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Hussein A. H. Salem, Mieczysław Cichoń, "On Solutions of Fractional Order Boundary Value Problems with Integral Boundary Conditions in Banach Spaces", Journal of Function Spaces, vol. 2013, Article ID 428094, 13 pages, 2013. https://doi.org/10.1155/2013/428094
On Solutions of Fractional Order Boundary Value Problems with Integral Boundary Conditions in Banach Spaces
The object of this paper is to investigate the existence of a class of solutions for some boundary value problems of fractional order with integral boundary conditions. The considered problems are very interesting and important from an application point of view. They include two, three, multipoint, and nonlocal boundary value problems as special cases. We stress on single and multivalued problems for which the nonlinear term is assumed only to be Pettis integrable and depends on the fractional derivative of an unknown function. Some investigations on fractional Pettis integrability for functions and multifunctions are also presented. An example illustrating the main result is given.
The theory of boundary value problems is one of the most important and useful branches of mathematical analysis. Boundary value problems of various types create a significant subject of several mathematical investigations and appear often in many applications, especially in solving numerous problems in physics and engineering. For example, heat conduction, chemical engineering, underground water flow, thermoelasticity, and plasma physics can be reduced to nonlocal problems with integral boundary conditions. For boundary value problems with integral boundary conditions and comments on their importance, we refer the reader to [1–3] and the references therein.
The class of boundary value problems with integral boundary conditions considered below contains as special cases numerous two, three, multipoint, and nonlocal boundary value problems. Such problems are mainly investigated when considering functions satisfying some conditions expressed in terms of the strong topology of a Banach space . We will investigate the case, when functions are not strongly continuous and strongly integrable. In this situation we need to introduce more general notion of a solution. We should note that the considered case seems to be a natural case and cover many particular cases considered for both the strong and weak topologies (cf. Lemma 19). A more general notion of solutions allows us to solve the problem under very general assumptions, not so restrictive as before (see our last section).
In contrast to the classical approach for the theory of boundary value problems, the theory for fractional order BVP's is still developing one and not satisfactorily described. It is caused by the fact that it is very difficult to find convenient and handy conditions ensuring the existence of solutions of several nonlinear boundary value problems of fractional order. In the considered case of a weak topology on our results form a relatively new branch of investigations.
As a pursuit of this, some sufficient conditions for the existence of solutions are presented for the following nonlinear -point boundary value problem of fractional type: where takes values in a Banach space and . Here for some and denotes the pseudo fractional differential operator of order (to be described later). We will assume that is a vector-valued Pettis integrable function on . We remark the following:(1)for real-valued functions with , we have problems studied in, for example, [4, 5], (2)for real-valued functions with and , we have problems studied in, for example, [6–10]; see also the references therein, (3)for real-valued functions and when the function is independent of the fractional derivatives, then we have problems studied in, for example, [1, 11–13], (4)in abstract spaces (for vector-valued functions) with and , we have problems studied in, for example, [14–19], (5)in abstract spaces with conditions related to the weak topology on and when the vector-valued function is independent of the fractional derivatives, then we have a problems studied in, for example, [20, 21].
In comparison with the existence results in the above list, our assumptions seem to be more natural. In contrast to earlier results, we drop the requirement that is a real-valued function independent of the fractional derivatives and we consider the case of vector-valued Pettis, but not necessarily Bochner, integrable functions. As we mentioned above, the assumptions in the existence theorem are expressed in terms of the weak topology. Such a result does not appear in the earlier literature and so it seems to be new. We collect all interesting properties for the fractional Pettis integral. Moreover, we are able also to start some studies for multivalued fractional boundary value problems with Pettis-integral boundary conditions and fractionally Pettis integrable multifunctions.
In the paper we stress also on comparison results for Pettis integrals and fractional Pettis integrals. This is also done for the multivalued integrals and seems to be interesting by itself, independently of applicability of our results. The properties of fractional integral operators on the spaces of Pettis integrable functions as well as on some of its subspaces are also investigated.
Finally, we remark that, in the Banach spaces, the existence of solutions of some boundary value problems of fractional orders has been considered in terms of Pettis integrals, for the first time, by Salem . In this paper, for clarity of proofs, we restrict ourselves to the case of reflexive spaces, but it is easy to extend our results for nonreflexive spaces by putting contraction hypothesis with respect to some measure of weak noncompactness and by using appropriate fixed point theorem (cf. ). Nevertheless, all auxiliary results in this paper are not restricted to reflexive spaces.
The question of proving the existence of solutions to the problem (1) reduces to proving the existence of solutions of a Fredholm integral equation. Since the space of all Pettis integrable functions is not complete (in general), we restrict our attention to the case of weakly continuous solution of the Fredholm integral equation (modeled off the problem (1)); hence we are ready to find the so-called pseudo-solutions of the problem (1) (cf. [22, 23]).
2. Preliminaries and Auxiliary Results
For the sake of the reader’s convenience here we collect a few facts which will be needed further on. Let . According to the custom , will denote the Banach space of real-valued measurable functions defined on . Let denote the Banach space of real-valued essentially bounded and measurable functions defined on . Through the paper, is considered to be a Banach space with norm and with its dual space . Moreover, let denote the space with its weak topology. By we will denote the Banach space of strongly continuous functions endowed with a standard , while denotes the space of all -valued Pettis integrable functions in the interval (see [24, 25] for the definition). Let us also recall that a function is said to be weakly-weakly sequentially continuous if takes each weakly convergent sequence in into weakly convergent sequence in . We point out that a bounded weakly measurable function need not to be Pettis integrable even if is reflexive. However, in reflexive Banach spaces, the weakly measurable function is Pettis integrable if and only if is Lebesgue integrable on for every .
Lemma 1. A convex subset of a normed space is closed if and only if it is weakly closed.
A simple consequence of the Hahn-Banach theorem is as follows.
Proposition 2. Let be a normed space with . Then there exits with and .
Now, we are in a position to recall a fixed point theorem being an extension of results from .
Theorem 3. Let be a Banach space with a nonempty, closed, convex, and weakly compact subset of . Assume that is weakly-weakly sequentially continuous. Then has a fixed point in .
We need to introduce some subspaces of the space of Pettis integrable functions on which are important in the sequel.
Definition 4. For , we define the class to be the class of all functions having for every . If , the added condition must be satisfied by each . The class is defined by
Remark 5. In a reflexive Banach space the set coincides with the space . This is due to the fact that in reflexive Banach spaces, the weakly measurable function is Pettis integrable if and only if is Lebesgue integrable on for every . In general, this is the space of Dunford integrable functions.
In the remaining part of this paper we let be fixed and is conjugated with ; that is, . The following results are due to Pettis (see [25, Theorem 3.4 and Corollary 3.41]).
Proposition 6. In order that could be in , it is necessary and sufficient that be Pettis integrable for every .
It is worthwhile to recall the following.
Definition 7. Let . The (left-sided) fractional Pettis-integral (shortly LS-FPI) of of order is defined by
In the above definition the sign “” denotes the Pettis integral. For further purpose, we define the right-sided fractional Pettis-integral (shortly RS-FPI) by
We will call a function fractionally Pettis integrable provided this integral exists as an element of (for arbitrary ).
We need to clarify the relations between Pettis integrability and fractional Pettis integrability. Similar results will be proved for classes . This will be important in our consideration, but it seems to be really interesting in itself. Here we restrict ourselves to the case of left-sided fractional Pettis-integrals.
To make the paper more expository, we will consider fractional Pettis integrability for both cases: and . The last case is more important in our paper, but the first one is necessary to compare our results with some earlier theorems.
Let us observe that such an integral is a convolution of a function for , for , and the function for , where outside the interval . Note that Pettis integrability of implies Pettis integrability of () and , so the convolution of Pettis integrable function with real-valued function can be properly defined. We start with an obvious observation that for
As a consequence of some properties of a convolution for the Pettis integral [28, Proposition 9], for arbitrary , we have the following.
Theorem 8. If is Pettis integrable, then (a) is defined almost a.e. on , (b) is fractionally Pettis integrable on , (c)if is Pettis integrable and strongly measurable, then is bounded, weakly continuous and
In the case , it is a well-known consequence of an inequality of Young that the linear fractional integral operators , send continuously into if satisfy (see ) (a deep result from interpolation theory implies that even is allowed if ). In particular, is compact for each . Moreover, for , the map is compact (see, e.g., [20, 30]).
The following results plays a major rule in our analysis.
Lemma 9. For any the operator takes into and is well defined.
Proof. Only the proof in case of the LS-FPI is given since the case of the RS-FPI is very similar.
It can be easily seen that if is weakly continuous, then . Since , , in the view of Proposition 6 and Theorem 8, we have that the function , , is Pettis integrable. Moreover, is well defined. To see this, we define by , . From the definition of fractional Pettis integrals and Theorem 8(c), we have for every that Since the function is continuous, is continuous. That is, is weakly continuous which finishes the proof.
Lemma 10. Let be a reflexive Banach space. For any and arbitrary , the operator takes into and is well defined.
Proof. Only the proof in case of the LS-FPI is given since the case of the RS-FPI is very similar.
Note first that, for , we have in the view of Proposition 6 that the function , , is Pettis integrable. That is, the operator makes sense. Further, is well defined. To see this, define by , . From the definition of fractional Pettis integrals, we have for every that Since , one have for every . One could see, by the properties of fractional integral operators on the Banach space , that for every . In particular, is weakly continuous. Since weak continuity implies weak measurability (see [26, page 73]), is weakly measurable. However, in reflexive Banach spaces, weakly measurable functions are Pettis integrable if and only if is Dunford integrable; that is, is Lebesgue integrable on for every . Thus is Pettis integrable. That is, .
Remark 11. Let us discuss some properties of and . Recall that for and for . The case of is trivial; that is, . When , we see that whenever . This means that and by the converse for the Young inequality takes into whenever . In particular, the space required for this property depends on . Let us note that need not be continuous as an operator from into (cf. [29, Remark 4.1.1]).
This means that for we have “uniform” estimations for all , but for the situation is more complicated (a weakly singular case).
Our consideration as well as Theorem 4.1.1 in  gives us a new property.
Lemma 12. For any and arbitrary the operator takes into , where with arbitrary .
As a consequence of Lemma 10, we are able to prove the following.
Lemma 13. Assume that and . Then,
Let us present the case .
Lemma 14. If , , and , then
Proof. Define the real-valued function by
Using the properties of fractional calculus in the Banach space (see, e.g., [20, 30]), we deduce that . Now, for we have, in the view of Lemma 10, that . Thanks to Proposition 6, the functions and are Pettis integrable on . That is, the integrals in both sides of (13) exist. Then there exists , such that By the definition of the Pettis integral, we have
By changing the order of integration results in Thus
The following Lemma is well known in the case , but to see that it also holds in the vector-valued case, we provide a proof.
Lemma 16. For we have for every weakly continuous function , , and
In particular, when , (20) means that the operator is defined in and that is the left-inverse of .
Proof. The first claim, that is, , follows from the fact that the integral of weakly continuous function is weakly continuous, then pseudo-differentiable with respect to the right endpoint of the integration interval. Let and with and . Then we have, in view of and Lemmas 9 and 10, that
Definition 17. A function is called pseudo-solution of the problem (1) if has fractional pseudo-derivative of order , , and satisfies
The following auxiliary Lemma will be needed in our techniques.
Lemma 18. If is a pseudo-solution to the problem then is a pseudo-solution for the problem (1).
Proof. Let be a pseudo-solution to the problem (23) and . As in the proof of Lemma 9 it follows that exists and the real function is continuous for every ; moreover Thus for every ; that is, . Further In the view of Lemma 16 we also have
If otherwise is not stated, we will assume from now that and .
To obtain the Hammerstein type integral equation modeled off the problem (23), we keep the boundary value problem (23) in mind and we formally put (cf. [11, Lemma 2.3]) In the view of Lemma 13, we obtain To facilitate our discussion, let be constant with the conjugate exponents . Suppose be a nonnegative real-valued function and satisfy the following assumptions: (1)for each , is weakly-weakly sequentially continuous, (2)for each , ,(3)for any and there exist a Pettis integrable function , function , and nondecreasing continuous function such that for a.e. and all .
Let us present two remarks about the above assumptions.(i)For the interesting discussion about the growth conditions for Pettis integrable functions of the above type see . For differential equations with Caputo fractional integrals (i.e., solutions in the space , ), where the problem of dominants for considered functions (Assumption (3)) was considered in . Nevertheless, in the paper by Lin rather strong boundedness conditions are investigated (dominants from or for or bounded functions for ) [32, Remark 2.3].(ii)For convenience of the readers, let us recall the following lemma describing particular sufficient conditions for Pettis integrability of [23, Lemma 15]. It is obvious that additional growth condition for allows us to characterize the functions from .
Lemma 19. Assume that is absolutely continuous and . Thus is Pettis integrable if at least one of the following cases holds: (a) satisfies Carathédory conditions; that is, is measurable, is continuous in , and there exists an integrable function such that for all and a.e. , (b) is weakly-weakly continuous and is a weakly sequentially complete space, (c) is weakly measurable, is weakly-weakly continuous in and is a WCG-space (weakly compactly generated space), (d) is strongly measurable and there exists a Young function such that and , (e) is strongly measurable and there exists such that for each (here ), (f) is strongly measurable, is weakly-weakly sequentially continuous in , and is bounded, (g) is strongly measurable, contains no copy of , and is bounded.
Now, we would like to pay our attention to solve (27) for by It follows that Therefore where Then (in account of Lemma 14), we have Here . Substituting into (27), one has Thus where . The Green function is given by with Since , the following can be easily seen.
Lemma 20. The map is continuous from to .
Remark 21. We point out that if is reflexive, it is not necessary to assume any compactness conditions on the nonlinearity of . This will be due to [33, Lemma 2] and the fact that a subset of reflexive Banach spaces is weakly compact if and only if it is weakly closed and norm bounded.
3. Weak Solutions of the Hammerstein Integral Equation
In this section, in the light of the Assumptions (1)–(3) imposed on , we proceed to obtain a result which relies on the fixed point Theorem 3 to ensure the existence of weak solution to the integral equation (35). For the sake of convenience, we introduce the following.
We need to explain why we consider continuous solutions. By the properties of the Pettis integral this should be weakly continuous function. Since is continuous and we impose (local) boundedness hypothesis for , our solutions are strongly continuous (cf. ). We restrict our attention to the space and then our integral operators will be defined on this space. In contrast to the case of weakly-weakly continuous functions , we need to replace the space endowed with its topology of weak uniform convergence by the space of (strongly) continuous functions with its weak topology. We will utilize in our proofs some characterization of its weak topology.
Now, we are in the position to state and prove the first existence result.
Theorem 23. Assume that , with and be a nonnegative real-valued function. If the Assumptions (1)–(3) hold along with then the integral equation (35) has at least one solution .
Proof. First of all, observe the expression of and note that the following implications:
hold. Consequently for any . Let
Consider the set of real numbers which satisfy the inequality
Then is bounded above; that is, there exists a constant with
To see this, suppose (42) is false. Then there exists a sequence with as and
Since for any sequences , we have . This contradicts (38). Then, for every the inequality
holds, which is in contradiction with and then contradicts (42).
Now, define the operator by We remark that for we have that, by Lemma 9, is weakly continuous, and consequently, (Assumption (2)). Since , for all , is Pettis integrable for all (thanks to Proposition 6), and thus the operator makes sense. Note that is well defined. To see this, let with . Since , we deduce that if , then and , where . Without loss of generality, assume . Then there exists (as a consequence of Proposition 2 ) with and .
Putting the Assumption (3) in mind, one can write the following chain of inequalities: Then
Therefore we deduce, in the view of Lemma 20, that maps into itself.
Let be the convex, closed and equicontinuous subset (required by Theorem 3). Define this set by We claim that restricted to the set maps this set into itself (i.e., ) and is weakly-weakly sequentially continuous. Once the claim is established, Theorem 3 guarantees the existence of a fixed point of . Hence the integral equation (35) has a solution in .
We start by showing that . To see this, take . Since , we deduce that , . The monotonicity of and the inequality (47) imply that
Now, without loss of generality, assume . Then there exists (consequence of Proposition 2) with and . By the Assumption (3), we obtain Therefore . Hence .
We need to prove now that is weakly-weakly sequentially continuous. Let us recall that the weak convergence in is exactly the weak pointwise convergence. Let be a sequence in weakly convergent to . Then in for each . Since is closed, by Lemma 1 we have .
Fix and note, in the view of Lebesgue dominated convergence theorem for the Pettis integral (see [31, 34]), that in . Let us recall that the topology on on equicontinuous subsets coincides with the topology of weak pointwise convergence. Since satisfies Assumption (1), we have converging weakly to ; hence again the Lebesgue dominated convergence theorem for Pettis integral yields converging weakly to in , but is an equicontinuous subset of , and then is weakly-weakly sequentially continuous. Applying now Theorem 3, we conclude that has a fixed point in , which completes the proof.
Let us present a multivalued problem:
Some basic results for multivalued boundary value problems with Pettis integrals are due to Maruyama , Azzam et al. , Azzam-Laouir and Boutana , and Satco . However these results are devoted to study the standard case and three-point boundary conditions. Our result is an essential extension for the previous ones.
By and we denote the family of all nonempty convex compact and nonempty convex weakly compact subsets of , respectively. For every nonempty convex bounded set the support function of is denoted by and defined on by , for each .
Definition 24. A multifunction with nonempty, closed values is weakly sequentially upper hemicontinuous if and only if for each is sequentially upper semicontinuous from into .
In the remaining part of the paper a multifunction is supposed to be Pettis integrable in the sense of Aumann.
Definition 25. The Aumann-Pettis integral of a multifunction is where denotes the set of all Pettis integrable selections of provided that this set is not empty.
Let us note that the multivalued Pettis integral can be defined by other methods. The above definition is the best choice for our consideration. This can be deduced from the following theorem.
Theorem 26 (see ). Let be measurable and scalarly integrable multifunction (i.e., the support functions are real-valued integrable functions). Then the following statements are equivalent:(a)the set is uniformly integrable,(b)every measurable selection of is Pettis integrable,(c)for every measurable subset of the Aumann-Pettis integral belongs to  and, for every , one has
Taking into account Theorem 8, we are able to add one more condition to the above theorem, which seems to be important in our consideration. Since , is continuous, and by taking arbitrary Pettis integrable selection we obtain Pettis integrability of .
Theorem 27. Each of the conditions from Theorem 26 implies the following:(d) for every the multifunction is fractionally Aumann-Pettis integrable; that is, belongs to .
Let us recall that we restrict ourselves to the case of the (left-sided) fractional Pettis-integral.
Note that for multivalued mappings we will utilize Kakutani's fixed point theorem (for continuity concepts see ).
Theorem 28 (see ). If is a nonempty weakly compact convex subset of and is sequentially weakly upper semi-continuous, then there exists a fixed point of ; that is, with .
An immediate consequence of the above theorems as well as our main theorems is the following result.
Theorem 29. Assume that is separable. Let with nonempty convex and weakly compact values satisfy the following: (a) is weakly sequentially upper hemicontinuous for each , (b) has a weakly measurable selection for each , (c) a.e. for some valued Pettis integrable multifunction .
Then there exists at least one pseudo-solution of the Cauchy problem (51) on .
We will follow the idea of the proof for the single-valued problem. Let us only sketch the main steps of the proof.
Note that the Assumption (c) implies weak compactness of and separability of the space are sufficient to replace reflexivity of (as announced in the preliminary part).
In this proof we need to define the multifunction by By [40, Lemma 3.2] our assumptions (a)–(c) imply that the set is nonempty for arbitrary and the multivalued Nemytskii operator is well defined, so by Theorem 27 the Aumann-Pettis integral of is nonempty too.
Let and .
For and we have . Then, by our assumptions, is Pettis uniformly integrable. Thus for arbitrary and there exists an appropriate and By uniform Pettis integrability of it follows that is an equicontinuous subset of . The property of the multivalued Pettis integral gives us the convexity of . Then is nonempty, convex, bounded, and equicontinuous in .
As is sequentially compact for the topology induced by the tensor product , the is closed. Since is convex, by Mazur's lemma (Lemma 1) is weakly closed. Thus by a weak version of Ascoli's theorem is weakly compact in .
As the set is strongly equicontinuous, then for each there exists such that for each and we have .
Then restricted to a ball with radius (as in the previous proof) has nonempty, closed, convex and weakly compact values. As a domain for we put . By repeating the proof from [40, Theorem 3.3] we are able to show that has weakly-weakly sequentially closed graph. Restricted to a weakly compact set an operatot is sequentially weakly upper semi-continuous. This means, that the Kakutani fixed point theorem (Theorem 28) gives us a fixed point of . The proof is complete.
4. Pseudo-Solutions to Fractional Order Boundary Value Problem
In this section, we are looking for sufficient conditions to ensure the existence of pseudo-solution to the boundary value problem (1) under the Pettis integrability assumption imposed on . In order to obtain the existence of solutions of the problem (1), we can make use of Theorem 23.
Proof. Firstly, we remark that, for any , we have (according to Proposition 6) that for , . Thus the integral boundary condition makes sense.
In account of Theorem 23 it can be easily seen that the integral equation (35) has a solution . Let be a weak solution of (35). Then
By Lemma 14 and using , a straightforward estimates show that Furthermore, we have Thus for any we have