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Wave equation, general solution, cylindrical symmetry

  1. Jul 28, 2009 #1
    I was interrested in the general solutions to the wave equation depending on only one spatial coordinate.

    For one linear coordinate, the general solution is:
    a f(x-ct) + b g(x+ct)​

    For one radial spherical coordinate, the general solution is:
    a f(r-ct)/r + b g(r+ct)/r​

    I thought that for a radial cylindrical coordinate, the solution would be:
    a f(r-ct)*Log(r) + b g(r+ct)*Log(r)​

    Yet I found this is not a solution since I got this residual for the wave equation:
    D²(f(r-ct)/r) = a (2+Log(r))*f'(r-ct)/r + b (2+Log(r))*g'(r+ct)/r​

    This would indicate that the cylindrical solution applies only to the static case.

    Any comment on this surprise?
    Did I do a mistake in the calculations, of does that instead mean something?
    Thanks.
     
    Last edited: Jul 28, 2009
  2. jcsd
  3. Aug 6, 2009 #2
    The complete wave equation in cylindrical coordinates is:

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

    But if you want a symmetry around the z-axis, you can suppose that the function is constant for [tex]\theta[/tex] and z, so you get the equation:

    [tex]\frac{\partial^2 u}{\partial t^2}=\frac{c^2}{\rho}\frac{\partial}{\partial \rho}\left(\rho\frac{\partial u}{\partial \rho}\right)[/tex]

    Which look like the spherical coordinates version but is much more difficult to solve. In fact, this is the wave equation in 2D polar coordinates.

    The solution to this equation uses Bessel's functions, which are represented by Taylor series, not a closed form. It's strange that the solution in 2D is much more complicated that the solution in 3D.
     
  4. Aug 7, 2009 #3
    (warning: my post does not really contribute anything meaningful to the discussion)
    I was under the impression that the solution to this equation was what defined bessel functions. I could be wrong, though.
     
  5. Aug 7, 2009 #4
    siyphsc: Yes, but I forgot to mention that this not gives a general solution, but only the steady state solution. So this says that the solution [tex]u(\rho,t)[/tex] is separable:

    [tex]u(\rho,t)=u_s(\rho)u_t(t)[/tex]

    Bessel's functions are modes in a cylindrical space for the radial component, like sinusoidal functions are modes in a 1D space.
     
  6. Dec 5, 2011 #5
    In Landau and Lifgarbagez, Fluid Mechanics book, it gives as a general solution to the wave equation with cylindrical symmetry (Section 71):

    \begin{equation}
    \psi = \int_{ct-\rho}^{ct+\rho}\frac{F(\xi)}{\sqrt{\rho^2-(\xi-ct)^2}}
    \end{equation}

    The way he derives this form is by integrating out the z-dependence of the spherically symmetric solutions in 3D. Note that F is a general function that is found by satisfying the initial conditions of the wave.
     
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