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It's well known that: Given a continuous function defined on the boundary of the disk, then there exists a unique harmonic function in the interior of the disk. What if we replace the disk by a cone?

Take $\Omega$ as $$z^2=x^2+y^2,x^2+y^2\leq1$$ for example.

What's the regularity of boundary function $f$ that can guarantee the existence of a harmonic function of $u$? If $u$ exists, what's the regularity of $u$? Does $u$ satisfy the Green's formula? i.e., $$\int_\Omega |\nabla u|^2dxdy=\int_{x^2+y^2=1}u\frac{\partial u}{\partial n}dS?$$

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Since $\Omega$ satisfies the exterior sphere condition, all the boundary points of $\Omega$ are regular. So for a continuous boundary function the solution is continuous in $\bar \Omega$. If the boundary function is piecewise $C^1$ then I think the Green formula should hold though I didn't check the details. –  Andrew Jan 12 at 12:20
    
@Andrew:For a disk, your answer is OK. But for a cone(It's not a $C^1$ manifold), it's not so obvious. –  jiangsaiyin Jan 12 at 12:46
    
Oops, my previous comment was for a cone. As for the surface, it is locally flat. And so the problem reduces to the Dirichlet problem in a circle sector with periodic conditions. May be it can be dealt with a standard approach via Fourier Series. –  Andrew Jan 12 at 13:26

1 Answer 1

If I understand correctly what a "harmonic function on the cone" is, there is no difference between the cone and the disc for the Dirichlet problem.

This cone is a surface equipped with a Riemannian metric (induced from $R^3$) with one singularity (at the vertex). The metric defines a conformal structure turning the cone minus vertex into a Riemann surface. It is easy to see that this Riemann surface is conformally equivalent to the punctured disc.

To write this conformal map explicitly, we first cut the cone along the segment which projects to $x\leq 0$, then "unwrap" onto the sector $|z|<R$, $|\arg z|<\alpha$, where $\alpha\in(0,\pi)$ is the total angle at the vertex, as seen on the surface of the cone, and then compose this with the conformal map $z\mapsto z^{\pi/\alpha}$, (principal branch) in the sector. The map evidently extends through the cut to the conformal map of the whole cone onto the whole $|z|<R^{\pi/\alpha}$, conformal except at the vertex, of course.

For the cone in the question, $R=\sqrt{2}$, and $\alpha=2\pi/\sqrt{2}$.

Now it is easy to solve the Dirichlet problem on the punctured disc (with data on the circle): the solution is the same as for the usual disc; it extends to the puncture by contnuity (removable singularity theorem for harmonic functions).

Thus there is no difference between this cone and the disc for the Dirichlet problem.

Concerning the Green formula, it also holds, but notation in the question seems strange: what is $dxdy$ and $dS$ ? One has to take the gradient with respect to the metric on the cone, and write the area element on the cone instead of $dxdy$. When all this is corrected, there will be a constant multiple in the inequality, depending on $\alpha$, the opening of the cone.

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It's evident that it is conformally equivalent to a sector. But what is the explicit map on the punctured disk? –  Andrew Jan 12 at 17:37
    
I inserted the explicit construction to my answer. –  Alexandre Eremenko Jan 12 at 20:04

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