Hindawi Publishing Corporation Mathematical Problems in Engineering Volume 2015, Article ID 837053, 10 pages http://dx.doi.org/10.1155/2015/837053
Research Article Study on H− Index of Stochastic Linear Continuous-Time Systems Yan Li,1 Tianliang Zhang,2 Xikui Liu,3 and Xiushan Jiang2 1
College of Electrical Engineering and Automation, Shandong University of Science and Technology, Qingdao, Shandong 266590, China College of Information and Control Engineering, China University of Petroleum (East China), Qingdao, Shandong 266590, China 3 College of Mathematics and Systems Science, Shandong University of Science and Technology, Qingdao, Shandong 266590, China 2
Correspondence should be addressed to Xikui Liu;
[email protected] Received 3 February 2015; Accepted 30 March 2015 Academic Editor: Ruihua Liu Copyright © 2015 Yan Li 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. This paper studies the H− index problem. We obtain a necessary and sufficient condition of H− index larger than 𝛾 > 0. A generalized differential equation is introduced and it is proved that its solvability and the feasibility of the H− index are equivalent. We extend the deterministic cases to the stochastic models. Our results can be used to fault detection filter analysis. Finally, the effectiveness of the proposed results is illustrated by an example.
1. Introduction It is well known that many control and filtering problems have been discussed based on a certain performance index of a system, such as H2 norm, H∞ norm, and H− index; see [1–9]. H∞ norm is the measure of the worst-case disturbance inputs on the controlled outputs [1–4]. The H− index is a measure of the minimum sensitivity of system outputs to system inputs. H∞ norm and H− index with specific application to fault detection filter have been carried out in [10–17]. To ensure robustness, H− index should be maximized and H∞ norm should be minimized. Using H− /H∞ performance can make certain that the residual signal is maximally sensitive to faults and highly robust to disturbance inputs; see [16, 17]. In [12], H− index was defined as the minimum nonzero singular value in zero frequency. In [10], the authors extended the results of [12] to all frequency range. By means of LMIs, a necessary and sufficient condition was given for the infinite frequency range. The case for finite frequency range was concluded through frequency weighting. In recent decades, a great deal of attention has been attracted to H− index in time domain. A fault residual generator was designed to maximize the fault sensitivity in the finite time
domain [16–20]. Based on H− index, results on optimal fault detection can be found in [17, 18] and the references. The lower bound of H− index for linear time-varying systems was proposed in [19, 20]. A sliding mode observer was designed for sensor fault diagnosis of nonlinear time-delay systems; see [21]. In [22], a fault-tolerant controller was projected to compensate nonlinear faults by using a fuzzy adaptive fault observer. Although there is much work on the H− index problem, to the best of our knowledge, very little work was concerned with the H− index in stochastic systems. In this paper, the H− index for stochastic linear continuous-time systems is discussed. The definition of the H− index is extended to the stochastic case. We present a necessary and sufficient condition of the H− index. A generalized differential equation is introduced and it is proved that its solvability and the feasibility of the H− index are equivalent. Comparing our results with the bounded real lemma [2, 9], it shows that the H− index is not completely dual to H∞ norm. The H− index discussed in this paper is only for tall or square systems. The reason for this is that H− index is zero for wide systems. But bounded real lemma for H∞ is applicable to any systems. Finally, the effectiveness of the given methods is illustrated by numerical example.
2
Mathematical Problems in Engineering
The outline of the paper is organized as follows. In Section 2, some efficient criteria are given for the H− index of stochastic linear systems in finite horizon. Section 3 contains an example provided to show the efficiency of the proposed results. Finally, we conclude this paper in Section 4. Notations. 𝑅 is the field of real numbers. 𝑅𝑚×𝑛 is the vector space of all 𝑚 × 𝑛 matrices with entries in 𝑅. S𝑛 (𝑅) is the set of all real symmetric matrices 𝑅𝑛×𝑛 . 𝐴 is the transpose of matrix 𝐴. 𝐴−1 is the inverse of 𝐴. Given positive semidefinite (positive definite) matrix 𝐴, we denote it by 𝐴 ≥ 0 (𝐴 > 0). 𝐸 is the mathematical expectation. 𝐼 is identity matrix. 0𝑛 is 𝑛×𝑛 zero matrix. L2F ([0, 𝑇], 𝑅𝑝 ) is the space of nonanticipative stochastic process 𝑦(𝑡) ∈ 𝑅𝑝 with respect to increasing 𝜎algebras F𝑡 (𝑡 ≥ 0) satisfying ‖𝑦(𝑡)‖2[0,𝑇] < ∞, where 𝑇
𝑇
‖𝑦(𝑡)‖2[0,𝑇] = 𝐸 ∫0 𝑦(𝑡) 𝑦(𝑡)𝑑𝑡 = 𝐸 ∫0 ‖𝑦(𝑡)‖2 𝑑𝑡. A square (wide or tall) system denotes a system when the number of inputs equals (is more than or less than) the outputs number.
Remark 2. If V is fault signal and 𝑧 is the residual, then the H− index describes the smallest fault sensitivity of system (1). In this paper, we suppose that system (1) is tall or square because the H− index is zero for wide system. Given 𝛾 > 0 and 0 < 𝑇 < ∞, let 𝛾
𝐽𝑇 (𝑥0 , V) = ‖𝑧 (𝑡)‖2[0,𝑇] − 𝛾2 ‖V (𝑡)‖2[0,𝑇] (3)
𝑇
2
= 𝐸 ∫ [𝑧 (𝑡) 𝑧 (𝑡) − 𝛾 V (𝑡) V (𝑡)] 𝑑𝑡. 0
We will study the following optimal problem: 𝛾
𝐽𝑇 (𝑥0 , V) .
min
In this section, we will discuss the H− index problem of stochastic linear continuous-time systems. We give a necessary and sufficient condition of the H− index larger than 𝛾 > 0 in finite horizon. Consider the following stochastic linear time-varying system G:
Remark 3. It can be shown that ‖G‖[0,𝑇] > 𝛾 is equivalent to − the following inequality 𝛾
𝐽𝑇 (0, V) = ‖𝑧 (𝑡)‖2[0,𝑇] − 𝛾2 ‖V (𝑡)‖2[0,𝑇]
𝑑𝑥 (𝑡) = [𝐴 (𝑡) 𝑥 (𝑡) + 𝐵 (𝑡) V (𝑡)] 𝑑𝑡
inf
0
∀V(𝑡) ∈ L2F ([0, 𝑇], 𝑅𝑝 ), V(𝑡) ≠ 0. Remark 4. When 𝑇 = ∞, (2) corresponds to the infinite horizon case. Lemma 5. Suppose 𝑃(𝑡) : [0, 𝑇] → S𝑛 (𝑅) is continuously differentiable, 𝑇 > 0. Then, for every 𝑥0 ∈ 𝑅𝑛 , V(𝑡) ∈ L2F ([0, 𝑇], 𝑅𝑝 ), 𝛾
𝐽𝑇 (𝑥0 , V) = 𝑥0 𝑃 (0) 𝑥0 − 𝐸 [𝑥 (𝑇) 𝑃 (𝑇) 𝑥 (𝑇)] 𝑇
𝑥 (𝑡)
0
V (𝑡)
+ 𝐸 ∫ [[
] M (𝑡, 𝑃 (𝑡)) [
𝑥 (𝑡) V (𝑡)
(6) ]] 𝑑𝑡,
‖𝑧 (𝑡)‖[0,𝑇]
V=0,𝑥 ̸ 0 =0 ‖V (𝑡)‖[0,𝑇]
V=0,𝑥 ̸ 0 =0
2
𝑥 (0) = 𝑥0 .
Definition 1. For stochastic system (1), given 0 < 𝑇 < ∞, its H− index in [0, 𝑇] is defined as
=
= 𝐸 ∫ [𝑧 (𝑡) 𝑧 (𝑡) − 𝛾 V (𝑡) V (𝑡)] 𝑑𝑡 > 0,
(1)
In the above, 𝜔(𝑡) is the one-dimensional standard Wiener process defined on the complete probability space (Ω, F, P), with the natural filter F𝑡 generated by 𝜔(𝑡) up to time 𝑡. Consider 𝑥(𝑡) ∈ 𝑅𝑛 , V(𝑡) ∈ L2F ([0, 𝑇], 𝑅𝑝 ), and 𝑧(𝑡) ∈ 𝑅𝑙 are the system state, control input, and regulated output, respectively. 𝐴(𝑡), 𝐵(𝑡), 𝐴 0 (𝑡), 𝐵0 (𝑡), 𝐶(𝑡), and 𝐷(𝑡) are coefficients with appropriate dimensions. For any 0 < 𝑇 < ∞ and (V(𝑡), 𝑥0 ) ∈ L2F ([0, 𝑇], 𝑅𝑝 ) × 𝑅𝑛 , there exists unique solution 𝑥(𝑡) = 𝑥(𝑡; V, 𝑥0 ) ∈ L2F ([0, 𝑇], 𝑅𝑛 ) of (1). The finite horizon stochastic H− index of system (1) can be stated as follows.
inf
(5)
𝑇
+ [𝐴 0 (𝑡) 𝑥 (𝑡) + 𝐵0 (𝑡) V (𝑡)] 𝑑𝜔 (𝑡) ,
= ‖G‖[0,𝑇] −
(4)
V∈L2F ([0,𝑇],𝑅𝑝 )
2. Finite Horizon Stochastic H− Index
𝑧 (𝑡) = 𝐶 (𝑡) 𝑥 (𝑡) + 𝐷 (𝑡) V (𝑡) ,
̇
𝑇
1/2
𝑇
1/2
{𝐸 ∫0 𝑧 (𝑡) 𝑧 (𝑡) 𝑑𝑡} {𝐸 ∫0 V (𝑡) V (𝑡) 𝑑𝑡}
where V(𝑡) ∈ L2F ([0, 𝑇], 𝑅𝑝 ).
(2) ,
𝑃(𝑡) 𝐾(𝑃(𝑡)) where M(𝑡, 𝑃(𝑡)) = [ 𝐿(𝑃(𝑡))+ ] ∈ S𝑛+𝑙 (𝑅), 𝐾(𝑃(𝑡)) 𝐻𝛾 (𝑃(𝑡))
𝐿 (𝑃 (𝑡)) = 𝑃 (𝑡) 𝐴 (𝑡) + 𝐴 (𝑡) 𝑃 (𝑡) + 𝐴 0 (𝑡) 𝑃 (𝑡) 𝐴 0 (𝑡) + 𝐶 (𝑡) 𝐶 (𝑡) ,
Mathematical Problems in Engineering
3
𝐾 (𝑃 (𝑡)) = 𝑃 (𝑡) 𝐵 (𝑡) + 𝐴 0 (𝑡) 𝑃 (𝑡) 𝐵0 (𝑡)
Ito’s formula to 𝑥(𝑡) 𝑃(𝑡)𝑥(𝑡) and taking expectations, we have that, for any 𝑇 > 0,
+ 𝐶 (𝑡) 𝐷 (𝑡) ,
𝐸 [𝑥 (𝑇) 𝑃 (𝑇) 𝑥 (𝑇)] − [𝑥 (0) 𝑃 (0) 𝑥 (0)] 𝑇
𝐻𝛾 (𝑃 (𝑡)) = 𝐵0 (𝑡) 𝑃 (𝑡) 𝐵0 (𝑡) + 𝐷 (𝑡) 𝐷 (𝑡) − 𝛾2 𝐼.
= 𝐸 ∫ 𝑑 [𝑥 (𝑡) 𝑃 (𝑡) 𝑥 (𝑡)] 𝑇
𝑥 (𝑡)
0
V (𝑡)
= 𝐸∫ [ 𝑛
L2F ([0, 𝑇], 𝑅𝑝 ),
and 𝑥(𝑡) = Proof. Let 𝑥0 ∈ 𝑅 , V(𝑡) ∈ 𝑥(𝑡; V, 𝑥0 ) denote the corresponding solution of (1). Applying
Q (𝑡, 𝑃 (𝑡)) = [ [
𝑥 (𝑡)
] Q (𝑡, 𝑃 (𝑡)) [
V (𝑡)
] 𝑑𝑡,
where
𝑃 (𝑡) 𝐴 (𝑡) + 𝐴 (𝑡) 𝑃 (𝑡) + 𝐴 0 (𝑡) 𝑃 (𝑡) 𝐴 0 (𝑡) + 𝑃̇ (𝑡) 𝑃 (𝑡) 𝐵 (𝑡) + 𝐴 0 (𝑡) 𝑃 (𝑡) 𝐵0 (𝑡) 𝐵 (𝑡) 𝑃 (𝑡) + 𝐵0 (𝑡) 𝑃 (𝑡) 𝐴 0 (𝑡)
𝐵0 (𝑡) 𝑃 (𝑡) 𝐵0 (𝑡)
].
(9)
]
Proof. By Lemma 5, for every V(𝑡) ∈ L2F ([0, 𝑇], 𝑅𝑝 ), V ≠ 0, 𝑥0 = 0, we conclude that
So 𝛾
(8)
0
(7)
𝑇
𝐽𝑇 (𝑥0 , V) = 𝐸 ∫ [‖𝐶𝑥 (𝑡) + 𝐷V (𝑡)‖2 − 𝛾2 ‖V (𝑡)‖2 ] 𝑑𝑡 0
+ 𝑥 (0) 𝑃 (0) 𝑥 (0)
𝛾 𝐽𝑇 (0, V)
− 𝐸 [𝑥 (𝑇) 𝑃 (𝑇) 𝑥 (𝑇)]
𝑥 (𝑡) ] 𝑑𝑡 ] Q (𝑡, 𝑃 (𝑡)) [ +𝐸∫ [ 0 V V (𝑡) (𝑡) ] ] [ [ 𝑇
𝑥 (𝑡)
(10)
(12)
𝑇
𝛾
𝑥 (𝑡)
𝑥 (𝑡)
By using completion of squares argument and the first equality in (11), we have
= 𝑥 (0) 𝑃 (0) 𝑥 (0) − 𝐸 [𝑥 (𝑇) 𝑃 (𝑇) 𝑥 (𝑇)] 𝑇
𝑥 (𝑡) ] 𝑑𝑡. ] M (𝑡, 𝑃𝑇 (𝑡)) [ = 𝐸∫ [ 0 V V (𝑡) (𝑡) [ ] [ ] 𝑇
𝐽𝑇 (0, V) = 𝐸 ∫ 𝑥 (𝑡) [𝐿 (𝑃𝑇 (𝑡)) + 𝑃𝑇̇ (𝑡) − 𝐾 (𝑃𝑇 (𝑡)) 0
𝑥 (𝑡)
] M (𝑡, 𝑃 (𝑡)) [ ] 𝑑𝑡, +𝐸∫ [ 0 V V (𝑡) (𝑡) ] ] [ [
−1
⋅ 𝐻𝛾 (𝑃𝑇 (𝑡)) 𝐾 (𝑃𝑇 (𝑡)) ] 𝑥 (𝑡) 𝑑𝑡 𝑇
which ends the proof.
−1
+ 𝐸 ∫ {[V (𝑡) + 𝐻𝛾 (𝑃𝑇 (𝑡)) 𝐾 (𝑃𝑇 (𝑡)) 𝑥 (𝑡)]
0
Below, we prove the following theorem which is necessary in this paper.
⋅ 𝐻𝛾 (𝑃𝑇 (𝑡))
Theorem 6. For (1) and some given 𝛾 > 0, if the following differential Riccati equation
⋅ [V (𝑡) + 𝐻𝛾 (𝑃𝑇 (𝑡)) 𝐾 (𝑃𝑇 (𝑡)) 𝑥 (𝑡)]} 𝑑𝑡
−1
𝑇
= 𝐸 ∫ {[V (𝑡) − V∗ (𝑡)] 𝐻𝛾 (𝑃𝑇 (𝑡))
𝐿 (𝑃 (𝑡)) + 𝑃̇ (𝑡) = 𝐾 (𝑃 (𝑡)) 𝐻𝛾 (𝑃 (𝑡))−1 𝐾 (𝑃 (𝑡)) , 𝐻𝛾 (𝑃 (𝑡)) > 0,
(13)
0
(11)
⋅ [V (𝑡) − V∗ (𝑡)]} 𝑑𝑡,
𝑃 (𝑇) = 0 > 𝛾. admits solution 𝑃𝑇 (𝑡) on [0, 𝑇], then ‖G‖[0,𝑇] −
where V∗ (𝑡) = −𝐻𝛾 (𝑃𝑇(𝑡))−1 𝐾(𝑃𝑇 (𝑡)) 𝑥(𝑡).
4
Mathematical Problems in Engineering 𝛾
𝛾
From 𝐻𝛾 (𝑃𝑇 (𝑡)) > 0, 𝐽𝑇 (0, V) ≥ 0, to show 𝐽𝑇 (0, V) > 0, we define the operator L: LV(𝑡) = V(𝑡) − V∗ (𝑡) with its realization: 𝑑𝑥 (𝑡) = (𝐴 (𝑡) 𝑥 (𝑡) + 𝐵 (𝑡) V (𝑡)) 𝑑𝑡
It is easy to see that (17) satisfies the following equation: 𝛾 𝑃𝐹̇ (𝑡) + [
= 0,
+ [𝐴 0 (𝑡) 𝑥 (𝑡) + 𝐵0 (𝑡) V (𝑡)] 𝑑𝜔 (𝑡) , 𝑥 (0) = 0, −1
(14)
𝛾
Lemma 7. Suppose 𝐹(𝑡) ∈ 𝐶[0, 𝑇] and 𝑃𝐹 (𝑡) is the solution of (18). Then if V(𝑡) ∈ L2F ([0, 𝑇], 𝑅𝑝 ), one obtains 𝛾
𝛾
𝐽𝑇 (𝑥0 , V + 𝐹𝑥𝐹 ) = 𝑥0 𝑃𝐹 (0) 𝑥0
−1
𝑑𝑥 (𝑡) = [𝐴 (𝑡) − 𝐵 (𝑡) 𝐻𝛾 (𝑃𝑇 (𝑡)) (𝐵 (𝑡) 𝑃𝑇 (𝑡)
𝑇
+ 𝐸 ∫ [V (𝑡) 𝐺 (𝑡) 𝑥𝐹 (𝑡) + 𝑥𝐹 (𝑡) 𝐺 (𝑡) V (𝑡) 0
+ 𝐵0 (𝑡) 𝑃𝑇 (𝑡) 𝐴 0 (𝑡) + 𝐷 (𝑡) 𝐶 (𝑡))] 𝑥 (𝑡) 𝑑𝑡
𝛾
+ [𝐴 0 (𝑡) − 𝐵0 (𝑡) 𝐻𝛾 (𝑃𝑇 (𝑡)) (𝐵 (𝑡) 𝑃𝑇 (𝑡)
(15)
where 𝑥𝐹 (𝑡) = 𝑥(𝑡, 𝐹(𝑡)𝑥𝐹 (𝑡) + V(𝑡), 𝑥0 ) is the solution of
(𝑡) 𝑃𝑇 (𝑡) 𝐴 0 (𝑡) + 𝐷 (𝑡) 𝐶 (𝑡))] 𝑥 (𝑡) 𝑑𝜔 (𝑡)
𝑑𝑥𝐹 (𝑡) = (𝐴 (𝑡) + 𝐵 (𝑡) 𝐹 (𝑡)) 𝑥𝐹 (𝑡) 𝑑𝑡
∗
+ 𝐵 (𝑡) (V (𝑡) − V (𝑡)) 𝑑𝑡 + 𝐵0 (𝑡) (V (𝑡) − V (𝑡)) 𝑑𝜔 (𝑡) ,
+ (𝐴 0 (𝑡) + 𝐵0 (𝑡) 𝐹 (𝑡)) 𝑥𝐹 (𝑡) 𝑑𝑤 (𝑡)
𝑥0 = 0,
(20)
+ 𝐵0 (𝑡) V (𝑡) 𝑑𝑤 (𝑡) + 𝐵 (𝑡) V (𝑡) 𝑑𝑡
where V(𝑡) = −𝐻𝛾 (𝑃𝑇(𝑡))−1 𝐾(𝑃𝑇 (𝑡)) 𝑥(𝑡) + (V(𝑡) − V∗ (𝑡)). We assume that 𝐻𝛾 (𝑃𝑇 (𝑡)) ≥ 𝜖𝐼, 𝜖 > 0, so there exists constant 𝐶0 > 0, such that 𝛾
𝐽𝑇 (0, V)
with 𝑥𝐹 (0) = 𝑥0 and
𝛾
𝛾
𝐺 (𝑡) = 𝐾 (𝑃𝐹 (𝑡)) + 𝐻𝛾 (𝑃𝐹 (𝑡)) 𝐹 (𝑡) .
(21)
As V = 0, then 𝑇
∗
𝛾
𝛾
0
2 ≥ 𝜖 V (𝑡) − V∗ (𝑡)[0,𝑇] = 𝜖 ‖LV (𝑡)‖2[0,𝑇]
𝛾
𝐽𝑇 (𝑥0 , 𝐹𝑥𝐹 ) = 𝑥0 𝑃𝐹 (0) 𝑥0 .
∗
= 𝐸 ∫ [V (𝑡) − V (𝑡)] 𝐻 (𝑃𝑇 (𝑡)) [V (𝑡) − V (𝑡)] 𝑑𝑡 (16)
𝐶0 ‖V (𝑡)‖2[0,𝑇]
𝛾
𝛾
𝐽𝑇 (𝑥0 , V + 𝐹𝑥𝐹 ) = 𝑥0 𝑃𝐹 (0) 𝑥0
> 0.
𝑇 𝑥𝐹 (𝑡) 𝛾 + 𝐸 ∫ {[ ] M (𝑡, 𝑃𝐹 (𝑡)) 𝐹 (𝑡) 𝑥𝐹 (𝑡) + V (𝑡) 0
> 𝛾. That is, ‖G‖[0,𝑇] − Now, we consider the following equation:
𝑥𝐹 (𝑡) 𝛾 ⋅[ ]} 𝑑𝑡 = 𝑥0 𝑃𝐹 (0) 𝑥0 𝐹 (𝑡) 𝑥𝐹 (𝑡) + V (𝑡)
𝑋̇ (𝑡) + 𝐿 (𝑋 (𝑡)) + 𝐾 (𝑋 (𝑡)) 𝐹 (𝑡) + 𝐹 (𝑡) 𝐾 (𝑋 (𝑡)) + 𝐹 (𝑡) 𝐻𝛾 (𝑋 (𝑡)) 𝐹 (𝑡) = 0,
(22)
Proof. In terms of Lemma 5 with 𝑃(𝑡) = 𝑃𝐹 (𝑡) and 𝐹(𝑡)𝑥𝐹 (𝑡)+ V(𝑡) for V(𝑡), 𝛾
≥
(19)
+ V (𝑡) 𝐻𝛾 (𝑃𝐹 (𝑡)) V (𝑡)] 𝑑𝑡,
−1
∗
(18)
𝑡 ∈ [0, 𝑇] ,
𝑃𝐹 (𝑇) = 0.
Then L−1 exists, which is determined by
+ 𝐵0
𝛾
𝛾
V (𝑡) − V∗ (𝑡) = V (𝑡) + 𝐻𝛾 (𝑃𝑇 (𝑡)) 𝐾 (𝑃𝑇 (𝑡)) 𝑥 (𝑡) .
𝛾
𝐿 (𝑃𝐹 (𝑡)) 𝐾 (𝑃𝐹 (𝑡)) 𝐼 ] [ ] ][ 𝛾 𝛾 𝛾 𝐹 (𝑡) 𝐾 (𝑃𝐹 (𝑡)) 𝐻 (𝑃𝐹 (𝑡)) 𝐹 (𝑡)
𝐼
𝑡 ∈ [0, 𝑇] ,
(17)
𝑋 (𝑇) = 0,
𝑇
]
𝐹 (𝑡)
0
𝛾
where 𝐹(𝑡) ∈ 𝐶[0, 𝑇] and this equation has unique solution 𝛾 𝑋(𝑡) = 𝑃𝐹 (𝑡), 𝑡 ∈ [0, 𝑇].
𝐼
+ 𝐸 ∫ {𝑥𝐹 (𝑡) [ 𝛾
𝛾
𝐿 (𝑃𝐹 (𝑡)) + 𝑃𝐹̇ (𝑡) 𝐾 (𝑃𝐹 (𝑡)) 𝐼 ⋅[ ][ ] 𝑥𝐹 (𝑡)} 𝑑𝑡 𝛾 𝛾 𝐾 (𝑃𝐹 (𝑡)) 𝐻𝛾 (𝑃𝐹 (𝑡)) 𝐹 (𝑡)
Mathematical Problems in Engineering
5
𝑇
that 𝐻𝜆 (𝑃𝐹𝜆 (𝑡)) ≥ 0. For any 𝑡0 ∈ [0, 𝑇), set 𝐹𝑡0 = 𝐹(𝑡 + 𝑡0 ), 𝑡 ∈ [0, 𝑇−𝑡0 ]. Let 𝑃𝐹𝜆𝑡 (𝑡) be the solution of (18) with 𝛾 replaced
+ 𝐸 ∫ {V (𝑡) 𝐺 (𝑡) 𝑥𝐹 (𝑡) + 𝑥𝐹 (𝑡) 𝐺 (𝑡) V (𝑡) 0
0
+ V (𝑡) 𝐻
𝛾
𝛾 (𝑃𝐹
(𝑡)) V (𝑡)} 𝑑𝑡 =
𝛾 𝑥0 𝑃𝐹
by 𝜆 and 𝐹 replaced by 𝐹𝑡0 on [0, 𝑇 − 𝑡0 ]. Then, 𝑃𝐹𝜆𝑡 (𝑡) =
(0) 𝑥0
0
𝑃𝐹𝜆 (𝑡 + 𝑡0 ), 𝑡 ∈ [0, 𝑇 − 𝑡0 ]. By (22), for any 𝑡0 ∈ [0, 𝑇), 𝑥0 ∈ 𝑅𝑛 ,
𝑇
+ 𝐸 ∫ {V (𝑡) 𝐺 (𝑡) 𝑥𝐹 (𝑡) + 𝑥𝐹 (𝑡) 𝐺 (𝑡) V (𝑡)
𝜆 𝑥0 𝑃𝐹𝜆 (𝑡0 ) 𝑥0 = 𝑥0 𝑃𝐹𝜆𝑡 (0) 𝑥0 = 𝐽𝑇−𝑡 (𝑥0 , 𝐹𝑡0 𝑥𝐹𝑡 ) 0
0
0
0
(28)
𝛾
+ V (𝑡) 𝐻𝛾 (𝑃𝐹 (𝑡)) V (𝑡)} 𝑑𝑡.
𝛾
0
(23) This means that (19) holds. Let V = 0 in (19); we obtain (22). 𝛾 Now we are in a position to prove that 𝐻𝛾 (𝑃𝐹 (𝑡)) is invertible for 𝑡 ∈ [0, 𝑇]. Lemma 8. For system (1), if ‖G‖[0,𝑇] > 𝛾 for some given 𝛾 > 0, − 𝛾 𝐹(𝑡) ∈ 𝐶[0, 𝑇], 𝑇 > 0, and 𝑃𝐹 (𝑡) satisfies (18). Then, 2
𝛾
𝐻𝛾 (𝑃𝐹 (𝑡)) ≥ [(‖G‖[0,𝑇] ) − 𝛾2 ] 𝐼 > 0, −
𝛾
and so 𝐻𝜆 (𝑃𝐹 (𝑡0 )) ≥ 𝐻𝜆 (𝑃𝐹𝜆 (𝑡0 )) ≥ 0. By continuity, 𝛾 𝐻𝛾 (𝑃𝐹 (𝑡)) ≥ 𝜌2 𝐼 for all 𝑡 ∈ [0, 𝑇]. As this holds for arbitrary 𝛾 2 𝜌 < (‖G‖[0,𝑇] )2 −𝛾2 , it follows that 𝐻𝛾 (𝑃𝐹 (𝑡)) ≥ [(‖G‖[0,𝑇] )2 − − − 𝛾2 ]𝐼 > 0. This completes the proof. 𝛾
Remark 9. When 𝑡 = 𝑇, (24) becomes 𝐻𝛾 (𝑃𝐹 (𝑇)) = 𝐷(𝑇) 𝐷(𝑇) − 𝛾2 𝐼 > 0. If system (1) is time-invariant, then
𝑡 ∈ [0, 𝑇] . (24)
𝛾
Proof. Let us first prove that 𝐻𝛾 (𝑃𝐹 (𝑡)) ≥ 0. Suppose this is false; then there exists 𝑡∗ ∈ [0, 𝑇), 𝑢 ∈ 𝑅𝑙 , ‖𝑢‖ = 1 such that 𝛾 𝑢 𝐻𝛾 (𝑃𝐹 (𝑡∗ ))𝑢 ≤ −𝜂 for some 𝜂 > 0. Then, for sufficiently small 𝛿 > 0, 𝜂 𝛾 𝑢 𝐻𝛾 (𝑃𝐹 (𝑡)) 𝑢 ≤ − , 2
𝛾
≤ 𝐽𝑇−𝑡0 (𝑥0 , 𝐹𝑡0 𝑥𝐹𝑡 ) = 𝑥0 𝑃𝐹 (𝑡0 ) 𝑥0 ,
𝑡 ∈ [𝑡∗ , 𝑡∗ + 𝛿] ⊂ [0, 𝑇] .
(25)
𝐷 𝐷 − 𝛾2 𝐼 > 0.
(29)
Remark 10. By the equality 𝐴(𝐼 − 𝐵𝐴)−1 = (𝐼 − 𝐴𝐵)−1 𝐴, we have that 𝐶 [𝐼 − 𝐷(𝐷 𝐷 − 𝛾2 𝐼)−1 𝐷 ]𝐶 = 𝐶 (𝐼 − 𝛾−2 𝐷𝐷 )−1 𝐶. If system (1) is time-invariant and square, by (29), −1
𝐶 [𝐼 − 𝐷 (𝐷 𝐷 − 𝛾2 𝐼) 𝐷 ] 𝐶 (30)
Define
−1
−2
= 𝐶 (𝐼 − 𝛾 𝐷𝐷 ) 𝐶 ≤ 0. ∗ ∗ {0, 𝑡 ∈ [0, 𝑡 ) ∪ (𝑡 + 𝛿, 𝑇] , V (𝑡) = { 𝑢, 𝑡 ∈ [𝑡∗ , 𝑡∗ + 𝛿] . {
(26)
Now, we present the following theorem which is important in this paper.
Using Lemma 7 with this V(𝑡) and 𝑥0 = 0, we can derive that 𝑥𝐹 (𝑡) = 0 for 𝑡 ∈ [0, 𝑡∗ ] and
Theorem 11. Suppose system (1) is time-invariant and square > 𝛾 for given 𝛾 ≤ 0. Then (11) has a and satisfies ‖G‖[0,𝑇] − unique solution 𝑃𝑇 (𝑡) ≤ 0 on [0, 𝑇] for every 𝑇 > 0. Moreover, 𝛾 𝐽𝑇 (𝑥0 , V) is minimized by the feedback control:
𝑇
𝛾 2 𝐽𝑇 (0, V) = 𝐸 ∫ [𝐶 (𝑡) 𝑥𝐹 (𝑡) + 𝐷 (𝑡) V (𝑡) 0
V∗ (𝑡) = 𝐹𝑇 (𝑡) 𝑥𝐹𝑇 (𝑡) ,
𝑇
− 𝛾2 ‖V (𝑡)‖2 ] 𝑑𝑡 = 𝐸 ∫ [V (𝑡) 𝐺 (𝑡) 𝑥𝐹 (𝑡) 0
𝛾
(27) 𝛾
+ 𝑥𝐹 (𝑡) 𝐺 (𝑡) V (𝑡) + V (𝑡) 𝐻𝛾 (𝑃𝐹 (𝑡)) V (𝑡)] 𝑑𝑡 𝑡∗ +𝛿
≤ 𝐸∫
𝑡∗
𝜂 (2 𝐺 (𝑡) 𝑢 𝑥𝐹 (𝑡) − ) 𝑑𝑡. 2
Since 𝑥𝐹 (𝑡) is continuous and 𝑥𝐹 (𝑡∗ ) = 0, (27) is negative. 𝛾 Moreover, the condition ‖G‖[0,𝑇] > 𝛾 implies 𝐽𝑇 (0, V) ≥ 0. − ∗ As a result, this is a contradiction. If 𝑡 = 𝑇, we can replace [𝑡∗ , 𝑡∗ + 𝛿] by [𝑇 − 𝛿, 𝑇] and use a similar proof. Next, let ‖G‖[0,𝑇] > (𝛾2 +𝜌2 )1/2 for any 𝜌 > 0 and 𝜆 = (𝛾2 + − 2 1/2 𝜌 ) . Replacing 𝛾 with 𝜆 in (18), we obtain the corresponding solution 𝑃𝐹𝜆 (𝑡). Applying the previous step, we can deduce
−1
(31)
𝐹𝑇 (𝑡) = − 𝐻 (𝑃𝑇 (𝑡)) 𝐾 (𝑃𝑇 (𝑡)) , where 𝑥𝐹𝑇 (𝑡) satisfies 𝑑𝑥𝐹𝑇 (𝑡) = (𝐴 + 𝐵𝐹𝑇 (𝑡)) 𝑥𝐹𝑇 (𝑡) 𝑑𝑡 + [𝐴 0 + 𝐵0 𝐹𝑇 (𝑡)] 𝑥𝐹𝑇 (𝑡) 𝑑𝑤 (𝑡) ,
(32)
𝑥𝐹𝑇 (0) = 𝑥0 and the optimal cost is min
𝛾
𝐽𝑇 (𝑥0 , V) = 𝑥0 𝑃𝑇 (0) 𝑥0 .
V∈L2F ([0,𝑇],𝑅𝑝 )
(33)
6
Mathematical Problems in Engineering
Proof. We prove that ‖G‖[0,𝑇] > 𝛾 implies the existence − of solution 𝑃𝑇(𝑡) of (11) on [0, 𝑇]. Using a contradiction argument, we suppose that (11) does not admit a solution. By the standard theory of differential equations, there exists unique solution 𝑃𝑇 (𝑡) backward in time on maximal interval [𝑇0 , 𝑇] (𝑇0 ≥ 0), and as 𝑡 → 𝑇0 , 𝑃𝑇 (𝑡) becomes unbounded. Let 0 < 𝛿 < 𝑇 − 𝑇0 , 𝑥(𝑇0 + 𝛿) = 𝑥𝑇0 ,𝛿 ∈ 𝑅𝑛 , 𝑄(𝑡) = 𝐵 𝑃𝑇 (𝑡) + 𝐵0 𝑃𝑇 (𝑡)𝐴 0 + 𝐷 𝐶, 𝑅(𝑡) = 𝐵0 𝑃𝑇(𝑡)𝐵0 + 𝐷 𝐷 − 𝛾2 𝐼, by completing the squares; then 𝐽𝛾 (𝑥, V, 𝑥𝑇0 ,𝛿 , 𝑇0 + 𝛿) = 𝐸 ∫
𝑇
[𝑧 (𝑡) 𝑧 (𝑡)
𝑇0 +𝛿
− 𝛾2 V (𝑡) V (𝑡)] 𝑑𝑡 = 𝐸 ∫
𝑇
𝑇0 +𝛿
2
− 𝛾 V (𝑡) V (𝑡)] 𝑑𝑡 + 𝐸 ∫
𝑇
𝑇0 +𝛿
𝑇0 +𝛿
− 𝛾2 V (𝑡) V (𝑡)} 𝑑𝑡 = 𝐸 ∫
𝑇0 +𝛿
−1
− 𝐷 (𝐷 𝐷 − 𝛾2 𝐼) 𝐷 ] 𝐶𝑥 (𝑡) 𝑑𝑡 + 𝐸 ∫
−1
𝑇
𝑇0 +𝛿
𝑇0 +𝛿
−1
𝑇
𝑇0 +𝛿
𝑥 (𝑡)
(34)
[V (𝑡)
where ̃V(𝑡) = −(𝐷 𝐷 − 𝛾2 𝐼)−1 𝐷 𝐶𝑥(𝑡). Considering (35), we get 𝑥𝑇 0 ,𝛿 𝑃𝑇 (𝑇0 + 𝛿) 𝑥𝑇0 ,𝛿 ≤ 𝐽𝛾 (𝑥, ̃V, 𝑥𝑇0 ,𝛿 , 𝑇0 + 𝛿) ≤ 0,
(38)
which implies that 𝑃𝑇 (𝑇0 + 𝛿) ≤ 0.
𝑄 (𝑡) 𝑥 (𝑡)] 𝑑𝑡 + 𝑥𝑇 0 ,𝛿 𝑃𝑇 (𝑇0
(39)
By linearity, the solution of (1) with initial state 𝑥𝑇0 ,𝛿 satisfies + 𝛿) 𝑥𝑇0 ,𝛿
𝑥 (𝑡, V, 𝑥𝑇0 ,𝛿 , 𝑇0 + 𝛿) = 𝑥 (𝑡, 0, 𝑥𝑇0 ,𝛿 , 𝑇0 + 𝛿)
[V (𝑡) + 𝑅 (𝑡)−1 𝑄 (𝑡) 𝑥 (𝑡)] 𝑅 (𝑡) [V (𝑡)
(40)
+ 𝑥 (𝑡, V, 0, 𝑇0 + 𝛿) . Suppose Φ(𝑡) is the solution of
+ 𝑅 (𝑡)−1 𝑄 (𝑡) 𝑥 (𝑡)] 𝑑𝑡 + 𝑥𝑇 0 ,𝛿 𝑃𝑇 (𝑇0 + 𝛿) 𝑥𝑇0 ,𝛿 .
𝐶 𝐶 + 𝐴 Φ (𝑡) + Φ (𝑡) 𝐴 + 𝐴0 Φ (𝑡) 𝐴 0 + Φ̇ (𝑡) = 0,
Obviously, min
(37)
⋅ 𝐶 [𝐼 − 𝐷 (𝐷 𝐷 − 𝛾2 𝐼) 𝐷 ] 𝐶𝑥 (𝑡) 𝑑𝑡 ≤ 0,
𝑇0 +𝛿
[V (𝑡)
+ (𝐷 𝐷 − 𝛾2 𝐼) 𝐷 𝐶𝑥 (𝑡)] (𝐷 𝐷 − 𝛾2 𝐼) [V (𝑡)
𝐽𝛾 (𝑥, ̃V, 𝑥𝑇0 ,𝛿 , 𝑇0 + 𝛿) = 𝐸 ∫
+ 𝑅 (𝑡)−1 𝑄 (𝑡) 𝑥 (𝑡)] 𝑅 (𝑡) [V (𝑡)
𝑇
𝑇
From Remark 10,
𝑥 (𝑡) [𝐶 𝐶
− 𝑄 (𝑡) 𝑅 (𝑡) 𝑄 (𝑡)] 𝑥 (𝑡) 𝑑𝑡 + 𝐸 ∫
= 𝐸∫
𝑥 (𝑡) 𝐶 [𝐼
−1
+ 𝐴 𝑃𝑇 (𝑡) + 𝑃𝑇 (𝑡) 𝐴 + 𝐴0 𝑃𝑇 (𝑡) 𝐴 0 + 𝑃𝑇̇ (𝑡)
+ 𝑅 (𝑡)
𝑇
(36)
𝑑 (𝑥 (𝑡) 𝑃𝑇 (𝑡) 𝑥 (𝑡)) 𝑑𝑡
𝑇0 +𝛿
−1
{[𝐶𝑥 (𝑡) + 𝐷V (𝑡)] [𝐶𝑥 (𝑡) + 𝐷V (𝑡)]
−1
𝑇
𝑇
+ (𝐷 𝐷 − 𝛾2 𝐼) 𝐷 𝐶𝑥 (𝑡)] 𝑑𝑡.
[𝑧 (𝑡) 𝑧 (𝑡)
+ 𝑥𝑇 0 ,𝛿 𝑃𝑇 (𝑇0 + 𝛿) 𝑥𝑇0 ,𝛿 = 𝐸 ∫
= 𝐸∫
(41)
Φ (𝑇) = 0, and we have
𝐽𝛾 (𝑥, V, 𝑥𝑇0 ,𝛿 , 𝑇0 + 𝛿)
V∈L2F ([𝑇0 +𝛿,𝑇],𝑅𝑝 )
(35)
= 𝐽𝛾 (𝑥, V∗ , 𝑥𝑇0 ,𝛿 , 𝑇0 + 𝛿) = 𝑥𝑇 0 ,𝛿 𝑃𝑇 (𝑇0 + 𝛿) 𝑥𝑇0 ,𝛿 ,
𝑇
𝑇0 +𝛿
− 𝛾2 V (𝑡) V (𝑡)] 𝑑𝑡
𝑇
= 𝑥𝑇 0 ,𝛿 Φ (𝑇0 + 𝛿) 𝑥𝑇0 ,𝛿 + 𝐸 ∫
𝑇0 +𝛿
where V∗ (𝑡) = −𝑅(𝑡)−1 𝑄(𝑡)𝑥(𝑡). Furthermore, we can see that 𝐽𝛾 (𝑥, V, 𝑥𝑇0 ,𝛿 , 𝑇0 + 𝛿) = 𝐸 ∫
𝐽𝛾 (𝑥, V, 𝑥𝑇0 ,𝛿 , 𝑇0 + 𝛿) − 𝐽𝛾 (𝑥, V, 0, 𝑇0 + 𝛿) [V (𝑡)
⋅ (𝐵0 Φ (𝑡) + 𝐷 𝐶) 𝑥 (𝑡, 0, 𝑥𝑇0 ,𝛿 , 𝑇0 + 𝛿)] 𝑑𝑡 [𝑧 (𝑡) 𝑧 (𝑡)
𝑇
+𝐸∫
𝑇0 +𝛿
(42)
[𝑥 (𝑡, 0, 𝑥𝑇0 ,𝛿 , 𝑇0 + 𝛿) (𝐵0 Φ (𝑡) + 𝐷 𝐶)
⋅ V (𝑡)] 𝑑𝑡.
Mathematical Problems in Engineering
7
Take ‖G‖[0,𝑇] ≥ (𝛾2 𝜖2 )1/2 , −
𝑇
−1 𝜖 (𝐵0 Φ (𝑡) + 𝐷 𝐶) 𝑥 (𝑡, 0, 𝑥𝑇0 ,𝛿 , 𝑇0 𝑇 +𝛿
−𝐸∫
0
V (𝑡)
2 + 𝛿) 𝑑𝑡 ≥ 𝑥𝑇 0 ,𝛿 Φ (𝑇0 + 𝛿) 𝑥𝑇0 ,𝛿
−1 {̃V (𝑡) = − (𝐷 𝐷 − 𝛾2 𝐼) 𝐷 𝐶𝑥 (𝑡) , 𝑡 ∈ [0, 𝑇0 + 𝛿] (43) ={ 𝑡 ∈ (𝑇0 + 𝛿, 𝑇] , {V (𝑡) ,
−𝐸∫
𝑇0 +𝛿
(45)
𝑇
𝛾
2
2
2
𝐽 (𝑥, V, 0, 𝑇0 + 𝛿) = 𝐸 ∫ (‖𝑧 (𝑡)‖ − 𝛾 ‖V (𝑡)‖ ) 𝑑𝑡
It is obvious that there exists constant 𝐶0 > 0 such that
0
−𝐸∫
0
2
2
𝑇 2 𝐶2 𝑥𝑇0 ,𝛿 ≥ 𝐸 ∫
2 𝑥 (𝑡, 0, 𝑥𝑇0 ,𝛿 , 𝑇0 + 𝛿) 𝑑𝑡. 𝑇 +𝛿
2
(‖𝑧 (𝑡)‖ − 𝛾 ‖̃V (𝑡)‖ ) 𝑑𝑡
𝑇
2
2
2
0
𝑇0 +𝛿
So, there is constant 𝐶1 > 0 such that
𝑥 (𝑡)
0
0
−1
0
(44)
−1
[̃V (𝑡) + (𝐷 𝐷 − 𝛾2 𝐼) 𝐷 𝐶𝑥 (𝑡)]
𝑥𝑇 0 ,𝛿 Φ (𝑇0 + 𝛿) 𝑥𝑇0 ,𝛿
⋅ (𝐷 𝐷 − 𝛾 𝐼) 2
−1
= −𝐸∫
⋅ [̃V (𝑡) + (𝐷 𝐷 − 𝛾 𝐼) 𝐷 𝐶𝑥 (𝑡)] 𝑑𝑡 𝑇
= 𝐸∫
𝑑 (𝑥 (𝑡) Φ (𝑡) 𝑥 (𝑡))
(48)
2 𝑥 (𝑡, 0, 𝑥𝑇0 ,𝛿 , 𝑇0 + 𝛿) 𝐶 𝑑𝑡 ≥ 0.
From (45), we have
≥ 𝜖2 ‖V (𝑡)‖2[𝑇 +𝛿,𝑇] .
2 𝐽𝛾 (𝑥, V, 𝑥𝑇0 ,𝛿 , 𝑇0 + 𝛿) ≥ − 𝐶1 𝑥𝑇0 ,𝛿 .
0
It follows that
(49)
In view of (35) and (39), it yields 𝑇
𝐽𝛾 (𝑥, V, 𝑥𝑇0 ,𝛿 , 𝑇0 + 𝛿) ≥ 𝐸 ∫
𝑇0 +𝛿
− 𝐶1 ≤ 𝑃𝑇 (𝑇0 + 𝛿) ≤ 0.
𝜖2 ‖V (𝑡)‖2 𝑑𝑡
𝑇
+ 𝑥𝑇 0 ,𝛿 Φ (𝑇0 + 𝛿) 𝑥𝑇0 ,𝛿 + 𝐸 ∫
𝑇0 +𝛿
[V (𝑡) (𝐵0 Φ (𝑡)
𝑃𝑇̇ (𝑡) + 𝐿 (𝑃𝑇 (𝑡)) + 𝐾 (𝑃𝑇 (𝑡)) 𝐹 (𝑡)
[𝑥 (𝑡, 0, 𝑥𝑇0 ,𝛿 , 𝑇0 + 𝛿) (𝐵0 Φ (𝑡) + 𝐷 𝐶)
⋅ V (𝑡)] 𝑑𝑡 =
𝑥𝑇 0 ,𝛿 Φ (𝑇0
(50)
So, 𝑃𝑇 (𝑇0 + 𝛿) can not become unbounded as 𝛿 → 0, which means that (11) has unique solution 𝑃𝑇 (𝑡) on [0, 𝑇]. Setting 𝐹(𝑡) = 𝐹𝑇(𝑡), 𝑡 ∈ [0, 𝑇], in (17), from (31), we obtain
+ 𝐷 𝐶) 𝑥 (𝑡, 0, 𝑥𝑇0 ,𝛿 , 𝑇0 + 𝛿)] 𝑑𝑡
𝑇0 +𝛿
𝑇
𝑇0 +𝛿
0
+𝐸∫
𝑇
𝑇0 +𝛿
≥ 𝐸 ∫ (‖𝑧 (𝑡)‖2 − 𝛾2 ‖V (𝑡)‖2 ) 𝑑𝑡 ≥ 𝜖2 ‖V (𝑡)‖2[0,𝑇]
𝑇
(47)
2 ≤ 𝐶1 𝑥𝑇0 ,𝛿 .
In addition,
2
𝑇
−1 2 𝜖 (𝐵0 Φ (𝑡) + 𝐷 𝐶) 𝑥 (𝑡, 0, 𝑥𝑇0 ,𝛿 , 𝑇0 + 𝛿) 𝑑𝑡 𝑇 +𝛿
𝐸∫
⋅ 𝐶 [𝐼 − 𝐷 (𝐷 𝐷 − 𝛾2 𝐼) 𝐷 ] 𝐶𝑥 (𝑡) 𝑑𝑡 𝑇0 +𝛿
(46)
0
= 𝐸 ∫ (‖𝑧 (𝑡)‖ − 𝛾 ‖V (𝑡)‖ ) 𝑑𝑡 − 𝐸 ∫
−𝐸∫
−1 𝜖 (𝐵0 Φ (𝑡) + 𝐷 𝐶) 𝑥 (𝑡, 0, 𝑥𝑇0 ,𝛿 , 𝑇0
2 + 𝛿) 𝑑𝑡.
and it is easy to show that
𝑇0 +𝛿
𝑇
+ 𝐹 (𝑡) 𝐾 (𝑃𝑇 (𝑡)) + 𝐹 (𝑡) 𝐻𝛾 (𝑃𝑇 (𝑡)) 𝐹 (𝑡) = 0.
Hence 𝑃𝑇 (𝑡) satisfies (17), or equivalently (18). So 𝛾
𝑃𝐹𝑇 (𝑡) = 𝑃𝑇 (𝑡) ,
𝑇
𝜖 [V + 𝛿) 𝑥𝑇0 ,𝛿 + 𝐸 ∫ 𝑇 +𝛿 0
2 − 𝜖−2 (𝐵0 Φ (𝑡) + 𝐷 𝐶) 𝑥 (𝑡, 0, 𝑥𝑇0 ,𝛿 , 𝑇0 + 𝛿)] 𝑑𝑡
(51)
𝑡 ∈ [0, 𝑇] .
(52)
By (31),
𝐺 (𝑡) = 𝐾 (𝑃𝑇 (𝑡)) + 𝐻𝛾 (𝑃𝑇 (𝑡)) 𝐹𝑇 (𝑡) = 0,
(53)
8
Mathematical Problems in Engineering > 𝛾, then ‖G‖[0,𝑇] > 𝛾. Using Theorem 12 to When ‖G‖[0,𝑇] − 𝛿− the modified data, it is easy to see that the following equation
and, in terms of Lemma 7, 𝛾
𝐽𝑇 (𝑥0 , V + 𝐹𝑇 𝑥) =
𝑇
𝑥0 𝑃𝑇 (0) 𝑥0
(54)
𝛾
+ 𝐸 ∫ [V (𝑡) 𝐻 (𝑃𝑇 (𝑡)) V (𝑡)] 𝑑𝑡.
𝑃 (𝑡) 𝐴 + 𝐴 𝑃 (𝑡) + 𝐴0 𝑃 (𝑡) 𝐴 0 + 𝐶 𝐶 + 𝛿2 𝐼 + 𝑃̇ (𝑡)
0
= (𝑃 (𝑡) 𝐵 + 𝐴0 𝑃 (𝑡) 𝐵0 + 𝐶 𝐷)
But by Lemma 8,
−1
2
𝛾
𝐻𝛾 (𝑃𝑇 (𝑡)) = 𝐻𝛾 (𝑃𝐹𝑇 (𝑡)) ⪰ [(‖G‖[0,𝑇] ) − 𝛾2 ] 𝐼 − ≻ 0,
⋅ (𝐵0 𝑃 (𝑡) 𝐵0 + 𝐷 𝐷 − 𝛾2 𝐼) (55)
𝑡 ∈ [0, 𝑇] .
∗
Hence, V (𝑡) = 𝐹𝑇 (𝑡)𝑥(𝑡) minimizes 𝛾 minV∈L2F ([0,𝑇],𝑅𝑝 ) 𝐽𝑇 (𝑥0 , V) = 𝑥0 𝑃𝑇(0)𝑥0 .
𝛾 𝐽𝑇 (𝑥0 , V)
and
Theorem 12. If system (1) is time-invariant and square, for given 𝛾 > 0, the following are equivalent: (i) Consider
> 𝛾.
(ii) The following equation
= (𝑃 (𝑡) 𝐵 + 𝐴0 𝑃 (𝑡) 𝐵0 + 𝐶 𝐷)
⋅ (𝑃 (𝑡) 𝐵 + 𝐴0 𝑃 (𝑡) 𝐵0
𝐵0 𝑃 (𝑡) 𝐵0 + 𝐷 𝐷 − 𝛾2 𝐼 > 0,
has unique solution 𝑃𝛿,𝑇 (𝑡) ≤ 0 on [0, 𝑇]. Moreover, 𝛾 minV∈L2F ([0,𝑇],𝑅𝑝 ) 𝐽𝑇,𝛿 (𝑥0 , V) = 𝑥0 𝑃𝛿,𝑇(0)𝑥0 . Now, we are to show what happens as 𝑇 increases. Theorem 14. If system (1) is time-invariant and square, ‖G‖[0,𝑇] > 𝛾 for some 𝛾 > 0. Then 𝑃𝑇 (𝑡) in (56) decreases as 𝑇 − increases for every 𝑡 ∈ [0, 𝑇].
V (𝜏) ∗
−1
+ 𝐶 𝐷) ,
∗ be Proof. Suppose 𝑇 > 𝑇, 𝑡 ∈ [0, 𝑇], and 𝑥0 ∈ 𝑅𝑛 . Let V𝑇−𝑡 optimal for 𝑥0 on [0, 𝑇 − 𝑡], and set
𝑃 (𝑡) 𝐴 + 𝐴 𝑃 (𝑡) + 𝐴0 𝑃 (𝑡) 𝐴 0 + 𝐶 𝐶 + 𝑃̇ (𝑡)
⋅ (𝐵0 𝑃 (𝑡) 𝐵0 + 𝐷 𝐷 − 𝛾2 𝐼)
(58)
𝑃 (𝑇) = 0
According to Theorems 6 and 11, we get the following theorem.
‖G‖[0,𝑇] −
⋅ (𝑃 (𝑡) 𝐵 + 𝐴0 𝑃 (𝑡) 𝐵0
(56)
+ 𝐶 𝐷) ,
{V𝑇−𝑡 (𝜏) , ={ 2 −1 {− (𝐷 𝐷 − 𝛾 𝐼) 𝐷 𝐶𝑥 (𝜏) ,
𝛾 𝑇−𝑡
𝑥0 𝑃𝑇 (𝑡) 𝑥0 = 𝑥0 𝑃𝑇−𝑡 (0) 𝑥0 ≤ 𝐽
𝑃 (𝑇) = 0 has unique solution 𝑃𝑇 (𝑡) ≤ 0 on [0, 𝑇]. Moreover, 𝛾 minV∈L2F ([0,𝑇],𝑅𝑝 ) 𝐽𝑇 (𝑥0 , V) = 𝑥0 𝑃𝑇(0)𝑥0 .
∗ = 𝐽𝑇−𝑡 (𝑥0 , V𝑇−𝑡 )+𝐸∫
Remark 13. For given 𝛾 > 0, if we replace 𝐵, 𝐶, 𝐷, and V(𝑡) 0𝑙×𝑛 with 𝐵𝛿 = [𝐵 0𝑛×𝑛 ], 𝐶𝛿 = [𝐶 𝛿𝐼𝑛 ] , 𝐷𝛿 = [ 0𝐷 ], and 𝑛×𝑙 0𝑛×𝑛
∗ )+𝐸∫ = 𝐽𝑇−𝑡 (𝑥0 , V𝑇−𝑡
𝛾
𝑇
𝛾
𝑇
0
𝑇−𝑡
𝑇−𝑡
(𝑥0 , V)
{‖𝑧 (𝜏)‖2 − 𝛾2 ‖V (𝜏)‖2 } 𝑑𝜏 {𝑥 (𝜏)
−1
⋅ 𝐶 [𝐼 − 𝐷 (𝐷 𝐷 − 𝛾2 𝐼) 𝐷 ] 𝐶𝑥 (𝜏)} 𝑑𝜏 𝑇−𝑡
+𝐸∫ (57)
𝑇−𝑡
𝑇−𝑡
V𝛿 (𝑡) = [V(𝑡) 0𝑛×𝑛 ] , respectively, and 𝑧(𝑡) with 𝑧𝛿 (𝑡) in (1), we deduce the corresponding H− index ‖G‖[0,𝑇] 𝛿− and
= 𝐸 ∫ {‖𝑧 (𝑡)‖2 − 𝛾2 ‖V (𝑡)‖2 + 𝛿2 𝐼} 𝑑𝑡.
𝜏 ∈ (𝑇 − 𝑡, 𝑇 − 𝑡] .
By the time invariance of 𝑃𝑇 (𝑡), 𝑃𝑇−𝑡 (0) = 𝑃𝑇(𝑡). Then,
𝐵0 𝑃 (𝑡) 𝐵0 + 𝐷 𝐷 − 𝛾2 𝐼 > 0,
𝛾 2 2 𝐽𝑇,𝛿 (𝑥0 , V) = 𝐸 ∫ {𝑧𝛿 (𝑡) − 𝛾2 V𝛿 (𝑡) } 𝑑𝑡 0
(59)
𝜏 ∈ [0, 𝑇 − 𝑡]
𝑇−𝑡
−1
{[V (𝜏) + (𝐷 𝐷 − 𝛾2 𝐼) 𝐷 𝐶𝑥 (𝜏)] −1
⋅ (𝐷 𝐷 − 𝛾2 𝐼) [V (𝜏) + (𝐷 𝐷 − 𝛾2 𝐼) 𝐷 𝐶𝑥 (𝜏)]} 𝑑𝜏
Mathematical Problems in Engineering 𝛾
∗ )+𝐸∫ = 𝐽𝑇−𝑡 (𝑥0 , V𝑇−𝑡
𝑇−𝑡
𝑇−𝑡
9 0
{𝑥 (𝜏)
−0.02 −0.04
−1
⋅ 𝐶 [𝐼 − 𝐷 (𝐷 𝐷 − 𝛾2 𝐼) 𝐷 ] 𝐶𝑥 (𝜏)} 𝑑𝜏
−0.06 −0.08
𝛾
∗ ≤ 𝐽𝑇−𝑡 (𝑥0 , V𝑇−𝑡 ) = 𝑥0 𝑃𝑇 (𝑡) 𝑥0 .
−0.1
(60)
−0.12
This means that 𝑃𝑇 (𝑡) decreases as 𝑇 increases for every 𝑡 ∈ [0, 𝑇].
−0.14 −0.16 −0.18
3. A Numerical Example Below, we give a numerical example to illustrate the rightness of Theorems 12 and 14. Example 1. In system (1), we consider a two-dimensional linear stochastic system with the following parameters: 0 1 ], 𝐴=[ 1 2
1 1 ], 2 3
(61)
1 0 ], 𝐷=[ 0 1
1 0 ]. 0 0
𝑃3 (𝑡) =
(𝑡) (𝑡)
(𝑡)
(62)
],
(𝑡)
for which their trajectories are shown in Figure 1. If we set 𝑡 = 1, then it yields 𝑃2 (1) = [ 𝑃3 (1) = [
2
P211 (t)
P311 (t)
P222 (t)
P322 (t)
P212 (t)
2.5
3
P312 (t)
Figure 1: The trajectories of 𝑃2 (𝑡) and 𝑃3 (𝑡).
In this paper, we have solved the H− index problem where both stochastic and deterministic perturbations are present. Necessary and sufficient condition for the lower bound of H− index is given by means of the solvability of a generalized differential equation. The proposed results are not completely dual to H∞ norm, and the effectiveness of the given methods is illustrated by numerical example.
Acknowledgments
𝑝211 (𝑡) 𝑝212 (𝑡) 𝑃2 (𝑡) = [ 12 ], 𝑝2 (𝑡) 𝑝222 (𝑡) 𝑝312 𝑝322
1.5 t
The authors declare that there is no conflict of interests regarding the publication of this paper.
Set 𝛾 = 0.5, 𝑇 = 2, 3; by solving (56), we can obtain the solutions of
𝑝311 [ 12 𝑝3
1
Conflict of Interests
1 2 𝐴0 = [ ], 2 1 𝐵0 = [
0.5
4. Conclusion
2 1 ], 𝐵=[ 1 3 𝐶=[
0
−0.0892 −0.0953 −0.0953 −0.1446 −0.1044 −0.0889 −0.0889 −0.1489
], (63) ].
It is easy to see that 𝑃2 (1) > 𝑃3 (1), which verifies the rightness of Theorem 14.
This work is supported by National Natural Science Foundation of China (Grants nos. 61174078, 61170054, and 61402265) and the Research Fund for the Taishan Scholar Project of Shandong Province of China.
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