Bài toán điều khiển được hệ phương trình rời rạc tuyến tính - Pdf 30

class="bi x0 y0 w1 h1"
class="bi x0 y1 w1 h2"
class="bi x0 y2 w2 h3"
class="bi x0 y3 w2 h4"

R
n
n
Z
+
A A
I
A

A
|A| A
F

(M) F M
A A
˙x(t) = t (t, x(t), u(t)) t ≥ 0,
x(k + 1) = f (k, x(k), u(k)) , k ∈ Z
+
,
x(.) u(.)
class="bi x0 y4 w2 h5"
class="bi x0 y4 w2 hd"
class="bi x0 y4 w4 he"
x
0
, x

, u(0)) , u(1)) , u(2))
. . . .
f (k, x(k), u(k))
f (k, x(k), u(k))
f (k, x(k), u(k)) = A(k)x(k) + B(k)u(k), k ∈ Z
+
.
x(k + 1) = A(k)x(k) + B(k)u(k), k ∈ Z
+
.
x(0) = x
0
u
k
= (u(0), u(1), . . . , u(k − 1)) ,
x(k) k > 0
x(k) = F (k, 0)x
0
+
k−1

s=0
F (k, s + 1)B(s)u(s),
F (k, s)
x(k + 1) = A(k)x(k), k ∈ Z
+
.
F (k, s)
F (k, s) = A(k − 1) × . . . × A(s), k ≥ s ≥ 0,
F (k, k) = I.

0
u(k)
x (k, x
0
, u) k > 0
x (k, x
0
, u) = F (k, 0)x
0
+
k−1

i=0
F (k, i + 1)B(i)u(i),
F (k, i)



F (k, i) = A(k − 1)A(k − 2) . . . A(i), k > i,
F (k, k) = I.
x
0
, x
1
∈ R
n
(x
0
, x
1

V (0) ⊂ R
n
V (0)
x
1
∈ R
n
k
1
> 0 (0, x
1
) k
1
0
x
0
∈ R
n
k
1
> 0 (x
0
, 0)
k
1
u
k
= (u(0), u(1), . . . , u(k − 1)) ∈ R
km
.


k>0
C
k
,

0
= {0} .
 = R
n
0 C = R
n
0 ∈ int 
0 0 ∈ int C
X
ρ(x, y) : X × X → R
+
ρ(x, y) = 0 x = y.
ρ(x, y) = ρ(y, x).
ρ(x, y) ≤ ρ(x, z) + ρ(z, y).
{x
n
} α > 0
N > 0 n, m > N ρ (x
n
, x
m
) < α
(X, ρ)
X

L(X, Y )
A = sup
x≤1
Ax .
Y = R A ∈ L(X, Y )
X X

X x

, x
x

(x) x

∈ X

x ∈ X
X M ⊂ X M
x, y ∈ M λ ∈ [0, 1]
λx + (1 − λ)y ∈ M.
M M M
X = R
n
M =

y =
n

i=1
a

∈ R, x
i
∈ M

.
X M ⊂ X
λM ⊂ M, ∀λ > 0.
M 0 ∈ M M
M = {δx : ∀δ > 0, ∀x ∈ M}.
M M
+
M
+
= {x

∈ X : x

, x ≥ ∀x ∈ M} .
M =

(x
1
, x
2
) ∈ R
2
: x
1
≥ 0, x
2

2
= 0

M
+
=

(x
1
, x
2
) ∈ R
2
: x
1
≤ 0, x
2
∈ R

.
M ⊂ X

¯
M

+
= M
+
M
+

i
A
j
= A
j
A
i
N
A
i
(int M) ⊂ int M, ∀i ∈ N.
M = X.
f

∈ M
+
⊂ X

{A

i
}
i∈N
A

i
f

= λ
i

n
, u(k) ∈ R
m
,
A(k), B(k) (n × n) (n × m)
C(k) = [F (k, k)B(k − 1), F (k, k − 1)B(k − 2), . . . , F (k, 1)B(0)] ; k ∈ Z
+
,
F (k, s)
x(k + 1) = A(k)x(k), k ∈ Z
+
.
k
0
≥ 1
C(k
0
) = n.
L
k
u
k
=
k−1

i=0
F (k, i + 1)B(i)u(i).
L
k
: R

k
0

R
k
0
m

= R
n
= 
k
0
,
(n × n)
D(k
0
) = C(k
0
)C

(k
0
)
D
−1
(k) x
1
∈ R
n

−1
(k
0
) x
1
= D (k
0
) D
−1
(k
0
) x
1
= x
1
.
0 x
1
∈ R
n

A, B
k
0
≥ 1

B, AB, , A
k
0
−1

(k) − x
2
(k) + u(k).
A =


1 1
1 −1


, B =


1
1


⇒ AB =


1 1
1 −1




1
1



(k + 1) = x
1
(k) − x
2
(k) + u(k).
A =


1 −1
1 −1


, B =


1
1


⇒ AB =


1 −1
1 −1




1
1

k
= [−F (k, 0)]

(
k
) ,
F

(M) M
F

(M) = {x ∈ R
n
: F x ∈ M} .
k
0
> 0
C
k
0
= R
n
[−F (k
0
, 0)]

(
k
0
) = R


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