Problem 33 - 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. $$ \frac{d x}{d t}=\frac{1}{2}-\frac{x^{2}}{x^{2}+1} $$

Step-by-Step Solution

Answer

The differential equation has two equilibria at \(x = 1\) and \(x = -1\), both stable.

Step 1: Rewrite the Differential Equation

The given differential equation is \(\frac{dx}{dt} = \frac{1}{2} - \frac{x^2}{x^2+1}\). First, rewrite it to identify the zero points, i.e., \( \frac{dx}{dt} = 0 \). This leads to the equation \( \frac{1}{2} = \frac{x^2}{x^2+1} \).

Step 2: Set Up the Equation for Equilibrium Points

Transform the equation \( \frac{1}{2} = \frac{x^2}{x^2+1} \) into an algebraic equation. Multiply both sides by \( x^2 + 1 \) to get \( \frac{1}{2}(x^2 + 1) = x^2 \). This simplifies to \( \frac{x^2}{2} + \frac{1}{2} = x^2 \).

Step 3: Solve for \(x^2\)

Subtract \( \frac{x^2}{2} \) from both sides to get \( \frac{1}{2} = \frac{x^2}{2}\). Simplifying this equation leads to \( x^2 = 1 \). Therefore, \( x = 1 \) or \( x = -1 \). These are the equilibrium solutions.

Step 4: Analyze Stability

To determine stability, evaluate \( \frac{d}{dx}\left(\frac{dx}{dt}\right) \) at the equilibria. The derivative \( \frac{d}{dx}\left(\frac{1}{2} - \frac{x^2}{x^2+1}\right) \) becomes essential. Calculate \( \frac{d}{dx}\left(\frac{1}{2} - \frac{x^2}{x^2+1}\right) \) at \( x = 1 \) and \( x = -1 \).

Step 5: Classify Equilibria Using the Derivative

Compute \( \frac{d}{dx}\left(\frac{dx}{dt}\right) = -\frac{2x}{(x^2+1)^2} \). Evaluate at \( x = 1 \) to get a negative value, indicating stability. Evaluate at \( x = -1 \) to also get a negative value, indicating stability at both points.

Key Concepts

Equilibrium Points in Differential Equations

In the context of differential equations, equilibrium points are values of the variable (in this scenario, "x") where the rate of change (\( \frac{dx}{dt} \)) is zero.
This means that at an equilibrium point, the system is not changing.
To determine equilibrium points, the equation is rearranged and set to zero, solving for the variable.

In this exercise, when we set\( \frac{dx}{dt} = 0 \), it basically translates to finding the zeros of the equation given by\( \frac{1}{2} - \frac{x^2}{x^2+1} = 0 \).This leads us to the equilibrium points, \( x = 1 \) and \( x = -1 \).
These are the points where changes stop, and the system may either remain constant or shift, depending on stability.

Stability Analysis of Equilibrium Points

Once the equilibrium points are found, it's vital to understand their behavior.
This is where stability analysis comes in. By analyzing the derivative of the original differential equation with respect to "x," we can infer stability around these points.

If the derivative is negative around an equilibrium point, that point is considered "stable."

• Stable points imply that small deviations from the equilibrium will diminish over time. The system automatically corrects to return to that point.
• If the derivative is positive, the equilibrium point is "unstable." This means a slight disturbance will cause the system to move further away, amplifying the deviation.

The derivative calculation\( \frac{d}{dx}\left(\frac{dx}{dt}\right) = -\frac{2x}{(x^2+1)^2} \)helps classify the equilibrium points.
In this exercise, both points \( x = 1 \) and \( x = -1 \) yield a negative derivative, indicating that both are stable.

Calculus in Differential Equations

Calculus forms the backbone of solving and understanding differential equations.
It provides tools like differentiation which are crucial for stability analysis of equilibrium points.
The given differential equation is analyzed by first setting the change rate \( \frac{dx}{dt} \) to zero to find where the system remains unchanged.

Differentiation was used in solving the stability of equilibrium points by computing\( \frac{d}{dx} \left(\frac{dx}{dt}\right)\).
This second derivative highlights how the system behaves when slightly disturbed from equilibrium.

• The logic of calculus allows us to predict the system's stability without needing to simulate it.
• By examining the change caused by small variations, we can foresee whether equilibrium states are maintained or disrupted.

The power of calculus simplifies complex dynamic systems, making it a powerful tool for mathematicians and scientists.