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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!

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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. –  Siminore Jul 3 '12 at 11:00
1  
Related: mathoverflow.net/questions/66418/… –  Willie Wong Jul 3 '12 at 11:22
    
@Siminore : Do you know if there is any paper about sufficient conditions for the first eigenvalue of $-\Delta +V$ to be $> 0$? –  Pacciu Jul 4 '12 at 14:08
    
You could start from these lecture notes. Some conditions are hidden there :-) math.nsysu.edu.tw/~amen/posters/pankov.pdf –  Siminore Jul 4 '12 at 14:28
    
I'm absolutely going to take a look at those notes! Thank you. –  Pacciu Jul 4 '12 at 14:51

1 Answer 1

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.

share|improve this answer
    
Great! Thank you Nick. –  Pacciu Jul 4 '12 at 11:54
    
Switched up an important minus sign. Acceleration should be in the opposite direction to the position, not the same. Everything else is still good though. –  Nick Alger Jul 4 '12 at 21:50

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