Theory and Modern Applications

# Gronwall-type integral inequalities with impulses on time scales

## Abstract

In this article, some Gronwall-type integral inequalities with impulses on time scales are investigated. Our results extend some known dynamic inequalities on time scales, unify and extend some continuous inequalities and their corresponding discrete analogues. Some applications of the main results are given in the end of this article.

AMS (MOS) Subject Classification 34D09; 34D99, 37M10, 35D05, 49K25, 90C46.

## Introduction

The theory of time scales, which has recently received a lot of attention, was initiated by Hilger  in his Ph.D. thesis in 1988 to contain both difference and differential calculus in a consistent way. Since then many authors have investigated the dynamic equations, the calculus of variations and the optimal control problem on time scales (see ). At the same time, a few papers have studied the theory of integral inequalities on time scales (see ).

In this article, we study some Gronwall-type integral inequalities on time scales, which extend some known dynamic inequalities on time scales, unify and extend some continuous inequalities and their corresponding discrete analogues. It is helpful in our result to study dynamic systems and optimal control problem on time scales.

## Preliminaries

For convenience, we present some important theorem on time scales in this section. The approach is based on the ideas in  and will be of fundamental importance in following results.

A time scale $T$ is a nonempty closed subset of R. The two most popular examples are $T=R$ and $T=Z$. Define the forward and backward jump operators $\sigma ,\rho :T\to T$ by

$\sigma \left(t\right)=inf\left\{s\in T|s>t\right\},\phantom{\rule{1em}{0ex}}\rho \left(t\right)=sup\left\{s\in T|s

where, in this definition, we write $sup\varnothing =infT\equiv a$ and $inf\varnothing =supT\equiv b$. A point $t\in T$ is said to be left-dense, left-scattered, right-dense, right-scattered if ρ(t) = t, ρ (t) < t, σ (t) = t, σ (t) > t, respectively. The forward (backward) graininess $\mu :T\to \left[0,+\infty \right)\left(\nu :T\to \left[0,+\infty \right)\right)$ is defined by μ(t) = σ(t) - t(v(t) = t - ρ(t)). Define $\Lambda =\left\{{t}_{i}\in T|i=1,2,\cdots \phantom{\rule{0.3em}{0ex}},n\right\}$, $P{C}_{lrd}\left(T,R\right)\left(P{C}_{rrd}\left(T,R\right)\right)=\left\{x:T\to R|x$ is continuous at right-dense point $t\in T\\Lambda$, x exists right limit at t Λ or left-dense point $t\in T$, x is left (right) continuous and exists right (left) limit at t Λ}. Endowed with norm

$||x|{|}_{PC}=max\left\{\underset{t\in T}{sup}||x\left(t+\right)||,\phantom{\rule{2.77695pt}{0ex}}\underset{t\in T}{sup}||x\left(t-\right)||\right\}.$

$\left\{P{C}_{lrd}\left(T,X\right),||\cdot |{|}_{PC}\right\}$ and $\left\{P{C}_{rrd}\left(T,X\right),||\cdot |{|}_{PC}\right\}$ are Banach spaces.

Theorem 2.1. (1) Let $f\in {L}^{1}\left(T,R\right)$. Then

$F\left(t\right)=\underset{\left[a,t\right)}{\int }f\left(\tau \right)\Delta \tau ,t\in T$

implies $F\in {C}_{rd}\left(T,R\right)$ is Δ-differentiable Δ-a.e. on $T$ and

$F\left(t\right)-F\left(s\right)=\underset{\left[s,t\right)}{\int }f\left(\tau \right)\Delta \tau for\phantom{\rule{2.77695pt}{0ex}}s,t\in T.$
1. (2)

If f and g are Δ-differentiable Δ-a.e. on $T$, then

$\underset{\left[s,t\right)}{\int }\left[{f}^{\Delta }\left(\tau \right)g\left(\tau \right)+{f}^{\sigma }\left(\tau \right){g}^{\Delta }\left(\tau \right)\right]\phantom{\rule{0.3em}{0ex}}\Delta \tau =f\left(t\right)g\left(t\right)-f\left(s\right)g\left(s\right)fors,t\in T.$

The exponential function e p on time scale plays a very important role for discussing dynamic equations on time scales. Define ${\Gamma }_{1}\left(T\right)=\left\{p\in {L}_{\mathsf{\text{loc}}}^{1}\left(T,R\right)|1+\mu \left(t\right)p\left(t\right)\ne 0\right\}$. For any $p,q\in {\Gamma }_{1}\left(T\right)$, define p q = p + q + μpq, $\ominus p=-\frac{p}{1+\mu p}$, $p\ominus q=\frac{p-q}{1+\mu q}$. Further, we can show that $p\oplus q,p\ominus q,\ominus p\in {\Gamma }_{1}\left(T\right)$. Define the generalized exponential function as follows:

${e}_{p}\left(t,s\right)=exp\left\{\underset{\left[s,t\right)}{\int }{\xi }_{\mu \left(\tau \right)}\left(p\left(\tau \right)\right)\Delta \tau \right\}.$

Theorem 2.2. Assume that $p,q\in {\Gamma }_{1}\left(T\right)$, then the following hold:

1. (1)

e 0(t, s) ≡ 1, e p (t, t) ≡ 1, e p (t, s)e p (s, r) = e p (t, r), e p (σ(t), s) = [1 + μ(t)p(t)]e p (t, s);

2. (2)

${e}_{p}\left(t,s\right)=\frac{1}{{e}_{p}\left(s,t\right)}={e}_{\ominus p}\left(s,t\right)$, e p (t, s)e q (t, s) = e pq (t, s), $\frac{{e}_{p}\left(t,s\right)}{{e}_{q}\left(t,s\right)}={e}_{p\ominus q}\left(t,s\right)$;

3. (3)

${e}_{p}\left(\cdot ,s\right)\in {C}_{rd}\left(T,R\right)$, (e p (·, s))Δ = p(·)e p (·, s),e p (s, ·))Δ = -p(·)e p (s, σ(·) Δ-a.e. on $T$.

## Main results

In this section, we deal with Gronwall-type integral inequalities with impulses on time scales. For convenience, we always assume that $f\in {L}^{1}\left(T,R\right)$, p i , ${q}_{i}\in {L}^{1}\left(T,{R}^{+}\right)$ with R + = [0, +∞), 0 < λ i < 1 (i = 1, 2, 3, 4), α ≥ 0, β k ≥ 0 (k = 1, 2, · · ·, n), ${x}_{t}=\underset{a\le s\le t}{sup}|x\left(s\right)|$, ${x}_{{\tau }_{b}}\underset{\tau \le s\le b}{sup}|x\left(s\right)|$ in the section.

Theorem A. (1) If $x\in {C}_{rd}\left(T,R\right)$ satisfies the following inequality

${x}^{\Delta }\left(t\right)\le {p}_{1}\left(t\right)x\left(t\right)\phantom{\rule{2.77695pt}{0ex}}+\phantom{\rule{2.77695pt}{0ex}}f\left(t\right)\phantom{\rule{2.77695pt}{0ex}}\Delta -a.e.\phantom{\rule{2.77695pt}{0ex}}onT$

then

$x\left(t\right)\le {e}_{{p}_{1}}\left(t,a\right)x\left(a\right)+\underset{\left[a,t\right)}{\int }{{e}_{p}}_{1}\left(t,\phantom{\rule{2.77695pt}{0ex}}\sigma \left(\tau \right)\right)f\left(\tau \right)\Delta \tau \phantom{\rule{2.77695pt}{0ex}}for\phantom{\rule{2.77695pt}{0ex}}all\phantom{\rule{2.77695pt}{0ex}}t\in T.$
1. (2)

If $x\in P{C}_{lrd}\left(T,{R}^{+}\right)$ satisfies the following inequality

$x\left(t\right)\le \alpha +\underset{\left[a,t\right)}{\int }{p}_{1}\left(\tau \right)x\left(\tau \right)\Delta \tau +\sum _{{t}_{k}

then

$x\left(t\right)\le \alpha \phantom{\rule{2.77695pt}{0ex}}\prod _{{t}_{k}
1. (3)

If $x\in P{C}_{lrd}\left(T,{R}^{+}\right)$ satisfies the following inequality

$x\left(t\right)\le \alpha +\underset{\left[a,t\right)}{\int }{p}_{1}\left(\tau \right)x\left(\tau \right)\Delta \tau +\underset{\left[a,t\right)}{\int }{q}_{1}\left(\tau \right){x}^{{\lambda }_{1}}\left(\sigma \left(\tau \right)\right)\Delta \tau +\sum _{{t}_{k}

then there is a constant M > 0 such that

$x\left(t\right)\le M,\phantom{\rule{2.77695pt}{0ex}}\forall t\in T.$
1. (4)

If $x\in P{C}_{lrd}\left(T,{R}^{+}\right)$ satisfies the following inequality

$\begin{array}{lll}\hfill x\left(t\right)\phantom{\rule{1em}{0ex}}\le \phantom{\rule{1em}{0ex}}& \alpha +\underset{\left[a,t\right)}{\int }{p}_{1}\left(\tau \right)x\left(\tau \right)\Delta \tau +\underset{\left[a,t\right)}{\int }{p}_{2}\left(\tau \right){x}_{\tau }\Delta \tau +\underset{\left[a,t\right)}{\int }{q}_{1}\left(\tau \right){x}^{{\lambda }_{1}}\left(\sigma \left(\tau \right)\right)\Delta \tau \phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ +\underset{\left[a,t\right)}{\int }{q}_{2}\left(\tau \right){x}_{\sigma \left(\tau \right)}^{{\lambda }_{2}}\Delta \tau +\sum _{{t}_{k}

then there is a constant M > 0 such that

$x\left(t\right)\le M,\phantom{\rule{2.77695pt}{0ex}}\forall t\in T.$
1. (5)

If $x\in P{C}_{lrd}\left(T,{R}^{+}\right)$ satisfies the following inequality

then there is a constant M > 0 such that

$x\left(t\right)\le M,\phantom{\rule{2.77695pt}{0ex}}\forall t\in T.$

Proof. (1) Note that ${p}_{1}\in {L}^{1}\left(T,{R}^{+}\right)$ implies ${p}_{1}\in {\Gamma }_{1}\left(T\right)$ and 1 + μ(t)p 1(t) > 0 for all $t\in T$. Now

${\left[x\left(t\right){e}_{\ominus {p}_{1}}\left(t,a\right)\right]}^{\Delta }=\left[{x}^{\Delta }\left(t\right)-{p}_{1}\left(t\right)x\left(t\right)\right]{e}_{\ominus {p}_{1}}\left(\sigma \left(t\right),a\right)\phantom{\rule{1em}{0ex}}\Delta -\mathsf{\text{a}}\mathsf{\text{.e}}\mathsf{\text{.}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{on}}\phantom{\rule{2.77695pt}{0ex}}T.$

Therefore,

$x\left(t\right){e}_{\ominus {p}_{1}}\left(t,a\right)-x\left(a\right)=\underset{\left[a,t\right)}{\int }\left[{x}^{\Delta }\left(\tau \right)-{p}_{1}\left(\tau \right)x\left(\tau \right)\right]{e}_{\ominus {p}_{1}}\left(\sigma \left(\tau \right),a\right)\Delta \tau \le \underset{\left[a,t\right)}{\int }f\left(\tau \right){e}_{\ominus {p}_{1}}\left(\sigma \left(\tau \right),a\right)\Delta \tau ,$

that is,

$x\left(t\right)\le {e}_{{p}_{1}}\left(t,a\right)x\left(a\right)+\underset{\left[a,t\right)}{\int }{e}_{{p}_{1}}\left(t,\sigma \left(\tau \right)\right)f\left(\tau \right)\Delta \tau \phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{all}}\phantom{\rule{2.77695pt}{0ex}}t\in T.$
1. (2)

Define

$y\left(t\right)=\alpha +\underset{\left[a,t\right)}{\int }{p}_{1}\left(\tau \right)x\left(\tau \right)\Delta \tau +\sum _{{t}_{k}

By Theorem 2.1, y is Δ-differential Δ-a.e. on $T$ and

$\left\{\begin{array}{cc}\hfill {y}^{\Delta }\left(t\right)={p}_{1}\left(t\right)x\left(t\right)\le {p}_{1}\left(t\right)y\left(t\right),\hfill & \hfill t\ne {t}_{k},\hfill \\ \hfill y\left(a\right)=\alpha ,y\left({t}_{k}+\right)=\left(1+{\beta }_{k}\right)y\left({t}_{k}\right).\hfill \end{array}\right\$

For t [a, t 1], it is obvious to

$y\left(t\right)\le \alpha {e}_{{p}_{1}}\left(t,a\right).$

Further, we have

$y\left(t\right)\le y\left({t}_{i}+\right)\phantom{\rule{2.77695pt}{0ex}}{e}_{{p}_{1}}\left(t,{t}_{i}\right)\le \alpha \left(1+{\beta }_{i}\right)\prod _{{t}_{k}<{t}_{i}}\left(1+{\beta }_{k}\right)\phantom{\rule{2.77695pt}{0ex}}{e}_{{p}_{1}}\left({t}_{i},a\right){e}_{{p}_{1}}\left(t,{t}_{i}\right)=\alpha \prod _{{t}_{k}

Thus,

$x\left(t\right)\le \alpha \prod _{{t}_{k}
1. (3)

Setting

$y\left(t\right)=\alpha +\underset{\left[a,t\right)}{\int }{p}_{1}\left(\tau \right)x\left(\tau \right)\Delta \tau +\underset{\left[a,t\right)}{\int }{g}_{1}\left(\tau \right){x}^{{\lambda }_{1}}\left(\sigma \left(\tau \right)\right)\Delta \tau +\sum _{{t}_{k}

then

$\left\{\begin{array}{cc}\hfill {y}^{\Delta }\left(t\right)={p}_{1}\left(t\right)x\left(t\right)+{g}_{1}\left(t\right){x}^{{\lambda }_{1}}\left(\sigma \left(t\right)\right)\le {p}_{1}\left(t\right)y\left(t\right)+{g}_{1}\left(t\right){y}^{{\lambda }_{1}}\left(\sigma \left(t\right)\right)\phantom{\rule{2.77695pt}{0ex}}\Delta -\mathsf{\text{a}}\mathsf{\text{.e}}\mathsf{\text{.}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{on}}\phantom{\rule{2.77695pt}{0ex}}T,\hfill & \hfill t\ne {t}_{k},\hfill \\ \hfill y\left(a\right)=\alpha ,y\left({t}_{k}+\right)=y\left({t}_{k}\right)+{\beta }_{k}x\left({t}_{k}\right)=\left(1+{\beta }_{k}\right)y\left({t}_{k}\right).\hfill \end{array}\right\$

Using the conclusion (1), we have

For $t\in T$, let

$h\left(t\right)=\alpha {e}_{{p}_{1}}\left(b,a\right)+{e}_{{p}_{1}}\left(b,a\right)\underset{\left[a,t\right)}{\int }{g}_{1}\left(\tau \right){y}^{{\lambda }_{1}}\left(\sigma \left(\tau \right)\right)\Delta \tau +{e}_{{p}_{1}}\left(b,a\right)\underset{\left[a,b\right)}{\int }{g}_{1}\left(\tau \right){y}^{{\lambda }_{1}}\left(\sigma \left(\tau \right)\right)\Delta \tau ,$

then h is monotone increasing function and $h\left(b\right)=2h\left(a\right)-\alpha {e}_{{p}_{1}}\left(b,a\right)$,

${h}^{\Delta }\left(t\right)={e}_{{p}_{1}}\left(b,a\right){g}_{1}\left(t\right){y}^{{\lambda }_{1}}\left(\sigma \left(t\right)\right)\le \prod _{k=1}^{n}{\left(2+{\beta }_{k}\right)}^{{\lambda }_{1}}{e}_{{p}_{1}}\left(b,a\right){g}_{1}\left(t\right){q}^{{\lambda }_{1}}\left(t\right),\phantom{\rule{2.77695pt}{0ex}}\forall t\in T.$

Δ-integrating from a to t, we obtain

${h}^{1-{\lambda }_{1}}\left(t\right)-{h}^{1-{\lambda }_{1}}\left(a\right)\le \rho \underset{\left[a,t\right)}{\int }{g}_{1}\left(\tau \right)\Delta \tau ,\phantom{\rule{2.77695pt}{0ex}}\forall t\in T,$

where

$\rho =\left(1-{\lambda }_{1}\right){e}_{{p}_{1}}\left(b,a\right)\prod _{k=1}^{n}{\left(2+{\beta }_{k}\right)}^{{\lambda }_{1}}.$

Now, we observe that

${\left(2h\left(a\right)-\alpha {e}_{{p}_{1}}\left(b,a\right)\right)}^{1-{\lambda }_{1}}-{h}^{1-{\lambda }_{1}}\left(a\right)\le \rho \underset{\left[a,b\right)}{\int }{g}_{1}\left(\tau \right)\Delta \tau .$

Letting

$\Gamma \left(z\right)={\left(2z-\alpha {e}_{{p}_{1}}\left(b,a\right)\right)}^{1-{\lambda }_{1}}-{z}^{1-{\lambda }_{1}},$

then $\Gamma \in C\left(\left[\frac{\alpha {e}_{{p}_{1}}\left(b,a\right)}{2},+\infty \right),R\right)$ and $\Gamma \left(\frac{\alpha {e}_{{p}_{1}}\left(b,a\right)}{2}\right)=-{\left(\frac{\alpha {e}_{{p}_{1}}\left(b,a\right)}{2}\right)}^{1-{\lambda }_{1}}<0$,

$\underset{z\to +\infty }{lim}\Gamma \left(z\right)=\underset{z\to +\infty }{lim}\frac{\Gamma \left(z\right)}{{z}^{1-{\lambda }_{1}}}{z}^{1-{\lambda }_{1}}=\underset{z\to +\infty }{lim}\left[{\left(2-\frac{\alpha {e}_{{p}_{1}}\left(b,a\right)}{z}\right)}^{1-{\lambda }_{1}}-1\right]{z}^{1-{\lambda }_{1}}=+\infty .$

Using the proof by contraction, one can show that there exists a constant M > 0 such that q(a) < M. Thus,

$x\left(t\right)\le y\left(t\right)\le h\left(a\right)\le M,\phantom{\rule{2.77695pt}{0ex}}\forall t\in T.$
1. (4)

Setting λ = max{λ 1, λ 2},

$z\left(t\right)=\left\{\begin{array}{cc}\hfill 1,\hfill & \hfill x\left(t\right)<1,\hfill \\ \hfill x\left(t\right),\hfill & \hfill x\left(t\right)\ge 1,\hfill \end{array}\right\p\left(t\right)={p}_{1}\left(t\right)+{p}_{2}\left(t\right),g\left(t\right)={g}_{1}\left(t\right)+{g}_{2}\left(t\right),\phantom{\rule{2.77695pt}{0ex}}\forall t\in T,$

we have

$x\left(t\right)\le \alpha +\underset{\left[a,t\right)}{\int }p\left(\tau \right){z}_{\tau }\Delta \tau +\underset{\left[a,t\right)}{\int }g\left(\tau \right){z}_{\sigma \left(\tau \right)}^{\lambda }\Delta \tau +\sum _{{t}_{k}

Furthermore,

${z}_{t}\le \alpha +\underset{\left[a,t\right)}{\int }p\left(\tau \right){z}_{\tau }\Delta \tau +\underset{\left[a,t\right)}{\int }g\left(\tau \right){z}_{\sigma \left(\tau \right)}^{\lambda }\Delta \tau +\sum _{{t}_{k}

By the conclusion (3), there is a constant M > 0 such that

$x\left(t\right)\le M,\phantom{\rule{2.77695pt}{0ex}}\forall t\in T.$
1. (5)

Setting λ = max{λ 1, λ 2, λ 3, λ 4},

$y\left(t\right)=\left\{\begin{array}{cc}\hfill 1,\hfill & \hfill {x}_{t}<1,\hfill \\ \hfill {x}_{t},\hfill & \hfill {x}_{t}\ge 1,\hfill \end{array}\right\\phantom{\rule{1em}{0ex}}p\left(t\right)={p}_{1}\left(t\right)+{p}_{2}\left(t\right),\phantom{\rule{1em}{0ex}}g\left(t\right)=\sum _{i=1}^{4}{g}_{i}\left(t\right),\phantom{\rule{2.77695pt}{0ex}}\forall t\in T.$

we have

$y\left(t\right)\le \alpha +\underset{\left[a,t\right)}{\int }p\left(\tau \right)y\left(\tau \right)\Delta \tau +\underset{\left[a,b\right)}{\int }g\left(\tau \right){y}^{\lambda }\left(\tau \right)\Delta \tau +\sum _{{t}_{k}

By the conclusions (3) and (4), we can show that the conclusion (5) is true. The proof is completed. □

In Theorem A, we give some Gronwall-type generalized integral inequalities on time scales. Next, we give some backward Gronwall-type generalized integral inequalities on time scales which can not be directly obtained from Gronwall inequalities.

Theorem B. (1) If $x\in {C}_{rd}\left(T,{R}^{+}\right)$ satisfies the following inequality

$x\left(t\right)\le f\left(t\right)+\underset{\left[t,b\right)}{\int }{p}_{1}\left(\tau \right){x}^{\sigma }\left(\tau \right)\Delta \tau ,$

then

$x\left(t\right)\le f\left(t\right)+\underset{\left[t,b\right)}{\int }{e}_{{p}_{1}}\left(\tau ,t\right){p}_{1}\left(\tau \right){f}^{\sigma }\left(\tau \right)\Delta \tau \phantom{\rule{2.77695pt}{0ex}}for\phantom{\rule{2.77695pt}{0ex}}t\in T.$
1. (2)

If $x\in P{C}_{rrd}\left(T,{R}^{+}\right)$ satisfies the following inequality

$x\left(t\right)\le \alpha +\underset{\left[t,b\right)}{\int }{p}_{1}\left(\tau \right){x}^{\sigma }\left(\tau \right)\Delta \tau +\sum _{{t}_{k}>t}{\beta }_{k}x\left({t}_{k}\right)$

then

$x\left(t\right)\le \alpha \prod _{{t}_{k}>t}\left(1+{\beta }_{k}\right){e}_{{p}_{1}}\left(b,t\right),\phantom{\rule{2.77695pt}{0ex}}\forall t\in T.$

(3) If $x\in P{C}_{rrd}\left(T,{R}^{+}\right)$ satisfies the following inequality

$x\left(t\right)\le \alpha +\underset{\left[t,b\right)}{\int }{p}_{1}\left(\tau \right){x}^{{\lambda }_{1}}\left(\tau \right)\Delta \tau +\underset{\left[t,b\right)}{\int }{g}_{1}\left(\tau \right){x}^{\sigma }\left(\tau \right)\Delta \tau ,$

then there is a constant M > 0 such that

$x\left(t\right)\le M,\phantom{\rule{2.77695pt}{0ex}}\forall t\in T.$

(4) If $x\in P{C}_{rrd}\left(T,{R}^{+}\right)$ satisfies the following inequality

$\begin{array}{lll}\hfill x\left(t\right)\phantom{\rule{1em}{0ex}}\le \phantom{\rule{1em}{0ex}}& \alpha +\underset{\left[t,b\right)}{\int }{p}_{1}\left(\tau \right){x}^{{\lambda }_{1}}\left(\tau \right)\Delta \tau +\underset{\left[t,b\right)}{\int }{p}_{2}\left(\tau \right){x}_{{\tau }_{b}}^{{\lambda }_{2}}\Delta \tau +\underset{\left[t,b\right)}{\int }{g}_{1}\left(\tau \right)x\left(\sigma \left(\tau \right)\right)\Delta \tau \phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ +\underset{\left[t,b\right)}{\int }{g}_{2}\left(\tau \right){x}_{\sigma {\left(\tau \right)}_{b}}\Delta \tau +\sum _{{t}_{k}>t}{\beta }_{k}x\left({t}_{k}\right),\phantom{\rule{2.77695pt}{0ex}}\forall t\in T,\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$

then there is a constant M > 0 such that

$x\left(t\right)\le M,\phantom{\rule{2.77695pt}{0ex}}\forall t\in T.$

(5) If $x\in P{C}_{rrd}\left(T,{R}^{+}\right)$ satisfies the following inequality

then there is a constant M > 0 such that

$x\left(t\right)\le M,\phantom{\rule{2.77695pt}{0ex}}\forall t\in T.$

Proof. (1) Define

$y\left(t\right)=\underset{\left[t,b\right)}{\int }{p}_{1}\left(\tau \right){x}^{\sigma }\left(\tau \right)\Delta \tau \phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}t\in T.$

Then y(b) = 0 and

${y}^{\Delta }\left(t\right)=-{p}_{1}\left(t\right){x}^{\sigma }\left(t\right)\ge -{p}_{1}\left(t\right){y}^{\sigma }\left(t\right)-{p}_{1}\left(t\right){f}^{\sigma }\left(t\right)\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\Delta -\mathsf{\text{a}}\mathsf{\text{.e}}\mathsf{\text{.}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{on}}\phantom{\rule{2.77695pt}{0ex}}T.$

Note that,

${\left[y\left(t\right){e}_{{p}_{1}}\left(t,b\right)\right]}^{\Delta }=\left[{y}^{\Delta }\left(t\right)+{p}_{1}\left(t\right){y}^{\sigma }\left(t\right)\right]{e}_{{p}_{1}}\left(t,b\right)\phantom{\rule{2.77695pt}{0ex}}\Delta -\mathsf{\text{a}}\mathsf{\text{.e}}\mathsf{\text{.}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{on}}\phantom{\rule{2.77695pt}{0ex}}T,$

therefore,

$-y\left(t\right){e}_{{p}_{1}}\left(t,b\right)\ge -\underset{\left[t,b\right)}{\int }{p}_{1}\left(\tau \right){f}^{\sigma }\left(\tau \right){e}_{{p}_{1}}\left(\tau ,b\right)\Delta \tau .$

Moreover, we obtain

$x\left(t\right)\le f\left(t\right)+\underset{\left[t,b\right)}{\int }{e}_{{p}_{1}}\left(\tau ,t\right){p}_{1}\left(\tau \right){f}^{\sigma }\left(\tau \right)\Delta \tau \phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}t\in T.$
1. (2)

Setting

$y\left(t\right)=\alpha +\underset{\left[t,b\right)}{\int }{p}_{1}\left(\tau \right){x}^{\sigma }\left(\tau \right)\Delta \tau +\sum _{{t}_{k}>t}{\beta }_{k}x\left({t}_{k}\right),\phantom{\rule{2.77695pt}{0ex}}\forall t\in T.$

then y is Δ-differential Δ-a.e. on $T$ and

$\left\{\begin{array}{cc}\hfill {y}^{\Delta }\left(t\right)=-{p}_{1}\left(t\right){x}^{\sigma }\left(t\right)\ge -{p}_{1}\left(t\right){y}^{\sigma }\left(t\right),\hfill & \hfill t\ne {t}_{k},\hfill \\ \hfill y\left({t}_{k}-\right)=y\left({t}_{k}\right)+{\beta }_{k}x\left({t}_{k}\right)\le \left(1+{\beta }_{k}\right)y\left({t}_{k}\right),\hfill \\ \hfill y\left(b\right)=\alpha .\hfill \end{array}\right\$

For t [t n , b], by the conclusion (1) we have

$y\left(t\right)\le \alpha {e}_{{p}_{1}}\left(b,t\right).$

When t (t n , t n+1], By Theorem 2.2 and the conclusion (1) we also obtain

$y\left(t\right)\le y\left({t}_{n}-\right){e}_{{p}_{1}}\left({t}_{n},t\right)\le \alpha \left(1+{\beta }_{n}\right){e}_{{p}_{1}}\left(b,t\right).$

Thus,

$x\left(t\right)\le \alpha \prod _{{t}_{k}>t}\left(1+{\beta }_{k}\right){e}_{{p}_{1}}\left(b,t\right),\phantom{\rule{2.77695pt}{0ex}}\forall t\in T.$
1. (3)

Setting γ = (α + 1)(β + 1), β = ∫[a,b) e g (τ, a)g(τ)Δτ, then

Letting

$h\left(t\right)=\gamma +\gamma \underset{\left[t,b\right)}{\int }{p}_{1}\left(\tau \right){x}^{\lambda }\left(\tau \right)\Delta \tau +\gamma \underset{\left[a,b\right)}{\int }{p}_{1}\left(\tau \right){x}^{{\lambda }_{1}}\left(\tau \right)\Delta \tau ,$

then h is monotone descending function and

${h}^{\Delta }\left(t\right)\ge -\gamma {p}_{1}\left(t\right){h}^{\lambda }\left(t\right).$

Δ-integrating from t to b, we obtain

${h}^{1-{\lambda }_{1}}\left(t\right)-{h}^{1-{\lambda }_{1}}\left(b\right)\le \left(1-{\lambda }_{1}\right)\gamma \underset{\left[t,b\right)}{\int }{p}_{1}\left(\tau \right)\Delta \tau .$

Therefore,

${\left(2h\left(b\right)-\gamma \right)}^{1-{\lambda }_{1}}-{h}^{1-{\lambda }_{1}}\left(b\right)\le \left(1-{\lambda }_{1}\right)\gamma \underset{\left[a,b\right)}{\int }{p}_{1}\left(\tau \right)\Delta \tau .$

Using the method of the conclusion (3) in Theorem A, one can show that there is a constant M > 0 such that

$x\left(t\right)\le M,\phantom{\rule{2.77695pt}{0ex}}\forall t\in T.$
1. (4)

For $t\in T$, define

$z\left(t\right)=\left\{\begin{array}{cc}\hfill 1,\hfill & \hfill x\left(t\right)<1,\hfill \\ \hfill x\left(t\right),\hfill & \hfill x\left(t\right)\ge 1,\hfill \end{array}\right\$

we have

$x\left(t\right)\le \alpha +\underset{\left[t,b\right)}{\int }p\left(\tau \right){z}_{{\tau }_{b}}\Delta \tau +\underset{\left[t,b\right)}{\int }g\left(\tau \right){z}_{\sigma {\left(\tau \right)}_{b}}^{\lambda }\Delta \tau +\sum _{{t}_{k}>t}{\beta }_{k}{z}_{{t}_{kb}},\phantom{\rule{2.77695pt}{0ex}}\forall t\in T,$

where λ = max{λ 1, λ 2}, p(t) = p 1(t) + p 2(t)g(t) = g 1(t) + g 2(t). Hence

${z}_{{t}_{b}}\le \alpha +\underset{\left[t,b\right)}{\int }p\left(\tau \right){z}_{{\tau }_{b}}\Delta \tau +\underset{\left[t,b\right)}{\int }g\left(\tau \right){z}_{\sigma {\left(\tau \right)}_{b}}^{\lambda }\Delta \tau +\sum _{{t}_{k}>t}{\beta }_{k}{z}_{{t}_{kb}},\phantom{\rule{2.77695pt}{0ex}}\forall t\in T,$

By the conclusion (3), there exists a constant M > 0 such that

$x\left(t\right)\le M,\phantom{\rule{2.77695pt}{0ex}}\forall t\in T.$
1. (5)

Setting λ = max{λ 1, λ 2, λ 3, λ 4},

$y\left(t\right)=\left\{\begin{array}{cc}\hfill 1,\hfill & \hfill {x}_{t}<1,\hfill \\ \hfill {x}_{t},\hfill & \hfill {x}_{t}\ge 1,\hfill \end{array}\right\\phantom{\rule{1em}{0ex}}p\left(t\right)={p}_{1}\left(t\right)+{p}_{2}\left(t\right),\phantom{\rule{1em}{0ex}}g\left(t\right)=\sum _{i=1}^{4}{g}_{i}\left(t\right),\phantom{\rule{2.77695pt}{0ex}}\forall t\in T.$

then we have

$y\left(t\right)\le \alpha +\underset{\left[t,b\right)}{\int }p\left(\tau \right)y\left(\tau \right)\Delta \tau +\underset{\left[a,b\right)}{\int }g\left(\tau \right){y}^{\lambda }\left(\tau \right)\Delta \tau +\sum _{{t}_{k}>t}{\beta }_{k}y\left({t}_{k}\right).$

Further, we also can prove that the conclusion (5) is hold. This completes the proof.

Remark 3.1: (1) If $T=R$, then the inequality established in Theorem A reduces to the inequality established by Peng and Wei in .

1. (2)

Using our main results, we can obtain many dynamic inequalities for some peculiar time scales. Due to limited space, their statements are omitted here.

## Application

In this section, we present some applications of Theorems A and B to investigate certain properties of solutions of the following impulsive dynamic integral equation

$\left\{\begin{array}{cc}\hfill {x}^{\Delta }\left(t\right)+p\left(t\right){x}^{\sigma }\left(t\right)=f\left(t,x\left(t\right),{x}^{\sigma }\left(t\right),\left(Sx\right)\left(t\right),\left(Wx\right)\left(t\right)\right),\hfill & \hfill t\notin \Lambda ,\hfill \\ \hfill \left(Sx\right)\left(t\right)=\underset{\left[a,t\right)}{\int }k\left(t,\tau \right)s\left(\tau ,x\left(\tau \right),{x}^{\sigma }\left(\tau \right)\right)\Delta \tau ,\hfill & \hfill t\in T,\hfill \\ \hfill \left(Wx\right)\left(t\right)=\underset{\left[a,b\right)}{\int }m\left(t,\tau \right)w\left(\tau ,x\left(\tau \right),{x}^{\sigma }\left(\tau \right)\right)\Delta \tau ,\hfill & \hfill t\in T\hfill \\ \hfill x\left({t}_{k}+\right)=x\left({t}_{k}\right)+{J}_{k}\left(x\left({t}_{k}\right)\right),\hfill & \hfill {t}_{k}\in \Lambda ,\hfill \\ \hfill x\left(a\right)={x}_{0},\hfill \end{array}\right\$
(4.1)

where $p\in {\Gamma }_{1}\left(T\right)$.

Definition 4.1: A function $x\in P{C}_{lrd}\left(T,R\right)$ is said to be a weak solution of (4.1), if x satisfies the impulsive integral equation

$x\left(t\right)={e}_{\ominus p}\left(t,a\right){x}_{0}+\underset{\left[a,t\right)}{\int }{e}_{\ominus p}\left(t,\tau \right)f\left(\tau ,x\left(\tau \right),{x}^{\sigma }\left(\tau \right),\left(Sx\right)\left(\tau \right),\left(Wx\right)\left(\tau \right)\right)\Delta \tau +\sum _{{t}_{k}

Suppose that:

[F] (1) k, $m\in {C}_{rd}\left(T×T,R\right)$, the functions $f:T×R×R×R×R\to R$, $s:T×R×R\to R$, $w:T×R×R\to R$ are measurable in $t\in T$ and locally Lipschitz continuous, that is, for any ρ > 0, there exists a constant L(ρ) > 0, for all x i , y i , z i , w i X, satisfying ||x i ||, ||y i ||, ||z i ||, ||w i || ≤ ρ (i = 1, 2), we have

1. (2)

There are constants 0 < λ i < 0 (i = 1, · · ·, 4) and function q 1, q 2, ${q}_{3}\in {L}^{1}\left(T,{R}^{+}\right)$ such that

2. (3)

There are constants β k ≥ 0 such that the mapping J k : RR (k = 12 · · · n) satisfies

$\parallel {J}_{k}\left(x\right)-{J}_{k}\left(y\right)\parallel \phantom{\rule{2.77695pt}{0ex}}\le {\beta }_{k}\parallel x-y\parallel ,\phantom{\rule{2.77695pt}{0ex}}\forall x,y\in X.$

Theorem C. Under assumption [F], if $p\in {\Gamma }_{1}\left(T\right)$, then the system (4.1) has a weak solution $x\in P{C}_{lrd}\left(T,R\right)$.

Proof. Define the operator H on $P{C}_{lrd}\left(T,R\right)$ given by

$\left(Hx\right)\left(t\right)=\underset{\left[a,t\right)}{\int }{e}_{\ominus p}\left(t,\tau \right)f\left(\tau ,x\left(\tau \right),{x}^{\sigma }\left(\tau \right),\left(Sx\right)\left(\tau \right),\left(Wx\right)\left(\tau \right)\right)\Delta \tau \phantom{\rule{2.77695pt}{0ex}}+\phantom{\rule{2.77695pt}{0ex}}{e}_{\ominus p}\left(t,a\right){x}_{0}+\sum _{{t}_{k}

We can first prove that $H:P{C}_{lrd}\left(T,R\right)\to P{C}_{lrd}\left(T,R\right)$ is continuous and compact.

Let $x\in \Upsilon \equiv \left\{x\in P{C}_{lrd}\left(T,R\right)|x=\delta \left(Hx\right),\phantom{\rule{2.77695pt}{0ex}}\delta \in \left[0,1\right]\right\}$. When δ ≠ 0, set $y=\frac{1}{\delta }x$, if not y = 0. Note that

By Gronwall inequality (5) in Theorem A, there is a constant M > 0 such that

$|y\left(t\right)|\phantom{\rule{2.77695pt}{0ex}}\le M,\phantom{\rule{2.77695pt}{0ex}}\forall t\in T.$

It follows by Leray-Schauder fixed point theorem, H has a fixed point in $P{C}_{rd}\left(T,R\right)$, that is, the impulsive integro-differential equation (4.1) has a weak solution $x\in P{C}_{rd}\left(T,R\right)$. □

For the following backward problem

$\left\{\begin{array}{c}\hfill {\phi }^{\Delta }\left(t\right)+p\left(t\right){\phi }^{\sigma }\left(t\right)=w\left(t,\phi \left(t\right),{\phi }^{\sigma }\left(t\right)\right),\phantom{\rule{1em}{0ex}}a\le t
(4.2)

we introduce the following assumption:

[W] (1) The function $w:T×R×R\to R$ is Δ-measurable in $t\in T$ and locally Lipschitz continuous, i.e. for all φ 1, φ 2, ψ 1, ψ 2 R, satisfying |φ 1|, |φ 2|, |ψ 1|, |ψ 2| ≤ ρ, we have

$|w\left(t,{\phi }_{1},{\psi }_{1}\right)-w\left(t,{\phi }_{2},{\psi }_{2}\right)|\phantom{\rule{2.77695pt}{0ex}}\le L\left(\rho \right)\left(|{\phi }_{1}-{\phi }_{2}|+|{\psi }_{1}-{\psi }_{2}|\right)\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{all}}\phantom{\rule{2.77695pt}{0ex}}t\in T.$
1. (2)

There exist a constant 0 < λ < 1 and a function $q\in {L}^{1}\left(T,\left[0,+\infty \right)\right)$ such that

$|w\left(t,\phi ,\psi \right)|\phantom{\rule{2.77695pt}{0ex}}\le q\left(t\right)\left(1+|\phi {|}^{\lambda }+|\psi |\right)\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{all}}\phantom{\rule{2.77695pt}{0ex}}\phi ,\psi \in R.$

We can prove the following result.

Theorem D. Let $p\in {\Gamma }_{1}\left(T\right)$. Under the assumption [W], the backward problem of the nonlinear dynamical equation (4.2) has a unique weak solution $\phi \in {C}_{rd}\left(T,R\right)$.

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## Acknowledgements

The purpose of this paper is to give some Gronwall type integral inequalities with impulses on time scales, which extend some known dynamic inequalities on time scales, unify and extend some continuous inequalities and their corresponding discrete analogues. It is helpful on our result to study dynamic systems and optimal control problem on time scales.

We would like to thank the referees very much for their valuable comments and suggestions. This work is supported by the National Natural Science Foundation of China under grant no. 10961009, the Fok Ying Tung Education Foundation under grant no. 121104 and Introducing Talents Foundation for the Doctor of Guizhou University under grant no. 2010031.

## Author information

Authors

### Corresponding author

Correspondence to Y Peng.

### Competing interests

The authors declare that they have no competing interests.

### Authors' contributions

In this paper, YP carried out the main theorem studies, participated in the sequence alignment and drafted the manuscript. YK carried out the Gronwall inequalities studies. MY carried out the backward Gronwall inequalities. LY participated in the Gronwall inequalities studies. RH present some applications on Gronwall inequality in dynamic systems. All authors read and approved the final manuscript.

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Peng, Y., Kang, Y., Yuan, M. et al. Gronwall-type integral inequalities with impulses on time scales. Adv Differ Equ 2011, 26 (2011). https://doi.org/10.1186/1687-1847-2011-26 