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Finding an Antiderivative in an Annulus

  1. Mar 21, 2008 #1
    [SOLVED] Finding an Antiderivative in an Annulus

    The problem statement, all variables and given/known data
    Let A be the annulus, A = {z : 0 <= r < |z| < R}. Suppose that g is an analytic function on A (and that g' is continuous) with the property that

    [tex]\int_{|z| = s} g(z) \, dz = 0[/tex]

    where r < s < R. Show that there is an analytic function G on A with G on A with G' = g throught A.

    The attempt at a solution
    My plan of attack is as follows:

    1. Show that for any piecewise smooth simple closed curve C in A, [itex]\int_C g(z) \, dz = 0[/itex].

    2. Conclude that g(z) is path-independent on A and so there is a function G on A such that if g(z = x + iy) = u(x, y) + iv(x, y), then Gx = u and Gy = y or more succinctly that G' = g.

    I'm having trouble with 1. If the interior of C is entirely contained in A, then by Cauchy's Theorem, the integral is 0. But what if C goes around the annulus? I was thinking of using Green's Theorem but I can't because A is open and so I can't integrate around its boundaries. But what if I apply Green's Theorem to the portion of A contained within |z| = s and C? Hmm...

    If |z| = s is contained inside C, then

    [tex]\int_C g(z) \, dz - \int_{|z|=s} g(z) \, dz = i\iint_B (g_x + ig_y) \, dxdy[/tex]

    The second term on the LHS is 0. The term on the RHS is 0 because g is analytic and by Cauchy-Riemann equations. That means the first term on the LHS is 0. Aha! Is correct?
     
  2. jcsd
  3. Apr 18, 2008 #2
    In order to better understand this problem, I think it would be best to simplify: suppose g is analytic inside D = {z : |z| < R}.

    D is simply-connected. By Cauchy's Theorem,

    [tex]\int_C g(z) \, dz = 0[/tex]

    That means the line integral of g(z) is path-independent. So if [itex]\gamma[/itex] is a piecewise smooth simple curve connecting points [itex]z_0[/itex] and [itex]z[/itex] in D, then

    [tex]\int_\gamma g(z) \, dz = G(z) - G(z_0)[/tex]

    by the Fundamental Theorem of Calculus, where G is the antiderivative of g or equiv. G' = g. Does this make sense? Would I be able to employ the same reasoning in the original problem?
     
  4. Apr 19, 2008 #3

    Dick

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    If C winds once around the hole in the annulus, just deform it into the circular contour |z|=s.
     
  5. Apr 20, 2008 #4
    Deform it how?

    Also, do you agree with my analysis in my first and second post?
     
  6. Apr 20, 2008 #5

    Dick

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    Define a new contour. Pick a point z0 on C. Connect z0 to |z|=s along a path D. Wind around the hole along |z|=s the opposite way that C winds around. Then follow D back to z0. If you draw the picture you can see that the curve as a whole doesn't wind around the hole. It's interior is simply connected (or can be broken into simply connected parts by breaking it where it crosses itself).
     
  7. Apr 20, 2008 #6
    Yes it does since I'm traveling around |z| = s.

    But the interior contains the hole |z| < r. I don't get.
     
  8. Apr 20, 2008 #7

    Dick

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    If the contour wraps around the hole one way then wraps around the other way and closes, then the hole isn't really 'in' the contour. The interior of the contour is simply connected. It can be contracted to a point. Wish I had a picture handy. Do you see what I mean?
     
  9. Apr 20, 2008 #8
    Perhaps there's a misunderstanding. Do you mean the contour from z0 around C (clockwise) back to z0, then through D onto |z| = s, then around |z| = s (counterclockwise) back to D, then through D back to z0? This is the only way I imagine the interior of the contour 'without the hole'.

    Is the interior simply connected just because you can't have a closed curved around it (because D prevents it)?
     
  10. Apr 20, 2008 #9

    Dick

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    That's exactly what I mean. The hole in the domain is no longer in the interior of the contour. It's simply connected because any loop in the domain can be contracted to a point without hitting the hole. So you can now apply Cauchy to the contour. You should be able to conclude from this that the integral around ANY closed contour in the annulus is zero.
     
  11. Apr 21, 2008 #10
    Sorry. I'm unable to conclude what you wrote. I need to show that the integral around C is 0. I don't understand how to conclude this from knowing that the interior of the contour is simply connected.

    I know that integrating around |z| = s is 0 and going twice through D back and forth makes that part of the integral 0. That just leaves the integral around C, which is what I need.
     
  12. Apr 21, 2008 #11

    Dick

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    The integral about the whole contour is 0, from Cauchy. And as you point out, only C contributes. So the integral around C is?
     
  13. Apr 21, 2008 #12
    What domain are you using? Are you applying Cauchy's Theorem using the interior of the contour as the domain? I don't think you can do that since domains are open sets.
     
  14. Apr 21, 2008 #13

    Dick

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    Why do you think the interior of the contour isn't open? It looks like a big fat letter C with the open ends almost closed but with a slit between them. Do I really have to find or make a picture? You described it pretty well.
     
  15. Apr 22, 2008 #14
    On the contrary. I know it is open. However, the contour does not form part of it and so you can't apply Cauchy's Theorem on the contour using it's interior as the domain.
     
  16. Apr 22, 2008 #15

    HallsofIvy

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    To make that a single path, imagine two radial lines, very close together, going between the two circles. You start at one of the lines, on the outer circle, go around the outer circle, counter-clockwise, to the second line, along that line to the inner circle, around the inner circle clockwise until you are back to the first line, the up that line to the outer circle. That gives you a single closed path. Is z= 0 inside or outside that path?

    But that gives you the same integral as the two circles because it does not change as you move the two straight lines closer together when they are "together", their integrals cancel.
     
  17. Apr 22, 2008 #16
    You've just described the so-called keyhole contour.

    z = 0 is at the center of the keyhole but is effectively outside of it.

    Oh really? That seems counterintuitive to me. However, how does that help me determine that the line integral around C is 0.
     
  18. Apr 25, 2008 #17
    I had an epiphany: one may safely apply Cauchy's Theorem to the keyhole contour because both the contour and its inside are in the annulus. Duh!

    Dick: You are right as always. Thank you very much.
     
  19. Apr 25, 2008 #18

    Dick

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    Glad you got it. But the MOST important thing about the keyhole interior is that it is simply connected. You see that too, right?
     
  20. Apr 25, 2008 #19
    Yes I do. As a matter of fact, it has to be simply-connected for me to apply Cauchy's Theorem.

    When I started out with this problem, I was dead set in trying to determine using elementary means, i.e. without Cauchy's Theorem, that the integral of g around C is 0. The idea of building a keyhole contour to apply Cauchy's Theorem never occurred to me. Your intuition is very good Dick.
     
  21. Apr 25, 2008 #20

    Dick

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    Thanks for the compliment. But I can't claim I created the idea. Somebody explained it to me once. And probably had a hard time doing it...
     
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