Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose we are trying to solve the Dirichlet problem in a possibly unbounded domain $\Omega \subseteq \mathbb R ^n$ with continuous prescribed boundary data $f$. When $\Omega$ is bounded, it is well known that provided $\partial \Omega$ is nice enough (e.g. for each point there is a sphere touching it only at that point), a solution exists. One makes use of Perron's method to show this, and the boundedness of $f$ on $\partial \Omega$ plays a seemingly crucial role. I am interested in the case where $\Omega$ is unbounded and $f$ is not necessarily bounded on the boundary. Supposing $\partial \Omega$ is nice as before, will the Dirichlet problem generally admit a solution? Are there counterexamples?

share|improve this question
add comment

1 Answer 1

up vote 2 down vote accepted

As long as the complement of $\Omega$ has interior points, you can use an inversion in a sphere in the complement (using the Kelvin transform) to reduce this to the case of bounded domains with not necessarily continuous or even bounded boundary data.

share|improve this answer
    
I see. If I'm understanding you correctly, one can start with a domain contained in the ball with its boundary passing through the ball's centre. Then we specify boundary data so that it's continuous everywhere except at the centre, and when inverting we are left with an infinite domain with continuous boundary data. So I suppose the answer will in general be no. –  user18063 Nov 25 '12 at 5:33
    
I just realized that inversions in spheres don't preserve harmonic functions in dimensions $\ne 2$, so the only case in which this works directly is $n=2$. In that case your interpretation is correct. –  Lukas Geyer Nov 25 '12 at 16:39
    
Oh, thanks, I'll edit the post accordingly. –  Lukas Geyer Nov 25 '12 at 23:14
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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