Let \(I=[0,T]\) and \((V,H,V^{*})\) be an evolution triple of spaces. Consider a nonlocal problems of evolution inclusions in the form
$$ \begin{aligned} & \dot{x}(t)+B\bigl(t,x(t)\bigr)+Dx(t)\in\mathcal{F}\bigl(t, x(t)\bigr)\quad \mbox{a.e. for }t\in I, \\ &x(0)=\tau(x), \end{aligned} $$
(3.1)
where \(B:I\times V\rightarrow V^{*}\) is a nonlinear operator, \(D:V\rightarrow V^{*}\) is a bounded linear operator, \(\tau :C(I,H)\rightarrow H\) is a continuous map, and \(\mathcal{F}:I\times H\rightarrow2^{V^{*}}\) is a multifunction to be given later. Some classical examples for τ are

(i)
\(\tau(x)=x(T)\);

(ii)
\(\tau(x)=x(T)\);

(iii)
\(\tau(x)=\frac{1}{T}\int_{0}^{T} x(t)\,dt\);

(iv)
\(\tau(x)=\sum_{i=1}^{n}\theta_{i}x(t_{i})\), where \(0\leq t_{1}<\cdots<t_{n}\leq T\) are arbitrary, but fixed and \(\sum_{i=1}^{n}\vert \theta _{i}\vert \leq1\).
We investigate the solutions of (3.1) in the weak sense. Let u be a solution of problem (3.1), we mean a function \(u\in W_{pq}(I)\) and there exists a function \(f(t)\in\mathcal{F}(t,u)\) such that
$$\bigl\langle \dot{u}(t),v\bigr\rangle _{*}+\bigl\langle B\bigl(t,u(t)\bigr),v\bigr\rangle _{*}+\bigl\langle Du(t),v\bigr\rangle _{*}=\bigl\langle f(t),v\bigr\rangle _{*}, $$
where \(u(0)=\tau(u)\) for all \(v \in V\) and almost all \(t \in I\).
From [1], the existence of solution \(x\in W_{pq}(I)\cap C(I,H)\) of problem (3.1) is obtained. Let S denote the solution set of problem (3.1) defined as
$$S=\bigl\{ x\in W_{pq}(I)\cap C(I,H): x \mbox{ is the solution of (3.1)}\bigr\} . $$
In this paper, we can study the first structural property of the solution set \(S\subset C(I,H)\) of (3.1). We show that S is an \(R_{\delta}\) set in \(C(I,H)\).
We need the following hypotheses on the data of (3.1).

(H1)
\(B:I\times V\rightarrow{V^{*}}\) is an operator such that

(i)
\(t\rightarrow B(t,u)\) is measurable;

(ii)
for almost all \(t\in I\), there exists a constant \(C_{1}>0\) such that
$$\bigl\langle B(t,u_{1})B(t,u_{2}),u_{1}u_{2} \bigr\rangle _{*}\geq C_{1}\Vert u_{1}u_{2}\Vert _{H}^{p} $$
for all \(u_{1},u_{2}\in V\), and the map \(s\mapsto\langle B(t,u+sz),y\rangle_{*}\) is continuous on \([0,1]\) for all \(u,y,z\in V\), and \(p>1\);

(iii)
there exist a constant \(C_{2}>0\), a function \(a(\cdot)\in L^{q}_{+}(I)\) where q is the conjugate of \(p>1\), and a nondecreasing continuous function \(\alpha(\cdot)\) such that \(\Vert B(t,u)\Vert _{V^{*}}\leq a(t)+C_{2}\alpha(\Vert u\Vert _{V})\) for all \(u\in V\), a.e. for \(t\in I\);

(iv)
there exist \(C_{3}>0\), \(C_{4}>0\), \(b(\cdot)\in L^{1}(t)\) such that
$$\begin{aligned} \bigl\langle B(t,u),u\bigr\rangle _{*}\geq{}& C_{3}\Vert u\Vert _{V}^{p}C_{4}\Vert u\Vert _{V}^{p1}\\ &{}+\frac{1}{2T}\bigl\Vert u(0)\bigr\Vert ^{2}b(t) \quad \mbox{a.e. }t\in I, \forall u\in V, 1< p\leq2, \end{aligned}$$
and
$$\bigl\langle B(t,u),u\bigr\rangle _{*}\geq C_{3}\Vert u\Vert _{V}^{p}C_{4}\Vert u\Vert _{V}^{p1}b(t) \quad \mbox{a.e. }t\in I, \forall u\in V, p>2. $$

(H2)

(i)
\(D: V\rightarrow V^{*}\) is a bounded linear selfadjoint operator such that \(\langle Du,u\rangle_{*}\geq0\), for all \(u\in V\), for almost all \(t\in I\);

(ii)
there exists a continuous function \(\tau:C(I,H)\rightarrow H\) such that
$$\bigl\Vert \tau(u_{1})\tau(u_{2})\bigr\Vert _{H}\leq \Vert u_{1}u_{2}\Vert _{C(I,H)}, \quad \forall u_{1},u_{2}\in C(I,H), $$
and \(\tau(0)=0\).
For every \(f\in L^{q}(I,H)\), consider the following nonlinear evolution equation:
$$ \begin{aligned} &\dot{u}(t)+B\bigl(t,u(t)\bigr)+Du(t)=f(t), \quad \mbox{a.e. for }t\in I, \\ &u(0)=\tau(u), \end{aligned} $$
(3.2)
where \(B:I\times V\rightarrow V^{*}\) is a nonlinear map, \(D:V\rightarrow V^{*}\) is a bounded linear map, \(\tau:C(I,H)\rightarrow H\) is a continuous map. If hypotheses (H1), (H2) hold, then for every \(f\in L^{q}(I,H)\), from [1], the problem (3.2) has a unique solution \(u\in W_{pq}(I)\subset C(I,H)\), moreover, \(\Vert u\Vert _{C(I,H)}\leq M\), for some constant \(M>0\). Let us define an operator \(L: W_{pq}(I)\rightarrow L^{q}(I,H)\) as \(L(u(t))=\dot {u}(t)+B(t,u(t))+Du(t)\) and \(u(0)=\tau(u)\). By Theorem 3.1 of [1], then L is onetoone and surjective. Thus the inverse mapping \(P:L^{q}(I,H)\rightarrow W_{pq}(I)\) is well defined. From (H1)(ii) and (H2), it is easy to deduce that \(L: W_{pq}(I)\rightarrow L^{q}(I,H)\) is continuous. So \(P:L^{q}(I,H)\rightarrow W_{pq}(I)\) is continuous. Because \(W_{pq}(I)\) embeds continuously into \(C(I,H)\) (see Proposition 23.23 of [17]), we see that \(P:L^{q}(I,H)\rightarrow C(I,H)\) is continuous. Let \(L^{q}(I,H)_{w}\) denote the LebesgueBochner space furnished with the weak topology.
Let us give a proposition which is essential for our results.
Proposition 3.1
If hypotheses (H1) and (H2) hold, then
\(P:L^{q}(I,H)_{w}\rightarrow C(I,H)\)
is sequentially continuous.
Proof
Let
$$W=\bigl\{ v\in L^{q}(I,H):\bigl\Vert v(t)\bigr\Vert _{H} \leq\psi(t) \mbox{ a.e. for }t\in I\bigr\} , $$
with \(\psi(t)\in L^{q}_{+}(I)\). Similar to the step 2 of Theorem 3.1 in [1], we will show that there is a prior bound to problem (3.2) for each \(f\in W\). Let \(f_{n}\in W\) for every \(n\geq1\). From the definition of W, \(\{f_{n}\}_{n\geq1}\) is uniformly bounded in \(L^{q}(I,H)\). By the DunfordPettis theorem, passing to a subsequence if necessary, we may assume that \(f_{n}\longrightarrow f\) weakly in \(L^{q}(I,H)\) for some \(f\in W\) as \(n\longrightarrow\infty\). For the possible solution \(u_{n}=P(f_{n})\), we have
$$\dot{u}_{n}(t)+B\bigl(t,u_{n}(t) \bigr)+Du_{n}(t)=f_{n}(t) \quad \mbox{a.e. for }t\in I. $$
It follows that
$$\langle\dot{u}_{n},u_{n}\rangle_{**}+ \bigl\langle B(t,u_{n}),u_{n}\bigr\rangle _{**}+ \langle Du_{n},u_{n}\rangle_{**}=\langle f_{n},u_{n}\rangle_{**}. $$
From (H1)(iv) and
$$\begin{aligned}& \int_{0}^{T}\bigl\langle \dot{u}_{n}(t),u_{n}(t) \bigr\rangle \,dt=\frac{1}{2}\bigl\Vert u_{n}(T)\bigr\Vert _{H}^{2}\frac{1}{2}\bigl\Vert u_{n}(0) \bigr\Vert _{H}^{2}, \\& \langle f_{n},u_{n}\rangle_{**}\leq \Vert f_{n}\Vert _{L^{q}(I,V^{*})}\Vert u_{n}\Vert _{L^{p}(I,V)}, \end{aligned}$$
we derive
$$C_{3}\Vert u_{n}\Vert _{L^{p}(I,V)}^{p}\leq C_{4}\Vert u_{n}\Vert _{L^{p}(I,V)}^{p1}+ \Vert f_{n}\Vert _{L^{q}(I,V^{*})}\Vert u_{n}\Vert _{L^{p}(I,V)}+\Vert b\Vert _{L^{1}}, $$
with \(p>2\), and
$$C_{3}\Vert u_{n}\Vert _{L^{p}(I,V)}^{p}\leq C_{4}\Vert u_{n}\Vert _{L^{p}(I,V)}^{p1}+ \Vert f_{n}\Vert _{L^{q}(I,V^{*})}\Vert u_{n}\Vert _{L^{p}(I,V)}+\Vert b\Vert _{L^{1}}+\Vert u_{n}\Vert ^{2}_{L^{p}(I,V)}, $$
with \(1< p\leq2\). Hence, we have, for all \(n\in\mathbb{N}\), \(\Vert u_{n}\Vert _{L^{p}(I,V)}\leq M\) where \(M>0\). From (3.2), it is easy to deduce that, for all \(n\in \mathbb{N}\), \(\Vert u_{n}\Vert _{L^{q}(I,V^{*})}\leq\overline{M}\) where \(\overline{M}>0\). So we see that \(\{u_{n}\}_{n\geq1}\) is uniformly bounded in \(W_{pq}(I)\). Due to the boundedness of the sequence \(\{u_{n}\}_{n\geq1}\subset W_{pq}(I)\), it follows that the sequence \(\{ u_{n}\}_{n\geq1}\subset L^{q}(I,V^{*})\) is uniformly bounded. So passing to a subsequence if necessary, we may assume that \(\dot {u}_{n}\longrightarrow v\) weakly in \(L^{q}(I,V^{*})\). Obviously, \(v=\dot{u}\) and \(u_{n}\longrightarrow u\) weakly in \(W_{pq}(I)\). Since the embedding \(V \rightarrow H\) is compact, the embedding \(W_{pq}(I)\rightarrow L^{p}(I,H)\) is compact. Thus, we have \(u_{n}\longrightarrow u\) in \(L^{p}(I,H)\). Since the operator B is hemicontinuous and monotone and D is a continuous linear operator, we have \(B(t,x_{n})\longrightarrow B(t,x)\), \(Dx_{n}\longrightarrow Dx\) weakly in \(L^{q}(I,V^{*})\) as \(n\longrightarrow\infty\). Therefore, we obtain \(\dot{u}(t)+B(t,u(t))+Du(t) = f(t)\). Since the embedding \(W_{pq}(I)\rightarrow C(I,H)\) is continuous, the sequence \(\{u_{n}\} _{n\geq1}\subset C(I,H)\) is uniformly bounded. Also, \(\{u_{n}\}_{n\geq1}\) is equicontinuous. By the ArzelaAscoli theorem, passing to a subsequence, we may assume \(u_{n}\longrightarrow u\) in \(C(I,H)\). Since \(\tau: C(I,H)\rightarrow H\) is continuous, we have \(u_{n}(0)=\tau(u_{n})\longrightarrow u(0)=\tau(u)\). Hence, \(u=P(f)\). Because of \(W_{pq}(I)\subset C(I,H)\), one has \(u_{n}u\in C(I,H)\). Note that
$$\bigl\langle \dot{u}_{n}(t)\dot{u}(t),u_{n}(t)u(t) \bigr\rangle \leq\bigl\langle f_{n}(t)f(t),u_{n}(t)u(t)\bigr\rangle . $$
Integrating the above inequality over \([0,t]\), by hypotheses (H1), (H2), for arbitrary \(t\leq T\) we find
$$\bigl\Vert u_{n}(t)u(t)\bigr\Vert _{H}^{2} \leq2 \int_{0}^{t}\bigl\langle f_{n}(s)f(s),u_{n}(s)u(s) \bigr\rangle \,ds+\bigl\Vert u_{n}(0)u(0)\bigr\Vert _{H}^{2}. $$
By
$$\begin{aligned}& \int_{0}^{t}\bigl\langle f_{n}(s)f(s),u_{n}(s)u(s) \bigr\rangle \,ds \\& \quad \leq2 \int_{0}^{T}\bigl\Vert f_{n}(t)\bigr\Vert _{H}\bigl\Vert u_{n}(t)u(t)\bigr\Vert _{H}\,dt+2 \int_{0}^{T}\bigl\Vert f(t)\bigr\Vert _{H}\bigl\Vert u_{n}(t)u(t)\bigr\Vert _{H} \,dt \\& \quad \leq4 \int_{0}^{T}\bigl\vert \psi(t)\bigr\vert \bigl\Vert u_{n}(t)u(t)\bigr\Vert _{H}\,dt, \end{aligned}$$
it follows that
$$\begin{aligned} \bigl\Vert u_{n}(t)u(t)\bigr\Vert _{H}^{2} &\leq 4 \int_{0}^{T}\bigl\vert \psi(t)\bigr\vert \bigl\Vert u_{n}(t)u(t)\bigr\Vert _{H}\,dt+\bigl\Vert u_{n}(0)u(0)\bigr\Vert _{H}^{2} \\ &\leq 4 \int_{0}^{T}\bigl\vert \psi(t)\bigr\vert \bigl\Vert u_{n}(t)u(t)\bigr\Vert _{H}\,dt+\bigl\Vert \tau(u_{n})\tau (u)\bigr\Vert _{H}^{2} \\ &\leq 4\Vert \psi \Vert _{L^{q}}\Vert u_{n}u \Vert _{L^{p}(I,H)}+\bigl\Vert \tau(u_{n})\tau (u)\bigr\Vert _{H}^{2} \\ &\longrightarrow 0 \end{aligned}$$
(3.3)
as \(n\longrightarrow\infty\).
Then we get
$$\max_{t\in I}\bigl\Vert u_{n}(t)u(t)\bigr\Vert _{H}\longrightarrow0 \quad \mbox{as } n\longrightarrow\infty. $$
So, \(u_{n}\longrightarrow u\) in \(C(I,H)\), i.e.
\(P:L^{q}(I,H)_{w}\longrightarrow C(I,H)\) is sequentially continuous. □
To prove such a structural result for the solution set of (3.1) we need the following hypotheses on \(\mathcal{F}(t,x)\):

(H3)
\(\mathcal{F}:I\times H\rightarrow2^{H}\) is a multifunction with closed and convex values such that

(i)
\((t,u)\rightarrow\mathcal{F}(t,u)\) is graph measurable;

(ii)
for almost all \(t\in I\), \(u\rightarrow\mathcal{F}(t,u)\) has a closed graph;

(iii)
there exist a function \(b_{1}(\cdot)\in L_{+}^{q}(I)\) and a constant \(C_{5}>0\) such that
$$\begin{aligned} \bigl\vert \mathcal{F}(t,u)\bigr\vert &=\sup\bigl\{ \Vert f\Vert _{H}:f\in\mathcal{F}(t,u)\bigr\} \\ &\leq b_{1}(t)+C_{5} \Vert u\Vert _{H}^{k1},\quad \forall x\in H,\mbox{ a.e. for } t \in I, \end{aligned}$$
where \(1\leq k< p\).
Theorem 3.1
Under assumptions (H1)(H3), S
is an
\(R_{\delta}\)
set in
\(C(I,H)\).
Proof
From the prior estimation executed in the proof of Theorem 3.1 in [1], without loss of generality, we may suppose that, for almost all \(t\in I\), all \(u\in H\), and all \(h(t)\in\mathcal{F}(t, u)\), we have \(\Vert h(t)\Vert _{H}\leq\psi(t)\) where \(\psi\in L_{+}^{q}(I)\). Applying Lemma 2.1, we can obtain a sequence of multifunctions \(\mathcal{F}_{n} : I\times H \rightarrow P_{fc}(H)\). For every \(n\geq1\), consider the following evolution inclusion:
$$ \begin{aligned} &\dot{u}(t)+B\bigl(t,u(t)\bigr)+Du(t)\in\mathcal{F}_{n}\bigl(t, u(t)\bigr), \quad \mbox{a.e. for }t\in I, \\ &u(0)=\tau(u). \end{aligned} $$
(3.4)
From Theorem 4.1 of [1], we see that the solution set \(S_{n}\subseteq W_{pq}(I)\subset C(I,H)\) of problem (3.4) is compact in \(C(I,H)\). The following two steps are needed to complete the proof.
Step 1. This set \(S_{n}\) is contractible.
Let \(w_{n}(t, u)\) be measurable selector of \(\mathcal{F}_{n}(t,u)\) generated by Lemma 2.1, the locally Lipschitz continuous in \(u\in H\). Fixing \(w_{n}(t,u)\in\mathcal{F}_{n}(t,u)\), let \(\widehat{u}=P(w_{n}(t,u))\in S_{n}\), where P is defined as before. Consider the following Cauchy problem:
$$ \begin{aligned} & \dot{z}(t)+B\bigl(t,z(t)\bigr)+Dz(t)=w_{n}(t, z)\quad \mbox{a.e. for }t\in [\delta T, T], \\ &z(\delta T)=\delta u(\delta T)+(1\delta)\widehat{u}(0). \end{aligned} $$
(3.5)
It is easy to check the existence of solution in (3.5) (see [20]). Because B is hemicontinuous and monotone, D is a continuous linear operator and \(w_{n}(t, u)\) be the locally Lipschitz continuous in \(u\in H\), the solution of (3.5) is unique. For every \(\delta\in[0,1]\) and every \(u\in S_{n}\), let \(z(\delta,u)(t)\in W_{pq}(I)\) be the unique solution of (3.5). In order for this set \(S_{n}\) to be contractible, for given \(\widehat {u}\in S_{n}\), we only need to find a continuous function \(\mu :[0,1]\times S_{n}\rightarrow S_{n}\) such that, for all \(u\in S_{n}\), we have \(\mu(0,u)(t)=\widehat{u}(t)\) and \(\mu(1,u)(t)=u(t)\). So we can define \(\mu:[0,1]\times S_{n}\rightarrow S_{n}\) by
$$\mu(\delta,u) (t)=\left ( \begin{matrix} \delta u(t)+(1\delta)\widehat{u}(0) &\quad \mbox{for } t\in[0,\delta T],\\ z(\delta,u)(t) &\quad \mbox{for } t\in(\delta T,T]. \end{matrix} \right ) $$
If \(\delta=0\), by (H1) and (H2), we know that \(z(t)=\widehat{u}(t)\) for every \(t\in I\). So \(\mu(0,u)(t)=\widehat{u}\). From the definition of \(\mu(\delta,u)(t)\), obviously, \(\mu(1,u)(t)=u(t)\) for every \(u(t)\in S_{n}\). We show that \(\mu(\delta,u)(t)\in S_{n}\) for each \((\delta, u)\in [0,1]\times S_{n}\). Note that for each \(u\in S_{n}\), there exists \(\widetilde{f}\in\mathcal{F}_{n}(t,u)\), such that \(u=P(\widetilde{f})\). Let
$$v(t)=\widetilde{f}\chi_{[0,\delta T]}(t)+w_{n}\chi_{(\delta T,T]}(t) \quad \mbox{for } t\in I, $$
where χ is the characteristic function. Since \(\mathcal{F}_{n}(t,u)\) is closed and convexvalued, from the definition of a decomposable value (see Section 2), we see that \(\mathcal{F}_{n}(t,u)\) is decomposable, from which one deduces \(v\in\mathcal{F}_{n}(t,u)\). Obviously, \(P(v)=u(t)\) for all \(t\in[0,\delta T]\) and \(P(v)=z(\delta,u)(t)\) for all \(t\in(\delta T, T]\), which means \(P(v)=\mu(\delta,u)\) for any \(\delta\in[0,1]\). Hence, \(\mu(\delta,u)\in S_{n}\).
To prove that \(S_{n}\) is contractible, we first note that \(\mu :[0,1]\times S_{n}\rightarrow S_{n}\) and \(\mu(0,u)(t)=\widehat{u}(t)\) and \(\mu(1,u)(t)=u(t)\) for every \(u\in S_{n}\). Next, it remains to show that \(\mu(\delta,u)(t)\) is continuous in \([0,1]\times C(I,H)\). To this aim, let \((\delta_{m},u_{m})\longrightarrow(\delta,u)\) in \([0,1]\times S_{n}\). Next, we will distinguish two distinct cases to proceed.
I:
\(\delta_{m}\geq\delta\)
for every
\(m\geq1\). Let \(v_{m}(t)=\mu(\delta_{m},u_{m})(t)\), for \(t\in I\). Obviously, \(v_{m}\in S_{n}\), \(m\geq1\). Due to the compactness of the solution set \(S_{n}\) in \(C(I,H)\), by passing to a subsequence if necessary, we may suppose that \(v_{m}\longrightarrow v\) in \(C(I,H)\) as \(m\longrightarrow\infty\). Next, we only need to check \(v(t)=\mu(\delta,u)(t)\). Clearly, \([0,\delta T]\subseteq[0,\delta_{m}T]\), by the definition of \(\mu(\delta,u)\), one has \(v_{m}(t)=\delta_{m} u_{m}(t)+(1\delta_{m})\widehat{u}(0)\) for \(0\leq t\leq \delta_{m} T\). As \((\delta_{m},u_{m})\rightarrow(\delta,u)\) in \([0,1]\times C(I,H)\), we obtain \(v(t)=\delta u(t)+(1\delta)\widehat{u}(0)\) for \(0\leq t\leq\delta T\). Also let \(y\in W_{pq}(I)\) be the unique solution of the following equation:
$$ \begin{aligned} & \dot{y}(t)+B\bigl(t,y(t)\bigr)+Dy(t)= w_{n}(t, v)\quad \mbox{a.e. for }t\in [\delta T, T], \\ &y(\delta T)=v(\delta T). \end{aligned} $$
(3.6)
Let \(m \geq N\) be large enough for some constant \(N > 0\), by (3.6), \(v_{m}(\cdot)\) satisfies
$$\begin{aligned} \dot{v}_{m}(t)+B\bigl(t,v_{m}(t)\bigr)+Dv_{m}(t)= w_{n}\bigl(t, v_{m}(t)\bigr)\quad \mbox{a.e. for } t\in [ \delta_{m} T, T]. \end{aligned}$$
(3.7)
Exploiting the monotonicity of the operator \(B(t,\cdot)\) and D, and combining (3.6) with (3.7), we find
$$\begin{aligned} &\bigl\langle \dot{y}(t)\dot{v}_{m}(t),y(t)v_{m}(t) \bigr\rangle \\ &\quad \leq \bigl\langle w_{n}\bigl(t,v(t)\bigr)w_{n} \bigl(t,v_{m}(t)\bigr),y(t)v_{m}(t)\bigr\rangle \quad \mbox{a.e. for } t\in [\delta_{N} T,T]. \end{aligned}$$
Taking the integral over the above inequality from \(\delta_{N}T\) to t (\(t\leq T\)), we have
$$\begin{aligned}& \bigl\Vert y(t)v_{m}(t)\bigr\Vert ^{2}_{H} \\& \quad \leq\bigl\Vert y(\delta_{N} T)v_{m}( \delta_{N} T)\bigr\Vert _{H}^{2}+2 \int_{\delta_{N} T}^{t}\bigl\Vert w_{n}\bigl(s,v(s) \bigr)w_{n}\bigl(s,v_{m}(s)\bigr)\bigr\Vert _{H}\bigl\Vert y(s)v_{m}(s)\bigr\Vert _{H} \,ds. \end{aligned}$$
By Brezis [21], p.157, we obtain
$$\bigl\Vert y(t)v_{m}(t)\bigr\Vert _{H}\leq\bigl\Vert y(\delta_{N} T)v_{m}(\delta_{N} T)\bigr\Vert _{H}+ \int_{\delta_{N} T}^{t}\bigl\Vert u_{n}\bigl(s,v(s) \bigr)u_{n}\bigl(s,v_{m}(s)\bigr)\bigr\Vert _{H}\,ds. $$
Passing to the limit as \(m\longrightarrow\infty\), and recalling that \(v_{m}\longrightarrow v\) in \(C(I,H)\) and \(w_{n}(t, u)\) is locally Lipschitz continuous in u, we get \(\Vert y(t)v(t)\Vert _{H}\leq \Vert y(\delta_{N} T)v(\delta _{N} T)\Vert _{H}\) for \(\delta_{N} T\leq t\leq T\). Since \(y(t)\) is the solution in (3.6), \(y\in C(I,H)\). So \(y(t)\) is continuous with respect to t. Finally, \(y(\delta_{N} T)\rightarrow v(\delta T)\) and \(v(\delta_{N} T)\rightarrow v(\delta T)\) in H as \(N\longrightarrow\infty\), in the limit we get \(y(t)=v(t)\) for every \(\delta T\leq t\leq T\). So
$$\begin{aligned} & \dot{v}(t)+B\bigl(t,v(t)\bigr)+Dv(t)=w_{n} \bigl(t, v(t)\bigr) \quad \mbox{a.e. for } t\in [\delta T, T], \\ & v(\delta T)=\delta u(\delta T)+(1\delta)\widehat{u}(0), \end{aligned} $$
hence \(v=\mu(\delta,u)\). Therefore \(\mu(\delta_{m},u_{m})\longrightarrow\mu (\delta,u)\) in \(C(I,H)\) as \((\delta_{m},u_{m})\longrightarrow(\delta,u)\).
II:
\(\delta_{m}\leq\delta\)
for every
\(m\geq1\). Keeping the notation as in case I, we have \([0,\delta_{m}T]\subseteq[0,\delta T]\). By the definition of \(\mu(\delta,u)\), one has \(v_{m}(t)=\delta_{m} u_{m}(t)+(1\delta_{m})\widehat{u}(0)\) for \(0\leq t\leq \delta_{m} T\). As \((\delta_{m},u_{m})\longrightarrow(\delta,u)\) in \([0,1]\times C(I,H)\), we get \(v(t)=\delta u(t)+(1\delta)\widehat {u}(0)\) for \(0\leq t\leq\delta T\). Also let \(y\in W_{pq}(I)\) be the unique solution of the following equation:
$$ \begin{aligned} & \dot{y}(t)+B\bigl(t,y(t)\bigr)+Dy(t)= w_{n}(t, v)\quad \mbox{a.e. for } t\in [\delta T, T], \\ &y(\delta T)=v(\delta T). \end{aligned} $$
(3.8)
Let \(m\geq N\) large enough for some constant \(N>0\), by (3.8), \(v_{m}(\cdot )\) satisfies
$$\begin{aligned} \dot{v}_{m}(t)+B\bigl(t,v_{m}(t)\bigr)+Dv_{m}(t)= w_{n}\bigl(t, v_{m}(t)\bigr)\quad \mbox{a.e. for } t\in [ \delta_{N} T, T]. \end{aligned}$$
(3.9)
Due to the monotonicity of the operator \(B(t,\cdot)\) and D, and combining (3.8) with (3.9), we have
$$\begin{aligned} &\bigl\langle \dot{y}(t)\dot{v}_{m}(t),y(t)v_{m}(t) \bigr\rangle \\ &\quad \leq \bigl\langle w_{n}\bigl(t,v(t)\bigr)w_{n} \bigl(t,v_{m}(t)\bigr),y(t)v_{m}(t)\bigr\rangle \quad \mbox{a.e. for } t\in[\delta T,T]. \end{aligned}$$
Integrating the above inequality over \([\delta T,t)\), \(t\leq T\), we have
$$\begin{aligned}& \bigl\Vert y(t)v_{m}(t)\bigr\Vert ^{2}_{H} \\& \quad \leq\bigl\Vert y(\delta T)v_{m}(\delta T)\bigr\Vert _{H}^{2}+2 \int_{\delta T}^{t}\bigl\Vert w_{n}\bigl(s,v(s) \bigr)w_{n}\bigl(s,v_{m}(s)\bigr)\bigr\Vert _{H}\bigl\Vert y(s)v_{m}(s)\bigr\Vert _{H} \,ds. \end{aligned}$$
As before, we obtain
$$\bigl\Vert y(t)v_{m}(t)\bigr\Vert _{H}\leq\bigl\Vert y(\delta T)v_{m}(\delta T)\bigr\Vert _{H}+ \int_{\delta T}^{t}\bigl\Vert w_{n}\bigl(s,v(s) \bigr)w_{n}\bigl(s,v_{m}(s)\bigr)\bigr\Vert _{H}\,ds. $$
Taking the limit as \(m\longrightarrow\infty\), and recalling that \(v_{m}\longrightarrow v\) in \(C(I,H)\) and \(w_{n}(t, x)\) is the locally Lipschitz continuous in \(x\in H\), we get \(\Vert y(t)v(t)\Vert _{H}\leq \Vert y(\delta T)v(\delta T)\Vert _{H}\) for \(\delta T\leq t\leq T\). Since \(y(t)\) is the solution in (3.8), \(y(\delta T)= v(\delta T)\). Finally, in the limit we get \(y(t)=v(t)\) for every \(\delta T\leq t\leq T\). So
$$\begin{aligned}& \dot{v}(t)+B\bigl(t,v(t)\bigr)+Dv(t)=w_{n} \bigl(t, v(t)\bigr)\quad \mbox{a.e. for } t\in [\delta T, T], \\& v(\delta T)=\delta u(\delta T)+(1\delta)\widehat{u}(0), \end{aligned}$$
hence \(v=\mu(\delta,u)\). Therefore \(\mu(\delta_{m},u_{m})\longrightarrow\mu (\delta,u)\) in \(C(I,H)\) as \((\delta_{m},u_{m})\longrightarrow(\delta,u)\).
In fact, we can always get a subsequence of \(\{\delta_{m}\}_{m\geq1}\) conforming to I or II. Thus we have established the continuity of \(\mu(\delta,u)\). Therefore, for every \(n\geq1\), \(S_{n}\subseteq C(I,H) \) is compact and contractible.
Step 2. \(S=\bigcap_{n\geq1}S_{n}\).
Evidently, \(S\subseteq\bigcap_{n\geq1}S_{n}\). Let \(u\in\bigcap_{n\geq1}S_{n}\). Then through definition \(u=P(f_{n})\), where \(f_{n}\in S^{q}_{\mathcal{F}_{n}(\cdot,u_{n}(\cdot))}\) (the set of all \(L^{q}(I,H)\)selectors of \(\mathcal{F}_{n}\)), \(n\geq1\). By passing to a subsequence if necessary, we may assume that \(f_{n}\longrightarrow f\) weakly in \(L^{q}(I,H)\). Then \(f\in S^{q}_{\mathcal{F}(\cdot,u(\cdot))}\) (see Theorem 3.2 of [1]). So \(u=P(f)\) with \(f\in S^{q}_{\mathcal{F}(\cdot,u(\cdot))}\) from which we conclude that \(S=\bigcap_{n\geq1}S_{n}\). Finally, by Steps 1 and 2, Hyman’s result [22] implies that S is an \(R_{\delta}\) set in \(C(I,H)\). □
An immediate conclusion of Theorem 3.1 is the following result for the multivalued problem (3.1).
Remark 3.1
Assume (H1)(H3), then for every \(t\in I\), \(S(t)=\{x(t)x\in S\}\) (the reachable set at time \(t\in I\)) is compact and connected in H.
We can establish an analogous result for the topological structure of a nonconvex evolution inclusion (i.e., \(\mathcal{F}(t,u)\) has nonconvex values) provided we strengthen our assumption on the continuity of \(\mathcal{F}(t,u)\). In fact, in this case we can see that the solution set is path connected.
To prove such a structural result for the solution set of (3.1) we need the following hypotheses on \(\mathcal{F}(t,x)\) and τ:
 (H4):

\(\mathcal{F}:I\times H\rightarrow P_{f}(H)\) is a multifunction such that

(i)
\(t\rightarrow\mathcal{F}(t,u)\) is measurable;

(ii)
for every \(u_{1},u_{2}\in H\), \(h(\mathcal{F}(t,u_{1}),\mathcal{F}(t,u_{2}))\leq z(t)\Vert u_{1}u_{2}\Vert _{H}\) a.e. with \(z(t)\in L_{+}^{q}(I)\);

(iii)
there exist a function \(b_{2}(\cdot)\in L_{+}^{q}(I)\) and a constant \(C_{5}>0\) such that
$$\bigl\vert \mathcal{F}(t,u)\bigr\vert =\sup\bigl\{ \Vert f\Vert _{H}:f\in\mathcal{F}(t,u)\bigr\} \leq b_{2}(t)+C_{5} \Vert u\Vert _{H}^{k1} \quad \forall u\in V\mbox{ a.e. for } t\in I $$
with \(1\leq k< p\).
 (H2)_{1}
:


(i)
\(D: V\rightarrow V^{*}\) is a bounded linear selfadjoint operator such that \(\langle Du,u\rangle_{*}\geq0\), for all \(u\in V\), a.e. for \(t\in I\);

(ii)
there exists a continuous function \(\tau:C(I,H)\rightarrow H\) such that
$$\bigl\Vert \tau(u_{1})\tau(u_{2})\bigr\Vert _{H}\leq\mathcal{L}\Vert u_{1}u_{2}\Vert _{C(I,H)}, \quad \forall u_{1},u_{2}\in C(I,H), $$
where \(\mathcal{L}\in[0,1)\), \(\tau(0)=0\).
Theorem 3.2
If hypotheses (H1), (H2)_{1}, and (H4) are satisfied, then
\(S\subseteq C(I, H)\)
is nonempty and path connected.
Proof
The prior estimation conducted in the three part of the proof of Theorem 3.1 in [1] is still valid here. So according to it, we see that without any loss of generality we may suppose that \(\vert \mathcal{F}(t,u)\vert \leq\psi(t)\) a.e. with \(\psi(t)\in L_{+}^{q}(I)\). Let \(W_{\alpha}=\{g\in L^{q}(I,H):\Vert g(t)\Vert _{H}\leq\psi (t)\mbox{ a.e. for }t\in I\}\), and the multifunction \(N:W_{\alpha}\rightarrow P_{f}(L^{q}(I,H))\) be defined by \(N(g)=S^{q}_{\mathcal{F}(\cdot,P(g)(\cdot))}\) (here \(P(g)\) is defined as before). Next, we present a new norm on \(L^{q}(I,H)\) defined by
$$\begin{aligned} \Vert g\Vert _{q}=\biggl( \int_{0}^{T}\exp^{r\int_{0}^{t}z^{q}(s)\,ds}\bigl\Vert g(t) \bigr\Vert _{H}^{q} \,dt\biggr)^{\frac{1}{q}}, \end{aligned}$$
where r is a number satisfying \(r>T^{\frac{q}{p}}\), \(p>1\), and q is the conjugate of p. It is easy to check that the norm \(\Vert \cdot \Vert _{q}\) is the equivalent of \(\Vert \cdot \Vert _{L^{q}(I,H)}\). Denote by \(d_{q}(\cdot,\cdot )\) and \(d_{\widehat{H}}(\cdot,\cdot)\) the distance function and Hausdorff metric, respectively, generated by \(\Vert \cdot \Vert _{q}\). Let \(f,g\in W_{\alpha}\) and \(v\in N(g)\). Let \(\epsilon>0\) and define
$$D_{\epsilon}(t)=\bigl\{ w\in\mathcal{F}\bigl(t,P(f)\bigr):\bigl\Vert v(t)w(t) \bigr\Vert _{H}\leq d\bigl(v(t),F\bigl(t,P(f) (t)\bigr)\bigr)+ \epsilon\bigr\} . $$
Let \(b(t,w)=d(v(t),\mathcal{F}(t,P(f)))\Vert vw\Vert _{H}+\epsilon\). Obviously, for every \(t\in I\), \(D_{\epsilon}(t)\neq\emptyset\) and \(\operatorname {\mathbf {Gr}}D_{\epsilon}=\{(t,w)\in \operatorname {\mathbf {Gr}}\mathcal{F}(t,P(f)(t)): b(t,w)\geq0\}\). In fact, because of (H4)(i) and (ii), \((t,u)\rightarrow\mathcal{F}(t,u)\) is measurable, thus \(t\rightarrow\mathcal{F}(t,P(f)(t))\) is measurable. So, \((t,w)\rightarrow b(t,w)\) is a Carathéodory function, thus jointly measurable. Hence \(\operatorname {\mathbf {Gr}}D_{\epsilon}\in{\mathcal{B}}(I)\times {\mathcal{B}}(H)\). Apply Aumann’s selection theorem to get \(w: I\rightarrow H\) measurable such that \(w(t)\in D_{\epsilon}(t)\) a.e. for \(t\in I\). Then we find
$$\begin{aligned} d_{q}\bigl(v,N(f)\bigr) \leq& \Vert vw \Vert _{q} \\ =& \biggl( \int_{0}^{T}\Vert vw\Vert _{H}^{q} \exp^{r\int_{0}^{t}z^{q}(s)\,ds}\,dt \biggr)^{\frac{1}{q}} \\ \leq&\biggl( \int_{0}^{T}\bigl[d\bigl(v(t),\mathcal{F}\bigl(t,P(f) (t)\bigr)\bigr)\bigr]^{q}\exp^{r\int _{0}^{t}z^{q}(s)\,ds}\,dt\biggr)^{\frac{1}{q}}+(T)^{\frac{1}{q}}\epsilon \\ \leq&\biggl( \int_{0}^{T}\bigl[h\bigl(\mathcal{F}\bigl(t,P(g) (t) \bigr),\mathcal{F}\bigl(t,P(f) (t)\bigr)\bigr)\bigr]^{q}\exp^{r\int_{0}^{t}z^{q}(s)\,ds} \,dt\biggr)^{\frac{1}{q}}+(T)^{\frac {1}{q}}\epsilon \\ \leq&\biggl( \int_{0}^{T}z^{q}(t)\bigl\Vert P(f) (t)P(g) (t)\bigr\Vert _{H}^{q}\exp^{r\int _{0}^{t}z^{q}(s)\,ds}\,dt \biggr)^{\frac{1}{q}}+(T)^{\frac{1}{q}}\epsilon. \end{aligned}$$
(3.10)
For every \(y\in C(I,H)\), consider the following differential inclusion:
$$ \begin{aligned} & \dot{x}(t)+B\bigl(t,x(t)\bigr)+Dx(t)=f(t),\quad \mbox{a.e. for }t\in I, \\ &x(0)=\tau(y), \end{aligned} $$
(3.11)
where \(f(t)\in\mathcal{F}(t, x(t))\). Let \(S(y)\) denote the solution set of (3.11), by Theorem 3.1 in [1], we have \(S(y)\neq\emptyset\). Since \(B(t,\cdot)\) is monotone, D is a continuous linear map, and \(P(f)(0)=P(g)(0)\) (\(P(f),P(g)\in S(y) \)), we have
$$\begin{aligned} \frac{1}{2}\bigl\Vert P(f) (t)P(g) (t)\bigr\Vert ^{2}_{H}\leq \int_{0}^{t}\bigl\Vert P(f)P(g)\bigr\Vert _{H}\bigl\Vert f(s)g(s)\bigr\Vert _{H}\,ds. \end{aligned}$$
(3.12)
Hence, we obtain \(\Vert P(f)(t)P(g)(t)\Vert _{H}\leq\int_{0}^{t}\Vert f(s)g(s)\Vert _{H}\,ds\) for all \(t\in I\) (see Brezis [21], p.157), then \(\Vert P(f)(t)P(g)(t)\Vert _{H}^{q}\leq T^{\frac {q}{p}}\int_{0}^{t}\Vert f(s)g(s)\Vert _{H}^{q}\,ds\). From (3.10), it follows that
$$\begin{aligned} d_{q}\bigl(v,N(f)\bigr) \leq& \biggl(T^{\frac{q}{p}} \int_{0}^{T}z^{q}(t) \int_{0}^{t}\bigl\Vert f(s)g(s)\bigr\Vert _{H}^{q}\,ds\exp^{r\int_{0}^{t}z^{q}(s)\,ds}\,dt\biggr)^{\frac{1}{q}}+(T)^{\frac {1}{q}}\epsilon \\ \leq&\biggl(\frac{1}{r}T^{\frac{q}{p}} \int_{0}^{T} \int_{0}^{t}\bigl\Vert f(s)g(s)\bigr\Vert _{H}^{q}\,ds\frac{d}{dt} \bigl(\exp^{r\int_{0}^{t}z^{q}(s)\,ds} \bigr)\,dt\biggr)^{\frac {1}{q}}+(T)^{\frac{1}{q}}\epsilon \\ \leq&\biggl(\frac{1}{r}T^{\frac{q}{p}} \int_{0}^{T}\exp^{r\int _{0}^{t}z^{q}(s)\,ds}\bigl\Vert f(t)g(t)\bigr\Vert _{H}^{q}\,dt\biggr)^{\frac{1}{q}}+(T)^{\frac{1}{q}}\epsilon \\ \leq&\biggl(\frac{1}{r}\biggr)^{\frac{1}{q}}T^{\frac{1}{p}}\bigl\Vert f(t)g(t)\bigr\Vert _{q}+(T)^{\frac {1}{q}} \epsilon. \end{aligned}$$
(3.13)
Let \(\epsilon\rightarrow0\) to get \(d_{q}(v,N(f))\leq(\frac{1}{r})^{\frac{1}{q}} T^{\frac{1}{p}}\Vert f(t)g(t)\Vert _{q}\). Exchanging the roles of f and g we also get \(d_{q}(u,N(g))\leq(\frac{1}{r})^{\frac{1}{q}}T^{\frac{1}{p}}\Vert f(t)g(t)\Vert _{q}\). Thus, finally, \(d_{\widehat{H}}(N(f)(t),N(g)(t))\leq (\frac{1}{r})^{\frac{1}{q}}T^{\frac{1}{p}}\Vert f(t)g(t)\Vert _{q}\) where \((\frac {1}{r})^{\frac{1}{q}}T^{\frac{1}{p}}<1\). Set \(\Theta=\{g\in W_{\alpha}:g\in N(g)\}\). From Nadler’s fixedpoint theorem [23] we get \(\Theta\neq\emptyset\), and then from Theorem 1 in [24] we see that Θ is an absolute retract in \(L^{q}(I,H)\). Since an absolute retract is path connected, we see that Θ is path connected. Then \(P(\Theta)\) is path connected in \(C(I,H)\). However, note that \(P(\Theta)=S(y)\), to conclude that \(S(y)\) is nonempty and path connected in \(C(I,H)\). For every \(y_{1}, y_{2}\in C(I,H)\), we see that there exist \(x_{1}(t)\in S(y_{1})\), \(x_{2}(t)\in S(y_{2})\) such that
$$\begin{aligned} d_{H}\bigl(S(y_{1}),S(y_{2})\bigr)= \bigl\Vert x_{1}(t)x_{2}(t)\bigr\Vert _{C(I,H)}. \end{aligned}$$
(3.14)
Note that
$$x_{i}(t)= \int_{0}^{t}\dot{x}_{i}(s)\,ds+\tau \bigl(y_{i}(t)\bigr), $$
for \(i=1,2\). By (3.11), one has
$$\begin{aligned}& x_{1}(t)x_{2}(t)+ \int_{0}^{t}\bigl(B\bigl(s,x_{1}(s) \bigr)B\bigl(s,x_{2}(s)\bigr)\bigr)\,ds+ \int _{0}^{t}\bigl(Dx_{1}(s)Dx_{2}(s) \bigr)\,ds \\& \quad =\tau(y_{1})\tau(y_{2}) \end{aligned}$$
(3.15)
for every \(y_{1}, y_{2}\in C(I,H)\). Taking the inner product over (3.15) with \(x_{1}(t)x_{2}(t)\), by (H1)(ii) and (H2)_{1}(i), then
$$\begin{aligned} \bigl\langle x_{1}(t)x_{2}(t),x_{1}(t)x_{2}(t) \bigr\rangle \leq\bigl\langle \tau \bigl(y_{1}(t)\bigr)\tau \bigl(y_{2}(t)\bigr),x_{1}(t)x_{2}(t)\bigr\rangle . \end{aligned}$$
It follows that
$$\begin{aligned} \Vert x_{1}x_{2}\Vert _{C(I,H)}\leq\bigl\Vert \tau(y_{1})\tau(y_{2})\bigr\Vert _{H}. \end{aligned}$$
By (H2)_{1}(ii), we have
$$\begin{aligned} \Vert x_{1}x_{2}\Vert _{C(I,H)}\leq\mathcal{L}\Vert y_{1}y_{2}\Vert _{C(I,H)}. \end{aligned}$$
From (3.14), one has
$$\begin{aligned} d_{H}\bigl(S(y_{1}),S(y_{2}) \bigr)\leq\mathcal{L}\Vert y_{1}y_{2}\Vert _{C(I,H)}. \end{aligned}$$
Since \(\mathcal{L}\in[0,1)\), from Nadler’s fixedpoint theorem [23] and Theorem 1 in [24], we know that \(S=\{y:y\in S(y)\}\) is an absolute retract in \(C(I,H)\), which shows that S is path connected in \(C(I,H)\). The proof is completed. □
Remark 3.2
In [25], it is proved that the set of extremal solutions of a differential inclusion in \({\mathbb{R}}^{N}\) is path connected based on the Baire category method.