Oscillation of first order linear differential equations with several non-monotone delays

Lemma 2.1

([6]). Let x(t) be an eventually positive solution of Eq. (1). Then

whereq_kis defined by(4).

Lemma 2.2

([10]). Assume that

andx(t) is an eventually positive solution of the first order delay differential inequality

whereP ∈ C([t₁, ∞), [0, ∞)). If 0 ≤ α^* ≤ 1e, then

Theorem 2.3

Assume that

and

where

andϵ ∈ (0, λ (ρ)). Then all solutions of Eq.(1)are oscillatory.

Proof

Assume the contrary, i.e., there exists a nonoscillatory solution x(t) of (1). Because of the linearity of (1), we assume that x(t) is eventually positive. Therefore, there exists a sufficiently large t₂ ≥ t₁ such that x(τ_k(t)) > 0, for all t ≥ t₂, k = 1, 2, …, n. Thus, equation (1) implies that x(t) is nonincreasing for all t ≥ t₂. Integrating (1) from τ_i(t) to t, we obtain

Also, dividing (1) by x(t) and integrating the resulting equation from τ_k(s) to g_k(t), we get

Substituting this into (10),

Multiplying the above equation by p_i(t), and taking the sum over i, it follows that

The substitution y(t)=e∫t0t∑ℓ=1npℓ(s)dsx(t), reduces this equation to

which in turn, by integrating from g(t) to t, leads to

Hence, the nonincreasing nature of y(t) implies that

for all t ≥ t₃ and some t₃ ≥ t₂.

On the other hand, using the nonincreasing nature of x(t), equation (1) implies that

Therefore, from [11, Lemma 2.1.2], we obtain lim inft→∞x(g(t))x(t)≥λ(ρ). Thus, for sufficiently small ϵ > 0, we have

for some t₄ ≥ t₃. This together with (12) implies that

for all t ≥ t₅, where t₅ ≥ t₄. That is

for all t ≥ t₅. Also, in view of Lemma 2.2, for sufficiently small ϵ^∗ > 0 and some t₆ ≥ t₅, inequality (13) leads to

Therefore,

Combining this inequality with (15), it follows that

Then

Notice that, by integrating (13) from g(t) to t and using the nonincreasing nature of x(t), we obtain

Now, letting ϵ^∗ → 0 in (16), we arrive at a contradiction with (9). The proof is complete. □

Remark 2.4

Theorem 2.3 is proved using the core idea of the proof of [12, Theorem 2.1] which is given for Equation (1) when n = 1. However, Theorem 2.3 produces a new oscillation criterion even for equations with only one non-monotone delay.

Using Lemma 2.1 instead of [11, Lemma 2.1.2], similar reasoning as in the proof of Theorem 2.3 implies the following result:

Theorem 2.5

Assume that

where

q_ℓ is defined by (4), ρ is defined as in Theorem 2.3 and ϵ_ℓ ∈ (0, λ (q_ℓ)). Then all solutions of Equation (1) oscillate.

Theorem 2.6

Assume that

where β_k is defined by (5) with 0 < β_k ≤ 1eand

ρ is defined as in Theorem 2.3 and ϵ ∈ (0, λ(ρ )). Then all solutions of Equation (1) oscillate.

Proof

Let x(t) be a nonoscillatory solution of (1). As usual, we assume that x(t) is an eventually positive solution. Substituting (14) into (11), we obtain

for all t ≥ t₂ and some t₂ ≥ t₁ where ϵ ∈ (0, λ (ρ)) and y(t)=e∫t0t∑ℓ=1npℓ(s)dsx(t) Integrating from g_j(t) to t and using the nonincreasing nature of y(t), we obtain

and some t₃ ≥ t₂. Using the relation between arithmetic and geometric mean, it follows that

Taking the product on both sides,

Therefore,

Since y(t)=e∫t0t∑ℓ=1npℓ(s)dsx(t), then

Substituting in (17),

On the other hand, the nonincreasing nature of x(t) and (1) imply that

and hence, by Lemma 2.2, it follows that

for some t₄ ≥ t₃ and sufficiently small ϵ^∗ > 0. This together with (18) leads to

for all t ≥ t₄. Consequently,

On the other hand, integrating (1) from g_k(t) to t, we get

Then, using the nonincreasing nature of x(t), we obtain

But applying Lemma 2.2 to (19), we obtain

Therefore, e∑k=1n∫gk(t)t∑ℓ=1npℓ(s)ds is bounded. Now, allowing ϵ^∗ → 0 in (20), it follows that

This contradiction completes the proof. □

The following example shows that Theorem 2.3 can be applied but conditions (2), (3) and (6) fail to apply as well as (8) when r = 1.

Example 2.7

Consider the first order delay differential equation

wherea=0.41.137π+2,b=1.784and

Clearly

Fig. 1

Equation (21) has the form (1) with p₁(t) = a(b + sin (t)), p₂(t) = a(b + cos (t)), τ₁(t) = t − π2and τ₂(t) = t − π2 − σ sin²(300t). Therefore, we can choose g₁(t) = g₂(t) = g(t) = t − π2and ϵ = 0.001. Then

Consequently,

LetI(t)=∫g(t)tQ(v)dv+c(ρ)e∫g(t)t∑i=12pi(s)ds.Then, the property that e^x ≥ ex for x ≥ 0 leads to

Now, using Maple, we get

ChooseTk=3π4+2πk,then

which means that