Consider the problem $x'=-x^3,x(0)=x_0.$
I was wondering if it's possible to prove that this problem has a global solution (on $[0,U],$ for every $U>0$) without solving it: considering $\psi:C([0,U],\mathbb{R}) \to C([0,U],\mathbb{R}),\psi(x)(r)=x_0-\int_0^r(x(q))^3dq$ for $r\in [0,U],x \in C([0,U],\mathbb{R})$ (space of continuous functions equipped with the supremum norm $\Vert \cdot \Vert_{\sup}$), define $\Vert x \Vert_\lambda=\sup_{r\in [0,U]}e^{-\lambda r}|x(r)|,C'=\{ y \in C([0,U],\mathbb{R}),\Vert y\Vert_{\sup} \leq R\},$ for convenient condition on $R$ and $U$ so that $C'$ is complete (for $\Vert \cdot \Vert_{\lambda}$) and invariant by $\psi,$ we can also choose $\lambda$ so that $\psi$ is a contraction on $C'$ and we can apply a fixed point argument to prove that a local solution exists (which make sense since $u\to -u^3$ is locally Lipchitz).
How to obtain a global solution? Is it possible to apply a fixed point argument to obtain it?