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Existence and Global Asymptotic Behavior of Positive Solutions for Nonlinear Fractional Dirichlet Problems on the Half-Line
We are interested in the following fractional boundary value problem: , , , , where , , is the standard Riemann-Liouville fractional derivative, and is a nonnegative continuous function on satisfying some appropriate assumptions related to Karamata regular variation theory. Using the Schauder fixed point theorem, we prove the existence and the uniqueness of a positive solution. We also give a global behavior of such solution.
Fractional differential equations arise in various fields of science and engineering such as control, porous media, electrochemistry, viscoelasticity, and electromagnetic. They also serve as an excellent tool for the description of hereditary properties of various materials and processes (see [1–3]). In consequence, the subject of fractional differential equations has been gaining much importance and attention. Most of the related results focused on developing the global existence and uniqueness of solutions on finite intervals (see [4–12]) and the references therein). However, to the best of our knowledge, there exist few articles dealing with the existence of solutions to fractional differential equations on the half-line; see, for instance, [13–21]. In , by using the recent Leggett-Williams norm-type theorem due to O’Regan and Zima, the author established the existence of positive solutions for fractional boundary value problems of resonance on infinite intervals. On the other hand, in , Su and Zhang studied the following fractional differential problem on the half-line by using Schauder’s fixed point theorem: where is the standard Riemann-Liouville fractional derivative (see Definition 8 below).
In , by means of the Leray-Schauder alternative theorem, Zhao and Ge proved the existence of solutions to the following boundary value problem: where and .
In this paper, we aim at studying the existence, uniqueness, and the exact asymptotic behavior of a positive solution to the following fractional boundary value problem: where , , and is a nonnegative continuous function on that may be singular at .
To state our result, we need some notations. We first introduce the following Karamata classes.
Definition 1. The class is the set of all Karamata functions defined on by for some and where and such that .
Definition 2. The class is the set of all Karamata functions defined on by where and such that .
It is easy to verify the following.
(i) A function is in if and only if is a positive function in , for some , such that .
(ii) A function is in if and only if is a positive function in such that .
Remark 4 (see ). Let be a function in , and then there exists such that for every and we have
As a typical example of function belonging to the class , we quote where are real numbers, ( times, and is a sufficiently large positive real number such that is defined and positive on , for some .
In the sequel, we denote by the set of nonnegative Borel measurable functions in and by the set of all functions such that is continuous on .
We also denote by the set of continuous functions on such that . It is easy to see that is a Banach space with the uniform norm .
For two nonnegative functions and defined on a set , the notation , means that there exists such that , for all .
Finally, for , we put .
Throughout this paper we assume that the function is nonnegative on and satisfies the following condition:
(H) such that where , , defined on , for some and satisfying In what follows, we put and we define the function on by where, for , and, for , Our main result is the following.
Theorem 5. Let , and assume (H). Then problem (3) has a unique positive solution satisfying, for ,
The content of this paper is organized as follows. In Section 2, we present some properties of the Green function of the operator on with Dirichlet conditions and . Next, we give some fundamental properties of the two Karamata classes and and we establish sharp estimates on some potential functions. In Section 3, exploiting the results of the previous section and using the Schauder fixed point theorem, we prove Theorem 5.
2.1. Fractional Calculus and Green Function
Definition 7. The Riemann-Liouville fractional integral of order of a function is given by provided that the right-hand side is pointwise defined on .
Definition 8. The Riemann-Liouville fractional derivative of order of a function is given by where provided that the right-hand side is pointwise defined on . Here means the integer part of the number .
So we have the following properties (see ).
Proposition 9. (1) Let and let , and then one has
(2) Let , and then where is the smallest integer greater than or equal to and .
Corollary 10. Let and assume that . Then, where is the smallest integer greater than or equal to and .
Lemma 11. Let and . The unique solution of is given by where is Green’s function for the boundary value problem (20).
Proof. We may apply Corollary 10 and Proposition 9 to reduce equation to an equivalent integral equation where . Hence the general solution of is By using and , we get Therefore, the unique solution of problem (20) is The proof is complete.
Next we give sharp estimates on the Green function . To this end, we need the following lemma.
Lemma 12. (i) For and , one has (ii) For one has
Proposition 13. The Green function defined by (22) satisfies
Proof. For we have Since for , then by applying Lemma 12 (i) with and , we obtain
From here on, we define the potential kernel on by
Corollary 14. Let and , and then the function belongs to if and only if the integral converges.
Proposition 15. Let and be a function such that the map is continuous and integrable on . Then is the unique solution in of the boundary value problem
Proof. From Corollary 14, the function is well defined in . Using Proposition 13 and Lemma 12 (ii), we get
This implies that is finite on . So by using Fubini’s theorem, we obtain
Observe that by considering the substitution , we obtain
Using this fact and (22) we deduce that
Now, assume that , and then by (36) we have
On the other hand, if and , we have
So combining (38) and (39), we obtain
This implies that
and , for .
Moreover, using Proposition 13 and the dominated convergence theorem, we deduce that Finally, we need to prove the uniqueness. Let be two solutions of (33) and put . Then and . Hence, it follows from Corollary 10 that . Using the fact that , we deduce that and therefore . The proof is complete.
2.2. Sharp Estimates on the Potential of Some Karamata Functions
We collect in this paragraph some properties of functions belonging to the Karamata class (resp., ) and we give estimates on some potential functions.
Theorem 17 (see [26, 27]). (a) Let and be a function in defined on . One has the following.(i)If , then diverges and .(ii)If , then converges and .
(b) Let and be a function in . One has the following.(i)If , then diverges and .(ii)If , then converges and .
The proof of the next lemma can be found in .
Lemma 18. Let be a function in defined on . Then one has
If further converges, then one has .
In the next lemma, we have the following properties related to the class . For the proof we refer to .
Lemma 19. Let be a function in . Then one has In particular, If further converges, then one has In particular,
Now, we put where and . We aim at giving sharp estimates on the potential function .
Proposition 20. Assume that defined on , for some and . Let and such that Then for where, for , and, for ,
Proof. Using Proposition 13 and Remark 4, we have
Case 1. Assume that .
By using (53), we deduce that On the other hand, Using Theorem 17 and hypothesis (53), we deduce that Hence, it follows by Lemma 18, Proposition 16, and hypothesis (53) that Combining (58) and (61) and using Proposition 16 and hypothesis (53), we deduce that, for , Case 2. Assume that .
By using (53), we deduce that On the other hand, Using again Theorem 17 and hypothesis (53), we deduce that Hence, it follows from Lemma 19 and hypothesis (53) that Combining (63) and (66) and using Proposition 16, hypothesis (53), and Remark 4, we deduce that, for , Now since the functions and are positive and continuous on , we deduce that, for , Finally, using (62), (67), and (68), we obtain the required result.
3. Proof of the Main Result
The next lemma will play a crucial role in the proof of Theorem 5
Lemma 21. Assume that the function satisfies (H) and put for . Then one has for
Proof. We recall that where, for , and, for , For , we have Using Proposition 20 with and and , we obtain for On the other hand, using again Proposition 20 and Remark 4, we get, for ,This completes the proof.
Proof of Theorem 5. From Lemma 21, there exists such that for each
Put and let In order to use a fixed point theorem, we define the operator on by For this choice of and using (76), we easily prove that for all and On the other hand, using Proposition 13 and Lemma 12 (ii), there exists such that for all , we have This implies that there exists such that, for each and , Now by hypothesis (H) and Theorem 17, the function is in , which implies that the family is uniformly bounded.
Using (80) and the fact that, for each , the function is in , we deduce that the family is equicontinuous in .
Hence, it follows by Ascoli’s theorem that is relatively compact in and therefore .
Next, we will prove the continuity of in the supremum norm. Let be a sequence in which converges to in . Using again (80) and Lebesgue’s theorem, we deduce that as , for .
Since is relatively compact in , then the pointwise convergence implies the uniform convergence. Thus we have proved that is a compact mapping from to itself.
Now, the Schauder fixed point theorem implies the existence of such that Put . Then and satisfies the equation Since the function is continuous and integrable on , then by Proposition 15, the function is a positive solution in of problem (3).
Finally, it remains to prove that is the unique positive solution in satisfying (14). To this end, assume that problem (3) has two positive solutions satisfying (14). Then there exists a constant such that This implies that the set is not empty. Let . Then and we have . It follows that and consequently This implies by Proposition 15 that . By symmetry, we obtain also . Hence and . Since , then and therefore .