Using an Initial Value in the Logistic Formula

From an Initial Population to a Prediction Formula

A logistic differential equation describes a population whose rate of change depends on both its current size and a carrying capacity. The explicit logistic solution contains a constant that must be determined from an initial value. Once that constant is known, the formula gives a population prediction at any chosen time in its domain.

For this tutorial, use the logistic model

where \(P(t)\) is the population at time \(t\), \(r>0\) is the growth-rate parameter, and \(K>0\) is the carrying capacity. If time is measured in days, for example, \(r\) has units of inverse days. The product \(r(t-t_0)\) is dimensionless, as required in the exponential.

The explicit solution, with its constant not yet determined, can be written as

Here \(C\) is a dimensionless constant, and \(t_0\) is a reference time. An initial condition specifies the population at that time: \(P(t_0)=P_0\). The main task is to use this information to find \(C\).

Finding the Constant from the Initial Condition

At the initial time, the exponential factor is \(e^0=1\). Substituting \(t=t_0\) and \(P(t_0)=P_0\) into the solution gives \(P_0=K/(1+C)\). Solving this equation for \(C\) produces the formula above.

Why This Substitution Works

The initial condition determines the constant because it fixes the value of the solution at one specific time. Substituting that time into the explicit formula gives an equation containing only the known initial population, the carrying capacity, and \(C\).

Multiply both sides by \(1+C\), then divide by \(P_0\), which is allowed when \(P_0>0\):

This derivation also shows why the initial time belongs in the exponent as \(t-t_0\): at the specified initial time, that difference is zero, so the exponential contributes exactly \(1\).

Checking the Formula Against the Differential Equation

After finding a constant, it is useful to check that the resulting expression both matches the initial condition and satisfies the logistic differential equation. The following calculation verifies the general explicit form wherever its denominator is nonzero.

Write \(D=1+C e^{-r(t-t_0)}\), so \(P=K/D\). By the chain rule,

On the other hand, substituting \(P=K/D\) into the right-hand side of the differential equation gives

The two expressions agree. Also, when \(t=t_0\), the formula gives \(P(t_0)=K/(1+C)\); with \(C=(K-P_0)/P_0\), this equals \(P_0\). Thus the formula satisfies both the differential equation and the initial condition, provided its denominator is nonzero.

Worked Example: A Prediction Three Days After the Initial Measurement

Worked Example: An Initial Value at a Nonzero Time

Special Initial Values and the Formula’s Domain

The formula \(C=(K-P_0)/P_0\) requires division by \(P_0\). If \(P_0=0\), use the equilibrium solution \(P(t)=0\) instead. If \(P_0=K\), then \(C=0\), and the explicit formula gives the constant solution \(P(t)=K\). These are equilibrium cases, so they do not require a nonzero denominator in the expression for \(C\).

For a population model with \(P_0>0\), \(K>0\), and \(r>0\), the formula is defined for every future time \(t\ge t_0\). Indeed, \(C=(K-P_0)/P_0>-1\). If \(C\ge 0\), then \(1+C e^{-r(t-t_0)}\ge 1\). If \(-1<C<0\), then for \(t\ge t_0\), \(0<e^{-r(t-t_0)}\le1\), so \(1+C e^{-r(t-t_0)}\ge1+C>0\). In either case the denominator is positive for future predictions.

A Common Mistake: Reversing the Ratio

The constant is the difference between carrying capacity and initial population, divided by the initial population. It is not generally the initial population divided by carrying capacity.

Using the Result Carefully

Before using a logistic formula to answer a contextual question, check that the initial condition and parameters are attached to the right quantities. In particular, distinguish the initial time \(t_0\) from elapsed time and distinguish the carrying capacity \(K\) from the initial population \(P_0\).

• Use \(P(t_0)=P_0\) to calculate \(C\), rather than guessing the constant.
• Use \(t-t_0\) in the exponent when the initial measurement is at a nonzero time.
• Keep \(r\) and time units consistent, so \(r(t-t_0)\) is dimensionless.
• Check the initial value by substituting \(t=t_0\) into the completed formula.
• For a population initially equal to zero, use the zero equilibrium solution rather than dividing by \(P_0\).
• Include population units in the prediction and round only at the final step.

These checks turn the explicit solution into a usable prediction rule: the model parameters and the initial measurement determine the constant, and the resulting formula can then be evaluated at the time of interest.