Solving a Nonlinear ODE Does this problem have a unique solution in $[t_0,t_1]$?
$$ \ddot x(t)=\alpha(1-t)\cos(x(t))$$ 
$$x(t_0)=0$$
$$ \dot x(t_1)=0$$
Thanks!
 A: Yes, by the Picard-Lindelöf theorem, or the Cauchy-Lipschitz theorem. To see this, write your second order ODE as a two-dimensional system of first order ODE's:
\begin{align}
 \frac{\text{d} x}{\text{d} t} &= y(t),\\
 \frac{\text{d} y}{\text{d} t} &= (1-t) \cos(x(t)).
\end{align}
You can now show that the function
\begin{equation}
f : (t_0,t_1) \times \mathbb{R}^2 \to \mathbb{R}^2,\quad f(t;x,y) = \begin{pmatrix} y \\ (1-t) \cos(x) \end{pmatrix}
\end{equation}
is Lipschitz continuous (on a certain set $U \subset \mathbb{R}^2$), by showing that there exists a constant $L$ such that
\begin{equation}
 \|f(t;x,y) - f(t;\hat{x},\hat{y})\| \leq L \left\| \begin{pmatrix} x \\ y \end{pmatrix} - \begin{pmatrix}\hat{x}\\ \hat{y} \end{pmatrix} \right\|
\end{equation}
for all $t \in (t_0,t_1)$ and all $(x,y),(\hat{x},\hat{y}) \in U$.
EDIT: The above is true for initial value problems. In your case, the question at hand is really one of existence. In other words: given $t_0$, $t_1$, does there exist $y_0,x_1 \in \mathbb{R}$ such that the orbit in the above ODE system associated to the initial condition $(x(t_0),y(t_0)) = (0,y_0)$ obeys $(x(t_1),y(t_1)) = (x_1,0)$? This question is a lot harder to answer.
A clue to the answer can be found in the observation that we can bound the right hand side of the original ODE:
\begin{equation}
 \left| \alpha (1-t) \cos x \right| \leq |\alpha(1-t)|.
\end{equation}
For second order ODE's with such a property, existence of a boundary value problem such as the one you described can be found in Theorem 2.1 and Corollary 2.2 in 
Lloyd K Jackson, Subfunctions and second-order ordinary differential inequalities, Advances in Mathematics 2(3) (1968), pp 307-363 [DOI].
