Convergence of the sequence of discretized solutions of a variational problem

**Dynamite** · June 20th 2009, 10:11 PM

Consider the Galerkin discretization of an abstract variational problem where the Hilbert space V is separable.
http://en.wikipedia.org/wiki/Galerki...stract_problem

Each of the subspaces Vn is generated by the first n terms of a sequence of elements of the separable Hilbert space V. This sequence is such that each of these subspaces

will be dense in V. The problem is proving that there is a subsequence of the bounded sequence of discretized solutions

, that converges weakly to the solution of the variational abstract problem

, then proving that the same subsequence converges strongly and finally proving that the sequence of discretized solutions

converges to the solution of the variational abstract problem

.

**Rebesques** · July 16th 2009, 11:51 PM

Let's see... Let $(u^{(n)})$ be the Galerkin approximation. Identify elements in each subspace $V_n$ with their natural injections into $V$ . Coercivity of $a$ implies that $c||u^{(n)}||^2\leq a(u^{(n)},u^{(n)})=f(u^{(n)})\leq ||f||||u^{(n)}||$ , which gives that $(u^{(n)})$ is bounded in $V$ and must therefore possess a subsequence, denoted by $(u^{(n)})$ also, to weakly converge to $u\in V$ . Since $a(\cdot,v)$ is linear, we have $a(u^{(n)},v)\rightarrow a(u,v)$ which means $u$ is also a solution of the initial problem. Now, to demonstrate the strong convergence of $u^{(n)}$ , notice that coercivity and Galerkin orthogonality imply $c||u^{(n)}-u||^2\leq a(u-u^{(n)},u-u^{(n)})=f(u-u^{(n)})=0(n)$ .

**Dynamite** · July 29th 2009, 08:58 PM

Thank you for the answer. I haven´t been able to prove as yet that, as you state in your answer, for any linear form

and for any

weakly convergent to u, we will always have that

. Can this be infered from the Lax Milgram lemma?

**Rebesques** · July 30th 2009, 01:37 AM

Not really. Just the definition of weak convergence.

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