TITLE: Proving existence of solutions for second-order differential equation $u''(t)=-w^2\sin(u(t))$
QUESTION [5 upvotes]: Let $w>0$ and $$u''(t)=-w^2\sin(u(t))$$ for $t\geq 0$. 
I'd like to see that the solutions to the system exist on the whole $\mathbb{R}_{+}$ and if an initial value is given, they are supposed to be unique.
Usually I'd translate a second-order system to a first-order one, by renaming $v(t):= u''(t)$ and then rewriting it into a first-order matrix differential equation, which would be linear and therefore I'd easily be able to draw conclusions about the existence of solutions, but this does not work here, because of the $\sin()$ function.
Another approach would be to immediately try to solve by seperating variables, i.e. 
$$ \int_0^t \frac{u''(s)}{\sin(u(s))}\,ds = \int_0^t -w^2 \, ds$$
but here the second order ruins the substitution ( or can you somehow integrate that? ).
I'm not looking for a solution, but a hint on how to approach these second order autonomous systems!

REPLY [5 votes]: You can rewrite
$$ \begin{cases}
u''(t) = -w^2 \sin(u(t)),& t\in \mathbb{R}_{\geq 0};\\
u(0)=u_0; \\
u'(0)=v_0.
\end{cases}$$
as
$$ \begin{cases}
\begin{pmatrix}
u(t) \\ v(t)
\end{pmatrix}' =
\begin{pmatrix}
v(t) \\ -w^2 \sin(u(t))
\end{pmatrix}; \\
\begin{pmatrix}
u(0) \\ v(0)
\end{pmatrix}
= \begin{pmatrix}
u_0 \\ v_0
\end{pmatrix}
\end{cases} $$
by setting $v(t)=u'(t)$. Note that the function
$$F: \mathbb{R}^2 \rightarrow \mathbb{R}^2, \  F( u, v)= (v,-w^2 \sin(u)) $$
is lipschitz (as $\sin$ has bounded derivative). Thus, by the Picard-Lindelöf Theorem our second Cauchy problem has a unique solution. As the second system is equivalent to the first one, it has a unique solution as well.