Unique solution differential equation proof Prove that there is a $\delta>0$ such that there is a unique solution of the differential equation $y'(t)=\sin(y(t))$ with $y(0)=1$ on the interval $[-\delta, \delta]$. How large can you choose $\delta$ to be?
Don't even know where to to start here... Our class has touched on local existence, but we haven't really done any example questions like this
 A: Please read about Picard-Lindelöf Theorem on the existence and uniqueness of solutions to an intial value problem in a Lipschitz (or locally Lipschitz) vector field. As soon as you understand the theorem, you will know what to do. A proof given below uses the contraction fixed point theorem, and reduce existence/uniqueness of solution to existence/uniqueness of fixed point.
See here.
Now with Picard-Lindelöf in hand, all you need to do is show the vector field is Lipschitz. Try it yourself and in the box is a sketch which is a spoiler.

 The function $f(y)=\sin y$ is bounded by $M=1$ in absolute value and satisfies the  Lipschitz condition with $L=1$. The height $b$ of the "safety rectangle" can be chosen arbitrarily large. So, the limiting factor in the theorem is $1/L$. 

Then try to find $\delta$ such that your vector field remain Lipschitz (with same constant), which turns out to be as large as you want.
A: Define
$$
F(x)=\int_1^x \frac{du}{\sin u}, \quad x\in(0,\pi).
$$
It is not hard to show that:

*

*$\lim_{x\to 0}F(x)=-\infty\,\,$ and
$\,\,\lim_{x\to \pi}F(x)=\infty$.


*$F: (0,\pi)\to\mathbb R$ is one-to-one and onto, $F'(x)>0$, for all $x\in(0,\pi),$ and hence strictly increasing, and $C^\infty$.


*Consequently $F$ possesses an inverse $\varphi: \mathbb R\to(0,\pi)$, which is also $C^\infty$ and satisfies $F\big(\varphi(t)\big)=t,$ for all $t\in\mathbb R$.


*Therefore, $\varphi(0)=F^{-1}(0)=1$ and
$$
F'\big(\varphi(t)\big)\varphi'(t)=1
\quad\Longrightarrow\quad \varphi'(t)=\frac{1}{F'\big(\varphi(t)\big)}=\sin\big(\varphi(t)\big),
$$
for all $t\in\mathbb R$. This means that the IVP
$$
x'=\sin(x), \quad x(0)=1,
$$
possesses a global solution.


*Uniqueness. The flux function $f(x)=\sin x$ is Lipschitz continuous, and by virtue of Picard-Lindelöf Theorem, our IVP enjoys global uniqueness.
