Two-point boundary value problem: $u'' = -\lambda \cos u$ Suppose that we have the following
$$
u'' = -\lambda \cos u
$$
with $u(0) = u(1) = 0$. How can it be shown that if $|\lambda|$ is sufficiently small
then the problem above has a unique solution?

The hint is to reformulate the problem as a nonlinear integral equation. So here is my attempt:
$$
u'(t) = C -\int_{0}^{t} \lambda \cos(u(s)) ds
$$
If we define the map $T$ as
$$
T(u(t)) = C -\int_{0}^{t} \lambda \cos(u(s)) ds
$$
we can write
$$
u' = T(u)
$$
With the contraction mapping theorem I could show that if $T(x) = x$ and $T$ is a 
contraction mapping then $T(x) = x$ has a unique solution. 
I suppose that if I integrate again:
$$
u(x) = C_2 + C_1 x - \int_{0}^{x} \int_{0}^{t} \lambda \cos(u(s)) ds dt
$$
but that is where I'm running out of ideas.
 A: I think you have the right idea, but you can calculate expressions for $C_1, C_2$ by using the boundary conditions. Integrating twice yields
$$ u(t) = u(0) + tu'(0) - \lambda \int_0^t \int_0^{\tau} \cos(u(s)) \, ds \, d\tau$$
We immediately eliminate $u(0) = 0$. If we evaluate at $t = 1$, we get
$$ 0 = u(1) = u'(0) - \lambda \int_0^1 \int_0^{\tau} \cos(u(s)) \, ds \, d\tau$$
Substituting everything, we get
$$ u(t) = t\lambda \int_0^1 \int_0^{\tau} \cos(u(s)) \, ds \, d\tau - \lambda \int_0^t \int_0^{\tau} \cos(u(s)) \, ds \, d\tau$$
Try to work with this integral equation.
A: While this is not the intended solution method, you may find another approach of interest.  I'll assume $\lambda > 0$ (the case $\lambda = 0$ is trivial, and $\lambda < 0$ can be obtained from $\lambda > 0$ by $u \to -u$).
The differential equation describes a nonlinear pendulum.  It is a Hamiltonian system, where the total energy $E = (u')^2/2 + \lambda \sin(u)$ is constant.  A part of the phase portrait is shown here:

The solutions in which $u$ is not monotonic are those for $-\lambda < E < \lambda$, and in order for $u$ to ever reach $0$ we need $E \ge 0$.  For $0 \le E < \lambda$ there is a trajectory that goes from $(u,u') = (0,\sqrt{2E})$ to $(0,-\sqrt{2E})$
to the right of $u=0$.  There are also trajectories that go from $(0,-\sqrt{2E})$ to $(0,\sqrt{2E})$ to the left of $u=0$, but these cover more than half a period, which is longer than $1$ if $\lambda$ is sufficiently small.  For $E$ near $0$ the time to get from $(0,\sqrt{2E})$ to $(0,-\sqrt{2E})$ is near $0$, while as $E \to \lambda$ the time goes to $+\infty$ because the trajectory passes near the fixed point $(\pi/2, 0)$.  By the Intermediate Value Theorem there is some $E \in (0,\lambda)$ for which the time is $1$, corresponding to a solution with $u(0) = u(1) = 0$.  It seems somewhat tricky to show that the time is an increasing function of $E$, which you would need for uniqueness.    
A: This is not an answer, but a suggestion. Your equations actually can be written in the form
$$
\frac{(\dot u)^2}{2}+\lambda \sin u=0,
$$
from which the implicit solution is given by
$$
\pm\frac {1}{\sqrt 2}\int_{u_0}^u\frac{du}{\sqrt{E-\lambda \sin u}}=t-t_0,
$$
where $E$ is a constant that determines the orbit.
