# Efficient Solution Criteria for a Coupled Fractional Laplacian System on Some Infinite Domains
## 3 Main Results
### Existences and Uniqueness Result
\]  
which implies that  
\[\|\mathbb{Y}_{2}\mathcal{U}-\mathbb{Y}_{2}\mathcal{U}\|_{\mathbb{Y}_{2}}\leq \mathcal{B}_{1}\big{\|}\mathcal{U}-\bar{\mathcal{U}}\big{\|}_{\mathbb{Y}_{2}}. \tag{20}\]  
Hence, using inequalities (19) and (20), we find  
\[ \begin{split}\|\mathbb{Y}(\mathcal{U},\mathcal{V})-\mathbb{Y}( \bar{\mathcal{U}},\bar{\mathcal{V}})\|_{\mathbb{M}}&=\|(\mathbb{Y }_{1}\mathcal{V},\mathbb{Y}_{2}\mathcal{U})-(\mathbb{Y}_{1}\bar{\mathcal{V}},\mathbb{Y}_{2}\bar{\mathcal{U}})\|_{\mathbb{M}}\\ &=\left\|(\mathbb{Y}_{1}\mathcal{V}-\mathbb{Y}_{1}\bar{\mathcal{V }}),\mathbb{Y}_{2}\mathcal{U}-\mathbb{Y}_{2}\bar{\mathcal{U}})\right\|_{ \mathbb{M}}\\ &=\max\{\|\mathbb{Y}_{1}\mathcal{V}-\mathbb{Y}_{1}\bar{\mathcal{ V}}\|_{\mathbb{Y}_{1}}\|_{\mathbb{Y}_{2}}\mathcal{U}-\mathbb{Y}_{2}\bar{ \mathcal{U}}\|_{\mathbb{Y}_{2}}\}\\ &\leq\max\{\mathcal{A}_{1}\big{\|}\mathcal{V}-\bar{\mathcal{V}} \big{\|}_{\mathbb{Y}_{1}},\mathcal{B}_{1}\big{\|}\mathcal{U}-\bar{\mathcal{U}} \big{\|}_{\mathbb{Y}_{2}}\}\\ &=\max\{\mathcal{B}_{1}\big{\|}\mathcal{U}-\bar{\mathcal{U}} \big{\|}_{\mathbb{Y}_{2}},\mathcal{A}_{1}\big{\|}\mathcal{V}-\bar{\mathcal{V}} \big{\|}_{\mathbb{Y}_{1}}\}\\ &\leq\mathcal{G}\|(\mathcal{U},\mathcal{V})-(\bar{\mathcal{U}}, \bar{\mathcal{V}})\|_{\mathbb{M}},\end{split}  \tag{21}\]  
where \(\mathcal{G}=\max\{\mathcal{A}_{1},\mathcal{B}_{1}\}\), and since \(\mathcal{G}\in[0,1)\), then the operator \(\mathbb{Y}\) is contractive. Hence, in view of the Banach fixed point result, we infer that \(\mathbb{Y}\) has only one fixed point \((\bar{\mathcal{U}},\bar{\mathcal{V}})\) in \(\mathbb{M}\) and acts a solution of the coupled \(\varphi\)-\(\mathbb{R}\mathbb{L}\) fractional Laplacian implicit differential system (1) on an infinite domain \(\mathcal{T}\times\mathcal{T}\). Moreover, for any \((\mathcal{U}_{0},\mathcal{V}_{0})\in\mathbb{M}\), \(\|(\mathcal{U}_{n},\mathcal{V}_{n})-(\bar{\mathcal{U}},\bar{\mathcal{V}})\|_{ \mathbb{M}}\to 0\), as \(n\to\infty\), where \((\mathcal{U}_{n},\mathcal{V}_{n})=\mathbb{Y}(\mathcal{U}_{n-1},\mathcal{V}_{n- 1}),n\in\mathbb{N}\). Furthermore, in view of (21), we obtain  
\[\|(\mathcal{U}_{n},\mathcal{V}_{n})-(\mathcal{U}_{n-1},\mathcal{V}_{n-1})\|_{ \mathbb{M}}\leq\mathcal{G}^{n-1}\|(\mathcal{U}_{1},\mathcal{V}_{1})-( \mathcal{U}_{0},\mathcal{V}_{0})\|_{\mathbb{M}},\quad n\in\mathbb{N}.\]  
Then, for any \(k\), and \(k>n\), one finds  
\[ \begin{split}\|(\mathcal{U}_{k},\mathcal{V}_{k})-(\mathcal{U}_{n },\mathcal{V}_{n})\|_{\mathbb{M}}&\leq\|(\mathcal{U}_{k}, \mathcal{V}_{k})-(\mathcal{U}_{k-1},\mathcal{V}_{k-1})\|_{\mathbb{M}}+\|( \mathcal{U}_{k-1},\mathcal{V}_{k-1})-(\mathcal{U}_{k-2},\mathcal{V}_{k-2})\|_{ \mathbb{M}}+\ldots\\ &\quad+\|(\mathcal{U}_{n+1},\mathcal{V}_{n+1})-(\mathcal{U}_{n}, \mathcal{V}_{n})\|_{\mathbb{M}}\\ &\leq\mathcal{G}^{n}\|(\mathcal{U}_{1},\mathcal{V}_{1})-(\mathcal{ U}_{0},\mathcal{V}_{0})\|_{\mathbb{M}}\sum_{r=0}^{k-n-1}\mathcal{G}^{r}.\end{split} \]  
This shows that, for \(k\to\infty\),  
\[\|(\bar{\mathcal{U}},\bar{\mathcal{V}})-(\mathcal{U}_{n},\mathcal{V}_{n})\|_{ \mathbb{M}}\leq\frac{\mathcal{G}^{n}}{1-\mathcal{G}}\|(\mathcal{U}_{1}, \mathcal{V}_{1})-(\mathcal{U}_{0},\mathcal{V}_{0})\|_{\mathbb{M}},\quad n\in \mathbb{N}.\]  
Hence, the proof is completed.