# Physical interpretation: weighted eigenvalues of the Laplacian with a potential

I'm a mathematician with only the basic knowledge of Physics, so my question may be trivial: in this case, mercy me. :-)

Let $\Omega \subseteq \mathbb{R}^N$ be a domain and let $V,m:\Omega \to \mathbb{R}$ be two measurable and sufficiently summable functions.

When one considers the eigenvalue problem for the operator $\mathcal{L}:=-\Delta +V$ w.r.t. the weight $m$, i.e.: $$\tag{P} \begin{cases} -\Delta u(x) + V(x)\ u(x) = \lambda\ m(x)\ u(x) &\text{, in } \Omega\\ u(x)=0 &\text{, on } \partial \Omega , \end{cases}$$ the function $V$ is usually called potential and the function $m$ is called weight.

Then, a weighted eigenvalue of $-\Delta +V$ w.r.t. $m$ is any number $\lambda \in \mathbb{R}$ s.t. (P) has at least one nontrivial weak solution $u\in H_0^1(\Omega)$, i.e.: $$\forall \phi \in C_c^\infty(\Omega),\quad \int_\Omega \nabla u\cdot \nabla \phi\ \text{d} x + \int_\Omega V\ u\ \phi\ \text{d} x = \lambda\ \int_\Omega m\ u\ \phi\ \text{d} x\; .$$

My questions are:

1. Is there any reasonable physical interpretation of those eigenvalues? And what is it?

2. Why have the functions $V$ and $m$ those names?

Moreover:

What about any possible physical meaning of the nonlinear weighted eigenvalues coming from the problem: $$\tag{Q} \begin{cases} -\Delta_p u(x) + V(x)\ |u(x)|^{p-2}\ u(x) = \lambda\ m(x)\ |u(x)|^{p-2}\ u(x) &\text{, in } \Omega\\ u(x)=0 &\text{, on } \partial \Omega , \end{cases}$$ where $1 < p < \infty$ and $\Delta_p$ is the $p$-laplacian (i.e., $\Delta_p u := \nabla \cdot (|\nabla u|^{p-2}\ \nabla u)$, which reduces to the usual laplacian when $p=2$)?

• I do not dare to give the answer. But I can point out why $V$ is named like that. To me, it seems that (P) is the Dirichlet problem associated to the time-independent Schrödinger equation $H u=E u$. $H$ contains the kinetic part, $(\hat{p})^2/2m$, which is essentialy the Laplacian you wrote in (P), and the potential part $V$; whence the notation and name potential in your problem. – c.p. Jul 1 '12 at 21:32
• (and being those equations generalizations of Schrödinger's, $\lambda$ is a genralization of the energy) – c.p. Jul 1 '12 at 23:48