Abstract and Applied Analysis

Volume 2012 (2012), Article ID 937908, 12 pages

doi:10.1155/2012/937908

Research Article

A New Fractional Integral Inequality with Singularity and Its Application

Qiong-Xiang Kong¹ and Xiao-Li Ding²

¹Department of Building Environment and Services Engineering, Xi'an Jiaotong University, Xi'an, Shaanxi 710049, China

²Department of Mathematical Sciences, Xi'an Jiaotong University, Xi'an, Shaanxi 710049, China

Received 3 March 2012; Revised 30 April 2012; Accepted 30 April 2012

Academic Editor: Bashir Ahmad

Copyright © 2012 Qiong-Xiang Kong and Xiao-Li Ding. 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.

We prove an integral inequality with singularity, which complements some known results. This inequality enables us to study the dependence of the solution on the initial condition to a fractional differential equation in the weighted space.

1. Introduction

Integral inequalities provide an excellent tool for the properties of solutions to differential equations, such as boundedness, existence, uniqueness, and stability (e.g., see [1–10]). For this reason, the study of integral inequalities has been emphasized by many authors. For example, in 1919, Gronwall in [11] proved a remarkable inequality which can be described by the following.

Suppose that satisfies the relation

where all the functions involved are continuous on the interval , , and . Consider

The inequality has attracted and continues to attract considerable attention in the literature. In 2007, Ye et al. [12] reported an integral inequality with singular kernel. The inequality can be stated as follows.

If , is a nonnegative and locally integrable on , is a nonnegative, nondecreasing continuous function on , and , where , is a positive constant. Further suppose that is nonnegative and locally integrable on with

Then

Besides the above-mentioned inequalities, there are still many inequalities (e.g., see [13–15]).

But in the analysis of the dependence of the solution on the initial condition of a fractional differential equation in the weighted space, the bounds provided by the existing inequalities are not adequate. So it is natural and necessary to seek new inequality in order to obtain our desired results. In this paper, we present a new integral inequality, and then apply our inequality to investigate the dependence of the solution on the initial condition of a fractional differential equations in the weighted space.

2. An Integral Inequality

In this section, our main aim is to establish an integral inequality with singularity. Before proceeding, we give some useful definitions and lemmas.

Definition 2.1 (see [14, 16]). The gamma function is defined by .

Definition 2.2 (see [14, 16]). The beta function is defined by .

The beta function is connected with gamma function by the following relation [3, 14]:

Lemma 2.3 (see [14]). Let ,  . Then the quotient expansion of two gamma functions at infinity can be represented as follows:

Lemma 2.4. Let ,  . Then one has

Proof. By Lemma 2.3, we have , which proves that as . The proof of this lemma is completed.

Based on Lemma 2.4, we can define a function.

Definition 2.5. Let ,  . Then the following definition:

is well defined, where is a positive constant, and .

Proof. We only need to show that the series in (2.4) is uniformly convergent for . By Lemma 2.4, we know that as . Since ,   as . This implies that the series in (2.4) is uniformly convergent for . It follows that the definition is well defined.

Lemma 2.6. Let ,   and . Then one has

Proof. Making the substitution and combining the relation (2.1), we obtain

The proof of this lemma is completed.

Now we can state the integral inequality.

Theorem 2.7. Let ,  ,  ,  , and let be a nonnegative, nondecreasing continuous function on ,  , where , is a positive constant. Further suppose that is nonnegative and is locally integrable on with

Then one has

Proof. For convenience, we define an operator

Then (2.7) can be rewritten in the form

Since and are nonnegative, it is easy to induce that

Let us prove that the following relation

holds for any , where , and as for each in .

Obviously, inequality (2.12) is valid for , due to . Suppose that the inequality is satisfied for any fixed . Let us verify that it is also satisfied for . We first prove the case . According to the induction hypothesis and Lemma 2.6, we have

which is estimated with the help of

So, for the case , inequality (2.12) is true for any . Now we prove the case . Similarly, according to the induction hypothesis and Lemma 2.6, we get

which is calculated with the help of

So, for the case , inequality (2.12) is true for any . Based on this analysis, we conclude that inequality (2.12) holds for any .

Next, we show that as . Now, we go back to inequality (2.12). For the case , we denote , where . Note that

Since , by Lemma 2.4, we obtain as . This implies that as . It follows that as for the case . For the case , we denote , where . Note that

Using the same arguments as above, we know that as . It follows that as for the case . So, it has as for the two cases and . This, together with (2.11), leads to .

Finally, we show that

where ,  ,  .

Obviously, inequality (2.19) is true for . Suppose that the inequality is satisfied for any fixed . Let us verify that it is also satisfied for . According to the induction hypothesis and Lemma 2.6, we obtain

This proves that inequality (2.19) is satisfied for any . In other words, we have proved that

where , , . By virtue of Definition 2.5, we can arrive at inequality (2.8) and the proof of this theorem is completed.

For the case in Theorem 2.7, we can obtain the following corollary, which can be found in [17].

Corollary 2.8. Let ,  ,  ,  . And suppose that is nonnegative and is locally integrable on () with

Then one has

For in Theorem 2.7, we can arrive at the following corollary, which can be found in [12].

Corollary 2.9. Let ,  ,   be a nonnegative, nondecreasing continuous function on , , where , is a positive constant. And suppose that is nonnegative and locally integrable on with

Then one has

3. Application

In this section, we will apply our established result to study the dependence of the solution on the initial condition of a fractional differential equation with the Riemann-Liouville derivative.

For the reader's convenience, we first recall several definitions of the Reimann-Liouville integral and derivative. From now on, we assume that is a finite positive constant, that is, .

Definition 3.1 (see [14, 16]). Let . The Riemann-Liouville integral of order is defined by

Definition 3.2 (see [14, 16]). Let . The Riemann-Liouville derivative of order is defined by

Now we consider the following initial value problem of the form

With regard to problem (3.3), the existence and uniqueness of the solution can be found in the book by Kilbas et al. [14]. For the completeness of this paper, we give the existence and uniqueness of the solution to (3.3) in the weighted space . The space consists of all functions such that , which turns out to be a Banach space when endowed with the norm .

Theorem 3.3 (see [14]). Let , and be a function such that for any , . Further assume that for any ,  , the following inequality

holds, where is a constant. Then there exists a unique solution to problem (3.3) in the space .

Theorem 3.4. Let , and be a function such that for any , . Further assume that for any , , the following inequality

holds, where is a constant. Assume that and are the solutions of problem (3.3) and

respectively. Then, for , one has

where are positive constants such that

Proof. The proof is rather technical. We first prove the case and . Since and are the solutions of (3.3) and (3.6), we have

Subtracting (3.10) from (3.9) and using the Lipschitz condition (3.5), we obtain

Taking into account that , we multiply at both sides of inequality (3.11) by to get

Denote . Then, (3.12) can be written as

Putting , , , , , we see that , , , , and is a nondecreasing continuous function due to . So, applying Theorem 2.7 to (3.13), we obtain

Next, we prove the case and . Notice that the Lipschitz condition (3.5) holds for each in . Since and , we can always choose two positive constants , such that

Condition (3.15) means that and . That is to say, if the Lipschitz condition (3.5) holds for each in , then we can always choose two positive constants ,   such that the following condition

holds for each in .

Subtracting (3.10) from (3.9) and using condition (3.16), we obtain

Multiplying on both sides of (3.17), we get

where is defined as before. Now, putting , , , , , we see that , , , , is a nondecreasing continuous function due to . So, applying Theorem 2.7 to (3.18), we have

Finally, we prove the case and . Since and , we can always choose two positive constants , such that

Condition (3.20) means that and . Using the same arguments as above, we can obtain that

where is defined as before. So the conclusion of this theorem is true.

From the proof of Theorem 3.4, we can see that the integral inequality in Theorem 2.7 is very useful. This demonstrates that our investigation is meaningful.

Acknowledgmets

This work was supported by the National Natural Science Foundation of China (11071192) and the International Science and Technology (S&T) Cooperation Program of China (2010DFA14700).

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