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Volume 2010, Article ID 586971, 8 pagesdoi:10.1155/2010/586971 Research Article Periodic Problem with a Potential Landesman Lazer Condition Petr Tomiczek Department of Mathematics, Unive

Volume 2010, Article ID 586971, 8 pages

doi:10.1155/2010/586971

Research Article

Periodic Problem with a Potential Landesman

Lazer Condition

Petr Tomiczek

Department of Mathematics, University of West Bohemia, Univerzitn´ ı 22, 306 14 Plzeˇn, Czech Republic

Correspondence should be addressed to Petr Tomiczek,tomiczek@kma.zcu.cz

Received 6 January 2010; Revised 30 June 2010; Accepted 22 September 2010

Academic Editor: Pavel Dr´abek

Copyrightq 2010 Petr Tomiczek This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

We prove the existence of a solution to the periodic nonlinear second-order ordinary differential

equation with damping ux  rxux  gx, ux  fx, u0  uT, u0  uT We

suppose thatT

0r xdx  0, the nonlinearity g satisfies the potential Landesman Lazer condition

and prove that a critical point of a corresponding energy functional is a solution to this problem

1 Introduction

Let us consider the nonlinear problem

ux  rxux  gx, ux  fx, x ∈ 0, T,

u 0  uT, u0  uT, 1.1

where r ∈ L10, T, the nonlinearity g : 0, T × R → R is a Caratheodory function and

f ∈ L10, T.

To state an existence result to 1.1 Amster 1 assumes that r is a nondecreasing

function see also 2 He supposes that the nonlinearity g satisfies the growth condition

gx, s − gx, t/s − t ≤ c1, c1 < λ1 for x ∈ 0, T, s, t ∈ R, s / t, where λ1 is the first eigenvalue of the problem−u  λu, u0  uT  0 and there exist a−, a such that

g|0,T×I a ≥T

0 p1xfxdx/p11 ≥ g| 0,T×I a− An interval I a is centered in a with the radius

δ1|a|  δ2 where δ1  λ1c1T/ λ1 − c1 < 1, 0 < δ2 and p1 is a solution to the problem

p1− rp1  k1 , k1∈ R with p10  p1T  1.

In3,4 authors studied 1.1 with a constant friction term rx  c and results with

repulsive singularities were obtained in5,6

In this paper we present new assumptions, we suppose that the friction term r has

zero mean value

T 0

r xdx  0, 1.2

the nonlinearity g is bounded by a L1 function and satisfies the following potential Landesman-Lazer conditionsee also 7,8

T 0



R x2G−xdx <

T 0



R x2f xdx <

T 0



R x2Gxdx, 1.3

where Gx, s s

0g x, tdt, Gx  lim inf s→ ∞G x, s/s, G−x  lim sup s→ −∞Gx, s/s and Rx  ex 1/2rξdξ

To obtain our result we use variational approach even if the linearization of the periodic problem1.1 is a non-self-adjoint operator

2 Preliminaries

Notation We will use the classical space C k 0, T of functions whose kth derivative is continuous and the space L p 0, T of measurable real-valued functions whose pth power

of the absolute value is Lebesgue integrable We denote H the Sobolev space of absolutely continuous functions u : 0, T → R such that u ∈ L20, T and u0  uT with the norm

u  T

0 u2x  u2xdx 1/2 By a solution to1.1 we mean a function u ∈ C10, T such that uis absolutely continuous, u satisfies the boundary conditions and1.1 is satisfied a.e

in0, T.

We denote Rx  ex 1/2rξdξ and we study1.1 by using variational methods We

investigate the functional J : H → R, which is defined by

J u  1

2

T 0



R2

u2

dx−

T 0



R2G x, u − R2fu

dx, 2.1

where

G x, s 

s 0

g x, t dt. 2.2

We say that u is a critical point of J, if

Ju, v  0 ∀v ∈ H. 2.3

We see that every critical point u ∈ H of the functional J satisfies

T 0



R2uv

dx−

T 0



R2

g x, u − fv

dx 0 2.4

for all v ∈ H.

Now we prove that any critical point of the functional J is a solution to1.1 mentioned above

Lemma 2.1 Let the condition 1.2 be satisfied Then any critical point of the functional J is a solution

to1.1.

Proof Setting v 1 in 2.4 we obtain

T 0



R2

g x, u − fdx  0. 2.5

We denote

Φx 

x 0



R t2

g t, ut − ftdt 2.6

then previous equality2.5 implies Φ0  ΦT  0 and by parts in 2.4 we have

T 0



R2u Φv

for all v ∈ H Hence there exists a constant c usuch that

R2u Φ  c u 2.8

on 0, T The condition 1.2 implies R0  RT  1 and from 2.8 we get u0 

R20u0  −Φ0  c u  −ΦT  c u  uT Using R2 R2r and differentiating equality

2.8 with respect to x we obtain

R2

u ru gx, u − f 0. 2.9

Thus u is a solution to1.1

We say that J satisfies the Palais-Smale condition PS if every sequence u n for which

J u n  is bounded in H and Ju n  → 0 as n → ∞ possesses a convergent subsequence.

To prove the existence of a critical point of the functional J we use the Saddle Point

Theorem which is proved in Rabinowitz9 see also 10

H⊕  H < ∞ and dim  H  ∞ Let

J : H → R be a functional such that J ∈ C1H, R and

H and a constant α such that J/∂D ≤ α,

b there is a constant β > α such that J/ H ≥ β,

c J satisfies the Palais-Smale condition (PS).

Then, the functional J has a critical point in H.

3 Main Result

We define

Gx  lim inf

s→ ∞

G x, s

s , G−x  lim sup

s→ −∞

G x, s

s . 3.1 Assume that the following potential Landesman-Lazer type condition holds:

T 0



R x2G−xdx <

T 0



R x2f xdx <

T 0



R x2Gxdx. 3.2

We also suppose that there exists a function qx ∈ L10, T such that

g x, s ≤ qx, x ∈ 0,T, s ∈ R. 3.3

Theorem 3.1 Under the assumptions 1.2, 3.2, 3.3, problem 1.1 has at least one solution.

Proof We verify that the functional J satisfies assumptions of the Saddle PointTheorem 2.2

on H, then J has a critical point u and due toLemma 2.1u is the solution to1.1

It is easy to see that J ∈ C1H, R Let  H  {u ∈ H :T

0 u xdx  0} then H  R ⊕  H

and dimH  ∞

In order to check assumptiona, we prove

lim

|s| → ∞ J s  −∞ 3.4

by contradiction Then, assume on the contrary there is a sequence of numberss n ⊂ R such that|s n | → ∞ and a constant c1satisfying

lim inf

n→ ∞ J s n  ≥ c1 3.5

From the definition of J and from3.5 it follows

lim inf

n→ ∞

T 0

R2

−Gx, s n   fs n



|s n| dx ≥ 0. 3.6

We note that from3.2 it follows there exist constants s , s−and functions Ax, A−x ∈

L10, T such that Ax ≤ Gx, s, Gx, s ≤ A−x for a.e x ∈ 0, T and for all s ≥ s, s ≤ s−,

respectively We suppose that for this moment s n → ∞ Using 3.6 and Fatou’s lemma we obtain

T 0



R x2f xdx≥

T 0



R x2Gxdx, 3.7

a contradiction to 3.2 We proceed for the case s n → −∞ Then assumption a of

b Now we prove that J is bounded from below on  H For u∈ H, we have

T 0



u2

dx  u2 3.8

and assumption3.3 implies

|Gx, s| ≤ qx|s|, x ∈ 0, T, s ∈ R. 3.9

Hence and due to compact imbedding H ⊂ C0, Tu C 0,T ≤ c2u we obtain

J u  1

2

T 0



R2

u2

dx−

T 0



R2G x, u − R2fu

dx

≥ 1

2xmin∈0,T R x2

T 0



u2

dx− max

x ∈0,T R x2

T 0

q   f|u|dx

≥ 1

2xmin∈0,T R x2u2− max

x ∈0,T R x2q

1f

1



c2u.

3.10

Since the function R is strictly positive equality3.10 implies that the functional J is bounded

from below

a constant α such that J/∂D ≤ α, and there is a constant β > α such that J/ H ≥ β.

In order to check assumptionc, we show that J satisfies the Palais-Smale condition.

First, we suppose that the sequenceu n  is unbounded and there exists a constant c3 such that







1 2

T 0



R2

un2

dx−

T 0



R2

G x, u n  − fu n



dx



 ≤c3, 3.11 lim

n→ ∞Ju n  0 3.12

Let w k  be an arbitrary sequence bounded in H It follows from 3.12 and the Schwarz inequality that





nlim→ ∞

k→ ∞

T 0



R2un wk

dx−

T 0



R2

g x, u n w k − fw k



dx









nlim→ ∞

k→ ∞

Ju n w k





 ≤nlim→ ∞

k→ ∞

Ju n · wk   0.

3.13

From3.3 we obtain

lim

n→ ∞

k→ ∞

T 0



R2g x, u n

u n w k− R2f

u nw k



dx  0. 3.14

Put v n  u n / u n  and w k  v nthen3.13, 3.14 imply

lim

n→ ∞

T 0



R2

v n2

dx  0. 3.15

Due to compact imbedding H ⊂ C0, T and 3.15 we have |v n | → d in C0, T, d > 0 Suppose that v n → d and set w k  v n − d in 3.13, we get

lim

n→ ∞

T 0



R2un v n

dx−

T 0



R2

g x, u n  − fv n − ddx  0. 3.16

Because the nonlinearity g is boundedassumption 3.3 and v n → d the second integral in

previous equality3.16 converges to zero Therefore

lim

n→ ∞

T 0



R2un v n

dx  0. 3.17

Now we divide3.11 by u n We get

lim

n→ ∞

 1 2

T 0



R2un v n

dx−

T 0

R2

G x, u n  − fu n



u n dx



 0. 3.18

Equalities3.17, 3.18 imply

lim

n→ ∞

T 0

R2



−G x, u n

u n  f



v n dx  0. 3.19

Because v n → d > 0, lim n→ ∞u n x  ∞ Using Fatou’s lemma and 3.19 we conclude

T 0



R x2f xdx≥

T 0



R x2Gxdx, 3.20

a contradiction to3.2 We proceed for the case v n → −d similarly This implies that the

sequenceu n  is bounded Then there exists u0 ∈ H such that u n  u0 in H, u n → u0 in

L20, T, C0, T taking a subsequence if it is necessary It follows from equality 3.13 that

lim

n→ ∞

m→ ∞

k→ ∞

T

0



R2u n − u mwk

dx−

T 0



R2

g x, u n  − gx, u mw k dx



 0. 3.21

The strong convergence u n → u0 in C0, T and the assumption 3.3 imply

lim

n→ ∞

m→ ∞

T 0



R2

g x, u n  − gx, u mu n − u mdx  0. 3.22

If we set w k  u n , w k  u min3.21 and subtract these equalities, then using 3.22 we have

lim

n→ ∞

m→ ∞

T 0



R2

un − u

m

2

dx  0. 3.23

Hence we obtain the strong convergence u n → u0 in H This shows that J satisfies the

Palais-Smale condition and the proof ofTheorem 3.1is complete

Acknowledgment

This work was supported by Research Plan MSM 4977751301

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Let w k  be an arbitrary sequence bounded in H It follows from 3.12 and the Schwarz inequality that