A Landesman Lazer-type condition for asymptotically linear second-order equations with a singularity

Transcription

1 Proceedings of the Royal Society of Edinburgh, 142A, , 212 A Landesman Lazer-type condition for asymptotically linear second-order equations with a singularity Alessandro Fonda Dipartimento di Matematica e Geoscienze, Università di rieste, Piazzale Europa 1, rieste, Italy (a.fonda@units.it) Maurizio Garrione SISSA International School for Advanced Studies, Via Bonomea 265, rieste, Italy (garrione@sissa.it) (MS received 21 January 211; accepted 27 October 211) We consider the -periodic problem x + g(t, x) =, x() = x( ), x () = x ( ), where g : [,] ], + [ R exhibits a singularity of a repulsive type at the origin, and an asymptotically linear behaviour at infinity. In particular, for large x, g(t, x) is controlled from both sides by two consecutive asymptotes of the -periodic Fučik spectrum, with possible equality on one side. Using a suitable Landesman Lazer-type condition, we prove the existence of a solution. 1. Introduction Del Pino et al. [6] considered the problem of finding -periodic positive solutions of the equation x + g(t, x) =, (1.1) where g : R ], + [ R is continuous, -periodic in its first variable and has a singularity of a repulsive type at the origin. hey proved the following result. heorem 1.1 (Del Pino et al. [6, theorem 1]). Assume that there exist an integer N and two constants ˆµ, µ such that ( ) 2 ( ) 2 Nπ g(t, x) g(t, x) (N + 1)π < ˆµ lim inf lim sup µ <, (1.2) x + x x + x uniformly for every t [,]. Moreover, suppose that there exist positive constants c, c, δ and ν 1 such that c c g(t, x) xν x ν for every t [,] and every x ],δ]. (1.3) hen equation (1.1) has a -periodic solution c 212 he Royal Society of Edinburgh

2 1264 A. Fonda and M. Garrione Assumption (1.2) seems to be a kind of non-resonance condition with respect to the set {( ) 2 kπ Σ =,k N}, whose elements correspond to the heights of the asymptotes of the curves in the Fučik spectrum (see, for example, [1]). hey also coincide with the eigenvalues of the associated Dirichlet problem on the interval [,]. As a direct consequence of theorem 1.1, the equation x 1 + βx = e(t), (1.4) xν where ν 1 and e(t) is continuous and -periodic, has a -periodic solution, provided that β> satisfies β/ Σ. We mention that the study of the existence of periodic solutions for equations with a singularity, like (1.1), was first considered by Lazer and Solimini in [14], and later investigated by many authors (see, for example, [2, 4, 5, 7 9, 11, 17, 19 23]). A few lines below the statement of their main theorem, del Pino et al. raised the problem of finding sufficient conditions on e(t) in order to ensure the existence of a -periodic solution for equation (1.4) in the case when β belongs to the set Σ. As a matter of fact, this seems to be a delicate problem for the following reason. As already noted in [6] (see also [1, 16]), if ν = 3, in the phase plane there is an isochronous centre for Steen s equation (see [18]): x 1 + βx =, (1.5) x3 for any choice of β>, since all the positive solutions have minimal period equal to π/ β. his fact is crucial when dealing with the -periodic problem associated with (1.5), in the case when β Σ. Indeed, it was shown in [16] using a phase plane analysis that, due to isochronicity, if one considers a forcing made by a -periodic chain of Dirac deltas, then all the solutions of the forced equation are unbounded, both in the past and in the future. he same type of situation arises for the equation x 1 x 3 + βx = ε sin( βt), for sufficiently small ε, as proved in [4, theorem 3]. However, in the case ν = 3, defining the (π/ β)-periodic function Φ(θ) = e(t) sin( β(t + θ)) dt, it was shown in [4] that if Φ has only simple zeros, and their number in [,π/ β[ is different from 2, then (1.4) has a -periodic solution. In particular, this is true assuming that Φ has a constant sign, which, in this context, corresponds to the so-called Landesman Lazer condition. he choice of ν = 3 is crucial in [4] for exploiting the isochronicity of the homogeneous equation. he presence of the repulsive singularity has some analogies with the problem of a bouncing particle, which was treated, for example, in [3, 12, 13, 15]. For instance,

3 concerning the simple model Second-order singular equations at resonance 1265 x + βx = e(t), x(t )= = x (t + )= x (t ), describing the behaviour of a particle bouncing elastically against the barrier {x = }, with β > belonging to the set Σ, it was proved in [3] that, defining the function Φ as above, the same type of result holds. In particular, as also shown in [15, theorem 4.2], this is true if e(t) sin( β(t + θ)) dt > for every θ [,]. (1.6) he aim of this paper is to propose a kind of Landesman Lazer condition in the case of a nonlinearity with a singularity like the one considered in [6], with the strict inequalities in (1.2) possibly replaced by equalities. We will succeed in replacing the upper inequality with an equality by adding a suitable Landesman Lazer-type condition on that side. Moreover, condition (1.3) will be replaced by a one-sided control on the function g(t, x). We note that, under our assumptions, we are far from perturbing an isochronous oscillator like the one considered in [4]. Our situation seems to be more delicate than the one studied in [4], or, for the bouncing problem, in [3, 15]. For example, dealing with equation (1.4) with β> belonging to Σ, and ν 3, it is not clear whether a condition like (1.6) is sufficient for the existence of -periodic solutions. In order to explain this, we appeal to the analogy between the perturbed isochronous Steen equation and the bouncing problem. When looking for a priori estimates in the elastic bouncing problem, it is clear that the elasticity implies that all the bumps of the limit function must have the same height. Unfortunately, this behaviour is not guaranteed for the limit function obtained for the singular periodic problem with lack of isochronicity (see remark 2.5). Hence, since we are unable to predict that the height of each bump will be the same, we will need a Landesman Lazer condition that refers to each bump separately. hus, defining sin( βt) if t ψ(t) = if t [, π β ], [ ] π β,, instead of (1.6) we will ask a stronger condition, namely, e(t)ψ(t + θ)dt> for every θ [,]. (1.7) Under this assumption, we will prove that equation (1.4) has a -periodic solution. However, the possibility of replacing condition (1.7) with (1.6) in our setting remains an open problem. Note that, following the arguments in [8], under the assumptions of theorem 2.1 it is possible to prove that the radially symmetric system z = g(t, z ) z z, z RM \{},

4 1266 A. Fonda and M. Garrione has infinitely many subharmonic solutions rotating around the origin. However, we will not discuss this argument further for reasons of brevity. 2. he main result Let us consider the -periodic problem } x + g(t, x) =, x() = x( ), x () = x ( ), (2.1) where g : [,] ], + [ R is an L -Carathéodory function, namely, t g(t, x) is measurable for every x>, x g(t, x) is continuous, for almost every t [,], for every compact interval [a, b] ], + [ there exists a constant C> such that g(t, x) C for almost every (a.e.) t [,] and every x [a, b]. Henceforth, when dealing with a solution x(t) of (2.1), in the Carathéodory sense, we will implicitly assume x(t) > for every t [,]. Moreover, we will assume that all the functions defined on [,] are extended by -periodicity to the whole real line. Define, for a fixed non-negative integer N, ( ) 2 Nπ µ N =, and set Let us state our main result. sin( µ N+1 t) if t ψ(t) = [ if t heorem 2.1. Assume the following hypotheses. (H1) here exists a constant ˆµ such that uniformly for almost every t [,]. (H2) here exists a constant ˆη such that [, ], N +1 N +1, g(t, x) lim inf ˆµ >µ N, x + x g(t, x) µ N+1 x +ˆη for a.e. t [,] and every x 1. Moreover, for every θ [,], lim sup(g(t, x) µ N+1 x)ψ(t + θ)dt<. (2.2) x + ].

5 Second-order singular equations at resonance 1267 (H3) here exist a constant δ> and a continuous function f : ],δ] R such that g(t, x) f(x) for a.e. t [,] and every x ],δ], with lim x + f(x) =, δ f(x)dx =. hen, problem (2.1) has a solution. Remark 2.2. heorem 2.1 clearly generalizes theorem 1.1, since (1.2) implies (H1) and (H2), while (1.3) implies (H3), being, in this case, the left-hand side of (2.2) equal to. Let us introduce, for λ [, 1], the family of functions ( g λ (t, x) =λg(t, x) + (1 λ) 1 ) x (µ N + µ N+1 )x. (2.3) Moreover, for R>1, let { } Λ R = x C([,]) 1 <x(t) <Rfor every t [,]. R Following the arguments in [8, 2], it is sufficient to prove that there exists an R>1 such that, for any λ [, 1], every solution of the problem } x + g λ (t, x) =, x() = x( ), x () = x (2.4) ( ), belongs to Λ R. We will prove this using the following two lemmas, the first of which gives the estimate for the maxima, and the second of which focuses on minima. Lemma 2.3. here exists a constant R>1 such that, for any λ [, 1], if x(t) is a solution of (2.4), then max x(t) <R. t [, ] Proof. By contradiction, assume that there exist (λ n ) n [, 1] and a sequence (x n ) n satisfying } x n + g λn (t, x n )=, x n () = x n ( ), x n() = x n( ), (2.5) with +. Up to a subsequence, we can assume λ n λ [, 1]. We will prove the following claims. Claim 1. here exists M>1 such that min t [, ] x n(t) M for every n N.

6 1268 A. Fonda and M. Garrione Proof of claim 1. By contradiction, assume that this is not the case. Namely, there exists a subsequence, still denoted by (x n ) n, such that x n (t) + uniformly in t [,]. Integrating (2.5), we see that g λn (t, x n (t)) dt =. On the other hand, in view of (H1), there exists d> such that g λn (t, x) 1 2 ˆµx for a.e. t [,] and every x d. Since, for n large, x n (t) d for every t [,], this is clearly impossible. Claim 2. For large n, (x n (t),x n(t)) makes exactly N + 1 clockwise revolutions around the point (1, ) when t varies from to. Proof of claim 2. We will use some arguments from [8], introducing, with the same notation, the function N : ], + [ R, defined as ( ) 1/2 1 N (u, v) = u 2 + u2 + v 2. Let ɛ > be such that ɛ < (N + 1)(N + 2). (2.6) It is possible to see [8, lemma 2] that there exist ζ > and a sufficiently large R 1 >M with the following property. If τ 1 <τ 2 are such that (x n (t),x n(t)) makes exactly one rotation in the phase plane around the point (1, ) when t varies from τ 1 to τ 2, and N (x n (t),x n(t)) R 1 for every t [τ 1,τ 2 ], then N +1 ɛ τ 2 τ 1 ζ< N. (2.7) Now choose a positive integer n such that ( ) n N +1 ɛ >. (2.8) Using [8, lemma 1], there exists R 2 >R 1 such that if, for some t 1 <t 2, and N (x n (t 1 ),x n(t 1 )) = R 1, N (x n (t 2 ),x n(t 2 )) = R 2, R 1 < N (x n (t),x n(t)) <R 2 for every t ]t 1,t 2 [, then (x n (t),x n(t)) makes at least n clockwise turns in the phase plane around (1, ) when t varies from t 1 to t 2. Since +, for every large n we will have max{n (x n (t),x n(t)) t [,]} >R 2. Consequently, x n (t) being -periodic, for sufficiently large n, it will be N (x n (t), x n(t)) >R 1 for every t [,], otherwise (x n (t),x n(t)) would make at least n turns around (1, ) in the time period, which is impossible due to (2.7) and (2.8).

7 Second-order singular equations at resonance 1269 Since N (x n (t),x n(t)) >R 1 for every t [,], enlarging R 1 if necessary, as shown in the proof of [8, lemma 2], (x n (t),x n(t)) clockwise rotates at least once around (1, ) in the time. aking into account (2.6) and (2.7), then, for sufficiently large n, (x n (t),x n(t)) will make, in the phase plane, strictly more than N and strictly less than N +2 turns around (1, ) in time. Since x n (t) is -periodic, this implies that (x n (t),x n(t)) will make exactly N + 1 such turns in the time. Claim 3. Setting v n = x n, the sequence (v n ) n converges uniformly, up to a subsequence, to a continuous function v. his function is not identically zero and it is twice continuously differentiable, except on a finite subset of [,]. Moreover, v satisfies v (t) +µ N+1 v(t) = for almost every t [,]. Proof of claim 3. For every n, the -periodic function v n solves v n + g λ n (t, x n (t)) =. (2.9) Multiplying both sides of equation (2.9) by v n and integrating between and, we have v n(t) 2 g λn (t, x n (t)) dt = v n (t)dt. By (H3), we can choose δ ], 1[ such that, for every n, g λn (t, x) for a.e. t [,] and every x ], δ]. For large n, it follows that g λn (t, x n (t)) v n (t)dt = {x n δ} C { δ<x n<1} {x n 1} {x n 1} (C + µ N+1 ) +1, µ N+1 x n (t)+ˆη for a suitable constant C> coming from the Carathéodory condition. his immediately gives an estimate of the L 2 -norm of v n which is independent of n. It follows that (v n ) n is bounded in H 1 (,), so there exists v H 1 (,) such that, up to a subsequence, v n v(weakly) in H 1 (,) and v n v uniformly. Let us still denote such a subsequence by (v n ) n. Since v n = 1 for every n, we have that v = 1. Hence, v is not identically zero. Moreover, it is clear that, since v is the uniform limit of positive continuous -periodic functions, it is a non-negative continuous -periodic function. We now prove that v (t) +µ N+1 v(t) = almost everywhere in [,]. o this end, define the open set Ω + = {t R v(t) > }, dt

8 127 A. Fonda and M. Garrione which is an at most countable union of open intervals, and consider a C 1 -function ϕ with compact support K ϕ contained in Ω +. Multiplying equation (2.9) by ϕ and integrating on K ϕ, we obtain v n(t)ϕ (t)dt = K ϕ = K ϕ K ϕ g λn (t, x n (t)) ϕ(t)dt g λn (t, x n (t)) v n (t)ϕ(t)dt. (2.1) x n (t) Observe now that, in view of the continuity of v (which is strictly greater than on K ϕ ) and recalling that K ϕ is compact and strictly contained in Ω +, we have that x n (t) + uniformly for every t K ϕ. By hypotheses (H1) and (H2), for every positive integer m, there exists a sufficiently large n m such that µ N < ˆµ 1 m g λ nm (t, x nm (t)) x nm (t) µ N m, (2.11) for almost every t K ϕ. We can assume the sequence (n m ) m to be strictly increasing. Considering the subsequence ( gλnm (t, x ) n m (t)) x nm (t) it then turns out that it is bounded in L 2 (K ϕ ), so that (up to a further subsequence) it converges weakly in L 2 (K ϕ ) to a function γ(t) which, in view of (2.11), satisfies, m µ N < ˆµ γ(t) µ N+1 for a.e. t K ϕ. (2.12) Since v n v uniformly and v n vin H 1 (,), passing to the limit in (2.1) then yields v (t)ϕ (t)dt = K ϕ γ(t)v(t)ϕ(t)dt. K ϕ By the uniqueness of the weak limit, γ(t) can be uniquely extended on the whole set Ω +, hence, we have v (t)ϕ (t)dt = γ(t)v(t)ϕ(t)dt Ω + Ω + for every C 1 -function ϕ with compact support in Ω +. his implies that v(t) is a weak solution of the equation v + γ(t)v = (2.13) in Ω +. Hence, v H 2 loc (Ω+ ), and satisfies (2.13) for almost every t Ω +. Moreover, v(t) is continuously differentiable at any point t Ω +. We now aim to prove that γ(t) =µ N+1 for almost every t Ω +. Observe first that, writing the equation as a first-order system, and taking into account that, for every n, the angular velocity of (v n,v n) has to be non-positive (see [8]), the set {t [,] x n (t) =R 1 } (which is non-empty since R 1 >M) is discrete. Moreover, returning to claim 2, we know that, in the phase plane, (x n (t),x n(t)) has to perform

9 Second-order singular equations at resonance 1271 exactly N + 1 turns around the point (1, ) in time. hen, in the interval [, 2 ], we can find α n 1 <β n 1 <α n 2 <β n 2 < <α n N+1 <β n N+1 <α n 1 + such that, setting α n N+2 = αn 1 +, and x n (t) >R 1 if t ]α n r,β n r [, r =1,...,N +1, x n (t) <R 1 if t ]β n r,α n r+1[, r =1,...,N +1. Since both the sequences (βr n ) n and (αr n ) n are bounded, there exist ξr and ξ r +, with r =1,...,N + 1 such that, up to subsequences, with α n r ξ r, β n r ξ + r, r =1,...,N +1, ξ 1 ξ+ 1 ξ 2 ξ+ 2 ξ N+1 ξ+ N+1 ξ 1 +. (2.14) By the estimates in [8, lemma 2], given any ɛ>, for sufficiently large n we have β n r α n r ɛ, r =1,...,N +1. N +1 Passing to the limit, since ɛ is arbitrary, we deduce Since, by (2.14), ξ + N+1 ξ 1 ξ + r = ξ r + N +1 ξ r + ξr, r =1,...,N +1. N +1, this implies For simplicity of notation, we set ξ r = ξ r and ξ r+1 = ξ+ r, r =1,...,N +1. v n (α n r )=v n (β n r )= R 1, for r =1,...,N + 1. Recalling that we deduce that v(ξ r ) = for every r =1,...,N + 1, since v is continuous and v n v uniformly. Let us now focus on an interval [, β] such that v> on ], β[, with v() = v( β) =. Since, on ], β[,v is continuously differentiable and satisfies (2.13) almost everywhere, writing we have v(t) =ρ(t) cos(θ(t)), which yields, in view of (2.12), v (t) =ρ(t) sin(θ(t)), θ (t) = γ(t)v(t)2 + v (t) 2 v(t) 2 + v (t) 2, θ (t) µ N+1 cos 2 θ(t) + sin 2 θ(t) 1 < θ (t) ˆµ cos 2 θ(t) + sin 2 θ(t).

10 1272 A. Fonda and M. Garrione Consequently, integrating between and β, we infer, taking into account that ˆµ >µ N, N +1 β < N. However, since v(ξ r ) = for every r =1,...,N + 1, and the points ξ r are equally distributed at a distance /(N + 1) from each other, then it must be the case that β = /(N + 1). his means that, whenever v becomes positive, it is forced to remain positive for a time exactly equal to /(N + 1). Recalling that v solves v + γ(t)v = on ], β[, we now pass to generalized polar coordinates by writing v(t) = 1 µn+1 ˆρ(t) cos(ˆθ(t)), Integrating ˆθ (t) on [, β], we have π = µ N+1 β from which we deduce that v (t) =ˆρ(t) sin(ˆθ(t)). γ(t)v(t) 2 + v (t) 2 µ N+1 v(t) 2 + v (t) 2 dt µ N+1 N +1 = π, γ(t) =µ N+1 his means, in particular, that Ω + ]ξ r,ξ r+1 [, on which we have for a.e. t [, β]. is the union of some intervals of the type v(t) =c r sin( µ N+1 (t ξ r )), where, for every r, the constants c r are in [, 1], and at least one of them is equal to 1. he claim is thus proved. Conclusion of the proof of lemma 2.3. Let us focus again on an interval [, β] such that v() =v( β) =, and v>on ], β[. Note first that, since γ(t) =µ N+1 almost everywhere in Ω +, it must be the case that λ n 1. Indeed, if it were λ n λ <1, taking any interval I ], β[, for every t I it would be, for large n, g λn (t, x n (t)) x n (t) ζ <µ N+1, for a suitable constant ζ and, taking the weak limit, we would reach the contradiction γ(t) <µ N+1 for almost every t I. By the previous discussion, we know that β = /(N + 1) and that, on ], β[, v solves the Dirichlet problem v + µ N+1 v =, v() ==v( β). Explicitly, for some C>, ( ) (N + 1)π v(t) =C sin (t ). (2.15)

11 Second-order singular equations at resonance 1273 Let (φ k ) k 1 be the orthonormal L 2 (, β)-basis made of the solutions of the Dirichlet problem φ k + µ k φ k =, φ k () ==φ k ( β), where µ k is the kth eigenvalue. Let us denote by, the scalar product in L 2 (, β) and by 2 the L 2 (, β)-norm. We write, for every n, the Fourier series of x n, x n = + k=1 x n,φ k φ k. It is useful to split such an expression as follows: x n = x n + x n, where while x n = x n,φ N+1 φ N+1, x n = x n,φ k φ k. k N+1 It is known that x n is L 2 -orthogonal to x n in view of the properties of the eigenfunctions φ k. Dividing x n by, we have v n = vn + vn with Since v = v, we have As v n = x n, v n = x n. v n = v n,v v 2 v v uniformly in t [, β]. we have that x n = + k=1 x n,φ k φ + = x n,φ k µ k φ k, k=1 (x n) =(x n), (x n) =(x n ). k Multiplying equation (2.5) by v n and integrating between and β, we then have β (x n) (t)v n(t)dt = β g λn (t, x n (t))v n(t)dt. (2.16)

12 1274 A. Fonda and M. Garrione Integrating twice by parts, we obtain β (x n) (t)v n(t)dt = β = β x n(t)(v n) (t)dt µ N+1 x n(t)v n(t)dt. (2.17) On the other hand, note that, by (H3), there exists ω ], min{δ, 1}[ such that g(t, x) f(x) for every x ],ω[. Define the two L -Carathéodory functions g(t, x) if x ], 1 2 ω[, 2 σ(t, x) = g(t, x)(ω x) ω if x [ 1 2ω, ω], if x ]ω, + [, and r(t, x) =g(t, x) µ N+1 x σ(t, x). Using (H2), together with the fact that σ(t, x), there exists a positive constant η such that As a consequence, β so that β r(t, x) η for a.e. t [,] and every x>. (2.18) g λn (t, x n (t))v n(t)dt β β = λ n σ(t, x n (t))vn(t)dt + λ n µ N+1 x n (t)vn(t)dt β + λ n r(t, x n (t))vn(t)dt + (1 λ n ) g λn (t, x n (t))v n(t)dt β β ( 1 ) x n (t) (µ N + µ N+1 )x n (t) vn(t)dt, β µ N+1 x n(t)vn(t)dt + λ n r(t, x n (t))vn(t)dt, in view of the fact that x n and v n are positive. Using (2.16), (2.17) and the fact that λ n 1, we deduce that, for sufficiently large n, β r(t, x n (t))v n(t)dt.

13 Second-order singular equations at resonance 1275 As v n v uniformly, by (2.18) the integrand is bounded from above by a positive constant. It is then possible to apply Fatou s lemma, so that β lim sup r(t, x n (t))vn(t)dt, n + whence, since lim n + v n(t) =v(t) for every t [, β], As β lim sup r(t, x n (t))v(t)dt. n + r(t, x) =g(t, x) µ N+1 x for a.e. t [,] and every x 1, and x n (t) + for every t ], β[, we deduce that β lim sup(g(t, x) µ N+1 x)v(t)dt. x + Since, in view of (2.15), v(t) =Cψ(t ) for every t [, β], this contradicts (2.2). We now prove the counterpart of lemma 2.3 for minima. Lemma 2.4. here exists a constant R ], 1[ such that, for any λ [, 1], if x(t) is a solution of (2.4), then min t [, ] x(t) >R. Proof. By contradiction, assume that there exist two sequences (λ n ) n [, 1] and (x n ) n, with x n (t) solving (2.4) for λ = λ n, such that min t [, ] x n (t). Let R>1 be as in lemma 2.3 and fix a sufficiently small ɛ>. As we have already recalled, by [8, lemma 2] there exists a sufficiently large R 1 > N (R, ) such that if (x n (t),x n(t)) makes exactly one rotation around the point (1, ) when t varies from τ 1 to τ 2, and N (x n (t),x n(t)) R 1 for every t [τ 1,τ 2 ], then τ 2 τ 1 N +1 ɛ. Now choose a positive integer n such that ( ) n N +1 ɛ >. Recalling again [8, lemma 1], as used in the proof of lemma 2.3, there exists R 2 >R 1 such that if for some t 1 <t 2, and N (x n (t 1 ),x n(t 1 )) = R 1, N (x n (t 2 ),x n(t 2 )) = R 2, R 1 < N (x n (t),x n(t)) <R 2 for every t ]t 1,t 2 [,

14 1276 A. Fonda and M. Garrione then (x n (t),x n(t)) turns at least n times around (1, ) in the phase plane when t varies from t 1 to t 2. As min t [, ] x n (t), for every large n, we will have max{n (x n (t),x n(t)) t [,]} >R 2. Since x n (t) is -periodic, and in view of the continuity of N (x n (t),x n(t)), for sufficiently large n it follows that N (x n (t),x n(t)) >R 1 for every t [,], otherwise (x n (t),x n(t)) would turn at least n times around (1, ) in time, which is impossible. Choosing t n such that x n ( t n ) = max t [, ] x n (t), being x n( t n ) = it would then be, for large n, N (x n ( t n ), ) >R 1, which, since x n ( t n ) <R, contradicts the inequality N (R, ) <R 1. Remark 2.5. As mentioned in 1, for the special case of equation (1.4), condition (2.2) reads as (1.7). Let us try to explain why we are not able to replace (1.7) with the more general assumption (1.6). We recall that, in [8], a curve Γ : [, + [ R 2 was constructed in order to control the solutions in the phase plane. One could try to estimate the height of the bumps of v(t) by using such a curve for the solutions x n (t). However, consider, for example, equation (1.5) and define g λ (t, x) as in (2.3). We would then have g λ (t, x) = 1 x 3 +(1 2 (µ N + µ N+1 )+ 1 2 λ(µ N+1 µ N ))x. Hence, the curve Γ (t), as it is constructed in [8], should use the level lines of the functions V 1 (u, v) = 1 ( ) 1 2 u 2 + µ N+1u 2 + v 2, V 2 (u, v) = 1 ( ) 1 2 u (µ N + µ N+1 )u 2 + v 2. A straightforward computation shows, however, that, denoting by P m =(u m, ) and P m+1 =(u m+1, ) two consecutive intersections of Γ (t) with the half-line {(u, v) R 2 u 1,v =}, it is u m+1 lim inf > 1. m + u m his fact makes us unable to prove that the height of the bumps of v(t), which is obtained as the limit of the normalized sequence (v n ) n, is always the same. his is why we did not manage, in our setting, to assume a Landesman Lazer condition like (1.6). Acknowledgements he authors acknowledge the support of GNAMPA through the project Soluzioni periodiche di alcune classi di equazioni differenziali ordinarie. References 1 S. Bolotin and R. S. Mackay. Isochronous potentials. In Localization and energy transfer in nonlinear systems, Proc. hird Conf. San Lorenzo de El Escorial, Madrid, Spain, June 22, pp (World Scientific, 23).

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