# Laplace equation in polars

1. Apr 1, 2009

### briteliner

1. The problem statement, all variables and given/known data

Laplace's equation in 2 dimensions may be written, using plane polar coordinates r, θ, as

Find all separable solutions of this equation which have the form V(r, θ)=R(r)S(θ), which are single valued for all r, θ. What property of the equation makes any linear combination of such solutions also a solution?

2. Relevant equations

3. The attempt at a solution
i get how to separate the variables and i am left with to expressions equaling a constant. but from there it doesnt make sense to me how to end up with 2 solutions

2. Apr 1, 2009

### latentcorpse

assume a seperable solution then sub it in and do the derivatives. then multiply thorugh by $\frac{r^2}{RS}$

if you make your constant $\lambda$ you should get a nice solution for $\lambda>0$ hint : let $R=r^m$ with m a constant. the others will be less pretty unless you have simplifying boundary conditions.

3. Apr 1, 2009

### briteliner

ok, got it thanks. the second part of the question which i didnt include before is:
A continuous potential V(r, θ) satisfies Laplace's equation everywhere except on the concentric circles r=a, r=b where b>a.
(i) Given that V(r=a, θ)=Vo(1+cos θ), and that V is finite as r-->infinity , find V in the region r less than or equal to a
(ii) given, separately, that V(r=0, θ)=2Vo and V is finite as r--> ∞, find V for r≥b

for (i), i dont really know how to select solutions for that V. my solutions arecombinations of sinmtheta cosmtheta r^m and r^-m. is there a way to do this through an expansion?

4. Apr 1, 2009

### latentcorpse

ok i think that the periodicity of the cos term means that the solutions for $\lambda<0$ are useless here and the $\lambda=0$ solutions will be trivial after you apply the b.c. that it must be finit as r goes to infinity.

so for the $\lambda>0$ solutions i have

$V=(Ar^n+Br^{-n})(C \cos{n \theta} + D \sin{n \theta})$

for this to be finite at infinity what can you say about the coefficient of $r^n$

5. Apr 1, 2009

### latentcorpse

then for finding V in the region r<a, your right to think of expansions, consider

$\frac{1}{2}c_0 + \sum_{n=0}^{\infty} r^{-m} (c_n \cos{n \theta} + d_n \sin{n \theta})$

equate this to the boundary condition you have and you should be able to work out the relative non-zero coefficients.

my answer : $V(r<a)=2V_0+\frac{aV_0}{r}$

6. Apr 1, 2009

### briteliner

Well Br^-n goes to 0 so A should be a constant or 1 so that the whole expression doesn't go to zero?

7. Apr 1, 2009

### latentcorpse

zero is finite though. so that's allowed. what is not allowed is have $\infty^n$. so what can you say about A and B now?

8. Apr 1, 2009

### briteliner

ok so should i then just throw out the solutions r^n and r^-n since they don't meet the initial condition?