Let's consider the heat equation on a $\Omega \subset \mathbb{R}^2$ manifold with a boundary $\Gamma$, with initial and boundary conditions \begin{align} \dot{u}(\mathbf{r}, t) &= \Delta u(\mathbf{r}, t)\quad \mathbf{r}\in\Omega\\ \lim_{\mathbf{r}\to \Gamma} u &= 0\\ \lim_{t\to 0}\left<T_u, \varphi\right> &= \left<\delta_\rho, \varphi\right>, \end{align} where $T\in D^\star$ is a distribution, $\varphi\in D$ is a test function, and $\rho\in \Omega$ is the initial concentration point.
We can convert this problem into an eigenvalue problem by substituting a solution $$u = \sum_{n=1}^\infty e^{-\lambda_n t} \psi_n(\mathbf{\rho})\psi_n(\mathbf{r}),$$ where $\psi_n$ and $\lambda_n$ are the eigenfunctions and eigenvalues respectively. The substitution will yield $$-\sum_{n=1}^\infty \lambda_n e^{-\lambda_n t}\psi_n(\mathbf{\rho})\psi_n(\mathbf{r}) = \sum_{n=1}^\infty e^{-\lambda_n t}\psi_n(\mathbf{\rho})\Delta \psi_n(\mathbf{r}).$$ For some $n$ we will get $$-\lambda_n \psi_n(\mathbf{r}) = \Delta \psi_n(\mathbf{r}).$$
The question is:
Do the eigenvalues $\lambda_n$ have a clear physical meaning in this scenario?
