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

    and

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

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


    b) Determine the stream function  \psi


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

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

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

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

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

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

      \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  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 v_x and v_y ....

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

    \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 \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

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

    Do the same for \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 v_x and v_y ....

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

    \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 \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\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\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

      v_{y} = axy^{2} + C

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

    where dO I go from here?

    Thanks
    Last edited by Tweety; December 6th 2010 at 12:01 PM.
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  7. #7
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    I think you're confusing yourself with the v's versus the \psi's. For part b, lose the v's entirely and work just with the function \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 v's versus the \psi's. For part b, lose the v's entirely and work just with the function \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  \psi

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

    and the same for  - \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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