The logistic differential equation
Master the logistic differential equation in Year 12 VCE Specialist Mathematics. 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. It sits in the Calculus area of study of the VCE Mathematics Study Design (VCAA), within the Differential equations topic of Unit 4.
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 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 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 \(P=M\).