Show Lipschitz continuity of ODE solution with respect to initial condition

Question

I am working on the nonlinear pendulum differential equation given by

$$ \theta'' = -g\sin \theta. $$

Let $t \geq 0$ be fix, I define for $\theta_0 \in (-\frac\pi2, \frac\pi2)$ a mapping $S$ by

$$ S(\theta_0) = \theta(t; \theta_0), $$

where $\theta(t; \theta_0)$ is the solution of the initial value problem for the ODE defined above with initial conditions $\theta'(0) = 0$ and $ \theta(0) = \theta_0$.

I believe the mapping $S$ is Lipschitz continuous, but is it true? If yes, how could I prove it?

Answer 1

Reading through some old lecture notes, it seems that the derivative of $S$ is as many times continuously differentiable as the RHS of the problem. (I can't link anything here unfortunately)

By changing $(-\frac\pi2, \frac\pi2)$ to $[-\frac\pi2, \frac\pi2]$, $S$ is now a $C^1$ function on a compact interval, thus Lipschitz continuous.

Answer 2

Multiplying your equation by $\theta'$ gives $\theta'\theta''=-g\theta'\sin\theta$ or

$\frac{d}{dt}\left(\frac{1}{2}\theta'^2-g\cos\theta\right)=0\Rightarrow\frac{1}{2}\theta'^2-g\cos\theta=c$

$\theta' = \pm\sqrt{2(c+g\cos\theta)}$

Every function that has a bounded derivative is Lipshitz continuous. This would be where I would start the proof. Notice that the square root is bounded $\forall\,\theta$.