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Help needed to find the flow potential function

  1. Nov 30, 2007 #1
    Here is the catch:

    We are given a 2-D velocity flow incompressible, irrotational of the form:

    --> ^ ^ ^ ^
    V = u i + v j = [4y -x(1+x)] i + y(2x+1) j

    and we are asked to find the flow potential which obeys the Laplace Eq. for 2-D incompressible, irrotational flow: dΦ = u dx + v dy

    in other words:

    ---- = u

    ---- = v

    I integrated the first one and then the second one and compared the two functions and combined the terms, but at the end the Φ does not satisfy the first equation only the second one.

    Another technique, I integrated the first function with respect to x and Φ is expressed as 4xy - x^2/2 - x^3/3 + f(y) = Φ (x,y)

    now I differentiate with respect to y and equate it to v:

    4x - f'(y) = y(2x+1) which solves to f(y) = xy^2 + y^2/2 -4xy + C

    Plug it in the above expression and get:

    Φ (x,y) = xy^2 + y^2/2 - x^2/2 - x^3/3 + C

    now the first partial diff. eq is not satisfied but the first is.

    Can someone explain what is wrong here?
  2. jcsd
  3. Nov 30, 2007 #2


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    Science Advisor

    What reason do you have to thinks such a potential exists?

    If there exist [itex]\phi[/itex] such that [itex]d\phi = udx+ vdy[/itex] (as long as u and v are differentiable) then it must be true that the mixed derivatives are equal:
    [tex]\frac{\partial^2 \phi}{\partial x \partial y}= \frac{\partial u}{\partial y}= \frac{\partial v}{\partial x}= \frac{\partial^2 \phi}{\partial y\partial x}[/tex]
    Here it is clear that that is not true: [itex](4y -x(1+x))_y= 4 \ne (y(2x+1))_x= 2y[/itex]. This is not an "exact differential" and there is no "flow potential".
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