Does the following initial value problem \begin{equation*} \begin{cases} u'(t) = f(u(t)), \\ u(0) = u_0 \end{cases} \quad \text{with} \quad f(u_1,u_2):=\sqrt{1+\sqrt{u_1^2+u_2^2}}\begin{pmatrix}u_1+u_2\\3 u_2-u_1\end{pmatrix} \quad \text{and} \quad u_0 := \begin{pmatrix} 1 \\ 0 \end{pmatrix} \end{equation*} have a global solution on $[0, \infty)$, where $f: \mathbb{R}^2 \to \mathbb{R}^2$?
Since the system is non-linear, we can't transform this system into a matrix and find eigenvalues etc, so I couldn't come up with a sensible answer. In our course (ODE) we discussed absolute continuity and Caratheodory solutions to equations but I couldn't think of how this could help.
The IVP does indeed have a global solution because the function $f$ is globally Lipschitz. Therefore the intervals of existence obtained by Picard's Theorem are all of $\mathbb{R}$. To see this, consider the linear and nonlinear parts separately. for the matrix part, we have $\max |\lambda|=2$ and for the nonlinear part, we have $\sup \left\|\nabla \left(\sqrt{1+\sqrt{u_1^2+u_2^2}}\right)\right\|=1/2$. Therefore $f$ has is globally Lipschitz with constant 1, so the global solution exists.