1 Introduction

The theory of differential equations driven by measures has been continuously growing over the last decade (e.g. [2, 4, 7,8,9, 21, 27]) since it offers a tool to study in a unified way several classical problems: first-order differential equations (in the case when the driving measure is absolutely continuous with respect to the Lebesgue measure), impulsive differential problems (when we take into consideration a measure which can be written as a sum of Lebesgue measure with a discrete measure) with no limitations on the impulse moments, dynamic equations on time scales (see [4, 8, 9]) and generalized differential equations (e.g. [16, 21, 28, 29]).

On the other hand, an equivalent formulation in terms of a notion of (Stieltjes) derivative with respect to a nondecreasing function is available (c.f. [22], see also [19]). This derivative, considered in [22] (even if the idea is not really new in literature, c.f. [31]) has found recent interesting applications in biology, population dynamics or chemistry (see [12, 13] or [23]).

At the same time, it is well known that differential problems with periodic boundary conditions have wide applicability in various areas of science.

Relying on these considerations, we focus on first order nonlinear periodic boundary value problems of the form (1):

$$\begin{aligned} \left\{ \begin{array}{l} u'_g(t)+b(t) u(t) =f(t,u(t)),\; t\in [0,T]\\ u(0)=u(T) \end{array} \right. \end{aligned}$$

involving the Stieltjes derivative with respect to a function \(g:[0,T]\rightarrow {\mathbb {R}}\) left-continuous and nondecreasing.

The maps \(b:[0,T]\rightarrow {\mathbb {R}}\) and \(f:[0,T]\times {\mathbb {R}}\rightarrow {\mathbb {R}}\) are supposed to be continuous at the continuity points of g.

In two steps (first, for the linear and then, applying Schaeffer’s fixed point theorem, for the general case), we prove that the specified problem possesses solutions and provide a useful integral characterization via a Green function.

We essentially use an idea borrowed from [12] which consists in modifying in an appropriate manner the involved functions at the discontinuity points of g to adjust the g-derivative at these points.

This is, as far as the authors know, the first existence result for periodic boundary value problems in such a general framework (involving Stieltjes derivatives).

In the same spirit as described above for measure differential problems, in particular when g is the sum of an absolutely continuous function with a sum of Heaviside functions, we cover the framework of periodic impulsive differential problems, therefore we can deduce new results for impulsive equations (studied, under different assumptions, e.g. in [3, 17, 33]). Moreover, the number of impulses can be not only finite, but countable and it can contain accumulation points (known as Zeno points in the theory of hybrid systems, e.g. [18]). Our outcome is also related to some existence results for periodic dynamic equations on time scales (see [5, 14] or [32]). We finally note that the same problem has also been studied in the framework of fractional differential equations (e.g. [1]) or in that of functional impulsive differential equations (as in [15]).

2 Notations and auxiliary results

A function \(u : [0,T] \rightarrow {\mathbb {R}}\) is said to be regulated ([10]) if there exist the right and left limits \(u(t+)\) and \(u(s-)\) at every points \(t \in [0,T)\) and \(s \in (0,T]\). The set of discontinuity points of a regulated function is at most countable ([25]) and any function of bounded variation (and also any continuous function) is regulated. Regulated functions are bounded and the space of these functions is a Banach space when endowed with the norm \(\Vert u\Vert _C=\sup _{t\in [0,T]}|u(t)|\).

A family \({\mathcal {A}}\) of regulated real-valued functions on [0, T] is called equiregulated if for every \({\overline{t}}\in [0,T]\) and every \(\varepsilon >0\) one can find \(\delta >0\) such that for any \(u\in {\mathcal {A}}\)

$$\begin{aligned} |u(t)-u({\overline{t}}-)|<\varepsilon ,\; \mathrm{for \; every}\; t\in ({\overline{t}}-\delta ,{\overline{t}}) \end{aligned}$$

and

$$\begin{aligned} |u(t)-u({\overline{t}}+)|<\varepsilon ,\; \mathrm{for \; every}\; t\in ({\overline{t}}, {\overline{t}}+\delta ). \end{aligned}$$

The notion of equiregulatedness is related to compactness in the space of regulated functions.

Lemma 1

([10, Corollary 2.4]) A set of regulated functions is relatively compact if and only if it is equiregulated and pointwise bounded.

This is a consequence of another interesting result.

Lemma 2

[10] Let \((f_n)_n\) be an equiregulated sequence of functions which converges pointwise to a function f. Then \((f_n)_n\) converges uniformly to f.

The following remark will be useful later:

Remark 3

Let \({\mathcal {A}}\) be a set of regulated functions. If there exists a regulated function \(\chi :[0,T]\rightarrow {\mathbb {R}}\) such that for every \(u\in {\mathcal {A}}\),

$$\begin{aligned} |u(t)-u(t')|\le |\chi (t)-\chi (t')|,\quad \forall \; 0\le t<t'\le T, \end{aligned}$$

then \({\mathcal {A}}\) is equiregulated.

Let \(g:[0,T]\rightarrow {\mathbb {R}}\) be a nondecreasing left-continuous function. In the whole paper, we deal with the Kurzweil–Stieltjes integral; to be allowed to use its properties, we recall below the basic facts concerning this type of integral.

Definition 4

[16, 20, 25, 26, 28] or [30] A function \(f:[0,T]\rightarrow {\mathbb {R}}\) is said to be Kurzweil–Stieltjes integrable with respect to \(g:[0,T]\rightarrow {\mathbb {R}}\) on [0, T] (or KS-integrable) if there exists \(\int _0^{T}f(s)\mathrm{{d}}g(s)\in {\mathbb {R}}\) such that, for every \(\varepsilon > 0\), there is a positive function \(\delta _{\varepsilon }\) on [0, T] with

$$\begin{aligned} \left| \sum _{i=1}^p f(\xi _i)(g(t_i)-g(t_{i-1}))-\int _{0}^{T} f(s)\mathrm{{d}}g(s) \right| <\varepsilon \end{aligned}$$

for every \(\delta _{\varepsilon }\)-fine partition \(\{([t_{i-1},t_i],\xi _i): \ i=1, \ldots ,p\}\) of [0, T].

A partition \(\{([t_{i-1},t_i],\xi _i): \ i=1, \ldots ,p\}\) is \(\delta _{\varepsilon }\)-fine if for all \(i=1, \ldots ,p\), \([t_{i-1},t_i] \subset \left]\xi _i-\delta _{\varepsilon }(\xi _i),\xi _i+\delta _{\varepsilon }(\xi _i) \right[\). The KS-integrability is preserved on all sub-intervals of [0, T].

In the particular case, where \(g(t)=t\) for every \(t\in [0,T]\), one finds the Henstock–Kurzweil integral (see [11]).

It is known that regulated functions are KS-integrable with respect to bounded variation functions and vice versa (see [30]). The properties of the primitive contained in the proposition below are important in what follows.

Proposition 5

([30, Proposition 2.3.16]) Let \(f:[0,T]\rightarrow {\mathbb {R}}\) be KS-integrable w.r.t. \(g:[0,T]\rightarrow {\mathbb {R}}\). If g is regulated, then so is the primitive \(F:[0,T]\rightarrow {{\mathbb {R}}}\), \(F(t)=\int _0^t f(s) \mathrm{{d}}g(s)\) and for every \(t \in [0,T)\) and \(s\in (0,T]\),

$$\begin{aligned} F(t+)-F(t){=}f(t)\left[ g(t+)-g(t) \right] \quad and \quad F(s)-F(s-){=}f(s)\left[ g(s)-g(s-) \right] . \end{aligned}$$

It follows that F is left-continuous, respectively, right-continuous at the points where g has the same property.

Moreover, when g is of bounded variation and f is bounded, F is of bounded variation as well.

Note that the Lebesgue–Stieltjes integrability of a function f (i.e., the abstract Lebesgue integrability w.r.t. the Stieltjes measure \(\mu _g\) generated by g, see [24, Example 6.14]) implies the Kurzweil–Stieltjes integrability. In the framework of a left-continuous nondecreasing function g, as a consequence of [20, Theorem 6.11.3] (see also [25, Theorem 8.1]), for \(t\in [0,T]\),

$$\begin{aligned} \int _0^t f(s)\mathrm{{d}}g(s)=\int _{[0,t]} f(s)\mathrm{{d}}\mu _g(s)-f(t)(g(t+)-g(t))=\int _{[0,t)} f(s)\mathrm{{d}}\mu _g(s). \end{aligned}$$

In [22], a notion of differentiability related to Stieltjes type integrals was introduced (following an idea in [31]).

Definition 6

Let \(g:[0,T]\rightarrow {\mathbb {R}}\) be a nondecreasing left-continuous function. The derivative with respect to g (or the g-derivative) of a function \(f:[0,T]\rightarrow {\mathbb {R}}\) at a point \({\overline{t}}\in [0,T]\) is given by

$$\begin{aligned} f'_g({\overline{t}})= & {} \lim _{t\rightarrow {\overline{t}}}\frac{f(t)-f({\overline{t}})}{g(t)-g({\overline{t}})}\quad \hbox { if} \ g \ \hbox {is continuous at } {\overline{t}},\\ f'_g({\overline{t}})= & {} \lim _{t\rightarrow {\overline{t}}+}\frac{f(t)-f({\overline{t}})}{g(t)-g({\overline{t}})}\quad \hbox { if } g \hbox { is discontinuous at } {\overline{t}}, \end{aligned}$$

provided the limit exists.

Define the following sets:

$$\begin{aligned} C_g= & {} \{t\in [0,T]: g \text{ is } \text{ constant } \text{ on } (t-\varepsilon ,t+\varepsilon ) \text{ for } \text{ some } \varepsilon>0\}\\ D_g= & {} \{t\in [0,T]: g(t+)-g(t)>0\}. \end{aligned}$$

It is worth mentioning that \(D_g\) is the set of atoms of the measure \(\mu _g\) and if \(t\in D_g\), the g-derivative \(f'_g(t)\) exists if and only if the sided limit \(f(t+)\) exists, and in this case

$$\begin{aligned} f'_g(t)=\frac{f(t+)-f(t)}{g(t+)-g(t)}. \end{aligned}$$

Note that Definition 6 has no meaning in \(C_g\). Anyway (see [22]), this set is not significant, in the sense that \(\mu _g(C_g)=0\).

What’s more, if \(C_g=\bigcup _{n\in {\mathbb {N}}}(u_n,v_n)\) is a disjoint decomposition of \(C_g\) and

$$\begin{aligned} N_g=\{u_n,v_n:n\in {\mathbb {N}} \}{\setminus } D_g, \end{aligned}$$

then Definition 6 has no meaning in \(N_g\) neither, but again \(\mu _g(N_g)=0\).

This type of derivative has already been used in solving various problems where abrupt changes (corresponding to discontinuity points of g) and dead times (corresponding to intervals where g is constant) are present, such as [12, 13, 23] or [27].

Fundamental Theorems of Calculus are essential when taking into account the connection between integrals and derivatives. Such a result for Kurzweil–Stieltjes integrals can be found in [22, Theorem 6.5]:

Theorem 7

Let \(g:[0,T]\rightarrow {\mathbb {R}}\) be a nondecreasing left-continuous function. If \(f:[0,T]\rightarrow {\mathbb {R}}\) is KS-integrable with respect to g and

$$\begin{aligned} F(t)=\int _0^t f(s)\,\mathrm{{d}}g(s),\quad t\in [0,T], \end{aligned}$$

then F is g-differentiable \(\mu _g\)-a.e. on [0, T] and \(F_g'=f\).

It would be useful to know more precisely at which points the differentiability w.r.t. g is achieved.

Proposition 8

Let \(f:[0,T]\rightarrow {\mathbb {R}}\) be KS-integrable w.r.t. g and let \(F(t)=\int _0^t f(s)\,\mathrm{{d}}g(s)\) be its primitive. Then F is g-differentiable (with the g-derivative equal to f(t)) at any point \(t\in [0,T]{\setminus } \left( C_g\cup N_g \right) \) where f is continuous.

Proof

Let \({\overline{t}}\in [0,T]{\setminus } \left( C_g\cup N_g \right) \) where f is continuous.

If \({\overline{t}}\notin D_g\), then

$$\begin{aligned} F'_g({\overline{t}})=\lim _{t\rightarrow {\overline{t}}}\frac{F(t)-F({\overline{t}})}{g(t)-g({\overline{t}})}. \end{aligned}$$

We can write

$$\begin{aligned} \lim _{t\rightarrow {\overline{t}},t>{\overline{t}}}\min _{\tau \in [{\overline{t}},t)} f(\tau )= & {} \lim _{t\rightarrow {\overline{t}},t>{\overline{t}}}\frac{\min _{\tau \in [{\overline{t}},t)} f(\tau )(g(t)-g({\overline{t}}))}{g(t)-g({\overline{t}})} \\\le & {} \lim _{t\rightarrow {\overline{t}},t>{\overline{t}}}\frac{F(t)-F({\overline{t}})}{g(t)-g({\overline{t}})} \end{aligned}$$

and

$$\begin{aligned} \lim _{t\rightarrow {\overline{t}},t>{\overline{t}}}\frac{F(t)-F({\overline{t}})}{g(t)-g({\overline{t}})}\le & {} \lim _{t\rightarrow {\overline{t}},t>{\overline{t}}}\frac{\max _{\tau \in [{\overline{t}},t)}f(\tau )(g(t)-g({\overline{t}}))}{g(t)-g({\overline{t}})}\\= & {} \lim _{t\rightarrow {\overline{t}},t>{\overline{t}}}\max _{\tau \in [{\overline{t}},t)} f(\tau ), \end{aligned}$$

whence, by the continuity of f at \({\overline{t}}\),

$$\begin{aligned} \lim _{t\rightarrow {\overline{t}},t>{\overline{t}}}\frac{F(t)-F({\overline{t}})}{g(t)-g({\overline{t}})}=f({\overline{t}}). \end{aligned}$$

Similarly, it can be proved that

$$\begin{aligned} \lim _{t\rightarrow {\overline{t}},t<{\overline{t}}}\frac{F(t)-F({\overline{t}})}{g(t)-g({\overline{t}})}=f({\overline{t}}), \end{aligned}$$

therefore

$$\begin{aligned} F'_g({\overline{t}})=f({\overline{t}}). \end{aligned}$$

On the other hand, if \({\overline{t}}\in D_g\), then

$$\begin{aligned} F'_g({\overline{t}})=\lim _{t\rightarrow {\overline{t}}, t>{\overline{t}}}\frac{F(t)-F({\overline{t}})}{g(t)-g({\overline{t}})} \end{aligned}$$

and, as before, it equals \(f({\overline{t}})\). \(\square \)

In the same line, we could define partial g-derivatives.

Definition 9

The partial g-derivative w.r.t. the variable t of a function \(f:[0,T]\times [0,T] \rightarrow {\mathbb {R}}\) at a point \(({\overline{t}},{\overline{s}})\in [0,T]\times [0,T]\) is given by

$$\begin{aligned} \frac{\partial f}{\partial _g t}({\overline{t}},{\overline{s}})= & {} \lim _{t\rightarrow {\overline{t}}}\frac{f(t,{\overline{s}})-f({\overline{t}},{\overline{s}})}{g(t)-g({\overline{t}})}\quad \hbox { if } g \hbox { is continuous at } {\overline{t}},\\ \frac{\partial f}{\partial _g t}({\overline{t}},{\overline{s}})= & {} \lim _{t\rightarrow {\overline{t}}+}\frac{f(t,{\overline{s}})-f({\overline{t}},{\overline{s}})}{g(t)-g({\overline{t}})}\quad \hbox { if } g \hbox { is discontinuous at } {\overline{t}} \end{aligned}$$

provided the limit exists.

Let us recall the chain rules for g-derivatives at the continuity points of g (borrowed from [22]):

Lemma 10

([22, Theorem 2.3]) Let f be a real function defined on a neighborhood of \(t\in {\mathbb {R}}{\setminus } D_g\) and h be another function defined on a neighborhood of f(t). Then

  1. 1.

    If \(h'(f(t))\) and \(f'_g(t)\) exist, then

    $$\begin{aligned} (h\circ f)'_g(t)=h'(f(t))\cdot f'_g(t). \end{aligned}$$
  2. 2.

    If \(h'_g(f(t))\), \(g'(f(t))\) and \(f'_g(t)\) exist, then

    $$\begin{aligned} (h\circ f)'_g(t)=h'_g(f(t))\cdot g'(f(t))\cdot f'_g(t). \end{aligned}$$

We deduce several auxiliary (technical) results concerning the partial derivative of a function with respect to another function:

Lemma 11

i) Let \(h:{\mathbb {R}}\rightarrow {\mathbb {R}}\) be differentiable and \(f:{\mathbb {R}}^2\rightarrow {\mathbb {R}}\) be partially g-differentiable w.r.t. t at some point (t, s) with \(t\in {\mathbb {R}}{\setminus } D_g\). Then \(h\circ f\) is partially g-differentiable w.r.t. t and

$$\begin{aligned} \frac{\partial (h\circ f)}{\partial _g t}(t,s)=h'(f(t,s))\cdot \frac{\partial f}{\partial _g t}(t,s). \end{aligned}$$

ii) Let \(h:{\mathbb {R}}^2\rightarrow {\mathbb {R}}\) be partially g-differentiable w.r.t. both arguments u, v and \(f,j:{\mathbb {R}}\rightarrow {\mathbb {R}}\) be g-differentiable at some point \(t\in {\mathbb {R}}{\setminus } D_g\) such that g is differentiable at f(t) and j(t). Then \(H(t)=h(f(t),j(t))\) is g-differentiable at t and

$$\begin{aligned} H'_g(t)=\frac{\partial h}{\partial _g u}(f(t),j(t))\cdot g'(f(t))\cdot f'_g(t)+\frac{\partial h}{\partial _g v}(f(t),j(t))\cdot g'(j(t))\cdot j'_g(t). \end{aligned}$$

Consequently, we can get the following version of Leibniz’s rule for g-differentiation of Stieltjes integrals depending on a parameter:

Lemma 12

Let \(g:[0,T]\rightarrow {\mathbb {R}}\) be nondecreasing and left-continuous and \(\omega :[0,T]\times [0,T]\rightarrow {\mathbb {R}}\) partially g-differentiable w.r.t. t be such that \(s\rightarrow \omega (t,s)\) is Kurzweil–Stieltjes integrable w.r.t. g. Then

$$\begin{aligned} J(t)=\int _{\phi (t)}^{\psi (t)} \omega (t,s)\mathrm{{d}}g(s),\; t\in [0,T] \end{aligned}$$

is g-differentiable at each point \(t\in [0,T]{\setminus } D_g\), where

  • t, \(\phi \) and \(\psi \) are g-differentiable,

  • g is differentiable at t, \(\phi (t)\) and \(\psi (t)\),

  • \(s\rightarrow \omega (t,s)\) is continuous at \(\phi (t)\) and \(\psi (t)\).

Besides, \(J'_g(t)=\)

$$\begin{aligned} \int _{\phi (t)}^{\psi (t)} \frac{\partial \omega }{\partial _g t}(t,s)\mathrm{{d}}g(s)+\omega (t,\psi (t))\cdot g'(\psi (t))\cdot \psi '_g(t)-\omega (t,\phi (t))\cdot g'(\phi (t))\cdot \phi '_g(t). \end{aligned}$$

Proof

Defining

$$\begin{aligned} {\tilde{J}}(u,v,t)=\int _u^v \omega (t,s)\mathrm{{d}}g(s) \end{aligned}$$

one notices that

$$\begin{aligned} J(t)={\tilde{J}}(\phi (t),\psi (t),t). \end{aligned}$$

By Lemma 11,

$$\begin{aligned} J'_g(t)= \frac{\partial {\tilde{J}}}{\partial _g u}\cdot g'(u)\cdot u'_g+\frac{\partial {\tilde{J}}}{\partial _g v}\cdot g'(v)\cdot v'_g+\frac{\partial {\tilde{J}}}{\partial _g t}\cdot g'(t)\cdot t'_g. \end{aligned}$$

Applying Proposition 8, whenever \(s\rightarrow \omega (t,s)\) is continuous at u (resp. at v), one gets

$$\begin{aligned} \frac{\partial {\tilde{J}}}{\partial _g u}=-\omega (t,u),\quad \frac{\partial {\tilde{J}}}{\partial _g v}=\omega (t,v), \end{aligned}$$

whence

$$\begin{aligned} J'_g(t)= & {} -\omega (t,\phi (t))\cdot g'(\phi (t))\cdot \phi '_g(t)+\omega (t,\psi (t))\cdot g'(\psi (t))\cdot \psi '_g(t)\\&+\int _{\phi (t)}^{\psi (t)} \frac{\partial \omega }{\partial _g t}(t,s)\mathrm{{d}}g(s)\cdot g'(t)\cdot t'_g. \end{aligned}$$

At the same time, by definition, at a continuity point of g,

$$\begin{aligned} t'_g=\lim _{t'\rightarrow t}\frac{t'-t}{g(t')-g(t)}, \end{aligned}$$

while

$$\begin{aligned} g'(t)=\lim _{t'\rightarrow t}\frac{g(t')-g(t)}{t'-t}, \end{aligned}$$

and so,

$$\begin{aligned} g'(t)\cdot t'_g=1. \end{aligned}$$

\(\square \)

A formula for the g-derivative of a product of two functions will also be necessary:

Lemma 13

Let \(f,h:[0,T]\rightarrow {\mathbb {R}}\) be g-differentiable at \({\overline{t}}\in D_g\). Then \(f\cdot h\) is also g-differentiable at \({\overline{t}}\) and

$$\begin{aligned} (f\cdot h)'_g({\overline{t}})=f'_g({\overline{t}})h({\overline{t}}+)+f({\overline{t}})h'_g({\overline{t}}). \end{aligned}$$

Proof

Since \({\overline{t}}\in D_g\),

$$\begin{aligned} (f\cdot h)'_g({\overline{t}})= & {} \lim _{t\rightarrow {\overline{t}},t>{\overline{t}}}\frac{(f\cdot h)(t)-(f\cdot h)({\overline{t}})}{g(t)-g({\overline{t}})}\\= & {} \lim _{t\rightarrow {\overline{t}},t>{\overline{t}}}\frac{(f(t)-f({\overline{t}}))h(t)+f({\overline{t}})(h(t)-h({\overline{t}}))}{g(t)-g({\overline{t}})}\\= & {} f'_g({\overline{t}})h({\overline{t}}+)+f({\overline{t}})h'_g({\overline{t}}). \end{aligned}$$

\(\square \)

3 Main results

Our goal is to provide existence of solutions for the boundary value problem with nonlinear right-hand side:

$$\begin{aligned} \left\{ \begin{array}{l} u'_g(t)+b(t) u(t) =f(t,u(t)),\; t\in [0,T]\\ u(0)=u(T). \end{array} \right. \end{aligned}$$

involving the g-derivative.

We impose the nonresonance condition

$$\begin{aligned} 1-b(t)\mu _g(\{t\})\ne 0,\quad \mathrm{for \; every\;} t\in [0,T]. \end{aligned}$$

Definition 14

A left-continuous, regulated function \(u:[0,T]\rightarrow {\mathbb {R}}\) is a solution of our problem if it is constant on any interval where g is constant, g-differentiable \(\mu _g\)-a.e. and it verifies the equality

$$\begin{aligned} u'_g(t)+b(t) u(t) =f(t,u(t)),\; \mu _g-a.e. \; on\; [0,T] \end{aligned}$$

and the condition

$$\begin{aligned} u(0)=u(T). \end{aligned}$$

3.1 Existence result for the linear problem

First, we study the linear periodic Stieltjes differential equation

$$\begin{aligned} \left\{ \begin{array}{l} u'_g(t)+b(t) u(t)=f(t),\; t\in [0,T]\\ u(0)=u(T) \end{array} \right. \end{aligned}$$
(2)

under the assumptions that \(b:[0,T]\rightarrow {\mathbb {R}}\) and \(f:[0,T]\rightarrow {\mathbb {R}}\) are continuous on \([0,T]{\setminus } D_g\).

To solve the problem (2), we must take into account the sign of \(1-b(t)\mu _g(\{t\})\).

As in [12], if \(b\in L^1_g([0,T]\), the set

$$\begin{aligned} D_g^-=\{t\in D_g: 1-b(t)\mu _g(\{t\})<0\} \end{aligned}$$

is finite since

$$\begin{aligned} \infty>\Vert b\Vert _{L^1_g}>\sum _{t\in D_g^-} b(t)\mu _g(\{t\})>\sum _{t\in D_g^-} 1. \end{aligned}$$

Denote by \(t_1< \cdots <t_k\) its elements and, for simplicity, let \(t_0=0\) and \(t_{k+1}=T\).

Let

$$\begin{aligned} \alpha (t)= \left\{ \begin{array}{l} 1,\; if \;0\le t\le t_1\\ (-1)^i, \; if \; t_i< t\le t_{i+1},\; i=1, \ldots ,k. \end{array} \right. \end{aligned}$$

To simplify the proof of the existence theorem, we will use the following lemma.

Lemma 15

Let \({\overline{t}}\in D_g\) and \(c,h:[0,T]\rightarrow {\mathbb {R}}\) be KS-integrable w.r.t. g. Then

  1. i)

    The function \(t\rightarrow e^{\int _0^t c(r)\mathrm{{d}}g(r)}\) is g-differentiable at \({\overline{t}}\) and

    $$\begin{aligned} \left( e^{\int _0^t c(r)\mathrm{{d}}g(r)} \right) '_g({\overline{t}})=e^{\int _0^{{\overline{t}}} c(r)\mathrm{{d}}g(r)}\cdot \frac{e^{c({\overline{t}})\mu _g(\{{\overline{t}}\})}-1}{\mu _g(\{{\overline{t}}\})}. \end{aligned}$$
  2. ii)

    1. The function

    $$\begin{aligned} F(t)=\frac{1}{\alpha (t)}\int _0^t \alpha (s)e^{-\int _s^t c(r)\mathrm{{d}}g(r)}h(s)\mathrm{{d}}g(s) \end{aligned}$$

    is g-differentiable at \({\overline{t}}\) and

    $$\begin{aligned} F'_g({\overline{t}})=\frac{e^{-c({\overline{t}})\mu _g(\{{\overline{t}}\})}-1}{\mu _g(\{{\overline{t}}\})}\cdot F({\overline{t}})+e^{-c({\overline{t}})\mu _g(\{{\overline{t}}\})}h({\overline{t}}) \end{aligned}$$

    if \({\overline{t}}\in D_g{\setminus } \{t_1, \ldots ,t_k\}\) and

    $$\begin{aligned} F'_g({\overline{t}})=\frac{-e^{-c({\overline{t}})\mu _g(\{{\overline{t}}\})}-1}{\mu _g(\{{\overline{t}}\})}\cdot F({\overline{t}})-e^{-c({\overline{t}})\mu _g(\{{\overline{t}}\})}h({\overline{t}}) \end{aligned}$$

    if \({\overline{t}}=t_i,i=1, \ldots ,k\).

    2. The function

    $$\begin{aligned} G(t)=\frac{1}{\alpha (t)}\int _t^T \alpha (s) e^{-\int _s^t c(r)\mathrm{{d}}g(r)}h(s)\mathrm{{d}}g(s) \end{aligned}$$

    is g-differentiable at \({\overline{t}}\) and

    $$\begin{aligned} G'_g({\overline{t}})=\frac{e^{-c({\overline{t}})\mu _g(\{{\overline{t}}\})}-1}{\mu _g(\{{\overline{t}}\})}\cdot G({\overline{t}})-e^{-c({\overline{t}})\mu _g(\{{\overline{t}}\})}h({\overline{t}}) \end{aligned}$$

    if \({\overline{t}}\in D_g{\setminus } \{t_1, \ldots ,t_k\}\) and

    $$\begin{aligned} G'_g({\overline{t}})=\frac{-e^{-c({\overline{t}})\mu _g(\{{\overline{t}}\})}-1}{\mu _g(\{{\overline{t}}\})}\cdot G({\overline{t}})+e^{-c({\overline{t}})\mu _g(\{{\overline{t}}\})}h({\overline{t}}) \end{aligned}$$

    if \({\overline{t}}=t_i,i=1, \ldots ,k\).

Proof

i) By the definition of the g-derivative at a discontinuity point and using Proposition 5,

$$\begin{aligned} \left( e^{\int _0^t c(r)\mathrm{{d}}g(r)} \right) '_g({\overline{t}})= & {} \frac{e^{\int _0^{{\overline{t}}+} c(r)\mathrm{{d}}g(r)}-e^{\int _0^{{\overline{t}}} c(r)\mathrm{{d}}g(r)}}{\mu _g(\{{\overline{t}}\})}\\= & {} \frac{e^{\int _0^{{\overline{t}}} c(r)\mathrm{{d}}g(r)}\left( e^{\int _{{\overline{t}}}^{{\overline{t}}+} c(r)\mathrm{{d}}g(r)}-1 \right) }{\mu _g(\{{\overline{t}}\})}\\= & {} e^{\int _0^{{\overline{t}}} c(r)\mathrm{{d}}g(r)}\cdot \frac{e^{c({\overline{t}})\mu _g(\{{\overline{t}}\})}-1}{\mu _g(\{{\overline{t}}\})}. \end{aligned}$$

ii). 1. By definition, at any point \({\overline{t}}\in D_g{\setminus } \{t_1, \ldots ,t_k\}\),

$$\begin{aligned} F'_g({\overline{t}})= & {} \frac{F({\overline{t}}+)-F({\overline{t}})}{\mu _g(\{{\overline{t}}\})}\\= & {} \frac{1}{\mu _g(\{{\overline{t}}\})}\left[ \frac{1}{\alpha ({\overline{t}}+)}\int _0^{{\overline{t}}+} \alpha (s)e^{-\int _s^{{\overline{t}}+} c(r)\mathrm{{d}}g(r)}h(s)\mathrm{{d}}g(s)\right. \\&- \left. \frac{1}{\alpha ({\overline{t}})}\int _0^{{\overline{t}}} \alpha (s)e^{-\int _s^{{\overline{t}}} c(r)\mathrm{{d}}g(r)}h(s)\mathrm{{d}}g(s)\right] \\= & {} \frac{1}{\mu _g(\{{\overline{t}}\})}\left[ \frac{1}{\alpha ({\overline{t}})}\int _0^{{\overline{t}}} \alpha (s)e^{-\int _s^{{\overline{t}}} c(r)\mathrm{{d}}g(r)}e^{-\int _{{\overline{t}}}^{{\overline{t}}+} c(r)\mathrm{{d}}g(r)}h(s)\mathrm{{d}}g(s)\right. \\&+ \frac{1}{\alpha ({\overline{t}})}\int _{{\overline{t}}}^{{\overline{t}}+} \alpha (s)e^{-\int _s^{{\overline{t}}+} c(r)\mathrm{{d}}g(r)}h(s)\mathrm{{d}}g(s)\\&- \left. \frac{1}{\alpha ({\overline{t}})}\int _0^{{\overline{t}}} \alpha (s)e^{-\int _s^{{\overline{t}}} c(r)\mathrm{{d}}g(r)}h(s)\mathrm{{d}}g(s)\right] . \end{aligned}$$

Using Proposition 5,

$$\begin{aligned} F'_g({\overline{t}})= & {} \frac{1}{\alpha ({\overline{t}})}\frac{e^{-c({\overline{t}})\mu _g(\{{\overline{t}}\})}-1}{\mu _g(\{{\overline{t}}\})}\cdot \int _0^{{\overline{t}}} \alpha (s)e^{-\int _s^{{\overline{t}}} c(r)\mathrm{{d}}g(r)}h(s)\mathrm{{d}}g(s)\\&+ \frac{1}{\alpha ({\overline{t}})}\frac{\alpha ({\overline{t}})e^{-c({\overline{t}})\mu _g(\{{\overline{t}}\})}h({\overline{t}})\mu _g(\{{\overline{t}}\})}{\mu _g(\{{\overline{t}}\})}\\= & {} \frac{e^{-c({\overline{t}})\mu _g(\{{\overline{t}}\})}-1}{\mu _g(\{{\overline{t}}\})}\cdot F({\overline{t}})+e^{-c({\overline{t}})\mu _g(\{{\overline{t}}\})}h({\overline{t}}). \end{aligned}$$

On the other hand, if \({\overline{t}}=t_i,i\in \{1, \ldots ,k\}\), then \(\alpha ({\overline{t}}+)=-\alpha ({\overline{t}})\), so

$$\begin{aligned} F'_g({\overline{t}})= & {} \frac{F({\overline{t}}+)-F({\overline{t}})}{\mu _g(\{{\overline{t}}\})}\\= & {} \frac{-e^{-c({\overline{t}})\mu _g(\{{\overline{t}}\})}-1}{\mu _g(\{{\overline{t}}\})}\cdot F({\overline{t}})-e^{-c({\overline{t}})\mu _g(\{{\overline{t}}\})}h({\overline{t}}). \end{aligned}$$

The g-derivative of G can be computed in a similar way. \(\square \)

Consider, following an idea used in [12] for initial value problems driven by measures,

$$\begin{aligned} {\tilde{b}}(t)= \left\{ \begin{array}{l} b(t),\; if \; t\in [0,T]{\setminus } D_g\\ \frac{-\mathrm{{log}}|1-b(t)\mu _g(\{t\})|}{\mu _g(\{t\})}, \; \mathrm{{if}} \; t\in D_g \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} {\tilde{f}}(t)=\frac{f(t)}{1-b(t)\mu _g(\{t\})}. \end{aligned}$$

Notice that \({\tilde{f}}(t)=f(t)\) whenever \(t\notin D_g\).

Suppose that

$$\begin{aligned} \sum _{t\in D_g}\left| \mathrm{{log}}|1-b(t)\mu _g(\{t\})| \right| <\infty . \end{aligned}$$
(3)

It implies that there exists a positive constant \(\delta \) such that

$$\begin{aligned} |1-b(t)\mu _g(\{t\})|>\delta ,\; \forall t\in D_g. \end{aligned}$$

Indeed, if we consider the countable set \(D_g\) written as a sequence \(({\tilde{t}}_n)_n\), then

$$\begin{aligned} \left| \mathrm{{log}}|1-b({\tilde{t}}_n)\mu _g(\{{\tilde{t}}_n\})| \right| \rightarrow 0\quad \mathrm{as}\; n\rightarrow \infty \end{aligned}$$

which means that

$$\begin{aligned} |1-b({\tilde{t}}_n)\mu _g(\{{\tilde{t}}_n\})|\rightarrow 1\quad \mathrm{as}\; n\rightarrow \infty \end{aligned}$$

therefore there exists a positive constant \(\delta \) such that

$$\begin{aligned} |1-b(t)\mu _g(\{t\})|>\delta ,\; \forall t\in D_g. \end{aligned}$$

What’s more, as \(D_g^-\) is finite, from one place onwards \({\tilde{t}}_n\in D_g{\setminus } D_g^-\), whence

$$\begin{aligned} 1-b({\tilde{t}}_n)\mu _g(\{{\tilde{t}}_n\})=|1-b({\tilde{t}}_n)\mu _g(\{{\tilde{t}}_n\})|\rightarrow 1, \end{aligned}$$

thus \(b({\tilde{t}}_n)\mu _g(\{{\tilde{t}}_n\})\rightarrow 0\) as \(n\rightarrow \infty \).

Lemma 16

If \(b,f:[0,T]\rightarrow {\mathbb {R}}\) are continuous on \([0,T]{\setminus } D_g\), then so are \({\tilde{b}}\) and \({\tilde{f}}\) under assumption (3).

Proof

Let \({\overline{t}}\in [0,T]{\setminus } D_g\) and let \(t_n\rightarrow {\overline{t}}\), \((t_n)_n\subset [0,T]\).

If \((t_n)_n \subset [0,T]{\setminus } D_g\), then \({\tilde{b}}(t_n)=b(t_n)\rightarrow b({\overline{t}})={\tilde{b}}({\overline{t}})\) by hypothesis.

If (at least on a subsequence) \(t_n \in D_g\), then

$$\begin{aligned}&{\tilde{b}}(t_n)=\frac{-\mathrm{{log}}|1-b(t_n)\mu _g(\{t_n\})|}{\mu _g(\{t_n\})}\\&=\frac{-\mathrm{{log}}|1-b(t_n)\mu _g(\{t_n\})|}{b(t_n)\mu _g(\{t_n\})}\cdot b(t_n)\rightarrow b({\overline{t}})={\tilde{b}}({\overline{t}}) \end{aligned}$$

since, by the previous discussion, \(b(t_n)\mu _g(\{t_n\})\rightarrow 0\) as \(n\rightarrow \infty \).

So, the continuity of \({\tilde{b}}\) on \([0,T]{\setminus } D_g\) is proved. Obviously, the continuity of \({\tilde{f}}\) on \([0,T]{\setminus } D_g\) can be proved similarly.\(\square \)

Theorem 17

Let \(g:[0,T]\rightarrow {\mathbb {R}}\) be a nondecreasing left-continuous function which, \(\mu _g\)-a.e. on the set of continuity points, has the property that g is differentiable and the identical function is g-differentiable.

Let \(b:[0,T]\rightarrow {\mathbb {R}}\) be LS-integrable w.r.t. g, continuous on \([0,T]{\setminus } D_g\) and \(f:[0,T]\rightarrow {\mathbb {R}}\) be continuous on \([0,T]{\setminus } D_g\), such that \({\tilde{f}}\) is KS-integrable w.r.t. g. Suppose condition (3) is fulfilled.

Then the function \(u:[0,T]\rightarrow {\mathbb {R}}\) given by

$$\begin{aligned} u(t)=\frac{1}{\alpha (t)}\int _0^T \alpha (s){\tilde{g}}(t,s){\tilde{f}}(s)\mathrm{{d}}g(s), \end{aligned}$$

where

$$\begin{aligned} {\tilde{g}}(t,s)=\frac{1}{\alpha (T) e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1} \left\{ \begin{array}{l} \alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)-\int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r)},\; if \;0\le s\le t\le T\\ e^{-\int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r)}, \; if \; 0\le t<s\le T \end{array} \right. \end{aligned}$$

is a solution of the periodic Stieltjes differential problem (2).

Proof

Note that hypothesis (3) ensures (together with the LS-integrability w.r.t. g of b) the LS-integrability of \({\tilde{b}}\) w.r.t. g.

Let \(t\in [0,T]{\setminus } D_g\) be a point where g is differentiable and the identical function is g-differentiable.

One can see that

$$\begin{aligned} u(t)= & {} \frac{1}{\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1} \left[ \frac{\alpha (T)}{\alpha (t)}\int _0^t \alpha (s)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)-\int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r)}\cdot {\tilde{f}}(s)\mathrm{{d}}g(s)\right. \\&+\left. \frac{1}{\alpha (t)}\int _t^T \alpha (s)e^{-\int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r)}\cdot {\tilde{f}}(s)\mathrm{{d}}g(s) \right] . \end{aligned}$$

By making use of Proposition 5, \(s\rightarrow \int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r)\) is continuous at t (because g is continuous at t), therefore \(s\rightarrow e^{-\int _s^t{\tilde{b}}(r)\mathrm{{d}}g(r)}\cdot \alpha (s){\tilde{f}}(s)\) is continuous at t.

We are able to apply Lemma 12 and use the fact that \(\alpha \) is constant on a neighborhood of t:

$$\begin{aligned} u'_g(t)= & {} \frac{1}{\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1} \\&\cdot \left[ \frac{\alpha (T)}{\alpha (t)}\int _0^t \alpha (s)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)-\int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r)}\cdot {\tilde{f}}(s)\cdot (-{\tilde{b}}(t))\mathrm{{d}}g(s)\right. \\&+ \frac{\alpha (T)}{\alpha (t)}e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)-\int _t^t {\tilde{b}}(r)\mathrm{{d}}g(r)}\cdot \alpha (t) {\tilde{f}}(t)\cdot g'(t)\cdot t'_g\\&+ \left. \frac{1}{\alpha (t)}\int _t^T e^{-\int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r)}\cdot \alpha (s){\tilde{f}}(s)\cdot (-{\tilde{b}}(t))\mathrm{{d}}g(s)\right. \\&\left. -\frac{1}{\alpha (t)}e^{-\int _t^t {\tilde{b}}(r)\mathrm{{d}}g(r)}\cdot \alpha (t){\tilde{f}}(t)\cdot g'(t)\cdot t'_g\right] \\= & {} -b(t)u(t)+f(t) \end{aligned}$$

since \(\int _t^t {\tilde{b}}(r)\mathrm{{d}}g(r)=0\) (t being a continuity point of g).

Note that when calculating the partial g-derivative w.r.t. the argument t of the function under the integral sign, we used Lemma 11.i) together with Proposition 8 (due to the continuity of \({\tilde{b}}\) at the point t), for instance

$$\begin{aligned} \frac{\partial }{\partial _g t} e^{-\int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r)}= & {} e^{-\int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r)} \cdot \frac{\partial (-\int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r))}{\partial _g t}\\= & {} e^{-\int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r)} \cdot (-{\tilde{b}}(t)). \end{aligned}$$

Consider next \(t\in D_g{\setminus } \{t_1,...,t_k\}\).

Denoting by

$$\begin{aligned} F(t)=\frac{1}{\alpha (t)}\int _0^t \alpha (s) e^{-\int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r)}{\tilde{f}}(s)\mathrm{{d}}g(s) \end{aligned}$$

respectively

$$\begin{aligned} G(t)=\frac{1}{\alpha (t)}\int _t^T \alpha (s) e^{-\int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r)}{\tilde{f}}(s)\mathrm{{d}}g(s) \end{aligned}$$

as in Lemma 15, we are able to write

$$\begin{aligned} u(t)=\frac{\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}}{\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1}\cdot F(t)+\frac{1}{\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1}\cdot G(t) \end{aligned}$$

and so we can compute the g-derivative:

$$\begin{aligned} u'_g(t)= & {} \frac{\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}}{\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1}\cdot F'_g(t)+\frac{1}{\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1}\cdot G'_g(t)\\= & {} \frac{\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}}{\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1}\cdot \left[ \frac{e^{-{\tilde{b}}(t)\cdot \mu _g(\{t\})}-1}{\mu _g(\{t \})}\cdot F(t)+e^{-{\tilde{b}}(t)\cdot \mu _g(\{t \})}\cdot {\tilde{f}}(t) \right] \\&+ \frac{1}{\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1}\cdot \left[ \frac{e^{-{\tilde{b}}(t)\cdot \mu _g(\{t\})}-1}{\mu _g(\{t \})}\cdot G(t)-e^{-{\tilde{b}}(t)\cdot \mu _g(\{t \})}\cdot {\tilde{f}}(t) \right] \\= & {} \frac{e^{-{\tilde{b}}(t)\mu _g(\{t\})}-1}{\mu _g(\{t\})} \cdot u(t)+e^{-{\tilde{b}}(t)\mu _g(\{t\})}\cdot {\tilde{f}}(t). \end{aligned}$$

It follows that

$$\begin{aligned} u'_g(t)= & {} \frac{e^{\mathrm{{log}}(1-b(t)\mu _g(\{t\}))}-1}{\mu _g(\{t\})}\cdot u(t)+e^{\mathrm{{log}}(1-b(t)\mu _g(\{t\}))}\cdot \frac{f(t)}{1-b(t)\mu _g(\{t\})}\\= & {} -b(t)\cdot u(t)+f(t). \end{aligned}$$

At each \(t=t_i,i=1,...,k\),

$$\begin{aligned} u'_g(t)= & {} \frac{\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}}{\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1}\cdot F'_g(t)+\frac{1}{\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1}\cdot G'_g(t)\\= & {} \frac{\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}}{\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1}\cdot \left[ \frac{-e^{-{\tilde{b}}(t)\cdot \mu _g(\{t\})}-1}{\mu _g(\{t \})}\cdot F(t)-e^{-{\tilde{b}}(t)\cdot \mu _g(\{t \})}\cdot {\tilde{f}}(t) \right] \\&+ \frac{1}{\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1}\cdot \left[ \frac{-e^{-{\tilde{b}}(t)\cdot \mu _g(\{t\})}-1}{\mu _g(\{t \})}\cdot G(t)+e^{-{\tilde{b}}(t)\cdot \mu _g(\{t \})}\cdot {\tilde{f}}(t) \right] \\= & {} \frac{-e^{-{\tilde{b}}(t)\mu _g(\{t\})}-1}{\mu _g(\{t\})} \cdot u(t)-e^{-{\tilde{b}}(t)\mu _g(\{t\})}\cdot {\tilde{f}}(t), \end{aligned}$$

so

$$\begin{aligned} u'_g(t)= & {} \frac{-e^{\mathrm{{log}}(-1+b(t)\mu _g(\{t\}))}-1}{\mu _g(\{t\})}\cdot u(t)-e^{\mathrm{{log}}(-1+b(t)\mu _g(\{t\}))}\cdot \frac{f(t)}{1-b(t)\mu _g(\{t\})}\\= & {} -b(t)\cdot u(t)+f(t). \end{aligned}$$

Finally,

$$\begin{aligned} u(0)= & {} \frac{1}{\alpha (0)}\int _0^T \alpha (s){\tilde{g}}(0,s){\tilde{f}}(s)\mathrm{{d}}g(s)\\= & {} \frac{1}{\alpha (T)e^{\int _0^T b(r)\mathrm{{d}}g(r)}-1}\int _0^T \alpha (s)e^{\int _0^s b(r)\mathrm{{d}}g(r)}\cdot {\tilde{f}}(s)\mathrm{{d}}g(s) \end{aligned}$$

and

$$\begin{aligned} u(T)= & {} \frac{1}{\alpha (T)}\int _0^T \alpha (s){\tilde{g}}(T,s){\tilde{f}}(s)\mathrm{{d}}g(s)\\= & {} \frac{1}{\alpha (T)e^{\int _0^T b(r)\mathrm{{d}}g(r)}-1}\int _0^T \alpha (s)e^{\int _0^T b(r)\mathrm{{d}}g(r)-\int _s^T b(r)\mathrm{{d}}g(r)}\cdot {\tilde{f}}(s)\mathrm{{d}}g(s)\\= & {} \frac{1}{\alpha (T)e^{\int _0^T b(r)\mathrm{{d}}g(r)}-1}\int _0^T \alpha (s)e^{\int _0^s b(r)\mathrm{{d}}g(r)}\cdot {\tilde{f}}(s)\mathrm{{d}}g(s); \end{aligned}$$

therefore,

$$\begin{aligned} u(0)=u(T). \end{aligned}$$

\(\square \)

Remark 18

If f is LS-integrable w.r.t. g, then the LS-integrability (thus, the KS-integrability as well) w.r.t. g of \({\tilde{f}}\) is a simple consequence of condition (3) since

$$\begin{aligned} |{\tilde{f}}(t)|\le \max \left( 1,\frac{1}{\delta }\right) \cdot |f(t)| \quad \forall t\in [0,T]. \end{aligned}$$

The reciprocal assertion is also valid:

Theorem 19

If a function \(u:[0,T]\rightarrow {\mathbb {R}}\) is a solution in the sense of Definition 14 of the periodic Stieltjes differential problem (2), then

$$\begin{aligned} u(t)=\frac{1}{\alpha (t)}\int _0^T \alpha (s){\tilde{g}}(t,s){\tilde{f}}(s)\mathrm{{d}}g(s),\quad \forall \; t\in [0,T]. \end{aligned}$$

Proof

Let u be a solution of problem (2).

Take first \(t\in [0,T]{\setminus } D_g\) a point where the equation is satisfied. Then since \(\alpha \) is constant in a neighborhood of t,

$$\begin{aligned}&\left( \alpha (t)e^{\int _0^t {\tilde{b}}(r)\mathrm{{d}}g(r)}u(t) \right) '_g \\&\quad = \alpha (t)e^{\int _0^t {\tilde{b}}(r)\mathrm{{d}}g(r)}{\tilde{b}}(t)\cdot u(t)+\alpha (t) e^{\int _0^t {\tilde{b}}(r)\mathrm{{d}}g(r)} \cdot u'_g(t)\\&\quad = \alpha (t)e^{\int _0^t {\tilde{b}}(r)\mathrm{{d}}g(r)}f(t)=\alpha (t)e^{\int _0^{t} {\tilde{b}}(r)\mathrm{{d}}g(r)}\cdot {\tilde{f}}(t). \end{aligned}$$

Let now \(t\in D_g{\setminus } \{t_1,...,t_k\}\). Applying Lemma 15, one gets

$$\begin{aligned}&\left( \alpha (t)e^{\int _0^t {\tilde{b}}(r)\mathrm{{d}}g(r)}u(t) \right) '_g \\&\quad = \alpha (t) u'_g(t)\cdot e^{\int _0^{t+} {\tilde{b}}(r)\mathrm{{d}}g(r)}+\alpha (t)u(t)\cdot e^{\int _0^t {\tilde{b}}(r)\mathrm{{d}}g(r)} \cdot \frac{e^{{\tilde{b}}(t)\mu _g(\{t\})}-1}{\mu _g(\{t\})}\\&\quad = \alpha (t)e^{\int _0^{t} {\tilde{b}}(r)\mathrm{{d}}g(r)}\cdot \left[ u'_g(t)\cdot e^{\int _t^{t+} {\tilde{b}}(r)\mathrm{{d}}g(r)} +u(t)\cdot \frac{e^{{\tilde{b}}(t)\mu _g(\{t\})}-1}{\mu _g(\{t\})}\right] \\&\quad = \alpha (t)e^{\int _0^{t} {\tilde{b}}(r)\mathrm{{d}}g(r)}\cdot \left( u'_g(t)\cdot e^{{\tilde{b}}(t) \mu _g(\{t\})}+u(t)\cdot \frac{e^{{\tilde{b}}(t)\mu _g(\{t\})}-1}{\mu _g(\{t\})} \right) \\&\quad = \alpha (t)e^{\int _0^{t} {\tilde{b}}(r)\mathrm{{d}}g(r)}\cdot \left( (-b(t)u(t)+f(t))\cdot \frac{1}{1-b(t)\mu _g(\{t\})}\right. \\&\qquad + \left. u(t)\cdot \frac{\frac{1}{1-b(t)\mu _g(\{t\})}-1}{\mu _g(\{t\})}\right) \\&\quad = \alpha (t)e^{\int _0^{t} {\tilde{b}}(r)\mathrm{{d}}g(r)}\cdot {\tilde{f}}(t). \end{aligned}$$

Also, at any point \(t=t_i, i\in \{1,...,k\}\),

$$\begin{aligned}&\left( \alpha (t)e^{\int _0^t {\tilde{b}}(r)\mathrm{{d}}g(r)}u(t) \right) '_g \\&\quad =\frac{\alpha (t+)e^{\int _0^{t+} {\tilde{b}}(r)\mathrm{{d}}g(r)}u(t+)-\alpha (t)e^{\int _0^t {\tilde{b}}(r)\mathrm{{d}}g(r)}u(t)}{\mu _g(\{t\})}\\&\quad = \frac{-\alpha (t)e^{\int _0^{t} {\tilde{b}}(r)\mathrm{{d}}g(r)}e^{\int _t^{t+}{\tilde{b}}(r)\mathrm{{d}}g(r)}u(t+)-\alpha (t)e^{\int _0^t {\tilde{b}}(r)\mathrm{{d}}g(r)}u(t)}{\mu _g(\{t\})}\\&\quad = \frac{-\alpha (t)e^{\int _0^{t} {\tilde{b}}(r)\mathrm{{d}}g(r)}e^{{\tilde{b}}(t)\mu _g(\{t\})}u(t+)-\alpha (t)e^{\int _0^t {\tilde{b}}(r)\mathrm{{d}}g(r)}u(t)}{\mu _g(\{t\})}.\\ \end{aligned}$$

Writing now

$$\begin{aligned} u(t+)=u'_g(t)\mu _g(\{t\})+u(t) \end{aligned}$$

and taking into account that

$$\begin{aligned} u'_g(t)=-b(t)u(t)+f(t) \end{aligned}$$

and the definition of \({\tilde{b}}\), one gets

$$\begin{aligned}&\left( \alpha (t)e^{\int _0^t {\tilde{b}}(r)\mathrm{{d}}g(r)}u(t) \right) '_g \\&\quad = \frac{-\alpha (t)e^{\int _0^{t} {\tilde{b}}(r)\mathrm{{d}}g(r)}e^{{\tilde{b}}(t)\mu _g(\{t\})}(u'_g(t)\mu _g(\{t\})+u(t))-\alpha (t)e^{\int _0^t {\tilde{b}}(r)\mathrm{{d}}g(r)}u(t)}{\mu _g(\{t\})}\\&\quad =\frac{ -\alpha (t)e^{\int _0^{t} {\tilde{b}}(r)\mathrm{{d}}g(r)}}{\mu _g(\{t\})} \left[ e^{{\tilde{b}}(t)\mu _g(\{t\})}((-b(t)u(t)+f(t))\mu _g(\{t\})+u(t))+u(t) \right] \\&\quad =\frac{ -\alpha (t)e^{\int _0^{t} {\tilde{b}}(r)\mathrm{{d}}g(r)}}{\mu _g(\{t\})} \left[ e^{{\tilde{b}}(t)\mu _g(\{t\})}((-b(t)\mu _g(\{t\})+1) u(t)\right. \\&\qquad +\left. e^{{\tilde{b}}(t)\mu _g(\{t\})}f(t)\mu _g(\{t\})+u(t) \right] \\&\quad =\frac{ -\alpha (t)e^{\int _0^{t} {\tilde{b}}(r)\mathrm{{d}}g(r)}}{\mu _g(\{t\})} \left[ e^{-\mathrm{{log}}(-1+b(t)\mu _g(\{t\})}((-b(t)\mu _g(\{t\})+1) u(t)\right. \\&\qquad +\left. e^{-\mathrm{{log}}(-1+b(t)\mu _g(\{t\})}f(t)\mu _g(\{t\})+u(t) \right] \\&\quad = \alpha (t)e^{\int _0^{t} {\tilde{b}}(r)\mathrm{{d}}g(r)}\cdot {\tilde{f}}(t). \end{aligned}$$

From here, the result is achieved by integrating on [0, t] w.r.t. g the equality

$$\begin{aligned} \left( \alpha (s)e^{\int _0^s {\tilde{b}}(r)\mathrm{{d}}g(r)}u(s) \right) '_g=\alpha (s)e^{\int _0^{s} {\tilde{b}}(r)\mathrm{{d}}g(r)}\cdot {\tilde{f}}(s) \end{aligned}$$

and imposing the boundary conditions as in the classical case of periodic differential problems (where \(g(t)=t\)). While integrating the g-derivative in the Kurzweil-Stieltjes sense we apply the fundamental theorem for this type of integral [12, Theorem 6.2] (this is possible since \(t\mapsto e^{\int _0^t {\tilde{b}}(r)\mathrm{{d}}g(r)}\) and u and \(\alpha \) are left-continuous and constant on any subinterval where g is constant). \(\square \)

Remark 20

If for every \(t\in [0,T]\)

$$\begin{aligned} 1-b(t)\mu _g(t)>0 \end{aligned}$$

then the set \(\{t\in D_g: 1-b(t)\mu _g(t)<0\}\) is empty and \(\alpha (t)=1\) on the whole interval; therefore, the calculus is much simpler. In this case, b can be assumed to be only KS-integrable w.r.t. g.

3.2 Existence result for the nonlinear problem

We go further to studying the nonlinear problem (1):

$$\begin{aligned} \left\{ \begin{array}{l} u'_g(t)+b(t) u(t) =f(t,u(t)),\; t\in [0,T]\\ u(0)=u(T). \end{array} \right. \end{aligned}$$

We shall apply Schaeffer’s fixed point theorem.

Theorem 21

Let S be a normed linear space and the operator \(A:S\rightarrow S\) be continuous and compact. If the set

$$\begin{aligned} \{x\in S: x=\lambda Ax \; \mathrm{for\; some}\; \lambda \in (0,1)\} \end{aligned}$$

is bounded, then the operator has a fixed point.

Theorem 22

Let \(g:[0,T]\rightarrow {\mathbb {R}}\) be a nondecreasing left-continuous function which, \(\mu _g\)-a.e. on the set of continuity points, has the property that g is differentiable and the identical function is g-differentiable.

Let \(b:[0,T]\rightarrow {\mathbb {R}}\) be LS-integrable w.r.t. g, continuous on \([0,T]{\setminus } D_g\) and suppose condition (3) is fulfilled.

Let \(f:[0,T]\times {\mathbb {R}}\rightarrow {\mathbb {R}}\) satisfy the following hypotheses:

  • f is continuous on \(([0,T]{\setminus } D_g)\times {\mathbb {R}}\) and for every \(t\in D_g\), \(f(t,\cdot )\) is continuous;

  • for every \(u\in {\mathbb {R}}\), \(f(\cdot ,u)\) is \(\mu _g\)-measurable;

  • there exists a function \({\overline{\phi }}\) KS-integrable w.r.t. g such that

    $$\begin{aligned} \left| f(t,u)\right| \le {\overline{\phi }}(t) \end{aligned}$$

    for every \(t\in [0,T], u\in {\mathbb {R}}\).

Then the periodic Stieltjes differential equation (1) possesses solutions.

Proof

Let \(S_g\) be the subspace of the space of regulated real-valued maps defined on [0, T] consisting of those functions which are continuous on \([0,T]{\setminus } D_g\).

Consider the operator \(A:S_g\rightarrow S_g\) given by

$$\begin{aligned} Au=\frac{1}{\alpha (t)}\int _0^T \alpha (s){\tilde{g}}(t,s){\tilde{f}}(s,u(s))\mathrm{{d}}g(s),\; t\in [0,T] \end{aligned}$$

with \({\tilde{g}}\) as in Theorem 17 and

$$\begin{aligned} {\tilde{f}}(t,u)=\frac{f(t,u)}{1-b(t)\mu _g(\{t\})}. \end{aligned}$$

It is well defined: \({\tilde{f}}(\cdot ,u(\cdot ))\) is KS-integrable w.r.t. g for each \(u\in S_g\) since it is \(\mu _g\)-measurable and

$$\begin{aligned} \left| {\tilde{f}}(t,u(t))\right| \le \phi (t)=\max \left( 1,\frac{1}{\delta }\right) \cdot {\overline{\phi }}(t)\; \forall t\in [0,T], u\in S_g. \end{aligned}$$

Also, whenever \(u\in S_g\), i.e. u is regulated and continuous on \([0,T]{\setminus } D_g\), Au has the same feature by Proposition 5 due to the fact that \(\alpha \) is constant in a neighborhood of \(t\in [0,T]{\setminus } D_g\), since

$$\begin{aligned} Au(t)= & {} \frac{1}{\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1}\\&\cdot \left[ \frac{\alpha (T)}{\alpha (t)}\int _0^t \alpha (s)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)-\int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r)}\cdot {\tilde{f}}(s,u(s))\mathrm{{d}}g(s)\right. \\&+\left. \frac{1}{\alpha (t)}\int _t^T \alpha (s)e^{-\int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r)}\cdot {\tilde{f}}(s,u(s))\mathrm{{d}}g(s) \right] . \end{aligned}$$

We shall see that A satisfies the hypotheses of Schaeffer’s fixed point theorem.

First, let us prove that it is continuous. Let \((u_n)_n \subset S_g\) converge uniformly to u. As f is continuous with respect to the second argument,

$$\begin{aligned} {\tilde{f}}(s,u_n(s))\rightarrow {\tilde{f}}(s,u(s)),\; \forall \; s\in [0,T]. \end{aligned}$$

We are now able to apply a dominated convergence result, e.g. [20, Theorem 6.8.6], since \(-\phi (s)\le {\tilde{f}}(s,u_{n}(s)) \le \phi (s)\) for every \(n\in {\mathbb {N}}\); we get, for all \(t\in [0,T]\),

$$\begin{aligned} \int _0^T \alpha (s){\tilde{g}}(t,s){\tilde{f}}(s,u_{n}(s))\mathrm{{d}}g(s)\rightarrow \int _0^T \alpha (s){\tilde{g}}(t,s){\tilde{f}}(s,u(s))\mathrm{{d}}g(s), \end{aligned}$$

so \(Au_n(t)\rightarrow Au(t)\).

We check, using Lemma 2, that the convergence is uniform.

Take \(0\le t<t'\le T\) .

We can see that for each n,

$$\begin{aligned} \left| Au_n(t)-Au_n(t')\right|\le & {} \left| \frac{1}{\alpha (t)}\int _0^T \alpha (s)({\tilde{g}}(t,s)-{\tilde{g}}(t',s)){\tilde{f}}(s,u_{n}(s))\mathrm{{d}}g(s) \right| \\&+\left| \left( \frac{1}{\alpha (t)}-\frac{1}{\alpha (t')}\right) \int _0^T \alpha (s){\tilde{g}}(t',s){\tilde{f}}(s,u_{n}(s))\mathrm{{d}}g(s) \right| . \end{aligned}$$

We note that \(|\alpha (t)|=1\) for each \(t\in [0,T]\), so we can see that

$$\begin{aligned}&\left| \frac{1}{\alpha (t)}\int _0^T \alpha (s)({\tilde{g}}(t,s)-{\tilde{g}}(t',s)){\tilde{f}}(s,u_{n}(s))\mathrm{{d}}g(s) \right| \\&\quad \le \frac{1}{\left| \alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1\right| }\\&\qquad \left[ \left| \alpha (T)\right. \right. \\&\qquad \left. \int _0^t \alpha (s)(e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)-\int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r)}-e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r) -\int _s^{t'} {\tilde{b}}(r)\mathrm{{d}}g(r)}){\tilde{f}}(s,u_{n}(s))\mathrm{{d}}g(s) \right| \\&\qquad + \left| \int _{t'}^T \alpha (s)(e^{-\int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r)}-e^{-\int _s^{t'} {\tilde{b}}(r)\mathrm{{d}}g(r)}){\tilde{f}}(s,u_{n}(s))\mathrm{{d}}g(s)\right| \\&\qquad + \left| \int _t^{t'} \alpha (s)(e^{-\int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r)}-\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)-\int _s^{t'} {\tilde{b}}(r)\mathrm{{d}}g(r)})\left. {\tilde{f}}(s,u_{n}(s))\mathrm{{d}}g(s)\right| \right] \\&\quad = \frac{1}{\left| \alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1\right| }\\&\qquad \left[ \left| \int _0^t \alpha (s)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)-\int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r)}(1-e^{-\int _t^{t'} {\tilde{b}}(r)\mathrm{{d}}g(r)}){\tilde{f}}(s,u_{n}(s))\mathrm{{d}}g(s) \right| \right. \\&\qquad +\left| \int _{t'}^T \alpha (s)e^{-\int _s^{t'} {\tilde{b}}(r)\mathrm{{d}}g(r)}(e^{\int _t^{t'} {\tilde{b}}(r)\mathrm{{d}}g(r)}-1){\tilde{f}}(s,u_{n}(s))\mathrm{{d}}g(s)\right| \\&\qquad +\left| \int _t^{t'} \alpha (s)(e^{-\int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r)}-\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)-\int _s^{t'} {\tilde{b}}(r)\mathrm{{d}}g(r)})\right. \\&\left. \left. \qquad {\tilde{f}}(s,u_{n}(s))\mathrm{{d}}g(s)\right| \right] . \end{aligned}$$

On the other hand,

$$\begin{aligned}&\left| \left( \frac{1}{\alpha (t)}-\frac{1}{\alpha (t')}\right) \int _0^T \alpha (s){\tilde{g}}(t',s){\tilde{f}}(s,u_{n}(s))\mathrm{{d}}g(s) \right| \\&\quad = |\alpha (t)-\alpha (t')|\left| \int _0^T\alpha (s){\tilde{g}}(t',s){\tilde{f}}(s,u_{n}(s))\mathrm{{d}}g(s)\right| \\&\quad \le \frac{|\alpha (t)-\alpha (t')|}{|\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1|}\left[ \left| \int _0^{t'} \alpha (s)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)-\int _s^{t'} {\tilde{b}}(r)\mathrm{{d}}g(r)}{\tilde{f}}(s,u_{n}(s))\mathrm{{d}}g(s) \right| \right. \\&\qquad + \left. \left| \int _{t'}^T \alpha (s)e^{-\int _s^{t'} {\tilde{b}}(r)\mathrm{{d}}g(r)}{\tilde{f}}(s,u_{n}(s))\mathrm{{d}}g(s) \right| \right] \\&\quad \le \frac{|\alpha (t)-\alpha (t')|}{|\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1|}\left[ \int _0^{t'}\left| e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)-\int _s^{t'} {\tilde{b}}(r)\mathrm{{d}}g(r)}{\tilde{f}}(s,u_{n}(s)) \right| \mathrm{{d}}g(s)\right. \\&\qquad +\left. \int _{t'}^T \left| e^{-\int _s^{t'} {\tilde{b}}(r)\mathrm{{d}}g(r)}{\tilde{f}}(s,u_{n}(s)) \right| \mathrm{{d}}g(s) \right] . \end{aligned}$$

But

$$\begin{aligned} \int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)-\int _s^t {\tilde{b}}(r)\mathrm{{d}}g(r)=\int _0^s {\tilde{b}}(r)\mathrm{{d}}g(r)+\int _t^T {\tilde{b}}(r)\mathrm{{d}}g(r) \end{aligned}$$

and

$$\begin{aligned} \int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)-\int _s^{t'} {\tilde{b}}(r)\mathrm{{d}}g(r)=\int _0^s {\tilde{b}}(r)\mathrm{{d}}g(r)+\int _{t'}^T {\tilde{b}}(r)\mathrm{{d}}g(r). \end{aligned}$$

The map \((s',s'')\in [0,T]\times [0,T]\rightarrow e^{\int _{s'}^{s''} {\tilde{b}}(s)\mathrm{{d}}g(s)}\) is regulated in both arguments, therefore bounded. If we note by

$$\begin{aligned} M=\sup _{(s',s'')\in [0,T]\times [0,T]} e^{\int _{s'}^{s''} {\tilde{b}}(s)\mathrm{{d}}g(s)}, \end{aligned}$$

we get

$$\begin{aligned}&\left| Au_n(t)-Au_n(t')\right| \\&\quad \le \frac{M}{\left| \alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1\right| } \left[ M\int _0^t\left| 1-e^{-\int _t^{t'} {\tilde{b}}(r)\mathrm{{d}}g(r)}\right| \cdot |{\tilde{f}}(s,u_{n}(s))|\mathrm{{d}}g(s)\right. \\&\qquad +\int _{t'}^T \left| e^{\int _t^{t'} {\tilde{b}}(r)\mathrm{{d}}g(r)}-1\right| \cdot |{\tilde{f}}(s,u_{n}(s))|\mathrm{{d}}g(s)\\&\qquad +\left. (1+M) \int _t^{t'} \left| {\tilde{f}}(s,u_{n}(s))\right| \mathrm{{d}}g(s)\right] \\&\qquad + \frac{|\alpha (t)-\alpha (t')|}{|\alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1|}\left[ M^2 \int _0^{t'} |{\tilde{f}}(s,u_n(s))|\mathrm{{d}}g(s)\right. \\&\qquad \left. +M\int _{t'}^T |{\tilde{f}}(s,u_n(s))|\mathrm{{d}}g(s)\right] , \end{aligned}$$

so

$$\begin{aligned}&\left| Au_n(t)-Au_n(t')\right| \\&\quad \le \frac{M}{\left| \alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1\right| }\left[ M\left| 1-e^{-\int _t^{t'} {\tilde{b}}(r)\mathrm{{d}}g(r)}\right| \cdot \int _0^T \phi (s)\mathrm{{d}}g(s)\right. \\&\qquad + \left| e^{\int _t^{t'} {\tilde{b}}(r)\mathrm{{d}}g(r)}-1\right| \cdot \left. \int _{0}^T \phi (s)\mathrm{{d}}g(s)+(1+M) \int _t^{t'} \phi (s)\mathrm{{d}}g(s)\right. \\&\qquad \left. +(M+1)|\alpha (t)-\alpha (t')|\int _0^T \phi (s)\mathrm{{d}}g(s)\right] . \end{aligned}$$

But

$$\begin{aligned} \left| 1-e^{-\int _t^{t'} {\tilde{b}}(r)\mathrm{{d}}g(r)}\right|= & {} \left| e^{\int _0^{t} {\tilde{b}}(r)\mathrm{{d}}g(r)}\left( e^{-\int _0^{t} {\tilde{b}}(r)\mathrm{{d}}g(r)}-e^{-\int _0^{t'} {\tilde{b}}(r)\mathrm{{d}}g(r)}\right) \right| \\\le & {} M\left| e^{-\int _0^{t} {\tilde{b}}(r)\mathrm{{d}}g(r)}-e^{-\int _0^{t'} {\tilde{b}}(r)\mathrm{{d}}g(r)}\right| \end{aligned}$$

and \(e^{-\int _0^{\cdot } {\tilde{b}}(r)\mathrm{{d}}g(r)}\) is regulated.

A similar calculus can be made for \(\left| e^{\int _t^{t'} {\tilde{b}}(r)\mathrm{{d}}g(r)}-1\right| \) , while

$$\begin{aligned} \int _t^{t'} \phi (s)\mathrm{{d}}g(s)=\int _0^{t'} \phi (s)\mathrm{{d}}g(s)-\int _0^{t} \phi (s)\mathrm{{d}}g(s) \end{aligned}$$

and \(\int _0^{\cdot } \phi (s)\mathrm{{d}}g(s)\) and \(\alpha \) are regulated.

Remark 3 yields now that the sequence is equiregulated whence, by Lemma 2, \((Au_n)_n\) converges uniformly to Au, i.e. the operator A is continuous.

Let us next prove that the operator is compact. Take \(B\subset S_g\) be a bounded set. Then, in the same manner as before, it can be seen that \(\{Au: u\in B\}\) is equiregulated.

It is pointwise bounded as well. Indeed, fix \(t\in [0,T]\). Then for every \(u\in B\),

$$\begin{aligned} \left| Au(t)\right|\le & {} \int _0^T \left| {\tilde{g}}(t,s)\right| \cdot \left| {\tilde{f}}(s,u(s))\right| \mathrm{{d}}g(s)\\\le & {} \frac{ \max (M,M^2) }{\left| \alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1\right| }\int _0^T |{\tilde{f}}(s,u(s))|\mathrm{{d}}g(s)\\\le & {} \frac{ \max (M,M^2) }{\left| \alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1\right| } \int _0^T \phi (s)\mathrm{{d}}g(s); \end{aligned}$$

therefore, the set is pointwise (in fact, even uniformly) bounded.

Lemma 1 implies that \(\{Au: u\in B\}\) is relatively compact, thus A is a compact operator.

Let us now see that the set

$$\begin{aligned} \{u\in S_g: u=\lambda Au \; \mathrm{for\; some}\; \lambda \in (0,1)\} \end{aligned}$$

is bounded.

Let u be an arbitrary element of this set. One can find \(\lambda \in (0,1)\) such that

$$\begin{aligned} u(t)=\frac{\lambda }{\alpha (t)} \int _0^T \alpha (s){\tilde{g}}(t,s){\tilde{f}}(s,u(s))\mathrm{{d}}g(s),\; t\in [0,T]. \end{aligned}$$

It follows, as before, that

$$\begin{aligned} \Vert u\Vert _C\le & {} \lambda \frac{\max (M,M^2) }{\left| \alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1\right| } \int _0^T \phi (s)\mathrm{{d}}g(s)\\\le & {} \frac{ \max (M,M^2) }{\left| \alpha (T)e^{\int _0^T {\tilde{b}}(r)\mathrm{{d}}g(r)}-1\right| } \int _0^T \phi (s)\mathrm{{d}}g(s) \end{aligned}$$

and the boundedness is achieved.

Schaeffer’s fixed point theorem yields that the operator has fixed points, which are solutions to our problem. This follows exactly as in Theorem 17 since \({\tilde{b}}\) and \({\tilde{f}}(\cdot ,u(\cdot ))\) satisfy the Stieltjes integrability conditions and, by Lemma 16, the continuity on \([0,T]{\setminus } D_g\). \(\square \)

Remark 23

The large applicability of our result is a motivation for the less general assumptions on the function f (comparing with the existence results available in particular cases, e.g. [3, 14, 17, 33] or [32]).

In counterbalance, this is, as far as the authors know, the first existence result for periodic Stieltjes differential boundary value problems (and, consequently, for periodic measure boundary value problems). In particular, new results can be deduced for dynamic boundary value problems on time scales or for periodic impulsive differential equations allowing a countable number of impulses (i.e. the impulse moments can accumulate) which cannot be studied through the theory of impulsive differential equations.