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Proof of an admissible Airy stress function

  1. Oct 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that a stress function of the form [itex]\phi(r,\theta) = f(r)\cos(2\theta)[/itex], where f(r) is a function of r only, satisfies the biharmonic equation and hence an admissible Airy stress function.


    2. Relevant equations
    [itex]
    g \equiv \nabla^2 \phi = \frac{\partial^2 \phi}{\partial r^2} + \frac{1}{r}\frac{\partial \phi}{\partial r} + \frac{1}{r^2}\frac{\partial^2 \phi}{\partial \theta}
    [/itex]

    [itex]
    \nabla^4 \phi = \nabla^2 g = \frac{\partial^2 g}{\partial r^2} + \frac{1}{r}\frac{\partial g}{\partial r} + \frac{1}{r^2}\frac{\partial^2 g}{\partial \theta}
    [/itex]


    3. The attempt at a solution

    Evaluate the harmonic equation:
    [itex]
    g = \frac{\partial^2f(r)}{\partial r^2}\cos(2\theta) + \frac{1}{r}\frac{\partial f(r)}{\partial r}\cos(2\theta) -\frac{4}{r^2}f(r)\cos(2\theta)
    [/itex]

    Evaluate the biharmonic equation:
    I will split the equation into three parts to better show the different derivatives


    [itex]
    \frac{\partial^2 g}{\partial r^2} = \left[\frac{\partial^4 f(r)}{\partial r^4} + \frac{\partial}{\partial r}\left(-\frac{1}{r^2}\frac{\partial f(r)}{\partial r} + \frac{1}{r}\frac{\partial^2f(r)}{\partial r^2} + \frac{8}{r^3}f(r) - \frac{4}{r^2}\frac{\partial f(r)}{\partial r}\right)\right]\cos(2\theta)
    [/itex]

    [itex]
    =\left[\frac{\partial^4 f(r)}{\partial r^4} + \frac{1}{r^2}\frac{\partial f(r)}{\partial r} - \frac{1}{r}\frac{\partial^2 f(r)}{\partial r^2} - \frac{1}{r^2}\frac{\partial^2 f(r)}{\partial r^2} + \frac{1}{r}\frac{\partial^3 f(r)}{\partial r^3} - \frac{24}{r^4}f(r) + \frac{8}{r^3}\frac{\partial f(r)}{\partial r} + \frac{8}{r^3}\frac{\partial f(r)}{\partial r} - \frac{4}{r^2}\frac{^2 f(r)}{\partial r^2}\right]\cos(2\theta)
    [/itex]

    [itex]
    \frac{1}{r}\frac{\partial g}{\partial r} = \left[\frac{1}{r}\left(\frac{\partial^3f(r)}{\partial r^3} - \frac{1}{r^2}\frac{\partial f(r)}{\partial r} + \frac{1}{r}\frac{\partial^2 f(r)}{\partial r^2} + \frac{8}{r^3}f(r) - \frac{4}{r^2}\frac{\partial f(r)}{\partial r}\right)\right]\cos(2\theta)
    [/itex]

    [itex]
    =\left[\frac{1}{r}\frac{\partial^3 f(r)}{\partial r^3} - \frac{1}{r^2}\frac{\partial f(r)}{\partial r} + \frac{1}{r^2}\frac{\partial^2 f(r)}{\partial r^2} + \frac{8}{r^4}f(r) - \frac{4}{r^3}\frac{\partial f(r)}{\partial r}\right]\cos(2\theta)
    [/itex]

    [itex]
    \frac{1}{r^2}\frac{\partial^2g}{\partial \theta^2} = \left[\frac{1}{r^2}\left(-4\frac{\partial^2f(r)}{\partial r^2} - \frac{4}{r}\frac{\partial f(r)}{\partial r} + \frac{16}{r^2}f(r)\right)\right]\cos(2\theta)
    [/itex]

    [itex]
    = \left[\frac{-4}{r^2}\frac{\partial^2f(r)}{\partial r^2} - \frac{4}{r^3}\frac{\partial f(r)}{\partial r} + \frac{16}{r^4}f(r)\right]\cos(2\theta)
    [/itex]

    Combining these terms:

    [itex]
    \nabla^4 \phi = \left[\frac{\partial^4 f(r)}{\partial r^4} - \frac{1}{r}\frac{\partial^2 f(r)}{\partial r^2} + \frac{2}{r}\frac{\partial^3 f(r)}{\partial r^3} - \frac{8}{r^2}\frac{\partial^2 f(r)}{\partial r^2} + \frac{8}{r^3}\frac{\partial f(r)}{\partial r}\right]\cos(2\theta)
    [/itex]

    The final form of the biharmonic equation above does not equal 0 (and for this problem it is supposed to). Is there anywhere I went wrong?
     
  2. jcsd
  3. Oct 20, 2013 #2

    pasmith

    User Avatar
    Homework Helper

    Why would you expect it to for arbitrary [itex]f[/itex]? Clearly you must impose some additional condition.

    Since [itex]\phi[/itex] is separable we have
    [tex]
    \nabla^2 \phi = \nabla^2(f(r) \cos 2\theta) =
    \left(\frac{1}r\frac{\partial}{\partial r} \left(r\frac{\partial f}{\partial r}\right) - 4\frac{f}{r^2}\right)\cos 2\theta = D_r^2(f)\cos 2 \theta[/tex]
    which is separable. Hence
    [tex]
    \nabla^4 \phi = \nabla^2(\nabla^2 \phi) = \nabla^2(D_r^2(f) \cos 2\theta)
    = D_r^2(D_r^2 f) \cos 2\theta = D_r^4 (f) \cos 2\theta
    [/tex]
    which again is separable, and it is this separability which allows us to impose a condition on [itex]f(r)[/itex] so that [itex]\nabla^4 \phi = 0[/itex] for all [itex]r[/itex] and [itex]\theta[/itex]: we must have [itex]D_r^4 f = 0[/itex].

    (I suspect your expression for [itex]D_r^4(f)[/itex] is wrong, because it is dimensionally inconsistent: I expect terms of the form [itex]\frac 1{r^n} \frac{\partial^m}{\partial r^m}[/itex] where [itex]n + m = 4[/itex].)
     
    Last edited: Oct 20, 2013
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