Chapter 7 ยท Introduction to Differential Equations

7.4 The Logistic Equation

Real populations cannot grow exponentially for ever. Adding a braking term gives the S-shaped logistic curve, which rises fastest at half the carrying capacity and levels off at $K$.

In Section 7.3 we saw that $\dfrac{dP}{dt}=kP$ gives exponential growth $P=P_0e^{kt}$. For a while this is a good model: bacteria in fresh broth, a new species released on an island, the first weeks of an epidemic, early sales of a new phone. Followed for long enough, though, the exponential model becomes absurd. A single bacterium dividing every $20$ minutes would, after only two days, have descendants weighing more than the Earth. In reality food, space, light or customers run out, and growth slows down.

The simplest way to build this into a model is to let the growth rate per individual fall as the population rises, from about $r$ when the population is small to zero when it reaches the largest size the environment can support. The result is the logistic equation. Ecologists use it for fish stocks and wildlife, epidemiologists for the spread of infections, and economists for the adoption of new technology. It is separable, so the methods of Section 7.3 solve it exactly. Even before we solve it, we can read off almost everything about its solutions from the equation itself.

Limits to Growth: the Logistic Model

Write the per-capita growth rate as $\dfrac{1}{P}\dfrac{dP}{dt}$. Exponential growth keeps it constant at $r$. For the logistic model we instead let it decrease linearly from $r$ (when $P$ is near $0$) to $0$ (when $P=K$): $$\frac{1}{P}\frac{dP}{dt}=r\left(1-\frac PK\right).$$ Multiplying by $P$ gives the logistic equation.

๐Ÿ“˜ Definition 7.10 โ€” The logistic equation
Let $r\gt0$ and $K\gt0$ be constants. The logistic differential equation is $$\frac{dP}{dt}=rP\left(1-\frac{P}{K}\right).$$ The constant $r$ is the intrinsic growth rate (the per-capita growth rate when the population is small) and $K$ is the carrying capacity (the largest population the environment can sustain in the long run). A solution $P(t)$ is called a logistic function.

Three regimes can be read straight from the right-hand side:

โœ๏ธ Example 1 โ€” Reading a logistic equation
A deer population in a reserve is modelled by $\dfrac{dP}{dt}=0.4P\left(1-\dfrac{P}{1000}\right)$, with $t$ in years. (a) State $r$ and $K$. (b) Find $\dfrac{dP}{dt}$ when $P=200$, $500$ and $1200$, and interpret each value.
(a) Comparing with Definition 7.10: $r=0.4$ per year and $K=1000$ deer.
(b) $$\begin{aligned}P=200:&\quad 0.4(200)(1-0.2)=64\ \text{deer/year}\\ P=500:&\quad 0.4(500)(1-0.5)=100\ \text{deer/year}\\ P=1200:&\quad 0.4(1200)(1-1.2)=-96\ \text{deer/year}\end{aligned}$$
Interpret At $200$ deer the herd grows by about $64$ deer per year. At $500$ deer it grows fastest, $100$ per year. (We will see why $K/2$ always gives the fastest growth.) At $1200$ deer the reserve is overcrowded and the herd shrinks by about $96$ per year.
Check The rate $0.4P-0.0004P^2$ is a downward parabola in $P$ with roots $0$ and $1000$, so its maximum is at the midpoint $P=500$. โœ“

Equilibria and the Phase Line

The logistic equation is autonomous: its right-hand side $f(P)=rP(1-P/K)$ depends on $P$ only, not on $t$. For autonomous equations a single picture, the graph of $f(P)$ against $P$, tells us the direction of motion at every population level. Where $f(P)\gt0$ the population increases (an arrow pointing right on the $P$-axis), and where $f(P)\lt0$ it decreases (arrow pointing left). The zeros of $f$ are the equilibrium solutions. The $P$-axis marked with these arrows and equilibria is the phase line of Section 7.2, laid horizontally here (rather than vertically) so that it sits under the graph of $f$.

For the logistic equation, $f(P)=0$ when $P=0$ or $P=K$. Between them $f\gt0$ and beyond $K$ it is negative, so the arrows point away from $0$ and towards $K$.

๐Ÿ“˜ Definition 7.11 โ€” Stable and unstable equilibria (recap of Section 7.2)
Let $c$ be an equilibrium of the autonomous equation $\dfrac{dy}{dt}=f(y)$, so $f(c)=0$. The equilibrium is stable if every solution that starts close enough to $c$, on either side, stays close to $c$ and tends to $c$ as $t\to\infty$. It is unstable if solutions starting arbitrarily close to $c$, on both sides, move away from it. If solutions approach $c$ from one side but move away on the other, $c$ is semi-stable.
๐Ÿ“ Theorem 7.9 โ€” The derivative test for equilibria (recap of Section 7.2)
Let $f$ have a continuous derivative, and let $f(c)=0$.
  • If $f^\prime(c)\lt0$, then $y\equiv c$ is a stable equilibrium of $\dfrac{dy}{dt}=f(y)$.
  • If $f^\prime(c)\gt0$, then $y\equiv c$ is an unstable equilibrium.
(If $f^\prime(c)=0$ the test gives no information.) Reason: if $f^\prime(c)\lt0$, then $f$ is decreasing through $0$ at $c$. So $f\gt0$ just below $c$ (solutions there increase towards $c$) and $f\lt0$ just above (solutions decrease towards $c$). The arrows on both sides point inwards.

For the logistic equation, $f(P)=rP-\dfrac rKP^2$ gives $f^\prime(P)=r-\dfrac{2r}{K}P$. Then $f^\prime(0)=r\gt0$, so $P=0$ is unstable: a few individuals grow away from extinction. And $f^\prime(K)=-r\lt0$, so $P=K$ is stable: every positive population is drawn towards the carrying capacity. Interactive 2 below shows this phase line live.

Solving the Logistic Equation

Suppose $P\ne0$ and $P\ne K$ (those are the equilibria). Separate the variables: $$\frac{dP}{P\left(1-\frac PK\right)}=r\,dt\quad\Longleftrightarrow\quad\frac{K\,dP}{P(K-P)}=r\,dt.$$ By partial fractions (Section 5.3), $\dfrac{K}{P(K-P)}=\dfrac1P+\dfrac{1}{K-P}$. (Check: the right side is $\dfrac{(K-P)+P}{P(K-P)}$.) Integrating, $$\ln|P|-\ln|K-P|=rt+C\quad\Longrightarrow\quad\left|\frac{P}{K-P}\right|=e^{C}e^{rt}\quad\Longrightarrow\quad\frac{P}{K-P}=Be^{rt},$$ where $B=\pm e^C$. At $t=0$ this gives $B=\dfrac{P_0}{K-P_0}$. Solving for $P$: $P=(K-P)Be^{rt}$, so $P\left(1+Be^{rt}\right)=KBe^{rt}$ and $$P=\frac{KBe^{rt}}{1+Be^{rt}}=\frac{K}{1+B^{-1}e^{-rt}}.$$ Writing $A=\dfrac1B=\dfrac{K-P_0}{P_0}$ gives the standard form.

๐Ÿ“ Theorem 7.10 โ€” Solution of the logistic equation
Let $r\gt0$, $K\gt0$ and $P_0\gt0$. The solution of $$\frac{dP}{dt}=rP\left(1-\frac PK\right),\qquad P(0)=P_0,$$ is $$P(t)=\frac{K}{1+Ae^{-rt}},\qquad A=\frac{K-P_0}{P_0},$$ equivalently $P(t)=\dfrac{KP_0}{P_0+(K-P_0)e^{-rt}}$. It is defined for all $t\ge0$ and $P(t)\to K$ as $t\to\infty$.
  • If $0\lt P_0\lt K$ then $A\gt0$ and $P$ increases to $K$ without ever reaching it.
  • If $P_0=K$ then $A=0$ and $P\equiv K$ (the equilibrium).
  • If $P_0\gt K$ then $-1\lt A\lt0$ and $P$ decreases to $K$.
โœ๏ธ Example 2 โ€” Solving a logistic initial-value problem
The deer reserve of Example 1 starts with $100$ deer: $\dfrac{dP}{dt}=0.4P\left(1-\dfrac{P}{1000}\right)$, $P(0)=100$. Find $P(t)$, the population after $5$ years, and the time when the herd reaches $900$ deer.
Formula $A=\dfrac{1000-100}{100}=9$, so by Theorem 7.10 $$P(t)=\frac{1000}{1+9e^{-0.4t}}.$$
After 5 years $P(5)=\dfrac{1000}{1+9e^{-2}}=\dfrac{1000}{1+1.2180}\approx451$ deer.
Reach 900 $\dfrac{1000}{1+9e^{-0.4t}}=900\ \Longrightarrow\ 1+9e^{-0.4t}=\dfrac{10}{9}\ \Longrightarrow\ e^{-0.4t}=\dfrac1{81}\ \Longrightarrow\ t=\dfrac{\ln81}{0.4}\approx11.0$ years.
Check The pure exponential model $100e^{0.4t}$ would predict $100e^{2}\approx739$ deer after 5 years and $8100$ after $11$ years, far above the logistic values. A numerical (RK4) solution of the differential equation gives $P(5)=450.853$, matching $\dfrac{1000}{1+9e^{-2}}=450.853$. โœ“

The inflection point: where growth is fastest

Differentiate the logistic equation with respect to $t$, using the chain rule on $f(P)=rP-\frac rKP^2$: $$\frac{d^2P}{dt^2}=\left(r-\frac{2r}{K}P\right)\frac{dP}{dt}=r\left(1-\frac{2P}{K}\right)\frac{dP}{dt}.$$ While $0\lt P\lt K$ we have $\dfrac{dP}{dt}\gt0$, so the sign of $P^{\prime\prime}$ is the sign of $1-\dfrac{2P}{K}$: positive (concave up) while $P\lt K/2$ and negative (concave down) once $P\gt K/2$.

๐Ÿ“ Theorem 7.11 โ€” Inflection point and maximum growth rate
Let $P(t)=\dfrac{K}{1+Ae^{-rt}}$ solve the logistic equation with $r\gt0$, $K\gt0$ and $0\lt P_0\lt\dfrac K2$ (so $A\gt1$). Then:
  • the graph of $P$ has exactly one inflection point, where $P=\dfrac K2$, at time $$t^{*}=\frac{\ln A}{r};$$
  • the growth rate $\dfrac{dP}{dt}$ is largest there, with maximum value $\dfrac{rK}{4}$.
Proof. $P=K/2$ exactly when $Ae^{-rt}=1$, i.e. $t=\ln A/r$, and $P^{\prime\prime}$ changes sign there as shown above. Also $\dfrac{dP}{dt}=rP\left(1-\frac PK\right)$ is a downward parabola in $P$ with roots $0$ and $K$, so its largest value occurs at the midpoint $P=K/2$: $r\cdot\frac K2\cdot\frac12=\frac{rK}{4}$. $\blacksquare$  (If $\frac K2\le P_0\lt K$, the curve is concave down for all $t\ge0$ and there is no inflection point after time $0$.)
โœ๏ธ Example 3 โ€” When is the herd growing fastest?
For the herd of Example 2, $P(t)=\dfrac{1000}{1+9e^{-0.4t}}$, find when the population grows fastest, the size of the herd then, and the maximum growth rate.
Time By Theorem 7.11, $t^{*}=\dfrac{\ln9}{0.4}\approx5.49$ years.
Size and rate At that moment $P=K/2=500$ deer and $\dfrac{dP}{dt}=\dfrac{rK}{4}=\dfrac{0.4\times1000}{4}=100$ deer per year, matching Example 1(b).
Interpret Before about $5.5$ years each year's increase is larger than the last (the curve bends upward). After that the increases shrink (the curve bends over towards $1000$). The curve is symmetric about the inflection point: it takes as long to grow from $100$ to $500$ as from $500$ to $900$, about $5.5$ years each.
Check $P(5.4931)=\dfrac{1000}{1+9e^{-2.1972}}=\dfrac{1000}{1+9\cdot\frac19}=500$. โœ“ And $11.0-5.49=5.49$, consistent with the symmetry.
Interactive 1 โ€” Logistic versus Exponential Growth

Both models start at $P_0$ with the same intrinsic rate $r$. The red dashed exponential $P_0e^{rt}$ races off the top, while the blue logistic curve bends over at the inflection point on the line $P=K/2$ and settles on the carrying capacity $K$. Try $P_0\gt K$, or $P_0$ between $K/2$ and $K$, and watch the inflection point disappear. The defaults are Examples 2 and 3.

time $t$โ€”
logistic $P(t)$โ€”
exponential $P_0e^{rt}$โ€”
logistic $dP/dt$โ€”
inflection time $t^{*}$โ€”
max rate $rK/4$โ€”

Fitting a Logistic Model to Data

In practice nobody hands us $r$ and $K$. We estimate them from measurements. Two situations are common:

โœ๏ธ Example 4 โ€” Estimating $r$ when $K$ is known
Biologists estimate that a lake can support $10\,000$ trout. They stock it with $2000$ trout, and one year later a survey counts $3000$. Assuming logistic growth, find $r$, predict the population after $2$ years, and find when it reaches $8000$.
Set up $K=10\,000$, $P_0=2000$, so $A=\dfrac{10\,000-2000}{2000}=4$ and $P(t)=\dfrac{10\,000}{1+4e^{-rt}}$.
Find $r$ $P(1)=3000$: $\;1+4e^{-r}=\dfrac{10}{3}\Rightarrow e^{-r}=\dfrac{7}{12}\Rightarrow r=\ln\dfrac{12}{7}\approx0.539$ per year.
Predict $e^{-2r}=\left(\tfrac{7}{12}\right)^2=\tfrac{49}{144}$, so $P(2)=\dfrac{10\,000}{1+4\cdot\frac{49}{144}}=\dfrac{10\,000}{2.3611}\approx4235$ trout.
Reach 8000 $1+4e^{-rt}=\dfrac{10}{8}\Rightarrow e^{-rt}=\dfrac1{16}\Rightarrow t=\dfrac{\ln16}{\ln(12/7)}\approx5.14$ years.
๐Ÿงญ Three equally spaced measurements: the reciprocal trick

Let $u=\dfrac1P$. From Theorem 7.10, $u(t)=\dfrac1K+\dfrac AK\,e^{-rt}$. If we measure at $t=0,\ T,\ 2T$ and put $q=e^{-rT}$, then with $\alpha=\dfrac1K$, $\beta=\dfrac AK$: $$u_0=\alpha+\beta,\qquad u_1=\alpha+\beta q,\qquad u_2=\alpha+\beta q^2.$$ Subtracting, $u_1-u_0=\beta(q-1)$ and $u_2-u_1=\beta q(q-1)$, so $$q=\frac{u_2-u_1}{u_1-u_0},\qquad \alpha=\frac{u_0u_2-u_1^2}{u_0+u_2-2u_1},\qquad K=\frac1\alpha,\qquad r=-\frac{\ln q}{T},\qquad A=Ku_0-1.$$ The data must actually look logistic: we need $0\lt q\lt1$ and $\alpha\gt0$, otherwise no logistic curve passes through the three points.

โœ๏ธ Example 5 โ€” Fitting $r$ and $K$ from three measurements
A yeast culture in a fermentation lab has biomass $10$ g, $25$ g and $50$ g at $t=0$, $2$ and $4$ hours. Fit a logistic model and predict the biomass at $t=8$ hours.
Reciprocals $u_0=0.1$, $u_1=0.04$, $u_2=0.02$ (per gram), with $T=2$ hours.
Fit $$q=\frac{0.02-0.04}{0.04-0.1}=\frac{-0.02}{-0.06}=\frac13,\qquad\alpha=\frac{(0.1)(0.02)-0.04^2}{0.1+0.02-0.08}=\frac{0.0004}{0.04}=0.01.$$ So $K=100$ g, $r=\dfrac{\ln3}{2}\approx0.549$ per hour, and $A=100(0.1)-1=9$.
Model $P(t)=\dfrac{100}{1+9e^{-0.549t}}=\dfrac{100}{1+9\cdot3^{-t/2}}$ g. At $t=8$: $3^{-4}=\tfrac1{81}$, so $P(8)=\dfrac{100}{1+\frac19}=90$ g.
Check $P(2)=\dfrac{100}{1+9/3}=25$ โœ“ and $P(4)=\dfrac{100}{1+9/9}=50$ โœ“. Notice $50=K/2$: the culture was at its inflection point at $t=4$ h, growing at its maximum rate $rK/4\approx13.7$ g/h.

Logistic Growth with Harvesting

Fisheries, forestry and wildlife management remove part of a population every year. The simplest model removes a constant amount $H$ per unit time: $$\frac{dP}{dt}=rP\left(1-\frac PK\right)-H.$$ On the phase-line picture this shifts the parabola down by $H$. Its maximum value, $\dfrac{rK}{4}-H$, tells the whole story:

โœ๏ธ Example 6 โ€” A fishery under quota
A fish stock (in tonnes) follows $\dfrac{dP}{dt}=0.5P\left(1-\dfrac{P}{8000}\right)-H$, with $t$ in years. (a) Find the maximum sustainable yield. (b) For a quota of $H=600$ tonnes per year, find the equilibria and classify them. What happens if a bad year leaves only $1200$ tonnes?
(a) $\dfrac{rK}{4}=\dfrac{0.5\times8000}{4}=1000$ tonnes per year, taken at a stock of $K/2=4000$ tonnes.
(b) Equilibria $0.5P-\dfrac{0.5}{8000}P^2=600$. Multiply by $-16\,000$: $P^2-8000P+9\,600\,000=0$, so $$P=\frac{8000\pm\sqrt{64\,000\,000-38\,400\,000}}{2}=\frac{8000\pm5059.6}{2},$$ giving $P_1\approx1470$ tonnes and $P_2\approx6530$ tonnes.
Classify $f(P)=0.5P(1-P/8000)-600$ has $f^\prime(P)=0.5-P/8000$. Then $f^\prime(1470)\approx0.316\gt0$, so $P_1$ is unstable, and $f^\prime(6530)\approx-0.316\lt0$, so $P_2$ is stable (Theorem 7.9).
Bad year $1200\lt P_1$, so $f(1200)=0.5(1200)(0.85)-600=-90\lt0$: the stock keeps falling and, if the quota is not cut, collapses. The safe policy is to fish well below $1000$ t/yr and to cut the quota when the stock drops. This is the lesson of several real fishery collapses.
Interactive 2 โ€” The Phase Line: $dP/dt$ against $P$

The parabola is $f(P)=rP(1-P/K)-H$ with $K=1000$. Arrows on the $P$-axis show which way the population moves. Filled dots are stable equilibria and hollow dots are unstable ones. The red dot is a population evolving in time from $P_0$, and its height on the parabola is its current growth rate. Start with $H=0$, then raise $H$: the parabola sinks, the equilibria close in, and above $rK/4$ they vanish.

time $t$ (years)โ€”
population $P$โ€”
$dP/dt$โ€”
equilibriaโ€”
max sustainable yield $rK/4$โ€”

Exercises

I
Parts 1โ€“5 refer to the fish-pond model $\dfrac{dP}{dt}=0.2P\left(1-\dfrac{P}{500}\right)$, with $t$ in months.
  • 1 State the intrinsic growth rate and the carrying capacity, with units.
  • 2 Find the equilibrium solutions and classify each as stable or unstable.
  • 3 Find $\dfrac{dP}{dt}$ when $P=100$.
  • 4 At what population is the growth fastest, and what is the fastest growth rate?
  • 5 Describe the solution with $P(0)=700$: is it increasing or decreasing, concave up or down, and what is its limit?
  • 6 Write $\dfrac{dP}{dt}=3P-0.01P^2$ in the form of Definition 7.10 and identify $r$ and $K$.
  • 7 Write $\dfrac{dP}{dt}=0.05P-0.0001P^2$ in logistic form and identify $r$ and $K$.
  • 8 Conceptual. Explain why a solution with $0\lt P(0)\lt K$ never reaches the value $K$, using (a) the solution formula and (b) the equilibrium solution $P\equiv K$.
1 $r=0.2$ per month; $K=500$ fish.
2 $P\equiv0$ and $P\equiv500$. With $f(P)=0.2P-0.0004P^2$, $f^\prime(P)=0.2-0.0008P$: $f^\prime(0)=0.2\gt0$ (unstable), $f^\prime(500)=-0.2\lt0$ (stable).
3 $0.2(100)\left(1-\tfrac{100}{500}\right)=20(0.8)=16$ fish per month.
4 At $P=K/2=250$ fish; the rate is $\dfrac{rK}{4}=\dfrac{0.2\times500}{4}=25$ fish per month.
5 Since $700\gt K$, $\dfrac{dP}{dt}\lt0$: the population decreases. Because $P\gt K\gt K/2$ and $P^\prime\lt0$, $P^{\prime\prime}=r\left(1-\frac{2P}{K}\right)P^\prime\gt0$: concave up. It tends to $500$ from above and never drops below it. Here $A=\frac{500-700}{700}=-\frac27$, so $P=\dfrac{500}{1-\frac27e^{-0.2t}}$.
6 $3P-0.01P^2=3P\left(1-\dfrac{0.01}{3}P\right)=3P\left(1-\dfrac{P}{300}\right)$: $r=3$, $K=300$.
7 $0.05P-0.0001P^2=0.05P\left(1-\dfrac{P}{500}\right)$: $r=0.05$, $K=500$.
8 (a) $P(t)=\dfrac{K}{1+Ae^{-rt}}$ with $A\gt0$, and $Ae^{-rt}\gt0$ for every $t$, so $P(t)\lt K$ always. (b) The right-hand side $f(P)$ is a polynomial, so initial-value problems have unique solutions. If $P(t_1)=K$ for some $t_1$, then both $P(t)$ and the constant solution $P\equiv K$ would pass through the point $(t_1,K)$, which is impossible unless $P\equiv K$. Solution curves of this equation cannot cross or touch.
II
Solving logistic equations. Parts 2โ€“4 refer to the solution of part 1.
  • 1 Solve $\dfrac{dP}{dt}=0.5P\left(1-\dfrac{P}{200}\right)$, $P(0)=20$.
  • 2 Find $P(4)$.
  • 3 When does the population reach $150$?
  • 4 Find the inflection time $t^{*}$, the population then, and the growth rate then.
  • 5 Solve $\dfrac{dP}{dt}=0.1P\left(1-\dfrac{P}{50}\right)$, $P(0)=100$, and find $P(10)$.
  • 6 Verify the partial-fraction identity $\dfrac{K}{P(K-P)}=\dfrac1P+\dfrac1{K-P}$ and use it to evaluate $\displaystyle\int\frac{dP}{P(10-P)}$.
  • 7 Verify by differentiation that $P(t)=\dfrac{K}{1+Ae^{-rt}}$ satisfies $\dfrac{dP}{dt}=rP\left(1-\dfrac PK\right)$ for every constant $A$.
  • 8 Solve $\dfrac{dy}{dt}=2y-2y^2$, $y(0)=0.1$.
Check: a numerical solver (RK4, red dots) uses only the differential equation. It must land on the formula (curve).
$t$โ€”
formulaโ€”
RK4 valueโ€”
differenceโ€”
1 $K=200$, $r=0.5$, $A=\dfrac{200-20}{20}=9$: $\;P(t)=\dfrac{200}{1+9e^{-0.5t}}$.
2 $P(4)=\dfrac{200}{1+9e^{-2}}\approx\dfrac{200}{2.2180}\approx90.2$.
3 $1+9e^{-0.5t}=\dfrac{200}{150}=\dfrac43\Rightarrow e^{-0.5t}=\dfrac1{27}\Rightarrow t=2\ln27\approx6.59$.
4 $t^{*}=\dfrac{\ln9}{0.5}=2\ln9\approx4.39$; then $P=100$ and $\dfrac{dP}{dt}=\dfrac{rK}{4}=\dfrac{0.5\times200}{4}=25$ per unit time.
5 $A=\dfrac{50-100}{100}=-\dfrac12$: $\;P(t)=\dfrac{50}{1-\frac12e^{-0.1t}}$. $P(10)=\dfrac{50}{1-0.5e^{-1}}\approx\dfrac{50}{0.81606}\approx61.3$. It decreases towards $50$.
6 $\dfrac1P+\dfrac1{K-P}=\dfrac{(K-P)+P}{P(K-P)}=\dfrac{K}{P(K-P)}$ โœ“. With $K=10$: $\displaystyle\int\frac{dP}{P(10-P)}=\frac1{10}\int\left(\frac1P+\frac1{10-P}\right)dP=\frac1{10}\left(\ln|P|-\ln|10-P|\right)+C=\frac1{10}\ln\left|\frac{P}{10-P}\right|+C$.
7 $P=K\left(1+Ae^{-rt}\right)^{-1}$ gives $P^\prime=K\left(1+Ae^{-rt}\right)^{-2}\cdot rAe^{-rt}$. Also $1-\dfrac PK=\dfrac{Ae^{-rt}}{1+Ae^{-rt}}$, so $rP\left(1-\dfrac PK\right)=r\cdot\dfrac{K}{1+Ae^{-rt}}\cdot\dfrac{Ae^{-rt}}{1+Ae^{-rt}}=\dfrac{rKAe^{-rt}}{\left(1+Ae^{-rt}\right)^2}=P^\prime$ โœ“.
8 $2y-2y^2=2y(1-y)$: logistic with $r=2$, $K=1$; $A=\dfrac{1-0.1}{0.1}=9$, so $y(t)=\dfrac{1}{1+9e^{-2t}}$.
III
Modelling with the logistic equation. Give units in every answer.
  • 1 A rumour. In a town of $5000$ people, the number $N$ who have heard a rumour satisfies $\dfrac{dN}{dt}=0.0002\,N(5000-N)$, $t$ in days, $N(0)=50$. Write this in logistic form, solve it, and find when half the town has heard.
  • 2 World population. Suppose world population follows a logistic model with $K=12$ billion and $r=0.025$ per year, with $6.1$ billion people in the year $2000$. Predict the population in $2050$, and find the year in which the growth rate was largest.
  • 3 A tilapia pond can support $2000$ fish. It is stocked with $200$ fish, and $3$ months later there are $500$. Find $r$ and the time when the pond holds $1800$ fish.
  • 4 A protected bird population is counted at $200$, $400$ and $600$ in three surveys $5$ years apart. Fit a logistic model (find $K$, $A$ and $r$).
  • 5 For the harvested model $\dfrac{dP}{dt}=0.4P\left(1-\dfrac{P}{1000}\right)-H$ ($t$ in years), find the maximum sustainable yield and the population at which it is taken.
  • 6 In part 5 take $H=64$. Find and classify the equilibria, and describe what happens when $P(0)=150$ and when $P(0)=300$.
  • 7 In part 5 take $H=120$. Show that the population collapses whatever its starting size.
  • 8 Challenge. Show that every solution of the logistic equation satisfies $\dfrac{d^2P}{dt^2}=r^2P\left(1-\dfrac PK\right)\left(1-\dfrac{2P}{K}\right)$, and use this to find where solution curves are concave up and concave down for all $P\gt0$.
1 $0.0002N(5000-N)=0.0002\cdot5000\,N\left(1-\frac{N}{5000}\right)=1\cdot N\left(1-\frac{N}{5000}\right)$: $r=1$ per day, $K=5000$ people. $A=\frac{5000-50}{50}=99$, so $N(t)=\dfrac{5000}{1+99e^{-t}}$. Half the town ($2500$) at $t^{*}=\ln99\approx4.6$ days.
2 $t$ = years after 2000, $A=\frac{12-6.1}{6.1}\approx0.9672$. $P(50)=\dfrac{12}{1+0.9672e^{-1.25}}\approx\dfrac{12}{1.2771}\approx9.40$ billion. Fastest growth when $P=K/2=6$ billion: $t^{*}=\dfrac{\ln A}{r}=\dfrac{\ln0.9672}{0.025}\approx-1.3$, i.e. around 1999. In this model the world passed its inflection point just before 2000. (The actual population in 2050 is projected by the UN to be nearer $9.7$ billion, a reminder that $r$ and $K$ are only estimates.)
3 $A=9$; $500=\dfrac{2000}{1+9e^{-3r}}\Rightarrow e^{-3r}=\dfrac13\Rightarrow r=\dfrac{\ln3}{3}\approx0.366$ per month. $1800$ fish: $9e^{-rt}=\dfrac19\Rightarrow e^{-rt}=\dfrac1{81}=\left(e^{-3r}\right)^4\Rightarrow t=12$ months.
4 $u_0=0.005$, $u_1=0.0025$, $u_2=\tfrac1{600}\approx0.0016667$. $q=\dfrac{u_2-u_1}{u_1-u_0}=\dfrac13$; $\alpha=\dfrac{u_0u_2-u_1^2}{u_0+u_2-2u_1}=0.00125$, so $K=800$ birds, $A=800(0.005)-1=3$, $r=\dfrac{\ln3}{5}\approx0.220$ per year. Check: $\dfrac{800}{1+3/3}=400$ โœ“, $\dfrac{800}{1+3/9}=600$ โœ“.
5 MSY $=\dfrac{rK}{4}=\dfrac{0.4\times1000}{4}=100$ animals per year, at $P=K/2=500$.
6 $0.4P-0.0004P^2=64\iff P^2-1000P+160\,000=0\iff(P-200)(P-800)=0$. $f^\prime(P)=0.4-0.0008P$: $f^\prime(200)=0.24\gt0$ (unstable threshold), $f^\prime(800)=-0.24\lt0$ (stable). From $P(0)=150\lt200$: $f\lt0$, so the population declines to extinction. From $P(0)=300$: $f\gt0$, so it grows and settles at $800$.
7 $0.4P\left(1-\frac{P}{1000}\right)\le100$ for all $P$ (its maximum), so $\dfrac{dP}{dt}\le100-120=-20$ animals per year. The population falls by at least $20$ per year, so it reaches $0$ within $P(0)/20$ years.
8 By the chain rule, $P^{\prime\prime}=f^\prime(P)P^\prime=r\left(1-\frac{2P}{K}\right)\cdot rP\left(1-\frac PK\right)$ โœ“. For $P\gt0$, the factors $1-P/K$ and $1-2P/K$ change sign at $K$ and $K/2$: concave up for $0\lt P\lt K/2$, concave down for $K/2\lt P\lt K$, concave up again for $P\gt K$.
IV
Applications and comparisons.
  • 1 An international tuna fishery is modelled with $r=0.3$ per year and $K=50\,000$ tonnes. What is the maximum sustainable yield, and at what stock level should it be taken?
  • 2 Show that, for any logistic solution with $0\lt P_0\lt K$, the time taken to grow from $K/10$ to $9K/10$ is $\dfrac{2\ln9}{r}$, independent of $K$ and $P_0$.
  • 3 Technology adoption. In a certain country $10\%$ of households used mobile banking in 2015 and $50\%$ in 2020. Assuming logistic growth towards $100\%$, find $r$ and predict the percentage in 2025.
  • 4 With $r=0.4$, $K=1000$ and $P_0=100$, compare the exponential prediction $P_0e^{rt}$ with the logistic prediction at $t=10$.
1 $\dfrac{rK}{4}=\dfrac{0.3\times50\,000}{4}=3750$ tonnes per year, at a stock of $K/2=25\,000$ tonnes.
2 $P=\dfrac{K}{10}\iff1+Ae^{-rt}=10\iff Ae^{-rt_1}=9$; $\;P=\dfrac{9K}{10}\iff Ae^{-rt_2}=\dfrac19$. Dividing, $e^{r(t_2-t_1)}=81$, so $t_2-t_1=\dfrac{\ln81}{r}=\dfrac{2\ln9}{r}\approx\dfrac{4.39}{r}$. Neither $K$ nor $A$ (hence $P_0$) appears.
3 In percent, $K=100$, $P_0=10$ (2015), $A=9$. In 2020: $50=\dfrac{100}{1+9e^{-5r}}\Rightarrow e^{-5r}=\dfrac19\Rightarrow r=\dfrac{\ln9}{5}\approx0.439$ per year. In 2025: $e^{-10r}=\dfrac1{81}$, so $P=\dfrac{100}{1+9/81}=90\%$.
4 Exponential: $100e^{4}\approx5460$. Logistic: $\dfrac{1000}{1+9e^{-4}}\approx\dfrac{1000}{1.1648}\approx858.5$. The exponential model is more than six times too large, and more than five times the carrying capacity.
๐Ÿ”ฌ Researchโ€“Teaching NexusResearch-tutored ยท optional, about 45 minutes

Research question. Does the spread of internet use in a country follow a logistic curve, and what "carrying capacity" does the data suggest?

Why it matters. Governments, telecom companies and development agencies plan networks, services and digital-inclusion programmes around adoption forecasts. Epidemiologists use the same S-curve thinking for the early spread of infections.

What to do.

  1. From the World Bank open data site, download "Individuals using the Internet (% of population)" for one country of your choice (for example Oman, Kenya, India, Brazil or the United Kingdom).
  2. Choose three equally spaced years (for example 2004, 2010, 2016) and use the reciprocal trick of Example 5 to compute $K$, $A$ and $r$.
  3. Plot your logistic curve together with all the data points in a spreadsheet. Where does it fit well, and where does it miss?
  4. Repeat with a different triple of years. How sensitive is the fitted $K$ to your choice?

Tools: Spreadsheet, World Bank open data (or any national statistics office).

Hand in: A one-page note with your fitted $K$, $r$ and $t^{*}$, a plot, and two sentences on whether a logistic model is believable for this country.

Going further. Compare two countries with very different incomes. Do they have similar $r$ but different $K$, or the other way round? What might explain this?

Instructor note: many countries give fitted $K$ between about $80\%$ and $100\%$ and $r$ around $0.2$โ€“$0.4$ per year. Triples taken before the inflection point give unstable or even impossible fits ($q$ outside $(0,1)$ or $\alpha\le0$), which is a good discussion point about fitting before the curve bends.

๐Ÿงช Self-check quiz

0/5 correct

Q1. What is the carrying capacity of $\dfrac{dP}{dt}=0.3P\left(1-\dfrac{P}{600}\right)$?

Q2. For $\dfrac{dP}{dt}=2P-0.01P^2$, the population grows fastest when $P=$

Q3. If a logistic population starts above its carrying capacity ($P_0\gt K$), then $P(t)$

Q4. The largest value of $\dfrac{dP}{dt}$ along a logistic solution with $0\lt P_0\lt K/2$ is

Q5. With constant harvesting, $\dfrac{dP}{dt}=rP\left(1-\dfrac PK\right)-H$ has a positive equilibrium exactly when

โ† 7.3 Separable EquationsChapter 8 โ€” Sequences and Series โ†’