I have the following problem to solve and am not certain of whether my attempt at the solution is correct.
Problem description:
Given an IVP $\dot{y} = \frac{1}{1+|y|},y(0)=y_0\in\mathbb{R}\forall t>0$
Check existence and uniqueness of solutions (local and global).
Attempt at solution
The first thing that comes to mind when hearing EE is the Theorem of Picard, which states, that given
an ODE $x'=f(t,x(t)$ the function $f$ is continuous w.r.t. $t$ and Lipschitz-Continuous w.r.t. $x$ in
a domain $U \subset \mathbb{R}\times\mathbb{R}$, where $(t_0,x_0)\in U, x(t_0)=x_0$ being the initial condition.
Then there exists an interval $I=[t_0-\epsilon, t_0+\epsilon]$ s.t. there exists a unique solution to the IVP
$x'(t)=f(t,x(t)),x(t_0)=x_0$.
Now I have $f(t,y(t))=f(y(t))$ and thus an autonomous ODE. Following holds: $||f(y(t))||_\infty \leq 1\quad\forall t\in I$ where $I$ is any interval. Generally $f$ is continuous $\forall t$
Now let $y,z$ be solutions of the ODE, then: $||f(y)-f(z)||=||\frac{|z|-|y|}{(1+|y|)(1+|z|)}||=(*)$
Now assuming $y(t),z(t)\lessgtr 0$ I can certainly get:
$(*) \leq \frac{max_I |z(t) -y(t)|}{min_I |(1+|y(t)|)(1+|z(t)|)|}\leq max_I |z(t) -y(t)| = ||z-y||$
where in the first inequality I use $||x|-|y||\leq |x-y|$.
Thus I have obtained a Lipschitz-constant and can use Picard's theorem to assert, that for any $y_0\neq 0$ there exists a unique solution to the IVP. My problem is, that in the equation above I have used the reverse triangle inequality to obtain the needed inequality for the L-constant. But if, say, $z(t)>0,y(t)<0$ then I get $||y+z||$ which does not help.
On the other hand, if I look at $f(y(t))$, then it is obvious, that $y(t)$ is increasing for all $t$ ($\dot{y}=f(y(t))\geq 0$) From this I would deduce, that $y(t)$ is unique given $y(0)=y_0$, globally.
Question: What am I missing to check (global)existence and uniqueness and what (if any) errors are present in the argumentation above.
