*Article* **Applications of Stieltjes Derivatives to Periodic Boundary Value Inclusions**

**Bianca Satco 1,\* and George Smyrlis <sup>2</sup>**


Received: 3 November 2020; Accepted: 27 November 2020; Published: 1 December 2020 -

**Abstract:** In the present paper, we are interested in studying first-order Stieltjes differential inclusions with periodic boundary conditions. Relying on recent results obtained by the authors in the single-valued case, the existence of regulated solutions is obtained via the multivalued Bohnenblust–Karlin fixed-point theorem and a result concerning the dependence on the data of the solution set is provided.

**Keywords:** periodic boundary value inclusion; Stieltjes derivative; Stieltjes integrals; Bohnenblust–Karlin fixed-point theorem; regulated function

#### **1. Introduction**

Allowing the study in a unique framework of many classical problems: ordinary differential or difference equations (in the case of an absolutely continuous measure—with respect to the Lebesgue measure—respectively of a discrete measure), impulsive differential problems (for a sum of Lebesgue measure with a discrete one), dynamic equations on time scales (see [1]) and generalized differential equations (e.g., [2,3]), it is clear why the theory of differential equations driven by measures has seen a significant growth (e.g., [1,4]).

Using a natural notion of Stieltjes derivative with respect to a non-decreasing function (c.f. [5], see also [6] or [7,8] for applications), measure-driven differential equations can be expressed, in an equivalent form, as a Stieltjes differential equation.

On the other hand, the set-valued setting covers a wider range of problems ([9–12], see also [13–15]), therefore passing from the single-valued to the multivalued case brings a real improvement.

Based on the results obtained in [4] for measure-driven differential equations with periodic boundary conditions, in the present paper we focus on nonlinear differential inclusions of the form:

$$\begin{cases} u\_\S'(t) + b(t)u(t) \in F(t, u(t)), \; t \in [0, T] \\ u(0) = u(T) \end{cases} \tag{1}$$

where *u <sup>g</sup>* denotes the Stieltjes derivative of the state *u* with respect to a left-continuous non-decreasing function *g* : [0, *T*] → R. This form is preferred since in many real-world problems the linear, respectively the nonlinear term has different practical meanings.

In the particular case of the identical function *g*, periodic differential problems have been widely considered in the literature; to mention only a few works, we refer to [16–18] for the single-valued setting and to [19,20] (without impulses) or [21,22] (allowing impulses) in the set-valued framework. As far as the authors know, periodic differential problems driven by a non-decreasing left-continuous function *g* have been studied only in the single-valued case in [4].

Applying Bohnenblust–Karlin set-valued fixed-point theorem, we prove that the specified problem (1) possesses solutions and characterize the solutions as Stieltjes integrals with an appropriate Green function.

We then study the dependence of the solution set of (1) on the data; specifically, we want to estimate the perturbation of the corresponding solution set if perturbations occur in the values of *b* and *F*. Such an estimation is provided in the case where the multifunction does not depend on the state.

New results for impulsive periodic inclusions (studied, e.g., in [21,22]) can be deduced by considering as function *g* the sum of an absolutely continuous function with step functions. Moreover, no restrictions are imposed on the number of impulses (it can be countable, so Zeno behavior is allowed).

Having in mind that the theory of measure-driven equations is equivalent, in most situations, with the theory of dynamic equations on time scales ([1], see also [23]), our study could be used to deduce new existence and dependence on the data results for periodic dynamic inclusions on time scales (see [24,25]).

The outline of the paper is as follows. After introducing the notations and recalling some necessary known facts, in Section 3 we present an existence result for the single-valued case and then we proceed to the main results in Section 4: we prove (for the multivalued setting) an existence result and also a result on the dependence of the solution set on the data.

#### **2. Notations and Known Facts**

A regulated map *<sup>u</sup>* : [0, *<sup>T</sup>*] <sup>→</sup> <sup>R</sup>*<sup>d</sup>* [26] is a map with right and left limits *<sup>u</sup>*(*t*+) and *<sup>u</sup>*(*s*−) at every point *t* ∈ [0, *T*) and *s* ∈ (0, *T*]. It is known that regulated functions have at most countably many discontinuities [27] and that the space *<sup>G</sup>*([0, *<sup>T</sup>*], <sup>R</sup>*d*) of regulated functions *<sup>u</sup>* : [0, *<sup>T</sup>*] <sup>→</sup> <sup>R</sup>*<sup>d</sup>* is a Banach space with respect to the norm *u<sup>C</sup>* <sup>=</sup> sup*t*∈[0,*T*] *u*(*t*).

A collection A ⊂ *<sup>G</sup>*([0, *<sup>T</sup>*], <sup>R</sup>*d*) is said to be equiregulated if the following conditions hold:

• for each *t* ∈ (0, *T*] and *ε* > 0, one can choose *δε*,*<sup>t</sup>* ∈ (0, *T*] such that for all *u* ∈ A

$$\|\|u(t') - u(t-)\|\| < \varepsilon, \text{ for every } t' \in (t - \delta\_{\varepsilon, t\_\prime} t)^2$$

• for each *t* ∈ [0, *T*) and *ε* > 0, one can choose *δε*,*<sup>t</sup>* ∈ (0, *T* − *t*] such that for all *u* ∈ A

*u*(*t* ) − *u*(*t*+) < *ε*, for every *t* ∈ (*t*, *t* + *δε*,*t*).

Let us recall an Ascoli-type result.

**Lemma 1.** ([26], Corollary 2.4) *A set* A ⊂ *<sup>G</sup>*([0, *<sup>T</sup>*], <sup>R</sup>*d*) *is relatively compact if and only if it is equiregulated and pointwise bounded.*

It is not difficult to check that:

**Remark 1.** *A set* A *of regulated functions is equiregulated if*

$$\|\|u(t) - u(t')\|\| \le |\chi(t) - \chi(t')|, \quad \forall \ 0 \le t < t' \le T, \quad \forall \ u \in \mathcal{A}$$

*for some regulated function χ* : [0, *T*] → R*.*

In the whole paper, *g* : [0, *T*] → R will be a non-decreasing left-continuous function and *μ<sup>g</sup>* the Stieltjes measure defined by *g*. Without any loss of generality, suppose *g*(0) = 0. We deal with the Kurzweil–Stieltjes integral; we recall below the basic facts concerning this integral.

**Definition 1.** (Refs [2,3,27,28] or [29]) *One says that <sup>f</sup>* : [0, *<sup>T</sup>*] <sup>→</sup> <sup>R</sup>*<sup>d</sup> is Kurzweil–Stieltjes integrable (or KS-integrable) with respect to <sup>g</sup>* : [0, *<sup>T</sup>*] <sup>→</sup> <sup>R</sup> *if there is <sup>T</sup>* <sup>0</sup> *<sup>f</sup>*(*s*)*dg*(*s*) <sup>∈</sup> <sup>R</sup>*<sup>d</sup> with the property that for every ε* > 0*, one can find δε* : [0, *T*] → R<sup>+</sup> *satisfying*

$$\left\| \left| \sum\_{i=1}^k f(\xi\_i) (g(t\_i) - g(t\_{i-1})) - \int\_0^T f(s) dg(s) \right| \right\| < \varepsilon$$

*for every δε-fine partition* {([*ti*−1, *ti*], *<sup>ξ</sup>i*) : *<sup>i</sup>* = 1, ..., *<sup>k</sup>*} *of* [0, *<sup>T</sup>*]. *( A partition* {([*ti*−1, *ti*], *<sup>ξ</sup>i*) : *<sup>i</sup>* = 1, ..., *<sup>k</sup>*} *of* [0, *<sup>T</sup>*] *is δε-fine iff* [*ti*−1, *ti*] ⊂ (*ξ<sup>i</sup>* − *δε*(*ξi*), *<sup>ξ</sup><sup>i</sup>* + *δε*(*ξi*)), *for all* <sup>1</sup> ≤ *<sup>i</sup>* ≤ *k).*

The well-known Henstock–Kurzweil integral (see [30–32]) is recovered in the case where *g* is the identical function and *d* = 1.

In general, the Lebesgue–Stieltjes integrability with respect to *g* (i.e., the abstract Lebesgue integrability with respect to the Stieltjes measure *μg*) yields the Kurzweil–Stieltjes integrability with respect to *g*. When *g* is left-continuous and non-decreasing, by ([28], Theorem 6.11.3) (or ([27], Theorem 8.1)),

$$\int\_0^t f(s)d\mathbf{g}(\mathbf{s}) = \int\_{[0,t]} f(\mathbf{s})d\mu\_{\mathcal{S}}(\mathbf{s}) - f(t)(\mathbf{g}(t+) - \mathbf{g}(t)) = \int\_{[0,t)} f(\mathbf{s})d\mu\_{\mathcal{S}}(\mathbf{s}), \,\forall t \in [0,T].$$

(Ref [29], Proposition 2.3.16) asserts that the KS-primitive *<sup>F</sup>* : [0, *<sup>T</sup>*] <sup>→</sup> <sup>R</sup>*d*, *<sup>F</sup>*(*t*) = *<sup>t</sup>* <sup>0</sup> *f*(*s*)*dg*(*s*) is regulated whenever *g* is regulated, it is left-continuous if *g* is left-continuous and for every *t* ∈ [0, *T*),

$$F(t+) - F(t) = f(t) \left[ \mathbf{g}(t+) - \mathbf{g}(t) \right].$$

Consequently, if *g* is continuous at some point, then *F* is also continuous.

To recall more properties of the primitive, we need a notion of (Stieltjes) derivative of a function with respect to another function, given in [5] (see also [33]).

**Definition 2.** *Let <sup>g</sup>* : [0, *<sup>T</sup>*] <sup>→</sup> <sup>R</sup> *be non-decreasing and left-continuous. The derivative of <sup>f</sup>* : [0, *<sup>T</sup>*] <sup>→</sup> <sup>R</sup>*<sup>d</sup> with respect to g (or the g-derivative) at the point t* ∈ [0, *T*] *is*

$$f\_{\mathcal{S}}'(t) = \lim\_{t' \to t} \frac{f(t') - f(t)}{\mathcal{g}(t') - \mathcal{g}(t)} \quad \text{if } \mathcal{g} \text{ is continuous at } t,$$

$$f\_{\mathcal{S}}'(t) = \lim\_{t' \to t+} \frac{f(t') - f(t)}{\mathcal{g}(t') - \mathcal{g}(t)} \quad \text{if } \mathcal{g} \text{ is discontinuous at } t,$$

*if the limit exists.*

The *g*-derivative has found interesting applications in solving real-world problems where periods of time where no activity occurs and instants with abrupt changes are both present, such as [7] or [8].

Define the following set:

$$D\_{\mathcal{S}} = \{ t \in [0, T] : \mathcal{g}(t+) - \mathcal{g}(t) > 0 \},$$

namely the collection of atoms of *μg*; remark that if *t* ∈ *Dg*, then

$$f'\_{\mathcal{S}}(t) = \frac{f(t+) - f(t)}{\mathbf{g}(t+) - \mathbf{g}(t)}.$$

There is a set where Definition 2 has no meaning, more precisely,

$$\mathcal{C}\_{\mathcal{S}} = \{ t \in [0, T] : \mathbb{g} \text{ is constant on } (t - \varepsilon, t + \varepsilon) \text{ for some } \varepsilon > 0 \}. $$

It is convenient, when working with the *g*-derivative, to also disregard the points of the set

$$N\_{\S} = \{ \mu\_n, \upsilon\_n : n \in \mathbb{N} \} \backslash D\_{\S'},$$

where *Cg* = ! *<sup>n</sup>*∈N(*un*, *vn*) is a pairwise disjoint decomposition of *Cg* (such a writing is possible due to the fact that *Cg* is open in the usual topology of the real line, see [5]).

To warrant this, take into account that *μg*(*Cg*) = *μg*(*Ng*) = 0 [5] and, when studying differential equations, the equation has to be satisfied *μg*-almost everywhere.

The connection between Stieltjes integrals and the Stieltjes derivative is given by Fundamental Theorems of Calculus ([5], Theorems 5.4, 6.2, 6.5).

**Theorem 1.** ([5], Theorem 6.5) *Let <sup>f</sup>* : [0, *<sup>T</sup>*] <sup>→</sup> <sup>R</sup>*<sup>d</sup> be KS-integrable with respect to the non-decreasing left-continuous function g* : [0, *T*] → R*. Then its primitive*

$$F(t) = \int\_0^t f(s) \, d\mathfrak{g}(s), \quad t \in [0, T]\_\prime$$

*is g-differentiable μg-a.e. in [0,T] and F <sup>g</sup>*(*t*) = *f*(*t*), *μg-a.e. in [0,T].*

As our aim is to study a differential inclusion, we end this section with basic notions of set-valued analysis (the reader is referred to [34,35] or [36]).

Let <sup>P</sup>*bc*(R*d*) be the space of all non-empty bounded, closed and convex subsets of <sup>R</sup>*<sup>d</sup>* endowed with the Hausdorff–Pompeiu distance

$$D(A, A') = \max(\varepsilon(A, A'), \varepsilon(A', A)),$$

where the (Pompeiu-) excess of the set *<sup>A</sup>* ∈ P*bc*(R*d*) over *<sup>A</sup>* ∈ P*bc*(R*d*) is given by

$$e(A, A') = \sup\_{a \in A} \inf\_{a' \in A'} ||a - a'||.$$

If *<sup>A</sup>* ∈ P*bc*(R*d*), denote by <sup>|</sup>*A*<sup>|</sup> <sup>=</sup> *<sup>D</sup>*(*A*, {0}) = sup*a*∈*<sup>A</sup> a*.

Let *X*,*Y* be Banach spaces and let *F* : *X* → P(*Y*) be a multimapping. *F* is said to be upper semicontinuous at *u*<sup>0</sup> ∈ *X* if for each *ε* > 0 there is *δε*,*u*<sup>0</sup> > 0 such that whenever *u* − *u*0 < *δε*,*u*<sup>0</sup> ,

$$F(u) \subset F(u\_0) + \varepsilon B\_{\prime}$$

*B* being the closed unit ball of *Y*.

Moreover, *<sup>F</sup>* has closed graph if for all (*un*)*n*∈<sup>N</sup> ⊂ *<sup>X</sup>*, (*vn*)*n*∈<sup>N</sup> ⊂ *<sup>Y</sup>* with

$$u\_{\mathfrak{n}} \to u \in X, \quad \upsilon\_{\mathfrak{n}} \to \upsilon \in \mathcal{Y}, \quad \upsilon\_{\mathfrak{n}} \in F(u\_{\mathfrak{n}}), \quad \mathfrak{n} \in \mathbb{N}\_{\prime}$$

we have *v* ∈ *F*(*u*).

#### **3. Preliminary Result—Existence Theory for the Single-Valued Problem**

In this section, relying on the theory in [4], we present an existence result for the linear Stieltjes differential equation with periodic boundary conditions

$$\begin{cases} \begin{array}{ll} u'\_{\mathcal{S}}(t) + b(t)u(t) = f(t), & \mu\_{\mathcal{S}} - \text{a.e. in } [0, T], \\ u(0) = u(T), \end{array} \end{cases} \tag{2}$$

where *g* : [0, *T*] → R is non-decreasing and left-continuous and *b* : [0, *T*] → R is a *μg*-measurable function satisfying the non-resonance condition:

$$1 - b(t)\mu\_{\mathcal{J}}(\{t\}) \neq 0, \quad \text{for every } t \in [0, T]. \tag{3}$$

**Definition 3.** *A function <sup>u</sup>* : [0, *<sup>T</sup>*] <sup>→</sup> <sup>R</sup>*<sup>d</sup> is a solution of problem* (2) *if it is left-continuous and regulated, constant on the intervals where g is constant, g-differentiable μg-a.e. in* [0, *T*] *satisfying*

$$
\mu'\_{\mathcal{S}}(t) + b(t)\mu(t) = f(t), \; \mu\_{\mathcal{S}} - a.e. \; \text{in } [0, T].
$$

*and*

$$
u(0) = 
u(T).$$

Let us remark that when *<sup>b</sup>* <sup>∈</sup> *<sup>L</sup>*1(*μg*), the following condition is fulfilled:

$$\sum\_{t \in D\_{\mathcal{S}}} \left| \log \left| 1 - b(t) \mu\_{\mathcal{S}}(\{t\}) \right| \right| < \infty. \tag{4}$$

Indeed, if *Dg* is countable, we note its elements by {˜*tn*}*n*∈<sup>N</sup> and we get

$$\sum\_{n=1}^{\infty} |b(\tilde{t}\_n)\mu\_{\mathcal{S}}(\{\tilde{t}\_n\})| \le ||b||\_{L^1(\mu\_{\mathcal{S}})} < \infty$$

which implies *<sup>b</sup>*(˜*tn*)*μg*({˜*tn*}) → 0 as *<sup>n</sup>* → <sup>∞</sup>. Then, since

$$\lim\_{n \to \infty} \frac{\left| \log \left| 1 - b(\overline{t}\_n) \mu\_{\overline{\mathcal{S}}}(\{\overline{t}\_n\}) \right| \right|}{\left| b(\overline{t}\_n) \mu\_{\overline{\mathcal{S}}}(\{\overline{t}\_n\}) \right|} = 1,\tag{5}$$

(4) comes from the Limit Comparison Criterion for the convergence of numerical series. If *Dg* is finite, then (4) is trivially fulfilled.

It turns out (see [4]) that for some positive constant *δ*,

$$|1 - b(t)\mu\_{\mathcal{S}}(\{t\})| > \delta, \ \forall t \in D\_{\mathcal{S}}\dots$$

Moreover, *t* → |*b*(*t*)*μg*({*t*})| is bounded on [0, *T*] since on [0, *T*] \ *Dg* it vanishes, while on *Dg* we may see that is obviously bounded if *Dg* is finite, respectively *<sup>b</sup>*(˜*tn*)*μg*({˜*tn*}) → 0 as *<sup>n</sup>* → <sup>∞</sup> if *Dg* is countable.

To solve the problem (2), the sign of 1 − *b*(*t*)*μg*({*t*}) has to be taken into account.

As in [7], if *<sup>b</sup>* <sup>∈</sup> *<sup>L</sup>*<sup>1</sup> *<sup>g</sup>*([0, *T*]), the set

$$D\_{\mathcal{S}}^{-} = \{ t \in D\_{\mathcal{S}} : 1 - b(t)\mu\_{\mathcal{S}}(\{t\}) < 0 \}$$

is finite since

$$\infty > \||b||\_{L\_{\mathcal{S}}^1} > \sum\_{t \in D\_{\mathcal{S}}^-} b(t)\mu\_{\mathcal{S}}(\{t\}) > \sum\_{t \in D\_{\mathcal{S}}^-} 1.$$

Denote by *t*<sup>1</sup> < ... < *tk* its elements and, for simplicity, let *t*<sup>0</sup> = 0 and *tk*<sup>+</sup><sup>1</sup> = *T*. Let

$$\mathfrak{a}(t) = \begin{cases} \begin{array}{c} 1, \ if \; 0 \le t \le t\_1 \\ (-1)^i, \; if \; t\_i < t \le t\_{i+1}, \; i = 1, \dots, k \end{array} \end{cases}$$

and

$$\tilde{b}(t) = \begin{cases} \begin{array}{c} b(t), \operatorname{if} \, t \in [0, T] \\ \frac{-\log|1 - b(t)\mu\_{\mathcal{S}}(\{t\})|}{\mu\_{\mathcal{S}}(\{t\})}, \operatorname{if} \, t \in D\_{\mathcal{S}}. \end{array} \end{cases}$$

Applying Theorem 1, the following existence result can be proved:

**Theorem 2.** *Let <sup>b</sup>* : [0, *<sup>T</sup>*] <sup>→</sup> <sup>R</sup> *be LS-integrable with respect to g, satisfying (3) and let <sup>f</sup>* : [0, *<sup>T</sup>*] <sup>→</sup> <sup>R</sup>*<sup>d</sup> be such that* ˜ *f*(*t*) = *<sup>f</sup>*(*t*) <sup>1</sup>−*b*(*t*)*μg*({*t*}) *is KS-integrable with respect to g. Denoting by*

$$\mathfrak{g}(t,s) = \frac{1}{a(T)e^{\int\_0^T \mathfrak{b}(r)d\mathfrak{g}(r)} - 1} \begin{cases} a(T)e^{\int\_0^T \mathfrak{b}(r)d\mathfrak{g}(r) - \int\_s^t \mathfrak{b}(r)d\mathfrak{g}(r)}, \text{ if } 0 \le s \le t \le T\\ e^{-\int\_s^t \mathfrak{b}(r)d\mathfrak{g}(r)}, \text{ if } 0 \le t < s \le T \end{cases}$$

*the function u* : [0, *<sup>T</sup>*] <sup>→</sup> <sup>R</sup>*d,*

$$
\mu(t) = \frac{1}{\alpha(t)} \int\_0^T \frac{\alpha(s)}{1 - b(s)\mu\_\S(\{s\})} \overline{g}(t, s) f(s) dg(s),
$$

*is a solution of problem* (2)*.*

**Proof.** Obviously, the LS-integrability of ˜ *b* with respect to *g* follows from condition (4) and the LS-integrability of *b*.

One can see that for all *t* ∈ [0, *T*],

$$\begin{split} u(t) &= \\ &= \frac{1}{a(T)e^{\int\_0^T \tilde{b}(r)d\mathfrak{g}(r)} - 1} \left[ \frac{a(T)}{a(t)} e^{\int\_0^T \tilde{b}(r)d\mathfrak{g}(r)} e^{-\int\_0^t \tilde{b}(r)d\mathfrak{g}(r)} \int\_0^t a(s) e^{\int\_0^s \tilde{b}(r)d\mathfrak{g}(r)} \cdot \tilde{f}(s) d\mathfrak{g}(s) \right] \\ &+ \frac{1}{a(t)} e^{-\int\_0^t \tilde{b}(r)d\mathfrak{g}(r)} \int\_t^T a(s) e^{\int\_0^s \tilde{b}(r)d\mathfrak{g}(r)} \cdot \tilde{f}(s) d\mathfrak{g}(s) \right]. \end{split} \tag{6}$$

Let *<sup>t</sup>* <sup>∈</sup> [0, *<sup>T</sup>*] \ *Dg* be a point where the maps · 0 ˜ *b*(*r*)*dg*(*r*) and · <sup>0</sup> *α*(*s*)*e s* 0 ˜ *<sup>b</sup>*(*r*)*dg*(*r*) · ˜ *f*(*s*)*dg*(*s*) are *g*-differentiable (we know that it happens *μg*-a.e.).

*Mathematics* **2020**, *8*, 2142

We notice that *α* is constant on a neighborhood of *t*, so, by the product differentiation rule (see [5], Proposition 2.2),

*u <sup>g</sup>*(*t*) = <sup>1</sup> *α*(*T*)*e T* 0 ˜ *<sup>b</sup>*(*r*)*dg*(*r*) <sup>−</sup> <sup>1</sup> · ) *α*(*T*) *<sup>α</sup>*(*t*) *<sup>e</sup> T* 0 ˜ *b*(*r*)*dg*(*r*) *e* <sup>−</sup> *<sup>t</sup>* 0 ˜ *b*(*r*)*dg*(*r*) (−˜ *<sup>b</sup>*(*t*)) *<sup>t</sup>* 0 *α*(*s*)*e s* 0 ˜ *<sup>b</sup>*(*r*)*dg*(*r*) · ˜ *f*(*s*)*dg*(*s*) <sup>+</sup> *<sup>α</sup>*(*T*) *<sup>α</sup>*(*t*) *<sup>e</sup> T* 0 ˜ *b*(*r*)*dg*(*r*) *e* <sup>−</sup> *<sup>t</sup>* 0 ˜ *b*(*r*)*dg*(*r*) *α*(*t*)*e t* 0 ˜ *<sup>b</sup>*(*r*)*dg*(*r*) · ˜ *f*(*t*) + 1 *α*(*t*) *e* <sup>−</sup> *<sup>t</sup>* 0 ˜ *b*(*r*)*dg*(*r*) (−˜ *<sup>b</sup>*(*t*)) *<sup>T</sup> t α*(*s*)*e s* 0 ˜ *<sup>b</sup>*(*r*)*dg*(*r*) · ˜ *f*(*s*)*dg*(*s*) + 1 *α*(*t*) *e* <sup>−</sup> *<sup>t</sup>* 0 ˜ *b*(*r*)*dg*(*r*) (−*α*(*t*)*e t* 0 ˜ *<sup>b</sup>*(*r*)*dg*(*r*) · ˜ *f*(*t*))\* <sup>=</sup> <sup>1</sup> *α*(*T*)*e T* 0 ˜ *<sup>b</sup>*(*r*)*dg*(*r*) <sup>−</sup> <sup>1</sup> · −˜ *b*(*t*) · ) *α*(*T*) *α*(*t*) *t* 0 *α*(*s*)*e T* 0 ˜ *<sup>b</sup>*(*r*)*dg*(*r*)− *<sup>t</sup> s* ˜ *b*(*r*)*dg*(*r*) ˜ *f*(*s*)*dg*(*s*) + 1 *α*(*t*) *T t α*(*s*)*e* <sup>−</sup> *<sup>t</sup> s* ˜ *b*(*r*)*dg*(*r*) ˜ *f*(*s*)*dg*(*s*) \* + [*α*(*T*)*e T* 0 ˜ *<sup>b</sup>*(*r*)*dg*(*r*) <sup>−</sup> <sup>1</sup>] ˜ *f*(*t*) <sup>=</sup> <sup>−</sup>˜ *<sup>b</sup>*(*t*) <sup>1</sup> *α*(*t*) *T* 0 *α*(*s*)*g*˜(*t*,*s*) ˜ *f*(*s*)*dg*(*s*) + ˜ *f*(*t*) <sup>=</sup> <sup>−</sup>˜ *b*(*t*)*u*(*t*) + ˜ *f*(*t*) = −*b*(*t*)*u*(*t*) + *f*(*t*) (recall that *t* ∈ [0, *T*] \ *Dg*).

When calculating the *g*-derivative of the exponential function, we used a chain rule ([5], Theorem 2.3) together with Theorem 1, namely:

$$\begin{aligned} \left(\boldsymbol{\varepsilon}^{-\int\_0^t \tilde{b}(r)d\boldsymbol{\xi}(r)}\right)'\_{\mathcal{S}} &= \left.\boldsymbol{\varepsilon}^{-\int\_0^t \tilde{b}(r)d\boldsymbol{\xi}(r)}\cdot \left(-\int\_0^t \tilde{b}(r)d\boldsymbol{\xi}(r)\right)'\_{\mathcal{S}}\right|\_{\mathcal{S}}\\ &= \left.\boldsymbol{\varepsilon}^{-\int\_0^t \tilde{b}(r)d\boldsymbol{\xi}(r)}\cdot (-\tilde{b}(t))\right|\_{\mathcal{S}}\end{aligned}$$

The equality *u <sup>g</sup>*(*t*) = −*b*(*t*)*u*(*t*) + *f*(*t*) at the points in *Dg* can be proved exactly as in ([4], Theorem 17).

**Remark 2.** *If we impose the LS-integrability with respect to g of f , then the LS-integrability (therefore, the KS-integrability) of <sup>f</sup>*(*t*) <sup>1</sup>−*b*(*t*)*μg*({*t*}) *comes from the inequality*

$$\frac{1}{|1 - b(t)\mu\_{\mathcal{S}}(\{t\})|} \le \max\left(1, \frac{1}{\delta}\right), \quad \forall t \in [0, T].$$

**Remark 3.** *The reciprocal assertion of Theorem 2 is also valid (see* [4]*, Theorem 19). Specifically, if b*, *g*˜, *f are as postulated in Theorem <sup>2</sup> and u* : [0, *<sup>T</sup>*] <sup>→</sup> <sup>R</sup>*<sup>d</sup> is a solution of* (2)*, then*

$$
\mu(t) = \frac{1}{a(t)} \int\_0^T \frac{a(s)}{1 - b(s)\mu\_\mathcal{S}(\{s\})} \overline{g}(t, s) f(s) dg(s), \quad t \in [0, T].
$$

**Remark 4.** *As seen in* [4]*, the application* (*s* ,*s*) ∈ [0, *T*] × [0, *T*] → *e <sup>s</sup> <sup>s</sup>* ˜ *<sup>b</sup>*(*s*)*dg*(*s*) *is regulated in both variables, therefore it is bounded. If*

$$\mathcal{M} = \sup\_{(s',s'') \in [0,T] \times [0,T]} \mathfrak{e}^{\int\_{s'}^{s''} \mathbb{B}(s)d\mathfrak{g}(s)}$$

*then from the definition of g it can easily be deduced that* ˜

$$|\bar{\mathfrak{z}}(t,s)| \le \frac{\max(M, M^2)}{\left| \alpha(T)e^{\int\_0^T \bar{b}(r)d\bar{\mathfrak{z}}(r)} - 1 \right|}, \ \forall s, t \in [0, T].$$

Obviously, if

$$1 - b(t)\mu\_{\mathcal{S}}(t) > 0 \quad \text{for all } t \in [0, T]\_{\prime}$$

then *α*(*t*) = 1 for every *t* ∈ [0, *T*], therefore the formulas and the computations become much simpler.

#### **4. Main Results**

*4.1. Existence of Solutions*

We aim to obtain the existence of solutions for the set-valued periodic boundary value problem (1):

$$\begin{cases} \boldsymbol{u}\_{\mathcal{S}}'(t) + b(t)\boldsymbol{u}(t) \in F(t, \boldsymbol{u}(t)), \quad \boldsymbol{\mu}\_{\mathcal{S}} - \text{a.e. in } [0, T],\\ \boldsymbol{u}(0) = \boldsymbol{u}(T). \end{cases}$$

The notion of solution adapted from the single-valued case (Definition 3) reads as follows.

**Definition 4.** *A function <sup>u</sup>* : [0, *<sup>T</sup>*] <sup>→</sup> <sup>R</sup>*<sup>d</sup> is a solution of problem* (1) *if it is left-continuous and regulated, constant on the intervals where g is constant, g-differentiable μg-a.e. in* [0, *T*] *and*

$$
\mu'\_{\mathcal{S}}(t) + b(t)\mu(t) = f(t).
$$

*with f*(*t*) ∈ *F*(*t*, *u*(*t*)), *μ<sup>g</sup>* − *a*.*e*. *in* [0, *T*]*.*

We shall apply the following fixed-point theorem for multivalued operators.

**Theorem 3.** *(Bohnenblust–Karlin) Let X be a Banach space,* M ⊂ *X be closed and convex and the operator A* : M→P(M) *with closed, convex values be upper semicontinuous such that A*(M) *is relatively compact. Then the operator has a fixed point.*

**Theorem 4.** *Let b* : [0, *T*] → R *be LS-integrable with respect to g and suppose that (3) is fulfilled. Let F* : [0, *<sup>T</sup>*] <sup>×</sup> <sup>R</sup>*<sup>d</sup>* → P*bc*(R*d*) *satisfy the following hypotheses:*


$$|F(t,u)| \le \overline{\phi}(t)$$

*for every t* <sup>∈</sup> [0, *<sup>T</sup>*], *<sup>u</sup>* <sup>∈</sup> <sup>R</sup>*d.*

*Then the Stieltjes differential inclusion* (1) *has solutions. Moreover, the solution set of* (1) *is C-bounded.*

**Proof.** Let *Xg* be the subspace of *G*([0, *T*], R*d*) consisting of the functions being continuous on [0, *T*] \ *Dg*.

Condition (4) together with the LS-integrability with respect to *g* of *b* imply that ˜ *b* has the same feature.

Following Remark 4, we note by

$$M = \sup\_{(s',s'') \in [0,T] \times [0,T]} e^{\int\_{s''}^{s''} \mathbb{B}(s)d\mathfrak{g}(s)}.$$

By condition (4), for every *t* ∈ [0, *T*],

$$\frac{1}{|1 - b(t)\mu\_{\mathcal{S}}(\{t\})|}|F(t, u)| \le \max\left(1, \frac{1}{\delta}\right) \cdot \overline{\phi}(t)$$

so we shall denote by

$$\phi(t) = \max\left(1, \frac{1}{\delta}\right) \cdot \overline{\phi}(t), \,\forall t \in [0, T].$$

Consider

$$\mathcal{M} = \left\{ \boldsymbol{\upsilon} \in \mathcal{X}\_{\mathcal{S}} : \|\boldsymbol{\upsilon}\|\_{\mathcal{C}} \le \frac{\max\left(M, M^2\right)}{\left|a(\boldsymbol{T})e^{\int\_0^T b(\boldsymbol{r})d\boldsymbol{g}(\boldsymbol{r})} - 1\right|} \int\_0^T \boldsymbol{\phi}(s)d\boldsymbol{g}(s) \right\}.$$

and the operator *A* : M→P(M) given, for each *u* ∈ M, by

$$Au = \left\{ v \in X\_{\mathcal{S}} : v(t) = \frac{1}{a(t)} \int\_0^T \frac{a(s)}{1 - b(s)\mu\_{\mathcal{S}}(\{s\})} \tilde{g}(t, s) f(s) dg(s) : f \in S\_{F(\cdot, \mu(\cdot))} \right\}$$

with *g*˜ as in Theorem 2 and

$$S\_{F(\cdot,\mu(\cdot))} = \left\{ f \in L^1(\mu\_{\mathcal{S}'} \mathbb{R}^d) : f(t) \in F(t, \mu(t)) \,\,\mu\_{\mathcal{S}} - \text{a.e.} \right\}.$$

*A* is well defined: for each *u* ∈ *Xg*, *SF*(·,*u*(·)) is non-empty and whenever *u* ∈ *Xg*, i.e., *u* is regulated and continuous on [0, *T*] \ *Dg*, each element of *Au* has the same feature. Indeed, we note that *α* is constant in a neighborhood of *t* ∈ [0, *T*] \ *Dg*, and writing each element of *Au* as in (6), by ([29], Proposition 2.3.16) we deduce that it is regulated and continuous on [0, *T*] \ *Dg*.

We next show that *u<sup>C</sup>* <sup>≤</sup> max(*M*,*M*2) *α*(*T*)*<sup>e</sup> T* 0 ˜ *b*(*r*)*dg*(*r*) −1 *<sup>T</sup>* <sup>0</sup> *φ*(*s*)*dg*(*s*) implies that every *v* ∈ *Au* satisfies the

same inequality.

Indeed, fix *t* ∈ [0, *T*]. Then every *v* ∈ *Au* is given (by the definition of the operator *A*) by some selection *f* of *F*(·, *u*(·)) and we can see, by Remark 4, that

$$\begin{aligned} \|v(t)\| &\leq \max\left(1, \frac{1}{\delta}\right) \int\_0^T |\bar{g}(t,s)| \cdot |f(s)| \, dg(s) \\ &\leq \max\left(1, \frac{1}{\delta}\right) \frac{\max(M, M^2)}{\left|a(T)e^{\int\_0^T \bar{b}(r)d\mathfrak{g}(r)} - 1\right|} \int\_0^T |f(s)| \, dg(s) \\ &\leq \max\left(1, \frac{1}{\delta}\right) \frac{\max(M, M^2)}{\left|a(T)e^{\int\_0^T \bar{b}(r)d\mathfrak{g}(r)} - 1\right|} \int\_0^T \overline{\Phi}(s) dg(s), \end{aligned}$$

whence

$$||v||\_{\mathcal{C}} \le \frac{\max(M, M^2)}{\left|\kappa(T)e^{\int\_0^T \mathbb{B}(r)d\mathbb{y}(r)} - 1\right|} \int\_0^T \phi(s)d\mathbb{y}(s).$$

Let us next check that the operator has closed, convex values.

Let *u* ∈ M. Obviously, *SF*(·,*u*(·)) is convex (recall that *F* has convex values), therefore, *Au* is convex as well.

To prove that it is closed, take (*vn*)*n*∈<sup>N</sup> ⊂ *Au* uniformly convergent to *<sup>v</sup>* ∈ M; specifically, for each *<sup>n</sup>* ∈ N, one can find *fn* ∈ *SF*(·,*u*(·)) such that

$$w\_n(t) = \frac{1}{a(t)} \int\_0^T \frac{a(s)}{1 - b(s)\mu\_\mathcal{S}(\{s\})} \overline{g}(t, s) f\_n(s) dg(s), \ t \in [0, T] \text{ and } v\_n \to v \text{ uniformly.} $$

One can see that

$$||f\_n(t)|| \le \overline{\phi}(t), \ \forall n \in \mathbb{N}, \ t \in [0, T]\_\prime$$

so there exists a subsequence (*fnk* )*k*∈<sup>N</sup> weakly *<sup>L</sup>*1(*μg*, <sup>R</sup>*d*) convergent to a function *<sup>f</sup>* <sup>∈</sup> *<sup>L</sup>*1(*μg*, <sup>R</sup>*d*) (Dunford-Pettis Theorem). In a classical way (Mazur's theorem and properties of norm-convergent sequences in *L*1(*μg*, R*d*)), a sequence of convex combinations tends pointwise *μ<sup>g</sup>* -a.e. to *f* , whence

$$f(\cdot) \in \mathcal{S}\_{F(\cdot, u(\cdot))^\*}$$

By a dominated convergence result (see [28], Theorem 6.8.7) applied for the components of (*fnk* )*k*∈N, *<sup>f</sup>* , one deduces that

$$\begin{aligned} v\_{n\_k}(t) &= \frac{1}{\mathfrak{a}(t)} \int\_0^T \frac{\mathfrak{a}(s)}{1 - b(s)\mu\_{\mathfrak{S}}(\{s\})} \mathfrak{g}(t, s) f\_{n\_k}(s) d\mathfrak{g}(s) \rightarrow \\ &\quad \frac{1}{\mathfrak{a}(t)} \int\_0^T \frac{\mathfrak{a}(s)}{1 - b(s)\mu\_{\mathfrak{S}}(\{s\})} \mathfrak{g}(t, s) f(s) d\mathfrak{g}(s) \end{aligned}$$

and so,

$$v(t) = \frac{1}{a(t)} \int\_0^T \frac{\kappa(s)}{1 - b(s)\mu\_{\mathcal{G}}(\{s\})} \overline{g}(t, s) f(s) dg(s), \ t \in [0, T], t$$

thus *Au* is closed.

We will prove that *A* satisfies the hypotheses of Theorem 3.

We check that *A*(M) is relatively compact, using Lemma 1.

Take 0 ≤ *t* < *t* ≤ *T*.

For each *u* ∈ M and each *v* ∈ *Au* (defined via a selection *f* of *F*(·, *u*(·)) LS-integrable with respect to *g*),

$$\begin{split} \left\| \left| v(t) - v(t') \right| \right\| &\leq \quad \left\| \left| \frac{1}{a(t)} \int\_{0}^{T} \frac{a(s)}{1 - b(s)\mu\_{\mathcal{S}}(\{s\})} (\bar{\boldsymbol{\mathcal{g}}}(t, s) - \bar{\boldsymbol{\mathcal{g}}}(t', s)) f(s)) d\boldsymbol{\mathfrak{g}}(s) \right\| \\ &\quad + \quad \left\| \left( \frac{1}{a(t)} - \frac{1}{a(t')} \right) \int\_{0}^{T} \frac{a(s)}{1 - b(s)\mu\_{\mathcal{S}}(\{s\})} \bar{\boldsymbol{\mathcal{g}}}(t', s) f(s)) d\boldsymbol{\mathfrak{g}}(s) \right\| . \end{split}$$

We note that |*α*(*t*)| = 1 for each *t* ∈ [0, *T*], so we can write

( ( ( ( 1 *α*(*t*) *T* 0 *α*(*s*) 1 − *b*(*s*)*μg*({*s*}) (*g*˜(*t*,*s*) − *g*˜(*t* ,*s*))*f*(*s*))*dg*(*s*) ( ( ( ( ≤ 1 *α*(*T*)*<sup>e</sup> T* 0 ˜ *<sup>b</sup>*(*r*)*dg*(*r*) <sup>−</sup> <sup>1</sup> )( ( ( (*α*(*T*) *t* 0 *α*(*s*) 1 − *b*(*s*)*μg*({*s*}) (*e T* 0 ˜ *<sup>b</sup>*(*r*)*dg*(*r*)− *<sup>t</sup> s* ˜ *<sup>b</sup>*(*r*)*dg*(*r*) <sup>−</sup> *<sup>e</sup> T* 0 ˜ *<sup>b</sup>*(*r*)*dg*(*r*)− *<sup>t</sup> s* ˜ *b*(*r*)*dg*(*r*) )*f*(*s*)*dg*(*s*) ( ( ( ( + ( ( ( ( *T t α*(*s*) 1 − *b*(*s*)*μg*({*s*}) (*e* <sup>−</sup> *<sup>t</sup> s* ˜ *<sup>b</sup>*(*r*)*dg*(*r*) <sup>−</sup> *<sup>e</sup>* <sup>−</sup> *<sup>t</sup> s* ˜ *b*(*r*)*dg*(*r*) )*f*(*s*)*dg*(*s*) ( ( ( ( + ( ( ( ( *t t α*(*s*) 1 − *b*(*s*)*μg*({*s*}) (*e* <sup>−</sup> *<sup>t</sup> s* ˜ *<sup>b</sup>*(*r*)*dg*(*r*) <sup>−</sup> *<sup>α</sup>*(*T*)*<sup>e</sup> T* 0 ˜ *<sup>b</sup>*(*r*)*dg*(*r*)− *<sup>t</sup> s* ˜ *b*(*r*)*dg*(*r*) )*f*(*s*)*dg*(*s*) ( ( ( ( \* <sup>=</sup> <sup>1</sup> *α*(*T*)*<sup>e</sup> T* 0 ˜ *<sup>b</sup>*(*r*)*dg*(*r*) <sup>−</sup> <sup>1</sup> )( ( ( ( *t* 0 *α*(*s*) 1 − *b*(*s*)*μg*({*s*}) *e T* 0 ˜ *<sup>b</sup>*(*r*)*dg*(*r*)− *<sup>t</sup> s* ˜ *b*(*r*)*dg*(*r*) (1 − *e* <sup>−</sup> *<sup>t</sup> t* ˜ *b*(*r*)*dg*(*r*) )*f*(*s*)*dg*(*s*) ( ( ( ( + ( ( ( ( *T t α*(*s*) 1 − *b*(*s*)*μg*({*s*}) *e* <sup>−</sup> *<sup>t</sup> s* ˜ *b*(*r*)*dg*(*r*) (*e t t* ˜ *<sup>b</sup>*(*r*)*dg*(*r*) <sup>−</sup> <sup>1</sup>)*f*(*s*)*dg*(*s*) ( ( ( ( + ( ( ( ( *t t α*(*s*) 1 − *b*(*s*)*μg*({*s*}) (*e* <sup>−</sup> *<sup>t</sup> s* ˜ *<sup>b</sup>*(*r*)*dg*(*r*) <sup>−</sup> *<sup>α</sup>*(*T*)*<sup>e</sup> T* 0 ˜ *<sup>b</sup>*(*r*)*dg*(*r*)− *<sup>t</sup> s* ˜ *b*(*r*)*dg*(*r*) )*f*(*s*)*dg*(*s*) ( ( ( ( \* .

On the other hand, again by |*α*(*t*)| = |*α*(*t* )| = 1,

$$\begin{aligned} & \left\| \left( \frac{1}{\alpha(t)} - \frac{1}{\alpha(t')} \right) \int\_0^T \frac{\alpha(s)}{1 - b(s)\mu\_{\mathcal{S}}(\{s\})} \bar{\mathfrak{g}}(t', s) f(s) d\mathfrak{g}(s) \right\|\_{\ast} \\ &= |\mathfrak{a}(t) - \mathfrak{a}(t')| \left\| \int\_0^T \frac{\mathfrak{a}(s)}{1 - b(s)\mu\_{\mathcal{S}}(\{s\})} \bar{\mathfrak{g}}(t', s) f(s) d\mathfrak{g}(s) \right\|\_{\ast} \end{aligned}$$

and using the definition of *g*˜ together with Remark 2 one gets

$$\begin{split} & \left\| \left( \frac{1}{a(t)} - \frac{1}{a(t')} \right) \int\_0^T \frac{a(s)}{1 - b(s)\mu\_{\mathcal{S}}(\{s\})} \bar{g}(t', s) f(s) dg(s) \right\| \\ & \leq \frac{|a(t) - a(t')|}{|a(T)e^{\int\_0^T b(r)d\mathfrak{g}(r)} - 1|} \left[ \left| \int\_0^{t'} \frac{a(s)}{1 - b(s)\mu\_{\mathcal{S}}(\{s\})} e^{\int\_0^T \bar{b}(r)d\mathfrak{g}(r) - \int\_s^{t'} \bar{b}(r)d\mathfrak{g}(r)} f(s) dg(s) \right| \right] \\ & + \left\| \int\_{t'}^T \frac{a(s)}{1 - b(s)\mu\_{\mathcal{S}}(\{s\})} e^{-\int\_s^{t'} \bar{b}(r)d\mathfrak{g}(r)} f(s) dg(s) \right\| \right] \\ & \leq \max\left(1, \frac{1}{\delta} \right) \frac{|a(t) - a(t')|}{|a(T)e^{\int\_0^T b(r)d\mathfrak{g}(r)} - 1|} \left[ \int\_0^{t'} \left\| e^{\int\_0^T \bar{b}(r)d\mathfrak{g}(r) - \int\_s^{t'} \bar{b}(r)d\mathfrak{g}(r)} f(s) \right\| \right] d\mathfrak{g}(s) \\ & + \int\_{t'}^T \left\| e^{-\int\_s^{t'} \bar{b}(r)d\mathfrak{g}(r)} f(s) \right\| \, d\mathfrak{g}(s) \right]. \end{split}$$

But

$$\mathfrak{e}\_{\mathfrak{e}}\int\_{0}^{T}\mathfrak{b}(r)d\mathfrak{g}(r)-\int\_{s}^{t}\mathfrak{b}(r)d\mathfrak{g}(r) \begin{array}{c} \displaystyle \leq \ M^{2} \end{array} \text{ and } \quad \mathfrak{e}\_{\mathfrak{e}}\int\_{0}^{T}\mathfrak{b}(r)d\mathfrak{g}(r)-\int\_{s}^{t'}\mathfrak{b}(r)d\mathfrak{g}(r) \begin{array}{c} \displaystyle \leq \ M^{2} \end{array}$$

We thus get

$$\begin{split} & \quad \|\boldsymbol{\varepsilon}(t) - \boldsymbol{\upsilon}(t')\| \\ & \leq \frac{M}{\left|\boldsymbol{a}(T)e^{\int\_{0}^{T}\boldsymbol{\delta}(\boldsymbol{r})d\boldsymbol{g}(\boldsymbol{r})} - 1\right|} \max\left(1, \frac{1}{\delta}\right) \left[M\int\_{0}^{t} \left|1 - e^{-\int\_{t}^{f}\boldsymbol{\delta}(\boldsymbol{r})d\boldsymbol{g}(\boldsymbol{r})}\right| \cdot \|\boldsymbol{f}(\boldsymbol{s})\| \|d\boldsymbol{g}(\boldsymbol{s})\| \right] \\ & + \int\_{t'}^{T} \left|e^{\int\_{t'}^{f'}\boldsymbol{\delta}(\boldsymbol{r})d\boldsymbol{g}(\boldsymbol{r})} - 1\right| \cdot \|\boldsymbol{f}(\boldsymbol{s})\| \|d\boldsymbol{g}(\boldsymbol{s}) \\ & + (1 + M) \int\_{t}^{t'} \|\boldsymbol{f}(\boldsymbol{s})\| \|d\boldsymbol{g}(\boldsymbol{s})\| \\ & + \frac{|\boldsymbol{a}(t) - \boldsymbol{a}(t')|}{|\boldsymbol{a}(T)e^{\int\_{0}^{T}\boldsymbol{\delta}(\boldsymbol{r})d\boldsymbol{g}(\boldsymbol{r})} - 1|} \max\left(1, \frac{1}{\delta}\right) \left[M^{2} \int\_{0}^{t'} \|\boldsymbol{f}(\boldsymbol{s})\| \|d\boldsymbol{g}(\boldsymbol{s})\| \right. \\ & \left. + M \int\_{t'}^{T} \|\boldsymbol{f}(\boldsymbol{s})\| \|d\boldsymbol{g}(\boldsymbol{s})\| \right]. \end{split}$$

so, taking into account the definition of *φ*, it follows that

$$\begin{split} & \left|| \boldsymbol{\varepsilon}(t) - \boldsymbol{\upsilon}(t') \right|| \\ & \leq \frac{M}{\left| \boldsymbol{a}(\boldsymbol{T}) \boldsymbol{e}\_{0}^{\boldsymbol{f}\_{0}^{\boldsymbol{f}} \boldsymbol{\delta}(\boldsymbol{r}) \boldsymbol{d} \boldsymbol{g}(\boldsymbol{r})} - 1 \right|} \left| \boldsymbol{M} \right| \boldsymbol{1} - \boldsymbol{e}^{-\int\_{t}^{\boldsymbol{f}'} \boldsymbol{\delta}(\boldsymbol{r}) \boldsymbol{d} \boldsymbol{g}(\boldsymbol{r})} \right| \cdot \int\_{0}^{T} \boldsymbol{\phi}(\boldsymbol{s}) d\boldsymbol{g}(\boldsymbol{s}) \\ & \quad + \left| \boldsymbol{e}\_{t}^{\int\_{t}^{\boldsymbol{f}'} \boldsymbol{\delta}(\boldsymbol{r}) \boldsymbol{d} \boldsymbol{g}(\boldsymbol{r})} - 1 \right| \cdot \int\_{0}^{T} \boldsymbol{\phi}(\boldsymbol{s}) d\boldsymbol{g}(\boldsymbol{s}) + (1 + M) \int\_{t}^{\boldsymbol{f}'} \boldsymbol{\phi}(\boldsymbol{s}) d\boldsymbol{g}(\boldsymbol{s}) \\ & + (M + 1) \left| \boldsymbol{a}(\boldsymbol{t}) - \boldsymbol{a}(\boldsymbol{t}') \right| \int\_{0}^{T} \boldsymbol{\phi}(\boldsymbol{s}) d\boldsymbol{g}(\boldsymbol{s}) \Big{]}. \end{split}$$

But

$$\left| \left| 1 - \mathfrak{e}^{-\int\_{t}^{t'} \mathbb{B}(r) d\mathfrak{g}(r)} \right| \leq M \left| \mathfrak{e}^{-\int\_{0}^{t} \mathbb{B}(r) d\mathfrak{g}(r)} - \mathfrak{e}^{-\int\_{0}^{t'} \mathbb{B}(r) d\mathfrak{g}(r)} \right| \right| $$

and similarly for *e t t* ˜ *<sup>b</sup>*(*r*)*dg*(*r*) <sup>−</sup> <sup>1</sup> , while

$$\int\_{t}^{t'} \phi(s) dg(s) = \int\_{0}^{t'} \phi(s) dg(s) - \int\_{0}^{t} \phi(s) dg(s).$$

Remark 1 yields now that the set *A*(M) is equiregulated.

The pointwise boundedness is immediate, therefore Lemma 1 implies that {*Au* : *u* ∈ M} is relatively compact.

Next, let us prove that *A* is upper semicontinuous. As *A*(M) is relatively compact, it suffices to verify that *A* has closed graph (see [36], Proposition 2.23).

Let (*un*)*n*∈<sup>N</sup> ⊂ M converge uniformly to *<sup>u</sup>* ∈ M and (*vn*)*n*∈<sup>N</sup> ⊂ M converge uniformly to *v* ∈ M be such that *vn* ∈ *Aun* for all *n* ∈ N.

One can find, for every *<sup>n</sup>* ∈ N, *fn* ∈ *SF*(·,*un*(·)) such that

$$w\_n(t) = \frac{1}{a(t)} \int\_0^T \frac{a(s)}{1 - b(s)\mu\_\mathcal{S}(\{s\})} \overline{g}(t, s) f\_n(s) dg(s), \ t \in [0, T].$$

As before,

$$\|\|f\_n(t)\|\| \le \overline{\Phi}(t), \ \forall n \in \mathbb{N}, \ t \in [0, T]\_\prime\prime$$

so there is a subsequence (*fnk* )*k*∈<sup>N</sup> convergent in the weak-*L*1(*μg*, <sup>R</sup>*d*) topology to a function *<sup>f</sup>* <sup>∈</sup> *<sup>L</sup>*1(*μg*, <sup>R</sup>*d*). It follows that a sequence of convex combinations of { *fnk* : *<sup>k</sup>* <sup>∈</sup> <sup>N</sup>} tends pointwise (*μg*-a.e.) to *f* . On the other hand, *F* is upper semicontinuous with respect to the second value also with closed values , thus it has closed graph with respect to the second value (see [36], Proposition 2.17). Combining these two facts, we may easily check that

$$f \in \mathcal{S}\_{F(\cdot,\mu(\cdot))} \cdot$$

By a dominated convergence result (see [28], Theorem 6.8.7) applied for the components of (*fnk* )*k*∈<sup>N</sup> and *<sup>f</sup>* , one deduces that the corresponding sequence of convex combinations of (*vnk* )*<sup>k</sup>* converges to

$$\frac{1}{\varkappa(t)} \int\_0^T \frac{\varkappa(s)}{1 - b(s)\mu\_{\mathcal{S}}(\{s\})} \overline{g}(t, s) f(s) dg(s)$$

whence

$$w(t) = \frac{1}{a(t)} \int\_0^T \frac{a(s)}{1 - b(s)\mu\_{\mathcal{S}}(\{s\})} \overline{g}(t, s) f(s) dg(s), \ t \in [0, T]$$

and consequently *v* ∈ *Au*.

Finally, Bohnenblust–Karlin fixed-point theorem yields that the operator has fixed points, which are solutions to problem (1) by Theorem 2.

#### *4.2. Dependence on the Data*

Let us now study in which manner the solution set of problem (1) depends on the data. For this purpose, we are forced to drop the dependence on the state of the right-hand side. To be more precise, if we consider functions *<sup>b</sup>*1, *<sup>b</sup>*<sup>2</sup> as in Theorem <sup>4</sup> and multifunctions *<sup>F</sup>*1, *<sup>F</sup>*<sup>2</sup> : [0, *<sup>T</sup>*] → P*bc*(R*d*) such that the considered problem has solutions, we are interested in finding the relation between the solution set S<sup>1</sup> of

$$\begin{cases} \ u\_{\S}'(t) + b\_1(t)\mu(t) \in F\_1(t), \; \mu\_{\S}-\text{a.e. in } [0,T],\\ \ u(0) = u(T) \end{cases}$$

and the solution set S<sup>2</sup> of

$$\begin{cases} \ u\_{\S}'(t) + b\_2(t)u(t) \in F\_2(t), \; \mu\_{\S}-\text{a.e. in } [0,T],\\ u(0) = u(T). \end{cases}$$

The perturbation of *b* shall be measured through

$$||b\_1 - b\_2||\_\mathcal{C} = \sup\_{t \in [0, T]} |b\_1(t) - b\_2(t)| \quad \text{or} \quad ||b\_1 - b\_2||\_{L^1} = \int\_0^T |b\_1(\mathbf{s}) - b\_2(\mathbf{s})| d\mathbf{y}(\mathbf{s})$$

while the perturbation of *F* through

$$D\_{\mathbb{C}}(F\_1, F\_2) = \sup\_{t \in [0, T]} D(F\_1(t), F\_2(t)) \quad \text{or} \quad D\_{L^1}(F\_1, F\_2) = \int\_0^T D(F\_1(s), F\_2(s)) d\mathfrak{g}(s).$$

Correspondingly, one can measure the distance between the *C*-bounded sets S1, S<sup>2</sup> of regulated functions in the following ways:

$$D\_{\mathbb{C}}(\mathcal{S}\_1, \mathcal{S}\_2) = \max(e\_{\mathbb{C}}(\mathcal{S}\_1, \mathcal{S}\_2), e\_{\mathbb{C}}(\mathcal{S}\_2, \mathcal{S}\_1))\_{\prime\prime}$$

where the Pompeiu-excess of the set S<sup>1</sup> over the set S<sup>2</sup> is defined by

$$e\_{\mathbb{C}}(\mathcal{S}\_1, \mathcal{S}\_2) = \sup\_{\boldsymbol{\mu} \in \mathcal{S}\_1} \inf\_{\boldsymbol{\mu'} \in \mathcal{S}\_2} ||\boldsymbol{\mu} - \boldsymbol{\mu'}||\_{\mathcal{C}}.$$

or

$$D\_{L^1}(\mathcal{S}\_1, \mathcal{S}\_2) = \max(e\_{L^1}(\mathcal{S}\_1, \mathcal{S}\_2), e\_{L^1}(\mathcal{S}\_2, \mathcal{S}\_1))\_\prime$$

where the excess of S<sup>1</sup> over S<sup>2</sup> is

$$\varrho\_{L^1}(\mathcal{S}\_1, \mathcal{S}\_2) = \sup\_{u \in \mathcal{S}\_1} \inf\_{u' \in \mathcal{S}\_2} ||u - u'||\_{L^1}.$$

Let us note that

**Remark 5.** *For any x*1, *x*<sup>2</sup> ∈ [*a*, *b*] ⊂ R*,*

*(i)*

$$|e^{\mathbf{x}\_1} - e^{\mathbf{x}\_2}| \le e^b |\mathbf{x}\_1 - \mathbf{x}\_2|.$$

*(ii) if a* > 0*,*

$$|\log|\mathbf{x}\_1| - \log|\mathbf{x}\_2|| \le \frac{1}{a} |\mathbf{x}\_1 - \mathbf{x}\_2|.$$

**Theorem 5.** *Let b*1, *b*<sup>2</sup> : [0, *T*] → R *be LS-integrable with respect to g and suppose that (3) is fulfilled for both b*<sup>1</sup> *and b*2*.*

*Let F*1, *<sup>F</sup>*<sup>2</sup> : [0, *<sup>T</sup>*] → P*bc*(R*d*) *satisfy the following hypotheses:*


$$|F\_i(t)| \le \overline{\phi}(t), \ i = 1, 2, \ \forall t \in [0, T].$$

*Then there exist positive constants Ci*, *i* = 1, 6 *such that for every u*<sup>1</sup> ∈ S1*, one can find u*<sup>2</sup> ∈ S<sup>2</sup> *satisfying, for all t* ∈ [0, *T*]*,*

$$\begin{aligned} & \|u\_1(t) - u\_2(t)\| \\ & \le \mathcal{C}\_1 \int\_0^T |b\_1(s) - b\_2(s)| d\mathfrak{g}(s) + \mathcal{C}\_2 |a\_1(T) - a\_2(T)| \\ & + \mathcal{C}\_3 \int\_0^T \overline{\mathfrak{g}}(s) |a\_1(s) - a\_2(s)| d\mathfrak{g}(s) + \mathcal{C}\_4 \int\_0^T |b\_1(s) - b\_2(s)| \overline{\mathfrak{g}}(s) d\mathfrak{g}(s) \\ & + \mathcal{C}\_5 \int\_0^T D(F\_1(s), F\_2(s)) d\mathfrak{g}(s) + \mathcal{C}\_6 |a\_1(t) - a\_2(t)|. \end{aligned}$$

**Proof.** Let *u*<sup>1</sup> ∈ S1. Then there exists a selection *f*<sup>1</sup> of *F*<sup>1</sup> which is LS-integrable with respect to *g* such that

$$\mu\_1(t) = \frac{1}{a\_1(t)} \int\_0^T \frac{a\_1(s)}{1 - b\_1(s)\mu\_\mathcal{g}(\{s\})} \tilde{g}\_1(t, s) f\_1(s) dg(s), \ t \in [0, T].$$

where

$$D\_{\mathcal{S}}^{1,-} = \{ t \in D\_{\mathcal{S}} : 1 - b\_1(t)\mu\_{\mathcal{S}}(\{t\}) < 0 \} = \{ t\_1^1, \dots, t\_k^1 \}$$

(with the obvious convention *t* 1 <sup>0</sup> = 0 and *t* 1 *<sup>k</sup>*+<sup>1</sup> = *T*),

$$a\_1(t) = \begin{cases} \ 1, \text{ if } 0 \le t \le t\_1^1 \\ \ (-1)^i, \text{ if } t\_i^1 < t \le t\_{i+1}^1, \ i = 1, \dots, k. \end{cases}$$

$$\tilde{b\_1}(t) = \begin{cases} \begin{array}{c} b\_1(t), \text{ if } t \in [0, T] \\ \frac{-\log|1 - b\_1(t)\mu\_{\mathcal{S}}(\{t\})|}{\mu\_{\mathcal{S}}(\{t\})}, \text{ if } t \in D\_{\mathcal{S}} \end{array} \end{cases}$$

and

$$\mathcal{g}\_1(t,s) = \frac{1}{a\_1(T)e^{\int\_0^T \tilde{b}\_1(r)d\mathfrak{g}(r)} - 1} \left\{ \begin{array}{ll} a\_1(T)e^{\int\_0^T \tilde{b}\_1(r)d\mathfrak{g}(r) - \int\_s^t \tilde{b}\_1(r)d\mathfrak{g}(r)}, \text{ if } 0 \le s \le t \le T\\ e^{-\int\_s^t \tilde{b}\_1(r)d\mathfrak{g}(r)}, \text{ if } 0 \le t < s \le T. \end{array} \right\}$$

By ([34], Corollary 8.2.13) we can choose *f*<sup>2</sup> as the *μg*-measurable selection of *F*<sup>2</sup> satisfying

*f*1(*t*) − *f*2(*t*) = *d*(*f*1(*t*), *F*2(*t*)), ∀ *t* ∈ [0, *T*]

so, by the very definition of the Pompeiu-Hausdorff distance,

$$\|f\_1(t) - f\_2(t)\| \le D(F\_1(t), F\_2(t)), \ \forall \ t \in [0, T].$$

Consider now the function *<sup>u</sup>*<sup>2</sup> : [0, *<sup>T</sup>*] <sup>→</sup> <sup>R</sup>*<sup>d</sup>* given by

$$
\mu\_2(t) = \frac{1}{a\_2(t)} \int\_0^T \frac{a\_2(s)}{1 - b\_2(s)\mu\_\mathcal{g}(\{s\})} \mathfrak{g}\_2(t, s) f\_2(s) d\mathcal{g}(s), \ t \in [0, T],
$$

where

$$D\_{\mathcal{J}}^{2,-} = \{ t \in D\_{\mathcal{J}} : 1 - b\_2(t)\mu\_{\mathcal{J}}(\{t\}) < 0 \} = \{t\_1^2, \dots, t\_l^2\},$$

(again, with the convention *t* 2 <sup>0</sup> = 0 and *t* 2 *<sup>l</sup>*+<sup>1</sup> = *T*),

$$\begin{aligned} a\_2(t) &= \begin{cases} 1, \text{ if } 0 \le t \le t\_1^2 \\ (-1)^i, \text{ if } t\_i^2 < t \le t\_{i+1}^2, \ i = 1, \dots, l\_i \end{cases} \\\\ \tilde{b}\_2(t) &= \begin{cases} \begin{array}{c} b\_2(t), \text{ if } t \in [0, T] \\ \frac{-\log|1 - b\_2(t)\mu\_{\mathcal{S}}(\{t\})|}{\mu\_{\mathcal{S}}(\{t\})}, \text{ if } t \in D\_{\mathcal{S}} \end{cases} \end{aligned} \end{aligned}$$

and

$$\mathfrak{g}\_2(t,s) = \frac{1}{a\_2(T)e^{\int\_0^T \tilde{b}\_2(r)d\mathfrak{g}(r)} - 1} \left\{ \begin{array}{ll} a\_2(T)e^{\int\_0^T \tilde{b}\_2(r)d\mathfrak{g}(r) - \int\_s^t \tilde{b}\_2(r)d\mathfrak{g}(r)}, \text{ if } 0 \le s \le t \le T\\ e^{-\int\_s^t \tilde{b}\_2(r)d\mathfrak{g}(r)}, \text{ if } 0 \le t < s \le T. \end{array} \right\}$$

Obviously, *u*<sup>2</sup> ∈ S<sup>2</sup> by Theorem 2. Let us see that it satisfies the requested inequality for some well-chosen constants *Ci*, *i* = 1, ..., 6.

First, we may write

*u*1(*t*) − *u*2(*t*) = ( ( ( ( 1 *α*1(*t*) *T* 0 *α*1(*s*) 1 − *b*1(*s*)*μg*({*s*}) *g*˜1(*t*,*s*)*f*1(*s*)*dg*(*s*) <sup>−</sup> <sup>1</sup> *α*2(*t*) *T* 0 *α*2(*s*) 1 − *b*2(*s*)*μg*({*s*}) *g*˜2(*t*,*s*)*f*2(*s*)*dg*(*s*) ( ( ( ( ≤ ( ( ( ( 1 *α*1(*t*) *T* 0 *α*1(*s*) 1 − *b*1(*s*)*μg*({*s*}) *g*˜1(*t*,*s*)*f*1(*s*)*dg*(*s*) <sup>−</sup> <sup>1</sup> *α*1(*t*) *T* 0 *α*2(*s*) 1 − *b*2(*s*)*μg*({*s*}) *g*˜2(*t*,*s*)*f*2(*s*)*dg*(*s*) ( ( ( ( + ( ( ( ( 1 *α*1(*t*) *T* 0 *α*2(*s*) 1 − *b*2(*s*)*μg*({*s*}) *g*˜2(*t*,*s*)*f*2(*s*)*dg*(*s*) <sup>−</sup> <sup>1</sup> *α*2(*t*) *T* 0 *α*2(*s*) 1 − *b*2(*s*)*μg*({*s*}) *g*˜2(*t*,*s*)*f*2(*s*)*dg*(*s*) ( ( ( (

Using the remark that |*α*1(*t*)| = 1 and also |*α*2(*t*)| = 1 for every *t* ∈ [0, *T*], we obtain

$$\begin{aligned} & \left\| \| u\_1(t) - u\_2(t) \| \\ & \le \left\| \int\_0^T \left( \frac{a\_1(s)}{1 - b\_1(s)\mu\_\mathcal{S}(\{s\})} \tilde{g}\_1(t, s) f\_1(s) - \frac{a\_2(s)}{1 - b\_2(s)\mu\_\mathcal{S}(\{s\})} \tilde{g}\_2(t, s) f\_2(s) \right) d\mathfrak{g}(s) \right\| \\ & + \left\| \left( a\_1(t) - a\_2(t) \right) \int\_0^T \frac{a\_2(s)}{1 - b\_2(s)\mu\_\mathcal{S}(\{s\})} \tilde{g}\_2(t, s) f\_2(s) d\mathfrak{g}(s) \right\| \end{aligned} \tag{7}$$

(please note that 1 *<sup>α</sup>*1(*t*) <sup>−</sup> <sup>1</sup> *α*2(*t*) <sup>=</sup> <sup>|</sup>*α*1(*t*) <sup>−</sup> *<sup>α</sup>*2(*t*)|, for every *<sup>t</sup>* <sup>∈</sup> [0, *<sup>T</sup>*]). Let us evaluate the first term of the sum (7):

$$\begin{split} & \left\| \left| \int\_{0}^{T} \left( \frac{a\_{1}(s)}{1 - b\_{1}(s)\mu\_{\mathcal{S}}(\{s\})} \tilde{g}\_{1}(t, s) f\_{1}(s) - \frac{a\_{2}(s)}{1 - b\_{2}(s)\mu\_{\mathcal{S}}(\{s\})} \tilde{g}\_{2}(t, s) f\_{2}(s) \right) d\mathcal{g}(s) \right| \right\| \\ & \leq \left\| \int\_{0}^{T} \left( \frac{a\_{1}(s)}{1 - b\_{1}(s)\mu\_{\mathcal{S}}(\{s\})} \tilde{g}\_{1}(t, s) - \frac{a\_{2}(s)}{1 - b\_{2}(s)\mu\_{\mathcal{S}}(\{s\})} \tilde{g}\_{2}(t, s) \right) f\_{1}(s) d\mathcal{g}(s) \right\| \\ & + \left\| \int\_{0}^{T} \frac{a\_{2}(s)}{1 - b\_{2}(s)\mu\_{\mathcal{S}}(\{s\})} \tilde{g}\_{2}(t, s) (f\_{1}(s) - f\_{2}(s)) d\mathcal{g}(s) \right\| \end{split}$$

and by Remark 2,

 *α*1(*s*) 1 − *b*1(*s*)*μg*({*s*}) *<sup>g</sup>*˜1(*t*,*s*) <sup>−</sup> *<sup>α</sup>*2(*s*) 1 − *b*2(*s*)*μg*({*s*}) *g*˜2(*t*,*s*) ≤ *α*1(*s*) 1 − *b*1(*s*)*μg*({*s*}) (*g*˜1(*t*,*s*) − *g*˜2(*t*,*s*)) + *α*1(*s*) <sup>1</sup> <sup>−</sup> *<sup>b</sup>*1(*s*)*μg*({*s*}) <sup>−</sup> *<sup>α</sup>*2(*s*) 1 − *b*2(*s*)*μg*({*s*}) *g*˜2(*t*,*s*) = *α*1(*s*) 1 − *b*1(*s*)*μg*({*s*}) (*g*˜1(*t*,*s*) − *g*˜2(*t*,*s*)) + *α*1(*s*) − *α*2(*s*)+(*α*2(*s*)*b*1(*s*) − *α*1(*s*)*b*2(*s*))*μg*({*s*}) (<sup>1</sup> <sup>−</sup> *<sup>b</sup>*1(*s*)*μg*({*s*}))(<sup>1</sup> <sup>−</sup> *<sup>b</sup>*2(*s*)*μg*({*s*})) *<sup>g</sup>*˜2(*t*,*s*) <sup>≤</sup> max 1, <sup>1</sup> *δ*1 |*g*˜1(*t*,*s*) − *g*˜2(*t*,*s*)| <sup>+</sup> max 1, <sup>1</sup> *δ*1 max 1, <sup>1</sup> *δ*2 |*g*˜2(*t*,*s*)|(|*α*1(*s*) − *α*2(*s*)| +|*α*1(*s*) − *α*2(*s*)|·|*b*2(*s*)*μg*({*s*})| + |*b*1(*s*) − *b*2(*s*)|*μg*({*s*})),

where *δ*<sup>1</sup> , *δ*<sup>2</sup> are the corresponding positive constants in Remark 2 for *b*<sup>1</sup> , *b*<sup>2</sup> respectively. Since the condition (4) is verified, |*b*2(*s*)*μg*({*s*})| is bounded, say by *m*2. Then

$$\begin{split} & \left\| \int\_{0}^{T} \left( \frac{\mathfrak{a}\_{1}(s)}{1 - b\_{1}(s)\mu\_{\mathcal{S}}(\{s\})} \tilde{g}\_{1}(t, s) - \frac{\mathfrak{a}\_{2}(s)}{1 - b\_{2}(s)\mu\_{\mathcal{S}}(\{s\})} \tilde{g}\_{2}(t, s) \right) f\_{1}(s) dg(s) \right\| \\ & \leq \max\left(1, \frac{1}{\delta\_{1}} \right) \int\_{0}^{T} |\tilde{g}\_{1}(t, s) - \tilde{g}\_{2}(t, s)| ||f\_{1}(s)|| dg(s) \\ & + \max\left(1, \frac{1}{\delta\_{1}} \right) \max\left(1, \frac{1}{\delta\_{2}} \right) \int\_{0}^{T} (1 + m\_{2}) |\tilde{g}\_{2}(t, s)| ||f\_{1}(s)|| |a\_{1}(s) - a\_{2}(s)| dg(s) \\ & + \max\left(1, \frac{1}{\delta\_{1}} \right) \max\left(1, \frac{1}{\delta\_{2}} \right) g(T) \int\_{0}^{T} |\tilde{g}\_{2}(t, s)| ||b\_{1}(s) - b\_{2}(s)|| ||f\_{1}(s)|| dg(s). \end{split}$$

We can also see, by the choice of *f*<sup>2</sup> that

$$\begin{aligned} & \left\| \int\_0^T \frac{a\_2(s)}{1 - b\_2(s)\mu\_{\mathcal{S}}(\{s\})} \bar{g}\_2(t, s) (f\_1(s) - f\_2(s)) dg(s) \right\| \\ & \le \max\left(1, \frac{1}{\delta\_2}\right) \int\_0^T |\bar{g}\_2(t, s)| \cdot D(F\_1(s), F\_2(s)) dg(s). \end{aligned}$$

We are now evaluating the second term of the sum (7):

$$\begin{aligned} & \left\| \left( \mathfrak{a}\_1(t) - \mathfrak{a}\_2(t) \right) \int\_0^T \frac{\mathfrak{a}\_2(s)}{1 - b\_2(s)\mu\_\mathcal{S}(\{s\})} \mathfrak{g}\_2(t, s) f\_2(s) d\mathfrak{g}(s) \right\| \\ & \le \max\left( 1, \frac{1}{\delta\_2} \right) |\mathfrak{a}\_1(t) - \mathfrak{a}\_2(t)| \int\_0^T ||\mathfrak{g}\_2(t, s) f\_2(s)|| d\mathfrak{g}(s). \end{aligned} $$

As in Remark 4, we denote by

$$M\_1 = \sup\_{\left(\mathfrak{s}',\mathfrak{s}''\right) \in \left[0,T\right] \times \left[0,T\right]} \mathfrak{e}^{\int\_{s''}^{s''} \tilde{b}\_1(s) d\mathfrak{g}(s)}\_{\mathfrak{s}'} $$

respectively

$$M\_2 = \sup\_{(s',s'') \in [0,T] \times [0,T]} \mathfrak{e}\_{s''}^{\int\_{s''}^{s''} \tilde{\nu}\_2(s) d\S(s)}$$

and so,

$$|\mathcal{g}\_1(t,s)| \le \frac{\max(M\_1, M\_1^2)}{\left|\alpha\_1(T)e^{\int\_0^T \bar{b}\_1(r)d\xi(r)} - 1\right|} = \overline{M}\_1.$$

respectively

$$|\mathcal{g}\_2(t,s)| \le \frac{\max(M\_2, M\_2^2)}{\left| \varkappa\_2(T) e^{\int\_0^T \tilde{\nu}\_2(r) d\xi(r)} - 1 \right|} = \overline{M}\_2.$$

It follows that

$$\begin{split} & \| u\_1(t) - u\_2(t) \| \\ & \le \max\left(1, \frac{1}{\delta\_1}\right) \int\_0^T |\tilde{g}\_1(t,s) - \tilde{g}\_2(t,s)| \| |f\_1(s)| \| d\mathcal{g}(s) \\ & + \max\left(1, \frac{1}{\delta\_1}\right) \max\left(1, \frac{1}{\delta\_2}\right) \int\_0^T (1+m\_2) |\tilde{g}\_2(t,s)| \| f\_1(s) \| |a\_1(s) - a\_2(s)| d\mathcal{g}(s) \\ & + \max\left(1, \frac{1}{\delta\_1}\right) \max\left(1, \frac{1}{\delta\_2}\right) \mathcal{g}(T) \int\_0^T |\tilde{g}\_2(t,s)| |b\_1(s) - b\_2(s)| \| f\_1(s) \| d\mathcal{g}(s) \\ & + \max\left(1, \frac{1}{\delta\_2}\right) \int\_0^T |\tilde{g}\_2(t,s)| \cdot D(F\_1(s), F\_2(s)) d\mathcal{g}(s) \\ & + \max\left(1, \frac{1}{\delta\_2}\right) |a\_1(t) - a\_2(t)| \int\_0^T \| \tilde{g}\_2(t,s) f\_2(s) \| d\mathcal{g}(s) \end{split}$$

so

$$\begin{split} \|u\_1(t) - u\_2(t)\| &\le \max\left(1, \frac{1}{\delta\_1}\right) \int\_0^T |\underline{g}\_1(t,s) - \underline{g}\_2(t,s)| \overline{\Phi}(s) dg(s) \\ &+ \max\left(1, \frac{1}{\delta\_1}\right) \max\left(1, \frac{1}{\delta\_2}\right) (1 + m\_2) \overline{M}\_2 \int\_0^T \overline{\Phi}(s) |a\_1(s) - a\_2(s)| dg(s) \\ &+ \max\left(1, \frac{1}{\delta\_1}\right) \max\left(1, \frac{1}{\delta\_2}\right) \overline{M}\_2 \underline{g}(T) \int\_0^T |b\_1(s) - b\_2(s)| \overline{\Phi}(s) dg(s) \\ &+ \max\left(1, \frac{1}{\delta\_2}\right) \overline{M}\_2 \int\_0^T D(F\_1(s), F\_2(s)) dg(s) \\ &+ \max\left(1, \frac{1}{\delta\_2}\right) \overline{M}\_2 |a\_1(t) - a\_2(t)| \int\_0^T \overline{\Phi}(s) dg(s). \end{split} \tag{8}$$

We are now evaluating the difference *g*˜1(*t*,*s*) − *g*˜2(*t*,*s*). It can be seen that

$$= \begin{cases} \frac{\mathcal{J}\_{1}(t,\mathbf{s}) - \mathcal{J}\_{2}(t,\mathbf{s})}{a\_{1}(T)c\_{0}^{\dagger\_{0}}b\_{1}(r)\varrho(\mathbf{r}) - \mathbf{1}} e^{\int\_{0}^{T}\mathbf{f}\_{1}(r)\mathbf{d}\varrho(\mathbf{r}) - \int\_{s}^{t}\mathbf{b}\_{1}(r)\mathbf{d}\varrho(\mathbf{r})} - \frac{a\_{2}(T)}{a\_{2}(T)c\_{0}^{\dagger\_{0}}b\_{2}(r)\varrho(\mathbf{r}) - \int\_{s}^{T}\mathbf{b}\_{2}(r)\mathbf{d}\varrho(\mathbf{r}) - \int\_{s}^{t}\mathbf{b}\_{2}(r)\mathbf{d}\varrho(\mathbf{r})} \\\\ \frac{1}{a\_{1}(T)c\_{0}^{\dagger\_{0}}b\_{1}(r)\varrho(\mathbf{r}) - 1}e^{-\int\_{s}^{t}\mathbf{b}\_{1}(r)\mathbf{d}\varrho(\mathbf{r})} - \frac{if \ 0 \le s \le t \le T \\\\ \frac{1}{a\_{2}(T)c\_{0}^{\dagger\_{0}}b\_{2}(r)\varrho(\mathbf{r}) - 1}e^{-\int\_{s}^{t}\mathbf{b}\_{2}(r)\mathbf{d}\varrho(\mathbf{r})} \end{cases}$$
 
$$if \ 0 \le t < s \le T.$$

In the first case (0 ≤ *s* ≤ *t* ≤ *T*),


and, by Remark 5(i),

$$\begin{split} & \left| \tilde{g}\_1(t,s) - \tilde{g}\_2(t,s) \right| \\ & \leq \frac{1}{|(a\_1(T)e^{\int\_0^{\tilde{l}} b\_1(r)d\xi(r)} - 1)(a\_2(T)e^{\int\_0^{\tilde{l}} b\_2(r)d\xi(r)} - 1)|} \\ & \left| M\_1 M\_2 \max(M\_1, M\_2) \left| \int\_s^t \tilde{b}\_1(r)d\xi(r) - \int\_s^t \tilde{b}\_2(r)d\xi(r) \right| \\ & + \max(M\_1^2, M\_2^2) \left| \int\_0^T \tilde{b}\_1(r)d\xi(r) - \int\_s^t \tilde{b}\_1(r)d\xi(r) - \int\_0^T \tilde{b}\_2(r)d\xi(r) + \int\_s^t \tilde{b}\_2(r)d\xi(r) \right| \\ & + |a\_1(T) - a\_2(T)|M\_2^2| \\ & \leq \frac{1}{|(a\_1(T)e^{\int\_0^{\tilde{l}} b\_1(r)d\xi(r)} - 1)(a\_2(T)e^{\int\_0^{\tilde{l}} b\_2(r)d\xi(r)} - 1)|} \\ & \leq \frac{1}{|(a\_1(T)e^{\int\_0^{\tilde{l}} b\_1(r)d\xi(r)} - 1)(a\_2(T)e^{\int\_0^{\tilde{l}} b\_2(r)d\xi(r)} - 1)|} \\ & \left| \left( M\_1 M\_2 \max(M\_1, M\_2) + \max(M\_1^2, M\_2^2) \right) \int\_0^T |\tilde{b}\_1(s) - \tilde{b}\_2(s)|d\xi(s) + |a\_1(T) - a\_2(T)|M\_2^2| \right) \end{split}$$

Similarly, in the second case (0 ≤ *t* < *s* ≤ *T*) it can be proved that

$$\begin{aligned} &|\tilde{g}\_1(t,s) - \tilde{g}\_2(t,s)| \\ &\le \frac{1}{|(a\_1(T)e^{\int\_0^T \tilde{b}\_1(r)d\xi(r)} - 1)(a\_2(T)e^{\int\_0^T \tilde{b}\_2(r)d\xi(r)} - 1)|} \\ &\left[ (M\_1M\_2 + \max(M\_1, M\_2)) \int\_0^T |\tilde{b}\_1(s) - \tilde{b}\_2(s)|d\xi(s) + |a\_1(T) - a\_2(T)|M\_1M\_2| \right]. \end{aligned}$$

.

*Mathematics* **2020**, *8*, 2142

Denoting by

$$\tilde{M}\_1 = \max\left(M\_1 M\_2 \max(M\_1, M\_2) + \max(M\_1^2, M\_2^2), M\_1 M\_2 + \max(M\_1, M\_2)\right)$$

respectively

$$
\bar{M}\_2 = \max \left( M\_{2'}^2, M\_1 M\_2 \right),
$$

we may say that for every *s*, *t* ∈ [0, *T*],

$$|\varrho\_1(t, \mathbf{s}) - \varrho\_2(t, \mathbf{s})| \le \tilde{M}\_1 \int\_0^T |\tilde{b}\_1(\mathbf{s}) - \tilde{b}\_2(\mathbf{s})| d\mathbf{g}(\mathbf{s}) + \tilde{M}\_2 |a\_1(T) - a\_2(T)|.\tag{9}$$

We use next Remark 5(ii) and the fact that from (4), any *t* ∈ *Dg* satisfies

$$|1 - b\_1(t)\mu\_{\mathcal{S}}(\{t\})| > \delta\_1 \qquad \text{and} \qquad |1 - b\_2(t)\mu\_{\mathcal{S}}(\{t\})| > \delta\_2$$

to see that for each *t* ∈ *Dg*,

$$\begin{aligned} |b\_1(t) - b\_2(t)| &= \left| \frac{-\log|1 - b\_1(t)\mu\_{\mathcal{S}}(\{t\})| + \log|1 - b\_2(t)\mu\_{\mathcal{S}}(\{t\})|}{\mu\_{\mathcal{S}}(\{t\})} \right| \\ &\leq \max\left(\frac{1}{\delta\_1}, \frac{1}{\delta\_2}\right) |b\_1(t) - b\_2(t)|. \end{aligned}$$

It is immediate that for each *t* ∈ [0, *T*],

$$|\bar{b}\_1(t) - \bar{b}\_2(t)| \le \max\left(1, \frac{1}{\delta\_1}, \frac{1}{\delta\_2}\right) |b\_1(t) - b\_2(t)|.\tag{10}$$

Finally, exploiting (8), (9), (10) we obtain that for all *t* ∈ [0, *T*],

$$\begin{split} & \| u\_1(t) - u\_2(t) \| \\ & \le \max\left( 1, \frac{1}{\delta\_1} \right) \int\_0^T \overline{\Phi}(s) d\mathfrak{g}(s) \left( \tilde{M}\_1 \int\_0^T |\tilde{b}\_1(s) - \tilde{b}\_2(s)| d\mathfrak{g}(s) + \tilde{M}\_2 |a\_1(T) - a\_2(T)| \right) \\ & + \max\left( 1, \frac{1}{\delta\_1} \right) \max\left( 1, \frac{1}{\delta\_2} \right) (1 + m\_2) \overline{M}\_2 \int\_0^T \overline{\Phi}(s) |a\_1(s) - a\_2(s)| d\mathfrak{g}(s) \\ & + \max\left( 1, \frac{1}{\delta\_1} \right) \max\left( 1, \frac{1}{\delta\_2} \right) \overline{M}\_2 \mathfrak{g}(T) \int\_0^T |b\_1(s) - b\_2(s)| \overline{\Phi}(s) d\mathfrak{g}(s) \\ & + \max\left( 1, \frac{1}{\delta\_2} \right) \overline{M}\_2 \int\_0^T D(F\_1(s), F\_2(s)) d\mathfrak{g}(s) \\ & + \max\left( 1, \frac{1}{\delta\_2} \right) \overline{M}\_2 |a\_1(t) - a\_2(t)| \int\_0^T \overline{\Phi}(s) d\mathfrak{g}(s) \end{split}$$

so

$$\begin{split} & \| u\_1(t) - u\_2(t) \| \\ & \leq \overline{M}\_1 \int\_0^T \overline{\Phi}(s) d\mathfrak{g}(s) \cdot \max\left( 1, \frac{1}{\delta\_1} \right) \cdot \max\left( 1, \frac{1}{\delta\_1}, \frac{1}{\delta\_2} \right) \cdot \int\_0^T |b\_1(s) - b\_2(s)| d\mathfrak{g}(s) \\ & + \overline{M}\_2 \max\left( 1, \frac{1}{\delta\_1} \right) \int\_0^T \overline{\Phi}(s) d\mathfrak{g}(s) \cdot |a\_1(T) - a\_2(T)| \\ & + (1 + m\_2) \overline{M}\_2 \max\left( 1, \frac{1}{\delta\_1} \right) \max\left( 1, \frac{1}{\delta\_2} \right) \cdot \int\_0^T \overline{\Phi}(s) |a\_1(s) - a\_2(s)| d\mathfrak{g}(s) \\ & + \overline{M}\_2 \mathfrak{g}(T) \max\left( 1, \frac{1}{\delta\_1} \right) \max\left( 1, \frac{1}{\delta\_2} \right) \cdot \int\_0^T |b\_1(s) - b\_2(s)| \overline{\Phi}(s) d\mathfrak{g}(s) \\ & + \overline{M}\_2 \max\left( 1, \frac{1}{\delta\_2} \right) \cdot \int\_0^T D(F\_1(s), F\_2(s)) d\mathfrak{g}(s) \\ & + \overline{M}\_2 \max\left( 1, \frac{1}{\delta\_2} \right) \int\_0^T \overline{\Phi}(s) d\mathfrak{g}(s) \cdot |a\_1(t) - a\_2(t)|. \end{split}$$

Denoting thus by

$$\begin{split} \mathbb{C}\_{1} &= \ \breve{M}\_{1} \int\_{0}^{T} \overline{\theta}(s) d\underline{g}(s) \cdot \max\left(1, \frac{1}{\delta\_{1}}\right) \cdot \max\left(1, \frac{1}{\delta\_{1}}, \frac{1}{\delta\_{2}}\right), \\ \mathbb{C}\_{2} &= \ \breve{M}\_{2} \max\left(1, \frac{1}{\delta\_{1}}\right) \int\_{0}^{T} \overline{\theta}(s) d\underline{g}(s), \; \mathbb{C}\_{3} = (1 + m\_{2}) \overline{M}\_{2} \max\left(1, \frac{1}{\delta\_{1}}\right) \max\left(1, \frac{1}{\delta\_{2}}\right), \\ \mathbb{C}\_{4} &= \ \overline{M}\_{2} \underline{g}(T) \max\left(1, \frac{1}{\delta\_{1}}\right) \max\left(1, \frac{1}{\delta\_{2}}\right), \\ \mathbb{C}\_{5} &= \ \overline{M}\_{2} \max\left(1, \frac{1}{\delta\_{2}}\right), \; \mathbb{C}\_{6} = \overline{M}\_{2} \max\left(1, \frac{1}{\delta\_{2}}\right) \int\_{0}^{T} \overline{\theta}(s) d\underline{g}(s) \end{split}$$

one gets the required inequality.

Consider now

$$\begin{array}{rcl}\overline{\mathsf{C}}\_{1} &=& \overline{\mathsf{M}}\_{1} \int\_{0}^{T} \overline{\Phi}(s) d\boldsymbol{\xi}(s) \cdot \max\left(1, \frac{1}{\delta\_{1}}, \frac{1}{\delta\_{2}}\right)^{2}, \\\overline{\mathsf{C}}\_{4} &=& \max(\overline{\mathsf{M}}\_{2}, \overline{\mathsf{M}}\_{1}) \underline{\mathsf{J}}(T) \max\left(1, \frac{1}{\delta\_{1}}\right) \max\left(1, \frac{1}{\delta\_{2}}\right), \\\overline{\mathsf{C}}\_{5} &=& \max(\overline{\mathsf{M}}\_{2}, \overline{\mathsf{M}}\_{1}) \max\left(1, \frac{1}{\delta\_{1}}, \frac{1}{\delta\_{2}}\right). \end{array}$$

**Corollary 1.** *Under the assumptions of Theorem 5, if for every t* ∈ [0, *T*]

$$1 - b\_1(t)\mu\_{\mathcal{S}}(t) > 0 \qquad \text{and} \qquad 1 - b\_2(t)\mu\_{\mathcal{S}}(t) > 0$$

*then:*

$$(i)$$

$$D\_{\mathbb{C}}(\mathcal{S}\_1, \mathcal{S}\_2) \le \left(\overline{\mathsf{C}}\_1 \underline{\mathcal{G}}(T) + \overline{\mathsf{C}}\_4 \int\_0^T \overline{\mathfrak{g}}(s) d\underline{\mathcal{g}}(s)\right) \cdot \|b\_1 - b\_2\|\_{\mathbb{C}} + \overline{\mathsf{C}}\_5 \underline{\mathcal{G}}(T) \cdot D\_{\mathbb{C}}(F\_1, F\_2).$$

*(ii) if φ is bounded,*

$$D\_{L^1}(\mathcal{S}\_1, \mathcal{S}\_2) \le \left(\overline{\mathcal{C}}\_1 + \overline{\mathcal{C}}\_4 \sup\_{t \in [0, T]} \overline{\phi}(t)\right) \mathcal{g}(T) \|b\_1 - b\_2\|\_{L^1} + \overline{\mathcal{C}}\_5 \mathcal{g}(T) D\_{L^1}(F\_1, F\_2).$$

**Proof.** Under the additional hypothesis on *b*<sup>1</sup> and *b*2, it can be seen that *α*1(*t*) = *α*2(*t*) = 1 on the whole interval and so, Theorem 5 yields that for every *u*<sup>1</sup> ∈ S<sup>1</sup> one can find *u*<sup>2</sup> ∈ S<sup>2</sup> such that for all *t* ∈ [0, *T*],

$$\begin{split} \|u\_1(t) - u\_2(t)\| &\quad \le \quad \mathbb{C}\_1 \int\_0^T |b\_1(s) - b\_2(s)| d\mathfrak{g}(s) + \mathbb{C}\_4 \int\_0^T |b\_1(s) - b\_2(s)| \overline{\mathfrak{g}}(s) d\mathfrak{g}(s) \\ &\quad + \quad \mathbb{C}\_5 \int\_0^T D(F\_1(s), F\_2(s)) d\mathfrak{g}(s). \end{split} \tag{11}$$

(i) By taking the supremum in (11) over *t* ∈ [0, *T*],

$$\|\|\boldsymbol{\mu}\_{1}-\boldsymbol{\mu}\_{2}\|\|\_{\boldsymbol{\mathcal{C}}} \leq \left(\mathsf{C}\_{1}\mathsf{g}(\boldsymbol{T}) + \mathsf{C}\_{4}\int\_{0}^{T} \overline{\boldsymbol{\Phi}}(\mathsf{s})d\boldsymbol{\mathcal{g}}(\boldsymbol{\mathsf{s}})\right) \cdot \|\|\boldsymbol{b}\_{1}-\boldsymbol{b}\_{2}\|\|\_{\boldsymbol{\mathcal{C}}} + \mathsf{C}\_{5}\mathsf{g}(\boldsymbol{T}) \cdot \boldsymbol{D}\_{\boldsymbol{\mathcal{C}}}(\mathsf{F}\_{1},\mathsf{F}\_{2}) .$$

By the definition of the Pompeiu-excess, it follows that

$$\mathcal{e}\_{\mathbb{C}}(\mathcal{S}\_1, \mathcal{S}\_2) \le \left( \mathbb{C}\_1 \mathbb{g}(T) + \mathbb{C}\_4 \int\_0^T \overline{\mathfrak{g}}(s) d\mathfrak{g}(s) \right) \cdot \|b\_1 - b\_2\|\_{\mathbb{C}} + \mathbb{C}\_5 \mathfrak{g}(T) \cdot D\_{\mathbb{C}}(F\_1, F\_2),$$

and, by interchanging the roles of S<sup>1</sup> and S2, one obtains the announced estimation.

(ii) If *φ* is bounded, the inequality (11) implies that

$$||\mu\_1(t) - \mu\_2(t)|| \le \left(\mathbb{C}\_1 + \mathbb{C}\_4 \sup\_{t \in [0, T]} \overline{\phi}(t)\right) ||b\_1 - b\_2||\_{L^1} + \mathbb{C}\_5 D\_{L^1}(F\_1, F\_2)$$

By integrating it with respect to *g* on [0, *T*] we get

$$\|\|u\_1 - u\_2\|\|\_{L^1} \le \left(\mathbb{C}\_1 + \mathbb{C}\_4 \sup\_{t \in [0, T]} \overline{\mathfrak{g}}(t)\right) \mathcal{g}(T) \|\|b\_1 - b\_2\|\|\_{L^1} + \mathbb{C}\_5 \mathfrak{g}(T) D\_{L^1}(F\_1, F\_2)$$

whence

$$\mathfrak{e}\_{L^{1}}(\mathcal{S}\_{1},\mathcal{S}\_{2}) \leq \left(\mathbb{C}\_{1} + \mathbb{C}\_{4} \sup\_{t \in [0,T]} \overline{\mathfrak{g}}(t)\right) \mathfrak{g}(T) \|b\_{1} - b\_{2}\|\_{L^{1}} + \mathbb{C}\_{5} \mathfrak{g}(T) D\_{L^{1}}(F\_{1}, F\_{2}).$$

and the inequality comes from interchanging the roles of S<sup>1</sup> and S2.

**Author Contributions:** Both authors have equally contributed to this work. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research received no external funding.

**Acknowledgments:** The authors are grateful to the three referees for their comments and suggestions.

**Conflicts of Interest:** The authors declare no conflict of interest.

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