$x'(t) = |y|, x(0) = 1$
$y'(t) = |x|, y(0) = 1$
How to show that the system has a global existence $\forall t \in (-\infty, \infty)$?
As $|y|,|x|$ are continuous over all of the real lines, so we can say that the solutions exist!, but I think it guarantees us the local existence if so how i can prove for global existence? Also the partial derivative $\frac{\partial{f_{1}}}{\partial{x}},\frac{\partial{f_{2}}}{\partial{y}} = 0$ implying boundedness and hence Lipschitz, ?
But is this correct? any stronger argument I should give?
Sketch: Observe \begin{align} \left|\frac{1}{2}\frac{d}{dt}\log|x(t)^2+y(t)^2| \right|=\left|\frac{x(t)x'(t)+y(t)y'(t)}{x(t)^2+y(t)^2}\right| \leq \frac{2|x(t)||y(t)|}{x(t)^2+y(t)^2} \leq 1 \end{align} then we have that \begin{align} \frac{1}{2}|\log|x(t)^2+y(t)^2|| \leq \frac{1}{2}\log 2+ \frac{1}{2}\int^t_0\left|\frac{d}{d\tau}\log|x(\tau)^2+y(\tau)^2| \right|\ d\tau \leq t+\frac{1}{2}\log 2. \end{align} Hence it follows \begin{align} x(t)^2+y(t)^2= \exp\left(\log|x(t)^2+y(t)^2| \right) \leq \exp\left(|\log|x(t)^2+y(t)^2|| \right) \leq 2e^{2t}. \end{align} Since $x(t), y(t)$ are bounded for every finite interval $[0, T]$ then the solution can be extended to a bigger time interval. Hence $x(t), y(t)$ have global solutions.