On Solvability Theorems of Second-Order Ordinary Differential Equations with Delay

Lemma 1. Let k ∈ N ∪ {0} and Γ be a nonnegative L¹(0,2π)-function such that for a.e. x ∈ (0,2π),   Γ(x) ≤ 2k + 1 with strict inequality on a positive measurable subset of (0,2π). Then there exists a constant K₁ > 0 such that

whenever p ∈ L¹(0,2π) with 0 ≤ p(x) ≤ Γ(x) for a.e. x ∈ (0,2π) and u ∈ W^2,1(0,2π) is a periodic function of period 2π with u(0) − u(2π) = u^′(0) − u^′(2π) = 0.

Proof. Just as in [20, Lemma 1], we can modify slightly the proof of [11, Lemma 2] or [1, Lemma 2.2] to obtain the fact that there exists a constant K₁ > 0 such that

whenever p ∈ L¹(0,2π) with 0 ≤ p(x) ≤ Γ(x) for a.e. x ∈ (0,2π) and u ∈ W^2,1(0,2π) with u(0) − u(2π) = u^′(0) − u^′(2π) = 0. Let us extend u(x) and p(x)  2π periodically in x to all of R and then use the same notations for the periodic extensions as for the original functions. In this case, we have $$ and Since $$,  p(x) ≥ 0 for a.e. x ∈ (0,2π), and $$ for all v, w ∈ W^2,1(0,2π) with v(0) − v(2π) = v^′(0) − v^′(2π) = 0 and w(0) − w(2π) = w^′(0) − w^′(2π) = 0, we have $$ and Combining (12) with (13), we have

Lemma 2. Let k ∈ N and Γ be a nonnegative L¹(0,2π)-function such that for a.e. x ∈ (0,2π),   Γ(x) ≤ 2k − 1 with strict inequality on a positive measurable subset of (0,2π). Then there exists a constant K₂ > 0 such that

whenever p ∈ L¹(0,2π) with 0 ≤ p(x) ≤ Γ(x) for a.e. x ∈ (0,2π) and u ∈ W^2,1(0,2π) is a periodic function of period 2π with u(0) − u(2π) = u^′(0) − u^′(2π) = 0. Here $$ and $$ for each v ∈ W^2,1(0,2π) with v(0) − v(2π) = v^′(0) − v^′(2π) = 0.

Theorem 3. Let k ∈ N ∪ {0} and g : (0,2π) × R → R be a Caratheodory function satisfying (H). Then for each h ∈ L¹(0,2π) problem (1) has a solution u, provided that either (8) with −1 < β < 0 or (9) with β = 0 holds.

Proof. Let α ∈ R be fixed and 0 < α < 2k + 1. We consider the boundary value problems

for 0 ≤ t ≤ 1, which becomes the original problem when t = 1. Since 0 < α < 2k + 1, we observe from Lemma 1 that (17) has only a trivial solution when t = 0. To apply the Leray-Schauder continuation method, it suffices to show that solutions to (17) for 0 < t < 1 have an a priori bound in H¹(0,2π). To this end, let θ : R → R be a continuous function such that 0 ≤ θ ≤ 1,  θ(u) = 0 for |u| ≤ r₀, and θ(u) = 1 for |u| ≥ 2r₀. We define $$, and g₂(x, u) = g(x, u)  −  g₁(x, u). Then g₁, g₂ : (0,2π)  ×  R → R are Caratheodory functions, such that for a.e. x ∈ (0,2π) and u ∈ R,  u ≠ 0 If u is a possible solution to (17) for some 0 < t < 1, then using (19), (20), and Lemma 1, we have which implies that for some constants C₁, C₂ > 0 independent of u. It remains to show that solutions to (17) for 0 < t < 1 have an a priori bound in H¹(0,2π). We argue by contradiction and suppose that there exists a sequence {u_n} of periodic functions with period 2π and a corresponding sequence {t_n} in (0,1) such that u_n is a solution to (17) with t = t_n and $$ for all n. Let $$; then $$ for all n ∈ N, and by (22) we have $$ as n → ∞. Since $$ and $$ for all n ∈ N, we have a bounded sequence {P_kv_n} in H¹(0,2π). For simplicity, we may assume that v_n converges to v in H¹(0,2π) for some v ∈ N(L) with $$. In particular, v_n → v in C[0,2π]. Clearly, v(·−x₀) ∈ N(L) and $$. It follows that u_n(x) → ∞ for each x ∈ R with v(x) > 0, and u_n(x)→−∞ for each x ∈ R with v(x) < 0. Since $$ and $$, we have Multiplying each side of (17) by P_kv_n(x − x₀), and then integrating them over [0,2π] when u = u_n and t = t_n, we get By (19) and the assumption of −1 < β ≤ 0, we have for a.e. x ∈ (0,2π). Combining (22) with (25), we get that $$ is bounded from below by an L¹(0,2π)-function independent of n. By (20) and the assumption of −1 < β ≤ 0, we have for a.e. x ∈ (0,2π), In particular, $$ is bounded from below by an L¹(0,2π)-function independent of n, which implies that $$ is also so, $$, and for all n ∈ N with u_n(x − x₀) ≠ 0. Here sign(w) = 1 if w > 0, sign(w) = 0 if w = 0, and sign(w) = −1 if w < 0. Applying Fatou’s lemma to the integral $$, we have which is a contradiction when either (8) with −1 < β < 0 or (9) with β = 0 is satisfied. Hence, the proof of this theorem is complete.

Theorem 4. Let k ∈ N and g : (0,2π) × R → R be a Caratheodory function satisfying (G). Then for each h ∈ L¹(0,2π) problem (2) has a solution u, provided that either (8) with −1 < β < 0 or (9) with β = 0 holds.