Problem 65 - Calculus for Biology and Medicine

Question

We consider differential equations of the form \(\frac{d x}{d t}=A x(t)\) where \(A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]\) Analyze the stability of the equilibrium \(\mathbf{( 0 , 0 ) , \text { and classify the }}\) equilibrium. \(A=\left[\begin{array}{rr}1 & 2 \\ -1 & -1\end{array}\right]\)

Step-by-Step Solution

Answer

The equilibrium \((0,0)\) is a stable node or sink since both eigenvalues have negative real parts.

Step 1: Find the Eigenvalues

To analyze stability, we need the eigenvalues of matrix \(A\). The matrix given is \(A=\begin{bmatrix} 1 & 2 \ -1 & -1 \end{bmatrix}\). The characteristic equation for a 2x2 matrix \(A\) is \(\det(A - \lambda I) = 0\), where \(\lambda\) is the eigenvalue and \(I\) is the identity matrix. For this matrix: \[(1-\lambda)(-1-\lambda) - 2(-1) = 0 \implies \lambda^2 + \lambda - 1 = 0\].

Step 2: Solve for Eigenvalues

Next, solve the quadratic equation \(\lambda^2 + \lambda - 1 = 0\) using the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=1\), and \(c=-1\). This gives: \[\lambda = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}\]. Thus, the eigenvalues are \(\lambda_1 = \frac{-1 + \sqrt{5}}{2}\) and \(\lambda_2 = \frac{-1 - \sqrt{5}}{2}\).

Step 3: Determine the Stability

For a 2x2 matrix system, the stability at the equilibrium \((0,0)\) is determined by the real parts of the eigenvalues. If all eigenvalues have negative real parts, the equilibrium is stable. If any eigenvalue has a positive real part, the equilibrium is unstable. Here, since both eigenvalues have negative real parts: \(\frac{-1 + \sqrt{5}}{2} < 0\) and \(\frac{-1 - \sqrt{5}}{2} < 0\), the equilibrium is stable.

Step 4: Classify the Equilibrium

Both eigenvalues are real and distinct, and they both have negative real parts making the equilibrium a stable node or sink. In this context, this means that trajectories starting close to \((0,0)\) will converge to \((0,0)\) as time approaches infinity.

Key Concepts

Eigenvalues

Eigenvalues play a crucial role in understanding the dynamics of differential equations, especially when it comes to systems characterized by matrices, like the one given in the exercise. To find eigenvalues, we calculate them from the matrix associated with the system. Specifically, for a 2x2 matrix like \(A = \begin{bmatrix} 1 & 2 \-1 & -1 \end{bmatrix} \), we derive the characteristic equation by using the formula: \(\det(A - \lambda I) = 0\). This involves subtracting \(\lambda\), a placeholder for eigenvalues, from every diagonal element, and then calculating the determinant of the resulting matrix. Once we solve the characteristic equation, we attain the eigenvalues, which in our example are \(\lambda_1 = \frac{-1 + \sqrt{5}}{2}\) and \(\lambda_2 = \frac{-1 - \sqrt{5}}{2}\). These eigenvalues provide insights into the behavior of solutions around the equilibrium point \((0,0)\).

Stability Analysis

Stability analysis is a fundamental part of evaluating differential equations and refers to how solutions behave when they are near an equilibrium point. In our scenario, the equilibrium point is \((0,0)\). The stability of this equilibrium depends on the real parts of the eigenvalues that we determined earlier.

• If all eigenvalues have negative real parts, the system is stable around the equilibrium. Solutions will return to equilibrium over time if perturbed.
• If any eigenvalue has a positive real part, the equilibrium is unstable. Solutions diverge away from the equilibrium.

In our exercise, both eigenvalues \(\lambda_1\) and \(\lambda_2\) have negative real parts, indicating that the system is stable. For students analyzing a different problem, remember to carefully evaluate the real part of each eigenvalue to predict long-term behavior.

Equilibrium Classification

The classification of equilibrium helps us understand the nature and long-term behavior of the solutions around an equilibrium point. With our matrix \(A \), we are focusing on \((0,0)\) as the equilibrium. In this context, the eigenvalues not only speak to stability but also allow us to classify the equilibrium:

• If eigenvalues are real and distinct with negative real parts, like \(\lambda_1\) and \(\lambda_2\) in the exercise, the equilibrium is a stable node. This indicates that trajectories move directly in towards the equilibrium.
• When eigenvalues are complex with negative real parts, it forms a stable focus, causing spiraling motion but eventual convergence.

For our example, as both eigenvalues are real, distinct, and negative, the system is characterized by a stable node, often called a sink. This means nearby solutions will converge to \((0,0)\) as time progresses.