# Infinitely Many Solutions for a Robin Boundary Value Problem

Authors: Aixia Qian; Chong Li
Journal: International Journal of Differential Equations
 (2010)

---

## Abstract

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.


---

## Body

## 1. Introduction

In this paper, we consider the following equation:(1.1)-Δu=f(x,u),inΩ,∂u∂n+b(x)u=0,on∂Ω,
where Ω is a bounded domain in ℝn with smooth boundary ∂Ω and 0≤b∈L∞(∂Ω). Denote(1.2)F(x,s)=∫0sf(x,t)dt,ℱ=f(x,s)s-2F(x,s),
and let λ1≤λ2≤⋯≤λj<⋯ be the eigenvalues of -Δ with the Robin boundary conditions. We assume that the following hold:(f1)
f∈C(Ω̅×ℝ), ∃q∈(2,2*) such that
(1.3)|f(x,s)|≤c(1+|s|q-1),where1≤s<2N/(N-2), N≥3. If N=1,2, let 2*=∞;(f2)
f(x,s)s≥0, lim|s|→+∞(f(x,s)s)/|s|2=+∞ uniformly for x∈Ω.(f3)
there existθ≥1,s∈[0,1] s.t.(1.4)θℱ(x,t)≥ℱ(x,st),(x,t)∈Ω×ℝ;(f4)
f(x,-t)=-f(x,t), (x,t)∈Ω×ℝ.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(AR)
∃μ>2, R>0 such that(1.5)x∈Ω,|s|≥R⇒0<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 [3](1.6)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 thatX is a Banach space, we say that J∈C1(X,ℝ) satisfies Cerami condition (C), if for all c∈ℝ:(i)
any bounded sequence{un}⊂X satisfying J(un)→c, J′(un)→0 possesses a convergent subsequence;(ii)
there existσ,R,β>0 s.t. for any u∈J-1([c-σ,c+σ]) with ∥u∥≥R, ∥J′(u)∥∥u∥≥β.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 [5]). There is the following Fountain theorem under condition (C).AssumeX=⨁j=1∞Xj¯, where Xj are finite dimensional subspace of X. For each k∈ℕ, let(1.7)Yk=⨁j=1kXj,Zk=⨁j≥kXj¯.
Denote Sρ={u∈X:∥u∥=ρ}.Proposition 1.2.
Assume thatJ∈C1(X,ℝ) satisfies condition (C), and J(-u)=J(u). For each k∈ℕ, there exist ρk>rk>0 such that(i)
bk:=infu∈Zk∩srkJ(u)→+∞, k→∞,(ii)
ak:=maxu∈Yk∩sρkJ(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 [6] proved multiple solutions for the problem (1.1) with Neumann boundary value condition; the author in [3] 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 [8], they showed the existence of one nontrivial solution for problem (1.1), while we get its infinitely many solutions under weaker conditions than [8].Remark 1.6.
In the work in [9], 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 [9]. Therefore, Theorem 1.3 applied to Dirichlet boundary value problem improves those results in [1, 2, 8, 9].


## 2. Preliminaries

Let the Sobolev spaceX=H1(Ω). Denote(2.1)∥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→ℝ:(2.2)J(u)=12∫Ω|∇u|2dx+12∫∂Ωb(x)u2dS-∫ΩF(x,u)dx.
Then by (f1), J is C1 and(2.3)〈J′(u),ϕ〉=∫Ω∇u∇ϕdx+∫∂Ωb(x)uϕdS-∫Ωf(x,u)ϕdx,u,ϕ∈X.
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 functionalJ satisfies condition (C) instead of condition (PS).Lemma 2.1.
Under (f1)–(f3), J satisfies condition (C).Proof.
For allc∈ℝ, we assume that {un}⊂X is bounded and
(2.4)J(un)→c,J′(un)→0,n→∞.
Going, if necessary, to a subsequence, we can assume that un⇀u in X, then
(2.5)∥un-u∥2=∫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,
(2.6)∥un-u∥2+∫∂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 imbeddingW1,2(Ω)↪Lγ(Ω)(1≤γ<2*) is compact, we have the right-hand side of (2.6) converges to 0. While ∫∂Ωb(x)(un-u)2dS≥0, we have ∥un-u∥2→0. It follows that un→u in X and J′(u)=0, that is, condition (i) of Definition 1.1 holds.
Next, we prove condition (ii) of Definition1.1, if not, there exist c∈ℝ and {un}⊂X satisfying, as n→∞(2.7)J(un)→c,∥un∥→∞,∥J′(un)∥∥un∥→0,
then we have
(2.8)limn→∞∫Ω(12f(x,un)un-F(x,un))dx=limn→∞(J(un)-12〈J′(un),un〉)=c.
Denotevn=un/∥un∥, then ∥vn∥=1, that is, {vn} is bounded in X, thus for some v∈X, we get
(2.9)vn⇀v,inX,vn→v,inL2(Ω),vn→v,a.e.  inΩ.
Ifv=0, define a sequence {tn}⊂ℝ as in [4]
(2.10)J(tnun)=maxt∈[0,1]J(tun).
If for some n∈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. x∈Ω that
(2.11)limn→∞∫ΩF(x,v̅n)dx=limn→∞∫ΩF(x,2mvn)dx=0.
Then for n large enough, by (2.9), (2.11), and ∫∂Ωb(x)vn2≥0, we have
(2.12)J(tnun)≥J(v̅n)=12∫Ω|∇v̅n|2dx+12∫∂Ωb(x)v̅n2dS-∫ΩF(x,v̅n)dx=2m∫Ω|∇vn|2dx+2m∫∂Ωb(x)vn2dS-∫ΩF(x,v̅n)dx=2m∥vn∥2-∫ΩF(x,v̅n)dx+2m∫∂Ωb(x)vn2dS-2m∫Ωvn2dx≥2m-∫ΩF(x,v̅n)dx≥m.
That is, limn→∞J(tnun)=+∞. Since J(0)=0 and J(un)→c, then 0<tn<1. Thus
(2.13)∫Ω|∇tnun|2dx+∫∂Ωb(x)(tnun)2dS-∫Ωf(x,tnun)tnundx=〈J′(tnun),tnun〉=tnddt|t=tnJ(tun)=0.
We see that
(2.14)12∫Ωf(x,tnun)tnundx-∫ΩF(x,tnun)dx=12∫Ω|∇tnun|2dx+12∫∂Ωb(x)(tnun)2dS-∫ΩF(x,tnun)dx=J(tnun).
From the aforementioned, we infer that
(2.15)∫Ω(12f(x,un)un-F(x,un))dx=12∫ΩF̃(x,un)dx≥12θ∫ΩF̃(x,tnun)dx=1θ∫Ω[12f(x,tnun)tnun-F(x,tnun)]dx→+∞,n→∞,
which contradicts (2.8).
Ifv≢0, by (2.7)
(2.16)∫Ω|∇un|2dx+∫∂Ωb(x)un2dS-∫Ωf(x,un)undx=〈J′(un,un)〉=o(1).
That is,
(2.17)∥un∥2-∫Ωf(x,un)undx-∫Ωun2dx+∫∂Ωb(x)un2dS=o(1)1-o(1)=∫Ωf(x,un)un∥un∥2dx+∫Ωun2∥un∥2dx-∫∂Ωb(x)un2∥un∥2dS.
Since limn→∞∫Ω(un2/∥un∥2)dx=limn→∞∫Ωvn2=|v|22 exists, and by vn⇀v in X (the weakly convergent sequence is bounded), we get
(2.18)∫∂Ωb(x)un2∥un∥2dS=∫∂Ωb(x)vn2dS≤C∥b∥L∞(∂Ω)∥vn∥2<∞,
where C is the constant of Sobolev Trace imbedding from H1(Ω)→L2(∂Ω), see [10]. We have
(2.19)1-o(1)≥∫Ωf(x,un)un∥un∥2dx-C̃=(∫v≠0+∫v=0)f(x,un)un|un|2|vn|2dx-C̃.
For x∈Ω′:={x∈Ω:v(x)≠0}, we get |un(x)|→+∞. Then by (f2)(2.20)f(x,un(x))un(x)|un(x)|2|vn(x)|2→+∞,n→∞.
By using Fatou lemma, since the Lebesgue measure |Ω′|>0,
(2.21)∫v≠0f(x,un)un|un|2|vn|2dx→+∞,n→∞.
On the other hand, by (f2), there exists γ>-∞, such that (f(x,s)s)/|s|2≥γ for (x,s)∈Ω×ℝ. Moreover,
(2.22)∫v=0∥vn∥2dx→0,n→∞.
Now, there is Λ>-∞ s.t.
(2.23)∫v=0f(x,un)un|un|2|vn|2dx≥γ∫v=0∥vn∥2dx≥Λ>-∞.
Together with (2.19) and (2.21), (2.23), it is a contradiction.
This proves thatJ satisfies condition (C).


## 3. Proof of Theorem1.3

We will apply the Fountain theorem of Proposition1.2 to the functional in (2.2). Let(3.1)Xj=ker(-Δ-λi),Yk=⨁j=1kXj,Zk=⨁j≥kXj¯,
then X=⨁j=1∞Xj¯. It shows that J∈C1(X,ℝ) by (f1) and satisfies condition (C) by Lemma 2.1.(i)
After integrating, we obtain from(f1) that there exist c1>0 such that(3.2)|F(x,u)|≤c1(1+|u|q).Let us defineβk=supu∈Zk∩S1|u|q. By [2, Lemma 3.8], we get βk→0 as k→∞. Since |u|2≤C(Ω)|u|q, let c=c1+(1/2)C(Ω), and rk=(cqβkq)1/2-q, then by (3.2), for u∈Zk with ∥u∥=rk, we have(3.3)J(u)=12∫Ω|∇u|2dx+12∫∂Ωb(x)u2dS-∫ΩF(x,u)dx≥12∥u∥2-c1|u|qq-c1|Ω|+12∫∂Ωb(x)u2dS-12∫Ωu2dx≥12∥u∥2-c1|u|qq-c1|Ω|-12|u|22≥12∥u∥2-c|u|qq-c1|Ω|≥(12-1q)(cqβkq)2/(2-q)-c1|Ω|.
Notice that βk→0 and q>2, we infer that(3.4)bk=infu∈Zk∩srkJ(u)→+∞,k→∞.(ii)
While(3.5)λ1=infu∈H1(Ω)∖{0}∫Ω|∇u|2dx+∫∂Ωα(x)u2dS∫Ωu2dx>0,
we can deduce that ∫Ω|∇u|2dx+∫∂Ωα(x)u2dS is the equivalent norm of ∥u∥2 in X. Since dimYk<+∞ and all norms are equivalent in the finite-dimensional space, there exists Ck>0, for all u∈Yk, we get(3.6)12∫Ω|∇u|2dx+12∫∂Ωb(x)u2dS=12∥u∥2≤Ck|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×ℝ, we obtain(3.7)F(x,s)≥2Ck|s|2-Mk.
It follows from (3.6), (3.7), for all u∈Yk that(3.8)J(u)=12∫Ω|∇u|2dx+12∫∂Ωb(x)u2dS-∫ΩF(x,u)dx=12∥u∥2-∫ΩF(x,u)dx≤-Ck|u|22+Mk|Ω|≤-12∥u∥2+Mk|Ω|.
Therefore, we get that for ρk large enough (ρk>rk),(3.9)ak=maxu∈Yk,∥u∥=ρkJ(u)≤0.
By Fountain theorem of Proposition 1.2, J has a sequence of critical points un∈X, such that J(un)→+∞ as n→∞, that is, (1.1) has infinitely many solutions.Remark 3.1.
By Theorem1.3, the following equation:
(3.10)-Δu=2ulog(1+|u|),inΩ,∂u∂n+b(x)u=0,on∂Ω,
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).



---

*Source: 548702-2009-11-22.xml*