Electronic Journal of Differential Equations, Vol. 2005(2005), No. 134, pp. 1–18.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
OSCILLATION CRITERIA FOR FIRST-ORDER NONLINEAR
NEUTRAL DELAY DIFFERENTIAL EQUATIONS
ELMETWALLY M. ELABBASY, TAHER S. HASSAN, SAMIR H. SAKER
Abstract. Oscillation criteria are obtained for all solutions of first-order nonlinear neutral delay differential equations. Our results extend and improve
some results well known in the literature. Some examples are considered to
illustrate our main results.
1. Introduction
In recent years, the literature on the oscillation of neutral delay differential equations has grown very rapidly. It is a relatively new field with interesting applications
in real world life problems. In fact, neutral delay differential equations appear in
modelling of the networks containing lossless transmission lines (as in high-speed
computers where the lossless transmission lines are used to interconnect switching
circuits), in the study of vibrating masses attached to an elastic bar, as the Euler
equation in some variational problems, in the theory of automatic control and in
neuro-mechanical systems in which inertia plays an important role. See Hale [17],
Driver [8], Brayton and Willoughby [6], Popove [32], and Boe and Chang [5], and
the references cited therein. Also this is evident by the number of references in the
recent books by Ladde et al. [14] and by Ladas [16].
We consider a general first-order nonlinear neutral delay differential equation
(x(t) − q(t)x(t − r))0 + f (t, x(τ (t))) = 0,
(1.1)
where for t ≥ t0
q, τ ∈ C([t0 , ∞), R+ ), q(t) 6= 1, r ∈ (0, ∞), τ (t) < t, lim τ (t) = ∞,
t→∞
f ∈ C([t0 , ∞) × R, R),
n Y
i
X
uf (t, u) ≥ 0,
1
→ ∞ as n → ∞.
q(tj )
i=1 j=1
(1.2)
(1.3)
(1.4)
we assume that the nonlinear function f (t, u) in (1.1) satisfies the following
conditions:
(H) There are a piecewise continuous function p : [t0 , ∞) → R+ = [0, ∞), a
function g ∈ C(R, R+ ), and a number ε0 > 0 such that
2000 Mathematics Subject Classification. 34K15, 34C10.
Key words and phrases. Oscillation; non-oscillation; neutral delay of differential equations.
c
2005
Texas State University - San Marcos.
Submitted August 2, 2005. Published November 30, 2005.
1
2
E. M. ELABBASY, T. S. HASSAN, S. H. SAKER
(i)
(ii)
(iii)
(iv)
(v)
EJDE-2005/134
g is nondecreasing on R+
g(−u)
g(u), limu→0 g(u) = 0,
R∞ =
−u
g(e
)du < ∞,
0
1
|f
(t,
u)
− p(t)u| ≤ p(t)g(u) for t ≥ t0 and 0 < |u| < ε0 ,
|u|
For each ϕ ∈ C([t0 , ∞), R) with limt→∞ ϕ(t) > 0,
Z ∞
Z ∞Z
f (t, ϕ(τ (t)))dt = ∞,
f (t, −ϕ(τ (t)))dt = −∞.
t0
t0
As usual a solution x(t) of equation (1.1) is said to be oscillatory if it has arbitrarily
large zeros in [t0 , ∞). Otherwise it is nonoscillatory and the equation (1.1) is called
oscillatory if every solution of this equation is oscillatory.
When q(t) = 0, (1.1) reduces to
x0 (t) + f (t, x(τ (t))) = 0,
which was studied by Tang and Shen [36]. They obtained some infinite integral
sufficient conditions for oscillations.
The oscillatory behavior of other neutral delay differential equations have been
investigated by many authors, see [1, 2, 3, 4, 9, 10, 12, 13, 14, 15, 16, 20, 21, 22,
23, 24, 25, 26, 27, 28, 29, 30, 31, 33, 37, 38, 39, 40] and references therein.
In recent papers Elabbasy and Saker [10], Kubiaczyk and Saker [22] obtained
an infinite integral conditions for oscillation of the linear neutral delay differential
equation
(x(t) − q(t)x(t − r))0 + p(t)x(t − τ ) = 0 .
Let δ(t) = max{τ (t) : t0 ≤ s ≤ t} and δ −1 (t) = min{s ≥ t0 : δ(s) = t}. Clearly, δ
and δ −1 are non-decreasing and satisfy
(A) δ(t) < t and δ −1 (t) > t
(B) δ(δ −1 (t)) = t and δ −1 (δ(t)) ≤ t.
Let δ −k (t) be defined on [t0 , ∞) by
δ −(k+1) (t) = δ −1 (δ −k (t)),
k = 1, 2, . . .
(1.5)
Throughout this paper, we use the sequence {pk }, of functions defined by
Z δ−1 (t)
p1 (t) =
p(s)ds, t ≥ t0 ,
t
Z
pk+1 (t) =
δ −1 (t)
p(s)pk (s)ds,
t ≥ t0 , k = 1, 2, . . .
t
Our main results are the following.
Theorem 1.1. Assume that (1.2), (1.4), (1.3), and (H) hold, and there exist a
bounded positive function σ(t) such that
Z t
1
B(s)ds > ,
(1.6)
e
τ (t)
and
Z
∞
t0
h
Z
p(t)σ(t) exp
t
τ (t)
p(s)ds −
i
σ(t) − 1 dt = ∞,
e
where B(t) = p(t)/σ(t). Then every solution of (1.1) oscillates.
(1.7)
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OSCILLATION CRITERIA
3
Theorem 1.2. Assume that (1.2), (1.4), (1.3) and (H) hold, and that
Z t
p(s)ds ≥ 0.
lim inf
t→∞
(1.8)
τ (t)
and suppose that there exists a positive integer n such that
Z ∞
p(t) ln(en−1 pn (t) + 1)dt = ∞.
(1.9)
t0
Then every solution of (1.1) oscillates.
Corollary 1.3. Assume that (1.2) (1.3), (1.4), (1.8) and (H) hold, and that
Z ∞
h
Z t
i
p(t) exp
p(s)dsBig) − 1 dt = ∞.
(1.10)
t0
τ (t)
Then every solution of (1.1) oscillates.
Corollary 1.4. Assume that (1.2), (1.3), (1.4), (1.8) and (H) hold, and that
Z ∞
Z δ−1 (t)
p(t) ln
p(s)ds + 1 dt = ∞.
t0
t
Then every solution of (1.1) oscillates.
Corollary 1.5. Assume that (1.2), (1.3), (1.4), (1.8) and (H) hold, and suppose
that there exists a positive integer n such that
Z ∞
p(t) ln(en pn (t))dt = ∞.
t0
Then every solution of (1.1) oscillates.
Note that if
Z
t
lim sup
t→∞
p(s)ds > 2,
τ (t)
then by Lemma 2.4 every solution of (1.1) oscillates. Thus, we will consider the
case
Z t
lim sup
p(s)ds ≤ 2.
t→∞
τ (t)
This implies that for some > 0 and large t,
Z t
p(s)ds ≤ 2 + .
τ (t)
Thus we have
lim inf pk (t) ≤ (2 + )k−1 lim inf
t→∞
t→∞
Z
δ −1 (t)
p(s)ds ≤ (2 + )k−1 lim inf
t
t→∞
Z
t
p(s)ds.
τ (t)
As a result, by Theorem 1.2 we have
Corollary 1.6. Assume that (1.2), (1.3), (1.4) and (H) hold, and that there exists
a positive integer n such that
lim inf pn (t) > 0.
t→∞
Then every solution of (1.1) oscillates.
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E. M. ELABBASY, T. S. HASSAN, S. H. SAKER
EJDE-2005/134
The proofs of the above Theorems and also some Lemmas to be used in these
proofs will be given in the next two sections. Some examples which illustrate and
the advantage of our results will be given in section 4.
2. Preliminary Lemmas
Lemma 2.1. Assume that (1.2), (1.4) and (1.3) hold. Let x(t) be an eventually
positive solution of 1.1 and set
z(t) = x(t) − q(t)x(t − r).
(2.1)
Then z(t) is eventually nonincreasing and positive function.
Proof. From (1.1)), (1.3), we have z 0 (t) = −f (t, x(τ (t))) ≤ 0 eventually. We prove
that z(t) is a positive function. If not, then there exist T ≥ t0 and α < 0 such that
z(t) < α for t ≥ T . Then from (2.1), we have x(t) < α + q(t)x(t − r) which implies
x(t + r) < α + q(t + r)x(t).
Now we choose k such that tk = t∗ + kr > T . Then x(tk+1 ) < α + q(tk+1 )x(tk ).
Applying this inequality by induction, it gives
n
n−i
n
h
i
X
Y
Y
x(tn ) < α 1 +
q(tn−j ) +
q(ti )x(tk ).
i=k+2 j=0
i=k+1
Now define qn and dn by
qn = 1 +
n−i
n
Y
X
q(tn−j ),
dn =
i=k+2 j=0
n
Y
q(ti ),
i=k+1
and let
sn =
n Y
i
X
1
.
q(tj )
i=1 j=1
Then
k+1
s∗n =
i
XY 1 qn
= sn −
q(tk+1 ) . . . q(t1 ) → ∞ as n → ∞,
dn
q(tj )
i=1 j=1
by condition (1.4). Using the above inequality,
x(tk ) αdn → −∞ as n → ∞,
x(tn ) < s∗n +
α
and this contradicts the assumption that x(t) > 0. Then z(t) must be positive
function. The proof is complete.
Note that the proof of Lemma 2.1 is similar to that in [7, Lemma 1]; we state it
here for the sake of completeness.
Lemma 2.2. Assume that (1.2), (1.3), (1.4) and (H) hold. Then every nonoscillatory solution of (1.1) converges to zero monotonically for large t as t → ∞.
Proof. Suppose that x(t) is a non-oscillatory solution of equation (1.1) which we
shall assume to be eventually positive [If x(t) is eventually negative the proof is
similar]. From Lemma 2.1, we have z(t) is eventually non-increasing and positive
function.
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OSCILLATION CRITERIA
5
Choose a t1 ≥ t0 such that x(t) > 0, z(t) > 0 for t ≥ t1 . It follows from equations
(1.1)-(1.3) and (H) that there exists t2 > t1 such that τ (t) ≥ t1 and z 0 (t) ≤ 0 for
t > t2 . Hence the following limits exist and
lim x(t) ≥ lim z(t) = α ≥ 0 .
t→∞
t→∞
If α > 0, then from (1.1) we have
Z
t
z(t) − z(t0 ) = −
f (t, x(τ (s)))ds.
t0
It follows from assumption (H)(v) that limt→∞ z(t) = −∞, which contradicts that
z(t) being positive function, then α = 0, from (1.2), we have limt→∞ x(t) = 0. The
proof of Lemma 2.2 is complete.
Lemma 2.3. Assume that (1.2), 1.3, (1.4) and (H) hold. If x(t) is a nonoscillatory solution of (1.1), then there exist A > 0, ε > 0 and T ∈ (0, ∞) such
that for t ≥ T ,
1Z t
p(s)ds + ε,
(2.2)
|x(t)| ≤ A exp −
2 T
Proof. We shall assume x(t) to be eventually positive [If x(t) is eventually negative
the proof is similar]. By Lemma 2.2, there exists t1 > 0 such that
0 < x(t) ≤ x(τ (t)) < ε for t ≥ t1 .
From (H), we find that for t ≥ t1
f (t, x(τ (t))) ≥ p(t)[1 − g(x(τ (t)))]x(τ (t)),
and limt→∞ x(t) = 0. By assumption (H), there exists T > t1 such that for t ≥ T ,
f (t, x(τ (t))) ≥
1
1
p(t)x(τ (t)) ≥ p(t)x(t),
2
2
and it follows from (1.1) that for t ≥ T ,
1
1
(x(t) − q(t)x(t − r))0 + p(t)x(t) ≤ 0, z 0 (t) + p(t)z(t) ≤ 0,
2
2
where z(t) = x(t) − q(t)x(t − r). This yields, for t ≥ T ,
h 1Z t
i
z(t) ≤ A exp −
p(s)ds ,
2 T
1Z t
|x(t)| ≤ A exp −
p(s)ds + ε,
2 T
where A = x(T ) − q(T )x(T − r).
Lemma 2.4. Assume that (1.2), 1.3, (1.4) and (H) hold. If equation (1.1)) has a
nonoscillatory solution, then
Z t
p(s)ds ≤ 2 and pk (t) ≤ 2k , k = 1, 2, . . .
(2.3)
τ (t)
eventually.
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E. M. ELABBASY, T. S. HASSAN, S. H. SAKER
EJDE-2005/134
Proof. Suppose that x(t) is a nonoscillatory solution of equation (1.1) which we
shall assume to be eventually positive [if x(t) is eventually negative the proof is
similar]. By Lemma 2.2, there exists T ≥ 0 such that
x(τ (t)) ≥ x(t) > 0
for t ≥ T,
1
(x(t) − q(t)x(t − r))0 + p(t)x(τ (t)) ≤ 0,
2
1
z 0 (t) + p(t)z(τ (t)) ≤ 0 for t ≥ T.
2
Integrating both sides from τ (t) to t yields
Z
1 t
p(s)z(τ (s))ds ≤ 0 for t ≥ T.
z(t) − z(τ (t)) +
2 τ (t)
(2.4)
By the decreasing nature of z(t) for large t and the increasing nature of τ (t), there
exists T1 ≥ T such that
Z t
1
z(t) − z(τ (t)) + z(τ (t))
p(s)ds ≤ 0 for t ≥ T1 .
2
τ (t)
Rt
Then, τ (t) p(s)ds ≤ 2.
Also, integrating both sides of equation (2.4) from t to δ −1 (t) yields
Z
1
−1
δ −1 (t)p(s)z(τ (s))ds ≤ 0 for t ≥ T.
z(δ (t)) − z(t) +
2 t
By the decreasing nature of z(t) for large t and the increasing nature of τ (t), there
exists T1 ≥ T such that
Z −1
1 δ (t)
−1
z(δ (t)) − z(t) +
p(s)ds z(τ (δ −1 (t))) ≤ 0 for t ≥ T1 .
2 t
or
Z −1
1 δ (t)
−1
z(δ (t)) − z(t) +
p(s)ds z(t) ≤ 0 for t ≥ T1 .
2 t
Then, we have
Z δ−1 (t)
p1 (t) =
p(s)ds ≤ 2.
t
By iteration we deduce, from this, that pk (t) ≤ 2k which shows that (2.3) holds for
t ≥ T1 . The proof of Lemma 2.4 is complete.
Lemma 2.5. Assume that (1.2), (1.3), (1.8), (1.4) and (H) hold. If x(t) is a
(t))
is well defined for large t
nonoscillatory solution of equation (1.1)), then z(τ
z(t)
and is bounded.
Proof. Suppose that x(t) is a nonoscillatory solution of equation (1.1) which we
shall assume to be eventually positive [if x(t) is eventually negative the proof is
similar]. By the same argument as in the proof of Lemma 2.3, there exists T > 0,
such that
x(τ (t)) ≥ x(t) > 0
for t ≥ T,
1
(x(t) − q(t)x(t − σ))0 + p(t)x(τ (t)) ≤ 0,
2
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OSCILLATION CRITERIA
7
1
z 0 (t) + p(t)z(τ (t)) ≤ 0 for t ≥ T.
2
The rest of the proof is similar to in [28, Lemma 5], and thus it is omitted.
3. Proofs of Theorems
Proof of Theorem 1.1. Assume that (1.1) has a nonoscillatory solution x(t)
which will be assumed to be eventually positive (if x(t) is eventually negative the
proof is similar). By Lemma 2.2, there exists t1 ≥ t0 such that
0 < x(t) ≤ x(τ (t)) < ε0 ,
g(x(τ (t))) < 1,
t ≥ t1 ,
(3.1)
where ε0 is given by assumption (H). From (3.1) and (H), we have
f (t, x(τ (t))) ≥ p(t)[1 − g(x(τ (t)))]x(τ (t)),
t ≥ t1 .
(3.2)
Set
σ(t)z(τ (t))
for t ≥ t1 .
z(t)
From Lemmas 2.1 and 2.2, ω(t) ≥ σ(t) for t ≥ t1 . From (1.1) and (3.2), we have
ω(t) =
z 0 (t)
+ B(t)ω(t)[1 − g(x(τ (t)))] ≤ 0,
z(t)
t ≥ t1 .
(3.3)
Let t2 > t1 be such that τ (t) ≥ t1 for t ≥ t2 . Integrating both sides of (3.3) from
τ (t) to t, we obtain
Z t
ω(t) ≥ σ(t) exp
B(s)ω(s)[1 − g(x(τ (s)))]ds , t ≥ t2 .
(3.4)
τ (t)
By (1.6), for t ≥ t2 , we have
Z t
Z
p(s)ds =
t
τ (t∗ )
δ(t)
Z
t∗
p(s)ds ≥
p(s)ds ≥ e−1 ,
(3.5)
τ (t∗ )
where t∗ ∈ [t0 , t] with τ (t∗ ) = δ(t). From (1.6) and (3.4)), we find that for t ≥ t2 ,
Z t
σ(t) ω(t) ≥ σ(t) exp
B(s)(ω(s) − σ(t))ds +
e
τ (t)
Z
t
Z t
σ(t) × exp
p(s)ds −
exp exp −
B(s)ω(s)g(x(τ (s)))ds
e
τ (t)
τ (t)
Z t
Z t
σ(t) ≥ σ(t) e
B(s)(ω(s) − σ(t))ds + σ(t) exp
p(s)ds −
e
δ(t)
τ (t)
Z t
× exp −
B(s)ω(s)g(x(τ (s)))ds .
τ (t)
Let υ(t) = ω(t) − σ(t) for t ≥ t1 . Then υ(t) ≥ 0 for t ≥ t1 , and so for t ≥ t2 ,
Z t
υ(t) − e
B(s)υ(s)ds
δ(t)
Z
≥ e
t
h
Z
B(s)υ(s)ds + σ(t) σ(t) exp
δ(t)
t
p(s)ds −
τ (t)
Z
t
× exp −
τ (t)
i
B(s)ω(s)g(x(τ (s)))ds − 1 ,
σ(t) e
8
E. M. ELABBASY, T. S. HASSAN, S. H. SAKER
EJDE-2005/134
that is, for t ≥ t2 ,
Z
t
B(t)υ(t) − B(t)e
B(s)υ(s)ds
δ(t)
Z
≥ B(t) e
t
h
Z
B(s)υ(s)ds + σ(t) σ(t) exp
δ(t)
Z
× exp −
t
p(s)ds −
τ (t)
t
σ(t) e
(3.6)
i
B(s)ω(s)g(x(τ (s)))ds − 1 .
τ (t)
By Lemmas 2.1–2.5, there exist T > t2 , A > 0, ε > 0 and M > 0 such that for
t ≥ T,
1 Z τ (t)
x(τ (t)) ≤ A exp −
p(s)ds + ε,
(3.7)
2 T
Z t
p(s)ds ≤ 2,
(3.8)
τ (t)
ω(t) ≤ σ(t)M,
σ(t) ≤ η.
(3.9)
Let
α(t) =
1
2
Z
t
p(s)ds,
t ≥ T.
T
Clearly, (1.6) implies that α(t) → ∞ as t → ∞. For t ≥ t2 , set
Z t
Z t
σ(t)
D(t) = p(t) e
B(s)υ(s)ds + σ(t) exp
p(s)ds −
|Big)
e
δ(t)
τ (t)
h
Z t
i
× 1 − exp −
B(s)ω(s)g(x(τ (s)))ds .
(3.10)
τ (t)
One can easily see that
0 ≤ 1 − e−c ≤ c for c ≥ 0.
It follows from (3.10) that for t ≥ t2 ,
Z t
Z
D(t) ≤ p(t) e
B(s)υ(s)ds + σ(t) exp
δ(t)
Z
(3.11)
t
p(s)ds −
τ (t)
t
×
σ(t) e
(3.12)
B(s)ω(s)g(x(τ (s)))ds.
τ (t)
Therefore,
D(t)
Z
≤ p(t) e
t
Z
B(s)υ(s)ds + σ(t) exp
δ(t)
t
τ (t)
p(s)ds
Z
t
B(s)ω(s)g(x(τ (s)))ds.
τ (t)
Let T ∗ > T be such that τ (τ (t)) ≥ T for t ≥ T ∗ and α(T ∗ ) > 2 + ln A. Set
M1 = e2 ηM [2e(M − 1) + η] and A1 = eA. Noting that
Z t
B(s)υ(s)ds + σ(t) ≤ 2e(M − 1) + η for t ≥ T.
e
δ(t)
EJDE-2005/134
OSCILLATION CRITERIA
9
from (3.7)–(3.9), (3.12), and assumption (H), we obtain N ≥ T ∗ ,
Z
N
D(t)dt
T a st
N
t
1 Z τ (t)
p(s)g A exp
p(s)ds + ε ds dt
2 T
τ (t)
T a st
Z N
Z t
Z
1Z s
1 s
= M1
p(t)
p(s)g A exp −
p(µ)dµ +
p(µ)dµ + ε ds dt
2 T
2 τ (s)
T a st
τ (t)
Z N
Z t
≤ M1
p(t)
p(s)g A1 e−α(s) + ε ds dt
Z
≤ M1
Z
p(t)
T a st
Z
τ (t)
N
α(t)
Z
= 2M1
g(A1 e−u + ε)du dt
p(t)
T a st
Z
α(τ (t))
N
α(t)
Z
= 2M1
g(A1 e−u + ε) du dt,
p(t)
T a st
Z
α(t)−β(t)
α(N )
≤ 4M1
α(T ∗ )
1
2
Z
t
p(s)ds
τ (t)
v
Z
g(A1 e−u + ε)du dv
p(t)
Z
β(t) =
v−1
α(N )
g(A1 e−u + ε)du
≤ 4M1
α(T ∗ )−1
Z
= 4M1
ln(A1 e−α(N ) +ε)−1
ln(A1 e1−α(T ∗ ) +ε)−1
Z
α(N )
g(e−u )
≤ 4M1
Z0 ∞
≤ 4M1
e−u
du
e−u − ε
g(e−u )
e−u
du
e−u − ε
g(e−u )du < ∞.
0
and
Z
∞
D(t)dt < ∞.
(3.13)
T
Substituting (3.10) into (3.6), for t ≥ t2 , we obtain
Z
t
B(t)υ(t) − eB(t)
B(s)υ(s)ds
δ(t)
Z
≥ p(t) e
t
h
Z
B(s)υ(s)ds + σ(t) exp
δ(t)
p(s)ds −
τ (t)
Z
i
σ(t) − 1 − D(t),
e
t
B(t)υ(t) − eB(t)
h
≥ p(t)σ(t) exp
t
B(s)υ(s)ds
δ(t)
Z t
τ (t)
p(s)ds −
i
σ(t) − 1 − D(t).
e
10
E. M. ELABBASY, T. S. HASSAN, S. H. SAKER
EJDE-2005/134
Integrating both sides from T ∗ to N > τ −1 (T ∗ ), we have
Z N
Z N
Z t
B(t)υ(t)dt − e
B(t)
B(s)υ(s) ds dt
T a st
Z
N
≥
T a st
δ(t)
t
Z N
i
σ(t) p(s)ds −
D(t)dt.
− 1 dt −
e
τ (t)
T a st
h
Z
p(t)σ(t) exp
T a st
(3.14)
By interchanging the order of integrations and by (3.5), we have
Z δ−1 (t)
Z δ(N )
Z t
Z N
B(s) ds dt
B(t)υ(t)
B(s)υ(s) ds dt ≥ e
B(t)
e
T a st
T∗
δ(t)
Z
t
(3.15)
δ(N )
≥
B(t)υ(t)dt.
T∗
From this and (3.14), it follows that
Z N
Z N
h
Z
B(t)υ(t)dt ≥
p(t)σ(t) exp
T a st
δ(N )
p(s)ds −
τ (t)
By (3.8) and (3.9),
Z N
Z
B(t)υ(t)dt ≤ (M − 1)
δ(N )
t
N
Z
N
h
Z
p(t)σ(t) exp
∞
p(s)ds −
Z N
i
σ(t) − 1 dt −
D(t)dt.
e
T a st
p(s)ds −
Z ∞
i
σ(t) − 1 dt −
D(t)dt,
e
T∗
t
τ (t)
h
Z
p(t)σ(t) exp
T∗
t
τ (t)
which together with (3.13) yields
Z ∞
h
Z
p(t)σ(t) exp
T∗
p(t)dt ≤ 2(M − 1),
τ (N )
T a st
This implies that
Z
2(M − 1) ≥
N
p(t)dt ≤ (M − 1)
δ(N )
and so by (3.16),
Z
2(M − 1) ≥
Z N
i
σ(t) − 1 dt −
D(t)dt.
e
T a st
(3.16)
t
p(s)ds −
τ (t)
i
σ(t) − 1 dt < ∞.
e
This contradicts (1.7) and so the proof is complete.
Proof of Theorem 1.2. Assume that (1.1) has a nonoscillatory solution x(t)
which will be assumed to be eventually positive (if x(t) is eventually negative the
proof is similar). By Lemma 2.1 and assumption (H), there exists t∗0 ≥ t0 such that
0 < x(t) ≤ x(δ(t)) ≤ x(τ (t)) < ε0 ,
g(x(τ (t))) < 1,
t ≥ t∗0 ,
where ε0 is given by assumption (H). (3.17) and (H) yield that for t ≥
f (t, x(τ (t))) ≥ p(t)[1 − g(x(τ (t)))]x(τ (t))
≥ p(t)[1 − g(x(τ (t)))]z(δ(t)),
(3.17)
t∗0 ,
(3.18)
and it follows from (1.1) that
z(δ(t))
z 0 (t)
+ p(t)
[1 − g(x(τ (t)))] ≤ 0,
z(t)
z(t)
t ≥ t∗0 .
(3.19)
EJDE-2005/134
OSCILLATION CRITERIA
11
By Lemmas 2.1–2.5, there exist T > t2 , A > 0, ε > 0 and M > 0 such that for
t ≥ T,
1 Z τ (t)
x(τ (t)) ≤ A exp −
p(s)ds) + ε,
(3.20)
2 T
Z t
Z t
p(s)ds ≤
p(s)ds ≤ 2, pk (t) ≤ 2k , k = 1, 2, . . . ,
(3.21)
δ(t)
τ (t)
z(δ(t))
z(τ (t))
≤
≤ M.
z(t)
z(t)
(3.22)
Let tk = δ −k (T ), k = 1, 2, . . . Clearly tk → ∞ as k → ∞. Set λ(t) = −z 0 (t)/z(t),
for t ≥ T . Then
Z t
z(δ(t))
λ(s)ds, t ≥ t1 ,
= exp
z(t)
δ(t)
and from (3.19), for t ≥ t1 , we have
Z t
z(δ(t))
λ(t) ≥ p(t) exp
λ(s)ds − p(t)g(x(τ (t)))
.
z(t)
δ(t)
(3.23)
It follows from (3.20)–(3.23) that for t ≥ t1 ,
λ(t)
Z
1 τ (t)
≥ p(t) exp
λ(s)ds − M p(t)g A exp −
p(s)ds + ε
2 T
δ(t)
Z t
1Z
T t p(s)ds + ε ,
≥ p(t) exp
λ(s)ds − M p(t)g A1 exp −
2
δ(t)
Z
t
where A1 = eA. By the inequality ec ≥ ec for c ≥ 0, we have for t ≥ t1 ,
Z t
1Z t
λ(t) ≥ ep(t)
λ(s)ds − M p(t)g A1 exp −
p(s)ds + ε .
2 T
δ(t)
(3.24)
(3.25)
Set
α(t) =
1
2
Z
t
p(s)ds,
t ≥ T,
(3.26)
T
and
λ0 (t) = λ(t),
Z
t ≥ T,
t
λk (t) = p(t)
λk−1 (s)ds,
t ≥ tk , k = 1, 2, . . . , n,
(3.27)
δ(t)
and
G0 (t) = 0,
Z
t ≥ T,
t
Gk (t) =ep(t)
Gk−1 (s)ds +M p(t)g(A1 exp(−α(t)) + ε),
(3.28)
δ(t)
for t ≥ tk , k = 1, 2, . . . , n. Clearly (1.8) implies that α(t) is nondecreasing on [T, ∞)
and α(t) → ∞ as t → ∞. By iteration we deduce from (3.25) that
λ(t) ≥ ek λk (t) − Gk (t),
t ≥ tk , k = 1, 2, . . . n − 1,
(3.29)
12
E. M. ELABBASY, T. S. HASSAN, S. H. SAKER
EJDE-2005/134
and so by (3.24),
n−1
Z
t
Z
λn−1 (s)ds exp −
λ(t) ≥ p(t) exp e
δ(t)
t
Gn−1 (s)ds − G1 (t),
(3.30)
δ(t)
for t ≥ tn . From (3.28), one can easily obtain
Z t
[Gk (s) − Gk−1 (s)]ds,
Gk+1 (t) − Gk (t) = ep(t)
(3.31)
δ(t)
for t ≥ tk+1 , k = 1, 2, . . . , n − 1. By (3.21), (3.26) and (3.28), for t ≥ t2 , we have
Z t
Z t
G1 (s)ds = M
p(s)g(A1 exp(−α(s)) + ε)ds
δ(t)
δ(t)
Z
α(t)
g(A1 e−u + ε)du
= 2M
(3.32)
α(δ(t))
Z
α(t)
g(A1 e−u + ε)du.
≤ 2M
α(t)−1
Thus, from (3.31), we get
Z t
Z
[G2 (t) − G1 (t) = ep(t)
G1 (s)ds ≤ 2eM p(t)
δ(t)
α(t)
g(A1 e−u + ε)du,
t ≥ t2 ,
α(t)−1
Z
t
G3 (t) − G2 (t) = ep(t)
[G2 (s) − G1 (s)]ds
δ(t)
Z
2
t
≤ 2e M p(t)
g(A1 e−u + ε)du ds
p(s)
δ(t)
Z
2
α(s)
Z
α(t)
α(s)−1
v
Z
g(A1 e−u + ε)du dv
= 4e M p(t)
α(δ(t))
≤ 4e2 M p(t)
Z
α(t)
α(t)−1
≤ 4e2 M p(t)
Z
v−1
Z
v
g(A1 e−u + ε)du dv
v−1
α(t)
g(A1 e−u + ε)du,
t ≥ t3 .
α(t)−2
By induction, one can prove in general that for k = 2, 3, . . . , n − 1,
Z α(t)
k−1
Gk (t) − Gk−1 (t) ≤ (2e)
(k − 2)!M p(t)
g(A1 e−u + ε)du,
t ≥ tk ,
α(t)−(k−1)
and so
Gn−1 (t) =
n−1
X
[Gk (t) − Gk−1 (t)]
k=1
≤ G1 (t) + M p(t)
n−1
X
k=2
(2e)k−1 (k − 2)!
Z
(3.33)
α(t)
α(t)−(k−1)
g(A1 e−u + ε)du,
EJDE-2005/134
OSCILLATION CRITERIA
13
for t ≥ tn−1 . By (3.21), (3.22) and (3.27), we obtain
Z t
z(δ(t)) λ(s)ds = p(t) ln
λ1 (t) = p(t)
z(t)
δ(t)
≤ p(t) ln M, t ≥ t1 ,
Z t
Z
λ2 (t) = p(t)
λ1 (s)ds ≤ p(t) ln M
δ(t)
t
p(s)ds
(3.34)
δ(t)
≤ 2p(t) ln M,
t ≥ t2 ,
...
λn−1 (t) ≤ 2n−2 p(t) ln M,
t ≥ tn−1 .
For t ≥ tn , set
Z
= p(t) exp en−1
D(t)
t
λn−1 (s)ds
h
Z
1 − exp −
δ(t)
t
i
Gn−1 (s)ds + G1 (t).
δ(t)
From (3.11), (3.21), (3.32), (3.33) and (3.34), we have
Z t
Z t
D(t) ≤ p(t) exp en−1
λn−1 (s)ds
Gn−1 (s)ds + G1 (t)
δ(t)
≤ G1 (t) + p(t) exp 2n−2 en−1 ln M
δ(t)
Z t
p(s)ds
δ(t)
Z
t
×
h
G1 (s) + M p(s)
δ(t)
n−1
X
α(s)
Z
(2e)k−1 (k − 2)!
i
g(A1 e−u + ε)du ds
α(s)−(k−1)
k=2
≤ G1 (t) + 2M p(t) exp((2e)n−1 ln M )
α(t)
Z
g(A1 e−u + ε)du
α(t)−1
+ M p(t) exp((2e)n−1 ln M )
Z t
Z
n−1
X
k−1
(2e)
(k − 2)!
×
p(s)
δ(t)
k=2
≤ G1 (t) + M1 p(t)
n−1
X
α(s)
g(A1 e−u + ε)du ds
α(s)−(k−1)
(2e)k−1 (k − 1)!
Z
α(t)
g(A1 e−u + ε)du,
t ≥ tn ,
α(t)−k
k=1
(3.35)
where M1 = 2M exp((2e)n−1 ln M ). Let T ∗ > tn be such that α(T ∗ ) > n + ln A1 .
It follows from (3.35) and (H) that
Z ∞
D(t)dt
T∗
Z
∞
≤
G1 (t)dt + M1
T∗
n−1
X
(2e)k−1 (k − 1)!
Z
∞
g(A1 e−u )du + 2M1
≤ 2M
α(T ∗ )
Z
T∗
−u
g(A1 e
α(T ∗ )
n−1
X
)du + 2M1
n−1
X
k=1
α(t)
g(A1 e−u ) du dt
α(t)−k
(2e)k−1 (k − 1)!
Z
∞
α(T ∗ )
k=1
∞
≤ 2M
Z
p(t)
k=1
Z
∞
k−1
(2e)
Z
Z
v
g(A1 e−u )du dv
v−k
∞
k!
α(T ∗ )−(k+1)
g(A1 e−u )du
14
E. M. ELABBASY, T. S. HASSAN, S. H. SAKER
Z
∞
∞
Z
g(e−u )du + 2M1
D(t)dt ≤ 2M
T∗
0
n−1
X
(2e)k−1 k!
EJDE-2005/134
∞
Z
g(e−u )du < ∞.
0
k=1
Since
n−1
Z
t
Z
λn−1 (s)ds exp −
p(t) exp e
= p(t) exp en−1
Gn−1 (s)ds − G1 (t)
δ(t)
δ(t)
t
Z
t
λn−1 (s)ds − D(t),
t ≥ tn ,
δ(t)
it follows from (3.30) that
Z
λ(t) ≥ p(t) exp en−1
t
λn−1 (s)ds − D(t),
t ≥ tn .
(3.36)
δ(t)
One can easily show that γex ≥ x + ln(γ + 1) for γ > 0, and so for t ≥ tn ,
Z t
pn (t)λ(t) ≥ p(t)e1−n (en−1 pn (t)) exp en−1
λn−1 (s)ds − pn (t)D(t)
δ(t)
t
Z
λn−1 (s)ds + e1−n p(t) ln(en−1 pn (t) + 1) − pn (t)D(t),
≥ p(t)
δ(t)
that is, for t ≥ tn ,
Z
pn (t)λ(t) − p(t)
t
λn−1 (s)ds ≥ e1−n p(t) ln(en−1 pn (t) + 1) − pn (t)D(t). (3.37)
δ(t)
For N > δ −n (T ∗ ), we have
Z N
Z
pn (t)λ(t)dt −
T a st
≥ e1−n
N
p(t)
T a st
Z
t
Z
λn−1 (s) ds dt
δ(t)
N
p(t) ln(en−1 pn (t) + 1)dt −
(3.38)
N
Z
pn (t)D(t)dt.
T∗
T a st
Let δ 1 (t) = δ(t), δ k+1 (t) = δ(δ k (t)), k = 1, 2, . . . , n. Then by interchanging the
order of integration, we have
Z N
Z t
Z δ(N )
Z δ−1 (t)
p(t)
λn−1 (s) ds dt ≥
λn−1 (t)
p(s) ds dt
T a st
T∗
δ(t)
Z
t
δ(N )
=
t
Z
p(t)p1 (t)
λn−2 (s) ds dt
T∗
Z
δ(t)
δ 2 (N )
≥
δ −1 (t)
Z
λn−2 (t)
T∗
Z δ2 (N )
=
p(s)p1 (s) ds dt
t
Z
t
p(t)p2 (t)
T∗
...
Z
≥
λn−3 (s) ds dt
δ(t)
δ n (N )
T∗
λ(t)pn (t)dt.
EJDE-2005/134
OSCILLATION CRITERIA
From this and (3.38), we have
Z
Z N
Z N
p(t) ln(en−1 pn (t) + 1)dt −
pn (t)λ(t)dt ≥ e1−n
T∗
δ n (N )
15
N
pn (t)D(t)dt, (3.39)
T a st
which together with (3.21) yields
Z N
Z N
Z
n
1−n
n−1
n
λ(t)dt ≥ e
p(t) ln(e
2
pn (t) + 1)dt − 2
δ n (N )
T a st
N
D(t)dt,
T a st
or
ln
x(δ n (N ))
≥ 2−n e1−n
x(N )
Z
N
p(t) ln(en−1 pn (t) + 1)dt −
Z
N
D(t)dt.
(3.40)
T∗
T a st
In view of (1.9) and (3), we have
x(δ n (N ))
= ∞.
N →∞
x(N )
lim
(3.41)
On the other hand, (3.22) implies that
x(δ n (N ))
x(δ 1 (N )) x(δ 2 (N ))
x(δ n (N ))
=
.
·
·
·
≤ M n.
x(N )
x(N ) x(δ 1 (N ))
x(δ n−1 (N ))
This contradicts (3.41) and completes the proof.
4. Examples
In this section we introduce some examples to illustrate our main results.
Example 4.1. Consider the neutral delay differential equation
0
5
x(t) − ( + sin t)x(t − π) + f (t, x(τ (t))) = 0, t ≥ 3.
(4.1)
2
For f (t, u) = p(t)f (u), with
(
u[1 + (1 + ln2 |u|)−1 ], u 6= 0,
f (u) =
(4.2)
0,
u = 0,


|u| > 1,
1,
(4.3)
g(u) = (1 + ln2 |u|)−1 , 0 < |u| ≤ 1,


0,
u = 0,
1
t
1
+
, τ (t) = ,
(4.4)
p(t) =
et ln 2 t ln t
2
R∞
with 3 p(t)dt = ∞. It is easily seen that condition (H) holds. We check that the
conditions (1.8) and (1.10) in Corollary 1.3 hold. In fact, for t ≥ 3,
Z t
Z t
1
1 1
ln 2 1
p(s)ds =
+
ds = − ln 1 −
≥ ,
t
t
es ln 2 s ln s
e
ln t
e
2
2
Rt
lim inf t→∞ t/2 p(s)ds = 1/e, and
Z ∞
Z
Z
Z
t
t
1
1
p(t) exp
p(s)ds −
− 1 dt ≥ ∞p(t)
p(s)ds − dt
e
e
3
t/2
3
t/2
Z ∞
1
1
ln 2
≥−
ln[1 −
]dt = ∞,
e ln 2 3 t
ln t
16
E. M. ELABBASY, T. S. HASSAN, S. H. SAKER
because
Z
EJDE-2005/134
∞
1
ln 2
dt = ∞ and lim (ln t) ln[1 −
] = − ln 2.
t→∞
t
ln
t
ln t
3
By Theorem 1.1 every solution of (4.1) oscillates.
Example 4.2. Consider the neutral delay differential equation
5
π 0
(4.5)
x(t) − ( + sin t)x(t − ) + f (t, x(τ (t))) = 0, t ≥ 3,
2
2
For
δ
t
1
p(t) = , τ (t) = , δ <
, λ > 1, f (t, u) = p(t)f (u),
t
λ
e ln λ
R∞
where f (u) and g(u) are defined by (4.2) and (4.3) and with 3 p(t)dt = ∞, and
Z t
Z t
δ
t
1
p(s)ds =
ds = δ ln t − ln
= δ ln λ <
λ
e
t/λ
t/λ s
It is easily seen that condition (H) holds. We check that the conditions (1.6) and
(1.10) in Theorem 1.2 hold. In fact, for t ≥ 3,
Z t
Z t
t
δ
lim inf
ds = δ(ln t − ln ) = δ ln λ > 0
p(s)ds = lim inf
t→∞
t→∞
s
λ
t/λ
t/λ
and
Z
∞
Z
t
∞
p(s)ds) − 1]dt ≥
p(t)[exp(
3
Z
t/λ
Z
p(t)(
3
Z
t
p(s)ds)dt
t/λ
∞
≥
3
δ 2 ln λ
dt = δ 2 ln λ(∞) = ∞
t
By Corollary 1.3 every solution of (4.5) oscillates.
Example 4.3. Consider the neutral delay differential equation
0
5
x(t) − ( + sin t)x(t − π) + f (t, x(τ (t))) = 0, t ≥ 3,
2
where
τ (t) = t − 1 and f (t, u) = [exp 3(sin t − 1) + |u|]1/3 u.
(4.6)
Let p(t) = exp(sin t) − 0.1 and g(u) = e2 |u|1/3 . It is easy to see that assumption
(H) holds. Clearly
Z t
1
lim inf
p(s)ds < ,
t→∞
e
t−1
Z ∞
Z ∞
Z t+1
Z t+1
p(t) ln
p(s)ds + 1 dt ≥
exp(sin t − 1) ln
exp(sin s)ds dt
0
t
0
By Jensen’s inequality,
Z ∞
Z
Z t+1
p(t) ln
p(s)ds + 1 dt ≥
0
t
t
∞
Z
t+1
exp(sin t − 1)
sin s ds dt
0
t
Z
2 sin 2−1 ∞
1
=
exp(sin t) sin(t + )dt.
e
2
0
EJDE-2005/134
OSCILLATION CRITERIA
17
Rt
On the other hand, it is easy to see that 0 exp(sin s) cos s ds is bounded and
Z 2π
exp(sin t) sin tdt > 0.
0
Thus
Z
∞
p(t) ln
0
Z
t+1
p(s)ds + 1 dt = ∞.
t
By Corollary 1.4, every solution of (4.6) oscillates.
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Elmetwally M. Elabbasy
Department of Mathematics, Faculty of Science, Mansoura University, Mansoura, 35516,
Egypt
E-mail address: emelabbasy@mans.edu.eg
Taher S. Hassan
Department of Mathematics, Faculty of Science, Mansoura University, Mansoura, 35516,
Egypt
E-mail address: tshassan@mans.edu.eg
Samir H. Saker
Department of Mathematics, Faculty of Science, Mansoura University, Mansoura, 35516,
Egypt
E-mail address: shsaker@mans.edu.eg