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Lipschitz condition |
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| Sep22-06, 12:13 PM | #1 |
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Lipschitz condition
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] and [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. |
| Sep23-06, 10:04 AM | #2 |
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anyone? help please. thanks!
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| Sep23-06, 11:30 AM | #3 |
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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. |
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