## 2 An Auxiliary Lemma
### Existence Results  
The first existence result for the problem (1) and (2) is based on the Leray-Schauder alternative (Lemma 3). **Theorem 1**. : _Assume that the condition \((A_{1})\) holds and that_  
\[\mathcal{O}_{1}\sigma_{1}+\mathcal{O}_{2}\theta_{1}<1\;\;\text{and}\;\; \mathcal{O}_{1}\sigma_{2}+\mathcal{O}_{2}\theta_{2}<1, \tag{19}\]  
_where \(\mathcal{O}_{i}\)\((i=1,2)\) are given by (17). Then there exists at least one solution for the problem (1) and (2) on \([a,b]\)._  
**Proof. ** In the first step, we show that the operator \(\mathcal{T}:\mathcal{P}\times\mathcal{P}\rightarrow\mathcal{P}\times\mathcal{P}\) defined by (14) is completely continuous. From the continuity of the functions \(f\)_and_\(g\), it follows that \(\mathcal{T}_{1}\) and \(\mathcal{T}_{2}\) are continuous and hence the operator \(\mathcal{T}\) is continuous. Let \(\Lambda\subset\mathcal{P}\times\mathcal{P}\) be bounded. Then, there exist positive constants \(Q_{f}\) and \(Q_{g}\) such that \(|f(t,u(t),v(t))|\leq Q_{f}\), and \(|g(t,u(t),v(t))|\leq Q_{g}\), \(\forall(u,v)\in\Lambda\). Then, for any \((u,v)\in\Lambda\), we obtain  
\[|\mathcal{T}_{1}(u,v)(t)| =\Bigg{|}\int_{a}^{t}\Bigg{(}\frac{1}{p(s)}\int_{a}^{s}f(\tau,u( \tau),v(\tau))d\tau\Bigg{)}ds\] \[\quad+\frac{1}{B}\Bigg{[}-\int_{a}^{b}\Bigg{(}\frac{1}{p(s)}\int_ {a}^{s}f(\tau,u(\tau),v(\tau))d\tau\Bigg{)}ds\] \[\quad+\sum_{j=1}^{m}\alpha_{j}\int_{a}^{\eta_{j}}\frac{1}{q(s)} \Bigg{(}\int_{a}^{s}g(\tau,u(\tau),v(\tau))d\tau\Bigg{)}ds\] \[\quad-\int_{a}^{b}\Bigg{(}\frac{\sum_{j=1}^{m}\alpha_{j}}{q(s)} \int_{a}^{s}g(\tau,u(\tau),v(\tau))d\tau\Bigg{)}ds\] \[\quad+\Bigg{(}\sum_{j=1}^{m}\alpha_{j}\Bigg{)}\Bigg{(}\sum_{k=1} ^{n}\beta_{k}\int_{a}^{\xi_{k}}\frac{1}{p(s)}\Bigg{(}\int_{a}^{s}f(\tau,u( \tau),v(\tau))d\tau\Bigg{)}ds\Bigg{)}\Bigg{]}\Bigg{|}\] \[\quad+Q_{g}\Bigg{\{}\frac{\sum_{j=1}^{m}\alpha_{j}}{|B\bar{q}|} \Bigg{[}\frac{(\eta_{j}-a)^{2}}{2}+\frac{(b-a)^{2}}{2}\Bigg{]}\Bigg{\}},\]which, on taking the norm for \(t\in[a,b]\), yields \(\|\mathcal{T}_{1}(u,v)\|\leq\mathcal{Q}_{f}L_{1}+\mathcal{Q}_{g}L_{2}\). Similarly, we have \(\|\mathcal{T}_{2}(u,v)\|\leq Q_{f}L_{3}+\mathcal{Q}_{g}L_{4}\), where \(L_{i}\) (\(i=1,\ldots,4\)) are given by (18). In consequence, we get  
\[\|\mathcal{T}(u,v)\|\leq\mathcal{Q}_{f}\mathcal{O}_{1}+\mathcal{Q}_{g} \mathcal{O}_{2},\]  
where \(\mathcal{O}_{i}\) (\(i=1,2\)) are given by (17). Hence, we deduce that the operator \(\mathcal{T}\) is uniformly bounded. Next, we prove that \(\mathcal{T}\) is an equicontinuous operator. For \(t_{1},t_{2}\in[a,b]\) with \(t_{1}<t_{2}\), we have  
\[|\mathcal{T}_{1}(u,v)(t_{2})-\mathcal{T}_{1}(u,v)(t_{1})|\] \[=\Bigg{|}\int_{a}^{t_{2}}\left(\frac{1}{p(s)}\int_{a}^{s}f(\tau,u (\tau),v(\tau))d\tau\right)ds-\int_{a}^{t_{1}}\left(\frac{1}{p(s)}\int_{a}^{s }f(\tau,u(\tau),v(\tau))d\tau\right)ds\Bigg{|}\] \[|\mathcal{T}_{1}(u,v)(t)| =\Bigg{|}\int_{a}^{t}\Bigg{(}\frac{1}{p(s)}\int_{a}^{s}f(\tau,u( \tau),v(\tau))d\tau\Bigg{)}ds\]0 \[|\mathcal{T}_{1}(u,v)(t)| =\Bigg{|}\int_{a}^{t}\Bigg{(}\frac{1}{p(s)}\int_{a}^{s}f(\tau,u( \tau),v(\tau))d\tau\Bigg{)}ds\]1 \[|\mathcal{T}_{1}(u,v)(t)| =\Bigg{|}\int_{a}^{t}\Bigg{(}\frac{1}{p(s)}\int_{a}^{s}f(\tau,u( \tau),v(\tau))d\tau\Bigg{)}ds\]2  
In a similar manner, one can find that  
\[|\mathcal{T}_{1}(u,v)(t)| =\Bigg{|}\int_{a}^{t}\Bigg{(}\frac{1}{p(s)}\int_{a}^{s}f(\tau,u( \tau),v(\tau))d\tau\Bigg{)}ds\]3  
Thus, the operator \(\mathcal{T}\) is equicontinuous. Finally, we verify that the set \(\Theta=\{(u,v)\in\mathcal{P}\times\mathcal{P}|(u,v)=e\mathcal{T}(u,v),\ 0<e<1\}\) is bounded. Let \((u,v)\in\Theta\).