Combining the foregoing continuation principle with the ScorzaDragonitype technique (cf. Proposition 2.2), we are ready to state the main result of the paper concerning the solvability and localization of a solution of the multivalued Dirichlet problem (2).
Theorem 3.1 Consider the Dirichlet b.v.p. (2). Suppose that $F:[0,T]\times E\times E\u22b8E$ is an upperCarathéodory mapping which is either globally measurable or quasicompact. Furthermore, let $K\subset E$ be a nonempty, open, convex, bounded subset containing 0 of a separable Banach space E satisfying the RadonNikodym property. Let the following conditions (${2}_{\mathrm{i}}$)(${2}_{\mathrm{iii}}$) be satisfied:
(${2}_{\mathrm{i}}$) $\gamma (F(t,{\mathrm{\Omega}}_{1}\times {\mathrm{\Omega}}_{2}))\le g(t)(\gamma ({\mathrm{\Omega}}_{1})+\gamma ({\mathrm{\Omega}}_{2}))$, for a.a. $t\in [0,T]$ and each ${\mathrm{\Omega}}_{1}\subset \overline{K}$, and each bounded ${\mathrm{\Omega}}_{2}\subset E$, where $g\in {L}^{1}([0,T],[0,\mathrm{\infty}))$ and γ is the Hausdorff m.n.c. in E.
(
${2}_{\mathrm{ii}}$)
For every non
empty,
bounded $\mathrm{\Omega}\subset E$,
there exists ${\nu}_{\mathrm{\Omega}}\in {L}^{1}([0,T],[0,\mathrm{\infty}))$ such that $\parallel F(t,x,y)\parallel \le {\nu}_{\mathrm{\Omega}}(t),$
(9)
for a.a. $t\in [0,T]$ and all $(x,y)\in \mathrm{\Omega}\times E$.
(
${2}_{\mathrm{iii}}$)
$(T+4){\parallel g\parallel}_{{L}^{1}([0,T],\mathbb{R})}<4.$
Furthermore,
let there exist $\epsilon >0$ and a function $V\in {C}^{2}(E,\mathbb{R})$,
i.
e.
a twice continuously differentiable function in the sense of Fréchet,
satisfying (H1)(H3) (
cf.
Proposition 2.3)
with Fréchet derivative $\dot{V}$ Lipschitzian in $\overline{B(\partial K,\epsilon )}$.
^{b} Let there still exist $h>0$ such that $\u3008{\ddot{V}}_{x}(v),v\u3009\ge 0,\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}x\in B(\partial K,h),v\in E,$
(10)
where ${\ddot{V}}_{x}(v)$ denotes the second Fréchet derivative of V at x in the direction $(v,v)\in E\times E$.
Finally,
let $\u3008{\dot{V}}_{x},w\u3009>0,$
(11)
for a.a. $t\in (0,T)$ and all $x\in \partial K$, $v\in E$, and $w\in F(t,x,v)$.
Then the Dirichlet b.v.p. (2) admits a solution whose values are located in $\overline{K}$. If, moreover, $0\notin F(t,0,0)$, for a.a. $t\in [0,T]$, then the obtained solution is nontrivial.
Proof Since the proof of this result is rather technical, it will be divided into several steps. At first, let us define the sequence of approximating problems. For this purpose, let
k be as in Proposition 2.3 and consider a continuous function
$\tau :E\to [0,1]$ such that
$\tau (x)=0$, for all
$x\in E\setminus B(\partial K,k)$, and
$\tau (x)=1$, for all
$x\in \overline{B(\partial K,\frac{k}{2})}$. According to Proposition 2.3 (see also Remark 2.1), the function
$\stackrel{\u02c6}{\varphi}:E\to E$, where
$\stackrel{\u02c6}{\varphi}(x)=\{\begin{array}{ll}\tau (x)\cdot \varphi (x)\cdot \parallel {\dot{V}}_{x}\parallel ,& \text{for all}x\in \overline{B(\partial K,k)},\\ 0,& \text{for all}x\in E\setminus \overline{B(\partial K,k)},\end{array}$
is well defined, continuous and bounded.
Since the mapping $(t,x,y)\u22b8F(t,x,y)$ has, according to Proposition 2.2, the ScorzaDragoni property, we are able to find a decreasing sequence ${\{{J}_{m}\}}_{m}$ of subsets of $[0,T]$ and a mapping ${F}_{0}:[0,T]\times E\times E\u22b8E\cup \{\mathrm{\varnothing}\}$ with compact, convex values such that, for all $m\in \mathbb{N}$,

$\mu ({J}_{m})<\frac{1}{m}$,

$[0,T]\setminus {J}_{m}$ is closed,

$(t,x,y)\u22b8{F}_{0}(t,x,y)$ is u.s.c. on $[0,T]\setminus {J}_{m}\times E\times E$,

${\nu}_{\overline{K}}$ is continuous in $[0,T]\setminus {J}_{m}$ (cf. e.g.[2]).
If we put $J={\bigcap}_{m=1}^{\mathrm{\infty}}{J}_{m}$, then $\mu (J)=0$, ${F}_{0}(t,x,y)\ne \mathrm{\varnothing}$, for all $t\in [0,T]\setminus J$, the mapping $(t,x,y)\u22b8{F}_{0}(t,x,y)$ is u.s.c. on $[0,T]\setminus J\times E\times E$ and ${\nu}_{\overline{K}}$ is continuous in $[0,T]\setminus J$.
For each
$m\in \mathbb{N}$, let us define the mapping
${F}_{m}:[0,T]\times E\times E\u22b8E$ with compact, convex values by the formula
${F}_{m}(t,x,y):=\{\begin{array}{ll}{F}_{0}(t,x,y)+{\nu}_{\overline{K}}(t)({\chi}_{{J}_{m}}(t)+\frac{1}{m})\stackrel{\u02c6}{\varphi}(x),& \text{for all}(t,x,y)\in [0,T]\setminus J\times E\times E,\\ {\nu}_{\overline{K}}(t)({\chi}_{{J}_{m}}(t)+\frac{1}{m})\stackrel{\u02c6}{\varphi}(x),& \text{for all}(t,x,y)\in J\times E\times E.\end{array}$
Let us consider the b.v.p.
$({P}_{m})\phantom{\rule{1em}{0ex}}\begin{array}{l}\ddot{x}(t)\in {F}_{m}(t,x(t),\dot{x}(t)),\phantom{\rule{1em}{0ex}}\text{for a.a.}t\in [0,T],\\ x(T)=x(0)=0.\end{array}\}$
Now, let us verify the solvability of problems
$({P}_{m})$. Let
$m\in \mathbb{N}$ be fixed. Since
${F}_{0}$ is globally u.s.c. on
$[0,T]\setminus J\times E\times E$,
${F}_{m}(\cdot ,x,y)$ is measurable, for each
$(x,y)\in E\times E$, and, due to the continuity of
$\stackrel{\u02c6}{\varphi}$,
${F}_{m}(t,\cdot ,\cdot )$ is u.s.c., for all
$t\in [0,T]\setminus J$. Therefore,
${F}_{m}$ is an upperCarathéodory mapping. Moreover, let us define the upperCarathéodory mapping
${H}_{m}:[0,T]\times E\times E\times E\times E\times [0,1]\u22b8E$ by the formula
$\begin{array}{c}{H}_{m}(t,x,y,u,v,\lambda )\hfill \\ \phantom{\rule{1em}{0ex}}\equiv {H}_{m}(t,u,v,\lambda )\hfill \\ \phantom{\rule{1em}{0ex}}:=\{\begin{array}{ll}\lambda {F}_{0}(t,u,v)+{\nu}_{\overline{K}}(t)({\chi}_{{J}_{m}}(t)+\frac{1}{m})\stackrel{\u02c6}{\varphi}(u),& \text{for all}(t,x,y,u,v,\lambda )\in [0,T]\setminus J\\ \phantom{\text{for all}(t,x,y,u,v,\lambda )\in}\times {E}^{4}\times [0,1],\\ {\nu}_{\overline{K}}(t)({\chi}_{{J}_{m}}(t)+\frac{1}{m})\stackrel{\u02c6}{\varphi}(u),& \text{for all}(t,x,y,u,v,\lambda )\in J\times {E}^{4}\times [0,1].\end{array}\hfill \end{array}$
Let us show that, when $m\in \mathbb{N}$ is sufficiently large, all assumptions of Proposition 2.4 (for $\phi (t,x,\dot{x}):={F}_{m}(t,x,\dot{x})$) are satisfied.
For this purpose, 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
${P}_{m}(q,\lambda )\phantom{\rule{1em}{0ex}}\begin{array}{l}\ddot{x}(t)\in {H}_{m}(t,q(t),\dot{q}(t),\lambda ),\phantom{\rule{1em}{0ex}}\text{for a.a.}t\in [0,T],\\ x(T)=x(0)=0,\end{array}\}$
and denote by ${\mathfrak{T}}_{m}$ the solution mapping which assigns to each $(q,\lambda )\in Q\times [0,1]$ the set of solutions of ${P}_{m}(q,\lambda )$.
ad (i) In order to verify condition (i) in Proposition 2.4, we need to show that, for each
$(q,\lambda )\in Q\times [0,1]$, the problem
${P}_{m}(q,\lambda )$ is solvable with a convex set of solutions. So, let
$(q,\lambda )\in Q\times [0,1]$ be arbitrary and let
${f}_{q,\lambda}(\cdot )$ be a measurable selection of
${H}_{m}(\cdot ,q(\cdot ),\dot{q}(\cdot ),\lambda )$, which surely exists (see,
e.g., [[
27], Theorem 1.3.5]). According to (
${2}_{\mathrm{ii}}$) and the definition of
${H}_{m}$, it is also easy to see that
${f}_{q,\lambda}\in {L}^{1}([0,T],E)$. The homogeneous problem corresponding to b.v.p.
${P}_{m}(q,\lambda )$,
$\begin{array}{l}\ddot{x}(t)=0,\phantom{\rule{1em}{0ex}}\text{for a.a.}t\in [0,T],\\ x(T)=x(0)=0,\end{array}\}$
(12)
has only the trivial solution, and therefore the singlevalued Dirichlet problem
$\begin{array}{l}\ddot{x}(t)={f}_{q,\lambda}(t),\phantom{\rule{1em}{0ex}}\text{for a.a.}t\in [0,T],\\ x(T)=x(0)=0\end{array}\}$
admits a unique solution
${x}_{q,\lambda}(\cdot )$ which is one of solutions of
${P}_{m}(q,\lambda )$. This is given, for a.a.
$t\in [0,T]$, by
${x}_{q,\lambda}(t)={\int}_{0}^{T}G(t,s){f}_{q,\lambda}(s)\phantom{\rule{0.2em}{0ex}}ds$, where
G is the Green function associated to the homogeneous problem (12). The Green function
G and its partial derivative
$\frac{\partial}{\partial t}G$ are defined by (
cf. e.g. [[
28], pp.170171])
$\begin{array}{c}G(t,s)=\{\begin{array}{ll}\frac{(sT)t}{T},& \text{for all}0\le t\le s\le T,\\ \frac{(tT)s}{T},& \text{for all}0\le s\le t\le T,\end{array}\hfill \\ \frac{\partial}{\partial t}G(t,s)=\{\begin{array}{ll}\frac{(sT)}{T},& \text{for all}0\le ts\le T,\\ \frac{s}{T},& \text{for all}0\le st\le T.\end{array}\hfill \end{array}$
Thus, the set of solutions of ${P}_{m}(q,\lambda )$ is nonempty. The convexity of the solution sets follows immediately from the definition of ${H}_{m}$ and the fact that problems ${P}_{m}(q,\lambda )$ are fully linearized.
ad (ii) Let
$\mathrm{\Omega}\subset E\times E\times E\times E$ be bounded. Then, there exists a bounded
${\mathrm{\Omega}}_{1}\subset E$ such that
$\mathrm{\Omega}\subset {\mathrm{\Omega}}_{1}\times {\mathrm{\Omega}}_{1}\times {\mathrm{\Omega}}_{1}\times {\mathrm{\Omega}}_{1}$ and, according to (
${2}_{\mathrm{ii}}$) and the definition of
${H}_{m}$, there exists
$\stackrel{\u02c6}{J}\subset [0,T]$ with
$\mu (\stackrel{\u02c6}{J})=0$ such that, for all
$t\in [0,T]\setminus (J\cup \stackrel{\u02c6}{J})$,
$(x,y,u,v)\in \mathrm{\Omega}$ and
$\lambda \in [0,1]$,
$\parallel {H}_{m}(t,u,v,\lambda )\parallel \le {\nu}_{{\mathrm{\Omega}}_{1}}(t)+2{\nu}_{\overline{K}}(t)\cdot \underset{x\in \overline{B(\partial K,k)}}{max}\parallel \stackrel{\u02c6}{\varphi}(x)\parallel .$
Therefore, the mapping ${H}_{m}(t,q(t),\dot{q}(t),\lambda )$ satisfies condition (ii) from Proposition 2.4.
ad (iii) Since the verification of condition (iii) in Proposition 2.4 is technically the most complicated, it will be split into two parts: (iii_{1}) the quasicompactness of the solution operator ${\mathfrak{T}}_{m}$, (iii_{2}) the condensity of ${\mathfrak{T}}_{m}$ w.r.t. the monotone and nonsingular m.n.c. α defined by (7).
ad (iii
_{1}) Let us firstly prove that the solution mapping
${\mathfrak{T}}_{m}$ is quasicompact. Since
${C}^{1}([0,T],E)$ is a complete metric space, it is sufficient to prove the sequential quasicompactness of
${\mathfrak{T}}_{m}$. 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
${x}_{n}\in {\mathfrak{T}}_{m}({q}_{n},{\lambda}_{n})$, for all
$n\in \mathbb{N}$. Then there exists, for all
$n\in \mathbb{N}$,
${k}_{n}(\cdot )\in {F}_{0}(\cdot ,{q}_{n}(\cdot ),{\dot{q}}_{n}(\cdot ))$ such that
${\ddot{x}}_{n}(t)={f}_{n}(t),\phantom{\rule{1em}{0ex}}\text{for a.a.}t\in [0,T],$
(13)
where
${f}_{n}(t)={\lambda}_{n}{k}_{n}(t)+{\nu}_{\overline{K}}(t)({\chi}_{{J}_{m}}(t)+\frac{1}{m})\stackrel{\u02c6}{\varphi}({q}_{n}(t)),$
(14)
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 $\mathrm{\Omega}\times \mathrm{\Omega}\subset E\times E$ such that $({q}_{n}(t),{\dot{q}}_{n}(t))\in \mathrm{\Omega}\times \mathrm{\Omega}$, for all $t\in [0,T]$ and $n\in \mathbb{N}$. Therefore, there exists, according to condition (${2}_{\mathrm{ii}}$), ${\nu}_{\mathrm{\Omega}}\in {L}^{1}([0,T],[0,\mathrm{\infty}))$ such that $\parallel {f}_{n}(t)\parallel \le \varpi (t)$, for every $n\in \mathbb{N}$ and a.a. $t\in [0,T]$, where $\varpi (t):={\nu}_{\mathrm{\Omega}}(t)+2{\nu}_{\overline{K}}(t)\cdot {max}_{x\in \overline{B(\partial K,\epsilon )}}\parallel \stackrel{\u02c6}{\varphi}(x)\parallel $.
Moreover, for every
$n\in \mathbb{N}$ and a.a.
$t\in [0,T]$,
${x}_{n}(t)={\int}_{0}^{T}G(t,s){f}_{n}(s)\phantom{\rule{0.2em}{0ex}}ds$
(15)
and
${\dot{x}}_{n}(t)={\int}_{0}^{T}\frac{\partial}{\partial t}G(t,s){f}_{n}(s)\phantom{\rule{0.2em}{0ex}}ds.$
(16)
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
$a:=\frac{T}{4}{\int}_{0}^{T}\varpi (s)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}b:={\int}_{0}^{T}\varpi (s)\phantom{\rule{0.2em}{0ex}}ds.$
Furthermore, for every
$n\in \mathbb{N}$ and a.a.
$t\in [0,T]$, we have
$\parallel {\ddot{x}}_{n}(t)\parallel \le \varpi (t).$
Hence, the sequences $\{{x}_{n}\}$ and $\{{\dot{x}}_{n}\}$ are bounded and $\{{\ddot{x}}_{n}\}$ is uniformly integrable.
For each
$t\in [0,T]$, the properties of the Hausdorff m.n.c. yield
$\begin{array}{rcl}\gamma \left({\{{f}_{n}(t)\}}_{n}\right)& \le & \gamma \left({\{{\lambda}_{n}{k}_{n}(t)\}}_{n}\right)+{\nu}_{\overline{K}}(t)({\chi}_{{J}_{m}}(t)+\frac{1}{m})\gamma \left({\left\{\stackrel{\u02c6}{\varphi}({q}_{n}(t))\right\}}_{n}\right)\\ \le & \gamma \left({\{{k}_{n}(t)\}}_{n}\right)+{\nu}_{\overline{K}}(t)({\chi}_{{J}_{m}}(t)+\frac{1}{m})\\ \times \gamma \left(\{\varphi ({q}_{n}(t))\parallel {\dot{V}}_{{q}_{n}(t)}\parallel :{q}_{n}(t)\in \overline{B(\partial K,\epsilon )}\}\right).\end{array}$
Since
${q}_{n}(t)\in \overline{K}$, for all
$t\in [0,T]$ and all
$n\in \mathbb{N}$, it follows from condition (
${2}_{\mathrm{i}}$) that, for a.a.
$t\in [0,T]$,
$\begin{array}{rcl}\gamma \left({\{{f}_{n}(t)\}}_{n}\right)& \le & g(t)(\gamma \left({\{{q}_{n}(t)\}}_{n}\right)+\gamma \left({\{{\dot{q}}_{n}(t)\}}_{n}\right))\\ +{\nu}_{\overline{K}}(t)({\chi}_{{J}_{m}}(t)+\frac{1}{m})\gamma \left(\{\varphi ({q}_{n}(t))\parallel {\dot{V}}_{{q}_{n}(t)}\parallel :{q}_{n}(t)\in \overline{B(\partial K,\epsilon )}\}\right)\\ \le & g(t)\underset{t\in [0,T]}{sup}(\gamma \left({\{{q}_{n}(t)\}}_{n}\right)+\gamma \left({\{{\dot{q}}_{n}(t)\}}_{n}\right))\\ +{\nu}_{\overline{K}}(t)({\chi}_{{J}_{m}}(t)+\frac{1}{m})\gamma \left(\{\varphi ({q}_{n}(t))\parallel {\dot{V}}_{{q}_{n}(t)}\parallel :{q}_{n}(t)\in \overline{B(\partial K,\epsilon )}\}\right).\end{array}$
Since the function
$x\to \varphi (x)\parallel {\dot{V}}_{x}\parallel $ is Lipschitzian on
$\overline{B(\partial K,\epsilon )}$ with some Lipschitz constant
$\stackrel{\u02c6}{L}>0$ (see Remark 2.1), we get
$\gamma \left({\{{f}_{n}(t)\}}_{n}\right)\le (g(t)+\stackrel{\u02c6}{L}{\nu}_{\overline{K}}(t)({\chi}_{{J}_{m}}(t)+\frac{1}{m}))\underset{t\in [0,T]}{sup}(\gamma \left({\{{q}_{n}(t)\}}_{n}\right)+\gamma \left({\{{\dot{q}}_{n}(t)\}}_{n}\right)).$
(17)
Since ${q}_{n}\to q$ and ${\dot{q}}_{n}\to \dot{q}$ in $C([0,T],E)$, we get, for all $t\in [0,T]$, $\gamma ({\{{q}_{n}(t)\}}_{n})=\gamma ({\{{\dot{q}}_{n}(t)\}}_{n})=0$, which implies that $\gamma ({\{{f}_{n}(t)\}}_{n})=0$, for all $t\in [0,T]$.
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 semihomogeneity of the Hausdorff m.n.c.,
$\gamma \left(\{G(t,s){f}_{n}(s)\}\right)\le G(t,s)\gamma \left(\{{f}_{n}(s)\}\right)=0,\phantom{\rule{1em}{0ex}}\text{for all}(t,s)\in [0,T]\times [0,T].$
(18)
Moreover, by means of (4) and (18),
$\gamma \left(\{{x}_{n}(t)\}\right)=\gamma \left(\{{\int}_{0}^{T}G(t,s){f}_{n}(s)\phantom{\rule{0.2em}{0ex}}ds\}\right)=0,\phantom{\rule{1em}{0ex}}\text{for all}t\in [0,T].$
By similar reasoning, we also get
$\gamma \left(\{{\dot{x}}_{n}(t)\}\right)=0,\phantom{\rule{1em}{0ex}}\text{for all}t\in [0,T],$
by which $\{{x}_{n}(t)\}$, $\{{\dot{x}}_{n}(t)\}$ are relatively compact, for all $t\in [0,T]$.
Moreover, since ${x}_{n}$ satisfies for all $n\in \mathbb{N}$ (13), $\{{\ddot{x}}_{n}(t)\}$ is relatively compact, for a.a. $t\in [0,T]$. Thus, according to [[23], Lemma III.1.30], 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 ${L}^{1}([0,T],E)$. According to the classical closure results (cf. e.g. [[27], Lemma 5.1.1]), $x\in {\mathfrak{T}}_{m}(q,\lambda )$, which implies the quasicompactness of ${\mathfrak{T}}_{m}$.
ad (iii
_{2}) In order to show that, for
$m\in \mathbb{N}$ sufficiently large,
${\mathfrak{T}}_{m}$ is
αcondensing with respect to the m.n.c.
α defined by (7), let us consider a bounded subset
$\mathrm{\Theta}\subset Q$ such that
$\alpha ({\mathfrak{T}}_{m}(\mathrm{\Theta}\times [0,1]))\ge \alpha (\mathrm{\Theta})$. Let
$\{{x}_{n}\}\subset {\mathfrak{T}}_{m}(\mathrm{\Theta}\times [0,1])$ be a sequence such that
$\begin{array}{c}\alpha \left({\mathfrak{T}}_{m}(\mathrm{\Theta}\times [0,1])\right)\hfill \\ \phantom{\rule{1em}{0ex}}=(\underset{t\in [0,T]}{sup}[\gamma \left({\{{x}_{n}(t)\}}_{n}\right)+\gamma \left({\{{\dot{x}}_{n}(t)\}}_{n}\right)],{mod}_{C}\left({\{{x}_{n}\}}_{n}\right)+{mod}_{C}\left({\{{\dot{x}}_{n}\}}_{n}\right)).\hfill \end{array}$
At first, let us show that the set
${\mathfrak{T}}_{m}(\mathrm{\Theta}\times [0,1])$ is bounded. If
$x\in {\mathfrak{T}}_{m}(\mathrm{\Theta}\times [0,1])$, then there exist
$q\in \mathrm{\Theta}$,
$\lambda \in [0,1]$ and
$k(\cdot )\in {F}_{0}(\cdot ,q(\cdot ),\dot{q}(\cdot ))$ such that
$x(t)={\int}_{0}^{T}G(t,s)f(s)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{2em}{0ex}}\dot{x}(t)={\int}_{0}^{T}\frac{\partial G(t,s)}{\partial t}f(s)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}\text{for all}t\in [0,T],$
with $f(t)=\lambda k(t)+{\nu}_{\overline{K}}(t)({\chi}_{{J}_{m}}(t)+\frac{1}{m})\stackrel{\u02c6}{\varphi}(q(t))$, for a.a. $t\in [0,T]$.
Since Θ is bounded, there exists
$\mathrm{\Omega}\subset E$ such that
$q(t)\in \mathrm{\Omega}$, for all
$q\in \mathrm{\Theta}$ and all
$t\in [0,T]$. Hence, according to (
${2}_{\mathrm{ii}}$), there exists
${\nu}_{\mathrm{\Omega}}\in {L}^{1}([0,T])$ such that
$\parallel k(t)\parallel \le {\nu}_{\mathrm{\Omega}}(t)$, for a.a.
$t\in [0,T]$. Consequently
$\begin{array}{rcl}{\parallel x(t)\parallel}_{E}& \le & \underset{(t,s)\in [0,1]\times [0,1]}{max}G(t,s)[{\int}_{0}^{T}{\nu}_{\mathrm{\Omega}}(s)\phantom{\rule{0.2em}{0ex}}ds+2\underset{x\in \overline{B(\partial K,k)}}{max}\parallel \stackrel{\u02c6}{\varphi}(x)\parallel {\int}_{0}^{T}{\nu}_{\overline{K}}(t)]\\ \le & \frac{T}{4}\parallel {\nu}_{\mathrm{\Omega}}\parallel +2\underset{x\in \overline{B(\partial K,k)}}{max}\parallel \stackrel{\u02c6}{\varphi}(x)\parallel \cdot \parallel {\nu}_{\overline{K}}\parallel .\end{array}$
Similarly,
$\begin{array}{rcl}{\parallel \dot{x}(t)\parallel}_{E}& \le & \underset{(t,s)\in [0,1]\times [0,1]}{max}\left\frac{\partial G(t,s)}{\partial}\right[{\int}_{0}^{T}\parallel k(s)\parallel \phantom{\rule{0.2em}{0ex}}ds+2\underset{x\in \overline{B(\partial K,k)}}{max}\parallel \stackrel{\u02c6}{\varphi}(x)\parallel {\int}_{0}^{T}{\nu}_{\overline{K}}(t)]\\ \le & \parallel {\nu}_{\mathrm{\Omega}}\parallel +2\underset{x\in \overline{B(\partial K,k)}}{max}\parallel \stackrel{\u02c6}{\varphi}(x)\parallel \cdot \parallel {\nu}_{\overline{K}}\parallel .\end{array}$
Thus, the set ${\mathfrak{T}}_{m}(\mathrm{\Theta}\times [0,1])$ is bounded.
Moreover, we can find $\{{q}_{n}\}\subset \mathrm{\Theta}$, $\{{\lambda}_{n}\}\subset [0,1]$ and $\{{k}_{n}\}$ satisfying, for a.a. $t\in [0,T]$, ${k}_{n}(t)\in {F}_{0}(t,{q}_{n}(t),{\dot{q}}_{n}(t))$, such that, for all $t\in [0,T]$, ${x}_{n}(t)$ and ${\dot{x}}_{n}(t)$ are defined by (15) and (16), respectively, where ${f}_{n}(t)$ is defined by (14).
By similar reasoning as in the part
ad (iii
_{1}), we obtain
$\gamma \left({\{{f}_{n}(t)\}}_{n}\right)\le (g(t)+\stackrel{\u02c6}{L}{\nu}_{\overline{K}}(t)({\chi}_{{J}_{m}}(t)+\frac{1}{m}))\underset{t\in [0,T]}{sup}(\gamma \left({\{{q}_{n}(t)\}}_{n}\right)+\gamma \left({\{{\dot{q}}_{n}(t)\}}_{n}\right)),$
for a.a.
$t\in [0,T]$, and that
$\parallel {f}_{n}(t)\parallel \le \parallel {k}_{n}(t)\parallel +2\cdot \underset{x\in \overline{B(\partial K,\epsilon )}}{max}\parallel \stackrel{\u02c6}{\varphi}(x)\parallel \cdot {\nu}_{\overline{K}}(t),\phantom{\rule{1em}{0ex}}\text{for a.a.}t\in [0,T]\text{and all}n\in \mathbb{N}.$
Since
${k}_{n}(t)\in {F}_{0}(t,{q}_{n}(t),{\dot{q}}_{n}(t))$, for a.a.
$t\in [0,T]$, and
${q}_{n}\in \mathrm{\Theta}$, for all
$n\in \mathbb{N}$, where Θ is a bounded subset of
${C}^{1}([0,T],E)$, there exists
$\mathrm{\Omega}\subset \overline{K}$ such that
${q}_{n}(t)\in \mathrm{\Omega}$, for all
$n\in \mathbb{N}$ and
$t\in [0,T]$. Hence, it follows from condition (
${2}_{\mathrm{ii}}$) that
$\parallel {f}_{n}(t)\parallel \le {\nu}_{\mathrm{\Omega}}(t)+2\cdot {\nu}_{\overline{K}}(t)\cdot \underset{x\in \overline{B(\partial K,\epsilon )}}{max}\parallel \stackrel{\u02c6}{\varphi}(x)\parallel ,\phantom{\rule{1em}{0ex}}\text{for a.a.}t\in [0,T].$
(19)
This implies $\parallel G(t,s){f}_{n}(t)\parallel \le G(t,s)({\nu}_{\mathrm{\Omega}}(t)+2\cdot {\nu}_{\overline{K}}(t)\cdot {max}_{x\in \overline{B(\partial K,\epsilon )}}\parallel \stackrel{\u02c6}{\varphi}(x)\parallel )$, for a.a. $t,s\in [0,T]$ and all $n\in \mathbb{N}$.
Moreover, by virtue of the semihomogeneity of the Hausdorff m.n.c., for all
$(t,s)\in [0,T]\times [0,T]$, we have
$\begin{array}{rcl}\gamma \left({\{G(t,s){f}_{n}(s)\}}_{n}\right)& \le & G(t,s)\gamma \left({\{{f}_{n}(s)\}}_{n}\right)\le \frac{T}{4}\gamma \left({\{{f}_{n}(s)\}}_{n}\right)\\ \le & \frac{T}{4}(g(t)+\stackrel{\u02c6}{L}{\nu}_{\overline{K}}(t)({\chi}_{{J}_{m}}(t)+\frac{1}{m}))\\ \times \underset{t\in [0,T]}{sup}(\gamma \left({\{{q}_{n}(t)\}}_{n}\right)+\gamma \left({\{{\dot{q}}_{n}(t)\}}_{n}\right)).\end{array}$
Let us denote
$\mathcal{S}:=\underset{t\in [0,T]}{sup}(\gamma \left({\{{q}_{n}(t)\}}_{n}\right)+\gamma \left({\{{\dot{q}}_{n}(t)\}}_{n}\right))$
and
${\mathcal{S}}^{\ast}:=\underset{t\in [0,T]}{sup}(\gamma \left({\{{x}_{n}(t)\}}_{n}\right)+\gamma \left({\{{\dot{x}}_{n}(t)\}}_{n}\right)).$
According to (4) and (15) we thus obtain for each
$t\in [0,T]$,
$\begin{array}{rcl}\gamma \left({\{{x}_{n}(t)\}}_{n}\right)& =& \gamma \left({\{{\int}_{0}^{T}G(t,s){f}_{n}(s)\phantom{\rule{0.2em}{0ex}}ds\}}_{n}\right)\\ \le & \frac{T}{4}({\parallel g\parallel}_{{L}^{1}}+\stackrel{\u02c6}{L}({\parallel {\nu}_{\overline{K}}\parallel}_{{L}^{1}({J}_{m})}+\frac{1}{m}{\parallel {\nu}_{\overline{K}}\parallel}_{{L}^{1}}))\mathcal{S}.\end{array}$
By similar reasonings, we can see that, for each
$t\in [0,T]$,
$\gamma \left({\{{\dot{x}}_{n}(t)\}}_{n}\right)\le ({\parallel g\parallel}_{{L}^{1}}+\stackrel{\u02c6}{L}({\parallel {\nu}_{\overline{K}}\parallel}_{{L}^{1}({J}_{m})}+\frac{1}{m}{\parallel {\nu}_{\overline{K}}\parallel}_{{L}^{1}}))\mathcal{S},$
when starting from condition (16). Subsequently,
${\mathcal{S}}^{\ast}\le \frac{T+4}{4}({\parallel g\parallel}_{{L}^{1}}+\stackrel{\u02c6}{L}({\parallel {\nu}_{\overline{K}}\parallel}_{{L}^{1}({J}_{m})}+\frac{1}{m}{\parallel {\nu}_{\overline{K}}\parallel}_{{L}^{1}}))\mathcal{S}.$
(20)
Since we assume that
$\alpha ({\mathfrak{T}}_{m}(\mathrm{\Theta}\times [0,1]))\ge \alpha (\mathrm{\Theta})$ and
${\{{q}_{n}\}}_{n}\subset \mathrm{\Theta}$, we get
$\mathcal{S}\le {\mathcal{S}}^{\ast}\le \frac{T+4}{4}({\parallel g\parallel}_{{L}^{1}}+\stackrel{\u02c6}{L}({\parallel {\nu}_{\overline{K}}\parallel}_{{L}^{1}({J}_{m})}+\frac{1}{m}{\parallel {\nu}_{\overline{K}}\parallel}_{{L}^{1}}))\mathcal{S}.$
Since we have, according to (
${2}_{\mathrm{iii}}$),
$\frac{T+4}{4}{\parallel g\parallel}_{{L}^{1}}<1$, we can choose
${m}_{0}\in \mathbb{N}$ such that, for all
$m\in \mathbb{N}$,
$m\ge {m}_{0}$, we have
$\frac{T+4}{4}({\parallel g\parallel}_{{L}^{1}}+\stackrel{\u02c6}{L}({\parallel {\nu}_{\overline{K}}\parallel}_{{L}^{1}({J}_{m})}+\frac{1}{m}{\parallel {\nu}_{\overline{K}}\parallel}_{{L}^{1}}))<1.$
Therefore, we get, for sufficiently large $m\in \mathbb{N}$, the contradiction $\mathcal{S}<\mathcal{S}$ which ensures the validity of condition (iii) in Proposition 2.4.
ad (iv) For all
$q\in Q$, the set
${\mathfrak{T}}_{m}(q,0)$ coincides with the unique solution
${x}_{m}$ of the linear system
$\begin{array}{l}\ddot{x}(t)={\nu}_{\overline{K}}(t)({\chi}_{{J}_{m}}(t)+\frac{1}{m})\stackrel{\u02c6}{\varphi}(q(t)),\phantom{\rule{1em}{0ex}}\text{for a.a.}t\in [0,T],\\ x(T)=x(0)=0.\end{array}\}$
According to (15) and (16), for all
$t\in [0,T]$,
${x}_{m}(t)={\int}_{0}^{T}G(t,s){\phi}_{m}(s)\phantom{\rule{0.2em}{0ex}}ds$
and
${\dot{x}}_{m}(t)={\int}_{0}^{T}\frac{\partial}{\partial t}G(t,s){\phi}_{m}(s)\phantom{\rule{0.2em}{0ex}}ds,$
where ${\phi}_{m}(t):={\nu}_{\overline{K}}(t)({\chi}_{{J}_{m}}(t)+\frac{1}{m})\stackrel{\u02c6}{\varphi}(q(t))$.
Since
${\parallel {\phi}_{m}\parallel}_{{L}^{1}}\le \underset{x\in \overline{B(\partial K,\epsilon )}}{max}\parallel \stackrel{\u02c6}{\varphi}(x)\parallel \cdot ({\parallel {\nu}_{\overline{K}}\parallel}_{{L}^{1}({J}_{m})}+\frac{{\parallel {\nu}_{\overline{K}}\parallel}_{{L}^{1}}}{m}),$
we have, for all
$t\in [0,T]$,
$\parallel {x}_{m}(t)\parallel \le \frac{{T}^{2}}{4}\cdot \underset{x\in \overline{B(\partial K,\epsilon )}}{max}\parallel \stackrel{\u02c6}{\varphi}(x)\parallel \cdot ({\parallel {\nu}_{\overline{K}}\parallel}_{{L}^{1}({J}_{m})}+\frac{{\parallel {\nu}_{\overline{K}}\parallel}_{{L}^{1}}}{m}).$
(21)
Let us now consider $r>0$ such that $rB\subset K$. Then it follows from (21) that we are able to find ${m}_{0}\in \mathbb{N}$ such that, for all $m\in \mathbb{N}$, $m\ge {m}_{0}$, and $t\in [0,T]$, $\parallel {x}_{m}\parallel \le r$. Therefore, for all $m\in \mathbb{N}$, $m\ge {m}_{0}$, ${\mathfrak{T}}_{m}(q,0)\subset IntQ$, for all $q\in Q$, which ensures the validity of condition (iv) in Proposition 2.4.
ad (v) The validity of the transversality condition (v) in Proposition 2.4 can be proven quite analogously as in [
16] (see pp.4043 in [
16]) with the following differences:

due to the Dirichlet boundary conditions, ${t}_{0}$ belongs to the open interval $(0,T)$,

since $A(t)=B(t)=0$, we have $p(t)={\nu}_{\overline{K}}(t)$.
In this way, we can prove that there exists ${m}_{0}\in \mathbb{N}$ such that every problem $({P}_{m})$, where $m\ge {m}_{0}$, satisfies all the assumptions of Proposition 2.4. This implies that every such $({P}_{m})$ admits a solution, denoted by ${x}_{m}$, with ${x}_{m}(t)\in \overline{K}$, for all $t\in [0,T]$. By similar arguments as in [16], but with the expression $Z(4Zk+1)$ replaced by $\frac{T}{4}$, according to condition (${2}_{\mathrm{ii}}$), we can obtain the result that there exists a subsequence, denoted as the sequence, and a function $x\in A{C}^{1}([0,T],E)$ such that ${x}_{m}\to x$ and ${\dot{x}}_{m}\to \dot{x}$ in $C([0,T],E)$ and also ${\ddot{x}}_{m}\rightharpoonup x$ in ${L}^{1}([0,T],E)$, when $m\to \mathrm{\infty}$. Thus, a classical closure result (see e.g. [[27], Lemma 5.1.1]) guarantees that x is a solution of (2) satisfying $x(t)\in \overline{K}$, for all $t\in [0,T]$, and the sketch of proof is so complete. □
The case when $F={F}_{1}+{F}_{2}$, with ${F}_{1}(t,\cdot ,\cdot )$ to be completely continuous and ${F}_{2}(t,\cdot ,\cdot )$ to be Lipschitzian, for a.a. $t\in [0,T]$, represents the most classical example of a map which is γregular w.r.t. the Hausdorff measure of noncompactness γ. The following corollary of Theorem 3.1 can be proved quite analogously as in [[3], Example 6.1 and Remark 6.1].
Corollary 3.1 Let $E=H$ be a separable Hilbert space and let us consider the Dirichlet b.
v.
p.:
$\begin{array}{l}\ddot{x}(t)\in {F}_{1}(t,x(t),\dot{x}(t))+{F}_{2}(t,x(t),\dot{x}(t)),\phantom{\rule{1em}{0ex}}\mathit{\text{for a.a.}}t\in [0,T],\\ x(0)=x(T)=0,\end{array}\}$
(22)
where
 (i)
${F}_{1}:[0,T]\times H\times H\u22b8H$ is an upper
Carathéodory,
globally measurable,
multivalued mapping and ${F}_{1}(t,\cdot ,\cdot ):H\times H\u22b8H$ is completely continuous,
for a.
a.
$t\in [0,T]$,
such that $\parallel {F}_{1}(t,x,y)\parallel \le {\nu}_{1}(t,{D}_{0}),$
for a.
a.
$t\in [0,T]$,
all $x\in H$ with $\parallel x\parallel \le {D}_{0}$,
where ${D}_{0}>0$ is an arbitrary constant,
${\nu}_{1}\in {L}^{1}([0,T],[0,\mathrm{\infty}))$,
and all $y\in H$,
 (ii)
${F}_{2}:[0,T]\times H\times H\u22b8H$
is a Carathéodory multivalued mapping such that
$\parallel {F}_{2}(t,0,0)\parallel \le {\nu}_{2}(t),\phantom{\rule{1em}{0ex}}\mathit{\text{for a.a.}}t\in [0,T],$
where ${\nu}_{2}\in {L}^{1}([0,T],[0,\mathrm{\infty}))$,
and ${F}_{2}(t,\cdot ,\cdot ):H\times H\u22b8H$ is Lipschitzian,
for a.
a.
$t\in [0,T]$,
with the Lipschitz constant Moreover,
suppose that  (iii)
there exists $R>0$ such that,
for all $x\in H$ with $\parallel x\parallel =R$,
$t\in (0,T)$,
$y\in H$ and $w\in {F}_{1}(t,x,y)+{F}_{2}(t,x,y)$,
we have
Then the Dirichlet problem (22) admits, according to Theorem 3.1, a solution $x(\cdot )$ such that $\parallel x(t)\parallel \le R$, for all $t\in [0,T]$.
Remark 3.1 For ${F}_{2}(t,x,y)\equiv 0$, the completely continuous mapping ${F}_{1}(t,x,y)$ allows us to make a comparison with classical singlevalued results recalled in the Introduction. Unfortunately, our ${F}_{1}$ in (i) (see also (${2}_{\mathrm{ii}}$) in Theorem 3.1) is the only mapping which is (unlike in [[3], Example 6.1 and Remark 6.1], where under some additional restrictions quite liberal growth restrictions were permitted) globally bounded w.r.t. $y\in H$. Furthermore, our sign condition in (iii) is also (unlike again in [[3], Example 6.1 and Remark 6.1], where under some additional restrictions the Hartmantype condition like (i_{H}) in the Introduction was employed) the most restrictive among their analogies in [6–13]. On the other hand, because of multivalued upperCarathéodory maps ${F}_{1}+{F}_{2}$ in a Hilbert space which are γregular, our result has still, as far as we know, no analogy at all.