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Solvability for an impulsive fractional multi-point boundary value problem at resonance
Boundary Value Problems volume 2014, Article number: 247 (2014)
Abstract
In this paper, we consider the fractional multi-point boundary value problem with impulse effects. As indicated by M Fečkan, Y Zhou and JR Wang in 2012, the concepts of piecewise continuous solutions used in most of the current literature about the impulsive differential equations of fractional order are not appropriate. Based on a new concept of a piecewise continuous solution, we establish sufficient conditions for the existence of the solutions for the boundary value problem at resonance by the continuation theorem of coincidence degree theory.
MSC: 26A33, 34A08, 34A37.
1 Introduction
In this paper, we are concerned with the existence of the impulsive differential equation of fractional order
where is the Caputo fractional derivative, , is continuous, () are continuous, , , , , , , , , , and , where and denote the right and left limit of at (), respectively. has a similar meaning for . Further, we assume that
and
which implies that the problem (1.1) is at resonance. 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]), and there are a large number of papers dealing with the fractional differential equations (see [4]–[17]) due to their various applications.
Impulsive differential equations have found its importance in realistic mathematical modeling of the phenomena in both the physical and the social sciences. The boundary value problem of impulsive differential equation has been investigated extensively in the literature; see [9], [10], [12], [15], [18], [19] and references therein.
There are some papers considering the fractional impulsive differential equations today. However, in [9], [15], the authors indicated that the concept of piecewise continuous solutions used in most of the current literature about the impulsive differential equations of fractional order are not appropriate.
Motivated by the papers [6], [8], [9], [11], [15], [17], [20], in this paper we deal with the problem (1.1). Based on the new concept of a piecewise continuous solution presented in [9], [15], we establish some existence results about the problem (1.1). As far as we know, there are few papers to deal with fractional differential equations with impulses under resonant conditions.
The rest of the paper is organized as follows. In Section 2, we introduce some notations, definitions, and preliminary facts that will be used in the remainder of the paper. In Section 3, applying the results listed in Section 2, we prove the existence of the solution for the problem (1.1) by the coincidence degree theory. Then an example is given in Section 4 to demonstrate the application of our results.
2 Preliminaries
First of all, we present the necessary definitions and fundamental facts on the fractional calculus theory. These can be found in [1], [3].
Definition 2.1
The Riemann-Liouville fractional integral of order of a function is given by
provided that the right-hand side is pointwise defined on .
Definition 2.2
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 .
Definition 2.3
The Caputo fractional derivative of order of a function is given by
where , , provided that the right-hand side is pointwise defined on .
Lemma 2.1
([16])
Let; then the differential equation
has solutions, , , .
Lemma 2.2
([16])
Let; then
for some, , where.
Lemma 2.3
If, , and, then we have
Now let us recall some notations as regards the coincidence degree continuation theorem.
Let X, Z be real Banach spaces. Consider the 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. 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 . If is an open bounded subset of X, the map N will be called L-compact on if is bounded and is compact. For ImQ is isomorphic to KerL, there exists an isomorphism . Then we shall give the coincidence degree continuation theorem, which is proved in [21].
Theorem 2.1
Let L be a Fredholm operator of index zero and N be L-compact on, where Ω is an open bounded subset of X. Suppose that the following conditions are satisfied:
-
(i)
for each ;
-
(ii)
for each ;
-
(iii)
, where is a continuous projection as above with and is any isomorphism.
Then the equationhas at least one solution in.
Set and , . Then we shall introduce some function spaces in the following:
equipped with the norm
and
equipped with the norm
Obviously, and are Banach spaces.
Define and with the norm
where , . It is easy to verify that is a Banach space.
Let
and
where , , .
Then the multi-point boundary value problem can be written
At the end of the section we give a method for determining the compactness of a set in X.
Lemma 2.4
is a relatively compact set in X if and only if the following conditions are satisfied:
-
(a)
is uniformly bounded, that is, there exists a constant , such that for each , .
-
(b)
, there exists , such that
for, , , .
Proof
Analogous to the proof of the Lemma 2.2 in [17], we can prove the results. Here, we omit the details. □
3 Main results
In this section, we will establish the existence theorem for the impulsive fractional differential equation (1.1). In order to prove our main results, we need the following lemmas.
Lemma 3.1
Suppose thatand. A functionis a solution of the impulsive differential equation of fractional order
if and only if
whereand, and
Proof
In view of Lemma 2.7 in [9] or Lemma 4.1 in [15], we can get the conclusions by applying Lemma 2.2 and Lemma 2.3. □
Lemma 3.2
Assume that and
Then L is a Fredholm mapping of index zero. Moreover,
and
Proof
Taking into account Lemma 2.1 and the definition of L, we can get the result (3.4) easily. Further, Lemma 3.1 implies (3.5). Now, let us focus on the proof that L is a Fredholm mapping of index zero.
Define an auxiliary operator as follows:
where . It is obvious that is a continuous linear mapping.
Take the mapping defined by
where and . Evidently, and
and is a continuous linear projector. In fact, for an arbitrary , we have
and
that is to say, is idempotent.
Let , where z is an arbitrary element in Z. Since and , we obtain . Take . Then can be written as , where , , for . Thus we have , which implies that and . Hence , where we denote θ the zero element in Z. Then we get .
Now, , and noting that ImL is closed in Z, L is a Fredholm mapping of index zero. □
Let be defined by
Clearly, is a linear continuous projector and
Also, proceeding as in the proof of Lemma 3.2, we can show that .
Consider the mapping :
where and .
Note that
and
Thus, , where .
Lemma 3.3
For each, we have
Proof
For each and , , we have
and
Hence,
□
Lemma 3.4
Letand () are continuous. Thenis completely continuous.
Proof
By virtue of Lemma 2.4, we can conclude that the claim of the lemma is true. □
Next, let us list the assumptions that will be used in the rest of the section.
(H1): is continuous, and there exist three nonnegative functions such that for all and , we have
(H2): There exist , , () such that
where , .
(H3):
where .
(H4): There exists a constant such that
for each satisfying .
(H5): There exists a positive constant S such that for any , if , then either
or else
Theorem 3.1
Let, , and () are continuous. Assume that the hypotheses (H1)-(H5) all hold. Then the problem (1.1) has at least one solution in domL.
Proof
The proof consists of four main steps as follows.
-
(1)
Set and proof of is bounded.
-
(2)
Set and proof of is bounded.
-
(3)
If (3.14) holds, we set ; if (3.15) holds, we set , where is a linear isomorphism defined as
(3.16)
Then proof of is bounded.
-
(4)
Let Ω be a bounded open set such that and prove that
Now, let us prove the steps one by one.
Step 1: Take , then and , so and . Hence, , that is, . From (H4), we have .
Again, for , we get
In view of , by (3.8) and Lemma 3.3, we have
Combining (3.17) and (3.18), we can obtain
From (H1), for each , we have
where we denote .
From (H2), for each , we get
and
where .
Combining (3.20), (3.21), and (3.22), we obtain
Thus, by (H3), (3.19), and (3.23), we can derive
which clearly shows that is bounded.
Step 2: Let , then and , , . Since , we have , that is, . Taking into account (H5), , which implies that is bounded.
Step 3: Without loss of generality, in the following part of the proof, we assume that (3.14) holds in (H5). Then we set . For , then we have , , , and . Thus,
Therefore, via (H5) and (3.14), we have , which shows that is bounded.
Step 4: Let Ω be a bounded open set such that . The operator N is L-compact on due to the fact that is bounded and is compact by Lemma 3.4. Then by Step 1 and Step 2, we have
-
(i)
for each ;
-
(ii)
for each .
Define , where I is the identity operator in X. According to the arguments in Step 3, we have
and therefore, via the homotopy property of degree, we obtain
which verifies the condition (iii) of Theorem 2.1. Then, applying Theorem 2.1, we conclude that the problem (1.1) has at least one solution in . The proof is complete. □
Corollary 3.1
If the conditions (H3) and (H4) are replaced by:
():
():There exists a constantsuch that
for eachsatisfying.
Then the problem (1.1) has at least one solution in domL.
Theorem 3.2
Let, (), and () are continuous. Suppose that (H2) and the following conditions are satisfied:
():
(H6):There exists a constant, such that for any, we have
whereand.
(H7):For, , .
(H8):
Then the problem (1.1) has at least one solution in domL.
Proof
In view of () and (H7), we have .
First, let us set . Hence, for each , we have , that is, . We claim that there exists such that , .
In fact, if we assume that , in view of , (H6), and (H8), we can get , which contradicts . Thus, the claim that there exists such that is true.
If , . If , we have
Since
we have
Thus,
Similar to the discussion in Theorem 3.1, we can get
Therefore, is bounded.
Also, , (H6), and (H8) imply that (H5) is satisfied. Then, proceeding with the proof of the theorem similar to the proof of Theorem 3.1, we can derive the conclusion. □
4 Examples
To illustrate our main results, we shall present an example.
Example 4.1
where
and
and , , , , , , , , , , , , , .
From the above conditions, we can find that (H2), (), and (H7) hold. Noting that , , , , for each , we have
where , and , so (H6) is satisfied. Also we can certify that (H8) holds, too.
To sum up the points which we have just indicated, by Theorem 3.2, we can conclude that the problem (4.1) has at least one solution.
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Acknowledgements
This research was supported by the Fundamental Research Funds for the Central Universities (2014QNA52) and Natural Science Foundation of Jiangsu Province of China (BK20140176). The research is also supported by the TianYuan Special Funds of the National Natural Science Foundation of China (Grant No. 11426211).
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Chen, Y., Lv, Z. & Xu, Z. Solvability for an impulsive fractional multi-point boundary value problem at resonance. Bound Value Probl 2014, 247 (2014). https://doi.org/10.1186/s13661-014-0247-7
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DOI: https://doi.org/10.1186/s13661-014-0247-7