Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lipschitz condition

  1. Sep 22, 2006 #1
    I need to prove that the function F is Lipschitz, using the
    [tex]\| \cdot \|_{1}[/tex] norm.

    that is, for
    [tex] t \in \mathbb{R}[/tex]
    [tex]y, z \in Y(t) \in \mathbb{R}^{2}[/tex]

    I must show that
    [tex]\|F(t, y) - F(t, z)\|_{1} < k|y-z|[/tex]

    F(t, Y(t)) is given as

    [tex]F(t, Y(t)) = \left( \begin{array}{cc} y' \\ \displaystyle{-\frac{g}{L}\sin(y)}\end{array} \right)[/tex]

    my only other given is that
    y"(t) = -g/L [sin y(t)]
    where g and L are constants.

    Now if my calculations are correct, I only need to show that the following is true:

    [tex]\|[\frac{g}{L}(\cos y(t) - \cos z(t)] - [\frac{-g}{L} (\sin y(t) - \sin z(t)] \|_{1} < K|y-z|[/tex]

    [tex]|\frac{g}{L}(\cos y(t) - \cos z(t)| + |-\frac{-g}{L} (\sin y(t) - \sin z(t)| < K|y-z|[/tex]

    however, I don't know how to prove the above inequality.
    I know that the absolute values of both cos and sin are less than or equal to one, but I don't know if that is helpful.
    Last edited: Sep 22, 2006
  2. jcsd
  3. Sep 23, 2006 #2
    anyone? help please. thanks!
  4. Sep 23, 2006 #3
    ok, please ignore this question,I think I may have solved it after I was able to prove |siny| <= |y| in the other thread.

    thanks for reading.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook