# PDE: Filling in missing steps.

#### Jester

MHF Helper
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?

#### NZJaguar

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.

#### NZJaguar

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
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.

Last edited:
• NZJaguar