Four 2—Nodal Solutions for Semilinear
Elliptic Equations in Finite Strip with Hole
Hsiu-Chuan Huang
Department of Mathematics
National Cheng Kung University
Tainan 701 Taiwan
e-mail: [email protected]
Tsung-fang Wu
Center for General Education
Southern Taiwan University of Technology
Tainan 710, Taiwan
e-mail: [email protected]
Abstract
In this paper, we study the decomposition of the filtration of the Nehari manifold via the variation of domain shape. We use this result to prove that the semilinear elliptic equation in a finite strip with hole has at least four 2—nodal solutions (solutions with precisely two nodal domains). Furthermore, we can describe the bump location of these solutions.
1
Introduction
Let N ≥ 2 and 2 < p < 2∗, where 2∗ = 2N
N −2 for N ≥ 3 and 2∗ = ∞ for
N = 2. Consider the nonlinear elliptic equation ½
−∆u + u = |u|p−2u in Ω,
u∈ H1
0(Ω),
where Ω is a smooth bounded domain in RN and H1
0(Ω) is the Sobolev space
in Ω with dual space H−1(Ω). Associated with equation (1), we consider the
energy functional J in H1 0(Ω) J(u) = 1 2 Z Ω (|∇u|2+ u2)−1 p Z Ω|u| p.
It is well known that the solutions of equation (1) are the critical points of the energy functional J and the equation (1) there exist infinitely many solutions (see Ambrosetti-Rabinowitz [1]).
That the number of solutions of equation (1) is affected by the shape of the domain Ω has been the focus of a great deal of research in recent years.
Let x = (x0, xN)∈ RN −1× R and O be a bounded smooth domain in RN −1.
Denote the N —ball BN(x
0; s) in RN, the infinite strip S and the finite strip
Sl,t as follows.
BN(x0; s) = {x ∈ RN | |x − x0| < s};
S = {(x0, xN)∈ RN | x0 ∈ O};
Sl,t = {(x0, xN)∈ S | l < xN < t}.
We should point out here that the precise definition of the finite strip
S−t,t are symmetric and convex domain in xN—axis, and has been smoothed
out at the corners. By the Rellich Compactness Theorem, there is a positive
solution for equation (1) in the finite strip S−t,t. Moreover, S−t,tis convex in
xN—axis. By the famous theorem of Gidas-Ni-Nirenberg [10], every positive
solution of equation (1) in S−t,tis axially symmetric in xN. Actually, Dancer
[7] proved that the positive solution of equation (1) in S−t,t for each t > 0
in R2 is unique. In Byeon [4] and Dancer [7], consider a perturbation of the
finite strip S−t,t, that is dumbbell type domain
D = BN((0,−t) ; r0)∪ S−t,t∪ BN((0, t) ; r0) for O ⊂ BN −1(0; r0) .
They proved that the equation (1) in D has at least three positive solutions,
for O is sufficiently close to a point x0 in RN −1. In Wang-Wu [15] and Wu
[17], consider another perturbation of the finite strip S−t,t,that is finite strip
with hole
Θt= S−t,t\ω,
where ω is a bounded domain in RN
with ω ⊂⊂ S−t0,t0 for some t0 > 0. They
proved that there exists t0 > t0 such that for t > t0, the equation (1) in Θt
In the aforementioned works, the authors considered positive solutions. For other situations, in Bartsch [2] obtained infinite nodal (change sign) so-lutions for equation (1) in bounded domains. In Furtado [8, 9], used the Ljusternik—Schnirelmann category constructed the number of 2—nodal solu-tions depends on the topology of bounded domain Ω. The definition of 2— nodal solution is: for a nontrivial solution u is such that the set {x ∈ Ω | u (x) 6= 0} has exactly two connected components, u is positive in one of them and nega-tive in the other (see Castro-Clapp [5] or Bartsch-Weth [3]). In Bartsch-Weth [3], proved that the equation (1) in a bounded domain Ω contains a large ball has three nodal solutions in which two 2—nodal solutions.
Motivated by the results of Bartsch-Weth [3] and Furtado [8, 9], we are interested in relating the geometry and topology of domain Ω with the num-ber of 2—nodal solutions. Now, we state our main result in this paper. Let
α (S) be a smallest positive Palais—Smale value in H01(S) for J (see Willem
[16, p.73] or Wang [14]). Then we have the following results.
Theorem 1 For each positive number ε ≤ p−2p α (S), there exists t0 > 0 such
that for t > t0,the equation (1) in Θt has four 2-nodal solutions u1,10 , u 1,2 0 , u 2,1 0 and u22 0 with Z Θi t ¯ ¯ ¯¡ui,j0 ¢+¯¯ ¯ p < ε and Z Θjt ¯ ¯ ¯¡ui,j0 ¢−¯¯ ¯ p < ε for all i, j = 1, 2 where u+ = max {u, 0} , u− = u− u+, Θ1 t = {(x0, xN)∈ Θt | xN > 0} and Θ2t ={(x0, xN)∈ Θt | xN < 0} .
Corollary 2 Suppose that the domain ω is axially symmetric in xN—axis.
Then there exists t0 > 0 such that for t > t0, the equation (1) in Θt possesses
at least two non—odd nodal solutions in xN—axis.
Among other interesting results, Bartsch-Weth [3] and Noussair-Wei [13] have considered the effect of domain topology on the existence of nodal so-lutions. Roughly speaking, if Ω has a "rich" topology, then the singular perturbation problem
½
−ε∆u + u = |u|p−2u in Ω
u = 0 on ∂Ω
This paper is organized as follows. In section 2, we describe various preliminaries. In section 3, we use the filtration of Nehari manifold to prove
that the equation (1) in Θt has at least four 2—nodal solutions provided that
t is sufficiently large.
2
Preliminary
In this section, we recall several known results will be used for later sections. First, we define the Palais—Smale (simply by (PS)) sequences, (PS)—values,
and (PS)—conditions in H1
0(Ω) for J as follows:
Definition 3 We define
(i) For β ∈ R, a sequence {un} is a (PS)β—sequence in H01(Ω) for J if
J(un) = β + o(1) and J0(un) = o(1) strongly in H−1(Ω) as n→ ∞;
(ii) β ∈ R is a (PS)—value in H1
0 (Ω) for J if there exists a (PS)β—sequence
in H01(Ω) for J;
(iii) J satisfies the (PS)β—condition in H01(Ω) if every (PS)β—sequence in
H1
0(Ω) for J contains a convergent subsequence.
For any β ∈ R, a (PS)β—sequence in H01(Ω) for J is bounded. Moreover,
a (PS)—value β should be nonnegative.
Lemma 4 Let β ∈ R and {un} be a (PS)β—sequence in H01(Ω) for J, then
there exists a c > 0 such that kunkH1 ≤ c for all n. Furthermore,
Z Ω (|∇un|2+ u2) = Z Ω |un|p+ o(1) = 2p p− 2β + o(1) and β ≥ 0.
Proof. See Willem [16].
Now, we consider the Nehari minimization problem
α(Ω) = inf
u∈M(Ω)J(u),
where M(Ω) = {u ∈ H01(Ω)\{0} | hJ0(u) , ui = 0}. Note that M(Ω)
con-tains every nonzero solution of equation (1) in Ω, α(Ω) > 0 and α (Ω1) ≥
α (Ω2) if Ω1 ⊂ Ω2 (see Wang-Wu[15] or Willem[16]). Moreover, we have the
Lemma 5 Let Ω be a bounded domain in RN. Then the (PS)
α(Ω)—condition
holds in H01(Ω) for J.
Lemma 6 If u ∈ H1
0 (Ω) is a nodal solution of equation (1) in Ω and J (u)≤
3α(Ω), then u is a 2—nodal solution.
Proof. Assume the contrary, without loss of generality, we may assume
that Ω\u−1(0) has three connected components A
1, A2 and A3 such that
Ω\u−1(0) = A1 ∪ A2 ∪ A3, u (z) > 0 for all z ∈ A1 ∪ A2 and u (z) < 0 for
all z ∈ A3. Define u+ = max{u, 0} and u− = u+ − u. Let vi(z) = u+(z)
for z ∈ Ai for i = 1, 2 and v3(z) = u−(z) for z ∈ A3. We note that every
solution u of the equation (1) is a C2—function on Ω. Hence, vj ∈ M (Ω) for
all j = 1, 2, 3 (see Müller-Pfeiffer [12, Lemma 1]). Moreover, J (u) = J (v1) + J (v2) + J (v3) > 3α(Ω),
which is a contradiction.
3
Existence of Four 2—Nodal Solutions
Throughout this section, let ω be a bounded domain in RN such that
ω ⊂⊂ S−t0,t0 for some t0 > 0.
We need the following notations:
Θ1t ={(x0, xN)∈ Θt | xN > 0} and Θ2t ={(x0, xN)∈ Θt | xN < 0} .
For positive numbers ε, δ, let
M (δ, Θt) = {u ∈ M (Θt) | J (u) ≤ α (S) + δ} ; M1(ε, δ, Θt) = ( u∈ M (δ, Θt) | Z Θ1 t |u|p < ε ) ; M2(ε, δ, Θt) = ( u∈ M (δ, Θt) | Z Θ2 t |u|p < ε ) ; Ni,j(ε, δ, Θt) = © u∈ H01(Θt) | u+∈ Mi(ε, δ, Θt) and u− ∈ Mj(ε, δ, Θt) ª for all i, j = 1, 2,
Lemma 7 For each positive number ε ≤ p−2p α (S), there exist positive num-bers δ (ε) , t (ε) such that for t > t (ε) , we have
(i) Mi(ε, δ (ε) , Θt)6= φ for all i = 1, 2;
(ii) M1(ε, δ (ε) , Θt)∩ M2(ε, δ (ε) , Θt) = φ;
(iii) M (δ (ε) , Θt) = M1(ε, δ (ε) , Θt)∪ M2(ε, δ (ε) , Θt) .
Proof. Our proof is almost the same as that in Wu [17, Lemma 3.3] and is
omitted here.
By Lemma 7, for each positive number ε ≤ p−2p α (S) there exist positive
numbers δ (ε) , t (ε) such that for t > t (ε) , Mi(ε, δ (ε) , Θt)6= φ for all i = 1, 2.
Moreover, we have the following results.
Lemma 8 Let δ (ε) , t (ε) > 0 as in Lemma 7, then there exists t0 ≥ t (ε)
such that for t > t0 we have
(i) infu∈Mi(ε,δ(ε),Θt)J (u) < α (S) +
1
2min{δ (ε) , α (S)} for all i = 1, 2;
(ii) Ni,j(ε, δ (ε) , Θt)6= φ for all i, j = 1, 2;
(iii) infu∈Ni,j(ε,δ(ε),Θt)J (u) < 2α (S) + min{δ (ε) , α (S)} for all i, j = 1, 2;
(iv) Ni,j(ε, δ (ε) , Θt) are disjoint.
Proof. (i) , (ii) and (iii) By the Lien-Tzeng-Wang [11, Lemma 2.2], we have
α³St 2,t ´ = α³S0,t 2 ´ & α (S) as t % ∞.
Thus, there exists t0 ≥ t (ε) such that
α ³ St 2,t ´ = α ³ S0,t 2 ´ < α (S) +1 2min{δ (ε) , α (S)}
for all t > t0.By Lemma 5, the equation (1) in S0,2t and in St2,t have positive
solutions u1 ∈ M ³ S0,t 2 ´ and u2 ∈ M ³ St 2,t ´ such that J (u1) = α ³ S0,t 2 ´ and J (u2) = α ³ St 2,t ´ . Set vi(x0, xN) = u1 ³ x0, (−1)i xN ´ and wi(x, y) = u2 ³ x0, (−1)ix N ´ . Clearly, vi, wi ∈ M (Θt) , J (vi) = J (wi) = α ³ S0,t 2 ´ < α (S) +1 2min{δ (ε) , α (S)} (2) and Z Θi t |vi| p = Z Θi t |wi| p = 0
for all i = 1, 2 and t > t0. We obtain vi, wi ∈ Mi(ε, δ (ε) , Θt) (3) and inf u∈Mi(ε,δ(ε),Θt) J (u) < α (S) +1 2min{δ (ε) , α (S)}
for all i = 1, 2 and t > t0. Let ui,j = vi − wj. By (2) and (3) we obtain
ui,j ∈ Ni,j(ε, δ (ε) , Θt) and
inf
u∈Ni,j(ε,δ(ε),Θt)
J (u)≤ J (ui,j) = J (vi) + J (wj) < 2α (S) + min{δ (ε) , α (S)}
for all i, j = 1, 2, · · · , m and t > t0.
(iv) Because the proof of every case is similar. Thus, we only need to prove
the case ”1, 1 and 1, 2”. Assume the contrary, then there exist t > t0 v0 ∈
N1,1(ε, δ (ε) , Θt)∩ N1,2(ε, δ (ε) , Θt) such that Z Θ1 t ¯ ¯v− 0 ¯ ¯p < ε and Z Θ2 t ¯ ¯v− 0 ¯ ¯p < ε. Since v−0 ∈ M (Θt) ,we have 2p p− 2α (Θt) ≤ Z Θt ¯ ¯v− 0 ¯ ¯p ≤ Z Θ1 t ¯ ¯v− 0 ¯ ¯p + Z Θ2 t ¯ ¯v− 0 ¯ ¯p < 2p p− 2α (S) , which is a contradiction.
Let Mi(ε, δ (ε) , Θt)and Ni,j(ε, δ (ε) , Θt)be denoted the closure of Mi(ε, δ (ε) , Θt)
and Ni,j(ε, δ (ε) , Θt) respectively, then we have the following result.
Lemma 9 Let δ (ε) , t0 > 0 as in Lemma 8, then for t > t0 we have
(i) Mi(ε, δ (ε) , Θt) = Mi(ε, δ (ε) , Θt) for all i = 1, 2;
(ii) Ni,j(ε, δ (ε) , Θt) = Ni,j(ε, δ (ε) , Θt) for all i, j = 1, 2.
Proof. (i)The proof of cases ”1” and ”2” are the similar arguments.
There-fore, we only need to prove the case ”1”. Suppose that u0 is a limit point of
M1(ε, δ (ε) , Θt) , then Z Θ1 t |u0| p ≤ ε ≤ p p − 2α (S)
and
J (u0)≤ α (S) + δ (ε) .
The fact that u0 ∈ M (δ (ε) , Θt). Since
M (δ (ε) , Θt) = M1(ε, δ (ε) , Θt)∪ M2(ε, δ (ε) , Θt) and M1(ε, δ (ε) , Θt)∩ M2(ε, δ (ε) , Θt) = φ. If RΘ1 t |u0| p = ε, then u0 ∈ M2(ε, δ (ε) , Θt) . We obtain 2p p− 2α (Θt) ≤ Z Θt |u0| p = Z Θ1 t |u0| p + Z Θ2 t |u0| p < 2p p− 2α (S) ,
which is a contradiction. Therefore, Mi(ε, δ (ε) , Θt) = Mi(ε, δ (ε) , Θt) for
all i = 1, 2.
(ii) By part (i) .
Now, we will to consider the minimization problem in Ni,j(ε, δ (ε) , Θt)
for J,
θi,j(ε, δ (ε) , Θt) = inf
u∈Ni,j(ε,δ(ε),Θt)
J (u) .
Clearly, θi,j(ε, δ (ε) , Θt)≥ 2α (Θt) for all i, j = 1, 2 and t > t0.Here, we will
use the idea of Clapp-Weth [6] to get the following results.
Lemma 10 For each v0 ∈ Ni,j(ε, δ (ε) , Θt) , there exist a map h : H01(Θt)→
R2 such that
(i) h¡s1v0++ s2v0−
¢
= (s1, s2) for s1, s2 ≥ 0,
(ii) h (u) = (1, 1) if and only if v0 ∈ Ni,j(ε, δ (ε) , Θt) .
Proof. Similarly to the method used in Clapp-Weth [6, Lemma 13].
Proposition 11 Let λ0 = 2α (S)+min{δ (ε) , α (S)}−θi,j(ε, δ (ε) , Θt) , then
for each λ ∈ (0, λ0) and μ > 0 there exists u0 ∈ H01(Θt) such that
(i) dist (u0, Ni,j(ε, δ (ε) , Θt))≤ μ,
(ii) J (u0)∈ [θi,j(ε, δ (ε) , Θt) , θi,j(ε, δ (ε) , Θt) + λ),
(iii) k∇J (u0)k ≤ max
n√ λ,λμo.
Proof. Fix v0 ∈ Ni,j(ε, δ (ε) , Θt) such that J (v0) < θi,j(ε, δ (ε) , Θt) + λ,
and fix l0 > 1 such that J
¡ l0v±0
¢
≤ 0. Let h : H01(Θt)→ R2 as in Lemma 10.
We put K = [0, l0]× [0, l0] and define
γ : K → H01(Θt) , γ (s1, s2) = s1v+0 + s2v−0.
Then h ◦ γ = id : K → K, in particular
deg (h◦ γ, K, (1, 1)) = 1. (4)
Notice also that
J (γ (s1, s2))≤ J (v0) < θi,j(ε, δ (ε) , Θt) + λ for every (s1, s2)∈ K. (5)
We now choose a Lipschitz continuous function χ : R → R such that 0 ≤ χ ≤
1, χ (s) = 1for s ≥ 0 and χ (s) = 0 for s ≤ −1. Then, since ∇J (u) ∈ H01(Θt)
for every u ∈ H1
0(Θt) , there is a global semiflow ϕ : [0, ∞) × H01(Θ (r)) →
H1
0(Θt) satisfying
½ ∂
∂tϕ (t, u) =−χ (J (ϕ (t, u))) ∇J (ϕ (t, u)) ,
ϕ (0, u) = u.
We will frequently write ϕt
in place of ϕ (t, ·) . Since J¡v±0¢< α (S) + min{δ (ε) , α (S)} and J¡l0v±0 ¢ ≤ 0, it follows that sup J (γ (∂K)) < α (S) + min{δ (ε) , α (S)} . Hence ¡ ϕt◦ γ¢(∂K)∩ N (Θt) = φ for every t ≥ 0
and, by Lemma 10, this implies ¡
h◦ ϕt◦ γ¢(y)6= (1, 1) for every y ∈ ∂K, t ≥ 0,
where N (Θt) = {u ∈ H01(Θt) | u± ∈ M (Θt)} . Equality (4) and the global
continuation principle of Leray—Schauder (see e.g. [18, p.629]) imply that there exists a connected subset Z ⊂ K × [0, 1] such that
(1, 1, 0)∈ Z ϕt(γ (s
1, s2))∈ N (Θt) for every (s1, s2, t)∈ Z
We put e Z =©ϕt(γ (s1, s2))∈ N (Θt) | (s1, s2, t)∈ Z ª . By inequality (5) , sup u∈ eZ
J (u) < θi,j(ε, δ (ε) , Θt) + λ < 2α (S) + min{δ (ε) , α (S)}
So, since Z is connected, we obtain that eZ ⊂ Ni,j(ε, δ (ε) , Θt) .We now pick
(¯s1, ¯s2, 1)∈ Z ∩ (K × {1}) and write
v1 := γ (¯s1, ¯s2) , v2 := ϕ1(v1) .
Then v2 ∈ eZ ⊂ Ni,j(ε, δ (ε) , Θt) . We distinguish two case.
Case 1. kϕt(v1)− v2k ≤ μ for all t ∈ [0, 1] . We choose t0 ∈ [0, 1] with
° °∇J¡ϕt0(v 1)¢°°= min 0≤t≤1 ° °∇J¡ϕt(v1)¢°°
and put u0 = ϕt0(v1) .Then
λ ≥ J (v1)− J (v2) =− Z 1 0 ∂ ∂tJ ¡ ϕt(v1) ¢ dt = Z 1 0 ° °∇J¡ϕt(v1)¢°° 2 dt≥ k∇J (u0)k2.
Hence u0 has the desired properties.
Case 2. There exists ¯t∈ [0, 1] such that °°ϕt¯(v 1)− v2 ° ° H1 > μ. Then let t1 = sup © t ≥ ¯t | °°ϕt(v1)− v2 ° ° H1 > μ ª . We choose t0 ∈ [t1, 1]with ° °∇J¡ϕt0(v 1)¢°°= min t1≤t≤1 ° °∇J¡ϕt(v1)¢°°
and put u0 = ϕt0(v1) .Then
μ≤ Z 1 t1 ° ° ° °∂t∂ ϕt(v1) ° ° ° ° dt ≤ Z 1 t1 ° °∇J¡ϕt(v1)¢°°dt and λ ≥ J¡ϕt1(v 1) ¢ − J (v2) = Z 1 t1 ° °∇J¡ϕt(v1)¢°° 2 dt ≥ k∇J (u0)k Z 1 t1 ° °∇J¡ϕt(v1)¢°°dt.
Corollary 12 For every t > t0, there exists a sequence {ui,jn } ⊂ H01(Θ (r))
such that (i) dist (ui,j
n , Ni,j(ε, δ (ε) , Θt))→ 0,
(ii) J (ui,j
n )→ θi,j(ε, δ (ε) , Θt) ,
(iii) J0(ui,jn ) = o(1) strongly in H−1(Θt) .
Now, we begin to show the proof of Theorem 1: Fix i, j ∈ {1, 2}
and t > t0. By Corollary 12, there exists a sequence {ui,jn } ⊂ H01(Θ (r))
such that dist (ui,j
n , Ni,j(ε, δ (ε) , Θt)) → 0, J (ui,jn ) → θi,j(ε, δ (ε) , Θt) and
J0(ui,j
n ) = o(1) strongly in H−1(Θt) . Then by the Rellich compactness
the-orem and Lemma 4 there exist subsequences {ui,j
n } and a nodal solution
ui,j0 such that ui,j n → u
i,j
0 strongly in H01(Θt) and J
¡
ui,j0 ¢ = θi,j(ε, δ (ε) , Θt).
Since θi,j(ε, δ (ε) , Θt) < 2α (S) +min{δ (ε) , α (S)} for all i, j ∈ {1, 2} . Thus,
by Lemmas 8 9 ui,j0 ∈ Ni,j(ε, δ (ε) , Θt) and ui,j0 are different. Moreover, by
Lemma 6 u1,10 , u1,20 , u2,10 and u22
0 are 2—nodal solutions of equation (1) in Θt.
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