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Math Help - Surfaces and Smoothness

  1. #1
    Super Member Aryth's Avatar
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    Surfaces and Smoothness

    Well, here's my problem. I've been at it all night and have made zero progress:

    Consider the real function

    f(x,y) = xy(x^2 + y^2)^{-N}, in the respective cases N = 2, 1, and \frac{1}{2}.

    a.) Show that in each case the function is differentiable ( \mathcal{C}^{\omega}) with respect to x, for any fixed y-value (and that the same holds with the roles of x and y reversed).

    Nevertheless, f is not smooth as a function of the pair (x,y).

    b.) Show this in the case N = 2 by demonstrating that the function is not even bounded in the neighborhood of the origin (0,0).

    c.) Show this in the case N = 1 by demonstrating that the function though bounded is not actually continuous as a function of (x,y).

    d.) Show this in the case N = \frac{1}{2} by showing that though the function is now continuous, it is not smooth along the line x = y.

    Thanks for any and all help. It's much appreciated.
    Last edited by Aryth; June 10th 2008 at 08:12 PM.
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  2. #2
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    Quote Originally Posted by Aryth View Post
    f(x,y) = xy(x^2 + y^2)^{-N}, in the respective cases N = 2, 1, and \frac{1}{2}.

    a.) Show that in each case the function is differentiable ( C^{\omega}) with respect to x, for any fixed y-value (and that the same holds with the roles of x and y reversed).
    You end up with xy_0(x^2+y_0)^{-N}. Thus, as long as y_0\not = 0 it would mean xy_0(x^2+y_0)^{-N} is defined eveywhere. And you can compute the derivatives g^{(n)}(x) where g(x) = f(x,y_0) or all orders n. Thus, g is \mathcal{C}^{\infty}.
    [quote]

    Nevertheless, f is not smooth as a function of the pair (x,y).

    b.) Show this in the case N = 2 by demonstrating that the function is not even bounded in the neighborhood of the origin (0,0).
    You need to show that for any A>0 we can find \epsilon > 0 so that there is a point \bold{x}\in \{ 0 < \sqrt{x^2+y^2} < \epsilon \} such that f(\bold{x}) > A. To show this think along the x-axis. Therefore, there is no way to define the function f to even a continous function at (0,0).

    c.) Show this in the case N = 1 by demonstrating that the function though bounded is not actually continuous as a function of (x,y).
    Show \lim_{(x,y)\to (0,0)} f(x,y) does not exist.

    d.) Show this in the case N = \frac{1}{2} by showing that though the function is now continuous, it is not smooth along the line x = y.
    Defining f(0,0) = 0 will make the function continous. It is continous for |f(x,y) - 0| = \left| \tfrac{xy}{\sqrt{x^2+y^2}} \right| \leq \tfrac{|xy|}{\sqrt{2|xy|}} = \tfrac{1}{\sqrt{2}} \cdot \sqrt{|xy|} \leq \tfrac{1}{2\sqrt{2}}\left( \sqrt{x^2+y^2} \right) < \epsilon if 0 < \sqrt{x^2+y^2} < \epsilon. Thus, the function is continous. To show it is not differenciable it is sufficient to show x^2\sqrt{x^2+x^2} is not differenciable at 0 (this is along the path y=x).
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  3. #3
    Super Member Aryth's Avatar
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    I just had a classic case of "thinking WAY too hard".

    That was much simpler than I intended...

    Thanks a lot.
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