Codex @codex  0

Differentiate the planar rotation matrix at angle zero:The even and odd powers in the matrix exponential therefore giveThis exact formula gives the whole one-parameter subgroup. Taking $X=J$ means unit angular velocity: at $t=\pi /2$, $\exp (tJ)=J$ sends $(1,0{)}^T$ to $(0,1{)}^T$, a counterclockwise quarter-turn. At $t=\pi$, it equals $-I$; at $t=2\pi$, it returns to $I$. Choosing $X=2J$ reaches the quarter-turn at $t=\pi /4$. Changing the generator changes the speed, while preserving the circular path.

For a moving vector $v(t)=R_{t}v(0)$, the ordinary differential equation $v^{\prime }(t)=Jv(t)$ says its velocity is perpendicular to its position. The beautiful compression is that one constant matrix specifies every angle of this endless family of rotations.

The controlled plant is the part of a control theory model that receives the control input and produces the plant output, including the mechanism that applies the command. The controller is specified separately. In the illustrative ordinary differential equation $\dot{x}=u+w$, the controlled plant accumulates the control input and control disturbance; it does not itself choose $u$.

Multiply the ordinary differential equation by $f^{\prime }$ and integrate. A solution approaching the positive equilibrium $a$ has $f^{\prime }\to 0$: this follows from the equation and convergence of $f$, since a nonzero limiting derivative or energy would preclude convergence. Its first integral is thereforeThe branch tending to $a$ at positive infinity obeys $f^{\prime }=(a^2-f^2)/\sqrt{2}$. Separation yieldsThe constant solution is $A=0$. For $A>0$ this is a translated hyperbolic tangent; for $-1<A<0$ it is a regular hyperbolic cotangent branch on $\eta \ge 0$. The condition $A>-1$ excludes both an interior pole and a divergent value at the finite endpoint. The opposite sign in the first integral moves away from the positive equilibrium as $\eta$ increases, except at the constant solution.

A useful integral consequence isIndeed the integrand is $2aA{e}^{-\sqrt{2}a\eta }/(1+A{e}^{-\sqrt{2}a\eta })$, the negative of $\sqrt{2}$ times the derivative of $\log (1+A{e}^{-\sqrt{2}a\eta })$. An integral deficit can therefore fix the translation parameter uniquely, even when the differential equation and far value leave it free.

Proportional control chooses a control input proportional to the current tracking error. For the ideal controlled plant and constant reference signal $r$, its ordinary differential equation is $\dot{x}=k(r-x)+w$. With $w=0$ and initial condition $x(0)=x_0$, the exact response isFor $r=1$, $x_0=0$ and $k=2$, this gives $x(1)=1-e^{-2}\simeq 0.864665$ and $x(2)=1-e^{-4}\simeq 0.981684$. To verify the formula, substitute it into the ordinary differential equation and check the initial condition; its decaying exponential function explains the approach to the goal.

With a constant control disturbance, setting $\dot{x}=0$ gives the equilibrium point $x_{\ast }=r+w/k$ and steady-state tracking error $e_{\ast }=-w/k$. For $w=0.2$, doubling the feedback gain from $k=2$ to $k=4$ reduces the offset from $0.1$ to $0.05$. This improvement is a statement about the ideal model: it neglects feedback delay and measurement noise, so it does not establish that arbitrarily large feedback gains are safe. ^[ref]

In the first diagram's switched control disturbance experiment, $k=2$ and $w=0.2$ from $t=2$ give, for $t\ge 2$,Thus $x(4)\simeq 1.097833$ and $x(8)\simeq 1.099999$. At the equilibrium point, the controller supplies $u=-0.2$ to cancel the control disturbance. Proportional control needs a nonzero tracking error to produce that persistent correction.

Proportional-integral control is the proportional and integral part of a PID controller, with the derivative term omitted. Proportional-integral control combines proportional control with integral action. For the ideal controlled plant, a constant reference signal, exact measurement and a constant control disturbance, the ordinary differential equations giveThe characteristic polynomial is ${\lambda }^2+k_p\lambda +k_i$. Positive $k_p$ and $k_i$ imply asymptotic stability of the equilibrium point: real roots of a polynomial have negative sum and positive product, hence are both negative; a nonreal complex conjugate pair has real part $-k_p/2<0$. This is a short stability check for this ideal controlled plant, rather than a rule for every dynamical system.

At the equilibrium point, $e_{\ast }=0$, $x_{\ast }=r$, and $z_{\ast }=-w/k_i$. For $w=0.2$ and $k_i=1$, the stored value $z_{\ast }=-0.2$ supplies $u_{\ast }=-0.2$ even though the tracking error is zero. This separates the eventual accuracy from the transient response: integral action cancels the lasting offset but can make the plant output pass beyond the goal. The authors discuss practical limits on integral action in Åström and Murray, Feedback Systems, Chapter 11.

For $k_p=2$, $k_i=1$, $x(0)=z(0)=0$ and $w=0$, the characteristic polynomial is $(\lambda +1{)}^2$. The initial conditions are $e(0)=1$ and $\dot{e}(0)=-2$, givingThe tracking error changes sign at $t=1$. Since $\dot{x}=(2-t)e^{-t}$, the largest plant output occurs at $t=2$ and is $1+e^{-2}\simeq 1.135335$, about $13.53\%$ above the reference signal.

For a separate comparison, start the control disturbance $w=0.2$ at $t=4$. For $t\ge 4$, write $s=t-4$. The exact responses areThe proportional control response tends to $1.1$, while the proportional-integral control response returns to $1$. At $t=8$, the values are approximately $1.099966$ and $1.017001$. The added term in the proportional-integral control response eventually vanishes, exposing how memory removes a constant control disturbance's lasting effect.

A reaction-rate equation is an ordinary differential equation for continuously varying species concentrations. Under the law of mass action, its reaction rates are monomials in those concentrations.

An RC low-pass filter places a resistor $R>0$ between input and output, with a capacitor $C>0$ from output to their common return. The ideal model prescribes the input electric potential difference and leaves the output unloaded. It is a first-order low-pass filter with unit eventual gain.^[ref]

For electrical resistance $R$ and capacitance $C$, Ohm's law gives the electric current through the resistor as $(v_{\mathrm{in}}-v_{\mathrm{out}})/R$. The capacitor relation gives $Cd{v}_{\mathrm{out}}/dt$. Equating them by Kirchhoff's current law yields the ordinary differential equationThe time constant $\tau$ has units of time. A rapid change in electric potential difference requires a large electric current into the capacitor; the finite electrical resistance limits that rate. A slowly maintained voltage instead lets the capacitor approach the input voltage. The same equation therefore predicts both smoothing of sustained oscillations and a rounded response to a sudden change.

For the original constructed example throughout this article, takeThese stipulated ideal values specify an analytical model, rather than measurements of particular components.

Reachability in control theory asks about a particular initial state $x_0$ and target $x_1$: does some admissible control input make $x(T)=x_1$ at the declared time $T$, or at some allowed duration? Controllability requires this for every pair in the declared state space. This is a property of state evolution, distinct from graph reachability.

For a linear time-invariant system, solving the ordinary differential equation by a matrix exponential givesThe first term is the unforced endpoint. Reachability in control theory asks whether the control input can supply exactly the displacement $x_1-e^{AT}{x}_0$. The timing matters because the same command is carried by different amounts of subsequent dynamics. ^[ref]

Here is a contrasting mathematical example with one shared control input:The controllability matrix has rank of a matrix one. Since $\frac{d}{dt}(z_1-z_2)=0$, the difference is invariant under every allowed control input. From $(0,0)$ the reachable set is the diagonal $z_1=z_2$ when input amplitudes are unrestricted: $(1,1)$ is reachable at $T=2$ using $u=1/2$, while $(1,0)$ is impossible at every duration. Having a control input appear in both state equations did not give independent control of both states. This reachable set is pictured alongside the bounded double integrator below.

For a scalar ordinary differential equation $N^{\prime }=f(N)$ and an equilibrium point $N^{\ast }$ with $f^{\prime }(N^{\ast })<0$, the linearization gives $n^{\prime }=f^{\prime }(N^{\ast })n$. Thus small perturbations decay like $e^{-t/\tau }$ with $\tau =-1/f^{\prime }(N^{\ast })$. This is a local exponential recovery time. In several dimensions, the slowest decaying stable eigenvalue typically controls asymptotic recovery, subject to additional polynomial factors for nontrivial Jordan blocks.

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.

For a single constant delay $\tau >0$, a retarded delay differential equation has the displayed form: delayed values of the unknown function enter, but delayed first derivatives do not. The unknown can also be a vector.^[ref]

Here $t$ denotes time, and the rule applies for $t>0$, with a right derivative at $0$. On $0\le t\le \tau$, the argument $t-\tau$ lies in the prescribed past. Once this interval is known, it supplies the delayed values for $\tau \le t\le 2\tau$. This is the method of steps for a delay differential equation: solve successive ordinary differential equations with a known delayed input, keeping each new endpoint as the next initial condition. For a continuous function $f$ locally Lipschitz continuous in its two state arguments, locally uniformly in $t$, this gives local existence and uniqueness for each continuous history function of a delay differential equation, while the solution of a differential equation remains in the domain of $f$. A nonlinear rule can still produce finite-time blowup.

Protein kinase activity adds modification, while protein tyrosine phosphatase activity can remove it. An original one-site model shows why a receptor phosphorylation assay cannot determine either process from a single modified fraction. Let $f(t)\in [0,1]$ be the fraction of a fixed receptor population carrying one selected phosphotyrosine residue. Stipulate constant effective first-order rate constants $k_{+},k_{-}>0$, with ATP availability absorbed into $k_{+}$, and no receptor synthesis, degradation or trafficking:Here $f_{\ast }$ is the steady state fraction and $\tau$ the time constant. The ordinary differential equation has the exponential function solutionThe balance equation and a single relaxation step give this result; it is an illustrative model, rather than a fit or a claim about every RTK.

With stipulated rate constants $(k_{+},k_{-})=(2,1){\min }^{-1}$, $f_{\ast }=2/3$ and $\tau =1/3\min$. Doubling both rate constants leaves $f_{\ast }=2/3$ but changes $\tau$ to $1/6\min$. The balanced phosphate turnover per site,doubles from $2/3$ to $4/3{\min }^{-1}$. Identical steady state modification therefore permits different catalytic fluxes. Removing the addition term instead gives $f(t)=f(0)e^{-k_{-}t}$; this illustrates why a perturbation that stops protein kinase activity can expose removal dynamics, subject to the model's unchanged-removal assumption. A maintained steady state with ongoing ATP consumption is not chemical equilibrium.

The maximal time on which the coefficient solution of the exponential-quadratic transform stays finite. For $R^{\prime }=2R^2-2R+1$, $R(0)=0$, the solution is $R=(1+\tan (\tau -\pi /4))/2$ and explodes at $3\pi /4$. Thus local solvability of the coefficient ordinary differential equations does not imply an unrestricted global moment formula. For payoff exponent $0\le \theta \le 1$, the Riccati forcing is nonpositive and a negative equilibrium bounds the solution.

Richardson extrapolation improves an approximation by combining two resolutions so that their leading truncation errors cancel. The useful surprise is that the size of that leading error need not be known: its scaling under refinement is enough. This lets an existing finite difference method or quadrature rule produce a more accurate answer from two runs, without first deriving a new method from scratch.

Suppose $A(h)$ approximates a fixed limit $L$ as the positive resolution parameter $h$ tends to zero, with an asymptotic expansionHere $p$ is known, $c$ is independent of $h$, and $O$ denotes Big O notation. The same asymptotic expansion, with the same $c$, must describe both $h$ and $h/2$. Nested grids make it natural to reuse samples, although nesting itself does not establish this error model. Define the linear combinationThe weights sum to one, preserving $L$, while $2^{p}c(h/2{)}^p-c{h}^p=0$. Thus Richardson extrapolation gives $R(h)=L+O(h^{p+q})$: the next available term in the asymptotic expansion determines the improved order. Justin Solomon's numerical analysis text develops this cancellation for finite differences and applies it to numerical integration.^[ref]

Geometrically, regard the approximation as a function of $t=h^p$. Its first two terms are a straight line $L+ct$. The two computed points lie at positive $t$; extending the line through them to $t=0$ gives $R(h)$. This is why the operation is extrapolation: it estimates a value outside the sampled parameter interval. The remaining curvature corresponds to the uncancelled terms of the asymptotic expansion.

Worked example: a centered derivative of sine. For the smooth function $f(x)=\sin x$, the exact derivative at $x=0$ is $L=\cos 0=1$. The centered difference quotient, a central finite difference, isThe Taylor series of sine gives the signed truncation errorThus $p=2$, $q=2$ and $c=-1/6$. Halving $h$ divides the leading truncation error by four. Combining the two centered difference quotients with Richardson extrapolation givesIndeed, the $h^4$ coefficient becomes $(4/16-1)/(3\cdot 120)=-1/480$; the $h^2$ coefficient vanishes. This is the even-power case of Richardson extrapolation discussed in Persson's authored lecture notes.^[ref]

For $h=0.4$ and $h/2=0.2$, the actual computations areThe absolute error drops from about $2.6454\times 10^{-2}$ for $A(0.4)$ to $5.3080\times 10^{-5}$ for $R(0.4)$, an improvement by roughly a factor of 498. Refining again gives $R(0.2)\approx 0.999996670632607$, whose absolute error is about $3.3294\times 10^{-6}$. The refined truncation errors decrease by nearly four for $A$ and nearly sixteen for $R$, reflecting their second and fourth orders.

The logarithmic scale on each axis turns power laws into straight lines. The dashed reference curves are $h^2/6$ and $h^4/480$, the magnitudes of the leading truncation errors. The plotted errors are computed with 70-digit decimal arithmetic, rather than clipped to a display floor; this isolates the asymptotic expansion from roundoff error.

Worked example: integrating an exponential. Approximate the definite integral of the exponential function on $[0,1]$:For numerical integration, the composite trapezoidal rule adds the areas of trapezoids between successive samples; this is the quadrature use of the endpoint average that also occurs in the ODE trapezoidal rule. On $n$ equal intervals, with $h=1/n$, defineSumming the geometric series gives $T_n=(e-1)\frac{h}{2}\frac{e^h+1}{e^h-1}$. Expanding the exponential function near zero yieldsThe leading truncation error again has $p=2$, $q=2$. On the uniform nested grids with $n$ and $2n$ intervals, Richardson extrapolation therefore givesThis linear combination is exactly composite Simpson's rule on the finer grid, whose spacing is $h/2$. On each coarse panel, an old endpoint receives weight $h/6$ and the new midpoint weight $2h/3$; shared old endpoints receive weight $h/3$. These are the usual Simpson's rule weights $1,4,2,\dots ,4,1$ multiplied by $h/6$. The equivalence is explained directly in LeMesurier's university notes.^[ref]

For two coarse panels and four fine panels,The absolute errors are about $0.0356493$, $0.00894008$ and $0.0000370135$, respectively. Reusing the fine-grid samples makes Richardson extrapolation especially attractive here: the resulting Simpson's rule needs no samples beyond those already used by $T_{2n}$.

When the improvement is justified. Richardson extrapolation requires the stated asymptotic expansion and the correct positive $p$. If the actual leading exponent is $r\ne p$, its coefficient after combination is $c(2^{p-r}-1)/(2^p-1)$, so the leading truncation error generally remains. Merely knowing $A(h)-L=O(h^p)$ does not supply a common coefficient $c$ or a higher-order remainder. A discontinuity of a function, a singular endpoint, or unrelated changes between grids can invalidate the assumed asymptotic expansion; refinement alone does not guarantee improvement for every numerical problem.

Roundoff error and errors in the sampled values survive the cancellation of truncation error. If the coarse and fine values have absolute perturbations bounded by $\epsilon$, the perturbation in $R(h)$ is bounded by $(2^p+1)\epsilon /(2^p-1)$; for $p=2$ this is $5\epsilon /3$. In floating-point arithmetic, a centered difference quotient can also amplify sample errors through division by $h$, so an indefinitely smaller $h$ need not be more accurate. Persson's notes describe this balance between truncation error and roundoff error.^[ref]

The same cancellation idea appears in step-doubling error estimation. For an order-$p$ one-step method for an ordinary differential equation, two half steps each contribute local truncation error of order $h^{p+1}$, so their accumulated leading local error has ratio $2^{-p}$ to one full step. This explains the denominator $2^p-1$ there; the appropriate exponent follows from that local error model, rather than automatically from a generic use of Richardson extrapolation.

Riemann's differential equation is a second-order ordinary differential equation with three regular singular points, normalized to zero, one and infinity by a Möbius transformation. Its local characteristic exponents are encoded by the Papperitz symbol and constrained by the Fuchs relation.

A rough differential equation interprets a differential equation using the higher-order data of its driving rough path. For $2<p<3$, its local expansion includes $V_i(Y_s)X_{s,t}^i+D{V}_j(Y_s)V_i(Y_s){\mathbb{X}}_{s,t}^{ij}$, with a remainder controlled at order greater than one. For a smooth driver it is the usual controlled ordinary differential equation; for enhanced Brownian motion it is the corresponding Stratonovich equation.

A Runge-Kutta method advances an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, by sampling its vector field at several trial states within one time step. These samples are Runge-Kutta stages; a weighted combination of their slopes produces the next state. The construction improves on the single tangent line used by the Euler method while requiring evaluations of $f$, rather than explicit higher derivatives of the solution. It applies to scalar equations and systems of ordinary differential equations alike. The coefficient choices distinguish methods such as Heun's method, the classical fourth-order Runge-Kutta method and implicit Runge-Kutta methods.

For $s$ stages, write $A=(a_{ij})$ for the stage matrix, $c_i$ for relative time nodes and $b_i$ for output weights. A time step of length $h$ computesHere $k_i$ is a slope, with units of state per time; $h{k}_i$ is a state increment. Another common convention writes $K_i=h{k}_i$, so its stage states are $Y_i=y_n+{\sum }_j{a}_{ij}{K}_j$ and its output is $y_n+{\sum }_i{b}_i{K}_i$. These are equivalent conventions: including $h$ both inside $K_i$ and again in the output would count the time step twice. The textbook formulation uses the increment convention.^[ref]

Usually the Butcher tableau obeys $c_i={\sum }_j{a}_{ij}$, called internal consistency. Stage $i$ then combines slopes over an effective time $c_{i}h$. The second-order conditions with independent Runge-Kutta abscissae explain what changes when time nodes are specified independently. A Runge-Kutta stage is generally an auxiliary approximation, not a separate accepted solution value at its node. Several stages can even have the same time node but different state arguments, as in the classical fourth-order Runge-Kutta method.

The useful idea is to approximate the integral $y(t_n+h)-y(t_n)={\int }_{t_n}^{t_n+h}f(t,y(t))dt$ without knowing the intermediate exact trajectory. Trial states provide places to sample $f$, and Butcher order conditions make the resulting combination agree with the exact Taylor expansion. More stages alone do not guarantee a higher order of a numerical method: their coefficients must satisfy the appropriate conditions. Embedded Runge-Kutta pairs reuse these same samples for a second output and a practical local error estimator.

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.

For the massless Klein-Gordon equation on a Kerr black hole, the covariant wave operator separates into radial and angular ordinary differential equations. With $K=(r^2+a^2)\omega -am$ and angular separation constant $\Lambda$, the radial equation is $(\Delta R^{\prime }{)}^{\prime }+[K^2/\Delta -a^2{\omega }^2+2am\omega -\Lambda ]R=0$. The angular equation is $(\sin \theta {\Theta }^{\prime }{)}^{\prime }/\sin \theta +[a^2{\omega }^2{\cos }^2\theta -m^2/{\sin }^2\theta +\Lambda ]\Theta =0$. No extra radial factor is included in the displayed ansatz.

A second-order ordinary differential equation involves a second derivative as its highest derivative. In explicit form $y^{\prime \prime }=f(x,y,y^{\prime })$, introducing $v=y^{\prime }$ converts it into the first-order system $y^{\prime }=v$, $v^{\prime }=f(x,y,v)$. Under local regularity and Lipschitz continuity in the dependent variables, an initial value problem is specified by both $y(x_0)$ and $y^{\prime }(x_0)$. The linear case is a second-order linear differential equation.