Problem 30 - Calculus for Biology and Medicine

Question

We consider differential equations of the form $$ \frac{d \mathbf{x}}{d t}=A \mathbf{x}(t) $$ where $$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] $$ The eigenvalues of A will be real, distinct, and nonzero. Analyze the stability of the equilibrium \((0,0)\), and classify the equilibrium according to whether it is a sink, a source, or a saddle point. $$ A=\left[\begin{array}{rr} -1 & 0 \\ 0 & -4 \end{array}\right] $$

Step-by-Step Solution

Answer

The equilibrium \((0,0)\) is stable and acts as a sink.

Step 1: Identify the System Matrix

The system matrix is given as \( A = \begin{bmatrix} -1 & 0 \ 0 & -4 \end{bmatrix} \). This matrix will be used to find the eigenvalues, which help in identifying stability and type of equilibrium.

Step 2: Compute the Eigenvalues

To find the eigenvalues \( \lambda \) of matrix \( A \), solve the characteristic equation \( \det(A - \lambda I) = 0 \). For matrix \( A = \begin{bmatrix} -1 & 0 \ 0 & -4 \end{bmatrix} \), the characteristic equation is \((-1 - \lambda)(-4 - \lambda) = 0\). Therefore, the eigenvalues are \( \lambda_1 = -1 \) and \( \lambda_2 = -4 \).

Step 3: Determine Stability from Eigenvalues

The eigenvalues \( \lambda_1 = -1 \) and \( \lambda_2 = -4 \) are both negative and real. This means that any solution \( \mathbf{x}(t) \) of the differential equation \( \frac{d \mathbf{x}}{d t}=A \mathbf{x}(t) \) will decay to zero as \( t \to \infty \). Therefore, the equilibrium point \((0,0)\) is stable.

Step 4: Classify the Equilibrium

A stable equilibrium with both eigenvalues negative indicates a sink. Thus, the equilibrium \((0,0)\) acts as a sink.

Key Concepts

Eigenvalues: Understanding Their Importance

Eigenvalues are a key concept in understanding differential equations. They provide crucial insight into the behavior of systems represented by matrices. When we have a system matrix like

• A = \( \begin{bmatrix} -1 & 0 \ 0 & -4 \end{bmatrix} \)

we need to compute its eigenvalues to analyze the system's properties. The eigenvalues are found by solving the characteristic equation, which in this case is

• \((-1 - \lambda)(-4 - \lambda) = 0\).

The roots of this equation, \( \lambda_1 = -1 \) and \( \lambda_2 = -4 \), are the eigenvalues. They help determine how solutions of the differential equation change over time.
In summary, eigenvalues are instrumental in assessing the dynamics of systems, predicting how solutions evolve, and understanding the tendencies towards stability or instability.

Stability Analysis: Decoding Equilibrium Behavior

Stability analysis is a process used to understand how solutions to differential equations respond over time. It often relies on the eigenvalues of the system matrix. In our case, the matrix

• A = \( \begin{bmatrix} -1 & 0 \ 0 & -4 \end{bmatrix} \)

produces eigenvalues \( \lambda_1 = -1 \) and \( \lambda_2 = -4 \). Both eigenvalues are negative, indicating stability in the system.
When eigenvalues are negative, the solutions of the matrix decay towards zero, meaning any disturbances in the system will diminish over time, leading the system back to equilibrium. Technically, solutions \( \mathbf{x}(t) \) decay to zero as time \( t \to \infty \), indicating a stable equilibrium.
Thus, in stability analysis, eigenvalues offer a roadmap for predicting long-term behavior of solutions to differential equations, helping determine if a system will remain stable or become chaotic.

Equilibrium Classification: Identifying System Types

Equilibrium classification helps us categorize the type of equilibrium point by using the signs of the eigenvalues. In a two-dimensional system, such as

• A = \( \begin{bmatrix} -1 & 0 \ 0 & -4 \end{bmatrix} \)

we look at the eigenvalues \( \lambda_1 = -1 \) and \( \lambda_2 = -4 \).
Here both eigenvalues are negative, indicating that the equilibrium point \((0,0)\) is a sink. In this context, a sink represents a system where trajectories eventually converge towards the equilibrium point as time passes.
Equilibrium classification helps determine if a point is a sink, representing a stable system, a source, indicating instability, or a saddle point, which can suggest more complex dynamics. In our exercise, a negative pair of eigenvalues classifies the equilibrium point as a sink, showing that the system stabilizes to \((0,0)\).