Laplace equation in spherical coordinates

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
7 replies · 2K views
Fgard
Messages
15
Reaction score
1

Homework Statement



Solve the Laplace equation inside a sphere, with the boundary condition:

\begin{equation}
u(3,\theta,\phi) = \sin(\theta) \cos(\theta)^2 \sin(\phi)
\end{equation}

Homework Equations


\begin{equation}
\sum^{\infty}_{l=0} \sum^{m}_{m=0} (A_lr^l + B_lr^{-l -1})P_l^m(\cos \theta) [S_m\sin(m\phi) + C_m\cos(m\phi)]
\end{equation}

The Attempt at a Solution


I have derived the general solution for the Laplace equation in spherical problems and that went okej. The problem is when I try to match it with the boundary conditions. If I use the method of identification it is imminently obvious that:
\begin{equation}
C_m=0, m=1 \text{ and that for} \quad l\neq1 \rightarrow A=0
\end{equation}
, m has to equal one so the arguments of sin is correct and A=0 so the solution does not blow up, when r approaches 0. Which means I have two different solutions: one where l=1 and another solution for all the rest of l´s. When l equals one the Legendre polynomials dose not match the boundary conditions so that solution can be discarded.

Which means that I have an equation that looks like:

\begin{equation}
\sum^{\infty}_{l=0} ( B_l3^{-l -1})P_l^1(\cos \theta) S_1\sin(\phi) = \sin(\theta) \cos(\theta)^2 \sin(\phi)
\end{equation}

When I look at table's for Legendre polynomials I do not find any that looks like it can solve my equation. I would really appreciate some guidance in the right direction.
 
Physics news on Phys.org
\begin{equation}
\sum^{\infty}_{l=0} \sum^{m}_{m=0} (A_lr^l + B_lr^{-l -1})P_l^m(\cos \theta) [S_m\sin(m\phi) + C_m\cos(m\phi)]
\end{equation}
Typo: what is the upper limit of the summation over ##m##?

You are dealing with the interior of a sphere. So, you don't need to worry about ##r## going to infinity. However, what happens to ##r^{-l-1}## at the center of the sphere?
 
The upper limit of the summation is suppose to be l.

I have singularity there, so the constant B has to be zero. Thanks.
 
No one that can help me?
 
Looks like now you just need to find the values of the ##A_l## coefficients by considering the boundary condition at r = 3. Look at a table of ##P_l^1(\cos \theta)## to see which ones can be used to obtain ##\sin \theta \cos^2 \theta##.
 
That is what I have trying to do, quite unsuccessfully so far. But I know now at least that this is the way to do it, so thank you.
 
Hint: Look at your boundary condition at r = 3 and compare to a table of associated Legendre polynomials.
 

Attachments

  • Assoc Legendre Polynomials.png
    Assoc Legendre Polynomials.png
    9 KB · Views: 565
Last edited:
I think I solved it. Thanks for all the help!