Electronic Journal of Differential Equations, Vol. 2005(2005), No. 53, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
THROUGHOUT POSITIVE SOLUTIONS OF SECOND-ORDER
NONLINEAR DIFFERENTIAL EQUATIONS
ZHENGUO ZHANG, CHUNJIAO WANG, QIAOLUAN LI, FANG LI
Abstract. In this paper, we consider the second-order nonlinear and the
nonlinear neutral functional differential equations
(a(t)x0 (t))0 + f (t, x(g(t))) = 0,
t ≥ t0
(a(t)(x(t) − p(t)x(t − τ ))0 )0 + f (t, x(g(t))) = 0,
t ≥ t0 .
Using the Banach contraction mapping principle, we obtain the existence of
throughout positive solutions for the above equations.
1. Introduction
Recently, there has been an increasing interest in the study of the oscillation
and nonoscillation of solutions of second-order ordinary and delay neutral differential and difference equations. Also eventually positive solutions and asymptotic
behavior of nonoscillatory solutions have been investigated widely. Delay differential equations play a very important role in many practical problems. The papers
[3, 4, 7, 8, 11, 12, 15] discuss the oscillation of second order differential and difference equations. The papers [1, 5] discuss the oscillation and non-oscillation criteria
for second order differential equations. Of course there is also the discussion of
the existence of eventually positive solutions, such as [10, 6, 13, 14]. But there are
relatively few which guarantee the existence of throughout positive solutions. The
paper [9] studies the positive solutions of the following second order non-neutral
ordinary differential equation
y 00 (t) + F (t, y(t)) = 0,
t≥a
where F : [a, ∞) × R → R is continuous and nonnegative. We have studied further and extended the results of Erik Wahlén [9] to the self-conjugate and neutral
functional differential equations. We obtain the existence of throughout positive
solutions by introducing a weighted norm (see [2, 9]) and using the Banach contraction mapping principle (see [2]).
2000 Mathematics Subject Classification. 34C10, 34K11.
Key words and phrases. Nonlinear differential equations; neutral term;
eventually positive solution; throughout positive solution.
c
2005
Texas State University - San Marcos.
Submitted December 13, 2004. Published May 19, 2005.
Supported by the Natural Science Foundation of Hebei Province and by the Main
Foundation of Hebei Normal University.
1
2
Z. ZHANG, C. WANG, Q. LI, F. LI
EJDE-2005/53
In this paper, we are concerned with existence of throughout positive solutions
for the following self-conjugate nonlinear differential equations
(a(t)x0 (t))0 + f (t, x(g(t))) = 0,
t ≥ t0
0 0
(a(t)(x(t) − p(t)x(t − τ )) ) + f (t, x(g(t))) = 0,
(1.1)
t ≥ t0
(1.2)
where a(t) > 0 is continuous; f (t, x) is continuous and satisfies f (t, x)x > 0 for
x 6= 0; g(t) is continuous, increasing and satisfies g(t) ≤ t, limt→∞ g(t) = ∞.
1.1. Definitions. A solution of differential equation is said to be oscillatory if it
has arbitrarily large zeros; otherwise it is said to be non-oscillatory.
A solution of differential equation is said to be eventually positive solution if
there exists some T ≥ t0 such that x(t) > 0 for all t ≥ T .
A solution of differential equation is said to be throughout positive solution if
x(t) > 0 for all t ≥ t0 .
Related Lemmas. To obtain our main results, we need the following lemma.
Lemma 1.1. Assume x(t) is bounded, limt→∞ p(t) = p, p 6= ±1,
z(t) = x(t) − p(t)x(t − τ ),
lim z(t) = l,
t→∞
then limt→∞ x(t) exists and limt→∞ x(t) = l/(1 − p).
Proof. (1) p ∈ (−∞, −1). Since x(t) is bounded, we get that lim supt→∞ x(t) =
M and lim inf t→∞ x(t) = m exist. Then there exists a sequence {tn } such that
limn→∞ x(tn − τ ) = M and
l = lim sup z(tn ) = lim sup(x(tn ) − p(tn )x(tn − τ )) ≥ m − pM .
n→∞
n→∞
Similarly there exists a sequence {t0n } such that limn→∞ x(t0n − τ ) = m and
l = lim inf z(t0n ) = lim inf (x(t0n ) − p(t0n )x(t0n − τ )) ≤ M − pm .
n→∞
n→∞
So we have M − pm ≥ m − pM , that is, (1 + p)M ≥ (1 + p)m. In view of 1 + p < 0,
we get M ≤ m. Hence M = m and limt→∞ x(t) exists. By the assumption, we
obtain limt→∞ x(t) = 1/(1 − p).
(2) p ∈ (−1, 0). Similarly, there exists a sequence {tn } such that limn→∞ x(tn ) =
M . Then there exists a sequence {t0n } such that limn→∞ x(t0n ) = m and
l = lim sup z(tn ) = lim sup(x(tn ) − p(tn )x(tn − τ )) ≥ M − pm,
n→∞
n→∞
l = lim inf z(t0n ) = lim inf (x(t0n ) − p(t0n )x(t0n − τ )) ≤ m − pM .
n→∞
n→∞
Therefore, M − pm ≤ m − pM , that is, (1 + p)M ≤ (1 + p)m. In view of 1 + p > 0,
we get M ≤ m. Hence M = m and limt→∞ x(t) exists. By the assumption, we
obtain limt→∞ x(t) = 1/(1 − p).
(3) p ∈ [0, 1). Similarly, there exists a sequence {tn } such that limn→∞ x(tn ) = M .
Then there exists a sequence {t0n } such that limn→∞ x(t0n ) = m and
l = lim sup z(tn ) = lim sup(x(tn ) − p(tn )x(tn − τ )) ≥ M (1 − p),
n→∞
n→∞
l = lim inf z(t0n ) = lim inf (x(t0n ) − p(t0n )x(t0n − τ )) ≤ m(1 − p) .
n→∞
n→∞
Therefore, M (1−p) ≤ m(1−p). In view of 1−p > 0 we get M ≤ m. Hence M = m
and limt→∞ x(t) exists. By the assumption, we obtain limt→∞ x(t) = 1/(1 − p).
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THROUGHOUT POSITIVE SOLUTIONS
3
(4) p ∈ (1, +∞). Similarly, there exists a sequence {tn } such that limn→∞ x(tn −
τ ) = M . Then there exists a sequence {t0n } such that limn→∞ x(t0n − τ ) = m and
l = lim sup z(tn ) = lim sup(x(tn ) − p(tn )x(tn − τ )) ≤ M (1 − p),
n→∞
l = lim inf
n→∞
n→∞
z(t0n )
= lim inf (x(t0n ) − p(t0n )x(t0n − τ )) ≥ m(1 − p) .
n→∞
Therefore, M (1 − p) ≥ m(1 − p). In view of 1 − p < 0 we get M ≤ m. Hence
M = m and limt→∞ x(t) exists. By the assumption, limt→∞ x(t) = l/(1 − p) which
completes the proof.
2. Main Results
In this section we give existence theorems of throughout positive solutions for
equations (1.1) and (1.2). First of all we need the following conditions:
Assume that the nonlinearity f satisfies a Lipschitz condition
|f (t, u) − f (t, v)| ≤ k(t)|u − v|,
for 0 ≤ u, v ≤ C and t ≥ t0 ,
(2.1)
where the constant C will be specified in the theorems below, and k(t) > 0 is a
continuous function satisfying
Z ∞
s
k(s)ds < ∞,
(2.2)
t0 a(s)
where a(s) = min{a(θ) : min{t0 − τ, g(t0 )} ≤ θ ≤ s}.
Theorem 2.1. For equation (1.1), we define the set
X = {u ∈ C 1 [t0 , ∞), 0 ≤ u(t) ≤ M, for t ≥ t0 ; u(t) = u(t0 ), for g(t0 ) ≤ t < t0 }.
Assume that for every u ∈ X,
Z ∞
Z ∞
ds
f (θ, u(g(θ)))dθ < M .
t0 a(s) s
(2.3)
Let conditions (2.1) and (2.2) hold for 0 ≤ u, v ≤ M . Assume further that there
exists a positive integer N > 1 such that 0 < Nl < 1, where l(N ) = max{ G(g(t))
G(t) ,
R∞ s
t ≥ t0 }, G(t) = exp(N t a(s) k(s)ds). Then equation (1.1) has a throughout
positive solution x(t) on [t0 , ∞) satisfying limt→∞ x(t) = M .
Proof. Define a mapping T on X as follows
(
R ∞ ds R ∞
f (θ, x(g(θ)))dθ
M − t a(s)
s
(T x)(t) =
(T x)(t0 )
t ≥ t0
g(t0 ) ≤ t < t0 .
(2.4)
From (2.3)
have 0 ≤ (T x)(t) ≤ M , so T X ⊆ X. From the assumption G(t) =
R ∞ we
s
exp(N t a(s)
k(s)ds), we introduce the norm k · k on X, kxk = supt≥t0 |x(t)|/G(t).
Note that X is closed with respect to this norm, and therefore we have a complete
metric space.
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Z. ZHANG, C. WANG, Q. LI, F. LI
EJDE-2005/53
We now show that T is a contraction mapping on X. For any x1 , x2 ∈ X, in
view of the assumptions we have
Z ∞
Z ∞
|(T x1 )(t) − (T x2 )(t)|
ds
1
|f (θ, x1 (g(θ))) − f (θ, x2 (g(θ)))|dθ
≤
G(t)
G(t) t a(s) s
Z ∞
Z ∞
1
ds
G(g(θ))k(θ)|x1 (g(θ)) − x2 (g(θ))|
≤
dθ
G(t) t a(s) s
G(g(θ))
Z ∞
Z ∞
ds
G(g(θ))
1
G(θ)k(θ)
≤
kx1 − x2 k
dθ
G(t)
a(s) s
G(θ)
Zt ∞
l
(s − t)G(s)k(s)
ds
≤
kx1 − x2 k
a(s)
G(t)
Zt ∞
l
sG(s)k(s)
≤
ds
kx1 − x2 k
a(s)
G(t)
Zt ∞
1
l
kx1 − x2 k
(− )G0 (s)ds
=
G(t)
N
t
G(t) − 1
=l
kx1 − x2 k
N G(t)
l
≤ kx1 − x2 k.
N
Since 0 < Nl < 1, T is a contraction mapping on X. Finally we use the Banach
fixed point theorem to deduce the existence of a unique fixed point in X,
Z ∞
Z ∞
ds
f (θ, x(g(θ)))dθ.
x(t) = (T x)(t) = M −
a(s) s
t
From (2.3) we know that x(t) > 0 for t ≥ t0 . Clearly x(t) satisfies
(a(t)x0 (t))0 + f (t, x(g(t))) = 0,
thus x(t) is a throughout positive solution of (1.1) and limt→∞ x(t) = M . The
proof is complete.
Now we discuss the equation (1.2).
Theorem 2.2. Assume that limt→∞ p(t) = p, where p ∈ [0, 1) and 0 < p(t) ≤ p.
Define
X = u ∈ C 1 [t0 , ∞), 0 ≤ u(t) ≤ M, for t ≥ t0 ; u(t) = u(t0 ),
for min{g(t0 ), t0 − τ } ≤ t < t0 .
Let condition (2.1) and (2.2) hold for 0 ≤ u, v ≤ M , and we replace (2.3) by
Z ∞
Z ∞
ds
f (θ, u(g(θ)))dθ < M (1 − p).
(2.5)
t0 a(s) s
Assume further there exists a positive integer N > 1 such that 0 < (p + N1 )l <
R∞ s
) G(g(t))
1, where l(N ) = max{ G(t−τ
G(t) , G(t) , t ≥ t0 }, G(t) = exp (N t a(s) k(s)ds).
Then equation (1.2) has a throughout positive solution x(t) on [t0 , ∞) satisfying
limt→∞ x(t) = M .
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THROUGHOUT POSITIVE SOLUTIONS
Proof. Define a mapping T on X as follows
(
R∞
M (1 − p) + p(t)x(t − τ ) − t
(T x)(t) =
(T x)(t0 )
ds
a(s)
R∞
s
f (θ, x(g(θ)))dθ
5
t ≥ t0
min{t0 − τ, g(t0 )} ≤ t ≤ t0 .
For t ≥ t0 , from (2.5) and p(t) ≤ p, we have 0 ≤ (T x)(t) ≤ M (1 − p) + pM = M ,
so T X ⊆ X. We introduce the norm k · k on X, kxk = supt≥t0 |x(t)|/G(t). Now
we show that T is a contraction mapping on X. For any x1 , x2 ∈ X, in view of the
assumptions we have
|(T x1 )(t) − (T x2 )(t)|
G(t)
|x1 (t − τ ) − x2 (t − τ )|
≤ p(t)
G(t)
Z ∞
Z ∞
1
ds
|f (θ, x1 (g(θ))) − f (θ, x2 (g(θ)))|dθ
+
G(t) t a(s) s
G(t − τ ) |x1 (t − τ ) − x2 (t − τ )|
≤ p(t)
G(t)
G(t − τ )
Z ∞
Z ∞
1
ds
|f (θ, x1 (g(θ))) − f (θ, x2 (g(θ)))|dθ
+
G(t) t a(s) s
Z ∞
Z ∞
ds
G(g(θ))k(θ)|x1 (g(θ)) − x2 (g(θ))|
1
dθ
≤ p l kx1 − x2 k +
G(t) t a(s) s
G(g(θ))
Z ∞
Z ∞
ds
G(g(θ))
1
≤ p l kx1 − x2 k +
kx1 − x2 k
G(θ)k(θ)
dθ
G(t)
a(s)
G(θ)
t
s
Z ∞
l
(s − t)G(s)k(s)
kx1 − x2 k
ds
≤ p l kx1 − x2 k +
G(t)
a(s)
t
Z ∞
l
sG(s)k(s)
≤ p l kx1 − x2 k +
kx1 − x2 k
ds
G(t)
a(s)
t
Z ∞
l
1
= p l kx1 − x2 k +
kx1 − x2 k
(− )G0 (s)ds
G(t)
N
t
G(t) − 1 l kx1 − x2 k
≤ p+
N G(t)
1
≤ (p + ) l kx1 − x2 k.
N
Since 0 < (p + N1 )l < 1, T is a contraction mapping on X. Finally we use the
Banach fixed point theorem to deduce the existence of a unique fixed point in X
Z ∞
Z ∞
ds
f (θ, x(g(θ)))dθ.
x(t) = (T x)(t) = M (1 − p) + p(t)x(t − τ ) −
a(s) s
t
From the condition (2.5) and p(t)x(t − τ ) ≥ 0 we know that x(t) > 0 for t ≥ t0 .
Clearly x(t) satisfies
(a(t)(x(t) − p(t)x(t − τ ))0 )0 + f (t, x(g(t))) = 0,
thus x(t) is a throughout positive solution of (1.2) and
lim (x(t) − p(t)x(t − τ )) = M (1 − p) .
t→∞
In view of the Lemma 1.1, limt→∞ x(t) = M which completes the proof.
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Z. ZHANG, C. WANG, Q. LI, F. LI
EJDE-2005/53
Theorem 2.3. Assume that limt→∞ p(t) = p where p ∈ (−1, 0) and p ≤ p(t) < 0
and define
Y = { u ∈ C 1 [t0 , ∞), 0 ≤ u(t) ≤ M (1 − p), fort ≥ t0 ; u(t) = u(t0 ),
for min{g(t0 ), t0 − τ } ≤ t < t0 }.
Let conditions (2.1) and (2.2) hold for 0 ≤ u, v ≤ M (1 − p). Assume that for every
u∈Y,
Z ∞
Z ∞
ds
f (θ, u(g(θ)))dθ < M (1 − p2 ).
(2.6)
t0 a(s) s
Assume further that there exists a positive integer N > 1 such that 0 < ( N1 −
R∞ s
) G(g(t))
p)l < 1, where l(N ) = max{ G(t−τ
G(t) , G(t) , t ≥ t0 }, G(t) = exp(N t a(s) k(s)ds).
Then equation (1.2) has a throughout positive solution x(t) on [t0 , ∞) satisfying
limt→∞ x(t) = M .
Proof. Define a mapping T on Y as follows
(
R∞
M (1 − p) + p(t)x(t − τ ) − t
(T x)(t) =
(T x)(t0 )
ds
a(s)
R∞
s
f (θ, x(g(θ)))dθ
t ≥ t0
min{t0 − τ, g(t0 )} ≤ t ≤ t0 .
Since p(t) < 0, we easily know that 0 ≤ (T x)(t) ≤ M (1 − p). So T X ⊆ X. We
introduce the norm k · k on Y , kxk = supt≥t0 |x(t)|/G(t). We now show that T is a
contraction mapping on Y . Similar to the proof of Theorem 2.2, for any x1 , x2 ∈ Y ,
in view of the assumptions we have
G(t − τ ) |x1 (t − τ ) − x2 (t − τ )|
|(T x1 )(t) − (T x2 )(t)|
≤ |p(t)|
G(t)
G(t)
G(t − τ )
Z ∞
Z ∞
1
ds
+
|f (θ, x1 (g(θ))) − f (θ, x2 (g(θ)))|dθ
G(t) t a(s) s
l
≤ |p| l kx1 − x2 k + kx1 − x2 k
N
1
= ( − p) l kx1 − x2 k.
N
Since 0 < ( N1 − p)l < 1, T is a contraction mapping on Y . Finally we use the
Banach fixed point theorem to deduce the existence of a unique fixed point in Y ,
Z ∞
Z ∞
ds
x(t) = (T x)(t) = M (1 − p) + p(t)x(t − τ ) −
f (θ, x(g(θ)))dθ.
a(s) s
t
Since x ∈ Y and p ≤ p(t) < 0, we have p(t)x(t − τ ) ≥ pM (1 − p). From the
inequality and the condition (2.6), we obtain
x(t) > M (1 − p) + pM (1 − p) − M (1 − p2 ) = 0.
Hence x(t) > 0 for t ≥ t0 . Substituting x(t) into (1.2), we know that x(t) is a
throughout positive solution of equation (1.2) and
lim (x(t) − p(t)x(t − τ )) = M (1 − p).
t→∞
In view of the Lemma 1.1, limt→∞ x(t) = M which completes the proof.
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THROUGHOUT POSITIVE SOLUTIONS
7
Theorem 2.4. Assume that limt→∞ p(t) = p where p ∈ (−∞, −1) and p(t) ≤ p.
Define
u ∈ C 1 [t0 , ∞), 0 ≤ u(t) ≤
for g(t0 ) ≤ t < t0
Z=
M (1 + |p|)
, for t ≥ t0 ; u(t) = u(t0 ),
|p|
where M is a positive constant. Let conditions (2.1) and (2.2) hold for 0 ≤ u, v ≤
M (1+|p|)
. Assume that for every u ∈ Z,
|p|
Z
∞
t0
ds
a(s)
Z
∞
f (θ, u(g(θ)))dθ <
s
M (p2 − 1)
.
|p|
(2.7)
1
Assume further there exists a positive integer N > 1 such that 0 < |p|
(1 + Nl ) < 1,
R
∞ s
where l(N ) = max{ G(g(t))
G(t) , t ≥ t0 }, G(t) = exp (N t a(s) k(s)ds). Then equation
(1.2) has a throughout positive solution x(t) on [t0 , ∞) satisfying limt→∞ x(t) = M .
Proof. Define a mapping T on Z as follows
R∞
1
M (1 − p) − x(t + τ ) − t+τ
−p(t+τ
)
(T x)(t) =
(T x)(t0 )
ds
a(s)
R∞
s
f (θ, x(g(θ)))dθ t ≥ t0
g(t0 ) ≤ t ≤ t0 .
. So T Z ⊆ Z. We introduce the norm
From (2.7), we have 0 ≤ (T x)(t) ≤ M (1+|p|)
|p|
k ·k on Z, kxk = supt≥t0 |x(t)|/G(t). We now show that T is a contraction mapping
on Z. For any x1 , x2 ∈ Z, in view of the assumptions we have
|(T x1 )(t) − (T x2 )(t)|
G(t)
−1
≤
|x1 (t + τ ) − x2 (t + τ )|
G(t + τ )p(t + τ )
Z ∞
Z ∞
−1
ds
+
|f (θ, x1 (g(θ))) − f (θ, x2 (g(θ)))|dθ
G(t)p(t + τ ) t+τ a(s) s
Z ∞
Z ∞
1
1
ds
G(g(θ))k(θ)|x1 (g(θ)) − x2 (g(θ))|
kx1 − x2 k +
dθ
≤
|p|
G(t)|p| t+τ a(s) s
G(g(θ))
Z ∞
Z ∞
1
1
ds
G(g(θ))
≤
kx1 − x2 k +
kx1 − x2 k
G(θ)k(θ)
dθ
|p|
G(t)|p|
a(s)
G(θ)
t+τ
s
Z ∞
1
l
(s − t − τ )G(s)k(s)
≤
kx1 − x2 k +
kx1 − x2 k
ds
|p|
G(t)|p|
a(s)
t+τ
Z ∞
1
l
sG(s)k(s)
≤
kx1 − x2 k +
kx1 − x2 k
ds
|p|
G(t)|p|
a(s)
t+τ
Z ∞
1
l
1
=
kx1 − x2 k +
kx1 − x2 k
(− )G0 (s)ds
|p|
G(t)|p|
N
t+τ
G(t + τ ) − 1
1
≤
1+l
kx1 − x2 k
|p|
N G(t)
1
l
≤
(1 + )kx1 − x2 k.
|p|
N
8
Z. ZHANG, C. WANG, Q. LI, F. LI
EJDE-2005/53
1
Since 0 < |p|
(1 + Nl ) < 1, T is a contraction mapping on Z. Finally we use the
Banach fixed point theorem to deduce the existence of a unique fixed point in Z,
x(t) = (T x)(t)
=
h
1
M (1 − p) − x(t + τ ) −
−p(t + τ )
Since x ∈ Z, we have x(t + τ ) ≤
(2.7), we obtain
x(t) >
M (1+|p|)
.
|p|
Z
∞
t+τ
ds
a(s)
Z
∞
i
f (θ, x(g(θ)))dθ .
s
From the inequality and the condition
1
M (1 + |p|) M (p2 − 1) M (1 − p) −
= 0.
−
−p(t + τ )
|p|
|p|
Hence x(t) > 0 for t ≥ t0 . Substituting x(t) into (1.2), we know that x(t) is a
throughout positive solution of (1.2) and
lim (x(t) − p(t)x(t − τ )) = M (1 − p).
t→∞
In view of Lemma 1.1, we have limt→∞ x(t) = M . The proof is complete.
Theorem 2.5. Assume that limt→∞ p(t) = p where p ∈ (1, +∞) and p(t) ≥ p.
Define
M (1 + p)
Ω = u ∈ C 1 [t0 , ∞), 0 ≤ u(t) ≤
, for t ≥ t0 ; u(t) = u(t0 ),
p
for g(t0 ) ≤ t < t0
where M is a positive constant. Let conditions (2.1) and (2.2) hold for 0 ≤ u, v ≤
M (1+p)
. We assume that for every u ∈ Ω
p
Z
∞
t0
ds
a(s)
Z
∞
f (θ, u(g(θ)))dθ ≤
s
p−1
M.
p
(2.8)
Assume further that there exists a positive integer N > 1 such that 0 < p1 (1 + Nl ) <
R∞ s
1, where l(N ) = max{ G(g(t))
G(t) , t ≥ t0 }, G(t) = exp(N t a(s) k(s)ds). Then (1.2)
has a throughout positive solution x(t) on [t0 , ∞) satisfying limt→∞ x(t) = M .
Proof. Define a mapping T on Ω as follows
(
(T x)(t) =
1
p(t+τ )
R∞
M (p − 1) + x(t + τ ) + t+τ
(T x)(t0 )
ds
a(s)
R∞
s
f (θ, x(g(θ)))dθ
t ≥ t0
g(t0 ) ≤ t ≤ t0 .
From (2.8), we have 0 ≤ (T x)(t) ≤ p+1
p M . So T Ω ⊆ Ω. We introduce the norm k·k
on Ω, kxk = supt≥t0 |x(t)|/G(t). We now show that T is a contraction mapping
on Ω. Similar to the proof of Theorem 2.4, for any x1 , x2 ∈ Ω, in view of the
EJDE-2005/53
THROUGHOUT POSITIVE SOLUTIONS
9
assumptions we have
|(T x1 )(t) − (T x2 )(t)|
G(t)
1
≤
|x1 (t + τ ) − x2 (t + τ )|
G(t + τ )p(t + τ )
Z ∞
Z ∞
1
ds
+
|f (θ, x1 (g(θ))) − f (θ, x2 (g(θ)))|dθ
G(t)p(t + τ ) t+τ a(s) s
1
l
≤ (1 + )kx1 − x2 k.
p
N
Since 0 < p1 (1 + Nl ) < 1, T is a contraction mapping on Ω. Finally we use the
Banach fixed point theorem to deduce the existence of a unique fixed point in Ω,
x(t) = (T x)(t)
=
1
M (p − 1) + x(t + τ ) +
p(t + τ )
Z
∞
t+τ
ds
a(s)
Z
∞
f (θ, x(g(θ)))dθ .
s
Because p > 1, that is M (p − 1) > 0, and all the other terms which are in the
expression of x(t) are nonnegative, we easily know that x(t) > 0 for t ≥ t0 . Substituting x(t) into (1.2), we know that x(t) is a throughout positive solution of
equation (1.2) and
lim (x(t) − p(t)x(t − τ )) = M (1 − p).
t→∞
In view of the Lemma 1.1 we have limt→∞ x(t) = M . The proof is complete.
3. Examples
Example 3.1. Consider the second order self-conjugate differential equation
(tx0 (t))0 +
4(t − 1)6
t6 (t − 2)3
x3 (t − 1) = 0,
t ≥ t0 = 6.
In our notation, a(t) = t, a(s) = 5, g(t) = t − 1, f (t, u) =
M = 1, k(t) =
4(t − 1)6 3
u . We choose
t6 (t − 2)3
12(t − 1)6
, N = 3. We know that for any 0 ≤ u, v ≤ 1,
t6 (t − 2)3
|f (t, u) − f (t, v)| = |
4(t − 1)6
12(t − 1)6
3
3
(u
−
v
)|
≤
|u − v|.
t6 (t − 2)3
t6 (t − 2)3
For any u, v ∈ X
Z
∞
t0
Z
∞
t0
ds
a(s)
Z
s
∞
1
s
k(s) ds =
a(s)
5
Z
∞
6
∞
12(s − 1)6
ds < ∞
s5 (s − 2)3
Z ∞
4
(θ − 1)6 (u(θ − 1))3
f (θ, u(g(θ))) dθ =
ds
dθ
s
θ6 (θ − 2)3
6
s
Z ∞
Z ∞
4
dθ
ds
≤
s
(θ
−
2)3
6
s
1 1 4
1
= + ln ≤ < 1 ,
4 2 6
4
Z
(3.1)
10
Z. ZHANG, C. WANG, Q. LI, F. LI
EJDE-2005/53
t0
Z 6 s 12(s − 1)6 s 12(s − 1)6 ds
=
exp
3
ds < 3.
6
3
6
3
t0 −1 t0 − 1 s (s − 2)
5 5 s (s − 2)
Thus the conditions in Theorem 2.1 are satisfied. So (3.1) has a throughout positive
solution x(t) on [t0 , ∞) and limt→∞ x(t) = 1. In fact, x(t) = 1− t12 is such a solution.
Z
l = exp N
Example 3.2. Consider the second-order neutral differential equation
1
2(t − 1)3 − t3 3
(x(t) − x(t − 1))00 +
x (t − 1) = 0,
2
(t − 1)3 (t − 2)3
t ≥ t0 = 13.
(3.2)
1
[2(t − 1)3 − t3 ]u3
. We
, g(t) = t − 1, f (t, u) =
2
(t − 1)3 (t − 2)3
3[2(t − 1)3 − t3 ]
choose M = 1, k(t) =
, N = 4. It is easy to show that for any
(t − 1)3 (t − 2)3
0 ≤ u, v ≤ 1,
Here a(t) = 1, a(s) = 1, p(t) =
|f (t, u) − f (t, v)| = |
For any u, v ∈ X
Z
∞
t0
Z
∞
t0
ds
a(s)
Z
2(t − 1)3 − t3
3[2(t − 1)3 − t3 ]
3
3
(u
−
v
)|
≤
|u − v|.
(t − 1)3 (t − 2)3
(t − 1)3 (t − 2)3
s
k(s) ds =
a(s)
Z
∞
3s
13
∞
Z
∞
2(s − 1)3 − s3
ds < ∞ ,
(s − 1)3 (s − 2)3
Z
f (θ, u(g(θ)))dθ =
s
13
∞
s
∞
2(θ − 1)3 − θ3
(u(θ − 1))3 dθds
(θ − 1)3 (θ − 2)3
Z
(θ − t)
=
Z13∞
2(θ − 1)3 − θ3
(u(θ − 1))3 dθ
(θ − 1)3 (θ − 2)3
2θ
dθ
(θ
−
2)3
13
24
1
=
< ,
121
2
Z 13
1
1 1
3[2(s − 1)3 − s3 ] (p + )l = ( + ) exp 4
s
ds < 1.
N
2 4
(s − 1)3 (s − 2)3
12
Thus the conditions in Theorem 2.2 are satisfied. So (3.2) has a throughout positive
solution x(t) on [t0 , ∞) and limt→∞ x(t) = 1. In fact, x(t) = 1− 1t is such a solution.
≤
Example 3.3. Consider the second-order self-conjugate neutral differential equation
(t − 1)3 3
t3 (t − 1)
0 0
(x(t)
+
2x(t
−
1))
+
x (t − 1) = 0, t ≥ t0 = 9. (3.3)
4((t − 1)3 + 2t3 )
t3 (t − 2)3
t3 (t − 1)
896
, a(s) = 1367
,
4((t − 1)3 + 2t3 )
(t − 1)3
27(t − 1)3
3
f (t, u) = 3
u
.
We
choose
that
M
=
1,
k(t)
=
, N = 3. Here
t (t − 2)3
4t3 (t − 2)3
3
1
we define Z = {u ∈ C [t0 , ∞) : 0 ≤ u(t) ≤ 2 , t ≥ t0 }. It is easy to show that for
any 0 ≤ u, v ≤ 23 ,
In our notation, p(t) = −2, g(t) = t−1, τ = 1, a(t) =
|f (t, u) − f (t, v)| = |
(t − 1)3
27(t − 1)3
3
3
(u
−
v
)|
≤
|u − v|.
t3 (t − 2)3
4t3 (t − 2)3
EJDE-2005/53
For any u, v ∈ Z,
Z
THROUGHOUT POSITIVE SOLUTIONS
11
Z
s
1367 ∞ 27(s − 1)3
ds < ∞ ,
k(s) ds =
896 9 4s2 (s − 2)3
t0 a(s)
Z ∞
Z ∞
ds
f (θ, u(g(θ)))dθ
t0 a(s) s
Z
Z ∞
4((t0 − 1)3 + 2t30 ) ∞
(θ − 1)3 (u(θ − 1))3
ds
≤
dθ
3
t0 (t0 − 1)
θ3 (θ − 2)3
9
s
Z ∞ Z ∞
12
dθ
≤
ds
t0 − 1 9
(θ − 2)3
s
3
3
=
< ,
28
2
Z 9 4((9 − 2)3 + 2(9 − 1)3 ) 27(s − 1)3
i
h
1
l
1
1
(1 + ) = 1 + exp 3
s
ds
< 1.
|p|
N
2
3
(9 − 1)3 (9 − 2)
4s2 (s − 2)3
8
Thus the conditions in Theorem 2.4 are satisfied. So (3.3) has a throughout positive
solution x(t) on [t0 , ∞) and limt→∞ x(t) = 1. In fact, x(t) = 1− t12 is such a solution.
∞
References
[1] M. M. A. El-Sheikh; Oscillation and nonoscillation criteria for second order nonlinear differential equations. J. Math. Anal. Appl. 179(1993) 14-27.
[2] L. H. Erbe; Qingkai. Kong; B. G. Zhang; Oscillation theory for functional differential equations. New York. Basel. Hong Kong, 1995.
[3] W. T. Li and J. R. Yan; Oscillation criteria for second order superlinear differential equations. Indina. J. Pure. Appl. Math. 28(6)(1997) 735-740.
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Appl. 217(1998) 1-14.
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equations with delays. Acta Sci. Nat. Univ. Norm. Hunan . Vol21. (1998) 12-18.
[7] S. H. Saker; Oscillation of second order nonlinear delay difference equations. Bull. Korean.
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[8] E. Thandapani, K. Ravi; Oscillation of second order half-linear difference equations. Applied
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[9] E. Wahlén; Positive solutions of second order differential equations. Nonlinear Analysis.
58(2004) 359-366.
[10] Jun Yang, Xin Ping Guan; Positive solution of a class of neutral delay difference equations.
Acta Mathematic Sinica. Vol. 44. No.3 (2001) 409-416.
[11] J. R. Yan; Oscillation of second order neutral functional differential equations. Appl. Math.
Comp. 83(1997) 27-41.
[12] J. R. Yan; The oscillation properties for the solution of the second order differential equation
with ”integral small” coefficient. Acta. Math. Sinica. 30(2)(1987) 206-215.
[13] X. J. Yang; Nonoscillation criteria for second order nonlinear differential equations. Appl.
Math. Comp. 131(2002) 125-131.
[14] J. S. Yu, M. P. Chen, H. Zhang; Oscillation and nonoscillation in neutral equations with
integrable coefficients. Comp. Math. Applic. Vol35. No.6 (1998) 65-71.
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Zhenguo Zhang
College of Mathematics and Information Science, Hebei normal University, Heibei, Shijiazhuang, 050016, China
E-mail address: Zhangzhg@mail.hebtu.edu.cn
12
Z. ZHANG, C. WANG, Q. LI, F. LI
EJDE-2005/53
Chunjiao Wang
College of Mathematics and Information Science, Hebei normal University, Heibei, Shijiazhuang, 050016, China
E-mail address: ccjjj0601@sina.com.cn
Qiaoluan Li
College of Mathematics and Information Science, Hebei normal University, Heibei, Shijiazhuang, 050016, China
E-mail address: qll71125@163.com
Fang Li
College of Mathematics and Information Science, Hebei normal University, Heibei, Shijiazhuang, 050016, China
E-mail address: lifanglucky@126.com