The first Dirichlet eigenvalue of a compact domain $\Omega\subset\mathbb{R}^n$ with smooth boundary is the smallest positive number for which there exists a non-trivial solution to $$ -\Delta\psi = \lambda\psi\ \ \ \text{in}\ \Omega\\ \psi=0\ \ \ \text{on}\ \partial\Omega. $$

I remember hearing that there is a very concrete interpretation of $\lambda_1(\Omega)$ in terms of (expected) boundary hitting times of random walks, but I can't seem to find this written down explicitly anywhere, and I can't seem to come up with it on my own.

I would greatly appreciate a reference, or if this is not true that would be helpful to know as well!

  • $\begingroup$ I remember seeing the required definition in Y.Kifer's Random Perturbation of Dynamical systems for a general elliptic operator. $\endgroup$ – Piyush Grover Mar 17 '14 at 20:32

The first eigenvalue is the exponential rate for hitting the boundary, i.e. $$ \lambda_1 = - \lim_{t\to\infty} \frac{1}{t} \log P_{u_0}(\tau > t) , $$ where $\tau := \inf\{X_t \in \partial \Omega\}$ and $\mathrm{law}\, X_0 = u_0$ for a Brownian motion $X_t$ killed at $\partial \Omega$. This probability is also given in terms of the density of $X_t$ which solves the pde $$\begin{align} \partial_t u &= \Delta u & \text{in } \Omega \\ u &= 0 &\text{on } \partial \Omega \\ u(t=0,\cdot) &= u_0(\cdot) . \end{align} $$ Therewith, the probability of not yet hit the boundary is the probability to be still inside of $\Omega$, that is $$ P_{u_0}(\tau > t ) = \int_\Omega u(t,x) \;\mathrm{d} x . $$ Assuming now that $0<\lambda_1 <\lambda_2 \leq \dots$, it follows by the semigroup representation of the solution to the pde $$ \int u(t,\cdot) = e^{-\lambda_1 t} \left( \langle u_0, \psi_1 \rangle \int \psi_1 + O(e^{-(\lambda_2-\lambda_1)t})\right), $$ where $\psi_1$ is the first eigenfunction. From here the claim follows.

  • $\begingroup$ How can we be sure that $\lambda_1$ in the above limit is finite? $\endgroup$ – rubikscube09 Oct 28 '19 at 16:21
  • $\begingroup$ On a compact domain, you have a Poincaré inequality and the inverse of the Poincaré constant gives a lower bound on $\lambda_1$ by the Rayleigh principle. On the other hand, an upper bound is also obtained by the Rayleigh principle provided that there exists a non-zero function in $f\in H_0^1(\Omega)$, that is Sobolev functions. For this any bump function supported inside of $\Omega$ will do it, if $\Omega$ has non-empty interior. $\endgroup$ – André Schlichting Mar 19 at 12:28

The First eigenvalue determine the behavior of the heat equation and then of the brownian motion, see reference therein http://www.actamath.com/Jwk_sxxb_en/EN/article/downloadArticleFile.do?attachType=PDF&id=6458

Edit: The new link is Free? Else I can send you the pdf.

Of course there is the marvelous article of Kac, where the role of the eigenvalue in the diffusion process is make so clearly https://www.math.ucdavis.edu/~saito/courses/ACHA.READ.F03/kac-drum.pdf

  • $\begingroup$ Hey Paul, thanks for the answer; the limit this paper presents on the first page doesn't seem to have any content, unless I'm misinterpreting the notation: the author claims $\lim_{t\rightarrow\infty} e^{-\lambda_1 t/2}P^x(x_t\in K)=0$, but $P^x\leq 1$ always, and for any compact domain of the type I described, $\lambda_1>0$. Am I misinterpreting something? $\endgroup$ – quick_q Mar 17 '14 at 20:11
  • $\begingroup$ @quick: I can't access the paper right now, but double check the sign of the operator being used, ie whether the author is discussing an eigenvalue of the Laplacian or its negative. $\endgroup$ – Nate Eldredge Mar 17 '14 at 23:40
  • $\begingroup$ @NateEldredge - thanks; they do say "the smallest eigenvalue of $-\Delta$" but you're definitely right - there's a sign error there. I found a paper, "Inequalities for Exit times and Eigenvalues of Balls", which has basically the same formula written. The author calls this "Kac's formula" with no reference, and I would really like to have a proof of this; searching "Kac's formula" always returns Feynman-Kac, which I don't think implies this limit... $\endgroup$ – quick_q Mar 18 '14 at 0:00
  • 2
    $\begingroup$ Ah, it can be derived from Feynman-Kac; thanks! I am willing to post a short sketch of the proof if anyone is interested. $\endgroup$ – quick_q Mar 18 '14 at 0:59

Your Answer

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

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