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How to solve the exterior problem on a ball with radius $r$ in the 3d space? I have to found u such that:

$\Delta u = 0$ in $B(0,r)^C$

Thanks!

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Where's the $r$ coming from? –  Cameron Buie Nov 2 '12 at 10:45
    
I gathered that much, but there's no $r$ in the expression for $u$, so I don't see why it should vanish as $r\to\infty$. Also, what is $S(0,a)$? –  Cameron Buie Nov 2 '12 at 11:31
    
I'm sorry this question is wrong. I want to know the solution 3d space for a ball B(0,a) with radius $a$ and center at point 0. $(r,\theta,\phi)$ are spherical coordinates. S is a sphere of radius $a$ with center in 0. –  TheStudent Nov 2 '12 at 16:41
    
You should clarify all of that in the body of the question, itself. That will "bump" the question up on the list, and someone should see it and hopefully be able to answer it. –  Cameron Buie Nov 2 '12 at 18:52

1 Answer 1

The solution to the Dirichlet problem uses the Green's function for the Laplacian. In 3D, this is $H(r) = -1/4\pi|r|$ (for vector $r$), such that $\nabla^2 H(r) = \delta(r)$, per the definition of the Green's function.

Some use of the various generalized Stokes theorems gives the following (when $\nabla^2 u = 0$):

$$u(r) = \oint \nabla'H(r-r') \cdot dS' \; u(r') - \oint H(r-r') \; dS' \cdot \nabla' u(r')$$

The choice of the Dirichlet Green's function enforces that $H(r-r') = 0$ on the boundary. The second integral vanishes, leaving only the first. You'll have to be careful about the orientation of $dS'$ (since this is an exterior problem), but otherwise, you should be able to exploit the symmetries of the problem to reduce the surface integral to something tractable. Good luck!

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Right. What do you think if I expand $H(r)$ in Legendre polynomials? Do you think is possible to simplify more than: $u(x)=C\sum(\frac{a}{\vert x \vert})^n \int_{S} g(\theta) P_n(cos(\mu)) d\omega$ ? Where $C$ is a constant. –  TheStudent Nov 2 '12 at 19:45
    
A Legendre expansion is appropriate, yes, and I don't see any obvious way to simplify this. Do you need an exact solution everywhere (one that will ultimately be expressed in these polynomials)? Or will you be able to reduce it down to the leading order contribution at long distances? –  Muphrid Nov 2 '12 at 19:55
    
I need an exact solution. The problem ask: Justify your answer using the maximum principle in the regions $\vert a \vert\leq x \leq R$. Do you have any idea? –  TheStudent Nov 2 '12 at 20:02
    
The maximum principle merely says that a harmonic function has to take on all values between boundary's minimum and maximum. I'm not really sure what that has to do with this problem other than maybe verifying the final answer. If you have not posted the exact text of the question, I suggest you do so. It feels like something is missing here. –  Muphrid Nov 2 '12 at 20:16
    
Now the question is complete! –  TheStudent Nov 2 '12 at 20:25

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