0
$\begingroup$

Let $V \subset H \subset V^*$ be a Gelfand triple, all Hilbert and separable spaces.

I consider the PDE with weak form: find $u \in L^2(0,T;V)$ with $u' \in L^2(0,T;V')$ such that $$\langle u'(t), v(t) \rangle + b(u(t),v(t)) = \langle f(t), v(t) \rangle$$ for appropriate bilinear form $b$.

Is there a way to obtain an uniform bound on the derivative of the Galerkin approximations, $u_n'$, in the space $L^2(0,T;V^*)$ without using a orthogonal and orthonormal basis $w_j$ for the spaces $V$ and $H$ respectively? So I am looking for techniques to get $$\lVert u'_n \rVert_{L^2(0,T;V^*)} < C$$ when the basis is not orthogonal/orthonormal.

(I know I can orthonormalise any basis but I don't want to do that).

$\endgroup$
3
  • $\begingroup$ Can you show me how it goes if you have orthonormal bases? $\endgroup$
    – timur
    Commented Jun 25, 2013 at 19:27
  • $\begingroup$ @timur The Galerkin approximation is $\langle u'_n(t), w_j \rangle + b(u_n(t), w_j(t)) = \langle f(t), w_j \rangle$ for all $j=1,...,N$. From this one can infer $$u'_n + Bu_n = P_nf$$ where $B$ is the operator that generates the bilinear form $b$ and $\langle P_n f , v \rangle := \langle f, P_n v \rangle$. Then in this equation we can get the bound since $Bu_n$ and $P_nf$ are bounded independently of $n$. $\endgroup$
    – user35613
    Commented Jun 25, 2013 at 19:35
  • 1
    $\begingroup$ Galerkin approximations depend only on the subspaces you choose, in your case the spaces spanned by the first $n$ basis functions. Try defining the operations $B$ and $P_n$ etc without using the basis functions. $\endgroup$
    – timur
    Commented Jun 26, 2013 at 1:19

0

You must log in to answer this question.

Browse other questions tagged .