# Question about Laplace(Potential) equation of disk.

1. May 6, 2010

### yungman

Laplace equation:

$$\frac{R''}{R} + \frac{1}{r}\frac{R'}{R} + \frac{1}{r^2}\frac{\Theta''}{\Theta} =0$$

Which give:

$$r^2\frac{R''}{R} + r\frac{R'}{R} - \lambda R =0 \;and\; \Theta '' + \lambda \Theta =0$$

$$\theta \;is\; 2\pi \;periodic\; \Rightarrow\; \Theta \;is\; 2\pi \;periodic\;$$

My question is: Why then $$\theta \;is\; 2\pi \;periodic\; \Rightarrow\;\lambda = n^2$$ where n= 0,1,2,3........

AND

$$\Theta = a_n cos(n\theta) + b_n sin(n\theta)$$

2. May 6, 2010

### LCKurtz

What do you mean by $\theta$ is periodic? It obviously isn't. Maybe you mean the solution is periodic in $\theta$?

If the solution is periodic in $\theta$ than

$$u(r,\theta) = u(r, \theta + 2\pi)\hbox{ and }u_\theta(r,\theta) = u_\theta(r,\theta+2\pi)$$

and those boundary conditions on your $\Theta$ will give you the eigenvalues and eigenfunctions.

As a side note, I find the your use of gigantic font size in your posts to be very annoying.

3. May 6, 2010

### yungman

Because $$0< \theta< 2\pi$$. which is periodic of $$2\pi$$.

Yes, $$u(r,\theta) = u(r, \theta + 2\pi)\hbox{ and }u_\theta(r,\theta) = u_\theta(r,\theta+2\pi)$$

4. May 6, 2010

### yungman

I want to clarify, I understand Eigen values. I want to know the physical meaning. I have been working on this myself and I come up with this explaination:

$$\Theta '' + \lambda \Theta =0$$

$$\theta \;is\; 2\pi \;periodic\; \Rightarrow\; \Theta \;is\; 2\pi \;periodic\;$$

Let $$\lambda = \mu^2$$

$$\mu=\frac{n(\hbox{ Radian in a Period})}{(\hbox {Length of Period(length or radian)})}$$

In this case, (Radian in a Period ) is $2\pi[/tex]. Length of the period is [itex] 2\pi[/tex]. Therefore [itex] \mu=\frac{n(2\pi)}{2\pi}=n$

$$\Theta = a_n cos(n\theta) + b_n sin(n\theta)$$

For period in length instead of radian in case of string with length of L and fixed at x=0 and x=L:

$$\mu=\frac{n(\hbox{ Radian in a Period})}{(\hbox {Length of Period(length or radian)})}=\frac{n\pi}{L}$$

$$\Theta = a_n cos(\frac{n\pi}{L}) + b_n sin(\frac{n\pi}{L})$$

Can anyone comment on my understanding?

Last edited: May 6, 2010
5. May 6, 2010

### LCKurtz

It is hard to tell what you understand from what you have written. Do you know how to solve the $\Theta$ boundary value problem to derive the $\mu_n$ and the sine and cosine eigenfunctions?

6. May 6, 2010

### yungman

Yes I do, I have no problem with the BVP:

$$\Theta '' +\lambda \Theta = 0$$ is a constant coef. which give:

$$\Theta(\theta)=acos(\sqrt{\lambda}\theta) +bsin(\sqrt{\lambda}\theta)$$

$$\lambda = \mu^2$$

From periodic,$$\lambda = \mu^2 = n^2$$

$$\Rightarrow \Theta_n(\theta)=a_n cos(n\theta) +b_n sin(n\theta)$$

All I want to do is to verify my understanding of the eigen value intepretation since this is respect to $$\theta$$ instead of respect to length on x-axis where $\mu_n=\frac{n\pi}{L}$.

I guess I am trying to define $\mu_n$ in English.

Last edited: May 7, 2010
7. May 8, 2010

### yungman

I did further digging. This basically boil down to the Fourier series expansions of function of arbitrary period:

For period of $2\pi$

$$\Theta = a_n cos(n\theta) + b_n sin(n\theta)$$ (1)

For function of arbitrary period of 2p:

$$\Theta = a_n cos(\frac{n\pi}{L}) + b_n sin(\frac{n\pi}{L})$$ (2)

I looked through a few text books, most don't even have any derivations. Only Asmar gave some sort of steps:

For f(x) is a function of period of T=2p, Let $g(x) = f(\frac{p}{\pi}x)$:

$$g(x+2\pi) = f(\frac{p}{\pi}(x+2\pi)}) = f(\frac{p}{\pi}x+2p) = f(\frac{p}{\pi}x)=g(x)$$

I still don't see the connection how this derive from (1) to (2).

8. May 8, 2010

### LCKurtz

Ok. You have f(x) of period 2p and you know how to calculate FS for functions of period 2π. So you let $g(x) = f(\frac p \pi x)$ so g has period 2π. So we know

$$g(x) \sim \sum a_n\cos{(nx)} + b_n\sin{(nx)}$$

where

$$b_n =\frac 1 \pi \int_{-\pi}^{\pi} g(x) \sin{(nx)}\ dx$$

and similarly for an. Now substitute $x = \frac \pi p t$ in the series:

$$g(\frac \pi p t) \sim \sum a_n\cos{(\frac {n\pi} p t)} + b_n\sin{(\frac {n\pi} p t)}$$

But $g(\frac \pi p t) = f(t)$ so we have

$$f(t) \sim \sum a_n\cos{(\frac {n\pi} p t)} + b_n\sin{(\frac {n\pi} p t)}$$

Now let's make the substitution $x =\frac \pi p t,\ dx =\frac \pi p dt$ in the formula for bn

$$b_n =\frac 1 \pi \int_{-\pi}^{\pi} g(x) \sin{(nx)}\ dx=\frac 1 \pi\int_{-p}^pg(\frac{\pi t} p )\ \frac \pi p\,\sin{\frac{n\pi t}p}\ dt =\frac 1 p\int_{-p}^p f(t)\sin{\frac{n\pi t}p}\ dt$$

The same thing works for the an. This shows how you can get the formulas for 2p from those for 2π. And, of course, putting p = π in these new formulas gets the old ones.

9. May 9, 2010

### yungman

Thanks, that really clear it up.

Have a nice day.