Problem 31 - Calculus for Biology and Medicine

Question

By breaking down each equation into two parts that you can sketch, determine how many equilib\mathrm{\\{} r i a ~ e a c h ~ d i f f e r e n t i a l ~ e q u a t i o n ~ h a s , ~ a n d ~ c l a s s i f y ~ t h e m ~ a s ~ s t a b l e ~ or unstable. You do not need to determine the location of the equilibria. $$ \frac{d x}{d t}=\frac{1}{x}-\frac{x}{x+1} \quad x>0 $$

Step-by-Step Solution

Answer

The differential equation has one equilibrium at \( x = 1 \), which is stable.

Step 1: Understand the Differential Equation

We need to find the equilibrium points for the differential equation \( \frac{dx}{dt} = \frac{1}{x} - \frac{x}{x+1} \). An equilibrium occurs when \( \frac{dx}{dt} = 0 \). So, set \( \frac{1}{x} - \frac{x}{x+1} = 0 \) and solve for \( x \).

Step 2: Rewrite the Equation for Equilibrium

Set \( \frac{1}{x} - \frac{x}{x+1} = 0 \). This equation can be rewritten by finding a common denominator, which gives: \( \frac{(x+1) - x^2}{x(x+1)} = 0 \). For the whole expression to be zero, the numerator must be zero.

Step 3: Solve the Numerator for Equilibrium Points

The numerator of the equation is \( 1 - x^2 = 0 \). Solving for \( x \) gives \( x^2 = 1 \), so \( x = 1 \) or \( x = -1 \). Since \( x > 0 \), we consider only \( x = 1 \).

Step 4: Determine Equilibrium Stability

To determine stability, examine the sign of \( \frac{dx}{dt} \) around \( x = 1 \). If \( x < 1 \), substitute a value (like \( x = 0.5 \)) into \( \frac{1}{x} - \frac{x}{x+1} \): it is positive. If \( x > 1 \), test a value (like \( x = 2 \)): it is negative. Thus, \( x = 1 \) is a stable equilibrium because \( \frac{dx}{dt} \) changes from positive to negative as \( x \) passes through 1.

Key Concepts

Equilibrium Points

In differential equations, equilibrium points, also known as steady states, are where the system's variables remain constant over time. They occur when the derivative (rate of change) is zero, which in this case means \( \frac{dx}{dt} = 0 \). To find these points, we solve the equation given in the differential equation. For the exercise, this involved simplifying \( \frac{1}{x} - \frac{x}{x+1} = 0 \).

• Set the equation to zero: \( \frac{1}{x} - \frac{x}{x+1} = 0 \).
• Finding a common denominator results in \( \frac{(x+1) - x^2}{x(x+1)} = 0 \).
• Focus on the numerator: \( 1 - x^2 = 0 \).

This gives potential equilibrium solutions at \( x = 1 \) and \( x = -1 \). However, considering the condition \( x > 0 \), only \( x = 1 \) qualifies as an equilibrium point in this context.

Stability Analysis

Stability analysis of equilibrium points helps us understand how a system behaves when it is slightly disturbed from an equilibrium. Because an equilibrium point not only serves as a solution but also offers insight into the system's dynamics. In the given example, we explored the behavior of the system around the identified equilibrium point \( x = 1 \).

• First, calculate \( \frac{dx}{dt} \) at points slightly less than 1 (e.g., \( x = 0.5 \)). The result is positive, indicating that the system moves towards \( x = 1 \) when \( x < 1 \).
• Next, compute \( \frac{dx}{dt} \) at points slightly greater than 1 (e.g., \( x = 2 \)). The outcome is negative, implying the system stabilizes back to \( x = 1 \) when \( x > 1 \).

This shift in sign from positive to negative demonstrates that \( x = 1 \) is a stable equilibrium. The system naturally returns to this point if there are small perturbations.

Mathematical Modeling

Mathematical modeling involves using equations to represent real-world systems, allowing us to predict behaviors and outcomes under various scenarios. Differential equations like the one in our exercise are foundational tools in modeling dynamic systems. They describe how quantities change over time, providing valuable insights into systems ranging from biological populations to engineering systems.
To effectively use mathematical models:

• Identify key variables and parameters that affect the system.
• Determine the relationships between variables using appropriate differential equations.
• Analyze equilibrium points to ascertain long-term behavior and stability.

In our differential equation, we model changes in \( x \) over time. By analyzing equilibrium points and stability, we understand how \( x \) behaves. This understanding can translate into making informed decisions or predictions based on the modeled system. Mathematical modeling's power lies in its ability to simplify complex real-world phenomena into manageable equations that offer deep insights.