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Proof of an admissible Airy stress function

  • Thread starter roldy
  • Start date
  • #1
232
1

Homework Statement


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.


Homework 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]


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?
 

Answers and Replies

  • #2
pasmith
Homework Helper
1,740
412

Homework Statement


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.

[snip]

The final form of the biharmonic equation above does not equal 0
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].)
 
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