Electronic Journal of Differential Equations, Monograph 06, 2004, (142 pages). ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu (login: ftp) PALAIS-SMALE APPROACHES TO SEMILINEAR ELLIPTIC EQUATIONS IN UNBOUNDED DOMAINS HWAI-CHIUAN WANG Abstract. Let Ω be a domain in RN , N ≥ 1, and 2∗ = ∞ if N = 1, 2, 2∗ = 2N N−2 if N > 2, 2 < p < 2∗. Consider the semilinear elliptic problem −∆u + u = |u|p−2u in Ω; u ∈ H1 0 (Ω). Let H1 0 (Ω) be the Sobolev space in Ω. The existence, the nonexistence, and the multiplicity of positive solutions are affected by the geometry and the topology of the domain Ω. The existence, the nonexistence, and the multiplicity of positive solutions have been the focus of a great deal of research in recent years. That the above equation in a bounded domain admits a positive solution is a classical result. Therefore the only interesting domains in which this equation admits a positive solution are proper unbounded domains. Such elliptic prob- lems are difficult because of the lack of compactness in unbounded domains. Remarkable progress in the study of this kind of problem has been made by P. L. Lions. He developed the concentration-compactness principles for solving a large class of minimization problems with constraints in unbounded domains. The characterization of domains in which this equation admits a positive so- lution is an important open question. In this monograph, we present various analyses and use them to characterize several categories of domains in which this equation admits a positive solution or multiple solutions. Contents 1. Introduction 2 2. Preliminaries 4 3. Palais-Smale Decomposition Theorems 25 4. Palais-Smale Values and Indexes of Domains 31 5. Palais-Smale Conditions 47 6. Symmetric Palais-Smale Conditions 57 7. Symmetric Palais-Smale Decomposition Theorems 60 8. Fundamental Properties, Regularity, and Asymptotic Behavior of Solutions 68 2000 Mathematics Subject Classification. 35J20, 35J25. Key words and phrases. Palais-Smale condition; index; decomposition theorem; achieved domain; Esteban-Lions domain; symmtric Palais-Smale condition. c©2004 Texas State University - San Marcos. Submitted September 17, 2004. Published September 30, 2004. 1 2 HWAI-CHIUAN WANG EJDE-2004/MON. 06 8.1. Fundamental Properties of Solutions 68 8.2. Regularity of Solutions 69 8.3. Asymptotic Behavior of Solutions 75 9. Symmetry of Solutions 79 9.1. Open Question: 83 10. Nonachieved Domains and Esteban-Lions Domains 83 11. Higher Energy Solutions 95 11.1. Existence Results 95 11.2. Dynamic Systems of Solutions 99 12. Achieved Domains 100 12.1. Open Question: 106 13. Multiple Solutions 106 13.1. Multiple Solutions for a Perturbed Equation 106 13.2. Symmetry Breaking in a Bounded Symmetry Domain 134 13.3. Multiple Solutions in Domains with Two Bumps 137 References 140 1. Introduction Let Ω be a domain in RN , N ≥ 1, and 2∗ = ∞ if N = 1, 2, 2∗ = 2N N−2 if N > 2, 2 < p < 2∗. Consider the semilinear elliptic problem −∆u+ u = |u|p−2u in Ω; u ∈ H1 0 (Ω). (1.1) Let H1 0 (Ω) be the Sobolev space in Ω. For the general theory of Sobolev spaces H1 0 (Ω), see Adams [2]. Associated with (1.1), we consider the energy functionals a, b and J for u ∈ H1 0 (Ω) a(u) = ∫ Ω (|∇u|2 + u2); b(u) = ∫ Ω |u|p; J(u) = 1 2 a(u)− 1 p b(u). As in Rabinowitz [64, Proposition B.10], a, b and J are of C2. It is well known that the solutions of (1.1) and the critical points of the energy functional J are the same. The existence, the nonexistence, and the multiplicity of positive solutions of (1.1) are affected by the geometry and the topology of the domain Ω. The existence, the nonexistence, and the multiplicity of positive solutions of (1.1) have been the focus of a great deal of research in recent years. That Equation (1.1) in a bounded domain admits a positive solution is a classical result. Gidas-Ni-Nirenberg [35] and Kwong [46] asserted that (1.1) in the whole space RN admits a “unique” positive spherically symmetric solution. Therefore the only interesting domains in which (1.1) admits a positive solution are proper unbounded domains. Such elliptic problems are difficult because of the lack of compactness in unbounded domains. Remarkable progress in the study of this kind of problem has been made by P. L. Lions [49] and [50]. EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 3 He developed the concentration-compactness principles for solving a large class of minimization problems with constraints in unbounded domains. The cornerstone is the paper of Esteban-Lions [33], in which they asserted : no any nonzeroal solutions H1 0 (Ω) for the (1.1) exist in an Esteban-Lions domain (see Definition 2.6.) The characterization of domains in which (1.1) admits a positive solution is an important open question. In this monograph, we present various analyses and use them to characterize several categories of domains in which (1.1) admits a positive solution or multiple solutions. In Section 2 we define the Palais-Smale (denoted by (PS))-sequences, (PS)- values, and (PS)-conditions. We study the properties of (PS)-values. We recall the classical compactness theorems such as the Lebesgue dominated convergence theo- rem and the Vitali convergence theorem. We then come to study (PS)-conditions: the modern concepts for compactness. In Section 3 we recall the (PS) decomposition theorems in RN of Lions [49] and the (PS) decomposition theorems in the infinite strip Ar of Lien-Tzeng-Wang [47]. In Section 4 we assert the four classical (PS)-values in Ω: the constrained maxi- mizing value, the Nehari minimizing value, the mountain pass minimax value, and the minimal positive (PS)-value are the same. We call any one of them the index of the functional J in the domain Ω. We also study in detail various indexes of the functional J in domain Ω. In Section 5 we use the indexes of the functional J in domains Ω to characterize the (PS)-conditions: we obtain a theorem in which eight conditions are equivalent to the (PS)-conditions. In Section 6 we establish y-symmetric (PS)-conditions. The development is in- teresting in its own right and will also be used to prove the multiplicity of nonzero solutions in Section 13. In Section 7 we present the y-symmetric (PS) decomposition theorems in the infinite strip Ar. In Section 8 we study the fundamental properties, regularity, and asymptotic behavior of solutions of (1.1). In Section 9 we use the asymptotic behavior of solutions developed in Section 8 and apply the “moving plane” method to prove the symmetry of positive solutions to (1.2) in the infinite strip Ar. Our approach is similar to those in Gidas-Ni- Nirenberg [34, Theorem 1] and [35, Theorem 2] but is more complicated. Finally we propose an open question—are positive solutions of (1.1) in the generalized infinite strip Sr unique up to a translation—? In Section 10 we characterize Esteban-Lions domains. We prove that proper large domains, Esteban-Lions domains, and some interior flask domains are nonachieved. Nonachieved domains may admit higher energy solutions. Berestycki conjectured that there is a positive solution of (1.1) in an Esteban-Lions domain with a hole. In Section 11 we answer the Berestycki conjecture affirmatively. We also study the dynamic system of those solutions. In Section 12 we assert that a bounded domain, a quasibounded domain, a periodic domain, some interior flask domains, some flat interior flask domains, canal domains, and manger domains are achieved. Finally we propose an open question: in Theorem 12.7, is s0 = r ? In Section 12 we prove that there is a ground state solution in an achieved domain. In Section 13 we prove that if we perturb (1.1), then we obtain three 4 HWAI-CHIUAN WANG EJDE-2004/MON. 06 nontrivial solutions of (1.2) or if we perturb the achieved domain by adding or removing a domain, then we obtain three positive solutions of (1.1). For the simplicity and the convenience of the reader, we present results for (1.1). As a matter of fact, our results also hold for more general semilinear elliptic equa- tions as follows: −∆u+ u = |u|p−2u+ h(z) in Ω; u ∈ H1 0 (Ω), (1.2) −∆u+ u = g(u) in Ω; u ∈ H1 0 (Ω), (1.3) −∆u = f(z, u) in Ω; u ∈ H1 0 (Ω). (1.4) Readers interested in other aspects of critical point theory may consult the fol- lowing books: Aubin-Ekeland [6], Brézis [14], Chabrowski [18], [19], Ghoussoub [37], Mawhin-Willem [56], Ni [58], Rabinowitz [64], Struwe [66], Willem [78], and Zei- dler [79]. For the study of semilinear elliptic equations in unbounded domains, we recommend the following articles: Ambrosetti-Rabinowitz [4], Benci-Cerami [11], Berestycki-Lions [13], Esteban-Lions [33], Lions [49], [50], Palais [59], and Palais- Smale [60]. I am grateful to Roger Temam for inviting me to visit Université de Paris-Sud in 1983, to Häım Brézis for introducing me to critical point theory in 1983, to Wei- Ming Ni for introducing me to the semilinear elliptic problems in 1987, to Henri Berestycki, Maria J. Esteban, and H. Attouch, and to Pierre-Louis Lions for giving me his preprints and for enlightening discussions. 2. Preliminaries Throughout this monograph, let X(Ω) be a closed linear subspace of H1 0 (Ω) with dual X−1(Ω) with the space X(Ω) satisfying the following three properties: (p1) If u ∈ X(Ω), then |u| ∈ X(Ω) (p2) If u ∈ X(Ω), then ξnu ∈ X(Ω) for each n = 1, 2, . . . , where ξ ∈ C∞([0,∞)) satisfies 0 ≤ ξ ≤ 1, ξ(t) = { 0 for t ∈ [0, 1]; 1 for t ∈ [2,∞), and ξn(z) = ξ( 2|z| n ). (2.1) (p3) If u ∈ X(Ω), then ηnu ∈ X(Ω) for each n = 1, 2, . . . , where η ∈ C∞c ([0,∞)) satisfies 0 ≤ η ≤ 1 and η(t) = { 1 for t ∈ [0, 1]; 0 for t ∈ [2,∞), and ηn(z) = η( 2|z| n ). (2.2) EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 5 n/2 n z 1 Figure 1. ξn(z). 1 n/2 n z Figure 2. ηn(z). Typical examples of X(Ω) are the whole space H1 0 (Ω) and the y-symmetric space H1 s (Ω). Let z = (x, y) ∈ RN−1 × R. In this monograph, we refer to three universal domains: the whole space RN , the infinite strip Ar, the infinite hole strip Ar1,r2 (in this case, N ≥ 3), and their subdomains: the ball BN (z0; s), the upper semi- strip Ar s, the interior flask domain Fr s, the infinite cone C, and the epigraph Π as follows. Ar = {(x, y) ∈ RN : |x| < r}; Ar s,t = {(x, y) ∈ Ar : s < y < t}; Ar s = {(x, y) ∈ Ar : s < y}; Ar\ω, where ω ⊂ Ar is a bounded domain; Ãr s = Ar\Ar s; BN (z0; s) = {z ∈ RN : |z − z0| < s}; Fr s = Ar 0 ∪BN (0; s); Ar1,r2 = {(x, y) ∈ RN : r1 < |x| < r2}; Ar1,r2 s,t = {(x, y) ∈ Ar1,r2 | s < y < t}; Ar1,r2 s = {(x, y) ∈ Ar1,r2 | s < y}; Ãr1,r2 s = Ar1,r2\Ar1,r2 s ; RN + = {(x, y) ∈ RN : 0 < y}; RN −ρ,ρ = {(x, y) ∈ RN : −ρ < y < ρ}; P+ = {(x, y) ∈ RN : y > |x|2}; P− = {(x,−y) : (x, y) ∈ P+}; 6 HWAI-CHIUAN WANG EJDE-2004/MON. 06 C = {(x, y) ∈ RN : |x| < y}; Π = {(x, y) ∈ RN : f(x) < y}, where f : RN−1 → R is a function. Definition 2.1. (i) We say that Ω is a large domain in RN if for any r > 0, z ∈ Ω exists such that B(z; r) ⊂ Ω; (i′) We say that Ω is a strictly large domain in RN if Ω contains an infinite cone of RN ; (ii) We call Ω a large domain in Ar if for any positive number m, a, b exist such that b− a = m and Ar a,b ⊂ Ω; (ii′) We call Ω a strictly large domain in Ar if Ω contains a semi-strip of Ar; (iii) We call Ω a large domain in Ar1,r2 if for any positive number m, a, b exist with a < b such that b− a = m and Ar1,r2 a,b ⊂ Ω; (iii′) We call Ω a strictly large domain in Ar1,r2 if Ω contains a semi-strip of Ar1,r2 . Let Ω be any one of RN , Ar, or Ar1,r2 . Then a strictly large domain in Ω is a large domain in Ω. Example 2.2. The infinite cone C, the upper semi-space RN + , the paraboloid P+, and the epigraph Π are strictly large domains in RN . Example 2.3. Ar s and Ar s \ D are strictly large domains in Ar, where s ∈ R and D ⊂ Ar s is a bounded domain. Example 2.4. Ar1,r2 s and Ar1,r2 s \ D are strictly large domains in Ar1,r2 , where s ∈ R and D ⊂ Ar1,r2 s is a bounded domain. There is a large domains in Ar which is not a strictly large domain in Ar. Example 2.5. Let Ω = Ar 0\ ∞ ∪ n=1 B(zn, r 4 ) where zn = (0, 0, . . . , 2n). Then Ω is a large domain in Ar which is not a strictly large domain in Ar. Figure 3. Large domains 1. Definition 2.6. A proper smooth unbounded domain Ω in RN is an Esteban-Lions domain if χ ∈ RN exists with ‖χ‖ = 1 such that n(z) · χ ≥ 0, and n(z) · χ 6≡ 0 on ∂Ω, where n(z) is the unit outward normal vector to ∂Ω at the point z. Example 2.7. An upper half strip Ar s, a lower half strip Ãr t , the epigraph Π, the infinite cone C, the upper half space RN + , and the paraboloid P+ are Esteban-Lions domains. We define the Palais-Smale (denoted by (PS)) sequences, (PS)-values, and (PS)- conditions in X(Ω) for J as follows. EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 7 Definition 2.8. (i) For β ∈ R, a sequence {un} is a (PS)β-sequence in X(Ω) for J if J(un) = β + o(1) and J ′(un) = o(1) strongly in X−1(Ω) as n→∞; (ii) β ∈ R is a (PS)-value in X(Ω) for J if there is a (PS)β-sequence in X(Ω) for J ; (iii) J satisfies the (PS)β-condition in X(Ω) if every (PS)β-sequence in X(Ω) for J contains a convergent subsequence; (iv) J satisfies the (PS)-condition in X(Ω) if for every β ∈ R, J satisfies the (PS)β- condition in X(Ω). A (PS)β-sequence in X(Ω) for J is a (PS)β-sequence in H1 0 (Ω) for J . Lemma 2.9. (i) For µ ∈ X−1(Ω), we can extend it to be µ ∈ H−1(Ω) such that ‖µ‖X−1 = ‖µ‖H−1 ; (ii) Let {un} be in X(Ω) and satisfy J ′(un) = o(1) strongly in X−1(Ω), then J ′(un) = o(1) strongly in H−1(Ω) as n→∞; (iii) If J ′(u) = 0 in X−1(Ω), then J ′(u) = 0 in H−1(Ω). Proof. (i) Since X(Ω) is a closed linear subspace of the Hilbert space H1 0 (Ω), we have H1 0 (Ω) = X(Ω)⊕X(Ω)⊥. Since µ is a bounded linear functional inX(Ω), by the Riesz representation theorem, there is a w ∈ X(Ω) such that µ(ϕ) = 〈w,ϕ〉H1 for each ϕ ∈ X(Ω) and ‖µ‖X−1 = ‖w‖H1 . Define µ(ϕ) = 〈w,ϕ〉H1 for each ϕ ∈ H1 0 (Ω). Note that 〈w, φ〉H1 = 0 for each φ ∈ X(Ω)⊥. For any v ∈ H1 0 (Ω) with ‖v‖H1 ≤ 1, vs ∈ X(Ω) and v⊥s ∈ X(Ω)⊥ exist such that v = vs + v⊥s . Then |µ(v)| = |〈w, v〉H1 | = |〈w, vs + v⊥s 〉H1 | = |〈w, vs〉H1 | ≤ ‖w‖H1 = ‖µ‖X−1 . Thus, ‖µ‖H−1 ≤ ‖µ‖X−1 . Moreover, ‖µ‖X−1 = sup{|µ(ϕ)| |ϕ ∈ X(Ω), ‖ϕ‖H1 ≤ 1} ≤ sup{|µ(ϕ)| |ϕ ∈ H1 0 (Ω), ‖ϕ‖H1 ≤ 1} ≤ ‖µ‖H−1 . Therefore, ‖µ‖H−1 = ‖µ‖X−1 . Part (ii) follows from part (i). Part (iii) follows form (i). � Bound and weakly convergence are the same. Lemma 2.10. Let Y be a normed linear space and un ⇀ u weakly in Y , then {un} is bounded in Y and ‖u‖ ≤ lim inf n→∞ ‖un‖. Lemma 2.11. Let un ⇀ u weakly in X(Ω). Then there exists a subsequence {un} such that: (i) {un} is bounded in X(Ω) and ‖u‖H1 ≤ lim infn→∞ ‖un‖H1 ; (ii) un ⇀ u, ∇un ⇀ ∇u weakly in L2(Ω), and un → u a.e. in Ω; (iii) ‖un − u‖2H1 = ‖un‖2H1 − ‖u‖2H1 + o(1). 8 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Proof. Part (i) follows from Lemma 2.10. (ii) For v ∈ (L2(Ω))N , define f(u) = ∫ Ω v∇u for u ∈ X(Ω), then |f(u)| ≤ ‖v‖L2‖∇u‖L2 ≤ ‖v‖L2‖u‖H1 . Thus, f is a bounded linear functional in X(Ω). By the Riesz representation theorem, w ∈ X−1(Ω) exists such that f(u) = 〈u,w〉H1 for u ∈ X(Ω). Hence, if un ⇀ u weakly in X(Ω), then f(un) → f(u), or∫ Ω (∇un)v → ∫ Ω (∇u)v for v ∈ (L2(Ω))N . Thus, ∇un ⇀ ∇u weakly in L2(Ω). Similarly, for v ∈ L2(Ω), define g(u) = ∫ Ω vu for u ∈ X(Ω), then we have un ⇀ u weakly in L2(Ω). Recall that the embedding X(Ω) ↪→ Lp loc(Ω) is compact. There is a subsequence {ui n} of {ui−1 n } and u in X(Ω) such that ui n → u in Lp(Ω ∩BN (0; i)) and a.e. in Ω ∩BN (0; i). Then we have un n → u a.e. in Ω. (iii) By the definition of weak convergence in X(Ω), we have∫ Ω (∇un∇u+ unu) = ∫ Ω (|∇u|2 + |u|2) + o(1). Therefore, ‖un − u‖2H1 = ∫ Ω |∇un −∇u|2 + ∫ Ω |un − u|2 = ∫ Ω (|∇un|2 + |un|2) + ∫ Ω (|∇u|2 + |u|2)− 2 ∫ Ω (∇un∇u+ unu) = ‖un‖2H1 − ‖u‖2H1 + o(1). � There is a sequence which converges weakly to zero. Lemma 2.12. For u ∈ H1(RN ) and {zn} in RN satisfying |zn| → ∞ as n → ∞, then u(z + zn) ⇀ 0 weakly in H1(RN ) as n→∞. Proof. For ε > 0, ϕ ∈ H1(RN ), and φ ∈ C1 c (RN ) exist such that ‖ϕ− φ‖H1 < ε/2(‖u‖H1 + 1). Let K = suppφ, then K is compact. We have 〈u(z + zn), φ(z)〉H1 = ∫ RN ∇u(z + zn)∇φ(z)dz + ∫ RN u(z + zn)φ(z)dz = ∫ K ∇u(z + zn)∇φ(z)dz + ∫ K u(z + zn)φ(z)dz ≤ ‖∇u(z + zn)‖L2(K)‖∇φ‖L2(K) + ‖u(z + zn)‖L2(K)‖φ‖L2(K) = o(1) as n→∞. Thus, for some N > 0 such that |〈u(z + zn), φ(z)〉H1 | < ε 2 for n ≥ N . In addition, 〈u(z + zn), ϕ(z)〉H1 = 〈u(z + zn), ϕ(z)− φ(z)〉H1 + 〈u(z + zn), φ(z)〉H1 ≤ ‖u(z + zn)‖H1(RN )‖ϕ(z)− φ(z)‖H1(RN ) EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 9 + 〈u(z + zn), φ(z)〉H1 ≤ ‖u(z)‖H1(RN )‖ϕ(z)− φ(z)‖H1(RN ) + ε 2 < ε for n ≥ N. Therefore, u(z + zn) ⇀ 0 weakly in H1(RN ). � Lemma 2.13. For u ∈ H1 0 (Ar) and {zn} in Ar satisfying |zn| → ∞ as n → ∞, then u(z + zn) ⇀ 0 weakly in H1 0 (Ar) as n→∞. The proof of this lemma is the same as the proof of Lemma 2.12. Therefore, we omit it. Bounded Lp(Ω) sequence admits interesting convergent properties. Lemma 2.14 (Brézis-Lieb Lemma). Suppose un → u a.e. in Ω and there is a c > 0 such that ‖un‖Lp(Ω) ≤ c for n = 1, 2, . . . . Then (i)‖un − u‖p Lp = ‖un‖p Lp − ‖u‖p Lp + o(1); (ii) |un − u|p−2(un − u)− |un|p−2un + |u|p−2u = o(1) in L p p−1 (Ω). Proof. (i) Let ϕ(t) = tp for t > 0, then ϕ′(t) = ptp−1 and |un − u|p − |un|p = ϕ(|un − u|)− ϕ(|un|) = ϕ′(t)(|un − u| − |un|), where t = (1 − θ)|un| + θ|un − u| ≤ |un| + |u| for some θ ∈ [0, 1]. Thus, by the Young inequality, for ε > 0 ||un − u|p − |un|p| ≤ p(|un|+ |u|)p−1|u| ≤ d(|un|p−1|u|) + d|u|p ≤ ε|un|p + cε|u|p. Thus, ||un − u|p − |un|p + |u|p| ≤ ε|un|p + (cε + 1)|u|p. We have ∫ Ω ||un − u|p − |un|p + |u|p| ≤ εcp + (cε + 1) ∫ Ω |u|p. Since ‖u‖Lp ≤ lim infn→∞ ‖un‖Lp ≤ c. For some δ > 0 |E| < δ implies ∫ E |u|p < ε. In addition, K in RN exists such that |K| <∞ and ∫ Kc |u|p < ε. Thus,∫ E ||un − u|p − |un|p + |u|p| ≤ (cp + cε + 1)ε,∫ Kc ||un − u|p − |un|p + |u|p| ≤ (cp + cε + 1)ε. Clearly, ||un − u|p − |un|p + |u|p| = o(1) a.e. in Ω. By Theorem 2.23 below,∫ Ω ||un − u|p − |un|p + |u|p| = o(1), or ‖un − u‖p Lp = ‖un‖p Lp − ‖u‖p Lp + o(1). (ii) Let ϕ(t) = |t|p−2t, then ϕ′(t) = (p − 1)|t|p−2. The proof is similar to part (i) � New (PS)-sequences can be produced as follows. Lemma 2.15. Let un ⇀ u weakly in X(Ω) and J ′(un) = −∆un + un − |un|p−2un = o(1) in X−1(Ω). Then (i) |un − u|p−2(un − u)− |un|p−2un + |u|p−2u = o(1) in X−1(Ω); (ii) J ′(ϕn) = −∆ϕn + ϕn − |ϕn|p−2ϕn = o(1) in X−1(Ω) where ϕn = un − u; (iii) if {un} is a (PS)β-sequence, then {ϕn} is a (PS)(β−J(u))-sequence. 10 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Proof. (i) By Lemma 2.14,∫ Ω ||un − u|p−2(un − u)− |un|p−2un + |u|p−2u| p p−1 = o(1). Now for ϕ ∈ H1(Ω), |〈|un − u|p−2(un − u)− |un|p−2un + |u|p−2u, ϕ〉| = | ∫ Ω εnϕ| ≤ ( ∫ Ω |εn| p p−1 ) p−1 p ( ∫ Ω |ϕ|p)1/p ≤ c‖εn‖ L p p−1 ‖ϕ‖H1 , where εn = |un − u|p−2(un − u)− |un|p−2un + |u|p−2u. Therefore, ‖|un − u|p−2(un − u)− |un|p−2un + |u|p−2u‖X−1 ≤ c‖εn‖ L p p−1 = o(1). (ii) Since J ′(un) = −∆un + un − |un|p−2un = o(1) in X(Ω) (2.3) and un ⇀ u, then by Lemma 2.11, we have J ′(u) = 0, or −∆u+ u− |u|p−2u = 0. (2.4) Now by part (i), (2.3), and (2.4), J ′(ϕn) = −∆ϕn + ϕn − |ϕn|p−2ϕn = −∆(un − u) + (un − u)− |un − u|p−2(un − u) = (−∆un + un − |un|p−2un)− (−∆u+ u− |u|p−2u) − (|un − u|p−2(un − u)− |un|p−2un + |u|p−2u) = o(1). (iii) Since un ⇀ u weakly in X(Ω) and {un} is a (PS)β-sequence, by Lemma 2.11, 2.14 and Theorem 2.28 below, a subsequence {un} exists such that a(ϕn) = a(un)− a(u) + o(1) and b(ϕn) = b(un)− b(u) + o(1). Thus, J(ϕn) = J(un)− J(u) + o(1) = β − J(u) + o(1). Therefore, by part (ii), {ϕn} is a (PS)(β−J(u))-sequence. � Define the concentration function of |un|2 in RN by Qn(t) = sup z∈RN ∫ z+BN (0;t) |un|2. Then we have the following concentration lemma. Lemma 2.16. Let {un} be bounded in H1(RN ) and for some t0 > 0, let Qn(t0) = o(1). Then (i) un = o(1) strongly in Lq(RN ) for 2 < q < 2∗; (ii) in addition, if un satisfies −∆un + un − |un|p−2un = o(1) in H−1(RN ), then un = o(1) strongly in H1(RN ). Proof. (i) Decompose RN into the family F0 = {P 0 i }∞i=1 of unit cubes P 0 i of edge 1. Continue to bisect the cubes to obtain the family Fm = {Pm i }∞i=1 of unit cubes Pm i of edge 1 2m . Let m0 satisfy √ N 1 2m0 < t0. For each i, let Bm0 i be a ball in RN with radius t0 such that the centers of Bm0 i and Pm0 i are the same. Then EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 11 Pm0 i ⊂ Bm0 i , RN = ∪∞i=1P m0 i and {Pm0 i }∞i=1 are nonoverlapping. Write Pi = Pm0 i , 2 < q < r < 2∗, and∫ RN |un|q = ∞∑ i=1 ∫ Pi |un|q = ∞∑ i=1 ∫ Pi |un|2(1−t)|un|rt ≤ ∞∑ i=1 (∫ Pi |un|2 )1−t(∫ Pi |un|r )t ≤ (Qn(t0))(1−t) ∞∑ i=1 (∫ Pi |un|r )t ≤ c(Qn(t0))(1−t) ∞∑ i=1 (∫ Pi |∇un|2 + u2 n)rt/2, where 0 < t < 1. Since rt 2 → q 2 > 1 as r → q, we may choose r satisfying 2 < q < r < 2∗ and s = rt 2 > 1. Recall that ‖{an}‖`s = ( ∞∑ n=1 |an|s)1/s ≤ ∞∑ n=1 |an| = ‖{an}‖`1 , `1 ⊂ `2 ⊂ · · · ⊂ `∞. Thus, ∞∑ i=1 (∫ Pi |∇un|2 + |un|2 )rt/2 ≤ ( ∞∑ i=1 ∫ Pi (|∇un|2 + |un|2) )s = (∫ RN (|∇un|2 + |un|2) )s = ‖un‖2s H1(RN ) ≤ c for n = 1, 2, . . . . Therefore,∫ RN |un|q ≤ c(Qn(t0))(1−t), or ∫ RN |un|q = o(1) as n→∞. (ii) In addition, if un satisfies −∆un + un − |un|p−2un = o(1) in H−1(RN ), (2.5) then {un} is bounded. Multiply Equation (2.5) by un and integrate it to obtain a(un) = b(un) + o(1). By part (i), b(un) = o(1). Thus, a(un) = o(1), or ‖un‖H1 = o(1) strongly in H1(RN ). � Lemma 2.17. Let {un} be bounded in H1 0 (Ar) and for some t0 > 0, Qr n(t0) = sup y∈R ∫ (0,y)+Ar −t0,t0 |un|2 = o(1). Then (i) un = o(1) strongly in Lq(Ar) for 2 < q < 2∗; (ii) In addition, if un satisfies −∆un + un − |un|p−2un = o(1) in H−1(Ar), 12 HWAI-CHIUAN WANG EJDE-2004/MON. 06 then un = o(1) strongly in H1 0 (Ar). The proof of the above lemma is the same as the proof of Lemma 2.16. We have a sufficient condition for a solution of (1.1) to be zero. Lemma 2.18. Let N ≥ 2. For c > 0, there is a δ > 0 such that if v ∈ H1 0 (Ω) solves (1.1) in Ω satisfying ‖v‖H1 ≤ c and ‖v‖L2 ≤ δ, then v ≡ 0. Proof. For 0 < t0 < 1 and p < q <∞, let γ = { 2t0 for n ≥ 3; qt0 for n = 2, and p = 2(1− t0)+γ. Since ‖v‖H1 ≤ c and ‖v‖L2 ≤ δ, multiply −∆v+ v = |v|p−2v by v and integrate it to obtain ‖v‖2H1 = ∫ Ω |v|p = ∫ Ω |v|2(1−t0)|v|γ ≤ ‖v‖2(1−t0) L2 ‖v‖γ Lγ/t0 ≤ dδ2(1−t0)‖v‖γ H1 . Thus, we have ‖v‖2H1 ≤ dδ2(1−t0)‖v‖γ H1 . (2.6) Suppose that ‖v‖H1 > 0. (i) Let γ − 2 ≥ 0. Note that 2(1− t0) > 0. By (2.6), we have 1 ≤ dδ2(1−t0)‖v‖γ−2 H1 ≤ dcγ−2δ2(1−t0). Let δ1 > 0 satisfy dcγ−2δ 2(1−t0) 1 < 1. If δ ≤ δ1, then 1 ≤ dcγ−2δ2(1−t0) ≤ dcγ−2δ 2(1−t0) 1 < 1, which is a contradiction. (ii) Let γ − 2 < 0. By (2.6), we have ‖v‖H1 ≤ δ 2(1−t0) 2−γ d 1 2−γ , since ‖v‖2H1 = ∫ Ω |v|p ≤ c1‖v‖p H1 , or 1 ≤ c1‖v‖p−2 H1 . Thus, we have 1 ≤ c1‖v‖p−2 H1 ≤ c2δ 2(1−t0)(p−2) 2−γ , where c2 = c1d p−2 2−γ > 0. Note that 2(1−t0)(p−2) 2−γ > 0. Let δ2 > 0 such that c2δ 2(1−t0)(p−2) 2−γ 2 < 1. If δ ≤ δ2, then 1 ≤ c2δ 2(1−t0)(p−2) 2−γ < 1, which is a contradiction. Take δ0 = min{δ1, δ2}, if δ ≤ δ0, from parts (i) and (ii), and we obtain ‖v‖H1 = 0 or v = 0. � Let ũ(z) = { u(z) for z ∈ Ω; 0 for z ∈ RN\Ω. Then we have the following characterization of a function in W 1,p 0 (Ω). EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 13 Lemma 2.19. Let Ω be a C0,1 domain in RN and u ∈ Lp(Ω) with 1 < p < ∞. Then the following are equivalent: (i) u ∈W 1,p 0 (Ω); (ii) there is a constant c > 0 such that | ∫ Ω u ∂ϕ ∂xi | ≤ c‖ϕ‖Lp , for each ϕ ∈ C1 c (RN ), i = 1, 2, . . . , N ; (iii) ũ ∈W 1,p 0 (RN ) and ∂ũ ∂zi = ∂̃u ∂zi . For the proof of this lemma, see Brézis [14, Proposition IX.18], Gilbarg-Trudinger [36, Theorem 7.25], and Grisvard [38, p26]. We recall the classical compactness theorems. The Lebesgue dominated conver- gence theorem is a well-known compactness theorem. Theorem 2.20 (Lebesgue Dominated Convergence Theorem). Suppose Ω is a domain in RN , {un}∞n=1 and u are measurable functions in Ω such that un → u a.e. in Ω. If ϕ ∈ L1(Ω) exists such that for each n |un| ≤ ϕ a.e. in Ω, then un → u in L1(Ω). The converse of the Lebesgue dominated convergence theorem fails. Example 2.21. For n = 1, 2, . . . , let un : R → R be defined by un(z) =  0 for z ≤ n; 2 for z = n+ 1/2n; 0 for z ≥ n+ 1/n; linear otherwise. R R 2 1 2 3 4 u1 2 3 u u 2+1/2 3+1/3 Figure 4. Counter example 1. We have ∫ R un(z)dz = 1 n <∞ for each n ∈ N. 14 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Hence, un → 0 a.e. in R and strongly in L1(R). Let ϕ : R → R satisfy |un| ≤ ϕ a.e. in R for each n ∈ N. Then ∞ = ∞∑ n=1 1 n = ∫ R ∞∑ n=1 un ≤ ∫ Rϕ. Consequently, ϕ /∈ L1(R). However, the generalized Lebesgue dominated convergence theorem is a neces- sary and sufficient result for L1 convergence. Theorem 2.22 (Generalized Lebesgue Dominated Convergence Theorem:). Sup- pose Ω is a domain in RN , {un}∞n=1 and u are measurable functions in Ω such that un → u a.e. in Ω. Then un → u in L1(Ω) if and only if {ϕn}∞n=1, ϕ ∈ L1(Ω) exist such that ϕn → ϕ a.e. in Ω, |un| ≤ ϕn a.e. in Ω for each n, and ϕn → ϕ in L1(Ω). Proof. (=⇒) Suppose that un → u in L1(Ω), take ϕn = |un| and ϕ = |u|, then ϕn → ϕ in L1(Ω). (⇐=) Suppose that a sequence of measurable functions {ϕn}∞n=1 and ϕ in Ω exist such that ϕn ∈ L1(Ω), ϕn → ϕ a.e. in Ω, |un| ≤ ϕn a.e. in Ω for each n, and ϕn → ϕ in L1(Ω). Applying the Fatou lemma, we have∫ Ω lim inf n→∞ (ϕn − un) ≤ lim inf n→∞ ∫ Ω (ϕn − un), or ∫ Ω u ≥ lim sup n→∞ ∫ Ω un. Applying the Fatou lemma again, we have∫ Ω lim inf n→∞ (ϕn + un) ≤ lim inf n→∞ ∫ Ω (ϕn + un), or ∫ Ω u ≤ lim inf n→∞ ∫ Ω un. Thus, ∫ Ω u = lim n→∞ ∫ Ω un. � Another necessary and sufficient result for L1 convergence is the Vitali conver- gence theorem. Theorem 2.23 (Vitali Convergence Theorem for L1(Ω)). Suppose Ω is a domain in RN , {un}∞n=1 in L1(Ω), and u ∈ L1(Ω). Then ‖un − u‖L1 → 0 if the following three conditions hold: (i) un → u a.e in Ω; (ii) (Uniformly integrable) For each ε > 0, a measurable set E ⊂ Ω exists such that |E| <∞ and ∫ Ec |un|dµ < ε for each n ∈ N, where Ec = Ω\E; (iii) (Uniformly continuous) For each ε > 0, δ > 0 exists such that |E| < δ implies∫ E |un|dµ < ε for each n ∈ N. Conversely, if ‖un − u‖L1 → 0, then conditions (ii) and (iii) hold and there is a subsequence {un} such that (i) holds. Furthermore, if |Ω| < ∞, then we can drop condition (ii). EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 15 Proof. Assume the three conditions hold. Choose ε > 0 and let δ > 0 be the cor- responding number given by condition (iii). Condition (ii) provides a measurable set E ⊂ Ω with |E| <∞ such that∫ Ec |un|dµ < ε for all positive integers n. Since |E| < ∞, we can apply the Egorov theorem to obtain a measurable set B ⊂ E with |E\B| < δ such that un converges uniformly to u on B. Now write∫ Ω |un − u|dµ = ∫ B |un − u|dµ+ ∫ E\B |un − u|dµ+ ∫ Ec |un − u|dµ. Since un → u uniformly in B, the first integral on the right can be made arbitrarily small for large n. The second and third integrals will be estimated with the help of the inequality |un − u| ≤ |un|+ |u|. From condition (iii), we have ∫ E\B |un|dµ < ε for all n ∈ N and the Fatou Lemma shows that ∫ E\B |u|dµ ≤ ε as well. The third integral can be handled in a similar way using condition (ii). Thus, it follows that ‖un − u‖L1 → 0. Now suppose ‖un − u‖L1 → 0. Then for each ε > 0, a positive integer n0 exists such that ‖un − u‖L1 < ε/2 for n > n0, and measurable sets A and B of finite measure exist such that∫ Ac |u|dµ < ε/2 and ∫ Bc |un|dµ < ε for n = 1, 2, . . . , n0. Minkowski’s inequality implies that ‖un‖L1(Ac) ≤ ‖un − u‖L1(Ac) + ‖u‖L1(Ac) < ε for n > n0. Then let E = A ∪ B to obtain the necessity of condition (ii). Similar reasoning establishes the necessity of condition (iii). Convergence in L1 implies convergence in measure. Hence, condition (i) holds for a subsequence. � There is a bounded sequence {un} in L1(R) that violates Theorem 2.23 condition (ii). Example 2.24. For n = 1, 2, . . . , let un : R → R be defined by un(z) =  0 for z ≤ n; 2 for z = n+ 1/2; 0 for z ≥ n+ 1; linear otherwise, then ∫ Run(z)dz = 1 for each n ∈ N. Clearly, {un} violates Theorem 2.23 (ii). There is a bounded sequence {un} in L1(R) that violates Theorem 2.23 condition (iii). 16 HWAI-CHIUAN WANG EJDE-2004/MON. 06 R R 2 1 2 3 4 u1 2 3u u 5 Figure 5. counter example violating Theorem 2.23 condition (ii). Example 2.25. For n = 1, 2, . . . , let un : R → R be defined by un(z) =  0 for z ≤ n; 2n for z = n+ 1/2n; 0 for z ≥ n+ 1/n; linear therwise. R R1 2 3 4 5 2 4 6 u u u 1 2 3 Figure 6. counter example violating Theorem 2.23 condition (iii). Then ∫ R un(z)dz = 1 for each n ∈ N. Clearly, {un} violates Theorem 2.23 condition (iii). Lemma 2.26. In the Vitali convergence theorem 2.23 condition (ii), the set E with |E| <∞ can be replaced by the condition that E is bounded. Proof. Let En = E ∩ BN (0;n) for n = 1, 2, . . . . Then E1 ⊂ E2 ⊂ · · · ↗ E. Thus |E1| ≤ |E2| ≤ · · · ↗ |E|. For δ > 0 as in Theorem 2.23 condition (iii), there is an EN such that |E\EN | < δ. Now∫ Ec N |un|dz = ∫ Ec |un|dz + ∫ E\EN |un|dz < 2ε EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 17 for each n ∈ N. � Lemma 2.27. Let Ω be a domain in RN , 1 ≤ r < q < s, and {un} in Lr(Ω)∩Ls(Ω). Suppose that either ‖un‖Lr = o(1) and ‖un‖Ls = O(1), or ‖un‖Lr = O(1) and ‖un‖Ls = o(1), then ‖un‖Lq = o(1). Proof. Note that q = (1− t)r + ts, 0 < t < 1, so by the Hölder inequality,∫ Ω |un|qdz ≤ ( ∫ Ω |un|rdz )1−t( ∫ Ω |un|sdz )t . Then the conclusion follows. � We recall the Sobolev embedding theorem as follows. Theorem 2.28 (Sobolev Embedding Theorem in Wm,p 0 (Ω))). Let m ∈ N and 1 ≤ p <∞. Then we have the following continuous injections. (i) If 1 p − m N > 0, then Wm,p 0 (Ω) ↪→ Lq(Ω), where q ∈ [p, p∗], 1 p∗ = 1 p − m N ; (ii) If 1 p − m N = 0, then Wm,p 0 (Ω) ↪→ Lq(Ω), where q ∈ [p,∞); (iii) If 1 p − m N < 0, then Wm,p 0 (Ω) ↪→ L∞(Ω). Moreover, if m − N p > 0 is not an integer, let k = [ m− N p ] and θ = m − N p − k (0 < θ < 1), then we have for u ∈Wm,p 0 (Ω) ‖Dβu‖L∞ ≤ c‖u‖W m,p for |β| ≤ k |u(x)− u(y)| ≤ c‖u‖W m,p |x− y|θ a.e. for x, y ∈ Ω. In particular, Wm,p 0 (Ω) ↪→ Ck,θ(Ω). For the proof ot the theorem above, see Gilbarg-Trudinger [36, p.164]. Definition 2.29. Ω satisfies a uniform interior cone condition if a fixed cone KΩ exists such that each x ∈ ∂Ω is the vertex of a cone KΩ(x) ⊂ Ω and congruent to KΩ. Theorem 2.30 (Sobolev Embedding Theorem in Wm,p(Ω)). Let Ω satisfy a uni- form interior cone condition, m ∈ N and 1 ≤ p < ∞. Then we have the following continuous injections. (i) If 1 p − m N > 0, then Wm,p(Ω) ↪→ Lq(Ω), where q ∈ [p, p∗] and 1 p∗ = 1 p − m N ; (ii) If 1 p − m N = 0, then Wm,p(Ω) ↪→ Lq(Ω), where q ∈ [p,∞); (iii) If 1 p − m N < 0, then Wm,p(Ω) ↪→ L∞(Ω). Moreover, if m− N p > 0 is not an integer, let k = [ m− N p ] and θ = m− N p − k (0 < θ < 1), then we have for u ∈Wm,p(Ω), ‖Dβu‖L∞ ≤ c‖u‖W m,p for β with |β| ≤ k |Dβu(x)−Dβu(y)| ≤ c‖u‖W m,p |x− y|θ a.e. for x, y ∈ Ω and |β| = k. In particular, Wm,p(Ω) ↪→ Ck,θ(Ω). For the proof of the theorem above, see Brézis [14, Cor. IX.13] and Gilbarg- Trudinger [36, Theorem 7.26]. 18 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Theorem 2.31 (Rellich-Kondrakov Theorem in Wm,p 0 (Ω)). Let Ω be a bounded domain, m ∈ N and 1 ≤ p <∞. Then we have the following compact injections. (i) If 1 p − m N > 0, then Wm,p 0 (Ω) ↪→ Lq(Ω), where q ∈ [1, p∗), 1 p∗ = 1 p − m N ; (ii) If 1 p − m N = 0, then Wm,p 0 (Ω) ↪→ Lq(Ω), where q ∈ [1,∞); (iii) If 1 p − m N < 0, then Wm,p 0 (Ω) ↪→ Ck(Ω), where m − N p > 0 is not an integer and k = [ m− N p ] . For the proof of the aboved theroem, see Gilbarg-Trudinger [36, Theorem 7.22]. Theorem 2.32 (Rellich-Kondrakov Theorem in Wm,p(Ω)). Let Ω be a bounded C0,1 domain in RN , m ∈ N and 1 ≤ p < ∞. Then we have the following compact injections. (i) If 1 p − m N > 0, then Wm,p(Ω) ↪→ Lq(Ω), where q ∈ [1, p∗), 1 p∗ = 1 p − m N ; (ii) If 1 p − m N = 0, then Wm,p(Ω) ↪→ Lq(Ω), where q ∈ [1,∞); (iii) If 1 p − m N < 0, then Wm,p(Ω) ↪→ Ck,β(Ω), where m− N p > 0 is not an integer, 0 < β < θ, k = [ m− N p ] , and θ = m− N p − k (0 < θ < 1). For the proof of the above theorem, see Brézis [14, p. 169] and Gilbarg-Trudinger [36, Theorem 7.26]. For the Sobolev spaceX(Ω), we can drop condition (iii) of the Vitali convergence theorem 2.23 through the interpolation results. Theorem 2.33 (Rellich-Kondrakov Theorem). Let Ω be a domain in RN of finite measure. Then the embedding X(Ω) ↪→ Lp(Ω) is compact. Proof. Let {un} be a bounded sequence in X(Ω), then by Lemma 2.11, a subse- quence {un} and u ∈ X(Ω) exist such that un → u a.e. in Ω. By the Egorov theorem, for ε > 0, a closed subset F in RN exists such that F ⊂ Ω, |Ω\F | < ε, and un → u uniformly in F . Thus,∫ F |un − u|p = o(1) as n→∞. For N > 2, we have∫ Ω\F |un − u|p ≤ (∫ Ω\F 1 )1/r(∫ Ω\F |un − u|ps )1/s ≤ |Ω\F |1/r (∫ Ω |un − u|ps )1/s ≤ c‖un − u‖p H1 |Ω\F |1/r < cε1/r, where ps = 2∗ and 1 r + 1 s = 1. For N = 2, take any s > 1 to obtain the above inequality. Hence, un → u strongly in Lp(Ω). � Theorem 2.34 (Vitali Convergence Theorem for X(Ω)). (i) Let Ω be a domain in RN of finite measure. Then the embedding X(Ω) ↪→ Lp(Ω) is compact; (ii) Let Ω be a domain in RN and let {un}∞n=1 be a sequence in X(Ω). Suppose that a constant c > 0 exists such that ‖un‖H1 ≤ c for each n and un → u a.e. in Ω. Then for each ε > 0, a measurable set E ⊂ Ω exists such that |E| < ∞ and∫ Ec |un|pdz < ε for each n ∈ N if and only if ‖un − u‖ Lp(Ω) = o(1). EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 19 Proof. Part (i) follows from Willem [78]. (ii) By the Fatou lemma, ∫ Ec |u|pdz ≤ ε. Since |E| <∞ and ‖un‖H1 ≤ c, by (i), there is a subsequence {un}∞n=1 satisfying∫ E |un − u|pdz = o(1). Therefore,∫ Ω |un − u|pdz = ∫ E∩Ω |un − u|pdz + ∫ Ec∩Ω |un − u|pdz = o(1). Now suppose ‖un − u‖Lp(Ω) = o(1). Then for each ε > 0, a positive integer n0 exists such that ‖un − u‖Lp(Ω) < ε1/p 2 for n > n0, and measurable sets A and B of finite measure exist such that∫ Ac |u|pdz < ε 2p and ∫ Bc |un|pdz < ε for n = 1, 2, . . . , n0. The Minkowski inequality implies ‖un‖Lp(Ac) ≤ ‖un − u‖Lp(Ac) + ‖u‖Lp(Ac) < ε1/p for n > n0. Then let E = A ∪B to obtain the conclusion. � Let Lp w(RN ) = {u ∈ Lp loc(RN ) : ∫ RN |u(z)|pw(z)dz <∞} be a weighted Lebesgue space, where the weight w is nonnegative with ‖u‖p Lp w(RN ) = ∫ RN |u(z)|pw(z)dz. We denote by Q(x, l) the cube of the form Q(x, l) = {y ∈ RN : |yj − xj | < l/2, j = 1, . . . , N}. Theorem 2.35 (Vitali Convergence Theorem for H1(RN )). The embedding of H1(RN ) into Lp w(RN ) is compact: (i) Let N > 2. Suppose that w ∈ L p+δ δ w (RN ), with 2 ≤ p < p + δ < 2∗ for some δ > 0, and lim |x|→∞ ∫ Q(x,l) w(z) p+δ δ dz = 0 (2.7) for some l > 0. Then H1(RN ) is compactly embedded in Lp w(RN ); (ii) Let N = 2 and suppose that w ∈ Ls w(RN ) for some s > 1 and lim |x|→∞ ∫ Q(x,l) w(z) s dz = 0 (2.8) for some l > 0. Then H1(RN ) is compactly embedded in Lp w(RN ) for every p ≥ 2; (iii) Let N = 1 and suppose that w ∈ L1 loc(RN ) and lim |x|→∞ ∫ Q(x,l) w(z)dz = 0 (2.9) for some l > 0. Then H1(RN ) is compact embedded in Lp w(RN ) for every p ≥ 2. Proof. (i) It suffices to show that for every ε > 0, a R > 0 exists such that ‖u− uχQ(0,R)‖Lp w(RN ) < ε (2.10) for each u ∈ H1(RN ) such that ‖u‖H1(RN ) ≤ 1, where χQ is the characteristic function of the cube. Indeed, let {un} be a bounded sequence in H1(RN ). We 20 HWAI-CHIUAN WANG EJDE-2004/MON. 06 assume that ‖un‖H1(RN ) ≤ 1 for all n ∈ N. Consequently, a subsequence {un} and a u ∈ H1(RN ) exist such that un ⇀ u in H1(RN ) and un → u in Lp(Q(0, R)). On the other hand, by (2.10), we have ‖un − u‖Lp w(RN\Q(0,R)) ≤ ‖un‖Lp w(RN\Q(0,R)) + ‖u‖Lp w(RN\Q(0,R)) ≤ 2ε. Combining this with the previous observation, it is easy to conclude that un → u in Lp w(RN ). To show (2.10), we cover RN with cubes Q(ẑ, 1), ẑ ∈ ZN . We may assume that (i) holds with l = 1. For η > 0, we use (2.7) to find a positive constant n0 such that ∫ Q w(z) p+δ δ dz < η for each Q = Q(ẑ, 1) outside Q(0, n0). By the Sobolev embedding theorem, for any u ∈ H1(RN ), a constant c > 0 exists such that ‖u‖Lp(Q) ≤ c‖u‖H1(Q) for all 2 ≤ p < 2∗. Thus, by the Hölder inequality, we have∫ Q |u|pwdz ≤ ( ∫ Q w p+δ δ dz ) δ p+δ ( ∫ Q |u|p+δdz ) p p+δ ≤ c′η1/s‖u‖p H1(Q) where c′ = cp/(p+δ). Now, choose c′η1/s < ε and add these inequalities over all Q(ẑ, 1) outside Q(0, n0) to obtain R = n0. (ii) and (iii) are similar to (i). � We define H1 r (Ω) = {u ∈ H1 0 (Ω) : u is radially symmetric}. Lemma 2.36. For N ≥ 2, every u ∈ H1 r (RN ) is equal to a continuous function U a.e. in RN\{0} such that for z 6= 0 |U(z)| ≤ ( 2 ωN )1/2|z| 1−N 2 ( ∫ |t|≥|z| |u(t)|2dt )1/4( ∫ |y|≥|z| |∇u(t)|2dt )1/4 , where ωN is the area of the unit ball in RN . Proof. Let ϕ ∈ C∞c (RN ) be a radially symmetric function. Then for 0 ≤ r <∞, rN−1ϕ(r)2 = ∫ r 0 (sN−1ϕ(s)2)′ds = (N − 1) ∫ r 0 sN−2ϕ(s)2ds+ 2 ∫ r 0 sN−1ϕ(s)ϕ′(s)ds. Thus, 0 = (N − 1) ∫ ∞ 0 sN−2ϕ(s)2ds+ 2 ∫ ∞ 0 sN−1ϕ(s)ϕ′(s)ds. Consequently, rN−1ϕ(r)2 ≤ (N − 1) ∫ ∞ 0 sN−2ϕ(s)2ds+ 2 ∫ r 0 sN−1ϕ(s)ϕ′(s)ds = −2 ∫ ∞ r sN−1ϕ(s)ϕ′(s)ds = ( −2 ωN ) ∫ |t|≥r ϕ(t)ϕ′(t)dt ≤ ( 2 ωN ) (∫ |t|≥r |ϕ(t)|2dt )1/2(∫ |t|≥r |∇ϕ(t)|2dt )1/2 . EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 21 For u ∈ H1 r (RN ), take a sequence {ϕn} radially symmetric in C∞c (RN ), such that ϕn → u in H1(RN ), then there is a subsequence {ϕn(r)} such that rN−1u(r)2 = lim n→∞ rN−1ϕn(r)2 ≤ lim n→∞ ( 2 ωN ) ( ∫ |t|≥r |ϕn(t)|2dt )1/2( ∫ |t|≥r |∇ϕn(t)|2dt )1/2 ≤ ( 2 ωN ) ( ∫ |t|≥r |u(t)|2dt )1/2( ∫ |t|≥r |∇u(t)|2dt )1/2 . Since u ∈ H1 r (RN ), it is a function in H1(R), and there is a continuous function U in R such that u = U a.e. and |U(z)| ≤ ( 2 ωN )1/2|z| 1−N 2 ( ∫ |t|≥|z| |u(t)|2dt )1/4( ∫ |t|≥|z| |∇u(t)|2dt )1/4 . � Let Θ be an annulus, say Θ = {z ∈ RN : 1 < |z|} with N ≥ 3. Theorem 2.37 (Rellich-Kondrakov Theorem for H1 r (Θ)). The embedding H1 r (Θ) ↪→ Lp(Θ) is compact. Proof. Let {un} be a bounded sequence in H1 r (Θ). Then a subsequence {un} exists such that un → u a.e. in Θ and un ⇀ u weakly in H1 0 (Θ). By Lemma 2.36, lim |z|→∞ un(z) = 0 uniformly in n and lim s→0 |s|p |s|2+|s|2∗ = 0. Thus, for ε > 0, there is a K > 0 such that if |z| ≥ K, for each n, we have |un(z)|p ≤ ε(|un(z)|2 + |un(z)|2 ∗ ), or ∫ Θc K |un|p ≤ cε, where ΘK = {z ∈ Θ : z| < K}. By the Fatou lemma,∫ Θc K |u|p ≤ cε. By the Rellich-Kondrakov compactness theorem, a subsequence {un} exists such that lim n→∞ ∫ ΘK |un − u|p = 0. Thus, lim n→∞ ∫ Θ |un − u|p = 0. � For any β ∈ R, a (PS)β-sequence in X(Ω) for J is bounded. Moreover, a (PS)- value β should be nonnegative. 22 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Lemma 2.38. Let β ∈ R and let {un} be a (PS)β-sequence in X(Ω) for J , then a positive sequence {cn(β)} exists such that ‖un‖H1 ≤ cn(β) ≤ c for each n and cn(β) = o(1) as n→∞ and β → 0. Furthermore, a(un) = b(un) + o(1) = 2p p− 2 β + o(1) and β ≥ 0. Proof. Since {un} is a (PS)β-sequence in X(Ω) for J , we have |β|+ δn + εn‖un‖H1 p ≥ J(un)− 1 p 〈J ′(un), un〉 = ( 1 2 − 1 p )‖un‖2H1 , where δn = o(1) and εn = o(1). Take cn(β) = 1 p− 2 (εn + √ ε2n + 2p(p− 2)(|β|+ δn)), then cn(β) = o(1) as n → ∞ and β → 0 and ‖un‖H1 ≤ c(β) ≤ c for each n. Since {un} is bounded, we have o(1) = 〈J ′(un), un〉 = a(un)− b(un), or β + o(1) = J(un) = 1 2 a(un)− 1 p b(un) = p− 2 2p a(un) + o(1). Therefore, a(un) = b(un) + o(1) = 2p p− 2 β + o(1). This implies β ≥ 0. � Lemma 2.39. Let {un} be in X(Ω)\{0} satisfying a(un) = b(un) + o(1) and let J(un) = β + o(1) with β > 0, then c > 0 exists such that ‖un‖H1 ≥ c for each n. Proof. Suppose that a subsequence {un} satisfies lim n→∞ ‖un‖H1 = 0. Then J(un) = o(1), but this contradicts β > 0. Thus, c > 0 exists such that ‖un‖H1 ≥ c for each n. � Let Ω be an unbounded domain and ξn as in (2.1), then we have the following lemma. Lemma 2.40. Let {un} be a (PS)β-sequence in X(Ω) for J such that∫ Ωn |un|p = o(1), where Ωn = Ω ∩BN (0;n). Then for any r ≥ 1, we have (i) ∫ Ω ξr n|un|p = ∫ Ω |un|p + o(1) = 2p p−2β + o(1); (ii) ∫ Ω ξr n(|∇un|2 + u2 n) = ∫ Ω ξr n|un|p + o(1) = 2p p−2β + o(1); (iii) ∫ Ω (ξr n − 1)unϕ = o(1)‖ϕ‖H1 for every ϕ ∈ X(Ω); (iv) | ∫ Ω (ξr n − 1)|un|p−2unϕ| = o(1)‖ϕ‖H1 for every ϕ ∈ X(Ω); (v) | ∫ Ω (ξr n − 1)∇un∇ϕ| = o(1)‖ϕ‖H1 for every ϕ ∈ X(Ω). EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 23 Proof. (i) Clearly, we have∫ Ω ξr n|un|p = ∫ Ω |un|p + o(1) = 2p p− 2 β + o(1). (ii) Let wn = ξr nun. Since {wn} is bounded in X(Ω), we have o(1) = 〈J ′(un), wn〉 = ∫ Ω (ξr n|∇un|2 + rξr−1 n un∇ξn · ∇un + ξr nu 2 n)− ∫ Ω ξr n|un|p. Note that |∇ξn(z)| ≤ c n and {un} is bounded in X(Ω), so∫ Ω ξr−1 n un∇ξn · ∇un = o(1). We conclude that∫ Ω ξr n(|∇un|2 + u2 n) = ∫ Ω ξr n|un|p + o(1) = 2p p− 2 β + o(1). Therefore, the results follow. (iii) By the Hölder and Sobolev inequalities, we have | ∫ Ω (ξr n − 1)unϕ| ≤ ( ∫ Ωn |un|2 )1/2( ∫ Ω |ϕ|2 )1/2 ≤ o(1)‖ϕ‖H1 . (iv) By the Hölder and Sobolev inequalities, we have∣∣∣∫ Ω (ξr n − 1)|un|p−2unϕ ∣∣∣ ≤ ( ∫ Ωn |un|p) p−1 p ( ∫ Ω |ϕ|p)1/p ≤ o(1)‖ϕ‖H1 . (v) By the hypothesis and part (i), we have o(1) = 〈J ′(un), wn〉 = 〈J ′(un), wn〉 − 〈J ′(un), un〉+ 〈J ′(un), un〉 = ∫ Ω (ξr n − 1)|∇un|2 + ∫ Ω (ξr n − 1)u2 n − ∫ Ω (ξr n − 1)|un|p + o(1) = ∫ Ω (ξr n − 1)|∇un|2 + o(1). Thus, ∣∣ ∫ Ω (ξr n − 1)|∇un|2 ∣∣ = ∫ Ω (1− ξr n)|∇un|2 = o(1). Therefore, by the Hölder inequality, | ∫ Ω (ξr n − 1)∇un∇ϕ| ≤ (∫ Ω (ξr n − 1)2|∇un|2 )1/2 ‖ϕ‖H1 ≤ (∫ Ω (1− ξr n)|∇un|2 )1/2 ‖ϕ‖H1 ≤ o(1)‖ϕ‖H1 . � Lemma 2.41. (i) Suppose that {un} is a sequence in X(Ω) satisfying un ⇀ 0 weakly in X(Ω), then there is a subsequence {un} in X(Ω) such that ∫ Ωn |un|p = o(1) as n→∞; 24 HWAI-CHIUAN WANG EJDE-2004/MON. 06 (ii) For any β > 0, suppose that {un} is a (PS)β-sequence in X(Ω) for J satisfying∫ Ωn |un|p = o(1) as n→∞, then {ξnun} is also a (PS)β-sequence in X(Ω) for J . Proof. (i) Since un ⇀ 0 weakly in X(Ω), there is a subsequence {un} such that un → u strongly in Lp loc(Ω), or there is a subsequence {un} such that∫ Ωn |un|p = o(1), where Ωn = Ω ∩BN (0;n). (ii) Let {un} be a (PS)β-sequence in X(Ω) for J satisfying ∫ Ωn |un|p = o(1) as n→∞. By Lemma 2.40, we have J(ξnun) = 1 2 ∫ Ω [ |∇(ξnun)|2 + (ξnun)2 ] − 1 p ∫ Ω |ξnun|p = 1 2 ∫ Ω [ |∇ξn|2u2 n + ξ2n(|∇un|2 + u2 n) + 2ξnun∇ξn∇un ] − 1 p ∫ Ω ξp n|un|p = 1 2 a(un)− 1 p b(un) + o(1) = β + o(1). Then for ϕ ∈ X(Ω), we have |〈J ′(ξnun), ϕ〉| = ∣∣〈J ′(ξnun), ϕ〉 − 〈J ′(un), ϕ〉+ 〈J ′(un), ϕ〉 ∣∣ = ∣∣ ∫ Ω (ξn∇un∇ϕ+ un∇ξn∇ϕ+ ξnunϕ− ξp−1 n |un|p−2unϕ) − 〈J ′(un), ϕ〉+ 〈J ′(un), ϕ〉 ∣∣ = ∣∣ ∫ Ω [ (ξn − 1)∇un∇ϕ+ (ξn − 1)unϕ− (ξp−1 n − 1)|un|p−2unϕ ] + 〈J ′(un), ϕ〉 ∣∣ ≤ o(1)‖ϕ‖H1 Thus, J ′(ξnun) = o(1). � Moreover, we have the following lemma. Lemma 2.42. Let {un} be a (PS)-sequence in H1 0 (Ω) for J satisfying un ⇀ 0 weakly in X(Ω) and let vn = ξnun. Then ‖un − vn‖H1 = o(1) as n→∞. Proof. Note that a(un − vn) = 〈un − vn, un − vn〉H1 = a(un) + a(vn)− 2〈un, vn〉H1 = 2a(un)− 2〈un, vn〉H1 + o(1). Thus, it suffices to show that a(un) = 〈un, vn〉H1 + o(1). We have 〈un, vn〉H1 = ∫ Ω ∇un∇vn + unvn = ∫ Ω ξn [ |∇un|2 + (un)2 ] + ∫ Ω un∇un∇ξn. EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 25 Note that |∇ξn| ≤ c n and {un} is a (PS)-sequence in H1 0 (Ω) for J , so∫ Ω un∇un∇ξn = o(1). Hence, 〈un, vn〉H1 = ∫ Ω ξn [ |∇un|2 + (un)2 ] + o(1). By Lemma 2.40 (i), (ii) and Lemma 2.41 (i), we have 〈un, vn〉H1 = ∫ Ω ξn [ |∇un|2 + (un)2 ] + o(1) = a(un) + o(1). � Bibliographical notes: The (PS)-sequences were originally introduced by Palais- Smale [60]. Lemma 2.10 is from Brézis [14, p. 35]. Lemma 2.11 is from Zeidler [79, II/A, p. 303]. Lemma 2.16 is from Bahri-Lions [10]. Lemma 2.19 is from Grisvard [38, p. 24]. 3. Palais-Smale Decomposition Theorems In this section, we present the Palais-Smale decomposition theorem in H1 0 (Ω) for J . This is the concentration-compactness method of P. L. Lions. Theorem 3.1 (Palais-Smale Decomposition Theorem in RN )). Let Ω be strictly large domain (see Definition 2.1) in RN and let {un} be a (PS)β-sequence in H1 0 (Ω) for J . Then there are a subsequence {un}, a positive integer m, sequences {zi n}∞n=1 in RN , a function ū ∈ H1 0 (Ω), and 0 6= wi ∈ H1(RN ) for 1 ≤ i ≤ m such that |zi n| → ∞, for i = 1, 2, . . . ,m, −∆ū+ ū = |ū|p−2ū in Ω, −∆wi + wi =| wi|p−2wi in RN , and un = ū+ m∑ i=1 wi(· − zi n) + o(1) strongly in H1(RN ), a(un) = a(ū) + m∑ i=1 a(wi) + o(1), b(un) = b(ū) + m∑ i=1 b(wi) + o(1), J(un) = J(ū) + m∑ i=1 J(wi) + o(1). In addition, if un ≥ 0, then ū ≥ 0 and wi ≥ 0 for each 1 ≤ i ≤ m. Proof. Step 0. Since {un} is a (PS)β-sequence in H1 0 (Ω) for J , by Lemma 2.38 there is a c > 0 such that ‖un‖H1 ≤ c. In the following proof of this theorem, we fix such a c. There is a subsequence {un} and a ū in H1 0 (Ω) such that un ⇀ ū weakly in H1 0 (Ω) and ū solves −∆ū+ ū = |ū|p−2ū in Ω. 26 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Suppose that un → ū strongly in H1 0 (Ω), then we have un = u + o(1) strongly in H1 0 (Ω), a(un) = a(u) + o(1), b(un) = b(u) + o(1), J(un) = J(u) + o(1). Step 1. Suppose that un 9 ū strongly in H1 0 (Ω). Let u1 n = un − ū for n = 1, 2, . . . . By Lemma 2.15, {u1 n} is a (PS)(β−J(ū))-sequence in H1 0 (Ω) for J . (1-0) ∫ BN (0;1) |w1 n(z)|2dz ≥ d1 2 for some constant d1 > 0 and n = 1, 2, . . . , where w1 n(z) = u1 n(z + y1 n) for some {y1 n} ⊂ RN : since {u1 n} is bounded, J ′(u1 n) = o(1), and u1 n 9 0 strongly in H1 0 (Ω). By Lemma 2.16 there is a subsequence {u1 n}, a constant d1 > 0 such that Q1 n = sup z∈RN ∫ z+BN (0;1) |u1 n|2 ≥ d1 for n = 1, 2, . . . . Take {y1 n} in RN such that∫ y1 n+BN (0;1) |u1 n(z)|2dz ≥ d1 2 . Let w1 n(z) = u1 n(z + y1 n), then∫ BN (0;1) |w1 n(z)|2dz ≥ d1 2 for n = 1, 2, . . . . (1-1) un(z) = ū(z) + w1 n(z − y1 n) in H1(RN ). (1-2) ‖w1 n‖H1(RN ) ≤ c for n = 1, 2, . . . and ‖w1‖H1 ≤ c, where w1 n ⇀ w1 weakly in H1(RN ): by Lemma 2.11 (iii), ‖w1 n‖2H1 = ‖u1 n‖2H1 = ‖un‖2H1 − ‖ū‖2H1 + o(1) ≤ c2 + o(1), we have ‖w1 n‖H1(RN ) ≤ c for n = 1, 2, . . . . Then there is a subsequence {w1 n} and a w1 in H1(RN ) such that w1 n ⇀ w1 weakly in H1(RN ). By Lemma 2.11 (i), we have ‖w1‖H1 ≤ lim inf n→∞ ‖w1 n‖H1 ≤ c. (1-3) {w1 n} is a (PS)(β−J(ū))-sequence in H1(RN ) for J : note that J ′(u1 n) = o(1) in H−1(Ω). Because Ω is a strictly large domain, (1-7) below and Theorem 2.35, we have for every ϕ ∈ H1 0 (RN ), 〈J ′(w1 n), ϕ〉 = ∫ RN ∇w1 n∇ϕ+ w1 nϕ− ∫ RN |w1 n|p−2w1 nϕ = o(1) Therefore, J ′(w1 n) = o(1) strongly in H−1(RN ). Moreover, we have J(w1 n) = J(u1 n(z + y1 n)) = J(u1 n) = (β − J(ū)) + o(1). (1-4) −∆w1 + w1 − |w1|p−2w1 = 0 in RN : by Theorem 5.6 (i) below. (1-5) w1 6≡ 0: by the Rellich-Kondrakov theorem 2.31 and (1− 0), we have∫ BN (0;1) |w1|2 = lim n→∞ ∫ BN (0;1) |w1 n|2 ≥ d1 2 , thus w1 6≡ 0. EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 27 (1-6) By (1− 2), (1− 4), (1− 5), and Lemma 2.18, δ > 0 exists such that ‖w1‖H1(RN ) ≥ ‖w1‖L2(RN ) > δ. Therefore, J(w1) = ( 1 2 − 1 p )a(w1) > ( 1 2 − 1 p )δ2 = δ′. (1-7) |y1 n| → ∞: otherwise, R > 0 exists such that y1 n +BN (0; 1) ⊂ BN (0;R) for n = 1, 2, . . . . Then by (1− 0), we have 0 = lim n→∞ ∫ BN (0;R) |u1 n|2 ≥ lim n→∞ ∫ y1 n+BN (0;1) |u1 n|2 ≥ d1 2 , which is a contradiction. (1-8) a(un) = a(ū) + a(w1 n) + o(1) : since un ⇀ ū weakly in H1(RN ), by Lemma 2.11 (iii), we have a(un)− a(ū) = a(un − ū) + o(1) = a(u1 n) + o(1) = a(w1 n) + o(1). Thus, a(un) = a(ū) + a(w1 n) + o(1). (1-9) b(un) = b(ū) + b(w1 n) + o(1): since un → ū a.e. in Ω and {un} is bounded in Lp(Ω), by Lemma 2.14 (i), we have b(un)− b(ū) = b(un − ū) + o(1) = b(u1 n) + o(1) = b(w1 n) + o(1). Thus, b(un) = b(ū) + b(w1 n) + o(1). (1-10) J(un) = J(ū) + J(w1 n) + o(1): by (1− 8) and (1− 9), we have J(un) = J(ū) + J(w1 n) + o(1). Step 2. Suppose that w1 n(z) 9 w1(z) strongly in H1(RN ). Let u2 n(z) = w1 n(z)− w1(z). We have u2 n ⇀ 0 weakly in H1(RN ) but u2 n 9 0 strongly in H1(RN ). (2-0) ∫ BN (0;1) |w2 n(z)|2dz ≥ d2 2 for some constant d2 > 0 and n = 1, 2, . . . , where w2 n(z) = u2 n(z + y2 n) for some {y2 n} ⊂ RN : since {u2 n} is bounded, J ′(u2 n) = o(1), and u2 n 9 0 strongly in H1(RN ), by Lemma 2.16 there are a subsequence {u2 n}, and a constant d2 > 0 such that Q2 n = sup z∈RN ∫ z+BN (0;1) |u2 n(z)|2dz ≥ d2 for n = 1, 2, . . . . For n = 1, 2, . . . , take {y2 n} in RN such that∫ y2 n+BN (0;1) |u2 n(z)|2dz ≥ d2 2 for n = 1, 2, . . . . Let w2 n(z) = u2 n(z + y2 n), then∫ BN (0;1) |w2 n(z)|2dz ≥ d2 2 for n = 1, 2, . . . . As in Step 1, we have the following results. (2-1) un(z) = ū(z) + w1(z − y1 n) + w2 n(z − y1 n − y2 n) in H1(RN ); (2-2) ‖w2 n‖H1 ≤ c for n = 1, 2, . . . and ‖w2‖H1 ≤ c, where w2 n ⇀ w2 weakly in H1(RN ); (2-3) {w2 n} is a (PS)-sequence in H1(RN ) for J ; (2-4) −∆w2 + w2 − |w2|p−2w2 = 0 in RN ; 28 HWAI-CHIUAN WANG EJDE-2004/MON. 06 (2-5) w2 6≡ 0; (2-6) ‖w2‖L2(RN ) > δ and J(w2) > δ′; (2-7) |y2 n| → ∞; (2-8) a(un) = a(ū) + a(w1) + a(w2 n) + o(1): since u2 n(z) = w1 n(z)− w1(z) ⇀ 0, we have a(w2 n) = a(u2 n) = a(w1 n)− a(w1) + o(1). Further by (1− 8), we have a(un)− a(ū) = a(w1 n) + o(1) = a(w1) + a(w2 n) + o(1). (2-9) b(un) = b(ū) + b(w1) + b(w2 n) + o(1); (2-10) J(un) = J(ū) + J(w1) + J(w2 n) + o(1). Continuing this process, we arrive at the m-th step. (m-0) ∫ BN (0;1) |wm n (z)|2dz ≥ dm 2 for some constant dm > 0 and n = 1, 2, . . . , where wm n (z) = um n (z + ym n ) for some {ym n } ⊂ RN ; (m-1) un(z) = ū(z) + m−1∑ i=1 wi(z − zi n) + wm n (z − zm n ) in H1(RN ), where zi n = y1 n + · · ·+ yi n for i = 1, 2, . . . ,m : since wm n (z) = um n (z + ym n ) = wm−1 n (z + ym n )− wm−1(z + ym n ), thus wm n (z) + wm−1(z + ym n ) = wm−1 n (z + ym n ). Continuing this way, we obtain wm n (z) + wm−1(z + ym n ) + · · ·+ w1(z + y2 n + · · ·+ ym n ) = w1 n(z + y2 n + · · ·+ ym n ) = u1 n(z + y1 n + y2 n + · · ·+ ym n ) (m-2) ‖wm n ‖H1 ≤ c for n = 1, 2, . . . and ‖wm‖H1 ≤ c, where wm n ⇀ wm weakly in H1(RN ); (m-3) {wm n } is a (PS)-sequence in H1(RN ) for J ; (m-4) −∆wm + wm − |wm|p−2wm = 0 in RN ; (m-5) wm 6≡ 0; (m-6) ‖wm‖L2(RN ) > δ and J(wm) > δ′; (m-7) |yi n| = |zi n − zi−1 n | → ∞ and |zi n| → ∞, for each i = 1, 2, . . . ,m : we show it by induction on i. For i = 1, |z1 n| = |y1 n| → ∞. Assume that |zi n| → ∞, for i = 1, 2, . . . , k, for some k < m. By Lemma 2.12, we have wi(z − zi n) ⇀ 0 weakly in H1(RN ) for i = 1, 2, . . . , k. We claim that |zk+1 n | → ∞. Otherwise, suppose that {zk+1 n } is bounded. Since ‖wk+1‖L2(RN ) > δ, R > 0 exists such that zk+1 n +BN (0;R) ⊂ BN (0; 2R) and ∫ BN (0;R) |wk+1(z)|2 ≥ ( δ 2 )2. EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 29 We have ( δ 2 )2 ≤ ∫ BN (0;R) ∫ BN (0;R) |wk+1(z)|2 = lim n→∞ ∫ BN (0;R) |u1 n(z + zk+1 n )|2dz ≤ lim n→∞ ∫ BN (0;2R) |u1 n(z)|2 = 0, which is a contradiction. By the induction hypothesis, we have |zi n| → ∞ for i = 1, 2, . . . ,m. (m-8) a(un) = a(ū) + ∑m−1 i=1 a(wi) + a(wm n ) + o(1); (m-9) b(un) = b(ū) + ∑m−1 i=1 b(wi) + b(wm n ) + o(1); (m-10) J(un) = J(ū) + ∑m−1 i=1 J(wi) + J(wm n ) + o(1). By the Archimedean principle, l ∈ N exists such that lδ2 > β. Then after step (l + 1), we obtain a(un) = a(ū) + a(w1) + a(w2) + · · ·+ a(wl) + a(wl+1 n ) + o(1). Since a(wl+1 n ) ≥ 0, a(ū) > 0, and a(wi) > δ2 for i = 1, 2, . . . , l, we have β + o(1) ≥ lδ2 > β, which is a contradiction. Therefore, there is an m ∈ N, such that wm n (z) = wm(z) + o(1) strongly in H1(RN ), wi n(z) = wi(z) + o(1) weakly, and wi n(z) 6= wi(z) + o(1) strongly in H1(RN ) for i = 1, 2, . . .m− 1. Then we have (sm-0) ∫ BN (0;1) |wm n (z)|2dz ≥ dm 2 for some constant dm > 0 and n = 1, 2, . . . , where wm n (z) = um n (z + ym n ) for some {ym n } ⊂ RN ; (sm-1) un(z) = ū(z) + m∑ i=1 wi(z − zi n) + o(1) strongly in H1(RN ), where zi n = y1 n + · · ·+ yi n for i = 1, 2, . . . ,m; (sm-2) ‖wm n ‖H1 ≤ c for n = 1, 2, . . . and ‖wm‖H1 ≤ c, where wm n ⇀ wm weakly in H1(RN ); (sm-3) {wm n } is a (PS)-sequence in H1(RN ) for J ; (sm-4) −∆wm + wm − |wm|p−2wm = 0 in RN ; (sm-5) wm 6≡ 0; (sm-6) ‖wm‖L2(RN ) > δ and J(wm) > δ′; (sm-7) |yi n| = |zi n − zi−1 n | → ∞ and |zi n| → ∞, for each i = 1, 2, . . . ,m; (sm-8) a(un) = a(ū) + ∑m i=1 a(w i) + o(1); (sm-9) b(un) = b(ū) + ∑m i=1 b(w i) + o(1); (sm-10) J(un) = J(ū) + m∑ i=1 J(wi) + o(1). Finally, suppose un ≥ 0 for n = 1, 2, . . . . Then (i) Since un ⇀ ū weakly in H1 0 (Ω). By Lemma 2.11 (ii), there is a subsequence {un} such that un → ū a.e. in Ω. Thus, ū ≥ 0. (ii) Since w1 n(z) = un(z+y1 n)−ū(z+y1 n) ⇀ w1(z) weakly inH1(RN ) and ū(z+y1 n) ⇀ 0 weakly in H1(RN ). Thus, un(z + y1 n) → w1(z) a.e. in Ω, or w1 ≥ 0. (iii) Continuing this process, we obtain wi ≥ 0 for each i = 1, 2, . . . ,m. � We have the following useful corollary. 30 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Corollary 3.2. Let Ω be a strictly large domain in RN . If {un} is a positive (PS)β-sequence in H1 0 (Ω) for J . (i) If β 6= jα(RN ) for each j ∈ N, then there is a positive solution u of (1.1) in Ω; (ii) If α(RN ) < β < 2α(RN ), then {un} contains a strongly convergent subsequence. Proof. By Theorem 3.1, we have J(un) = J(u) + ∑ i = 1mJ(wi) + o(1). By Corollary 3.6 below, the positive solutions of (1.1) in RN are unique, and we obtain J(wi) = α(RN ) for each i. Thus, we have β = J(u) +mJ(wi) + o(1). (i) If β 6= jα(RN ) for each j ∈ N, then J(u) 6= 0, or u 6= 0. By Theorem 5.6 (i) below, there is a positive solution u of (1.1) in Ω. (ii) Recall that we always have β ≥ α(Ω) ≥ α(RN ). Suppose that m ≥ 1 and α(RN ) < β < 2α(RN ), then J(u) 6= 0 or J(u) ≥ α(Ω). Thus, 2α(RN ) > β + o(1) = J(u) +mα(RN ) ≥ (m+ 1)α(RN ). This is a contradiction. Hence, m = 0. By the proof of Theorem 3.1, we have un = u+ o(1) strongly in H1 0 (Ω). � Remark 3.3. Note that if we replace a strictly large domain by a domain in Theorem 3.1, then the theorem may fail. Let Ar 0 be an upper semi-strip with sufficiently large r, then α(RN ) < α(Ar 0) < 2α(RN ). By the Esteban-Lions theorem 10.7, (1.1) in Ar 0 admits only trivial solution, but if Theorem 3.1 holds, by Corollary 3.2, (1.1) in Ar 0 admits a positive solution, a contradiction. Definition 3.4. A domain Θ in RN is a periodic domain if a partition {Qm}∞m=0 of Θ and points {zm}∞m=1 in RN exist, satisfying the following conditions: (i) {zm}∞m=1 forms a subgroup of RN ; (ii) Q0 is bounded; (iii) Qm = zm +Q0 for each m. Typical examples of periodic domains are the infinite strip Ar, the infinite hollow strip Ar1,r2 , and the whole space RN . Similarly, we have the Palais-Smale decomposition theorem in H1 0 (Ω) for J in a periodic domain in Θ ⊂ RN . Theorem 3.5 (Palais-Smale Decomposition Theorem in a Periodic Domain). Let Ω be a strictly large domain in Θ and let {un} be a positive (PS)β-sequence in H1 0 (Ω) for J . Then there are a subsequence {un}, a positive integer m, a subsequence {zi n}∞n=1 of {zm}∞m=1 in Θ, and a function ū ∈ H1 0 (Ω), and 0 6= wi ∈ H1(Θ), for 1 ≤ i ≤ m such that (i) |zi n| →,∞ for i = 1, 2, . . . ,m; (ii) −∆ū+ ū =| ū |p−2 ū in Ω; (iii) −∆wi + wi = |wi|p−2wi in Θ; (iv) un = ū+ ∑m i=1 w i(· − zi n) + o(1) strongly in H1(Θ); (v) a(un) = a(ū) + ∑m i=1 a(w i) + o(1); (vi) b(un) = b(ū) + ∑m i=1 b(w i) + o(1); EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 31 (vii) J(un) = J(ū) + ∑m i=1 J(wi) + o(1). In addition, if un ≥ 0, then ū ≥ 0 and wi ≥ 0 for each 1 ≤ i ≤ m. Proof. The proof is similar to those of Theorem 3.1: see Lien-Tzeng-Wang [47]. Note that instead of Qn = sup z∈RN ∫ z+BN (0;1) |un(z)|2dz we use Qr n = sup y∈R ∫ (0,y)+Ar −1,1 |un(z)|2dz, where Ar −1,1 = {(x, y) ∈ Ar | − 1 < y < 1}. � Corollary 3.6. Let Θ be a periodic domain in RN and let Ω be a strictly large domain in Θ, and let {un} be a positive (PS)β-sequence in H1 0 (Ω) for J . Suppose that the only positive solutions in Θ are ground state solutions. (i) If β 6= jα(Θ) for each j ∈ N, then there is a positive solution u of (1.1) in Ω; (ii) If α(Θ) < β < 2α(Θ), then {un} contains a strongly convergent subsequence. The proof of the above corollary is same as the proof of Corollary 3.2. Bibliographical notes: Theorem 3.1 is from Lions [49, Lemma 19] and Struwe [66]. Theorem 3.5 is from Lien-Tzeng-Wang [47, Theorem 4.1]. 4. Palais-Smale Values and Indexes of Domains In this section, we prove that four classical important (PS)-values in X(Ω) for J are the same. Then any one of them is called the index of a domain. The index of a domain Ω is important in studying the existence of solutions of (1.1) in Ω. (A) Consider the constrained maximization problem αγ(Ω) = ( 1 2 − 1 p )γ(Ω) 2p 2−p , where γ(Ω) = sup{b(u) | u ∈ X(Ω), a(u) = 1}. By the Sobolev embedding theorem, we have αγ(Ω) > 0. Moreover, αγ(Ω) is a (PS)-value in X(Ω) for J . Theorem 4.1. αγ(Ω) is a (PS)-value in X(Ω) for J . Proof. Let {un} in X(Ω) be a maximizing sequence of γ(Ω). Then a(un) = 1 for n = 1, 2, . . . , and ∫ Ω |un|p = γ(Ω)p + o(1) as n→∞. Let vn = γ(Ω) p 2−pun for each n = 1, 2, . . . . Then we have a(vn) = ∫ Ω (|∇vn|2 + v2 n) = γ(Ω) 2p 2−p for each n = 1, 2, . . . , b(vn) = ∫ Ω |vn|p = γ(Ω) 2p 2−p + o(1) as n→∞, and J(vn) = 1 2 a(vn)− 1 p b(vn) = ( 1 2 − 1 p )γ(Ω) 2p 2−p + o(1) as n→∞ = αγ(Ω) + o(1) as n→∞. 32 HWAI-CHIUAN WANG EJDE-2004/MON. 06 For each n = 1, 2, . . . and ϕ ∈ X(Ω), denote ln(ϕ) = ∫ Ω |vn|p−2vnϕ. Let φ ∈ X(Ω) satisfy ‖φ‖H1 = 1. Then γ(Ω) ≥ ‖φ‖Lp and |ln(φ)| = ∣∣∫ Ω |vn|p−2vnφ ∣∣ ≤ (∫ Ω |vn|p )(p−1)/p(∫ Ω |φ|p )1/p ≤ γ(Ω) 2p−2 2−p γ(Ω) + o(1) = γ(Ω) p 2−p + o(1) as n→∞. Thus, ‖ln‖X−1 ≤ γ(Ω) p 2−p + o(1) as n→∞. Furthermore, ln ( vn ‖vn‖H1 ) = ∫ Ω |vn|p ‖vn‖H1 = γ(Ω)2p/(2−p) γ(Ω)p/(2−p) + o(1) = γ(Ω) p 2−p + o(1) as n→∞. We conclude that ‖ln‖X−1 = γ(Ω) p 2−p + o(1) as n→∞. Since ln is a continuous linear functional in X(Ω), by the Riesz representation theorem, for each n, wn ∈ X(Ω) exists such that ln(ϕ) = 〈wn, ϕ〉H1 = ∫ Ω (∇wn · ∇ϕ+ wnϕ) for each ϕ ∈ X(Ω), and ‖wn‖H1 = ‖ln‖X−1 . Since 〈wn, vn〉H1 = ln(vn) = ∫ Ω |vn|p = γ(Ω) 2p 2−p + o(1) as n→∞, we obtain ‖vn − wn‖2H1 = 〈vn, vn〉H1 − 2〈vn, wn〉H1 + 〈wn, wn〉H1 = ‖vn‖2H1 − 2〈vn, wn〉H1 + ‖wn‖2H1 = γ(Ω) 2p 2−p − 2γ(Ω) 2p 2−p + γ(Ω) 2p 2−p + o(1) = o(1) as n→∞. For ϕ ∈ X(Ω) satisfying ‖ϕ‖H1 = 1, we have 〈J ′(vn), ϕ〉 = ∫ Ω (∇vn · ∇ϕ+ vnϕ)− ∫ Ω |vn|p−2vnϕ = 〈vn, ϕ〉H1 − 〈wn, ϕ〉H1 = 〈vn − wn, ϕ〉H1 , so |〈J ′(vn), ϕ〉| ≤ ‖vn − wn‖H1 . We conclude that J ′(vn) = o(1) strongly in X−1(Ω) as n→∞. � EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 33 (B) Consider the Nehari minimizing problem αM(Ω) = inf v∈M(Ω) J(v), where M(Ω) = {u ∈ X(Ω)\{0} : a(u) = b(u)}. Note that M(Ω) contains every nonzero solution of (1.1). Consider the unit sphere U(Ω) and the zero energy manifold Z(Ω), where U(Ω) = {u ∈ X(Ω) : ‖u‖H1 = 1}, Z(Ω) = {u ∈ X(Ω)\{0} : 1 2 a(u) = 1 p b(u)}. αM(Ω) > 0 is a consequence of part (i) of the following lemma. Part (ii) of the following lemma will be used later in Lemma 4.6 and Theorem 4.12. Lemma 4.2. (i) There is a bijective C1,1 map m from U(Ω) to M(Ω). Moreover, M(Ω) is path-connected and a constant c > 0 exists such that for u ∈ M(Ω), ‖u‖H1 ≥ c and J(u) ≥ c; (ii) There is a bijective C1,1 map z from U(Ω) to Z(Ω). Moreover, Z(Ω) is path- connected and a constant c′ > 0 exists such that for u ∈ Z(Ω), ‖u‖H1 ≥ c′. Proof. (i) For t ≥ 0, u ∈ U(Ω), let hu(t) = J(tu) = 1 2 t2 − 1 p tpb(u). Then h′u(t) = t − tp−1b(u). We take uniquely su ∈ R+ such that su > 0, suu ∈ M(Ω), and 0 = h′u(su). For v ∈ U(Ω), a sv ∈ R+ exists such that svv ∈ M(Ω): that is 〈J ′(svv), svv〉 = s2v − sp vb(v) = 0. Consider the function g(t, u) : R+ ×U(Ω) → R defined by g(t, u) = 〈J ′(tu), tu〉 = t2a(u)− tpb(u). Note that g(sv, v) = 〈J ′(svv), svv〉 = 0. Thus, ∂g ∂t (t, u) ∣∣ (sv,v) = 2sv − psp−1 v b(v) = sv(2− p) < 0. By the implicit function theorem, a neighborhood W of v in U(Ω) and a unique function t ∈ C1,1 exist such that t : W → R+, t(v) = sv, g(t(u), u) = 0 for all u ∈ W. Therefore, for each v ∈ U(Ω), t : U(Ω) → R+ and m : U(Ω) → M(Ω), t, m ∈ C1,1 exist such that t(v) = sv, m(v) = svv. Clearly, t and m are injective. For each u ∈ M(Ω), write u = svv, where sv = ‖u‖H1 and v = u ‖u‖H1 ∈ U(Ω). Since m(v) = u, m is surjective. Since U(Ω) is path-connected, M(Ω) is path-connected. Note that u ∈ M(Ω), so J ′(u) = 0, or s2v = ∫ Ω sp v|v|p. By the Sobolev embedding theorem, we have s2v = ∫ Ω sp v|v|p ≤ dsp v, or c ≤ sv, where d and c are two positive constants. Therefore, ‖u‖H1 = ‖svv‖H1 = sv ≥ c for u ∈ M(Ω). (ii) The proof is similar to part (i). � 34 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Theorem 4.3. Let β > 0 and let {un} in X(Ω)\{0} be a sequence for J such that J(un) = β + o(1) and a(un) = b(un) + o(1). Then there is a sequence {sn} in R+ such that sn = 1+ o(1), {snun} is in M(Ω) and J(snun) = β+ o(1). In particular, if {un} is a (PS)β-sequence for J , then there is a sequence {sn} in R+ such that {snun} is in M(Ω) and there is also a (PS)β-sequence in X(Ω) for J . Proof. By Lemma 4.2, there is a sequence {sn} in R+ such that {snun} is in M(Ω) : s2na(un) = sp nb(un) for each n, because a(un) = b(un) + o(1) and J(un) = β + o(1) imply sn = 1 + o(1). Therefore, J(snun) = β + o(1). The last part follows from Lemma 2.38. � A minimizing sequence {un} in M(Ω) of αM(Ω) is a (PS)αM(Ω)-sequence inX(Ω) for J . Theorem 4.4. Let {un} be in X(Ω). Then {un} is a (PS)αM(Ω)-sequence for J if and only if J(un) = αM(Ω) + o(1) and a(un) = b(un) + o(1). In particular, every minimizing sequence {un} in M(Ω) of αM(Ω) is a (PS)αM(Ω)-sequence in X(Ω) for J . In particular, αM(Ω) is a (PS)αM(Ω)−value in X(Ω) for J . Proof. Suppose {un} is a (PS)αM(Ω)-sequence in X(Ω) for J . By Lemma 2.38, we have a(un) = b(un) + o(1). Conversely, let {un} satisfy J(un) = αM(Ω) + o(1) and a(un) = b(un) + o(1). Then we have a(un) = 2p p− 2 αM(Ω) + o(1) as n→∞. (4.1) For n = 1, 2, . . . , denote ln(ϕ) = ∫ Ω |un|p−2unϕ for ϕ ∈ X(Ω). (4.2) Let φ ∈ U(Ω). By Lemma 4.2, t > 0 exists such that tφ ∈ M(Ω) : ‖tφ‖2H1 = ‖tφ‖p Lp ; we conclude that t = ‖φ‖ −p p−2 Lp and αM(Ω) ≤ ( 1 2 − 1 p )‖tφ‖2H1 = p− 2 2p t2 = p− 2 2p ‖φ‖ −2p p−2 Lp . Therefore, ‖φ‖Lp ≤ ( 2p p−2αM(Ω)) 2−p 2p . For each n, |ln(φ)| = ∣∣ ∫ Ω |un|p−2unφ ∣∣ ≤ ( ∫ Ω |un|p ) p−1 p ( ∫ Ω |φ|p )1/p ≤ ( 2p p− 2 αM(Ω)) p−1 p ( 2p p− 2 αM(Ω)) 2−p 2p + o(1) = ( 2p p− 2 αM(Ω) )1/2 + o(1) as n→∞, we have ‖ln‖X−1 ≤ ( 2p p− 2 αM(Ω))1/2 + o(1) as n→∞. (4.3) EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 35 Furthermore, by (4.2), we have ln( un ‖un‖H1 ) = ∫ Ω |un|p ‖un‖H1 = (b(un))1/2 + o(1) = ( 2p p− 2 αM(Ω))1/2 + o(1) as n→∞ (4.4) By (4.3) and (4.4), we conclude that ‖ln‖X−1 = ( 2p p− 2 αM(Ω))1/2 + o(1) as n→∞. By the Riesz representation theorem, for each n, wn ∈ X(Ω) exists such that, for each ϕ ∈ X(Ω), ln(ϕ) = 〈wn, ϕ〉H1 = ∫ Ω (∇wn · ∇ϕ+ wnϕ), and ‖wn‖H1 = ‖ln‖X−1 = ( 2p p− 2 αM(Ω))1/2 + o(1). (4.5) Consequently, 〈wn, un〉H1 = ln(un) = ∫ Ω |un|p = 2p p− 2 αM(Ω) + o(1). (4.6) By (4.1), (4.5), and (4.6), we obtain ‖un − wn‖2H1 = 〈un, un〉H1 − 2〈un, wn〉H1 + 〈wn, wn〉H1 = ‖un‖2H1 − 2〈un, wn〉H1 + ‖wn‖2H1 = 2p p− 2 αM(Ω)− 2 2p p− 2 αM(Ω) + 2p p− 2 αM(Ω) + o(1) = o(1) as n→∞. For ϕ ∈ U(Ω), we have 〈J ′(un), ϕ〉 = ∫ Ω (∇un · ∇ϕ+ unϕ)− ∫ Ω |un|p−2unϕ = 〈un, ϕ〉H1 − 〈wn, ϕ〉H1 = 〈un − wn, ϕ〉H1 , so ‖J ′(un)‖X−1 ≤ ‖un − wn‖H1 = o(1). We conclude that J ′(un) = o(1) strongly in X−1(Ω) as n→∞. � If u achieves αM(Ω), then u is a nonzero solution of (1.1). Theorem 4.5. Let u ∈ M(Ω) be such that J(u) = minv∈M(Ω) J(v). Then u is a nonzero solution of (1.1). Proof. Set g(v) = a(v)− b(v) for v ∈ X(Ω). Note that 〈g′(u), u〉 = (2− p)a(u) 6= 0. Since the minimum of J is achieved at u and is constrained in M(Ω), by the Lagrange multiplier theorem, λ ∈ R exists such that J ′(u) = λg′(u) in X(Ω). Thus, 0 = 〈J ′(u), u〉 = λ〈g′(u), u〉, or λ = 0. Thus, J ′(u) = 0. Hence, u is a weak solution of (1.1) such that J(u) = αM(Ω). � 36 HWAI-CHIUAN WANG EJDE-2004/MON. 06 (C) Consider the mountain pass minimax problem αΓ(Ω) = inf g∈Γ(Ω) max t∈[0,1] J(g(t)), where e 6= 0, J(e) = 0, and Γ(Ω) = {g ∈ C([0, 1], X(Ω)) : g(0) = 0, g(1) = e}. Then αΓ(Ω) > 0 is a consequence of the following lemma. Lemma 4.6. A ball B(0; r) in X(Ω), c > 0, and e /∈ B(0; r) exist such that J(e) = 0 and minv∈∂B(0;r) J(v) ≥ c. Proof. By Lemma 4.2 (ii), for each u ∈ U(Ω), there is a t > 0 such that J(tu) = 0. Let e = tu, then J(e) = 0. Since for each v ∈ X(Ω)\{0} J(v) = 1 2 a(v)− 1 p b(v), by the Sobolev inequality, there is a constant c1 > 0 such that b(v) ≤ c1a(v)p/2, and we have J(v) ≥ a(v){1 2 − c1 p a(v) p−2 2 }. Take r > 0 such that e /∈ B(0; r) and 1 2 − c1 p r p−2 ≥ 1 4 , then for ‖v‖H1 = r, we have J(v) ≥ c, where c = 1 4r 2. � We require the following lemma. Theorem 4.7 (Ekeland variational principle). Let M be a complete metric space with metric d and let F : M → R∪ {+∞} be lower semi-continuous, bounded from below, and 6≡ ∞. Then for any ε > 0 and λ > 0, and any u ∈M with F (u) ≤ inf M F + ε, there is an element v ∈M such that F (v) ≤ F (u), d(u, v) ≤ 1 λ , F (w) + ελd(v, w) > F (v) for w 6= v. Proof. It is sufficient to prove our assertion for λ = 1. The general case is obtained by replacing d by an equivalent metric λd. We define the relation on M : w ≤ v ⇐⇒ F (w) + εd(v, w) ≤ F (v). It is easy to see that this relation define a partial ordering on M . We now construct inductively a sequence {um} as follows: u0 = u; also assuming that um has been defined, we set Sn = {w ∈M | w ≤ un} and choose un+1 ∈ Sn so that F (un+1) ≤ inf Sn F + 1 n+ 1 . EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 37 Since un+1 ≤ un, Sn+1 ⊂ Sn, and by the lower semicontinuity of F , Sn is closed. We now show that diam Sn → 0. Indeed, if w ∈ Sn+1, then w ≤ un+1 ≤ un and consequently, εd(w, un+1) ≤ F (un+1)− F (w) ≤ inf Sn F + 1 n+ 1 − inf Sn F = 1 n+ 1 . This estimate implies diamSn+1 ≤ 2 ε(n+ 1) and our claim follows. The fact that M is complete implies that ∩n≥0Sn = {v} for some v ∈M . In particular, v ∈ S0, that is, v ≤ u0 = u. Hence, F (v) ≤ F (u)− εd(u, v) ≤ F (u). Moreover, d(u, v) ≤ ε−1(F (u)− F (v)) ≤ ε−1 ( inf M F + ε− inf M F ) = 1. To complete the proof we must show w ≤ v implies w = v. If w ≤ v, then w ≤ un for each integer n ≥ 0, that is w ∈ ∩n≥0Sn = {v}. � Lemma 4.8. Let Γ(Ω) be the complete metric space with the usual L∞ distance d and J ∈ C1(X(Ω),R). Then for each ε > 0 and each f ∈ Γ(Ω) such that max s∈[0,1] J(f(s)) ≤ αΓ(Ω) + ε, (4.7) v ∈ X(Ω) exists such that αΓ(Ω)− ε ≤ J(v) ≤ max s∈[0,1] J(f(s)), dist(v, f([0, 1])) ≤ ε1/2, |J ′(v)| ≤ ε1/2. Proof. Without loss of generality, we can assume that 0 < ε < αΓ(Ω). (4.8) Let f ∈ Γ(Ω) satisfy the condition (4.7). We define the function Φ : Γ(Ω) → R by Φ(g) = max s∈[0,1] J(g(s)). Then (i) Φ is bounded below: Φ(g) ≥ αΓ(Ω) > 0. (ii) Φ is continuous at each g ∈ Γ(Ω) : since J is continuous on the compact set K = g([0, 1]), for each ε > 0, u ∈ K, there is a δu > 0 such that if w ∈ B(u; δu) is an open ball in X(Ω), then |J(w) − J(u)| < 1 2ε. Since K is compact, finite values B(ui; δui), i = 1, . . . , n, exist such that K ⊂ B(u1; δu1 2 ) ∪ · · · ∪B(un; δun 2 ). Take δ = min{ δu1 2 , . . . , δun 2 }. Let k ∈ Γ(Ω) satisfy ‖k − g‖L∞ < δ. For each s ∈ [0, 1], we have |k(s)− g(s)| < δ, 38 HWAI-CHIUAN WANG EJDE-2004/MON. 06 or g(s) ∈ B(ui; δui 2 ), k(s) ∈ B(ui; δui ). Thus |J(k(s))− J(g(s))| < ε, or |Φ(k)− Φ(g)| ≤ ε. The Ekeland variational principle (Theorem 4.7) implies the existence of h ∈ Γ(Ω) such that Φ(h) ≤ Φ(f) ≤ αΓ(Ω) + ε, max s∈[0,1] |h(s)− f(s)| ≤ ε1/2, and Φ(g) > Φ(h)− ε 1 2 d(h, g) whenever g ∈ Γ(Ω) and g 6= h. (4.9) Let A = {s ∈ [0, 1] : αΓ(Ω)− ε ≤ J(h(s))}, then A is nonempty since αΓ(Ω)− ε < αΓ(Ω) = inf g∈Γ(Ω) max s∈[0,1] J(g(s)) ≤ max s∈[0,1] J(h(s)). Note that for s ∈ A, |J ′(h(s))| ≤ ε1/2, if and only if |〈J ′(h(s)), v〉| ≤ ε1/2 for v ∈ U(Ω), if and only if 〈J ′(h(s)), v〉 ≥ −ε1/2 for v ∈ U(Ω). We claim that there is some s ∈ A satisfying |J ′(h(s))| ≤ ε1/2. If this is not the case, then for each s ∈ A, vs ∈ U(Ω) exists such that 〈J ′(h(s)), vs〉 < −ε1/2. By the continuity of J ′, δs > 0 and an open ball Bs in [0, 1] containing s exist such that for t ∈ Bs and u ∈ X(Ω) with |u| ≤ δs, we have 〈J ′(h(t) + u), vs〉 < −ε1/2. (4.10) Since A is compact, a finite subcovering Bs1 , Bs2 . . .Bsk of A exists. We define the Lipschitz continuous functions, for each j = 1, 2, . . . , k, ψj : [0, 1] → [0, 1] by ψj(t) = { dist(t, Bc sj )/ ∑k i=1 dist(t, Bc si ) for t ∈ A; 0 for t /∈ ∪k i=1Bsi . Then k∑ j=1 ψj(t) = 1for t ∈ A; ‖ k∑ j=1 ψj(t)vsj ‖H1 ≤ 1 for t ∈ A. Let δ = min{δs1 , . . . δsk } and let ψ : [0, 1] → [0, 1] be a continuous function such that ψ(t) = { 1 if J(h(t)) ≥ αΓ(Ω); 0 if J(h(t)) ≤ αΓ(Ω)− ε, and let g ∈ C([0, 1], X(Ω)) be defined by g(t) = h(t) + δψ(t) k∑ j=1 ψj(t)vsj . EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 39 It follows from (4.8) that, for t ∈ {0, 1}, we have J(h(t)) = 0 < αΓ(Ω) − ε, or ψ(t) = 0. Consequently, g(0) = h(0) = 0 and g(1) = h(1) = e, that is, g ∈ Γ(Ω). The mean value theorem and (4.10) imply that, for each t ∈ A, there is some 0 < τ < 1 for which J(g(t))− J(h(t)) = 〈J ′(h(t) + τδψ ( t) k∑ j=1 ψj(t)vsj ) , δψ(t) k∑ j=1 ψj(t)vsj 〉 = δψ(t) k∑ j=1 ψj(t)〈J ′ ( h(t) + τδψ(t) k∑ j=1 ψj(t)vsj ) , vsj 〉 ≤ −ε1/2δψ(t). (4.11) Thus J(g(t)) ≤ J(h(t))− ε1/2δψ(t) ≤ J(h(t)). If t /∈ A, then ψ(t) = 0 and hence J(g(t)) = J(h(t)). Let t̄ ∈ [0, 1] satisfy J(g(t)) = Φ(g), then we obtain J(h(t)) ≥ J(g(t)) ≥ αΓ(Ω), so that t ∈ A and ψ(t) = 1. By (4.11), we obtain J(g(t))− J(h(t)) ≤ −ε1/2δ and in particular Φ(g) + ε1/2δ ≤ J(h(t)) ≤ Φ(h), so that g 6= h. However, by the definition of g, we have d(g, h) ≤ δ and Φ(g) + ε1/2d(g, h) ≤ Φ(h) which contradicts (4.9). The proof is complete . � αΓ(Ω) is a (PS)-value in X(Ω) for J . Theorem 4.9. Under the conditions of Lemma 4.8, for each minimizing sequence {fk} ⊂ Γ(Ω) such that Φ(fk) = max s∈[0,1] J(fk(s)) = αΓ(Ω) + o(1), there is a (PS)-sequence {vk} in X(Ω) for J satisfying J(vk) = αΓ(Ω) + o(1), dist(vk, fk([0, 1])) = o(1), J ′(vk) = o(1) strongly in X−1(Ω) as k →∞. In particular, αΓ(Ω) is a (PS)-value in X(Ω) for J . Proof. We define εk = max s∈[0,1] J(fk(s)) − αΓ(Ω) if max s∈[0,1] J(fk(s)) − αΓ(Ω) > 0 and εk = 1 k in the other case. Then we apply Lemma 4.8 to εk and fk: αΓ(Ω)− εk ≤ J(vk) ≤ max s∈[0,1] J(fk(s)) ≤ αΓ(Ω) + εk, dist(vk, fk([0, 1])) ≤ ε 1/2 k , |J ′(vk)| ≤ ε 1 2 k for each k > 0. 40 HWAI-CHIUAN WANG EJDE-2004/MON. 06 This completes the proof. � (D) Consider the infimum of positive (PS)-values in X(Ω) for J : αP(Ω) = inf β∈P(Ω) β, where P(Ω) is the set of all positive (PS)-values in X(Ω) for J . That αP(Ω) > 0 is a consequence of the following theorem. Theorem 4.10. There is a β0 > 0 such that β ≥ β0 for every positive (PS)-value β in X(Ω) for J . Proof. Let {un} be a (PS)β-sequence in X(Ω) for J with β > 0. By Lemma 2.38, a positive sequence {cn(β)} exists such that cn(β) = o(1) as n→∞, β → 0, and a(un) ≤ cn(β)2. (4.12) By the Sobolev embedding theorem, there is a constant d > 0 such that b(un) ≤ da(un)p/2. (4.13) By Lemma 2.38, (4.12), and (4.13), we have o(1) = a(un)− b(un) ≥ a(un) [ 1− dcn(β)p−2 ] . Take β0 > 0 and n0 > 0 such that if β < β0 and n ≥ n0, then 1− dcn(β)p−2 > 1 2 . Consequently, a(un) = b(un) = o(1), or J(un) = o(1). Thus, β ≥ β0. � αP(Ω) is a (PS)-value in X(Ω) for J . Theorem 4.11. αP(Ω) ∈ P(Ω). Proof. For each n ∈ N, take un ∈ X(Ω) and cn ∈ P(Ω) such that |cn − αP(Ω)| < 1 n , |J(un)− cn| < 1 n , ‖J ′(un)‖X−1 < 1 n . Then J(un) = αP(Ω) + o(1) and J ′(un) = o(1). Thus, αP(Ω) ∈ P(Ω). � The following theorem is very useful. Theorem 4.12. Let β be a positive (PS)-value in X(Ω) for J . Then (i) β ≥ αγ(Ω); (ii) β ≥ αM(Ω); (iii) β ≥ αΓ(Ω); (iv) β ≥ αP(Ω). Proof. Let {un} be a nonzero (PS)β-sequence in X(Ω) for J with β > 0. By Lemma 2.38, we have J(un) = β + o(1), a(un)− b(un) = o(1). (i) Let wn = un(a(un))− 1 2 , then a(wn) = 1 and b(wn) = a(un)−p/2b(un) ≤ γ(Ω)p. Thus, a(un) ≥ γ(Ω)2p/(2−p) + o(1), or β ≥ ( 1 2 − 1 p )γ(Ω)2p/(2−p) = αγ(Ω). (ii) By Theorem 4.3, there is a sequence {sn} in R+ such that {snun} ⊂ M(Ω) and J(snun) = β + o(1). Therefore, β ≥ αM(Ω). (iii) By Theorem 4.3 and Lemma 4.2 (ii), there are sequences {sn} and {tn} in R+ such that {snun} ⊂ M(Ω), {tnun} ⊂ Z(Ω), and J(snun) = β + o(1). Since the EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 41 manifold Z(Ω) is path-connected, there is a path ζn in Z(Ω) that connects tnun to e. Let γ′n be the line segment connecting 0 and tnun and the path γn = γ′n ∪ ζn, then αΓ(Ω) ≤ max 0≤t≤1 J(γn(t)) = J(snun) = β + o(1). Thus, β ≥ αΓ(Ω). (iv) Clearly, β ≥ αP(Ω). � By Theorems 4.1, 4.4, 4.9, 4.11, and 4.12, we have the following theorem. Theorem 4.13. αγ(Ω) = αM(Ω) = αΓ(Ω) = αP(Ω). Definition 4.14. By Theorem 4.13, we conclude that the positive (PS)-values αγ(Ω), αΓ(Ω), αM(Ω), and αP(Ω) in X(Ω) for J are the same. Any one of them is called the index of J in X(Ω) and denoted by αX(Ω). By the definition of αM(Ω), if u is a nonzero solution of Equation (1.1), then u ∈ M(Ω). Thus, J(u) ≥ αM(Ω) = αX(Ω). We say that a nonzero solution u of Equation (1.1) in X(Ω) is a ground state solution if J(u) = αX(Ω), and is a higher energy solution if J(u) > αX(Ω). Remark 4.15. We denote αX(Ω) by α(Ω) for X(Ω) = H1 0 (Ω) and by αs(Ω) for X(Ω) = H1 s (Ω) (see Definition 6.1). Remark 4.16. By Theorem 8.2, a ground state solution in X(Ω) is of constant sign. Note that if u is a solution of (1.1), then −u is also a solution of (1.1). By the maximum principle, if u is a nonzero and nonnegative solution of (1.1), then u is positive. From now on, by a ground state solution in X(Ω), we mean a positive solution of (1.1). Definition 4.17. We say that a domain Ω in RN is an achieved domain if there is a solution u in H1 0 (Ω) of (1.1) such that J(u) = α(Ω), by Remark 4.16, we may assume that u be positive. Otherwise, we say that Ω is a nonachieved domain. Theorem 4.18. (i) If Ω is a large domain in RN , then α(Ω) = α(RN ); (ii) If Ω is a large domain in Ar, then α(Ω) = α(Ar); (iii) If Ω is a large domain in Ar1,r2 , then α(Ω) = α(Ar1,r2). Proof. It suffices to prove part (i). Let w ∈ H1(RN ) be a ground state solution of Equation (1.1) satisfying a(w) = ∫ RN (|∇w|2 + w2) = b(w) = ∫ RN |w|p = ( 2p p− 2 )α(RN ). For rn → ∞, take {zn} ⊂ Ω such that BN (zn; rn) ⊂ Ω. Consider the cut-off function η ∈ C∞c ([0,∞)) as in (2.2), and for each n, let wn(z) = η( 2|z − zn| rn )w(z − zn). Then wn ∈ H1 0 (Ω) and a(wn) = ∫ Ω (|∇wn|2 + w2 n) = ( 2p p− 2 )α(RN ) + o(1), b(wn) = ∫ Ω |wn|p = ( 2p p− 2 )α(RN ) + o(1) as n→∞. Thus, J(wn) = α(RN ) + o(1), 42 HWAI-CHIUAN WANG EJDE-2004/MON. 06 a(wn) = b(wn) + o(1) as n→∞. By Theorem 4.4, {wn} is a (PS)α(RN )-sequence in H1 0 (Ω) for J . Therefore, α(Ω) ≤ α(RN ). Clearly, α(RN ) ≤ α(Ω), thus we have α(Ω) = α(RN ). � Theorem 4.19. Let Ω be a large domain in RN . If β is a positive (PS)-value in H1 0 (Ω) for J , then mβ is also a positive (PS)-value in H1 0 (Ω) for J , where m = 2, 3, . . . . Proof. It suffices to prove the case m = 2. First embed H1 0 (Ω) into H1(RN ). Let {un} be a (PS)β-sequence in H1 0 (Ω). Then by Lemma 2.38, there is a constant c > 0 such that, for each n, a(un) ≤ c and b(un) ≤ c. For rn → ∞, since Ω\BN (0; 5rn) is also a large domain in RN , zn ∈ Ω exists such that BN (zn; 2rn) ⊂ Ω and∫ BN (0;rn)c |∇un|2 + u2 n < 1 n and ∫ BN (0;rn)c |un|p < 1 n . Note that |zn| ≥ 5rn. Let ηn(z) = η( |z|rn ), where η is as in (2.2), vn(z) = ηn(z)un(z) and wn(z) = vn(z − zn). Then we have |∇ηn| ≤ 2 rn and suppwn ⊂ BN (zn; 2rn). (i) J(vn) = β + o(1): note that |∇vn|2 = |ηn|2|∇un|2 + |∇ηn|2|un|2 + 2ηnun∇ηn∇un. Thus, for z ∈ BN (0; rn), we have |∇vn| = |∇un| and∫ Ω |∇vn|2 = ∫ BN (0;rn) |∇vn|2 + ∫ BN (0;2rn)\BN (0;rn) |∇vn|2 = ∫ BN (0;rn) |∇un|2 + o(1) = ∫ Ω |∇un|2 + o(1). Similarly, we have∫ Ω |vn|2 = ∫ Ω |un|2 + o(1), ∫ Ω |vn|p = ∫ Ω |un|p + o(1). Thus, J(vn) = J(un) + o(1) = β + o(1). Clearly, for each n, J(wn) = J(vn), and hence J(wn) = β + o(1). (ii) J(vn +wn) = 2β + o(1) : since the supports of vn and wn are disjoint, we have a(vn + wn) = ∫ Ω |∇(vn + wn)|2 + (vn + wn)2 = ∫ Ω |∇vn|2 + v2 n + ∫ Ω |∇wn|2 + w2 n + 2 ∫ Ω ∇vn∇wn + 2 ∫ Ω vnwn = a(vn) + a(wn). Now,∫ Ω |vn + wn|p − |vn|p − |wn|p = ∫ BN (0;2rn) |vn + wn|p − |vn|p − |wn|p + ∫ BN (0;2rn)c∩Ω |vn + wn|p − |vn|p − |wn|p = 0. EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 43 Thus, b(vn + wn) = ∫ Ω |vn + wn|p = ∫ Ω |vn|p + ∫ Ω |wn|p = b(vn) + b(wn). Hence, J(vn + wn) = 1 2 a(vn + wn)− 1 p b(vn + wn) = J(vn) + J(wn) = 2β + o(1). (iii) ‖J ′(vn + wn)‖ = o(1) : for ϕ ∈ C∞c (Ω), we have 〈J ′(vn), ϕ〉 = ∫ BN (0;rn) un(∇ηn) · ∇ϕ+ ∫ BN (0;rn) ηn(∇un) · ∇ϕ+ ηnunϕ − ∫ BN (0;rn) |ηnun|p−2ηnunϕ+ o(1) = ∫ BN (0;rn) ∇un(z) · ∇ϕ(z) + un(z)ϕ(z) − ∫ BN (0;rn) |un|p−2unϕ(z) + o(1) = 〈J ′(un), ϕ〉+ o(1). Thus, ‖J ′(vn)‖H−1 = o(1). Similarly, ‖J ′(wn)‖H−1 = o(1). We have∫ Ω |vn + wn|p−2(vn + wn)ϕ− |vn|p−2vnϕ− |wn|p−2wnϕ = ∫ BN (0;2rn) |vn + wn|p−2(vn + wn)ϕ− |vn|p−2vnϕ− |wn|p−2wnϕ + ∫ BN (0;2rn)c∩Ω |vn + wn|p−2(vn + wn)ϕ− |vn|p−2vnϕ− |wn|p−2wnϕ = 0. Now for ϕ ∈ C∞c (Ω), we have 〈J ′(vn + wn), ϕ〉 = ∫ Ω ∇(vn + wn)∇ϕ+ (vn + wn)ϕ − ∫ Ω |vn + wn|p−2(vn + wn)ϕ = ∫ Ω ∇vn∇ϕ+ vnϕ+ ∫ Ω ∇wn∇ϕ+ wnϕ − ∫ Ω |vn|p−2vnϕ− ∫ Ω |wn|p−2wnϕ = 〈J ′(vn), ϕ〉+ 〈J ′(wn), ϕ〉. Therefore, ‖J ′(vn + wn)‖H−1 = o(1). We conclude that J(vn + wn) = 2β + o(1), J ′(vn + wn) = o(1) strongly in H−1(Ω). � The following theorem has a proof similar to that of Theorem 4.19. 44 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Theorem 4.20. Let Ω be a large domain in Ar. If β is a positive (PS)-value in H1 0 (Ω) for J , then mβ is also a positive (PS)-value in H1 0 (Ω) for J , where m = 2, 3, . . . . Lemma 4.21. The set P(Ω) is closed. The proof of this lemma is similar to the proof of Theorem 4.11; so we omit it. By Lemma 4.2, J(M(Ω)) is bounded below away from zero. Actually for any domain Ω in RN , J(M(Ω)) is unbounded above. Theorem 4.22. If Ω is a domain in RN , then J(M(Ω)) = (α(Ω),∞) for a nonachieved domain Ω and J(M(Ω)) = [α(Ω),∞) for an achieved domain Ω. Proof. (i) Suppose that Ω is bounded. By Struwe [66, p.116 Theorem 6.6], an unbounded sequence {un} exists in M(Ω) for J . Since J(un) = ( 1 2 − 1 p )a(un) and M(Ω) is path connected, then we have J(M(Ω)) = [α(Ω),∞). (ii) Let Ω be an unbounded domain and Ω1 be a bounded domain in Ω. Then M(Ω1) ⊂ M(Ω) and α(Ω) ≤α(Ω1). By part (i), we have [α(Ω1),∞) = J(M(Ω1)) ⊂ J(M(Ω)). Since M(Ω) is path connected, the result follows. � Theorem 4.23. Let Ω be an Esteban-Lions domain as well as a large domain in RN . Then we have P(Ω) = {α(Ω), 2α(Ω), 3α(Ω), . . . }. Proof. By Theorem 4.19, P(Ω) ⊃ {α(Ω), 2α(Ω), 3α(Ω), . . . }. Suppose that a (PS)β-sequence {un} exists for J , where kα(Ω) < β < (k + 1)α(Ω) for some k ∈ N. By the Palais-Smale decomposition theorem 3.1, Equation (1.1) has a nonzero solution. This contradicts Theorem 10.7. � By Lemma 2.38, if {un} is a (PS)β−sequence in H1 0 (Ω) for J , then a(un) = b(un) + o(1) = 2p p−2β + o(1). By Theorems 4.22 and 4.23, we have: Lemma 4.24. Let Ω be an Esteban-Lions domain as well as a large domain in RN . For each β and m = 0, 1, . . . , satisfying mα(Ω) < β < (m + 1)α(Ω), then there is a sequence {un} in H1 0 (Ω) for J satisfying a(un) = b(un) + o(1) = 2p p− 2 β + o(1) but J ′(un) 9 0 strongly in H−1(Ω). Let Ω be an unbounded domain in RN and Ωn = Ω ∩BN (0; rn), then we have the following theorem. Theorem 4.25. αX(Ωn) = αX(Ω) + o(1) as n→∞. Proof. Suppose that {un} in X(Ω) is a minimizing sequence in M(Ω) of αX(Ω), then by Lemma 2.38, {un} is bounded in X(Ω). Let {rn} be a sequence of strictly increasing positive integers such that rn ≥ n,∫ Ω∩{|z|≥ rn 2 } |∇un|2 + |un|2 < 1 n (4.14) and ∫ Ω∩{|z|≥ rn 2 } |un|p < 1 n . (4.15) EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 45 Define ηn(z) = η( 2|z| rn ), where η is as in (2.2). Then ηnun ∈ X(Ωn) ⊂ X(Ω). From the inequalities (4.14) and (4.15), we obtain a(ηnun) = a(un) + o(1) and b(ηnun) = b(un) + o(1). Therefore, we have J(ηnun) = J(un) + o(1) = αX(Ω) + o(1). and a(ηnun) = b(ηnun) + o(1). By Theorem 4.3, there is a sequence {sn} in R+ such that sn = 1+ o(1), {snηnun} is in M(Ω) and J(snηnun) = αX(Ω) + o(1). Note that J(snηnun) ≥ αX(Ωn) > αX(Ω). Hence, αX(Ωn) = αX(Ω) + o(1). � Let Ω be a domain containing zero in RN . For δ > 0, we define δΩ = {δz | z ∈ Ω}. Then we have the following theorem. Theorem 4.26. (i) limδ→∞ α(δΩ) = α(RN ); (ii) Let δΩ be achieved for each δ > 0, then limδ→0+α(δΩ) = ∞. Proof. (i) By Theorem 12.5 below, there is a ground state solution u in H1(RN ) such that a(u) = b(u) and J(u) = α(RN ). A sequence {δn} exists such that δn →∞ and BN (0;n) ⊂ δnΩ. Consider the cut-off function η and ηn(z) = η( 2|z| n ) for n = 1, 2, . . . . Let un(z) = ηn(z)u(z), then un(z) ∈ H1 0 (BN (0;n)), J(un) = α(RN ) + o(1), and a(un) = b(un) + o(1) as n → ∞. By Theorem 4.3, there is a sequence {sn} in R+ such that sn = 1 + o(1), {snun} is in M(BN (0;n)) and J(snun) = α(RN ) + o(1). Then we have lim n→∞ α(δnΩ) ≤ lim n→∞ α(BN (0;n)) ≤ lim n→∞ J(snun) = α(RN ) + o(1). However, α(RN ) ≤ α(δΩ) for each δ > 0. Thus, limδ→∞ α(δΩ) = α(RN ). (ii) Let u be a ground state solution of Equation (1.1) in H1 0 (δΩ), then a(u) = b(u) and J(u) = α(δΩ). Set v(z) = u(δz), then v ∈ H1 0 (Ω). Note that a(u) = ∫ δΩ (|∇u(z)|2 + u(z)2)dz = δN−2 ∫ Ω |∇v|2 + δN ∫ Ω v2, and (b(u))2/p = ( ∫ δΩ |u(z)|p)2/p = δ 2N p ( ∫ Ω |v|p) 2 p . Therefore, by the Sobolev continuous embedding theorem, a(u) ≥ δN−2 ∫ Ω |∇v|2 ≥ cδN−2( ∫ Ω |v|p) 2 p ≥ cδN−2− 2N p (b(u))2/p. That is, (a(u)) p−2 p ≥ cδN−2− 2N p . Thus, J(u) = ( 1 2 − 1 p )a(u) ≥ c( 1 2 − 1 p )(δN−2− 2N p )p/(p−2). Therefore, α(δΩ) ≥ c( 1 2 − 1 p )(δN−2− 2N p )p/(p−2). Since p < 2N N−2 , we have N − 2 − 2N p < 0. We conclude that limδ→0+ α(δΩ) = ∞. � As a corollary of Theorem 4.26, we have 46 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Theorem 4.27. (i) limr→∞ α(BN (0; r)) = α(RN ); (ii) limr→0+ α(BN (0; r)) = ∞. Using the same argument as for the proof of Theorem 4.26, we obtain the fol- lowing theorem. Theorem 4.28. (i) limt→∞ α(Ar −t,t) = α(Ar); (ii) limt→0+ α(Ar −t,t) = ∞. Let Ω = O × Rl, where O is a bounded domain in Rm, m ≥ 1, n ≥ 1. With the same argument of the proof in Theorem 4.26, we get Theorem 4.29. (i) limδ→∞ α(δΩ) = α(RN ); (ii) limδ→0+ α(δΩ) = ∞. We have the following continuity property. Theorem 4.30. We have lim δ→1 α(δAr) = α(Ar). Proof. (i). lim δ→1+ α(δAr) = α(Ar) : let 1 < δ < 2. By Theorem 5.7 (ii) below, α(2Ar) < α(δAr) < α(Ar). In addition, α(δAr) is increasing as δ is decreasing. Let c ≡ limδ→1+ α(δAr), then c ≤ α(Ar). We claim that c ≥ α(Ar). By Theorem 12.5 below, a ground state solution un in M((1 + 1 n )Ar) exists such that a(un) = b(un) and J(un) = α ( (1 + 1 n ) Ar) for each n ∈ N. Moreover, we have a(un) = b(un) = ( 2p p−2 )c + o(1) and J(un) = α((1 + 1 n )Ar) = c+ o(1). Define vn(z) = un((1 + 1 n )z) ∈ H1 0 (Ar). Since a(un) = (1 + 1 n )N−2 ∫ Ar |∇vn|2dz + (1 + 1 n )N ∫ Ar v2 ndz and b(un) = (1 + 1 n )N ∫ Ar |vn|pdz, then {vn} is bounded in H1 0 (Ar). Thus, a(vn) = b(vn)+ o(1) and J(vn) = c+ o(1). By Theorem 4.3, sn > 0 exists such that snvn ∈ M(Ar), sn = 1 + o(1), and J(snvn) = c+ o(1). Hence, c ≥ α(Ar). We conclude the proof. (ii) lim δ→1− α(δAr) = α(Ar) : by Theorem 12.5, u ∈ M(Ar) satisfies a(u) = b(u) and J(u) = α(Ar). Let vn(x, y) = u((1 + 1 n )x, y) for n ∈ N. Then vn(x, y) ∈ H1 0 ( n n+1A r), a(vn) = a(u)+o(1), and b(vn) = b(u)+o(1). Thus a(vn) = b(vn)+o(1) as n→∞ and J(vn) = 1 2 a(vn)− 1 p b(vn) = J(u) + o(1). By Theorem 4.3, for each n, there is an sn > 0 such that snvn ∈ M( n n+1A r), sn = 1 + o(1) and J(snvn) = J(u) + o(1) as n → ∞. Moreover, J(snvn) ≥ α( n n+1A r) > α(Ar) for each n ∈ N. By the squeeze theorem, lim δ→1− α(δAr) = α(Ar). � Bibliographical notes: The constrained maximization problem αγ(Ω) is a clas- sical problem. Theorem 4.1 is from Lien-Tzeng-Wang [47]. The Nehari minimizing problem αM(Ω) was first studied by Nehari [57]. Theorem 4.3 is from Chen-Wang [26, p.158]. For Theorem 4.4, Stuart [67] proved that there is a (PS)αM(Ω)-sequence. EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 47 However, Chen-Wang [26, Lemma 2.1] proved that every minimizing sequence in M(Ω) for αM(Ω) is a (PS)αM(Ω)-sequence. The mountain pass minimax problem αΓ(Ω) is originally from the mountain pass lemmas 4.8 and 4.9 by Ambrosetti- Rabinowitz [4] and the new version is from Brézis-Nirenberg [15]. Theorem 4.13 is due to Willem [78] and Wang [71, p.4241]. Theorem 2.7 is from Lien-Tzeng-Wang [47, Lemma 2.5]. 5. Palais-Smale Conditions The Palais-Smale conditions are conditions for compactness. They are useful in ascertaining the existence of solutions of (1.1). In this section, we assert that eight related (PS)-conditions in X(Ω) for J are actually equivalent. Theorem 5.1. The (PS)αX(Ω)-condition for J holds in a bounded domain Ω. In particular, there is a ground state solution of (1.1) in a bounded domain Ω. Proof. Let {un} be a (PS)αX(Ω)-sequence in X(Ω) for J , by Lemma 2.38, {un} is bounded and J(un) = αX(Ω) + o(1), a(un) = b(un) + o(1). Take a subsequence {un} and u ∈ X(Ω) such that un ⇀ u weakly in X(Ω). By the compactness theorem, un → u strongly in Lp(Ω). Suppose u = 0, then b(un) = o(1). Thus, a(un) = o(1) and J(un) = o(1), contradicting that α(Ω) > 0. By Theorem 5.6, u is a ground state solution in X(Ω) for J and un → u strongly in X(Ω). � The (PS)αX(Ω)-condition holds in unbounded domains and quasibounded do- mains. Definition 5.2. A domain Ω is quasibounded if lim z∈Ω,|z|→∞ dist(z, ∂Ω) = 0. Example 5.3. (i) Let f, g : RN−1 → R be two functions of C1, f ≤ g, lim |x|→∞ (g(x)− f(x)) = 0, and Ω = {z = (x, y) ∈ RN−1 × R : f(x) < y < g(x)}. Then Ω is a quasibounded domain; (ii) Let A be the domain in RN−1 × R with a hypersurface boundary. For each t ∈ R, let At = {(x, y) ∈ A : y = t} be the section of A at t. If limt→∞ diamAt = 0, then A is a quasibounded domain. Theorem 5.4. (i) Let Ω be a C1 quasibounded domain, then the embedding H1 0 (Ω) → Lq(Ω) is compact, where 2 < q < 2∗; (ii) The (PS)αX(Ω)-condition holds for J in a C1 quasibounded domain Ω, and there is a ground state solution of (1.1) in a C1 quasibounded domain. Proof. (i) By Adams [2]. (ii) Similar to the proof of Theorem 5.1. � Theorem 5.5. Let m ≥ 1, k ≥ 2, ω be a smooth bounded open set in Rm, and let E = ω × Rk. Denote by (x, y) a generic point in Rm × Rk and consider the space H1 s (E) consisting of functions in H1 0 (E) that are spherically symmetric in y. Then the Sobolev embedding from H1 s (E) into Lq(E) is compact for every q ∈ (2, 2N N−2 ) with N = m+ k. 48 HWAI-CHIUAN WANG EJDE-2004/MON. 06 The proof of this theorem can be found in Wang [72] and Lions [50]. As a consequence of Lemma 2.38, for each (PS)β-sequence {un} in X(Ω) for J , there is a subsequence {un} and a u in X(Ω) such that un ⇀ u weakly in X(Ω). Then u is a solution of Equation (1.1). Theorem 5.6. (i) Let {un} be a (PS)αX(Ω)-sequence in X(Ω) for J and u in X(Ω) satisfying un ⇀ u weakly in X(Ω). Then u is a solution of Equation (1.1); (ii) Let {un} be a (PS)αX(Ω)-sequence in X(Ω) for J and u in X(Ω) such that un ⇀ u weakly in X(Ω) and u is nonzero. Then u is a ground state solution of Equation (1.1) and un → u strongly in X(Ω); (iii) The (PS)αX(Ω)-condition holds for J if and only if for each (PS)αX(Ω)− se- quence {un} in X(Ω) for J , there is a subsequence {un}and a nonzero u in X(Ω) such that un ⇀ u weakly in X(Ω). Proof. (i) By Lemmas 2.38, 2.11, and Theorem 2.33, there is a subsequence {un} such that un ⇀ u weakly in X(Ω), a.e. in Ω, and strongly in Lq `oc(Ω) where 1 ≤ q < 2∗. By Lemma 2.11, we obtain, for φ ∈ X(Ω) ∩ C∞c (Ω), suppφ = K,∫ Ω ∇un · ∇φ→ ∫ Ω ∇u · ∇φ,∫ Ω unφ→ ∫ Ω uφ. Let gn = |un|p−1φ and g = |u|p−1φ . Then ||un|p−2unφ| ≤ gn for each n, gn → g a.e. By the Rellich-Kondrakov theorem 2.33, gn → g in L1(K). By Theorem 2.22,∫ Ω |un|p−2unφ→ ∫ Ω |u|p−2uφ. Thus, 〈J ′(un), φ〉 → 〈J ′(u), φ〉 for each φ ∈ X(Ω)∩C∞c (Ω). Since 〈J ′(un), φ〉 = o(1), for each φ ∈ X(Ω)∩C∞c (Ω), we have J ′(u) = 0 inX−1(Ω). Therefore, u is a solution of Equation (1.1). (ii) By part (i), u is a nonzero solution of Equation (1.1), hence u ∈ M(Ω) and J(un) = 1 2 a(un)− 1 p b(un) = αX(Ω) + o(1), 〈J ′(un), un〉 = a(un)− b(un) = o(1). Thus, a(un) = 2p p− 2 αX(Ω) + o(1). (5.1) Since a is weakly lower semicontinuous, we have αX(Ω) ≤ J(u) = ( 1 2 − 1 p )a(u) ≤ ( 1 2 − 1 p ) lim inf n→∞ a(un) = αX(Ω), or J(u) = αX(Ω). By Lemma 8.2 below, u is of constant sign. Recall that if w is a solution of Equation (1.1), then −w is also a solution of (1.1). By the maximum principle, we may assume that u is positive. Let pn = un− u. By Lemma 2.11 and 2.14, we have J(pn) = J(un)− J(u) + o(1) = o(1). By Lemma 2.15, {pn} is a Palais-Smale sequence for J , thus 〈J ′(pn), pn〉 = o (1). Similar to (5.1), we have a(pn) = 2p p− 2 J(pn) + o(1) = o(1). EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 49 Thus, un → u strongly in X(Ω). (iii) follows by part (ii). � Let Ω1 $ Ω2 and αi X = αX(Ωi) for i = 1, 2, then clearly α2 X ≤ α1 X . If α2 X = α1 X , then we have the following useful results. Theorem 5.7. Let Ω1 $ Ω2 and J : X(Ω2) → R be the energy functional. Suppose that α2 X = α1 X . Then (i) α1 X does not admit any ground state solution; (ii) J does not satisfy the (PS)α1 X -condition; (iii) J does not satisfy the (PS)α2 X -condition. Proof. (i) Suppose that α1 X admits a ground state solution u ∈ M(Ω1) ⊂ M(Ω2) such that J(u) = α1 X . Then we have J(u) = α1 X = α2 X = minv∈M(Ω2) J(v). By Lemma 4.5 and Theorem 5.6 (ii), (iii), u is a ground state solution of (1.1) in Ω2. Thus, u > 0 in Ω2, which contradicts the fact that u ∈ X(Ω1). (ii) By part (i) and Theorem 5.6 (ii), (iii). (iii) Let {un} in X(Ω1) satisfy J(un) → α1 X and J ′(un) → 0 strongly in X−1(Ω1). By Theorem 4.3, {sn} in R+ exists such that sn = 1+o(1), wn = snun ∈ M(Ω1) and J(wn) = α1 X+o(1) and J ′(wn) = o(1) strongly inX−1(Ω1). Since M(Ω1) ⊂ M(Ω2), {wn} ⊂ M(Ω2) and J(wn) = α2 X + o(1). By Theorem 4.4, we have J(wn) = α2 X + o(1), J ′(wn) = o(1) strongly in X−1(Ω2). Suppose that J satisfies the (PS)α2 X -condition. Then there is a subsequence {wn} and a w ∈ X(Ω2) satisfying wn → w strongly in X(Ω2) and J(w) = α2 X . Hence, w 6= 0. By Lemma 8.2 below and the maximum principle, w is a ground state solution of (1.1) in Ω2. Since {wn} ⊂ M(Ω1) and wn → w strongly in X(Ω2), we have w = 0 in (Ω1)c. This contradicts the fact that w is a positive solution of Equation (1.1) in Ω2. Thus, J does not satisfy the (PS)α2 X -condition. � Compare Theorem 5.7 for X(Ω) = H1 0 (Ω) and the following results. Corollary 5.8. Let E be either RN , or Ar, or Ar1,r2 and Ω a proper large domain of E. Then there is no any ground state solution of (1.1) in Ω. The proof of this corollary follows by Theorem 5.7 (ii) and Theorem 4.18. Theorem 5.9. Let X(Ω) = H1 0 (Ω). We have: (i) α(Ar) > α(Θ) for each domain Θ % Ar; (ii) J does not satisfy (PS)α(Ar)-condition. Proof. (i) Since the infinite strip Ar is a periodic domain, by Theorem 12.5 below, there is a ground state solution u0 ∈ M(Ar) such that J(u0) = α(Ar). The result follows from Theorem 5.7. (ii) Let un = u0(x, y + n) for each n. Since Ar is a periodic domain, we have un ∈ H1 0 (Ar) for each n ∈ N, J(un) = α(Ar) and a(un) = b(un). By Theorem 4.4, {un} is a (PS)α(Ar)-sequence for J . Moreover, for ϕ ∈ C∞c (Ar) and K =suppϕ, we have 〈un, ϕ(z)〉H1 = 〈u0(x, y + n), ϕ(z)〉H1 50 HWAI-CHIUAN WANG EJDE-2004/MON. 06 = ∫ Ar ∇u0(x, y + n)∇ϕ(z)dz + ∫ Ar u0(x, y + n)ϕ(z)dz = ∫ K ∇u0(x, y + n)∇ϕ(z)dz + ∫ K u0(x, y + n)ϕ(z)dz = o(1) as n→∞. For ε > 0, φ ∈ H1 0 (Ar), there is ϕ ∈ C∞c (Ar) such that ‖φ− ϕ‖H1 < ε/(‖u‖H1 + 1) and 〈un, φ(z)〉H1 = 〈un, φ(z)− ϕ(z)〉H1 + 〈un, ϕ(z)〉H1 ≤ ‖un‖H1‖φ(z)− ϕ(z)‖H1 + 〈un, ϕ(z)〉H1 < ε as n→∞. Thus, un ⇀ 0 weakly in H1 0 (Ar). Suppose that J satisfies the (PS)α(Ar)-condition, then there is a subsequence {un} such that un → 0 strongly in H1 0 (Ar). This contradicts α(Ar) > 0. Therefore, J does not satisfy the (PS)α(Ar)-condition. � From now on αX(Ω) is simply denoted by αX . Let {un} in X(Ω) be a (PS)αX - sequence for J in Ω. Clearly, {un} is bounded in X(Ω). Then a subsequence {un} and u ∈ X(Ω) exist such that un ⇀ u weakly in X(Ω) a.e. in Ω, and strongly in Lp loc(Ω). Define a∞ = lim R→∞ lim sup n→∞ ∫ Ω∩{R<|z|} (|∇un|2 + u2 n), b∞ = lim R→∞ lim sup n→∞ ∫ Ω∩{R<|z|} |un|p. We have the following results. Lemma 5.10. Let {un} be a (PS)αX -sequence in X(Ω) for J , then a subsequence {un} exists such that (i) lim supn→∞ ∫ Ω |∇un|2 + u2 n = ( ∫ Ω |∇u|2 + u2) + a∞; (ii) lim supn→∞ ∫ Ω |un|p = ∫ Ω |u|p + b∞; (iii) a∞ = b∞ and αX = J∞ + J(u), where J∞ = (p−2 2p )b∞. Proof. Since {un} is a (PS)αX -sequence in X(Ω) for J , by Lemma 2.38, {un} is bounded in X(Ω). A subsequence {un} and u ∈ X(Ω) exist such that un ⇀ u weakly in X(Ω), a.e. in Ω, and strongly in Lp loc(Ω). (i) Let ηR(z) = η( 2|z| R ), then∫ Ω |∇un|2 + u2 n = ∫ Ω∩{|z| 0 o(1) = 〈J ′(un), ηRun〉 = ∫ Ω ηR(|∇un|2 + u2 n) + un∇ηR∇un − ∫ Ω ηR|un|p. Thus, ∫ Ω ηR ( |∇un|2 + u2 n ) + un∇ηR∇un − ∫ Ω ηR|un|p = o(1). Then, for each R > 0, − | ∫ Ω un∇ηR∇un|+ ∫ Ω∩{|z| 0 lim sup n→∞ ∫ Ω ηR(|∇un|2 + u2 n) ≤ c R + ∫ Ω∩{|z| 0 lim sup n→∞ ∫ Ω |un|p = ∫ Ω∩{|z| 0. (5.12) Let vn = ξnun. Then vn ∈ X(Ω). By (5.12), we have b(vn) = ∫ Ω ξp n|un|p = 2p p− 2 αX(Ω) + o(1). (5.13) 54 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Since {ξ2nun} is bounded in X(Ω), we have o(1) = 〈J ′(un), ξ2nun〉 = ∫ Ω (ξ2n|∇un|2+2ξnun∇ξn·∇un+ξ2nu 2 n)− ∫ Ω ξ2n|un|p. (5.14) By |∇ξn(z)| ≤ c n and (5.11), we have∫ Ω ξnun∇ξn · ∇un = o(1). By (5.13) and (5.14) we have a(vn) = ∫ Ω ξ2n(|∇un|2 + u2 n) = b(vn) + o(1) = 2p p− 2 αX(Ω) + o(1). Thus J(vn) = 1 2 a(vn)− 1 p b(vn) = 1 2 2p p− 2 αX(Ω)− 1 p 2p p− 2 αX(Ω) + o(1) = αX(Ω) + o(1). Since J(vn) = αX(Ω) + o(1) and a(vn) = b(vn) + o(1), by Theorem 4.4, {vn} is a (PS)αX(Ω)-sequence in X(Ω) for J . (iii) =⇒ (i) Let {un} be a (PS)αX(Ω)-sequence in X(Ω) for J such that {ξnun} is also a (PS)αX(Ω)-sequence in X(Ω) for J . Let vn = ξnun. Claim: vn ⇀ 0 as n → ∞. For φ ∈ C1 c (Ω) and K = supp φ, K ⊂ Ω is compact and there is an n0 such that vn(z) = 0 in K for all n ≥ n0. We have 〈vn(z), φ(z)〉H1 = ∫ Ω ∇vn(z)∇φ(z)dz + ∫ Ω vn(z)φ(z)dz = 0 for all n ≥ n0. By Lemma 2.11, there is a C > 0 such that ‖vn(z)‖H1 ≤ C. For ε > 0, ϕ ∈ H1 0 (Ω), φ ∈ C1 c (Ω) exists such that ‖ϕ− φ‖H1 < ε/2(C + 1) Moreover, 〈vn(z), ϕ(z)〉H1 = 〈vn(z), ϕ(z)− φ(z)〉H1 + 〈vn(z), φ(z)〉H1 ≤ ‖vn(z)‖H1‖ϕ(z)− φ(z)‖H1 + 〈vn(z), φ(z)〉H1 ≤ C‖ϕ(z)− φ(z)‖H1 < ε for n ≥ n0. This implies vn ⇀ 0 weakly in H1 0 (Ω). Therefore, by Lemma 5.6 J does not satisfy the (PS)αX(Ω)-condition in X(Ω). � For k ≥ 1, i = 1, 2, . . . , k, let Ω be an unbounded domain and let Ωi be a proper domain in Ω such that Ω = ∪k i=1Ωi, Ωi ∩ Ωj is bounded, and at least one of Ωi is unbounded. Let αX = αX(Ω) and αi X = αX(Ωi), then M = {u ∈ X(Ω)\{0} | a(u) = b(u)}, Mi = {u ∈ H1 0 (Ωi)\{0} | a(u) = b(u)} for i = 1, 2, . . . , k. EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 55 Since X(Ωi) ⊂ X(Ω) and Mi ⊂ M, for i = 1, 2, . . . , k, αX ≤ min{α1 X , α 2 X , . . . , α k X}. Let Ω be an unbounded domain in RN and Ω0 $ Ω with the indexes αX = αX(Ω) and α0 X = αX(Ω0). Then we have αX ≤ α0 X . Let Ω̃n = Ω \ BN (0;n); M̃n = {u ∈ H1 0 (Ω̃n)\{0} | a(u) = b(u)}; α̃n X = α(Ω̃n) = inf u∈M̃n J(u). Theorem 5.12. The following properties are equivalent: (i) J satisfies the (PS)αX -condition; (ii) For every (PS)αX -sequence {un} in X(Ω) for J , there are a subsequence {un} and u 6= 0 in X(Ω) such that un → u strongly in X(Ω); (iii) For every (PS)αX -sequence {un} in X(Ω) for J , there are c > 0, a subsequence {un}, and positive integers K and n0 such that for each n ≥ n0, we have∫ Ω∩{|z| 0, there is a measurable set E such that |E| < ∞ and∫ Ec |un|pdz < ε for each n ∈ N; (v) αX < α̃n X for each n ∈ N; (vi) αX < min{α1 X , α 2 X , . . . , α k X}; (vii) J∞ < αX ; (viii) αX < α0 X for each proper subdomain Ω0 of Ω. Proof. (i) =⇒ (ii) Suppose that J satisfies the (PS)αX −condition. Let {un} be a (PS)αX -sequence in X(Ω) for J . Then there are a subsequence {un} and a u in X(Ω) such that un → u strongly in X(Ω). We conclude that J(u) = αX > 0. Thus, u 6= 0. (ii) =⇒ (iii) Suppose that {un} is a (PS)αX -sequence in X(Ω) for J that has a sub- sequence {un} and u 6= 0 in X(Ω) such that un ⇀ u weakly in X(Ω). By Theorem 5.6, limn→∞ ∫ Ω |un|p = ∫ Ω |u|p. Take K > 0 and c > 0 with ∫ Ω∩{|z| 0 exists such that ∫ Ω∩{|z| 0, a subsequence {un}, positive integers K and n0 such that for each n ≥ n0, we have∫ Ω∩{|z| 0, there is a set E such that |E| <∞ and ∫ Ec |un|pdz < ε for each n ∈ N. (iv) =⇒ (v) For every (PS)αX -sequence {un} in X(Ω) for J , there is a subsequence {un} such that for ε > 0, there is a set E such that |E| < ∞ and ∫ Ec |un|pdz < ε 56 HWAI-CHIUAN WANG EJDE-2004/MON. 06 for each n ∈ N. Then {un} is bounded and there is a subsequence {un} and a u in X(Ω) such that un → u a.e. in Ω. By Theorem 2.34, un → u in Lp(Ω). Note that αX + o(1) = J(un) = ( 1 2 − 1 p )b(un) + o(1) = ( 1 2 − 1 p )b(u) + o(1). Thus, u 6= 0. By Theorem 5.6, J satisfies the (PS)αX -condition in Ω. Suppose that α̃n0 X = αX for some n0 ∈ N, by Theorem 5.7 J does not satisfy the (PS)αX -condition in Ω, which is a contradiction. Hence, we have αX < α̃n X for each n. (v) =⇒ (vi) On the contrary, suppose that αX = min{α1 X , α 2 X , . . . , α k X}, say αX = α1 X . Since Ω1 $ Ω, by Theorem 5.7, J does not satisfy the (PS)αX -condition in Ω. By Theorem 5.6, there is a (PS)αX -sequence {un} such that un ⇀ 0 weakly in X(Ω). There is a subsequence {un} and a sequence {Ωn} such that∫ Ωn |un|p = o(1). Let ξn be as in (2.1) and vn = ξnun. By Lemma 2.41, we have J(vn) = αX + o(1). J ′(vn) = o(1) strongly in X−1(Ω). Then by Theorem 4.3, there is a sequence {sn} in R+ such that wn = snvn, {wn} ∈ M̃n, and J(wn) = J(vn)+o(1) = αX +o(1). Note that α̃n X ≤ J(wn) for each n ∈ N. Hence, limn→∞ α̃n X ≤ αX . Since Ω ⊃ Ω̃n ⊃ Ω̃n+1, we have αX ≤ α̃n X ≤ α̃n+1 X for each n ∈ N. Then we can conclude that αX = α̃n X for each n ∈ N, which is a contradiction. (vi) =⇒ (vii) Let {un} be a (PS)αX -sequence in X(Ω). Then a subsequence {un} and a u in X(Ω) exist such that un ⇀ u weakly in X(Ω). By Lemma 5.10, αX = J∞ + J(u). On the contrary, suppose that J∞ = αX , and we have u = 0. Thus, un ⇀ 0 weakly in X(Ω). There are a subsequence {un} and a sequence {Ωn} such that ∫ Ωn |un|p = o(1). Let vn = ξnun. By Lemma 2.41, {vn} is a (PS)αX -sequence in X(Ω). Since Ωi∩Ωj is bounded for i 6= j, n0 > 0, vn = 0 in BN (0;n0) exists for n ≥ n0, where BN (0;n0) ⊃ Ωi ∩ Ωj for i 6= j. Set vn = v1 n + v2 n + · · · + vk n, where vi n ∈ H1 0 (Ωi), and for i = 1, 2, . . . , k, vi n(z) = { vn(z) for z ∈ Ωi; 0 otherwise. As in the proof of Lemma 2.41, we obtain J ′(vi n) = o(1) strongly in X−1(Ω) for i = 1, 2, . . . , k. Assume J(vi n) = ci + o(1) for i = 1, 2, . . . , k. Since J(un) = αX + o(1), we have c1 + c2 + · · · + ck = αX . Since ci are (PS)- values in X(Ω) for J , by Lemma 2.38, they are nonnegative. There is at least one of the ci that is positive, say c1 > 0. By Theorem 4.10, c1 ≥ α1 X , thus αX ≥ c1 ≥ α1 X . This proves αX ≥ min{α1 X , α 2 X , . . . , α k X}. We conclude that αX = min{α1 X , α 2 X , . . . , α k X}. (vii) =⇒ (viii) Let {un} be a (PS)α-sequence in Ω. Then a subsequence {un} and EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 57 a u in X(Ω) exist such that un ⇀ u weakly in X(Ω). By Lemma 5.10, J(u) = αX − J∞. Suppose that J∞ < αX , then u 6= 0. By Theorem 5.6, J satisfies (PS)αX -condition in X(Ω). By Theorem 5.7, αX < α0 X for each proper subdomain Ω0 of Ω. (viii) =⇒ (i) Suppose that J does not satisfy the (PS)αX -condition in Ω. By Lemma 2.41 and Theorem5.7, {un} exists in X(Ω) and is a (PS)αX -sequence for J such that {ξnun} also is a (PS)αX -sequence in X(Ω)for J . Let vn = ξnun, then J(vn) = α+ o(1), J ′(vn) = o(1) strongly in X−1(Ω). By Theorem 4.3, there is a sequence {sn} in R+ such that wn = snvn, {wn} ∈ M(Ω\BN (0; n 2 )) and J(wn) = J(vn) + o(1) = αX + o(1). n0 > 0 exists such that Ω \ BN (0;n0) $ Ω. Let Ω0 = Ω\BN (0;n0). Then wn ∈ M(Ω0) for n ≥ n0. Since M(Ω0) ⊂ M(Ω) and J(wn) = αX + o(1), thus, α0 X = αX , which is a contradiction. � Bibliographical notes: Theorem 5.6 is from Chen-Lee-Wang [24, Lemma 19]. Theorem 5.12 is from Chen-Lin-Wang [25, Theorem 23]. 6. Symmetric Palais-Smale Conditions In this section, we focus on the symmetric Palais-Smale conditions which will be used in Section 13. Definition 6.1. (i) Suppose that (x, y) ∈ Ω if and only if (x,−y) ∈ Ω, then we call Ω a y-symmetric domain ; (ii) Let Ω be a y-symmetric domain and Θ be a y-symmetric bounded domain in RN . If two disjoint subdomains Ω1 and Ω2 of Ω exist such that (x, y) ∈ Ω2 if and only if (x,−y) ∈ Ω1, Ω\Θ̄ = Ω1 ∪ Ω2, then we say that Ω is separated by Θ; (iii) Let Ω be a y-symmetric domain in RN . If a function u : Ω → R satisfies u(x, y) = u(x,−y) for (x, y) ∈ Ω, then we call u a y-symmetric (axially symmetric) function; (iv) Let Ω be a y-symmetric domain in RN and denote the space H1 s (Ω) by the H1-closure of the space {u ∈ C∞0 (Ω) : u is y-symmetric}. Remark 6.2. (i) Note that H1 s (Ω) is a closed linear subspace of H1 0 (Ω). Let H−1 s (Ω) be the dual space of H1 s (Ω); (ii) Let Ω be a y-symmetric domain in Ar and let BN (0; r+1) be a N−ball. Then clearly Ω is separated by BN (0; r + 1). Example 6.3. (i) For each ρ > 0, let Ω = (RN\Kρ)∪Ar. Then Ω is a y-symmetric large domain in RN separated by a bounded domain Ar ρ; (ii) Let Ω = [P++(0, R 2 )]∪BN (0;R)∪ [P−−(0, R 2 )], then Ω is a y-symmetric large domain in RN separated by the bounded domain BN (0;R). Theorem 6.4. (i) αs(BN (0;R)) = α(BN (0;R)); (ii) αs(RN ) = α(RN ); 58 HWAI-CHIUAN WANG EJDE-2004/MON. 06 (iii) αs(Ar −t,t) = α(Ar −t,t); (iv) αs(Ar) = α(Ar). Proof. By Lien-Tzeng-Wang [47] and Theorem 12.3 below, there is a ground state solution of (1.1) in BN (0;R), RN , Ar −t,t, and Ar. By Gidas-Ni-Nirenberg [34] and [35] and Chen-Chen-Wang [23], every positive solution of (1.1) in BN (0;R), RN , Ar −t,t, and Ar is y-symmetric. � The following symmetric results are required to assert our main result. Theorem 6.5. Suppose that Ω is a y-symmetric large domain in RN separated by a y-symmetric bounded domain, then αs(Ω) ≤ 2α(Ω). Proof. First, by Lien-Tzeng-Wang [47] and Gidas-Ni-Nirenberg [35], there is a pos- itive solution u0 of Equation (1.1) with radial symmetry such that J(u0) = α(RN ). Since Ω is a y-symmetric proper large domain in RN , for n = 1, 2, . . . , sequences {zn} and {rn} exist such that BN (zn; rn) ⊂ Ω and rn → ∞ as n → ∞. Let ηn(z) = η( 2|z−zn| rn ) as in (2.2), and un(z) = ηn(z)u0(z − zn). Then un(z) ∈ H1 0 (Ω), and J(un) = J(u0) + o(1) = α(RN ) + o(1) a(un) = b(un) + o(1). By Theorem 4.4 and Theorem 4.18, {un} is a (PS)α(Ω)-sequence in H1 0 (Ω) for J . Moreover, if we let wn = un(x,−y), then wn is also a (PS)α(Ω)-sequence in H1 0 (Ω) for J such that suppwn ∩ suppun = ∅ and {un + wn} ⊂ H1 s (Ω). We have a(un + wn) = ∫ Ω |∇(un + wn)|2 + (un + wn)2 = ∫ Ω |∇un|2 + u2 n + ∫ Ω |∇wn|2 + w2 n + 2 ∫ Ω ∇un∇wn + 2 ∫ Ω unwn = a(un) + a(wn), and b(un + wn) = ∫ Ω |un + wn|p = ∫ Ω |un|p + ∫ Ω |wn|p = b(un) + b(wn). Hence, J(un + wn) = 1 2 a(un + wn)− 1 p b(un + wn) = J(un) + J(wn) = 2α(Ω) + o(1). Moreover, for ϕ ∈ C∞c (Ω) with y-symmetry, we have |〈J ′(un + wn), ϕ〉| = big| ∫ Ω ∇(un + wn)∇ϕ+ (un + wn)ϕ− ∫ Ω |un + wn|p−2(un + wn)ϕ ∣∣ = ∣∣∫ Ω ∇un∇ϕ+ unϕ+ ∫ Ω ∇wn∇ϕ+ wnϕ− ∫ Ω |un|p−2unϕ− ∫ Ω |wn|p−2wnϕ ∣∣ = |〈J ′(un), ϕ〉|+ |〈J ′(wn), ϕ〉| ≤ ‖J ′(un)‖H−1 + ‖J ′(wn)‖H−1 . EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 59 Therefore, ‖J ′(un + wn)‖H−1 s = o(1). We conclude that {un + wn} is a (PS)2α(Ω)- sequence in H1 s (Ω) for J . By Theorems 4.12 and 4.13, αs(Ω) ≤ 2α(Ω). � We have the following symmetric Palais-Smale condition. Theorem 6.6. Suppose that Ω is a y-symmetric large domain in RN separated by a y-symmetric bounded domain. Then αs(Ω) < 2α(Ω) if and only if J satisfies the (PS)αs(Ω)-condition in H1 s (Ω). Proof. Let αs(Ω) < 2α(Ω). Suppose J does not satisfy the (PS)αs(Ω)-condition. By Theorem 5.11, a (PS)αs(Ω)-sequence {un} in H1 s (Ω) for J exists such that {ξnun} is also a (PS)αs(Ω)-sequence in H1 s (Ω) for J , where ξn is as at (2.1). Let wn = ξnun, then by Lemma 2.9, we obtain J(wn) = αs(Ω) + o(1), J ′(wn) = o(1) in H−1(Ω). (6.1) Since Ω is a y-symmetric domain in RN separated by a bounded domain Q, n0 > 0, exists such that wn = 0 in Q for n ≥ n0 and two disjoint subdomains Ω1 and Ω2 of Ω exist such that (x, y) ∈ Ω2 if and only if (x,−y) ∈ Ω1, Ω\Q = Ω1 ∪ Ω2. Note that, for n ≥ n0, wn = w1 n +w2 n and w1 n(x, y) = w2 n(x,−y), where for i = 1, 2, wi n(x) = { wn(x) for x ∈ Ωi, 0 for x /∈ Ωi. Then wi n ∈ H1 0 (Ωi). We obtain J(w1 n) = J(w2 n) and αs(Ω) + o(1) = J(wn) = J(w1 n) + J(w2 n) = 2J(wi n) for i = 1, 2, or J(wi n) = 1 2 αs(Ω) + o(1) for i = 1, 2. By (6.1), we have J ′(wi n) = o(1) in H1 0 (Ωi) for i = 1, 2. Therefore 1 2αs(Ω) is a (PS)-value in H1 0 (Ω) for J . By Theorems 4.12 and 4.13, 1 2 αs(Ω) ≥ α(Ωi). Since Ω and Ωi are large domains of RN , by Theorem 4.18, we have α(Ωi) = α(RN ) = α(Ω). Thus αs(Ω) ≥ 2α(Ω), which is a contradiction. Conversely, suppose that J satisfies the (PS)αs(Ω)-condition in H1 s (Ω). By Theorem 6.5, we have αs(Ω) ≤ 2α(Ω). Suppose that αs(Ω) = 2α(Ω). By the definition of the large domain in RN , we may take a domain Ω̃ = Ω \ BN (0; r̃) for some r̃ > 0 such that Ω̃ $ Ω, and Ω̃ is a proper y-symmetric large domain in RN separated by a y- symmetric bounded domain. By Theorem 5.7, we have 2α(RN ) = 2α(Ω) = αs(Ω) < αs(Ω̃). By Theorem 6.5, αs(Ω̃) ≤ 2α(Ω̃) = 2α(RN ). Thus, 2α(RN ) < 2α(RN ), which is a contradiction. � 60 HWAI-CHIUAN WANG EJDE-2004/MON. 06 As a consequence of Theorem 6.6, we have the following result. Theorem 6.7. If Ω is a y-symmetric large domain in RN separated by a y- symmetric bounded domain, then α(Ω) < αs(Ω). Proof. Since Ω $ RN , we have αs(RN ) ≤ αs(Ω). Assume that αs(RN ) = αs(Ω). Then by Theorem 5.7, J does not satisfy the (PS)αs(Ω)-condition in H1 s (Ω) for J . By Theorem 4.18, α(RN ) = α(Ω), by Theorem 6.4, αs(RN ) = α(RN ), and by Theorem 6.6, 2α(Ω) ≤ αs(Ω). We conclude that 2α(RN ) = 2α(Ω) ≤ αs(Ω) = αs(RN ) = α(RN ), which is a contradiction. � Consider the y-symmetric large domain ΩR in RN separated by a y-symmetric bounded domain, where ΩR = [ P++(0, R 2 ) ] ∪ BN (0;R) ∪ [ P−−(0, R 2 ) ] . Then we have the following existence result. Theorem 6.8. An R0 > 0 exists such that for R ≥ R0, there is a positive y- symmetric solution of (1.1) in ΩR. Proof. By Lien-Tzeng-Wang [47], α(BN (0, R)) is strictly decreasing as R is strictly increasing and α(BN (0, R)) ↘ α(RN ) as R→∞. By Theorem 6.4, α(BN (0, R)) = αs(BN (0, R)) for each R. Thus, there is a R0 > 0 such that αs(ΩR) ≤ αs(BN (0, R)) < 2α(RN ) = 2α(ΩR) for each R ≥ R0. By Theorem 6.5 and Theorem 6.6, there is a y-symmetric positive solution of Equation (1.1) in ΩR for each R ≥ R0. � Bibliographical notes: The results of this section are from Wang-Wu [74]. 7. Symmetric Palais-Smale Decomposition Theorems In this section, we present the symmetric Palais-Smale decomposition theorem in Ar. Lemma 7.1. Let Θ1 ⊂ Θ2 ⊂ Θ3 ⊂ . . . , where ∞ ∪ n=1 Θn = Ar. If fn(z) = { gn(z)− hn(z), for z ∈ Θn, 0, otherwise, fn → f a.e., gn ≥ 0, and hn → 0 a.e., then f ≥ 0. Proof. For z ∈ Ar, we have z ∈ Θm for some m ∈ N, then z ∈ Θm+i for i = 0, 1, 2, . . . . Since gm+i(z) = fm+i(z) + hm+i(z), fm+i → f a.e., hm+i → 0 a.e. for i→∞, hence gm+i → f a.e., and since gm+i ≥ 0, we have f ≥ 0. � Theorem 7.2 (Symmetric Palais-Smale Decomposition Theorem in Ar). Let {un} be a (PS)β-sequence in H1 s (Ar) for J . Then there are a subsequence {un}, a positive integer m, sequences {z̃i,j n }∞n=1 in Ar, a function ū ∈ H1 s (Ar), and 0 6= wi,j ∈ H1 0 (Ar) for 1 ≤ i ≤ m, j = 1, 2 such that wi,1(x, y) = wi,2(x,−y), |z̃i,j n | → ∞ for i = 1, 2, . . . ,m, −∆ū+ ū =| ū |p−2 ū in Ar, EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 61 −∆wi,j + wi,j =| wi,j |p−2wi,j in Ar, and un = ū+ 2∑ j=1 m∑ i=1 wi,j(· − z̃i,j n ) + o(1) strongly in H1 0 (Ar), a(un) = a(ū) + 2 m∑ i=1 a(wi,j) + o(1) for some j = 1, 2, b(un) = b(ū) + 2 m∑ i=1 b(wi,j) + o(1) for some j = 1, 2, J(un) = J(ū) + 2 m∑ i=1 J(wi,j) + o(1) for some j = 1, 2. In addition, if un ≥ 0, then ū ≥ 0 and wi,j ≥ 0 for each 1 ≤ i ≤ m, j = 1, 2. Proof. Step 0. Since {un} is a (PS)β-sequence in H1 s (Ar) for J , by Lemma 2.38 there is a c > 0 such that ‖un‖H1 ≤ c. In the following proof of this theorem, we fix the value of c. There is a subsequence {un} and a ū in H1 s (Ar) such that un ⇀ ū weakly in H1 s (Ar) and ū solves −∆ū+ ū = |ū|p−2ū in Ar. Step 1. Suppose that un 9 ū strongly in H1 s (Ar). Let u1 n = un − ū for n = 1, 2, . . . . By Lemma 2.15, {u1 n} is a (PS)(β−J(ū))-sequence in H1 s (Ar) for J . Let v1 n = ξnu 1 n, where ξn are as in 2.1. Note that u1 n ⇀ 0 weakly in H1 s (Ar) and u1 n 9 0 strongly in H1 s (Ar). By Lemma 2.41, {v1 n} is also a (PS)(β−J(ū))-sequence in H1 s (Ar) for J . Moreover, J(u1 n) = J(v1 n) + o(1), v1 n ⇀ 0 weakly in H1 s (Ar) and v1 n 9 0 strongly in H1 s (Ar). Let K = 2([r] + 1) and QK = Ar ∩BN (0;K). Then v1 n = 0 in QK for n ≥ 2K. Two disjoint strictly large domains Ω1 = Ar K and Ω2 = Ãr −K in Ar exist such that (x, y) ∈ Ω2 if and only if (x,−y) ∈ Ω1, Ω\QK = Ω1 ∪ Ω2. For j = 1, 2, let v1,j n (z) = { v1 n(z) for z ∈ Ωj , 0 for z /∈ Ωj . Then v1,j n ∈ H1 0 (Ωj), v1,1 n (x, y) = v1,2 n (x,−y), v1 n = v1,1 n +v1,2 n , and J(v1,1 n ) = J(v1,2 n ). We claim that {v1,j n } is a (PS) 1 2 (β−J(ū))-sequence in H1 0 (Ωj) for J . In fact, J(v1,j n ) = 1 2 J(v1 n) = 1 2 J(u1 n) + o(1) = 1 2 [β − J(ū)] + o(1), and for ϕ ∈ C∞c (Ωj), ‖ϕ‖H1 = 1, we have |〈J ′(v1,j n ), ϕ〉| = |〈J ′(v1 n), ϕ〉| ≤ ‖J ′(v1 n)‖H−1‖ϕ‖H1 . Thus, ‖J ′(v1,j n )‖H−1 ≤ ‖J ′(v1 n)‖H−1 = o(1). 62 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Note that v1 n ⇀ 0 weakly in H1 0 (Ω) and v1 n 9 0 strongly in H1 0 (Ω), so we have v1,j n ⇀ 0 weakly in H1 0 (Ωj) and v1,j n 9 0 strongly in H1 0 (Ωj). (1-0) ∫ Ar −1,1 |w1,j n (z)|2dz ≥ d1 2 for some constant d1 > 0, n = 1, 2, . . . , and j = 1, 2, where w1,j n (z) = v1,j n (z + z1,j n ) for some {z1,j n } ⊂ Ar : for j = 1, 2, since {v1,j n } is bounded, J ′(v1,j n ) = o(1), and v1,j n 9 0 strongly in H1 0 (Ωj). By Lemma 2.17, there is a subsequence {v1,j n }, and a constant d1 > 0 such that Qr,1 n = sup y∈R ∫ (0,y)+Ar −1,1 |v1,j n (z)|2dz ≥ d1 for n = 1, 2, . . . . For n = 1, 2, . . . , take z1,1 n = (0, y1 n) and z1,2 n = (0,−y1 n) in Ar such that∫ z1,j n +Ar −1,1 |v1,j n (z)|2dz ≥ d1 2 for n = 1, 2, . . . . Let w1,j n (z) = v1,j n (z + z1,j n ) then ∫ Ar −1,1 |w1,j n (z)|2dz ≥ d1 2 for n = 1, 2, . . . . (1-1) un(z) = ū(z) + ∑2 j=1 w 1,j n (z − z1,j n ) + o(1) strongly in H1 0 (Ar). By Lemma 2.42, we have the following equalities in the strong sense in H1 0 (Ar) 2∑ j=1 w1,j n (z − z1,j n ) = 2∑ j=1 v1,j n (z) = v1 n(z) = u1 n(z) + o(1) = un(z)− u(z) + o(1), or un(z) = u(z) + 2∑ j=1 w1,j n (z − z1,j n ) + o(1) strongly in H1 0 (Ar). (1-2) ‖w1,j n ‖H1 ≤ c for n = 1, 2, . . . , and ‖w1,j‖H1 ≤ c, where w1,j n ⇀ w1,j weakly in H1 0 (Ar) for j = 1, 2. By Lemma 2.11(iii), ‖w1,j n ‖2H1 = ‖v1,j n ‖2H1 = 1 2 ‖v1 n‖2H1 = 1 2 ‖u1 n‖2H1 + o(1) = 1 2 (‖un‖2H1 − ‖ū‖2H1) + o(1) ≤ 1 2 c2 + o(1), we have ‖w1,j n ‖H1 0 (Ar) ≤ c for n = 1, 2, . . . . Then there is a subsequence {w1,j n } and w1,j in H1 0 (Ar) such that w1,j n ⇀ w1,j weakly in H1 0 (Ar). In addition, w1,1(x, y) = w1,2(x,−y). By Lemma 2.11 (i), we have ‖w1,j‖H1 ≤ lim inf n→∞ ‖w1,j n ‖H1 ≤ c for j = 1, 2. (1-3) {w1,j n } is a (PS) 1 2 (β−J(ū))-sequence in H1 0 (Ar) for J : note that J ′(v1,j n ) = o(1) in H−1(Ωj). Because Ωj is a half infinite strip, (1 − 7) below and Theorem 2.35, we have for every ϕ ∈ H1 0 (Ar), 〈J ′(w1,j n ), ϕ〉 = ∫ Ar ∇w1,j n ∇ϕ+ w1,j n ϕ− ∫ Ar |w1,j n |p−2w1,j n ϕ = o(1). EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 63 Therefore, J ′(w1,j n ) = o(1) in H−1(Ar). Moreover, we have J(w1,j n ) = J(v1,j n (z + z1,j n )) = 1 2 J(v1 n) = 1 2 (β − J(ū)) + o(1). (1-4) −∆w1,j + w1,j − |w1,j |p−2w1,j = 0 in Ar : by Theorem 5.6 (i). (1-5) w1,j 6≡ 0 : by the Rellich-Kondrakov theorem 2.31 and (1− 0), we have∫ Ar −1,1 |w1,j |2 = lim n→∞ ∫ Ar −1,1 |w1,j n |2 ≥ d1 2 , thus w1,j 6≡ 0. (1-6) By (1-2), (1-4), (1-5), and Lemma 2.18, there is a δ > 0 such that ‖w1,j‖H1 0 (Ar) ≥ ‖w1,j‖L2(Ar) > δ. Therefore, J(w1,j) = ( 1 2 − 1 p )a(w1,j) > ( 1 2 − 1 p )δ2 = δ′. (1-7) |z1,j n | → ∞ : otherwise, there is a R > 0 such that z1,j n + Ar −1,1 ⊂ Ar −R,R for n = 1, 2, . . . . Then by (1-0), we have 0 = lim n→∞ ∫ Ar −R,R |v1,j n |2 ≥ lim n→∞ ∫ z1,j n +Ar −1,1 |v1,j n |2 ≥ d1 2 , which is a contradiction. (1-8) a(un) = a(ū)+2a(w1,j n )+o(1) for j = 1, 2 : since un ⇀ ū weakly in H1 0 (Ar), by Lemma 2.11(iii), a(un)− a(ū) = a(un − ū) + o(1) = a(u1 n) + o(1) = a(v1 n) + o(1) = a(v1,1 n ) + a(v1,2 n ) + o(1) = a(w1,1 n ) + a(w1,2 n ) + o(1) = 2a(w1,j n ) + o(1) for j = 1, 2, thus, a(un) = a(ū) + 2a(w1,j n ) + o(1) for j = 1, 2. (1-9) b(un) = b(ū) + 2b(w1,j n ) + o(1) for j = 1, 2 : since un → ū a.e. in Ω and {un} is bounded in Lp(Ω), by Lemma 2.14(i), we have b(un)− b(ū) = b(un − ū) + o(1) = b(u1 n) + o(1) = b(v1 n) + o(1) = b(v1,1 n ) + b(v1,2 n ) + o(1) = b(w1,1 n ) + b(w1,2 n ) + o(1) = 2b(w1,j n ) + o(1) for j = 1, 2 thus b(un) = b(ū) + 2b(w1,j n ) + o(1) for j = 1, 2. 64 HWAI-CHIUAN WANG EJDE-2004/MON. 06 (1-10) J(un) = J(ū) + 2J(w1,j n ) + o(1) for j = 1, 2 : by (1 − 8), (1 − 9) and J(w1,1) = J(w1,2), we have J(un) = J(ū) + J(w1,1 n ) + J(w1,2 n ) + o(1) = J(ū) + 2J(w1,j n ) + o(1) for j = 1, 2. Step 2. Suppose that w1,j n (z) 9 w1,j(z) strongly in H1 0 (Ar). Let v2,j n (z) = w1,j n (z)− w1,j(z). We have v2,j n ⇀ 0 weakly in H1 0 (Ar) but v2,j n 9 0 strongly in H1 0 (Ar). (2-0) ∫ Ar −1,1 |w2,j n (z)|2dz ≥ d2 2 for some constant d2 > 0, n = 1, 2, . . . , and j = 1, 2, where w2,j n (z) = v2,j n (z+z2,j n ) for some {z2,j n } ⊂ Ar: for j = 1, 2, since {v2,j n } is bounded, J ′(v2,j n ) = o(1), and v2,j n 9 0 strongly in H1 0 (Ar), by Lemma 2.17, there is a subsequence {v2,j n }, a constant d2 > 0 such that Qr,2 n = sup y∈R ∫ (0,y)+Ar −1,1 |v2,j n (z)|2dz ≥ d2 for n = 1, 2, . . . . For n = 1, 2, . . . , take z2,1 n = (0, y2 n) and z2,2 n = (0,−y2 n) in Ar such that∫ z2,j n +Ar −1,1 |v2,j n (z)|2dz ≥ d2 2 for n = 1, 2, . . . . Let w2,j n (z) = v2,j n (z + z2,j n ), then∫ Ar −1,1 |w2,j n (z)|2dz ≥ d2 2 for n = 1, 2, . . . . As in Step 1, we have the following results. (2-1) un(z) = ū(z)+ ∑2 j=1 w 1,j(z−z1,j n )+ ∑2 j=1 w 2,j n (z−z1,j n −z2,j n )+o(1) strongly in H1 0 (Ar) ; (2-2) ‖w2,j n ‖H1 ≤ c for n = 1, 2, . . . and ‖w2,j‖H1 ≤ c, where w2,j n ⇀ w2,j weakly in H1 0 (Ar) for j = 1, 2; (2-3) {w2,j n } is a (PS)-sequence in H1 0 (Ar) for J ; (2-4) −∆w2,j + w2,j − |w2,j |p−2w2,j = 0 in Ar; (2-5) w2,j 6≡ 0; (2-6) ‖w2,j‖L2(Ar) > δ and J(w2,j) > δ′; (2-7) |z2,j n | → ∞; (2-8) a(un) = a(ū) + 2a(w1,j) + 2a(w2,j n ) + o(1) : since v2,j n (z) = w1,j n (z)− w1,j(z) ⇀ 0, we have a(w2,j n ) = a(v2,j n ) = a(w1,j n )− a(w1,j) + o(1). Further, by (1− 8), we have a(un)− a(ū) = a(w1,1 n ) + a(w1,2 n ) + o(1) = 2a(w1,j) + 2a(w2,j n ) + o(1). (2-9) b(un) = b(ū) + 2b(w1,j) + 2b(w2,j n ) + o(1); (2-10) J(un) = J(ū) + 2J(w1,j) + 2J(wj n) + o(1). Continuing this process, we arrive at the m-th step EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 65 (m-0) ∫ Ar −1,1 |wm,j n (z)|2dz ≥ dm 2 for some constant dm > 0, n = 1, 2, . . . and j = 1, 2, where wm,j n (z) = vm,j n (z + zm,j n ) for some {zm,j n } ⊂ Ar; (m-1) un(z) = ū(z)+ ∑2 j=1 m−1∑ i=1 wi,j(z− z̃i,j n )+ ∑2 j=1 w m,j n (z− z̃m,j n )+o(1) strongly in H1 0 (Ar), where z̃i,j n = z1,j n + · · ·+ zi,j n for i = 1, 2, . . . ,m and j = 1, 2; (m-2) ‖wm,j n ‖H1 ≤ c for n = 1, 2, . . . and ‖wm,j‖H1 ≤ c, where wm,j n ⇀ wm,j weakly in H1 0 (Ar); (m-3) {wm,j n } is a (PS)-sequence in H1 0 (Ar) for J ; (m-4) −∆wm,j + wm,j − |wm,j |p−2wm,j = 0 in Ar; (m-5) wm,j 6≡ 0; (m-6) ‖wm,j‖L2(Ar) > δ and J(wm,j) > δ′; (m-7) |zi,j n | = |z̃i,j n − z̃i−1,j n | → ∞ and |z̃i,j n | → ∞, for each i = 1, 2, . . . ,m: we show this by induction on i. For i = 1, |z̃i,j n | = |z1,j n | → ∞. Assume that |z̃i,j n | → ∞, for i = 1, 2, . . . , k, for some k < m. By Lemma 2.12, we have wi,j(z − z̃i,j n ) ⇀ 0 weakly in H1 0 (Ar) for i = 1, 2, . . . , k. We claim that |z̃k+1,j n | → ∞. Otherwise, {z̃k+1,j n } is bounded. Since ‖wk+1,j‖L2(Ar) > δ, R > 0 exists such that z̃k+1,j n +Ar −R,R ⊂ Ar −2R,2R and ∫ Ar −R,R |wk+1,j |2 ≥ ( δ 2 )2. We have ( δ 2 )2 ≤ ∫ Ar −R,R |wk+1,j |2 = lim n→∞ ∫ Ar −R,R |v1,j n (z + z̃k+1,j n )|2 ≤ lim n→∞ ∫ Ar −2R,2R |v1,j n (z)|2 = 0, which is a contradiction. By the induction hypothesis, we have |z̃i,j n | → ∞ for i = 1, 2, . . . ,m. (m-8) a(un) = a(ū) + 2 ∑m−1 i=1 a(wi,j) + 2a(wi,j n ) + o(1); (m-9) b(un) = b(ū) + 2 ∑m−1 i=1 b(wi,j) + 2b(wi,j n ) + o(1); (m-10) J(un) = J(ū) + 2 ∑m−1 i=1 J(wi,j) + 2J(wi,j n ) + o(1). By the Archimedean principle, k ∈ N exists such that kδ2 > β. Take l = [k 2 ]+ 1, then after step (l + 1), we obtain a(un) = a(ū) + 2a(w1,j) + 2a(w2,j) + · · ·+ 2a(wl,j) + 2a(wl+1,j n ) + o(1). Since a(wl+1,j n ) ≥ 0, a(ū) > 0, and a(wi,j) > δ2 for i = 1, 2, . . . , l, we have β+o(1) ≥ 2lδ2 > kδ2 > β, which is a contradiction. Therefore, there is an m ∈ N, such that wm,j n (z) = wm,j(z) + o(1) strongly in H1 0 (Ar), wi,j n (z) = wi,j(z) + o(1) weakly in H1 0 (Ar), and wi,j n (z) 6= wi,j(z) + o(1) strongly in H1 0 (Ar) for i = 1, 2, . . . ,m − 1. Then we have (sm-0) ∫ Ar −1,1 |wm,j n (z)|2dz ≥ dm 2 for some constant dm > 0, n = 1, 2, . . . , and j = 1, 2, where wm,j n (z) = vm,j n (z + zm,j n ) for some {zm,j n } ⊂ Ar; 66 HWAI-CHIUAN WANG EJDE-2004/MON. 06 (sm-1) un(z) = ū(z) + ∑2 j=1 m∑ i=1 wi,j(z − z̃i,j n ) + o(1) strongly in H1 0 (Ar), where z̃i,j n = z1,j n + · · ·+ zi,j n , for i = 1, 2, . . . ,m and j = 1, 2; (sm-2) ‖wm,j n ‖H1 ≤ c for n = 1, 2, . . . and ‖wm,j‖H1 ≤ c, where wm,j n ⇀ wm,j weakly in H1 0 (Ar); (sm-3) {wm,j n } is a (PS)-sequence in H1 0 (Ar) for J ; (sm-4) −∆wm,j + wm,j − |wm,j |p−2wm,j = 0 in Ar; (sm-5) wm,j 6≡ 0; (sm-6) ‖wm,j‖L2(Ar) > δ and J(wm,j) > δ′; (sm-7) |zi,j n | = |z̃i,j n − z̃i−1,j n | → ∞ and |z̃i,j n | → ∞, for each i = 1, 2, . . . ,m; (sm-8) a(un) = a(ū) + 2 m∑ i=1 a(wi,j) + o(1); (sm-9) b(un) = b(ū) + 2 m∑ i=1 b(wi,j) + o(1); (sm-10) J(un) = J(ū) + 2 m∑ i=1 J(wi,j) + o(1). Finally, suppose un ≥ 0 for n = 1, 2, . . . . Then (i) Since un ⇀ ū weakly in H1 0 (Ω). By Lemma 2.11(ii), there is a subsequence {un} such that un → ū a.e. in Ω. Thus, ū ≥ 0. (ii) For j = 1, 2, let z ∈ Ωj − z1,j n , then w1,j n (z) = v1,j n (z + z1,j n ) = v1 n(z + z1,j n ) = ξn(z + z1,j n )u1 n(z + z1,j n ). Thus, w1,j n (z) = { ξn(z + z1,j n )u1 n(z + z1,j n ) if z ∈ Ωj − z1,j n 0 otherwise. Let hn(z) = ξn(z + z1,j n )ū(z + z1,j n ) and gn(z) = ξn(z + z1,j n )un(z + z1,j n ) ≥ 0. Since w1,j n (z) ⇀ w1,j(z) weakly in H1 0 (Ar) and hn(z) ⇀ 0 weakly in H1 0 (Ar), we have w1,j n (z) → w1,j(z) a.e. in Ar and hn(z) → 0 a.e. in Ar. By Lemma 7.1, we have w1,j ≥ 0. (iii) In fact, we have w2,j n (z) = v2,j n (z + z2,j n ) = w1,j n (z + z2,j n )− w1,j(z + z2,j n ). By Lemma 2.12, w1,j(z + z2,j n ) ⇀ 0 weakly in H1 0 (Ar). Moreover, w2,j n (z) ⇀ w2,j(z) weakly in H1 0 (Ar), hence w1,j n (z + z2,j n ) ⇀ w2,j(z) weakly in H1 0 (Ar), since w1,j n (z + z2,j n ) = { ξn(z + z̃2,j n )u1 n(z + z̃2,j n ) if z ∈ Ωj − z1,j n , 0 otherwise. Similar to (ii), w2,j ≥ 0. (iv) Continuing this process, we obtain wi,j ≥ 0 for each i = 1, 2, . . . ,m. � Similarly, we have Theorem 7.3 (Symmetric Palais-Smale Decomposition Theorem in RN ). Let Ω be a y-symmetric large domain in RN separated by a y-symmetric bounded domain, and let {un} ⊂ H1 s (Ω) be a (PS)β-sequence in H1 0 (Ω) for J . Then there are a subsequence {un}, a positive integer m, sequences {z̃i,j n }∞n=1 in RN , a function ū ∈ H1 s (Ω), and 0 6= wi,j ∈ H1(RN ) for 1 ≤ i ≤ m, j = 1, 2 such that wi,1(x, y) = wi,2(x,−y), |z̃i,j n | → ∞ for i = 1, 2, . . . ,m, EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 67 −∆ū+ ū =| ū |p−2 ū in Ω, −∆wi,j + wi,j =| wi,j |p−2wi,j in RN , and un = ū+ m∑ i=1 wi,1(· − z̃i,1 n ) + m∑ i=1 wi,2(· − z̃i,2 n ) + o(1) strongly in H1(RN ), a(un) = a(ū) + 2 m∑ i=1 a(wi,j) + o(1) for some j = 1, 2, b(un) = b(ū) + 2 m∑ i=1 b(wi,j) + o(1) for some j = 1, 2, J(un) = J(ū) + 2 m∑ i=1 J(wi,j) + o(1) for some j = 1, 2. In addition, if un ≥ 0, then ū ≥ 0 and wi,j ≥ 0 for each 1 ≤ i ≤ m, j = 1, 2. Corollary 7.4. If Ω = Ar\ω, where ω is an axially symmetric bounded set in Ar and {un} is a (PS)β-sequence in H1 s (Ω) for J and 0 < β < 2α(Ar), then the sequence {un} contains a strongly convergent subsequence and there is a positive solution u of (1.1) in Ω. Proof. By Theorem 7.2, we have J(un) = J(ū) + 2 m∑ i=1 J(wi,j) + o(1) for some j = 1, 2. Note that J(wi,j) ≥ α(Ar) and J(ū) ≥ 0. If m ≥ 1, then we have 2α(Ar) > β + o(1) = J(un) ≥ J(ū) + 2mα(Ar), which is a contradiction. Thus, m = 0. By Theorem 7.2, we have un = ū+ o(1) strongly in H1 0 (Ar). Since J(ū) = β and β > 0, we have ū 6= 0. � Corollary 7.5. If Ω is a y-symmetric strictly large domain in RN separated by a y-symmetric bounded domain, and {un} is a positive (PS)β-sequence in H1 s (Ω) for J , with 0 < β < 3α(RN ) but β 6= 2α(RN ), then {un} contains a strongly convergent subsequence, and there is a positive solution ū of Equation (1.1) in Ω. In particular, if {un} is a (PS)β-sequence in H1 s (Ω) for J with 0 < β < 2α(RN ), then {un} contains a strongly convergent subsequence. Proof. By Theorem 7.3, we have J(un) = J(ū) + 2 m∑ i=1 J(wi,j) + o(1), for some j = 1, 2. By the uniqueness of the positive solution for Equation (1.1) in RN , we have J(wi,j) = α(RN ) and J(ū) ≥ 0. If m ≥ 1, then 3α(RN ) > β + o(1) = J(un) = J(ū) + 2mα(RN ), thus, m = 0, 1. Suppose that m = 1, then 2α(RN ) 6= β + o(1) = J(un) = J(ū) + 2α(RN ), 68 HWAI-CHIUAN WANG EJDE-2004/MON. 06 which implies that J(ū) 6= 0 or J(ū) > 0, or J(ū) ≥ α(RN ). Therefore, 3α(RN ) > β + o(1) = J(un) ≥ J(ū) + 2mα(RN ) ≥ 3mα(RN ), which is a contradiction. Thus, m = 0. By Theorem 7.3 again, we have un = ū+ o(1) strongly in H1 0 (RN ). Since J(ū) = β and β > 0, we have ū 6= 0. � Bibliographical notes: The results of this section are from Wang-Wu [77]. 8. Fundamental Properties, Regularity, and Asymptotic Behavior of Solutions In this section we study the fundamental properties, regularity, and asymptotic behavior of solutions of (1.1). 8.1. Fundamental Properties of Solutions. Theorem 8.1. Let u ∈ H1 0 (Ω) be a positive symmetric and radially decreasing solution of (1.1). Then u(0) ≥ 1. Proof. Since u is symmetric and radially decreasing, we have −∆u(0) ≥ 0. Thus, u(0) ≤ −∆u(0) + u(0) = up−1(0), or u(0) ≥ 1. � A ground state solution in X(Ω) is of constant sign. Lemma 8.2. Let u in X(Ω) be a solution of (1.1) that changes sign , and let αX(Ω) be the index of J in Ω. Then J(u) > 2αX(Ω). Proof. Let u− = max{−u, 0}. Then u− is nonzero. Multiply (1.1) by u− and integrate to obtain ∫ Ω ∇u∇u− + ∫ Ω uu− = ∫ Ω |u|p−2uu−. Consequently, ∫ Ω |∇u−|2 + ∫ Ω |u−|2 = ∫ Ω |u−|p. Thus, u− ∈ M(Ω) and hence J(u−) ≥ αX(Ω). Suppose that J(u−) = αX(X). By Theorem 4.5, u− is a nonzero solution of (1.1). By the maximum principle, u = u−, which contradicts the sign assumption on u. Thus J(u−) > αX(Ω). Similarly, J(u+) > αX(Ω), where u+ = max{u, 0}. Thus, J(u) = J(u+) + J(u−) > 2αX(Ω). � The positive solution of Equation (1.1) in RN is unique. Theorem 8.3. (i) There is a ground state solution of Equation (1.1) in RN ; (ii) The only positive solutions of Equation (1.1) in RN are ground state solutions; (iii) Every positive ground state solution ū ∈ H1(RN ) of Equation (1.1) is spheri- cally symmetric about some point x0 in RN , ū′(r) < 0 for r = |x− x0|, and lim r→∞ r N−1 2 erū(r) = γ > 0, EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 69 lim r→∞ r N−1 2 erū′(r) = −γ; (iv) The positive solution of Equation (1.1) in RN is unique. Proof. (i) By Lien-Tzeng-Wang [47]. For (ii) and (iii) see Gidas-Ni-Nirenberg [35]. For (iv) See Kwong [46]. � 8.2. Regularity of Solutions. In addition to the study of (1.1), we also study Equation (1.2), a perturbation of (1.1): associated with (1.2). We consider the energy functionals Jh for u ∈ H1 0 (Ω) : Jh(u) = 1 2 a(u)− 1 p b(u)− ∫ Ω hu. We let J0 = J . We first recall some fundamental estimates for elliptic equations. Let us first consider the classical Cβ−setting: Schauder estimates. Theorem 8.4. Let Ω be a bounded C2,β−domain, h ∈ Cβ(Ω). Then the Dirichlet problem (1.2) has a unique classical solution u ∈ C2,β(Ω). For the proof of the above theorem see Gilbarg-Trudinger [36, Theorem 6.14]. We have the following Kato regularity. Theorem 8.5. Let Ω be a domain in RN and let f : Ω×R → R be a Caratheodory function such that for almost every z ∈ Ω |f(z, u)| ≤ a(z)(1 + |u|) with a nonnegative function a ∈ L N/2 loc (Ω). In addition, let u ∈ H1 loc(Ω) be a weak solution of Equation (1.4). Then u ∈ Lq loc(Ω) for any 1 ≤ q <∞. If a ∈ LN/2(Ω)∩ L2(Ω) and u ∈ H1 0 (Ω), then u ∈ Lq(Ω) for any 2 ≤ q <∞. Proof. Recall that if u ∈ H1 loc(Ω) is a weak solution of Equation (1.4) in Ω, then u satisfies ∫ Ω ∇u∇ϕ = ∫ Ω f(z, u)ϕ for each ϕ ∈ C∞c (Ω). Choose η ∈ C∞c (Ω) and for s ≥ 0, L ≥ 1, let ϕ = ϕs,L = umin{|u|2s, L2}η2 ∈ H1 0 (Ω), with supp η = F ⊂⊂ Ω. Note that ∇ϕ = ∇umin{|u|2s, L2}η2 + 2sχ{|u|s≤L}|u|2s−2u2∇uη2 + 2umin{|u|2s, L2}η∇η. Testing Equation (1.4) with ϕ, we obtain∫ Ω |∇u|2 min{|u|2s, L2}η2 + s 2 ∫ {|u(x)|s≤L} |∇(|u|2)|2|u|2s−2η2 ∫ Ω 2u∇umin{|u|2s, L2}η∇η = ∫ Ω f(·, u)umin{|u|2s, L2}η2. Note that 2xy = 2 √ 1/2x √ 2y ≤ 1 2 x2 + 2y2, 70 HWAI-CHIUAN WANG EJDE-2004/MON. 06 if |u| ≤ 1 we have (1 + |u|)|u|min{|u|2s, L2} ≤ 2, and if |u| ≥ 1, we have (1 + |u|)|u|min{|u|2s, L2} ≤ 2|u|2 min{|u|2s, L2}. Thus, |a|(1 + |u|)|u|min{|u|2s, L2}η2 ≤ 2|a‖u|2 min{|u|2s, L2}η2 + 2|a|η2. Hence,∫ Ω |∇u|2 min{|u|2s, L2}η2 + s 2 ∫ {|u(x)|s≤L} |∇(|u|2)|2|u|2s−2η2 ≤ −2 ∫ Ω u∇umin{|u|2s, L2}η∇η + ∫ Ω |a|(1 + |u|)|u|min{|u|2s, L2}η2 ≤ 1 2 ∫ Ω |∇u|2 min{|u|2s, L2}η2 + 2 ∫ Ω |u|2 min{|u|2s, L2}|∇η|2 + 2 ∫ Ω a|u|2 min{|u|2s, L2}η2 + 2 ∫ Ω aη2. (8.1) Suppose that u ∈ L2s+2 loc (Ω) for some s ≥ 0. Let cs = ∫ F |u|2s+2 <∞; d = max{1,max |∇η|2}; ε(M) = (∫ {a≥M} |a|N/2 )2/N = o(1). We have ∫ Ω |u|2 min{|u|2s, L2}|∇η|2 ≤ d ∫ F |u|2s+2 = dcs, and ∫ Ω a|u|2 min{|u|2s, L2}η2 = ∫ {a≥M}∩F a|u|2 min{|u|2s, L2}η2 + ∫ {a 0 (independent of u) such that ‖u‖W 2,q(Ω) ≤ c‖ −∆u+ u‖Lq(Ω) for each u ∈W 2,q(Ω) ∩W 1,q 0 (Ω), 1 < q <∞. The proof of the above theorem can be found in Gilbarg-Trudinger [36, Lemma 9.17]. Theorem 8.8. Let k ≥ 0 be an integer and Ω be a Ck+1,1 domain in RN and h ∈W k,q(Ω) for some q, 1 < q <∞. If u ∈W 2,q loc (Ω) solves −∆u+ u = h(z) in Ω, then u ∈W k+2,q(Ω). The proof of the above theorem can be found in Gilbarg-Trudinger [36, Theorem 9.19]. Theorem 8.9. Let k ≥ 1 be an integer and let Ω be a Ck+1,β domain in RN and h ∈ Ck−1,β(Ω) for some q, 1 < q <∞. If u ∈ W 2,q loc (Ω) solves −∆u+ u = h(z) in Ω, then u ∈ Ck+1,β(Ω). The proof of the above theorem can be found in Gilbarg-Trudinger [36, Theorem 9.19]. We also have the following existence and uniqueness theorem for the Dirichlet problem for weak solutions: L2-theory. Its proof can be found in Gilbarg-Trudinger [36, Corollary 8.2]. Theorem 8.10. Let u ∈ H1 0 (Ω) be a weak solution of the equation −∆u + u = 0 in Ω. Then u = 0 in Ω. Applying these results, we have the following theorem. Theorem 8.11. Let Ω be a C1,1 domain in RN . Let u ∈ H1 0 (Ω) be a weak solution of (1.2) in Ω. (i) If h ∈ LN/2(Ω) ∩ L2( Ω), then u ∈ Lq(Ω) for every q ∈ [2,∞). Furthermore, ‖u‖Lq(Ω) ≤ p(‖u‖H1(Ω)), where p(‖u‖H1(Ω)) is a polynomial of ‖u‖H1(Ω) with real powers; (ii) Let h ∈ LN/2 ∩ Ls(Ω) ∩ L2(Ω) for some s, s > N and θ = 2 − N s − [2 − N s ], then 0 < θ < 1, u ∈ C1,θ(Ω) ∩W 2,s(Ω), and ‖u‖L∞(Ω) ≤ ‖u‖C1,θ(Ω) ≤ c‖u‖W 2,s(Ω); (iii) Let h ∈ Cθ(Ω) ∩ LN/2(Ω) ∩ Ls(Ω) ∩ L2(Ω) for s as defined in (ii), then u ∈ C2(Ω). Proof. (i) N = 1, 2 follows from the Sobolev inequality. Suppose that N ≥ 3. For d ≥ 0, l ≥ 1, let ϕ = ϕd,l = umin{|u|2d, l2}, then ∇ϕ = min{|u|2d, l2}∇u+ χ{|u|d≤l}2d|u|2d∇u. Since u ∈ H1 0 (Ω), we have∫ Ω |∇ϕ|2 = ∫ Ω |min{|u|2d, l2}∇u+ χ{|u|d≤l}2d|u|2d∇u|2 EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 73 ≤ 2 (∫ Ω |min{|u|2d, l2}∇u|2 + |χ{|u|d≤l}2d|u|2d∇u|2 ) ≤ 2l4 ∫ Ω |∇u|2 + 8d2l4 ∫ {|u|d≤l} |∇u|2 ≤ c ∫ Ω |∇u|2 <∞. Clearly, ϕ ∈ L2(Ω). Thus, ϕ ∈ H1 0 (Ω). Note that we have hϕ ≤ |h||u| for |u| ≤ 1 and hϕ ≤ |h||u|2 min{|u|2d, l2} for |u| > 1. Suppose u ∈ L2d+p. Multiplying and integrating (1.2) with ϕ, we have∫ Ω ∇u∇ϕ = − ∫ Ω uϕ+ ∫ Ω |u|p−2uϕ+ ∫ Ω hϕ ≤ − ∫ Ω |u|2 min{|u|2d, l2}+ ∫ Ω |u|2d+p + ∫ Ω |h||u|+ ∫ Ω |h||u|2 min{|u|2d, l2} ≤ ‖u‖2d+p L2d+p + ‖h‖L2‖u‖L2 + ∫ {h 0 be such that (1 + d 2 )Sε(M) < 1 2 . Then we have ‖∇(umin{|u|d, l})‖2L2 ≤ (2 + d)(‖u‖2d+p L2d+p + ‖h‖L2‖u‖L2 +M‖u‖α(2d+2) L2 ‖u‖(1−α)(2d+2) L2d+p ). By (8.7), we have∫ {|u|d≤l} |∇(|u|d+1)|2 ≤ ∫ Ω |∇(umin{|u|d, l})|2 ≤ (2 + d)(‖u‖2d+p L2d+p + ‖h‖L2‖u‖L2 +M‖u‖α(2d+2) L2 ‖u‖(1−α)(2d+2) L2d+p ). Letting l→∞, we obtain ‖∇(|u|d+1)‖2L2 ≤ (2 + d) ( ‖u‖2d+p L2d+p + ‖h‖L2‖u‖L2 +M‖u‖α(2d+2) L2 ‖u‖(1−α)(2d+2) L2d+p ) . Since ‖u‖2(d+1) L(d+1)2∗ = ‖|u|d+1‖2L2∗ ≤ S‖∇(|u|d+1)‖2L2 , we have ‖u‖2(d+1) L(d+1)2∗ EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 75 ≤ S(2 + d) [ ‖u‖2d+p L2d+p + ‖h‖L2‖u‖L2 +M‖u‖α(2d+2) L2 ‖u‖(1−α)(2d+2) L2d+p ] . Let d0 = 0 and 2di + p = (di−1 + 1)2∗ for i ≥ 1. Since p < 2∗, then d1 > 0 and di ≥ ( 2∗ 2 )i−1d1. Hence, limn→∞ di = ∞ and ‖u‖L2di+p = ‖u‖ L(di−1+1)2∗ ≤ pi−1(‖u‖L2di−1+p). By iterating, we conclude that ‖u‖L2di+p ≤ pi(‖u‖H1). By the interpolation property, for every q ∈ [2,∞), we have u ∈ Lq that satisfies ‖u‖Lq ≤ pq(‖u‖H1). (ii) Let g(z, u) = |u|p−2u + h(z), since (p − 1)s > s > N , then by (i), |u|p−2u ∈ Ls(Ω) ∩ L2(Ω). Thus, g(z, u) ∈ Ls(Ω) ∩ L2(Ω). By Theorem 8.6, the Dirichlet problem −∆v + v = g(z, u) in Ω, v ∈W 1,s 0 (Ω) ∩H1 0 (Ω), has a unique strong solution v ∈W 2,s(Ω) ∩W 1,s 0 (Ω) ∩H1 0 (Ω) and ‖v‖W 2,s(Ω) ≤ c‖g(z, u)‖Ls(Ω). Since v ∈ H1 0 (Ω) satisfies Equation (1.3), we have for each ϕ ∈ C∞c (Ω),∫ Ω (∇v∇ϕ+ vϕ) = ∫ Ω g(z, u)ϕ. Thus u and v satisfy weakly −∆v + v = g(z, u) in Ω, −∆u+ u = g(z, u) in Ω. Let w = v − u. Then −∆w + w = 0 in Ω. By Theorem 8.10, w = 0, or u = v ∈ W 2,s(Ω) ∩W 1,s 0 (Ω) ∩H1 0 (Ω) and ‖u‖W 2,s(Ω) ≤ c‖g(z, u)‖Ls(Ω). Now s > N and θ = 2 − N s − [2 − N s ]. Then by the Sobolev embedding theorem 2.30, u ∈ C1,θ(Ω) and ‖u‖L∞(Ω) ≤ ‖u‖C1,θ(Ω) ≤ c‖u‖W 2,s(Ω). (iii) By (ii), we know that u ∈ C1,θ(Ω). Since u is bounded, then |u|p−2u ∈ Cθ(Ω). By Theorem 8.9, we have u ∈ C2,θ(Ω). � 8.3. Asymptotic Behavior of Solutions. By Theorem 8.11, we obtain the fol- lowing three results about asymptotic behavior of solutions. We define the general- ized infinite strip Sr = Bm(0; r)×Rn, where N ≥ 4, m ≥ 2, n ≥ 1 and m+n = N . Let λ1 be the first eigenvalue of −∆ in Bm(0; r) with the Dirichlet problem, and φ1 the corresponding positive eigenfunction to λ1. Theorem 8.12. (i) Let h ∈ LN/2(Sr)∩Ls(Sr)∩L2(Sr) for s > N . If u in H1 0 (Sr) is a weak solution of (1.2) in Sr, then u ∈ C1(Sr) and lim |y|→∞ u(x, y) = 0 uniformly inx ∈ Bm(0; r); 76 HWAI-CHIUAN WANG EJDE-2004/MON. 06 (ii) Let h ∈ LN/2(Fr s) ∩ Ls(Fr s) ∩ L2(Fr s) for s > N . If u in H1 0 (Fr s) is a weak solution of (1.2) in Fr s, then u ∈ C1(Fr s) and lim y→∞ u(x, y) = 0 uniformly inx ∈ Bm(0; r); (iii) Let Ω be an unbounded domain in RN , h ∈ LN/2(Ω) ∩ Ls(Ω) ∩ L2(Ω) for s > N . If u in H1 0 (Ω) is a weak solution of (1.2) in Ω, then u ∈ C1(Ω) and lim|z|→∞ u(z) = 0. Proof. (i) By Theorem 8.11, u ∈ C1(Sr)∩W 2,s(Sr). For each t > 0, apply Theorem 8.11 (ii) to obtain ‖u‖L∞(Sr t ) ≤ c‖u‖W 2,N (Sr t ), where Sr t = {z = (x, y) ∈ Sr : |y| > t} . Since ‖u‖W 2,N (Sr t ) = o(1) as t → ∞, we obtain lim |y|→∞ u(x, y) = 0 uniformly in x ∈ Bm(0; r). The proofs of (ii) and (iii) are similar to (i). � By Lien-Tzeng-Wang [47], there is a positive solution of (1.1) in Sr. Such a solution admits exponential decay in y. Theorem 8.13. Let u be a positive solution of Equation (1.1) in Sr. Then for every 0 < δ < 1 + λ1 γ > 0 and β > 0 exist such that γφ1(x)e− √ 1+λ1+δ |y| ≤ u(z) ≤ βφ1(x)e− √ 1+λ1−δ |y| for z = (x, y) ∈ Sr. Proof. By Theorem 8.11 (iii), u ∈ C2(Ω). (i) γφ1(x)e− √ 1+λ1+δ |y| ≤ u(z) for z = (x, y) ∈ Sr : define wδ(z) = φ1(x)e− √ 1+λ1+δ |y| for z = (x, y) ∈ Bm(0; r)× Rn = Sr. For 0 < δ < 1 + λ1, take R > 0 such that δ − √ 1+λ1+δ(n−1) |y| ≥ 0 for |y| ≥ R (for n = 1, take R = 1). Set γ = inf z∈Sr, |y|≤R u(x, y) wδ(x, y) . Note that wδ(x, y) and u(x, y) are radially symmetric in x and y, and decreasing in |y| for a fixed x. Thus u(x, y) wδ(x, y) = u(x, y) φ1(x) e √ 1+λ1+δ|y| ≥ u(x,Re1) φ1(x) , for |y| ≤ R, x ∈ Bm(0; r), where e1 = (1, 0, . . . , 0) ∈ Rn. Therefore γ = inf z∈Sr, |y|≤R u(z) wδ(z) ≥ inf x∈Bm(0;r) u(x,Re1) φ1(x) = inf x∈L u(x,Re1) φ1(x) , where L is a fixed diameter of Bm(0; r). Note that u(x,Re1) φ1(x) > 0 for x ∈ L. Furthermore, for each x0 ∈ ∂L ⊂ ∂Bm(0; r), take a small ball B1 in Bm(0; r) such that x0 ∈ ∂B1. Note that φ1(x) > 0 for x ∈ B1, φ1(x0) = 0, and φ1(x) ∈ C2(Bm(0; r)). Then by Lemma 9.1 below, ∂φ1 ∂ν (x0) < 0, where ν is the outward unit normal vector of B1 at x0. Let u1(x) = u(x,Re1), and for each z1 = (x0, Re1) ∈ ∂L×Rn ⊂ ∂Sr, take a small ball B2 in Sr such that z1 ∈ ∂B2. Note that u(z) > 0 EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 77 for z ∈ B2, and u(z1) = 0. By Theorem 8.12, u(x) ∈ C2(Sr), then by Lemma 9.1 below, ∂u ∂ν̄ (z1) < 0, where ν̄ = (ν, 0) is the outward unit normal vector of B2 at z1. Thus, ∂u1 ∂ν (x0) = ∇u1(x0) · ν = ∇u(z1) · ν̄ = ∂u ∂ν̄ (z1) < 0. By L’Hôpital’s rule, we have lim x∈Bm(0;r) x→x0 normally u(x,Re1) φ1(x) = lim h→0− u1(x0 + hν) φ1(x0 + hν) = ∂u1 ∂ν (x0) ∂φ1 ∂ν (x0) > 0. Define u(x,Re1) φ1(x0) = lim x∈Bm(0;r) x→x0 normally u(x,Re1) φ1(x) . Thus u(x,Re1) φ1(x) > 0 for x ∈ L. Since u(x,Re1) φ1(x) : L→ R is continuous, we have γ≥ inf x∈L̄ u(x,Re1) φ1(x) > 0. Let v(z) = γwδ(z) for z ∈ Sr, then we have v(z) ≤ u(z) for z ∈ Sr, |y| ≤ R. For z ∈ Sr, |y| > R, we have −∆(u− v)(z) + (u− v)(z) = (−∆u(z) + u(z)) + (∆v(z)− v(z)) = up−1 + v(z)(δ − √ 1 + λ1 + δ(n− 1) |y| ) ≥ 0. (For n = 1, we only consider the domain {z ∈ Sr| y > R}.) Since u(z)−v(z) ≥ 0 on ∂A∪ ∂Sr where A = {z ∈ Sr : |y| > R}, by Lemma 9.3 below, we have v(z) ≤ u(z) for z = (x, y) ∈ Sr, |y| > R. Thus, we conclude that v(z) ≤ u(z) for z ∈ Sr, or γφ1(x)e− √ 1+λ1+δ |y| ≤ u(z). (ii) u(z) ≤ βφ1(x)e− √ 1+λ1−δ |y|, for z = (x, y) ∈ Sr : for 0 < δ < 1 + λ1. By Theorem 8.12, lim |y|→∞ u(x, y) = 0 uniformly in x. Take R′ > 0 such that up−2 ≤ δ 2+[λ1] for |y| ≥ R′, x ∈ Bm(0; r). Define wδ(z) = φ1(x)e− √ 1+λ1−δ |y| for z = (x, y) ∈ Bm(0; r)× Rn=Sr; β = sup z∈Sr, |y|≤R′ u(z) wδ(z) > 0; µ(z) = βwδ(z) for z ∈ Sr. Fix a diameter L of Bm(0; r). We have u(z) wδ(z) = u(x, y) φ1(x) e √ 1+λ1−δ |y| ≤ u(x, 0) φ1(x) e √ 1+λ1−δ R′ , for |y| ≤ R′, x ∈ Bm(0; r), and β = sup z∈Sr, |y|≤R′ u(z) wδ(z) ≤ sup x∈Bm(0;r) u(x, 0) φ1(x) e √ 1+λ1−δ R′ = sup x∈L u(x, 0) φ1(x) e √ 1+λ1−δ R′ . Similarly to part (i), for x0 ∈ ∂L ⊂ ∂Bm(0; r), we have lim x∈Bm(0;r) x→x0 normally u(x, 0) φ1(x) = lim h→0− u0(x0 + hν) φ1(x0 + hν) = ∂u0 ∂ν (x0) ∂φ1 ∂ν (x0) <∞, 78 HWAI-CHIUAN WANG EJDE-2004/MON. 06 where ν is the outward unit normal vector of Bm(0; r) at x0 and u0(x) = u(x, 0). Define u(x0, 0) φ1(x0) = lim x∈Bm(0;r) x→x0 normally u(x, 0) φ1(x) <∞. Thus, β≤ sup x∈L u(x, 0) φ1(x) e √ 1+λ1−δ R′ ≤ sup x∈L̄ u(x, 0) φ1(x) e √ 1+λ1−δ R′ <∞. Therefore, µ(z) ≥ u(z) for z ∈ Sr, |y| ≤ R′. For z ∈ Sr, |y| > R′ we have −∆(u− µ)(z) + (u− µ)(z) = (−∆u(z) + u(z)) + (∆µ(z)− µ(z)) = up−1(z) + ( − δ − √ 1 + λ1 − δ(n− 1) |y| ) µ(z) ≤ δ 2 + [λ1] (u− µ)(z), therefore, −∆(u− µ)(z) + (1− δ 2 + [λ1] )(u− µ)(z) ≤ 0. Since 1 − δ 2+[λ1] > 0, u(z) − µ(z) ≤ 0 on ∂B, where B = {z ∈ Sr : |y| > R′}, by Lemma 9.3 below, u(z) − µ(z) ≤ 0 in B. Thus, we conclude that u(z) ≤ µ(z) for z ∈ Sr. � We similarly present the asymptotic behavior of each solution of (1.1) in the interior flask domains Fr s, where s > r. Theorem 8.14. Let u be a positive solution of (1.1) in Fr s. Then for any 0 < δ < 1 + λ1, γ > 0, β > 0, and R > s exist such that for z = (x, y) ∈ Ar R, γφ1(x)e− √ 1+λ1 y ≤ u(z) ≤ βφ1(x)e− √ 1+λ1−δ y. Proof. (i) γφ1(x)e− √ 1+λ1 y ≤ u(z) for z = (x, y) ∈ Ar R : by Theorem 8.12 (ii), lim y→∞ u(x, y) = 0 uniformly in x, where (x, y) ∈ Fr s. For 0 < δ < 1 + λ1, take R > s such that up−2(x, y) ≤ δ 2+[λ1] for y ≥ R. In the remaining proofs, we fix such R. Define w(z) = φ1(x)e− √ 1+λ1 y for z = (x, y) ∈ Ar s. Set γ = inf z∈Ar s, y=R u(x, y) w(x, y) . Similarly to Theorem 8.13, γ > 0. Let v(z) = γw(z) for z ∈ Ar s, then we have v(z) ≤ u(z) for z ∈ Ar s, y = R. For z ∈ Ar R, we have −∆(u− v)(z) + (u− v)(z) = (−∆u(z) + u(z)) + (∆v(z)− v(z)) = up−1 ≥ 0. Since u − v ≥ 0 on ∂Ar R, by the strong maximum principle, we have v(z) ≤ u(z) for z = (x, y), z ∈ Ar R. (ii) u(z) ≤ βφ1(x)e− √ 1+λ1−δ y, for z = (x, y) ∈ Ar R : define wδ(z) = φ1(x)e− √ 1+λ1−δ y for z = (x, y) ∈ Ar s; β = sup z∈Ar s, y=R u(z) wδ(z) > 0; EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 79 µ(z) = βwδ(z) for z ∈ Ar s. Similarly to Theorem 8.13, β <∞. Therefore, u(z) ≤ µ(z) for z ∈ Ar s, y = R. For z ∈ Ar s, y ≥ R, we have −∆(u− µ)(z) + (u− µ)(z) = (−∆u(z) + u(z)) + (∆µ(z)− µ(z)) = up−1(z)− δµ(z) ≤ δ 2 + [λ1] (u− µ)(z); therefore, −∆(u− µ)(z) + (1− δ 2 + [λ1] )(u− µ)(z) ≤ 0. Since 1− δ 2+[λ1] > 0 and u− µ ≤ 0 on ∂Ar R, by the strong maximum principle, we obtain u(z) ≤ µ(z) for z ∈ Ar R. � Bibliographical notes: Theorem 8.2 is from Benci-Cerami [11]. Theorem 8.3 is from Lien-Tzeng-Wang [47] and Gidas-Ni-Nirenberg [35].The asymptotic behavior results are from Wang [71] and Chen-Chen-Wang [23]. 9. Symmetry of Solutions We use the asymptotic behavior of solutions developed in Section 8 and apply the “moving plane” method to prove the symmetry of solutions to (1.1) in the infinite strip Ar. Our approach is similar to those in Gidas-Ni-Nirenberg [34, Theorem 1] and [35, Theorem 2] but is more complicated. Before proving our main results, we first establish a version of the Hopf boundary point lemma and the strong maximum principle that will be used in our case. Lemma 9.1 (Hopf Boundary Point Lemma). Let Ω be a domain (possibly un- bounded) in Rn. Let L be a differential operator given by Lu = n∑ i,j=1 aij(x)Diju+ n∑ i=1 bi(x)Diu+ c(x)u, aij(x) = aji(x), which is uniformly elliptic, |bi(x)| λ(x) and c(x) λ(x) are uniformly bounded, where λ(x) is the minimum eigenvalue of [aij(x)]. Assume Lu ≤ 0. Let x0 ∈ ∂Ω satisfy: (i) u is continuous at x0; (ii) u(x0) < u(x) for all x ∈ Ω; (iii) A ball B ⊂ Ω exists with x0 ∈ ∂B. Suppose that one of the following conditions holds: (i) u(x0) < 0 and c(x) ≤ 0; (ii) u(x0) = 0; (iii) c(x) ≡ 0. If the outer normal derivative ∂u ∂ν (x0) of u at x0 exists, then ∂u ∂ν (x0) < 0. For the proof of the above lemma, see Gilbarg-Trudinger [36, Lemma 3.4]. Lemma 9.2 (Hopf Boundary Point Lemma). Let Ω be a domain (possibly un- bounded) in Rn. Let L be a differential operator given by Lu = n∑ i,j=1 aij(x)Diju+ n∑ i=1 bi(x)Diu+ c(x)u, aij(x) = aji(x), 80 HWAI-CHIUAN WANG EJDE-2004/MON. 06 which is uniformly elliptic, |bi(x)| λ(x) and c(x) λ(x) are uniformly bounded, where λ(x) is the minimum eigenvalue of [aij(x)]. Assume Lu ≥ 0. Let x0 ∈ ∂Ω satisfy: (i) u is continuous at x0; (ii) u(x0) > u(x) for all x ∈ Ω; (iii) A ball B ⊂ Ω exists with x0 ∈ ∂B. Suppose that one of the following conditions holds: (i) u(x0) > 0 and c(x) ≤ 0; (ii) u(x0) = 0; (iii) c(x) ≡ 0. If the outer normal derivative ∂u ∂ν (x0) of u at x0 exists, then ∂u ∂ν (x0) > 0. For the proof of the above lemma, see Gilbarg-Trudinger [36, Lemma 3.4]. Lemma 9.3 (Strong Maximum Principle). Let L be uniformly elliptic, c = 0 and Lu ≥ 0 (≤ 0) in a domain Ω (not necessarily bounded). Then if u achieves its maximum (minimum) in the interior of Ω, u is constant. If c ≤ 0 and c/λ is bounded, then u cannot achieve a nonnegative maximum (nonpositive minimum) in the interior of Ω unless it is constant. For the proof of the above lemma, see Gilbarg-Trudinger [36, Theorem 3.5]. We define the generalized infinite strip by Sr = Bm(0; r) × Rn, where m ≥ 2, n ≥ 1, and m+ n = N , and suppose that (g1) g(u) > 0 as u > 0; (g2) g(u) = O(up) as u→ 0 for some p > 1. Now we consider the equation −∆u+ u = g(u) + h(z) in Sr, u > 0 in Sr, u = 0 on ∂Sr, lim |y| → ∞u(x, y) = 0 uniformly in x ∈ Bm(0; r). (9.1) We apply the “moving plane” method to prove the symmetry of solutions of (9.1). Theorem 9.4. Assume that g ∈ C1 satisfies (g1) and h is radially symmetric in x and y and strictly decreasing in |x| and |y| . Let u(x, y) be a C2 solution of Equation (9.1). Then u is radially symmetric in x and in y; that is to say, u(x, y) = u(|x|, |y|). Part I: u is radially symmetric in y. Notation: Sθ = {(x, y1, y2, . . . , yn) ∈ Sr | x ∈ Bm(0; r), y1 = θ}; Γθ = {(x, y1, y2, . . . , yn) ∈ Sr | x ∈ Bm(0; r), y1 < θ}. For any (x, y) ∈ Sr, set (x, yθ) = (x, 2θ − y1, y2, . . . , yn); that is, (x, yθ) is the reflection of (x, y) with respect to Sθ; Let Θ be the collection of all θ ∈ R such that the following statements hold: u(x, y) < u(x, yθ) for all (x, y) ∈ Γθ, uy1(x, y) > 0 on Sθ. Lemma 9.5. There exists θ0 > 0 such that either (−∞,−θ0] ⊂ Θ or u(x, y) ≡ u(x, y−θ0) in Γ−θ0 . EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 81 Proof. Given θ < 0, set wθ(x, y) = u(x, y) − u(x, yθ) for (x, y) ∈ Γθ, and wθ(x, y) satisfies ∆wθ(x, y) + cθ(x, y)wθ(x, y) = h(x, yθ)− h(x, y) ≥ 0, (9.2) where cθ(x, y) = g(u(x,y))−g(u(x,yθ)) u(x,y)−u(x,yθ) − 1 = g′(ξθ) − 1 and ξθ is in between u(x, y) and u(x, yθ). We claim that θ0 > 0 exists such that if θ ≤ −θ0, then wθ(x, y) ≤ 0 in Γθ. Otherwise, suppose wθ(x, y) > 0 for some (x, y) ∈ Γθ. Since lim |y| → ∞wθ(x, y) = 0 uniformly in x, wθ(x, y) achieves its maximum at (xθ, yθ) ∈ Γθ. Then ∇wθ(xθ, yθ) = 0, quad{wθ ij(xθ, yθ)} ≤ 0. Note that by (g2), lim t→ 0+g′(t) = 0. Take t0 > 0 such that if 0 < t ≤ t0, then g′(t) < 1. Since lim y1 → −∞u(x, y) = 0, we can choose θ0 > 0 such that if y1 ≤ −θ0, then u(x, y) ≤ t0, therefore, g′(u(x, y)) < 1. For θ ≤ −θ0, (xθ, yθ) ∈ Γθ, then ∆wθ(xθ, yθ) ≤ 0, cθ(x, y)wθ(xθ, yθ) = (g′(ξθ)− 1)wθ(xθ, yθ) < 0, contradicting (9.2). As a consequence of the maximum principle and the Hopf boundary point lemma, either w−θ0(x, y) ≡ 0 in Γ−θ0 or wθ(x, y) < 0 in Γθ and wθ y1 (x, y) > 0 for (x, y) ∈ Sθ for θ ≤ −θ0, or uy1(x, y) > 0 for (x, y) ∈ Sθ. � Lemma 9.6. If (−∞, θ] ⊂ Θ, then there exists ε > 0 such that [θ, θ + ε) ⊂ Θ. Proof. Suppose instead that a decreasing sequence θk → θ and a sequence {(xk, yk)} of points in Γθk exist such that wθk(xk, yk) = u(xk, yk) − u(xk, yθk) > 0, where (xk, yθk) is the reflection of (xk, yk) with respect to Sθk . There is a subsequence {(xk, yk)} such that xk → x as k → ∞. Two possibilities may arise, as shown in Case 1 and 2. Case 1. |yk| → ∞. As shown in Lemma 9.5, we assume wθk(x̃k, ỹk) = max (x, y) ∈ Γθk wθk(x, y), ∇wθk(x̃k, ỹk) = 0, {wθk ij (x̃k, ỹk)} ≤ 0. (9.3) From lim |y| → ∞u(x, y) = 0, as in Lemma 9.5, we obtain a contradiction. Case 2. yk → y. We have (xk, yk) → (x, y) ∈ Γθ , thus wθ(x, y) ≥ 0. Clearly (x, y) /∈ Γθ. If (x, y) ∈ Sθ , then uy1(x, y) < 0, which contradicts θ ∈ Θ. Moreover, (x, y) /∈ ∂Sr ∩ Γθ. Note that wθ(x, y) satisfies Equation (9.2), and by the Hopf boundary point lemma, we obtain ∂ ∂νw θ(x, y) > 0. On the other hand, by taking limits in (9.3), we obtain ∇wθ(x, y) = 0, which is a contradiction. We conclude that either Case 1 or Case 2 is impossible. � Proof of Part I. Let σ = sup{θ ∈ R : (−∞, θ) ⊂ Θ}. Then σ /∈ Θ. If not, by Lemma 9.6 we would have [σ, σ + ε) ⊂ Θ, which contradicts the definition of σ. We claim that σ = 0. Suppose instead that σ ∈ (−∞, 0). By continuity, u(x, y) ≤ u(x, yσ) for all (x, y) ∈ Γσ, then by the maximum principle, we have u(x, y) ≡ u(x, yσ) for all (x, y) ∈ Γσ. This implies that h(x, y) = h(x, yσ) for all (x, y) ∈ Γσ, which is a contradiction. This proves u(x, y) is symmetric with respect to the hyperplane y1 = 0 for all (x, y) ∈ Sr. By reversing the y1 axis, we conclude that u(x, y) is symmetric with respect to the hyperplane S0. Since the y1 direction can be chosen arbitrarily, we conclude that u(x, y) is radially symmetric in Rn. � 82 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Part II: u is radially symmetric in Bm(0; r). Notation: Tλ = {(x, y) = (x1, x2, . . . , xN−1, y) ∈ Sr : x1 = λ}; Σλ = {(x, y) = (x1, x2, . . . , xN−1, y) ∈ Sr : x1 < λ}. For any (x, y) = (x1, x2, . . . , xN−1, y) ∈ Sr, set (xλ, y) = (2λ− x1, x2, . . . , xN−1, y), that is, (xλ, y) is the reflection of (x, y) with respect to Tλ. Let Λ be the collection of all λ ∈ (−r, 0) such that the following statements hold: u(x, y) < u(xλ, y) for all (x, y) ∈ Σλ, ux1(x, y) > 0 on Tλ. Lemma 9.7. For some δ such that 0 < δ < r, (−r,−r + δ) ⊂ Λ. Proof. Given λ ∈ (−r, 0), set vλ(x, y) = u(x, y) − u(xλ, y) for (x, y) ∈ Σλ, then vλ(x, y) = 0 for (x, y) ∈ Tλ, and vλ(x, y) satisfies ∆vλ(x, y) + cλ(x, y)vλ(x, y) = h(xλ, y)− h(x, y) ≥ 0, (9.4) where cλ(x, y) = g(u(x,y))−g(u(xλ,y)) u(x,y)−u(xλ,y) − 1 = g′(ζλ)− 1 where ζλ is in between u(x, y) and u(xλ, y). We claim that if −r < λ < −r + δ, then vλ(x, y) ≤ 0 in Σλ. Otherwise, suppose λ exists such that −r < λ < −r + δ, vλ(x, y) > 0 for some (x, y) ∈ Σλ. Since lim |y| → ∞vλ(x, y) = 0 uniformly in x, vλ(x, y) achieves its maximum at (xλ, yλ) ∈ Σλ. Then ∇vλ(xλ, yλ) = 0, {vλ ij(xλ, yλ)} ≤ 0. Note that by (g2), lim t→ 0+g′(t) = 0. Take t0 > 0 such that if 0 < t ≤ t0, then g′(t) < 1. Since lim |x| → r−u(x, y) = 0, we can choose δ, 0 < δ < r such that u(x, y) ≤ t0 whenever r − δ < |x| < r. For −r < λ < −r + δ, we have ∆vλ(xλ, yλ) ≤ 0, cλ(x, y)vλ(xλ, yλ) = (g′(ζλ)− 1)vλ(xλ, yλ) < 0, which contradicts (9.4). Therefore, for −r < λ < −r + δ, vλ(x, y) ≤ 0 in Σλ. By applying the maximum principle and the Hopf boundary point lemma, for −r < λ < −r + δ, we obtain vλ(x, y) < 0 in Σλ and vλ x1 (x, y) > 0 for (x, y) ∈ Tλ. Hence, ux1(x, y) > 0 for (x, y) ∈ Tλ. Then (−r,−r + δ) ⊂ Λ. � Lemma 9.8. If (−r, λ] ⊂ Λ, then there is a τ > 0 such that [λ, λ+ τ) ⊂ Λ. Proof. Suppose that a decreasing sequence λk → λ and a sequence {(xk, yk)} of points in Σλk exist such that vλk(xk, yk) = u(xk, yk) − u(xλk , yk) > 0, where (xλk , yk) is the reflection of (xk, yk) with respect to Tλ. There is a subsequence {(xk, yk)} such that xk → x ∈ Bm(0; r). Two possibilities may arise as shown in Case 1 and 2: Case 1. |yk| → ∞. As shown in Lemma 9.7, we assume vλk(x̃k, ỹk) = max (x, y) ∈ Σλk vλk(x, y), ∇vλk(x̃k, ỹk) = 0, {vλk ij (x̃k, ỹk)} ≤ 0. From lim |y| → ∞ u(x, y) = 0, as in Lemma 9.7, we obtain a contradiction. Case 2. yk → y. We have (xk, yk) → (x, y) ∈ Σλ. Thus, vλ(x, y) ≥ 0. Clearly (x, y) /∈ Σλ. If (x, y) ∈ Tλ then ux1(x, y) < 0, which contradicts λ ∈ Λ. Moreover, EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 83 (x, y) /∈ ∂Sr ∩ Σλ, since if (x, y) ∈ ∂Sr ∩ Σλ then 0 = u(x, y) ≥ u(xλ, y) > 0, a contraction. We conclude that either Case 1 or Case 2 is impossible. � Proof of Part II. Let µ = sup{λ ∈ (−r, 0) : (−r, λ) ⊂ Λ}. Then µ /∈ Λ. If not, by Lemma 9.8, we would have [µ, µ + ε) ⊂ Λ, which contradicts the definition of µ. We claim that µ = 0. Suppose instead that µ ∈ (−r, 0). By continuity we have u(x, y) ≤ u(xµ, y) for all (x, y) ∈ Σµ. Then by the maximum principle we have u(x, y) ≡ u(xµ, y) for all (x, y) ∈ Σµ, which is impossible. Thus µ = 0. By reversing the x1 axis, we conclude that u(x, y) is symmetric with respect to the hyperplane T0 and ux1(x, y) < 0 for x1 > 0. Since the x1 direction can be chosen arbitrarily, we conclude that u(x, y) is radially symmetric in Bm(0; r). � Corollary 9.9. Assume that g ∈ C1 satisfies property (g1) and h(x) ≡ 0. Let u(x, y) be a C2 solution of Equation (9.1). Then u is radially symmetric in x and in y; that is to say, u(x, y) = u(|x|, |y|). Proof of Part I. u is radially symmetric in y. Similarly to Theorem 9.4, let σ1 = sup{θ ∈ R | (−∞, θ) ⊂ Θ}. Note that σ1 is not necessarily zero. Similarly, u(x, y) is symmetric with respect to the hyperplane y1 = σ1 for all (x, y) ∈ Sr. The same conclusion holds for the other coordinate direction, and we conclude that u is symmetric about each of n planes yj = σj and ∇u = 0 only at their intersection. We may now take their intersection as the origin. The same argument may be applied to any unit direction γ and we infer that u is symmetric about some plane Bm(0; r)× {y ∈ R : y · γ = c(γ) = const.} At the point on this plane where u achieves its maximum we have ∇u = 0 (since the derivative normal to the plane is zero at every point of the plane). It follows that c(γ) = 0. Thus u is symmetric about every plane through the origin, that is, u is radially symmetric in y. In addition, we also conclude that uρ < 0 where ρ = |y|. Proof of Part II: Since u is radially symmetric in Bm(0; r): the proof is similarly to the proof of Theorem 9.4. � 9.1. Open Question: Are positive solutions of (1.1) in the generalized infinite strip by Sr unique up to a translation? Bibliographical notes: The results of this section are from Chen-Chen-Wang [23]. 10. Nonachieved Domains and Esteban-Lions Domains In this section we also characterize Esteban-Lions domains. We prove that proper large domains, Esteban-Lions domains, and some interior flask domains are nonachieved. Theorem 10.1. Let Ω2 be one of Ar, Ar1,r2 , and RN , and Ω1 a proper large domain of Ω2. Then α1 = α2, J does not satisfy the (PS)α1-condition, and Equation (1.1) does not admit any ground state solution in Ω1. In particular, a proper large domain Ω1 of Ω2 is nonachieved. Proof. Since Ω1 is a proper large domain of Ω2, by Theorem 4.18, α1 = α2. Then by Theorem 5.7 (i) and (ii), J does not satisfy the (PS)α1-condition, and Equation (1.1) does not admit any ground state solution in Ω1. � 84 HWAI-CHIUAN WANG EJDE-2004/MON. 06 The only solution in an Esteban-Lions domain is trivial. The following lemmas are required. The first lemma is from Protter [62]. Lemma 10.2. Let u be a C2 real-valued function in Ω. Suppose that δ0, ρ0 > 0 and z0 ∈ Ω exist such that 0 < δ0 < ρ0 < 1 and u(z) = 0 in {z ∈ RN : δ0 ≤ |z − z0| ≤ ρ0}c. Then m0 > 0 and c > 0 exist such that if m ≥ m0, then∫ Ω ρ−2m−2e2ρ−m u2dz ≤ c m4 ∫ Ω ρm+2e2ρ−m (∆u)2dz, where ρ = |z − z0|. Proof. Without loss of generality, we assume that z0 = 0, so that ρ = |z|. Let v = eρ−m u. By the inequality (A+B + C)2 ≥ 2B(A+ C), we have (∆u)2 = [ e−ρ−m ∆v + 2∇v∇(e−ρ−m ) + v∆(e−ρ−m ) ]2 = [e−ρ−m ∆v + 2me−ρ−m ρ−m−2 N∑ i=1 zi ∂v ∂zi +mvρ−m−2e−ρ−m (mρ−m −m− 2 +N)]2 ≥ 4me−2ρ−m ρ−m−2 N∑ i=1 zi ∂v ∂zi [ ∆v +mvρ−m−2(mρ−m −m− 2 +N) ] and ρm+2e2ρ−m (∆u)2 ≥ 4m N∑ i=1 zi ∂v ∂zi [ ∆v +m2vρ−2m−2 − (m+ 2−N)mvρ−m−2 ] . Thus, ∫ {δ0≤|z|≤ρ0} ρm+2e2ρ−m (∆u)2dz ≥ (I) + (II)− (III), where (I) = 4m ∑ i = 1N ∫ {δ0≤|z|≤ρ0} zi ∂v ∂zi ∆v dz , (II) = 4m3 ∑ i = 1N ∫ {δ0≤|z|≤ρ0} ρ−2m−2vzi ∂v ∂zi dz , (III) = 4m2 ∑ i = 1N ∫ {δ0≤|z|≤ρ0} ρ−m−2(m+ 2−N)vzi ∂v ∂zi dz . We claim that (i) (I) = 2m(N − 2) ∫ {δ0≤|z|≤ρ0} |∇v|2dz > 0; (ii) There exists m1 > 0 such that (II) ≥ m4 ∫ {δ0≤|z|≤ρ0} ρ−2m−2v2dz for m ≥ m1; EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 85 and (iii) for 0 < ε < 1, there exists m2 > 0 such that (III) ≤ εm4 (∫ {δ0≤|z|≤ρ0} ρ−2m−2v2dz ) for m ≥ m2. (i) Since ∂v ∂zi = 0 on ∂{δ0 ≤ |z| ≤ ρ0} for all i = 1, 2, . . . , N and N∑ i=1 ∫ {δ0≤|z|≤ρ0} zi ∂v ∂zi 4vdz = N∑ i=1 N∑ j=1 ∫ {δ0≤|z|≤ρ0} zi ∂v ∂zi ∂2v ∂z2 j dz. For i 6= j, by integration by parts∫ {δ0≤|z|≤ρ0} zi ∂v ∂zi ∂2v ∂z2 j dz = − ∫ {δ0≤|z|≤ρ0} ∂v ∂zj ∂ ∂zj (zi ∂v ∂zi )dz = − ∫ {δ0≤|z|≤ρ0} zi ∂v ∂zj ∂2v ∂zi∂zj dz = ∫ {δ0≤|z|≤ρ0} [ ( ∂v ∂zj )2 + zi ∂v ∂zj ∂2v ∂zi∂zj ] dz, we have ∫ {δ0≤|z|≤ρ0} zi ∂v ∂zj ∂2v ∂zi∂zj dz = −1 2 ∫ {δ0≤|z|≤ρ0} ( ∂v ∂zj )2dz. Thus, N∑ j=1 j 6=i ∫ {δ0≤|z|≤ρ0} zi ∂v ∂zi ∂2v ∂z2 j dz = 1 2 N∑ j=1 j 6=i ∫ {δ0≤|z|≤ρ0} ( ∂v ∂zj )2dz = 1 2 ∫ {δ0≤|z|≤ρ0} |∇v|2 − ( ∂v ∂zi )2dz. For i = j,∫ {δ0≤|z|≤ρ0} zi ∂v ∂zi ∂2v ∂z2 j dz = − ∫ {δ0≤|z|≤ρ0} [ zi ∂v ∂zi ∂2v ∂z2 i + ( ∂v ∂zi )2 ] dz, then we have ∫ {δ0≤|z|≤ρ0} zi ∂v ∂zi ∂2v ∂z2 j dz = −1 2 ∫ {δ0≤|z|≤ρ0} ( ∂v ∂zi )2dz. Hence, N∑ i=1 ∫ {δ0≤|z|≤ρ0} zi ∂v ∂zi 4v dz = N∑ i=1 (1 2 ∫ {δ0≤|z|≤ρ0} |∇v|2dz − ∫ {δ0≤|z|≤ρ0} ( ∂v ∂zi )2dz ) = ( N 2 − 1) ∫ |∇v|2 > 0. (ii) m1 > 0 exists such that 4m4 + 4m3− 2m3N ≥ m4 for m ≥ m1. Since v = 0 on ∂{δ0 ≤ |z| ≤ ρ0}, we have∫ {δ0≤|z|≤ρ0} ρ−2m−2vzi ∂v ∂zi dz 86 HWAI-CHIUAN WANG EJDE-2004/MON. 06 = − ∫ {δ0≤|z|≤ρ0} ∂ ∂zi (ρ−2m−2vzi)vdz = (2m+ 2) ∫ {δ0≤|z|≤ρ0} ρ−2m−4v2z2 i dz − ∫ {δ0≤|z|≤ρ0} ρ−2m−2vzi ∂v ∂zi dz − ∫ {δ0≤|z|≤ρ0} ρ−2m−2v2dz and ∫ {δ0≤|z|≤ρ0} ρ−2m−2vzi ∂v ∂zi dz = (m+ 1) ∫ {δ0≤|z|≤ρ0} ρ−2m−4v2z2 i dz − ∫ {δ0≤|z|≤ρ0} ρ−2m−2v2dz. Thus, for m ≥ m1 we have (II) = 4m3 N∑ i=1 ∫ {δ0≤|z|≤ρ0} ρ−2m−2vzi ∂v ∂zi dz = 4m3(m+ 1) N∑ i=1 ∫ {δ0≤|z|≤ρ0} ρ−2m−4v2z2 i dz − 2m3 N∑ i=1 ∫ {δ0≤|z|≤ρ0} ρ−2m−2v2dz = (4m4 + 4m3 − 2m3N) ∫ {δ0≤|z|≤ρ0} ρ−2m−2v2dz ≥ m4 ∫ {δ0≤|z|≤ρ0} ρ−2m−2v2dz for m ≥ m1. (iii) Since ρ0 < 1, for 0 < ε < 1 there is an m2 > 0 such that 2m2(m+ 2−N)2 ≤ 2m4 for m ≥ m2,∫ ρ−m−2v2dz ≤ ε 2 ∫ ρ−2m−2v2dz for m ≥ m2. Similarly to part (ii), we have∫ {δ0≤|z|≤ρ0} ρ−m−2vzi ∂v ∂zi dz = (m+ 2) ∫ {δ0≤|z|≤ρ0} ρ−m−4v2z2 i dz − ∫ {δ0≤|z|≤ρ0} ρ−m−2vzi ∂v ∂zi dz − ∫ {δ0≤|z|≤ρ0} ρ−m−2v2dz and N∑ i=1 ∫ {δ0≤|z|≤ρ0} ρ−m−2vzi ∂v ∂zi dz = 1 2 (m+ 2−N) ∫ {δ0≤|z|≤ρ0} ρ−m−2v2dz. Thus, for m ≥ m2 we have (III) = 4m2(m+ 2−N) N∑ i=1 ∫ {δ0≤|z|≤ρ0} ρ−m−2vzi ∂v ∂zi dz = 2m2(m+ 2−N)2 ∫ {δ0≤|z|≤ρ0} ρ−m−2v2dz EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 87 ≤ m4 ∫ {δ0≤|z|≤ρ0} ρ−m−2v2dz ≤ εm4 ∫ {δ0≤|z|≤ρ0} ρ−2m−2v2dz. Take m0 = max{m1,m2}. For m ≥ m0, since u ∈ C2 and u = 0 on the set {z ∈ RN : δ0 ≤ |z| ≤ ρ0}c, we have∫ Ω ρm+2e2ρ−m (∆u)2dz = ∫ {δ0≤|z|≤ρ0} ρm+2e2ρ−m (∆u)2dz ≥ (I) + (II)− (III) ≥ m4( ∫ {δ0≤|z|≤ρ0} ρ−2m−2v2dz)− εm4( ∫ {δ0≤|z|≤ρ0} ρ−2m−2v2dz) = (1− ε)m4( ∫ {δ0≤|z|≤ρ0} ρ−2m−2v2dz) = (1− ε)m4( ∫ Ω ρ−2m−2v2dz). Hence, for m ≥ m0, we have∫ ρ−2m−2e2ρ−m u2dz ≤ c m4 ∫ ρm+2e2ρ−m (∆u)2dz. � Recall the uniqueness of the analytic function: suppose that f is analytic in a domain Ω in R2. If f(zn) = 0 for some sequence {zn} of distinct points such that zn → z0 ∈ Ω, then f ≡ 0 in Ω. We know that f = u + iv, where u and v are harmonic functions, but the uniqueness of harmonic functions is not elegant, taking the form below. Let δ(z0) = dist(z0, ∂Ω). Then we have the following uniqueness result (see Heinz [39]). Theorem 10.3. Let u be a C2 real-valued function on Ω. Suppose that u(z) = 0 is in the neighborhood of a point z0 ∈ Ω and that M > 0 exists such that (∆u)2 ≤Mu2 for any z ∈ Ω. Then u(z) = 0 for any z ∈ Ω. Proof. Let R = min{ 1 2 , 1 4δ(z0)} and Φ(t) ∈ C2 c ([0,∞)) satisfy Φ(t) = { 1 for 0 ≤ t ≤ R, 0 for 3 2R ≤ t <∞. Let ũ(z) = u(z)Φ(|z− z0|). Note that 2R < δ(z0) and BN (z0; 2R) ⊂ Ω. Thus, ũ(z) is well-defined on BN (z0; 2R) and ũ(z) ∈ C2 c ( BN (z0; 2R) ) . We also have ũ(z) = { u(z) in {z ∈ RN : |z − z0| ≤ R}, 0 in {z ∈ RN : |z − z0| ≥ 3 2R}, with 3 2R < 1. By Lemma 10.2, m0 > 0 and c0 > 0 exist such that R−2m−2 − c0Mm−4Rm+2 > 0, 88 HWAI-CHIUAN WANG EJDE-2004/MON. 06 and for all m ≥ m0∫ ρ−2m−2e2ρ−m ũ2dz ≤ c0 m4 ∫ ρm+2e2ρ−m (∆ũ)2dz. We have R−2m−2 ∫ {|z−z0|≤R} e2ρ−m u2dz ≤ ∫ {|z−z0|≤R} ρ−2m−2e2ρ−m u2dz ≤ ∫ {|z−z0|≤2R} ρ−2m−2e2ρ−m ũ2dz ≤ c0m −4 ∫ {|z−z0|≤2R} ρm+2e2ρ−m (∆ũ)2dz ≤ c0Mm−4 ∫ {|z−z0|≤R} ρm+2e2ρ−m u2dz + c0m −4 ∫ {R≤|z−z0|≤2R} ρm+2e2ρ−m (∆ũ)2dz ≤ c0Mm−4Rm+2 ∫ {|z−z0|≤R} e2ρ−m u2dz + c0m −4(2R)m+2e2R−m ∫ {R≤|z−z0|≤2R} (∆ũ)2dz, or (R−2m−2 − c0Mm−4Rm+2) ∫ {|z−z0|≤R} e2ρ−m u2dz ≤ c0m −4(2R)m+2e2R−m ∫ {R≤|z−z0|≤2R} (∆ũ)2dz. Since ũ(z) ∈ C2 c (BN (z0; 2R)), we have (R−2m−2 − c0Mm−4Rm+2)e2R−m ∫ {|z−z0|≤R} u2dz ≤ (R−2m−2 − c0Mm−4Rm+2) ∫ {|z−z0|≤R} e2ρ−m u2dz ≤ c0m −4(2R)m+2e2R−m ∫ {R≤|z−z0|≤2R} (∆ũ)2dz ≤ c0m −4(2R)m+2e2R−m C. Thus, ∫ {|z−a|≤R} u2(z)dz ≤ c0m −4(2R)m+2C R−2m−2 − c0Mm−4Rm+2 . Let m→∞, then u(z) = 0 for any z ∈ BN (z0;R). We claim that u(z̃) = 0 for any z̃ ∈ Ω. For z̃ ∈ Ω, let h : [0, 1] → Ω be a path satisfying h(0) = z0, h(1) = z̃. Since h and δ are continuous functions, then A = min {1 2 , 1 4 inf 0≤t≤1 δ(h(t)) } > 0. EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 89 Since {h(t) : 0 ≤ t ≤ 1} is a compact set, 0 = t0 < t1 < t2 < · · · < tn = 1 exist such that {h(t)|0 ≤ t ≤ 1} ⊂ n ∪ i=0 BN (h(ti); 1 3 A) and BN ( h(ti); 1 3 A ) ∩BN ( h(ti+1); 1 3 A ) 6= ∅. For i = 1, 2, . . . , n, let zi = h(ti) and Ri = min{ 1 2 , 1 4δ(zi)}, then zi ∈ BN (zi−1;A) ⊂ BN (zi−1;Ri). Applying the same process to z0, z1 ,. . . , zn, we conclude that u(z) = 0 on ∪n i=0B N (zi;Ri), in particular, u(z̃) = 0. � Let us denote the linear space of k times weakly differentiable functions by W k(Ω). Clearly Ck(Ω) ⊂W k(Ω). For p ≥ 1 and a nonnegative integer k, we define W k,p(Ω) = {u ∈ W k(Ω) : Dαu ∈ Lp(Ω) for all |α| ≤ k}. We have the following lemma. Lemma 10.4. Let 1 ≤ p <∞, R > 0, and u ∈ W 2,p(Ω ∩ BN (0;R)) be a solution of −4u = f(u) in Ω; u = 0 on ∂Ω, (10.1) where f is locally Lipschitz continuous on R, and Ω is a smooth unbounded domain. Let F (t) = ∫ t 0 f(s)ds, and assume that ∇u ∈ L2(Ω) and F (u) ∈ L1(Ω). Then a sequence {Rk} ↗ ∞ exists such that (i) (Pohozaev identity)∫ Ω {NF (u) + (1− N 2 )|∇u|2}dz = lim k→∞ 1 2 ∫ ∂Ω∩BN (0;Rk) (z · ν(z))|∇u|2ds (ii) lim k→∞ ∫ ∂Ω∩BN (0;Rk) νi(z)|∇u|2ds = 0 for each 1 ≤ i ≤ N, where ν(z) = (ν1(z), ν2(z), . . . , νN (z)) is the outward unit normal vector at z. Proof. (i) See Esteban-Lions [33, Proposition I.1] . (ii) Let BR = BN (0;R). Then we multiply (10.1) by ∂u ∂zi and use integration by parts over Ω ∩BR to obtain∫ Ω∩BR (−4u) ∂u ∂zi = ∫ Ω∩BR f(u) ∂u ∂zi = ∫ Ω∩BR ∂F (u) ∂zi = ∫ ∂(Ω∩BR) F (u)νids = ∫ Ω∩∂BR F (u) zi |z| ds. Note that ∇u = ∂u ∂ν ν on ∂Ω and ν = z |z| on ∂BR. We use the Green first identity and integrate by parts to obtain∫ Ω∩BR (−4u) ∂u ∂zi = ∫ Ω∩BR ∇u∇( ∂u ∂zi )− ∫ ∂(Ω∩BR) ∂u ∂ν ∂u ∂zi ds = ∑ j = 1N ∫ Ω∩BR ∂u ∂zj ∂2u ∂zi∂zj − ∫ ∂(Ω∩BR) ∂u ∂ν ∂u ∂zi ds 90 HWAI-CHIUAN WANG EJDE-2004/MON. 06 = ∑ j = 1N 1 2 ∫ Ω∩BR ∂ ∂zi ( ∂u ∂zj )2 − ∫ ∂(Ω∩BR) ∂u ∂ν ∂u ∂zi ds = ∑ j = 1N 1 2 ∫ ∂(Ω∩BR) ( ∂u ∂zj )2νids− ∫ ∂(Ω∩BR) ∂u ∂ν ∂u ∂zi ds = 1 2 ∫ ∂Ω∩BR |∇u|2νids+ 1 2 ∫ Ω∩∂BR |∇u|2 zi |z| ds − ∫ ∂Ω∩BR ( ∂u ∂ν )2νids− ∫ Ω∩∂BR ∂u ∂ν ∂u ∂zi ds = −1 2 ∫ ∂Ω∩BR |∇u|2νids+ 1 2 ∫ Ω∩∂BR |∇u|2 zi |z| ds − ∫ Ω∩∂BR ∂u ∂ν ∂u ∂zi ds. Thus, we have∣∣− 1 2 ∫ ∂Ω∩BR |∇u|2νids ∣∣ = ∣∣ ∫ Ω∩∂BR [ F (u) zi |z| − 1 2 |∇u|2 zi |z| + ∂u ∂ν ∂u ∂zi ] ds ∣∣ ≤ ∫ Ω∩∂BR ( |F (u)|+ 1 2 |∇u|2 + |∇u|2 ) ds. Since F (u) ∈ L1(Ω) and ∇u ∈ L2(Ω), ∞ > ∫ Ω (|F (u)|+ 3 2 |∇u|2)dz = lim R→∞ ∫ Ω∩BR (|F (u)|+ 3 2 |∇u|2)dz = ∫ ∞ 0 rN−1 [∫ Ω∩∂Br (|F (u)|+ 3 2 |∇u|2)ds ] dr = ∫ ∞ 0 rN−1M(r)dr, where M(r) = ∫ Ω∩∂Br (|F (u)|+ 3 2 |∇u| 2)ds. Suppose that for every sequence Rk → ∞ we have M(Rk) 9 0 as k →∞, that is, a subsequence {Rk} and c > 0 exist such thatM(Rk) → c as n→∞. Then for sufficiently large k we obtainM(Rk) ≥ c 2 > 0. Thus, ∫ ∞ 0 rN−1M(r)dr ≥ c 2 ∫ ∞ k rN−1dr = ∞. This is a contradiction. We conclude that Rk →∞ exists such that lim k→∞ ∫ Ω∩∂BRk (|F (u)|+ 3 2 |∇u|2)ds = 0, that is, lim k→∞ ∫ Ω∩∂BRk |∇u|2νids = 0 for 1 ≤ i ≤ N. � Lemma 10.5. Let Lu = aij(z)Diju + bi(z)Diu + c(z)u and u ∈ W 2,p loc (Ω) be a solution of the elliptic equation Lu = f in a domain Ω, where the coefficients of L belong to Ck−1,α(Ω), f ∈ Ck−1,α(Ω) with 1 < p < ∞, k ≥ 1, 0 < α < 1. Then u ∈ Ck+1,α(Ω). EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 91 The proof of the lemma above can be found in Gilbarg-Trudinger [36, Theorem 9.19]. Theorem 10.6. Let Ω be an Esteban-Lions domain in RN with χ as in Definition 2.6, and let f be a locally Lipschitz-continuous functional on R with f(0) = 0. If u ∈ C2(Ω) is a solution of −4u = f(u) in Ω; u = 0 on ∂Ω, and satisfies ∇u ∈ L2(Ω), F (u) ∈ L1(Ω) (with F (t) = ∫ t 0 f(s)ds), then u ≡ 0 in Ω. Proof. By Lemma 10.4 (ii), a sequence Rk →∞ as k →∞ exists such that lim k→∞ ∫ ∂Ω∩BN (0;Rk) (ν(z) · χ)|∇u|2ds = 0. This immediately yields (ν(z) · χ)|∇u| = 0 on ∂Ω. Since Ω is an Esteban-Lions domain, then z0 ∈ ∂Ω exists such that ν(z0) · χ > 0. Thus, δ > 0 exists such that ν(z) · χ > 0 for z ∈ ∂Ω ∩BN (z0; δ). Then we have ∇u = 0 on ∂Ω ∩BN (z0; δ). Let ũ(z) = { u(z) for z ∈ Ω; 0 for z ∈ BN (z0; δ)\Ω. x y Figure 7. Esteban-Lions domain with a small ball. We claim that ũ is twice weakly differentiable. Since u ∈ C2(Ω) and u = 0 on ∂Ω, for each ϕ ∈ C1 c (Ω ∪BN (z0; δ)), we have∫ Ω∪BN (z0;δ) ũ ∂ϕ ∂zi = ∫ Ω u ∂ϕ ∂zi = − ∫ Ω ∂u ∂zi ϕ+ ∫ ∂Ω uϕνids = − ∫ Ω∪BN (z0;δ) viϕ, where vi(z) = { ∂u ∂zi for z ∈ Ω; 0 for z ∈ BN (z0; δ)\Ω. 92 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Thus, vi = ∂ũ ∂zi . Similarly, we have wij = ∂2ũ ∂zi∂zj , where wij(z) = { ∂2u ∂zi∂zj for z ∈ Ω; 0 for z ∈ BN (z0; δ)\Ω. Therefore, ũ ∈W 2,2 loc (Ω). Next, since u ∈ C2(Ω) and by the Green first identity, for each ϕ ∈ C∞c (Ω ∪BN (z0; δ)), we have∫ Ω∪BN (z0;δ) ∇ũ∇ϕ = ∫ Ω ∇u∇ϕ = − ∫ Ω (4u)ϕ+ ∫ ∂Ω ∂u ∂ν ϕds = ∫ Ω f(u)ϕ+ ∫ ∂Ω∩BN (z0;δ) ∂u ∂ν ϕds+ ∫ ∂Ω\BN (z0;δ) ∂u ∂ν ϕds = ∫ Ω∪BN (z0;δ) f(ũ)ϕ. Hence, ũ is a weak solution of −4ũ = f(ũ) in Ω ∪BN (z0; δ); ũ = 0 on ∂(Ω ∪BN (z0; δ)). We claim that f ◦ ũ is locally Hölder continuous with exponent α in Ω∪BN (z0; δ), where 0 < α < 1. That is, for each compact set K ⊂ Ω ∪BN (z0; δ), we must show that a constant C(K) > 0 exists such that |f ◦ ũ(z)− f ◦ ũ(z)| ≤ C(K)|z − z|α (10.2) for all z, z ∈ K. Since f is locally Lipschitz with f(0) = 0, and u ∈ C2(Ω), for z, z ∈ K, we have (i) (10.2) holds for z, z ∈ {BN (z0; δ)\Ω} ∩K ; (ii) if z ∈ Ω and z ∈ {BN (z0; δ)\Ω}∩K, then ẑ ∈ ∂Ω exists such that |z−ẑ| ≤ |z−z|. Thus, |f ◦ ũ(z)− f ◦ ũ(z)| ≤ |f ◦ ũ(z)− f ◦ ũ(ẑ)|+ |f ◦ ũ(ẑ)− f ◦ ũ(z)| ≤ C1(K)|ũ(z)− ũ(ẑ)| ≤ C2(K)|z − ẑ| ≤ C2(K)|z − z| ≤ C2(K)|z − z|1−α|z − z|α ≤ C3(K)|z − z|α; (iii) it is clear that both z and z are in Ω ∩K. By Lemma 10.5, we have ũ ∈ C2(Ω∪BN (z0; δ)). Finally, since f is locally Lipschitz and f(0) = 0, then (4ũ)2 = |f(ũ)|2 ≤ C(K)(ũ(z))2 on each compact subset K of Ω ∪ BN (z0; δ). By Theorem 10.3, ũ ≡ 0 on K, and ũ ≡ 0 on Ω ∪ BN (z0; δ). Otherwise, if there is a z ∈ Ω such that ũ(z) 6= 0, then a bounded domain Ω1 exists such that z ∈ Ω1 ⊂ Ω1 ⊂ Ω ∪ BN (z0; δ), and by the previous argument, ũ ≡ 0 on Ω1, which is a contradiction. Hence, u ≡ 0. � Esteban-Lions [33, Theorem I.1] proved the following result. EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 93 Theorem 10.7. Equation (1.1) in an Esteban-Lions domain Ω does not admit any nontrivial solution. In particular, an Esteban-Lions domain is a nonachieved domain. We have the following lemma. Lemma 10.8. (i) Esteban-Lions domains are invariant under any rigid motions; (ii) If Ω is an Esteban-Lions domain, then Ω c is also an Esteban-Lions domain. Proof. (i) Clearly. (ii) Note that n(x) · χ = −n(x) · −χ. � In fact, in a star-shaped domain Ω ⊂ RN , there is a fixed point z0 ∈ Ω such that the segment zz0 is contained in Ω for each z ∈ Ω. We assert that an Esteban-Lions domain Ω is an infinite star-shaped domain in the sense that it is v−convex for some direction v: if z1, z2 ∈ Ω with z1− z2 = tv for some t ∈ R, then the segment z1z2 is contained in Ω. This is a consequence of the following lemma. Lemma 10.9. An Esteban-Lions domain Ω with χ as in Definition 2.6 in RN is χ−convex. Proof. Let Ω be an Esteban-Lions domain in RN . Without loss of generality, we may assume that χ = (0, . . . , 0,−1) ∈ RN satisfying n(z) · χ ≥ 0, and n(z) · χ 6≡ 0 for each z ∈ ∂Ω. For each z1 ∈ Ω, we claim that {z1 − λχ|λ ≥ 0} ⊂ Ω. Otherwise, set λ0 = inf{λ ≥ 0|z1 − λχ 6∈ Ω}. Then λ0 > 0 and the point z0 = z1 − λ0χ ∈ ∂Ω. There are only two possibilities: (i) the curve ∂Ω is transverse with the xN−axis at z0, and then n(z0) · χ < 0; (ii) the curve ∂Ω is tangent with the xN−axis at z0, and then z ∈ ∂Ω near z0 exists such that n(z) · χ < 0. Both (i) and (ii) contradict the definition of χ. We conclude that an Esteban-Lions domain Ω in RN is χ−convex. � In R2, let an upper semi-strip Ω1, a first quadrant Ω2, a second quadrant Ω3, and an upper half plane Ω4 be defined as follows: Ω1 = {z = (x, y) ∈ R2 : a < x < b, f1(x) < y}; Ω2 = {z = (x, y) ∈ R2 : a < x <∞, f2(x) < y}; Ω3 = {z = (x, y) ∈ R2 : −∞ < x < b, f3(x) < y}; Ω4 = {z = (x, y) ∈ R2 : −∞ < x <∞, f4(x) < y}, where f1 : (a, b) → R, f2 : (a,∞) → R, f3 : (−∞, b) → R, and f4 : (−∞,∞) → R are smooth functions with single values. Figure 8. Esteban-Lions domains 1. As a consequence of Lemma 10.9, we have the following two lemmas. 94 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Figure 9. Esteban-Lions domains 2. Lemma 10.10. An Esteban-Lions domain in R2 must be one of the following types: an upper half strip Ω1, a first quadrant Ω2, a second quadrant Ω3, or an upper half plane Ω4. Lemma 10.11. An Esteban-Lions domain Ω with χ as in Definition 2.6 in RN is a large domain if and only if the projection Ω̃ of Ω in the hyperplane RN−1 that is perpendicular to χ, is a large domain in RN−1. Proof. Suppose that Ω is a large domain in RN . Then for r > 0, x ∈ Ω exists such that BN (x, r) ⊂ Ω which implies B̃N (x, r) ⊂ Ω̃, where B̃N (x, r) and Ω̃ are the projections of BN (x, r) and Ω, respectively. On the other hand, suppose Ω̃ is a large domain in RN−1. Then for r > 0, x ∈ Ω exists such that B̃N (x, r) ⊂ Ω̃, which means that for any ỹ ∈ B̃N (x, r), y ∈ Ω exists and ỹ is the projection of y. By Lemma 10.9, λ > 0 exists such that {ỹ − χt : t ≥ λ} ⊂ Ω. Let λ̄ = inf ỹ ∈ B̃N (x, r){λ > 0 : ỹ − χt ∈ Ω for t ≥ λ}, then BN (z, r) ⊂ Ω, where z = x̃ − (r + 1 + λ̄)χ. Thus, Ω is a large domain in RN . � In R3, there is an Esteban-Lions domain that is not a large domain. Example 10.12. Let Ω be a domain in R3 such that Ω contains the point (0, 0, 1) with the boundary ∂Ω = {((u+ 2) cos v, sin v, u) ∈ R3 | 0 ≤ v ≤ 2π, u ≥ 0}. Then Ω is an Esteban-Lions domain in R3 with χ = (0, 0,−1), but it is not a large domain in either R3 or Ar. x y z Figure 10. Esteban-Lion domain but not a large domain. EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 95 Proof. Let p(u, v) = ((u+ 2) cos v, sin v, u). Then pv = (−(u+ 2) sin v, cos v, 0) and pu = (cos v, 0, 1). For x ∈ ∂Ω, we have the outer normal vector n(x) = pv × pu = (cos v, (u + 2) sin v,− cos2 v), and χ · n(x) = cos2 v ≥ 0. Therefore, we have χ · n(x) ≥ 0. Clearly, B(x; 2) * Ω for any x ∈ Ω, and thus, Ω is not a large domain in R3. � Let Fr s = Ar 0 ∪BN (0; s) be an interior flask domains. Interior flask domains are achieved for large s, but are nonachieved for small s. By Theorem 12.7 below, we have the following result. Theorem 10.13. s0 > 0 exists such that the interior flask domains Fr s nonachieved if s < s0. Bibliographical notes: Theorem 10.6 is from Esteban-Lions [33]. 11. Higher Energy Solutions Nonachieved domains may admit higher energy solutions. The Berestycki con- jecture states: there is a positive solution of Equation (1.1) in an Esteban-Lions domain with a hole. We answer this conjecture affirmatively. The Berestycki conjecture is based on some historical and physical reasons: sup- pose that no solution of an equation exists in a domain Θ. If we break the symmetry of the domain Θ by adding a ball to it or by removing a ball, then the same equa- tion in the perturbed domain admits a solution. (i) Pohozaev [61] proved that the Dirichlet problem ∆u+ u N+2 N−2 = 0 in a ball does not have any nontrivial solution. However, if we remove a small ball, then Coron [28] proved that there is a positive solution. (ii) Some turbulence equations in a ball do not admit any nontrivial solutions. Lions-Zuazua [48] added a small ball on the boundary to break the symmetry, and proved that the equation then has a nontrivial solution. One description of the phenomenon is that if we add a small bump to the surface of a plane, then the turbulence will be controlled. We assert the existence of higher energy solutions of Equation (1.1) in Ω, thus answering the Berestycki conjecture affirmatively. Then we study the dynamic systems of solutions. Our results in this paper are still true for any one of the above known four Esteban-Lions domains. For simplicity, however,we study only the upper half strip with a hole Ω. We also believe that the analyses and the results in this paper will be helpful for studying the existence of solutions of equations in unbounded domains. 11.1. Existence Results. For h > r and B =BN ((0, h); r/2), let Ω = Ωh = (Ar 0 ∪ BN (0; r))\B be the upper half strip with a hole. By Theorem 10.1, there are no ground state solutions of Equation (1.1) in Ω. However, in this section, we prove that a positive higher energy solution of Equation (1.1) exists in Ω. Let u be a ground state solution of Equation (1.1) in Ar, h = (0, h) ∈ Ar and φ : Ar→ [0, 1], a C∞ cut-off function such that 0 ≤ φ ≤ 1 and φ(z) =  0 textforz ∈ B ∪ (Ar\(Ar 0 ∪BN (0; r))), 1 for z ∈ (Ar 0 ∪BN (0; r))\(BN (0; 2 3r) ∪{z = (x, y) ∈ Ar|y ≤ r}), 96 HWAI-CHIUAN WANG EJDE-2004/MON. 06 I = {0} × [−r 2 , r 2 ], Ih = h+ I, vt(z) = φ(z)ū(z − t− 2h) for z ∈ Ar, t ∈ I. Then vt ∈ H1 0 (Ω). Furthermore, we have the following lemma. Lemma 11.1. For t ∈ I or t ∈ Ih, where t = h+ t, then (i) ‖vt(z)− ū(z − t− 2h)‖Lp(Ar) = o(1) as h→∞; (ii) ‖vt(z)− ū(z − t− 2h)‖H1(Ar) = o(1) as h→∞; (iii) J(vt) = α(Ar) + o(1) as h→∞. Proof. (i) ‖vt(z)− ū(z − t− 2h)‖p Lp(Ar) = ∫ Ar |φ(z)− 1|p|ū(z − t− 2h)|p ≤ ∫ (Ar h+r)c |ū(z − t− 2h)|p = o(1) as h→∞. (ii) We have ‖vt(z)− ū(z − t− 2h)‖2H1(Ar) = ‖(φ(z)− 1)ū(z − t− 2h)‖2H1(Ar) ≤ c( 1 r2 + 1) ∫ (Ar h+r)c (|∇ū(z − t− 2h)|2 + |ū(z − t− 2h)|2) = o(1) as h→∞. (iii) This follows from (i), (ii), Theorem 12.5, and α(Ar) = J(u) = 1 2 a(ū)− 1 p b(ū) = 1 2 a(vt)− 1 p b(vt) + o(1) = J(vt) + o(1). � From Lemma 11.1, since ‖u‖2H1(Ar) = ‖u‖p Lp(Ar), we have ‖vt‖2H1(Ar) = ‖u‖2H1(Ar) + o(1) as h→∞, ‖vt‖p Lp(Ar) = ‖u‖p Lp(Ar) + o(1) as h→∞. Therefore, ‖vt‖2H1(Ar) = ‖vt‖p Lp(Ar) + o(1) as h→∞. By Lemma 4.2, there exists λt > 0 such that ut = λtvt in M: ‖ut‖2H1(Ar) = ‖ut‖p Lp(Ar). Therefore, λt → 1 as h→∞, or J(ut) = α(Ar) + o(1) as h→∞. For u ∈ H1 0 (Ω), define the center of mass function by j(u) = ‖u‖−p Lp(Ar) ∫ Ar (h+ r 2 z |z| )|u(x, y)|pdxdy. Let β0 = inf{J(u) : u ∈ M(Ω), u ≥ 0, j(u) = h}. Proposition 11.2. α(Ar) = α(Ω) < β0. EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 97 Proof. By Theorem 10.1, α(Ar) = α(Ω). Clearly α(Ar) ≤ β0. Suppose α(Ar) = β0. By Theorem 4.4, there is a sequence {uk} in M(Ω), uk ≥ 0, j(uk) = h for each k, such that J(uk) = α(Ar) + o(1) as k →∞, J ′(uk) = o(1) strongly in H−1(Ω) as k →∞. By Theorem 3.5, there is an unbounded sequence {(0, yk)} in Ar such that uk(x, y) = u(x, y − yk) + o(1) strongly in H1 0 (Ar), where u is a ground state solution of Equation (1.1) in Ar. Assume (h+ r 2 (0,yk) |(0,yk)| ) = ς + o(1) as k →∞, where ς ∈ ∂Ih. Then by the Lebesgue Dominated Convergence Theorem, we have h = j(uk) = ‖uk‖−p Lp(Ar) ∫ Ar (h+ r 2 z |z| )|uk(x, y)|pdxdy = ‖u‖−p Lp(Ar) ∫ Ar (h+ r 2 (x, y + yk) |(x, y + yk)| )|u(x, y)|pdxdy + o(1) = ς + o(1) as k →∞, which is a contradiction. Therefore α(Ar) = α(Ω) < β0. � Let V = {u ∈ M(Ω) : u ≥ 0}; Γ = {k : Ih → V continuous : k(t) = ut for t ∈ ∂I}; β1 = inf k∈Γ max t∈Ih J(k(t)). Proposition 11.3. There is an h0 > 0 such that for h ≥ h0, (i) α(Ar) < J(ut) < β0+α(Ar) 2 < β0, for t ∈ I; (ii) α(Ar) < J(ut) < 2 p−2 p α(Ar), for t ∈ I; (iii) (j ◦ ut, t) > 0, for t ∈ ∂I. Proof. (i) and (ii) follow from Theorem 10.1, Proposition 11.2, and Lemma 11.1. (iii) c1, c2 > 0 exist such that c1 ≤ ‖φ(z)ū(z− t−2h)‖Lp(Ar) ≤ c2. For t ∈ ∂I with z + t+ 2h 6= 0, we have ( z + t+ 2h |z + t+ 2h| , t) = |z + t+ 2h| − 1 |z + t+ 2h| (z + t+ 2h, z + 2h) ≥ |z + t+ 2h| − |z + 2h| ≥ |t| − 2|z + 2h| = r 2 − 2|z + 2h|. Then there are constants c3, c4, c5, c6, h0 > 0 such that for h ≥ h0 (j(ut), h+ t) = ‖ut(z)‖−p Lp(Ar) ∫ Ar (h+ r 2 z |z| , h+ t)|ut(z)|pdz = c3 ∫ Ar (h+ r 2 z |z| , h+ t)|φ(z)ū(z − t− 2h)|pdz = c3 ∫ Ar (h+ r 2 z + t+ 2h |z + t+ 2h| , h+ t)|φ(z + t+ 2h)ū(z)|pdz ≥ c3(h2c5 − hc4c5 − hc5 − 2c6 − 4hc5) > 0 since h ≥ h0, 98 HWAI-CHIUAN WANG EJDE-2004/MON. 06 where ∫ Ar |φ(z + t + 2h)ū(z)|pdz ≥ c5 and ∫ Ar |z‖φ(z + t + 2h)ū(z)|pdz ≥ c6. By Theorem 8.13, c4 <∞. � Proposition 11.4. For h ≥ h0, we have α(Ar) < β0 = β1 < 2 p−2 p α(Ar). Proof. We claim that: (i) β0 = β1 : for any k ∈ Γ, consider the homotopy H(λ, t) : [0, 1]× Ih → RN defined by H(λ, t) = (1− λ)j(k(t)) + λi(t), where i denotes the identity map. Note that j(k(t)) = j(ut) for t ∈ ∂Ih. By Proposition 11.3 (iii), H(λ, t) 6= h for t ∈ ∂Ih and λ ∈ [0, 1]. Therefore deg(j ◦ k, Ih, h) = deg(i, Ih, h) = 1. t0 ∈ Ih exists such that j(k(t0)) = h. Hence, for each k ∈ Γ, β0 = inf{J(u) : u ∈ M, u ≥ 0, j(u) = h} ≤ J(k(t0)) ≤ max t∈Ih J(k(t)). We have β0 ≤ β1. On the other hand, by Proposition 11.3 (i), for t ∈ I, we have ut ∈ V and J(ut) < β0. Thus, maxt∈I J(ut) ≤ β0, or β1 ≤ β0. (ii) β1 < 2 p−2 p α(Ar) : by Proposition 11.3 (ii), J(ut) < 2 p−2 p α(Ar) for t ∈ I. Thus max t∈I J(ut) < 2 p−2 p α(Ar). We have β1 < 2 p−2 p α(Ar). By Proposition 11.3 (i), we have α(Ar) < β0 = β1 < 2 p−2 p α(Ar). � Now we assert that there is a higher energy solution of Equation (1.1) in Ω. Theorem 11.5. Suppose that the positive solution of Equation (1.1) in the infinite strip Ar is unique up to y-translations. h0 > 0 exists such that if h ≥ h0, then there is a positive higher energy solution v of Equation (1.1) in the upper half strip with a hole Ω such that α(Ar) < J(v) < 2 p−2 p α(Ar). Proof. Note that β0 = inf{J(u) : u ∈ M, u ≥ 0, j(u) = h}. Take a minimizing sequence {uk} in M: J(uk) → β0 as k → ∞. By Theorem 4.4, {uk} is a (PS)β0- sequence for J : J(uk) → β0 and J ′(uk) → 0 as k → ∞. By Theorem 3.5, an integer ` ≥ 0, and sequences {zi k}, where zi k = (0, yi k) ∈ Ar for 1 ≤ i ≤ ` exist, such that for some subsequence {uk}, there are u0 ∈ H1 0 (Ω), u0 ≥ 0 in Ω, ui ∈ H1 0 (Ar), and ui > 0 in Ar, 1 ≤ i ≤ `, satisfying uk(z) = u0(z) + [u1(z − z1 k) + u2(z − z2 k) + . . . + u`(z − z` k)] + o(1) strongly inH1 0 (Ar), −4u0 + u0 = (u0)p−1 in Ω, −4ui + ui = (ui)p−1 in Ar, 1 ≤ i ≤ `, EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 99 J(uk) = J(u0) + ∑̀ i=1 J(ui) + o(1) as k →∞. Suppose that the solution of (1.1) in the infinite strip Ar is unique up to y- translations. From Theorems 3.5 and 9.4, we find that ui are the same and J(ui) = α(Ar) for i = 1, 2, . . . , l. Therefore β0 = J(u0) + lα(Ar). Since α(Ar) < β0 < 2 p−2 p α(Ar). We conclude that u0 is nonzero and l = 0. Thus, there is a positive higher energy solution v = u0 of Equation (1.1) in the upper half strip with a hole Ω such that α(Ar) < J(v) = β0 < 2 p−2 p α(Ar). � 11.2. Dynamic Systems of Solutions. As in Subsection 1, for k = 1, 2, . . . , define Ωk = (Ar 0 ∪BN (0; r)) \BN ((0, h); 1 k ), where h ≥ 2h0, 1 k < r. Then Ωk is an increasing sequence such that (Ar 0 ∪BN (0; r)) \ {0} = ∪∞k=1Ωk. By Theorem 11.5, we have, for each k, a positive solution uk ∈ H1 0 (Ωk) of −∆uk + uk = up−1 k in Ωk satisfying α(Ar) < J(uk) < 2 p−2 p α(Ar) . Lemma 11.6. If uk ⇀ u weakly in H1 0 ((Ar 0 ∪BN (0; r))) as k →∞, then u ≡ 0. Proof. For ϕ ∈ C∞0 ((Ar 0 ∪BN (0; r))), we have∫ (Ar 0∪BN (0;r)) uk(−∆ϕ+ ϕ) = ∫ (Ar 0∪BN (0;r)) (−∆uk + uk)ϕ = ∫ (Ar 0∪BN (0;r)) up−1 k ϕ. Let k →∞, and we obtain∫ (Ar 0∪BN (0;r)) u(−∆ϕ+ ϕ) = ∫ (Ar 0∪BN (0;r)) up−1ϕ. Thus, −∆u + u = up−1 in (Ar 0 ∪ BN (0; r)). By Theorems 3.5 and 9.4, u ≡ 0, or uk ⇀ 0 weakly in H1 0 ((Ar 0 ∪BN (0; r))) as k →∞. � We have the following dynamic systems of solutions {uk}: Theorem 11.7. |∇uk|2dz = cδ0 + o(1) for some positive number c. Proof. Let uk ⇀ u weakly in H1 0 ((Ar 0∪BN (0; r))) as k →∞, µk = |∇uk|2dz = µ+ o(1) weak∗, and νk = |uk|pdz = ν + o(1) weak∗. Then by the second concentration lemma (see Lions [50, Lemma I.1, p.24]), there exist {aj}∞j=1, {bj}∞j=1 in R+ such that mp/(p−2) + o(1) = ‖uk‖2H1(Ωk) = ∫ (Ar 0∪BN (0;r)) dµk = ∫ (Ar 0∪BN (0;r)) dµ+ o(1) ≥ ‖u‖2H1(Ωk) + ∑ j aj + o(1) 100 HWAI-CHIUAN WANG EJDE-2004/MON. 06 ≥ m(‖u‖2Lp + ∑ j bj 2 p ) + o(1) ≥ m(‖u‖p Lp + ∑ j bj) 2 p + o(1) = m (∫ (Ar 0∪BN (0;r)) dν )2/p + o(1) = mp/(p−2) + o(1). By Lemma 11.6, u = 0. Thus, only one of the aj is different from 0, say a1 = c > 0, aj = 0, j = 2, 3, . . . . Thus, |∇uk|2dz = cδz1 + o(1). Clearly, z1 = 0. � Similarly, let {wk} be the solutions of Equation (1.1) in the interior flask domains Fr s, where s > s0. Then we have dynamic systems of {wk} as follows: Theorem 11.8. Let w be a ground state solution of Equation (1.1) in RN . Then wk → w strongly in H1(RN ) as k →∞. Proof. Note that J(wk) = α(RN ) + o(1), J ′(wk) = o(1) as k →∞. By Theorem 3.1, we prove that an integer ` ≥ 0 and sequences {zi k} ⊂ RN for 0 ≤ i ≤ ` exist such that for some subsequence {wk}, there are wi ∈ H1(RN ), wi > 0 in RN for 0 ≤ i ≤ `, satisfying wk(z) = w0(z) + [w1(z − z1 k) + w2(z − z2 k) + . . . + w`(z − z` k)] + o(1) strongly inH1(RN ), −4wi + wi = (wi)p−1 textinRN , 0 ≤ i ≤ `, J(wk) = ∑̀ i=0 J(wi) + o(1) as k →∞. Then, since J(wk) = α(RN ) + o(1), we conclude that wk(z) = w(z) + o(1) strongly in H1(RN ). � Bibliographical notes: The results of this section are from Wang [71]. 12. Achieved Domains In this section we assert that the bounded domains, the quasibounded domains, the periodic domains, some interior flask domains, some flat interior flask domains, some canal domains, and some manger domains are achieved. We begin with the following lemma. Lemma 12.1. γ(Ω) is achieved if and only if α(Ω) is achieved. Proof. Recall that α(Ω) = ( 1 2 − 1 p )γ(Ω) 2p 2−p . Suppose that there is a u ∈ H1 0 (Ω) such that J(u) = α(Ω), 〈J ′(u), u〉 = a(u)− b(u) = 0. Then we have a(u)( 1 p− 1 2 ) = γ(Ω). Let v = u ‖u‖H1 . Then ‖v‖Lp = b(u)1/p a(u)1/2 = a(u) 1 p− 1 2 = γ(Ω). EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 101 Thus, γ(Ω) is achieved by v. On the other hand, let γ(Ω) be achieved by some function u where a(u) = ‖u‖2H1 = 1 and b(u) = ‖u‖p Lp = γ(Ω)p. By the Lagrange multiplier theorem there is a λ such that b′(u) = λa′(u). It is easy to see that λ = p 2γ(Ω)p, so we have b′(u) = p 2 γ(Ω)pa′(u). This implies γ(Ω)−p( ∫ |u|p−2uϕ) = ( ∫ ∇u∇ϕ+ uϕ). Thus, u is a weak solution of −∆u+ u = γ(Ω)−p|u|p−2u. Let v = γ(Ω) p 2−pu. Then −∆v + v = |v|p−2v. We have a(v) = b(v) = γ(Ω) 2p 2−p , 〈J ′(v), ϕ〉 = 0 for each ϕ ∈ C∞c (Ω), and J(v) = ( 1 2 − 1 p )γ(Ω) 2p 2−p = α(Ω). � Remark 12.2. Note that if u is a ground state solution for J in Ω, then u solves the semilinear elliptic (1.1) and J(u) = α(Ω). By the Kato regularity, Lp−regularity and Schauder regularity, the ground state solution u of (1.1) is classical. Theorem 12.3. A bounded domain Ω is an achieved domain. For the proof of this theorem follows from Theorem 5.1. An unbounded domain may be achieved. Theorem 12.4. A C1 quasibounded domain is achieved. The statment of this theorem follows from Theorem 5.4. A periodic domain in RN is achieved. In Theorem 9.1, we proved that if a (PS)α- sequence for J admits a nonzero weak limit u, then u is a ground state solution for J . However, even though the weak limit is zero we can still obtain a ground state solution for J if the domain is periodic. Theorem 12.5. A periodic domain in RN is achieved. In particular, there is a ground state solution of Equation (1.1) in Ar, Ar1,r2 , and RN . Proof. It suffices to prove the case Ω = Ar. Let {un} be a (PS)α(Ar)-sequence such that J(un) = α(Ar) + o(1), J ′(un) = o(1). By Lemma 2.38, there are a subsequence {un} and a u ∈ H1 0 (Ar) such that un ⇀ u weakly in H1 0 (Ar). Suppose that u is nonzero, then by Theorem 5.6, we are done. Suppose that un ⇀ 0 weakly in H1 0 (Ar). Since α(Ω) is positive, we have un 9 0 strongly in H1 0 (Ω). By Lemma 2.16, there is a subsequence {un}, and a constant α > 0 such that for n = 1, 2, . . . , Qn = sup y∈R ∫ (0,y)+Ar −2,2 |un(z)|2dz > α > 0. 102 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Take {zn} in Ar, where zn = (0, yn) such that ∫ zn+Ar −2,2 |un(z)|2dz ≥ α/2, and let wn(z) = un(z + zn). Then for n = 1, 2, . . . ,∫ Ar −2,2 |wn(z)|2dz = ∫ zn+Ar −2,2 |un(z)|2dz ≥ α/2, ‖wn‖H1(Ar) = ‖un‖H1(Ar) ≤ c, so w ∈ H1 0 (Ar) exists such that wn ⇀ w weakly in H1 0 (Ar). Clearly, {wn} is a (PS)-sequence in H1 0 (Ar) for J . By Theorem 2.31,∫ Ar −2,2 |w|2 = lim n→∞ ∫ Ar −2,2 |wn|2 ≥ α/2, so w 6≡ 0. By Theorem 5.6, there is a ground state solution of Equation (1.1) in H1 0 (Ω). � Moreover, R is an achieved domain: there is a classical solution u of the Equation u′′ = u− γ(R)−1|u|p−2u. (12.1) By Berestycki-Lions [13], such a solution is unique. The solution can be constructed as follows by routine computations. Theorem 12.6. With µ = 2/(p− 2), we have u(r) = (pγ(R) 2 )µ 2 {cosh (r/µ)}−µ; γ(R) = [ (2µ+ 1)Γ(2µ) µΓ(µ)2 ]µ−1 ( µ 4 )(µ+ 1)− p 2 , to solve Equation (12.1). In particular, R is an achieved domain. Next we present achieved domains from the perturbations of nonachieved do- mains. By Theorem 10.7, the upper half strip Ar 0 and the upper half hollow strip Ar1,r2 0 are nonachieved. However, the perturbed domains of Ar 0 and Ar1,r2 0 may be achieved. Let Fr s = Ar 0 ∪ BN (0; s) be an interior flask domain. Interior flask domains are achieved for large s, but are nonachieved for small s. Theorem 12.7. s0 > 0 exists such that Equation (1.1) has a ground state solution in Fr s if s > s0, but does not have any ground state solution if s < s0. In particular, the interior flask domains Fr s are achieved if s > s0, while Fr s are nonachieved if s < s0. Proof. By Theorem 12.5, the infinite strip Ar admits a ground state solution. Then by Theorem 5.7 (ii), α(Ar) > α(RN ). By Theorem 4.18 (ii), we have α(Ar) = α(Ar 0) and by Theorem 4.27, lims→∞ α(BN (0; s)) = α(RN ). Take s large enough so that α(BN (0; s)) < α(Ar) = α(Ar 0). By Theorem 5.1, there is a ground state solution of Equation (1.1) in BN (0; s). Then by Theorem 5.7 (ii), we have α(Fr s) < α(BN (0; s)). We conclude that α(Fr s) < α(BN (0; s)) < α(Ar 0), EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 103 or α(Fr s) < min{α(BN (0; s)), α(Ar 0)}. By the equivalence of (i) and (vi) in Theorem 5.12, Equation (1.1) has a ground state solution in Fr s for large s. If Equation (1.1) has a ground state solution in Fr s1 and s1 < s2, then Fr s2 = Fr s1 ∪ BN (0; s2). By Theorem 5.1 and Theorem 5.7 (ii), α(Fr s2 ) < α(BN (0; s2)) and α(Fr s2 ) < α(Fr s1 ). By the equivalence of (i) and (vi) in Theorem 5.12, Equation (1.1) has a ground state solution in Fr s2 . Let s0 = inf{s > r : Equation (1.1) has a ground state solution in Fr s}. We then conclude that Equation (1.1) has a ground state solution in Fr s if s > s0, and Equation (1.1) does not have any ground state solution in Fr s if s < s0. � Remark 12.8. In Theorem 12.7, we have asserted that the interior flask domains Fr s = Ar 0 ∪BN (0; s) are achieved if s > s0. In fact, if we replace Ar 0 ∪BN (0; s) by Ar 0 ∪Ω, where Ω is a bounded domain containing BN (0; s), the theorem still holds. For δ > 0, there is a ε(δ) > 0 such that a flat interior flask domain Ωε is an achieved domain, where Eε = {(x, y) ∈ RN : (x, εy) ∈ B(0; r + δ)}; Ωε = Ar 0 ∪ Eε. Theorem 12.9. Given δ > 0, ε0 > 0 exists such that if ε ≤ ε0, then the flat interior flask domain Ωε is an achieved domain. Proof. By Theorem 12.5, the infinite strip Ar admits a ground state solution. Since Ar ( Ar+δ, by Theorem 9.4 we have α(Ar+δ) < α(Ar). Since Eε ⊂ Ar+δ and limε→0 α(Eε) = α(Ar+δ), there is an ε0 > 0 such that if ε ≤ ε0, then α(Eε) < α(Ar). If ε, ε ≤ ε0 is fixed, a large N ∈ N exists such that α((Ω̃ε)N ) = α(Ar N ) = α(Ar). Thus, α(Ωε) ≤ α(Eε) < α(Ar) = α((Ω̃ε)N ). By Theorem 9.5, a ground state solution u of (1.1) exists. By Theorem 12.1, Ωε is an achieved domain. � Fix a number 1 5 l 5 N − 1 and write RN = Rl × RN−l, so that a generic z ∈ RN is written as z = (x, y) with x ∈ Rl and y ∈ RN−l. Let Ω be a domain in RN . For y ∈ RN−l, we denote by Ωy ⊂ Rl the y-section of Ω, that is, Ωy = {x ∈ Rl| (x, y) ∈ Ω}. We consider the following canal properties: (Ω1) Ω is a smooth domain in RN and the sections Ωy are contained in a bounded set for each y ∈ RN−l; (Ω2) there is a smooth domain O in Rl such that O ⊂ Ωy for each y ∈ RN−l; (Ω3) for each δ > 0 there is an M > 0 such that Ωy ⊂ {x ∈ Rl|dist(x,O) < δ} for each |y| = M. Theorem 12.10. Assume that Ω satisfies (Ω1), (Ω2) and (Ω3). Then Equation (1.1) admits a ground state solution in the canal domain Ω. 104 HWAI-CHIUAN WANG EJDE-2004/MON. 06 R R M O l N-l -M Figure 11. perturbed infinite strip domain. Proof. (i) N − l = 1 : (Ω2) gives Ô = O × RN−l $ Ω. By Theorem 12.5, there is a ground state solution of (1.1) in Ô. By Theorem 5.7 (ii), we have α(Ω) < α(Ô). (12.2) Let Ω+ = {z = (x, y) ∈ Ω : y > −1}, Ω− = {z = (x, y) ∈ Ω : y < 1}, Ô+ = {z = (x, y) ∈ Ô : y > −1}, Ô− = {z = (x, y) ∈ Ô : y < 1}. Then Ω = Ω+ ∪ Ω and−Ô = Ô+ ∪ Ô. Moreover, both Ω+ ∩ Ω− and T, Ô+ ∩ Ô are bounded. Since Ô+ ⊂ Ω+, then α(Ω+) ≤ α(Ô+) = α(Ô). (a) Suppose that α(Ω+) = α(Ô+) = α(Ô). By (12.2), we have α(Ω) < α(Ω+). (b) Suppose that α(Ω+) < α(Ô+) = α(Ô). By Theorem 4.30, lim δ→1 α(δÔ) = α(Ô), and a δ0 > 1 exists such that α(Ω+) < α(δ0Ô). From (Ω3), there is n0 > 0 such that Ω+\BN (0;n0) ⊂ δ0Ô. Thus, α(δ0Ô) ≤ α(Ω+\BN (0;n)) for n ≥ n0. Therefore, α(Ω+) < α(Ω+\BN (0;n)) for n ≥ n0. From the proof of Theorem 5.12, if we EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 105 assume that (v) holds for each n ≥ n0, then we obtain (i). Thus J satisfies the (PS)α(Ω+)-condition. By Theorem 5.7 (i), α(Ω) < α(Ω+). From Case (i) and Case (ii), we conclude that α(Ω) < α(Ω+). Similarly, we have α(Ω) < α(Ω−). Finally, we have α(Ω) < min{α(Ω+), α(Ω−)}. By the equivalence of (i) and (vi) in Theorem 5.12, J satisfies the (PS)α(Ω)- condition. By Theorem 5.6 (iii), (1.1) admits a ground state solution in Ω. (ii) N − l ≥ 2 : (Ω2) gives Ô = O × RN−l $ Ω, and by Theorem 12.5 and 5.7 (ii), we have α(Ω) < α(Ô). By Lemma 4.30, lim γ→1 α(γÔ) = α(Ô). Thus, take a γ close to 1 such that α(Ω) < α(γÔ). By (Ω3), there is an n0 > 0 such that Ω\BN (0;n) ⊂ γÔ for n ≥ n0. Hence, α(γÔ) ≤ α̃n for n ≥ n0. Thus α(Ω) < α̃n for each n ≥ n0. The result follows from the proof of (v) to (i) in Theorem 5.12. � Assume z = (x, y) ∈ RN−1 × R for N ≥ 3. For r2 > r1 > 0 and t > 0, we consider a manger domain Dt = Ar1,r2 0 ∪Ar2 0,t. We have the following result. Theorem 12.11. t̃ ≥ 0 exists such that Equation (1.1) admits a ground state solution in the manger domain Dt if t > t̃, and does not admit any ground state solution in Dt if t < t̃. Proof. (i) We claim that there is a t0 > 0 such that Equation (1.1) admits a ground state solution in Dt if t ≥ t0. Method (I): let {un} be a (PS)α(Dt)-sequence in H1 0 (Dt) for J . By Lemma 5.10 (iii), we have a∞ = b∞ and J∞ ≤ α(Dt), where J∞ = p−2 2p b∞. We claim that there is a t0 > 0 such that J∞ < α(Dt) for t ≥ t0. On the contrary, suppose that J∞ = α(Dtn) for a sequence {tn} such that tn →∞ as t→∞. Then J∞ = α(Dt) for each t ≥ t1. Let ξ(z) be as in (2.1) and ξR(z) = ξ( 2|z| R ). Then there is an R0 > 0 such that ξRun ∈ H1 0 (Ar1,r2 0 ) for R ≥ R0. Let λn = ( a(ξRun) b(ξRun) )1/(p−2). Then we have a(λnξRun) = b(λnξRun). For R ≥ R0, we have J(λnξRun) ≥ α(Ar1,r2 0 ), or ( 1 2 − 1 p ) a(ξRun)p/(p−2) b(ξRun)2/(p−2) ≥ α(Ar1,r2 0 ). Letting R→∞ and n→∞ and using a∞ = b∞, we obtain α(Dt) = J∞ = p− 2 2p b∞ ≥ α(Ar1,r2 0 ). Thus we have α(Dt) ≥ α(Ar1,r2 0 ) for each t > t1. (12.3) 106 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Since Ar2 0 and Ar1,r2 0 are a large domain of Ar2 and Ar1,r2 , respectively, then by Theorem 4.18, we have α(Ar2 0 ) = α(Ar2) and α(Ar1,r2 0 ) = α(Ar1,r2). Moreover, by Theorem 12.5, α(Ar2) < α(Ar1,r2). We conclude that α(Ar2 0 ) < α(Ar1,r2 0 ). (12.4) By Theorem 4.25, we have α((Ar2 0 )n) = α(Ar2 0 ) + o(1). Hence, there is an n0 > 0 such that α(Ar2 0 ) < α((Ar2 0 )n) < α(Ar1,r2 0 ) for n ≥ n0. Then we have α(Dn) ≤ α((Ar2 0 )n) < α(Ar1,r2 0 ) for n ≥ n0. This contradicts (12.3). Thus, there is a t0 > 0 such that J∞ < α(Dt) for t ≥ t0. By the equivalence of (i) and (vii) in Theorem 5.12, Equation (1.1) admits a ground state solution in Dt for t ≥ t0. Method (II): Since Ar2 0 and Ar1,r2 0 are large domains of Ar2 and Ar1,r2 , respec- tively, then by Theorem 4.18, we have α(Ar2 0 ) = α(Ar2) and α(Ar1,r2 0 ) = α(Ar1,r2). Moreover, by Theorem 12.5, α(Ar2) < α(Ar1,r2). We conclude that α(Ar2 0 ) < α(Ar1,r2 0 ). By Theorem 4.25, we have α((Ar2 0 )n) = α(Ar2 0 ) + o(1). Hence, there is an n0 > 0 such that α(Ar2 0 ) < α((Ar2 0 )n) < α(Ar1,r2 0 ) for n ≥ n0. Then we have α(Dn) < α((Ar2 0 )n) < α(Ar1,r2 0 ) for n ≥ n0. By the equivalence of (i) and (vi) in Theorem 5.12, Equation (1.1) admits a ground state solution in Dn for n ≥ n0. (ii) If Equation (1.1) admits a ground state solution in Dt2 and t2 < t3, then Dt3 = Dt2 ∪Ar2 0,t3 . By Theorem 5.1 and Theorem 5.6 (ii), α(Dt3) < α(Ar2 0,t3 ) and α(Dt3) < α(Dt2). By the equivalence of (i) and (vi) in Theorem 5.12, Equation (1.1) admits a ground state solution in Dt3 . Let t̃ = inf{t > 0 : Equation (1.1) has a ground state solution in Dt}. Then t̃ ≥ 0 such that Equation (1.1) admits a ground state solution in Dt if t > t̃ and does not admit any ground state solution in Dt if t < t̃. � 12.1. Open Question: in Theorem 12.7, is s0 = r? Bibliographical notes: Theorem 12.5 is from Lien-Tzeng-Wang [47]. Theorem 12.7 is from Chen-Lee-Wang [24] and Chen-Wang [26]. Theorem 12.7 is from Lien- Tzeng-Wang [47], Chen- Lee-Wang [24, Lemma 19], and Chen-Wang [26, Propo- sition 2.10]. Theorem 12.10 is from del Pino-Felmer [31]. Theorem 12.11 is from Chabrowski [20]. 13. Multiple Solutions In Section 12 we prove that there is a ground state solution in an achieved domain. In this section we prove that if we perturb (1.1) or perturb the achieved domain by adding or taking out a domain, then we obtain multiple solutions. 13.1. Multiple Solutions for a Perturbed Equation. EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 107 13.1.1. Introduction. Let N ≥ 2 and 2 < p < 2∗, where 2∗ = 2N N−2 for N ≥ 3, and 2∗ = ∞ for N = 2. Consider the semilinear elliptic Equation (1.2) −∆u+ u = |u|p−2u+ h(z) in Ω; u ∈ H1 0 (Ω), where Ω is a domain in RN and 0´6≡h(z) ∈ L2(Ω). Associated with (1.2), we consider the functionals a, b, and Jh, for u ∈ H1 0 (Ω), a(u) = ∫ Ω (|∇u|2 + u2); b(u) = ∫ Ω |u|p; Jh(u) = 1 2 a(u)− 1 p b(u)− ∫ Ω hu. By Rabinowitz [64, Proposition B.10.], a, b, and Jh are from C2. It is well-known that the solutions of (1.2) and the critical points of the energy functional Jh are the same. When Ω is bounded, Equation (1.1) has been studied by many mathemati- cians: see, for instance, Bahri-Berestycki [7], Bahri-Lions [10], Rabinowitz [63], and Tanaka [69]. Suppose that h is nonnegative, small and exponential decay. Us- ing the concentration-compactness principle of P. L. Lions [49] and [50] to prove that Equation (1.2) has at least two positive solutions, Cao-Zhu [17], Hirano [40], and Zhu [80] studied Equation (1.2) in RN , Hsu-Wang [41] in an exterior strip domain, and Wang [71] in an upper semi-strip with hole. In this section, we generalize the results of Zhu [80] and Cao-Zhu [17], by relaxing the assumptions of the function h and the domain Ω, to obtain two nonzero solutions of (1.1). By adding the exponential decay of the function h, we obtain three nonzero solutions of (1.1). The main ingredients of the proofs are from Adachi-Tanaka [1], Cao-Zhu [17], Tarantello [70], and Wang [71]. 13.1.2. Existence of (PS)-Sequences. We define the Palais-Smale (denoted by (PS)) sequences, (PS)-values, and (PS)-conditions in H1 0 (Ω) for Jh as follows. Definition 13.1. (i) For β ∈ R, a sequence {un} is a (PS)β-sequence in H1 0 (Ω) for Jh if Jh(un) = β + o(1) and J ′h(un) = o(1) strongly in H−1(Ω) as n→∞; (ii) β ∈ R is a (PS)-value in H1 0 (Ω) for Jh if there is a (PS)β-sequence in H1 0 (Ω) for Jh; (iii) Jh satisfies the (PS)β-condition in H1 0 (Ω) if every (PS)β-sequence in H1 0 (Ω) for Jh contains a convergent subsequence; (iv) Jh satisfies the (PS)-condition in H1 0 (Ω) if for every β ∈ R, Jh satisfies the (PS)β-condition in H1 0 (Ω). Lemma 13.2. If u ∈ H1 0 (Ω)\{0}, then(a(u)p/2 b(u) ) 1 p−2 ≥ ( 2p p− 2 )1/2 α(Ω)1/2. Proof. By Lemma 4.2, we can take λ > 0 such that λu ∈ M(Ω), and then the computations is routine. � 108 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Throughout this section, we assume that h(z) satisfies 0 < ‖h‖L2 < d(p, α), where d(p, α) = (p− 2)( 1 p− 1 ) p−1 p−2 ( 2p p− 2 )1/2α(Ω)1/2. (13.1) Let Mh = {u ∈ H1 0 (Ω)\{0} : 〈J ′h(u), u〉 = 0}, and M0 = M. Define ψ(u) = 〈J ′h(u), u〉 = a(u)− b(u)− ∫ Ω hu. Then Lemma 13.3. For each u ∈ Mh, we have 〈ψ′(u), u〉 = a(u)− (p− 1)b(u) 6= 0. Proof. For u ∈ Mh, we have a(u) − b(u) − ∫ Ω hu = 0. Then 〈ψ′(u), u〉 = 2a(u) − pb(u)− ∫ Ω hu = a(u)− (p− 1)b(u). We claim that 〈ψ′(u), u〉 6= 0 for u ∈ Mh. Let I : Mh → R be given by I(u) = K(p)( a(u)p−1 b(u) ) 1 p−2 − ∫ Ω hu, where K(p) = (p− 2)( 1 p−1 ) p−1 p−2 . Then by Lemma 13.2, we have I(u) = K(p) (a(u)p−1 b(u) ) 1 p−2 − ∫ Ω hu ≥ K(p) (a(u)p−1 b(u) ) 1 p−2 − ‖h‖L2‖u‖H1 = ‖u‖H1(K(p) (a(u)p/2 b(u) ) 1 p−2 − ‖h‖L2) ≥ ‖u‖H1(K(p) ( 2p p− 2 )1/2 α(Ω) 1 2 − ‖h‖L2) = ‖u‖H1(d(p, α)− ‖h‖L2) for u ∈ Mh. Thus, I(u) ≥ ‖u‖H1(d(p, α)− ‖h‖L2) > 0 for each u ∈ Mh. (13.2) Suppose that there is a w ∈ Mh such that 〈ψ′(w), w〉 = 0. Then we have a(w) = (p− 1)b(w) and ∫ Ω hw = a(w)− b(w) = (p− 2)b(w). Now 0 < I(w) = K(p) (a(w)p−1 b(w) ) 1 p−2 − ∫ Ω hw = ( 1 p− 1 ) p−1 p−2 (p− 2) ( (p− 1)p−1b(w)p−1 b(w) ) 1 p−2 − (p− 2)b(w) = 0, which contradicts (13.2). Thus, we conclude that 〈ψ′(u), u〉 6= 0 for each u ∈ Mh. � By Lemma 13.3, we can decompose Mh into M+ h and M− h , where M+ h = {u ∈ Mh : a(u)− (p− 1)b(u) > 0}; M− h = {u ∈ Mh : a(u)− (p− 1)b(u) < 0}. Consider the Nehari minimization problems for Equation (1.2). Let αh(Ω) = inf u∈Mh Jh(u); EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 109 α+ h (Ω) = inf u∈M+ h Jh(u); α−h (Ω) = inf u∈M− h Jh(u). Let αh(Ω) = α(Ω), Mh = M(Ω) and Jh(u) = J(u) for h = 0. For each u ∈ H1 0 (Ω)\{0}, we write tmax = [ a(u) (p− 1)b(u) ] 1 p−2 . Clearly, tmax > 0. Note that Jh(tu) = 1 2 t2a(u)− 1 p tpb(u)− t ∫ Ω hu; d dt Jh(tu) = ta(u)− tp−1b(u)− ∫ Ω hu; d2 dt2 Jh(tu) = a(u)− (p− 1)tp−2b(u); 〈J ′h(tu), tu〉 = t2a(u)− tpb(u)− t ∫ Ω hu. (13.3) The following lemma is required. Lemma 13.4. For each u ∈ H1 0 (Ω)\{0}, (i) there is a unique t− = t−(u) > tmax > 0 such that t−u ∈ M− h and Jh(t−u) = maxt≥tmax Jh(tu); (ii) t−(u) is a continuous function for nonzero u ; (iii) M− h = {u ∈ H1 0 (Ω)\{0} : 1 ‖u‖H1 t−( u ‖u‖H1 ) = 1}; (iv) if ∫ Ω hu > 0, then there is a unique 0 < t+ = t+(u) < tmax such that t+u ∈ M+ h and Jh(t+u) = min0≤t≤t− Jh(tu). Proof. (i) Fix u ∈ H1 0 (Ω)\{0}. Let s(t) = ta(u)− tp−1b(u) for t ≥ 0. Then s(0) = 0, s(t) → −∞ as t→∞, and s(t) is concave and achieves its maximum at tmax. Furthermore, by Lemma 13.2, we have s(tmax) = ( a(u)p−1 (p− 1)b(u) ) 1 p−2 − ( a(u) (p− 1)b(u) 1 p−1 ) p−1 p−2 = ‖u‖H1(p− 2)( 1 p− 1 ) p−1 p−2 [ a(u) p 2 b(u) ] 1 p−2 ≥ ‖u‖H1(p− 2)( 1 p− 1 ) p−1 p−2 ( 2p p− 2 ) 1 2α(Ω)1/2 = ‖u‖H1d(p, α), or s(tmax) ≥ ‖u‖H1d(p, α) (13.4) Case (a) : ∫ Ω hu ≤ 0. There is a unique t− > tmax such that s(t−) = ∫ Ω hu and s′(t−) < 0. Now, a(t−u)− (p− 1)b(t−u) = (t−)2 [ a(u)− (p− 1)(t−)p−2b(u) ] = (t−)2s′(t−) < 0, 110 HWAI-CHIUAN WANG EJDE-2004/MON. 06 and 〈J ′h(t−u), t−u〉 = (t−)2a(u)− (t−)pb(u)− t− ∫ Ω hu = t− [ t−a(u)− (t−)p−1b(u)− ∫ Ω hu ] = t− [ s(t−)− ∫ Ω hu ] = 0. Thus, t−u ∈ M− h , since for t > tmax, we have a(tu)− (p− 1)b(tu) < 0; d2 dt2 Jh(tu) = 1 t2 [a(tu)− (p− 1)b(tu)] < 0; d dt Jh(tu) = ta(u)− tp−1b(u)− ∫ Ω hu = 0 if t = t−. Thus, Jh(t−u) = maxt≥tmax Jh(tu). Case (b) : ∫ Ω hu > 0. By (13.4), s(0) = 0 < ∫ Ω hu ≤ ‖h‖L2‖u‖H1 < ‖u‖H1d(p, α) ≤ s(tmax), there are unique t+ and t− such that 0 < t+ < tmax < t−, s(t+) = ∫ Ω hu = s(t−) s′(t+) > 0 > s′(t−). This case is similar to Case (a): We have t+u ∈ M+ h , t−u ∈ M− h , and Jh(t−u) ≥ Jh(tu) ≥ Jh(t+u) for each t ∈ [t+, t−], and Jh(t+u) ≤ Jh(tu) for each t ∈ [0, t+]. Thus, Jh(t−u) = max t≥tmax Jh(tu), Jh(t+u) = min 0≤t≤t− Jh(tu). (ii) By the uniqueness of t−(u) and the extremity property of t−(u), t−(u) is a continuous function for nonzero u. (iii) For u ∈ M− h , let v = u ‖u‖H1 . By part (i), there is a unique t−(v) > 0 such that t−(v)v ∈ M− h or t−( u ‖u‖H1 ) 1 ‖u‖H1 u ∈ M− h . Since u ∈ M− h , we have t−( u ‖u‖H1 ) 1 ‖u‖H1 = 1, implying M− h ⊂ { u ∈ H1 0 (Ω)\{0} : 1 ‖u‖H1 t−( u ‖u‖H1 ) = 1 } . Conversely, let u ∈ H1 0 (Ω)\{0} such that 1 ‖u‖H1 t−( u ‖u‖H1 ) = 1. Then t−( u ‖u‖H1 ) u ‖u‖H1 ∈ M− h . Thus, M− h = { u ∈ H1 0 (Ω)\{0} : 1 ‖u‖H1 t−( u ‖u‖H1 ) = 1 } . (iv) By Case (b) of part (i). � EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 111 We have the following results. Lemma 13.5. (i) For each u ∈ M+ h , we have ∫ Ω hu > 0 and Jh(u) < 0. In particular, αh(Ω) ≤ α+ h (Ω) < 0; (ii) Jh is coercive and bounded below on Mh; (iii) For each minimizing sequence {un} in Mh for Jh, we have 0 < lim sup n→∞ |〈ψ′(un), un〉| <∞. Proof. (i) For each u ∈ M+ h , a(u)− (p− 1)b(u) > 0 and a(u) = b(u)+ ∫ Ω hu. Thus,∫ Ω hu = a(u)− b(u) > (p− 2)b(u) > 0, and Jh(u) = 1 2 a(u)− 1 p b(u)− ∫ Ω hu = ( 1 2 − 1 p )b(u)− 1 2 ∫ Ω hu < p− 2 2p b(u)− p− 2 2 b(u) = − (p− 1)(p− 2) 2p b(u) < 0. Then αh(Ω) = inf u∈Mh Jh(u) ≤ inf u∈M+ h Jh(u) = α+ h (Ω) < 0. (ii) For u ∈ Mh, we have a(u)− b(u)− ∫ Ω hu = 0. Then Jh(u) = ( 1 2 − 1 p )a(u)− (1− 1 p ) ∫ Ω hu ≥ ( 1 2 − 1 p )‖u‖2H1 − (1− 1 p )‖h‖L2‖u‖H1 = ( 1 2 − 1 p ) ( ‖u‖H1 − p− 1 p− 2 ‖h‖L2 )2 − 1 2p(p− 2) [(p− 1)‖h‖L2 ]2 ≥ − 1 2p(p− 2) [(p− 1)‖h‖L2 ]2 . Thus, Jh is coercive and bounded below on Mh. (iii) Let {un} be a minimizing sequence in Mh for Jh. Since Jh is coercive on Mh, we can assume {un} is bounded in Mh. By the Sobolev embedding theo- rem, a c > 0 exists such that |〈ψ′(un), un〉| = |a(un) − (p − 1)b(un)| ≤ c. Thus, lim supn→∞ |〈ψ′(un), un〉| <∞. Suppose that there is a minimizing sequence {wn} in Mh for Jh such that 〈ψ′(wn), wn〉 = o(1). Since Jh is a continuous function with Jh(0) = 0, by part (i), αh(Ω) < 0. We claim that there is a δ > 0 such that ‖wn‖H1 > δ for each n. Otherwise, a subsequence {wn} exists such that ‖wn‖H1 = o(1). Then Jh(wn) = o(1), which is a contradiction. Since ‖wn‖H1 > δ for each n and o(1) = 〈ψ′(wn), wn〉 = a(wn)− (p− 1)b(wn), (13.5) there is a γ > 0 such that b(wn) ≥ γ for each n, and(a(wn)p−1 b(wn) ) 1 p−2 = (p− 1) p−1 p−2 b(wn) + o(1). 112 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Since wn ∈ Mh, by (13.2) and (13.5), we have I(wn) ≥ ‖wn‖H1(d(p, α)− ‖h‖L2) ≥ δ(d(p, α)− ‖h‖L2) and ∫ Ω hwn = a(wn)− b(wn) = (p− 2)b(wn) + o(1). Now we have 0 < δ(d(p, α)− ‖h‖L2) ≤ I(wn) = ( 1 p− 1 ) p−1 p−2 (p− 2) (a(wn)p−1 b(wn) ) 1 p−2 − ∫ Ω hwn = ( 1 p− 1 ) p−1 p−2 (p− 2)(p− 1) p−1 p−2 b(wn)− (p− 2)b(wn) + o(1) = o(1), which is a contradiction. � Lemma 13.6. Let u be in Mh such that Jh(u) = min v∈Mh Jh(v) = αh(Ω). Then (i) ∫ Ω hu > 0; (ii) u is a solution of Equation (1.2) in Ω. Proof. (i) By Lemma 13.5 (i), we have 0 > Jh(u) = ( 1 2 − 1 p )a(u)− (1− 1 p ) ∫ Ω hu. Thus, ∫ Ω hu > 0. (ii) By Lemma 13.3 〈ψ′(v), v〉 = a(v)− (p− 1)b(v) 6= 0 for each v ∈ Mh. Since Jh(u) = minv∈Mh Jh(v), by the Lagrange multiplier theorem, there is a λ ∈ RN such that J ′(u) = λψ′(u) in H−1(Ω). Then we have 0 = 〈J ′h(u), u〉 = λ〈ψ′(u), u〉. Thus, λ = 0 and J ′h(u) = 0 in H−1(Ω). Therefore, u is a solution of Equation (1.2) in Ω with Jh(u) = αh(Ω). � The following Lemma is required to prove the existence of the (PS)αh(Ω)- se- quence for Jh. Lemma 13.7. Given u ∈ Mh, then a δ > 0 and a differentiable functional l : B(0; δ) ⊂ H1 0 (Ω) → R+ exist such that l(0) = 1, l(v)(u − v) ∈ Mh for v ∈ B(0; δ) and 〈l′(v), ϕ〉 ∣∣ (l,v)=(1,0) = 〈ψ′(u), ϕ〉 〈ψ′(u), u〉 for ϕ ∈ C∞c (Ω). Proof. For u ∈ Mh, let G : R×H1 0 (Ω) → R be given by G(l, v) = ψ(l(u− v)). Note that G(1, 0) = ψ(u) = 〈J ′h(u), u〉 = 0. Then by Lemma 13.3 DlG(1, 0) = ∂ ∂l [ l2a(u− v)− |l|pb(u− v)− l ∫ Ω h(u− v) ]∣∣ (1,0) EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 113 = [ 2la(u− v)− p|l|p−2lb(u− v)− ∫ Ω h(u− v) ]∣∣ (1,0) = 2a(u)− pb(u)− (a(u)− b(u)) = a(u)− (p− 1)b(u) 6= 0. By the implicit function theorem, there exist δ > 0 and a differentiable functional l : B(0; δ) ⊂ H1 0 (Ω) → R such that l(0) = 1 and G(l(v), v) = 0 for v ∈ B(0; δ), Thus, l(v)(u− v) ∈ Mh for v ∈ B(0; δ). Moreover, ϕ ∈ C∞c (Ω) Dϕl(v)|(1,0) = 〈l′(v), ϕ〉|(1,0) = −Gv(l, v) Gl(l, v) |(1,0) = − −2 ∫ Ω ∇u∇ϕ+ uϕ+ p ∫ Ω |u|p−2uϕ+ ∫ Ω hϕ a(u)− (p− 1)b(u) = 〈ψ′(u), ϕ〉 〈ψ′(u), u〉 . � Proposition 13.8. (i) A (PS)αh(Ω)-sequence {un} exists in Mh for Jh; (ii) A (PS)α+ h (Ω)-sequence {un} exists in M+ h for Jh; (iii) A (PS)α−h (Ω)-sequence {un} exists in M− h for Jh. Proof. (i) Let {vn} be a minimizing sequence in Mh for Jh. Since Jh is continuous and bounded below on Mh, by the Ekeland variational principle we have a mini- mizing sequence {un} in Mh such that (a) Jh(un) ≤ Jh(vn) < αh(Ω) + 1 n2 ; (b) ‖un − vn‖H1 = o(1); (c) Jh(w) ≥ Jh(un)− 1 n‖un − w‖H1 for each w ∈ Mh. Assume that there is an n0 > 0 such that ‖J ′h(un)‖H−1 > 0 for n ≥ n0, otherwise we are done. For n ≥ n0, by the Riesz representation theorem, a unique φn ∈ H1 0 (Ω) exists such that ‖φn‖H1 = 1 and 〈 J ′h(un) ‖J ′h(un)‖H−1 , ϕ〉 = 〈φn, ϕ〉H1 for each ϕ ∈ H1 0 (Ω). Let tn(ε) = ln(εφn). Applying Lemma 13.7, we have wε = tn(ε) [un − εφn] ∈ Mh. Now, t′n(0) = lim ε→0 ln(εφn)− ln(0) ε = 〈l′n(0), φn〉 = 〈ψ′(un), φn〉 〈ψ′(un), un〉 . By Lemma 13.5 (iii), we have 0 < lim supn→∞ |〈ψ′(un), un〉| <∞. Thus there is a subsequence {un} and c1 > 0 such that |〈ψ′(un), un〉| ≥ c1. By the Hölder inequality and ‖φn‖H1 = 1, we obtain |〈ψ′(un), φn〉| = |2〈un, φn〉H1 − p ∫ Ω |un|p−2unφn − ∫ Ω hφn| ≤ 2‖un‖H1 + p‖un‖p−1 Lp ‖φn‖Lp + ‖h‖L2 114 HWAI-CHIUAN WANG EJDE-2004/MON. 06 ≤ c1 + ‖h‖L2 . Thus, |t′n(0)| ≤ c2 for each n ≥ n0. Moreover, ε0 and c3 > 0 exist such that for ε ≤ ε0 ‖un − wε‖H1 ε = 1 ε ‖(1− tn(ε))un + εtn(ε)φn‖H1 ≤ [ |tn(0)− tn(ε)| ε ‖un‖H1 + tn(ε)‖φn‖H1 ] = |t′n(0)|‖un‖H1 + 1 + o(1) as ε→ 0. ≤ c3. Note that {un} is bounded in H1 0 (Ω), so ‖t0un + (1− t0)wε − un‖H1 = ‖(1− t0)(tn(ε)− 1)un − ε(1− t0)tn(ε)φn‖H1 ≤ (1− t0)|tn(ε)− 1|‖un‖H1 + ε(1− t0)tn(ε) = o(1) as ε→ 0, Since Jh ∈ C1, we have 1 ε |〈J ′h(t0un + (1− t0)wε), (un − wε)〉 − 〈J ′h(un), (un − wε)〉| ≤ ‖J ′h(t0un + (1− t0)wε)− J ′h(un)‖H−1 ‖un − wε‖H1 ε = o(1) as ε→ 0. Thus, 〈J ′h(t0un + (1− t0)wε), (un − wε)〉 = 〈J ′h(un), (un − wε)〉+ o(ε), where o(ε) ε → 0 as ε→ 0. By condition (c) and the mean value theorem, we have 1 n ‖un − wε‖H1 ≥ Jh(un)− Jh(wε) = 〈J ′h(t0un + (1− t0)wε), (un − wε)〉, where t0 ∈ (0, 1). Then we obtain 1 n ‖un − wε‖H1 ≥ 〈J ′h(un), (un − wε)〉+ o(ε). (13.6) We divide (13.6) by ε > 0 and obtain c3 n ≥ 1 nε ‖un − wε‖H1 ≥ (1− tn(ε)) ε 〈J ′h(un), un〉+ tn(ε)〈J ′h(un), φn〉+ o(ε) ε = (1− tn(ε)) ε 〈J ′h(un), un〉+ tn(ε)‖J ′h(un)‖H−1 + o(ε) ε . Since {un} ⊂ Mh, let ε→ 0 to obtain 0 ≥ −t′n(0)〈J ′h(un), un〉+ ‖J ′h(un)‖H−1 = ‖J ′h(un)‖H−1 , which is a contradiction. Therefore, ‖J ′h(un)‖H−1 → 0 as n→∞. (ii) and (iii) can be proved similarly. � EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 115 13.1.3. Existence of a Local Minimum. By Proposition 13.8 (i), there is a (PS)αh(Ω)- sequence {un} in Mh for Jh. Then we have the following (PS)αh(Ω)-condition. Proposition 13.9. Let {un} in Mh be a (PS)αh(Ω)-sequence for Jh. Then a sub- sequence {un} and u0 in H1 0 (Ω) exist such that un → u0 strongly in H1 0 (Ω). Fur- thermore, u0 is a solution of Equation (1.2) such that Jh(u0) = αh(Ω). Proof. By Lemma 13.5 (ii), {un} is bounded in H1 0 (Ω). Take a subsequence {un} and u0 ∈ H1 0 (Ω) such that un ⇀ u0 weakly in H1 0 (Ω). By Lemma 2.11, u0 is a nonzero solution of Equation (1.2) in Ω. Since Jh(un) = 1 2 a(un)− 1 p b(un)− ∫ Ω hun = αh(Ω) + o(1), 〈J ′h(un), un〉 = a(un)− b(un)− ∫ Ω hun = o(1). (13.7) By (13.7), we have ( 1 2 − 1 p )a(un)− (1− 1 p ) ∫ Ω hun = αh(Ω) + o(1). Since the functional a is weakly lower semicontinuous and ∫ Ω hun → ∫ Ω hu0 as n→∞, we have αh(Ω) ≤ Jh(u0) = ( 1 2 − 1 p )a(u0)− (1− 1 p ) ∫ Ω hu0 ≤ ( 1 2 − 1 p ) lim inf n→∞ a(un)− (1− 1 p ) lim n→∞ ∫ Ω hun = lim inf n→∞ [ ( 1 2 − 1 p )a(un)− (1− 1 p ) ∫ Ω hun ] = αh(Ω) or Jh(u0) = αh(Ω). Let pn = un − u0. By Lemma 2.11 and 2.14, we have Jh(pn) = 1 2 a(pn)− 1 p b(pn)− ∫ Ω hpn = 1 2 a(un)− 1 2 a(u0)− 1 p b(un) + 1 p b(u0)− ∫ Ω hun + ∫ Ω hu0 + o(1) = Jh(un)− Jh(u0) + o(1) = o(1) . (13.8) By Lemma 2.11, 2.14, ∫ Ω hpn = o(1) and u0 is a solution of Equation (1.2), so 〈J ′h(pn), pn〉 = a(pn)− b(pn)− ∫ Ω hpn = a(un)− a(u0)− b(un) + b(u0)− ∫ Ω hun + ∫ Ω hu0 + o(1) = 〈J ′h(un), un〉 − 〈J ′h(u0), u0〉 = o(1) . (13.9) Thus, by (13.8), (13.9) and ∫ Ω hpn = o(1), we have p− 2 2p a(pn) = o(1) or un → u0 strongly in H1 0 (Ω). � 116 HWAI-CHIUAN WANG EJDE-2004/MON. 06 The following result is required to prove that u0 is the unique critical point of Jh(u) in B(r0). Lemma 13.10. Let r0 = ( 1 p−1 ) 1 p−2 ( 2p p−2 )1/2α(Ω) 1 2 . Then (i) M+ h ⊂ B(r0) = {u ∈ H1 0 (Ω) : ‖u‖H1 < r0}; (ii) Jh(u) is strictly convex in B(r0). Proof. (i) If u ∈ M+ h , then a(u) > (p− 1)b(u) and a(u) = b(u) + ∫ Ω hu. Thus, a(u) < 1 p− 1 a(u) + ‖h‖L2‖u‖H1 . This implies ‖u‖H1 < ( p− 1 p− 2 )‖h‖L2 < ( p− 1 p− 2 )(p− 2)( 1 p− 1 ) p−1 p−2 ( 2p p− 2 ) 1 2α(Ω)1/2 = ( 1 p− 1 ) 1 p−2 ( 2p p− 2 )1/2α(Ω)1/2 = r0. (ii) Similarly to Adachi-Tanaka [1], we have J ′′h (u)(v, v) = a(v)− (p− 1) ∫ Ω |u|p−2v2 for all v ∈ H1 0 (Ω). Thus, by Lemma 13.2 for u ∈ H1 0 (Ω)\{0}[a(u)p/2 b(u) ] 1 p−2 ≥ ( 2p p− 2 )1/2α(Ω)1/2, then J ′′h (u)(v, v) ≥ a(v)− (p− 1)‖u‖p−2 Lp ‖v‖2Lp ≥ a(v)− (p− 1) [ a(u) p−2 2 ( p− 2 2p ) p−2 2 α(Ω)− (p−2)2 2p ] × [ a(v)( p− 2 2p ) p−2 p α(Ω) −(p−2) p ] ≥ a(v) [ 1− (p− 1)( 2p p− 2 α(Ω)) 2−p 2 ‖u‖p−2 H1 ] > 0 for u ∈ B(r0). Thus, J ′′h (u) is positive definite for u ∈ B(r0) and Jh is strictly convex in B(r0). � By Proposition 13.9, a solution u0 ∈ Mh of Equation (1.2) exists such that Jh(u0) = αh(Ω). Furthermore, we have the following theorem. Theorem 13.11. (i) u0 ∈ M+ h and Jh(u0) = α+ h (Ω) = αh(Ω); (ii) u0 is the unique critical point of Jh(u) in B(r0), where r0 is as in Lemma 13.10; (iii) Jh(u0) is a local minimum in H1 0 (Ω). Proof. (i) By Lemma 13.6 (i), ∫ Ω hu0 > 0. We claim that u0 ∈ M+ h . Otherwise, if u0 ∈ M− h , then by Lemma 13.4 a unique t−(u0) = 1 > t+(u0) > 0 exists such that t+(u0)u0 ∈ M+ h and αh(Ω) ≤ α+ h (Ω) ≤ Jh(t+(u0)u0) < Jh(u0) = αh(Ω), EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 117 which is a contradiction. Since u0 ∈ M+ h , α+ h (Ω) ≤ Jh(u0) = αh(Ω) ≤ α+ h (Ω), that is, Jh(u0) = α+ h (Ω). (ii) By part (i) and Lemma 13.10. (iii) Since u0 ∈ Mh, by Lemma 13.7, δ1 > 0 and a differentiable functional l(w) > 0 exist such that l(0) = 1 and l(w)(u0 + w) ∈ Mh for ‖w‖H1 < δ1. (13.10) By Lemma 13.4, 1 = t+(u0) < tmax(u0), (13.11) l(0) = 1, and the continuity of tmax, δ2 exists with δ1 > δ2 > 0, such that l(w) < tmax(u0 + w) for ‖w‖H1 < δ2. (13.12) By (13.10) and (13.12 ), we have l(w)(u0 + w) ∈ M+ h for ‖w‖H1 < δ2. By Lemma 13.4, for 0 < s < tmax(u0 + w), we have Jh(u0) = α+ h (Ω) ≤ Jh(l(w)(u0 + w)) ≤ Jh(s(u0 + w)). By the continuity of tmax and (13.11), δ exists with δ2 > δ > 0 such that 1 < tmax(u0 + w) for ‖w‖H1 < δ. Thus, we can take s = 1 to obtain Jh(u0 + w) ≥ Jh(u0) for ‖w‖H1 < δ. Hence, Jh(u0) is a local minimum in H1 0 (Ω). � Theorem 13.12. (i) If ∫ Ω h|u0| > 0, then u0 is a nonnegative solution of Equation (1.1) in Ω. Moreover, a positive solution of equation (1.2 ) exists for h � 0; (ii) If ∫ Ω h|u0| < 0, then u0 is a nonpositive solution of Equation (1.2) in Ω. Moreover, a negative solution of Equation (1.2) exists for h � 0; (iii) If ∫ Ω h|u0| = 0, then u0 is a solution of Equation (1.2) in Ω that changes sign. Proof. (i) If ∫ Ω h|u0| > 0, then by Lemma 13.4 tmax(|u0|) > t+(|u0|) > 0 exists such that t+(|u0|)|u0| ∈ M+ h . Since tmax(|u0|) = tmax(u0) > 1, we have αh(Ω) = α+ h (Ω) ≤ Jh(t+(|u0|)|u0|) ≤ Jh(|u0|) ≤ Jh(u0) = αh(Ω), or Jh(|u0|) = αh(Ω). By Lemma 13.6 (ii) and Lemma 13.11 (ii), u0 = |u0|. Thus, we can take u0 ≥ 0. Moreover, if h � 0, we apply the maximum principle and obtain u0 > 0. (ii) The proof is similar to (i). (iii) Let u+ 0 = max{u0, 0} and u−0 = max{−u0, 0}. Since ∫ Ω h|u0| = 0, then∫ Ω hu+ 0 + ∫ Ω hu−0 = 0. By Lemma 13.6 (i), we have ∫ Ω hu+ 0 > 0 and ∫ Ω hu0 < 0. Thus, u+ 0 � 0 and u−0 � 0. Hence, u0 is a solution of equation (1.2) in Ω that changes sign. � 13.1.4. Existence of Two Solutions. Let u0 be the local minimum for Jh in H1 0 (Ω) in Theorem 13.11. Then we have the following restricted (PS)-condition. Proposition 13.13. If {un} is a (PS)β-sequence in H1 0 (Ω) for Jh with β < αh(Ω)+ α(Ω), then a subsequence {un} and u0 in H1 0 (Ω) exist such that un → u0 strongly in H1 0 (Ω) and Jh(u0) = β. 118 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Proof. Let {un} be a (PS)β-sequence in H1 0 (Ω) for Jh. By Lemma 13.5 (ii), {un} is bounded. As in the proof of Proposition 13.9, a subsequence {un} and a solution u0 of Equation (1.2) exist such that un ⇀ u0 weakly in H1 0 (Ω). Suppose that un 9 u0 strongly in H1 0 (Ω). Let pn = un − u0 for n = 1, 2, . . . . By Lemma 2.11 and 2.14, we have a(pn) = a(un)− a(u0) + 0(1), b(pn) = b(un)− b(u0) + 0(1). (13.13) Since a(un) − b(un) − ∫ Ω hun = o(1), a(u0) − b(u0) − ∫ Ω hu0 = 0 and un ⇀ u0 weakly in H1 0 (Ω), we have a(pn) = a(un)− a(u0) + 0(1) = b(un) + ∫ Ω hun − b(u0)− ∫ Ω hu0 + o(1) = b(pn) + o(1). Since pn 9 0, we have J(pn) = 1 2 a(pn)− 1 p b(pn) = ( 1 2 − 1 p )a(pn) + o(1) > 0. By Theorem 4.3, a sequence {sn} in R+ exists such that {snpn} in M(Ω) and J(snpn) = J(pn) + o(1). Thus, by (13.13) and un ⇀ u0 weakly in H1 0 (Ω), we have α(Ω) ≤ J(snpn) = J(pn) + o(1) = Jh(pn) + o(1) = Jh(un)− Jh(u0) + o(1) = β − Jh(u0) + o(1) < αh(Ω) + α(Ω)− Jh(u0) + o(1). Then αh(Ω) > Jh(u0) ≥ αh(Ω), which is a contradiction. Thus, un → u0 strongly in H1 0 (Ω). � Throughout this section, let Ω be an achieved domain in RN . Lemma 13.14. Let u be a positive solution of Equation ( 1.1) in Ω such that J(u) = α(Ω) and u0 is the local minimum in Theorem 13.11. Then (i) If ∫ Ω h|u0| > 0, then we have sup t≥0 Jh(u0 + tu) < Jh(u0) + α(Ω). (ii) If ∫ Ω h|u0| < 0, then we have sup t≥0 Jh(u0 + t(−u)) < Jh(u0) + α(Ω). EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 119 Proof. (i) Since ∫ Ω h|u0| > 0, by Theorem 13.12 (i), u0 is a nonnegative solution of Equation (1.2). Let f(s) = sp−1 and F (u) = ∫ u 0 f(s)ds = 1 pb(u), then Jh(u0 + tu) = 1 2 a(u0 + tu)− 1 p b(u0 + tu)− ∫ Ω h(u0 + tu) = 1 2 [a(u0) + a(tu) + 2〈u0, tu〉H1 ]− 1 p b(u0 + tu)− ∫ Ω h(u0 + tu) = Jh(u0) + J(tu) + 〈u0, tu〉H1 + 1 p [b(u0) + b(tu)− b(u0 + tu)]− ∫ Ω htu = Jh(u0) + J(tu) + t( ∫ Ω up−1 0 u+ hu)− ∫ Ω htu + 1 p [b(u0) + b(tu)− b(u0 + tu)] = Jh(u0) + J(tu)− ∫ Ω { ∫ tu 0 [f(u0 + s)− f(s)− f(u0)]ds } . (13.14) For v > 0 and w > 0, we have f(v+w) = (v+w)p−1 = (v+w)p−2v+(v+w)p−2w > vp−1 +wp−1 = f(v)+ f(w). (13.15) Thus, Jh(u0 + tu) ≤ Jh(u0)+J(tu). Since J(tu) → −∞ as t→∞, there is a t0 > 0 such that Jh(u0 + tu) < Jh(u0) for t ≥ t0. Hence, sup t≥0 Jh(u0 + tu) = sup 0≤t≤t0 Jh(u0 + tu). Let g1(t) = Jh(u0 + tu) for t ≥ 0. By the continuity of g1(t), given ε = 1 2α(Ω) > 0 there is a t0 > t1 > 0 such that g1(t) < g1(0) + 1 2α(Ω) for 2t1 > t ≥ 0. Then sup 0≤t≤t1 Jh(u0 + tu) ≤ Jh(u0) + 1 2 α(Ω) < Jh(u0) + α(Ω). Now, it only remains to show that sup t1≤t≤t0 Jh(u0 + tu) < Jh(u0) + α(Ω). Let g2(t) = J(tu) for t ≥ 0, then g′2(t) = ta(u)− tp−1b(u) and g′′2 (t) = a(u)− (p− 1)tp−2b(u). There is a unique t = [a(u) b(u) ]1/(p−2) = 1 such that g′2(t) = 0 and g′′2 (t) < 0. Thus, g2(t) has an absolute maximum at t = 1. Therefore, sup t≥0 J(tu) = J(u) = α(Ω). By (13.14), (13.15), we obtain sup t1≤t≤t0 Jh(u0 + tu) ≤ Jh(u0) + α(Ω)− inf t1≤t≤t0 ∫ Ω { ∫ tu 0 [f(u0 + s)− f(s)− f(u0)] ds } < Jh(u0) + α(Ω). 120 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Thus, sup t≥0 Jh(u0 + tu) < Jh(u0) + α(Ω). (ii) Since ∫ Ω h|u0| < 0, by Theorem 13.12 (ii), u0 is a nonpositive solution of Equation (1.2). Let f(s) = |s|p−2s and F (u) = ∫ u 0 f(s)ds = 1 pb(u). Then Jh(u0 − tu) = 1 2 a(u0 − tu)− 1 p b(u0 − tu)− ∫ Ω h(u0 − tu) = 1 2 [a(u0) + a(tu) + 2〈u0,−tu〉H1 ] − 1 p b(−u0 + tu)− ∫ Ω h(u0 − tu) = Jh(u0) + J(tu) + 〈u0,−tu〉H1 + 1 p [b(u0) + b(tu)− b(−u0 + tu)] + ∫ Ω htu = Jh(u0) + J(tu)− t( ∫ Ω |u0|p−2u0u+ hu) + ∫ Ω htu + 1 p [b(−u0) + b(tu)− b(−u0 + tu)] = Jh(−u0) + J(tu) + ∫ Ω | − u0|p−2(−u0)tu + 1 p [b(u0) + b(tu)− b(−u0 + tu)] = Jh(u0) + J(tu)− ∫ Ω { ∫ tu 0 [f(−u0 + s)− f(s)− f(−u0)] ds } . Similarly to part (i), we have sup t≥0 Jh(u0 + t(−u)) < Jh(u0) + α(Ω). � Theorem 13.15. If ∫ Ω h|u0| 6= 0, where u0 is the local minimum in Theorem 13.11, then Equation (1.2) has two solutions u0 ∈ M+ h , u0 ∈ M− h such that Jh(u0) = α+ h < α−h = Jh(u0). Moreover, if h � 0 (� 0), then Equation (1.2) has at least two positive (negative) solutions in Ω. Proof. For u ∈ H1 0 (Ω) with ‖u‖H1 = 1, by Lemma 13.4 there is a unique t−(u) > 0 such that t−(u)u ∈ M− h and Jh(t−(u)u) = max t≥tmax Jh(tu). By Lemma 13.4 (ii) and (iii), we have that t−(u) is a continuous function for nonzero u and M− h = { u ∈ H1 0 (Ω)\{0} : 1 ‖u‖H1 t−( u ‖u‖H1 ) = 1 } . Let A1 = { u ∈ H1 0 (Ω)\{0} : 1 ‖u‖H1 t−( u ‖u‖H1 ) > 1 } ∪ {0} A2 = { u ∈ H1 0 (Ω)\{0} : 1 ‖u‖H1 t−( u ‖u‖H1 ) < 1 } . Then H1 0 (Ω)\M− h = A1 ∪A2. For each u ∈ M+ h , we have 1 < tmax(u) < t−(u). EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 121 Since t−(u) = 1 ‖u‖H1 t−( u ‖u‖H1 ), then M+ h ⊂ A1. In particular, u0 ∈ A1. We claim that a t0 > 0 exists such that u0 + t0ū ∈ A2. First, we find a constant c > 0 such that 0 < t−( u0+tu ‖u0+tu‖H1 ) < c for each t ≥ 0. Otherwise, a sequence {tn} exists such that tn → ∞ and t−( u0+tnu ‖u0+tnu‖H1 ) → ∞ as n → ∞. Let vn = u0+tnu ‖u0+tnu‖H1 . Since t−(vn)vn ∈ M− h ⊂ Mh and by the Lebesgue dominated convergence theorem, b(vn) = 1 ‖u0 + tnu‖p H1 ∫ Ω (u0 + tnu)p = 1 ‖u0 tn + u‖p H1 ∫ Ω ( u0 tn + u)p → ∫ Ω up ‖u‖p H1 as n→∞. We have Jh(t−(vn)vn) = 1 2 [ t−(vn) ]2 − 1 p [ t−(vn) ]p b(vn)− t−(vn) ∫ Ω hvn → −∞ as n→∞. However, Jh is bounded below on Mh, which is a contradiction. Let t0 = ( 2p (p− 2)α(Ω) |c2 − a(u0)| )1/2 + 1. Then ‖u0 + t0u‖2H1 = a(u0) + t20( p− 2 2p )α(Ω) + o(1) > c2 > [ t− ( u0 + t0u ‖u0 + t0u‖H1 )]2 , that is, u0 + t0u ∈ A2. Define a path γ(s) = u0 + st0u for s ∈ [0, 1] where t0 > 1, then γ(0) = u0 ∈ A1, γ(1) = u0 + t0u ∈ A2. Since 1 ‖u‖H1 t−( u ‖u‖H1 ) is a continuous function for nonzero u and γ([0, 1]) is con- nected, a s0 ∈ (0, 1) exists such that u0 + s0t0u ∈ M− h . Thus, by Lemma 13.12 and Theorem 13.14 we have α−h ≤ Jh(u0 + s0t0u) ≤ max s∈[0,1] Jh(γ(s)) < Jh(u0) + α(Ω) for ∫ Ω h|u0| > 0. Similarly, we also have α−h < Jh(u0) + α(Ω) for ∫ Ω h|u0| < 0. By Proposition 13.8 (iii), a sequence {un} in M− h exists such that Jh(un) = α−h (Ω) + o(1), J ′h(un) = o(1) strongly in H−1(Ω). Then by Proposition 13.13, a subsequence {un} and u0 ∈ Mh exist such that un → u0 strongly in H1 0 (Ω), u0 is a solution of Equation (1.2), and Jh(u0) = α−h (Ω). By the Sobolev continuous embedding theorem, we have un → u0 in Lp(Ω). Thus, a(u0)− (p− 1)b(u0) ≤ 0. 122 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Then u0 ∈ M− h . This implies u0 6= u0. We must now show that Jh(u0) = αh < α−h = Jh(u0). Otherwise, assume that Jh(u0) = α−h = Jh(u0) = αh. By Lemma 13.6 (i) we have∫ Ω hu0 > 0. By Lemma 13.4, t+(u0) > 0 exists such that t+(u0)u0 ∈ M+ h and α+ h (Ω) ≤ Jh(t+(u0)u0) < Jh(u0) ≤ αh ≤ α−h , which is a contradiction. Finally, if h � 0, by Lemma 13.4 t−(|u0|) > 0 exists such that t−(|u0|)|u0| ∈ M− h , t−(|u0|) > tmax(|u0|) = tmax(u0) and α−h (Ω) ≤ Jh(t−(|u0|)|u0|) ≤ Jh(t−(|u0|)u0) ≤ max t≥tmax(u0) Jh(tu0) = Jh(u0) = α−h (Ω). Thus, Jh(t−(|u0|)|u0|) = Jh(t−(|u0|)u0) = α−h (Ω). We conclude that ∫ Ω hu0 = ∫ Ω h|u0|. Let u0 + = max{u0, 0} and u0 − = max{−u0, 0}, then ∫ Ω hu0 − = 0. Since h � 0 and u0 − ≥ 0, we have u0 − = 0. Hence, u0 ≥ 0. By the maximum principle, u0 > 0. � Remark 13.16. By Theorems 13.11 and 13.15, there is a unique solution u0 of Equation (1.2) in Ω such that Jh(u0) = α+ h (Ω) = αh(Ω). Bibliographical notes: The results of this section are from Lin-Wang-Wu [55]. 13.1.5. Three Solutions. Throughout this section, we consider a C1,1 domain Ω to be Ar, RN , Ar\D, or RN\D, where D is a bounded domain in RN and assume that h ∈ LN 2 (Ω) ∩ Ls(Ω) ∩ L2(Ω) for some s > N , 0 < ‖h‖L2 < d(p, α), and 0 ≤ h(z) ≤ c exp(−(1 + ε)|z|) for any z ∈ Ω, for some positive constants c and ε, where d(p, α) is defined as in (13.1). Then we have the following lemma. Lemma 13.17. Let u be a positive solution of the Equation (1.2). Then for any 0 < δ < min{ε, 1}, positive constants c1δ, c 2 δ and R exist such that for |z| ≥ R c1δ exp(−(1 + δ)|z|) ≤ u(z) ≤ c2δ exp(−(1− δ)|z|). Proof. By Lemma 8.11 and 8.12, we have u ∈ W 2,s(Ω) ∩ C1,θ(Ω) for some θ, 0 < θ < 1 and lim |z|→∞ u(z) = 0. Take R1 > 0 such that D ⊂ BN (0;R1). For any 0 < δ < min{ε, 1}, we choose R2 > R1 > 0 such that (1 + δ)− √ 1 + δ(N − 1) |z| ≥ 1 for |z| ≥ R2. (13.16) Let β = √ 1 + δ and v1(z) = µ exp(−β(|z| − R2)), where µ = min|z|=R2 u(z) > 0. Then min|z|=R2(u− v1)(z) ≥ 0. By (13.16), for |z| > R2 4(u− v1)(z) = u− |u|p−2u− h(z)− ( β2 − β(N − 1) |z| ) v1 EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 123 ≤ u− ( β2 − β(N − 1) |z| ) v1 ≤ (u− v1)(z). By Lemma 9.3, for |z| > R2, u(z)− v1(z) ≥ min |z|=R2 (u− v1)(z) ≥ 0. Thus, we have u(z) ≥ v1(z) = µ exp(−β(|z| −R2)) = µ exp(R2 √ 1 + δ) exp(−β|z|) ≥ c1δ exp(−(1 + δ)|z|) for |z| ≥ R2. (13.17) We know that there exist positive numbers ε, c such that 0 ≤ h(z) ≤ c exp(−(1 + ε)|z|) for any z ∈ Ω. For any 0 < δ < min{ε, 1}, by (13.17), there is R3 > R2 > 0 such that δ 2 u(z) ≥ h(z) for |z| ≥ R3. (13.18) Since lim|z|→∞ u(z) = 0, there is R > R3 > 0 such that 1− up−2 ≥ 1− δ 2 for |z| ≥ R. (13.19) Let γ = √ 1− δ and v2(z) = ν exp(−γ(|z| − R)), where ν = max|z|=R u(z) > 0. Thus min |z| = R(v2 − u)(z) ≥ 0. By (13.18) and (13.19), for |z| > R 4(v2 − u)(z) = (γ2 − γ(N − 1) |z| )v2(z)− u+ |u|p−2u+ h(z) ≤ γ2v2 − (1− δ 2 )u+ h(z) = (1− δ)(v2(z)− u(z))− δ 2 u+ h(z) ≤ (1− δ)(v2(z)− u(z)). By Lemma 9.3, for |z| > R v2(z)− u(z) ≥ min |z|=R (v2 − u)(z) ≥ 0. Thus, we have u(z) ≤ v2(z) = ν exp(−γ(|z| −R)) = ν exp(R √ 1− δ) exp(−γ|z|) ≤ c2δ exp(−(1− δ)|z|) for |z| ≥ R. � By Lien-Tzeng-Wang [47], there is a positive ground state solution ū of the Equation (1.1) in RN such that J(ū) = α(RN ). By Gidas-Ni-Nirenberg [35], we have that u is radially symmetric about 0 in RN . Similarly to Lemma 13.17, for any δ′ > 0, positive constants c1δ′ and c2δ′ exist such that c1δ′ exp(−(1 + δ′)|z|) ≤ u(z) ≤ c2δ′ exp(−(1− δ′)|z|) for z ∈ RN . 124 HWAI-CHIUAN WANG EJDE-2004/MON. 06 By Lemma 13.17, there is a R > 0 such that D ⊂ B(0;R). For such R, let ψR : RN → [0, 1] be a C∞−function on RN such that 0 ≤ ψR ≤ 1, ψR(z) = { 1 for |z| ≥ R+ 1; 0 for |z| ≤ R. For z ∈ RN , we define vz(z) = ψR(z)ū(z − z). Clearly, vz(z) ∈ H1 0 (Ω). Lemma 13.18. (i) a(vz) = b(vz) + o(1) as |z| → ∞; (ii) J(vz) = α(Ω) + o(1) as |z| → ∞; (iii) vz ⇀ 0 weakly in H1 0 (Ω) as |z| → ∞. Proof. (i-1) a(vz) = a(ū) + o(1) as |z| → ∞ : since u ∈ H1 0 (RN ), we have ‖vz(z)− ū(z − z)‖2H1 = ∫ RN |∇[(ψR(z)− 1)ū(z − z)]|2dz + ∫ RN |(ψR(z)− 1)ū(z − z)|2dz ≤ 2 ∫ RN |∇(ψR(z)− 1)|2|ū(z − z)|2dz + 2 ∫ RN |(ψR(z)− 1)|2|∇ū(z − z)|2dz + ∫ {|z|≤R+1} |ū(z − z)|2dz ≤ 2 ∫ {R≤|z|≤R+1} |ū(z − z)|2dz + 2 ∫ {|z|≤R+1} (|∇ū(z − z)|2 + |ū(z − z)|2)dz = o(1). Thus, a(vz) = a(ū(z − z)) + o(1) = a(ū) + o(1) as |z| → ∞. (i-2) b(vz) = b(ū) + o(1) as |z| → ∞: since u ∈ H1 0 (Ω), we have ‖vz(z)− ū(z − z)‖p Lp = ∫ RN |(ψR(z)− 1)|p|ū(z − z)|pdz ≤ ∫ {|z|≤R+1} |ū(z − z)|pdz = o(1). Thus, b(vz) = b(ū(z − z)) + o(1) = b(ū) + o(1) as |z| → ∞. By (i-1), (i-2)and that ū is a solution of the Equation (1.1), we have a(vz) = a(ū) + o(1) = b(ū) + o(1) = b(vz) + o(1) as |z| → ∞. (ii) By Lemma 4.18, J(vz) = J(ū)+o(1) = α(RN )+o(1) = α(Ω)+o(1) as |z| → ∞. (iii) For φ ∈ C1 c (Ω) with K = suppφ, then K ⊂ Ω is compact. |〈vz, φ〉H1 | = ∣∣∫ Ω ∇vz(z)∇φ(z)dz + ∫ Ω vz(z)φ(z)dz ∣∣ = ∣∣∫ K ∇ [ψR(z)ū(z − z)]∇φ(z)dz + ∫ K ψR(z)ū(z − z)φ(z)dz ∣∣ ≤ ‖∇ [ψR(z)ū(z − z)] ‖L2(K)‖∇φ‖L2(K) + ‖ψR(z)ū(z − z)‖L2(K)‖φ‖L2(K) = o(1) as |z| → ∞. EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 125 By part (i), there is a c > 0 such that ‖vz‖H1 ≤ c. For ε > 0 and ϕ ∈ H1 0 (Ω), there exist φ ∈ C1 c (Ω) and l0 > 0 such that ‖ϕ− φ‖H1 < ε/2c |〈vz, φ〉H1 | < ε/2 for |z| ≥ l0. Thus, 〈vz, ϕ〉H1 = 〈vz, ϕ− φ〉H1 + 〈vz, φ〉H1 ≤ ‖vz‖H1‖ϕ− φ‖H1 + 〈vz, φ〉H1 < c‖ϕ− φ‖H1 + ε 2 < ε for |z| ≥ l0. Therefore, vz ⇀ 0 weakly in H1 0 (Ω) as |z| → ∞. � Lemma 13.19. Let D be a domain in RN . If f : D → R satisfies∫ D |f(z)eσ|z||dz <∞ for some σ > 0, then ( ∫ D f(z)e−σ|z−z|dz)eσ|z| = ∫ D f(z)eσ 〈z,z〉 |z| dz + o(1) as |z| → ∞. Proof. Since σ|z| ≤ σ|z|+ σ|z − z|, we have |f(z)e−σ|z−z|eσ|z|| ≤ |f(z)eσ|z||. Since −σ|z − z| + σ|z| = σ 〈z,z〉 |z| + o(1) as |z| → ∞, then the lemma follows from Theorem 2.20. � Since Ω here is nonachieved, we need a delicate result. Lemma 13.20. l0 > 0 exists such that for |z| ≥ l0 sup t≥0 Jh(u0 + tvz) < Jh(u0) + α(Ω), where u0 is the local minimum in Theorem 13.12. Proof. Since Jh is continuous in H1 0 (Ω) and {vz} is bounded in H1 0 (Ω), there is t0 > 0 such that for 0 ≤ t < t0 and each vz ∈ H1 0 (Ω) Jh(u0 + tvz) < Jh(u0) + α(Ω). Thus, we only need to show that there exists l0 > 0 such that for |z| ≥ l0 sup t≥t0 Jh(u0 + tvz) < Jh(u0) + α(Ω). First, we observe that if a ≥ 0 and b ≥ 0, then there is c = c(p) > 0 independent of a and b such that (a+ b)p ≥ ap + bp + p(ap−1b+ abp−1)− cap/2bp/2. Hence, we get∫ Ω (u0 + tvz)pdz ≥ ∫ Ω (up 0 + (tvz)p + ptup−1 0 vz + pu0(tvz)p−1)dz − c ∫ Ω u p/2 0 (tvz)p/2dz. 126 HWAI-CHIUAN WANG EJDE-2004/MON. 06 By Lemma 4.2 and Theorem 4.3, there exists a t0(vz) > 0 such that t0(vz)vz ∈ M(Ω) and max t ≥ 0J(tvz) = J(t0(vz)vz) = α(Ω) + o(1) as |z| → ∞. (13.20) By (13.20) and Lemma 13.18, we deduce for t ≥ t0 Jh(u0 + tvz) = 1 2 ∫ Ω [ |∇(u0 + tvz)|2 + (u0 + tvz)2 ] dz − 1 p ∫ Ω (u0 + tvz)pdz − ∫ Ω h(u0 + tvz)dz ≤ Jh(u0) + α(Ω)− t p 2 ( t p−2 2 0 ∫ Ω vp−1 z u0dz − c p ∫ Ω u p/2 0 v p 2 z dz ) + o(1). Let 0 < δ′ < min{ε, p−2 p+2}, then (1−δ′)p 2 − (1 + δ′) > 0. We choose 0 < δ < δ′ < (p− 2) + (p− 1)δ. By Lemma 13.17, 13.18 and u0 is a positive solution of the Equation (1.2), we have∫ Ω vp−1 z u0dz = ∫ Ω (∇vz∇u0 + vzu0)dz + o(1) = ∫ Ω (vzu p−1 0 + hvz)dz + o(1) = ∫ Ω vzu p−1 0 dz + o(1) ≥ c1 ∫ {|z|≥R+1} e(−(1+δ′)|z−z|)e(p−1)(−(1+δ)|z|)dz + o(1), and ∫ Ω u p/2 0 v p/2 z dz ≤ c2 ∫ {|z|≥R} e(− (1−δ′)p 2 |z−z|)e(− (1−δ)p 2 |z|)dz, where c1 = c1(δ, δ′) and c2 = c2(δ, δ′). Since∫ {|z|≥R+1} e(p−1)(−(1+δ)|z|)e(1+δ′)|z|dz <∞ and ∫ {|z|≥R} e(− (1−δ)p 2 |z|)e (1−δ′)p 2 |z|dz <∞, by Lemma 13.19, we deduce that as |z| → ∞∫ Ω vp−1 z u0dz ≥ c1 ∫ {|z|≥R+1} e(−(1+δ′)|z−z|)e(p−1)(−(1+δ)|z|)dz + o(1) = c1 { ∫ {|z|≥R+1} e(p−1)(−(1+δ)|z|)e(1+δ′) 〈z,z〉 |z| dz + o(1) } e−(1+δ′)|z| and ∫ Ω u p/2 0 v p/2 z dz ≤ c2 ∫ {|z|≥R} e(− (1−δ′)p 2 |z−z|)e(− (1−δ)p 2 |z|)dz = c2 { ∫ {|z|≥R} e(− (1−δ)p 2 |z|)e (1−δ′)p 2 〈z,z〉 |z| dz + o(1) } e− (1−δ′)p 2 |z|. EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 127 Thus,∫ Ω vp−1 z u0dz∫ Ω u p 2 0 v p/2 z dz ≥ c1{ ∫ {|z|≥R+1} e (p−1)(−(1+δ)|z|)e(1+δ′) 〈z,z〉 |z| dz + o(1)} c2{ ∫ {|z|≥R} e (− (1−δ)p 2 |z|)e (1−δ′)p 2 〈z,z〉 |z| dz + o(1)} e( (1−δ′)p 2 −(1+δ′))|z| which approaches ∞ as |z| → ∞. Then l0 > 0 exists such that for |z| ≥ l0 t p−2 2 0 ∫ Ω vp−1 z u0dz − c p ∫ Ω u p/2 0 v p/2 z dz > 0. Hence, we have that for |z| ≥ l0, sup t≥t0 Jh(u0 + tvz) < Jh(u0) + α(Ω). � Take the sequence {tn} ⊂ R such that |tn| ↗ +∞ as n → +∞. For n ∈ N, we define vn(z) = ψR(z)ū(z − tnz), where z is a unit vector in RN . Clearly, vn ∈ H1 0 (Ω). Remark 13.21. There exists n0 > 0 such that for n ≥ n0 sup t≥0 Jh(u0 + tvn) < Jh(u0) + α(Ω) uniformly in z, where u0 is the local minimum in Theorem 13.12. We use the notation: For c ∈ R, [Jh ≤ c] = {u ∈ M− h : Jh(u) ≤ c}. In this section, we show for a sufficiently small σ > 0 cat([Jh ≤ αh(Ω) + α(RN )− σ]) ≥ 2. (13.21) To prove (13.21), we need some preliminaries. Recall the definition of Lusternik- Schnirelman category. Definition 13.22. (i) For a topological space X, we say a non-empty, closed subset A ⊂ X is contractible to a point in X if and only if there exists a continuous mapping η : [0, 1]×A→ X such that for some x0 ∈ X η(0, x) = x for all x ∈ A, η(1, x) = x0 for all x ∈ A. (ii) We define cat(X) = min { k ∈ N : there exist closed subsets A1, . . . , Ak ⊂ X such that Aj is contractible to a point in X for all j and ∪k j=1 Aj = X}. 128 HWAI-CHIUAN WANG EJDE-2004/MON. 06 When we do not have finitely many closed subsets A1, . . . , Ak ⊂ X such that Aj is contractible to a point in X for all j and k ∪ j=1 Aj = X, we say cat(X) = ∞. For fundamental properties of Lusternik-Schnirelman category, we refer to Am- brosetti [2] and Schwartz [66]. Here we use the following property: Theorem 13.23. Suppose that X is a Hilbert manifold and Ψ ∈ C1(X,R). Assume that there are c0 ∈ R and k ∈ N, (i) Ψ(x) satisfies the (PS)c-condition for c ≤ c0 ; (ii) cat({x ∈ X : Ψ(x) ≤ c0}) ≥ k. Then Ψ(x) has at least k critical points in {x ∈ X; Ψ(x) ≤ c0}. Theorem 13.24. Let N ≥ 1, SN−1 = {x ∈ RN ; |x| = 1}, and let X be a topological space. Suppose that there are two continuous maps F : SN−1 → X, G : X → SN−1 such that G ◦ F is homotopic to the identity map of SN−1, that is, a continuous map ζ : [0, 1]× SN−1 → SN−1 exists such that ζ(0, x) = (G ◦ F )(x) for each x ∈ SN−1, ζ(1, x) = x for each x ∈ SN−1. Then cat(X) ≥ 2. Proof. We argue indirectly and suppose that cat(X) = 1, that is, that X is con- tractible to a point in itself. Thus, a continuous map η : [0, 1]×X → X exists such that for some x0 ∈ X η(0, x) = x for all x ∈ X, η(1, x) = x0 for all x ∈ X. Consider a homotopy β : [0, 1]× SN−1 → SN−1 defined by β(s, x) = G(η(s, F (x))). Then β(0, x) = (G ◦ F )(x) for all x ∈ X, β(1, x) = G(x0) for all x ∈ X. Thus G ◦ F is homotopic to a constant map. However,by assumption, G ◦ F is homotopic to the identity. Thus SN−1 is contractible to a point in SN−1, which is a contradiction. Therefore cat(X) ≥ 2. � Let A1 = { u ∈ H1 0 (Ω)\{0} : 1 ‖u‖H1 t−( u ‖u‖H1 ) > 1 } ∪ {0} A2 = { u ∈ H1 0 (Ω)\{0} : 1 ‖u‖H1 t−( u ‖u‖H1 ) < 1}. Lemma 13.25. We have the following results: (i) H1 0 (Ω)\M− h = A1 ∪A2; (ii) M+ h ⊂ A1; (iii) t0 > 1 and n1 ≥ n0 exist such that u0 + t0vn ∈ A2 for each n ≥ n1, where n0 is defined as in Remark 13.21; (iv) a sequence {sn} ⊂ (0, 1) exists such that u0 + snt0vn ∈ M− h for each n ≥ n1. EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 129 Proof. (i) By Lemma 13.4 (iii), we have M− h = { u ∈ H1 0 (Ω)\{0} : 1 ‖u‖H1 t−( u ‖u‖H1 ) = 1 } . Then H1 0 (Ω)\M− h = A1 ∪A2. (ii) For each u ∈ M+ h , we have 1 < tmax(u) < t−(u). Since t−(u) = 1 ‖u‖H1 t−( u ‖u‖H1 ), then M+ h ⊂ A1. In particular, u0 ∈ A1. (iii) There is a constant c > 0 such that 0 < t−( u0+tvn ‖u0+tvn‖H1 ) < c for each t ≥ 0 and each n ∈ N. Otherwise, a sequence {tn} and a subsequence {vn} exist such that t−( u0+tnvn ‖u0+tnvn‖H1 ) → ∞ as n → ∞. Let wn = u0+tnvn ‖u0+tnvn‖H1 . We claim that b(wn) is bounded below away from zero. Case (a) : tn = c0 + o(1) as n→∞, where c0 > 0. By Lemma 13.18, we have a(vn) = b(vn) + o(1) = 2p p− 2 α(Ω) + o(1). Thus, b(wn) = 1 ‖u0 tn + vn‖p H1 ∫ Ω ( u0 tn + vn)p ≥ b(vn) 2p−1 ( ‖u0 tn ‖p H1 + ‖vn‖p H1 ) = 2p p−2α(Ω) 2p−1 (‖u0‖p H1 cp 0 + ( 2p p−2α(Ω) )p/2) + o(1). Case (b) : tn →∞ as n→∞. The proof is similar to Case (a). Case (c) : tn = o(1) as n→∞. By Lemma 13.18, we have ‖u0 + tnvn‖2H1 = ‖u0‖2H1 + t2n‖vn‖2H1 + 2tn〈vn, u0〉H1 = ‖u0‖2H1 + o(1). Thus, b(wn) ≥ 1 ‖u0 + tnvn‖p H1 ∫ Ω up 0 = 1 ‖u0‖p H1 ∫ Ω up 0 + o(1). From Case (a), (b) and (c), b(wn) is bounded below away from zero. Since t−(wn)wn ∈ M− h ⊂ Mh, we have Jh(t−(wn)wn) = 1 2 [t−(wn)]2 − 1 p [t−(wn)]pb(wn)− t−(wn) ∫ Ω hwn which approaches −∞ as n→∞. However, Jh is bounded below on Mh, which is a contradiction. Let t0 = ( p− 2 2pα(Ω) |c2 − a(u0)|)1/2 + 1, then ‖u0 + t0vn‖2H1 = a(u0) + t20( 2p p− 2 )α(Ω) + o(1) > c2 + o(1) ≥ [ t−( u0 + t0vn ‖u0 + t0vn‖H1 ) ]2 + o(1). 130 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Thus, there is an n1 ≥ n0, where n0 is defined as in Remark 13.21, such that, or n ≥ n1, 1 ‖u0 + t0vn‖H1 t−( u0 + t0vn ‖u0 + t0vn‖H1 ) < 1, or u0 + t0vn ∈ A2. (iv) Define a path γn(s) = u0 + st0vn for s ∈ [0, 1] and each n ≥ n1where t0 > 1, then γn(0) = u0 ∈ A1, γn(1) = u0 + t0vn ∈ A2. Since 1 ‖u‖H1 t−( u ‖u‖H1 ) is a continuous function for nonzero u and γn([0, 1]) is con- nected, a sequence {sn} ⊂ (0, 1) exists such that u0 + snt0vn ∈ M− h . � Define a map Fn : SN−1 → H1 0 (Ω) by, for z ∈ SN−1, Fn(z)(z) = u0(z) + snt0vn(z) for n ≥ n1, where vn(z) = ψR(z)u(z− tnz) and n1 is defined as in Lemma 13.25. Then we have the following proposition. Proposition 13.26. A sequence {σn} ⊂ R+ exists such that Fn(SN−1) ⊂ [ Jh ≤ αh(Ω) + α(RN )− σn ] . Proof. By Lemma 13.25 (iv) and Remark 13.21, we have that for each n ≥ n1 u0 + snt0vn ∈ M− h and Jh(u0 + snt0vn) ≤ αh(Ω) + α(RN ) − σn, the conclusion holds. � For c > 0, we define bc(u) = ∫ Ω c|u|p; Ic(u) = 1 2 a(u)− 1 p bc(u); MIc = {u ∈ H1 0 (Ω)\{0} : 〈I ′c(u), u〉 = 0}. Recall that a unique t− = t−(u) > 0 and t1 = t1(u) > 0 exist such that t−u ∈ M− h and t1u ∈ M(Ω). Lemma 13.27. For u ∈ Σ = {u ∈ H1 0 (Ω) | ‖u‖H1 = 1}, we have the following results: (i) a unique tc(u) > 0 exists such that tc(u)u ∈ MIc and max t≥0 Ic(tu) = Ic(tc(u)u) = ( 1 2 − 1 p )bc(u)− 2 p−2 ; (ii) for 0 < µ < 1, d1(µ) > 0 exists such that for ‖h‖L2 < d1(µ) Jh(t−u) ≥ (1− µ) p p−2 J(t0u)− 1 2µ ‖h‖2L2 . Proof. (i) For each u ∈ Σ, let f(t) = Ic(tu) = 1 2 t 2 − 1 p t pbc(u), then f(t) → −∞ as t→∞, f ′(t) = t− tp−1bc(u) and f ′′(t) = 1− (p− 1)tp−2bc(u). Let tc(u) = [ 1 bc(u) ] 1 p−2 > 0. EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 131 Then f ′(tc(u)) = 0, tc(u)u ∈ MIc and (tc(u))2f ′′(tc(u)) = a(tc(u)u)− (p− 1)bc(tc(u)u) = (2− p)(tc(u))2a(u) < 0. Thus, a unique tc(u) > 0 exists such that tc(u)u ∈ MIc and max t≥0 Ic(tu) = Ic(tc(u)u) = ( 1 2 − 1 p )bc(u)− 2 p−2 . (ii) Let c = 1 1−µ , tc = t 1 1−µ > 0 and t1 = t1(u) > 0 such that tcu ∈ MIc and t1u ∈ M(Ω). For µ ∈ (0, 1), we have | ∫ Ω htcudz| ≤ ‖tcu‖H1‖h‖L2 ≤ µ 2 ‖tcu‖2H1 + 1 2µ ‖h‖2L2 . Then by part (i), sup t≥0 Jh(tu) ≥ Jh(tcu) ≥ 1− µ 2 ‖tcu‖2H1 − 1 p b(tcu)− 1 2µ ‖h‖2L2 = (1− µ) [1 2 ‖tcu‖2H1 − 1 (1− µ)p ∫ Ω |tcu|p ] − 1 2µ ‖h‖2L2 = (1− µ)I 1 1−µ (tcu)− 1 2µ ‖h‖2L2 = (1− µ) p p−2 ( 1 2 − 1 p )b(u)− 2 p−2 − 1 2µ ‖h‖2L2 = (1− µ) p p−2 J(t1u)− 1 2µ ‖h‖2L2 ≥ (1− µ) p p−2α(Ω)− 1 2µ ‖h‖2L2 . For µ ∈ (0, 1), there exists d1(µ) > 0 such that for ‖h‖L2 < d1(µ) sup t≥0 Jh(tu) > 0. By Lemma 13.4, there exists t− = t−(u) > 0 such that t−u ∈ M− h and sup t≥0 Jh(tu) = Jh(t−u). Thus, for ‖h‖L2 < d1(µ), Jh(t−u) ≥ (1− µ) p p−2 J(t1u)− 1 2µ ‖h‖2L2 . � Lemma 13.28. A δ0 > 0 exists such that if u ∈ M(Ω) and J(u) ≤ α(RN ) + δ0, then ∫ RN z |z| (|∇u|2 + u2)dz 6= 0. Proof. If not, a sequence {un} ⊂ M(Ω) exists such that J(un) = α(RN )+ o(1) and∫ RN z |z| (|∇un|2 + u2 n)dz = 0. 132 HWAI-CHIUAN WANG EJDE-2004/MON. 06 By Theorem 4.4, {un} is a (PS)α(RN )-sequence in H1 0 (Ω) for J . By Theorem 4.18, infv∈M(Ω) J(v) = α(Ω) = α(RN ) is not achieved. Let u be the unique positive solution of Equation (1.1) in RN . It follows from Theorem 3.1 that a sequence {zn} exists in RN such that |zn| → ∞ as n→∞ and un(z) = u(z − zn) + o(1) strongly in H1(RN ). Assume zn |zn| → z0 as n → ∞, where z0 is a unit vector in RN . Then by Theorem 2.20, we have 0 = ∫ RN z |z| (|∇un|2 + u2 n)dz = ∫ RN z + zn |z + zn| (|∇u|2 + u2)dz + o(1) = ( 2p p− 2 )z0α(RN ) + o(1), which is a contradiction. � Lemma 13.29. d0 > 0 exists such that for ‖h‖L2 < d0, we have∫ RN z |z| (|∇u|2 + u2)dz 6= 0, for u ∈ [ Jh < αh(Ω) + α(RN ) ] . Proof. For u ∈ [ Jh < αh(Ω) + α(RN ) ] , then u/‖u‖H1 ∈ Σ. There exists a t0 > 0 such that t0u/‖u‖H1 ∈ M. By Lemma 13.27 (ii), we have for any µ ∈ (0, 1) and ‖h‖L2 < d1(µ) J( t0u ‖u‖H1 ) ≤ (1− µ)− p p−2 ( Jh(u) + 1 2µ ‖h‖2L2 ) . (13.22) Since αh(Ω) < 0, we have [ Jh < αh(Ω) + α(RN ) ] ⊂ [ Jh < α(RN ) ] . Thus by (13.22), we have, for u ∈ [ Jh < αh(Ω) + α(RN ) ] , J( t0u ‖u‖H1 ) ≤ (1− µ)− p p−2 ( α(RN ) + 1 2µ ‖h‖2L2 ) . Take µ ∈ (0, 1) such that d1(µ) > d0 > 0 and δ0 > 0 exist such that for ‖h‖L2 < d0 J( t0u ‖u‖H1 ) ≤ α(RN ) + δ0. (13.23) Since t0u/‖u‖H1 ∈ M, by Lemma 13.28 and (13.23)∫ RN z |z| (|∇( t0u ‖u‖H1 )|2 + ( t0u ‖u‖H1 )2)dz 6= 0, or, ∫ RN z |z| (|∇u|2 + (u)2)dz 6= 0. � EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 133 By Lemma 13.29 ∫ RN z |z| (|∇u|2 + |u|2)dz 6= 0 for all u ∈ [ Jh < αh(Ω) + α(RN ) ] . We define G : [ Jh < αh(Ω) + α(RN ) ] → SN−1 by G(u) = ∫ RN z |z| (|∇u|2 + |u|2)dz�| ∫ RN z |z| (|∇u|2 + |u|2)dz|. Proposition 13.30. For n ≥ n1 and ‖h‖L2 < d0, the map G ◦ Fn : SN−1 → SN−1 is homotopic to the identity. Proof. We define ζn(θ, z) : [0, 1]× SN−1 → SN−1 by ζn(θ, z) =  G((1− 2θ)Fn(z) + 2θu(z − tnz)) for θ ∈ [0, 1/2); G ( u ( z − tn 2(1−θ)z )) for θ ∈ [1/2, 1); z for θ = 1. Since h ∈ L2(Ω) ∩ LN 2 (Ω) ∩ Ls(Ω) for some s > N , then by Theorem 8.11 we have u0 ∈ C1,θ(Ω). First, we claim that limθ→1− ζn(θ, z) = z and limθ→ 1 2 − ζn(θ, z) = G(u(z − tnz)). (a) lim θ→1− ζn(θ, z) = z : since∫ RN z |z| (∣∣∇u(z − tn 2(1− θ) z )∣∣2 + u ( z − tn 2(1− θ) z )2)dz = ∫ RN z + tn 2(1−θ)z |z + tn 2(1−θ)z| (|∇u(z)|2 + u(z)2)dz = ( 2p p− 2 )α(RN )z + o(1) as θ → 1−, then lim θ→1− ζn(θ, z) = z. (b) lim θ→ 1 2 − ζn(θ, z) = G(u(z − tnz)) : since u and u0 ∈ C1,θ(Ω), then ‖(1− 2θ)Fn(z) + 2θu(z − tnz)‖H1 = ‖u(z − tnz)‖+ o(1) as θ → 1 2 − . By the continuity of G, we obtain lim θ→ 1 2 − ζn(θ, z) = G(u(z − tnz)). Thus, ζn(θ, z) ∈ C([0, 1]× SN−1, SN−1) and ζn(0, z) = G(Fn(z)) for all z ∈ SN−1, ζn(1, z) = z for all z ∈ SN−1, provided n ≥ n1 and ‖h‖L2 < d0. This completes the proof. � Thus we have the following theorem. Theorem 13.31. Jh(u) has at least two critical points in[ Jh < αh(Ω) + α(RN ) ] . 134 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Proof. Applying Theorem 13.24 and Proposition 13.30, we have for sufficiently large n ≥ n1 and ‖h‖L2 < d0, cat( [ Jh ≤ αh(Ω) + α(RN )− σn ] ) ≥ 2. By Proposition 13.13 and Theorem 13.23, the theorem holds. � By Theorem 13.11, there is a nontrivial solution in M+ h and by Theorem 13.31, there are two nontrivial solutions in M− h . Therefore, we have Theorem 13.32. Let the domain Ω to be Ar, RN , Ar\D, or RN\D, where D is a C1,1 bounded domain in RN and assume that h ∈ L N 2 (Ω) ∩ Ls(Ω) ∩ L2(Ω) for some s > N , 0 < ‖h‖L2 < d(p, α), and h � 0 0 ≤ h(z) ≤ c exp(−(1 + ε)|z|) for any z ∈ Ω and for some positive constants c, ε, where d(p, α) is defined as in (13.1). Then there are three positive solutions of equation (1.2). Bibliographical notes: The results of this section are from Adachi-Tanaka [1]. 13.2. Symmetry Breaking in a Bounded Symmetry Domain. The main purpose of this sectionis to present the breaking of symmetry by a perturbation of the finite strip Ar −t,t. Let 0 < r1 < r and consider the finite strip with a hole, Ωt = Ar −t,t \ BN ((x, 0); r1). We prove that t0 > 0 exists such that for t ≥ t0, Equation (1.1) on Ωt has three positive solutions, one of which is y-symmetric while the other two are nonaxially symmetric. 13.2.1. Existence of Three Solutions. Example 13.33 (y-symmetric large domain). (i) For 0 < r1 < r and x in BN−1(0; r + r1), consider the infinite strip with holes Ωl = Ar \ [ BN ((x, l); r1) ∪BN ((x,−l); r1) ] for some l > 0. Then Ωl is a y-symmetric large domain in Ar; (ii) Let 0 < r1 < r and 0 < yn+1 = nyn for n = 1, 2, . . . . Consider the infinite strip with infinite holes D = Ar \ { ∞ ∪ n=1 [ BN ((0, yn); r1) ∪BN ((0,−yn); r1) ]} . Then D is a y-symmetric large domain in Ar. Proposition 13.34. αs(Ar −t,t) = α(Ar −t,t) and αs(Ar) = α(Ar). Proof. By Gidas-Ni-Nirenberg [34] and Chen-Chen-Wang [23], every positive solu- tion of (1.1) in a finite strip Ar −t,t and in an infinite strip Aris y-symmetric. � The following symmetric results are required to assert our main result. Theorem 13.35. (i) Suppose that Ω is a proper y-symmetric large domain in Ar and J does not satisfy the (PS)αs(Ω)-condition in H1 s (Ω). Then αs(Ω) ≥ 2α(Ar); (ii) If Ω is a proper y-symmetric large domain in Ar, then α(Ar) < αs(Ω). EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 135 Proof. (i) Suppose that J does not satisfy the (PS)αs(Ω)-condition in H1 s (Ω). By Lemma 2.41, a subsequence {un} exists such that J(ξnun) = αs(Ω) + o(1) and J ′(ξnun) = o(1) in H−1 s (Ω), where ξn is defined as in (2.1). Let wn = ξnun. Then by Lemma 2.9, we obtain J(wn) = αs(Ω) + o(1), J ′(wn) = o(1) in H−1(Ω). (13.24) Since Ω $ Ar, K > 0 exists such that wn = 0 in QK , and two disjoint large domains Ω1 and Ω2 in Ar exist such that (x, y) ∈ Ω2 if and only if (x,−y) ∈ Ω1, Ω\QK = Ω1 ∪ Ω2 where QK = Ω ∩BN (0;K). Let wi n(x) = { wn(x) for x ∈ Ωi, 0 for x /∈ Ωi, for i = 1, 2. Then wi n ∈ H1 0 (Ωi), w1 n(x, y) = w2 n(x,−y), wn = w1 n + w2 n, and J(w1 n) = J(w2 n). Moreover, we have αs(Ω) + o(1) = J(wn) = J(w1 n) + J(w2 n) = 2J(wi n) for i = 1, 2, or J(wi n) = 1 2 αs(Ω) + o(1) for i = 1, 2. By (13.24), we have J ′(wi n) = o(1) in H−1(Ωi) for i = 1, 2. Therefore, 1 2αs(Ω) is a positive (PS)-value in H1 0 (Ωi) for J . By Lemma 4.12, Theorem 4.13, and Definition 4.14, 1 2 αs(Ω) ≥ αM(Ωi) = α(Ωi). Since Ωi is a large domain in Ar, by Lemma 4.18 (ii), we have α(Ωi) = α(Ar). Thus, αs(Ω) ≥ 2α(Ar). (ii) Clearly, we have α(Ar) ≤ αs(Ω). Assume that α(Ar) = αs(Ω), then by Theorem 5.7, J does not satisfy the (PS)αs(Ω)-condition in H1 s (Ω) for J . By Lemma 13.35, 2α(Ar) ≤ αs(Ω) = α(Ar), which is a contradiction. � Theorem 13.36. Let 0 < r1 < r, and for each t > 0 Θt = Ar \ [ BN ((x, t+ r1); r1) ∪BN ((x,−(t+ r1)); r1) ] . Then t0 > 0 exists such that αs(Θt) < 2α(Ar) for all t ≥ t0. In particular, there is a y-symmetric positive ground state solution of Equation (1.1) in Ωt. Proof. By Lien-Tzeng-Wang [47], α(Ar −t,t) is strictly decreasing as t is strictly increasing and α(Ar −t,t) ↘ α(Ar) as t→∞. Thus, there is a t0 > 0 such that α(Ar −t,t) < 2α(Ar) for each t ≥ t0. By Proposition 13.34, α(Ar −t,t) = αs(Ar −t,t) for each t. Thus, αs(Ar −t,t) < 2α(Ar) for each t ≥ t0. Since Θt ⊃ Ar −t,t for each t ≥ t0. Therefore, we have αs(Θt) ≤ αs(Ar −t,t) for each t ≥ t0. We then conclude that αs(Θt) ≤ αs(Ar −t,t) < 2α(Ar) for each t ≥ t0. 136 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Since Θt is a y-symmetric large domain in Ar, by Theorem 13.35, J satisfies the (PS)αs(Ω)-condition in H1 s (Ω), or telse here is a y-symmetric positive ground state solution of Equation (1.1) in Θt for each t ≥ t0. � x y t -t r r1 x y t -t r 1r Figure 12. the finite strip with a hole Let 0 < r1 < r and consider the finite strip with a hole Ωt = Ar −t,t\BN ((x, 0); r1). Then we have the following assertion. Theorem 13.37. t0 > 0 exists such that for t ≥ t0, Equation (1.1) on Ωt has three positive solutions of which one is y-symmetric and the other two are nonaxially symmetric. Proof. Let Ω = Ar \ BN ((x, 0); r1). Then Ω is a y-symmetric large domain in Ar. By Theorem 13.35, we have α(Ar) < αs(Ω). By Lien-Tzeng-Wang [47], (1.1) admits a ground state solution in Ar 0,t and in Ar, and α(Ar 0,t) is strictly decreasing as t is strictly increasing and α(Ar 0,t) ↘ α(Ar) as t→∞. Take t1 > 0 such that for t ≥ t1, α(Ar) < α(Ar 0,t) < αs(Ω). (13.25) Note that Ar r1,t1+r1 $ Ωt $ Ar for t ≥ t0 = t1 + r1. By Theorem 5.7, we conclude that α(Ar) < α(Ωt) < α(Ar r1,t1+r1 ). (13.26) By Lien-Tzeng-Wang [47], if Ω is a domain of RN , then α(Ω) is invariant by rigid motions. Thus, α(Ar r1,t1+r1 ) = α(Ar 0,t1). (13.27) EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 137 Therefore, by (13.25)-(13.27) α(Ar) < α(Ωt) < α(Ar 0,t1) < αs(Ω). (13.28) Since Ωt ⊂ Ω, we have αs(Ω) ≤ αs(Ωt). (13.29) By (13.28) and (13.29), we obtain α(Ωt) < αs(Ωt). (13.30) By Theorem 12.3, there are a y−symmetry solution u1 and a solution u2 of Equation (1.1) in domain Ωt such that J(u1) = αs(Ωt), J(u2) = α(Ωt). By Theorem 5.7, we may take u1 and u2 to be positive. Let u3(x, y) = u2(x,−y). Then u3 is the third solution. By (13.30), u1, u2 and u3 are different. Moreover, u1 is a y-symmetric solution while both u2 and u3 are nonaxially symmetric solutions of Equation (1.1) in domain Ωt. � Bibliographical notes: The results of this section are from Wang-Wu [74]. 13.3. Multiple Solutions in Domains with Two Bumps. That the existence of solutions of (1.1) 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 solution of (1.1) in a bounded domain. For general unbounded domains Ω, because of the lack of compactness, the existence of solutions of Equa- tion (1.1) is an important open question. Recently, there has been some progress in determining the existence and multiplicity of solutions as follows: Bahri-Lions [13], Coti Zelati [28], Chabrowski [20], Chen-Lee-Wang [24], Chen-Wang[26] , Chen-Lin- Wang [25], Lien-Tzeng-Wang [47], del Pino-Felmer [30], [31], and Wang [71] used the (PS)−theory to treat the existence of solutions of (1.1). Byeon [16], Chen-Ni- Zhou [22], Dancer [29], and Wang-Wu [74] asserted the existence of three positive solutions of semilinear elliptic equations in a dumbbell domain. Jimbo [43] and [44] asserted the existence of solutions depending on the width of the corridor of the dumbbell. In this section we assert that there is a R0 > 0 such that for R > R0 Equation (1.1) on the two bumps domain DR has three positive solutions in which one is y-symmetric and other two are nonaxially symmetric. (see Theorem 13.41). Since finite dumbbell is a two bumps domain, the results of Byeon [16], Chen-Ni-Zhou [22], and Dancer [29] are the consequences of our Theorem 13.41. 13.3.1. Existence of Three Solutions. We have the following results. Theorem 13.38. (i) The bounded domains in RN are the achieved domains in RN ; (ii) The C1 quasi-bounded domains are the achieved domains in RN ; (iii) RN is an achieved domain in RN ; (iv) The periodic domains in RN are the achieved domains. In particular, the infinite strip Ar is an achieved domain in RN ; (v) s0 > 0 exists such that Fr s is an achieved domain in RN if s > s0. 138 HWAI-CHIUAN WANG EJDE-2004/MON. 06 Proof. (i) By Theorem 12.3. (ii) By Theorem 5.4. (iii) and (iv) follows from Lien-Tzeng-Wang [47]. (v) By Theorem 12.7. � Throughout this section, let Θ be a proper achieved y-symmetric domain in RN bounded in the x−direction, such as bounded domains and the infinite strip Ar. For R > 0, let Ω1 R and Ω2 R be two bounded domains in RN such that R = dist{0,Ω1 R}, Ω2 R = {(x, y) : (x,−y) ∈ Ω1 R}. and the y-symmetric domain DR = Ω1 R ∪Θ ∪ Ω2 R. We call DR a two-bumps domain. Here are some examples of two-bumps domains. Example 13.39. (i) For t > R > r > 0. The bounded dumbbell domain DR is a two-bumps domain, where DR = BN ((0,−t), r) ∪Ar −t,t ∪BN ((0, t), r); (ii) For t > R > r > 0. The unbounded dumbbell domain DR is a two-bumps domain, where DR = BN ((0,−t), r) ∪Ar ∪BN ((0, t), r); (iii) For t > R > r > 0. The curved dumbbell domain DR = Ω1 R ∪Θ∪Ω2 R is a two bumps domain, where Ω1 R and Ω2 R are two bounded domains in RN such that R = dist{0,Ω1 R}, Ω2 R = {(x, y) : (x,−y) ∈ Ω1 R}. and Θ is a curved bounded y-symmetric domain in RN . Then we have the following assertion. Theorem 13.40. Let DR be a two-bumps domain, where DR = Ω1 R ∪Θ∪Ω2 R, and let Θ be a proper achieved y-symmetric domain in RN bounded in the x−direction. Then we have, for all R > 0, (i) α(Θ) ≥ α(DR) > α(RN ); (ii) J satisfies the (PS)αX(DR)-condition in X(DR). Proof. By Theorem 5.7 and Theorem 12.3, it suffices to assume that Θ is un- bounded. (i) Since Θ ⊂ DR $ RN , we have α(Θ) ≥ α(DR) ≥ α(RN ). Suppose that α(DR) = α(RN ), by Theorem 5.7, J does not satisfy the (PS)α(DR)-condition. By Theorem 5.11, a sequence {un} in H1 0 (DR) exists such that {un} and {ξnun} are the (PS)α(DR)-sequences for J , where ξn is defined as in (2.1). Let wn = ξnun. Then J(wn) = α(DR) + o(1), J ′(wn) = o(1) in H−1(DR). Since DR = Ω1 R ∪Θ ∪Ω2 R is a y-symmetric domain in RN separated by a bounded domain, n0 > 0 exists such that for n ≥ n0, wn ∈ H1 0 (Θ), J(wn) = α(DR) + o(1), and a(wn) = b(wn) + o(1). By Theorem 4.3, there is a sequence {sn} in R+ such EJDE-2004/MON. 06 PALAIS-SMALE APPROACHES 139 that {snwn} is in M(Ω) and {snwn} is a (PS)α(DR)-sequence in X(Ω) for J . Thus α(Θ) ≤ α(DR). We then conclude that α(Θ) = α(DR) = α(RN ). However, since Θ is a proper achieved y-symmetric domain in RN , by Theorem 5.7, α(Θ) > α(RN ). This is a contradiction. Thus, α(Θ) ≥ α(DR) > α(RN ) for all R > 0. (ii) It suffices to prove the case X(DR) = H1 0 (DR). Since Ω1 R, Θ, and Ω2 R are achieved, by Theorem 5.7, α(DR) < min{α(Ω1 R), α(Θ), α(Ω2 R)}. By Theorem 5.12, J satisfies the (PS)α(DR)-condition in H1 0 (DR). � We apply Theorems 6.7 and 13.40 to prove the following result. Theorem 13.41. There is an R0 > 0 such that for R > R0 Equation (1.1) on DR has three positive solutions, of which one is y-symmetric and other two are nonaxially symmetric. Proof. Take ρ > 0 such that Ω = (RN \ RN −ρ,ρ) ∪ Θ is connected. Then Ω is a y-symmetric large domain in RN separated by a bounded domain. By Theorem 6.7, we have α(RN ) = α(Ω) < αs(Ω). By Lien-Tzeng-Wang [47], α(BN (0, R)) is strictly decreasing as R is strictly increasing and α(BN (0, R)) ↘ α(RN ) as R→∞. Take R0 > 0, such that for R > R0 α(RN ) < α(BN (0, R)) < αs(Ω). (13.31) Since BN ((xR, yR), R) $ DR $ RN , by Theorem 5.7 and Theorem 13.40, we con- clude that α(RN ) < α(DR) < α(BN ((xR, yR), R)) = α(BN (0, R)). (13.32) Therefore, by (13.31) and (13.32) and DR ⊂ Ω, we have α(DR) < α(BN (0, R)) < αs(Ω) ≤ αs(DR). (13.33) Thus, α(DR) < αs(DR). (13.34) By Theorem 12.3, there are a y−symmetry positive solution u1 and a positive solution u2 of Equation (1.1) in domain DR for R > R0 such that J(u1) = αs(DR), J(u2) = α(DR). Let u3(x, y) = u2(x,−y), then u3 is the third positive solution. By (13.34), u1, u2, and u3 are different. Moreover, u1 is a y-symmetric positive solution while both u2, and u3 are nonaxially symmetric positive solutions of (1.1) in domain DR. � Bibliographical notes: The results of this section are from Wang-Wu [74]. 140 HWAI-CHIUAN WANG EJDE-2004/MON. 06 References [1] S. Adachi and K. Tanaka, Four positive solutions for the semilinear elliptic equation: −4u+ u = a(x)up + f(x) in RN , Calc. Var. 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Differential Equations, 92 (1991), 163-178. Department of Mathematics, National Tsing Hua University, Hsinchu, Taiwan E-mail address: hwang@mail.math.nthu.edu.tw 1. Introduction 2. Preliminaries 3. Palais-Smale Decomposition Theorems 4. Palais-Smale Values and Indexes of Domains 5. Palais-Smale Conditions 6. Symmetric Palais-Smale Conditions 7. Symmetric Palais-Smale Decomposition Theorems 8. Fundamental Properties, Regularity, and Asymptotic Behavior of Solutions 8.1. Fundamental Properties of Solutions 8.2. Regularity of Solutions 8.3. Asymptotic Behavior of Solutions 9. Symmetry of Solutions 9.1. Open Question: 10. Nonachieved Domains and Esteban-Lions Domains 11. Higher Energy Solutions 11.1. Existence Results 11.2. Dynamic Systems of Solutions 12. Achieved Domains 12.1. Open Question: 13. Multiple Solutions 13.1. Multiple Solutions for a Perturbed Equation 13.2. Symmetry Breaking in a Bounded Symmetry Domain 13.3. Multiple Solutions in Domains with Two Bumps References