Existence of Solutions for Fractional Differential Inclusions with Antiperiodic Boundary Conditions

We study the existence of solutions for a class of fractional differential inclusions with anti-periodic boundary conditions. The main result of the paper is based on Bohnenblust- Karlins fixed point theorem. Some applications of the main result are also discussed.

In some cases and real world problems, fractional-order models are found to be more adequate than integer-order models as fractional derivatives provide an excellent tool for the description of memory and hereditary properties of various materials and processes. The mathematical modelling of systems and processes in the fields of physics, chemistry, aerodynamics, electro dynamics of complex medium, polymer rheology, and so forth, involves derivatives of fractional order. In consequence, the subject of fractional differential equations is gaining much importance and attention. For details and examples, see [1–14] and the references therein.

Antiperiodic boundary value problems have recently received considerable attention as antiperiodic boundary conditions appear in numerous situations, for instance, see [15–22].

Differential inclusions arise in the mathematical modelling of certain problems in economics, optimal control, and so forth. and are widely studied by many authors, see [23–27] and the references therein. For some recent development on differential inclusions, we refer the reader to the references [28–32].

Chang and Nieto [33] discussed the existence of solutions for the fractional boundary value problem:

In this paper, we consider the following fractional differential inclusions with antiperiodic boundary conditions

where denotes the Caputo fractional derivative of order , Bohnenblust-Karlin fixed point theorem is applied to prove the existence of solutions of (1.2).

Let denote a Banach space of continuous functions from into with the norm Let be the Banach space of functions which are Lebesgue integrable and normed by

Now we recall some basic definitions on multivalued maps [34, 35].

Let be a Banach space. Then a multivalued map is convex (closed) valued if is convex (closed) for all The map is bounded on bounded sets if is bounded in for any bounded set of (i.e., . is called upper semicontinuous (u.s.c.) on if for each the set is a nonempty closed subset of , and if for each open set of containing there exists an open neighborhood of such that . is said to be completely continuous if is relatively compact for every bounded subset of If the multivalued map is completely continuous with nonempty compact values, then is u.s.c. if and only if has a closed graph, that is, imply In the following study, denotes the set of all nonempty bounded, closed, and convex subset of . has a fixed point if there is such that

Let us record some definitions on fractional calculus [8, 11, 13].

Definition 2.1.

For a function the Caputo derivative of fractional order is defined as

where denotes the integer part of the real number and denotes the gamma function.

Definition 2.2.

The Riemann-Liouville fractional integral of order for a function is defined as

provided the right-hand side is pointwise defined on

Definition 2.3.

The Riemann-Liouville fractional derivative of order for a function is defined by

provided the right-hand side is pointwise defined on

In passing, we remark that the Caputo derivative becomes the conventional derivative of the function as and the initial conditions for fractional differential equations retain the same form as that of ordinary differential equations with integer derivatives. On the other hand, the Riemann-Liouville fractional derivative could hardly produce the physical interpretation of the initial conditions required for the initial value problems involving fractional differential equations (the same applies to the boundary value problems of fractional differential equations). Moreover, the Caputo derivative for a constant is zero while the Riemann-Liouville fractional derivative of a constant is nonzero. For more details, see [13].

For the forthcoming analysis, we need the following assumptions:

(A₁)let be measurable with respect to for each , u.s.c. with respect to for a.e. , and for each fixed the set is nonempty,

(A₂)for each there exists a function such that for each with , and

where depends on For example, for we have and hence If then is not finite.

Definition 2.4 ([16, 33]).

A function is a solution of the problem (1.2) if there exists a function such that a.e. on and

which, in terms of Green's function , can be expressed as

where

Here, we remark that the Green's function for takes the form (see [22])

Now we state the following lemmas which are necessary to establish the main result of the paper.

Lemma 2.5 (Bohnenblust-Karlin [36]).

Let be a nonempty subset of a Banach space , which is bounded, closed, and convex. Suppose that is u.s.c. with closed, convex values such that and is compact. Then G has a fixed point.

Lemma 2.6 ([37]).

Let be a compact real interval. Let be a multivalued map satisfying and let be linear continuous from then the operator is a closed graph operator in

Theorem 3.1.

Suppose that the assumptions and are satisfied, and

Then the antiperiodic problem (1.2) has at least one solution on

Proof.

To transform the problem (1.2) into a fixed point problem, we define a multivalued map as

Now we prove that satisfies all the assumptions of Lemma 2.6, and thus has a fixed point which is a solution of the problem (1.2). As a first step, we show that is convex for each For that, let Then there exist such that for each we have

Let Then, for each we have

Since is convex ( has convex values), therefore it follows that

In order to show that is closed for each let be such that in Then and there exists a such that

As has compact values, we pass onto a subsequence to obtain that converges to in Thus, and

Hence

Next we show that there exists a positive number such that where Clearly is a bounded closed convex set in for each positive constant If it is not true, then for each positive number , there exists a function with and

On the other hand, in view of , we have

where we have used the fact that

Dividing both sides of (3.8) by and taking the lower limit as , we find that which contradicts (3.1). Hence there exists a positive number such that

Now we show that is equicontinuous. Let with Let and then there exists such that for each we have

Using (3.8), we obtain

Obviously the right-hand side of the above inequality tends to zero independently of as Thus, is equicontinuous.

As satisfies the above assumptions, therefore it follows by Ascoli-Arzela theorem that is a compact multivalued map.

Finally, we show that has a closed graph. Let and We will show that By the relation we mean that there exists such that for each

Thus we need to show that there exists such that for each

Let us consider the continuous linear operator so that

Observe that

Thus, it follows by Lemma 2.6 that is a closed graph operator. Further, we have Since therefore, Lemma 2.6 yields

Hence, we conclude that is a compact multivalued map, u.s.c. with convex closed values. Thus, all the assumptions of Lemma 2.6 are satisfied and so by the conclusion of Lemma 2.6, has a fixed point which is a solution of the problem (1.2).

Remark 3.2.

If we take where is a continuous function, then our results correspond to a single-valued problem (a new result).

Applications

As an application of Theorem 3.1, we discuss two cases in relation to the nonlinearity in (1.2), namely, has (a) sublinear growth in its second variable (b) linear growth in its second variable (state variable). In case of sublinear growth, there exist functions such that for each In this case, For the linear growth, the nonlinearity satisfies the relation for each In this case and the condition (3.1) modifies to In both the cases, the antiperiodic problem (1.2) has at least one solution on

Examples

(a) We consider and in (1.2). Here, Clearly satisfies the assumptions of Theorem 3.1 with (condition (3.1). Thus, by the conclusion of Theorem 3.1, the antiperiodic problem (1.2) has at least one solution on

(b) As a second example, let be such that and in (1.2). In this case, (3.1) takes the form As all the assumptions of Theorem 3.1 are satisfied, the antiperiodic problem (1.2) has at least one solution on

The authors are grateful to the referees for their valuable suggestions that led to the improvement of the original manuscript. The research of V. Otero-Espinar has been partially supported by Ministerio de Educacion y Ciencia and FEDER, Project MTM2007-61724, and by Xunta de Galicia and FEDER, Project PGIDIT06PXIB207023PR.

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