KHADEMLOO Somayehand MOHSENHI Rahelh

Department of Basic Sciences,Babol Noushirvani University of Technology,Babol, Iran.

Received 30 May 2013;Accepted 22 November 2013

Multiple Positive Solutions for Semilinear Elliptic EquationsInvolvingSubcriticalNonlinearitiesinRN

KHADEMLOO Somayeh?and MOHSENHI Rahelh

Department of Basic Sciences,Babol Noushirvani University of Technology,Babol, Iran.

Received 30 May 2013;Accepted 22 November 2013

.In this paper,we study how the shape of the graph of a(z)affects on the number of positive solutions of

We prove for large enough λ,μ＞0,there exist at least k+1 positive solutions of the this semilinear elliptic equations where 1≤q＜2＜p＜2?=2N/(N?2)forN≥3.

AMS Subject Classifications:35J20,35J25,35J65

Chinese Library Classifications:O175.8,O175.25

Sobolev spaces;semilinear elliptic equations;critical exponent;Nehari manifold; Palais-Smale condition.

1 Introduction

For N≥3,1≤q＜2＜p＜2?=2N/(N?2),we suppose the semilinear elliptic equations

where λ,μ＞0.Suppose a,b and h satisfy the following conditions:

(a1)a is a positive continuous function in RNand lim|z|→∞a(z)=a∞＞0.

(a2)There are k points a1,a2,···,akin RNsuch that a(ai)=amax=maxz∈RNa(z);for 1≤i≤k and a∞＜amax.(h1)h∈Lpp?q(RN)∩L∞(RN)and h?0.

(b1)b is a bounded and positive continuous function in RN.

Forμ=1,λ=0,a(z)=b(z)=1 for all z∈RN,we assume the semilinear elliptic equation

where

and the energy functional

We consider the semilinear elliptic equation

have been studied by Huei-li Lin[1](b(z)=1,μ=1 and for N≥3,1≤q＜2＜p＜2?= 2N/(N?2))andshestudiedtheeffectofthecoefficient a(z)ofthesubcritical nonlinearity in RN,Ambrosetti[2](a≡1 and 1＜q＜2＜p≤2?=2N/(N?2)and Wu[3](a∈C(?)and changes sign,1＜q＜2＜p＜2?).They showed that this equation has at least two positive solutions for small enough λ＞0.In[4],Hsu and Lin have studied that there are four positive solutions of the general cases

for small enough λ＞0.

Inthis paper,we studythe existenceand multiplicity ofpositive solutionsoftheequation(Eλ,μ)in RN.By the change of variables

Eq.(Eλ,μ)is converted to

Based on Eq.(Eε,λ),we consider the C1-functional Jε,λ,for u∈H1(RN).

where

is the norm in H1(RN).In fact that d=max{1,b(εz)}then‖u‖H≤‖u‖b≤d‖u‖H,i.e.,‖u‖bis an equivalent norm by‖u‖H.We know that the nonnegative weak solutions of Eq.(Eε,λ)are equivalent to the critical points of Jε,λ.Here we study the existence and multiplicity of positive solutions of Eq.(Eε,λ)in RN.

We organize this paper in this way.In Section 2,we apply the argument of Tarantello [5]to divide the Neharimanifold Mε,λinto twopartsand.?InSection3,we show that the existence of a positive ground state solution u0of Eq.(Eε,λ).In Section 4, there are at least k critical pointsfor 1≤i≤k.Let

2 Main results

Theorem 2.1.Under assumptions a1and h1,if

(a)

where‖h‖#is the norm in Lp?q(RN),then Eq.(Eε,λ)accepts at least a positive ground state solution,(see Theorem 3.4).

(b)Under assumptions a1,a2and h1,if λ is large enough,then Eq.(Eλ,μ)archives at least k+1 positive solutions,(see Theorem 4.10).

For the semilinear elliptic equations

if a=amaxand.We define the energy functional

then γmax=infu∈?Imax(u).

Lemma 2.1.We have

Proof.If

then

Definition 2.1.We define the Palais-Smale(denoted by(PS))-sequences,(PS)-value,and(PS)-conditi

ons in H1(RN)for Jε,λas follows. (i)For β∈R,a sequence{un}is a(PS)β-sequence in H1(RN)for Jε,λif Jε,λ(un)=β+on(1)and J′ε,λ(un)=on(1)strongly in H?1(RN)as n?→∞,where H?1(R)Nis the dual space ofH1(RN); (ii)β∈R is a(PS)-value in H1(RN)for Jε,λif there is a(PS)β-sequence in H1(R)for Jε,λ; (iii)Jε

,λsatisfy the(PS)β-condition in H1(RN)if every(PS)β-sequence in H1(RN)for Jε,λ includ

es a convergent subsequence.

Next,since Jε,λis not bounded form below in H1(RN),we consider the Nehari manifold

where

Notice Mε,λincludes all nonnegative solutions of Eq.(Eλ,μ).

Lemma 2.2.The energy functional Jε,λis coercive and bounded from below on Mε,λ.

Proof.For u∈Mε,λ,the Holder inequality(p1=p/(p?q),p2=p/q)and the Sobolev embedding we get

where

i.e,we have that Jε,λis coercive and bounded from below on Mε,λ.

Definition 2.2.Define

Under assumptions for u∈Mε,λ,we get

We apply the method in Tarantello[5],suppose

Lemma 2.3.Under assumptions a1,a2and h1,if 0＜λ＜Λ,then M0ε,λ=?. Proof.On the contrary,there is a number λ0∈R and 0＜λ0＜Λ such that M0λ0=?.Then for u∈,by(2.2),we have

By the Holder and the Sobolev embedding theorem,we obtain

Thus,

This makes a contradiction.

Lemma 2.4.Suppose that u is a local minimizer for Jε,λon Mε,λand u∈M0ε,λ.Then J′ε,λ(u)=0 in H?1(RN).

Proof.See[6,Theorem 2.3].

Lemma 2.5.For each u∈M+ε,λ,we have

For every u∈Mε+,λ?Mε,λ,by(2.2),we apply the Holder inequality(p1=p/(p?q),p2= p/q)to obtain that

This completes the proof.

Lemma 2.6.For each u∈,we have Proof.For every u∈,by(2.2),we have that

This completes the proof.

So Jε,λ(u)≥d0＞0 for some d0=d0(ε,p,q,S,λ,‖h‖#,amax).

For u∈H1(RN){0} and u+ /≡0, let

Lemma 2.8.For every u∈H1(RN){0}and u+/≡0,we have that,if

then there is a unique positive number.

Proof.For every u∈H1(RN){0}and u+/≡0,define

Clearly,we get that k(0)=0 and k(l)→?∞as l→∞since

then k′(l)=0,k′(l)＞0 for 0＜l＜l,and k′(l)＜0 for l＞l.Thus,k(l)get its maximum at l. Furthermore,by the Sobolev embedding theorem,we have that

There is a unique positive number l?=l?(u)＞l such thatR

and k′(l?)＞0.Then

and

Lemma 2.9.If 0＜λ＜Λ and,then there is unique positive number l+= l+(u)＜l＜l?=l?(u)such that,and

It follows that there are unique positive number l+=l+(u)and l?=l?(u)such that

We also have that

for every l∈[l+,l?],and Jε,λ(l+u)≤Jε,λ(lu)for every l∈[0,l].Hence,

This completes the proof.

Define

Lemma 2.10..

Then

By the definition αε,λand α+ε,λ,we conclude that αε,λ≤α+ε,λ＜0.

Lemma 2.11.Iffor some d0=d0(ε,λ,p,q,S,‖h‖#).

Proof.See[4,Lemma 2.5].

Lemma 2.12.We conclude

3 Existence of a ground state solution

At first,we show that Jε,λsatisfy the(PS)β-condition in H1(RN)for β∈(?∞,γmax?C0λ2?2q),whereLemma 3.1.Under some assumptions a1,a2,h1and 0＜λ＜Λ.If{un}is a(PS)β-sequence in H1(RN)for Jε,λwith un?u weakly in H1(RN),then J′ε,λ(u)=0 in H?1(RN).

Proof.Suppose{un}be a(PS)β-sequence in H1(RN)for Jε,λsuch that Jε,λ(un)=β+on(1)

and Jε′,λ(un)=on(1)in H?1(RN).Then

then

where dn=on(1)as n→∞.It follows that{un}is bounded in H1(RN).Furthermorethere are a subsequence{un}and u∈H1(RN)such that

Lemma 3.2.Under some assumptions a1,a2,h1and 0＜λ＜Λ.If{un}is a(PS)β-sequence in

H1(RN)for Jε,λwith un?u weakly in H1(RN),,where

Hence,by the Young inequality.

This completes the proof.

Lemma 3.3.Assume that a,b and h satisfy a1and h1.If 0＜λ＜Λ.Then Jε,λsatisfy the(PS)β-

Proof.Suppose{un}be a(PS)β-sequence in H1(RN)for Jε,λsuch that and.Then it follows that{un}is bounded in H1(RN).Moreover,there are a subsequence{un}and u∈H1(RN)such thatin H?1(RN). un?u weakly in H1(RN),un→u a.e.in RN,un?u strongly infor every 1≤s＜2?.Next,claim that

Using the Brezis-Lieb lemma to get

For every σ＞0,there is r＞0 so that

By the Holder inequality and the Sobolev embedding theorem,we get

{un}is bounded in H1(RN)and un→u in.Applying a1and, we get that

Let pn=un?u.Suppose pn?0 strongly in H1(RN).By(3.1),(3.2),we conclude that

also

then

By Theorem 4.3 in Wang[7],there is a sequence{sn}?R+such that

It follows that

which is a contradiction.Hence,un→u strongly in H1(RN).

Theorem 3.1.Under some assumptions a1,a2,h1and 0＜λ＜Λ,then there is at least onepositive ground state solution u0of Eq.(Eε,λ)in RN.Moreover,we have that u0∈M+ε,λand

Proof.There is a minimizing sequence{un}?Mε,λfor Jε,λsuch that

By Lemma 3.2(i),there is a subsequence{un}and u0∈H1(RN).We claim that

On the contrary that u0∈,we get that

Otherwise,

It follows that

that contradicts to αε,λ＜0.By Lemma 2.11(ii),then there are positive numbers l+＜l＜ l?=1 such thatand that is a contradiction.Hence,

This completes the proof.

4 Existence of multiple solutions

From this time,we assume that a and h satisfy a1,a2and h1.Suppose w∈H1(RN)be the positive ground state solution of Eq.(E0)in RNfor a≡amax.

(i)w∈L∞(RN)∩C2,θ

loc(RN)for some 0＜θ＜1 and lim|z|→∞w(z)=0.

(ii)For every ε＞0,there are positive numbers C1,Cε2and Cε3such that for all

and

For 1≤i≤k,we define Clearly,.By Lemma 2.11(ii)there is a unique numberso that,where 1≤i≤k.

Lemma 4.1.There is a number t0＞0 such that for 0≤t＜t0and every ε＞0,we have that

Proof.For every ε＞0,we have

Since Jε,λis continuous in H1(RN),is uniformly bounded in H1(RN)for every ε＞0 and γmax＞0 there is t0＞0 such that for 0≤t≤t0and every ε＞0

This completes the proof.

Lemma 4.2.There are positive numbers t1and ε1such that for every t＞t1and ε＜ε1,we have that

Proof.There is an r0＞0 such that a(z)≥amax/2 for z∈BN(ai:r0)uniformly in i.Then is ε1＞0 such that for ε＜ε1

Thus,there is t1＞0 such that for every t＞t1and ε＜ε1

This completes the proof.

Lemma 4.3.Suppose that a1,a2,and h1hold.If 0＜λ＜qΛ/2,then

Proof.By Lemma 4.1 we just try to indicate

uniformly in i;we learn that supt≥0Imax(tw)=γmax.For t0≤t≤t1,we get

Since

as ε→0+uniformly in i.And

then

uniformly in i.

Remark 4.1.Applying the results of Lemma 4.3,we can conclude that

Since there is ε0＞0 such that

where

By Lemma 4.3,there is tε＞0 such that tεwε＞0∈Mε,λfor every 1≤i≤k.

Lemma 4.4.There is 0＜ε0≤ε0such that if ε＜ε0,then Qε((ti ε)?wiε)∈Kρ02for every 1≤i≤k.Proof.Since

There is ε0＞0 such that

This completes the proof.

Lemma 4.5.There is a number δ＞0 such that if u∈? and Imax(u)≤γmax+δ then Qε(u)∈for every 0＜ε＜ε0.

Proof.On thecontrary,thereexistthesequences{εn}?R+and{un}∈? suchthat εn→0+. Iεn(un)=γmax(＞0)+on(1)as n→∞and Qεn(un)/∈Kρ02for all n∈N.It is not difficult to find that{un}is bounded in H1(RN).Suppose that

strongly in Lp(RN).Since

then

That is a contradiction.Then

Thus un?0 strongly in Lp(RN).Also the concentration-compactness principle(see Wang[7,Lemma 2.16],then there is a fixed d0＞0 and a sequence{zn}?RNsuch that

Suppose νn(z)=un(z+zn)then there a subsequence{νn}and ν∈H1(RN)such that νn?ν weakly in H1(RN).Using the same computation in Lemma 2.11.There is a sequence{snmax}?R+such that νn=snmaxνn∈? and

as n→∞.

Weconclude that aconvergentsubsequence{snmax}satisfy snmax→s0＞0.Thenthereare subsequences{νn}and ν∈H1(RN)such that νn?ν(=s0ν)weakly in H1(RN).By(4.2), thenν/=0.Furthermore,wecan obtain that νn→ν stronglyin H1(RN),and Imax(ν)=γmax. Now,we try to indicate that there is a subsequence{zn}={εnzn}such that zn→z0∈K.

(i)Claim that the sequence{zn}is bounded in RN.On the contrary,assume that |zn|→∞,then

that is a contradiction.

(ii)Claim that z0∈K.On the contrary,assume that z0/∈K,that is a(z0)＜amax.Then using the above argument to obtain that

that is a contradiction.Since νn?ν/=0 in H1(RN),we have that a

Hence,there is a number δ＞0 such that if u∈? and Imax(u)≤γmax+δ.Then Qε(u)∈

for every c＜ε0.Choosing 0＜δ0＜δ such that

This completes the proof.

Lemma4.6.and,then there isanumberΛ?＞0sothatfor every 0＜ε＜Λ?.

Proof.We apply the same computation in Lemma 2.11 to obtain that there is a unique positive number

that is

and

that is

Moreover,we have that Jε,λis coercive on Mε,λ,thenfor some C1and C2(independent of u).Next,we claim thatfor some C3(independent of u). On the contrary,there is a sequenceso that(1)as n→∞.By(2.3)

that is a contradiction.Thus,for some C＞0(independent of u).Now,we get that

Form the above inequality,we conclude that

Hence,there is 0＜Λ?≤ε0such that for 0＜ε≤Λ?

By Lemma 4.6,we get

Applying the above lemma,we get that

By Lemmas 4.3,4.4,and Eq.(4.3),there every 0＜ε?＜Λ?.So that

This completes the proof.

Lemma 4.7.Given,then there is an η＞0 and differentiable functional l:B(0;η)? H1(RN)→R+such that

and

Proof.See Cao and Zhou[8].

Lemma 4.8.For each 1≤i≤k,there is a(PS)βiε,λ-sequence.

Proof.See[1,Lemma 4.7].

Theorem 4.1.According to a1,a2,h1,there is a positive number(ε?)?2such that for λ,μ＞(ε?)?2,Eq.(Eλ,μ)has k+1 positive solution in RN.

Proof.We know that there is a(PS)βiε,λ-sequencefor Jε,λfor every 1≤i≤k,and(4.5).Since Jε,λsatisfy the(PS)β-condition for β, then Jε,λhas at least k critical points in

for 0＜ε≤ε?.It follows that Eq.(Eλ,μ) has k nonnegative solution in RN.Applying the maximum principle and Theorem 3.4, Eq.(Eε,λ)has k+1 positive solution in RN.

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?Corresponding author.Email addresses:s.khademloo@nit.ac.ir(S.Khademloo),rm omran78@yahoo.com (R.Mohsenhi)