hfill \end{array}$$

(14)

Set $\rho =\mu -\zeta$ and $\sigma =\nu -\eta$, and by applying the adequate expansion of the upper bound of integrals to $\mu \to \infty$ and $\nu \to \infty$, Equation (14) can be written as

$$\begin{array}{ccc}\hfill L_{\mu }{G}_{\alpha }\left[\varphi \left(\mu ,\nu \right)\ast \ast \phi \left(\mu ,\nu \right)\right]& =& s^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{e}^{-\zeta \mu -\frac{\eta }{s}}\phi (\mu -\rho ,\nu -\sigma )d\zeta d\eta \underset{-\zeta }{\overset{\infty }{\int }}\underset{-\eta }{\overset{\infty }{\int }}{e}^{-p\rho -\frac{\sigma }{s}}\varphi (\rho ,\sigma )d\rho d\sigma ,\hfill \end{array}$$

(15)

where the functions $\varphi \left(\mu ,\nu \right)$ and $\phi \left(\mu ,\nu \right)$ equal zero at $\mu$<$0,$$\nu$< 0; therefore, it follows with respect to the lower limit of integrations that

$$L_{\mu }{G}_{\alpha }\left[\varphi \left(\mu ,\nu \right)\ast \ast \phi \left(\mu ,\nu \right)\right]=\frac{1}{s^{\alpha }}\left(s^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{e}^{-\zeta \mu -\frac{\eta }{s}}\phi (\zeta ,\eta )d\zeta d\eta \right)\left(s^{\alpha }\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{e}^{-p\rho -\frac{\sigma }{s}}\varphi (\rho ,\sigma )d\rho d\sigma \right).$$

It is thus easy to see that

$$s^{\alpha }{L}_{\mu }{G}_{\alpha }\left[\varphi \left(\mu ,\nu \right)\ast \ast \phi \left(\mu ,\nu \right)\right]={\varphi }_{\alpha }\left(p,s\right){\phi }_{\alpha }\left(p,s\right).$$

□

Theorem 4.

If $G_{\alpha }$-Laplace Transform of the function $f(\mu ,\nu )$ is given by $L_{\mu }{G}_{\alpha }\left[f(\mu ,\nu )\right]=F_{\alpha }\left(p,s\right),$ then $G_{\alpha }$-Laplace Transform of $\frac{{\partial }^{n}f\left(\mu ,\nu \right)}{\partial {\nu }^n}$ and $\frac{{\partial }^{n}f\left(\mu ,\nu \right)}{\partial {\mu }^n}$ are given as

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

(16)

and

$$L_{\mu }{G}_{\alpha }\left[\frac{{\partial }^{n}f\left(\mu ,\nu \right)}{\partial {\nu }^n}\right]=\frac{F_{\alpha }\left(p,s\right)}{s^n}-s^{\alpha }\sum _{k=1}^n\frac{1}{s^{n-k}}{L}_{\mu }\left[\frac{{\partial }^{k-1}f\left(\mu ,0\right)}{\partial {\nu }^{k-1}}\right].$$

(17)

Proof. 

Now, by substituting $n=1$ in Equation (16):

$$L_{\mu }{G}_{\alpha }\left[\frac{\partial f\left(\mu ,\nu \right)}{\partial \mu }\right]=s^{\alpha }\underset{0}{\overset{\infty }{\int }}{e}^{-\frac{\nu }{s}}\left[\underset{0}{\overset{\infty }{\int }}{e}^{-p\mu }\frac{\partial f\left(\mu ,\nu \right)}{\partial \mu }d\mu \right]d\nu ,$$

(18)

first, we calculate the integral inside bracket to obtain

$$\begin{array}{ccc}\hfill \underset{0}{\overset{\infty }{\int }}{e}^{-p\mu }\frac{\partial f\left(\mu ,\nu \right)}{\partial \mu }d\mu & =& {\left[e^{-p\mu }f\left(\mu ,\nu \right)\right]}_0^{\infty }+p\underset{0}{\overset{\infty }{\int }}{e}^{-p\mu }f\left(\mu ,\nu \right)d\mu \hfill \\ & =& pF\left(p,\nu \right)-f\left(0,\nu \right),\hfill \end{array}$$

and by substituting in Equation (18), we have

$$L_{\mu }{G}_{\alpha }\left[\frac{\partial f\left(\mu ,\nu \right)}{\partial \mu }\right]=s^{\alpha }\underset{0}{\overset{\infty }{\int }}{e}^{-\frac{\nu }{s}}\left[pF\left(p,\nu \right)-f\left(0,\nu \right)\right]d\nu ,$$

then

$$L_{\mu }{G}_{\alpha }\left[\frac{\partial f\left(\mu ,\nu \right)}{\partial \mu }\right]=p{F}_{\alpha }\left(p,s\right)-G_{\alpha }\left[f\left(0,\nu \right)\right],$$

(19)

at $n=2$. In a similar way, one can easily see that

$$L_{\mu }{G}_{\alpha }\left[\frac{{\partial }^{2}f\left(\mu ,\nu \right)}{\partial {\mu }^2}\right]=p^2{F}_{\alpha }\left(p,s\right)-p{G}_{\alpha }\left[f\left(0,\nu \right)\right]-G_{\alpha }\left[\frac{\partial f\left(0,\nu \right)}{\partial \mu }\right],$$

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

$$\begin{array}{ccc}\hfill L_{\mu }{G}_{\alpha }\left[\frac{{\partial }^{m}f\left(\mu ,\nu \right)}{\partial {\mu }^m}\right]& =& p^m{F}_{\alpha }\left(p,s\right)-p^{m-1}{G}_{\alpha }\left[f\left(0,\nu \right)\right]-p^{m-2}{G}_{\alpha }\left[\frac{\partial f\left(0,\nu \right)}{\partial \mu }\right]\hfill \\ & & -\dots -p{G}_{\alpha }\left[\frac{{\partial }^{m-2}f\left(0,\nu \right)}{\partial {\mu }^{m-2}}\right]-G_{\alpha }\left[\frac{{\partial }^{m-1}f\left(0,\nu \right)}{\partial {\mu }^{m-1}}\right],\hfill \end{array}$$

(20)

hold for $\frac{{\partial }^0}{\partial }=1$, and now, we show that

$$\begin{array}{ccc}\hfill L_{\mu }{G}_{\alpha }\left[\frac{{\partial }^{m+1}f\left(\mu ,\nu \right)}{\partial {\mu }^{m+1}}\right]& =& p^{m+1}{F}_{\alpha }\left(p,s\right)-p^m{G}_{\alpha }\left[f\left(0,\nu \right)\right]-p^{m-1}{G}_{\alpha }\left[\frac{\partial f\left(0,\nu \right)}{\partial \mu }\right]\hfill \\ & & -\dots -p{G}_{\alpha }\left[\frac{{\partial }^{m-1}f\left(0,\nu \right)}{\partial {\mu }^{m-1}}\right]-G_{\alpha }\left[\frac{{\partial }^{m}f\left(0,\nu \right)}{\partial {\mu }^m}\right],\hfill \end{array}$$

(21)

and by the notion of $n=1$, we have

$$\begin{array}{ccc}\hfill L_{\mu }{G}_{\alpha }\left[\frac{{\partial }^{m+1}f\left(\mu ,\nu \right)}{\partial {\mu }^{m+1}}\right]& =& p\left[\begin{array}{c}{p}^m{F}_{\alpha }\left(p,s\right)-p^{m-1}{G}_{\alpha }\left[f\left(0,\nu \right)\right]-p^{m-2}{G}_{\alpha }\left[\frac{\partial f\left(0,\nu \right)}{\partial \mu }\right]\\ \dots -G_{\alpha }\left[\frac{{\partial }^{m-1}f\left(0,\nu \right)}{\partial {\mu }^{m-1}}\right]\end{array}\right]\hfill \\ & & -G_{\alpha }\left[\frac{{\partial }^{m}f\left(0,\nu \right)}{\partial {\mu }^m}\right],\hfill \end{array}$$

of which, the formula inside the bracket is

$$L_{\mu }{G}_{\alpha }\left[\frac{{\partial }^{m}f\left(\mu ,\nu \right)}{\partial {\mu }^m}\right]=p^m{F}_{\alpha }\left(p,s\right)-p^{m-1}{G}_{\alpha }\left[f\left(0,\nu \right)\right]-p^{m-2}{G}_{\alpha }\left[\frac{\partial f\left(0,\nu \right)}{\partial \mu }\right]\dots -G_{\alpha }\left[\frac{{\partial }^{m-1}f\left(0,\nu \right)}{\partial {\mu }^{m-1}}\right].$$

Therefore,

$$L_{\mu }{G}_{\alpha }\left[\frac{{\partial }^{m+1}f\left(\mu ,\nu \right)}{\partial {\mu }^{m+1}}\right]=p{L}_{\mu }{G}_{\alpha }\left[\frac{{\partial }^{m}f\left(\mu ,\nu \right)}{\partial {\mu }^m}\right]-G_{\alpha }\left[\frac{{\partial }^{m}f\left(0,\nu \right)}{\partial {\mu }^m}\right].$$

Hence, Equation (16) can be written as follows:

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

For Equation (17), by substituting $n=1$ in Equation (16),

$$L_{\mu }{G}_{\alpha }\left[\frac{\partial f\left(\mu ,\nu \right)}{\partial \nu }\right]=\underset{0}{\overset{\infty }{\int }}{e}^{-p\mu }\left[s^{\alpha }\underset{0}{\overset{\infty }{\int }}{e}^{-\frac{\nu }{s}}\frac{\partial f\left(\mu ,\nu \right)}{\partial \nu }d\nu \right]d\mu ,$$

we calculate the integral inside bracket

$$\begin{array}{ccc}\hfill s^{\alpha }\underset{0}{\overset{\infty }{\int }}{e}^{-\frac{\nu }{s}}\frac{\partial f\left(\mu ,\nu \right)}{\partial \nu }d\nu & =& s^{\alpha }{\left[e^{-\frac{\nu }{s}}f\left(\mu ,\nu \right)\right]}_0^{\infty }+s^{\alpha }\underset{0}{\overset{\infty }{\int }}{e}^{-\frac{\nu }{s}}f\left(\mu ,\nu \right)d\nu \hfill \\ & =& -s^{\alpha }f(\mu ,0)+\frac{1}{s}{F}_{\alpha }\left(\mu ,s\right).\hfill \end{array}$$

Therefore,

$$\begin{array}{ccc}\hfill L_{\mu }{G}_{\alpha }\left[\frac{\partial f\left(\mu ,\nu \right)}{\partial \nu }\right]& =& -s^{\alpha }\underset{0}{\overset{\infty }{\int }}{e}^{-p\mu }f(\mu ,0)d\mu +\frac{1}{s}\underset{0}{\overset{\infty }{\int }}{e}^{-p\mu }{F}_{\alpha }\left(\mu ,s\right)d\mu \hfill \\ \hfill L_{\mu }{G}_{\alpha }\left[\frac{\partial f\left(\mu ,\nu \right)}{\partial \nu }\right]& =& \frac{1}{s}{F}_{\alpha }\left(p,s\right)-s^{\alpha }f(p,0).\hfill \end{array}$$

Now, assume that $n=m,$ Equation (17) is then correct for some $m.$ Thus,

$$L_{\mu }{G}_{\alpha }\left[\frac{{\partial }^{m}f\left(\mu ,\nu \right)}{\partial {\nu }^m}\right]=\frac{F_{\alpha }\left(p,s\right)}{s^m}-s^{\alpha }\sum _{k=1}^m\frac{1}{s^{m-k}}{L}_{\mu }\left[\frac{{\partial }^{k-1}f\left(\mu ,0\right)}{\partial {\nu }^{k-1}}\right],$$

$$\begin{array}{ccc}\hfill L_{\mu }{G}_{\alpha }\left[\frac{{\partial }^{m}f\left(\mu ,\nu \right)}{\partial {\nu }^m}\right]& =& \frac{F_{\alpha }\left(p,s\right)}{s^m}-\frac{s^{\alpha }}{s^{m-1}}{L}_{\mu }\left[f\left(\mu ,0\right)\right]-\frac{s^{\alpha }}{s^{m-2}}{L}_{\mu }\left[\frac{\partial f\left(\mu ,0\right)}{\partial \nu }\right]\hfill \\ & & -\dots -s^{\alpha -1}{L}_{\mu }\left[\frac{{\partial }^{m-2}f\left(\mu ,0\right)}{\partial {\nu }^{m-2}}\right]-s^{\alpha }{L}_{\mu }\left[\frac{{\partial }^{m-1}f\left(\mu ,0\right)}{\partial {\nu }^{m-1}}\right],\hfill \end{array}$$

allows us to indicate

$$\begin{array}{ccc}\hfill L_{\mu }{G}_{\alpha }\left[\frac{{\partial }^{m+1}f\left(\mu ,\nu \right)}{\partial {\nu }^{m+1}}\right]& =& \frac{F_{\alpha }\left(p,s\right)}{s^{m+1}}-\frac{s^{\alpha }}{s^m}{L}_{\mu }\left[f\left(\mu ,0\right)\right]-\frac{s^{\alpha }}{s^{m-1}}{L}_{\mu }\left[\frac{\partial f\left(\mu ,0\right)}{\partial \nu }\right]\hfill \\ & & -\dots -s^{\alpha -1}{L}_{\mu }\left[\frac{{\partial }^{m-1}f\left(\mu ,0\right)}{\partial {\nu }^{m-1}}\right]-s^{\alpha }{L}_{\mu }\left[\frac{{\partial }^{m}f\left(\mu ,0\right)}{\partial {\nu }^m}\right],\hfill \end{array}$$

and by the notion of $n=1$, we have

$$\begin{array}{ccc}\hfill L_{\mu }{G}_{\alpha }\left[\frac{{\partial }^{m+1}f\left(\mu ,\nu \right)}{\partial {\nu }^{m+1}}\right]& =& \frac{1}{s}\left[\begin{array}{c}\frac{F_{\alpha }\left(p,s\right)}{s^m}-\frac{s^{\alpha }}{s^{m-1}}{L}_{\mu }\left[f\left(\mu ,0\right)\right]-\frac{s^{\alpha }}{s^{m-2}}{L}_{\mu }\left[\frac{\partial f\left(\mu ,0\right)}{\partial \nu }\right]\\ -\dots -s^{\alpha -1}{L}_{\mu }\left[\frac{{\partial }^{m-2}f\left(\mu ,0\right)}{\partial {\nu }^{m-2}}\right]-s^{\alpha }{L}_{\mu }\left[\frac{{\partial }^{m-1}f\left(\mu ,0\right)}{\partial {\nu }^{m-1}}\right]\end{array}\right]\hfill \\ & & -s^{\alpha }{L}_{\mu }\left[\frac{{\partial }^{m}f\left(\mu ,0\right)}{\partial {\nu }^m}\right]\hfill \\ \hfill L_{\mu }{G}_{\alpha }\left[\frac{{\partial }^{m+1}f\left(\mu ,\nu \right)}{\partial {\nu }^{m+1}}\right]& =& \frac{1}{s}{L}_{\mu }{G}_{\alpha }\left[\frac{{\partial }^{m}f\left(\mu ,\nu \right)}{\partial {\nu }^m}\right]-s^{\alpha }{L}_{\mu }\left[\frac{{\partial }^{m}f\left(\mu ,0\right)}{\partial {\nu }^m}\right].\hfill \end{array}$$

Thus, the theorem is correct at an arbitrary natural number k. Hence, Equation (17) is correct. □

3. ${\mathit{G}}_{\mathbf{\alpha }}$-Laplace Transform Decomposition Method Applied to Coupled Burgers’ Equation

In this section, we discuss the solutions of two problems by applying $G_{\alpha }$-Laplace Transform decomposition method:

The first problem: Regular Burgers’ equation is given by

$${\varphi }_{\nu }-{\varphi }_{\mu \mu }+\varphi {\varphi }_{\mu }=f\left(\mu ,\nu \right),$$

(22)

subject to

$$\varphi \left(\mu ,0\right)=f_1\left(\mu \right),$$

(23)

for $\nu ,\mu >0$. Here, $f(\mu ,\nu )$ and $f_1\left(\mu \right)$ are given functions. By taking $G_{\alpha }$-Laplace Transform for both sides of Equation (22) and Laplace transform for Equation (23), we obtain

$${\mathsf{\Psi }}_{\alpha }(p,s)=s^{\alpha +1}{F}_1\left(p\right)+s{F}_{\alpha }(p,s)+s{L}_{\mu }{G}_{\alpha }\left[{\varphi }_{\mu \mu }-\varphi {\varphi }_{\mu }\right].$$

(24)

By utilizing the inverse $G_{\alpha }$-Laplace Transform for Equation (22), we have

$$\begin{array}{ccc}\hfill \varphi \left(\mu ,\nu \right)& =& f_1\left(\mu \right)+L_p^{-1}{G}_{\alpha }^{-1}\left[s{F}_{\alpha }(p,s)\right]\hfill \\ & & +L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\frac{{\partial }^2}{\partial {\mu }^2}\varphi \right]\right]\hfill \\ & & -L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\varphi {\varphi }_{\mu }\right]\right].\hfill \end{array}$$

(25)

The $G_{\alpha }$-Laplace Transform decomposion method (GLTDM) supposes the solution $\varphi \left(\mu ,\nu \right)$ can be expanded into infinite series as

$$\varphi \left(\mu ,\nu \right)=\sum _{n=0}^{\infty }{\varphi }_n\left(\mu ,\nu \right).$$

(26)

We can present Adomian’s polynomials $A_n$, respectively, as follows:

$$N\left(\varphi \right)=A_n=\sum _{n=0}^{\infty }{\varphi }_n{\varphi }_{n\mu },$$

(27)

where, the Adomian polynomials for the nonlinear term $\varphi {\varphi }_{\mu }$ are given by

$$\begin{array}{ccc}\hfill A_0& =& {\varphi }_0{\varphi }_{0\mu }\hfill \\ \hfill A_1& =& {\varphi }_0{\varphi }_{1\mu }+{\varphi }_1{\varphi }_{0\mu },\hfill \\ \hfill A_2& =& {\varphi }_0{\varphi }_{2\mu }+{\varphi }_1{\varphi }_{1\mu }+{\varphi }_2{\varphi }_{0\mu },\hfill \\ \hfill A_3& =& {\varphi }_0{\varphi }_{3\mu }+{\varphi }_1{\varphi }_{2\mu }+{\varphi }_2{\varphi }_{1\mu }+{\varphi }_3{\varphi }_{0\mu },\hfill \\ \hfill A_4& =& {\varphi }_0{\varphi }_{4\mu }+{\varphi }_1{\varphi }_{3\mu }+{\varphi }_2{\varphi }_{2\mu }+{\varphi }_3{\varphi }_{1\mu }+{\varphi }_4{\varphi }_{0\mu }.\hfill \end{array}$$

(28)

By substituting Equation (26) into Equation (25), we have

$$\begin{array}{ccc}\hfill \sum _{n=0}^{\infty }{\varphi }_n\left(\mu ,\nu \right)& =& f_1\left(\mu \right)+L_p^{-1}{G}_{\alpha }^{-1}\left[s{F}_{\alpha }(p,s)\right]\hfill \\ & & +L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\frac{{\partial }^2}{\partial {\mu }^2}\sum _{n=0}^{\infty }{\varphi }_n\right]\right]\hfill \\ & & -L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\sum _{n=0}^{\infty }{A}_n\right]\right].\hfill \end{array}$$

(29)

Upon comparing both sides of the Equations (26) and (29), we obtain the following iterative algorithm:

$${\varphi }_0=f_1\left(\mu \right)+L_p^{-1}{G}_{\alpha }^{-1}\left[s{F}_{\alpha }(p,s)\right].$$

The value of the rest component ${\varphi }_{n+1}$, where $n\ge 0$, is determined by utilizing the following relation:

$${\varphi }_{n+1}=L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\frac{{\partial }^2}{\partial {\mu }^2}{\varphi }_n\right]\right]-L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[A_n\right]\right].$$

Here, we applied the inverse $G_{\alpha }$-Laplace Transform to each term on the right-hand side of the above equation, with respect to p and s, to obtain the corresponding expressions for each term.

To examine this method for one-dimensional Burgers’ equations, we use the following example:

Example 4.

Consider that one-dimensional Burgers’ equation is given by

$${\varphi }_{\nu }-{\varphi }_{\mu \mu }+\varphi {\varphi }_{\mu }=0,$$

(30)

subject to the condition

$$\varphi \left(\mu ,0\right)=2\mu .$$

(31)

By taking $G_{\alpha }$-Laplace Transform for both sides of Equation (30) and Laplace transform for Equation (31), we obtain

$${\mathsf{\Psi }}_{\alpha }(p,s)=\frac{2s^{\alpha +1}}{p^2}+s{L}_{\mu }{G}_{\alpha }\left[{\varphi }_{\mu \mu }-\varphi {\varphi }_{\mu }\right],$$

(32)

and by using Equation (29), we have

$$\begin{array}{ccc}\hfill \sum _{n=0}^{\infty }{\varphi }_n\left(\mu ,\nu \right)=2\mu & & +L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\frac{{\partial }^2}{\partial {\mu }^2}\sum _{n=0}^{\infty }{\varphi }_n\right]\right]\hfill \\ & & -L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\sum _{n=0}^{\infty }{A}_n\right]\right],\hfill \end{array}$$

(33)

where $A_n$ is given by Equation (28). On matching both sides of Equations (29) and (33), we have

$${\varphi }_0=2\mu .$$

Overall, the recursive link is given by

$${\varphi }_{n+1}=L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\frac{{\partial }^2}{\partial {\mu }^2}{\varphi }_n\right]\right]-L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[A_n\right]\right],$$

where $n\ge 0.$ At $n=0,$

$$\begin{array}{ccc}\hfill {\varphi }_1& =& L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\frac{{\partial }^2}{\partial {\mu }^2}{\varphi }_0\right]\right]-L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[A_0\right]\right]\hfill \\ & =& L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\frac{{\partial }^2}{\partial {\mu }^2}{\varphi }_0\right]\right]-L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[{\varphi }_0{\varphi }_{0\mu }\right]\right]\hfill \\ & =& -L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[4\mu \right]\right]=-4\mu \nu .\hfill \end{array}$$

In a similar manner, at $n=1,$ we have

$$\begin{array}{ccc}\hfill {\varphi }_2& =& L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\frac{{\partial }^2}{\partial {\mu }^2}{\varphi }_1\right]\right]-L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[A_1\right]\right]\hfill \\ & =& L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\frac{{\partial }^2}{\partial {\mu }^2}{\varphi }_1\right]\right]-L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[{\varphi }_0{\varphi }_{1\mu }+{\varphi }_1{\varphi }_{0\mu }\right]\right]\hfill \\ & =& -L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[16\mu \nu \right]\right]\hfill \\ & =& 8\mu {\nu }^2.\hfill \end{array}$$

In a similar way, at $n=2,$ we obtain

$$\begin{array}{ccc}\hfill {\varphi }_3& =& L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\frac{{\partial }^2}{\partial {\mu }^2}{\varphi }_2\right]\right]-L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[A_2\right]\right]\hfill \\ & =& L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\frac{{\partial }^2}{\partial {\mu }^2}{\varphi }_2\right]\right]-L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[{\varphi }_0{\varphi }_{2\mu }+{\varphi }_1{\varphi }_{1\mu }+{\varphi }_2{\varphi }_{0\mu }\right]\right]\hfill \\ & =& -L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[48\mu {\nu }^2\right]\right]=-L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\frac{96s^{\alpha +4}}{p^2}\right]\right]\hfill \\ & =& -16\mu {\nu }^3.\hfill \end{array}$$

Upon applying Equation (26), the convergent solutions are thus specified by

$$\varphi \left(\mu ,\nu \right)={\varphi }_0+{\varphi }_1+{\varphi }_2+{\varphi }_3+\dots =2\mu \left(1-2\nu +{\left(2\nu \right)}^2-{\left(2\nu \right)}^3+{\left(2\nu \right)}^4-\begin{array}{c}\hfill \dots \end{array}\right),$$

and therefore, the delicate solution becomes

$$\varphi \left(\mu ,\nu \right)=\frac{2\mu }{1+2\nu }$$

The second problem: Consider the following one-dimensional Burgers’ equations

$$\begin{array}{ccc}\hfill {\varphi }_{\nu }-{\varphi }_{\mu \mu }+a\varphi {\varphi }_{\mu }+{\left(\varphi \phi \right)}_{\mu }& =& f\left(\mu ,\nu \right)\hfill \\ \hfill {\phi }_{\nu }-{\phi }_{\mu \mu }+b\phi {\phi }_{\mu }+{\left(\varphi \phi \right)}_{\mu }& =& h\left(\mu ,\nu \right),\hfill \end{array}$$

(34)

subject to

$$\begin{array}{ccc}\hfill \varphi \left(\mu ,0\right)& =& f_1\left(\mu \right)\hfill \\ \hfill \phi \left(\mu ,0\right)& =& h_1\left(\mu \right).\hfill \end{array}$$

(35)

On using $G_{\alpha }$-Laplace Transform and the characteristic of the differentiation of Laplace transform, we have

$$\begin{array}{ccc}\hfill {\mathsf{\Psi }}_{\alpha }(p,s)& =& s^{\alpha +1}{F}_1\left(p\right)+s{F}_{\alpha }(p,s)+s{L}_{\mu }{G}_{\alpha }\left[{\varphi }_{\mu \mu }-a\varphi {\varphi }_{\mu }-{\left(\varphi \phi \right)}_{\mu }\right]\hfill \\ \hfill {\mathsf{\Phi }}_{\alpha }(p,s)& =& s^{\alpha +1}{H}_1\left(p\right)+s{H}_{\alpha }(p,s)+s{L}_{\mu }{G}_{\alpha }\left[{\phi }_{\mu \mu }-b\phi {\phi }_{\mu }-{\left(\varphi \phi \right)}_{\mu }\right].\hfill \end{array}$$

(36)

The following step in the $G_{\alpha }$-Laplace Transform decomposition method illustrates the solution of the series given below:

$$\varphi \left(\mu ,\nu \right)=\sum _{n=0}^{\infty }{\varphi }_n\left(\mu ,\nu \right),\phi \left(\mu ,\nu \right)=\sum _{n=0}^{\infty }{\phi }_n\left(\mu ,\nu \right),$$

(37)

where the nonlinear operators are determined by

$$B_n=\sum _{n=0}^{\infty }{\phi }_n{\phi }_{n\mu },C_n=\sum _{n=0}^{\infty }{\phi }_n{\varphi }_{n\mu },D_n=\sum _{n=0}^{\infty }{\varphi }_n{\phi }_{n\mu },$$

(38)

and the terms $\phi {\phi }_{\mu }$, $\phi {\varphi }_{\mu }$, and $\varphi {\phi }_{\mu }$ are determined by

$$\begin{array}{cc}\hfill B_0& ={\phi }_0{\phi }_{0\mu }\hfill \\ \hfill B_1& ={\phi }_0{\phi }_{1\mu }+{\phi }_1{\phi }_{0\mu },\hfill \\ \hfill B_2& ={\phi }_0{\phi }_{2\mu }+{\phi }_1{\phi }_{1\mu }+{\phi }_2{\phi }_{0\mu },\hfill \\ \hfill B_3& ={\phi }_0{\phi }_{3\mu }+{\phi }_1{\phi }_{2\mu }+{\phi }_2{\phi }_{1\mu }+{\phi }_3{\phi }_{0\mu }.\hfill \end{array}$$

(39)

$$\begin{array}{cc}\hfill C_0& ={\phi }_0{\varphi }_{0\mu },\hfill \\ \hfill C_1& ={\phi }_0{\varphi }_{1\mu }+{\phi }_1{\varphi }_{0\mu },\hfill \\ \hfill C_2& ={\phi }_0{\varphi }_{2\mu }+{\phi }_1{\varphi }_{1\mu }+{\phi }_2{\varphi }_{0\mu },\hfill \\ \hfill C_3& ={\phi }_0{\varphi }_{3\mu }+{\phi }_1{\varphi }_{2\mu }+{\phi }_2{\varphi }_{1\mu }+{\phi }_3{\varphi }_{0\mu }.\hfill \end{array}$$

(40)

And

$$\begin{array}{cc}\hfill D_0& ={\varphi }_0{\phi }_{0\mu }\hfill \\ \hfill D_1& ={\varphi }_0{\phi }_{1\mu }+{\varphi }_1{\phi }_{0\mu },\hfill \\ \hfill D_2& ={\varphi }_0{\phi }_{2\mu }+{\varphi }_1{\phi }_{1\mu }+{\varphi }_2{\phi }_{0\mu },\hfill \\ \hfill D_3& ={\varphi }_0{\phi }_{3\mu }+{\varphi }_1{\phi }_{2\mu }+{\varphi }_2{\phi }_{1\mu }+{\varphi }_3{\phi }_{0\mu }.\hfill \end{array}$$

(41)

Implement the inverse $G_{\alpha }$-Laplace Transform into Equation (36) and, using Equations (27), (37), and (38), we have

$$\begin{array}{ccc}\hfill \sum _{n=0}^{\infty }{\varphi }_n\left(\mu ,\nu \right)& =& f_1\left(\mu \right)+L_p^{-1}{G}_{\alpha }^{-1}\left[s{F}_{\alpha }(p,s)\right]\hfill \\ & & +L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\sum _{n=0}^{\infty }{\varphi }_{n\mu \mu }\right]\right]\hfill \\ & & -L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\sum _{n=0}^{\infty }{A}_n+C_n+D_n\right]\right],\hfill \end{array}$$

(42)

and

$$\begin{array}{ccc}\hfill \sum _{n=0}^{\infty }{\phi }_n\left(\mu ,\nu \right)& =& h_1\left(\mu \right)+L_p^{-1}{G}_{\alpha }^{-1}\left[s{H}_{\alpha }(p,s)\right]\hfill \\ & & +L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\sum _{n=0}^{\infty }{\phi }_{n\mu \mu }\right]\right]\hfill \\ & & -L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\sum _{n=0}^{\infty }{B}_n+C_n+D_n\right]\right]\hfill \end{array}$$

(43)

provides our desired recursive relation in this way

$$\begin{array}{ccc}\hfill {\varphi }_0& =& f_1\left(\mu \right)+L_p^{-1}{G}_{\alpha }^{-1}\left[s{F}_{\alpha }(p,s)\right],\hfill \\ \hfill {\phi }_0& =& h_1\left(\mu \right)+L_p^{-1}{G}_{\alpha }^{-1}\left[s{H}_{\alpha }(p,s)\right],\hfill \end{array}$$

(44)

with the rest terms presented by

$${\varphi }_{n+1}=L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[{\varphi }_{n\mu \mu }-A_n-C_n-D_n\right]\right],$$

(45)

and

$${\phi }_{n+1}=L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[{\phi }_{n\mu \mu }-B_n-C_n-D_n\right]\right].$$

(46)

To explain this method for the coupled Burgers’ equation problem, we check the next examples

Example 5.

([21]). Consider the following one-dimensional homogeneous Burgers’ equations:

$$\begin{array}{ccc}\hfill {\varphi }_{\nu }-{\varphi }_{\mu \mu }-2\varphi {\varphi }_{\mu }+{\varphi }_{\mu }\phi +\varphi {\phi }_{\mu }& =& 0\hfill \\ \hfill {\phi }_{\nu }-{\phi }_{\mu \mu }-2\phi {\phi }_{\mu }+{\varphi }_{\mu }\phi +\varphi {\phi }_{\mu }& =& 0,\hfill \end{array}$$

(47)

subject to

$$\varphi \left(\mu ,0\right)=sin\mu ,\phi \left(\mu ,0\right)=sin\mu .$$

(48)

By using the mentioned method, we receive

$${\varphi }_0=sin\mu ,{\phi }_0=sin\mu ,$$

(49)

$${\varphi }_{n+1}=L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[{\varphi }_{n\mu \mu }+2{\varphi }_{n\mu }{\varphi }_n-{\varphi }_{n\mu }{\phi }_n-{\varphi }_n{\phi }_{n\mu }\right]\right],$$

(50)

and

$${\phi }_{n+1}=L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[{\phi }_{n\mu \mu }+2{\phi }_{n\mu }{\phi }_n-{\varphi }_{n\mu }{\phi }_n-{\varphi }_n{\phi }_{n\mu }\right]\right].$$

(51)

The following terms are presented, wherein $n\ge 0.$ At $n=0,$ we have

$$\begin{array}{cc}\hfill {\varphi }_1& =L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[{\varphi }_{0\mu \mu }+2{\varphi }_{0\mu }{\varphi }_0-{\varphi }_{0\mu }{\phi }_0-{\varphi }_0{\phi }_{0\mu }\right]\right]\hfill \\ \hfill & =L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[-sin\mu +2sin\mu cos\mu -2sin\mu cos\mu \right]\right]\hfill \\ \hfill {\varphi }_1& =L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[-sin\mu \right]\right],\hfill \\ \hfill {\varphi }_1& =L_p^{-1}{G}_{\alpha }^{-1}\left[-\frac{s^{\alpha +2}}{p^2+1}\right]=-\nu sin\mu ,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\phi }_1& =L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[{\phi }_{0\mu \mu }+2{\phi }_{0\mu }{\phi }_0-{\varphi }_{0\mu }{\phi }_0-{\varphi }_0{\phi }_{0\mu }\right]\right]\hfill \\ \hfill & =L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[-sin\mu +2sin\mu cos\mu -2sin\mu cos\mu \right]\right]\hfill \\ \hfill {\phi }_1& =L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[-sin\mu \right]\right],\hfill \\ \hfill {\phi }_1& =L_p^{-1}{G}_{\alpha }^{-1}\left[-\frac{s^{\alpha +2}}{p^2+1}\right]=-\nu sin\mu .\hfill \end{array}$$

In the same way at $n=1,$ we obtain

$$\begin{array}{cc}\hfill {\varphi }_2& =L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[{\varphi }_{1\mu \mu }+2\left({\varphi }_0{\varphi }_{1\mu }+{\varphi }_1{\varphi }_{0\mu }\right)\right]\right]\hfill \\ \hfill & -L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\left({\phi }_0{\varphi }_{1\mu }+{\phi }_1{\varphi }_{0\mu }\right)+\left({\varphi }_0{\phi }_{1\mu }+{\varphi }_1{\phi }_{0\mu }\right)\right]\right]\hfill \\ \hfill & =L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\nu sin\mu \right]\right],\hfill \\ \hfill {\varphi }_2& =L_p^{-1}{G}_{\alpha }^{-1}\left[\frac{s^{\alpha +3}}{p^2+1}\right]=\frac{{\nu }^2}{2!}sin\mu ,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\phi }_2& =L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[{\phi }_{1\mu \mu }+2\left({\phi }_0{\phi }_{1\mu }+{\phi }_1{\phi }_{0\mu }\right)\right]\right]\hfill \\ \hfill & -L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\left({\phi }_0{\varphi }_{1\mu }+{\phi }_1{\varphi }_{0\mu }\right)+\left({\varphi }_0{\phi }_{1\mu }+{\varphi }_1{\phi }_{0\mu }\right)\right]\right]\hfill \\ \hfill & =L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[\nu sin\mu \right]\right],\hfill \\ \hfill {\phi }_2& =L_p^{-1}{G}_{\alpha }^{-1}\left[\frac{s^{\alpha +3}}{p^2+1}\right]=\frac{{\nu }^2}{2!}sin\mu .\hfill \end{array}$$

Similarly, we obtain the remaining terms as follows:

$${\varphi }_3=L_p^{-1}{G}_{\alpha }^{-1}\left[-\frac{s^{\alpha +4}}{p^2+1}\right]=-\frac{{\nu }^3}{3!}sin\mu ,{\phi }_3=L_p^{-1}{G}_{\alpha }^{-1}\left[-\frac{s^{\alpha +4}}{p^2+1}\right]=-\frac{{\nu }^3}{3!}sin\mu .$$

We keep the same style to obtain the approximate solutions

$$\begin{array}{ccc}\hfill \varphi \left(\mu ,\nu \right)& =& {\varphi }_0+{\varphi }_2+{\varphi }_3+\dots =\left(1-\nu +\frac{{\nu }^2}{2!}-\frac{{\nu }^3}{3!}+\dots \right)sin\mu ,\hfill \\ \hfill \phi \left(\mu ,\nu \right)& =& {\phi }_0+{\phi }_2+{\phi }_3+\dots =\left(1-\nu +\frac{{\nu }^2}{2!}-\frac{{\nu }^3}{3!}+\dots \right)sin\mu .\hfill \end{array}$$

Therefore, the perfect solutions become

$$\varphi \left(\mu ,\nu \right)=e^{-\nu }sin\mu ,\phi \left(\mu ,\nu \right)=e^{-\nu }sin\mu .$$

The result we have reached is similar to the one in [21].

In the next example, we apply our technique to solve the nonhomogenous coupled system of Burgers’ equation.

Example 6.

Consider the following one-dimensional Burgers’ equations:

$$\begin{array}{ccc}\hfill {\varphi }_{\nu }-{\varphi }_{\mu \mu }-2\varphi {\varphi }_{\mu }+{\varphi }_{\mu }\phi +\varphi {\phi }_{\mu }& =& -\mu e^{-\nu }\hfill \\ \hfill {\phi }_{\nu }-{\phi }_{\mu \mu }-2\phi {\phi }_{\mu }+{\varphi }_{\mu }\phi +\varphi {\phi }_{\mu }& =& -\mu e^{-\nu },\hfill \end{array}$$

(52)

subject to

$$\begin{array}{ccc}\hfill \varphi \left(\mu ,0\right)& =& \mu \hfill \\ \hfill \phi \left(\mu ,0\right)& =& \mu .\hfill \end{array}$$

(53)

By using the aforesaid method, we obtain

$${\varphi }_0=\mu e^{-\nu },{\phi }_0=\mu e^{-\nu },$$

(54)

$${\varphi }_{n+1}=L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[{\varphi }_{n\mu \mu }-{\varphi }_{n\mu }{\varphi }_n-{\varphi }_{n\mu }{\phi }_n-{\varphi }_n{\phi }_{n\mu }\right]\right],$$

(55)

and

$${\phi }_{n+1}=L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[{\phi }_{n\mu \mu }-{\phi }_{n\mu }{\phi }_n-{\varphi }_{n\mu }{\phi }_n-{\varphi }_n{\phi }_{n\mu }\right]\right].$$

(56)

The subsequent terms are presented by

$$\begin{array}{cc}\hfill {\varphi }_1& =L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[{\varphi }_{0\mu \mu }-{\varphi }_{0\mu }{\varphi }_0-{\varphi }_{0\mu }{\phi }_0-{\varphi }_0{\phi }_{0\mu }\right]\right]\hfill \\ \hfill & =L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[0+2\mu e^{-2\nu }-2\mu e^{-2\nu }\right]\right]\hfill \\ \hfill {\varphi }_1& =0,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\phi }_1& =L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[{\phi }_{0\mu \mu }-{\phi }_{0\mu }{\phi }_0-{\varphi }_{0\mu }{\phi }_0-{\varphi }_0{\phi }_{0\mu }\right]\right]\hfill \\ \hfill & =L_p^{-1}{G}_{\alpha }^{-1}\left[s{L}_{\mu }{G}_{\alpha }\left[0+2\mu e^{-2\nu }-2\mu e^{-2\nu }\right]\right]\hfill \\ \hfill {\phi }_1& =0.\hfill \end{array}$$

In a similar manner, we have

$${\varphi }_2=0,{\varphi }_3=0,{\varphi }_4=0,\dots ,$$

and

$${\phi }_2=0,{\phi }_3=0,{\phi }_4=0,\dots$$

Hence, the solution series is given by

$$\varphi (\mu ,\nu )=\mu e^{-\nu },\phi (\mu ,\nu )=\mu e^{-\nu }.$$

4. Conclusions

In this study, a strong method called the G-Laplace transform was examined to find the exact solution of the Burgers’ equation and coupled Burgers’ equation. Moreover, some of the theorems of the properties of our technique are presented, and the uniqueness of our solutions is proved.