86
.pdfChapter II. Absolute stability and Aizerman’s problem of multidimensional regulated systems
Then in the sector [0, 0 ],
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> 0, |
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i = 1, m
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arbitrarily small numbers, Iserman’s problem |
has a solution, |
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0 = diag( 01, , 0m ), |
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0 = diag( 01, , 0m ) the limit value of the matrix |
from the Hurwitz condition for the matrix A B 0S.
Proof. From (2.28) when we have the condition (2.29), we get
where found
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due to the fact that |
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then |
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)W y(t)]dt |
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< . |
As matrix |
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( y(t))dt = |
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where V1( y(t)) = |
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t I. As surface V1 |
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trajectories, then repeating the proofs of Theorem 5, we get |
lim y(t) = 0. |
As follows |
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from [14] the sufficient condition for the absence of whole trajectories |
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V1( y) = y T1 y = 0 |
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y(t) = 2 y |
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(t)T [ Ay(t) B ( (t))] |
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t I = [0, ). |
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It can be shown that for the system (2.10) the given inequality is performed. The theorem is proved.
As follows from Lemma 4, equation (2.10) can be represented as
H y = A1H y A2 H y ( ), |
H y = A3H y A4 H y, |
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(2.30)
where A11 = ( A1, A2 ),
m m, |
m (n m), |
(n |
lows from equality
A12 = ( A3 , A4 ),
m) m, |
(n m) (n |
A1,
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A4 |
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A2 , A3 , |
are matrices of orders |
respectively. The equation (2.30) fol-
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H1 y |
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A3 |
A 2 A4
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n m,m |
( ), ( ) .0
91
Lectures on the stability of the solution of an equation with differential inclusions
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Theorem 7. Let |
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of Theorem 6 |
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be satisfied, where |
the matrix |
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0, |
T11 |
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surface |
V1 ( y) = y |
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0 y = 0 does not contain whole |
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T1 = H0T11 H0 |
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trajectories, the matrix |
A4 |
is a Hurwitz matrix. |
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Then in the sector |
[0, 0 ], |
0 = 0 , |
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is a diagonal matrix of order |
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m m with arbitrarily small elements, Iserman’s problem has a solution. |
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Proof. As |
follows |
from the |
condition |
of |
the |
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Theorem, the |
inequality holds |
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< ) , where V1 |
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0 y, T11 = T11 |
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V (y) > 0, y, |
y R |
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V (0) = 0, |
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tain whole trajectories. Then according to the statement of the Theorem lim |
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y(t) = 0. |
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Consider |
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If the |
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of order |
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(n m) n m is a Hurwitz matrix, then when |
lim |
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y(t) = 0, |
limit |
lim H y(t) = 0. |
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As y(t) = (H0 y(t), H1 y(t)), |
t I , |
then |
lim y(t) = 0, |
, 0 . |
The theorem is proved. |
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It should be noted that matrices 1, N1, |
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2 , S |
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R = R* 0, = * |
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W * ) 0. |
As follows from Theorem 6, Iserman’s |
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problem has a solution. Consequently, Theorem 6 gives necessary and sufficient con-
ditions of absolute stability, and matrices |
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expand areas of absolute stability in |
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space of parameters.
For multidimensional regulated systems with limited nonlinearities we get improper integral I1 I2 < without any additional requirement on values of constructive parameters of the system.
Lecture 17.
The solution of model problems in the general case
The constructiveness of the proposed method for solving the Iserman’s problem is shown in examples.
1. Iserman’s problem for a one-dimensional system.
The equation of a one-dimensional system has the form
x |
= x |
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= 2x |
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1.03x |
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0.03x |
0.75 ( ), |
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(2.31)
For this example matrices A, B, S are equal:
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A = |
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92
Chapter II. Absolute stability and Aizerman’s problem of multidimensional regulated systems
the equation (2.31) in vector form can be written as
x = Ax B ( ), = Sx, ( ) |
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А. Nonsingular transformation. The characteristic polynomial of the matrix A is equal
( ) =| I |
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A |= |
1.04 |
0.98 = a |
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a2 = 1.04. |
Matrix |
A is a Hurwitz matrix. |
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Define the vectors |
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As a result, we get |
2 = (0,1, 3), |
3 = (1,1, |
transformation
B = 1, |
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3). |
Matrix of |
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1111 |
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RRR=R=||==||||||, ,,,, ,,, ||=|||=||=|= 1111 |
1111 |
1111, ,|,,R||R|R|=R|=|=|=44440.0.0.0. |
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1 1112 2223 333 |
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1111 3333 3333 |
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Consequently, the rank of the matrix
R
is equal to
n = 3.
Matrices
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Matrices
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S = SR 1 = ( 34 1 14 1, 1 14 1 14 1 , 1 ),
A11 |
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93
Lectures on the stability of the solution of an equation with differential inclusions