Problem 80 - Calculus for Biology and Medicine

Question

Find all equilibria, and, by calculating the eigenvalue of the differential equation, determine which equilibria are stable and which are unstable. $$ \frac{d y}{d t}=y(y-1)(y-2) $$

Step-by-Step Solution

Answer

Equilibrium points are at \( y = 0, 1, 2 \); \( y = 1 \) is stable, \( y = 0 \) and \( y = 2 \) are unstable.

Step 1: Find Equilibrium Points

Equilibrium points occur when the derivative is zero, i.e., \( \frac{dy}{dt} = 0 \). Given the equation \( \frac{dy}{dt} = y(y-1)(y-2) \), set it equal to zero: \( y(y-1)(y-2) = 0 \). Therefore, the solutions are \( y = 0 \), \( y = 1 \), and \( y = 2 \). These are the equilibrium points.

Step 2: Calculate Derivative for Stability Analysis

To determine stability, examine the sign of the derivative of \( \frac{dy}{dt} \) with respect to \( y \). The derivative is \( \frac{d}{dy}(y(y-1)(y-2)) = (y-1)(y-2) + y(1)(y-2) + y(y-1)(1) = 3y^2 - 6y + 2 \).

Step 3: Find Eigenvalue at Each Equilibrium

Substitute each equilibrium point into the derivative from Step 2: \( y = 0: 3(0)^2 - 6(0) + 2 = 2 \), \( y = 1: 3(1)^2 - 6(1) + 2 = -1 \), \( y = 2: 3(2)^2 - 6(2) + 2 = 2 \). Therefore, the eigenvalues are \( \lambda = 2, -1, 2 \) for \( y = 0, 1, 2 \) respectively.

Step 4: Determine Equilibria Stability

The stability of each equilibrium point is determined by the sign of the eigenvalue: \( y = 0 \), eigenvalue \( \lambda = 2 > 0 \), unstable; \( y = 1 \), eigenvalue \( \lambda = -1 < 0 \), stable; \( y = 2 \), eigenvalue \( \lambda = 2 > 0 \), unstable.

Key Concepts

Differential Equations

Differential equations are mathematical expressions that relate a function with its derivatives. They are fundamental in modeling how quantities change over time or space. In this particular exercise, we explore a simple differential equation: \[ \frac{dy}{dt} = y(y-1)(y-2) \]This equation suggests how the value of \( y \) changes as time \( t \) progresses. Differential equations are used extensively to model real-world phenomena, like population growth, heat transfer, or even the motion of planets. They come in various forms, such as ordinary differential equations (ODEs) where derivatives are with respect to one independent variable, or partial differential equations (PDEs), which involve derivatives with respect to multiple variables.

In solving differential equations, finding equilibrium points is crucial. These are the values where the rate of change (or derivative) is zero – suggesting that the system is in a stable state with no change over time. In our example, the equilibrium points are found by solving \( y(y-1)(y-2) = 0 \), giving us points \( y = 0, 1, \) and \( 2 \). Each point tells us about potential long-term behavior of the system.

Stability Analysis

Stability analysis is the process of determining whether small disturbances or deviations from equilibrium points will diminish over time (making it stable), or whether they will grow, leading to an unstable system. In simpler terms, it helps predict whether a system will settle into a steady state or if it will spiral out of control.

To perform stability analysis on the given differential equation, we first calculate the derivative of \( \frac{dy}{dt} \) with respect to \( y \). This is known as the Jacobian in more complex systems, and it provides insight into how sensitive the system is around the equilibrium points. For the equation:\[ \frac{d}{dy}(y(y-1)(y-2)) = 3y^2 - 6y + 2 \]By substituting each equilibrium point \( y = 0, 1, \) and \( 2 \) into the derived expression, we determine how the system behaves. This will inform us if the equilibrium points are stable, meaning small changes will remain small, or unstable, where changes will grow.

Eigenvalues

Eigenvalues play a key role in determining stability within the context of differential equations. When analyzing a system's stability at its equilibrium points, we often look at the eigenvalues of the equation's linearization around these points. An eigenvalue essentially gives us information about the direction and rate at which disturbances grow or decay.

For the differential equation given by:\[ \frac{dy}{dt} = y(y-1)(y-2) \]The derivative of \( \, y(y-1)(y-2) \) yielded an expression \( 3y^2 - 6y + 2 \). At each equilibrium point, this expression evaluates to an eigenvalue. Here's what it means in practice:

• For \( y = 0 \), the eigenvalue is 2, indicating instability since positive eigenvalues make deviations grow.
• For \( y = 1 \), the eigenvalue is -1, signaling stability because negative eigenvalues mean disturbances shrink over time.
• For \( y = 2 \), like at \( y = 0 \), the eigenvalue is 2, leading to instability.

Understanding these values is crucial for predicting the behavior of the system as it helps in deciding whether the system will return to equilibrium after perturbation or not.