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. $$ A=\left[\begin{array}{ll} 1 & 3 \\ -2 & 1 \end{array}\right] $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Equilibrium Point

For the differential equation \(\frac{d \mathbf{x}}{dt} = A \mathbf{x}(t)\), the equilibrium point can be found by setting \(\frac{d \mathbf{x}}{dt} = 0\). This leads to \(A \mathbf{x}(t) = 0\). The trivial solution \(\mathbf{x} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\) always satisfies this equation, indicating that the equilibrium is at \((0,0)\).

Step 2: Calculate the Eigenvalues

Determine the eigenvalues of the matrix \(A\) by solving the characteristic equation \(\text{det}(A - \lambda I) = 0\). Here, \(I\) is the identity matrix. For the given \(A = \begin{bmatrix} 1 & 3 \ -2 & 1 \end{bmatrix}\), the characteristic equation is \(\text{det}\left(\begin{bmatrix} 1-\lambda & 3 \ -2 & 1-\lambda \end{bmatrix}\right) = (1-\lambda)^2 + 6 = 0\).

Step 3: Solve the Characteristic Equation

Expand and simplify the characteristic equation: \((1-\lambda)^2 + 6 = 0 \Rightarrow \lambda^2 - 2\lambda + 7 = 0\). Use the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = -2, c = 7\). The discriminant \(b^2 - 4ac = -24\), indicating complex roots.

Step 4: Determine Eigenvalues as Complex Numbers

Substitute into the quadratic formula: \(\lambda = \frac{2 \pm \sqrt{-24}}{2}\). Simplifying, we get \(\lambda = 1 \pm i\sqrt{6}\). These are the complex conjugate eigenvalues.

Step 5: Analyze the Stability

For complex eigenvalues \(\lambda = \alpha \pm i\beta\), where \(\alpha > 0\), the equilibrium is an unstable spiral. Since \(\lambda = 1 \pm i\sqrt{6}\), \(\alpha = 1 > 0\). Therefore, the equilibrium \((0,0)\) is an unstable spiral.

Key Concepts

Eigenvalues

When we analyze a system of differential equations, the eigenvalues can tell us a lot about the behavior of the system. Eigenvalues are scalars associated with a matrix that satisfy the characteristic equation \( \text{det}(A - \lambda I) = 0 \). Here, \( A \) is our matrix, \( \lambda \) represents the eigenvalues, and \( I \) is the identity matrix.
Understanding eigenvalues involves solving this characteristic equation, which can be a polynomial. The degree of the polynomial corresponds to the number of eigenvalues, which may be real or complex.
In the current exercise, solving the characteristic equation resulted in the eigenvalues \( \lambda = 1 \pm i\sqrt{6} \), which are complex numbers. Complex eigenvalues often come in conjugate pairs, and this indicates an oscillatory or spiral behavior in the system's solutions.

• Real parts: Influence the growth or decay (stability) of solutions.
• Imaginary parts: Indicate oscillatory behavior.

Understanding these values helps predict the system's future states.

Stability Analysis

Stability analysis in systems of differential equations is crucial to understand the long-term behavior of solutions. Specifically, it tells us whether solutions that start near an equilibrium point will stay close, move toward, or move away from it over time.
We determine stability by examining the eigenvalues of the system's matrix. In this exercise, we see complex eigenvalues \( \lambda = 1 \pm i\sqrt{6} \). The critical part here is the real component of the eigenvalues, which is \( \alpha = 1 \).

• If the real part \( \alpha \) is positive (\( \alpha > 0 \)), the system is unstable. Solutions move away from the equilibrium point.
• If \( \alpha = 0 \), the system can be either stable or exhibit neutral stability (like a center, with no exponential growth or decay).
• If \( \alpha < 0 \), the system is stable. Solutions approach the equilibrium point.

Since \( \alpha = 1 > 0 \) in this exercise, the equilibrium at \((0,0)\) is an unstable spiral. Solutions will exhibit a spiral motion moving away from this point.

Complex Conjugates

Complex numbers play a pivotal role in systems of differential equations, especially when dealing with oscillations and rotations in the system's behavior. When eigenvalues are complex, they usually appear in conjugate pairs, meaning if \( \lambda = a + bi \) is an eigenvalue, then \( \lambda = a - bi \) will also be an eigenvalue.
These pairs are essential in determining the type of stability and nature of equilibrium points in linear systems. Here's how they affect analysis:

• The real part of the eigenvalue (\(a\)) determines stability: Positive values lead to instability.
• The imaginary part (\(b\)) relates to the frequency of oscillation.

In our exercise, the eigenvalues were \( \lambda = 1 \pm i\sqrt{6} \), indicating a spiral nature due to their imaginary part \( i\sqrt{6} \). The real part \(1\) confirms it as unstable, leading to solutions spiraling outward. Recognizing these complex conjugate pairs allows for a deeper understanding of how the system behaves around an equilibrium point, especially in two dimensions where such dynamics are common.