Let

$I=[0,T]$, consider the following evolution inclusions:

$\begin{array}{r}\dot{x}+A(t,x)+Bx\in F(t,x),\phantom{\rule{1em}{0ex}}\text{a.e.}I,\\ x(0)=\phi (x),\end{array}$

(3.1)

where $A:I\times V\to {V}^{\ast}$ is a nonlinear map, $B:V\to {V}^{\ast}$ is a bounded linear map, $\phi :H\to H$ is a continuous map and $F:I\times H\to {2}^{{V}^{\ast}}$ is a multifunction satisfying some conditions mentioned later.

**Definition 3.1** A function

$x\in {W}_{pq}(I)$ is called a solution to the problem (3.1) iff

$\u3008\dot{x}(t),v\u3009+\u3008A(t,x(t)),v\u3009+\u3008Bx(t),v\u3009=\u3008f(t),v\u3009,$

where $x(0)=\phi (x)$, $f(t)\in F(t,x(t))$ for all $v\in V$ and almost all $t\in I$.

We will need the following hypotheses on the data problem (3.1).

(H1)

$A:I\times V\to {V}^{\ast}$ is an operator such that

- (i)
$t\to A(t,x)$ is measurable;

- (ii)
for each

$t\in I$, the operator

$A(t,\cdot ):V\to {V}^{\ast}$ is uniformly monotone and hemicontinuous,

*that is*, there exists a constant

${C}_{1}>0$ (independent of

*t*) such that

$\u3008A(t,{x}_{1})-A(t,{x}_{2}),{x}_{1}-{x}_{2}\u3009\ge {C}_{1}{\parallel {x}_{1}-{x}_{2}\parallel}_{H}^{p}$

for all

${x}_{1},{x}_{2}\in V$, and the map

$s\mapsto \u3008A(t,x+sz),y\u3009$ is continuous on

$[0,1]$ for all

$x,y,z\in V$;

- (iii)
there exist a constant ${C}_{2}>0$, a nonnegative function $a(\cdot )\in {L}_{q}(I)$ and a nondecreasing continuous function $\eta (\cdot )\in {L}_{q}(I)$ such that ${\parallel A(t,x)\parallel}_{{V}^{\ast}}\le a(t)+{C}_{2}\eta ({\parallel x\parallel}_{V})$ for all $x\in V$, a.e. on *I*;

- (iv)
there exist

${C}_{3}>0$,

${C}_{4}>0$,

$b(\cdot )\in {L}_{1}(I)$ such that

$\u3008A(t,x),x\u3009\ge {C}_{3}{\parallel x\parallel}_{V}^{p}-{C}_{4}{\parallel x\parallel}_{V}^{p-1}+\frac{1}{2T}{\parallel x(0)\parallel}^{2}-b(t)\phantom{\rule{1em}{0ex}}\text{a.e.}I,\mathrm{\forall}x\in V,$

or

$\u3008A(t,x),x\u3009\ge {C}_{3}{\parallel x\parallel}_{V}^{p}-{C}_{4}{\parallel x\parallel}_{V}^{p-1}-b(t)\phantom{\rule{1em}{0ex}}\text{a.e.}I,\mathrm{\forall}x\in V,p2.$

(H2)

$F:I\times H\to {P}_{k}({V}^{\ast})$ is a multifunction such that

- (i)
$(t,x)\to F(t,x)$ is graph measurable;

- (ii)
for almost all $t\in I$, $x\to F(t,x)$ is LSC;

- (iii)
there exist a nonnegative function

${b}_{1}(\cdot )\in {L}_{q}(I)$ and a constant

${C}_{5}>0$ such that

$|F(t,x)|=sup\{{\parallel f\parallel}_{{V}^{\ast}}:f\in F(t,x)\}\le {b}_{1}(t)+{C}_{5}{\parallel x\parallel}_{H}^{k-1}\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in V\text{a.e.}I,$

where $1\le k<p$.

(H3)

- (i)
$B:V\to {V}^{\ast}$ is a bounded linear self-adjoint operator such that $(Bx,x)\ge 0$ for all $x\in V$, a.e. on *I*;

- (ii)
there exists a continuous function

$\phi :{L}_{p}(I,H)\to H$ such that

$\parallel \phi (u)-\phi (v)\parallel \le {\parallel u-v\parallel}_{C(I,H)}\phantom{\rule{1em}{0ex}}\mathrm{\forall}u,v\in C(I,H),$

and $\phi (0)=0$.

It is convenient to rewrite the system (3.1) as an operator equation in

${W}_{pq}(I)$. For

$x\in X$, we get

$\begin{array}{c}A(x)(t)=A(t,x),\phantom{\rule{2em}{0ex}}B(x)(t)=Bx(t),\hfill \\ F(x)(t)=F(t,x(t)),\phantom{\rule{1em}{0ex}}t\in I.\hfill \end{array}$

It follows from Theorem 30.A of Zeidler [

23] that the operator

$A:X\to {X}^{\ast}$ is bounded, monotone, hemicontinuous and coercive. By using the same technique, one can show that the operator

$F:{L}_{p}(I,H)\to {X}^{\ast}$ is bounded and satisfies

$|F(t,x)|=sup\{{\parallel f\parallel}_{{X}^{\ast}}:f\in F(t,x)\}\le \stackrel{\u02c6}{{M}_{1}}+\stackrel{\u02c6}{{M}_{2}}{\parallel x\parallel}_{{L}_{p}(I,H)}^{k-1}$

for some constants $\stackrel{\u02c6}{{M}_{1}},\stackrel{\u02c6}{{M}_{2}}>0$ and all $x\in {L}_{p}(I,H)$.

We define

$Lu=\dot{u},\phantom{\rule{2em}{0ex}}D(L)=\{u\in {W}_{pq}(I):u(0)=\xi \in H\},$

(3.2)

where

$\dot{u}$ stands for the generalized derivative of

*u*,

*i.e.*,

${\int}_{0}^{T}\dot{u}(t)v(t)\phantom{\rule{0.2em}{0ex}}dt=-{\int}_{0}^{T}u(t)\dot{v}(t)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{1em}{0ex}}\mathrm{\forall}v(\cdot )\in {C}_{0}^{\mathrm{\infty}}(I).$

For the proofs of main results, we need the following lemma.

**Lemma 3.1** *Let* $V\subseteq H\subseteq {V}^{\ast}$ *be an evolution triple and let* $X={L}_{p}(I,V)$, *where* $1<p<\mathrm{\infty}$ *and* $0<T<\mathrm{\infty}$. *Then the linear operator* $L:D(L)\subseteq X\to {X}^{\ast}$ *defined by* (3.2) *is maximal monotone*.

*Proof* In the sequel we will show that

*L* is maximal monotone. To prove this, suppose that

$(v,w)\in X\times {X}^{\ast}$ and

$0\le \u3008w-Lu,v-u\u3009\phantom{\rule{1em}{0ex}}\mathrm{\forall}u\in D(L).$

We have to show that

$v\in D(L)$ and

$w=Lv$,

*i.e.*,

$w=\dot{v}$. Due to the arbitrariness of

*u*, we choose

$u=\varphi z+\xi $, where

$\varphi \in {C}_{0}^{\mathrm{\infty}}(I)$,

$\xi ={u}_{0}$ and

$z\in V$. Then

$\dot{u}=\dot{\varphi}z$, so

$\u300aLu,u\u300b=0$. From

$\u3008w-Lu,v-u\u3009\ge 0$, we obtain that

$0\le \u300aw,v-\xi \u300b-{\int}_{0}^{T}\u3008\dot{\phi}v+\phi w,z\u3009\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{1em}{0ex}}\mathrm{\forall}z\in V.$

By the arbitrariness of

*z*, one has that

${\int}_{0}^{T}(\dot{\phi}v+\phi w)\phantom{\rule{0.2em}{0ex}}dt=0\phantom{\rule{1em}{0ex}}\mathrm{\forall}\phi \in {C}_{0}^{\mathrm{\infty}}(I).$

Hence,

$w=\dot{v}$. Since

$v\in {W}_{pq}(I)$, then

$w\in {X}^{\ast}$. It remains to show that

$v\in D(L)$. Using the integration by parts formula for functions in

${W}_{pq}(I)$ (see Zeidler [

23], Proposition 23.23), we obtain from (3.2) that

$0\le \u300a\dot{v}-\dot{u},v-u\u300b=\frac{1}{2}({\parallel v(T)-u(T)\parallel}^{2}-{\parallel v(0)-u(0)\parallel}^{2})\phantom{\rule{1em}{0ex}}\mathrm{\forall}u\in D(L).$

(3.3)

Choose a set of functions ${({a}_{n})}_{n\ge 1}$ in *H* such that $T{a}_{n}\to v(T)-\xi $ as $n\to \mathrm{\infty}$. For $\xi \in H$, let $u(t)=t{a}_{n}+\xi $, then $u\in D(L)$. By (3.3), we have $v(0)=u(0)=\xi $ as $n\to \mathrm{\infty}$. Hence, $v\in D(L)$. This completes the proof. □

**Theorem 3.1** *If hypotheses* (H1), (H2) *and* (H3) *hold*, *the problem* (3.1) *has at least one solution*.

*Proof* The process of proof is divided into four parts.

*Step 1.* We claim that the equation

$\begin{array}{r}\dot{x}+A(t,x)+Bx=f(t)\phantom{\rule{1em}{0ex}}\text{a.e.}I,\\ x(0)=(1-\u03f5)\phi (x)\end{array}$

(3.4)

has only one solution.

Firstly, for every

$\u03f5\in (0,1]$,

$y\in X$ and

$f\in {X}^{\ast}$, we claim that the equation

$\begin{array}{r}\dot{x}+A(t,x)+Bx=f(t)\phantom{\rule{1em}{0ex}}\text{a.e.}I,\\ x(0)=(1-\u03f5)\phi (y)\end{array}$

(3.5)

has only one solution. By (H1) and (H3), it is easy to check that

$(A+B)$ is bounded, monotone, hemicontinuous and coercive. Moreover, by Lemma 3.1,

*L* is a linear maximal monotone operator. Therefore,

$R(L+A+B)={V}^{\ast}$,

*i.e.*,

$L+(A+B)$ is surjective (see [

23], p.868]). The uniqueness is clear. Hence, for the Cauchy problem (3.5) has a unique

${x}_{y}(t)\in {W}_{pq}(I)$. By

${W}_{pq}(I)\subset C(I,H)$, then the operator

$P:{W}_{pq}(I)\to {W}_{pq}(I)$ is defined as follows:

$P(y)={\int}_{0}^{t}\dot{x}(s)\phantom{\rule{0.2em}{0ex}}ds+(1-\u03f5)\phi (y).$

By (3.5), we have

$P({y}_{1})-P({y}_{2})+{\int}_{0}^{t}A(s,{x}_{1})-A(s,{x}_{2})\phantom{\rule{0.2em}{0ex}}ds=(1-\u03f5)\phi ({y}_{2})-(1-\u03f5)\phi ({y}_{1})$

for all

${y}_{1},{y}_{2}\in {W}_{pq}(I)$. Take an inner product over (3.5) with

${x}_{1}-{x}_{2}$, then

$(P({y}_{1})-P({y}_{2}),{x}_{1}-{x}_{2})+{\int}_{0}^{t}(A(s,{x}_{1})-A(s,{x}_{2}),{x}_{1}-{x}_{2})\phantom{\rule{0.2em}{0ex}}ds=(1-\u03f5)(\phi ({y}_{2})-\phi ({y}_{1}),{x}_{1}-{x}_{2}).$

By (H1)(ii), we have

${\parallel P({y}_{1})-P({y}_{2})\parallel}^{2}\le (1-\u03f5)\parallel \phi ({y}_{2})-\phi ({y}_{1})\parallel \parallel {x}_{1}-{x}_{2}\parallel .$

Hence,

$\begin{array}{rcl}{\parallel P({y}_{1})-P({y}_{2})\parallel}_{C(I,H)}& \le & (1-\u03f5){\parallel \phi ({y}_{2})-\phi ({y}_{1})\parallel}_{C(I,H)}\\ \le & (1-\u03f5){\parallel {y}_{2}-{y}_{1}\parallel}_{C(I,H)}.\end{array}$

(3.6)

Invoking the Banach fixed point theorem, the operator *P* has only one fixed point ${x}_{\u03f5}=P({x}_{\u03f5})$, *i.e.*, ${x}_{\u03f5}$ is the uniform solution of (3.4).

Therefore, we define ${L}_{\u03f5}:{W}_{pq}(I)\to {X}^{\ast}$ as ${L}_{\u03f5}x=\dot{x}+A(t,x)+Bx$ and $x(0)=(1-\u03f5)\phi (x)$. By Step 1, we have ${L}_{\u03f5}:{W}_{pq}(I)\to {X}^{\ast}$ is one-to-one and surjective, and so ${L}_{\u03f5}^{-1}:{X}^{\ast}\to {W}_{pq}(I)$ is well defined.

*Step 2.* ${L}_{\u03f5}^{-1}:{X}^{\ast}\to {L}_{p}(I,H)$ is completely continuous.

We only need to show that ${L}_{\u03f5}^{-1}$ is continuous and maps a bounded set into a relatively compact set. We claim that ${L}_{\u03f5}:{W}_{pq}(I)\to {X}^{\ast}$ is continuous. In fact, let ${\{{x}_{n}\}}_{n\ge 1}\subset {W}_{pq}(I)$ such that ${x}_{n}\to x$ as $n\to \mathrm{\infty}$. From (H1)(ii) and (H3), we infer that ${x}_{n}\to x$, $A({x}_{n})\to A(x)$, $B{x}_{n}\to Bx$, a.e. *I* as $n\to \mathrm{\infty}$. Obviously, $\phi ({x}_{n})\to \phi (x)$. Therefore, ${L}_{\u03f5}:{W}_{pq}(I)\to {X}^{\ast}$ is continuous and ${L}_{\u03f5}^{-1}$ is continuous.

Let

$K\subset {X}^{\ast}$ be a bound set, for any

$f\in K$, there is

*a priori* bound in

${W}_{pq}(I)$ for the possible solution

$x(t)={L}_{\u03f5}^{-1}f$ of (3.4). Then

$\dot{x}+A(t,x)+Bx=f(t)\phantom{\rule{1em}{0ex}}\text{a.e.}I.$

(3.7)

It follows that

$\u300a\dot{x},x\u300b+\u300aAx,x\u300b+\u300aBx,x\u300b=\u300af(t),x\u300b.$

(3.8)

By (H1)(iv),

$\u300aAu,u\u300b\ge {C}_{3}{\parallel u\parallel}_{X}^{p}-{C}_{4}{\parallel u\parallel}_{X}^{p-1}+\frac{1}{2}{\parallel u(0)\parallel}^{2}-{\parallel b\parallel}_{{L}^{1}},$

or

$\u300aAu,u\u300b\ge {C}_{3}{\parallel u\parallel}_{X}^{p}-{C}_{4}{\parallel u\parallel}_{X}^{p-1}-{\parallel b\parallel}_{{L}^{1}}$

with

$p>2$. But

$\begin{array}{c}\u300a\dot{u},u\u300b={\parallel u(T)\parallel}^{2}-{\parallel u(0)\parallel}^{2},\hfill \\ \u300af,u\u300b\le {\parallel f\parallel}_{{X}^{\ast}}{\parallel u\parallel}_{X}.\hfill \end{array}$

Therefore,

${C}_{3}{\parallel x\parallel}_{X}^{p}\le {C}_{4}{\parallel x\parallel}_{X}^{p-1}+{\parallel f\parallel}_{{X}^{\ast}}{\parallel x\parallel}_{X}+{\parallel b\parallel}_{{L}^{1}},$

or

${C}_{3}{\parallel x\parallel}_{X}^{p}\le {C}_{4}{\parallel x\parallel}_{X}^{p-1}+{\parallel f\parallel}_{{X}^{\ast}}{\parallel x\parallel}_{X}+{\parallel b\parallel}_{{L}^{1}}+{\parallel x\parallel}_{X}^{2},$

with $p>2$. So, there exists an ${M}_{1}>0$ such that ${\parallel x\parallel}_{X}\le {M}_{1}$. Because of the boundedness of operators *A*, *B*, we obtain that there exists an ${M}_{2}>0$ such that ${\parallel \dot{x}\parallel}_{{X}^{\ast}}\le {M}_{2}$. Hence, ${\parallel x\parallel}_{{W}_{pq}}\le M$ for some constant $M>0$. Therefore, we have ${L}_{\u03f5}^{-1}(K)$ is bounded in ${W}_{pq}(I)$. But ${W}_{pq}(I)$ is compactly embedded in ${L}_{p}(I,H)$. Therefore, ${L}_{\u03f5}^{-1}(K)$ is relatively compact in ${L}_{p}(I,H)$.

Let $\stackrel{\u02c6}{N}:{L}_{p}(I,H)\to {2}^{{X}^{\ast}}$ be a multivalued Nemitsky operator corresponding to *F* and $\stackrel{\u02c6}{N}$ was defined by $\stackrel{\u02c6}{N}(x)=\{v\in {X}^{\ast}:v(t)\in F(t,x(t))\}$ a.e. on *I*.

*Step 3.* $\stackrel{\u02c6}{N}(\cdot )$ has nonempty, closed, decomposable values and is LSC.

The closedness and decomposability of the values of

$\stackrel{\u02c6}{N}(\cdot )$ are easy to check. For the nonemptiness, note that if

$x\in {L}_{p}(I,H)$, by the hypothesis (H2)(i),

$(t,x)\to F(t,x)$ is graph measurable, so we apply Aumann’s selection theorem and obtain a measurable map

$v:I\to {V}^{\ast}$ such that

$v(t)\in F(t,x(t))$ a.e. on

*I*. By the hypothesis (H2)(iii),

$v\in {X}^{\ast}$. Thus, for every

$x\in {L}_{p}(I,H)$,

$\stackrel{\u02c6}{N}(x)\ne \mathrm{\varnothing}$. To prove the lower semicontinuity of

$\stackrel{\u02c6}{N}(\cdot )$, we only need to show that every

$u\in {X}^{\ast}$,

$x\to d(u,\stackrel{\u02c6}{N}(x))$ is a USC

${R}_{+}$-valued function. Note that

$\begin{array}{rcl}d(u,\stackrel{\u02c6}{N}(x))& =& inf\{{\parallel u-v\parallel}_{{X}^{\ast}}:v\in \stackrel{\u02c6}{N}(x)\}\\ =& inf\{{\left[{\int}_{0}^{T}{\parallel u(t)-v(t)\parallel}_{{V}^{\ast}}^{q}\phantom{\rule{0.2em}{0ex}}dt\right]}^{1/q}:v\in \stackrel{\u02c6}{N}(x)\}\\ =& {\{{\int}_{0}^{T}inf\{{\parallel u(t)-v(t)\parallel}_{{V}^{\ast}}^{q}:v\in \stackrel{\u02c6}{N}(x)\}\phantom{\rule{0.2em}{0ex}}dt\}}^{1/q}\\ =& {\left\{{\int}_{0}^{T}{\left[d(u(t),F(t,x(t)))\right]}^{q}\phantom{\rule{0.2em}{0ex}}dt\right\}}^{1/q}\end{array}$

(see Hiai and Umegaki [

27] Theorem 2.2). We will show that for every

$\lambda \ge 0$, the superlevel set

${U}_{\lambda}=\{x\in {L}_{p}(I,H):d(u,\stackrel{\u02c6}{N}(x))\ge \lambda \}$ is closed in

${L}_{p}(I,H)$. Let

${\{{x}_{n}\}}_{n\ge 1}\subseteq {U}_{\lambda}$ and assume that

${x}_{n}\to x$ in

${L}_{p}(I,H)$. By passing to a subsequence if necessary, we may assume that

${x}_{n}(t)\to x(t)$ a.e. on

*I* as

$n\to \mathrm{\infty}$. By (H2)(ii),

$x\to d(u(t),F(t,x))$ is an upper semicontinuous

${R}_{+}$-valued function. So, via Fatou’s lemma, we have

$\begin{array}{rcl}{\lambda}^{q}& \le & \overline{lim}{\left[d(u,\stackrel{\u02c6}{N}({x}_{n}))\right]}^{q}\\ =& \overline{lim}{\int}_{0}^{T}{\left[d(u(t),F(t,{x}_{n}(t)))\right]}^{q}\phantom{\rule{0.2em}{0ex}}dt\\ \le & {\int}_{0}^{T}\overline{lim}{\left[d(u(t),F(t,{x}_{n}(t)))\right]}^{q}\phantom{\rule{0.2em}{0ex}}dt\\ \le & {\int}_{0}^{T}{\left[d(u(t),F(t,x(t)))\right]}^{q}\phantom{\rule{0.2em}{0ex}}dt={\left[d(u,\stackrel{\u02c6}{N}(x))\right]}^{q}.\end{array}$

Therefore, $x\in {U}_{\lambda}$ and this proves the LSC of $\stackrel{\u02c6}{N}(\cdot )$. By Lemma 2.2, we obtain a continuous map $f:{L}_{p}(I,H)\to {X}^{\ast}$ such that $f(x)\in \stackrel{\u02c6}{N}(x)$. To finish our proof, we need to solve the fixed point problem: $x={L}_{\u03f5}^{-1}\circ f(x)$.

Since the embedding $V\to H$ is compact, the embedding ${W}_{pq}(I)\to {L}_{p}(I,H)$ is compact. That is, ${x}_{n}\to x$ in ${L}_{p}(I,H)$ whenever ${x}_{n}\rightharpoonup x$ in ${W}_{pq}(I)$. By using the above relation and the continuity of *f*, we have $f({x}_{n})\to f(x)$ in ${X}^{\ast}$ whenever ${x}_{n}\rightharpoonup x$ in ${W}_{pq}(I)$. So, ${L}_{\u03f5}^{-1}\circ f:{L}_{p}(I,H)\to {L}_{p}(I,H)$ is compact.

*Step 4.* We claim that the set $\mathrm{\Gamma}=\{x\in {L}_{p}(I,H):x=\sigma {L}_{\u03f5}^{-1}\circ f(x),\sigma \in (0,1)\}$ is bounded.

Let

$x\in \mathrm{\Gamma}$, then we have

${L}_{\u03f5}(\frac{x}{\sigma})=f(x)$. Note that

$\u300a\dot{\frac{x}{\sigma}},\frac{x}{\sigma}\u300b+\u300aA\left(\frac{x}{\sigma}\right),\frac{x}{\sigma}\u300b+\u300aB\frac{x}{\sigma},\frac{x}{\sigma}\u300b=\u300af(x),\frac{x}{\sigma}\u300b.$

By (H1)(iv) and (H3)(i), one has that

$\u300aAu,u\u300b+\u300aBx,x\u300b\ge {C}_{3}{\parallel u\parallel}_{X}^{p}-{C}_{4}{\parallel u\parallel}_{X}^{p-1}+\frac{1}{2}{\parallel u(0)\parallel}^{2}-{\parallel b\parallel}_{{L}^{1}},$

(3.9)

or

$\u300aAu,u\u300b+\u300aBx,x\u300b\ge {C}_{3}{\parallel u\parallel}_{X}^{p}-{C}_{4}{\parallel u\parallel}_{X}^{p-1}-{\parallel b\parallel}_{{L}^{1}}$

(3.10)

with

$p>2$. By using the integration by parts formula, we have

$\begin{array}{r}\u300aA\left(\frac{x}{\sigma}\right),\frac{x}{\sigma}\u300b+\u300aB\frac{x}{\sigma},\frac{x}{\sigma}\u300b\\ \phantom{\rule{1em}{0ex}}=\u300af(x),\frac{x}{\sigma}\u300b-\u300a\dot{\frac{x}{\sigma}},\frac{x}{\sigma}\u300b\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{\sigma}{\left\{{\int}_{0}^{T}{\parallel f(t,x)\parallel}_{V\ast}^{q}\phantom{\rule{0.2em}{0ex}}dt\right\}}^{1/q}{\left\{{\int}_{0}^{T}{\parallel x\parallel}_{V}^{p}\phantom{\rule{0.2em}{0ex}}dt\right\}}^{1/p}+\frac{1}{2{\sigma}^{2}}{\parallel \phi (x)\parallel}^{2}\\ \phantom{\rule{1em}{0ex}}\le \frac{1}{\sigma}{\left\{{\int}_{0}^{T}{(h(t)+{\parallel x\parallel}_{H}^{k-1})}^{q}\phantom{\rule{0.2em}{0ex}}dt\right\}}^{1/q}{\parallel x\parallel}_{X}+\frac{1}{2{\sigma}^{2}}{\parallel \phi (x)\parallel}^{2}\\ \phantom{\rule{1em}{0ex}}\le \frac{2}{\sigma}{\{{\int}_{0}^{T}{|h(t)|}^{q}+{\parallel x\parallel}_{H}^{q(k-1)}\phantom{\rule{0.2em}{0ex}}dt\}}^{1/q}{\parallel x\parallel}_{X}+\frac{1}{2{\sigma}^{2}}{\parallel \phi (x)\parallel}^{2}\\ \phantom{\rule{1em}{0ex}}\le \frac{2}{\sigma}\{{\left({\int}_{0}^{T}{|h(t)|}^{q}\right)}^{1/q}+{\left({\int}_{0}^{T}{\parallel x\parallel}_{H}^{q(k-1)}\phantom{\rule{0.2em}{0ex}}dt\right)}^{1/q}\}{\parallel x\parallel}_{X}+\frac{1}{2{\sigma}^{2}}{\parallel \phi (x)\parallel}^{2}\\ \phantom{\rule{1em}{0ex}}\le {\gamma}_{1}{\parallel x\parallel}_{X}+{\gamma}_{2}{\parallel x\parallel}_{X}^{k}+\frac{1}{2{\sigma}^{2}}{\parallel \phi (x)\parallel}^{2},\end{array}$

(3.11)

where

${\gamma}_{1},{\gamma}_{2}>0$. By (3.9), (3.10) and (3.11), if

$1\le k<p$, then we have

${\parallel x\parallel}_{X}^{p}\le {\beta}_{1}{\parallel x\parallel}_{X}^{p-1}+{\beta}_{2}{\parallel x\parallel}_{X}^{k}+{\beta}_{3}{\parallel x\parallel}_{X}+{\beta}_{4}{\parallel b\parallel}_{{L}^{1}}.$

(3.12)

If

$1\le k<p$,

$p>2$, then we have

${\parallel x\parallel}_{X}^{p}\le {\beta}_{1}{\parallel x\parallel}_{X}^{p-1}+{\beta}_{2}{\parallel x\parallel}_{X}^{k}+{\beta}_{3}{\parallel x\parallel}_{X}+{\beta}_{4}{\parallel b\parallel}_{{L}^{1}}+{\beta}_{5}{\parallel x\parallel}_{X}^{2}.$

(3.13)

Thus, by virtue of the inequalities (3.12) and (3.13), we can find a constant

${M}_{1}>0$ such that

${\parallel x\parallel}_{X}\le {M}_{1}$ for all

$x\in \mathrm{\Gamma}$. From the boundedness of operators

*A*,

*B* and

*f*, and the continuous embedding

$X\to {L}_{p}(I,H)$, we obtain

${\parallel A(x)\parallel}_{{X}^{\ast}}\le {M}_{2}$,

${\parallel Bx\parallel}_{{X}^{\ast}}\le {M}_{3}$ and

${\parallel f(x)\parallel}_{{X}^{\ast}}\le {M}_{4}$ for some constants

${M}_{2}>0$,

${M}_{3}>0$,

${M}_{4}>0$ and all

$x\in \mathrm{\Gamma}$. Therefore,

${\parallel \dot{x}\parallel}_{{X}^{\ast}}\le {\parallel A(x)\parallel}_{{X}^{\ast}}+{\parallel Bx\parallel}_{{X}^{\ast}}+{\parallel f(x)\parallel}_{{X}^{\ast}}\le {M}_{2}+{M}_{3}+{M}_{4}$

(3.14)

for all $x\in \mathrm{\Gamma}$.

It follows from (3.14) that ${\parallel x\parallel}_{{W}_{pq}}\le {\parallel x\parallel}_{X}+{\parallel \dot{x}\parallel}_{{X}^{\ast}}\le \stackrel{\u02c6}{M}$ for some constant $\stackrel{\u02c6}{M}>0$. Hence, Γ is a bounded subset of ${W}_{pq}(I)$. So, Γ is a bounded subset of ${L}_{p}(I,H)$ since the embedding ${W}_{pq}(I)\to {L}_{p}(I,H)$ is compact.

Invoking the Leray-Schauder theorem, one has that there exists an

${x}_{\u03f5}\in {W}_{pq}(I)$ such that

${x}_{\u03f5}={L}_{\u03f5}^{-1}f({x}_{\u03f5})$,

*i.e.*,

${x}_{\u03f5}$ is a solution of the following problem:

$\begin{array}{r}{\dot{x}}_{\u03f5}+A(t,{x}_{\u03f5})+B{x}_{\u03f5}=f({x}_{\u03f5}),\\ f({x}_{\u03f5})\in F(t,{x}_{\u03f5})\phantom{\rule{1em}{0ex}}\text{a.e.}I,\\ {x}_{\u03f5}(0)=(1-\u03f5)\phi ({x}_{\u03f5}).\end{array}$

(3.15)

Let

${({\u03f5}_{n})}_{n\ge 1}\subset (0,1)$ and

${\u03f5}_{n}\to 0$. For every

$n\in N$, there exists an

${x}_{n}\in {W}_{pq}(I)$ which is a solution of the following equations:

$\begin{array}{r}{\dot{x}}_{n}+A(t,{x}_{n})+B{x}_{n}=f({x}_{n})\phantom{\rule{1em}{0ex}}\text{a.e.}I,\\ f({x}_{n})\in F(t,{x}_{n}),\\ {x}_{n}(0)=(1-{\u03f5}_{n})\phi ({x}_{n}).\end{array}$

(3.16)

By Step 3, we have that

${\{{x}_{n}\}}_{n\ge 1}$ is uniformly bounded. By the boundedness of the sequence

${\{{x}_{n}\}}_{n\ge 1}\subset {W}_{pq}(I)$, it follows that the sequence

${\{{\dot{x}}_{n}\}}_{n\ge 1}\subset {X}^{\ast}$ is uniformly bounded and passing to subsequence if necessary, we may assume that

${\dot{x}}_{n}\rightharpoonup u$ in

${X}^{\ast}$. Evidently,

$u=\dot{x}$ and

${x}_{n}\rightharpoonup x$ in

${W}_{pq}(I)$. Since the embedding

${W}_{pq}(I)\hookrightarrow {L}_{p}(I,H)$ is compact, then

${x}_{n}\to x$ in

${L}_{p}(I,H)$. Hence, from the hypothesis (H2)(ii), we obtain

$f({x}_{n})\to f(x)$ and

$f(x)\in F(t,x)$. Since the operator

*A* is hemicontinuous and monotone and

*B* is a continuous linear operator, thus

$A({x}_{n})\rightharpoonup A(x)$,

$B{x}_{n}\rightharpoonup Bx$ in

${X}^{\ast}$ as

$n\to \mathrm{\infty}$. Therefore, we obtain

$\dot{x}+A(x)+Bx=f$,

$f\in F(t,x)$ a.e. on

*I*. Since

${x}_{n}(t)\to x(t)$ in

${L}_{p}(I,H)$ and

$\phi :{L}_{p}(I,H)\to H$ is continuous, then we have

${x}_{n}(0)=(1-{\u03f5}_{n})\phi ({x}_{n})\to \phi (x)=x(0).$

Hence, *x* is a solution of (3.1). The proof is completed. □

Next, we consider the convex case, the assumption on *F* is as follows:

(H4)

$F:I\times H\to {P}_{kc}({V}^{\ast})$ is a multifunction such that

- (i)
$(t,x)\to F(t,x)$ is graph measurable;

- (ii)
for almost all $t\in I$, $x\to F(t,x)$ has a closed graph; and (H2)(iii) hold.

**Theorem 3.2** *If hypotheses* (H1), (H3) *and* (H4) *hold*, *the problem* (3.1) *has at least one solution*; *moreover*, *the solution set is weakly compact in* ${W}_{pq}(I)$.

*Proof* The proof is as that of Theorem 3.1. So, we only present those particular points where the two proofs differ.

In this case, the multivalued Nemistsky operator

$\stackrel{\u02c6}{N}:{L}_{p}(I,H)\to {2}^{{X}_{w}^{\ast}}$ has nonempty closed, convex values in

${X}^{\ast}$ and is USC from

${L}_{p}(I,H)$ into

${X}^{\ast}$ furnished with the weak topology (denoted by

${X}_{w}^{\ast}$). The closedness and convexity of the values of

$\stackrel{\u02c6}{N}(\cdot )$ are clear. To prove the nonemptiness, let

$x\in {L}_{p}(I,H)$ and

${\{{s}_{n}\}}_{n\ge 1}$ be a sequence of step functions such that

${s}_{n}(t)\to x(t)$ in

*H* and

${\parallel {s}_{n}(t)\parallel}_{H}\le {\parallel x(t)\parallel}_{H}$ a.e. on

*I*. Then by virtue of the hypothesis (H4)(i), for every

$n\ge 1$,

$t\to F(t,{s}_{n}(t))$ admits a measurable selector

${v}_{n}(t)$. From the hypothesis (H4)(iii), we have that

${\parallel {v}_{n}\parallel}_{{X}^{\ast}}\le \stackrel{\u02c6}{{M}_{1}}+\stackrel{\u02c6}{{M}_{2}}{\parallel x\parallel}_{{L}_{p}(I,H)}^{k-1}$, so

${\{{v}_{n}(t)\}}_{n\ge 1}\subseteq {X}^{\ast}$ is uniformly integrable. So, by the Dunford-Pettis theorem, and by passing to a subsequence if necessary, we may assume that

${v}_{n}\to v$ weakly in

${X}^{\ast}$. Then from Theorem 3.1 in [

28], we have

$v(t)\in \overline{conv}\phantom{\rule{0.2em}{0ex}}\overline{lim}{\{{v}_{n}(t)\}}_{n\ge 1}\subseteq \overline{conv}\phantom{\rule{0.2em}{0ex}}\overline{lim}F(t,{s}_{n}(t))\subseteq F(t,x(t))\phantom{\rule{1em}{0ex}}\text{a.e. on}I,$

the last inclusion being a consequence of the hypothesis (H4)(ii). So, $v\in \stackrel{\u02c6}{N}(x)$, which means that $\stackrel{\u02c6}{N}(\cdot )$ is nonempty.

Next, we show that

$\stackrel{\u02c6}{N}(\cdot )$ is USC from

${L}_{p}(I,H)$ into

${X}_{w}^{\ast}$. Let Ξ be a nonempty and weakly closed subset of

${X}^{\ast}$. Obviously, it is sufficient to show that the set

${\stackrel{\u02c6}{N}}^{-1}(\mathrm{\Xi})=\{x\in {L}_{p}(I,H):\stackrel{\u02c6}{N}(x)\cap \mathrm{\Xi}\ne \varphi \}$

is closed. Let

${\{{x}_{n}\}}_{n\ge 1}\subseteq {\stackrel{\u02c6}{N}}^{-1}(\mathrm{\Xi})$ and assume

${x}_{n}\to x$ in

${L}_{p}(I,H)$. Passing to a subsequence, we can get that

${x}_{n}(t)\to x(t)$ a.e. on

*I*. Let

${f}_{n}\in \stackrel{\u02c6}{N}({x}_{n})\cap \mathrm{\Xi}$,

$n\ge 1$. Then by virtue of the hypothesis (H4)(iii), we have

${\parallel {f}_{n}\parallel}_{{X}^{\ast}}\le \stackrel{\u02c6}{{M}_{1}}+\stackrel{\u02c6}{{M}_{2}}{\parallel x\parallel}_{{L}_{p}(I,H)}^{k-1}.$

So, by the Dunford-Pettis theorem, we may assume that

${f}_{n}\to f\in \mathrm{\Xi}$ in

${X}_{w}^{\ast}$. As before, we have

$f(t)\in \overline{conv}\phantom{\rule{0.2em}{0ex}}\overline{lim}{\{{f}_{n}(t)\}}_{n\ge 1}\subseteq \overline{conv}\phantom{\rule{0.2em}{0ex}}\overline{lim}F(t,{x}_{n}(t))\subseteq F(t,x(t))\phantom{\rule{1em}{0ex}}\text{a.e. on}I,$

then $f\in \stackrel{\u02c6}{N}(x)\cap \mathrm{\Xi}$, *i.e.*, ${\stackrel{\u02c6}{N}}^{-1}(\mathrm{\Xi})$ is closed in ${L}_{p}(I,H)$. This proves the upper semicontinuity of $\stackrel{\u02c6}{N}(\cdot )$ from ${L}_{p}(I,H)$ into ${X}_{w}^{\ast}$.

We consider the following fixed point problem:

$x\in {L}_{\u03f5}^{-1}\circ \stackrel{\u02c6}{N}(x).$

Recalling that

${L}_{\u03f5}^{-1}:{X}^{\ast}\to {L}_{p}(I,H)$ is completely continuous, we see that

${L}_{\u03f5}^{-1}\circ \stackrel{\u02c6}{N}:{L}_{p}(I,H)\to {P}_{kc}({L}_{p}(I,H))$ is USC and maps bounded sets into relatively compact sets. We easily check that

${\mathrm{\Gamma}}_{1}=\{x\in {L}_{p}(I,H):x\in \lambda {L}_{\u03f5}^{-1}\circ \stackrel{\u02c6}{N}(x),\lambda \in (0,1)\}$

is bounded, as a proof of Theorem 3.1. Invoking the Leray-Schauder fixed point theorem, one has that there exists an

${x}_{\u03f5}\in {W}_{pq}(I)$ such that

${x}_{\u03f5}\in {L}_{\u03f5}^{-1}\circ \stackrel{\u02c6}{N}({x}_{\u03f5})$,

*i.e.*,

${x}_{\u03f5}$ is a solution of the following problem:

$\begin{array}{r}{\dot{x}}_{\u03f5}+A(t,{x}_{\u03f5})+B{x}_{\u03f5}\in F(t,{x}_{\u03f5})\phantom{\rule{1em}{0ex}}\text{a.e.}I,\\ {x}_{\u03f5}(0)=(1-\u03f5)\phi ({x}_{\u03f5}).\end{array}$

(3.17)

Let

${({\u03f5}_{n})}_{n\ge 1}\subset (0,1)$ and

${\u03f5}_{n}\to 0$. For every

$n\in N$, there exists an

${x}_{n}\in {W}_{pq}(I)$ which is a solution of the following problem:

$\begin{array}{r}{\dot{x}}_{n}+A(t,{x}_{n})+B{x}_{n}={f}_{n}(t)\phantom{\rule{1em}{0ex}}\text{a.e.}I,\\ {f}_{n}(t)\in F(t,{x}_{n}),\\ {x}_{n}(0)=(1-{\u03f5}_{n})\phi ({x}_{n}).\end{array}$

(3.18)

By Step 3,

${\{{x}_{n}\}}_{n\ge 1}$ is uniformly bounded. By the boundedness of the sequence

${\{{x}_{n}\}}_{n\ge 1}\subset {W}_{pq}(I)$, it follows that the sequence

${\{{\dot{x}}_{n}\}}_{n\ge 1}\subset {X}^{\ast}$ is uniformly bounded and, passing to subsequence if necessary, we may assume that

${\dot{x}}_{n}\rightharpoonup \dot{x}$ in

${X}^{\ast}$. Thus,

$A({x}_{n})\rightharpoonup A(x)$,

$B{x}_{n}\rightharpoonup Bx$ in

${X}^{\ast}$ as

$n\to \mathrm{\infty}$. Evidently, there exists

${f}_{n}\in N({x}_{n})$, by virtue of the hypothesis (H4)(iv), we have that

${\parallel {f}_{n}\parallel}_{{X}^{\ast}}\le \stackrel{\u02c6}{{M}_{1}}+\stackrel{\u02c6}{{M}_{2}}{\parallel x\parallel}_{{L}_{p}(I,H)}^{k-1}$, so

${\{{f}_{n}(t)\}}_{n\ge 1}\subseteq {X}^{\ast}$ is uniformly integrable. So, by the Dunford-Pettis theorem and by passing to a subsequence if necessary, we may assume that

${f}_{n}\to f$ weakly in

${X}^{\ast}$. Therefore, we obtain

$\dot{x}+A(x)+Bx=f$,

$f\in F(t,x)$ a.e. on

*I*. Since

${x}_{n}(t)\to x(t)$ in

${L}_{p}(I,H)$ and

$\phi :{L}_{p}(I,H)\to H$ is continuous, then we have

${x}_{n}(0)=(1-{\u03f5}_{n})\phi ({x}_{n})\to \phi (x)=x(0).$

Hence, evidently

*x* is a solution of (3.1). As in the proof of Theorem 3.1, we have that

$|S|=sup\{{\parallel x\parallel}_{{W}_{pq}}:x\in S\}\le \stackrel{\u02c6}{M}$, for some

$\stackrel{\u02c6}{M}>0$. So,

$S\subseteq {W}_{pq}(I)$ is uniformly bounded. So, by the Dunford-Pettis theorem and by passing to a subsequence if necessary, we may assume that

${x}_{n}\to x$ weakly in

${W}_{pq}(I)$. As before, we have

${L}_{\u03f5}(x)(t)\in \overline{conv}\phantom{\rule{0.2em}{0ex}}\overline{lim}{\{{L}_{\u03f5}{x}_{n}(t)\}}_{n\ge 1}\subseteq \overline{conv}\phantom{\rule{0.2em}{0ex}}\overline{lim}F(t,{x}_{n}(t))\subseteq F(t,x(t))\phantom{\rule{1em}{0ex}}\text{a.e. on}I.$

Clearly, $x(0)=\phi (x)$, then $x\in S$. Thus, *S* is weakly compact in ${W}_{pq}(I)$. □