# Laplace Operator

1. Dec 13, 2011

### tsamocki

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

The Laplace operator Δ is defined by: Δ=

Show in polar coordinates r and Θ, that the Laplace operator takes the following form:

2. Relevant equations

x=rcos(Θ), y=rsin(Θ), r ≥ 0, Θ ∈ [0,2∏]

3. The attempt at a solution

It seems simple; convert the x and y in terms of polar coordinates, and differentiate twice. However, when i attempt to differentiate twice, in respect to r and theta, my r disappears and the trig function simply reverts.

2. Dec 13, 2011

### I like Serena

Can you show your work where you differentiate the first time?

3. Dec 14, 2011

### tsamocki

Sorry for the delayed response, been studying for other finals.

Here is where i am stuck.

For the x partial derivative:

$\partial$ / $\partial$x= -rsinθ+cosθ

For the y partial derivative:

$\partial$ / $\partial$y= sinθ+rcosθ

I know this is incorrect because i was informed by my professor that there is a nested chain ruled inside a product rule in the second differentiation (or something similar). d/dr d/dx +d/dθ d/dx is the format for the x partial derivative, and just replacing x with y gives you the format for the y partial derivative.

Where am i going wrong?

4. Dec 15, 2011

### I like Serena

What did you differentiate? That seems to be missing.
And how did you get this result?

Yes, the chain rule for multivariable differentiation is:
$${\partial \over \partial x}f(u(x,y), v(x,y))={\partial f \over \partial u}{\partial u\over \partial x}+{\partial f \over \partial v}{\partial v \over \partial x}$$

You should do something similar to f(x(r,θ), y(r,θ))...