# Wild Laplacian!

1. Nov 9, 2005

### quasar987

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.

2. Nov 9, 2005

### Tide

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.?

3. Nov 9, 2005

### HallsofIvy

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.

4. Nov 9, 2005

### quasar987

Yeah that's the "second way" I was refering to! Why do you call "backward" the first way? It's just as good imo. It's not guarenteed that you'll be able to isolate the cartesian laplancian, but if you can, shouldn't it give the right answer too ?!

5. Nov 9, 2005

### quasar987

Ok, after working out the algebra completely and correcting a few mistake I had made in the second way (the one advetised by HallsofIvy), it works. But I'm still very interested to know why the "backwards" way doesn't work.