Adaptative high-gain extended Kalman filter and applications

tel-00559107, version 1 - 24 Jan 2011
B.2 Proofs of the Technical Lemmas
B.2 Proofs of the Technical Lemmas
Lemma 97 ([38]) Let {x (t) > 0,t≥ 0} ⊂ R n be absolutely continuous, and satisfying:
dx(t)
dt ≤−k1x + k2x √ x,
for almost all t>0, for k1, k2 > 0. Then, if x (0) < k2 1
4k2 , x(t) ≤ 4 x (0) e
2
−k1t .
Proof.
We make three successive change of variables: y = √ x, z = 1
y and w(t) =e− k 1 2 t z(t).
Then all the quantities y(t) z(t), w(t) are positive and absolutely continuous on any finite
time interval [0,T]. We have
˙y = 1
2 √ x
k2
˙x ≤ −k1
2 y + 2 y2
˙z = − 1
y2 ˙y ≥ k1 k2
2 z − 2
˙w = − k1
2 e− k1 t − 2 z(t)+e k1 t k2
2 ˙z ≥ − 2 e− k1 t 2
Moreover, w(0) = 1 √ . Then, for almost all t>0,
x(0)
If 1
√x(0) − k2
k1
w(t) ≥
1
� x(0) − k2
k1
+ k2
e
k1
− k1 t 2
> 0, then w(t) > 0 and we can go backwards in the previous inequalities:
w(t) ≥ √x(0) 1 − k2
�
1 − e k1
− k1 t 2 �
z(t) ≥ e k �
1 t 2 √ 1 −
x(0) k2
�
k1
+ k1
y(t) ≤ 1
2
.
x(t) ≤
Hence, if x(0) ≤ k2 1
4k2 , or 1 −
2
� x(0) k2
k1
e k �
1
2
t
√x(0)
1
− k �
2 + k1
k1 2
x(0)e−k1t �
1− √ x(0) k �2 2
k1
1 ≥ 2 , then:
x(t) ≤ 4x(0)e −k1t
Lemma 98 ([38])
Consider �b (˜z) − �b (˜x) − �b ∗ (˜z)˜ε that appears in the inequality (3.12) (omitting to write u
in ˜ �
�
b) and suppose θ ≥ 1. Then ��b (˜z) − �b (˜x) − �b ∗ �
�
(˜z)˜ε � ≤ Kθn−1 |˜ε| 2 , for some K>0.
Proof.
Let us denote ε = z − x and consider a smooth expression E(z, x) of the form:
E(z, x) =f(z) − f(x) − df (z)ε,
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