0
$\begingroup$

We have the problem: $$ -u''(x) + u(x) = f(x) ,\quad \quad x \in [0,L] $$ $$u(0) = 0 $$ $$u'(L) + u(L) = 4 $$ I then put it into variational form (hopefully correctly done) with introduction of $v \in T(I) = \{v \in C^1 : v(0) =0 \} $. $$u(L)v(L)+\int_0^L u'v\,' \, dx + \int_0^L uv\,dx = \int_0^L fv\,dx+ 4v(L) $$ We have bilinear operator as $$\therefore a(u,v) = u(L)v(L)+\int_0^L u'v\,' \, dx + \int_0^L uv\,dx $$ as well as $$ f(v) = \int_0^L fv\,dx+ 4v(L) $$ I am not sure on how to go about proving/showing that the variational form has at most one solution. The Lax-Milgram Theorem gets thrown around a lot in proving uniqueness. I do not know how to use it in application though. Please guide me or recommend a path/resource that I must follow in order to show this.

Thank you for your time.

$\endgroup$
1
$\begingroup$

You need to satisfy the conditions of Lax-Milgram lemma. Proving we’ll-posedness of a weak formulation is written in literally any textbook about the finite element method. Briefly, you need to show that your bilinear form is symmetric and coercive on the correct functional space and that the righthand side is a bounded linear functional in the same space.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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