Problem 29 - Calculus for Biology and Medicine

Question

In Problems 29-42, 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 equilib\mathrm{\\{} r i u m ~ a c c o r d i n g ~ t o ~ w h e t h e r ~ i t ~ i s ~ a ~ s i n k , ~ a ~ s o u r c e , ~ o r ~ a ~ s a d d l e ~ point. \(A=\left[\begin{array}{ll}2 & 1 \\ 0 & 3\end{array}\right]\)

Step-by-Step Solution

Answer

The equilibrium is a source, as both eigenvalues are positive.

Step 1: Understand the Scenario

You are given a differential equation of the form \(\frac{d \mathbf{x}}{d t}=A \mathbf{x}(t)\) with \(A\) as a matrix. The task involves analyzing the stability of the equilibrium point \((0, 0)\) and classifying it.

Step 2: Find Eigenvalues

To find the eigenvalues of matrix \(A\), solve the characteristic equation: \(\det(A - \lambda I) = 0\). For the given matrix \(A=\begin{bmatrix} 2 & 1 \ 0 & 3 \end{bmatrix}\), calculate \(\det\left(\begin{bmatrix} 2-\lambda & 1 \ 0 & 3-\lambda \end{bmatrix}\right)\). This results in \((2-\lambda)(3-\lambda) - 0 = 0\). Simplifying, we get \(\lambda^2 - 5\lambda + 6 = 0\).

Step 3: Solve for Eigenvalues

Solve the quadratic equation \(\lambda^2 - 5\lambda + 6 = 0\) using the quadratic formula: \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -5\), \(c = 6\). Thus, \(\lambda = \frac{5 \pm \sqrt{1}}{2}\), resulting in \(\lambda_1 = 3\) and \(\lambda_2 = 2\).

Step 4: Classify Stability and Equilibrium

Since both eigenvalues \(\lambda_1 = 3\) and \(\lambda_2 = 2\) are real and positive, vector field trajectories move away from the equilibrium. This implies that the equilibrium \((0, 0)\) is unstable and known as a source.

Key Concepts

Eigenvalues

Eigenvalues are incredibly important in understanding the behavior of differential equations, like the one given in the problem. In this context, they help us determine how solutions to the differential equation behave over time. The eigenvalues of a matrix are special numbers that allow us to see how a system changes. They come from the characteristic equation, which is based on the matrix of the system itself. To find them, you solve this equation by setting the determinant of matrix \( (A - \lambda I) \) equal to zero, where \( I \) is the identity matrix.In our specific exercise, we computed the eigenvalues for matrix \[A=\begin{bmatrix} 2 & 1 \ 0 & 3 \end{bmatrix}\], and found them to be \( \lambda_1 = 3 \) and \( \lambda_2 = 2 \). These values are crucial as they reveal if the solutions to the differential equation are stable or not.

Stability Analysis

Stability analysis helps us figure out if small disturbances or changes will die out or get bigger over time. It is an essential tool to predict the future behavior of a system at equilibrium. We conduct stability analysis by looking at the signs of the eigenvalues determined earlier. - If all eigenvalues are negative, the system tends to return to equilibrium, indicating stability.- If all are positive, like in our exercise ( \( \lambda_1 = 3 \) and \( \lambda_2 = 2 \)), solutions move away from equilibrium, making the system unstable.- Mixed signs imply that the equilibrium point is a saddle point. This means it's stable along some directions but unstable in others.In our case, since both eigenvalues are positive, the disturbance grows, indicating that our equilibrium point (0,0) is unstable. Our system will not settle back to equilibrium after a slight perturbation.

Equilibrium Classification

Equilibrium classification involves identifying the type of equilibrium in the system, usually categorized as a sink, a source, or a saddle point.We classify the equilibrium by the nature of its stability:- **Source**: An equilibrium is a source if all eigenvalues are positive, just like in our exercise. This means that any disturbance will cause the system to move away from the equilibrium point. In simple terms, think of it like a hilltop where any small push sends an object rolling away.- **Sink**: If all eigenvalues were negative, the equilibrium would be a sink. This implies stability as disturbance fades away over time, dragging the system back to equilibrium.- **Saddle Point**: If the eigenvalues contain both positive and negative values, the equilibrium is a saddle point. This creates areas of stability along some directions and instability along others, allowing disturbances to grow or shrink depending on the direction.Aligning with the results from our stability analysis, we conclude that the equilibrium at (0,0) is a source because the eigenvalues \( \lambda_1 = 3 \) and \( \lambda_2 = 2 \) are both positive.