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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 positive $m,n$, this mutual repression system can display bistability for some $\lambda$ exactly when both $m>1$ and $n>1$. If either exponent is at most one, it has a unique stable equilibrium for every $\lambda$. To prove the less obvious case $m\le 1<n$, put $a=x/(1+x)$ and $b=y/(1+y)$ at equilibrium. Equality of the two expressions for $\lambda$ gives $a\le b(1-b{)}^{n-1}$. The loop gain is then bounded by $mn{b}^2(1-b{)}^{n-1}$, whose maximum is $4mn(n-1{)}^{n-1}/(n+1{)}^{n+1}<1$. Swap the two components for the other case. If both exponents exceed one, sufficiently large $\lambda$ yields two stable equilibria with $(x,y)\sim (\lambda ,{\lambda }^{1-m})$ and $(x,y)\sim ({\lambda }^{1-n},\lambda )$. The weaker condition $mn\le 1$ is sufficient to exclude multistability, but does not exhaust all exclusions when the same production prefactor is imposed.

// Target: mathematical-biology.bigb

Ice-ocean albedo feedback is the ocean-surface case of ice-albedo feedback: loss of bright sea ice, often covered by snow, exposes darker water, lowers surface albedo and increases absorbed sunlight. The extra stored ocean heat content can promote further melting and delay refreezing. Reversing the sequence can amplify cooling. This is a positive climate feedback, meaning that it reinforces the initial change; the word positive does not mean beneficial, and it does not guarantee runaway warming.

For unchanged downward solar irradiance $S_{\downarrow }$ on a newly exposed opaque patch, the extra absorbed heat flux isFor illustrative local surface albedos $0.60$ and $0.10$, and $S_{\downarrow }=200\mathrm{W}{\mathrm{m}}^{-2}$, the patch absorbs an additional $100\mathrm{W}{\mathrm{m}}^{-2}$. This is a local sunlight calculation, not a whole-planet radiative forcing: newly exposed area, illumination, clouds and season determine its global contribution. With no incident sunlight this particular heating pathway vanishes. NASA's notes describe the contrast between reflective ice and snow and darker water. ^[ref]

To isolate how a positive climate feedback competes with Planck feedback, retain the assumptions of the zero-dimensional planetary energy balance and prescribe the smooth broadband Bond albedoThe exponential function makes a smooth transition from a bright cold state near $0.6$ to a darker warm state near $0.3$. The midpoint $240\mathrm{K}$ and width $3\mathrm{K}$ are chosen to illustrate the dynamics. They are not a seawater freezing law, an observed global relation, or a calibration of modern Earth. No evolving ice thickness or latent heat is represented.

An equilibrium satisfies $F(T_{\ast })=0$. For a small perturbation $\delta T$, linear stability follows fromSince $C>0$, $F^{\prime }<0$ gives an asymptotically stable equilibrium, while $F^{\prime }>0$ gives an unstable equilibrium. With $x=(T-240\mathrm{K})/(3\mathrm{K})$, the derivative isThus the optical term $-S{\alpha }^{\prime }/4$ is destabilizing, while the thermal radiation term $-4\sigma T^3$ is stabilizing. Their sum, not the optical term alone, decides the linear stability.

For the same $S=1360\mathrm{W}{\mathrm{m}}^{-2}$ and Stefan-Boltzmann constant used above, the bisection method on the intervals $(220,223)$, $(238,241)$ and $(253,255)$, with temperature in kelvins, gives:

Unrounded numerical zeros of a function have $|F(T_{\ast })|<10^{-10}\mathrm{W}{\mathrm{m}}^{-2}$; the table rounds them for readability. In the positive-temperature domain the sign of $F$ alternates between heating, cooling, heating and cooling across these three equilibria. The middle equilibrium separates the basins of attraction of the two stable equilibria. For example, a starting temperature of $235\mathrm{K}$ cools toward the cold state, whereas $245\mathrm{K}$ warms toward the warm state: this is bistability at identical external illumination.

The same model illustrates hysteresis. Its equilibrium curve can be written as $S(T)=4\sigma T^4/[1-\alpha (T)]$. The two turning points, where $4[1-\alpha (T)]+T{\alpha }^{\prime }(T)=0$, are saddle-node bifurcations: approximately $(T,S)=(232.637\mathrm{K},1567.801\mathrm{W}{\mathrm{m}}^{-2})$ and $(245.893\mathrm{K},1250.481\mathrm{W}{\mathrm{m}}^{-2})$. For illumination strictly between these thresholds there are two stable equilibria and one unstable equilibrium. Under sufficiently slow increase of $S$, the cold branch ends at the upper threshold; under sufficiently slow decrease, the warm branch ends at the lower threshold. The ensuing warming and cooling transitions differ, so the state depends on its history. Finite sweep rates can delay the transitions; these thresholds describe this model's static equilibrium curve, not predicted tipping points for modern Earth.

The striking feature is that a smooth albedo law can produce distinct long-time states and abrupt transitions, although no discontinuity was inserted into the optical law. It is useful for understanding why positive climate feedback can amplify a response while stabilizing thermal radiation still limits it. Here $0.3<\alpha (T)<0.6$ bounds the incoming heating, whereas outgoing thermal radiation grows as $T^4$; high enough temperatures therefore cool rather than run away indefinitely.

Historically, M. I. Budyko's 1969 paper coupled changing albedo from expanding ice cover to a latitude-dependent temperature and heat transfer calculation, showing how that coupling amplified the effect of changed sunlight. ^[ref] William D. Sellers's 1969 paper constructed an annual energy-budget model in 10-degree latitude belts, including planetary albedo, atmospheric infrared transmission and atmospheric and oceanic exchange. ^[ref] These primary papers provide historical examples of ice-sensitive energy-budget modeling; the smooth single-temperature example above is a separate illustration, not a reproduction of either paper's equations or predictions.

Coexistence of multiple stable long-time states at the same parameter values. Bistability is the case of two states; distinct basins of attraction allow a system to remember its initial state or a sufficiently large perturbation.

// Target: mathematical-biology.bigb

Direction arises from switching between mutually incompatible states, with feedback providing memory and regulated destruction resetting the system. Different transitions use related logic, but a steep response, a bistable switch and an irreversible molecular event are not identical. The cell cycle also requires checkpoints and once-per-cycle licensing so that the ordered biochemical states produce a correct sequence of physical events.

At the G1/S transition, growth signals promote cyclin-dependent kinase activity and relieve RB-mediated restraint on E2F. E2F-dependent expression of additional cell-cycle components can reinforce the transition. More generally, activation of a kinase together with removal of its inhibitor creates positive feedback. In yeast, multisite phosphorylation of Sic1 permits recognition by the Cdc4 ubiquitin-ligase system and degradation of an inhibitor of S-phase kinase. This supplies a threshold that prevents weak or premature signals from launching the next programme.^[ref] The detailed regulators vary across organisms and biological cell types; the conserved point is a coordinated transition from an inhibited state to one that reinforces its own activation.

Mitotic entry gives an experimentally accessible example of bistability. cyclin B binding increases the available CDK1 complex, but Wee1-dependent inhibitory phosphorylation can keep it inactive. Cdc25 removes the inhibitory phosphate. Active CDK1 promotes Cdc25 activity and suppresses Wee1, giving a positive-feedback loop and a double-negative loop. Over a suitable input range, both a low-activity and a high-activity state can be stable. If cyclin input rises from interphase, activation occurs at a higher threshold than the input required to maintain a pre-existing mitotic state. These distinct thresholds constitute hysteresis and prevent small fluctuations from repeatedly reversing a committed transition.

In egg extracts, nondegradable cyclin could be supplied while CDK activity was approached from different initial states. The same cyclin input then supported different persistent activities depending on history, directly distinguishing hysteresis from a single steep but history-independent response.^[ref] These are extract measurements, not universal numerical thresholds for every intact biological cell. A useful conceptual representation iswhere $u$ represents effective activating input. hysteresis supplies memory within this interval; it does not mean that the switch can never be reset.

Reset is achieved by coupling activation to delayed negative feedback and proteolysis. Mitotic kinase activity helps prepare activation of the anaphase-promoting complex with Cdc20, while the spindle checkpoint prevents productive progression until attachment conditions are met. Once this restraint is satisfied, APC/C-dependent ubiquitination promotes destruction of securin and cyclin B. securin loss allows separase to cleave cohesin, making sister-chromatid separation possible. Experiments identifying cohesin cleavage connected regulated proteolysis to a physical transition that simple reversal of a phosphorylation state cannot undo.^[ref] Chromatids cannot be returned to a pre-anaphase chromosome merely by reducing separase activity.

cyclin B destruction lowers CDK1 activity, allowing phosphatases to reverse mitotic phosphorylation and promoting an interphase state. Activation of anaphase-promoting complex with Cdh1 at low kinase activity can help keep cyclins low, while high kinase activity opposes that state. Thus low-CDK and high-CDK programmes are mutually antagonistic. Cyclin synthesis eventually provides a new activating input. Early experiments in extracts showed the importance of cyclin synthesis and destruction, including failure of normal exit with nondegradable cyclin.^[ref] A bistable activation module combined with an appropriate delayed reset can generate an ordered oscillation; a bistable module held at constant input need not oscillate on its own.

replication licensing enforces another essential directionality. At low CDK activity, origin recognition complex with Cdc6 and Cdt1 loads MCM2-7 double hexamers, establishing competence to replicate. In S phase, activating kinases promote origin firing while high CDK activity and additional controls prevent renewed licensing. In animal biological cells, geminin inhibition of Cdt1 and regulated degradation help separate these states. Purified-protein reconstitution demonstrated loading of the replication-helicase precursor as a distinct biochemical event.^[ref] Separating licensing from firing allows many origins to be prepared but prevents fired regions from being copied repeatedly during the same cycle. Lowering mitotic kinase activity must eventually restore licensing competence for the next cycle, rather than permit uncontrolled rereplication during the current S phase.

Checkpoint pathways modify or hold these switches when DNA is unreplicated, damaged or chromosomes are improperly attached; they do not supply the entire oscillator. Energy-consuming phosphorylation cycles, ubiquitination, proteolysis and resynthesis make the system a driven process, while feedback stabilizes distinct states. Thresholds provide commitment, hysteresis provides memory, destruction and cleavage enforce selected transitions, and licensing/checkpoints couple those transitions to one correct round of replication and segregation.

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.

Using the constants as printed, with volume flux measured in Sverdrups, givesIn particular the two small expansion coefficients should not be silently replaced by different oceanographic values.

For a positive steady volume flux, the equation in 2(a) is $q^2-Aq+BE=0$. For a negative steady volume flux, it is $q^2-Aq-BE=0$, with only its negative root admissible. Therefore for $E>0$ the branches areThe upper positive branch $q_{+}$ is stable, the lower positive branch $q_{-}$ is unstable, and the reversed branch $q_{\mathrm{rev}}$ is stable under the usual salt budget dynamics with fixed $\Delta T$. Thus there is bistability below the positive-branch fold. At the fold the two positive equilibria merge with a zero restoring rate; the point is not an ordinary asymptotically stable interior point of either branch. The negative algebraic root of the positive-$q$ quadratic, or positive root of the negative-$q$ quadratic, must never be added as extra solutions.

At $E=0.3\mathrm{Sv}$, the three steady volume fluxes areThe present thermally dominated, northward-upper-flow Atlantic meridional overturning circulation corresponds to $q_{+}\simeq 17.7\mathrm{Sv}$. These algebraic data alone also allow a stable reversed state, so selecting the present branch uses the stated circulation direction. On the large positive branch, the predicted salinity contrast is $\Delta S=E{S}_0/q_{+}\simeq 0.594\mathrm{g}\mathrm{k}{\mathrm{g}}^{-1}$, consistent with the reference-salinity approximation. The weaker branches give larger contrasts, approximately $3.16$ and $4.20\mathrm{g}\mathrm{k}{\mathrm{g}}^{-1}$, so the approximation is less accurate there.

The saddle-node bifurcation occurs atFor $E>E_c$, only the reversed steady volume flux remains. At $E=E_c$ there are still two distinct values, the double positive root and the negative root; “single-valued” holds strictly above this threshold.

Slowly increasing $E$ from the present value causes the stable positive volume flux to weaken. Its salinity contrast then increases because salt exchange is weaker, reducing the cold-box density advantage further: this is the salt-advection feedback. At the fold the positive steady state disappears, producing an abrupt transition to the reversed branch in the quasistatic limit. At the same $E_c$ that branch hasFor a finite slow ramp there can be a small dynamical delay near the fold. In this model subsequently reducing $E$ while keeping it positive follows the stable reversed branch instead of automatically restoring the positive branch, an example of hysteresis. The two small branches approach $q=0$ as $E\to 0^{+}$. At $E=0$ this is an equal-density, no-exchange equilibrium with $\Delta S=\alpha \Delta T/\beta =3.75\mathrm{g}\mathrm{k}{\mathrm{g}}^{-1}$, whereas the positive branch tends to $q=A$ with $\Delta S=0$. One must use the original density relation at this endpoint, rather than divide by $|q|=0$.

The reversal is thermally opposed but salinity driven: freshwater makes the cold box sufficiently fresh that the warm, salty box becomes denser. A real weakening would reduce the northward ocean heat flux and alter regional climate and ocean ventilation. The fold and full reversal here describe the ideal two-box overturning model; its numerical threshold is not a forecast for the real ocean, whose other controls appear in 2(c).

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.

At a positive equilibrium of the bounded Hill model, the mutual repression stability criterion becomesThus $mn>1$ is necessary. Because the same $\lambda$ is imposed in both synthesis functions, it is not sufficient for arbitrary unequal positive exponents. The additional equilibrium constraint isHere is an exact criterion for all the printed exponent choices, including nonintegers. If $m>1$ and $n>1$, multistability is possible: for large $\lambda$ the intermediate equilibrium satisfies $x\sim {\lambda }^{(m-1)/(mn-1)}$, $y\sim {\lambda }^{(n-1)/(mn-1)}$, so both increase without bound and $K\to mn>1$. These asymptotics follow by balancing $x{y}^m\sim \lambda$ and $y{x}^n\sim \lambda$; the limiting two algebraic equations have a nonsingular log-Jacobian when $mn\ne 1$, so a nearby exact equilibrium exists for large $\lambda$. If both exponents are at most one, $K<1$ everywhere and multistability is impossible.

For the remaining case, interchange $x,y$ if necessary so $0<m\le 1<n$. For each $x>0$, let $Y(x)>0$ be the unique solution ofwhich exists because the left side is strictly decreasing; for $m=1$ the right side always exceeds one. DefineThen the exact answer for this remaining range isThis is a one-variable test depending only on $m,n$, with no unknown $\lambda$. To prove it, parameterize all equilibria by $\lambda (x)=x(1+Y(x{)}^m)$. Implicit differentiation givesThe denominator is positive, and $\lambda (x)$ goes from zero to infinity as $x$ goes from zero to infinity. A negative-slope portion therefore forces two folds and an interval of three equilibria, with two attracting branches. If $K\le 1$ everywhere, $\lambda$ is strictly increasing apart from possible isolated stationary points, so there is only one equilibrium for each $\lambda$. This is the common-strength Hill repression criterion.

For illustration and a closed test on the boundary $m=1$, $Y=x/(1+x^n-x)$ andobtained at $x^n=(n+1)/(n-1)$. In particular $(m,n)=(1,2)$ has $mn>1$ but maximum $3\sqrt{3}/8<1$, so it cannot be multistable. For positive integer exponents the allowed choices simplify to both exponents at least two, or one equal to one and the other at least four. For arbitrary real exponents the preceding exact criterion is needed; replacing it by $mn>1$ would discard the common-strength constraint.

The unbounded power laws behave differently. Combining their equilibrium equations gives $x^{1-mn}={\lambda }^{1-m}$. If $mn\ne 1$, there is exactly one positive equilibrium,Its slope product is exactly $mn$, so it is attracting if $mn<1$ and a saddle if $mn>1$, but there are never two isolated attracting equilibria. If $mn=1$, equilibria exist only when ${\lambda }^{1-m}=1$; when this holds there is a whole curve of equilibria, with a zero tangent eigenvalue and transverse eigenvalue $-2$. This includes $m=n=1$, $xy=\lambda$, for every $\lambda$. A neutrally degenerate continuum is not bistability between isolated attractors. Thus these power laws do not generate ordinary multistability; boundedness was essential to the outer-root argument in part (a).

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.

For this equal-strength power-law mutual repression, the logarithmic gains and their product at equilibrium areThe mutual repression stability criterion immediately excludes multistability if $mn\le 1$, since $P<mn$ at every finite positive equilibrium. The common prefactor $\lambda$ gives a stronger exclusion: if either exponent is at most one, multistability is impossible for every $\lambda$, including some cases with $mn>1$.

To prove the additional cases, suppose $m\le 1<n$. At equilibrium the two expressions for $\lambda$ implyPut $a=x/(1+x)$ and $b=y/(1+y)$. Because $m\le 1$, the left side is at least $a$, so $a\le b(1-b{)}^{n-1}$. It follows thatThe maximum in the second inequality occurs at $b=2/(n+1)$. Its strict bound follows from $4n/(n+1{)}^2\le 1$, $m\le 1$, and $((n-1)/(n+1){)}^{n-1}<1$. Interchanging the components proves the case $n\le 1<m$; the case with both exponents at most one was already covered by $mn\le 1$. Every root of $F(x)=f(g(x))-x$ therefore has $F^{\prime }<0$. Since $F$ begins positive and ends negative, there is exactly one root: two downward crossings would require another crossing with nonnegative derivative between them. The unique equilibrium is stable.

For completeness, neither exponent can be relaxed above one in this exclusion. If $m>1$ and $n>1$, take $\lambda$ sufficiently large. On $x\in [\lambda /2,\lambda ]$, $g(x)=O({\lambda }^{1-m})\to 0$, so $f(g(x))=\lambda (1+o(1))$ maps that interval into itself. It has a fixed point withIt is a stable equilibrium. By exchanging the components, there is a second stable fixed point with $y\sim \lambda$ and $x\sim {\lambda }^{1-n}$. These two regions are disjoint for large $\lambda$, proving bistability for some production strength. Hence the full answer isFor example, $m=1/2,n=3$ has $mn>1$ but still a unique stable equilibrium for all $\lambda$. Thus the product-only necessary condition for strong feedback is not sufficient when both production prefactors are constrained to be equal.

Use all coefficients exactly as printed and use Sv consistently for transports. The present salt contrast and density-driven friction coefficient areThusThe large numerical value of $k$ reflects division by a small dimensionless density contrast, about $2.07\times 10^{-6}$ with the supplied effective expansion coefficients. It is not the actual overturning transport, which remains $15$ Sv.

For $q>0$, the steady equation is a quadratic with rootsThe critical freshwater forcing in a two-box overturning model occurs where the two roots coalesce:There is also a negative-flow branch for every $F>0$:At the present forcing, the three roots are $15.000$, $6.774$ and $-3.950$ Sv. The prescribed present state is therefore the upper positive branch, not an arbitrary choice among the roots.

Stability needs a salinity evolution law, not just the algebraic steady equation. For fixed temperature contrast and reference salinity, the standard salt-contrast dynamics can be writtenwhere $V_{\mathrm{eff}}>0$ summarizes the reservoir volumes. The linear slope at a positive root is $(A-2q)/V_{\mathrm{eff}}$: the upper branch is stable and the lower branch unstable. At a negative root it is $(2q-A)/V_{\mathrm{eff}}<0$, so the reversed branch is stable. The two stable states below the fold have different attraction basins, demonstrating bistability. Their stability is conditional on this fixed-temperature, fixed-reference-salinity closure.

As forcing increases quasistatically from $0.5$ Sv, the normal branch weakens and reaches a saddle-node bifurcation at $F_c$. Beyond it no positive steady circulation remains, so the reduced dynamics move toward the stable reversed branch. The reversal is salinity dominated: the fresh cold reservoir is no longer the denser one. Recovery under a later forcing change can depend on the system's state and feedback history.

network resilience depends on which function must persist, which disturbances are applied and how the network responds over time. Local interaction motifs, global architecture and physical operating constraints can protect different properties, often with tradeoffs. A graph that remains connected is not necessarily a functioning regulatory system, and a system that changes its output during a stimulus has not necessarily failed.

It helps to distinguish network robustness, meaning persistence of a specified property under perturbation, from network resilience, which also considers recovery after a disturbance. Relevant performance measures include the maintained output, transient deviation, recovery time, surviving connected component, transport capacity and cost. Disturbances can be brief or persistent, small or destructive, random or targeted, independent or correlated. Consequently, resilience is not one scalar attribute that can be read off a degree distribution.

At the local scale, negative feedback can oppose fluctuations, while network redundancy can provide an alternative route when one component fails. A feed-forward loop can filter stimuli by duration, and positive feedback can stabilize a switched state through bistability. These roles are conditional: a switch that remembers a transient signal can be useful in a developmental decision but undesirable in a homeostatic controller. feedback delay or excessive gain can destabilize a corrective loop; redundancy does little against a disturbance shared by every supposedly independent backup. One must analyze the signs, strengths, kinetics and operating ranges of the interactions, not just count motifs.

Bacterial chemotaxis supplies a concrete cellular example. In Escherichia coli, an attractant change affects receptor-associated kinase activity and hence downstream phosphorylation controlling swimming behavior. Slower receptor methylation and demethylation provide a feedback memory of the stimulus. In an appropriate operating range, the output responds transiently but approximately returns to its earlier level even while the new stimulus persists: robust perfect adaptation. This preserves sensitivity to subsequent changes rather than holding the entire molecular state fixed. The control interpretation identifies conditions under which adaptation does not require precise tuning of every reaction constant.^[ref]

The essential logic can be isolated without pretending to reproduce every reaction. If $y$ is the output and $z$ an internal controller state, supposeWhenever the closed system reaches a stable steady state with an unsaturated integrator, $\dot{z}=0$ forces $y=y_{\ast }$. The controller accumulates the error, so a constant disturbance need not leave a persistent output offset. This is integral action. The equation demonstrates the mechanism, not an unconditional guarantee: the surrounding feedback must be stable, the required control action must be attainable, and the effective set point must remain meaningful under the parameter changes considered.

The example also clarifies what is and is not robust. Returning kinase activity to a set point does not imply an unchanged response time, noise level or swimming trajectory. Strong or sustained stimuli can exhaust the receptor's modification range; deleting an essential feedback component can remove adaptation; correlated changes in controller parameters can shift its set point. Cellular resilience therefore combines biochemical organization with finite resource and timescale constraints. A useful experiment measures the response and recovery after stimuli of different amplitudes, while independently perturbing feedback components, rather than inferring perfect adaptation from one endpoint measurement.

At the global scale, several architectural features matter. Alternative paths and disjoint routes reduce dependence on one component. network modularity can confine damage, although inter-module bridges may then become critical bottlenecks. Short paths improve communication but may also permit rapid spread of a disturbance. Highly connected hubs can support efficient coordination while concentrating vulnerability. Directed links, edge weights and capacities change which routes are usable; a path in an unweighted undirected projection need not carry a signal or the required load in the real system.

The Internet illustrates these distinctions for a real-world network. Nodes can represent routers or autonomous systems, with edges representing different kinds of connectivity; conclusions should state the chosen resolution. In a heterogeneous network, random deletion is likely to hit one of many low-degree nodes, whereas targeted deletion of high-degree nodes can disrupt a large fraction of the remaining paths. The classic attack-tolerance study demonstrated this contrast in model networks and mapped communication networks.^[ref] Its lesson is a vulnerability conditional on the topology and attack rule, not that every network with an apparent power-law tail is invulnerable to arbitrary failures.

Actual Internet resilience also uses routing protocols, spare capacity, replicated services and operational recovery. Those are dynamical and engineering mechanisms additional to the static graph. Removing a node may reroute traffic through surviving links and overload them; failures can then cascade even while many topological paths remain. Conversely, route recomputation may restore service after an initially severe disruption. cascading failure therefore requires a load and dependency model, while geographically correlated failures require an account of shared physical infrastructure. Maintaining connectivity, throughput and low latency are different objectives.

The cellular and communication examples share a design principle but not a single universal architecture. A local controller can reject a persistent input error; a global alternative route can bypass a failed connection. Neither guarantees protection against every perturbation, and both incur costs in energy, components or spare capacity. A sound comparison specifies the disturbance ensemble, the measured function, the recovery interval and the intervention used to test the mechanism. Resilience is produced by matched topology and dynamics, and protection against one class of stimulus can create vulnerability to another.

Yes: there is a quasistatic hysteresis loop. Equilibrium points lie on the mathematical model parameter curve $a=-P(N)$. Its two saddle-node bifurcations occur at the stationary points computed in part (a), at mathematical model parameter valuesBetween these values there are two attracting branches separated by a repelling branch. Starting below $a_{\mathrm{down}}$ and increasing $a$ slowly, a solution follows the lower attracting branch. This branch persists through the bistability interval and ends only at $a_{\mathrm{up}}$, where it meets the repelling branch in a saddle-node bifurcation. The solution then moves to the upper attracting branch.

If the direction of mathematical model parameter change is reversed, the upper branch persists until $a_{\mathrm{down}}$, where a different saddle-node bifurcation forces the downward transition. The two transition values differ, and both lie inside the allowed range $[-1/2,1/2]$. Thus the biological population selected at an intermediate mathematical model parameter value depends on its prior branch. This is the equilibrium point hysteresis shown in the branch diagram; finite-rate passage can additionally delay the jumps near the saddle-node bifurcations.

With three distinct positive roots of an equation, the negative-leading cubic polynomial has the sign patternEquivalently its derivative is negative at the first root of an equation, positive at the middle root of an equation, and negative at the final root of an equation. The phase line arrows therefore point toward the two outer equilibrium points and away from the middle one:The middle equilibrium point separates the two basin of attraction in the bistability mathematical model parameter interval.

The four nonnegative equilibria are $0,a_1,a_2,a_3$. Since all factors are positive below $a_1$, the sign of $h(N)$ on consecutive intervals isThe phase line therefore points away from zero, towards $a_1$, away from $a_2$, and towards $a_3$. Equivalently $h^{\prime }(0)=a_1{a}_2{a}_3>0$, and at each simple positive root $a_i$, $h^{\prime }(a_i)=-a_i{\prod }_{j\ne i}(a_j-a_i)$. Their signs alternate negative, positive, negative.

Hence extinction and $a_2$ are unstable, while $a_1$ and $a_3$ are asymptotically stable. Initial positive populations below $a_2$ tend to $a_1$; those above $a_2$ tend to $a_3$; an initial population exactly at $a_2$ remains there. This is bistability with a threshold separating the two basins.

Positive feedback closes a causal loop whose net effect reinforces an initiating change. A loop with two inhibitory links can therefore have positive feedback. In signal transduction, a product that increases the activity of the enzyme making it illustrates reinforcement; the consequences depend on the full mathematical model, not just the loop sign. See signaling feedback.

A biological example is cell-cycle activation feedback. Cyclin B binds Cdc2, whose activity promotes activation of Cdc25; Cdc25 removes inhibitory phosphorylation from Cdc2, permitting further activation. Cdc2 also inhibits Wee1, an enzyme that inhibits Cdc2 through phosphorylation. That second return contains two inhibitions and therefore also has positive feedback. These are regulatory paths, not a claim that every connection is a single direct molecular encounter. ^[ref]

Reinforcement can preserve a decision after its initiating stimulus weakens: returning from the high state may require a lower input than entering it. The cell-cycle hysteresis experiment below tests this history dependence at matched cyclin B concentrations. A steep response alone is insufficient evidence of bistability, and finite observation times can make slow relaxation look like memory. A proposed switch must specify its input, readout and observation interval.

The replicator equation says that a strategy grows in frequency when its expected payoff exceeds the population's current mean payoff. It is an ordinary differential equation for a composition, rather than for total population size. The subtraction of the mean is what makes relative success matter: even a type growing in absolute number can shrink as a fraction if other types grow faster.

Let $x_i(t)$ be the frequency of pure strategy $i$, with $n$ available pure strategies andIn a symmetric bimatrix game, let the payoff matrix $A$ have entry $a_{ij}$ equal to the payoff to strategy $i$ against $j$. The other player's payoff in that encounter is $a_{ji}$: symmetry between player roles does not require $A=A^T$. Uniform random encounters give the expected payoffs and their population meanAfter choosing the time unit so that the growth-rate scale is one, the continuous-time replicator equation isHere a payoff is measured on a scale proportional to net growth per unit time, not merely as an arbitrary preference ranking. More generally one can specify locally Lipschitz frequency-dependent payoffs $f_i(x)$ directly; the same equation with $\bar{f}={\sum }_i{x}_i{f}_i$ then defines a unique frequency trajectory. The linear pairwise model and its reproductive interpretation are given in Taylor and Jonker (1978), sections 2–3; the formulas here also follow by the following direct normalization.

Suppose the number $N_i$ of type $i$ obeys ${\dot{N}}_i=N_i[b(t)+f_i(x)]$, where the baseline $b(t)$ is identical for all types. Set $N={\sum }_i{N}_i>0$ and $x_i=N_i/N$. The quotient rule givesThus the common baseline cancels. Adding the same constant to every entry of $A$ leaves the replicator equation unchanged; multiplying $A$ by a positive constant changes only the speed along trajectories. The same payoff matrix can therefore describe frequency evolution at different growth-rate baselines and time scales.

The probability simplex is invariant in forward time. On ${\sum }_i{x}_i=1$,Normalization alone would not exclude negative frequencies. Nonnegativity follows as well: along a solution,An absent type remains absent, and a positive frequency remains positive at every finite time. For the fixed payoff matrix model, bounded rates on the compact probability simplex give global forward solutions. Frequencies can still tend to zero as $t\to \infty$. Each face is invariant, so this model has no spontaneous introduction of new pure strategies; mutation or outside introductions would require additional terms. For two present types the logarithm of their frequency ratio satisfies $\frac{d}{dt}\log (x_i/x_j)=f_i-f_j$, directly expressing the accumulated payoff advantage.

For two pure strategies, write the composition as $(x,1-x)$, where $x$ is the frequency of strategy 1, and letTheir payoff difference and the resulting scalar ordinary differential equation areIn the interior $0<x<1$, the sign of the payoff difference determines the direction of motion. The endpoints are always equilibrium points, and an additional interior equilibrium point exists precisely when the affine payoff difference has a zero strictly between them. At a simple interior zero $x_{\ast }$, $F^{\prime }(x_{\ast })=x_{\ast }(1-x_{\ast })\Delta f^{\prime }(x_{\ast })$: a negative value gives asymptotic stability, a positive value gives instability. At the endpoints $F^{\prime }(0)=b-d$ and $F^{\prime }(1)=c-a$, with stability understood relative to $[0,1]$. A zero derivative requires examining the actual flow rather than this strict sign test.

For an original coordination game, imagine two conventions that reward matching, with convention 1 giving the larger matching reward. ChooseThe crossing is $x_{\ast }=2/5$. Below it, convention 1 loses frequency; above it, convention 1 gains frequency. The endpoint equilibrium points $0$ and $1$ have one-sided asymptotic stability, while $2/5$ is unstable: the corresponding derivatives are $-2,-3$ and $6/5$, respectively. Every initial $0<x(0)<2/5$ tends to $0$, every initial $2/5<x(0)<1$ tends to $1$, and exactly $x(0)=2/5$ stays there. Even the convention with the larger matching reward can disappear if it starts below the threshold. This is bistability, not selection of a universally best convention.

The compositions giving a symmetric Nash equilibrium in this coordination game are the two pure conventions and the mixed composition $(2/5,3/5)$. Both pure conventions are evolutionarily stable strategies because a mutant convention has strictly lower payoff against the pure resident. The mixed symmetric Nash equilibrium is not an evolutionarily stable strategy: against that resident both conventions tie, but after a pure convention-1 mutant is introduced its payoff advantage is $5(x-2/5)>0$. Equal payoffs at a symmetric Nash equilibrium need not provide resistance to invasion or asymptotic stability.

For an original coexistence model, choose payoffs making each convention advantageous when sufficiently rare:Here $x_{\ast }=3/4$ attracts every initial composition with both types present: $\dot{x}>0$ below it and $\dot{x}<0$ above it. The interior derivative is $F^{\prime }(3/4)=-3/4$, while $F^{\prime }(0)=3$ and $F^{\prime }(1)=1$ make both endpoints unstable to introduction of the other type. The endpoints nevertheless remain stationary if the other type is exactly absent; the attraction statement excludes those boundary initial states.

The sole symmetric Nash equilibrium in this coexistence model is $p=(3/4,1/4)$. It is also an evolutionarily stable strategy. For any different mixed strategy $q=(y,1-y)$, the first comparison in the two-condition criterion for evolutionary stability ties because $Ap=(3/2,3/2)$. Its second comparison is strictly positive:Thus restoring frequency feedback can protect a mixture even though all present types earn equal payoffs at the mixture. Coexistence here means different individuals using different pure strategies; it does not require every individual to randomize with a mixed strategy.

Frequency evolution need not increase the mean payoff. In the coexistence example $\bar{f}=5x-4x^2$, whose maximum occurs at $x=5/8$, whereas the attracting composition is $3/4$. For $5/8<x<3/4$, the frequency $x$ increases while $\frac{d}{dt}\bar{f}=(5-8x)\dot{x}<0$. Individual payoff advantages drive the equation; maximizing collective mean payoff is a different optimization problem.

More generally, a rest point only equalizes payoffs among present types: $x_i>0$ implies $(Ax{)}_i=x^{T}Ax$. A symmetric Nash equilibrium additionally requires $(Ax{)}_i\le x^{T}Ax$ for every absent type. All vertices are rest points even when an absent type could invade. For finite symmetric bimatrix games, a rest point with Lyapunov stability on the full probability simplex is a symmetric Nash equilibrium; an interior trajectory that converges has a symmetric Nash equilibrium as its limit. These are conditional results, not a convergence theorem for every trajectory. See Cressman and Tao (2014), Theorem 1.

The rock paper scissors payoff matrix gives an explicit obstruction to universal convergence:Since $A^T=-A$, $x^{T}Ax=0$, and in the interior $\frac{d}{dt}\log (x_1{x}_2{x}_3)={\sum }_i(Ax{)}_i=0$. Thus $x_1{x}_2{x}_3$ is conserved. Its maximum on the probability simplex occurs only at $(1/3,1/3,1/3)$, the unique symmetric Nash equilibrium, so an interior trajectory starting elsewhere cannot converge to that composition. Every other interior product level is a closed curve without a rest point, giving a periodic orbit. The existence of these cycles is also discussed in Cressman and Tao (2014), Example 1. The useful lesson is to inspect the specified payoff feedback and dynamical system, rather than infer convergence from the existence of an evolutionarily stable strategy or a Nash equilibrium.

A circulation change alters the salinity contrast and hence the density difference that supports the circulation. In a thermally dominated positive branch, weaker exchange removes the imposed salinity contrast less effectively, making the high-latitude reservoir fresher and weakening the circulation further. This positive feedback permits bistability and a saddle-node bifurcation.

Saturating positive feedback limits the reinforcing rate as activity increases. Here is an explicitly synthetic ordinary differential equation for normalized nonnegative input $u$, response $x$, and return strength $\beta$:The return is an increasing, nonproportional function of $x$, while removal is proportional to $x$. Time and activities are dimensionless variables. At $x=0$ the vector field points into the nonnegative domain, and $dx/dt\le u+\beta -x$ prevents unbounded growth there. This makes the toy useful for asking about a bounded switch; it supplies no molecular mechanism for the chosen rate law.

At an equilibrium point, $f=0$. For a simple equilibrium point, the sign of the derivativedetermines local asymptotic stability. If $\beta =1$, the maximum reinforcing slope is $3\sqrt{3}/8<1$, so $f$ decreases strictly: every $u\ge 0$ has one asymptotically stable equilibrium point despite positive feedback.

Three equilibrium points become possible only if the reinforcing slope exceeds removal's slope somewhere. Its maximum is $3\sqrt{3}\beta /8$, at $x=1/\sqrt{3}$. Consequently this toy requiresfor two distinct folds. This is a necessary threshold for a bistability interval in this toy, not a universal biochemical number. At a fold, both $f=0$ and $f_x=0$, givingFor $\beta =1.8$, the folds are at $(x,u)\simeq (0.350003,0.153563)$ and $(0.887240,0.094405)$. Between these input values there are two asymptotically stable equilibrium points separated by an unstable one. At $u=0.12$, the three activities are approximately $0.171331$, $0.621230$ and $1.127439$; the low and high ones are stable.

A sufficiently slow increasing sweep follows the low branch to $u_{\mathrm{on}}\simeq 0.153563$ and then jumps upward. A decreasing sweep follows the high branch to $u_{\mathrm{off}}\simeq 0.094405$ and then jumps downward: hysteresis. These are limiting quasistatic thresholds, with relaxation between input changes and negligible perturbations that could cause earlier switching. Finite sweep rates and fluctuations can alter observed thresholds. A switchlike response with only one stable equilibrium point does not have this equilibrium hysteresis mechanism.