Abstract
The paper deals with the secondorder Dirichlet boundary value problem with one statedependent impulse
Proofs of the main results contain a new approach to boundary value problems with statedependent impulses which is based on a transformation to a fixed point problem of an appropriate operator in the space . Sufficient conditions for the existence of solutions to the problem are given here. The presented approach can be extended to more impulses and to other boundary conditions.
MSC: 34B37, 34B15.
Keywords:
impulsive differential equation; statedependent impulses; Dirichlet problem; secondorder ODE1 Introduction
Differential equations involving impulse effects appear as a natural description of observed evolution phenomena of several real world problems. We refer to the monographs [13].
Most papers in the literature on impulsive boundary value problems concern the case with fixed moments of impulsive effects. Papers dealing with statedependent impulses, called also impulses at variable times, focus their attention on initial value problems or periodic problems. Such papers investigate the existence, stability or asymptotic properties of solutions of initial value problems [48] or solvability of autonomous periodic problems [9,10] and nonautonomous ones [1115]. We can also find papers investigating other boundary value problems with statedependent impulses through some initial value problems for multivalued maps [16,17].
In this paper we provide a new approach to boundary value problems with statedependent impulses based on a construction of proper sets and operators and the topological degree arguments. Unlike previous existing results, our approach enables us to find simple existence conditions for data functions and it can be used for other regular (and also singular) problems. We demonstrate it on the secondorder Dirichlet boundary value problem with one statedependent impulse
where we assume
and
Under assumptions (4)(8), we prove the solvability of problem (1)(3). In particular, we transform problem (1)(3) to a fixed point problem for a proper operator in the space . This approach can be also used for other types of boundary conditions and it can be easily extended to more impulses.
Here, we denote by the set of all continuous functions on the interval J, by the set of all functions having continuous derivatives on the interval J and by the set of all Lebesgue integrable functions on J. For a compact interval J, we consider the linear space of functions from or equipped, respectively, with the norms
In this paper we work with the linear space , where , equipped with the norm
It is wellknown that the mentioned normed spaces are Banach spaces. Recall that for , a function satisfies the Carathéodory conditions on (we write ) if
• for each compact set , there exists a function such that for a.e. and each .
We say that is a solution of problem (1)(3), if z is continuous on , there exists unique such that , and have absolutely continuous first derivatives, z satisfies equation (1) for a.e. and fulfills conditions (2), (3).
2 Operators
In this section we assume that (4)(8) are fulfilled. We introduce sets and operators corresponding to problem (1)(3) and prove their properties which are needed for an application of the LeraySchauder degree theory. Let us consider K of (7) and define the set
Lemma 1For each, there exists a uniquesuch that
Proof Let us take an arbitrary . Obviously, the constant is a solution of the equation
i.e., is a root of the function
From (8) it follows , . According to (8) and the definition of , we get
Therefore, σ is strictly decreasing on and hence it has exactly one root in . □
where fulfills (9). The next lemma provides an important result about the continuity of which is fundamental for our approach.
Lemma 2The functionalis continuous on.
Proof Let us consider , for such that in . Let us denote
By Lemma 1, and , where and , respectively. According to (8), we get , for and
We will prove that . Let us take an arbitrary . Since and (cf. (10)), we can find and such that
From (11) it follows the existence of such that
for each . By Lemma 1 and the continuity of , it follows that for . □
Further, consider of (8) and define sets and Ω by
and
Finally, define an operator by , where
and G is the Green function of the problem , , that is,
Lemma 3The operator ℱ is compact on.
Proof First, we will prove the continuity of the operator ℱ. Let us choose , for such that
Let us denote , , , for each . We will prove that in . For each , we get by (13)(15)
and
Since
we get
By (16), there exists a compact set such that , for each and . Consequently, by (4), there exists such that
for a.e. , then due to the Lebesgue dominated convergence theorem, it follows that
as . Since , the absolute continuity of the Lebesgue integral yields
Further, we have for the inequality
for each and the same is true for . The continuity of , and I imply that
as uniformly w.r.t. . Therefore, converges to x in . Similar arguments can be applied to the sequence .
Now we will prove that is relatively compact. The boundedness of implies the existence of and such that for all ,
and
Therefore, by (13), we get
We have proved that the set is bounded in . We now show that the set is equicontinuous on . For a.e. and all , we have
As a result, for each , there exists such that for each satisfying , the inequality
holds for all . Consequently, is relatively compact in by the ArzelàAscoli theorem. □
Lemma 4Letbe a fixed point of ℱ. Then the function
is a solution of problem (1)(3).
Proof Let be such that , that is,
Let us consider the function z defined in (17). Hence, , ,
and by Lemma 1,
In addition, by (17), is a unique point in satisfying (20). Put , . Due to (19) and (20), we get . Further,
Therefore, σ is strictly decreasing on , which yields for . Consequently, is a unique point in satisfying (20).
Further, we get
Therefore,
Finally,
Since
we have
□
3 Main result
Here, using the LeraySchauder degree theory, we prove our main result about the solvability of problem (1)(3). To this end, we will need the following lemma on a priori estimates.
Lemma 5Assume (4)(8). Then for anyand any solutionof the equation
the implication
holds.
Proof Let us choose and let satisfy (21), i.e.,
. Since , it follows that and therefore , , and . There are two possibilities as follows.
Case A. Let . Then and from (15) and (24), it follows
which implies, due to (6) and (8),
Case B. Let . From (24), (14), (6), (5) and (8), it follows
This inequality together with (7) implies
which is a contradiction.
For , the solution of (21) is , and it clearly belongs to Ω. □
Theorem 6Assume (4)(8). Then the operator ℱ has a fixed point in Ω.
Proof According to Lemma 5, the operator is a homotopy. Therefore,
and consequently the equation
has a solution in Ω. This solution is a fixed point of the operator ℱ. □
Theorem 7Assume (4)(8). Then problem (1)(3) has a solutionzsuch that
Proof From Theorem 6 it follows that there exists a fixed point of the operator ℱ. Lemma 4 yields that the function z defined in (17) (with ) is a solution of problem (1)(3). Estimates (25) follow from (17) and from the definitions of Ω and (cf. (12) and (8)). □
Remark 8 Let us note that assumption (7) follows from the condition
4 Examples
In this section we demonstrate that Theorem 7 can be applied to sublinear, linear and superlinear problems.
Example 9 (Sublinear problem)
Let us consider problem (1)(3) with
that is, f and I are sublinear in x. Then assumptions (5) and (6) are valid for
Since
Remark 8 yields that condition (7) is satisfied for any sufficiently large K. In particular, let us put
If we choose , we see that (7) holds. Then by (8), we have
For instance, if we choose and put
we can check that conditions (8) are satisfied in both cases. Therefore, by Theorem 7, the corresponding problem (1)(3) has at least one solution.
Note that (27) shows that γ need not be monotonous.
Example 10 (Linear problem)
Let us consider problem (1)(3) with f and I having the linear behavior in x and put
Then assumptions (5) and (6) are valid for
Since
Theorem 7 can be applied, due to Remark 8, under the additional assumption
If (28) holds, then for any sufficiently large K, condition (7) is satisfied. By (8), we have , and problem (1)(3) has a solution for any γ satisfying (8). Consequently, if γ is given by (26) or (27), problem (1)(3) is solvable.
Example 11 (Superlinear problem) Let us consider problem (1)(3) with f and I superlinear in x. Put, for example,
Then assumptions (5) and (6) are valid for
It holds
By virtue of (7), Theorem 7 can be applied provided there exists such that
Let us search K in the interval . Then and it holds
Consequently, each fulfilling the equation
satisfies (30) as well. Put, for example, , . Then we get that for inequality (30) holds. Consequently, (8) gives and the corresponding problem (1)(3) is solvable for any γ satisfying (8). In particular, γ given by (26) or (27) can be considered in this case as well.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
Both authors contributed equally to the manuscript and read and approved the final manuscript.
Acknowledgements
Dedicated to Jean Mawhin on the occasion of his 70th birthday.
The authors would like to thank the anonymous referees for their valuable comments and suggestions. This work was supported by the grant Matematické modely a struktury, PrF_ 2012_ 017.
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