1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Complex Analysis => Fluid Flow

  1. Mar 10, 2006 #1

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I'm struggling with this question right now:

    Let the complex velocity potential [itex] \Omega(z) [/itex] be defined implicitly by

    [tex] z = \Omega + e^{\Omega} [/tex]

    Show that this map corresponds to (some kind of fluid flow, shown in a diagram, not important).

    For background,

    [tex] \Omega = \Phi + i\Psi [/tex]

    where Phi is the velocity potential:

    [tex] \mathbf{v} = \nabla\Phi [/tex]

    and Psi is the harmonic conjugate of Phi (therefore it is the streamfunction of the fluid flow.

    My first thought was that I need to find the level curves of the streamfunction in order to find out what kind of flow this is. But before I can do that, I need to solve for Omega explicitly. THAT's where I'm stumped. Any suggestions on a strategy or approach?
     
  2. jcsd
  3. Mar 10, 2006 #2

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    Well, you can perhaps find a power series expansion for [itex]\Omega[/itex]. You have:

    [tex]I = (I + \exp)\circ\Omega[/tex]

    [tex](I + \exp)^{-1} = \Omega[/tex]

    I think the best way to compute the left side is to look at the power series expansion of

    [tex]\frac{1}{1 - x}[/tex]

    where x is replaced with -exp. I can't think of a way to get a closed form solution out of this, but at least it will give you something to work with, hopefully enough to prove the other things you need to prove. Note given any thing like Z = X + iY, it's easy to express Y solely in terms if Z, and similarly X only in terms of Z

    X = (Z + Z*)/2

    Where Z* is the complex conjugate of Z.
     
    Last edited: Mar 10, 2006
  4. Mar 10, 2006 #3

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I'm really not sure what's going on here.
     
  5. Mar 10, 2006 #4

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    Sorry, I is the identity function. The left is I, so I(z) = z. The right is (I + exp)oQ, so

    [(I + exp)oQ](z)
    = (I + exp)(Q(z))
    = I(Q(z)) + exp(Q(z))
    = Q(z) + exp(Q(z))

    I'm using Q for [itex]\Omega[/itex]. Anyways, I don't think the thing I suggested will work...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?