Combining the continuation principle with the bound sets technique, we are ready to state the main result of the paper concerning the solvability and localization of a solution of the multivalued Dirichlet problem (1).

**Theorem 5.1** *Consider the Dirichlet b*.*v*.*p*. (1), *where* $F:[0,T]\times E\times E\u22b8E$ *is an upper*-*Carathéodory multivalued mapping*. *Assume that* $K\subset E$ *is an open*, *convex set containing * 0. *Furthermore*, *let the following conditions be satisfied*:

(${5}_{i}$) $\gamma (F(t,{\mathrm{\Omega}}_{1}\times {\mathrm{\Omega}}_{2}))\le g(t)(\gamma ({\mathrm{\Omega}}_{1})+\gamma ({\mathrm{\Omega}}_{2}))$ *for a*.*a*. $t\in [0,T]$ *and each bounded* ${\mathrm{\Omega}}_{1},{\mathrm{\Omega}}_{2}\subset E$, *where* $g\in {L}^{1}([0,T],[0,\mathrm{\infty}))$ *and* *γ* *is the Hausdorff measure of noncompactness in E*.

(

${5}_{ii}$)

*For every nonempty*,

*bounded set* $\mathrm{\Omega}\subset E\times E$,

*there exists* ${\nu}_{\mathrm{\Omega}}\in {L}^{1}([0,T],[0,\mathrm{\infty}))$ *such that* $\parallel F(t,x,y)\parallel \le {\nu}_{\mathrm{\Omega}}(t)$

(18)

*for a*.*a*. $t\in [0,T]$ *and all* $(x,y)\in \mathrm{\Omega}$,

(${5}_{iii}$) $(T+4){\parallel g\parallel}_{{L}^{1}([0,T],[0,\mathrm{\infty}))}<2$.

*Finally*, *let there exist a function* $V\in {C}^{1}(E,\mathbb{R})$ *with a locally Lipschitzian Fréchet derivative* $\dot{V}$ *satisfying conditions* (H1), (H2), *and at least one of conditions* (7), (8) *for a suitable* $\epsilon >0$, *all* $x\in \overline{K}\cap B(\partial K,\epsilon )$, $t\in (0,T)$, $y\in E$, $\lambda \in (0,1)$ *and* $w\in \lambda F(t,x,y)$. *Then the Dirichlet b*.*v*.*p*. (1) *admits a solution whose values are located in* $\overline{K}$.

*Proof* Let us define the closed set

$S={S}_{1}$ by

$S:=\{x\in A{C}^{1}([0,T],E):x(T)=x(0)=0\}$

and let the set *Q* of candidate solutions be defined as $Q:={C}^{1}([0,T],\overline{K})$. Because of the convexity of *K*, the set *Q* is closed and convex.

For all

$q\in Q$ and

$\lambda \in [0,1]$, consider still the associated fully linearized problem

and denote by $\mathfrak{T}$ a solution mapping which assigns to each $(q,\lambda )\in Q\times [0,1]$ the set of solutions of $P(q,\lambda )$. We will show that the family of the above b.v.p.s $P(q,\lambda )$ satisfies all assumptions of Proposition 3.1.

In this case, $\phi (t,x,\dot{x})=F(t,x,\dot{x})$ which, together with the definition of $P(q,\lambda )$, ensures the validity of (6).

ad (i) In order to verify condition (i) in Proposition 3.1, we need to show that for each

$(q,\lambda )\in Q\times [0,1]$, the problem

$P(q,\lambda )$ is solvable with a convex set of solutions. So, let

$(q,\lambda )\in Q\times [0,1]$ be arbitrary and let

${f}_{q}(\cdot )$ be a strongly measurable selection of

$F(\cdot ,q(\cdot ),\dot{q}(\cdot ))$. The homogeneous problem corresponding to b.v.p.

$P(q,\lambda )$,

$\begin{array}{l}\ddot{x}(t)=0\phantom{\rule{1em}{0ex}}\text{for a.a.}t\in [0,T],\\ x(T)=x(0)=0,\end{array}\}$

(19)

has only the trivial solution, and therefore the single-valued Dirichlet problem

$\begin{array}{l}\ddot{x}(t)=\lambda {f}_{q}(t)\phantom{\rule{1em}{0ex}}\text{for a.a.}t\in [0,T],\\ x(T)=x(0)=0\end{array}\}$

admits a unique solution

${x}_{q,\lambda}(\cdot )$ which is one of solutions of

$P(q,\lambda )$. This is given, for a.a.

$t\in [0,T]$, by

${x}_{q,\lambda}(t)={\int}_{0}^{T}G(t,s)\lambda {f}_{q}(s)\phantom{\rule{0.2em}{0ex}}ds$, where

*G* is the Green function associated to the homogeneous problem (19). The Green function

*G* and its partial derivative

$\frac{\partial}{\partial t}G$ are defined by (

*cf.*,

*e.g.*, [[

12], pp.170-171])

$\begin{array}{r}G(t,s)=\{\begin{array}{cc}\frac{(s-T)t}{T}\hfill & \text{for all}0\le t\le s\le T,\hfill \\ \frac{(t-T)s}{T}\hfill & \text{for all}0\le s\le t\le T,\hfill \end{array}\\ \frac{\partial}{\partial t}G(t,s)=\{\begin{array}{cc}\frac{(s-T)}{T}\hfill & \text{for all}0\le t\le s\le T,\hfill \\ \frac{s}{T}\hfill & \text{for all}0\le s\le t\le T.\hfill \end{array}\end{array}$

Thus, the set of solutions of $P(q,\lambda )$ is nonempty. The convexity of the solution sets follows immediately from the properties of a mapping *F* and the fact that problems $P(q,\lambda )$ are fully linearized.

ad (ii) Assuming that $H:[0,T]\times E\times E\times E\times E\times [0,1]\u22b8E$ is defined by $H(t,x,y,q,r,\lambda ):=\lambda F(t,q,r)$, condition (ii) in Proposition 3.1 is ensured directly by assumption (5
).

ad (iii) Since the verification of condition (iii) in Proposition 3.1 is technically the most complicated, it will be subdivided into two parts: (iii_{1}) the quasi-compactness of the solution operator $\mathfrak{T}$, (iii_{2}) the condensity of $\mathfrak{T}$ w.r.t. the monotone and nonsingular (*cf.* Lemma 2.1) m.n.c. *μ* defined by (4).

ad (iii

_{1}) Let us firstly prove that the solution mapping

$\mathfrak{T}$ is quasi-compact. Since

${C}^{1}([0,T],E)$ is a metric space, it is sufficient to prove the sequential quasi-compactness of

$\mathfrak{T}$. Hence, let us consider the sequences

$\{{q}_{n}\},\{{\lambda}_{n}\},{q}_{n}\in Q$,

${\lambda}_{n}\in [0,1]$ for all

$n\in \mathbb{N}$ such that

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

${C}^{1}([0,T],E)$ and

${\lambda}_{n}\to \lambda $. Moreover, let

${x}_{n}\in \mathfrak{T}({q}_{n},{\lambda}_{n})$ for all

$n\in \mathbb{N}$. Then there exists, for all

$n\in \mathbb{N}$,

${f}_{n}(\cdot )\in F(\cdot ,{q}_{n}(\cdot ),{\dot{q}}_{n}(\cdot ))$ such that

${\ddot{x}}_{n}(t)={\lambda}_{n}{f}_{n}(t)\phantom{\rule{1em}{0ex}}\text{for a.a.}t\in [0,T],$

(20)

and that ${x}_{n}(T)={x}_{n}(0)=0$.

Since ${q}_{n}\to q$ and ${\dot{q}}_{n}\to \dot{q}$ in $C([0,T],E)$, there exists a bounded $\mathrm{\Omega}\subset E\times E$ such that $({q}_{n}(t),{\dot{q}}_{n}(t))\in \mathrm{\Omega}$ for all $t\in [0,T]$ and $n\in \mathbb{N}$. Therefore, there exists, according to condition (${5}_{ii}$), ${\nu}_{\mathrm{\Omega}}\in {L}^{1}([0,T],[0,\mathrm{\infty}))$ such that $\parallel {f}_{n}(t)\parallel \le {\nu}_{\mathrm{\Omega}}(t)$ for every $n\in \mathbb{N}$ and a.a. $t\in [0,T]$.

Moreover, for every

$n\in \mathbb{N}$ and a.a.

$t\in [0,T]$,

${x}_{n}(t)={\lambda}_{n}{\int}_{0}^{T}G(t,s){f}_{n}(s)\phantom{\rule{0.2em}{0ex}}ds$

and

${\dot{x}}_{n}(t)={\lambda}_{n}{\int}_{0}^{T}\frac{\partial}{\partial t}G(t,s){f}_{n}(s)\phantom{\rule{0.2em}{0ex}}ds.$

Thus,

${x}_{n}$ satisfies, for every

$n\in \mathbb{N}$ and a.a.

$t\in [0,T]$,

$\parallel {x}_{n}(t)\parallel \le a$ and

$\parallel {\dot{x}}_{n}(t)\parallel \le b$, where

$a:=\frac{T}{4}{\int}_{0}^{T}{\nu}_{\mathrm{\Omega}}(s)\phantom{\rule{0.2em}{0ex}}ds$

and

$b:={\int}_{0}^{T}{\nu}_{\mathrm{\Omega}}(s)\phantom{\rule{0.2em}{0ex}}ds.$

Furthermore, for every

$n\in \mathbb{N}$ and a.a.

$t\in [0,T]$, we have

$\parallel {\ddot{x}}_{n}(t)\parallel \le {\nu}_{\mathrm{\Omega}}(t).$

Hence, the sequences $\{{x}_{n}\}$ and $\{{\dot{x}}_{n}\}$ are bounded and $\{{\ddot{x}}_{n}\}$ is uniformly integrable.

Since the sequences

$\{{q}_{n}\}$,

$\{{\dot{q}}_{n}\}$ are converging, we obtain, in view of (

${5}_{i}$),

$\gamma \left(\{{f}_{n}(t)\}\right)\le g(t)(\gamma \left(\{{q}_{n}(t)\}\right)+\gamma \left(\{{\dot{q}}_{n}(t)\}\right))=0$

for a.a. $t\in [0,T]$, which implies that $\{{f}_{n}(t)\}$ is relatively compact.

For all

$(t,s)\in [0,T]\times [0,T]$, the sequence

$\{G(t,s){f}_{n}(s)\}$ is relatively compact as well since, according to the semi-homogeneity of the Hausdorff m.n.c.,

$\gamma \left(\{G(t,s){f}_{n}(s)\}\right)\le |G(t,s)|\gamma \left(\{{f}_{n}(s)\}\right)=0\phantom{\rule{1em}{0ex}}\text{for all}(t,s)\in [0,T]\times [0,T].$

(21)

Moreover, by means of (2), (3), (21) and the semi-homogeneity of the Hausdorff m.n.c.,

$\gamma \left(\{{x}_{n}(t)\}\right)\le \gamma \left(\bigcup _{\lambda \in [0,1]}\lambda \{{\int}_{0}^{T}G(t,s){f}_{n}(s)\phantom{\rule{0.2em}{0ex}}ds\}\right)\le \gamma \left(\{{\int}_{0}^{T}G(t,s){f}_{n}(s)\phantom{\rule{0.2em}{0ex}}ds\}\right)=0.$

By similar reasonings, we can also get

$\gamma \left(\{{\dot{x}}_{n}(t)\}\right)=0,$

by which $\{{x}_{n}(t)\}$, $\{{\dot{x}}_{n}(t)\}$ are relatively compact for a.a. $t\in [0,T]$. Moreover, since ${x}_{n}$ satisfies for all $n\in \mathbb{N}$ equation (20), $\{{\ddot{x}}_{n}(t)\}$ is relatively compact for a.a. $t\in [0,T]$. Thus, according to Lemma 2.2, there exist a subsequence of $\{{\dot{x}}_{n}\}$, for the sake of simplicity denoted in the same way as the sequence, and $x\in {C}^{1}([0,T],E)$ such that $\{{\dot{x}}_{n}\}$ converges to $\dot{x}$ in $C([0,T],E)$ and $\{{\ddot{x}}_{n}\}$ converges weakly to $\ddot{x}$ in ${L}^{1}([0,T],E)$. Therefore, the mapping $\mathfrak{T}$ is quasi-compact.

ad (iii

_{2}) In order to show that

$\mathfrak{T}$ is

*μ*-condensing, where

*μ* is defined by (4), we will prove that any bounded subset

$\mathrm{\Theta}\subset Q$ such that

$\mu (\mathfrak{T}(\mathrm{\Theta}\times [0,1]))\ge \mu (\mathrm{\Theta})$ is relatively compact. Let

${\{{x}_{n}\}}_{n}\subset \mathfrak{T}(\mathrm{\Theta}\times [0,1])$ be a sequence such that

Then we can find

${\{{q}_{n}\}}_{n}\subset \mathrm{\Theta}$,

${\{{f}_{n}\}}_{n}$ satisfying

${f}_{n}(t)\in F(t,{q}_{n}(t),{\dot{q}}_{n}(t))$ for a.a.

$t\in [0,T]$ and

${\{{\lambda}_{n}\}}_{n}\subset [0,1]$ such that for all

$t\in [0,T]$,

${x}_{n}(t)={\lambda}_{n}{\int}_{0}^{T}G(t,s){f}_{n}(s)\phantom{\rule{0.2em}{0ex}}ds$

(22)

and

${\dot{x}}_{n}(t)={\lambda}_{n}{\int}_{0}^{T}\frac{\partial}{\partial t}G(t,s){f}_{n}(s)\phantom{\rule{0.2em}{0ex}}ds.$

(23)

In view of (

${5}_{i}$), we have, for all

$t\in [0,T]$,

Since ${\{{q}_{n}\}}_{n}\subset \mathrm{\Theta}$ and Θ is bounded in ${C}^{1}([0,T],E)$, by means of (${5}_{ii}$), we get the existence of ${\nu}_{\mathrm{\Theta}}\in {L}^{1}([0,T],[0,\mathrm{\infty}))$ such that $\parallel {f}_{n}(t)\parallel \le {\nu}_{\mathrm{\Theta}}(t)$ for a.a. $t\in [0,T]$ and all $n\in \mathbb{N}$. This implies $\parallel G(t,s){f}_{n}(t)\parallel \le |G(t,s)|{\nu}_{\mathrm{\Theta}}(t)$ for a.a. $t,s\in [0,T]$ and all $n\in \mathbb{N}$.

Moreover, by virtue of the semi-homogeneity of the Hausdorff m.n.c., for all

$(t,s)\in [0,T]\times [0,T]$, we have

According to (2), (3) and (22), we so obtain for each

$t\in [0,T]$,

$\begin{array}{rcl}\gamma \left(\{{x}_{n}(t),n\in \mathbb{N}\}\right)& \le & \gamma \left(\{{\int}_{0}^{T}G(t,s){f}_{n}(s)\phantom{\rule{0.2em}{0ex}}ds,n\in \mathbb{N}\}\right)\\ \le & 2\frac{T}{4}{\parallel g\parallel}_{{L}^{1}}\underset{t\in [0,T]}{sup}(\gamma \left(\{{q}_{n}(t),n\in \mathbb{N}\}\right)+\gamma \left(\{{\dot{q}}_{n}(t),n\in \mathbb{N}\}\right))\\ =& \frac{T}{2}{\parallel g\parallel}_{{L}^{1}}\mathcal{S},\end{array}$

where

$\mathcal{S}:=\underset{t\in [0,T]}{sup}(\gamma \left(\{{q}_{n}(t),n\in \mathbb{N}\}\right)+\gamma \left(\{{\dot{q}}_{n}(t),n\in \mathbb{N}\}\right)).$

By the similar reasonings, we can obtain that for each

$t\in [0,T]$,

$\gamma \left(\{{\dot{x}}_{n}(t),n\in \mathbb{N}\}\right)\le 2{\parallel g\parallel}_{{L}^{1}}\mathcal{S},$

when starting from condition (23). Subsequently,

$\gamma \left(\{{x}_{n}(t),n\in \mathbb{N}\}\right)+\gamma \left(\{{\dot{x}}_{n}(t),n\in \mathbb{N}\}\right)\le \frac{T+4}{2}{\parallel g\parallel}_{{L}^{1}}\mathcal{S},$

yielding

$\underset{t\in [0,T]}{sup}(\gamma \left(\{{x}_{n}(t),n\in \mathbb{N}\}\right)+\gamma \left(\{{\dot{x}}_{n}(t),n\in \mathbb{N}\}\right))\le \frac{T+4}{2}{\parallel g\parallel}_{{L}^{1}}\mathcal{S}.$

(24)

Since

$\mu (\mathfrak{T}(\mathrm{\Theta}\times [0,1]))\ge \mu (\mathrm{\Theta})$ and

${\{{q}_{n}\}}_{n}\subset \mathrm{\Theta}$, we so get

and, in view of (24) and (

${5}_{iii}$), we have that

$\underset{t\in [0,T]}{sup}(\gamma \left(\{{q}_{n}(t),n\in \mathbb{N}\}\right)+\gamma \left(\{{\dot{q}}_{n}(t),n\in \mathbb{N}\}\right))=0.$

Inequality (24) implies that

$\underset{t\in [0,T]}{sup}(\gamma \left(\{{x}_{n}(t),n\in \mathbb{N}\}\right)+\gamma \left(\{{\dot{x}}_{n}(t),n\in \mathbb{N}\}\right))=0.$

(25)

Now, we show that both the sequences

$\{{x}_{n}\}$ and

$\{{\dot{x}}_{n}\}$ are equi-continuous. Let

$\tilde{\mathrm{\Theta}}\subset E$ be such that

${q}_{n}(t)\in \tilde{\mathrm{\Theta}}$ and

${\dot{q}}_{n}(t)\in \tilde{\mathrm{\Theta}}$ for all

$n\in \mathbb{N}$ and

$t\in [0,T]$. Thus, we get that

$\parallel {\ddot{x}}_{n}(t)\parallel ={\lambda}_{n}\parallel {f}_{n}(t)\parallel \le {\nu}_{\tilde{\mathrm{\Theta}}}(t)$, where

${\nu}_{\tilde{\mathrm{\Theta}}}\in {L}^{1}([0,T],[0,\mathrm{\infty}))$ comes from (

${5}_{ii}$), and so

${\{{\ddot{x}}_{n}\}}_{n}$ is uniformly integrable. This implies that

${\{{\dot{x}}_{n}\}}_{n}$ is equi-continuous. Moreover, according to (23), we obtain that

$\parallel {\dot{x}}_{n}(t)\parallel \le {\int}_{0}^{T}{\nu}_{\tilde{\mathrm{\Theta}}}(s)\phantom{\rule{0.2em}{0ex}}ds$

for all

$n\in \mathbb{N}$ and

$t\in [0,T]$, implying that

${\{{\dot{x}}_{n}\}}_{n}$ is bounded; consequently, also

${\{{x}_{n}\}}_{n}$ is equi-continuous. Therefore,

${mod}_{C}(\{{x}_{n}\})={mod}_{C}(\{{\dot{x}}_{n}\})=0.$

In view of (25), we have so obtained that

$\mu \left(\mathfrak{T}(\mathrm{\Theta}\times [0,1])\right)=(0,0).$

Hence, also $\mu (\mathrm{\Theta})=(0,0)$ and since *μ* is regular, we have that Θ is relatively compact. Therefore, condition (iii) in Proposition 3.1 holds.

ad (iv) For all $q\in Q$, the problem $P(q,0)$ has only the trivial solution. Since $0\in K$, condition (iv) in Proposition 3.1 is satisfied.

ad (v) Let

${q}_{\ast}\in Q$ be a solution of the b.v.p.

$P({q}_{\ast},\lambda )$ for some

$\lambda \in (0,1)$,

*i.e.*, a fixed point of the solution mapping

$\mathfrak{T}$. In view of conditions (7), (8) (see Proposition 4.1),

*K* is, for all

$\lambda \in (0,1)$, a bound set for the problem

$\begin{array}{l}{\ddot{q}}_{\ast}(t)\in \lambda F(t,{q}_{\ast}(t),{\dot{q}}_{\ast}(t)),\phantom{\rule{1em}{0ex}}\text{for a.a.}t\in [0,T],\\ x(T)=x(0)=0.\end{array}\}$

This implies that ${q}_{\ast}\notin \partial Q$, which ensures condition (v) in Proposition 3.1. □

If the mapping $F(t,x,y)$ is globally u.s.c. in $(t,x,y)$ (*i.e.*, a Marchaud map), then we are able to improve Theorem 5.1 in the following way.

**Theorem 5.2** *Consider the Dirichlet b*.*v*.*p*. (1), *where* $F:[0,T]\times E\times E\u22b8E$ *is an upper semicontinuous mapping with compact*, *convex values*. *Assume that* $K\subset E$ *is an open*, *convex set containing* 0. *Moreover*, *let conditions* (${5}_{i}$), (${5}_{ii}$), (${5}_{iii}$) *from Theorem * 5.1 *be satisfied*.

*Furthermore*, *let there exist a function* $V\in {C}^{1}(E,\mathbb{R})$ *with a locally Lipschitz Frechét derivative* $\dot{V}$ *satisfying* (H1) *and* (H2). *Moreover*, *let*, *for all* $x\in \partial K$, $t\in (0,T)$, $\lambda \in (0,1)$ *and* $y\in E$ *satisfying* (11), *condition* (12) *hold for all* $w\in \lambda F(t,x,y)$. *Then the Dirichlet b*.*v*.*p*. (1) *admits a solution whose values are located in* $\overline{K}$.

*Proof* The verification is quite analogous as in Theorem 5.1 when just replacing the usage of Proposition 4.1 by Proposition 4.2. □