Abstract
A class of nonlinear multipoint boundary value problems for singular fractional differential equations is considered. By means of a coupled fixed point theorem on ordered sets, some results on the existence and uniqueness of positive solutions are obtained.
1. Introduction
Fractional calculus is used to formulate different phenomena in physics, biology, medicine, and so forth. For more details on the applications of fractional calculus, we refer the reader to [1–4]. On the other hand, some basic theory for the initial value problems of fractional differential equations involving Riemann-Liouville differential operator has been discussed by Lakshmikantham [5–7], Bai and Lü [8], El-Sayed et al. [9, 10], Bai [11, 12], Zhang [13], and so forth.
Very recently, Liang and Zhang [14] considered the following m-point boundary value problem: where is the Riemann-Liouville fractional derivative, satisfies , and is continuous and nondecreasing with respect to the second variable. Under some hypotheses, using a fixed point theorem on ordered sets, the authors established the existence and uniqueness of solution to such problem.
Motivated by the abovementioned work, in this paper, we deal with the following multipoint boundary value problem: where is continuous and ( is singular at ).
2. Preliminaries
The following preliminaries will be useful later.
Definition 1. The Riemann-Liouville fractional derivative of order of a continuous function is given by where , denotes the integer part of number , provided that the right side is pointwise defined on . Here, is the Euler gamma function defined by
Definition 2. The Riemann-Liouville fractional integral of order of a given function is defined by provided that the right side is defined on .
From the definition of the Riemann-Liouville derivative, we can obtain the following statement.
Lemma 3 (see [15]). Let . If one assumes that , then the fractional differential equation has , , as unique solutions, where is the smallest integer greater than or equal to .
Lemma 4 (see [15]). Assume that with a fractional derivative of order that belongs to . Then for some , , where is the smallest integer greater than or equal to .
Lemma 5 (see [14]). Assume that . If , then the boundary value problem has a unique solution where
It is clear that, for all , the function . For , we have ; then . Hence for one has . The following properties of will be used later.
Lemma 6 (see [14]). The following properties hold.(i)is a nonnegative continuous function on ;(ii) is strictly increasing for all .
Let be a partially ordered set endowed with a metric . Let be a given mapping.
Definition 7. One says that is directed if, for every , there exists such that , and there exists such that , .
Definition 8. We say that is regular if the following conditions hold. )if is a nondecreasing sequence in such that , then for all ;() if is a decreasing sequence in such that , then for all .
Example 9. Let , be the set of real continuous functions on . We endow with the standard metric given by We define the partial order on by Let . For , that is, , for all , we have and . For , that is, , for all , we have and . This implies that is directed. Now, let be a nondecreasing sequence in such that as , for some . Then, for all , is a nondecreasing sequence of real numbers converging to . Thus, we have , for all , that is, for all . Similarly, if is a decreasing sequence in such that as , for some , we get that for all . Then, we proved that is regular.
Definition 10 (see [16]). An element is called a coupled fixed point of if
Definition 11 (see [16]). One says that has the mixed monotone property if
Denote by the set of functions satisfying the following. is continuous; is nondecreasing;.
The following two lemmas are fundamental for the proof of our main result.
Lemma 12 (see [17]). Let be a partially ordered set and suppose that there exists a metric on such that is a complete metric space. Let be a mapping having the mixed monotone property on such that for all with and , where . Suppose also that is regular and there exist such that Then, has a coupled fixed point . Moreover, if and are the sequences in defined by then
Lemma 13 (see [17]). Adding to the hypotheses of Lemma 12 the condition is directed; one obtains uniqueness of the coupled fixed point. Moreover, one has the equality .
3. Main Result
Let Banach space be endowed with the norm . We define the partial order on by In Example 9, we proved that with the classic metric given by satisfies the following properties: is directed and is regular.
Define the closed cone by where denotes the zero function.
Definition 14. One says that is a coupled lower and upper solution to (2)-(3) if, for all , one has
The main result of this paper is the following.
Theorem 15. Suppose that the following conditions hold. (i) satisfies with for ;(ii) is continuous, ;(iii)there exists such that is continuous on for all ;(iv)there exists
such that for all with , and ,
where is nondecreasing, ;(v)equations (2)-(3) has a coupled lower and upper solution .
Then, (I)the boundary value problem (2)-(3) has a unique positive solution ;(II)the sequences and defined by
converge uniformly to .
Proof. Suppose that is a solution to the boundary value problem (2)-(3). Then, from Lemma 5, we have
for all .
Consider the operator defined by
for all , for all . From (iii) and Lemma 6, we have that .
Let such that and . From (25), we have
Since the operator is linear and increasing with respect to the function , we deduce that
This implies that has the mixed monotone property with respect to the partial order given by (20).
In the sequel, we denote
Let such that and . For all , using (25) and Lemma 6, we have
Thus, for all such that and , we have
where
Now, let . We have
where denotes the beta function. Recall that beta and gamma functions satisfy the following properties:
Thus we have
Using the same computation as above, we can show that, for all , we have
Now, (37) and (38) give us that
where
But the maximum of the above function depending in is attained at . Then
Now, it follows from (33), (41), and condition (iv) that
with and .
Finally, taking , we have from condition (v) that and .
Now, from Lemmas 12 and 13, there exists a unique such that ; that is, is the unique positive solution to (2)-(3). The convergence of the sequences and to follows immediately from (19). This makes end to the proof.
Example 16. Consider the fractional boundary value problem
In this case, we have
Note that is continuous on and . Let and . For all with , and , we have
On the other hand, we have
Now, since , , and are nonnegative continuous functions, it is easy to prove that is a lower solution to (43)-(44). Moreover, for all , we use (37), (38), and (41) to obtain
Then, an upper solution to (43)-(44) will be any constant satisfying
Finally, from Theorem 15, Problems (43) and (44) have a unique positive solution.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Authors’ Contributions
All authors contributed equally and significantly to writing this paper. All authors read and approved the final manuscript.
Acknowledgments
This project was supported by King Saud University, Deanship of Scientific Research, College of Science Research Center.