# Homework Help: The Laplace Equation in Polar Coordinates

1. Oct 2, 2008

### thejinx0r

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

$$\frac{\partial^2f}{\partial x^2}+\frac{\partial ^2f}{\partial y^2}= 0$$

2. Relevant equations

Show that the equation above is equal to:
$$\frac{\partial^2f}{\partial r^2}+\frac{1}{r^2} \frac{\partial ^2f}{\partial \theta^2} + \frac{1}{ r} \frac{\partial f}{\partial r}= 0$$

3. The attempt at a solution

So, let $$f=(r,\theta)$$
$$r = \sqrt{x^2+y^2}$$
$$\theta = tan^{-1}(y/x)$$
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 ...

Last edited by a moderator: May 3, 2017
2. Oct 2, 2008

### gabbagabbahey

One of your LaTeX images contained an error and didn't load properly. Was it supposed to be:

$$\frac{\partial f}{\partial r}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial r}$$

If so, then just use this again:

$$\frac{\partial ^2 f}{\partial r^2}= \frac{\partial }{\partial r} \left( \frac{\partial f}{\partial r} \right)=\frac{\partial}{\partial x} \left( \frac{\partial f}{\partial r} \right) \cdot \frac{\partial x}{\partial r}+\frac{\partial }{\partial y} \left( \frac{\partial f}{\partial r} \right) \cdot \frac{\partial y}{\partial r}$$

Substitute $\frac{\partial f}{\partial r}$ into this, and compute $\frac{\partial x}{\partial r}$ and $\frac{\partial y}{\partial r}$.

3. Oct 3, 2008

### thejinx0r

So I fixed up the first post.

But it does help me do the second derivative and shows that I sort of did write it out wrong

:)