On Some New Impulsive Integral Inequalities

Abstract

We establish some new impulsive integral inequalities related to certain integral inequalities arising in the theory of differential equalities. The inequalities obtained here can be used as handy tools in the theory of some classes of impulsive differential and integral equations.

1. Introduction

Differential and integral inequalities play a fundamental role in global existence, uniqueness, stability, and other properties of the solutions of various nonlinear differential equations; see [1–4]. A great deal of attention has been given to differential and integral inequalities; see [1, 2, 5–8] and the references given therein. Motivated by the results in [1, 5, 7], the main purpose of this paper is to establish some new impulsive integral inequalities similar to Bihari's inequalities.

Let , , and , then we introduce the following spaces of function:

is continuous for , , and exist, and  

is continuously differentiable for , , and exist, and

To prove our main results, we need the following result (see [1, Theorem 1.4.1]).

Lemma 1.1.

Assume that

the sequencesatisfies, with;

andis left-continuous at, ;

for,

(1.1)

where , and are constants.

Then,

(1.2)

2. Main Results

In this section, we will state and prove our results.

Theorem 2.1.

Let , , and be constants. If

(2.1)

for , then

(2.2)

for .

Proof.

Define a function by

(2.3)

where is an arbitrary small constant. For , differentiating (2.3) and then using the fact that , we have

(2.4)

and so

(2.5)

For , we have ; thus . Let ; it follows that

(2.6)

From Lemma 1.1, we obtain

(2.7)

Now by using the fact that in (2.7) and then letting , we get the desired inequality in (2.2). This proof is complete.

Theorem 2.2.

Let and be constants, and let be a nonnegative constant. If

(2.8)

for , then

(2.9)

for .

Proof.

This proof is similar to that of Theorem 2.1; thus we omit the details here.

Theorem 2.3.

Let , and be constants. If

(2.10)

for , then

(2.11)

for , where

(2.12)

Proof.

Let be an arbitrary small constant, and define a function by

(2.13)

Let ; similar to the proof of Theorem 2.1, we have

(2.14)

Set ; then , and so from (2.14) we get that . Thus, for ,

(2.15)

and for ,

(2.16)

and so . By Lemma 1.1, we have

(2.17)

Let , then we obtain

(2.18)

where is defined in (2.12). Substituting (2.18) into (2.14), we have

(2.19)

Applying Lemma 1.1 again, we obtain

(2.20)

Now using and letting , we get the desired inequality in (2.11).

Theorem 2.4.

Let , and be constants. If

(2.21)

for , then

(2.22)

for .

Proof.

Set

(2.23)

where is an arbitrary small constant; then is nondecreasing. Let , then it follows for that

(2.24)

since is nondecreasing. Also, for , we have . Applying Lemma 1.1, we obtain

(2.25)

Now by using the fact that in (2.25) and letting , we get the inequality (2.22).

Remark 2.5.

If , then (2.1), (2.8), (2.10), and (2.21) have no impulses. In this case, it is clear that Theorems 2.2-2.3 improve the corresponding results of [5, Theorem 1].

Theorem 2.6.

Let , for , and be constants. Let be a nondecreasing function with , for , and , for ; here , for . If

(2.26)

for , then for

(2.27)

where

(2.28)

(2.29)

(2.30)

Proof.

We first assume that and define a function by the right-hand side of (2.26). Then, , and is nondecreasing. For ,

(2.31)

and for . As , from (2.31) we have

(2.32)

and so

(2.33)

Now assume that for , we have

(2.34)

Then, for , it follows from (2.32) that . Using , we arrive at

(2.35)

From the supposition of , we see that

(2.36)

If , then

(2.37)

Otherwise, we have

(2.38)

This implies, by induction hypothesis, that

(2.39)

Thus, (2.35) and (2.39) yield, for ,

(2.40)

and so

(2.41)

Using (2.41) in , we have the required inequality in (2.27).

If is nonnegative, we carry out the above procedure with instead of , where is an arbitrary small constant, and by letting , we obtain (2.27). The proof is complete.

Remark 2.7.

If , then and the inequality in (2.27) is true for .

An interesting and useful special version of Theorem 2.6 is given in what follows.

Corollary 2.8.

Let , and be as in Theorem 2.6 . If

(2.42)

for , then

(2.43)

for , where is defined by (2.28).

Proof.

Let in Theorem 2.6. Then, (2.26) reduces to (2.42) and

(2.44)

Consequently, by Theorem 2.6, we have

(2.45)

This proof is complete.

3. Application

Example 3.1.

Consider the integrodifferential equations

(3.1)

where with and are continuous; is continuous at and exist and ; are constants with . Here, we assume that the solution of (3.1) exists on . Multiplying both sides of (3.1) by and then integrating them from 0 to , we obtain

(3.2)

We assume that

(3.3)

where . From (3.2) and (3.3), we obtain

(3.4)

Now applying Theorem 2.3, we have

(3.5)

where

(3.6)

for all . The inequality (3.5) gives the bound on the solution of (3.1).