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

Step-by-Step Solution

Answer

The equilibrium \((0,0)\) is a stable spiral.

Step 1: Compute the Eigenvalues

To analyze the stability, we first need to compute the eigenvalues of the matrix \(A\). The eigenvalues \(\lambda\) of a matrix are found by solving the characteristic equation: \(\det(A - \lambda I) = 0\). For our matrix \(A\), this translates to: \[\begin{vmatrix}-1 - \lambda & 1 \-3 & 1 - \lambda\end{vmatrix} = 0\]Calculating the determinant, we find:\(-1 - \lambda)(1 - \lambda) - (-3)(1) = 0\), which simplifies to:\(\lambda^2 + \lambda + 2 = 0\).

Step 2: Solve for Eigenvalues

Now, solve the quadratic equation \(\lambda^2 + \lambda + 2 = 0\) using the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b = 1\), and \(c = 2\).Plug in the values:\[\lambda = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times 2}}{2 \times 1} = \frac{-1 \pm \sqrt{1 - 8}}{2}\]\[= \frac{-1 \pm \sqrt{-7}}{2} = \frac{-1 \pm i\sqrt{7}}{2}\]The eigenvalues are complex conjugates: \(\lambda = \frac{-1}{2} \pm \frac{i\sqrt{7}}{2}\).

Step 3: Analyze the Real Part of Eigenvalues

The stability of the equilibrium is determined by the real part of the eigenvalues. The real parts are both \(-1/2\), which are negative. Negative real parts indicate that the equilibrium is a stable spiral.

Step 4: Conclude the Equilibrium Classification

Since the real part of the eigenvalues is negative and they are complex conjugates with non-zero imaginary parts, the equilibrium point \((0,0)\) is classified as a stable spiral.

Key Concepts

Eigenvalues

In the context of differential equations, eigenvalues play a crucial role in determining the behavior of solutions. To find the eigenvalues of a matrix, we use the characteristic equation. For a matrix \( A \) like the one in our example, the characteristic equation is given by the determinant of \( (A - \lambda I) = 0 \), where \( \lambda \) represents the eigenvalues and \( I \) is the identity matrix.
Calculating the determinant for our given matrix \( A = \begin{bmatrix} -1 & 1 \ -3 & 1 \end{bmatrix} \), we obtain a quadratic equation: \( \lambda^2 + \lambda + 2 = 0 \). This leads us to solve for \( \lambda \) using the quadratic formula.
The solution yields complex conjugates: \( \lambda = \frac{-1}{2} \pm \frac{i\sqrt{7}}{2} \). These eigenvalues, having both real and imaginary parts, suggest a spiral nature around the equilibrium, with the sign of the real part indicating stability.

Stability Analysis

Stability analysis helps us understand if solutions tend to remain close to equilibrium over time. It uses the real parts of the eigenvalues to determine the nature of this stability.
For the differential system with our matrix \( A \), the real part of the eigenvalues \( \lambda = \frac{-1}{2} \pm \frac{i\sqrt{7}}{2} \) is \(-1/2\). Since the real part is negative, it signals that trajectories will spiral inwards towards the equilibrium point as time progresses, indicating a stable spiral.
Therefore, when eigenvalues have negative real parts, the system's responses to perturbations diminish over time, signifying that such an equilibrium point is stable.

Equilibrium Classification

Classifying equilibrium points in relation to their stability and dynamic behavior is essential. In our analysis, the equilibrium point is classified based on the characteristics of the eigenvalues.
For our exercise, we found complex conjugate eigenvalues with negative real parts. This configuration describes a stable spiral.
Generally,

• A stable spiral occurs when eigenvalues have negative real parts with non-zero imaginary parts.
• An unstable spiral presents itself when eigenvalues have positive real parts.
• A center happens when eigenvalues are purely imaginary, showing neither inward nor outward spiraling, leading to neutral stability.

Therefore, in systems like ours, the eigenvalue analysis directly informs how we classify the equilibrium, emphasizing its tendency to naturally return to, or stay near, the equilibrium point over time.