Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2011, Article ID 693890, 23 pages
doi:10.1155/2011/693890
Research Article
Existence of Nonoscillatory Solutions
for a Third-Order Nonlinear Neutral Delay
Differential Equation
Zeqing Liu,1 Lin Chen,1 Shin Min Kang,2 and Sun Young Cho3
1
Department of Mathematics, Liaoning Normal University, Dalian, Liaoning 116029, China
Department of Mathematics and RINS, Gyeongsang National University,
Jinju 660-701, Republic of Korea
3
Department of Mathematics, Gyeongsang National University, Jinju 660-701, Republic of Korea
2
Correspondence should be addressed to Shin Min Kang, smkang@gnu.ac.kr
Received 4 April 2011; Accepted 28 June 2011
Academic Editor: Elena Braverman
Copyright q 2011 Zeqing Liu et al. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
The aim of this paper is to study the solvability of a third-order nonlinear neutral delay differential
equation of the form {αtβtxt ptxt − τ } ft, xσ1 t, xσ2 t, . . . , xσn t 0,
t ≥ t0 . By using the Krasnoselskii’s fixed point theorem and the Schauder’s fixed point theorem, we
demonstrate the existence of uncountably many bounded nonoscillatory solutions for the above
differential equation. Several nontrivial examples are given to illustrate our results.
1. Introduction and Preliminaries
In recent years, the study of the oscillation, nonoscillation, asymptotic behaviors and
existence of solutions for various kinds of first- and second-order neutral delay differential
equations and systems of differential equations have attracted much attention, for example,
see 1–12 and the references therein. Dorociaková and Olach 2 discussed the existence
of nonoscillatory solutions and asymptotic behaviors for the first-order delay differential
equation
x t ptxt qtxτt 0,
t ≥ 0.
1.1
2
Abstract and Applied Analysis
Elbert 3 and Huang 5 established a few oscillation and nonoscillation criteria for the
second-order linear differential equation
x t qtxt 0,
t ≥ 0,
1.2
where q ∈ C0, ∞, R . Tang and Liu 10 studied the existence of bounded oscillation for
the second-order linear delay differential equation of unstable type
x t ptxt − τ,
1.3
t ≥ t0 ,
where τ > 0, p ∈ Ct0 , ∞, R and pt /
≡ 0 on any interval of length τ. In view of the
Banach fixed point theorem, Kulenović and Hadžiomerspahić 7 deduced the existence of
a nonoscillatory solution for the second-order linear neutral delay differential equation with
positive and negative coefficients
xt cxt − τ Q1 txt − σ1 − Q2 txt − σ2 0,
t ≥ t0 ,
1.4
where c ∈ R\{−1, 1}, τ > 0, σ1 , σ2 ∈ 0, ∞, Q1 , Q2 ∈ Ct0 , ∞, R . Qin et al. 9 and Yang
et al. 11 developed several oscillation criteria for the second-order differential equation
rt xt ptxt − τ
qtfxt − δ 0,
t ≥ t0 ,
1.5
where τ and δ are nonnegative constants, r, p, q ∈ Ct0 , ∞, R, and f ∈ CR, R. By
utilizing the Krasnoselskii’s fixed point theorem, Zhou 12 discussed the existence of
nonoscillatory solutions of the second-order nonlinear neutral differential equation
m
rt xt ptxt − τ
Qi tfxi t − σi 0,
t ≥ t0 ,
1.6
i1
where m ≥ 1 is an integer, τ > 0, σi ≥ 0, r, p, Qi ∈ Ct0 , ∞, R, and fi ∈ CR, R for
i ∈ {1, 2, . . . , m}.
However, the existence of nonoscillatory solutions of third-order neutral differential
equations received much less attention, moreover, the results in 7, 11, 12 only figured out
the existence of a nonoscillatory solution of 1.3–1.5, respectively.
Motivated by the papers mentioned above, in this paper, we investigate the following
third-order nonlinear neutral delay differential equation
ft, xσ1 t, xσ2 t, . . . , xσn t 0,
αt βt xt ptxt − τ
t ≥ t0 , 1.7
where n ≥ 1 is an integer, τ > 0, α, β ∈ Ct0 , ∞, R \ {0}, p ∈ Ct0 , ∞, R, and f ∈
Ct0 , ∞ × Rn , R. By applying the Krasnoselskii’s fixed point theorem and the Schauder’s
fixed point theorem, we obtain some sufficient conditions for the existence of uncountably
many bounded nonoscillatory solutions of 1.7.
Abstract and Applied Analysis
3
Throughout this paper, we assume that R −∞, ∞, R 0, ∞, Ct0 , ∞, R
denotes the Banach space of all continuous and bounded functions on t0 , ∞ with the norm
x supt≥t0 |xt| for each x ∈ Ct0 , ∞, R and
AN, M {x ∈ Ct0 , ∞, R : N ≤ xt ≤ M, t ≥ t0 }
for M > N > 0.
1.8
It is easy to see that AN, M is a bounded closed and convex subset of Ct0 , ∞, R.
By a solution of 1.7, we mean a function x ∈ Ct1 − τ, ∞, R for some t1 ≥ t0
such that xt ptxt − τ, βtxt ptxt − τ and αtβtxt ptxt − τ are
continuously differentiable in t1 , ∞ and such that 1.7 is satisfied for t ≥ t1 . As is
customary, a solution of 1.7 is called oscillatory if it has arbitrarily large zeros, and otherwise
it is said to be nonoscillatory.
Definition 1.1 see 6. A family F of functions in Ct0 , ∞, R is equicontinuous on t0 , ∞
if for any ε > 0, the interval t0 , ∞ can be decomposed into a finite number of subintervals
I1 , I2 , . . . , In such that
fx − f y ≤ ε,
∀f ∈ F, x, y ∈ Ii , i ∈ {1, 2, . . . , n}.
1.9
Lemma 1.2 see Krasnoselskii’s fixed point theorem, 4. Let X be a Banach space. Let Ω be a
bounded closed convex subset of X and S1 and S2 mappings from Ω into X such that S1 x S2 y ∈ Ω
for every pair x, y ∈ Ω. If S1 is a contraction and S2 is completely continuous, then the equation
S1 x S2 x x has at least one solution in Ω.
Lemma 1.3 see Schauder’s fixed point theorem, 4. Let Ω be a nonempty closed convex subset
of a Banach space X. Let S : Ω → Ω be a continuous mapping such that SΩ is a relatively compact
subset of X. Then S has at least one fixed point in Ω.
2. Main Results
Now we study those conditions under which 1.7 possesses uncountably many bounded
nonoscillatory solutions.
Theorem 2.1. Assume that there exist constants M, N, c1 , c2 , T0 and a function h ∈ Ct0 , ∞, R satisfying
min{c1 , c2 } ≥ 0,
c1 c2 < 1,
0 < N < 1 − c1 − c2 M,
−c2 ≤ pt ≤ c1 , t ≥ T0 > t0 ;
ft, u1 , u2 , . . . , un ≤ ht, t ∈ t0 , ∞,
ui ∈ N, M,
∞ ∞ ∞
hu
du ds dv < ∞.
αsβv
t0
v
s
2.1
2.2
i ∈ {1, . . . , n};
Then 1.7 possesses uncountably many bounded nonoscillatory solutions in AN, M.
2.3
2.4
4
Abstract and Applied Analysis
Proof. Set k ∈ c1 M N, 1 − c2 M. From 2.4, we pick up T > T0 such that
∞ ∞ ∞
T
v
s
hu
du ds dv < min{1 − c2 M − k, k − c1 M − N}.
αsβv
2.5
Define two mappings S1k and S2k : AN, M → Ct0 , ∞, R by
S1k xt S2k xt ⎧
∞ ∞ ∞
⎪
⎨
⎪
⎩
t
v
s
⎧
⎨k − ptxt − τ,
t ≥ T,
⎩S xT ,
1k
t0 ≤ t < T,
2.6
1
fu, xσ1 u, . . . , xσn udu ds dv,
αsβv
t ≥ T,
t0 ≤ t < T
S2k xT ,
2.7
for x ∈ AN, M.
Firstly, we prove that S1k AN, M S2k AN, M ⊆ AN, M. By 2.1–2.3 and
2.5–2.7, we get that
S1k xt S2k y t ≤ k c2 M ∞ ∞ ∞
T
v
s
hu
du ds dv
αsβv
≤ k c2 M min{1 − c2 M − k, k − c1 M − N}
≤ M,
x, y ∈ AN, M, t ≥ T,
∞ ∞ ∞
hu
du ds dv
S1k xt S2k y t ≥ k − c1 M −
αsβv
T
v
s
2.8
≥ k − c1 M − min{1 − c2 M − k, k − c1 M − N}
≥ N,
x, y ∈ AN, M, t ≥ T,
which infer that S1k AN, M S2k AN, M ⊆ AN, M for any x, y ∈ AN, M.
Secondly, we show that S1k is a contraction mapping. By 2.1, 2.2, and 2.6, we
deduce that
S1k xt − S1k y t ≤ ptxt − τ − yt − τ ≤ c1 c2 x − y
2.9
for x, y ∈ AN, M and t ≥ T , which gives that
S1k x − S1k y ≤ c1 c2 x − y.
2.10
Abstract and Applied Analysis
5
Thirdly, we show that S2k is completely continuous. Let x ∈ AN, M and {xm }m≥1 ⊆
AN, M with limm → ∞ xm x. By 2.7, we obtain that
|S2k xm t − S2k xt|
∞ ∞ ∞
1
fu, xm σ1 u, . . . , xm σn u
≤
αsβv
T
v
s
−fu, xσ1 u, . . . , xσn udu ds dv,
2.11
t ≥ T, m ≥ 1.
Using 2.3 and 2.4, we conclude that
fu, xm σ1 u, . . . , xm σn u − fu, xσ1 u, . . . , xσn u ≤ 2hu,
u ∈ T, ∞, m ≥ 1,
∞
fu, xm σ1 u, . . . , xm σn u − fu, xσ1 u, . . . , xσn udu ≤ 2
s
∞
hudu,
s
s ∈ T, ∞, m ≥ 1,
∞ ∞
v
1 fu, xm σ1 u, . . . , xm σn u − fu, xσ1 u, . . . , xσn udu ds
αs
s
∞ ∞
hu
≤2
du ds, v, s ∈ T, ∞, m ≥ 1.
αs
v
s
2.12
In light of 2.11–2.12,
fu, xm σ1 u, . . . , xm σn u − fu, xσ1 u, . . . , xσn u −→ 0
as m −→ ∞,
2.13
and the Lebesgue dominated convergence theorem, we conclude that
lim S2k xm − S2k x 0,
2.14
m → ∞
which means that S2k is continuous in AN, M.
Now we show that S2k is completely continuous. By virtue of 2.3, 2.4, and 2.7, we
get that
S2k x ≤
∞ ∞ ∞
T
v
s
hu
du ds dv,
αsβv
x ∈ AN, M.
2.15
6
Abstract and Applied Analysis
That is, S2k AN, M is uniform bounded. It follows from 2.4 that for each ε > 0, there
exists T ∗ > T such that
∞ ∞ ∞
T∗
v
s
hu
ε
du ds dv < .
αsβv
2
2.16
In view of 2.3, 2.7, and 2.16, we infer that
|S2k xt2 − S2k xt1 |
∞ ∞ ∞
∞ ∞ ∞
hu
hu
du ds dv du ds dv
≤
αsβv
αsβv
t2
v
s
t1
v
s
< ε,
2.17
x ∈ AN, M, t2 > t1 ≥ T ∗ .
From 2.3 and 2.7, we get that
|S2k xt2 − S2k xt1 |
t2 ∞ ∞
hu
≤
du ds dv,
αsβv
s
t1 v
x ∈ AN, M, T ≤ t1 ≤ t2 ≤ T ∗ ,
2.18
which together with 2.4 ensures that there exists δ > 0 satisfying
|S2k xt2 − S2k xt1 | < ε,
x ∈ AN, M, t1 , t2 ∈ T, T ∗ with |t2 − t1 | < δ.
2.19
Clearly,
|S2k xt2 − S2k xt1 | 0 < ε,
x ∈ AN, M, t0 ≤ t1 ≤ t2 ≤ T.
2.20
That is, S2k AN, M is equicontinuous on t0 , ∞. Consequently, S2k is completely
continuous. By Lemma 1.2, there is x0 ∈ AN, M such that S1 x0 S2 x0 x0 , which is a
bounded nonoscillatory solution of 1.7.
Lastly, we demonstrate that 1.7 possesses uncountably many bounded nonoscilla k2 . For each j ∈ {1, 2},
tory solutions in AN, M. Let k1 , k2 ∈ c1 M N, 1 − c2 M and k1 /
we choose Tj > T0 and the mappings Sjk1 and Sjk2 satisfying 2.5–2.7 with k and T replaced
by kj and Tj , respectively, and some T3 > max{T1 , T2 } such that
∞ ∞ ∞
T3
v
s
hu
|k1 − k2 |
du ds dv <
.
αsβv
2
2.21
Obviously, there are x, y ∈ AN, M such that S1k1 x S2k1 x x and S1k2 y S2k2 y y,
respectively. That is, x and y are two bounded nonoscillatory solutions of 1.7 in AN, M.
Abstract and Applied Analysis
7
In order to prove that 1.7 possesses uncountably many bounded nonoscillatory solutions in
AN, M, we prove only that x /
y. In fact, by 2.6 and 2.7, we gain that for t ≥ T3
xt k1 − ptxt − τ ∞ ∞ ∞
t
yt k2 − ptyt − τ v
s
∞ ∞ ∞
t
v
s
1
fu, xσ1 u, . . . , xσn udu ds dv,
αsβv
1
f u, yσ1 u, . . . , yσn u du ds dv,
αsβv
2.22
which together with 2.1–2.3 imply that
xt − yt ≥ |k1 − k2 | − ptxt − τ − yt − τ −
∞ ∞ ∞
1
αsβv
t
v
s
× fu, xσ1 u, . . . , xσn u − f u, yσ1 u, . . . , yσn u du ds dv
≥ |k1 − k2 | − c1 c2 x − y
−2
∞ ∞ ∞
T3
v
s
hu
du ds dv,
αsβv
t ≥ T3 ,
2.23
which means that
x − y ≥
1
1 c1 c2
|k1 − k2 | − 2
∞ ∞ ∞
T3
v
s
hu
du ds dv
αsβv
> 0,
2.24
by 2.1 and 2.21. That is, x /
y. This completes the proof.
Theorem 2.2. Assume that there exist constants M, N, c1 , c2 , T0 and a function h ∈ Ct0 , ∞, R satisfying 2.3, 2.4 and
0 < 1 − c2 N < 1 − c1 M;
0 ≤ c2 ≤ pt ≤ c1 < 1,
t ≥ T 0 > t0 .
2.25
2.26
Then 1.7 possesses uncountably many bounded nonoscillatory solutions in AN, M.
Proof. Set k ∈ c1 M N, M c2 N. It follows from 2.4 that there exists T > T0 such that
∞ ∞ ∞
T
v
s
hu
du ds dv < min{M c2 N − k, k − c1 M − N}.
αsβv
2.27
8
Abstract and Applied Analysis
Define two mappings S1k and S2k : AN, M → Ct0 , ∞, R by 2.6 and 2.7,
respectively. By virtue of 2.3, 2.6, 2.7, 2.26, and 2.27, we obtain that
S1k xt S2k y t ≤ k − ptxt − τ ∞ ∞ ∞
T
v
s
hu
du ds dv
αsβv
≤ k − c2 N min{M c2 N − k, k − c1 M − N}
x, y ∈ AN, M, t ≥ T,
∞ ∞ ∞
S1k xt S2k y t ≥ k − ptxt − τ −
2.28
≤ M,
T
v
s
hu
du ds dv
αsβv
≥ k − c1 M − min{M c2 N − k, k − c1 M − N}
≥ N,
2.29
x, y ∈ AN, M, t ≥ T,
which yield that S1k AN, M S2k AN, M ⊆ AN, M.
By a similar argument used in the proof of Theorem 2.1, we gain that S1k is a
contraction mapping S2k is completely continuous, and 1.7 possesses uncountably many
nonoscillatory solutions. This completes the proof.
Theorem 2.3. Assume that there exist constants M, N, c1 , c2 , T0 and a function h ∈ Ct0 , ∞, R satisfying 2.3, 2.4, 2.25, and
−1 < −c1 ≤ pt ≤ −c2 ≤ 0,
t ≥ T 0 > t0 .
2.30
Then 1.7 possesses uncountably many bounded nonoscillatory solutions in AN, M.
Proof. Set k ∈ 1 − c2 N, 1 − c1 M. It follows from 2.4 that there exists T > T0 satisfying
∞ ∞ ∞
T
v
s
hu
du ds dv < min{1 − c1 M − k, k − 1 − c2 N}.
αsβv
2.31
Let S1k and S2k : AN, M → Ct0 , ∞, R be defined by 2.6 and 2.7, respectively. In
view of 2.3, 2.6, 2.7, 2.30, and 2.31, we obtain that
S1 xt S2 y t ≤ k − ptxt − τ ∞ ∞ ∞
T
v
s
hu
du ds dv
αsβv
≤ k c1 M min{1 − c1 M − k, k − 1 − c2 N}
≤ M,
x, y ∈ AN, M, t ≥ T,
∞ ∞ ∞
S1 xt S2 y t ≥ k − ptxt − τ −
T
v
s
hu
du ds dv
αsβv
≥ k c2 N − min{1 − c1 M − k, k − 1 − c2 N}
≥ N,
x, y ∈ AN, M, t ≥ T,
which mean that S1k AN, M S2k AN, M ⊆ AN, M.
2.32
Abstract and Applied Analysis
9
The rest of the proof is similar to the proof of Theorem 2.1 and is omitted. This
completes the proof.
Theorem 2.4. Assume that there exist constants M and N with M > N > 0 and a function h ∈
Ct0 , ∞, R satisfying 2.3 and 2.4. If pt 1 for each t ∈ t0 , ∞, then 1.7 possesses
uncountably many bounded nonoscillatory solutions in AN, M.
Proof. Set k ∈ N, M. It follows from 2.4 that there exists T ≥ t0 τ satisfying
∞ ∞ ∞
T
v
s
hu
du ds dv < min{M − k, k − N}.
αsβv
2.33
Define a mapping Sk : AN, M → Ct0 , ∞, R by
Sk xt ⎧
∞ t2iτ
⎪
⎨k
⎪
⎩
i1
t2i−1τ
∞ ∞
v
s
1
fu, xσ1 u, . . . , xσn udu ds dv, t ≥ T,
αsβv
t0 ≤ t < T
2.34
Sk xT ,
for x ∈ AN, M.
Firstly, we prove that Sk AN, M ⊆ AN, M. By virtue of 2.3, 2.33 and 2.34,
we obtain that
Sk xt ≤ k ∞ t2iτ
i1
≤k
Sk xt ≥ k −
s
t2i−1τ
∞ ∞ ∞
T
≥ N,
v
v
s
hu
du ds dv
αsβv
hu
du ds dv
αsβv
x ∈ AN, M, t ≥ T,
∞ ∞
∞ t2iτ
i1
≥k−
t2i−1τ
∞ ∞ ∞
T
≤ M,
∞ ∞
v
s
v
s
hu
du ds dv
αsβv
hu
du ds dv
αsβv
x ∈ AN, M, t ≥ T,
which imply that Sk AN, M ⊆ AN, M.
2.35
10
Abstract and Applied Analysis
Secondly, we show that Sk is continuous in AN, M. Let x ∈ AN, M and {xm }m≥1 ⊆
AN, M with limm → ∞ xm x. By 2.34, we get that
|Sk xm t − Sk xt|
∞ ∞
∞ t2iτ
≤
i1
≤
t2i−1τ
∞ ∞ ∞
T
v
s
v
1
fu, xm σ1 u, . . . , xm σn u
αsβv
−fu, xσ1 u, . . . , xσn udu ds dv
s
1
fu, xm σ1 u, . . . , xm σn u
αsβv
−fu, xσ1 u, . . . , xσn udu ds dv,
t ≥ T, m ≥ 1.
2.36
In view of 2.12, 2.13, 2.36, and the Lebesgue dominated convergence theorem, we
deduce that
lim Sk xm − Sk x 0,
2.37
m → ∞
which means that Sk is continuous in AN, M.
Thirdly, we show that Sk AN, M is relatively compact. From 2.3, 2.33, and
2.34, we gain that
Sk x ≤ k ∞ ∞ ∞
T
v
s
hu
du ds dv ≤ 2M,
αsβv
x ∈ AN, M,
2.38
which means that Sk AN, M is uniform bounded.
Let ε > 0. It follows from 2.4 that there exist T > T ∗ > T such that
∞ ∞ ∞
T∗
v
s
∞ ∞ ∞
v
T
s
hu
ε
du ds dv < ,
αsβv
2
2.39
ε
hu
du ds dv < .
αsβv
4
2.40
By 2.3, 2.34, and 2.39, we deduce that
|Sk xt2 − Sk xt1 | ≤
t 2iτ
∞
2
t2 2i−1τ
i1
≤2
∞ ∞ ∞
< ε,
T∗
v
s
t1 2iτ
t1 2i−1τ
∞ ∞
v
hu
du ds dv
αsβv
x ∈ AN, M, t2 > t1 ≥ T ∗ .
s
hu
du ds dv
αsβv
2.41
Abstract and Applied Analysis
11
Choose an integer z ≥ 1 with T 2z 1τ ≥ T , and put
∞ ∞
hu
du ds : v ∈ T τ, T ∗ 2z − 1τ ,
αsβv
v
s
∞ ∞
hu
du ds : v ∈ T 2τ, T ∗ 2zτ ,
B max
αsβv
v
s
ε
ε
,
.
δ min
1 4Az 1 4Bz
A max
2.42
Equation 2.4 means that max{A, B} < ∞. It follows from 2.3, 2.34, and 2.40–2.42
that for x ∈ AN, M, t1 , t2 ∈ T, T ∗ with t1 ≤ t2 ≤ t1 δ
|Sk xt2 − Sk xt1 |
t1 2iτ t2 2iτ
z
≤
−
i1
t2 2i−1τ
t1 2i−1τ
1
fu, xσ1 u, . . . , xσn udu ds dv
×
αsβv
v
s
t 2iτ
t1 2iτ ∞ ∞
∞
2
hu
du ds dv
αsβv
t2 2i−1τ
t1 2i−1τ
v
s
iz1
t2 2iτ
t1 2i−1τ t1 2i−1τ
z
≤
i1
t2 2i−1τ
t1 2i−1τ
t1 2iτ
∞ ∞
1
fu, xσ1 u, . . . , xσn udu ds dv
×
αsβv
v
s
∞ ∞
2
∞ ∞ ∞
v
T
≤
2.43
z
t 2i−1τ
i1
t1 2i−1τ
2
s
hu
du ds dv
αsβv
t2 2iτ ∞ ∞
t1 2iτ
≤ Azt2 − t1 Bzt2 − t1 v
s
ε
hu
du ds dv αsβv
2
ε
2
< ε.
It is not difficult to verify that
|Sk xt2 − Sk xt1 | 0 < ε,
x ∈ AN, M, t0 ≤ t1 ≤ t2 ≤ T.
2.44
12
Abstract and Applied Analysis
Therefore Sk AN, M is equicontinuous on t0 , ∞, and consequently Sk is relatively
compact. By Lemma 1.3, there is x0 ∈ AN, M such that Sx0 x0 , which together with
2.34 yields that for t ≥ T τ
x0 t k ∞ t2iτ
t2i−1τ
i1
x0 t − τ k ∞ ∞
v
s
1
fu, x0 σ1 u, . . . , x0 σn udu ds dv,
αsβv
∞ t2i−1τ ∞ ∞
i1
t2i−1τ
v
s
2.45
1
fu, x0 σ1 u, . . . , x0 σn udu ds dv,
αsβv
which mean that
x0 t x0 t − τ 2k ∞ ∞ ∞
t
v
s
1
fu, x0 σ1 u, . . . , x0 σn udu ds dv. 2.46
αsβv
It is easy to show that x0 is a bounded nonoscillatory solution of 1.7.
Finally, we demonstrate that 1.7 possesses uncountably many bounded nonoscilla k2 . For each j ∈ {1, 2}, we pick up a
tory solutions in AN, M. Let k1 , k2 ∈ N, M and k1 /
positive integer Tj ≥ t0 τ and the mapping Skj satisfying 2.33 and 2.34, where k and T
are replaced by kj and Tj , respectively, and some T3 > max{T1 , T2 } satisfying 2.21. Clearly,
there are x and y ∈ AN, M such that Sk1 x x and Sk2 y y, respectively. That is, x and y
are bounded nonoscillatory solutions of 1.7 in AN, M. By 2.34 we get that for t ≥ T3
xt k1 ∞ t2iτ
i1
yt k2 ∞ ∞
t2i−1τ
∞ t2iτ
i1
v
s
∞ ∞
t2i−1τ
v
s
1
fu, xσ1 u, . . . , xσn udu ds dv,
αsβv
1
f u, yσ1 u, . . . , yσn u du ds dv,
αsβv
2.47
which together with 2.3 and 2.21 yield that
x − y ≥ |k1 − k2 | − 2
∞ ∞ ∞
T3
v
s
hu
du ds dv > 0,
αsβv
2.48
which implies that x / y. This completes the proof.
Theorem 2.5. Assume that there exist constants M and N with M > N > 0 and a function h ∈
Ct0 , ∞, R satisfying 2.3 and
∞ ∞ ∞ ∞
i1
t0 iτ
v
s
hu
du ds dv < ∞.
αsβv
2.49
If pt −1 for each t ∈ t0 , ∞, then 1.7 possesses uncountably many bounded nonoscillatory
solutions in AN, M.
Abstract and Applied Analysis
13
Proof. Set k ∈ N, M. It follows from 2.49 that there exists T > t0 satisfying
∞ ∞ ∞ ∞
T iτ
i1
v
hu
du ds dv < min{M − k, k − N}.
αsβv
s
2.50
Define a mapping Sk : AN, M → Ct0 , ∞, R by
Sk xt ⎧
∞ ∞ ∞ ∞
⎪
⎨k −
⎪
⎩
i1
tiτ
v
s
1
fu, xσ1 u, . . . , xσn udu ds dv, t ≥ T,
αsβv
t0 ≤ t < T
2.51
Sk xT ,
for x ∈ AN, M.
Firstly, we prove that Sk AN, M ⊆ AN, M. By 2.3, 2.50, and 2.51, we obtain
that
Sk xt ≤ k ∞ ∞ ∞ ∞
tiτ
i1
Sk xt ≥ k −
v
s
∞ ∞ ∞ ∞
tiτ
i1
v
s
hu
du ds dv ≤ M,
αsβv
x ∈ AN, M, t ≥ T,
hu
du ds dv ≥ N,
αsβv
x ∈ AN, M, t ≥ T,
2.52
which imply that SAN, M ⊆ AN, M.
Secondly, we show that Sk is continuous in AN, M. Let x ∈ AN, M and {xm }m≥1 ⊆
AN, M with limm → ∞ xm x. By 2.51, we get that
|Sk xm t − Sk xt|
∞ ∞ ∞ ∞
≤
i1
T iτ
v
s
1
fu, xm σ1 u, . . . , xm σn u
αsβv
−fu, xσ1 u, . . . , xσn udu ds dv,
t ≥ T, m ≥ 1.
2.53
Equation 2.53 together with 2.3, 2.49 and the Lebesgue dominated convergence theorem
yields that
lim Sk xm − Sk x 0,
2.54
m → ∞
that is, Sk is continuous in AN, M.
Thirdly, we show that Sk AN, M is relatively compact. From 2.3, 2.50, and
2.51, we obtain that
Sk x ≤ k ∞ ∞ ∞ ∞
i1
T iτ
v
s
hu
du ds dv ≤ 2M,
αsβv
x ∈ AN, M,
2.55
14
Abstract and Applied Analysis
which means that Sk AN, M is uniform bounded. It follows from 2.49 that, for each
ε > 0, there exists T ∗ > T such that
∞ ∞ ∞ ∞
T ∗ iτ
i1
v
s
ε
hu
du ds dv < .
αsβv
2
2.56
Notice that 2.3, 2.51, and 2.56 yield that
∞ ∞ ∞ ∞
|Sk xt2 − Sk xt1 | ≤ 2
i1
< ε,
T ∗ iτ
v
s
hu
du ds dv
αsβv
2.57
∗
x ∈ AN, M, t2 > t1 ≥ T .
Choose a positive integer z satisfying T zτ ≥ T ∗ . Put
D max
∞ ∞
v
s
hu
∗
du ds : v ∈ T τ, T zτ ,
αsβv
ε
.
δ
1 2Dz
2.58
By 2.3, 2.51, and 2.58, we gain that
|Sk xt2 − Sk xt1 |
∞ ∞ ∞
∞
∞
1
fu, xσ1 u, . . . , xσn udu ds dv
−
i1
αsβv
t2 iτ
t1 iτ
v
s
≤
∞ t2 iτ ∞ ∞
i1
≤
t1 iτ
v
s
z t2 iτ ∞ ∞
i1
t1 iτ
v
≤ Dzt2 − t1 s
∞ t2 iτ ∞ ∞
hu
hu
du ds dv du ds dv
αsβv
αsβv
s
iz1 t1 iτ v
∞ ∞ ∞ ∞
i1
< ε,
hu
du ds dv
αsβv
T ∗ iτ
v
s
2.59
hu
du ds dv
αsβv
x ∈ AN, M, t1 , t2 ∈ T, T ∗ with t1 ≤ t2 ≤ t1 δ.
Clearly,
|Sk xt2 − Sk xt1 | 0 < ε,
x ∈ AN, M, t0 ≤ t1 ≤ t2 ≤ T.
2.60
That is, Sk AN, M is equicontinuous on t0 , ∞, and Sk is relatively compact. The rest
argument is similar to the proof of Theorem 2.4 and is omitted. This completes the proof.
Abstract and Applied Analysis
15
Theorem 2.6. Assume that there exist constants M, N, c1 , c2 , T0 and a function h ∈ Ct0 , ∞, R satisfying 2.3, 2.4, and
c12 − c2
c 2 − c1
N< 2
M;
c1
c2
c2 ≤ pt ≤ c1 , t ≥ T0 ≥ t0 τ.
1 < c2 < c1 < c22 ,
0<
2.61
2.62
Then 1.7 possesses uncountably many bounded nonoscillatory solutions in AN, M.
Proof. Set k ∈ c1 N c1 /c2 M, c2 M c2 /c1 N. It follows from 2.4 that there exists T > T0
such that
∞ ∞ ∞
T
v
s
hu
c2
c2
du ds dv < min c2 M N − k, k − M − c2 N .
αsβv
c1
c1
2.63
Define two mappings S1k and S2k : AN, M → Ct0 , ∞, R by
⎧
⎪
⎨
k
xt τ
−
,
pt
τ
pt τ
S1k xt ⎪
⎩S1k xT ,
S2k xt ⎧
⎪
⎨
⎪
⎩
1
pt τ
∞ ∞ ∞
tτ
v
s
t ≥ T,
t0 ≤ t < T,
1
fu, xσ1 u, . . . , xσn udu ds dv,
αsβv
2.64
t ≥ T,
t0 ≤ t < T
2.65
S2k xT ,
for x ∈ AN, M.
Firstly, we show that S1k AN, M S2k AN, M ⊆ AN, M. From 2.3, 2.61–
2.65, we get that
S1k xt S2k y t
≤
k N
1
c2
c2
−
min c2 M N − k, k − M − c2 N
c2 c1 c2
c1
c1
≤ M,
x, y ∈ AN, M, t ≥ T,
S1k xt S2k y t
c2
c2
k M 1
−
− min c2 M N − k, k − M − c2 N
≥
c1 c2 c2
c1
c1
≥ N,
x, y ∈ AN, M, t ≥ T,
That is, S1k AN, M S2k AN, M ⊆ AN, M.
2.66
16
Abstract and Applied Analysis
Secondly, by 2.61, 2.62, and 2.64, we conclude that
S1k x − S1k y ≤ 1 x − y,
c2
x, y ∈ AN, M,
2.67
which implies that S1k is a contraction mapping in AN, M.
Thirdly, we show that S2k is completely continuous. Let x ∈ AN, M and {xm }m≥1 ⊆
AN, M be such that xm → x as m → ∞. By 2.61, 2.62, and 2.65, we gain that
|S2k xm t − S2k xt|
∞ ∞ ∞
1
1
fu, xm σ1 u, . . . , xm σn u
≤
pt τ tτ v
αsβv
s
−fu, xσ1 t, . . . , xσn tdu ds dv
1 ∞ ∞ ∞
1
fu, xm σ1 u, . . . , xm σn u
≤
c2 T
αsβv
v
s
−fu, xσ1 u, . . . , xσn udu ds dv, t ≥ T, m ≥ 1.
2.68
In view of 2.12, 2.13, 2.68, and the Lebesgue dominated convergence theorem, we obtain
that
lim S2k xm − S2k x 0,
m → ∞
2.69
that is, S2k is continuous in AN, M.
For each ε > 0, it follows from 2.4 that there exists T ∗ ≥ T satisfying
∞ ∞ ∞
T∗
v
s
c2 ε c22 ε
hu
.
du ds dv < min
,
αsβv
2 4c1
2.70
From 2.3, 2.61, 2.62, 2.65, and 2.70, we gain that
|S2k xt2 − S2k xt1 |
1 ∞ ∞ ∞ hu
1 ∞ ∞ ∞ hu
du ds dv du ds dv
≤
c2 t2 τ v
αsβv
c2 t1 τ v
αsβv
s
s
2 ∞ ∞ ∞ hu
du ds dv
≤
c2 T ∗ v
αsβv
s
< ε,
x ∈ AN, M, t2 > t1 ≥ T ∗ .
2.71
Abstract and Applied Analysis
17
By 2.3, 2.61, 2.62, 2.65, and 2.70, we obtain that, for x ∈ AN, M, T ≤ t1 ≤ t2 ≤ T ∗ ,
|S2k xt2 − S2k xt1 |
t2 τ ∞ ∞
1
hu
≤
du ds dv
pt2 τ t1 τ v
αsβv
s
∞ ∞ ∞
1
hu
1
−
du ds dv
pt2 τ pt1 τ t1 τ v
αsβv
s
1 t2 τ ∞ ∞ hu
2c1 ∞ ∞ ∞ hu
≤
du ds dv 2
du ds dv,
c2 t1 τ v
αsβv
αsβv
c2 T ∗ v
s
s
2.72
which means that there exists δ > 0 such that
x ∈ AN, M, t1 , t2 ∈ T, T ∗ with |t2 − t1 | < δ.
|S2k xt2 − S2k xt1 | < ε,
2.73
It is easy to verify that
|S2k xt2 − S2k xt1 | 0 < ε,
x ∈ AN, M, t0 ≤ t1 ≤ t2 ≤ T.
2.74
That is, S2k AN, M is equicontinuous on t0 , ∞, and S2k is completely continuous. By
Lemma 1.2, there is x0 ∈ AN, M such that S1k x0 S2k x0 x0 , which is a bounded
nonoscillatory solution of 1.7.
Lastly, we demonstrate that 1.7 possesses uncountably many nonoscillatory
k2 . For each j ∈ {1, 2},
solutions. Let k1 , k2 ∈ c1 N c1 /c2 M, c2 M c2 /c1 N and k1 /
we choose Tj > t0 and contraction mappings Sjk1 and Sjk2 satisfying 2.63–2.65 with k and
T replaced by kj and Tj , respectively, and some T3 > max{T1 , T2 } satisfying
∞ ∞ ∞
T3
v
s
c2 |k1 − k2 |
hu
du ds dv <
.
αsβv
2c1
2.75
Obviously, there are x, y ∈ AN, M such that S1k1 x S2k1 x x and S1k2 y S2k2 y y,
respectively, and x and y are two bounded nonoscillatory solutions of 1.7 in AN, M. By
2.64 and 2.65, we gain that for t ≥ T3
xt xt τ
k1
−
pt τ pt τ
∞ ∞ ∞
1
1
fu, xσ1 u, . . . , xσn udu ds dv,
pt τ tτ v
αsβv
s
yt τ
k2
−
yt pt τ pt τ
∞ ∞ ∞
1
1
f u, yσ1 u, . . . , yσn u du ds dv,
pt τ tτ v
αsβv
s
2.76
18
Abstract and Applied Analysis
which together with 2.3, 2.61, and 2.62 mean that
xt − yt ≥
≥
1
1
xt − τ − yt − τ
|k1 − k2 | −
pt τ
pt τ
∞ ∞ ∞
1
1
−
pt τ tτ v
αsβv
s
× fu, xσ1 u, . . . , xσn u − f u, yσ1 u, . . . , yσn u du ds dv
1
1
|k1 − k2 | − x − y
c1
c2
∞ ∞ ∞
2
hu
−
du ds dv,
c 2 T3 v
αsβv
s
t ≥ T3 ,
2.77
which together with 2.75 yields that
x − y ≥
c2
1 c2
1
2
|k1 − k2 | −
c1
c2
∞ ∞ ∞
T3
v
s
hu
du ds dv
αsβv
> 0,
2.78
that is, x /
y. This completes the proof.
Theorem 2.7. Assume that there exist constants M, N, c1 , c2 , T0 and a function h ∈ Ct0 , ∞, R satisfying 2.3, 2.4, and
0 < c1 − 1N < c2 − 1M;
−∞ < −c1 ≤ pt ≤ −c2 < −1,
t ≥ T0 ≥ t0 τ.
2.79
2.80
Then 1.7 possesses uncountably many bounded nonoscillatory solutions in AN, M.
Proof. Set k ∈ c1 − 1N, c2 − 1M. It follows from 2.4 that there exists T > T0 such that
∞ ∞ ∞
T
v
s
hu
c2 k N
du ds dv < min c2 − 1M − k,
− c2 N .
αsβv
c1
2.81
Define two mappings S1k and S2k : AN, M → Ct0 , ∞, R by
S1k xt S2k xt ⎧
⎪
⎨
⎪
⎩
1
pt τ
∞ ∞ ∞
tτ
S2k xT ,
v
s
⎧
⎪
⎨
⎪
⎩
−k
xt τ
−
,
pt τ pt τ
t ≥ T,
S1k xT ,
t0 ≤ t < T,
1
fu, xσ1 u, . . . , xσn udu ds dv,
αsβv
t ≥ T,
t0 ≤ t < T
2.82
Abstract and Applied Analysis
19
for x ∈ AN, M. From 2.3, 2.80–2.82, we obtain that
S1k xt S2k y t
k M 1
c2 k N
min c2 − 1M − k,
− c2 N
c2 c2 c2
c1
≤
≤ M,
x, y ∈ AN, M, t ≥ T,
S1k xt S2k y t
k N
1
c2 k N
≥
− min c2 − 1M − k,
− c2 N ,
c1 c1 c2
c1
2.83
≥ N,
x, y ∈ AN, M, t ≥ T
which implies that S1k AN, M S2k AN, M ⊆ AN, M. The rest of the proof is similar
to the proof of Theorem 2.6 and is omitted. This completes the proof.
3. Examples
In this section we construct seven examples as applications of the results presented in
Section 2.
Example 3.1. Consider the following third-order nonlinear neutral delay differential equation:
⎧ ⎨ t3
⎩ et
t
ln1 t
1
et
x2
⎫
⎬
3 sin t2 − 2 1
xt − τ
xt ⎭
7
2
2t
1 3t
x2 t −
1
t
3.1
0,
t ≥ t0 ,
where τ > 0 and t0 > 0 are fixed. Let n 2, and let M and N be two positive constants with
M > 7N and
t3
αt t ,
e
c1 t
,
βt ln1 t
4
,
7
c2 ft, u, v 2
,
7
σ1 t u2 v2
,
et
3 sin t2 − 2 1
pt ,
7
2t2
,
1 3t
1
σ2 t t − ,
t
ht 2M2
,
et
3.2
t, u, v ∈ t0 , ∞ × R2 .
Obviously, 2.1–2.3 hold. On the other hand,
∞ ∞ ∞
t0
v
s
hu
du ds dv M2
αsβv
∞
t0
ln1 v
dv < ∞.
v3
3.3
20
Abstract and Applied Analysis
That is, 2.4 holds. Thus Theorem 2.1 means that 1.7 possesses uncountably many bounded
nonoscillatory solutions in AN, M.
Example 3.2. Consider the following third-order nonlinear neutral delay differential equation:
⎧
⎨ 2t 13 ⎩
t
2
4t 4t 5 xt 2t2
xt − τ
1 3t2
⎫
⎬
√
2x t1
0,
⎭
t2
t ≥ t0 ,
3.4
where τ > 0 and t0 > 0 are fixed. Let n 1, and let M and N be two positive constants with
3t20 1M > 3t20 1N and
αt c1 2t 13
,
t
2
,
3
ft, u c2 2u
,
t2
βt 4t2 4t 5,
2t20
,
2
1 3t0
σ1 t √
pt t 1,
2t2
,
1 3t2
ht 2M
,
t2
3.5
t, u ∈ t0 , ∞ × R.
Obviously, 2.3, 2.25, and 2.26 hold. Moreover,
∞ ∞ ∞
t0
v
s
2M
1
1π
hu
du ds dv −
− arc tan2t0 1 < ∞.
αsβv
16 22t0 1 2 2
3.6
Hence 2.4 holds. Thus Theorem 2.2 ensures that 1.7 possesses uncountably many bounded
nonoscillatory solutions in AN, M.
Example 3.3. Consider the following third-order nonlinear neutral delay differential equation:
e
t−t2
1/3 −3t − 1
xt − τ
t 1
xt 5t 2
4
tx t3
et2 1 x2 t2 2t
0,
t ≥ 0,
3.7
where τ > 0 is fixed and t0 0. Let n 2, and let M and N be two positive constants with
4M > 5N and
2
αt et−t ,
c2 1
,
2
ft, u, v 1/3
βt t4 1
,
σ1 t tt 2,
tv
,
et2 1 u2 pt −3t − 1
,
5t 2
σ2 t t3 ,
t, u, v ∈ t0 , ∞ × R2 .
ht c1 3
,
5
Mt
,
1 N 2 et2
3.8
Abstract and Applied Analysis
21
Clearly, 2.3, 2.25, and 2.30 hold, and
∞ ∞ ∞
t0
v
s
M
hu
du ds dv αsβv
21 N 2 ∞
−1/3
e−v t4 1
dv < ∞,
3.9
t0
hence 2.4 holds. Therefore Theorem 2.3 guarantees that 1.7 possesses uncountably many
bounded nonoscillatory solutions in AN, M.
Example 3.4. Consider the following third-order nonlinear neutral delay differential equation:
x 2 t2 − t x t 2 − t 3
t3 1
1
0,
xt xt − τ
t2 arc tan t t
t4
t ≥ t0 ,
3.10
where τ > 0 and t0 > 0 are fixed. Let n 1, and let M and N be two positive constants with
M > N and
αt t3 1
,
arc tan t
t2
βt M2 M 3
,
ht t4
1
, pt 1,
t
σ1 t t2 − t,
u2 u 3
ft, u ,
t4
3.11
t, u ∈ t0 , ∞ × R.
Obviously, 2.3 holds. Furthermore,
∞ ∞ ∞
t0
v
s
hu
M2 M 3
du ds dv αsβv
3
∞
t0
varc tan v
dv < ∞,
v3 1
3.12
that is, 2.4 holds. Thus Theorem 2.4 ensures that 1.7 possesses uncountably many
bounded nonoscillatory solutions in AN, M.
Example 3.5. Consider the following third-order nonlinear neutral delay differential equation:
1
x2 t 1/t ln1 t
0,
t3 xt − xt − τ
t
t3
t ≥ t0 ,
3.13
where τ > 0 and t0 > 0 are fixed. Let n 1, and let M and N be two positive constants with
M > N and
αt t3 ,
βt M2
ht 3 ,
t
1
,
t
pt −1,
u2
ft, u 3 ,
t
σ1 t t 1
ln1 t,
t
t, u ∈ t0 , ∞ × R.
3.14
22
Abstract and Applied Analysis
Obviously, 2.3 holds. Notice that
∞ ∞ ∞ ∞
i1
t0 iτ
v
s
∞
M2 hu
1
du ds dv < ∞,
αsβv
16 i1 t0 iτ2
3.15
that is, 2.49 holds. Hence Theorem 2.5 implies that 1.7 possesses uncountably many
bounded nonoscillatory solutions in AN, M.
Example 3.6. Consider the following third-order nonlinear neutral delay differential equation:
t2 5π
t
2
2/3
arc tan t
t2 2 2
1 ln t xt xt − τ
0,
2
t
2
e
e x t 1 1
t ≥ t0 ,
3.16
where τ > 0 and t0 > 0 are fixed. Let n 1, and let M and N be two positive constants with
5/29π − 5/2N < 325π/4 − 3M and
αt pt t2
,
et2
βt 1 ln2 t,
5π
arc tan t2/3 ,
2
c1 3π, c2 5π
,
2
t
,
1et2
3.17
π
hu
1
du ds dv − arc tanln t0 < ∞,
2
αsβv
2N 1 2
3.18
ft, u σ1 t t2 1,
t
,
u2 1et2
ht N 2
t, u ∈ t0 , ∞ × R.
Clearly, 2.3, 2.61, and 2.62 hold, and
∞ ∞ ∞
t0
v
s
that is, 2.4 holds also. Thus Theorem 2.6 ensures that 1.7 possesses uncountably many
bounded nonoscillatory solutions in AN, M.
Example 3.7. Consider the following third-order nonlinear neutral delay differential equation
√
tx2 t2 1
2
1 t 1
1
xt sin t −
0,
4 cosln t − 7 xt − τ
2
t
t
t
e
et2
t ≥ t0 ,
3.19
Abstract and Applied Analysis
23
where τ > 0 and t0 > 0 are fixed. Let n 1, and let M and N be two positive constants with
M > 11N and
t2 1
1
,
βt ,
t
et2
1
pt sin t −
4 cosln t − 7,
t
αt ht tM2
,
et2
ft, u tu2
,
et2
c1 12, c2 2,
σ1 t !
t2 1,
3.20
t, u ∈ t0 , ∞ × R.
Obviously, 2.3, 2.79, and 2.80 hold. Note that
∞ ∞ ∞
t0
v
s
M2 π
hu
du ds dv − arc tan t0 < ∞,
αsβv
4 2
3.21
that is, 2.4 holds. It follows from Theorem 2.7 that 1.7 possesses uncountably many
bounded nonoscillatory solutions in AN, M.
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