Problem 10 - Thomas Calculus

Question

The autonomous differential equations in Exercises \(9-12\) represent models for population growth. For each exercise, use a phase line analysis to sketch solution curves for \(P(t),\) selecting different starting values \(P(0) .\) Which equilibria are stable, and which are unstable? $$\frac{d P}{d t}=P(1-2 P)$$

Step-by-Step Solution

Answer

The stable equilibrium is at \(P = \frac{1}{2}\), and the unstable equilibrium is at \(P = 0\).

Step 1: Identify Equilibria

To find the equilibrium points, set the derivative of the function to zero: \(\frac{d P}{d t} = P(1-2P) = 0\). This implies either \(P = 0\) or \(1 - 2P = 0\). Solving \(1 - 2P = 0\), we get \(P = \frac{1}{2}\). Thus, there are two equilibria: \(P = 0\) and \(P = \frac{1}{2}\).

Step 2: Determine Stability of Equilibria

To determine the stability, evaluate the derivative of \(f(P) = P(1-2P)\): \(f'(P) = 1 - 4P\). For \(P = 0\), \(f'(0) = 1\), which is positive, making this equilibrium unstable. For \(P = \frac{1}{2}\), \(f'(\frac{1}{2}) = 1 - 4(\frac{1}{2}) = -1\), which is negative, making this equilibrium stable.

Step 3: Draw the Phase Line

Draw a vertical line representing the \(P\)-axis with two equilibria at \(P=0\) (unstable) and \(P=\frac{1}{2}\) (stable). In regions: \(P < 0\), \(f(P) < 0\) and points move to the left, and for \(0 < P < \frac{1}{2}\), \(f(P) > 0\), points move to the right, and for \(P > \frac{1}{2}\), \(f(P) < 0\), points move to the left.

Step 4: Sketching Solution Curves

For different initial conditions \(P(0)\): if \(P(0) < 0\), solutions decrease further; if \(0 < P(0) < \frac{1}{2}\), solutions increase towards \(\frac{1}{2}\); if \(P(0) = \frac{1}{2}\), solutions remain constant; and if \(P(0) > \frac{1}{2}\), solutions decrease towards \(\frac{1}{2}\). This confirms the earlier stability analysis.

Key Concepts

Population Growth Models

Population growth models help us understand how populations change over time. They are often represented by differential equations. These equations model how a certain quantity, in this case, a population size, changes with time. For this exercise, our differential equation is given by\[\frac{dP}{dt} = P(1 - 2P) \ .\]This model suggests that the change in population over time depends on both the current population size \(P\) and a factor \((1 - 2P)\), which could represent the effect of environmental limits or available resources.

In general, population growth models can be used to:

• Predict future population sizes based on current conditions.
• Analyze how populations reach equilibrium points, where the growth rate becomes zero.
• Understand stability by exploring how small changes in population size affect its long-term behavior.

A crucial part of these models is identifying equilibrium points, where the population doesn't change because growth is balanced by limitations.

Phase Line Analysis

Phase line analysis is a valuable technique used to study differential equations, particularly when analyzing the stability and direction of solutions. In our context, we focus on the population model\[\frac{dP}{dt} = P(1 - 2P) \ .\]To use phase line analysis, we first identify the equilibrium points by setting the derivative equal to zero, giving us \(P = 0\) and \(P = \frac{1}{2}\). These points indicate where the population may stabilize.

Next, we create a phase line diagram. This diagram is a simple visual representation of equilibrium points and the stability of these points. Here's how:

• Draw a vertical line representing all possible values of \(P\) from negative to positive infinity.
• Mark the equilibria on this line: \(P=0\) and \(P=\frac{1}{2}\).
• Determine the direction of arrows between these points:
  • For \(P < 0\), the derivative is negative, so solutions move left.
  • For \(0 < P < \frac{1}{2}\), the derivative is positive, so solutions move right.
  • For \(P > \frac{1}{2}\), the derivative is negative, so solutions move left again.

Phase line analysis helps visualize how solutions behave for different initial conditions and whether they approach or diverge from equilibrium points.

Stability of Equilibria

Understanding the stability of equilibria in a system is critical for predicting long-term behavior of a model. For our differential equation\[\frac{dP}{dt} = P(1 - 2P) \ ,\]we identify two equilibria: \(P = 0\) and \(P = \frac{1}{2}\). The concept of stability determines if these points are attractive (stable) or repulsive (unstable).To assess stability, we calculate the derivative of the function with respect to \(P\), which is \[ f'(P) = 1 - 4P \ .\]We then evaluate this derivative at each equilibrium:

• At \(P = 0\): \(f'(0) = 1\), positive, indicating this point is unstable. If the population starts near zero, it will diverge away.
• At \(P = \frac{1}{2}\): \(f'\left(\frac{1}{2}\right) = -1\), negative, meaning this point is stable. Populations near \(\frac{1}{2}\) will tend to stay or return to this equilibrium.

Stable equilibria act like attractors, pulling solutions toward them, while unstable equilibria push solutions away. This analysis is crucial for predicting the evolution of population dynamics under different initial conditions.