(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

[tex]\frac{\partial^2f}{\partial x^2}+\frac{\partial ^2f}{\partial y^2}= 0[/tex]

2. Relevant equations

Show that the equation above is equal to:

[tex]\frac{\partial^2f}{\partial r^2}+\frac{1}{r^2} \frac{\partial ^2f}{\partial \theta^2} + \frac{1}{ r} \frac{\partial f}{\partial r}= 0[/tex]

3. The attempt at a solution

So, let [tex]f=(r,\theta)[/tex]

[tex] r = \sqrt{x^2+y^2}[/tex]

[tex]\theta = tan^{-1}(y/x) [/tex]

then by the chain rule, partial f partial r is:

http://www.texify.com/img/%5CLARGE%5C%21%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20r%7D%20%5Cfrac%7Bx%7D%7B%20%5Csqrt%7Bx%5E2%20%2B%20y%5E2%7D%20%7D%20%2B%20%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20r%7D%20%5Cfrac%7By%7D%7B%20%5Csqrt%7Bx%5E2%20%2B%20y%5E2%7D%20%7D.gif [Broken]

And I have something similar for partial f partial theta,

but I'm not sure if what I'm doing is right...

Because, I applied the chain rule to get the top part, (which I do not know how to write a matrix in latex :S).

But I've never applied the chain rule twice to get a second order differential.

So I'm stuck there. I tried to replace df/dr with d^f/dr^2 and similarly with the theta and multiplying it by the gaussian matrix (dr/dx dr/dy ; dt/dx dt/dy) [dt = d theta].

But then I realized that I would not get the 3rd term because of the way I just changed df/dr with d^f/dr^2 which is probably wrong on my part ...

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: The Laplace Equation in Polar Coordinates

**Physics Forums | Science Articles, Homework Help, Discussion**