Problem 35 - 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. $$ A=\left[\begin{array}{rr} 6 & -4 \\ -3 & 5 \end{array}\right] $$

Step-by-Step Solution

Answer

The equilibrium \((0,0)\) is a source, as both eigenvalues are positive.

Step 1: Find the Eigenvalues

To determine the stability of the equilibrium \((0,0)\), we first need to find the eigenvalues of the matrix \(A\). The eigenvalues \(\lambda\) are found from the characteristic equation \(\det(A - \lambda I) = 0\), where \(I\) is the identity matrix \(\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\). So we calculate the determinant: \[\det\left(\begin{bmatrix} 6 & -4 \ -3 & 5 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\right) = \det\begin{bmatrix} 6 - \lambda & -4 \ -3 & 5 - \lambda \end{bmatrix}\] This simplifies to: \[(6 - \lambda)(5 - \lambda) - (-4)(-3) = \lambda^2 - 11\lambda + 18 - 12 = 0\] \[\lambda^2 - 11\lambda + 6 = 0\]

Step 2: Solve the Characteristic Equation

Now solve the quadratic equation \(\lambda^2 - 11\lambda + 6 = 0\) using the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -11\), and \(c = 6\). \[\lambda = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} = \frac{11 \pm \sqrt{121 - 24}}{2} = \frac{11 \pm \sqrt{97}}{2}\] Thus, the eigenvalues are \(\lambda_1 = \frac{11 + \sqrt{97}}{2}\) and \(\lambda_2 = \frac{11 - \sqrt{97}}{2}\).

Step 3: Evaluate the Eigenvalues

Both eigenvalues are real and distinct. To classify the stability, note the sign of the eigenvalues: if both eigenvalues are positive, the equilibrium is a source; if both are negative, it is a sink; if one is positive and one is negative, it is a saddle point. Since \(\lambda_1 > 0\) and \(\lambda_2 > 0\), the equilibrium \((0,0)\) is a source.

Key Concepts

Stability Analysis

Stability analysis is a crucial concept in understanding dynamical systems described by differential equations. It involves evaluating how solutions to these equations behave over time, especially near equilibrium points. When analyzing the stability of an equilibrium point, such as \( (0,0) \) in our matrix differential equation, we are determining whether small perturbations or changes will grow or diminish over time.
For linear systems of differential equations, the stability of an equilibrium is related directly to the eigenvalues of the system's matrix. In simple terms:

• If all eigenvalues have negative real parts, the equilibrium is stable (or asymptotically stable), and solutions tend to come back to the equilibrium point over time.
• If any eigenvalue has a positive real part, even if just one, the equilibrium is unstable, meaning solutions tend to move away from the point.

These outcomes help predict whether a system will return to equilibrium after being slightly disturbed, which is essential in various fields such as biology, physics, and economics.

Eigenvalues

Eigenvalues are fundamental to the study of differential equations and linear algebra. They provide insights into the behavior of systems described by matrices. To find the eigenvalues of a matrix, you solve the characteristic equation \( \det(A - \lambda I) = 0 \), where \( A \) is your matrix and \( I \) is the identity matrix.
In our example, we computed the eigenvalues \( \lambda \) from matrix:\[A=\begin{bmatrix}6 & -4 \-3 & 5\end{bmatrix}\]By solving the characteristic equation for this matrix, we derived the eigenvalues \( \lambda_1 = \frac{11 + \sqrt{97}}{2} \) and \( \lambda_2 = \frac{11 - \sqrt{97}}{2} \).

• The real and distinct nature of these eigenvalues is significant because it allows us to classify the equilibrium point straightforwardly.
• Real eigenvalues indicate the system's solutions are primarily influenced by exponential growth or decay, crucial for predicting long-term behavior.

Understanding eigenvalues is thus critical in stability analysis and the classification of equilibrium points in dynamical systems.

Equilibrium Classification

Classifying equilibria based on eigenvalues helps us understand how different types of behavior manifest in a system. Equilibrium points can generally be termed as 'sink,' 'source,' or 'saddle point,' each describing different stability properties:

• Sink: All eigenvalues have negative real parts. The system is stable around this point, and perturbations die out over time, drawing solutions back to the equilibrium.
• Source: All eigenvalues have positive real parts. The equilibrium is unstable, and any disturbance grows over time, pulling the system away.
• Saddle Point: Eigenvalues are mixed; some positive, some negative. The system is unstable, but the behavior is more complex, with some trajectories approaching while others diverge.

In our example, both eigenvalues are positive, so the equilibrium \( (0,0) \) is classified as a source. This means the system will diverge from the equilibrium point upon perturbation, indicative of an intrinsically unstable system. This classification is vital in many fields, helping engineers, biologists, and economists to design systems with desired stability characteristics.