Laplace problem for eccentric spheres

[gparker267]

Homework Statement

General solution for eccentric spheres, smaller sphere (radius, b) completely embedded within larger sphere of radius c. The centers of both spheres lie on z-axis, distance a, apart (note: c>b+a). Problem is symmetric, so consider θ=[0,∏], r=[0,c]. The inner sphere is grounded while the outer sphere is held at a potential f2(θ).

Homework Equations

General solution (see Morse & Feshbach, 1953)is given in coordinates of larger sphere as:

V(r,θ)=Ʃ1((r/a)^s×Ʃ2(A_n×(a/b)^n×[(-1)^(n-s)n!/s!(n-s)!]-
(a/r)^(s+1)×Ʃ3(A_n×(b/a)^(n+1)×[s!/n!(s-n)!])P_s(cosθ)
Limits of summations
Ʃ1, s=0 to s=s_max (can use s_max=50)
Ʃ2, n=s to n=n_max (can use n_max=50)
Ʃ3, n=0 to s=s

The Attempt at a Solution

I have substituted the relevant boundary conditions into the general solution and computed the unknown coefficients A_n. For the special case when the offset,a is zero,the solution behaves well (and converges nicely to prescribed boundary conditions, BCs). However, in cases where the offset is greater than zero, the solution blows up ( and does not converge to BCs). Any ideas on what might be going wrong?

 

[soothsayer]

Usually, with these boundary condition problems, when you get something blowing up to infinity when it shouldn't, it means you are forgetting to throw out some general solution (perhaps some term in V(r,θ) ) that doesn't obey the boundary conditions. You might look at your Legendre polynomial, P_s(cosθ). You have to compute V for different ranges of r and θ, and some solutions that work for certain ranges won't work for others. For example, if a general solution were e^{ir} + e^{-ir}, you'd want to throw away the e^{ir} when evaluating the solution for c < r (i.e: r → ∞) because that solution would blow up. Similarly, you need to think about what solutions are appropriate for r → 0, and get rid of the ones that aren't, which I suspect could be an issue here.

Can't say for sure that this is your problem: I don't have to resources to go into this specific problem with depth right now, but that's a general issue with PDEs that you have to work around in physics.

 

[gparker267]

Thanks for the feedback!
I have looked at the Legendre polynomials, I am using the unnormalized polynomials. The problem is (a/r)^(s+1) term in V(r,θ) which diverges in the region r<a, for all θ values.

V(r,θ)=Ʃ1(((r/a)^s×Ʃ2(A_n×(a/b)^n×[(-1)^(n-s)n!/s!(n-s)!])-(a/r)^(s+1)×Ʃ3(A_n×(b/a)^(n+1)×[s!/n!(s-n)!]))P_s(cosθ)

Limits of summations
Ʃ1, s=0 to s=s_max (can use s_max =50)
Ʃ2, n=s to n=n_max (can use n_max=50)
Ʃ3, n=0 to s=s

 

[soothsayer]

Yeah, the \left( \frac{a}{r}\right)^{s+1} term is exactly the kind of term I am talking about. That needs to be taken out when evaluating V(r, \theta ) in a region containing r = 0

 

[gparker267]

Actually, the radial domain of the solution is r ε [b,c], so r=0, is already excluded because a singularity exists there. As for the formualtion, (a/r)^(s+1) cannot be removed because I am seeking for a solution in a region between the two spheres. Any other ideas?

 

[soothsayer]

Are you sure (a/r)^(s+1) can't still be removed? It does diverge in the r ε [b,c] region. It seems like for the region within r = a (and not within the inner, grounded sphere) you would need to throw it out and have a different solution for the rest of the region. Basically, a new boundary condition? I don't think it would cause a discontinuity, since (a/r)^(s+1) just converges to 1 at r =a, so V should be equal on either side of the boundary.

EDIT: Nevermind. I might be wrong about that. I'm giving this problem a try in depth, but it might be a bit beyond my pay grade, if I'm to be honest. I think the answer does lie in enforcing another condition to remove divergent terms in the general solution, but I'm not clear on how to go about it.

 

[gparker267]

Thanks for following up! Yes, the trick lies in enforcing a condition that removes the divergent terms (however, the (a/r)^(s+1) term is still part of the solution). I intend to shift the range of radii (from r=[b,c] to r=[a,a+c]) and see how the solution responds.

 

[soothsayer]

No doubt, the fact that we have a radius, as well as some measurement along the z axis, complicates things. Might not be a bad idea.