Generalized Chain Rules and Applications to Stieltjes Differential and Integral Equations

Abstract

We present new generalizations of the chain rule, which involve Stieltjes derivatives and integrals. The results are subsequently used to obtain the power rules for two generalizations of the exponential function, and to investigate Bernoulli-type equations with Stieltjes derivatives and integrals.

1 Introduction

The classical chain rule for real functions f and h states that if f is differentiable at $t\in {\mathbb {R}}$ and h is differentiable at f(t), then the composite function $h\circ f$ is differentiable at t, and its derivative is

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

(1.1)

The goal of the present paper is to investigate several generalizations of this rule.

The first generalization consists in replacing an ordinary derivative by the Stieltjes derivative with respect to a left-continuous nondecreasing function g; the definition of this derivative will be recalled in Sect. 2. Two formulas of this type, namely

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

and

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

are already available in the literature (see [7, Theorem 2.3] and [9, Proposition 3.15 and Remark 3.16]). The disadvantage of these formulas is that they do not apply if t is a discontinuity point of g. To avoid this problem, we will present another pair of formulas, namely,

$$\begin{aligned} (h\circ f)'_g(t)=f'_g(t)\int _0^1 h'(f(t)+\tau f'_g(t)\Delta ^+ g(t))\,\textrm{d}\tau , \end{aligned}$$

(1.2)

and

$$\begin{aligned} (h\circ f)'_g(t)=f'_g(t)\int _0^1 h'_g(f(t)+\tau f'_g(t)\Delta ^+ g(t))g'(f(t)+\tau f'_g(t)\Delta ^+ g(t))\,\textrm{d}\tau ,\nonumber \\ \end{aligned}$$

(1.3)

where $\Delta ^+g(t)=g(t+)-g(t)$; these formulas are valid even if g is discontinuous at t, and their proofs together with precise assumptions are given in Sect. 3.

Another possible generalization of the chain rule is as follows: we observe that if h and f are continuously differentiable on [a, b], it is possible to rewrite Eq. (1.1) in the integral form

$$\begin{aligned} h(f(b))-h(f(a))=\int _a^b \left( h\circ f\right) '(t)\,\textrm{d}t=\int _a^b h'(f(t))f'(t)\,\textrm{d}t =\int _a^b h'(f(t))\,\textrm{d}f(t). \end{aligned}$$

The Stieltjes integral on the right-hand side no longer involves the derivative of f, and this formula does not require differentiability but merely continuity of f. The precise conditions on h and f depend on the choice of the integral, see e.g. [14, Corollary 1.2 and Eq. (6.3)]. In the same interesting paper, Rimas Norvaiša also considered the case when f is discontinuous, and obtained the formula

$$\begin{aligned} h(f(b))-h(f(a)) =\int _a^b h'(f(t))\,\textrm{d}f(t)&+\sum _{t\in (a,b]}\left( \Delta ^-(h\circ f)(t)-(h'\circ f)\Delta ^-f(t)\right) \\&+\sum _{t\in [a,b)}\left( \Delta ^+(h\circ f)(t)-(h'\circ f)\Delta ^+f(t)\right) , \end{aligned}$$

provided that h is continuously differentiable and f has bounded variation; see [14, Lemma 3.1]. In Sect. 7 of the present paper, we will provide an independent proof of the alternative chain rule

$$\begin{aligned} h(f(b))-h(f(a))=\int _a^b\left( \int _0^1 h'(\tau f(t+)+(1-\tau )f(t-))\,\textrm{d}\tau \right) \,\textrm{d}f(t),\nonumber \\ \end{aligned}$$

(1.4)

which is closer in spirit to the formula (1.2) (their relation is explained in Remark 7.3). The proof relies on some basic properties of Stieltjes integrals, which are recalled in Sect. 6.

The three formulas (1.2), (1.3), (1.4) are inspired by the chain rule on time scales (measure chains), which was introduced by Christian Pötzsche in [15] (see also [2, Theorem 1.90]), and which can be obtained as a special case of our results.

The remaining sections of the paper contain several applications of the generalized chain rules. In Sect. 4, we will focus on the g-exponential function $t\mapsto \exp _g(p,t)$, $t\in [a,b]$, which is defined as the solution of the Stieltjes differential equation

$$\begin{aligned} x'_g(t)=p(t)x(t),\quad x(a)=1. \end{aligned}$$

In particular, we will discuss the product rule

$$\begin{aligned}\exp _g(p,t)\cdot \exp _g(q,t)=\exp _g(p\oplus q, t)\end{aligned}$$

and the power rule

$$\begin{aligned} (\exp _g(p,t))^\alpha =\exp _g(\alpha \odot p,t), \end{aligned}$$

(1.5)

where $\oplus$ and $\odot$ are certain operations that reduce to ordinary addition and multiplication if g is continuous, but otherwise take into account the jumps of g. Along the way, we will introduce two classes of regressive functions, which provide a natural setting for the study of the operations $\oplus$ and $\odot$. Moreover, we will establish some basic properties of these operations. This part is inspired by the results for dynamic equations on time scales (see [1, Section 2.5]).

Similarly, in Sect. 8, we will deal with the generalized exponential function $t\mapsto e_{\textrm{d}P}(t,t_0)$, $t\in [a,b]$, which is defined as the solution of the Stieltjes integral equation

$$\begin{aligned} x(t)=1+\int _{t_0}^t x(s)\,\textrm{d}P(s),\quad t\in [a,b]. \end{aligned}$$

The product rule for the generalized exponential function is already available in the literature (see [12, Theorem 3.3] or [13, Theorem 8.5.6]). Thus, we will restrict ourselves to the power rule

$$\begin{aligned} e_{\textrm{d}P}(t,t_0)^{\alpha }=e_{\textrm{d}(\alpha *P)}(t,t_0), \end{aligned}$$

(1.6)

where $\alpha *P$ is a certain function that reduces to $\alpha P$ if P is continuous, but otherwise takes into account the jumps of P. The proofs of both power rules (1.5) and (1.6) are based on the corresponding generalized chain rules where the outer function is $h(x)=x^{\alpha }$.

The final topic of this paper are Bernoulli-type equations. It is well known that the classical Bernoulli differential equation

$$\begin{aligned} y'(t)=(p(t)+f(t)y(t)^{\alpha })y(t), \end{aligned}$$

where $\alpha \in {\mathbb {R}}\setminus \{0\}$, can be converted to the linear equation

$$\begin{aligned} z'(t)=-\alpha p(t)z(t)-\alpha f(t) \end{aligned}$$

using the substitution $z(t)=y(t)^{-\alpha }$. Conversely, one can define the Bernoulli equation as the differential equation that is obtained from the linear differential equation

$$\begin{aligned} z'(t)=p(t)z(t)+f(t) \end{aligned}$$

(we abandon the factor $-\alpha$, which can be incorporated into p and f) using the substitution $y(t)=z(t)^{-1/\alpha }$.

In Sect. 5, we will investigate what happens if we instead begin with the linear Stieltjes differential equation

$$\begin{aligned} z'_g(t)=p(t)z(t)+f(t). \end{aligned}$$

Performing again the substitution $y(t)=z(t)^{-1/\alpha }$, we will obtain the Bernoulli equation with Stieltjes derivatives, and provide an explicit solution formula. The choice $\alpha =1$ yields the logistic equation with Stieltjes derivatives, which was introduced in [10]. We will also discuss a slight modification of this procedure, where we begin with the adjoint linear Stieltjes differential equation introduced in [11].

Similarly, in Sect. 9, we begin with the linear Stieltjes integral equation

$$\begin{aligned} z(t)=z(t_0)+\int _{t_0}^t \left( p(s)z(s)+f(s)\right) \,\textrm{d}g(s), \end{aligned}$$

perform the substitution $y(t)=z(t)^{-1/\alpha }$, and obtain the Stieltjes-integral version of the Bernoulli equation. The results of Sects. 5 and 9 are again based on the generalized chain rules with the outer function $h(x)=x^{\alpha }$. They are inspired by a similar treatment of the Bernoulli equation in the context of dynamic equations on time scales (see [1, Section 2.6]).

Certain parts of the paper can be read independently. Readers interested mainly in the chain rules themselves can focus only on Sects. 3 and 7 (and perhaps consult the preliminaries in Sects. 2 and 6). Readers who are concerned only with Stieltjes derivatives and Stieltjes differential equations are invited to check Sects. 2–5, while readers interested only in Stieltjes integral equations might skip to Sects. 6–9.

Some basic familiarity with Stieltjes derivatives and Stieltjes integral will be helpful, but we have tried to make the paper self-contained by recalling the necessary basic facts in Sects. 2 and 6.

2 Preliminaries on Stieltjes Derivatives

Let $g:{\mathbb {R}}\rightarrow {\mathbb {R}}$ be a nondecreasing and left-continuous function. We shall denote by $\mu _g$ the Lebesgue–Stieltjes measure associated to g given by

$$\begin{aligned} \mu _g([a,b))=g(a)-g(b),\quad a,b\in {\mathbb {R}},\ a<b; \end{aligned}$$

for more information on measure theory and the Lebesgue–Stieltjes measure, we refer the reader to [3, 16, 17]. We denote by ${\mathcal {L}}^1_{g}(X,{\mathbb {R}})$ the set of Lebesgue–Stieltjes $\mu _g$-integrable functions on a $\mu _g$-measurable set X with values in ${\mathbb {R}}$, whose integral we denote by

$$\begin{aligned} \int _X f(s)\,\textrm{d}\mu _g(s),\quad f\in {\mathcal {L}}^1_{g}(X,{\mathbb {R}}). \end{aligned}$$

Throughout this paper, we shall talk about properties holding g-almost everywhere in a set X, or holding for g-almost all (or simply, g-a.a.) $x\in X$, as a simplified way to express that they hold $\mu _g$-almost everywhere in X or for $\mu _g$-almost all $x\in X$, respectively.

Define

$$\begin{aligned} C_g&=\{ t \in {\mathbb {R}} \, : \, g \, \text{ is } \text{ constant } \text{ on } \, (t-\varepsilon ,t+\varepsilon ) \, \text{ for } \text{ some }\, \varepsilon>0 \},\\ D_g&=\{ t \in {\mathbb {R}} \, : \, \Delta ^+ g(t)>0\}, \end{aligned}$$

where $\Delta ^+g(t)=g(t+)-g(t)$, $t\in {\mathbb {R}}$. Observe that, as pointed out in [7], $\mu _g(\{t\})>0$ for each $t\in D_g$, whereas the set $C_g$ has null $\mu _g$-measure. Furthermore, $C_g$ is open in the usual topology, so it can be uniquely expressed as the countable union of open disjoint intervals, say

$$\begin{aligned} C_g=\bigcup _{n\in {\mathbb {N}}} (a_n,b_n), \end{aligned}$$

(2.1)

for some $a_n, b_n\in [-\infty ,+\infty ]$, $n\in {\mathbb {N}}$. With this notation, we also define

$$\begin{aligned}{} & {} N_g^-=\{a_n\in {\mathbb {R}}: n\in {\mathbb {N}}\}\setminus D_g,\quad N_g^+=\{b_n\in {\mathbb {R}}:n\in {\mathbb {N}}\}\setminus D_g,\quad \\{} & {} N_g=N_g^-\cup N_g^+. \end{aligned}$$

We now introduce the Stieltjes derivative of a real-valued function as in [7, 8].

Definition 2.1

Let $f:{\mathbb {R}}\rightarrow {\mathbb {R}}$ and $t\in {\mathbb {R}}{\setminus } C_g$. We define the Stieltjes derivative, or g-derivative, of f at t as follows, provided the corresponding limit exists:

$$\begin{aligned} f'_g(t)=\left\{ \begin{array}{rl} \displaystyle \lim _{s \rightarrow t}\frac{f(s)-f(t)}{g(s)-g(t)},\quad &{} t\not \in D_{g}\cup N_g,\\ \displaystyle \lim _{s\rightarrow t-}\frac{f(s)-f(t)}{g(s)-g(t)},\quad &{} t\in N_g^-,\\ \displaystyle \lim _{s\rightarrow t+}\frac{f(s)-f(t)}{g(s)-g(t)},\quad &{} t\in D_{g}\cup N_g^+, \end{array} \right. \end{aligned}$$

In that case, we say that f is g-differentiable at t.

Remark 2.2

It is important to note that, as explained in [8, Remark 2.2], for $t\in N_g$ we have

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

as the domain of the quotient function gives the corresponding one-sided limit. Furthermore, since g is a regulated function, it follows that the g-derivative of f at a point $t\in D_g$ exists if and only if $f(t+)$ exists and, in that case,

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

Remark 2.3

Observe that the definition of Stieltjes derivative presented above excludes the points of $C_g$. A possible definition for such points can be given as follows ([4, Definition 3.7]): if $t\in (a_n,b_n)\subset C_g$ for $a_n,b_n$ as in (2.1), then we define $f'_g(t)$ as

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

provided the limit exists (in other words, $f'_g(t)=f'_g(b_n)$ for the corresponding $b_n$). In this paper we will restrict ourselves to the definition of derivative given in Definition 2.1, despite the fact that it is less general. This is because, later, we shall study differential equations which need only be satisfied g-almost everwhere on an interval so, since $\mu _g(C_g)=0$, it is not essential to have the derivative defined at such points. Nevertheless, occasional remarks for the derivative on the points of $C_g$ will be presented when they are relevant.

For some basic properties of the Stieltjes derivative such as linearity or the product and quotient rule, we refer the reader to [7,8,9].

Finally, we present the concept of g-absolute continuity introduced in [7], as well as some of its properties. For simplicity, we introduce such concept as part of the following result, [7, Proposition 5.4].

Theorem 2.4

Let $F:[a,b]\rightarrow {\mathbb {R}}$. The following conditions are equivalent:

• 1. The function F is g–absolutely continuous on [a, b] according to the following definition: for every $\varepsilon >0$, there exists $\delta >0$ such that for every open pairwise disjoint family of subintervals $\{(a_n,b_n)\}_{n=1}^m$ satisfying

  $$\begin{aligned} \sum _{n=1}^m (g(b_n)-g(a_n))<\delta , \end{aligned}$$

  we have

  $$\begin{aligned} \sum _{n=1}^m |F(b_n)-F(a_n)|<\varepsilon . \end{aligned}$$

• 2. The function F satisfies the following conditions:

  • (i) there exists $F'_g(t)$ for g-a.a. $t\in [a,b)$;

  • (ii) $F'_g\in {\mathcal {L}}^1_{g}([a,b),{\mathbb {R}})$;

  • (iii) for each $t\in [a,b]$,

    $$\begin{aligned} F(t)=F(a)+\int _{[a,t)}F'_g(s)\,\textrm{d}\mu _g(s). \end{aligned}$$

Remark 2.5

In what follows, we denote by $\mathcal{A}\mathcal{C}_g([a,b],{\mathbb {R}})$ the set of g-absolutely continuous functions in [a, b] with values on ${\mathbb {R}}$.

The following result is a direct consequence of [5, Propositions 3.2 and 5.5] and it contains some basic properties of g-absolutely continuous functions that will be useful for the work ahead.

Proposition 2.6

If $f\in \mathcal{A}\mathcal{C}_g([a,b],{\mathbb {R}})$, then f is bounded on [a, b]. Moreover:

• The map f is continuous from the left at every $t\in (a,b]$.

• If g is continuous at $t\in [a,b)$, then so is f.

• If g is constant on some $[c,d]\subset [a,b]$, then so is f.

The set of g-absolutely continuous functions is fundamental for the study of differential equations in the context of Stieltjes derivatives. In particular, if we consider initial value problems of the form

$$\begin{aligned} x'_g(t)=F(t,x(t)),\quad \quad x(t_0)=x_0, \end{aligned}$$

(2.2)

with $t_0,T,x_0\in {\mathbb {R}}$, $T>0$, and $F:[t_0,t_0+T]\times {\mathbb {R}}\rightarrow {\mathbb {R}}$, we have the following definition of solution, see [5, Definition 5.7].

Definition 2.7

Given $\tau \in (0,T]$, a solution of (2.2) on $[t_0,t_0+\tau ]$ is function $x\in \mathcal{A}\mathcal{C}_g([t_0,t_0+\tau ],{\mathbb {R}})$ such that $x(t_0)=x_0$ and

$$\begin{aligned} x'_g(t)=F(t,x(t)),\quad g\text{-a.a. } t\in [t_0,t_0+\tau ). \end{aligned}$$

3 Chain Rule for Stieltjes Derivatives

The chain rule is a fundamental tool in the study of derivatives. In the context of the Stieltjes derivatives, there are two versions of the chain rule for derivatives for the points of ${\mathbb {R}}{\setminus } (C_g\cup D_g)$, nevertheless no information is available for the points of $D_g$. The aim of this section is to establish a formulation of the chain rule that also includes these points.

We start by recalling the two versions of the chain rule for Stieltjes derivatives mentioned in the introduction, which can be found in [9, Proposition 3.15]. The next result is a refinement of [7, Theorem 2.3, statement 2] as the formulation presented there was missing condition (3.1) which, as indicated in [9, Remark 3.16], is necessary for the second formulation to be true.

Proposition 3.1

Let f be a real-valued function defined in a neighborhood of $t\in {\mathbb {R}}\setminus (C_g\cup D_g)$ and let h be another real-valued function defined in a neighborhood of f(t). Then:

• 1. If f is g-differentiable at t and h is differentiable (in the usual sense) at f(t), then $h\circ f$ is g-differentiable at t and

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

• 2. If f and h are g-differentiable at t and f(t), respectively; g is differentiable (in the usual sense) at f(t) and

  $$\begin{aligned} g(s)\not =g(f(t)),\quad \text{ for } \text{ all } s\in {\mathbb {R}},\ s\not =f(t), \end{aligned}$$

  (3.1)

  then $h\circ f$ is g-differentiable at t and

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

In what follows, we shall provide alternative formulations of the two versions of the chain rule above that can be applied to all the points except those of $C_g$, i.e., all the points where we can consider the Stieltjes derivative. We start by extending the result in which the inner function is differentiable in the Stieltjes sense and the outer one is differentiable in the usual sense, thus generalizing the first version of the chain rule in Proposition 3.1. This result and its proof are inspired by an existing version of the chain rule for the $\Delta$-derivative on time scales, see [2, Theorem 1.90].

Theorem 3.2

Let f be a real-valued function defined in a neighborhood of $t\in {\mathbb {R}}\setminus C_g$, denoted by $U_t$; and $h\in \mathcal C^1(I,{\mathbb {R}})$, where I denotes the closed convex hull of $f(U_t)$. If f is g-differentiable at t, then so is $h\circ f$ and

$$\begin{aligned} (h\circ f)'_g(t)=f'_g(t)\int _0^1 h'(f(t)+\tau f'_g(t)\Delta ^+ g(t))\,\textrm{d}\tau . \end{aligned}$$

(3.2)

Proof

Let $t\in {\mathbb {R}}\setminus C_g$. Observe that the expression inside the integral of (3.2) makes sense. Indeed, if $t\not \in D_g$ we have that $f(t)+\tau f'_g(t)\Delta ^+ g(t)=f(t)\in I$ for all $\tau \in [0,1]$; while whenever $t\in D_g$, Remark 2.2 ensures that

$$\begin{aligned} f(t)+\tau f'_g(t)\Delta ^+ g(t)=\tau f(t+)+(1-\tau )f(t),\quad \tau \in [0,1]; \end{aligned}$$

so we have that these values belong to I for all $\tau \in [0,1]$ since they represent the line segment connecting f(t) and $f(t+)$, both of which belong to I by definition.

If $t\not \in D_g$, then $\Delta ^+ g(t)=0$ and Proposition 3.1 ensures that $h\circ f$ is g-differentiable at t and, furthermore,

$$\begin{aligned} (h\circ f)'_g(t)=&f'_g(t)h'(f(t))=f'_g(t)\int _0^1h'(f(t))\,\textrm{d}\tau \\ =&f'_g(t)\int _0^1h'(f(t)+\tau f'_g(t)\Delta ^+ g(t))\,\textrm{d}\tau . \end{aligned}$$

On the other hand, if $t\in D_g$, since h is continuous at $f(t+)$, we have that

$$\begin{aligned} \lim _{s\rightarrow t+} (h\circ f)(s)=h\left( \lim _{s\rightarrow t+}f(s)\right) =h(f(t+)). \end{aligned}$$

Hence, since $(h\circ f)(t+)$ exists, so does $(h\circ f)'_g(t)$ (see Remark 2.2) and

$$\begin{aligned} (h\circ f)'_g(t)=\frac{\Delta ^+(h\circ f)(t)}{\Delta ^+ g(t)}. \end{aligned}$$

Now, if $f(t+)=f(t)$, then $\Delta ^+ f(t)=\Delta ^+ (h\circ f)(t)=0$ which implies that $f'_g(t)=(h\circ f)'_g(t)=0$ so (3.2) is trivially satisfied. Otherwise, we must have that $f(t+)\not =f(t)$ and, in that case,

$$\begin{aligned} (h\circ f)'_g(t)=\frac{\Delta ^+(h\circ f)(t)}{\Delta ^+ g(t)}=\frac{\Delta ^+f(t)}{\Delta ^+ g(t)}\frac{\Delta ^+(h\circ f)(t)}{\Delta ^+ f(t)}=f'_g(t)\frac{\Delta ^+(h\circ f)(t)}{\Delta ^+ f(t)}. \end{aligned}$$

Now, since $h\in {\mathcal {C}}^1(I,{\mathbb {R}})$, we have that

$$\begin{aligned} \Delta ^+(h\circ f)(t)&=\int _{f(t)}^{f(t+)}h'(r)\,\textrm{d}r\nonumber \\&=\Delta ^+f(t)\int _0^1h'(f(t)+\tau \Delta ^+f(t))\,\textrm{d}\tau \nonumber \\&=\Delta ^+f(t)\int _0^1h'(f(t)+\tau f'_g(t)\Delta ^+g(t))\,\textrm{d}\tau , \end{aligned}$$

(3.3)

where the last equality follows from Remark 2.2. Hence,

$$\begin{aligned} (h\circ f)'_g(t)=f'_g(t)\frac{\Delta ^+(h\circ f)(t)}{\Delta ^+ f(t)}=f'_g(t)\int _0^1 h'(f(t)+\tau f'_g(t)\Delta ^+ g(t))\,\textrm{d}\tau , \end{aligned}$$

which finishes the proof. $\square$

Remark 3.3

Looking at the proof of Theorem 3.2, it is easy to see that it is possible to relax the hypotheses on the map h. In particular, if $t\not \in D_g$ or $t\in D_g$ and $f(t+)=f(t)$, it is enough to assume that h is differentiable (in the usual sense) at f(t); whereas for the remaining case ($t\in D_g$, $f(t+)\not =f(t)$), it suffices to assume that h is absolutely continuous (in the usual sense) on the interval with end points f(t) and $f(t+)$ to ensure that the equality (3.3) holds. Here, we have opted for this formulation to provide the reader with a simple and comprehensive statement of the chain rule that applies to all points in ${\mathbb {R}}\setminus C_g$.

Remark 3.4

It is possible to extend Theorem 3.2 by including the points of $C_g$ for the definition of derivative mentioned in Remark 2.3 using the notation introduced in [4, Proposition 3.9]: given $t\in {\mathbb {R}}$, denote

$$\begin{aligned} t^*=\left\{ \begin{array}{ll} t,\quad &{} t\not \in C_g,\\ b_n,\quad &{} t\in (a_n,b_n)\subset C_g, \end{array} \right. \end{aligned}$$

where $a_n, b_n$ are as in (2.1). In that case, given $t\in {\mathbb {R}}$, if f is real-valued function defined in a neighborhood of $t^*$ containing t, denoted by $U_t^*$; f is g-differentiable at t, and $h\in {\mathcal {C}}^1(I,{\mathbb {R}})$, where I denotes the closed convex hull of $f(U_t^*)$, then $h\circ f$ is g-differentiable at t and

$$\begin{aligned} (h\circ f)'_g(t)=f'_g(t)\int _0^1 h'(f(t^*)+\tau f'_g(t)\Delta ^+ g(t^*))\,\textrm{d}\tau . \end{aligned}$$

As we mentioned before, Theorem 3.2 serves as a generalization of the first statement of Proposition 3.1 to the points of $D_g$. It is natural, then, to wonder if it is possible to extend the second statement in Proposition 3.1 in some way to also include the discontinuity points of the map g. This is the content of the following result.

Theorem 3.5

Let f be a real-valued function defined in a neighborhood of $t\in {\mathbb {R}}\setminus C_g$, denoted by $U_t$; and denote by I the closed convex hull of $f(U_t)$. If g is continuously differentiable (in the usual sense) and injective in I, f is g-differentiable at t; h is g-differentiable at every $s\in I$ and $h'_g$ is continuous (in the usual sense) on I, then $h\circ f$ is g-differentiable at t and

$$\begin{aligned} (h\circ f)'_g(t)=f'_g(t)\int _0^1 h'_g(f(t)+\tau f'_g(t)\Delta ^+ g(t))g'(f(t)+\tau f'_g(t)\Delta ^+ g(t))\,\textrm{d}\tau . \end{aligned}$$

Proof

First observe that it is enough to show that

$$\begin{aligned} h'(s)=h'_g(s)g'(s),\quad s\in I, \end{aligned}$$

(3.4)

to prove the result. Indeed, if (3.4) holds, then the hypotheses guarantee that $h\in {\mathcal {C}}^1(I,{\mathbb {R}})$ and so Theorem 3.2 guarantees that $h\circ f$ is g-differentiable at t and

$$\begin{aligned} (h\circ f)'_g(t)&=f'_g(t)\int _0^1 h'(f(t)+\tau f'_g(t)\Delta ^+ g(t))\,\textrm{d}\tau \\&=f'_g(t)\int _0^1 h'_g(f(t)+\tau f'_g(t)\Delta ^+ g(t))g'(f(t)+\tau f'_g(t)\Delta ^+ g(t))\,\textrm{d}\tau , \end{aligned}$$

where the last equality follows from the fact that $\{ f(t)+f'_g(t)\Delta ^+ g(t): \tau \in [0,1]\}\subset I$ (this can be deduced following a similar reasoning to the one used in the proof of Theorem 3.2).

In order to prove (3.4), suppose that s is an interior point of I. Since g is injective on I, we have that $I\cap C_g=\emptyset$; while the fact that g is continuous on I guarantees that $I\cap D_g=\emptyset$. Hence, since h is g-differentiable at s, by definition, the following limit exists and

$$\begin{aligned} \lim _{r\rightarrow s}\frac{h(r)-h(s)}{g(r)-g(s)}=h'_g(s). \end{aligned}$$

Similarly, since g is differentiable (in the usual sense) at s, the following limit exists and

$$\begin{aligned} \lim _{r\rightarrow s}\frac{g(r)-g(s)}{r-s}=g'(s). \end{aligned}$$

Now, since g is injective on I, it holds that

$$\begin{aligned} \frac{h(r)-h(s)}{r-s}=\frac{h(r)-h(s)}{g(r)-g(s)}\frac{g(r)-g(s)}{r-s},\quad r\in I, r\not =s. \end{aligned}$$

Hence,

$$\begin{aligned} \lim _{r\rightarrow s}\frac{h(r)-h(s)}{r-s}= & {} \lim _{r\rightarrow s}\left( \frac{h(r)-h(s)}{g(r)-g(s)}\frac{g(r)-g(s)}{r-s}\right) \\= & {} \lim _{r\rightarrow s}\frac{h(r)-h(s)}{g(r)-g(s)}\lim _{r\rightarrow s}\frac{g(r)-g(s)}{r-s}=h'_g(s)g'(s), \end{aligned}$$

which shows that h is differentiable at t and (3.4) holds. The proof of (3.4) for $s=\min I$ and $s=\max I$ is analogous considering the corresponding one-sided limits and, thus, we omit it. $\square$

Remark 3.6

Note that the formulation of Theorem 3.5 allows the point t at which f is g-differentiable to belong to $D_g$ even though the map g is continuously differentiable in I since, in general, $t\not \in I$. Furthermore, the fact that g is injective in I is fundamental for Theorem 3.5 as it guarantees that I contains no points of $C_g$ which, in turn, allows us to consider the Stieltjes derivative of h for all the points of I. Moreover, it is also possible to obtain a formulation of the result including the elements of $C_g$ in a similar fashion to Remark 3.4: given $t\in {\mathbb {R}}$, if f is real-valued function defined in a neighborhood of $t^*$ containing t, denoted by $U_t^*$; $g\in {\mathcal {C}}^1(I,{\mathbb {R}})$ where I denotes the closed convex hull of $f(U_t^*)$; f is g-differentiable at t; h is g-differentiable at every $s\in I$ and $h'_g$ is continuous (in the usual sense) on I, then $h\circ f$ is g-differentiable at t and

$$\begin{aligned} (h\circ f)'_g(t)=f'_g(t)\int _0^1 h'_g(f(t^*)+\tau f'_g(t)\Delta ^+ g(t^*))g'(f(t^*)+\tau f'_g(t)\Delta ^+ g(t^*))\,\textrm{d}\tau . \end{aligned}$$

As a final note to this section, we use Theorem 3.2 to compute the Stieltjes derivatives of the different real powers of a given function. This is fundamental for the work in the following sections, in particular, for the study of certain differential equations. For that reason, we present the result in the context of g-absolutely continuous functions.

Lemma 3.7

Let $f\in \mathcal{A}\mathcal{C}_g([a,b],{\mathbb {R}})$, $\alpha \in {\mathbb {R}}$, and consider the map $f^\alpha :[a,b]\rightarrow {\mathbb {R}}$ defined as

$$\begin{aligned} f^\alpha (t)=(f(t))^\alpha ,\quad t\in [a,b]. \end{aligned}$$

• 1. If $\alpha \in {\mathbb {N}}$, the map $f^\alpha$ is well-defined, g-absolutely continuous on [a, b] and, furthermore,

  $$\begin{aligned} (f^\alpha )'_g(t)=\alpha f'_g(t)\int _0^1(f(t)+\tau f'_g(t)\Delta ^+ g(t))^{\alpha -1}\,\textrm{d}\tau ,\quad g\text{-a.a. } t\in [a,b). \end{aligned}$$

  (3.5)

  If, in addition, $f(t)\not =0$ for g-almost all $t\in [a,b]$, then

  $$\begin{aligned} \frac{(f^\alpha )'_g(t)}{f^\alpha (t)}=\alpha \frac{f'_g(t)}{f(t)}\int _0^1\left( 1+\tau \frac{f'_g(t)}{f(t)}\Delta ^+ g(t)\right) ^{\alpha -1}\textrm{d}\tau ,\quad g\text{-a.a. } t\in [a,b). \end{aligned}$$

  (3.6)

• 2. If $\alpha \in {\mathbb {R}}\setminus {\mathbb {N}}$ and there exists $M>0$ such that $f(t)>M$ for all $t\in [a,b]$ then, the map $f^\alpha$ is well-defined, g-absolutely continuous on [a, b] and (3.5) and (3.6) hold.

• 3. If $\alpha \in {\mathbb {R}}\setminus {\mathbb {N}}$, $f(t)>0$ for all $t\in [a,b]$ and

  $$\begin{aligned} f(t)+f'_g(t)\Delta ^+ g(t)\not =0,\quad t\in [a,b]\cap D_g, \end{aligned}$$

  (3.7)

  then, the map $f^\alpha$ is well-defined, g-absolutely continuous on [a, b] and (3.5) and (3.6) hold.

Proof

First note that, under the corresponding hypotheses in 1–3, the map $f^\alpha$ is well-defined, so we only need to show that it is g-absolutely continuous and (3.5) and (3.6) hold.

First, we prove 1 and 2. If $\alpha =0$ the result holds trivially, so we assume $\alpha \not =0$. Since $f\in \mathcal{A}\mathcal{C}_g([a,b],\mathbb R)$, Proposition 2.6 ensures there exists $K>0$ such that $f([a,b])\subset [-K,K]$ and, in particular, if $\alpha \in {\mathbb {R}}\setminus {\mathbb {N}}$, the hypotheses ensure that $f([a,b])\subset [M,K]\subset (0,+\infty )$. Now (3.5) follows from Theorem 3.2 by noting that the map $h(t)=t^\alpha$ is continuously differentiable on $(0,+\infty )$ if $\alpha \in \mathbb R{\setminus }{\mathbb {N}}$; and for $\alpha \in {\mathbb {N}}$, it is continuously differentiable on ${\mathbb {R}}$. Basic computations are enough to obtain (3.6) from (3.5). Finally, $f^\alpha \in \mathcal{A}\mathcal{C}_g([a,b],{\mathbb {R}})$ as it is the composition of a g-absolutely continuous map, f, with h, which satisfies a Lipschitz condition on $[-K,K]$ for $\alpha \in \mathbb N$ and on [M, K] for $\alpha \in {\mathbb {R}}{\setminus } {\mathbb {N}}$, see [5, Proposition 5.3].

Finally, 3 is a particular case of 2. Indeed, if $f\in \mathcal{A}\mathcal{C}_g([a,b],{\mathbb {R}})$ is positive function, it follows from Proposition 2.6 that $f(t-)>0$, $t\in (a,b]$, and $f(t+)>0$, $t\in [a,b)\setminus D_g$. Meanwhile, for $t\in [a,b)\cap D_g$, the definition of derivative (see Remark 2.2) and (3.7) ensure that $f(t+)=f(t)+f'_g(t)\Delta ^+ g(t)\not =0$. This means that the conditions of [10, Lemma 2.7] are satisfied and, thus, there exists $M>0$ such that $f(t)=|f(t)|>M$ for all $t\in [a,b]$, so 3 follows from 2. $\square$

4 Operations with the g-Exponential Function

In the context of ordinary differential equations, given $p\in {\mathcal {L}}^1([a,b],{\mathbb {R}})$, the map $\exp (p,\cdot ):[a,b]\rightarrow {\mathbb {R}}$ defined as

$$\begin{aligned} \exp (p,t)=\exp \left( \int _a^t p(s)\, \textrm{d}s\right) ,\quad t\in [a,b], \end{aligned}$$

is the unique solution of

$$\begin{aligned} x'(t)=p(t)x(t),\quad \quad x(a)=1. \end{aligned}$$

Given the particular properties of the exponential map, we have that

$$\begin{aligned} \exp (p+q,\cdot )&=\exp (p,\cdot )\exp (q,\cdot ),\quad \quad \quad p,q\in {\mathcal {L}}^1([a,b],{\mathbb {R}}), \end{aligned}$$

(4.1)

$$\begin{aligned} \exp (\alpha p,\cdot )&=(\exp (p,\cdot ))^\alpha ,\quad \quad \quad \quad \quad \ p\in {\mathcal {L}}^1([a,b],{\mathbb {R}}),\ \alpha \in \mathbb R. \end{aligned}$$

(4.2)

In other words, $\exp (p,\cdot )\exp (q,\cdot )$ and $(\exp (p,\cdot ))^\alpha$ are, respectively, the solutions of

$$\begin{aligned} x'(t)&=(p(t)+q(t))x(t),\quad \quad x(a)=1,\\ x'(t)&= \alpha p(t)x(t),\quad \quad \quad \quad \quad x(a)=1. \end{aligned}$$

For differential equations with Stieltjes derivatives there exists a generalization of the exponential map above, called the g-exponential map. For the explicit formulas, we refer the reader to [8, Theorem 3.4] or, more generally, [4, Theorem 4.3]. For the purposes of this paper, it is enough to consider the following definition of the g-exponential map.

Definition 4.1

Given $p \in {{\mathcal {L}}}^1_g([a,b),{\mathbb {R}})$, we define the g-exponential map on [a, b] associated to p, denoted by $\exp _g(p,\cdot )$, as the unique solution (in the sense of Definition 2.7) on [a, b] of the homogeneous linear initial value problem

$$\begin{aligned} x'_g(t)=p(t)x(t),\quad x(a)=1. \end{aligned}$$

(4.3)

Remark 4.2

Note that [5, Theorem 7.3] ensures that (4.3) has, indeed, a unique solution. Furthermore, given [8, Theorem 3.4], we see that if $p \in {\mathcal L}^1_g([a,b),{\mathbb {R}})$ is such that

$$\begin{aligned} 1+p(t)\Delta ^+ g(t)\not =0,\quad \text{ for } \text{ all } t\in [a,b)\cap D_g, \end{aligned}$$

(4.4)

then $\exp _g(p,t)\not =0$ for all $t\in [a,b]$ and, if in addition,

$$\begin{aligned} 1+p(t)\Delta ^+ g(t)>0,\quad \text{ for } \text{ all } t\in [a,b)\cap D_g, \end{aligned}$$

then [8, Theorem 3.2] implies that $\exp _g(p,t)>0$ for all $t\in [a,b]$.

Standard computations with the explicit expression of the g-exponential map show that (4.1) and (4.2) do not hold in this context, so the idea behind this section is to define two operations, $\oplus$ and $\odot$, such that

$$\begin{aligned} \exp _g(p\oplus q,\cdot )&=\exp _g(p,\cdot )\exp _g(q,\cdot ),\quad \quad \quad \ p,q\in {\mathcal {L}}^1_g([a,b],{\mathbb {R}}), \end{aligned}$$

(4.5)

$$\begin{aligned} \exp _g(\alpha \odot p,\cdot )&=(\exp _g(p,\cdot ))^\alpha ,\quad \quad \quad \quad \quad \quad p\in {\mathcal {L}}^1_g([a,b],{\mathbb {R}}),\ \alpha \in \mathbb R. \end{aligned}$$

(4.6)

Equivalently, we want the maps $\exp _g(p,\cdot )\exp _g(q,\cdot )$ and $(\exp _g(p,\cdot ))^\alpha$ to be, respectively, the solutions of

$$\begin{aligned} x'_g(t)&=(p\oplus q)(t)x(t),\quad \quad \quad x(a)=1,\nonumber \\ x'_g(t)&= (\alpha \odot p)(t)x(t),\quad \quad \quad x(a)=1. \end{aligned}$$

(4.7)

If we expect $x(t)=(\exp _g(p,\cdot ))^\alpha$ to be the solution of (4.7), we can use Lemma 3.7 to see that, under certain conditions,

$$\begin{aligned} (\alpha \odot p)(t)=\frac{x'_g(t)}{x(t)}=\alpha \frac{(\exp _g(p,\cdot ))'_g(t)}{\exp _g(p,t)}\int _0^1\left( 1+\tau \frac{(\exp _g(p,\cdot ))'_g(t)}{\exp _g(p,t)}\Delta ^+ g(t)\right) ^{\alpha -1}\textrm{d}\tau , \end{aligned}$$

so, using the fact that $\exp _g(p,\cdot )$ is the unique solution of (4.3), we can obtain the explicit expression of $\alpha \odot p$. A similar reasoning using the product rule (cf. [8, Proposition 2.5, (i)]) instead of Lemma 3.7 can be used to obtain the expression for $\oplus$. Alternatively, we can deduce it directly from (4.5) and the following result containing some basic properties of the g-exponential map, see [4, Proposition 4.6].

Proposition 4.3

Let $p, q \in {{\mathcal {L}}}^1_g([a,b),{\mathbb {R}})$. Then:

• 1. The map $p+q+pq\Delta ^+g$ belongs to ${{\mathcal {L}}}^1_g([a,b),{\mathbb {R}})$ and, furthermore,

  $$\begin{aligned} \exp _g(p,t)\exp _g(q,t)=\exp _g\left( p+q+pq\Delta ^+g,t\right) ,\quad t\in [a,b]. \end{aligned}$$

• 2. For each $n\in {\mathbb {N}}$ and $t\in [a,b]$, $(\exp _g(p,t))^n=\exp _g(p_n,t)$, where

  $$\begin{aligned} p_n(t)=np(t)+\sum _{k=2}^n \left( {\begin{array}{c}n\\ k\end{array}}\right) p(t)^k\Delta ^+ g(t)^{k-1},\quad t\in [a,b]. \end{aligned}$$

• 3. For each $n\in {\mathbb {N}}$ and $t\in [a,b]$ such that $\exp _g(p,t)\not =0$, $(\exp _g(p,t))^{-n}=\exp _g(q_n,t)$, where

  $$\begin{aligned} q_n(t)=-\frac{p_n(t)}{1+p_n(t)\Delta ^+g(t)}. \end{aligned}$$

With these ideas in mind, we have the following definition of the operations $\oplus$ and $\odot$ on the set of regressive functions, i.e., the set of functions for which the corresponding g-exponential map is defined and does not vanish at any point. The following definitions are inspired by the corresponding definitions in [1, 2].

Definition 4.4

Consider the sets

$$\begin{aligned} {\mathcal {R}}_g([a,b),{\mathbb {R}})&=\left\{ p \in {\mathcal L}^1_g([a,b),{\mathbb {R}}) : 1+p(t)\Delta ^+ g(t)\not =0 \text{ for } \text{ all } t\in [a,b)\cap D_g\right\} ,\\ {\mathcal {R}}_g^+([a,b),{\mathbb {R}})&=\left\{ p \in {\mathcal L}^1_g([a,b),{\mathbb {R}}) : 1+p(t)\Delta ^+ g(t)>0 \text{ for } \text{ all } t\in [a,b)\cap D_g\right\} , \end{aligned}$$

which we shall refer to as the set of regressive functions on [a, b) and the set of positively regressive functions on [a, b), respectively.

We define the regressive sum on ${\mathcal {R}}_g([a,b),\mathbb R)$ as the binary operation $\oplus :\mathcal R_g([a,b),{\mathbb {R}})\times {\mathcal {R}}_g([a,b),{\mathbb {R}})\rightarrow \mathcal R_g([a,b),{\mathbb {R}})$ given by

$$\begin{aligned} (p\oplus q) (t)= p(t)+q(t)+p(t)q(t)\Delta ^+ g(t),\quad t\in [a,b). \end{aligned}$$

Similarly, we define the regressive scalar multiplication on ${\mathcal {R}}_g^+([a,b),{\mathbb {R}})$ as the map $\odot :\mathbb R\times {\mathcal {R}}_g^+([a,b),{\mathbb {R}})\rightarrow \mathcal R_g^+([a,b),{\mathbb {R}})$ given by

$$\begin{aligned} (\alpha \odot p)(t)=\alpha p(t)\int _0^1(1+\tau p(t)\Delta ^+ g(t))^{\alpha -1}\,\textrm{d}\tau ,\quad t\in [a,b). \end{aligned}$$

(4.8)

Remark 4.5

Observe that, in general, $\odot$ can only be defined on $\mathcal R_g^+([a,b),{\mathbb {R}})$ as for $\alpha \in {\mathbb {R}}{\setminus } \mathbb N$, it is necessary to have that $1+\tau p(t)\Delta ^+ g(t)>0$ for all $\tau \in [0,1]$ for the function in the integral in (4.8) to be well-defined. Nevertheless, it is possible to consider the regressive scalar multiplication on $\mathcal R_g([a,b),{\mathbb {R}})$ as the map $\odot :{\mathbb {N}}\times {\mathcal {R}}_g([a,b),{\mathbb {R}})\rightarrow {\mathcal {R}}_g([a,b),{\mathbb {R}})$ given by (4.8).

A similar definition of regressive function can be found in [6, Section 2.2], where they do not require the map p to be $\mu _g$-integrable, but instead impose the condition

$$\begin{aligned} \sum _{t\in [a,b)\cap D_g} |\log |1+p(t)\Delta ^+g(t)||<+\infty . \end{aligned}$$

(4.9)

However, as pointed out in [8, Lemma 3.1], if p is a $\mu _g$-integrable map satisfying (4.4), we have that (4.9) is satisfied. In that sense, our definition of regressive functions is more restrictive. Nevertheless, given the objectives of this section, we are interested in considering the g-exponential maps of regressive functions, which is why impose the integrability condition instead of (4.9).

In the following result we show that the class of regressive functions is, in fact, closed under regressive sums and regressive scalar multiplications.

Proposition 4.6

The operations $\oplus$ and $\odot$ are well-defined.

Proof

In order to prove that the operations are well-defined, we need to show that $p\oplus q\in {\mathcal {R}}_g([a,b),{\mathbb {R}})$ for $p,q\in {\mathcal {R}}_g([a,b),{\mathbb {R}})$, and $\alpha \odot p\in \mathcal R_g^+([a,b),{\mathbb {R}})$ for $\alpha \in {\mathbb {R}}$ and $p\in \mathcal R_g^+([a,b),{\mathbb {R}})$.

Let $p,q\in {\mathcal {R}}_g([a,b),{\mathbb {R}})$ and let us show that $p\oplus q\in {\mathcal {R}}_g([a,b),{\mathbb {R}})$. First, statement (i) in Proposition 4.3 guarantees that $p\oplus q\in {\mathcal L}^1_g([a,b),{\mathbb {R}})$.

Now, since $p,q\in {\mathcal {R}}_g([a,b),{\mathbb {R}})$, given $t\in [a,b)\cap D_g$,

$$\begin{aligned} 1+(p\oplus q)(t)\Delta ^+ g(t)&=1+p(t)\Delta ^+ g(t)+q(t)\Delta ^+ g(t)+p(t)q(t)(\Delta ^+ g(t))^2\nonumber \\&=(1+p(t)\Delta ^+ g(t))(1+q(t)\Delta ^+ g(t))\not =0, \end{aligned}$$

(4.10)

which proves that $p\oplus q\in {\mathcal {R}}_g([a,b),{\mathbb {R}})$.

Consider $p\in {\mathcal {R}}_g^+([a,b),{\mathbb {R}})$ and $\alpha \in {\mathbb {R}}$. Let us show that $\alpha \odot p\in \mathcal R_g^+([a,b),{\mathbb {R}})$. First, note that for each $t\in [a,b)$, $1+p(t)\Delta ^+ g(t)>0$, so

$$\begin{aligned} 1+\tau p(t)\Delta ^+g(t)\ge \tau +\tau p(t)\Delta ^+g(t)=\tau (1+p(t)\Delta ^+g(t))\ge 0,\quad \tau \in [0,1], \end{aligned}$$

which shows that the integral in the definition of $\alpha \odot p$ is well-defined. Furthermore, it can be explicitly computed for each $t\in [a,b)$, which allows us to rewrite

$$\begin{aligned} (\alpha \odot p)(t) = {\left\{ \begin{array}{ll} \alpha p(t), &{}\text {if } \,t \in [a,b)\backslash D_g,\\ \frac{(1+p(t)\Delta ^+ g(t))^{\alpha }-1}{\Delta ^+ g(t)}, &{}\text {if}\, t \in [a,b)\cap D_g. \end{array}\right. } \end{aligned}$$

(4.11)

Hence, since $\alpha \odot p=\alpha p$ on $[a,b)\setminus D_g$ and $D_g$ is countable, it follows that $\alpha \odot p$ is $\mu _g$-measurable. Moreover, since $p\in \mathcal R_g^+([a,b),{\mathbb {R}})$, given $t\in [a,b)\cap D_g$,

$$\begin{aligned} 1+(\alpha \odot p)(t)\Delta ^+g(t)&=1+\frac{(1+p(t)\Delta ^+ g(t))^{\alpha }-1}{\Delta ^+ g(t)}\Delta ^+ g(t)\\&=(1+p(t)\Delta ^+ g(t))^{\alpha }>0, \end{aligned}$$

so it only remains to show that $\alpha \odot p\in {\mathcal L}^1_g([a,b),{\mathbb {R}})$ to prove that $\alpha \odot p\in \mathcal R_g^+([a,b),{\mathbb {R}})$. Observe that

$$\begin{aligned} \int _{[a,b)}|(\alpha \odot p)(s)|\,\text {d}\mu _g(s)=&\int _{[a,b)\setminus D_g}|(\alpha \odot p)(s)|\,\text {d}\mu _g(s)+\int _{[a,b)\cap D_g}|(\alpha \odot p)(s)|\,\text {d}\mu _g(s)\nonumber \\ =&\int _{[a,b)\setminus D_g}|\alpha p(s)|\,\text {d}\mu _g(s)\nonumber \\ {}&+\int _{[a,b)\cap D_g}\left| \frac{(1+p(s)\Delta ^+ g(s))^{\alpha }-1}{\Delta ^+ g(s)}\right| \,\text {d}\mu _g(s)\nonumber \\ =&|\alpha |\int _{[a,b)\setminus D_g}|p(s)|\,\text {d}\mu _g(s)\nonumber \\ {}&+\sum _{t\in [a,b)\cap D_g}\left| \frac{(1+p(t)\Delta ^+ g(t))^{\alpha }-1}{\Delta ^+ g(t)}\right| \Delta ^+g(t)\nonumber \\ =&|\alpha |\int _{[a,b)\setminus D_g}|p(s)|\,\text {d}\mu _g(s)+\sum _{t\in [a,b)\cap D_g}\left| (1+p(t)\Delta ^+ g(t))^{\alpha }-1\right| , \end{aligned}$$

(4.12)

so, since $p\in {{\mathcal {L}}}^1_g([a,b),{\mathbb {R}})$, it is enough to prove that the sum in (4.12) is finite to show that $\alpha \odot p$ is $\mu _g$-integrable on [a, b).

First, since $p\in {\mathcal {L}}^1_g([a,b),{\mathbb {R}})$, the map $p(t)\Delta ^+ g(t)$ is bounded by some constant, say $M>0$; and, moreover, the sum $\sum _{t\in [a,b)\cap D_g}|p(t)\Delta ^+ g(t)|$ is finite (cf. [11, Lemma 3.4]). Hence, if we consider the set $P_g=\{t\in [a,b)\cap D_g: p(t)\Delta ^+ g(t)\ge -1/2\}$, we necessarily have that $D_g{\setminus } P_g$ must be finite. Thus, in order to prove that the sum in (4.12) is finite, it suffices to show that the sum $\sum _{t\in [a,b)\cap P_g}\left| (1+p(t)\Delta ^+ g(t))^{\alpha }-1\right|$ is finite.

Observe that $1+p(t)\Delta ^+ g(t)\in [1/2, 1+M]$ for $t\in P_g$. At the same time, the map $h(x)=x^\alpha$ is Lipschitz continuous on $[1/2, 1+M]$ so, if $L>0$ denotes the corresponding Lipschitz constant,

$$\begin{aligned} \sum _{t\in [a,b)\cap P_g}\left| (1+p(t)\Delta ^+ g(t))^{\alpha }-1\right|\le & {} L\sum _{t\in [a,b)\cap P_g}\left| p(t)\Delta ^+ g(t)\right| \\\le & {} L\sum _{t\in [a,b)\cap D_g}\left| p(t)\Delta ^+ g(t)\right| <+\infty , \end{aligned}$$

which finishes the proof of the result. $\square$

Remark 4.7

Observe that it follows from (4.10) that if $p,q\in {\mathcal {R}}^+_g([a,b),{\mathbb {R}})$, then $p\oplus q\in \mathcal R^+_g([a,b),{\mathbb {R}})$.

Next, we verify that the operations $\oplus$ and $\odot$ have the properties we desired, namely, (4.5) and (4.6). To that end, we note that, under the corresponding hypotheses, (3.6) can be rewritten in terms of $\odot$ as

$$\begin{aligned} \frac{(f^\alpha )'_g(t)}{f^\alpha (t)}=\left( \alpha \odot \frac{f'_g}{f}\right) \hspace{-0.1cm}(t). \end{aligned}$$

(4.13)

Proposition 4.8

Let $p,q\in {\mathcal {R}}_g([a,b),{\mathbb {R}})$. Then:

• (i) For each $t\in [a,b)$,

  $$\begin{aligned}\exp _g(p,t)\cdot \exp _g(q,t)=\exp _g(p\oplus q, t).\end{aligned}$$

• (ii) If, in addition $p\in {\mathcal {R}}_g^+([a,b),{\mathbb {R}})$, then, for each $\alpha \in {\mathbb {R}}$,

  $$\begin{aligned}(\exp _g(p,t))^\alpha =\exp _g(\alpha \odot p,t),\quad t\in [a,b).\end{aligned}$$

Proof

First, observe that (i) follows directly from Proposition 4.3, so we shall focus on proving (ii). To that end, let $p\in \mathcal R_g^+([a,b),{\mathbb {R}})$ and $\alpha \in {\mathbb {R}}$. In that case, Proposition 4.6 ensures that $\alpha \odot p\in {\mathcal {R}}_g^+([a,b),{\mathbb {R}})$ so, by definition, $\exp _g(\alpha \odot p, \cdot )$ is the unique solution on [a, b] of

$$\begin{aligned} x'_g(t)=(\alpha \odot p)(t)x(t),\quad x(a)=1. \end{aligned}$$

(4.14)

Hence, it suffices to show that $x(t)=(\exp _g(p,t))^\alpha$ is a solution of (4.14) as, in that case, the equality for all points in [a, b] follows from the uniqueness of solution.

First, note that $x(a)=(\exp _g(p,a))^\alpha =1$. Furthermore, by definition, $\exp _g(p,\cdot )\in \mathcal{A}\mathcal{C}_g([a,b],{\mathbb {R}})$ and, as pointed out in Remark 4.2, $\exp _g(p,t)>0$ for $t\in [a,b]$. Moreover, for each $t\in [a,b]\cap D_g$, we have that

$$\begin{aligned} \exp _g(p,t)+(\exp _g(p,\cdot ))'_g(t)\Delta ^+ g(t)= & {} \exp _g(p,t)+p(t)\exp _g(p,t)\Delta ^+ g(t)\\= & {} \exp _g(p,t)(1+p(t)\Delta ^+g(t))>0, \end{aligned}$$

so Lemma 3.7 guarantees that $x\in \mathcal{A}\mathcal{C}_g([a,b],{\mathbb {R}})$ and

$$\begin{aligned} \frac{x'_g(t)}{x(t)}=\left( \alpha \odot \frac{(\exp _g(p,\cdot ))'_g}{\exp _g(p,\cdot )}\right) \hspace{-0.1cm}(t),\quad g\text{-a.a. } t\in [a,b). \end{aligned}$$

Equivalently, for g-a.a. $t\in [a,b)$,

$$\begin{aligned} x'_g(t)=x(t)\left( \alpha \odot \frac{(\exp _g(p,\cdot ))'_g}{\exp _g(p,\cdot )}\right) \hspace{-0.1cm}(t)=x(t)\left( \alpha \odot \frac{p\exp _g(p,\cdot )}{\exp _g(p,\cdot )}\right) \hspace{-0.1cm}(t)=(\alpha \odot p)(t)x(t), \end{aligned}$$

which finishes the proof of the result. $\square$

Remark 4.9

Statement (ii) in Proposition 4.8 is a generalization of statements 2 and 3 in Proposition 4.3, where similar properties are presented for integer powers of g-exponential map. First, we look at 2 in Proposition 4.3. Note that, in general, (iii) in Proposition 4.8 requires $p\in \mathcal R_g^+([a,b),{\mathbb {R}})$, however, since we are considering $n\in {\mathbb {N}}$, the same equality holds for $p\in \mathcal R_g([a,b),{\mathbb {R}})$, see Remark 4.5. Hence, all that is left to do is to show that, for $n\in {\mathbb {N}}$, $n\odot p=p_n$. Using (4.11), we see that for $t\in [a,b]{\setminus } D_g$, $(n\odot p)(t)=np(t)=p_n(t)$; while for $t\in [a,b]\cap D_g$,

$$\begin{aligned} (n\odot p)(t)&=\frac{(1+p(t)\Delta ^+g(t))^n-1}{\Delta ^+g(t)}=\frac{\sum _{k=0}^n\left( {\begin{array}{c}n\\ k\end{array}}\right) p(t)^k\Delta ^+ g(t)^{k} -1}{\Delta ^+g(t)}\\&=\frac{1+np(t)\Delta ^+g(t)+\sum _{k=2}^n\left( {\begin{array}{c}n\\ k\end{array}}\right) p(t)^k\Delta ^+ g(t)^{k}-1}{\Delta ^+g(t)}\\&=np(t)+\sum _{k=2}^n \left( {\begin{array}{c}n\\ k\end{array}}\right) p(t)^k\Delta ^+ g(t)^{k-1}=p_n(t), \end{aligned}$$

which shows that $n\odot p=p_n$ for each $n\in {\mathbb {N}}$. The reasoning for 3 in Proposition 4.3 is similar and we omit it.

The rest of this section is devoted to the study of the properties of the operations $\oplus$ and $\odot$, as well as the algebraic structure of the set of regressive functions. We begin by looking at the regressive sum.

Theorem 4.10

The pair $({\mathcal {R}}_g([a,b),{\mathbb {R}}), \oplus )$ is an abelian group, where the identity element is the map 0 and the inverse element of an element $p\in {\mathcal {R}}_g([a,b),{\mathbb {R}})$ is given by

$$\begin{aligned} \ominus p(t)=-\frac{p(t)}{1+p(t)\Delta ^+ g(t)},\quad t\in [a,b). \end{aligned}$$

(4.15)

Proof

In order to show that $({\mathcal {R}}_g([a,b),{\mathbb {R}}), \oplus )$ is an abelian group, it is enough to show that it is a group as the commutativity is straightforward from the definition.

Let $p,q,r\in {\mathcal {R}}_g([a,b),{\mathbb {R}})$. Then, for $t\in [a,b)$,

$$\begin{aligned} ((p\oplus q)\oplus r)(t)=&\,((p+q+pq\Delta ^+ g)\oplus r)(t)\\ =&\,p(t)+q(t)+p(t)q(t)\Delta ^+ g(t)+r(t)\\ {}&+(p(t)+q(t)\,+p(t)q(t)\Delta ^+ g(t))r(t)\Delta ^+g(t)\\ =&\,p(t)+q(t)+r(t)+q(t)r(t)\Delta ^+g(t)\\ {}&+p(t)(q(t)+r(t)\,+q(t)r(t)\Delta ^+ g(t))\Delta ^+g(t)\\ =&\,(p\oplus (q+r+qr\Delta ^+ g))(t)= (p\oplus (q\oplus r))(t), \end{aligned}$$

which proves the associativity property.

Next, it is clear that the map 0 belongs to $\mathcal R_g([a,b),{\mathbb {R}})$ and, furthermore,

$$\begin{aligned} (p\oplus 0)(t)=p(t)+0(t)+p(t)0(t)\Delta ^+ g(t)=p(t),\quad t\in [a,b), \end{aligned}$$

which shows that 0 is the identity element.

Now, consider $\ominus p$ given by (4.15). Observe that [11, Lemma 3.4] ensures that $\ominus p\in {\mathcal {L}}^1_g([a,b),{\mathbb {R}})$. Furthermore, given $t\in [a,b)\cap D_g$,

$$\begin{aligned} 1+\ominus p(t)\Delta ^+g(t)=1-\frac{p(t)}{1+p(t)\Delta ^+ g(t)}\Delta ^+ g(t)=\frac{1}{1+p(t)\Delta ^+ g(t)}\not =0,\nonumber \\ \end{aligned}$$

(4.16)

so $\ominus p\in {\mathcal {R}}_g([a,b),{\mathbb {R}})$. Moreover, for $t\in [a,b)$, we have that

$$\begin{aligned} (p\oplus (\ominus p))(t)&=p(t)+\ominus p(t)+p(t)(\ominus p(t))\Delta ^+ g(t)\\&=p(t)-\frac{p(t)}{1+p(t)\Delta ^+ g(t)}-p(t)\frac{p(t)}{1+p(t)\Delta ^+ g(t)}\Delta ^+g(t)\\&=\frac{p(t)+p(t)^2\Delta ^+g(t)-p(t)-p(t)^2\Delta ^+ g(t)}{1+p(t)\Delta ^+ g(t)}=0, \end{aligned}$$

which proves that $\ominus p$ is, indeed, the inverse element of p. $\square$

Remark 4.11

Given the definition of inverse element in $(\mathcal R_g([a,b),{\mathbb {R}}), \oplus )$ in (4.15), we define the regressive difference on ${\mathcal {R}}_g([a,b),{\mathbb {R}})$ as the map $\ominus : {\mathcal {R}}_g([a,b),{\mathbb {R}})\times \mathcal R_g([a,b),{\mathbb {R}})\rightarrow {\mathcal {R}}_g([a,b),{\mathbb {R}})$ given by

$$\begin{aligned} (p\ominus q)(t):= & {} \, (p\oplus (\ominus q))(t)\\= & {} p(t)-\frac{q(t)}{1+q(t)\Delta ^+ g(t)}-\,\frac{p(t)q(t)}{1+q(t)\Delta ^+ g(t)}\Delta ^+ g(t),\quad t\in [a,b). \end{aligned}$$

Observe that, in the particular case that $p\in \mathcal R_g^+([a,b),{\mathbb {R}})$, it follows from (4.16) that $\ominus p\in {\mathcal {R}}_g^+([a,b),{\mathbb {R}})$. Furthermore, it is clear from (ii) in Proposition 4.3 that $(\exp _g(p,\cdot ))^{-1}=\exp _g(\ominus p, \cdot )$ and, as a consequence,

$$\begin{aligned} \frac{\exp _g(p,\cdot )}{\exp _g(q,\cdot )}&=\exp _g\left( p \ominus q, \cdot \right) . \end{aligned}$$

We now move on to the study of the algebraic properties of the regressive scalar multiplication on the set of positively regressive functions. To that end, we present the following technical lemma containing some properties of the g-exponential map that, to the best of our knowledge, are not available in the literature. The proof of this result is based on an analogous result for the exponential map on time scales, see [1, Lemma 2.45].

Lemma 4.12

Let $p,q \in {{\mathcal {R}}}_g([a,b),{\mathbb {R}})$. Then:

• (i) If $\exp _g(p,t)=1$, $t\in [a,b]$, then $p(t)=0$ for g-a.a. $t\in [a,b)$.

• (ii) If $\exp _g(p,t)=\exp _g(q,t)$, $t\in [a,b]$, then $p(t)=q(t)$ for g-a.a. $t\in [a,b)$.

Proof

First, let $p \in {{\mathcal {R}}}_g([a,b),{\mathbb {R}})$ be such that $\exp _g(p,t)=1$, $t\in [a,b]$. In that case, by definition of g-derivative we have that $(\exp _g(p,\cdot ))'_g(t)=0$ for all $t\in [a,b)\setminus C_g$, while, by definition of g-exponential map, it is the unique solution on [a, b] of

$$\begin{aligned} x'_g(t)=p(t)x(t),\quad x(a)=1. \end{aligned}$$

Therefore, combining these two facts, it follows that, for g-a.a. $t\in [a,b]$,

$$\begin{aligned} p(t)=p(t)\exp _g(p,t)=(\exp _g(p,\cdot ))'_g(t)=0, \end{aligned}$$

as we needed to prove (i). Now, in order to prove (ii), let us consider $p, q \in {{\mathcal {R}}}_g([a,b),{\mathbb {R}})$ such that $\exp _g(p,t)=\exp _g(q,t)$, $t\in [a,b]$. In that case, thanks to Proposition 4.3, we have that

$$\begin{aligned} 1= & {} \frac{\exp _g(p,t)}{\exp _g(q,t)}=\exp _g\left( p \ominus q, t\right) \\= & {} \exp _g\left( p-\frac{q}{1+q\Delta ^+ g}-p\frac{q}{1+q\Delta ^+ g},t\right) =\exp _g\left( \frac{p-q}{1+q\Delta ^+ g},t\right) , \end{aligned}$$

so, it follows from (i) that $p(t)=q(t)$ for g-a.a. $t\in [a,b]$. $\square$

Remark 4.13

Observe that the equalities in Lemma 4.12 cannot be guaranteed for every point of the interval [a, b] as a consequence of considering the Lebesgue–Stieltjes integral of the function p in the definition of the g-exponential map. Indeed, take $t_0\not \in D_g$ and consider the map $p:[a,b]\rightarrow {\mathbb {R}}$ defined as

$$\begin{aligned} p(t) = {\left\{ \begin{array}{ll} 0, &{}{}\text{ if } \,t\not =t_0,\\ 1, &{}{}\text{ if } \,t=t_0. \end{array}\right. } \end{aligned}$$

In this case, $\mu _g(\{t_0\})=0$ so it follows that $\exp _g(p,t)=1$, $t\in [t_0,t_0+T]$, despite p not vanishing everywhere in $[t_0,t_0+T]$.

We are finally in position to prove some properties of the regressive scalar multiplication.

Proposition 4.14

Let $p,q\in {\mathcal {R}}_g^+([a,b),{\mathbb {R}})$ and $\alpha ,\beta \in {\mathbb {R}}$. Then:

• (i) $1\odot p=p$ and $0\odot p=0$.

• (ii) $\alpha \odot (\ominus p)=(-\alpha )\odot p$.

• (iii) $\alpha \odot (\beta \odot p)=(\alpha \beta )\odot p$.

• (iv) $\alpha \odot (p\oplus q)=(\alpha \odot p)\oplus (\alpha \odot q)$.

• (v) $(\alpha +\beta )\odot p=(\alpha \odot p)\oplus (\beta \odot p)$.

Proof

It is clear that (i) follows directly from the definition, so we shall focus on proving (ii)–(v). Given the properties in Proposition 4.8, we have that, for each $t\in [a,b]$,

$$\begin{aligned} \exp _g(\alpha \odot (\ominus p), t)&=(\exp _g(\ominus p,t))^\alpha =\left( \frac{1}{\exp _g(p,t)}\right) ^\alpha \\ {}&=(\exp _g(p,t))^{-\alpha }=\exp _g((-\alpha )\odot p,t),\\ \exp _g(\alpha \odot (\beta \odot p), t)&=(\exp _g(\beta \odot p, t))^\alpha =((\exp _g(p, t))^\beta )^\alpha \\ {}&=(\exp _g(p, t))^{\alpha \beta }=\exp _g((\alpha \beta )\odot p,t),\\ \exp _g(\alpha \odot (p\oplus q), t)&=(\exp _g(p\oplus q, t))^\alpha =(\exp _g(p, t)\exp _g(q, t))^\alpha \\ {}&=\exp _g(\alpha \odot p, t)\exp _g(\alpha \odot p, t)=\exp _g((\alpha \odot p)\oplus (\alpha \odot q), t),\\ \exp _g((\alpha +\beta )\odot p, t)&=(\exp _g(p, t))^{\alpha +\beta }=(\exp _g(p, t))^{\alpha }(\exp _g(p, t))^{\beta }\\ {}&=\exp _g(\alpha \odot p, t)\exp _g(\beta \odot p, t)=\exp _g((\alpha \odot p)\oplus (\beta \odot p), t). \end{aligned}$$

Therefore, (ii) in Lemma 4.12 guarantees that equalities (ii)–(v) hold g-almost everywhere in [a, b] and, in particular, since $\mu _g(\{t\})>0$ for each $t\in [a,b]\cap D_g$, we know that the equalities hold on $[a,b]\cap D_g$. For $t\in [a,b]{\setminus } D_g$, keeping in mind that $\Delta ^+ g(t)=0$ and given the expression of $\oplus$ and (4.11), we see that for such points, the operations $\oplus$ and $\odot$ agree with the usual sum and product on the real line, so equalities (ii)–(v) for the points of $[a,b]\setminus D_g$ now follow from the properties of the usual sum and product on ${\mathbb {R}}$, which finishes the proof. $\square$

Remark 4.15

Note that Theorem 4.10 and Proposition 4.14 guarantee that the set $\mathcal R_g^+([a,b),{\mathbb {R}})$ with the operations $\oplus$ and $\odot$ is a real vector space.

Remark 4.16

It follows from (i) and (v) in Proposition 4.14 that $0=(\alpha -\alpha )\odot p=(\alpha \odot p)\oplus ((-\alpha )\odot p)$. Thus,

$$\begin{aligned} \ominus (\alpha \odot p)=(-\alpha )\odot p=\alpha \odot (\ominus p), \end{aligned}$$

where the last equality is given by (ii) in Proposition 4.14. Therefore, we have that

$$\begin{aligned} \alpha \odot (p\ominus q)=&\,\alpha \odot (p\oplus (\ominus q))=(\alpha \odot p)\oplus (\alpha \odot (\ominus q))\nonumber \\ =&\,(\alpha \odot p)\oplus (\ominus (\alpha \odot q))=(\alpha \odot p)\ominus (\alpha \odot q). \end{aligned}$$

(4.17)

5 Bernoulli Equations in the Context of Differential Equations with Stieltjes Derivatives

Recently, in [10], the authors defined and studied the logistic equation in the context of differential equations with Stieltjes derivatives. In this setting, the logistic equation was defined as the equation that, after the usual change of variables, leads to a linear equation with Stieltjes derivatives. This idea can be observed in the following result which combines the information available in [8, Theorem 3.5] and [11, Proposition 3.10], as well as Theorems 3.4 and 3.8 and Propositions 3.6 and 3.9 in [10].

Theorem 5.1

Let $x_0,y_0\in {\mathbb {R}}{\setminus }\{0\}$, $p\in \mathcal R_g([t_0,t_0+T),{\mathbb {R}})$, $f\in {\mathcal {L}}^1_g([t_0,t_0+T),\mathbb R)$ and define

$$\begin{aligned} \phi (t)&=\frac{1}{x_0}+\int _{[t_0,t)}\frac{f(s)}{1+p(s)\Delta ^+ g(s)}\exp _g(p,s)^{-1}\,\textrm{d}\mu _g(s),&t\in [t_0,t_0+T],\\ \varphi (t)&=\frac{1}{y_0}+\int _{[t_0,t)}f(s)\exp _g(p,s)\,\textrm{d}\mu _g(s),&t\in [t_0,t_0+T]. \end{aligned}$$

• 1. The map $x(t)=\exp _g(p,t)\phi (t)$, $t\in [t_0,t_0+T]$, is the unique solution on $[t_0,t_0+T]$ of the linear equation

  $$\begin{aligned} x'_g(t)=p(t)x(t)+f(t),\quad x(t_0)=\frac{1}{x_0}. \end{aligned}$$

  (5.1)

• 2. If there exists $\tau \in (0,T]$ such that $\phi (t)\not =0$ for $t\in [t_0,t_0+\tau ]$ and

  $$\begin{aligned} \phi (t)\not =-\frac{f(t)\Delta ^+ g(t)}{1+p(t)\Delta ^+g(t)}\exp _g(p,t)^{-1},\quad t\in [t_0,t_0+\tau ]\cap D_g, \end{aligned}$$

  (5.2)

  then, the map $x(t)=(\exp _g(p,t)\phi (t))^{-1}$, $t\in [t_0,t_0+\tau ]$, is a solution on $[t_0,t_0+\tau ]$ of the logistic equation with Stieltjes derivatives,

  $$\begin{aligned} x'_g(t)=-\frac{p(t)+f(t)x(t)}{1+\Delta ^+g(t)(p(t)+f(t)x(t))}x(t),\quad x(t_0)=x_0. \end{aligned}$$

  (5.3)

• 3. The map $y(t)=\exp _g(p,t)^{-1}\varphi (t)$, $t\in [t_0,t_0+T]$, is the unique solution on $[t_0,t_0+T]$ of the adjoint linear equation,

  $$\begin{aligned} y'_g(t)=-\frac{p(t)}{1+p(t)\Delta ^+g(t)}y(t)+\frac{f(t)}{1+p(t)\Delta ^+g(t)},\quad y(t_0)=\frac{1}{y_0}. \end{aligned}$$

  (5.4)

• 4. If there exists $\tau \in (0,T]$ such that $\varphi (t)\not =0$ for $t\in [t_0,t_0+\tau ]$ and

  $$\begin{aligned} \varphi (t)\not =-f(t)\exp _g(p,t),\quad t\in [t_0,t_0+\tau ]\cap D_g, \end{aligned}$$

  (5.5)

  then, the map $y(t)=\exp _g(p,t)(\varphi (t))^{-1}$, $t\in [t_0,t_0+\tau ]$, is a solution on $[t_0,t_0+\tau ]$ of the adjoint logistic equation with Stieltjes derivatives,

  $$\begin{aligned} y'_g(t)=\frac{p(t)-f(t)y(t)}{1+\Delta ^+g(t)f(t)y(t)}y(t),\quad y(t_0)=y_0. \end{aligned}$$

  (5.6)

In particular, Theorem 5.1 shows that the solutions of (5.3) and (5.6) are, respectively, the multiplicative inverses of the solutions of (5.1) and (5.4). This is a direct consequence of the change of variables used in [10], which is based on the quotient rule, [8, Proposition 2.5, (ii)].

With this idea in mind, and taking into consideration that the logistic equation in the context of ordinary differential equations is a particular case of a Bernoulli differential equation, in this section we aim to propose a formulation for Bernoulli equations in the context of Stieltjes derivatives in a similar fashion to the logistic equation in [10], using the chain rule instead of the quotient rule. Specifically, we will define Bernoulli type equations with parameter $\alpha$ in the context of Stieltjes differential equations as those equations which, after the change of variables $u(t)=x(t)^{-\alpha }$, $\alpha \in {\mathbb {R}}\setminus \{0\}$, yield a linear equation in u, namely (5.1). In that case, we have that $x(t)=u(t)^{-1/\alpha }$ so, under the corresponding conditions, we can use Lemma 3.7 (expressed in terms of $\odot$ as in (4.13)) to study $x'_g(t)$, from which we would have

$$\begin{aligned} \frac{x'_g(t)}{x(t)}&=\left( \left( -\frac{1}{\alpha }\right) \odot \frac{u'_g}{u}\right) \hspace{-0.1cm}(t) =\left( \left( -\frac{1}{\alpha }\right) \odot \frac{pu+f}{u}\right) \hspace{-0.1cm}(t)\\&=\left( \left( -\frac{1}{\alpha }\right) \odot \frac{px^{-\alpha }+f}{x^{-\alpha }}\right) \hspace{-0.1cm}(t) =\left( \left( -\frac{1}{\alpha }\right) \odot (p+fx^\alpha )\right) \hspace{-0.1cm}(t). \end{aligned}$$

Therefore, we shall refer to differential problems of the form

$$\begin{aligned} x'_g(t)=\left( \left( -\frac{1}{\alpha }\right) \odot (p+fx^\alpha )\right) \hspace{-0.1cm}(t)x(t),\quad x(t_0)=x_0 \end{aligned}$$

(5.7)

as Bernoulli differential problems with Stieltjes derivatives with parameter $\alpha \in {\mathbb {R}}\setminus \{0\}$. Naturally, (5.7) can be regarded as a particular case of (2.2) and, therefore, the concept of solution in Definition 2.7 can still be considered. Nevertheless, it is important to note that with this definition, we are implicitly assuming that $p+fx^\alpha \in {\mathcal {R}}_g^+([t_0,t_0+\tau ),{\mathbb {R}})$, as this is necessary in order to be able to consider the $\odot$ product in (5.7).

Naturally, for $\alpha =-1$ (5.7) is just a linear equation of the form (5.1); whereas when $\alpha =1$, for $t\in [t_0,t_0+T]$,

$$\begin{aligned} \left( \left( -\frac{1}{\alpha }\right) \odot (p+fx^\alpha )\right) \hspace{-0.1cm}(t) ={} & {} \left( (-1)\odot (p+fx)\right) \hspace{-0.1cm}(t)=\left( \ominus \left( p+fx\right) \right) \hspace{-0.1cm}(t)\\ ={} & {} -\frac{p(t)+f(t)x(t)}{1+\Delta ^+g(t)(p(t)+f(t)x(t))}, \end{aligned}$$

which shows that (5.7) coincides with (5.3). Furthermore, when $D_g=\emptyset$ (and in particular, when $g=\textrm{Id}$), it follows from (4.11) that $\odot$ is just the standard scalar multiplication, so in that case, the right-hand side of (5.7) coincides with the corresponding right-hand side of a Bernoulli equation in the sense of ordinary differential equations.

The following result includes an explicit solution for (5.7). This result is, to some extent, a more general version of Theorem 5.1, statement 2; and its proof shares some similarities with the proof of [10, Theorem 3.4].

Theorem 5.2

Let $\alpha \in {\mathbb {R}}\!\setminus \{0\}$, $x_0\in {\mathbb {R}}^+$, $p\in {\mathcal {R}}_g([t_0,t_0+T),{\mathbb {R}})$ and $f\in \mathcal L^1_g([t_0,t_0+T),{\mathbb {R}})$ and define

$$\begin{aligned} \phi _\alpha (t)=\frac{1}{x_0^\alpha }+\int _{[t_0,t)}\frac{f(s)}{1+p(s)\Delta ^+ g(s)}\exp _g(p,s)^{-1}\,\textrm{d}\mu _g(s),\quad t\in [t_0,t_0+T].\nonumber \\ \end{aligned}$$

(5.8)

If there exists $\tau \in (0,T]$ such that $\exp _g(p,t)\phi _\alpha (t)>0$ for $t\in [t_0,t_0+\tau ]$ and

$$\begin{aligned} \phi _\alpha (t)\not =-\frac{f(t)\Delta ^+ g(t)}{1+p(t)\Delta ^+g(t)}\exp _g(p,t)^{-1},\quad t\in [t_0,t_0+\tau ]\cap D_g, \end{aligned}$$

(5.9)

then, the map $x(t)=(\exp _g(p,t)\phi _\alpha (t))^{-1/\alpha }$, $t\in [t_0,t_0+\tau ]$, is a solution of (5.7) on $[t_0,t_0+\tau ]$.

Proof

Denote

$$\begin{aligned} \Phi _\alpha (t)=\exp _g(p,t)\phi _\alpha (t),\quad t\in [t_0,t_0+\tau ]. \end{aligned}$$

With this notation, we have that $x=\Phi _\alpha ^{-1/\alpha }$ and, since by hypothesis $\Phi _\alpha (t)>0$, $t\in [t_0,t_0+\tau )$, x is well-defined and $x(t)>0$, $t\in [t_0, t_0+\tau ]$. Furthermore, statement 1 in Theorem 5.1 guarantees that $\Phi _\alpha$ is the solution of

$$\begin{aligned} (\Phi _\alpha )'_g(t)=p(t)\Phi _\alpha (t)+f(t),\quad \Phi (t_0)=\frac{1}{x_0^{\alpha }}. \end{aligned}$$

Note that, in particular, we have that $x(t_0)=(\Phi _\alpha (t_0))^{-1/\alpha }=x_0$, so all that is left to do is to check that $x\in \mathcal{A}\mathcal{C}_g([t_0,t_0+\tau ],{\mathbb {R}})$ and

$$\begin{aligned} x'_g(t)=\left( \left( -\frac{1}{\alpha }\right) \odot (p+fx^\alpha )\right) \hspace{-0.1cm}(t)x(t),\quad g\text{-a.a. } t\in [t_0,t_0+\tau ). \end{aligned}$$

Observe that $\Phi _\alpha \in \mathcal{A}\mathcal{C}_g([t_0,t_0+\tau ],\mathbb R)$, and, moreover, for each $t\in [t_0,t_0+\tau ]\cap D_g$,

$$\begin{aligned} \Phi _\alpha (t)+(\Phi _\alpha )'_g(t)\Delta ^+ g(t)&=\Phi _\alpha (t)+(p(t)\Phi _\alpha (t)+f(t))\Delta ^+ g(t)\\&=\Phi _\alpha (t)(1+p(t)\Delta ^+ g(t))+f(t)\Delta ^+ g(t)\\&=\phi _\alpha (t)\exp _g(p,t)(1+p(t)\Delta ^+ g(t))+f(t)\Delta ^+ g(t)\not =0, \end{aligned}$$

since we have that (5.9) holds. Thus, Lemma 3.7 guarantees that $x\in \mathcal{A}\mathcal{C}_g([t_0,t_0+\tau ],{\mathbb {R}})$ and, furthermore, for g-a.a. $t\in [t_0,t_0+\tau )$,

$$\begin{aligned} \frac{x'_g(t)}{x(t)}&=\left( \left( -\frac{1}{\alpha }\right) \odot \left( \frac{(\Phi _\alpha )'_g}{\Phi _\alpha }\right) \right) \hspace{-0.1cm}(t)=\left( \left( -\frac{1}{\alpha }\right) \odot \left( \frac{\Phi _\alpha p+f}{\Phi _\alpha }\right) \right) \hspace{-0.1cm}(t)\\&=\left( \left( -\frac{1}{\alpha }\right) \odot \left( p+f\Phi _\alpha ^{-1}\right) \right) \hspace{-0.1cm}(t) =\left( \left( -\frac{1}{\alpha }\right) \odot \left( p+fx^{\alpha }\right) \right) \hspace{-0.1cm}(t), \end{aligned}$$

from which the result follows. $\square$

Remark 5.3

Observe that, although it is not explicitly shown in the proof of the Theorem 5.2, we have that $p+fx^\alpha \in \mathcal R_g^+$ as a direct consequence of Lemma 3.7 and the fact that $(\Phi _\alpha )'_g/\Phi _\alpha =p+fx^\alpha$.

Remark 5.4

Condition (5.9) is essential to ensure that the map x is g-differentiable. Note that this condition is only relevant when $D_g\not =\emptyset$, which explains why it is not required for Bernoulli ordinary differential equations. On the other hand, requiring $\Phi _\alpha >0$ on an interval is necessary for x to be well-defined, so this condition can be found in similar results for ODEs. Naturally, if $-1/\alpha \in {\mathbb {N}}$, there is no need to require such condition (or any other) for the solution to be well-defined; while for $1/\alpha \in {\mathbb {N}}$, it is enough to ask for $\Phi _\alpha \not =0$ on an interval for the solution to be well-defined which, keeping in mind that $p\in \mathcal R_g([t_0,t_0+T),{\mathbb {R}})$, means that it is only necessary to impose that $\phi _\alpha \not =0$ on an interval. Similarly, the assumption that $x_0\in {\mathbb {R}}^+$ is necessary for the definition of $\phi _\alpha$ as it includes the expression $x_0^{-\alpha }$, so whenever $\alpha \in {\mathbb {N}}$, it is only necessary that $x_0\in {\mathbb {R}}{\setminus }\{0\}$; while for $-\alpha \in {\mathbb {N}}$ no condition is required. These observations are key to see that Theorem 5.2 is equivalent to statement 2 in Theorem 5.1 when (5.7) coincides with (5.3), i.e., for $\alpha =1$.

The following result is a consequence of Theorem 5.2 and it shows that, if the map p belongs to the set of positively regressive functions, then some of the conditions required in Theorem 5.2 might be ignored and, furthermore, the solution of (5.7) in Theorem 5.2 might be expressed in an alternative way.

Corollary 5.5

Let $\alpha \in {\mathbb {R}}\setminus \{0\}$, $x_0\in {\mathbb {R}}^+$ and $f\in {\mathcal {L}}^1_g([t_0,t_0+T),{\mathbb {R}})$. If there exists $\tau \in (0,T]$ such that $p\in {\mathcal {R}}_g^+([t_0,t_0+\tau ),\mathbb R)$; and the map $\phi _\alpha$ in (5.8) is such that $\phi _\alpha (t)>0$, $t\in [t_0,t_0+\tau ]$, and (5.9) holds; then, the function

$$\begin{aligned} x(t)=\exp _g\left( \frac{-1}{\alpha }\odot p,t\right) \phi _\alpha (t)^{-1/\alpha }, \quad t\in [t_0,t_0+\tau ], \end{aligned}$$

(5.10)

is a solution of (5.7) on $[t_0,t_0+\tau ]$.

Proof

First, observe that, since $p\in \mathcal R_g^+([t_0,t_0+\tau ),{\mathbb {R}})$, we have that $\exp _g(p,t)>0$ for all $t\in [t_0,t_0+T]$, see Remark 4.2. Hence, given the hypotheses, it is clear that $\exp _g(p,t)\phi _\alpha (t)>0$ for $t\in [t_0,t_0+\tau ]$ and, since (5.9) holds, we have that the conditions of Theorem 5.2 are satisfied. Thus,

$$\begin{aligned} x(t)=(\exp _g(p,t)\phi _\alpha (t))^{-1/\alpha }, \quad t\in [t_0,t_0+\tau ], \end{aligned}$$

is a solution on $[t_0,t_0+\tau ]$ of (5.7). Now, (5.10) follows from statement (iii) in Proposition 4.8. $\square$

In a similar fashion to [10], we can also consider the adjoint version of the Bernoulli equation with Stieltjes derivatives as the equation for which the change of variables $v(t)=y(t)^{-\alpha }$, $\alpha \in {\mathbb {R}}\setminus \{0\}$, yields the adjoint version of the linear equation, i.e. (5.4), which can be rewritten, under suitable conditions, as

$$\begin{aligned} v'_g(t)&=-\frac{p(t)}{1+p(t)\Delta ^+ g(t)}v(t)+\frac{f(t)}{1+p(t)\Delta ^+ g(t)}\\&=f(t)-\frac{p(t)}{1+p(t)\Delta ^+ g(t)}v(t)-\frac{f(t)p(t)\Delta ^+ g(t)}{1+p(t)\Delta ^+ g(t)}\\&=\left( \frac{f(t)}{v(t)}-\frac{p(t)}{1+p(t)\Delta ^+ g(t)}-\frac{f(t)}{v(t)}\frac{p(t)\Delta ^+ g(t)}{1+p(t)\Delta ^+ g(t)}\right) \hspace{-0.08cm}v(t)=\left( \frac{f}{v}\ominus p\right) \hspace{-0.1cm}(t)v(t). \end{aligned}$$

Under the mentioned transformation, we would have that $y(t)=v(t)^{-1/\alpha }$ so Lemma 3.7 would ensure that

$$\begin{aligned} \frac{y'_g(t)}{y(t)}&=\left( \left( -\frac{1}{\alpha }\right) \odot \frac{v'_g}{v}\right) \hspace{-0.1cm}(t) =\left( \left( -\frac{1}{\alpha }\right) \odot \frac{\left( \frac{f}{v}\ominus p\right) v}{v}\right) \hspace{-0.1cm}(t)=\left( \left( -\frac{1}{\alpha }\right) \odot \left( \frac{f}{v}\ominus p\right) \right) \hspace{-0.1cm}(t)\\&=\left( \frac{1}{\alpha }\odot \left( \ominus \left( \frac{f}{v}\ominus p\right) \right) \right) \hspace{-0.1cm}(t)=\left( \frac{1}{\alpha }\odot \left( p\ominus \frac{f}{v}\right) \right) \hspace{-0.1cm}(t)=\left( \frac{1}{\alpha }\odot \left( p\ominus (fy^\alpha )\right) \right) \hspace{-0.1cm}(t), \end{aligned}$$

provided certain conditions are satisfied. Hence, we define the adjoint Bernoulli differential problem with Stieltjes derivatives with parameter $\alpha \in {\mathbb {R}}\setminus \{0\}$ as the initial value problem

$$\begin{aligned} y'_g(t)=\left( \frac{1}{\alpha }\odot \left( p\ominus (fy^\alpha )\right) \right) \hspace{-0.1cm}(t)y(t),\quad y(t_0)=y_0. \end{aligned}$$

(5.11)

Note that (5.11) is another particular case of (2.2) so the concept of solution in Definition 2.7 is still valid (in this case, we implicitly assume that $p\ominus (fy^\alpha )\in {\mathcal {R}}_g^+([t_0,t_0+\tau ),{\mathbb {R}})$ for $\odot$ in (5.11) to be well-defined). Furthermore, similarly to (5.7), it is easy to see that for $\alpha =-1$, (5.11) coincides with (5.4); while for $\alpha =1$, (5.6) is a particular case of (5.11). Moreover, (5.11) is a particular case of (5.7) whenever $p\in {\mathcal {R}}_g([t_0,t_0+T],{\mathbb {R}})$ as it can be rewritten as

$$\begin{aligned} y'_g(t)=\left( \left( -\frac{1}{\alpha }\right) \odot (P+Fy^\alpha )\right) \hspace{-0.1cm}(t)y(t),\quad y(t_0)=y_0, \end{aligned}$$

(5.12)

where the maps $P,F:[t_0,t_0+T]\rightarrow {\mathbb {R}}$ are defined as

$$\begin{aligned} P(t)=-\frac{p(t)}{1+p(t)\Delta ^+ g(t)},\quad F(t)=\frac{f(t)}{1+p(t)\Delta ^+ g(t)}, \end{aligned}$$

(5.13)

since it holds that $P+Fy^\alpha =fy^\alpha \ominus p$ and, as a consequence,

$$\begin{aligned} \left( \left( -\frac{1}{\alpha }\right) \odot (P+Fy^\alpha )\right){} & {} =\left( \left( -\frac{1}{\alpha }\right) \odot (fy^\alpha \ominus p)\right) \\{} & {} =\left( \frac{1}{\alpha }\odot (\ominus (fy^\alpha \ominus p))\right) =\left( \frac{1}{\alpha }\odot ((p\ominus fy^\alpha ))\right) . \end{aligned}$$

Hence, we can use Theorem 5.2 and Corollary 5.5 to obtain a solution of (5.11) under the corresponding hypotheses, which leads to the following result. Note that the first statement is a generalization of statement 4 in Theorem 5.1 in the same way that Theorem 5.2 is an extension of statement 2 in Theorem 5.1.

Theorem 5.6

Let $\alpha \in {\mathbb {R}}\!\setminus \{0\}$, $y_0\in {\mathbb {R}}^+$, $p\in {\mathcal {R}}_g([t_0,t_0+T),{\mathbb {R}})$ and $f\in \mathcal L^1_g([t_0,t_0+T),{\mathbb {R}})$ and define

$$\begin{aligned} \varphi _\alpha (t)=\frac{1}{y_0^\alpha }+\int _{[t_0,t)}f(s)\exp _g(p,s)\,\textrm{d}\mu _g(s),\quad t\in [t_0,t_0+T]. \end{aligned}$$

• 1. If there exists $\tau \in (0,T]$ such that $\exp _g(p,t)\varphi _\alpha (t)>0$ for $t\in [t_0,t_0+\tau ]$ and

  $$\begin{aligned} \varphi _\alpha (t)\not =-f(t)\exp _g(p,t),\quad t\in [t_0,t_0+\tau ]\cap D_g, \end{aligned}$$

  (5.14)

  then, the map $y(t)=(\exp _g(p,t)\varphi _\alpha (t)^{-1})^{1/\alpha }$, $t\in [t_0,t_0+\tau ]$, is a solution of (5.11) on $[t_0,t_0+\tau ]$.

• 2. If there exists $\tau \in (0,T]$ such that $p\in {\mathcal {R}}_g^+([t_0,t_0+\tau ),{\mathbb {R}})$, $\varphi _\alpha (t)>0$ for all $t\in [t_0,t_0+\tau ]$, and (5.14) holds; then, the function

  $$\begin{aligned} y(t)=\exp _g\left( \frac{1}{\alpha }\odot p,t\right) \varphi _\alpha (t)^{-1/\alpha }, \quad t\in [t_0,t_0+\tau ], \end{aligned}$$

  is a solution of (5.7) on $[t_0,t_0+\tau ]$.

As a final comment to this section, note that if we consider (5.11) under the assumption that p is a positively regressive function, we can use (4.17) to rewrite the equation as

$$\begin{aligned} y'_g(t)=\left( \left( \frac{1}{\alpha }\odot p\right) \ominus \left( \frac{1}{\alpha }\odot (fy^\alpha )\right) \right) \hspace{-0.1cm}(t)y(t),\quad y(t_0)=y_0. \end{aligned}$$

Hence, the previous problem is equivalent to

$$\begin{aligned} y'_g(t)=\left( q\ominus \left( \frac{1}{\alpha }\odot (fy^\alpha )\right) \right) \hspace{-0.1cm}(t)y(t),\quad y(t_0)=y_0, \end{aligned}$$

(5.15)

for $q\in {\mathcal {R}}_g^+([t_0,t_0+T),{\mathbb {R}})$, which is the equivalent Stieltjes differential equation of the corresponding Bernoulli equation in the time scales setting, see [1, Equation 2.42]. For more information on the relation between Stieltjes differential equations and dynamic equations on time scales we refer the reader to [5, Section 8.3], [7, Section 3], [9, Section 3.3.3], [10, Section 4] and [11, Section 4].

The following result is a direct consequence of statement 2 of Theorem 5.6 and it is a generalization of [1, Theorem 2.48].

Corollary 5.7

Let $\alpha \in {\mathbb {R}}\setminus \{0\}$, $x_0\in {\mathbb {R}}^+$, $q\in {\mathcal {R}}_g^+([t_0,t_0+T),{\mathbb {R}})$, $f\in \mathcal L^1_g([t_0,t_0+T),{\mathbb {R}})$ and define

$$\begin{aligned} \psi _\alpha (t)=\frac{1}{y_0^\alpha }+\int _{[t_0,t)}f(s)\exp _g(q,s)^\alpha \,\textrm{d}\mu _g(s),\quad t\in [t_0,t_0+T]. \end{aligned}$$

If there exists $\tau \in (0,T]$ such that $\psi _\alpha (t)>0$ for all $t\in [t_0,t_0+\tau ]$ and

$$\begin{aligned} \psi _\alpha (t)\not =-f(t)\exp _g(q,t)^\alpha ,\quad t\in [t_0,t_0+\tau ]\cap D_g, \end{aligned}$$

then, the function $y(t)=\exp _g\left( q,t\right) \psi _\alpha (t)^{-1/\alpha }$, $t\in [t_0,t_0+\tau ]$, is a solution on $[t_0,t_0+\tau ]$ of (5.15).

6 Preliminaries on Stieltjes Integrals

In the rest of the paper, we will deal with Stieltjes integrals of the form $\int _a^b f\,\textrm{d}g$, where $f:[a,b]\rightarrow {\mathbb {R}}$ is referred to as the integrand, and $g:[a,b]\rightarrow {\mathbb {R}}$ is the integrator. All integrals will be understood in the Kurzweil–Stieltjes sense (or, equivalently, in the Perron–Stieltjes sense). The definition of this integral is based on Riemann–Stieltjes sums of the form $\sum _{i=1}^m f(\xi _i)(g(t_i)-g(t_{i-1}))$, where $a=t_0<\cdots <t_m=b$ is a partition of [a, b], and $\xi _i\in [t_{i-1},t_i]$ for each $i\in \{1,\ldots ,m\}$. The precise definition can be found e.g. in [13, Definition 6.2.2], but it is not too important for our purposes. We will need just a few basic properties of the integral:

• If one of the functions f, g has bounded variation and the other is regulated, then $\int _a^b f\,\textrm{d}g$ exists. Moreover, if we let $F(t)=\int _a^t f\,\textrm{d}g$ for all $t\in [a,b]$, then F is regulated and its one-sided limits are

  $$\begin{aligned} F(t+)&=F(t)+f(t)\Delta ^+g(t),\quad t\in [a,b), \end{aligned}$$

  (6.1)
  $$\begin{aligned} F(t-)&=F(t)-f(t)\Delta ^-g(t),\quad t\in (a,b]. \end{aligned}$$

  (6.2)

  For proofs of these statements, see [13, Theorem 6.3.11 and Corollary 6.5.5]. Here and in the rest of the paper, we use the notation

  $$\begin{aligned} \Delta ^+g(t)&=g(t+)-g(t),\\ \Delta ^-g(t)&=g(t)-g(t-),\\ \Delta g(t)&=g(t+)-g(t-). \end{aligned}$$

• The substitution theorem for the Kurzweil–Stieltjes integral says that if $h:[a,b]\rightarrow {\mathbb {R}}$ is a bounded function and $f,g:[a,b]\rightarrow {\mathbb {R}}$ are such that $\int _a^b f\,\textrm{d}g$ exists, then

  $$\begin{aligned} \int _a^b h(x)\,\textrm{d}\left( \int _a^x f(z)\,\textrm{d}g(z)\right) =\int _a^b h(x) f(x)\,\textrm{d}g(x), \end{aligned}$$

  whenever either side of the equation exists (see [13, Theorem 6.6.1]).

• If $f:[a,b]\rightarrow {\mathbb {R}}$ is zero except on a countable set $\{t_1,t_2,\ldots \}\subset [a,b]$, $\sum _i f(t_i)$ is absolutely convergent, and $g:[a,b]\rightarrow {\mathbb {R}}$ is regulated, then

  $$\begin{aligned} \int _a^b f\,\textrm{d}g=f(a)\Delta ^+g(a)+\sum _{x\in (a,b)}f(x)\Delta g(x)+f(b)\Delta ^-g(b), \end{aligned}$$

  (6.3)

  where the infinite series is absolutely convergent (see [18, Proposition 2.12]).

• If $f:[a,b]\rightarrow {\mathbb {R}}$ is regulated, $g:[a,b]\rightarrow {\mathbb {R}}$ has bounded variation and its continuous part (cf. [13, Definition 2.6.2]) vanishes, then

  $$\begin{aligned} \int _a^b f\,\textrm{d}g=f(a)\Delta ^+g(a)+\sum _{x\in (a,b)}f(x)\Delta g(x)+f(b)\Delta ^-g(b), \end{aligned}$$

  (6.4)

  where the infinite series is absolutely convergent (see [13, Theorem 6.3.13]).

• If g is left-continuous, nondecreasing, and the Lebesgue–Stieltjes integral $\int _{[a,b)}f\,\textrm{d}\mu _g$ exists, then the Kurzweil–Stieltjes integral $\int _a^b f\,\textrm{d}g$ exists as well, and has the same value (see [13, Theorem 6.12.3]).

All remaining properties of the integral will be recalled whenever they are needed.

7 Chain Rule for Stieltjes Integrals

We now return to the topic of generalized chain rules. Recalling the Stieltjes-integral version

$$\begin{aligned} h(f(b))-h(f(a)) =\int _a^b h'(f(t))\,\textrm{d}f(t) \end{aligned}$$

presented in the introduction, our goal is to generalize this formula to the case when h remains continuously differentiable but f has merely bounded variation and need not be continuous. We begin with the case when f is a step function.

Lemma 7.1

If $h:{\mathbb {R}}\rightarrow {\mathbb {R}}$ is continuously differentiable and $f:[a,b]\rightarrow {\mathbb {R}}$ is a step function, then

$$\begin{aligned} h(f(b))-h(f(a))=\int _a^b\left( \int _0^1 h'(\tau f(t+)+(1-\tau )f(t-))\,\textrm{d}\tau \right) \,\textrm{d}f(t), \end{aligned}$$

with the convention that $f(a-)=f(a)$ and $f(b+)=f(b)$.

Proof

Since f is a step function, there exists a partition $a=\alpha _0<\cdots <\alpha _m=b$ such that f is a constant function on each interval $(\alpha _{j-1},\alpha _j)$. Hence,

$$\begin{aligned} (h\circ f)(b)-(h\circ f)(a)=(h\circ f)(b+)-(h\circ f)(a-)=\sum _{j=0}^m\Delta (h\circ f)(\alpha _j). \end{aligned}$$

For each $j\in \{1,\ldots ,m-1\}$, we calculate

$$\begin{aligned} \Delta (h\circ f)(\alpha _j)&=\lim _{\delta \rightarrow 0+}((h\circ f)(\alpha _j+\delta )-(h\circ f)(\alpha _j-\delta ))=\lim _{\delta \rightarrow 0+}\int _{f(\alpha _j-\delta )}^{f(\alpha _j+\delta )}h'(s)\,\textrm{d}s\\&=\lim _{\delta \rightarrow 0+}\left( (f(\alpha _j+\delta )-f(\alpha _j-\delta ))\int _0^1 h'(\tau f(\alpha _j+\delta )+(1-\tau )f(\alpha _j-\delta ))\,\textrm{d}\tau \right) \\&=\Delta f(\alpha _j)\lim _{\delta \rightarrow 0+}\left( \int _0^1 h'(\tau f(\alpha _j+\delta )+(1-\tau )f(\alpha _j-\delta ))\,\textrm{d}\tau \right) \\&=\Delta f(\alpha _j)\int _0^1 \lim _{\delta \rightarrow 0+}\left( h'(\tau f(\alpha _j+\delta )+(1-\tau )f(\alpha _j-\delta ))\right) \,\textrm{d}\tau \\&=\Delta f(\alpha _j)\int _0^1 h'(\tau f(\alpha _j+)+(1-\tau )f(\alpha _j-))\,\textrm{d}\tau . \end{aligned}$$

To interchange the order of the limit and the integral, we have used the bounded convergence theorem (whose assumptions are satisfied because $h'$ is bounded on bounded subsets of ${\mathbb {R}}$). In a similar way, one can prove that the relation

$$\begin{aligned} \Delta (h\circ f)(\alpha _j)=\Delta f(\alpha _j)\int _0^1 h'(\tau f(\alpha _j+)+(1-\tau )f(\alpha _j-))\,\textrm{d}\tau \end{aligned}$$

also holds for $j=0$ (with the convention that $f(a-)=f(a)$) and for $j=m$ (with the convention that $f(b+)=f(b)$). Thus, we get

$$\begin{aligned} (h\circ f)(b)-(h\circ f)(a)&=\sum _{j=0}^m \left( \int _0^1 h'(\tau f(\alpha _j+)+(1-\tau )f(\alpha _j-))\,\textrm{d}\tau \right) \Delta f(\alpha _j)\\&=\int _a^b \left( \int _0^1 h'(\tau f(\alpha _j+)+(1-\tau )f(\alpha _j-))\,\textrm{d}\tau \right) \,\textrm{d}f(t), \end{aligned}$$

where the last equality follows from the fact that $\int _a^b H\,\textrm{d}f=\sum _{j=0}^m H(\alpha _j)\Delta f(\alpha _j)$ for an arbitrary function $H:[a,b]\rightarrow {\mathbb {R}}$ (see e.g. [13, Theorem 6.3.13]). $\square$

We now extend the formula from Lemma 7.1 to functions of bounded variation.

Theorem 7.2

If $f:[a,b]\rightarrow {\mathbb {R}}$ has bounded variation, $I=[\inf _{x\in [a,b]} f(x),\sup _{x\in [a,b]} f(x)]$, and $h:I\rightarrow {\mathbb {R}}$ is continuously differentiable, then

$$\begin{aligned} h(f(b))-h(f(a))=\int _a^b\left( \int _0^1 h'(\tau f(t+)+(1-\tau )f(t-))\,\textrm{d}\tau \right) \,\textrm{d}f(t), \end{aligned}$$

with the convention that $f(a-)=f(a)$ and $f(b+)=f(b)$.

Proof

We extend the function h to the whole real line; the values on ${\mathbb {R}}\setminus I$ can be chosen in an arbitrary way (because the identity we want to prove does not depend on the values outside I), provided that the extended function is differentiable on ${\mathbb {R}}$.

Since the function f has bounded variation, there exist nondecreasing functions $f^1,f^2:[a,b]\rightarrow {\mathbb {R}}$ such that $f=f^1-f^2$. Also, for each $i\in \{1,2\}$, there exists a sequence of nondecreasing step functions $\{f^i_n\}_{n=1}^\infty$ which is uniformly convergent to $f^i$. Without loss of generality, we can assume that these sequences are such that

$$\begin{aligned} f^i(a)\le f_n^i(a)\le f_n^i(b)\le f^i(b) \end{aligned}$$

for all $n\in {\mathbb {N}}$ and $i\in \{1,2\}$. Therefore,

$$\begin{aligned} \mathop {\text {var}}(f^i_n,[a,b])=f^i_n(b)-f^i_n(a)\le f^i(b)-f^i(a), \quad n\in {\mathbb {N}},\quad i\in \{1,2\}. \end{aligned}$$

Consequently, by letting $f_n=f^1_n-f^2_n$ for all $n\in {\mathbb {N}}$, we obtain a sequence of step functions $\{f_n\}_{n=1}^\infty$, which is uniformly convergent to f, and its members have uniformly bounded variation.

Using the convention that $f_n(a-)=f_n(a)$ and $f_n(b+)=f_n(b)$ for each $n\in {\mathbb {N}}$, note that $f_n(t-)\rightrightarrows f(t-)$ and $f_n(t+)\rightrightarrows f(t+)$ with respect to $t\in [a,b]$ (see [13, Lemma 4.2.3]). Therefore,

$$\begin{aligned} \tau f_n(t+)+(1-\tau )f_n(t-)\rightrightarrows \tau f(t+)+(1-\tau )f(t-), \end{aligned}$$

where the convergence is uniform with respect to $(\tau ,t)\in [0,1]\times [a,b]$, and consequently (since $h'$ is continuous, and therefore uniformly continuous on compact sets),

$$\begin{aligned} h'(\tau f_n(t+)+(1-\tau )f_n(t-))\rightrightarrows h'(\tau f(t+)+(1-\tau )f(t-)) \end{aligned}$$

with respect to $(\tau ,t)\in [0,1]\times [a,b]$. It follows that

$$\begin{aligned} \int _0^1 h'(\tau f_n(t+)+(1-\tau )f_n(t-))\,\textrm{d}\tau \rightrightarrows \int _0^1h'(\tau f(t+)+(1-\tau )f(t-))\,\textrm{d}\tau , \end{aligned}$$

where the convergence is uniform with respect to $t\in [a,b]$. Finally, using Lemma 7.1 and the uniform convergence theorem for integrals whose integrators have uniformly bounded variation (see [13, Theorem 6.8.8]), we get

$$\begin{aligned} h(f(b))-h(f(a))&=\lim _{n\rightarrow \infty }\left( h(f_n(b))-h(f_n(a))\right) \\&=\lim _{n\rightarrow \infty }\int _a^b\left( \int _0^1 h'(\tau f_n(t+)+(1-\tau )f_n(t-))\,\textrm{d}\tau \right) \,\textrm{d}f_n(t)\\&=\int _a^b\left( \int _0^1 h'(\tau f(t+)+(1-\tau )f(t-))\,\textrm{d}\tau \right) \,\textrm{d}f(t). \end{aligned}$$

$\square$

Remark 7.3

If f is left-continuous, the formula from Theorem 7.2 reduces to

$$\begin{aligned} h(f(b))-h(f(a))=\int _a^b\left( \int _0^1 h'(f(t)+\tau \Delta ^+ f(t))\,\textrm{d}\tau \right) \,\textrm{d}f(t). \end{aligned}$$

Furthermore, if g is a left-continuous nondecreasing function, f is g-absolutely continuous and $f_g'$ is the g-derivative of f, then $f(t)=f(a)+\int _{a}^t f_g'(s)\,\textrm{d}g(s)$ and $\Delta ^+f(t)=f_g'(t)\Delta ^+g(t)$. In this case, using the substitution theorem, we get

$$\begin{aligned} h(f(b))-h(f(a))=\int _a^b\left( \int _0^1 h'(f(t)+\tau f_g'(t)\Delta ^+ g(t))\,\textrm{d}\tau \right) f_g'(t)\,\textrm{d}g(t), \end{aligned}$$

which is in agreement with the chain rule for Stieltjes derivatives obtained in Theorem 3.2.

As a corollary of Theorem 7.2, we obtain the following generalization of the formula $\frac{\textrm{d}}{\textrm{d}t}z(t)^{\alpha }=\alpha z(t)^{\alpha -1}z'(t)$.

Corollary 7.4

Suppose that $z:[a,b]\rightarrow {\mathbb {R}}$ has bounded variation, and one of the following conditions holds:

• $\alpha \in {\mathbb {N}}$.

• $\alpha \in {\mathbb {R}}\setminus {\mathbb {N}}$, $z(t)>0$ for all $t\in [a,b]$, $z(t-)>0$ for all $t\in (a,b]$, and $z(t+)>0$ for all $t\in [a,b)$.

Then

$$\begin{aligned} z(b)^{\alpha }-z(a)^{\alpha }=\int _a^b \left( \int _0^1 \alpha (\tau z(s+)+(1-\tau )z(s-))^{\alpha -1}\,\textrm{d}\tau \right) \,\textrm{d}z(s), \end{aligned}$$

(7.1)

with the convention that $z(a-)=z(a)$ and $z(b+)=z(b)$.

Proof

Eq. (7.1) follows from Theorem 7.2 by choosing $h(x)=x^{\alpha }$. If $\alpha \in {\mathbb {N}}$, then h is continuously differentiable on ${\mathbb {R}}$. If $\alpha \in \mathbb R{\setminus }{\mathbb {N}}$, then h is continuously differentiable on $(0,\infty )$. According to the assumption, z is positive. In fact, it is bounded away from zero, for otherwise (using compactness of [a, b]) it would be possible to find a $t\in [a,b]$ with $z(t+)=0$ or $z(t-)=0$ (cf. [10, Lemma 2.7]). Hence, the interval $I=[\inf _{x\in [a,b]} z(x),\sup _{x\in [a,b]} z(x)]$ is contained in $(0,\infty )$. $\square$

Remark 7.5

Note that if z is continuous at s, then the inner integral in (7.1) equals $\alpha z(s)^{\alpha -1}$. Otherwise, it can be evaluated using the formula

$$\begin{aligned} \int _0^1 \alpha (\tau \gamma +(1-\tau )\beta )^{\alpha -1}\,\textrm{d}\tau ={\left\{ \begin{array}{ll} \alpha \beta ^{\alpha -1}&{} \text{ if } \beta =\gamma ,\\ \frac{\gamma ^\alpha -\beta ^{\alpha }}{\gamma -\beta }&{} \text{ if } \beta \ne \gamma \end{array}\right. } \end{aligned}$$

(7.2)

whenever $\beta ,\gamma \in {\mathbb {R}}$ and $\alpha \in {\mathbb {N}}$ or $\beta ,\gamma >0$ and $\alpha \in {\mathbb {R}}{\setminus }{\mathbb {N}}$.

If z is a differentiable function with $z'=w$ being nonzero, then $\frac{\textrm{d}}{\textrm{d}t}z(t)^{\alpha }=\alpha z(t)^{\alpha }\frac{w(t)}{z(t)}$. We will now obtain a more general version of this formula applicable in the situation when $z(t)=z(t_0)+\int _{t_0}^t w(s)\,\textrm{d}g(s)$ for all $t\in [a,b]$. For this purpose, we introduce the following notation.

Definition 7.6

Given $\alpha \in {\mathbb {R}}$, $t_0,t\in [a,b]$ and functions $p,g:[a,b]\rightarrow {\mathbb {R}}$, we denote

$$\begin{aligned} (\alpha \odot _{g,t_0,t}p)(s)=\alpha p(s)\int _0^1 (1+p(s)(\tau \Delta g(s)-\Delta ^-g(s)))^{\alpha -1}\,\textrm{d}\tau \end{aligned}$$

whenever the right-hand side makes sense, and with the convention that $\Delta g(s)=\Delta ^-g(s)$ if $s=\max (t_0,t)$, and $\Delta g(s)=\Delta ^+g(s)$, $\Delta ^-g(s)=0$ if $s=\min (t,t_0)$.

Remark 7.7

A straightforward calculation reveals that

$$\begin{aligned}{} & {} (\alpha \odot _{g,t_0,t}p)(s)\\{} & {} ={\left\{ \begin{array}{ll} \frac{(1+p(s)\Delta ^+ g(s))^{\alpha }-(1-p(s)\Delta ^- g(s))^{\alpha }}{\Delta g(s)}&{} \text{ if } \Delta g(s)\ne 0,\\ \alpha p(s)\left( 1-p(s)\Delta ^- g(s)\right) ^{\alpha -1}=\alpha p(s)\left( 1+p(s)\Delta ^+ g(s)\right) ^{\alpha -1}&{} \text{ if } \Delta g(s)=0, \end{array}\right. } \end{aligned}$$

with the same convention concerning the jumps of g as in Definition 7.6.

Theorem 7.8

Suppose that $z:[a,b]\rightarrow {\mathbb {R}}$ has bounded variation and one of the following conditions holds:

• $\alpha \in {\mathbb {N}}$, $z(t)\ne 0$ for all $t\in [a,b]$.

• $\alpha \in {\mathbb {R}}\setminus {\mathbb {N}}$, $z(t)> 0$ for all $t\in [a,b]$, $z(t-)>0$ for all $t\in (a,b]$, and $z(t+)>0$ for all $t\in [a,b)$.

If $z(t)=z(t_0)+\int _{t_0}^t w(s)\,\textrm{d}g(s)$ for all $t\in [a,b]$, then

$$\begin{aligned} z(t)^{\alpha }-z(t_0)^{\alpha }=\int _{t_0}^t z(s)^{\alpha }\left( \alpha \odot _{g,t_0,t}\frac{w}{z}\right) \hspace{-0.1cm}(s)\,\textrm{d}g(s),\quad t\in [a,b]. \end{aligned}$$

(7.3)

Proof

Using Eq. (7.1), the relations $z(s+)=z(s)+w(s)\Delta ^+g(s)$ and $z(s-)=z(s)-w(s)\Delta ^-g(s)$, as well as the substitution theorem, we obtain

$$\begin{aligned} z(t)^{\alpha }-z(t_0)^{\alpha }&=\int _{t_0}^t \left( \int _0^1 \alpha \left( z(s)+w(s)(\tau \Delta g(s)-\Delta ^-g(s))\right) ^{\alpha -1}\,\textrm{d}\tau \right) \,\textrm{d}z(s)\\&=\int _{t_0}^t z(s)^{\alpha -1}\left( \int _0^1 \alpha \left( 1+\frac{w(s)}{z(s)}(\tau \Delta g(s)-\Delta ^-g(s))\right) ^{\alpha -1}\,\textrm{d}\tau \right) w(s)\,\textrm{d}g(s), \end{aligned}$$

with the convention that $\Delta g(s)=\Delta ^-g(s)$ if $s=\max (t_0,t)$, and $\Delta g(s)=\Delta ^+g(s)$, $\Delta ^-g(s)=0$ if $s=\min (t,t_0)$; this proves that Eq. (7.3) holds. $\square$

The next result is essentially a special case of Theorem 7.8 with $w(s)=p(s)z(s)$; however, we formulate it as a standalone result since in this case, we can omit the requirement $z(t)\ne 0$ for all $t\in [a,b]$ if $\alpha \in {\mathbb {N}}$.

Theorem 7.9

Suppose that $z:[a,b]\rightarrow {\mathbb {R}}$ has bounded variation, and one of the following conditions holds:

• $\alpha \in {\mathbb {N}}$.

• $\alpha \in {\mathbb {R}}\setminus {\mathbb {N}}$, $z(t)> 0$ for all $t\in [a,b]$, $z(t-)>0$ for all $t\in (a,b]$, and $z(t+)>0$ for all $t\in [a,b)$.

If $z(t)=z(t_0)+\int _{t_0}^t p(s)z(s)\,\textrm{d}g(s)$ for all $t\in [a,b]$, then

$$\begin{aligned} z(t)^{\alpha }-z(t_0)^{\alpha }=\int _{t_0}^t z(s)^{\alpha }\left( \alpha \odot _{g,t_0,t}p\right) \hspace{-0.05cm}(s)\,\textrm{d}g(s),\quad t\in [a,b]. \end{aligned}$$

(7.4)

Proof

Similarly to the proof of Theorem 7.8, we use Eq. (7.1), the relations $z(s+)=z(s)+p(s)z(s)\Delta ^+g(s)$ and $z(s-)=z(s)-p(s)z(s)\Delta ^-g(s)$, as well as the substitution theorem to obtain

$$\begin{aligned} z(t)^{\alpha }-z(t_0)^{\alpha }=\int _{t_0}^t z(s)^{\alpha -1}\left( \int _0^1 \alpha \left( 1+p(s)(\tau \Delta g(s)-\Delta ^-g(s))\right) ^{\alpha -1}\,\textrm{d}\tau \right) p(s)z(s)\,\textrm{d}g(s), \end{aligned}$$

with the convention that $\Delta g(s)=\Delta ^-g(s)$ if $s=\max (t_0,t)$, and $\Delta g(s)=\Delta ^+g(s)$, $\Delta ^-g(s)=0$ if $s=\min (t,t_0)$; this proves Eq. (7.4). $\square$

8 Powers of the Generalized Exponential Function

The simplest linear Stieltjes integral equation has the form

$$\begin{aligned} z(t)=z(t_0)+\int _{t_0}^t z(s)\,\textrm{d}P(s),\quad t\in [a,b], \end{aligned}$$

(8.1)

where $P:[a,b]\rightarrow {\mathbb {R}}$ has bounded variation. It is well known that if $1+\Delta ^+P(t)\ne 0$ for all $t\in [a,t_0)$ and $1-\Delta ^-P(t)\ne 0$ for all $t\in (t_0,b]$, then for each choice of the initial value $z(t_0)=z_0$, the equation has a unique solution $z:[a,b]\rightarrow {\mathbb {R}}$ (see e.g. [12, Theorem 2.7] or [13, Theorem 8.5.1]). Because of linearity, it suffices to focus on the case $z_0=1$; the corresponding solution is called the generalized exponential function and denoted by $t\mapsto e_{\textrm{d}P}(t,t_0)$. An explicit formula for the generalized exponential function is available in [13, Theorem 8.5.4].

Let $\textrm{BV}([a,b],{\mathbb {R}})$ be the class of all real functions on [a, b] having bounded variation, and denote

$$\begin{aligned} {\mathcal {P}}_{t_0}([a,b],{\mathbb {R}})&=\{P\in \text {BV}([a,b],{\mathbb {R}}): 1+\Delta ^+P(t)\ne 0 \text { for } t\in [a,t_0) \\ {}&\qquad \text {and } 1-\Delta ^-P(t)\ne 0 \text { for } t\in (t_0,b]\},\\ {\mathcal {P}}^+([a,b],{\mathbb {R}})&=\{P\in \text {BV}([a,b],{\mathbb {R}}): 1+\Delta ^+P(t)>0 \text { for } t\in [a,b)\\ {}&\qquad \text {and } 1-\Delta ^-P(t)> 0 \text { for } t\in (a,b]\}. \end{aligned}$$

The meaning of the class ${\mathcal {P}}_{t_0}([a,b],{\mathbb {R}})$ is clear – it contains all functions P for which it is possible to define $t\mapsto e_{\textrm{d}P}(t,t_0)$. In general, $e_{\textrm{d}P}$ need not be positive; it can vanish or become negative. However, if we assume that $P\in {\mathcal {P}}^+([a,b],{\mathbb {R}})$, then $e_{\textrm{d}P}(t,t_0)>0$ for all $t\in [a,b]$; see [12, Theorem 3.5] or [13, Theorem 8.5.9].

There is a product rule for generalized exponentials: For each pair $P,Q\in {\mathcal {P}}_{t_0}([a,b],{\mathbb {R}})$, there is a function $R\in {\mathcal {P}}_{t_0}([a,b],{\mathbb {R}})$ such that

$$\begin{aligned} e_{\textrm{d}P}(t,t_0)e_{\textrm{d}Q}(t,t_0)=e_{\textrm{d}R}(t,t_0),\quad t\in [a,b]. \end{aligned}$$

If P, Q are continuous, then $R=P+Q$; the general case is more complicated, and an explicit formula for R can be found in [12, Theorem 3.3] or [13, Theorem 8.5.6].

On the other hand, the analogue of the power rule for generalized exponentials is not yet available, and the goal of the present section is to obtain this formula. Given a function $P\in \mathcal P_{t_0}([a,b],{\mathbb {R}})$ and a number $\alpha \in {\mathbb {R}}$, we would like to find a function denoted by $\alpha *P$ such that

$$\begin{aligned} e_{\textrm{d}P}(t,t_0)^{\alpha }=e_{\textrm{d}(\alpha *P)}(t,t_0),\quad t\in [a,b]. \end{aligned}$$

In other words, if $z(t)=e_{\textrm{d}P}(t,t_0)$, we are looking for a function $\alpha *P$ such that

$$\begin{aligned} z(t)^{\alpha }=1+\int _{t_0}^t z(s)^{\alpha }\,\textrm{d}(\alpha *P)(s),\quad t\in [a,b]. \end{aligned}$$

(8.2)

Observe that if P is continuously differentiable with $P'=p$, it suffices to take $\alpha *P=\alpha P$, because $\frac{\textrm{d}}{\textrm{d}t}z(t)^{\alpha }=\alpha z(t)^{\alpha -1}z'(t)=z(t)^{\alpha }\alpha p(t)$. However, if P is discontinuous, the formula for $\alpha *P$ is more complicated. To guess the correct formula, suppose there exists a function $\alpha *P$ such that (8.2) holds. Then, using the relations (6.1) and (6.2), we necessarily have

$$\begin{aligned} z(t+)^{\alpha }-z(t)^{\alpha }=\Delta ^+z(t)^{\alpha }=z(t)^{\alpha }\Delta ^+(\alpha *P)(t),\\ z(t)^{\alpha }-z(t-)^{\alpha }=\Delta ^-z(t)^{\alpha }=z(t)^{\alpha }\Delta ^-(\alpha *P)(t). \end{aligned}$$

On the other hand, Eq. (8.1) and the relations (6.1) and (6.2) imply

$$\begin{aligned} z(t+)=(1+\Delta ^+P(t))z(t) \text{ and } z(t-)=(1-\Delta ^- P(t))z(t),\quad t\in [a,b]. \end{aligned}$$

(8.3)

Hence, we obtain the following formulas for the jumps of $\alpha *P$:

$$\begin{aligned} (1+\Delta ^+P(t))^{\alpha }-1=\Delta ^+(\alpha *P)(t), \end{aligned}$$

(8.4)

$$\begin{aligned} 1-(1-\Delta ^- P(t))^{\alpha }=\Delta ^-(\alpha *P)(t). \end{aligned}$$

(8.5)

Consequently,

$$\begin{aligned} (1+\Delta ^+P(t))^{\alpha }-(1-\Delta ^- P(t))^{\alpha }=\Delta (\alpha *P)(t). \end{aligned}$$

The idea behind the formula for $\alpha *P$ presented in the next theorem is now simple: We begin with $\alpha P$, subtract its jumps and replace them by the jumps calculated above.

If $\alpha \in {\mathbb {R}}\setminus {\mathbb {N}}$, we have to assume that $P\in {\mathcal {P}}^+([a,b],{\mathbb {R}})$ to ensure that $e_{\textrm{d}P}$ remains positive.

Theorem 8.1

Suppose that $P:[a,b]\rightarrow {\mathbb {R}}$ has bounded variation, $t_0\in [a,b]$, and either $\alpha \in {\mathbb {N}}$ and $P\in \mathcal P_{t_0}([a,b],{\mathbb {R}})$, or $\alpha \in {\mathbb {R}}\setminus {\mathbb {N}}$ and $P\in {\mathcal {P}}^+([a,b],{\mathbb {R}})$. If $z(t)=e_{\textrm{d}P}(t,t_0)$ for all $t\in [a,b]$, then

$$\begin{aligned} z(t)^{\alpha }=1+\int _{t_0}^t z(s)^{\alpha }\,\textrm{d}(\alpha *P)(s),\quad t\in [a,b], \end{aligned}$$

(8.6)

where $\alpha *P:[a,b]\rightarrow {\mathbb {R}}$ is given by

$$\begin{aligned} (\alpha *P)(s)={\left\{ \begin{array}{ll} \alpha P(s)-\alpha \left( \Delta ^+P(t_0)+\sum \limits _{\tau \in (t_0,s)}\Delta P(\tau )+\Delta ^-P(s)\right) \\ +\sum \limits _{\tau \in (t_0,s)}\left( (1+\Delta ^+ P(\tau ))^{\alpha }-(1-\Delta ^-P(\tau ))^{\alpha }\right) \\ +(1+\Delta ^+P(t_0))^{\alpha }-(1-\Delta ^-P(s))^{\alpha },&{}s>t_0,\\ \alpha P(t_0),&{}s=t_0,\\ \alpha P(s)+\alpha \left( \Delta ^+P(s)+\sum \limits _{\tau \in (s,t_0)}\Delta P(\tau )+\Delta ^-P(t_0)\right) \\ +\sum \limits _{\tau \in (s,t_0)}\left( (1-\Delta ^-P(\tau ))^{\alpha }-(1+\Delta ^+ P(\tau ))^{\alpha }\right) \\ +(1-\Delta ^-P(t_0))^{\alpha }-(1+\Delta ^+P(s))^{\alpha },&{}s<t_0. \end{array}\right. }\qquad \end{aligned}$$

(8.7)

The function $\alpha *P$ has bounded variation, and its jumps are given by the relations (8.4) and (8.5).

Moreover, $(\alpha *P)\in {\mathcal {P}}_{t_0}([a,b],{\mathbb {R}})$, and $e_{\textrm{d}P}(t,t_0)^{\alpha }=e_{\textrm{d}(\alpha *P)}(t,t_0)$ holds for all $t\in [a,b]$.

Proof

Throughout the proof, we will assume that $\alpha \ne 0$; the case $\alpha =0$ is straightforward to check.

The fact that P has bounded variation implies that

$$\begin{aligned} V=\sum _{t\in [a,b)}|\Delta ^+ P(\tau )|+\sum _{t\in (a,b]}|\Delta ^- P(\tau )|<\infty \end{aligned}$$

(8.8)

(see [13, Corollary 2.3.8]). We will also need the fact that

$$\begin{aligned} \sum _{t\in [a,b)}|(1+\Delta ^+P(t))^{\alpha }-1|+\sum _{t\in (a,b]}|1-(1-\Delta ^- P(t))^{\alpha }|<\infty . \end{aligned}$$

(8.9)

To see this, let $D\subset [a,b]$ be the set of all discontinuity points of P, and $D'\subset D$ the set of all points $t\in [a,b]$ satisfying $\Delta ^+P(t)\ge -1/2$ and $\Delta ^-P(t)\le 1/2$. For each $t\in D'$, the values $1+\Delta ^+P(t)$ and $1-\Delta ^- P(t)$ lie between 1/2 and $1+V$. The function $x\mapsto x^{\alpha }$ is Lipschitz continuous on $[1/2,1+V]$; if L is the corresponding Lipschitz constant, we have

$$\begin{aligned}{} & {} \sum _{t\in D'}|(1+\Delta ^+P(t))^{\alpha }-1|+\sum _{t\in D'}|1-(1-\Delta ^- P(t))^{\alpha }|\\{} & {} \quad \le L \sum _{t\in D'}|\Delta ^+P(t)|+L \sum _{t\in D'}|\Delta ^-P(t)|<\infty . \end{aligned}$$

Since $D\setminus D'$ is finite (see [13, Corollary 4.1.7]), it follows that (8.9) holds.

Suppose first that $t\in [t_0,b]$. Consider the functions

$$\begin{aligned} P^B(s)&=\sum _{\tau \in [t_0,b)}\Delta ^+P(\tau )\chi _{(\tau ,b]}(s)+\sum _{\tau \in (t_0,b]}\Delta ^-P(\tau )\chi _{[\tau ,b]}(s),\quad s\in [t_0,b],\\ (\alpha *P)^C(s)&=\alpha (P(s)-P^B(s)),\quad s\in [t_0,b],\\ (\alpha *P)^B(s)&=\sum _{\tau \in [t_0,b)}((1+\Delta ^+P(\tau ))^{\alpha }-1)\chi _{(\tau ,b]}(s)\\&\quad +\sum _{\tau \in (t_0,b]}(1-(1-\Delta ^-P(\tau ))^{\alpha })\chi _{[\tau ,b]}(s),\quad s\in [t_0,b]. \end{aligned}$$

Note that (8.8) ensures that $P^B$ is well-defined. Furthermore, it is a step function with countably many steps, has bounded variation, and its jumps coincide with the jumps of P; in fact, it is the jump part of P. Consequently, $P-P^B$ as well as $(\alpha *P)^C$ are continuous and have bounded variation; in fact, $(\alpha *P)^C$ is the continuous part of $\alpha P$. Similarly, because of (8.9), the definition of $(\alpha *P)^B$ makes sense. It is a step function with countably many steps, has bounded variation, and its jumps are given by

$$\begin{aligned} \Delta ^+(\alpha *P)^B(s)=(1+\Delta ^+P(s))^{\alpha }-1, \end{aligned}$$

(8.10)

$$\begin{aligned} \Delta ^-(\alpha *P)^B(s)=1-(1-\Delta ^- P(s))^{\alpha }. \end{aligned}$$

(8.11)

Next, observe that

$$\begin{aligned} (\alpha *P)^C(s)+(\alpha *P)^B(s)=(\alpha *P)(s), \end{aligned}$$

where $\alpha *P$ is given by (8.7). Therefore, $\alpha *P$ has bounded variation. Since $(\alpha *P)^C$ is continuous, the jumps of $(\alpha *P)$ coincide with the jumps of $(\alpha *P)^B$; this means that the relations (8.4) and (8.5) hold.

If $\alpha \in {\mathbb {R}}\setminus {\mathbb {N}}$, the condition $P\in {\mathcal {P}}^+([a,b],{\mathbb {R}})$ ensures that z(t), $z(t+)$ and $z(t-)$ are positive for all $t\in [a,b]$.

Let $t\in (t_0,b]$ be fixed. According to Theorem 7.9 with $g(s)=P(s)$ and $p(s)=1$, we have

$$\begin{aligned}&z(t)^\alpha -1=z(t)^\alpha -z(t_0)^\alpha =\int _{t_0}^t z(s)^{\alpha }\left( \alpha \odot _{P,t_0,t}1\right) (s)\,\text {d}P(s)\nonumber \\ {}&=\int _{t_0}^{t} z(s)^{\alpha }I(s)\,\text {d}P(s), \end{aligned}$$

(8.12)

where

$$\begin{aligned} I(s)=\int _0^1 \alpha (1+\xi \Delta P(s)-\Delta ^- P(s))^{\alpha -1}\,\textrm{d}\xi ,\quad s\in [t_0,t], \end{aligned}$$

with the convention that $\Delta P(t)=\Delta ^-P(t)$ and $\Delta ^-P(t_0)=0$, or equivalently

$$\begin{aligned} I(s)= & {} \int _0^1 \alpha (\xi (1+\Delta ^+P(s)\chi _{[t_0,t)}(s))\\{} & {} +(1-\xi )(1-\Delta ^- P(s)\chi _{(t_0,t]}(s))^{\alpha -1}\,\textrm{d}\xi ,\quad s\in [t_0,t]. \end{aligned}$$

If P is continuous at s, then $I(s)=\alpha$. Since P has bounded variation, this happens for all s with countably many exceptions. Thus, it makes sense to write

$$\begin{aligned} z(t)^\alpha -1=\alpha \int _{t_0}^{t} z(s)^{\alpha }\,\textrm{d}P(s)+\int _{t_0}^{t} z(s)^{\alpha }(I(s)-\alpha )\,\textrm{d}P(s). \end{aligned}$$

(8.13)

The function $s\mapsto z(s)^{\alpha }(I(s)-\alpha )$ is regulated and zero with countably many exceptions. Thus, using the formula (6.3), we get

$$\begin{aligned}&\int _{t_0}^{t} z(s)^{\alpha }(I(s)-\alpha )\,\textrm{d}P(s)=z(t_0)^{\alpha }(I(t_0)-\alpha )\Delta ^+P(t_0)\\&\qquad \ +\sum _{\tau \in (t_0,t)}z(\tau )^{\alpha }(I(\tau )-\alpha )\Delta P(\tau )+z(t)^{\alpha }(I(t)-\alpha )\Delta ^-P(t)\\&\qquad =z(t_0)^{\alpha }I(t_0)\Delta ^+P(t_0)+\sum _{\tau \in (t_0,t)}z(\tau )^{\alpha }I(\tau )\Delta P(\tau )+z(t)^{\alpha }I(t)\Delta ^-P(t)\\&\qquad \ -\alpha \left( z(t_0)^{\alpha }\Delta ^+P(t_0)+\sum _{\tau \in (t_0,t)}z(\tau )^{\alpha }\Delta P(\tau )+z(t)^{\alpha }\Delta ^-P(t)\right) . \end{aligned}$$

Using the formula (6.4), we can rewrite the expression on the last line as an integral:

$$\begin{aligned}{} & {} z(t_0)^{\alpha }\Delta ^+P(t_0)+\sum _{\tau \in (t_0,t)}z(\tau )^{\alpha }\Delta P(\tau )+z(t)^{\alpha }\Delta ^-P(t)\nonumber \\{} & {} =z(t_0)^{\alpha }\Delta ^+P^B(t_0)+\sum _{\tau \in (t_0,t)}z(\tau )^{\alpha }\Delta P^B(\tau )+z(t)^{\alpha }\Delta ^-P^B(t)\nonumber \\{} & {} =\int _{t_0}^t z(s)^{\alpha }\,\textrm{d}P^B(s). \end{aligned}$$

(8.14)

To investigate the terms involving the function I, observe that Eq. (7.2) implies

$$\begin{aligned} (\gamma -\beta )\int _0^1 \alpha (\xi \gamma +(1-\xi )\beta )^{\alpha -1}\,\textrm{d}\xi = \gamma ^\alpha -\beta ^{\alpha } \end{aligned}$$

whenever $\beta ,\gamma \in {\mathbb {R}}$ and $\alpha \in {\mathbb {N}}$ or $\beta ,\gamma >0$ and $\alpha \in {\mathbb {R}}{\setminus }{\mathbb {N}}$. It follows that

$$\begin{aligned} I(t_0)\Delta ^+P(t_0)&=\left( \int _0^1 \alpha (\xi (1+\Delta ^+P(t_0))+(1-\xi ))^{\alpha -1}\,\text {d}\xi \right) \Delta ^+P(t_0) \\ {}&\quad =(1+\Delta ^+P(t_0))^{\alpha }-1,\\ I(t)\Delta ^-P(t)&=\left( \int _0^1 \alpha (\xi +(1-\xi )(1-\Delta ^- P(t))^{\alpha -1}\,\text {d}\xi \right) \Delta ^-P(t) \\ {}&\quad =1-(1-\Delta ^- P(t))^{\alpha },\\ I(\tau )\Delta P(\tau )&=\left( \int _0^1 \alpha (\xi (1+\Delta ^+P(\tau ))+(1-\xi )(1-\Delta ^- P(\tau ))^{\alpha -1}\,\text {d}\xi \right) \Delta P(\tau )\\ {}&=(1+\Delta ^+P(\tau ))^{\alpha }-(1-\Delta ^- P(\tau ))^{\alpha } \end{aligned}$$

for all $\tau \in (t_0,t)$. The expressions on the right-hand sides correspond, respectively, to $\Delta ^+(\alpha *P)^B(t_0)$, $\Delta ^-(\alpha *P)^B(t)$, and $\Delta (\alpha *P)^B(\tau )$. Thus, using again the formula (6.4), we see that

$$\begin{aligned} z(t_0)^{\alpha }I(t_0)\Delta ^+P(t_0){} & {} +\sum _{\tau \in (t_0,t)}z(\tau )^{\alpha }I(\tau )\Delta P(\tau )+z(t)^{\alpha }I(t)\Delta ^-P(t) \nonumber \\{} & {} =\int _{t_0}^t z(s)^{\alpha }\,\textrm{d}(\alpha *P)^B(s). \end{aligned}$$

(8.15)

Using Eq. (8.14) and Eq. (8.15) and returning back to Eq. (8.13), we get

$$\begin{aligned} z(t)^\alpha -1&=\alpha \int _{t_0}^{t} z(s)^{\alpha }\,\textrm{d}P(s)-\alpha \int _{t_0}^t z(s)^{\alpha }\,\textrm{d}P^B(s)+\int _{t_0}^t z(s)^{\alpha }\,\textrm{d}(\alpha *P)^B(s)\\&=\int _{t_0}^t z(s)^{\alpha }\,\textrm{d}(\alpha *P)^C(s)+\int _{t_0}^t z(s)^{\alpha }\,\textrm{d}(\alpha *P)^B(s)=\int _{t_0}^t z(s)^{\alpha }\,\textrm{d}(\alpha *P)(s), \end{aligned}$$

i.e., we have proved that Eq. (8.6) holds for all $t\in (t_0,b]$.

The case $t\in [a,t_0)$ is similar. The functions $P^B$, $(\alpha *P)^C$, $(\alpha *P)^B$ are now given by

$$\begin{aligned} P^B(s)&=-\sum _{\tau \in [a,t_0)}\Delta ^+P(\tau )\chi _{[a,\tau ]}(s)-\sum _{\tau \in (a,t_0]}\Delta ^-P(\tau )\chi _{[a,\tau )}(s),\quad s\in [a,t_0],\\ (\alpha *P)^C(s)&=\alpha (P(s)-P^B(s)),\quad s\in [a,t_0],\\ (\alpha *P)^B(s)&=-\sum _{\tau \in [a,t_0)}((1+\Delta ^+P(\tau ))^{\alpha }-1)\chi _{[a,\tau ]}(s)\\&\quad -\sum _{\tau \in (a,t_0]}(1-(1-\Delta ^-P(\tau ))^{\alpha })\chi _{[a,\tau )}(s),\quad s\in [t_0,b]. \end{aligned}$$

As in the previous part of the proof, one finds that

$$\begin{aligned} z(t)^\alpha -1=-(z(t_0)^\alpha -z(t)^\alpha )=-\int ^{t_0}_{t} z(s)^{\alpha }I(s)\,\textrm{d}P(s), \end{aligned}$$

where I is now given by

$$\begin{aligned} I(s)= & {} \int _0^1 \alpha (\xi (1+\Delta ^+P(s)\chi _{[t,t_0)}(s))\\{} & {} +(1-\xi )(1-\Delta ^- P(s)\chi _{(t,t_0]}(s))^{\alpha -1}\,\textrm{d}\xi ,\quad s\in [t,t_0]. \end{aligned}$$

The remaining calculations are straightforward adaptations of those in the previous part and, thus, we omit them.

The relations (8.4) and (8.5) imply that

$$\begin{aligned} (1+\Delta ^+(\alpha *P)(t))&= (1+\Delta ^+P(t))^{\alpha }\ne 0,\\ (1-\Delta ^-(\alpha *P)(t))&= (1-\Delta ^-P(t))^{\alpha }\ne 0 \end{aligned}$$

for all $t\in [a,b]$, and therefore $(\alpha *P)\in \mathcal P_{t_0}([a,b],{\mathbb {R}})$. Since $e_{\textrm{d}(\alpha *P)}$ is the unique solution of Eq. (8.6), we see that $e_{\textrm{d}P}(t,t_0)^{\alpha }=z(t)^{\alpha }=e_{\textrm{d}(\alpha *P)}(t,t_0)$. $\square$

Remark 8.2

Using the definition of $\alpha *P$ with $\alpha =-1$, a simple calculation reveals that

$$\begin{aligned} (-1*P)(s)={\left\{ \begin{array}{ll} -P(s)+\sum \limits _{\tau \in [t_0,s)}\frac{(\Delta ^+P(\tau ))^2}{1+\Delta ^+P(\tau )} -\sum \limits _{\tau \in (t_0,s]}\frac{(\Delta ^-P(\tau ))^2}{1-\Delta ^-P(\tau )},&{} s\ge t_0,\\ -P(s)-\sum \limits _{\tau \in [s,t_0)}\frac{(\Delta ^+P(\tau ))^2}{1+\Delta ^+P(\tau )} +\sum \limits _{\tau \in (s,t_0]}\frac{(\Delta ^-P(\tau ))^2}{1-\Delta ^-P(\tau )},&{} s\le t_0. \end{array}\right. } \end{aligned}$$

This result agrees with [12, Theorem 3.4] (see also [13, Theorem 8.5.8]), where the previous function is denoted by $\ominus P$, and it is shown that $(e_{{\textrm{d}}P}(t,t_0))^{-1}=e_{{\textrm{d}}(\ominus P)}(t,t_0)$.

Let us briefly explain the connection between the results of the present section and those obtained earlier in Sect. 4 for the g-exponential function. Consider the linear Stieltjes integral equation

$$\begin{aligned} z(t)=1+\int _{t_0}^t p(s)z(s)\,\textrm{d}g(s),\quad t\in [a,b], \end{aligned}$$

(8.16)

where $g:[a,b]\rightarrow {\mathbb {R}}$ is nondecreasing and $p:[a,b]\rightarrow {\mathbb {R}}$ is Kurzweil–Stieltjes integrable with respect to g.

It follows from the substitution theorem that Eq. (8.16) is a special case of Eq. (8.1) with

$$\begin{aligned} P(t)=\int _{t_0}^t p(s)\,\textrm{d}g(s),\quad t\in [a,b]. \end{aligned}$$

(8.17)

To ensure that P has bounded variation, we will assume that p is absolutely integrable with respect to g (see [13, Section 6.7]). Note that the relations (6.1) and (6.2) imply

$$\begin{aligned} \Delta ^+P(t)=p(t)\Delta ^+g(t),\quad \Delta ^-P(t)=p(t)\Delta ^-g(t) \end{aligned}$$

for all $t\in [a,b]$. Hence, we have $P\in \mathcal P_{t_0}([a,b],{\mathbb {R}})$ if and only if

$$\begin{aligned} 1+p(t)\Delta ^+g(t)\ne 0 \text{ for } t\in [a,t_0) \text{ and } 1-p(t)\Delta ^-g(t)\ne 0 \text{ for } t\in (t_0,b]. \end{aligned}$$

Under these conditions, Eq. (8.16) has the unique solution $z(t)=e_{\textrm{d}P}(t,t_0)$. If we in addition assume that

$$\begin{aligned} 1+p(t)\Delta ^+g(t)>0 \text{ for } t\in [a,b) \text{ and } 1-p(t)\Delta ^-g(t)> 0 \text{ for } t\in (a,b], \end{aligned}$$

then $P\in {\mathcal {P}}^+([a,b],{\mathbb {R}})$, and $z(t)>0$ for all $t\in [a,b]$.

In the special case when g is left-continuous and $t_0=a$, the solution z coincides with the g-exponential function introduced earlier, i.e., we have

$$\begin{aligned} e_{\textrm{d}P}(t,t_0)=\exp _g(p,t),\quad t\in [a,b]. \end{aligned}$$

(8.18)

The product rule

$$\begin{aligned} \exp _g(p,t)\exp _q(p,t)=\exp _g(p\oplus q,t),\quad t\in [a,b], \end{aligned}$$

(8.19)

where

$$\begin{aligned} (p\oplus q)(t)=p(t)+q(t)+p(t)q(t)\Delta ^+g(t), \end{aligned}$$

can be derived from the product rule for generalized exponential functions (see [12, Theorem 3.3] or [13, Theorem 8.5.6]).

To derive the power rule, we use Theorem 7.9, which implies that the function $z(t)=\exp _g(p,t)$ satisfies

$$\begin{aligned} z(t)^{\alpha }=1+\int _{t_0}^t z(s)^{\alpha }\left( \alpha \odot _{g,t_0,t}p\right) \hspace{-0.05cm}(s)\,\textrm{d}g(s),\quad t\in [a,b]. \end{aligned}$$

(8.20)

In the left-continuous case, the definition of $\alpha \odot _{g,t_0,t}p$ reduces to (cf. Remark 7.7)

$$\begin{aligned} (\alpha \odot _{g,t_0,t}p)(s)={\left\{ \begin{array}{ll} \frac{(1+p(s)\Delta ^+ g(s))^{\alpha }-1}{\Delta ^+ g(s)}&{} \text{ if } \Delta ^+g(s)\ne 0,\\ \alpha p(s)&{} \text{ if } \Delta ^+ g(s)=0, \end{array}\right. } \end{aligned}$$

with the convention that $\Delta ^+ g(s)=0$ if $s=\max (t_0,t)$. However, since the integrator in Eq. (8.20) is left-continuous, the value of the integrand for $s=\max (t_0,t)$ does not matter, and we obtain

$$\begin{aligned} z(t)^{\alpha }=1+\int _{t_0}^t z(s)^{\alpha }\left( \alpha \odot p\right) \hspace{-0.05cm}(s)\,\textrm{d}g(s),\quad t\in [a,b], \end{aligned}$$

(8.21)

where

$$\begin{aligned} (\alpha \odot p)(s)={\left\{ \begin{array}{ll} \frac{(1+p(s)\Delta ^+ g(s))^{\alpha }-1}{\Delta ^+ g(s)}&{} \text{ if } \Delta ^+g(s)\ne 0,\\ \alpha p(s)&{} \text{ if } \Delta ^+ g(s)=0. \end{array}\right. } \end{aligned}$$

This is in agreement with the definition of $\odot$ given in Sect. 4, cf. Eq. (4.11).

Equation (8.21) already implies the power rule

$$\begin{aligned} \exp _g(p,t)^{\alpha }=\exp _g(\alpha \odot p,t),\quad t\in [a,b], \end{aligned}$$

because both sides of this identity satisfy the same integral equation.

9 Integral version of the Bernoulli Equation

To discover the integral version of the Bernoulli equation, we begin with the linear Stieltjes integral equation

$$\begin{aligned} z(t)=z(t_0)+\int _{t_0}^t \left( p(s)z(s)+f(s)\right) \,\textrm{d}g(s). \end{aligned}$$

Suppose that z is a solution satisfying $z(t)>0$, $z(t+)>0$ and $z(t-)>0$ for all $t\in [a,b]$. We will find an integral equation for the function $y(t)=z(t)^{-1/\alpha }$, where $\alpha \in \mathbb R{\setminus }\{0\}$. Using Theorem 7.8 with $w(s)=p(s)z(s)+f(s)$, we get

$$\begin{aligned} z(t)^{-\frac{1}{\alpha }}-z(t_0)^{-\frac{1}{\alpha }} =\int _{t_0}^t \left( \frac{-1}{\alpha }\odot _{g,t_0,t} \left( p+\frac{f}{z}\right) \right) \hspace{-0.1cm}(s)z(s)^{-\frac{1}{\alpha }}\,\textrm{d}g(s). \end{aligned}$$

Rewriting the equation in terms of $y(s)=z(s)^{-\frac{1}{\alpha }}$, we obtain the following integral version of the Bernoulli equation:

$$\begin{aligned} y(t)-y(t_0)=\int _{t_0}^t \left( \frac{-1}{\alpha }\odot _{g,t_0,t} \left( p+{f}y^{\alpha }\right) \right) \hspace{-0.1cm}(s)y(s)\,\textrm{d}g(s),\quad t\in [a,b]. \end{aligned}$$

According to Remark 7.7, we have

$$\begin{aligned}{} & {} \left( \frac{-1}{\alpha }\odot _{g,t_0,t} \left( p+{f}y^{\alpha }\right) \right) \hspace{-0.1cm}(s)\\{} & {} ={\left\{ \begin{array}{ll} \frac{(1+\left( p(s)+f(s)y(s)^{\alpha }\right) \Delta ^+ g(s))^{-\frac{1}{\alpha }}-(1-\left( p(s)+f(s)y(s)^{\alpha }\right) \Delta ^- g(s))^{-\frac{1}{\alpha }}}{\Delta g(s)},&{} \Delta g(s)\ne 0,\\ \frac{-1}{\alpha }\left( p(s)+f(s)y(s)^{\alpha }\right) \left( 1-\left( p(s)+f(s)y(s)^{\alpha }\right) \Delta ^- g(s)\right) ^{-\frac{1}{\alpha }-1},&{} \Delta g(s)=0, \end{array}\right. } \end{aligned}$$

with the convention that $\Delta g(s)=\Delta ^-g(s)$ if $s=\max (t,t_0)$, and $\Delta ^-g(s)=0$ and $\Delta g(s)=\Delta ^+g(s)$ if $s=\min (t,t_0)$.

In the special case when $\alpha =1$, the previous formula reduces to

$$\begin{aligned}{} & {} \left( -1\odot _{g,t_0,t} \left( p+{f}y\right) \right) \hspace{-0.05cm}(s)\\{} & {} \qquad =-\frac{p(s)+f(s)y(s)}{\left( 1+\left( p(s)+f(s)y(s)\right) \Delta ^+g(s)\right) \left( 1-\left( p(s)+f(s)y(s)\right) \Delta ^-g(s)\right) }, \end{aligned}$$

no matter whether $\Delta g(s)$ is zero or not. Therefore, the Bernoulli equation corresponding to $\alpha =1$ has the form

$$\begin{aligned}{} & {} y(t)-y(t_0)\\{} & {} \qquad =-\int _{t_0}^t \frac{y(s)(p(s)+f(s)y(s))}{\left( 1+\left( p(s)+f(s)y(s)\right) \Delta ^+g(s)\right) \left( 1-\left( p(s)+f(s)y(s)\right) \Delta ^-g(s)\right) } \,\textrm{d}g(s), \end{aligned}$$

which agrees with the Stieltjes-integral version of the logistic equation introduced in [10, Section 7].