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Harmonic oscillators at resonance, perturbed by a non-linear friction force

This note is an addendum to the results of P.O. Frederickson and A.C. Lazer [1], and A.C. Lazer [4] on periodic oscillations, with linear part at resonance. We show that a small modification of the argument in [4] provides a more general result. It turns out that things are different for the corresponding Dirichlet boundary value problem.


Introduction
We are interested in the existence of 2π periodic solutions to the problem (x = x(t)) x ′′ + f (x)x ′ + n 2 x = e(t) .
Here e(t) ∈ C(R) satisfies e(t + 2π) = e(t) for all t, f (u) ∈ C(R), n ≥ 1 is an integer. The linear part, x ′′ + n 2 x = e(t), is at resonance, with the null space spanned by cos nt and sin nt. Define F (x) = x 0 f (t) dt. We assume that the finite limits F (∞) and F (−∞) exist, and F (−∞) < F (x) < F (∞) for all x . The following theorem was proved in case n = 1 by A.C. Lazer [4], based on P.O. Frederickson and A.C. Lazer [1]. The paper [1] was the precursor to the classical works of E.M. Landesman and A.C. Lazer [3], and A.C. Lazer and D.E. Leach [3].
is necessary and sufficient for the existence of 2π periodic solution of (1.1).
We provide a proof for all n, by modifying the argument in [4].
Remarkably, things are different for the corresponding Dirichlet boundary value problem, for which we derive a necessary condition for the existence of solutions, but show by a numerical computation that this condition is not sufficient. Observe that the condition (1.3) depends on n, unlike the condition in A.C. Lazer and D.E. Leach [3].
Proof of the Theorem 1.1: 1. Necessity. Given arbitrary numbers a and b, we can find a δ ∈ [0, 2π), so that We multiply (1.1) by a cos nt, then by b sin nt, integrate and add the results Using that x(t) is a 2π periodic solution, and Lemma 2.2, we have On the right in (2.1) we have the scalar product of the vector (A n , B n ) and an arbitrary unit vector. The condition (1.3) follows.

A boundary value problem
Consider the Dirichlet problem The linear part has a kernel spanned by sin t. Denote A = π 0 e(t) sin t dt. Then from (3.1) We conclude that is a necessary condition for the existence of solutions.
It is natural to ask if the condition (3.2) is sufficient for the existence of solutions. The following numerical computations indicate that the answer is No.
Example We have solved the problem Here F (±∞) = ±1, and so the necessary condition for the existence of solutions is |A| < 2. Writing the solution as x(t) = ξ sin t + X(t), with π 0 X(t) sin t dt = 0, for each value of ξ we compute the value of A for which the problem (3.3) has a solution with the first harmonic equal to ξ, and that solution x(t), see P. Korman [2] for more details. (I.e., we compute the solution curve (A, x(t))(ξ).) In Figure 1 we