
#1
Aug1210, 01:59 PM

P: 745

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 coordinates 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
Aug1210, 06:07 PM

HW Helper
P: 3,309

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 



#3
Aug1610, 06:50 PM

P: 745

Can you offer any assistance please. 



#4
Aug1610, 08:17 PM

HW Helper
P: 3,309

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




#5
Aug1810, 06:01 PM

P: 745

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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