I am trying to derive the wave equation for a membrane of general shape from Hamilton's principle: $$W(u)=\int_{t_1}^{t_2}(T(u,t)-U(u,t))dt$$ and $$\frac{d}{ds}W(u+s\phi)|_{s=0} =0$$ The kinetic energy $$T(u,t)=\rho/2\int_\Omega d^2x \dot{u}^2(x,t)$$ and the potential energy is given by $$\sigma/2 \int_\Omega d^2x ||\nabla u||^2$$ If I plug these last two definitions into the equation describing Hamilton's principle I get: $$\int_{t_1}^{t_2}\int_\Omega\left[\rho \dot{u}\dot{\phi}-\sigma \left(\partial_{x_1} u \partial_{x_1}\phi +\partial_{x_2} u \partial_{x_2}\right)\right]dtd^2x$$ Manipulating the first part is easy: By using partial integration wrt $t$ I can get rid of the first derivative of $\phi$ wrt $t$ but I am not so sure about the second part. I got a hint to use Gauss's integral theorem, but I don't know how to use it here.


1 Answer 1


I do not think Gauss theorem is necessary to solve the equation. We can have the answer just with integration by part: \begin{align} \int_{\Omega}{(\partial_{x_1} u\partial_{x_1} \phi+\partial_{x_2} u \partial_{x_2} \phi)d^2x} & = \int{\bigg(\int{(\partial_{x_1} u)d\phi\big|_{x_2}}\bigg) dx_2}+\int{\bigg(\int{(\partial_{x_2} u)d\phi\big|_{x_1}}\bigg) dx_1} \\ & = -\int_{\Omega}{\bigg({\partial^2 u \over \partial x_1^2}\bigg)\phi d^2x}-\int_{\Omega}{\bigg({\partial^2 u \over \partial x_2^2}\bigg)\phi d^2x} \\ & = -\int_{\Omega}{(\nabla^2u)\phi d^2x} \end{align} If we plug this back in the equation derived from the Hamiltonian's principle, we have $$\int_{t_1}^{t_2}{\int_{\Omega}{(-\rho\ddot{u}+\sigma\nabla^2u)\phi d^2x}dt}=0$$ This gives us the wave equation $-\rho\ddot{u}+\sigma\nabla^2u=0$.

Edit: The folowing is the updated version of how Gauss theorem can be applied to the problem. I did not think of the function $\nabla \cdot (\phi \nabla u)$ in my original consideration.

At last, let us take a look how Gauss theorem can be applied to the problem. Since $$\nabla \cdot (\phi \nabla u) = \nabla\phi \cdot \nabla u +\phi\nabla^2u = \partial_{x_1} u\partial_{x_1} \phi+\partial_{x_2} u \partial_{x_2} \phi+\phi\nabla^2u$$ the integral $\int_{\Omega}{(\partial_{x_1} u\partial_{x_1} \phi+\partial_{x_2} u \partial_{x_2} \phi)d^2x}$ is \begin{align} \int_{\Omega}{(\partial_{x_1} u\partial_{x_1} \phi+\partial_{x_2} u \partial_{x_2} \phi)d^2x} & = \int_{\Omega}{(\nabla \cdot (\phi \nabla u)-\phi\nabla^2u)d^2x} \\ & = \int_{\partial \Omega}{(\phi \nabla u) \cdot dl}-\int_{\Omega}{(\nabla^2u)\phi d^2x} \end{align} The first term, $\int_{\partial \Omega}{\phi \nabla u \cdot dl}$ is $0$ since $\phi=0$ on the boundary, and we have $$\int_{\Omega}{(\partial_{x_1} u\partial_{x_1} \phi+\partial_{x_2} u \partial_{x_2} \phi)d^2x}=-\int_{\Omega}{(\nabla^2u)\phi d^2x}$$

  • $\begingroup$ That's a really good answer! Thanks! $\endgroup$
    – eeqesri
    Aug 6 at 16:29
  • 1
    $\begingroup$ The answer has been updated since I found a way to apply Gauss theorem without integration by part. $\endgroup$
    – Andy Chen
    Aug 6 at 17:20

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