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Stream Function Confusion

  1. Mar 13, 2010 #1
    In Batchelor's text (2000) on page 76, the stream function is defined as

    [tex]
    \psi - \psi_0 = \int\left(u dy - v dx\right)
    [/tex]

    where [itex] \psi_0 [/itex] is a constant

    Now I begin with a simple function for [itex]u[/itex] where

    [tex]
    u = x^3
    [/tex]

    From mass conservation,

    [tex]
    \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0
    [/tex]

    [tex]
    3x^2 + \frac{\partial v}{\partial y} = 0
    [/tex]

    [tex]
    v = -3x^2y
    [/tex]

    Plugging this into the equation for the stream function

    [tex]
    \psi - \psi_0 = \int\left(u dy - v dx\right)
    [/tex]

    [tex]
    \psi - \psi_0 = \int\left(x^3 dy + 3x^2y dx\right)
    [/tex]

    [tex]
    \psi - \psi_0 = \int\left(x^3 dy\right) + \int\left(3x^2y dx\right)
    [/tex]

    [tex]
    \psi - \psi_0 = x^3y + x^3y + C
    [/tex]

    [tex]
    \psi - \psi_0 = 2x^3y + C
    [/tex]

    Now using the equations for [itex] u [/itex] and [itex] v [/itex],

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

    [tex]
    u = \frac{\partial (2x^3y + C - \psi_0)}{\partial y}
    [/tex]

    [tex]
    u = 2x^3
    [/tex]

    [tex]
    v = -\frac{\partial \psi}{\partial x}
    [/tex]

    [tex]
    v = -\frac{\partial (2x^3y + C - \psi_0)}{\partial x}
    [/tex]

    [tex]
    v = -6x^2y
    [/tex]

    I seems like the initial [itex]v=-3x^2y[/itex] and [itex]u=x^3[/itex] are off from the recalculated [itex]v = -6x^2y[/itex] and [itex]u = 2x^3[/itex] by a factor of two. Am I doing something wrong? Thanks.
     
  2. jcsd
  3. Mar 14, 2010 #2
    Hello Ryan.

    First off I recommend revising partial differentiation and partial differential equations.

    In general with ordinary differential equations we add an arbitrary constant when we integrate them.

    With partials you add an arbitrary function.

    So your solution of the continuity equation should be

    [tex]\begin{array}{l}
    \frac{{\partial v}}{{\partial y}}\quad = \quad - 3{x^2} \\
    v\quad = \quad - 3{x^2}y\quad + \quad P(x) \\
    \end{array}[/tex]

    Where P(x) is an arbitrary function of x alone.

    You can see that if this is so then differentiation with respect to y will yield the same result, for any P(x) whatsoever.

    Now you need more information to evalutate P(x)

    I suggest you read on a few pages and try the condition for irrotational flow.
    I'm afraid Batchelor makes a bit of a meal of it but the condition boils down to

    [tex]\frac{{\partial u}}{{\partial y}}\quad = \quad \frac{{\partial v}}{{\partial x}}[/tex]
     
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