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Math Help - PDE: Filling in missing steps.

  1. #1
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    PDE: Filling in missing steps.

    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.

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  2. #2
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    Quote Originally Posted by NZJaguar View Post
    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?
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  3. #3
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    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.
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  4. #4
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    Anyone? Danny don't leave me now!

    Or even if any of you think that it doesn't make sense that would be helpful! =)
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  5. #5
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    You have \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, \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} = r\frac{\partial^2 v_\theta}{\partial r^2}+ \frac{\partial v_\theta}{\partial r}.

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

    Subtracting \frac{v_\theta}{r^2} from that gives the second line: \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:
    \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} = \frac{1}{r}\frac{\partial v_\theta}{\partial r}- \frac{v_\theta}{r^2}
    and, of course, if we know that v_\theta is a function of r only, we can change the partial derivatives symbols to common derivatives: \partial to d.
    Last edited by HallsofIvy; June 1st 2010 at 03:13 AM.
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