I've already posted this question on Physics.SE, but I thougth it could be useful to ask also here.

No problem if moderators will ask me to cancel this thread... But, please, have mercy! :-D

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 $\mathcal{L}$ 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, I heard that the $p$-laplacian (i.e., $\Delta_p u := \operatorname{div} (|\nabla u|^{p-2}\ \nabla u)$, which reduces to the usual laplacian when $p=2$) can be used to model nonlinear elasticity or something like that; therefore I have also the following question:

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 $1 < p < \infty$?

Many thanks in advance, guys!

  • $\begingroup$ In the linear case, the spectrum of $-\Delta + V(\cdot )$ is strictly connected to the study of standing waves for Schrödinger equations. There are hundreds of papers about this. For the $p$-Laplace operator, the problem is nonlinear and mostly open. As far as I know, it is chiefly a mathematical problem. $\endgroup$ – Siminore Jul 3 '12 at 11:00
  • 1
    $\begingroup$ Related: mathoverflow.net/questions/66418/… $\endgroup$ – Willie Wong Jul 3 '12 at 11:22
  • $\begingroup$ @Siminore : Do you know if there is any paper about sufficient conditions for the first eigenvalue of $-\Delta +V$ to be $> 0$? $\endgroup$ – Pacciu Jul 4 '12 at 14:08
  • $\begingroup$ You could start from these lecture notes. Some conditions are hidden there :-) math.nsysu.edu.tw/~amen/posters/pankov.pdf $\endgroup$ – Siminore Jul 4 '12 at 14:28
  • $\begingroup$ I'm absolutely going to take a look at those notes! Thank you. $\endgroup$ – Pacciu Jul 4 '12 at 14:51

The left hand side $-F=-\Delta u + Vu$ models force in a material where points try to pull their neighbors towards their local value in a spring-like manner, but also get pulled down by an external force that increases linearly with displacement (for example, other springs or long range gravity).

Now suppose $m$ is understood as a mass (density), and consider Newton's law $F=ma=mu_{tt}$. We see that solving $-\Delta u + Vu=\lambda m u$ is finding modes such that $$-u_{tt}=\lambda u.$$ In other words, modes that will stay the same shape, but simply grow (complex-)exponentially in time.

Here is a 1-dimensional diagram:

enter image description here

Edit: To clarify, the extension to the p-laplacean, $\nabla \cdot |\nabla u|^{p-2} \nabla u=\nabla \cdot k(u,x) \nabla u$ models a material where the force of molecules pulling on their neighbors is p-nonlinear in the displacement gradient. In other words, the "springs" in the above diagram are not ideal.

  • $\begingroup$ Great! Thank you Nick. $\endgroup$ – Pacciu Jul 4 '12 at 11:54

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.