Let us consider at first the scalar Dirichlet problem for differential equations involving a dry friction (4), where *a*, *b*, *c*, \(x_{0}\), \(x_{T}\) and \(T >0\) are real constants and \(p\colon J \rightarrow \mathbb{R}\), \(J=[0,T]\), is a Lebesgue integrable function.

Since the function \(\operatorname{sgn}(\cdot)\) is discontinuous in the spatial variable, problem (4) need not have a *Carathéodory solution*, *i.e.* a function \(x \colon J \rightarrow\mathbb {R}\) with an absolutely continuous derivative, satisfying (4), for almost all \(t \in J\). Therefore, we need another notion of an appropriate solution, namely the one in the sense of Filippov. For this goal, we use the concept of the Filippov-like regularization (see [15]) of spatially discontinuous maps. More precisely, applying Definition 2 to the right-hand side involving spatial discontinuities, we can speak about a *solution in the sense of Filippov* of the original problem, provided it is a Carathéodory solution of a multivalued problem with a Filippov-like regularized right-hand side.

In our situation, the discontinuous function to be regularized is the function *signum*. On the basis of the Filippov-like regularization of \(\operatorname{sgn}(\cdot)\), we obtain the multivalued mapping *Signum* defined in (3), *i.e.*
\(\operatorname{Sgn}(\cdot)\).

One can readily check that the *Signum* mapping is u.s.c. with compact and convex values. Hence, after the described Filippov-like regularization, problem (4) with a discontinuous function \(\operatorname{sgn}(\cdot)\) becomes multivalued, *i.e.* (5).

By a *Filippov solution* of (4), we understand a function \(x(\cdot)\colon J \rightarrow\mathbb{R}\) with absolutely continuous derivative, satisfying problem (5), almost everywhere on *J*.

For our needs, it will be convenient to consider still the problem involving more multivalued terms, namely (9), where \(F_{1} \colon\mathbb{R}\multimap\mathbb{R}\), \(F_{2} \colon\mathbb {R}\multimap\mathbb{R}\) are u.s.c. maps with compact, convex values and \(P \colon J \multimap\mathbb{R}\) is an Aumann integrable mapping, \(J=[0,T]\).

Making the change of variables \(y(t) = x(t) - v(t)\), where \(v(t) = \frac{x_{T}-x_{0}}{T} t + x_{0}\), we can immediately see that \(x(t)\) is a Carathéodory solution of (9) if and only if \(y(t)\) satisfies

$$ \left . \textstyle\begin{array}{l} y''(t)+a (y'(t)+v'(t))+b (y(t)+v(t)) \\ \quad \in P(t)+F_{1}(y(t)+v(t)) +F_{2}(y'(t)+v'(t))-c \operatorname{Sgn} (y'(t)+v'(t) ),\quad \text{for a.a. } t \in J, \\ y(0)=0,\qquad y(T)=0. \end{array}\displaystyle \right \} $$

(12)

One can also easily check that \(|v(t)| \leq k_{2} := \max\{|x_{0}|, |x_{T}|\} \), for all \(t\in J\), and \(|v'(t)|= k_{3} :=\frac{|x_{T}-x_{0}|}{T}\), for all \(t \in J\).

Our aim at this moment is to prove the existence and localization theorems for problem (9). Since the desired sufficient conditions will tendentiously take the form of growth restrictions, let us start with the problem involving the truncated maps of \(F_{1}\), \(F_{2}\), *i.e.*

$$ \left . \textstyle\begin{array}{l} y''(t)+a y'(t)+b y(t) \\ \quad \in P(t)-a v'(t)-b v(t)+F_{1}^{*}(y(t)+v(t)) \\ \qquad {}+F_{2}^{*}(y'(t)+v'(t))-c \operatorname{Sgn} (y'(t)+v'(t) ),\quad \text{for a.a. } t \in J, \\ y(0)=0, \qquad y(T)=0, \end{array}\displaystyle \right \} $$

(13)

where

$$\begin{aligned}& F_{1}^{*}(z) := \left \{ \textstyle\begin{array}{l@{\quad}l} F_{1}(z), & \text{for } |z| \leq D, \\ F_{1}(D \operatorname{sgn}(z)), & \text{for } |z| \geq D, \end{array}\displaystyle \right . \\& F_{2}^{*}(z) := \left \{ \textstyle\begin{array}{l@{\quad}l} F_{2}(z), & \text{for } |z| \leq D, \\ F_{2}(D \operatorname{sgn}(z)), & \text{for } |z| \geq D. \end{array}\displaystyle \right . \end{aligned}$$

Obviously, \(F_{1} \vert_{\overline{B}_{D}} \colon\overline{B}_{D} \multimap\mathbb{R}\) and \(F_{2} \vert_{\overline{B}_{D}} \colon \overline{B}_{D} \multimap\mathbb{R}\) are u.s.c. maps with compact and convex values, where \(\overline{B}_{D} := \{ z \in\mathbb{R}\mid|z| \leq D\}\) is the closed ball, constant *D* is such that \(D = D_{0}+k_{2}+k_{3}\), where \(D_{0}>0\) is a suitable constant which will be specified later, and \(k_{2}\), \(k_{3}\) are defined above. Thus, the same is true for \(F_{1}^{*}, F_{2}^{*} \colon\mathbb{R}\multimap\mathbb{R}\) defined as above.

Observe that, in this way, \(F_{1}^{*}(z+v(t)) = F_{1}(z+v(t))\), for \(|z| \leq D_{0}\), and \(F_{2}^{*}(z+v'(t)) = F_{2}(z+v'(t))\), for \(|z| \leq D_{0}\).

Hence, let us find sufficient conditions for the solvability of (13). We distinguish two cases in order to separate formally a linear differential operator and a multivalued perturbation.

In this section, we will deal with the problem formally written in the following way:

$$ \left . \textstyle\begin{array}{l} y''(t) \in F(t,y(t),y'(t)),\quad \text{for a.a. } t \in J, \\ y(0)=0, \qquad y(T)=0, \end{array}\displaystyle \right \} $$

(14)

where \(F(t,y(t),y'(t)):=P(t)+F_{1}^{*}(y(t)+v(t))+F_{2}^{*}(y'(t)+v'(t))-a (y'(t)+v'(t)) -b (y(t)+v(t))-c \operatorname{Sgn} (y'(t)+v'(t) )\), *i.e.*
\(F\colon J \times\mathbb {R}\times\mathbb{R}\multimap\mathbb{R}\) is a u-Carathéodory multivalued mapping and \(L y(t) := y''(t)\).

For a solvability of the multivalued nonlinear problem (14), we use the Schauder linearization device. Thus, we parametrize the right-hand side (r.h.s.) *F* in order to have a one-parameter family of linear problems. Let

$$Q:=\bigl\{ u \in C^{1}(J,\mathbb{R}),\|u\|_{C^{1}}\leq D_{0}\bigr\} $$

be the set of candidate solutions, where \(\|u\|_{C^{1}}:=\sup\{ |u(t)|+|u'(t)|, t \in J\}\) and \(D_{0}>0\) is a suitable constant.

Then, for each \(q \in Q\), we get a fully linearized problem

$$ \left . \textstyle\begin{array}{l} y''(t)\in F_{q}(t), \quad \text{for a.a. } t \in J, \\ y(0) = 0,\qquad y(T)=0, \end{array}\displaystyle \right \} $$

(15)

where \(F_{q}(t)=P(t)+F_{1}^{*}(q(t)+v(t))+F_{2}^{*}(q'(t)+v'(t))-a q'(t)-a v'(t) -b q(t)-b v(t)-c \operatorname{Sgn} (q'(t)+v'(t) )\). Obviously, \(F_{q} \colon J\multimap\mathbb{R}\) is, for every \(q \in Q\), an Aumann integrable function of *t*.

From Lemma 3, the existence follows of at least one measurable selection \(f_{q} \subset F_{q}\) of the multivalued composition \(F_{q}(t)\). Thus, we can consider the single-valued linear Dirichlet problem

$$ \left . \textstyle\begin{array}{l} y''(t) = f_{q}(t), \quad \text{for a.a. } t \in J, \\ y(0) = 0, \qquad y(T)=0. \end{array}\displaystyle \right \} $$

(16)

The homogeneous problem associated with the problem (16), *i.e.*

$$ \left . \textstyle\begin{array}{l} y''(t) = 0, \quad \text{for a.a. } t \in J, \\ y(0) = 0,\qquad y(T)=0, \end{array}\displaystyle \right \} $$

(17)

has only a trivial solution. Hence, it follows from the Fredholm alternative (see Proposition 3) that problem (16) has a unique Carathéodory solution \(y(\cdot)\), which takes the form

$$y(t)= \int_{0}^{T} G_{1} (t,s) f_{q}(s) \,\mathrm{d}s, $$

where \(G_{1} \) is the Green’s function of problem (17), *i.e.*

$$ G_{1}(t,s)= \left \{ \textstyle\begin{array}{l@{\quad}l} \frac{t (s-T)}{T}, & \text{for all } 0 \leq t \leq s \leq T, \\ \frac{s (t-T)}{T}, & \text{for all } 0\leq s \leq t \leq T. \end{array}\displaystyle \right . $$

(18)

One can easily check that \(\vert G_{1}(t, s) \vert \leq \frac{T}{4}\), for all \(t, s \in[0, T]\).

In fact, since the multivalued composition \(F_{q}(t)\) is, for every \(q \in Q\), obviously measurable, according to Proposition 1, we can even write

$$y(t) \in\int_{0}^{T} G_{1}(t, s) \overline{\bigcup_{n\in\mathbb{N}} f_{n, q} (s)} \, \mathrm{d}s, $$

where \(\{f_{n, q}(t) \subset F_{q}(t) \}_{n \in\mathbb {N}}\) is a sequence of measurable selections of \(F_{q}\) and the integral is understood in the sense of Aumann.

Observe that because of

$$ \frac{\partial G_{1}}{\partial t}(t,s)= \left \{ \textstyle\begin{array}{l@{\quad}l} \frac{(s-T)}{T}, & \text{for all } 0 \leq t \leq s \leq T, \\ \frac{s}{T}, & \text{for all } 0\leq s \leq t \leq T, \end{array}\displaystyle \right . $$

(19)

we also have

$$y'(t)\in\int_{0}^{T} \frac{\partial G_{1} }{\partial t}(t,s) \overline {\bigcup_{n \in\mathbb{N}} f_{n, q}(s)} \,\mathrm{d}s, $$

where the integral is again understood in the sense of Aumann. Moreover, \(\vert \frac{\partial G_{1}}{\partial t} (t, s)\vert \leq1\), for all \(t, s \in[0, T]\).

Hence, denoting by \(\varphi \colon Q \multimap C^{1}(J, \mathbb{R})\) the solution operator of (15), where

$$\varphi:= \int_{0}^{T} G_{1} (t,s) F_{q}(s) \,\mathrm{d}s= \int_{0}^{T} G_{1}(t, s) \overline{\bigcup_{n \in\mathbb{N}} f_{n, q}(s)} \,\mathrm{d}s, $$

instead of the existence of the solution of differential problem (15), we can equivalently investigate the existence of a fixed point of the multivalued operator *φ*. For this purpose, we will apply the Kakutani-Ky Fan-type fixed point theorem (Proposition 2). In this way, we will prove the following theorem.

### Theorem 1

*Let*
*a*, *b*, *c*, \(x_{0}\), \(x_{T}\)
*and*
\(T>0\)
*be real constants such that*

$$ \frac{4}{T(T+4)} > \max\bigl\{ \vert a\vert , |b|\bigr\} . $$

(20)

*Assume that*
\(P \colon J \multimap\mathbb{R}\)
*is an Aumann integrable multivalued mapping and*
\(F_{1} \vert_{\overline{B}_{D}} \colon\overline{B}_{D} \multimap \mathbb{R}\), \(F_{2} \vert_{\overline{B}_{D}} \colon\overline{B}_{D} \multimap\mathbb{R}\)
*are u*.*s*.*c*. *multivalued mappings with convex and compact values*, *where*
\(D=D_{0}+k_{2}+k_{3}\), \(D_{0}>0\)
*is still supposed to be a suitable constant such that*

$$ D_{0} \geq\Delta_{1}(D_{0}), $$

(21)

*where*

$$ \Delta_{1} (D_{0}):= \frac{ [\mathcal{P}+T (M_{1}(D_{0})+M_{2}(D_{0})+|a|k_{3}+|b|k_{2}+|c|) ](T+4)}{4-k_{1} T(T+4)} $$

(22)

*and*

$$\begin{aligned}& M_{1}(D_{0}) := \max_{|z| \leq D_{0}+k_{2}+k_{3}}\bigl\vert F_{1}(z)\bigr\vert ,\qquad M_{2}(D_{0}):=\max _{|z| \leq D_{0}+k_{2}+k_{3}}\bigl\vert F_{2}(z)\bigr\vert , \\& k_{1} := \max\bigl\{ \vert a\vert ,|b|\bigr\} , \qquad k_{2} := \max\bigl\{ \vert x_{0}\vert , |x_{T}| \bigr\} , \qquad k_{3} :=\frac{|x_{T}-x_{0}|}{T}, \\& \mathcal{P} := \sup_{p \subset P} \biggl\{ \int_{0}^{T} \bigl\vert p(t)\bigr\vert \, \mathrm{d}t \Bigm| p \subset P \textit{ is a Lebesgue integrable selection of }P \biggr\} . \end{aligned}$$

*Then problem* (9) *admits a solution*
\(x(\cdot)\)
*such that*

$$\max_{t \in J}\bigl\{ \bigl\vert x(t)\bigr\vert +\bigl\vert x'(t)\bigr\vert \bigr\} \leq D. $$

### Proof

In order to check all the assumptions of Proposition 2, we will proceed in four steps.

(i) Since problem (16) is uniquely solvable, the set \(\varphi(Q)\) is nonempty.

(ii) Let us prove that the set \(\varphi(Q)\), *i.e.* the set of solutions of (15), is relatively compact. According to the well known Arzelá-Ascoli lemma, the set of solutions is relatively compact in \(C^{1}(J, \mathbb{R})\) if and only if it is uniformly bounded and equi-continuous, both in the \(C^{1}\)-norm.

(a) Let us show that the set of solutions of (15) is uniformly bounded in \(C^{1}(J, \mathbb{R})\). Let \(u(\cdot)\) be a solution of (15) and \(f_{q} \subset F_{q}\) be a measurable selection of \(F_{q}\). We could see that such a measurable selection exists and that \(u(\cdot)\) takes the form

$$u(t)=\int_{0}^{T} G_{1}(t,s) f_{q}(s) \, \mathrm{d}s. $$

It is obvious that

$$\begin{aligned}& \max_{|z| \leq D_{0}}\bigl\vert F_{1}\bigl(z + v(t)\bigr) \bigr\vert = \max_{|z| \leq D_{0}} \bigl\vert F_{1}^{*} \bigl(z+v(t)\bigr)\bigr\vert \leq M_{1}(D_{0}), \\& \max_{|z| \leq D_{0}}\bigl\vert F_{2}\bigl(z+v'(t) \bigr)\bigr\vert = \max_{|z| \leq D_{0}} \bigl\vert F_{2}^{*} \bigl(z+v'(t)\bigr)\bigr\vert \leq M_{2}(D_{0}), \end{aligned}$$

where \(v(t)= \frac{x_{T}-x_{0}}{T} t +x_{0}\) and \(v'(t)=\frac{x_{T}-x_{0}}{T}\).

By means of Lemma 3 and in view of (18), (19), for any \(t \in J\), we obtain the following estimate:

$$\begin{aligned} \bigl\vert u(t)\bigr\vert +\bigl\vert u'(t)\bigr\vert =& \biggl\vert \int_{0}^{T} G_{1}(t,s) f_{q}(s) \,\mathrm{d}s\biggr\vert +\biggl\vert \int _{0}^{T} \frac{\partial G_{1}}{\partial t}(t,s) f_{q}(s) \,\mathrm{d}s\biggr\vert \\ \leq& \biggl(\frac{T}{4}+1 \biggr) \int_{0}^{T} \bigl\vert f_{q}(s) \bigr\vert \,\mathrm{d}s \\ \leq& \biggl( \frac{T}{4}+1 \biggr)\int_{0}^{T} \bigl\vert p(s)\bigr\vert \\ &{}+\bigl\vert f_{1}^{*}\bigl(q(s)+v(s)\bigr)\bigr\vert +\bigl\vert f_{2}^{*}\bigl(q'(s)+v'(s)\bigr)\bigr\vert +\bigl\vert a\bigl(q'(s)+ v'(s)\bigr)\bigr\vert \\ &{}+\bigl\vert b \bigl(q(s)+ v(s)\bigr)\bigr\vert +\bigl\vert c \operatorname{Sgn}_{\mathrm{Sel}}\bigl(q'(s)+v'(s)\bigr) \bigr\vert \,\mathrm{d}s \\ \leq& \biggl(\frac{T}{4}+1 \biggr) \biggl[ \mathcal {P}+T \bigl(M_{1}(D_{0})+M_{2}(D_{0})+|a| k_{3}+|b| k_{2}+|c|\bigr) \\ &{}+ k_{1} \int_{0}^{T}\bigl\vert q(s)\bigr\vert +\bigl\vert q'(s)\bigr\vert \,\mathrm{d}s \biggr], \\ \leq& \biggl(\frac{T}{4}+1 \biggr) \bigl[ \mathcal {P}+T \bigl(M_{1}(D_{0})+M_{2}(D_{0})+|a| k_{3}+|b|k_{2}+|c| +k_{1} D_{0}\bigr) \bigr], \end{aligned}$$

where \(f_{1}^{*} \subset F_{1}^{*}\), \(f_{2}^{*} \subset F_{2}^{*}\), \(p \subset P\) and \(\operatorname{Sgn}_{\mathrm{Sel}} \subset\operatorname{Sgn}\) are the related measurable selections.

Since this estimate holds in the same way for all \(q \in Q\), this already means that the solutions \(u(\cdot)\) of (15), *i.e.* the set \(\varphi(Q)\), are uniformly bounded in the \(C^{1}\)-norm.

Moreover, according to (20), (22), if there exists a positive constant \(D_{0}\) such that \(D_{0} \geq\Delta_{1}(D_{0})\), then the set \(\varphi(Q)\) satisfies \(\varphi(Q) \subset Q\).

(b) Now, let us show that the elements of the set \(\varphi(Q)\), *i.e.* the solutions \(u(\cdot)\) of (15), are equi-continuous in the \(C^{1}\)-norm. For any \(t_{1}, t_{2} \in J\) with \(t_{1}< t_{2}\), we have by means of Lemma 3 and in view of (19)

$$\begin{aligned} \bigl\vert u(t_{2})-u(t_{1})\bigr\vert =& \biggl\vert \int_{t_{1}}^{t_{2}} u'(t) \, \mathrm{d}t \biggr\vert = \biggl\vert \int_{t_{1}}^{t_{2}} \int_{0}^{T} \frac{\partial G_{1}}{\partial t}(t, s) f_{q}(s) \,\mathrm{d}s \,\mathrm{d}t\biggr\vert \\ \leq& \biggl\vert \int_{t_{1}}^{t_{2}} \int _{0}^{T} \biggl\vert \frac{\partial G_{1}}{\partial t}(t, s) \biggr\vert \bigl\vert f_{q}(s)\bigr\vert \,\mathrm{d}s \, \mathrm{d}t\biggr\vert \leq\biggl\vert \int_{t_{1}}^{t_{2}} \int_{0}^{T} \bigl\vert f_{q}(s)\bigr\vert \,\mathrm{d}s \,\mathrm{d}t\biggr\vert \\ \leq& (t_{2}-t_{1}) \bigl[ \mathcal{P} + \bigl( M_{1}(D_{0})+M_{2}(D_{0})+k_{1} D_{0}+|a| k_{3}+|b| k_{2}+|c| \bigr)T \bigr]. \end{aligned}$$

Furthermore, we have still

$$\begin{aligned} \bigl\vert u'(t_{2})-u'(t_{1}) \bigr\vert =&\biggl\vert \int_{t_{1}}^{t_{2}}u''(t) \,\mathrm{d}t\biggr\vert = \biggl\vert \int_{t_{1}}^{t_{2}} f_{q}(t) \,\mathrm{d}t\biggr\vert \leq \biggl\vert \int _{t_{1}}^{t_{2}}\bigl\vert f_{q}(t)\bigr\vert \,\mathrm{d}t\biggr\vert \\ \leq& (t_{2}-t_{1}) \bigl[ M_{1}(D_{0})+M_{2}(D_{0})+k_{1} D_{0}+|a| k_{3}+|b| k_{2}+|c| \bigr] \\ &{}+\int_{t_{1}}^{t_{2}}\bigl\vert p(t)\bigr\vert \, \mathrm{d}t. \end{aligned}$$

Therefore, the solutions \(u(\cdot)\) of (15) are equi-continuous in the \(C^{1}\)-norm. Summing up (a) and (b), the elements of the set \(\varphi(Q)\) are relatively compact in the \(C^{1}\)-norm, as claimed.

(iii) We will show that the operator *φ* is u.s.c. In view of Lemma 2, and since *φ* was shown to be compact, it is sufficient to show that the graph \(\Gamma_{\varphi}\) is closed. Let \(\{(q_{k}, u_{k})\} \subset\Gamma_{\varphi}\) be a sequence such that \(\{(q_{k}, q'_{k}, u_{k})\} \rightarrow(q, q', u)\), where \(q \in Q\). For all \(k \in\mathbb{N}\) and a.a. \(t \in J\), the sequence \(\{u_{k}'\}\) is bounded and \(|u_{k}''(t)|\leq |p(t)|+M_{1}(D_{0})+M_{2}(D_{0})+k_{1} D_{0}+|c|+|a| k_{3}+|b| k_{2}\), for a.a. \(t \in J\). The sequence \(\{w_{k}:=u'_{k}\}\) satisfies all the assumptions of Lemma 4.

Thus, applying Lemma 4 to the sequence \(\{w_{k}:=u'_{k}\}\), we find that there exists a subsequence of \(\{u_{k}'\}\), for the sake of simplicity denoted in the same way as the sequence, which converges uniformly to \(u'\) on *J* and such that \(\{u''_{k}\}\) converges weakly to \(u''\) in \(L^{1}(J,\mathbb{R})\).

If we set \(z_{k}:=(u_{k}, w_{k})\), then \(z'_{k}= (u_{k}', w_{k}')=(u_{k}', u_{k}'')\rightarrow(u', u'')\), weakly in \(L^{1}(J, \mathbb{R}^{2})\). Let us consider the system

$$z_{k}'(t) \in H\bigl(t, q_{k}(t), q_{k}'(t)\bigr),\quad \mbox{for a.a. } t \in J, $$

where \(z_{k}'(t) = (u_{k}'(t), w_{k}'(t))\) and \(H(t, q_{k}(t), q_{k}'(t)) = (w_{k}, F_{q_{k}} (t))\).

Applying Lemma 5, for \(f_{i}:=z'_{k}\), \(f := (u', u'')\), \(x_{i}:=(q_{k}, q'_{k})\), it follows that

$$\bigl(u'(t), u''(t)\bigr) \in H\bigl(t, q(t), q'(t)\bigr), \quad \mbox{for a.a. } t \in J, $$

*i.e.*
\(u''(t) \in F_{q}(t)\), for a.a. \(t \in J\).

The set \(\varphi(Q)\) is relatively compact and graph \(\Gamma_{\varphi }\) is closed. Therefore, the mapping *φ* is u.s.c., compact and, in particular, with compact values.

(iv) Finally, we will show that the mapping *φ* has convex values.

Let \(u_{1}\), \(u_{2}\) be two distinct solutions of problem (15) associated with measurable selections \(f_{1,q}, f_{2,q} \subset F_{q}\). Then, for all \(t \in J\), we have

$$\begin{aligned}& u_{1}(t) = \int_{0}^{T} G_{1} (t,s) f_{1,q}(s) \,\mathrm{d}s, \\& u_{2}(t) = \int_{0}^{T} G_{1} (t,s) f_{2,q}(s) \,\mathrm{d}s. \end{aligned}$$

Let \(\lambda\in[ 0, 1 ]\) be arbitrary. Since *F* is convex-valued, the same is true for \(F_{q}\), and subsequently

$$f_{q}(t)=\lambda f_{1,q}(t)+(1-\lambda) f_{2,q}(t) $$

must be a measurable selection of \(F_{q}\), *i.e.*
\(f_{q} \subset F_{q}\). We have

$$\begin{aligned} u(t) =&\lambda u_{1}(t)+(1-\lambda) u_{2}(t) \\ =& \lambda\int_{0}^{T} G_{1} (t,s) f_{1,q}(s) \,\mathrm{d}s + (1-\lambda) \int_{0}^{T} G_{1} (t,s) f_{2,q}(s) \,\mathrm{d}s \\ =&\int_{0}^{T} G_{1} (t,s) \bigl[ \lambda f_{1,q}(s) + (1-\lambda) f_{2,q}(s)\bigr] \,\mathrm{d}s = \int_{0}^{T} G_{1}(t, s) f_{q}(s) \,\mathrm{d}s. \end{aligned}$$

Thus, \(u(\cdot)\) must also be a solution of (15), by which the mapping *φ* has convex values.

After all, applying Proposition 2, we obtain the existence of a fixed point of the multivalued mapping *φ* which represents a solution of problem (13). However, because of the definitions of \(F_{1}^{*}\), \(F_{2}^{*}\), such a solution must be a solution of problem (12) as well as (9). □

### Example 1

As an illustrative example of a Dirichlet problem satisfying rather implicit conditions (20)-(22), we can consider (9), where

$$\begin{aligned}& |a|=|b|=\frac{1}{4} \quad \Longrightarrow\quad k_{1}= \frac{1}{4}, \\& x_{0}=x_{T}=\frac{1}{4} \quad \Longrightarrow\quad k_{2}=\frac{1}{4},\qquad k_{3}=0, \\& T=2(\sqrt{2}-1) \quad \Longrightarrow \quad \frac{4}{T(T+4)}=1>\frac{1}{4}=k_{1} \quad (\mbox{see}\ (20)). \end{aligned}$$

In this way, conditions (21), (22) take, for \(D_{0}=\frac{1}{4}\), the form:

$$\frac{\mathcal{P}}{2(\sqrt{2}-1)}+ \biggl[ M_{1} \biggl(\frac{1}{4} \biggr) + M_{2} \biggl(\frac{1}{4} \biggr)+|c| \biggr] \leq \frac{1}{8}. $$

Taking, furthermore, \(\mathcal{P}\leq\frac{\sqrt{2}-1}{16}\) and \(|c|=\frac{1}{32}\), we get for \(F_{1}(z):=f_{1}z^{m}\) and \(F_{2}(z):=f_{2}z^{n}\) the inequality:

$$|f_{1}| \biggl(\frac{1}{2} \biggr)^{m}+|f_{2}| \biggl(\frac{1}{2} \biggr)^{n}\leq\frac{1}{16}. $$

Hence, if \(|f_{1}|,|f_{2}|\leq\frac{1}{16}\) and \(m,n\geq1\), then we arrive at (21), (22), because for \(|f_{1}|=|f_{2}|=\frac{1}{16}\) and \(m=n\), we have

$$\biggl(\frac{1}{2} \biggr)^{m-1}\leq1 $$

which holds for all \(m\geq1\).

After all, under these assumptions, problem (9) admits a solution \(x(\cdot)\) such that

$$\max_{t\in[0,T]}\bigl\{ \bigl\vert x(t)\bigr\vert +\bigl\vert x'(t)\bigr\vert \bigr\} \leq\frac{1}{2}. $$

In fact, the same is true for (9), where \(F_{1}\), \(F_{2}\) are as above for \(z\in[-\frac{1}{2},\frac{1}{2}]\) and can be arbitrary outside of the interval \([-\frac{1}{2},\frac{1}{2}]\).

### Remark 1

If, additionally, a multivalued mapping *P* is essentially bounded, then condition (20) can be replaced, for the same conclusion for (9), by

$$\frac{8}{T (T+4)} > \max\bigl\{ \vert a\vert , |b|\bigr\} , $$

and (21) with \(\Delta_{1}(D_{0})\) in (22) can be replaced by \(\widetilde{D}_{0} \geq\Delta'_{1}(\widetilde{D}_{0}) \), where

$$\Delta'_{1} (\widetilde{D}_{0}):= \frac{ [\mathcal{P}_{1}+ M_{1}(\widetilde{D}_{0})+M_{2}(\widetilde{D}_{0})+|a| k_{3}+|b|k_{2}+|c| ]T (T+4 )}{8-k_{1}T (T+4 )}, $$

\(\mathcal{P}_{1}:=\operatorname{ess}\sup_{t \in J} | P(t)|\), because

$$\int_{0}^{T} \bigl\vert G_{1}(t,s) \bigr\vert \,\mathrm{d}s\leq\frac{T^{2}}{8}\quad \text{and}\quad \int _{0}^{T} \biggl\vert \frac{\partial G_{1}}{\partial t}(t,s)\biggr\vert \,\mathrm{d}s\leq\frac{T}{2}. $$

Thus, the same conclusion holds for (9) with *D* replaced by \(\widetilde{D}=\widetilde{D}_{0}+k_{2}+k_{3}\).

### Remark 2

If, additionally, a multivalued mapping \(F_{2}\) is bounded, *i.e.*
\(M_{2} := \max_{z \in\mathbb{R}} |F_{2}(z)|\), and \(a=0\), we can still improve the estimate for a solution of (9). Condition (20) can then be replaced by \(\frac{4}{T^{2}} > |b|\), provided there still exists a positive constant \(D_{1}\) such that

$$D_{1} \geq\frac{T [\mathcal{P} + (M_{1}(D_{1})+M_{2}+|b| k_{2} +|c| ) T ]}{4-T^{2} |b|}. $$

Under these assumptions, problem (9) has a solution \(x(\cdot )\) such that

$$\begin{aligned}& \max_{t \in J} \bigl\vert x(t)\bigr\vert \leq D_{1}+k_{2}, \\& \max_{t \in J} \bigl\vert x'(t)\bigr\vert \leq D_{2}+k_{3} := \mathcal{P} + T \bigl[M_{1}(D_{1})+M_{2}+|b| (D_{1}+ k_{2})+|c| \bigr]+k_{3}, \end{aligned}$$

where \(M_{1}(D_{1}):=\max_{|z|\leq D_{1}+k_{2}}|F_{1}(z)|\), \(k_{2}:=\max\{|x_{0}|, |x_{T}|\}\), \(k_{3}:=\frac{|x_{T}-x_{0}|}{T}\).

Similarly if, additionally, a multivalued mapping \(F_{1}\) is bounded, *i.e.*
\(M_{1}:= \max_{z \in\mathbb{R}} |F_{1}(z)|\), and \(b=0\), then condition (20) can be replaced by \(\frac{1}{T}>|a|\), provided there still exists a positive constant \(D_{2}\) such that

$$D_{2} \geq\frac{\mathcal{P} +T (M_{1}+M_{2}(D_{2})+|a| k_{3} +|c| ) }{1-T |a|}, $$

where \(M_{2}(D_{2}):=\max_{|z|\leq D_{2}+k_{3}}|F_{2}(z)|\). Here, problem (9) has a solution \(x(\cdot)\) such that

$$\begin{aligned}& \max_{t \in J} \bigl\vert x(t)\bigr\vert \leq D_{1}+k_{2}:=\frac{T}{4} \bigl[\mathcal {P}+T \bigl(M_{1}+M_{2}(D_{2})+|a| (D_{2}+k_{3})+|c| \bigr) \bigr]+k_{2}, \\& \max_{t \in J} \bigl\vert x'(t)\bigr\vert \leq D_{2}+k_{3}. \end{aligned}$$