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Viscous Burgers

We now consider the viscous Burgers' equation, a standard nonlinear transport equation. We also consider only one iteration of the BFN algorithm:

\begin{displaymath}\begin{array}{rl} (F) & \left\{ \begin{array}{rcl} \partial_t...
...etilde{u}\vert _{t=T} &=& u(T), \end{array} \right. \end{array}\end{displaymath} (3.30)

with the same notations as before. The observations $ u_{obs}$ also satisfy the forward Burgers' equation:

$\displaystyle \left\{ \begin{array}{rcl} \partial_t u_{obs} -\nu \partial_{xx} ...
...rt _{x=0}=u\vert _{x=1}&=&0,\\ u\vert _{t=0} &=& u_{obs}^0. \end{array} \right.$ (3.31)

Then, the following result holds true [29]:

Theorem 3.4   If $ K(t,x)\not\equiv 0$ , then the BFN iteration (3.30) for viscous Burgers' equation, with observations satisfying (3.31), is ill-posed, even when $ K(t,x)$ is constant: there does not exist, in general, a solution $ (u,\widetilde{u})$ .

In the particular case where $ K\equiv K'\equiv 0$ , the backward problem is ill-posed in the sense of Hadamard (the solution does not depend continuously on the data), but it has a unique solution if the final condition $ \widetilde{u}\vert _{t=T}$ is set to a final solution of the direct equation. Moreover, in this particular case, the backward solution is exactly equal to the forward one: $ \widetilde{u}(t) = u(t)$ for all $ t\in[0,T]$ . The main result is the following [29]:

Proposition 3.1   If $ K\equiv K'\equiv 0$ , then problem (3.30) is well-posed in the sense of Hadamard, and there exists a unique solution $ (u,\widetilde{u})$ . Moreover, $ u=\widetilde{u}$ .

The BFN algorithm is then ill-posed (except if $ K\equiv K'\equiv 0$ ) when applied to a viscous Burgers' equation, as there does not exist a solution to the backward problem. However, from the numerical point of view, the BFN algorithm has been successfully applied to this model [21]. This phenomenon is probably due to the fact that we numerically solve a discrete problem and not the exact continuous one.


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