---
title: Autonomous Equations, Phase Lines, and Population Dynamics
module: First-Order Equations
moduleNumber: 2
lessonNumber: 4
order: 204
summary: >
  An autonomous equation y' = f(y) can be analyzed qualitatively without being
  solved. Its constant solutions are the zeros of f, and the sign of f between
  them fixes whether nearby solutions rise or fall, which the phase line records
  as a column of arrows. The logistic and threshold models, constant- and
  effort-proportional harvesting, and the properties nonlinear equations lose all
  follow from this reading.
topics: [First-Order Equations]
draft: false
sources:
  - book: Boyce
    ref: "§2.5 Autonomous Differential Equations and Population Dynamics; §2.4 Differences Between Linear and Nonlinear Equations"
  - book: Simmons
    ref: "Ch. 1 §4 Growth, Decay, Chemical Reactions, and Mixing"
---

An equation is **autonomous** when the independent variable does not appear
explicitly:

$$
\frac{\d y}{\d t} = f(y).
$$

The slope at a point depends only on $y$, not on $t$, so the direction field is
constant along every horizontal line and the qualitative behavior is determined on
the $y$-axis alone. Autonomous equations are separable and so solvable in
principle, but their stability and long-time behavior can be read straight off $f$
without integrating.[^boyce-25] This qualitative reading is the starting point of
the [qualitative theory](/differential-equations/nonlinear/phase-plane-autonomous-stability)
used when equations cannot be solved at all.

## Equilibria and the phase line

The simplest solutions are constant. A constant $y = y^\ast{}$ solves $y' = f(y)$
precisely when $f(y^\ast{}) = 0$, since then $\d y/\d t = 0$ for all $t$.

> **Definition (Equilibrium / critical point).** A value $y^\ast{}$ with
> $f(y^\ast{}) = 0$ is an equilibrium solution, or critical point, of
> $y' = f(y)$. The constant function $y(t) = y^\ast{}$ neither grows nor decays.

Between consecutive equilibria $f$ keeps one sign, so every solution there is
monotonic. Where $f(y) > 0$ the solution increases; where $f(y) < 0$ it decreases.
Encoding this on the $y$-axis gives the **phase line**: mark the critical points,
then draw an arrow pointing up on intervals where $f > 0$ and down where $f < 0$.

> **Definition (Stability).** An equilibrium $y^\ast{}$ is **asymptotically
> stable** if nearby solutions approach it as $t \to \infty$, and **unstable** if
> nearby solutions move away. It is **semistable** if solutions approach from one
> side and depart on the other.

The arrows read off stability directly: an equilibrium with inflowing arrows on
both sides is stable, one with outflowing arrows on both sides is unstable. A
calculus test says the same thing through the slope of $f$.

> **Theorem (Linearized stability).** If $f(y^\ast{}) = 0$, then $y^\ast{}$ is
> asymptotically stable when $f'(y^\ast{}) < 0$ and unstable when
> $f'(y^\ast{}) > 0$.

$$
% caption: The phase line marks equilibria as dots and flow direction as arrows;
% arrows converging on a point signal stability, diverging arrows instability.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % vertical phase line
  \draw[black, thick] (0,-2.6) -- (0,2.9);
  \node[black, anchor=south] at (0,2.9) {$y$};
  % equilibria (only the two critical points 0 and K)
  \fill[acc] (0,-1.8) circle (2.4pt);
  \fill[acc] (0,1.8) circle (2.4pt);
  \node[black, anchor=west] at (0.25,-1.8) {$y = 0$ (unstable)};
  \node[black, anchor=west] at (0.25,1.8) {$y = K$ (stable)};
  % flow arrows: below 0 down (away), between 0 and K up (toward K), above K down (toward K)
  \draw[->, black, thick] (0,-1.2) -- (0,-0.2);
  \draw[->, black, thick] (0,0.4) -- (0,1.2);
  \draw[->, black, thick] (0,2.6) -- (0,2.05);
  \draw[->, black, thick] (0,-2.1) -- (0,-2.55);
  \node[black, anchor=west] at (0.25,0.1) {f\/low toward $K$};
\end{tikzpicture}
$$

A double root of $f$ produces the third stability type, where the flow has the
same direction on both sides of the equilibrium.

> **Worked example.** The equation $\dfrac{\d y}{\d t} = k(1 - y)^2$ with $k > 0$
> has the single equilibrium $y = 1$, a double root of $f$. Since $f(y) \ge 0$
> everywhere, every solution is increasing: below $y = 1$ it rises toward the
> equilibrium, above $y = 1$ it rises away from it, so $y = 1$ is semistable.
> Separating,
>
> $$
> \frac{\d y}{(1 - y)^2} = k\,\d t
> \qquad\Longrightarrow\qquad
> \frac{1}{1 - y} = kt + \frac{1}{1 - y_0}.
> $$
>
> Solving for $y$,
>
> $$
> y(t) = 1 - \frac{1 - y_0}{1 + kt\,(1 - y_0)}.
> $$
>
> For $y_0 < 1$ the denominator grows and $y \to 1^-$; for $y_0 > 1$ it reaches
> zero at $t = 1/[k(y_0 - 1)]$, where the solution blows up.

$$
% caption: The phase line of y' = k(1-y)^2 has one equilibrium at y = 1 with
% upward flow on both sides: solutions below approach it, solutions above leave
% it, the signature of a semistable point.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (0,-2.6) -- (0,2.9);
  \node[black, anchor=south] at (0,2.9) {$y$};
  \fill[acc] (0,0.4) circle (2.4pt);
  \node[acc, anchor=west] at (0.25,0.4) {$y = 1$ (semistable)};
  \draw[->, black, thick] (0,-2.1) -- (0,-1.1);
  \draw[->, black, thick] (0,-0.7) -- (0,0.05);
  \draw[->, black, thick] (0,0.9) -- (0,1.7);
  \draw[->, black, thick] (0,2.15) -- (0,2.7);
  \node[black, anchor=west] at (0.25,-1.5) {approach from below};
  \node[black, anchor=west] at (0.25,1.55) {depart above};
\end{tikzpicture}
$$

## Logistic growth

The exponential model $y' = ry$ predicts unbounded growth, which no real
population sustains. Replacing the constant rate $r$ by a rate that falls as $y$
rises gives the **logistic equation**, introduced by Verhulst in 1838:

$$
\frac{\d y}{\d t} = r\left(1 - \frac{y}{K}\right)y,
$$

with $r > 0$ the intrinsic growth rate and $K > 0$ the **carrying capacity**. The
right side $f(y) = r(1 - y/K)y$ is a downward parabola with zeros at $y = 0$ and
$y = K$. Between them $f > 0$, so populations grow; above $K$, $f < 0$, so they
decline. The phase line has $y = 0$ unstable and $y = K$ stable: any positive
starting population converges to $K$.

$$
% caption: The logistic rate f(y) is a parabola positive between its roots; the
% solution curves it generates grow toward the carrying capacity K, with an
% inflection at K/2 where f peaks.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  % left panel: f(y) vs y parabola
  \begin{scope}
    \draw[->, black] (-0.2,0) -- (3.0,0) node[right] {$y$};
    \draw[->, black] (0,-0.3) -- (0,2.2) node[above] {$f(y)$};
    \draw[acc, very thick] plot[domain=0:2.4, samples=40] (\x, {1.7*(\x/2.4)*(1 - \x/2.4)*4});
    \fill[black] (0,0) circle (1.8pt);
    \fill[black] (2.4,0) circle (1.8pt);
    \node[black, anchor=north] at (2.4,-0.05) {$K$};
    \node[black, anchor=north] at (1.2,-0.05) {$\tfrac{K}{2}$};
  \end{scope}
  % right panel: solutions
  \begin{scope}[xshift=4.6cm]
    \draw[->, black] (-0.2,0) -- (3.4,0) node[right] {$t$};
    \draw[->, black] (0,-0.3) -- (0,2.6) node[above] {$y$};
    \draw[acc, thick, dashed] (0,2.1) -- (3.2,2.1);
    \node[acc, anchor=south east] at (3.2,2.1) {$y = K$};
    % S-curve from below
    \draw[black, thick] plot[domain=0:3.2, samples=60] (\x, {2.1/(1 + 5*exp(-2.1*\x))});
    % curve from above
    \draw[black, thick] plot[domain=0:3.2, samples=60] (\x, {2.1 + 0.35*exp(-2.1*\x)});
    \node[black, anchor=west] at (0.15,0.3) {$y_0 < K$};
  \end{scope}
\end{tikzpicture}
$$

The shape of the solution follows from the sign of $f$ and $f'$. Differentiating,
$y'' = f'(y)\,f(y)$, so a solution is concave up where $f$ and $f'$ agree in sign
and concave down where they disagree. For the logistic curve this gives an
**inflection point** at $y = K/2$, where the growth rate $f$ is largest.

> **Worked example.** Solve the logistic equation
> $\dfrac{\d y}{\d t} = r\left(1 - \dfrac{y}{K}\right)y$ by separation. Writing the
> right side as $\tfrac{r}{K}y(K - y)$ and separating,
>
> $$
> \frac{\d y}{y(K - y)} = \frac{r}{K}\,\d t.
> $$
>
> Partial fractions split the left side,
> $\dfrac{1}{y(K-y)} = \dfrac{1}{K}\!\left(\dfrac{1}{y} + \dfrac{1}{K-y}\right)$, so
>
> $$
> \left(\frac{1}{y} + \frac{1}{K - y}\right)\d y = r\,\d t
> \qquad\Longrightarrow\qquad
> \ln\frac{y}{K - y} = rt + c_0,
> $$
>
> valid for $0 < y < K$. Exponentiating and applying $y(0) = y_0$ fixes
> $e^{c_0} = y_0/(K - y_0)$; solving for $y$ then gives
>
> $$
> y(t) = \frac{y_0 K}{y_0 + (K - y_0)e^{-rt}}
> \;\xrightarrow{\,t \to \infty\,}\; K.
> $$
>
> The graph is an S-curve rising to the carrying capacity, with inflection at
> $y = K/2$.

However small the nonlinear term, solutions of the logistic equation stay bounded,
while solutions of the linear $y' = ry$ diverge: an arbitrarily small nonlinearity
changes the entire long-time behavior.

## Thresholds

Reversing the sign of the logistic right side gives a **threshold** model:

$$
\frac{\d y}{\d t} = -r\left(1 - \frac{y}{T}\right)y,
\qquad r, T > 0.
$$

Now $f$ is an upward-opening parabola. The equilibria are again $y = 0$ and
$y = T$, but the stability swaps: $y = 0$ is stable and $y = T$ is unstable. Below
the threshold $T$ the population declines to extinction; above it, the population
grows without bound, in fact reaching infinity in finite time. The critical value
$T$ separates two qualitatively opposite outcomes.

$$
% caption: In the threshold model the critical population T is unstable: below it
% solutions collapse to extinction, above it they escape upward, with T a wall
% between the two outcomes.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-0.2,0) -- (5.0,0) node[right] {$t$};
  \draw[->, black] (0,-0.2) -- (0,3.0) node[above] {$y$};
  % threshold line T at height 1.5
  \draw[acc, thick, dashed] (0,1.5) -- (4.8,1.5);
  \node[acc, anchor=south east] at (4.8,1.5) {$y = T$ (unstable)};
  % below threshold: decay to 0
  \draw[black, thick] plot[domain=0:4.6, samples=50] (\x, {1.3*exp(-0.6*\x)});
  % above threshold: escape upward
  \draw[black, thick] plot[domain=0:2.6, samples=50] (\x, {1.5 + 0.25*exp(0.85*\x)});
  \node[black, anchor=west] at (0.15,0.35) {$y_0 < T$: extinction};
  \node[black, anchor=west] at (1.5,2.75) {$y_0 > T$: blow-up};
\end{tikzpicture}
$$

Combining the two effects — a threshold below and a carrying capacity above —
gives the logistic-with-threshold model
$y' = -r\left(1 - \tfrac{y}{T}\right)\left(1 - \tfrac{y}{K}\right)y$ with
$0 < T < K$. It has three equilibria: $y = 0$ and $y = K$ stable, $y = T$
unstable. A population starting below $T$ dies out; above $T$ it climbs to the
carrying capacity $K$. This structure is thought to describe the passenger pigeon,
which bred successfully only in large concentrations and collapsed to extinction
once its numbers fell below the breeding threshold.

## Harvesting

Suppose the logistic population is harvested. Two policies give different
equilibrium structures.

- **Effort-proportional (Schaefer model).** Catching at rate $Ey$ replaces the
  equation with $y' = r(1 - y/K)y - Ey$. For $E < r$ the equilibria are $y_1 = 0$
  (unstable) and $y_2 = K(1 - E/r)$ (stable). The **sustainable yield** is
  $Y = E\,y_2 = EK(1 - E/r)$, maximized at $E = r/2$, giving the maximum
  sustainable yield $Y_m = rK/4$ at population $K/2$.
- **Constant-rate.** Catching at a fixed rate $h$ gives $y' = r(1 - y/K)y - h$.
  For $h < rK/4$ there are two equilibria $y_1 < y_2$, with $y_1$ unstable and
  $y_2$ stable. As $h$ rises they approach each other, merging at $y = K/2$ when
  $h = rK/4$, and for $h > rK/4$ no equilibrium exists and the population crashes
  to zero.

> **Worked example.** Harvest the logistic population at a constant rate $h$:
> $\dfrac{\d y}{\d t} = r\left(1 - \dfrac{y}{K}\right)y - h$. The equilibria solve
> $r(1 - y/K)y = h$, which rearranges to
>
> $$
> y^2 - Ky + \frac{Kh}{r} = 0
> \qquad\Longrightarrow\qquad
> y = \frac{K}{2}\left(1 \pm \sqrt{1 - \frac{4h}{rK}}\,\right).
> $$
>
> Two equilibria exist when $h < rK/4$, the lower unstable and the upper stable.
> Taking $r = 1$, $K = 4$, and $h = \tfrac34$ gives $y^2 - 4y + 3 = 0$, so
> $y_1 = 1$ (unstable) and $y_2 = 3$ (stable): a population above $1$ settles at
> $3$, one below $1$ collapses. At the critical rate $h = rK/4$ the roots merge at
> $y = K/2$, and beyond it no equilibrium remains.

The constant-rate case is a **bifurcation**: a qualitative change in the number of
equilibria as the parameter $h$ passes the critical value $rK/4$, where the stable
and unstable equilibria merge into a single semistable point.

$$
% caption: As the constant harvest rate h rises, the stable and unstable
% equilibria of the population move together, collide at h = rK/4, and vanish —
% a bifurcation beyond which the population cannot persist.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-0.2,0) -- (5.2,0) node[right] {$h$};
  \draw[->, black] (0,-0.2) -- (0,3.0) node[above] {equilibrium $y$};
  % upper (stable) branch descending, lower (unstable) branch rising, meeting at (h*, K/2)
  \draw[very thick] plot[domain=0:3.6, samples=50] (\x, {1.4 + 1.2*sqrt(1 - \x/3.6)});
  \draw[very thick, dash pattern=on 3pt off 2pt] plot[domain=0:3.6, samples=50] (\x, {1.4 - 1.2*sqrt(1 - \x/3.6)});
  \fill[acc] (3.6,1.4) circle (2.2pt);
  \node[acc, anchor=west] at (3.65,1.55) {$h = \tfrac{rK}{4}$};
  \draw[black, dashed] (3.6,0) -- (3.6,1.4);
  \node[black, anchor=north] at (3.6,-0.05) {critical rate};
  \node[anchor=south west] at (0.1,2.5) {stable};
  \node[anchor=north west] at (0.1,0.35) {unstable};
\end{tikzpicture}
$$

## Linear versus nonlinear equations

The
[linear equation](/differential-equations/first-order/linear-first-order-integrating-factors)
$y' + p(t)y = g(t)$ has three properties the nonlinear case generally lacks.[^boyce-24]

- **General solution.** A linear equation has a solution containing one arbitrary
  constant that captures _all_ solutions. A nonlinear equation may have extra
  solutions no value of the constant reaches. For $y' = y^2$ the family
  $y = -1/(t + c)$ misses the constant solution $y = 0$ entirely.
- **Interval of existence.** For a linear equation the interval is read off the
  coefficients in advance. For a nonlinear equation it can depend on the initial
  value: $y' = y^2$, $y(0) = y_0 > 0$ has solution $y = y_0/(1 - y_0 t)$, which
  blows up at $t = 1/y_0$ — a singularity set by the data, invisible in the
  equation.
- **Explicit form.** Linear equations give explicit solutions; nonlinear ones
  usually give only an implicit relation, and often not even that.

What both share is the geometric consequence of uniqueness: under the
[existence-uniqueness hypotheses](/differential-equations/first-order/existence-uniqueness-euler),
distinct solution curves cannot cross, because a crossing point would be an
initial condition with two solutions. This is why logistic solutions approach the
line $y = K$ but never touch it in finite time.

## Summary

- An **autonomous** equation $y' = f(y)$ is analyzed on the phase line: equilibria
  are zeros of $f$, and the sign of $f$ between them sets the flow direction.
- **Stability** follows from the arrows, or equivalently from $f'(y^\ast{})$:
  negative slope means stable, positive means unstable.
- **Logistic** growth carries every positive population to the carrying capacity
  $K$; a **threshold** model makes $T$ a dividing line between extinction and
  escape; **harvesting** at a constant rate bifurcates the equilibria at
  $h = rK/4$.
- Nonlinear equations lose the linear guarantees of a catch-all general solution,
  a predictable interval of existence, and an explicit formula.

The stability arguments here rest on solutions being unique, which the
[existence and uniqueness theorem](/differential-equations/first-order/existence-uniqueness-euler)
guarantees under continuity of $f$ and $\partial f/\partial y$.

[^boyce-25]: **Boyce**, _Elementary Differential Equations_, §2.5 — Autonomous Differential Equations and Population Dynamics: the phase line, logistic and threshold models, carrying capacity, and the harvesting analyses (Schaefer and constant-rate).
[^boyce-24]: **Boyce**, _Elementary Differential Equations_, §2.4 — Differences Between Linear and Nonlinear Equations: general solutions, initial-value-dependent intervals of existence, implicit solutions, and the non-crossing of integral curves.