4
$\begingroup$

For a second order PDE (lets say the Laplace equation), is there a problem with specifying neumann boundary conditions, which instead of being specified in the direction normal to the boundary are instead specified in some other direction.

For example, could one specify the derivative in the direction of the boundary.

Would this lead to a unique solution?

This seems like a stupid question, but I couldn't find any information on it.

$\endgroup$

2 Answers 2

3
$\begingroup$

Suppose for definiteness we work with Laplace's equation $\triangledown^{2}u =0$ on the unit disk in $R^{2}$. Assuming things are somewhat smooth, suppose one specified the tangential, instead of the normal, derivative of $u$, i.e. specified $\partial{u} / \partial{\theta}$ on the unit circle. Picking any point $\theta_{0}$ on the circle, one could integrate in $\theta$ to obtain $u(\theta)$ on the circle:

$u(\theta) = \int_{\theta_{0}}^{\theta} \partial{u}/\partial{\theta}(\alpha) d \alpha + u(\theta_{0})$.

So a tangential derivative really specifies a Dirichlet boundary condition. The additive constant in the above integral merely corresponds to a constant solution in the disk, i.e. it shifts the solution $u$ by a constant amount.

You can in general freely specify Dirichlet or Neumann conditions, but not both. So take your pick, a tangential or a normal derivative for the boundary condition.

$\endgroup$
1
  • $\begingroup$ We love those phat checks! $\endgroup$ Jan 6, 2011 at 21:28
3
$\begingroup$

There is an extensive literature on oblique derivative boundary conditions. A Google search with the keyword "oblique derivative" will get you started.

$\endgroup$

Your Answer

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

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