0
$\begingroup$

I am studying introduction to Multigrid methods. In all tutorials, authors write that eigenvectors of Laplacian (1D, finite difference) are given as

$w_k(x_i) = \sin(k \pi x_i),$

where $x_i$ is a discretization point. Can anybody point me to literature which derives this formula? Or can anybody sketch the derivation?

$\endgroup$
2
  • 3
    $\begingroup$ At some level this result is just intuitive, because sines/cosines are the (continuous) eigenfunctions of the (continuous) d^2/dx^2 operator. The choice of boundary conditions will restrict you to one family or the other. Armed with that intuition you should be able to guess and check this result precisely, by just inserting those wk's into the (discrete) eigenstatement. You should be able to find a trigonometric identity that lets you reduce/equate [-1 2 -1] ' [wk(i-1) wk(0) wk(i+1)] = lambda*wk(0) .. probably some flavor of angle sum/difference identity(s). $\endgroup$ Mar 14, 2022 at 18:30
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Mar 14, 2022 at 20:17

2 Answers 2

2
$\begingroup$

You can find a discussion about the topic (and the derivation) in chapter V of:

R. Courant and D. Hilbert, Methods of Mathematical Physics, vol. I (1989). doi:10.1002/9783527617210

$\endgroup$
6
$\begingroup$

They're on Wikipedia, for instance, in a page with the slightly unclear name of "Eigenvalues and eigenvectors of the second derivative".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.