Proof of an admissible Airy stress function

[roldy]

Homework Statement

Show that a stress function of the form $\phi(r,\theta) = f(r)\cos(2\theta)$, where f(r) is a function of r only, satisfies the biharmonic equation and hence an admissible Airy stress function.

Homework Equations

$g \equiv \nabla^2 \phi = \frac{\partial^2 \phi}{\partial r^2} + \frac{1}{r}\frac{\partial \phi}{\partial r} + \frac{1}{r^2}\frac{\partial^2 \phi}{\partial \theta}$

$\nabla^4 \phi = \nabla^2 g = \frac{\partial^2 g}{\partial r^2} + \frac{1}{r}\frac{\partial g}{\partial r} + \frac{1}{r^2}\frac{\partial^2 g}{\partial \theta}$

The Attempt at a Solution

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

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


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

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

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

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

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

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

Combining these terms:

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

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?

 

[pasmith]

Homework Statement

Show that a stress function of the form $\phi(r,\theta) = f(r)\cos(2\theta)$, 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 $f$? Clearly you must impose some additional condition.

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

(I suspect your expression for $D_r^4(f)$ is wrong, because it is dimensionally inconsistent: I expect terms of the form $\frac 1{r^n} \frac{\partial^m}{\partial r^m}$ where $n + m = 4$.)