0
$\begingroup$

In the following model of a membrane with a mass particle in it, why does the integral represents the elastic energy of the system?

Let $\Omega$ be an open connected region (the membrane) in $\Re^2$, $x\in \Omega$ and $u(x)$ be the profile of the membrane with $u=0$ at $\partial\Omega$. If $P$ is a unit mass particle we put in the membrane in position $q$, then $$\Delta u=\delta_{q} $$ is satisfied, where $\delta_{q}$ is the Dirac measure. The problem of finding the equilibrium position of the particle $P$ can be reduced to find the function u that minimizes the energy functional: $$ E(u,q)=\frac{1}{2}\int_{\Omega}|\nabla u(x)|^2dx +u(q)$$ which is the sum of the elastic energy and the gravitational energy (considering all physical constants equal to $1$).

$\endgroup$
  • 1
    $\begingroup$ It is probably a small $u$ approximation to the area functional. $$\verb/Area/(u) - \verb/Area/(u \equiv 0) = \int_\Omega \left( \sqrt{1 + |\nabla u|^2} - 1\right) dx \approx \frac12 \int_\Omega |\nabla u|^2 dx$$ $\endgroup$ – achille hui Apr 27 at 4:38
  • $\begingroup$ Perhaps you can try the physics site. For what it's worth, this post there is probably an overkill for your inquiry. Might or might not be helpful to you. $\endgroup$ – Lee David Chung Lin Apr 27 at 6:33
  • $\begingroup$ Crossposted to physics.stackexchange.com/q/476334/2451 $\endgroup$ – Qmechanic Apr 27 at 21:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.