0
$\begingroup$

enter image description here ______________________________________________________________ Conversely, in a different approach using Green's 1st identity, I showed that by choosing v ≡ 1, that the compatibility condition is derived, and so if u is harmonic on Ω, then the integral of the normal derivative over the boundary is equal to zero.

enter image description here ______________________________________________________________ However, for circles/balls lying within the closure of the domain (i.e. intersection with the boundary), do I need to use the mean value principle with Green's theorem to prove that u is a harmonic function in Ω?

$\endgroup$
2
$\begingroup$

Since every open set can be expresses as a countable union of open balls you don't have to deal with intersection of the closure of an open set $\Omega$ with balls.

$\endgroup$
4
  • $\begingroup$ Any other suggestions on how to approach this problem? $\endgroup$ – Blasius Boundary Layer Nov 4 '15 at 13:50
  • $\begingroup$ what else do you looking for? $\endgroup$ – Michael Medvinsky Nov 4 '15 at 13:53
  • $\begingroup$ Ok, so to prove that u is harmonic in Ω, we must show that u satisfies the Laplacian on the boundary. By the given condition, u and its first two derivatives are continuous in Ω. $\endgroup$ – Blasius Boundary Layer Nov 4 '15 at 20:47
  • 1
    $\begingroup$ In general, the definition of harmonic function is that applying Laplacian on it you get zero. $\endgroup$ – Michael Medvinsky Nov 4 '15 at 20:52
1
$\begingroup$

To show that u satisfies the Laplacian, we use Green's 1st identity to obtain the inside-outside theorem equality, setting another arbitrary function, v, contained in the same space as u.

Then, we set v = 1 and obtain an new (simplified) expression of Green's first identity. Hence, for an arbitrary circle B lying in Ω closure, with the compatibility condition = 0, then the integral of the Laplacian = 0. Thus, it follows that the Laplacian = 0 and so the second condition of u being harmonic in Ω is satisfied. enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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