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Complex analysis: having partials is the same as being well defined?

  1. Feb 25, 2008 #1
    Complex analysis: having partials is the same as being "well defined?"

    My professor proved this theorem in class and I don't know if I even wrote it down correctly in my notes. I don't have access to the book so I need to know if this makes sense. Here is the theorem:

    Under these conditions:
    [tex]R[/tex] is a simply connected region in [tex]\mathbb{C}[/tex] and

    p,q : [tex]R \rightarrow \mathbb{R}[/tex] continuous.

    [tex]\sigma - rect[/tex] is a rectangle in [tex]\mathcal{R}[/tex].

    The existence of a function U on [tex]R[/tex] such that:

    [tex]\frac{\partial U}{\partial x}=p, \frac{\partial U}{\partial y}=q[/tex]

    [tex]\Longleftrightarrow \displaystyle\oint_{\sigma - rect} pdx + qdy =0[/tex]

    ---
    QUESTIONS:

    1. My professor said that having the partials for p and q is the same thing as saying that U is well defined. Why is this the case?

    2. Then he said U would be well defined if it was path independent, well, if we know it's path independent then, of course [tex]\displaystyle\oint_{\sigma - rect} pdx + qdy =0[/tex]. So, am I right to think of path independence as a consequence of this proof, not a condition? (I know this seems obvious, but I keep losing track of what we "know" when we start the proof and what we are trying to prove so, I want to get this absolutely clear!)
     
    Last edited: Feb 25, 2008
  2. jcsd
  3. Feb 25, 2008 #2

    HallsofIvy

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    If the partials are continuous, and satisfy
    [tex]\frac{\partial p}{\partial y}= \frac{\partial q}{\partial x}[/tex]
    , the "mixed second derivative" condition, which for the real and imaginary parts of the function of a complex variable is the same as saying the function is analytic then that is true.
     
  4. Feb 25, 2008 #3
    Okay, I remember that from last semester... That's the Cauchy Rieman condition (or half of it) -- we didn't mention that at all or anything about being analytic--- except I think if U has continuous partials for its real and imaginary parts, it must be analytic... right?

    Uggg.. this is so confusing-- If I could get a coherent question together I'd email my professor for help since he has a low tolerance level for hand holding...
     
  5. Feb 25, 2008 #4

    Are you saying that I left out a condition? Do I need to say that U is analytic in R in addition to the other things I mentioned in the set-up?
     
  6. Feb 27, 2008 #5
    I found out the answer and I thought I'd share. It turns out that the way that we proved the Cauchy's theorem (It's from the book by Alfors) started out by splitting the complex function in to two real funtions with the form pdx + qdy.

    That's why I only had "half" of the C-R equations.
     
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