In this section, we investigate the existence for BVP (1). For convenience we list the hypothesis.
- (\(H_{1}\)):
-
\(f:[0,T]\times \mathbf{R} \rightarrow \mathbf{R} \) is continuous.
- (\(H_{2}\)):
-
There exists a constant \(L>0 \) such that
$$ \bigl\vert f(t,x_{1})-f(t,x_{2}) \bigr\vert \leq L \vert x_{1}-x_{2} \vert $$
for all \(x_{1}, x_{2}\in \mathbf{R}\) and \(t\in [0,T]\).
- (\(H_{3}\)):
-
There exists \(u\in C([0,T],\mathbf{R}^{+})\) such that
$$ \bigl\vert f(t,x) \bigr\vert \leq u(t) $$
for each \(t\in [0,T]\) and \(x\in \mathbf{R}\).
- (\(H_{4}\)):
-
There exist a continuous function \(\phi_{1}(t)\in C([0,T], \mathbf{R}^{+})\) and a nondecreasing function \(\phi_{2}\in C([0,T], \mathbf{R}^{+})\) such that
$$ \bigl\vert f(t,x) \bigr\vert \leq \phi_{1}(t)\phi_{2} \bigl( \vert x \vert \bigr) $$
for each \(t\in [0,T]\) and \(x\in \mathbf{R}\).
Theorem 3.1
Assume that (\(H_{1}\)) and (\(H_{2}\)) hold. If
then the boundary value problem (1) has a unique solution in
\(C([0,T],\mathbf{R})\).
Proof
We define an operator \(S: C([0,T],\mathbf{R})\rightarrow C([0,T],\mathbf{R})\) by
$$\begin{aligned} Sx(t)=\frac{\gamma_{3}}{\varTheta }t^{\alpha -1}+ \int_{0}^{T} G_{1}(t,s)x(s) \,ds+ \int_{0}^{T} G_{2}(t,s)f \bigl(s,x(s) \bigr) \,ds, \end{aligned}$$
for each \(t\in (0,T)\).
By Lemma 2.3, \(x\in C([0,T],\mathbf{R})\) is a solution of problem (1) if and only if x is a fixed point of S. We now prove that S has a fixed point. Taking \(x_{1},x_{2}\in C([0,T],\mathbf{R})\) arbitrary, according to (\(H_{2}\)), we find
$$\begin{aligned}& \bigl\vert Sx_{1}(t)-Sx_{2}(t) \bigr\vert \\& \quad \leq \int_{0}^{T} \bigl\vert G_{1}(t,s) \bigr\vert \bigl\vert x_{1}(s)-x _{2}(s) \bigr\vert \,ds + \int_{0}^{T} \bigl\vert G_{2}(t,s) \bigr\vert \bigl\vert f \bigl(s,x_{1}(s) \bigr)-f \bigl(s,x_{2}(s) \bigr) \bigr\vert \,ds \\& \quad \leq \Vert x_{1}-x_{2} \Vert \int_{0}^{T} \bigl\vert G_{1}(t,s) \bigr\vert \,ds+L \Vert x_{1}-x _{2} \Vert \int_{0}^{T} \bigl\vert G_{2}(t,s) \bigr\vert \,ds \\& \quad \leq (M_{1}+LM_{2}) \Vert x_{1}-x_{2} \Vert , \end{aligned}$$
and hence
$$ \Vert Sx_{1}-Sx_{2} \Vert \leq (M_{1}+LM_{2}) \Vert x_{1}-x_{2} \Vert . $$
Since \(M_{1}+LM_{2}<1\), S is a contraction. By Banach contraction principle, S has a unique fixed point in \(C([0,T],\mathbf{R})\), i.e., the boundary value problem (1) has a unique solution. This completes the proof. □
Theorem 3.2
Assume that (\(H_{1}\)) and (\(H_{3}\)) hold. If
then the boundary value problem (1) has at least one solution in
\(C([0,T],\mathbf{R})\).
Proof
We define mappings A and B from \(C((0,T), \mathbf{R})\) into itself by
$$\begin{aligned} Ax(t)=\frac{\gamma_{3}}{\varTheta }t^{\alpha -1}+ \int_{0}^{T} G_{1}(t,s)x(s) \,ds \end{aligned}$$
and
$$\begin{aligned} Bx(t)= \int_{0}^{T} G_{2}(t,s)f \bigl(s,x(s) \bigr) \,ds. \end{aligned}$$
It is easy to verify that B is continuous on \(C([0,T],\mathbf{R})\), since \(G_{2}\) and f are continuous. Further, since \(M_{1}<1\), we can choose a constant \(r>0\) large enough such that \(M_{1}+\frac{M_{2} \Vert u \Vert }{r}+\frac{\gamma_{3} T^{\alpha -1}}{r\varTheta }<1\), i.e., \(rM_{1}+M _{2} \Vert u \Vert +\frac{\gamma_{3} T^{\alpha -1}}{\varTheta }< r \), where u belongs to \(C([0,T],{\mathbf{R}}^{+})\) such that \(\vert f(t,x) \vert \leq u(t)\), according to (\(H_{3}\)). Set \(B_{r}=\{x\in C((0,T),\mathbf{R}): \Vert x \Vert \leq r\}\). Then \(B_{r}\) is nonempty, bounded, closed and convex. Moreover, for any \(x,y\in B_{r}\) and \(t\in [0,T] \), we have
$$\begin{aligned} \bigl\vert Ax(t) \bigr\vert \leq & \biggl\vert \frac{\gamma_{3} t^{\alpha -1}}{\varTheta } \biggr\vert + \int_{0} ^{T} \bigl\vert G_{1}(t,s) \bigr\vert \bigl\vert x(s) \bigr\vert \,ds \\ \leq & rM_{1}+\frac{\gamma_{3} T^{\alpha -1}}{\varTheta } \end{aligned}$$
and
$$\begin{aligned} \bigl\vert By(t) \bigr\vert \leq & \int_{0}^{T} \bigl\vert G_{2}(t,s) \bigr\vert \bigl\vert f \bigl(s,x(s) \bigr) \bigr\vert \,ds \\ \leq & M_{2} \Vert u \Vert . \end{aligned}$$
Thus, \(\vert Ax(t)+By(t) \vert \leq rM_{1}+M_{2} \Vert u \Vert +\frac{\gamma_{3} T^{ \alpha -1}}{\varTheta }< r \), i.e., \(Ax+By\in B_{r}\).
Next we prove that A is a contraction. In fact, for any \(x, y\in C((0,T), \mathbf{R})\) and \(t\in [0, 1]\),
$$\begin{aligned} \bigl\vert Ax(t)-Ay(t) \bigr\vert \leq \int_{0}^{T} \bigl\vert G_{1}(t,s) \bigr\vert \bigl\vert x(t)-y(t) \bigr\vert \,ds \leq M _{1} \Vert x-y \Vert . \end{aligned}$$
It follows that \(\Vert Ax-Ay \Vert \leq M_{1} \Vert x-y \Vert \). Since \(M_{1}<1\), we know that A is a contraction.
Finally, we have to show that B is compact. Take any bounded subset \(U\subseteq C([0,T],\mathbf{R}) \). Then there exists a constant \(r_{0}>0\) such that \(U=\{u\in U: \Vert u \Vert \leq r_{0}\}\). We prove that BU is bounded and equicontinuous. In fact, for any \(x\in U \), we have
$$\begin{aligned} \bigl\vert Bx(t) \bigr\vert \leq \int_{0}^{T} \bigl\vert G_{2}(t,s) \bigr\vert \bigl\vert f \bigl(s,x(s) \bigr) \bigr\vert \,ds \leq M_{2} \Vert u \Vert \leq M_{2}r_{0} \end{aligned}$$
for each \(t\in (0,T)\). Hence BU is bounded. Further, for any \(0\leq t_{1}< t_{2}\leq T\) and \(x\in U \), we have
$$\begin{aligned} \bigl\vert Bx(t_{2})-Bx(t_{1}) \bigr\vert =& \biggl\vert \int_{0}^{T} G_{2}(t_{2},s)f \bigl(s,x(s) \bigr)- \int _{0}^{T} G_{2}(t_{1},s)f \bigl(s,x(s) \bigr) \,ds \biggr\vert \\ \leq & \biggl\vert \int_{0}^{T} \frac{\mu }{\varTheta \lambda \varGamma (\alpha - \gamma_{1})} \bigl(t_{1}^{\alpha -1}-t_{2}^{\alpha -1} \bigr) (T-s)^{\alpha -\gamma _{1}-1}f \bigl(s,x(s) \bigr) \,ds \biggr\vert \\ & {} + \biggl\vert \int_{0}^{\eta }\frac{1}{\varTheta \lambda \varGamma (\alpha +\gamma _{2})} \bigl(t_{1}^{\alpha -1}-t_{2}^{\alpha -1} \bigr) (\eta -s)^{\alpha +\gamma _{2}-1}f \bigl(s,x(s) \bigr) \,ds \biggr\vert \\ & {} + \biggl\vert \int_{0}^{t_{1}} \frac{1}{\lambda \varGamma (\alpha )} \bigl[(t_{2}-s)^{ \alpha -1}-(t_{1}-s)^{\alpha -1} \bigr]f \bigl(s,x(s) \bigr) \,ds \biggr\vert \\ & {} + \biggl\vert \int_{t_{1}}^{t_{2}} \frac{1}{\lambda \varGamma (\alpha )}(t_{2}-s)^{ \alpha -1}f \bigl(s,x(s) \bigr) \,ds \biggr\vert \\ \leq & \frac{\mu (t_{2}^{\alpha -1}-t_{1}^{\alpha -1})}{\varTheta \lambda \varGamma (\alpha -\gamma_{1})} \int_{0}^{T} (T-s)^{\alpha -\gamma _{1}-1} \,ds \Vert u \Vert \\ & {} +\frac{t_{2}^{\alpha -1}-t_{1}^{\alpha -1}}{\varTheta \lambda \varGamma (\alpha +\gamma_{2})} \int_{0}^{\eta }(\eta -s)^{\alpha +\gamma_{2}-1} \,ds \Vert u \Vert \\ & {} +\frac{1}{\lambda \varGamma (\alpha )} \int_{0}^{t_{1}} \bigl\vert (t_{2}-s)^{ \alpha -1}-(t_{1}-s)^{\alpha -1} \bigr\vert \,ds \Vert u \Vert \\ & {} +\frac{1}{\lambda \varGamma (\alpha )} \int_{t_{1}}^{t_{2}} (t_{2}-s)^{ \alpha -1} \,ds \Vert u \Vert \\ =& \frac{\mu T^{\alpha -\gamma_{1}}(t_{2}^{\alpha -1}-t_{1}^{\alpha -1})}{\varTheta \lambda \varGamma (\alpha -\gamma_{1}+1)}r_{0} \\ & {} +\frac{\eta^{\alpha +\gamma_{2}}(t_{2}^{\alpha -1}-t_{1}^{\alpha -1})}{ \varTheta \lambda \varGamma (\alpha +\gamma_{2}+1)}r_{0} \\ & {} +\frac{1}{\lambda \varGamma (\alpha +1)} \bigl[t_{2}^{\alpha }-t_{1}^{ \alpha }+2(t_{2}-t_{1})^{\alpha } \bigr]r_{0}. \end{aligned}$$
We can see from the above inequality that \(\vert Bx(t_{2})-Bx(t_{1}) \vert \rightarrow 0\) as \(t_{2}-t_{1}\rightarrow 0\), and the convergence is independent on x, \(t_{1}\) and \(t_{2}\). This shows that BU is equicontinuous. An employment of Arzelà–Ascoli theorem shows that B is compact. Now, we apply Krasnoselskii’s fixed point theorem on operator A and B to deduce that there exists at least one x such that \(Ax+Bx=x\), which is the solution of the boundary value problem (1). □
Theorem 3.3
Assume that (\(H_{1}\)) and (\(H_{4}\)) hold. If
$$ M_{1}+M_{2} \Vert \phi_{1} \Vert \limsup _{r\rightarrow \propto } \frac{\phi_{2}(r)}{r}< 1, $$
then the boundary value problem (1) has at least one solution in
\(C([0,T],\mathbf{R})\).
Proof
Define a mapping \(S: C([0,T],\mathbf{R})\rightarrow C([0,T], \mathbf{R})\) by
$$ Sx(t)=\frac{\gamma_{3}}{\varTheta }t^{\alpha -1}+ \int_{0}^{T} G_{1}(t,s)x(s) \,ds+ \int_{0}^{T} G_{2}(t,s)f \bigl(s,x(s) \bigr) \,ds $$
for \(t\in [0,T]\). It is easy to prove that S is continuous. In order to apply Schauder’s fixed point theorem, we only need to show that S is compact.
Firstly, we take any bounded subset \(Q\subseteq C([0,T],\mathbf{R})\). Then there exists q (\(q>0\)) satisfying that \(Q\subseteq B_{q}=\{x \in C([0,T],\mathbf{R}): \Vert x \Vert \leq q\}\). Notice that \(B_{q}\) is a closed convex and bounded subset. For any \(x\in B_{q}\), we have
$$\begin{aligned} \bigl\vert Sx(t) \bigr\vert \leq & \biggl\vert \frac{\gamma_{3}t^{\alpha -1}}{\varTheta } \biggr\vert + \int_{0}^{T} \bigl\vert G_{1}(t,s) \bigr\vert \bigl\vert x(s) \bigr\vert \,ds+ \int_{0}^{T} \bigl\vert G_{2}(t,s) \bigr\vert \bigl\vert f \bigl(s,x(s) \bigr) \bigr\vert \,ds \\ \leq & \frac{\gamma_{3}T^{\alpha -1}}{\varTheta }+ \Vert x \Vert \int_{0}^{T} \bigl\vert G _{1}(t,s) \bigr\vert \,ds+ \Vert \phi_{1} \Vert \phi_{2} \bigl( \Vert x \Vert \bigr) \int_{0}^{T} \bigl\vert G_{2}(t,s) \bigr\vert \,ds, \\ \leq & \frac{\gamma_{3}T^{\alpha -1}}{\varTheta }+TM_{1}q+TM_{2} \Vert \phi _{1} \Vert \phi_{2}(q), \end{aligned}$$
which means \(SB_{q}\), and therefore SQ, is uniformly bounded in \(C((0,T),\mathbf{R})\). For \(0\leq t_{1}< t_{2}\leq T\) and any \(x\in B_{q}\),
$$\begin{aligned}& \begin{aligned} \bigl\vert Sx(t_{1})-Sx(t_{2}) \bigr\vert &\leq \frac{\gamma_{3}}{\varTheta } \bigl(t_{2}^{\alpha -1}-t_{1}^{\alpha -1} \bigr)+ \biggl\vert \int_{0}^{T} G_{1}(t_{2},s)x(s)- \int_{0}^{T} G _{1}(t_{1},s)x(s) \,ds \biggr\vert \\ & \quad {} + \biggl\vert \int_{0}^{T} G_{2}(t_{2},s)x(s)- \int_{0}^{T} G_{2}(t_{1},s)x(s) \,ds \biggr\vert \\ &= P_{1}+P_{2}+P_{3}; \end{aligned} \\ & P_{1}=\frac{\gamma_{3}}{\varTheta } \bigl(t_{2}^{\alpha -1}-t_{1}^{\alpha -1} \bigr), \\ & \begin{aligned} P_{2} &= \biggl\vert \int_{0}^{T} G_{1}(t_{2},s)x(s)- \int_{0}^{T} G_{1}(t_{1},s)x(s) \,ds \biggr\vert \\ &\leq \biggl\vert \int_{0}^{T} \frac{\mu }{\varTheta \lambda \varGamma (\alpha - \beta -\gamma_{1})} \bigl(t_{2}^{\alpha -1}-t_{1}^{\alpha -1} \bigr) (T-s)^{\alpha -\beta -\gamma_{1}-1}x(s) \,ds \biggr\vert \\ & \quad {} + \biggl\vert \int_{0}^{\eta }\frac{1}{\varTheta \lambda \varGamma (\alpha -\beta + \gamma_{2})} \bigl(t_{2}^{\alpha -1}-t_{1}^{\alpha -1} \bigr) (\eta -s)^{\alpha - \beta +\gamma_{2}-1}x(s) \,ds \biggr\vert \\ & \quad {} + \biggl\vert \int_{0}^{t_{1}} \frac{1}{\lambda \varGamma (\alpha -\beta )} \bigl[(t _{2}-s)^{\alpha -\beta -1}-(t_{1}-s)^{\alpha -\beta -1} \bigr]x(s) \,ds \biggr\vert \\ & \quad {} + \biggl\vert \int_{t_{1}}^{t_{2}} \frac{1}{\lambda \varGamma (\alpha -\beta )}(t _{2}-s)^{\alpha -\beta -1}x(s) \,ds \biggr\vert \\ &\leq \frac{\mu (t_{2}^{\alpha -1}-t_{1}^{\alpha -1})}{\varTheta \lambda \varGamma (\alpha -\beta -\gamma_{1})} \int_{0}^{T} (T-s)^{\alpha -\beta -\gamma_{1}-1} \,ds \Vert x \Vert \\ & \quad {} +\frac{t_{2}^{\alpha -1}-t_{1}^{\alpha -1}}{\varTheta \lambda \varGamma (\alpha -\beta +\gamma_{2})} \int_{0}^{\eta }(\eta -s)^{\alpha -\beta +\gamma_{2}-1} \,ds \Vert x \Vert \\ & \quad {} +\frac{1}{\lambda \varGamma (\alpha -\beta )} \int_{0}^{t_{1}} \bigl\vert (t_{2}-s)^{ \alpha -\beta -1}-(t_{1}-s)^{\alpha -\beta -1} \bigr\vert \,ds \Vert x \Vert \\ & \quad {} +\frac{1}{\lambda \varGamma (\alpha -\beta )} \int_{t_{1}}^{t_{2}} (t _{2}-s)^{\alpha -\beta -1} \,ds \Vert x \Vert \\ &=\quad \frac{\mu (t_{2}^{\alpha -1}-t_{1}^{\alpha -1})T^{\alpha -\beta - \gamma_{1}}q}{\varTheta \lambda \varGamma (\alpha -\beta -\gamma_{1})} \\ & \quad {} +\frac{(t_{2}^{\alpha -1}-t_{1}^{\alpha -1}) \eta^{\alpha -\beta -\gamma_{1}}q}{\varTheta \lambda \varGamma (\alpha - \beta +\gamma_{2})} \\ & \quad {} +\frac{q}{\lambda \varGamma (\alpha -\beta )} \bigl[t_{2}^{\alpha -\beta }-t _{1}^{\alpha -\beta }+2(t_{2}-t_{1})^{\alpha -\beta } \bigr], \end{aligned} \\ & P_{3} = \biggl\vert \int_{0}^{T} G_{2}(t_{2},s)f \bigl(s,x(s) \bigr)- \int_{0}^{T} G_{2}(t _{1},s)f \bigl(s,x(s) \bigr) \,ds \biggr\vert \\ & \hphantom{P_{3}} \leq \biggl\vert \int_{0}^{T} \frac{\mu }{\varTheta \lambda \varGamma (\alpha - \gamma_{1})} \bigl(t_{1}^{\alpha -1}-t_{2}^{\alpha -1} \bigr) (T-s)^{\alpha -\gamma _{1}-1}f \bigl(s,x(s) \bigr) \,ds \biggr\vert \\ & \hphantom{P_{3}} \quad {} + \biggl\vert \int_{0}^{\eta }\frac{1}{\varTheta \lambda \varGamma (\alpha +\gamma _{2})} \bigl(t_{1}^{\alpha -1}-t_{2}^{\alpha -1} \bigr) (\eta -s)^{\alpha +\gamma _{2}-1}f \bigl(s,x(s) \bigr) \,ds \biggr\vert \\ & \hphantom{P_{3}} \quad {} + \biggl\vert \int_{0}^{t_{1}} \frac{1}{\lambda \varGamma (\alpha )} \bigl[(t_{2}-s)^{ \alpha -1}-(t_{1}-s)^{\alpha -1} \bigr]f \bigl(s,x(s) \bigr) \,ds \biggr\vert \\ & \hphantom{P_{3}}\quad {} + \biggl\vert \int_{t_{1}}^{t_{2}} \frac{1}{\lambda \varGamma (\alpha )}(t_{2}-s)^{ \alpha -1}f \bigl(s,x(s) \bigr) \,ds \biggr\vert \\ & \hphantom{P_{3}}\leq \frac{\mu (t_{2}^{\alpha -1}-t_{1}^{\alpha -1})}{\varTheta \lambda \varGamma (\alpha -\gamma_{1})} \biggl[ \int_{0}^{T} (T-s)^{\alpha - \gamma_{1}-1} \,ds \biggr] \Vert \phi_{1} \Vert \phi_{2} \bigl( \Vert x \Vert \bigr) \\ & \hphantom{P_{3}}\quad {} +\frac{(t_{2}^{\alpha -1}-t_{1}^{\alpha -1})}{\varTheta \lambda \varGamma (\alpha +\gamma_{2})} \biggl[ \int_{0}^{\eta }(\eta -s)^{\alpha +\gamma _{2}-1} \,ds \biggr] \Vert \phi_{1} \Vert \phi_{2} \bigl( \Vert x \Vert \bigr) \\ & \hphantom{P_{3}}\quad {} +\frac{1}{\lambda \varGamma (\alpha )} \biggl[ \int_{0}^{t_{1}} \bigl\vert (t_{2}-s)^{ \alpha -1}-(t_{1}-s)^{\alpha -1} \bigr\vert \,ds \biggr] \Vert \phi_{1} \Vert \phi_{2} \bigl( \Vert x \Vert \bigr) \\ & \hphantom{P_{3}}\quad {} +\frac{1}{\lambda \varGamma (\alpha )} \int_{t_{1}}^{t_{2}} (t_{2}-s)^{ \alpha -1} \,ds \Vert \phi_{1} \Vert \phi_{2} \bigl( \Vert x \Vert \bigr) \\ & \hphantom{P_{3}}= \frac{\mu T^{\alpha -\gamma_{1}}(t_{2}^{\alpha -1}-t_{1}^{\alpha -1})}{\varTheta \lambda \varGamma (\alpha -\gamma_{1}+1)} \Vert \phi_{1} \Vert \phi _{2}(q) \\ & \hphantom{P_{3}}\quad {} +\frac{\eta^{\alpha +\gamma_{2}}(t_{2}^{\alpha -1}-t_{1}^{\alpha -1})}{ \varTheta \lambda \varGamma (\alpha +\gamma_{2}+1)} \Vert \phi_{1} \Vert \phi_{2}(q) \\ & \hphantom{P_{3}}\quad {} +\frac{1}{\lambda \varGamma (\alpha +1)} \bigl[t_{2}^{\alpha }-t_{1}^{ \alpha }+2(t_{2}-t_{1})^{\alpha } \bigr] \Vert \phi_{1} \Vert \phi_{2}(q). \end{aligned}$$
It is trivial that \(P_{1}\), \(P_{2} \) and \(P_{3}\) all tend to 0 as \(t_{2}-t_{1}\rightarrow 0\). Hence, \(\vert Sx(t_{2})-Sx(t_{1}) \vert \rightarrow 0\) (\(t_{2}-t_{1}\rightarrow 0\)). Notice that the convergence is independent on x, \(t_{1}\) and \(t_{2}\). It follows that \(SB_{q}\) is equicontinuous. An application of the Arzelà–Ascoli Theorem yields that \(SB_{q}\) is compact. Therefore, we have shown that S maps bounded subsets into compact subset, i.e., S is a compact mapping.
Now, from the condition \(M_{1}+M_{2} \Vert \phi_{1} \Vert \lim_{r\rightarrow \propto }\sup\frac{\phi_{2}(r)}{r}<1\), we can choose a positive r large enough, such that
$$ M_{1}+M_{2} \Vert \phi_{1} \Vert \frac{\phi_{2}(r)}{r}+\frac{\gamma_{3}T^{ \alpha -1}}{r\varTheta }< 1, $$
i.e., \(M_{1}r+M_{2} \Vert \phi_{1} \Vert \phi_{2}(r)+\frac{\gamma_{3} T^{\alpha -1}}{\varTheta }< r\). Hence, we can take \(C>0\) such that \(M_{1}C+M_{2} \Vert \phi_{1} \Vert \phi_{2}(C)+\frac{\gamma_{3} T^{\alpha -1}}{\varTheta }< C\). Let \(U=\{x\in C((0,T),\mathbf{R}): \Vert x \Vert \leq C\}\). Then \(S:\overline{U} \rightarrow C((0,T),\mathbf{R})\) is compact and continuous. If there exist \(\lambda \in (0,1)\) and \(x\in \overline{U}\) such that \(x=\lambda Sx\), then, for every \(t\in (0,T)\),
$$\begin{aligned} \bigl\vert x(t) \bigr\vert = &\bigl\vert \lambda Sx(t) \bigr\vert \leq \bigl\vert Sx(t) \bigr\vert \\ \leq & \biggl\vert \frac{\gamma_{3}t^{\alpha -1}}{\varTheta } \biggr\vert + \biggl\vert \int_{0}^{T} G_{1}(t,s)x(s) \,ds \biggr\vert + \biggl\vert \int_{0}^{T} G_{2}(t,s)f \bigl(s,x(s) \bigr) \,ds \biggr\vert \\ \leq &\frac{\gamma_{3} T^{\alpha -1}}{\varTheta }+M_{1}C+M_{2} \Vert \phi _{1} \Vert \phi_{2} \bigl( \Vert x \Vert \bigr) \\ \leq & M_{1}C+M_{2} \Vert \phi_{1} \Vert \phi_{2}(C)+\frac{\gamma_{3} T^{ \alpha -1}}{\varTheta }. \end{aligned}$$
It then follows that \(C= \Vert x \Vert \leq M_{1}C+M_{2} \Vert \phi_{1} \Vert \phi_{2}(C)+\frac{ \gamma_{3} T^{\alpha -1}}{\varTheta }< C\), which contradict to the fact that \(x\in \overline{U}\). Thus, \(x\neq \lambda Sx\) for each \(\lambda \in (0,1)\) and \(x\in \overline{U}\). By the Leray Schauder alternative, S has at least one fixed point, which is the solution to the boundary value problem (1). This completes the proof. □