Proving Lipschitz Condition for F with $\| \cdot \|_{1}$ Norm

island-boy
Messages
93
Reaction score
0
I need to prove that the function F is Lipschitz, using the
\| \cdot \|_{1} norm.

that is, for
t \in \mathbb{R}
and
y, z \in Y(t) \in \mathbb{R}^{2}

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

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

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

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:

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

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

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:
Physics news on Phys.org
anyone? help please. thanks!
 
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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top