OSCILLATION IN NEUTRAL PARTIAL FUNCTIONAL DIFFERENTIAL EQUATIONS AND INEQUALITIES

Xilin Fu^{1} and Jianhong Wu^{2}

1 Department of Mathematics, Shandong Normal University, Jinan 250014, P. R. China 2 Department of Mathematics and Statistics, York University North York, Ontario, M3J 1P3, Canada

Abstract: We derive some sufficient conditions for certain classes of ordinary differential inequalities of neutral type with distributed delay not to have eventually positive or negative solutions. These, together with the technique of spatial average, the Green’s Theorem and Jensen’s inequality, yield some sufficient conditions for all solutions of a class of neutral partial functional differential equations to be oscillatory. An example is given to illustrate the result. AMS (MOS) Subject Classification. 34K40

1. INTRODUCTION

Consider the following scalar neutral partial functional differential equation

\begin{matrix}\frac{∂}{∂t}[u(x, t) − ∑_{i=1}^{m} λ_{i}(t)u(x, t − r_{i})] + p(x, t)u(x, t) + ∫_{α}^{β} q(x, t, s)F [u(x, h(t, s))]dσ(s) \\ = a(t)Δu(x, t) + ∑_{j=1}^{n} a_{j}(t)Δu(x, τ_{j}(t)), (x, t) ∈ Ω × [0, ∞) \end{matrix}

(1)

subject to either the Dirichlet boundary condition

u(x, t) = 0, (x, t) ∈ ∂Ω × [0, ∞)  (2)

or the Neumann boundary condition

\frac{∂}{∂ν} u(x, t) + r(x, t)u(x, t) = 0, (x, t) ∈ ∂Ω × [0, ∞),  (3)

where Ω is a bounded domain in R^{N} with piecewise smooth boundary ∂Ω, Δ is the Lapalcian operator in R^{N}, r_{i} are given positive constants, λ_{i} ∈ C(R^{+}; R^{+}) for all 1 ≤ i ≤ m, p ∈ C(Ω̄ × R^{+}; R^{+}), q ∈ C(Ω̄ × R^{+} × J; R^{+}), J = [α, β] with two given reals α ≤ β, F ∈ C(R; R), h ∈ C(R^{+} × J; R) and h(t, s) is nondecreasing with respect to t and s, respectively, with h(t, s) < t for all t ∈ R^{+} and s ∈ J and lim_{t→∞} min_{s∈J} h(t, s) = ∞, σ : J → R is nondecreasing, a, a_{j} ∈ C(R^{+}; R^{+}), j = 1 ⋯, n, τ_{j} ∈ C(R^{+}; R^{+}) is nondecreasing, τ_{j}(t) < t and lim_{t→∞} τ_{j}(t) = ∞ for all 1 ≤ j ≤ n, ν is the unit exterior normal vector to ∂Ω and r ∈ C(∂Ω × R^{+}; R^{+}).

We say a solution of (1)subject to either (2) or (3) is oscillatory in the domain Ω×R^{+} if for every positive real τ there exists a point (x_{0}, t_{0}) ∈ Ω×[τ, ∞) such that u(x_{0}, τ_{0}) = 0.

Our goal is to give sufficient conditions guaranteeing all solutions of the problem (1)-(2) or (1)-(3) are oscillatory. One of the typical results we will prove is as follows:

Theorem 1.1. Assume that

F : R → R is odd, convex and inf_{y>0} F (y)/y > 0;  (4)

∑_{i=1}^{m} λ_{i}(t) = 1, and there is i_{0} such that lim_{t→∞} λ_{i_{0}}(t) = λ_{0} > 0.  (5)

∫ ^{∞} ∫_{α}^{β} min_{x∈Ω} q(x, s, χ)dσ(χ)ds = ∞.  (6)

Then every solution of (1) subject to either (2) or (3) is oscillatory in Ω × R^{+}.

Our approach is to show that if u is a solution of (1) subject to, for example, the Neumann boundary condition, then

U(t) = \frac{1}{∫_{Ω} ϕ(x)ds} ∫_{Ω} u(x, t)ϕ(x)de  (7)

with large t satisfies the scalar functional differential inequality of neutral type

\frac{d}{dt}[U(t) − ∑_{i=1}^{m} U(t − r_{i})] + ∫_{α}^{β} min_{x∈Ω̄} q(x, t, χ)F [U(h(t, χ))]dσ(χ) ≤ 0,  (8)

where ϕ is a positive eigenfunction associated with the smallest eigenvalue of the Dirichlet problem

Δu + μu = 0 in Ω and u|_{∂Ω} = 0.  (9)

This enables us to apply our general sufficient conditions for scalar differential inequalities of neutral type not to have eventually positive or negative solutions.

The rest of this paper is organized as follows: in Section 2, we collect several technical lemmas concerning the operator

z(t) = y(t) − ∑_{j=1}^{m} λ_{i}(t)y(t − r_{i})  (10)

and the relation between oscillations and convergence for a class of ordinary differntial inequalities of neutral type. Section 3 contains our main results on sufficient conditions for the ordinary differential inequalities of neutral type under consideration not to have eventually positive solutions. Finally, in Section 4, we apply these general results to prove Theorem 1.1 and other general results for all solutions of the partial functional differential equation (1) subject to certain boundary conditions to be oscillatory.

2. TECHNICAL LEMMAS

We consider the following first order nonlinear differential inequality with distributed deviating arguments

\frac{d}{dt}[y(t) − ∑_{i=1}^{m} λ_{i}(t)y(t − r_{i})] + ∫_{α}^{β} Q(t, s)F [y(h(t, s))]dσ(s) ≤ 0,  (11)

where r_{i} are given positive constants, λ_{i} ∈ C(R^{+}; R^{+}) for all 1 ≤ i ≤ m, Q ∈ C(R^{+} × J; R^{+}), J = [α, β] with two given reals α ≤ β, F ∈ C(R; R), h ∈ C(R^{+}×J; R) and h(t, s) is nondecreasing with respect to t and s, respectively, with h(t, s) < t for all t ∈ R^{+} and s ∈ J and lim_{t→∞} min_{s∈J} h(t, s) = ∞, and σ : J → R is nondecreasing.

We will derive some sufficient conditions for inequality (11) not to have any eventually positive solutions. Note that if we let F^{∗}(y) = −F (−y) for all y ∈ R, then any solution y of (11) yields a solution −y for the differntial inequality

\frac{d}{dt}[y(t) − ∑_{i=1}^{m} λ_{i}(t)y(t − r_{i})] + ∫_{α}^{β} Q(t, s)F^{∗}[y(h(t, s))]dσ(s) ≥ 0.  (12)

Therefore, one should easily obtain similar sufficient conditions for (12) not to have any eventually negative solutions. This also shows all of the sufficient conditions stated below for inequality (11) not to have eventually positive solutions can be modified to those sufficient conditions for all solutions of the following neutral functional differntial equation

\frac{d}{dt}[y(t) − ∑_{i=1}^{m} λ_{i}(t)y(t − r_{i})] + ∫_{α}^{β} Q(t, s)F [y(h(t, s))]dσ(s) = 0  (13)

to be oscillatory (i.e., to have arbitrarily large zeros).

LEMMA 2.1. Assume that

F (y) ≥ 0 for any y > 0  (14)

and that there exist i_{0} ∈ {1, ⋯, m} such that

lim_{t→∞} ∑_{i=1}^{m} λ_{i}(t) = λ ∈ (0, 1], lim_{t→∞} λ_{i_{0}}(t) = λ_{0} > 0.  (15)

Consider a bounded and eventually positive solution y of inequality (11) and let

z(t) = y(t) − ∑_{i=1}^{m} λ_{i}(t)y(t − r_{i}), t ∈ R^{+}.  (16)

Then we have (i) lim_{t→∞} z(t) = 0 in case where λ = 1; (ii) lim_{t→∞} y(t) exists in case where λ < 1.

Proof. Clearly, z is bounded on R^{+} and for large t ≥ 0, we have

z′(t) ≤ − ∫_{α}^{β} Q(t, s)F [y(h(t, s))]dσ(s) ≤ 0.

Therefore, lim_{t→∞} z(t) = z(∞) exists.

On the other hand, since y is bounded on R^{+}, we can find two sequences of nonnegative reals {t_{n}} and {t_{n}^{∗}} such that

lim_{n→∞} t_{n} = lim_{n→∞} t_{n}^{∗} = ∞

and

lim_{n→∞} y(t_{n}) = lim sup_{t→∞} y(t) = A, lim_{n→∞} y(t_{n}^{∗}) = lim inf_{t→∞} y(t) = B.

For each ϵ > 0 there is a positive integer N such that for all n ≥ N, we have

λ_{i_{0}}(t_{n}) > 0, λ_{i_{0}}(t_{n}^{∗}) > 0

and

y(t_{n} − r_{i}) ≤ A + ϵ, y(t_{n}^{∗} − r_{i}) ≥ B − ϵ, i = 1, ⋯, m.

Therefore, for n ≥ N, we get

\begin{aligned}y(t_{n} − r_{i_{0}}) & = \frac{1}{λ_{i_{0}}(t_{n})}[y(t_{n}) − z(t_{n}) − ∑_{i=1,i≠i_{0}}^{m} λ_{i}(t_{n})y(t_{n} − r_{i})] \\ & ≥ \frac{1}{λ_{i_{0}}(t_{n})}[y(t_{n}) − z(t_{n}) − (A + ϵ) ∑_{i=1,i≠i_{0}}^{m} λ_{i}(t_{n})] \end{aligned}

and

\begin{aligned}y(t_{n}^{∗} − r_{i_{0}}) & = \frac{1}{λ_{i_{0}}(t_{n}^{∗})}[y(t_{n}^{∗}) − z(t_{n}^{∗}) − ∑_{i=1,i≠i_{0}}^{m} λ_{i}(t_{n}^{∗})y(t_{n}^{∗} − r_{i})] \\ & ≤ \frac{1}{λ_{i_{0}}(t_{n}^{∗})}[y(t_{n}^{∗}) − z(t_{n}^{∗}) − (B − ϵ) ∑_{i=1,i≠i_{0}}^{m} λ_{i}(t_{n}^{∗})]. \end{aligned}

Taking n → ∞, we obtain

A ≥ \frac{1}{λ_{0}} [A − z(∞) − (A + ϵ)(λ − λ_{0})]

and

B ≤ \frac{1}{λ_{0}} [B − z(∞) − (B − ϵ)(λ − λ_{0})].

Consequently,

A(1 − λ) ≤ z(∞) ≤ B(1 − λ).  (17)

In case where λ = 1, we immediately obtain from (17) that lim_{t→∞} z(t) = 0, establishing the conclusion (i).

In case where λ < 1, we obtaion from (17) that A ≤ B. This yields A = B and completes the proof for (ii).

LEMMA 2.2. Assume condition (15) holds with λ < 1, and assume

inf_{y>0} F (y)/y = K > 0.  (18)

If

∫ ^{∞} ∫_{α}^{β} Q(t, s)dσ(s)dt = ∞,  (19)

then every eventually positive solution of inequality (11) tends to zero as t → ∞.

Proof: Consider an eventually positive solution y of (11) and set z as in (16). Since

z′(t) ≤ − ∫_{α}^{β} Q(t, s)F [y(h(t, s))]dσ(s) ≤ −K ∫_{α}^{β} Q(t, s)y(h(t, s))dσ(s) ≤ 0

for all sufficiently large t, z is eventually decreasing. Hence, there exists t_{1} ≥ 0 such that for t ≥ t_{1} and s ∈ J,

z(t) ≤ z(h(t, s)) = y(h(t, s)) − ∑_{i=1}^{m} λ_{i}(h(t, s))y(h(t, s) − r_{i}) ≤ y(h(t, s)).

Thus, there exists T ≥ t_{1} so that for all t ≥ T, one has

0 ≥ z′(t) + ∫_{α}^{β} Q(t, s)F [y(h(t, s))]dσ(s)
≥ z′(t) + K ∫_{α}^{β} Q(t, s)y(h(t, s))dσ(s)
≥ z′(t) + Kz(t) ∫_{α}^{β} Q(t, s)dσ(s)

from which it follows that

z(t) ≤ z(T)e^{− ∫_{T}^{t} ∫_{α}^{β} KQ(t,s)dσ(s)dt}, t ≥ T.

Set

μ = max{z(T), 0}.

Then,

\begin{aligned}y(t) & = z(t) + ∑_{i=1}^{m} λ_{i}(t)y(t − r_{i}) \\ & ≤ ∑_{i=1}^{m} λ_{i}(t)y(t − r_{i}) + μe^{− ∫_{T}^{t} ∫_{α}^{β} KQ(t,s)dσ(s)dt}, t ≥ T. \end{aligned}

We are going to show that y is bounded on R^{+}. By way of contradiction, if y is unbounded, then there exists a sequence {t_{n}} ⊂ [T, ∞) such that lim_{n→∞} t_{n} = ∞ and

lim_{n→∞} y(t_{n}) = ∞, y(t_{n}) = max_{T ≤t_{n}≤t_{n}} y(t).  (20)

Now we have

\begin{aligned}y(t_{n}) & ≤ ∑_{i=1}^{m} λ_{i}(t_{n})y(t_{n} − r_{i}) + μe^{− ∫_{T}^{tn} ∫_{α}^{β} KQ(t,s)dσ(s)dt} \\ & ≤ y(t_{n}) ∑_{i=1}^{m} λ_{i}(t_{n}) + μe^{− ∫_{T}^{tn} ∫_{α}^{β} KQ(t,s)dσ(s)dt}. \end{aligned}

Since λ < 1, we can find some positive integer N such that

∑_{i=1}^{m} λ_{i}(t_{n}) < 1 for n ≥ N.

Thus for n ≥ N we have

y(t_{n}) ≤ \frac{μ}{1 − ∑_{i=1}^{m} λ_{i}(t_{n})}e^{− ∫_{T}^{tn} ∫_{α}^{β} KQ(t,s)dσ(s)dt}.

But as n → ∞, the right hand side tends to \frac{μ}{1−λ} · 0 due to condition (19). This yields a contradiction to lim_{n→∞} y(t_{n}) = ∞.

So, y is bounded and hence, Lemma 2.1 ensures that lim_{t→∞} y(t) = η exists, which implies that z is bounded as well. We claim that η = 0. Otherwise, η > 0 and y(t) ≥ η/2 for large t. Therefore, there exists T^{∗} ≥ T such that

\begin{aligned}0 & ≥ z′(t) + ∫_{α}^{β} Q(t, s)F [y(h(t, s))]dσ(s) \\ & ≥ z′(t) + \frac{η}{2} ∫_{α}^{β} KQ(t, s)dσ(s) \end{aligned}

for all t ≥ T^{∗}, and hence

z(t) ≤ z(T^{∗}) − \frac{η}{2} ∫_{T^{∗}}^{t} ∫_{α}^{β} KQ(t, s)dσ(s)dt

which, due to the condition (19), leads to a contradiction to the boundedness of z. This completes the proof.

In what follows, we are going to state three results from [2,3] which will be used in oue work. Consider the retarded differential inequality

X′(t) + ∫_{α}^{β} P (t, s)X(h(t, s))dσ(s) ≤ 0,  (21)

where P ∈ C(R^{+} × J; R^{+}).

LEMMA 2.3. Assume that (H0) there exists a function g ∈ C(R^{+} ×J, (0, ∞)) such that g(g(t, s), s) = h(t, s); g(t, s) is nondecreasing with respect to t ≥ 0 and s ∈ J respectively; t > g(t, s) > h(t, s) and lim_{t→∞} min_{s∈J} g(t, s) = ∞; (H1) σ is not right continuous at C_{i}(i = 0, 1, ⋯ ν) with α < C_{1} < C_{2} < ⋯ < C_{ν} < β and lim inf_{t→∞} ∫_{g(t,χ^{∗})}^{t} ∫_{C_{j}}^{χ^{∗}} P (η, s)dσ(s)dη > 0 for all χ^{∗} ∈ (C_{j}, C_{j+1}], j = 0, ⋯ ν, where

C_{0} = α, C_{ν+1} = β, σ(α − 0) = σ(α), σ(β + 0) = σ(β);

(H2) lim inf_{t→∞} ∫_{h(t,β)}^{t} ∫_{α}^{β} P (η, s)dσ(s)dsdη > \frac{1}{e}. Then (21) does not have eventually positive solutions.

LEMMA 2.4. Assume that conditions (H0) and (H2) are satisfied and (H3) lim inf_{t→∞} ∫_{g(t,β)}^{t} ∫_{α}^{β} P (η, s)dσ(s)dη > 0. Then (21) does not have eventually positive solutions.

LEMMA 2.5. Assume that (H1) and (H2) are satisfied. Then (21) has no eventually positive solutions if one of the following conditions is satisfied: (H4) lim inf_{t→∞} ∫_{h(t,α_{i})}^{t} ∫_{α}^{β} P (η, s)dσ_{i}(s)dη > \frac{1}{e} for some i ∈ \{1, ⋯ ν\}, here we assume σ is continuous at α = α_{0} < α_{1} < α_{2} < ⋯ < α_{ν} = β and set

σ_{i}(s) = \{ \begin{matrix}σ(s), s ∈ [α_{i−1}, α_{i}], \\ σ(α_{i−1}), s < α_{i−1}, \\ σ(α_{i}), s > α_{i}. \end{matrix}

(H5) ∫_{α}^{β} ln[∫_{α}^{β} lim inf_{t→∞} ∫_{h(t,χ)}^{t} P (θ, η)dθdσ(χ)]dσ(η) + ∫_{α}^{β} dσ(η) > 0; (H6)

∫_{α}^{β} ∫_{α}^{β}[lim inf_{t→∞} ∫_{h(t,η)}^{t} P (θ, χ)dθ]^{\frac{1}{2}}
[lim inf_{t→∞} ∫_{h(t,χ)}^{t} |P (θ, η)dθ]^{\frac{1}{2}} dσ(χ)dσ(η)
> \frac{1}{e}.

3. MAIN RESULTS

THEOREM 3.1 Assume that (15), (18) and (19) are satisfied. In addition, assume that

∑_{i=1}^{m} λ_{i}(t) = 1, t ≥ t_{0} ≥ 0.  (22)

Then (11) has no eventually positive solution. Proof: Assume that (11) has an eventually positive solution y. Then there exists t_{1} ≥ t_{0} ≥ 0 so that y(t) > 0 for all t ≥ t_{1} ≥ t_{0}. Set

z(t) = y(t) − ∑_{i=1}^{m} λ_{i}(t)y(t − r_{i}), t ≥ 0.

Since there exists t_{2} ≥ t_{1} such that for t ≥ t_{2},

z′(t) ≤ − ∫_{α}^{β} Q(t, s)F [y(h(t, s))]dσ(s) ≤ −K ∫_{α}^{β} Q(t, s)y(h(t, s))dσ(s) ≤ 0,

z is nonincreasing on [t_{2}, ∞).

We first consider the case where there exists t_{3} ≥ t_{2} such that z(t) < 0 for all t ≥ t_{3}. In this case, we have

z(t) ≤ z(t_{3}) < 0, t ≥ t_{3}.  (23)

We want to show that y is bounded. If this is not true, then there exists a sequence {s_{n}} ⊂ [t_{4}, ∞), t_{4} = t_{3} + max_{1≤i≤m} r_{i}, such that s_{n} → ∞ and y(s_{n}) → ∞ as n → ∞, and y(t_{n}) = max_{t_{3}≤t≤s_{n}} y(t) for all n ≥ 1. Using (22) and (23), we get

y(s_{n}) = z(s_{n}) + ∑_{i=1}^{m} λ_{i}(s_{n})y(s_{n} − r_{i})
≤ z(t_{3}) + y(s_{n}) ∑_{i=1}^{m} λ_{i}(s_{n})
= z(t_{3}) + y(s_{n}),

from which we get z(t_{3}) ≥ 0, a contradiction to (23). Hence, y is bounded and thus, z is bounded. By Lemma 2.1, we have lim_{t→∞} z(t) = 0, which contradicts the fact that z′(t) ≤ 0 and z(t) < 0 for all t ≥ t_{4}.

We next consider the remaining possible case that there exists T ≥ t_{1} such that z(t) > 0 for all t ≥ T. Then we have

y(t) > ∑_{i=1}^{m} λ_{i}(t)y(t − r_{i}), t ≥ T.  (24)

If lim inf_{t→∞} y(t) = 0, then there exists a sequence {s_{n}} ⊂ [T, ∞) such that lim_{n→∞} s_{n} = ∞ and

lim_{n→∞} y(s_{n}) = 0 y(s_{n}) = min_{t_{1}≤t≤s_{n}} y(t), n ≥ 1.

Using (22) and (24), we obtain the following contradiction

y(s_{n}) > ∑_{i=1}^{m} λ_{i}(s_{n})y(s_{n} − r_{i})
≥ y(s_{n}) ∑_{i=1}^{m} λ_{i}(s_{n})
= y(s_{n}), n ≥ 1.

Hence, lim inf_{t→∞} y(t) > 0. Therefore, there exist T^{∗} ≥ T and a constant M > 0 such that

y(t) ≥ M, y(h(t, s)) > M for all t ≥ T^{∗}, s ∈ J.

Thus,

z′(t) ≤ −K ∫_{α}^{β} Q(t, s)y(h(t, s))dσ(s) ≤ −KM ∫_{α}^{β} Q(t, s)dσ(s), t ≥ T^{∗}.

This yields

z(t) ≤ z(T^{∗}) − KM ∫_{T^{∗}}^{t} ∫_{α}^{β} Q(η, χ)dσ(χ)dη → −∞

as t → ∞, a contradiction to z(t) > 0 for t ≥ T^{∗}. This completes the proof.

THEOREM 3.2 Assume that (18) and (H0) are satisfied. In addition, we assume that

∑_{i=1}^{m} λ_{i}(t) ≤ 1 for large t;  (25)

and

ρ := inf_{(t,s)∈R^{+}×J} Q(t, s) > 0  (26)

and either (H1^{∗}) σ is not right continuous at C_{i}(i = 0, 1, ⋯ ν) with α < C_{1} < C_{2} < ⋯ < C_{ν} < β and lim inf_{t→∞} ∫_{g(t,χ^{∗})}^{t} ∫_{C_{j}}^{χ^{∗}} Q(η, χ)dσ(χ)dη > 0 for χ^{∗} ∈ (C_{j}, C_{j+1}], j = 0, ⋯, ν, where

C_{0} = α, C_{ν+1} = β, σ(α − 0) = σ(α), σ(β + 0) = σ(β); or

(H2^{∗}) lim inf_{t→∞} ∫_{h(t,β)}^{t} ∫_{α}^{β} Q(η, χ)dσ(χ)dη > \frac{1}{Ke}. Then (11) has no eventually positive solutions. Proof: Suppose that y is an eventually positive solution of (11). Then there exists t_{0} ≥ 0 such that for all t ≥ t_{0}, s ∈ J and 1 ≤ i ≤ m,

y(t) > 0, y(t − r_{i}) > 0, y(h(t, s)) > 0, y(h(t, s) − r_{i}) > 0.

Let

z(t) = y(t) − ∑_{i=1}^{m} λ_{i}(t)y(t − r_{i}).

We have

z(h(t, s)) ≤ y(h(t, s)), t ≥ t_{0}, s ∈ J.

From (11) and (18), we have

z′(t) + ∫_{α}^{β} KQ(t, s)z(h(t, s))dσ(s) ≤ 0, t ≥ t_{0}.  (27)

We shall show that z(t) > 0 for sufficiently large t. In fact, from (11) and (26) it follows that

z′(t) ≤ −ρ ∫_{α}^{β} Ky(h(t, s))dσ(s), t ≥ t_{0}.

Thus, lim_{t→∞} z(t) = z(∞) ∈ [−∞, ∞). If z(∞) = −∞, then from the definition of z and (25) it follows that y is unbounded. Hence there exists t_{1} ≥ t_{0} with z(t_{1}) < 0 and y(t_{1}) = max_{t_{0}≤t≤t_{1}} y(t). But we have

z(t_{1}) = y(t_{1}) − ∑_{i=1}^{m} λ_{i}(t_{1})y(t_{1} − r_{i}) ≥ y(t_{1})[1 − ∑_{i=1}^{m} λ_{i}(t_{1})] ≥ 0.

This yields a contradiction. Hence z(∞) > −∞. Using (26), we get

ρK ∫_{t_{1}}^{t} ∫_{α}^{β} z(h(η, χ))dσ(χ)dη
≤ ρK ∫_{t_{1}}^{t} ∫_{α}^{β} y(h(η, χ))dσ(χ)dη
≤ ∫_{t_{1}}^{t} ∫_{α}^{β} KQ(η, χ)y(h(η, χ))dσ(χ)dη
≤ − ∫_{t_{1}}^{t} z′(s)ds = z(t_{1}) − z(t) ≤ z(t_{1}) − z(∞) < ∞.

Since z′(t) < 0 for t ≥ t_{0} and h(t, s) ≤ t for all s ∈ J, we have t_{2} ≥ t_{0} such that

z(h(t, s)) > z(t) for all t ≥ t_{2}, s ∈ J.

Then it follows that

ρK ∫_{α}^{β} dσ(χ) ∫_{t_{2}}^{t} z(s)ds < ∞,

i.e., z ∈ L^{1}((t_{2}, ∞)) and hence, lim_{t→∞} z(t) = 0. As z is strictly decreasing, there must be t_{3} ≥ t_{1} such that z(t) > 0 for all t ≥ t_{3}. Therefore, z is an eventually positive solution of (27). However, Lemma 2.3 ensures that (27) has no eventually positive solutions. This contradiction shows that (11) does not have eventually positive solution. This completes the proof.

One of the key steps in the above proof is to show that if y is an eventually positive solution of (11), then z is an eventually positive solution (27). Using the same technique and Lemmas 2.4 and 2.5, we can also establieh the following two results:

THEOREM 3.3. Assume that conditions (18), (25), (26), (H0) and (H2^{∗}) are satisfied. Also assume that (H3)^{∗} lim inf_{t→∞} ∫_{g(t,β)}^{t} ∫_{α}^{β} Q(η, χ)dσ(χ)dη > 0. Then (11) has no eventually positive solutions.

THEOREM 3.4. Assume that conditions (18), (25), (26), (H0) and (H1^{∗}) are satisfied. Then (11) has no eventually positive solutions if one of the following conditions holds: (H4)^{∗} lim inf_{t→∞} ∫_{h(t,α_{i})}^{t} ∫_{α}^{β} Q(η, s)dσ_{i}(s)dη > \frac{1}{Ke} for some i ∈ \{1, ⋯ ν\}, here we assume σ is continuous at α = α_{0} < α_{1} < α_{2} < ⋯ < α_{ν} = β and set

σ_{i}(s) = \{ \begin{matrix}σ(s), s ∈ [α_{i−1}, α_{i}], \\ σ(α_{i−1}), s < α_{i−1}, \\ σ(α_{i}), s > α_{i}. \end{matrix} ;

(H5)^{∗} ∫_{α}^{β} ln[∫_{α}^{β} lim inf_{t→∞} ∫_{h(t,χ)}^{t} KQ(η, θ)dηdσ(χ)]dσ(θ) + ∫_{α}^{β} dσ(s) > 0; (H6)^{∗} ∫_{α}^{β} ∫_{α}^{β}[lim inf_{t→∞} ∫_{h(t,χ)}^{t} KQ(η, χ)dη]^{\frac{1}{2}} [lim inf_{t→∞} ∫_{h(t,χ)}^{t} KQ(η, θ)dη]^{\frac{1}{2}} dσ(η)dσ(θ) > \frac{1}{e}.

We now consider the case where λ < 1.

THEOREM 3.5. Assume that conditions (18), (15), (26), (H0), (H1^{∗}) and (H2^{∗}) are satisfied and that λ < 1. Then (11) has no eventually positive solutions and (1.2) has no eventually negative solutions. Proof: Let y be an eventually positive solution of (11). We can assume that there exists t_{0} ≥ 0 such that y(t) > 0 for all t ≥ t_{0} ≥ 0. Note that condition (H2) implies (19), thus Lemma 2.2 implies that lim_{t→∞} y(t) = 0. Therefore, for z(t) = y(t) − ∑_{i=1}^{m} λ_{i}(t)t(t − r_{i}), we have lim_{t→∞} z(t) = 0. From (11) and (26) it follows that there exists t_{1} ≥ t_{0} such that

z′(t) ≤ − ∫_{α}^{β} Q(t, s)F [y(h(t, s))]dσ(s) ≤ −ρK ∫_{α}^{β} y(h(t, s))dσ(s) < 0

for all t ≥ t_{1} ≥ t_{0}. This shows that z is strictly increasing on [t_{1}, ∞). Thus

z(t) > 0 for all t ≥ t_{1}.  (28)

Choose t_{2} ≥ t_{1} so that z(h(t, s)) ≤ y(h(t, s)) for all t ≥ t_{2} ≥ t_{1}. Then we have

z′(t) + ∫_{α}^{β} KQ(t, s)z(h(t, s))dσ(s) ≤ 0, t ≥ t_{2}.  (29)

By Lemma 2.3, we see that (29) has no eventually positive solutions, which contradicts (28). This completes the proof.

Using the same argument and Lemmas 2.2 and 2.5, we can also get THEOREM 3.6. Assume that all conditions (18), (15), (26), (H0), (H2^{∗}) and (H6^{∗}) are satisfied and that λ < 1, Then (11) has no eventually positive solution.

4. OSCILLATIONS FOR NEUTRAL PARTIAL FDES

We begin with the proof of Theorem 1.1:

Assume, by way of contradiction, that (1)-(2) has a solution u such that for some τ > 0, u(x, t) has no zeros in Ω × [τ, ∞). Since F is odd we may assume, without loss of generality, that u(x, t) > 0 for (x, t) ∈ Ω × [τ, ∞). Then there exists t_{1} > τ such that

u(x, t − r_{i}) > 0, u(x, τ_{j}(t)) > 0, x ∈ Ω, t ≥ t_{1}, 1 ≤ i ≤ m, 1 ≤ j ≤ n

and

u(x, h(x, t)) > 0, x ∈ Ω, t ≥ t_{1}, s ∈ J.

Multiplying both sides of (1) by the positive eigenfunction associated with the smallest eigenvalue μ_{0} > 0 of the Dirichlet problem (9) and integrating with respect to x over the domain Ω yield

\begin{matrix}\frac{d}{dt}[∫_{Ω} u(x, t)ϕ(x)dx − ∑_{i=1}^{m} λ_{i}(t) ∫_{Ω} u(x, t − r_{i})ϕ(x)ds] + ∫_{Ω} p(x, t)u(x, t)ϕ(x)dx \\ + ∫_{Ω} ∫_{α}^{β} q(x, t, χ)F [u(x, h(t, χ))]ϕ(x)dσ(χ)dx \\ = a(t) ∫_{Ω} δuϕ(x) + ∑_{i=1}^{n} a_{j}(t) ∫_{Ω} Δu(x, τ_{j}(t))ϕ(x)ds, t ≥ t_{1}. \end{matrix} (30)

It is easy to see that

\begin{matrix}∫_{Ω} ∫_{α}^{β} q(x, t, χ)F [u(x, h(t, χ))]ϕ(x)dσ(χ)dx \\ = ∫^{β} ∫_{Ω} q(x, t, χ)F [u(x, h(t, χ))]ϕ(x)dxdσ(χ). \end{matrix}  (31)

Using Green’s Theorem, we obtain

\begin{matrix}∫_{Ω} Δu(x, t)ϕ(x)dx \\ = ∫_{∂Ω}[ϕ(x)\frac{∂}{∂ν} u(x, t) − u(x, t)\frac{∂}{∂ν} ϕ(x)]dS + ∫_{Ω} u(x, t)Δϕ(x)dx \\ = −μ_{0} ∫_{Ω} u(x, t)ϕ(x)dx, t ≥ t_{1}, \end{matrix}  (32)

and

\begin{matrix}∫_{Ω} Δu(x, τ_{j}(t))ϕ(x)dx \\ = ∫_{∂Ω}[ϕ(x)\frac{∂}{∂ν} u(x, τ_{j}(t)) − u(x, τ_{j}(t))\frac{∂}{∂ν} ϕ(x)]dS + ∫_{Ω} u(x, τ_{j}(t))Δϕ(x)dx \\ = −μ_{0} ∫_{Ω} u(x, τ_{j}(t))ϕ(x)dx, t ≥ t_{1}, 1 ≤ j ≤ n, \end{matrix}  (33)

where dS is the surface integral element on ∂Ω. As F is convex, we can apply Jensen’s inequality to get

∫_{Ω} F [u(x, h(t, χ))]ϕ(x)dx ≥ ∫_{Ω} ϕ(x)dx · F [\frac{1}{∫_{Ω} ϕ(x)dx} ∫_{Ω} u(x, h(t, χ))ϕ(x)dx]. (34)

Combining (30)-(34) yields

\frac{d}{dt}[∫_{Ω} u(x, t)ϕ(x)dx − ∑_{i=1}^{m} λ_{i}(t) ∫_{Ω} u(x, t − r_{i})ϕ(x)dx] + ∫_{Ω} p(x, t)u(x, t)ϕ(x)dx
+[∫_{Ω} ϕ(x)dx] ∫_{α}^{β} min_{x∈Ω̄} q(x, t, χ)F (_{\frac{1}{∫_{Ω} ϕ(x)dx}} ∫_{Ω} u(x, h(t, χ))ϕ(x)dx)dσ(χ)
≤ −μ_{0}a(t) ∫_{Ω} u(x, t)ϕ(x)dx − μ_{0} ∑_{j=1}^{n} a_{j}(t)u(x, τ_{j}(t))ϕ(x)dx ≤ 0, t ≥ t_{1}.

(35)

So, the function

U(t) = \frac{1}{∫_{Ω} ϕ(x)dx} ∫_{Ω} u(x, t)ϕ(x)dx, t ≥ t_{1}

satisfies the following scalar ordinary neutral differntial inequality

\frac{d}{dt}[U(t) − ∑_{i=1}^{m} λ_{i}(t)U(t − r_{i})] + ∫_{α}^{β} min_{x∈Ω̄} q(x, t, χ)F [U(h(t, χ))]dσ(χ) ≤ 0, t ≥ t_{1}. (36)

Theorem 3.1 shows that (36) can not have eventually positive solution, a contradiction to U(t) > 0 for all t ≥ t_{1}. So, all solutions of (1)-(2) are oscillatory.

Now we assume, again by way of contradiction, that (1)-(3) has a solution which is positive in Ω × [τ, ∞) for some τ > 0. Then there exists t_{1} > τ such that

u(x, t − r_{i}) > 0, u(x, τ_{j}(t)) > 0, x ∈ Ω, t ≥ t_{1}, 1 ≤ i ≤ m, 1 ≤ j ≤ n

and

u(x, h(x, t)) > 0, x ∈ Ω, t ≥ t_{1}, s ∈ J.

By Green’s Theorem, we have

∫_{Ω} Δudx = ∫_{∂Ω} \frac{∂u}{∂ν}dS = − ∫_{∂Ω} r(x, t)u(x, t)dS ≤ 0,  (37)

and

∫_{Ω} Δu(x, τ_{j}(t))dx = ∫_{∂Ω} \frac{∂}{∂ν} u(x, τ_{j}(t))dS = − ∫_{∂Ω} r(x, τ_{j}(t))u(x, τ_{j}(t))dS ≤ 0, 1 ≤ j ≤ n.

(38)

Using Jensen’s inequality, we get

∫_{Ω} F [u(x, h(t, χ))]dx ≥ |Ω|F [\frac{1}{|Ω|} ∫_{Ω} u(x, h(t, χ))dx], t ≥ t_{1}.  (39)

Integrating (1) with respect to x over Ω and applying (37)-(40) yield

\frac{d}{dt}[∫_{Ω} u(x, t)dx − ∑_{i=1}^{m} λ_{i}(t) ∫_{Ω} u(x, t − r_{i})dx] + ∫_{Ω} p(x, t)u(x, t)dx
+|Ω| ∫_{α}^{β} min_{x∈Ω̄} q(x, t, χ)F [\frac{1}{|ω|} ∫_{Ω} u(x, h(t, χ))dx]dσ(χ)
≤ −a(t) ∫_{∂Ω} r(x, t)u(x, t)dS − ∑_{j=1}^{n} a_{j}(t) ∫_{∂Ω} r(x, τ_{j}(t))u(x, τ_{j}(t))dS ≤ 0, t ≥ t_{1}.

Therefore, the function

V (t) = \frac{1}{|Ω|} ∫_{Ω} u(x, t)dx, t ≥ t_{1}

satisfies the following scalar ordinary neutral differential inequality

\frac{d}{dt}[V (t) − ∑_{i=1}^{m} λ_{i}(t)V (t − r_{i})] + ∫_{α}^{β} min_{x∈Ω̄} F [V (x, h(t, χ))]dσ(χ) ≤ 0, t ≥ t_{1}.  (40)

By Theorem 3.1, (40) can not have an eventually positive solution, a contradiction to the fact that V (t) > 0 for all t ≥ t_{1}. This completes the proof.

For the purpose, we give the following Example. Consider

\begin{matrix}\frac{∂}{∂t}[u(x, t) − e^{−π}u(x, t − \frac{π}{2}) − (1 − e^{−π})u(x, t − π)] + (\frac{t}{2} + e^{−π})u(x, t) \\ + ∫_{−π}^{−π/2} 3tu(x, t + χ)dχ \\ = \frac{5}{2}t\frac{∂^{2}}{∂x^{2}} u(x, t − \frac{π}{2}) + (e^{−π} + 4t)\frac{∂^{2}}{∂x^{2}} u(x, t − \frac{3π}{2}), \\ (x, t) ∈ (0, π) × (0, ∞) \end{matrix}  (41)

subject the Dirichlet boundary condition

u(0, t) = u(π, t) = 0, t > 0.  (42)

It is easy to verify all hopotheses of Theorem 1.1 for this case, and thus all solutions of (41)-(42) are oscillatory.

Using Theorems 3.2-3.5, we can get other sets of sufficient conditions for all solutions of (1)-(2) or (1)-(3) to be oscillatory. We will only state the results, since the proof is similar to that for Theorem 1.1.

THEOREM 4.1. Assume that (4), (25) and (H0) are satisfied. In addition, we assume that

ρ := inf_{(t,s)∈R^{+}×J,x∈Ω̄} q(x, t, x) > 0  (43)

and either (H1^{∗∗}) σ is not right continuous at C_{i}(i = 0, 1, ⋯ ν) with α < C_{1} < C_{2} < ⋯ < C_{ν} < β and lim inf_{t→∞} ∫_{g(t,χ^{∗})}^{t} ∫_{C_{j}}^{χ^{∗}} min x ∈ Ω̄q(x, η, χ)dσ(χ)dη > 0 for χ^{∗} ∈ (C_{j}, C_{j+1}], j = 0, ⋯, ν, where

C_{0} = α, C_{ν+1} = β, σ(α − 0) = σ(α), σ(β + 0) = σ(β); or

(H2^{∗∗}) lim inf_{t→∞} ∫_{h(t,β)}^{t} ∫_{α}^{β} min_{x∈Ω̄} q(x, η, χ)dσ(χ)dη > \frac{1}{Ke}. Then every solution of (1)-(2) or (1)-(3) is oscillatory.

THEOREM 4.2. Assume that conditions (4), (25), (43), (H0) and (H2^{∗∗}) are satisfied. Also assume that (H3)^{∗∗} lim inf_{t→∞} ∫_{g(t,β)}^{t} ∫_{α}^{β} min_{x∈Ω̄} q(x, η, χ)Q(η, χ)dσ(χ)dη > 0. Then every solution of (1)-(2) or (1)-(3) is oscillatory.

THEOREM 4.3. Assume that conditions (4), (25), (43), (H0) and (H1^{∗∗}) are satisfied. Then every solution of 91)-(2) or (1)-(3) is oscillatory if one of the following conditions holds: (H4)^{∗∗} lim inf_{t→∞} ∫_{h(t,α_{i})}^{t} ∫_{α}^{β} min_{x∈Ω̄} q(x, η, s)dσ_{i}(s)dη > \frac{1}{Ke} for some i ∈ \{1, ⋯ ν\}, here we assume σ is continuous at α = α_{0} < α_{1} < α_{2} < ⋯ < α_{ν} = β and set

σ_{i}(s) = \{ \begin{matrix}σ(s), s ∈ [α_{i−1}, α_{i}], \\ σ(α_{i−1}), s < α_{i−1}, \\ σ(α_{i}), s > α_{i}. \end{matrix}

(H5)^{∗∗} ∫_{α}^{β} ln[∫_{α}^{β} lim inf_{t→∞} ∫_{h(t,χ)}^{t} K min_{x∈Ω̄}(x, η, θ)dηdσ(χ)]dσ(θ) + ∫_{α}^{β} dσ(s) > 0; (H6)^{∗∗}

∫_{α}^{β} ∫_{α}^{β}[lim inf_{t→∞} ∫_{h(t,χ)}^{t} K min_{x∈Ω̄}(x, η, χ)dη]^{\frac{1}{2}}
[lim inf_{t→∞} ∫_{h(t,χ)}^{t} K min_{x∈Ω̄}(x, η, θ)dη]^{\frac{1}{2}} dσ(η)dσ(θ)
> \frac{1}{e}.

THEOREM 4.4. Assume that conditions (4), (15), (43), (H0), (H1^{∗∗}) and (H2^{∗∗}) are satisfied and that λ < 1. Then every solution of (1)-(2) or (1)-(3) is oscillatory.

ACKNOWLEDGEMENTS

The research of the first author is partially supported by the National Science Foundation of People’s Republic of China and the research of the second author is partially supported by Natural Sciences and Engineering Research Council of Canada and the Alexander von Humboldt Foundation of Germany.

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