Question

Suppose that a certain population obeys the logistic equation \(d y / d t=r y[1-(y / K)]\). (a) If \(y_{0}=K / 3\), find the time \(\tau\) at which the initial population has doubled. Find the value of \(\tau\) corresponding to \(r=0.025\) per year. (b) If \(y_{0} / K=\alpha,\) find the time \(T\) at which \(y(T) / K=\beta,\) where \(0<\alpha, \beta<1 .\) Observe that \(T \rightarrow \infty\) as \(\alpha \rightarrow 0\) or as \(\beta \rightarrow 1 .\) Find the value of \(T\) for \(r=0.025\) per year, \(\alpha=0.1\) and \(\beta=0.9 .\)

Short answer

Also, what is the time T in years at which \(y(T) / K=\beta\), when \(y_{0} / K=\alpha,\) for given values of \(r=0.025\) per year, \(\alpha=0.1\), and \(\beta=0.9\)? Solution: Based on our calculations, the time τ at which the initial population has doubled is around \(\frac{27.73}{K}\) years. For the time T at which \(y(T) / K=\beta\) when \(y_{0} / K=\alpha\), the solution is approximately \(\frac{205.17}{K}\) years with given values of \(r=0.025\) per year, \(\alpha=0.1\), and \(\beta=0.9\).

Step-by-step solution

Solve the logistic differential equation

To solve the logistic differential equation, we first need to separate variables: $$\frac{dy}{y(1-\frac{y}{K})} = r dt$$ Now, we can use partial fractions to rewrite the left-hand side of the equation: $$\frac{1}{y(1-\frac{y}{K})} = \frac{A}{y} + \frac{B}{1-\frac{y}{K}}$$ Equate the numerators: $$1 = A(1-\frac{y}{K}) + B \cdot y$$ By comparing the coefficients, we obtain \(A=-B\) and \(B=-\frac{1}{K}\). Thus, the equation becomes: $$\frac{dy}{y(1-\frac{y}{K})} = -\frac{1}{K} \left(\frac{1}{y} - \frac{1}{1-\frac{y}{K}} \right) r dt$$ Now we can integrate both sides: $$\left[-\frac{1}{K} \ln(|y(1-\frac{y}{K})|) \right] = rt + C$$

Find the time τ when the initial population has doubled

To find the time τ, we first need to determine the constant C. We can do this by inputing the initial conditions: \(y_{0}=K / 3\) and \(t=0\). Solve for C: $$-\frac{1}{K} \ln \left(\frac{K}{3} \cdot \frac{2K}{3} \right) = C$$ Now, we need to find the time at which the population has doubled, i.e., \(y = 2 y_{0}\). Substitute the equation: $$-\frac{1}{K} \ln \left(\frac{2K}{3} \cdot \frac{K}{3} \right) = r\tau + C$$ Subtract the equation for C from the equation above: $$r\tau = \frac{1}{K} \ln \left( \frac{\frac{2K}{3} \cdot \frac{K}{3}}{\frac{K}{3} \cdot \frac{2K}{3}} \right)$$ Now we can plug in the given value for r: $$\tau = \frac{1}{0.025K} \ln \left( \frac{\frac{2K}{3} \cdot \frac{K}{3}}{\frac{K}{3} \cdot \frac{2K}{3}} \right)$$ After solving, we obtain: $$\tau = \frac{1}{0.025K} \ln(2) \approx \frac{27.73}{K}$$

Find the time T at which \(y(T) / K=\beta\) when \(y_{0} / K=\alpha\)

For this step, we need to change the initial condition to \(y_{0} / K=\alpha.\) Then, we can find the time T when \(y(T) / K=\beta\). First, let's find the constant C: $$C = -\frac{1}{K} \ln(\alpha K(\frac{K - y_{0}}{K}))$$ Now, we set the new condition \(y(T) / K=\beta\): $$-\frac{1}{K} \ln \left(\beta K \cdot (K-\beta K) \right) = rT + C$$ Subtract the equation for C from the equation above: $$rT = \frac{1}{K} \ln \left( \frac{\alpha K(\frac{K - y_{0}}{K})}{\beta K \cdot (K-\beta K)} \right)$$ With given values, \(r=0.025\) per year, \(\alpha=0.1\), and \(\beta=0.9\), we obtain: $$T = \frac{1}{0.025K} \ln \left( \frac{0.1K \cdot (0.9K)}{0.9K \cdot (0.1K)} \right)$$ After solving this, we obtain: $$T = \frac{1}{0.025K} \ln(81) \approx \frac{205.17}{K}$$

Key concepts

Population Dynamics

Population dynamics explores how populations change over time and what factors contribute to these changes. In the context of the logistic equation, this involves the growth of a population in an environment with limited resources.
The logistic model assumes that a population will expand rapidly when resources are plentiful. However, as resources become scarce, the growth rate decreases, stabilizing the population.
This can be represented mathematically by:

• Initial rapid growth
• Slowing as resources deplete
• Equilibrium at carrying capacity (K)

The logistic equation is given by \( \frac{dy}{dt} = r y \left(1-\frac{y}{K}\right) \) where:

• \(y\) is the population size
• \(r\) is the intrinsic growth rate
• \(K\) is the carrying capacity

This model helps in understanding how a population can grow towards its carrying capacity over time, representing realistic limitations on growth in an ecosystem.

Differential Equations

Differential equations are a class of mathematical equations used to describe how a quantity changes with another variable, typically time. In population dynamics, we use them to understand how populations evolve.
The logistic equation is a type of differential equation known as a first-order nonlinear ordinary differential equation. This equation can model real-world phenomena like population growth limited by resources.
To solve a differential equation like the logistic equation \( \frac{dy}{dt} = r y \left(1-\frac{y}{K}\right) \), we often separate variables or use integrating techniques.
Thus, understanding differential equations is crucial to predicting the future populations based on current data and understanding the factors influencing change.

Partial Fractions

Partial fractions is a mathematical technique used to decompose a complex rational expression into simpler fractions that are easier to integrate.
This method is particularly useful when solving integral problems in differential equations, like the logistic equation's separation of variables step.
During the solution process of the logistic equation, we encounter an expression \( \frac{1}{y\left(1-\frac{y}{K}\right)} \). This can be decomposed using partial fractions into:

• \( \frac{A}{y} + \frac{B}{1-\frac{y}{K}} \)

By finding suitable values for constants \(A\) and \(B\), the integration becomes straightforward, helping us to solve the differential equation.
This technique leverages the known forms and simplifications to solve the complex logistic growth equation efficiently, providing a deeper insight into population growth dynamics.