# Lecture Notes

math121b:04-03

## 2020-04-03, Friday

Today we consider PDE problem in cylindrical coordinate and spherical coordinate.

### Cylindrical Coordinate

Recall that Laplacian in cylindrical coordinate $r, \theta, z$ is written as $$\Delta u = \frac{1}{r} \d_r(r \d_r(u)) + \frac{1}{r^2} \d_\theta^2 u + \d_z^2 u.$$ We shall look for eigenfunctions of $\Delta$ of the following form $$u(r,\theta, z) = R( r ) \Theta(\theta) Z(z),$$ Then we have $$\frac{1}{u}\Delta u = \frac{1}{R} \frac{1}{r} \d_r(r \d_r(R)) + \frac{1}{\Theta} \frac{1}{r^2} \d_\theta^2 \Theta + \frac{1}{Z} \d_z^2 Z.$$ such that $$\Theta^{-1} \d_\theta^2 \Theta = \lambda_\theta$$ $$\frac{1}{R} \frac{1}{r} \d_r(r \d_r(R)) + r^{-2} \lambda_\theta = \lambda_r$$ $$\frac{1}{Z} \d_z^2 Z = \lambda_z.$$

eigenvalue problem for $\Theta$.
$$\lambda_\theta = -n^2, \quad \Theta(\theta) = \cos(n \theta), \sin(n \theta)$$

eigenvalue problem for $R( r)$.
Assume $\lambda_\theta = -n^2$, then we get $$r \d_r(r \d_r(R)) + (-\lambda_r r^2 - n^2) R = 0.$$

Compare with Bessel equation (12.2) from Boas $$x(xy')' + (x^2 - p^2) y = 0 \Rightarrow y(x) = a J_p(x) + b Y_p(x).$$ And notice that we want $R(0)$ finite, and $|Y_p(0)| = \infty$, we see $$R( r) = \begin{cases} J_n(\sqrt{-\lambda_r} r) & \lambda_r \neq 0 \cr r^n & \lambda_r = 0 \cr \end{cases}$$ Note that, the $\sqrt{-\lambda_r}$ has an ambiguity of $\pm 1$ sign, but $J_n(x)=(-1)^n J_n(-x)$, hence you can take either sign for the square root.