9.4 The Logistic Equation

The simplest model of population growth is \({dy}/dt=ky\), according to which populations grow exponentially. This may be true over short periods of time, but it is clear that no population can increase without limit. Therefore, population biologists use a variety of other differential equations that take into account environmental limitations to growth such as food scarcity and competition between species. One widely used model is based on the logistic differential equation: \[ \begin{equation} \label{10.popmod.logsol} \boxed{\bbox[#FAF8ED,5pt]{\frac{dy}{dt} = ky \left(1-\frac{y}A \right),\qquad}}\tag {1} \end{equation} \] Here \(k>0\) is the growth constant, and \(A>0\) is a constant called the carrying capacity. Figure 9.19 shows a typical \(S\)-shaped solution of Eq. (1). As in the previous section, we also denote \({dy}/dt\) by \(\dot{y}\).

The slope field in Figure 9.20 shows clearly that there are three families of solutions, depending on the initial value \(y_0=y(0)\).

• If \(y_0>A\), then \(y(t)\) is decreasing and approaches \(A\) as \(t\to\infty\).
• If \(0< y_0<A\), then \(y(t)\) is increasing and approaches \(A\) as \(t\to\infty\).
• If \(y_0<0\), then \(y(t)\) is decreasing and \(\lim\limits_{t\to t_{b}-} y(t) = -\infty\) for some time \(t_b\).

Equation (1) also has two constant solutions: \(y=0\) and \(y=A\). They correspond to the roots of \(ky(1-y/A) = 0\), and they satisfy Eq. (1) because \(\dot{y}=0\) when \(y\) is a constant. Constant solutions are called equilibrium or steady-state solutions. The equilibrium solution \(y=A\) is a stable equilibrium because every solution with initial value \(y_0\) close to \(A\) approaches the equilibrium \(y=A\) as \(t\to \infty\). By contrast, \(y=0\) is an unstable equilibrium because every nonequilibrium solution with initial value \(y_0\) near \(y=0\) either increases to \(A\) or decreases to \(-\infty\).

Having described the solutions qualitatively, let us now find the nonequilibrium solutions explicitly using separation of variables. Assuming that \(y\ne 0\) and \(y\ne A\), we have \[ \begin{align} \frac{dy}{dt} &= ky \left(1-\dfrac{y}A \right)\notag\\ \frac{dy}{y \left(1-y/A \right)} &= k\,dt\tag {2}\\ \int \left(\frac{1}{y} - \frac{1}{y-A}\right)dy &= \int k\, dt \label{10.popmod.partfrac}\notag\\ \ln |y| - \ln |y-A| &= kt+C\notag\\ \left|\frac{y}{y-A}\right| &=e^{kt+C} \quad\Rightarrow\quad \frac{y}{y-A} =\pm e^Ce^{kt}\notag \end{align} \] Since \(\pm e^{C}\) takes on arbitrary nonzero values, we replace \(\pm e^{C}\) with \(C\) (nonzero): \[ \begin{equation} \label{10.popmod.partlythere} \frac{y}{y-A} = Ce^{kt} \tag {3} \end{equation} \] For \(t=0\), this gives a useful relation between \(C\) and the initial value \(y_0=y(0)\): \[ \begin{equation} \label{10.popmod.loginit} \frac{y_0}{y_0-A} =C \tag {4} \end{equation} \] To solve for \(y\), multiply each side of Eq. (3) by \((y-A)\): \[ \begin{align*} y &= (y-A)Ce^{kt} \\ y(1-Ce^{kt}) &= - AC e^{kt}\\ y&= \frac{ AC e^{kt}}{ Ce^{kt}-1 } \end{align*} \] As \(C\ne 0\), we may divide by \(Ce^{kt}\) to obtain the general nonequilibrium solution: \[ \boxed{\bbox[#FAF8ED,5pt]{\frac{dy}{dt} = ky \left(1 - \frac{y}{A}\right),\qquad y= \frac{A}{1- e^{-kt}/C}}} \tag {5} \]

9.4.1 Summary

• The logistic equation and its general nonequilibrium solution (\(k>0\) and \(A>0\)): \[ \frac{dy}{dt} =ky\left(1-\frac{y}A\right),\qquad y= \frac{A}{1 - e^{-kt}/C}, \qquad\text{or equivalently}\qquad \frac{y}{y-A} = Ce^{kt} \]
• Two equilibrium (constant)solutions:
  • - \(y=0\) is an unstable equilibrium.
  • - \(y=A\) is a stable equilibrium.
• If the initial value \(y_0 = y(0)\) satisfies \(y_0 > 0\), then \(y(t)\) approaches the stable equilibrium \(y = A\); that is, \(\lim_{t\to\infty} y(t) = A\).