Remark:

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Remark:

For the special case $K(t,x)=K(x)=K\mathbbm{1}_{[a,b]}(x)$ , where

is a constant and

is a sub-interval of

, we have

$\displaystyle w(T,\psi(T,x)) = w(0,x) \exp \left( -K \chi(x)-\displaystyle\int_0^T \partial_x u_{obs}(\sigma,\psi(\sigma,x))d\sigma \right),$

(3.45)

where

$\displaystyle \chi(x) = \displaystyle\int_0^T \mathbbm{1}_{Supp(K)}(\psi(\sigma,x)) d\sigma$

(3.46)

is the time during which the characteristic curve $\psi(\sigma,x)$ with foot

of equation (3.39-F) with

lies in the support of

The system is then observable if and only if the function $\chi$ has a non-zero lower bound, i.e. $m := \displaystyle \min_{x} \chi(x) > 0$ , the observability being defined by (see e.g. [93]):

$% latex2html id marker 7176 $\displaystyle \exists C, \forall u \textrm{ soluti... ...with } K=0,\quad \Vert u(T,.)\Vert^2 \le C\int_0^T \Vert K(.)u(s,.)\Vert^2\,ds.$$

(3.47)

In this case, proposition 3.2 proves the global exponential decrease of the error, provided

is larger than $\displaystyle \frac{MT}{m}$ , where

is defined by equation (3.41).

From this remark, we can easily deduce that if for each iteration, both in the forward and backward integrations, the observability condition is satisfied, then the algorithm converges and the error decreases exponentially to 0 . Note that this is not a necessary condition, as even if $\chi(x)=0$ , the last exponential of equation (3.45) is bounded.

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