Problem 55 - 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. \(A=\left[\begin{array}{rr}1 & 2 \\ -5 & -3\end{array}\right]\)

Step-by-Step Solution

Answer

The equilibrium \((0,0)\) is a stable spiral because the real parts of the eigenvalues are negative and they have non-zero imaginary parts.

Step 1: Determine the Eigenvalues

To analyze the stability, we first need to determine the eigenvalues of the matrix \( A \). The characteristic equation is given by \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix and \( \lambda \) represents the eigenvalues. For the given matrix \( A = \begin{pmatrix} 1 & 2 \ -5 & -3 \end{pmatrix} \), the characteristic polynomial is \( \lambda^2 + 2\lambda + 7 = 0 \). Solving for \( \lambda \), we find the eigenvalues to be \( \lambda_1, \lambda_2 = -1 \pm 3i \).

Step 2: Analyze the Real Part of the Eigenvalues

The stability and type of equilibrium depend on the real part of the eigenvalues. We found \( \lambda_1 = -1 + 3i \) and \( \lambda_2 = -1 - 3i \). Both eigenvalues have a real part equal to \(-1\).

Step 3: Determine the Stability and Type of Equilibrium

Since the real part of both eigenvalues is negative (\(-1\)), the equilibrium point \((0,0)\) is stable. Additionally, because the imaginary part is non-zero \((\pm 3i)\), the system behaves as a spiral. The equilibrium is therefore classified as a stable spiral.

Key Concepts

Understanding Eigenvalues

Eigenvalues are essential for solving and understanding systems of differential equations. When given a matrix like \( A \), the eigenvalues tell us a lot about the system's characteristics and behavior over time. An eigenvalue \( \lambda \) is a special number related to a square matrix, indicating a dilation factor in the direction of a corresponding eigenvector. The eigenvalues are found by solving the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. A matrix's eigenvalues can be real or complex, and they determine the nature of equilibrium points in the system. For example, in our given matrix \( A = \begin{pmatrix} 1 & 2 \ -5 & -3 \end{pmatrix} \), solving the characteristic polynomial \( \lambda^2 + 2\lambda + 7 = 0 \) reveals eigenvalues \( \lambda_1, \lambda_2 = -1 \pm 3i \). Here, the complex eigenvalues indicate the presence of oscillatory (spiral-like) behavior in the system.

Stability Analysis in Differential Equations

Stability analysis is crucial in determining whether the system consistently remains near an equilibrium point over time. For linear differential equations, the stability of equilibrium points directly depends on the eigenvalues' real parts.

• If the real part of any eigenvalue is positive, the system is unstable.
• If all real parts are negative, the system is stable, and any trajectory starting close to the equilibrium remains near or converges to it.
• If the real part is zero, the equilibrium might be stable (center) or unstable; further analysis is needed.

In our case, the eigenvalues \( \lambda_1 \) and \( \lambda_2 \) have real parts equal to \(-1\). This negative value indicates stability, meaning trajectories around the equilibrium point \((0,0)\) eventually remain close or converge to the equilibrium. Thus, the system is stable in nature.

Equilibrium Classification

Equilibrium classification involves determining the type of behavior a linear system exhibits around its equilibrium points. For systems with complex eigenvalues, like in our example, the classification depends on the real and imaginary parts of these eigenvalues.

• Stable Spiral: When the real part of eigenvalues is negative and the imaginary part is non-zero, trajectories spiral toward the equilibrium. This is the case for our system, with \( \lambda_1 = -1 + 3i \) and \( \lambda_2 = -1 - 3i \), making the equilibrium a stable spiral.
• Unstable Spiral: If the real part is positive, trajectories spiral away, classifying it as an unstable spiral.
• Center: A purely imaginary eigenvalue (real part is zero) can suggest oscillation without spiraling inward or outward, known as a center.

The classification aids in visualizing system responses and predicting long-term behavior near the equilibrium. Understanding these classifications is key for applications, like engineering and physics, where stability insights are fundamental for system design and control.