Abstract
By using fixed point results on cones, we study the existence of solutions for the singular nonlinear fractional boundary value problem
where is an integer, , , , , f is an Caratheodory function, and may be singular at value 0 in one dimension of its space variables x, y, z. Here, ^{c}D stands for the Caputo fractional derivative.
Keywords:
boundary value problem; fixed point; fractional differential equation; Green function; regularization; singular1 Introduction
Fractional differential equations (see, for example, [16] and references therein) started to play an important role in several branches of science and engineering. There are some works about existence of solutions for the nonlinear mixed problems of singular fractional boundary value problem (see, for example, [711] and [12]). Also, there are different methods for solving distinct fractional differential equations (see, for example, [1318] and [19]). By using fixed point results on cones, we focus on the existence of positive solutions for a nonlinear mixed problem of singular fractional boundary value problem. For the convenience of the reader, we present some necessary definitions from fractional calculus theory (see, for example, [20]). The Caputo derivative of fractional order α for a function is defined by
Let . As you know, denotes the space of functions, whose qth powers of modulus are integrable on , equipped with the norm . We consider the sup norm
on the space . Also, is the set of absolutely continuous functions on . Let B be a subset of . A function is called an Caratheodory function whenever the realvalued function on is measurable for all , the function is continuous for almost all , and for each compact set , there exists a function such that for almost all and . Consider the nonlinear fractional boundary value problem
where is an integer, , , , and . We say that the function is a positive solution for the problem whenever on , is a function in , and u satisfies the boundary conditions almost everywhere on . In this paper, we suppose that f is an Caratheodory function on , where , there exists a positive constant m such that for almost all and , f satisfies the estimate
where are positive and nonincreasing, and are positive, w is nondecreasing in all its variables, , , , and . Since we suppose that problem (∗) is singular, that is, may be singular at the value 0 of its space variables x, y, z, we use regularization and sequential techniques for the existence of positive solutions of the problem. In this way, for each natural number n define the function by
It is easy to see that each is an Caratheodory function on , ,
and
for almost all and all . In 2012, Agarwal et al. proved the following result.
Lemma 1.1[7]
Letand. Then we havefor alland
2 Main results
Now, we are ready to investigate the problem in regular and singular cases. First, we give the following result.
Lemma 2.1Let. Then the boundary value problem
is equivalent to the fractional integral equation, where
Proof From and the boundary conditions, we obtain
By properties of the Caputo derivative, we get
By using the boundary conditions and , we get and
Thus,
This completes the proof. □
Put and . It is easy to check that the Green function G in the last result belongs to , for all ,
for all . Consider the Banach space with the norm and the cone
For each natural number n, define the operator on P by
Now, we prove that is a completely continuous operator (see [2]).
Lemma 2.2The operatoris a completely continuous operator.
Proof Let . Then, and . Now, define for almost all . Then and for almost all . By using the properties of fractional integral , it is easy to see that , and
for all . This implies that and on . Consequently, maps P into P. In order to prove that is a continuous operator, let be a convergent sequence in P and . Thus, uniformly on for . Since
we get and uniformly on . Also, we have on , and so . Now, put
Then, it is easy to see that for almost all , and there exists such that for almost all and all . Since is an Caratheodory function, is bounded in , and is bounded in . Therefore, uniformly on . Since is convergent on ,
uniformly on . Hence, is a continuous operator. Now, we have to show that for each bounded sequence in P, the sequence is relatively compact in . Choose a positive constant k such that and for all m. Note that and for all m, where . But we have
and
for all and m. This implies that is bounded in . Also, we have
for all , where . Hence, is equicontinuous on . Thus, is relatively compact in by the ArzelaAscoli theorem. Hence, is a completely continuous operator. □
We need the following result (see [2] and [21]).
Lemma 2.3[21]
LetYbe a Banach space, Pa cone inYandandbounded open balls inYcentered at the origin with. Suppose thatis a completely continuous operator such thatfor allandfor all. ThenThas a fixed point in.
Theorem 2.4For each natural numbern, problem (∗) has a solutionsuch that, andfor all.
Proof Let . It is sufficient to show that has a fixed point in P with the desired conditions. In this way, note that
and so . Put . Then for all . If , then
and
for all and , because w is nondecreasing in all its variables. Since , , and
and
there exists a positive constant L such that
for all . Thus, for all with . Put . Then for all . By using last result, has a fixed point in . But and
for all . This completes the proof. □
Now, we give our last result.
Theorem 2.5Problem (∗) has a solutionusuch that, andfor all.
Proof By using Theorem 2.4, one gets that for each natural number n, problem (∗) has a solution with the desired conditions. Thus, , and for all and n. Also, we have . Suppose that
Then
for almost all and n. Since , we get
and
We show that is bounded on . Let . Note that
and
Thus, for all , where . Also, we have
Since
and
we get for all n, where , and also . On the other hand, there exists a positive constant L such that for all , and so for all n. Thus, for almost all and all n, we have , where
Note that . We show that is equicontinuous on . Let and . Then
and so
Hence, is equicontinuous on . Since is a bounded sequence in , by using the ArzelaAscoli theorem, without loss of generality, we can assume that is convergent in . Let . Then, it is easy to see that , and uniformly converges to on . Thus, converges to in . Hence,
for almost all . Since , by using the dominated convergence theorem on the relation
we get for all . This completes the proof. □
2.1 Examples for the problem
Example 2.1 Let , for almost all t in . Suppose that
on , , whenever and whenever , , , and . Then Theorem 2.5 guarantees that problem (∗) has a positive solution.
Example 2.2 Consider the nonlinear mixed problem of singular fractional boundary value problem
via boundary value conditions , and , where . Let
Then the map f is singular at , and f satisfies the desired conditions, where whenever and whenever , , , , , and . Then Theorem 2.5 guarantees that this problem has a positive solution.
3 Conclusions
One of the most interesting branches is obtaining solutions of singular fractional differential via boundary value problems. Having these things in mind, we study the existence of solutions for a singular nonlinear fractional boundary value problem. Two illustrative examples illustrate the applicability of the proposed method. It seems that the obtained results could be extended to more general functional spaces. Finally, note that all calculations in proofs of the results depend on the definition of the fractional derivative.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors have equal contributions. All authors read and approved the final manuscript.
Acknowledgements
Research of the second and third authors was supported by Azarbaijan Shahid Madani University. Also, the authors express their gratitude to the referees for their helpful suggestions, which improved the final version of this paper.
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