Multiple Solutions for Biharmonic Equations with Asymptotically Linear Nonlinearities

Volume 2010, Article ID 241518,11pages doi:10.1155/2010/241518

Research Article

Multiple Solutions for Biharmonic Equations with

Asymptotically Linear Nonlinearities

Ruichang Pei

1_{Center for Nonlinear Studies, Northwest University, Xi’an 710069, China} 2_{Department of Mathematics, Tianshui Normal University, Tianshui 741001, China}

Correspondence should be addressed to Ruichang Pei,[email protected]

Received 26 February 2010; Revised 2 April 2010; Accepted 22 April 2010

Academic Editor: Kanishka Perera

Copyrightq2010 Ruichang Pei. 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.

The existence of multiple solutions for a class of fourth elliptic equation with respect to the resonance and nonresonance conditions is established by using the minimax method and Morse theory.

1. Introduction

Consider the following Navier boundary value problem:

Δ2_{ux fx, u, inΩ,}

u Δu0 in∂Ω,

1.1

whereΩis a bounded smooth domain inRN _{N >4, andfx, tsatisfies the following:}

H₁f ∈C1Ω×R,R, fx,0 0, fx, tt≥0 for allx∈Ω,t∈R;

H₂lim|t| →0fx, t/t f0, lim|t| → ∞fx, t/t luniformly forx∈Ω, wheref0andl

are constants;

In view of the conditionH₂, problem1.1is called asymptotically linear at both zero and infinity. Clearly,u0 is a trivial solution of problem1.1. It follows fromH₁andH₂

that the functional

Iu 1

2

Ω|Δu| 2_dx−

ΩFx, udx 1.2

is ofC2_{on the spaceH}1

0Ω∩H2Ωwith the norm

u:

Ω|Δu|

2_dx1/2_{. 1.3}

Under the condition H₂, the critical points of I are solutions of problem 1.1. Let 0 < λ1 < λ2 < · · · < λk < · · · be the eigenvalues of Δ2, H2Ω ∩ H₀1Ω and φ1x > 0

be the eigenfunction corresponding to λ1. Let Eλk denote the eigenspace associated to λk.

Throughout this paper, we denoted by| · |_ptheLp_Ωnorm.

Iflin the above conditionH₂is an eigenvalue ofΔ2_{, H}2_Ω∩H1

0Ω,then problem

1.1 is called resonance at infinity. Otherwise, we call it non-resonance. A main tool of seeking the critical points of functionalIis the mountain pass theoremsee1–3. To apply this theorem to the functionalIin1.2, usually we need the following condition1, that is, for someθ >2 andM >0,

AR

0< θFx, s≤fx, ss for a.e. x∈Ω, |s|> M. 1.4

It is well known that the conditionARplays an important role in verifying that the functionalI has a “mountain pass” geometry and a relatedP S_c sequence is bounded in

H2_Ω∩H1

0Ωwhen one uses the mountain pass theorem.

If fx, t admits subcritical growth and satisfies AR condition by the standard argument of applying mountain pass theorem, we known that problem1.1has nontrivial solutions. Similarly, lasefx, tis of critical growthsee, e.g.,4–7and their references.

It follows from the condition AR that lim|t| → ∞Fx, t/t2 ∞ after a simple

computation. That is,fx, tmust be superlinear with respect tot at infinity. Noticing our condition H₂, the nonlinear term fx, t is asymptotically linear, not superlinear, with respect totat infinity, which means that the usual conditionARcannot be assumed in our case. If the mountain pass theorem is used to seek the critical points ofI, it is difficult to verify that the functionalIhas a “mountain pass” structure and theP S_csequence is bounded.

In8, Zhou studied the following elliptic problem:

−Δufx, u, u∈H1

0Ω, 1.5

where the conditions onfx, tare similar toH₁andH₂.He provided a valid method to verify theP Ssequence of the variational functional, for the above problem is bounded in

H1

To the author’s knowledge, there seems few results on problem 1.1 when fx, t

is asymptotically linear at infinity. However, the method in 8cannot be applied directly to the biharmonic problems. For example, for the Laplacian problem, u ∈ H1

0Ω implies |u|, u , u− ∈ H01Ω, where u maxu,0, u− max−u,0.We can use u or u− as a

test function, which is helpful in proving a solution nonnegative. While for the biharmonic problems, this trick fails completely sinceu∈H2

0Ωdoes not implyu , u− ∈H02Ω see11,

Remark 2.1.10. As far as this point is concerned, we will make use of the methods in12to discuss in the followingLemma 2.3. In this paper we consider multiple solutions of problem 1.1in the cases of resonance and non-resonance by using the mountain pass theorem and Morse theory. At first, we use the truncated skill and mountain pass theorem to obtain a positive solution and a negative solution of problem1.1under our more general condition H₁andH₂with respect to the conditionsH1andH3in8. In the course of proving

existence of positive solution and negative solution, the monotonicity conditionH2of8

on the nonlinear termfis not necessary, this point is very important because we can directly prove existence of positive solution and negative solution by using Rabinowitz’s mountain pass theorem. That is, the proof of our compact condition is more simple than that in 8. Furthermore, we can obtain a nontrivial solution when the nonlinear termf is resonance or non-resonance at the infinity by using Morse theory.

2. Main Results and Auxiliary Lemmas

Let us now state the main results.

Theorem 2.1. Assume that conditionsH₁andH₂hold,f0 < λ1, andl ∈ λk, λk 1for some

k≥2; then problem1.1has at least three nontrivial solutions.

Theorem 2.2. Assume that conditionsH₁)–(H₃hold, f0 < λ1,andl λk for somek ≥ 2; then

problem1.1has at least three nontrivial solutions.

Consider the following problem:

Δ2_{uf x, u, x∈Ω,}

u|_∂Ω Δu|_∂Ω 0,

2.1

where

f x, t

⎧ ⎨ ⎩

fx, t, t >0,

0, t≤0.

2.2

Define a functionalI :H2_Ω∩H1

0Ω → Rby

I u 1

2

Ω|Δu| 2_dx−

ΩF x, udx, 2.3

whereF x, t ₀tf x, sds,and thenI ∈C2_H2_Ω∩H1

Lemma 2.3. I satisfies the (PS) condition.

Proof. Let{un} ⊂H2_Ω∩H1

0Ωbe a sequence such that|Iun| ≤c, < Iun, φ >→ 0 as

n → ∞.Note that

Iun, φ

ΩΔunΔφdx−

Ωf x, unφdxo

φ 2.4

for allφ ∈H2Ω ∩ H₀1Ω.Assume that|un|2 is bounded, takingφ un in2.4. ByH₂,

there existsc >0 such that|f x, unx| ≤c|unx|,a.e.x∈Ω.Sounis bounded inH2Ω ∩

H1

0Ω. If|un|2 → ∞,asn → ∞,setvn un/|un|2, and then|vn|2 1. Takingφ vn in

2.4, it follows thatvnis bounded. Without loss of generality, we assume thatvn v in

H2_Ω∩H1

0Ω, and thenvn → vinL2Ω. Hence,vn → va.e. inΩ. Dividing both sides of

2.4by|un|2, we get

ΩΔvnΔφdx−

Ω

f x, un

|un|₂ φdxo φ

|un|₂

, ∀φ∈H2Ω∩H₀1Ω. 2.5

Then for a.e.x∈Ω, we deduce thatf x, un/|un|2 → lv asn → ∞,wherev max{v,0}.

In fact, whenvx>0,byH₂we have

unx vnx|un|2−→ ∞,

f x, un |un|₂

f x, un

un

vn−→lv.

2.6

Whenvx 0, we have

f x, un

|un|₂ ≤c|vn| −→0. 2.7

Whenvx<0, we have

unx vnx|un|2−→ −∞,

f x, un |un|₂ 0.

2.8

Sincef x, un/|un|2 ≤c|vn|, by2.5and the Lebesgue dominated convergence theorem, we

arrive at

ΩΔvΔφdx−

Ωlv φdx0, for any φ∈H

2_Ω∩H1

Choosingφφ1, we deduce that

l

Ωv φ1dxλ1

Ωvφ1dx. 2.10

Notice that

Ωv φ1dx −

Ωvφ1dx

Ω−

−vφ1dx≥0, 2.11

whereΩ−{x∈Ω:vx<0}.

Now we show that there is a contradiction in both cases of|Ω₋|0 and|Ω₋|>0.

Case 1. Suppose|Ω−| 0,thenvx ≥ 0 a.e. inΩ.Byvx/≡0 we have_Ωvφ1dx > 0.Thus

2.11implies that

l

Ωvφ1dxl

Ωv φ1dxλ1

Ωvφ1dx 2.12

which contradicts tol > λ1.

Case 2. Suppose|Ω₋| > 0,then_Ω

−−vφ1dx > 0,and

Ωv φ1dx >

Ωvφ1dx.It follows from

2.11that

l

Ωv φ1dxλ1

Ωvφ1dx < λ1

Ωv φ1dx 2.13

which contradicts tol > λ1if

Ωv φ1dx >0 and contradicts to 0/<0 if

Ωv φ1dx0.

Lemma 2.4. Letφ1be the eigenfunction corresponding toλ1withφ11. Iff0 < λ1< l, then

athere existρ, β >0such thatI u≥βfor allu∈H2_Ω∩H1

0Ωwithuρ;

bI tφ1 −∞ast → ∞.

Proof. ByH₁andH₂, ifl∈λ1, ∞, for anyε >0, there existAAε≥0 andBBε

such that for allx, s∈Ω×R,

F x, s≤ 1

2

f0 ε

s2 Asp 1, 2.14

F x, s≥ 1

2l−εs

2_{−B, 2.15}

Chooseε >0 such thatf0 ε < λ1.By2.14, the Poincar´e inequality, and the Sobolev

inequality, we get

I u 1

2

Ω|Δu| 2_dx−

ΩF x, udx ≥ 1

2

Ω|Δu| 2_dx−1

2

Ω

f0 ε

u2 A|u|p 1dx

≥ 1

2

1−f0 ε

λ1

u2−cup 1.

2.16

So, partaholds if we chooseuρ >0 small enough.

On the other hand, ifl∈λ1, ∞,takeε >0 such thatl−ε > λ1. By2.15, we have

I u≤ 1

2u

2₋l−ε

2 |u|

2

2 B|Ω|. 2.17

Sincel−ε > λ1andφ11, it is easy to see that

I tφ1

≤ 1

2

1− l−ε

λ1

t2 B|Ω| −→ −∞ ast−→ ∞, 2.18

and partbis proved. Lemma 2.5. Let H2_{Ω ∩ H}1

0Ω V ⊕ W, where V Eλ1 ⊕ Eλ2 ⊕ · · · ⊕ Eλk. Iff satisfies H₁)–(H₃,then

ithe functionalIis coercive onW, that is,

Iu−→ ∞ as u −→ ∞, u∈W 2.19

and bounded from below onW; iithe functionalIis anticoercive onV.

Proof. Foru∈W, byH₂, for anyε >0, there existsB1B1εsuch that for allx, s∈Ω×R,

Fx, s≤ 1

2l εs

2 _B

1. 2.20

So we have

Iu 1

2

Ω|Δu| 2_dx−

ΩFx, udx

≥ 1

2

Ω|Δu| 2_dx−1

2l ε|u|

2

2−B1|Ω|

≥ 1

2

1−l ε

λk 1

u2−B1|Ω|.

2.21

ii We firstly consider the casel λk. WriteGx, t Fx, t−1/2λkt2, gx, t

fx, t−λkt. ThenH₂andH₃imply that

lim

|t| → ∞

gx, tt−2Gx, t−∞, _2.22

lim

|t| → ∞

2Gx, t

t2 0. 2.23

It follows from2.22that for everyM >0, there exists a constantT >0 such that

gx, tt−2Gx, t≤ −M, ∀t∈R, |t| ≥T, a.e. x∈Ω. 2.24

Forτ >0,we have

d dτ

Gx, τ τ2

gx, ττ−2Gx, τ

τ3 . 2.25

Integrating2.25overt, s⊂T, ∞, we deduce that

Gx, s s2 −

Gx, t t2 ≤

M

2

1

s2 −

1

t2

. 2.26

Lets → ∞and use2.23; we see thatGx, t≥M/2,fort∈R, t≥T,a.e.x∈Ω.A similar argument shows thatGx, t≥M/2,fort∈R, t≤ −T,a.e.x∈Ω. Hence

lim

|t| → ∞Gx, t−→ ∞, a.e. x∈Ω. 2.27

By2.27, we get

Iv 1

2

Ω|Δv| 2_dx−

ΩFx, vdx

1

2

Ω|Δv| 2_dx−1

2λk

Ωv 2_dx−

ΩGx, vdx

≤ −δv−2−

ΩGx, vdx−→ −∞

2.28

forv∈Vwithv → ∞, wherev− ∈Eλ1⊕Eλ2⊕ · · · ⊕Eλk−1.

In the case ofλk < l < λk 1, we do not need the assumptionH3and it is easy to see

Lemma 2.6. Ifλk< l < λk 1, thenIsatisfies the (PS) condition.

Proof. Let{un} ⊂H2_Ω∩H1

0Ωbe a sequence such that|Iun| ≤c, < Iun, φ >→ 0. One

has

Iun, φ

ΩΔunΔφdx−

Ωfx, unφdxo

φ 2.29

for allφ ∈ H2_Ω∩H1

0Ω.If|un|2 is bounded, we can takeφ un. ByH2, there exists a

constantc > 0 such that |fx, unx| ≤ c|unx|,a.e.x ∈ Ω.Soun is bounded in H2Ω∩

H1

0Ω. If|un|2 → ∞, as n → ∞,set vn un/|un|2, and then |vn|2 1. Takingφ vn

in2.29, it follows thatvnis bounded. Without loss of generality, we assumevn vin

H2_Ω∩H1

0Ω, and thenvn → vinL2Ω. Hence,vn → va.e. inΩ. Dividing both sides of

2.29by|un|2, we get

ΩΔvnΔφdx−

Ω

fx, un

|un|₂ φdxo φ

|un|₂

for anyφ∈H2Ω∩H₀1Ω. 2.30

Then for a.e.x∈Ω, we havefx, un/|un|2 → lvasn → ∞.In fact, ifvx/0,byH2, we

have

|unx||vnx||un|2 −→ ∞,

fx, un |un|₂

fx, un

un

vn−→lv.

2.31

Ifvx 0, we have

fx, un

|un|₂ ≤c|vn| −→0. 2.32

Since|fx, un|/|un|2 ≤ c|vn|, by 2.30and the Lebesgue dominated convergence theorem,

we arrive at

ΩΔvΔφdx−

Ωlvφdx0, ∀φ∈H

2_Ω∩H1

0Ω. 2.33

It is easy to see thatv /≡0. In fact, ifv≡0, then|v|20 contradicts to limn→ ∞|vn|2 |v|2 1.

Hence,lis an eigenvalue ofΔ2_{, H}2_Ω∩H1

0Ω. This contradicts our assumption.

Lemma 2.7. Suppose thatlλkandfsatisfiesH3. Then the functionalIsatisfies the (C) condition

which is stated in [13].

Proof. Supposeun∈H2Ω∩H₀1Ωsatisfies

In view ofH₂, it suffices to prove thatunis bounded inH2Ω∩H₀1Ω. Similar to the proof

ofLemma 2.6, we have

ΩΔvΔφdx−

Ωlvφdx0, ∀φ∈H

2_Ω∩H1

0Ω. 2.35

Thereforev /≡0 is an eigenfunction ofλk, then|unx| → ∞for a.e.x ∈ Ω. It follows from

H₃that

lim

n→ ∞

fx, unxunx−2Fx, unx

−∞ 2.36

holds uniformly inx∈Ω, which implies that

Ω

fx, unun−2Fx, un

dx−→ −∞ asn−→ ∞. 2.37

On the other hand,2.34implies that

2Iun− Iun, un

−→2c asn−→ ∞. 2.38

Thus

Ω

fx, unun−2Fx, un

dx−→2c asn−→ ∞, 2.39

which contradicts to2.37. Henceunis bounded.

It is well known that critical groups and Morse theory are the main tools in solving elliptic partial differential equation. Let us recall some results which will be used later. We refer the readers to the book14for more information on Morse theory.

LetHbe a Hilbert space, letI ∈C1_{H,Rbe a functional satisfying thePScondition}

or C condition, let HqX, Y be the qth singular relative homology group with integer

coefficients. Letu0 be an isolated critical point of I with Iu0 c, c ∈ R,and letUbe a

neighborhood ofu0. The group

CqI, u0:HqIc∩U, Ic∩U\ {u0}, q∈Z 2.40

is said to be theqth critical group ofIatu0, whereIc{u∈H:Iu≤c}.

LetK : {u ∈ H : Iu 0}be the set of critical points ofI anda < infIK; the critical groups ofIat infinity are formally defined bysee15

CqI,∞:HqH, Ia, q∈Z. 2.41

Proposition 2.8see15. Assume that H H_∞⊕H_∞−, I is bounded from below onH_∞ and

Iu → −∞asu → ∞withu∈H_∞−. Then

CkI,∞0, if kdimH∞− <∞. 2.42

3. Proof of the Main Results

Proof ofTheorem 2.1. By Lemmas2.32.4and the mountain pass theorem, the functionalI has a critical pointu1satisfyingI u1≥β. SinceI 0 0,u1/0, and by the maximum principle,

we getu1>0. Henceu1is a positive solution of the problem1.1and satisfies

C1I , u1/0, u1>0. 3.1

Using the results in14, we obtain

CqI, u1 Cq

I_C1 0Ω, u1

Cq

I |_C1 0Ω, u1

CqI , u1 δq1Z. 3.2

Similarly, we can obtain another negative critical pointu2ofIsatisfying

CqI, u2 δq,1Z. 3.3

Sincef0< λ1,the zero function is a local minimizer ofI, and then

CqI,0 δq,0Z. 3.4

On the other hand, by Lemmas2.52.6andProposition 2.8, we have

CkI,∞0. 3.5

HenceIhas a critical pointu3satisfying

CkI, u30. 3.6

Since k ≥ 2, it follows from 3.2–3.6 that u1, u2, and u3 are three different nontrivial

solutions of problem1.1.

Proof ofTheorem 2.2. By Lemmas2.52.7and theProposition 2.8, we can prove the conclusion 3.5. The other proof is similar to that ofTheorem 2.1.

Acknowledgments

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