A class of nonlinear fractional multipoint boundary value problems at resonance is considered in this article. The existence results are obtained by the method of the coincidence degree theory of Mawhin. An example is given to illustrate the results.
Keywords:coincidence degree; fractional differential equation; resonance; multipoint boundary conditions
The subject of fractional calculus has gained considerable popularity during the past decades, due mainly to its frequent appearance in a variety of different areas such as physics, aerodynamics, polymer rheology, etc. (see [1-3]). Many methods have been introduced for solving fractional differential equations (FDEs for short in the remaining), such as the Laplace transform method, the iteration method, the Fourier transform method, etc. (see ).
Recently, there have been many works related to the existence of solutions for multipoint boundary value problems (BVPs for short in the remaining) at nonresonance of FDEs (see [5-11]). Motivated by the above articles and recent studies on FDEs (see [12-19]), we consider the existence of solutions for a nonlinear fractional multipoint BVPs at resonance in this article.
In , Zhang and Bai considered the following fractional three-point boundary value problems at resonance:
where is a natural number; is a real number; and are the standard Riemann-Liouville derivative and integral respectively; is continuous; ; are given constants such that . In their article, they made the operator and got . In , Bai discussed fractional m-point boundary value problems at resonance with the case of .
In 2010, Bai and Jiang studied the fractional differential equation of boundary value problems at resonance with the case of respectively (see [18,19]), and we can see that they obtained the results by the assumption that a specific algebraic expression is not equal to zero; for example,
is referred to as a condition in . We will show that the assumption like above is not necessary.
In this article, we will use the coincidence degree theory to study the existence of solutions for a nonlinear FDEs at resonance which is given by
with boundary conditions
where ; ; ; ; with satisfying Carathéodory conditions; and are the standard Riemann-Liouville fractional derivative and fractional integral respectively.
BVPs (1.1)-(1.2) being at resonance means that the associated linear homogeneous equation with boundary conditions (1.2) has as a nontrivial solution, where , .
We will always suppose that the following conditions hold:
The rest of this article is organized as follows: In Section 2, we give some definitions, lemmas and notations. In Section 3, we establish theorems of existence result for BVPs (1.1)-(1.2). In Section 4, we give an example to illustrate our result.
Definition 2.1 The Riemann-Liouville fractional integral of order of a function is given by
where is the Gamma function, provided the right side is pointwise defined on .
Definition 2.2 The Riemann-Liouville fractional derivative of order of a function is given by
where , provided the right side is pointwise defined on .
Definition 2.3 ()
We say that the map satisfies Carathéodory conditions with respect to if the following conditions are satisfied:
(i) for each , the mapping is Lebesgue measurable;
(ii) for almost every , the mapping is continuous on ;
(iii) for each , there exists a such that, for a.e. and every , we have .
Lemma 2.4 ()
Assume , , then . And, for all , , we have that
Lemma 2.5 ()
Let , and assume that , then the following equality holds almost everywhere on ,
Now, we briefly recall some notations and an abstract existence result, which can be found in . Let YZ be real Banach spaces, be a Fredholm map of index zero, and be continuous projectors such that
It follows that is invertible. We denote the inverse by . If Ω is an open bounded subset of Y such that , the map will be called L-compact on Ω if is bounded and is compact.
Lemma 2.6 ()
LetLbe a Fredholm operator of index zero andNbeL-compact on . The equation has at least one solution in if the following conditions are satisfied:
(i) for each ;
(ii) for each ;
where is a projection such that and is a any isomorphism.
In this article, we use the Banach space with the norm .
Lemma 2.7 ()
Given and , for any , ( ), we can define a Banach space
with the norm defined by .
Lemma 2.8 ()
is a sequentially compact set if and only ifEis uniformly bounded and equicontinuous. Here, a uniform bound means that there exists a constant with each , such that
and equicontinuation means that there exists a with for any , and , such that
In this article, let with the norm and with the norm . Define the operator by
where . Define the operator by
Thus, BVP (1.1) can be written as for each .
3 Main results
First, let us introduce the following notations for convenience, with setting and with ,
Then, let us make some assumptions which will be used throughout the article.
(H1) There exist functions and a constant such that for all , ,
(H2) For any , , there exists a constant such that if , then either
(H3) For any , , there exists a constant such that if , then either
Theorem 3.1If conditions (C), (H1)-(H3) hold, then BVPs (1.1)-(1.2) have at least one solution provided that .
In order to obtain our main result, we first present and prove Lemmas 3.2-3.8. Now, let us define operators ( ) as follows:
Lemma 3.2If condition (C) holds andLis defined by (2.1), then
Proof By (2.1) and Lemma 2.5, has a solution
Combining with the condition (1.2), we get .
Suppose , then there exists such that , i.e., , . By Lemma 2.5, we have
Then in view of condition (C), (1.2) and Lemma 2.4, x satisfies
On the other hand, suppose and it satisfies (3.1), let , then , , i.e., . Therefore, we obtain that
Lemma 3.3If condition (C) holds, then there exist two constants and with such that .
Proof From , we obtain that for any nonnegative integer l, there exists such that . If else, we obtain that , , .
If , we have
It is equal to
Since the determinant of coefficients is not equal to zero, we have that ( ), which is a contradiction to condition (C).
If , we get
Similarly, we can deduce that the determinant of coefficients is not equal to zero, so we have that ( ), which is a contradiction to condition (C). Thus, there exists such that .
Similarly, from , we have that there exists a constant such that
we shall prove that S is a finite set. If else, there exists a strict increasing sequence such that
Since , we have . Thus,
which is a contradiction to (3.2). Therefore, there exists two constants and with such that . □
Lemma 3.4If the condition (C) holds andLis defined by (2.1), thenLis a Fredholm operator of index zero. Define the linear operator with , then it is the inverse ofL. Furthermore, we have
Proof For each and with , define operator by
It is clear that . It follows from (3.4), the definition of and that
similarly, we can derive that
Hence, for each and , it follows from the (3.3)-(3.6) that
Furthermore, Q is a continuous linear projector.
For each , we have , i.e., . On the other hand, for each , we have that
However, the determinant of coefficients is as follows
then we have ( ), i.e., . Thus, .
Take any in the type , obviously, and , so . For any with , by Lemma 3.2, we have
but the determinant of coefficients is as follows
we can deduce that . Hence, . Furthermore, we get . Therefore, , which means that L is a Fredholm operator of index zero.
Let operator and
It is easy to calculate that ; furthermore, P is a continuous linear projector. Obviously
It is clear that .
For any , in view of the definition of operators Kp and L, we have . On the other hand, if , we have , . Therefore, by Lemma 2.5 and definitions of operators and L, we know that , which implies that . By the definition of , we have
It follows from Lemma 2.4 that
Then, we have
By the definition of the norm in space Y, we get . □
Lemma 3.5Assume is an open bounded subset such that , andNis defined by (2.2), thenNisL-compact on .
Proof In order to prove N is L-compact, we only need to prove that is bounded and is compact. Since the function f satisfies Carathéodory conditions and , for each , there exists a such that, for a.e. and every , we have . By the definition of operators Q and on the interval , it is easy to get that and are bounded. Thus, there exists a constant with each , such that .
For all , , , we have
Since is uniformly continuous on and , so and are equicontinuous. By Lemma 2.8, we get that is completely continuous. □
Lemma 3.6Suppose (H1)-(H3) hold, then the set is bounded.
Proof Taking any , then we have , which yields and , i.e., for all . It follows from (H2) and (H3) that there exists such that . Then we can get that
Furthermore, we have that, with setting ,
By (3.7)-(3.9) and Lemma 2.4, we have that
As before, for any , we have and . From Lemma 3.4 and for each , we can get
Furthermore, we have
By (H1) and the definition of N, we have
where . Since and hold true, we can get that
which yield that
Furthermore, from the previous inequalities, we know that
Since , there exist constants such that
Therefore, is bounded. □
Lemma 3.7Suppose (H2) and (H3) hold, then the set is bounded.
Proof For any and , then and . By (H2), we get that , then we have . By (H3), we have that , thus . Therefore, is bounded. □
Lemma 3.8If the first parts of (H2) and (H3) hold, then the set is bounded.
Proof Taking any and , we have . For all , we define the isomorphism by
By the definition of the set , we can get that
If , we have
By the first parts of (H2) and (H3), similar to the proof of Lemma 3.7, then
Therefore, is bounded.
If , we have .
If , we get that and , similar to the proof of Lemma 3.7, is bounded. If else, we have that and . It contradicts (3.10), thus is bounded. □
Remark 3.9 If the other parts of (H2) and (H3) hold, then the set is bounded.
Now with Lemmas 3.2-3.8 in hands, we can begin to prove our main result - Theorem 3.1.
Proof of Theorem 3.1 Assume that Ω is a bounded open set of Y with . By Lemma 3.5, N is L-compact on . Then by Lemmas 3.6 and 3.7, we have
(i) for every ;
(ii) for every .
Finally, we will prove that (iii) of Lemma 2.6 is satisfied. We let I as the identity operator in the Banach space Y and , according to Lemma 3.8 (or Remark 3.9) we know that for all , . By the homotopic property of degree, we have
so (iii) of Lemma 2.6 is satisfied.
Consequently, by Lemma 2.6, the equation has at least one solution in . Namely, BVPs (1.1)-(1.2) have at least one solution in the space Y. □
According to Theorem 3.1, we have the following corollary.
Corollary 3.10Suppose that (H1) is replaced by the following condition,
(H4) there exist functions and a constant such that for all , ,
and the others in Theorem 3.1 are not changed, then BVPs (1.1)-(1.2) have at least one solution.
4 An example
Example Consider the following boundary value problem for all :
Let , and , , , , , , , . We can get that the condition (C) holds, i.e., , , . Moreover,
Thus, we have
Taking , , , , , , , , , , , , we can calculate that (H1)-(H3) hold. Furthermore, we can get
By Corollary 3.10, the BVP (4.1) has at least one solution in .
The authors declare that they have no competing interests.
NX designed all the steps of proof in this research and also wrote the article. WBL suggested many good ideas in this article. LSX helped to draft the first manuscript and gave an example to illustrate our result. All authors read and approved the final manuscript.
The authors would like to acknowledge the anonymous referee for many helpful comments and valuable suggestions on this article. This work is sponsored by Fundamental Research Funds for the Central Universities (2012LWB44).
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