A Generalized Henry-Type Integral Inequality and Application to Dependence on Orders and Known Functions for a Fractional Differential Equation

Abstract

We discuss on integrable solutions for a generalized Henry-type integral inequality in which weak singularity and delays are involved. Not requiring continuity or differentiability for some given functions, we use a modified iteration argument to give an estimate of the unknown function in terms of the multiple Mittag-Leffler function. We apply the result to give continuous dependence of solutions on initial data, derivative orders, and known functions for a fractional differential equation.

1. Introduction

Since Gronwall [1] and Bellman [2] discussed the integral inequalities respectively, there have been made many generalizations, some of which were applied to existence, uniqueness, boundedness, and stability of solutions and invariant manifolds for differential equations and integral equations. In 1956 Bihari [3] discussed a nonlinear version of the integral inequality where the given function is continuous, nondecreasing, and positive definite (i.e., for all and if and only if ) on . In 2000 this result was generalized by Lipovan [4] to the delay case where is a continuously differentiable and nondecreasing function from to such that . In 2005 Agarwal et al. [5] investigated the integral inequality of a finite sum where the monotonicity of is not required but has stronger monotonicity than for each . In recent years some new generalizations were given in, for example, [6–8]. Many results on integral inequalities can be found in Pachpatte’s monograph (see [9]).

Among various types of integral inequalities, integral inequalities with weakly singular kernels are important tools in the discussion of reaction-diffusion equations and fractional differential equations. As shown in [10], the integral is singular on the line ; it is referred to be weakly singular if it is singular and for all . In fractional differential equations we need to consider the following Riemann-Liouville derivative operator and integral operator (see [11]) in which the singular kernels and are included, respectively. Another example can be seen from the Cauchy problem of the evolution equation in a Banach space , where is a sectorial operator and is locally Hölder continuous with Hölder index . From the variation-constant formula (see [12]), we encounter the following: and therefore the singular kernel is the estimation of its solution. For this reason the integral inequality with a weakly singular kernel where , and both and are nonnegative and locally integrable, was considered in Henry’s book [12] in 1981 and the estimate where is the Gamma function, was given by an iteration approach. Following Henry’s idea, in 1994 Sano and Kunimatsu [13] extended Henry’s result to a more general integral inequality where , . Inequality (9) was also extended by Ye et al. [14] in 2007 by replacing the constant with a nonnegative, nondecreasing, bounded, and continuous function .

Another idea for inequalities with a weakly singular kernel was introduced by Medved [15] in 1997 for the following Henry-Bihari type integral inequality: where . This inequality is of Bihari’s form (3) with a weakly singular kernel. He applied the well-known Hölder inequality to separate the unknown from the singular kernel, that is, where are certain constants such that , so that the inserted exponential factor makes the singular integral convergent as and the inequality is reduced to the classic Bihari’s form (3). In 2002 Ma and Yang [16] improved Medved’s method and gave an estimation to the Volterra-type integral inequality with a more general form of weakly singular kernel where . Recently Ma and Pečarić [17] also employed the separation approach to discuss another weakly singular integral inequality under the condition .

In this paper we investigate the following integral inequality of finite sum: in which weakly singular kernels and delays are involved. Not requiring continuity of or differentiability of , we give an estimate for locally integrable in terms of the multiple Mittag-Leffler function (see [18]). We prefer Henry’s iteration approach in our proof because the approach does not reduce the problem to the classic one so that no more continuity and differentiability are required. Our result generalizes the works made in [14–17] in some sense because of the more general form (16). Finally, we apply the result to give continuous dependence of solutions on initial data, derivative orders, and known functions for a fractional differential equation.

2. Main Results

For constants ,  such that , consider inequality (16), where and are given nonnegative functions and satisfy the following hypotheses:(H₁) is a locally integrable function on , that is, Lebesgue integrable on every compact subset of , and the integrations in (16) are bounded by replacing with in (16);(H₂)every ,  , is continuous on ;(H₃)every , , is continuous and strictly increasing such that for all ;(H₄)every , , is continuously differentiable on such that and for all .

Theorem 1. Suppose that (H₁)–(H₄) hold and , . Then, every nonnegative and locally integrable function which satisfies (16) on and the integrations in (16) bounded has the estimate where , .

We leave the proof of this theorem to next section. In what follows, we express the estimate of series form in terms of the multiple Mittag-Leffler function (see [18]).

The original Mittag-Leffler function was proposed as an extension of the exponential function by Mittag-Leffler ([19]) in 1903. An extension of two parameters was proposed by Wiman [20] in 1905. Later, two Mittag-Leffler functions with three parameters were given separately by Prabhakar [21] and Kilbas and Saigo [22]. In 1996 Hadid and Luchko [18] generalized the function into the multiple form where , ,   and These generalized Mittag-Leffler functions have been treated as significant special functions since they played an important role in computing fractional calculus and solving fractional differential and integral equations modeled in physics, chemistry, biology, engineering, and applied sciences (see monographs [11, 23]). We have the following inequality: In fact, we can show that By Euler’s definition on Gamma function (see [24]), we have It follows that It is easy to know that . Then, we can prove by induction that Obviously, (26) is true for . Assume that (26) is true for . We get since implying that (26) holds for . Thus, from (26) we see that (23) holds. By (23), we have Therefore, (22) is proved.

Corollary 2. Suppose that the hypotheses of Theorem 1 hold and additionally that is continuous on . Then, where and and are defined in (18) and (20).

Proof. Starting from (17), we have It follows from (22) that The corollary is proved.

3. Proof of Theorem

Let consist of all locally integrable nonnegative functions on such that for arbitrarily given . We can verify that is a linear space, due to the linearity of integration. Define a linear operator on as where . We claim that is self-mapping on . In fact, for all and given , , implying that , the set of all continuous and nonnegative functions on . We know that . For all , from the continuity of and , we know that they are all locally bounded. We also know that are integrable on by (H₃), (H₄) and . It follows that , implying that . Then, inequality (16) can be simplified as , from which we can prove by induction that We claim that for all integers , where are well-defined on since and are continuous and on . In fact, (35) is true for . Assume that (35) is true for . Then It follows that where Note that by (H₄). It implies that for all by the monotonicity of . It follows from (39) that is a nondecreasing function. Hence, by (H₄). On the other hand, from (H₃) and (H₄) we see that , implying that . With the change of variables , it follows from (38) that Letting denote and substituting (39) in (40), we get by interchanging the order of integration. In (41) we observe that where and by (H₄). Thus, from (37) we see that inequality (35) holds for and the claimed (35) is proved.

For every and each , define a sequence of linear operators by where since from the supposition of Theorem 1 and are continuous on . It implies from (35) that In what follows we will prove that for all and arbitrarily given , converges. Define where and is a constant. By Euler’s definition on Gamma function (see [24]), , we know that and therefore is strictly decreasing on . Thus, for all integers , , , and satisfying that , we have , where , implying that Multiplying the above inequality by , we get that Let , and . Then (48) implies that When , we have where

Let Then is convergent on by the ratio test ([25, pages 66-67]) because where we note that is a continuous function on and is a Stirling’s approximation of as , as known in [26, pages 59]. It implies that is convergent on . The case of can be proved similarly. Then, is convergent for and arbitrarily given . Passing to the limit as in (34), by (44) we get Since ’s are chosen arbitrarily, by interchanging with in (54), we obtain the estimate (17) and complete the proof of the theorem.

4. Application to Dependence

Recently, increasing interest was given to fractional differential equations (see monographs [11, 23]). In this section we consider the Cauchy problem of the general fractional differential equation where , , is defined as in the Introduction, particularly, denotes and denotes identity map, and , . This system includes the system as a special case, which was considered in [27] and corresponds to the Basset problem when , a classical problem in fluid dynamics concerning the unsteady motion of a particle accelerating in a viscous fluid under the action of gravity (see [28]). We will give continuous dependence for solutions of (55) associated with the initial condition (56) on the derivative orders ’s, the initial data , and the functions and .

Proposition 3. Suppose that each is strictly increasing such that and each satisfies that , , where , , and are constant. Then, the Cauchy problem (55)-(56) has a unique solution on for a certain constant and the solution depends continuously on , ’s, ’s, and ’s for all .

Proof. The Cauchy problem (55)-(56) is equivalent to the integral equation Define a sequence such that and where . We first claim that all are well-defined and continuous in , such that . In fact, it is true for . Suppose that it is true for some . Then, is also well defined by (59) and continuous in . Moreover, where and is a positive constant such that The existence of is guaranteed by the intermediate theorem for continuous functions. Thus, the claim is proved by induction for all .
The convergence of the sequence is equivalent to the convergence of the series . We claim that for all and all integers . It can be checked easily for . Suppose that it is true for some . By changing into in (63), we get from that Combining with (59), we have Thus, (63) holds for and the claim is proved for all by induction. From (63), for we have where , , and . By the ratio test ([25, pp. 66-67]), we get that is convergent, implying that the series and therefore the sequence are uniformly convergent in . Let , which is well-defined and continuous in such that . One can check that is a continuous solution of the Cauchy problem (55)-(56) in .
Next, we prove the continuous dependence of the solution. Consider the Cauchy problem where , , is strictly increasing such that , and is Lipschitzian in the second variable with the Lipschitz constant , . As above, we similarly obtain a solution of the Cauchy problem (67)-(68). Similar to (58), we have It follows that where , and . Thus, applying the inequality given in Corollary 2 to (70), we obtain that where .
Note from (71) that contains and , which will be estimated with and , respectively. In (71) we have in (61). Since and are both functions in , and for all , we see that is in . By the Lagrange mean value theorem (see [25, page 108]), function satisfies the Lipschitz condition uniformly with respect to ; that is, there exists a constant such that On the other hand, from the equivalent integral equation of (67) we get where Choosing and in (74), we obtain Summarily, from (71), (73) and (75) we get It follows from (72) that that is, the solution of the Cauchy problem (55)-(56) depends continuously on , , and .
As a complement, we note from (79) that in if , , , and . This implies the uniqueness of the solution.

Conflict of Interests

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

Acknowledgment

The author is very grateful to Professor Weinian Zhang for his valuable suggestions.