Existence and multiplicity of solutions for a fourth-order elliptic equation

  • Fanglei Wang1Email author and

    Affiliated with

    • Yukun An2

      Affiliated with

      Boundary Value Problems20122012:6

      DOI: 10.1186/1687-2770-2012-6

      Received: 26 August 2011

      Accepted: 17 January 2012

      Published: 17 January 2012

      Abstract

      This article is concerned with the existence and multiplicity of nontrival solutions for a fourth-order elliptic equation

      Δ 2 u - M Ω | u | 2 d x Δ u = f ( x , u ) , in Ω , u = Δ u = 0 , on Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equa_HTML.gif

      by using the mountain pass theorem.

      Keywords

      fourth-order elliptic equation nontrivial solutions mountain pass theorem

      1 Introduction

      In this article we study the existence of nontrivial solutions for the fourth-order boundary value problem
      Δ 2 u - M Ω | u | 2 d x Δ u = f ( x , u ) , in Ω , u = Δ u = 0 , on Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equ1_HTML.gif
      (1)

      where Ω ⊂ R N is a bounded smooth domain, f : Ω × RR and M : RR are continuous functions. The existence and multiplicity results for Equation (1) are considered in [13] by using variational methods and fixed point theorems in cones of ordered Banach space with space dimension is one.

      On the other hand, The four-order semilinear elliptic problem
      Δ 2 u + c Δ u = f ( x , u ) , in Ω , u = Δ u = 0 , on Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equ2_HTML.gif
      (2)

      arises in the study of traveling waves in a suspension bridge, or the study of the static deflection of an elastic plate in a fluid, and has been studied by many authors, see [410] and the references therein.

      Inspired by the above references, the object of this article is to study existence and multiplicity of nontrivial solution of a fourth-order elliptic equation under some conditions on the function M(t) and the nonlinearity. The proof is based on the mountain pass theorem, namely,

      Lemma 1.1. Let E be a real Banach space, and IC1(E, R) satisfy (PS)-condition. Suppose
      1. (1)
        There exist ρ > 0, α > 0 such that
        I | B ρ I ( 0 ) + α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equb_HTML.gif
         
      where B p = {uE|∥u∥ ≤ ρ}.
      1. (2)
        There is an eE and ∥e∥ > ρ such that
        I ( e ) I ( 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equc_HTML.gif
         
      Then I(u) has a critical value c which can be characterized as
      C = inf γ Γ max u γ ( [ 0 , 1 ] ) I ( u ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equd_HTML.gif

      where Γ = {γC([0, 1],E)|γ(0) = 0,γ(1) = e}.

      The article is organized as follows: Section 2 is devoted to giving the main result and proving the existence of nontrivial solution of Equation (1). In Section 3, we deal with the multiplicity results of Equation (1) whose nonlinear term is asymptotically linear at both zero and infinity

      2 Main result I

      Theorem 2.1. Assume the function M(t) and the nonlinearity f(x, t) satisfying the following conditions:

      (H1) M(t) is continuous and satisfies
      M ( t ) > m 0 , t > 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equ3_HTML.gif
      (3)
      for some m0 > 0. In addition, that there exist m' > m0 and t0 > 0, such that
      M ( t ) = m , t > t 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equ4_HTML.gif
      (4)

      (H2) f(x, t) ∈ C(Ω × R); f(x, t) ≡ 0, ∀x ∈ Ω, t ≤ 0, f(x, t) ≥ 0, ∀x ∈ Ω, t > 0;

      (H3) |f(x, t)| ≤ a(x) + b|t| p , ∀tR and a.e. x in Ω, where a(x) ∈ L q (Ω), bR

      and 1 < p < N + 4 N - 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq1_HTML.gif if N > 4 and 1 < p < ∞ if N ≤ 4 and 1 q + 1 p = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq2_HTML.gif;

      (H4) f(x, t) = o(|t|) as t → 0 uniformly for x ∈ Ω ;

      (H5) There exists a constant Θ > 2 and R > 0, such that
      Θ F ( x , s ) s f ( x , s ) , | s | R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Eque_HTML.gif

      Then Equation (1) has at least one nonnegative solution.

      Let Ω ⊂ R N be a bounded smooth open domain, H = H 2 ( Ω ) H 0 1 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq3_HTML.gif be the Hilbert space equipped with the inner product
      ( u , v ) - Ω ( Δ u Δ v + u v ) d x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equf_HTML.gif
      and the deduced norm
      | | u | | 2 = Ω | Δ u | 2 d x + Ω | u | 2 d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equg_HTML.gif
      Let λ1 be the positive first eigenvalue of the following second eigenvalue problem
      - Δ v = λ v , in Ω , v = 0 , on Ω . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equh_HTML.gif
      Then from [4], it is clear to see that Λ1 = λ1(λ1 - c) is the positive first eigenvalue of the following fourth-order eigenvalue problem
      Δ 2 u + c Δ u = λ u , in Ω , Δ u = u = 0 , on Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equi_HTML.gif
      where c < λ1. By Poincare inequality, for all uH, we have
      | | u | | 2 Λ 1 | | u | | L 2 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equ5_HTML.gif
      (5)
      A function uH is called a weak solution of Equation (1) if
      Ω Δ u Δ v d x + M Ω | u | 2 d x Ω u v d x = Ω f ( x , u ) v d x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equj_HTML.gif
      holds for any vH. In addition, we see that weak solutions of Equation (1) are critical points of the functional I : HR defined by
      I ( u ) = 1 2 Ω | Δ u | 2 d x + 1 2 M ^ Ω | u | 2 d x - Ω F ( x , u ) d x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equk_HTML.gif

      where M ^ ( t ) = 0 t M ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq4_HTML.gif and F(x, t) = ∫ f(x, t)dt. Since M is continuous and f has subcritical growth, the above functional is of class C1 in H. We shall apply the famous mountain pass theorem to show the existence of a nontrivial critical point of functional I(u).

      Lemma 2.2. Assume that (H1)-(H5) hold, then I(u) satisfies the (PS)-condition.

      Proof. Let {u n } ⊂ H be a (PS)-sequence. In particular, {u n } satisfies
      I ( u n ) C , and I ( u n ) , u n 0 as n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equ6_HTML.gif
      (6)

      Since f(x, t) is sub-critical by (H3), from the compactness of Sobolev embedding and, following the standard processes we know that to show that I verifies (PS)-condition it is enough to prove that {u n } is bounded in H. By contradiction, assume that ∥u n ∥ → +∞.

      Case I. If Ω | u n | 2 d x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq5_HTML.gif is bounded, Ω | Δ u n | 2 d x + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq6_HTML.gif. We assume that there exist a constant K > 0 such that Ω | u n | 2 d x K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq7_HTML.gif. By (H1), it is easy to obtain that m ̃ = max t [ 0 , K ] M ( t ) > m 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq8_HTML.gif. Set l 1 = min { 1 , m 0 } , l 2 = max { 1 , m ̃ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq9_HTML.gif. Then, from

      (H1), (H3), and (H5), we have
      I ( u n ) - l 1 2 l 2 I ( u n ) u n = 1 2 Ω | Δ u n | 2 d x + 1 2 M ^ Ω | u n | 2 d x - Ω F ( x , u n ) d x - l 1 2 l 2 Ω | Δ u n | 2 d x + M Ω | u n | 2 d x Ω | u n | 2 d x + l 1 2 l 2 Ω f ( x , u n ) u n d x 1 2 l 1 | | u n | | 2 + Ω l 1 2 l 2 f ( x , u n + ) u n - F ( x , u n + ) d x 1 2 l 1 | | u n | | 2 + | | u n | | R l 1 2 l 2 f ( x , u n + ) u n + - F ( x , u n + ) d x - C 1 1 2 l 1 | | u n | | 2 + l 1 2 l 2 | | u n | | R f ( x , u n + ) u n + - 2 l 2 l 2 F ( x , u n + ) d x - C 1 1 2 l 1 | | u n | | 2 + l 1 2 l 2 | | u n | | R f ( x , u n + ) u n + - Θ F ( x , u n + ) d x - C 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equl_HTML.gif
      On the other hand, it is easy to obtain that
      I ( u n ) - l 1 2 l 2 I ( u n ) u n C + C | | u n | | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equm_HTML.gif
      Then, from above, we can have
      | | u | | 2 C + C | | u n | | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equn_HTML.gif

      which contradicts ∥u n ∥ → +∞. Therefore {u n } is bounded in H.

      Case II. if Ω | Δ u n | 2 d x + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq6_HTML.gif. By (H1), let l2 = max{1, m'}, we also can obtain that {u n } is bounded in H.

      This lemma is completely proved.

      Lemma 2.3. Suppose that (H1)-(H5) hold, then we have
      1. (1)

        there exist constants ρ > 0, α > 0 such that I | B ρ α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq10_HTML.gif with B p = {uHu∥ ≤ ρ};

         
      2. (2)

        I(1) → -∞ as t → +∞.

         
      Proof. By (H1)-(H4), we see that for any ε > 0, there exist constants C 1 > 0, C2 such that for all (x, s) ∈ Ω × R, one have
      F ( x , s ) 1 2 ε s 2 + C 1 s p + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equ7_HTML.gif
      (7)
      Choosing ε > 0 small enough, we have
      I ( u ) = 1 2 Ω | Δ u | 2 d x + 1 2 M ^ Ω | u | 2 d x - Ω F ( x , u ) d x 1 2 Ω | Δ u | 2 d x + 1 2 m 0 Ω | u | 2 d x - Ω F ( x , u ) d x 1 2 l 1 | | u | | 2 - ε 2 | | u | | L 2 2 - C 1 | | u | | L p + 1 p + 1 1 2 ( l 1 - ε ) | | u | | 2 - C 3 | | u | | p + 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equo_HTML.gif

      by (3), (5), (7) and the Sobolev inequality. So, part 1 is proved if we choose ∥uρ > 0 small enough.

      On the other hand, we have
      I ( u ) = 1 2 Ω | Δ u | 2 d x + 1 2 M ^ Ω | u | 2 d x - Ω F ( x , u ) d x 1 2 Ω | Δ u | 2 d x + 1 2 m 1 Ω | u | 2 d x - Ω F ( x , u ) d x 1 2 l 2 | | u | | 2 - | | u | | Θ Θ + C 4 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equp_HTML.gif
      using (4) and (H5). Hence,
      I ( t φ 1 ) 1 2 l 2 t 2 | | φ 1 | | 2 - t Θ | | φ 1 | | Θ Θ + C 4 - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equq_HTML.gif

      as t → +∞ and part 2 is proved.

      Proof of Theorem 2.1. From Lemmas 2.2 and 2.3, it is clear to see that I(u) satisfies the hypotheses of Lemma 1.1. Therefore I(u) has a critical point.

      3 Existence result II

      Theorem 3.1. Assume that (H1) holds. In addition, assume the following conditions are hold:

      (H6) f(x, t)t ≥ 0 for x ∈ Ω, tR;

      (H7) lim t 0 f ( x , t ) t = α , lim | t | + f ( x , t ) t = β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq11_HTML.gif, uniformly in a.e x ∈ Ω, where α min { 1 , m 0 } < λ 1 ( λ 1 + m ) < β < + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq12_HTML.gif.

      Then Equation (1) has at least two nontrivial solutions, one of which is positive and the other is negative.

      Let u+ = max{u, 0}, u- = min{u, 0}. Consider the following problem
      Δ 2 u - M Ω | u | 2 d x Δ u = f + ( x , u ) , in Ω , u = Δ u = 0 , on Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equ8_HTML.gif
      (8)
      where
      f + ( x , t ) = f ( x , t ) if t 0 , 0 , if t < 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equr_HTML.gif
      Define the corresponding functional I+ : HR as follows:
      I + ( u ) = 1 2 Ω | Δ u | 2 d x + 1 2 M ^ Ω | u | 2 d x - Ω F + ( x , u ) d x , u H , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equs_HTML.gif

      where F + ( x , u ) = 0 u f + ( x , t ) d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq13_HTML.gif. Obviously, I+Cl(H, R). Let u be a critical point of I+ which implies that u is the weak solution of Equation (8). Futhermore, by the weak maximum principle it follows that u ≥ 0 in Ω. Thus u is also a solution of Equation (1).

      Similarly, we also can define
      f - ( x , t ) = f ( x , t ) if t 0 , 0 , if t < 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equt_HTML.gif
      and
      I - ( u ) = 1 2 Ω | Δ u | 2 d x + 1 2 M ^ Ω | u | 2 d x - Ω F - ( x , u ) d x , u H , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equu_HTML.gif

      where F - ( x , u ) = 0 u f - ( x , t ) d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq14_HTML.gif. Obviously, I-C1(H, R). Let u be a critical point of I- which implies that u is the weak solution of Equation (1) with I-(u) = I(u).

      Lemma 3.2. Assume that (H1), (H6), and (H7) hold, then I± satisfies the (PS) condition.

      Proof. We just prove the case of I+. The arguments for the case of I- are similar. Since Ω is bounded and (H7) holds, then if {u n } is bounded in H, by using the Sobolve embedding and the standard procedures, we can get a convergent subsequence. So we need only to show that {u n } is bounded in H.

      Let {u n } ⊂ H be a sequence such that
      I + ( u n ) c , I + ( u n ) 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equ9_HTML.gif
      (9)
      By (H7), it is easy to see that
      | f + ( x , s ) s | C ( 1 + | s | 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equv_HTML.gif
      Now, (9) implies that, for all ϕH, we have
      Ω Δ u n Δ ϕ d x + M Ω | u n | 2 d x Ω u n ϕ d x = Ω f + ( x , u n ) ϕ d x 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equ10_HTML.gif
      (10)
      Set ϕ = u n , we have
      min { 1 , m 0 } | | u n | | 2 Ω | Δ u n | 2 d x + M Ω | u n | 2 d x Ω | u n | 2 d x = Ω f + ( x , u n ) u n d x + I + ( u n ) , u n Ω f + ( x , u n ) u n d x + o ( 1 ) | | u n | | C + C | | u n | | L 2 2 + o ( 1 ) | | u n | | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equ11_HTML.gif
      (11)
      Next, we will show that | | u n | | L 2 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq15_HTML.gif is bounded. If not, we may assume that ∥u n L 2 → +∞ as n → +∞. Let ω n = u n | | u n | | L 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq16_HTML.gif, then | | ω n | | L 2 = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq17_HTML.gif. From (11), we have
      | | ω n | | 2 o ( 1 ) + C + o ( 1 ) | | u n | | L 2 | | u n | | | | u n | | L 2 = o ( 1 ) + C + o ( 1 ) | | ω n | | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equw_HTML.gif
      thus {ω n } is bounded in H. Passing to a subsequence, we may assume that there exists ωH with | | ω | | L 2 = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq18_HTML.gif such that
      ω n ω , weakly in H , n + , ω n ω , strongly in L 2 ( Ω ) , n + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equx_HTML.gif
      On the other hand, | | u n | | L 2 + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq19_HTML.gif as n → +∞, by Poincare inequality, it is easy to know that Ω | Δ u n | 2 d x + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq6_HTML.gif as n → +∞. Thus by (H1), the function M Ω | u n | 2 d x = m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq20_HTML.gif. So as n → +∞, by (10), we have
      Ω Δ ω Δ ϕ d x + m Ω ω ϕ d x Ω β ω + ϕ d x = 0 , ϕ H . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equ12_HTML.gif
      (12)
      Then ωH is a weak solution of the equation
      Δ 2 ω - m Δ ω = β ω + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equy_HTML.gif
      The weak maximum principle implies that ω = ω+ ≥ 0. Choosing ϕ (x) = φ1(x) > 0, which is the corresponding eigenfunctions of λ1. From (10), we get
      Ω Δ ω Δ φ 1 d x + m Ω ω φ 1 d x = β Ω ω + φ 1 d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equ13_HTML.gif
      (13)
      On the other hand, we can easily see that Λ = λ1(λ1 + m') is the eigenvalue of the problem
      Δ 2 u + m Δ u = Λ u , in Ω , Δ u = u = 0 , on Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equz_HTML.gif
      and the corresponding eigenfunction is still φ1(x). If ω(x) > 0, we also have
      Ω Δ ω Δ φ 1 d x + m Ω ω φ 1 d x = Λ Ω ω + φ 1 d x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equ14_HTML.gif
      (14)

      which follows that ω ≡ 0 by Λ < β But this conclusion contradicts | | ω | | L 2 = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq18_HTML.gif.

      Hence {u n } is bounded in H.

      Now we prove that the functionals I± has a mountain pass geometry.

      Lemma 3.3. Assume that (H1), (H7) hold, then we have
      1. (1)

        there exists ρ, R > 0 such that I±(u) > R, if ∥u∥ = ρ;

         
      2. (2)

        I±(u) are unbounded from below.

         
      Proof. By (H7), for any ε > 0, there exists C 1 > 0, C2 > 0 such that ∀(x, s) ∈ Ω × R, we have
      F ( x , s ) 1 2 ( α + ε ) s 2 + C 1 s p + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equ15_HTML.gif
      (15)
      and
      F ( x , s ) 1 2 ( β + ε ) s 2 + C 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equ16_HTML.gif
      (16)

      where 2 < p < 2 * = 2 N N - 2 N > 2 , + N 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_IEq21_HTML.gif

      We just prove the case of I+. The arguments for the case of I- are similar. Let ϕ = 1. When t is sufficiently large, by (16) and (H1), it is easy to see that
      I + ( t φ 1 ) = 1 2 Ω | Δ ( t φ 1 ) | 2 d x + 1 2 M ^ Ω | ( t φ 1 ) | 2 d x - Ω F + ( x , t φ 1 ) d x 1 2 Ω | Δ ( t φ 1 ) | 2 d x + 1 2 m Ω | ( t φ 1 ) | 2 d x - Ω 1 2 ( β - ε ) ( t φ 1 ) 2 - C 2 d x = t 2 2 Ω | Δ φ 1 | 2 d x + m Ω | Δ φ 1 | 2 d x - ( β - ε ) Ω φ 1 2 d x + C 2 | Ω | = t 2 2 [ Λ - ( β - ε ) ] | | φ 1 | | L 2 + C 2 | Ω | - , a s t + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equaa_HTML.gif
      On the other hand, by (17), (H1), the Poincare inequality and the Sobolve embedding, we have
      I + ( u ) = 1 2 Ω | Δ u | 2 d x + 1 2 M ^ Ω | u | 2 d x - Ω F + ( x , u ) d x 1 2 min { 1 , m 0 } | | u | | - α + ε 2 Ω | u | 2 d x - C 1 Ω | u | p + 1 d x 1 2 min { 1 , m 0 } - α + ε 2 Λ | | u | | - C 4 | | u | | p + 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equab_HTML.gif

      where C4 is a constant. Choosing ∥u∥ = ρ small enough, we can obtain I+(u) ≥ R > 0 if ∥u∥ = ρ.

      Proof of Theorem 3.1. From Lemma 3.3, it is easy to see that there exists eH with ∥e∥ > ρ such that I±(e) < 0.

      Define
      P = { γ : [ 0 , 1 ] H : γ is continuous and γ ( 0 ) = 0 , γ ( 1 ) = e } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equac_HTML.gif
      and
      c ± = inf γ P max t [ 0 , 1 ] I ± ( γ ( t ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equad_HTML.gif
      From Lemma 3.3, we have
      I ± ( 0 ) = 0 , I ± ( e ) < 0 , I ± ( u ) | B ρ R > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-6/MediaObjects/13661_2011_Article_107_Equae_HTML.gif

      Moreover, by Lemma 3.2, the functions I± satisfies the (PS)-condition. By Lemma 1.1, we know that c+ is a critical value of I+ and there is at least one nontrivial critical point in H corresponding to this value. This critical in nonnegative, then the strong maximum principle implies that is a positive solution of Equation (1). By an analogous way we know there exists at least one negative solution, which is a nontrivial critical point of I- Hence, Equation (1) admits at least a positive solution and a negative solution.

      Declarations

      Acknowledgements

      The authors' would like to thank the referees for valuable comments and suggestions for improving this article.

      Authors’ Affiliations

      (1)
      College of Science, Hohai University
      (2)
      Department of Mathematics, Nanjing University of Aeronautics and Astronautics

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      © Wang and An; licensee Springer. 2012

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