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# Quasilinear boundary value problem with impulses: variational approach to resonance problem

Pavel Drábek12 and Martina Langerová2*

Author Affiliations

1 Department of Mathematics, University of West Bohemia, Univerzitní 22, Plzeň, 306 14, Czech Republic

2 NTIS, University of West Bohemia, Univerzitní 22, Plzeň, 306 14, Czech Republic

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Boundary Value Problems 2014, 2014:64  doi:10.1186/1687-2770-2014-64

We dedicate this paper to Professor Ivan Kiguradze for his merits in the theory of differential equations.

 Received: 5 December 2013 Accepted: 7 March 2014 Published: 24 March 2014

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

### Abstract

This paper deals with the resonance problem for the one-dimensional p-Laplacian with homogeneous Dirichlet boundary conditions and with nonlinear impulses in the derivative of the solution at prescribed points. The sufficient condition of Landesman-Lazer type is presented and the existence of at least one solution is proved. The proof is variational and relies on the linking theorem.

MSC: 34A37, 34B37, 34F15, 49K35.

##### Keywords:
quasilinear impulsive differential equations; Landesman-Lazer condition; variational methods; critical point theory; linking theorem

### 1 Introduction

Let be a real number. We consider the homogeneous Dirichlet boundary value problem for one-dimensional p-Laplacian

(1)

where is a spectral parameter and , , is a given right-hand side.

Let be given points and let , , be given continuous functions. We are interested in the solutions of (1) satisfying the impulse conditions in the derivative

(2)

For the sake of brevity, in further text we use the following notation:

For this problem is considered in [1] where the necessary and sufficient condition for the existence of a solution of (1) and (2) is given. In fact, in the so-called resonance case, we introduce necessary and sufficient conditions of Landesman-Lazer type in terms of the impulse functions , , and the right-hand side f. They generalize the Fredholm alternative for linear problem (1) with .

In this paper we focus on a quasilinear equation with and look just for sufficient conditions. We point out that there are principal differences between the linear case () and the nonlinear case (). In the linear case, we could benefit from the Hilbert structure of an abstract formulation of the problem. It could be treated using the topological degree as a nonlinear compact perturbation of a linear operator. However, in the nonlinear case, completely different approach must be chosen in the resonance case. Our variational proof relies on the linking theorem (see [2]), but we have to work in a Banach space since the Hilbert structure is not suitable for the case .

It is known that the eigenvalues of

(3)

are simple and form an unbounded increasing sequence whose eigenspaces are spanned by functions such that has evenly spaced zeros in , , and . The reader is invited to see [[3], p.388], [[4], p.780] or [[5], pp.272-275] for further details. See also Example 1 below for more explicit form of and .

Let ,  , in (1). This is the nonresonance case. Then, for any , there exists at least one solution of (1). In the case , this solution is unique. In the case , the uniqueness holds if , but it may fail for certain right-hand sides if . See, e.g., [6] (for ) and [7] (for ).

The same argument as that used for in [[1], Section 3] for the nonresonance case yields the following existence result for the quasilinear impulsive problem (1), (2).

Theorem 1 (Nonresonance case)

Let,  , , , be continuous functions which are-subhomogeneous at ±∞, that is,

Then (1), (2) has a solution for arbitrary.

Variational approach to impulsive differential equations of the type (1), (2) with was used, e.g., in paper [8]. The authors apply the mountain pass theorem to prove the existence of a solution for . Our Theorem 1 thus generalizes [[8], Theorem 5.2] in two directions. Firstly, it allows also (, ) and, secondly, it deals with quasilinear equations (), too.

Let for some . This is the resonance case. Contrary to the linear case (), there is no Fredholm alternative for (1) in the nonlinear case (). If , then

is the sufficient condition for solvability of (1), but it is not necessary if . Moreover, if but f is ‘close enough’ to , problem (1) has at least two distinct solutions. The reader is referred to [3] or [9] for more details. It appears that the situation is even more complicated for , (see, e.g., [10]).

In the presence of nonlinear impulses which have certain asymptotic properties (to be made precise below), we show that the fact might still be the sufficient condition for the existence of a solution to (1) (with ) and (2). For this purpose we need some notation. Let denote evenly spaced zeros of , let and denote the union of intervals where or , respectively. We arrange , , into three sequences: , , ; , , ; , . Obviously, we have and . Assume that , i.e., . The impulse condition (2) can be written in an equivalent form

(4)

We assume that , ; ; , are continuous, bounded functions and there exist limits , . We consider the following Landesman-Lazer type conditions: either

(5)

or

(6)

Our main result is the following.

Theorem 2 (Resonance case)

Letfor somein (1). Let the nonlinear bounded impulse functions, , and the right-hand sidesatisfy either (5) or (6). Then (1), (2) has a solution.

The result from Theorem 2 is illustrated in the following special example.

Example 1 It follows from the first integral associated with the equation in (3) that the eigenvalues and the eigenfunctions of (3) have the form

where and , , , , , , see [[3], p.388]. Let us consider in (1) and , , , , , in (2). Since , , condition (6) reads as follows:

### 2 Functional framework

We say that u is the classical solution of (1), (2) if the following conditions are fulfilled:

, , is absolutely continuous in , ;

• the equation in (1) holds a.e. in and ;

• one-sided limits , exist finite and (2) holds.

We say that is a weak solution of (1), (2) if the integral identity

(7)

holds for any function .

Integration by parts and the fundamental lemma in calculus of variations (see [[11], Lemma 7.1.9]) yields that every weak solution of (1), (2) is also a classical solution and vice versa. Indeed, let u be a weak solution of (1), (2), (the space of smooth functions with a compact support in , ), elsewhere in , then

Since v is arbitrary, we have for a.e. . Then is absolutely continuous in and

(8)

for a.e. , . Taking now arbitrary, integrating by parts in the first integral in (7) and using (8), we get

and hence also (2) follows. Similarly, we show that every classical solution is a weak solution at the same time.

Let with the norm , be the dual of X and be the duality pairing between and X. For , we set

Then, for , we have

Lemma 1The operatorshave the following properties:

(A) is-homogeneous, odd, continuously invertible, andfor any.

(B) is-homogeneous, odd and compact.

(J) is bounded and compact.

By the linearity of, is a fixed element.

Proof See [[12], Lemma 10.3, p.120]. □

With this notation in hands we can look for (classical) solutions of (1), (2) either as for solutions of the operator equation

(9)

or, alternatively, as for critical points of the functional ,

(10)

As mentioned already above, in the nonresonance case (, ), we can use the Leray-Schauder degree argument and prove the existence of a solution of the equation (9) exactly as in [[1], proof of Thm. 1]. Note that the -subhomogeneous condition on is used here instead of the sublinear condition imposed on in [1] and the proof of Theorem 1 follows the same lines. For this reason we skip it and concentrate on the resonance case ( for some ) in the next section.

### 3 Resonance problem, variational approach

We use the following definition of linked sets and the linking theorem (cf.[13]).

Definition 1 Let ℰ be a closed subset of X and let Q be a submanifold of X with relative boundary ∂Q. We say that ℰ and ∂Q link if

(i) and

(ii) for any continuous map such that , there holds .

(See [[14], Def. 8.1, p.116].)

Suppose thatsatisfies the Palais-Smale condition. Consider a closed subsetand a submanifoldwith relative boundary∂Q, and let. Suppose thatand∂Qlink in the sense of Definition 1, and

Thenis a critical value of ℱ.

(See [[14], Thm. 8.4, p.118].)

The purpose of the following series of lemmas is to show that the hypotheses of Theorem 3 are satisfied provided that either (5) or (6) holds. From now on we assume that (for some ) in (1).

Lemma 2If either (5) or (6) is satisfied, thensatisfies the Palais-Smale condition.

Proof Suppose that such that and in . We must show that has a subsequence that converges in X. We prove first that is a bounded sequence. We proceed via contradiction and suppose that and consider . Without loss of generality, we can assume that there is such that (weakly) in X (X is a reflexive Banach space). Since

dividing through by , we have

By the boundedness of we know that . We also have . By the compactness of we get in . Thus in X by Lemma 1(A). It follows that .

We assume and remark that a similar argument follows if . Next we estimate

(11)

Our assumption yields

(12)

and the Cauchy-Schwarz inequality implies

(13)

where denotes the norm in . It follows from (11)-(13) that

Dividing through by and writing , where

, we get

(14)

Since as , we obtain from (14):

(15)

Recall that X embeds compactly in , so, without loss of generality, we assume that , , as . Hence, for , which implies as well as as by an application of the l’Hospital rule to . Notice that by the boundedness of we have

if is a zero point of for some . Thus, passing to the limit in (15) as , we get

which contradicts (5) or (6). Hence is bounded.

By compactness there is a subsequence such that and converge in (see Lemma 1(B), (J)). Since by our assumption, we also have that converges in . Finally, converges in X by Lemma 1(A). The proof is finished. □

With the Palais-Smale condition in hands, we can turn our attention to the geometry of the functional ℱ. To this end we have to find suitable sets which link in the sense of Definition 1. Actually, we use the sets constructed in [13] and explain that they fit with the hypotheses of Theorem 3 if either (5) or (6) is satisfied.

Consider the even functional

and the manifold

For any , let , where represents the unit sphere in . Next we define

(16)

It is proved in [[15], Section 3] that is a sequence of eigenvalues of homogeneous problem (3). It then follows from the results in [16] that this sequence exhausts the set of all eigenvalues of (3) with the properties described in Section 1.

Now consider the functions for , where is a characteristic function of the interval , and let

Observe that is symmetric and is homeomorphic to the unit sphere in . Moreover, for , we have

Notice that the second equality holds thanks to the fact

for , , while the third one follows from the p-homogeneity of B. Thus and so . A similar computation then shows that for all . For a given , we let

Then is homeomorphic to the closed unit ball in . For a given , we denote by

a super-level set, and

The existence of a pseudo-gradient vector field with the following properties is proved in [[13], Lemma 6] (cf. [[14], pp.77-79] and [[2], p.55]).

Lemma 3For, there isand a one-parameter family of homeomorphismssuch that

(i) ifor if;

(ii) is strictly decreasing intifand;

(iii) ;

(iv) .

An important fact is that the flow η ‘lowers’ and ‘raises’ if we modify them as follows:

and

Then, by Lemma 3 and the definition of , we have

for with equality if and only if for some . Similarly,

for with equality if and only if for some .

It is proved in [[13], Lemma 7] that the couple and satisfies condition (ii) from Definition 1. It is also proved in [[13], Lemma 8] that the couple and satisfies the same condition. To show that also other hypotheses of Theorem 3 are satisfied, we need some technical lemmas.

Lemma 4If (6) is satisfied, then there existandsuch thatfor anyand.

Proof We proceed via contradiction and assume that there exist and such that

(17)

Since is compact, we may assume, without loss of generality, that in for some .

If , then there exists such that

Hence, there exists such that for any we have

This implies

for . However, this contradicts (17).

If , we still have

and so

for all . The boundedness of , , and uniform convergence as (due to continuous embedding ) then yield

by the first inequality in (6). This contradicts (17) again. Notice that by the boundedness of we have

if is a zero point of for some . The case is proved similarly using the second inequality in (6). □

Lemma 5If (6) is satisfied, then there existssuch that

(18)

Proof There exists such that for any we have

By Lemma 4 there exists such that for all and we have

Thus there exists such that

for all , . In particular, for all and (18) is proved. □

Now we can finish the proof of Theorem 2 under assumption (6). Indeed, it follows from (18) that and thus the hypotheses of Theorem 3 hold with and . It then follows that ℱ has a critical point and hence (1), (2) has a solution.

Next we show that the sets and satisfy the hypotheses of Theorem 3 if (5) is satisfied.

The principal difference consists in the fact that, in contrast with , the set is not compact. That is why one more technical lemma is needed.

Lemma 6For any, there existssuch that

(19)

for. (Hereis the ball inXcentered atwith radius .)

Proof We note that the pseudo-gradient flow η from Lemma 3 is constructed as a solution of the initial value problem , , where

is a locally Lipschitz continuous symmetric pseudo-gradient vector field associated with E on and is a smooth function such that for u satisfying and for u satisfying or .

Let and . Without loss of generality, we may assume that . Let be such that . Observe that there is a constant such that for we have

Hence for . Since E satisfies the Palais-Smale condition on (see [[13], Lemma 2]), there exists such that for all . Then

for all . The last but one inequality holds due to the following property of :

(see [14] and [2]). We also used the fact that for . Hence

with . □

The following lemma is a counterpart of Lemma 4 in the case of condition (5).

Lemma 7If (5) is satisfied, then there existandsuch thatfor anyand.

Proof We proceed via contradiction and assume that there exist and such that

(20)

If there is such that for all k large enough, then Lemma 6 leads to the estimate

contradicting (20). Thus it must be as . If as , we still have

and so

for all . Similar arguments as in the proof of Lemma 4 lead to

by the second inequality in (5). This contradicts (20) again. The case as is proved similarly but using the first inequality in (5). □

Lemma 8If (5) is satisfied, then there existssuch that

(21)

Proof By Lemma 7 there exists such that for all and we have

Hence, there exists such that for any we have

On the other hand, for any and , we get

Thus, there exists such that, for ,

and (21) is proved. □

It follows that the sets and satisfy the hypotheses of Theorem 3 if (5) is satisfied. The proof of Theorem 2 is thus completed.

Final remark Reviewers of our manuscript suggested to include some recent references on impulsive problems. Variational approach to impulsive problems can be found, e.g., in [17-21]. The last reference deals with the p-Laplacian with the variable exponent . Singular impulsive problems are treated in [22-24]. Impulsive problems are still ‘hot topic’ attracting the attention of many mathematicians and the bibliography on that topic is vast.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

Both authors read and approved the final manuscript.

### Acknowledgements

This research was supported by Grant 13-00863S of the Grant Agency of Czech Republic and by the European Regional Development Fund (ERDF), project ‘NTIS - New Technologies for the Information Society’, European Centre of Excellence, CZ.1.05/1.1.00/02.0090.

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