Example 4: Logistic growth

Ordinary differential equations: Separable differential equations

Example 4: Logistic growth

The first-order ODE \[\frac{\dd y}{\dd t}=r\cdot y\cdot \left(1-\frac{y}{a}\right)\] where \(r\) and \(a\) are positive constants, can be rewritten in differential form: \[\frac{a}{y\,(y-a)}\,\dd y=-r\,\dd t\] Integrating the left- and right-hand side \[\int\frac{a}{y\, (y-a)}\,\dd y=-r \int \dd t\] is doable. The result of computing the integral on the right-hand side is \(-r\cdot t\). The integral on the left-hand side can be computed by partial fraction decomposition of the integrand: we can decompose the integrand (check this!) as \[\frac{a}{y\, (y-a)}=\frac{1}{y-a}-\frac{1}{y}\] Thus: \[\int\frac{a}{y\, (y-a)}\,\dd y = \ln|y-a|-\ln|y| + C=\ln\left|\frac{y-a}{y}\right|+ C\] Equality of the two integration results up to some constant leads after some algebraic manipulation to the following formula: \[\frac{y-a}{y}=-c\cdot e^{-r\cdot t}\] for some constant \(c\). Why is it useful to place a minus sign in the right-hand side becomes obvious when we isolate the variable \(y\): \[\begin{aligned} \frac{y-a}{y}=-c\cdot e^{-r\cdot t} &\implies 1-\frac{a}{y}=-c\cdot e^{-r\cdot t} \\ \\ &\implies \frac{a}{y} = 1+ c\cdot e^{-r\cdot t}\\ \\ &\implies y = \frac{a}{1+ c\cdot e^{-r\cdot t}}\end{aligned}\] We have now reached the classical formula for a logistic function.

The constant \(c\) is determined by the inital value: If \(y(0)=\alpha\) then \(\alpha=\dfrac{a}{1+c}\), that is, \(c\,\alpha = a-\alpha\) (check this!). Thus: \[\begin{aligned}y &= \frac{a}{1+c\cdot e^{-r\cdot t}}\\ \\ &= \frac{a\cdot \alpha }{\alpha+c\cdot \alpha\cdot e^{-r\cdot t}}\\ \\ &= \frac{a\cdot\alpha}{\alpha+(a-\alpha)e^{-r\cdot t}}\\ \\ &= \frac{a}{1+\left(\dfrac{a}{\alpha}-1\right)e^{-r\cdot t}}\end{aligned}\]

In other words, we have proven the following theorem.

The general solution of the initial value problem \[\frac{\dd y}{\dd t}=r\cdot y\cdot \left(1-\frac{y}{a}\right),\quad y(0)=\alpha\] where \(r\) and \(a\) are positive constants and \(-\infty<\alpha<\infty\), is \[y(t) = \frac{a}{1+\left(\dfrac{a}{\alpha}-1\right) e^{-r\, t}}\]

By the way, this example also illustrates that the maximum existence interval is not always equal to the whole real line \(\mathbb{R}\), but also can be restricted to an interval. Because this interval depends on the initial value \(\alpha\) we denote the existence-inverval as \(I_\alpha\). In this example, there are the following intervals: \[\begin{array}{lll}
\alpha>a: & I_\alpha=(a^{-},\infty), & {\displaystyle \lim_{t\downarrow a^{-}}}\;y(t)=\infty \text{ and } {\displaystyle \lim_{t\rightarrow \infty}}\;y(t)=a,\\ \\
\alpha=a: & I_\alpha=\mathbb{R}, & y(t)=a \text{ for all }t,\\ \\ \\
0\lt\alpha\lt a: & I_\alpha=\mathbb{R}, & {\displaystyle \lim_{t\rightarrow -\infty}\;}y(t)=0 \text{ and } {\displaystyle \lim_{t\rightarrow \infty}}\;y(t)=a,\\ \\
\alpha=0: & I_\alpha=\mathbb{R}, & y(t)=0 \text{ for all }t,\\ \\
\alpha\lt 0: & I_\alpha=(-\infty,a^{+}), & {\displaystyle \lim_{t\rightarrow -\infty}}\;y(t)=0 \text{ and } {\displaystyle \lim_{t\uparrow a^{+}}}\;y(t)=-\infty, \\
\end{array}\] where \[\begin{aligned}a^{-}&= -\frac{1}{r}\ln\left(\frac{\alpha}{\alpha-a}\right) \text{ for }\alpha\gt a,\\ a^{+}&= -\frac{1}{r}\ln\left(\frac{\alpha}{\alpha-a}\right) \text{ for }\alpha\lt 0.\end{aligned}\]

In short, only for \(0\lt y(0)\lt a\) there exist logistic solution curves (for all times \(t\)); under other constraints, there are either constant solutions or solutions that go to plus or minus infinity when time reaches a certain endpoint.