Problem 60 - Calculus for Biology and Medicine

Question

We consider differential equations of the form \(\frac{d x}{d t}=A x(t)\) where \(A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]\) Analyze the stability of the equilibrium \(\mathbf{( 0 , 0 ) , \text { and classify the }}\) equilibrium. \(A=\left[\begin{array}{ll}2 & 2 \\ 2 & 1\end{array}\right]\)

Step-by-Step Solution

Answer

The equilibrium at \((0,0)\) is an unstable node.

Step 1: Write the System of Differential Equations

The problem provides a matrix \( A \) from which we derive the system of differential equations: \[ \frac{dx}{dt} = 2x + 2y, \quad \frac{dy}{dt} = 2x + y. \] These equations describe how the quantities \( x(t) \) and \( y(t) \) change over time.

Step 2: Set Up the Characteristic Equation

To analyze stability, we need to find the eigenvalues of matrix \( A \). First, form the characteristic equation by calculating \( \text{det}(A - \lambda I) = 0 \), where \( I \) is the identity matrix. The matrix form is: \[ A - \lambda I = \begin{bmatrix} 2 - \lambda & 2 \ 2 & 1 - \lambda \end{bmatrix}. \]

Step 3: Compute the Determinant

Calculate the determinant of the matrix \( A - \lambda I \): \[ \text{det}(A - \lambda I) = (2-\lambda)(1-\lambda) - (2)(2). \] This simplifies to \[ \lambda^2 - 3\lambda - 2. \]

Step 4: Solve the Characteristic Equation

Solve the quadratic equation \( \lambda^2 - 3\lambda - 2 = 0 \) for \( \lambda \). Using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we find: \[ \lambda = \frac{3 \pm \sqrt{3^2 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2}. \]

Step 5: Analyze the Eigenvalues

The eigenvalues are \( \lambda_1 = \frac{3 + \sqrt{17}}{2} \) and \( \lambda_2 = \frac{3 - \sqrt{17}}{2} \). Both eigenvalues are real and the larger one is positive, indicating an unstable node at the equilibrium point \((0,0)\).

Step 6: Classify the Equilibrium

Since both eigenvalues are real and distinct, and one is positive, the equilibrium is an unstable node. The node is asymmetric as the eigenvalues have different magnitudes.

Key Concepts

Stability Analysis

Stability analysis helps us understand the behavior of equilibrium points in dynamic systems, such as differential equations. For the differential equation given by \(\frac{dx}{dt} = 2x + 2y, \frac{dy}{dt} = 2x + y\), with equilibrium at \((0,0)\), stability analysis is crucial to predict system behavior.

We analyze stability by examining the eigenvalues of the matrix \(A\), derived from the linearized system at the equilibrium. If all eigenvalues have negative real parts, the equilibrium is stable. If any eigenvalue has a positive real part, the equilibrium is unstable.

In our case, as per the solution, one eigenvalue is positive, indicating that the system will not return to equilibrium if disturbed. This conclusion points to an unstable equilibrium at the origin.

Eigenvalues

Eigenvalues are pivotal in analyzing systems of differential equations. They are solutions to the characteristic equation derived from \(A - \lambda I = 0\). By finding the eigenvalues, we understand crucial aspects of the equilibrium points.

In our example involving the matrix \( A = \begin{bmatrix} 2 & 2 \ 2 & 1 \end{bmatrix} \), we compute the eigenvalues by solving the characteristic equation \( \lambda^2 - 3\lambda - 2 = 0 \). This gives us the eigenvalues \( \lambda_1 = \frac{3 + \sqrt{17}}{2} \) and \( \lambda_2 = \frac{3 - \sqrt{17}}{2} \).

The significance of these eigenvalues lies in their sign. Since \( \lambda_1 \) is positive, it highlights the instability at the equilibrium point \((0,0)\). Analyzing eigenvalues is thus essential for stability insights and dynamic predictions.

Equilibrium Classification

Equilibrium classification tells us more about the nature of stability and the behavior of the system near equilibrium points. In this exercise, we found that the equilibrium point \((0,0)\) is classified as an **unstable node**.

This classification arises because we have two distinct real eigenvalues, \( \lambda_1 = \frac{3 + \sqrt{17}}{2} \) and \( \lambda_2 = \frac{3 - \sqrt{17}}{2} \).

With one eigenvalue being positive and the other being less positive implies dominance of instability in the system. The equilibrium point \((0,0)\) will not remain in its position if disturbed, affirming our classification. Understanding classification helps predict how a system will behave under different conditions and guides in control strategies.

Characteristic Equation

The characteristic equation is fundamental in linear systems analysis. It emerges when we determine the eigenvalues of a matrix \(A\) related to a differential equation. The equation encapsulates how matrix transformations influence vector behaviors, especially near equilibrium points.

For the matrix \( A = \begin{bmatrix} 2 & 2 \ 2 & 1 \end{bmatrix} \), we set up the equation by calculating \( \text{det}(A - \lambda I) \). This gives us \( \lambda^2 - 3\lambda - 2 = 0 \). Solving this provides the eigenvalues that determine system stability.

The characteristic equation provides a clear pathway to understand stability without solving entire differential equations. It is a crucial tool that simplifies the analysis and helps to avert complex computations while retaining rich qualitative understanding.