= Logistic differential equation
{title2=$N'=rN(1-N/K)$}
= Logistic equation
{synonym}
The <logistic differential equation> is the <nonlinear ordinary differential equation>
$$
\frac{dN}{dt}=rN\left(1-\frac{N}{K}\right),\qquad r>0,\quad K>0,\quad N(0)=N_0\geq0.
$$
As 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.
\Image[/logistic-growth-trajectories.png]
{title=Logistic trajectories approach the carrying capacity from below or above}
{description=Here $K=1000$ and $r=0.4$ per day. The <initial conditions> are $N_0=100,500,1200$; dashed horizontal lines show the two <equilibrium points> $N=0$ and $N=K$. Only the $N_0=100$ trajectory has a future <inflection point>, marked at $N=500$ and $t\approx5.49306$ days. The other marked values are $N(5)\approx450.85306$ and $N(10)\approx858.48645$ for that same trajectory.}
For $N>0$, dividing the <derivative> by abundance gives the instantaneous per-capita rate:
$$
\frac{N'}{N}=r\left(1-\frac{N}{K}\right).
$$
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 $N'$, 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 https://ocw.mit.edu/courses/18-03sc-differential-equations-fall-2011/9893a2556a1d21717cd31bbf6b29fe9a_MIT18_03SCF11_s10_1text.pdf[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 $f'(0)=r>0$ and $f'(K)=-r<0$: 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 $t\geq0$ is
$$
N(t)=\frac{K}{1+\left(\frac{K}{N_0}-1\right)e^{-rt}}.
$$
The 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 $z'=-rz$, 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>:
$$
f(N)=\frac{rK}{4}-\frac{r}{K}\left(N-\frac{K}{2}\right)^2,
\qquad \max_{N\geq0}f(N)=\frac{rK}{4}\quad\text{at }N=\frac{K}{2}.
$$
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.
\Image[/logistic-growth-rate.png]
{title=Total logistic growth is greatest at half the carrying capacity}
{description=With $K=1000$ and $r=0.4$ per day, the maximum is $100$ individuals per day at $N=500$. The <function graph> crosses zero at the <equilibrium points> $N=0$ and $N=1000$; beyond $1000$ the rate is negative. The lower <phase line> has arrows toward increasing abundance below $K$ and toward decreasing abundance above $K$, with a hollow mark for the <unstable equilibrium> and a filled mark for the <asymptotically stable> <equilibrium point>.}
The <chain rule> connects this maximum to the shape of the time trajectory through its <second derivative>:
$$
N''=f'(N)N'=r\left(1-\frac{2N}{K}\right)N'.
$$
For $0<N_0<K/2$, the trajectory initially has <convexity> and then <concavity>, with its one future <inflection point> at
$$
t_{1/2}=\frac1r\ln\left(\frac{K}{N_0}-1\right),\qquad N(t_{1/2})=\frac{K}{2}.
$$
For $N_0=K/2$, 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 https://openstax.org/books/calculus-volume-2/pages/4-4-the-logistic-equation[OpenStax's logistic-equation solution and inflection calculation]{ref}.
For the plotted example with $K=1000$, $r=0.4\,\mathrm{day}^{-1}$ and $N_0=100$, time $t$ measured in days gives
$$
N(t)=\frac{1000}{1+9e^{-0.4t}},\qquad N(5)\approx450.85306,\qquad N(10)\approx858.48645.
$$
The initial total rate is $N'(0)=0.4\cdot100\cdot0.9=36$ 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 $t_{1/2}=\ln9/0.4\approx5.49306$ 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 $p'=mp-np^2$, 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.https://www.neo-classical-physics.info/uploads/3/4/3/6/34363841/verhulst_-_law_of_population.pdf[Verhulst's 1838 paper, translated by D. H. Delphenich]{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>.
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