CUBO A Mathematical Journal Vol.10, N o ¯ 03, (21–41). October 2008 Multiple Solutions for Doubly Resonant Elliptic Problems Using Critical Groups Ravi P. Agarwal Department of Mathematical Sciences, Florida Institute of Technology, Melbourne 32901-6975, FL, U.S.A email: agarwal@fit.edu Michael E. Filippakis Department of Mathematics, National Technical University, Zografou Campus, Athens 15780, Greece email: mfilip@math.ntua.gr Donal O’Regan Department of Mathematics, National University of Ireland, Galway, IRELAND email: donal.oregan@nuigalway.ie and Nikolaos S. Papageorgiou Department of Mathematics, National Technical University, Zografou Campus, Athens 15780, Greece email: npapg@math.ntua.gr ABSTRACT We consider a semilinear elliptic equation, with a right hand side nonlinearity which may grow linearly. Throughout we assume a double resonance at infinity in the spectral interval [λ1,λ2]. In this paper, we can also have resonance at zero or even double 22 Ravi P. Agarwal et al. CUBO 10, 3 (2008) resonance in the order interval [λm,λm+1], m ≥ 2. Using Morse theory and in particular critical groups, we prove two multiplicity theorems. RESUMEN Nosotros consideramos una ecuación semilinear eliptica con una no-linealidad la cual puede crecer linealmente. Asumimos una doble resonancia en infinito en el intervalo espectral [λ1,λ2]. En este art́ıculo, podemos también tener resonancia en cero o incluso doble resonancia en el intervalo ordenado [λm,λm+1], m ≥ 2. Usando teoria de Morse y en particular grupos cŕıticos, provamos dos teoremas de mulplicidad. Key words and phrases: Double resonance, C-condition, critical groups, critical point of moun- tain pass-type, Poincare-Hopf formula. Math. Subj. Class.: 35J20, 35J25. 1 Introduction Let Z ⊆ RN be a bounded domain with a C2-boundary ∂Z. We consider the following semilinear elliptic problem: { −△x(z) = λ1x(z) + f(z,x(z)) a.e. on Z, x|∂Z = 0. } (1.1) Here λ1 > 0 is the principal eigenvalue of (−△,H 1 0 (Z)). Assume that lim |x|→∞ f(z,x) x = 0 uniformly for a.a. z ∈ Z. (1.2) The problem (1.1) is resonant at infinity with respect to the principal eigenvalue λ1 > 0. Res- onant problems, were first studied by Landesman-Lazer [7], who assumed a bounded nonlinearity and introduced the well-known sufficient asymptotic solvability conditions, which carry their name (the LL-conditions for short). We can be more general and instead of (1.2), assume only that lim inf |x|→∞ f(z,x) x and lim sup |x|→∞ f(z,x) x belong in the interval [0,λ2 − λ1] uniformly for a.a. z ∈ Z, with λ2 (λ2 > λ1) being the second eigenvalue of (−△,H1 0 (Z)). In this more general setting, the nonlinearity f(z,x) need not be bounded. This more general situation was examined by Berestycki-De Figueiredo [2], Landesman-Robinson-Rumbos [8], Nkashama [11], Robinson [13],[14], Rumbos [15] and Su [16]. From these works, Berestycki-De Figueiredo [2], Nkashama [11], Robinson [13] and Rumbos [15], prove existence theorems in a double resonance setting (i.e. asymptotically at ±∞, we have CUBO 10, 3 (2008) Multiple Solutions for Doubly ... 23 complete interaction of the ”slope” f (z,x) x with both ends of the spectral interval [0,λ2 − λ1]; see Berestycki-De Figueiredo [2] who coined the term ”double resonance” and Robinson [13]) or in a one-sided resonance setting (i.e. the ”slope” f (z,x) x is not allowed to cross λ2 − λ1; see Nkashama [11] and Rumbos [15]). Multiplicity results were proved by Landesman-Robinson-Rumbos [8] (one- sided resonant problems) and by Robinson [14] and Su [16] (doubly resonant problems). In this paper, we extend the work of Landesman-Robinson-Rumbos [8] and partially extend and complement the works of Robinson [14] and Su [16], by covering cases which are not included in their multiplicity results. 2 Mathematical background We start by recalling some basic facts about the following weighted linear eigenvalue problem: { −△u(z) = ̂λm(z)u(z) a.e. on Z, u|∂Z = 0, ̂λ ∈ R. } (2.1) Here m ∈ L∞(Z)+ = {m ∈ L ∞(Z) : m(z) ≥ 0 a.e. on Z}, m 6= 0 (the weight function). By an eigenvalue of (2.1), we mean a real number ̂λ, for which problem (2.1) has a nontrivial solution u ∈ H1 0 (Z). It is well-known (see for example Gasinski-Papageorgiou [5]), that problem (2.1) (or equivalently that (−△,H1 0 (Z),m)), has a sequence {̂λk(m)}k≥1 of distinct eigenvalues, ̂λ1(m) > 0 and ̂λk(m) → +∞ as k → +∞. Moreover, ̂λ1(m) > 0 is simple (i.e. the corresponding eigenspace E(̂λ1) is one-dimensional). Also we can find an orthonormal basis {un}n≥1 ⊆ H 1 0 (Z) ∩ C∞(Z) for the Hilbert space L2(Z) consisting of eigenfunctions corresponding to the eigenvalues {̂λk(m)}k≥1. Note that {un}n≥1 is also an orthogonal basis for the Hilbert space H 1 0 (Z). Moreover, since by hypothesis ∂Z is a C2-manifold, then un ∈ C 2(Z) for all n ≥ 1. For every k ≥ 1, by E(̂λk) we denote the eigenspace corresponding to the eigenvalue ̂λk(m). This space has the so-called ”unique continuation property”, namely, if u ∈ E(̂λk) is such that it vanishes on a set of positive measure, then u(z) = 0 for all z ∈ Z. We set Hk = k ⊕ i=1 E(̂λi) and ̂Hk+1 = ⊕ i≥k+1 E(λi) = H ⊥ k , k ≥ 1. We have the orthogonal direct sum decomposition H 1 0 (Z) = Hk ⊕ ̂Hk+1. Using these spaces, we can have useful variational characterizations of the eigenvalues {̂λk(m)}k≥1 using the Rayleigh quotient. Namely we have: ̂λ1(m) = min [ ‖Du‖2 2 ∫ Z mu2dz : u ∈ H1 0 (Z),u 6= 0 ] . (2.2) 24 Ravi P. Agarwal et al. CUBO 10, 3 (2008) In (2.2) the minimum is attained on E(̂λ1)\{0}. By u1 ∈ C 2 0 (Z), we denote the principal eigenfunction satisfying ∫ Z mu2 1 dz = 1. For k ≥ 2, we have ̂λk(m) = max [ ‖Du‖2 2 ∫ Z mu2dz : u ∈ Hk,u 6= 0 ] (2.3) = min [ ‖Dû‖2 2 ∫ Z mû2dz : û ∈ ̂Hk, û 6= 0 ] . (2.4) In (2.3) (resp.(2.4)), the maximum (resp.minimum) is attained on E(̂λk). From these varia- tional characterizations of the eigenvalues and the unique continuation property of the eigenspaces E(̂λk), we see that the eigenvalues {̂λk(m)}k≥1 have the following strict monotonicity property: ”If m1,m2 ∈ L ∞(Z)+, m1(z) ≤ m2(z) a.e. on Z and m1 6= m2, then ̂λk(m2) < ̂λk(m1) for all k ≥ 1.” If m ≡ 1, then we simply write λk for all k ≥ 1 and we have the full-spectrum of (−△,H 1 0 (Z)). Let H be a Hilbert space and ϕ ∈ C1(H). We say that ϕ satisfies the ”Cerami condition” (the C-condition for short), if the following is true:”every sequence {xn}n≥1 ⊆ H such that |ϕ(xn)| ≤ M1 for some M1 > 0, all n ≥ 1 and (1 + ‖xn‖)ϕ ′(xn) → 0 in H ∗ as n → ∞, has a strongly convergent subsequence”. This condition is a weakened version of the well-known Palais-Smale condition (PS-condition for short). Bartolo-Benci-Fortunato [1], showed that the C-condition suffices to prove a deformation theorem and from this produce minimax expressions for the critical values of the functional ϕ. For every c ∈ R, let ϕc = {x ∈ X : ϕ ≤ c} (the sublevel set at c of ϕ), K = {x ∈ x : ϕ′(x) = 0} (the set of critical points of ϕ) and Kc = {x ∈ K : ϕ(x) = c} (the critical points of ϕ at level c). If X is a Hausdorff topological space and Y a subspace of it, for every integer n ≥ 0, by Hn(X,Y ) we denote the n th-relative singular homology group with integer coefficients. The critical groups of ϕ at an isolated critical point x0 ∈ H with ϕ(x0) = c, are defined by Cn(ϕ,x0) = Hn(ϕ c ∩ U, (ϕc ∩ U)\{x0}), where U is a neighborhood of x0 such that K ∩ϕ c ∩U = {x0}. By the excision property of singular homology theory, we see that the above definition of critical groups, is independent of U (see for example Mawhin-Willem [10]). Suppose that −∞ < inf ϕ(K). Choose c < inf ϕ(K). The critical groups at infinity, are defined by Ck(ϕ,∞) = Hk(H,ϕ c) for all k ≥ 0. CUBO 10, 3 (2008) Multiple Solutions for Doubly ... 25 If K is finite, then the Morse-type numbers of ϕ, are defined by Mk = ∑ x∈K rankCk(ϕ,x). The Betti-type numbers of ϕ, are defined by βk = rankCk(ϕ,∞). By Morse theory (see Chang [4] and Mawhin-Willem [10]), we have m ∑ k=0 (−1)m−kMk ≥ m ∑ k=0 (−1)m−kβk and ∑ k≥0 (−1)kMk = ∑ k≥0 (−1)kβk. From the first relation, we deduce that βk ≤ Mk for all k ≥ 0. Therefore, if βk 6= 0 for some k ≥ 0, then ϕ must have a critical point x ∈ H and the critical group Ck(ϕ,x) is nontrivial. The second relation (the equality), is known as the ”Poincare-Hopf formula”. Finally, if K = {x0}, then Ck(ϕ,∞) = Ck(ϕ,x0) for all k ≥ 0. 3 Multiplicity of solutions The hypotheses on the nonlinearity f(z,x) are the following: H(f): f : Z × R → R is a function such that f(z, 0) = 0 a.e. on Z and (i) for all x ∈ R, z → f(z,x) is measurable; (ii) for almost all z ∈ Z, f(z, ·) ∈ C1(R); (iii) |f′x(z,x)| ≤ c(1 + |x| r), r < 4 N−2 ,c > 0. (iv) 0 ≤ lim inf |x|→∞ f (z,x) x ≤ lim sup |x|→∞ f (z,x) x ≤ λ2 − λ1 uniformly for a.a. z ∈ Z; (v) suppose that ‖xn‖ → ∞, (i) if ‖x 0 n ‖ ‖xn‖ → 1, xn = x 0 n + x̂n with x 0 n ∈ E(λ1) = H1, x̂n ∈ ̂H2, then there exist γ1 > 0 and n1 ≥ 1 such that ∫ Z f(z,xn(z))x 0 n(z)dz ≥ γ1 for all n ≥ n1; (ii) if ‖x 0 n ‖ ‖xn‖ → 1, xn = x 0 n + x̂n with x 0 n ∈ E(λ2), x̂n ∈ W = E(λ2) ⊥, then there exist γ2 > 0 and n ≥ 1 such that ∫ Z (f(z,xn(z)) − (λ2 − λ1)xn(z))x 0 n(z)dz ≤ −γ2 for all n ≥ n2; 26 Ravi P. Agarwal et al. CUBO 10, 3 (2008) (vi) if F(z,x) = ∫ x 0 f(z,s)ds, then there exist η ∈ L∞(Z) and δ > 0, such that η(z) ≤ 0 a.e. on Z with strict inequality on a set of positive measure and F(z,x) ≤ η(z) 2 x 2 for a.a. z ∈ Z and all |x| ≤ δ. Remark 3.1. Hypothesis H(f)(iv) implies that asymptotically at ±∞, we have double resonance. Hypothesis H(f)(v) is a generalized LL-condition. Similar conditions can be found in the works of Landesman-Robinson-Rumbos [8], Robinson [13],[14] and Su [16]. Consider a C2-function x → F(x) which in a neighborhood of zero equals x4 − sinx2, while for |x| large (say |x| ≥ M > 0), F(x) = c|x| 3 2 ,c > 0. If f(x) = F ′(x), then f ∈ C1(R) satisfies hypothesis H(f) above. To verify the generalized LL-condition in hypothesis H(f)(v), we use Lemma 2.1 of Su-Tang [17]. Similarly we can consider if near the origin, F(x) = 1 2 x2 − tan−1x2 or F(x) = −cosx2. This second case is interesting because then f(x) = 2xsinx2 and f′(x) = 2sinx2+4x2cosx2. So f′(0) = 0. This example, which is covered by hypotheses H(f), illustrates that our framework of analysis incorporates also problems with resonance at zero with respect to λ1 > 0 (double-double resonance). This is not possible in the setting of Landesman-Robinson-Rumbos [8] (see Theorem 2 in [8]). Also such a potential function is not covered by the multiplicity results of Robinson [14] (theorem 2) and Su [16] (Theorem 2). We consider the Euler functional for problem (1.1), ϕ : H1 0 (Z) → R defined by ϕ(x) = 1 2 ‖Dx‖2 2 − λ1 2 ‖x‖2 2 − ∫ Z F(z,x(z))dz for all x ∈ H1 0 (Z). It is well-known that ϕ ∈ C2(H1 0 (Z)) and if by 〈·, ·〉 we denote the duality brackets for the pair (H1 0 (Z),H−1(Z) = H1 0 (Z)∗), we have 〈ϕ′(x),y〉 = ∫ Z (Dx,Dy)RNdz − λ1 ∫ Z xydz − ∫ Z f(z,x(z))y(z)dz and ϕ′′(x)(u,v) = ∫ Z (Du,Dv)RNdz − λ1 ∫ Z uvdz − ∫ Z f ′(z,x(z))u(z)v(z)dz for all x,y,u,v ∈ H1 0 (Z). Proposition 3.2. If hypotheses H(f) hold then ϕ satisfies the C-condition. Proof. Let {xn}n≥1 ⊆ H 1 0 (Z) be a sequence such that (1 + ‖xn‖)ϕ ′(xn) → 0 as n → ∞. We will show that {xn}n≥1 ⊆ H 1 0 (Z) is bounded. We argue indirectly. Suppose that {xn}n≥1 ⊆ H 1 0 (Z) is unbounded. We may assume that ‖xn‖ → ∞. Let yn = xn ‖xn‖ , n ≥ 1. By passing to a suitable subsequence if necessary, we may assume that yn w → y in H1 0 (Z), yn → y in L 2(Z), yn(z) → y(z) a.e. on Z and |yn(z)| ≤ k(z) a.e. on Z, for all n ≥ 1, with k ∈ L 2(Z)+. CUBO 10, 3 (2008) Multiple Solutions for Doubly ... 27 Hypotheses H(f)(iii) and (iv), imply that |f(z,x)| ≤ a(z) + c|x| for a.a. z ∈ Z, all x ∈ R, with a ∈ L∞(Z)+,c > 0, ⇒ |f(z,xn(z))| ‖xn‖ ≤ a(z) ‖xn‖ + c|yn(z)| for a.a. z ∈ Z, all n ≥ 1, (3.1) ⇒ { f(·,xn(·)) ‖xn‖ } n≥1 ⊆ L2(Z) is bounded. Thus we may assume that f(·,xn(·)) ‖xn‖ w → h in L2(Z) as n → ∞. For every ε > 0 and n ≥ 1, we set C+ε,n = {z ∈ Z : xn(z) > 0, −ε ≤ f(z,xn(z)) xn(z) ≤ λ2 − λ1 + ε} and C−ε,n = {z ∈ Z : xn(z) < 0, −ε ≤ f(z,xn(z)) xn(z) ≤ λ2 − λ1 + ε} Note that xn(z) → +∞ a.e. on {y > 0} and xn(z) → −∞ a.e. on {y < 0}. Then by virtue of hypothesis H(f)(iv), we have χ C + ε,n (z) → χ{y>0}(z) and χC−ε,n (z) → χ{y<0}(z) a.e. on Z. Using the dominated convergent theorem, we see that ‖(1 − χ C + ε,n ) f(·,xn(·)) ‖xn‖ ‖L2({y>0}) → 0 and ‖(1 − χ C − ε,n ) f(·,xn(·)) ‖xn‖ ‖L2({y<0}) → 0 as n → ∞. It follows that χ C + ε,n (·) f(·,xn(·)) ‖xn‖ w → h in L({y > 0}) and χ C − ε,n (·) f(·,xn(·)) ‖xn‖ w → h in L({y < 0}) as n → ∞. From the definitions of the sets C+ε,n and C − ε,n we have −εyn(z) ≤ f(z,xn(z)) ‖xn‖ = f(z,xn(z)) xn(z) yn(z) ≤ (λ2 − λ1 + ε)yn(z) a.e. on C + ε,n and −εyn(z) ≥ f(z,xn(z)) ‖xn‖ = f(z,xn(z)) xn(z) yn(z) ≥ (λ2 − λ1 + ε)yn(z)a.e. on C − ε,n. 28 Ravi P. Agarwal et al. CUBO 10, 3 (2008) Passing to the limit as n → ∞, using Mazur’s lemma and recalling that ε > 0 is arbitrary, we obtain 0 ≤ h(z) ≤ (λ2 − λ1)y(z) a.e. on {y > 0} (3.2) and 0 ≥ h(z) ≥ (λ2 − λ1)y(z) a.e. on {y < 0}. (3.3) Moreover, from (3.1) it is clear that h(z) = 0 a.e. on {y = 0}. (3.4) From (3.2), (3.3) and (3.4), it follows that h(z) = g(z)y(z) a.e. on Z, where g ∈ L∞(Z)+, 0 ≤ g(z) ≤ λ2 − λ1 a.e. on Z. Recall that by 〈·, ·〉 we denote the duality brackets for the pair (H1 0 (Z),H−1(Z)). Let A ∈ L(H1 0 (Z),H−1(Z)) be defined by 〈A(x),y〉 = ∫ Z (Dx,Dy)RNdz for all x,y ∈ H 1 0 (Z). Also let N : L2(Z) → L2(Z) be the Nemitskii operator corresponding to the nonlinearity f(z,x), i.e. N(x)(·) = f(·,x(·)) for all x ∈ L2(Z). Because of (3.1), by Krasnoselskii’s theorem, we know that N is continuous and bounded. Moreover, exploiting the compact embedding of H1 0 (Z) into L2(Z), we see that N is completely continuous (hence compact too) as a map from H1 0 (Z) into L2(Z) (see for example Gasinski- Papageorgiou [5], pp.267-268). We have ϕ′(xn) = A(xn) − λ1xn − N(xn) for all n ≥ 1. From the choice of the sequence {xn}n≥1 ⊆ H 1 0 (Z), we know that |〈ϕ′(xn),v〉| ≤ εn for all v ∈ H 1 0 (Z) with εn ↓ 0, ⇒ ∣ ∣ ∣ ∣ 〈A(yn) − λ1yn − N(xn) ‖xn‖ ,v〉 ∣ ∣ ∣ ∣ ≤ εn ‖xn‖ for all n ≥ 1. (3.5) Let v = yn − y ∈ H 1 0 (Z), n ≥ 1. Then ∣ ∣ ∣ ∣ 〈A(yn),yn − y〉 − λ1 ∫ Z yn(yn − y)dz − ∫ Z N(xn) ‖xn‖ (yn − y)dz ∣ ∣ ∣ ∣ ≤ εn ‖xn‖ for all n ≥ 1. (3.6) CUBO 10, 3 (2008) Multiple Solutions for Doubly ... 29 Evidently ∫ Z yn(yn − y)dz → 0 and ∫ Z N(xn) ‖xn‖ (yn − y) → 0 as n → ∞. So from (3.6), we infer that 〈A(yn),yn − y〉 → 0. (3.7) We have A(yn) w → A(y) in H−1(Z). From (3.7) it follows that 〈A(yn),yn〉 → 〈A(y),y〉, ⇒‖Dyn‖2 → ‖Dy‖2. Also Dyn w → Dy in L2(Z, RN). Since the Hilbert space L2(Z, RN) has the Kadec-Klee property, we deduce that Dyn → Dy in L 2(Z, RN) ⇒ yn → y in H 1 0 (Z), i.e. ‖y‖ = 1, y 6= 0. We return to (3.5) and we pass to the limit as n → ∞. We obtain 〈A(y) − λ1y − gy,v〉 = 0 for all v ∈ H 1 0 (Z), ⇒A(y) = (λ1 + g)y in H −1(Z), ⇒ − △y(z) = (λ1 + g(z))y(z) a.e. on Z, y|∂Z = 0. (3.8) We distinguish three cases for problem (3.8) depending on where the function g ∈ L∞(Z)+ stands in the interval [0,λ2 − λ1]. Case 1: g(z) = 0 a.e. on Z. Then from (3.8), we have − △y(z) = λ1y(z) a.e. on Z, y|∂Z = 0, ⇒y ∈ E(λ1), y 6= 0. We consider the orthogonal direct sum decomposition H1 0 (Z) = E(λ1) ⊕ ̂H2, ̂H2 = E(λ1) ⊥. Then for every n ≥ 1, we have xn = x 0 n + x̂n and x 0 n ∈ E(λ1), x̂n ∈ ̂H2. We have yn = y 0 n + ŷn, with y0n = x0n ‖xn‖ ∈ E(λ1) and ŷn = x̂n ‖xn‖ ∈ ̂H2 for all n ≥ 1. 30 Ravi P. Agarwal et al. CUBO 10, 3 (2008) Since y ∈ E(λ1), ‖y‖ = 1, we have ‖x0n‖ ‖xn‖ → 1 as n → ∞. Recall that ∣ ∣ ∣ ∣ 〈A(xn),v〉 − λ1 ∫ Z xnvdz − ∫ Z N(xn)vdz ∣ ∣ ∣ ∣ ≤ εn for all v ∈ H 1 0 (Z). Let v = x0n ∈ H 1 0 (Z). We have ∣ ∣ ∣ ∣ ‖Dx0n‖ 2 2 − λ1‖x 0 n‖ 2 2 − ∫ Z f(z,xn(z))x 0 n(z)dz ∣ ∣ ∣ ∣ ≤ εn, ⇒ ∫ Z f(z,xn(z))x 0 n(z)dz ≤ εn (see (2.2)) for all n ≥ 1. (3.9) But by virtue of hypothesis H(f)(v) 0 < γ1 ≤ ∫ Z f(z,x(z))x0n(z)dz for all n ≥ n1. (3.10) Comparing (3.9) and (3.10), we reach a contradiction. Case 2: g(z) = λ2 − λ1 a.e. on Z. In this case, from (3.8) we have − △y(z) = λ2y(z) a.e. on Z, y|∂Z = 0, ⇒y ∈ E(λ2), y 6= 0. Now we consider the orthogonal direct sum decomposition H1 0 (Z) = E(λ2) ⊕ W, with W = E(λ2) ⊥. Then xn = x 0 n + x̂n with x 0 n ∈ E(λ2), x̂n ∈ W, n ≥ 1. Since y ∈ E(λ2), ‖y‖ = 1, we have ‖x0n‖ ‖xn‖ → 1 as n → ∞. (3.11) We have |〈A(xn),v〉 − λ1 ∫ Z xnvdz − ∫ Z f(z,xn(z))v(z)dz| ≤ εn for all v ∈ H1 0 (Z), with εn ↓ 0. CUBO 10, 3 (2008) Multiple Solutions for Doubly ... 31 Let v = x0n. Then ∣ ∣ ∣ ∣ ‖Dx0n‖ 2 2 − λ1‖x 0 n‖ 2 2 − ∫ Z f(z,xn(z))x 0 n(z)dz ∣ ∣ ∣ ∣ ≤ εn, ⇒ ∣ ∣ ∣ ∣ ‖Dx0n‖ 2 2 − λ2‖x 0 n‖ 2 2 − ∫ Z (f(z,xn(z)) − (λ2 − λ1)xn(z))x 0 n(z)dz ∣ ∣ ∣ ∣ ≤ εn, ⇒ ∫ Z (f(z,xn(z)) − (λ2 − λ1)xn(z))x 0 n(z)dz ≥ −εn (see (2.3) and (2.4)). (3.12) But again hypothesis H(f)(v) implies 0 > −γ2 ≥ ∫ Z (f(z,xn(z)) − (λ2 − λ1)xn(z))x 0 n(z)dz for all n ≥ n2. (3.13) Comparing (3.12) and (3.13) we reach a contradiction. Case 3: 0 ≤ g(z) ≤ λ2 − λ1 a.e. on Z with g 6= 0, g 6= λ2 − λ1. Note that λ1 ≤ λ1 + g(z) ≤ λ2 a.e. on Z and the inequalities are strict on sets (in general different) of positive measure. Exploiting the strict monotonicity property of the eigenvalues of (−△,H1 0 (Z), m) on the weight function m (see Section 2), we have ̂λ1(λ1 + g) < ̂λ1(λ1) = 1 and ̂λ2(λ1 + g) > ̂λ2(λ2) = 1. Combining this with (2.2), we see that y = 0, a contradiction to the fact that ‖y‖ = 1. So in all these cases we have reached a contradiction. This means that {xn}n≥1 is bounded and so we may assume (at least for a subsequence) that xn w → x in H1 0 (Z), xn → x in L 2(Z), xn(z) → x(z) a.e. on Z and |xn(z)| ≤ k(z) a.e. on Z for all n ≥ 1, with k ∈ L 2(Z)+. Recall that ∣ ∣ ∣ ∣ 〈A(xn),xn − x〉 − λ1 ∫ Z xn(xn − x)dz − ∫ Z f(z,xn(z))(xn − x)dz ∣ ∣ ∣ ∣ ≤ εn. Since ∫ Z xn(xn − x)dz → 0 and ∫ Z f(z,xn(z))(xn − x)(z)dz → 0 as n → ∞, we obtain 〈A(xn),xn − x〉 → 0 as n → ∞. We know that A(xn) w → A(x) in H−1(Z). So as before, via the Kadec-Klee property of H1 0 (Z), we conclude that xn → x in H 1 0 (Z). This proves that ϕ satisfies the C-condition. 32 Ravi P. Agarwal et al. CUBO 10, 3 (2008) In the sequel, we will need the following simple lemma: Lemma 3.3. If β ∈ L∞(Z), β(z) ≤ λ1 a.e. on Z and the inequality is strict on a set of positive measure, then there exists ξ1 > 0 such that ψ(x) = ‖Dx‖2 2 − ∫ Z β(z)x(z)2dz ≥ ξ1‖Dx‖ 2 2 for all x ∈ H1 0 (Z). Proof. From (2.2), we see that ψ ≥ 0. Suppose that the lemma is not true. Exploiting the 2- homogeneity of ψ, we can find {xn}n≥1 ⊆ H 1 0 (Z) such that ‖Dxn‖2 = 1 for all n ≥ 1 and ψ(xn) ↓ 0 as n → ∞. By Poincare’s inequality {xn}n≥1 ⊆ H 1 0 (Z) is bounded. So we may assume that xn w → x in H1 0 (Z), xn → x in L 2(Z), xn(z) → x(z) a.e. on Z and |xn(z)| ≤ k(z) a.e. on Z for all n ≥ 1, with k ∈ L 2(Z)+. From the weak lower semicontinuity of the norm functional, we have ‖Dx‖2 2 ≤ lim inf n→∞ ‖Dxn‖ 2 2 , while from the dominated convergence theorem, we have ∫ Z β(z)xn(z) 2dz → ∫ Z β(z)x(z)2dz as n → ∞. Hence ψ(x) ≤ lim inf n→∞ ψ(xn) = 0, (3.14) ⇒‖Dx‖2 2 ≤ ∫ Z β(z)x(z)2dz ≤ λ1‖x‖ 2 2 , ⇒‖Dx‖2 2 = λ1‖x‖ 2 2 (see (2.2)), ⇒x = 0 or x = ±u1 with u1 ∈ E(λ1). If x = 0, then ‖Dxn‖2 → 0, a contradiction to the fact that ‖Dxn‖2 = 1 for all n ≥ 1. If x = ±u1, then |x(z)| > 0 for all z ∈ Z and so from the first inequality in (3.9) and the hypothesis on β, we have ‖Dx‖2 2 < λ1‖x‖ 2 2 , a contradiction to (2.2). Using this lemma, we prove the following proposition. Proposition 3.4. If hypotheses H(f) hold, then the origin is a local minimizer of ϕ. CUBO 10, 3 (2008) Multiple Solutions for Doubly ... 33 Proof. Let δ > 0 be as in hypothesis H(f)(vi) and consider the closed ball B C 1 0 δ = {x ∈ C 1 0 (Z) : ‖x‖ C1 0 (Z) ≤ δ}. By virtue of hypothesis H(f)(vi), for every x ∈ B C 1 0 δ , we have F(z,x(z)) ≤ η(z) 2 x(z)2 for a.a. z ∈ Z. (3.15) Thus, for all x ∈ B C 1 0 δ , we have ϕ(x) = 1 2 ‖Dx‖2 2 − λ1 2 ‖x‖2 2 − ∫ Z F(z,x(z))dz ≥ 1 2 ‖Dx‖2 2 − 1 2 ∫ Z (λ1 + η(z))x(z) 2 dz (see (3.15)) ≥ ξ1 2 ‖Dx‖2 2 (apply Lemma 3.3 with g = λ1 + η ∈ L ∞(Z)) ≥ 0 = ϕ(0). (3.16) From (3.16) we see that x = 0 is a local C1 0 (Z)-minimizer of ϕ. But then from Brezis-Nirenberg [3], we have that x = 0 is a local H1 0 (Z)-minimizer of ϕ. We may assume that the origin is an isolated critical point of ϕ or otherwise we have a sequence of nontrivial solutions for problems (1.1). Then from the description of the critical groups at an isolated local minimizer (see Chang [4], p.33 and Mawhin-Willem [10], p.175), we have: Corollary 3.5. If hypotheses H(f) hold, then Ck(ϕ, 0) = δk,0Z for all k ≥ 0. In the next proposition, we produce the first nontrivial solution for problem (1.1). Proposition 3.6. If hypotheses H(f) hold then problem (1.1) has a nontrivial solution x0 ∈ C 1 0 (Z) and x0 is a critical point of ϕ of mountain pass-type. Proof. Recall that x = 0 is an isolated local minimum of ϕ. So we can find ρ0 > 0 such that ϕ|∂Bρ0 > 0. (3.17) Let u1 ∈ C 1 0 (Z) be the L2(Z)-normalized principal eigenfunction of (−△, H1 0 (Z)) and let t > 0. For 0 < β0 < t, via the mean value theorem, we have F(z,tu1(z)) = F(z,β0u1(z)) + ∫ t β0 f(z,µu1(z))u1(z)dµ a.e. on Z. (3.18) Integrating over Z and using Fubini’s theorem, we obtain ∫ Z F(z,tu1(z))dz = ∫ Z F(z,β0u1(z))dz + ∫ t β0 1 µ ∫ Z f(z,µu1(z))µu1(z)dzdµ. 34 Ravi P. Agarwal et al. CUBO 10, 3 (2008) Choosing β0 > 0 large, because of hypothesis H(f)(v), we have ∫ Z f(z,µu1(z))µu1(z)dz ≥ γ1 > 0 for all µ ∈ [β0, t]. (3.19) From (3.18) and (3.19), we obtain ∫ Z F(z,tu1(z))dz ≥ ∫ Z F(z,β0u1(z))dz + ∫ t β0 γ1 µ dµ for β0 > 0 large, ⇒ ∫ Z F(z,tu1(z))dz ≥ ∫ Z F(z,β0u1(z))dz + γ1(lnt − lnβ0). (3.20) So from (3.20) it follows that − ∫ Z F(z,tu1(z))dz → −∞ as t → +∞. Hence ϕ(tu1) = − ∫ Z F(z,tu1(z))dz → −∞ as t → +∞ (see (2.2)). Therefore for t > 0 large, we have ϕ(tu1) < ϕ(0) = 0 < inf ∂Bρ0 ϕ = c. This fact together with Proposition 3.2, permit the use of the mountain pass theorem (see Bartolo- Benci-Fortunato [1]), which gives x0 ∈ H 1 0 (Z) such that ϕ ′(x0) = 0 and ϕ(0) = 0 < c ≤ ϕ(x0). (3.21) From (3.21), we deduce that x0 6= 0. From the equality in (3.21), we have A(x0) = λ1x0 + N(x0), ⇒ − △x0(z) = λ1x0(z) + f(z,x0(z)) a.e. on Z, x0|∂Z = 0. Thus x0 ∈ H 1 0 (Z) is a nontrivial solution of problem (1.1) and from regularity theory (see for example Gasinski-Papageorgiou [5], pp.737-738), we have x0 ∈ C 1 0 (Z). Let d = ϕ(x0) and assume without loss of generality that Kd is discrete (otherwise we have a whole sequence of nontrivial solutions for problem (1.1)). Then invoking Theorem 1 of Hofer [6], we can say that x0 ∈ C 1 0 (Z) is a critical point of ϕ which is of mountain pass-type. From the description of the critical groups for a critical point of a mountain pass-type (see Chang [4], p.91 and Mawhin-Willem [10], pp.195-196), we have: Corollary 3.7. If hypotheses H(f) hold and x0 ∈ C 1 0 (Z) is the nontrivial solution of (1.1) obtained in Proposition 3.6, then Ck(ϕ,x0) = δk,1Z for all k ≥ 0. CUBO 10, 3 (2008) Multiple Solutions for Doubly ... 35 In the next proposition, we determine the critical groups of ϕ at infinity. To do this, we will need the following slight generalization of Lemma 2.4 of Perera-Schechter [12]. Lemma 3.8. If H is a Hilbert space, {ϕt}t∈[0,1] is a one-parameter family of C 1(H)-functions such that ϕ′t and ∂tϕt are both locally Lipschitz in u ∈ H and there exists R > 0 such that inf[(1 + ‖u‖)‖ϕ′t(u)‖ : t ∈ [0, 1],‖u‖ > R] > 0 and inf[ϕt(u) : t ∈ [0, 1],‖u‖ ≤ R] > −∞, then Ck(ϕ0,∞) = Ck(ϕ1,∞) for all k ≥ 0. Proof. Let ξ < inf[ϕt(u) : t ∈ [0, 1],‖u‖ ≤ R]. Let h(t; u) (t ∈ [0, 1],u ∈ ϕ ξ 0 ) be the flow generated by the Cauchy problem · h(t) = − ∂tϕt(h(t)) ‖ϕ′t(h(t))‖ 2 ϕ′t(h(t)) a.e. on R+, h(0) = u. We have d dt ϕt(h(t)) = 〈ϕ ′ t(h(t)), · h(t)〉 + ∂tϕt(h(t)) = 0 for all t ≥ 0, ⇒ϕt(h(t)) = ϕ0(u) for all t ≥ 0. Since u ∈ ϕa 0 , we have ϕt(h(t)) ≤ ξ and so ‖h(t)‖ > R for all t ≥ 0. This then by virtue of the hypothesis of the lemma, implies that this flow exists for all t ≥ 0 (see Bartolo-Benci-Fortunato [1]). It can be reversed, if we replace ϕt with ϕ1−t. Therefore h(1) is a homeomorphism of ϕ ξ 0 and ϕ ξ 1 and so Ck(ϕ0,∞) = Hk(H,ϕ ξ 0 ) ∼= Hk(H,ϕ ξ 1 ) = Ck(ϕ1,∞). Proposition 3.9. If hypotheses H(f)(i) → (v) hold, then Ck(ϕ,∞) = δk,1Z for all k ≥ 0. Proof. Let 0 < σ < λ2 − λ1 and consider the following one-parameter C 2-functions on the Hilbert space H1 0 (Z) : ϕt(x) = 1 2 ‖Dx‖2 2 − λ1 + σ 2 ‖x‖2 2 − t ∫ Z (F(z,x(z)) − σx(z))dz for all x ∈ H1 0 (Z). We claim that we can find R > 0 such that inf[(1 + ‖u‖)‖ϕ′t(u)‖ : t ∈ [0, 1], ‖u‖ > R] > 0. (3.22) 36 Ravi P. Agarwal et al. CUBO 10, 3 (2008) Suppose that this is not possible. Then we can find tn → t ∈ [0, 1] and ‖un‖ → ∞ such that ϕ′tn (un) → 0 in H −1(Z) as n → ∞. Let yn = un ‖un‖ , n ≥ 1. By passing to a suitable subsequence if necessary, we may assume that yn w → y in H−1(Z), yn → y in L 2(Z), yn(z) → y(z) a.e. on Z, and |yn(z)| ≤ k(z) for a.a. z ∈ Z, all n ≥ 1, with k ∈ L 2(Z). We have ∣ ∣ ∣ ∣ 〈 ϕ′tn (un) ‖un‖ ,v〉 ∣ ∣ ∣ ∣ ≤ εn for all v ∈ H ( 0 Z), with εn ↓ 0 (see (3.22)) ⇒ ∣ ∣ ∣ ∣ 〈A(yn),v〉 − (λ1 + σ) ∫ Z ynvdz − tn ∫ Z N(un) ‖un‖ vdz + tnσ ∫ Z ynvdz ∣ ∣ ∣ ∣ ≤ εn (3.23) From the proof of Proposition 3.2, we know that N(un) ‖un‖ w → h = gy in L2(Z) with g ∈ L∞(Z)+, 0 ≤ g(z) ≤ λ2 − λ1 a.e. on Z. Moreover, arguing as in that proof, we can also show that yn → y in H 1 0 (Z), hence ‖y‖ = 1, i.e. y 6= 0. So, if we pass to the limit as n → ∞ in (3.23), we obtain 〈A(y),v〉 = (λ1 + σ) ∫ Z yvdz + t ∫ Z (g + σ)yvdz for all v ∈ H1 0 (Z), ⇒A(y) = (λ1 + (1 − t)σ + tg)y. (3.24) As in the proof of Proposition 3.2, we consider three distinct possibilities for the weight function m = λ1 + (1 − t)σ + tg ∈ L ∞(Z)+. Case 1: t = 1 and g = 0. From (3.24), we have A(y) = λ1(y), ⇒ − △y(z) = λ1y(z) a.e. on Z, y|∂Z = 0, ⇒y ∈ E(λ1), y 6= 0. So, if un = u 0 n + ûn with u 0 n ∈ E(λ1), ûn ∈ ̂H2 = E(λ1) ⊥, n ≥ 1, then ‖u0n‖ ‖un‖ → 1 as n → ∞. (3.25) CUBO 10, 3 (2008) Multiple Solutions for Doubly ... 37 We have ∣ ∣ ∣ ∣ 〈A(un),v〉 − (λ1 + σ) ∫ Z unvdz − tn ∫ Z N(un)vdz + tnσ ∫ Z unvdz ∣ ∣ ∣ ∣ ≤ εn for all v ∈ H1 0 (Z). Let v = u0n ∈ E(λ1). We obtain ∣ ∣ ∣ ∣ ‖Du0n‖ 2 2 − (λ1 + σ)‖u 0 n‖ 2 2 − tn ∫ Z f(z,un(z))u 0 n(z)dz + tnσ‖u 0 n‖ 2 2 ∣ ∣ ∣ ∣ ≤ εn. (3.26) Since u0n ∈ E(λ1), we know that ‖Du 0 n‖ 2 2 = λ1‖u 0 n‖ 2 2 . Also because of (3.25) and hypothesis H(f)(v), we have ∫ Z f(z,un(z))u 0 n(z)dz ≥ γ1 for all n ≥ n1. Then from (3.26), we obtain (1 − tn)σ‖u 0 n‖ 2 2 + tnγ1 ≤ εn for all n ≥ n1, ⇒tnγ1 ≤ εn for all n ≥ n1. Since tn → t = 1 and εn ↓ 0, in the limit as n → ∞, we obtain 0 < γ1 ≤ 0, a contradiction. Case 2: t = 1 and g = λ2 − λ1. From (3.24), we have A(y) = λ2y, ⇒ − △y(z) = λ2y(z) a.e. on Z, y|∂Z = 0, ⇒y ∈ E(λ2), y 6= 0. Now we write un = u 0 n + ûn with u 0 n ∈ E(λ2) and ûn ∈ W = E(λ2) ⊥. We have ‖u0n‖ ‖un‖ → 1 as n → ∞. (3.27) Recall that ∣ ∣ ∣ ∣ 〈A(un),v〉 − (λ1 + σ) ∫ Z unvdz − tn ∫ Z N(un)vdz + tnσ ∫ Z unvdz ∣ ∣ ∣ ∣ ≤ εn for all v ∈ H1 0 (Z). 38 Ravi P. Agarwal et al. CUBO 10, 3 (2008) Let v = u0n ∈ E(λ2). We obtain ∣ ∣ ∣ ∣ ‖Du0n‖ 2 2 − tnλ2‖u 0 n‖ 2 2 − (1 − tn)(λ1 + σ)‖u 0 n‖ 2 2 − tn ∫ Z (f(z,un(z)) − (λ2 − λ1)un(z))u 0 n(z)dz ∣ ∣ ∣ ∣ ≤ εn. (3.28) Note that tnλ2 + (1 − tn)(λ1 + σ) < λ2 and so 0 < ‖Du0n‖ 2 2 − (tnλ2 + (1 − tn)(λ1 + σ))‖u 0 n‖ 2 2 . (3.29) In addition because of (3.27) and hypothesis H(f)(v), we have ∫ Z (f(z,un(z)) − (λ2 − λ1)un(z))u 0 n(z)dz ≤ −γ2 < 0 for all n ≥ n2. (3.30) Using (3.29) and (3.30) in (3.28), we obtain tnγ2 ≤ εn for all n ≥ n2. Passing to the limit as n → ∞ and recalling that tn → 1 and ε ↓ 0, we get 0 < γ2 ≤ 0, again a contradiction. Case 3: t 6= 1 or 0 ≤ g(z) ≤ λ2 − λ1 a.e. on Z with g 6= 0 and g 6= λ2 − λ1. From (3.24), we have A(y) = (λ1 + ̂ξ)y, y 6= 0 with ̂ξ = (1 − t)σ + tg ∈ L ∞(Z)+, ⇒ − △y(z) = (λ1 + ̂ξ(z))y(z) a.e. on Z, y|∂Z = 0. (3.31) Note that since t 6= 1 or (g 6= 0 and g 6= λ2 − λ1), we have λ1 ≤ λ1 + ̂ξ(z) ≤ λ2 a.e. on Z, λ1 6= λ1 + ̂ξ and λ2 6= λ1 + ̂ξ. Hence from the strict monotonicity of the eigenvalues on the weight function, we infer that ̂λ1(λ1 + ̂ξ) < ̂λ1(λ1) = 1 and ̂λ2(λ1 + ̂ξ). (3.32) Using (3.32) in (3.31), we infer that y = 0, a contradiction to the fact that ‖y‖ = 1. So in all three cases we have reached a contradiction and this means that there exists R > 0 for which (3.22) is valid. Also it is clear, that due to hypotheses H(f)(iii), (iv), we have inf[ϕt(u) : t ∈ [0, 1], ‖u‖ ≤ R] > −∞. CUBO 10, 3 (2008) Multiple Solutions for Doubly ... 39 So we can apply Lemma 3.8 and have that Ck(ϕ0,∞) = Ck(ϕ,∞) for all k ≥ 0. (3.33) Note that ϕ0(x) = 1 2 ‖Dx‖2 2 − λ1 + σ 2 ‖x‖2 2 and ϕ1(x) = ϕ(x) for all x ∈ H 1 0 (Z). Since 0 < σ < λ2 − λ1, the only critical point of ϕ0 is u = 0. Hence Ck(ϕ0,∞) = Ck(ϕ, 0) for all k ≥ 0. (3.34) Moreover, from Proposition 2.3 of Su [16], we have Ck(ϕ0, 0) = δk,1Z for all k ≥ 0. (3.35) From (3.33), (3.34) and (3.35), we conclude that Ck(ϕ,∞) = δk,1Z for all k ≥ 0. Now we are ready for the first multiplicity theorem. Theorem 3.10. If hypotheses H(f) hold, then problem (1.1) has at least two nontrivial solutions x0,v0 ∈ C 1 0 (Z). Proof. One nontrivial solution x0 ∈ C 1 0 (Z), exists by virtue of Proposition 3.6. Suppose that {0,x0} are the only critical points of ϕ. Then using Corollaries 3.5, 3.7, 3.9 and the Poincare-Hopf formula, we have (−1)0 + (−1)1 = (−1)1, a contradiction. So there exists a third critical point v0 6= x0, v0 6= 0. Evidently v0 is a solution of (1.1) and by regularity theory, we have v0 ∈ C 1 0 (Z). We have another multiplicity result by modifying hypothesis H(f)(vi). So the new hypotheses on the nonlinearity f(z,x) are the following: H(f)′: f : Z × R → R is a function such that f(z, 0) = 0 a.e. on Z, hypotheses H(f)′(i) → (v) are the same as hypotheses H(f)(i) → (v) respectively and (vi) there exist m ≥ 2 and δ > 0 such that λm − λ1 ≤ f(z,x) x ≤ λm+1 − λ1 for a.a. z ∈ Z and all 0 < |x| ≤ δ. 40 Ravi P. Agarwal et al. CUBO 10, 3 (2008) Remark 3.11. Hypotheses H(f)′(iv) and (vi) imply that we can have double resonance both at infinity and at zero. A double-double resonance situation. Theorem 3.12. If hypotheses H(f)′ hold, then problem (1.1) has at least two nontrivial solutions x0,v0 ∈ C 1 0 (Z). Proof. Because of hypothesis H(f)′(vi) and Proposition 1.1 of Li-Perera-Su [9], we have Ck(ϕ, 0) = δk,dZ, (3.36) where d =sum of multiplicities of {λk} m k=1 = dimHm ≥ 2, since m ≥ 2. Also from Proposition 3.9, we know that Ck(ϕ,∞) = δk,1Z. (3.37) So there exists a critical point x0 of ϕ such that C1(ϕ,x0) 6= 0. (3.38) Comparing this with (3.36), we infer that x0 6= 0. Moreover, due to (3.38) x0 is of mountain pass type and so C1(ϕ,x0) = δk,1Z. (3.39) If {0,x0} are the only critical points of ϕ, then from (3.36), (3.37) and (3.39) and the Poincare-Hopf formula, we have (−1)d + (−1)1 = (−1)1, ⇒(−1)d = 0, a contradiction. So there exists a second nontrivial critical point v0 of ϕ. Evidently x0,v0 ∈ H 1 0 (Z) are nontrivial solutions of problem (1.1). From regularity theory, we conclude that x0,v0 ∈ C 1 0 (Z). Remark 3.13. Theorem 3.12 above partially extends Theorem 3 of Robinson [14] and also Theorem 2 of Su [16]. Received: February 2008. Revised: April 2008. References [1] P. Bartolo, V. Benci and D. Fortunato, Abstract critical point theorems to some nonlinear problems with strong resonance at infinity, Nonlin. Anal., 7 (1983), 981–1012. [2] H. Berestycki and D. De Figueiredo, Double resonance and semilinear elliptic problems, Comm. PDE, 6 (1981), 91–120. 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[17] J-B.Su and C.L. Tang, Multiplicity results for semilinear eliptic equations with resonance at higher eigenvalues, Nonlin. Anal., 44 (2001), 311–321. N03