I am looking to understand the conditions under which one can expect "reasonably" accurate solution to leading eigenvalues/eigenvectors of a second order differential operator posed on the real line.

Specifically, consider the following "Smoluchowski" equation on $\mathbb{R}$: Let $U(x)$ be a potential function (such as $x^2$ or $x^4$), then the probability distribution of velocity of a noisy particle moving under this potential in high friction limit satisfies (the linear advection diffusion equation):

$\partial_t(p(x,t))=L[p](x,t)$, where

$L[p](x)=\nabla((\nabla U)p)+(\sigma^2/2)\Delta p$

There is always a "decay at infinity" type of condition that accompanies such equations.

My question is: how does this decay at infinity condition translate into practical computational "rules of thumb" ?

For example, if I wanted to obtain the spectrum of L, I would need to truncate the domain to some finite length. It is known that the invariant distribution (i.e. the 'principle' eigenfunction) decays rapidly away from 0, and I believe other eigenfunctions do so too. Hence, So my question are:

1). When can I claim that my computation on a finite domain gives me accurate information about the top few eigenvalues ?

2). What boundary condition is most appropriate in the numerical computation (Dirichlet or Neumann) ?

  • $\begingroup$ Is your domain $(-\infty, \infty)$? If so, can you do a transformation, say $\tanh: (-\infty,\infty)\to (-1,1)$? $\endgroup$ – Hans Jan 28 '18 at 4:29
  • $\begingroup$ @Hans That sounds like a good idea. Do you have any reference where similar thing has been studied ? $\endgroup$ – mystupid_acct Jan 28 '18 at 6:30
  • $\begingroup$ I do not have any reference but just thought it was natural. $\endgroup$ – Hans Jan 28 '18 at 7:22
  • $\begingroup$ Have you tried it? I am curious how it works out. $\endgroup$ – Hans Jan 28 '18 at 16:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.