Adaptative high-gain extended Kalman filter and applications

tel-00559107, version 1 - 24 Jan 2011
2. β 0}: it has positive measure.
Suppose it is not the case: therefore x decreases from time α to time β. This
implies that x(α) >x(β) which is in contradiction with ⋆. Therefore the
measure of F is strictly positive.
However for any τ ∈]α,β [⊂ E we have x(τ) > 2b
a : −aX2 +2bX < 0 since x(τ) > 2b
a .
Then ˙x(τ) < 0 and τ /∈ F , which means that F is not of positive measure. This gives
us a contradiction: either α = 0 or E = ∅ (first item at the beginning of the proof).
Consider α = 0: E = [0, τ1[, τ1 > 0 and x(t) ≤ max (x(0), 2b
a ), for all τ ≥ τ1. For
τ ∈ [0, τ1[:
� �
2b
˙x(τ) ≤ a − x(τ) x(τ) < 0,
a
which means that x(τ) is decreasing on [0, τ1[ (x(τ) 2b
2b
a , then x(τ) > a for τ ∈ [0, τ1[ as above. It means that:
� �
2b
˙x ≤ a − x(τ) x(τ).
a
We rewrite it:
Consider
or in other terms
Integration with respect to time gives:
�
x
ln
x − 2b
�
a
− ˙x(τ)
a � 2b
a − x� ≥ 1.
x
� �
d ln
x
x − 2b
��
a
= − 2b
a dx
x(x − 2b
a )
� �
a d
2b dτ
ln
x
x − 2b
��
a
=
− ˙x
(x − 2b ≥ a.
a )x
≥ a 2b
�
τ + ln
a
Therefore, since x0 > 2b
2b
a , and x(τ) > a for τ