Let us introduce the space \(X=\{u(t)|u(t)\in C([0,T], {\mathbb {R}})\}\) endowed with the norm \(\|u\|=\sup\{|u(t)|, t\in[0,T]\}\). Obviously, \((X,\|\cdot\|)\) is a Banach space. Then the product space \((X\times X, \|(u,v)\|)\) is also a Banach space equipped with the norm \(\|(u,v)\|=\| u\|+\|v\|\).
In view of Lemma 2.5, we define the operator \(T: X\times X\to X\times X\) by
$$T(u,v) (t)=\left ( \textstyle\begin{array}{@{}c@{}} T_{1}(u,v)(t) \\ T_{2}(u,v)(t) \end{array}\displaystyle \right ), $$
where
$$\begin{aligned} \begin{aligned} &T_{1}(u,v) (t) \\ &\quad = \frac{\mu_{2}}{1-\lambda_{2}\mu_{2}} \biggl(\frac{\lambda_{1}T(\mu _{1}\lambda_{2}+1)}{1-\lambda_{1}\mu_{1}}+\lambda_{2} t \biggr)B_{2f} +\frac{\lambda_{2}}{1-\lambda_{2}\mu_{2}} \biggl(\frac{T(\mu_{1}+\mu _{2})\lambda_{1}}{1-\lambda_{1}\mu_{1}}+t \biggr)A_{2g} \\ &\qquad {} +\frac{\lambda_{1}}{1-\lambda_{1}\mu_{1}}(A_{1g}+ \mu_{1} B_{1f})+ \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f\bigl(s,x(s), y(s) \bigr)\,ds, \end{aligned} \end{aligned}$$
and
$$\begin{aligned}& T_{2}(u,v) (t) \\& \quad = \frac{\mu_{2}}{1-\lambda_{2}\mu_{2}} \biggl(\frac{T\mu_{1}(\lambda _{1}+\lambda_{2})}{1-\lambda_{1}\mu_{1}}+t \biggr)B_{2f}+ \frac{\lambda_{2}}{1-\lambda_{2}\mu_{2}} \biggl(\frac{T\mu_{1}(\lambda _{1}\mu_{2}+1)}{1-\lambda_{1}\mu_{1}}+\mu_{2}t \biggr)A_{2g} \\& \qquad {} +\frac{\mu_{1}}{1-\lambda_{1}\mu_{1}}(\lambda_{1}A_{1g}+B_{1f})+ \int _{0}^{t}\frac{(t-s)^{\beta-1}}{\Gamma(\beta)}g\bigl(s,x(s), y(s) \bigr)\,ds, \\& A_{1g} = \int_{0}^{T}\frac{(T-s)^{\beta-1}}{\Gamma(\beta)}g\bigl(s, x(s), y(s) \bigr)\,ds, \qquad B_{1f}= \int_{0}^{T}\frac{(T-s)^{\alpha-1}}{\Gamma(\alpha )}f\bigl(s, x(s), y(s) \bigr)\,ds, \\& A_{2g} = \int_{0}^{T}\frac{(T-s)^{\beta-2}}{\Gamma(\beta-1)}g\bigl(s, x(s), y(s) \bigr)\,ds,\qquad B_{2f}= \int_{0}^{T}\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}f\bigl(s, x(s), y(s) \bigr)\,ds. \end{aligned}$$
For convenience, we put
$$\begin{aligned}& M_{1} = \frac{|\mu_{2}|}{|1-\lambda_{2}\mu_{2}|} \biggl(\frac{|\lambda _{1}|(|\mu_{1}||\lambda_{2}|+1)}{|1-\lambda_{1}\mu_{1}|}+| \lambda_{2}| \biggr)\frac{T^{\alpha}}{\Gamma(\alpha)} \\& \hphantom{M_{1} ={}}{}+ \biggl(\frac{|\lambda_{1}||\mu_{1}|}{|1-\lambda_{1}\mu_{1}|}+1 \biggr)\frac{T^{\alpha}}{\Gamma(\alpha+1)}, \end{aligned}$$
(3.1)
$$\begin{aligned}& M_{2} = \frac{|\lambda_{2}|}{|1-\lambda_{2}\mu_{2}|} \biggl(\frac{(|\mu _{1}|+|\mu_{2}|)|\lambda_{1}|}{|1-\lambda_{1}\mu_{1}|}+1 \biggr) \frac {T^{\beta}}{\Gamma(\beta)}+\frac{|\lambda_{1}|}{|1-\lambda_{1}\mu _{1}|}\frac{T^{\beta}}{\Gamma(\beta+1)}, \end{aligned}$$
(3.2)
$$\begin{aligned}& M_{3} = \frac{|\mu_{2}|}{|1-\lambda_{2}\mu_{2}|} \biggl(\frac{|\mu _{1}|(|\lambda_{1}|+|\lambda_{2}|)}{|1-\lambda_{1}\mu_{1}|}+1 \biggr) \frac {T^{\alpha}}{\Gamma(\alpha)}+\frac{|\mu_{1}|}{|1-\lambda_{1}\mu _{1}|}\frac{T^{\alpha}}{\Gamma(\alpha+1)}, \end{aligned}$$
(3.3)
$$\begin{aligned}& M_{4} = \frac{|\lambda_{2}|}{|1-\lambda_{2}\mu_{2}|} \biggl(\frac{|\mu _{1}|(|\lambda_{1}||\mu_{2}|+1)}{|1-\lambda_{1}\mu_{1}|}+| \mu_{2}| \biggr)\frac {T^{\beta}}{\Gamma(\beta)} \\& \hphantom{M_{4} ={}}{}+ \biggl(\frac{|\lambda_{1}||\mu_{1}|}{|1-\lambda_{1}\mu_{1}|} +1 \biggr)\frac{T^{\beta}}{\Gamma(\beta+1)}. \end{aligned}$$
(3.4)
In the first result, we prove the existence and uniqueness of solutions of boundary value problem (1.1)-(1.2) via Banach’s contraction principle.
Theorem 3.1
Assume that:
- (H1):
-
\(f,g:[0,T]\times\mathbb{R}\times\mathbb {R}\rightarrow\mathbb{R}\)
are continuous functions and there exist positive constants
\(\ell_{1}\)
and
\(\ell_{2}\)
such that for all
\(t\in[0,T]\)
and
\(x_{i},y_{i}\in\mathbb{R}\), \(i=1,2\), we have
$$\begin{aligned}& \bigl\vert f(t,x_{1},x_{2})-f(t,y_{1},y_{2}) \bigr\vert \leq\ell_{1} \bigl( \vert x_{1}-y_{1} \vert + \vert x_{2}-y_{2} \vert \bigr), \\& \bigl\vert g(t,x_{1},x_{2})-g(t,y_{1},y_{2}) \bigr\vert \leq\ell_{2} \bigl( \vert x_{1}-y_{1} \vert + \vert x_{2}-y_{2} \vert \bigr). \end{aligned}$$
If
$$(M_{1}+M_{3})\ell_{1}+(M_{2}+M_{4}) \ell_{2}< 1, $$
where
\(M_{i}\), \(i=1,2,3,4\), are given by (3.1)-(3.4), then system (1.1)-(1.2) has a unique solution.
Proof
Define \(\sup_{t\in[0,T]}f(t,0,0)=N_{1}<\infty\) and \(\sup_{t\in[0,T]}g(t,0,0)=N_{2}<\infty\) and \(r>0\) such that
$$r>\frac{(M_{1}+M_{3})N_{1}+(M_{2}+M_{4})N_{2}}{1-(M_{1}+M_{3})\ell_{1}-(M_{2}+M_{4})\ell_{2}}. $$
We show that \(TB_{r}\subset B_{r}\), where \(B_{r}=\{(u,v)\in X\times X: \| (u,v)\|\le r\}\).
By assumption (H1), for \((u,v)\in B_{r}\), \(t\in[0,T]\), we have
$$\begin{aligned} \bigl\vert f\bigl(t,u(t),v(t)\bigr) \bigr\vert \leq& \bigl\vert f \bigl(t,u(t),v(t)\bigr)-f(t,0,0) \bigr\vert + \bigl\vert f(t,0,0) \bigr\vert \\ \leq& \ell_{1} \bigl( \bigl\vert u(t) \bigr\vert + \bigl\vert v(t) \bigr\vert \bigr)+N_{1} \\ \leq&\ell_{1} \bigl( \Vert u \Vert + \Vert v \Vert \bigr)+N_{1} \leq \ell_{1} r + N_{1}, \end{aligned}$$
and
$$ \bigl\vert g\bigl(t,u(t),v(t)\bigr) \bigr\vert \leq\ell_{2} \bigl( \Vert u \Vert + \Vert v \Vert \bigr)+N_{2} \leq \ell_{2} r +N_{2}, $$
which lead to
$$\begin{aligned} \bigl\vert T_{1}(u, v) (t) \bigr\vert \le& \frac{|\mu_{2}|}{|1-\lambda_{2}\mu_{2}|} \biggl(\frac{|\lambda _{1}|(|\mu_{1}||\lambda_{2}|+1)}{|1-\lambda_{1}\mu_{1}|}+|\lambda_{2}| \biggr)\frac{T^{\alpha}}{\Gamma(\alpha)}( \ell_{1} r + N_{1}) \\ &{}+\frac{|\lambda_{2}|}{|1-\lambda_{2}\mu_{2}|} \biggl(\frac{(|\mu _{1}|+|\mu_{2}|)|\lambda_{1}|}{|1-\lambda_{1}\mu_{1}|}+1 \biggr)\frac {T^{\beta}}{\Gamma(\beta)}( \ell_{2} r +N_{2}) \\ &{}+\frac{|\lambda_{1}|}{|1-\lambda_{1}\mu_{1}|} \biggl(\frac{T^{\beta }}{\Gamma(\beta+1)}( \ell_{2} r +N_{2})+ |\mu_{1}| \frac{T^{\alpha}}{\Gamma(\alpha+1)}(\ell_{1} r + N_{1}) \biggr) \\ &{}+\frac{T^{\alpha}}{\Gamma(\alpha+1)}(\ell_{1} r + N_{1}) \\ =& (M_{1}\ell_{1}+M_{2}\ell_{2})r+M_{1}N_{1}+M_{2}N_{2}. \end{aligned}$$
Hence
$$\bigl\Vert T_{1}(u,v) \bigr\Vert \le(M_{1} \ell_{1}+M_{2}\ell_{2})r+M_{1}N_{1}+M_{2}N_{2}. $$
In the same way, we can obtain that
$$\bigl\Vert T_{2}(u,v) \bigr\Vert \le(M_{3} \ell_{1}+M_{4}\ell_{2})r+M_{3}N_{1}+M_{4}N_{2}. $$
Consequently,
$$\bigl\Vert T(u,v) \bigr\Vert \le\bigl[(M_{1}+M_{3}) \ell_{1}+(M_{2}+M_{4})\ell _{2} \bigr]r+(M_{1}+M_{3})N_{1}+(M_{2}+M_{4})N_{2} \le r. $$
Now, for \((u_{2},v_{2}), (u_{1},v_{1})\in X\times X\) and for any \(t\in[0,T]\), we get
$$\begin{aligned}& \bigl\vert T_{1}(u_{2},v_{2}) (t)-T_{1}(u_{1},v_{1}) (t) \bigr\vert \\& \quad \le \frac{|\mu_{2}|}{|1-\lambda_{2}\mu_{2}|} \biggl(\frac{|\lambda _{1}|(|\mu_{1}||\lambda_{2}|+1)}{|1-\lambda_{1}\mu_{1}|}+|\lambda_{2}| \biggr)\frac{T^{\alpha}}{\Gamma(\alpha)}\ell_{1}\bigl( \Vert u_{2}-u_{1} \Vert + \Vert v_{2}-v_{1} \Vert \bigr) \\& \qquad {} +\frac{|\lambda_{2}|}{|1-\lambda_{2}\mu_{2}|} \biggl(\frac{(|\mu _{1}|+|\mu_{2}|)|\lambda_{1}|}{|1-\lambda_{1}\mu_{1}|}+1 \biggr) \frac {T^{\beta}}{\Gamma(\beta)}\ell_{2}\bigl( \Vert u_{2}-u_{1} \Vert + \Vert v_{2}-v_{1} \Vert \bigr) \\& \qquad {} +\frac{|\lambda_{1}|}{|1-\lambda_{1}\mu_{1}|}\frac{T^{\beta}}{\Gamma (\beta+1)}\ell_{2}\bigl( \Vert u_{2}-u_{1} \Vert + \Vert v_{2}-v_{1} \Vert \bigr) \\& \qquad {} + \frac{|\lambda_{1}||\mu_{1}|}{|1-\lambda_{1}\mu_{1}|} \frac{T^{\alpha}}{\Gamma(\alpha+1)}\ell_{1}\bigl( \Vert u_{2}-u_{1} \Vert + \Vert v_{2}-v_{1} \Vert \bigr) \\& \qquad {} +\frac{T^{\alpha}}{\Gamma(\alpha+1)}\ell_{1}\bigl( \Vert u_{2}-u_{1} \Vert + \Vert v_{2}-v_{1} \Vert \bigr) \\& \quad = ( M_{1} \ell_{1}+ M_{2} \ell_{2}) \bigl( \Vert u_{2}-u_{1} \Vert + \Vert v_{2}-v_{1} \Vert \bigr), \end{aligned}$$
and consequently we obtain
$$ \bigl\Vert T_{1}(u_{2},v_{2})-T_{1}(u_{1},v_{1}) \bigr\Vert \le( M_{1} \ell _{1}+ M_{2} \ell_{2}) \bigl( \Vert u_{2}-u_{1} \Vert + \Vert v_{2}-v_{1} \Vert \bigr). $$
(3.5)
Similarly,
$$ \bigl\Vert T_{2}(u_{2},v_{2}) (t)-T_{2}(u_{1},v_{1}) \bigr\Vert \le( M_{3} \ell_{1}+ M_{4} \ell_{2}) \bigl( \Vert u_{2}-u_{1} \Vert + \Vert v_{2}-v_{1} \Vert \bigr). $$
(3.6)
It follows from (3.5) and (3.6) that
$$\bigl\Vert T(u_{2},v_{2})-T(u_{1},v_{1}) \bigr\Vert \le\bigl[(M_{1}+M_{3})\ell_{1}+(M_{2}+M_{4}) \ell_{2}\bigr]\bigl( \Vert u_{2}-u_{1} \Vert + \Vert v_{2}-v_{1} \Vert \bigr). $$
Since \((M_{1}+M_{3})\ell_{1}+(M_{2}+M_{4})\ell_{2}<1\), therefore, T is a contraction operator. So, by Banach’s fixed point theorem, the operator T has a unique fixed point, which is the unique solution of problem (1.1)-(1.2). This completes the proof. □
The second result is based on the Leray-Schauder alternative.
Lemma 3.2
Leray-Schauder alternative [28], p.4
Let
\(F: E\to E\)
be a completely continuous operator (i.e., a map restricted to any bounded set in
E
is compact). Let
$${\mathcal {E}}(F)=\bigl\{ x\in E: x=\lambda F(x) \textit{ for some } 0< \lambda< 1 \bigr\} . $$
Then either the set
\({\mathcal {E}}(F)\)
is unbounded or
F
has at least one fixed point.
Theorem 3.3
Assume that:
- (H3):
-
\(f,g:[0,T]\times\mathbb{R}\times\mathbb {R}\rightarrow\mathbb{R}\)
are continuous functions and there exist real constants
\(k_{i},\gamma_{i}\geq0\) (\(i=0,1,2\)) and
\(k_{0}>0\), \(\gamma_{0}>0 \)
such that
\(\forall x_{i}\in\mathbb{R}\) (\(i=1,2\)), we have
$$\begin{aligned}& \bigl\vert f(t,x_{1},x_{2}) \bigr\vert \leq k_{0} + k_{1} \vert x_{1} \vert +k_{2} \vert x_{2} \vert , \\& \bigl\vert g(t,x_{1},x_{2}) \bigr\vert \leq \gamma_{0}+\gamma_{1} \vert x_{1} \vert + \gamma_{2} \vert x_{2} \vert . \end{aligned}$$
If
$$(M_{1}+M_{3})k_{1}+(M_{2}+M_{4}) \gamma_{1}< 1 \quad \textit {and}\quad (M_{1}+M_{3})k_{2}+(M_{2}+M_{4}) \gamma_{2}< 1, $$
where
\(M_{i}\), \(i=1,2,3,4\), are given by (3.1)-(3.4), then system (1.1)-(1.2) has at least one solution.
Proof
First we show that the operator \(T:X\times X\to X\times X\) is completely continuous. By the continuity of functions f and g, the operator T is continuous.
Let \(\Omega\subset X\times X\) be bounded. Then there exist positive constants \(L_{1}\) and \(L_{2}\) such that
$$\bigl\vert f\bigl(t, u(t), v(t)\bigr) \bigr\vert \le L_{1}, \qquad \bigl\vert h\bigl(t, u(t), v(t)\bigr) \bigr\vert \le L_{2}, \quad \forall (u,v)\in\Omega. $$
Then, for any \((u,v)\in\Omega\), we have
$$\begin{aligned} \bigl\vert T_{1}(u, v) (t) \bigr\vert \le& \frac{|\mu_{2}|}{|1-\lambda_{2}\mu_{2}|} \biggl(\frac{|\lambda_{1}|(|\mu _{1}||\lambda_{2}|+1)}{|1-\lambda_{1}\mu_{1}|}+|\lambda_{2}| \biggr)\frac {T^{\alpha}}{\Gamma(\alpha)}L_{1} \\ &{}+\frac{|\lambda_{2}|}{|1-\lambda_{2}\mu_{2}|} \biggl(\frac{(|\mu _{1}|+|\mu_{2}|)|\lambda_{1}|}{|1-\lambda_{1}\mu_{1}|}+1 \biggr)\frac {T^{\beta}}{\Gamma(\beta)}L_{2} \\ &{}+\frac{|\lambda_{1}|}{|1-\lambda_{1}\mu_{1}|} \biggl(\frac{T^{\beta }}{\Gamma(\beta+1)}L_{2}+ | \mu_{1}| \frac{T^{\alpha}}{\Gamma(\alpha+1)}L_{1} \biggr)+ \frac{T^{\alpha }}{\Gamma(\alpha+1)}L_{1} \\ =& M_{1}L_{1}+M_{2}L_{2}, \end{aligned}$$
which implies that
$$\bigl\Vert T_{1}(u,v) \bigr\Vert \le M_{1}L_{1}+M_{2}L_{2}. $$
Similarly, we get
$$\bigl\Vert T_{2}(u,v) \bigr\Vert \le M_{3}L_{1}+M_{4}L_{2}. $$
Thus, it follows from the above inequalities that the operator T is uniformly bounded, since \(\|T(u,v)\|\le (M_{1}+M_{3})L_{1}+(M_{2}+M_{4})L_{2}\).
Next, we show that T is equicontinuous. Let \(t_{1}, t_{2} \in[0,T]\) with \(t_{1}< t_{2}\). Then we have
$$\begin{aligned}& \bigl\vert T_{1}\bigl(u(t_{2}),v(t_{2}) \bigr)-T_{1}\bigl(u(t_{1}),v(t_{1})\bigr) \bigr\vert \\& \quad \le \frac{L_{1}|\lambda_{2}||\mu_{2}|}{|1-\lambda_{2}\mu _{2}|}(t_{2}-t_{1})+ \frac{L_{2}|\lambda_{2}|}{|1-\lambda_{2}\mu_{2}|}(t_{2}-t_{1}) \\& \qquad {}+L_{1} \biggl\vert \frac{1}{\Gamma(\alpha)} \int_{0}^{t_{2}} (t_{2}-s)^{\alpha-1} \,ds-\frac{1}{\Gamma(\alpha)} \int_{0}^{t_{1}} (t_{1}-s)^{\alpha-1} \,ds \biggr\vert \\& \quad \le \frac{L_{1}|\lambda_{2}||\mu_{2}|}{|1-\lambda_{2}\mu _{2}|}(t_{2}-t_{1})+ \frac{L_{2}|\lambda_{2}|}{|1-\lambda_{2}\mu_{2}|}(t_{2}-t_{1}) \\& \qquad {}+L_{1} \biggl\{ \frac{1}{\Gamma(\alpha)} \int_{0}^{t_{1}} \bigl[(t_{2}-s)^{\alpha-1}-(t_{1}-s)^{\alpha-1} \bigr] \,ds+ \frac{1}{\Gamma(q)} \int _{t_{1}}^{t_{2}}(t_{2}-s)^{q-1} \,ds \biggr\} \\& \quad \le \frac{L_{1}|\lambda_{2}||\mu_{2}|}{|1-\lambda_{2}\mu _{2}|}(t_{2}-t_{1})+ \frac{L_{2}|\lambda_{2}|}{|1-\lambda_{2}\mu _{2}|}(t_{2}-t_{1}) \\& \qquad {}+\frac{L_{1}}{\Gamma(\alpha+1)} \bigl[2(t_{2}-t_{1})^{\alpha}+\bigl|t_{2}^{\alpha}-t_{1}^{\alpha}\bigr| \bigr]. \end{aligned}$$
Analogously, we can obtain
$$\begin{aligned} \begin{aligned} &\bigl\vert T_{2}\bigl(u(t_{2}),v(t_{2}) \bigr)-T_{2}\bigl(u(t_{1}),v(t_{1})\bigr) \bigr\vert \\ &\quad \le\frac{L_{1}|\mu_{2}|}{|1-\lambda_{2}\mu_{2}|}(t_{2}-t_{1})+ \frac {L_{2}|\lambda_{2}||\mu_{2}|}{|1-\lambda_{2}\mu_{2}|}(t_{2}-t_{1})+\frac {L_{2}}{\Gamma(\beta+1)} \bigl[2(t_{2}-t_{1})^{\beta}+ \bigl\vert t_{2}^{\beta}-t_{1}^{\beta}\bigr\vert \bigr]. \end{aligned} \end{aligned}$$
Therefore, the operator \(T(u,v)\) is equicontinuous, and thus the operator \(T(u,v)\) is completely continuous.
Finally, it will be verified that the set \({\mathcal {E}}=\{(u,v)\in X\times X| (u,v)=\lambda T(u,v), 0\le\lambda\le1\}\) is bounded. Let \((u,v)\in{\mathcal {E}}\), with \((u,v)=\lambda T(u,v)\). For any \(t\in [0,T]\), we have
$$u(t)=\lambda T_{1}(u,v) (t), \qquad v(t)=\lambda T_{2}(u,v) (t). $$
Then
$$\begin{aligned} \bigl\vert u(t) \bigr\vert \le& M_{1}\bigl(k_{0}+k_{1} \vert u \vert +k_{2} \vert v \vert \bigr)+M_{2}\bigl( \gamma_{0}+\gamma_{1} \vert u \vert + \gamma_{2} \vert v \vert \bigr) \\ =&M_{1}k_{0}+M_{2}\gamma_{0}+(M_{1}k_{1}+M_{2} \gamma_{1}) \vert u \vert +(M_{1}k_{2}+M_{2} \gamma_{2}) \vert v \vert \end{aligned}$$
and
$$\begin{aligned} \bigl\vert v(t) \bigr\vert \le& M_{3}\bigl(k_{0}+k_{1} \vert u \vert +k_{2} \vert v \vert \bigr)+M_{4} \bigl(\gamma_{0}+\gamma_{1} \vert u \vert + \gamma_{2} \vert v \vert \bigr) \\ =&M_{3}k_{0}+M_{4}\gamma_{0}+(M_{3}k_{1}+M_{4} \gamma_{1}) \vert u \vert +(M_{3}k_{2}+M_{4} \gamma_{2}) \vert v \vert . \end{aligned}$$
Hence we have
$$\|u\|\le M_{1}k_{0}+M_{2}\gamma_{0}+(M_{1}k_{1}+M_{2} \gamma_{1})\|u\| +(M_{1}k_{2}+M_{2} \gamma_{2})\|v\| $$
and
$$\|v\|\le M_{3}k_{0}+M_{4}\gamma_{0}+(M_{3}k_{1}+M_{4} \gamma_{1})\|u\| +(M_{3}k_{2}+M_{4} \gamma_{2})\|v\|, $$
which imply that
$$\begin{aligned} \|u\|+\|v\| \le& (M_{1}+M_{3})k_{0}+ (M_{2}+M_{4})\gamma_{0} +\bigl[(M_{1}+M_{3})k_{1}+(M_{2}+M_{4}) \gamma_{1}\bigr] \| u\| \\ &{}+\bigl[(M_{1}+M_{3})k_{2}+(M_{2}+M_{4}) \gamma_{2}\bigr]\|v\|. \end{aligned}$$
Consequently,
$$\bigl\Vert (u,v) \bigr\Vert \le\frac{(M_{1}+M_{3})k_{0}+ (M_{2}+M_{4})\gamma_{0}}{M_{0}}, $$
where \(M_{0}=\min\{1-[(M_{1}+M_{3})k_{1}+(M_{2}+M_{4})\gamma_{1}], 1-[(M_{1}+M_{3})k_{2}+(M_{2}+M_{4})\gamma_{2}]\}\), which proves that \({\mathcal {E}}\) is bounded. Thus, by Lemma 3.2, the operator T has at least one fixed point. Hence boundary value problem (1.1)-(1.2) has at least one solution. The proof is complete. □
Example 3.4
Consider the following system of fractional boundary value problem:
$$ \textstyle\begin{cases} {}^{\mathrm{c}}D^{3/2}x(t) = \frac{1}{4(t+2)^{2}}\frac{|x(t)|}{1+|x(t)|}+1+\frac{1}{32}\sin ^{2}y(t)+\frac{1}{\sqrt{t^{2}+1}},\quad t\in[0,1], \\ {}^{\mathrm{c}}D^{3/2}y(t) = \frac{1}{32 \pi}\sin(2 \pi x(t))+\frac{|y(t)|}{16(1+|y(t)|)}+\frac{1}{2},\quad t\in[0,1], \\ x(0)= \frac{1}{2}y(1), \qquad x'(0)=\frac{1}{3}y'(1), \\ y(0)= \frac{4}{5}x(1), \qquad y'(0)=\frac{3}{4}x'(1). \end{cases} $$
(3.7)
Here \(\alpha=\beta=3/2\), \(\lambda_{1}=1/2\), \(\lambda_{2}=1/3\), \(\mu_{1}=4/5\), \(\mu_{2}=3/4\), \(f(t,u,v)= \frac{1}{4(t+2)^{2}}\frac{|u|}{1+|u|}+1+\frac {1}{32}\sin^{2}v\), and \(h(t,u,v)=\frac {1}{32\pi}\sin(2\pi u)+\frac {|v|}{16(1+|v|)}+\frac{1}{2}\). With the given data, we find that \(M_{1}\approx2.8216701\), \(M_{2}\approx1.4892147\), \(M_{3}\approx3.3860044\), \(M_{4}\approx2.5499538\). Note that \(|f(t,u_{1},u_{2})-f(t,v_{1},v_{2})| \le\frac{1}{16} |u_{1}-u_{2}|+\frac{1}{16}|v_{1}-v_{2}|\), \(|g(t,u_{1},u_{2})-g(t,v_{1},v_{2})| \le\frac{1}{16} |u_{1}-u_{2}|+\frac{1}{16}|v_{1}-v_{2}|\), and \((M_{1}+M_{3})\ell_{1}+(M_{2}+M_{4})\ell_{2} \approx0.6404276<1\). Thus all the conditions of Theorem 3.1 are satisfied, and consequently, its conclusion applies to problem (3.7).