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Real part of a analytic function

  1. Sep 28, 2016 #1
    1. The problem statement, all variables and given/known data


    Show that xux + yuy is the real part of an analytic function if u(x,y) is.
    To which analytic function is the real part of u = Re (f(z))?

    2. Relevant equations
    What I know about analytic functions is Cauchy-Riemann condition
    (∂u/∂x) =(∂v/∂y) and (∂y/∂y)=-(∂v/∂x)

    I know actually Harmonic functions and Laplace equation (2-dim) but I don't know if I need it here:
    (∂2φ/∂x2) + (∂2φ/∂y2) =0


    3. The attempt at a solution

    I say that there is a analytic function : f(z)=u(x,y)+iv(x,y)
    (∂u/∂x)=ux+xuxx+yuxy =(∂v/∂y)
    (∂u/∂y)=xuxy+uy+yuyy=-(∂v/∂x)
    But further , should I integrate to find v(x,y) ?!

    Am I in right path ?!
     
  2. jcsd
  3. Sep 28, 2016 #2
    If I say there is a function F(z)=U(x,y)+iV(x,y)
    and
    U(x,y) = x (∂u/∂x) + y(∂u/∂y)
    Can I use Cauchy-Riemann to find V(x,y)?

    Is this right ?!
     
  4. Sep 28, 2016 #3
    Please can somebody help me ?!
     
  5. Sep 28, 2016 #4

    Mark44

    Staff: Mentor

    In the second equation, it should be (∂u/∂y)=-(∂v/∂x)
    For the first part, I don't think you are. The statement to prove is: If u(x, y) is the real part of an analytic function f(z), then Re(f(z)) = xux + yuy.
     
  6. Sep 28, 2016 #5
    OK if I say f(z)= U(x,y) + i V(x,y), then U(x,y) = x (∂u/∂x) + y(∂u/∂y) ?! (U is a new function and is not the same like u)
    But how can I use it ?! It's not easy to use Cauchy-Riemann to find V
     
  7. Sep 28, 2016 #6

    Mark44

    Staff: Mentor

    What you're calling U above is, I believe, the same as u in ∂u/∂x. If they are the same, you should use just one letter and be consistent.
    The situation as I see it is that f(z) = f(x + iy) = u(x, y) + iv(x, y).

    If I'm wrong in this assumption, what's the meaning of u?


     
  8. Sep 28, 2016 #7

    Mark44

    Staff: Mentor

    Is this the exact wording of the problem?

    It suggests to me that if f(z) = f(x + iy) = u(x, y) + iv(x, y), then u(x, y) = xux(x, y) + yuy(x, y). I don't see why that would be true, but I have to admit it's been a long time since I took a class in complex analysis.
     
  9. Sep 28, 2016 #8
    OK somebody told me F(z) = U(x,y)+iV(x,y) and U(x,y) =xux + yuy
    But if I use Cauchy -Riemann , it's gonna be very hard to find V(x,y) .... I can come with more post and show why
     
  10. Sep 28, 2016 #9
    OK thank you anyway
     
  11. Sep 28, 2016 #10
    ∂V/∂x = (-∂u/∂y) - x(∂2/∂x∂y) -y(∂2/∂y2)

    ∂V/∂y = y(∂u/∂x) +x(∂2u/∂x2)+y(∂2u/∂x∂y)

    But I don't know how I can integrate to find V
     
  12. Sep 28, 2016 #11

    Mark44

    Staff: Mentor

    These parts in the first line above don't make sense:


    This -- ∂2/∂x∂y -- and the other one are operators. You have to indicate what you are taking the partial of, like you did in the second line of the first quote above.
     
    Last edited: Sep 28, 2016
  13. Sep 28, 2016 #12

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Basically, you start with some analytic ##f(z) = f(x+iy) = u(x,y) + i v(x,y)## and are then asked to show that there is an analytic ##F(z)## such that ##F(x+iy) = [x u_x + y u_y]+ i V(x,y)## for some ##V##. Apply Cauchy-Riemann to the pair ##U(x,y) = x u_x(x,y) + y u_y(x,y)## and ##V(x,y)##, so see what can be said about ##V## in terms of ##u## and ##v##. In particular, can you show that an appropriate ##V## actually exists?
     
  14. Sep 29, 2016 #13
    Cauchy-Riemann : ∂U/∂x = ∂V/∂y and ∂U/∂y = - ∂V/∂x

    ∂V/∂x = (-∂u/∂y) - x(∂2/∂x∂y) -y(∂2/∂y2)

    ∂V/∂y = y(∂u/∂x) +x(∂2u/∂x2)+y(∂2u/∂x∂y)

    But I don't know how I find V in this equation!

    I know that the U(x,y) is harmonic (∂2U/∂x2) +(∂2U/∂y2) =0
    But that gives me nothing to find V!
     
  15. Sep 29, 2016 #14

    Mark44

    Staff: Mentor

    You could integrate both sides of the first equation above with respect to x, and integrate the second equation with respect to y.Keep in mind that when you integrate Vx with respect to x, you don't end up with an arbitrary constant, you end up with an arbitrary function of y alone, say g(y). Similar idea when you integrate the second equation with respect to y -- you end up with an arbitrary function of x alone, say f(x).

    This will give you two "views" of V, which you can set equal. I believe that's what Ray was suggesting.
     
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