96 Z. Zhang et al. / Neural Networks 25 (2012) 94–105
methods and techniques in order to obtain the global asymptotic
result of the equilibrium point for system (1.5).
The paper is organized as follows. In Section 2, a novel sufficient
condition is derived for the existence and uniqueness of the
equilibrium point for system (1.5). In Section 3, a novel sufficient
condition is derived for the global asymptotic stability of system
(1.5). In Section 4, an example and its simulation are given to show
the effectiveness of the obtained results. Finally, in Section 5, some
conclusions are provided.
For the sake of convenience, we introduce some notations and
definitions as follows.
For any matrix A, A
T
stands for the transpose of A and A
−1
denotes the inverse of A. If A is a symmetric matrix, A >
0 (A ≥ 0) means that A is positive definite (positive semidefinite).
Similarly, A < 0 (A ≤ 0) means that A is negative definite
(negative semidefinite). λ
M
(A), λ
m
(A) denote the maximum and
the minimum eigenvalue of a square matrix A. |A| = (|a
ij
|), |x| =
(|x
1
|, . . . , |x
m
|)
T
, ‖A‖ =
λ
M
(A
T
A). Let R
m
be an m-dimensional
Euclidean space, which is endowed with a norm ‖.‖ and inner
product (., .). Given column vector x = (x
1
, x
2
, . . . , x
m
)
T
∈ R
m
,
the norm is the Euclidean vector norm, i.e., ‖x‖ =
∑
m
i=1
x
2
i
1
2
.
Definition 1.1. A point (x
∗
, y
∗
)
T
∈ R
m
× R
m
is said to be an
equilibrium point of system (1.5) if
a
i
(x
∗
i
)
b
i
(x
∗
i
) −
m
−
j=1
s
ij
f
j
(x
∗
j
, y
∗
j
) + I
i
= 0, i = 1, 2, . . . , m,
c
j
(y
∗
j
)
d
j
(y
∗
j
) −
n
−
i=1
t
ji
g
i
(x
∗
i
, y
∗
i
) + J
j
= 0, j = 1, 2, . . . , m,
where x
∗
= (x
∗
1
, x
∗
2
, . . . , x
∗
m
)
T
, y
∗
= (y
∗
1
, y
∗
2
, . . . , y
∗
m
)
T
.
Lemma 1 (Forti & Tesi, 1995). Let H : R
2m
→ R
2m
be continuous.
Assume that H satisfies the following conditions
1. H(u) is injective on R
2m
,
2. ‖H(u)‖ → ∞ as ‖u‖ → ∞.
Then H is a homeomorphism.
Lemma 2. If a > 0, b > 0, β > 2, then
a
2
1
+ a
2
2
2
1
2
≤
a
β
1
+ a
β
2
2
1
β
.
Proof. Set αβ = 2 (0 < α < 1), then
2
−
i=1
a
2
i
2
=
2
−
i=1
1
2
α
a
αβ
i
1
2
1−α
≤
2
−
i=1
a
β
i
2
α
2
−
i=1
1
2
1−α
=
2
−
i=1
a
β
i
2
α
.
Hence
2
−
i=1
a
2
i
2
1
2
≤
2
−
i=1
a
β
i
2
α
2
=
2
−
i=1
a
β
i
2
1
β
.
Lemma 3. If a > 1, x > 0, then
x
a
≥ ax + 1 − a.
Proof. Set f (x) = x
a
− ax, x > 0. Then f
′
(x) = a(x
a−1
− 1). Since
a > 1, when 0 < x < 1, f
′
(x) < 0 and when x > 1, f
′
(x) > 0.
Hence, f (1) = 1 − a = min f (x). Namely x > 0, f (x) > 1 − a.
Lemma 4. If a > 0, b > 0, then
0 < p < 1, (a + b)
p
< a
p
+ b
p
.
Proof. This is a well-known inequality and its proof is omitted.
Lemma 5. If a > 0, b > 0,
1
p
+
1
q
= 1, then
ab ≤
a
p
p
+
b
q
q
.
Proof. This is a well-known inequality and its proof is omitted.
Lemma 6. If k > 1,
1
k
+
1
k
′
= 1, a
i
, b
i
∈ R, then
m
−
i=1
|a
i
| |b
i
| ≤
m
−
i=1
|a
i
|
k
1
k
m
−
i=1
|b
i
|
k
′
1
k
′
.
Proof. This is a well-known inequality and its proof is omitted.
Lemma 7. Assume that m and r are two positive integers, x
i
, y
i
∈ R,
then for i = 1, 2, . . . , m,
m
−
i=1
(|x
i
|
2r−1
+ |y
i
|
2r−1
) < A
∗
m
−
i=1
|x
i
|
2r
+ |y
i
|
2r
2r−1
2r
,
where A
∗
= m
1
2r
∑
2r
k=0
|C
k
2r
|
1
2r−1
max
k
2r
,
2r−k
2r
2r−1
2r
.
Proof. By using Lemmas 4 and 5, we obtain
m
−
i=1
(|x
i
|
2r−1
+ |y
i
|
2r−1
)
=
m
−
i=1
|x
i
|
2r−1
+
m
−
i=1
|y
i
|
2r−1
2r
2r−1
2r−1
2r
=
2r
−
k=0
C
k
2r
m
−
i=1
|x
i
|
2r−1
k
m
−
i=1
|y
i
|
2r−1
(2r−k)
1
2r−1
2r−1
2r
≤
2r
−
k=0
|C
k
2r
|
1
2r−1
m
−
i=1
|x
i
|
2r−1
k
2r−1
m
−
i=1
|y
i
|
2r−1
2r−k
2r−1
2r−1
2r
≤
2r
−
k=0
|C
k
2r
|
1
2r−1
k
2r
m
−
i=1
|x
i
|
2r−1
2r
2r−1
+
2r − k
2r
m
−
i=1
|y
i
|
2r−1
2r
2r−1
2r−1
2r
. (1.6)
Using Lemma 6, we have
m
−
i=1
|x
i
|
2r−1
≤
m
−
i=1
1
1
2r
m
−
i=1
(|x
i
|
2r−1
)
2r
2r−1
2r−1
2r
= m
1
2r
m
−
i=1
|x
i
|
2r
2r−1
2r
(1.7)
and
剩余11页未读，继续阅读
weixin_38750644
- 粉丝: 5
- 资源: 907
我的内容管理
展开
最新资源
- 李兴华Java基础教程：从入门到精通
- U盘与硬盘启动安装教程：从菜鸟到专家
- C++面试宝典：动态内存管理与继承解析
- C++ STL源码深度解析：专家级剖析与关键技术
- C/C++调用DOS命令实战指南
- 神经网络补偿的多传感器航迹融合技术
- GIS中的大地坐标系与椭球体解析
- 海思Hi3515 H.264编解码处理器用户手册
- Oracle基础练习题与解答
- 谷歌地球3D建筑筛选新流程详解
- CFO与CIO携手：数据管理与企业增值的战略
- Eclipse IDE基础教程：从入门到精通
- Shell脚本专家宝典：全面学习与资源指南
- Tomcat安装指南：附带JDK配置步骤
- NA3003A电子水准仪数据格式解析与转换研究
- 自动化专业英语词汇精华：必备术语集锦
资源上传下载、课程学习等过程中有任何疑问或建议，欢迎提出宝贵意见哦~我们会及时处理！
点击此处反馈
