If $\alpha$ > 0, and $y(x)$ for $x \in \mathbb R$ be a solution of the initial value problem $$y^\prime(x)=|1-\sin(y(x))|^{\alpha}, y(0)=0$$ How do I know for which $\alpha\gt 0$, $y(x)$ is unique?
I tried $$y'(x) = \sqrt{(1-\sin(y(x)))^2}$$ and then $$\int_{0}^{y(x)} \frac 1 {(1-\sin(s))^{\alpha /2}} \,ds $$ Bur I am not sure how to precede.
The map
$$y \mapsto \lvert 1- \sin y \rvert^\alpha$$ is continuously differentiable on the interval $I=(-\pi/2, \pi/2)$. Therefore it is locally Lipschitz and we can apply Picard-Lindelöf theorem around zero and conclude that the given Initial Value Problem has a unique local solution satisfying $y(0)=0$.