- #1

island-boy

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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.

[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.

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