0
$\begingroup$

Polyharmonic equations, to my understanding, are defined as:

$$\Delta ^k u = 0$$

i.e. one repeatedly applies the laplace operator to the function a certain number of times and the result must be 0.

The system can be discretized quite easily using finite differences or in the case I care about, triangle meshes, using the discrete Laplace-Beltrami operator.

However, this system is incomplete, we must also have boundary conditions, else there is a trivial solution yielded by $u = 0$.

So, let's say I am given a mesh $M$. I can construct a matrix $L$ through the use of the laplace beltrami oerator. But this is only possible at interior (not boundary) vertices.

Let's say I know which vertices in my mesh are fixed, what do I do to construct an appropriate linear system to solve the polyharmonic problem in the discrete setting?

$\endgroup$

1 Answer 1

1
$\begingroup$

This is, ultimately, a modeling question: The boundary conditions you choose are the ones that match the physical situation you are trying to describe. For example, the biharmonic equation ($\Delta^2$) describes the vertical deflection of thin plates, and among the boundary conditions you can choose are the "simply supported" case (the edge of the plate just rests on a support) and the "clamped" case (the edge of the plate is clamped). In one case, you prescribe $u$ and $\Delta u$ at the boundary, in the other you prescribe $u$ and $\partial u/\partial n$.

Which one of these is correct depends of course on what you want to do.

$\endgroup$
3
  • $\begingroup$ I am trying to replicate equation 5 in this paper: cs.toronto.edu/~jacobson/seminar/botsch-and-kobbelt-2004.pdf $\endgroup$
    – Makogan
    Sep 20, 2023 at 18:51
  • 1
    $\begingroup$ I have to admit that I don't understand what they do in that paper without spending more time reading through it :-( $\endgroup$ Sep 20, 2023 at 23:05
  • $\begingroup$ That is fair, thank you for the help still. $\endgroup$
    – Makogan
    Sep 20, 2023 at 23:33

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.