Theorem 3.1
Assume that the following conditions hold.
 (H1):

\(F:J\times R\rightarrow P_{cv,cp}(R)\)
is an
\(L^{1}\)Carathéodory multivalued map.
 (H2):

Functions
\(\alpha,\beta\in PC(J,R)\cap AC(J',R)\)
are related lower and upper solutions of problem (1.1)–(1.3), which are given in Definition 2.2
and satisfy
\(\alpha(t)\leq\beta(t)\), \(t\in J\).
 (H3):

\(I_{k}\in C(R,R)\), \(k=1,\ldots,m\).
 (H4):

g
is a continuous singlevalued map in
\((x,y)\in [\alpha(0),\beta(0)]\times[\alpha(b),\beta(b)]\)
and nondecreasing in
\(y\in[\alpha(b),\beta(b)]\).
 (H5):

A
is the infinitesimal generator of a linear bounded semigroup
\(T(t)\), \(t\geq0\), and there exists
\(M>0\)
such that
\(\ T(t)\_{L(R)}\leq M\).
 (H6):

For
\(x(t)<\alpha(t)\), \(v\in F(t,\alpha(t))\), \(t\in J\), and
\(Ax(t)+ v(t)\geq A\alpha(t)+v_{1}(t)\), and for
\(x(t)>\beta(t)\), \(v\in F(t,\beta(t))\), \(t\in J\), and
\(Ax(t)+ v(t)\leq A\beta(t)+v_{2}(t)\), where
\(v_{1}, v_{2}\in L^{1}(J,R)\)
satisfy (2.1)–(2.4).
Then system (1.1)–(1.3) has at least one solution
x
such that
\(\alpha(t)\leq x(t)\leq\beta(t)\)
for all
\(t\in J\).
Proof
We transform (1.1)–(1.3) into a fixed point problem. Consider the modified problem
$$ \textstyle\begin{cases} x'(t)+x(t)\in Ax(t)+F_{1}(t,x(t)),\quad t\in J',\\ \Delta x(t_{k})=I_{k}(\tau(t_{k}, x(t_{k}))),\quad k=1,\ldots,m,\\ x(0)=\tau(0,x(0)g(\tau(0,x),\tau(b,x))), \end{cases} $$
(3.1)
where \(F_{1}(t,x)=F(t,\tau(t,x))+\tau(t,x)\), and \(\tau: C(J,R)\rightarrow C(J,R)\) is defined by
$$\tau(t,x)= \textstyle\begin{cases} \beta(t),& x(t)>\beta(t),\\ x(t),& \alpha(t)\leq x(t)\leq\beta(t),\\ \alpha(t),& x(t)< \alpha(t). \end{cases} $$
Evidently, if x is a solution of (3.1), \(\alpha(t)\leq x(t)\leq\beta (t)\), and \(\alpha(0)\leq x(0)g(\tau(0,x), \tau(T,x))\leq\beta(0)\), then x is a solution of (1.1)–(1.3).
By Lemma 2.10 we have that a solution of (3.1) is a fixed point of the operator \(N:PC(J,R)\rightarrow P(PC(J,R))\) defined by
$$\begin{aligned} N(x) =& \biggl\{ h\in PC(J,R): h(t)=T(t)x(0)+ \int ^{t}_{0}T(ts)\bigl[v(s)+\tau(s, x)x(s)\bigr] \,ds \\ &{} +\sum_{0< t_{k}< t}T(tt_{k})I_{k} \bigl(\tau\bigl(t_{k},x(t_{k})\bigr)\bigr), v\in S_{F,\tau (t,x)} \biggr\} , \end{aligned}$$
where
$$S_{F,\tau(t,x)}=\bigl\{ v\in L^{1}(J,R):v(t)\in F\bigl(t,\tau(t,x) \bigr)\mbox{ for a.e. } t\in J\bigr\} . $$
Note that, for each \(x\in C(J,R)\), \(S_{F,x}\) is nonempty (see [14]), so \(S_{F,\tau(t,x)}\) is nonempty.
Next, we will show that N has a fixed point by applying Lemma 2.5. The proof will be given in several steps. We first show that N is a completely continuous multivalued map, upper semicontinuous with convex closed values.
Step 1.
\(N(x)\) is convex for each \(x\in PC(J,R)\).
Indeed, if \(h_{1}\), \(h_{2}\) belong to \(N(x)\), then there exist \(\overline {v}_{1}, \overline{v}_{2}\in S_{F,\tau(t,x)}\) such that
$$\begin{aligned} h_{i}(t) =&T(t)x(0)+ \int^{t}_{0}T(ts)\bigl[\overline{v}_{i}(s)+ \tau(s, x)x(s)\bigr]\,ds \\ & +\sum_{0< t_{k}< t}T(tt_{k})I_{k} \bigl(\tau \bigl(t_{k},x(t_{k})\bigr)\bigr),\quad i=1,2. \end{aligned}$$
Let \(0\leq l\leq1\). Then, for each \(t\in J\), we have
$$\begin{aligned} \bigl[l h_{1}+(1l)h_{2}\bigr](t) =&T(t)x(0)+\sum _{0< t_{k}< t}T(tt_{k})I_{k}\bigl(\tau \bigl(t_{k},x(t_{k})\bigr)\bigr) \\ &{} + \int^{t}_{0}T(ts)\bigl[l\overline{v}_{1}(s)+(1l) \overline{v}_{2}(s)+\tau (s, x)x(s)\bigr]\,ds. \end{aligned}$$
Since \(S_{F,\tau(t,x)}\) is convex (because F has convex values in (H1)), then \(l h_{1}+(1l)h_{2}\in N(x)\), so \(N(x)\) is convex.
Step 2.
N is completely continuous.
First, we show that N maps bounded sets into bounded sets in \(PC(J,R)\). Let q be a positive constant, \(B_{q}=\{x\in PC(J,R): \x\ _{PC}< q\}\) be a bounded set, and \(x\in B_{q}\). Then for each \(h\in N(x)\), there exists \(v\in S_{F,\tau(t,x)}\) such that
$$\begin{aligned} h(t) =&T(t)x(0)+ \int^{t}_{0}T(ts)\bigl[v(s)+\tau(s, x)x(s)\bigr] \,ds \\ &{} +\sum_{0< t_{k}< t}T(tt_{k})I_{k} \bigl(\tau \bigl(t_{k},x(t_{k})\bigr)\bigr). \end{aligned}$$
(3.2)
Noting the boundary condition of (3.1) and the definition of τ, we have
$$\begin{aligned} &\alpha(0)\leq x(0)\leq\beta(0), \end{aligned}$$
(3.3)
$$\begin{aligned} &\alpha(t)\leq\tau(t,x)\leq\beta(t). \end{aligned}$$
(3.4)
Let \(\rho_{1}=\max(q,\sup_{t\in J}\alpha(t),\sup_{t\in J}\beta (t))\). Then \(\tau(t,x)\leq\rho_{1}\). By (H1) there exists \(\varphi _{\rho_{1}}\in L^{1}(J,[0,+\infty))\) such that
$$ \sup\bigl\{ \vert v \vert : v\in F\bigl(t,\tau(t,x)\bigr)\bigr\} \leq \varphi_{\rho_{1}}(t). $$
(3.5)
If \(x\in B_{q}\), then there exist \(c_{k}>0\), \(k=1,\ldots,m\), such that \(I_{k}(\tau(t_{k},x(t_{k})))\leq c_{k}\), since \(I_{k}\) are continuous in (H3) and (3.4). So, with (3.3), (3.5), and (H5), we have
$$\begin{aligned} \bigl\vert h(t) \bigr\vert \leq& \bigl\Vert T(t) \bigr\Vert _{L(R)} \bigl\vert x(0) \bigr\vert + \int_{0}^{d} \bigl\Vert T(ts) \bigr\Vert _{L(R)}\bigl[ \bigl\vert v(s) \bigr\vert + \bigl\vert \tau(s, x) \bigr\vert + \bigl\vert x(s) \bigr\vert \bigr]\,ds \\ &{} +\sum_{0< t_{k}< t} \bigl\Vert T(tt_{k}) \bigr\Vert _{L(R)} \bigl\vert I_{k}\bigl(\tau \bigl(t_{k},x(t_{k})\bigr)\bigr) \bigr\vert \\ \leq& M\max\bigl( \bigl\vert \alpha(0) \bigr\vert , \bigl\vert \beta(0) \bigr\vert \bigr)+M \Vert \varphi_{\rho_{1}} \Vert _{L^{1}}+Md( \rho_{1}+q) \\ &{} +M\sum^{m}_{k=1}c_{k}:=K, \end{aligned}$$
and thus \(\N(x)\_{PC}\leq K\).
Second, we prove that N maps bounded sets into quasiequicontinuous sets of \(PC(J,R)\). Let \(u_{1},u_{2}\in J_{k}\), \(k=0,1,\ldots,m\), \(u_{1}< u_{2}\), \(x\in B_{q}\), and \(h\in N(x)\). Then
$$\begin{aligned} \bigl\vert h(u_{2})h(u_{1}) \bigr\vert \leq& \bigl\vert T(u_{2})T(u_{1}) \bigr\vert \max\bigl( \bigl\vert \alpha(0) \bigr\vert , \bigl\vert \beta (0) \bigr\vert \bigr) \\ &{} + \int _{0}^{u_{1}} \bigl\vert T(u_{2}s)T(u_{1}s) \bigr\vert \bigl(\varphi_{\rho_{1}}(s)+\rho _{1}+q\bigr)\,ds \\ &{} + \int _{u_{1}}^{u_{2}}M\bigl(\varphi_{\rho_{1}}(s)+ \rho_{1}+q\bigr)\,ds \\ &{} +\sum_{0< t_{k}< u_{1}} \bigl\vert T(u_{2}t_{k})T(u_{1}t_{k}) \bigr\vert c_{k}+\sum_{u_{1}< t_{k}< u_{2}}Mc_{k}. \end{aligned}$$
As \(u_{2}\rightarrow u_{1}\), the righthand side of this inequality tends to zero since \(T(t)\) is strongly continuous. This proves that \(N(B_{q})\) is quasiequicontinuous. By Lemma 2.8, N is completely continuous and therefore a condensing map.
Step 3.
N has a closed graph.
Let \(x_{n}\rightarrow x^{\ast}\), \(h_{n}\in N(x_{n})\), and \(h_{n}\rightarrow h^{\ast}\). We will prove that \(h^{\ast}\in N(x^{\ast })\). Since \(h_{n}\in N(x_{n})\), there exist \(v_{n}\in S_{F,\tau (t,x_{n})}\) such that
$$\begin{aligned} h_{n}(t) =&T(t)x_{n}(0)+ \int^{t}_{0}T(ts)\bigl[v_{n}(s)+\tau(s, x_{n})x_{n}(s)\bigr]\,ds \\ &{} +\sum_{0< t_{k}< t}T(tt_{k})I_{k} \bigl(\tau\bigl(t_{k},x_{n}(t_{k})\bigr)\bigr). \end{aligned}$$
Next, we need prove that there exists \(v^{\ast}\in S_{F,\tau(t,x^{\ast })}\) such that, for each \(t\in J\),
$$\begin{aligned} h^{\ast}(t) =&T(t)x^{\ast}(0)+ \int^{t}_{0}T(ts)\bigl[v^{\ast }(s)+\tau \bigl(s, x^{\ast}\bigr)x^{\ast}(s)\bigr]\,ds \\ &{} +\sum_{0< t_{k}< t}T(tt_{k})I_{k} \bigl(\tau\bigl(t_{k},x^{\ast}(t_{k})\bigr)\bigr). \end{aligned}$$
Since \(x_{n}\rightarrow x^{\ast}\), \(h_{n}\rightarrow h^{\ast}\), and \(I_{k}\in C(R,R)\) in (H3), by the definition of τ we have
$$ \begin{aligned}[b] &\biggl\ h_{n}(t)T(t)x_{n}(0) \int^{t}_{0}T(ts)\bigl[\tau(s, x_{n})x_{n}(s) \bigr]\,ds \\ &\quad {} \sum_{0< t_{k}< t}T(tt_{k})I_{k} \bigl(\tau \bigl(t_{k},x_{n}(t_{k})\bigr)\bigr) \\ &\quad {}  \biggl[h^{\ast}(t)T(t)x^{\ast}(0) \int ^{t}_{0}T(ts)\bigl[\tau\bigl(s, x^{\ast}\bigr)x^{\ast}(s)\bigr]\,ds \\ &\quad {} \sum_{0< t_{k}< t}T(tt_{k})I_{k} \bigl(\tau\bigl(t_{k},x^{\ast }(t_{k})\bigr)\bigr) \biggr]\biggr\ _{PC}\rightarrow0 \end{aligned} $$
(3.6)
as \(n\rightarrow\infty\). Consider the linear continuous operator \(\varGamma : L^{1}(J,R)\rightarrow C(J,R)\) defined by
$$v\mapsto\varGamma(v) (t)= \int^{t}_{0}T(ts)v(s)\,ds. $$
Note that \(S_{F,\tau(t,x)}\) is nonempty, so by Lemma 2.4, \(\varGamma\circ S_{F}\) is a closed graph operator. Moreover,
$$ \begin{aligned}[b] &h_{n}(t)T(t)x_{n}(0) \int^{t}_{0}T(ts)\bigl[\tau(s, x_{n})x_{n}(s) \bigr]\,ds \\ &\quad {} \sum_{0< t_{k}< t}T(tt_{k})I_{k} \bigl(\tau \bigl(t_{k},x_{n}(t_{k})\bigr)\bigr)\in \varGamma(S_{F,\tau(t,x_{n})}). \end{aligned} $$
(3.7)
Since \(x_{n}\rightarrow x^{\ast}\), by (3.6) and (3.7) there exists \(v^{\ast}\in S_{F,\tau(t,x^{\ast})}\) satisfying
$$\begin{aligned} &h^{\ast}(t)T(t)x^{\ast}(0) \int^{t}_{0}T(ts)\bigl[\tau\bigl(s, x^{\ast} \bigr)x^{\ast}(s)\bigr]\,ds \\ &\quad {} \sum_{0< t_{k}< t}T(tt_{k})I_{k} \bigl(\tau\bigl(t_{k},x^{\ast }(t_{k})\bigr)\bigr)= \int^{t}_{0}T(ts)v^{\ast}(s)\,ds. \end{aligned}$$
As a consequence of Steps 1 to 3, N is a completely continuous multivalued upper semicontinuous map with convex closed values.
Step 4. The set \(\Re=\{x\in PC(J,R):\lambda x\in N(x) \mbox{ for some } \lambda>1\}\) is bounded.
Let \(x\in\Re\). Then \(\lambda x\in N(x)\) for some \(\lambda>1\). Thus, for each \(t\in J\),
$$\begin{aligned} x(t) =&\lambda^{1} \biggl[T(t)x(0)+ \int^{t}_{0}T(ts)\bigl[v(s)+\tau (s, x)x(s)\bigr] \,ds \\ &{} +\sum_{0< t_{k}< t}T(tt_{k})I_{k} \bigl(\tau \bigl(t_{k},x(t_{k})\bigr)\bigr) \biggr] \end{aligned}$$
for some \(v\in S_{F,\tau(t,x)}\). Let \(\rho_{2}=\max(\sup_{t\in J}\alpha (t),\sup_{t\in J}\beta(t))\). From (3.4) it follows that \(\tau (t,x)\leq\rho_{2}\). By (H1) there exists \(\varphi_{\rho_{2}}\in L^{1}(J,[0,+\infty))\) such that
$$ \sup\bigl\{ \vert v \vert : v\in F\bigl(t,\tau(t,x)\bigr)\bigr\} \leq \varphi_{\rho_{2}}(t). $$
(3.8)
Since \(I_{k}\in C(R,R)\) in (H3) and (3.4), there exist \(c'_{k}>0\), \(k=1,\ldots,m\), such that \(I_{k}(\tau(t_{k}, x(t_{k})))\leq c'_{k}\). So, by (3.3) and (3.8), for each \(t\in J\), we have
$$\begin{aligned} \bigl\vert x(t) \bigr\vert \leq& M \biggl[ \bigl\vert x(0) \bigr\vert + \int_{0}^{t}\bigl[ \bigl\vert v(s) \bigr\vert + \bigl\vert \tau(s, x) \bigr\vert + \bigl\vert x(s) \bigr\vert \bigr]\,ds + \sum_{0< t_{k}< t} \bigl\vert I_{k}\bigl(\tau \bigl(t_{k},x(t_{k})\bigr)\bigr) \bigr\vert \biggr] \\ \leq& M \Biggl[\max\bigl( \bigl\vert \alpha(0) \bigr\vert , \bigl\vert \beta (0) \bigr\vert \bigr)+ \Vert \varphi_{\rho_{2}} \Vert _{L^{1}}+b\rho_{2} + \int _{0}^{t} \bigl\vert x(s) \bigr\vert \,ds+ \sum^{m}_{k=1}c'_{k} \Biggr]. \end{aligned}$$
Set
$$K_{0}= M \Biggl[\max\bigl( \bigl\vert \alpha(0) \bigr\vert , \bigl\vert \beta(0) \bigr\vert \bigr)+ \Vert \varphi_{\rho _{2}} \Vert _{L^{1}}+b\rho_{2} +\sum^{m}_{k=1}c'_{k} \Biggr]. $$
Using Gronwall’s lemma (see [18], p. 36), for each \(t\in J\), we have
$$\bigl\vert x(t) \bigr\vert \leq K_{0} e^{Mt}. $$
So,
$$\Vert x \Vert _{PC}\leq K_{0} e^{Mb}. $$
This shows that the set ℜ is bounded. As a consequence of Lemma 2.5, we deduce that N has a fixed point, which is a solution of problem (3.1).
Step 5. The solution x of (3.1) satisfies
$$ \alpha(t)\leq x(t)\leq\beta(t),\quad t\in J, $$
(3.9)
and
$$ \alpha(0)\leq x(0)g\bigl(\tau(0,x),\tau(b,x)\bigr)\leq\beta(0). $$
(3.10)
We first prove (3.9). Let x be a solution of (3.1). We prove that \(x(t)\leq\beta(t)\) for all \(t\in J\). Suppose that \(x\beta\) attains a positive maximum on J at \(s_{0}\). As (3.3), we consider the only possible case \(s_{0}\in(0,T]\). Then there exists \(s_{1}\in(0,s_{0})\), \(s_{1}\neq t_{k}\ (k=1,2,\ldots,m)\), such that
$$0< x(t)\beta(t)\leq x(s_{0})\beta(s_{0}),\quad t \in[s_{1},s_{0}]. $$
So, \(\tau(t, x)=\beta(t)\) for \(t\in[s_{1},s_{0}]\), and there exists \(v\in F(t,\beta(t))\) such that
$$\begin{aligned} \beta(s_{0})\beta(s_{1}) \leq& x(s_{0})x(s_{1})\\ =& \int ^{s_{0}}_{s_{1}}\bigl[Ax(s)+v(s)+\beta(s)x(s)\bigr] \,ds \\ < & \int ^{s_{0}}_{s_{1}}\bigl[Ax(s)+v(s)\bigr]\,ds. \end{aligned}$$
By (H6) and Definition 2.2 we have
$$\begin{aligned} \beta(s_{0})\beta(s_{1}) \leq& \int ^{s_{0}}_{s_{1}}\bigl[Ax(s)+v(s)\bigr]\,ds\leq \int^{s_{0}}_{s_{1}}\bigl[A\beta (s)+v_{2}(s) \bigr]\,ds \\ \leq& \int ^{s_{0}}_{s_{1}}\beta'(s)\,ds= \beta(s_{0})\beta(s_{1}). \end{aligned}$$
This is a contradiction. Consequently, \(x(t)\leq\beta(t)\) for all \(t\in J\).
Similarly, we can prove that \(\alpha(t)\leq x(t)\) on J. This shows that (3.9) holds.
Finally, we prove that the solution x of (3.1) satisfies (3.10). Suppose that
$$ \alpha(0)>x(0)g\bigl(\tau(0,x),\tau(b,x)\bigr). $$
(3.11)
Then by the boundary condition of (3.1) and the definition of τ we have
$$ x(0)=\alpha(0). $$
(3.12)
By (3.9) and the definition of τ we have
$$ \tau(0,x)=x(0),\qquad \tau(b,x)=x(b). $$
(3.13)
From (3.11) to (3.13) we get
$$g\bigl(\alpha(0),x(b)\bigr)=g\bigl(\tau(0,x),\tau(b,x)\bigr)>0. $$
Since g is nondecreasing in the second variable in (H4) and \(x(b)\leq \beta(b)\), we have
$$g\bigl(\alpha(0),\beta(b)\bigr)\geq g\bigl(\alpha(0),x(b)\bigr)>0, $$
which contradicts \(g(\alpha(0),\beta(b))\leq0\) in Definition 2.2. So, we have
$$ \alpha(0)\leq x(0)g\bigl(\tau(0,x),\tau(b,x)\bigr). $$
(3.14)
Analogously, we can prove that
$$ x(0)g\bigl(\tau(0,x),\tau(b,x)\bigr)\leq\beta(0). $$
(3.15)
Inequalities (3.14) and (3.15) show that (3.10) holds.
According to Steps 1 to 5, the solution x of (3.1) is also a solution of (1.1)–(1.3). The proof is complete. □
Remark 3.2
If \(g(x(0),x(b))=x(0)+x(b)\) in (1.1)–(1.3), that is, \(x(0)=x(b)\), which satisfies (H4), then (1.1)–(1.3) become an antiperiodic boundary value problem for semilinear impulsive differential inclusions.