Problem 36 - 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 real, distinct, and nonzero. Analyze the stability of the equilibrium \((0,0)\), and classify the equilibrium according to whether it is a sink, a source, or a saddle point. $$ A=\left[\begin{array}{rr} -1 & 3 \\ 2 & -5 \end{array}\right] $$

Step-by-Step Solution

Answer

The equilibrium point \((0, 0)\) is a stable sink because both eigenvalues are negative.

Step 1: Find the Eigenvalues

The eigenvalues of matrix \( A \) determine the stability of the equilibrium. To find the eigenvalues, solve the characteristic equation \( \text{det}(A - \lambda I) = 0 \). For matrix \( A = \begin{bmatrix} -1 & 3 \ 2 & -5 \end{bmatrix} \), this becomes: \[ \text{det}(\begin{bmatrix} -1 - \lambda & 3 \ 2 & -5 - \lambda \end{bmatrix}) = 0 \]. Expanding the determinant gives \[ (-1 - \lambda)(-5 - \lambda) - (3)(2) = 0 \] which simplifies to \( \lambda^2 + 6\lambda + 1 = 0 \).

Step 2: Solve the Characteristic Equation

The characteristic equation \( \lambda^2 + 6\lambda + 1 = 0 \) needs to be solved to find \( \lambda \). We use the quadratic formula: \[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Here, \( a = 1, b = 6, \) and \( c = 1 \). Substitute these values in to get: \[ \lambda = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-6 \pm \sqrt{32}}{2} = \frac{-6 \pm 4\sqrt{2}}{2} \]. Thus, the eigenvalues are \( \lambda_1 = -3 + 2\sqrt{2} \) and \( \lambda_2 = -3 - 2\sqrt{2} \).

Step 3: Analyze Stability

Determine the sign of the eigenvalues to assess the stability of the equilibrium point. Both \( \lambda_1 = -3 + 2\sqrt{2} \) and \( \lambda_2 = -3 - 2\sqrt{2} \) are negative since \( -3 + 2\sqrt{2} \approx -0.1715 \) and \( -3 - 2\sqrt{2} \approx -5.828 \). Since both eigenvalues are negative, the equilibrium point \((0,0)\) is stable.

Step 4: Classify the Equilibrium

Because both eigenvalues are negative, the system is stable and behaves as a sink. A sink implies all solutions of the differential equation approach the equilibrium point \((0,0)\) as \(t \rightarrow \infty\).

Key Concepts

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They play a crucial role in modeling and understanding dynamic systems, such as the growth of populations, the motion of a pendulum, or the change in investment over time.
They are often written in the form \( \frac{d \mathbf{x}}{d t}=A \mathbf{x}(t) \), where \( \mathbf{x}(t) \) is a vector of functions dependent on time \( t \).
Such equations are fundamental in stability analysis, as they allow us to predict the behavior of a system over time.

In our example, the matrix \( A \) determines how the system changes. By analyzing \( A \), we can understand the system's stability, which is crucial when scrutinizing equilibrium points.
This leads us to examine the eigenvalues. But first, what are eigenvalues, and how do they tie into this process?

Eigenvalues

Eigenvalues are scalar values that provide significant insights into the characteristics of a linear transformation or matrix. Specifically, they demonstrate how much vectors are scaled during a transformation associated with the matrix.
We find eigenvalues by solving the characteristic equation \( \det(A - \lambda I) = 0 \), where \( A \) is the matrix, \( I \) is the identity matrix, and \( \lambda \) represents the eigenvalues.
In the context of differential equations, these values play a critical role in determining whether the system will reach stability or continue to evolve chaotically.

• If both eigenvalues are negative, as found in our problem, the system tends toward a stable equilibrium, indicating a situation known as a 'sink'.
• If both were positive, it would imply instability, commonly referred to as a 'source'.
• A combination of one positive and one negative eigenvalue results in a 'saddle point', suggesting a direction-dependent stability.

By understanding eigenvalues, one can predict the system's future behavior and categorize its equilibrium points.

Equilibrium Point Classification

Equilibrium points in a system described by differential equations are solutions where the system remains constant if undisturbed.
Classification of these points depends significantly on the signs and nature of the eigenvalues derived from the system's matrix \( A \).

In our example, we have deduced that the eigenvalues are negative, leading to a stable equilibrium point at \((0,0)\).

• When both eigenvalues are negative, the point is classified as a 'sink', meaning all system paths eventually converge to this equilibrium over time, regardless of initial conditions.
• On the other hand, if the eigenvalues were positive, the equilibrium would be classified as a 'source', where trajectories diverge away from the point.
• Mixed-sign eigenvalues lead to a 'saddle point', where stability depends on the route chosen within the system's trajectory.

This classification is crucial in engineering, physics, and other fields, as it predicts how systems will respond to perturbations and adapt over time.