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Harmonic Function

by billiards
Tags: function, harmonic
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billiards
#1
Aug12-10, 01:59 PM
P: 748
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.
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lanedance
#2
Aug12-10, 06:07 PM
HW Helper
P: 3,307
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
billiards
#3
Aug16-10, 06:50 PM
P: 748
Quote Quote by lanedance View Post
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
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.

Can you offer any assistance please.

lanedance
#4
Aug16-10, 08:17 PM
HW Helper
P: 3,307
Harmonic Function

none of the intergal limits or integrations variables depend on r, so you can directly differentiate under the integral sign
billiards
#5
Aug18-10, 06:01 PM
P: 748
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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