Problem 32 - 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. \(A=\left[\begin{array}{rr}1 & 1 \\ 2 & -1\end{array}\right]\)

Step-by-Step Solution

Answer

The equilibrium (0,0) is a saddle point.

Step 1: Find the Eigenvalues of Matrix A

To determine the stability, we need to find the eigenvalues of matrix \(A\). The eigenvalues can be found by solving the characteristic equation given by \(\det(A - \lambda I) = 0\), where \(A = \begin{bmatrix} 1 & 1 \ 2 & -1 \end{bmatrix}\) and \(I\) is the identity matrix. So, we have:\[\det\left(\begin{bmatrix} 1 - \lambda & 1 \ 2 & -1 - \lambda \end{bmatrix}\right) = 0\]Simplifying, we get:\[(1-\lambda)(-1-\lambda) - 2 = 0\]\[\lambda^2 + \lambda - 1 - 2 = 0\]\[\lambda^2 + \lambda - 3 = 0\]

Step 2: Solve the Characteristic Equation

The characteristic equation \(\lambda^2 + \lambda - 3 = 0\) is a quadratic equation. We solve it using the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 1\), \(c = -3\). Thus,\[\lambda = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1}\]\[\lambda = \frac{-1 \pm \sqrt{1 + 12}}{2}\]\[\lambda = \frac{-1 \pm \sqrt{13}}{2}\]So, the eigenvalues are \(\lambda_1 = \frac{-1 + \sqrt{13}}{2}\) and \(\lambda_2 = \frac{-1 - \sqrt{13}}{2}\).

Step 3: Determine the Nature of the Equilibrium

The nature of the equilibrium at the origin \((0,0)\) depends on the eigenvalues. If both are negative, the equilibrium is a sink. If both are positive, it's a source. If one is positive and one is negative, it is a saddle point.In this problem, \(\lambda_1 = \frac{-1 + \sqrt{13}}{2} > 0\) and \(\lambda_2 = \frac{-1 - \sqrt{13}}{2} < 0\). Therefore, the origin is a saddle point.

Key Concepts

Differential Equations

Differential equations play a fundamental role in modeling changes in various fields such as physics, engineering, and economics. In this case, the differential equation \( \frac{d \mathbf{x}}{d t}=A \mathbf{x}(t) \) is considered. It involves a matrix \( A \) that affects a state vector \( \mathbf{x}(t) \), which changes over time. These types of differential equations are called linear differential equations because of the linear relationship between the vector's rate of change and the vector itself.
For our particular problem, the matrix \( A \) has real, distinct, and nonzero eigenvalues, which influences how the solutions behave.
Understanding how \( A \) influences \( \mathbf{x}(t) \) is crucial, as it tells us how the state of a system evolves, allowing us to predict its future dynamics.

Equilibrium Stability

Equilibrium stability helps us understand how a system behaves when it is disturbed from a state of balance or equilibrium. An equilibrium point, such as \( (0,0) \) in this problem, represents a state where the system does not change unless it is subjected to external disturbances. The stability of this point can be analyzed using eigenvalues derived from the system's matrix \( A \).
Depending on the eigenvalues:

• If both are negative, the system naturally returns to equilibrium, indicating a sink.
• If both are positive, the system moves away, indicating a source.
• If they differ in sign, the system moves in a combination of both directions, indicating a saddle point.

In our exercise, we discover that the equilibrium at \( (0,0) \) is unstable due to the presence of one positive and one negative eigenvalue, leading us towards the concept of saddle points.

Saddle Point Analysis

Saddle point analysis provides insight into a system's directional behavior near an equilibrium. A saddle point occurs when the equilibrium is unstable along at least one direction, typically because one eigenvalue is positive and the other is negative. This can be visualized in a physical landscape as a saddle shape, which curves upward in one direction and downward in the perpendicular direction.
For our matrix \( A \), with eigenvalues \( \lambda_1 = \frac{-1 + \sqrt{13}}{2} \) and \( \lambda_2 = \frac{-1 - \sqrt{13}}{2} \), we see that \( \lambda_1 > 0 \) and \( \lambda_2 < 0 \).
This mix of positive and negative provides the classic dynamics of a saddle point, where perturbations cause the system to move away in some directions but might return to equilibrium in others.