Harmonic Function

  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. lanedance

    lanedance 3,307
    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. 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. lanedance

    lanedance 3,307
    Homework Helper

    none of the intergal limits or integrations variables depend on r, so you can directly differentiate under the integral sign
     
  6. 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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