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