Problem 38 - 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. $$ A=\left[\begin{array}{rr} -3 & 1 \\ 1 & -2 \end{array}\right] $$

Step-by-Step Solution

Answer

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

Step 1: Find the Characteristic Equation

To determine the stability of the equilibrium at \((0,0)\), we need the eigenvalues of matrix \(A\). Find the characteristic equation of \(A\). The characteristic equation is given by: \[\det(A - \lambda I) = 0\]where \(I\) is the identity matrix of the same size as \(A\). So, we have:\[\det\left(\begin{bmatrix} -3 & 1 \ 1 & -2 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\right) = \det\left(\begin{bmatrix} -3-\lambda & 1 \ 1 & -2-\lambda \end{bmatrix}\right)\]This reduces to:\[(-3-\lambda)(-2-\lambda) - (1)(1) = \lambda^2 + 5\lambda + 5 = 0\]

Step 2: Solve for Eigenvalues

Solve the quadratic equation for \(\lambda\): \[\lambda^2 + 5\lambda + 5 = 0\]The solutions are given by the quadratic formula: \[\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a = 1\), \(b = 5\), and \(c = 5\). So:\[\lambda = \frac{-5 \pm \sqrt{5^2 - 4 \times 1 \times 5}}{2 \times 1} = \frac{-5 \pm \sqrt{25 - 20}}{2} = \frac{-5 \pm \sqrt{5}}{2}\]Thus, the eigenvalues are \(\lambda_1 = \frac{-5 + \sqrt{5}}{2}\) and \(\lambda_2 = \frac{-5 - \sqrt{5}}{2}\). Both eigenvalues are real.

Step 3: Analyze Stability

Examine the signs of the eigenvalues \(\lambda_1\) and \(\lambda_2\) to determine stability. Since both eigenvalues are real and negative (\(\lambda_1 < 0\) and \(\lambda_2 < 0\)), the equilibrium at \((0,0)\) is stable. Specifically, this means all trajectories in the phase plane will approach the origin over time.

Step 4: Classify the Equilibrium

With both eigenvalues being negative, the equilibrium point is classified as a sink. This is because all solutions will tend towards the equilibrium at \((0,0)\), indicating that the system dissipates energy, moving state vectors toward zero over time.

Key Concepts

Differential Equations

Differential equations are mathematical expressions involving derivatives, expressing how a certain quantity changes with respect to another. In this context, we have a system of linear differential equations represented by \( \frac{d\mathbf{x}}{dt} = A \mathbf{x}(t) \). Here, \( \mathbf{x}(t) \) represents a vector function that depends on time \( t \), while \( A \) is a matrix affecting how \( \mathbf{x}(t) \) evolves over time. For the given matrix \( A = \begin{bmatrix} -3 & 1 \ 1 & -2 \end{bmatrix} \), we're interested in understanding how solutions behave near the origin, \((0,0)\). These systems often represent physical processes such as motion, electrical circuits, or population dynamics. Understanding the solutions means finding the type of equilibrium and how the system responds when perturbed from this state. In our exercise, the critical tool for this analysis is the eigenvalues of the matrix \( A \).

Eigenvalues help decipher patterns of growth, decay, or oscillation in the system, crucial for conducting

Stability Analysis

Stability analysis involves studying how a system responds to small disturbances. To do this, we need the eigenvalues of the matrix \( A \), derived from its characteristic equation. This equation, \( \det(A - \lambda I) = 0 \), leads us to find potential eigenvalues \( \lambda \) that describe the system's behavior.

For our example, the characteristic equation derived was \( \lambda^2 + 5\lambda + 5 = 0 \). Solving this using the quadratic formula yielded two real and distinct eigenvalues \( \lambda_1 = \frac{-5 + \sqrt{5}}{2} \) and \( \lambda_2 = \frac{-5 - \sqrt{5}}{2} \). Both eigenvalues are negative, a key indicator in determining stability.

Since both eigenvalues are negative, any small displacement from the origin leads the system back to equilibrium. The equilibrium at \((0,0)\) is stable, meaning that over time, solutions tend to return to the origin without oscillating or diverging away.

Equilibrium Classification

After analyzing the stability, the next step is to classify the equilibrium point based on the characteristics of the eigenvalues. Our goal is to decide whether it is a sink, a source, or a saddle point.

A **sink** is characterized by all eigenvalues being negative, indicating that the system's trajectories move towards the equilibrium point over time, like water pouring into a drain. This is precisely what we have in our exercise, given both eigenvalues are negative. Thus, the system at \((0,0)\) acts as a sink.

In contrast, a **source** would have all positive eigenvalues, causing trajectories to move away from the equilibrium, suggesting instability. A **saddle point**, on the other hand, presents an unstable node with a mix of positive and negative eigenvalues, demonstrating attraction along certain directions and repulsion along others. Therefore, due to the negative eigenvalues, the equilibrium at \((0,0)\) is correctly classified as a sink, showing consistent stability.