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Harmonic Function

  1. Aug 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Derive that:

    [tex]\left[r\frac{\partial\overline{f}}{\partial r}\right]}^{r=R}_{r=0}=0[/tex]


    2. Relevant equations

    I have taken the Laplacian [tex]\nabla^{2}f=0[/tex] for a disk in cylindrical co-ordinates and have found that:

    [tex]\int^{R}_{0}\int^{2\pi}_{0} \left[\frac{\partial}{\partial r}(r\frac{\partial f}{\partial r})\right] d\varphi dr=0[/tex]

    And the definition of the average of the function around the circle of radius r is provided:

    [tex]\overline{f}(r)\equiv\frac{1}{2\pi}\int^{2\pi}_{0}f(r,\varphi)d\varphi[/tex]

    3. The attempt at a solution

    This ones seems to have me stumped.

    I've tried setting

    [tex]\int^{2\pi}_{0} \left[\frac{\partial}{\partial r}(r\frac{\partial f}{\partial r})\right] d\varphi = \int^{2\pi}_{0}f(r,\varphi)d\varphi = 2\pi\overline{f}(r)[/tex]

    But that didn't seem to be fruitful.

    I've tried expanding

    [tex]\left[\frac{\partial}{\partial r}(r\frac{\partial f}{\partial r})\right]d\varphi = \frac{\partial f}{\partial r}d\varphi + r\frac{\partial^{2} f}{\partial r^{2}}d\varphi [/tex]

    That looks a little bit like a Taylor series but I don't know what to do with it. I've been playing around with the algebra but can't seem to find my break through.
     
  2. jcsd
  3. Aug 12, 2010 #2

    lanedance

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    Homework Helper

    i think you've pretty much got it, just need to work backwards

    so you have
    [tex]\overline{f}(r)=\frac{1}{2\pi}\int^{2\pi}_{0} f(r,\phi)d\phi[/tex]

    differentiate that whole expression w.r.t. r, multiply by r then intergate over r from 0 to R and see what you end up with
     
  4. Aug 16, 2010 #3
    Thanks lanedance. That seems like a good approach, the reason I didn't try it was because I'm afraid I don't know how to differentiate the expression w.r.t. r. :redface:

    Can you offer any assistance please.
     
  5. Aug 16, 2010 #4

    lanedance

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    Homework Helper

    none of the intergal limits or integrations variables depend on r, so you can directly differentiate under the integral sign
     
  6. Aug 18, 2010 #5
    Thanks lanedance, I didn't know about differentiating under the integral sign, good stuff!

    I think I have the answer, would appreciate feedback as this is not 100% comfortable stuff for me.

    So far i get:

    1) Use the definition of [tex]\overline{f}(r)\equiv\frac{1}{2\pi}\int^{2\pi}_{0}f(r,\varphi)d\varphi[/tex]

    to find [tex](r\frac{\partial \overline{f}}{\partial r}) = \frac{1}{2\pi}\int^{2\pi}_{0} r\frac{\partial f(r,\varphi)}{\partial r} d\varphi[/tex] in terms of [tex]f[/tex]

    2) Note that

    [tex]\left[r\frac{\partial\overline{f}}{\partial r}\right]}^{r=R}_{r=0}=\int^{R}_{0} \frac{\partial}{\partial r} (r\frac{\partial \overline{f}}{\partial r})dr = \frac{1}{2\pi}\int^{R}_{0}\int^{2\pi}_{0} \frac{\partial}{\partial r} (r\frac{\partial f(r,\varphi)}{\partial r}) d\varphi dr[/tex]

    3) Sub in the Laplacian expression [tex]\int^{R}_{0}\int^{2\pi}_{0} \left[\frac{\partial}{\partial r}(r\frac{\partial f}{\partial r})\right] d\varphi dr=0[/tex]

    To find that

    [tex]\left[r\frac{\partial\overline{f}}{\partial r}\right]}^{r=R}_{r=0}=0[/tex]


    QED?
     
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