& Applications
Volume 1, Number 4 (2009), 497–515
MULTIPLICITY OF 2–NODAL SOLUTIONS FOR A SEMILINEAR ELLIPTIC EQUATION
TSUNG-FANGWU (Communicated by C. Alves)
Abstract. In this paper, we consider the multiplicity of 2-nodal solutions of semilinear elliptic
equations. Using the generalized barycenter map, we prove that existence of multiple 2-nodal solutions for semilinear elliptic equations in some domains with hole.
1. Introduction
In this paper, we study the multiplicity of 2-nodal solutions of semilinear elliptic equations of the form
−Δu + u = |u|p−2u++ |u|q−2u− in Ω,
u= 0 on ∂Ω, (Ep,q)
whereΩ is a domain in RN, 2 < p,q < 2∗2∗= 2N
N−2 if N 3,2∗= ∞ if N = 2
,u+= max{0,u} and u−= min{u,0}. Associated with equation (E
p,q), we consider the
energy functional J in the Sobolev space H1 0(Ω), J(u) =12 u2−1 p Ωu +pdx−1 q Ωu −qdx, where u = Ω |∇u|2+ u2dx1/2 is a standard norm in H1
0(Ω). It is well-known that the functional J ∈ C2
H1 0(Ω),R
and the solutions of equation(Ep,q) in Ω are the critical points of the energy functional
J in H1
0(Ω) (see Ambrosetti-Rabinowitz [1] and Willem [23]).
Generally, a standard technique to find the one sign solutions of equation (Ep,q)
inΩ is using the Nehari minimization problems:
α±(Ω) = inf v∈M±(Ω)J(v), Mathematics subject classification (2000): 35J60, 35J20, 35J25.
Keywords and phrases: semilinear elliptic equations, multiple nodal solutions, Palais-Smale.
c
, Zagreb
where
M±(Ω) = u∈ H1
0(Ω)\{0} |
J(u),u = 0,±u 0 .
Note that α±(Ω) are positive numbers and α±(Ω1) α±(Ω2) if Ω1⊂ Ω2 (see Willem [23]). Furthermore, we call a nonzero critical point u0 of J is a least en-ergy positive (or negative) solution of equation (Ep,q) in Ω if u0> 0 (or < 0) and J(u0) =α+(Ω) (orα−(Ω)).
That the existence of one sign solutions of equation(Ep,q) is affected by the shape of the domain Ω has been the focus of a great deal of research in recent years. By the Rellich compactness theorem, it is easy to obtain a one sign solution of equation (Ep,q) in bounded domains. For general unbounded domains Ω, because of the lack of compactness, the issue of existence of one sign solutions of equation(Ep,q) in Ω is very difficult and unclear. Indeed, a by now classical result of Esteban-Lions [14] states that for unbounded domains satisfying the condition: there exists χ∈ RN,||χ|| = 1 such that n(z)·χ 0 and n(z)·χ≡ 0 on∂Ω, where n(z) is the unit outward normal vector to∂Ω at the point z, equation (Ep,q) does not admit any nontrivial solution. Recently, there has benn some progress for the existence and multiplicity of one sign solutions of equation (Ep,q) in unbounded domains (see Ambrosetti-Rabinowitz [1], Bahri-Lions [2], Benci-Cerami [5], Berestycki-Lions [6], Lions [20], Lien-Tzeng-Wang [19], Del Pino-Felmer [12, 13] and Wu [24, 25], etc.). Furthermore, if Ω = RN, then equation (Ep,q) has a unique positive solution (see Kwong [18]). For the equation (Ep,q) in exterior domainΩ, we can see that the Mountain Pass value is equal to the first level of breaking down of Palais-Smale condition (see Benci-Cerami [5]) and we cannot get a positive solution through the Mountain Pass Theorem (i.e. equation(Ep,q) does not admit any least energy solution). However, Benci-Cerami [5] and Bahri-Lions [2] showed the existence of at least one positive solution of equation (Ep,q) in exterior domainΩ.
In the works mentioned above, the authors considered one sign solutions. For other situations, Wang [21] proved the existence of a nodal (sign-changing) solution if the domainΩ is bounded and the operator is −Δ rather than −Δ + 1 for equation (Ep,q). Bartsch [3] obtained infinite nodal solutions for equation (Ep,q) in bounded domains. Furtado [15, 16] used the Ljusternik-Schnirelmann category and showed that the number of 2-nodal solutions of equation(Ep,q) depends on the topology and the symmetries of a symmetric bounded domainΩ. A 2-nodal solution is a nontrivial so-lution u such that the set{x ∈ Ω : u(x) = 0} has exactly two connected components, and u is positive in one of them and negative in the other (see Castro-Clapp [8] or Bartsch-Weth [4]). Bartsch-Weth [4] proved that equation(Ep,q) in a bounded domain Ω that contains a large ball has three nodal solutions in which two are 2-nodal solu-tions. Huang-Wu [17] proved that equation(Ep,q) in a finite strip with hole has at least four 2-nodal solutions. Wu [26], proved that equation(Ep,q) in an infinite upper strip with m−holes has at least m2 2-nodal solutions.
Motivated by the results of [4, 17, 26], we are interested in the relation between the “holes” of domain and the number of 2-nodal solutions of equation(Ep,q). Before stating our main results, we need the following notations. Denote the infinite strip A,
the upper half strip A+and the finite strip A(s,l) as follows:
A= (x,xN) ∈ Θ × R : Θ is a bounded domain in RN−1 ; A+= {(x,xN) ∈ A : xN > 0};
A(s,l) = {(x,xN) ∈ A : s < xN< l}. Then we have the following result.
THEOREM1.1. There exists a positive number l0 such that for l> l0 and the
bounded domainΩ(l) satisfy A(−l,l) ⊂ Ω(l) ⊂ A, equation (Ep,q) in Ω(l) has at least two 2-nodal solutions.
Let Ω(l) be a bounded domain as in Theorem 1.1 and letω ⊂ A be a nonempty open set such that A− l, l\ω is a domain in RN for some l> 0. Then [17] proved that equation (Ep,q) in Ω(l)\ω has at least four 2-nodal solutions if l sufficiently large. Here we will use the generalized barycenter map to improve the result of [17]. Our result is the following theorem.
THEOREM1.2. There is l0> l such that for l > l0 equation (Ep,q) in Ω(l)\ω has at least six 2-nodal solutions.
Next, we consider the upper infinite strip with m-holes Dm(l) = A+\[∪mi=1(ω+ (0,il))].
By Wu [26], we know that equation(Ep,q) in Dm(l) has at least m22-nodal solutions if l sufficiently large. Here we can show that existence of more than m22-nodal solutions. Our result is the following theorem.
THEOREM1.3. There exists l0> lsuch that for l > l0, equation (Ep,q) in Dm(l) has at least m× (m + 1) 2-nodal solutions.
This paper is organized as follows. In Section 2, we set up preliminaries. In Sections 3-5, we complete the proofs of our Theorems 1.1-1.3.
2. Preliminaries
In this section, we recall several known results will be used in later section. First, we define the Palais-Smale (denoted by (PS)) sequences in H1
0(Ω) for J as follows.
DEFINITION2.1. For β ∈ R, a sequence {un} is a (PS)β-sequence in H01(Ω)
for J if J(un) =β+ o(1) and J(un) = o(1) strongly in H−1(Ω) as n → ∞. Now, we consider the minimization problems
α±(Ω) = inf
where M±(Ω) = u∈ H1 0(Ω)\{0} | J(u),u = 0,±u 0 and N(Ω) = u∈ H1 0(Ω) | u+∈ M+(Ω),u−∈ M−(Ω) . Clearly,α+(Ω) +α−(Ω) θ(Ω). We need the following definition.
DEFINITION2.2. (i) The domain Ω is called a large domain in A if Ω ⊂ A and
for any n> 0 there exist s < l such that l − s = n and A(s,l) ⊂ Ω;
(ii) The domain Ω is called a strictly large domain in A if Ω is a large domain in A andΩ = A.
Note that the infinite strip A is a large domain in itself and the upper half strip with m-holes Dm(l) is a strictly large domain in A for all l > 0. Furthermore, by Lien-Tzeng-Wang [19, Lemma 2.5] we have the following result.
LEMMA2.3. IfΩ is a large domain in A, thenα±(Ω) =α±(A). Furthermore, if Ω is a strictly large domain in A, then equation (Ep,q) in Ω does not admits any least energy one sign solution.
LEMMA2.4. If u is a nodal solution of the equation (Ep,q) in Ω and J(u) <
α+(Ω) +α−(Ω) + min{α+(Ω),α−(Ω)}, then u is a 2-nodal solution of equation (Ep,q) in Ω.
Proof. The proof is similar to that of Proposition 3.1 in Furtado [15] (or see Bartsch-Weth [4].
Now, we recall the generalized barycenter map (cf. Bartsch-Weth [4, Theorem 2.1] and Cerami-Passaseo [10]) given byΦ : LpRN\{0} → RN a continuous map satisfying
Φ(u(z −ζ)) =ζ+ Φ(u) and Φu◦ A−1= AΦ(u)
for everyζ ∈ RN, every orthogonal N× N matrix A and every u ∈ LPRN\{0}. Since Lp(A) ⊂ LpRNand the infinite strip A is only translation invariant and sym-metric on xN-axis. Therefore, we may redefine a new generalized barycenter map h : Lp(A)\{0} → R such that forξ ∈ R and u ∈ Lp(A)\{0}, we have
hux,xN−ξ=ξ+ hux,xNand hux,−xN= −hux,xN. (1) Then we have the following result.
LEMMA2.5. Let Ω be a large domain in A. Then for each positive number L there exists a positive numberδ(L) such that for u ∈ N(Ω) with J (u) θ(A)+δ(L) we have either h(u+) − h(u−) > L or h(u−) − h(u+) > L. Furthermore, if Ω is a strictly large domain in A, then:
(i) the function u+ satisfies either h(u+) > L or h(u+) < −L; (ii) the function u− satisfies either h(u−) > L or h(u−) < −L.
Proof. Suppose otherwise, then there exist L0> 0 and a sequence {un} ⊂ N(Ω) such that J(un) =θ(A)+ o(1),
h(u+
n) − h(u−n) L0 and h(u−n) − h(u+n) L0. (2) By Lemma 2.3 and Wu [26, Theorem 1.2]
lim
n→∞J(un) =θ(A) =α
+(A)+α−(A)
and
J(un) = Ju+n+ Ju−nα+(A) +α−(A).
This implies limn→∞J(u±n) =α±(A). Since u±n ∈ M±(Ω) ⊂ M±(A), by Wang-Wu [22, Lemma 7], {u±
n} are (PS)α±(A)-sequences in H01(A) for J. Clearly, {u±n} are
bounded sets in H1
0(A). Then by using a similar argument as in Lien-Tzeng-Wang [19, Theorem 4.1], there exist R,C0> 0 and {y±n} ⊂ R with |y±n| → ∞ such that
A(−R,R)−(0,y±n)
u±n2dx C0for all n∈ N.
Moreover, by Lien-Tzeng-Wang [19, Theorem 4.1] and Chen-Chen-Wang [9], equation (Ep,q) in A has a positive solution u+0 and a negative solution u−0 such that u±0 are axially symmetric in xN-axis and
u± n − u±0
x,xN− y±
nH1→ 0 as n → ∞. (3)
Now we will show that |y+
n− y−n| → ∞ as n → ∞. Suppose otherwise, then we can assume that y+
n− y−n → y0for some y0∈ R. Then by (3) 0= Au + nru−nsdx= Au + 0 x,xN− y+ nru−0 x,xN− y− nsdx+ o(1) = Au + 0 x,xNru−0 x,xN− y−n+ y+nsdx+ o(1) = Au − 0 x,xNru− 0 x,xN+ y0sdx+ o(1), which is a contradiction, where rp+s
q= 1. Thus, |y+n− y−n| → ∞ as n → ∞. Moreover, by(3) we can conclude h(u+ n) = h(u+0 x,xN− y+ n ) + o(1) = h(u+ 0) + y+n+ o(1) (4) and h(u− n) = h(u−0 x,xN− y− n ) + o(1) = h(u− 0) + y−n+ o(1). (5) This implies h(u+ n) − h(u−n) → ∞ as n → ∞
this contradicts(2). Next, for Ω is a strictly large domain in A, then by (4) and (5) we only need to prove that
y+
The proofs of two cases are similar argument. Therefore, we only need to prove the case |y+
n| → ∞ as n → ∞. Suppose otherwise, then {y+n} is a bounded sequence in R. Without loss of generality, we may assume that y+
n → y0 for some y0∈ R. Since u+n ∈ H1 0(Ω). Then u+nx,xN+ y+n ∈ H1 0 Ω −0,y+n . By(3) and limn→∞[Ω − (0,y+
n)] → [Ω − (0,y0)] we obtain [Ω − (0,y0)] = A and u+0 ∈ H1
0(Ω − (0,y0)) which contradicts the fact that u+0 is a positive solution of equation (Ep.q) in A.
3. Proof of Theorem 1.1
For positive numbers L,δ and the domain Ω ⊂ A, we denote N(δ,Ω) = {u ∈ N(Ω) : J (u) θ(A) +δ}, Ni(L,δ,Ω) =
u∈ N(δ,Ω) : (−1)ig(u) > L, where g(u) = h(u+) − h(u−). Then we have the following results.
LEMMA3.1. Let L andδ(L) be positive numbers as in Lemma 2.5. Then there exists a positive number l0 such that for l> l0 and the bounded domainΩ(l) satisfy A(−l,l) ⊂ Ω(l) ⊂ A, we have:
(i) Ni(L,δ(L),Ω(l)) = /0 for all i = 1,2;
(ii) N(δ(L),Ω(l)) = N1(L,δ(L),Ω(l)) ∪ N2(L,δ(L),Ω(l)); (iii) N1(L,δ(L),Ω(l)) ∩ N2(L,δ(L),Ω(l)) = /0.
Proof. By N(Ω(l)) ⊂ N(A) for all l > 0 and Lemma 2.5, we only need to prove that there exists l0> 0 such that Ni(L,δ(L),Ω(l)) = /0 for all l > l0 and i= 1,2. By Lien-Tzeng-Wang [19, Lemma 2.2], α±A−l 2 , l 2 α±(A) as l ∞. Then there exists l0> L such that
α±A−l 2 , l 2 <α±(A) +δ(L) 2 for all l> l0.
Moreover, by Ambrosetti-Rabinowitz [1] and Chen-Chen-Wang [9], equation (Ep,q) in A(−l/2,l/2) has a positive solution v+∈ M+(A(−l/2,l/2)) and a negative solu-tion v−∈ M−(A(−l/2,l/2)) such that J (v±) =α±(A(−l/2,l/2)) and v±(x,−y) = v±(x,y). Clearly, h(v±) = 0. Setting v+i x,xN= v+x,xN− (−1)il 2 , v− i x,xN= v−x,xN+ (−1)i l 2 for i= 1,2.
From the translation invariance of the functional in xN-axis we get that v±i ∈ M±(Ω(l)) and h(v± i ) = ±(−1) il 2 . Setting vi= v+
i + v−i , we obtain vi∈ N(Ω(l)),(−1)ig(vi) = l > L and J(vi) <α+(A)+α−(A) +δ(L) =θ(A)+δ(L) for all l > l
0. This implies Ni(L,δ(L),Ω(l)) = /0 for all l > l0.
Furthermore, we have the following results.
LEMMA3.2. Let l0> 0 as in Lemma 3.1. Then for each l > l0, we have that Ni(L,δ(L),Ω(l)) are closed sets.
Proof. The proofs of cases “i= 1” and “i = 2” are similar argument. There-fore, we only need to prove the case “i= 1”. Suppose that u0 is a limits point of N1(L,δ(L),Ω(l)), then −g(u0) L and J (u0) θ(A) +δ(L). This implies u0∈ N(δ(L),Ω(l)). If −g(u0) = L, then by Lemma 3.1 u0∈ N2(L,δ(L),Ω(l)). We obtain
−L = g(u0) > L
which is a contradiction. Thus, g(u0) < L, and so u0∈ N1(L,δ(L),Ω(l)). Therefore, Ni(L,δ(L),Ω(l)) are closed sets.
Now we consider the minimization problem in Ni(L,δ(L),Ω(l)) for J
θi(Ω(l)) = inf u∈Ni(L,δ(L),Ω(l))
J(u) for all i = 1,2. Then we have the following result.
LEMMA3.3. For each v0∈ Ni(L,δ(L),Ω(l)) there is a map Φ : H01(Ω(l)) → R2 such that:
(i) Φt1v+0 +t2v−0
= (t1,t2) for t1,t2 0; (ii) Φ(u) = (1,1) if and only if u ∈ N(Ω(l)).
Proof. Similarly to the method used in Clapp-Weth [11, Lemma 13]. The next result is a variant of Proposition 14 in Clapp-Weth [11], and its proof follows from the arguments of applying the Leray-Schauder continuation principle. Let
b=θ(A) + min α+(A),α−(A),δ(L) and distH(u,D) = inf u − v : v ∈ D ⊂ H01(Ω)
. Then we have the following results.
PROPOSITION3.4. Letλ0= b−θi(Ω(l)). Then for eachλ ∈ (0,λ0) and μ> 0
there exists u0∈ H01(Ω(l)) such that: (i) distH(u0,Ni(L,δ(L),Ω(l))) μ; (ii) J (u0) ∈ [θi(Ω(l)),θi(Ω(l)) +λ); (iii) ∇J (u0) max√λ,λμ
.
Proof. The proofs of cases “i= 1” and “i = 2” are similar argument. Therefore, we only need to prove the case “i= 1”. Fix v0∈ N1(L,δ(L),Ω(l)) such that J (v0) <
θ1(Ω(l)) +λ, and fix d0> 1 such that Jd0v±0
0. Let Φ : H1
0(Ω(l)) → R2 be as in Lemma 3.3. We put K= [0,d0] × [0,d0] and define
η: K→ H1
0(Ω(l)),η(s1,s2) = s1v+0+ s2v−0. ThenΦ ◦η= id : K → K, in particular
deg(Φ ◦η,K,(1,1)) = 1. (6)
Notice also that
J(η(s1,s2)) J (v0) <θ1(Ω(l)) +λ for all(s1,s2) ∈ K. (7) We now choose a Lipschitz continuous functionχ:R → R such that 0 χ 1,χ(s) = 1 for s 0 and χ(s) = 0 for s −1. Then since J ∈ C2H1
0(Ω(l)),R , there is a semiflowϕ:[0,∞) × H1 0(Ω(l)) → H01(Ω(l)) satisfying ∂
∂tϕ(t,u) = −χ(J (ϕ(t,u)))∇J (ϕ(t,u)),
ϕ(0,u) = u.
We will frequently writeϕt in place ofϕ(t,·). Since Jv±0<α+(A) +α−(A) and Jd
0v±0
0, it follows that
sup J(η(∂K)) <α+(A) +α−(A) =θ(A).
Hence
ϕt◦η(∂K) ∩ N(Ω(l)) = /0 for all t 0 and, by Lemma 3.3, this implies
Φ ◦ϕt◦η(y) = (1,1) for all y ∈∂K,t 0.
Equality(6) and the global continuation principle of Leray-Schauder (see e.g. Zeidler [27, p.629]) imply that there exists a connected subset Z⊂ K × [0,1] such that
(1,1,0) ∈ Z; ϕt(η(s1,s2)) ∈ N(Ω(l)) for all (s1,s2,t) ∈ Z; Z∩ (K × {1}) = /0. We put Z = ϕt(η(s 1,s2)) ∈ N(Ω(l)) : (s1,s2,t) ∈ Z . By inequality(7) and Lemma 3.1 (ii), we obtain
Since Z is connected, we obtain that Z⊂ N1(L,δ(L),Ω(l)). We now pick (s1,s2,1) ∈ Z∩ (K × {1}) and write
v1:=η(s1,s2),v2:=ϕ1(v1). Then v2∈ Z ⊂ N1(L,δ(L),Ω(l)). We distinguish two case.
Case 1. ϕt(v1) − v2 μ for all t∈ [0,1]. We choose t0∈ [0,1] with ∇Jϕt0(v
1) = min 0t1∇Jϕ
t(v 1) and put u0=ϕt0(v1). Then
λ J (v1) − J (v2) = − 1 0 ∂ ∂tJ ϕt(v 1)ds = 1 0 ∇Jϕ t(v 1)2dt ∇J (u0)2. Hence u0has the desired properties.
Case 2. There exists t∈ [0,1] such thatϕt(v1)− v2 >μ. Then let t1= sup t t :ϕt(v1) − v2 >μ . We choose t0∈ [t1,1] with ∇Jϕt0(v 1) = min t1t1∇Jϕ t(v 1) and put u0=ϕt0(v1). Then
μ 1 t1 ∂∂tϕt(v 1)dt 1 t1 ∇Jϕ t(v 1)dt and λ Jϕt1(v 1)− J (v2) = 1 t1 ∇Jϕ t(v 1)2dt ∇J (u0) 1 t1 ∇Jϕ t(v 1)dt.
We conclude that∇J (u0) λμ. Thus, u0 has the desired properties.
COROLLARY3.5. For each l> l0 there exists a sequence
u(i)n ⊂ H1 0(Ω(l)) such that:
(i) distHu(i)n ,Ni(K,δ(K),Ω(l))
→ 0; (ii) Ju(i)n
→θi(Ω(l)) <θ(A) + min{α+(A),α−(A),δ(K)}; (iii) J(u(i)
We begin to show the proof of Theorem 1.1 for l> l0. Then by Corollary 3.5, there exist sequences u(i)n ⊂ H01(Ω(l)) such that (i)−(iii) in Corollary 3.5 are hold. Then by the Rellich compactness theorem and Lemma 3.2 there exist subsequences
u(i)n and a nodal solution u(i)0 ∈ Ni(K,δ(K),Ω(l)) such that u(i)n → u(i)0 strongly in H1
0(Ω(l)) and J
u(i)0 =θi(Ω(l)). Thus, by Lemmas 3.1 u(1)0 and u(2)0 are different. Since
θi(Ω(l)) <θ(A) + min α+(A),α−(A),δ(K) ,
by Lemma 2.4 u(1)0 and u(2)0 are 2-nodal solutions of equation(Ep,q) in Ω(l). 4. Proof of Theorem 1.2
Throughout this paper, we let ω⊂ A be a nonempty open set such that the set A− l, l\ω is a domain in RN for some l> 0. Then A\ω is a strictly large domain in A. Furthermore, by Lemma 2.5 we have the following result.
LEMMA4.1. For each positive number L there exists a positive number δ(L) such that for u∈ N(A\ω) with J (u) θ(A)+δ(L) we have:
(i) either h(u+) − h(u−) > L or h(u−) − h(u+) > L;
(ii) the function u+ satisfies either h(u+) > L or h(u+) < −L; (iii) the function u−satisfies either h(u−) > L or h(u−) < −L.
For positive numbers L,δ, we denote:
N1(L,l) = u∈ N(δ,D(l)) : h(u+) > L and h(u−) < −L , N2(L,l) = u∈ N(δ,D(l)) : h(u+) < −L and h(u−) > L , N3(L,l) = u∈ N(δ,D(l)) : h(u+) > L,h(u−) > L and g(u) > L , N4(L,l) = u∈ N(δ,D(l)) : h(u+)) > L,h(u−) > L and − g(u) > L , N5(L,l) = u∈ N(δ,D(l)) : h(u+) < −L,h(u−) < −L and g(u) > L , N6(L,l) = u∈ N(δ,D(l)) : h(u+) < −L,h(u−) < −L and − g(u) > L , where g(u) = h(u+) − h(u−). Then we have the following result.
LEMMA4.2. For each positive number L, there exist positive numbersδ(L) and l0 such that for l> l0:
(i) Ni(L,l) = /0 for all i = 1,2,...,6; (ii) Ni(L,l) ∩ Nj(L,l) = /0 for all i = j; (iii) N(δ,D(l)) = ∪6
i=1 Ni(L,l).
Proof. Since N(D(l)) ⊂ N(A\ω) for all l > l. By Lemma 4.1, we only need to prove that there exists l0> l such that Ni(L,l) = /0 for all l > l0 and i= 1,2,...,6.
Moreover, the proofs of all cases are similar argument. Thus, we only need to prove the case “ N1(L,l) = /0”. By Lien-Tzeng-Wang [19, Lemma 2.2], we have
α±A−l 2 + l, l 2 − l α±(A) as l ∞. Thus, there exists l0> max 2L, l such that
α±A−l 2 + l, l 2 − l <α±(A)+δ(L) 2 for all l> l0.
Moreover, by Ambrosetti-Rabinowitz [1] and Chen-Chen-Wang [9], equation(Ep,q) in A− l/2 + l,l/2 − l has a positive solution v+∈ M+A− l/2 + l,l/2 − land a negative solution v−∈ M−A− l/2 + l,l/2 − lsuch that
Jv±=α±A−l 2 + l,
l 2 − l
and v±(x,−y) = v±(x,y). Clearly, h(v±) = 0. Setting u+(x,y) = v+(x,y − l/2) and u−(x,y) = v−(x,y + l/2). From the translation invariance of the functional in y-axis we get that u±∈ M±(D(l)) and h(u±) = ±l/2. Setting u = u++ u−, we obtain u∈ N(D(l)),h+(u+) > L and h−(u−) < −L. Moreover
J(u) <α+(A) +α−(A)+δ(L) =θ(A) +δ(L) for all l > l 0. This implies N1(L,l) = /0 for all l > l0.
Similar to the argument in Lemma 3.2 we have the following result.
LEMMA4.3. Let l0> 0 as in Lemma 3.1. Then for each l > l0, we have that Ni(L,l) is closed set for all i = 1,2...,6.
Now we consider the minimization problem in Ni(L,l) for J
θi(D(l)) = inf u∈ Ni(L,l)
J(u) for all i = 1,2...,6.
Similar to the method used in the proof of Proposition 3.4, we can get the following result.
PROPOSITION4.4. For each l> l0 there exist sequences u(i)n ⊂ H01(D(l)) such that:
(i) distHu(i)n , Ni(L,l)
→ 0; (ii) Ju(i)n
→ θi(D(l)) <θ(A) + min{α+(A),α−(A),δ(L)}; (iii) J(u(i)
We begin to show the proof of Theorem 1.2 for l> l0. By Proposition 4.4, there exist sequence u(i)n ⊂ H01(D(l)) such that (i) − (iii) in Proposition 4.4 are hold. Then by the Rellich compactness theorem and Lemma 4.3 there exist subsequences
u(i)n and nodal solutions u(i)0 ∈ Ni(L,l) such that u(i)n → u(i)0 strongly in H01(D(l)) and Ju(i)0 = θi(D(l)) for all i = 1,2...,6. Thus, by Lemmas 4.2 u(1)0 ,u(2)0 ,...,u(6)0 are different. Since
θi(D(l)) <θ(A) + min α+(A),α−(A),δ(K) ,
by Lemma 2.4 u(1)0 ,u(2)0 ,...,u(6)0 are 2-nodal solutions of equation(Ep,q) in D(l). 5. Proof of Theorem 1.3
For each i∈ {1,2,...,m} and l > 2 l we denote the set B(i,l) as follows: B(i,l) = {y ∈ R : y = (i − 1)l or y = il}. Furthermore, we denote: M+i (l) = u∈ M+(Dm(l)) : (i − 1)l < h(u) < il , ∂M+i (l) = u∈ M+(Dm(l)) : h(u) ∈ B(i,l) , M−i (l) = u∈ M−(Dm(l)) : (i − 1)l < h(u) < il , ∂M−i (l) = u∈ M−(Dm(l)) : h(u) ∈ B(i,l) , Ni, j(l) = u∈ H1 0(Dm(l)) : u+∈ M+i (ε,t) and u−∈ M−j (ε,t) , ∂Ni, j(l) = u∈ H01(Dm(l)) : u+∈ M+i (l),u−∈ M−j (l) and either u+∈∂M+i (l) or u−∈∂M− j (l) ,
where M±i (l) = M±i (l) ∪∂M±i (l). Then we have the following result. LEMMA5.1. For each l> 2 l, we have that Ni, j(l) are mutually disjoint. Proof. The proofs of all cases are similar argument. Thus, we only need to prove the case “1,1” and “1,2”. Suppose otherwise, then there exists v0∈ H01(Dm(l)) such that v0∈ N1,1(l) ∩ N1,2(l). Then
0< h(v−0) < l and l < h(v−0) < 2l, this contradicts A(0,l) ∩ A(l,2l) = /0 for all l > 2 l.
Define the minimization problems in Ni, j(l) and ∂Ni, j(l) for J,
γi, j(l) = inf v∈Ni, j(l)
J(v) and γi, j(l) = inf v∈∂Ni, j(l)
J(v).
LEMMA5.2. For each positive numberσ min{α+(A),α−(A),δ(L)} there exists l1> 2 l such that Ni, j(l) = /0 and γi, j(l) <θ(A) +σ for all i, j = 1,2,...,m and l> l1.
Furthermore, we have the following result.
LEMMA5.3. There exist positive numbersδ and l2> 2 lsuch that for each i, j ∈ {1,2,...,m} we have
γi, j(l) >θ(A) +δ for all l l2.
Proof. Fix i, j ∈ {1,2,...,m}. Suppose otherwise, then there exist ln→ ∞ as n→ ∞ and {un} ⊂∂Ni, j(l) such that:
J(un) →θ(A), (8) Dm(ln) ∇u+ n2+ u+n2dx= Dm(ln) u+ npdx, (9) Dm(ln) ∇u− n2+ u−n2dx= Dm(ln) u− nqdx (10) and either u+ n ∈∂M+i (ln) or u−n ∈∂M−j (ln). Since u±n ∈ M±(Dm(ln)) ⊂ M±(A+) ⊂ M±(A),θ(A) =α+(A) +α−(A) and J (un) = J (u+
n) + J (u−n). By (8) and Wang-Wu [22, Lemma 7]{u±
n} are (PS)α±(A)-sequences in H01(A) for J. Moreover, by (10)
and the Sobolev imbedding theorem, there exists c> 0 such that
A+∇u ± n2+ u±n2dx> c for all n.
From the concentration compactness principle of Lions [20] (or see [19, Theorem 4.1]), there exist R> 0,d > 0 and {(0,y±
n)} ⊂ RN−1× R+such that A(−R,R)+(0,y+n) u+ npdx d and A(−R,R)+(0,y−n) u− nqdx d for all n. Without loss of generality, we may assume that u+
n ∈∂M+i (ln) that is, h+(u+n) ∈ B(i,ln). Set un(x,xN) = u+
n(x,xN+ y+n). From the translation invariance of the func-tional in xN-axis, we get that{un} satisfies
un∈ M+A+−0,y+ n
⊂ M+(A) (11)
and is (PS)α+(A)-sequences in H01(A) for J. Then there exist a subsequence {un} and a nonnegative function u0∈ H01(A) such that
un u0 weakly in H01(A) as n → ∞, un→ u0 a.e. in A as n→ ∞, and A(−R,R)|un| pdx→ A(−R,R)|u0| pdx d as n → ∞. (12)
This implies u0 is a nonzero nonnegative solution of equation (Ep,q) in A. By the Fatou lemma α+(A) J (u 0) = 1 2 − 1 p A|u0| pdx liminf 1 2 − 1 p A|un| pdx=α+(A),
and so J(u0) =α+(A). Moreover, by the maximum principle and Chen-Chen-Wang [9], u0 is a positive solution of equation (Ep,q) in A and is axially symmetric in y-axis. Set wn= un− u0, Since {un} is uniformly bounded, by Br´ezis-Lieb lemma [7]
we obtain A|wn| pdx= A|un| pdx− A|u0| pdx+ o(1). (13)
Moreover, un u0 weakly in H01(A) we have
wn2= un2− u02+ o(1). (14) Then wn2= A|wn| pdx+ o(1) and so 1 2 − 1 p
wn2= J (wn) = J (un) − J (u0) + o(1) = o(1).
This implies un→ u0 strongly in H01(A) as n → ∞. We will show that y+n → ∞ as n→ ∞. Suppose otherwise, then {y+
n} is a bounded sequence in R or there exists a subsequence{y+
n} such that y+n → −∞ as n → ∞. If {y+n} is a bounded sequence in R. Without loss of generality, we may assume that y+
n → y0. Since u+n ∈ M+(Dm(ln)) ⊂ M+(A+) and un(x,y) = u+ n(x,xN+ y+n), by (11) and lim n→∞ A+−0,y+ n → A+ y0 = x,xN∈ A : xN> −y0 we have u0∈ M+A+y0
which contradicts the fact that u0 is a positive solution of equation(Ep,q) in A. If y+
n → −∞ as n → ∞, then there exists n0∈ N such that
A+−0,y+ n
∩ A(−R,R) = /0 for all n n0,
this implies un≡ 0 on A(−R,R) for all n n0which contradicts(12). Thus y+n → ∞ as n→ ∞. Moreover, un→ u0strongly in H01(A) as n → ∞, we have
h(un) = h(u0) + o(1) (15)
and so h(u+ n) = h unx,xN+ y+ n = h(u0)− y+n + o(1). (16) This implies dist(y+
n,B(i,ln)) → 0 as n → ∞. By passing to a subsequence, we may assume that one of the following cases occurs:
(a) |iln− y+
n| → 0 as n → ∞ for a subsequence; (b) |(i − 1)ln− y+
n| → 0 as n → ∞ for a subsequence. In case(a) since y+
n → ∞ as n → ∞ and ln→ ∞ as n → ∞. Then (i − 1)ln−y+n → −∞ as n → ∞ and (i + 1)ln− y+ n → ∞ as n → ∞, this implies lim n→∞ Dm(ln) −0,y+n→ A\ω. Moreover, un→ u0strongly in H01(A) as n → ∞ and
un∈ H01
Dm(ln) −0,y+n
.
Thus, u0∈ M+(A\ω) which contradicts to the fact that u0 is a positive solution of equation(Ep,q) in A.
In case(b) : since dist(y+
n,B(i,l)) → 0 as n → ∞ and ln→ ∞ as n → ∞, we may assume iln− y+n → ∞ as n → ∞. If i = 1, then (i − 1)ln− y+n → −∞ as n → ∞, this is a contradiction. For i 2. Then (i − 2)ln− y+
n → −∞ as n → ∞, this implies lim n→∞ Dm(ln) −0,y+n → A\ω. Moreover, un→ u0strongly in H01(A) as n → ∞ and
un∈ H01
Dm(ln) −0,y+n
.
Thus, u0∈ M+(A\ω) which contradicts to the fact that u0 is a positive solution of equation(Ep,q) in A. Therefore, we have completed our proof.
By Lemmas 5.2, 5.3, there exists l0> 2 l such that for l > l0
γi, j(l) < min θ(A) + min α+(A),α−(A),δ(L) , γi, j(l) (17) for all i, j ∈ {1,2,...,m}. Similar to the method used in the proof of Lemma 3.1, we can get the following result.
LEMMA5.4. Let L and δ(L) be positive numbers as in Lemma 2.5. Then for each i∈ {1,2,...,m} and l > l0, we have:
(i) N(k)i,i (l) = /0 for all k = 1,2; (ii) Ni,i(l) = N(1)i,i (l) ∪ N(2)i,i (l);
(iii) N(1)i,i (l) ∩ N(2)i,i (l) = /0, where N(k)i,i (l) =u∈ Ni,i(l) : (−1)kg(u) > L.
Similar to the method used in the proof of Proposition 3.4 (or see Wu [26, Propo-sition 3.6]), we can get the following result.
PROPOSITION5.5. For each l> l0 there exist sequences u(i, j)n ⊂ H1 0(Dm(l)) such that:
(ii) J(u(i, j)n ) →γi, j(l); (iii) J(u(i, j)
n ) = o(1) strongly in H−1(Dm(l));
(iv) dist(h((u(i, j)n )+),((i − 1)l,il)) → 0 and dist(h((u(i, j)n )−),(( j − 1)l, jl)) → 0. Then we have the following result.
THEOREM5.6. For each L> 0, there exists l0> 0 such that for l > l0, equation (Ep,q) in Dm(l) has m22-nodal solutions u(i, j)
0
big}i, j∈{1,2,...,m} with u(i, j)0 ∈ Ni, j(l) for all i, j ∈ {1,2,...,m}.
Proof. It follows from Proposition 5.5 that there exists l0> 2 l such that for each l> l0 and i, j ∈ {1,2,...,m} we can find a sequence u(i, j)n ⊂ H01(Dm(l)) such that (i) − (iv) in Proposition 5.5 hold. Since u(i, j)n is bounded in H01(Dm(l)), we have
u(i, j)n + and u(i, j)n − are also bounded in H01(Dm(l)) and u(i, j)n + 2 = Dm(l) u(i, j)n + p dx+ o(1), and u(i, j)n − 2 = Dm(l) u(i, j)n − q dx+ o(1).
Thus, there exist a subsequence u(i, j)n and u(i, j)0 in H01(Dm(l)) such that u(i, j)n ui, j0 ; u(i, j)n ±u(i, j)0 ± weakly in H01(Dm(l)) and
u(i, j)n → u(i, j)0 ;u(i, j)n ±→u(i, j)0 ±a.e. in Dm(l).
Moreover, u(i, j)0 is a solution of equation(Ep,q) in Dm(l). We will show thatu(i, j)0 ±≡ 0. If not, then we may assume that u(i, j)0 +≡ 0. Since distu(i, j)n ,Ni, j(l)→ 0 as n→ ∞ andγi, j(l) > 0, we deduce from the Sobolev imbedding theorem that
u(i, j)n + >ν> 0 for some constantν and for all n.
Applying the concentration-compactness principle of P. L. Lions [20], there are positive constants R,c0 and a unbounded sequence{y+n} ⊂ R such that
A(−R,R)
u(i, j)n +x,xN+ y+n p
dx c0 for n sufficiently large. (18) Set u(i, j)n (x,xN) =u(i, j)n +(x,y + y+n). Since
u(i, j)n is bounded in H01(A), we may assume that there exists u(i, j)0 ∈ H1
0(A) such that u(i, j)n u(i, j)0 weakly in H01(A). From (18) we have u(i, j)0 0 and u(i, j)0 ≡ 0 in A. Set vn= u(i, j)n − u(i, j)0 . We distinguish between two cases:
Case I : vn → 0 as n → ∞;
Case II : vn c for large n and for some constant c > 0.
Assume Case I, we employ the argument in Lemma 5.3 to obtain that there exists c0> 0 such that
disthu(i, j)n +,((i − 1)l,il)
> c0for large n, which is a contradiction.
In Case II, we notice first that J(u(i, j)
n ) → 0 strongly in H−1(A) and distu(i, j)n ,Ni, j(l)→ 0 as n → ∞ implies
A ∇ u(i, j) 0 2+ u(i, j)0 2dx − A u (i, j) 0 pdx= o(1) (19) and A ∇ u(i, j) n 2+ u(i, j)n 2 dx− A u (i, j) n pdx= o(1). (20)
By(19),(20) and Brezis-Lieb lemma [7] we obtain
A |∇vn|2+ v2 n dx+ A|vn| pdx= o(1).
Since vn c for large n, is is easy to find a sequence {sn} ⊂ R+ with sn→ 1 as n→ ∞ such that snvn∈ M+(A), and so
1 2 A |∇vn|2+ v2 n dx−1p A|vn| pdxα+(A) + o(1). Similarly 1 2 A ∇ u(i, j) 0 2+ u(i, j)0 2dx−1 p A u (i, j) 0 pdxα+(A)+ o(1) and 1 2 A ∇ u(i, j)n −2+u(i, j)n −2 dx−1q Au (i, j) n −qdxα−(A) + o(1). Thus by Brezis-Lieb lemma [7] we have
Ju(i, j)n = 12 A ∇ u(i, j)n +2+ u(i, j)n +2 dx−1p A u (i, j) n −pdx +12 A ∇u(i, j)n −2+u(i, j)n −2 dx−1q Au (i, j) n −qdx = 12 A |∇vn|2+ v2 n dx−1p A|vn| pdx +12 A ∇ u(i, j) 0 2 + u(i, j)0 2dx−1p A u (i, j) 0 pdx +12 A ∇ u(i, j)n −2+u(i, j)n −2 dx−1q Au (i, j) n −qdx+ o(1) 2α+(A)+α−(A) + o(1)
and so
lim n→∞J
u(i, j)n =γi, j(l) θ(A) +α+(A), (21)
this contradicts (17). Next we will show that u(i, j)n → u(i, j)0 strongly in H01(Dm(l)). Using Case II we can prove that we result, otherwise, we may use a similar argument as above to reach the contradiction (21). Finally, we will show that u(i, j)0 ∈ Ni, j(l). Since
disthu(i, j)n +,((i − 1)l,il)
→ 0 and disthu(i, j)n −,(( j − 1)l, jl)
→ 0, we have u(i, j)0 ∈ Ni, j(l). Moreover, Ju(i, j)0 =γi, j(l) < γi, j(l) and so u(i, j)0 /∈∂Ni, j(l). Thus, u(i, j)0 ∈ Ni, j(l) and so u(i, j)0 i, j∈{1,2,...,m} are different. This completes the proof.
THEOREM5.7. For each L> 0, there exists l0> 0 such that for l > l0 and i∈ {1,2,...,m}, equation (Ep,q) in Dm(l) has two 2-nodal solutions u(i,i,1)0 and u(i,i,2)0 such that u(i,i,k)0 ∈ Ni,i(l) for all k = 1,2.
Proof. Similar to the method used in the proof of Proposition 3.4 (or see Wu [26, Proposition 3.6]), there exists a sequence u(i,i,k)n ⊂ H01(Dm(l)) such that:
(i) distHu(i,i,k)n ,N(k)i,i (l)
→ 0; (ii) Ju(i,i,k)n
→γi,i,k(l) = infu∈N(k)
i,i(l)J(u);
(iii) J(u(i,i,k)n ) = o(1) strongly in H−1(Dm(l)); (iv) disthu(i,i,k)n ±,((i − 1)l,il)
→ 0 and (−1)kgu(i,i,k)n > L.
Then we may use a similar argument as in Theorem 5.6, there exist subsequences
u(i,i,k)n and u(i,i,k)0 ∈ N(k)i,i (l) such that u(i,i,k)n → u(i,i,k)0 strongly in H01(Dm(l)). Fur-thermore, u(i,i,1)0 and u(i,i,2)0 are 2-nodal solutions of equation (Ep,q) in Dm(l). By Lemma 5.4, u(i,i,1)0 and u(i,i,2)0 are different.
The proof of Theorem 1.3 follows by a combination of the results of Theorems 5.6 and 5.7 and so, we have equation(Ep,q) in Dm(l) has at least m × (m + 1) 2-nodal solutions.
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(Received November 4, 2008) Tsung-fang Wu
Department of Applied Mathematics National University of Kaohsiung Kaohsiung 811 Taiwan e-mail:[email protected]
Differential Equations & Applications www.ele-math.com
