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Moving Infinite Plate at a Constant Velocity

  1. Mar 8, 2007 #1
    Hi all!

    I am required to find the velocity distribution of the flow around an infinite plate that suddenly starts moving with a constant speed U_o. The solution has already been worked out, but I still do not understand all of it. The part that is perplexing to me is where they use the transformation variable to rewrite the terms (del u/del t) and (del^2 u/del y^2) with respect to eta.

    Simple rearrangement of the terms in eta certainly does not yield the equation in (del u/del t), although this is true for (del^2 u/del y^2). I would think that there would be more of a consistency?

    Please help! Thanks!
     

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  3. Mar 8, 2007 #2

    AlephZero

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    It's just the chain rule for differentiation

    del u/del t = (del u/del eta)(del eta/del t)

    and a similar equation for del^2 u/del y^2
     
  4. Mar 8, 2007 #3
    Why of course! Chain Rule, so elementary yet I didn't even consider it. Thanks!!

    This brings up one more question:

    For (del^2 eta/del y^2), why is it that the answer is the square of the first derivative of eta instead of the second derivative of eta? That is,

    (del eta/del y) = 1/(2*sqrt(nu*t))

    => (del^2 eta/del y^2) = 0


    But according to the solution,

    (del^2 eta/del y^2) = [(del eta/del y)]^2 = 1/(4*nu*t)
     
  5. Mar 9, 2007 #4

    AlephZero

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    [tex]\frac{\partial}{\partial y} = \frac{\partial \eta}{\partial y}\frac{\partial}{\partial \eta}[/tex]

    [tex]\frac{\partial^2}{\partial y^2} =
    \frac{\partial \eta}{\partial y} \frac{\partial}{\partial \eta}
    \left( \frac{\partial \eta}{\partial y} \frac {\partial}{\partial \eta} \right)[/tex]
     
    Last edited: Mar 9, 2007
  6. Mar 10, 2007 #5
    I'm sorry, but I'm not familiar with your notation. For the expression for
    [tex]\frac{\partial}{\partial y}[/tex], you used [tex]\frac{\partial}{\partial \eta}[/tex]. What function are you taking the partial (with respect to eta)??

    Also, for your expression of [tex]\frac{\partial^2}{\partial y^2}[/tex], are you just restating what I had posted in my second question? I'm looking for why the partial of second order (that is, [tex]\frac{\partial^2 \eta}{\partial y^2}[/tex]) is not zero, but rather just the square of the the partial of the first order.

    Thank you for your patience!
     
    Last edited: Mar 10, 2007
  7. Mar 10, 2007 #6

    AlephZero

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    I was using [tex]\frac{\partial}{\partial y}[/tex] as an operator, that is something that can operate on any function.

    If you don't like that standard calculus notation, then

    Differentiate u once:

    [tex]\frac{\partial u}{\partial y} = \frac{\partial \eta}{\partial y}\frac{\partial u}{\partial \eta}[/tex]

    Then differentiate again:

    [tex]\frac{\partial^2 u}{\partial y^2} = \frac{\partial \eta}{\partial y} \partial \left( \frac{\partial \eta}{\partial y} \frac {\partial u}{\partial \eta} \right) / \partial \eta} [/tex]

    So, because [tex]\frac{\partial \eta}{\partial y}[/tex] is only a function of [tex]\nu[/tex] and t so it's just a constant when differentiating with respect to [tex]\eta[/tex]

    [tex]\frac{\partial^2 u}{\partial y^2} =
    \frac{\partial \eta}{\partial y}
    \frac{\partial \eta}{\partial y}
    \frac {\partial^2 u}{\partial \eta^2}[/tex]
     
    Last edited: Mar 10, 2007
  8. Mar 10, 2007 #7

    AlephZero

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    A simple example to cut out all the partial differentials, if they are confusing you:

    Let y = ax, where a is a constant. Then dy/dx = a

    For any function u(x), [tex]\frac{du}{dx} = \frac{dy}{dx} \frac{du}{dy} = a \frac{du}{dy}[/tex]

    [tex]\frac{d^2u}{dx^2} = \frac{d}{dx}\left(\frac{du}{dx}\right) = \frac{dy}{dx} \frac{d}{dy}\left(a \frac{du}{dy}\right)[/tex]
    [tex] = a^2 \frac{d^2u}{dy^2}[/tex]
     
    Last edited: Mar 10, 2007
  9. Mar 11, 2007 #8
    AlephZero, THANK YOU!

    Your explanantion with the example was amazing! I truly appreciate your patience with me, as I realize that my math is a little rusty (since I just started school again recently).

    Once again, many thanks!
     
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