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A body ejecting mass at constant rate $Q>0$ and relative exhaust speed $V>0$, against quadratic drag $\alpha u^2$, has remaining mass $m=m_0-Qt$. A short-interval momentum balance gives $m\dot{u}=QV-\alpha u^2$. From rest, separation of variables givesFor final mass $m_f>0$, this formula applies until $t=(m_0-m_f)/Q$. The zero-drag limit is the rocket equation $u_f=V\log (m_0/m_f)$. Two dimensionless variables are $\alpha V/Q$ and $m_0/m_f$.

A feedback delay is a time interval between an event and its returning influence. Multiple biochemical steps can introduce an effective lag; a fixed delay is an idealization, not a universal description of signal transduction. To isolate timing, let $x$ be a signed displacement from an equilibrium point and use normalized dimensionless variables:The local removal is $-x$ and the returning influence is negative at every delay. At $d=0$ the ordinary differential equation reduces to the decay already illustrated under signaling feedback. For $d>0$ a delay differential equation requires a history on $[-d,0]$, not just an initial condition at one instant. Negative values of $x$ are displacements, not negative chemical concentrations.

An exponential function $x=e^{\lambda t}$ gives the characteristic equation of a delay differential equationPutting $\lambda =i\omega$ and equating real and imaginary parts gives $\cos (\omega d)=-1/2$ and $\omega =2\sin (\omega d)$. The first crossing with $\omega >0$ therefore occurs atFor this particular mathematical model, the equilibrium point is asymptotically stable for $0\le d<d_c$. No nonnegative real delayed feedback characteristic root can satisfy the equation; the first loss of asymptotic stability is the indicated pair of imaginary delayed feedback characteristic roots. Differentiating the characteristic equation of a delay differential equation along a root yieldsThus the pair moves into growth as the delay increases through $d_c$. The neutral mode at $d_c$ is a periodic function, but its amplitude depends on history: this linear mathematical model does not select an attracting limit cycle.

The figure plots $x(t)=e^{at}\cos (bt)$ using a delayed feedback characteristic root $\lambda =a+ib$ and that same formula on the history interval. At $d=0.7$ the displayed mode decays; at $d=1.5$ it grows. These are exact modes up to numerical root accuracy, not step responses with identical histories. Molecular bounds and a changed return law would be needed to interpret what happens after an unstable biological response becomes large.

A dimensionless quantity has zero exponent for every base physical dimension. It can be formed by dividing two quantities with the same dimensions, such as $x=r/R$ or $q=m/M$. Its value is independent of the choice of units for those dimensions. The quantities $G{M}^2/R$ and $Lt$ both have dimensions of energy, so $RLt/(G{M}^2)$ is dimensionless. A dimensionless variable is such a quantity used as an unknown or independent variable in an equation.

Homeostasis is the maintenance of usable internal conditions despite disturbances. It concerns selected regulated physiological variables, rather than keeping every part of an organism unchanged: core body temperature may remain near its usual range while skin temperature, sweating and blood pressure change. Regulation makes a relatively dependable environment for biological cells out of a changing external world. ^[ref]

A useful causal description separates a physiological sensor, a physiological integrator and a physiological effector. The return from the physiological effector's action to the sensed regulated physiological variable closes a feedback control loop. In negative feedback, that return opposes the disturbance. The insulin and glucagon articles already trace the blood glucose example; synaptic integration explains how several inputs can jointly influence a neuron. These are implementations and ingredients, not a requirement that every regulatory loop use one anatomical control center. ^[ref]

Homeostasis is useful precisely because internal activity can change: a nearly steady temperature can conceal a large increase in heat transfer. Conversely, observing a flat trace alone does not identify a regulatory mechanism. The environment might simply be constant, or the measuring instrument too insensitive; a perturbation and an independent response readout provide stronger evidence. The skin-temperature perturbation experiment below makes that distinction tangible.

The historical motivation was the puzzle of stability in changing living material. In his 1932 The Wisdom of the Body, Walter Bradford Cannon emphasized that an organism exchanges material with its surroundings and continually replaces its own structure, yet maintains relatively steady internal conditions. This authored excerpt presents homeostasis as active compensation for external and internal disturbances, rather than a static body. ^[ref]

Here is an explicitly synthetic mathematical model of a warming disturbance and its compensatory cooling. It is a teaching analogy for thermoregulation, not a measured human heat budget. Write $x=(T-T_r)/\Delta T$, where $T$ is a lumped internal temperature, $T_r$ is a fixed reference and $\Delta T>0$ is a chosen temperature scale. Let $d$ be an additional heating rate and $u$ an additional cooling rate, both scaled to the same reference rate. All three are dimensionless variables, while $\tau >0$ is a time constant in the chosen time units. The ordinary differential equation isThe term $-x$ idealizes increased passive heat transfer with increased internal temperature, around the baseline. The term $-u$ opposes warming through proportional control with feedback gain $k$. This local signed model does not treat a negative $u$ as negative sweating: it would represent decreased cooling relative to baseline. Changes of the environment and multiple body compartments are omitted.

For a constant disturbance $d=D$ and initial condition $x(0)=0$, the exponential function solution isSubstitution checks the ordinary differential equation directly. Increasing $k$ reduces the lasting offset and, in this instantaneous model, shortens the response time constant to $\tau /(1+k)$. It does not make $x_{\ast }$ zero for a nonzero disturbance and finite $k$: a persistent cooling command requires a persistent displacement under proportional control. At the equilibrium point, the extra heating is balanced by passive loss $x_{\ast }$ plus active cooling $u_{\ast }$. Zero change of temperature means balanced rates, rather than zero rates, consistent with conservation of energy.

The first diagram uses $D=1$, $\tau =4$, and a pulse from $t=2$ to $t=10$, comparing $k=0$ with $k=3$. The latter approaches $x_{\ast }=1/4$ during the pulse, compared with $1$ without active feedback control. After the pulse ends, each displacement decays as $x(10)e^{-(1+k)(t-10)/\tau }$. These are computed curves, not physiological measurements. The $k=0$ curve still has passive loss; it is not a claim that an unregulated body warms without limit.

Real physiological effectors take time to respond. To isolate that issue, set $\tau =1$ and replace the instantaneous command by another ordinary differential equation:Here the disturbance remains $1$ from $t=0$ onward and the effector time constant is also $1$. The equilibrium point is still $(x_{\ast },u_{\ast })=(1/5,4/5)$, but the characteristic polynomial is $(\lambda +1{)}^2+4$. Its eigenvalues $-1\pm 2i$ give decaying oscillations and asymptotic stability. The exact response isIn particular, $\dot{x}=e^{-t}\cos (2t)$, so the first maximum occurs at $t=\pi /4$, with $x=1/5+(2/5)e^{-\pi /4}\simeq 0.3824$. The response exceeds its eventual displacement while remaining above its baseline. The return has the same negative sign as before; the extra state changes the timing. A finite effector time constant is distinct from the fixed time lag in the delayed negative feedback model.

The logistic differential equation is the nonlinear ordinary differential equationAs a mathematical model of population dynamics, it treats abundance $N$ as a continuous nonnegative quantity. The constant $r$ has units of inverse time, while the carrying capacity $K$ has the same units as $N$. The useful idea is to combine growth resembling exponential growth at low abundance with a slowing approach to a finite abundance. The three trajectories below already show an important distinction: starting above the carrying capacity gives a decreasing curve, whereas starting below it gives an increasing one.

For $N>0$, dividing the derivative by abundance gives the instantaneous per-capita rate:It decreases linearly with $N$: it approaches $r$ at low abundance, is $r/2$ at $N=K/2$, vanishes at the carrying capacity, and is negative above it. This per-capita rate differs from the total rate $N^{\prime }$, which also counts how many individuals contribute. It is an instantaneous rate, rather than a fixed percentage increase over a whole day. The linear per-capita assumption and the resulting qualitative picture are developed in MIT 18.03SC's logistic-model notes.

Writing $f(N)=rN(1-N/K)$, the equilibrium points satisfy $f(N)=0$, so the constant solutions of a differential equation are $N(t)=0$ and $N(t)=K$. The phase line points upward for $0<N<K$ and downward for $N>K$. Also $f^{\prime }(0)=r>0$ and $f^{\prime }(K)=-r<0$: the criterion for fixed point stability for an autonomous differential equation makes zero an unstable equilibrium and $K$ an asymptotically stable equilibrium point. Every positive initial condition approaches $K$ as $t\to \infty$; $N_0=0$ stays exactly zero. Instability of zero means that a small positive population grows away from it, not that a zero population spontaneously appears.

For every $N_0>0$, the unique solution of a differential equation for all $t\ge 0$ isThe denominator is positive on this domain, including when $N_0>K$. For $0<N_0<K$, this is a translated and scaled logistic function; its extension to all real times is a sigmoid function. For $N_0>K$ the coefficient of the exponential function is negative, and the future curve instead decreases toward $K$. No nonconstant positive solution of a differential equation crosses $K$ or reaches it at finite time. A short derivation uses the dimensionless variable $z=K/N-1$: the chain rule gives $z^{\prime }=-rz$, so exponential decay yields $z(t)=(K/N_0-1)e^{-rt}$. The substitution divides by $N$, so the zero equilibrium point must be included separately.

The total growth rate is a downward-opening quadratic function of abundance. Completing the square gives its maximum without solving the differential equation:The balance is beautiful: more individuals contribute to growth, but each contributes less, and the strongest total increase occurs halfway to the carrying capacity. This is a maximum over possible abundances. A trajectory starting above $K/2$ never passes through that abundance in its future.

The chain rule connects this maximum to the shape of the time trajectory through its second derivative:For $0<N_0<K/2$, the trajectory initially has convexity and then concavity, with its one future inflection point atFor $N_0=K/2$, this point lies at the initial time, with concavity for $t>0$. For $K/2<N_0<K$, the inflection point belongs to the past extension, so the future trajectory has concavity throughout. Above $K$, the decreasing trajectory has convexity and no inflection point. The two constant solutions of a differential equation have no change of concavity. Thus an entire logistic function has an S shape, but its observed future segment need not show both bends. See also ^[ref].

For the plotted example with $K=1000$, $r=0.4{\mathrm{day}}^{-1}$ and $N_0=100$, time $t$ measured in days givesThe initial total rate is $N^{\prime }(0)=0.4\cdot 100\cdot 0.9=36$ individuals per day, and the initial per-capita rate is $0.36{\mathrm{day}}^{-1}$. The fastest total increase is $rK/4=100$ individuals per day, reached at $t_{1/2}=\ln 9/0.4\approx 5.49306$ days. At that time the per-capita rate has already fallen to $0.2{\mathrm{day}}^{-1}$.

These parameters are illustrative choices, rather than an empirical fit. The equation specifies deterministic evolution under a phenomenological assumption: it summarizes how net growth depends on abundance without separately modelling every process that affects it. The carrying capacity is a parameter for a specified setting, not a universal constant for all populations. In his 1838 paper "Notice sur la loi que la population suit dans son accroissement", Correspondance mathématique et physique 10, pp. 113–121, Pierre-François Verhulst proposed $p^{\prime }=mp-n{p}^2$, equivalent to this equation with $r=m$ and $K=m/n$. He also noted that different assumed growth-retarding terms could describe the observations similarly while giving very different limiting populations. This is an early illustration of why a good-looking curve alone does not establish a unique long-term prediction.^[ref]

The logistic differential equation describes continuous time. The logistic map instead advances in discrete steps and can exhibit a period-doubling cascade and Devaney chaos; the monotone positive trajectories here cannot. Their connection through the Forward Euler method is explained under Euler discretization of logistic growth, where a large step can introduce behaviour absent from the original differential equation.

A metabotropic receptor couples ligand recognition to intracellular signal transduction, which can regulate a separate ion channel, alter enzyme activity or change gene expression. In the usual neurotransmitter context, metabotropic receptors are G protein-coupled receptors: receptor activation is relayed by a G protein. There need not be a change in membrane potential at all. ^[ref]

Two routes should be distinguished. An activated G protein can regulate an ion channel locally at the cell membrane, without an obligatory diffusible second messenger. Alternatively, it can regulate an enzyme that changes the availability of a second messenger, such as cAMP or inositol trisphosphate. Downstream proteins then change channel opening, phosphorylation or other cellular activities. The receptor, G protein, second messenger and final target are distinct components; an activated G protein is a relay protein, rather than itself a small-molecule second messenger.^[ref]

Do not turn this useful chemical synapse classification into the claim that every receptor is a GPCR, or that every indirect response must pass through a G protein. For example, the insulin receptor is an enzyme-linked cell-surface receptor whose intracellular catalytic activity initiates protein phosphorylation. Intracellular receptors supply another category. “Metabotropic” commonly names the GPCR branch in neuroscience; the broader idea of biochemical coupling does not make all non-channel receptors structurally identical. ^[ref]

Separating recognition from execution allows signal amplification and branching. A persistently activated receptor can activate several relay proteins; a regulated enzyme can produce many second messenger molecules. As an explicitly invented count example, suppose one activated receptor activates four relay proteins during one interval, and each relay activates a distinct enzyme that produces fifteen second messenger molecules during the next interval. With sufficient substrates and no loss, this gives $4\times 15=60$ molecules. Those counts do not specify a physiological gain or response duration. A directly gated ion channel also passes many ions while open, so amplification is not an exclusive property of metabotropic receptors: always name the input and output being compared. ^[ref]

A particularly instructive electrophysiology experiment separated a chemical trigger from the ion channel carrying its eventual electric current. In 1987, Hirono, Ito and Sugiyama injected messenger RNA from rat brain into frog egg cells, then tested acquired responses to neurotensin and muscarinic acetylcholine stimulation. Under voltage clamp at $-60\mathrm{mV}$, neurotensin elicited transient, oscillatory inward electric currents after a delay of several tens of seconds. This delay refers to that response and preparation. Reversal potential measurements implicated increased permeability to chloride anions, rather than a pore specific to the applied ligand. Both neurotensin and acetylcholine responses were inhibited by pertussis toxin, which interferes with susceptible G proteins, and by intracellular EGTA, which binds free calcium. Injecting inositol trisphosphate mimicked the responses; a large prior dose of inositol trisphosphate reduced subsequent responses to either extracellular signal. These interventions implicate a shared intracellular relay and calcium-dependent channel opening.^[ref]

The informative comparison with the nicotinic acetylcholine receptor experiment is mechanistic: an extracellular ligand can produce an inward electric current either by directly gating the pore or by mobilizing an intracellular second messenger and stored calcium that activate another pore. Intracellular inositol trisphosphate bypasses reception, whereas EGTA tests the need for free calcium. However, neither inhibition nor mimicry alone identifies every molecular connection: a perturbation can affect several targets, and an injected mediator can activate a pathway without proving that it is the only natural route. The egg cell's endogenous machinery and the mixed messenger RNA preparation also limit extrapolation to an intact neuron. The experiment establishes functional coupling in its preparation, rather than the exact molecular composition or timing of every metabotropic receptor.

Even G protein-dependent channel regulation need not involve a lengthy second messenger cascade. Pfaffinger and colleagues reported in 1985 that acetylcholine-evoked potassium channel activation in embryonic chick atrial cells required intracellular GTP, a guanine-containing nucleotide, and was abolished by pertussis-toxin pretreatment. Changes in cyclic nucleotides, including cAMP, did not account for the channel response. The report discusses intrinsic activation delays of roughly $30$–$100\mathrm{ms}$, illustrating relatively rapid metabotropic receptor action. These observations support a G protein coupling step without cyclic-nucleotide second messengers; they do not by themselves exclude every possible intermediary or prove the specific subunit-channel interaction.^[ref]

A small timing model makes the extra steps visible without inventing an experimental trace. Let $u(t)$ jump from zero to one at $t=0$ and remain on. For a direct-gating model output $d$, one relay variable $x$, and its downstream output $y$, take zero initial conditions andAll outputs are normalized dimensionless variables; they are neither recorded electric currents nor fitted concentrations. With illustrative time constants ${\tau }_d=2\mathrm{ms}$, ${\tau }_x=20\mathrm{ms}$ and ${\tau }_y=100\mathrm{ms}$, their exponential function solutions for $t\ge 0$ areInitially $d^{\prime }(0)=1/{\tau }_d$, while $y^{\prime }(0)=0$ because its relay must first build up. Both outputs eventually approach one. Equal normalized final values do not imply equal biological amplitudes or absence of signal amplification; normalization deliberately removes that question.

The model's useful feature is the delayed buildup created by sequential relaxation, not its arbitrary numbers. Actual kinetics also depend on ligand delivery, relay deactivation, neurotransmitter clearance and the final target. A local G protein route can act promptly, while an ionotropic receptor can remain active during continued exposure or feed a longer-lasting intracellular response. The beauty of metabotropic receptor coupling is its flexibility: the same recognition event can adjust electrical responsiveness and other cellular tasks through different targets. To classify an observed response, combine timing with molecular identity and controlled perturbations.

Rescaling dimensional quantities by chosen characteristic scales produces dimensionless variables. The remaining dimensionless ratios identify independent parameters and possible asymptotic regimes.

In the simple pendulum equation, choose the time scale $t_{\ast }=\sqrt{\ell /g}$ and put $\tau =t/t_{\ast }$. The chain rule gives $\ddot{\theta }=(g/\ell )d^2\theta /d{\tau }^2$, so the equation becomes $d^2\theta /d{\tau }^2+\sin \theta =0$. This removes $\ell$ and $g$ from the equation, while the initial conditions still include angular amplitude and dimensionless initial angular speed. A solution can therefore be calculated once and rescaled to many physical lengths and gravity values with the same dimensionless initial data.

For a separated weighted density, write $P_n=e^{\gamma t}p(\sigma )$. Its eigenvalue equation isSet $p=\psi \exp [-{\sigma }^2/(2\kappa \tau )]$. Differentiation eliminates the first derivative of $\psi$, giving the tilted Ornstein-Uhlenbeck oscillator transformationComplete the square and introduce the dimensionless variable $x=(\sigma -\kappa n{\tau }^2)/\sqrt{\kappa \tau }$. The equation becomesComparing with the quantum harmonic oscillator gives $2E=1+\kappa n^2{\tau }^3-2\gamma \tau$. Its energies $E_m=m+1/2$ therefore correspond to ${\gamma }_m=\kappa n^2{\tau }^2/2-m/\tau$. The dominant nonnegative separated mode is the ground state, which is nodeless. ConsequentlyFor an independent check, $\log [\tilde{B}(t)/B_0]={\int }_0^t\tilde{\sigma }(s)ds$ is a centered Gaussian random variable with varianceThe exponential moment of a Gaussian linear functional gives $\mathbb{E}[B^n]=B_0^n\exp [n^{2}V(t)/2]$, whose long-time growth rate is the same ${\gamma }_n$. This exact Ornstein-Uhlenbeck multiplicative amplification formula also distinguishes the finite-time transient from the asymptotic exponential law. A general transient can contain sign-changing excited eigenfunctions in its expansion while the total weighted density remains nonnegative.

During ejection, the remaining mass is $m(t)=m_o+m_w-Qt$, and the water is exhausted at $t_c=m_w/Q$. Consider the momentum of the octopus and the small portion of water expelled during $dt$. Initially it is $mu$. At the end of this interval, the remaining body has mass $m-Qdt$ and velocity $u+du$, while the expelled mass $Qdt$ has laboratory velocity $u-V$, to first order. Hence the total final momentum isThe external impulse from quadratic drag is $-\alpha u^{2}dt$. Equating this to the change in total momentum, and dividing by $dt$, proves the equation of motionThe outgoing water's momentum flux is essential: applying $d(mu)/dt$ to the remaining body alone would omit it.

For dimensional analysis, the independent physical inputs are $m_o,m_w,Q,V,\alpha$, with dimensionsThe scales $m_o$, $m_o/Q$ and $V{m}_o/Q$ fix units of mass, time and length. In these units, the only remaining independent dimensionless variables areFor example, with $s=Qt/m_o$ and $w=u/V$, the equation of motion becomes $(1+\lambda -s)dw/ds=1-\mu w^2$, with $w(0)=0$ and ejection ending at $s=\lambda$. Thus the dimensionless terminal value depends only on $\lambda ,\mu$, proving $u_c=Vf(\lambda ,\mu )$.

For its explicit value, put $m_0=m_o+m_w$ and $U=\sqrt{QV/\alpha }$. The initially stationary solution stays below $U$, since the acceleration vanishes at $U$ and is positive below it. Separation of variables givesThe two integrals yieldThis is constant mass-loss propulsion with quadratic drag. At $t_c$, the remaining mass is $m_o$, givingThus $f(\lambda ,\mu )={\mu }^{-1/2}\tanh (\sqrt{\mu }\log (1+\lambda ))$, verifying the required dimensional form. The limit as drag tends to zero is $V\log (1+\lambda )$, the ordinary rocket equation, while the positive-drag answer is below the terminal velocity $U$.

The dimensions are $[m]=\mathsf{M}$, $[\alpha ]=\mathsf{M}{\mathsf{T}}^{-1}$, $[u_0]=\mathsf{L}{\mathsf{T}}^{-1}$, and $[g]=\mathsf{L}{\mathsf{T}}^{-2}$. Under the stated model, $m,\alpha ,u_0,g$ are the only parameters determining the ascent time. If $m^p{\alpha }^q{u}_0^r{g}^s$ is dimensionless, then $p+q=0$, $r+s=0$, and $-q-r-2s=0$, leaving one independent dimensionless variable. One convenient choice isThus dimensional analysis fixes the form, but Newton's second law must determine the function $f$.

Take upward velocity as positive. Throughout ascent, Newton's second law with linear drag gives $m\dot{v}=-mg-\alpha v$. An integrating factor or direct solution of this linear ordinary differential equation givesThe first zero of this velocity is the highest point, soThis vertical ascent under linear drag has $T<u_0/g$ for $u_0>0$, since $\log (1+\lambda )<\lambda$. The limit $\alpha \to 0$ recovers the no-drag ascent time $u_0/g$; $u_0=0$ gives $T=0$.

Take $z$ positive downwards. The leading axial velocity of a slender viscous thread is independent of radius. Incompressible flow then gives $u_r=-r{w}^{\prime }/2$, while steady mass conservation gives $(a^{2}w{)}^{\prime }=0$. The Newtonian fluid stress tensor has ${\sigma }_{rr}=-p-\mu w^{\prime }$ and ${\sigma }_{zz}=-p+2\mu w^{\prime }$. The Young–Laplace equation, neglecting axial curvature at slender order, gives ${\sigma }_{rr}=-p_e-\gamma /a$ and hence $p=p_e+\gamma /a-\mu w^{\prime }$.

The axial cut transmits both the integrated Newtonian fluid stress tensor and the circumferential surface tension. After subtracting the ambient pressure, its tensile force isThe coefficient three is the Trouton ratio for axial extension. The last term is essential: the capillary tensile force of a slender thread is $\pi \gamma a$, not $2\pi \gamma a$, because the capillary pressure subtracts half the interfacial line contribution. Axial force balance is $T^{\prime }+\pi \rho g{a}^2=0$. Together the governing equations areDefine the dimensionless variablesThen $a=a_0{W}^{-1/2}$ and the axial equation becomesFor the increasing power-law branch, all three terms have the same dependence when $\alpha =2$. Substitution leaves $1-6k^2-\Gamma k=0$. ThusThis remains valid at $\Gamma =0$, when $k=1/\sqrt{6}$. The other root is negative and describes a decreasing local branch up to its singular endpoint, not a thread accelerating downwards indefinitely. Positive surface tension decreases $k$, so it slows the fall at a given height on this branch. Thinning reduces the capillary tensile force $\pi \gamma a$. Its negative axial derivative supplies an upward capillary force on a thread segment, balancing part of the weight. The viscous tensile force therefore supplies a smaller remaining balance; on the quadratic family this reduces the stretching rate.

The condition $W(0)=1$ alone does not select a unique solution of this second-order ordinary differential equation. The boxed power law is the requested verified branch, with tensile force tending to zero at large height; an additional nozzle tension or downstream condition is required to select it physically. The slender, negligible-inertia approximation must remain valid over the height range considered.

For this chemical reaction network, the law of mass action gives the reaction rates are $k_1{s}^{2}e$, $k_{2}c$ and $k_{3}c$. Their stoichiometry givesHere the convention for $k_1$ absorbs any combinatorial factor for two identical reactants. Direct differentiation gives the two conservation lawsEliminating $e$ and $p$ leavesFor positive $s_0,e_0,k_1$, use the dimensionless variables $u=s/s_0$, $v=c/e_0$ and $\tau =k_1{e}_0{s}_{0}t$. With prime marks denoting differentiation in $\tau$, this becomeswhere

Introduce the dimensionless variablesUsing $P=\Sigma H^2{\Omega }_z^2$, vertical hydrostatic equilibrium becomes the parameter-free equationwith normalizations $\int \tilde{\rho }d\tilde{z}=1$ and $\int \tilde{p}d\tilde{z}=1$. For an isothermal atmosphere, $p=c_s^2\rho$, and these conventions give the Gaussian distribution

With the dimensionless variables $x=r/r_s$ and $y=u/c_s$, mass conservation and part (c) giveSubstitution into the Bernoulli function, together with $GM/r=2c_s^2/x$, eliminates every dimensional parameter and gives the transcendental equationEquivalently,orThe transonic branch passes through $(x,y)=(1,1)$.

Differentiating the equilibrium velocity givesOn the other hand, the definition of vortensity and $\Sigma ={\Sigma }_0{e}^{\sigma }$ giveMultiplying by $2/\Omega$, and introducing the dimensionless variablesyields

SetThen $\dot{L}=(L_0/t_0)f^{\prime }$ and $\ddot{L}=(L_0/t_0^2)f^{\prime \prime }$. Every term in the scaling equation has the common dimensions and magnitude ${\rho }^2{\sigma }^3/{\eta }^4$. Cancelling that factor yields the dimensionless equationor equivalently

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.

Signaling adaptation is a decline of a response toward its earlier level while the initiating signal remains present. Specify which response is measured and its baseline: complete signaling adaptation means eventual return to that baseline; partial signaling adaptation leaves a lasting displacement. The continuing signal distinguishes signaling adaptation from simply switching off the stimulus. Signaling adaptation describes behavior, while signaling feedback describes a causal arrangement, so neither implies the other.

By reducing a persistent response to background ligand concentration, signaling adaptation can make a biological cell sensitive to changes rather than only to absolute levels. Resetting the response helps leave room for another change over a useful range of ligand concentrations. This is valuable in chemotaxis: swimming bacteria encounter changing chemical conditions, and their signal transduction influences whether a run continues or a tumble occurs. Signaling adaptation helps separate a change encountered during motion from a lasting background. This usefulness does not imply exact return or equal sensitivity at every possible concentration. ^{[ref] [ref]}

Here is a solvable incoherent feedforward loop toy that exhibits complete signaling adaptation without signaling feedback. The input $s(t)$ jumps from $0$ to $1$ at $t=0$ and remains $1$, a Heaviside step function. Let $x$ be a fast activating branch, $a$ a slower inhibitory branch, and $r=x-a$ a readout. The activities $s,x,a,r$ are dimensionless variables; physical time is in seconds. Choose time constants ${\tau }_x=1$ second and ${\tau }_a=5$ seconds, with initial conditions $x(0)=a(0)=0$:Both branches read $s$ directly. Neither branch depends on $r$, and $a$ does not change $x$: the subtraction only defines the readout. Thus the slow opposing influence is feedforward, with no closed feedback control loop.

For $t\ge 0$, solving the two ordinary differential equations givesWriting $t$ for the numerical time in seconds, these are $x=1-e^{-t}$, $a=1-e^{-t/5}$ and $r=e^{-t/5}-e^{-t}$. Initially $r=0$, then $x$ rises faster than $a$, producing a pulse. At long times $x$ and $a$ both approach $1$, so $r$ returns to $0$ despite sustained $s=1$. For this upward step $e^{-t/5}\ge e^{-t}$, hence $r(t)\ge 0$. The striking feature of this signaling adaptation toy is that both branches remain active while the readout becomes quiet: the continuing background and the response to its arrival are separable.

Find the pulse maximum by differentiating the exponential functions. With numerical time in seconds, the derivative isThe stationary condition is $e^{-4t/5}=1/5$, so the natural logarithm givesBefore this time the derivative is positive; afterwards it is negative. Together with $r(0)=0$ and ${\lim }_{t\to \infty }r(t)=0$, this proves that it is the unique pulse maximum. As another numerical check, at $t=10$ seconds $x\simeq 0.999955$, $a\simeq 0.864665$ and $r\simeq 0.135290$: the slow branch is still catching up.

The subtraction $r=x-a$ is a mathematical readout, not a claim that an actual chemical concentration equals this difference for arbitrary inputs. Nonnegativity was established only for the specified upward step; other input histories can give a negative signed readout. Complete signaling adaptation here also relies on equal eventual branch gains: changing the readout to $r=x-0.9a$ leaves $r_{\ast }=0.1$ for $s=1$. This toy illustrates cancellation and different time constants; it does not establish molecular implementation or robustness to arbitrary parameter changes. Ma and colleagues' 2009 study separately analyzed biochemical mathematical models with negative feedback and incoherent feedforward loops, identifying additional conditions for signaling adaptation in those models. ^[ref]

A historical connection to feedback control comes from Yi, Huang, Simon and Doyle's 2000 analysis of a bacterial chemotaxis model. They identified integral action in its regulation of receptor activity: accumulated deviations alter a slower adjustment, which in turn changes the measured activity. At a stationary state, an adjustment that integrates a nonzero error cannot remain constant, explaining how zero error can follow. Their conclusion depended on stated biochemical assumptions and stable behavior; it was not a claim that every biological cell uses that mechanism. A transient pulse alone, as the incoherent feedforward loop toy shows, cannot identify the underlying signaling feedback architecture. ^[ref]