Zeitschrift für Analysis und ihre Anwendungen

Journal for Analysis and its Applications Volume 18 (1999), No. 4, 1107–1116

Oscillations of Certain Delay Integro-Differential Equations

Y. H. Wang and B. G. Zhang

Abstract. In this paper, the oscillation of solutions of certain delay integro-differential systems are considered. Sufficient conditions for the non-existence of non-oscillatory solutions for these systems are obtained. Comparison results are obtained also.

Keywords: Integro-differential systems, oscillation, non-oscillation, comparison theorems AMS subject classification: 34 A 15

1. Introduction

Recently the oscillation in systems of functional-differential equations has received attention in the literature [1 - 6]. In this paper we consider the delay integro-differential system

x′_{i}(t) + ∑_{j=1}^{n} ∫_{0}^{t} D_{i,j}(t − s)x_{j}(s − τ) ds = 0  (i = 1, 2, …, n)  (1)

and

x′_{i}(t) + ∑_{j=1}^{n} ∫_{0}^{t} D_{i,j}(t − s)x_{j}(s − τ) ds = f_{i}(t)  (i = 1, 2, …, n)  (2)

with initial condition

x(s) = ϕ(s)  (s ∈ [−τ, 0])  (3)

where τ > 0, D_{i,j} ∈ C(ℝ_{+}, ℝ_{+}) and ϕ ∈ C([−τ, 0]). System (1) with n = 1 has been considered in [7].

By a solution of (1) - (3) we mean a vector x = (x_{1}, …, x_{n})^{T}, which is continuous on [−τ, ∞) and continuously differentiable and satisfies (1) for t ≥ 0 and such that (3) holds. By a solution of (1) on [T, ∞) we mean a vector x ∈ ℝ^{n} which is continuous on [−τ, ∞) and continuously differentiable and satisfies (1) for t ≥ T. A vector x ∈ ℝ^{n} is said to be non-oscillatory, if x_{i}(t) ≠ 0 on [−τ, ∞) (i = 1, 2, …, n), and to be positive if every component of x is positive on [−τ, ∞).

Y. H. Wang: Yantai University, Dept. Math., Yantai, Shandong, 264005, China B. G.Zhang: Ocean University of Qingdao, Dept. Appl. Math., Qingdao, 266003, China The research was supported by NNSF of China ISSN 0232-2064 / $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag Berlin

We will reduce the non-existence of non-oscillatory solutions of (1) to the same for the scalar integro-differential inequality

y′(t) + ∫_{0}^{t} D(t − s)y(s − τ) ds ≤ 0  (4)

where

D(t) = \{ \begin{matrix}D_{11}(t) & for n = 1 \\ min_{1≤j≤n}[D_{jj}(t) − ∑_{i=1,i≠j}^{n} |D_{i,j}(t)|] & for n > 1 \end{matrix} (t ≥ 0).

By a solution of (4) on [T, ∞) we mean a function y ∈ C([−τ, ∞), ℝ) ∩ C^{1}([T, ∞), ℝ) which satisfies (4) on [T, ∞) where T ≥ 0. A solution y of (4) is called positive, if y > 0 on [−τ, +∞).

2. Main results

Let us start with the following theorem.

Theorem 1. If (4) has no positive solutions, then (1) has no non-oscillatory solutions.

Proof. Suppose the contrary, let x be a non-oscillatory solution of (1) with x_{i}(t) ≠ 0 (t ≥ −τ; i = 1, 2, …, n) and set δ_{i} = \frac{x_{i}(t)}{|x_{i}(t)|}. Then w = (w_{1}, w_{2}, …, w_{n})^{T} = (δ_{1}x_{1}, …, δ_{n}x_{n})^{T} satisfies w_{i}(t) > 0 (t ≥ −τ) and

w′_{i}(t) + ∑_{j=1}^{n} ∫_{0}^{t} D̄_{i,j}(t − s)w_{j}(s − τ) ds = 0  (t ≥ 0)  (5)

where D̄_{i,j} = \frac{δ_{i}}{δ_{j}} D_{i,j}. Clearly, D̄_{i,i} = D_{i,i} and |D̄_{i,j}| = |D_{i,j}| (i ≠ j). Define y(t) = ∑_{i=1}^{n} w_{i}(t) > 0 (t ≥ −τ). Summing (5), we have

0 = ∑_{i=1}^{n} w′_{i}(t) + ∑_{i=1}^{n} ∑_{j=1}^{n} ∫_{0}^{t} D̄_{i,j}(t − s)w_{j}(s − τ) ds

= y′(t) + ∑_{j=1}^{n} ∫_{0}^{t} D̄_{j,j}(t − s)w_{j}(s − τ) ds + ∑_{i,j=1,i≠j}^{n} ∫_{0}^{t} D̄_{i,j}(t − s)w_{j}(s − τ) ds

≥ y′(t) + ∑_{j=1}^{n} ∫_{0}^{t} (D̄_{j,j}(t − s) − ∑_{i,j=1,i≠j}^{n} |D̄_{i,j}(t − s)|)w_{j}(s − τ) ds

≥ y′(t) + ∫_{0}^{t} D(t − s)y(s − τ) ds

which is a contradiction Theorem 2. Assume that D_{ii}(t) ≢ 0 on [0, τ]. If

z′_{i}(t) + ∑_{j=1}^{n} ∫_{0}^{t} D_{i,j}(t − s)z_{j}(s − τ) ds ≤ 0  (i = 1, 2, …, n)  (6)

has a positive solution z(t) on [−τ, ∞), then (1) has a positive solution x(t) with 0 < x(t) ≤ z(t) (t ≥ 0) and lim_{t→∞} x(t) = 0.

Proof. Define the set

X = {x ∈ C([−τ, ∞), ℝ^{n})| 0 ≤ x(t) ≤ z(t) (t ≥ −τ), x(t) > 0 on [−τ, 0]}

and an operator S on X by

\begin{matrix}(Sx_{i})(t) = \\ \{ & \begin{matrix}∑_{j=1}^{n} ∫_{t}^{∞}∫_{0}^{u} D_{i,j}(u − s)x_{j}(s − τ)dsdu & if t ≥ 0 \\ (∑_{j=1}^{n} ∫_{0}^{∞}∫_{0}^{u} D_{i,j}(u − s)x_{j}(s − τ)dsdu)(1 − \frac{|t|}{τ})\frac{z_{i}(t)}{z_{i}(0)} + z_{i}(t)\frac{|t|}{τ} & if −τ ≤ t ≤ 0 \end{matrix} \end{matrix} (7)

for i = 1, 2, …, n. Clearly, (Sx_{i})(t) ≤ z_{i}(t) (t ≥ −τ) and (Sx_{i})(t) > 0 on [−τ, 0], i.e. SX ⊂ X. Define the sequences {y_{m}(t)}, y_{m} ∈ ℝ^{n} by

\begin{matrix}y_{0} & = z \\ y_{m} & = Sy_{m−1} (m ∈ ℕ). \end{matrix} \}

Then

y_{0} ≥ y_{1} ≥ … ≥ y_{m}(t) ≥ … (t ≥ −τ)

and hence lim_{m→∞} y_{m}(t) = y(t) (t ≥ −τ) exists. It is easy to see that y(t) is a solution of (1) for t ≥ 0 and y(t) ≤ z(t). We claim that y(t) > 0 (t ≥ 0). From (7) and the condition D_{i,i}(t) ≢ 0 on [0, τ] we have y(t) > 0 on [−τ, 0]. If ξ is the first zero of y_{i} on [0, ∞), i.e. y_{i}(t) > 0 (0 ≤ t < ξ) and y_{i}(ξ) = 0, then

0 = y_{i}(ξ)

= ∑_{j=1}^{n} ∫_{ξ}^{∞}∫_{0}^{u} D_{i,j}(u − s)y_{j}(s − τ) dsdu

≥ ∫_{ξ}^{∞}∫_{0}^{u} D_{ii}(u − s)y_{i}(s − τ) dsdu

which implies that ∫_{0}^{u} D_{i,i}(u−s)y_{i}(s−τ) ds = 0, which contradicts the fact that D_{i,i} ≢ 0 on [0, τ] and y_{i}(s) > 0 (s ∈ [−τ, 0]). This contradiction proves the Theorem From Theorem 2 we can derive a comparison theorem. Corollary 1. If D_{i,j}(t) ≤ E_{i,j}(t) (t ≥ 0) and

y′_{i}(t) + ∑_{j=1}^{n} ∫_{0}^{t} E_{i,j}(t − s)y_{j}(s − τ) ds = 0 (i = 1, 2, …, n)

has a positive solution y(t), then (1) has a positive solution x(t) with x(t) ≤ y(t). Proof. In fact, we have

y′_{i}(t) + ∑_{j=1}^{n} ∫_{0}^{t} D_{i,j}(t − s)y_{j}(t − τ) ds ≤ y′_{i}(t) + ∑_{j=1}^{n} ∫_{0}^{t} E_{i,j}(t − s)y_{j}(t − τ) ds = 0.

By Theorem 2, (1) has a positive solution Corollary 2. Let d(t) = max_{1≤i≤n} ∑_{j=1}^{n} D_{i,j}(t). If there exists λ > 0 such that

−λ + e^{λτ} ∫_{0}^{∞}d(s)e^{λs}ds ≤ 0,

then system (1) has a positive solution x with

0 < x_{i}(t) ≤ e^{−λt}  (t ≥ 0; i = 1, 2, …, n).  (8)

Proof. In fact, let z_{i}(t) = e^{−λt} (i = 1, 2, …, n). Then

z′_{i}(t) + ∑_{j=1}^{n} ∫_{0}^{t} D_{i,j}(t − s)z_{j}(s − τ) ds

= z′_{i}(t) + ∑_{j=1}^{n} ∫_{0}^{t} D_{i,j}(s)z_{j}(t − s − τ) ds

= e^{−λt}[ − λ + ∑_{j=1}^{n} e^{λτ} ∫_{0}^{t} D_{i,j}(s)e^{λs} ds]

≤ e^{−λt}[ − λ + e^{λτ} ∫_{0}^{t} d(s)e^{λs}ds]

≤ 0.

By Theorem 2, (1) has a solution x, which satisfies (8)

Now we want to show a sufficient condition for the non-existence of positive solutions of (4). Let

\begin{aligned}D_{1}(t) & = ∫_{t−τ}^{t} ∫_{0}^{u} D(u − s) dsdu \\ D_{n}(t) & = ∫_{t−τ}^{t} ∫_{0}^{u} D(u − s)D_{n−1}(s) dsdu (n ≥ 2) \end{aligned}

and

\begin{aligned}p_{0} & = min_{0≤t≤nτ} D_{n}(t) \\ p_{k} & = min_{knτ≤t≤(k+1)nτ} D_{n}(t) (k ≥ 1). \end{aligned}

Lemma 1. Assume that

D_{n}(t) ≥ \frac{1}{e^{n}} .  (9)

Let y(t) be a positive solution of (4). Then

\frac{y(t − τ)}{y(t)} ≤ 4e^{2n}.  (10)

Proof. Integrating (4), we have

y(t) − y(t − τ) + ∫_{t−τ}^{t} ∫_{0}^{u} D(u − s)y(s − τ) dsdu ≤ 0.  (11)

Thus

\begin{aligned}y(t − τ) & > ∫_{t−τ}^{t} ∫_{0}^{u} D(u − s)y(s − τ) dsdu \\ & ≥ y(t − τ) ∫_{t−τ}^{t} ∫_{0}^{u} D(u − s) dsdu \\ & = D_{1}(t)y(t − τ). \end{aligned}  (12)

Substituting (12) into (11) we obtain y(t − τ) > D_{2}(t)y(t − τ). In general, we have

y(t − τ) > y(t − τ)D_{n}(t).  (13)

Assumption (9) implies that there exists t^{∗} ∈ [t − τ, t] such that

∫_{t−τ}^{t^{∗}} ∫_{0}^{u} D(u − s)D_{n−1}(s) dsdu ≥ \frac{1}{2e^{n}} and ∫_{t^{∗}}^{t} ∫_{0}^{u} D(u − s)D_{n−1}(s) dsdu ≥ \frac{1}{2e^{n}} .

Integrating (4), we have

y(t^{∗}) − y(t − τ) + ∫_{t−τ}^{t^{∗}} ∫_{0}^{u} D(u − s)y(s − τ) dsdu ≤ 0.  (14)

In view of (13) and (14), we obtain

y(t − τ) > ∫_{t−τ}^{t^{∗}} ∫_{0}^{u} D(u − s)y(s − τ) dsdu

≥ ∫_{t−τ}^{t^{∗}} ∫_{0}^{u} D(u − s)D_{n−1}(s)y(s − τ) dsdu

≥ \frac{y(t^{∗} − τ)}{2e^{n}}.

Similarly, we have y(t^{∗}) > \frac{y(t−τ)}{2e^{n}}. Combining the above two inequalities we have \frac{y(t^{∗}−τ)}{y(t^{∗})} < (2e^{n})^{2} and the proof is complete

Lemma 2. Assume that (9) holds. Let y(t) be a positive solution of (4) and define

M = min_{n(m−1)τ≤t≤mnτ} \frac{y(t − τ)}{y(t)}  and  N = min_{mnτ≤t≤(m+1)nτ} \frac{y(t − τ)}{y(t)}.

Then M > 1 and

N ≥ exp(e^{n−1}Mp_{m}) ≥ exp (\frac{M}{e} ) ≥ M.  (15)

Proof. M > 1 is obvious. Dividing (4) by y(t) and integrating it we have

\frac{y(t − τ)}{y(t)} ≥ exp ∫_{t−τ}^{t} \frac{1}{y(s)} ∫_{0}^{s} D(s − u)y(u − τ) duds.  (16)

Let

d_{k} = min_{(mn+k−1)τ≤t≤(mn+k)τ}{D_{n}(t)}
N_{k} = min_{(mn+k−1)τ≤t≤(mn+k)τ} \frac{y(t − τ)}{y(t)}
M_{l} = min_{((m−1)n+l)τ≤t≤(mn+l)τ} \frac{y(t − τ)}{y(t)}

for k, l = 1, 2, …, n. By definition, p_{m} ≤ d_{k} (k = 1, 2, …, n) and N = min_{1≤k≤n} N_{k}.

Oscillations of Integro-Differential Equations 1113

For t ∈ [mnτ, (mn + 1)τ], from (16) and the inequality e^{x} ≥ ex for x > 0, we have

\frac{y(t − τ)}{y(t)} ≥ exp ∫_{t−τ}^{t} ∫_{0}^{s_{1}}D(s_{1} − u)\frac{y(u − τ)}{y(u)} duds_{1}

≥ exp ∫_{t−τ}^{t} ∫_{0}^{s_{1}}D(s_{1} − u) exp ∫_{u−τ}^{u} ∫_{0}^{s_{2}}D(s_{2} − u_{1})\frac{y(u_{1} − τ)}{y(u_{1})} du_{1}ds_{2}duds_{1}

≥ exp (e ∫_{t−τ}^{t} ∫_{0}^{s_{1}}D(s_{1} − u) ∫_{u−τ}^{u} ∫_{0}^{s_{2}}D(s_{2} − u_{1})\frac{y(u_{1} − τ)}{y(u_{1})}du_{1}ds_{2}duds_{1})

≥ ⋮

≥ exp (e^{n−1} ∫_{t−τ}^{t} ∫_{0}^{s_{1}}D(s_{1} − u) ∫_{u−τ}^{u} ∫_{0}^{s_{2}}D(s_{2} − u_{2}) …

… ∫_{u_{n−2}−τ}^{u_{n−2}} ∫_{0}^{s_{n}}D(s_{n} − u_{n−1})\frac{y(u_{n−1} − τ)}{y(u_{n−1})} du_{n−1}ds_{n} … duds_{1})

≥ exp (e^{n−1} ∫_{t−τ}^{t} ∫_{0}^{s_{1}}D(s_{1} − u) ∫_{u−τ}^{u} ∫_{0}^{s_{2}}D(s_{2} − u_{2}) …

… ∫_{u_{n−2}−τ}^{u_{n−2}} \frac{y(s_{n} − τ)}{y(s_{n})} ∫_{0}^{s_{n}}D(s_{n} − u_{n−1}) du_{n−1}ds_{n} … duds_{1})

where s_{n} ∈ [(m − 1)nτ, (mn + 1)τ]. In view of the inequality exp(\frac{x}{e}) > x for x ≠ e we have

N_{1} ≥ exp (e^{n−1} min(M, N_{1})d_{1})

≥ exp (e^{n−1} min(M, N_{1})p_{m})

≥ exp (\frac{min(M, N_{1})}{e})

> min(M, N_{1}).

Hence min(M, N_{1}) = M. Therefore

N_{1} ≥ exp(e^{n−1}Mp_{m}) ≥ exp (\frac{M}{e} ) ≥ M

and M_{1} ≥ min(M, N_{1}) ≥ M. Similar to the above, we can prove that

N_{k} ≥ exp(e^{n−1}Mp_{m}) ≥ exp (\frac{M}{e} ) ≥ M (k = 1, 2, …, n).

1114 Y. H. Wang and B. G. Zhang

Therefore

N = min_{1≤k≤n} N_{k} ≥ exp(e^{n−1}Mp_{m}) ≥ exp (\frac{M}{e} ) ≥ M

and the proof is complete Theorem 3. Assume that (9) holds and

∑_{i=1}^{∞} (p_{i} − \frac{1}{e^{n}} ) = ∞.  (17)

Then (4) has no positive solutions.

Proof. Suppose the contrary, let y(t) be a positive solution of (4). Define the sequence {N̄_{i}} by

N̄_{i} = min_{(k+i−1)nτ≤t≤(k+i)nτ} \frac{y(t − τ)}{y(t)}  (i ≥ 0).

By Lemma 2, we have N̄_{i} > 1 and

N̄_{i+1} ≥ exp(e^{n−1}N̄_{i}p_{k+i})

≥ exp (\frac{N̄_{i}}{e} ) exp (N̄_{i}(e^{n−1}p_{k+i} − \frac{1}{e}))

≥ exp (\frac{N̄_{i}}{e} )

> N̄_{i}.

(18)

Therefore {N̄_{i}} is increasing. On the other hand, by Lemma 1, {N̄_{i}} is bounded. Hence lim_{i→∞} N̄_{i} = N̄ exists. From the last inequality, we have N̄ ≥ exp(\frac{N̄}{e}) > N̄ if N̄ ≠ e which implies that N̄ = e. From (18),

N̄_{i+1} ≥ N̄_{i}(1 + e^{n−1}N̄_{i}(p_{k+i} − e^{−n})).

Hence

N̄_{i+1} − N̄_{i} ≥ e^{n−1}N̄_{i}^{2}(p_{k+i} − e^{−n})

and

N̄_{i+2} − N̄_{i+1} ≥ e^{n−1}N̄_{i+1}^{2}(p_{k+i+1} − e^{−n}) > e^{n−1}N̄_{i}^{2}(p_{k+i+1} − e^{−n}).

Summing up the above inequality, we get

e − N̄_{i} > e^{n−1}N̄_{i}^{2} ∑_{j=i}^{∞}(p_{k+j} − e^{−n})

for some k, which contradicts (17)

Oscillations of Integro-Differential Equations 1115

Corollary 3. If there exists a positive integer n such that

lim inf_{t→∞} D_{n}(t) > \frac{1}{e^{n}} ,  (19)

then (4) has no positive solutions.

In fact, (19) implies (17). Combining Theorem 1 and Corollary 3 we have the following result.

Corollary 4. If (19) holds, then (1) has no non-oscillatory solutions.

Now we consider the forced system (2).

Theorem 4. Let F (t) = ∑_{i=1}^{n} δ_{i}f_{i}(t) with δ_{i} = ±1 and F (t) = h′(t), h_{+}(t) = max(h(t), 0) ≢ 0, h_{−}(t) = max(−h(t), 0) ≢ 0, and

∫_{0}^{∞}∫_{0}^{t} D(t − s)h_{+}(s − τ) dsdt = ∞.  (20)

Then (2) has no non-oscillatory solutions.

Proof. Suppose the contrary, let {x_{i}(t)} (i = 1, 2, …, n) be a non-oscillatory solution of (2). Then we have

\begin{aligned}δ_{i}x′_{i}(t) + ∑_{j=1}^{n} ∫_{0}^{t} D_{i,j}(t − s)δ_{i}x_{j}(s − τ) ds & = δ_{i}f_{i}(t) \\ w′_{i}(t) + ∑_{j=1}^{n} ∫_{0}^{∞}D_{i,j}(t − s)δ_{i}δ_{j}^{−1}w_{j}(s − τ) ds & = δ_{i}f_{i}(t). \end{aligned}

That is,

w′_{i}(t) + ∑_{i=1}^{n} ∫_{0}^{t} D̄_{i,j}(t − s)w_{j}(s − τ) ds = δ_{i}f_{i}(t).

Summing the above equation, we obtain

\begin{aligned}∑_{i=1}^{n} w′_{i}(t) & + ∑_{j=1}^{n} ∫_{0}^{t} D̄_{i,j}D̄_{j,j}(t − s)w_{j}(s − τ) ds \\ & + ∑_{i,j=1,i≠j}^{n} ∫_{0}^{t} D̄_{i,j}(t − s)w_{j}(s − τ) ds = ∑_{i=1}^{n} δ_{i}f_{i}(t). \end{aligned}

Hence

y′(t) + ∑_{j=1}^{n} ∫_{0}^{t} (D̄_{j,j}(t − s) − ∑_{i,j=1,i≠j} |D̄_{i,j}(t − s)|)w_{j}(s − τ) ds ≤ ∑_{i=1}^{n} δ_{i}f_{i}(t)

and hence

y′(t) + ∫_{0}^{t} D(t − s)y(s − τ) ds ≤ F (t).

Thus,

(y(t) − h(t))′ + ∫_{0}^{t} D(t − s)y(s − τ) ds ≤ 0.  (21)

If y(t) > 0 eventually, then y(t) − h(t) is non-increasing. There are two possible cases:

(i) y(t) − h(t) ≤ 0 eventually and

(ii) y(t) − h(t) ≥ 0 eventually.

For the case (i), y(t) ≤ h(t) eventually, which contradicts the positivity of y. Therefore, the case (ii) holds. Hence y(t) ≥ h_{+}(t) eventually. From this and (21), we obtain

(y(t) − h(t))′ + ∫_{0}^{t} D(t − s)h_{+}(s − τ) ds ≤ 0.  (22)

This together with condition (20) lead to a contradiction and the proof is completed

Remark 1. (2) has no positive solution, if (20) holds, where F (t) = ∑_{i=1}^{n} f_{i}(t).

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[3] Kong, Q. and H. L. Freedman: Oscillation in delay differential systems. Diff. Int. Equ. 6 (1993), 1325 – 1336.

[4] Qingkai Kong: Oscillation for systems of functional differential equations. J. Math. Anal. Appl. 198 (1996), 608 – 619.

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Received 05.01.1999