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When solving Poisson's equation on the unit square $\Omega$ with homogeneous Dirichlet boundary conditions for $x=0$ and Robin-type conditions at the rest of the boundary, $$ \begin{cases} -\Delta u = 0 &\quad \text{ in }\Omega,\\ u = 0 &\quad\text{ on } \Gamma_{\text{left}},\\ \alpha(u-1) = \mathbf{n}\cdot\nabla u &\quad\text{ on } \Gamma_{\text{rest}}, \end{cases} $$ the result will will of course very much depend on the value of $\alpha$.

For $\alpha\gg 1$, the Robin conditions will essentially enforce $u\approx 1$ on $\Gamma_\text{rest}$, and act as homogeneous Neumann conditions if $\alpha\ll 1$. The behavior for intermediate values is less clear.

Here are some numerical experiments for $\alpha\in\{10^0, 10^1, 10^2, 10^3\}$.

alpha1e0

alpha1e1

alpha1e3

alpha1e4

The oscillations near the boundary stand out. Is there a physical explanation for it or any way this curious behavior could be explained from the equations?

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    $\begingroup$ As the answer by JLC suggests, you are using the wrong sign. $\endgroup$ – Guido Kanschat Sep 20 '13 at 6:54
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Is the sign accurate? If so, you may have an issue since your var form should be

$(\nabla u, \nabla v) - \langle n\cdot \nabla u, v\rangle_{\Gamma_{\rm rest}} = 0$

substituting the Robin condition in gives

$(\nabla u, \nabla v) - \alpha\langle u, v\rangle_{\Gamma_{\rm rest}} = -\langle 1, v\rangle_{\Gamma_{\rm rest}}$

which can mess with your coercivity if $\alpha$ is too large. We've used $\alpha < 0$ in our modeling cases, but I'm not sure what your problem needs.

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