# Nonlinear integral inequalities with delay for discontinuous functions and their applications

## Abstract

This paper investigates integral inequalities with delay for discontinuous functions involving two nonlinear terms. We do not require the classes and ȷ in Gallo and Piccirillo’s paper (Bound. Value Probl. 2009:808124, 2009). Our main results can be applied to generalize Gallo and Piccirillo’s results and Iovane’s results (Nonlinear Anal., Theory Methods Appl. 66:498-508, 2007). Examples to show the bounds of solutions of an impulsive differential equation are also given, which can not be estimated by Gallo and Piccirillo’s results.

MSC:26D15, 26D20.

## 1 Introduction

The Gronwall-Bellman integral inequalities and their various linear and nonlinear generalizations, involving continuous or discontinuous functions, play very important roles in investigating different qualitative characteristics of solutions for differential equations and impulsive differential equations such as existence, uniqueness, continuation, boundedness, continuous dependence of parameters, stability, attraction, practical stability. The literature on inequalities for continuous functions and their applications is vast (see [18]). Recently, more attention has been paid to generalizations of Gronwall-Bellman’s results for discontinuous functions (see [917]) and their applications (see [11, 18, 19]). Among them, one of the important things is that Samoilenko and Perestyuk [17] studied the following inequality

$u\left(x\right)\le c+{\int }_{{x}_{0}}^{x}f\left(s\right)u\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+\sum _{{x}_{0}<{x}_{i}
(1.1)

about the nonnegative piecewise continuous function $u\left(x\right)$, where c, ${\beta }_{i}$ are nonnegative constants, $f\left(s\right)$ is a positive function, and ${x}_{i}$ are the first kind discontinuity points of the function $u\left(x\right)$. Then Borysenko [20] considered

$\begin{array}{r}u\left(x\right)\le c+{\int }_{{x}_{0}}^{x}f\left(s\right){u}^{m}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+\sum _{{x}_{0}<{x}_{i}0,m\ne 1,\\ u\left(x\right)\le c+{\int }_{{x}_{0}}^{x}f\left(s\right){u}^{m}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+\sum _{{x}_{0}<{x}_{i}0,m\ne 1.\end{array}$
(1.2)

He replaced the constant c by a positive monotonously nondecreasing function $a\left(x\right)$, and also estimated the inequalities

$\begin{array}{r}u\left(x\right)\le a\left(x\right)+{\int }_{{x}_{0}}^{x}f\left(s\right)u\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+\sum _{{x}_{0}<{x}_{i}0,\\ u\left(x\right)\le a\left(x\right)+{\int }_{{x}_{0}}^{x}f\left(s\right){u}^{m}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+\sum _{{x}_{0}<{x}_{i}0,m\ne 1.\end{array}$
(1.3)

In 2005, he [18] generalized the inequalities above from one integral to two integrals with a form

$\begin{array}{rl}u\left(x\right)\le & c+{\int }_{{x}_{0}}^{x}q\left(s\right)u\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+{\int }_{{x}_{0}}^{x}q\left(s\right){\int }_{{x}_{0}}^{s}g\left(\tau \right){u}^{m}\left(\tau \right)\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds\\ +\sum _{{x}_{0}<{x}_{i}0.\end{array}$
(1.4)

In 2007, Iovane [21] investigated the inequalities with delay

$\begin{array}{r}u\left(x\right)\le a\left(x\right)+{\int }_{{x}_{0}}^{x}f\left(s\right)u\left(b\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+\sum _{{x}_{0}<{x}_{i}0,\\ u\left(x\right)\le a\left(x\right)+{\int }_{{x}_{0}}^{x}f\left(s\right){u}^{m}\left(b\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+\sum _{{x}_{0}<{x}_{i}0,m\ne 1.\end{array}$
(1.5)

Later, Gallo and Piccirillo [22] further discussed

$u\left(x\right)\le a\left(x\right)+h\left(x\right){\int }_{{x}_{0}}^{x}f\left(s\right)w\left(u\left(b\left(s\right)\right)\right)\phantom{\rule{0.2em}{0ex}}ds+\sum _{{x}_{0}<{x}_{i}
(1.6)

with a general nonlinear term $w\left(u\right)$ of u. They assumed that $w\in \mathrm{\wp }$ or $w\in ȷ$, where the class consists of all nonnegative, nondecreasing and continuous functions $w\left(u\right)$ on $\left[0,\mathrm{\infty }\right)$ such that $w\left(0\right)=0$ and $w\left(\alpha u\right)\le w\left(\alpha \right)w\left(u\right)$ for all $\alpha >0$ and $u\ge 0$, and the class ȷ consists of all positive, nondecreasing and continuous functions $w\left(u\right)$ on $\left(0,\mathrm{\infty }\right)$ such that $w\left(0\right)=0$ and $w\left({\alpha }^{-1}u\right)>{\alpha }^{-1}w\left(u\right)$ for all $\alpha \ge 1$ and $u>0$. The classes and ȷ allow a reduction of $a\left(t\right)$ to the case of a constant ${a}_{0}$ by dividing $a\left(x\right)$ if $a\left(x\right)$ is a positive and nondecreasing function. Actually, when we study behaviors of solutions of impulsive differential equations, $a\left(x\right)$ may not be a nondecreasing function, and w may not satisfy the condition $w\in \mathrm{\wp }$ or $w\in ȷ$. For example, $w\left(u\right)={e}^{u}$ does not belong to the class and ȷ for any $\alpha >1$ and large $u>0$. Thus, it is interesting to avoid such conditions.

Motivated by this observation, in this paper, we consider the following much more general inequality

$\begin{array}{rl}u\left(x\right)\le & a\left(x\right)+{\int }_{{b}_{1}\left({x}_{0}\right)}^{{b}_{1}\left(x\right)}{f}_{1}\left(x,s\right){w}_{1}\left(u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+{\int }_{{b}_{2}\left({x}_{0}\right)}^{{b}_{2}\left(x\right)}{f}_{2}\left(x,s\right){w}_{2}\left(u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ +\sum _{{x}_{0}<{x}_{i}0\end{array}$
(1.7)

with two nonlinear terms ${w}_{1}\left(u\right)$ and ${w}_{2}\left(u\right)$ of u, where we do not restrict ${w}_{1}$ and ${w}_{2}$ to the class or the class ȷ. We also show that many integral inequalities for discontinuous functions such as (1.3), (1.4) and (1.6) can be reduced to the form of (1.7). Our main result is applied to estimate the bounds of solutions of an impulsive ordinary differential equation.

## 2 Main results

Consider (1.7), and assume that

(C1) ${w}_{1}\left(x\right)$ and ${w}_{2}\left(x\right)$ are continuous and nondecreasing functions on $\left[0,\mathrm{\infty }\right)$ and are positive on $\left(0,\mathrm{\infty }\right)$ such that $\frac{{w}_{2}\left(x\right)}{{w}_{1}\left(x\right)}$ is nondecreasing;

(C2) $a\left(x\right)$ is defined on $\left[{x}_{0},\mathrm{\infty }\right)$ and $a\left({x}_{0}\right)\ne 0$; ${\beta }_{i}$ is a nonnegative constant for any positive integer i;

(C3) ${f}_{1}\left(x,s\right)$ and ${f}_{2}\left(x,s\right)$ are continuous and nonnegative functions on $\left[{x}_{0},\mathrm{\infty }\right)×\left[{x}_{0},\mathrm{\infty }\right)$;

(C4) ${b}_{1}\left(x\right)$ and ${b}_{2}\left(x\right)$ are continuously differentiable and nondecreasing such that ${x}_{0}\le {b}_{1}\left(x\right)\le x$ and ${x}_{0}\le {b}_{2}\left(x\right)\le x$ on $\left[{x}_{0},\mathrm{\infty }\right)$;

(C5) For $x\in \left[{x}_{0},\mathrm{\infty }\right)$, $u\left(x\right)$ is nonnegative and piecewise-continuous with the first kind of discontinuities at the points ${x}_{i}:{x}_{0}<{x}_{1}<\cdots$ , where i is a nonnegative integer and ${lim}_{i\to \mathrm{\infty }}{x}_{i}=\mathrm{\infty }$.

Let ${W}_{j}\left(u\right)={\int }_{{\stackrel{˜}{u}}_{j}}^{u}\frac{dz}{{w}_{j}\left(z\right)}$ for $u\ge {\stackrel{˜}{u}}_{j}$ and $j=1,2$, where ${\stackrel{˜}{u}}_{j}$ is a given positive constant. Clearly, ${W}_{j}$ is strictly increasing so its inverse ${W}_{j}^{-1}$ is well defined, continuous and increasing in its corresponding domain.

Theorem 2.1 Suppose that (C k ) ($k=1,\dots ,5$) hold, and $u\left(x\right)$ satisfies (1.7) for a positive constant m. Let ${u}_{i}\left(x\right)=u\left(x\right)$ for $x\in \left[{x}_{i},{x}_{i+1}\right)$. Then the estimate of $u\left(x\right)$ is recursively given by for $x\in \left[{x}_{i},{x}_{i+1}\right)$, $i=0,1,2,\dots$ ,

${u}_{i}\left(x\right)\le {W}_{2}^{-1}\left[{W}_{2}\circ {W}_{1}^{-1}\left({W}_{1}\left({r}_{i+1}\left(x\right)\right)+{\int }_{{b}_{1}\left({x}_{i}\right)}^{{b}_{1}\left(x\right)}{\stackrel{˜}{f}}_{1}\left(x,s\right)\phantom{\rule{0.2em}{0ex}}ds\right)+{\int }_{{b}_{2}\left({x}_{i}\right)}^{{b}_{2}\left(x\right)}{\stackrel{˜}{f}}_{2}\left(x,s\right)\phantom{\rule{0.2em}{0ex}}ds\right],$
(2.1)

where

$\begin{array}{c}\begin{array}{r}{\stackrel{˜}{f}}_{j}\left(x,s\right)=\underset{{x}_{0}\le \tau \le x}{max}{f}_{j}\left(\tau ,s\right),\phantom{\rule{1em}{0ex}}j=1,2,\phantom{\rule{2em}{0ex}}{r}_{1}\left(x\right)=\underset{{x}_{0}\le \tau \le x}{max}|a\left(\tau \right)|,\end{array}\hfill \\ \begin{array}{rl}{r}_{i+1}\left(x\right)=& {r}_{1}\left(x\right)+\sum _{k=1}^{i}{\int }_{{b}_{1}\left({x}_{k-1}\right)}^{{b}_{1}\left({x}_{k}\right)}{\stackrel{˜}{f}}_{1}\left(x,s\right){w}_{1}\left({u}_{k-1}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ +\sum _{k=1}^{i}{\int }_{{b}_{2}\left({x}_{k-1}\right)}^{{b}_{2}\left({x}_{k}\right)}{\stackrel{˜}{f}}_{2}\left(x,s\right){w}_{2}\left({u}_{k-1}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+\sum _{k=1}^{i}{\beta }_{k}{u}_{k-1}^{m}\left({x}_{k}-0\right),\end{array}\hfill \end{array}$
(2.2)

provided that

$\begin{array}{r}{W}_{1}\left({r}_{i+1}\left(x\right)\right)+{\int }_{{b}_{1}\left({x}_{i}\right)}^{{b}_{1}\left(x\right)}{\stackrel{˜}{f}}_{1}\left(x,s\right)\phantom{\rule{0.2em}{0ex}}ds\le {\int }_{{\stackrel{˜}{u}}_{1}}^{\mathrm{\infty }}\frac{dz}{{w}_{1}\left(z\right)},\\ {W}_{2}\circ {W}_{1}^{-1}\left({W}_{1}\left({r}_{i+1}\left(x\right)\right)+{\int }_{{b}_{1}\left({x}_{i}\right)}^{{b}_{1}\left(x\right)}{\stackrel{˜}{f}}_{1}\left(x,s\right)\phantom{\rule{0.2em}{0ex}}ds\right)+{\int }_{{b}_{2}\left({x}_{i}\right)}^{{b}_{2}\left(x\right)}{\stackrel{˜}{f}}_{2}\left(x,s\right)\phantom{\rule{0.2em}{0ex}}ds\le {\int }_{{\stackrel{˜}{u}}_{2}}^{\mathrm{\infty }}\frac{dz}{{w}_{2}\left(z\right)}.\end{array}$
(2.3)

The proof is given in Section 3.

Remark 2.1 (1) If ${w}_{j}$ satisfies ${\int }_{{\stackrel{˜}{u}}_{j}}^{\mathrm{\infty }}\frac{dz}{{w}_{j}\left(z\right)}=\mathrm{\infty }$ for $j=1,2$, then i in Theorem 2.1 can be any nonzero integer. [6] pointed out that different choices of ${\stackrel{˜}{u}}_{j}$ in ${W}_{j}$ do not affect our results for $j=1,2$. If $a\left(x\right)\equiv 0$, then define ${W}_{1}\left(0\right)=0$, and (2.1) is still true.

1. (2)

Take ${b}_{1}\left(x\right)=x$, $a\left(x\right)=c$, ${f}_{1}\left(t,s\right)=f\left(s\right)$, ${f}_{2}\left(t,s\right)=0$, ${w}_{1}\left(u\right)=u$ and $m=1$. Hence, (1.7) becomes (1.1). It is easy to check that ${W}_{1}\left(u\right)=ln\frac{u}{{\stackrel{˜}{u}}_{1}}$ and ${W}_{1}^{-1}\left(u\right)={\stackrel{˜}{u}}_{1}{e}^{u}$. From Theorem 2.1, we know that for $x\in \left[{x}_{i},{x}_{i+1}\right)$,

${u}_{i}\left(x\right)\le {r}_{i+1}\left(x\right){e}^{{\int }_{{x}_{i}}^{x}f\left(s\right)\phantom{\rule{0.2em}{0ex}}ds}$

with

${r}_{i+1}\left(x\right)=c+\sum _{k=1}^{i}{\int }_{{x}_{k-1}}^{{x}_{k}}f\left(s\right){u}_{k-1}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+\sum _{k=1}^{i}{\beta }_{k}{u}_{k-1}\left({x}_{k}-0\right).$

Hence,

$\begin{array}{r}{r}_{1}\left(x\right)=c,\phantom{\rule{2em}{0ex}}{u}_{0}\left(x\right)\le c{e}^{{\int }_{{x}_{0}}^{x}f\left(s\right)\phantom{\rule{0.2em}{0ex}}ds},\\ {r}_{2}\left(x\right)=c+{\int }_{{x}_{0}}^{{x}_{1}}f\left(s\right){u}_{0}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+{\beta }_{1}{u}_{0}\left({x}_{1}-0\right)\\ \phantom{{r}_{2}\left(x\right)}\le c+{\int }_{{x}_{0}}^{{x}_{1}}f\left(s\right)c{e}^{{\int }_{{x}_{0}}^{s}f\left(\tau \right)\phantom{\rule{0.2em}{0ex}}d\tau }\phantom{\rule{0.2em}{0ex}}ds+c{\beta }_{1}{e}^{{\int }_{{x}_{0}}^{{x}_{1}}f\left(s\right)\phantom{\rule{0.2em}{0ex}}ds}\\ \phantom{{r}_{2}\left(x\right)}=c+c{e}^{{\int }_{{x}_{0}}^{s}f\left(\tau \right)\phantom{\rule{0.2em}{0ex}}d\tau }{|}_{{x}_{0}}^{{x}_{1}}+c{\beta }_{1}{e}^{{\int }_{{x}_{0}}^{{x}_{1}}f\left(s\right)\phantom{\rule{0.2em}{0ex}}ds}=c\left(1+{\beta }_{1}\right){e}^{{\int }_{{x}_{0}}^{{x}_{1}}f\left(s\right)\phantom{\rule{0.2em}{0ex}}ds},\\ {u}_{1}\left(x\right)\le c\left(1+{\beta }_{1}\right){e}^{{\int }_{{x}_{0}}^{x}f\left(s\right)\phantom{\rule{0.2em}{0ex}}ds}.\end{array}$

After recursive calculations, we have for $x\ge {x}_{0}$

$u\left(x\right)\le c{\mathrm{\Pi }}_{{x}_{0}<{x}_{k}

which is same as the one in [17].

1. (3)

Clearly, (1.2) and (1.3) are special cases of (1.7). If ${b}^{\prime }\left(x\right)>0$ on $\left[{x}_{0},\mathrm{\infty }\right)$, then (1.6) can be rewritten as

$u\left(x\right)\le a\left(x\right)+h\left(x\right){\int }_{b\left({x}_{0}\right)}^{b\left(x\right)}\frac{f\left({b}^{-1}\left(s\right)\right)}{{b}^{\prime }\left({b}^{-1}\left(s\right)\right)}w\left(u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+\sum _{{x}_{0}<{x}_{i}

Let ${f}_{1}\left(x,s\right)=h\left(x\right)\frac{f\left({b}^{-1}\left(s\right)\right)}{{b}^{\prime }\left({b}^{-1}\left(s\right)\right)}$ and ${f}_{2}\left(x,s\right)\equiv 0$, the inequality above is same as (1.7). Similarly, (1.5) can also be reduced to (1.7).

Consider the inequality

$\begin{array}{rl}u\left(x\right)\le & a\left(x\right)+{\int }_{{x}_{0}}^{x}{g}_{1}\left(x,s\right){\int }_{{x}_{0}}^{s}{h}_{1}\left(s,\tau \right){w}_{1}\left(u\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds\\ +{\int }_{{x}_{0}}^{x}{g}_{2}\left(x,s\right){\int }_{{x}_{0}}^{s}{h}_{2}\left(s,\tau \right){w}_{2}\left(u\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds+\sum _{{x}_{0}<{x}_{i}
(2.4)

which looks more complicated than (1.7).

Corollary 2.1 Suppose that (C1)-(C3) and (C5) hold, and that the functions ${g}_{j}$ and ${h}_{j}$ ($j=1,2$) are both nonnegative and continuous on $\left[{x}_{0},\mathrm{\infty }\right)×\left[{x}_{0},\mathrm{\infty }\right)$. If (2.4) holds, then for $x\in \left[{x}_{i},{x}_{i+1}\right)$, $i=0,1,2,\dots$ ,

$\begin{array}{rl}{u}_{i}\left(x\right)\le & {W}_{2}^{-1}\left[{W}_{2}\circ {W}_{1}^{-1}\left({W}_{1}\left({r}_{i+1}\left(x\right)\right)+{\int }_{{x}_{i}}^{x}{\int }_{s}^{x}\underset{{x}_{0}\le \tau \le x}{max}{g}_{1}\left(\tau ,v\right){h}_{1}\left(v,s\right)\phantom{\rule{0.2em}{0ex}}dv\phantom{\rule{0.2em}{0ex}}ds\right)\\ +{\int }_{{x}_{i}}^{x}{\int }_{s}^{x}\underset{{x}_{0}\le \tau \le x}{max}{g}_{2}\left(\tau ,v\right){h}_{2}\left(v,s\right)\phantom{\rule{0.2em}{0ex}}dv\phantom{\rule{0.2em}{0ex}}ds\right],\end{array}$
(2.5)

where ${r}_{i+1}$ and its related functions are defined as in Theorem 2.1 by replacing ${f}_{j}\left(x,s\right)$ with ${\int }_{s}^{x}{max}_{{x}_{0}\le \tau \le x}{g}_{j}\left(\tau ,v\right){h}_{j}\left(v,s\right)\phantom{\rule{0.2em}{0ex}}dv$, $j=1,2$.

Proof Because ${f}_{j}$, ${h}_{j}$ and ${w}_{j}$ are continuous, we have

$\begin{array}{r}{\int }_{{x}_{0}}^{x}{g}_{j}\left(x,s\right){\int }_{{x}_{0}}^{s}{h}_{j}\left(s,\tau \right){w}_{j}\left(u\left(\tau \right)\right)\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}={\int }_{{x}_{0}}^{x}{w}_{j}\left(u\left(\tau \right)\right){\int }_{\tau }^{x}{g}_{j}\left(x,s\right){h}_{j}\left(s,\tau \right)\phantom{\rule{0.2em}{0ex}}ds\phantom{\rule{0.2em}{0ex}}d\tau \\ \phantom{\rule{1em}{0ex}}={\int }_{{x}_{0}}^{x}{w}_{j}\left(u\left(s\right)\right){\int }_{s}^{x}{g}_{j}\left(x,\tau \right){h}_{j}\left(\tau ,s\right)\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds\le {\int }_{{x}_{0}}^{x}{f}_{j}\left(x,s\right){w}_{j}\left(u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds,\end{array}$

where ${f}_{j}\left(x,s\right):={\int }_{s}^{x}{max}_{{x}_{0}\le \tau \le x}{g}_{j}\left(\tau ,v\right){h}_{j}\left(v,s\right)\phantom{\rule{0.2em}{0ex}}dv$. Then (2.4) is reduced to

$\begin{array}{rl}u\left(x\right)\le & a\left(x\right)+{\int }_{{x}_{0}}^{x}{f}_{1}\left(x,s\right){w}_{1}\left(u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+{\int }_{{x}_{0}}^{x}{f}_{2}\left(x,s\right){w}_{2}\left(u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ +\sum _{{x}_{0}<{x}_{i}

which is just the form of (1.7), if we take ${b}_{j}\left(x\right)=x$ for $j=1,2$. Note that for fixed s, the function ${f}_{j}\left(x,s\right)$ is increasing in x. So ${\stackrel{˜}{f}}_{j}\left(x,s\right):={max}_{{t}_{0}\le \tau \le x}{f}_{j}\left(\tau ,s\right)={f}_{j}\left(x,s\right)$. By Theorem 2.1, for $x\in \left[{x}_{i},{x}_{i+1}\right)$, $i=0,1,2,\dots$ ,

$\begin{array}{rl}{u}_{i}\left(x\right)\le & {W}_{2}^{-1}\left[{W}_{2}\circ {W}_{1}^{-1}\left({W}_{1}\left({r}_{i+1}\left(x\right)\right)+{\int }_{{x}_{i}}^{x}{\int }_{s}^{x}\underset{{x}_{0}\le \tau \le x}{max}{g}_{1}\left(\tau ,v\right){h}_{1}\left(v,s\right)\phantom{\rule{0.2em}{0ex}}dv\phantom{\rule{0.2em}{0ex}}ds\right)\\ +{\int }_{{x}_{i}}^{x}{\int }_{s}^{x}\underset{{x}_{0}\le \tau \le x}{max}{g}_{2}\left(\tau ,v\right){h}_{2}\left(v,s\right)\phantom{\rule{0.2em}{0ex}}dv\phantom{\rule{0.2em}{0ex}}ds\right].\end{array}$

□

Remark 2.2 Using the same way, we can change inequality (1.4) into the form of (1.7) with $a\left(x\right)=c$, ${f}_{1}\left(x,s\right)=q\left(s\right)$, ${f}_{2}\left(x,s\right)=g\left(s\right){\int }_{s}^{x}q\left(\tau \right)\phantom{\rule{0.2em}{0ex}}d\tau$, ${w}_{1}\left(u\right)=u$ and ${w}_{2}\left(u\right)={u}^{m}$.

## 3 Proof of Theorem 2.1

Obviously, ${r}_{1}\left(x\right)$ is positive and nondecreasing in x, and ${\stackrel{˜}{f}}_{j}\left(x,s\right)$ is nonnegative and nondecreasing in x for each fixed s and $j=1,2$. They satisfy ${r}_{1}\left(x\right)\ge a\left(x\right)$ and ${\stackrel{˜}{f}}_{j}\left(x,s\right)\ge {f}_{j}\left(x,s\right)$.

We first consider $x\in \left[{x}_{0},{x}_{1}\right)$, and we have from (1.7) and (2.2)

$\begin{array}{rl}u\left(x\right)& \le a\left(x\right)+{\int }_{{b}_{1}\left({x}_{0}\right)}^{{b}_{1}\left(x\right)}{f}_{1}\left(x,s\right){w}_{1}\left(u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+{\int }_{{b}_{2}\left({x}_{0}\right)}^{{b}_{2}\left(x\right)}{f}_{2}\left(x,s\right){w}_{2}\left(u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \le {r}_{1}\left(x\right)+{\int }_{{b}_{1}\left({x}_{0}\right)}^{{b}_{1}\left(x\right)}{f}_{1}\left(x,s\right){w}_{1}\left(u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+{\int }_{{b}_{2}\left({x}_{0}\right)}^{{b}_{2}\left(x\right)}{f}_{2}\left(x,s\right){w}_{2}\left(u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds.\end{array}$
(3.1)

Take any fixed $T\in \left({x}_{0},{x}_{1}\right)$, and we investigate the following inequality

$u\left(x\right)\le {r}_{1}\left(T\right)+{\int }_{{b}_{1}\left({x}_{0}\right)}^{{b}_{1}\left(x\right)}{\stackrel{˜}{f}}_{1}\left(T,s\right){w}_{1}\left(u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+{\int }_{{b}_{2}\left({x}_{0}\right)}^{{b}_{2}\left(x\right)}{\stackrel{˜}{f}}_{2}\left(T,s\right){w}_{2}\left(u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds$
(3.2)

for $x\in \left[{x}_{0},T\right]$, where ${\stackrel{˜}{f}}_{1}$ and ${\stackrel{˜}{f}}_{2}$ are defined in (2.2). Let

$z\left(x\right)={\int }_{{b}_{1}\left({x}_{0}\right)}^{{b}_{1}\left(x\right)}{\stackrel{˜}{f}}_{1}\left(T,s\right){w}_{1}\left(u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+{\int }_{{b}_{2}\left({x}_{0}\right)}^{{b}_{2}\left(x\right)}{\stackrel{˜}{f}}_{2}\left(T,s\right){w}_{2}\left(u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds$

and $z\left({x}_{0}\right)=0$. Hence, $u\left(x\right)\le {r}_{1}\left(T\right)+z\left(x\right)$. Clearly, $z\left(x\right)$ is a nonnegative, nondecreasing and differentiable function for $x\in \left({x}_{0},T\right]$. Moreover, ${b}_{j}\left(x\right)$ is differentiable and nondecreasing in $x\in \left[{x}_{0},T\right]$ for $j=1,2$. Thus, ${b}_{j}^{\prime }\left(x\right)\ge 0$ for $x\in \left[{x}_{0},T\right]$. Since ${w}_{1}$ and ${w}_{2}$ are nondecreasing, $z\left(x\right)+{r}_{1}\left(T\right)>0$ and ${b}_{j}\left(x\right)\le x$ for $x\in \left[{x}_{0},T\right]$, we have

$\begin{array}{rcl}\frac{{z}^{\prime }\left(x\right)}{{w}_{1}\left(z\left(x\right)+{r}_{1}\left(T\right)\right)}& \le & \frac{{b}_{1}^{\prime }\left(x\right){\stackrel{˜}{f}}_{1}\left(T,{b}_{1}\left(x\right)\right){w}_{1}\left(u\left({b}_{1}\left(x\right)\right)\right)}{{w}_{1}\left(z\left(x\right)+{r}_{1}\left(T\right)\right)}+\frac{{b}_{2}^{\prime }\left(x\right){\stackrel{˜}{f}}_{2}\left(T,{b}_{2}\left(x\right)\right){w}_{2}\left(u\left({b}_{2}\left(x\right)\right)\right)}{{w}_{1}\left(z\left(x\right)+{r}_{1}\left(T\right)\right)}\\ \le & \frac{{b}_{1}^{\prime }\left(x\right){\stackrel{˜}{f}}_{1}\left(T,{b}_{1}\left(x\right)\right){w}_{1}\left(z\left({b}_{1}\left(x\right)\right)+{r}_{1}\left(T\right)\right)}{{w}_{1}\left(z\left(x\right)+{r}_{1}\left(T\right)\right)}\\ +\frac{{b}_{2}^{\prime }\left(x\right){\stackrel{˜}{f}}_{2}\left(T,{b}_{2}\left(x\right)\right){w}_{2}\left(z\left({b}_{2}\left(x\right)\right)+{r}_{1}\left(T\right)\right)}{{w}_{1}\left(z\left(x\right)+{r}_{1}\left(T\right)\right)}\\ \le & \frac{{b}_{1}^{\prime }\left(x\right){\stackrel{˜}{f}}_{1}\left(T,{b}_{1}\left(x\right)\right){w}_{1}\left(z\left(x\right)+{r}_{1}\left(T\right)\right)}{{w}_{1}\left(z\left(x\right)+{r}_{1}\left(T\right)\right)}\\ +\frac{{b}_{2}^{\prime }\left(x\right){\stackrel{˜}{f}}_{2}\left(T,{b}_{2}\left(x\right)\right){w}_{2}\left(z\left({b}_{2}\left(x\right)\right)+{r}_{1}\left(T\right)\right)}{{w}_{1}\left(z\left(x\right)+{r}_{1}\left(T\right)\right)}\\ \le & {b}_{1}^{\prime }\left(x\right){\stackrel{˜}{f}}_{1}\left(T,{b}_{1}\left(x\right)\right)+\frac{{b}_{2}^{\prime }\left(x\right){\stackrel{˜}{f}}_{2}\left(T,{b}_{2}\left(x\right)\right){w}_{2}\left(z\left({b}_{2}\left(x\right)\right)+{r}_{1}\left(T\right)\right)}{{w}_{1}\left(z\left({b}_{2}\left(x\right)\right)+{r}_{1}\left(T\right)\right)}.\end{array}$

Integrating both sides of the inequality above, from ${x}_{0}$ to x, we obtain

$\begin{array}{rcl}{W}_{1}\left(z\left(x\right)+{r}_{1}\left(T\right)\right)& \le & {W}_{1}\left({r}_{1}\left(T\right)\right)+{\int }_{{x}_{0}}^{x}{b}_{1}^{\prime }\left(s\right){\stackrel{˜}{f}}_{1}\left(T,{b}_{1}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ +{\int }_{{x}_{0}}^{x}{b}_{2}^{\prime }\left(s\right){\stackrel{˜}{f}}_{2}\left(T,{b}_{2}\left(s\right)\right)\varphi \left(z\left({b}_{2}\left(s\right)\right)+{r}_{1}\left(T\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \le & {W}_{1}\left({r}_{1}\left(T\right)\right)+{\int }_{{b}_{1}\left({x}_{0}\right)}^{{b}_{1}\left(x\right)}{\stackrel{˜}{f}}_{1}\left(T,s\right)\phantom{\rule{0.2em}{0ex}}ds+{\int }_{{b}_{2}\left({x}_{0}\right)}^{{b}_{2}\left(x\right)}{\stackrel{˜}{f}}_{2}\left(T,s\right)\varphi \left(z\left(s\right)+{r}_{1}\left(T\right)\right)\phantom{\rule{0.2em}{0ex}}ds\end{array}$

for ${x}_{0}, where $\varphi \left(x\right)=\frac{{w}_{2}\left(x\right)}{{w}_{1}\left(x\right)}$, or equivalently,

$\xi \left(x\right)\le {W}_{1}\left({r}_{1}\left(T\right)\right)+{\int }_{{b}_{1}\left({x}_{0}\right)}^{{b}_{1}\left(x\right)}{\stackrel{˜}{f}}_{1}\left(T,s\right)\phantom{\rule{0.2em}{0ex}}ds+{\int }_{{b}_{2}\left({x}_{0}\right)}^{{b}_{2}\left(x\right)}{\stackrel{˜}{f}}_{2}\left(T,s\right)\varphi \left({W}_{1}^{-1}\left(\xi \left(s\right)\right)\right)\phantom{\rule{0.2em}{0ex}}ds\triangleq {z}_{1}\left(x\right),$

where

$\xi \left(x\right)={W}_{1}\left(z\left(x\right)+{r}_{1}\left(T\right)\right).$

It is easy to check that $\xi \left(x\right)\le {z}_{1}\left(x\right)$, ${z}_{1}\left({x}_{0}\right)={W}_{1}\left({r}_{1}\left(T\right)\right)$ and ${z}_{1}\left(x\right)$ is differentiable, positive and nondecreasing on $\left({x}_{0},T\right]$. Since $\varphi \left({W}_{1}^{-1}\left(u\right)\right)$ is nondecreasing from the assumption (C1), we have by (2.3)

$\begin{array}{r}\frac{{z}_{1}^{\prime }\left(x\right)}{\varphi \left({W}_{1}^{-1}\left({z}_{1}\left(x\right)\right)\right)}\\ \phantom{\rule{1em}{0ex}}\le \frac{{\stackrel{˜}{f}}_{1}\left(T,{b}_{1}\left(x\right)\right){b}_{1}^{\prime }\left(x\right)}{\varphi \left({W}_{1}^{-1}\left({z}_{1}\left(x\right)\right)\right)}+\frac{{\stackrel{˜}{f}}_{2}\left(T,{b}_{2}\left(x\right)\right)\varphi \left({W}_{1}^{-1}\left(\xi \left({b}_{2}\left(x\right)\right)\right)\right){b}_{2}^{\prime }\left(x\right)}{\varphi \left({W}_{1}^{-1}\left({z}_{1}\left(x\right)\right)\right)}\\ \phantom{\rule{1em}{0ex}}\le \frac{{\stackrel{˜}{f}}_{1}\left(T,{b}_{1}\left(x\right)\right){b}_{1}^{\prime }\left(x\right)}{\varphi \left({W}_{1}^{-1}\left({z}_{1}\left(x\right)\right)\right)}+\frac{{\stackrel{˜}{f}}_{2}\left(T,{b}_{2}\left(x\right)\right)\varphi \left({W}_{1}^{-1}\left({z}_{1}\left({b}_{2}\left(x\right)\right)\right)\right){b}_{2}^{\prime }\left(x\right)}{\varphi \left({W}_{1}^{-1}\left({z}_{1}\left(x\right)\right)\right)}\\ \phantom{\rule{1em}{0ex}}\le \frac{{\stackrel{˜}{f}}_{1}\left(T,{b}_{1}\left(x\right)\right){b}_{1}^{\prime }\left(x\right)}{\varphi \left({W}_{1}^{-1}\left({W}_{1}\left({r}_{1}\left(T\right)\right)+{\int }_{{b}_{1}\left({x}_{0}\right)}^{{b}_{1}\left(x\right)}{\stackrel{˜}{f}}_{1}\left(T,s\right)\phantom{\rule{0.2em}{0ex}}ds\right)\right)}+{\stackrel{˜}{f}}_{2}\left(T,{b}_{2}\left(x\right)\right){b}_{2}^{\prime }\left(x\right).\end{array}$
(3.3)

Note that

$\begin{array}{rl}{\int }_{{x}_{0}}^{x}\frac{{z}_{1}^{\prime }\left(s\right)}{\varphi \left({W}_{1}^{-1}\left({z}_{1}\left(s\right)\right)\right)}\phantom{\rule{0.2em}{0ex}}ds& ={\int }_{{x}_{0}}^{x}\frac{{w}_{1}\left({W}_{1}^{-1}\left({z}_{1}\left(s\right)\right)\right){z}_{1}^{\prime }\left(s\right)}{{w}_{2}\left({W}_{1}^{-1}\left({z}_{1}\left(s\right)\right)\right)}\phantom{\rule{0.2em}{0ex}}ds={\int }_{{W}_{1}^{-1}\left({z}_{1}\left({x}_{0}\right)\right)}^{{W}_{1}^{-1}\left({z}_{1}\left(x\right)\right)}\frac{\phantom{\rule{0.2em}{0ex}}du}{{w}_{2}\left(u\right)}\\ ={W}_{2}\circ {W}_{1}^{-1}\left({z}_{1}\left(x\right)\right)-{W}_{2}\circ {W}_{1}^{-1}\left({z}_{1}\left({x}_{0}\right)\right)\\ ={W}_{2}\circ {W}_{1}^{-1}\left({z}_{1}\left(x\right)\right)-{W}_{2}\left({r}_{1}\left(T\right)\right).\end{array}$

Integrating both sides of inequality (3.3), from ${x}_{0}$ to x, we obtain

$\begin{array}{r}{W}_{2}\circ {W}_{1}^{-1}\left({z}_{1}\left(x\right)\right)-{W}_{2}\left({r}_{1}\left(T\right)\right)\\ \phantom{\rule{1em}{0ex}}={\int }_{{x}_{0}}^{x}\frac{{z}_{1}^{\prime }\left(s\right)}{\varphi \left({W}_{1}^{-1}\left({z}_{1}\left(s\right)\right)\right)}\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}\le {\int }_{{x}_{0}}^{x}\frac{{\stackrel{˜}{f}}_{1}\left(T,{b}_{1}\left(s\right)\right){b}_{1}^{\prime }\left(s\right)}{\varphi \left({W}_{1}^{-1}\left({W}_{1}\left({r}_{1}\left(T\right)\right)+{\int }_{{b}_{1}\left({x}_{0}\right)}^{{b}_{1}\left(s\right)}{\stackrel{˜}{f}}_{1}\left(T,\tau \right)\phantom{\rule{0.2em}{0ex}}d\tau \right)\right)}\phantom{\rule{0.2em}{0ex}}ds+{\int }_{{x}_{0}}^{x}{\stackrel{˜}{f}}_{2}\left(T,{b}_{2}\left(s\right)\right){b}_{2}^{\prime }\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}\le {W}_{2}\circ {W}_{1}^{-1}\left({W}_{1}\left({r}_{1}\left(T\right)\right)+{\int }_{{b}_{1}\left({x}_{0}\right)}^{{b}_{1}\left(x\right)}{\stackrel{˜}{f}}_{1}\left(T,s\right)\phantom{\rule{0.2em}{0ex}}ds\right)-{W}_{2}\left({r}_{1}\left(T\right)\right)+{\int }_{{b}_{2}\left({x}_{0}\right)}^{{b}_{2}\left(x\right)}{\stackrel{˜}{f}}_{2}\left(T,s\right)\phantom{\rule{0.2em}{0ex}}ds.\end{array}$

Thus,

${W}_{2}\circ {W}_{1}^{-1}\left({z}_{1}\left(x\right)\right)\le {W}_{2}\circ {W}_{1}^{-1}\left({W}_{1}\left({r}_{1}\left(T\right)\right)+{\int }_{{b}_{1}\left({x}_{0}\right)}^{{b}_{1}\left(x\right)}{\stackrel{˜}{f}}_{1}\left(T,s\right)\phantom{\rule{0.2em}{0ex}}ds\right)+{\int }_{{b}_{2}\left({x}_{0}\right)}^{{b}_{2}\left(x\right)}{\stackrel{˜}{f}}_{2}\left(T,s\right)\phantom{\rule{0.2em}{0ex}}ds.$

We have by (2.3)

$\begin{array}{rl}u\left(x\right)& \le z\left(x\right)+{r}_{1}\left(T\right)\le {W}_{1}^{-1}\left(\xi \left(x\right)\right)\le {W}_{1}^{-1}\left({z}_{1}\left(x\right)\right)\\ \le {W}_{2}^{-1}\left[{W}_{2}\circ {W}_{1}^{-1}\left({W}_{1}\left({r}_{1}\left(T\right)\right)+{\int }_{{b}_{1}\left({x}_{0}\right)}^{{b}_{1}\left(x\right)}{\stackrel{˜}{f}}_{1}\left(T,s\right)\phantom{\rule{0.2em}{0ex}}ds\right)+{\int }_{{b}_{2}\left({x}_{0}\right)}^{{b}_{2}\left(x\right)}{\stackrel{˜}{f}}_{2}\left(T,s\right)\phantom{\rule{0.2em}{0ex}}ds\right].\end{array}$

Since the inequality above is true for any $x\in \left[{x}_{0},T\right]$, we obtain

$u\left(T\right)\le {W}_{2}^{-1}\left[{W}_{2}\circ {W}_{1}^{-1}\left({W}_{1}\left({r}_{1}\left(T\right)\right)+{\int }_{{b}_{1}\left({x}_{0}\right)}^{{b}_{1}\left(T\right)}{\stackrel{˜}{f}}_{1}\left(T,s\right)\phantom{\rule{0.2em}{0ex}}ds\right)+{\int }_{{b}_{2}\left({x}_{0}\right)}^{{b}_{2}\left(T\right)}{\stackrel{˜}{f}}_{2}\left(T,s\right)\phantom{\rule{0.2em}{0ex}}ds\right].$

Replacing T by x yields

$u\left(x\right)\le {W}_{2}^{-1}\left[{W}_{2}\circ {W}_{1}^{-1}\left({W}_{1}\left({r}_{1}\left(x\right)\right)+{\int }_{{b}_{1}\left({x}_{0}\right)}^{{b}_{1}\left(x\right)}{\stackrel{˜}{f}}_{1}\left(x,s\right)\phantom{\rule{0.2em}{0ex}}ds\right)+{\int }_{{b}_{2}\left({x}_{0}\right)}^{{b}_{2}\left(x\right)}{\stackrel{˜}{f}}_{2}\left(x,s\right)\phantom{\rule{0.2em}{0ex}}ds\right].$
(3.4)

This means that (2.1) is true for $x\in \left[{x}_{0},{x}_{1}\right)$ and $i=0$ if replace $u\left(x\right)$ with ${u}_{0}\left(x\right)$.

For $i=1$ and $x\in \left[{x}_{1},{x}_{2}\right)$, (1.7) becomes

$\begin{array}{rl}u\left(x\right)\le & {r}_{1}\left(x\right)+{\int }_{{b}_{1}\left({x}_{0}\right)}^{{b}_{1}\left({x}_{1}\right)}{f}_{1}\left(x,s\right){w}_{1}\left({u}_{0}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+{\int }_{{b}_{2}\left({x}_{0}\right)}^{{b}_{2}\left({x}_{1}\right)}{f}_{2}\left(x,s\right){w}_{2}\left({u}_{0}\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ +{\beta }_{1}{u}_{0}^{m}\left({x}_{1}-0\right)+{\int }_{{b}_{1}\left({x}_{1}\right)}^{{b}_{1}\left(x\right)}{f}_{1}\left(x,s\right){w}_{1}\left(u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+{\int }_{{b}_{2}\left({x}_{1}\right)}^{{b}_{2}\left(x\right)}{f}_{2}\left(x,s\right){w}_{2}\left(u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \le & {r}_{2}\left(x\right)+{\int }_{{b}_{1}\left({x}_{1}\right)}^{{b}_{1}\left(x\right)}{f}_{1}\left(x,s\right){w}_{1}\left(u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+{\int }_{{b}_{2}\left({x}_{1}\right)}^{{b}_{2}\left(x\right)}{f}_{2}\left(x,s\right){w}_{2}\left(u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds,\end{array}$
(3.5)

where the definition of ${r}_{2}\left(x\right)$ is given in (2.2). Note that the estimate of ${u}_{0}\left(x\right)$ is known. Equation (3.5) is same as (3.1) if replace ${r}_{1}\left(x\right)$ and ${x}_{0}$ by ${r}_{2}\left(x\right)$ and ${x}_{1}$. Thus, by (3.4), we have

$u\left(x\right)\le {W}_{2}^{-1}\left[{W}_{2}\circ {W}_{1}^{-1}\left({W}_{1}\left({r}_{2}\left(x\right)\right)+{\int }_{{b}_{1}\left({x}_{1}\right)}^{{b}_{1}\left(x\right)}{\stackrel{˜}{f}}_{1}\left(x,s\right)\phantom{\rule{0.2em}{0ex}}ds\right)+{\int }_{{b}_{2}\left({x}_{1}\right)}^{{b}_{2}\left(x\right)}{\stackrel{˜}{f}}_{2}\left(x,s\right)\phantom{\rule{0.2em}{0ex}}ds\right].$
(3.6)

This implies that (2.1) is true for $x\in \left[{x}_{1},{x}_{2}\right)$ and $i=1$ if replace $u\left(x\right)$ by ${u}_{1}\left(x\right)$.

Assume that (2.1) is true for $x\in \left[{x}_{i},{x}_{i+1}\right)$, i.e.,

${u}_{i}\left(x\right)\le {W}_{2}^{-1}\left[{W}_{2}\circ {W}_{1}^{-1}\left({W}_{1}\left({r}_{i+1}\left(x\right)\right)+{\int }_{{b}_{1}\left({x}_{i}\right)}^{{b}_{1}\left(x\right)}{\stackrel{˜}{f}}_{1}\left(x,s\right)\phantom{\rule{0.2em}{0ex}}ds\right)+{\int }_{{b}_{2}\left({x}_{i}\right)}^{{b}_{2}\left(x\right)}{\stackrel{˜}{f}}_{2}\left(x,s\right)\phantom{\rule{0.2em}{0ex}}ds\right]$
(3.7)

for $x\in \left[{x}_{i},{x}_{i+1}\right)$.

For $x\in \left[{x}_{i+1},{x}_{i+2}\right)$, (1.7) becomes

$\begin{array}{rl}u\left(x\right)\le & a\left(x\right)+{\int }_{{b}_{1}\left({x}_{0}\right)}^{{b}_{1}\left(x\right)}{f}_{1}\left(x,s\right){w}_{1}\left(u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds+{\int }_{{b}_{2}\left({x}_{0}\right)}^{{b}_{2}\left(x\right)}{f}_{2}\left(x,s\right){w}_{2}\left(u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ +\sum _{{x}_{0}<{x}_{i+1}
(3.8)

where we use the fact that the estimate of $u\left(x\right)$ is already known for $x\in \left[{x}_{0},{x}_{i+1}\right)$ by the assumption (3.7). Again (3.8) is same as (3.1) if replace ${r}_{1}\left(x\right)$ and ${x}_{0}$ by ${r}_{i+2}\left(x\right)$ and ${x}_{i+1}$. Thus, by (3.4), we have

$u\left(x\right)\le {W}_{2}^{-1}\left[{W}_{2}\circ {W}_{1}^{-1}\left({W}_{1}\left({r}_{i+2}\left(x\right)\right)+{\int }_{{b}_{1}\left({x}_{i+1}\right)}^{{b}_{1}\left(x\right)}{\stackrel{˜}{f}}_{1}\left(x,s\right)\phantom{\rule{0.2em}{0ex}}ds\right)+{\int }_{{b}_{2}\left({x}_{i+1}\right)}^{{b}_{2}\left(x\right)}{\stackrel{˜}{f}}_{2}\left(x,s\right)\phantom{\rule{0.2em}{0ex}}ds\right].$

This yields that (2.1) is true for $x\in \left[{x}_{i+1},{x}_{i+2}\right)$ if replace $u\left(x\right)$ by ${u}_{i+1}\left(x\right)$. By induction, we know that (2.1) holds for $x\in \left[{x}_{i},{x}_{i+1}\right)$ for any nonnegative integer i. This completes the proof of Theorem 2.1.

## 4 Applications

Consider the following impulsive differential equation

$\begin{array}{r}\frac{dy}{dx}=F\left(x,y\right),\phantom{\rule{1em}{0ex}}x\ne {x}_{i},\\ \mathrm{\Delta }y{|}_{x={x}_{i}}={I}_{i}\left(y\right),\end{array}$
(4.1)

where $y\in {\mathbf{R}}^{n}$, $F:{\mathbf{R}}^{n+1}\to {\mathbf{R}}^{n}$, ${I}_{i}:{\mathbf{R}}^{n}\to {\mathbf{R}}^{n}$ ($i=1,2,\dots$), $x\ge {x}_{0}\ge 0$, ${lim}_{i\to \mathrm{\infty }}{x}_{i}=\mathrm{\infty }$, ${x}_{i-1}<{x}_{i}$ for all $i=1,2,\dots$ .

Assume that

1. (1)

$\parallel F\left(x,y\right)\parallel \le {h}_{1}\left(x\right)\parallel y\parallel +{h}_{2}\left(x\right){e}^{\parallel y\parallel }$, where ${h}_{1}$, ${h}_{2}$ are nonnegative and continuous on $\left[{x}_{0},\mathrm{\infty }\right)$;

2. (2)

$\parallel {I}_{i}\left(y\right)\parallel \le {\beta }_{i}{\parallel y\parallel }^{m}$, where ${\beta }_{i}$ and m are nonnegative constants.

The solution of (4.1) with an initial value $y\left({x}_{0}\right)={y}_{0}$ is given by

$y\left(x\right)={y}_{0}+{\int }_{{x}_{0}}^{x}F\left(s,y\right)\phantom{\rule{0.2em}{0ex}}ds+\sum _{{x}_{0}<{x}_{i}
(4.2)

which implies that

$\parallel y\left(x\right)\parallel \le \parallel {y}_{0}\parallel +{\int }_{{x}_{0}}^{x}\left({h}_{1}\left(s\right)\parallel y\parallel +{h}_{2}\left(s\right){e}^{\parallel y\parallel }\right)\phantom{\rule{0.2em}{0ex}}ds+\sum _{{x}_{0}<{x}_{i}
(4.3)

Let

$\begin{array}{r}u\left(x\right)=\parallel y\left(x\right)\parallel ,\phantom{\rule{2em}{0ex}}a\left(x\right)\equiv \parallel {y}_{0}\parallel ,\phantom{\rule{2em}{0ex}}{b}_{1}\left(x\right)=x,\phantom{\rule{2em}{0ex}}{b}_{2}\left(x\right)=x,\\ {f}_{1}\left(x,s\right)={h}_{1}\left(s\right),\phantom{\rule{2em}{0ex}}{f}_{2}\left(x,s\right)={h}_{2}\left(s\right),\phantom{\rule{2em}{0ex}}{w}_{1}\left(u\right)=u,\phantom{\rule{2em}{0ex}}{w}_{2}\left(u\right)={e}^{u},\end{array}$

so (4.3) is same as (1.7). It is easy to obtain for any positive constants ${\stackrel{˜}{u}}_{1}$ and ${\stackrel{˜}{u}}_{2}$

$\begin{array}{r}{r}_{1}\left(x\right)\equiv \parallel {y}_{0}\parallel ,\phantom{\rule{2em}{0ex}}{\stackrel{˜}{f}}_{1}\left(x,s\right)={h}_{1}\left(s\right),\phantom{\rule{2em}{0ex}}{\stackrel{˜}{f}}_{2}\left(x,s\right)={h}_{2}\left(s\right),\phantom{\rule{2em}{0ex}}{W}_{1}\left(u\right)={\int }_{{\stackrel{˜}{u}}_{1}}^{u}\frac{dz}{{w}_{1}\left(z\right)}=ln\frac{u}{{\stackrel{˜}{u}}_{1}},\\ {W}_{1}^{-1}\left(u\right)={\stackrel{˜}{u}}_{1}{e}^{u},\phantom{\rule{2em}{0ex}}{W}_{2}\left(u\right)={\int }_{{\stackrel{˜}{u}}_{2}}^{u}\frac{dz}{{w}_{2}\left(u\right)}={e}^{-{\stackrel{˜}{u}}_{2}}-{e}^{-u},\phantom{\rule{2em}{0ex}}{W}_{2}^{-1}\left(u\right)=-ln\left({e}^{-{\stackrel{˜}{u}}_{2}}-u\right),\\ {r}_{i+1}\left(x\right)=\parallel {y}_{0}\parallel +\sum _{k=1}^{i}{\int }_{{x}_{k-1}}^{{x}_{k}}{h}_{1}\left(s\right){u}_{k-1}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds+\sum _{k=1}^{i}{\int }_{{x}_{k-1}}^{{x}_{k}}{h}_{2}\left(s\right){e}^{{u}_{k-1}\left(s\right)}\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{{r}_{i+1}\left(x\right)=}+\sum _{k=1}^{i}{\beta }_{k}{u}_{k-1}^{m}\left({x}_{k}-0\right).\end{array}$

Thus, for any nonnegative integer i and $x\in \left({x}_{i},{x}_{i+1}\right)$

${u}_{i}\left(x\right)\le -ln\left({e}^{-{r}_{i+1}\left(x\right){e}^{{\int }_{{x}_{i}}^{x}{h}_{1}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds}}-{\int }_{{x}_{i}}^{x}{h}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds\right),$

provided that

${e}^{-{r}_{i+1}\left(x\right){e}^{{\int }_{{x}_{i}}^{x}{h}_{1}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds}}-{\int }_{{x}_{i}}^{x}{h}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}ds>0.$

Remark 4.1 From (4.3), we know that ${w}_{2}\left(u\right)={e}^{u}$. Clearly, ${w}_{2}\left(2u\right)={e}^{2u}\le {w}_{2}\left(2\right){w}_{2}\left(u\right)={e}^{2}{e}^{u}$ does not hold for large $u>0$. Thus, ${w}_{2}\left(u\right)$ does not belong to the class . Again ${w}_{2}\left(\frac{u}{2}\right)={e}^{\frac{u}{2}}\ge \frac{1}{2}{w}_{2}\left(u\right)=\frac{1}{2}{e}^{u}$ does not hold for large $u>0$, so ${w}_{2}\left(u\right)$ does not belong to the class ȷ. Hence, the results in [22] can not be applied to inequality (4.3).

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## Acknowledgements

This work was supported by the Project of Department of Education of Guangdong Province, China (No. 2012KJCX0074), the PhD Start-up Fund of the Natural Science Foundation of Guangdong Province, China (No. S2011040000464), the China Postdoctoral Science Foundation-Special Project (No. 201104077), the Scientific Research Foundation for the Returned Overseas Chinese Scholars, State Education Ministry (No. (2012)940), the Natural Fund of Zhanjiang Normal University (No. LZL1101), and the Doctoral Project of Zhanjiang Normal University (No. ZL1109).

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Correspondence to Shengfu Deng.

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Mi, Y., Deng, S. & Li, X. Nonlinear integral inequalities with delay for discontinuous functions and their applications. J Inequal Appl 2013, 430 (2013). https://doi.org/10.1186/1029-242X-2013-430