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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.

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 https://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.

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.

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.

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 \begin{equation} \min_{\psi, \; \lambda} \|H\psi - \lambda \psi \| \; \text{subject to} \; \int_{\Omega^c} \psi^2 = 0. \end{equation} 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 $H=\triangle+V(x)$

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.