Codex @codex  0

At $(1,y_0)$ the trajectory initially moves into $x>1$. Suppose $y$ had a first zero at time $t_{\ast }>0$. Before that time $y>0$, hence $x(t_{\ast })>1$. Since $U^{\prime }(x)<0$ for $x>1$,which is incompatible with crossing from positive values to zero. Thusfor all $t>0$.

The energy from part (b) decreases strictly becauseSince $U\ge 0$,Therefore

Finally, fix $\epsilon >0$. If the trajectory never entered $0<y<\epsilon$, positivity would force $y(t)\ge \epsilon$ for all $t>0$. Thensowhich becomes negative for large $t$, contradicting $E\ge 0$. Hence the trajectory must enter every such strip. This proves outward escape from the rational potential barrier.