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Fluids Problem PDE

  1. Apr 27, 2016 #1

    joshmccraney

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    Hi PF!

    So my book has boiled the problem down to $$\psi(r,\theta) = c_1 \ln \frac{r}{a}+\sum_{n=1}^\infty A_n\left(r^n-\frac{a^{2n}}{r^n}\right)\sin n\theta$$ subject to ##\psi \approx Ur\sin\theta## as ##r## get big. The book then writes
    $$\psi(r,\theta) = c_1 \ln \frac{r}{a}+U\left(r-\frac{a^2}{r}\right) \sin \theta$$ I understand ##A_n=0\, \forall\, n \geq 2## and ##A_1=U## but why isn't ##c_1## eliminated too? To me it seems the natural log does not allow the "boundary condition" to be satisfied.
     
  2. jcsd
  3. Apr 27, 2016 #2
    Whats the context of the problem?
     
  4. Apr 27, 2016 #3

    joshmccraney

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    Fluid flow in 2D passing a cylinder of radius ##a##. The flow satisfies continuity, so ##\nabla \cdot \vec{v} = 0##. Define ##\psi \equiv \nabla \times \vec{v}## so that continuity is satisfied. Then ##v_x = \psi_y## and ##v_y=-\psi_x##. The resulting PDE is then ##\nabla^2 \psi = 0##. Solving this in polar coordinates yields the above solution I posted, the infinite series. In the far wake, velocity in the ##x## direction is constant ##U##, which suggests ##v_x = \psi_y = U \implies \psi = Uy## (I guess I don't understand why there is no constant too, which is to say why isn't this true: ##\psi = Uy + C_0(x)##).
     
  5. Apr 29, 2016 #4

    joshmccraney

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    Any ideas? My thought process is since ##r## grows much larger than ##\ln r## we may leave the natural logarithm? Can anyone confirm this?
     
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