Journal of Applied Mathematics and Stochastic Analysis, 16:3 (2003), 257-265.
c
Printed in the USA 2003
by North Atlantic Science Publishing Company
ON AN EVEN ORDER NEUTRAL
DIFFERENTIAL INEQUALITY
PEIGUANG WANG
Hebei University
College of Electronic and Information
Baoding 071002, PR of China
E-mail: pgwang@mail.hbu.edu.cn
and
LOKENATH DEBNATH
University of Texas-Pan American1
Department of Mathematics
Edinburg, TX 78539 USA
E-mail: debnath@panam.edu
(Received July, 2002; Revised November, 2002)
In this paper, we prove new results related to the nonexistence criteria for eventually
positive solutions of certain even order neutral differential inequality with distributed
deviating arguments.
Keywords: Differential Inequality, Neutral Type, Distributed Deviating Arguments,
Eventually Postive Solution.
AMS (MOS) subject classification: 34K11, 34K40.
1
Introduction
In order to make this paper self-contained, we introduce the following definition.
Definition 1: The function f (t) is said to be eventually zero if there exists a
sufficiently large tµ such that f (t) ≡ 0 holds for t ≥ tµ .
This paper is concerned with nonexistence conditions of eventually positive solutions
of the even order neutral differential inequality with distributed deviating arguments
Z b
p(t, ξ)f (x[g(t, ξ)])dσ(ξ) ≤ 0,
t ≥ t0 ,
(1)
[x(t) + c(t)x(t − τ )](n) +
a
in which τ > 0 is a constant, n is an even positive integer; c(t) ∈ C(I, R), 0 ≤ c(t) ≤ 1,
and p(t, ξ) ∈ C(I × J, R+ ) is not eventually zero on any Iµ × J, I = [t0 , ∞), J = [a, b],
1 This
work was supported by the Faculty Research Council of the University of Texas-Pan American
257
258
P. WANG and L. DEBNATH
Iµ = [tµ , ∞), tµ ≥ t0 , R+ = [0, ∞). Furthermore, we assume that g(t, ξ) ∈ C(I × J, R)
d
is nondecreasing with respect to t and ξ, respectively, dt
g(t, a) exists, g(t, ξ) ≤ t for
ξ ∈ J, and lim inf {g(t, ξ)} = ∞; f (x) ∈ C(R, R), and xf (x) > 0 (x 6= 0); σ(ξ) ∈ (J, R)
t→∞,ξ∈J
is nondecreasing; integral of inequality (1) is in Lebesgue-Stieltjes sense.
Recently, Li and Cui [1] have obtained some results dealing with a class of even order
neutral differential inequalities with applications. On the other hand, Liu and Fu [2]
have studied nonlinear differential inequality with distributed deviating arguments and
their applications. These authors provided some results on nonexistence conditions of
eventually positive solutions of inequality (1). For example,
Theorem A:(See [1]) If 0 ≤ c(t) ≤ 1, and
Z tZ
t0
b
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ)ds = ∞,
a
then inequality (1) has no eventually positive solutions.
Theorem B:(See ([2]) Assume that f (−x) = −f (x), x ∈ (0, ∞), and
f (x)
≥ λ,
x
x ∈ (0, ∞),
for some constant
λ > 0.
(2)
If there exists a monotonically increasing function ϕ(t) ∈ C 0 (I, (0, ∞)) such that
Z
t
[λϕ(s)
t0
Z
b
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ) − Aϕ0 (s)]ds = ∞
a
for any number A > 0, then inequality (1) has no eventually positive solutions.
The purpose of this paper is to obtain two new results related to the nonexistence
criteria for eventually positive solutions of inequality (1). In the established nonexistence
criteria, there is a general class of function H(t, s) as the parameter function. By
choosing a different function H(t, s), we are able to derive some useful corollaries.
Definition 2: The solution x(t) ∈ C (n) (I, R) of inequality (1) is said to be eventually positive if there exists a sufficiently large positive number T ≥ t0 such that the
inequality x(t) > 0 holds for t ≥ T .
To develop the nonexistence criteria of eventually positive solutions of inequality
(1), we first need the following Lemmas:
Lemma 1: (See [1]) Assume that x(t) is an eventually positive solution of inequality
(1). Let
y(t) = x(t) + c(t)x(t − τ ).
(3)
Then there exists a t1 ≥ t0 such that
y(t) > 0, y 0 (t) > 0, y (n−1) (t) > 0 and y (n) (t) ≤ 0,
t ≥ t1 .
Lemma 2: (See [3]) Let x(n) (t) ∈ C(I, R+ ). If x(n) (t) is eventually of one sign for
all large t, and x(n) (t) × x(n−1) (t) ≤ 0 for t1 > t0 , then there exists a constant θ ∈ (0, 1)
such that for sufficiently large t, there exists a constant Mθ > 0 satisfying
|u0 (t/2)| ≥ Mθ tn−2 |u(n−1) (t)|.
Even Order Neutral Differential Inequality
2
259
Main Results
The following theorems provide sufficient conditions leading to nonexistence of eventually positive solutions for inequality (1).
Theorem 1: Assume that the condition of Theorem B holds, and there exist functions H(t, s) ∈ C 0 (D; R), h(t, s) ∈ C(D; R), with D = {(t, s)|t ≥ s ≥ t0 } satisfying
(H1 ) H(t, t) = 0, t ≥ t0 ; H(t, s) > 0, t > s ≥ t0 ;
p
(H2 ) Ht 0 (t, s) ≥ 0, Hs 0 (t, s) ≤ 0, and −Hs 0 (t, s) = h(t, s) H(t, s), (t, s) ∈ D.
If
lim sup
t→∞
1
H(t, t0 )
Z
t
H(t, s)
Z
t0
t→∞
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ)ds = ∞,
(4)
a
and
lim sup
b
1
H(t, t0 )
Z
t
t0
h2 (t, s)
g n−2 (s, a)g 0 (s, a)
ds < ∞,
(5)
then inequality (1) has no eventually positive solutions.
Proof: Assume to the contrary that x(t) is an eventually positive solution of inequality (1). Then from limt→∞, inf ξ∈J {g(t, ξ)} = ∞, there exists a t1 ≥ t0 such that
x(t) > 0, x(t − τ ) > 0 and x[g(t, ξ)] > 0 for t ≥ t1 and ξ ∈ J. From (2) and (3),
inequality (1) can be written as
0 ≥
y (n) (t) +
Z
b
p(t, ξ)f (x[g(t, ξ)])dσ(ξ)
(6)
a
≥
y (n) (t) + λ
Z
b
p(t, ξ){y[g(t, ξ)] − c[g(t, ξ)]x[g(t, ξ) − τ ]}dσ(ξ).
a
From Lemma 1, y 0 (t) > 0 and y(t) ≥ x(t), t ≥ t1 , hence y[g(t, ξ)] ≥ y[g(t, ξ) − τ ] ≥
x[g(t, ξ) − τ ]. Thus
y (n) (t) + λ
Z
b
p(t, ξ){1 − c[g(t, ξ)]}y[g(t, ξ)]dσ(ξ) ≤ 0,
t ≥ t1 .
(7)
a
Furthermore, in view of g(t, ξ) being nondecreasing with respect to ξ, we have
y (n) (t) + λy[g(t, a)]
Z
b
p(t, ξ){1 − c[g(t, ξ)]}dσ(ξ) ≤ 0,
t ≥ t2 .
(8)
a
Let
z(t) =
y (n−1) (t)
y[ g(t,a)
2 ]
.
(9)
d
d
g(t, a) exists, we obtain y 0 [g(t, a)] = dy
Then z(t) ≥ 0. Since dt
dg dt g(t, a). Furthermore,
(n)
from Lemma 1, y (t) ≤ 0, and in view of g(t, ξ) being nondecreasing with respect to ξ,
g(t, ξ) ≤ t for ξ ∈ J, we obtain y (n−1) (t) ≤ y (n−1) [g(t, a)] ≤ y (n−1) [ g(t,a)
2 ]. Thus, from
260
P. WANG and L. DEBNATH
Lemma 2, we have
z 0 (t) =
y (n) (t)
y[ g(t,a)
2 ]
g(t,a)
−
1 y (n−1) (t)y 0 [ 2 ]g 0 (t, a)
2
y 2 [ g(t,a) ]
2
(10)
≤
y
(n)
(t)
y[ g(t,a)
2 ]
−
Mθ n−2
g
(t, a)g 0 (t, a)z 2 (t),
2
Furthermore, from y 0 (t) > 0 and (8), for t ≥ t2 , we obtain
Z b
Mθ n−2
g
z 0 (t) ≤ −λ
p(t, ξ){1 − c[g(t, ξ)]}dσ(ξ) −
(t, a)g 0 (t, a)z 2 (t).
2
a
(11)
Integrating by parts for any t > T ≥ t1 , and using the properties (H1 ) and (H2 ), we
have
Z b
Z t
H(t, s)
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ)ds
λ
T
a
Z
Z t
Mθ t
H(t, s)z 0 (s)ds −
H(t, s)g n−2 (s, a)g 0 (s, a)z 2 (s)ds
≤−
2 T
T
Z
Z t
Mθ t
H(t, s)dz(s) −
H(t, s)g n−2 (s, a)g 0 (s, a)z 2 (s)ds
=−
2 T
T
Z t
p
h(t, s) H(t, s)z(s)ds
= H(t, T )z(T ) −
T
Z t
Mθ
H(t, s)g n−2 (s, a)g 0 (s, a)z 2 (s)ds
−
2 T
Z
1 t hp
Mθ H(t, s)g n−2 (s, a)g 0 (s, a)z(s)
= H(t, T )z(T ) −
2 T
#2
Z t
h2 (t, s)
h(t, s)
p
ds,
ds +
+
n−2
(s, a)g 0 (s, a)
Mθ g n−2 (s, a)g 0 (s, a)
T 2Mθ g
which implies that
Z t"
Z
λH(t, s)
#
h2 (t, s)
ds
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ) −
2Mθ g n−2 (s, a)g 0 (s, a)
T
a
Z
1 t hp
≤ H(t, T )z(T ) −
Mθ H(t, s)g n−2 (s, a)g 0 (s, a)z(s)
2 T
#2
h(t, s)
+p
ds.
(12)
Mθ g n−2 (s, a)g 0 (s, a)
b
Furthermore, in view of (H2 ), for t1 ≥ t0 , we have H(t, t1 ) ≤ H(t, t0 ). Thus
#
Z t"
Z b
h2 (t, s)
ds
λH(t, s)
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ) −
2Mθ g n−2 (s, a)g 0 (s, a)
t1
a
≤ H(t, t1 )z(t1 ) ≤ H(t, t0 )z(t1 )
Even Order Neutral Differential Inequality
1
H(t, t0 )
=
1
H(t, t0 )
Z t"
261
#
h2 (t, s)
λH(t, s)
ds
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ) −
2Mθ g n−2 (s, a)g 0 (s, a)
t0
a
Z t1 Z t "
Z b
+
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ)
λH(t, s)
Z
t0
b
t1
2
a
h (t, s)
ds
2Mθ g n−2 (s, a)g 0 (s, a)
Z
Z t1
H(t, s) b
≤ z(t1 ) +
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ)ds
t0 H(t, t0 ) a
Z t1 Z b
≤ z(t1 ) +
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ)ds.
−
(13)
a
t0
It follows from (13) that
1
lim sup
H(t,
t0 )
t→∞
≤ lim sup
t→∞
1
H(t, t0 )
Z
Z
t
b
λH(t, s)
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ)
t0
a
"
Z t
Z b
λH(t, s)
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ)
t0
2
a
h (t, s)
ds
2Mθ g n−2 (s, a)g 0 (s, a)
Z t
1
h2 (t, s)
+ lim sup
ds
n−2
(s, a)g 0 (s, a)
t→∞ H(t, t0 ) t0 g
Z t1 Z b
≤ z(t1 ) +
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ)ds
−
t0
a
+ lim sup
t→∞
1
H(t, t0 )
Z
t
t0
h2 (t, s)
g n−2 (s, a)g 0 (s, a)
ds
< ∞,
which contradicts (4). Therefore, the proof of Theorem 1 is complete.
Remark 1: From Theorem 1, we can establish various sufficient conditions by means
of the choices of parameter function H(t, s). For example, choosing H(t, s) = (t−s)m−1 ,
m−3
t ≥ s ≥ t0 , in which m > 2 is an integer, we obtain h(t, s) = (m−1)(t−s) 2 , t ≥ s ≥ t0 .
From Theorem 1, we have
Corollary 1: If there exists an integer m > 2 such that
lim sup
t→∞
Z
1
tm−1
t
(t − s)m−1
t0
lim sup
t→∞
Z
b
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ)ds = ∞,
(14)
a
1
tm−1
Z
t
t0
(m − 1)2 (t − s)m−3
ds < ∞,
g n−2 (s, a)g 0 (s, a)
then inequality (1) has no eventually positive solutions.
If
Z t
Z b
1
lim sup
H(t, s)
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ)ds < ∞,
t→∞ H(t, t0 ) t0
a
(15)
(16)
262
P. WANG and L. DEBNATH
we have the following result:
Theorem 2: Assume that the conditions of Theorem 1 and (16) hold. If Ht0 (t, s) is
nondecreasing, and there exists a function ϕ(t) ∈ C(I, R) satisfying
Z t"
1
lim inf
t→∞ H(t, t0 )
−
1
t→∞ H(t, t0 )
lim
Z
λH(t, s)
Z
u
b
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ)
a
h2 (t, s)
ds ≥ ϕ(u), u ≥ t0 ,
2Mθ g n−2 (s, a)g 0 (s, a)
(17)
t
H(t, s)g n−2 (s, a)g 0 (s, a)ϕ2+ (s)ds = ∞, ϕ+ (s) = max{ϕ(s), 0}, (18)
s≥t0
t0
then inequality (1) has no eventually positive solutions.
Proof: Assume to the contrary that x(t) is an eventually positive solution of inequality (1). Then from the proof of Theorem 1, there exists a t1 ≥ t0 such that
Z
z 0 (t) ≤ −λ
b
p(t, ξ){1 − c[g(t, ξ)]}dσ(ξ) −
a
Mθ n−2
g
(t, a)g 0 (t, a)z 2 (t).
2
(19)
Thus
λ
≤
Z
−
t
H(t, s)
t1
Z t
Z
b
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ)ds
a
H(t, s)z 0 (s)ds −
t1
= H(t, t1 )z(t1 ) −
Mθ
−
2
Z
Mθ
2
Z
t
H(t, s)g n−2 (s, a)g 0 (s, a)z 2 (s)ds
t1
Z tp
H(t, s)h(t, s)z(s)ds
t1
t
H(t, s)g n−2 (s, a)g 0 (s, a)z 2 (s)ds
(20)
t1
and
λ
Z
t
H(t, s)
t1
Z
b
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ)ds
a
1
≤ H(t, t1 )z(t1 ) −
2
Z t np
Mθ g n−2 (s, a)g 0 (s, a)H(t, s)z(s)
t1
)2
h(t, s)
+p
ds
Mθ g n−2 (s, a)g 0 (s, a)
Z t
h2 (t, s)
ds
+
n−2 (s, a)g 0 (s, a)
t1 2Mθ g
Z t
h2 (t, s)
ds.
≤ H(t, t1 )z(t1 ) +
n−2
(s, a)g 0 (s, a)
t1 2Mθ g
(21)
Even Order Neutral Differential Inequality
263
Furthermore, for t > u ≥ t0 , we have
#
Z t"
Z b
h2 (t, s)
1
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ) −
ds
λH(t, s)
H(t, t0 ) u
2Mθ g n−2 (s, a)g 0 (s, a)
a
≤
H(t, u)
z(u).
H(t, t0 )
(22)
From (17) and (H2 ), we conclude that
Z t"
Z b
1
λH(t, s)
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ)
ϕ(u) ≤
H(t, t0 ) u
a
h2 (t, s)
−
ds
2Mθ g n−2 (s, a)g 0 (s, a)
H(t, u)
≤
z(u) ≤ z(u),
H(t, t0 )
(23)
which implies that
ϕ2+ (u) ≤ z 2 (u).
(24)
Let
v(t)
w(t)
1
H(t, t0 )
=
1
H(t, t0 )
=
Z tp
H(t, s)h(t, s)z(s)ds
Z
t1
t
Mθ
H(t, s)g n−2 (s, a)g 0 (s, a)z 2 (s)ds.
2
t1
Then, from (20), we find
v(t)+w(t) ≤
1
H(t, t1 )
z(t1 )−
H(t, t0 )
H(t, t0 )
Z
t
λH(t, s)
t1
Z
b
p(s, ξ){1−c[g(s, ξ)]}dσ(ξ)ds. (25)
a
It follows from (17) that
lim inf
t→∞
1
H(t, t0 )
Z
t
λH(t, s)
Z
u
b
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ)ds ≥ ϕ(u).
a
Furthermore, we obtain
lim sup
t→∞
1
H(t, t0 )
Z
1
− lim inf
t→∞ H(t, t0 )
t
λH(t, s)
t1
Z
t
t1
Z
b
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ)ds
a
2
h (t, s)
ds ≥ ϕ(t1 ).
2Mθ g n−2 (s, a)g 0 (s, a)
It turns out from (26) and (16)
lim inf
t→∞
1
H(t, t0 )
Z
t
t1
h2 (t, s)
2Mθ
g n−2 (s, a)g 0 (s, a)
ds < ∞.
(26)
264
P. WANG and L. DEBNATH
Thus, there exists a sequence {tn }∞
1 in [t1 , ∞) such that limn→∞ tn = ∞ that satisfies
Z tn
1
h2 (tn , s)
lim
ds < ∞.
(27)
n→∞ H(tn , t0 ) t
2Mθ g n−2 (s, a)g 0 (s, a)
1
Result (27) implies that
lim sup{v(t) + w(t)}
t→∞
1
≤ z(t1 ) − lim inf
t→∞ H(t, t0 )
Z
t
λH(t, s)
t1
Z
b
p(s, ξ){1 − c[g(s, ξ)]}dσ(ξ)ds
a
4
≤ z(t1 ) − ϕ(t1 ) = M.
(28)
Then, for any sufficiently large n, we have
v(tn ) + w(tn ) < M1 ,
(29)
where M1 > M , M and M1 are constant. According to the definition of w(t), we have
Z t
Mθ (Ht0 (t, s)H(t, t0 ) − Ht0 (t, t0 )H(t, s)) n−2
(s, a)g 0 (s, a)z 2 (s)ds.
g
w0 (t) =
2H 2 (t, t0 )
t1
Since Ht0 (t, s) is nondecreasing and (H2 ) holds, we have w0 (t) ≥ 0, thus, w(t) is increasing, and lim w(t) = l exists, where l is finite or infinite. In the case of lim w(tn ) = ∞.
t→∞
n→∞
Consequently, it follows from (29) that
lim v(tn ) = −∞,
n→∞
and
(30)
M1
v(tn )
+1<
.
w(tn )
w(tn )
Thus, for any 0 < ε < 1 and sufficiently large n, we have
v(tn )
< ε − 1 < 0.
w(tn )
(31)
On the other hand, by using the Schwartz inequality, for t ≥ t1 , we obtain
Z tn
2
p
1
0 ≤ v 2 (tn ) = 2
H(tn , s)h(tn , s)z(s)ds
H (tn , t0 )
t1
Z tn
Mθ
1
n−2
0
2
H(tn , s)g
≤
(s, a)g (s, a)z (s)ds
H(tn , t0 ) t1 2
Z tn
2h2 (tn , s)
1
ds
×
H(tn , t0 ) t1 Mθ g n−2 (s, a)g 0 (s, a)
Z tn
2h2 (tn , s)
1
ds.
= w(tn )
n−2
H(tn , t0 ) t1 Mθ g
(s, a)g 0 (s, a)
Then,
0≤
1
v 2 (tn )
≤
w(tn )
H(tn , t0 )
Z
tn
t1
Mθ
2h2 (tn , s)
ds.
n−2
g
(s, a)g 0 (s, a)
(32)
Even Order Neutral Differential Inequality
265
It follows from (27) that
v 2 (tn )
< ∞.
n→∞ w(tn )
(33)
0 ≤ lim
In view of (31), we obtain
v(tn )
v 0 (tn )
= lim 0
≤ ε − 1 < 0,
n→∞ w(tn )
n→∞ w (tn )
lim
and then,
v 2 (tn )
2v(tn )v 0 (tn )
v 0 (tn )
= lim
= 2 lim v(tn ) lim 0
= ∞,
0
n→∞ w(tn )
n→∞
n→∞
n→∞ w (tn )
w (tn )
lim
which contradicts (33). Thus, we have lim w(t) = l < ∞. Furthermore, according to
t→∞
(24), we conclude that
1
t→∞ H(t, t0 )
lim
1
t→∞ H(t, t0 )
≤ lim
Z
Z
t
H(t, s)g n−2 (s, a)g 0 (s, a)ϕ2+ (s)ds
t1
t
H(t, s)g n−2 (s, a)g 0 (s, a)z 2 (s)ds =
t1
2
lim w(t) < ∞,
Mθ t→∞
(34)
which implies that
1
H(t, t0 )
lim
t→∞
=
≤
Z
t
H(t, s)g n−2 (s, a)g 0 (s, a)ϕ2+ (s)ds
t0
Z t1 Z t 1
H(t, s)g n−2 (s, a)g 0 (s, a)ϕ2+ (s)ds
+
lim
t→∞ H(t, t0 )
t0
t1
Z t1
2
H(t, s)g n−2 (s, a)g 0 (s, a)ϕ2+ (s)ds +
lim w(t) < ∞.
M
θ t→∞
t0
The latter contradicts (18). Therefore, the proof of Theorem 2 is complete.
References
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