- #1
thejinx0r
- 27
- 0
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 ...
Last edited by a moderator: