1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Existence of improper integral using polar transform (Munkres Analysis on Manifolds)

  1. Jun 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Let U be the open set in [itex]R^2[/itex] consisting of all x with (Euclidean norm) [itex] ||x|| < 1.[/itex] Let [itex] f(x,y) = 1/(x^2 + y^2)[/itex] for [itex] (x,y) \not = 0.[/itex] Determine whether [itex] f [/itex] is integrable over [itex] R^2 - \overline{U}; [/itex] if so, evaluate it.

    2. Relevant equations
    [itex]g:R^2 \rightarrow R^2 [/itex] is the polar coordinate transformation defined by [itex] g(r, \theta ) = (r\cos \theta , r\sin \theta ). [/itex]


    3. The attempt at a solution
    My thought is to use the sequence of sets [itex] A_n = \{(r,\theta )| 1<r<n, 0<\theta < 2\pi \}[/itex] in the polar plane whose infinite union will equal [itex] R^2 - \overline{U} [/itex], where I can then show that [itex]\int _{A_N} f [/itex] is unbounded, implying [itex] \int _{R^2 - \overline{U}} f[/itex] does not exist. However, in order to use the polar coordinate transform, I need to show that [itex]g [/itex] is a diffeomorphism. However, I'm not sure how to show that g is bijective. If anyone has any advice, I would be very grateful. Thanks!
     
  2. jcsd
  3. Jun 12, 2012 #2
    Re: Existence of improper integral using polar transform (Munkres Analysis on Manifol

    I don't think the integral exists first of all, because you have 1/r2 integrated against dA=rdθdr, so you are integrating (1/r)dr, which diverges near 0 (or to ∞, by the way; it's symmetrically divergent in a sense).

    also, i think you meant r goes from 1/n to 1.

    To get the diffeomorphism you'll need an open set, r in (1/n,1) and θ in (0,π). You could give θ a larger region, just has to be smaller than [0,2π). But it will be easier to show the diffeomorphism on a smaller region, and the integral is symmetric enough, so you could just take half the region (or smaller).

    To show it is one-to-one, how about showing if cos(θ)=cos(θ'), then θ=θ' (for θ in (0,π)). To work with (0,2π), you'll need to make sure you understand what the inverse trig functions do with some care.

    EDIT:

    if I think about it, if sin(θ)>0 means θ is in (0,π), sin(θ)=0 means θ=π, and sin(θ)<0 means θ is in (π,2π). Then use cos to get 1-1 on each region, and stir the logic to get a neat statement.
     
  4. Jun 12, 2012 #3
    Re: Existence of improper integral using polar transform (Munkres Analysis on Manifol

    Assuming your approach will work, I'm just going to talk about the polar coordinate transformation. First off, [itex]g:R^2\rightarrow R^2[/itex] is not a diffeomorphism. Think about it; in the polar plane, you describe a point by its radius and an angle. However, to remain consistent, we throw out the point at the origin (because the origin could be assigned any angle), and we restrict our angles to being in some interval of length [itex]2\pi[/itex] (since otherwise, any point could be described by an infinite number of angles). Customarily, we choose [itex](-\pi,\pi][/itex] (or [itex](-\pi,-\pi)[/itex] if we are interested in differentiability, as we require open sets).

    So really, we can make g a diffeomorphism by choosing the domain and codomain carefully. Since your set [itex]\overline{U}[/itex] contains the origin, and we are only interested in the behaviour of your function and its integral over [itex]R^2-\overline{U}[/itex], this is OK.

    I claim that [itex]g:(0,\infty)\times(-\pi,\pi)\rightarrow R^2\backslash\{0\}[/itex] defined by [itex]g(r,\theta)=(r\cos\theta,r\sin\theta)[/itex] is a diffeomorphism which is continuous on [itex](0,\infty)\times(-\pi,\pi][/itex].

    As for proving it is a bijection, simply find the inverse! I then leave it to you to check anything else.
     
  5. Jun 12, 2012 #4
    Re: Existence of improper integral using polar transform (Munkres Analysis on Manifol

    Simply finding the inverse is not enough, for instance does the inverse imply diffeomorphism on (0,3π)? Since your proof proves something false, there must be an error. Finding the inverse may not get you to the proof any faster. Of course, it was my first thought as well, but then I worried about that larger interval business.

    EDIT: Now I'm not sure we don't need the inverse, it may help with the r part. You may have to play around with it to find the most efficient/elegant proof.
     
  6. Jun 12, 2012 #5
    Re: Existence of improper integral using polar transform (Munkres Analysis on Manifol

    Certainly not; can't have a diffeomorphism without a bijection, and [itex](0,3\pi)[/itex] is too big for that. I only intended on ever explaining how one could turn g into a bijection.

    Not sure what you mean by false... One can define the inverse in terms of the arctan function. Once you know that, you can just check that g has no critical points, and you have a diffeomorphism. I would link to something here as a source, but I'm not at 10 posts yet :P

    Edit: I'm at 10 posts now. Here's a link: http://math.bard.edu/belk/math461/MultivariableCalculus.pdf
     
    Last edited: Jun 12, 2012
  7. Jun 13, 2012 #6
    Re: Existence of improper integral using polar transform (Munkres Analysis on Manifol

    So, I believe I need to show that given the set [itex] A_N = \{(r, \theta)| 1 < r < N, 0 < \theta < 2\pi \} [/itex] in the [itex] (r, \theta) [/itex] plane and the set [itex] U_N = \{(x,y)| 1<x^2 + y^2 < N \text{ and } x<0 \text{ when } y=0\}[/itex] in the [itex] (x,y) [/itex] plane, then [itex]g:A_N \rightarrow U_N [/itex] is a diffeomorphism.

    To show that it is one-to-one, assume that [itex]g(r, \theta ) = g(s, \tau ) [/itex]. Then, [itex] (r\cos \theta , r\sin \theta ) = (s\cos \tau , s\sin \tau ), [/itex], which implies that [itex]r^2 \cos ^2 \theta + r^2 \sin ^2 \theta = s^2 \cos \theta + s^2 \sin \theta [/itex], which of course implies that [itex] r^2 = s^2[/itex], or that [itex] r = s [/itex] since both are restricted to be positive numbers. Then, we can easily show that [itex] \theta = \tau [/itex], completing the proof that [itex] g [/itex] is one-to-one.

    The proof that [itex] g [/itex] is onto comes from the inverse function [itex] g^{-1}[/itex] defined as [itex] g^{-1}(x,y) = (\sqrt{x^2 + y^2}, \arctan({y/x}))[/itex].

    Hence, [itex] g [/itex] is bijective, and moreover [itex] Det(Dg(r, \theta )) = r[/itex] for all [itex](r, \theta ) [/itex] which is non-zero, so [itex] g [/itex] is a diffeomorphism.

    Now that [itex] g [/itex] is a diffeomorphism, I can invoke the change of variables theorem to show that [itex] \int _{U_N} f = \int _{A_N} 1/(r^2 \cos ^2 \theta + r^2 \sin ^2 \theta )r[/itex], which is unbounded as [itex] N \rightarrow \infty[/itex]. Hence, the integral does not exist.

    How does this proof look? Thanks again for the help!
     
    Last edited: Jun 13, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Existence of improper integral using polar transform (Munkres Analysis on Manifolds)
Loading...