Logistic growth, read without solving it

A logistic differential equation has the form

\[\frac{dy}{dt} = ky(a-y)\]

with \(k>0\). The equation describes growth that is initially reinforced by the size of the population and eventually limited by a carrying capacity, and a surprising amount can be read directly from the differential equation without solving it.

The two factors

The factor \(y\) makes the growth rate proportional to the current population, and the factor \(a-y\) slows the growth as the population approaches \(a\). The equilibrium solutions occur where \(\tfrac{dy}{dt}=0\), so \(y=0\) and \(y=a\) are equilibria.

For a population starting between them, \(0<y<a\), both factors are positive and the population increases. As \(y\) approaches \(a\), the factor \(a-y\) approaches zero and the growth slows, and the value \(a\) is the carrying capacity.

Growth is fastest halfway up

The growth rate can be viewed as a function of \(y\), namely \(R(y) = ky(a-y) = k(ay-y^2)\), which is a downward-opening parabola. Its maximum occurs at \(y=\tfrac a2\), and at that point, \(R\left(\tfrac a2\right) = \tfrac{ka^2}{4}\). So a logistic population grows fastest when it reaches half its carrying capacity, and that same point is the inflection point of the solution curve.

Concavity from the equation

Differentiate the differential equation with respect to time and substitute \(y'=ky(a-y)\):

\[\begin{aligned} y'' &= k y'(a-2y)\\ &= k^2y(a-y)(a-2y) \end{aligned}\]

For a solution with \(0<y<a\), the first three factors except \(a-2y\) are positive, so the sign of \(y''\) is determined by \(a-2y\). When \(y<\tfrac a2\), the solution is concave up, and when \(y>\tfrac a2\), it is concave down. The change occurs at half the carrying capacity, exactly where the growth rate is largest.

The visualization fixes the carrying capacity at 100 and lets the initial value and growth constant vary. Changing \(k\) changes the time scale, and a larger \(k\) makes the population move through the same stages more quickly. It does not change the carrying capacity or the population size at which growth is fastest.

Reading an expanded equation

A logistic equation may not be presented in factored form. Suppose \(\tfrac{dP}{dt} = 2P-0.02P^2\), which factors as \(\tfrac{dP}{dt} = 0.02P(100-P)\), and now the structure is visible.

The carrying capacity is \(100\), the fastest growth occurs at \(P=50\), and the maximum growth rate is \(0.02(50)(50)=50\). If \(P\) is measured in thousands of fish and \(t\) in years, that means the population grows fastest at 50 thousand fish, at a rate of 50 thousand fish per year. The units come from the context, not from the algebra alone.

Initial values outside the usual range

Most population examples begin with \(0<y_0<a\). The differential equation also tells us what happens outside that range. If \(y>a\), then \(a-y<0\), so \(y'<0\) and the solution decreases toward the carrying capacity. If \(y=0\) or \(y=a\), the solution remains constant, and these conclusions come from the sign of the differential equation itself.