Problem 1 - Calculus for Biology and Medicine

Question

Suppose that $$\frac{d y}{d x}=y(2-y)$$ (a) Find the equilibria of this differential equation. (b) Graph \(d y / d x\) as a function of \(y\), and use your graph to discuss the stability of the equilibria. (c) Compute the eigenvalues associated with each equilibrium, and discuss the stability of the equilibria.

Step-by-Step Solution

Answer

Equilibria are at \( y = 0 \) (unstable) and \( y = 2 \) (stable).

Step 1: Set Up Equilibrium Conditions

To find the equilibria of the differential equation \( \frac{dy}{dx} = y(2-y) \), we need to set the right-hand side to zero: \[ y(2-y) = 0 \].This gives us potential equilibria where the value of \( dy/dx \) becomes zero.

Step 2: Solve for Equilibria

Solve the equation from Step 1: \[ y(2-y) = 0 \].This results in two equilibria: \( y = 0 \) and \( y = 2 \).

Step 3: Plot \( dy/dx \) as a Function of \( y \)

Graph \( \frac{dy}{dx} = y(2-y) \) as a function of \( y \). The roots \( y = 0 \) and \( y = 2 \) are where \( \frac{dy}{dx} = 0 \). The expression is positive for \( 0 < y < 2 \) and negative for \( y < 0 \) or \( y > 2 \).

Step 4: Determine Stability From the Graph

From the graph in Step 3, if an equilibrium value is approached from either side and \( \frac{dy}{dx} \) leads away, it is unstable; if it leads towards the equilibrium, it is stable. Thus, \( y = 0 \) is unstable, and \( y = 2 \) is stable.

Step 5: Linearize Near Each Equilibrium to Find Eigenvalues

Linearize the differential equation around each equilibrium point to find stability through eigenvalues. Use the linear approximation:\[ \frac{d}{dy}(y(2-y)) = 2 - 2y \]. Evaluate this derivative at each equilibrium.

Step 6: Compute Eigenvalues for \( y = 0 \)

Substitute \( y = 0 \) into the derivative: \[ 2 - 2(0) = 2 \].The positive eigenvalue indicates \( y = 0 \) is unstable.

Step 7: Compute Eigenvalues for \( y = 2 \)

Substitute \( y = 2 \) into the derivative: \[ 2 - 2(2) = -2 \].The negative eigenvalue indicates \( y = 2 \) is stable.

Key Concepts

Equilibrium Points

Equilibrium points are fundamental in the study of differential equations as they represent steady-state solutions. These are the points where the system does not change, meaning the rate of change, or derivative, is zero. For the differential equation \( \frac{d y}{d x} = y(2-y) \), equilibrium points occur where \( y(2-y) = 0 \).
Solving this equation, we find two equilibria: \( y = 0 \) and \( y = 2 \). At these points, the system is in balance and no longer changes with respect to \( x \).

Equilibrium points offer insights into the behavior of the system—indicating where forces balance and where no net change occurs. These poise states are crucial for predicting long-term trends in dynamic systems. Understanding where these points lie is the first step in analyzing the system's overall stability and behavior.

Stability Analysis

Once equilibrium points are identified, the next crucial step is to assess their stability. Stability analysis involves examining whether small perturbations (or changes) will cause the system to return to equilibrium or diverge away from it.
To understand this concept, imagine perturbing the system slightly at the equilibrium point:

• If small changes cause the system to gravitate back toward equilibrium, the point is considered stable.
• If small changes veer the system away, the point is unstable.

For our specific equation, we can examine the graph of \( \frac{d y}{d x} = y(2-y) \) against \( y \):

• At \( y = 0 \), \( \frac{d y}{d x} \) is positive for \( y > 0 \) and negative for \( y < 0 \), driving away from \( y=0 \). This makes \( y = 0 \) unstable.
• At \( y = 2 \), the opposite occurs, with \( \frac{d y}{d x} \) leading back to the equilibrium point, showing stability.

Stability analysis helps predict system responses to disturbances and determine whether solutions will naturally persist or shift over time.

Linearization

Linearization simplifies analyzing non-linear differential equations by approximating them near their equilibrium points. It involves expanding the original equation into a linear format using derivatives, allowing for easier stability assessment through algebraic methods like eigenvalue calculation.
For \( \frac{d y}{d x} = y(2-y) \), we linearize around the equilibrium points \( y = 0 \) and \( y = 2 \). This is achieved by differentiating the right-hand side, \( \,\frac{d}{dy}(y(2-y)) = 2 - 2y \, \).

• For \( y = 0 \), substitute \( y \) with 0 to find that the slope of the linearized equation, or derivative, is 2: \( 2 - 2(0) = 2 \).
• For \( y = 2 \), substitution yields a slope of \( -2 \): \( 2 - 2(2) = -2 \).

With these linear approximations, we effectively transform the behavior of complex systems into a simpler one around equilibria, making analysis more straightforward. This technique is fundamental in understanding the local stability of differential systems.

Eigenvalues

Eigenvalues are central in determining the stability of equilibrium points through linearization. After linearizing a system, eigenvalues emerge from the derivatives calculated around these equilibrium points.
Consider the linearized form \( \, 2-2y \, \). The computed values give insight:

• Substituting \( y = 0 \) yields an eigenvalue of 2. A positive eigenvalue indicates the system is unstable since disturbances grow exponentially.
• In the case of \( y = 2 \), the eigenvalue is \(-2\), a negative number showing stability as it attracts solutions back after small disturbances.

Eigenvalues assess how deviations from equilibrium evolve. Positive eigenvalues suggest divergence and instability, while negative ones imply convergence and stability. By evaluating these values, one can predict long-term behavior, ensuring a firm grasp of the system's dynamics.