Non-Normal derivative boundary conditions for a PDE - MathOverflow

Non-Normal derivative boundary conditions for a PDE

2011-01-06T05:06:37Z

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.

Answer by drbobmeister for Non-Normal derivative boundary conditions for a PDE

2011-01-06T07:41:14Z

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.

Answer by Michael Renardy for Non-Normal derivative boundary conditions for a PDE

2011-01-06T15:35:06Z

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