Fifth Mississippi State Conference on Differential Equations and Computational Simulations, Electronic Journal of Differential Equations, Conference 10, 2003, pp. 79–88. ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu ftp ejde.math.swt.edu (login: ftp)

OSCILLATION OF THIRD ORDER FUNCTIONAL DIFFERENTIAL EQUATIONS WITH DELAY

TUNCAY CANDAN & RAJBIR S. DAHIYA

Abstract. We consider third order functional differential equations with discrete and continuous delay. We then develop several theorems related to the oscillatory behavior of these differential equations.

1. Introduction

Our goal in this paper is to study functional differential equations of the form

(b(t)(a(t)x′(t))′)′ + ∑_{i=1}^{m} q_{i}(t)f(x(σ_{i}(t))) = h(t),  (1.1)

where a, b, h ∈ C([t_{0}, ∞), ℝ), a(t), b(t) > 0, f : ℝ → ℝ continuous, σ_{i}(t) → ∞, as t → ∞, i = 1, 2, …, m, and

(b(t)(a(t)x′(t))′)′ + ∫_{c}^{d} q(t, ξ)f(x(σ(t, ξ)))dξ = 0,  (1.2)

where a, b ∈ C([t_{0}, ∞), ℝ), f ∈ C(ℝ, ℝ). The oscillations of solutions of third order equations were studied by Rao and Dahiya [8], Tantawy [9], Waltman [10] and Zafer and Dahiya [11]. The results in this paper for equation (1.1) are more general comparing to Zafer and Dahiya [11]. The results for equation (1.2) are essentially new.

As is customary, a solution of equations (1.1) and (1.2) is called oscillatory if it has arbitrarily large zeros, otherwise it is called nonoscillatory. The solution of equations (1.1) and (1.2) is called almost oscillatory if it is oscillatory or lim_{t→∞} x^{(i)}(t) = 0, i = 0, 1, 2.

2. Main Results

Oscillatory behavior of third order differential equations with discrete delay. Assume that xf(x) > 0, x ≠ 0, q_{i}(t) ≥ 0 is not identically zero in any half line of the form (τ, ∞) for some τ ≥ 0, i = 1, 2, …, m and σ_{i}(t) < t, σ′_{i}(t) > 0, i=1,2,…,m, b′(t) ≥ 0, and

1991 Mathematics Subject Classification. 34K11.

Key words and phrases. Oscillation theory.

\begin{matrix}c \\ ◯ \end{matrix}2003 Southwest Texas State University.

Published February 28, 2003.

∫ ^{∞} \frac{dt}{b(t)} = ∞, ∫ ^{∞} \frac{dt}{a(t)} = ∞.  (2.1)

Theorem 2.1. Let f(x) = x and h(t) = 0. Suppose that there exist a differentiable function p ∈ C([t_{0}, ∞), ℝ), p(t) > 0 such that

∫ ^{∞} [q(t)p(t) − \frac{b(t)(p′(t))^{2}}{∑_{i=1}^{m} \frac{(σ_{i}(t)−T)}{a(σ_{i}(t))}σ′_{i}(t)4p(t)}]dt = ∞,  (2.2)

where q(t) = min{q_{1}(t), q_{2}(t), …, q_{m}(t)}, for every T ≥ 0, and that

∫_{σ(t)}^{t} [ ∫_{σ(t)}^{r} \frac{1}{a(u)}du (∫_{u}^{r} \frac{1}{b(v)}dv) ] ∑_{i=1}^{m} q_{i}(r)dr > 1,  (2.3)

where σ(t) = max{σ_{1}(t), σ_{2}(t), …, σ_{m}(t)}. Then the equation (1.1) is oscillatory.

Proof. Let x(t) be a non-oscillatory solution of (1.1). Assume x(t) is eventually positive. Since σ_{i}(t) → ∞ as t → ∞ for i = 1, 2, …, m, there exist a t_{1} ≥ t_{0} such that x(t) > 0 and x(σ_{i}(t)) > 0 for t ≥ t_{1}. From (1.1), we have

(b(t)(a(t)x′(t))′)′ = − ∑_{i=1}^{m} q_{i}(t)x(σ_{i}(t)).  (2.4)

Since q_{i}(t) is not negative and x(σ_{i}(t)) > 0 is positive for t ≥ t_{1}, the right-hand side becomes non-positive. Therefore, we have

(b(t)(a(t)x′(t))′)′ ≤ 0

for t ≥ t_{1}. Thus, x(t), x′(t), (a(t)x′(t))′ are monotone and eventually one-signed. Now we want to show that there is a t_{2} ≥ t_{1} such that for t ≥ t_{2}

(a(t)x′(t))′ > 0.  (2.5)

Suppose this is not true, then (a(t)x′(t))′ ≤ 0. Since q_{i}(t), i = 1, 2, …, m are not identically zero and b(t) > 0, it is clear that there is t_{3} ≥ t_{2} such that b(t_{3})(a(t_{3})x′(t_{3}))′ < 0. Then, for t > t_{3} we have

b(t)(a(t)x′(t))′ ≤ b(t_{3})(a(t_{3})x′(t_{3}))′ < 0.  (2.6)

Dividing (2.6) by b(t) and then integrating between t_{3} and t, we obtain

a(t)x′(t) − a(t_{3})x′(t_{3}) < b(t_{3})(a(t_{3})x(t_{3})′)′ ∫_{t_{3}}^{t} \frac{1}{b(s)}ds.  (2.7)

Letting t → ∞ in (2.7), and because of (2.1) we see that a(t)x′(t) → −∞ as t → ∞. Thus there is a t_{4} ≥ t_{3} such that a(t_{4})x′(t_{4}) < 0. Using (a(t)x′(t))′ ≤ 0, we have for t ≥ t_{4}

a(t)x′(t) ≤ a(t_{4})x′(t_{4}).  (2.8)

If we divide (2.8) by a(t) and integrate from t_{4} to t with t → ∞, the right-hand side becomes negative. Thus, we have x(t) → −∞. But this is a contradiction x(t) being eventually positive and therefore it proves that (2.5) holds. Now we consider two cases.

Suppose x′(t) is eventually positive, say x′(t) > 0 for t ≥ t_{2}. Define the function z(t) by

z(t) = \frac{b(t)(a(t)x′(t))′}{∑_{i=1}^{m} x(σ_{i}(t))} p(t).

It is obvious that z(t) > 0 for t ≥ t_{2} and z′(t) is

z′(t) = −\frac{∑_{i=1}^{m} q_{i}(t)x(σ_{i}(t))}{∑_{i=1}^{m} x(σ_{i}(t))}p(t) + \frac{p′(t)}{p(t)}z(t) − \frac{∑_{i=1}^{m} x′(σ_{i}(t))σ′_{i}(t)}{∑_{i=1}^{m} x(σ_{i}(t))}z(t).

Then

z′(t) ≤ −q(t)p(t) + \frac{p′(t)}{p(t)}z(t) − \frac{∑_{i=1}^{m} x′(σ_{i}(t))σ′_{i}(t)}{∑_{i=1}^{m} x(σ_{i}(t))}z(t),  (2.9)

where q(t) = min{q_{1}(t), q_{2}(t), …, q_{m}(t)}. On the other hand, using (b(t)(a(t)x′(t))′)′ ≤ 0, b′(t) ≥ 0 and (2.5), we can find that

(a(t)x′(t))′′ ≤ 0.  (2.10)

Using (2.10) and the equality

a(t)x′(t) = a(T)x′(T) + ∫_{T}^{t}(a(s)x′(s))′ds,  (2.11)

we have

a(t)x′(t) ≥ (t − T)(a(t)x′(t))′  (2.12)

for T ≥ t_{2}. Since (a(t)x′(t))′ is non-increasing, we obtain

a(σ_{i}(t))x′(σ_{i}(t)) ≥ (σ_{i}(t) − T)(a(t)x′(t))′ for i = 1, 2, …, m.  (2.13)

Multiplying both sides of (2.13) by

\frac{σ′_{i}(t)}{a(σ_{i}(t))}

and taking the summation from 1 to m, we have

∑_{i=1}^{m} σ′_{i}(t)x′(σ_{i}(t)) ≥ ∑_{i=1}^{m} \frac{(σ_{i}(t) − T)}{a(σ_{i}(t))}σ′_{i}(t)(a(t)x′(t))′.  (2.14)

Then, using (2.14) in (2.9), it follows that

z′(t) ≤ −q(t)p(t) + \frac{p′(t)}{p(t)}z(t) − \frac{∑_{i=1}^{m} \frac{(σ_{i}(t)−T)}{a(σ_{i}(t))}σ′_{i}(t)}{b(t)p(t)}z^{2}(t),

and completing the square will leads to

z′(t) ≤ −q(t)p(t) + \frac{b(t)(p′)^{2}(t)}{∑_{i=1}^{m} \frac{(σ_{i}(t)−T)}{a(σ_{i}(t))}σ′_{i}(t)4p(t)}.  (2.15)

Integrating (2.15) between T and t and letting t → ∞, we see that lim_{t→∞} z(t) = −∞. This contradicts z(t) being eventually positive.

If x′(t) is eventually negative. We integrate (1.1) from t to ∞ and since

b(t)(a(t)x′(t))′ > 0,

we have

−b(t)(a(t)x′(t))′ + ∫_{t}^{∞} ∑_{i=1}^{m} q_{i}(r)x(σ_{i}(r))dr ≤ 0.  (2.16)

Now integrating (2.16) from t to ∞ after dividing by b(t) and using a(t)x′(t) < 0, will lead to

a(t)x′(t) + ∫_{t}^{∞} ( ∫_{t}^{r} \frac{1}{b(u)}du) ∑_{i=1}^{m} q_{i}(r)x(σ_{i}(r))dr ≤ 0.  (2.17)

Dividing (2.17) by a(t) and integrating again from t to ∞ gives

∫_{t}^{∞} [ ∫_{t}^{r} \frac{1}{a(u)}du( ∫_{u}^{r} \frac{1}{b(v)}dv)] ∑_{i=1}^{m} q_{i}(r)x(σ_{i}(r))dr ≤ x(t).  (2.18)

Replacing t by σ(t) in (2.18), where σ(t) = max{σ_{1}(t), σ_{2}(t), …, σ_{m}(t)}, will give

∫_{σ(t)}^{t} [ ∫_{σ(t)}^{r} \frac{1}{a(u)}du( ∫_{u}^{r} \frac{1}{b(v)}dv)] ∑_{i=1}^{m} q_{i}(r)x(σ_{i}(r))dr ≤ x(σ(t)).  (2.19)

Using the fact that σ_{i}(t) < t and x(t) is decreasing in (2.19), we obtain

∫_{σ(t)}^{t} [ ∫_{σ(t)}^{r} \frac{1}{a(u)}du( ∫_{u}^{r} \frac{1}{b(v)}dv)] ∑_{i=1}^{m} q_{i}(r)dr ≤ 1.

This is a contradiction to (2.3). Therefore, the proof is complete

Example 2.2. Consider the following functional differential equation

(e^{−t}x′)′′ + ∑_{i=1}^{2}(2i − 1)e^{−t}x(t − (i + 1)π) = 0.

Now a(t) = e^{−t}, b(t) = 1, q_{1}(t) = e^{−t}, q_{2}(t) = 3e^{−t} σ_{1}(t) = t − 2π, σ_{2}(t) = t − 3π, p(t) = e^{t}.

We can easily see that the conditions of Theorem 2.1 are satisfied. It is easy to verify that x(t) = cost is a solution of this problem.

Theorem 2.3. Let f′(x) ≥ λ for some λ > 0, and h(t) = 0. Suppose that there exist a differentiable function p ∈ C([t_{0}, ∞), ℝ), p(t) > 0 such that

∫ ^{∞} [q(t)p(t) − \frac{b(t)(p′(t))^{2}}{∑_{i=1}^{m} \frac{(σ_{i}(t)−T)}{a(σ_{i}(t))}σ′_{i}(t)4λp(t)}]dt = ∞,  (2.20)

where q(t) = min{q_{1}(t), q_{2}(t), …, q_{m}(t)}, for every T ≥ 0, and that

lim sup_{t→∞} ∫_{σ(t)}^{t} [ ∫_{σ(t)}^{r} \frac{1}{a(u)}du( ∫_{u}^{r} \frac{1}{b(v)}dv)] ∑_{i=1}^{m} q_{i}(r)dr = ∞,  (2.21)

where σ(t) = max{σ_{1}(t), σ_{2}(t), …, σ_{m}(t)}. Then the equation (1.1) is oscillatory.

Proof. The beginning part of the proof is similar to the proof of Theorem 2.1 until we reach at two possible cases. Suppose x′(t) is eventually positive. Then, we can define

z(t) = \frac{b(t)(a(t)x′(t))′}{∑_{i=1}^{m} f(x(σ_{i}(t)))}p(t) > 0.

It is obvious that z(t) > 0 for t ≥ t_{2} and z′(t) is

z′(t) = −\frac{∑_{i=1}^{m} q_{i}(t)f(x(σ_{i}(t)))}{∑_{i=1}^{m} f(x(σ_{i}(t)))}p(t)

+\frac{p′(t)}{p(t)}z(t) − \frac{∑_{i=1}^{m} f′(x(σ_{i}(t)))x′(σ_{i}(t))σ′_{i}(t)}{∑_{i=1}^{m} f(x(σ_{i}(t)))}z(t).

Then

z′(t) ≤ −q(t)p(t) + \frac{p′(t)}{p(t)}z(t) − \frac{∑_{i=1}^{m} f′(x(σ_{i}(t))x′(σ_{i}(t))σ′_{i}(t)}{∑_{i=1}^{m} f(x(σ_{i}(t)))}z(t),  (2.22)

where q(t) = min{q_{1}(t), q_{2}(t), …, q_{m}(t)}. On the other hand, since (b(t)(a(t)x′(t))′)′ ≤ 0, (2.5) holds and b′(t) ≥ 0, we can obtain

(a(t)x′(t))′′ ≤ 0.  (2.23)

Using (2.23) and the equality

a(t)x′(t) = a(T)x′(T) + ∫_{T}^{t}(a(s)x′(s))′ds  (2.24)

will lead to

a(t)x′(t) ≥ (t − T)(a(t)x′(t))′.  (2.25)

Now using non-increasing nature of (a(t)x′(t))′, we obtain

a(σ_{i}(t))x′(σ_{i}(t)) ≥ (σ_{i}(t) − T)(a(t)x′(t))′ for i = 1, 2, …, m.  (2.26)

Multiplying both sides of (2.26) by

\frac{σ′_{i}(t)}{a(σ_{i}(t))}

and taking the summation from 1 to m, we have

∑_{i=1}^{m} σ′_{i}(t)x′(σ_{i}(t)) ≥ ∑_{i=1}^{m} \frac{(σ_{i}(t) − T)}{a(σ_{i}(t))}σ′_{i}(t)(a(t)x′(t))′.  (2.27)

Then, using (2.27) in (2.22), it follows that

z′(t) ≤ −q(t)p(t) + \frac{p′(t)}{p(t)}z(t) − λ\frac{∑_{i=1}^{m} \frac{(σ_{i}(t)−T)}{a(σ_{i}(t))}σ′_{i}(t)}{b(t)p(t)}z^{2}(t),

and then completing the square leads to

z′(t) ≤ −q(t)p(t) + \frac{b(t)(p′(t))^{2}}{λ ∑_{i=1}^{m} \frac{(σ_{i}(t)−T)}{a(σ_{i}(t))}σ′_{i}(t)4p(t)}.  (2.28)

Integrating (2.28) between T to t and letting t → ∞, we see that lim_{t→∞} z(t) = −∞. This contradicts z(t) being eventually positive.

If x′(t) is eventually negative and proceeding as in the proof of Theorem 2.1 we will end up with

∫_{σ(t)}^{∞} [ ∫_{σ(t)}^{r} \frac{1}{a(u)}du( ∫_{u}^{r} \frac{1}{b(v)}dv)] ∑_{i=1}^{m} q_{i}(r)f(x(σ_{i}(r)))dr ≤ x(σ(t)),

where σ(t) = max{σ_{1}(t), σ_{1}(t), …, σ_{n}(t)}. Thus we have

∫_{σ(t)}^{t} [ ∫_{σ(t)}^{r} \frac{1}{a(u)}du( ∫_{u}^{r} \frac{1}{b(v)}dv)] ∑_{i=1}^{m} q_{i}(r)f(x(σ_{i}(r)))dr ≤ x(σ(t)).  (2.29)

Using the fact that σ_{i}(t) < t, f(x) is increasing and x(t) is decreasing in (2.29), we obtain

∫_{σ(t)}^{t} [ ∫_{σ(t)}^{r} \frac{1}{a(u)}du( ∫_{u}^{r} \frac{1}{b(v)}dv)] ∑_{i=1}^{m} q_{i}(r)dr ≤ \frac{x(σ(t))}{f(x(σ(t)))}.

Since x(t) is decreasing and positive, it is approaching a finite non-negative number as t → ∞. In view of (2.21) and the last equation, it is not possible that lim_{t→∞} x(t) > 0. Suppose lim_{t→∞} x(t) = 0, then

lim_{t→∞} \frac{x(σ(t))}{f(x(σ(t)))} = lim_{t→∞} \frac{1}{f′(x(σ(t)))} = \frac{1}{f′(0)} ≤ \frac{1}{λ}.

This is a contradiction to (2.21) . Therefore, the proof is complete.

Theorem 2.4. Suppose that f′(x) ≥ λ for some λ > 0 and

lim sup_{t→∞} ∫_{σ(t)}^{t} [ ∫_{σ(t)}^{r} \frac{1}{a(u)}du( ∫_{u}^{r} \frac{1}{b(v)}dv)] ∑_{i=1}^{m} q_{i}(r)dr = ∞.  (2.30)

In addition, suppose that there exist a continuously differentiable function p ∈ C([t_{0}, ∞), ℝ), p(t) > 0 and an oscillatory function ψ(t) such that

∫ ^{∞} [q(t)p(t) − \frac{b(t)(p′(t))^{2}}{∑_{i=1}^{m} \frac{(σ_{i}(t)−T)}{a(σ_{i}(t))}σ′_{i}(t)4λdp(t)}]dt = ∞  (2.31)

for some d ∈ (0, 1) and for every T ≥ 0, and

(b(t)(a(t)ψ′(t))′)′ = h(t), lim_{t→∞} ψ^{(i)}(t) = 0, i = 0, 1, 2.  (2.32)

Then the equation (1.1) is almost oscillatory.

Proof. Let x(t) be a non-oscillatory solution of (1.1). Without loss of generality we may assume that x(t) is eventually positive. Consider

y(t) = x(t) − ψ(t).  (2.33)

Obviously y(t) is eventually positive, otherwise, x(t) < ψ(t) and it is a contradiction with oscillatory behavior of ψ(t). We know that,

(b(t)(a(t)y′(t))′)′ ≤ 0.  (2.34)

Proceeding as in the proof of Theorem 2.1, there is a t_{1} ≥ 0 such that for t ≥ t_{1}

(a(t)y′(t))′ > 0 and (a(t)y′(t))′′ ≤ 0.

Consider again two cases. Suppose that y′(t) is eventually positive, then y(t) is increasing and eventually positive. On the other hand, since ψ(t) → 0 as t → ∞ and y(t) = x(t) − ψ(t), there exists a t_{2} ≥ t_{1} such that

x(σ_{i}(t)) ≥ dy(σ_{i}(t)) for t ≥ t_{2} and d ∈ (0, 1), i = 1, 2, …, m.

Since f is an increasing function, we obtain

f(x(σ_{i}(t))) ≥ f(dy(σ_{i}(t))) for t ≥ t_{2}, i = 1, 2, …, m.

Define z(t) by

z(t) = \frac{b(t)(a(t)y′(t))′}{∑_{i=1}^{m} f(dy(σ_{i}(t)))}p(t),

then obviously z(t) > 0 for t ≥ t_{2} and z′(t) is

z′(t) = −\frac{∑_{i=1}^{m} q_{i}(t)f(x(σ_{i}(t)))}{∑_{i=1}^{m} f(dy(σ_{i}(t)))}p(t) + \frac{p′(t)}{p(t)}z(t)

−d\frac{∑_{i=1}^{m} f′(dy(σ_{i}(t)))y′(σ_{i}(t))σ′_{i}(t)}{∑_{i=1}^{m} f(dy(σ_{i}(t)))}z(t).

Then, using f′(x) ≥ λ > 0, we obtain

z′(t) ≤ −q(t)p(t) + \frac{p′(t)}{p(t)}z(t) − dλ\frac{∑_{i=1}^{m} y′(σ_{i}(t))σ′_{i}(t)}{∑_{i=1}^{m} f(dy(σ_{i}(t)))}z(t),  (2.35)

where q(t) = min{q_{1}(t), q_{2}(t), …, q_{m}(t)}. We can now show that

∑_{i=1}^{m} σ′_{i}(t)y′(σ_{i}(t)) ≥ ∑_{i=1}^{m} \frac{(σ_{i}(t) − T)}{a(σ_{i}(t))}σ′_{i}(t)(a(t)y′(t))′  (2.36)

as in proof of Theorem 2.1. Using (2.35) and (2.36), we have

z′(t) ≤ −q(t)p(t) + \frac{p′(t)}{p(t)}z(t) − dλ\frac{∑_{i=1}^{m} \frac{(σ_{i}(t)−T)}{a(σ_{i}(t))}σ′_{i}(t)}{b(t)p(t)}z^{2}(t).

Completing the square in the above equation leads to

z′(t) ≤ −q(t)p(t) + \frac{b(t)(p′)^{2}(t)}{∑_{i=1}^{m} \frac{(σ_{i}(t)−T)}{a(σ_{i}(t))}σ′_{i}(t)4λdp(t)}.  (2.37)

Integrating (2.37) from T to t and letting t → ∞, we see that lim_{t→∞} z(t) = −∞. This contradicts z(t) being eventually positive.

Now suppose y′(t) is eventually negative. Since y is eventually positive and decreasing, lim_{t→∞} y(t) = c, where c is a nonnegative number. Therefore, lim_{t→∞} x(t) = c. Integrating (1.1) three times as we did in the proof of Theorem 2.1, we will end up with

∫_{σ(t)}^{∞} [ ∫_{σ(t)}^{r} \frac{1}{a(u)}du( ∫_{u}^{r} \frac{1}{b(v)}dv)] ∑_{i=1}^{m} q_{i}(r)f(x(σ_{i}(r)))dr ≤ y(t),

where σ(t) = max{σ_{1}(t), σ_{1}(t), …, σ_{n}(t)}. Thus we have

∫_{σ(t)}^{t} [ ∫_{σ(t)}^{r} \frac{1}{a(u)}du( ∫_{u}^{r} \frac{1}{b(v)}dv)] ∑_{i=1}^{m} q_{i}(r)f(x(σ_{i}(r)))dr ≤ y(t).  (2.38)

Hence, we conclude that lim inf_{t→∞} x(t) = 0. But x(t) is monotone, so we have

lim_{t→∞} x(t) = 0. Thus c = 0 and by (2.32) and (2.33) lim_{t→∞} x^{(i)}(t) = 0, i = 0, 1, 2, which means that x(t) is almost oscillatory. This completes the proof.

Oscillatory behavior of third order differential equations with continuous deviating arguments. Suppose that the following conditions hold unless stated otherwise

(a) a(t) > 0, b(t) > 0, b′(t) ≥ 0, ∫^{∞} \frac{dt}{a(t)} = ∞, ∫^{∞} \frac{dt}{b(t)} = ∞,
(b) q(t, ξ) ∈ C([t_{0}, ∞) × [c, d], ℝ), q(t, ξ) > 0,
(c) \frac{f(x)}{x} ≥ ϵ > 0, for x ≠ 0, ϵ is a constant,

(d) σ(t, ξ) ∈ C([t_{0}, ∞) × [c, d], ℝ), σ(t, ξ) < t, ξ ∈ [c, d], σ(t, ξ) is nondecreasing with respect to t and ξ and

lim_{t→∞} min_{ξ∈[c,d]} σ(t, ξ) = ∞.

Theorem 2.5. If

∫_{t_{1}}^{∞} ∫_{c}^{d} q(s, ξ)dξds = ∞  (2.39)

and

ϵ ∫_{g(t)}^{t} [ ∫_{g(t)}^{r} \frac{1}{a(u)}du( ∫_{u}^{r} \frac{1}{b(v)}dv)] ∫_{c}^{d} q(r, ξ)dξdr > 1,  (2.40)

where g(t) = σ(t, d). Then the equation (1.2) is oscillatory.

Proof. Suppose that x(t) is non-oscillatory solution of (1.2). Without loss of generality we may assume that x(t) is eventually positive. (If x(t) is eventually negative solution, it can be proved by the same arguments). From (1.2), we have

(b(t)(a(t)x′(t))′)′ = − ∫_{c}^{d} q(t, ξ)f(x(σ(t, ξ)))dξ.  (2.41)

Proceeding as in the proof of Theorem 2.1, we have

(b(t)(a(t)x′(t))′)′ ≤ 0,
(a(t)x′(t))′ > 0 and (a(t)x′(t))′′ ≤ 0

for large enough t. Thus, x(t), x′(t) and (a(t)x′(t))′ are monotone and eventually one-signed. From condition (c),

f(x(σ(t, ξ))) ≥ ϵx(σ(t, ξ)) > 0.

Therefore,

0 ≥ (b(t)(a(t)x′(t))′)′ + ϵ ∫_{c}^{d} q(t, ξ)x(σ(t, ξ))dξ.  (2.42)

Now consider again two cases.

Suppose that x′(t) is eventually positive, say x′(t) > 0 for t > t_{2}. Now we can choose a constant k > 0 such that x(k) > 0. By (d), there exist a sufficiently large T such that σ(t, ξ) > k for t > T, ξ ∈ [c, d]. Therefore,

x(σ(t, ξ)) ≥ x(k).

Thus,

(b(t)(a(t)x′(t))′)′ + ϵx(k) ∫_{c}^{d} q(t, ξ)dξ ≤ 0.  (2.43)

Integrating this last equation from t_{1} to t, we get

b(t)(a(t)x′(t))′ ≤ b(t_{1})(a(t_{1})x′(t_{1}))′ − ϵx(k) ∫_{t_{1}}^{t} ∫_{c}^{d} q(s, ξ)dξds.  (2.44)

Taking the limit of both sides as t → ∞ and using (2.39), the last inequality above leads to a contradiction to (a(t)x′(t))′ > 0. Now suppose x′(t) is eventually negative. Proceeding as in the proof of Theorem 2.1 and integrating equation (1.2) three times, we get

∫_{t}^{∞} [ ∫_{t}^{r} \frac{1}{a(u)}du( ∫_{u}^{r} \frac{1}{b(v)}dv)] ∫_{c}^{d} q(r, ξ)f(x(σ(r, ξ)))dξdr ≤ x(t)  (2.45)

Using (c) in (2.45), we obtain

∫_{t}^{∞} [ ∫_{t}^{r} \frac{1}{a(u)}du( ∫_{u}^{r} \frac{1}{b(v)}dv)]ϵ ∫_{c}^{d} q(r, ξ)x(σ(r, ξ))dξdr ≤ x(t).  (2.46)

Replace t by g(t) in (2.46), where g(t) = σ(t, d), then we have

ϵ ∫_{g(t)}^{t} [ ∫_{g(t)}^{r} \frac{1}{a(u)}du( ∫_{u}^{r} \frac{1}{b(v)}dv)] ∫_{c}^{d} q(r, ξ)x(σ(r, ξ))dξdr ≤ x(g(t)). (2.47)

Since x(t) is decreasing and positive,

ϵ ∫_{g(t)}^{t} [ ∫_{g(t)}^{r} \frac{1}{a(u)}du( ∫_{u}^{r} \frac{1}{b(v)}dv)] ∫_{c}^{d} q(r, ξ)dξdr ≤ 1.

This is a contradiction to (2.40). Therefore, the proof is complete.

Example 2.6. Consider the following functional differential equation

x′′′ + ∫_{2/7π}^{1/2π} \frac{2e^{−1/ξ}}{ξ^{2}} x(t − \frac{1}{ξ})dξ = 0

so that a(t) = 1, b(t) = 1, f(x) = x, q(t, ξ) = \frac{2e^{−1/ξ}}{ξ^{2}} , σ(t, ξ) = t − \frac{1}{ξ}. We can easily see that the conditions of Theorem 2.5 are satisfied. It is easy to verify that x(t) = e^{−t} sin t is a solution of this problem.

Theorem 2.7. Suppose (2.40) holds. In addition to that suppose there exist p ∈ C([t_{0}, ∞), ℝ), p(t) > 0 such that

∫ ^{∞} [Γ(t)p(t) − \frac{a(σ(t, c))b(t)(p′(t))^{2}}{(σ(t, c) − T)σ′(t, c)4p(t)}]dt = ∞,  (2.48)

where Γ(t) = ϵ ∫_{c}^{d} q(t, ξ)dξ. Then the equation (1.2) is oscillatory.

Proof. Suppose that x(t) is non-oscillatory solution of (1.2). We can assume that x(t) is eventually positive. The case of x(t) is eventually negative can be proved by the same arguments. Proceeding as in the proof of Theorem 2.1, we have

(b(t)(a(t)x′(t))′)′ ≤ 0,
(a(t)x′(t))′ > 0 and (a(t)x′(t))′′ ≤ 0.

Thus, x(t), x′(t) and (a(t)x′(t))′ are monotone and eventually one-signed. From condition (c),

f(x(σ(t, ξ))) ≥ ϵx(σ(t, ξ)) > 0.
(b(t)(a(t)x′(t))′)′ + ϵ ∫_{c}^{d} q(t, ξ)x(σ(t, ξ))dξ ≤ 0.  (2.49)

If x′(t) is eventually positive, then we can define

z(t) = \frac{b(t)(a(t)x′(t))′}{x(σ(t, c))} p(t).

It is obvious that z(t) > 0 for t ≥ t_{2} and z′(t) is

z′(t) = \frac{(b(t)(a(t)x′(t))′)′}{x(σ(t, c))} p(t) + \frac{p′(t)}{p(t)}z(t) − \frac{x′(σ(t, c))σ′(t, c)}{x(σ(t, c))}z(t).  (2.50)

From proof of Theorem 2.1, we have

a(t)x′(t) ≥ (t − T)(a(t)x′(t))′.

Since (a(t)x(t))′ is non-increasing, we have

a(σ(t, c))x′(σ(t, c)) ≥ (σ(t, c) − T)(a(t)x′(t))′,

then

x′(σ(t, c)) ≥ \frac{(σ(t, c) − T)(a(t)x′(t))′}{a(σ(t, c))} .  (2.51)

Plug (2.51) in (2.50), then we obtain

z′(t) = \frac{(b(t)(a(t)x′(t))′)′}{x(σ(t, c))} p(t) + \frac{p′(t)}{p(t)}z(t) − \frac{(σ(t, c) − T)σ′(t, c)}{p(t)b(t)a(σ(t, c)))}z^{2}(t).

Completing the square leads to

z′(t) ≤ −Γ(t)p(t) + \frac{b(t)a(σ(t, c))(p′(t))^{2}}{(σ(t, c) − T)σ′(t, c))4p(t)}.  (2.52)

Integrating (2.52) from T to t and letting t → ∞, we see that lim_{t→∞} z(t) = −∞. This contradicts z(t) being eventually positive.

If x′(t) is eventually negative, the proof is exactly the same as in the second part of the proof of previous Theorem.

References

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[8] V. Sree Hari Rao and R. S. Dahiya, Properties of solutions of a class of third-order linear differential equations, Period. Math. Hungarica 20(3) (1989), 177-184.

[9] A. S. Tantawy, “Oscillation of Third and Higher Order Functional Differential Equations, Ph.D Thesis, ISU (1989).

[10] P. Waltman, Oscillation criteria for third-order nonlinear differential equations, Pasific J. Math. 18 (1966), 385-389.

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Tuncay Candan Department of Mathematics, Iowa State University Ames, IA 50011, USA

E-mail address: tuncayca@iastate.edu

Rajbir S. Dahiya Department of Mathematics, Iowa State University Ames, IA 50011, USA

E-mail address: dahiya@math.iastate.edu