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 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. \(A=\left[\begin{array}{rr}-1 & 3 \\ 2 & 4\end{array}\right]\)

Step-by-Step Solution

Answer

The equilibrium \((0,0)\) is a saddle point and is unstable.

Step 1: Find the Characteristic Equation

To find the eigenvalues of matrix \( A \), we must first calculate its characteristic equation. The characteristic equation is found using \( \text{det}(A - \lambda I) = 0 \). For the given matrix \[A = \begin{bmatrix}-1 & 3 \ 2 & 4\end{bmatrix},\]the characteristic equation is:\[\left|\begin{array}{cc} -1 - \lambda & 3 \ 2 & 4 - \lambda\end{array}\right| = 0\] \[\Rightarrow (-1 - \lambda)(4 - \lambda) - (3)(2) = 0.\]

Step 2: Simplify the Characteristic Equation

Now, simplify the equation obtained in Step 1: \[(-1 - \lambda)(4 - \lambda) - 6 = 0.\] This becomes: \[\lambda^2 - 3\lambda - 10 = 0.\] This is a quadratic equation in \(\lambda\).

Step 3: Solve the Quadratic Equation for Eigenvalues

Solve the quadratic equation \(\lambda^2 - 3\lambda - 10 = 0\) using the quadratic formula, \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): Here, \(a=1\), \(b=-3\), and \(c=-10\). Plug these into the quadratic formula: \[ \lambda = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1}\] \[ \lambda = \frac{3 \pm \sqrt{9 + 40}}{2}\] \[ \lambda = \frac{3 \pm \sqrt{49}}{2}\] \[ \lambda = \frac{3 \pm 7}{2}\] Thus, the eigenvalues are \(\lambda_1 = 5\) and \(\lambda_2 = -2\).

Step 4: Analyze Stability Using Eigenvalues

Examine the sign of the eigenvalues:- \(\lambda_1 = 5\) is positive.- \(\lambda_2 = -2\) is negative.In the case where eigenvalues have opposite signs, the equilibrium point at \((0,0)\) is a saddle point, which is unstable.

Step 5: Classify the Equilibrium

Since one eigenvalue is positive and the other is negative, the equilibrium point \((0,0)\) is a saddle point. This implies that it is neither a sink nor a source, as it describes an unstable solution where trajectories are moving away in at least one direction.

Key Concepts

Eigenvalues

In the context of differential equations, eigenvalues play a crucial role in determining the behavior of dynamic systems. They are essentially special scalars associated with a square matrix, often denoted by \( \lambda \). For the system described by the equation \( \frac{d \mathbf{x}}{d t} = A \mathbf{x}(t) \), the matrix \( A \) has eigenvalues that are solutions to the characteristic equation \( \text{det}(A - \lambda I) = 0 \).

To find these eigenvalues for a given 2x2 matrix, like the one in the original exercise, you solve the characteristic polynomial derived from\(|A - \lambda I| = 0\). This involves substituting \( \lambda \) into the determinant of \( A \), subtracting it along the diagonal, and simplifying. In our case, this resulted in the quadratic equation \( \lambda^2 - 3\lambda - 10 = 0 \), with solutions \( \lambda_1 = 5 \) and \( \lambda_2 = -2 \).

Understanding eigenvalues is key to predicting how solutions to the differential equations evolve over time, especially in relation to stability.

Stability Analysis

Stability analysis helps us determine the behavior of equilibrium points in dynamic systems. For linear systems, the eigenvalues of the system matrix \( A \) provide insights into whether solutions converge to equilibrium (stable), diverge to infinity (unstable), or exhibit neutral behaviors.

An equilibrium point is called stable if perturbations from the point lead to trajectories that return to it over time. In contrast, an equilibrium point is deemed unstable if any perturbation causes the system to move away to some extent. If all eigenvalues are real and negative, the equilibrium is stable. If any eigenvalue is positive, the equilibrium is unstable.

In the given exercise, with eigenvalues \( \lambda_1 = 5 \) and \( \lambda_2 = -2 \), the signs are critical. The presence of a positive eigenvalue indicates that there are directions along which solutions grow without bound, suggesting the equilibrium at \((0,0)\) is unstable.

Saddle Point

A saddle point is a type of equilibrium in dynamical systems that is characterized by its unstable nature, generally in systems with real and distinct eigenvalues. Specifically, it occurs when one eigenvalue is positive and the other is negative, as seen with the matrix \( A \) in the exercise with \( \lambda_1 = 5 \) and \( \lambda_2 = -2 \).

At a saddle point, trajectories of the system will move away from the equilibrium in one direction while possibly approaching in another, resulting in an overall unstable condition. This means that while some paths contract towards the point, others diverge away from it, creating a 'saddle-like' contour in the phase space.

The saddle point behavior implies it cannot act as a source (where all trajectories move away) or a sink (where all paths converge to the point). This dual nature leads to complex dynamics depending on the initial conditions of the system.