Concept

Logistic population growth

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Statement

Let \(N(t)\) be the size of a single closed, unstructured, well-mixed population, treated as a differentiable function of continuous time, and suppose that the per-capita growth rate declines affinely with density, \(N^{-1}\,dN/dt=r\left(1-N/K\right)\), with the intrinsic rate of increase \(r\gt 0\) and the carrying capacity \(K\gt 0\) both constant, the response to density instantaneous, and no immigration, emigration, demographic stochasticity or Allee effect. Then \(N\) obeys the logistic equation \(dN/dt=rN\left(1-N/K\right)\), whose solution through any \(N_0\gt 0\) exists, is unique, and is \(N(t)=K\left[1+\left(\left(K-N_0\right)/N_0\right)e^{-rt}\right]^{-1}\). The state \(N=K\) is globally asymptotically stable on \(\left(0,\infty\right)\) with linear relaxation rate \(r\), the state \(N=0\) is unstable with rate \(+r\); for \(0\lt N_0\lt K\) the trajectory is strictly increasing, and for \(0\lt N_0\lt K/2\) it is sigmoid, with a single inflection at \(N=K/2\), at which the absolute growth rate attains its maximum \(rK/4\) — a trajectory started at or above \(K/2\) is concave throughout and has no inflection; and the logit \(\ln\left[N/\left(K-N\right)\right]\) is exactly linear in \(t\) with slope \(r\).

Why it matters

Exponential growth is the correct model of a population that is not limited by anything, and no population is not limited by anything. The logistic equation is the minimal repair: it is what you get by adding the single leading correction to the per-capita rate and stopping there. That minimality is its point. It has exactly two parameters, both with clean operational meanings — \(r\) is measured from the early, uncrowded phase, \(K\) from the plateau — and it converts them into a closed-form trajectory, a unique stable equilibrium, a rate of return to it, and a density at which production is maximal. Every one of those is a quantity an ecologist, a fisheries manager or a microbiologist actually needs, and none of them is available from the exponential model at all.

It is also the load-bearing single-species component of most of the rest of population ecology. putting this equation’s density-dependent term in place of the exponential prey term of predator–prey dynamics is exactly what converts that model’s neutral cycles into a damped approach to a stable coexistence point; the Lotka–Volterra competition equations underlying Competition and the niche are the logistic with one extra term per competitor; r- and K-selection takes its two parameters as the axes of a life-history classification and so inherits whatever they do or do not mean; the sigmoid of the bacterial growth curve is its exponential and stationary phases; and the epidemic curve of the SIR model is logistic exactly in the limit of negligible recovery. Because population density feeds back on fitness, the equation is also where density-dependent selection enters evolutionary theory: the sign of selection on a life-history trait can reverse between the \(N\ll K\) and \(N\approx K\) regimes, which is the formal content of the claim that natural selection maximises different things in growing and in saturated populations. Practically, the same two-parameter fit sets fishing quotas, forecasts a culture’s harvest time, and produces the extinction thresholds used in conservation planning — and, as the worked examples below show, some of those uses are far more dangerous than they look.

Hypotheses

The population is closed, unstructured and well mixed, and \(N\) is large enough to treat as a continuous, differentiable variable.“Unstructured” means one number suffices: no age, size, sex or spatial classes. An age-structured population obeys \(r\) only asymptotically, once the stable age distribution is reached, and shows damped oscillations of period roughly one generation on the way there — transients that the logistic cannot produce. Continuity is a large-\(N\) idealisation: at \(N=3\) the equation predicts fractional individuals and, worse, assigns zero probability to the extinction that a real population of three faces every generation.

The per-capita growth rate is affine and strictly decreasing in current density: \(g(N)=r\left(1-N/K\right)\).This is the entire content of the model and it is an assumption, not a derivation. It is a first-order Taylor truncation of an arbitrary smooth declining \(g\) (Step 2), so it is defensible near any equilibrium and indefensible far from one. The theta-logistic \(g=r\left(1-\left(N/K\right)^{\theta}\right)\) and the Gompertz form \(g=r\ln\left(K/N\right)\) fit sigmoid data equally well and place the inflection at \(K\left(1+\theta\right)^{-1/\theta}\) and \(K/e\) instead of \(K/2\).

The response to density is instantaneous: growth at time \(t\) depends on \(N(t)\), not on \(N\) at any earlier time.Real crowding acts through resource depletion, maternal condition and juvenile recruitment, all of which take time. Replacing \(N(t)\) by \(N(t-\tau)\) in the bracket gives Hutchinson’s delay logistic, which is stable only while \(r\tau\lt\pi/2\) and beyond that oscillates permanently (Fails without; Problem 4). No smooth overshoot of \(K\) is possible in the undelayed model, so any observed overshoot falsifies this hypothesis and not the others.

\(r\) and \(K\) are constants of the environment and the organism over the interval modelled.Seasonality, weather and resource pulses make \(K\) a function of time, and a population tracking a fluctuating \(K\) with relaxation time \(1/r\) lags it by roughly \(1/r\) and never sits on it. When \(K(t)\) varies faster than \(1/r\) the population responds to a time-average; when it varies more slowly the population tracks it quasi-statically. Neither regime is the model derived here.

The per-capita rate is maximal as \(N\to 0^{+}\): there is no Allee effect and no positive density dependence at any density.Mate limitation, cooperative defence, satiation of predators and obligate group foraging all make \(g\) increase over some low-density range. A strong Allee effect, \(g=r\left(N/A-1\right)\left(1-N/K\right)\), inserts an unstable threshold at \(N=A\) below which the population goes deterministically extinct — a qualitative change of phase portrait, not a quantitative correction, and the reason logistic-based recovery targets are unsafe for depleted social species.

Dynamics are deterministic: no demographic or environmental noise.The logistic is the mean-field limit of a density-dependent birth–death process with per-capita birth rate \(b\) and per-capita death rate \(d+cN\) (Discussion), and that process has \(N=0\) as its only absorbing state. Extinction is certain in the stochastic model and impossible in the deterministic one; the deterministic \(K\) is a quasi-stationary mode, and the mean time to extinction from it grows roughly exponentially in \(K\), which is why the idealisation is harmless for a chemostat and dangerous for a population of forty whales.

Proof

Steps 1–3 set up the equation and show in what sense it is forced rather than chosen. Steps 4–7 solve it exactly, twice, and settle existence and uniqueness. Steps 8–11 read the qualitative content off the solution: monotonicity, the inflection, the maximum of production, and the stability of both equilibria. Steps 12–13 give the two forms in which the result is actually used — the logit linearisation for fitting, and the scaling law for elapsed time between two densities.

\[ g(N)\;\equiv\;\frac{1}{N}\frac{dN}{dt},\qquad \left[g\right]=\mathrm{time}^{-1},\qquad g\equiv r\ \Longrightarrow\ N(t)=N_0e^{rt} \]

Definitions, and the model being repaired. The per-capita growth rate \(g\) is the demographic primitive: it is births minus deaths per individual per unit time, it has dimensions of inverse time, and it is the quantity a life table estimates. Density independence is the statement that \(g\) is constant, and it integrates immediately to unbounded exponential growth. Note what is and is not wrong with that: nothing, over any interval short compared with \(1/r\) and at densities far below any limit. A

\[ g(N)=g(0)+g'(0)N+O\left(N^2\right),\qquad r\equiv g(0)\gt 0,\qquad g'(0)\lt 0,\qquad K\equiv-\frac{g(0)}{g'(0)}\ \Longrightarrow\ g(N)=r\left(1-\frac{N}{K}\right)+O\left(N^{2}\right) \]

Why this particular correction and not another. Expand any smooth, strictly decreasing per-capita rate about zero density and truncate at first order; the two Taylor coefficients are then renamed. So \(r\) is defined as the per-capita rate extrapolated to vanishing density — not the rate ever observed — and \(K\) is defined as the density at which the linear extrapolation of \(g\) crosses zero, which coincides with the true zero of \(g\) only when the truncation is exact. The logistic is therefore the unique two-parameter model consistent with the leading behaviour of every model in its class, which is exactly the sense in which it is a null model rather than a mechanism. B

\[ \frac{dN}{dt}=f(N)\equiv rN\left(1-\frac{N}{K}\right)=rN-\frac{r}{K}N^{2},\qquad \left[r\right]=\mathrm{time}^{-1},\ \left[K\right]=\mathrm{individuals},\ \left[f\right]=\mathrm{individuals}\;\mathrm{time}^{-1} \]

Multiply through by \(N\). The quadratic form is the more revealing one: growth is a linear birth term minus a term proportional to the number of pairwise encounters, which is what a mass-action account of interference or resource depletion produces. Both terms are rates of individuals per unit time, so the equation is dimensionally homogeneous, and \(r/K\) carries dimensions of (individual \(\times\) time)\(^{-1}\) — it, not \(K\), is the primitive coefficient of density dependence. A

\[ u\equiv\frac{N}{K},\qquad \tau\equiv rt\ \Longrightarrow\ \frac{du}{d\tau}=u\left(1-u\right) \]

Non-dimensionalise before solving. Both parameters are absorbed: \(K\) sets the unit of population and \(1/r\) the unit of time, so there is no free parameter left and every logistic trajectory in the universe is the same curve, differing only in how the axes are labelled. This is the reason a single fitted \(r\) predicts every timescale in the problem — doubling time, time to the inflection, time from ten to ninety per cent of \(K\) — with no further information. B

\[ v\equiv\frac{1}{N}\ \Longrightarrow\ \frac{dv}{dt}=-\frac{1}{N^{2}}\frac{dN}{dt}=-\frac{r}{N}\left(1-\frac{N}{K}\right)=-rv+\frac{r}{K} \]

The substitution that makes the equation linear. The logistic is a Bernoulli equation with exponent two, and \(v=N^{-1}\) is the standard change of variable for it. The resulting equation for the reciprocal density — equivalently, for the resource or space per individual — is linear with constant coefficients, which is why the whole problem is solvable in closed form while almost no other nonlinear population model is. B

\[ \frac{d}{dt}\left(ve^{rt}\right)=\frac{r}{K}e^{rt}\ \Longrightarrow\ v(t)=\frac{1}{K}+\left(\frac{1}{N_0}-\frac{1}{K}\right)e^{-rt}\ \Longrightarrow\ N(t)=\frac{K}{1+Ae^{-rt}},\qquad A\equiv\frac{K-N_0}{N_0} \]

Integrating factor \(e^{rt}\), then invert. The constant is fixed by \(v(0)=1/N_0\). Read the middle expression demographically: the reciprocal density relaxes exponentially, at rate \(r\), from its initial value to \(1/K\) — so it is space per individual, not population size, that decays exponentially towards its limit. The final form is the logistic function; \(A\) is dimensionless and equals \(\left(K-N_0\right)/N_0\), the initial ratio of unfilled to filled capacity. A

\[ f\in C^{\infty}\left(\mathbb{R}\right)\ \Longrightarrow\ f\ \text{locally Lipschitz}\ \Longrightarrow\ \text{the solution through}\ N_0\ \text{is unique};\qquad 0\lt N_0\lt K\ \Longrightarrow\ A\gt 0\ \Longrightarrow\ 1+Ae^{-rt}\gt 1\ \ \forall t\ge 0 \]

Existence, uniqueness and invariance. \(f\) is a polynomial, hence locally Lipschitz, so Picard–Lindelöf gives a unique maximal solution; the expression in Step 6 satisfies the equation and the initial condition, so it is that solution. Because \(f(0)=f(K)=0\), the constant solutions \(N\equiv 0\) and \(N\equiv K\) are trajectories, and uniqueness forbids any other trajectory from crossing them: the interval \(\left(0,K\right)\) is forward and backward invariant, which is the rigorous version of “a population below \(K\) never overshoots it”. C

\[ 0\lt N\lt K\ \Longrightarrow\ \frac{dN}{dt}\gt 0;\qquad N\gt K\ \Longrightarrow\ A\in\left(-1,0\right),\ 1+Ae^{-rt}\in\left(1+A,1\right)\subset\left(0,1\right),\ \frac{dN}{dt}\lt 0;\qquad \lim_{t\to\infty}N(t)=K \]

Monotonicity and the limit, in both directions of approach. The sign of \(dN/dt\) is the sign of \(1-N/K\), so trajectories rise from below and fall from above, and in neither case do they cross. In the closed form, \(e^{-rt}\to 0\) whatever the sign of \(A\), so the denominator tends to one; the approach is monotone, not oscillatory, and \(K\) is reached only in the limit — never at any finite time. A

\[ \frac{d^{2}N}{dt^{2}}=f'(N)\frac{dN}{dt}=r^{2}N\left(1-\frac{N}{K}\right)\left(1-\frac{2N}{K}\right)\ \Longrightarrow\ \frac{d^{2}N}{dt^{2}}=0\ \text{on}\ \left(0,K\right)\ \iff\ N=\frac{K}{2} \]

The inflection, by the chain rule applied to an autonomous equation. On \(\left(0,K\right)\) the first two factors are positive, so the curvature has the sign of \(1-2N/K\): the trajectory is convex below \(K/2\) and concave above it. This single sign change is what makes the curve sigmoid, and it is a theorem about the model, not an observation — a fitted logistic must place its inflection at exactly half of its own asymptote, which is a falsifiable prediction and the cleanest way to reject the logistic in favour of Gompertz or the theta-logistic. B

\[ \max_{0\le N\le K}f(N)=f\left(\frac{K}{2}\right)=r\cdot\frac{K}{2}\cdot\frac{1}{2}=\frac{rK}{4};\qquad t^{\ast}=\frac{1}{r}\ln A=\frac{1}{r}\ln\!\left(\frac{K-N_0}{N_0}\right) \]

Maximum production, and when it happens. \(f\) is a downward parabola with roots \(0\) and \(K\), so its vertex is at the midpoint; the value \(rK/4\) is the largest number of individuals per unit time the population can produce, and it is the theoretical maximum sustainable yield. The time of the inflection follows from setting \(N\left(t^{\ast}\right)=K/2\), i.e. \(Ae^{-rt^{\ast}}=1\); it is positive only if \(N_0\lt K/2\), and if the population starts above half capacity there is no inflection at all and growth decelerates from the outset. A

\[ f'(N)=r\left(1-\frac{2N}{K}\right):\qquad f'(0)=+r,\qquad f'(K)=-r;\qquad \xi\equiv N-K\ \Longrightarrow\ \frac{d\xi}{dt}=-r\xi+O\left(\xi^{2}\right)\ \Longrightarrow\ \xi(t)\approx\xi_0e^{-rt} \]

Stability, and the rate that goes with it. Linearising about each equilibrium gives eigenvalues \(+r\) at \(N=0\) (unstable; a single immigrant grows) and \(-r\) at \(N=K\) (stable). Both are hyperbolic, so the linearisation determines the local phase portrait, and Step 8 upgrades local to global: \(K\) is globally asymptotically stable on \(\left(0,\infty\right)\). The same number \(r\) therefore sets three things at once — the speed of growth when rare, the resilience time \(1/r\) after a perturbation at \(K\), and the width of the sigmoid — which is a strong and testable structural claim of the model. B

\[ \frac{1}{u\left(1-u\right)}=\frac{1}{u}+\frac{1}{1-u}\ \Longrightarrow\ \frac{d}{d\tau}\ln\!\left(\frac{u}{1-u}\right)=1\ \Longrightarrow\ \ln\!\left(\frac{N}{K-N}\right)=\ln\!\left(\frac{N_0}{K-N_0}\right)+rt \]

The same solution obtained by separation, in the form used for fitting. Partial fractions turn the separated integral into two logarithms whose difference is the logit of \(u\). The logit of the occupied fraction is therefore exactly linear in time with slope \(r\) and intercept fixed by \(N_0\) — the logistic analogue of a Lineweaver–Burk plot. Given an estimate of \(K\) from the plateau, \(r\) is then a linear regression, and curvature in that plot is direct evidence against the affine hypothesis of Step 2. B

\[ t_2-t_1=\frac{1}{r}\ln\!\left[\frac{N_2}{K-N_2}\cdot\frac{K-N_1}{N_1}\right];\qquad N\left(t^{\ast}+s\right)+N\left(t^{\ast}-s\right)=\frac{K}{1+e^{-rs}}+\frac{K}{1+e^{rs}}=K \]

Elapsed time between two densities, and the exact symmetry of the curve. Subtracting two copies of Step 12 eliminates \(N_0\) entirely, so the time to go from one density to another depends only on \(r\), \(K\) and the two densities — the standard tool for predicting harvest or recovery times. The second identity says the logistic curve has odd symmetry about its inflection point: the time spent climbing from \(\alpha K\) to \(K/2\) equals the time from \(K/2\) to \(\left(1-\alpha\right)K\). Real growth curves are usually left-skewed relative to this, which is one more reason the theta-logistic with \(\theta\lt 1\) often fits better. C

Result

\[ \frac{dN}{dt}=rN\left(1-\frac{N}{K}\right)\qquad\Longleftrightarrow\qquad N(t)=\frac{K}{1+\left(\dfrac{K-N_0}{N_0}\right)e^{-rt}},\qquad \left(\frac{dN}{dt}\right)_{\max}=\frac{rK}{4}\ \ \text{at}\ \ N=\frac{K}{2} \]

Reading. Per-capita growth falls linearly from \(r\) at vanishing density to zero at \(K\), so total growth is a downward parabola in \(N\): nearly exponential while the population is rare, maximal at half capacity, and vanishing as capacity is filled. The population approaches \(K\) monotonically and asymptotically from either side, with deviations decaying like \(e^{-rt}\), and the fraction occupied has a logit that is exactly linear in time with slope \(r\).

Scope. Units: \(N\) and \(K\) in individuals (or individuals per unit area, volume or biomass — any consistent measure of abundance); \(r\) per unit time; \(dN/dt\) in individuals per unit time; \(A\), \(u=N/K\) and \(\tau=rt\) dimensionless. Holds for one closed, unstructured, well-mixed, deterministic population with constant \(r\) and \(K\), an instantaneous and affinely decreasing per-capita rate, and no Allee effect. It is a leading-order model near an equilibrium, and its three sharp predictions — inflection at exactly \(K/2\), maximum production exactly \(rK/4\), and return rate exactly \(r\) — are the first things to test before using it quantitatively.

Corollaries & converses

Early doubling time. While \(N\ll K\) the bracket is within \(N/K\) of one, so \(N\approx N_0e^{rt}\) and the doubling time is \(\ln 2/r\). At \(N=0.1K\) the logistic rate is already ten per cent below exponential, which sets the practical ceiling on the range usable for estimating \(r\) by log-linear regression.
The whole sigmoid takes \(\left(2/r\right)\ln 9\) to traverse. From Step 13 with \(N_1=0.1K\) and \(N_2=0.9K\), the time from ten to ninety per cent of capacity is \(\left(2\ln 9\right)/r\approx 4.394/r\), independent of \(K\) and of \(N_0\). Two populations with the same \(r\) trace the same S-curve at the same speed however different their capacities.
Maximum sustainable yield, and the trap in it. Harvesting at constant quota \(h\) gives equilibria \(N_{\pm}=\tfrac{K}{2}\left[1\pm\sqrt{1-4h/(rK)}\right]\): a stable one at \(N_+\) and an unstable extinction threshold at \(N_-\). The two collide in a saddle-node bifurcation at \(h=rK/4\), so the maximum sustainable yield is attained exactly at the point where the safety margin is zero (Example 2).
Effort control is safer than quota control. With harvest proportional to abundance, \(dN/dt=rN\left(1-N/K\right)-qEN\), the equilibrium \(N^{\ast}=K\left(1-qE/r\right)\) is stable for every \(qE\lt r\) and slides continuously to zero. The same peak yield \(rK/4\) occurs at \(qE=r/2\), but overshooting the effort erodes the stock instead of collapsing it (Problem 3).
The equation solves its own inverse problem. Three abundances at equal spacing determine both parameters exactly, because \(1/N-1/K\) is geometric in \(t\): \(1/K=\left(v_1v_3-v_2^{2}\right)/\left(v_1+v_3-2v_2\right)\) with \(v_i=1/N_i\), and \(r\) follows from the common ratio (Problem 2).
Density dependence of selection. Compare two types with \(g_i=r_i\left(1-N/K_i\right)\). If they differ in one parameter only, nothing reverses: a larger \(r\) at equal \(K\), or a larger \(K\) at equal \(r\), gives the higher \(g\) at every density. Selection can therefore change sign only under a trade-off, \(r_1\gt r_2\) with \(K_1\lt K_2\), and then the two lines cross once, at \(N_c=\left(r_1-r_2\right)/\left(r_1/K_1-r_2/K_2\right)\in\left(0,K_1\right)\): the high-\(r\) type is favoured below \(N_c\), the high-\(K\) type above it. That crossing is the entire formal basis of r- and K-selection — and it evaporates if \(r\) and \(K\) are not independently variable, which mechanistically they are not (Problem 5).
Converse fails: sigmoid does not imply logistic. Gompertz, theta-logistic, Richards and Weibull growth curves are all monotone and sigmoid, and over a typical noisy time series they are close to indistinguishable. What distinguishes the logistic is the location of the inflection at \(K/2\) and the symmetry of Step 13; a fit that reproduces the plateau but places the inflection at \(0.37K\) is Gompertz evidence, not logistic evidence.
Converse fails: a plateau does not imply a carrying capacity. A population held at constant abundance by a predator, a pathogen or a harvester is at an equilibrium of a larger system in which \(f(N)\gt 0\); calling that plateau \(K\) and inferring \(rK/4\) as its production overestimates the true surplus by an amount equal to the removal rate.

Fails without

Instantaneous response dropped — the delayed regime: Hutchinson’s equation \(dN/dt=rN\left(1-N(t-\tau)/K\right)\) linearises about \(K\) to \(d\xi/dt=-r\xi\left(t-\tau\right)\), whose characteristic equation \(\lambda=-re^{-\lambda\tau}\) has a pair of roots crossing the imaginary axis at \(r\tau=\pi/2\). Below that threshold the approach to \(K\) is damped and oscillatory rather than monotone; above it, a Hopf bifurcation produces a permanent limit cycle of period \(4\tau\) at onset. A population with \(r=0.9\,\mathrm{yr}^{-1}\) tolerates a lag of only \(1.75\) years before cycling begins (Problem 4). Delayed density dependence through juvenile recruitment is the standard explanation for cyclic ungulate and insect populations, and it is invisible in the undelayed model, which cannot overshoot \(K\) at all.
Continuous time dropped — the discrete, chaotic regime: for non-overlapping generations, \(N_{t+1}=N_t+rN_t\left(1-N_t/K\right)\) is conjugate to the logistic map with parameter \(\lambda=1+r\) under \(N=\left(\left(1+r\right)K/r\right)y\). The fixed point \(N^{\ast}=K\) has multiplier \(1-r\), so it is stable only for \(0\lt r\lt 2\); at \(r=2\) it period-doubles, and the cascade accumulates at \(r\approx 2.570\), beyond which the dynamics are chaotic — bounded, aperiodic, and sensitive to initial conditions, with no approach to \(K\) whatever. The continuous logistic, being a one-dimensional autonomous flow, is incapable of any of this: its trajectories are monotone by Step 8. Whether a population cycles can therefore depend on nothing more than whether its generations overlap.
Positive density dependence at low density — the Allee regime: with \(dN/dt=rN\left(N/A-1\right)\left(1-N/K\right)\) and \(0\lt A\lt K\), the origin becomes stable and \(N=A\) is an unstable threshold. Extinction is now the fate of any population knocked below \(A\), the return rate at \(K\) changes from \(r\) to \(r\left(K/A-1\right)\), and maximum sustainable yield is no longer \(rK/4\). Passenger-pigeon-scale social species, broadcast spawners diluted below fertilisation range, and plants dependent on a specialist pollinator all sit in this regime, and a logistic-based recovery plan for them systematically overstates safety.
Constant environment dropped — the tracking regime: if \(K=K(t)\) varies on timescale \(T\), the population tracks it with lag \(\sim 1/r\). For \(T\gg 1/r\) the approximation \(N\approx K(t)\) is quasi-static and usable; for \(T\ll 1/r\) the population responds to a weighted average and, because \(f\) is concave in \(K\) at fixed \(N\), the long-run mean abundance falls below the mean of \(K\) — environmental variance costs abundance. Estimating \(K\) from a plateau observed in one good year and applying it to a bad one is the commonest way this failure appears in practice.
Determinism dropped — the small-population regime: in the birth–death process with rates \(bN\) and \(\left(d+cN\right)N\), \(N=0\) is absorbing, so extinction occurs with probability one. The deterministic \(K\) becomes a quasi-stationary distribution with mean near \(K\) and variance of order \(K\), and the expected time to extinction grows roughly exponentially in \(K\) — astronomically long for a chemostat at \(K=10^{9}\), and of order decades for a vertebrate population at \(K=50\). The deterministic model gives no warning of this because it has removed the only mechanism that produces it.

Common errors

“\(K\) is a property of the environment.” \(K=-g(0)/g'(0)\) is a ratio of two demographic coefficients of a particular species in a particular environment (Step 2). Anything that raises adult mortality lowers \(r\) and therefore lowers \(K\) in the same proportion; two species in the identical habitat have different \(K\). The mechanistic derivation of Problem 5 makes this explicit: doubling the total resource changes both \(r\) and \(K\), and by different factors.
“\(r\) is the population’s growth rate.” \(r\) is the per-capita rate extrapolated to zero density, and it is never observed directly. The observed per-capita rate is \(r\left(1-N/K\right)\), which is smaller by a factor that is already \(0.5\) at the inflection. Fitting \(\ln N\) against \(t\) over data extending past \(0.1K\) biases \(r\) downwards.
“Growth is fastest when the population is smallest.” Two different maxima are being confused. The per-capita rate is maximal as \(N\to 0\); the absolute rate \(dN/dt\) is maximal at \(N=K/2\) and tends to zero at both ends (Step 10). Harvesters care about the second, evolutionary biologists usually about the first.
“The population reaches \(K\) at time \(t^{\ast}\)” — or reaches it at all. \(t^{\ast}\) is the inflection, where the population is at exactly \(K/2\). Convergence to \(K\) is asymptotic: \(K-N\) decays like \(e^{-rt}\) and vanishes at no finite time (Step 8). “Time to carrying capacity” is only ever shorthand for time to a stated fraction of it.
“The curve is S-shaped, so growth is logistic.” Sigmoidicity is generic; the logistic makes the far stronger claims that the inflection sits at \(K/2\), that production peaks at \(rK/4\), and that the curve is odd-symmetric about the inflection (Step 13). Test those, not the shape.
“Fit \(N\) against \(t\) by eye and read off \(K\) from the last few points.” The plateau is the least informative part of the record: there \(dN/dt\approx 0\), so the data constrain \(K\) but carry almost no information about \(r\). Estimate \(K\) from the plateau, then obtain \(r\) as the slope of the logit (Step 12), where every point contributes.
“Harvest at maximum sustainable yield to be both efficient and safe.” At \(h=rK/4\) the stable and unstable equilibria have merged (Corollaries), \(f'\left(K/2\right)=0\), and recovery from a downward fluctuation is no longer exponential but algebraic. Any error in \(r\), \(K\) or the catch tips the stock onto a trajectory to zero. Managing at MSY is managing exactly at a bifurcation point.
“A negative \(dN/dt\) above \(K\) means the model has broken down.” It does not: \(N\gt K\) is a perfectly good part of the phase line, the solution of Step 6 remains valid with \(A\in\left(-1,0\right)\), and the population descends monotonically to \(K\). What the model genuinely cannot produce is a population that rises above \(K\) from below.

Discussion

Pierre-François Verhulst published the equation in 1838, explicitly as a correction to the geometric growth of Malthus’s Essay on the Principle of Population of 1798 — the same essay that, read by Darwin and by Wallace, supplied the surplus of offspring over resources that makes natural selection inevitable. Verhulst’s work was then forgotten for eighty years and rediscovered by Raymond Pearl and Lowell Reed in 1920, who fitted it to United States census data and read off an asymptote just under two hundred million — a figure the United States passed in the 1960s, which is the standard cautionary tale about extrapolating a fitted \(K\) beyond the range of the data. Georgii Gause’s Paramecium and yeast cultures in the early 1930s gave the first clean experimental sigmoids, and the same experiments produced the competition results that became the principle of competitive exclusion.

The logistic can be derived rather than postulated, and the derivations sharpen what its parameters mean. From below, it is the mean-field limit of a birth–death process with per-capita birth rate \(b\) and per-capita death rate \(d+cN\): the expected change is \(N\left(b-d-cN\right)\), so \(r=b-d\) and \(K=\left(b-d\right)/c\). From the side, it is what an explicit consumer–resource model collapses to when the resource is a conserved pool partitioned between free and sequestered forms and uptake is linear in free resource, which is the low-concentration limit of Michaelis–Menten kinetics (Problem 5). Both derivations deliver the same lesson: \(K\) is a composite, \(K=r/c\), and the natural parameters are \(r\) and the density-dependence coefficient \(c\), not \(r\) and \(K\). This matters because the \(r\)–\(K\) parameterisation makes the two look independently adjustable, and a great deal of life-history theory has been built on treating them that way.

The model’s statistical behaviour is worse than its mathematics. In the theta-logistic family, \(\theta\) and \(r\) are strongly correlated in the likelihood, so time series of the length ecologists actually have — a few decades of annual counts, with observation error comparable to process error — typically cannot distinguish \(\theta=1\) from \(\theta=0.3\) or \(\theta=3\), and estimates of \(\theta\) from such data are known to be badly biased. Since \(\theta\) controls where the inflection sits and therefore what fraction of \(K\) maximises yield, this is not a cosmetic uncertainty: the harvest policy that follows from a fitted model can be wrong by a factor of two while the fit itself looks excellent. Fitting is further compromised because logistic residuals are autocorrelated by construction — the state at \(t+1\) is a deterministic function of the state at \(t\) — so ordinary least squares on the logit understates the standard errors, and state-space methods separating process from observation noise are the current standard. The honest summary is that the logistic is an excellent conceptual instrument, a good interpolator, and an unreliable extrapolator.

Common misconceptions. That a population at \(K\) is static: \(K\) is a dynamic equilibrium at which births equal deaths, with turnover continuing at rate \(b\) per individual per unit time, exactly as in the equilibrium of island biogeography, where immigration balances extinction while species identities keep changing. That density dependence must act on deaths: it may act on births, on emigration, or on any combination, and the logistic cannot tell you which — only \(c=r/K\) is identifiable from abundance data, and separating its demographic components requires vital rates, not counts. And that the model is about resources at all: nothing in Step 2 mentions food. Territoriality, disease transmission, cannibalism and interference all produce a per-capita rate declining in density, and every one of them yields the same equation.

Worked examples

Example 1. A closed \(50\ \mathrm{mL}\) culture of a ciliate is inoculated from a small stock and sampled by counting fixed aliquots. The counts are \(21.7\), \(238.1\) and \(474.1\) individuals \(\mathrm{mL}^{-1}\) at \(t=2\), \(6\) and \(10\) days, and the culture is left running until it plateaus at \(500\) individuals \(\mathrm{mL}^{-1}\). Estimate \(r\) and the inoculum density, then predict the maximum rate of production, the day of fastest growth, and the density at day \(8\).

\[ K=500\ \mathrm{mL}^{-1}\ \text{(observed plateau)},\qquad y(t)\equiv\ln\!\left(\frac{N}{K-N}\right)=y_0+rt \]

Symbols before numbers. Step 12 says the logit of the counts is exactly linear in time with slope \(r\), so with \(K\) taken from the plateau the two remaining unknowns enter linearly and the estimation problem is an ordinary regression. Both \(N\) and \(K-N\) are in \(\mathrm{mL}^{-1}\), so \(y\) is dimensionless and \(r\) comes out in \(\mathrm{day}^{-1}\). A

\[ y(2)=\ln\frac{21.7}{478.3}=-3.0929,\qquad y(6)=\ln\frac{238.1}{261.9}=-0.0953,\qquad y(10)=\ln\frac{474.1}{25.9}=2.9072 \]

Transform the three counts. Already the numbers are informative: the successive differences are \(2.998\) and \(3.003\) over equal four-day intervals, so the points are collinear to within the rounding of the counts themselves. That collinearity is the test of the model, and it has passed; curvature here would have sent us to Gompertz or the theta-logistic instead. B

\[ \hat r=\frac{\sum\left(t_i-\bar t\right)\left(y_i-\bar y\right)}{\sum\left(t_i-\bar t\right)^{2}}=\frac{\left(-4\right)\left(-2.9993\right)+0+\left(4\right)\left(3.0008\right)}{16+0+16}=\frac{24.000}{32}=0.7500\ \mathrm{day}^{-1} \]

Least-squares slope, with \(\bar t=6\ \mathrm{day}\) and \(\bar y=-0.0937\). The estimate is \(r=0.750\ \mathrm{day}^{-1}\) to three figures. Sanity check on magnitude: a doubling time of \(\ln 2/r=0.924\ \mathrm{day}\) in the uncrowded phase is entirely ordinary for a ciliate feeding on bacteria. A

\[ \hat y_0=\bar y-\hat r\,\bar t=-0.0937-4.5000=-4.5937\ \Longrightarrow\ \frac{N_0}{K-N_0}=e^{-4.5937}=0.010115\ \Longrightarrow\ N_0=\frac{0.010115\,K}{1+0.010115}=5.01\ \mathrm{mL}^{-1} \]

Back-transform the intercept. Solving \(N_0/\left(K-N_0\right)=e^{y_0}\) for \(N_0\) gives \(N_0=Ke^{y_0}/\left(1+e^{y_0}\right)\) — the logistic function again, which it must be, since \(N_0=N(0)\). The inoculum is recovered as \(5.0\) individuals \(\mathrm{mL}^{-1}\) purely by extrapolation, and \(A=\left(K-N_0\right)/N_0=98.9\), which is \(99.0\) to the precision the counts support. B

\[ \left(\frac{dN}{dt}\right)_{\max}=\frac{rK}{4}=\frac{\left(0.750\ \mathrm{day}^{-1}\right)\left(500\ \mathrm{mL}^{-1}\right)}{4}=93.8\ \mathrm{mL}^{-1}\,\mathrm{day}^{-1}\ \ \text{at}\ \ N=250\ \mathrm{mL}^{-1} \]

Maximum production, from Step 10. Units multiply correctly: an inverse time times a density gives a density per unit time. In the whole \(50\ \mathrm{mL}\) culture that is \(4.69\times10^{3}\) new individuals per day at peak, and it is the largest standing harvest the culture can sustain if cropped back to half capacity each time. A

\[ t^{\ast}=\frac{1}{r}\ln\!\left(\frac{K-N_0}{N_0}\right)=\frac{\ln 99.0}{0.750\ \mathrm{day}^{-1}}=\frac{4.5951}{0.750}=6.13\ \mathrm{day} \]

The inflection, from Step 10. Consistency check against the raw data: the observation at \(t=6\ \mathrm{day}\) is \(238.1\ \mathrm{mL}^{-1}\), just below \(K/2=250\ \mathrm{mL}^{-1}\), so the culture crosses half capacity a little after day six, as the formula says. A

\[ N(8)=\frac{K}{1+Ae^{-rt}}=\frac{500}{1+99.0\,e^{-6.00}}=\frac{500}{1+99.0\left(0.0024788\right)}=\frac{500}{1.2454}=401\ \mathrm{mL}^{-1} \]

Prediction at a day not sampled, from the Result. Compare the exponential model with the same \(r\) and \(N_0\): \(5.0\,e^{6}=2.02\times10^{3}\ \mathrm{mL}^{-1}\), four times the capacity of the flask. By day eight the density-dependent brake has removed eighty per cent of the growth the unconstrained model predicts. B

\[ \Delta t_{10\%\to 90\%}=\frac{2\ln 9}{r}=\frac{4.394}{0.750\ \mathrm{day}^{-1}}=5.86\ \mathrm{day} \]

The width of the sigmoid, from Step 13 with \(N_1=0.1K\), \(N_2=0.9K\). It depends on \(r\) alone: a culture with the same \(r\) but a hundred-fold larger flask would take the same \(5.86\) days to cross that range, merely starting later. This is the practical content of the non-dimensionalisation in Step 4, and it is what makes \(r\) — not \(K\) — the parameter that sets every timescale in the experiment. C

\[ \hat r=0.750\ \mathrm{day}^{-1},\quad K=500\ \mathrm{mL}^{-1},\quad \hat N_0=5.01\ \mathrm{mL}^{-1};\qquad t^{\ast}=6.13\ \mathrm{day},\quad \left(\frac{dN}{dt}\right)_{\max}=93.8\ \mathrm{mL}^{-1}\mathrm{day}^{-1},\quad N(8)=401\ \mathrm{mL}^{-1} \]

Reading. Three counts and a plateau determine the whole trajectory. The logit of the occupied fraction rises by exactly \(0.750\) per day, the culture crosses half capacity on day \(6.1\), produces at most \(94\) individuals \(\mathrm{mL}^{-1}\,\mathrm{day}^{-1}\) at that moment, and spends \(5.9\) days crossing from a tenth to nine tenths of capacity.

Scope. Only the interval sampled is warranted. The linearity of the three logits tests the affine hypothesis over \(4\) to \(95\) per cent of \(K\) and says nothing about behaviour below that range, where an Allee effect or a lag phase of adjustment to the medium would show up, nor about the eventual senescence and death phase of a closed batch culture, which the logistic does not model at all: after the plateau, waste accumulation drives \(N\) back down and no constant \(K\) can describe it.

Example 2. A fish stock is assessed at \(r=0.40\ \mathrm{yr}^{-1}\) and \(K=8.0\times10^{5}\ \mathrm{t}\) of spawning biomass. (a) Find the maximum sustainable yield and the biomass at which it occurs. (b) A quota of \(7.5\times10^{4}\ \mathrm{t\,yr^{-1}}\) — nominally a cautious \(94\) per cent of MSY — is set. Find the equilibria, classify them, and state the resilience at each. (c) The stock is surveyed at \(2.9\times10^{5}\ \mathrm{t}\) with the quota unchanged. How long until it is gone? (d) Under an immediate moratorium, how long to rebuild from \(3.5\times10^{5}\) to \(4.5\times10^{5}\ \mathrm{t}\)?

\[ \mathrm{MSY}=\frac{rK}{4}=\frac{\left(0.40\ \mathrm{yr}^{-1}\right)\left(8.0\times10^{5}\ \mathrm{t}\right)}{4}=8.0\times10^{4}\ \mathrm{t\,yr^{-1}}\ \ \text{at}\ \ N=\frac{K}{2}=4.0\times10^{5}\ \mathrm{t} \]

Part (a), directly from Step 10. Biomass is a legitimate abundance measure provided \(r\) was estimated on the same scale; the surplus-production models used in stock assessment are exactly this equation written in tonnes. A

\[ rN\left(1-\frac{N}{K}\right)=h\ \Longrightarrow\ N^{2}-KN+\frac{hK}{r}=0\ \Longrightarrow\ N_{\pm}=\frac{K}{2}\left[1\pm\sqrt{1-\frac{4h}{rK}}\right] \]

Part (b), in symbols first. A constant quota is subtracted from the parabola, so equilibria are where the horizontal line \(h\) cuts it. The discriminant \(1-4h/(rK)\) is non-negative exactly when \(h\le rK/4\): there is no equilibrium at all above MSY, which is the mathematical statement that an over-quota fishery has nowhere to settle. B

\[ \frac{4h}{rK}=\frac{4\left(7.5\times10^{4}\right)}{\left(0.40\right)\left(8.0\times10^{5}\right)}=0.9375,\quad \sqrt{1-0.9375}=0.250\ \Longrightarrow\ N_{+}=5.0\times10^{5}\ \mathrm{t},\quad N_{-}=3.0\times10^{5}\ \mathrm{t} \]

Numbers. A six per cent reduction below MSY buys an equilibrium only \(25\) per cent above the MSY biomass, and it creates an extinction threshold at \(3.0\times10^{5}\ \mathrm{t}\) — only \(40\) per cent below the operating point. The gap \(N_+-N_-=K\sqrt{1-4h/(rK)}\) has infinite slope in \(h\) at \(h=rK/4\), so the last few per cent of the approach to MSY costs most of the margin; conversely, buying margin back is slow — halving the distance from MSY widens the gap by only \(\sqrt{2}\). B

\[ \frac{d}{dN}\left[f(N)-h\right]=r\left(1-\frac{2N}{K}\right):\qquad N_{+}:\ 0.40-0.50=-0.10\ \mathrm{yr}^{-1};\qquad N_{-}:\ 0.40-0.30=+0.10\ \mathrm{yr}^{-1} \]

Classification and resilience, from Step 11 (the constant \(h\) drops out on differentiating). \(N_+\) is stable with relaxation time \(1/0.10=10\ \mathrm{yr}\) — already four times slower than the unharvested \(1/r=2.5\ \mathrm{yr}\), because harvesting flattens the effective restoring force. \(N_-\) is unstable with the same rate: a stock displaced below it departs, doubling its distance from the threshold every \(\ln 2/0.10=6.9\ \mathrm{yr}\). At \(h=\mathrm{MSY}\) both eigenvalues would be zero and recovery would be algebraic rather than exponential. B

\[ \frac{dN}{dt}=-\frac{r}{K}\left(N-N_{-}\right)\left(N-N_{+}\right)\ \Longrightarrow\ T=\int_{0}^{N_a}\frac{K\,dN}{r\left(N_{-}-N\right)\left(N_{+}-N\right)}=\frac{K}{r\left(N_{+}-N_{-}\right)}\ln\!\left[\frac{N_{+}-N_a}{N_{-}-N_a}\cdot\frac{N_{-}}{N_{+}}\right] \]

Part (c), in symbols. Factorising the harvested quadratic through its two roots and separating gives an integral done by partial fractions, \(\left[\left(N_--N\right)\left(N_+-N\right)\right]^{-1}=\left(N_+-N_-\right)^{-1}\left[\left(N_--N\right)^{-1}-\left(N_+-N\right)^{-1}\right]\). Below \(N_-\) both factors in the denominator are positive and \(dN/dt\lt 0\), so the stock declines; unlike the unharvested model, it reaches zero in finite time, because \(dN/dt\to-h\lt 0\) as \(N\to 0\). C

\[ \frac{K}{r\left(N_{+}-N_{-}\right)}=\frac{8.0\times10^{5}}{\left(0.40\right)\left(2.0\times10^{5}\right)}=10\ \mathrm{yr},\qquad T=10\ln\!\left[\frac{2.1\times10^{5}}{1.0\times10^{4}}\cdot\frac{3.0}{5.0}\right]=10\ln 12.6=25.3\ \mathrm{yr} \]

Numbers. A stock only \(3.4\) per cent below its threshold is extinguished in twenty-five years by a quota advertised as six per cent below MSY. The decline is slow at first — \(dN/dt=-1.05\times10^{3}\ \mathrm{t\,yr^{-1}}\) at the survey, just over a third of a per cent of the stock per year, indistinguishable from survey noise — and accelerates to \(-7.5\times10^{4}\ \mathrm{t\,yr^{-1}}\) at the end. This is the characteristic signature of a fold bifurcation crossed unnoticed. B

\[ t=\frac{1}{r}\ln\!\left[\frac{N_2}{K-N_2}\cdot\frac{K-N_1}{N_1}\right]=\frac{1}{0.40}\ln\!\left[\frac{4.5}{3.5}\cdot\frac{4.5}{3.5}\right]=\frac{2\ln\left(1.2857\right)}{0.40}=1.26\ \mathrm{yr} \]

Part (d), from Step 13 with \(h=0\). Rebuilding across the most productive part of the curve is fast — fifteen months — because the stock is near \(K/2\) where surplus production is greatest. Rebuilding the same absolute increment near the top of the curve, from \(7.0\times10^{5}\) to \(7.5\times10^{5}\ \mathrm{t}\), takes \(\left(1/0.40\right)\ln\left[\left(7.5/0.5\right)\left(1.0/7.0\right)\right]=1.91\ \mathrm{yr}\), and the last stretch to \(K\) takes forever. A

\[ \mathrm{MSY}=8.0\times10^{4}\ \mathrm{t\,yr^{-1}}\ \text{at}\ 4.0\times10^{5}\ \mathrm{t};\quad h=7.5\times10^{4}:\ N_{+}=5.0\times10^{5}\ \mathrm{t}\ \left(\lambda=-0.10\ \mathrm{yr}^{-1}\right),\ N_{-}=3.0\times10^{5}\ \mathrm{t}\ \left(\lambda=+0.10\ \mathrm{yr}^{-1}\right);\quad T_{\text{collapse}}=25.3\ \mathrm{yr},\quad t_{\text{rebuild}}=1.26\ \mathrm{yr} \]

Reading. Constant-quota harvesting converts a globally stable population into a bistable one: a stable stock at \(5.0\times10^{5}\ \mathrm{t}\), an extinction threshold at \(3.0\times10^{5}\ \mathrm{t}\), and nothing to stop the fall once the threshold is crossed. Recovery under a moratorium is rapid where production is high; collapse under an unchanged quota is slow enough to be mistaken for noise for a decade.

Scope. A deterministic surplus-production model with perfectly known \(r\) and \(K\) and a quota taken exactly. Real assessments carry uncertainty of tens of per cent in both parameters, and since the threshold \(N_-\) depends on \(r\) and \(K\) through \(\sqrt{1-4h/(rK)}\), a twenty per cent overestimate of \(r\) can move it by more than the entire margin computed here. Recruitment variability, an Allee effect at low spawning biomass, and age structure all make the true situation worse, never better; none of them is in this model.

Problems

A population of grazing mammals introduced to an island has \(r=0.15\ \mathrm{yr}^{-1}\) and \(K=1200\) individuals, starting from \(N_0=100\). Find (a) the doubling time in the early phase, (b) the year of fastest growth and the growth rate then, (c) the population at \(t=10\ \mathrm{yr}\), and (d) the population the exponential model would have predicted at that time. Comment on when the two models first differ by more than ten per cent.
Solution
(a) While \(N\ll K\), \(N\approx N_0e^{rt}\), so the doubling time is \(t_2=\ln 2/r=0.6931/0.15=4.62\ \mathrm{yr}\).
(b) Fastest growth is at the inflection, \(N=K/2=600\) individuals (Step 10). Its timing is \(t^{\ast}=\left(1/r\right)\ln\left[\left(K-N_0\right)/N_0\right]=\left(1/0.15\right)\ln\left(1100/100\right)=\ln\left(11\right)/0.15=2.3979/0.15=16.0\ \mathrm{yr}\). The rate there is \(rK/4=\left(0.15\right)\left(1200\right)/4=45\) individuals per year.
(c) \(A=\left(K-N_0\right)/N_0=11.0\), so \(N(10)=1200/\left(1+11.0\,e^{-1.5}\right)\). With \(e^{-1.5}=0.22313\), the denominator is \(1+2.4544=3.4544\) and \(N(10)=347\) individuals.
(d) Exponential: \(N_0e^{rt}=100\,e^{1.5}=448\) individuals. The logistic is \(22.5\) per cent lower.
Divergence, in symbols first. Using \(K=N_0\left(1+A\right)\),
\[ \frac{N(t)}{N_0e^{rt}}=\frac{K}{N_0e^{rt}\left(1+Ae^{-rt}\right)}=\frac{1+A}{e^{rt}+A}. \]
Setting this to \(0.90\) gives \(e^{rt}+A=\left(1+A\right)/0.90=13.33\), so \(e^{rt}=2.333\), \(rt=0.8473\) and \(t=5.65\ \mathrm{yr}\). At that time \(N=210\) and \(N_0e^{rt}=233\). The models part company after less than six years — about \(1.2\) early doubling times — while the population is still only \(17.5\) per cent of \(K\). The general rule follows from Step 2: the fractional discrepancy in the per-capita rate is exactly \(N/K\), so a ten per cent error in rate arrives at ten per cent of carrying capacity, and a ten per cent error in the trajectory shortly after.
A population is censused three times at equal intervals of \(h=2\ \mathrm{yr}\), giving \(N_1=150.0\), \(N_2=324.2\) and \(N_3=566.0\) individuals. Assuming logistic growth, derive an estimator for \(K\) and \(r\) from three equally spaced counts and evaluate it. Why does this method fail if the intervals are unequal, and what is the practical weakness of the estimator?
Solution
Derivation. By Step 6 the reciprocal \(v=1/N\) obeys \(v(t)=1/K+\left(v_0-1/K\right)e^{-rt}\). Writing \(c=1/K\) and \(\rho=e^{-rh}\), equal spacing makes \(v_i-c\) a geometric sequence: \(v_2-c=\rho\left(v_1-c\right)\) and \(v_3-c=\rho\left(v_2-c\right)\). Eliminating \(\rho\) gives \(\left(v_2-c\right)^{2}=\left(v_1-c\right)\left(v_3-c\right)\). The \(c^{2}\) terms cancel, leaving \(v_2^{2}-2cv_2=v_1v_3-c\left(v_1+v_3\right)\), hence
\[ c=\frac{v_1v_3-v_2^{2}}{v_1+v_3-2v_2},\qquad K=\frac{1}{c},\qquad \rho=\frac{v_2-c}{v_1-c},\qquad r=-\frac{\ln\rho}{h}. \]
Evaluation. \(v_1=1/150.0=6.66667\times10^{-3}\), \(v_2=1/324.2=3.08451\times10^{-3}\), \(v_3=1/566.0=1.76678\times10^{-3}\), all in individuals\(^{-1}\).
Numerator: \(v_1v_3-v_2^{2}=1.177853\times10^{-5}-9.51420\times10^{-6}=2.26433\times10^{-6}\).
Denominator: \(v_1+v_3-2v_2=8.43345\times10^{-3}-6.16902\times10^{-3}=2.26443\times10^{-3}\).
So \(c=2.26433\times10^{-6}/2.26443\times10^{-3}=9.99956\times10^{-4}\) and \(K=1.00\times10^{3}\) individuals.
Common ratio: \(\rho=\left(3.08451-0.99996\right)/\left(6.66667-0.99996\right)=2.08455/5.66671=0.36786\), so \(r=-\ln\left(0.36786\right)/2=1.00007/2=0.500\ \mathrm{yr}^{-1}\).
Check. Forward-substituting \(K=1000\), \(r=0.500\), \(N_1=150\) into the Result gives \(N(2)=324.2\) and \(N(4)=566.0\), reproducing the data.
Why equal spacing is needed. The whole method rests on \(v_i-c\) being a geometric sequence, which requires the exponentials \(e^{-rt_i}\) to be in constant ratio — true only when the \(t_i\) are equally spaced. With unequal spacing the three equations still determine \(c\), \(\rho\) and \(v_0\), but no longer by eliminating \(\rho\) linearly; one must solve a transcendental equation numerically or fit by least squares.
Weakness. The estimator is a ratio of two small differences. Substituting \(v_i=c+d\rho^{\,i-1}\) with \(d=v_1-c\) shows that the numerator is \(cd\left(1-\rho\right)^{2}\) and the denominator \(d\left(1-\rho\right)^{2}\): both vanish quadratically as \(\rho\to 1\), so the relative error in \(c\) is amplified by a factor of order \(\left(1-\rho\right)^{-2}\), which blows up whenever the sampling interval is short compared with \(1/r\). Even at the comfortable spacing here (\(\rho=0.368\)) the amplification is about fivefold: perturbing \(N_2\) by one per cent, to \(327.4\), gives \(v_2=3.05437\times10^{-3}\), numerator \(2.44940\times10^{-6}\), denominator \(2.32472\times10^{-3}\), hence \(c=1.05362\times10^{-3}\) and \(K=949\) — a five per cent error in \(K\) from a one per cent error in one count, and a one per cent error the other way gives \(K=1062\). Three points are the minimum, never the right number; in practice one fits all the data by the logit regression of Step 12 or by a state-space method that separates process from observation error.
Compare constant-quota and constant-effort harvesting for a stock with \(r=0.60\ \mathrm{yr}^{-1}\) and \(K=5.0\times10^{4}\ \mathrm{t}\), where effort-based harvest is \(qEN\) with catchability \(q=1.0\times10^{-4}\) per boat-year. (a) Find the equilibrium biomass and the sustained yield as functions of effort. (b) Find the effort giving maximum yield, and that yield. (c) Compute the equilibrium and yield at \(E=4000\) boat-years, and explain the result. (d) What is the effort at which the stock is driven to zero, and how does the approach differ from the constant-quota case?
Solution
(a) Setting \(rN\left(1-N/K\right)=qEN\) and cancelling \(N\gt 0\) gives \(r\left(1-N/K\right)=qE\), so \(N^{\ast}=K\left(1-qE/r\right)\). Sustained yield is \(Y=qEN^{\ast}=qEK\left(1-qE/r\right)\). Note the structure: the yield is a downward parabola in the dimensionless effort \(qE/r\), just as production is a parabola in \(N\).
(b) \(dY/d\left(qE\right)=K\left(1-2qE/r\right)=0\) gives \(qE=r/2\), i.e. \(E=r/\left(2q\right)=0.60/\left(2\times10^{-4}\right)=3.0\times10^{3}\) boat-years. Then \(N^{\ast}=K/2=2.5\times10^{4}\ \mathrm{t}\) and \(Y=\left(0.30\right)\left(2.5\times10^{4}\right)=7.5\times10^{3}\ \mathrm{t\,yr^{-1}}\), which equals \(rK/4\) as it must — the same peak of the same production curve, reached by a different control.
(c) At \(E=4.0\times10^{3}\), \(qE=0.40\ \mathrm{yr}^{-1}\), so \(N^{\ast}=5.0\times10^{4}\left(1-0.40/0.60\right)=5.0\times10^{4}/3=1.67\times10^{4}\ \mathrm{t}\) and \(Y=0.40\times1.67\times10^{4}=6.67\times10^{3}\ \mathrm{t\,yr^{-1}}\). Thirty-three per cent more effort yields eleven per cent less fish, and leaves the stock a third lower: past MSY, effort and catch move in opposite directions, which is the standard signature of an overcapitalised fishery.
(d) \(N^{\ast}=0\) when \(qE=r\), i.e. \(E=r/q=6.0\times10^{3}\) boat-years. Stability: for the effort model \(d/dN\left[rN\left(1-N/K\right)-qEN\right]=r-qE-2rN/K\), which at \(N^{\ast}\) equals \(-\left(r-qE\right)\lt 0\) for all \(qE\lt r\). So every equilibrium below the collapse effort is stable, and \(N^{\ast}\) slides continuously to zero as effort rises: there is no threshold, no bistability, and no fold. Under a constant quota, by contrast, the stable and unstable equilibria collide at \(h=rK/4\) and the stock collapses discontinuously from \(K/2\) to zero. The price of the safety is that the yield near collapse comes from a nearly exhausted stock, and that catch rates fall as the stock falls, which quotas are designed to prevent.
A herbivore population responds to density with a one-generation delay, so that \(dN/dt=rN\left(t\right)\left[1-N\left(t-\tau\right)/K\right]\). (a) Linearise about \(N=K\) and obtain the characteristic equation. (b) Find the condition on \(r\tau\) for stability, and the period of oscillation at onset. (c) For \(r=0.90\ \mathrm{yr}^{-1}\), find the critical lag. (d) If the true lag is \(\tau=2.0\ \mathrm{yr}\), what behaviour is expected, and with what period?
Solution
(a) Put \(N=K+\xi\) with \(\left|\xi\right|\ll K\). Then \(1-N\left(t-\tau\right)/K=-\xi\left(t-\tau\right)/K\), so \(d\xi/dt=r\left(K+\xi\right)\left[-\xi\left(t-\tau\right)/K\right]=-r\xi\left(t-\tau\right)+O\left(\xi^{2}\right)\). Substituting the trial solution \(\xi=e^{\lambda t}\) gives \(\lambda e^{\lambda t}=-re^{\lambda\left(t-\tau\right)}\), hence the characteristic equation \(\lambda=-re^{-\lambda\tau}\). Unlike the undelayed case this is transcendental and has infinitely many roots.
(b) Stability is lost when a pair of roots crosses the imaginary axis, so set \(\lambda=i\omega\) with \(\omega\gt 0\): \(i\omega=-r\left[\cos\left(\omega\tau\right)-i\sin\left(\omega\tau\right)\right]\). Real part: \(0=-r\cos\left(\omega\tau\right)\), so \(\omega\tau=\pi/2\) (the first crossing). Imaginary part: \(\omega=r\sin\left(\omega\tau\right)=r\). Combining, the boundary is \(r\tau=\pi/2\), with stability for \(r\tau\lt\pi/2\) and a Hopf bifurcation to a limit cycle above it. At onset \(\omega=r\), so the period is \(T=2\pi/\omega=2\pi/r=4\tau\), since \(\tau=\pi/\left(2r\right)\) there.
(c) \(\tau_{c}=\pi/\left(2r\right)=3.1416/\left(2\times0.90\right)=1.75\ \mathrm{yr}\).
(d) With \(\tau=2.0\ \mathrm{yr}\), \(r\tau=1.80\gt\pi/2=1.571\), so the equilibrium is unstable and the population settles onto a sustained limit cycle rather than approaching \(K\). Near onset the period is \(2\pi/r=6.98\ \mathrm{yr}\), close to \(4\tau_c=6.98\ \mathrm{yr}\); with \(r\tau\) fifteen per cent above threshold the realised period is somewhat longer than this and the amplitude is large, so the linear estimate should be read as a lower bound of roughly seven years. The qualitative point is decisive: the undelayed logistic cannot oscillate at all (Step 8), so an observed cycle of a few generations around a stable mean is evidence for delayed density dependence, not against density dependence.
Derive the logistic equation from an explicit resource model, and use it to test whether \(r\) and \(K\) are independent parameters. A closed algal culture contains total nitrogen \(T=200\ \mu\mathrm{mol\,L^{-1}}\), partitioned between free nitrogen \(R\) and nitrogen locked in cells at a fixed cell quota \(c=0.50\ \mu\mathrm{mol}\) per \(10^{6}\) cells. Cells divide at a rate proportional to free nitrogen, \(\varepsilon a R\) with \(\varepsilon a=6.0\times10^{-3}\ \mathrm{L\,\mu mol^{-1}\,day^{-1}}\), and die at rate \(m=0.30\ \mathrm{day}^{-1}\). (a) Show the model is exactly logistic and identify \(r\) and \(K\). (b) Evaluate them. (c) Find the free nitrogen remaining at equilibrium and comment. (d) The medium is prepared at \(T=400\ \mu\mathrm{mol\,L^{-1}}\) instead. Recompute \(r\) and \(K\), and state the consequence for life-history theory.
Solution
(a) Conservation of nitrogen in a closed culture gives \(R=T-cN\), with \(N\) in \(10^{6}\) cells \(\mathrm{L}^{-1}\) so that \(cN\) is in \(\mu\mathrm{mol\,L^{-1}}\). The population equation is \(dN/dt=N\left(\varepsilon aR-m\right)=N\left[\varepsilon a\left(T-cN\right)-m\right]=N\left(\varepsilon aT-m\right)\left[1-\dfrac{\varepsilon acN}{\varepsilon aT-m}\right]\), which is the logistic with
\[ r=\varepsilon aT-m,\qquad K=\frac{\varepsilon aT-m}{\varepsilon ac}=\frac{r}{\varepsilon ac}. \]
The density-dependence coefficient of Step 3 is \(r/K=\varepsilon ac\), the product of uptake efficiency and cell quota — and it, unlike \(K\), is independent of the resource supply. Growth requires \(r\gt 0\), i.e. \(T\gt m/\left(\varepsilon a\right)\).
(b) \(r=\left(6.0\times10^{-3}\right)\left(200\right)-0.30=1.20-0.30=0.90\ \mathrm{day}^{-1}\). \(K=0.90/\left[\left(6.0\times10^{-3}\right)\left(0.50\right)\right]=0.90/\left(3.0\times10^{-3}\right)=300\), i.e. \(3.0\times10^{2}\times10^{6}=3.0\times10^{8}\) cells \(\mathrm{L}^{-1}\).
(c) \(R^{\ast}=T-cK=200-\left(0.50\right)\left(300\right)=200-150=50\ \mu\mathrm{mol\,L^{-1}}\). Equivalently, from \(\varepsilon aR^{\ast}=m\), \(R^{\ast}=m/\left(\varepsilon a\right)=0.30/6.0\times10^{-3}=50\ \mu\mathrm{mol\,L^{-1}}\) — which contains no \(T\) at all. The equilibrium free-resource level is a property of the consumer alone, not of the supply: this is the \(R^{\ast}\) rule, and it is what makes resource competition predictable, since the species with the lowest \(R^{\ast}\) draws the resource below every rival’s break-even point and excludes it (Competition and the niche). Note also that \(K\) is not the density at which resource runs out: a quarter of the nitrogen is still dissolved at equilibrium, and the population stops because uptake at that concentration only balances death.
(d) With \(T=400\ \mu\mathrm{mol\,L^{-1}}\): \(r=\left(6.0\times10^{-3}\right)\left(400\right)-0.30=2.40-0.30=2.10\ \mathrm{day}^{-1}\), and \(K=2.10/\left(3.0\times10^{-3}\right)=700\), i.e. \(7.0\times10^{8}\) cells \(\mathrm{L}^{-1}\). Doubling the resource multiplied \(r\) by \(2.33\) and \(K\) by \(2.33\) as well — they moved together, in exactly the ratio \(K=r/\left(\varepsilon ac\right)\) fixed by the two consumer traits.
Consequence: \(r\) and \(K\) are not independent axes here. Any environmental change acting through \(T\) moves both proportionally, and only changes in \(\varepsilon a\) or \(c\) — genuinely heritable consumer traits — can move them apart, by changing \(\varepsilon ac\). A life-history theory built on trading “high \(r\)” against “high \(K\)” (r- and K-selection) therefore needs the trade-off to sit in the underlying traits, not in the fitted parameters; a population with a larger \(r\) and a larger \(K\) than another is not a contradiction, it is the ordinary consequence of a richer medium. This is the central mechanistic objection to the \(r\)–\(K\) parameterisation, and it is why modern treatments write the equation as \(dN/dt=N\left(r-cN\right)\) with \(c\) as the primitive.