Hindawi Publishing Corporation Boundary Value Problems Volume 2008, Article ID 217636, 10 pages doi:10.1155/2008/217636 Research Article Existence Result for a Class of Elliptic Systems with Indefinite Weights in R2 Guoqing Zhang1 and Sanyang Liu2 1 2 College of Sciences, University of Shanghai for Science and Technology, Shanghai 200093, China Department of Applied Mathematics, Xidian University, Xi’an 710071, China Correspondence should be addressed to Guoqing Zhang, zgqw2001@sina.com.cn Received 31 October 2007; Accepted 4 March 2008 Recommended by Zhitao Zhang We obtain the existence of a nontrivial solution for a class of subcritical elliptic systems with indefinite weights in R2 . The proofs base on Trudinger-Moser inequality and a generalized linking theorem introduced by Kryszewski and Szulkin. Copyright q 2008 G. Zhang and S. Liu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 1. Introduction In this paper, we study the existence of a nontrivial solution for the following systems of two semilinear coupled Poisson equations P −Δu u gx, v, x ∈ R2 , −Δv v fx, u, x ∈ R2 , 1.1 where fx, t and gx, t are continuous functions on R2 × R and have the maximal growth on t which allows to treat problem P variationally, Δ is the Laplace operator. Recently, there exists an extensive bibliography in the study of elliptic problem in RN 1–6. As dimensions N ≥ 3, in 1998, de Figueiredo and Yang 5 considered the following coupled elliptic systems: −Δu u gx, v, x ∈ RN , −Δv v fx, u, x ∈ RN , 1.2 2 Boundary Value Problems where f, g are radially symmetric in x and satisfied the following Ambrosetti-Rabinowitz condition: t fx, sds ≥ c|t|2δ1 , 0 t gx, sds ≥ c|t|2δ2 , ∀t ∈ R, 1.3 0 and for some δ1 > 0, δ2 > 0. They obtained the decay, symmetry, and existence of solutions for problem 1.2. In 2004, Li and Yang 6 proved that problem 1.2 possesses at least a positive solution when the nonlinearities fx, t and gx, t are “asymptotically linear” at infinity and “superlinear” at zero, that is, 1 limt→∞ fx, t/t l > 1, limt→∞ gx, t/t m > 1, uniformly in x ∈ RN ; 2 limt→0 fx, t/t limt→0 gx, t/t 0, uniformly with respect to x ∈ RN . In 2006, Colin and Frigon 1 studied the following systems of coupled Poission equations with critical growth in unbounded domains: ∗ −Δu |v|2 −2 v, 1.4 ∗ −Δv |u|2 −2 u, where 2∗ 2N/N − 2 is critical Sobolev exponent, u, v ∈ D01,2 Ω∗ and Ω∗ RN \ E with E a∈ZN a ω∗ for a domain containing the origin ω∗ ⊂ ω∗ ⊂ B0, 1/2. Here, B0, 1/2 denotes the open ball centered at the origin of radius 1/2. The existence of a nontrivial solution was obtained by using a generalized linking theorem. As it is well known in dimensions N ≥ 3, the nonlinearities are required to have polynomial growth at infinity, so that one can define associated functionals in Sobolev spaces. Coming to dimension N 2, much faster growth is allowed for the nonlinearity. In fact, the Trudinger-Moser estimates in N 2 replace the Sobolev embedding theorem used in N ≥ 3. In dimension N 2, Adimurth and Yadava 7, de Figueiredo et al. 8 discussed the solvability of problems of the type −Δu fx, u, u 0, x ∈ Ω, 1.5 x ∈ ∂Ω, where Ω is some bounded domain in R2 . Shen et al. 9 considered the following nonlinear elliptic problems with critical potential: Δu − μ u 2 fx, u, |x| log R/|x| u 0, x∈Ω 1.6 x ∈ ∂Ω, and obtained some existence results. In the whole space R2 , some authors considered the following single semilinear elliptic equations: −Δu V xu fx, u, x ∈ R2 . 1.7 G. Zhang and S. Liu 3 As the potential V x and the nonlinearity fx, t are asymptotic to a constant function, Cao 10 obtained the existence of a nontrivial solution. As the potential V x and the nonlinearity fx, t are asymptotically periodic at infinity, Alves et al. 11 proved the existence of at least one positive weak solution. Our aim in this paper is to establish the existence of a nontrivial solution for problem P in subcritical case. To our knowledge, there are no results in the literature establishing the existence of solutions to these problems in the whole space. However, it contains a basic difficulty. Namely, the energy functional associated with problem P has strong indefinite quadratic part, so there is not any more mountain pass structure but linking one. Therefore, the proofs of our main results cannot rely on classical min-max results. Combining a generalized linking theorem introduced by Kryszewski and Szulkin 12 and Trudinger-Moser inequality, we prove an existence result for problem P. The paper is organized as follows. In Section 2, we recall some results and state our main results. In Section 3, main result is proved. 2. Preliminaries and main results Consider the Hilbert space 13 H 1 R2 u ∈ L2 R2 , ∇u ∈ L2 R2 , 2.1 and denote the product space Z H 1 R2 × H 1 R2 endowed with the inner product: u, v, φ, ψ R2 ∇u∇φ uφdx R2 ∇v∇ψ vψdx, ∀φ, ψ ∈ Z. 2.2 If we define Z {u, u ∈ Z}, Z − {v, −v ∈ Z}. 2.3 It is easy to check that Z Z ⊕ Z − , since 1 1 u, v u v, u v u − v, v − u. 2 2 2.4 Let us denote by P resp., Q the projection of Z on to Z resp., Z − , we have 2 2 1 1 u v, u v − 1 1 u − v, v − u 2 2 2 2 2 1 |∇u|2 |∇v|2 2∇u∇v dx |u| |v|2 2uv dx 2 2 4 R R 2 2 2 2 − |∇u| |∇v| − 2∇u∇v dx − |u| |v| − 2uv dx 1 P u, v2 − Qu, v2 2 R2 R2 R2 ∇u∇v uvdx. 2.5 4 Boundary Value Problems Now, we define the functional Iu, v 2 ∇u∇v uvdx − R P u, v2 2 R 2 Qu, v2 − 2 Fx, u Gx, v dx 2.6 − ϕu, v, ∀u, v ∈ Z, where ϕu, v 2 Fx, u Gx, v dx. 2.7 R Let z0 ∈ Z \ {0} and let R > r > 0, we define M z z− λz0 : z− ∈ Z − , z ≤ R, λ ≥ 0 , M0 z z− λz0 : z− ∈ Z − , z R and λ ≥ 0 or z ≤ R and λ ≥ 0 , N z ∈ Z : z r . 2.8 Here, we assume the following condition: H1 f, g ∈ CR2 × R, R; H2 limt→0 fx, t/t limt→0 gx, t/t 0 uniformly with respect to x ∈ R2 ; H3 there exist μ > 2 and η > 0 such that 0 < μFx, t ≤ tfx, t, 0 < μGx, t ≤ tgx, t, ∀|t| ≥ η. 2.9 Lemma 2.1 see 12, 14. Assume (H1), (H2), and (H3), and suppose 1 Iz 1/2P z2 −Qz2 −ϕz, where ϕ ∈ C1 Z, R is sequentially lower semicontinuous, bounded below, and ∇ϕ is weakly sequentially continuous; 2 there exist z0 ∈ Z \ {0}, α > 0, and R > r > 0, such that inf Iz ≥ α > 0, N sup Iz ≤ 0. 2.10 M0 Then, there exist c > 0 and a sequence zn ⊂ Z such that Izn −→ c, I zn −→ 0, as n −→ ∞. 2.11 Moreover, c ≥ α. Theorem 2.2. Under the assumptions (H1), (H2), and (H3), if f and g has subcritical growth (see definition below), problem (P) possesses a nontrivial weak solution. G. Zhang and S. Liu 5 In the whole space R2 , do Ó and Souto 15 proved a version of Trudinger-Moser inequality, that is, i if u ∈ H 1 R2 , β > 0, we have 2 exp β|u|2 − 1 dx < ∞; 2.12 R ii if 0 < β < 4π and |u|L2 R2 ≤ c, then there exists a constant c2 c1 c, β such that sup R2 |∇u|L2 R2 ≤1 exp β|u|2 − 1 dx < c2 . 2.13 Definition 2.3. We say fx, t has subcritical growth at ∞, if for all β > 0, there exists a positive constant c3 such that fx, t ≤ c3 exp βt2 , ∀x, t ∈ R2 × 0, ∞. 2.14 3. Proof of Theorem 2.2 In this section, we will prove Theorem 2.2. under our assumptions and 2.14, there exist cε > 0, β > 0 such that 2 Fx, t, Gx, t ≤ t ε cε exp βt2 − 1 , 2 ∀ε > 0, ∀t ∈ R. 3.1 Then, we obtain Fx, u, Gx, v ∈ L2 R2 , ∀u, v ∈ H 1 R2 . 3.2 Therefore, the functional Iu, v is well defined. Furthermore, using standard arguments, we obtain the functional Iu, v is C1 functional in Z and I u, vφ, ψ R2 ∇u∇ψ uψdx − R2 R2 ∇v∇φ vφ dx fx, uφ gx, vψ dx, 3.3 ∀φ, ψ ∈ Z. Consequently, the weak solutions of problem P are exactly the critical points of Iu, v in Z. Now, we prove that the functional Iu, v satisfied the geometry of Lemma 2.1. Lemma 3.1. There exist r > 0 and α > 0 such that infN Iu, u ≥ α > 0. Proof. By 2.14 and assumption H2, there exists cε > 0 such that Fx, t, Gx, t ≤ t2 ε cε t3 exp βt2 − 1 , 2 ∀t ∈ R, 3.4 6 Boundary Value Problems and thus on N, we have Iu, u ≥ R 2 ≥ ≥ R2 R2 2 2 εu2 cε u3 exp βu2 − 1 dx R |∇u| u dx − 2 2 |∇u| u dx − ε 2 |∇u|2 u2 dx − ε u dx − cε 2 R2 R2 u dx u2 dx − cε u3 R2 R2 R2 1/2 6 exp 2 exp βu2 − 1 dx 2 βu − 1 dx 1/2 1/2 . 3.5 So, by the Sobolev embedding theorem and 2.12, we can choose r > 0 sufficiently small, such that Iu, u ≥ α > 0, whenever u r. 3.6 Lemma 3.2. There exist u0 , u0 ∈ Z \ {0} and R > r > 0 such that supM0 I ≤ 0. Proof. 1 By assumption H3, we have on Z − Iu, u R2 |∇u|2 u2 dx − R2 Fx, u Gx, −u dx ≤ 0 3.7 because Fx, t ≥ 0, Gx, t ≥ 0 for any x, t ∈ R2 × R. 2 Assumption H3 implies that there exist c4 > 0, c5 > 0 such that Fx, t, Gx, t ≥ c4 tμ − c5 , ∀t ∈ R. 3.8 Now, we choose u0 , u0 ∈ Z \ {0} such that u0 , u0 r, then I −v, v λ u0 , u0 λ2 R2 − ≤− 2 |∇u0 |2 u20 dx − R2 |∇v|2 v2 dx Fλu0 v G λu0 − v dx 3.9 R R2 |∇u|2 u2 dx c λ2 − λμ . Because μ > 2, it follows that for w ∈ M0 Iw −→ −∞, whenever w −→ ∞, and so, taking R > r large, we get supM0 I ≤ 0. 3.10 G. Zhang and S. Liu 7 Proof of Theorem 2.2. By Lemma 3.1, there exist r > 0 and α > 0 such that infN Iu, u ≥ α > 0. By Lemma 3.2, there exist u0 , u0 ∈ Z \{0} and R > r > 0 such that supM0 I ≤ 0. Since Z Z ⊕Z − , we have Iu, v ∇u∇v uvdx − Fx, u Gx, v dx R2 P u, v2 2 R2 − Qu, v2 2 3.11 − ϕu, v, ∀u, v ∈ Z. From 2.14, 3.1, and assumption H3, ϕu, v ∈ C1 , ϕu, v ≥ 0 and ϕu, v is sequentially lower semicontinuous by Z ⊂ L2loc R2 × L2loc R2 and Fatou’s lemma; ∇ϕ is weakly sequentially continuous. Thus, by Lemma 2.1 there exists a sequence un , vn ⊂ Z such that Iun , vn −→ c ≥ α, I un , vn −→ 0. 3.12 Claim 3.3. There is c < ∞, such that un , vn ≤ c for any n. Indeed, from 3.12, we obtain that the sequence un , vn ⊂ Z satisfies I un , vn c δn , I un , vn φ, ψ εn un , vn , as n −→ ∞, 3.13 where φ, ψ ∈ {un , vn }, δn → 0, εn → 0 as n → ∞. Taking φ, ψ {un , vn } in 3.13 and assumption H3, we have 2 f x, un un g x, vn vn dx R ≤2 2 ≤ μ F x, un G x, vn dx 2c 2δn εn un , vn R2 2 3.14 fx, un un g x, vn vn dx C 2δn εn un , vn , R where C depends only on c and η in assumption H3. Since μ > 2, we have 1 − 2/μ > 0, and thus 2 1− f x, un un g x, vn vn dx ≤ C 2δn εn un , vn , ∀n ∈ N. 3.15 2 μ R On the other hand, let φ, ψ vn , 0, φ, ψ 0, un in 3.13, we obtain 2 vn − εn vn ≤ 2 f x, un vn dx, 2 un − εn un ≤ R g x, vn un dx. 3.16 un g x, vn dx εn . un 3.17 R 2 that is, vn ≤ vn f x, un dx εn , vn 2 R un ≤ 2 R 8 Boundary Value Problems Now, we recall the following inequality see 7, Lemma 2.4: ⎧ 2 1/2 n 1/4 ⎪ ⎨ e − 1 mlog m , n ≥ 0, m ≥ e , mn ≤ ⎪ ⎩ e n2 − 1 1 m2 , n ≥ 0, 0 ≤ m ≤ e1/4 . 2 3.18 Let n vn /vn and m fx, un /c3 , where c3 is defined in 2.14, we have f x, un vn vn 2 exp c3 − 1 dx dx ≤ c3 c3 vn vn R2 R2 1/2 f x, un f x, un c3 dx log c3 c3 {x∈R2 ,fx,un /c3 ≥e1/4 } 2 f x, un dx. c3 c3 {x∈R2 ,fx,un /c3 ≤e1/4 } By 2.12, we have R2 3.19 exp vn /vn 2 − 1dx < ∞. By 2.14, we have log fx, t c3 1/2 ≤ β1/2 t. 3.20 Hence, we have f x, un vn 1/2 dx ≤ c6 β c3 f x, un un dx 2 2 c3 vn R R 3.21 for some positive constant c6 . So we have vn ≤ c6 β1/2 2 f x, un un dx εn . 3.22 g x, vn vn dx εn 3.23 R Using a similar argument, we obtain un ≤ c7 β1/2 R2 for some positive constant c7 . Combining 3.22 and 3.23, we have un , vn ≤ c8 1 δn εn un , vn εn 3.24 for some positive constant c8 , which implies that un , vn ≤ c. Thus, for a subsequence still denoted by un , vn , there is u0 , v0 ∈ Z such that un , vn −→ u0 , v0 weakly in Z, as n −→ ∞, un , vn −→ u0 , v0 in Lsloc R2 × Lsloc R2 for s ≥ 1, as n −→ ∞, un x, vn x −→ u0 x, v0 x , almost every, in R2 , as n −→ ∞. 3.25 G. Zhang and S. Liu 9 1 2 2 Then, there exists hx ∈ 2H R such that |un x| ≤ h, ∀x ∈ R , ∀n ∈ N. From 2.12 and 2.14, we have R2 expβh x − 1dx < c, this implies f x, un φdx −→ f x, u0 φdx, as n −→ ∞. 3.26 R2 Similarly, we can obtain R 2 R2 g x, vn ψdx −→ 2 g x, v0 ψdx, as n −→ ∞. 3.27 R From these, we have I un , vn φ, ψ 0, so u0 , v0 is weak solution of problem P. Claim 3.4. u0 , v0 is nontrivial. By contradiction, since fx, t has subcritical growth, from 2.14 and Hölder inequality, we have f x, un un dx ≤ c un exp βu2n − 1 dx R2 R2 ≤c R2 q |un | dx 1/q R2 exp βqu2n − 1 dx where 1/q 1/q 1. Choosing suitable β and q, we have exp βqu2n − 1 dx ≤ c. R2 f x, un un dx ≤ c R2 q un dx , 3.29 R2 Then, we obtain 3.28 1/q 1/q . 3.30 Since un → 0 in Lq R2 , as n → ∞, this will lead to f x, un un dx −→ 0, R2 Similarly, we have R2 g x, vn vn dx −→ 0, Using assumption H3, we obtain F x, un dx −→ 0, R 2 2 as n −→ ∞. 3.31 as n −→ ∞. 3.32 G x, vn dx −→ 0, This together with I un , vn un , vn → 0, we have ∇un ∇vn un vn dx −→ 0, R2 Thus, we see that as n −→ ∞. 3.33 R I un , vn −→ 0, as n −→ ∞. as n −→ ∞. 3.34 3.35 which is a contradiction to Iun , vn → c ≥ α > 0, as n → ∞. 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