Qualitative Theory of Functional Differential and Integral EquationsView this Special Issue
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Existence and Estimates of Positive Solutions for Some Singular Fractional Boundary Value Problems
We establish the existence and uniqueness of a positive solution for the following fractional boundary value problem, with the conditions , , where , , and is a nonnegative continuous function on that may be singular at or . We also give the global behavior of such a solution.
Recently, the theory of fractional differential equations has been developed very quickly and the investigation for the existence of solutions of these differential equations has attracted considerable attention of researchers in the last few years (see [1–11] and the references therein).
Fractional differential equations arise in various fields of science and engineering such as control, porous media, electrochemistry, viscoelasticity, and electromagnetism. They also serve as an excellent tool for the description of hereditary properties of various materials and processes (see [12–14]). In consequence, the subject of fractional differential equations is gaining much importance. Motivated by the surge in the development of this subject, we consider the following singular Dirichlet problem: where , , and is a nonnegative continuous function on that may be singular at or . Then we study the existence and exact asymptotic behavior of positive solutions for this problem.
We recall that, for a measurable function , the Riemann-Liouville fractional integral and the Riemann-Liouville derivative of order are, respectively, defined by provided that the right hand sides are pointwise defined on . Here and means the integer part of the number and is the Euler Gamma function.
Several results are obtained for fractional differential equations with different boundary conditions, but none of them deal with the existence of a positive solution to problem (1).
Our aim in this paper is to establish the existence and uniqueness of a positive solution for problem (1) with a precise asymptotic behavior, where is the set of all functions such that is continuous on .
To state our result, we need some notations. We will use to denote the set of Karamata functions defined on by for some , where and such that . It is clear that a function is in if and only if is a positive function in such that For two nonnegative functions and defined on a set , the notation , , means that there exists such that for all . We denote by for and by the set of all nonnegative measurable functions on .
Throughout this paper, we assume that is nonnegative on and satisfies the following condition:
such that for where , , satisfying In the sequel, we introduce the function defined on by where Our main result is the following.
Theorem 1. Let and assume that satisfies . Then problem (1) has a unique positive solution satisfying for ,
Remark 2. Note that, for , we have This implies in particular that, for and , the solution blows up at and for , .
This paper is organized as follows. Some preliminary lemmas are stated and proved in the next section, involving some already known results on Karamata functions. In Section 3, we give the proof of Theorem 1.
2. Technical Lemmas
Lemma 3. The following hold.(i)Letting and , then one has (ii)Let and . Then one has , , and .
Example 4. Let be a positive integer. Let , , and be a sufficiently large positive real number such that the function is defined and positive on , for some , where times. Then .
Lemma 5. Let and be a function in defined on . One has the following:(i)if , then diverges and ;(ii)if , then converges and .
Lemma 6. Let be defined on . Then one has If further converges, then one has
Proof. We distinguish two cases.
Case 1. We suppose that converges. Since the function is nonincreasing in , for some , it follows that, for each , we have It follows that . So we deduce (13).
Now put Using that , we obtain This implies that So (14) holds.
Case 2. We suppose that diverges. We have, for some , This implies that This proves (13) and completes the proof.
Lemma 8. Given and , then the unique continuous solution of is given by where is Green's function for the boundary value problem (23).
Proof. Since , then is a solution of the equation . Hence . Consequently there exist two constants such that . Using the fact that and , we obtain and . So
In the following, we give some estimates on the Green function . So, we need the following lemma.
Lemma 9. For and one has
Proposition 10. On , one has(i); (ii) there exist two constants
Proof. (i) For we have
Since for , then by applying Lemma 9 with and , we obtain
(ii) Using the following inequalities for , we deduce the result from (i).
As a consequence of Proposition 10, we obtain the following.
Corollary 11. Let and put for . Then
Proposition 12. Given and such that the function is continuous and integrable on , then is the unique solution in of the following boundary value problem:
Proof. From Corollary 11, the function is defined on and by Proposition 10, we have
which implies that is bounded on . Now, using Fubini's theorem, we have
On the other hand, we have
Now, suppose that ; then we have
By considering the substitution , we obtain
Moreover if and , we have .
So, it follows that This implies that Moreover, using part of Proposition 10 and the dominated convergence theorem, we conclude that and .
Finally, we prove the uniqueness. Let be two solutions of (33) and put . Then and . Hence, it follows that . Using the fact that , we conclude that and so .
In the sequel, we assume that and and we put where satisfy So, we aim to give some estimates on the potential function .
We define the Karamata functions , by
Then, we have the following.
Proposition 13. For ,
Proof. Using Proposition 10, we have For , we have . So, using Lemma 5 and hypothesis (42), we deduce that Now, we have which implies by Lemma 5 that Hence, it follows by Lemma 6 and hypothesis (42) that, for , we get That is Now, for , we use again Lemma 5 and hypothesis (42) to deduce that Hence, it follows from Lemma 3 that, for , we get That is This together with (51) implies that, for , we have
3. Proof of Theorem 1
In order to prove Theorem 1, we need the following Lemma.
Lemma 14. Assume that the function satisfies and put for . Then one has, for ,
Proof. Put and . Then for , we have Let , , , and . Then, using Proposition 13, we obtain by a simple computation that
Proof of Theorem 1. From Lemma 14, there exists such that, for each ,
Put and let In order to use a fixed point theorem, we denote and we define the operator on by For this choice of , we can easily prove that, for , we have and .
Now, we have Since the function is continuous on and by Proposition 10, Corollary 11, and Lemma 14, the function is integrable on , we deduce that the operator is compact from to itself. It follows by the Schauder fixed point theorem that there exists such that . Put . Then and satisfies the equation Since the function is continuous and integrable on , then by Proposition 12, the function is a positive continuous solution of problem (1).
Finally, let us prove that is the unique positive continuous solution satisfying (9). To this aim, we assume that (1) has two positive solutions satisfying (9) and consider the nonempty set and put . Then and we have . It follows that and consequently which implies by Proposition 12 that . By symmetry, we also obtain that . Hence, and . Since , then and consequently .
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
This paper was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. The authors, therefore, acknowledge with thanks DSR technical and financial support.
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