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Standard Galerkin method reduces the problem

Find $u\in V$ such that $a(u,v) = f(v)$ for all $v \in V$, where $V$ is Hilbert space, $a$ is bilinear form and $f\in V^*$.

to a finite dimensional problem by introducing a $n$-dimensional subspace $V_n\subset V$, then we look for an approximation $u_n$ of the solution $u$ such that

Find $u_n \in V_n$ such that $a(u_n,v) = f(v)$ for all $v \in V_n$.


I would like to replace $V_n$ by a $n$-dimensional manifold $\mathcal{M}_n$ in $V$. So the reduced problem would be

Find $u_n\in \mathcal{M}_n$ such that $a(u_n,v)=f(v)$ for all $v\in T_{u_n}\mathcal{M}_n$, where $T_{u_n}\mathcal{M}_n$ is tangent space to the manifold $\mathcal{M}_n$ at the point $u_n$.

What do we know about this problem? What are the conditions on $\mathcal{M}_n$ for existence of $u_n$? Does $u_n$ converge to $u$ as we make $\mathcal{M}_n$ bigger and bigger ($n\rightarrow \infty$)? What is this method called?(I called it Manifold Galerkin method)

Can you please point me to the literature where they discuss this problem?

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  • $\begingroup$ The answers to the following scicomp question may be helpful, though the question is more computationally oriented than the question here: scicomp.stackexchange.com/questions/19007/… $\endgroup$ – Nick Alger Aug 26 '16 at 18:15
  • $\begingroup$ @NickAlger Sorry but it is not helpful, the question is very different from mine and only the name of the question is similar. Please read my question a little bit more carefully. $\endgroup$ – tom Aug 26 '16 at 18:20
  • $\begingroup$ I did read your question carefully. My comment above is a helpful link to additional related information, not an answer $\endgroup$ – Nick Alger Aug 26 '16 at 18:28

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