Quadratic escape desingularization

For $\dot{u}=u(\mu +u-v)$, $\dot{v}=\lambda u^2$ in $u>0$, the time change $d\tau /dt=u$ gives the affine linear system $u_{\tau }=\mu +u-v$, $v_{\tau }=\lambda u$. Shifting $w=v-\mu$ removes the forcing. Its eigenvalues are $(1\pm \sqrt{1-4\lambda })/2$. For $\lambda >1/4$, putting $\omega =\sqrt{4\lambda -1}/2$ givesFor fixed $\mu >0$ and initial $h,v_0$ tending to zero, the first return to small positive $u$ occurs after approximately $\pi /\omega =2\pi /\sqrt{4\lambda -1}$. This is a half-turn, not a full turn. At $\lambda =1/4$, $u=e^{\tau /2}[h+(\mu +h/2-v_0)\tau ]$ does not return and grows without bound. The loss of return diagnoses escape to infinity rather than a finite-amplitude homoclinic orbit. Physical time obeys $dt=d\tau /u$ and need not share the divergent desingularized travel time.