# Laplace problem-analytically

In which we show that the method of separation of variables gives an analytic solution to Laplace's equation.

Our prototypical problem is shown at right: We have an electric field which, at infinity is pointing uniformly in the $\uv{z}$ direction. (Perhaps we're between the flat plates of a large capacitor...) We insert a spherical conductor in the field. Now what's the electric field?

To solve this, we note that...

• The whole (conducting) sphere is at the same potential, and
• this problem has azimuthal symmetry--that is, the solution must be independent of the $\phi$ coordinate.
• Laplace's equation holds outside of the sphere.
• Inside the sphere, we know that the field is zero.

The Laplace equation (see your inside front cover) in spherical coordinates, assuming azimuthal symmetry, is...

$$\grad^2 V = \frac{1}{r^2}\frac{\del}{\del r}\left(r^2 \frac{\del V}{\del r}\right)+\frac{1}{r^2 \sin \theta}\frac{\del}{\del \theta}\left(\sin \theta \frac{\del V}{\del \theta}\right)=0$$.

## Separation of variables

The method of separation of variables consists in guessing that the solutions to a partial differential equation consist of the products of functions, each of which is a function of just one coordinate.

Though it seems, erm, less than obvious that this will work, this approach to solving partial diff.eq.s actually has a pretty good track record!

Since we're using spherical polar coordinates, let's guess that the solution is the product of two functions: $$V(\theta,r) = \Theta(\theta)R(r)$$.

Substituting this in to Laplace's equation gives... $$\begineq\grad^2 V = 0 &=& \frac{1}{r^2}\frac{\del}{\del r}$r^2 \frac{\del \Theta(\theta)R(r)}{\del r}$+\frac{1}{r^2 \sin \theta}\frac{\del}{\del \theta}$\sin \theta \frac{\del \Theta(\theta)R(r)}{\del \theta}$\\ &=&\frac{\Theta(\theta)}{r^2}\frac{d}{d r}$r^2 \frac{dR(r)}{d r}$+\frac{R(r)}{r^2 \sin \theta}\frac{d}{d \theta}$\sin \theta \frac{d \Theta(\theta)}{d \theta}$\endeq$$

Multiplying through by $r^2/(\Theta(\theta)R(r))$:

$$\frac{1}{R(r)}\frac{d}{d r}$r^2 \frac{dR(r)}{d r}$= -\frac{1}{\Theta(\theta) \sin \theta}\frac{d}{d \theta}$\sin \theta \frac{d \Theta(\theta)}{d \theta}$.$$

The only way for this to hold, since $\theta$ and $r$ are independent variables, is if the left chunk (only a function of $r$) and the right chunk (only a function of $\theta$) are each equal to a constant--the same constant, called the separation constant $k$. So this one equation is really two equations.

#### $\Theta(\theta)$ equation

$$-\frac{1}{\Theta(\theta) \sin \theta}\frac{d}{d \theta}$\sin \theta \frac{d \Theta(\theta)}{d \theta}$=k=l(l+1).$$

Taking the $\theta$ equation first... It turns out that solutions are only possible for certain values of the separation constant. These are $k=l(l+1)$ where $l>=0$ is an integer. Re-arranging our equation for $\Theta(\theta)$ we get: $$\frac{1}{\sin \theta}\frac{d}{d \theta}$\sin \theta \frac{d \Theta(\theta)}{d \theta}$ + l(l+1)\Theta(\theta) = 0.$$

The solutions that work for physical situations are the Legendre polynomials $P_l(x)$: