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I have a doubt about Lax-Milgram theorem, you can see here: http://mathworld.wolfram.com/Lax-MilgramTheorem.html.

I'm studying weak problems, and I have a bilineal form that it isn't coercive, can i conlude that exists a solution of the PDE but it isn`t unique?

Thank you in advance.

My attempt or my thoughts: I think that if the bilineal form is symmetric it is true because we can relate it with the minimization problem of a functional.

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  • $\begingroup$ Interesting question. Can you give more details? I think you need the bilinear form the be coercive. $\endgroup$ – Niklas Feb 28 '17 at 15:06
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No, this is not true. Let us define the bilinear form $a(u,v) := 0$, which is not coercive. Then, $$a(u,v) = \ell(v) \quad\forall v$$ does not have a solution for $\ell \ne 0$.

You also cannot use minimization of $\frac12 a(u,u) - \ell(u)$, since this problem will not have a minimizer (unless $\ell = 0$).

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  • $\begingroup$ Thank you very much for your answer, I haven´t thought that it has been very helpful. And is this positive defined?? $\endgroup$ – Skullgreymon Feb 28 '17 at 19:33
  • $\begingroup$ What do you mean by 'positive defined'? $\endgroup$ – gerw Mar 1 '17 at 9:24
  • $\begingroup$ $a(u,u)>0 \quad \forall u$ $\endgroup$ – Skullgreymon Mar 1 '17 at 16:07
  • $\begingroup$ No, my bilinear form is not positive defined. $\endgroup$ – gerw Mar 1 '17 at 20:27

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