# Harmonic Function

1. Aug 12, 2010

### billiards

1. The problem statement, all variables and given/known data

Derive that:

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

2. Relevant equations

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

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

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

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

3. The attempt at a solution

This ones seems to have me stumped.

I've tried setting

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

But that didn't seem to be fruitful.

I've tried expanding

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

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. Aug 12, 2010

### lanedance

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

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

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. Aug 16, 2010

### billiards

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.

4. Aug 16, 2010

### lanedance

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

5. Aug 18, 2010

### billiards

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

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

2) Note that

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

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

To find that

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

QED?