Abstract
The paper investigates a fixed point problem in the space which is connected to boundary value problems with state-dependent impulses of the form , a.e. , , . Here, the impulse instants are determined as solutions of the equations , . We assume that , , the vector function f satisfies the Carathéodory conditions on , the impulse functions , , are continuous on , and the barrier functions , , are continuous on . The operator ℓ is an arbitrary linear and bounded operator on the space of left-continuous regulated on vector valued functions and is represented by the Kurzweil-Stieltjes integral. Provided the data functions f and are bounded, transversality conditions which guarantee that this fixed point problem is solvable are presented. As a result it is possible to realize the construction of a solution of the above impulsive problem.
MSC: 34B37, 34B10, 34B15.
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1 Introduction
In the literature most of impulsive boundary value problems deals with impulses at fixed times. This is the case that moments, where impulses act in state variables, are known (cf. Section 2). The theory of these impulsive problems is widely developed and presents direct analogies with methods and results for problems without impulses. Important texts in this area are [1]–[6].
A different situation arises, when impulse moments satisfy a predetermined relation between state and time variables, see e.g.[7]–[12]. This case, which is represented by state-dependent impulses, is studied here, where we are interested in a system of n () nonlinear ordinary differential equations of the first order with state-dependent impulses and general linear boundary conditions on the interval . The main reason that boundary value problems with state-dependent impulses are developed significantly less than those with impulses at fixed moments is that new difficulties with an operator representation of the problem appear when examining state-dependent impulses (cf. Section 4). Therefore almost all existence results for boundary value problems with state-dependent impulses have been reached for periodic problems which can be transformed to fixed point problems of corresponding Poincaré maps in . Hence, the difficulties with the construction of a functional space and an operator have been cleared in the periodic case. See e.g.[13]–[16]. Other types of boundary value problems with state-dependent impulses have been studied very rarely, see [17], [18].
In this paper we construct and investigate a fixed point problem in some subset of the product space and we provide conditions for its solvability (cf. Section 4 and Theorem 14). The existence of such fixed point allows us to construct a solution of the system of differential equations
subject to the state-dependent impulse conditions
and the general linear boundary condition
For nonzero impulse functions , , this solution is discontinuous on and, since discontinuity points , , are not fixed and depend on the solution through (2), such a solution has to be searched in the space ; see the notation below. Note that conditions which guarantee the solvability of problem (1)-(3) have not been known before. Some results for special cases of problem (1)-(3) can be found in our previous papers [19]–[24].
In what follows we use this notation. Let . By we denote the set of all matrices of the type with real valued coefficients equipped with the matrix norm
Let denote the transpose of . Let be the set of all n-dimensional column vectors , where , , and . The (vector) norm of is a special case of the norm of , i.e. it has the form
It is well known that
By , (with ), we denote the set of all mappings , , with continuous components, respectively. By , , , , , we denote the sets of all mappings whose components are, respectively, essentially bounded functions, Lebesgue integrable functions, left-continuous regulated functions, continuous functions and functions with bounded variation on the interval . Let us note that the norm in the linear space is taken as
especially, in
We will make use of the Sobolev space , which is the linear space of vector functions, whose components are absolutely continuous having essentially bounded first derivatives on , equipped with the norm
By we denote the set of all mappings satisfying the Carathéodory conditions on the set . Finally, by we denote the characteristic function of the set .
Note that a mapping is left-continuous regulated on if for each and each
is a linear space and equipped with the sup-norm it is a Banach space (see [25], Theorem 3.6). In particular, we set
A mapping satisfies the Carathéodory conditions on if
is measurable for all ,
is continuous for a.e. ,
for each compact set there exists a function such that for a.e. and each .
Throughout we assume that
Now let us define a solution of problem (1)-(3).
Definition 1
A mapping is a solution of problem (1)-(3) if for each there exists a unique such that
, the restrictions are absolutely continuous, z satisfies (1) for a.e. and fulfills conditions (2) and (3).
2 Problem with impulses at fixed times
In this section we summarize results from the paper [23], where we investigated boundary value problems having impulses at fixed times. This is the case that the barrier functions in (2) are constant functions, i.e. there exist satisfying such that
and each solution of the problem crosses i th barrier at the same time instant for .
In [23], the following boundary value problem was investigated:
where
In order to get an operator representation of this problem (cf. Theorem 4) the Green’s matrix is constructed.
Definition 2
([23], Definition 7)
A mapping is the Green’s matrix of the problem
if
-
(a)
is continuous on and on for each ,
-
(b)
for each ,
-
(c)
for any the mapping
is a unique solution of the problem
Lemma 3
([23], Lemma 8)
Assume (8). Problem (10) has a unique solution if and only if
where Y is a fundamental matrix of the system of differential equations in (9). If (11) is valid, then there exists a Green’s matrix of problem (9), which is in the form
where H is defined by
and it has the following properties:
-
(i)
G is bounded on ,
-
(ii)
is absolutely continuous on and for each and its columns satisfy the differential equation from (9) a.e. on ,
-
(iii)
for each ,
-
(iv)
for each and
Theorem 4
([23], Theorem 11)
Let (8) and (11) be satisfied and let G be given by (12) with H of (13). Thenis a fixed point of an operatordefined by
for, if and only if z is a solution of problem (5)-(7). Moreover, the operator ℱ is completely continuous.
Similar results can be found also in [26], Chapter 6].
Remark 5
As in [23], we denote
i.e.
Remark 6
In the present paper we need the Green’s matrix of problem (9) for . Therefore and . The existence of the Green’s matrix is then equivalent with the regularity of K, i.e. with the assumption . If this is satisfied, then H from (13) is given by the formula
and the Green’s matrix takes the form
In this case the matrix functions , from Remark 5 are written as
3 Transversality conditions
Here we formulate conditions which guarantee that each possible solution of problem (1)-(3) in some region, which will be specified later (cf. (21)), crosses each barrier at the unique impulse point , . Consider positive real numbers , , and denote
We assume that
Further we choose positive real numbers , , and assume that
Under conditions (14)-(17), which we call transversality conditions, we define the set
In Section 4 we define an operator (cf. (26)) whose fixed point is used for the construction of a solution z of problem (1)-(3) (cf. (28)). In order to get a correct definition of we need to describe intersection point t of a function with the barriers , . These intersection points are roots of the functions , and their uniqueness is stated in Lemma 7.
Lemma 7
Let, be given by (14), and let, and, , , satisfy (15), (16) and (17). Finally, let ℬ be given by (18). Then for eachthe functions
are continuous and decreasing onand they have unique roots in the interval, i.e. forthere exists a unique solution of the equation
Proof
Let , . By (15),
are valid. This together with the fact that σ is continuous on shows that σ has at least one root in . Now, we will prove that σ is decreasing, by a contradiction. Let , be such that
i.e.
This contradicts (17). Therefore (19) has a unique solution. □
According to Lemma 7, for and , there exists a unique point which is an intersection point of the graph of v with the graph of the barrier . Therefore we define a functional by
where is a solution of (19), i.e. a unique root of the function from Lemma 7, for .
Since solutions are affected by impulses at the points , the functionals , , are used in the definition of the operator (cf. (26)), it is important to prove their properties which are presented in Lemma 8 and Corollary 9 and which are necessary for the compactness of (cf. Lemma 13).
Lemma 8
Let the assumptions of Lemma 7be satisfied. Then for eachthere exists a constantsuch that for every
Proof
Let , . Let us denote
Then from (16) and (18) we get
Subtracting the second term from the right-hand side of the inequality we obtain
and using (17) we arrive at
which is the desired inequality. □
Corollary 9
Let the assumptions of Lemma 7be satisfied. Then the functionals, , which are given by (20), are continuous onin the norm of.
4 Fixed point problem
The main result of this section is contained in Theorem 11, where we present a connection between a (discontinuous) solution z of problem (1)-(3) and a fixed point of some operator which operates on ordered -tuples of absolutely continuous vector functions. We work with the product space
where for we write and , . The sequence of elements of X is denoted as ; and the sequence of its k th components as . The space X is equipped with the norm
It is well known that X is a Banach space. For the construction of a fixed point problem we need the set
where ℬ is defined in (18) with constants , , , satisfying the assumptions of Lemma 7.
Now, assume that the matrix K from (4) fulfills
and consider an operator defined by
for , , where
and , , are continuous functionals from Corollary 9. Here , , Y, take values from Remark 6. Then , for . Assume in addition that f is essentially bounded, that is,
Then the operator maps to X. Unfortunately, is not compact on . We can overcome this obstacle by redefining the operator by means of an operator given by
where , , and are defined by (24). As we will show this will be enough for our needs (cf. Theorem 11).
Remark 10
The important property of the operator is that for we have
Let us note that for the operator satisfies this identity only on the interval , because
This fact obstructs the compactness of the operator in X.
Consider from (14), and assume
Then we are ready to prove the following theorem.
Theorem 11
Let the assumptions of Lemma 7and conditions (22), (25) and (27) hold. Ifis a fixed point of the operator, then a function z defined by
is a solution of problem (1)-(3). Here, , are continuous functionals from Corollary 9.
Proof
Let ℬ be defined by (18) and . Further, let be a fixed point of the operator . Then for each we have and hence, by Lemma 7, there exists a unique solution of the equation . Due to (15) the inequalities
are valid. Let us consider z defined by (28). We will prove that z is a fixed point of the operator ℱ from Theorem 4, taking
Let us denote
Let us choose and consider . Then
Of course,
Let be such that . Then and therefore Remark 5 yields
Let be such that (such i exists only if ). Then and Remark 5 gives
These facts imply that
Consequently, by virtue of Theorem 4, z is a solution of problem (5)-(7) with and (29). The function z satisfies (1) a.e. on and fulfills the boundary condition (3). In addition, since z fulfills the impulse conditions (6) with , and in place of , where , , we see that z fulfills (2). It remains to prove that are the only instants at which the function z crosses the barriers , respectively. To this aim, due to (15) and (28) it suffices to prove that
Choose an arbitrary and consider from Lemma 7 for , i.e.
Since z fulfills (2) we have
and according to (27) we get
Since is decreasing on we have
□
5 Existence results
Properties of the operator which is defined by (23), (24), and (26), in particular its compactness and the existence of its fixed point, will be proved in this section. Then the existence of a solution of problem (1)-(3) will follow (cf. Theorem 15). Besides the conditions from Section 4 we assume in addition that
Here is from (14) and , , are from (15).
Lemma 12
Let the assumptions of Lemma 7and conditions (22), (25), (27), (30), (31), and (32) be fulfilled. Letbe defined by (23), (24), and (26). Then for eachthere existssuch that eachsatisfy
Proof
Consider and denote
where is defined in (23). Let us choose a fixed .
Step 1. According to Remark 10 we have
Denote (cf. (24))
By Lemma 8, we have
Choose an arbitrary . By (35), there exists such that for each
For we have
and therefore, by (26),
Then, using (25) and (34), we get
Due to (35) and (36) there exists such that for each
It remains to discuss the expression . We have
Step 2. Treating the first term on the right-hand side of equality (39) we have
The function G is bounded on ; it follows from (31) that there exists such that for each
In view of Remark 6
and therefore, by (25) and (36), there exists such that for each
Similarly, since G is bounded on and f fulfills (25), we can find satisfying
Consequently, by (36), there exists such that for each
Step 3. Finally we discuss the second and third term on the right-hand side of equality (39). According to Remark 6, we have
Therefore, due to the uniform continuity of , , on (cf. (4) and (14)), the uniform continuity of V on , (cf. (32) and (15)) and by (36), there exists such that for each
Relations (37), (38), (40), (41), (42), (43), and (44) imply (33). □
Lemma 13
Let the assumptions of Lemma 12be fulfilled. Then the operatordefined by (23), (24), and (26) is compact on.
Proof
First, we prove the continuity of . Choose . Then there exists such that each satisfy (33). Since , , each satisfy
Now, we prove the relative compactness of the set . Let be a sequence of elements from the set . Then there exists a sequence such that for every . Since , we have (cf. (18))
for each , . This implies
The Arzelà-Ascoli theorem and the diagonalization principle give the existence of a subsequence which is convergent in the -norm. Let us denote it as . Then, by Lemma 12, for each there exist and such that for each , the inequality holds, and consequently, by (33),
Therefore there exists a subsequence which is convergent in X. □
Theorem 14
Assume that (25) and (30) hold and that numbers, , , satisfy
Define sets, ℬ and Ω by (14), (18), and (21), respectively, and assume that conditions (15), (16), (17), (27), (31), and (32) hold. Then the operatorhas a fixed point in.
Proof
It suffices to show that . Let and , (cf. (23) and (26)). That is and , where for . Choose , . Having in mind (24), we get by (23), (26), (45), and Remark 6
Therefore
From (25) and Remark 10 we have
which yields, due to (45),
Consequently, by virtue of (18), for , that is, . □
Theorems 11 and 14 give an existence result for problem (1)-(3).
Theorem 15
Under the assumptions of Theorem 14problem (1)-(3) has at least one solution z such that
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This work was supported by the grant No. 14-06958S of the Grant Agency of the Czech Republic.
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Rachůnková, I., Tomeček, J. Fixed point problem associated with state-dependent impulsive boundary value problems. Bound Value Probl 2014, 172 (2014). https://doi.org/10.1186/s13661-014-0172-9
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DOI: https://doi.org/10.1186/s13661-014-0172-9