AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 10.1155/2014/623726 623726 Research Article A Note on Gronwall’s Inequality on Time Scales Lin Xueru Xia Yonghui 1 College of Mathematics and Computer Science Fuzhou University Fuzhou 350002 China fzu.edu.cn 2014 172014 2014 29 05 2014 06 06 2014 1 7 2014 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.

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  and references cited therein. A time scale T is an arbitrary nonempty closed subset of the real numbers R. The forward and backward jump operators are defined by σ(t):=inf{sT:s>t},ρ(t):=sup{sT:s>t}. A point tT, t>infT, is said to be left dense if ρ(t)=t and right dense if t<infT and σ(t)=t. The mapping μ:TR+ defined by μ(t)=σ(t)-t is called graininess. A function g:TR is said to be rd-continuous provided g is continuous at right-dense points. The set of all such rd-continuous functions is denoted by Crd(T,R). A function p:TR is regressive provided 1+μ(t)p(t)0 for tT. Denote R+(T,R):={pCrd(T,R):1+μ(t)p(t)>0}.

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 y,fCrdR(T,R), p(t)R+(T,R), and p(t)0. Then (1)y(t)f(t)+t0ty(s)p(s)Δs,tT, implies (2)y(t)f(t)+t0tep(t,σ(s))f(s)p(s)Δs,tT.

Taking f(t)α, a classical version of Gronwall’s inequality follows (see [2, Corollary 6.7, pp 257]).

Theorem B.

Let pR+(T,R), p(t)0, yCrdR(T,R), and αR. Then (3)y(t)α+t0ty(s)p(s)Δs,tT, implies (4)y(t)αep(t,t0),tT.

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

Theorem 1.

Let -pR+(T,R) and yCrdR(T,R). Suppose that p(t)0, y(t)0, and α>0. Then (5)y(t)α+|t0ty(s)p(s)Δs|,tT, implies (6)y(t){αep(t,t0),fort[t0,+)T,αe-p(t,t0),fort(-,t0]T.

Remark 2.

Note that, for t(-,t0]T, inequality (5) reduces to (7)y(t)α-t0ty(s)p(s)Δs, which is different from inequality (3) in Theorem B. Since Theorem B requires p(t)0, 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 . Let z(t)=t0ty(s)p(s)Δs. Then z(t0)=0 and (8)zΔ=y(t)p(t)zΔ[f(t)+z(t)]p(t)=p(t)z(t)+p(t)f(t). By comparing theorem and variation of constants formula, (9)z(t)t0tep(t,σ(s))f(s)p(s)Δs, and hence Theorem A follows in view of y(t)f(t)+z(t).

Now we try to adopt the same idea used in  to estimate inequality (7). Let z(t)=t0ty(s)p(s)Δs. Then z(t0)=0 and (10)zΔ=y(t)p(t)[f(t)-z(t)]p(t)=-p(t)z(t)+p(t)f(t)=-p(t)zσ+(1+μ(t)p(t))p(t)f(t). By comparing theorem and variation of constants formula, we have (11)z(t)t0tep(t,σ(s))f(s)p(s)Δs, which implies (12)-z(t)-t0tep(t,σ(s))f(s)p(s)Δs. If we were to use the same idea as in , we should combine (12) with (13)y(t)f(t)-z(t). However, on one side, y(t); on the other side, f(t)-z(t). 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 [<xref ref-type="bibr" rid="B2">2</xref>]).

Assume g:TX is delta differentiable on T. Assume further that f:XX is continuously differentiable. Then fg:TX is delta differentiable and satisfies (14)(fg)Δ(t)={01f(g(t)+hμ(t)gΔ(t))dh}gΔ(t).

Lemma 4.

Suppose that g:TR+ is positive delta differentiable on T and gΔ(t)/g(t) is regressive. Then ξμ(t)(gΔ(t)/g(t)) is a preantiderivative of function Log[g(t)], where ξh(z)=(1/h)Log(1+zh) and Log is the principal logarithm function.

Proof.

Let f(x)=Logx. Obviously, f:R+R is continuous on R+. To prove Lemma 4, it suffices to show that [Log[g(t)]]Δ=ξμ(t)(gΔ(t)/g(t)). In fact, by using Lemma 3, we have (15)[Log[g(t)]]Δ=(fg)Δ(t)={01f(g(t)+hμ(t)gΔ(t))dh}gΔ(t)={011g(t)+hμ(t)gΔ(t)dh}gΔ(t)={1μ(t)gΔ(t)Log[g(t)+hμ(t)gΔ(t)]|h=0h=1}gΔ(t)={1μ(t){Log[g(t)+μ(t)gΔ(t)]-Log[g(t)]}ifμ(t)0,gΔ(t)g(t)ifμ(t)=0={1μ(t){Logg(t)+μ(t)gΔ(t)g(t)}ifμ(t)0,gΔ(t)g(t)ifμ(t)=0=1μ(t)Log{1+μ(t)gΔ(t)g(t)}=ξμ(t)(gΔ(t)g(t)).

Proof of Theorem <xref ref-type="statement" rid="thm1.1">1</xref>.

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

Case 1. For  t[t0,+)T, in this case, we have (16)y(t)α+|t0ty(s)p(s)Δs|=α+t0ty(s)p(s)Δs,fort[t0,+)T. Hence, it is easy to conclude that y(t)αep(t,t0) for t[t0,+)T.

Case 2. For t(-,t0]T, let  z(t)=t0ty(s)p(s)Δs. For any s[t,t0]T, we have (17)y(s)α+|t0sy(τ)p(τ)Δτ|y(s)=α-t0sy(τ)p(τ)Δτ=α-z(s). Noting that y0,p0,α>0, we have α-z(s)>0. Thus, we have (18)y(s)α-z(s)1. Multiplied by -p(s) on both sides of the above inequality, it follows that (19)-y(s)p(s)α-z(s)-p(s), or (20)[α-z(s)]Δα-z(s)-p(s). Since -pR+, -p[α-z(s)]Δ/(α-z(s))R+. Using the fact that ξμ(t)(z) is nondecreasing with respect to z for zR+, we have (21)ξμ(s)[(α-z(s))Δα-z(s)]ξμ(s)[-p(s)]. An integration of the above inequality over [t,t0]T leads to (22)tt0ξμ(s)[(α-z(s))Δα-z(s)]Δstt0ξμ(s)[-p(s)]Δs. It follows from Lemma 4 that (23)Log[α-z(s)]|tt0tt0ξμ(s)[-p(s)]Δs, or  (24)Logα-Log[α-z(t)]tt0ξμ(t)[-p(s)]Δs, which leads to (25)α-z(t)αexp(-tt0ξμ(s)[-p(s)]Δs)=αexp(t0tξμ(s)[-p(s)]Δs)=αe-p(t,t0). Therefore, y(t)α-z(t)e-p(t,t0) for t(-,t0]T. 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: (26)xΔ=A(t)x. Let X(t,t0,x0) and X(t,t0,x~0) be two solutions of (26) satisfying the initial conditions X(t0)=x0 and X(t0)=x~0, respectively.

Theorem 5.

Suppose that A(t) is bounded on T. Then one has (27)X(t,t0,x0)-X(t,t0,x~0){x0-x~0ep1(t,t0),fort[t0,+)T,x0-x~0e-p1(t,t0),fort(-,t0]T, where p1(t)M.

Proof.

Integrating (7) over [t0,t], we have (28)X(t,t0,x0)=x0+t0t[A(s)X(s,t0,x0)+f(s,X(s,t0,x0))]Δs. Denoting M=suptTA(t), simple computation leads us to (29)X(t,t0,x0)-X(t,t0,x~0)x0-x~0+M|t0tX(s,t0,x0)-X(s,t0,x~0)Δs|. By Theorem 1, it follows from (29) that (30)X(t,t0,x0)-X(t,t0,x~0){x0-x~0ep1(t,t0),fort[t0,+)T,x0-x~0e-p1(t,t0),fort(-,t0]T.

Remark 6.

One can see that, for the case t(-,t0]T, (29) reduces to (31)X(t,t0,x0)-X(t,t0,x~0)x0-x~0-Mt0t[X(s,t0,x0)-X(s,t0,x~0)]Δs. As you see, Theorem B cannot be used to (31) because the essential condition in Theorem B is p(t)0.

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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