Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Drumhead PDE

  1. Sep 10, 2006 #1
    I've searched through about 5 math books but don't know how to start this one:

    I have a drumskin of radius a, and small transverse oscillations of amplitude:

    [tex] \nabla^2 z = \frac{1}{c^2}\frac{\partial^2 z }{dt^2} [/tex]

    Ok, so I can write the normal mode as
    [tex]z=Z(\rho)cos(\omega t) [/tex]


    1) If I want to find the differential equation for [tex]z=Z(\rho) [/tex], do I just plug the second equation into the first, but use the polar coordinate version of nabla?

    2) If I want to obtain an estimate for the lowest normal mode frequency using a trial function of form [tex]a^{\nu}-\rho^{\nu}[/tex] with [tex]\nu[/tex] an adjustable parameter...where do I start???

    Last edited: Sep 10, 2006
  2. jcsd
  3. Sep 11, 2006 #2


    User Avatar

    Staff: Mentor

    Answer to #1 is yes, one needs the polar version of nabla.

    It's been decades since I solved this problem, so I have to refresh my memory on it's solution, but one should get a Bessel's function, J(r), IIRC, and one looks for the first zero for the fundamental mode.
  4. Sep 11, 2006 #3
    Thanks astronuc! The polar version works out nicely.

    I'm still not sure how to use the trial function. Is the suggested trial function something I should use as z and plug into the equation?


  5. Sep 17, 2006 #4
    Ok, so writing out nabla in polar I get:

    [tex]\frac{1}{\rho}\frac{\partial}{\partial \rho}(\rho \frac{\partial z }{\partial \rho})+\frac{1}{\rho^2} \frac{\partial^2 z}{\partial \phi^2}+\frac{\partial^2 z}{\partial t^2} = \frac{1}{c^2}\frac{\partial^2 z}{\partial t^2}[/tex]

    I am trying to find a normal mode independant of azimuthal angle which is why I let the phi derivative go to zero.

    after some re-arranging I get:

    [tex]\rho^2 Z''+\rho Z'+\rho^2 \omega^2(\frac{1}{c^2}-1)Z=0[/tex]

    now, this kind of looks like Bessels equation, except the last term should look like [tex](\rho^2-\nu^2)Z[/tex]

    with [tex]\nu[/tex] some real number

    I can't quite make the jump from my equation to Bessels...can someone give me any ideas?

  6. Sep 19, 2006 #5


    User Avatar

    Staff: Mentor

    Yes, this is fine. The assumes no azimuthal dependency.

    The temporal term on the left side is not correct. Nabla is spatial only. So, the correct equation should be,

    [tex]\frac{1}{\rho}\frac{\partial}{\partial \rho}(\rho \frac{\partial z }{\partial \rho})\, =\, \frac{1}{c^2}\frac{\partial^2 z}{\partial t^2}[/tex]

    Now we can try a trial function, [tex]Z(\rho)\,T(t)[/tex]

    [tex]\frac{1}{\rho}\frac{\partial}{\partial \rho}(\rho \frac{\partial (Z(\rho)\,T(t)) }{\partial \rho})\, =\, \frac{1}{c^2}\frac{\partial^2 (Z(\rho)\,T(t))}{\partial t^2}[/tex]

    which then yields

    [tex]\frac{1}{\rho}\frac{\partial}{\partial \rho}(\rho \frac{\partial Z(\rho) }{\partial \rho})T(t)\,=\,\frac{1}{c^2}\frac{\partial^2 T(t)}{\partial t^2}Z(\rho)[/tex].

    Dividing through by [tex]Z(\rho)\,T(t)[/tex]

    [tex]\frac{1}{\rho}\frac{\partial}{\partial \rho}(\rho \frac{\partial Z(\rho) }{\partial \rho})\,\frac{1}{Z(\rho)}\,=\,\frac{1}{c^2}\frac{\partial^2 T(t)}{\partial t^2}\,\frac{1}{T(t)} = -\lambda^2[/tex]

    which then give two equations.

    [tex]\frac{1}{\rho}\frac{d}{d \rho}(\rho \frac{d Z(\rho) }{d \rho})\,\frac{1}{Z(\rho)}\,=\, -\lambda^2[/tex]


    [tex]\frac{1}{c^2}\frac{d^2 T(t)}{dt^2}\,\frac{1}{T(t)} = -\lambda^2[/tex]

    The second equation can be written

    [tex]\frac{d^2 T(t)}{dt^2}\,+\,\omega^2\,T(t)\,=\,0[/tex]

    and the other equation can be written as

    [tex]\frac{1}{\rho}\frac{d}{d \rho}(\rho \frac{d Z(\rho) }{d \rho})\,\frac{1}{Z(\rho)}\,=\, -\lambda^2[/tex]

    which becomes

    [tex]\frac{d}{d \rho}(\rho \frac{d Z(\rho) }{d \rho})\,+\,{\rho}\lambda^2\,Z(\rho)\,=\,0[/tex]

    The last equation should simplify to

    [tex]\rho^2 Z''+\rho Z'+\rho^2 \lambda^2Z\,=\,0[/tex]

    Then letting [tex] x\,=\lambda\rho[/tex]

    One can write

    [tex]x^2 Z''(x)\,+\,x Z'(x)\,+\,x^2Z\,=\,0[/tex]

    which is Bessel's equation of order zero.
    Last edited: Sep 19, 2006
  7. Sep 19, 2006 #6
    mmmmm....lovely. :)

    Thanks astronuc. Thanks for pointing out that the nabla term was spatial only. I 'knew' that...but obviously it didn't surface from the deep cobwebs of my brain! I probably would have been stumped on this for ages without that!

    The solutions makes a lot of sense. I saw that you used separation of variables and Bessels equation came out nicely!

    Thanks again.

    PS - cool beard!
  8. Oct 4, 2006 #7
    Thanks again for the help.

    Now, I have to find an estimate for the lowest normal more using the adjustable parameter:


    Here is my effort, please tell me what you think:

    Our equation reads:

    [tex]\rho^2 Z'' + \rho Z'+\rho\frac{\omega^2}{c^2}Z=0[/tex]

    Let: [tex]Z=a^\nu-\rho^\nu[/tex]

    Plugging into our original equation we get:

    [tex]-\nu^2\rho^2\rho^{\nu-2}-\rho\nu\rho^{\nu-1}+\rho^2\frac{\omega^2}{c^2}(a^\nu - \rho^\nu)=0[/tex]

    ...some lines of algebra...


    Is this the correct way to get the normal mode? It makes sense to me but seems like its a little to simple.
    Last edited: Oct 4, 2006
  9. Oct 14, 2006 #8
    anyone care to take a stab?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Drumhead PDE
  1. Prerequisites for PDE ? (Replies: 10)

  2. Solutions to this PDE? (Replies: 4)

  3. Tough pde (Replies: 21)

  4. Linearity of PDE (Replies: 4)