Question

In each exercise, consider the linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\). Since \(A\) is a constant invertible \((2 \times 2)\) matrix, \(\mathbf{y}=\mathbf{0}\) is the unique (isolated) equilibrium point. (a) Determine the eigenvalues of the coefficient matrix \(A\). (b) Use Table \(6.2\) to classify the type and stability characteristics of the equilibrium point at the phase-plane origin. If the equilibrium point is a node, designate it as either a proper node or an improper node. $$ \mathbf{y}^{\prime}=\left[\begin{array}{ll} -1 & 8 \\ -1 & 5 \end{array}\right] \mathbf{y} $$

Short answer

Answer: The equilibrium point at the phase-plane origin is a saddle point.

Step-by-step solution

Determine the eigenvalues of matrix A

We are given the matrix \(A = \begin{bmatrix} -1 & 8 \\ -1 & 5 \end{bmatrix}\). To find the eigenvalues, we need to solve the characteristic equation: $$\det(A - \lambda I) = 0$$ where \(\lambda\) represents the eigenvalues and \(I\) is the identity matrix. Now, let's calculate the determinant of \((A - \lambda I)\): $$\det\left(\begin{bmatrix} -1-\lambda & 8 \\ -1 &5-\lambda \end{bmatrix}\right) = (-1-\lambda)(5-\lambda) - (-1)(8) = \lambda^2 -4\lambda - 3$$ Next, we will solve the quadratic equation to find the eigenvalues: $$\lambda^2 - 4\lambda - 3=0$$

Solve the quadratic equation to find the eigenvalues

We can use the quadratic formula to solve the equation \(\lambda^2 - 4\lambda - 3=0\). The quadratic formula is given by: $$\lambda = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ In our case, \(a=1\), \(b=-4\), and \(c=-3\). Substituting the values, we get: $$\lambda = \frac{4\pm\sqrt{(-4)^2-4(1)(-3)}}{2(1)} = \frac{4\pm\sqrt{28}}{2}$$ So, the eigenvalues are: $$\lambda_1 = \frac{4+\sqrt{28}}{2}, \quad \lambda_2 = \frac{4-\sqrt{28}}{2}$$

Classify the type and stability characteristics of the equilibrium point

Now that we have the eigenvalues, we can classify the equilibrium point at the phase-plane origin using Table 6.2. 1. If both eigenvalues are real and positive, the equilibrium point is an unstable (improper) node. 2. If both eigenvalues are real and negative, the equilibrium point is a stable (improper) node. 3. If eigenvalues have different signs, the equilibrium point is a saddle point. 4. If both eigenvalues are complex with positive real parts, the equilibrium point is an unstable spiral. 5. If both eigenvalues are complex with negative real parts, the equilibrium point is a stable spiral. In our case, since both eigenvalues are real and have different signs (one positive and one negative): $$\lambda_1 > 0 \quad \text{and} \quad \lambda_2 < 0$$ So, the equilibrium point at the phase-plane origin is a saddle point.

Key concepts

Linear Differential Equations

Linear differential equations form the basis of many problems and applications in physics, engineering, and mathematics. These equations involve an unknown function and its derivatives and are linear in terms of both. In the simplest case, a linear differential equation has the form \( y' = ay + b \) where \( a \) and \( b \) are constants, and \( y' \) denotes the derivative of \( y \) with respect to a variable, usually time. The solution to such equations describes dynamic systems that evolve over time.

When dealing with systems of linear differential equations, we often represent them in matrix form as \( \mathbf{y}' = A\mathbf{y} \) where \( A \) is a matrix of coefficients, and \( \mathbf{y} \) is a vector of unknown functions. The matrix \( A \) plays a crucial role because its eigenvalues determine the system's behavior. Such systems can model mechanical systems, electronic circuits, population dynamics, and much more. By analyzing the eigenvalues of the matrix, we gain insight into the stability and type of equilibrium points of the system, which are pivotal for understanding the long-term behavior of the system.

Equilibrium Point Classification

An equilibrium point in a system of differential equations is a point where the system is at rest or unchanged over time. In the context of our equation \( \mathbf{y}' = A\mathbf{y} \) with constant coefficients, the unique equilibrium point is \( \mathbf{y} = \mathbf{0} \), meaning that when the system reaches this point, it will stay there indefinitely unless perturbed. Classification of these points is critical, as they indicate the long-term behavior of the system surrounding these points.

To classify equilibrium points, we first find the eigenvalues of the matrix \( A \) as these provide significant information on the system's dynamics. Based on the eigenvalues we can typically classify equilibrium points into nodes (proper or improper), spirals, foci, or centers, with further characterization into stable, unstable, or semi-stable, depending on the nature of the eigenvalues. Stable points attract nearby trajectories whereas unstable points repel them. When a system exhibits eigenvalues of varying signs, we encounter what is known as a saddle point, which repels in some directions and attracts in others. These classifications help in predicting how the system reacts to small disturbances around the equilibrium.

Phase-Plane Analysis

Phase-plane analysis provides a geometric visualization of the behavior of a system of two first-order autonomous differential equations. It allows us to study the trajectories of the system over time, which are curves in the plane that represent the state of the system. By plotting these trajectories, also known as phase paths or phase curves, we get a 'phase portrait' that depicts the behavior around equilibrium points.

To conduct a phase-plane analysis, we need the eigenvalues and eigenvectors of the matrix \( A \) associated with the system. The eigenvalues help classify the type of equilibrium points and the eigenvectors give the direction of the phase paths. For example, in the given exercise, the computation of the eigenvalues led to the conclusion that the equilibrium point is a saddle point, which can be visualized as having trajectories approaching the equilibrium along one eigendirection and diverging along another, depicting different system behaviors. Phase-plane plots are invaluable tools for engineers and scientists, for they provide a way to predict the possible outcomes of a system under various initial conditions without having to solve the differential equations analytically.