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A note on singular two-dimensional fractional coupled Burgers’ equation and triple Laplace Adomian decomposition method

Abstract

The present article focuses on how to find the exact solutions of the time-fractional regular and singular coupled Burgers’ equations by applying a new method that is called triple Laplace and Adomian decomposition method. Furthermore, the proposed method is a strong tool for solving many problems. The accuracy of the method is considered through the use of some examples, and the results obtained are compared with those of the existing methods in the literature.

Introduction

Fractional Burgers’ equation has received significant attention. The solution of this equation becomes very important for mathematical and physical phenomena. This equation has been discovered to explain different kinds of events, for example, it is a mathematical model of turbulence and approximate theory of a flow over a shock wave traveling in a viscous fluid [1, 2]. Local fractional homotopy analysis method has been explained in [3, 4]. The authors in [5] applied the new semianalytical method which is called the homotopy analysis Shehu transform method to solve multidimensional fractional diffusion equations. The researchers in [6, 7] examined the numerical solutions of three-dimensional Burger’s equation and Riccati differential equations by applying Laplace decomposition methods. From a recent couple of years, important dedication has been given to Laplace decomposition method and its modifications for studying mathematical model [8, 9]. In the literature several authors have suggested different types of approximation and exact methods for solving fractional Burger’s equation [1012]. The authors in [13] applied the variational iteration method to obtain Burger’s equation. In [14] the Laplace decomposition method (LDM) is suggested to solve the two-dimensional nonlinear Burgers’ equations. The authors in [15] used the double Laplace decomposition methods to solve singular Burgers’ equation and coupled Burgers’ equations. In this study, we propose a new hybrid triple Laplace Adomian decomposition method to obtain exact solutions of the time-fractional regular and singular coupled Burgers’ equations. Finally, two examples are provided to illustrate the suggested approach.

Here we recall some definitions, notation of Laplace transform, and fractional calculus facts which are useful in this article.

Definition 1

([16])

Let f be a function of three variables x, y, and t, where \(x,y,t>0\). The triple Laplace transform of f is defined by

$$ L_{x}L_{y}L_{t} \bigl[ f(x,y,t) \bigr] =F(p,q,s)= \int _{0}^{\infty } \int _{0}^{\infty } \int _{0}^{\infty }e^{-px-qy-st}f(x,y,t)\,dt \,dy \,dx, $$

where p, q, s are Laplace variables. Further, the triple Laplace transform of the partial derivatives is denoted by

$$\begin{aligned}& L_{x}L_{y}L_{t} \bigl[ u_{x} ( x,y,t ) \bigr] = pU(p,q,s)-U(0,q,s), \\& L_{x}L_{y}L_{t} \bigl[ u_{t} ( x,y,t ) \bigr] = sU(p,q,s)-U(p,q,0). \end{aligned}$$

Similarly, the triple Laplace transform for the second partial derivatives with respect to x, y, and t are defined as follows:

$$\begin{aligned}& L_{x}L_{y}L_{t} \bigl[ u_{tt} ( x,y,t ) \bigr] = p^{2}U(p,q,s)-pU(0,q,s)- \frac{\partial U(0,q,s)}{\partial x}, \\& L_{x}L_{y}L_{t} \bigl[ u_{yy} ( x,y,t ) \bigr] = q^{2}U(p,q,s)-qU(p,0,s)- \frac{\partial U(p,0,s)}{\partial y}, \\& L_{x}L_{y}L_{t} \bigl[ u_{tt} ( x,y,t ) \bigr] = s^{2}U(p,q,s)-sU(p,q,0)- \frac{\partial U(p,q,0)}{\partial t}. \end{aligned}$$

The inverse triple Laplace transform \(L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} [ F ( p,q,s ) ] =f(x,y,t)\) is defined as in [16] by the complex triple integral formula

$$ f(x,y,t)=\frac{1}{2\pi i} \int _{e-i\infty }^{e+i\infty }e^{px}\,dp \frac{1}{2\pi i} \int _{c-i\infty }^{c+i\infty }e^{qy}\,dq \frac{1}{2\pi i} \int _{d-i\infty }^{d+i\infty }e^{st}\,ds. $$

Definition 2

([1719])

The Caputo time-fractional derivative operator of order \(\alpha >0\) is defined by

$$ D_{t}^{\alpha }u(r,t)= \textstyle\begin{cases} \frac{\partial ^{m}u(r,t)}{\partial t^{m}}, & \mbox{for }m=\alpha \in \mathbb{N}; \\ \frac{1}{\varGamma ( m-\alpha ) }\int _{0}^{t} ( t- \tau )^{m-\alpha -1} \frac{\partial ^{m}u(r,\tau )}{\partial \tau ^{m}}\,d\tau , & m-1< \alpha < m. \end{cases} $$

In the following theorem, we present the triple Laplace transforms of the partial fractional Caputo derivatives.

Theorem 1

([20])

Letα, \(\beta ,\gamma >0\), \(n-1<\alpha \leq n\), \(m-1<\beta \leq m\), \(r-1< \gamma \leq r\), and\(n,m,p\in \mathbb{N} \), so that\(f\in C^{l} ( \mathbb{R} ^{+}\times \mathbb{R} ^{+}\times \mathbb{R} ^{+} )\), \(l=\max \{n,m,p\}\), \(f^{ ( l ) }\in L_{1} [ ( 0,a ) \times ( 0,b ) \times ( 0,c ) ] \)for any\(a, b, c >0\), \(\vert f(x,y,t) \vert \leq we^{x\tau _{1}+y\tau _{2}+t \tau _{3}}\), \(x >a>0\), \(y>b>0\), and\(t>c>0\). Then the triple Laplace transform of Caputo’s fractional derivatives\(D_{t}^{\alpha }u(x,y,t)\), \(D_{t}^{\alpha }u(x,y,t)\)and\(D_{t}^{\alpha }u(x,y,t)\)are given by

$$\begin{aligned}& L_{x}L_{y}L_{t} \bigl[ D_{t}^{\alpha }u(x,y,t) \bigr] \\& \quad =s^{\alpha }U ( p,q,s ) -\sum_{i=0}^{n-1}s^{\alpha -1-i}L_{y}L_{t} \bigl[ D_{t}^{i}u ( x,y,0 ) \bigr] ,\quad n-1< \alpha < n, \end{aligned}$$
(1.1)
$$\begin{aligned}& L_{x}L_{y}L_{t} \bigl[ D_{y}^{\beta }u(x,y,t) \bigr] \\& \quad =q^{\beta }U ( p,q,s ) -\sum_{j=0}^{m-1}q^{\beta -1-j}L_{y}L_{t} \bigl[ D_{y}^{j}u ( x,0,t ) \bigr] ,\quad m-1< \beta < m, \end{aligned}$$
(1.2)
$$\begin{aligned}& L_{x}L_{y}L_{t} \bigl[ D_{x}^{\gamma }u(x,y,t) \bigr] \\& \quad =p^{\gamma }U ( p,q,s ) -\sum_{k=0}^{r-1}p^{\gamma -1-k}L_{y}L_{t} \bigl[ D_{x}^{k}u ( 0,y,t ) \bigr] ,\quad r-1< \gamma < r. \end{aligned}$$
(1.3)

Below, we establish the relation between Mittag-Leffler function and Laplace transform, which will be useful in this paper. The Mittag-Leffler function is defined by the following series representation:

$$ E_{\beta } ( z ) =\sum_{k=0}^{\infty } \frac{z^{k}}{\varGamma ( \beta k+1 ) },\quad z\in \mathbb{C} , \Re (\beta )>0, $$
(1.4)

the Mittag-Leffler function with two parameters is defined by

$$ E_{\beta ,\gamma } ( z ) =\sum_{k=0}^{\infty } \frac{z^{k}}{\varGamma ( \beta k+\gamma ) },\quad z\in \mathbb{C} , \Re (\alpha )>0, $$
(1.5)

see [21]. If we put \(\beta =1\) in Eq. (1.5), we obtain Eq. (1.4). It follows from Eq. (1.5) that

$$\begin{aligned}& E_{1,1} ( z ) =\sum_{k=0}^{\infty } \frac{z^{k}}{\varGamma ( k+1 ) }=\sum_{k=0}^{\infty } \frac{z^{k}}{k!}=e^{z}, \end{aligned}$$
(1.6)
$$\begin{aligned}& E_{1,2} ( z ) =\sum_{k=0}^{\infty } \frac{z^{k}}{\varGamma ( k+2 ) }=\sum_{k=0}^{\infty } \frac{z^{k}}{ ( k+1 ) !}=\frac{1}{z}\sum_{k=0}^{\infty } \frac{z^{k+1}}{ ( k+1 ) }= \frac{e^{z}-1}{z}, \end{aligned}$$
(1.7)

and

$$ E_{1,3} ( z ) =\sum_{k=0}^{\infty } \frac{z^{k}}{\varGamma ( k+3 ) }=\sum_{k=0}^{\infty } \frac{z^{k}}{ ( k+2 ) !}=\frac{1}{z^{2}}\sum_{k=0}^{\infty } \frac{z^{k+2}}{ ( k+2 ) }= \frac{e^{z}-1-1}{z^{2}}, $$
(1.8)

in general,

$$ E_{1,m} ( z ) =\frac{1}{z^{m-1}} \Biggl[ e^{z}-\sum _{k=0}^{m-2} \frac{z^{k}}{k!} \Biggr] . $$
(1.9)

In the following, we introduce the Laplace transforms of some Mittag-Leffler functions which are useful in this work:

$$\begin{aligned}& L_{x}L_{y}L_{t} \bigl[ x^{2}t^{\alpha }E_{1,\alpha +1}(t) \bigr] = \frac{2!}{p^{3}qs^{\alpha } ( s-1 ) }, \\& L_{x}L_{y}L_{t} \bigl[ t^{\alpha }E_{1,\alpha +1}(t) \bigr] = \frac{1}{pqs^{\alpha } ( s-1 ) }, \\& L_{x}L_{y}L_{t} \bigl[ t^{2\alpha }E_{1,2\alpha +1}(t) \bigr] = \frac{1}{pqs^{2\alpha } ( s-1 ) }. \end{aligned}$$

Analysis of the triple Laplace decomposition method

The main objective of this section is to address the use of triple Laplace Adomian decomposition method (TLADM) for solving two-dimensional time-fractional coupled Burger’s equation. We consider two-dimensional time-fractional coupled Burger’s equation in the form:

$$ \begin{aligned} &D_{t}^{\alpha }u+uu_{x}+vu_{y} = \frac{1}{\Re } ( u_{xx}+u_{yy} ) ,\quad x,y,t>0, \\ &D_{t}^{\alpha }v+uv_{x}+vv_{y} = \frac{1}{\Re } ( v_{xx}+v_{yy} ) , \quad x,y,t>0, \\ &\quad n-1 < \alpha < n; \end{aligned} $$
(2.1)

with the initial condition

$$ u(x,y,0)=f_{1}(x,y),\qquad v(x,y,0)=g_{1}(x,y), $$

where \(D_{t}^{\alpha }=\frac{\partial ^{\alpha }}{\partial t^{\alpha }}\) is the fractional Caputo derivative, is the Reynolds number, and the velocity components are given by \(u(x,y,t)\) and \(v(x,y,t)\) in the x and y direction, respectively. For the purpose of finding the solution of Eq. (2.1), we apply triple Laplace Adomian decomposition method as follows:

Step 1. Taking the triple Laplace transform for Eq. (2.1), we obtain

$$ \begin{aligned} &\begin{aligned} s^{\alpha }L_{x}L_{y}L_{t} \bigl[ u(x,y,t) \bigr] -s^{\alpha -1}U ( p,q,0 ) ={}& {-}L_{x}L_{y}L_{t} ( uu_{x}+vu_{y} ) \\ &{}+L_{x}L_{y}L_{t} \biggl( \frac{1}{\Re } ( u_{xx}+u_{yy} ) \biggr), \end{aligned} \\ &\begin{aligned} s^{\alpha }L_{x}L_{y}L_{t} \bigl[ v(x,y,t) \bigr] -s^{\alpha -1}U ( p,q,0 ) ={}& {-}L_{x}L_{y}L_{t} ( uv_{x}+vv_{y} ) \\ &{} +L_{x}L_{y}L_{t} \biggl( \frac{1}{\Re } ( v_{xx}+v_{yy} ) \biggr). \end{aligned} \end{aligned} $$
(2.2)

Step 2. Now, employing the differentiation property of the Laplace transform, we have

$$\begin{aligned}& \begin{aligned}[b] L_{x}L_{y}L_{t} \bigl[ u(x,y,t) \bigr] ={}& \frac{1}{s}F_{1} ( p,q ) -\frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( uu_{x}+vu_{y} ) \\ &{} +\frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \biggl( \frac{1}{\Re } ( u_{xx}+u_{yy} ) \biggr) , \end{aligned} \\& \begin{aligned} L_{x}L_{y}L_{t} \bigl[ v(x,y,t) \bigr] ={}& \frac{1}{s}G_{1} ( p,q ) -\frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( uv_{x}+vv_{y} ) \\ &{} +\frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \biggl( \frac{1}{\Re } ( v_{xx}+v_{yy} ) \biggr) . \end{aligned} \end{aligned}$$
(2.3)

Step 3. By implementing the triple inverse Laplace transformation of Eq. (2.3), we obtain

$$ \begin{aligned} &\begin{aligned} u(x,y,t) ={}& L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s}F_{1} ( p,q ) \biggr) -L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( uu_{x}+vu_{y} ) \biggr) \\ &{} +L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \biggl( \frac{1}{\Re } ( u_{xx}+u_{yy} ) \biggr) \biggr) , \end{aligned} \\ &\begin{aligned} v(x,y,t) ={}& L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s}G_{1} ( p,q ) \biggr) -L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( uv_{x}+vv_{y} ) \biggr) \\ &{} +L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \biggl( \frac{1}{\Re } ( v_{xx}+v_{yy} ) \biggr) \biggr) . \end{aligned} \end{aligned} $$
(2.4)

Step 4. The Laplace Adomian decomposition solution functions \(u(x,y,t)\) and \(v(x,y,t)\) are given by the infinite series

$$ u(x,y,t)=\sum_{n=0}^{\infty }u_{n}(x,y,t), \qquad v(x,y,t)= \sum_{n=0}^{\infty }v_{n}(x,y,t), $$
(2.5)

further, the nonlinear terms \(uu_{x}\), \(v\frac{\partial u}{\partial y}\), \(u \frac{\partial v}{\partial x}\), and \(v\frac{\partial v}{\partial y}\) are given by:

$$ uu_{x}=\sum_{n=0}^{\infty }A_{n}, \qquad vu_{y}= \sum_{n=0}^{\infty }B_{n}, \qquad uv_{x}= \sum_{n=0}^{ \infty }C_{n},\qquad vv_{y}= \sum_{n=0}^{\infty }D_{n}, $$
(2.6)

and, by substituting Eqs. (2.5) and (2.6) into Eq. (2.4), we get

$$\begin{aligned} \sum_{m=0}^{\infty }u_{n}(x,y,t) =&L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s}F_{1} ( p,q ) \biggr) \\ &{}-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \Biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \Biggl( \sum_{n=0}^{\infty } ( A_{n}+B_{n} ) \Biggr) \Biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \Biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \Biggl( \frac{1}{\Re } \Biggl( \sum_{n=0}^{ \infty } ( u_{xxn}+u_{yyn} ) \Biggr) \Biggr) \Biggr) \end{aligned}$$
(2.7)

and

$$\begin{aligned} \sum_{n=0}^{\infty }v_{n}(x,y,t) =&L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s}G_{1} ( p,q ) \biggr) . \\ &{}-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \Biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \Biggl( \sum_{n=0}^{\infty } ( C_{n}+D_{n} ) \Biggr) \Biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \Biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \Biggl( \frac{1}{\Re } \Biggl( \sum_{n=0}^{ \infty } ( v_{xxn}+v_{yyn} ) \Biggr) \Biggr) \Biggr) . \end{aligned}$$
(2.8)

Step 5. Using Laplace Adomian decomposition method, we introduce the recursive relations and get:

$$ \begin{aligned} &u_{0}(x,y,t) = L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s}F_{1} ( p,q ) \biggr), \\ &v_{0}(x,y,t) = L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s}G_{1} ( p,q ) \biggr) , \end{aligned} $$
(2.9)

and the remainder components \(u_{n+1}\) and \(v_{n+1}\), for \(n\geq 0\), are given by

$$\begin{aligned} u_{n+1}(x,y,t) =&-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \bigl( ( A_{n}+B_{n} ) \bigr) \biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \biggl( \frac{1}{\Re } ( u_{xxn}+u_{yyn} ) \biggr) \biggr) \end{aligned}$$
(2.10)

and

$$\begin{aligned} v_{n+1}(x,y,t) =&-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( C_{n}+D_{n} ) \biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \biggl( \frac{1}{\Re } ( v_{xx}+v_{yy} ) \biggr) \biggr) , \end{aligned}$$
(2.11)

where a few first terms of the Adomian polynomials \(A_{n}\), \(B_{n}\), \(C_{n}\), and \(D_{n}\) are given by

$$\begin{aligned}& \begin{aligned} &A_{0} = u_{0}u_{0x},\qquad A_{1}=u_{0}u_{1x}+u_{1}u_{0x}, \\ &A_{2} = u_{0}u_{2x}+u_{1}u_{1x}+u_{2}u_{0x}, \\ &A_{3} = u_{0}u_{3x}+u_{1}u_{2x}+u_{2}u_{1x}+u_{3}u_{0x}, \end{aligned} \end{aligned}$$
(2.12)
$$\begin{aligned}& \begin{aligned} &B_{0} = v_{0}u_{0y},\qquad B_{1}=v_{0}u_{1y}+v_{1}u_{0y}, \\ &B_{2} = v_{0}u_{2y}+v_{1}u_{1y}+v_{2}u_{0y}, \\ &B_{3} = v_{0}u_{3y}+v_{1}u_{2y}+v_{2}u_{1y}+v_{3}u_{0y}, \end{aligned} \end{aligned}$$
(2.13)
$$\begin{aligned}& \begin{aligned} &C_{0} = u_{0}v_{0x},\qquad C_{1}=u_{0}v_{1x}+u_{1}v_{0x}, \\ &C_{2} = u_{0}v_{2x}+u_{1}v_{1x}+u_{2}v_{0x}, \\ &C_{3} = u_{0}v_{3x}+u_{1}v_{2x}+u_{2}v_{1x}+u_{3}v_{0x}, \end{aligned} \end{aligned}$$
(2.14)
$$\begin{aligned}& \begin{aligned} &D_{0} = v_{0}v_{0y},\qquad D_{1}=v_{0}v_{1y}+v_{1}v_{0y}, \\ &D_{2} = v_{0}v_{2y}+v_{1}v_{1y}+v_{2}v_{0y}, \\ &D_{3} = v_{0}v_{3y}+v_{1}v_{2y}+v_{2}v_{1y}+v_{3}v_{0y}, \end{aligned} \end{aligned}$$
(2.15)

and where \(L_{x}L_{y}L_{t}\) is the triple Laplace transform with respect to x, y, t and the triple inverse Laplace transform with respect to p, q, s is denoted by \(L_{p}^{-1}L_{q}^{-1}L_{s}^{-1}\). We supposed that the triple inverse Laplace transform with respect to p, q, and s exists in Eqs. (2.9), (2.10), and (2.11).

In the next numerical example, the suggested method is applied to a two-dimensional time-fractional coupled Burger’s equation when \(\Re =1 \) as follows:

Example 1

Consider a two-dimensional nonlinear Burger’s differential equation

$$ \begin{aligned} &D_{t}^{\alpha }u+uu_{x}+vu_{y} = u_{xx}+u_{yy}, \quad x,y,t>0, \\ &D_{t}^{\alpha }v+uv_{x}+vv_{y} = v_{xx}+v_{yy}, \quad x,y,t>0, \\ &\quad n-1 < \alpha < n; \end{aligned} $$
(2.16)

subject to the condition

$$ u(x,y,0)=x+y, \qquad v(x,y,0)=x-y. $$

As reported by the above steps, we have

$$ \begin{aligned} &\begin{aligned} u(x,y,t) = {}&x+y-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( uu_{x}+vu_{y} ) \biggr) \\ &{} +L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \biggl( \frac{1}{2} ( u_{xx}+u_{yy} ) \biggr) \biggr) , \end{aligned} \\ &\begin{aligned} v(x,y,t) ={}& x-y-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( uv_{x}+vv_{y} ) \biggr) \\ &{} +L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( v_{xx}+v_{yy} ) \biggr) . \end{aligned} \end{aligned} $$
(2.17)

The zeroth components \(u_{0}\) and \(v_{0}\), recommended by Adomian method, always contain initial condition and the source term, so we set

$$ u_{0}=x+y,\qquad v_{0}=x-y. $$

The other components \(u_{n+1}\), \(v_{n+1}\), \(n\geq 0\) are given by the relations

$$\begin{aligned} u_{n+1} =&-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \bigl( ( A_{n}+B_{n} ) \bigr) \biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( u_{xxn}+u_{yyn} ) \biggr) \end{aligned}$$
(2.18)

and

$$\begin{aligned} v_{n+1} =&-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( C_{n}+D_{n} ) \biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( v_{xxn}+v_{yyn} ) \biggr) . \end{aligned}$$
(2.19)

By taking \(n=0\) in Eqs. (2.18) and (2.19), we get

$$\begin{aligned} u_{1} =&-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \bigl( ( A_{0}+B_{0} ) \bigr) \biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( u_{0xx}+u_{0yy} ) \biggr) \\ =&-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( 2x ) \biggr) \\ =&-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{2}{p^{2}s^{\alpha +1}} \biggr) \\ =&\frac{-2xt^{\alpha }}{\varGamma ( \alpha +1 ) } \end{aligned}$$

and

$$\begin{aligned} v_{1} =&-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( C_{0}+D_{0} ) \biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( v_{0xx}+v_{0yy} ) \biggr) \\ =&-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( 2y) \biggr) \\ =&\frac{-2yt^{\alpha }}{\varGamma ( \alpha +1 ) }, \end{aligned}$$

similarly, when \(n=1\), we have

$$\begin{aligned} u_{2} =&-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \bigl( ( u_{0}u_{1x}+u_{1}u_{0x}+v_{0}u_{1y}+v_{1}u_{0y} ) \bigr) \biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \biggl( \frac{1}{2} ( u_{1xx}+u_{1yy} ) \biggr) \biggr) \\ =&-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \biggl( \frac{-t^{\alpha }}{\varGamma ( \alpha +1 ) } ( 2x+2y ) \biggr) \biggr) \\ =&L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{2\alpha +1}} \biggl( \frac{4}{p^{2}q}+\frac{4}{pq^{2}} \biggr) \biggr) \\ =&\frac{4 ( x+y ) t^{2\alpha }}{\varGamma ( 2\alpha +1 ) } \end{aligned}$$

and

$$\begin{aligned} v_{2} =&-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( D_{1}+E_{1} ) \biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( v_{1xx}+v_{1yy} ) \biggr) \\ =&L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \biggl( \frac{-t^{\alpha }}{\varGamma ( \alpha +1 ) } ( -4x+4y ) \biggr) \biggr) \\ =&L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{2\alpha +1}} \biggl( \frac{-4}{p^{2}q}+\frac{4}{pq^{2}} \biggr) \biggr) \\ =&\frac{4 ( x-y ) t^{2\alpha }}{\varGamma ( 2\alpha +1 ) }, \end{aligned}$$

when \(n=2\), we have

$$\begin{aligned} u_{3} =&-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( u_{0}u_{2x}+u_{1}u_{1x}+u_{2}u_{0x}+v_{0}u_{2y}+v_{1}u_{1y}+v_{2}u_{0y} ) \biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \biggl( \frac{1}{2} ( u_{2xx}+u_{2yy} ) \biggr) \biggr) \\ =&-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \biggl( \biggl( 16x+ \frac{4x\varGamma ( 2\alpha +1 ) }{ ( \varGamma ( \alpha +1 ) ) ^{2}} \biggr) \frac{t^{2\alpha }}{\varGamma ( 2\alpha +1 ) } \biggr) \biggr) \\ =&-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{16}{p^{2}qs^{3\alpha +1}}+\frac{4\varGamma ( 2\alpha +1 ) }{ ( \varGamma ( \alpha +1 ) ) ^{2}p^{2}qs^{3\alpha +1}} \biggr) \\ =& \biggl( -16x- \frac{4x\varGamma ( 2\alpha +1 ) }{ ( \varGamma ( \alpha +1 ) ) ^{2}} \biggr) \frac{t^{3\alpha }}{\varGamma ( 3\alpha +1 ) }, \end{aligned}$$

similarly,

$$\begin{aligned} v_{3} =&-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( D_{2}+E_{2} ) \biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \biggl( \frac{1}{2} ( v_{2xx}+v_{2yy} ) \biggr) \biggr) \\ =&-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \biggl( \biggl( 16y+ \frac{4y\varGamma ( 2\alpha +1 ) }{ ( \varGamma ( \alpha +1 ) ) ^{2}} \biggr) \frac{t^{2\alpha }}{\varGamma ( 2\alpha +1 ) } \biggr) \biggr) \\ =& \biggl( -16y- \frac{4y\varGamma ( 2\alpha +1 ) }{ ( \varGamma ( \alpha +1 ) ) ^{2}} \biggr) \frac{t^{3\alpha }}{\varGamma ( 3\alpha +1 ) }, \end{aligned}$$

hence, the solution of Eq. (2.16) is given by

$$\begin{aligned}& u(x,y,t) = \sum_{n=0}^{\infty }u_{n}=u_{0}+u_{1}+u_{2}+u_{3}+ \cdots , \\& u(x,y,t) = x+y- \frac{2xt^{\alpha }}{\varGamma ( \alpha +1 ) }+ \frac{4 ( x+y ) t^{2\alpha }}{\varGamma ( 2\alpha +1 ) }- \biggl( 16x+ \frac{4x\varGamma ( 2\alpha +1 ) }{ ( \varGamma ( \alpha +1 ) ) ^{2}} \biggr) \frac{t^{3\alpha }}{\varGamma ( 3\alpha +1 ) }+ \cdots , \end{aligned}$$

and

$$\begin{aligned}& v(x,y,t) = \sum_{n=0}^{\infty }v_{n}=v_{0}+v_{1}+v_{2}+v_{3}+ \cdots , \\& v(x,y,t) = x-y- \frac{2yt^{\alpha }}{\varGamma ( \alpha +1 ) }+ \frac{4 ( x-y ) t^{2\alpha }}{\varGamma ( 2\alpha +1 ) }+ \biggl( -16y- \frac{4y\varGamma ( 2\alpha +1 ) }{ ( \varGamma ( \alpha +1 ) ) ^{2}} \biggr) \frac{t^{3\alpha }}{\varGamma ( 3\alpha +1 ) }+ \cdots , \end{aligned}$$

at \(\alpha =1\) the solution of above equation becomes

$$\begin{aligned}& \begin{aligned} u(x,y,t) ={}&x+y-2xt+2 ( x+y ) t^{2}-4xt^{3}+4 ( x+y ) t^{4}-8xt^{5} \\ &{}+8 ( x+y ) t^{6}-16xt^{7}+16 ( x+y ) t^{8}+ \cdots \\ ={}&x \bigl( 1+2t^{2}+4t^{4}+8t^{6}+\cdots \bigr) +y \bigl( 1+2t^{2}+4t^{4}+8t^{6}+ \cdots \bigr) \\ &{}-2xt \bigl( 1+2t^{2}+4t^{4}+8t^{6}+ \cdots \bigr), \end{aligned} \\& u(x,y,t) =\frac{ ( x+y-2xt ) }{1-2t^{2}}, \end{aligned}$$

and

$$\begin{aligned}& \begin{aligned} v(x,y,t) ={}&x-y- \frac{2yt^{\alpha }}{\varGamma ( \alpha +1 ) }+ \frac{4 ( x-y ) t^{2\alpha }}{\varGamma ( 2\alpha +1 ) }+ \biggl( -16y- \frac{4y\varGamma ( 2\alpha +1 ) }{ ( \varGamma ( \alpha +1 ) ) ^{2}} \biggr) \frac{t^{3\alpha }}{\varGamma ( 3\alpha +1 ) }+ \cdots \\ ={}& ( x-y ) -2yt+2 ( x-y ) t^{2}-4yt^{3}+4 ( x-y ) t^{4}-8yt^{5}+8 ( x-y ) t^{6} \\ &{}-16yt^{7}+16 ( x-y ) t^{8}+\cdots \\ ={}&x \bigl( 1+2t^{2}+4t^{4}+8t^{6}+\cdots \bigr) -y \bigl( 1+2t^{2}+4t^{4}+8t^{6}+ \cdots \bigr) \\ &{}-2yt \bigl( 1+2t^{2}+4t^{4}+8t^{6}+ \cdots \bigr), \end{aligned} \\& v(x,y,t) =\frac{ ( x-y-2yt ) }{1-2t^{2}}. \end{aligned}$$

We obtained the same results as in [14].

Triple Laplace Adomian decomposition method and singular two-dimensional fractional coupled Burgers’ equation

The principal algorithm of the triple Laplace decomposition method will be applied to singular two-dimensional fractional coupled Burgers’ equation of the form

$$ \begin{aligned} &D_{t}^{\alpha }u+\frac{1}{x}uu_{x}+ \frac{1}{y}vu_{y}-\frac{1}{x} ( xu_{x} ) _{x}-\frac{1}{y} ( yu_{y} ) _{y} = f ( x,y,t ) , \\ &D_{t}^{\alpha }v+\frac{1}{x}uv_{x}+ \frac{1}{y}vv_{y}-\frac{1}{x} ( xv_{x} ) _{x}-\frac{1}{y} ( yv_{y} ) _{y} = g ( x,y,t ) , \\ &\quad x,y,t > 0, \end{aligned} $$
(3.1)

associated with the initial condition

$$ u(x,y,0)=f_{1}(x,y), \qquad v(x,y,0)=g_{1}(x,y), $$
(3.2)

where \(D_{t}^{\alpha }=\frac{\partial ^{\alpha }}{\partial t^{\alpha }}\) is the fractional Caputo derivative and \(\frac{1}{x} ( xu_{x} ) _{x}\), \(\frac{1}{y} ( yu_{y} ) _{y}\) are called Bessel operators, \(u(x,y,t)\) and \(v(x,y,t)\) are the velocity components to be determined, \(f ( x,y,t ) \); \(g ( x,y,t ) \); \(f_{1}(x,y)\) and \(g_{1}(x,y)\) are known functions. In order to obtain the solution of Eq. (3.1), we use the following steps:

Step 1. Multiplying both sides of Eq. (3.1) by xy, we have

$$ \begin{aligned} &xyD_{t}^{\alpha }u+yuu_{x}+xvu_{y}-y ( xu_{x} ) _{x}-x ( yu_{y} ) _{y} = xyf ( x,y,t ) , \\ &xyD_{t}^{\alpha }v+yuv_{x}+xvv_{y}-y ( xv_{x} ) _{x}-x ( yv_{y} ) _{y} = xyg ( x,y,t ) , \\ &\quad x,y,t > 0, \end{aligned} $$
(3.3)

Step 2. Applying the triple Laplace transform to both sides of Eq. (3.3), we obtain

$$\begin{aligned} \frac{\partial ^{2}}{\partial p\partial q} \bigl( s^{\alpha }U ( p,q,s ) -s^{\alpha -1}U ( p,q,0 ) \bigr) =&L_{x}L_{y}L_{t} \bigl( y ( xu_{x} ) _{x}+x ( yu_{y} ) _{y} \bigr) \\ &{}-L_{x}L_{y}L_{t} ( yuu_{x}+xvu_{y} ) \\ &{}+L_{x}L_{y}L_{t} \bigl( xyf ( x,y,t ) \bigr) \end{aligned}$$
(3.4)

and

$$\begin{aligned} \frac{\partial ^{2}}{\partial p\partial q} \bigl( s^{\alpha }V ( p,q,s ) -s^{\alpha -1}V ( p,q,0 ) \bigr) =&L_{x}L_{y}L_{t} \bigl( y ( xv_{x} ) _{x}+x ( yv_{y} ) _{y} \bigr) \\ &{}-L_{x}L_{y}L_{t} ( yuv_{x}+xvv_{y} ) \\ &{}+L_{x}L_{y}L_{t} \bigl( xyg ( x,y,t ) \bigr) . \end{aligned}$$
(3.5)

Now, using the differentiation property of the Laplace transform yields

$$ \begin{aligned} &\begin{aligned} \frac{\partial ^{2}}{\partial p\partial q}U ( p,q,s ) ={}& \frac{1}{s}F_{1} ( p,q ) -\frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( yuu_{x}+xvu_{y} ) \\ &{} +\frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \bigl( y ( xu_{x} ) _{x}+x ( yu_{y} ) _{y} \bigr) \\ &{} +\frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \bigl( xyf ( x,y,t ) \bigr) , \end{aligned} \\ &\begin{aligned} \frac{\partial ^{2}}{\partial p\partial q}V ( p,q,s ) = {}&\frac{1}{s}G_{1} ( p,q ) -\frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} ( yuv_{x}+xvv_{y} ) \\ &{} +\frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \bigl( y ( xv_{x} ) _{x}+x ( yv_{y} ) _{y} \bigr) \\ &{} +\frac{1}{s^{\alpha }}L_{x}L_{y}L_{t} \bigl( xyg ( x,y,t ) \bigr) . \end{aligned} \end{aligned} $$
(3.6)

Step 3. Integrating both sides of Eq. (3.6) from 0 to p and from 0 to q with respect to p and q, respectively, we have

$$\begin{aligned} U ( p,q,s ) =&\frac{1}{s} \int _{0}^{p} \int _{0}^{q} \biggl( \frac{\partial ^{2}}{\partial p\partial q}F_{1} ( p,q ) \biggr) \,dq\,dp \\ &{}+\frac{1}{s^{\alpha }} \int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} \bigl( y ( xu_{x} ) _{x}+x ( yu_{y} ) _{y} \bigr) \bigr) \,dq\,dp \\ &{}-\frac{1}{s^{\alpha }} \int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} ( yuu_{x}+xvu_{y} ) \bigr) \,dq\,dp \\ &{}+\frac{1}{s^{\alpha }} \int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} \bigl( xyf ( x,y,t ) \bigr) \bigr) \,dq\,dp \end{aligned}$$
(3.7)

and

$$\begin{aligned} V ( p,q,s ) =&\frac{1}{s} \int _{0}^{p} \int _{0}^{q} \biggl( \frac{\partial ^{2}}{\partial p\partial q}G_{1} ( p,q ) \biggr) \,dq\,dp \\ &{}+\frac{1}{s^{\alpha }} \int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} \bigl( y ( xv_{x} ) _{x}+x ( yv_{y} ) _{y} \bigr) \bigr) \,dq\,dp \\ &{}-\frac{1}{s^{\alpha }} \int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} ( yuv_{x}+xvv_{y} ) \bigr) \,dq\,dp \\ &{}+\frac{1}{s^{\alpha }} \int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} \bigl( xyg ( x,y,t ) \bigr) \bigr) \,dq\,dp. \end{aligned}$$
(3.8)

Step 4. By taking the triple inverse Laplace transformation of Eqs. (3.7) and (3.8), we obtain

$$\begin{aligned} u ( x,y,t ) =&L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s}\int _{0}^{p} \int _{0}^{q} \biggl( \frac{\partial ^{2}}{\partial p\partial q}F_{1} ( p,q ) \biggr) \,dq\,dp \biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} \bigl( y ( xu_{x} ) _{x}+x ( yu_{y} ) _{y} \bigr) \bigr) \,dq\,dp \biggr) \\ &{}-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} ( yuu_{x}+xvu_{y} ) \bigr) \,dq\,dp \biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} \bigl( xyf ( x,y,t ) \bigr) \bigr) \,dq\,dp \biggr) \end{aligned}$$
(3.9)

and

$$\begin{aligned} v ( x,y,t ) =&L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s}\int _{0}^{p} \int _{0}^{q} \biggl( \frac{\partial ^{2}}{\partial p\partial q}G_{1} ( p,q ) \biggr) \,dq\,dp \biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} \bigl( y ( xv_{x} ) _{x}+x ( yv_{y} ) _{y} \bigr) \bigr) \,dq\,dp \biggr) \\ &{}-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} ( yuv_{x}+xvv_{y} ) \bigr) \,dq\,dp \biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} \bigl( xyg ( x,y,t ) \bigr) \bigr) \,dq\,dp \biggr) . \end{aligned}$$
(3.10)

Step 5. Substituting Eqs. (2.5) and (2.6) into Eqs. (3.9) and (3.10), we get

$$\begin{aligned} \sum_{n=0}^{\infty }u_{n}(x,y,t) =&L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s} \int _{0}^{p} \int _{0}^{q} \biggl( \frac{\partial ^{2}}{\partial p\partial q}F_{1} ( p,q ) \biggr) \,dq\,dp \biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \Biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \Biggl( L_{x}L_{y}L_{t} \Biggl( y \Biggl( x\sum_{n=0}^{\infty }u_{n}{}_{x} \Biggr) _{x} \\ &{}+x \Biggl( y\sum_{n=0}^{\infty }u_{n}{}_{y} \Biggr) _{y} \Biggr) \Biggr) \,dq\,dp \Biggr) \\ &{}-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \Biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \Biggl( L_{x}L_{y}L_{t} \Biggl( y\sum_{n=0}^{\infty }A_{n}+x \sum_{n=0}^{ \infty }B_{n} \Biggr) \Biggr) \,dq\,dp \Biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} \bigl( xyf ( x,y,t ) \bigr) \bigr) \,dq\,dp \biggr) \end{aligned}$$

and

$$\begin{aligned} \sum_{n=0}^{\infty }v_{n}(x,y,t) =&L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s} \int _{0}^{p} \int _{0}^{q} \biggl( \frac{\partial ^{2}}{\partial p\partial q}G_{1} ( p,q ) \biggr) \,dq\,dp \biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \Biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \Biggl( L_{x}L_{y}L_{t} \Biggl( y \Biggl( x\sum_{n=0}^{\infty }v_{n}{}_{x} \Biggr) _{x} \\ &{}+x \Biggl( y\sum_{n=0}^{\infty }v_{n}{}_{y} \Biggr) _{y} \Biggr) \Biggr) \,dq\,dp \Biggr) \\ &{}-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \Biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \Biggl( L_{x}L_{y}L_{t} \Biggl( y\sum_{n=0}^{\infty }C_{n}+x \sum_{n=0}^{ \infty }D_{n} \Biggr) \Biggr) \,dq\,dp \Biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} \bigl( xyg ( x,y,t ) \bigr) \bigr) \,dq\,dp \biggr) . \end{aligned}$$

Step 6. Using Laplace Adomian decomposition method, we introduce the recursive relations and get

$$\begin{aligned} u_{0}(x,y,t) =&L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s}\int _{0}^{p} \int _{0}^{q} \biggl( \frac{\partial ^{2}}{\partial p\partial q}F_{1} ( p,q ) \biggr) \,dq\,dp \biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} \bigl( xyf ( x,y,t ) \bigr) \bigr) \,dq\,dp \biggr) \end{aligned}$$
(3.11)

and

$$\begin{aligned} v_{0}(x,y,t) =&L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s}\int _{0}^{p} \int _{0}^{q} \biggl( \frac{\partial ^{2}}{\partial p\partial q}G_{1} ( p,q ) \biggr) \,dq\,dp \biggr) \\ &{}+L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} \bigl( xyg ( x,y,t ) \bigr) \bigr) \,dq\,dp \biggr) , \end{aligned}$$
(3.12)

the other components \(u_{n+1}\) and \(v_{n+1}\), for \(n\geq 0\), are given by

$$\begin{aligned} u_{n+1}(x,y,t) =&L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \Biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \Biggl( L_{x}L_{y}L_{t} \Biggl( y \Biggl( x\sum_{n=0}^{\infty }u_{n}{}_{x} \Biggr) _{x} \\ &{}+x \Biggl( y\sum_{n=0}^{\infty }u_{n}{}_{y} \Biggr) _{y} \Biggr) \Biggr) \,dq\,dp \Biggr) \\ &{}-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \Biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \Biggl( L_{x}L_{y}L_{t} \Biggl( y\sum_{n=0}^{\infty }A_{n}+x \sum_{n=0}^{ \infty }B_{n} \Biggr) \Biggr) \,dq\,dp \Biggr) \end{aligned}$$
(3.13)

and

$$\begin{aligned} v_{n+1}(x,y,t) =&L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \Biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \Biggl( L_{x}L_{y}L_{t} \Biggl( y \Biggl( x\sum_{n=0}^{\infty }v_{n}{}_{x} \Biggr) _{x} \\ &{}+x \Biggl( y\sum_{n=0}^{\infty }v_{n}{}_{y} \Biggr) _{y} \Biggr) \Biggr) \,dq\,dp \Biggr) \\ &{}-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \Biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \Biggl( L_{x}L_{y}L_{t} \Biggl( y\sum_{n=0}^{\infty }C_{n}+x \sum_{n=0}^{ \infty }D_{n} \Biggr) \Biggr) \,dq\,dp \Biggr) , \end{aligned}$$
(3.14)

where \(L_{x}L_{y}L_{t}\) is the triple Laplace transform with respect to x, y, t and the triple inverse Laplace transform with respect to p, q, s is denoted by \(L_{p}^{-1}L_{q}^{-1}L_{s}^{-1}\). We assumed that the triple inverse Laplace transform with respect to p, q, and s exists for Eqs. (3.11), (3.12), (3.13), and (3.14). In the following example we apply the triple Laplace Adomian decomposition method to solve singular two-dimensional time-fractional coupled Burgers’ equations.

Example 2

Consider singular two-dimensional time-fractional coupled Burgers’ equations given by

$$ \begin{aligned} &D_{t}^{\alpha }u+\frac{1}{x}uu_{x}+ \frac{1}{y}vu_{y}-\frac{1}{x} ( xu_{x} ) _{x}-\frac{1}{y} ( yu_{y} ) _{y} = \bigl( x^{2}-y^{2} \bigr) e^{t}, \\ &D_{t}^{\alpha }v+\frac{1}{x}uv_{x}+ \frac{1}{y}vv_{y}-\frac{1}{x} ( xv_{x} ) _{x}-\frac{1}{y} ( yv_{y} ) _{y} = \bigl( x^{2}-y^{2} \bigr) e^{t}, \\ &\quad x,y,t > 0, \end{aligned} $$
(3.15)

with the initial condition

$$ u(x,y,0)=x^{2}-y^{2},\qquad v(x,y,0)=x^{2}-y^{2}. $$

As stated by the above steps, we have

$$\begin{aligned} u ( x,y,t ) =&x^{2}-y^{2}+\frac{1}{s^{\alpha }} \int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} \bigl( y ( xu_{x} ) _{x}+x ( yu_{y} ) _{y} \bigr) \bigr) \,dq\,dp \\ &{}-\frac{1}{s^{\alpha }} \int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} ( yuu_{x}+xvu_{y} ) \bigr) \,dq\,dp \\ &{}+x^{2}t^{\alpha }E_{1,\alpha +1}(t)-y^{2}t^{\alpha }E_{1,\alpha +1}(t) \end{aligned}$$
(3.16)

and

$$\begin{aligned} v ( x,y,t ) =&x^{2}-y^{2}+\frac{1}{s^{\alpha }} \int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} \bigl( y ( xv_{x} ) _{x}+x ( yv_{y} ) _{y} \bigr) \bigr) \,dq\,dp \\ &{}-\frac{1}{s^{\alpha }} \int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} ( yuv_{x}+xvv_{y} ) \bigr) \,dq\,dp \\ &{}+x^{2}t^{\alpha }E_{1,\alpha +1}(t)-y^{2}t^{\alpha }E_{1,\alpha +1}(t). \end{aligned}$$
(3.17)

By applying Eqs. (3.11), (3.12), (3.13), and (3.14), we obtain

$$ \begin{aligned} &u_{0}(x,y,t) = x^{2}-y^{2}+x^{2}t^{\alpha }E_{1,\alpha +1}(t)-y^{2}t^{ \alpha }E_{1,\alpha +1}(t), \\ &v_{0}(x,y,t) = x^{2}-y^{2}+x^{2}t^{\alpha }E_{1,\alpha +1}(t)-y^{2}t^{ \alpha }E_{1,\alpha +1}(t), \end{aligned} $$
(3.18)

and the other components \(u_{n+1}\) and \(v_{n+1}\), for \(n\geq 0\), are given by

$$\begin{aligned} u_{n+1}(x,y,t) =&L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} \bigl( y ( xu_{n}{}_{x} ) _{x}+x ( yu_{n}{}_{y} ) _{y} \bigr) \bigr) \,dq\,dp \biggr) \\ &{}-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} ( yA_{n}+xB_{n} ) \bigr) \,dq\,dp \biggr) \end{aligned}$$
(3.19)

and

$$\begin{aligned} v_{n+1}(x,y,t) =&L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} \bigl( y ( xv_{n}{}_{x} ) _{x}+x ( yv_{n}{}_{y} ) _{y} \bigr) \bigr) \,dq\,dp \biggr) \\ &{}-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} ( yC_{n}+xD_{n} ) \bigr) \,dq\,dp \biggr) , \end{aligned}$$
(3.20)

where a few first terms of the Adomian polynomials \(A_{n}\), \(B_{n}\), \(C_{n}\), and \(D_{n}\) are given by Eqs. (2.12), (2.13), (2.14), and (2.15), respectively.

By substituting \(n=0\) into Eqs. (3.19) and (3.20), we get

$$\begin{aligned}& \begin{aligned} u_{1}(x,y,t) ={}&L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} \bigl( y ( xu_{0}{}_{x} ) _{x}+x ( yu_{0}{}_{y} ) _{y} \bigr) \bigr) \,dq\,dp \biggr) \\ &{}-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} ( yA_{0}+xB_{0} ) \bigr) \,dq\,dp \biggr) , \end{aligned} \\& u_{1}(x,y,t) =0, \end{aligned}$$

and

$$\begin{aligned}& \begin{aligned} v_{1}(x,y,t) ={}&L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} \bigl( y ( xv_{0}{}_{x} ) _{x}+x ( yv_{0}{}_{y} ) _{y} \bigr) \bigr) \,dq\,dp \biggr) \\ &{}-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} ( yC_{0}+xD_{0} ) \bigr) \,dq\,dp \biggr) , \end{aligned} \\& v_{1}(x,y,t) =0. \end{aligned}$$

In the same manner for \(n=1\), we obtain that

$$\begin{aligned}& \begin{aligned} u_{2}(x,y,t) ={}&L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} \bigl( y ( xu_{1}{}_{x} ) _{x}+x ( yu_{1}{}_{y} ) _{y} \bigr) \bigr) \,dq\,dp \biggr) \\ &{}-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} ( yA_{1}+xB_{1} ) \bigr) \,dq\,dp \biggr) , \end{aligned} \\& u_{2}(x,y,t) =0, \end{aligned}$$

and

$$\begin{aligned}& \begin{aligned} v_{2}(x,y,t) ={}&L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} \bigl( y ( xv_{1}{}_{x} ) _{x}+x ( yv_{1}{}_{y} ) _{y} \bigr) \bigr) \,dq\,dp \biggr) \\ &{}-L_{p}^{-1}L_{q}^{-1}L_{s}^{-1} \biggl( \frac{1}{s^{\alpha }}\int _{0}^{p} \int _{0}^{q} \bigl( L_{x}L_{y}L_{t} ( yC_{1}+xD_{1} ) \bigr) \,dq\,dp \biggr) , \end{aligned} \\& v_{2}(x,y,t) =0. \end{aligned}$$

The solution of Eq. (3.15) is given by

$$\begin{aligned}& u(x,y,t) = u_{0}+u_{1}+u_{2}+\cdots +u_{n}, \\& v(x,y,t) = v_{0}+v_{1}+v_{2}+\cdots +v_{n}. \end{aligned}$$

Hence, the exact solution is given by

$$\begin{aligned}& u(x,y,t) = x^{2}-y^{2}+x^{2}t^{\alpha }E_{1,\alpha +1}(t)-y^{2}t^{ \alpha }E_{1,\alpha +1}(t), \\& v(x,y,t) = x^{2}-y^{2}+x^{2}t^{\alpha }E_{1,\alpha +1}(t)-y^{2}t^{ \alpha }E_{1,\alpha +1}(t), \end{aligned}$$

where E denotes Mittag-Leffler function. Setting \(\alpha =1\) in Eq. (3.15), we get that the exact solution of the singular two-dimensional coupled Burgers’ equation

$$\begin{aligned}& D_{t}u+\frac{1}{x}uu_{x}+ \frac{1}{y}vu_{y}-\frac{1}{x} ( xu_{x} ) _{x}-\frac{1}{y} ( yu_{y} ) _{y} = \bigl( x^{2}-y^{2} \bigr) e^{t},\\& D_{t}v+\frac{1}{x}uv_{x}+ \frac{1}{y}vv_{y}-\frac{1}{x} ( xv_{x} ) _{x}-\frac{1}{y} ( yv_{y} ) _{y} = \bigl( x^{2}-y^{2} \bigr) e^{t},\\& \quad x,y,t > 0, \end{aligned}$$

with the initial condition

$$ u(x,y,0)=x^{2}-y^{2},\qquad v(x,y,0)=x^{2}-y^{2}, $$

is given by

$$ u(x,y,t)= \bigl( x^{2}-y^{2} \bigr) e^{t} ,\qquad v(x,y,t)= \bigl( x^{2}-y^{2} \bigr) e^{t}. $$

Conclusion

In this study, the triple Laplace transform and Adomian decomposition have been successfully combined to obtain a new powerful method named a hybrid triple Laplace Adomian decomposition method (TLADM). This method has been used to solve regular and singular coupled Burgers’ equations. By applying this method on some examples, we have obtained new efficient relations to solve our problems. It allows more realistic series solution that converges very quickly to the true solution.

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The authors would like to extend their sincere appreciation to the Deanship of Scientific Research at King Saud University for its funding of this Research group No. RG-1440-030.

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Eltayeb, H., Bachar, I. A note on singular two-dimensional fractional coupled Burgers’ equation and triple Laplace Adomian decomposition method. Bound Value Probl 2020, 129 (2020). https://doi.org/10.1186/s13661-020-01426-0

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Keywords

  • Double and triple Laplace transform
  • Inverse double and triple Laplace transform
  • Singular fractional coupled Burgers’ equation
  • Mittag-Leffler function
  • Single Laplace transform
  • Decomposition methods