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Proof.
The satisfaction of (15.33) means that the set
,
ζ
,
δ
1
) is included in the
set
S
(
u
0
) defined as in (15.22). Thus, one can conclude that for any
x
S
1
(
V
∈S
,
ζ
,
δ
1
(
V
1
)
K
the nonlinearity
x
) satisfies the sector condition (15.23). Moreover, the satisfac-
tion of relations (15.34) and (15.35) means that the set
φ
(
S
,
ζ
,
δ
1
(
V
1
) is included in
the set
1
), the constraints C2
and C3 are respected. Consider a positive definite function
V
(
x
) (
V
(
x
)
>
Ω
(
x
) defined by (15.19). Thus, for any
x
∈S
1
(
V
,
ζ
,
δ
= 0).
We want to prove that the time-derivative of this function satisfies (15.29) along the
trajectories of the closed-loop system (15.21) for all admissible nonlinearity
0,
∀
x
x
),
all admissible uncertain vector
z
and all disturbances satisfying (15.11). Hence, us-
ing Lemma 15.1, one gets
φ
(
K
V
(
x
)
−
ω
ω
≤
V
(
x
)
x
)
(
−
ω
ω
.
−
2
φ
(
K
φ
(
K
x
)+
G
x
)
Thus, the satisfaction of relation (15.32) implies that
V
(
x
)
x
)
M
(
−
ω
ω
<
−
2
φ
(
K
φ
(
K
x
)+
G
x
)
0
along the trajectories of system (15.21). Relations (15.30) and (15.31) are then satis-
fied. Therefore, when
= 0, one gets
V
(
x
)
ω
<
0,
∀
x
(0)
∈S
1
(
V
,
ζ
,
δ
1
).When
ω
= 0,
the closed-loop trajectories of system (15.21) remain bounded in
S
1
(
V
,
ζ
,
δ
1
),
for any
x
(0)
∈ S
0
(
V
,
ζ
) and any disturbance satisfying (15.11). One can con-
6
;
V
(
x
)
1
ζ
1
δ
1
}
clude that the gain
K
and the sets
S
1
(
V
,
ζ
,
δ
1
)=
{
x
∈
ℜ
≤
+
and
≤
ζ
−
1
6
;
V
(
x
)
S
0
(
V
,
ζ
)=
{
x
∈
ℜ
}
are solutions to Problem 15.1.
Theorem 15.1 provides a sufficient condition to solve the control gain design. How-
ever, such a condition appears not really constructive to exhibit a suitable function
V
(
x
),again
K
1
. The idea then consists in considering a
quadratic function for
V
(
x
),as
V
(
x
)=
x
Px
,
P
=
P
>
and two scalars
ζ
and
δ
0. Thus, by exploiting the de-
scription of the closed-loop system through a polytopic model (15.28), the following
proposition derived from Theorem 15.1 can be stated.
6
×
6
,R
1
∈
Proposition 15.1.
If there exist symmetric positive definite matrices W
∈
ℜ
3
×
3
, a diagonal positive matrix S
3
×
3
, two matrices Y
3
×
6
3
×
6
,
ℜ
∈
ℜ
∈
ℜ
and Z
∈
ℜ
δ
1
satisfying
1
three positive scalars
ε
,
ζ
and
⎡
⎣
⎤
⎦
A
1
j
+
B
1
Y
+
Y
B
1
+
R
R
1
R
W
A
1
j
W
+
(
u
1(3)
(1 +
2
)+
2
)
R
R
+
ε
β
β
u
1(2)
B
2
j
R
W
−
R
1
B
3
I
3
<
0
,
R
W
0
−
ε
I
6
B
−
S
Z
00
−
2
S
B
4
j
R
000
−
I
4
0
0
0
0
B
5
j
−
ε
I
3
(15.36)
1
stands for symmetric blocks.
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