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R E S E A R C H

Open Access

Multiple solutions for a fourth-order

nonlinear elliptic problem which is

superlinear at

+

and linear at

Ruichang Pei

1,2*

and Jihui Zhang

2

*Correspondence: [email protected] 1School of Mathematics and

Statistics, Tianshui Normal University, Tianshui, 741001, P.R. China

2School of Mathematics and

Computer Sciences, Nanjing Normal University, Nanjing, 210097, P.R. China

Abstract

We consider a semilinear fourth-order elliptic equation with a right-hand side nonlinearity which exhibits an asymmetric growth at +∞and at –∞. Namely, it is linear at –∞and superlinear at +∞. Combining variational methods with Morse theory, we show that the problem has at least two nontrivial solutions, one of which is negative.

Keywords: fourth-order elliptic boundary value problems; multiple solutions; critical groups; Morse theory

1 Introduction

Consider the following Navier boundary value problem:

⎧ ⎨ ⎩

u(x) +cu=f(x,u) in,

u=u=  on, ()

whereis the biharmonic operator, andis a bounded smooth domain inRN (N> ),

andc<λ the first eigenvalue of –inH().

The conditions imposed onf(x,t) are as follows:

(H) fC(¯ ×R,R),f(x, ) = for allxandf(x,t) < for allt< and allx; (H) there existr∈(,p∗)andA,B> such that|ft(x,t)| ≤A+B|t|r– for allx, and

t∈R, wherep∗=NN–, ifN> ;

(H) limt→–∞f(xt,t)=luniformly forx, wherelis a nonnegative constant;

(H) there existβ,ξ∈Rsuch that forF(x,t) =

t

f(x,s)ds, we have lim

t→–∞sup

F(x,t) –f(x,t)tβ uniformly for allx, ()

lim

t→+∞

F(x,t)

t = +∞ uniformly for allx ()

and

lim

t→+∞inf

f(x,t)t– F(x,t)≥ξ uniformly for allx; ()

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(H) there existϑ,ϑ∈L∞()+and an integerk≥such that

λkϑ(x)≤ϑ(x)≤λk+ for allx, ()

the first and the last inequality are strict on sets (not necessary the same) of positive measure, and

ϑ(x)≤ft(x, ) =lim t→

f(x,t)

tϑ(x) uniformly for allx. ()

In view of the conditions (H) and equation () in (H), it is clear that for allx,f(x,t)

is linear at –∞and superlinear at +∞. Clearly,u=  is a trivial solution of problem (). It follows from (H) and (H) that the functional

I(u) =  

|u|–c|∇u|dx

F(x,u)dx ()

isCon the spaceH()∩H() with the norm

u:=

|u|c|∇u|dx

 

,

whereF(x,t) =tf(x,s)ds. Under the condition (H), the critical points ofIare solutions

of problem (). Let  <λ<λ<· · ·<λk<· · · be the eigenvalues of (+c,H()∩ H

()) andφ(x) >  be the eigenfunction corresponding toλ. In fact,λ=λ∗(λ∗–c). Let

Eλk denote the eigenspace associated withλk. Throughout this article, we denoted by| · |p

theLp() norm andE=H()∩H(). The aim of this paper is to prove a multiplicity theorem for problem () when the nonlinearity termf(x,t) exhibits an asymmetric be-havior ast∈Rapproaches +∞and –∞. In the past, some authors studied the following elliptic problem:

–u=f(x,u), uH() () with asymmetric nonlinearities by using the Fučík spectrum of the operator (–,H

()).

This approach requires thatf(x,t) exhibits linear growth at both +∞and –∞and that the limitslimt→±∞f(x,t)

t exist and belong toR. See the works of Các [], Dancer and Zhang [],

Magalhães [], de Paiva [], Schechter [] and the references therein. Equations with non-linearities which are superlinear in one direction and linear in the other were investigated by Arcoya and Villegas [] and Perera []. They let the nonlinearityf(x,t) be line at –∞ and satisfy the Ambrosetti-Rabinowitz condition at +∞. Particularly, it is worth noticing paper []. The authors relax several of the above restrictions on the nonlinearityf(x,t). Their nonlinearity is only measurable inx. The limit ast→–∞off(xt,t) need not ex-ist and the growth at –∞can be linear or sublinear. Furthermore, their nonlinearityf(x,t) does not satisfy the famous AR-condition. They use the truncated skill of first order weak derivative to verify the (PS) condition and obtain multiple solutions for problem () by combining variational methods and Morse theory.

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method in [] cannot be applied directly to the biharmonic problems. For example, for the Laplacian problem,uH

() implies|u|,u+,u–∈H(), whereu+=max(u, ),

u–=max(–u, ). We can useu+oru–as a test function, which is helpful in proving a

so-lution nonnegative. While for the biharmonic problems, this trick fails completely since

uH

() does not implyu+,u–∈H() (see [, Remark ..] and [, ]). As far as

this point is concerned, we will make use of the new methods to overcome it.

This fourth-order semilinear elliptic problem can be considered as an analogue of a class of second-order problems which have been studied by many authors. In [], there was a survey of results obtained in this direction. In [], Micheletti and Pistoia showed that (P) admits at least two solutions by a variation of linking iff(x,u) is sublinear. Chipot []

proved that the problem (P) has at least three solutions by a variational reduction method

and a degree argument. In [], Zhang and Li showed that (P) admits at least two

nontriv-ial solutions by Morse theory and local linking iff(x,u) is superlinear and subcritical onu. In this article, under the guidance of [], we consider multiple solutions of problem () with the asymmetric nonlinearity by using variational methods and Morse theory.

2 Main result and auxiliary lemmas

Let us now state the main result.

Theorem . Assume conditions(H)-(H)hold.If l<λ,then problem()has at least two

nontrivial solutions.

Lemma . Under the assumptions of Theorem.,then I satisfies the(PS)condition.

Proof Let{un} ⊂Ebe a sequence such that for everynN,

|un|–c|∇un|

dx

F(x,un)dx

c, ()

(unv–c∇un∇v)dx

f(x,un)v dx

εnv, vE, ()

wherec>  is a positive constant and{εn} ⊂R+is a sequence which converges to zero. By

a standard argument, in order to prove that{un}has a convergence subsequence, we have

to show that it is a bounded sequence. To do this, we argue by contradiction assuming that for a subsequence, denoted by{un}, we have

un →+∞ asn→ ∞.

Without loss of generality we can assumeun>  for alln∈Nand definezn=uunn.

Ob-viously,zn= ∀n∈Nand then it is possible to extract a subsequence (denoted also by {zn}) such that

znz inE, ()

znz inL(), ()

zn(x)→z(x), a.e.x, ()

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wherez∈EandqL(). Dividing both sides of inequality () byun, we obtain

(znv–c∇zn∇v)dx

f(x,un)

un

v dxεn unv

for allvE.

Passing to the limit we deduce from equation () that

lim

n→∞

f(x,un)

un v dx=

(zvcz∇v)dx ()

for allvE.

Now we claim thatz(x)≤ a.e.x. To verify this, let us observe that by choosing

v=zin equation () we have

lim

n→∞

+ f(x,un)

un

zdx< +∞, ()

where+={x|z

(x) > }. But, on the other hand, from (H) and equation () in (H),

we have

f(x,un(x)) un

z(x)≥

lun(x) –K

un

z(x)≥

lq(x) –K

z(x), a.e.x

for some positive constantK> . Moreover, usinglimn→∞un(x) = +∞a.e.x+,

equa-tion () and the superlinearity off, we also deduce

lim

n→∞

f(x,un(x)) un

z(x) = lim

n→∞

f(x,un(x)) un

zn(x)z(x) = +∞, a.e.x+.

Therefore, if|+|>  we will obtain by Fatou’s lemma that

lim

n→∞

+

f(x,un(x)) un

z(x)dx= +∞

which contradicts inequality (). Thus|+|=  and the claim is proved.

Clearly,z(x)≡, by (H), there existsC>  such that |f(|xu,unn)| | ≤Cfor a.e.x. By

using Lebesgue dominated convergence theorem in equation (), we have

(zv–c∇z∇v)dx

lzv dx=  ()

for allvE. This contradictsl<λ. Lemma . Let E=VW,where V=⊕⊕ · · · ⊕Eλk.If k≥is an integer,ϑ

L∞()+,ϑ(x)≤λk+a.e.onand the inequality is strict on a set of positive measure,then

there existsγ > such that

u–

ϑudxγu

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Proof We claim that there exists a constantϑ<  such that

ϑ(x)udxϑu ()

for alluW. In fact, if not, there exists a sequence{un}such that

ϑ(x)|un|dx>  –

n

un

for allnN, which impliesun=  for alln. By the homogeneity of the above inequality,

we may assume thatun=  and

ϑ(x)|un|dx>  –

n ()

for alln. It follows from the weak compactness of the unit ball ofW that there exists a subsequence, say{un}, such thatunweakly converges touinW. Now Sobolev’s embedding

theorem suggests that{un}converges touinL(). From inequality () we obtain

ϑ(x)|u|dx≥.

Moreover one has

≥ u≥λk+|u|≥

ϑ(x)|u|dx≥.

Hence we have

u=λ

k+|u|

and

λk+–ϑ(x)

udx= 

which implies thatuEλk+\ {}andu=  on a positive measure subset. It contradicts

the unique continuation property of the eigenfunction.

3 Computation of the critical groups

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 book [] for more information on Morse theory.

LetHbe a Hilbert space andIC(H,R) be a functional satisfying the (PS) condition or (C) condition, andHq(X,Y) be theqth singular relative homology group with integer

coefficients. Letube an isolated critical point ofIwithI(u) =c,c∈R, andUbe a

neigh-borhood ofu. The group

Cq(I,u) :=Hq

IcU,IcU\ {u}

, qZ

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LetK:={u∈H:I(u) = }be the set of critical points ofIanda<infI(K), the critical groups ofIat infinity are formally defined by []

Cq(I,∞) :=Hq

H,Ia, qZ.

From the deformation theorem, we see that the above definition is independent of the particular choice ofc<infI(K). Ifc<infI(K) then

Cq(I,∞) :=Hq

HIa, qZ. ()

For the convenience of our proof, we first recall two interesting results and prove two important propositions.

Proposition .[] Under(H),if uE:=H()∩H()is an isolated critical point of

I,then C∗(I,u)∼=C∗(I|C (),u).

Proposition . [] If D ⊂ DE⊂E ⊂X and for some integer k ≥ we have

Hk(E,D)= and Hk(E,D) = ,then either Hk+(E,E)= or Hk–(D,D)= . Proposition . If the assumptions of Theorem.hold,then

Ck(I,∞) =  for all integers k≥.

Proof Under the guidance of [] and [], we begin to prove this result. LetI=I|C(¯).

Indeed, it follows from above Proposition . thatIandI have same critical set. Since

C(¯) is dense inE, invoking Proposition  of Palais [], we have

Hk

E,I˙a=Hk

C(¯),I˙a for allaRand all integersk≥. () From equations () and (), we see that in order to prove the proposition, it suffices to show that

Hk

C(¯),Ia=  for alla<  with|a|large and all integersk≥. () In order to prove equation (), we proceed as follows. We define the sets

∂Bc=uC(¯) :uC (¯)= 

and

∂Bc,+=u∂Bc:u(x) >  for somex. Consider the maph+: [, ]×∂Bc,+→∂Bc,+defined by

h+(t,u) =

( –t)u+

( –t)u+C(¯)

for all (t,u)∈[, ]×∂Bc,+.

Clearly,h+is a continuous homotopy andh(,u) =φfor allx∂Bc,+. Therefore,∂Bc,+is

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By equation () in (H), given anyγ > , we can findC=C(γ) >  such that

F(x,t)≥γt

for allxand alltC. ()

Similarly, from condition (H), and by choosingC>  even bigger if necessary, we observe

that there is a numberγ>  such that

F(x,t)≥–γ  t

for allxand alltC(γ). ()

Moreover, by condition (H), we have

F(x,t)≤C for allxand all|t| ≤C(γ) ()

for someC> .

Letu∂Bc,+. By inequalities (), (), and (), for allt>  we have

I(tu) =t

u

F(x,tu)dx

=t

u

tuC

F(x,tu)dx

tu≤–C

F(x,tu)dx

|tu|≤C

F(x,tu)dx

t

  +

γλ

uγ

tuC udx

+C||. ()

Recalling thatγ>  is arbitrary, from (), we have

I(tu)→–∞ ast→–∞. ()

Using formula () in condition (H), we see that there exist constantsξandM>  such

that

f(x,t)t– F(x,t)≥ξ for allxand alltM. ()

By (H) and formula () in condition (H), we have

F(x,t) –f(x,t)tC for allxand allt<M () for someC> . By inequalities () and (), for anyuEwe have

F(x,t) –f(x,t)tdxC, ()

whereCis a positive constant. Leti:C

(¯)→Ebe the continuous embedding map. Let

·,·denote the duality brackets for the pair (C(¯)∗,C(¯)). We letI=Ii, and so

I(u) =iIi(u) for alluC(¯),

d dtI(tu) =

I(tu),u=tu–

f(x,tu)u dx≤ t

I(tu) +C∗.

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Then, from equation (), we obtain

d

dtI(tu) <  for allt>  large such thatI(tu) < – C

 . ()

From conditions (H) and (H), we see that given> , we can findM>  such that

F(x,t)≤ (l+)t

+M for allxand allt. ()

Using inequality (), we have

I(u)≥ 

ul|u| –|u|

MCuM

foru∈–C+, whereC+is defined as

uC() :u(x)≥ for allx

andC>  is a positive constant. SoI|–C+is coercive, thus we findC∗∗>  such thatI|C+≥

C∗∗. We pick

a<min

C

 , –C

∗∗,inf

∂BcI

.

Then inequality () implies that we can findk(u) >  such that

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

I(tu) >a ift∈[,k(u)),

I(tu) =a ift=k(u),

I(tu) <a ift>k(u).

Moreover, the implicit function theorem implies thatkC(∂Bc

,+, [, +∞)).

By the choice ofa, we have

Ia=tu:u∂Bc,+,tk(u). ()

We define the setE+={tu:u∂Bc,+,t≥}. The maphˆ+: [, ]×E+→E+defined by

ˆ

h+(s,tu) =

⎧ ⎨ ⎩

( –s)tu+sk(u)u if ≤t<k(u),

tu iftk(u), s∈[, ], ()

is a continuous deformation ofE+,hˆ+(,E+)⊂Iaandhˆ+(s,·)|Ia=id|Iafor alls∈[, ] (see

equations () and ()). Therefore,Ia is a strong deformation retract ofE+. Hence we

have

Hk

C(¯),Ia=Hk

C(¯),E+

=Hk

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Recalling that in the first part of the proof, we established that∂Bc,+is contractible. This yields

Hk

C(¯),∂Bc,+=  for all integersk≥.

Combining with equation () leads to equation (), which completes the proof.

Proposition . If the assumptions of Theorem.hold,then Cd(I, )= ,

where d=dimV (V being defined in Lemma.).

Proof By condition (H), given> , we can findδ∗>  such that

 

ϑ(x) –

t≤F(x,t) for allxand all|t| ≤δ∗. ()

SinceVis finite dimensional, all norms are equivalent. Thus we can findρ>  small such that

u ≤ρ ⇔ u∞≤δ∗ ()

for alluV. Taking inequalities () and () into account, for alluVwithu ≤ρwe have

I(u)≤ 

λkϑ(x)

udx+ |u|

. ()

Similar to the proof of Lemma ., there existsC>  such that

I(u)≤(–C+)u≤ ()

for alluVandu ≤ρ.

On the other hand, for given> , it follows from (H) and (H) that

F(x,t)≤ϑ(x) +t

+C|t|r ()

for allxandtR. By () and Lemma ., we have

I(u)≥Cu–Cur ()

for alluW. From inequality (), we infer that forρsmall enough we have

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4 Proof of the main result

Proof of Theorem. We consider the following problem:

u+cu=f

–(x,u), x,

u|∂=u|∂= , where

f–(x,t) =

f(x,t), t< , , t≥.

Define a functionalI–:E=H()∩H()→Rby

I–(u) =

 

|u|c|∇u|dx

F–(x,u)dx,

whereF–(x,t) =

t

f–(x,s)ds, thenI–∈C(E,R). Obviously, by conditions (H) and (H),

we know thatI–is coercive and boundedness from below. Thus we can findv∈Esuch

that

I–(v) =infI–=:m–. ()

Next, we claim thatv= . By condition (H), given∈(,λkλ), there existsδ>  such

that  

ϑ(x) –

t≤F–(x,t) for allx,t∈[–δ, ]. ()

Forssmall enough, it follows from inequality () that

I–(–) =

s

λ–

F–(x, –)dx

s

(λ–λk+) < ,

and thus, by equation (),I–(v)≤I–(–) < , sov= . From condition (H) and strong

maximum principle, we havev<  and

Ck(I–,v) =δk,Z for all integersk≥.

Sincevis an interior point of –C+, from Proposition ., we know

Ck(I,v) =Ck(I–,v) =δk,Z for all integersk≥. ()

Letθ∈R,>  be such thatθ<m–=I(v) < –. We consider the sublevel sets

IIE.

Suppose that  andvare the only critical points ofI. Otherwise, we have a second

non-trivial smooth solution and so we are done. By Proposition ., we have

Hk

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We know thatIsatisfies the (PS) condition (see Lemma .). Hence choosing>  small enough, we have

Hd

I,I=C

d(I, )=  ()

(see Proposition .). Because of equations () and (), using Proposition ., we obtain

Hd+

E,I=  or Hd–

I,= .

IfHd+(E,I)= , then there is a critical pointv∗∈EofIsuch that

Iv∗>>  and v∗= ,v.

IfHd–(I,)= , then there is a critical pointv∗∈EofIsuch that

Cd–

I,v∗= . ()

Sinced≥, from equations () and (), we see thatv∗=v. It is obvious thatv∗= .

Thereforevandv∗are two solutions of problem ().

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

The authors read and approved the final manuscript.

Acknowledgements

The authors would like to thank the referees for valuable comments and suggestions in improving this article. This study was supported by the National NSF (Grant No. 11101319) of China and Planned Projects for Postdoctoral Research Funds of Jiangsu Province (Grant No. 1301038C).

Received: 14 August 2013 Accepted: 16 December 2013 Published:10 Jan 2014

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