IJDEInternational Journal of Differential Equations1687-96511687-9643Hindawi Publishing Corporation54870210.1155/2010/548702548702Review ArticleInfinitely Many Solutions for a Robin Boundary Value ProblemQianAixia1LiChong2ZouWenming1School of Mathematic SciencesQufu Normal UniversityQufu Shandong 273165Chinaqfnu.edu.cn2Institute of MathematicsAMSSAcademia SinicaBeijing 100080Chinasinica.edu201011112009201029082009071120092010Copyright © 2010This 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.

By combining the embedding arguments and the variational methods, we obtain infinitely many solutions for a class of superlinear elliptic problems with the Robin boundary value under weaker conditions.

1. Introduction

In this paper, we consider the following equation:

-Δu=f(x,u),inΩ,un+b(x)u=0,onΩ, where Ω is a bounded domain in n with smooth boundary Ω and 0bL(Ω). Denote

F(x,s)=0sf(x,t)dt,=f(x,s)s-2F(x,s), and let λ1λ2λj< be the eigenvalues of -Δ with the Robin boundary conditions. We assume that the following hold:

fC(Ω̅×), q(2,2*) such that |f(x,s)|c(1+|s|q-1),

where 1s<2N/(N-2), N3. If N=1,2, let 2*=;

f(x,s)s0, lim|s|+(f(x,s)s)/|s|2=+ uniformly for xΩ.

there exist θ1,s[0,1] s.t.

θ(x,t)(x,st),(x,t)Ω×;

f(x,-t)=-f(x,t), (x,t)Ω×.

Because of (f2), (1.1) is usually called a superlinear problem. In [1, 2], the author obtained infinitely many solutions of (1.1) with Dirichlet boundary value condition under (f1), (f4) and

μ>2, R>0 such that

xΩ,|s|R0<μF(x,s)f(x,s)s.

Obviously, (f2) can be deduced form (AR). Under (AR), the (PS) sequence can be deduced bounded. However, it is easy to see that the example 

f(x,t)=2tlog(1+|t|) does not satisfy (AR), while it satisfies the aforementioned conditions (take θ=1 in (f3)). (f3) is from [3, 4].

We need the following condition (C), see [3, 5, 6].

Definition 1.1.

Assume that X is a Banach space, we say that JC1(X,) satisfies Cerami condition (C), if for all c:

any bounded sequence {un}X satisfying J(un)c, J(un)0 possesses a convergent subsequence;

there exist σ,R,β>0 s.t. for any uJ-1([c-σ,c+σ]) with uR, J(u)uβ.

In the work in [2, 7], the Fountain theorem was obtained under the condition (PS). Though condition (C) is weaker than (PS), the well-known deformation theorem is still true under condition (C) (see ). There is the following Fountain theorem under condition (C).

Assume X=j=1Xj¯, where Xj are finite dimensional subspace of X. For each k, let

Yk=j=1kXj,Zk=jkXj¯. Denote Sρ={uX:u=ρ}.

Proposition 1.2.

Assume that JC1(X,) satisfies condition (C), and J(-u)=J(u). For each k, there exist ρk>rk>0 such that

bk:=infuZksrkJ(u)+, k,

ak:=maxuYksρkJ(u)0.

Then J has a sequence of critical points un, such that J(un)+ as n.

As a particular linking theorem, Fountain theorem is a version of the symmetric Mountain-Pass theorem. Using the aforementioned theorem, the author in  proved multiple solutions for the problem (1.1) with Neumann boundary value condition; the author in  proved multiple solutions for the problem (1.1) with Dirichlet boundary value condition. In the present paper, we also use the theorem to give infinitely many solutions for problem (1.1). The main results are follows.

Theorem 1.3.

Under assumptions (f1)–(f4), problem (1.1) has infinitely many solutions.

Remark 1.4.

In the work in [1, 2], they got infinitely many solutions for problem (1.1) with Dirichlet boundary value condition under condition (AR).

Remark 1.5.

In the work in , they showed the existence of one nontrivial solution for problem (1.1), while we get its infinitely many solutions under weaker conditions than .

Remark 1.6.

In the work in , they also obtained infinitely many solutions for problem (1.1) with Dirichlet boundary value condition under stronger conditions than the aforementioned (f2) and (f3) above. Furthermore, function (1.6) does not satisfy all conditions in . Therefore, Theorem 1.3 applied to Dirichlet boundary value problem improves those results in [1, 2, 8, 9].

2. Preliminaries

Let the Sobolev space X=H1(Ω). Denote

u=(Ω(|u|2+u2)dx)1/2 to be the norm of u in X, and |u|q the norm of u in Lq(Ω). Consider the functional J:X:

J(u)=12Ω|u|2dx+12Ωb(x)u2dS-ΩF(x,u)dx. Then by (f1), J is C1 and

J(u),ϕ=Ωuϕdx+Ωb(x)uϕdS-Ωf(x,u)ϕdx,u,ϕX. The critical point of J is just the weak solution of problem (1.1).

Since we do not assume condition (AR), we have to prove that the functional J satisfies condition (C) instead of condition (PS).

Lemma 2.1.

Under (f1)–(f3), J satisfies condition (C).

Proof.

For all c, we assume that {un}X is bounded and J(un)c,J(un)0,n. Going, if necessary, to a subsequence, we can assume that unu in X, then un-u2=Ω(|(un-u)|2+(un-u)2)dx=J(un)-J(u),un-u+Ω-b(x)(un-u)2dS+Ω[(un-u)2+(f(x,un)-f(x,u))(un-u)]dx. that is, un-u2+Ωb(x)(un-u)2dS=Ω(|(un-u)|2+(un-u)2)dx=J(un)-J(u),un-u+Ω[(un-u)2+(f(x,un)-f(x,u))(un-u)]dx.

Since the Sobolev imbedding W1,2(Ω)Lγ(Ω)(1γ<2*) is compact, we have the right-hand side of (2.6) converges to 0. While Ωb(x)(un-u)2dS0, we have un-u20. It follows that unu in X and J(u)=0, that is, condition (i) of Definition 1.1 holds.

Next, we prove condition (ii) of Definition 1.1, if not, there exist c and {un}X satisfying, as nJ(un)c,un,J(un)un0, then we have limnΩ(12f(x,un)un-F(x,un))dx=limn(J(un)-12J(un),un)=c.

Denote vn=un/un, then vn=1, that is, {vn} is bounded in X, thus for some vX, we get vnv,inX,vnv,inL2(Ω),vnv,a.e.  inΩ.

If v=0, define a sequence {tn} as in  J(tnun)=maxt[0,1]J(tun). If for some n, there is a number of tn satisfying (2.10), we choose one of them. For all m>0, let v̅n=2mvn, it follows by vn(x)v(x)=0 a.e. xΩ that limnΩF(x,v̅n)dx=limnΩF(x,2mvn)dx=0. Then for n large enough, by (2.9), (2.11), and Ωb(x)vn20, we have J(tnun)J(v̅n)=12Ω|v̅n|2dx+12Ωb(x)v̅n2dS-ΩF(x,v̅n)dx=2mΩ|vn|2dx+2mΩb(x)vn2dS-ΩF(x,v̅n)dx=2mvn2-ΩF(x,v̅n)dx+2mΩb(x)vn2dS-2mΩvn2dx2m-ΩF(x,v̅n)dxm. That is, limnJ(tnun)=+. Since J(0)=0 and J(un)c, then 0<tn<1. Thus Ω|tnun|2dx+Ωb(x)(tnun)2dS-Ωf(x,tnun)tnundx=J(tnun),tnun=tnddt|t=tnJ(tun)=0. We see that 12Ωf(x,tnun)tnundx-ΩF(x,tnun)dx=12Ω|tnun|2dx+12Ωb(x)(tnun)2dS-ΩF(x,tnun)dx=J(tnun). From the aforementioned, we infer that Ω(12f(x,un)un-F(x,un))dx=12ΩF̃(x,un)dx12θΩF̃(x,tnun)dx=1θΩ[12f(x,tnun)tnun-F(x,tnun)]dx+,n, which contradicts (2.8).

If v0, by (2.7) Ω|un|2dx+Ωb(x)un2dS-Ωf(x,un)  undx=J(un,un)=o(1). That is, un2-Ωf(x,un)undx-Ωun2dx+Ωb(x)un2dS=o(1)1-o(1)=Ωf(x,un)unun2dx+Ωun2un2dx-Ωb(x)un2un2dS. Since limnΩ(un2/un2)dx=limnΩvn2=|v|22 exists, and by vnv in X (the weakly convergent sequence is bounded), we get Ωb(x)un2un2dS=Ωb(x)vn2dSCbL(Ω)vn2<, where C is the constant of Sobolev Trace imbedding from H1(Ω)L2(Ω), see . We have 1-o(1)Ωf(x,un)unun2dx-C̃=(v0+v=0)f(x,un)un|un|2|vn|2dx-C̃. For xΩ:={xΩ:v(x)0}, we get |un(x)|+. Then by (f2)f(x,un(x))un(x)|un(x)|2|vn(x)|2+,n. By using Fatou lemma, since the Lebesgue measure |Ω|  >0, v0f(x,un)un|un|2|vn|2dx+,n. On the other hand, by (f2), there exists γ>-, such that (f(x,s)s)/|s|2γ for (x,s)Ω×. Moreover, v=0vn2dx0,n. Now, there is Λ>- s.t. v=0f(x,un)un|un|2|vn|2dxγv=0vn2dxΛ>-. Together with (2.19) and (2.21), (2.23), it is a contradiction.

This proves that J satisfies condition (C).

3. Proof of Theorem <xref ref-type="statement" rid="thm1.1">1.3</xref>

We will apply the Fountain theorem of Proposition 1.2 to the functional in (2.2). Let

Xj=ker(-Δ-λi),Yk=j=1kXj,Zk=jkXj¯, then X=j=1Xj¯. It shows that JC1(X,) by (f1) and satisfies condition (C) by Lemma 2.1.

After integrating, we obtain from (f1) that there exist c1>0 such that

|F(x,u)|c1(1+|u|q).

Let us define βk=supuZkS1|u|q. By [2, Lemma 3.8], we get βk0 as k. Since |u|2C(Ω)|u|q, let c=c1+(1/2)C(Ω), and rk=(cqβkq)1/2-q, then by (3.2), for uZk with u=rk, we have

J(u)=12Ω|u|2dx+12Ωb(x)u2dS-ΩF(x,u)dx12u2-c1|u|qq-c1|Ω|+12Ωb(x)u2dS-12Ωu2dx12u2-c1|u|qq-c1|Ω|-12|u|2212u2-c|u|qq-c1|Ω|(12-1q)(cqβkq)2/(2-q)-c1|Ω|. Notice that βk0 and q>2, we infer that

bk=infuZksrkJ(u)+,k.

While

λ1=infuH1(Ω){0}Ω|u|2dx+Ωα(x)u2dSΩu2dx>0, we can deduce that Ω|u|2dx+Ωα(x)u2dS is the equivalent norm of u2 in X. Since dimYk<+ and all norms are equivalent in the finite-dimensional space, there exists Ck>0, for all uYk, we get

12Ω|u|2dx+12Ωb(x)u2dS=12u2Ck|u|22. Next by (f2), there is Rk>0 such that F(x,s)2Ck|s|2 for |s|Rk. Take Mk := max{0, inf|s|RkF(x,s)}, then for all (x,s)Ω×, we obtain

F(x,s)2Ck|s|2-Mk. It follows from (3.6), (3.7), for all uYk that

J(u)=12Ω|u|2dx+12Ωb(x)u2dS-ΩF(x,u)dx=12u2-ΩF(x,u)dx-Ck|u|22+Mk|Ω|-12u2+Mk|Ω|. Therefore, we get that for ρk large enough (ρk>rk),

ak=maxuYk,u=ρkJ(u)0. By Fountain theorem of Proposition 1.2, J has a sequence of critical points unX, such that J(un)+ as n, that is, (1.1) has infinitely many solutions.

Remark 3.1.

By Theorem 1.3, the following equation: -Δu=2ulog(1+|u|),inΩ,un+b(x)u=0,onΩ, has infinitely many solutions, while the results cannot be obtained by [1, 2, 8, 9]

Remark 3.2.

In the next paper, we wish to consider the sign-changing solutions for problem (1.1).

Acknowledgments

We thank the referee for useful comments. C. Li is supported by NSFC (10601058, 10471098, 10571096). This work was supported by the Chinese National Science Foundation (10726003), the National Science Foundation of Shandong (Q2008A03), and the Foundation of Qufu Normal University.

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