86
.pdfChapter II. Absolute stability and Aizerman’s problem of multidimensional regulated systems
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2 0 |
( ) 2 0, |
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(2.152) |
From (2.152) it follows that improper integral (v. (2.149))
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From here taking into account inequalities (2.141), (2.143) we have
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I2 = [H |
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A11 y(t)] |
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0 y(t) |
A11 y(t)] |
2 0 |
0 y(t) |
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[H |
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[S11H |
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S12H y(t)] dt = |
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y(t) A11 y(t)] |
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where matrices |
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3 of |
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(n 2m) (n 2m) |
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formula (2.150). The theorem is proved. |
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Lemma 18. Let the conditions of the lemmas 13–15 be satisfied, matrix |
A |
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1 |
the
be
Hurwitz matrix, the function |
( ) |
1 |
. Then for any matrices |
N |
, N |
2 |
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(n 2m) (n m), |
(n 2m) (n m) |
respectively, the improper integral |
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[ y |
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(t)P y |
(t) |
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(t)P y(t) y |
(t)P y(t)]dt = 0 |
, |
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where |
P = N H |
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2 12 |
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12 . |
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Lemma’s proof directly follows from the identity (2.146).
of orders
(2.153)
Lecture 24.
Absolute stability and Iserman’s problem of multidimensional systems in a critical case
Absolute stability. Based on the above results on the estimation of improper integrals, and also Lemmas 13–15 we can formulate conditions of absolute stability of equilibrium state of the system (2.126), (2.127).
We introduce the following notation
R = P = H * |
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1H |
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N H , |
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151
Lectures on the stability of the solution of an equation with differential inclusions
0 = 2 2 P2 = H0* 1 S12 A12 H0* S11 1 A11 2H0* 2 0 1 A11
H0* 2 S11H0 H0* 2 S12H1 H1N1* N2 A12 ,
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W = |
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P = A11 |
S11 A12 |
A11 |
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A11 |
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A11 2 S11H0 A11 2 S12 H1 N1 A12 . |
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In particular when matrix |
N1 |
= A11 1 S12 A11 2T |
A12 K1, |
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W = A11 |
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A11 |
A12 K A12 0 |
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where |
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> 0, A |
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are matrices of orders |
(n m) (n m), m (n m) |
, respectively. |
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Then improper integral
(2.154)
then the matrix
(2.155)
A12. |
Here |
K |
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(t)R y(t) y |
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(t) y(t) |
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( i ) 1i d i |
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y |
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. (2.156)
Theorem 18. Let the conditions of the lemmas 13–17 be satisfied, matrix |
A |
be a |
1 |
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R |
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Hurwitz matrix, the function ( ) 1, and let, moreover, matrices |
of orders |
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0 |
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(n 2m) (n 2m), (n 2m) (n 2m) |
such that: 1) |
R R |
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; 2) |
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improper integral
I
m i ( )
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i=1 i (0)
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= |
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40 |
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(2.157) |
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where matrices R0 , 0 , |
W0 are determined by the formula (2.154), 1, |
2 0 are |
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any diagonal matrices |
of orders m m, N1, N2 |
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any matrices |
of orders |
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respectively. |
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Proof. As matrix 0 = 0* , then |
y* (t) y(t) = |
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[ |
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* (t) y(t)], t I , conse- |
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quently, improper integral (v. Lemma 2.17)
152
Chapter II. Absolute stability and Aizerman’s problem of multidimensional regulated systems
=
1 2
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y* (t) y(t)]dt = |
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<
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(2.158)
by virtue of the assessment |
| y( ) | c |
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0 < c2 |
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< . |
From (2.156) taking |
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into account (2.158) we get |
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< |
. (2.159) |
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hypothesis |
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matrix |
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R0 0. Then |
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y |
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(t)[ |
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(R |
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R |
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0, |
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t,t I. |
Consequently, from (2.159) we |
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get inequality |
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(t)W y(t)dt |
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[ y |
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(t)R y(t) y |
* |
(t)W |
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y(t)]dt < . |
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40 |
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The theorem is proved. |
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Theorem 19. Let the conditions of |
lemmas |
13–17 be satisfied, matrices |
A , |
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1 |
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A |
( ), |
0 |
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be Hurwitz matrices, the function |
( ) |
1 |
and conditions 1), |
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2) |
of Theorem 18 be satisfied, and let, moreover, matrix |
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W W |
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> 0. |
Then the |
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equilibrium state of the system (2.126), (2.127) is absolutely stable. |
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If in addition diagonal matrix |
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where |
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> 0 |
is a diagonal matrix |
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with arbitrarily small elements, then Iserman’s problem has a solution in the sector
[E, 0 ] .
Proof. As all conditions of Theorem 18 are satisfied, then we get inequality (2.157). Consequently, improper integral
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40 |
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y* (t)W y(t)dt = |
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y* (t)[ |
(W |
W * )]y(t)dt = |
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y* (t)T y(t)dt < , |
(2.160) |
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153
Lectures on the stability of the solution of an equation with differential inclusions
where T0 = 12 (W0 W0* ) > 0.
y Rn 2m , |
y 0, |
V (0) = 0, |
Let the function |
V ( y) = y*T y. |
Then V ( y) > 0, |
y, |
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where | y(t) | c2 , |
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t, |
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t I = [0; ). We show that lim y(t) = 0. |
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t |
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Suppose the contrary, |
i.e. |
lim y(t) 0. |
Then there exists |
a sequence |
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t |
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tk ,tk > 0,tk when k such that |
| y(tk ) | > 0, k = 1,2,... We choose |
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tk 1 tk 1 > 0, k = 1,2,... |
As |
y(t),t I |
is continuously |
differentiable, |
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| y(t) | c2 ,| y(t) | c3 ,t I, |
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then |
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| y(t) y(tk ) | c3 | t tk |, t,t I |
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V ( y(t))dt |
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V ( y) > 0, y R |
n 2m |
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V ( y(t))dt, |
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then
where |
| y(t) |=| y(t |
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) y(t) y(t |
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t, t [t |
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]. |
We can always choose the value |
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> 0 |
such that the |
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value |
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> 0. |
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So |
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, |
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t [t |
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As function |
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V ( y) = y T y |
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is continuous on compact set |
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then there is a number |
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m > 0 |
such as |
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min |
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V ( y) = m . Then the value of the integral |
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0 |y| c2 |
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V ( y(t))dt 1m, |
k = 1,2,... |
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Consequently, |
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V ( y(t))dt = y* (t)W0 y(t)dt |
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V ( y(t))dt lim k 1m = . |
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154
Chapter II. Absolute stability and Aizerman’s problem of multidimensional regulated systems
This contradicts the condition (2.160). Consequently,
lim y(t) = 0. t
From
z(t) = K |
1 |
y(t), |
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K is a nonsingular matrix, we have lim z(t) = 0, , |
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A1( ), 0 are |
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1. Hence taking into account that matrices A1, |
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Hurwitz matrices, lim z(t,0, z0 , ) = 0, |
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1 , according to definition 7 we |
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t |
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get that the equilibrium state of the system (2.126), (2.127) is absolutely stable. Notice
that if |
= |
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2 |
, |
then Iserman’s problem has a solution in the sector |
[E, |
0 |
]. |
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0 |
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The theorem is proved. |
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From theorems 18, 19 it follows that in cases when matrices A1, A1( ), E 0 |
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are |
Hurwitz |
matrices, |
( ) |
, |
and |
the following conditions are satisfied: |
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1) |
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R R |
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0; |
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W W |
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> 0 |
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2) |
= |
3) |
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the equilibrium state of multidimensional |
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0 |
1 |
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controlled system (2.126), (2.127) in a critical case is absolutely stable.
Remark 3. As follows from the proof of theorem 19, the equilibrium state of the
system (2.126), (2.127) is absolutely stable and in case, when |
T |
= |
1 |
(W |
W |
* |
) 0, |
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surface |
V ( y) = y T y |
does not contain whole trajectories. In |
this |
case, integral |
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0 |
curves "stitch" |
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surface |
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V ( y) = 0. |
Notice that surface |
V ( y) = 0 |
does not contain |
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whole trajectories, if |
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y |
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(t)T y(t) 0, |
t I = [0, ). |
This condition for regulated |
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systems with nonlinearities |
( ) |
1 is performed. |
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It should |
be |
noted |
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that as |
a result |
of nonsingular |
transformation |
the |
identity |
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(2.142) is obtained. This identity is used to determine the improper integral |
I |
3 . The |
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improper |
integral |
I |
3 |
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depends |
on |
arbitrary |
matrices |
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N |
, N |
2 |
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of |
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orders |
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1 |
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(n 2m) (n m), (n 2m) (n m) |
respectively. Matrices |
R |
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, ,W |
depend on |
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0 |
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0 |
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0 |
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matrices |
N , N |
2 |
. As |
follows |
from the condition |
of |
theorems |
18, |
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19 |
matrices |
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1 |
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1, 2 0, |
N1, |
N2 |
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* |
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0 |
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* |
W0 |
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* |
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ensure the fulfillment of conditions |
R0 R0 0, |
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= 0 |
, |
W0 > 0. |
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Consequently, matrices |
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N |
, N |
2 |
allow us to significantly expand the area of absolute |
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1 |
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stability of system parameters in space. In the method of A.I. Lurie and in the method of V.M. Popov we have no improper integral I3 , consequently, we have no matrices
N1, N2 . The presence of matrices N1, N2 allowed us to solve the Iserman’s problem. Iserman’s problem. The question arises: Is it possible to distinguish a class of
multidimensional regulated systems, by selecting a feedback matrix |
S |
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for which |
1 |
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Iserman’s problem has a solution? For this class of multidimensional regulated systems, the obtained results are necessary and sufficient conditions of absolute stability.
155
Lectures on the stability of the solution of an equation with differential inclusions
As follows from Theorem 16 and Lemma 2.17 he improper integral
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(t)R y(t) |
y |
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(t) y(t) |
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m |
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i ( i ) 1i d i |
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i=1 |
(0) |
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i |
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y |
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< |
(t)W1 y(t)]dt =,
(2.161)
where |
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R = P = H |
* |
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N |
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S11H |
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H , |
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P = H * |
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A H*N* N |
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S12 |
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A12 , |
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11 |
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W1 = 3 P3 = A11* 1 S12 A12 |
N1 A12 . |
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(2.162) |
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In |
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when |
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N1 = A11 1 S12 |
A12 K, |
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then |
W1 |
= A12 K A12 |
0, |
where |
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K = K |
* |
> 0 matrix of order (n m) (n m). |
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Theorem 20. Let the conditions of the lemmas 13–17 be satisfied, matrices |
A , |
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1 |
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A ( ), |
E |
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2 |
be Hurwitz matrices, the function |
( ) |
, |
and let |
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besides: |
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1) |
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diagonal |
matrix |
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1 = diag ( 11,..., 1m ), |
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N |
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N |
2 |
, |
S1 |
= (S11, S12 ) of |
orders |
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(n 2m) (n m), |
(n 2m) (n m), |
m (n 2m) |
such that |
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R R |
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0, |
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(2.163) |
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where matrices |
R , |
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are determined by the formula (2.162); |
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2) |
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Matrix |
T |
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(W W |
> 0 |
(or |
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surface |
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V ( y) = y Ty = 0 does |
not |
contain |
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whole trajectories), where matrix |
W |
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1 is determined by formula (2.162). |
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Then in the sector |
[E, ], |
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> 0 |
is a diagonal matrix of order |
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m m with arbitrarily small positive elements, Iserman’s problem has a solution,
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where 0 |
= diag( 01,..., 0m ) is the |
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limitation |
value of |
the matrix |
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0 |
= diag ( 01,..., 0m ) , found from the Hurwitz condition |
for the matrix |
A1 ( 0 ). |
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Proof. From (2.161) with account for (2.163) we have |
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[ y* (t)R y(t) y* (t)W y(t)]dt = |
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I 5 = |
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m i |
( ) |
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d |
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y* (t) 1 y(t)]dt < , |
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= i ( i ) 1i d i |
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0 dt |
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156
Chapter II. Absolute stability and Aizerman’s problem of multidimensional regulated systems
due to the fact that | y( ) | c |
< , |
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| y(0) | c |
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< , |
As T = |
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1 |
(W |
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T1 0 ), then from (2.164) taking into account (2.163) we get |
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I |
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(t)W y(t)dt = |
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(t)T y(t)dt |
I |
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< |
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y |
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5 |
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t, |
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(t)[ |
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due to the fact that |
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(t)R1 y(t) = y |
2 |
R1 )]y(t) 0, |
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W1 )*
I.
> 0 (or
(2.165)
Let the
function
V |
(0) = |
1 |
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V1
0,
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* |
( y) = y T y, |
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1 |
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1 |
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(V y(t))dt < |
0 |
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y ,
R |
n 2m |
. |
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where
Then
| y(t)
V ( y) |
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, |
2 |
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> 0, | y(t)
y, |
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3 |
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yt,
R |
n 2m |
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t I |
= |
y 0, [0, ).
Further, repeating the proofs of Theorem 5, we get
lim y(t) = 0. t
Matrix
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0 |
= |
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there |
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( ) |
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0 |
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is |
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= |
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2 is determined from the Hurwitz condition |
for the matrix |
A |
( |
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matrix |
S |
1 of order |
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m (n 2m) |
such |
that |
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, |
E |
= |
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2 |
solution of |
linear |
system |
z |
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0 |
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= |
). |
Thus, |
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for any A1( )z,
E |
0 |
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is asymptotically stable, for any
( ) |
, |
1 |
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lim |
y(t) = lim Kz(t) = 0. |
t |
t |
In this case, as follows from definition 9 Iserman’s problem has a solution. The theorem is proved.
As follows from Lemma 16, equation (2.137) can be represented as
H y = A11 y(t) ( ) = N H y N H y(t) ( ), |
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0 |
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11 |
0 |
12 |
1 |
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where equality
A11 |
= (N |
, N |
12 |
), |
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11 |
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H y |
= |
1 |
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A12 = |
A12 y(t) = |
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(N |
22 |
, N |
23 |
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N ) .
22 |
H |
y |
0 |
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The
N |
23 |
H |
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1 |
equation
y ,
(2.166)
(2.166)
follows from
H y |
N |
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y = |
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y |
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H |
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N |
23 |
H |
y |
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1 |
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n m,m |
( ), ( ) ,
1
where |
N11, |
N12 , |
N22 , |
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N23 |
are |
matrices |
of orders |
m m, |
m (n m), |
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(n m) m, |
(n m) (n m) respectively. |
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Theorem |
21. |
Let |
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the |
conditions |
of Theorem 20 |
be satisfied, |
where |
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T = |
1 |
(W W * ) = H *T H |
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0, |
T |
= T * |
> 0 |
is a matrix of order m m, |
surface |
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1 |
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0 |
11 |
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157
Lectures on the stability of the solution of an equation with differential inclusions
V ( y) = y* H *T H |
0 |
y = 0 does not |
contain whole |
trajectories, |
and matrix |
N |
23 |
of |
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1 |
0 |
11 |
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order |
(n m) (n m) is a Hurwitz matrix. |
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Then in the sector [E, 0 ], 0 |
= |
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2 , 2 > 0 is a diagonal matrix of order |
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0 |
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m m with arbitrarily small positive elements, Iserman’s problem has a solution, |
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where |
0 = diag( 01,..., 0m ) is a |
limit value of |
the matrix |
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0 |
= diag ( |
01 |
,..., |
0m |
) |
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found from the Hurwitz condition for matrix A1 ( 0 ).
Proof. As follows from the condition of the Theorem, equality (2.165) holds,
where |
* |
H |
* |
T |
= T |
* |
> 0. |
Function |
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* |
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V1 ( y) = y |
0T11H0 y, |
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V ( y) = y T y > 0, |
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11 |
11 |
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1 |
11 |
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y, y = H0 y 0,V1(0) = 0 trajectories. Then, as
lim |
y(t) = lim H |
0 y(t) = 0. |
t |
t |
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order |
(n m) (n m) |
is |
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lim |
H y(t) = 0. |
As |
y(t) = ( |
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1 |
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t |
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is proved.
when |
y = 0, |
surface |
V ( y) = 0 |
does not contain whole |
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1 |
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in |
the proof |
of |
Theorem 19, we have |
Consider the second equation from (2.166). If |
N |
23 of |
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a |
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Hurwitz matrix, then when |
lim |
H |
0 |
y(t) = 0, |
limit |
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y(t), H y(t)), |
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then lim |
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The theorem |
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Lecture 25.
The solution to the model problem in a critical case
The equation of a regulated system has the form
x |
= x ( ), x |
= x , x = ( ), = x x |
x |
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( ) |
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= { ( ) C(R |
, R ) | ( ) = ( ), ( ) } |
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, R ) | 0 ( ) |
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( ) = { ( ) C(R |
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(0) |
= 0,| ( ) | * , R1}, |
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(2.167) |
where > 0 is an arbitrarily small |
number, |
= x , |
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= x2 , L1 = 1, L2 = 0, |
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S = 1, T1 = 1, T2 = 1. |
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For this example matrices |
A1 = A1( ), |
B1, |
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S1 , are equal to: |
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1 1 |
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A1 = A1 ( ) = |
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= ( 1 |
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, B1 |
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, S1 |
1 ). |
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The equation (2.167) in the vector form has the form
158
Chapter II. Absolute stability and Aizerman’s problem of multidimensional regulated systems
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z = A1z B1 ( ), |
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= S1z, z(0) = z0 ,| z0 |< , ( ) 1,t I = [0, ) , |
(2.168) |
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where z = (x1, x2 , x3 ). If the matrix |
A1 = A1( ) |
is a Hurwitz matrix, the function |
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( ) |
, (0) = 0 |
only when |
= 0 , |
then |
the system |
(2.168) has the only |
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equilibrium state. |
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Characteristic polynomial of matrix |
A = A ( ) |
is equal to |
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( ) =| I |
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A |= a a a |
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where |
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= 1 ( |
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= 1 |
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matrix |
A |
( ) |
is a Hurwitz matrix, if: |
1 |
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А. Nonsingular transformation.
( |
1 |
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a |
= . |
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< 0, |
1 |
< 0 |
for any |
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Define the vectors |
1 |
As follows from
> 0 .
R |
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( ) |
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R |
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from the condition
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B = 1, |
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1 |
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*B = 0,
2 1
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B = 0. |
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1 |
As a result, we get
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= (1,1,0), |
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* = (1,0, 1),
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= (1,1, 1). |
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Matrices of transformation
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0 |
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R* = |
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K 1 = |
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Then |
matrices |
A1, |
B1, |
S |
1 are equal |
to: |
A1 ( ) = KA K |
1 |
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S |
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y = Kz, |
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where |
z = K y . |
Then |
equation |
(2.168) |
by |
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ransformation is reduced to |
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y |
= A1 ( ) y B ( ), |
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= S |
y, ( ) |
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, y(0) = y |
,| y |< ,t I = [0, ) |
. |
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The equation (2.169) can be written as
B1 |
= KB |
, |
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nonsingular
(2.169)
y |
= ( |
) y [ 1 ( )]y |
2 |
[ ( |
)]y |
( ), |
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= y y |
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= ( |
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2 |
( |
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= ( 1 |
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1) y1 ( 1 1)( y1 |
y2 y3 ) ( 1 |
1 |
1 )( y2 ) . (2.170) |
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159
Lectures on the stability of the solution of an equation with differential inclusions
Б. Solution properties. Characteristic polynomial
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( ) =| I |
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A1( ) |= a ( , ) a |
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( , ) a ( , ), |
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where |
a ( , ) =1 ( )( |
), |
a ( , ) = ( )( |
1 |
), |
a ( , ) = ( ) . |
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Notice that as value > 0 is an arbitrarily small number, then from (2.170) we |
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have |
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y |
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2 |
( ), y |
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y |
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= y , ( (t)) = y (t) y (t), |
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(t) = ( |
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( )( y |
y |
2 |
y |
) ( |
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)( y |
(t)) |
, (2.171) |
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where characteristic polynomial |
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( ) = a ( ) |
a |
( ) a |
( ) |
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(2.172) |
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where |
a ( ) = 1 ( |
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a ( ) = ( |
1 |
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a ( ) = . |
Limit value |
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0 is determined from the condition of Hurwitz of polynomial |
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В. Improper integrals. As follows from Theorem 16 the improper integral
where
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(t) 1 y(t) |
= [ y |
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( (t)), (t),t I |
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I |
= |
( (t)) (t)dt = |
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0 |
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( ) |
y |
* |
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* |
(t) 3 y(t)]dt = |
( (t)) 1 d < , |
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(t) 2 y(t) y |
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(0) |
are denoted by identities from (2.171), matrices
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1 ( 1 1 ) |
0 |
0 |
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1 ( 1 1 ) |
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1 ) |
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1 ( 1 1 ) |
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Improper integral
160