Open Access

Some new generalized retarded inequalities for discontinuous functions and their applications

Journal of Inequalities and Applications20162016:7

https://doi.org/10.1186/s13660-015-0943-6

Received: 30 June 2015

Accepted: 13 December 2015

Published: 4 January 2016

Abstract

In this paper, some new generalized retarded inequalities for discontinuous functions are discussed, which are effective in dealing with the qualitative theory of some impulsive differential equations and impulsive integral equations. Compared with some existing integral inequalities, these estimations can be used as tools in the study of differential-integral equations with impulsive conditions.

Keywords

retarded differential-integral equation global existence estimation impulsive equation

1 Introduction

In analyzing the impulsive phenomenon of a physical system governed by certain differential and integral equations, one often needs some kinds of inequalities, such as Gronwall-like inequalities; these inequalities and their various linear and nonlinear generalizations are crucial in the discussion of the existence, uniqueness, boundedness, stability, and other qualitative properties of solutions of differential and integral equations (see [112] and references therein). In [1], Lipovan studied the inequality with delay (\(b(t)\leq t\), \(b(t)\to\infty\) as \(t\to\infty\))
$$u(t)\leq c+ \int_{t_{0}}^{t}f(s)w\bigl(u(s)\bigr)\,ds+ \int_{b(t_{0})}^{b(t)}g(s)w\bigl(u(s)\bigr)\,ds,\quad t_{0}< t< t_{1}, $$
in [2], Agarwal et al. investigated the retarded Gronwall-like inequality
$$u(t)\leq a(t)+\sum_{i=1}^{n} \int_{b_{i}(t_{0})}^{b_{i}(t)} f_{i}(t, s)w_{i} \bigl(u(s)\bigr)\,ds, $$
in 2004, Borysenko [3] obtained the explicit bound to the unknown function of the following integral inequality with impulsive effect:
$$u(t)\leq a(t)+ \int_{t_{0}}^{t} f(s)u(s)\,ds+\sum _{t_{0}< t_{i}< t}{\alpha _{i}}u^{r}(t_{i}-0), $$
in 2007, Iovane [4] studied the following integral inequalities:
$$\begin{aligned}& u(t)\leq a(t)+ \int_{t_{0}}^{t} f(s)u\bigl(\lambda(s)\bigr)\,ds+\sum _{t_{0}< t_{i}< t}{\alpha _{i}}u^{r}(t_{i}-0), \\& u(t)\leq a(t)+q(t) \biggl[ \int_{t_{0}}^{t}f(s)u\bigl(\alpha(s)\bigr)\,ds+ \int _{t_{0}}^{t}f(s) \int_{t_{0}}^{s} g(t)u\bigl(\tau(t)\bigr)\,dt\,ds \\& \hphantom{u(t)\leq{}}{}+\sum _{t_{0}< t_{i}< t}{\alpha _{i}}u^{r}(t_{i}-0) \biggr], \end{aligned}$$
in 2012, Wang and Li [5] gave the upper bound of solutions for the nonlinear inequality
$$v^{p}(t)\leq A_{0}(t)+\frac{p}{p-q} \int_{t_{0}}^{t}f(s)v^{q}\bigl(\tau(s)\bigr) \,ds+\sum_{t_{0}< t_{i}< t}{\alpha_{i}}v^{q}(t_{i}-0), $$
in 2013, Yan [6] considered the following inequality:
$$\begin{aligned} u(t) \leq &a(t)+ \int_{t_{0}}^{t}f(t, s)u\bigl(\alpha(s)\bigr)\,ds + \int_{t_{0}}^{t}f(t, s) \biggl( \int_{t_{0}}^{s}g(s, \lambda)u\bigl(\tau(\lambda ) \bigr)\,d\lambda \biggr)\,ds \\ &{}+q(t)\sum_{t_{0}< t_{i}< t}{ \alpha_{i}}u^{r}(t_{i}-0), \end{aligned}$$
and gave an upper bound estimation. Because of the fundamental importance, over the years, many generalizations and analogous results have been established. However, the bounds given on such inequalities are not directly applicable in the study of some complicated retarded inequalities for discontinuous functions. It is desirable to establish new inequalities of the above type, which can be used more effectively in the study of certain classes of retarded nonlinear differential and integral equations. So in this paper, the following new integral inequalities are presented:
$$\begin{aligned}& u(t)\leq a(t)+\sum_{i=1}^{N} \int_{t_{0}}^{t}g_{i}(s)u\bigl( \phi_{i}(s)\bigr)\,ds+\sum_{j=1}^{L} \int_{t_{0}}^{t}b_{j}(s) \int_{t_{0}}^{s} c_{j}(\theta)u \bigl(w_{j}(\theta )\bigr)\,d\theta \,ds, \end{aligned}$$
(1)
$$\begin{aligned}& u^{p}(t)\leq a_{0}(t)+ \frac{p}{p-q}\sum _{i=1}^{N} \int_{t_{0}}^{t}g_{i}(s)u^{q}\bigl( \phi _{i}(s)\bigr)\,ds+\sum_{j=1}^{L} \int_{t_{0}}^{t}b_{j}(s) \int_{t_{0}}^{s} c_{j}(\theta )u^{q}\bigl(w_{j}(\theta)\bigr)\,d\theta \,ds \\& \hphantom{u^{p}(t)\leq{}}{}+ \sum_{t_{0}< t_{i}< t}\beta_{i}u^{q}(t_{i}-0), \end{aligned}$$
(2)
$$\begin{aligned}& u^{p}(t)\leq a_{0}(t)+ q_{1}(t)\sum _{i=1}^{N} \int_{t_{0}}^{t}g_{i}(t, s)u^{q} \bigl(\phi _{i}(s)\bigr)\,ds \\& \hphantom{u^{p}(t)\leq{}}{}+ \sum_{j=1}^{L} \int_{t_{0}}^{t}b_{j}(t, s) \int_{t_{0}}^{s} c_{j}(s, \theta)u^{q}\bigl(w_{j}(\theta)\bigr)\,d\theta \,ds+q_{2}(t)\sum_{t_{0}< t_{i}< t} \beta_{i}u^{q}(t_{i}-0). \end{aligned}$$
(3)
We give the explicit upper bounds estimation of unknown function of these new inequalities, some applications of these inequalities in impulsive differential equations are also involved.

2 Main results

We consider the inequality (1) first.

Theorem 2.1

Suppose that for \(t_{0}\in{\mathbb{R}}\) and \(t_{0}\leq t< \infty\), the functions \(u(t)\), \(a(t)\), and \(g_{i}(t)\), \(b_{j}(t)\), \(c_{j}(t) \) (\(1\leq i\leq N\), \(1\leq j\leq L\)) are positive and continuous functions on \([t_{0}, \infty)\), and \(c_{j}(t)\) are nondecreasing functions on \([t_{0}, \infty)\). Moreover, \(\phi_{i}(t)\), \(w_{j}(t)\) are continuous functions on \([t_{0}, \infty)\) and \(t_{0}\leq\phi _{i}(t)\leq t\), \(t_{0}\leq w_{j}(t)\leq t\) for \(1\leq i\leq N\) and \(1\leq j\leq L\). Then the inequality (1) implies that
$$ u(t)\leq a(t)+\exp \biggl( \int_{t_{0}}^{t} Q_{1}(s)\,ds \biggr) \biggl( \int_{t_{0}}^{t} Y(s)\exp\biggl(- \int_{t_{0}}^{s} Q_{1}(\tau)\,d\tau\biggr)\,ds \biggr), $$
(4)
where
$$Q_{1}(t)=L+\sum_{i=1}^{N}g_{i}(t)+ \sum_{j=1}^{L}b_{j}(t)c_{j}^{1}(t), $$
and \(c_{j}^{1}(t)=\max\{c_{j}(t), 1\}\).

Proof

Let \(a(t)+z_{1}(t)\) denote the function on the right-hand side of inequality (1). Obviously \(z_{1}(t)\) is a positive and increasing function, and it satisfies \(u(t)\leq a(t)+z_{1}(t)\),
$$\begin{aligned} \frac{dz_{1}(t)}{dt} =&\sum_{i=1}^{N}g_{i}(t)u \bigl(\phi_{i}(t)\bigr)+\sum_{j=1}^{L}b_{j}(t) \int_{t_{0}}^{t}c_{j}(\theta)u \bigl(w_{j}(\theta)\bigr)\,d\theta \\ \leq&\sum_{i=1}^{N}g_{i}(t) \bigl(a\bigl(\phi_{i}(t)\bigr)+z_{1}\bigl( \phi_{i}(t)\bigr) \bigr) \\ &{}+ \sum_{j=1}^{L}b_{j}(t) \int_{t_{0}}^{t}c_{j}(\theta) \bigl(a \bigl(w_{j}(\theta )\bigr)+z_{1}\bigl(w_{j}( \theta)\bigr) \bigr)\,d\theta. \end{aligned}$$
(5)
Let
$$\begin{aligned}& Y(t)=\sum_{i=1}^{N}g_{i}(t)a \bigl(\phi_{i}(t)\bigr)+\sum_{j=1}^{L}b_{j}(t) \int _{t_{0}}^{t}c_{j}(\theta)a \bigl(w_{j}(\theta)\bigr)\,d\theta, \end{aligned}$$
(6)
$$\begin{aligned}& z_{2}(t)=\sum_{i=1}^{N}g_{i}(t)z_{1}(t)+ \sum_{j=1}^{L}b_{j}(t) \int _{t_{0}}^{t}c_{j}(\theta)z_{1} \bigl(w_{j}(\theta)\bigr)\,d\theta. \end{aligned}$$
(7)
Obviously, \(\frac{dz_{1}(t)}{dt}\leq Y(t)+z_{2}(t)\). Let \(c_{j}^{1}(t)=\max\{c_{j}(t), 1\}\). We can obtain
$$ z_{2}(t)\leq \Biggl(\sum_{i=1}^{N}g_{i}(t)+ \sum_{j=1}^{L}b_{j}(t)c_{j}^{1}(t) \Biggr) \Biggl(z_{1}(t)+\sum_{j=1}^{L} \int_{t_{0}}^{t}z_{1}\bigl(w_{j}( \theta)\bigr)\,d\theta \Biggr). $$
(8)
Let
$$z_{3}(t)=z_{1}(t)+\sum_{j=1}^{L} \int_{t_{0}}^{t}z_{1}\bigl(w_{j}( \theta)\bigr)\,d\theta, $$
we note that \(z_{3}(t)\) is a positive and nondecreasing function on I with \(z_{3}(t_{0})=0\), and \(z_{1}(t)\leq z_{3}(t)\), which satisfies
$$\begin{aligned} \frac{dz_{3}(t)}{dt}&=\frac{dz_{1}(t)}{dt}+\sum _{j=1}^{L}z_{1}\bigl(w_{j}(t) \bigr) \\ &\leq Y(t)+ \Biggl(\sum_{i=1}^{N}g_{i}(t)+ \sum_{j=1}^{L}b_{j}(t)c_{j}^{1}(t) \Biggr)z_{3}(t)+\sum_{j=1}^{L}z_{3} \bigl(w_{j}(t)\bigr) \\ &\leq \Biggl(L+\sum_{i=1}^{N}g_{i}(t)+ \sum_{j=1}^{L}b_{j}(t)c_{j}^{1}(t) \Biggr)z_{3}(t)+Y(t) \\ &=Q_{1}(t)z_{3}(t)+Y(t). \end{aligned}$$
(9)
Consider the initial value problem of the differential equation
$$ \left \{ \textstyle\begin{array}{l} \frac{dz_{4}(t)}{dt}=Q_{1}(t)z_{4}(t)+Y(t), \\ z_{4}(t_{0})=0. \end{array}\displaystyle \right . $$
(10)
The solution of equation (10) is
$$ z_{4}(t)=\exp \biggl( \int_{t_{0}}^{t}Q_{1}(s)\,ds \biggr) \biggl( \int _{t_{0}}^{t}Y(s)\exp \biggl(- \int_{t_{0}}^{s}Q_{1}(\tau)\,d\tau \biggr) \,ds \biggr). $$
(11)
Then by comparison of the differential inequality, we have \(z_{3}(t)\leq z_{4}(t)\), so
$$ u(t)\leq a(t)+\exp \biggl( \int_{t_{0}}^{t}Q_{1}(s)\,ds \biggr) \biggl( \int _{t_{0}}^{t}Y(s)\exp \biggl(- \int_{t_{0}}^{s}Q_{1}(\tau)\,d\tau \biggr) \,ds \biggr). $$
(12)
This completes the proof. □

Now, we consider the inequality (2).

Theorem 2.2

Suppose that \(g_{i}(t)\), \(b_{j}(t)\), \(\phi_{i}(t)\), \(w_{j}(t)\) are defined as those in Theorem  2.1, \(p>q>0\), \(t_{0}< t_{1}< t_{2}<\cdots\), \(\beta_{i}\geq0\), \(a_{0}(t)\) is continuous and nondecreasing function on \([t_{0}, t_{1})\) and \(a_{0}(t)\geq1\). \(u(t)\) is a piecewise continuous nonnegative function on \([t_{0}, \infty)\) with only the first discontinuous points \(t_{i}\), \(i=1, 2, \ldots\) . Then, for all \(t\in I_{k}\) and \(I_{k}=[t_{k-1}, t_{k}]\), we obtain
$$ u(t)\leq\tau_{k}^{q^{-1}}(t), $$
(13)
where
$$\begin{aligned}& \tau_{k}(t)= \biggl\{ a_{k-1}(t) \exp \biggl( \int_{t_{k-1}}^{t}Q_{2}(s)\,ds \biggr) \biggr\} ^{\frac{q}{p}} \\& \hphantom{\tau_{k}(t)={}}{}\times \biggl[1+L\biggl(1-\frac{q}{p}\biggr) \bigl(1+a_{k-1}(t)\bigr)^{\frac{p}{p-q}} \\& \hphantom{\tau_{k}(t)={}}{}\times\int _{t_{k-1}}^{t}\exp \biggl(- \int_{t_{k-1}}^{s}\biggl(1-\frac{q}{p} \biggr)Q_{2}(\tau)\,d\tau \biggr)\,ds \biggr]^{\frac{q}{p-q}}, \\& a_{k}(t)= a_{0}(t)+\frac{p}{p-q}\sum _{j=1}^{k} \int_{t_{j-1}}^{t_{j}}\sum_{i=1}^{N}g_{i}(s) \tau_{j}\bigl(\phi_{i}(s)\bigr)\,ds \\& \hphantom{a_{k}(t)={}}{}+ \sum_{j=1}^{k} \int_{t_{j-1}}^{t_{j}}\sum_{m=1}^{L}b_{m}(s) \int _{t_{0}}^{s}c_{m}(\theta) \tau_{j}\bigl(w_{m}(\theta)\bigr)\,d\theta \,ds+\sum _{j=1}^{k}\beta _{j} \tau_{j}(t_{j}),\quad t\in I_{k}. \end{aligned}$$
(14)

Proof

Let \(V=u^{q}\). The inequality (2) is equivalent to
$$\begin{aligned} V^{\frac{p}{q}}(t) \leq&a_{0}(t)+\frac{p}{p-q}\sum _{i=1}^{N} \int _{t_{0}}^{t}g_{i}(s)V\bigl( \phi_{i}(s)\bigr)\,ds \\ &{}+\sum_{j=1}^{L} \int_{t_{0}}^{t}b_{j}(s) \int_{t_{0}}^{s}c_{j}(\theta)V \bigl(w_{j}(\theta )\bigr)\,d\theta \,ds \\ &{}+\sum _{t_{0}< t_{i}< t}\beta_{i}u^{q}(t_{i}-0), \quad \forall t \in [0, \infty ). \end{aligned}$$
(15)
Let \(I_{i}=[t_{i-1}, t_{i}]\), \(i=1, 2, \ldots \) . First, we consider the following inequality on \(I_{1}\):
$$\begin{aligned} V^{\frac{p}{q}}(t) \leq&a_{0}(t)+\frac{p}{p-q}\sum _{i=1}^{N} \int _{t_{0}}^{t}g_{i}(s)V\bigl( \phi_{i}(s)\bigr)\,ds \\ &{}+\sum_{j=1}^{L} \int_{t_{0}}^{t}b_{j}(s) \int_{t_{0}}^{s}c_{j}(\theta)V \bigl(w_{j}(\theta )\bigr)\,d\theta \,ds. \end{aligned}$$
(16)
For \(T\in I_{1}\) and \(t\in[t_{0}, T)\), let
$$\begin{aligned} Y_{1}(t) =&a_{0}(T)+\frac{p}{p-q}\sum _{i=1}^{N} \int_{t_{0}}^{t}g_{i}(s)V\bigl( \phi_{i}(s)\bigr)\,ds \\ &{}+\sum_{j=1}^{L} \int_{t_{0}}^{t}b_{j}(s) \int_{t_{0}}^{s}c_{j}(\theta)V \bigl(w_{j}(\theta )\bigr)\,d\theta \,ds. \end{aligned}$$
(17)
Obviously \(Y_{1}(t)\geq1\), and \(V^{\frac{p}{q}}(t)\leq Y_{1}(t)\), so \(V(t)\leq Y_{1}^{\frac{q}{p}}(t)\), and
$$\begin{aligned} \frac{dY_{1}(t)}{dt}&=\frac{p}{p-q}\sum _{i=1}^{N}g_{i}(t)V\bigl(\phi _{i}(t)\bigr)+\sum_{j=1}^{L}b_{j}(t) \int_{t_{0}}^{t}c_{j}(\theta)V \bigl(w_{j}(\theta )\bigr)\,d\theta \\ &\leq\frac{p}{p-q}\sum_{i=1}^{N}g_{i}(t)Y_{1}^{\frac{q}{p}}(t)+ \sum_{j=1}^{L}b_{j}(t) \int_{t_{0}}^{t}c_{j}(\theta)Y_{1}^{\frac{q}{p}} \bigl(w_{j}(\theta )\bigr)\,d\theta \\ &\leq \Biggl(\frac{p}{p-q}\sum_{i=1}^{N}g_{i}(t)+ \sum_{j=1}^{L}b_{j}(t)c_{j}^{1}(t) \Biggr) \Biggl(Y_{1}(t)+\sum_{j=1}^{L} \int_{t_{0}}^{t}Y_{1}^{\frac{q}{p}} \bigl(w_{j}(\theta )\bigr)\,d\theta \Biggr). \end{aligned}$$
(18)
Let
$$Y_{2}(t)=Y_{1}(t)+\sum_{j=1}^{L} \int_{t_{0}}^{t}Y_{1}^{\frac{q}{p}} \bigl(w_{j}(\theta )\bigr)\,d\theta. $$
Then \(Y_{1}(t)\leq Y_{2}(t)\), and differentiating \(Y_{2}(t)\) implies
$$\begin{aligned} \frac{dY_{2}(t)}{dt}&= \frac{dY_{1}(t)}{dt}+\sum _{j=1}^{L}Y_{1}^{\frac{q}{p}} \bigl(w_{j}(t)\bigr) \\ &\leq \Biggl(\frac{p}{p-q}\sum _{i=1}^{N}g_{i}(t)+\sum _{j=1}^{L}b_{j}(t)c_{j}^{1}(t) \Biggr)Y_{2}(t)+LY_{2}^{\frac{q}{p}}(t). \end{aligned}$$
(19)
Let
$$Q_{2}(t)=\frac{p}{p-q}\sum_{i=1}^{N}g_{i}(t)+ \sum_{j=1}^{L}b_{j}(t)c_{j}^{1}(t). $$
Considering \(\frac{dY_{3}(t)}{dt}=Q_{2}(t)Y_{3}(t)+LY_{3}^{\frac{q}{p}}(t)\), we obtain \(Y_{3}^{\frac{-q}{p}}(t)\frac{dY_{3}(t)}{dt}=Q_{2}(t)Y_{3}^{1-\frac {q}{p}}(t)+L\). Denote \(R(t)=Y_{3}^{\frac{p-q}{p}}(t)\), we have \(Y_{3}(t)=R^{\frac{p}{p-q}}(t)\). Furthermore,
$$ \left \{ \textstyle\begin{array}{l} \frac{dR(t)}{dt}= (Q_{2}(t)R(t)+L)(1-\frac{q}{p}), \\ R(t_{0})= a_{0}^{\frac{p-q}{p}}(T). \end{array}\displaystyle \right . $$
(20)
Then we get
$$\begin{aligned} R(t) =&\exp \biggl[\biggl(1-\frac{q}{p}\biggr) \int_{t_{0}}^{t}Q_{2}(s)\,ds \biggr] \\ &{}\times \biggl[a_{0}^{\frac{p-q}{p}}(T)+ L\biggl(1- \frac{q}{p}\biggr) \int_{t_{0}}^{t}\exp \biggl(- \int_{t_{0}}^{s}\biggl(\frac {p-q}{p} \biggr)Q_{2}(\tau)\,d\tau \biggr)\,ds \biggr]. \end{aligned}$$
(21)
Then
$$\begin{aligned} Y_{3}(t) =&a_{0}(T)\exp \biggl( \int_{t_{0}}^{t}Q_{2}(s)\,ds \biggr) \\ &{}\times \biggl[1+L\biggl(1-\frac{q}{p}\biggr)a_{0}^{\frac{p}{p-q}}(T) \int _{t_{0}}^{t}\exp \biggl(- \int_{t_{0}}^{s} \biggl(1-\frac{q}{p} \biggr)Q_{2}(\tau)\,d\tau \biggr)\,ds \biggr]^{\frac{p}{p-q}}. \end{aligned}$$
(22)
By comparison of the differential inequality, we have \(Y_{2}(t)\leq Y_{3}(t)\). Moreover, \(V^{\frac{p}{q}}(t)\leq Y_{2}(t)\) implies \(V(t)\leq Y_{3}^{\frac {q}{p}}(t)\), and this inequality is equivalent to
$$\begin{aligned} V(t) \leq& \biggl[a_{0}(T)\exp \biggl( \int_{t_{0}}^{t}Q_{2}(s)\,ds \biggr) \biggr]^{\frac{q}{p}} \\ &{}\times \biggl[1+L\biggl(1-\frac{q}{p}\biggr)a_{0}^{\frac{p}{p-q}}(T) \int _{t_{0}}^{t}\exp \biggl(- \int_{t_{0}}^{s} \biggl(1-\frac{q}{p} \biggr)Q_{2}(\tau)\,d\tau \biggr)\,ds \biggr]^{\frac{q}{p-q}}. \end{aligned}$$
(23)
Letting \(t=T\), where T is a positive constant chosen arbitrarily, we get
$$\begin{aligned} \begin{aligned}[b] V(T)\leq{}& \biggl[a_{0}(T)\exp \biggl( \int_{t_{0}}^{T}Q_{2}(s)\,ds \biggr) \biggr]^{\frac{q}{p}} \\ &{}\times \biggl[1+L\biggl(1-\frac{q}{p}\biggr)a_{0}^{\frac{p}{p-q}}(T) \int_{t_{0}}^{T} \exp \biggl(- \int_{t_{0}}^{s}\biggl(1-\frac{q}{p} \biggr)Q_{2}(\tau)\,d\tau \biggr)\,ds \biggr]^{\frac{q}{p-q}}. \end{aligned} \end{aligned}$$
(24)
Obviously,
$$\begin{aligned} V(t) \leq& \biggl[a_{0}(t)\exp \biggl( \int_{t_{0}}^{t}Q_{2}(s)\,ds \biggr) \biggr]^{\frac{q}{p}} \\ &{}\times \biggl[1+L\biggl(1-\frac{q}{p}\biggr)a_{0}^{\frac{p}{p-q}}(t) \int_{t_{0}}^{t} \exp \biggl(- \int_{t_{0}}^{s}\biggl(1-\frac{q}{p} \biggr)Q_{2}(\tau)\,d\tau \biggr)\,ds \biggr]^{\frac{q}{p-q}} \\ =&\tau_{1}(t). \end{aligned}$$
(25)
For all \(t\in I_{2}\), we can obtain the following estimation by (15) and (25):
$$\begin{aligned} V^{\frac{p}{q}}(t) \leq& a_{0}(t)+\frac{p}{p-q} \sum_{i=1}^{N} \int _{t_{0}}^{t}g_{i}(s)V\bigl( \phi_{i}(s)\bigr)\,ds \\ &{}+\sum_{j=1}^{L} \int_{t_{0}}^{t}b_{j}(s) \int_{t_{0}}^{s}c_{j}(\theta)V \bigl(w_{j}(\theta )\bigr)\,d\theta \,ds+\beta_{1}V(t_{1}-0) \\ \leq& a_{0}(t)+\frac{p}{p-q}\sum_{i=1}^{N} \int_{t_{0}}^{t_{1}}g_{i}(s)\tau _{1} \bigl(\phi_{i}(s)\bigr)\,ds +\sum_{j=1}^{L} \int_{t_{0}}^{t_{1}}b_{j}(s) \int_{t_{0}}^{s}c_{j}(\theta)\tau \bigl(w_{j}(\theta)\bigr)\,d\theta \,ds \\ &{}+\beta_{1}\tau_{1}(t_{1})+\frac{p}{p-q} \sum_{i=1}^{N} \int _{t_{1}}^{t}g_{i}(s)V\bigl( \phi_{i}(s)\bigr)\,ds \\ &{}+\sum_{j=1}^{L} \int_{t_{1}}^{t}b_{j}(s) \int_{t_{0}}^{s}c_{j}(\theta)V \bigl(w_{j}(\theta )\bigr)\,d\theta \,ds \\ =&a_{1}(t)+\frac{p}{p-q}\sum_{i=1}^{N} \int_{t_{1}}^{t}g_{i}(s)V\bigl( \phi_{i}(s)\bigr)\,ds \\ &{}+\sum_{j=1}^{L} \int_{t_{1}}^{t}b_{j}(s) \int_{t_{0}}^{s}c_{j}(\theta)V \bigl(w_{j}(\theta )\bigr)\,d\theta \,ds. \end{aligned}$$
(26)
Since it has the same style as (16), we can use the same ways to obtain the estimation as (25). Therefore
$$\begin{aligned} V(t) \leq& \biggl[a_{1}(t)\exp \biggl( \int_{t_{1}}^{t}Q_{2}(s)\,ds \biggr) \biggr]^{\frac{q}{p}} \\ &{}\times \biggl[1+L\biggl(1-\frac{q}{p}\biggr)a_{1}^{\frac{p}{p-q}}(t) \int_{t_{1}}^{t} \exp \biggl(- \int_{t_{1}}^{s}\biggl(1-\frac{q}{p} \biggr)Q_{2}(\tau)\,d\tau \biggr)\,ds \biggr]^{\frac{q}{p-q}}. \end{aligned}$$
(27)
Let \(\tau_{2}(t)\) denote the function of the right-hand side of (26), which is a positive and nondecreasing function on \(I_{2}\). Using mathematical induction, \(\forall k\in Z\), when \(\forall t\in I_{k}\), the estimation is obtained. We have
$$\begin{aligned} V(t) \leq& \biggl[a_{k-1}(t)\exp \biggl( \int_{t_{k-1}}^{t}Q_{2}(s)\,ds \biggr) \biggr]^{\frac{q}{p}} \\ &{}\times \biggl[1+L\biggl(1-\frac{q}{p}\biggr)a_{k-1}^{\frac{p}{p-q}}(t) \int_{t_{k-1}}^{t} \exp \biggl(- \int_{t_{k-1}}^{s}\biggl(1-\frac{q}{p} \biggr)Q_{2}(\tau)\,d\tau \biggr)\,ds \biggr]^{\frac{q}{p-q}}. \end{aligned}$$
(28)
This completes the proof. □

We consider the inequality (3) now.

Theorem 2.3

Suppose \(\phi_{i}(t)\), \(w_{j}(t)\), \(a_{0}(t)\), p, q are defined as those in Theorem  2.2. \(g_{i}(t, s)\), \(b_{j}(t, s)\), \(c_{j}(t, s)\) are nondecreasing functions with their two variables. \(q_{1}(t)\), \(q_{2}(t)\) are continuous and nondecreasing functions on \([t_{0}, \infty)\) and positive on \([t_{0}, \infty )\) and \(u(t)\) is a piecewise continuous nonnegative function on \([t_{0}, \infty)\) with only the first discontinuous points \(t_{i}\), \(i=1, 2, \ldots\) , and satisfying (3). Then, for all \(t\in I_{k}\),
$$ u(t)\leq R_{k}^{q^{-1}}(t), $$
(29)
where
$$ \begin{aligned} &R_{k}(t)= \biggl[a_{k-1}(t)q(t) \exp \biggl( \int_{t_{k-1}}^{t}\bigl(\tilde {Q}_{1}(s)+ \tilde{B}_{1}(s)\bigr)\,ds \biggr) \biggr]^{\frac{q}{p}} \\ &\hphantom{R_{k}(t)={}}{}\times \biggl[1+\biggl(1-\frac{q}{p}\biggr) \int_{t_{k-1}}^{t}\exp\biggl( \int _{t_{k-1}}^{s} \biggl(\frac{q}{p}-1\biggr) \bigl(\tilde{Q}_{1}(\tau)+\tilde{B}_{1}(\tau)\bigr)\,d\tau \biggr)\,ds \biggr]^{\frac{q}{p-q}}, \\ &a_{k-1}(t)= a_{0}(t)q(t)\Biggl\{ 1+\sum _{i=1}^{k} \int_{t_{k-1}}^{t_{k}}g_{i}(t, s)\biggl[ \frac{R_{i}(\phi_{i}(s))}{a_{0}(s)}\biggr]\,ds \\ &\hphantom{a_{k-1}(t)={}}{}+\sum_{j=1}^{L} \int_{t_{k-1}}^{t_{k}}\biggl[\frac{b_{j}(t, s)}{a_{0}(s)}\biggr] \biggl( \int_{t_{k-1}}^{s}c_{j}(s, \theta)R_{j}\bigl(w_{j}(\theta)\bigr)\,d\theta \biggr)\,ds + \sum_{i=1}^{k}\beta_{i} \frac{R_{i}(t_{i}-0)}{a(t_{i}-0)}\Biggr\} , \\ &\tilde{Q}_{1}(t)= \sum_{i=1}^{N}g_{i}(t, s)\frac{[q(\phi_{i}(t))a_{0}(\phi _{i}(t))]^{\frac{q}{p}}}{a_{0}(t)}, \\ &\tilde{B}_{1}(t)= \sum_{j=1}^{L}b_{j}(t, s)c_{j}(t, s)\frac {[q(w_{j}(t))a_{0}(w_{j}(t))]^{\frac{q}{p}}}{a_{0}(t)}. \end{aligned} $$
(30)

Proof

Let \(v=u^{q}\), so the inequality (3) is equivalent to
$$\begin{aligned} v^{\frac{p}{q}}(t) \leq& a_{0}(t)+q_{1}(t) \sum_{i=1}^{N} \int_{t_{0}}^{t}g_{i}(t, s)v\bigl( \phi_{i}(s)\bigr)\,ds \\ &{} +\sum_{j=1}^{L} \int_{t_{0}}^{t}b_{j}(t, s) \int_{t_{0}}^{s}c_{j}(s, \theta )v \bigl(w_{j}(\theta)\bigr)\,d\theta \,ds +q_{2}(t)\sum _{t_{0}< t_{i}< t}\beta_{i}v(t_{i}-0). \end{aligned}$$
(31)
Note that \(w(t)= v^{\frac{p}{q}}(t)\), then from (31), we get
$$\begin{aligned} \frac{w(t)}{a_{0}(t)} \leq&1+q_{1}(t)\sum _{i=1}^{N} \int_{t_{0}}^{t}g_{i}(t, s)\biggl[ \frac{v(\phi_{i}(s))}{a_{0}(s)}\biggr]\,ds \\ &{}+\sum_{j=1}^{L} \int_{t_{0}}^{t}\biggl[\frac{b_{j}(t, s)}{a_{0}(s)}\biggr] \int _{t_{0}}^{s}c_{j}(s, \theta)v \bigl(w_{j}(\theta)\bigr)\,d\theta \,ds \\ &{}+q_{2}(t)\sum _{t_{0}< t_{i}< t}\beta_{i}\frac{v(t_{i}-0)}{a_{0}(t_{i}-0)}. \end{aligned}$$
(32)
Moreover, with the assumption that \(q(t)=\max\{q_{1}(t), q_{2}(t)\}+1\), we see
$$\begin{aligned} \frac{w(t)}{a_{0}(t)} \leq& q(t)\Biggl\{ 1+\sum _{i=1}^{N} \int_{t_{0}}^{t}g_{i}(t, s)\biggl[ \frac{v(\phi_{i}(s))}{a_{0}(s)}\biggr]\,ds \\ & +\sum_{j=1}^{L} \int_{t_{0}}^{t}\biggl[\frac{b_{j}(t, s)}{a_{0}(s)}\biggr] \int _{t_{0}}^{s}c_{j}(s, \theta)v \bigl(w_{j}(\theta)\bigr)\,d\theta \,ds +\sum _{t_{0}< t_{i}< t}\beta_{i}\frac{v(t_{i}-0)}{a_{0}(t_{i}-0)}\Biggr\} . \end{aligned}$$
(33)
Let
$$\begin{aligned} \Gamma_{1}(t) =&1+\sum_{i=1}^{N} \int_{t_{0}}^{t}g_{i}(t, s)\biggl[ \frac{v(\phi _{i}(s))}{a_{0}(s)}\biggr]\,ds \\ &{}+\sum_{j=1}^{L} \int_{t_{0}}^{t}\biggl[\frac{b_{j}(t, s)}{a_{0}(s)}\biggr] \int _{t_{0}}^{s}c_{j}(s, \theta)v \bigl(w_{j}(\theta)\bigr)\,d\theta \,ds +\sum _{t_{0}< t_{i}< t}\beta_{i}\frac{v(t_{i}-0)}{a_{0}(t_{i}-0)}. \end{aligned}$$
(34)
Then \(\Gamma_{1}(t)\) is a positive and nondecreasing function on I with \(\Gamma_{1}(t_{0})=1\) and
$$ \frac{w(t)}{a_{0}(t)}\leq q(t)\Gamma_{1}(t), w(t)\leq q(t) \Gamma_{1}(t)a_{0}(t), $$
(35)
so \(v(t)\leq(q(t)\Gamma_{1}(t)a_{0}(t))^{\frac{q}{p}}\). Applying (31) to (35), we obtain
$$\begin{aligned} \Gamma_{1}(t) \leq&1+\sum_{i=1}^{N} \int_{t_{0}}^{t}g_{i}(t, s) \frac{[q(\phi _{i}(s))a_{0}(\phi_{i}(s))]^{\frac{q}{p}}}{a_{0}(s)}{\Gamma_{1}}^{\frac {q}{p}}(s)\,ds \\ &{} +\sum_{j=1}^{L} \int_{t_{0}}^{t}\biggl[\frac{b_{j}(t, s)}{a_{0}(s)}\biggr] \biggl( \int_{t_{0}}^{s}c_{j}(s, \theta)\bigl[q \bigl(w_{j}(\theta)\bigr)a_{0}\bigl(w_{j}(\theta) \bigr)\bigr]^{\frac{q}{p}} {\Gamma_{1}}^{\frac{q}{p}}(\theta) \,d\theta \biggr)\,ds \\ &{}+\sum_{t_{0}< t_{i}< t}\beta_{i} \frac{[q(t_{i}-0)a_{0}(t_{i}-0)\Gamma _{1}(t_{i}-0)]^{\frac{q}{p}}}{a_{0}(t_{i}-0)}. \end{aligned}$$
(36)
Let \(I_{i}=[t_{i-1}, t_{i})\), first, we consider the condition under which, for all t in \([t_{0}, t_{1})\), we have
$$\begin{aligned} \Gamma_{1}(t) \leq&1+\sum_{i=1}^{N} \int_{t_{0}}^{t}g_{i}(t, s) \frac{[q(\phi _{i}(s))a_{0}(\phi_{i}(s))]^{\frac{q}{p}}}{a_{0}(s)}{\Gamma_{1}}^{\frac {q}{p}}(s)\,ds \\ &{} +\sum_{j=1}^{L} \int_{t_{0}}^{t}\biggl[\frac{b_{j}(t, s)}{a_{0}(s)}\biggr] \biggl( \int _{t_{0}}^{s}c_{j}(s, \theta)\bigl[q \bigl(w_{j}(\theta)\bigr)a_{0}\bigl(w_{j}(\theta) \bigr)\bigr]^{\frac{q}{p}} {\Gamma_{1}}^{\frac{q}{p}}(\theta) \,d\theta \biggr)\,ds. \end{aligned}$$
(37)
For all \(t\in[t_{0}, T)\), where \(T\in I_{1}\), we get
$$\begin{aligned} \Gamma_{1}(t) \leq&1+\sum_{i=1}^{N} \int_{t_{0}}^{t}g_{i}(T, s) \frac{[q(\phi _{i}(s))a_{0}(\phi_{i}(s))]^{\frac{q}{p}}}{a_{0}(s)}{\Gamma_{1}}^{\frac {q}{p}}(s)\,ds \\ &{} +\sum_{j=1}^{L} \int_{t_{0}}^{t}\biggl[\frac{b_{j}(T, s)}{a_{0}(s)}\biggr] \biggl( \int _{t_{0}}^{s}c_{j}(s, \theta)\bigl[q \bigl(w_{j}(\theta)\bigr)a_{0}\bigl(w_{j}(\theta) \bigr)\bigr]^{\frac{q}{p}} {\Gamma_{1}}^{\frac{q}{p}}(\theta) \,d\theta \biggr)\,ds. \end{aligned}$$
(38)
Let \(\Gamma_{2}(t)\) denote the function on the right-hand side of (38), which is a positive and nondecreasing function on \(I_{1}\) with \(\Gamma_{2}(t_{0})= 1\), and \(\Gamma_{1}(t)\leq\Gamma_{2}(t)\). For all \(t\in [t_{0}, T)\), differentiating \(\Gamma_{2}(t)\),
$$\begin{aligned} \frac{d\Gamma_{2}(t)}{dt} \leq&\sum_{i=1}^{N}g_{i}(T, t)\frac{[q(\phi _{i}(t))a_{0}(\phi_{i}(t))]^{\frac{q}{p}}}{a_{0}(t)}{\Gamma_{2}}^{\frac{q}{p}}(t) \\ &{} +\sum_{j=1}^{L}\biggl[ \frac{b_{j}(T, t)}{a_{0}(t)}\biggr] \int_{t_{0}}^{t}c_{j}(s, \theta )\bigl[q \bigl(w_{j}(\theta)\bigr)a_{0}\bigl(w_{j}(\theta) \bigr)\bigr]^{\frac{q}{p}} {\Gamma_{2}}^{\frac{q}{p}}(\theta) \,d\theta. \end{aligned}$$
(39)
Let
$$\begin{aligned}& Q(t)=\sum_{i=1}^{N}g_{i}(T, t) \frac{[q(\phi_{i}(t))a_{0}(\phi_{i}(t))]^{\frac {q}{p}}}{a_{0}(t)}, \\& B(t)=\sum_{j=1}^{L}b_{j}(T, t)c_{j}(T, t)\frac{[q(w_{j}(t))a_{0}(w_{j}(t))]^{\frac {q}{p}}}{a_{0}(t)}. \end{aligned}$$
Since \(c_{j}\), q, \(a_{0}\) are nondecreasing functions, we can estimate (40) further to obtain
$$ \frac{d\Gamma_{2}(t)}{dt}\leq Q(t){\Gamma_{2}}^{\frac{q}{p}}(t)+B(t) \int _{t_{0}}^{t}{\Gamma_{2}}^{\frac{q}{p}}( \theta)\,d\theta. $$
(40)
Moreover, we can get
$$\begin{aligned} \frac{d\Gamma_{2}(t)}{dt}&\leq\bigl(Q(t)+B(t)\bigr) \biggl( \Gamma_{2}^{\frac {q}{p}}(t)+ \int_{t_{0}}^{t}\Gamma_{2}^{\frac{q}{p}}( \theta)\,d\theta \biggr) \\ &\leq \bigl(Q(t)+B(t)\bigr) \biggl(\Gamma_{2}(t)+ \int_{t_{0}}^{t}{\Gamma_{2}}^{\frac {q}{p}}( \theta)\,d\theta \biggr). \end{aligned}$$
(41)
Let \(\Gamma_{3}(t)=\Gamma_{2}(t)+\int_{t_{0}}^{t}\Gamma_{2}^{\frac{q}{p}}(\theta )\,d\theta\). We see that \(\Gamma_{3}(t)\) satisfies \(\Gamma_{2}(t)\leq\Gamma _{3}(t)\), and differentiating \(\Gamma_{3}(t)\), we can obtain
$$\begin{aligned} \frac{d\Gamma_{3}(t)}{dt}&= \frac{d\Gamma_{2}(t)}{dt}+{\Gamma_{2}}^{\frac {q}{p}}(t) \\ &\leq \bigl(Q(t)+B(t)\bigr)\Gamma_{3}(t)+{\Gamma_{3}}^{\frac{q}{p}}(t). \end{aligned}$$
(42)
Consider
$$ \left \{ \textstyle\begin{array}{l} \frac{d\Gamma_{4}(t)}{dt}= (Q(t)+B(t))\Gamma_{4}(t)+\Gamma _{4}^{\frac{q}{p}}(t), \\ \Gamma_{4}(t_{0})= 1. \end{array}\displaystyle \right . $$
(43)
Since (43) is a Bernoulli equation, we compute it to obtain
$$\begin{aligned} \Gamma_{4}(t) =& \exp \biggl( \int_{t_{0}}^{t}\bigl(Q(s)+ B(s)\bigr)\,ds \biggr) \\ &{}\times \biggl[1+\biggl(1-\frac{q}{p}\biggr) \int_{t_{0}}^{t}\exp\biggl( \int_{t_{0}}^{s} -\biggl(1-\frac{q}{p}\biggr) \bigl(Q(\tau)+B(\tau)\bigr)\,d\tau\biggr)\,ds \biggr]^{\frac{p}{p-q}}. \end{aligned}$$
(44)
Then by comparison of the differential inequality, we have \(\Gamma _{3}(t)\leq\Gamma_{4}(t)\). Therefore,
$$\begin{aligned} v(t) \leq& \biggl(a_{0}(t)q(t)\exp \biggl( \int_{t_{0}}^{t}\bigl(Q(s)+B(s)\bigr)\,ds \biggr) \biggr)^{\frac{q}{p}} \\ &{}\times \biggl[1+\biggl(1-\frac{q}{p}\biggr) \int_{t_{0}}^{t}\exp\biggl( \int_{t_{0}}^{s} -\biggl(1-\frac{q}{p}\biggr) \bigl(Q(\tau)+B(\tau)\bigr)\,d\tau\biggr)\,ds \biggr]^{\frac{q}{p-q}}. \end{aligned}$$
(45)
By taking \(t=T\) in the above inequality, and noticing the definitions of \(\tilde{Q}_{1}(t)\) and \(\tilde{B}_{1}(t)\), we get
$$\begin{aligned} v(t) \leq& \biggl[a_{0}(t)q(t)\exp \biggl( \int_{t_{0}}^{t}\bigl(\tilde{Q}_{1}(s)+\tilde {B}_{1}(s)\bigr)\,ds \biggr) \biggr]^{\frac{q}{p}} \\ &{}\times \biggl[1+ \int_{t_{0}}^{t}\biggl(1-\frac{q}{p}\biggr)\exp \biggl( \int_{t_{0}}^{t} -\biggl(1-\frac{q}{p}\biggr) \bigl(\tilde{Q}_{1}(s)+\tilde{B}_{1}(s)\bigr)\,ds\biggr) \biggr]^{\frac{q}{p-q}}. \end{aligned}$$
(46)
Let \(R_{1}(t)\) denote the function on the right-hand side of (46). When \(t\in I_{2}\), we obtain
$$\begin{aligned} v^{\frac{p}{q}}(t) \leq& a_{0}(t)q(t)\Biggl\{ 1+\sum _{i=1}^{N} \int _{t_{0}}^{t_{1}}g_{i}(t, s)\biggl[ \frac{R_{1}(\phi_{i}(s))}{a_{0}(s)}\biggr]\,ds \\ &{}+\sum_{j=1}^{L} \int_{t_{0}}^{t_{1}}\biggl[\frac{b_{j}(t, s)}{a_{0}(s)}\biggr] \biggl( \int_{t_{0}}^{s}c_{j}(s, \theta)R_{1}\bigl(w_{j}(\theta)\bigr)\,d\theta \biggr)\,ds+ \sum_{i=1}^{k}\beta_{i} \frac{R_{i}(t_{i}-0)}{a(t_{i}-0)}\Biggr\} \\ &{}+a_{0}(t)q(t) \Biggl\{ \sum_{i=1}^{N} \int_{t_{1}}^{t}g_{i}(t, s)\biggl[ \frac{v_{(}\phi _{i}(s))}{a_{0}(s)}\biggr]\,ds \\ &{}+\sum_{j=1}^{L} \int_{t_{1}}^{t}\biggl[\frac{b_{j}(t, s)}{a_{0}(s)}\biggr] \biggl( \int _{t_{0}}^{s}c_{j}(s, \theta)v \bigl(w_{j}(\theta)\bigr)\,d\theta \biggr)\,ds \Biggr\} . \end{aligned}$$
(47)
Let
$$\begin{aligned} a_{1}(t) =&a_{0}(t)q(t)\Biggl\{ 1+\sum _{i=1}^{N} \int_{t_{0}}^{t_{1}}g_{i}(t, s)\biggl[ \frac {R_{1}(\phi_{i}(s))}{a_{0}(s)}\biggr]\,ds \\ &{}+\sum_{j=1}^{L} \int_{t_{0}}^{t_{1}}\biggl[\frac{b_{j}(t, s)}{a_{0}(s)}\biggr] \biggl( \int_{t_{0}}^{s}c_{j}(s, \theta)R_{1}\bigl(w_{j}(\theta)\bigr)\,d\theta \biggr)\,ds \\ &{}+ \sum_{i=1}^{k}\beta_{i} \frac{R_{i}(t_{i}-0)}{a(t_{i}-0)}\Biggr\} . \end{aligned}$$
(48)
Obviously, \(a_{1}(t)\geq a_{0}(t)\). Since \(q(t)\geq1\), we can go further to obtain
$$\begin{aligned} v^{\frac{p}{q}}(t) \leq& a_{1}(t)q(t) \Biggl\{ 1+\sum _{i=1}^{N} \int _{t_{1}}^{t}g_{i}(t, s)\biggl[ \frac{v_{(}\phi_{i}(s))}{a_{0}(s)}\biggr]\,ds \\ &{}+\sum_{j=1}^{L} \int_{t_{1}}^{t}\biggl[\frac{b_{j}(t, s)}{a_{0}(s)}\biggr] \biggl( \int_{t_{0}}^{s}c_{j}(s, \theta)v \bigl(w_{j}(\theta)\bigr)\,d\theta \biggr)\,ds \Biggr\} . \end{aligned}$$
(49)
Since (49) has the same style as (38), we can use the same solution to deal with it, finally the estimation of the unknown function in the inequality (3) is obtained. We have
$$\begin{aligned} v(t) \leq& \biggl(a_{k-1}(t)q(t)\exp \biggl( \int_{t_{k-1}}^{t}\bigl(\tilde {Q}_{1}(s)+ \tilde{B}_{1}(s)\,ds\bigr) \biggr) \biggr)^{\frac{q}{p}} \\ &{}\times \biggl[1+\biggl(1-\frac{q}{p}\biggr) \int_{t_{k-1}}^{t}\exp\biggl(-\biggl(1-\frac {q}{p} \biggr) \int_{t_{k-1}}^{s}\bigl(\tilde{Q}_{1}(\tau)+ \tilde{B}_{1}(\tau)\bigr)\,d\tau \biggr)\,ds \biggr]^{\frac{q}{p-q}}. \end{aligned}$$
(50)
Let \(R_{k}(t)\) denote the function on the right-hand side, and
$$\begin{aligned} a_{k-1}(t) =&a_{0}(t)q(t)\Biggl\{ 1+ \sum _{i=1}^{k} \int_{t_{k-1}}^{t_{k}}g_{i}(t, s)\biggl[ \frac{R_{i}(\phi _{i}(s))}{a_{0}(s)}\biggr]\,ds \\ &{}+\sum_{j=1}^{L} \int_{t_{k-1}}^{t_{k}}\biggl[\frac{b_{j}(t, s)}{a_{0}(s)}\biggr] \biggl( \int_{t_{k-1}}^{s}c_{j}(s, \theta)R_{j}\bigl(w_{j}(\theta)\bigr)\,d\theta \biggr)\,ds+ \sum_{i=1}^{k}\beta_{i} \frac{R_{i}(t_{i}-0)}{a(t_{i}-0)}\Biggr\} . \end{aligned}$$
(51)
So we obtain
$$u(t)\leq R _{k}^{q^{-1}}(t). $$
This proves Theorem 2.3. □

3 Applications

In this section we will apply our Theorem 2.1 and Theorem 2.2 to discuss the following differential-integral equation and retarded differential equation for discontinuous functions, respectively. We present the following propositions.

Proposition 3.1

Consider the following equation:
$$ \left \{ \textstyle\begin{array}{l} \frac{dx(t)}{dt}=H(t, x(\phi_{1}(t)), \ldots, x(\phi_{N}(t)), \int_{0}^{t}K(s, x(w_{1}(s)))\,ds, \ldots, \int_{0}^{t}K(s, x(w_{L}(s)))\,ds), \\ x(0)=x_{0}, \quad \forall t \in I=[0, \infty), \end{array}\displaystyle \right . $$
(52)
where the function K is in \(C(R\times R, R_{+})\) and \(\phi_{i}(t)\leq t\), \(w_{j}(t)\leq t\), for \(t>0\), H satisfies the following condition:
$$\begin{aligned}& \biggl\vert H\biggl(t, u_{1}, u_{2}, \ldots, u_{N}, \int_{0}^{t}K(s, v_{1})\,ds, \ldots, \int_{0}^{t}K(s, v_{L})\,ds\biggr)\biggr\vert \\& \quad \leq\sum_{i=1}^{N}g_{i}(t)u_{i}+ \sum_{j=1}^{L}b_{j}(t) \int_{0}^{t}c_{j}(\theta )v_{j}\, d\theta, \end{aligned}$$
(53)
where \(g_{i}(t)\), \(b_{j}(t)\), \(c_{j}(t)\), \(w_{j}(t)\) are defined as in Theorem  2.1. If
$$\int_{t_{0}}^{t} Q_{1}(s)\,ds< \infty,\qquad \int_{t_{0}}^{t} Y(s)\exp\biggl(- \int_{t_{0}}^{s} Q_{1}(\tau)\,d\tau\biggr) \,ds< \infty. $$
Then all the solutions of equation (53) exist on I and for all t in \(I=[0, \infty)\), and they satisfy the following estimate:
$$ \bigl\vert x(t)\bigr\vert \leq x_{0}+\exp \biggl( \int_{t_{0}}^{t} Q_{1}(s)\,ds \biggr) \biggl( \int_{t_{0}}^{t} Y(s)\exp\biggl(- \int_{t_{0}}^{s} Q_{1}(\tau)\,d\tau\biggr)\,ds \biggr). $$
(54)

Proof

Integrating both sides of equation (52) from 0 to t, we get
$$\begin{aligned} x(t) =& x_{0}+ \int_{0}^{t}H\biggl(s, x\bigl(\phi_{1}(s) \bigr), \ldots, x\bigl(\phi_{N}(s)\bigr), \int _{0}^{s}K\bigl(\tau, x\bigl(w_{1}( \tau)\bigr)\bigr)\,d\tau, \ldots, \\ &\int_{0}^{s}K\bigl(\tau, x\bigl(w_{L}( \tau)\bigr)\bigr)\,d\tau\biggr)\,ds. \end{aligned}$$
(55)
Using the conditions (52) and (53), we can obtain
$$ \bigl\vert x(t)\bigr\vert \leq x_{0}+ \sum _{i=1}^{N} \int_{t_{0}}^{t} g_{i}(s)x\bigl( \phi_{i}(s)\bigr)\,ds+\sum_{j=1}^{L} \int_{t_{0}}^{t}b_{j}(s) \int_{t_{0}}^{s} c_{j}(\theta)x \bigl(w_{j}(\theta )\bigr)\,d\theta \,ds. $$
(56)
Applying Theorem 2.1 to (56), we can obtain the estimate. □

Proposition 3.2

Consider the impulsive differential system
$$ \left \{ \textstyle\begin{array}{l} \frac{dx^{p}(t)}{dt}=H(t, x(\phi_{1}(t)), \ldots, x(\phi _{N}(t)), \int_{0}^{t}K(s, x(w_{1}(s)))\,ds, \ldots, \int_{0}^{t}K(s, x(w_{L}(s)))\,ds), \\ \Delta(x)|_{t=t_{0}}= \beta_{i}x^{q}(t_{i}-0), \\ x(0)=c, \quad \forall t \in I=[0, \infty), \end{array}\displaystyle \right . $$
(57)
where \(\phi_{i}(t)\), \(w_{j}(t) \) are defined as in Theorem  2.2 and the function K is in \(C(R\times R, R_{+})\). Furthermore, H satisfies
$$\begin{aligned}& \biggl\vert H\biggl(t, x\bigl(\phi_{1}(t)\bigr), \ldots, x\bigl(\phi_{N}(t)\bigr), \int_{0}^{t}K\bigl(s, x\bigl(w_{1}(s) \bigr)\bigr)\,ds, \ldots, \int_{0}^{t}K\bigl(s, x\bigl(w_{L}(s) \bigr)\bigr)\,ds\biggr)\biggr\vert \\& \quad \leq \sum_{i=1}^{N}g_{i}(t)x^{q} \bigl(\phi_{i}(t)\bigr)+\sum_{j=1}^{L}b_{j}(s) \int _{0}^{s}c_{j}(t)x^{q} \bigl(w_{j}(t)\bigr)\, dt. \end{aligned}$$
(58)
Then all the solutions of equation (57) exist on I and satisfy \(|x(t)|\leq\tau_{k}(t)\) for all \(t\in I_{k}\), where \(\tau_{k}(t)\) is defined as in Theorem  2.2.

Proof

Integrating (57) we obtain
$$\begin{aligned} \bigl\vert x^{p}(t)\bigr\vert \leq& c^{p}+ \int_{0}^{t}H \biggl(s, x\bigl(\phi_{1}(s) \bigr), \ldots, x\bigl(\phi_{N}(s)\bigr), \int_{0}^{s}K\bigl(\tau, x\bigl(w_{1}( \tau)\bigr)\bigr)\,d\tau, \ldots, \\ & \int_{0}^{s}K\bigl(\tau, x\bigl(w_{L}( \tau )\bigr)\bigr)\,d\tau \biggr)\,ds \\ &{} +\sum_{t_{0}< t_{i}< t}\beta_{i}u^{q}(t_{i}-0), \quad \forall t\in I. \end{aligned}$$
(59)
Furthermore, we get
$$\begin{aligned} \bigl\vert x^{p}(t)\bigr\vert \leq& c^{p}+\sum_{i=1}^{N} \int_{0}^{t}g_{i}(s)x^{q}\bigl( \phi_{i}(s)\bigr)\,ds \\ &{} +\sum_{j=1}^{L} \int_{0}^{t}b_{j}(s) \int_{0}^{s}c_{j}(\theta)x^{q} \bigl(w_{j}(\theta )\bigr)\,d\theta \,ds+\sum _{t_{0}< t_{i}< t}\beta_{i}u^{q}(t_{i}-0). \end{aligned}$$
(60)
Then we use Theorem 2.2 to obtain the estimation. □

Declarations

Acknowledgements

The authors sincerely thank the referees for their constructive suggestions and corrections. This project is supported by the NNSF of China (Grants 11171178 and 11271225), NSF of Shandong Province (ZR2015PA005).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
School of Mathematical Sciences, Qufu Normal University
(2)
Department of Mathematics, Jining University

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