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Drawing flow lines of an irrotational and sourceless flow

  1. Mar 17, 2012 #1
    1. The problem statement, all variables and given/known data

    Draw flow lines of f(x+iy)=x^2-y^2+2ixy

    2. Relevant equations

    n/a

    3. The attempt at a solution

    I know the flow is irrotational and sourceless by Cauchy Theorem but I don't know how to draw the flow lines... I'm thinking I might have to parameterize the equation and go from there? Any help is appreciated
     
  2. jcsd
  3. Mar 17, 2012 #2
    This is strange. f(z)=z^2 is certainly analytic, but curl[x^2-y^2,2xy]=4y in the z direction, certainly not irrotational. Did I make a mistake?
     
  4. Mar 17, 2012 #3
    I used Cauchy Theorem to integrate around closed curve (unit circle) which is equal to 0... I thought this meant that it was irrotational and sourceless (at least locally), maybe not globally though. The book says to integrate the conjugate.. so (x-iy)^2 and if it is 0 then it is irrotational and sourceless..
     
  5. Mar 17, 2012 #4
    I guess f(z) integrated along a curve on the complex plane is different from the vector field [real{f(x,y)}, imag{f(x,y)}] integrated along a curve in the R2 plane. So your sourceless and irrotational assertion are both incorrect.
     
  6. Mar 17, 2012 #5
    what did I do wrong? and how should I go about it?
     
  7. Mar 17, 2012 #6
    let f(z)=u(x,y)+iv(x,y), f(z)dz=(u+iv)(dx+idy)=udx-vdy+i(vdx+udy), while [u,v].[dx,dy]=udx+vdy, obviously different. However, Conjugate(f(z))dz seem to recover the vector field integral correctly...
     
  8. Mar 18, 2012 #7
    so what happens if we do use the conjugate? it is sourceless and irrotational right?
     
  9. Mar 18, 2012 #8
    ok so I went back and tried to do it using green's theorem.. but then it shows that the function is not irrotational and not sourceless.. so something must be wrong because it must be 1 or both of these... i'm guessing maybe I missed a negative sign somewhere or something..
     
  10. Mar 18, 2012 #9
    looks like it is neither irrotational nor sourceless.. so no streamlines are necessary (i guess)

    is this correct? Or can I still draw streamlines?
     
    Last edited: Mar 18, 2012
  11. Mar 18, 2012 #10
    You can certainly draw streamlines. If there's curl. You'll see closed loops. If there's divergence, you'll see streamlines converging into a sink or diverging from a source. They can coexist. A streamline is nothing but a curve whose tangent is in the same direction as the vector field at the same point. You can parameterize the curve and solve some DE for it, but I suspect there's an easier way for your problem, because it is a homework problem, I suppose?
     
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