- #1

thejinx0r

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## Homework Statement

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

## Homework 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]

## 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

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 ...

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