1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
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

  1. Oct 2, 2008 #1
    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 ...
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Oct 2, 2008 #2


    User Avatar
    Homework Helper
    Gold Member

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

    [tex]\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}[/tex]

    If so, then just use this again:

    [tex]\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}[/tex]

    Substitute [itex]\frac{\partial f}{\partial r}[/itex] into this, and compute [itex]\frac{\partial x}{\partial r}[/itex] and [itex]\frac{\partial y}{\partial r}[/itex].
  4. Oct 3, 2008 #3
    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

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook