Show that the IVP:
$y'= \sin(t*y(t))$
$y(0)=y_{0}$
$y_{0}>0 $
a)It's well defined $\forall t>0$
b)It's integrable?, at least theoretically.
Help please, i been trying whith the Lipschitz theorem
$ |f(t,y_{1})-f(t,y_{2})|\leq|\sin(t*y_{1})-\sin(t*y_{2})|\leq|ty_{1}-ty_{2}|=t|y_{1}-y_{2}|$
but i need a Lipschitz constant, and t isn't a constant.
Consider
$$\dot{y} = f(t,y(t))$$
The Picard Lindelof theorem gives local existence and uniqueness of a $C^1$ solution on time some interval provided your vector field $f$ is locally Lipschitz in $y$ and continuous in $t$. Fix some $\delta>0$, then you have that
$$|f(t,y_1) - f(t,y_2)| \leq \delta |y_1 - y_2|, \forall t\in [-\delta,\delta]$$
Therefor you have local existence on $[-\epsilon,\epsilon]$ for some $\epsilon(\delta) := \epsilon >0$ and any inital data $y_0$. Next observe that
$$|\dot{y}| \leq 1 \implies |y(t)| \leq y_0 + t $$
Therefor you can extend your solution by setting $y_1 = y(\epsilon)$ and by the above $|y_1|\leq |y_0| +\epsilon <\infty $. Re-applying the theorem with new initial data $y(\epsilon) = y_1$ you obtain uniqueness and existence on $[-\epsilon, 2\epsilon]$. You can continue this procedure (and the reverse) to obtain global existence on $\mathbb{R}$
Note that in general, your solution will exist for as long as you can show it remains bounded. More precisely for $f$ locally Lipschitz in $y$ and continuous in $t$, your solution will exist on $(T_-, T_+)$, where $$T_- : = \inf \{ t\in\mathbb{R} \ | \ ||u(t)||_{\mathbb{R}^n} <\infty \}$$ $$T_+ : = \sup \{ t\in\mathbb{R} \ | \ ||u(t)||_{\mathbb{R}^n} <\infty \}$$