The logistic differential equation
Master the logistic differential equation in Year 12 Specialist Mathematics for Queensland (QCAA). The model \(\dfrac{dP}{dt}=kP\left(1-\dfrac{P}{M}\right)\) describes population growth that is limited by a carrying capacity \(M\), producing the familiar S-shaped growth curve.
You will learn to read the growth constant \(k\) and capacity \(M\) from the equation, find the equilibria \(P=0\) and \(P=M\), locate the fastest growth at \(P=\dfrac{M}{2}\) with maximum rate \(\dfrac{kM}{4}\), and interpret the solution \(P(t)=\dfrac{M}{1+Ae^{-kt}}\) — a key model in Unit 4 rates of change and differential equations.
Theory
The logistic differential equation \(\dfrac{dP}{dt}=kP\left(1-\dfrac{P}{M}\right)\) models growth that is limited by a carrying capacity \(M\). In Year 12 Specialist Mathematics (QCAA, Queensland) you read \(k\) and \(M\) from the equation, find the equilibria \(P=0\) and \(P=M\), locate the fastest growth at \(P=\tfrac{M}{2}\), and interpret the S-shaped solution \(P(t)=\dfrac{M}{1+Ae^{-kt}}\).
The logistic differential equation describes a population whose growth slows as it fills up its environment. It is written \(\dfrac{dP}{dt}=kP\left(1-\dfrac{P}{M}\right)\), where \(P\) is the population at time \(t\).
The constant \(k>0\) is the growth constant: it sets how fast the population grows when it is small. The constant \(M>0\) is the carrying capacity — the largest population the environment can sustain. The bracket \(\left(1-\dfrac{P}{M}\right)\) is the limiting factor: it is close to \(1\) when \(P\) is small (near-exponential growth) and shrinks to \(0\) as \(P\to M\), so growth stalls.
An equilibrium (constant) solution is a population that does not change, so \(\dfrac{dP}{dt}=0\). Setting the right-hand side to zero gives \(P=0\) and \(P=M\). Here \(P=M\) is stable (the population settles there) and \(P=0\) is unstable (any small population moves away from it).
The solution curve is an S-shaped (sigmoid) curve: it rises from a small \(P_0\), grows fastest at the inflection point \(P=\dfrac{M}{2}\), then levels off towards the horizontal asymptote \(P=M\). The provided solution is \(P(t)=\dfrac{M}{1+Ae^{-kt}}\), with the constant \(A=\dfrac{M-P_0}{P_0}\) fixed by the initial population \(P_0\).
The logistic model for a population \(P\) at time \(t\), with growth constant \(k\) and carrying capacity \(M\):
The equilibrium solutions (where the rate is zero) and the population of fastest growth:
The provided solution of the logistic equation, with the constant \(A\) fixed by the initial population \(P_0\):
How to work with a logistic model
- Read \(k\) and \(M\) from \(\dfrac{dP}{dt}=kP\left(1-\dfrac{P}{M}\right)\): \(M\) is the number in the bracket \(\left(1-\dfrac{P}{M}\right)\); \(k\) is the coefficient. If the equation is factored as \(aP(b-P)\), first rewrite it as \(ab\,P\left(1-\dfrac{P}{b}\right)\).
- Find the equilibria by setting \(\dfrac{dP}{dt}=0\): this gives \(P=0\) (unstable) and \(P=M\) (stable).
- Locate the fastest growth at the inflection \(P=\dfrac{M}{2}\); the maximum rate there is \(\dfrac{kM}{4}\). To find the rate at any \(P\), substitute that \(P\) into the equation.
- Use the solution \(P(t)=\dfrac{M}{1+Ae^{-kt}}\): find \(A=\dfrac{M-P_0}{P_0}\) from \(P_0\), then substitute a value of \(t\) (or solve for \(t\)) and interpret in context.
Compare with the standard form \(\dfrac{dP}{dt}=kP\left(1-\dfrac{P}{M}\right)\):
| \(\dfrac{dP}{dt}\) | \(=\) | \(0.08P\left(1-\dfrac{P}{1500}\right)\) |
| \(1-\dfrac{P}{M}\) | \(=\) | \(1-\dfrac{P}{1500}\) |
| \(M\) | \(=\) | \(1500\) |
| \(k\) | \(=\) | \(0.08\) |
The carrying capacity is \(M=1500\) and the growth constant is \(k=0.08\).
Equilibria occur where the rate of change is zero:
| \(0.5P\left(1-\dfrac{P}{600}\right)\) | \(=\) | \(0\) |
| \(P\) | \(=\) | \(0\) |
| \(1-\dfrac{P}{600}\) | \(=\) | \(0 \Rightarrow P=600\) |
Growth is fastest at the inflection, halfway to the carrying capacity:
| \(P_{\text{fastest}}\) | \(=\) | \(\dfrac{M}{2}\) |
| \(=\) | \(\dfrac{600}{2}\) | |
| \(=\) | \(300\) |
The equilibria are \(P=0\) and \(P=600\); growth is fastest at \(P=300\).
The maximum rate occurs at the inflection \(P=\dfrac{M}{2}=500\); substitute it in:
| \(P\) | \(=\) | \(\dfrac{1000}{2}=500\) |
| \(\dfrac{dP}{dt}\) | \(=\) | \(0.4\times 500\left(1-\dfrac{500}{1000}\right)\) |
| \(=\) | \(0.4\times 500\times 0.5\) | |
| \(=\) | \(100\) |
Check against the shortcut \(\dfrac{kM}{4}\):
| \(\dfrac{kM}{4}\) | \(=\) | \(\dfrac{0.4\times 1000}{4}\) |
| \(=\) | \(100\) |
The maximum rate of spread is \(100\) animals per day.
Read \(k\) and \(M\) from the equation:
| \(M\) | \(=\) | \(2000\) |
| \(k\) | \(=\) | \(0.03\) |
Find \(A\) from the initial population using \(A=\dfrac{M-P_0}{P_0}\):
| \(A\) | \(=\) | \(\dfrac{2000-250}{250}\) |
| \(=\) | \(\dfrac{1750}{250}\) | |
| \(=\) | \(7\) |
Substitute \(M\), \(A\) and \(k\) into the solution:
| \(P(t)\) | \(=\) | \(\dfrac{2000}{1+7e^{-0.03t}}\) |
The particular solution is \(P(t)=\dfrac{2000}{1+7e^{-0.03t}}\).
Common pitfalls
Frequently asked questions
What is the logistic differential equation?
It is \(\dfrac{dP}{dt}=kP\left(1-\dfrac{P}{M}\right)\), a model for population growth that is limited by a carrying capacity \(M\). The growth constant \(k\) sets the early growth rate, and the factor \(\left(1-\dfrac{P}{M}\right)\) slows growth as \(P\) approaches \(M\).
What is the carrying capacity?
The carrying capacity \(M\) is the largest population the environment can sustain. It is the non-zero equilibrium of the logistic equation, and the solution curve rises towards the horizontal asymptote \(P=M\) without exceeding it.
What are the equilibrium solutions of the logistic equation?
They are \(P=0\) and \(P=M\), found by setting \(\dfrac{dP}{dt}=0\). The value \(P=M\) is stable (the population settles there) and \(P=0\) is unstable (any small population grows away from it).
Where does a logistic population grow fastest?
At \(P=\dfrac{M}{2}\), half the carrying capacity — the inflection point of the S-curve. The maximum growth rate there is \(\dfrac{kM}{4}\).
How do you find A in the logistic solution?
Use the initial population \(P_0\): since \(P(0)=\dfrac{M}{1+A}\), rearranging gives \(A=\dfrac{M-P_0}{P_0}\).
How is logistic growth different from exponential growth?
Exponential growth \(\dfrac{dP}{dt}=kP\) increases without bound. Logistic growth includes the factor \(\left(1-\dfrac{P}{M}\right)\), so growth slows as \(P\) nears the carrying capacity and the population levels off at \(P=M\).