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Let $H = \triangle + V(x) : \mathbb{R}^2 \rightarrow \mathbb{R}^2$. I am interested in domain decomposition for an eigenproblem involving $H$.

The lowest 1000 eigenfunctions of $H$, $\psi_i$, can be partitioned using a region, $\Omega \subset \mathbb{R}^2$, such that each $\psi_i$ localizes either inside of $\Omega$ or outside of $\Omega$. $\Omega$ is not a subspace of $\mathbb{R}^2$ as it may be an oddly shaped region.

Label the inner eigenfunctions $\psi_i^{in}$ and the outer ones $\psi_i^{out}$. There's only about 10 $\psi_i^{in}$s. Given $\Omega$, my goal is to efficiently compute the $\psi_i^{in}$.

One way to find the $\psi_i^{in}$ would be to discretize, compute all 1000 $\psi_i$s, and then partition. This is what I do now (5-point stencil for $\triangle$ on a $10^3 \times 10^3$ grid). The problem is that this requires diagonalizing over a 1000 dimensional space in order to get 10 eigenvectors. It seems like there would be a cheaper way to compute the $\psi_i^{in}$.

Edit: I reposted to http://scicomp.stackexchange.com/questions/1396/efficiently-computing-a-few-localized-eigenvectors#comment2200_1396 and hopefully clarified the problem statement.

Edit I think I can solve this if I can at least figure a way to solve \begin{equation} \max \psi^T H \psi \text{ subject to } P\psi = \psi \text{ and } \psi^T \psi = 1 \end{equation} where $P$ is projection onto the space of functions localized over $\Omega$. My guess is that this will end up looking like power iterations with a projection step built in between matrix applies. If this is doable then something like inverse iteration should be doable which will give me what I want.

5 It was confusing before. It's still confusing.

Let $H = \in triangle + V(x) : \mathbb{R}^{1000 mathbb{R}^2 \times 1000}$ be symmetric positive definite. rightarrow \mathbb{R}^2$. I am interested in domain decomposition for an eigenproblem involving$H$. The lowest 100 eigenvectors 1000 eigenfunctions of$H$,$ \psi_i$, \psi_i$, can be partitioned using a region, $\Omega \subset \mathbb{R}^{1000}$, mathbb{R}^2$, such that each$\psi_i$localizes either inside of$\Omega$or outside of$\Omega$.$\Omega$is not a subspace of$\mathbb{R}^2$as it may be an oddly shaped region. Label the inner eigenvectors eigenfunctions$\psi_i^{in}$and the outer ones$\psi_i^{out}$. There's only about 10$\psi_i^{in}$s. Given$\Omega$, my goal is to efficiently compute the$\psi_i^{in}$. One way to find the$\psi_i^{in}$would be to discretize, compute all 100 1000$\psi_i$s \psi_i$s, and then partition. This is what I do now (5-point stencil for $\triangle$ on a $10^3 \times 10^3$ grid). The problem is that this requires diagonalizing over a 100 1000 dimensional space in order to get 10 eigenvectors. It seems like there would be a cheaper way to compute the $\psi_i^{in}$.

I suppose another way to do it would be to solve\min_{\psi, \; \lambda} \|H\psi - \lambda \psi \| \; \text{subject to} \; \int_{\Omega^c} \psi^2 = 0.Which I imagine is a pretty standard way to rewrite eigenvalue problems but I don't know where to look for more on this.

Edit: Replace 1000 by 100000. I can't cop out of this with a full svd!

Edit two and additional question: I'm very interested in this idea to restart Lanczos. It seems like that should work but don't we need some kind of localizing convergence'' result? Has anyone seen something like this before?

Edit three: I think we can get "localizing convergence" by looking at the convergence behavior of the power method. Here, we see that $x_k \rightarrow \psi$ along a direction orthogonal to $\psi$. Thus, if $\psi$ is in the span of $\Omega$ and the angle between some sequence $y_k$ and $\Omega^\bot$ is increasing then we can infer that $y_k$ is not converging reposted to $\psi$ http://scicomp.stackexchange.com/questions/1396/efficiently-computing-a-few-localized-eigenvectors#comment2200_1396 and we can throw it out. Maybe. I still need to look more at the proof in hopefully clarified the case of subspacesproblem statement.

Edit four: it's mp since $H=\triangle+V(x)$

Let $H \in \mathbb{R}^{1000 \times 1000}$ be symmetric positive definite. The lowest 100 eigenvectors of $H$, $\psi_i$, can be partitioned using a region, $\Omega \subset \mathbb{R}^{1000}$, such that each $\psi_i$ localizes either inside of $\Omega$ or outside of $\Omega$. Label the inner eigenvectors $\psi_i^{in}$ and the outer ones $\psi_i^{out}$. There's only about 10 $\psi_i^{in}$s. Given $\Omega$, my goal is to efficiently compute the $\psi_i^{in}$.

One way to find the $\psi_i^{in}$ would be to compute all 100 $\psi_i$s and then partition. The problem is that this requires diagonalizing over a 100 dimensional space in order to get 10 eigenvectors. It seems like there would be a cheaper way to compute the $\psi_i^{in}$.

I suppose another way to do it would be to solve $$\min_{\psi, \; \lambda} \|H\psi - \lambda \psi \| \; \text{subject to} \; \int_{\Omega^c} \psi^2 = 0.$$ Which I imagine is a pretty standard way to rewrite eigenvalue problems but I don't know where to look for more on this.

Edit: Replace 1000 by 100000. I can't cop out of this with a full svd!

Edit two and additional question: I'm very interested in this idea to restart Lanczos. It seems like that should work but don't we need some kind of localizing convergence'' result? Has anyone seen something like this before?

Edit three: I think we can get "localizing convergence" by looking at the convergence behavior of the power method. Here, we see that $x_k \rightarrow \psi$ along a direction orthogonal to $\psi$. Thus, if $\psi$ is in the span of $\Omega$ and the angle between some sequence $y_k$ and $\Omega^\bot$ is increasing then we can infer that $y_k$ is not converging to $\psi$ and we can throw it out. Maybe. I still need to look more at the proof in the case of subspaces.

Edit four: it's mp since $H=\triangle+V(x)$

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