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Question and answer on 'not exact' forms

  1. Sep 26, 2006 #1
    The question is:
    Show that on R^2\{0} (without zero),
    let w=(xdy-ydx)/(x^2+y^2) and show (a) closed (b) not exact.

    (a) is straightforward,
    and for (b), the following is the solution lecturer provided.
    Firstly convert to polar coordinates letting x=rcos(p) y=rsin(p) where p is supposed to be angle, and get w=dp.
    And then! (from here on I don't understand) suppose w=df where f is smooth on R^2\{0}. On R dp=df implies d(p-f) is constant. This contradicts that f smooth as in any neighbourhood of (1,0) there exists p1,p2 in R^2\{0} such that f(p1)-f(p2)>pi (as in pi=3.14...)so f is not continuous.

    Could someone 'explain' what he's doing in this solution? I don't understand how exterior derivatives got to do with these functions being constant etc.
    I appreciate your help :)
     
  2. jcsd
  3. Sep 26, 2006 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Do you mean continuous (not "constant")?

    There is a general theorem that, as long as we are working with differential forms defined on a simply connected set, then any closed form must be exact.
    An exact form is one which is the differential of some other form. A closed form is one whose differential is 0. It is easy to show that an exact form is always closed: d(da)= 0 for any differential form a. The converse, that closed forms are exact, is only true on simply connected domains.

    Your instructor's suggestion is exactly right. In polar coordinates, x= rcos(p) so dx= cos(p)dx- r sin(p)dp and y= r sin(p) so dy= sin(p)dr+ rcos(p)dy. Then xdy= rsin(p)cos(p)dx+ r2cos2(p)dy while ydx= r sin(p)cos(p)dr- r2sin2(p)dp.
    Then xdy- ydx= r2(cos2(p)dp+ sin2(p)dp= r2dp. Since the denominator, x2+ y2= r2, w= dp in polar coordinates. What isn't that an exact differential, then? because neither p nor dp is defined at r= 0 so dw is not an exact differential on any region containing the origin.

    The problem is that dw itself is not defined at the origin and so for any region containing the origin, its domain is not simply connected.
     
  4. Sep 26, 2006 #3
    Thank you for your explanation, but all the things you wrote down I already understand. My specific question was interpreting the solution that lecturer wrote down, I'll write down again,

    suppose w=df where f is smooth on R^2\{0}. On R dp=df implies d(p-f) is constant. This contradicts that f smooth as in any neighbourhood of (1,0) there exists p1,p2 in R^2\{0} such that f(p1)-f(p2)>pi (as in pi=3.14...)so f is not continuous.

    and no, it's not 'continuous' instead of 'constant' and THAT's why I don't understand it you see?
     
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