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Putting $\epsilon =0$ gives the local regular perturbation result $y_0=(x^2+1)/2$. It cannot describe the whole half-line: its derivative grows like $x$, making $\epsilon y^{\prime }$ order one at $x=O({\epsilon }^{-1})$.

The useful exact change of variables isDirect substitution turns the equation into two autonomous differential equations,The initial values here are $z(0)=w(0)=\epsilon$. Near the origin, linearizing this system gives eigenvalues $1$ and $-1-\epsilon$. Equivalently, neglecting only the derivative-square nonlinearity in the original equation gives the two modes $x^2$ and $x^{-\epsilon }$. Their matching coefficients areThus the growing derivative has $w\sim Cx$. Keeping these unexpanded linear-mode coefficients is convenient for matching; it does not assert that the linearized solution is the exact nonlinear one.

At the distinguished limit $X=Cx=O(1)$, set $R=\log X$. To leading order the derivative equation becomes $w_R=w(1-w)$, the logistic differential equation. Matching $w\sim e^R$ as $R\to -\infty$ selectsThen $z_R+z=w$, and the growing-mode matching $z\sim X/2$ givesThis proves the logistic crossover in a weakly nonlinear Euler equation. In particular the nonlinear-region leading solution is $y\sim H(Cx)/(\epsilon C)$, or ${\epsilon }^{-2}H(\epsilon x)$ with the constants taken at their lowest order.

A composite asymptotic expansion retaining the small-distance stable mode and satisfying the initial value exactly isSince $H(X)=X^2/2+O(X^3)$, this matches $A{x}^2+B{x}^{-\epsilon }$ when $Cx\ll 1$ and hence $(x^2+1)/2$ at fixed $x$ as $\epsilon \to 0$. Its derivative at one differs from the prescribed value only by $O(\epsilon )$, within leading-order accuracy. For $Cx\gg 1$ it gives $y\sim x/\epsilon$. Thus the composite covers the initial region, the nonlinear transition and the arbitrarily large-$x$ regime; uniform leading accuracy is understood relative to the growing solution, not as an absolute error independent of unbounded $x$.

There is no later hidden growth regime. The exact positive system keeps $z,w$ in $[0,1+\epsilon {]}^2$, because its vector field points inward at every boundary. Its positive equilibrium point is $z=w=1+\epsilon$, with Jacobian matrix having trace $-2-3\epsilon$ and determinant $1+\epsilon$. The origin has no attracting direction inside the positive quadrant, while the divergence $-\epsilon -2w<0$ excludes cycles by the Bendixson-Dulac theorem. The Poincaré-Bendixson theorem therefore gives convergence to the positive stable equilibrium. At fixed small positive $\epsilon$, the exact far-field slope is $y/x\to (1+\epsilon )/\epsilon$, whose difference from $1/\epsilon$ is only relative $O(\epsilon )$, consistent with the global leading composite.

Within the positive-amplitude wedge the reduced periodic orbit equilibrium has amplitude trace ${\mu }_1$ and positive determinant $4r_{\ast }^2$. At ${\mu }_1=0,{\mu }_2<0$ the transverse pair is imaginary, so the possible secondary event is a Hopf bifurcation of the amplitude equations, corresponding to an invariant torus in the full system.

It is degenerate. Set ${\mu }_2=-a^2$ and ${\mu }_1=0$. The first integral of the quadratic fold-Hopf amplitude flow isThe amplitude equations have the Hamiltonian system form $\dot{r}=F_x$, $\dot{x}=-F_r$, so $\dot{F}=0$. At $(r,x)=(a,0)$, $F$ has a strict local maximum $2a^3/3$ and negative-definite Hessian. Every nearby regular level is a closed amplitude periodic orbit, rather than one isolated orbit selected by nonlinear damping. Adding the rotating phase gives an entire family of invariant tori; the full motion is quasiperiodic or periodic according to the phase advance during one amplitude period.

For ${\mu }_1\ne 0$, the amplitude vector field has constant divergence ${\mu }_1$. The Bendixson-Dulac theorem rules out closed amplitude trajectories in the physical simply connected half-plane $r>0$. Hence there is no ordinary branch of isolated invariant tori on either side in this quadratic system. The secondary event has vanishing radial saturation coefficients; its imaginary linear pair alone does not establish a nondegenerate torus bifurcation.