Abstract
We establish novel results on the existence of impulsive problems for fractional differential equations with functional boundary value conditions at resonance with . Our results are based on the degree theory due to Mawhin, which requires appropriate Banach spaces and suitable projection schemes.
1. Introduction
Impulsive differential equations serve as basic models to study the dynamics of processes that are subject to sudden changes in their states. Recent development in this field has been motivated by many applied problems arising in control theory, population dynamics, and medicines. For some recent works on the theory of impulsive differential equations, we refer the reader to [1–3].
Fractional differential equations have proved valuable and effective tools in the modeling of many phenomena in various fields of biology, medicine, mechanics, engineering, viscoelasticity, and so forth; see [4, 5]. In order to improve the mathematical modeling of several concepts arising in these areas, many researchers have paid a considerable attention to the subject of impulsive fractional differential equations in the recent literature [6–13].
A boundary value problem is said to be at resonance if the corresponding homogeneous boundary value problem has a nontrivial solution. In [14, 15], the authors studied the existence of solutions for functional boundary value problems with a linear differential operator by using Mawhin continuation theorem [16]. In [15], N.Kosmatov and Jiang extended the results of [14] to new resonance scenarios.
To the best of the author knowledge, the solvability of the functional boundary value problems for impulsive fractional differential equation at resonance has not been well studied till now. We will fill this gap in the literature. Inspired by the above excellent results [14–24], in this paper, we investigate the following boundary value problem for impulsive fractional equations at resonance:where , are continuous functions with and are continuous linear functionals with the resonance condition: . and , with , , , , .
2. Preliminaries
For the convenience of the reader, we introduce the theoretical foundations of the method.
Definition 1 (see [4, 5]). The Riemann-Liouville fractional integrals of order of a function is given by where the right side is pointwise defined on .
Definition 2 (see [4, 5]). The Caputo fractional derivatives of order of a function is given bywhere denotes the integer part of number , and this derivative is called the right side which is pointwise defined on .
Lemma 3 (see [4, 5]). Let . If or , then the fractional differential equation (or ) has solutionwhere , and .
Lemma 4 (see [4, 5]). (1) Let ; if or , then one haswhere , and ;
(2) The equality holds for every and .
Definition 5 (see [16]). Let , be real Banach spaces and let be a linear operator. is said to be the Fredholm operator of index zero provided that (i) is a closed subset of ;(ii).
Let , be continuous projectors such that , , , and . It follows that is reversible. We denote the inverse of the mapping by (generalized inverse operator of ). If is an open bounded subset of such that , the mapping will be called on , if and are continuous and compact.
Theorem 6 (see [16] Mawhin continuation theorem). Let be a Fredholm operator of index zero and is -compact on . Assume that the following conditions are satisfied:(i) for every ;(ii) for every ;(iii), where is a continuous projection such that . Then the equation has at least one solution in .
In the following, we use the classical Banach space
, exist with with .
with the norm .
with the norm .
Let endowed with the norm . Then and are Banach spaces.
Define the operators as follows: where For convenience, let . So problem (1) becomes
3. Main Results
In order to obtain the main results, we will always suppose the following condition holds.
The functionals are linear continuous with respective norms , that is, and satisfy , where .
There must exist a function such thatand , where is introduced in .
Let satisfy the following conditions:(1) is measurable for each fixed ; is continuous for a.e..(2), and there must exist such that .
In addition,
Remark 7. We show that there must be a satisfying .
For convenience, set , and Then, there must exist such that . If not, . That is . By , . Thus, for every polynomial . Since is dense in X, we can get . That is a contradiction (since and are linearly independent on ).
There must be some such that . For these , take , we have and , where is beta function. Obviously, . Thus, there exists satisfying .
Lemma 8. Assume that holds; then is a Fredholm operator of index zero. Moreover,and
Proof. It is easy to see thatFor , if , thenFrom (12) and Lemma 4, we havewhere , are two arbitrary constants. Substituting the boundary condition into (13), one hasBased on the condition , we haveConversely, if (15) holds, settingthen it is easy to check that is a solution of (12) and satisfies .
Combining the above arguments, we obtain (10).
Define as follows:where is introduced in .
Hence,by and , which implies the operator is a projector.
Now, It is obvious that and . Moreover, it follows from that . Thus, and . Observe that is a closed subspace of and is a Fredholm operator of index zero.
Take as follows:It is easy to check that , and it is also elementary to confirm the identity . So, .
Lemma 9. The mapping defined byis the inverse of .
Proof. Obviously, for all . For , it is clear that
For each , noting that , where , we get From , we obtain , i.e., .
Thus,The next lemma provides norm-estimates needed for the main results.
Lemma 10. For ,where and .
Proof. It is easy to see that so thatBy representation of and observing that due to , similarly, Finally, (23) follows from above arguments.
Lemma 11. Assume hold, is an open bounded set, and For any , there exist a function and two positive constants such that , Then is compact on .
Proof. Since is bounded, there exists a constant such that .
Denote ,Since hence is bounded.
For , we have This, together with and means thatSince wherethen is uniformly bounded.
For each Then andFor , by the mean value theorem of differentiation, we have and absolute continuity of integral. Thus, we obtain that is equicontinuous. By the Arzela-Ascoli theorem, is completely continuous. The proof is completed.
Theorem 12. Assume and the following conditions hold: There exists a constant such that if for , thenThere exists a constant such that either or if , where .For any , there exist nonnegative functions , nonnegative constants and positive constants such that where Then the functional boundary value problem (1) has at least one solution in X.
In order to prove Theorem 12, we show three lemmas.
Lemma 13. The set is bounded if and are satisfied.
Proof. For , , i.e.,By , there exists a constant such that
It follows from thatand, subsequently,which impliessinceLet , where and . Then since , . As in the proof of Lemma 10,and as in (23),where , . Now, , so and Recall that , where is introduced for the sake of brevity. Hencethat is, Thus,for . By (49), (53), and ,And, then, , by , i.e., is bounded.
Lemma 14. The set is bounded if and hold.
Proof. Let , and then and . By , we have , so, ; that is, is bounded.
Define by . Obviously, . If , then , where is introduced in , and Thus, is an isomorphism.
Lemma 15. The set is bounded if conditions and are satisfied, where is defined above and
Proof. Suppose that , we have , and If , we have By Lemma 14, . If , then . If , we suppose , and then Hence, which contradicts . So is bounded.
Proof of Theorem 12. Set as a bounded open subset of such that By Lemma 11, we know that is L-compact on Then according to the above arguments, we have(i), for every ;(ii), for every ;At last we will prove that (iii) of Theorem 6 is satisfied. If is the identity map, we define the homotopy . If for every , by the homotopic property of degree, we have The assumption (iii) of Theorem 6 is verified and the proof is completed.
Then by Theorem 6, the functional boundary value problem (1) has at least one solution in X. The proof of Theorem 12 is also completed.
In the next result the inequality is replaced with , which will lead to slight modifications of the proof of Lemma 13, we recall that .
Theorem 16. Assume that , and the following conditions hold.There exists a constant such that if , thenFor any , there exist nonnegative functions , nonnegative constants , and positive constants such that where Then the boundary value problem (1) has at least one solution.
Proof. As in the proof of Lemma 13, by , there exists a constant such that Since , where and , dimilarly, . As in the proof of Lemma 13,where Again, (49) holds. Now, , so, by (63),As in the proof of Lemma 13, .
Hence Since , we haveThus,for .
By (63) and (67),where , by . Then, i.e., is bounded. The rest of the proof repeats that of Theorem 12, which completes the proof of Theorem 16.
Data Availability
No data were used to support this study.
Conflicts of Interest
The authors declare that they have no competing interests.
Authors’ Contributions
All results belong to B. Sun, W. Jiang, and S. Zhang.
Acknowledgments
This work was supported by the Natural Science Foundation of Hebei Province (A2018208171) and the Natural Science Foundation of China (11671181).