Unique global solution of $x'=-\sin(x)$

Question

I have the following problem:

Prove that the problem $x'=-\sin(x)$, $x(0)=1$ has only one global solution.

Considering $f(t,x)=-\sin(x)$, I've used the Picard-Lindelöf theorem to prove that there is only one local solution. For the unicity of the global solution, I've found that is enough to prove that $|f(t,x)|\leq a|x|^2+b$ for some $a,b\in\mathbb{R}$, then the domain of the solution is $\mathbb{R}$. I used $a=b=1$. However, where does this result come from?

That is a clear way to prove that there exists a unique global solution of an IVP?

Answer 1

You get $|\dot x|\le 1$ so that $|x(t)|\le|x(0)|+|t|$ which immediately tells you that the solution is bounded at all finite times and thus can be indefinitely extended.

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As there is a global Lipschitz constant $1$ it immediately follows from one of the variants of the Picard-Lindelöf theorem that there is a unique solution with domain $\Bbb R$.

Theorem: If $I$ is an interval and $f:I\times\Bbb R^n \to\Bbb R^n$ satisfies a global Lipschitz condition in the second argument on the indicated domain, any IVP $\dot x=f(t,x)$, $x(t_0)=x_0$, $t_0\in I$ has a unique solution $x:I\to\Bbb R^n$.