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`Abstract and Applied AnalysisVolume 2014 (2014), Article ID 623726, 4 pageshttp://dx.doi.org/10.1155/2014/623726`
Research Article

## A Note on Gronwall’s Inequality on Time Scales

College of Mathematics and Computer Science, Fuzhou University, Fuzhou 350002, China

Received 29 May 2014; Accepted 6 June 2014; Published 1 July 2014

Copyright © 2014 Xueru Lin. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

This paper gives a new version of Gronwall’s inequality on time scales. The method used in the proof is much different from that in the literature. Finally, an application is presented to show the feasibility of the obtained Gronwall’s inequality.

#### 1. Introduction and Motivation

Recently, an interesting field of research is to study the dynamic equations on time scales, which have been extensively studied. For example, one can see [117] and references cited therein. A time scale is an arbitrary nonempty closed subset of the real numbers . The forward and backward jump operators are defined by . A point , , is said to be left dense if and right dense if and . The mapping defined by is called graininess. A function is said to be rd-continuous provided is continuous at right-dense points. The set of all such rd-continuous functions is denoted by . A function is regressive provided for . Denote .

One of important topics is the differential inequalities on time scales. A nonlinear version of Gronwall’s inequality is presented in [2, Theorem 6.4, pp 256]. This version is stated as follows.

Theorem A. Let , , and . Then implies

Taking , a classical version of Gronwall’s inequality follows (see [2, Corollary 6.7, pp 257]).

Theorem B. Let , , , and . Then implies

This paper presents a new version of Gronwall’s inequality as follows.

Theorem 1. Let and . Suppose that , , and . Then implies

Remark 2. Note that, for , inequality (5) reduces to which is different from inequality (3) in Theorem B. Since Theorem B requires , we see that Theorem B cannot be applied to (7). Moreover, the method used to prove Theorem A cannot be used to prove Theorem 1. To explain this, recall the proof of Theorem A in [2]. Let . Then and By comparing theorem and variation of constants formula, and hence Theorem A follows in view of .

Now we try to adopt the same idea used in [2] to estimate inequality (7). Let . Then and By comparing theorem and variation of constants formula, we have which implies If we were to use the same idea as in [2], we should combine (12) with However, on one side, ; on the other side, . These two inequalities cannot lead us anywhere.

Therefore, some novel proof is employed to prove Theorem 1. One can see the detailed proof in the next section.

#### 2. Proof of Main Result

Before our proof of Theorem 1, we need some lemmas.

Lemma 3 (chain rule [2]). Assume is delta differentiable on . Assume further that is continuously differentiable. Then is delta differentiable and satisfies

Lemma 4. Suppose that is positive delta differentiable on and is regressive. Then is a preantiderivative of function , where and is the principal logarithm function.

Proof. Let . Obviously, is continuous on . To prove Lemma 4, it suffices to show that . In fact, by using Lemma 3, we have

Proof of Theorem 1. To prove Theorem 1, we divide it into two cases.

Case 1. For  , in this case, we have Hence, it is easy to conclude that for .

Case 2. For , let  . For any , we have Noting that , we have . Thus, we have Multiplied by on both sides of the above inequality, it follows that or Since , . Using the fact that is nondecreasing with respect to for , we have An integration of the above inequality over leads to It follows from Lemma 4 that or  which leads to Therefore, for . This completes the proof of Theorem 1.

#### 3. An Application

Inequality (5) has many potential applications. For instance, it can be used to study the property of the solutions to the dynamic systems. Consider the following linear system: Let and be two solutions of (26) satisfying the initial conditions and , respectively.

Theorem 5. Suppose that is bounded on . Then one has where

Proof. Integrating (7) over , we have Denoting , simple computation leads us to By Theorem 1, it follows from (29) that

Remark 6. One can see that, for the case , (29) reduces to As you see, Theorem B cannot be used to (31) because the essential condition in Theorem B is .

#### Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgment

This work was supported by JB12254.

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