Problem 31 - 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. \(A=\left[\begin{array}{rr}-2 & 2 \\ 2 & 1\end{array}\right]\)

Step-by-Step Solution

Answer

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

Step 1: Find Eigenvalues of Matrix A

To determine the stability of the system, first compute the eigenvalues of matrix \(A\). The matrix \(A\) is \(\begin{bmatrix}-2 & 2 \ 2 & 1\end{bmatrix}\). The characteristic equation is given by \(\det(A - \lambda I) = 0\), where \(I\) is the identity matrix and \(\lambda\) are the eigenvalues. This yields:\[\det\begin{bmatrix} -2-\lambda & 2 \ 2 & 1-\lambda \end{bmatrix}=0\]Expanding the determinant, we have:\[(\lambda+2)(\lambda-1) - 4 = 0\]\[\lambda^2 + \lambda - 6 = 0\]Factoring yields:\[(\lambda - 2)(\lambda + 3) = 0\]The eigenvalues are \(\lambda_1 = 2\) and \(\lambda_2 = -3\).

Step 2: Analyze Signs of Eigenvalues

The stability of the equilibrium point \((0,0)\) depends on the signs of the eigenvalues. If all eigenvalues are negative, the equilibrium is a sink (stable). If all are positive, it's a source (unstable). If one is positive and one is negative, it's a saddle point (unstable). Here, \(\lambda_1 = 2 > 0\) and \(\lambda_2 = -3 < 0\).

Step 3: Classify the Equilibrium Point

Since one eigenvalue is positive and the other is negative, the equilibrium point \((0,0)\) is classified as a saddle point. It is an unstable equilibrium because trajectories near the equilibrium will be repelled away in some directions while being attracted in others.

Key Concepts

Stability Analysis

When analyzing differential equations, stability analysis helps us understand the behavior of solutions near equilibrium points. For a system described by the equation \( \frac{d \mathbf{x}}{dt} = A \mathbf{x}(t) \), the matrix \( A \) gives insights into how the system behaves around these points.
The key to stability lies in the eigenvalues of the matrix \( A \).- If all eigenvalues are negative, every solution converges to the equilibrium as time progresses, making the equilibrium stable or a sink.- If all eigenvalues are positive, solutions diverge, so the equilibrium is unstable or a source.- If there are both positive and negative eigenvalues, the behavior is mixed, leading to instability, known as a saddle point.In our specific example, the eigenvalues were \( \lambda_1 = 2 \) (positive) and \( \lambda_2 = -3 \) (negative), indicating a saddle point scenario.

Equilibrium Classification

Equilibrium classification involves determining the nature of an equilibrium point by examining the eigenvalues of the system matrix. Given these scenarios:- **Sink:** If all eigenvalues are negative, the equilibrium is a stable point where solutions eventually settle. This corresponds to damping behavior where, over time, the system doesn't exhibit exponential growth but rather dies down.- **Source:** If all eigenvalues are positive, the equilibrium behaves as an unstable point. Any small perturbation will grow, indicating that solutions "escape" rapidly.- **Saddle Point:** This occurs when some eigenvalues are negative and some are positive. The mixed signs cause some directions to attract and others to repel trajectories from the equilibrium.In our case, one positive and one negative eigenvalue means the system's equilibrium at \((0,0)\) is a saddle point and thus unstable.

Saddle Point

A saddle point in the context of differential equations is a type of equilibrium point characterized by having directions of both attraction and repulsion. It arises when there are both positive and negative eigenvalues.- **Behavior:** Near a saddle point, trajectories will tend to move away from it in some directions and toward it in others.- **Example:** For our matrix \( A \), having \( \lambda_1 = 2 > 0 \) and \( \lambda_2 = -3 < 0 \) illustrates this clearly. This imbalance indicates instability because while one direction might seem to "calm," the other opposes it by pushing trajectories outward.Understanding saddle points is crucial for predicting long-term behavior in dynamical systems as it indicates the presence of inherent system instability.