Here, with the aid of fixed point theory, we demonstrate the existence and uniqueness for the solution of the considered problem. Now, by the aid of Eq. (27), one can have

$$ {}_{0}^{\mathit{ABC}} D_{t}^{\alpha} \bigl[ u ( x,y,t ) \bigr] =\mathcal{G} ( x,y,t, u ). $$

(35)

With the help of Eq. (35) and Theorem 2, one can get

$$ u ( x,y,t ) -u ( x,y,0 ) = \frac{ ( 1-\alpha )}{\mathcal{B} ( \alpha )} \mathcal{G} ( x,y,t,u ) + \frac{\alpha}{\mathcal{B} ( \alpha ) \varGamma ( \alpha )} \int_{0}^{t} \mathcal{G} ( x,y,\zeta,u ) ( t- \zeta )^{\alpha-1}\, d\zeta. $$

(36)

### Theorem 3

*The kernel*\(\mathcal{G}\)*satisfies the Lipschitz condition and also contraction if it satisfies*\(0 \leq ( \mu \Delta^{2} - \Delta+ ( a^{2} + b^{2} +ab ) -1 ) <1\).

### Proof

Let us choose two functions *u* and \(u_{1}\) to verify the required condition, therefore we get

$$\begin{aligned}[b] \bigl\Vert \mathcal{G} ( x,y,t, u ) -\mathcal{G} ( x,y,t, u_{1} ) \bigr\Vert &= \bigl\Vert \mu\Delta^{2} \bigl[ u ( x,y,t ) -u ( x,y, t_{1} ) \bigr] - \Delta \bigl[ u ( x,y,t ) -u ( x,y, t_{1} ) \bigr] \\ &\quad + \bigl[ u^{3} ( x,y,t ) - u^{3} ( x,y, t_{1} ) \bigr] - \bigl[ u ( x,y,t ) -u ( x,y, t_{1} ) \bigr] \bigr\Vert \\ &\leq \bigl\Vert \mu\Delta^{2} - \Delta+ \bigl( a^{2} + b^{2} +ab \bigr) -1 \bigr\Vert \bigl\Vert u ( x,y,t ) -u(x,y, t_{1} ) \bigr\Vert \\ &\leq \bigl( \mu\Delta^{2} - \Delta+ \bigl( a^{2} + b^{2} +ab \bigr) -1 \bigr) \bigl\Vert u ( x,y,t ) -u ( x,y, t_{1} ) \bigr\Vert .\end{aligned} $$

(37)

Since *u* and \(u_{1}\) are bounded and here \(a= \Vert u \Vert \) and \(b= \Vert u_{1} \Vert \). Inserting \(\eta=\mu\Delta^{2} - \Delta + ( a^{2} + b^{2} +ab ) -1\) in Eq. (37), we get

$$ \bigl\Vert \mathcal{G} ( x,y,t, u ) -\mathcal{G} ( x,y,t, u_{1} ) \bigr\Vert \leq\eta \bigl\Vert u ( x,y,t ) -u(x,y, t_{1} ) \bigr\Vert . $$

(38)

It is clear that the Lipschitz condition is achieved for \(\mathcal{G}_{1}\). Moreover, if \(0 \leq ( \mu\Delta^{2} - \Delta+ ( a^{2} + b^{2} +ab ) -1 ) <1\), then it leads to contraction. Now, the recursive form is present for the above relation and initial condition as follows:

$$ u_{n} ( x,y,t ) = \frac{ ( 1-\alpha )}{\mathcal{B} ( \alpha )} \mathcal{G} ( x,y,t, u_{n-1} ) + \frac{\alpha}{\mathcal{B} ( \alpha ) \varGamma ( \alpha )} \int_{0}^{t} \mathcal{G} ( x,y,\zeta, u_{n-1} ) ( t-\zeta )^{\alpha-1} \,d\zeta $$

(39)

and

$$ u ( x,y,0 ) = u_{0} ( x,y,t ). $$

(40)

The difference between successive terms is defined as follows:

$$\begin{aligned}[b] \phi_{n} ( x,y,t ) &= u_{n} ( x,y,t ) - u_{n-1} ( x,y,t )\\& = \frac{ ( 1-\alpha )}{\mathcal{B} ( \alpha )} \bigl( \mathcal{G}_{1} ( x,y,t, u_{n-1} ) -\mathcal{G} ( x,y,t, u_{n-2} ) \bigr)\\&\quad + \frac{\alpha}{\mathcal{B} ( \alpha ) \varGamma ( \alpha )} \int_{0}^{t} \mathcal{G} ( x,y,\zeta, u_{n-1} ) ( t-\zeta )^{\alpha-1}\, d\zeta. \end{aligned} $$

(41)

Notice that

$$ u_{n} ( x,y,t ) = \sum_{i=1}^{n} \phi_{i} ( x,y,t ). $$

(42)

By employing the norm on Eq. (41) and considering Eq. (38), we have

$$ \bigl\Vert \phi_{n} ( x,y,t ) \bigr\Vert \leq \frac{ ( 1-\alpha )}{\mathcal{B} ( \alpha )} \eta \bigl\Vert \phi_{(n-1)} ( x,y,t ) \bigr\Vert + \frac{\alpha}{\mathcal{B} ( \alpha ) \varGamma ( \alpha )} \eta \int_{0}^{t} \bigl\Vert \phi_{(n-1)} ( x,y,\zeta ) \bigr\Vert \,d\zeta. $$

(43)

□

With the aid of foregoing results, we prove the following theorems.

### Theorem 4

*For the projected system* (27), *the solution will exist and be unique for a particular*\(t_{0}\)*such that*

$$\frac{ ( 1-\alpha )}{\mathcal{B} ( \alpha )} \eta+ \frac{\alpha}{\mathcal{B} ( \alpha ) \varGamma ( \alpha )} \eta< 1. $$

### Proof

Let \(u (x,y, t )\) be a bounded function sustaining the Lipschitz condition. With the assistance of Eq. (43), we get

$$ \bigl\Vert \phi_{i} ( x,y,t ) \bigr\Vert \leq \bigl\Vert u_{n} ( x,y,0 ) \bigr\Vert \biggl[ \frac{ ( 1-\alpha )}{ \mathcal{B} ( \alpha )} \eta+ \frac{\alpha}{\mathcal{B} ( \alpha ) \varGamma ( \alpha )} \eta \biggr]^{n}. $$

(44)

This proves the continuity and existence of the achieved solution. Further, to verify that the above equation is the solution for Eq. (27), we consider

$$ u ( x,y,t ) -u ( x,y,0 ) = u_{n} ( x,y,t ) - \mathcal{K}_{n} ( x,y,t ). $$

(45)

Then, we consider achieving the required result

$$\begin{aligned}[b] \bigl\Vert \mathcal{K}_{n} ( x,y,t ) \bigr\Vert &=\biggl\Vert \frac{ ( 1-\alpha )}{\mathcal{B} ( \alpha )} \bigl( \mathcal{G} ( x,y,t,u ) -\mathcal{G}( x,y,t, u_{n-1} ) \bigr) \\ &\quad+ \frac{\alpha}{\mathcal{B} ( \alpha ) \varGamma ( \alpha )} \int_{0}^{t} ( t-\zeta )^{\mu-1} \bigl( \mathcal{G} ( x,y,\zeta,u ) -\mathcal{G} ( x,y,\zeta, u_{n-1} ) \bigr) \,d\zeta\biggr\Vert \\ &\leq\frac{ ( 1-\alpha )}{\mathcal{B} ( \alpha )} \bigl\Vert \mathcal{G} ( x,y,\zeta,u ) -\mathcal{G} ( x,y,\zeta, u_{n-1} ) \bigr\Vert \\ &\quad+ \frac{\alpha}{\mathcal{B} ( \alpha ) \varGamma ( \alpha )} \int_{0}^{t} \bigl\Vert \mathcal{G} ( x,y, \zeta,u ) -\mathcal{G} ( x,y,\zeta, u_{n-1} ) \bigr\Vert \,d\zeta \\ &\leq\frac{ ( 1-\alpha )}{\mathcal{B} ( \alpha )} \eta_{1} \Vert u- u_{n-1} \Vert + \frac{\alpha}{\mathcal{B} ( \alpha ) \varGamma ( \alpha )} \eta_{1} \Vert u- u_{n-1} \Vert t.\end{aligned} $$

(46)

Similarly, at \(t_{0}\) we can obtain

$$ \bigl\Vert \mathcal{K}_{n} ( x,y,t ) \bigr\Vert \leq \biggl( \frac{ ( 1-\alpha )}{\mathcal{B} ( \alpha )} + \frac{\alpha t_{0}}{ \mathcal{B} ( \alpha ) \varGamma ( \alpha )} \biggr)^{n+1} \eta^{n+1} M. $$

(47)

We have from Eq. (47), for *n* tending to ∞, \(\Vert\mathcal{K}_{n} ( x,y,t ) \Vert\) approaches 0. Now, it is important to present the uniqueness for the obtained solution. Suppose that \(u^{*} ( t )\) is a different solution, then one can get

$$\begin{aligned}[b] u ( x,y,t ) - u^{*} ( x,y,t ) &= \frac{ ( 1-\alpha )}{\mathcal{B} ( \alpha )} \bigl( \mathcal{G} ( x,y,t,u ) -\mathcal{G} \bigl( x,y,t, u^{*} \bigr) \bigr) \\&\quad+ \frac{\alpha}{\mathcal{B} ( \alpha ) \varGamma ( \alpha )} \int_{0}^{t} \bigl( \mathcal{G} ( x,y,\zeta,u ) -\mathcal{G} \bigl( x,y,\zeta, u^{*} \bigr) \bigr) \,d\zeta.\end{aligned} $$

(48)

With the help of properties of the norm, Eq. (48) reduces to

$$\begin{aligned}[b] \bigl\Vert u ( x,y,t ) - u^{*} ( x,y,t ) \bigr\Vert &= \biggl\Vert \frac{ ( 1-\alpha )}{\mathcal{B} ( \alpha )} \bigl( \mathcal{G} ( x,y,t,u ) -\mathcal{G} \bigl( x,y,t, u^{*} \bigr) \bigr) \\&\quad+ \frac{\alpha}{\mathcal{B} ( \alpha ) \varGamma ( \alpha )} \int_{0}^{t} \bigl( \mathcal{G} ( x,y,\zeta,u ) -\mathcal{G} \bigl( x,y,\zeta, u^{*} \bigr) \bigr) \,d\zeta \biggr\Vert \\&\leq\frac{ ( 1-\alpha )}{\mathcal{B} ( \alpha )} \eta \bigl\Vert u ( x,y,t ) - u^{*} ( x,y,t ) \bigr\Vert \\&\quad + \frac{\alpha}{\mathcal{B} ( \alpha ) \varGamma ( \alpha )} \eta t \bigl\Vert u ( x,y,t ) - u^{*} ( x,y,t ) \bigr\Vert . \end{aligned} $$

(49)

On simplification

$$ \bigl\Vert u ( x,y,t ) - u^{*} ( x,y,t ) \bigr\Vert \biggl( 1- \frac{ ( 1-\alpha )}{\mathcal{B} ( \alpha )} \eta- \frac{\alpha}{\mathcal{B} ( \alpha ) \varGamma ( \alpha )} \eta t \biggr) \leq0. $$

(50)

From the above condition, it is clear that \(u ( t ) = u^{*} ( t )\), if

$$ \biggl( 1- \frac{ ( 1-\alpha )}{\mathcal{B} ( \alpha )} \eta- \frac{\alpha}{\mathcal{B} ( \alpha ) \varGamma ( \alpha )} \eta t \biggr) \geq0. $$

(51)

Hence, Eq. (51) shows our required result. □

### Theorem 5

*Let*\(( B [ 0, T ], \Vert \cdot \Vert )\)*be a Banach space*. *Suppose that*\(u_{n} ( x,y,t )\)*and*\(u ( x,y,t )\)*define**B*, *if*\(0<\lambda<1\), *then the series solution presented in Eq*. (26) *tends towards the solution of Eq*. (10).

### Proof

Let \(\{ \mathcal{S}_{n} \}\) be a partial sum of Eq. (26), then we need to show that \(\{ \mathcal{S}_{n} \}\) is a Cauchy sequence in \(( B [ 0, T ], \Vert \cdot \Vert )\). Then

$$\begin{aligned} \bigl\Vert \mathcal{S}_{n+1} ( x,y,t ) - \mathcal{S}_{n} ( x,y,t ) \bigr\Vert & = \bigl\Vert u_{n+1} ( x,y,t ) \bigr\Vert \\ &\leq\lambda \bigl\Vert u_{n} ( x,y,t ) \bigr\Vert \\ &\leq\lambda^{2} \bigl\Vert u_{n-1} ( x,y,t ) \bigr\Vert \leq\cdots\leq\lambda^{n+1} \bigl\Vert u_{0} ( x,y,t ) \bigr\Vert .\end{aligned} $$

For every \(n, m\in N\) (\(m\leq n\)), we have

$$\begin{aligned}[b] \Vert \mathcal{S}_{n} - \mathcal{S}_{m} \Vert & = \bigl\Vert ( \mathcal{S}_{m+1} - \mathcal{S}_{m} ) + \cdots+ ( \mathcal{S}_{n-1} - \mathcal{S}_{n-2} ) + ( \mathcal{S}_{n} - \mathcal{S}_{n-1} ) \bigr\Vert \\ &\leq \Vert \mathcal{S}_{m+1} - \mathcal{S}_{m} \Vert +\cdots+ \Vert \mathcal{S}_{n-1} - \mathcal{S}_{n-2} \Vert + \Vert \mathcal{S}_{n} - \mathcal{S}_{n-1} \Vert \\ &\leq \bigl( \lambda^{n} + \lambda^{n-1} +\cdots+ \lambda^{m+1} \bigr) \Vert u_{0} \Vert \\ &\leq\lambda^{m+1} \bigl( 1+\lambda+\cdots++ \lambda^{n-m-2} + \lambda^{n-m-1} \bigr) \Vert u_{0} \Vert \\ &\leq \biggl( \frac{1- \lambda^{n-m}}{1-\lambda} \biggr) \lambda^{m+1} \Vert u_{0} \Vert .\end{aligned} $$

(52)

But \(0<\lambda< 1\), therefore \(\lim_{n, m\rightarrow\infty} \Vert \mathcal{S}_{n} - \mathcal{S}_{m} \Vert =0\). Therefore, \(\{ \mathcal{S}_{n} \}\) is the Cauchy sequence. Hence, it shows the above-mentioned result. □

### Theorem 6

*The series solution for Eq*. (10) *is defined in* (26), *then the maximum absolute error is*

$$\Biggl\Vert u ( x,y,t ) - \sum_{n=0}^{M} u_{n} (x,y,t ) \Biggr\Vert \leq\frac{\lambda_{1}^{M+1}}{1- \lambda_{1}} \bigl\Vert u_{0} ( x,y,t ) \bigr\Vert . $$

### Proof

With the assistance of Eq. (52) we have

$$\bigl\Vert u ( x,y,t ) - \mathcal{S}_{n} \bigr\Vert = \lambda^{m+1} \biggl( \frac{1- \lambda^{n-m}}{1-\lambda} \biggr) \bigl\Vert u_{0} ( x,y,t ) \bigr\Vert . $$

But \(0<\lambda<0\Rightarrow1- \lambda^{n-m} <1\). Hence, we have

$$\Biggl\Vert u ( x,y,t ) - \sum_{n=0}^{M} u_{n} (x,t ) \Biggr\Vert \leq\frac{\lambda^{M+1}}{1-\lambda} \bigl\Vert u_{0} ( x,y,t ) \bigr\Vert . $$

This proves the required result. □