86
.pdfChapter 1. Absolute stability and Aizerman’s problem of one-dimensional regulated systems
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n 1 |
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(t) (t)] = * (t)N (t), t I , |
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where (t) = ( (t), y1 (t), , yn 1 (t)). Then improper integral |
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I1 = |
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( )d |
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due to the fact that |
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(t) c2 , |
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t I , |
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where |
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N |
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is a constant matrix of order |
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(n 2) (n 2). |
As follows from lemma 9 |
(Q = N ), |
the following equality holds |
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(t)N (t) = N |
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(t) N y |
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(t) |
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(t) |
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[ y |
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(t)F y(t)], |
t I. |
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n 1 |
n 1 |
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1 1 |
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dt |
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Then the ratios (1.62), (1.63) follow from (1.60), (1.61), where |
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[ y |
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(t)F y(t)]dt = y |
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(t)F y(t) | |
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due to limited |
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y(0) |
and |
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The theorem is proved. |
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Notice that |
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( ) |
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(0) |
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( )d = c1. |
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I1 = ( ) dt = |
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( )d = |
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(0) |
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( ) |
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Then for |
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0, |
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c , |
if |
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< 0, |
then |
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2 I1 |
=| 2 |
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| ( I1 ) | 2 | c1. |
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Consequently, for any value 2 |
value |
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I |
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c . |
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Theorem 11. Let the conditions of lemmas 6, 7 be satisfied, matrices |
A, |
A1 be |
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Hurwitz matricies, function ( ) 1. |
Then for any value |
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1 |
> 0 |
along the solution |
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of the system (1.51) improper integral |
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1 2 ( (t))]dt = |
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I |
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(1.64) |
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= [M 2 |
(t) M |
1 |
y 2 (t) M |
n 1 |
y 2 |
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(t)]dt l |
2 |
0, |
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l |
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= y* (t)F y(t) | = y* ( )F y( ) y* (0)F y(0), |
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(1.65) |
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41
Lectures on the stability of the solution of an equation with differential inclusions
where Mi |
= Mi ( 1 ), 1 > 0, i = 0, n 1, |
F2 |
is a constant matrix of order (n 1) ( |
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N0 0. |
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Proof. From inclusion ( ) 1 it follows that |
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, R . |
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Then for any value 1 > 0 |
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an equality holds ( ) 1 > 1 0 |
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Hence it follows that along the solution of the system (1.51) the inequality holds
n 1)
R |
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1 |
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where
and M
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( (t)) (t) |
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( (t)) (t) |
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( (t)) = ( |
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n |
y |
n 1 |
) 1 |
1 1 ( a y |
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0 |
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is a constant matrix of order |
(n |
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(t)) 0, |
(t) 0, t, t I , |
(1.66) |
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a y |
a |
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) |
1 |
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n |
n 1 |
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0 1 |
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a |
n |
y |
n 1 |
)2 |
= * |
(t)M (t), t I , |
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2) (n 2). |
Next applying lemma 8, where |
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Q = M , t I.
we get
* |
2 |
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(t) M |
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y |
2 |
(t) |
d |
* |
(t)F y(t)], |
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(t)M (t) = M |
(t) M y |
n 1 |
n 1 |
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dt |
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Then the ratios (1.64), (1.65) follow from (1.60), (1.61), (1.66). The theorem is proved.
Theorem 12. Let the conditions of the lemmas be satisfied 6, 7, matrices
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be Hurwitz matrices, function ( ) 1. Then for any values |
0 |
1 |
n 1 |
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the solution of the system (1.51) improper integral |
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y (t) |
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(t)]2 dt = |
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= [ (t) |
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y |
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= [ 2 |
(t) y2 (t) |
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y2 |
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(t)]dt l |
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0, |
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A, A1
along
(1.67)
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l = y* |
(t)F y(t) | = y *( )F y( ) y* (0)F y(0), | l |
3 |
|< , |
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(1.68) |
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where ( |
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, , |
n 1 |
), i = 0, n 1, |
F is a constant matrix of order (n 1) (n 1) |
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i |
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The |
proof of |
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the |
theorem is |
similar |
to |
the |
proof of |
theorem |
4, |
where |
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[ |
y |
n 1 |
y |
n 1 |
]2 = * (t) (t), |
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is |
a constant matrix |
of |
order |
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0 |
1 1 |
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(n 2) (n 2).
42
Chapter 1. Absolute stability and Aizerman’s problem of one-dimensional regulated systems
Absolute stability. Based on the results described above, conditions for the absolute stability of the equilibrium position can be formulated for the system (1.41), (1.42).
Theorem 13. Let the following conditions hold:
1. matrices A1, A1( ), 0 < 0 < 0 are Hurwitz matrices, the function
( ) 1 , |
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is an arbitrarily small number; |
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2. there is a vector |
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* |
R |
n 1 |
such that |
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B = 0, A B = 0, , An 1B = 0, An B 0; |
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3. rank of the matrix |
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is equal to n 1; |
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* , A* * |
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4. estimates (1.62), (1.64), (1.67) are true, and let besides: |
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0, |
2 |
N M |
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0, |
2 |
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> 0, |
i = 2, n 1. |
(1.69) |
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Then the equilibrium state of the system (1.41), (1.42) is absolutely stable. If in
addition the value
the sector |
[ , |
0 |
] |
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0 |
= 0 , where > 0 is an arbitrarily small number, then in |
Iserman’s problem has a solution.
Proof. Let the conditions 1) - 4) of the theorem hold. Then
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I |
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= |
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(t) ( |
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[( |
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l |
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(1.70)
where |
| c |
|< , |
| l1 |< , | l2 |< , |
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1 |
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inequalities (1.63), (1.65), (1.68). From (1.70) taking into account
l |
3 |
|< , |
2 |
is any number, |
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1 |
, |
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the first inequality from (1.69), we get
due to
where |
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1 |
2 Ni Mi i
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( 2 Ni Mi i ) yi2 |
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i=1 |
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0, |
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> 0, |
i = 2, n 1. |
Hence in case when |
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> 0, |
i = 1, n 1, |
we get |
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[( |
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43
Lectures on the stability of the solution of an equation with differential inclusions
Then as follows from lemma 10 the limit
by Lemma 7 the limit |
lim z(t) = 0. |
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t |
lim y |
i |
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t |
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As |
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(t) = 0, |
i = |
matrices
1, n 1.
A |
= |
1 |
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Consequently,
A ( ), |
A |
( ), |
1 |
1 |
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0 < |
0 |
< |
0 |
are Hurwitz matrices, then according to Definition 4, a trivial |
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solution |
z(t) 0, |
t |
I |
of the |
system (1.41), (1.42) |
is absolutely |
stable, |
where |
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z(t) = (x(t), (t)), |
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t I. |
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Consider |
the case |
when |
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2 N1 M1 |
1 = 0. In |
this |
case as |
proved |
above |
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lim |
y (t) = 0, |
i = 2, n 1. |
It |
remains to |
prove that |
lim y1 |
(t) = 0. |
As |
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* |
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, |
2 |
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t,
t I ,
then function
( (t)),
t I
is uniformly continuous in
t,
t I.
Then function
y |
= a y (t) a y |
(t) A B ( (t)), |
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n |
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n 1 |
0 1 |
1 2 |
1 |
1 |
t I
is
uniformly continuous by t t I. Then according to
lim y |
n 1 |
(t) = 0 |
entails |
t |
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due the
lim t
to uniform continuity Barbalat lemma [14;
y |
n 1 |
(t) = 0. |
Then |
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y |
(t), |
i |
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§ |
21, |
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when
i = 1, n
Lemma
t
1, |
( (t)), |
1] equality we get
0 = a |
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lim y (t) A |
B lim ( (t)), |
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n |
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t |
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lim (t) = |
0 |
lim y (t). |
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t |
t |
1 |
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Hence, taking into
account that |
a |
= a |
( ) =| A ( ) | |
is an arbitrarily small number, |
y1 (t), |
0 |
0 |
1 |
limited function, we get |
0 = lim ( |
0 |
y (t)). |
As |
( ) = 0 |
only when |
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1 |
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lim y |
(t) = 0. |
Notice that equality to zero of the sum |
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2 |
N M |
1 |
= 0 |
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t |
1 |
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1 |
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t I |
is a |
= 0, |
then |
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is one of
the features of a simple critical case compared to the main case from [6–9].
In case |
0 |
= 0 , > 0 is an arbitrarily small number, then all conditions of |
definition 6 are satisfied, consequently in the sector |
[ , |
0 |
] |
Iserman’s problem |
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has a solution. The theorem is proved.
Lecture 9.
Aizerman’s problem. The solution of the model problem for a simple critical case
The question arises: is it possible to single out a class of regulated systems for which Iserman’s problem has a solution, without resorting to checking the conditions of absolute stability from Theorem 13.
Theorem 14. Let the following conditions hold:
1)matrices A1
( ) 1, > 0
1 |
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0 |
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0 < 0 < |
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, A |
( ), |
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is an arbitrarily small number,
are Hurwitz matrices, the function
A1 = A1( );
2) there is a vector * Rn 1 such that
B1 = 0, A1B1 = 0, , A1n 1B1 = 0, A1n B1 0;
44
Chapter 1. Absolute stability and Aizerman’s problem of one-dimensional regulated systems
3) rank of the matrix R = |
* |
* |
* |
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*n |
* |
is equal to |
n 1; |
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4) the condition (1.62) is satisfied. There |
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is a |
feedback vector S Rn 1 and |
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number 2 such that 2 N0 |
0, 2 N1 0, 2 N2 |
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> 0, , 2 Nn 1 > 0. |
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Then in the sector |
[ , |
0 |
], |
0 = 0 |
, > 0 |
is an arbitrarily small number, |
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Iserman’s problem has a |
solution, |
the |
value |
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0 |
is a limiting value of Hurwitz |
matrices
A1 ( ) =
A B S, |
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1 |
1 |
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0 |
. |
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Proof. Let the conditions of the theorem hold. Then the improper integral
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= |
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N |
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(t) N y |
(t) |
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N |
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y |
(t) |
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( (t)) (t)dt = |
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Hence taking we get
2 |
N |
n |
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into
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y |
2 |
(t)]dt l |
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account that |
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2 |
2 |
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|<
, 0,
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2 |
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2 |
0, , |
N |
n 1 |
> 0, |
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(t)dt < , |
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(t)dt < , |
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(t)dt < , i = 2, n 1. |
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N |
N y |
N y |
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1 |
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Consider two cases: 1) case when 2 Ni > 0, i = 1, n 1;
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2 |
N |
i |
> 0, i = 2, n 1. |
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In the first case we have the inequality
2) case when
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N = 0, |
2 |
1 |
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N |
y |
2 |
(t)dt < , |
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i |
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i |
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Consequently, |
lim yi (t) = 0, |
i |
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= 1, n 1, |
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t |
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conditions of Hurwitz matrices |
A1 |
( ). Then |
blem has a solution.
In the second case we have the inequality
2 Ni yi2 (t)dt < ,
i = 1, n 1.
the value |
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0 |
is |
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0 |
= |
0 |
, > |
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i = 2, n 1.
determined from the
0 |
and Iserman’s pro- |
0
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Then lim yi (t) = 0, i = 2, n 1. Next repeating the proofs of theorem 6 from the |
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t |
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condition lim y |
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n 1 |
(t) = 0, |
we get lim y (t) = 0. |
Then in the sector [ , |
0 |
] |
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t |
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t 1 |
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Iserman’s problem has a solution. The fulfillment of condition 4) is enough to check for the values 2 = 1, or 2 = 1. The theorem is proved.
45
Lectures on the stability of the solution of an equation with differential inclusions
Solution of model problem. The effectiveness of the proposed method for determining the conditions of absolute stability and solving the Aizerman problem will be shown on examples.
Example 1. The differential equation of the regulated system has a form
x1 = x1 x2 , x2 = 3x1 2x2 ( ), = ( ), = 3x1 2x2 2x3 , (1.71)
where ( ) = ( ), ( ) 1. For this example, the source data is as follows
1 |
1 |
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0 |
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D = ( 3, 2), |
E = 2, n = 2. |
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A = |
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, |
B = |
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3 |
2 |
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1 |
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1.The characteristic
( ) =| I |
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A |= 1. |
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2 |
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2 |
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Matrix
polynomial of the matrix It is easy to check that matrix
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A |
has a form |
A |
is Hurwitz matrix. |
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1 |
1 |
0 |
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0 |
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A = |
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3 3 |
2 2 |
2 |
, |
B = |
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1 |
, |
S = ( 3, 2, 2). |
1 |
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3 |
2 |
2 |
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Function |
z(t), |
t I |
is a solution of the differential equation |
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z = |
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= Sz. |
Characteristic |
polynomial of the matrix |
A1 ( ) |
A ( )z B ( ), |
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1 |
1 |
is |
equal to |
((())=|)=| III AAA((())|=)|= (1(1 ))) 222... |
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333 |
222 |
1 |
11 |
3 |
33 |
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For Hurwitz of the matrix
A |
( ) |
1 |
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it is
necessary |
and |
sufficient, that 1 |
> 0, 2 > 0, (1 ) > 0. Hence, it |
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matrix A1 ( ) |
is a Hurwitz matrix |
with an arbitrarily small |
> 0. |
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a = 2 , |
a1 = 1 , a2 = 1, > 0 is an arbitrarily small number. |
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0 |
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follows that Coefficients
String vector
= ( |
, |
, |
) = ( 2,1, 1), |
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3 |
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A1 = (1,0,0),
A |
2 |
= (1, 1,0), |
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0. |
A B = 1 |
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* 1 |
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R = ( |
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where
| R |= 1,
rangR = 3.
Vectors
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* |
* |
* |
*2 |
* |
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1 |
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form a basis in R3. Based on
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ratios |
A = R* A R* 1 , B = R* B , |
S = SR* 1 from (1.71), we get |
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1 |
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y1 |
= y2 , y2 |
= y3 , y3 = a0 y1 |
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a1 y2 a3 y3 ( ), = S y = 2y1 y2 , (1.72) |
46
Chapter 1. Absolute stability and Aizerman’s problem of one-dimensional regulated systems
where
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A = |
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B = |
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S = (2, 1, 0). |
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2. From (1.72) it follows that identities are true:
( (t)) = (t) y |
(t) y |
(t), (t) = 2y (t) y |
(t), (t) = 2y |
(t) y |
(t), |
t I, |
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3 |
1 |
2 |
2 |
3 |
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where (t) = y3 (t), t I. As > 0 is an arbitrarily small number, then we accept a0 = 0, a1 = 1, a2 = 1. As rangR = 3, then the systems (1.71), (1.72) are equivalent
and from |
y(t) = ( y1 (t), y2 (t), y3 (t)) 0 |
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when |
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z(t) = (x1(t), x2 (t), (t)) 0 |
when t . |
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3. Let us calculate improper integrals |
I |
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2 |
, |
I |
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1 |
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a) Improper integral |
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= |
( (t)) (t)dt = |
[N |
2 |
(t) N |
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y |
2 |
(t) |
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d |
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where |
N |
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= 0, |
N2 |
= 2, |
N |
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= 1, |
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0 |
3 |
l |
= |
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dt |
[F (t)]dt, |
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t 0 it follows
N y |
2 |
(t)]dt l |
= c1 |
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3 |
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( ) |
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c1 = |
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1 |
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(0) |
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that
|< ,
F (t) = |
3 |
y |
2 |
(t) |
1 |
y |
2 |
(t), |
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t I.
Notice that for any number
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the product
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2 |
I |
1 |
= |
2 |
c1, |
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c1 |< . |
2 |
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b) Improper integral
where
I
M
2
0
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2 |
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= |
[ ( (t)) (t) |
1 |
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( (t))]dt = |
[M |
2 |
(t) M y |
2 |
(t) |
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M |
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y2 (t) M |
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y |
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(t)]dt l |
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0, | l |
2 |
|< , |
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= 0, M 2 |
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1, |
M3 |
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1 |
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= |
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[F |
(t)]dt. |
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= 1 0 |
1 |
= 1 0 |
= 1 0 |
2 |
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c) Improper integral
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I |
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1 |
y |
2 |
y |
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3 |
y ]2 dt = [ 2 |
y2 |
y2 |
y2 ]dt l 0, | l |< , |
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where 0 |
= 02 , 1 |
= 12 , 2 |
= 22 |
2 1 3 , 3 |
= 32 |
2 0 2 , l |
= |
d |
[F (t)]dt. |
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47
Lectures on the stability of the solution of an equation with differential inclusions
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4. Let us denote limit value 0 from Hurwitz matrices A1( ). |
Characteristic |
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equation for matrices |
A1( ) has a form |
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1 ( ) =| I3 A1 ( ) |= 3 2 (1 ) 2 = 0. |
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Limit value |
0 |
is determined from condition |
(1 ) 2 > 0. |
Hence, it fol- |
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lows that limit value |
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0 |
= 1. So, matrix |
A ( ) is a Hurwitz matrix, if |
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1 |
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> 0 is an arbitrarily small number. |
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5. Absolute stability. Conditions of theorem 6.1 (v. (1.69)) will be written as:
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N |
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(1.73)
Solving a system of algebraic equations (1.73) we find the limit value |
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0max |
= 1 |
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when 3 = 0, 2 = 2 0 , |
1 = |
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lim y2 |
(t) = 0, |
lim y3 (t) = 0. |
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0 > 0, 2 = |
2 |
0 . Consequently, |
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t |
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It remains to prove that |
lim y |
(t) = 0. As |
y |
= y |
2 |
, |
lim y |
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(t) = 0, then lim y (t) = y |
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= const. |
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Let us show that |
y |
1 |
= 0. |
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Notice that |
y |
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= a |
y (t) a y |
(t) a y |
(t) ( (t)), t I |
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is a uniformly |
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3 |
0 |
1 |
1 |
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3 |
3 |
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continuous function. Then, according to the Barbalat lemma [14; |
§ |
21, Lemma 1] |
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condition |
lim |
y |
(t) = 0, |
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entails |
lim y |
3 |
(t) = 0. |
Passing |
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lim y |
3 |
(t) = a |
0 |
lim y (t) a |
lim y |
2 |
(t) a lim y |
3 |
(t) lim ( (t)) |
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t |
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to the limit
we get
0 = a0 y lim ( (t)),
t
lim (t) = 2lim y |
(t) lim y |
2 |
(t) = 2 y |
. |
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t |
t |
1 |
t |
1 |
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Hence, taking into
account that |
a = 2 , > 0 is |
an arbitrarily small number, y (t), |
t I is a limited |
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lim y (t) = y |
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= 0. |
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function, we get lim (2 y |
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As ( ) = 0, only when = 0, we get |
1 |
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Thus, equilibrium states x* = 0, * = 0 of the system (1.71) are absolutely stable |
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in the sector [ , |
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= 1 , > 0, > 0 is an arbitrarily small number. |
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0 |
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(0;1) |
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Then in the |
sector |
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Iserman’s problem has a solution, |
0max = 0 = 1. |
Consequently, the necessary and sufficient condition for absolute stability is obtained.
6. We study absolute stability of equilibrium states of the system (1.71) |
by |
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~ |
~ |
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applying the theorem of V.M. Popov [15]. The ransfer function from input |
to |
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( ) |
the output ~ is equal to
48
Chapter 1. Absolute stability and Aizerman’s problem of one-dimensional regulated systems
|
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p 2 |
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~ |
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W ( p) = S( A pI |
) |
1 |
B = |
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= |
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, |
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2 |
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~ |
~ |
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3 |
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p( p |
p |
1) |
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( ) |
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where |
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~ |
~ |
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are function images (t), |
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p is a complex variable. |
( p), ( p) |
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respectively. It is easy to verify that |
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y ( p) = |
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According to the frequency condition of V.M. Popov [15] the value mined from the condition
( (t)), t I |
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1) |
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Re(1 i q)W (i ) 0, < < , |
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where |
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W (i ) = ReW (i ) iImW (i ), |
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ReW (i ) = |
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ImW (i ) = |
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Modified frequency response |
W* (i ) = X iY , |
X = ReW (i ), |
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For variables |
X ,Y |
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frequency condition of |
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V.M. Popov |
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> 0 for all |
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0 < . As |
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X = qY 0 |
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X = |
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0 < , |
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then the value = |
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0 for which Y = 0, |
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1 = 2/3. |
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value |
X ( 1) = |
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33/7. |
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where the value 0 = 1 is obtained by the method proposed above.
To obtain a modified frequency response we calculate the values X ( ), Y ( ) :
1. |
= 0, |
X (0) = 3, |
Y (0) = 2; |
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2.= 0.1, (0.1) = −3.406, (0.1) = 1.868;
3.= 0.4, (0.4) = −4.474, (0.4) = 1.053;
4.= 0.6, (0.6) = −4.737, (0.6) = 0.263;
5.= 2/3, (2/3) = −4.714, (2/3) = 0;
6.= 0.7, (0.7) = −4.683, (0.7) = −0.126;
7.= 0.8, (0.8) = −4.523, (0.8) = −1.476;
8.= 1, X (1) = 4, Y (1) = 1;
9.= , X ( ) = 0, Y ( ) = 0;
49
Lectures on the stability of the solution of an equation with differential inclusions
Based on these data, we built modified frequency response and found values
1 = 33/7,
0
q = 0. Consequently, the value 0 according to the frequency condition
of V.M. Popov is equal to 0 = 337 = 0.212 and the value 0 found by the proposed method is equal to 0 = 1 , > 0 is an arbitrarily small number. This means that
the proposed method allows us to outline, in the space of system parameters, the area of absolute stability wider than that given by the known methods (method of A.I. Lurie, method of V.M. Popov). It is shown, that the frequency condition of V.M. Popov is not effective for the study of absolute stability of regulated systems in a simple critical case with limited nonlinearity.
Aizerman’s problem. The solution to Aizerman’s problem was obtained from
the boundedness of the improper integral |
I1 |
along the solution of the system. Notice |
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Consequently, |
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where | c1 l1 | | c1 | | l1 |< .
Example 2. We consider the equations of the following kind
̇ = |
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− 2 + ( ), ̇= ( ), = −0.2 |
+ 0.4 |
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This |
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1 = (−0.2 |
0.4 |
−0.5). Matrix |
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50