Example of application-relevant PDEs not satisfying conditions of Lax-Milgram theorem

Question

In every elementary course on Finite Element Methods, Numerical Methods for PDEs, Simulation, and alike, probably, the reasoning goes something like

application motivated PDE $\to$ weak form $\to$ discretization

and this is justified by showing that the weak form satisfies the conditions of the Lax-Milgram theorem or a generalization thereof.

I am looking for PDEs that are relevant to any kind of application and which either lead to bilinear forms that are either

not continuous or/and
not coercive

or that do not lead to a bilinear form at all.

Please don't resent my lack of knowledge about $\rm inf$-$\rm sup$ conditions.

By relevant to any kind of application I mean that the PDE should have any meaning besides serving as a counterexample to the conditions of Lax-Milgram.

PS: please also provide a short argument of why your counter example actually is one.

Gustave · Answer 1 · 2018-06-08T16:33:53.600

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Try this $$\Delta u+u=f$$ with f in $L^2(0,1)$ and Dirichlet boundary conditions.

edited Jun 08 '18 at 16:33

answered Jun 07 '18 at 23:22

Gustave

1,523
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What does the notation $\Delta^2$ mean to you? – rehctawrats Jun 08 '18 at 08:25
I'm sorry, I made a mistake. – Gustave Jun 08 '18 at 16:33
@Gustave , are you familiar with a simple example of a physical system governed by this PDE and/or a simple way to show that Lax-Milgram fails for it? – aghostinthefigures Jun 11 '18 at 16:25

score 0 · Answer 2 · answered Jun 08 '18 at 15:55

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Try this:

Any employment for the Varignon parallelogram?

answered Jun 08 '18 at 15:55

Han de Bruijn

17,070

Example of application-relevant PDEs not satisfying conditions of Lax-Milgram theorem

2 Answers2