Laplacian in 1D, 2D, or 3D

Sparse (1-3)D Laplacian on a rectangular grid with exact eigenpairs.

Description

The code computes the exact eigenpairs of (1-3)D negative Laplacian on a rectangular finite-difference grid for combinations of Dirichlet, Neumann, and Periodic boundary conditions using explicit formulas from
http://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors_of_the_second_derivative

The code can also compute the sparse matrix itself, using Kronecker sums of 1D Laplacians. For more information on tensor sums, see
http://en.wikipedia.org/wiki/Kronecker_sum_of_discrete_Laplacians

Example, compute everything for 3D negative Laplacian with mixed boundary conditions:
[lambda,V,A] = laplacian([100,45,55],{'DD' 'NN' 'P'}, 20);
Compute only the eigenvalues:
lambda = laplacian([100,45,55],{'DD' 'NN' 'P'}, 20);
Compute the matrix only:
[~,~,A] = laplacian([100,45,55],{'DD' 'NN' 'P'});

GNU OCTAVE compatible.

This code is a part of the BLOPEX eigensolver package, see
http://en.wikipedia.org/wiki/BLOPEX
or go directly to
http://code.google.com/p/blopex/

Copyright owners: Bryan C. Smith and Andrew V. Knyazev

• MATLAB release: MATLAB 7.11 (2010b)

[Thomas Clark]

This works extremely well, extremely flexibly and with no fuss.

I'm SO upset I didn't find this earlier. I've basically coded an alternative, already, but it isn't close to being as good.

Thank you very much!

[Mr Smart]

 

[Marios Karaoulis]

Hi, useful add on, but one question.
Let's say we have a 3x3 grid, then why all the element on the diagonal are 4? Should be something like this
[2 -1 0 -1 0 0 0 0 0
 -1 3 -1 0 -1 0 0 0
 0 -1 2 0 0 -1 0 0 0 ....

...]

Only line 5 should have 4 on it's diagonal.

[Andrew Knyazev]

Re: Marios Karaoulis

It depends on boundary conditions. Your example apparently uses Neumann boundary condition as is http://en.wikipedia.org/wiki/Laplacian_matrix#As_an_approximation_to_the_negative_continuous_Laplacian
while the code default us the Dirichlet boundary conditions. To get your matrix, please use

[~,~,A]=laplacian([3 3],{'NN' 'NN'})

full(A) shows what you want:

     2 -1 0 -1 0 0 0 0 0
    -1 3 -1 0 -1 0 0 0 0
     0 -1 2 0 0 -1 0 0 0
    -1 0 0 3 -1 0 -1 0 0
     0 -1 0 -1 4 -1 0 -1 0
     0 0 -1 0 -1 3 0 0 -1
     0 0 0 -1 0 0 2 -1 0
     0 0 0 0 -1 0 -1 3 -1
     0 0 0 0 0 -1 0 -1 2

[Marios Karaoulis]

Thanks for the reply Andrew, you are right.