Problem 19 - Calculus for Biology and Medicine

Question

For make vector field plots of each of the differential equations. Find any equilibria of each differential equation and use your vector field plot to classify whether each equilibrium is stable or unstable. $$ \frac{d x}{d t}=x-x^{3} $$

Step-by-Step Solution

Answer

Equilibria are at \( x = 0 \) (stable), \( x = 1 \) (unstable), and \( x = -1 \) (unstable).

Step 1: Identify the Differential Equation

We are given the differential equation \( \frac{dx}{dt} = x - x^3 \). Our task is to find the equilibria and classify them using a vector field plot.

Step 2: Find the Equilibria

To find the equilibria, we set \( \frac{dx}{dt} = 0 \).\[ x - x^3 = 0 \]This implies the equation \( x(1 - x^2) = 0 \). Solving this gives the equilibria points: \( x = 0 \), \( x = 1 \), and \( x = -1 \).

Step 3: Draw the Vector Field Plot

We plot the vector field of \( \frac{dx}{dt} = x - x^3 \) by sketching arrows to represent the direction and magnitude of the rate of change \( \frac{dx}{dt} \) against \( x \).- For \( x < -1 \) and \( x > 1 \), \( \frac{dx}{dt} > 0 \), the arrows point right (indicating increasing \( x \)).- For \( -1 < x < 0 \) and \( 0 < x < 1 \), \( \frac{dx}{dt} < 0 \), the arrows point left (indicating decreasing \( x \)).

Step 4: Classify the Equilibria

Using the vector field plot:- At \( x = 0 \), arrows point towards \( x = 0 \) from both sides, making it a stable equilibrium.- At \( x = 1 \), arrows point away from \( x = 1 \) on both sides, making it an unstable equilibrium.- At \( x = -1 \), arrows also point away from \( x = -1 \) on both sides, making it an unstable equilibrium.

Key Concepts

Vector Field Plots

A vector field plot is a visual representation of a differential equation. It shows the direction and magnitude of the rate of change, which can help us understand how solutions to the equation behave over time. When you create a vector field plot for the equation \( \frac{dx}{dt} = x - x^3 \), it involves sketching arrows on a graph. These arrows indicate how the value of \( x \) changes with respect to time for different starting values of \( x \).

The length of the arrow represents the speed of change, with longer arrows indicating faster changes. In the context of our equation:

• For \( x < -1 \) and \( x > 1 \), the function \( x - x^3 > 0 \), meaning the arrows point to the right, indicating that \( x \) will increase.
• For \( -1 < x < 0 \) and \( 0 < x < 1 \), the function \( x - x^3 < 0 \), meaning the arrows point to the left, indicating that \( x \) will decrease.

These visual cues are essential as they give us insight into the behavior of the system without solving the equation explicitly for every initial condition.

Equilibria

Finding equilibria in a differential equation means identifying points where the rate of change is zero. These are the points at which the system is at rest, and no further changes occur unless the system is disturbed. For our given differential equation \( \frac{dx}{dt} = x - x^3 \), we find the equilibria by setting \( \frac{dx}{dt} = 0 \).

This leads us to the equation \( x(1 - x^2) = 0 \). Solving this gives the equilibria at \( x = 0 \), \( x = 1 \), and \( x = -1 \). Each equilibrium represents a constant solution to the differential equation, where the forces driving change are balanced.

The nature of the equilibria—whether they are stable or unstable—will dictate how the system responds to small disturbances at these points. This is why identifying and analyzing equilibria is critical in differential equations.

Stability Analysis

Stability analysis involves determining the behavior of a system near its equilibrium points. To classify the equilibria as stable or unstable, we use the vector field plot to observe the directions of the arrows around these points.

In our equation \( \frac{dx}{dt} = x - x^3 \), we found three equilibrium points: \( x = 0 \), \( x = 1 \), and \( x = -1 \). Here's how we analyze their stability:

• For \( x = 0 \): Arrows point towards the equilibrium from both sides, meaning any small disturbance will decay, bringing the system back to the equilibrium. Thus, \( x = 0 \) is a stable equilibrium.
• For \( x = 1 \) and \( x = -1 \): Arrows point away from the equilibrium on both sides, meaning any small disturbance will grow, moving the system away from the equilibrium. Therefore, both \( x = 1 \) and \( x = -1 \) are unstable equilibria.

Understanding stability is crucial for predicting how a system will behave over time, especially when subjected to small external perturbations. Stable equilibria tend to attract trajectories, while unstable ones repel them, drastically changing the dynamic outcome.