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Viết đề tài giá sinh viên – ZALO:0973.287.149-
TEAMLUANVAN.COM
ĐẠI HỌC THÁI NGUYÊN
TRƯỜNG ĐẠI HỌC SƯ PHẠM
––––––––––––––––––––
NGUYỄN MINH TRANG
BÀI TOÁN ỔN ĐỊNH HÓA
HỆ PHƯƠNG TRÌNH VI PHÂN
PHI TUYẾN CÓ TRỄ
Chuyên ngành: Toán giải tích
Mã số: 60.46.01.02
LUẬN VĂN THẠC SĨ TOÁN HỌC
NGƯỜI HƯỚNG DẪN KHOA HỌC: GS. TSKH. VŨ NGỌC PHÁT
THÁI NGUYÊN - 2016

2. i
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L i cam oan
T i xin cam oan n i dung trong lu n v n Th c s chuy n ng nh To n gi i t ch
v i t i "B i to n n nh h a h ph ng tr nh vi ph n phi tuy n
c tr " c ho n th nh b i nh n th c c a t i, kh ng tr ng l p v i lu n
v n, lu n n v c c c ng tr nh c ng b .
Th i Nguy n, th ng 4 n m 2016
Ng i vi t Lu n v n
Nguy n Minh Trang

3. ii
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L i c m n
T i xin b y t l ng bi t n t i GS. TSKH V Ng c Ph t, ng i nh
h ng ch n t i v t n t nh h ng d n, cho t i nh ng nh n x t qu b u
t i c th ho n th nh lu n v n.
T i c ng xin b y t l ng bi t n ch n th nh t i ph ng Sau i h c, c c
th y c gi o d y cao h c chuy n ng nh To n gi i t ch tr ng i h c s
ph m - i h c Th i Nguy n gi p v t o i u ki n cho t i trong su t
qu tr nh h c t p v nghi n c u khoa h c.
Nh n d p n y t i c ng xin g i l i c m n ch n th nh t i gia nh, b n b
lu n ng vi n, c v , t o m i i u ki n thu n l i cho t i trong su t qu
tr nh h c t p.
Th i Nguy n, th ng 4 n m 2016
Ng i vi t lu n v n
Nguy n Minh Trang

4. iii
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M c l c
L i cam oan i
L i c m n ii
M c l c ii
M u 1
M t s k hi u vi t t t 3
1 C s to n h c 4
1.1 H ph ng tr nh vi ph n i u khi n . . . . . . . . . . . . . . . 4
1.2 B i to n n nh h a . . . . . . . . . . . . . . . . . . . . . . . 5
1.2.1 Ph ng ph p h m Lyapunov . . . . . . . . . . . . . . . 8
1.2.2 B i to n n nh h a . . . . . . . . . . . . . . . . . . . 9
1.3 C c b b tr . . . . . . . . . . . . . . . . . . . . . . . . . 9
2 B i to n n nh h a h ph ng tr nh vi ph n phi tuy n c
tr 11

5. iv
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2.1 H ph ng tr nh vi ph n c tr ..............................................11
2.2 H ph ng tr nh vi ph n phi tuy n t n m c tr....................14
2.3 H ph ng tr nh vi ph n phi tuy n kh ng t n m c tr...........27
K t lu n chung 45
T i li u tham kh o 46

6. 1
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M u
Trong l thuy t nh t nh c c h ng l c, b i to n n nh v n nh
h a c vai tr r t quan tr ng. S nghi n c u b i to n n nh h th ng tr
th nh m t h ng nghi n c u kh ng th thi u trong l thuy t ph ng tr nh
vi ph n, l thuy t h th ng v ng d ng.
T nh n nh l m t trong nh ng t nh ch t quan tr ng c a l thuy t nh
t nh c c h ng l c v c s d ng nhi u trong c c l nh v c c h c, v t
l to n, k thu t,... N i m t c ch h nh t ng, m t h th ng c g i l n
nh t i tr ng th i c n b ng n o n u c c nhi u nh c a c c d ki n ho c
c u tr c ban u c a h th ng kh ng l m cho h th ng thay i nhi u so
v i tr ng th i c n b ng . S nghi n c u b i to n n nh h th ng c
b t u t cu i th k XIX b i nh to n h c V. Lyapunov v n nay tr
th nh m t h ng nghi n c u kh ng th thi u trong l thuy t ph ng tr nh
vi ph n, l thuy t h th ng v ng d ng. T nh ng n m 60 c a th k XX,
song song v i s ph t tri n c a l thuy t i u khi n v do nhu c u nghi n
c u c c t nh ch t nh t nh c a h th ng i u khi n, ng i ta b t u nghi n
c u t nh n nh c c h i u khi n d ng ẋ(t) = f(t, x(t), u(t)), t ≥ 0(0.1) b i
to n n nh h a c a h l t m h m i u khi n ng c: u(t, x) = h(t, x) sao

7. 2
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cho h ng l c ẋ(t) = f(t, x(t), h(t, x(t))) = F(t, x(t)) l n nh ho c n
nh ti m c n t i tr ng th i c n b ng. Trong c c b i to n n nh h a t ng
qu t, h i u khi n (0.1) th ng c m h nh h a v i c c t c ng c a i u
khi n ng c, c a c c nhi u i u khi n v quan s t,... Nh v y m c ch c a
v n n nh h a m t h th ng i u khi n l t m c c h m i u khi n ng c
sao cho h th ng cho ng v i i u khi n tr th nh h th ng n nh
c t i tr ng th i c n b ng. C s to n h c c a b i to n n nh h a l l
thuy t n nh Lyapunov. D a tr n nh ng k t qu bi t c a t nh n nh
Lyapunov ng i ta nghi n c u, ph t tri n v ng d ng v o gi i b i to n
n nh h a c c h th ng i u khi n.
N i dung c a b n lu n v n c tr nh b y trong hai ch ng.
Ch ng 1 tr nh b y c s to n h c h ph ng tr nh vi ph n i u khi n,
ph ng ph p h m Lyapunov trong l thuy t n nh, b i to n n nh h a
v c c b li n quan.
Ch ng 2 tr nh b y b i to n h ph ng tr nh vi ph n phi tuy n c tr , h
ph ng tr nh vi ph n phi tuy n t n m c tr , h ph ng tr nh vi ph n phi
tuy n kh ng t n m c tr .

8. 3
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√
M t s k hi u vi t t t
R+
T p h p c c s th c kh ng m.
Rn
Kh ng gian Euclid n chi u.
< x, y > ho c xT
y T ch v h ng c a 2 v ct x, y.
ǁxǁ Chu n v ct Euclid c a x.
Rn×r
Kh ng gian c c ma tr n n × r chi u.
AT
Ma tr n chuy n v c a A.
I Ma tr n ng nh t.
λ(A) Gi tr ri ng c a A.
λmax(A) = max{Reλ : λ ∈ λ(A)}.
η(A) Chu n ph c a ma tr n c x c nh b i:
η(A) = λmax(AT A).
µ(A) o c a ma tr n A x c nh b i :
1
µ(A) = λ
2
max(A + AT
).
L2([0, t], Rn
) Kh ng gian kh t ch b c 2 tr n [0, t] gi tr trong Rn
.
A ≥ 0 Ma tr n x c nh kh ng m.
A > 0 Ma tr n x c nh d ng.
C([−h, 0], Rn
) Kh ng gian c c h m li n t c tr n [−h, 0] gi tr trong Rn
.
ǁxtǁ = sups∈[−h,0]ǁx(t + s)ǁ.
BM +
(0, ∞) T p h p c c h m ma tr n x c nh kh ng m
v b ch n tr n [0, ∞).

9. 4
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Ch ng 1
C s to n h c
Ch ng n y tr nh b y m t s ki n th c c s to n h c v h ph ng tr nh
vi ph n i u khi n, ph ng ph p h m Lyapunov, b i to n n nh h a v
c c b b tr . N i dung ch ng n y c tr nh b y t t i li u [1], [2].
1.1 H ph ng tr nh vi ph n i u khi n
H ph ng tr nh i u khi n m t b i ph ng tr nh vi ph n hay r i r c
d ng:
ẋ(t) = f(t, x(t), u(t)), t ≥ 0,
x(k + 1) = f(k, x(k), u(k)), k = 0, 1, 2, ...
trong x(t)(x(k)) ∈ Rn
l v ct tr ng th i, u(t)(u(k)) ∈ Rm
, n ≥ m, l
v ct i u khi n v h m f (t, x, u) : R+
× Rn
× Rm
→ Rn
. C c i t ng i u
khi n trong c c m h nh i u khi n h ng l c c m t nh nh ng d
li u u v o c t c ng quan tr ng, m c n y ho c m c kh c, c th
l m nh h ng n s v n h nh u ra c a h th ng. Nh v y, ta hi u m t

10. 5
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x˙= f(t, x, u)
h th ng i u khi n l m t m h nh to n h c c m t b i ph ng tr nh
to n h c bi u th s li n h v o - ra :
u(t) → → x(t).
M t trong nh ng m c ch ch nh c a b i to n i u khi n h th ng l t m
i u khi n ( u v o) sao cho h th ng ( u ra) c nh ng t nh ch t m ta
mong mu n. Th ng th ng, vi c chuy n m t h th ng c i u khi n t v
tr n y sang v tr kh c c th th c hi n b ng nhi u ph ng ph p d i t c
ng b i c c i u khi n kh c nhau. C n c v o nh ng m c ch c th c a
h th ng u ra ng i ta x c nh c c b i to n i u khi n kh c nhau: b i
to n i u khi n c, b i to n n nh h a, b i to n i u khi n t i u, v.v...
Trong lu n v n n y ch ng ta ch x t b i to n n nh h a.
1.2 B i to n n nh h a
B i to n n nh h a l b i to n n nh ( n nh Lyapunov) c c h i u
khi n. Do c s to n h c c a b i to n n nh h a l l thuy t n nh
Lyapunov. D a tr n nh ng k t qu bi t c a t nh n nh Lyapunov ng i
ta nghi n c u, ph t tri n v ng d ng v o gi i b i to n n nh h a c c
h th ng i u khi n. T nh n nh l m t trong nh ng t nh ch t quan tr ng
c a l thuy t nh t nh c c h ng l c v c s d ng nhi u trong c c l nh
v c c h c, v t l to n,... N i m t c ch h nh t ng, m t h th ng c g i
l n nh t i m t tr ng th i c n b ng n o n u c c nhi u nh c a c c d
ki n ho c c u tr c ban u c a h th ng kh ng l m cho h th ng thay i

11. 6
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ǁ − ǁ
nhi u so v i tr ng th i c n b ng . S nghi n c u b i to n n nh h th ng
c b t u t cu i th k XIX b i nh to n h c V. Lyapunov v n nay
tr th nh m t h ng nghi n c u kh ng th thi u trong l thuy t ph ng
tr nh vi ph n, l thuy t h th ng v ng d ng. Tr c ti n ta ph i t b i
to n n nh ( n nh Lyapunov) cho h ng l c kh ng c i u khi n. X t
h ph ng tr nh vi ph n
ẋ(t) = f(t, x(t)), t ≥ 0. (1.1)
nh ngh a 1.2.1. Nghi m x(t) c a h (1.1) g i l n nh n u v i m i
s ϵ > 0, t0 ≥ 0 s t n t i δ > 0 (ph thu c ϵ, t0) sao cho b t k nghi m
y(t), y(t0) = y0 c a h th a m n ǁy0 − x0ǁ < 0 th s nghi m ng b t ng
th c
ǁy(t) − x(t)ǁ < ϵ, ∀t ≥ t0.
N i c ch kh c, nghi m x(t) l n nh khi m i nghi m kh c c a h c gi tr
ban u g n v i gi tr ban u c a x(t) th v n g n n trong su t
th i gian t ≥ t0.
nh ngh a 1.2.2. Nghi m x(t) c a h (1.1) g i l n nh ti m c n n u n
l n nh v c m t s δ > 0 sao cho v i ǁy0 − x0ǁ < δ th
lim y(t) x(t) = 0.
t→∞
Ngh a l , nghi m x(t) l n nh ti m c n n u n n nh v m i nghi m y(t)
kh c c gi tr ban u y0 g n v i gi tr ban u x0 s ti n g n t i x(t) khi
t ti n t i v c ng.

12. 7
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ǁ ǁ
Nh n x t r ng b ng ph p bi n i (x − y) ›→ z, (t − t0) ›→ τ h ph ng tr nh
(1.1) s c a v d ng
z˙ = F(τ, z), (1.2)
trong F (τ, 0) = 0 v khi s n nh c a m t nghi m x(t) n o c a
h (1.1) s c a v nghi n c u t nh n nh c a nghi m 0 c a h (1.2).
ng n g n, t nay ta s n i h (1.2) l n nh thay v o n i nghi m 0 c a
h l n nh. Do t b y gi ta x t h (1.1) v i gi thi t h c nghi m 0,
t c l , f(t, 0) = 0, t ∈ R+
. Ta n i :
H (1.1) l n nh n u v i b t k ϵ > 0, t0 ∈ R+
s t n t i s δ > 0 (ph
thu c v o ϵ, t0 ) sao cho b t k nghi m x(t): x(t0) = x0 th a m n ǁx0ǁ < δ
v i m i t ≥ t0 th ǁx(t)ǁ < ϵ, ∀t ≥ 0.
H (1.1) l n nh ti m c n n u h l n nh v c m t s δ > 0 sao cho
n u ǁx0ǁ < δ th
lim x(t) = 0.
t→∞
nh ngh a 1.2.3. H (1.1) l n nh m n u t n t i c c s M > 0, δ > 0
sao cho m i nghi m c a h (1.1) v i x(t0) = x0 th a m n
ǁx(t)ǁ ≤ Me−δ(t−t0)
ǁx0ǁ, ∀t ≥ t0.
i u n y c ngh a l nghi m 0 c a h kh ng nh ng n nh ti m c n m
nghi m c a n ti n t i 0 nhanh v i t c theo h m s m .
B i to n n nh h a c a h (1.1) l t m h m i u khi n (c th ph
thu c v o bi n tr ng th i m ng i ta th ng g i l h m i u khi n ng c):

13. 8
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+ n
∂x
u(t) = h(t, x(t)) sao cho h ng:
ẋ(t) = f(t, x(t), h(t, x(t))) = F(t, x(t))
l n nh ti m c n (ho c n nh m ). Nh v y m c ch c a v n n
nh h a h th ng i u khi n l t m c c h m i u khi n ng c sao cho h
th ng cho ng v i i u khi n tr th nh h th ng n nh.
1.2.1 Ph ng ph p h m Lyapunov
gi i b i to n n nh c c h phi tuy n ng i ta hay d ng ph ng ph p
h m Lyapunov. Ph ng ph p n y d a v o s t n t i c a m t l p h m tr n
c bi t g i l h m Lyapunov m t nh n nh c a h c th tr c ti p qua
d u c a o h m theo nghi m (h m v ph i) c a h cho.
X t h ph ng tr nh vi ph n phi tuy n (1.1). Tr c h t ta x t l p h m K
l t p c c h m li n t c t ng ch t a(.) : R+
→ R+
, a(0) = 0. H m V (t, x) :
R+
× Rn
→ R g i l h m Lyapunov n u:
i) V (t, x) l h m x c nh d ng theo ngh a
∃a(.) ∈ K : V (t, x) ≥ a(ǁxǁ), ∀(t, x) ∈ R × R .
ii) o h m theo nghi m l kh ng m:
∂V
Df V (t, x) =
∂t
+
∂V
f(t, x) ≤ 0, ∀(t, x) ∈ R+
× Rn
.
Tr ng h p V (t, x) l h m Lyapunov v th a m n th m i u ki n:
iii) ∃a(.) ∈ K : V (t, x) ≤ b(ǁxǁ), ∀(t, x) ∈ R+
× Rn
.
iv) ∃γ(.) ∈ K : Df V (t, x) ≤ −γ(ǁxǁ), ∀x ∈ Rn
0,
th g i l h m Lyapunov ch t.

14. 9
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nh l 1.2.4. N u h phi tuy n kh ng d ng (1.1) c h m Lyapunov th h
l n nh. N u h m Lyapunov l ch t th h l n nh ti m c n.
1.2.2 B i to n n nh h a
X t h i u khi n m t b i h ph ng tr nh vi ph n
ẋ(t) = f(t, x(t), u(t)), t ≥ 0,
(1.3)
x(0) = x0.
nh ngh a 1.2.5. H (1.3) g i l n nh h a c n u t n t i h m h(x) :
Rn
→ Rm
sao cho h ng :
x(t) = f(t, x(t), h(x(t))), t ≥ 0,
l n nh ti m c n (ho c n nh m ). H m u(t) = h(x(t)) th ng g i l
h m i u khi n ng c.
Tr ng h p h (1.3) l h tuy n t nh x˙= Ax + Bu th h l n nh ho
c n u t n t i ma tr n K sao cho h ng ẋ(t) = (A +BK)x(t) l n nh
ti m c n, ho c n i c ch kh c l n u ma tr n (A + BK) l n nh (t.l. gi
tr ph n th c c a c c gi tr ri ng c a ma tr n l m).
1.3 C c b b tr
B 1.3.1. ( B t ng th c ma tr n Cauchy).

15. 10
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X ZT
∫ ∫
≥
∫
∫ ∫
(i) Gi s S ∈ Rn×n
l m t ma tr n i x ng x c nh d ng v Q ∈ Rn×n
,
ta c :
2 < Qy, x > − < Sy, y >≤< QS−1
QT
x, x >, ∀y, x ∈ Rn
.
(ii) Gi s N ∈ Rn×n
l m t ma tr n i x ng x c nh d ng, ta c :
±2xT
y ≤ xT
Nx + yT
N−1
y, ∀x, y ∈ Rn
.
B 1.3.2. Cho ma tr n h ng Z = ZT
> 0 b t k v c c i l ng h, h, 0 <
h < h sao cho c c t ch ph n sau l x c nh th ta c :
(i)
t 1 t
xT
(s)Zx(s)ds (
h
t
x(s)ds)T
Z( x(s)ds);
t−h
(ii)
−
∫h ∫
t
t−h
xT
(τ )Zx(τ )dτds ≥
2
(
h2 − h2
t−h
−
∫h ∫t
x(τ )dτds)T
−h t
Z( x(τ)dτds.
−h t+s
B 1.3.3. ( B Schur)
−h t+s −h t+s
Cho c c ma tr n X, Y, Z, trong Y = Y T
> 0, X = XT
, ta c
X + ZT
Y −1
Z < 0 ⇔
Z −Y
< 0.

16. 11
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ǁ ǁ
ǁ ǁ
Ch ng 2
B i to n n nh h a h ph ng tr nh
vi ph n phi tuy n c tr
Ch ng n y tr nh b y b i to n n nh h a h ph ng tr nh vi ph n phi
tuy n t n m c tr v h ph ng tr nh vi ph n phi tuy n kh ng t n m c
tr . N i dung tr nh b y t t i li u [3], [4].
2.1 H ph ng tr nh vi ph n c tr
Cho r ∈ R, r > 0, C([a, b], Rn
) l kh ng gian c c h m li n t c nh x t
[a,b] v o Rn
v i chu n
trong φ ∈ C[a, b].
φ = sup
t∈[a,b]
ǁφ(t)ǁ,
N u [a, b] = [−r, 0], ta t C = C([−r, 0], Rn
) v i chu n trong C
φ c = sup
t∈[−r,0]
ǁφ(t)ǁ.

17. 12
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V i t0 ∈ R, A ≥ 0 v x ∈ C([t0 −r, t0 +A], Rn
), th v i b t k t ∈ [t0, t0 +A]
ta x c nh c xt ∈ C nh sau: xt(θ) = x(t + θ), −r < θ < 0.
D ng t ng qu t c a h ph ng tr nh vi ph n c tr l
ẋ(t) = f(t, xt), (2.1)
trong x(t) ∈ Rn
, f : R × C → Rn
, xt : C → Rn
.
T (2.1) ta th y r ng o h m c a bi n tr ng th i x t i t ph thu c v o
t v x(s) v i t − r ≤ s < t. Nh v y, x c nh c tr ng th i x(t) th i
i m t ≥ t0 ta c n bi t tr ng th i ban u tr n kho ng d i r, t c l
xt0 = φ, (2.2)
trong φ ∈ C. Hay x(t0 + φ) = φ(θ), −r ≤ θ ≤ 0.
V i m t s A > 0, m t h m x c g i l nghi m c a (2.1) tr n [t0 −r, t0 +
A] n u trong kho ng n y x l h m li n t c v th a m n RFDE (Retarded
Functional Differential Equation) (2.1) v (t, xt) n m trong mi n x c nh
c a h m f . N u nghi m c ng th a m n i u ki n ban u (2.2) ta n i n l
nghi m c a ph ng tr nh RFDE (2.1) v i i u ki n ban u (2.2), hay n
gi n l nghi m t i (t0, φ).
Ta k hi u x(t0, φ, f) khi c n ch r nghi m c a ph ng tr nh (2.1) v i
i u ki n ban u (t0, φ). Gi tr c a x(t0, φ, f ) t i t k hi u x(t; t0, φ, f ). N u
kh ng g y nh m l n ta s t m qu n i f v vi t x(t0, φ) ho c x(t; t0, φ).
T ng t nh ph ng tr nh vi ph n th ng ta c ng c c ng th c nghi m
d ng t ch ph n c a h (2.1) v (2.2) l
xt0 = φ

18. 13
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∫
+
ǁ ǁ
f t
∂x
t t
t
x(t) = φ(0) +
t0
f(s, xs)ds, t ≥ t0.
Ta c ng c nh ngh a v n nh cho h ph ng tr nh vi ph n phi tuy n
c tr t ng t nh cho h ph ng tr nh vi ph n th ng.
nh ngh a 2.1.1. Nghi m x(t) = 0 c a h (2.1) c g i l n nh n u
∀ϵ > 0, ∀t0 > 0 u t n t i δ = δ(t0, ϵ) > 0 sao cho m i nghi m x(t, t0, φ)
th a m n ǁxt0ǁ < 0 th
ǁx(t, t0, φ)ǁ ≤ ϵ, ∀t ≥ t0.
nh ngh a 2.1.2. Nghi m x(t) = 0 c a h (2.1) c g i l n nh ti m
c n n u n n nh v h n n a v i b t k t0 > 0, ∀ϵ > 0, t n t i δa =
δa(t0, ϵ) > 0 sao cho m i nghi m x(t, t0, φ) c a h th a m n ǁxt0ǁ < δa th
lim x(t, t0, φ) = 0.
t→t0
gi i b i to n n nh h ph ng tr nh vi ph n c tr ta s d ng ph ng
ph p h m Lyapunov nh v i ph ng tr nh vi ph n th ng kh ng c tr : H m
V (t, xt) : R+
× C → R g i l h m Lyapunov n u:
i) V (t, xt) l h m x c nh d ng theo ngh a
∃a(.) ∈ K : V (t, xt) ≥ a(ǁxǁ), ∀(t, xt) ∈ R × C.
ii) o h m theo nghi m l kh ng m:
∂V
D V (t, x ) = +
∂V
f(t, x ) ≤ 0, ∀(t, x ) ∈ R+
× C.
∂t

19. 14
Tải tài liệu tại sividoc.com
(2.3)
Tr ng h p V (t, xt) l h m Lyapunov v th a m n th m i u ki n:
iii) ∃a(.) ∈ K : V (t, xt) ≤ b(ǁxtǁ), ∀(t, xt) ∈ R+
× C.
iv) ∃γ(.) ∈ K : Df V (t, xt) ≤ −γ(ǁxtǁ), ∀x ∈ Rn
0,
th g i l h m Lyapunov ch t.
Ta c ti u chu n n nh cho h ph ng tr nh vi ph n c tr sau:
nh l 2.1.3. N u t n t i h m Lyapunov V (t, xt) v c c s λ1, λ2 > 0 sao
cho th a m n c c i u ki n sau:
(i) λ1ǁx(t)ǁ2
≤ V (t, xt) ≤ λ2ǁxtǁ2
,
(ii) V˙ (t,xt) ≤ 0,
th m i nghi m x(t) c a h l b ch n.
∃N > 0 : ǁx(t, φ)ǁ ≤ Nǁφǁ, ∀t ≥ 0.
H n n a, n u i u ki n (ii) c thay th b i
(iii) ∃λ3 > 0 : V˙ (t, xt) ≤ −2λ3V (t, xt)
th nghi m 0 c a h l to n b s n nh h m s m
∃N > 0 : ǁx(t, φ)ǁ ≤ Nǁφǁe−λ3t
, ∀t ≥ 0.
2.2 H ph ng tr nh vi ph n phi tuy n t n m c tr
X t h ph ng tr nh vi ph n phi tuy n t n m c tr
ẋ(t) = Ax(t) + Dx(t − h(t)) + f(x(t)) + g(x(t − h(t)) + Bu(t), t ≥ 0,
x(t) = φ(t), t ∈ [−h2, 0],

20. 15
Tải tài liệu tại sividoc.com
trong x(t) ∈ Rn
l v ct tr ng th i, u(t) ∈ Rm
l v ct i u khi n, A, D v
B l c c ma tr n c s chi u t ng ng th ch h p v φ(t) ∈ C1
([−h2, 0]; Rn
)
l h m ban u v i chu n ǁφǁC1 = max{supt∈[−h2,0]ǁφ(t)ǁ, supt∈[−h2,0]ǁφ̇(t)ǁ}.
H m tr h(t), th a m n i u ki n sau:
0 < h1 ≤ h(t) ≤ h2, t ≥ 0, (2.4)
h1, h2 l c c h ng s tr cho tr c.
nh ngh a 2.2.1. Cho α > 0. H (2.3) v i u(t) = 0 l α - n nh n u t n
t i m t s d ng β > 0 sao cho m i nghi m x(t, φ) c a h th a m n i u
ki n sau:
ǁx(t, φ)ǁ ≤ βe−αt
ǁφǁC1, ∀t ∈ R+
.
nh ngh a 2.2.2. Cho α > 0. H (2.3) l α - n nh h a n u t n t i h m
i u khi n ng c u(t) = Kx(t) sao cho h ng
ẋ(t) = (A + BK)x(t) + Dx(t − h(t)) + f(x(t)) + g(x(t − h(t))), (2.5)
l α - n nh.
ph t bi u nh l ta a v o c c k hi u (v n t t) sau:
λ =λmin(P −1
),
A =λmax(P −1
) + h1λmax(P −1
Q1P −1
) + h2λmax(P −1
Q2P −1
)
+ (h2 − h1)λmax(P −1
Q3P −1
)
+
1
h2
[λ (P −1
S P −1
) + λ (P −1
S P −1
)]
2 1 max 1 max 3

21. 16
Tải tài liệu tại sividoc.com
−
h2
−
6 1
6 2 1
2 1
77 1 2 1 2
2 1
2 2 1
2
2
2
1
2
1 2 2 −1 −1 −1
+
2
(h2 − h1)[λmax(P S2P ) + λmax(P −1S4P )]
+
1
h3
λ (P −1
R P −1
) +
1
(h3
− h3
)λ (P −1
R P −1
),
E11 =AP + PAT
+ BY + Y T
BT
+ 2αP + 2I + Q1 + Q2
+ h1S3 + (h2 − h1)S4
1
e−2αh1
S
h1
1
— 2e−4αh1
R1 2
h2 − h1
e−4αh2
R ,
h2 + h1 2
E22 = − (1 − δ)e−2αh2
Q2
2
e−2αh2
S ,
h2 − h1
E33 = −e−2αh2
Q3
1
e−2αh2
S ,
h2 − h1
E44 =e−2αh1
Q3 − e−2αh1
Q1
1
e−2αh1
S
h1
1
1
e−2αh2
S ,
h2 − h1
E55 =
1
e−2αh1
S
h1
3
2
e−4αh1
R ,
1
E66 =
1
e−2αh2
S
h2 − h1
2
e−4αh2
R ,
h2
− h2
E =h S
1 2
+ (h − h )S + h R
1 2 2
+ (h − h )R + 2I − 2P,
E14 =
1
e−2αh1
S
h1
1, E15 =
2
e−4αh1
R
h1
1, E16
2
=
h2 + h1
e−4αh2
R2,
E17 =PAT
+ Y T
BT
, E23 = E24
1
=
h2 − h1
e−2αh2
S2.
nh l 2.2.3. Cho α > 0, h2 > h1 > 0, 0 ≤ δ < 1. H (2.3) l α - n nh
h a n u t n t i ma tr n i x ng x c nh d ng P, Q1, Q2, Q3, S1, S2, S3, S4,
− −
−
−
− −
−
−
max 1 max 2
1 1 2
4

22. 17
Tải tài liệu tại sividoc.com
2
Σ
+
2
R1, R2, b t k ma tr n Y sao cho b t ng th c ma tr n (LMI) sau th a m n:
E11 DP 0 E14 E15 E16 E17 a2
PFT
0
∗ E22 E23 E24 0 0 PDT
0 d2
PGT
∗ ∗ E33 0 0 0 0 0 0
∗ ∗ ∗ E44 0 0 0 0 0
∗ ∗ ∗ ∗ E55 0 0 0 0
∗ ∗ ∗ ∗ ∗ E66 0 0 0
∗ ∗ ∗ ∗ ∗ ∗ E77 0 0
∗ ∗ ∗ ∗ ∗ ∗ ∗ −1
a2
I 0
< 0. (2.6)
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ −1
d2
I
H n n a, h m i u khi n ng c c x c nh b i :
u(t) = Y P−1
x(t), t ≥ 0
v nghi m x(t, φ) c a h th a m n
ǁx(t, φ)ǁ ≤
Ch ng minh. Ch ng ta k hi u
a
e−αt
λ
ǁφǁC1, t ∈ R .
Qi = P−1
QiP−1
, i = 1, 2, 3, Sj = P−1
SjP−1
, j = 1, 2, 3, 4,
R1 = P−1
R1P−1
, R2 = P−1
R2P−1
.
X t h m Lyapunov-Krasovskii nh sau:
5
V (t, xt) = Vi(t, xt), t ≥ 0,
i=1
r

23. 18
Tải tài liệu tại sividoc.com
∫
∫
∫
∫
∫
∫
∫
V1(t, xt) =xT
(t)P −1
x(t),
t t
V2(t, xt) = e2α(s−t)
xT
(s)Q1x(s)ds +
t−h1 t−h(t)
t
∫
−h1
e2α(s−t)
xT
(s)Q2x(s)ds
+ e2α(s−t)
xT
(s)Q3x(s)ds,
t−h2
0 t
V3(t, xt) = e2α(τ−t)
ẋT
(τ)S1ẋ(τ)dτds +
∫
−h1∫t
e2α(τ−t)
ẋT
(τ)S2ẋ(τ)dτds,
−h1 t+s −h2 t+s
0
V4(t, xt) =
t
e2α(τ−t)
xT
(τ )S3x(τ )dτds +
∫
−h1∫t
e2α(τ−t)
xT
(τ )S4x(τ )dτds,
−h1 t+s −h2 t+s
0
V5(t, xt) =
∫0 ∫t
e2α(τ+s−t)
ẋT
(τ)R1ẋ(τ)dτdsdθ
−h1 θ t+s
∫
−h1∫0 ∫t
e2α(τ+s−t)
ẋT
(τ)R2ẋ(τ)dτdsdθ.
−h2 θ t+s
L y o h m hai v c a V1(t, xt) k o nghi m c a h (2.5) c
V̇1(t, xt) =xT
(t)[P−1
(A + BK) + (A + BK)T
P−1
]x(t)
+ 2xT
(t)P−1
Dx(t − h(t)) + 2xT
(t)P −1
f(t, x(t))
+ 2xT
(t)P−1
g(t, x(t − h(t))). (2.7)
Ta c
2xT
(t)P−1
f(t, x(t)) ≤ xT
(t)P −1
P−1
x(t) + fT
(t, x(t))f(t, x(t))
≤ xT
(t)P −1
P −1
x(t) + a2
xT
(t)FT
Fx(t) (2.8)
+

24. 19
Tải tài liệu tại sividoc.com
∫
∫
t
v
2xT
(t)P −1
g(t, x(t − h(t)))
≤ xT
(t)P −1
P −1
x(t) + gT
(t, x(t − h(t)))g(t, x(t − h(t)))
≤ xT
(t)P −1
P −1
x(t) + d2
xT
(t − h(t))GT
Gx(t − h(t)). (2.9)
p d ng i u ki n (2.7) -(2.9), ch ng ta c
V̇1(t, xt) =xT
(t)[P−1
(A + BK) + (A + BK)T
P−1
+ 2P−1
P−1
+ a2
FT
F]x(t) + d2
xT
(t − h(t))GT
Gx(t − h(t)). (2.10)
Ti p t c l y o h m 2 v c a Vi(t, xt), i = 2, 3, 4, 5 theo nghi m c a h (2.5),
ta c
V̇2(t, xt) ≤ − 2αV2(t, xt) + xT
(t)[Q1
+ Q2]x(t) − e−2αh1
xT
(t − h1)Q1x(t − h1)
— (1 − δ)e−2αh2
xT
(t − h(t))Q2x(t − h(t))
+ e−2αh1
xT
(t − h1)Q3x(t − h1)
— e−2αh2
xT
(t − h2)Q3x(t − h2); (2.11)
V̇3(t, xt) ≤ − 2αV3(t, xt) + ẋT
(t)[h1S1 + (h2 − h1)S2]ẋ(t)
— e−2αh1
ẋT
(s)S1ẋ(s)ds
t−h1
— e−2αh2
t−h1
ẋT
(s)S2ẋ(s)ds; (2.12)
t−h2

25. 20
Tải tài liệu tại sividoc.com
∫
∫
∫
∫ ∫
∫
1 1 2 1 2
2 1 1 +
2
( 2 1 2
1
1
t
V̇4(t, xt) ≤ −2αV4(t, xt) + xT
(t)[h1S3 + (h2 − h1)S4]x(t)
— e−2αh1
xT
(s)S3x(s)ds
t−h1
— e−2αh2
t−h1
xT
(s)S4x(s)ds; (2.13)
t−h2
T 1 2
V (t, x ) ≤ −2αV (t, x ) + x˙ (t)[ h R h2
− h2
)R ]ẋ(t)
5 t 5 t
2 1
∫0 t
1 +
2
( 2 1 2
— e−4αh1
−h1 t+θ
ẋT
(s)R1ẋ(s)dsdθ
— e−4αh2
−h1 t
ẋT
(s)R2ẋ(s)dsdθ. (2.14)
−h2 t+θ
T c c i u ki n (2.10) n (2.14) suy ra
V˙ (t, xt) + 2αV (t, xt)
≤ xT
(t)[P −1
(A + BK) + (A + BK)T
P −1
+ 2αP−1
+ 2P−1
P −1
+ a2
FT
F + Q1 + Q2 + h1S3 + (h2 − h1)S4]x(t)
+ xT
(t − h(t))[d2
GT
G − (1 − δ)e−2αh2
Q2]x(t − h(t))
+ xT
(t − h1)[e−2αh1
Q3 − e−2αh1
Q1]x(t − h1)
+ xT
(t − h2)[−e−2αh2
Q3]x(t − h2)
1
+ ẋT
(t)[h S + (h − h )S + h2
R h2
− h2
)R ]ẋ(t)
+ 2xT
(t)P −1
Dx(t − h(t)) − e−2αh1
t
ẋT
(s)S1ẋ(s)ds
t−h1

26. 21
Tải tài liệu tại sividoc.com
∫ ∫
∫ ∫
∫ ∫
∫
1
∫
2 1
∫ ∫
∫
2 1
∫
t
— e−2αh2
t−h1
ẋT
(s)S2ẋ(s)ds − e−2αh1
t−h2
t
xT
(s)S3x(s)ds
t−h1
— e−2αh2
t−h1 0
xT
(s)S4x(s)ds − e−4αh1
t
ẋT
(s)R1ẋ(s)dsdθ
t−h2
−h1 t
−h1 t+θ
— e−4αh2
−h2 t+θ
ẋT
(s)R2ẋ(s)dsdθ. (2.15)
p d ng b (1.3.2) cho c c nh gi sau:
— e−2αh1
t−h1
ẋT
(s)S1ẋ(s)ds
1
≤ −
h
[x(t) − x(t − h1
t
∫
−h1
)]T
e−2αh1
S 1[x(t) − x(t − h1 )] (2.16)
— e−2αh2
ẋT
(s)S2ẋ(s)ds
t−h2
t−h(t) t−h1
= e−2αh2
t−h2
ẋT
(s)S2ẋ(s)ds − e−2αh2
∫
t−h(t)
ẋT
(s)S2ẋ(s)ds
t−h(t) t−h(t)
1
≤ −
h − h
( ẋ(s)ds)T
e−2αh2
S2( ẋ(s)ds)
t−h2 t−h2
1
(
h2 − h1
t
∫
−h1
t−h1
ẋ(s)ds)T
e−2αh2
S2( ẋ(s)ds)
t−h(t) t−h(t)
1
= −
h − h
[x(t − h(t)) − x(t − h2
)]T
e−2αh2
S 2[x(t − h(t)) − x(t − h2)]
−

27. 22
Tải tài liệu tại sividoc.com
2 1
∫
∫
∫
∫
∫
∫ ∫
∫
h
∫
h
∫ ∫
∫ ∫
∫
−
∫
∫ ∫ ∫
−
1
—
h − h
[x(t − h1 ) − x(t − h(t))]T
e−2αh2
S 2[x(t − h1 ) − x(t − h(t))],
(2.17)
— e−2αh1
t
xT
(s)S3x(s)ds
t−h1
t
1
≤ −
h1
t
x(s)ds)T
e−2αh1
S3( x(s)ds)
t−h1
t
1
t−h1
∫t
= (
h1
t−h1
x(θ)dθ)T
e−2αh1
S3(
t−h1
x(θ)dθ), (2.18)
— e−2αh2
t−h1
xT
(s)S4x(s)ds
t−h2
1
t
∫
−h1 t−h1
≤ −
h2 − h1
(
t−h2
x(s)ds)T
e−2αh2
S4(
t−h2
x(s)ds)
1
= (
h2 − h1
t−h1
x(θ)dθ)T
e−2αh2
S4(
t−h1
x(θ)dθ), (2.19)
0
— e−4αh1
t−h2
t
ẋT
(s)R1ẋ(s)dsdθ
t−h2
−h1 t+θ
0
2
≤ − 2 (
t 0
ẋ(s)dsdθ)T
e−4αh1
R1(
t
ẋ(s)dsdθ)
−h1 t+θ
2
= 2 [h1x(t)
1
t
t−h1
−h1 t+θ
x(θ)dθ]T
e−4αh1
R1[h1x(t) −
t
t−h1
x(θ)dθ]; (2.20)
— e−4αh2
−h1 t
ẋT
(s)R2ẋ(s)dsdθ
−h2 t+θ
1
−
(
−

28. 23
Tải tài liệu tại sividoc.com
∫ ∫
∫
∫
∫
2
≤ −
h2 − h2 (
∫
−h1∫t −h1 t
ẋ(s)dsdθ)T
e−4αh2
R2( ẋ(s)dsdθ)
2 1
−h2 t+θ −h2 t+θ
2
=
h2
− h2 [(h2 − h1)x(t) −
t−h1
x(θ)dθ]T
2 1
t−h2
t
∫
−h1
× e−4αh2
R2[(h2 − h1)x(t) − x(θ)dθ]. (2.21)
t−h2
H n n a, t (2.5) ta c
2ẋT
(t)P−1
[(A + BK)x(t) + Dx(t − h(t)) − ẋ(t)] + 2ẋT
(t)P−1
f(t, x(t))
+ 2ẋT
(t)P−1
g(t, x(t − h(t))) = 0.
M t kh c ta ch r ng
2ẋT
(t)P−1
f(t, x(t)) ≤ ẋT
(t)P−1
P−1
ẋ(t) + fT
(t, x(t))f(t, x(t))
≤ ẋT
(t)P−1
P−1
ẋ(t) + a2
xT
(t)FT
Fx(t),
2ẋT
(t)P−1
g(t, x(t − h(t)))
≤ ẋT
(t)P−1
P−1
ẋ(t) + gT
(t, x(t − h(t)))g(t, x(t − h(t)))
(2.22)
(2.23)
≤ ẋT
(t)P−1
P−1
ẋ(t) + d2
xT
(t − h(t))GT
Gx(t − h(t)) (2.24)
K t h p c c i u ki n (2.15 - 2.24) cho ta
V˙ (t, xt) + 2αV (t, xt) ≤ ξT
(t)Ωξ(t), t ≥ 0, (2.25)
trong
t
ξ(t) = [x(t), x(t − h(t)), x(t − h2), x(t − h1), x(θ)dθ, (
t−h1
x(θ)dθ), ẋ(t)]T
,
t−h1 t−h2
−

29. 24
Tải tài liệu tại sividoc.com
— −
e S
− 2
−
77 1 1
2 1 1 +
2
( 2 1
∗
∗
2
1
2
−1
Ω11 P−1
D 0 Ω14 Ω15 Ω16 Ω17
∗
Ω = ∗ ,
∗
∗
Ω11 =P −1
(A + BK) + (A + BK)T
P −1
+ 2αP −1
+ 2P −1
P −1
+ 2a2
FT
F + Q1 + Q2 + h1S3 + (h2 — h1)S4
1
e−2αh1
S
h1
1
— 2e−4αh1
R1 2
h2 − h1
e−4αh2
R ,
h2 + h1 2
Ω22 =2d2
GT
G − (1 − δ)e−2αh2
Q
2 2αh2
,
2
h2 − h1 2
Ω33 = − e−2αh2
Q3
1
e−2αh2
S
h2 − h1
2,
Ω44 =e−2αh1
Q3 − e−2αh1
Q1 1
e−2αh1
S
h1
1
1
e−2αh2
S ,
h2 − h1
Ω55
=
1
e−2αh1
S
h1
3
2
e−4αh1
R ,
1
Ω66 =
1
e−2αh2
S
h2 − h1
2
e−4αh2
R ,
h2
− h2
Ω =h S + (h — h )S
2 1
+
1
h2
R h2
− h2
)R + 2P −1
P −1
− 2P −1
,
Ω14 =
1
e−2αh1
S
h1
1, Ω15 =
2
e−4αh1
R ,
h1
1
−
−
−
− −
− h
−
1 2 2 2
4
1
Ω22 Ω23 Ω24 0 0 DT
P
∗ Ω33 0 0 0 0
∗ ∗ Ω44 0 0 0
∗ ∗ ∗ Ω55 0 0
∗ ∗ ∗ ∗ Ω66 0
∗ ∗ ∗ ∗ ∗ Ω77

30. 25
Tải tài liệu tại sividoc.com
C
C
Ω16
2
=
h2 + h1
e−4αh2
R2 , Ω17 = (A + BK)T
P −1
,
Ω23 = Ω24
1
=
h2 − h1
e−2αh2
S2.
B y gi , nh n hai v tr i v ph i c a Ω v i nh n t
v t
diag{P, P, P, P, P, P, P }
K = Y P−1
(2.26)
v s d ng B Schur (B 1.3.3), ta nh n c i u ki n Ω < 0 l t ng
ng v i i u ki n (2.6). Do , t i u ki n (2.6), ch ng ta c
V˙ (t, xt) + 2αV (t, xt) ≤ 0, ∀t ≥ 0. (2.27)
L y t ch ph n 2 v c a (2.27) t 0 n t ta c
V (t, xt) ≤ V (0, x0)e−2αt
, t ≥ 0.
B ng m t s t nh to n n gi n, ch ng ta c
V (t, xt) ≥ λmin(P −1
)ǁx(t)ǁ2
= λǁx(t)ǁ2
, ∀t ≥ 0
v
n n suy ra
V (0, x0) ≤ Aǁφǁ2
1,
λǁx(t, φ)ǁ2
≤ V (t, xt) ≤ V (0, x0)e−2αt
≤ Ae−2αt
ǁφǁ2
1,

31. 26
Tải tài liệu tại sividoc.com
−0.2577 13.2139
1
1.7401 12.6133
2
0.0983 0.9995
3
1.0057 4.3166
1 2
3
0.2863 44.7318
4
−2.5985 14.6220 −0.2157 9.2005
v khi nghi m x(t, φ) c a h th a m n
ǁx(t, φ)ǁ ≤
A
e−αt
λ ǁφǁC1, ∀t ≥ 0,
i u n y ch ng t h ng l α - n nh. nh l c ch ng minh xong.
V d 2.2.4. Cho h (2.3), A = 1 0
T
, B = ,
D =
−2 −0.1
, F = G = I,
0 −1
0 1
0 1.1
v h m tr bi n thi n h(t) = 0.1 + 0.3sin2 5
t. H m nhi u phi tuy n th a
3
m n (2.3) v (2.4) , G = F = I, a = 0.1 v d = 0.1. D d ng ki m
tra ma tr n A v A + D kh ng n nh. Cho α = 0.2 v kho ng tr v i
h1 = 0.1, h2 = 0.4, δ = 0.5. B ng vi c s d ng LMI Toolbox in Matlab, (2.6)
th nh l 2.2.3 th a m n v i c c nghi m sau:
P =
11.3967 −0.2577
, Q =
0.8090 1.7401
,
Q =
0.0974 0.0983
, Q =
0.6348 1.0057
,
S =
19.2787 −2.5985
, S =
20.5908 −0.2157
,
S =
6.8511 0.2863
, S =
0.5965 −0.0064
,
−0.0064 5.2520

32. 27
Tải tài liệu tại sividoc.com
ǁ ǁ
1 2
−7.8298 76.8225 −1.8201 14.1727
R =
19.0366 −7.8298
, R =
1.5744 −1.8201
,
Y = 0.8193 −20.1793
V y, h l 0.2 - n nh v nghi m c a h ng th a m n
ǁx(t.φ)ǁ ≤ 1.2776e−0.2t
ǁφǁ, t ≥ 0.
Khi h m i u khi n ng c l
u(t) = Y P −1
x(t) = [0.0374 − 1.5264], t ≥ 0.
2.3 H ph ng tr nh vi ph n phi tuy n kh ng t n m
c tr
X t h phi tuy n kh ng t n m c tr d ng:
ẋ(t) =A(t)x(t) + A1(t)x(t − h(t)) + B(t)u(t)
+ f(t, x(t), x(t − h(t)), u(t)), t ≥ 0,
x(t) =φ(t), t ∈ [−h, 0], h ≥ 0,
(2.28)
trong x(t) ∈ Rn
l tr ng th i, u(t) ∈ Rm
l i u khi n, A(t),A1(t) ∈ Rn×n
,
B(t) ∈ Rn×m
l ma tr n h m s li n t c tr n R+
, φ ∈ C([−h, 0], Rn
) l h m
ban u c chu n φ = sup
s∈[−h,0]
ǁφ(s)ǁ, h(t) l h m s b ch n th a m n:
0 ≤ h(t) ≤ h, ḣ(t) ≤ δ < 1, ∀t ≥ 0,

33. 28
Tải tài liệu tại sividoc.com
α
v f(t, x, y, u) : [0, ∞) × Rn
× Rn
× Rm
→ Rn
l h m phi tuy n th a m n
i u ki n t ng :
∃a, b, c > 0 : ǁf(t, x, y, u)ǁ ≤ aǁxǁ + bǁyǁ + cǁuǁ, (2.29)
v i ∀(x, y, u) ∈ Rn
× Rn
× Rm
.
nh ngh a 2.3.1. Cho α > 0. H (2.28) c g i l α - n nh h a c
n u t n t i m t h m i u khi n ng c u(t) = g(x(t)) v m t s N > 0 sao
cho m i nghi m x(t, φ) c a h ng:
ẋ(t) =A(t)x(t) + A1(t)x(t − h(t)) + B(t)g(x(t))
+ f(t, x(t)), x(t − h(t)), g(x(t)), t ≥ 0,
x(t) =φ(t), t ∈ [−h, 0], h ≥ 0,
th a m n nh gi m sau:
ǁx(t, φ)ǁ ≤ Ne−αt
ǁφǁ, ∀t ≥ 0.
nh l 2.3.2. Cho α > 0. Gi thi t r ng t n t i c c s d ng β, ϵ1, ϵ2 v
ma tr n h m s P ∈ BM+
(0, ∞) th a m n ph ng tr nh vi ph n Riccati sau:
P˙ (t) + AT
(t)P(t) + P(t)Aα(t) − P(t)Q(t)P(t) + ϵI = 0 (RDE1)
Khi h (2.1) l α - n nh h a n u c c i u ki n sau th a m n:
ϵ2
a <
2(p + β)
(i1)
1
b <
(p + β)eαh
ϵ1(1 − δ)[ϵ2 − 2a(p + β)]
3
(i2)
r

34. 29
Tải tài liệu tại sividoc.com
−
2
2
c <
ϵ2 − 2a(p + β)
(p + β)2ǁBǁ ϵ1
3b2e2αh
(i3)
ǁBǁ(1 − δ)
H n n a, h m i u khi n ng c c cho b i
u(t) = −
1
BT
(t)[P (t) − βI]x(t), t ≥ 0,
v nghi m x(t, φ) th a m n i u ki n
ǁx(t, φ)ǁ ≤ Ne−αt
ǁφǁ, ∀t ≥ 0.
Ch ng minh. t bi n tr ng th i m i:
y(t) = eαt
x(t), ∀t ≥ 0. (2.30)
v p d ng i u khi n ng c u(t) = K(t)x(t), K(t) = −
1
BT
(t)[P (t) −
βI], h ng c a ph ng tr nh (2.28) l
ẏ(t) =[Aα(t) + B(t)K(t)]y(t) + Ā1,α(t)y(t − h(t))
+ eαt
f(t, e−αt
y(t), e−α(t−h(t))
)y(t − h(t)), ũ(t)), t ≥ 0,
y(t) = eαt
φ(t), t ∈ [−h, 0], h ≥ 0,
(2.31)
ũ(t) = e−αt
K(t)y(t). i v i h (2.31), ch ng ta x t h m Lyapunov-
Krasovskii sau
V (t, yt) = V1(.) + V2(.) + V3(.),
∫t
V1(.) =< P (t)y(t), y(t) >, V2(.) = βǁy(t)ǁ2
, V3(.) = ϵ1
t−h(t)
ǁy(s)ǁ ds,
D d ng th y r ng
βǁy(t)ǁ2
≤ V (t, yt) ≤ (p + β + ϵh)ǁytǁ2
, ∀t ≥ 0. (2.32)
2

35. 30
Tải tài liệu tại sividoc.com
α
α
α
α
2 2
α
α
3
3
3
ng n g n ta k hi u
f
¯(.) := eαt
f(t, e−αt
y(t), e−α(t−h(t))
y(t − h(t)), ũ(t)).
L y o h m V (t, yt) theo t d c theo nghi m y(t) c a h (2.31) ch ng ta c
V̇1 + V̇2 = < Ṗ(t)y(t), y(t) > +2 < P(t)ẏ(t), y(t) > +2β < ẏ(t), y(t) >
= < [P˙ + AT
P + PAαPBBT
[P − βI])y(t), y(t)] >
+ 2 < P(t)Ā1,α(t)y(t − h(t)), y(t) > +2 < P(t)f
¯(.), y(t) >
+ β < (Aα(t) + AT
(t) − B(t)BT
(t)[P (t) − βI])y(t), y(t) >
+ 2β < Ā1,α(t)y(t − h(t)), y(t) > +2β < f
¯(.), y(t) >
=< [P˙ (t) + AT
(t)P(t) + P(t)Aα(t) − P(t)B(t)BT
(t)P(t)]y(t), y(t) >
+ < [β2
B(t)BT
(t) + β(Aα(t) + AT
(t))]y(t), y(t) >
+ 2 < P(t)Ā1,α(t)y(t − h(t)), y(t) > +2β < Ā1,α(t)y(t − h(t)), y(t) >
+ 2 < [P(t) + βI]f¯(.), y(t) >
V̇3 =ϵ1ǁy(t)ǁ2
− ϵ1(1 − ḣ(t))ǁy(t − h(t))ǁ2
≤ ϵ1ǁy(t)ǁ − ϵ1(1 − δ)ǁy(t − h(t))ǁ .
Do ch ng ta c
V˙ (t,yt) = < [P˙ + AT
P + PAα − PBBT
P + ϵ1I]y(t), y(t) >
+ < [β2
B(t)BT
(t) + β(Aα(t) + AT
(t))]y(t), y(t) >
+ 2 < PĀ1,α y(t − h(t)), y(t) > −
ϵ1(1 − δ)
< y(t − h(t)), y(t − h(t)) >
+ 2β < Ā1,α y(t − h(t)), y(t) > −
ϵ1(1 − δ)
< y(t − h(t)), y(t − h(t)) >
+ 2 < [P(t) + βI]f¯(.), y(t) > −
ϵ1(1 − δ)
ǁy(t − h(t))ǁ2
.

36. 31
Tải tài liệu tại sividoc.com
3
3
2 αh 2 2
α
α
1
S d ng B 1.3.1, ta c
2 < P(t)Ā1,α (t)y(t − h(t)), y(t) >≤
ϵ1(1 − δ)
< y(t − h(t)), y(t − h(t)) >
3
+
ϵ1(1 − δ)
< PĀ1,α
¯T
1,αPy(t), y(t) >
ϵ1(1 − δ)
3
< y(t − h(t)), y(t − h(t)) >
3
+
ϵ1(1 − δ)
< PA1,α
T
1,αPy(t), y(t) >,
2β < Ā1,α (t)y(t − h(t)), y(t) >≤d
ϵ1(1 − δ)
< y(t − h(t)), y(t − h(t)) >
3β2
¯ ¯T
+
ϵ (1 − δ)
< A1,α(t)A1,α(t)y(t), y(t) >
ϵ1(1 − δ)
3
3β2
< y(t − h(t)), y(t − h(t)) >
T
L y (2.29) th v o, ta c
+
ϵ (1 − δ)
< A1,α(t)A1,α(t)y(t), y(t) > .
2 < [P(t)+βI]f
¯(.), y(t) >≤ 2(p+β)(aǁy(t)ǁ+beαh
ǁy(t−h(t))ǁ+cǁũ(t)ǁ)ǁy(t)ǁ
≤ 2a(p + β)ǁy(t)ǁ + 2b(p + β)e ǁy(t − h(t))ǁǁy(t)ǁ + c(p + β) ǁBǁǁy(t)ǁ .
,
V˙ (t, yt) ≤ < [P˙ + AT
P + PAα − PQP + ϵ1I]y(t), y(t) >
+ < [β2
B(t)BT
(t) + β(Aα(t) + AT
(t))
+ [2a(p + β) + c(p + β)2
ǁBǁ]ǁy(t)ǁ2
+ 2b(p + β)eαh
ǁy(t − h(t))ǁǁy(t)ǁ
ϵ1(1 − δ) 2 3β2
T
—
3
ǁy(t − h(t))ǁ +
ϵ (1 − δ)
A1,α(t)A1,α(t)]y(t), y(t) >
A
≤
A
≤
1
1

37. 32
Tải tài liệu tại sividoc.com
3
ǁ ǁ
α
1,α
α
2
1
S d ng B 1.3.1 l n n a, ch ng ta c
2b(p + β)eαh
ǁy(t − h(t))ǁǁy(t)ǁ ≤
ϵ1(1 − δ)
ǁy(t − h(t))ǁ2
+
3
b2
(p + β)2
e2αh
y(t) 2
,
ϵ1(1 − δ)
v
< B(t)BT
(t)y(t), y(t) >≤ ǁBǁ2
ǁy(t)ǁ2
,
< (Aα(t) + AT
(t))y(t), y(t) >≤ 2µ(Aα)ǁy(t)ǁ2
,
Do
< A1,α(t)AT
(t)y(t), y(t) >≤ η2
(A1,α)ǁy(t)ǁ2
.
V˙ (t, xt) ≤< [P˙ (t) + AT
(t)P(t) + P(t)Aα(t) − P(t)Q(t)P(t) + ϵI]y(t), y(t) >
−[ϵ2 − 2a(p + β) −
ϵ
b2
(1 − δ)
(p + β)2
e2αh
— c(p + β) ǁBǁ]ǁy(t)ǁ .
V P(t) l nghi m c a (RDE1), ch ng ta c
V˙ (t,xt) ≤ [ϵ2
3 2 2 2αh 2 2
— 2a(p + β) −
ϵ (1 − δ)
b (p + β) e − c(p + β) ǁBǁ]ǁy(t)ǁ .
Do , p d ng c c i u ki n (i1, i2, i3), ta c
V˙ (t, yt) ≤ 0, ∀t ≥ 0. (2.33)
H n n a, theo t nh b ch n c a nghi m y(t, φ) c a h (2.31)
∃N > 0 : ǁy(t, φ)ǁ ≤ Nǁφǁ, ∀t ≥ 0.
n n tr l i nghi m x(t, φ) c a h (2.28) b ng ph p bi n i (2.30), ch ng ta
c
ǁx(t, φ)ǁ ≤ Nǁφǁe−αt
, ∀t ≥ 0,
1
3 2

38. 33
Tải tài liệu tại sividoc.com
1
0
−
2 cost
suy ra s n nh m c a h ng (2.28). x c nh h s n nh N, t ch
ph n hai v c a (2.33) t 0 n t, ch ng ta c
V (t, yt) ≤ V (0, y0), ∀t ≥ 0.
Ngo i ra, t (2.32) suy ra
βǁy(t)ǁ2
≤ V (t, yt) ≤ V (0, y0).
T
V (0, y0) ≤ (p + β + hϵ1)ǁφǁ2
.
V
ǁy(t)ǁ ≤ Nǁφǁ.
n n khi ta tr l i ph p bi n i x(t) ch ng ta c
ǁx(t, φ)ǁ ≤ Ne−αt
ǁφǁ, ∀t ≥ 0.
nh l c ch ng minh xong.
V d 2.3.3. X t h i u khi n phi tuy n kh ng t n m c tr :
ẋ(t) = A(t)x(t) + A1(t)x(t − h(t)) + B(t)u(t) + f(t, x(t), x(t − h(t)), u(t)),
v i h m ban u φ(t) ∈ C([−
1
, 0], R2
), v h m tr : h(t) =
1
v
2 2sin2(
t
)
2
a(t) 1
2
√
3
e
1
−2 sint 0
−1 b(t)

39. 34
Tải tài liệu tại sividoc.com
1
√
3
e
A(t) =
, A (t) = 1

40. 35
Tải tài liệu tại sividoc.com
11
2 4
—
2
−
√
3
sint 0
(t) =
B(t) = cos2
t + 2 0
0
1
sin2
t + 1
2
1
x (t) sin[x2
1
(t − h(t))] − x2 (t − h(t))sin[tx1
(t − h(t))]
f(t, .) = 4
1
8
u2(t)cos[tx(t)]
4
1 1
a(t cos4
t + 4cos2t + 4)e−t
4et
,
) = (
2 —
2
−
b(t) =
1
(
1
sin4
t − cos2t + 1)e−t
1
4et
.
i u ki n t ng c a h m ǁf(t, x(t), x(t − h(t)), u(t))ǁ c nh gi l
1 1 1
ǁf(t, x(t), x(t − h(t)), u(t))ǁ ≤
4
ǁx(t)ǁ +
8
ǁx(t − h(t))ǁ +
4
ǁu(t)ǁ,
trong : a =
1
, b =
1
, c =
1
, h =
1
, δ =
1
. Cho α = 1, ch ng ta c
4 8 4 2 2
2
α 1,α
−1 b(t) + 1
1
0 √
3
cost
2
L y β =
1
, ϵ
4 1 = 2, ϵ2
µ(Aα) = 1, ǁBǁ = 3, η(A1,α) =
= 4 + , ϵ = 8 v
16
√
3
,
Q(t) = cos4
(t) + 4cos2t + 4 0
,
0
1
sin4
t − cos2t + 1
4
, A
a(t) + 1 1
(t) =
A
1

41. 36
Tải tài liệu tại sividoc.com
Th nghi m c a (RDE1) c x c nh l
e−t 0
P(t) =
0 e−t
≥ 0, ∀t ∈ R+
,
v ch ng ta c th ki m tra c t t c c c i u ki n c a nh l (2.3.2), do
h l α− n nh h a, v h m i u khi n ng c l :
u(t) = 1 (1 − 4e−t
)(cos2
t + 2) 0 x(t).
8 0 (1 − 4e−t
)(
1
sin2
t + 1)
Ch r ng nghi m c a (RDE1) kh ng ph thu c v o c c s a, b, c c a h mnhi
u phi tuy n f(.), tuy nhi n nh l (2.3.2) ch ng v i c c nhi u n y
nh ( v theo c c i u ki n i1 − i3). nh l d i, b ng vi c c i ti n c c
ho n thi n h m Lyapunov-Krasovskii ch ng ta s thu c k t qu t t h n
nh l (2.3.2) s ng v i c c s nhi u phi tuy n t y ( t.l. c c s a, b, c
kh ng c n nh ).
V i c c s d ng α, β, h, ϵi, i = 1, 2, 3, a, b, c ch ng ta t
Pβ(t) = P (t) + βI,
γ = ϵ−1
a2
+
3
b2
+ c2
,
ϵ1e−2αh(1 − δ)
µ(A) = sup µ(A(t)),
t∈R+
η(A1) = sup η(A1(t)),
t∈R+
Q(t) =
3
B(t)BT
(t) −
3
A (t)AT
(t) − γI,
4 ϵ1e−2αh(1 − δ)
2
3
1 1

42. 37
Tải tài liệu tại sividoc.com
s
2
2
M = p + β + hϵ1 + 2h2
ϵ2
, N =
M
,
β
ϵ = 2αβ + β2
γ + ϵ1 + ϵ2he2αh
+ ϵ3 + 3β2
ǁBǁ2
+2βµ(A) +
ϵ1
3β2
η
e−2αh(1 − δ)
2
(A1).
nh l 2.3.4. Cho α > 0. Gi thi t r ng t n t i c c s d ng β, ϵ1, ϵ2, ϵ3 v
m t ma tr n h m s P ∈ BM +
(0, ∞) th a m n ph ng tr nh vi ph n Riccati
sau y:
P˙ (t) + AT
(t)P (t) + P (t)A(t) − P (t)Q(t)P (t) + 2(α + βγ)P (t) + ϵI = 0.
(RDE2)
Th h (2.1) l α - n nh h a c v i h m i u khi n ng c:
u(t) = −
1
BT
(t)[P (t) − 2βI]x(t).
H n n a, nghi m x(t, φ) th a m n i u ki n
ǁx(t, φ)ǁ ≤ Ne−αt
ǁφǁ, ∀t ≥ 0.
Ch ng minh. Gi s u(t) = K(t)x(t), K(t) = −1
BT
(t)[P (t) − 2βI], t ≥
0. V i h ng (2.28), ch ng ta x t h m Lyapunov-Krasovskii sau
V (t, xt) = V1 + V2 + V3 + V4,
y
2
V1 =< P(t)x, x >, V2 = βǁx(t)ǁ ,
∫t
V3 = ϵ1 e2α(s−t)
ǁx(s)ǁ2
ds,
t−h(t)

43. 38
Tải tài liệu tại sividoc.com
T
∫ ∫
0 t
V4 = e2α(s+h−t)
ǁx(s)ǁ2
dsdr.
D d ng th y r ng
−h t+r−h(t+r)
λ1ǁx(t)ǁ2
≤ V (t, xt) ≤ λ2ǁxtǁ2
, ∀t ≥ 0,
v i c c h ng s d ng λ1, λ2.
L y o h m c a V (t, xt) theo t d c theo nghi m x(t), ch ng ta c
V̇1 + V̇2 = < Ṗ(t)x(t), x(t) > +2 < P(t)ẋ(t), x(t) > +2β < ẋ(t), x(t) >
= < (P˙ (t) + AT
(t)P(t) + P(t)A(t)
— P (t)B(t)B (t)[P (t) − 2βI])x(t), x(t) >
+ 2 < P (t)A1(t)x(t − h(t)), x(t) >
+ 2 < P (t)f(t, x(t)x(t − h(t)), u(t)), x(t) >
+ β < (A(t) + AT
(t) − B(t)BT
(t)[P (t) − 2βI])x(t), x(t) >
+ 2β < A1x(t − h(t)), x(t) > +2β < f(t, x(t)x(t − h(t)), u(t)), x(t) >
= < [P˙ + AT
P + PA − PBBT
P ]x(t), x(t) >
+ < βP (t)B(t)BT
(t)x(t), x(t) > + < [2β2
B(t)BT
(t)
+ β(A(t) + AT
(t))]x(t), x(t) > +2 < P(t)A1(t)x(t − h(t)), x(t) >
+ 2β < A1(t)x(t − h(t)), x(t) >
+ 2 < Pβ(t)f(t, x(t)x(t − h(t)), u(t)), x(t) >
V̇3(t, xt) = − 2αV3(t, xt) + ϵ1ǁx(t)ǁ2
− ϵ1e−2αh(t)
(1 − ḣ(t))ǁx(t − h(t))ǁ2
≤ −2αV3(t, xt) + ϵ1ǁx(t)ǁ2
− ϵ1e−2αh
(1 − δ)ǁx(t − h(t))ǁ2

44. 39
Tải tài liệu tại sividoc.com
2
2αh 2
∫0
V̇4(t, xt) = − 2αV4(t, xt) + ϵ2he2αh
ǁx(t)ǁ2
— ϵ2e−2αh
(1 − δ)
−h
ǁx(t + s − h(t + s))ǁ ds
, ta c
≤ −2αV4(t, xt) + ϵ2he ǁx(t)ǁ .
V̇ (t, xt)+2αV (t, xt) = V̇1(t, x) + 2αV1(t, x) + V̇2(t, x) + 2αV2(t, x)
+ V̇3(t, xt) + 2αV3(t, xt) + V̇4(t, xt) + 2αV4(t, xt)
≤< [P˙ (t) + AT
(t)P(t) + P(t)A(t)
— P (t)B(t)BT
(t)P (t)]x(t), x(t) >
+ < βP (t)B(t)BT
(t)x(t), x(t) >
+ < [2β2
B(t)BT
(t) + β(A(t) + AT
(t))]x(t), x(t) >
+ 2 < P (t)A1(t)x(t − h(t)), x(t) >
+ 2β < A1(t)x(t − h(t)), x(t) >
+ 2 < Pβ(t)f(t, x(t)x(t − h(t)), u(t)), x(t) >
+ 2α[< P (t)x(t), x(t) > +βǁx(t)ǁ2
]
+ ϵ1ǁx(t)ǁ2
− ϵ1e−2αh
(1 − δ)ǁx(t − h(t))ǁ2
+ ϵ2he2αh
ǁx(t)ǁ2
≤< [P˙ (t) + AT
(t)P (t) + P (t)A(t) − P (t)B(t)BT
(t)P (t)
+ 2αP (t)]x(t), x(t) >

45. 40
Tải tài liệu tại sividoc.com
−
1
−
1
−
1
−
1
−
+ < βPBBT
x(t), x(t) > + < [2β2
BBT
+ β(A(t)
+ AT
(t))]x(t), x(t) > +2 < P(t)A1(t)x(t − h(t)), x(t) >
ϵ1e−2αh
(1 − δ)
—
3
< x(t − h(t)), x(t − h(t)) >
ϵ e−2αh
(1 δ)
—
3
< x(t − h(t)), x(t − h(t)) >
+ 2 < Pβ(t)f(t, x(t)x(t − h(t)), u(t)), x(t) >
+ (2αβ + ϵ1 + ϵ2he2αh
)ǁxǁ2
+ 2β < A1x(t − h(t)), x >
ϵ e−2αh
(1 δ) 2
—
3
ǁx(t − h(t))ǁ . (2.34)
S d ng B (1.3.1), ta c
2 < P (t)A1(t)x(t − h(t)), x(t) > −
ϵ e−2αh
(1 δ)
3
< x(t − h(t))x(t − h(t)) >
3 T
≤
ϵ1e−2αh(1 − δ)
< P (t)A1(t)A1 (t)P (t)x(t), x(t) >,
(2.35)
2β < A1(t)x(t − h(t)), x(t) > −
ϵ e−2αh
(1 δ)
3
< x(t − h(t)), x(t − h(t)) >
3β2
T
L y (2.29) th v o ta c
≤
ϵ1e−2αh(1 δ)
< A1(t)A1 (t)x(t), x(t) > .
(2.36)
2 < Pβ(t)f(t, x(t)x(t − h(t)), u(t)), x(t) >
≤ 2ǁxT
(t)Pβ(t)ǁǁf(t, x(t)x(t − h(t)), u(t)), x(t)ǁ
≤ 2aǁxT
(t)Pβ(t)ǁǁx(t)ǁ + 2bǁxT
(t)Pβ(t)ǁǁx(t − h(t))ǁ
+ 2cǁxT
(t)Pβ(t)ǁǁu(t)ǁ (2.37)

46. 41
Tải tài liệu tại sividoc.com
1
−
1
4
1 1
Ti p t c s d ng B 1.3.1 ch ng ta c
2 < Pβ(t)f(t, x(t)x(t − h(t)), u(t)), x(t) >
≤ ϵ3
−1
a2
ǁxT
(t)Pβ (t)ǁ2
+ ϵ3
2 3 2 T
ǁx(t)ǁ +
ϵ e−2αh(1 − δ)
b ǁx (t)Pβ (t)ǁ2
ϵ1e−2αh
(1 − δ) 2 2 T 2 2
+
3
ǁx(t − h(t))ǁ + c ǁx (t)Pβ(t)ǁ + ǁu(t)ǁ (2.38)
≤< [ϵ−1
a2
+
3
b2
+ c2
]P2
(t)x(t), x(t) >
ϵ1e−2αh(1 − δ)
2 ϵ1e−2αh
(1 − δ) 2
+ ϵ3ǁx(t)ǁ +
3
ǁx(t − h(t))ǁ
+ < [P (t) − 2βI]B(t)BT
(t)[P (t) − 2βI]x(t), x(t) >
≤ γ < [P
2
(t) + 2βP (t) + β2
I]x(t), x(t) >
2 ϵ1e−2αh
(1 − δ) 2
+ ϵ3ǁx(t)ǁ +
3
ǁx(t − h(t))ǁ
+ < [
1
P (t)B(t)BT
(t)P (t) βP (t)B(t)BT
(t) + β2
B(t)BT
(t)]x(t), x(t) > .
4
V y, t c c i u ki n (2.34) - (2.38) ta c
V˙ (t, xt) + 2αV (t, xt)
≤< [P˙ (t) + AT
(t)P(t) + P(t)A(t) + 2(α + βγ)P(t)]x(t), x(t) >
(2.39)
— < [3
P (t)B(t)BT
(t)P(t) −
3
P (t)A (t)AT
(t)P (t)
4
ϵ1e−2αh(1 − δ)
−γP 2
(t)]x(t), x(t) > + < [3β2
B(t)BT
(t) + β(A(t) + AT
(t))
3β2
T
+
ϵ e−2αh(1 − δ)
A1(t)A1 (t)]x(t), x(t) >
+(2αβ + β2
γ + ϵ1 + ϵ2he2αh
+ ϵ3)ǁx(t)ǁ2
.
Ch r ng
< B(t)BT
(t)x(t), x(t) >≤ ǁBǁ2
ǁx(t)ǁ2
,
< (A(t) + AT
(t))x(t), x(t) >≤ 2µ(A)ǁx(t)ǁ2
,
3 β
1

47. 42
Tải tài liệu tại sividoc.com
1
β
th
< A1(t)AT
(t)x(t), x(t) >≤ η2
(A1)ǁx(t)ǁ2
,
V˙ (t, xt) + 2αV (t, xt) ≤< [P˙ (t) + AT
(t)P(t) + P(t)A(t)
−P (t)Q(t)P (t) + 2(α + βγ)P (t)]x(t), x(t) >
+[2αβ + β2
γ + ϵ1 + ϵ2he2αh
+ ϵ3 + 3β2
ǁBǁ2
+2βµ(A) +
ϵ1
3β2
η
e−2αh(1 − δ)
(A1)]ǁx(t)ǁ
≤< [P˙ (t) + AT
(t)P(t) + P(t)A(t) − P(t)Q(t)P(t)
+2(α + βγ)P (t) + ϵI]x(t), x(t) > .
T P(t) l nghi m c a (RDE2), ch ng ta c
V˙ (t, xt) + 2αV (t, xt) ≤ 0, ∀t ≥ 0.
V v y
V˙ (t, xt) ≤ −2αV (t, xt), ∀t ≥ 0, (2.40)
S d ng B 1.3.3, h l α - n nh. t m h s n nh N, l y t ch
ph n hai v c a (2.39) t 0 n t, ch ng ta c
V (t, xt) ≤ V (0, x0)e−2αt
, ∀t ≥ 0.
M t kh c, t nh gi b ch n c a V (t, xt) ta c
βǁx(t)ǁ2
≤ V (t, xt), ∀t ≥ 0,
suy ra
ǁx(t, φ)ǁ ≤
s
V (0, x0)
e−αt
, ∀t ≥ 0.
2 2

48. 43
Tải tài liệu tại sividoc.com
2
2 2 2
2
1
∫ ∫
H n n a, ta c
0 0
V (0, x0) ≤ (p + β + hϵ1)ǁφǁ2
+ ϵ2 e2α(s+h)
ǁx(s)ǁ2
dsdr,
n n
−h r−h(r)
≤ ((p + β + hϵ1)ǁφǁ + 2h ϵ2ǁφǁ
≤ Mǁφǁ ,
ǁx(t, φ)ǁ ≤ Ne−αt
ǁφǁ, ∀t ≥ 0.
nh l c ch ng minh xong.
Ch r ng c c s a, b, c trong i u ki n t ng c a h m nhi u phi tuy n
ch xu t hi n trong (RDE2) ch kh ng c n th a m n c c r ng bu c nh
i1 − i3, do h l n nh h a c v i c c nhi u phi tuy n t y .
V d 2.3.5. X t h phi tuy n kh ng t n m v i tr bi n thi n:
x˙= A(t)x(t) + A1(t)x(t − h(t)) + B(t)u(t) + f(t, x(t), x(t − h(t)), u(t)),
v i h m tr ban u φ ∈ C([−
1
, 0], R2
) v h m tr bi n thi n h(t) nh trong
V d 2.3.3 v
2x2(t)sin[tx2
(t − h(t))] − 2u1(t)cos[t2
x(t)]
f(t, x(t), x(t − h(t)), u(t)) =
√
2x1(t − h(t))sin[x(t)x1(t − h(t))]
A(t) =
a(t) 1
,
−1 b(t)

49. 44
Tải tài liệu tại sividoc.com
−
2 2
2
3
√ sint 0
A1(t) = 2
√
2cost
,
B(t) = 2(sin2
t + 3) 0
,
2
0 √
3
(cos t + 4)
a(t) = e−4t
(
3
sin4
t +
11
) − 5e4t
,
b(t) =
1
e−4t
cos4
t 5e4t
.
2
c l ng ǁf(t, x(t), x(t − h(t)), u(t))ǁ cho
ǁf(t, x(t), x(t − h(t)), u(t))ǁ ≤ 2ǁx(t)ǁ +
√
2ǁx(t − h(t))ǁ + 2ǁu(t)ǁ.
h =
1
, δ =
1
, a = 2, b =
√
2, c = 2,
2 2
3
V i α = 1, cho
µ(A) = 2, ǁBǁ = 8, η(A1) = √
2
.
1 3e 991
Ta c
β =
16
, ϵ1 =
2
, ϵ2 = (
64e
− 3), ϵ3 = 1,
Q(t) =
γ = 16, ϵ = 10,
3sin4
t + 11 0
0 cos4
t
Nghi m c a (RDE2) c t m b i
e−4t 0
P(t) =
0 e−4t ≥ 0, ∀t ∈ R+
.
0

50. 45
Tải tài liệu tại sividoc.com
Khi , h l n nh ho c v i h m i u khi n ng c l
(sin2
t + 3)(
1
− e−4t
) 0
u(t) =
8
1
0 √
3
(cos2
t + 4)(
1
8
— e−4t)
x(t).

51. 46
Tải tài liệu tại sividoc.com
K t lu n chung
Lu n v n tr nh b y c c c v n sau
• Tr nh b y v hi u c c kh i ni m h ph ng tr nh vi ph n i u khi n,
h ph ng tr nh vi ph n c tr , ph ng ph p h m Lyapunov trong vi c
gi i b i to n n nh h a v c c b li n quan.
• Tr nh b y c c ti u chu n v t nh n nh h a h ph ng tr nh vi ph n c
tr : h ph ng tr nh vi ph n phi tuy n t n m c tr , h ph ng tr nh
vi ph n phi tuy n kh ng t n m c tr v i c c ch ng minh chi ti t v
v d minh h a.

52. 47
Tải tài liệu tại sividoc.com
T i li u tham kh o
T i li u Ti ng Vi t
[1] V Ng c Ph t, (2001), Nh p m n l thuy t i u khi n to n h c, NXB
i H c Qu c Gia, H N i.
T i li u Ti ng Anh
[2] J. Hale and S.M. Lunel (2003), Introduction to Functional Differential
Equations, Springer, New York.
[3] VT. Tai and V.N. Phat (2009), Global exponential stabilization of non-
autonomous diferential equations via Riccati equations, Nonlinear Func-
tional Analysis and Applications, 14, pp. 245-260.
[4] M.V. Thuan, V.N. Phat, T.L. Fernando and H. Trinh (2014), Exponen-
tial stabilization of time-varying delay systems with nonlinear perturba-
tions, IMA J. Contr. Inform. 31, pp. 441-464.