Problem 50 - 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. \(A=\left[\begin{array}{ll}1 & -4 \\ 1 & -1\end{array}\right]\)

Step-by-Step Solution

Answer

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

Step 1: Find the Characteristic Equation

To determine the stability, we first need to find the eigenvalues of matrix \( A \). The eigenvalues are solutions to the characteristic equation \(|A - \lambda I| = 0\), where \( I \) is the identity matrix. For matrix \( A \), this becomes:\[\begin{vmatrix} 1-\lambda & -4 \1 & -1-\lambda \end{vmatrix} = 0\].This simplifies to \((1-\lambda)(-1-\lambda) - (-4)(1) = 0\). Expanding this equation, we get the characteristic equation: \(\lambda^2 - \lambda + 5 = 0\).

Step 2: Solve the Characteristic Equation

The next step is to solve the characteristic equation \(\lambda^2 - \lambda + 5 = 0\) using the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -1\), and \(c = 5\). Calculating the discriminant \(b^2 - 4ac = (-1)^2 - 4(1)(5) = 1 - 20 = -19\), which is negative, indicating complex roots. The eigenvalues are \(\lambda = \frac{1 \pm \sqrt{-19}i}{2}\), or \(\lambda = \frac{1}{2} \pm \frac{\sqrt{19}}{2}i\).

Step 3: Analyze the Eigenvalues

The eigenvalues \(\lambda = \frac{1}{2} \pm \frac{\sqrt{19}}{2}i\) are complex with a real part \(\frac{1}{2}\), which is positive. The presence of a positive real part means that the real parts of the eigenvalues are greater than zero. In terms of stability, if the real part of the eigenvalues is greater than zero, the system is unstable. Since the imaginary part indicates oscillation, we determine that the equilibrium point is not just unstable, but specifically an unstable spiral.

Key Concepts

Characteristic Equation

In differential equations, when you are tasked with determining the stability of a system, the first critical step is to find the characteristic equation. This equation is derived from the matrix form of the differential equation system. For example, consider a 2x2 matrix \( A \) related to our differential equation. The characteristic equation is formed by setting the determinant of \( A - \lambda I \) equal to zero. This looks like \(|A - \lambda I| = 0\), where \( I \) is the identity matrix.
To solve for \( \lambda \), you substitute into the determinant and simplify the expression. In our problem, the matrix \( A \) is \( \begin{bmatrix} 1 & -4 \ 1 & -1 \end{bmatrix} \). The characteristic equation becomes \((1-\lambda)(-1-\lambda) - (-4)(1) = 0\), simplifying further to \(\lambda^2 - \lambda + 5 = 0\).

• The solutions to this equation are termed as eigenvalues.
• These eigenvalues are pivotal in analyzing the system’s behavior.
• The process highlights the relationship between matrix algebra and differential equations.

Understanding the characteristic equation is fundamental as it directly influences the stability analysis of the system.

Complex Eigenvalues

Once the characteristic equation is established, the next step is to solve it. If solving yields complex numbers, the eigenvalues are known as complex eigenvalues. These are usually in a form such as \( \lambda = a \pm bi \), where \( a \) and \( b \) are real numbers.
In our solution, the eigenvalues of \( A \) are determined by solving the equation \( \lambda^2 - \lambda + 5 = 0 \). The discriminant \( b^2 - 4ac \) for this equation is negative, \( -19 \), indicating complex roots. Here, the eigenvalues actually are \( \lambda = \frac{1}{2} \pm \frac{\sqrt{19}}{2}i \). They consist of:

• A real part: \( \frac{1}{2} \), which dictates growth or decay.
• An imaginary part: \( \frac{\sqrt{19}}{2}i \), which signifies oscillatory motion.

Complex eigenvalues lead to rotational dynamics in the system, often informing us about spirals or circular motions around an equilibrium point. This is a typical scenario in oscillating systems, showing how real-life phenomena are modeled in differential equations.

Stability Analysis

Stability analysis in the context of differential equations involves understanding how the system behaves over time. Specifically, this refers to how trajectories behave as time tends to infinity—do they converge to a point, diverge, or oscillate?
With complex eigenvalues like \( \lambda = \frac{1}{2} \pm \frac{\sqrt{19}}{2}i \), the stability of the system predominantly depends on the real component. In our scenario, the positive real part \( \frac{1}{2} \) indicates an inherent instability, as it leads to exponential growth of solutions.

• If the real part is positive, the trajectory spirals outward, leading to an unstable spiral.
• If zero, the system remains stable, marking it a center.
• If negative, spirals inward, indicating a stable spiral.

Here, since the real part is positive, the solution paths repeatedly spiral away from the equilibrium. Thus, the system classifies as an unstable spiral, demonstrating how eigenvalues determine dynamic behavior and stability properties.