Why Look for Equilibria?
A differential equation describes how a quantity changes. In a population model, the quantity \(P(t)\) is the population at time \(t\), and its derivative \(dP/dt\) is the population’s instantaneous rate of change.
An equilibrium is a value of the quantity at which that rate is zero. If a population begins at such a value, the model predicts that it remains there. For the logistic model, two values make the rate of change vanish: zero population and the carrying capacity.
The Logistic Equation and Its Equilibria
The standard logistic differential equation is
where \(r>0\) is the growth-rate constant and \(K>0\) is the carrying capacity. The model is autonomous: the rate depends on the current population \(P\), not explicitly on time \(t\). For a population measured in individuals, \(K\) is measured in individuals and \(r\) has units of reciprocal time.
To find the equilibria of the logistic equation, set its right-hand side equal to zero and factor:
Since \(r>0\), the product is zero when \(P=0\) or \(1-P/K=0\). The second equation gives \(P=K\). Thus the two equilibrium values are \(0\) and \(K\).
Why Those Two Values Make Sense
The factors in the model explain the two equilibria. The factor \(P\) makes the rate zero when \(P=0\). With no population present, this model has no mechanism that creates individuals from nothing, so it predicts no change from zero.
The other factor, \(1-P/K\), measures the limiting effect represented by the carrying capacity. At \(P=K\), this factor is zero. The model represents growth as being offset by the population’s limitation at that level, so the net rate of change is zero.
These are statements about the model’s rate at particular values. They do not, by themselves, describe how solutions starting at other values behave over time. Here the question is specifically which constant populations satisfy the equation.
A Short Derivation
We can justify the equilibrium test directly. Suppose \(P(t)=C\) is constant. By the derivative rule for a constant function, \(dP/dt=0\). Substituting this function into \(dP/dt=f(P)\) therefore requires \(0=f(C)\).
Conversely, if \(f(C)=0\), then the constant function \(P(t)=C\) has derivative zero, and the right-hand side evaluated at \(P=C\) is also zero for every \(t\). It satisfies the differential equation. Therefore, for an autonomous equation, the constant solutions are exactly the constant values that make its right-hand side zero.
Applying this reasoning to the logistic equation gives the two roots of \(rP(1-P/K)\). Each root is not merely an algebraic answer: it defines a constant function that can be checked in the original differential equation.
Worked Examples
Worked Example: Finding Equilibria with Given Parameters
Find the equilibrium solutions of
where \(t\) is measured in weeks and \(P\) is a population count.
Set the right-hand side equal to zero:
Because \(0.4\ne 0\), the product is zero if \(P=0\) or \(1-P/800=0\). Solving the second equation gives \(P/800=1\), so \(P=800\). The equilibrium functions are therefore \(P(t)=0\) and \(P(t)=800\).
Check both in the original equation. For \(P=0\), the right-hand side is \(0.4(0)(1-0/800)=0\), equal to the derivative of the constant zero function. For \(P=800\), it is \(0.4(800)(1-800/800)=320(0)=0\), equal to the derivative of the constant function \(800\). Both are verified.
Worked Example: Interpreting the Carrying Capacity
A model for a fish population is
where \(P\) is measured in fish and \(t\) in years. Determine the population values for which the model predicts no instantaneous change.
Set the rate equal to zero:
The constant \(0.06\) is nonzero. The first factor gives \(P=0\). The second factor gives \(1-P/1200=0\), hence \(P/1200=1\) and \(P=1200\). Thus the model predicts zero instantaneous change at zero fish and at 1200 fish.
Substitute \(P=0\): the right-hand side is \(0.06(0)(1)=0\). Substitute \(P=1200\): it is \(0.06(1200)(1-1)=72(0)=0\). Each matches the derivative of the corresponding constant function. The units also agree: \(0.06\) is per year, so multiplying it by a population gives a population-change rate per year.
Worked Example: A Quadratic Form of the Logistic Model
Find the equilibria of
Factor the right-hand side:
Set this product to zero. The first factor gives \(P=0\). The second gives \(0.3-0.0005P=0\), so \(0.0005P=0.3\), and
The equilibrium functions are \(P(t)=0\) and \(P(t)=600\). Check \(P=0\): \(0.3(0)-0.0005(0)^2=0\). Check \(P=600\): \(0.3(600)-0.0005(600)^2=180-0.0005(360000)=180-180=0\). Both constant functions satisfy the equation.
This form can be rewritten as \(0.3P(1-P/600)\), so its carrying capacity is 600. The factorization makes the zero-population equilibrium and the carrying-capacity equilibrium visible.
A Common Pitfall: Dividing Away an Equilibrium
In solving a differential equation by separation, a learner might divide the logistic equation by \(P\), perhaps to prepare for separating variables. But this division is valid only when \(P\ne 0\). It excludes \(P=0\) from the equation being considered, even though zero is an equilibrium.
For example, starting from \(dP/dt=rP(1-P/K)\) and dividing by \(P\) gives
which applies only where \(P\ne 0\). A conclusion based only on this divided equation cannot rule out \(P(t)=0\). The correction is to find the constant solutions from the original equation first, or to check separately any values excluded by division.
Why Equilibria Matter in a Population Model
Equilibria identify population levels at which the model predicts no instantaneous change. They also give the two central reference values built into the logistic equation: extinction, represented by zero, and the limiting population level, represented by \(K\). Finding them is an algebraic step, but it has a direct interpretation in the context of the model.
The equilibrium calculation is also a useful check on the equation itself. If a proposed logistic model has a positive growth-rate parameter and a positive carrying capacity, its factored rate must vanish at both \(P=0\) and \(P=K\). If substitution does not produce zero at either value, there is likely an algebraic or modeling error.
Finding equilibria is distinct from deciding what happens to solutions that start elsewhere. The roots tell us where the rate is zero; additional analysis is needed to describe the direction of change or behavior over time. For this lesson, the essential result is the identification and verification of the constant solutions.
Check Your Understanding
For each question, use the right-hand side of the differential equation and verify any proposed equilibrium by substitution.
- Find the equilibrium solutions of \(\frac{dP}{dt}=0.2P(1-P/500)\).
- In the standard logistic equation, which factor makes the rate zero at \(P=0\), and which makes it zero at \(P=K\)?
- For \(\frac{dP}{dt}=0.4P-0.002P^2\), factor the right-hand side and find both equilibrium values.
- Why is dividing the logistic equation by \(P\) not a valid way to show that \(P=0\) is not an equilibrium?
- State the condition on \(C\) for a constant function \(P(t)=C\) to solve an autonomous equation \(\frac{dP}{dt}=f(P)\).
