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Thread: Partial derivatives, equation help

  1. #1
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    Partial derivatives, equation help

    The velocity components for a flow field are as follows;

    $\displaystyle \displaystyle v_{x} = \frac{\partial \psi}{ \partial y} = a( x^{2} - y^{2}) $

    and

    $\displaystyle \displaystyle v_{y} = -\frac{\partial \psi}{\partial x} = -2axy $

    a) Prove that it statifies the continuity equation ; $\displaystyle \displaystyle \frac{\partial v_{x}}{\partial x} + \frac{\partial v_{y}}{\partial y} = 0 $


    b) Determine the stream function $\displaystyle \psi $


    Not sure if my method for part 'a' is correct, I just started rearranging the equations

    $\displaystyle \displaystyle v_{x} = \frac{\partial \psi}{ \partial y} $

    $\displaystyle \displaystyle v_{x} \partial y = \partial \psi $

    sub this into $\displaystyle \displaystyle v_{y} = -\frac{\partial \psi}{\partial x} $

    $\displaystyle \displaystyle v_{y} = -\frac{v_{x} \partial y}{\partial x} $

    $\displaystyle \displaystyle \frac{ v_{y}}{\partial y} = -\frac{v_{x}}{\partial x} $

    $\displaystyle \displaystyle \frac{ v_{y}}{\partial y} + \frac{v_{x}}{\partial x} = 0 $


    It's not the exact desired equation but I am not sure how to get the $\displaystyle v_{x} and v_{y} $ as partial derivatives?

    Also any help on part 'b', do I integrate?

    Thank you.
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  2. #2
    Senior Member yeKciM's Avatar
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    you need to find $\displaystyle v_x $ and $\displaystyle v_y $ ....

    $\displaystyle \displaystyle v_{x} = \frac{\partial \psi}{ \partial y} \Rightarrow v_x = \int a(x^2-y^2) dx $

    $\displaystyle \displaystyle v_{y} = -\frac{\partial \psi}{\partial x} = -2axy \Rightarrow v_y = \int -2axy dy $

    then you will do partial derivation on v_x and v_y to do this

    $\displaystyle \displaystyle \frac{ v_{y}}{\partial y} + \frac{v_{x}}{\partial x} = 0$
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  3. #3
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    I think you're making it more complicated than it needs to be. Just compute

    $\displaystyle \dfrac{\partial v_{x}}{\partial x}=2ax.$

    Do the same for $\displaystyle \dfrac{\partial v_{y}}{\partial y}.$ The result is immediate.
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  4. #4
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    Quote Originally Posted by yeKciM View Post
    you need to find $\displaystyle v_x $ and $\displaystyle v_y $ ....

    $\displaystyle \displaystyle v_{x} = \frac{\partial \psi}{ \partial y} \Rightarrow v_x = \int a(x^2-y^2) dx $

    $\displaystyle \displaystyle v_{y} = -\frac{\partial \psi}{\partial x} = -2axy \Rightarrow v_y = \int -2axy dy $

    then you will do partial derivation on v_x and v_y to do this

    $\displaystyle \displaystyle \frac{ v_{y}}{\partial y} + \frac{v_{x}}{\partial x} = 0$

    Thank you, but I dont understand why you are integrating? How does that make v_{x} and v_{y} into partial derivatives?
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  5. #5
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    For the integration, just follow the general idea for solving exact ODE's. That is, start with

    $\displaystyle \displaystyle\frac{\partial\psi}{\partial y}=a(x^{2}-y^{2})\quad\rightarrow\quad \psi=ax^{2}y-a\frac{y^{3}}{3}+f(x)\quad\rightarrow\quad\frac{\p artial\psi}{\partial x}=2axy+f'(x).$

    Compare this with the other expression you have, and go from there.
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  6. #6
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    Quote Originally Posted by Ackbeet View Post
    For the integration, just follow the general idea for solving exact ODE's. That is, start with

    $\displaystyle \displaystyle\frac{\partial\psi}{\partial y}=a(x^{2}-y^{2})\quad\rightarrow\quad \psi=ax^{2}y-a\frac{y^{3}}{3}+f(x)\quad\rightarrow\quad\frac{\p artial\psi}{\partial x}=2axy+f'(x).$

    Compare this with the other expression you have, and go from there.
    after integrating I get

    $\displaystyle v_{y} = axy^{2} + C $

    $\displaystyle v_{x} = ax^{2}y - \frac{ay^{3}}{3} + C $

    where dO I go from here?

    Thanks
    Last edited by Tweety; Dec 6th 2010 at 12:01 PM.
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  7. #7
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    I think you're confusing yourself with the $\displaystyle v$'s versus the $\displaystyle \psi$'s. For part b, lose the $\displaystyle v$'s entirely and work just with the function $\displaystyle \psi.$ Does that clear away a few things?
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  8. #8
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    Quote Originally Posted by Ackbeet View Post
    I think you're confusing yourself with the $\displaystyle v$'s versus the $\displaystyle \psi$'s. For part b, lose the $\displaystyle v$'s entirely and work just with the function $\displaystyle \psi.$ Does that clear away a few things?

    I am not on part 'b' yet, I still can't do part 'a'/

    or is my orginal post correct for part a?
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  9. #9
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    No, you can't throw around partial differentials like you can total differentials. Your entire approach is, I'm afraid, doomed to failure. Just follow my hint in post # 3.
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  10. #10
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    ^^ oh right, now I get it. I really was over complicating it.

    Thanks for your help
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  11. #11
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    So for part 'b' to find the function $\displaystyle \psi $

    I have to integrate $\displaystyle \frac{\partial \psi}{\partial y} = a(x^{2} - y^{2} ) $

    and the same for $\displaystyle - \frac{\partial \psi}{\partial x} = -2axy $

    ?
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  12. #12
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    For part b, follow the hint in post # 5.
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