# Lax-Milgram theorem

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?

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.

• Interesting question. Can you give more details? I think you need the bilinear form the be coercive. – Niklas Feb 28 '17 at 15:06

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 $\ell = 0$).

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