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Hindawi Publishing Corporation
International Journal of Differential Equations
Volume 2010, Article ID 548702, 9 pages
doi:10.1155/2010/548702
Review Article
Infinitely Many Solutions for a Robin Boundary
Value Problem
Aixia Qian1 and Chong Li2
1
2
School of Mathematic Sciences, Qufu Normal University, Qufu Shandong 273165, China
Institute of Mathematics, AMSS, Academia Sinica, Beijing 100080, China
Correspondence should be addressed to Aixia Qian, qaixia@amss.ac.cn and
Chong Li, lichong@amss.ac.cn
Received 29 August 2009; Accepted 7 November 2009
Academic Editor: Wenming Zou
Copyright q 2010 A. Qian and C. Li. 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.
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 fx, u,
∂u
bxu 0,
∂n
in Ω,
1.1
on ∂Ω,
where Ω is a bounded domain in Rn with smooth boundary ∂Ω and 0 ≤ b ∈ L∞ ∂Ω. Denote
Fx, s s
fx, tdt,
F fx, ss − 2Fx, s,
1.2
0
and let λ1 ≤ λ2 ≤ · · · ≤ λj < · · · be the eigenvalues of −Δ with the Robin boundary conditions.
We assume that the following hold:
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f1 f ∈ CΩ × R, ∃q ∈ 2, 2∗ such that
fx, s ≤ c 1 |s|q−1 ,
1.3
where 1 ≤ s < 2N/N − 2, N ≥ 3. If N 1, 2, let 2∗ ∞;
f2 fx, ss ≥ 0, lim|s| → ∞ fx, ss/|s|2 ∞ uniformly for x ∈ Ω.
f3 there exist θ ≥ 1, s ∈ 0, 1 s.t.
θFx, t ≥ Fx, st,
x, t ∈ Ω × R;
1.4
f4 fx, −t −fx, 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
x ∈ Ω,
|s| ≥ R ⇒ 0 < μFx, s ≤ fx, ss.
1.5
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
fx, t 2t log1 |t|
1.6
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 J ∈ C1 X, R satisfies Cerami
condition C, if for all c ∈ R:
i any bounded sequence {un } ⊂ X satisfying Jun → c, J un → 0 possesses a
convergent subsequence;
ii there exist σ, 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.
Assume X ∞
j1 Xj , where Xj are finite dimensional subspace of X. For each k ∈ N,
let
Yk Denote Sρ {u ∈ X : u ρ}.
k
j1
Xj ,
Zk j≥k
Xj .
1.7
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3
Proposition 1.2. Assume that J ∈ C1 X, R satisfies condition (C), and J−u Ju. For each
k ∈ N, there exist ρk > rk > 0 such that
i bk : infu∈Zk ∩srk Ju → ∞, k → ∞,
ii ak : maxu∈Yk ∩sρk Ju ≤ 0.
Then J has a sequence of critical points un , such that Jun → ∞ 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 space X H 1 Ω. Denote
u Ω
|∇u|2 u2 dx
1/2
2.1
to be the norm of u in X, and |u|q the norm of u in Lq Ω. Consider the functional J : X → R:
Ju 1
2
Ω
|∇u|2 dx 1
2
bxu2 dS −
∂Ω
Ω
Fx, udx.
2.2
Then by f1 , J is C1 and
J u, φ Ω
∇u∇φdx bxuφdS −
∂Ω
Ω
fx, uφdx,
The critical point of J is just the weak solution of problem 1.1.
u, φ ∈ X.
2.3
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International Journal of Differential Equations
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 ∈ R, we assume that {un } ⊂ X is bounded and
Jun −→ c,
J un −→ 0,
n −→ ∞.
Going, if necessary, to a subsequence, we can assume that un
un − u
2
Ω
u in X, then
|∇un − u|2 un − u2 dx
− bxun − u2 dS
J un − J u, un − u 2.5
∂Ω
2.4
Ω
un − u2 fx, un − fx, u un − u dx.
that is,
un − u
2
bxun − u2 dS ∂Ω
Ω
|∇un − u|2 un − u2 dx
J un − J u, un − u
un − u2 fx, un − fx, u un − u dx.
2.6
Ω
Since the Sobolev imbedding W 1,2 Ω → Lγ Ω 1 ≤ γ < 2∗ is compact, we have the
right-hand side of 2.6 converges to 0. While ∂Ω bxun − u2 dS ≥ 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 Definition 1.1, if not, there exist c ∈ R and {un } ⊂ X
satisfying, as n → ∞
Jun −→ c,
un −→ ∞,
J un un −→ 0,
2.7
then we have
lim
n→∞ Ω
1
1
fx, un un − Fx, un dx lim Jun − J un , un
n→∞
2
2
c.
2.8
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Denote vn un / un , then vn 1, that is, {vn } is bounded in X, thus for some v ∈ X,
we get
vn
v,
in X,
vn −→ v,
in L2 Ω,
vn −→ v,
a.e. in Ω.
2.9
If v 0, define a sequence {tn } ⊂ R as in 4
Jtn un max Jtun .
2.10
t∈ 0,1
If for some n ∈ N, there is a number of tn satisfying 2.10, we choose one of them. For all
√
m > 0, let vn 2 mvn , it follows by vn x → vx 0 a.e. x ∈ Ω that
lim
n→∞ Ω
Fx, vn dx lim
n→∞ Ω
Then for n large enough, by 2.9, 2.11, and
1
2
Jtn un ≥ Jvn 2m
Ω
∂Ω
Ω
|∇vn |2 dx |∇vn |2 dx 2m
∂Ω
2
2m vn
−
Ω
≥ 2m −
Ω
1
2
√
F x, 2 mvn dx 0.
2.11
bxvn2 ≥ 0, we have
∂Ω
bxv2n dS −
bxvn2 dS −
Fx, vn dx 2m
∂Ω
Ω
Ω
Fx, vn dx
Fx, vn dx
bxvn2 dS
2.12
− 2m
Ω
vn2 dx
Fx, vn dx ≥ m.
That is, limn → ∞ Jtn un ∞. Since J0 0 and Jun → c, then 0 < tn < 1. Thus
Ω
|∇tn un |2 dx bxtn un 2 dS −
∂Ω
J tn un , tn un
Ω
fx, tn un tn un dx
d
tn Jtun 0.
dt ttn
2.13
We see that
1
2
Ω
fx, tn un tn un dx −
1
2
Ω
1
|∇tn un | dx 2
Ω
Fx, tn un dx
2.14
2
2
bxtn un dS −
∂Ω
Ω
Fx, tn un dx Jtn un .
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International Journal of Differential Equations
From the aforementioned, we infer that
Ω
1
fx, un un − Fx, un dx
2
1
1
tn un dx
Fx, un dx ≥
Fx,
2 Ω
2θ Ω
1
1
fx, tn un tn un − Fx, tn un dx −→ ∞,
θ Ω 2
2.15
n −→ ∞,
which contradicts 2.8.
If v ≡
/ 0, by 2.7
Ω
|∇un |2 dx ∂Ω
bxu2n dS −
Ω
fx, un un dx J un , un o1.
2.16
That is,
un
2
−
Ω
fx, un un dx −
1 − o1 fx, un un
Ω
2
un
Ω
u2n dx
dx ∂Ω
u2n
Ω
un
2
bxu2n dS o1
bxu2n
dx −
∂Ω
un
2
2.17
dS.
Since limn → ∞ Ω u2n / un 2 dx limn → ∞ Ω vn2 |v|22 exists, and by vn
convergent sequence is bounded, we get
bxu2n
∂Ω
un
2
v in X the weakly
dS ∂Ω
bxvn2 dS ≤ C b
L∞ ∂Ω
vn
2
< ∞,
2.18
where C is the constant of Sobolev Trace imbedding from H 1 Ω → L2 ∂Ω, see 10 . We
have
1 − o1 ≥
fx, un un
Ω
un
2
dx − C
v/
0
fx, un un
v0
|un |
2
|vn |2 dx − C.
2.19
For x ∈ Ω : {x ∈ Ω : vx /
0}, we get |un x| → ∞. Then by f2 fx, un xun x
|un x|2
|vn x|2 −→ ∞,
n −→ ∞.
2.20
By using Fatou lemma, since the Lebesgue measure |Ω | > 0,
fx, un un
v
/0
|un |2
|vn |2 dx −→ ∞,
n −→ ∞.
2.21
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On the other hand, by f2 , there exists γ > −∞, such that fx, ss/|s|2 ≥ γ for x, s ∈ Ω × R.
Moreover,
vn 2 dx −→ 0,
n −→ ∞.
2.22
vn 2 dx ≥ Λ > −∞.
2.23
v0
Now, there is Λ > −∞ s.t.
fx, un un
|un |2
v0
2
|vn | dx ≥ γ
v0
Together with 2.19 and 2.21, 2.23, it is a contradiction.
This proves that J satisfies condition C.
3. Proof of Theorem 1.3
We will apply the Fountain theorem of Proposition 1.2 to the functional in 2.2. Let
Xj ker−Δ − λi ,
then X ∞
j1 Xj . It shows that J
Yk k
Zk X,
j1 j
j≥k
Xj ,
3.1
∈ 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
|Fx, u| ≤ c1 1 |u|q .
3.2
Let us define βk supu∈Zk ∩S1 |u|q . By 2, Lemma 3.8 , we get βk → 0 as k → ∞. Since
q
|u|2 ≤ CΩ|u|q , let c c1 1/2CΩ, and rk cqβk 1/2−q , then by 3.2, for u ∈ Zk with
u rk , we have
1
Ju 2
1
|∇u| dx 2
Ω
1
≥ u
2
≥
1
u
2
2
2
2
−
q
c1 |u|q
bxu dS −
2
∂Ω
1
− c1 |Ω| 2
Fx, udx
1
bxu dS −
2
∂Ω
1
q
− c1 |u|q − c1 |Ω| − |u|22
2
1
q
u 2 − c|u|q − c1 |Ω|
2
1 1 q 2/2−q
−
≥
− c1 |Ω|.
cqβk
2 q
≥
Ω
2
Ω
u2 dx
3.3
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International Journal of Differential Equations
Notice that βk → 0 and q > 2, we infer that
bk inf Ju −→ ∞,
u∈Zk ∩srk
k −→ ∞.
3.4
ii While
λ1 inf
Ω |∇u|
u∈H 1 Ω\{0}
2
dx ∂Ω αxu2 dS
> 0,
u2 dx
Ω
3.5
we can deduce that Ω |∇u|2 dx ∂Ω αxu2 dS is the equivalent norm of u 2 in X. Since
dim Yk < ∞ and all norms are equivalent in the finite-dimensional space, there exists Ck > 0,
for all u ∈ Yk , we get
1
2
1
|∇u| dx 2
Ω
2
bxu2 dS ∂Ω
1
u
2
2
≤ Ck |u|22 .
3.6
Next by f2 , there is Rk > 0 such that Fx, s ≥ 2Ck |s|2 for |s| ≥ Rk . Take Mk : max{0,
inf|s|≤Rk Fx, s}, then for all x, s ∈ Ω × R, we obtain
Fx, s ≥ 2Ck |s|2 − Mk .
3.7
It follows from 3.6, 3.7, for all u ∈ Yk that
1
2
bxu dS −
Fx, udx
|∇u| dx 2 ∂Ω
Ω
Ω
1
2
Fx, udx
u −
2
Ω
1
Ju 2
≤
2
−Ck |u|22
Mk |Ω|
1
u
2
Mk |Ω|.
≤−
2
3.8
Therefore, we get that for ρk large enough ρk > rk ,
ak max
u∈Yk , u ρk
Ju ≤ 0.
3.9
By Fountain theorem of Proposition 1.2, J has a sequence of critical points un ∈ X, such that
Jun → ∞ as n → ∞, that is, 1.1 has infinitely many solutions.
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Remark 3.1. By Theorem 1.3, the following equation:
− Δu 2u log1 |u|,
∂u
bxu 0,
∂n
in Ω,
3.10
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.
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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