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 equilibrium according to whether it is a sink, a source, or a saddle 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.

Step 1: Finding the Eigenvalues of the Matrix A

To analyze stability, we first need to find the eigenvalues of matrix \( A \). The characteristic equation for a matrix \( A \) is given by \( \det(A - \lambda I) = 0 \). For our matrix \( A = \left[\begin{array}{cc} 4 & -1 \ 5 & -1 \end{array}\right] \), we write: \[\det\begin{vmatrix} 4 - \lambda & -1 \ 5 & -1 - \lambda \end{vmatrix} = 0\] Calculating this determinant:\[(4 - \lambda)(-1 - \lambda) - (-1)(5) = 0\]Expanding, we get:\[\lambda^2 - 3\lambda - 9 = 0\]

Step 2: Solving the Characteristic Equation

We solve the quadratic equation \( \lambda^2 - 3\lambda - 9 = 0 \) using the quadratic formula:\[\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \( a = 1 \), \( b = -3 \), and \( c = -9 \).Substitute these values into the formula:\[\lambda = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-9)}}{2 \cdot 1}\]Calculate the discriminant:\[\sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}\]Therefore, the eigenvalues are:\[\lambda_1 = \frac{3 + 3\sqrt{5}}{2}, \quad \lambda_2 = \frac{3 - 3\sqrt{5}}{2}\]

Step 3: Analyzing Stability and Classification of Equilibrium

The eigenvalues \( \lambda_1 \) and \( \lambda_2 \) are real and distinct. To determine the stability and classify the equilibrium:- If both eigenvalues are positive, the equilibrium is an unstable node (source).- If both are negative, the equilibrium is a stable node (sink).- If they have opposite signs, the equilibrium is a saddle point.Here, \( \lambda_1 \) is positive and \( \lambda_2 \) is negative. Thus, the equilibrium point \((0, 0)\) is a saddle point.

Key Concepts

Eigenvalues

Eigenvalues are a crucial part of analyzing the behavior of differential equations involving matrices. They are numerical values that reveal important information about the system's dynamics. In our scenario, the differential equation is expressed as \( \frac{d \mathbf{x}}{dt} = A \mathbf{x}(t) \), where the matrix \( A \) is given.

To find the eigenvalues of a matrix, you need to solve its characteristic equation \( \det(A - \lambda I) = 0 \). Here, \( \lambda \) represents the eigenvalues we are trying to determine, and \( I \) is the identity matrix.

• This process involves calculating the determinant of the matrix \( A - \lambda I \).
• The resulting polynomial equation is solved to find the values of \( \lambda \).

In our example, solving the quadratic equation \( \lambda^2 - 3\lambda - 9 = 0 \) using the quadratic formula, provides two distinct eigenvalues: \[ \lambda_1 = \frac{3 + 3\sqrt{5}}{2} \] \[ \lambda_2 = \frac{3 - 3\sqrt{5}}{2} \] These distinct and real eigenvalues help determine the subsequent stability of the system.

Stability Analysis

Stability analysis is an important step when working with differential equations. It helps us understand how a system behaves over time, particularly concerning its equilibrium points. In our context, stability analysis examines whether small deviations from an equilibrium point will dissipate or amplify.

The results depend on the eigenvalues of the matrix \( A \). When we:

• Have all eigenvalues with negative real parts, the system returns to equilibrium over time, meaning it is stable.
• Have all positive, the system moves further away — indicating instability.
• See a mix of positive and negative eigenvalues, the system is neither — it signifies a saddle point.

In our analysis, we found one positive and one negative eigenvalue, confirming that the equilibrium point is a saddle point. This implies that the system's behavior is unstable because initial small deviations along certain directions will grow over time.

Saddle Point

A saddle point in a system like ours occurs when the equilibrium is neither simply stable nor unstable. It is characterized by having eigenvalues of opposite signs. This results in behavior where:

• Motion along certain trajectories leads back towards the equilibrium (indicating stability along those paths).
• Other directions cause deviation away from the equilibrium (showing instability along these paths).

The mix of behavior is akin to a saddle, where some directions seem to "sink" while others "rise."

In our example, with \( \lambda_1 \) positive and \( \lambda_2 \) negative, the system at \((0, 0)\) resembles such a saddle shape. In practical scenarios, saddle points often depict systems with sensitive dependence on initial conditions, which might escalate small perturbations into significant changes.

Equilibrium Point

An equilibrium point in a differential equation is where the system does not change over time — i.e., the rate of change is zero. Mathematically, this means that when \( \mathbf{x}(t) = 0 \), then \( \frac{d \mathbf{x}}{dt} = 0 \). This point is vital as it helps determine the overall nature of the system's dynamics.

Equilibrium points can be classified into several types based on the eigenvalues derived from the system's matrix \( A \):

• Stable nodes (or sinks) occur when all eigenvalues are negative.
• Unstable nodes (or sources) occur when they are positive.
• Saddle points, as discussed earlier, arise with mixed sign eigenvalues.

For our specific case, the equilibrium point \((0,0)\) was identified via the system's dynamics equations. Given the eigenvalues' signs, \((0,0)\) acts as a saddle point, suggesting identified motion tendencies in different directions.