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Question about Laplace(Potential) equation of disk.

  1. May 6, 2010 #1
    Laplace equation:

    [tex]\frac{R''}{R} + \frac{1}{r}\frac{R'}{R} + \frac{1}{r^2}\frac{\Theta''}{\Theta} =0 [/tex]

    Which give:

    [tex]r^2\frac{R''}{R} + r\frac{R'}{R} - \lambda R =0 \;and\; \Theta '' + \lambda \Theta =0 [/tex]

    [tex]\theta \;is\; 2\pi \;periodic\; \Rightarrow\; \Theta \;is\; 2\pi \;periodic\;[/tex]



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

    AND

    [tex] \Theta = a_n cos(n\theta) + b_n sin(n\theta)[/tex]
     
  2. jcsd
  3. May 6, 2010 #2

    LCKurtz

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

    If the solution is periodic in [itex]\theta[/itex] than

    [tex]u(r,\theta) = u(r, \theta + 2\pi)\hbox{ and }u_\theta(r,\theta) = u_\theta(r,\theta+2\pi)[/tex]

    and those boundary conditions on your [itex]\Theta[/itex] 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.
     
  4. May 6, 2010 #3
    Because [tex]0< \theta< 2\pi[/tex]. which is periodic of [tex]2\pi[/tex].

    Yes, [tex]u(r,\theta) = u(r, \theta + 2\pi)\hbox{ and }u_\theta(r,\theta) = u_\theta(r,\theta+2\pi)[/tex]
     
  5. May 6, 2010 #4
    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:

    [tex]\Theta '' + \lambda \Theta =0 [/tex]

    [tex]\theta \;is\; 2\pi \;periodic\; \Rightarrow\; \Theta \;is\; 2\pi \;periodic\;[/tex]

    Let [tex]\lambda = \mu^2[/tex]

    [tex]\mu=\frac{n(\hbox{ Radian in a Period})}{(\hbox {Length of Period(length or radian)})}[/tex]

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

    [tex] \Theta = a_n cos(n\theta) + b_n sin(n\theta)[/tex]




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

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

    [tex] \Theta = a_n cos(\frac{n\pi}{L}) + b_n sin(\frac{n\pi}{L})[/tex]

    Can anyone comment on my understanding?
     
    Last edited: May 6, 2010
  6. May 6, 2010 #5

    LCKurtz

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    It is hard to tell what you understand from what you have written. Do you know how to solve the [itex]\Theta[/itex] boundary value problem to derive the [itex]\mu_n[/itex] and the sine and cosine eigenfunctions?
     
  7. May 6, 2010 #6
    Yes I do, I have no problem with the BVP:

    [tex]\Theta '' +\lambda \Theta = 0 [/tex] is a constant coef. which give:

    [tex] \Theta(\theta)=acos(\sqrt{\lambda}\theta) +bsin(\sqrt{\lambda}\theta)[/tex]

    [tex]\lambda = \mu^2[/tex]

    From periodic,[tex] \lambda = \mu^2 = n^2[/tex]

    [tex]\Rightarrow \Theta_n(\theta)=a_n cos(n\theta) +b_n sin(n\theta)[/tex]

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

    I guess I am trying to define [itex]\mu_n[/itex] in English.
     
    Last edited: May 7, 2010
  8. May 8, 2010 #7
    I did further digging. This basically boil down to the Fourier series expansions of function of arbitrary period:

    For period of [itex]2\pi[/itex]

    [tex] \Theta = a_n cos(n\theta) + b_n sin(n\theta)[/tex] (1)



    For function of arbitrary period of 2p:

    [tex] \Theta = a_n cos(\frac{n\pi}{L}) + b_n sin(\frac{n\pi}{L})[/tex] (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 [itex]g(x) = f(\frac{p}{\pi}x) [/itex]:

    [tex]g(x+2\pi) = f(\frac{p}{\pi}(x+2\pi)}) = f(\frac{p}{\pi}x+2p) = f(\frac{p}{\pi}x)=g(x)[/tex]

    I still don't see the connection how this derive from (1) to (2).
     
  9. May 8, 2010 #8

    LCKurtz

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

    [tex]g(x) \sim \sum a_n\cos{(nx)} + b_n\sin{(nx)}[/tex]

    where

    [tex]b_n =\frac 1 \pi \int_{-\pi}^{\pi} g(x) \sin{(nx)}\ dx[/tex]

    and similarly for an. Now substitute [itex]x = \frac \pi p t[/itex] in the series:

    [tex]g(\frac \pi p t) \sim \sum a_n\cos{(\frac {n\pi} p t)} + b_n\sin{(\frac {n\pi} p t)}[/tex]

    But [itex]g(\frac \pi p t) = f(t)[/itex] so we have

    [tex]f(t) \sim \sum a_n\cos{(\frac {n\pi} p t)} + b_n\sin{(\frac {n\pi} p t)}[/tex]

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

    [tex]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[/tex]

    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.
     
  10. May 9, 2010 #9
    Thanks, that really clear it up.

    Have a nice day.
     
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