Consider the following assumptions:
A(t) is a bounded linear operator on \(\mathbb {R}\) where t→A(t) is continuous in the strong operator topology and N= max{|A(t)|:t∈J}.
\(F:J\times \mathbb {R} \rightarrow P(\mathbb {R}) \) is L1-Carathéodory and has nonempty, compact, and convex values;
ϕ:J→J, ϕ(t)≤t is a deviated continuous function.
ψ:J→J, ψ(t)≥t is an advanced continuous function.
There exists a constant ν>0 such that
$$\frac{\nu~ \Gamma(1+\alpha) }{T^{\alpha}\left(\nu N +\|\varkappa\|_{L^{1}}\right)\left[ ~1+|\gamma|\left(|\lambda|\sum_{j=1}^{n} b_{j}+ \sum_{k=1}^{m} a_{k}\right) ~\right]} > 1. $$
where \(~ \gamma : = \left (\sum _{k=1}^{m} a_{k} -\lambda \sum _{j=1}^{n} b_{j}\right)^{-1}.\)
We need the following lemma to define a mild solution for Problem (1).
Lemma 2
Let \( \sum _{k=1}^{m} a_{k} \neq \lambda \sum _{j=1}^{n} b_{j} \). For a given single-valued function \(\mu \in C(J,\mathbb {R})\), the solution of the nonlocal problem
$$\begin{array}{*{20}l} \left\{\begin{array}{l} {~}^{c}D^{\alpha} u(t) = A(t)u(t) + \mu(t)~~ a.e.~~ t\in J; \\ \sum_{k=1}^{m} a_{k} u(\phi(\tau_{k})) = \lambda \sum_{j=1}^{n} b_{j} u(\psi (\eta_{j})),~a_{k}, b_{j} >0. \end{array}\right. \end{array} $$
(5)
can be expressed by the integral equation
$$ {\begin{aligned} u(t)&~=~ \gamma \lambda \sum_{j=1}^{n} b_{j} \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s) ds + \gamma \lambda \sum_{j=1}^{n} b_{j} \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} \mu(s)~ ds \\&\quad - \gamma \sum_{k=1}^{m} a_{k} \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s) ds - \gamma \sum_{k=1}^{m} a_{k} \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} \mu(s)~ ds \\&\quad + \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s) ds + \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} \mu(s)~ ds. \end{aligned}} $$
(6)
Proof
Let u(t) be a solution for Problem (5). Operating Iα on both sides of the fractional differential equation of Problem (5), we get
$$\begin{array}{*{20}l} u(t)=u(0) + \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s) ds + \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} \mu(s)~ ds. \end{array} $$
(7)
Putting t=ϕ(τk) in (7), we obtain
$$\begin{array}{*{20}l} u(\phi (\tau_{k}))= u(0) \,+\, \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s) ds \,+\, \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} \mu(s)~ ds, \end{array} $$
then
$$\begin{array}{*{20}l} \sum_{k=1}^{m} a_{k} u(\phi(\tau_{k})) & ~=~~ \sum_{k=1}^{m} a_{k} u(0) + \sum_{k=1}^{m} a_{k} \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s)~ ds \\&\qquad+ \sum_{k=1}^{m} a_{k} \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} \mu(s)~ ds. \end{array} $$
(8)
Putting t=ψ(ηj) in (7), we obtain
$$ u(\psi (\eta_{j}))=u(0) + \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s) ~ds + \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} \mu(s)~ ds, $$
then
$$\begin{array}{*{20}l} \lambda \sum_{j=1}^{n} b_{j} u(\psi (\eta_{j})) &~=~~ \lambda \sum_{j=1}^{n} b_{j} u(0) + \lambda \sum_{j=1}^{n} b_{j} \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s)~ ds \\&\qquad + \lambda \sum_{j=1}^{n} b_{j} \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} \mu(s)~ ds. \end{array} $$
(9)
From (8), (9), and the nonlocal condition of Problem (5), we get
$$ {\begin{aligned} u(0)&= \gamma \lambda \sum_{j=1}^{n} b_{j} \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s)~ ds + \gamma \lambda \sum_{j=1}^{n} b_{j} \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} \mu(s)~ ds\\ &~~ - \gamma \sum_{k=1}^{m} a_{k} \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s)~ ds - \gamma \sum_{k=1}^{m} a_{k} \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} \mu(s)~ ds. \end{aligned}} $$
(10)
Substituting (10) into (7), we get the required. □
Definition 5
A function \(u\in C(J,\mathbb {R}) \) is said to be a mild solution for Problem (1), if
$$ \sum_{k=1}^{m} a_{k}~ u(\phi(\tau_{k})) = \lambda \sum_{j=1}^{n} b_{j}~ u(\psi (\eta_{j})),~a_{k}, b_{j} >0 $$
and there exists a function \(f\in L^{1}(J,\mathbb {R})\) such that f(t)∈F(t,u(t))a.e. on J, and u(t) satisfies the integral equation
$$ {\begin{aligned} u(t)&~=~ \gamma \lambda \sum_{j=1}^{n} b_{j} \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s)~ ds + \gamma \lambda \sum_{j=1}^{n} b_{j} \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} f(s)~ ds \\&\quad - \gamma \sum_{k=1}^{m} a_{k} \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s)~ ds - \gamma \sum_{k=1}^{m} a_{k} \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} f(s)~ ds \\&\quad + \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s)~ ds + \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} f(s)~ ds. \end{aligned}} $$
(11)
In view of the nonlinear alternative fixed point theorem of Leray-Schauder type (Theorem 2), we will discuss the existence of at least one mild solution for Problem (1).
Theorem 3
Let the assumptions (H1)- (H5) be satisfied and \( \sum _{k=1}^{m} a_{k} \neq \lambda \sum _{j=1}^{n} b_{j} \). Then, Problem (1) has at least one mild solution \( u \in C(J,\mathbb {R})\).
Proof
Since assumption (H2) is satisfied, then by applying Theorem 1, there exists a single-valued selection \(f\in S^{1}_{F,u }\). So, consider the set-valued operator \(K: C(J,\mathbb {R}) \rightarrow P(C(J,\mathbb {R})) \) such that
$$ {\begin{aligned} K u(t)= \left\lbrace g\in C(J,\mathbb{R}): g(t) = \left\{\begin{array}{l} ~~ \gamma \lambda \sum_{j=1}^{n} b_{j} \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s)~ds \\ + \gamma \lambda \sum_{j=1}^{n} b_{j} \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} f(s)~ds\\- \gamma \sum_{k=1}^{m} a_{k} \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s)~ds \\- \gamma \sum_{k=1}^{m} a_{k} \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} f(s)~ds\\+ \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s) ds + \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} f(s)~ds \end{array}\right.,~f\in S^{1}_{F,u} \right\rbrace. \end{aligned}} $$
(12)
That is, for each \(g \in Ku,~u\in C(J,\mathbb {R})\), there exists \(f\in S^{1}_{F,u }\) such that
$$ {\begin{aligned} g(t)&~=~ \gamma \lambda \sum_{j=1}^{n} b_{j} \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s)~ds + \gamma \lambda \sum_{j=1}^{n} b_{j} \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} f(s)~ds \\&\quad - \gamma \sum_{k=1}^{m} a_{k} \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s)~ds - \gamma \sum_{k=1}^{m} a_{k} \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} f(s)~ds \\&\quad + \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s)~ds + \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} f(s)~ds. \end{aligned}} $$
(13)
Now, we will show that K satisfies the hypotheses of Theorem 2.
The proof will be given in 4 steps:Step 1: (K is convex for each \( u\in C(J,\mathbb {R}) \))Using assumption (H2), we have F has a convex values, then the set of selections \(S^{1}_{F,u }\) is convex. Therefore, the step is obvious.Step 2: (K is a completely continuous operator)Firstly, we have to show that K maps bounded sets into bounded sets in \( C(J,\mathbb {R}) \).
For a positive number r, let \(B_{r} \subset C(J,\mathbb {R})\) be defined as
$$ B_{r}= \left\lbrace u \in C(J,\mathbb{R}): \| u \| \leq r \right\rbrace. $$
(14)
Clearly, Br is a bounded ball in \(C(J,\mathbb {R})\).
Using (13) with applying (H3) and (H4), we get
$$\begin{array}{*{20}l} g(t)&~\leq~ \gamma \lambda \sum_{j=1}^{n} b_{j} \int_{0}^{\eta_{j}} \frac{(\eta_{j}-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s)~ds + \gamma \lambda \sum_{j=1}^{n} b_{j} \int_{0}^{\eta_{j}} \frac{(\eta_{j}-s)^{\alpha-1}}{\Gamma (\alpha)} f(s)~ ds \\&\quad - \gamma \sum_{k=1}^{m} a_{k} \int_{0}^{\tau_{k}} \frac{(\tau_{k}-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s) ds - \gamma \sum_{k=1}^{m} a_{k} \int_{0}^{\tau_{k}} \frac{(\tau_{k}-s)^{\alpha-1}}{\Gamma (\alpha)} f(s) ~ds \\& \quad+ \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s)~ ds + \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} f(s)~ ds, \end{array} $$
(15)
then for u∈Br,
$$ {\begin{aligned} \| g \| & \leq\left(r N+\|\varkappa\|_{L^{1}}\right) |\lambda \gamma| \sum_{j=1}^{n} b_{j} \int_{0}^{\eta_{j}} \frac{(\eta_{j}-s)^{\alpha-1}}{\Gamma (\alpha)} ds + \left(r N+\|\varkappa\|_{L^{1}}\right) |\gamma| \sum_{k=1}^{m} a_{k} \int_{0}^{\tau_{k}} \frac{(\tau_{k}-s)^{\alpha-1}}{\Gamma (\alpha)} ds \\& \quad + \left(r N+\|\varkappa\|_{L^{1}}\right) \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} ds \\ & \leq~ \frac{T^{\alpha}\left(r N+\|\varkappa\|_{L^{1}}\right)}{\Gamma(1+\alpha)} \left[ ~1+|\gamma|\left(|\lambda|\sum_{j=1}^{n} b_{j}+ \sum_{k=1}^{m} a_{k}\right) ~\right]. \end{aligned}} $$
Thus, K maps bounded sets into bounded sets in \( C(J,\mathbb {R}).\)
Secondly, we have to prove that K maps bounded sets into equicontinuous sets of \(C(J,\mathbb {R})\).
Let 0<t1<t2<T, for each g∈Ku and u∈Br.
$$\begin{array}{*{20}l} g(t_{2})-g(t_{1}) &=~~ \int_{0}^{t_{2}} \frac{(t_{2}-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s)~ds -\int_{0}^{t_{1}} \frac{(t_{1}-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s)~ds \\ &\quad+ \int_{0}^{t_{2}} \frac{(t_{2}-s)^{\alpha-1}}{\Gamma (\alpha)} f(s) ds - \int_{0}^{t_{1}} \frac{(t_{1}-s)^{\alpha-1}}{\Gamma (\alpha)} f(s)~ds \\ & =~~ \int_{t_{1}}^{t_{2}} \frac{(t_{2}-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s)~ds + \int_{t_{1}}^{t_{2}} \frac{(t_{2}-s)^{\alpha-1}}{\Gamma (\alpha)} f(s)~ds \\&\quad+ \int_{0}^{t_{1}} \left(\frac{(t_{2}-s)^{\alpha-1}}{\Gamma (\alpha)} - \frac{(t_{1}-s)^{\alpha-1}}{\Gamma (\alpha)}\right) A(s)u(s)~ds \\&\quad+ \int_{0}^{t_{1}} \left(\frac{(t_{2}-s)^{\alpha-1}}{\Gamma (\alpha)} - \frac{(t_{1}-s)^{\alpha-1}}{\Gamma (\alpha)}\right) f(s)~ds, \end{array} $$
then
$$ {\begin{aligned} \lvert g(t_{2})-g(t_{1}) \rvert & ~\leq~~N \int_{t_{1}}^{t_{2}} \frac{(t_{2}-s)^{\alpha-1}}{\Gamma (\alpha)} \lvert u(s) \rvert~ds + \int_{t_{1}}^{t_{2}} \frac{(t_{2}-s)^{\alpha-1}}{\Gamma (\alpha)} \lvert f(s) \rvert~ds \\ &\quad + N\int_{0}^{t_{1}} \left(\frac{(t_{2}-s)^{\alpha-1}}{\Gamma (\alpha)} - \frac{(t_{1}-s)^{\alpha-1}}{\Gamma (\alpha)}\right) \lvert u(s) \rvert~ds \\ &\quad+ \int_{0}^{t_{1}} \left(\frac{(t_{2}-s)^{\alpha-1}}{\Gamma (\alpha)} - \frac{(t_{1}-s)^{\alpha-1}}{\Gamma (\alpha)}\right) \lvert f(s) \rvert~ ds \\ & \leq~~ \left(rN +\|\varkappa\|_{L^{1}}\right) \left(\int_{t_{1}}^{t_{2}} \frac{(t_{2}-s)^{\alpha-1}}{\Gamma (\alpha)}~ds + \int_{0}^{t_{1}} \left(\frac{(t_{2}-s)^{\alpha-1}}{\Gamma (\alpha)} - \frac{(t_{1}-s)^{\alpha-1}}{\Gamma (\alpha)}\right) ds \right) \\& \leq ~~ \frac{ 1}{\Gamma (\alpha+1)} \left(rN +\|\varkappa\|_{L^{1}}\right) \lvert t_{2}^{\alpha} - t_{1}^{\alpha} \rvert. \end{aligned}} $$
(16)
Obviously, the right-hand side tends to 0 independently of u∈Br as t2 tends to t1. That is, K maps bounded sets into equicontinuous sets of \(C(J,\mathbb {R})\). Applying Arzela-Ascoli theorem, we have that K is a completely continuous operator.Step 3: (K is an upper semicontinuous operator)From (H2) and step 2, K is completely continuous with nonempty compact values. So, we have to show that K has a closed graph to be an upper semicontinuous operator.
Let \(u_{n} \rightarrow \check {u},~g_{n}\rightarrow \check {g} \) where gn∈Kun. We need to prove that \(\check {g} \in K\check {u}\). Associated with gn∈Kun, there exists \(f_{n} \in S^{1}_{F,u_{n}}\) such that for each t∈J, we have
$$ {\begin{aligned} g_{n}(t)&~=~ \gamma \lambda \sum_{j=1}^{n} b_{j} \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u_{n}(s)~ds + \gamma \lambda \sum_{j=1}^{n} b_{j} \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} f_{n}(s)~ds \\&\quad - \gamma \sum_{k=1}^{m} a_{k} \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u_{n}(s)~ds - \gamma \sum_{k=1}^{m} a_{k} \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} f_{n}(s)~ds \\&\quad + \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u_{n}(s)~ds + \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} f_{n}(s)~ds. \end{aligned}} $$
(17)
We want to prove that there exists \(\check {f} \in S^{1}_{F,\check {u}}\) such that for all t∈J, we have
$$ {{\begin{aligned} \check{g}(t)&~=~ \gamma \lambda \sum_{j=1}^{n} b_{j} \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)\check{u}(s)~ds + \gamma \lambda \sum_{j=1}^{n} b_{j} \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} \check{f}(s)~ds \\&\quad - \gamma \sum_{k=1}^{m} a_{k} \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)\check{u}(s)~ds - \gamma \sum_{k=1}^{m} a_{k} \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} \check{f}(s)~ds \\&\quad + \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)\check{u}(s)~ds + \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} \check{f}(s)~ds. \end{aligned}}} $$
(18)
In view of Lemma 1, define the continuous linear operator \(\Omega :L^{1}(J,\mathbb {R}) \rightarrow C(J,\mathbb {R})\) as
$$ {{\begin{aligned} \Omega(f)(t) &=~~ \gamma \lambda \sum_{j=1}^{n} b_{j} \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s)~ds + \gamma \lambda \sum_{j=1}^{n} b_{j} \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} f(s)~ds \\&\quad - \gamma \sum_{k=1}^{m} a_{k} \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s)~ds - \gamma \sum_{k=1}^{m} a_{k} \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} f(s)~ds \\&\quad + \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} A(s)u(s)~ds + \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} f(s)~ds. \end{aligned}}} $$
(19)
Thus,
$$\begin{array}{*{20}l} |g_{n}(t)- \check{g}(t)|&\leq~~ |\gamma \lambda|N \sum_{j=1}^{n} b_{j} \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} |u_{n}(s)-\check{u}(s)|~ds \\&\quad + |\gamma \lambda| \sum_{j=1}^{n} b_{j} \int_{0}^{\psi (\eta_{j})} \frac{(\psi (\eta_{j})-s)^{\alpha-1}}{\Gamma (\alpha)} |f_{n}(s)-\check{f}(s)|~ds \\&\quad + |\gamma|N \sum_{k=1}^{m} a_{k} \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} |u_{n}(s)-\check{u}(s)|~ds \\&\quad + |\gamma| \sum_{k=1}^{m} a_{k} \int_{0}^{\phi (\tau_{k})} \frac{(\phi (\tau_{k})-s)^{\alpha-1}}{\Gamma (\alpha)} |f_{n}(s)-\check{f}(s)|~ds \\&\quad + N \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} |u_{n}(s)-\overline{u}(s)|~ds + \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)} |f_{n}(s)-\check{f}(s)|~ds. \end{array} $$
(20)
That is, \(g_{n}(t)\rightarrow \check {g}(t)\) as n→∞. Applying Lemma 1, we have \(\Omega ~o~ S^{1}_{F,u}\) as a closed graph operator in \(C(J,\mathbb {R}) \times C(J,\mathbb {R}) \) and \(g_{n}\in \Omega (S^{1}_{F,u_{n}})\). Since \(u_{n}\rightarrow \check {u}\), then we have \(\check {g}(t)\) satisfies (18) for some \(\check {f}\in S^{1}_{F,\check {u}}\). Therefore, K is an upper semicontinuous operator.Step 4: (There exists an open set \(\Lambda \subset C(J,\mathbb {R})\) with u∉Ku for ϱ∈(0,1) and u∈∂Λ)Let ϱ∈(0,1) and u∈ϱ Ku. Then, there exists \(f\in L^{1}(J,\mathbb {R})\) with \(f\in S^{1}_{F,u}\), such that, for t∈J, we have g∈Ku satisfies (13). As in proving step 2, we get
$$\begin{array}{*{20}l} \| g \| \leq~ \frac{T^{\alpha}\left(N\|u\| +\|\varkappa\|_{L^{1}}\right)}{\Gamma(1+\alpha)} \left[ ~1+|\gamma|\left(|\lambda|\sum_{j=1}^{n} b_{j}+ \sum_{k=1}^{m} a_{k}\right) ~\right]. \end{array} $$
Thus,
$$\begin{array}{*{20}l} \frac{\Gamma(1+\alpha) \|u\|}{T^{\alpha}\left(N\|u\| +\|\varkappa\|_{L^{1}}\right)\left[ ~1+|\gamma|\left(|\lambda|\sum_{j=1}^{n} b_{j}+ \sum_{k=1}^{m} a_{k}\right) ~\right]} \leq 1. \end{array} $$
In view of (H5), there exists ν such that ν≠∥u∥. Set
$$ \Lambda:=\{u\in C(J,\mathbb{R}): \|u\|<\nu+1\}. $$
From the definition of Λ, there is no u∈∂Λ such that u∈ϱ Ku for some ϱ∈(0,1). Further, we have that the operator \(K:\overline {\Lambda } \rightarrow P_{cp,cv}(B_{r})\) is an upper semicontinuous and completely continuous operator.
Consequently, by the nonlinear alternative of Leray-Schauder type (Theorem 2), we deduce that K has a fixed point \(u\in \overline {\Lambda }\) which is a mild solution for Problem (1). □
To prove the existence of at least one mild solution for Problem (2), we give the following lemma.
Lemma 3
Let \( \sum _{k=1}^{m} a_{k} \neq \lambda \sum _{j=1}^{n} b_{j} \). For a given single-valued function \(y\in C(J,\mathbb {R})\), the solution of the nonlocal problem
$$\begin{array}{*{20}l} \left\{\begin{array}{l} {~}^{c}D^{\alpha} u(t) = A(t)u(t) + y(t)~~ a.e.~~ t\in J;\\ ~\int_{0}^{T} u(\phi(s)) ds = \lambda \int_{0}^{T} u(\psi(s)) ds \end{array}\right. \end{array} $$
(21)
can be expressed by the integral equation
$$ {\begin{aligned} u(t)&~=~\beta \lambda \int_{0}^{T} \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta ds + \beta \lambda \int_{0}^{T} \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} y(\theta) d\theta ds \\& \quad - \beta \int_{0}^{T} \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta ds -\beta \int_{0}^{T} \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} y(\theta) d\theta ds\\&\quad + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} y(\theta) d\theta, \end{aligned}} $$
(22)
where \(~\beta = \frac {1}{T(1-\lambda)}\) and λ≠1.
Proof
Let u(t) be a solution for Problem (21). Operating Iα on both sides of the fractional differential equation of Problem (21), we get
$$\begin{array}{*{20}l} u(t)=u(0) + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} y(\theta) d\theta. \end{array} $$
(23)
Putting t=ϕ(s) in (23), we obtain
$$\begin{array}{*{20}l} u(\phi (s))= u(0) + \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta + \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} y(\theta) d\theta, \end{array} $$
then
$$\begin{array}{*{20}l} \int_{0}^{T} u(\phi(s)) ds & ~=~ \int_{0}^{T} u(0) ds + \int_{0}^{T}\int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta ds \\&\quad+ \int_{0}^{T} \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} y(\theta) d\theta ds. \end{array} $$
(24)
Putting t=ψ(s) in (23), we obtain
$$\begin{array}{*{20}l} u(\psi (s))\,=\, u(0) \,+\, \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta \,+\, \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} g(\theta,u(\theta)) d\theta, \end{array} $$
then
$$\begin{array}{*{20}l} \lambda\int_{0}^{T} u(\psi(s)) ds & ~=~\lambda \int_{0}^{T} u(0) ds + \lambda \int_{0}^{T}\int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta ds \\&\quad+\lambda \int_{0}^{T} \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} y(\theta) d\theta ds,~\lambda \neq 1. \end{array} $$
(25)
From (24), (25), and the nonlocal condition of Problem (21), we get
$$ {\begin{aligned} u(0)&= \beta \lambda \int_{0}^{T} \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta ds + \beta \lambda \int_{0}^{T} \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} y(\theta) d\theta ds \\& \quad - \beta \int_{0}^{T} \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta ds -\beta \int_{0}^{T} \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} y(\theta) d\theta ds. \end{aligned}} $$
(26)
where λ≠1 and \( \beta = \frac {1}{T(1-\lambda)}\).
Substituting (26) into (23), we get (22). Then, the proof is completed. □
Definition 6
A function \(u\in C(J,\mathbb {R}) \) is said to be a mild solution for Problem (2), if
$$\int_{0}^{T} u(\phi(s)) ds = \lambda \int_{0}^{T} u(\psi(s)) ds, $$
and there exists a function \(f\in L^{1}(J,\mathbb {R})\) such that f(t)∈F(t,u(t))a.e. on J, and u(t) satisfies the integral equation
$$ {\begin{aligned} u(t)&~=~\beta \lambda \int_{0}^{T} \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta ds + \beta \lambda \int_{0}^{T} \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} f(\theta) d\theta ds \\& \quad - \beta \int_{0}^{T} \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta ds -\beta \int_{0}^{T} \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} f(\theta) d\theta ds\\&\quad + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} f(\theta) d\theta. \end{aligned}} $$
As in proving Theorem 3, we infer the existence of at least one mild solution for Problem (2).
Theorem 4
Let the assumptions (H1)– (H4) be satisfied. Then, Problem (2) has at least one solution \(u \in C(J,\mathbb {R})\) if there exists a positive constant ϖ such that
$$ \frac{\varpi~\Gamma(2+\alpha) }{T^{\alpha} (2+\alpha) \left(\varpi~N +||\varkappa||_{L_{1}} \right)} > 1. $$
(27)
Proof
Since the assumption (H2) be satisfied, then by applying Theorem 1, there exists a single-valued selection \(f\in S^{1}_{F,u }\). So, consider the set-valued operator \(Q: C(J,\mathbb {R}) \rightarrow P(C(J,\mathbb {R})) \) such that
$$ {\begin{aligned} Q u(t)= \left\lbrace h\in C(J,\mathbb{R}): h(t) = \left\{\begin{array}{l} ~~ \beta \lambda \int_{0}^{T} \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta ds\\ + \beta \lambda \int_{0}^{T} \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} f(\theta) d\theta ds \\ - \beta \int_{0}^{T} \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta ds \\-\beta \int_{0}^{T} \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} f(\theta) d\theta ds\\ + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} f(\theta) d\theta. \end{array}\right.,~f\in S^{1}_{F,u} \right\rbrace. \end{aligned}} $$
(28)
We have to show that Q satisfies the hypotheses of Theorem 2.Step 1:In view of (H2), F has convex values, then \(S^{1}_{F,u }\) is a convex set. So, Q is convex for each \( u\in C(J,\mathbb {R}) \).Step 2:Consider the bounded ball \(B_{\sigma } \subset C(J,\mathbb {R})\) such that
$$ B_{\sigma}= \left\lbrace u \in C(J,\mathbb{R}): \| u \| \leq \sigma,~\sigma >0 \right\rbrace. $$
(29)
Then, for each h∈Qu, u∈Bσ, there exists \(f\in S^{1}_{F,u }\), such that
$$ {\begin{aligned} h(t)&~=~\beta \lambda \int_{0}^{T} \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta ds + \beta \lambda \int_{0}^{T} \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} f(\theta) d\theta ds \\& \quad - \beta \int_{0}^{T} \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta ds -\beta \int_{0}^{T} \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} f(\theta) d\theta ds\\&\quad + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} f(\theta) d\theta. \end{aligned}} $$
(30)
Applying (H3) and (H4),
$$ {\begin{aligned} h(t)&~\leq~ ~\beta \lambda \int_{0}^{T} \int_{0}^{s} \frac{(s-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta ds + \beta \lambda \int_{0}^{T} \int_{0}^{s} \frac{(s-\theta)^{\alpha-1}}{\Gamma (\alpha)} f(\theta) d\theta ds \\& \qquad - \beta \int_{0}^{T} \int_{0}^{s} \frac{(s-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta ds -\beta \int_{0}^{T} \int_{0}^{s} \frac{(s-\theta)^{\alpha-1}}{\Gamma (\alpha)} f(\theta) d\theta ds\\&\qquad + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} f(\theta) d\theta \\&~\leq~ \beta (\lambda-1) \int_{0}^{T} \int_{0}^{s} \frac{(s-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta ds + \beta (\lambda-1) \int_{0}^{T} \int_{0}^{s} \frac{(s-\theta)^{\alpha-1}}{\Gamma (\alpha)} f(\theta) d\theta ds \\&\qquad + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} f(\theta) d\theta. \end{aligned}} $$
(31)
For u∈Br, we obtain
$$\begin{array}{*{20}l} | h(t) | & \leq~\left(\sigma N+||\varkappa||_{L_{1}} \right) \left(|\beta (\lambda-1)| \int_{0}^{T} \int_{0}^{s} \frac{(s-\theta)^{\alpha-1}}{\Gamma (\alpha)} d\theta ds+ \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} ~ d\theta\right) \\ &\leq~\frac{T^{\alpha} \left(\sigma N+||\varkappa||_{L_{1}} \right)}{ \Gamma (1+\alpha)} \left(\frac{ T|\beta (\lambda-1)|}{1+\alpha}+1\right)\\ &\leq~ \frac{ T^{\alpha} ~(2+\alpha) ~\left(\sigma N+||\varkappa||_{L_{1}} \right) }{ \Gamma (2+\alpha)}. \end{array} $$
(32)
Thus, Q maps bounded sets into bounded sets in \( C(J,\mathbb {R}).\)
Let 0<t1<t2<T, for each h∈Qu and u∈Bσ. Using (30), as in proving (16), we get
$$\begin{array}{*{20}l} \lvert h(t_{2})-h(t_{1}) \rvert \leq ~\frac{ \sigma N+ \|\varkappa\|_{L^{1}} }{\Gamma (\alpha+1)} ~\lvert t_{2}^{\alpha} - t_{1}^{\alpha} \rvert ~\rightarrow 0 ~~\text{as}~~t_{2}\rightarrow t_{1}. \end{array} $$
Therefore, Q maps bounded sets into equicontinuous sets of \(C(J,\mathbb {R})\). Applying Arzela-Ascoli theorem, we have that Q is a completely continuous operator.Step 3:Let \(u_{n} \rightarrow \check {u},~h_{n}\rightarrow \check {h} \) where hn∈Qun. We have to prove that \(\check {h} \in Q\check {u}\).
Associated with hn∈Qun, there exists \(f_{n} \in S^{1}_{F,u_{n}}\) such that for each t∈J, we have
$$ {\begin{aligned} h_{n}(t)&~=~\beta \lambda \int_{0}^{T} \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u_{n}(\theta) d\theta ds + \beta \lambda \int_{0}^{T} \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} f_{n}(\theta) d\theta ds \\& \quad - \beta \int_{0}^{T} \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u_{n}(\theta) d\theta ds -\beta \int_{0}^{T} \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} f_{n}(\theta) d\theta ds\\&\quad + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u_{n}(\theta) d\theta + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} f_{n}(\theta) d\theta \\&~\leq~ ~\beta \lambda \int_{0}^{T} \int_{0}^{s} \frac{(s-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u_{n}(\theta) d\theta ds + \beta \lambda \int_{0}^{T} \int_{0}^{s} \frac{(s-\theta)^{\alpha-1}}{\Gamma (\alpha)} f_{n}(\theta) d\theta ds \\& \quad - \beta \int_{0}^{T} \int_{0}^{s} \frac{(s-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u_{n}(\theta) d\theta ds -\beta \int_{0}^{T} \int_{0}^{s} \frac{(s-\theta)^{\alpha-1}}{\Gamma (\alpha)} f_{n}(\theta) d\theta ds\\&\quad + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u_{n}(\theta) d\theta + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} f_{n}(\theta) d\theta \end{aligned}} $$
We want to prove that there exists \(\check {f} \in S^{1}_{F,\check {u}}\), such that for all t∈J, we get
$$ {\begin{aligned} \check{h}(t)&~=~\beta \lambda \int_{0}^{T} \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)\check{u}(\theta) d\theta ds + \beta \lambda \int_{0}^{T} \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} \check{f}(\theta) d\theta ds \\& \quad - \beta \int_{0}^{T} \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)\check{u}(\theta) d\theta ds -\beta \int_{0}^{T} \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} \check{f}(\theta) d\theta ds\\&\quad + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)\check{u}(\theta) d\theta + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} \check{f}(\theta) d\theta. \end{aligned}} $$
(33)
In view of Lemma 1, define the continuous linear operator \(\Phi :L^{1}(J,\mathbb {R}) \rightarrow C(J,\mathbb {R})\) as
$$ {\begin{aligned} \Phi(f)(t) &=~~ \beta \lambda \int_{0}^{T} \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta ds + \beta \lambda \int_{0}^{T} \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} f(\theta) d\theta ds \\& \quad - \beta \int_{0}^{T} \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta ds -\beta \int_{0}^{T} \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} f(\theta) d\theta ds\\&\quad + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} A(\theta)u(\theta) d\theta + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} f(\theta) d\theta. \end{aligned}} $$
(34)
Thus,
$$\begin{array}{*{20}l} |h_{n}(t)- \check{h}(t)|&\leq~~N|\lambda \beta| \int_{0}^{T} \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} |u_{n}(\theta)-\check{u}(\theta)|~ d\theta ds\\&\quad + |\lambda \beta| \int_{0}^{T} \int_{0}^{\psi (s)} \frac{(\psi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} |f_{n}(\theta)-\check{f}(\theta)| ~ d\theta ds \\& \quad + N|\beta| \int_{0}^{T} \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} |u_{n}(\theta)-\check{u}(\theta)| d\theta ds\\&\quad +|\beta| \int_{0}^{T} \int_{0}^{\phi (s)} \frac{(\phi (s)-\theta)^{\alpha-1}}{\Gamma (\alpha)} |f_{n}(\theta)-\check{f}(\theta)| ~d\theta ds\\&\quad +N \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} |u_{n}(\theta)-\check{u}(\theta)| d\theta + \int_{0}^{t} \frac{(t-\theta)^{\alpha-1}}{\Gamma (\alpha)} |f_{n}(\theta)-\check{f}(\theta)|~ d\theta. \end{array} $$
(35)
That is, \(h_{n}(t)\rightarrow \check {h}(t)\) as n→∞. Lemma 1 gives that \(\Phi ~o~ S^{1}_{F,u}\) is a closed graph operator in \(C(J,\mathbb {R}) \times C(J,\mathbb {R}) \) and \(h_{n}\in \Phi (S^{1}_{F,u_{n}})\). Since \(u_{n}\rightarrow \check {u}\), then \(\check {h}(t)\) satisfies (34) for some \(\check {f}\in S^{1}_{F,\check {u}}\).
Therefore, Q is an upper semicontinuous operator.Step 4:Let ϱ∈(0,1) and u∈ϱ Qu. Then, there exists \(f\in L^{1}(J,\mathbb {R})\) with \(f\in S^{1}_{F,u}\), such that, for t∈J, we have h∈Qu satisfies (30). As in proving (32), we get
$$\begin{array}{*{20}l} \| h \| \leq~ \frac{ T^{\alpha} ~(2+\alpha) \left(N\|u\|+||\varkappa||_{L_{1}} \right) }{ \Gamma (2+\alpha)}. \end{array} $$
Thus,
$$\begin{array}{*{20}l} \frac{\Gamma(2+\alpha) ~\|u\|}{T^{\alpha} (2+\alpha) \left(N \|u\|+||\varkappa||_{L_{1}} \right)} \leq 1. \end{array} $$
In view of (27), there exists ϖ such that ϖ≠∥u∥. Set
$$\Upsilon:=\{u\in C(J,\mathbb{R}): \|u\|<\varpi+1\}. $$
It is clear that there is no u∈∂Υ such that u∈ϱ Qu for some ϱ∈(0,1). Further, we get that the operator \(Q:\overline {\Upsilon } \rightarrow P_{cp,cv}(B_{\sigma })\) is an upper semicontinuous and completely continuous operator.
Consequently, by the nonlinear alternative of Leray-Schauder type, we deduce that Q has a fixed point \(u\in \overline {\Upsilon }\) which is a mild solution for Problem (2). □
Finally, we give an illustrative example.
Example 1
Consider the following fractional differential inclusion
$$\begin{array}{*{20}l} {~}^{c}D^{0.4} -0.01\sin u(t) \in \left[\frac{|u|^{3}}{|u|^{3}+3}+4t^{3}+5,~\frac{|u|}{|u|+1}+2t+1 \right]~~a.e.~~t\in [0,1]. \end{array} $$
(36)
Here, \(\alpha =\frac {2}{5}\), \(A(t)u(t)= \frac {1}{100}\sin u(t)\), \(N=\frac {1}{100}\), T=1, J=[0,1] and \(F:J\times \mathbb {R}\rightarrow P(\mathbb {R})\) is a set-valued map given by
$$u\rightarrow F(t,u)= \left[\frac{|u|^{3}}{|u|^{3}+3}+4t^{3}+5,~\frac{|u|}{|u|+1}+2t+1 \right]. $$
For f∈F(t,u), we have
$$|f|\leq \max\left\lbrace\frac{|u|^{3}}{|u|^{3}+3}+t^{3},~\frac{|u|}{|u|+1}+1 \right\rbrace \leq 10,~u\in \mathbb{R}. $$
Then,
$$\|F(t,u)\|_{P}=\sup\left\lbrace |v|: v\in F(t,u),~u\in \mathbb{R}\right\rbrace \leq 10 ~~\text{and}~~\|\varkappa\|_{L^{1}}=10. $$
Consider the deviated-advanced nonlocal condition:
$$\begin{array}{*{20}l} \sum_{k=1}^{2}4^{-k} u(t_{k}^{2})=\frac{9}{4} \sum_{j=1}^{3}2^{-j} u(\sqrt{t_{j}}~). \end{array} $$
(37)
Clearly, \(m=2,~n=3,~a_{k}= 4^{-k},~\sum _{k=1}^{m}a_{k}= \frac {5}{16},~b_{j}=2^{-j},~ \sum _{j=1}^{n}b_{j}=\frac {7}{8},~\lambda =\frac {9}{4}, \gamma =-\frac {32}{53},~ \phi :J\rightarrow J~\) is defined by t↦t2, and ψ:J→J is defined by \(t\mapsto \sqrt {t}\).
Applying assumption (H5), we get ν>27.5321. Therefore, all conditions of Theorem 3 are satisfied. So, there exists at least one solution for Problem (36) with condition (37).
Now, consider the deviated-advanced nonlocal integral condition:
$$\begin{array}{*{20}l} \int_{0}^{1} u(s^{2})~ds=\frac{9}{4}\int_{0}^{1} u(\sqrt{s}~)~ds. \end{array} $$
(38)
Here, \(\lambda =\frac {9}{4}>1\), then \(\beta =-\frac {4}{5}\). In view of (27), there exists ϖ>19.7017. Therefore, all conditions of Theorem 4 are satisfied. So, there exists at least one solution for problem (36) with the integral condition (38).
Remark 1
For ak=tk−tk−1, τk∈[tk−1,tk]⊂J and bj=tj−tj−1, ηj∈[tj−1,tj]⊂J, the nonlocal condition for Problem (1) will be \(\sum _{k=1}^{m} (t_{k}-t_{k-1})~ u(\phi (\tau _{k})) = \lambda \sum _{j=1}^{n} (t_{j}-t_{j-1}) ~u(\psi (\eta _{j})). \) For \(u \in C(J,\mathbb {R})\), we have \(\lim \limits _{m\rightarrow \infty } \sum _{k=1}^{m} (t_{k}-t_{k-1})~ u(\phi (\tau _{k})) = \lambda \lim \limits _{n\rightarrow \infty } \sum _{j=1}^{n} (t_{j}-t_{j-1})~ u(\psi (\eta _{j})) \) which can be transformed into the nonlocal integral condition \( \int _{0}^{T} u(\phi (s)) ds = \lambda \int _{0}^{T} u(\psi (s)) ds~\) of Problem (2).