Problem 23 - 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. $$ \frac{d x}{d t}=\frac{x^{2}-x}{x^{2}+1} $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Differential Equation

The given differential equation is \( \frac{d x}{d t} = \frac{x^{2}-x}{x^{2}+1} \). This is an ordinary differential equation that describes how the variable \( x \) changes with respect to time \( t \). Our task is to find the equilibria of this equation and determine their stability.

Step 2: Find Equilibria

The equilibria of the differential equation occur where \( \frac{d x}{d t} = 0 \). Set \( \frac{x^2-x}{x^2+1} = 0 \) and solve for \( x \). The numerator must be zero, so we solve \( x(x-1) = 0 \). This gives us the equilibrium points \( x = 0 \) and \( x = 1 \).

Step 3: Analyze Stability at x=0

To classify the stability of the equilibrium at \( x = 0 \), consider the sign of \( \frac{d x}{d t} \) near this point. Test values just below and above \( x = 0 \) (e.g., \( x = -0.1 \) and \( x = 0.1 \)). If \( \frac{d x}{d t} > 0 \) for \( x < 0 \) and \( \frac{d x}{d t} < 0 \) for \( x > 0 \), then it's stable. Here, try calculating the expressions: \( \frac{(-0.1)^2 - (-0.1)}{(-0.1)^2 + 1} \) and \( \frac{(0.1)^2 - 0.1}{(0.1)^2 + 1} \).

Step 4: Analyze Stability at x=1

Similarly, analyze stability at \( x = 1 \). Again, when testing \( x = 0.9 \) and \( x = 1.1 \), if \( \frac{d x}{d t} < 0 \) to the left \( (x = 0.9) \) and \( \frac{d x}{d t} > 0 \) to the right \( (x = 1.1) \), then it is unstable. Calculate \( \frac{(0.9)^2 - 0.9}{(0.9)^2 + 1} \) and \( \frac{(1.1)^2 - 1.1}{(1.1)^2 + 1} \).

Step 5: Create a Vector Field Plot

Create a plot of the vector field by evaluating \( \frac{x^2-x}{x^2+1} \) for a range of \( x \) values. This gives you a visual representation of how \( x \) changes over time, helping to see where \( x \) moves towards or away from the equilibrium points.

Step 6: Interpret the Vector Field Plot

Using the vector field plot, verify the stability of equilibria points determined algebraically. A stable equilibrium will have vectors pointing towards it from both directions, whereas an unstable equilibrium will have vectors pointing away.

Key Concepts

Vector Field

A vector field provides a visual representation of a differential equation, showing the direction and magnitude of change for different states of the system. In the context of our exercise, the vector field will visually describe how the variable \(x\) evolves with time, according to the given differential equation:\[\frac{d x}{d t} = \frac{x^2-x}{x^2+1}\]Creating a vector field involves plotting vectors whose directions are determined by the derivative \(\frac{d x}{d t}\) and whose lengths indicate the rate of change. To interpret this vector field:

• If vectors point towards an equilibrium point, it indicates that the system tends to settle to this state, suggesting a stable equilibrium.
• Vectors pointing away suggest that the system tends to diverge from the equilibrium, indicating instability.

Thus, vector fields are essential tools for understanding the dynamics described by a given differential equation. They give both qualitative insights into the movement of solutions and insight into the nature of equilibria.

Equilibria

Equilibria in a differential equation refer to the solutions where the system remains in a constant state over time. For the given differential equation, equilibrium occurs where the derivative \(\frac{d x}{d t}\) equals zero. This indicates no change in \(x\) with respect to time.In our exercise:

• Equilibrium points can be found by solving the equation \( \frac{x^2-x}{x^2+1} = 0 \).
• This requires setting the numerator \(x^2-x\) equal to zero, yielding the solutions \(x=0\) and \(x=1\).

An equilibrium point serves as a potential resting position where, if the system reaches this point, it neither moves away nor approaches. Understanding equilibria is vital for analyzing how systems behave or evolve over time.

Stability Analysis

Stability analysis is the process of determining whether equilibria are stable or unstable. It is crucial to understanding how a system behaves when slightly perturbed from its equilibrium state.For our differential equation:

• Assess stability around each equilibrium point by evaluating \(\frac{d x}{d t}\) at points immediately adjacent.
• For \(x=0\), check values such as \(x=-0.1\) and \(x=0.1\). If \( \frac{d x}{d t} > 0 \) for \( x < 0 \) and \( \frac{d x}{d t} < 0 \) for \( x > 0 \), the equilibrium is stable.
• At \(x=1\), testing nearby values like \(x=0.9\) and \(x=1.1\) will reveal stability. Here, \( \frac{d x}{d t} < 0 \) when moving left and \( \frac{d x}{d t} > 0 \) moving right indicates instability.

A stable equilibrium returns the system to equilibrium after minor disruptions, while an unstable equilibrium means the system moves away when slightly perturbed.