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.