# Thread: Integration help - Biharmonic Equation

1. ## Integration help - Biharmonic Equation

Hi I'm having trouble expanding the following equation for the axially symmetric biharmonic equation in polar co-ordinates.

$\displaystyle \left ( \frac{d^2}{dr^2} + \frac{d}{rdr} \right ) \left ( \frac{d^2\phi}{dr^2} + \frac{d\phi}{rdr} \right ) = 0$

When I expanded it I got:
$\displaystyle \frac{d^4\phi}{dr^2} + \frac{2d\phi}{r^3dr} + \frac{d^3\phi}{rdr^3} + \frac{d^3\phi}{rdr^3} - \frac{d\phi}{r^3dr} + \frac{d^2\phi}{r^2dr^2}$

$\displaystyle = \frac{d^4\phi}{dr^2} + \frac{2d^3\phi}{rdr^3} + \frac{d^2\phi}{r^2dr^2} + \frac{d\phi}{r^3dr}$

But it should be:

$\displaystyle = \frac{d^4\phi}{dr^2} + \frac{2d^3\phi}{rdr^3} - \frac{d^2\phi}{r^2dr^2} + \frac{d\phi}{r^3dr}$

Any help would be much appreciated regards,

Callum

2. Originally Posted by CallumJ
Hi I'm having trouble expanding the following equation for the axially symmetric biharmonic equation in polar co-ordinates.

$\displaystyle \left ( \frac{d^2}{dr^2} + \frac{d}{rdr} \right ) \left ( \frac{d^2\phi}{dr^2} + \frac{d\phi}{rdr} \right ) = 0$

When I expanded it I got:
$\displaystyle \frac{d^4\phi}{dr^2} + \frac{2d\phi}{r^3dr} + \frac{d^3\phi}{rdr^3} + \frac{d^3\phi}{rdr^3} - \frac{d\phi}{r^3dr} + \frac{d^2\phi}{r^2dr^2}$

$\displaystyle = \frac{d^4\phi}{dr^2} + \frac{2d^3\phi}{rdr^3} + \frac{d^2\phi}{r^2dr^2} + \frac{d\phi}{r^3dr}$

But it should be:

$\displaystyle = \frac{d^4\phi}{dr^2} + \frac{2d^3\phi}{rdr^3} - \frac{d^2\phi}{r^2dr^2} + \frac{d\phi}{r^3dr}$

Any help would be much appreciated regards,

Callum
$\displaystyle \frac{d}{dr} \left( \frac{1}{r} \frac{d \phi}{dr} \right) = -\frac{1}{r^2} \frac{d \phi}{dr} + \frac{1}{r} \frac{d^2 \phi}{dr^2}$.

Therefore:

$\displaystyle \frac{d^2}{dr^2} \left( \frac{1}{r} \frac{d \phi}{dr} \right) = \frac{2}{r^3} \frac{d \phi}{dr} - \frac{2}{r^2} \frac{d^2 \phi}{dr^2} + \frac{1}{r} \frac{d^3 \phi}{dr^3}$

and

$\displaystyle \frac{1}{r} \frac{d}{dr} \left( \frac{1}{r} \frac{d \phi}{dr} \right) = -\frac{1}{r^3} \frac{d \phi}{dr} + \frac{1}{r^2} \frac{d^2 \phi}{dr^2}$.

Correctly substitute and simpify and you'll get the correct answer.

3. Originally Posted by CallumJ

But it should be:

$\displaystyle = \frac{d^4\phi}{dr^2} + \frac{2d^3\phi}{rdr^3} - \frac{d^2\phi}{r^2dr^2} + \frac{d\phi}{r^3dr}$

Any help would be much appreciated regards,

Callum
Should it not be:

$\displaystyle \frac{d^4\phi}{dr^4} + \frac{2d^3\phi}{rdr^3} - \frac{d^2\phi}{r^2dr^2} + \frac{d\phi}{r^3dr}=0$

Then what, multiply throughout by $\displaystyle r^4$, Euler, then Bingo-bango or no?