# Wild Laplacian!

Homework Helper
Gold Member
I'm supposed to prove the laplacian in cylindrical coord. is what it is. I tried tackling the problem in two ways and none work! I have no idea what's the matter. The first way is to calculate d²f/dr² , d²f/dO² and d²f/dz² and isolate d²f/dx² , d²f/dy² and d²f/dz². In cylindrical coord.,

x=rcosO
y=rsinO
z=z

(I'm not writting partial derivatives cuz in latex it's a pain in the @)

$$df/dr = df/dx dx/dr + df/dy dy/dr$$
$$\Rightarrow d^2f/dr^2 = d^2f/dx^2 dx/dr + df/dx d^2x/dr^2 + d^2f/dy^2 dy/dr + df/dy d^2y/dr^2$$
$$d^2f/dr^2 = d^2f/dx^2 cos^2\theta +d^2f/dy^2 sin^2\theta$$

In the same way,

$$d^2f/d\theta^2 = r^2 (d^2f/dx^2 d^2f/dx^2)$$

Hence, it would seem that the laplacian in cylindrical is just (1/r²)d²f/dO² + d^2f/dz^2.

Tide
Homework Helper
Have you tried

$$\frac {\partial}{\partial x} = \frac {\partial r}{\partial x} \frac {\partial}{\partial r} + \frac {\partial \theta}{\partial x} \frac {\partial}{\partial \theta}$$

etc.?

HallsofIvy
Homework Helper
You are doing it backwards!
You know that the Laplacian is
$$\frac{\partial^2U}{/partial x^2}+ \frac{\partial^2U}{\partial y^2}+ \frac{\partial^2U}{\partial z^2}$$
in Cartesian coordinates so you should be writing
$$\frac{\partialU}{\partial z}=\frac{\partial U}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial U}{\partial \theta}\frac{\partial \theta}{\partial x}+ \frac{\partial U}{\partial z}\frac{\partial z}{\partial x}$$
etc. so that you can substitute into that.