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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 an unit mass particle we put in the membrane in position q, then $$\Delta u=\delta_{q} $$ is satisfied, where $\delta_{q}$ is the shifted Dirac distribution. 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).

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  1. More generally, let there be given a potential $$V[u]~=~\int_{\Omega} d^2x ~{\cal V}(u(x), \nabla u(x),x)\tag{A}$$ with potential density $${\cal V}(u(x), \nabla u(x),x)~=~\frac{1}{2}|\nabla u(x)|^2-f_{\rm ext}(x)u(x).\tag{B}$$ In eq. (B) the first term is the elastic potential energy density (for sufficiently small displacements $u$). This is explained in any good textbook on continuum mechanics. Alternatively, see the last chapter of Goldstein, Classical Mechanics.

    Moreover in eq. (B), $f_{\rm ext}(x)$ is an external force density, cf. e.g. this Phys.SE post. In OP's case $$ f_{\rm ext}(x)~=~-\delta^2(x-q)\tag{C}$$ is the (shifted) Dirac delta distribution

  2. The force density is the functional derivative $$ f(x)~=~-\frac{\delta V[u]}{\delta u(x)}~=~\nabla^2u(x)+f_{\rm ext}(x). \tag{D}$$ Equilibrium points are stationary points.

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