17. Applications of Differential Equations

f. Logistic Growth

Exponential growth is a reasonable model for population growth when there are unlimited resources available. However, there is often an upper limit \(M\) to the size of the population because the environment cannot support a larger population. In that case, the rate of increase of the population \(P(t)\) is often jointly proportional to the population \(P(t)\) (as in exponential growth) and to the difference between the population and its maximum \(M-P(t)\). Thus the population satisfies \[ \dfrac{dP}{dt}=kP(M-P) \] As \(P\) gets bigger, the rate \(\dfrac{dP}{dt}\) gets bigger. As \(P\) gets closer to \(M\), the quantity \(M-P\) gets smaller, and the rate \(\dfrac{dP}{dt}\) gets smaller. This growth model is called logistic growth.

Find the general solution of the logistic growth equation: \[ \dfrac{dP}{dt}=kP(M-P) \] Find the asymptotic value of the solution.

We start by separating variables: \[ \int \dfrac{dP}{P(M-P)}=\int k\,dt \] The integral on the right is \[ \int k\,dt=kt+C \] The integral on the left is computed using the partial fraction expansion: (Remember \(M\) is a constant. Check it by adding them up.) \[ \dfrac{1}{P(M-P)}=\dfrac{1}{MP}+\dfrac{1}{M(M-P)} \] Thus \[\begin{aligned} \int \dfrac{dP}{P(M-P)} &=\dfrac{1}{M}\int \dfrac{dP}{P}+\dfrac{1}{M}\int \dfrac{dP}{(M-P)} \\ &=\dfrac{1}{M}\ln|P|-\dfrac{1}{M}\ln|M-P| \\ &=\ln\left|\dfrac{P}{M-P}\right|^{1/M} \end{aligned}\] Equating the left and right integrals, we have \[ \ln\left|\dfrac{P}{M-P}\right|^{1/M}=kt+C \] We exponentiate, raise both sides to the power \(M\) and remove the absolute values: \[\begin{aligned} \left|\dfrac{P}{M-P}\right|^{1/M}&=e^C e^{kt} \\ \left|\dfrac{P}{M-P}\right|&=e^{MC}e^{Mkt} \\ \dfrac{P}{M-P}&=Ae^{Mkt} \qquad \text{where} \quad A=\pm e^{MC} \end{aligned}\] We finally need to solve for \(P\). We multiply both sides by \(M-P\) and distribute: \[ P=MAe^{Mkt}-PAe^{Mkt} \] We add \(PAe^{Mkt}\) to both sides and factor the left: \[ P\left(1+Ae^{Mkt}\right)=MAe^{Mkt} \] Finally, we divide by \(1+Ae^{Mkt}\): \[ P(t)=\dfrac{MAe^{Mkt}}{1+Ae^{Mkt}} \] This is the logistic growth model. An alternate form is: \[ P(t)=\dfrac{M}{1+A^{-1}e^{-Mkt}} \] If we let \(B=A^{-1}\) then the solution is: \[ P(t)=\dfrac{M}{1+Be^{-Mkt}} \] For large times, the population approaches: \[\begin{aligned} \lim_{t\rightarrow\infty}P(t) &=\lim_{t\rightarrow\infty}\dfrac{M}{1+Be^{-Mkt}}=M \end{aligned}\]

Notice that the constant in the exponent of the logistic growth model is \(Mk\) and not just \(k\). There are three unknown constants: \(M\), \(k\) or \(Mk\) and \(A\) or \(B\). To find them, we will need three data points.

If you add flour to a yeast culture at a steady rate, then the culture reaches a steady state size. So it satisfies the logistic growth model. You start a yeast culture with \(P(0)=2000\) cells of yeast and add a cup of flour every \(10\) min. After \(40\) min, there are \(P(40)=3000\) cells of yeast. After \(80\) min, there are \(P(80)=4000\) cells of yeast. About how many cells of yeast are there after \(120\) min? What is the steady state number of yeast cells in this culture?

We need to find the constants, \(M\), \(Mk\) and \(A\), in the logistic growth model \(P(t)=\dfrac{MAe^{Mkt}}{1+Ae^{Mkt}}\) or equivalently in the equation \(\dfrac{P}{M-P}=Ae^{Mkt}\). Our \(3\) data points are at \(t=0\), \(t=40\) and \(t=80\):
\(P(0)=2000\): \[ \dfrac{2000}{M-2000}=A \qquad \qquad \qquad (1) \] \(P(40)=3000\): \[ \dfrac{3000}{M-3000}=Ae^{40Mk} \qquad \qquad (2) \] \(P(80)=4000\) \[ \dfrac{4000}{M-4000}=Ae^{80Mk} \qquad \qquad (3) \] The equation (1) gives a formula for \(A\) which we substitute into equations (2) and (3): \[ \dfrac{3000}{M-3000}=\dfrac{2000}{M-2000}e^{40Mk} \qquad \Longrightarrow \qquad e^{40Mk}=\dfrac{3(M-2000)}{2(M-3000)} \] \[ \dfrac{4000}{M-4000}=\dfrac{2000}{M-2000}e^{80Mk} \qquad \Longrightarrow \qquad e^{80Mk}=\dfrac{2(M-2000)}{M-4000} \] To eliminate \(Mk\), we square the first equation and equate it to the second equation: \[ e^{80Mk}=\left(\dfrac{3(M-2000)}{2(M-3000)}\right)^2=\dfrac{2(M-2000)}{M-4000} \] Cross multiplying and cancelling, this last equation can be written as \[\begin{aligned} 3^2(M-2000)^2(M-4000)&=2^3(M-2000)(M-3000)^2 \\ 9(M-2000)(M-4000)&=8(M-3000)^2 \end{aligned}\] Putting everything on one side of the equation and expanding we get \[ 9(M-2000)(M-4000)-8(M-3000)^2=0 \] \[ M^2-6000M=0 \] Consequently, \[ M=6000 \] To find \(A\) and \(Mk\), we plug \(M\) back into equations (1) and (2): \[ A=\dfrac{2000}{6000-2000} =\dfrac{1}{2} \] and \[ \dfrac{1}{2}e^{40Mk}=\dfrac{3000}{6000-3000}=1 \] so that \[ Mk=\dfrac{1}{40}\ln2\approx .017 33 \] Finally we plug the three constants back into the logistic growth model: \[ P(t)=\dfrac{M}{1+A^{-1}e^{-Mkt}} =\dfrac{6000}{1+2e^{-(t/40)\ln2}} \]