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For $A,B,\rho >0$, these Lotka-Volterra equations describe two populations limited by their own density and depressed by the other population. The nonnegative quadrant is invariant. Positive equilibria are intersections of the non-axis nullclines $u+Av=a$ and $Bu+v=b$; their existence and stability distinguish stable coexistence, competitive exclusion, and bistability. Positive solutions have no periodic orbits: multiplying the vector field by the Dulac function $1/(uv)$ makes its divergence $-1/v-\rho /u<0$, so the Bendixson-Dulac theorem applies. Boundedness, boundary equilibria, and their stable manifolds must still be checked before concluding convergence to one particular equilibrium.

For $\dot{v}=v(\alpha -v)(v-1)-w+I$, $\dot{w}=\beta v-\gamma w$, set $k=\beta /\gamma$. Intersections of the nullclines solve $I=h(v)=v^3-(1+\alpha )v^2+(\alpha +k)v$. Three simple equilibria are possible precisely when $k<M=(1-\alpha +{\alpha }^2)/3$ and $h(u_{+})<I<h(u_{-})$, where $u_{\pm }=(1+\alpha \pm \sqrt{1-\alpha +{\alpha }^2-3k})/3$. To prove this, observe that $h^{\prime }=k-M+3(v-(1+\alpha )/3{)}^2$ has two real zeros exactly when $k<M$; monotonicity on the three resulting intervals counts the intersections. At either endpoint there are a double and a simple root, and at $k=M$, $I=h((1+\alpha )/3)$ there is one triple root.

A gas-budget model in which inflow competes with star formation, gas return and outflows. A zero instantaneous gas derivative defines a balance nullcline; it need not define a fixed point of the evolving stellar and black-hole masses.

In a system where gas and stellar masses both evolve, the locus ${\dot{M}}_g=0$ gives the instantaneous gas-balance nullcline. Motion along the other state variables generally changes the equilibrium reservoir, so the nullcline is not itself an exact constant-gas trajectory.

For the first-order ordinary differential equation $y^{\prime }=f(x,y)$ defined on an open subset $D$ of the plane, an isocline of slope $k$ is the level setHere $k$ is a constant. At every point of this isocline, the direction field prescribes a tangent vector proportional to $(1,k)$ for a solution of a differential equation. It does not generally prescribe the tangent line of the isocline itself. The level set can have several components or fail to be a smooth curve, and points outside $D$ are excluded.

For example, take $y^{\prime }=x-y$ on the whole plane. Its isoclines are the parallel straight linesEach line has slope $1$, while the direction field along it has slope $k$. Thus the isocline $y=x$ carries horizontal segments even though the line itself rises. A function graph of a solution of a differential equation follows the segments, rather than following this line. Only the isocline with $k=1$, namely $y=x-1$, is also a solution of a differential equation.^[ref]

The first-order linear differential equation has the exact familyThe initial condition $y(x_0)=y_0$ selects $A=(y_0-x_0+1)e^{x_0}$ and hence a unique solution of a differential equation for every real $x$. The picture shows $y(0)=-1,0,1,2$. For $A=0$ the solution of a differential equation stays on the $k=1$ isocline; for other $A$ it moves through different isoclines. These formulas check the pictured trajectories by direct differentiation, independently of how the direction field was sampled.

The zero-slope isocline $f(x,y)=0$ is called a nullcline. Below the nullcline $y=x$ in this example, $y^{\prime }>0$; above it, $y^{\prime }<0$. For $A>0$, the solution of a differential equation meets it at $x=\log A$, where $y^{\prime }=0$ and $y^{\prime \prime }=1$, a strict local minimum. The nullcline is therefore a useful guide to turning points, although it is not itself a solution of a differential equation. In the equivalent planar autonomous differential equation $\dot{x}=1$, $\dot{y}=x-y$, it is the nullcline for the second component; there is no equilibrium point, because the first component never vanishes.

This makes isoclines useful even when no explicit solution of a differential equation is available: organize a direction field by common slopes, read where function graphs rise or fall, and identify possible turning points on nullclines.

The quadratic-Hill mutual repression switch is explicit: for $n=2$, steady state requires $Y(1+Z^2)=\gamma$ and $Z(1+Y^2)=\gamma$. Subtraction givesThus either $Y=Z=x$, giving the unique positive cubic root $x^3+x-\gamma =0$, or $YZ=1$. In the latter case, $Y+1/Y=\gamma$, henceThe non-symmetric pair is real, positive and distinct exactly when $\gamma >2$. For $0<\gamma <2$ only the symmetric solution exists. Eliminating one variable gives the source's hinted factorization,with no singular denominator for real concentrations. The cubic is strictly increasing since $3Y^2+1>0$, so it contributes only the symmetric state.

One can verify the switch interpretation without a lengthy stability analysis. At symmetry the Jacobian matrix eigenvalues are $-1\pm 2x^2/(1+x^2)$: symmetry is stable for $x<1$ and becomes a saddle for $x>1$, corresponding to $\gamma <2$ and $\gamma >2$. At either asymmetric equilibrium the eigenvalues are $-1\pm 2/\gamma$, both negative when $\gamma >2$. ThereforeAt $\gamma =2$, all three branches meet at $(1,1)$ with a zero linear eigenvalue, the switching threshold rather than two distinct robust memory states. The lower panels of the original figure in solution show representative nullclines on either side.

The nullclines areOn $y=\beta$, away from the equilibrium points, the vector field is vertical. Its crossing direction is downwards for $x<-1$, upwards for $-1<x<0$, downwards for $0<x<1$, and upwards for $x>1$, because $\dot{y}=x(x^2-1)$. On each of $x=-1,0,1$, the vector field is horizontal: it points right when $y>\beta$ and left when $y<\beta$.

Throughout each complementary region, $\dot{x}$ has the sign of $y-\beta$, and $\dot{y}$ has the four signs just listed. These signs orient the level sets from part (ii). In particular, the inner periodic orbits are clockwise. The saddle level is $h=1/4$, whose separatrices satisfyFor $-1<x<1$, the upper arc $y-\beta =(1-x^2)/\sqrt{2}$ goes from $(-1,\beta )$ to $(1,\beta )$; the lower arc goes back from $(1,\beta )$ to $(-1,\beta )$. These two heteroclinic orbits form a closed boundary around the family of periodic orbits. The remaining branches of the same saddle level extend outwards. The sketch shows the surrounding nonclosed trajectories as well as the nullclines and equilibrium points.

Set $z=y-\beta$. Then the autonomous planar system becomeswhich has no $\beta$ dependence. ThereforeMore precisely, replacing $\beta$ by $\beta +\delta$ sends every trajectory $(x(t),y(t))$ to $(x(t),y(t)+\delta )$, with the same time parametrization. The equilibrium points, nullclines, periodic orbits and separatrices all undergo this same translation. Their types and orientations are independent of $\beta$, so varying $\beta$ produces no bifurcation.

The fixed-point conditions are $y(y^2-1)=0$ and $x=y^2$. They give $(0,0)$ and $(1,\pm 1)$. The Jacobian matrix isAt $(1,1)$ its eigenvalues are $-1\pm \sqrt{3}$; at $(1,-1)$ they are $1\pm \sqrt{3}$. Both equilibria are saddle equilibria. At the origin the eigenvalues are $\pm i$. This linear calculation alone does not distinguish a nonlinear center from a weak focus.

To settle the origin and the global connections, setUsing $2a_{\pm }^2-2a_{\pm }-1=0$, direct differentiation gives ${\dot{F}}_{\pm }=-2a_{\pm }y{F}_{\pm }$. Hence each parabola $F_{\pm }=0$ is invariant and, away from them, the following is a first integral:The two parabolas intersect precisely at the saddle equilibria. Their arcs between $y=-1$ and $y=1$ enclose the lensInside it $F_{+}>0$, $F_{-}<0$. The origin is a critical point of $\log I$ with negative-definite Hessian $-2\sqrt{3}$ times the identity. It is therefore a strict local maximum of this first integral, proving a nonlinear center equilibrium. The regular interior levels are closed periodic orbits, circulating anticlockwise. In fact the only critical point of $I$ inside the lens is the origin, so these levels fill the lens up to its boundary.

On $F_{+}=0$, $\dot{y}=(a_{+}-1)(y^2-1)$, so the interior arc runs from the upper saddle equilibrium to the lower. On $F_{-}=0$ it runs from the lower to the upper. Together they form a heteroclinic cycle. Their outer branches supply the remaining stable and unstable separatrices.

For the full exterior portrait, the two invariant parabolas partition the rest of the plane into four regions. Where both $F_{\pm }>0$, $\dot{y}>0$ everywhere; where both are negative, $\dot{y}<0$. In the region above both saddle equilibria, $y>1$ and $\dot{x}>0$; in the corresponding lower region, $y<-1$ and $\dot{x}<0$. These exterior trajectories are unbounded and cannot cross the separatrices. The nullclines are $y=0,\pm 1$ for $\dot{x}=0$ and $x=y^2$ for $\dot{y}=0$.

At an equilibrium $(v_{\ast },k{v}_{\ast })$, set $p=f^{\prime }(v_{\ast })$. The Jacobian matrix, its trace and its determinant areThe eigenvalues and their discriminant areThus the complete local linear stability test isIn the last two cases, $\mathcal{D}>0$ gives a node, while $\mathcal{D}<0$ gives a focus; the sign of $T$ distinguishes inward and outward trajectories. At $\mathcal{D}=0$, a nonzero repeated eigenvalue gives a degenerate node. At $p=\gamma <k$ the eigenvalues are purely imaginary, so linear stability analysis alone does not decide asymptotic stability; with the required crossing and nondegeneracy this is a Hopf bifurcation. At $p=k$ there is a zero eigenvalue, requiring a nonlinear test and generically marking the saddle-node bifurcation in the preceding answer.

For a unique simple equilibrium, $p<k$, so it can be a sink equilibrium or a source equilibrium. The number of equilibria alone does not decide stability. If $k\le \gamma$, the unique simple equilibrium is always a sink equilibrium. If $\gamma <k$, it can become a source equilibrium where $\gamma <p<k$. The same conclusion follows particularly simply if $\gamma >M$: the trace is negative at every nonsaddle equilibrium.

For three simple equilibria ordered by their $v$ coordinates, the middle one has $p>k$ and is a saddle equilibrium; the two outer ones have $p<k$. When both outer equilibria are sink equilibria, each has an attracting neighborhood, giving bistability. In the usual two-basin portrait, the stable manifold of the saddle equilibrium separates their basins. Each branch of its unstable manifold initially leaves towards a different side; its eventual destination requires a global argument. If $k\le \gamma$, both outer equilibria must be sink equilibria. If $\gamma <k$, either or both outer equilibria can instead be source equilibria; the same test $p\lessgtr \gamma$ applies separately to each. The three lower panels illustrate all three possibilities up to exchanging left and right.

The arrows in the phase plane follow directly from the nullclines: below $w=f(v)+I$ the horizontal component points right, and above it left; below $w=kv$ the vertical component points up, and above it down. Near a stable focus trajectories spiral inward; near an unstable focus they spiral outward. Near a saddle equilibrium they approach along its stable manifold and leave along its unstable manifold. These directions, together with the local tests, determine the sketches.

There is also a useful global conclusion when the unique equilibrium is a source equilibrium. Withthe leading term in $v$ is $-\beta v^4$, so sufficiently large ellipses are trapping regions. A small repelling neighborhood of the hyperbolic source equilibrium can be removed to form an invariant annulus. The Poincaré-Bendixson theorem then guarantees at least one periodic orbit in that annulus. The numerical sketch illustrates the resulting sustained oscillation; this argument does not assert uniqueness of a limit cycle. A stable equilibrium can instead represent a resting membrane potential, with perturbations producing damped relaxation; near a threshold, a sufficiently large perturbation can produce a large excursion before recovery. Global basin boundaries and coexistence of periodic orbits need more information than the number and local stability of equilibria.

Put $k=\beta /\gamma$. The nullclines of the space-clamped FitzHugh-Nagumo model areThe first nullcline is a downward-leading cubic polynomial, shifted vertically by $I$; the second nullcline is a line through the origin with slope $k>0$. Their intersections satisfyWriting $M=(1-\alpha +{\alpha }^2)/3$, differentiation givesIf $k>M$, $h$ increases strictly, so there is exactly one equilibrium. If $k=M$, $h$ is still strictly increasing, although its derivative vanishes at one point, so the equilibrium is again unique. If $0<k<M$, defineHere $h$ has a local maximum at $u_{-}$ and a local minimum at $u_{+}$; thereforeAt either endpoint there are two distinct equilibria, one a double root. These are the saddle-node bifurcation boundaries omitted by the generic one-or-three wording. At $k=M$ and $I=h((1+\alpha )/3)$ there is one triple root. This is the Cubic FitzHugh-Nagumo nullcline criterion.

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.

At any equilibrium, the first balance gives $u=\mu$ and the second gives $f(\mu ,v)=0$. For every fixed $u>0$,andThe intermediate value theorem and strict monotonicity give precisely one positive root $v=g(u)$ for each $u>0$. Since $f_v\ne 0$, the implicit function theorem makes this a smooth nullcline. For every $\mu >0$, therefore,is the unique positive equilibrium. Graphically, draw the vertical line $u=\mu$ through the given graph: it meets the graph once, even on its steep descending section. This vertical line is the condition for simultaneous stationarity; the full $u$ nullcline at finite $\epsilon$ is actually $f(u,v)=(\epsilon /\gamma )(\mu -u)$.

Differentiate the identity $f(u,g(u))=0$ using the chain rule:Substitution into the trace givesBecause $f_v<0$, the exact instability condition isFor fixed $\delta >0$ and fixed equilibrium location, the correction tends to zero with $\epsilon$. To leading order this isThus instability requires the descending nullcline to be steeper than a line of slope $-1/\gamma$. The sign of the finite-$\epsilon$ correction makes the exact unstable interval slightly smaller. Since $\delta =0.001$ can make $|f_v|$ small, the approximation quantitatively requires $\epsilon /|f_v|\ll 1$ at the point in question; merely comparing $\epsilon$ with one can be insufficient.

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.

The term $p$ represents prey reproduction and $-p^2/a$ limits growth by available food, giving carrying capacity $a$ in the absence of hunters. The loss $-ph/a$ represents encounters between prey and hunters. For hunters, $hp/(8b)$ is prey-supported growth and $-h/8$ is mortality; their per-capita growth changes sign at prey population $b$. The coordinate axes are invariant, and positive initial populations remain positive.

For $a=1$, the physical equilibrium points are the origin, $(1,0)$, and $(b,1-b)$ when $b<1$. The Jacobian matrix isAt $(0,0)$ its eigenvalues are $1$ and $-1/8$, so the origin is always a saddle equilibrium, attracting along the hunter-only axis and repelling along the prey-only axis. At $(1,0)$ the eigenvalues areIt is a saddle equilibrium for $0<b<1/2$ and an asymptotically stable node for $b>1$.

For $0<b<1/2$, the coexistence Jacobian matrix isIts trace is negative and determinant positive. Its discriminant $b^2-(1-b)/2=(2b-1)(b+1)/2$ is negative, so $(b,1-b)$ is an asymptotically stable focus. Interior trajectories approach it in damped, counterclockwise oscillations in the $(p,h)$ plane. Both populations ultimately coexist.

To justify the global interior fate for this logistic predator-prey model, let $h_{\ast }=1-b$ and use the Lyapunov functionIt is nonnegative and its sublevel sets are compact inside the positive quadrant. Differentiation gives the explicit cancellationThe only invariant subset on which this vanishes is $p=b,h=h_{\ast }$: remaining at $p=b$ requires $\dot{p}=b(1-b-h)=0$. LaSalle's invariance principle therefore proves that every strictly positive initial state converges to this coexistence point.

For $b>1$, the formal coexistence point has negative $h$ and is outside the physical state space. The only physical equilibria are the saddle equilibrium at the origin and the stable prey-only node. Since $\dot{p}\le p(1-p)$, positive prey has $\mathrm{limsup}p\le 1$. Eventually $p<b$ with a fixed margin, so hunter population decays exponentially. When $h$ is sufficiently small, comparison with $\dot{p}=p(1-\epsilon -p)$ gives $\mathrm{liminf}p\ge 1-\epsilon$ for every $\epsilon >0$. Thus $h\to 0$ and $p\to 1$ for positive initial prey: hunters die out and prey reaches carrying capacity.

On the boundary, hunter-only initial data approach the origin; prey-only positive data approach $(1,0)$ in either parameter range. The sketches show representative values in the two requested ranges. Dashed and dotted lines are the prey and hunter nullclines; filled dots are attractors and open dots are saddle equilibria.

The nullclines are $x=f(y)$ and $y=g(x)$, both decreasing curves in the positive quadrant. To their left/right the horizontal velocity is respectively positive/negative; below/above the second nullcline the vertical velocity is positive/negative. Bounded positive synthesis functions give an inward-pointing sufficiently large rectangle, while the axes point into the quadrant. More directly, put $H(x)=f(g(x))-x$: $H(0)>0$, and $H(x)<0$ for large enough $x$. Continuity therefore gives a positive equilibrium.

At an equilibrium, the Jacobian matrix is $J=\left(\begin{array}{cc}-1& f^{\prime }(y_{\ast })\\ g^{\prime }(x_{\ast })& -1\end{array}\right)$, with eigenvalues $-1\pm \sqrt{f^{\prime }(y_{\ast })g^{\prime }(x_{\ast })}$. Its logarithmic-slope product isbecause $f(y_{\ast })=x_{\ast }$ and $g(x_{\ast })=y_{\ast }$. Thus a product exceeding one makes this equilibrium a saddle and gives $H^{\prime }(x_{\ast })>0$. Immediately on its left $H<0$, and immediately on its right $H>0$. Since $H(0)>0$ and $H$ is eventually negative, there must be other roots on both sides, crossing in the attracting direction. For simple roots those have $H^{\prime }<0$, hence two negative Jacobian eigenvalues: the saddle separates at least two stable steady-state alternatives. The divergence is $-2$, so periodic attractors cannot replace these equilibria in a bounded planar region. Nongeneric flat crossings or intervals of equilibria require a nonlinear stability interpretation; the strict saddle inequality alone does not guarantee that every other root is hyperbolic. This is the usual mutual repression stability criterion behind bistability.

The conclusion also has a nonlinear trapping justification. Choose $l<u$ on the left of the saddle with $H(l)>0>H(u)$; such endpoints exist by the signs just proved. The rectangle $[l,u]\times [g(u),g(l)]$ is forward invariant: on the vertical edges, $f(y)-l\ge f(g(l))-l>0$ and $f(y)-u\le f(g(u))-u<0$, while the horizontal-edge velocities point inward or are tangent. The same construction on the right of the saddle gives a second disjoint trapping rectangle. Each contains a nonempty compact attracting invariant set, obtained by intersecting its nested forward images; strict vertical inwardness also rules out invariant boundary trajectories. If a downward crossing is isolated, endpoints can be taken arbitrarily close to it, proving nonlinear stability, and the absence of periodic orbits makes nearby trajectories converge to that equilibrium. Intervals of equilibria instead give attracting alternatives consisting of families, so bounded monotone functions without nondegeneracy do not necessarily produce two isolated hyperbolic attractors.

The fixed points are $(0,0)$ and $(\pm 1,1)$. The Jacobian matrix isAt the origin its eigenvalues are $1,-1$, so it is a saddle with stable direction the y-axis. At either other fixed point, the characteristic polynomial is ${\lambda }^2+\lambda +2$, giving eigenvalues $(-1\pm i\sqrt{7})/2$. Thus both are stable foci. The nullclines are $x=0$, $y=1$ and $y=x^2$. In the right half-plane $x$ grows below $y=1$ and decreases above it; in the left half-plane these horizontal arrows reverse. Vertically, $y$ increases below $y=x^2$ and decreases above it.

A global argument determines the requested omega-limit sets, rather than guessing from the local phase portrait. For $x\ne 0$, use the quadratic-logarithmic Lyapunov function for a feedback systemIts sublevel sets are compact and bounded away from $x=0$, so trajectories starting in either half-plane exist for all positive time and stay in such a set. Since $L$ decreases, it tends to a limit, and is constant on the omega-limit set. Every point of that set therefore has $y=1$. Invariance then requires $\dot{y}=x^2-1=0$, so only the corresponding focus can occur. Also $x(t)=x(0)\exp ({\int }_0^t(1-y(s))ds)$ preserves its sign.

ConsequentlyOn $x=0$, the exact solution is $y(t)=y(0)e^{-t}$. The phase portrait below shows this stable axis, the nullclines and the spiral attraction in each half-plane.

Put $g(z)=s(az)=m{a}^2{z}^2/({\sigma }^2+a^2{z}^2)$. The nullclines areAny steady state has nonnegative coordinates because $g\ge 0$. For $a\ne 0$, $g$ is strictly increasing on the positive half-line. If a steady state had $x>y$, then $x=g(y)<g(x)=y$, a contradiction. The case $y>x$ is identical, so all steady states lie on the diagonal. Write $x=y=z$. Besides $z=0$, the equation $z=g(z)$ givesand henceThus three distinct steady states exist exactly when $|a|m>2\sigma$, with $a\ne 0$ and the positive parameter assumptions stated above. For positive excitatory coupling this is $am>2\sigma$. At equality the two positive states merge; below it only the origin remains. If $a=0$, the origin is again the only steady state.

Both nullclines begin at the origin and approach the limiting activity $m$ along their saturating direction; they are reflections of each other across $x=y$. Flow points to the right where $x<g(y)$ and to the left where $x>g(y)$. It points upwards where $y<g(x)$ and downwards where $y>g(x)$. The plot shows these nullclines and typical trajectories for the parameters of the next part.

The origin and upper diagonal state attract nearby trajectories, while the lower positive state is a saddle. Its stable manifold separates the two attracting regions: trajectories in each basin of attraction approach its stable node, giving bistability. The diagonal is invariant, and on it the scalar activity increases between $z_{-}$ and $z_{+}$ and decreases between zero and $z_{-}$ or above $z_{+}$. Activities remain nonnegative and bounded because $0\le g\le m$. The vector field has constant negative divergence $-2/\tau$, so the Bendixson-Dulac criterion also rules out periodic orbits in the physical quadrant. The network is a switch between persistent activity levels, rather than an oscillator.

Assume the decreasing functions are continuously differentiable so their local gains are defined. In the positive quadrant, the nullclines are $x=f(y)$ and $y=g(x)$, both decreasing. The phase plane arrows point right when $x<f(y)$ and left when $x>f(y)$; they point up when $y<g(x)$ and down when $y>g(x)$. Both axes point into the quadrant. Since $f,g$ are bounded, sufficiently large $x$ or $y$ decreases, giving a bounded forward-invariant rectangle containing all equilibria and eventually containing each trajectory. Thus mutual repression can have one attracting intersection or several intersections separating different basins of attraction.

An equilibrium obeys $x_{\ast }=f(y_{\ast })$, $y_{\ast }=g(x_{\ast })$. Its Jacobian matrix and eigenvalues areThe product is nonnegative. Hence an intersection is a stable equilibrium with two negative real eigenvalues when $f^{\prime }{g}^{\prime }<1$, a saddle equilibrium when $f^{\prime }{g}^{\prime }>1$, and nonhyperbolic when $f^{\prime }{g}^{\prime }=1$. This is the mutual repression stability criterion. To see why an intervening saddle normally accompanies multistability, eliminate $y$ and set $F(x)=f(g(x))-x$. Bounded positivity gives $F(0)>0$ and $F(x)<0$ for large $x$, whileTwo downward, stable crossings require an upward crossing between them. For simple roots that middle crossing has $f^{\prime }{g}^{\prime }>1$ and is a saddle equilibrium. Its stable manifold separates the attracting states. Thus robust multistability requires the loop gain to exceed one at an intervening equilibrium, with the stable equilibria themselves having gain below one. A tangency at gain one is the threshold where equilibria can merge; degenerate crossings require higher-order analysis. A gain exceeding one somewhere is necessary for multiple intersections, but by itself does not prove that a particular pair of nullclines has multiple stable intersections.

Define the logarithmic repression sensitivitiesBoth are negative. At equilibrium $f(y_{\ast })=x_{\ast }$ and $g(x_{\ast })=y_{\ast }$, so their product is exactly $f^{\prime }(y_{\ast })g^{\prime }(x_{\ast })$. The condition can therefore be expressed asEquivalently, the product of the magnitudes of the two negative logarithmic gains must exceed one at that saddle equilibrium. The divergence of the vector field is identically $-2$, so the Bendixson-Dulac criterion excludes periodic orbits in the positive quadrant. The relevant stable long-time states in this bounded phase plane are thus equilibria, rather than attracting limit cycles. The following sketches illustrate the single-attractor and two-attractor possibilities; the saddle in the second panel lies on the diagonal, with its stable diagonal separating the two basins.