Suppose we have the ODE $$x''=-a\sin{x}.$$
Then let $$x'=y$$ and $$y'=-a\sin{x}.$$ So $$\mathbf X = \begin{pmatrix} y \\ -a\sin{x} \end{pmatrix}.$$ Im confused about how to show a Lipschitz condtion and what interval to use. So the way I have been trying is to take $${||\mathbf X(t,x_1)-\mathbf X(t,x_2)||}= { \sqrt {(y_1-y_2)^2 +(-a\sin{(x_1)}-(-a)\sin{(x_2)})^2} } $$
Then I have: $${ \sqrt {(y_1-y_2)^2 +(-a\sin{(x_1)}-(-a)\sin{(x_2)})^2} } \over{\sqrt {(x_1-x_2)^2+(y_1-y_2)^2}} .$$
This is where Im stuck as to show that a Lipschitz constant exists.
Tomas is correct in the comments : A more direct proof :
Let $x_1' = x_2$. Let $x_2' = -a\sin{x_1}$. Let's show first that $\sin$ is lipschitz. We have $$ |\sin(x)-\sin(y)| = |2\sin(\frac{x-y}{2})\cos(\frac{x+y}{2})| \leq 2\frac{1}{2}|x-y| = |x-y|$$
Now for our system, we have
$$|X(x,t) - X(y,t)| = \sqrt{(x_2-y_2)^2 + (-a\sin x_1 + a\sin y_1)^2}$$ $$ \leq\sqrt{(x_2 - y_2)^2 + a^2(x_1-y_1)^2} \ \text{from above}$$ $$ \leq max(a^2,1)|x-y|$$
So we have Lipschitz (on bounded domains)