Given the Cauchy Problem: $ \left\{ \begin{array}{@{}l} x'(t) = \sin (x(t)),\ t\in\mathbb{R}\\ x(0)=x_0 \in ]0,\pi[ \end{array} \right. $
I try to prove that $x(t) \in ]0,\pi[ $
What I did: Using Cauchy-Lipschitz theorem, I proved that the Cauchy Problem has a unique solution on $\mathbb{R}$
I also proved that the solution cannot be constant since no $k \in \mathbb{Z}$ verifies $x_0=k\pi$.
I don't know to follow-up from there. any help much appreciated.
On
We have $$\frac{dx}{dt}=\sin x,$$ so $$\frac{dx}{\sin x}=dt$$ and the equation is easily solvable in the form $t=\varphi(x).$ You can now find a domain of $\varphi$ taking into account the initial condition.
On
Below is the explicit solution. Probably that will help you investigate the values of $x(t)$. $$x'(t)=\sin x(t)\Rightarrow x''(t)=\cos x(t)\cdot x'(t)=\cos x(t)\sin x(t)$$ On the other hand $$x'(t)^2=\sin x(t) x'(t)\Rightarrow \int x'(t)^2\,dt=\int \sin x(t) x'(t)\,dt$$ But $$\int \sin x(t) x'(t)\,dt=\int\sin x(t)\,dx(t)=-\cos x(t)+C$$ and after several steps of integration by parts we get $$\int x'(t)^2\,dt=\int x'(t)\,d x(t)=x'(t)x(t)-\int x(t)x''(t)\,dt\\ =x'(t)x(t)-\int \cos x(t) x(t)x'(t)\,dt=x'(t)x(t)-\int\cos x(t)x(t)\,d x(t)\\ =x'(t)x(t)-\frac{1}{2}x^2(t)\cos x(t)-\frac{1}{2}\int x^2(t)\sin x(t)\,dt\\=x'(t)x(t)-\frac{1}{2}x^2(t)\cos x(t)-\frac{1}{2}\int x^2(t)\,dx(t)\\ =x(t)\sin x(t)-\frac{1}{2}x^2(t)\cos x(t)-\frac{1}{6}x^3(t)$$ Overall we get $$x(t)\sin x(t)-\frac{1}{2}x^2(t)\cos x(t)-\frac{1}{6}x^3(t)=-\cos x(t)+C$$ where $C$ is uniquely determined by $x(0)=x_0$ and equals $$C=x_0\sin x_0-\frac{1}{2}x^2_0\cos x_0-\frac{1}{6}x^3_0+\cos x_0$$
It is an easy consequence of uniqueness of solutions. The constant functions $0$ and $\pi$ are solutions of the equation. By uniqueness, the graph of the solution with $x(0)\in(0,\pi)$ cannot cross the graph of those constant solutions (otherwise there would be two constant solutions through the same point.) This implies that $0<x(t)<\pi$.