Abstract

Motivated by the Hilfer fractional derivative (which interpolates the Riemann-Liouville derivative and the Caputo derivative), we consider a new type of fractional derivative (which interpolates the Hadamard derivative and its Caputo counterpart). We prove the well-posedness for a basic Cauchy type fractional differential equation involving this kind of derivative. This is established in an appropriate underlying space after proving the equivalence of this problem with a certain corresponding Volterra integral equation.

1. Introduction

In this work, we are concerned with the Hadamard derivative Its Caputo counterpart is where (see [15]). Here, we consider the following fractional derivative This type of fractional derivative interpolates the Hadamard fractional derivative () and the Caputo-Hadamard fractional derivative (). It has been introduced recently in [6]. In introducing this new fractional derivative we were motivated by the Hilfer fractional derivative of order and type (see [7]) which interpolates the Riemann-Liouville derivative and the Caputo derivative.

We study the existence and uniqueness of solutions of a basic fractional differential equation with an appropriate initial condition in a suitable underlying space after proving the equivalence of this problem with a corresponding Volterra integral equation. In addition to that, we discuss the stability of solutions for a large and important class of nonlinearities. We find that solutions decay to zero at a logarithmic rate as time goes to infinity. To this end, we prove an inequality (which is important by itself).

The literature is very rich in works on well-posedness for fractional differential equations [817] (see also the books [1, 2, 4, 5] and the survey paper [18]) to cite but a few. The Hadamard fractional derivative may be found in the books [1, 2, 46]. Differential equations involving such a derivative and others have been treated in [2, 3]. In contrast with the well-posedness, the stability issue and the long time behavior is not well studied [6, 12, 1923].

The rest of the paper is organized as follows: the next section contains some material needed in our proofs. The different fractional derivatives as well as the new one are defined there. In Section 3, we present our problem and prove an existence and uniqueness result after establishing the equivalence of the differential problem with its corresponding integral equation. Section 4 is devoted to a stability result.

2. Preliminaries

In this section we present some definitions, lemmas, properties, and notation which will be used in our theorems later.

Definition 1 (see [2]). Let () be a finite interval and , we introduce the weighted space of continuous functions on In the space , we define the norm

Definition 2 (see [2]). Let be the -derivative, for we denote by () the Banach space of functions which have continuous -derivatives on up to order and have the derivative of order on such that with the norm When , we set

Definition 3 (see [2]). Let be a finite or infinite interval of the half-axis and let . The Hadamard left-sided fractional integral of order is defined by provided that the integral exists. When , we set

Definition 4 (see [2]). Let    be a finite or infinite interval of the half-axis and let . The Hadamard right-sided fractional integral of order is defined by provided that the integral exists. When , we set

Definition 5 (see [2]). The left-sided Hadamard fractional derivative of order () on is defined by that is, In particular, when we have

Definition 6 (see [2]). The right-sided Hadamard fractional derivative of order () on is defined by that is, In particular, when we have

Definition 7. Let be a finite interval of the half-axis . The fractional derivative of order () on defined by where , is called the Hadamard-Caputo fractional derivative of order .

In the rest of the paper we shall assume when considering an interval .

Lemma 8 (see [2]). Let and let and be real numbers such that The following embeddings hold: with In particular, with

Lemma 9 (see [2]). If , and , then In particular, if and , then the Hadamard fractional derivative of a constant is not equal to zero: when .

Lemma 10 (see [2]). Let ,   and . If , then, for holds at any point . When this relation is valid at any point .

Lemma 11 (see [2]). Let and . The equality is valid for for any .

Theorem 12 (see [2]). Let and . Also let be the Hadamard fractional integral of order of the function .
If and , then holds at any point . If and , then the relation holds at any point .

Lemma 13 (see [2]). Let   and .(a)If , then the fractional integration operator is bounded from into : where In particular, is bounded in .(b)If , then the fractional integration operator is bounded from into : where In particular, is bounded in .

Lemma 14 (see [2]). The Hadamard fractional integration operator of order is a mapping from to and where .

Lemma 15 (see [17]). Let and . Then, and

Lemma 16 (see [24]). Let , be two continuous, positive functions defined on , , and be a continuous monotonic nondecreasing function such that and for . If is a positive differentiable function on that satisfies then we have for the values of , for which the right-hand side is well-defined, where

Theorem 17 (Banach fixed point theorem [2]). Let be a non-empty complete metric space, let , and let be a map such that, for every , the relation holds. Then, the operator has a unique fixed point .
Furthermore, if is the sequence of operators defined by then, for any , the sequence converges to the above fixed point .

Theorem 18 (Young’s inequality). If and are nonnegative real numbers and and are positive real numbers such that then we have Equality holds if and only if .

Finally, we refer the reader to the nice treatments of Hadamard-type fractional calculus in [25, 26]

3. Existence and Uniqueness for an FDE with Hilfer-Hadamard Fractional Derivative

In this section we discuss the existence, uniqueness and the stability of solutions of the Cauchy type problem (46) (below) with Hilfer-Hadamard fractional derivative.

Definition 19 (Hilfer-Hadamard fractional derivative (HHFD)). The left sided fractional derivative of order , and type with respect to is defined by for functions for which the expression on the right hand side exists, where is the Hadamard fractional derivative (Definition 5).

This new fractional derivative (introduced for the first time in [6]) may be viewed as interpolating the Hadamard fractional derivative and the Hadamard-Caputo fractional derivative. Indeed for this derivative (46) reduces to the Hadamard fractional derivative (Definition 5) and when , we recover the Hadamard-Caputo fractional derivative (Definition 7).

We will study the existence and uniqueness for the Cauchy type problem We consider the underlying spaces defined by where and . It is clear that for .

Here, and are weighted spaces of continuous functions on defined by Our investigations are based on reducing the fractional differential problem to a Volterra integral equation of the second kind: and then using the Banach fixed point theorem.

3.1. Equivalence of the Cauchy Type Problem and the Volterra Integral Equation

Here, we prove the equivalence of the Cauchy type problem (46) and the nonlinear Volterra integral equation (50) in the sense that, if satisfies one of them, then it also satisfies the other one. To establish this result, we assume that the function belongs to for any . We need the following lemma.

Lemma 20. Let , , and . If , then is continuous on and

Proof. Since then is continuous on and on we have for some positive constant . Therefore, and by using Lemma 9 (with ) we have As , we obtain the result.

Theorem 21. Let where and . Let be a function such that for any with .
If , then satisfies the (CFDP) (46) if and only if satisfies the (IE) (50).

Proof. First we prove the necessity. Let be a solution of problem (46). We want to prove that is also a solution of the integral equation (50). By the definition of the space (relation (48)) we have Moreover, by Lemma 13(b) we have since . Then, by Definition 2, we have Thus, we can apply Theorem 12 (with replaced by ) to get or where comes from the initial condition in (46). By our hypothesis , since , Lemma 13(a) and (b) we see that the integral for and for . Applying the operator to both sides of (46) we get We can sum up the exponents by Lemma 10 to get or From (58) and (61) we obtain which is (50), and hence the necessity is proved.
Now, we prove the sufficiency. Let satisfy (50), then exists and . Applying the operator to both sides of the last identity we find By using Lemma 10, Definition 5, Lemma 23 and the hypothesis , we have From (64) and the fact that , we obtain that Next, applying the operator to both sides of (64) we get that is, By virtue of and and Definition 2, we have (see the first part of the proof, or Lemma 13(b), for the continuity of for ). Then, Theorem 12 allows us to write Lemma 20 implies that because . Hence, the relation (69) reduces to Now, we show that the initial condition in (46) also holds. To this end we apply the operator to both sides of (50): and use the Lemma 8 (with replaced by and by ) and the Lemma 9 to obtain In (73), taking the limit as , we obtain mentioned above . Therefore, the sufficiency is proved, which completes the proof of Theorem 21.

3.2. Existence and Uniqueness of a Solution

In this section we establish the existence of a unique solution to the Cauchy type problem (46) in the space defined in (47) above under the conditions of Theorem 21 and an additional Lipschitz condition for some positive constant and every .

Theorem 22. Let where . Assume that is a function such that for any with and is Lipschitz continuous with respect to its second variable. Then, there exists a unique solution for the Cauchy type problem (46) in the space .

Proof. First we prove the existence of a unique solution in the space . According to Theorem 21, it suffices to prove the existence of a unique solution to the nonlinear Volterra integral equation (50).
Let us select in such that where is the Lipschitz constant. We start by proving that a unique solution to (50) exists on the interval (. It is easy to see that the space is a complete metric space when equipped with the distance given by The integral equation (50) takes the form where with We claim that maps into itself. Indeed, given by (80) is clearly in . Also, since for any with (), then, by Lemma 13(a) and (b), the integral in the right-hand side of (79) belongs to for and to for . Since , by Lemma 16 the right-hand side of (79) belongs to .
Our second claim is that is a contraction; that is, This follows from (79), Lemma 13(a), and the fact that Our assumption (76) allows us to apply the Banach fixed point theorem to obtain a unique solution to (50) on the interval .
This solution is the limit of a convergent sequence : where is any function in and Let us take with defined by (80). If we denote by then, clearly, Next, we consider the interval . From (50) we have where is defined by and is a known function. We note that .
We want to prove the existence of a unique solution of (50) on the interval . For this, we also use Banach fixed point theorem for the space where satisfies The space is a complete metric space with the distance given by The integral equation (87) may be written shortly as where the operator (again denoted by ) is given by As in the first part of this proof, since and for any , then, , for any , then by Lemma 14, we deduce that the integral in the right-hand side of (92) also belongs to , and hence .
Moreover, using the Lipschitz condition and applying the Lemma 14, we find This, together with our assumption , shows that is a contraction and therefore from Theorem 17, there exists a unique solution to (50) on the interval . Notice that . Further, Theorem 17 guarantees that this solution is the limit of a convergent sequence : where is any function in , which we can pick defined by (88). Therefore, where If , we consider the interval , where , such that and Using the same arguments as above, we derive that there exists a unique solution to (50) on the interval . If , then we continue the process until we reach a solution to (50), , and (), where and Assume that (for otherwise, take ). Then, divide the length of the interval by . Let be that quotient. It appears that for and is reached after a finite number of steps, . Then, there exists a unique solution to (50) on the interval .
Putting together the solutions in and and taking into account the Lemma 15, we obtain that there exists a unique solution to the Volterra integral equation (50) on the whole interval . Hence, is the unique solution to the Cauchy-type problem (46).
It remains to show that such a unique solution is actually in . To this end we need to prove that . Let us recall that our is a limit of the sequence , where ; that is, with a certain choice of on each subinterval . Indeed, this is a consequence of the construction adopted, the initial values are selected in the space (see (80) and (88)) and the operator maps this space into itself (see argument right after (80) and (92)). As for the convergence in that space it has been proved in (83) and (95).
If ,then we can take . Since , then by (46), the Lipschitz condition and Lemma 8, we have In virtue of (99) and (100), it follows that We entail from this relation that if , This latter property holds in as much as and for any .
Consequently, . This completes the proof of Theorem 22.

4. Stability

In this section, we consider the weighted Cauchy-type problem where is the Hilfer-Hadamard fractional derivative (HHFD) of order and type and (the set of all real numbers except ). It is interesting to note that, using an argument similar to the one in the proof of Lemma 3.5 in [2] (see also Lemma 3.2), we can prove that the initial condition in (102) and the one in (46) are equivalent.

We will assume the following hypotheses on the function :

(F*) is a continuous (nonlinear) function on and is such that where is a continuous (nonnegative) function on .

We first prove the following inequality.

Lemma 23. If , then for any , we have where is a positive constant independent of .

Proof. Let us denote by the left-hand side of the inequality in the Lemma. We consider the change of variable, then and . It follows that or Observe that, for and we have . Also since and the Gamma function is increasing in we have or . Moreover, (), and hence or Therefore, for we get For and such that we have This means that Consequently, Let , we see that Thus, As a result, .
For , it is clear that holds and we proceed in the same manner to conclude that for such that where . The proof is complete.

Let and be conjugate exponents; that is, , and let and , where . If and , then and . We denote by the positive real number

Theorem 24. Suppose that satisfies and . If for some , then, for any solution of Problem (102), there exists a positive constant such that , , , where .

Proof. Let us consider the Volterra integral equation associated to problem (102). Multiplying both sides of (119) by and using the assumption on we get Let denote the left-hand side of (120). The insertion of the term inside the integral gives Now, the Hölder inequality with exponents and yields Since , and , , , we may apply Lemma 9 (with replaced by , replaced by and replaced by ) to get where is the constant appearing in Lemma 9 corresponding to the present exponents. That is, Combining (122) and (124) we entail that where . Multiplying both sides of (126) by , we obtain Let denote the left-hand side of (127). The insertion of the term inside the integral gives Raising both sides of (129) to the power we get Let us set Then, clearly , and by differentiation Moreover, it is clear that is a continuous, nonnegative and nondecreasing function in .
Now, we would like to estimate the right hand side of (132) in term of . From (130) and (131) we entail that Raising both sides of (133) to the power we get The substitution of (134) in (132) yields Applying Lemma 16 (with ) we infer that Let us set then where we have used the fact that . Since , , , then and . That is, As long as In particular, if , then for some positive constant for all , and thus, from (129), we see that and then for some positive constant . This yields that for . The proof is complete.

Acknowledgment

The authors are very grateful for the financial support and the facilities provided by King Fahd University of Petroleum and Minerals through Project no. 101003.