Nodal type bound states of Schrödinger equations via invariant set and minimax methods

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1 J. Differential Equations 214 ( Nodal type bound states of Schrödinger equations via invariant set and minimax methods Zhaoli Liu a, Francois A. van Heerden b,1, Zhi-Qiang Wang b, a Department of Mathematics, Capital Normal University, Beijing , PR China b Department of Mathematics andstatistics, Utah State University, Logan, Utah 84322, USA Received 11 May 2004 Available online 14 October 2004 Abstract For nonlinear Schrödinger equations in the entire space we present new results on invariant sets of the gradient flows of the corresponding variational functionals. The structure of the invariant sets will be built into minimax procedures to construct nodal type bound state solutions of nonlinear Schrödinger type equations Elsevier Inc. All rights reserved. Keywords: Invariant sets; Bound state nodal solutions; Nonlinear Schrödinger equations; Minimax method 1. Introduction The principal project of this paper is to investigate the structure of invariant sets of the associated gradient flows for nonlinear Schrödinger equations in the entire space, and in conjunction with minimax method to construct nodal type bound state solutions. In particular, we shall study how the structure of global invariant sets depends upon the local behavior of the flow near the trivial critical point 0. The novelty of our work is to discover a new family of invariant sets when there is a hyperbolic structure near the trivial critical point 0. As applications we shall provide multiplicity results of nodal type bound state solutions for nonlinear Schrödinger type equations. More precisely, Corresponding author. address: wang@math.usu.edu (Zhi-Qiang Wang). 1 Current address: Department of Mathematics, University of Utah, Salt Lake City, UT 84112, USA /$ - see front matter 2004 Elsevier Inc. All rights reserved. doi: /j.jde

2 Z. Liu et al. / J. Differential Equations 214 ( as a model problem we consider the existence of nodal (sign-changing) solutions for nonlinear time-independent Schrödinger equations of the form Δu + V λ (x)u = f(x,u) in R N, (1.1) which satisfy u(x) 0as x. This type of equations arise also from study of standing wave solutions of time-dependent nonlinear Schrödinger equations. The potential function V λ (x) := λg(x) + 1 satisfies the following conditions: (V 1 )g C(R N, R) satisfies g 0 and Ω := int(g 1 (0)) is nonempty. (V 2 ) There exist M 0 > 0 and r 0 > 0 such that ( ) lim m {x R N : x y r 0 } {x R N : g(x) M 0 } = 0, y where m denotes the Lebesgue measure on R N. (V 3 ) Ω = g 1 (0) and Ω is locally Lipschitz. As λ, V λ has a steep potential well, and we are interested in finding solutions trapped in the potential well. Under the above conditions, the linear operator Δ + V λ may have a finite number of eigenvalues below the infimum of the essential spectrum [3]. Obviously, these eigenvalues (bound states to the linear problem), except the first one, have nodal eigenfunctions. We shall show that under suitable asymptotically linear perturbations f(x,u), the nonlinear problem (1.1) has multiple bound state nodal solutions resembling the nodal structure of the linear problem. In order to state our conditions on f we introduce the following eigenvalue problem on Ω (cf. [25, Proposition A.1]): } Δu + u = νu in Ω (1.2) u = 0 on Ω. Let the eigenvalues of this problem be denoted by 0 < ν 1 < ν 2 < ν 3 <, which will occasionally be written as 0 < μ 1 < μ 2 μ 3, counting their multiplicity. Using the convention ν 0 :=, we make the following assumptions on f: (f 0 )f C(R N R, R). (f 1 ) f (x, s) is odd in s. (f 2 ) There exists an m N 0 := {0, 1, 2,...} such that uniformly in x R N, ν m < lim inf s 0 f(x,s) f(x,s) lim sup < ν m+1. s s 0 s (f 3 ) There exists an n N 0, n = m such that uniformly in x R N, ν n < lim inf s f(x,s) f(x,s) lim sup < ν n+1. s s s

3 360 Z. Liu et al. / J. Differential Equations 214 ( (f 4 ) There exists a c 0 > 0 such that f(x,s) c 0 s for all (x, s). (f 5 ) There exists an L 0 such that f(x,s) + Ls is increasing in s. Let dim(ν i ) denote the dimension of the eigenspace associated with the eigenvalue ν i. Set d k := k dim(ν i ) and d 0 := 0. i=1 Theorem 1.1. Assume (V 1 ) (V 3 ) and (f 0 ) (f 5 ). Then there exists a Λ > 0 such that for all λ Λ, Eq. (1.1) has at least d m d n pairs of nodal solutions provided min{m, n} 1, andat least d m d n 1 pairs of nodal solutions if min{m, n} =0. Remark 1.2. In the case min{m, n} =0, there is also a pair of signed solutions u>0, u <0. Remark 1.3. Though the existence result is stated and will be proved in detail only for the model Eq. (1.1) with the nonlinearity given above, the approach we shall take will be useful for more general types of nonlinearities. The understanding of invariant sets for gradient flows has been the center issue in dealing with nodal solutions. The key new ingredient we provide here is the construction and the structure of a new family of invariant sets of the associated gradient flows (this is done in Section 4), which, to our knowledge, is the first nontrivial construction of invariant sets near a saddle critical point. By using a combination of invariant sets method and minimax method as is done in this paper, many multiplicity results on nodal solutions for bounded domains like in [1,4,14] can be generalized to nonlinear Schrödinger equations in the entire space, for example, superlinear problems with a saddle point at 0, asymptotically linear problems with resonance, nonlinear eigenvalue problems, etc. We leave the precise statements to interested readers. We finish the section with some historical comments on related work and methods involved, and outlining in more detail our approach. In the case where f is assumed to be superlinear, nonlinear Schrödinger type equations have received a lot of attention in the past. It is only recently that nonlinear Schrödinger type equations with asymptotically linear terms have been studied. In [22,23] Stuart and Zhou studied radially symmetric problem. More general situations were considered in [8,11,12,15,26]. In most of these papers, the potential function is either periodic or autonomous at infinity. Asymptotically linear problems with potentials in this paper have been studied in [24,25], in which multiple solutions were constructed without giving nodal information about the solutions. On the other hand, results like Theorem 1.1 for problems in bounded domains have been given in [14]. In this paper, we shall construct nodal solutions for Eq. (1.1) by building upon the general idea of combining invariant sets with minimax method (which has been very successful for bounded domain problems, e.g., [1,4,14,18]) and by developing new techniques which will overcome difficulties for unbounded domain problems. To

4 Z. Liu et al. / J. Differential Equations 214 ( describe our basic idea and approach, let us introduce some notations. It is well known that weak solutions of (1.1) correspond to critical points of I λ (u) := 1 ( u 2 + V λ u 2 )dx F (x, u) dx, 2 R N R N in H λ := {u H 1 (R N ) : u λ < }. Here F(x,u) := u 0 f(x,s)ds and λ is the norm induced by the inner product (u, v) λ := ( u v + V λ uv) dx. R N The main idea of this paper is to construct certain invariant sets of the gradient flow associated to the energy functional I λ so that all positive and negative solutions are contained in these invariant sets and that minimax procedures can be used to construct nodal critical points of the energy functional outside these invariant sets. As a byproduct we give more information of the dynamical nature of the gradient flow. This type of idea has been used successfully for elliptic problems on bounded domains (c.f. [1,4,9,13,14,18]). In general, the cone of positive (and negative) functions in the Sobolev space is invariant under the gradient flow. However, these cones have empty interior, and it is very difficult to build a deformation in relation to these cones and to construct critical points outside of these cones using a minimax method. In bounded domains, the dense subspace C0 1( Ω) of H0 1 (Ω) has been used, since the cone of positive (and negative) functions in C0 1( Ω) have nonempty interior. We remark also that an interesting approach without using the cones structure was used in [6] for bounded domain problems but may not be suitable for getting multiple nodal critical points for even functionals (see [5] for more references). For problems in R N, there is no known replacement for C0 1( Ω) since C0 1(RN ) has no interior points either. This has been the major obstacle for generalizing many results on nodal solutions from the bounded domain case to the entire space case. For a superlinear problem, this has been done recently [2]. However, it turns out that the existence of these invariant sets depend in a subtle way on the behavior of the functional near 0. In [2], under the condition that 0 is a strict local minimum critical point, it was shown that a neighborhood, in the Sobolev space norm, of the cone of positive (and negative) functions is an invariant set. When 0 is a saddle point (as in most of the cases of this paper) no neighborhood of the positive and negative cones can be invariant sets anymore. The main project of this paper is to develop new techniques to construct invariant sets for the case of 0 being a saddle point. Obviously, our construction can be used for more general type of problems. The paper is organized as follows. Section 2 contains some preliminary technical results. Section 3 is devoted to the construction of a minimal positive solution and a maximal negative solution to (1.1). This is essential for the construction of invariant sets containing all positive and negative solutions to (1.1), so all the solutions obtained outside of these invariant sets by minimax procedures are nodal solutions. In Section 4, we construct new invariant sets which depend upon the local behavior of the energy

5 362 Z. Liu et al. / J. Differential Equations 214 ( functional I λ near 0. Results in Sections 3 and 4 are also interesting for their own sake, and will be used in Section 5 for the proof of the existence results, Theorems 5.5 and 5.8, which together give Theorem Preliminaries In the following, B R will denote the ball in R N centered at zero with radius R, B c R := RN \B R. The space L p (R N ) will be denoted by L p. We first state the following useful technical result: Lemma 2.1 (Van Heerden and Wang [25]). Assume (V 1 ) (V 2 ). Then, for any ε > 0 there exist R>0 and Λ > 0 such that u 2 L 2 (B c R ) ε u 2 λ, for all u H λ and λ Λ. For any given elements ψ 1,...,ψ k of H λ,weset U λ (ψ 1,...,ψ k ) := inf u H λ { u 2 λ : u L 2 = 1, (u, ψ i ) λ = 0 for i = 1,...,k}. For k N, we define the spectral values of Δ + V λ by the kth Rayleigh quotient μ λ k = sup U λ (ψ 1,...,ψ k ). ψ 1,...,ψ k 1 H λ For any m N, [3, Corollary 2.2] asserts that for λ sufficiently large, the operator Δ + V λ has at least d m = m i=1 dim(ν i ) eigenvalues μ λ 1,...,μλ d m. The corresponding eigenfunctions are denoted by e1 λ,...,eλ d m with ek λ L2 = 1. As a consequence of [24, Lemma 2.5], we conclude that: Lemma 2.2. Assume (V 1 ) (V 3 ). Then μ λ k μ k as λ for all k N. Proof. Fix any k N. First we have μ λ k μ k as a simple minimax description of the eigenvalues (see [20, Section XIII.1], for instance). According to [24, Lemma 2.5], the limit μ k := lim λ μ λ k is an eigenvalue of (1.2). The proof of [24, Lemma 2.5] shows the weak limit e k := lim λ ek λ in H λ is an eigenfunction of (1.2) corresponding to μ k and e k L 2 = 1. Thus ek λ converges to e k strongly in L 2 as λ. For any i, j {1,...,k} with i = j, (ei λ,eλ j ) L 2 = 0 implies (e i,e j ) L 2 (Ω) = 0. As a consequence, μ k μ k. This together with μλ k μ k gives the result.

6 Z. Liu et al. / J. Differential Equations 214 ( Set E k := span{e 1,...,e dk }, E 0 := {0} and E k (λ) := span{e λ 1,...,eλ d k }, E 0 (λ) := {0}. Let E k and (E k (λ)) denote the orthogonal complement of E k and E k (λ) in H λ, respectively. Recall that (u n ) H λ is a Palais Smale ((PS) for short) sequence of I λ if I λ (u n ) is bounded and I λ (u n) 0. I λ is said to satisfy the (PS)-condition if any such sequence contains a convergent subsequence. The following lemma is a more general version of a result in [24,25], and the proof here is different from, and simpler than that in [24,25]. Lemma 2.3. Assume (V 1 ) (V 3 ), (f 0 ),(f 3 ) and (f 4 ). Then there exists a Λ > 0 such that the functional I λ satisfies the (PS)-condition for all λ Λ. Proof. We only consider the case n 1; the case n = 0 is similar and simpler. Take ν n <b 1 b 2 < ν n+1 and T>0 such that for u T, b 1 f (x, u)/u b 2. By Lemma 2.2, we can take Λ 1 > 0 such that for λ Λ 1 { } b 1 min μ λ 1, 1 b 2 d n μ λ a >0 d n +1 for some a>0. By Lemma 2.1, there exist Λ 2 > 0 and R 0 > 0 such that for λ Λ 2, u 2 L 2 (B c R 0 ) ε 0 u 2 λ, u H λ, (2.1) where 0 < ε 0 < min { ( ) a 2, 1} 2(α + c 0 ) (2.2) and α = b 1 b 2 /(b 2 b 1 ) and c 0 is from (f 4 ). Now let (u n ) be a (PS) sequence for I λ with λ max{λ 1, Λ 2 }. Let Z n := {x : u n (x) T }. Writing u n = v n + w n with v n E n (λ) and w n (E n (λ)) and taking

7 364 Z. Liu et al. / J. Differential Equations 214 ( inner product of I λ (u n) and v n w n, we see that o(1) u n λ = v n 2 λ w n 2 λ f(x, u n ) (vn 2 Z n u w2 n )dx f(x, u n )(v n w n ) n Zn c v n 2 λ w n 2 λ b 1 vn 2 + b 2 wn 2 f(x, u n )(v n w n ) Z n Z n Zn c = v n 2 λ w n 2 λ b 1 v 2 R N n + b 2 w 2 R N n + b 1 v 2 Zn c n b 2 w 2 Zn c n f(x, u n )(v n w n ) Z c n ( Next, we claim b 1 μ λ d n 1 ) v n 2λ ( f(x, u n )(v n w n ). Zn c b 1 Z c n 1 b 2 μ λ d n +1 v 2 n b 2 Z c n ) w n 2 λ + b 1 w 2 n α Z c n u 2 n. Z c n v 2 n b 2 Z c n w 2 n To see this we choose β = (α b 1 )/α = α/(α + b 2 ). Expanding u 2 n = v2 n + w2 n + 2v nw n and using Hölder s inequality, we get b 1 v 2 n b 2w 2 n αu2 n (b 1 α + αβ)v 2 n + ( α + α/β b 2)w 2 n = 0. On the other hand, Z c n f(x, u n )(v n w n ) ( Z c n ) 1/2 ( ) 1/2 f(x,u n ) 2 (v n w n ) 2 Zn c c 0 ( C(R0 ) + ε 0 u n λ ) un λ. Combining these we get, with C(R 0 ) a constant depending only on R 0, a u n 2 λ o(1) u n λ + α(c(r 0 ) + ε 0 u n 2 λ ) + c 0C(R 0 ) u n λ + c 0 ε0 u n 2 λ,

8 Z. Liu et al. / J. Differential Equations 214 ( which gives the bound of the sequence. Assume that, without loss of any generality, u n uin H λ and u n u in L 2 loc (RN ) for some u H λ, which is a solution to the problem Δu + V λ (x)u = f(x, u) in R N. Taking inner product of I λ (u n) and u n u, noting that I λ (u) = 0, and using (f 4), we have for any R R 0, Thus Since u n u 2 λ = o(1) + R N (f (x, u n ) f (x, u))(u n u) dx o(1) + c 0 B c R ( u n + u ) u n u dx + f(x, u n ) f(x, u) u n u dx B R o(1) + c 0 B c R u n u 2 dx + 2c 0 B c R + f(x, u n ) f(x, u) u n u dx B R o(1) u n u 2 λ + 2c 0 u n u λ ( B c R + f(x, u n ) f(x, u) u n u dx. B R u n u 2 λ o(1) + 4c 0 u n u λ ( B c R u 2 dx u u n u dx u 2 dx ) 1/2 +2 f(x, u n ) f(x, u) u n u dx. B R lim R ( B c R u 2 dx) 1/2 = 0 ) 1/2

9 366 Z. Liu et al. / J. Differential Equations 214 ( and lim f(x, u n ) f(x, u) u n u dx = 0 n B R for all R>0, we obtain lim n u n u λ = 0, as required. The following lemma is a variant of [18, Lemma 3.2]. It can be proved in a similar way as the proof of [18, Lemma 3.2] (cf. [18, Lemma 2.5]). Lemma 2.4. Let H be a Hilbert space, D 1 and D 2 be two closedconvex subsets of H, and I C 1 (H, R). Suppose I (u) = u A(u) and A(D i ) D i for i = 1, 2. Then there exists a pseudo gradient vector field V of I in the form V (u) = u B(u) with B satisfying B(D i ) D i for i = 1, 2. Moreover, B(D i ) int(d i ) if A(D i ) int(d i ) for i = 1, 2, andv is oddif I is even andd 1 = D 2. Here recall that V is a pseudo gradient vector field of I if V C(H, H), V H \K is locally Lipschitz continuous with K := {u H : I (u) = 0}, and (I (u), V (u)) 1 2 I (u) 2 and V (u) 2 I (u) for all u H. 3. Minimal positive solution and maximal negative solution In this section, we will construct a minimal positive solution and a maximal negative solution to problem (1.1). This is essential for the construction of invariant sets containing all the positive (and the negative) solutions in Section 4. In a bounded domain case, this can be done by an iterative procedure (see [17], for instance). Unfortunately, the argument used in bounded domain case does not work with our problem due to the lack of compactness of the operator. We will use variational arguments to construct these solutions. With respect to the norm λ,wehavei λ (u) = u K λf(,u) where K λ : H λ H λ denotes the inverse operator of Δ + V λ : K λ := ( Δ + V λ ) 1. In view of (f 5 ), we may assume that f(x, s) is increasing in s. Otherwise, we just replace the norm λ on H λ with the equivalent one u 2 L,λ := R N ( u 2 + (V λ (x) + L) u 2 )dx

10 Z. Liu et al. / J. Differential Equations 214 ( and replace f(x,u) with f(x,u)+ Lu. What follows for the case L = 0 work just as well for the case L>0. Throughout this section we will assume ( f 2 ) lim inf s 0 f (x, s)/s > ν 1 uniformly in x. The main result of this section is as follows. Theorem 3.1. Assume (V 1 ) (V 3 ), (f 0 ), ( f 2 ), (f 4 ), (f 5 ), andassume that there exists a positive solution w + anda negative solution w to (1.1). Then there exists a minimal positive solution u + anda maximal negative solution u to (1.1). Proof. We only prove the existence of u +. Define ˆ f(x,s) = f(x,w + (x)) f(x,s) f(x,w (x)) if s>w + (x), if w (x) s w + (x), if s<w (x). We consider the following cut-off problem: Δu + V λ (x)u = f(x,u) ˆ in R N }, u H λ. (3.1) We claim that for all solutions u of (3.1), it holds w (x) u(x) w + (x) for all x R N. Seeking a contradiction, suppose Z := {x R N : u(x) > w + (x)} = for some solution u of (3.1). Then Z is an open subset of R N and on Z: Δu + V λ u = f(x,w + ) = Δw + + V λ w +, which implies that Δ(u w + ) + V λ (u w + ) = 0. Since u(x), w + (x) 0 as x, we conclude by the maximum principle that u(x) = w + (x) for all x Z, a contradiction. In a similar fashion, we see that w (x) u(x) for all x R N. Now consider the energy functional associated with (3.1): ˆ I λ (u) := 1 2 u 2 λ R N ˆF (x, u) dx, where ˆF(x,u) = u ˆ 0 f(x,s)ds. The above discussion shows that any critical point of Iˆ λ is a (weak) solution of the original problem (1.1). From the definition of Iˆ λ and f ˆ, it follows that Iˆ λ (u) 1 2 u 2 λ C ( 1 w + + w ) u dx 1 R N 2 u 2 λ C ( 1 w + L 2 + w ) L 2 u λ.

11 368 Z. Liu et al. / J. Differential Equations 214 ( Here and in the sequel, C i are constants. Thus Iˆ λ is coercive. Next we verify the (PS)-condition for Iˆ λ. Suppose (u n ) H λ satisfy Iˆ λ (u n ) c for some number c, and I ˆ λ (u n) 0. Then (u n ) is bounded in H λ. Thus, up to a subsequence, u n uin H λ and u n u in L 2 loc (RN ) with u a solution of (1.1). To obtain the strong convergence in H λ, we observe that and for any R>0, B c R ( ) ( ) I ˆ λ (u n)u n = u n 2 λ u 2 λ f(x,u ˆ n )u n f ˆ(x, u)u dx R N ( ) f(x,u ˆ n )u n f ˆ(x, u)u dx C 2 B c R ( w + + w ) ( u n + u )dx C 3 ( w + L 2 (B c R ) + w L 2 (B c R ) ). Thus u n λ u λ which implies the strong convergence. Recall that e 1 is the first eigenfunction of (1.2). We may assume e 1 > 0. By ( f 2 ), there exist s > 0 and δ > 0 such that f (x, s)/s > ν 1 + δ for 0 < s s. Choose R 1 > 0 and s 0 > 0 such that w + (x) < s for x R 1 and s 0 e 1 (x) < w + (x) for x R 1. We claim that s 0 e 1 (x) < w + (x) for all x R N. Indeed, if X := { x R N : s 0 e 1 (x)>w + (x) } =, then X BR c 1 Ω and we have Δ(s 0 e 1 ) + V λ s 0 e 1 = ν 1 s 0 e 1 in X, (3.2) Δw + + V λ w + = f(x,w + ) in X, (3.3) s 0 e 1 = w + on X, (3.4) s 0 e 1 ν w+ ν on X, (3.5) where ν denotes the outer unit normal to X. Multiplying (3.3) with (s 0 e 1 ) and subtracting it from (3.2) multiplied with w + yields ( (s0 e 1 )Δw + w + Δ(s 0 e 1 ) ) ( dx = s 0 e 1 ν1 w + f(x,w + ) ) dx. X X Since for x X, f(x,w + )>(ν 1 + δ)w +, we see that ( s 0 e 1 ν1 w + f(x,w + ) ) dx < (s 0 e 1 )( δw + )dx <0. X X

12 Z. Liu et al. / J. Differential Equations 214 ( However, (3.4), (3.5) and the divergence theorem imply ( (s0 e 1 )Δw + w + Δ(s 0 e 1 ) ) dx = X X a contradiction. Thus, for 0 <s s 0, s 0 e 1 ( w + ν s 0 Iˆ λ (se 1 ) = I λ (se 1 ) = ν 1s 2 e1 2 2 dx F(x,se 1 )dx. Ω Ω ) e 1 dσ 0, ν Decreasing s 0 if necessary, we may assume F(x,se 1 (x)) > (ν 1 + δ)s 2 e1 2 (x)/2 for all 0 <s s 0 and x R N. Then, Iˆ λ (se 1 )< ν 1s 2 e1 2 2 dx (ν 1 + δ)s 2 e1 2 dx < 0, Ω 2 Ω for all 0 <s s 0. The fact that 0 and w + are solutions of (1.1) and f is increasing in s imply that 0 K λ f(x,u) w + if 0 u w +. For 0 <s s 0 and t 0, consider the following initial value problem: d dt ηt s = V(ηt s ), η 0 s = se 1, } (3.6) where V is a pseudo gradient vector field of Iˆ λ in the form V = I B with B satisfying B([0, w + ]) [0, w + ] and [0, w + ]={u H λ : 0 u w + }; the existence of V is guaranteed by Lemma 2.4 in which D 1 =[0, w + ] and D 2 =. Let [0, T s ) denote the existence interval of (3.6). Since Iˆ λ is coercive and satisfy the (PS)-condition, a standard argument shows that there exists a sequence tn s T s such that η ts n s u s for some u s H λ satisfying I λ (u s )<0 and I λ (us ) = 0. Since B([0, w + ]) [0, w + ] and 0 se 1 w +, a theorem of Brezis Martin [7, Chapter 1, Theorem 6.3] implies that 0 η t s (x) w+ (x) for all x R N,0 t <T s and 0 <s s 0. Thus 0 <u s 0(x) w + (x) for all x R N. Choose s 1 (0,s 0 /2) such that 0 s 1 e 1 (x) u s 0(x) for all x R N. Then the above discussion implies 0 η t s 1 (x) u s 0(x) for all t [0, T s1 ) and there exists a positive solution u s 1 of (1.1) intheω-limit set of η t s 1 satisfying 0 <u s 1(x) u s 0(x) for all x R N. Here a different pseudo gradient vector field V = I B may be used in order that B([0, u s 0]) [0, u s 0]. Repeating the above process infinitely many times, we obtain a decreasing sequence of positive numbers (s n ) and positive solutions (u s n) of (1.1) such that s n 0, Iˆ λ (u s n)<0, 0 <... u s n (x) u s n 1 u s 0 (x) w + (x) for all x R N

13 370 Z. Liu et al. / J. Differential Equations 214 ( and u s n lies in the ω-limit set of η t s n. Obviously, (u s n) is a (PS)-sequence. The (PS)- condition, together with the monotonicity of (u s n) imply that u s n u + in H λ and u s n(x) u + (x) for all x R N and for some nonnegative solution u + of Eq. (1.1). Now we prove that u + is a positive solution of (1.1). Arguing indirectly, we assume u + = 0. Then u s n 0 in H λ. (3.7) Choose a natural number k such that 0 <N 2 3k <4. Set γ i = 1 + i/k for i = 0, 1,...,k. For any R>0, from the elliptic L p loc estimates [10, Theorem 9.11], we have u s n W 2,2N/(N 2) (B γk 1 R) C 4 ( f(x,u s n ) L 2N/(N 2) (B γk R) + us n L 2N/(N 2) (B γk R) C 5 u s n L 2N/(N 2) (B γk R) C 6 u s n λ, which implies ) u s n L 2N/(N 5) (B γk 1 R) C 7 u s n λ, where C 7 depends only on N, R and f, but is independent of n. Repeating the above k times, we obtain, u s n W 2,2N/(N 2 3k) (B R ) C 8 u s n λ. Since 2N/(N 2 3k) > N/2, the embedding W 2,2N/(N 2 3k) (B R ) C(B R ) is continuous. Thus u s n L (B R ) C 9 u s n λ. (3.8) Since u s 0 solves (1.1), for the number s given above there exists a R 2 > 0 such that u s 0 <s for all x R 2. From (3.7) and (3.8) there exists an N > 0 such that u s n(x)<s for all x R 2 and n N. So, u s n (x)<s for all x R N and n N. Let e λ 1 > 0 denote the first eigenfunction of the operator Δ + V λ, i.e. Δe λ 1 + V λe λ 1 = μλ 1 eλ 1 on R N.

14 Z. Liu et al. / J. Differential Equations 214 ( Recall that μ λ 1 ν 1 for all λ 0. Finally, multiplying Δu s n + V λ u s n = f(x,u s n ) with e λ 1 and integrating, we obtain (for n N ), μ λ 1 u s n e λ R N 1 dx = f(x,u s n )e λ R N 1 dx (μλ 1 + δ) u s n e λ R N 1 dx, a contradiction. To complete the proof, it remains to show that u + constructed above is indeed a minimal positive solution. Since u + (x) > 0 on R N, from the above discussion we can choose an N 1 > 0 such that s n e 1 (x) u + (x) for all n N 1 and x R N. Thus η t s n (x) u + (x) for all n N 1, t 0 and x R N, which implies u s n (x) u + (x) for all n N 1 and x R N. Since (u s n(x)) is decreasing in n and u s n(x) u + (x) for all x R N, u s n (x) = u + (x) for all n N 1 and x R N. Now, let u 1 be another positive solution of (1.1). Fixing an n 1 N 1 such that s n1 e 1 (x) u 1 (x), wehaveη t s n1 (x) u 1 (x) for all t 0 and therefore u + (x) = u s n 1 (x) u 1 (x) for all x R N. This shows that u + is the minimal positive solution of problem (1.1). Remark 3.2. We note that the above proof is independent of λ. Thus the conclusion of Theorem 3.1 holds true for a large class of potentials. For instance, we can obtain the following more general result: Assume V C(R N, R), V is bounded below, and f satisfies (f 0 ), (f 4 ), and (f 5 ). Assume in addition that uniformly in x lim inf s 0 f(x, s) s > μ 1, where μ 1 := inf u H 1 (R N )\{0} R N ( u 2 + Vu 2 )dx R N u 2. dx

15 372 Z. Liu et al. / J. Differential Equations 214 ( Then the problem Δu + V(x)u= f (x, u), u H 1 0 (RN ) has a minimal positive (maximal negative) solution provided that it has a positive (negative) solution. We close this section with a sufficient condition for the existence of a positive (and negative) solution to (1.1). From Theorem 3.1 this would imply the existence of a minimal positive (maximal negative) solution. Theorem 3.3. Assume (V 1 ) (V 3 ), (f 0 ), ( f 2 ), (f 4 ) anduniformly in x R N lim sup s f(x,s) s < ν 1. Then, for λ sufficiently large, problem (1.1) has a positive anda negative solution. Proof. We only prove the existence of a positive solution. We first show that, for λ sufficiently large, I λ is coercive. By assumption, there exist a δ > 0 and C 1 > 0 such that F(x,s) 1 2 (ν 1 δ)s 2 + C 1 for all s R. For any R>0, I λ (u) = 1 2 u 2 λ B F(x,u)dx F(x,u)dx R BR c 1 2 u 2 λ ν 1 δ u 2 dx C 1 B R C 2 u 2 dx 2 B R BR c 1 2 u 2 λ ν 1 δ 2μ λ u 2 λ C 1 B R C 2 u 2 dx 1 BR c δ u 2 4ν λ C 1 B R C 2 u 2 dx, 1 for λ sufficiently large, say λ Λ 0 (Here we used Lemma 2.2). From Lemma 2.1, there exist R 1 > 0 and Λ 1 Λ 0 such that u 2 dx δ u 2 BR c 8C 2 ν λ, 1 1 B c R

16 for all λ Λ 1. It follows that Z. Liu et al. / J. Differential Equations 214 ( I λ (u) δ u 2 8ν λ C 1 B R1, 1 for all λ Λ 1, i.e. I λ is coercive. Fix an s such that I λ (se 1 )<0, the existence of such an s was showed in the proof of Theorem 3.1. Consider the initial value problem d dt ηt = V(η t ), η 0 = se 1. } Here V is a pseudo gradient vector field of I λ in the form V = I B with B satisfying B(P + λ ) P + λ, where P + λ := {u H λ : u 0} is the positive cone in H λ. The existence of such a V is guaranteed by (f 4 ) and by Lemma 2.4 (by (f 4 ), we may assume that f(x,s) 0 for all s 0, otherwise we need just to replace V λ and f(x,s) with V λ + c 0 and f(x,s) + c 0 s, respectively). Since I λ is coercive, a standard argument shows that there exists a positive sequence t n such that I λ (η t n ) I λ (se 1 )<0 for all n, (3.9) I λ (ηt n ) 0. (3.10) According to Lemma 2.3, passing to a subsequence if necessary we may assume η t n u in H λ, where u is a solution of (1.1). Since B(P + λ ) P + λ and se 1 P + λ, ηt n P + λ. Therefore u P + λ. The fact that I λ(u) I λ (η t n) I λ (se 1 )<0 implies u = 0. Then by the maximum principle, u(x) > 0, for all x R N. Remark 3.4. Under stronger assumptions, the existence of a positive and a negative solution was obtained in [24]. 4. Invariant sets of the gradient flow In order to construct nodal solutions we need to isolate the signed solutions (positive and negative solutions) into certain invariant sets. This section is devoted to this purpose. Here we discover that it is the behavior of I λ near the trivial critical point 0 that plays an important role for the structure of these invariant sets. This will be done by distinguishing the two opposite cases: lim sup s 0 f (x, s)/s < ν 1 and lim inf s 0 f (x, s)/s > ν 1. Define A(u) := K λ f(,u)= ( Δ + V λ ) 1 f(,u), u H λ and d dt ηt (u) = η t (u) + B(η t } (u)) η 0 (u) = u,

17 374 Z. Liu et al. / J. Differential Equations 214 ( where B is related to A via Lemma 2.4 in which D 1 and D 2 will be constructed in Theorem 4.1 and Theorem 4.2. This section is concerned with the construction of these sets which are invariant under the flow η t (u) such that all positive (and negative) solutions are contained in these invariant sets. Recall that a subset W H λ is an invariant set with respect to η if, for any u W, η t (u) W for all t>0. Denote P ± λ ={u H λ :±u 0}. For any M H λ and ε > 0, M ε denotes the closed ε-neighborhood of M, i.e. M ε := {u H λ : dist λ (u, M) ε}. The following result shows that a neighborhood of P ± λ is an invariant set if 0 is a local minimum critical point of the functional. This result was proved in [2] for a superlinear problem, but the proof covers our case as well. We quote the proof from [2] for the readers convenience. We also note that this result is essentially λ-independent as long as f s (x, 0) <μλ 1. Theorem 4.1 (Bartsch et al. [2]). Assume (V 1 ) (V 3 ), (f 0 ), (f 4 ), and (f 5 ). If lim sup s 0 f (x, s)/s < ν 1 then there exist ε 0 > 0 and Λ > 0 such that A((P ± λ ) ε) int((p ± λ ) ε) for all 0 < ε ε 0, λ Λ, and η t ((P ± λ ) ε) int((p ± λ ) ε) for all t > 0, 0 < ε ε 0, and λ Λ. Proof. Foru H λ, we denote v = Au and u + = max{0,u}, u = min{0,u}. Note that for any u H λ and 2 p 2 := 2N/(N 2), u L p = inf w P + λ u w L p. (4.1) Since, v 2 λ = (v, v ) λ = ( v v + V λ vv )dx = R N f (x, u)v dx, R N

18 Z. Liu et al. / J. Differential Equations 214 ( the fact that v + P + λ and v v+ = v, implies dist λ (v, P + λ ) v λ v 2 λ R N f(x,u )v dx. (4.2) Under the assumption lim sup s 0 f (x, s)/s < ν 1 there exist a δ > 0 and C 1 > 0 such that f(x,s) (ν 1 δ)s + C 1 s 2 2 s for s 0. Thus R N f(x,u )v dx R N [ (ν 1 δ)u + C 1 u 2 2 u ] v dx (ν 1 δ) u L 2 v L 2 + C 1 u 2 1 L 2 1 v L 2 1. (4.3) From the Sobolev imbedding and (4.1) (4.3), dist λ (v, P + λ ) v λ ν 1 δ μ λ 1 inf w P + λ u w L 2 v λ + C 2 inf w P + λ u w 2 1 L 2 1 v λ, which implies (if v λ = 0), dist λ (v, P + λ ) ν 1 δ μ λ 1 ν 1 δ μ λ 1 inf w P + λ inf w P + λ u w L 2 + C 2 u w λ + C 3 inf w P + λ inf w P + λ u w 2 1 L 2 1 u w 2 1 λ = ν 1 δ μ λ dist λ (u, P + λ ) + C ( 3 distλ (u, P + λ )) Therefore, there exist ε 0 > 0 and Λ > 0 such that if dist λ (u, P + λ ) ε 0 and λ Λ then dist λ (v, P + λ )<dist λ(u, P + λ ). The first conclusion in Theorem 4.1 is proved. The second conclusion is a consequence of the first as shown in [18] via Lemma 2.4. Next, we consider the case lim inf s 0 f (x, s)/s > ν 1. We first note that any neighborhoods of the positive (and negative) cones are no longer invariant sets of the gradient

19 376 Z. Liu et al. / J. Differential Equations 214 ( flow. We give a new construction here. Choose δ > 0 and s 0 > 0 such that f(x,s) s > ν 1 + δ μ λ 1 + δ, (4.4) for 0 <s s 0 and x R N. Let e λ 1 H λ be the eigenfunction associated with the eigenvalue μ λ 1 (we assume λ to be sufficiently large) such that eλ 1 > 0 and max x R N eλ 1 s 0. According to Theorem 3.1 and its proof, we may also assume that w (x) e λ 1 (x) and eλ 1 (x) w+ (x), for all x R N, all negative solutions w and positive solutions w + of (1.1). Define D ± λ := {u H λ :±u e λ 1 }. From the above discussion, all positive solutions and negative solutions to (1.1) are contained in D + λ and D λ, respectively. We show in the following that under the condition lim inf s 0 f (x, s)/s > ν 1, suitable neighborhoods of these sets are invariant sets. This result is λ-dependent and holds only for λ large. Theorem 4.2. Assume (V 1 ) (V 3 ), (f 0 ), (f 4 ), and (f 5 ). If lim inf s 0 f (x, s)/s > ν 1, then there exist ε 0 > 0 and Λ > 0 such that A((D ± λ ) ε) int((d ± λ ) ε) for all 0 < ε ε 0, λ Λ, and η t ((D ± λ ) ε) int((d ± λ ) ε) for all t > 0, 0 < ε ε 0, and λ Λ. Proof. We only prove the result for the positive sign, the other case follows analogously. For u H λ we denote v = Au and v 1 = max{e λ 1,v}. Then dist λ (v, D + λ ) v v 1 λ which implies dist λ (v, D + λ ) v v 1 λ v v 1 2 λ.

20 Note that v 1 = e λ 1 if v 1 = v. Since Z. Liu et al. / J. Differential Equations 214 ( v v 1 2 λ = (v eλ 1,v v 1) λ = (v e λ R N 1 ) (v v 1) + V λ (v e1 λ )(v v 1)dx ) = ( Δ(v e1 λ ) + V λ(v e1 λ ) (v v 1 )dx = R N R N ( ) f(x,u) μ λ 1 eλ 1 (v v 1 )dx, we have dist λ (v, D + λ ) v v 1 λ R N (μ λ 1 eλ 1 f(x,u) ) (v 1 v) dx. For the number s 0 givenin(4.4), if u>s 0, then f(x,u) f(x,s 0 )>(μ λ 1 + δ)s 0 (μ λ 1 + δ)eλ 1. Therefore, since v v 1, we conclude that dist λ (v, D + λ ) v v 1 λ u(x) s 0 (μ λ 1 eλ 1 f(x,u) ) (v 1 v) dx. Since f(x,u) (μ λ 1 + δ)u for 0 u s 0 and f(x,u) c 0 u for u 0, dist λ (v, D + λ ) v v 1 λ ) (μ λ 1 eλ 1 (μλ 1 + δ)u (v 1 v) dx + 0 u(x) s 0 = I 1 + I 2. u(x)<0 ( ) μ λ 1 eλ 1 c 0u (v 1 v) dx Note that I 1 0 u(x) μ λ 1 eλ 1 (x)/(μλ 1 +δ) (μ λ 1 eλ 1 (μλ 1 + δ)u ) (v 1 v) dx (μ λ 1 + δ) 0 u(x) μ λ 1 eλ 1 (x)/(μλ 1 +δ) (e λ 1 u)(v 1 v) dx

21 378 Z. Liu et al. / J. Differential Equations 214 ( and I 2 (μ λ 1 + c 0) (e1 λ u)(v 1 v) dx. u(x)<0 Denoting C = max{μ λ 1 + δ, μλ 1 + c 0}, wehave dist λ (v, D + λ ) v v 1 λ C Θ λ(eλ 1 u)(v 1 v) dx := CE λ, (4.5) where { } Θ λ := x R N : u(x) μλ 1 μ λ 1 + δeλ 1 (x). For any R>0, we write E λ = (e Θ λ 1 λ u) (v 1 v) dx + B R Since, on Θ λ B R, Θ λ B c R e1 λ u δ μ λ 1 + δeλ 1 δ R (e λ 1 u) (v 1 v) dx := E λ 1 + Eλ 2. for some δ R > 0, there exists a C 1 = C 1 (R) > 0 such that E1 λ C 1(R) e Θ λ 1 λ u 2 1 v 1 v dx B R The fact u(x) e λ 1 (x) on Θλ implies C 1 (R) e λ 1 u 2 1 L 2 (Θ λ B R ) v 1 v L 2 C 1 (R) e λ 1 u 2 1 L 2 (Θ λ B R ) v v 1 λ. (4.6) e λ 1 u L p (Θ λ ) = inf w u w D + L p (Θ λ ), λ for all u H λ and 2 p 2. Then, by the Sobolev imbedding and (4.6), E λ 1 C 2(R) ( dist λ (u, D + λ )) 2 1 v1 v λ. (4.7)

22 Z. Liu et al. / J. Differential Equations 214 ( From Lemma 2.1, there exist R 0 > 0 and Λ 0 > 0 such that E λ 2 eλ 1 u L 2 (Θ λ B c R 0 ) v 1 v L 2 inf u w w D + L 2 (Θ λ BR c ) v 1 v λ 0 λ 1 2C dist λ(u, D + λ ) v 1 v λ, (4.8) for all λ Λ 0, where C is the number from (4.5). Combining (4.5) (4.8) we conclude that ( 1 dist λ (v, D + λ ) C 2C dist λ(u, D + λ ) + C 2(R 0 ) ( ) dist λ (u, D + λ )) 2 1, for all v/ D + λ ε 0, it holds and λ Λ 0. Therefore there exists an ε 0 > 0 such that if dist λ (u, D + λ ) dist λ (Au, D + λ ) = dist λ(v, D + λ )<dist λ(u, D + λ ). Once again, the second conclusion follows directly from [18], via Lemma Proof of Theorem 1.1 In order to prove our main result Theorem 1.1, we need to use Ljusternik Schnirelman type minimax results. In general, results of this type do not give the nodal information of the solutions. We need to have the order structure and the invariant sets of the gradient flow from the previous section built into the minimax arguments. We need to distinguish two cases: d m >d n and d n >d m. Though the results are similar for the two cases, the minimax arguments used are quite different and the structure of the invariant sets are also quite different. These two cases will be dealt with in Sections 5.2 and 5.3, with some technical preparations in Section A deformation lemma in the presence of invariant sets Let us start with a more abstract setting. Consider I C 1 (X, R) where X is a Banach space. Let V be a pseudo gradient vector field of I such that V is odd if I is even, and consider d dt } σ(t, u) = V(σ), σ(0,u) = u X. (5.1)

23 380 Z. Liu et al. / J. Differential Equations 214 ( A subset W X is an invariant set with respect to σ if, for any u W, σ(t, u) W for all t 0. The following lemma is a variant of [14, Lemma 2.4]. Lemma 5.1. Assume I satisfies the (PS)-condition, and c R is fixed. Assume W = W int(w) is an invariant subset w.r.t. σ such that σ(t, W) int(w) for t>0. Define Kc 1 := K c W, Kc 2 := K c (X\W), where K c := {u X : I (u) = 0, I (u) = c}. Let δ > 0 be such that (Kc 1) δ W where ( Kc 1 ) δ ={u X : dist(u, K1 c )<δ}. Then there exists an ε 0 > 0 such that for any 0 < ε < ε 0, there exists η C([0, 1] X, X) satisfying: (i) η(t, u) = u for t = 0 or u/ I 1 ([c ε 0,c+ ε 0 ]) \ ( Kc 2 ) δ. (ii) η ( 1,I c+ε W \ ( Kc 2 (iii) η(t, ) is a homeomorphism of X for t [0, 1]. (iv) η(t, u) u δ, for any (t, u) [0, 1] X. (v) I(η(t, )) is non-increasing. (vi) η(t, W) W for any t [0, 1]. (vii) η(t, ) is oddif I is even andif W is symmetric with respect to 0. ) 3δ) I c ε W and η(1,i c+ε W) I c ε W if K 2 c =. Proof. Due to the (PS) condition, we may choose ε 0 > 0 such that I (u) I (u) 8ε 0 δ, for any u I 1 ([c ε 0,c+ ε 0 ]) \ (K c ) δ. Set X 1 := I 1 ([c ε 0,c+ ε 0 ]) \ (K 2 c ) δ.for anyfixed0< ε < ε 0, define X 2 = I 1 ([c ε,c+ ε]) \ (K 2 c ) 2δ and Consider d dt ψ(u) = dist(u, X \ X 1 ) dist(u, X \ X 1 ) + dist(u, X 2 ). u))v (ξ(t, u)) ξ(t, u) = ψ(ξ(t,, ξ(0,u)= u. 1 + V(ξ(t, u)) Then ξ(t, u) is well-defined and continuous on R X and we claim that η(t, u) := ξ(δt,u) has all the properties in the lemma. (i), (iii), (v), (vi) and (vii) are easily checked. For (iv) we note that η(t, u) u = 1 0 η δ (τ, u)dτ = ξ (τ, u)dτ δ. To show (ii), we suppose by contradiction that η(1,u) I c ε W for some u I c+ε W\ ( Kc 2 ) 3δ. Then η(t, u) I c ε W for all 0 t 1. We see that η(t, u) 0

24 Z. Liu et al. / J. Differential Equations 214 ( (K 1 c ) δ for 0 t 1 since (K 1 c ) δ W, and η(t, u) (K 2 c ) 2δ for 0 t 1 due to (iv) and u (K 2 c ) 3δ. Thus, for 0 t 1, η(t, u) I 1 ([c ε,c+ ε]) \ ( (K 2 c ) 2δ (K c ) δ ), which implies, for 0 t 1, ψ(η(t, u)) = 1 and I (η(τ, u)) I (η(τ, u)) 8ε 0 δ. Then 1 I(η(1, u)) = I (u) + 0 d 1 I(η(τ, u))dτ c + ε dτ 0 δ I (η(τ, u)) 2 4(1 + I dτ <c ε, (η(τ, u)) ) a contradiction. We shall also need the notion of genus (c.f. [19,21]). Set Σ λ := {A H λ \{0} :A is closed and A = A} and let γ(a) denote the genus of A, which is defined as the least integer n such that there exists an odd continuous map σ : A S n 1. We refer to [21] for the following properties of genus. Proposition 5.2. Let A, B Σ λ, and h C(H λ,h λ ) be an odd map. Then (i) A B γ(a) γ(b). (ii) γ(a B) γ(a) + γ(b). (iii) γ(a) γ(h(a)). (iv) If A is compact, there exists an N Σ λ such that A int(n) N and γ(a) = γ(n). (v) If F is a linear subspace of H λ with dimf = n, A F is bounded, open and symmetric, and 0 A, then γ( F A) = n. Recall in Section 4, D ± λ ={u H λ ±u e1 λ} with eλ 1 > 0 the first eigenfunction satisfying max e1 λ <s 0 and s 0 givenin(4.4). All positive (negative, resp.) solutions of problem (1.1) are contained in D + λ (D λ, resp.) and (D± λ ) ε are invariant sets if ε ε 0. We need the following result. Lemma 5.3. For any ρ > 0, let B ρ := {u H λ : u λ ρ}. Then ) dist λ ( B ρ (E 1 (λ)),d + λ D λ > 0. Proof. Assume for the contrary, there exists (u n ) D + λ, (v n) B ρ (E 1 (λ)), such that dist λ (u n,v n ) 0. Then (u n,e λ 1 ) λ = (u n v n,e λ 1 ) λ + (v n,e λ 1 ) λ 0asn.

25 382 Z. Liu et al. / J. Differential Equations 214 ( But since u n e1 λ,wehave (u n,e1 λ ) λ = μ λ 1 (u n,e1 λ ) L 2 μλ 1 (e λ R N 1 )2 dx = 0, a contradiction. Remark 5.4. We note that the above result is not true when D ( ± λ is replaced by P ± λ. That is, for any ρ > 0, it can be proved that dist λ Bρ (E 1 (λ)),p + λ P ) λ = The proof of Theorem 1.1 the case d m >d n We consider the case d m >d n first. In this case we have m 1 since ν n+1 ν m < lim inf s 0 We state the result more precisely here. f(x,s) f(x,s) lim sup < ν m+1. s s 0 s Theorem 5.5. Assume (V 1 ) (V 3 ) and (f 0 ) (f 5 ). Then there exists a Λ > 0 such that for all λ Λ, Eq. (1.1) has at least d m d n (d m d n 1, resp.) pairs of nodal solutions having negative critical values provided n 1 (n = 0, resp.). The following two lemmas are standard and their special cases were proved in [24,25]. Lemma 5.6. Assume (V 1 ) (V 2 ), (f 0 ), (f 2 ), and (f 4 ) with m = 0. Then there exist a ρ > 0 such that for all λ 0, sup I λ < 0. E m B ρ Proof. By(f 2 ) and (f 4 ), there exist δ > 0, C 1 > 0, and p (2, 2 ) such that for all x and u, F(x,u) (ν m + δ)u 2 /2 C 1 u p. Here 2 = 2N/(N 2) for N>2 and 2 =+ for N = 2. For u E m, the inequality u 2 λ ν m u 2 and the Sobolev L inequality imply 2 I λ (u) 1 2 u 2 λ (ν m + δ) u 2 dx C 1 u p dx R N R N δ 2ν m u 2 λ + C 2 u p λ, which gives the result.

26 Z. Liu et al. / J. Differential Equations 214 ( Lemma 5.7. Assume (V 1 ) (V 3 ), (f 0 ), (f 3 ), and (f 4 ). Then there exists a Λ > 0 such that for all λ Λ, dim E n (λ) = d n and inf (E n (λ)) I λ >. Proof. By(f 3 ), there exist δ > 0 and T>0such that for u T, F(x,u) (ν n+1 2δ)u 2 /2. Choose Λ > 0 such that for all λ Λ, μ dn +1 ν n+1 δ by Lemma 2.2. Let u (E n (λ)) with λ Λ and denote Z := {x : u(x) T }. For any R>0, the inequality u 2 λ μλ d n +1 u 2 implies L 2 I λ (u) 1 2 u 2 λ B R Z c F(x,u)dx 1 2 c 0 δ 2μ λ d n +1 u 2 λ C 1(R) 1 2 c 0 u 2 L 2 (B c R ). u 2 dx μλ dn +1 δ BR c Zc 2 u 2 dx Z Then using Lemma 2.1 gives the result. Proof of Theorem 5.5. Throughout the proof, we fix a λ sufficiently large such that I λ satisfies the (PS)-condition (Lemma 2.3) and all other relevant results hold. By Theorem 4.2 we may choose an ε > 0 small such that (D + λ ) ε (D λ ) ε = and W := (D + λ ) ε (D λ ) ε is an invariant set of the pseudo gradient flow. Recall that int(w) contains all positive and negative solutions. For j = 2,...,d m, define Γ j := {A Σ λ : γ(a) j} and c j := inf A Γ j sup I λ (u), A S where S := H λ \W. For any A Σ λ with γ(a) 2, we have A S = (since, if A S =, A W,butγ(W ) = 1, a contradiction). Thus c j for j = 2,...,d m can be defined. First we assume d n = 0 and we consider c j for j = 2,...,d m. As a consequence of Lemmas 5.6 and 5.7, <c 2 c 3 c dm < 0. We claim that if c := c j = c j+1 = = c j+k for some 2 j j + k d m with k 0 then γ(k c S) k + 1. Since Kc 2 = K c S is compact, there exists a closed neighborhood N in H λ with Kc 2 int(n) N such that γ(n) = γ(k2 c ). Without loss of generality, we may assume N = ( Kc 2 ) 3δ with δ > 0 satisfying ( Kc 1 ) δ W. Then there exists an ε 0 > 0 such that for 0 < ε < ε 0, there exists an η C([0, 1] H λ,h λ ) satisfying (i) (vii) of Lemma 5.1. Choose A Γ j+k such that sup I λ c + ε, A S

27 384 Z. Liu et al. / J. Differential Equations 214 ( Then η ( 1,A\ ( Kc 2 ) ) 3δ I c ε λ W and we claim Otherwise, γ ( η ( 1,A\ ( Kc 2 ) 3δ)) j and c = c j a contradiction. Now, ( ( ( ) )) γ η 1,A\ Kc 2 j 1. 3δ sup I λ η(1,a\(kc 2 ) 3δ ) S ( I c ε λ sup W ) S I λ sup I λ c ε, I c ε λ j + k γ(a) γ(a\int(n)) + γ(n) γ(η(1,a\(k 2 c ) 3δ)) + γ(k 2 c ) j 1 + γ(k 2 c ), which implies γ(kc 2) k + 1. Since all the solutions in K2 c are nodal solutions, the result is true in the case d n = 0. If d n 1, we may consider all c j for j = d n + 1,...,d m. By Lemmas 5.6 and 5.7, <c dn +1 c dm < 0. The same argument as above works in this case The proof of Theorem 1.1 the case d n >d m In this case we have n 1 and we prove the following result here which gives the other part of Theorem 1.1. Theorem 5.8. Assume (V 1 ) (V 3 ) and (f 0 ) (f 5 ). Then there exists a Λ > 0 such that for all λ Λ, Eq. (1.1) has at least d n d m (d n d m 1, resp.) pairs of nodal solutions having positive critical values provided m 1 (m = 0, resp.). We need the following two lemmas which are generalizations of similar results in [24]. Lemma 5.9. Assume (V 1 ) (V 2 ), (f 0 ), (f 3 ), and (f 4 ) with n = 0. Then there exists a R>0 such that for all λ 0 and B c R := H λ\b R, sup I λ < 0. E n BR c

28 Z. Liu et al. / J. Differential Equations 214 ( Proof. By(f 3 ), there exist δ > 0 and T > 0 such that for u T, F(x,u) (ν n + δ)u 2 /2. Let u E n and denote Z := {x : u(x) T }. For any R>0, I λ (u) 1 2 u 2 λ (c 0 + ν n + δ) u 2 dx ν n + δ u 2 dx Z c 2 R N δ 2ν n u 2 λ + C 1(R) (c 0 + ν n + δ) u 2 L 2 (B c R ), this together with Lemma 2.1 leads to the result. Lemma Assume (V 1 ) (V 2 ), (f 0 ), (f 2 ), and (f 4 ). Then there exist a Λ > 0 and ρ > 0 such that for all λ Λ, dim E m (λ) = d m and inf (E m (λ)) B ρ I λ > 0. Proof. There exist δ > 0 and C 1 > 0 such that F(x,u) (ν m+1 2δ)u 2 /2 + C 1 u p for some p (2, 2 ). Choose Λ > 0 such that for λ > Λ, μ λ d m +1 > ν m+1 δ. Then for λ Λ and u (E m (λ)), I λ (u) 1 2 u 2 λ μλ d m +1 δ 2 δ 2μ λ d m +1 u 2 λ C 2 u p λ, R N u 2 C 1 u p L p which gives the result. Proof of Theorem 5.8. We need to distinguish two subcases here: (i) d m 1 and (ii) d m = 0. Let us consider (i) first. Again, by Theorem 4.2 we may choose an ε small enough such that W := (D + λ ) ε (D λ ) ε is an invariant set of the gradient flow and all positive and negative solutions are contained in int(w ). Set S = H λ \ W. Wehavetousea different family of sets for the minimax procedure here. We essentially follow [19]. Define, G := {h C(B R E n,h λ ) : h is odd and h = id on B R E n }, where R>0 is given by Lemma 5.9. Note that G =, since id G. Set Γ j := { h ( B R E n \Y ) : h G, Y Σ λ and γ(y ) d n j }

29 386 Z. Liu et al. / J. Differential Equations 214 ( for j {2,...,d n }. From [19], Γ j possess the following properties: (1 ) Γ j = for all j {2,...,d n }. (2 ) Γ j+1 Γ j for j {2,...,d n 1}. (3 )Ifσ C(H λ,h λ ) is odd and σ = id on B R E n, then σ : Γ j Γ j for all j {2,...,d n } (i.e. σ(a) Γ j if A Γ j ). (4 )IfA Γ j, Z Σ λ, γ(z) s <j and j s 2, then A\Z Γ j s. Now, for j = d m + 1,...,d n, we define c j := inf sup I λ. A Γ j A S From [19, Proposition 9.23] for A Γ j with j d m + 1, A B ρ (E m (λ)) =. Since d m 1, by Lemma 5.3, B ρ (E m (λ)) S. Thus, for j d m + 1 and A Γ j, A S =, and from Lemma 5.10 we conclude that c j inf I λ α > 0. B ρ (E m (λ)) Then from the definition of c j and (2 ) we have 0 < α c dm +1 c dn <. We claim that if c := c j = = c j+k for some d m + 1 j j + k d n with k 0 then γ(k c S) k + 1. This also shows that each c j is a critical value. Since 0 / K c, and K 2 c = K c S is compact, we may choose N such that K2 c int(n) N and γ(k2 c ) = γ(n). Ifγ(K2 c ) k, wehaveγ(n) k. By Lemma 5.1 there exist ε > 0 and η C([0, 1] H λ,h λ ) such that η(1, ) is odd, η(1,u)= u for u I c 2ε, and ( ) η 1,I c+ε λ W \N I c ε λ W. We may assume c 2ε > 0. Choose A Γ j+k such that sup I λ c + ε. A S Then by (4 ) above A\N Γ j. As a consequence of Lemma 5.9, η(1,u) = u for u B R E n, and we have η(1, A\N) Γ j by (3 ). Then c sup I λ η(1,a\n) S ( I c ε λ sup W I ) λ c ε, S

30 Z. Liu et al. / J. Differential Equations 214 ( a contradiction. Therefore γ(k c S) k + 1 and I λ has at least d n d m pairs of nodal critical points. Next we consider case (ii): d m = 0. We have to use a different invariant set since 0 is a local minimum. From Theorem 4.1, we may choose an ε > 0 sufficiently small such that (P ± λ ) ε are invariant sets. Set W := (P + λ ) ε (P λ ) ε and S := H λ \W. We define c j := inf A Γ j sup I λ, A S for j = 2,...,d n. We need to show that for any A Γ j and j = 2,...,d n, A S = so that cj are well defined, and c 2 α > 0. Consider the attracting domain of 0 in H λ : Q := {u H λ : σ(t, u) 0, as t }. Note that Q is an invariant set. We claim that for A Γ j holds with j = 2,...,d n,it A S Q =. (5.2) This proves both A S = and c 2 α > 0, since B ρ Q and inf Q I λ inf Bρ I λ α > 0 by Lemma To prove (5.2), let A = h(b R E n \Y) with γ(y ) d n j and j 2. Define O := {u B R E n : h(u) Q}. Then O is a bounded open set with 0 O and O B R E n. Thus, from the Borsuk Ulam theorem γ( O) = d n and by the continuity of h, h( O) Q. It follows that γ( O\Y) j, h( O\Y) A Q and therefore γ(a Q) j. Since γ(w Q) = 1, which follows from (P + λ ) ε (P λ ) ε Q =, we conclude that γ(a S Q) γ(a Q) γ(w Q) 1, which proves (5.2). Thus cj are well defined for j = 2,...,d n and 0 < α c2 c 3 cd n <. Proceeding as for the case d m > 0 we have: if cj = c j+1 = = c j+k for 2 j j + k d n with k 0 then γ(k c j S ) k + 1. Therefore, I λ has at least d n d m 1 pairs of nodal critical points. The proof is similar and we omit it. Then the proof for the case (ii) is finished, and therefore the proof of the main Theorem 1.1 is complete.

31 388 Z. Liu et al. / J. Differential Equations 214 ( Related results Our method allows us to work on several other problems in the entire space R N and we mention some results here with sketch of the proofs. First we observe that additional information on the signed solutions will produce stronger existence results. In case d n >d m 1, we have d n d m pairs of nodal solutions. The following result gives additional nodal solutions to those in Theorem 1.1 in the presence of a pair of signed solutions. Theorem 6.1. Assume (V 1 ) (V 3 ) and (f 0 ) (f 5 ). Assume that there exists a positive solution w (so w is a negative solution). If d n >d m 2, Eq. (1.1) has at least d n 1 pairs of nodal solutions with d n d m pairs having positive critical values and d m 1 pairs having negative critical values. Sketch of the proof. Using Theorem 1.1 we get d n d m pairs of nodal solutions having positive critical values. Due to the presence of a pair of signed solutions we can also get d m 1 pairs of nodal solutions having negative critical values. This is done by modifying f(x,u) to a new function f(x,u) ˆ as in Section 3 and consider Eq. (3.1) in Section 3. It is easy to check the conditions of Theorem 1.1 are satisfied with d n = 0 in this case, so applying Theorem 1.1 we get d m 1 pairs of nodal solutions with negative critical values. Next, we consider the following problem. Δu + V(x)u= f (x, u), in R N, (6.1) which satisfy u(x) 0as x. The potential function V satisfies (V 1 )V C(R N, R) satisfies inf V V 0 > 0. (V 2 ) lim x V(x)=+. Under these conditions the linear operator is compact and has discrete spectrum only, i.e., σ( Δ + V) is given by 0 < ν 1 < ν 2 < with the dimension of each eigenvalue dim(ν k )<. Thus this is somewhat simpler case than the one we have considered. We state the conditions and a similar result and leave the details to the readers. We may assume the more general condition than (V 2 ) (V 2 ) There exists r 0 > 0 such that for any M>0 ( ) lim m {x R N : x y r 0 } {x R N : V(x) M} = 0. y Again, set d k := k dim(ν i ) and d 0 := 0. i=1