Problem 42 - 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 equilib\mathrm{\\{} r i u m ~ a c c o r d i n g ~ t o ~ w h e t h e r ~ i t ~ i s ~ a ~ s i n k , ~ a ~ s o u r c e , ~ o r ~ a ~ s a d d l e ~ point. \(A=\left[\begin{array}{ll}4 & -1 \\ 5 & -1\end{array}\right]\)

Step-by-Step Solution

Answer

The equilibrium \((0,0)\) is a saddle point, indicating instability.

Step 1: Calculate the Characteristic Polynomial

To find the eigenvalues of matrix \( A \), we start by calculating the characteristic polynomial: \[\text{det}(A - \lambda I) = \begin{vmatrix} 4-\lambda & -1 \ 5 & -1-\lambda \end{vmatrix}\]Solving the determinant, we get: \[(4-\lambda)(-1-\lambda) - (-1)(5) = \lambda^2 - 3\lambda - 9\]

Step 2: Solve for Eigenvalues

The next step is solving the characteristic polynomial for eigenvalues:\[\lambda^2 - 3\lambda - 9 = 0\]Using the quadratic formula, \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a=1, b=-3, c=-9 \) gives us:\[\lambda = \frac{3 \pm \sqrt{9 + 36}}{2} = \frac{3 \pm \sqrt{45}}{2} = \frac{3 \pm 3\sqrt{5}}{2}\]The eigenvalues are \( \lambda_1 = \frac{3 + 3\sqrt{5}}{2} \) and \( \lambda_2 = \frac{3 - 3\sqrt{5}}{2} \).

Step 3: Analyze Stability Using Eigenvalues

We determined that the eigenvalues are real, distinct, and nonzero: \( \lambda_1 = \frac{3 + 3\sqrt{5}}{2} > 0 \) and \( \lambda_2 = \frac{3 - 3\sqrt{5}}{2} < 0 \).Since one eigenvalue is positive and the other is negative, the equilibrium point at \( (0,0) \) is a saddle point. This indicates an unstable equilibrium.

Key Concepts

Eigenvalues

Eigenvalues are a fundamental concept in linear algebra and play a crucial role in understanding differential equations. Think of them as special numbers that provide insight into the behavior of a matrix when applied to a linear transformation. For the matrix \( A \), the eigenvalues represent the scaling factors of the linear transformation described by \( A \), which affect the equilibrium points.
To find these eigenvalues, we solve the characteristic equation derived from the matrix \( A \). In our example, the eigenvalues are real and distinct, which usually suggests more straightforward outcomes in terms of stability analysis. They directly help in determining whether a system is stable, unstable, or exhibits certain predictable behaviors over time.

Stability Analysis

Stability analysis involves examining whether solutions to a differential equation converge to an equilibrium (stable), move away from it (unstable), or do something else. Our matrix \( A \) has eigenvalues \( \lambda_1 \) and \( \lambda_2 \) that are real and distinct with different signs. This indicates that our system's equilibrium at \( (0,0) \) is unstable because at least one eigenvalue is positive.
When performing stability analysis, the sign of the eigenvalues is critical:

• If all eigenvalues are negative, the system is stable (sink).
• If all are positive, the system is unstable (source).
• If they have different signs, like in our case, the system is unstable and is usually a saddle point.

This informs us how solutions behave over time and whether they move towards or away from equilibrium.

Saddle Point

A saddle point is a specific type of equilibrium in differential equations where the solution behaves unusually due to competing effects. In mathematical terms, this occurs when the eigenvalues of the system have different signs.
Imagine riding a horse with a saddle: one can comfortably sit forward or backward (indicating stability in one direction), but any side movements (unstable directions) would cause you to fall off. Similarly, in the context of our example, the equilibrium at \( (0,0) \) is a saddle point because one eigenvalue is positive, and the other is negative.
Consequently, trajectories in the phase space tend to move away from the equilibrium in one direction, indicating instability, while simultaneously moving towards it in another direction. This dual behavior is hallmark of saddle points in such systems.

Characteristic Polynomial

The characteristic polynomial is a polynomial equation obtained from a square matrix used to find eigenvalues. It is created by setting the determinant of the matrix \( A - \lambda I \) to zero, where \( \lambda \) represents the eigenvalues, and \( I \) is the identity matrix.
In calculating the characteristic polynomial for our matrix \[ A = \begin{bmatrix} 4 & -1 \ 5 & -1 \end{bmatrix} \], we derived the equation \[ \lambda^2 - 3\lambda - 9 = 0 \]. Solving this polynomial using the quadratic formula gives us the system's eigenvalues.
This step is crucial because it directly impacts stability analysis and determines the type of equilibrium. Understanding how to derive and solve the characteristic polynomial provides the foundation for analyzing the system's dynamic behavior.