3
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ As the answer by JLC suggests, you are using the wrong sign. $\endgroup$ Sep 20, 2013 at 6:54

1 Answer 1

4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.