PDE: Filling in missing steps.

May 2010
3
0
Hi everyone,

I'm a new poster! I'm studying biomedical engineering and we're currently looking at fluid mechanics. In solving a particular problem I'm having trouble following the worked solutions, which I have uploaded as an image file below.

Your patient help would be most appreciated!

Thanks.

 

Jester

MHF Helper
Dec 2008
2,470
1,255
Conway AR
Hi everyone,

I'm a new poster! I'm studying biomedical engineering and we're currently looking at fluid mechanics. In solving a particular problem I'm having trouble following the worked solutions, which I have uploaded as an image file below.

Your patient help would be most appreciated!

Thanks.

Which part? The first step or the second?
 
May 2010
3
0
Both of them. I suspect there's something sneaky going on when expanding those brackets. As for turning it into an ODE I'm at a loss. Thanks.
 
May 2010
3
0
Anyone? Danny don't leave me now!

Or even if any of you think that it doesn't make sense that would be helpful! =)
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
You have \(\displaystyle \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial v_\theta}{\partial r}\right)- \frac{V_\theta}{r^2}= 0\)

Now, by the product rule, \(\displaystyle \frac{\partial}{\partial r}\left(r\frac{\partial v}{\partial r}\right)= \frac{\partial r}{\partial r}\frac{\partial v_\theta}{\partial r}+ r \frac{\partial^2 v_\theta}{\partial r^2}\)\(\displaystyle = r\frac{\partial^2 v_\theta}{\partial r^2}+ \frac{\partial v_\theta}{\partial r}\).

\(\displaystyle \frac{1}{r}\) times that gives
\(\displaystyle \frac{\partial^2 v_\theta}{\partial r^2}+ \frac{1}{r}\frac{\partial v_\theta}{\partial r}\)

Subtracting \(\displaystyle \frac{v_\theta}{r^2}\) from that gives the second line:\(\displaystyle \frac{\partial^2 v_\theta}{\partial r^2}+ \frac{1}{r}\frac{\partial v_\theta}{\partial r}-\frac{v_\theta}{r^2}\)

To see that this is the same as the third line, use the quotient rule:
\(\displaystyle \frac{\partial}{\partial r}\frac{v_\theta}{r}= \frac{r\frac{\partial v_\theta}{\partial r}- v_\theta\frac{\partial r}{\partial r}}{r^2}\)\(\displaystyle = \frac{1}{r}\frac{\partial v_\theta}{\partial r}- \frac{v_\theta}{r^2}\)
and, of course, if we know that \(\displaystyle v_\theta\) is a function of r only, we can change the partial derivatives symbols to common derivatives: \(\displaystyle \partial\) to d.
 
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