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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!

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  • $\begingroup$ I remember seeing the required definition in Y.Kifer's Random Perturbation of Dynamical systems for a general elliptic operator. $\endgroup$
    – Piyush Grover
    Commented Mar 17, 2014 at 20:32

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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.

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  • $\begingroup$ How can we be sure that $\lambda_1$ in the above limit is finite? $\endgroup$
    – rubikscube09
    Commented Oct 28, 2019 at 16:21
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    $\begingroup$ 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
    Commented Mar 19, 2021 at 12:28
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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

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  • $\begingroup$ 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
    Commented Mar 17, 2014 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
    Commented Mar 17, 2014 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
    Commented Mar 18, 2014 at 0:00
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    $\begingroup$ 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
    Commented Mar 18, 2014 at 0:59

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