I have a variational problem of the form

$$ E(u) = \int_{\Omega} F(x,y,u,u_x,u_y)dxdy $$

which leads me

$$ \frac{dE}{du} = \int_{\Omega} \left( \frac{\partial F}{\partial u} - \frac{d}{dx}\frac{\partial F}{\partial u_x} - \frac{d}{dy}\frac{\partial F}{\partial u_y} \right)h dxdy + \int_{\Gamma} \left(F_{u_x}dy - F_{u_y}dx \right)h $$

Assuming the increment function $h$ is not fixed,it can be arbitrary, how do I get from this term

$$ \int_{\Gamma} \left(F_{u_x}dy - F_{u_y}dx \right)h = 0 $$

the boundary conditions of my problem when I try to apply gradient descent? Namely how do I fill the system

$$ \left\{ \begin{array}{l} \frac{\partial u}{\partial t} = - \frac{\partial F}{\partial u} + \frac{d}{dx}\frac{\partial F}{\partial u_x} + \frac{d}{dy}\frac{\partial F}{\partial u_y} \\ \text{boundary conditions?} \end{array} \right. $$

I should end up having the Von Neumann boundary conditions, but I'm slightly confused how to pass from the line integral to the boundary conditions.

My guess is that we should pose $F_{u_x} = 0, F_{u_y} = 0$ on the boundary, I can't prove it though.

Thank you

  • $\begingroup$ $h=h(x)$? if so, not clear in your question $\endgroup$ – phdmba7of12 Aug 3 '18 at 16:28
  • $\begingroup$ $h=0$ on the boudary. $\endgroup$ – Rafa Budría Aug 3 '18 at 16:36
  • 1
    $\begingroup$ @phdmba7of12, yes it's h=h(x) $\endgroup$ – user8469759 Aug 3 '18 at 17:04
  • 1
    $\begingroup$ @Rafa, it's not 0. $\endgroup$ – user8469759 Aug 4 '18 at 13:19

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