In this section, we will present some preliminaries and lemmas that will be used in the proof of our main results.
Definition 2.1
([33, 34])
The Riemann-Liouville fractional integral of order \(\alpha > 0\) is given by
$$I^{\alpha}u(t)=\frac{1}{\Gamma(\alpha)} \int^{t}_{0}(t-s)^{\alpha-1}u(s)\,ds, $$
where \(n-1< \alpha< n\), provided that the right-hand side is pointwise defined on \((0,+\infty)\).
Definition 2.2
([33, 34])
For an \((n-1)\)-times absolutely continuous function \(u:[0,\infty)\to\mathbb{R}\), the Caputo derivative of fractional order α is defined as
$${}^{\mathrm{c}}\mathcal{D}^{\alpha}u(t)=\frac{1}{\Gamma(n-\alpha)} \int ^{t}_{0}(t-s)^{n-\alpha-1}u^{(n)}(s) \,ds, \quad n-1< \alpha< n, n=[\alpha]+1, $$
where \([\alpha]\) denotes the integer part of the real number α.
Lemma 2.1
For any
\(h,g\in L(0,1)\cap C(0,1)\), the system consisting of the equations
$$ \bigl({}^{\mathrm{c}}\mathcal{D}^{p}+ \lambda_{1} {}^{\mathrm{c}}\mathcal{D}^{p-1}\bigr)u(t)=h(t), \qquad \bigl({}^{\mathrm{c}}\mathcal{D}^{q}+\lambda_{2} {}^{\mathrm{c}}\mathcal{D}^{q-1}\bigr)v(t)=g(t),\quad t\in(0,1) $$
(2.1)
and the BCs
$$ \left \{ \textstyle\begin{array}{l} u(0)=u'(0)=0, \qquad u(1)=av(\xi), \\ v(0)=v'(0)=0,\qquad v(1)=bu(\eta), \end{array}\displaystyle \right . $$
(2.2)
has a unique integral representation
$$\begin{aligned}& u(t)= A_{1}(t) \biggl\{ a\bigl(\lambda_{2} \xi-1+e^{-\lambda_{2}\xi}\bigr) \biggl[ \int_{0}^{1}e^{-\lambda _{2}(1-s)}(Qg) (s)\,ds -b \int_{0}^{\eta}e^{-\lambda_{1}(\eta-s)}(Ph) (s)\,ds \biggr] \\& \hphantom{u(t)={}}{}-\bigl(\lambda_{2}-1+e^{-\lambda_{2}}\bigr) \biggl[a \int_{0}^{\xi}e^{-\lambda_{2}(\xi-s)}(Qg) (s)\,ds - \int_{0}^{1}e^{-\lambda_{1}(1-s)}(Ph) (s)\,ds \biggr] \biggr\} \\& \hphantom{u(t)={}}{}+ \int _{0}^{t}e^{-\lambda_{1}(t-s)}(Ph) (s) \,ds, \end{aligned}$$
(2.3)
$$\begin{aligned}& v(t)= A_{2}(t) \biggl\{ \bigl(\lambda_{1}-1+e^{-\lambda_{1}} \bigr) \biggl[ \int_{0}^{1}e^{-\lambda _{2}(1-s)}(Qg) (s)\,ds -b \int_{0}^{\eta}e^{-\lambda_{1}(\eta-s)}(Ph) (s)\,ds \biggr] \\& \hphantom{v(t)={}}{}-b\bigl(\lambda_{1}\eta-1+e^{-\lambda_{1}\eta}\bigr) \biggl[a \int_{0}^{\xi}e^{-\lambda _{2}(\xi-s)}(Qg) (s)\,ds - \int_{0}^{1}e^{-\lambda_{1}(1-s)}(Ph) (s)\,ds \biggr] \biggr\} \\& \hphantom{v(t)={}}{}+ \int _{0}^{t}e^{-\lambda_{2}(t-s)}(Qg) (s) \,ds, \end{aligned}$$
(2.4)
where
$$\begin{aligned}& A_{1}(t)=\frac{1}{\Lambda}\bigl(\lambda_{1}t-1+e^{-\lambda_{1}t} \bigr),\qquad A_{2}(t)=\frac{1}{\Lambda}\bigl(\lambda_{2}t-1+e^{-\lambda_{2}t} \bigr), \end{aligned}$$
(2.5)
$$\begin{aligned}& (Ph) (s)= \int_{0}^{s}\frac{(s-\tau)^{p-2}}{\Gamma(p-1)}h(\tau)\,d\tau,\qquad (Qg) (s)= \int_{0}^{s}\frac{(s-\tau)^{q-2}}{\Gamma(q-1)}g(\tau)\,d \tau. \end{aligned}$$
(2.6)
Proof
Solving (2.1), we obtain
$$\begin{aligned}& u(t)=c_{11}e^{-\lambda_{1}t}+\frac{c_{12}}{\lambda_{1}}\bigl(1-e^{-\lambda _{1}t} \bigr)+\frac{c_{13}}{\lambda_{1}^{2}}\bigl(\lambda_{1}t-1+e^{-\lambda_{1}t}\bigr) + \int_{0}^{t}e^{-\lambda_{1}(t-s)}(Ph) (s) \,ds, \end{aligned}$$
(2.7)
$$\begin{aligned}& v(t)=c_{21}e^{-\lambda_{2}t}+\frac{c_{22}}{\lambda_{2}}\bigl(1-e^{-\lambda _{2}t} \bigr)+\frac{c_{23}}{\lambda_{2}^{2}}\bigl(\lambda_{2}t-1+e^{-\lambda_{2}t}\bigr) + \int_{0}^{t}e^{-\lambda_{2}(t-s)}(Qg) (s) \,ds, \end{aligned}$$
(2.8)
where \(c_{ij}\) (\(1\leq i\leq2\), \(1\leq j\leq3\)) are constants to be determined. In the following, we determine \(c_{ij}\) (\(1\leq i\leq 2\), \(1\leq j\leq3\)), so that \(u(t)\) and \(v(t)\) satisfy (2.2). By BCs (2.2), we obtain
$$ c_{11}= c_{12}=0, \qquad c_{21}= c_{22}=0, $$
(2.9)
and
$$\begin{aligned}& \frac{\lambda_{1}-1+e^{-\lambda_{1}}}{\lambda_{1}^{2}}c_{13}-\frac{a(\lambda _{2}\xi-1+e^{-\lambda_{2}\xi})}{\lambda_{2}^{2}}c_{23} \\& \quad =a \int_{0}^{\xi}e^{-\lambda_{2}(\xi-s)}(Qg) (s)\,ds - \int_{0}^{1}e^{-\lambda_{1}(1-s)}(Ph) (s) \,ds, \end{aligned}$$
(2.10)
$$\begin{aligned}& \frac{b(\lambda_{1}\eta-1+e^{-\lambda_{1}\eta})}{\lambda_{1}^{2}}c_{13}-\frac {\lambda_{2}-1+e^{-\lambda_{2}}}{\lambda_{2}^{2}}c_{23} \\& \quad = \int_{0}^{1}e^{-\lambda_{2}(1-s)}(Qg) (s)\,ds -b \int_{0}^{\eta}e^{-\lambda_{1}(\eta-s)}(Ph) (s) \,ds. \end{aligned}$$
(2.11)
Note that
$$\begin{aligned} \begin{vmatrix} \frac{\lambda_{1}-1+e^{-\lambda_{1}}}{\lambda_{1}^{2}} & - \frac {a(\lambda_{2}\xi-1+e^{-\lambda_{2}\xi})}{\lambda_{2}^{2}} \\ \frac{b(\lambda_{1}\eta-1+e^{-\lambda_{1}\eta})}{\lambda_{1}^{2}} & -\frac{\lambda_{2}-1+e^{-\lambda_{2}}}{\lambda_{2}^{2}} \end{vmatrix} =& \frac{ab(\lambda_{1}\eta-1+e^{-\lambda_{1}\eta})(\lambda_{2}\xi -1+e^{-\lambda_{2}\xi})}{\lambda_{1}^{2}\lambda_{2}^{2}} \\ &{}-\frac{(\lambda_{1}-1+e^{-\lambda_{1}})(\lambda_{2}-1+e^{-\lambda_{2}})}{\lambda _{1}^{2}\lambda_{2}^{2}}=\frac{\Lambda}{\lambda_{1}^{2}\lambda_{2}^{2}}\triangleq \mathfrak{B} \neq0. \end{aligned}$$
Thus, the system (2.10)-(2.11) has a unique solution for \(c_{13}\) and \(c_{23}\). By Cramer’s rule and simple calculations, it follows that
$$\begin{aligned}& c_{13}= \frac{1}{\mathfrak{B}}\frac{a(\lambda_{2}\xi-1+e^{-\lambda_{2}\xi })}{\lambda_{2}^{2}} \biggl[ \int_{0}^{1}e^{-\lambda_{2}(1-s)}(Qg) (s)\,ds -b \int_{0}^{\eta}e^{-\lambda_{1}(\eta-s)}(Ph) (s)\,ds \biggr] \\& \hphantom{c_{13}={}}{}-\frac{1}{\mathfrak{B}}\frac{\lambda_{2}-1+e^{-\lambda_{2}}}{\lambda _{2}^{2}} \biggl[a \int_{0}^{\xi}e^{-\lambda_{2}(\xi-s)}(Qg) (s)\,ds - \int_{0}^{1}e^{-\lambda_{1}(1-s)}(Ph) (s)\,ds \biggr], \end{aligned}$$
(2.12)
$$\begin{aligned}& c_{23}= \frac{1}{\mathfrak{B}}\frac{\lambda_{1}-1+e^{-\lambda_{1}}}{\lambda _{1}^{2}} \biggl[ \int_{0}^{1}e^{-\lambda_{2}(1-s)}(Qg) (s)\,ds -b \int_{0}^{\eta}e^{-\lambda_{1}(\eta-s)}(Ph) (s)\,ds \biggr] \\& \hphantom{c_{23}={}}{}-\frac {1}{\mathfrak{B}}\frac{b(\lambda_{1}\eta-1+e^{-\lambda_{1}\eta})}{\lambda_{1}^{2}} \biggl[a \int_{0}^{\xi}e^{-\lambda_{2}(\xi-s)}(Qg) (s)\,ds - \int_{0}^{1}e^{-\lambda_{1}(1-s)}(Ph) (s)\,ds \biggr]. \end{aligned}$$
(2.13)
Substituting (2.9) and (2.12) in (2.7), one has
$$\begin{aligned} u(t) =&A_{1}(t) \biggl\{ a\bigl(\lambda_{2} \xi-1+e^{-\lambda_{2}\xi}\bigr) \biggl[ \int_{0}^{1}e^{-\lambda _{2}(1-s)}(Qg) (s)\,ds -b \int_{0}^{\eta}e^{-\lambda_{1}(\eta-s)}(Ph) (s)\,ds \biggr] \\ &{}-\bigl(\lambda_{2}-1+e^{-\lambda_{2}}\bigr) \biggl[a \int_{0}^{\xi}e^{-\lambda_{2}(\xi-s)}(Qg) (s)\,ds - \int_{0}^{1}e^{-\lambda_{1}(1-s)}(Ph) (s)\,ds \biggr] \biggr\} \\ &{}+ \int _{0}^{t}e^{-\lambda_{1}(t-s)}(Ph) (s)\,ds. \end{aligned}$$
So (2.3) holds. Similarly, substituting (2.9) and (2.13) in (2.8) we can get (2.4). This completes the proof of the lemma. □
Lemma 2.2
([35])
For any
\(h,g\in L(0,1)\cap C(0,1)\), we have
$$\begin{aligned}& \begin{aligned} &\biggl\vert \int_{0}^{\eta}e^{-\lambda_{1}(\eta-s)}(Ph) (s)\,ds\biggr\vert \leq\frac{\eta ^{p}}{\lambda_{1}\Gamma(p)}\bigl(1-e^{-\lambda_{1}\eta}\bigr)\|h\|, \\ &\biggl\vert \int_{0}^{t}e^{-\lambda_{1}(t-s)}(Ph) (s)\,ds\biggr\vert \leq\frac{1}{\lambda _{1}\Gamma(p)}\bigl(1-e^{-\lambda_{1}}\bigr)\|h\|, \end{aligned} \end{aligned}$$
(2.14)
$$\begin{aligned}& \begin{aligned} &\biggl\vert \int_{0}^{\xi}e^{-\lambda_{2}(\xi-s)}(Qg) (s)\,ds\biggr\vert \leq\frac{\xi ^{q}}{\lambda_{2}\Gamma(q)}\bigl(1-e^{-\lambda_{2}\xi}\bigr)\|g\|, \\ &\biggl\vert \int_{0}^{t}e^{-\lambda_{2}(t-s)}(Qg) (s)\,ds\biggr\vert \leq\frac{1}{\lambda _{2}\Gamma(q)}\bigl(1-e^{-\lambda_{2}}\bigr)\|g\|. \end{aligned} \end{aligned}$$
(2.15)
Let \(X=C[0,1]\), then \(X\times X\) is a Banach space with the norm
$$\bigl\Vert (u,v)\bigr\Vert _{1}:=\| u\|+\| v\|,\qquad \| u\| = \max_{0\leq t\leq1}\bigl\vert u(t)\bigr\vert ,\qquad \| v\| = \max _{0\leq t\leq1}\bigl\vert v(t)\bigr\vert , $$
for any \((u,v)\in X\times X\).
In view of Lemma 2.1, we define the operator \(T:X\times X\to X\times X\) by
$$T(u,v)=\bigl(T_{1}(u,v),T_{2}(u,v)\bigr), $$
where operators \(T_{i}:X\times X\to X\) (\(i=1,2\)) are defined by
$$\begin{aligned}& T_{1}(u,v) (t) \\& \quad = A_{1}(t) \biggl\{ a\bigl( \lambda_{2}\xi-1+e^{-\lambda_{2}\xi}\bigr) \biggl[ \int_{0}^{1}e^{-\lambda _{2}(1-s)}Q(u,v) (s)\,ds -b \int_{0}^{\eta}e^{-\lambda_{1}(\eta-s)}P(u,v) (s)\,ds \biggr] \\& \qquad {}-\bigl(\lambda_{2}-1+e^{-\lambda_{2}}\bigr) \biggl[a \int_{0}^{\xi}e^{-\lambda_{2}(\xi -s)}Q(u,v) (s)\,ds - \int_{0}^{1}e^{-\lambda_{1}(1-s)}P(u,v) (s)\,ds \biggr] \biggr\} \\& \qquad {}+ \int _{0}^{t}e^{-\lambda_{1}(t-s)}P(u,v) (s)\,ds \end{aligned}$$
(2.16)
and
$$\begin{aligned}& T_{2}(u,v) (t) \\& \quad = A_{2}(t) \biggl\{ \bigl( \lambda_{1}-1+e^{-\lambda_{1}}\bigr) \biggl[ \int_{0}^{1}e^{-\lambda _{2}(1-s)}Q(u,v) (s)\,ds -b \int_{0}^{\eta}e^{-\lambda_{1}(\eta-s)}P(u,v) (s)\,ds \biggr] \\& \qquad {}-b\bigl(\lambda_{1}\eta-1+e^{-\lambda_{1}\eta}\bigr) \biggl[a \int_{0}^{\xi}e^{-\lambda _{2}(\xi-s)}Q(u,v) (s)\,ds - \int_{0}^{1}e^{-\lambda_{1}(1-s)}P(u,v) (s)\,ds \biggr] \biggr\} \\& \qquad {}+ \int _{0}^{t}e^{-\lambda_{2}(t-s)}Q(u,v) (s)\,ds, \end{aligned}$$
(2.17)
with
$$\begin{aligned}& P(u,v) (s)= \int_{0}^{s}\frac{(s-\tau)^{p-2}}{\Gamma(p-1)}f_{1}\bigl( \tau,u(\tau ),v(\tau)\bigr)\,d\tau, \end{aligned}$$
(2.18)
$$\begin{aligned}& Q(u,v) (s)= \int_{0}^{s}\frac{(s-\tau)^{q-2}}{\Gamma(q-1)}f_{2}\bigl( \tau,u(\tau ),v(\tau)\bigr)\,d\tau. \end{aligned}$$
(2.19)
For the sake of convenience, we set
$$\begin{aligned}& A_{1}=\sup_{t\in[0,1]}\bigl\vert A_{1}(t) \bigr\vert , \qquad A_{2}=\sup_{t\in[0,1]}\bigl\vert A_{2}(t)\bigr\vert , \end{aligned}$$
(2.20)
$$\begin{aligned}& M_{1} = \frac{A_{1}[|ab|(\lambda_{2}\xi-1+e^{-\lambda_{2}\xi})(1-e^{-\lambda_{1}\eta })\eta^{p-1}+(\lambda_{2}-1+e^{-\lambda_{2}})(1-e^{-\lambda_{1}})] +(1-e^{-\lambda_{1}})}{\lambda_{1}\Gamma(p)}, \\& M_{2} =\frac{A_{1}[|a|(\lambda_{2}\xi-1+e^{-\lambda_{2}\xi})(1-e^{-\lambda _{2}})+(\lambda_{2}-1+e^{-\lambda_{2}})|a|(1-e^{-\lambda_{2}\xi})\xi ^{q-1}]}{\lambda_{2}\Gamma(q)}, \\& M_{1}' = \frac{A_{2}[(\lambda_{1}-1+e^{-\lambda_{1}})|b|(1-e^{-\lambda_{1}\eta})\eta ^{p-1}+|b|(\lambda_{1}\eta-1+e^{-\lambda_{1}\eta})(1-e^{-\lambda _{1}})]}{\lambda_{1}\Gamma(p)}, \\& M_{2}' =\frac{A_{2}[(\lambda_{1}-1+e^{-\lambda_{1}})(1-e^{-\lambda _{2}})+|ab|(\lambda_{1}\eta-1+e^{-\lambda_{1}\eta})(1-e^{-\lambda_{2}\xi})\xi^{q-1}] +(1-e^{-\lambda_{2}})}{\lambda_{2}\Gamma(q)}, \end{aligned}$$
(2.21)
and
$$ (P{\mathbf{1}}) (s)= \int_{0}^{s}\frac{(s-\tau)^{p-2}}{\Gamma(p-1)}\,d\tau, \qquad (Q{ \mathbf{1}}) (s)= \int_{0}^{s}\frac{(s-\tau)^{q-2}}{\Gamma(q-1)}\,d\tau. $$
(2.22)
Lemma 2.3
The operator
T: \(X\times X\rightarrow X\times X\)
is a completely continuous.
Proof
By continuity of the functions \(f_{i}\) (\(i=1,2\)), the operator T is continuous.
Let \(\Omega\subset X\times X\) be bounded. Then there exist constants \(L_{i}>0\) (\(i=1,2\)) such that
$$\bigl\vert f_{i}\bigl(t,u(t),v(t)\bigr)\bigr\vert \leq L_{i}, \quad \forall (u,v)\in\Omega, i=1,2. $$
Then for any \((u,v)\in\Omega\), it follows from (2.16), (2.14), (2.15), (2.20), and (2.22) that
$$\begin{aligned}& \bigl\vert T_{1}(u,v) (t)\bigr\vert \\& \quad \leq A_{1} \biggl\{ |a|\bigl(\lambda_{2}\xi-1+e^{-\lambda_{2}\xi} \bigr) \biggl[L_{2} \int _{0}^{1}e^{-\lambda_{2}(1-s)}(Q{\mathbf{1}}) (s) \,ds +|b|L_{1} \int_{0}^{\eta}e^{-\lambda_{1}(\eta-s)}(P{\mathbf{1}}) (s)\,ds \biggr] \\& \qquad {} +\bigl(\lambda_{2}-1+e^{-\lambda_{2}}\bigr) \biggl[|a|L_{2} \int_{0}^{\xi}e^{-\lambda_{2}(\xi -s)}(Q{\mathbf{1}}) (s)\,ds +L_{1} \int_{0}^{1}e^{-\lambda_{1}(1-s)}(P{\mathbf{1}}) (s)\,ds \biggr] \biggr\} \\& \qquad {}+L_{1} \int_{0}^{t}e^{-\lambda_{1}(t-s)}(P{{\mathbf{1}}}) (s) \,ds \\& \quad \leq A_{1} \biggl[|a|\bigl(\lambda_{2} \xi-1+e^{-\lambda_{2}\xi}\bigr) \biggl(\frac{L_{2}(1-e^{-\lambda _{2}})}{\lambda_{2}\Gamma(q)} +\frac{|b|L_{1}(1-e^{-\lambda_{1}\eta})\eta^{p-1}}{\lambda_{1}\Gamma(p)} \biggr) \\& \qquad {} +\bigl(\lambda_{2}-1+e^{-\lambda_{2}}\bigr) \biggl( \frac{|a|L_{2}(1-e^{-\lambda_{2}\xi})\xi ^{q-1}}{\lambda_{2}\Gamma(q)} +\frac{L_{1}(1-e^{-\lambda_{1}})}{\lambda_{1}\Gamma(p)} \biggr) \biggr] \\& \qquad {}+\frac {L_{1}(1-e^{-\lambda_{1}})}{\lambda_{1}\Gamma(p)} \\& \quad \leq L_{1} \frac{A_{1}[|ab|(\lambda_{2}\xi-1+e^{-\lambda_{2}\xi})(1-e^{-\lambda_{1}\eta })\eta^{p-1}+(\lambda_{2}-1+e^{-\lambda_{2}})(1-e^{-\lambda_{1}})] +(1-e^{-\lambda_{1}})}{\lambda_{1}\Gamma(p)} \\& \qquad {}+L_{2}\frac{A_{1}[|a|(\lambda_{2}\xi-1+e^{-\lambda_{2}\xi})(1-e^{-\lambda _{2}})+(\lambda_{2}-1+e^{-\lambda_{2}})|a|(1-e^{-\lambda_{2}\xi})\xi ^{q-1}]}{\lambda_{2}\Gamma(q)}, \end{aligned}$$
(2.23)
which implies that
$$ \bigl\Vert T_{1}(u,v)\bigr\Vert \leq L_{1}M_{1}+L_{2}M_{2}, $$
(2.24)
where \(M_{1}\), \(M_{2}\) are given by (2.21).
By (2.17), (2.14), (2.15), (2.20), (2.22), and proceeding as in (2.23), we can obtain
$$ \bigl\Vert T_{2}(u,v)\bigr\Vert \leq L_{1}M_{1}'+L_{2}M_{2}', $$
(2.25)
where \(M_{1}'\), \(M_{2}'\) are given by (2.21).
Combining (2.24) with(2.25), we obtain
$$\bigl\Vert T(u,v)\bigr\Vert _{1}=\bigl\Vert T_{1}(u,v)\bigr\Vert +\bigl\Vert T_{2}(u,v)\bigr\Vert \leq(L_{1}M_{1}+L_{2}M_{2})+ \bigl(L_{1}M_{1}'+L_{2}M_{2}' \bigr)=M, $$
which implies that the operator T is uniformly bounded.
Next, we show that T is equicontinuous. For any \(t_{1},t_{2}\in[0,1]\) with \(t_{1}\leq t_{2}\), noticing (2.22) then we have
$$\begin{aligned}& \bigl\vert T_{1}(u,v) (t_{2})-T_{1}(u,v) (t_{1})\bigr\vert \\& \quad \leq \biggl\vert \bigl[A_{1}(t_{2})-A_{1}(t_{1}) \bigr] \biggl\{ a\bigl(\lambda_{2}\xi-1+e^{-\lambda_{2}\xi}\bigr) \\& \qquad {}\times \biggl[L_{2} \int _{0}^{1}e^{-\lambda_{2}(1-s)}(Q{\mathbf{1}}) (s) \,ds -bL_{1} \int_{0}^{\eta}e^{-\lambda_{1}(\eta-s)}(P{\mathbf{1}}) (s)\,ds \biggr] \\& \qquad {}-\bigl(\lambda_{2}-1+e^{-\lambda_{2}}\bigr) \biggl[aL_{2} \int_{0}^{\xi}e^{-\lambda_{2}(\xi -s)}(Q{\mathbf{1}}) (s)\,ds -L_{1} \int_{0}^{1}e^{-\lambda_{1}(1-s)}(P{\mathbf{1}}) (s)\,ds \biggr]\biggr\} \biggr\vert \\& \qquad {}+L_{1}\biggl\vert \int_{0}^{t_{1}}\bigl(e^{-\lambda_{1}(t_{2}-s)}-e^{-\lambda _{1}(t_{1}-s)} \bigr) (P{\mathbf{1}}) (s)\,ds + \int_{t_{1}}^{t_{2}}e^{-\lambda_{1}(t_{2}-s)}(P{\mathbf{1}}) (s)\,ds \biggr\vert . \end{aligned}$$
Analogously, we can obtain the following inequalities:
$$\begin{aligned}& \bigl\vert T_{2}(u,v) (t_{2})-T_{2}(u,v) (t_{1})\bigr\vert \\& \quad \leq \biggl\vert \bigl[A_{2}(t_{2})-A_{2}(t_{1}) \bigr] \biggl\{ \bigl(\lambda_{1}-1+e^{-\lambda_{1}}\bigr) \\& \qquad {}\times \biggl[L_{2} \int_{0}^{1}e^{-\lambda _{2}(1-s)}(Q{\mathbf{1}}) (s)\,ds -bL_{1} \int_{0}^{\eta}e^{-\lambda_{1}(\eta-s)}(P{\mathbf{1}}) (s)\,ds \biggr] \\& \qquad {}-b\bigl(\lambda_{1}\eta-1+e^{-\lambda_{1}\eta}\bigr) \biggl[aL_{2} \int_{0}^{\xi }e^{-\lambda_{2}(\xi-s)}(Q{\mathbf{1}}) (s)\,ds -L_{1} \int_{0}^{1}e^{-\lambda_{1}(1-s)}(P{\mathbf{1}}) (s)\,ds \biggr]\biggr\} \biggr\vert \\& \qquad {}+L_{2}\biggl\vert \int_{0}^{t_{1}}\bigl(e^{-\lambda_{2}(t_{2}-s)}-e^{-\lambda _{2}(t_{1}-s)} \bigr) (Q{\mathbf{1}}) (s)\,ds + \int_{t_{1}}^{t_{2}}e^{-\lambda_{2}(t_{2}-s)}(Q{\mathbf{1}}) (s)\,ds \biggr\vert . \end{aligned}$$
Since for any fixed \(s\in[0,1]\), the functions \(e^{-\lambda_{i}(t-s)}\), \(A_{i}(t)\) (\(i=1,2\)) are uniformly continuous on the interval on [0,1], we can conclude that the operator \(T(u,v)\) is equicontinuous. Thus the operator \(T(u,v)\) is completely continuous. The proof is completed. □
Now we state a well-known fixed point theorem, which is needed to prove the existence of solutions for system (1.1).
Lemma 2.4
(Leray-Schauder alternative [36])
Let
E
be a Banach space. Assume that
\(T:E\to E\)
be a completely continuous operator. Let
$$V=\{x\in E|x=\mu Tx \textit{ for some } 0< \mu< 1\}. $$
Then either the set
V
is unbounded, or
T
has at least one fixed point.