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Math Help - Convergence of the sequence of discretized solutions of a variational problem

  1. #1
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    Convergence of the sequence of discretized solutions of a variational problem

    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 .
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    Super Member Rebesques's Avatar
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    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).
    Last edited by Rebesques; July 17th 2009 at 01:29 AM.
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    Thank you for the answer. I havenīt 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?
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  4. #4
    Super Member Rebesques's Avatar
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    Not really. Just the definition of weak convergence.
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