Autonomous System: Finding all orbits connected to a fixed point

Question

Given the autonomous system $$ \begin{pmatrix} \dot{x}\\ \dot{y} \end{pmatrix} = \begin{pmatrix} y\\ |x| \end{pmatrix} $$ I found that all orbits in the left halfplane are circular $$ \frac{d}{dt}\| (x,y) \|^2 = 0 \quad (x \leq 0) $$ Those in the first quadrant are growing in modulus $$ \frac{d}{dt}\| (x,y) \|^2 = 4xy > 0 \quad (x,y > 0) $$ while those in the forth quadrant are decreasing $$ \frac{d}{dt}\| (x,y) \|^2 = 4xy < 0 \quad (x > 0, y < 0) $$ Question: How can I prove that the only orbits approaching the fixed point (0,0) for $t \to \infty$ or $t \to -\infty$ are the diagonals in the first and forth quadrant in an elementary ODE-style way (no Poincaré-Bendixon or so)?

Answer 1

Arrgh, I found a first integral $$ E(x,y) = y^2 - \mathrm{sgn}(x)x^2 $$ which is constant along every orbit. Any solution approaching $(0,0)$ will satisfy $$ E(x_0, y_0) = \lim_t E(x(t), y(t)) = E(0,0) = 0 $$ and so we have $$ x_0 \geq 0 \wedge ( y_0 = -x_0 \vee y_0 = +x_0 ) $$ and using the monotonicity of the modulus we find the solution.