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For a system $\epsilon \dot{x}=F(x,y)$, $\dot{y}=G(x,y)$ with fast variable $x$ and slow variable $y$, the critical manifold is the set of equilibria of the fast problem $dx/ds=F(x,y)$ with $y$ frozen. It is attracting where the fast Jacobian matrix has eigenvalues with negative real parts and repelling where those real parts are positive. A smooth graph can cease to be attracting at a fold point of a critical manifold.

For initial data $u(x,0)=1+x^3/4$ on $0\le x\le 1$, the characteristic surface has $x=s{e}^r$, $y=3s^2(1-e^{-2r})/2$, $u=1+s^3(e^{-3r}-3e^{-r}/4)$. Its projection determinant is $3s^2{e}^r(2e^{-2r}-1)$, so the sheet connected to the initial data loses local invertibility at $r=\log 2/2$. The regular sheet has $r<\log 2/2$ and projects to $0<x<\sqrt{2}$, $3(1-x^{-2})/2<y<3x^2/8$. Its upper boundary is a fold point of a critical manifold caustic. Solving the inverse yields $s^2=[x^2+\sqrt{x^4-(8/3)x^{2}y}]/2$; the other square-root branch lies on the folded sheet and cannot simply be joined as a unique smooth solution. The initial point $s=0$ is already characteristic, illustrating the need to inspect endpoints as well as the interior of a Cauchy curve.

Near a nondegenerate fold point of a critical manifold crossed with nonzero slow speed, the local balance can be written $\epsilon \dot{x}=y-x^2$, $\dot{y}=-1$ up to nonzero scalings and higher-order terms. A time interval $T$ gives $y=O(T)$; balancing $x^2$ with $y$ gives $x=O(T^{1/2})$, and balancing $\epsilon x/T$ with $T$ gives $T=O({\epsilon }^{2/3})$, $x=O({\epsilon }^{1/3})$. This layer delays departure from the fold relative to an instantaneous singular jump. It differs from the order-$\epsilon$ fast motion away from the fold.

For one fast and one slow variable, a nondegenerate fold point of a critical manifold satisfies $F=F_x=0$ and $F_{xx}{F}_y\ne 0$. An attracting and a repelling branch of fast equilibria meet there. A slow trajectory reaching the fold generally leaves by a rapid jump when its slow drift crosses the fold transversely. The fold is defined relative to the fast direction: if fast trajectories are slanted in the plotted coordinates, their tangency to the critical manifold need not be an extremum of its plotted height.

Use dimensionless variables $X=X_{\ast }x$, $Y=Y_{\ast }y$, $Z=Z_{\ast }z$, $t=T\tau$, withSubstitution gives the requested system, with dots now meaning differentiation with respect to $\tau$, andIn particular the same scale for $Y$ makes the first two coefficients of the cubic equal in magnitude.

Its derivative is $f^{\prime }(y)=1-2y+\gamma y^2$. For $0<\gamma <1$ its two simple zeros areThe cubic increases below $y_{-}$, decreases between the two fold points of a critical manifold, and increases above $y_{+}$. It has a local maximum at $y_{-}$ and a local minimum at $y_{+}$.

After a rapidly decaying transient response, $z\simeq f(y)$ when its relaxation is faster than the other variables. The reduced autonomous system isFor $ϵ\ll 1$, the fast $y$ dynamics at fixed $x$ attracts to the branches where $f^{\prime }(y)>0$ and repels from the middle branch. On the low branch $y<y_{-}<\alpha$, $x$ increases slowly. At its fold point of a critical manifold $x=f(y_{-})$, the attracting low branch disappears and $y$ jumps upward, with nearly constant $x$, to the high branch. On that branch $y>y_{+}>\alpha$, $x$ decreases slowly. At $x=f(y_{+})$ it jumps back to the low branch. These slow arcs and fast jumps form the singular relaxation cycle. For small positive $ϵ$, and parameters bounded away from the fold point of a critical manifold-endpoint cases, the nearby attracting periodic motion has the same pattern. The equilibrium $(f(\alpha ),\alpha )$ is on the repelling branch: its Jacobian matrix has determinant $1/ϵ>0$ and trace $-f^{\prime }(\alpha )/ϵ>0$.

The limiting landing values are $3/\gamma -2y_{-}$ after the upward jump and $3/\gamma -2y_{+}$ after the downward jump, obtained from the third root of $f(y)=f(y_{\pm })$. Thus the two vertical segments in the phase plane plot really connect roots of the same fast equation.

If $\gamma >1$, $f^{\prime }$ is strictly positive everywhere, so the cubic is monotone and has no fold points of a critical manifold. The reduced equilibrium is asymptotically stable. In factThe only invariant trajectory in the zero-derivative set is the equilibrium, so bounded reduced solutions converge to it; relaxation oscillations disappear.

The stronger condition $\boxed{\delta \ll ϵ}$ matters during a fast jump: $y$ then varies on the $ϵ$ time scale, not on the slow unit scale. Only if $z$ adjusts faster still may one keep $z\simeq f(y)$ through that jump. Merely imposing $\delta \ll 1$ would not justify this reduction on all pieces of the cycle. This is a three-time-scale cubic relaxation oscillator.

Take $\gamma >{\gamma }_c$ and let $\epsilon \to 0^{+}$ with the other parameters fixed. The nullclines are the cubic polynomial graph $w=f(v)$ and the line $w=\gamma v$. In the phase plane, $v$ increases below the cubic polynomial graph and decreases above it; $w$ increases below the line and decreases above it. The fast ordinary differential equation, in time $s=t/\epsilon$, freezes $w$ and gives $dv/ds=f(v)-w$.

At $w=0$, the fast roots are $0,a,1$. Their linear stability follows from $f^{\prime }(0)=-a<0$, $f^{\prime }(a)=a(1-a)>0$ and $f^{\prime }(1)=a-1<0$. Thus a fixed $0<v^{\ast }<a$ falls towards the resting branch, whereas a fixed $a<v^{\ast }<1$ rises towards $v=1$. If $v^{\ast }>1$, the first fast segment instead settles down towards $1$, followed by the same recovery excursion. The threshold $a$ is the leading singular perturbation threshold; finite-$\epsilon$ corrections matter for initial values approaching $a$ with $\epsilon$.

LetThese are the roots of $f^{\prime }(v)=0$. The outer branches $v<v_L$ and $v>v_R$ of the critical manifold are attracting under fast motion; the middle branch is repelling. The full excursion has four stages.

For this cubic polynomial, the landing point is explicit. Since $f(v)-f(v_R)=-(v-v_R{)}^2(v-v_J)$, comparison of the quadratic coefficients givesThus a graph of $v(t)$ has a rapid rise, a slowly falling elevated plateau, a rapid fall through zero to a negative undershoot, and a slow approach back to zero. The figure shows both the phase plane and a computed time trace. The smaller initial perturbation stays near the resting branch and produces no elevated plateau.

The four-stage description concerns the singular limit. Close to a simple fold point of a critical manifold, balancing the local quadratic activation term with the slow drift gives $\Delta v=O({\epsilon }^{1/3})$ and $\Delta t=O({\epsilon }^{2/3})$ for fixed parameters. These layers shrink compared with the slow phases and do not alter the leading time integrals below.

On an attracting branch of the critical manifold, $w=f(v)$ to leading order. The chain rule therefore givesThe initial fast segment ends at $v=1$, and the first slow phase ends at the right fold point of a critical manifold, $v_R$. HenceOn this interval $f^{\prime }(v)<0$ and $\gamma v-f(v)>0$. The integrand is negative, but the upper limit is smaller than the lower limit, so $T_1>0$. The numerator tends to zero at the fold while the denominator remains positive; the leading integral is finite. Fast segments and the fold delay give vanishing corrections as $\epsilon \to 0^{+}$.

In a fast-slow singular perturbation $ϵ\dot{x}=f(x,y)$, $\dot{y}=g(x,y)$, the critical manifold is ${\mathcal{C}}_0=\{f=0\}$. A slow manifold for positive $ϵ$ is a nearby locally invariant manifold on which the rapid transient has relaxed and the remaining motion occurs on the original slow time. These two manifolds generally differ: invariance must account for motion of $y$, not just set its instantaneous fast equilibrium.

For example, consider $ϵ\dot{x}=y-x$, $\dot{y}=-y$. A graph $x=h_{ϵ}(y)$ is invariant only ifFor $0<ϵ<1$, $h_{ϵ}(y)=y/(1-ϵ)$ satisfies this identity, whereas the critical manifold $x=y$ does not. The transverse displacement $z=x-y/(1-ϵ)$ obeys $\dot{z}=-z/ϵ$ and decays rapidly. Thus the slow graph is displaced by $O(ϵ)$ on bounded $y$ intervals and attracts with an explicitly separated time scale.

Persistence of a smooth slow graph requires suitable smoothness and uniform normal hyperbolicity on the region considered, usually a compact piece of the critical manifold. Negative real parts of the fast linearization give attraction; near a fold point of a critical manifold, this spectral condition fails and the simple reduction can break down.

For $f(y)=y-y^2+\gamma y^3/3$ with $0<\gamma <1$, the fold points of a critical manifold are $y_{\pm }=(1\pm \sqrt{1-\gamma })/\gamma$. When $\delta \ll ϵ\ll 1$, the fastest variable satisfies $z\simeq f(y)$ even during jumps of the intermediate variable. At fixed $x$, the outer branches of $x=f(y)$ attract the fast $y$ dynamics because $f^{\prime }(y)>0$; its middle branch repels. If $y_{-}<\alpha <y_{+}$, the slow equation $\dot{x}=\alpha -y$ drives the low branch to its fold point of a critical manifold and the high branch back to the other fold point of a critical manifold. Fast jumps at almost fixed $x$ close a relaxation cycle. This explains why a reduction justified on the slow arcs must also be checked on the jump time scale. For $\gamma >1$, the cubic is strictly increasing; the reduced energy $V=[x-f(\alpha ){]}^2/2+ϵ(y-\alpha {)}^2/2$ has $\dot{V}=-(y-\alpha )[f(y)-f(\alpha )]\le 0$, precluding this oscillation mechanism.