Existence and Multiplicity of Solutions for a Class of Semilinear Elliptic Equations 1
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1 Journal of Mathematical Analysis and Applications 257, (2001) doi: /jmaa , available online at on Existence and Multiplicity of Solutions for a Class of Semilinear Elliptic Equations 1 Shui-Qiang Liu Department of Mathematics, Shaoyang Teachers College, Shaoyang , People s Republic of China and Chun-Lei Tang Department of Mathematics, Southwest Normal University, Chongqing , People s Republic of China Submitted by J. McKenna Received March 6, 2000 The existence and multiplicity results are obtained for solutions of a class of the Dirichlet problem for semilinear elliptic equations by the least action principle and the minimax methods, respectively Academic Press Key Words: Dirichlet problem; resonance; critical point; critical growth; the least action principle; the minimax methods. 1. INTRODUCTION AND MAIN RESULTS Consider the Dirichlet boundary value problem u = f x u+hx for a.e. x u = 0on (1) where R N N 1 is a bounded domain, f R R is a Carathéodory function, that is, f x t is measurable in x for every t R 1 Project supported by the National Natural Science Foundation of China and Project 00C024 supported by the Educative Department of Hunan Province X/01 $35.00 Copyright 2001 by Academic Press All rights of reproduction in any form reserved.
2 322 liu and tang and continuous in t for a.e. x, and h L 2. Letλ k k = 1 2 be the kth distinct eigenvalue of the eigenvalue problem u = λu in u = 0on and Eλ k k = 1 2 be the eigenspace corresponding to λ k. Set Fx t t 0 f x s ds for all t R and a.e. x. In this paper, we obtain an existence theorem by the least action principle for problem (1) in the critical growth case and a multiplicity result is obtained by using the minimax methods in the critical point theory, and in particular, a three-critical-point theorem proposed by Brezis and Nirenberg [4] in the subcritical growth case. The main results are the following theorems. Theorem 1. Suppose that there exist a constant C 1 > 0 and a real function γ L 1 such that f x t C 1 t γx (2) for all t R and a.e. x, where 2 2N/N 2 if N 3 and 2 may be replaced by any number in 1 + if N = 1 or 2, and that Fx t 1 2 λ 1t 2 (3) as t uniformly for a.e. x. Assume that h L 2 satisfies that hxψx dx = 0 (4) where ψ is the normalized eigenfunction corresponding to λ 1, ψx > 0 for all x. Then problem (1) has at least one solution in H0 1, where H1 0 is a Hilbert space with the norm given by ( ) 1 u = u 2 2 dx Theorem 2. Suppose that h = 0 (5) and that there exist C 2 > 0 and 2 <p<2n/n 2 for N 3(2 <p< +, for N = 1 2) such that f x t C 2 t p (6)
3 solutions for elliptic equations 323 for all t R and a.e. x. Assume that (3) holds and that there exists an integer m 1b>0, and δ>0 such that λ m f x t t λ m+1 b (7) for all 0 < t δ and a.e. x. Then problem (1) has at least two nonzero solutions in H 1 0. Many papers deal with the case that f x t lim = λ t t 1 (8) uniformly for a.e. x (see [1, 3, 5, 8, 10, 11, 13, 17 20] and their references), which is called resonant at the first eigenvalue. Note that (8) implies that for a.e. x, where A x =λ 1 A x lim sup t for a.e. x. The more general case that A x λ 1 2Fx t t 2 for a.e. x is considered by many authors (see [6, 7, 9, 14] and the references therein). The existence results are given for problem (1) in [1, 3, 5 11, 13, 14, 17 20]. In [2, 12, 21] some multiplicity theorems are obtained by using the topological degree technique and the variational methods, respectively. Except for [7, 14], the linear case is only treated. Reference [14] allows subcritical growth on f x t and in [7] the critical growth on f x t is considered. Remark 1 There are functions f x t and hx satisfying our Theorem 1 and not satisfying those in [1 3, 5 14, 16 20]. In fact, let f x t =λ 1 t 2t 1 + t t 2 1 cos t 2 and h L 2 satisfying (4). Then f x t does not satisfy the theorems in [1 3, 5, 6, 8 14, 16 20] for it is growing critically. Moreover f x t does not satisfy the theorems in [7] yet because A x =λ 1 in this case, { } inf v H 1 0 v L 2 =1 v 2 dx v 0 A xv dx = 0
4 324 liu and tang (see Theorem 1 in [7]). But Fx t = 1 2 λ 1t 2 ln1 + t 2 +sin t 2 satisfies (2) and (3). Hence f x t and hx satisfy our Theorem 1. Remark 2 There are functions f x t satisfying our Theorem 2 and not satisfying those in [2, 12, 21]. In fact, let { λm t t δ f x t = C 3 λ 1 t 2t/1 + t 2 +pt p 2 t cos t p t δ, where C 3 = λ m λ 1 2/1 + δ 2 +pδ p 2 cos δ p 1 and 2 <p<2n/ N 2 for N 3(2 <p<+, for N = 1 2). Then f x t satisfies our Theorem 2. But this f x t does not satisfy the theorems in [2, 12, 21] for it is not linearly growing. Let ϕu = PROOF OF THEOREMS u 2 dx Fx u dx hu dx for u H0 1. In a way similar to Theorem 1.4 in [15] we can prove that ϕ C 1 H0 1R. It is well known that u H1 0 is a solution of problem (1) if and only if u is a critical point of ϕ. Foru H0 1, set ( ) ū = u ψdx ψ ũ = u ū Then one has ū Eλ 1 ũ Eλ 1 Proof of Theorem 1 First there exist a real function g L 1, and G CR R which is subadditive, that is, Gs + t Gs+Gt (9) for all s t R, and coercive, that is, Gt + (10) as t, and satisfies that Gt t+4 (11)
5 for all t R, such that solutions for elliptic equations 325 Fx t 1 2 λ 1t 2 Gt+gx (12) for all t R and a.e. x. In fact, since Fx t 1 2 λ 1t 2 as t uniformly for a.e. x, there exists a sequence of positive integers (n k ) with n k+1 > 2n k for all positive integer k such that Fx t 1 2 λ 1t 2 k (13) for all t n k and a.e. x. Letn 0 = 0 and define Gt =k t n k 1 n k n k 1 (14) for n k 1 t <n k, where k N. By the definition of G we have for n k 1 t <n k. It follows that for all t R and a.e. x, where k + 2 Gt k + 3 (15) Fx t 1 2 λ 1t 2 Gt+gx gx = C 1 2 n2 1 + n 1γx+ λ 1 2 n2 1 + n 1+4 In fact, by (2) one has Fx t 1 2 λ 1t 2 C 1 2 t2 + γxt+ λ 1 +t (16) 2 t2 for all t R and a.e. x. For every t R there exists k N such that In the case that k = 1 we have n k 1 t <n k Fx t 1 2 λ 1t 2 C 1 2 n2 1 + γxn 1 + λ 1 2 n2 1 + n 1 = gx 4 Gt+gx for all t n 1 and a.e. x by (16) and (15). In the case that k 2, one has Fx t 1 2 λ 1t 2 k 1 k + 3+gx Gt+gx for all n k 1 t n k and a.e. x by (13) and (15).
6 326 liu and tang It is obvious that G is continuous and coercive. Moreover one has Gt t+4 for all t R. In fact, for every t R there exists k N such that which implies that n k 1 t <n k Gt k + 3 n k t+4 for all t R by (15) and the fact that n k k for all integers k 0. Now we only need to prove the subadditivity of G. Let and m = maxk j. Then we have Hence we obtain, by (15), n k 1 s <n k n j 1 t <n j s + t s+t <n k + n j 2n m <n m+1 Gs + t m + 4 k j + 2 Gs+Gt which shows that G is subadditive. Second, the functional Gv dx is coercive on Eλ 1. If not, there exist C 0 > 0 and a sequence v n in Eλ 1 such that v n and Gv n dx C 0, which implies that lim sup n Gv n dx C 0 < + (17) It follows from the first part of the proof of Lemma 3.2 in [3] that for every α>0 there exists m α > 0 such that measx vx <m α v <α for all v Eλ 1. For the convenience of the reader we state another proof. If it does not hold, there exist α 0 > 0 and a sequence v n n=1 Eλ 1 such that { meas x v n x < 1 } n v n α 0 for all n, which implies that v n 0 for all n. By the homogeneity of the above inequality we may assume that v n =1 and { meas x v n x < 1 } α n 0
7 solutions for elliptic equations 327 for all n. It follows from the compactness of the unit sphere of Eλ 1 that there exists a subsequence, say v n, such that v n converges to some v in Eλ 1. Hence v 0 and v n v 0asn by the equivalence of the norms on the finite-dimensional space Eλ 1. For every positive integer m, there exists N 2m such that v n v < 1/2m whenever n N. Hence one has { x v N x < 1 } { x vx < 1 } N m which follows from the inequality vx v N v +v N x 1 2m + 1 N 1 m for all x with v N x < 1/N. Thus we have { meas x vx < 1 } α m 0 for all m, which implies that v = 0 on a positive measure subset. It contradicts the unique continuation property of the eigenfunction. Let A n =x v n x m α v n Then one has meas\a n <α. For every β>0, there exists M>0 such that Gt β for all t M by the coercivity of G. Let For x A n, one has B n =x v n x M v n x m α v n which implies that A n B n for large n. Now we have, by (11), Gv n dx β meas B n M + 4 meas \B n β meas A n M + 4 meas \A n βmeas α M + 4α for large n. Hence one has lim inf Gv n dx βmeas α M + 4α n
8 328 liu and tang Letting α 0, we obtain lim inf n By the arbitarity of β we have lim inf n Gv n dx β meas Gv n dx =+ which contradicts (17). Third, the functional ϕ is coercive. In fact, from (12), (9), and (11) one obtains Fx u dx 1 2 λ 1 u 2 dx Gu dx + gx dx Gū G ũ dx + gx dx Gū dx +ũ L 1 + 4meas + Gū dx + C 4 ũ+1 for all u H0 1 and some C 4 = C + 4 meas + gx dx where C is a positive constant in Sobolev s inequality, gx dx u L 1 Cu u L 2 Cu for all u H0 1. Hence we have ϕu= 1 u 2 dx 1 ( 2 2 λ 1 u 2 dx 1 ũ 2 dx λ 1 1 ( 1 λ ) 1 2 λ 2 Fxudx 1 2 λ 1 ũ 2 dx+ Gūdx C 4 ũ+1 ũ 2 dx C 5 ũ+1+ Gūdx ) u 2 dx hudx hudx for all u H 1 0 and some C 5 = C 4 + Ch L 2, which implies that ϕ is coercive by the coercivity of the functional Gv dx on Eλ 1 and the fact that u 2 =ū 2 +ũ 2
9 solutions for elliptic equations 329 At last, the functional ϕ is weakly lower semicontinuous. Indeed, from (12) and (15) we obtain Fx t 1 2 λ 1t 2 + gx for all t R and a.e. x. In a way similar to the first part of the proof of Theorem 1 in [7], one can prove that ϕ is weakly lower semicontinuous. It follows from the least action principle (see, e.g., Theorem 1.1 in [15]) that ϕ has a minimum. Hence problem (1) has at least one solution in H 1 0, which completes our proof. Now we prove Theorem 2. For the convenience of the reader we state a three-critical-point theorem proposed by Brezis and Nirenberg (see Theorem 4 in [4]). Lemma 1 [4]. Let X be a Banach space with a direct sum decomposition X = X 1 X2 with dim X 2 < and let ϕ be a C 1 function on X with ϕ0 =0, satisfying the (PS) condition. Assume that, for some δ 0 > 0, and ϕv 0 for v X 1 with v δ 0 ϕv 0 for v X 2 with v δ 0 Assume also that ϕ is bounded from below and inf X ϕ<0. Then ϕ has at least two nonzero critical points. Proof of Theorem 2. Let X 2 be a finite-dimensional subspace of X = H0 1 given by X 2 = Eλ 1 Eλ m and let X 1 = X2. Then there exists δ 0 > 0 such that ϕu 0 ϕu 0 In fact, from (7) one obtains for all u X 2 with u δ 0 for all u X 1 with u δ 0 λ m t 2 tf x t λ m+1 bt 2 for all t <δand a.e. x, which implies that λ m t 2 s tf x st λ m+1 bt 2 s (18) for all 0 < s 1 t < δ and a.e. x. Noting that Fx t = 1 0 tf x stds, we obtain 1 2 λ mt 2 Fx t 1 2 λ m+1 bt 2
10 330 liu and tang for all t <δand a.e. x. By the equivalence of the norms on the finite-dimensional space X 2 there exists C 6 > 0 such that u C 6 u for all u X 2. Hence one has ϕu 1 u 2 dx λ m for all u X 2 with u δ/c 6 by (5). By (6) we have Fx t C 2 p 1 t p +t for all t R and a.e. x. Thus one has u 2 dx 0 Fx t C 2 p 1 + δ 1 p t p = C7 t p for all t δ and a.e. x. Hence we obtain, by (7), Fx t 1 2 λ m+1 bt 2 + C 7 t p for all t R and a.e. x. It follows from (5) that ϕu 1 u 2 dx λ m+1 b u 2 dx C 7 u p L p b u 2 C 2λ 7 C p u p m+1 0 for all u X 1 with u b/2λ m+1 C 7 C p 1/p 2, where C>0 is a constant such that u L p Cu for all u H0 1 by Sobolev s embedding theorem. It is obvious that ϕ is a C 1 -function with ϕ0 =0. By the proof of Theorem 1, ϕ is coercive and bounded from below. Thus ϕ satisfies the (PS) condition by the subcritical growth of f x t and the coercivity of ϕ. In the case that inf X ϕ<0, Theorem 2 follows from Lemma 1. In the case that inf X ϕ 0, by (18) one has ϕu =inf X ϕ = 0 for all u X 2 with u δ, which implies that all u X 2 with u δ are solutions of problem (1). Therefore Theorem 2 is proved. ACKNOWLEDGMENT The authors thank the referee for valuable suggestions.
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