5
$\begingroup$

What are the techniques on inverse problems to discover the distribution of parameters from a discrete set of values? For instance, I know that my domain where the PDE is defined is made up of materials A, B and C, but I do not know how they are distributed inside the domain. Working with discrete optimization algorithms seems unfeasible. What are the techniques to solve such a problem?

Edit: I will describe the problem for just two materials, as it seems that the strategy is different than for multimaterials, according to Clason's comments (bang-bang vs multibang)

$$ \underset{u \in U}{\text{min}} \frac{1}{2} \lVert F(u) - y^{\delta} \rVert_Y \\ $$ where $$ U = \{ u \in X : u \in \{ u_1, u_2 \} ~\text{pointwise}\}\\ F: X \rightarrow Y $$ $X$ and $Y$ are function spaces and $y^{\delta} \in Y$

$\endgroup$
10
  • 3
    $\begingroup$ That's one question where I can't resist mentioning my own work, because that was exactly the motivation for it... These slides (hopefully) give a decent overview, and there's code on my web page (look for papers involving "multibang"). $\endgroup$ Jun 17, 2017 at 18:00
  • $\begingroup$ Thanks for the reference. Is the problem much easier if there are two materials only? That's what it seems like from slide # 3. $\endgroup$
    – balborian
    Jun 18, 2017 at 1:35
  • $\begingroup$ Yes, the case of binary materials is well-known and in the classical control literature is called "bang-bang control" (hence the term "multibang" for more than two materials). In this case, you can just use the convex control constraints and don't need an additional penalty. $\endgroup$ Jun 18, 2017 at 9:01
  • $\begingroup$ What do you mean by convex control constraint? $\endgroup$
    – balborian
    Jun 18, 2017 at 17:46
  • 1
    $\begingroup$ Because that is non-convex, and hence not weakly lower semi-continuous, so you can't prove that a solution exist by standard techniques (and in fact, there are counter-examples). But while the constraints will not enforce such solutions (and neither will the multibang penalty), it turns out that in many cases you still obtain them. That's the price you pay for the efficient numerical solution (there ain't no free lunch...) $\endgroup$ Jun 19, 2017 at 6:30

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.