Problem 46 - 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 complex conjugates. Analyze the stability of the equilibrium \((0,0)\), and classify the equilibrium according to whether it is a stable spiral, an unstable spiral, or a center. \(A=\left[\begin{array}{rr}1 & -2 \\ 2 & 1\end{array}\right]\)

Step-by-Step Solution

Answer

The equilibrium \((0,0)\) is an unstable spiral because the eigenvalues have positive real parts.

Step 1: Calculate Eigenvalues

To determine the stability of the equilibrium, we need to find the eigenvalues of matrix \(A\). The eigenvalues \(\lambda\) are found by solving the characteristic equation: \[ \det(A - \lambda I) = 0 \] where \(I\) is the identity matrix. For matrix \(A = \begin{bmatrix} 1 & -2 \ 2 & 1 \end{bmatrix}\), the equation becomes: \[ \begin{vmatrix} 1 - \lambda & -2 \ 2 & 1 - \lambda \end{vmatrix} = (1-\lambda)^2 + 4 = \lambda^2 - 2\lambda + 5 = 0 \]

Step 2: Solve for Eigenvalues

Solve the quadratic equation \(\lambda^2 - 2\lambda + 5 = 0 \) using the quadratic formula: \[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \(a = 1\), \(b = -2\), and \(c = 5\). This gives: \[ \lambda = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2} = 1 \pm 2i \] This shows that the eigenvalues are \(1 + 2i\) and \(1 - 2i\).

Step 3: Analyze Eigenvalues for Stability

The eigenvalues \(1 + 2i\) and \(1 - 2i\) have a real part \(\text{Re}(\lambda) = 1\). Since the real part is positive, the equilibrium point \((0,0)\) is unstable. Because the eigenvalues are complex conjugates, this classifies the equilibrium as an unstable spiral.

Key Concepts

Eigenvalues

Eigenvalues are crucial in the study of differential equations, especially when analyzing stability. Think of eigenvalues as numbers that reveal important traits about a matrix and, consequently, about the dynamics of a system. For a given matrix, eigenvalues can tell us how small changes in initial conditions might evolve over time.
In our original exercise, we have matrix \[A = \begin{bmatrix}1 & -2 \2 & 1 \\end{bmatrix},\]and the eigenvalues are computed by solving the characteristic equation\[\det(A - \lambda I) = 0.\]
This is how we found the eigenvalues \(1 + 2i\) and \(1 - 2i\). Notice that they are complex conjugates, which means they can significantly influence the behavior of the solutions to the differential equation. Understanding these complex numbers helps in predicting how the solution behaves over time, like whether it spirals or stays steady.

Stability Analysis

Stability analysis is a core concept in differential equations, focusing on understanding whether solutions tend toward or away from an equilibrium point over time.
In our example, stability analysis helps us determine if the point \((0,0)\) is stable or unstable based on the eigenvalues we calculated. Here’s how it works:

• If the real parts of all eigenvalues are negative, the equilibrium is stable—solutions will tend toward it over time.
• If any eigenvalue has a positive real part, as in our exercise where the real part is \(1\), the equilibrium is unstable—solutions move away.
• If the real parts of the eigenvalues are zero, you might have a "center," indicating solutions could circle around without settling down or moving away.

This technique helps predict the system’s long-term behavior, ensuring that we have a nuanced understanding of how solutions behave as time progresses.

Equilibrium Points

An equilibrium point in the context of differential equations represents a steady state where the system doesn't change over time. For the differential equation \(\frac{d \mathbf{x}}{d t}=A \mathbf{x}(t)\), an equilibrium point is a solution where the derivatives (or rates of change) become zero. In our exercise, the point \((0,0)\) is analyzed as an equilibrium point.
Finding equilibria often revolves around setting the equation to zero and solving it. For linear systems like ours, this means that at the equilibrium, we have \(A\mathbf{x} = 0\).
In practical terms, equilibrium points help us figure out where a system might "rest" without any external changes. By understanding whether an equilibrium is stable or unstable, we can predict if the system tends to return to this restful state or moves away, helping in practical scenarios like predicting population dynamics or electrical circuits. These insights not only teach us about theoretical situations but also apply to real-world phenomena.