I'm studying the membrane problem. We want to find the vertical displacement of a membrane, this position is the solution of a boundary problem which involves the Poisson equation. And it is also equivalent to minimize the Dirichlet energy, this functional is: $$E(u)=\frac{1}{2}\int_{\Omega}\mid\nabla u(x)\mid^{2}dx.$$

Can anyone tell me how it is deduced the equation (Poisson) or the energy functional? What physical meaning have both?

Here there is more information about the problem, this is the obstacle problem which is related to this: https://en.wikipedia.org/wiki/Obstacle_problem


$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \delta E & = \int_{\Omega}\pars{\delta\verts{\nabla\mrm{u}}^{2}}\,\dd x = \int_{\Omega}\pars{\verts{\nabla\mrm{u} + \nabla\delta\mrm{u}}^{2} -\verts{\nabla\mrm{u}}^{2}}\,\dd x = \overbrace{\int_{\Omega}2\nabla\mrm{u}\cdot\nabla\delta\mrm{u}\,\dd x} ^{\ds{+\ \mbox{terms of order}\ \pars{\delta\mrm{u}}^{2}}} \\[5mm] & = \int_{\Omega}\bracks{2\nabla\cdot\pars{\delta\mrm{u}\nabla\mrm{u}} - 2\,\delta\mrm{u}\nabla^{2}\mrm{u}}\,\dd x = 2\int_{\partial\Omega}\delta\mrm{u}\nabla\mrm{u}\cdot\dd\mathbf{S} - 2\int_{\Omega}\delta\mrm{u}\nabla^{2}\mrm{u}\,\dd x = 0 \end{align}

$\ds{\mrm{u}}$ is 'fixed' at $\ds{\partial\Omega}$ such that the first integral vanishes out and $\ds{\nabla^{2}\mrm{u} = 0}$

Poisson requires a constraint as $\ds{\Phi = \int_{\Omega}\rho\,\mrm{u}\,\dd x}$. It yields ( via Lagrange Multipliers ) to an additional term like $\ds{\mu\int_{\Omega}\delta\mrm{u}\,\rho\,\dd x}$ which leads to ( Poisson ) $\ds{\nabla^{2}\mrm{u} = -\mu\rho}$.

  • 1
    $\begingroup$ I didn't understand anything, I don't know so much about physics. $\endgroup$ – Skullgreymon Jun 26 '17 at 18:00

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