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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

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