This article deals with the differential equations of fractional order on the half-line. By the recent Leggett-Williams norm-type theorem due to O’Regan and Zima, we present some new results on the existence of positive solutions for the fractional boundary value problems at resonance on unbounded domains.
MSC: 26A33, 34A08, 34A34.
Keywords:fractional order; half-line; coincidence degree; at resonance
In this article, we are concerned with the fractional differential equation
where is the Riemann-Liouville fractional derivative, , and satisfies the following condition: (H) = is continuous and for each , there exists satisfying and , such that
The problem (1.1) happens to be at resonance in the sense that the kernel of the linear operator is not less than one-dimensional under the boundary value conditions.
Fractional calculus is a generalization of the ordinary differentiation and integration. It has played a significant role in science, engineering, economy, and other fields. Some books on fractional calculus and fractional differential equations have appeared recently (see [1-3]); furthermore, today there is a large number of articles dealing with the fractional differential equations (see [4-15]) due to their various applications.
In , the researchers dealt with the existence of solutions for boundary value problems of fractional order of the form
where and is continuous. The results are based on the fixed point theorem of Schauder combined with the diagonalization method.
In , Su and Zhang studied the following fractional differential equations on the half-line using Schauder’s fixed point theorem
Employing the Leray-Schauder alternative theorem, in , Zhao and Ge considered the fractional boundary value problem
However, the articles on the existence of solutions of fractional differential equations on the half-line are still few, and most of them deal with the problems under nonresonance conditions. And as far as we know, recent articles, such as [4,6,7], investigating resonant problems are on the finite interval.
Motivated by the articles [16-20], in this article we study the differential equations (1.1) under resonance conditions on the unbounded domains. Moreover, we have successfully established the existence theorem by the recent Leggett-Williams norm-type theorem due to O’Regan and Zima. To our best knowledge, there is no article dealing with the resonant problems of fractional order on unbounded domains by the theorem.
The rest of the article is organized as follows. In Section 2, we give the definitions of the fractional integral and fractional derivative, some results about fractional differential equations, and the abstract existence theorem. In Section 3, we obtain the existence result of the solution for the problem (1.1) by the recent Leggett-Williams norm-type theorem. Then, an example is given in Section 4 to demonstrate the application of our result.
First of all, we present some fundamental facts on the fractional calculus theory which we will use in the next section.
The Riemann-Liouville fractional integral of order of a function is given by
provided that the right-hand side is pointwise defined on .
The Riemann-Liouville fractional derivative of order of a continuous function is given by
where , provided that the right-hand side is pointwise defined on .
Assume that . If , then
Lemma 2.2 ()
Assume that , . Then we have
for some , , whereNis the smallest integer greater than or equal toν.
Let X, Z be real Banach spaces. Consider an operation equation
where is a linear operator, is a nonlinear operator. If and ImL is closed in Z, then L is called a Fredholm mapping of index zero. And if L is a Fredholm mapping of index zero, there exist linear continuous projectors and such that , and , . Then it follows that is invertible. We denote the inverse of this map by . For ImQ is isomorphic to KerL, there exists an isomorphism .
It is known that the coincidence equation is equivalent to
A nonempty convex closed set is called a cone if
(i) for all and ;
(ii) implies .
Note that C induces a partial order ⪯ in X by
The following lemma is valid for every cone in a Banach space.
LetCbe a cone in the Banach spaceX. Then for every , there exists a positive number such that
for all .
Let be a retraction, i.e., a continuous mapping such that for all . Denote
LetCbe a cone inXand let , be open bounded subsets ofXwith and . Assume that: 1∘ = Lis a Fredholm operator of index zero;; 2∘ = is continuous and bounded and is compact on every bounded subset ofX;; 3∘ = for all and ;; 4∘ = γmaps subsets of into bounded subsets ofC;; 5∘ = , where stands for the Brouwer degree;; 6∘ = there exists such that for , where and is such that for every ;; 7∘ = ;; 8∘ = ..Then the equation has a solution in the set .
with the norm
equipped with the norm
Remark 2.1 It is easy for us to prove that and are Banach spaces.
Then the multi-point boundary value problem (1.1) can be written by
Definition 2.3 is called a solution of the problem (1.1) if and u satisfied Equation (1.1).
Lemma 2.4 is a relatively compact set inXif and only if the following conditions are satisfied:
(a) is uniformly bounded, that is, there exists a constant such that for each , .
(b) The functions from are equicontinuous on any compact subinterval of , that is, let J be a compact subinterval of , then , there exists such that for , ,
(c) The functions from are equiconvergent, that is, given , there exists such that
for , .
3 Main results
In this section, we will present the existence theorem for the fractional differential equation on the half-line. In order to prove our main result, we need the following lemmas.
Lemma 3.1Let . Then is the solution of the following fractional differential equation:
Proof In view of Lemmas 2.1 and 2.2, we can certify the conclusion easily, so we omit the details here. □
Lemma 3.2The operatorLis a Fredholm mapping of index zero. Moreover,
Proof It is obvious that Lemma 3.1 implies (3.1) and (3.2). Now, let us focus our minds on proving that L is a Fredholm mapping of index zero.
where . Evidently, , , and is a continuous linear projector. In fact, for an arbitrary , we have
that is to say, is idempotent.
Let , where is an arbitrary element. Since and , we obtain that . Take , then can be written as , , for . Since , by (3.2), we get that , which implies that , and then . Therefore, , thus, .
Now, , and observing that ImL is closed in Z, so L is a Fredholm mapping of index zero. □
Let be defined by
It is clear that is a linear continuous projector and
Also, proceeding with the proof of Lemma 3.2, we can show that .
Consider the mapping
Thus, , where .
Define the linear isomorphism as
Thus, is given by
Then, it is easy to verify that
Now, we state the main result on the existence of the positive solutions to the problem (1.1) in the following.
Theorem 3.1Let satisfy the condition (H). Assume that there exist six nonnegative functions ( ), ( ) and such that
where , , and is defined by (3.12), is bounded on , , , ,
Then the problem (1.1) has at least one positive solution in domL.
Proof For the simplicity of notation, we denote
Consider the cone
where , . Clearly, and are an open bounded set of X.
Step 1: In view of Lemma 3.2, the condition 1∘ of Theorem 2.1 is fulfilled.
Step 2: By virtue of Lemma 2.4, we can get that is continuous and bounded and is compact on every bounded subset of X, which ensures that the assumption 2∘ of Theorem 2.1 holds.
Step 3: Suppose that there exist and such that .
From (3.7) and (3.8), we get that
On account of the fact that
and considering (3.14) and (3.15), we have
By (3.9), (3.10) and (3.13), we obtain that
which is a contradiction to . Therefore, 3∘ is satisfied.
Step 4: Let , then we can verify that is a retraction and 4∘ holds.
Step 5: Let , then . Inspired by Aijun and Wang , we set
where and .
Define homeomorphism by , then
It is obvious that implies that by (3.8) and (3.11).
Take , then . Suppose that , , then we have that . Also, in view of (3.8),
It is a contradiction. Besides, if , then , which is impossible. Hence, for , , .
which shows that 5∘ is true.
Step 6: Let , then we have
And we can take .
Let such that
For , we have that
Therefore, combining (3.6), (3.8) and (3.11), we get that
Thus, for all . So, 6∘ holds.
Step 7: For , from (3.8) and (3.11), we have
which implies that . Hence, 7∘ holds.
Step 8: For , by (3.6), (3.8) and (3.11), we obtain that
Thus, , that is, 8∘ is satisfied.
Hence, applying Theorem 2.1, the problem (1.1) has a positive solution in the set . □
To illustrate our main result, we will present an example.
where , and for ,
It is easy for us to certify that f satisfies the condition (H).
for , where
Evidently, satisfies (3.11).
Meanwhile, by simple computation we can get that
Thus, to sum up the points which we have just indicated, by Theorem 3.1, we can conclude that the problem (4.1) has at least one positive solution.
The authors declare that they have no competing interests.
All the authors typed, read, and approved the final manuscript.
This project is supported by the Hunan Provincial Innovation Foundation For Postgraduate (NO. CX2011B079) and the National Natural Science Foundation of China (NO. 11171351).
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