Physics Asked by Jakub Korsak on January 12, 2021
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}inOmega
lim_{mathbf{r}to Gamma} u &= 0
lim_{tto 0}left<T_u, varphiright> &= left<delta_rho, varphiright>,
end{align}
where $Tin D^star$ is a distribution, $varphiin D$ is a test function, and $rhoin Omega$ 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?
The n'th eigenvalue relates to the time decay constant for an initial condition that is directly proportional to the n'th eigenfunction.
Answered by Chet Miller on January 12, 2021
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