Autonomous Equations, Phase Lines, and Population Dynamics

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

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.¹ This qualitative reading is the starting point of the qualitative theory used when equations cannot be solved at all.

Equilibria and the phase line

The simplest solutions are constant. A constant $y=y^{\ast }$ solves $y^{\prime }=f(y)$ precisely when $f(y^{\ast })=0$, since then $\mathbf{d}y/\mathbf{d}t=0$ for all $t$.

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 $f<0$.

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 $f$.

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

Logistic growth

The exponential model $y^{\prime }=ry$ 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:

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 $K$.

The shape of the solution follows from the sign of $f$ and $f^{\prime }$. Differentiating, $y^{\prime \prime }=f^{\prime }(y)f(y)$, so a solution is concave up where $f$ and $f^{\prime }$ 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.

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

Thresholds

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

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.

Combining the two effects — a threshold below and a carrying capacity above — gives the logistic-with-threshold model $y^{\prime }=-r\left(1-\frac{y}{T}\right)\left(1-\frac{y}{K}\right)y$ 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 $y^{\prime }=r(1-y/K)y-Ey$. 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 $y^{\prime }=r(1-y/K)y-h$. 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.

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.

Linear versus nonlinear equations

The linear equation$y^{\prime }+p(t)y=g(t)$ has three properties the nonlinear case generally lacks.²

• 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 $y^{\prime }=y^2$ 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: $y^{\prime }=y^2$, $y(0)=y_0>0$ 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, 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 $y^{\prime }=f(y)$ 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 $f^{\prime }(y^{\ast })$: 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 guarantees under continuity of $f$ and $\partial f/\partial y$.