Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 781983, 6 pages
http://dx.doi.org/10.1155/2013/781983
Research Article
ℎ-Stability of Linear Matrix Differential Systems
Peiguang Wang1 and Xiaowei Liu2
1
2
College of Electronic and Information Engineering, Hebei University, Baoding, Hebei 071002, China
College of Mathematics and Computer Science, Hebei University, Baoding, Hebei 071002, China
Correspondence should be addressed to Peiguang Wang; pgwang@mail.hbu.edu.cn
Received 2 December 2012; Accepted 1 February 2013
Academic Editor: Yong Hong Wu
Copyright © 2013 P. Wang and X. Liu. 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.
This paper investigates the stability problem of linear matrix differential systems and gives some sufficient conditions of ℎ-stability
for linear matrix system and its associated perturbed system by using the Kronecker product of matrices. An example is also worked
out to illustrate our results.
1. Introduction
The theory of stability in the sense of Lyapunov is well known
and is used in the real world. It is obvious that, in applications,
asymptotic stability is more important than stability because
the desirable feature is to know the size of the region of
asymptotic stability. However, when we study the asymptotic
stability, it is not easy to work with nonexponential types of
stability. In recent years, Medina and Pinto [1, 2] extended
the study of exponential stability to a variety of reasonable
systems called ℎ-systems. They introduced the notion of ℎstability with the intention of obtaining results about stability
for a weakly stable system (at least, weaker than those given
exponential stability and the uniform Lipschitz stability)
under some perturbations. Choi et al. [3] investigated ℎstability for the nonlinear differential systems by employing
the notion of 𝑡∞ -similarity and the Lyapunov functions. And
then, Choi et al. [4–6] also characterized the ℎ-stability in
variation for nonlinear difference systems via 𝑛∞ -similarity
and the Lyapunov functions and obtained some results
related to stability for the perturbations of nonlinear difference systems.
However, as far as the author’s scope, there are few
discussions and results for matrix differential systems. In this
paper, we shall investigate the ℎ-stability problem for linear
matrix differential systems by employing the Kronecker product of matrices which can be found in Lakshmikantham
and Deo’s monograph [7]. Some preliminaries are presented
in Section 2. A theorem is given in this section, which is
important to complete the main results of this paper. In
Section 3, sufficient conditions for the ℎ-stability are given for
linear matrix system and its associated perturbed system. An
example is also worked out at the end of this paper.
2. Preliminaries
Consider the linear matrix differential equation
𝑋󸀠 = 𝐴 (𝑡) 𝑋 + 𝑋𝐵 (𝑡) ,
𝑋 (𝑡0 ) = 𝑋0
(1)
and its associated perturbed system
𝑌󸀠 = 𝐴 (𝑡) 𝑌 + 𝑌𝐵 (𝑡) + 𝑅 (𝑡, 𝑌) ,
𝑌 (𝑡0 ) = 𝑋0 ,
(2)
where 𝐴, 𝐵 ∈ 𝐶[𝑅+ , 𝑅𝑛×𝑛 ], 𝑅 ∈ 𝐶[𝑅+ ×𝑅𝑛×𝑛 , 𝑅𝑛×𝑛 ], 𝑅(𝑡, 0) ≡ 0,
and 𝑋, 𝑌 ∈ 𝑅𝑛×𝑛 .
Now, we introduce the vec (⋅) operator which maps an 𝑚 ×
𝑛 matrix 𝑃 = (𝑝𝑖𝑗 ) onto the vector composed of the rows of 𝑃
𝑇
vec (𝑃) = (𝑝11 , . . . , 𝑝1𝑛 , 𝑝21 , . . . , 𝑝2𝑛 , . . . , 𝑝𝑚1 , . . . , 𝑝𝑚𝑛 ) .
(3)
Let us begin by defining the Kronecker product of matrices.
2
Abstract and Applied Analysis
Definition 1 (see [7]). If 𝑃 ∈ 𝑅𝑐×𝑑 , 𝑄 ∈ 𝑅𝑚×𝑛 , then the
Kronecker product of 𝑃 and 𝑄, 𝑃 ⊗ 𝑄 ∈ 𝑅𝑐𝑚×𝑑𝑛 , is defined
by the matrix
Lemma 3 (see [7]). Assume that 𝑥(𝑡, 𝑡0 , 𝑥0 ) is the solution of
(7) through (𝑡0 , 𝑥0 ), which exists for 𝑡 ≥ 𝑡0 , then
𝑝11 𝑄 𝑝12 𝑄 ⋅ ⋅ ⋅ 𝑝1𝑑 𝑄
𝑝21 𝑄 𝑝22 𝑄 ⋅ ⋅ ⋅ 𝑝2𝑑 𝑄
.
.
.
. )
(
𝑃⊗𝑄=
.
.
.
.
.
𝑝𝑐1 𝑄 𝑝𝑐2 𝑄 ⋅ ⋅ ⋅ 𝑝𝑐𝑑 𝑄
𝑥 (𝑡, 𝑡0 , 𝑥0 ) = [∫ Φ (𝑡, 𝑡0 , 𝑠𝑥0 ) 𝑑𝑠]𝑥0 ,
(
0
(4)
Among the main properties of this product presented in
[8], we recall the following useful ones:
(1) vec (𝑃𝑋𝑄) = (𝑃 ⊗ 𝑄𝑇 )vec (𝑋),
(2) vec (𝑃𝑋 + 𝑋𝑄) = (𝑃 ⊗ 𝐼 + 𝐼 ⊗ 𝑄𝑇 )vec (𝑋),
𝑦󸀠 = (𝐴 ⊗ 𝐼 + 𝐼 ⊗ 𝐵𝑇 ) 𝑦 + 𝑟 (𝑡, 𝑦) ,
where Φ(𝑡, 𝑡0 , 𝑥0 ) = 𝜕𝑥(𝑡, 𝑡0 , 𝑥0 )/𝜕𝑥0 .
Theorem 4. Assume that 𝑥(𝑡, 𝑡0 , 𝑥0 ) is the solution of (5) for
𝑡 ≥ 𝑡0 , let
𝐺 (𝑡, 𝑡0 , 𝑥0 ) = 𝐴 ⊗ 𝐼 + 𝐼 ⊗ 𝐵𝑇 .
where 𝑃, 𝑋, 𝑄, and 𝐼 ∈ 𝑅𝑛×𝑛 and 𝐼 is an identity matrix.
Then, the equivalent vector differential systems of (1) and
(2) can be written as
𝑥 (𝑡0 ) = 𝑥0 ,
𝜑󸀠 = 𝐺 (𝑡, 𝑡0 , 𝑥0 ) 𝜑,
(5)
2
where 𝑥 = vec (𝑋), 𝑦 = vec (𝑌), 𝑟 = vec (𝑅), 𝑟 ∈ 𝐶(𝑅+ × 𝑅𝑛 ,
2
𝑅𝑛 ), and 𝑟(𝑡, 0) ≡ 0.
In order to investigate ℎ-stability of linear matrix equation
and its associated perturbed system, we need to consider the
following systems and their properties. The techniques and
results are similar to those of [7].
Consider the linear differential system
(7)
where 𝑃 is an 𝑛 × 𝑛 continuous matrix and its perturbation
󸀠
𝑦 = 𝑃 (𝑡)𝑦 + 𝐹 (𝑡, 𝑦) ,
+
𝑛
𝑦 (𝑡0 ) = 𝑥0 ,
Φ (𝑡, 𝑡0 , 𝑥0 ) = 𝑊 (𝑡, 𝑡0 ) ⊗ 𝑍𝑇 (𝑡, 𝑡0 ) ,
𝑊󸀠 = 𝐴 (𝑡) 𝑊,
𝑊 (𝑡0 ) = 𝐼,
(15)
𝑍󸀠 = 𝑍𝐵 (𝑡) ,
𝑍 (𝑡0 ) = 𝐼,
(16)
respectively.
(ii) Any solution of (2) satisfies the integral equation
𝑌 (𝑡, 𝑡0 , 𝑋0 ) = 𝑋 (𝑡, 𝑡0 , 𝑋0 )
𝑡
+ ∫ 𝑊 (𝑡, 𝑠) 𝑅 (𝑠, 𝑌 (𝑠, 𝑡0 , 𝑋0 )) 𝑍 (𝑡, 𝑠) 𝑑𝑠,
(8)
where 𝐹 ∈ 𝐶[𝑅 × 𝑅 , 𝑅 ]. Suppose that the solution 𝑥(𝑡, 𝑡0 ,
𝑥0 ) of (7) exists for all 𝑡 ≥ 𝑡0 . The fundamental matrix solution Φ(𝑡, 𝑡0 , 𝑥0 ) of (7) is given by [7]
(9)
and Φ(𝑡0 , 𝑡0 , 𝑥0 ) = 𝐼.
We are now in a position to give the Alekseev formula,
which connects the solutions of (7) and (8).
Lemma 2 (see [7]). If 𝑥(𝑡, 𝑡0 , 𝑥0 ) is the solution of (7) and
exists for 𝑡 ≥ 𝑡0 , any solution 𝑦(𝑡, 𝑡0 , 𝑥0 ) of (8), with 𝑦(𝑡0 ) = 𝑥0 ,
satisfies the integral equation
𝑡0
(17)
for 𝑡 ≥ 𝑡0 .
Proof. (i) It is obvious that Φ(𝑡, 𝑡0 , 𝑥0 ) = 𝜕𝑥(𝑡, 𝑡0 , 𝑥0 )/𝜕𝑥0
exists and is the fundamental matrix solution of the variational equation
𝜑󸀠 = 𝐺 (𝑡, 𝑡0 , 𝑥0 ) 𝜑,
(18)
such that Φ(𝑡0 , 𝑡0 , 𝑥0 ) = 𝐼.
Furthermore, we get
𝜑󸀠 = 𝐺 (𝑡, 𝑡0 , 𝑥0 ) 𝜑 = [𝐴 ⊗ 𝐼 + 𝐼 ⊗ 𝐵𝑇 ] 𝜑,
(19)
with the initial value
𝑦 (𝑡, 𝑡0 , 𝑥0 ) = 𝑥 (𝑡, 𝑡0 , 𝑥0 )
𝑡
+ ∫ Φ (𝑡, 𝑠, 𝑦 (𝑠, 𝑡0 , 𝑥0 )) 𝐹 (𝑠, 𝑦 (𝑠, 𝑡0 , 𝑥0 ))𝑑𝑠,
𝑡0
(10)
for 𝑡 ≥ 𝑡0 , where Φ(𝑡, 𝑡0 , 𝑥0 ) = 𝜕𝑥(𝑡, 𝑡0 , 𝑥0 )/𝜕𝑥0 .
(14)
where 𝑊(𝑡, 𝑡0 ) and 𝑍(𝑡, 𝑡0 ) are solutions of
𝑛
𝜕𝑥 (𝑡, 𝑡0 , 𝑥0 )
Φ (𝑡, 𝑡0 , 𝑥0 ) =
𝜕𝑥0
(13)
such that Φ(𝑡0 , 𝑡0 , 𝑥0 ) = 𝐼, and therefore
𝑦 (𝑡0 ) = 𝑥0 ,
𝑥 (𝑡0 ) = 𝑥0 ,
(12)
Then one has the following.
(i) Φ(𝑡, 𝑡0 , 𝑥0 ) = 𝜕𝑥(𝑡, 𝑡0 , 𝑥0 )/𝜕𝑥0 exists and is the fundamental matrix solution of the variational equation
(6)
𝑥󸀠 = 𝑃 (𝑡) 𝑥,
(11)
The following theorem gives an analog of the variation of
parameters formula for the solution of (2).
)
𝑥󸀠 = (𝐴 ⊗ 𝐼 + 𝐼 ⊗ 𝐵𝑇 ) 𝑥,
1
𝜑 (𝑡0 , 𝑡0 , 𝑥0 ) = 𝑒,
𝑒 = vec (𝐼) ,
(20)
which has the solution
𝜑 (𝑡, 𝑡0 , 𝑥0 ) = (𝑊 (𝑡, 𝑡0 ) ⊗ 𝑍𝑇 (𝑡, 𝑡0 )) 𝑒,
(21)
Abstract and Applied Analysis
3
where 𝑊 and 𝑍 are the solutions of (15) and (16), respectively,
and 𝐼 is the 𝑛 × 𝑛 identity matrix.
Therefore,
Φ (𝑡, 𝑡0 , 𝑥0 ) = 𝑊 (𝑡, 𝑡0 ) ⊗ 𝑍𝑇 (𝑡, 𝑡0 ) .
(22)
(ii) Employing Lemma 2 and substituting for Φ the righthand side of (22), we get
𝑦 (𝑡, 𝑡0 , 𝑥0 ) = 𝑥 (𝑡, 𝑡0 , 𝑥0 )
𝑡
+ ∫ [𝑊 (𝑡, 𝑠) ⊗ 𝑍𝑇 (𝑡, 𝑠)] 𝑟 (𝑠, 𝑦 (𝑠, 𝑡0 , 𝑥0 ))𝑑𝑠
𝑡0
(23)
for 𝑡 ≥ 𝑡0 , where 𝑦(𝑡, 𝑡0 , 𝑥0 ) is any solution of (6).
Now, we define 𝑋(𝑡, 𝑡0 , 𝑋0 ), 𝑌(𝑡, 𝑡0 , 𝑋0 ), and 𝑅(𝑡, 𝑌) by
𝑥 = vec (𝑋), 𝑦 = vec (𝑌), and 𝑟 = vec (𝑅). Thus, we have
that
Lemma 7 (see [4]). The linear system
𝑥󸀠 = 𝐴 (𝑡) 𝑥,
𝑥 (𝑡0 ) = 𝑥0 ,
is ℎ𝑆 if and only if there exist a constant 𝑐 ≥ 1 and a positive
continuous bounded function ℎ defined on 𝑅+ such that for
every 𝑥0 in 𝑅𝑛 ,
󵄩󵄩
󵄩
−1
󵄩󵄩Φ (𝑡, 𝑡0 , 𝑥0 )󵄩󵄩󵄩 ≤ 𝑐ℎ (𝑡) ℎ (𝑡0 )
Theorem 8. The solution 𝑋 = 0 of (1) is ℎ𝑆 if and only if the
solution 𝑥 = 0 of (5) is ℎ𝑆.
Proof. Necessity. Since the solution 𝑋 = 0 of (1) is ℎ𝑆, there
exist 𝑐 ≥ 1, 𝛿 > 0, and a positive bounded continuous function ℎ on 𝑅+ such that
󵄩 󵄩
󵄩󵄩
󵄩
−1
󵄩󵄩𝑋 (𝑡, 𝑡0 , 𝑋0 )󵄩󵄩󵄩 ≤ 𝑐 󵄩󵄩󵄩𝑋0 󵄩󵄩󵄩 ℎ (𝑡) ℎ (𝑡0 )
+ ∫ 𝑊 (𝑡, 𝑠) 𝑅 (𝑠, 𝑌 (𝑠, 𝑡0 , 𝑋0 )) 𝑍 (𝑡, 𝑠) 𝑑𝑠,
𝑡0
(24)
for 𝑡 ≥ 𝑡0 , where 𝑋(𝑡, 𝑡0 , 𝑋0 ) is the unique solution of (1) for
𝑡 ≥ 𝑡0 .
The proof is completed.
𝑋󸀠 = 𝐴 (𝑡) 𝑋 + 𝑋𝐵 (𝑡) ,
𝑋 (𝑡0 ) = 𝑋0 ,
(31)
It follows that
𝑚×𝑛
Definition 5. A generalized matrix valued norm from 𝑅
𝑅+ is a mapping ‖ ⋅ ‖ : 𝑅𝑚×𝑛 → 𝑅+ such that
[vec (𝑋)]󸀠 = (𝐴 ⊗ 𝐼 + 𝐼 ⊗ 𝐵𝑇 ) vec (𝑋) ,
to
(32)
Thus,
󵄩󵄩
󵄩 󵄩
󵄩
󵄩󵄩𝑥 (𝑡, 𝑡0 , 𝑥0 )󵄩󵄩󵄩 = 󵄩󵄩󵄩vec (𝑋 (𝑡, 𝑡0 , 𝑋0 ))󵄩󵄩󵄩
󵄩
󵄩 󵄩
󵄩
= 󵄩󵄩󵄩𝑋 (𝑡, 𝑡0 , 𝑋0 )󵄩󵄩󵄩 ≤ 𝑐 󵄩󵄩󵄩𝑥0 󵄩󵄩󵄩 ℎ (𝑡) ℎ−1 (𝑡0 ) .
(a) ‖𝑋‖ ≥ 0, ‖𝑋‖ = 0 if and only if 𝑋 = 0,
(b) ‖𝜆𝑋‖ = |𝜆|‖𝑋‖, 𝜆 is a constant,
(c) ‖𝑋 + 𝑌‖ ≤ ‖𝑋‖ + ‖𝑌‖.
Definition 6. The zero solution of (1) is said to be
(hS) ℎ-stability if there exist 𝑐 ≥ 1, 𝛿 > 0, and a positive
bounded continuous function ℎ on 𝑅+ such that
(25)
for 𝑡 ≥ 𝑡0 ≥ 0 and ‖𝑋0 ‖ ≤ 𝛿, ℎ−1 (𝑡0 ) = 1/ℎ(𝑡0 ).
(hSV) ℎ-stability in variation if there exist 𝑐1 , 𝑐2 ≥ 1, 𝛿 > 0,
and a positive bounded continuous function ℎ on 𝑅+
satisfying
󵄩
󵄩󵄩
−1
󵄩󵄩𝑊 (𝑡, 𝑡0 )󵄩󵄩󵄩 ≤ 𝑐1 ℎ (𝑡) ℎ (𝑡0 ) ,
󵄩󵄩󵄩𝑍 (𝑡, 𝑡0 )󵄩󵄩󵄩 ≤ 𝑐2 ℎ (𝑡) ℎ−1 (𝑡0 ) ,
󵄩
󵄩
(30)
then we obtain
We firstly give some notions.
󵄩 󵄩
󵄩󵄩
󵄩
−1
󵄩󵄩𝑋 (𝑡, 𝑡0 , 𝑋0 )󵄩󵄩󵄩 ≤ 𝑐 󵄩󵄩󵄩𝑋0 󵄩󵄩󵄩 ℎ (𝑡) ℎ (𝑡0 ) ,
(29)
for every 𝑡 ≥ 𝑡0 ≥ 0, ‖𝑋0 ‖ ≤ 𝛿, where 𝑋(𝑡, 𝑡0 , 𝑋0 ) is the solution of (1), satisfying
vec (𝑋󸀠 ) = vec (𝐴 (𝑡) 𝑋 + 𝑋𝐵 (𝑡)) .
3. Main Results
(28)
for all 𝑡 ≥ 𝑡0 ≥ 0, where 𝐴(𝑡) is an 𝑛 × 𝑛 continuous matrix and
Φ(𝑡, 𝑡0 , 𝑥0 ) is a fundamental matrix of (27).
𝑌 (𝑡, 𝑡0 , 𝑋0 ) = 𝑋 (𝑡, 𝑡0 , 𝑋0 )
𝑡
(27)
(26)
provided ‖𝑋0 ‖ ≤ 𝛿, where 𝑊(𝑡, 𝑡0 ) and 𝑍(𝑡, 𝑡0 ) are
given in Theorem 4.
(33)
Sufficiency. It can be easily proved by the same method. The
proof is completed.
Theorem 9. The solution 𝑋 = 0 of (1) is ℎ𝑆 if and only if
there exist a constant 𝑐 ≥ 1 and a positive continuous bounded
function ℎ defined on 𝑅+ such that for every 𝑋0 in 𝑅𝑛×𝑛 ,
󵄩󵄩
󵄩
󵄩󵄩𝑊 (𝑡, 𝑡0 ) ⊗ 𝑍𝑇 (𝑡, 𝑡0 )󵄩󵄩󵄩 ≤ 𝑐ℎ (𝑡) ℎ−1 (𝑡0 )
󵄩
󵄩
(34)
for all 𝑡 ≥ 𝑡0 ≥ 0.
Proof. Sufficiency. Following Lemma 3 and Theorem 4, we
have
1
𝑥 (𝑡, 𝑡0 , 𝑥0 ) = [∫ Φ (𝑡, 𝑡0 , 𝑠𝑥0 )𝑑𝑠] 𝑥0 ,
0
𝑇
Φ (𝑡, 𝑡0 , 𝑥0 ) = 𝑊 (𝑡, 𝑡0 ) ⊗ 𝑍 (𝑡, 𝑡0 ) .
(35)
4
Abstract and Applied Analysis
2
(ii) ‖𝑥‖ ≤ 𝑉(𝑡, 𝑥) ≤ 𝑐‖𝑥‖, (𝑡, 𝑥) ∈ 𝑅+ × 𝑅𝑛 , 𝑐 ≥ 1,
It follows that
2
1
𝑇
𝑥 (𝑡, 𝑡0 , 𝑥0 ) = [∫ 𝑊 (𝑡, 𝑡0 ) ⊗ 𝑍 (𝑡, 𝑡0 )𝑑𝑠]𝑥0 .
0
(36)
Hence,
1
󵄩
󵄩󵄩
󵄩 󵄩 󵄩 󵄩
𝑇
󵄩󵄩𝑥 (𝑡, 𝑡0 , 𝑥0 )󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥0 󵄩󵄩󵄩 ∫ 󵄩󵄩󵄩󵄩𝑊 (𝑡, 𝑡0 ) ⊗ 𝑍 (𝑡, 𝑡0 )󵄩󵄩󵄩󵄩 𝑑𝑠
0
󵄩 󵄩
≤ 𝑐 󵄩󵄩󵄩𝑥0 󵄩󵄩󵄩 ℎ (𝑡) ℎ−1 (𝑡0 ) , 𝑡 ≥ 𝑡0 .
(iii) 𝐷+ 𝑉(5) (𝑡, 𝑥) ≤ ℎ󸀠 (𝑡)ℎ−1 (𝑡)𝑉(𝑡, 𝑥), (𝑡, 𝑥) ∈ 𝑅+ × 𝑅𝑛 .
Then, the solution 𝑋 = 0 of (1) is ℎ𝑆.
Proof. Let 𝑥(𝑡, 𝑡0 , 𝑥0 ) be the solution of (5). As a consequence
of (iii), we obtain
(37)
𝑉 (𝑡, 𝑋 (𝑡, 𝑡0 , 𝑥0 )) ≤ 𝑉 (𝑡0 , 𝑥0 ) exp ∫
𝑡0
Therefore, the solution 𝑥 = 0 of (5) is ℎ𝑆. By Theorem 8, it
implies that the solution 𝑋 = 0 of (1) is ℎ𝑆.
Necessity. If the solution 𝑋 = 0 of (1) is ℎ𝑆, then the solution
𝑥 = 0 of (5) is ℎ𝑆 using Theorem 8. By Lemma 7, we have
󵄩
󵄩󵄩
−1
(38)
󵄩󵄩Φ (𝑡, 𝑡0 , 𝑥0 )󵄩󵄩󵄩 ≤ 𝑐ℎ (𝑡) ℎ (𝑡0 ) .
From Theorem 4, we obtain that
Φ (𝑡, 𝑡0 , 𝑥0 ) = 𝑊 (𝑡, 𝑡0 ) ⊗ 𝑍𝑇 (𝑡, 𝑡0 ) ,
= 𝑉 (𝑡0 , 𝑥0 ) ℎ (𝑡) ℎ (𝑡0 ) .
From the condition (ii), we have
󵄩 󵄩
󵄩
󵄩󵄩
−1
󵄩󵄩𝑥 (𝑡, 𝑡0 , 𝑥0 )󵄩󵄩󵄩 ≤ 𝑐 󵄩󵄩󵄩𝑥0 󵄩󵄩󵄩 ℎ (𝑡) ℎ (𝑡0 ) ,
By Theorem 8, we can easily get that the solution 𝑋 = 0 of (1)
is ℎ𝑆. The proof is completed.
𝑡
𝑡
𝑡0
𝑡0
𝑥 (𝑡) − ∫ 𝑘 (𝑠, 𝑥 (𝑡))𝑑𝑠 ≤ 𝑦 (𝑡) − ∫ 𝑘 (𝑠, 𝑦 (𝑡)) 𝑑𝑠,
1
𝛿
(46)
𝑡 ≥ 𝑡0 ≥ 0
for 𝑥, 𝑦 ∈ 𝐶([𝑡0 , ∞), 𝑅𝑛 ). If 𝑥(𝑡0 ) < 𝑦(𝑡0 ), then 𝑥(𝑡) < 𝑦(𝑡) for
all 𝑡 ≥ 𝑡0 ≥ 0.
Theorem 13. Assume that 𝑋 = 0 of (1) is ℎ𝑆𝑉 with the nonincreasing function ℎ1 and ℎ2 . Consider the scalar differential
equation
𝑢󸀠 = 𝑐𝑙 (𝑡, 𝑢) ,
× [𝑉 (𝑡 + 𝛿, 𝑥 + 𝛿 (𝐴 ⊗ 𝐼 + 𝐼 ⊗ 𝐵𝑇 ) 𝑥)
𝑢 (𝑡0 ) = 𝑢0 ,
𝑤ℎ𝑒𝑟𝑒 𝑐 ≥ 1.
(47)
Suppose that
‖𝑅 (𝑡, 𝑌)‖ ≤ 𝑙 (𝑡, ‖𝑌‖) ,
−𝑉 (𝑡, 𝑥) ] ,
(41)
2
for (𝑡, 𝑥) ∈ 𝑅+ × 𝑅𝑛 and for the solution 𝑥(𝑡) = 𝑥(𝑡, 𝑡0 , 𝑥0 ) of
(5),
𝛿→0
(45)
(40)
Next, we offer sufficient conditions for the ℎ-stability of
linear matrix differential systems by using the Lyapunov functions.
Defining the Lyapunov functions
𝐷+ 𝑉 (𝑡, 𝑥) = lim sup
𝑡 ≥ 𝑡0 ≥ 0.
Lemma 12 (see [9]). Suppose that 𝑘(𝑡, 𝑥) ∈ 𝐶(𝑅+ × 𝑅𝑛 , 𝑅𝑛 ) is
strictly increasing in 𝑥 for 𝑡 ≥ 𝑡0 ≥ 0 with the property
Corollary 10. If the zero solution of (1) is ℎ𝑆𝑉, then the zero
solution of (1) is ℎ𝑆.
𝛿→0
(44)
−1
Now, we examine the properties of the perturbed linear
matrix differential system.
This completes the proof.
𝐷+ 𝑉(5) (𝑡, 𝑥) = lim sup
ℎ󸀠 (𝑠)
𝑑𝑠
ℎ (𝑠)
(39)
Thus,
󵄩󵄩
󵄩
󵄩󵄩𝑊 (𝑡, 𝑡0 ) ⊗ 𝑍𝑇 (𝑡, 𝑡0 )󵄩󵄩󵄩 ≤ 𝑐ℎ (𝑡) ℎ−1 (𝑡0 ) .
󵄩
󵄩
𝑡
1
[𝑉 (𝑡 + 𝛿, 𝑥 (𝑡 + 𝛿)) − 𝑉 (𝑡, 𝑥)] .
𝛿
(42)
(48)
where 𝑙 ∈ 𝐶(𝑅+ × 𝑅+ , 𝑅+ ) is strictly increasing in 𝑢 for each
fixed 𝑡 ≥ 𝑡0 ≥ 0 with 𝑙(𝑡, 0) = 0.
If 𝑢 = 0 is ℎ𝑆, then the solution 𝑌 = 0 of (2) is also ℎ𝑆,
whenever 𝑢0 = 𝑐‖𝑌0 ‖.
Proof. By Theorem 4, the solutions of (1) and (2) with the
same initial values are related by
𝑌 (𝑡, 𝑡0 , 𝑌0 ) = 𝑋 (𝑡, 𝑡0 , 𝑌0 )
Then, it is well known that
𝑡
𝐷+ 𝑉(5) (𝑡, 𝑥) = 𝐷+ 𝑉 (𝑡, 𝑥) ,
+ ∫ 𝑊 (𝑡, 𝑠) 𝑅 (𝑠, 𝑌 (𝑠, 𝑡0 , 𝑌0 )) 𝑍 (𝑡, 𝑠)𝑑𝑠.
(43)
𝑡0
+
(49)
if 𝑉(𝑡, 𝑥) is the Lipschitzian in 𝑥 for each 𝑡 ∈ 𝑅 .
Theorem 11. Suppose that ℎ(𝑡) is a positive bonded continuously differentiable function on 𝑅+ . Furthermore, assume that
there exists a function 𝑉(𝑡, 𝑥) satisfying the following properties:
2
(i) 𝑉 ∈ 𝐶(𝑅+ × 𝑅𝑛 , 𝑅+ ), and 𝑉(𝑡, 𝑥) is Lipschitzian in 𝑥
for each 𝑡 ∈ 𝑅+ ,
Then, we have
𝑡
󵄩 󵄩
󵄩
󵄩󵄩
󵄩󵄩𝑌 (𝑡, 𝑡0 , 𝑌0 )󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑋 (𝑡, 𝑡0 , 𝑌0 )󵄩󵄩󵄩 + ∫ ‖𝑊 (𝑡, 𝑠)‖
𝑡
0
󵄩
󵄩
× 󵄩󵄩󵄩𝑅 (𝑠, 𝑌 (𝑠, 𝑡0 , 𝑌0 ))󵄩󵄩󵄩 ‖Z (𝑡, 𝑠)‖ 𝑑𝑠.
(50)
Abstract and Applied Analysis
5
From Corollary 10, it easily follows that
󵄩󵄩
󵄩 󵄩
󵄩
−1
󵄩󵄩𝑌 (𝑡, 𝑡0 , 𝑌0 )󵄩󵄩󵄩 ≤ 𝑐1 󵄩󵄩󵄩𝑌0 󵄩󵄩󵄩 ℎ1 (𝑡) ℎ1 (𝑡0 )
𝑡
+ ∫ 𝑐2 𝑐3 [ℎ2 (𝑡) ℎ2−1 (𝑠)]
Then, by Gronwall’s inequality, we get
󵄩 󵄩
󵄩
󵄩󵄩
−1
󵄩󵄩𝑌 (𝑡, 𝑡0 , 𝑋0 )󵄩󵄩󵄩 ≤ 𝑐 󵄩󵄩󵄩𝑋0 󵄩󵄩󵄩 ℎ (𝑡) ℎ (𝑡0 )
∞
2
𝑡0
󵄩
󵄩
× 󵄩󵄩󵄩𝑅 (𝑠, 𝑌 (𝑠, 𝑡0 , 𝑌0 ))󵄩󵄩󵄩𝑑𝑠
2
× exp (∫ 𝑐1 𝑐2 [ℎ1 (𝑡) ℎ1−1 (𝑠)] 𝛾 (𝑠) 𝑑𝑠)
𝑡0
(51)
󵄩 󵄩
≤ 𝑀 󵄩󵄩󵄩𝑋0 󵄩󵄩󵄩 ℎ (𝑡) ℎ−1 (𝑡0 ) ,
(56)
𝑡
󵄩 󵄩
≤ 𝑐 󵄩󵄩󵄩𝑌0 󵄩󵄩󵄩 + 𝑐 ∫ 𝑙 (𝑠, ‖𝑌 (𝑠)‖) 𝑑𝑠,
𝑡
∞
0
where 𝑐 = max{𝑐1 , 𝑐2 𝑐3 }. Since ℎ1 (𝑡) and ℎ2 (𝑡) are nonincreasing, we obtain
4. Example
𝑡
󵄩
󵄩󵄩
󵄩󵄩𝑌 (𝑡, 𝑡0 , 𝑌0 )󵄩󵄩󵄩 − 𝑐 ∫ 𝑙 (𝑠, ‖𝑌 (𝑠)‖)𝑑𝑠
𝑡0
𝑡
󵄩 󵄩
≤ 𝑐 󵄩󵄩󵄩𝑌0 󵄩󵄩󵄩 = 𝑢0 = 𝑢 (𝑡) − ∫ 𝑐𝑙 (𝑠, 𝑢 (𝑠))𝑑𝑠.
(52)
𝑡0
By Lemma 12, we have ‖𝑌(𝑡)‖ < 𝑢(𝑡) for all 𝑡 ≥ 𝑡0 ≥ 0. Since
𝑢 = 0 of (47) is ℎ𝑆,
In this section, we give a simple but illustrative example. Consider the matrix differential equation
−1 1
−1 0
𝑋󸀠 (𝑡) = (
) 𝑋 (𝑡) + 𝑋 (𝑡) (
),
0 −2
0 −1
1 0
𝑋 (0) = (
).
0 1
‖𝑌 (𝑡)‖ < 𝑢 (𝑡) ≤ 𝑐4 𝑢0 ℎ (𝑡) ℎ−1 (𝑡0 )
󵄩 󵄩
= 𝑐4 𝑐 󵄩󵄩󵄩𝑌0 󵄩󵄩󵄩 ℎ (𝑡) ℎ−1 (𝑡0 )
󵄩 󵄩
= 𝑀 󵄩󵄩󵄩𝑌0 󵄩󵄩󵄩 ℎ (𝑡) ℎ−1 (𝑡0 ) ,
2
where 𝑀 = max{𝑐 exp(∫𝑡 𝑐1 𝑐2 [ℎ1 (𝑡)ℎ1−1 (𝑠)] 𝛾(𝑠)𝑑𝑠)}.
0
The proof is completed.
Then, we can obtain the following equations:
Theorem 14. Assume that
−1 1
) 𝑊,
0 −2
𝑊 (0) = 𝐼,
(58)
−1 0
),
𝑍󸀠 = 𝑍 (
0 −1
𝑍 (0) = 𝐼.
(59)
𝑊󸀠 = (
𝑐4 ≥ 1, 𝑀 = 𝑐4 𝑐 ≥ 1.
(53)
This completes the proof.
The solutions of (58) and (59) are
(i) the zero solution of (1) is ℎ𝑆𝑉,
(ii) ‖𝑅(𝑡, 𝑌(𝑡))‖ ≤ 𝛾(𝑡)‖𝑌(𝑡)‖ provided that 𝛾(𝑡) > 0 and
∞
∫𝑡 𝛾(𝑡)𝑑𝑡 < ∞ for 𝑡0 ≥ 0.
𝑒−𝑡 𝑒−𝑡 − 𝑒−2𝑡
),
𝑊 (𝑡) = (
0
𝑒−2𝑡
𝑍 (𝑡) = (
𝑒−𝑡 0
),
0 𝑒−𝑡
(60)
0
Then, the solution 𝑌 = 0 of (2) is ℎ𝑆.
Proof. By Theorem 4, the solutions of (1) and (2) with the
same initial values are related by
𝑌 (𝑡, 𝑡0 , 𝑋0 ) = 𝑋 (𝑡, 𝑡0 , 𝑋0 )
𝑡
+ ∫ 𝑊 (𝑡, 𝑠) 𝑅 (𝑠, 𝑌 (𝑠, 𝑡0 , 𝑋0 )) 𝑍 (𝑡, 𝑠)𝑑𝑠.
𝑡0
(54)
𝑒−2𝑡 0 𝑒−2𝑡 − 𝑒−3𝑡
0
−2𝑡
−2𝑡
0
𝑒
−
𝑒−3𝑡
0
𝑒
).
𝑊 (𝑡) ⊗ 𝑍𝑇 (𝑡) = (
−3𝑡
0
0
𝑒
0
−3𝑡
0
0
0
𝑒
(61)
Thus, we have
𝑡
𝑡0
󵄩 󵄩
× ‖𝑍 (𝑡, 𝑠)‖ 𝑑𝑠 ≤ 𝑐 󵄩󵄩󵄩𝑋0 󵄩󵄩󵄩 ℎ (𝑡) ℎ−1 (𝑡0 )
Acknowledgments
2
𝑡0
󵄩
󵄩
× 𝛾 (𝑠) 󵄩󵄩󵄩𝑌 (𝑠, 𝑡0 , 𝑋0 )󵄩󵄩󵄩 𝑑𝑠.
(62)
where ℎ(𝑡) = 𝑒−2𝑡 , 𝑐 = 5, ‖𝑊(𝑡) ⊗ 𝑍𝑇 (𝑡)‖ = 4𝑒−2𝑡 , ‖𝐷‖ =
𝑚×𝑛
, and 𝐼 is an identity matrix. So, from
∑𝑚,𝑛
𝑖,𝑗 |𝑑𝑖𝑗 |, 𝐷 ∈ 𝑅
Theorem 9, we can conclude that the solution 𝑋 = 0 of (57)
is ℎ𝑆.
󵄩
󵄩
+ ∫ ‖𝑊 (𝑡, 𝑠)‖ 󵄩󵄩󵄩𝑅 (𝑠, 𝑌 (𝑠, 𝑡0 , 𝑋0 ))󵄩󵄩󵄩
+ ∫ 𝑐1 𝑐2 [ℎ1 (𝑡) ℎ1−1 (𝑠)]
respectively.
Then,
󵄩
󵄩󵄩
󵄩󵄩𝑊 (𝑡) ⊗ 𝑍𝑇 (𝑡)󵄩󵄩󵄩 ≤ 𝑐ℎ (𝑡) ℎ−1 (0) ,
󵄩
󵄩
The assumptions (i) and (ii) yield
󵄩󵄩
󵄩 󵄩
󵄩
󵄩󵄩𝑌 (𝑡, 𝑡0 , 𝑋0 )󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑋 (𝑡, 𝑡0 , 𝑋0 )󵄩󵄩󵄩
𝑡
(57)
(55)
The authors would like to thank the reviewers for their valuable suggestions and comments. This paper is supported by
the National Natural Science Foundation of China (10971045
and 11271106).
6
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