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For a fast-slow system $ϵ\dot{x}=F(x,y)$, $\dot{y}=G(x,y)$ with $0<ϵ\ll 1$, the rescaled time $\tau =t/ϵ$ gives $x_{\tau }=F$, $y_{\tau }=ϵG$. Away from zeros of $F$, an order-one change of $x$ takes order-one fast time but only order-$ϵ$ original time, while $y$ changes little. Setting $ϵ=0$ in these rescaled equations freezes $y$ and produces the leading fast subsystem. This limit differs from the slow reduction $F=0$ in the original time; matching the two describes a relaxation oscillation.

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.

On the attracting right branch, $x>1$ and $\dot{y}=-x<0$, so the motion goes from $(2,2/3)$ towards the fold $(1,-2/3)$. At essentially fixed $y=-2/3$ it jumps to the other attracting root $x=-2$. The left branch then carries it to $(-1,2/3)$, where it jumps right to $x=2$. This gives the clockwise limiting relaxation oscillation shown in the preceding figure.

Along a slow branch $y=h(x)$, differentiation gives $(x^2-1)\dot{x}=-x$. The right-branch time isReflection symmetry gives the same time on the left. The fast jumps take a time tending to zero as $ϵ\to 0^{+}$, as do the fold-transition corrections at leading order. HenceThis is a leading asymptotic period, not an exact finite-$ϵ$ equality.

The voltage nullcline is $g(v,n)=I^{\ast }$ and the recovery nullcline is $n=n_{\infty }(v)$. In the relevant region $g_n>0$, so below the voltage curve $\dot{v}>0$ and above it $\dot{v}<0$. Below the recovery curve $\dot{n}>0$, while above it $\dot{n}<0$.

For small $ϵ$, the outer voltage branches with $g_v>0$ attract fast voltage motion; the middle branch with $g_v<0$ repels it. Arrange the intersections so the equilibrium is on that middle branch, with recovery increasing along the high-voltage attracting branch and decreasing along the low-voltage attracting branch. The trajectory then drifts towards a fold, jumps rapidly to the other attracting branch at nearly constant $n$, and repeats. This is a relaxation oscillation.

The local linear stability calculation supports the construction. At an equilibrium,A sufficiently steep recovery curve gives positive determinant, while $g_v<-ϵ$ gives positive trace. Hence the equilibrium is a repeller. The bounded physical rectangle can be chosen forward invariant, and a small neighborhood of the repeller removed to give a trapping annulus. The Poincaré-Bendixson theorem then supplies a periodic orbit. These conditions can be realized because the recovery function and the applied current control the intersection and its slope; oscillation is possible, not forced for every choice.

For the illustration, ${\gamma }_K=0.1$, $H=0.8$, $v_K=0.2$, $ϵ=0.01$, $I^{\ast }=0$, $M(v)=(1+e^{-25(v-0.2)}{)}^{-1}$ and $n_{\infty }(v)=0.8/(1+e^{-20v})$. The rectangle $[-0.2,1]\times [0,0.8]$ has inward-pointing flow, and the displayed numerical trajectory realizes the folded-branch mechanism.

The nullclines are $w=g(v)$ and $v=0$. The two outer pieces of $g$ have slope $-1$; the middle piece has positive slopeThe curve is continuous, with folds $(\alpha ,-\alpha )$ and $(\beta ,1-\beta )$. The unique equilibrium is $(0,-\alpha /(\beta -\alpha ))$. Its Jacobian matrix has trace $\mu /ϵ>0$ and determinant $1/ϵ>0$, so it is unstable. The outer branches attract the fast voltage motion.

Let $w_L=-\alpha$ and $w_R=1-\beta$. The hypothesis $\beta -\alpha <1$ gives $0<w_L<w_R<1$. The singular relaxation oscillation follows this route:

The jumps are almost horizontal in the phase plane, because $w$ changes negligibly on the fast $O(ϵ)$ time. Below $w=g(v)$, voltage increases; above it, voltage decreases; $w$ increases on the right of $v=0$ and decreases on the left. These signs provide a closed trapping region around the indicated route for sufficiently small positive $ϵ$, with the unstable equilibrium excluded, giving a periodic orbit by the planar return-map or Poincaré-Bendixson theorem argument. The continuous piecewise linear field is locally Lipschitz, so solution uniqueness is available at the corners.

On the descending part of the graph, mark the two tangency points whereChoose the crossings that enclose the portion with $g^{\prime }(u)<-1/\gamma$. For fixed $\mu$ strictly between these values and sufficiently small $\epsilon$, the unique equilibrium is a repelling equilibrium by part (e). The endpoints here are leading singular limit values; at finite $\epsilon$ the boundaries instead solve $f_v-\gamma f_u=\epsilon$ and approach these tangencies when the crossings are nondegenerate.

There is a necessary qualification: such an interval exists only when $\gamma$ is large enough that the descending graph actually attains slopes below $-1/\gamma$. If $M={\max }_{u>0}[-g^{\prime }(u)]$, one needs $\gamma M>1$. The supplied graph has a very steep descending section, but an arbitrarily small positive $\gamma$ need not satisfy this condition. For $\delta =0.001$, numerical evaluation gives $M\simeq 511.49$, so the leading threshold is $\gamma \simeq 0.0019551$. A graph alone does not remove this parameter restriction.

The fast motion is especially informative. In fast time $s=\tau /\epsilon$,At leading order $z=u+\gamma v$ is constant, so fast paths are straight lines with slope $dv/du=-1/\gamma$. Along such a line, the fast eigenvalue at the nullcline isIt is negative on the two outer attracting segments and positive on the intervening repelling segment. The tangencies defining ${\mu }_{\pm }$ are consequently the fold points of a critical manifold when it is expressed in the coordinates $(z,v)$. They are not generally the maximum and minimum of $g$ itself.

For ${\mu }_{-}<\mu <{\mu }_{+}$, slow motion on an attracting segment followsOn the left attracting segment $u<{\mu }_{-}$ this moves towards increasing $u$; on the right attracting segment $u>{\mu }_{+}$ it moves towards decreasing $u$. Each segment ends at a tangency and is followed by a fast jump along a line of slope $-1/\gamma$ to the other attracting segment. This constructs the limiting relaxation oscillation. The figure uses $\gamma =0.01$, for which ${\mu }_{-}\simeq 0.62970$ and ${\mu }_{+}\simeq 1.01157$; the marked endpoints change with $\gamma$.

To justify existence of a periodic orbit of the full planar system, rather than just draw a formal cycle, first establish a compact trapping region. The positive quadrant is forward invariant: at $u=0$, $\dot{u}=\mu +\gamma \delta v/\epsilon >0$, and at $v=0$, $\dot{v}=u^2/[\epsilon (1+u^2)]\ge 0$. MoreoverChoose $V>1/\delta$ above the initial store value. Since $f(u,v)\ge -1-\delta V$ when $0\le v\le V$,Thus the rectangle $0\le u\le U$, $0\le v\le V$ is a positively invariant set if $U>\mu +\gamma (1+\delta V)/\epsilon$ and the initial point lies inside it. Every forward solution is bounded. There are no equilibria on its axes for $\mu >0$, and its sole interior equilibrium is a hyperbolic equilibrium point with both eigenvalues having positive real parts in the strict unstable range. A sufficiently small neighborhood of this repelling equilibrium has outward flow; deleting it leaves a compact trapping region with no equilibria. The Poincaré-Bendixson theorem then supplies a periodic orbit.

Hence periodic solutions exist throughout the strict unstable interval, interpreted with the finite-$\epsilon$ correction above. This argument establishes existence, without asserting uniqueness, the criticality of possible endpoint Hopf bifurcations, or absence of coexisting periodic orbits outside the interval.

For Hopf thresholds in a two-pool calcium model, with the unique equilibrium parametrized by $u=\mu$, locate changes of stability by solving the two equationsIf there are two simple roots ${\mu }_{-}<{\mu }_{+}$ with positive trace between them, the equilibrium is a repeller throughout that interval, since its determinant stays positive. At a nondegenerate crossing a Hopf bifurcation is possible, with linear frequency $[-f_v/ϵ{]}^{1/2}$. A complete classification of criticality requires a nonlinear coefficient, not merely the trace calculation. At leading singular order the thresholds approach the folds determined by $f=0$ and $f_v-\gamma f_u=0$, equivalently $V^{\prime }=-1/\gamma$.

Periodic solutions are also supported globally in the positive quadrant. The axes point inward for positive input, $v^{\prime }\le (\beta -\delta v)/ϵ$ bounds $v$, and the $u$ equation then bounds $u$ by a linear restoring term plus a bounded source. A large rectangle can thus be chosen forward invariant. For an initial point distinct from the repelling unique equilibrium, its omega-limit set cannot contain that equilibrium or a boundary equilibrium. The Poincaré-Bendixson theorem supplies a periodic orbit. In the strong separation regime it follows the two attracting branches and jumps between them, a relaxation oscillation.

The interval requires the maximum of $f_v-\gamma f_u$ along the critical manifold to exceed $ϵ$; the listed scale hierarchy alone is not a substitute for that test for arbitrary remaining constants. As an explicit example, $\beta =0.1$, $\delta =10^{-5}$, $ϵ=5\times 10^{-4}$ and $\gamma =0.1$ give approximately ${\mu }_{-}=0.22618$ and ${\mu }_{+}=0.37350$. At $\mu =0.3$ the trace is positive, whereas at $0.03$ and $1$ it is negative. The figures use this example and label their singular-limit portions accordingly.

Use $W=w-I$. The nullclines become $W=f(v)$ and $W=\gamma v-I$, while the slow equation becomes $W_t=\gamma v-W-I$. The cubic has foldswith $W_{\ast }=f(v_{\ell })$ its local minimum and $W^{\ast }=f(v_r)$ its local maximum. Let $V_{-}(W)$ and $V_{+}(W)$ be the left and right attracting roots of $f(v)=W$. A line whose intersection is on the repelling middle branch permits a relaxation oscillation: upward slow travel on the right branch from $W_{\ast }$ to $W^{\ast }$, a fast jump to the left branch, downward slow travel to $W_{\ast }$, and a return jump. For sufficiently large fixed $\gamma$ this occurs for currents between the fold values $\gamma v_{\ell }-f(v_{\ell })$ and $\gamma v_r-f(v_r)$, away from the small-$ϵ$ onset layers. The straight line then crosses each outer branch beyond the relevant slow excursion, so the slow velocities have the required signs.

Neglecting the fast-jump durations, integrate $dt=dW/(\gamma V_{\pm }(W)-W-I)$ along the two slow segments. Their opposite orientations giveThe printed period omits the current shift. With the source's definition that $V_{\pm }$ solve $f(v)=w$, both denominators must retain $-I$ in the shifted coordinate. Equivalently, in the physical recovery coordinate the bounds are $I+W_{\ast },I+W^{\ast }$ and the integrand uses $\gamma V_{\pm }(w-I)-w$. The printed formula is therefore not the leading slow period for the displayed $I>0$ equations unless its notation is changed or an additional approximation justifying the omitted term is supplied.

For $\mu \gg 1$, trajectories rapidly approach the cubic nullclineand then move slowly along its attracting outer branches $|x|>1$. For $0<F<1$, the equilibrium lies on the repelling middle branch. Slow motion to a fold, a rapid jump to the opposite outer branch, and repetition produce a relaxation oscillation. For $F>1$, the equilibrium lies on the attracting right branch; trajectories undergo any necessary fast jump and then drift to that equilibrium, so no relaxation cycle remains.

For $ϵ\dot{v}=g(v)-w$, $\dot{w}=v$, let $g=-v$ below $\alpha$, $g=1-v$ above $\beta$, and join the endpoints by an increasing line, with $-1<\alpha <0<\beta <1$ and $\beta -\alpha <1$. The slow branches obey $\dot{w}=-w$ and $\dot{w}=1-w$ between $w_L=-\alpha$ and $w_R=1-\beta$. Their travel times are $\log [(1-\beta )/(-\alpha )]$ and $\log [(1+\alpha )/\beta ]$. Fast horizontal jumps contribute vanishing time as $ϵ\to 0$, giving the displayed relaxation oscillation period.

A short optical pulse can draw on previously stored excitation. In the original calculation below, $C=100$, $K=20$, $\beta =10^{-4}$, $u(0)=s(0)=0$, and $r=3$ for $0\le x<4$, followed by $r=0$. The cavity field initially remains small while pumping fills the upper state. Once sufficient gain develops, the photon population grows and depletes the stored inversion. Continuing pump action can refill it, giving relaxation oscillations in this simple model. This is not a fit to Maiman's recorded output.

The equations expose the excitation budget underlying conservation of energy. Adding them cancels the stimulated exchange:Pumping supplies excitations; nonlasing decay and cavity loss remove them. Integrating gives the boundOnly part of the cavity loss may be useful output; passive absorption is also a loss. The pulse's instantaneous power can draw on stored excitation, while its total extracted energy still obeys the integrated budget. The model also preserves $0\le u\le 1$ and $s\ge 0$: at each boundary its derivatives point inward or along the boundary.

There is a further energy cost in the pump cycle. For one absorbed pump photon producing at most one emitted laser photon, the ideal energy-conversion bound isChosen wavelengths $550\mathrm{nm}$ and $694.3\mathrm{nm}$ give about $0.7922$ before other losses. The missing energy can leave through nonradiative relaxation and heating. Actual efficiency also depends on how much pump light is absorbed, unwanted decays and optical losses; a population model alone cannot predict it.