Problem 9 - Differential Equations and Linear Algebra

Question

Characterize the equilibrium point for the system \(\mathbf{x}^{\prime}=A \mathbf{x}\) and sketch the phase portrait. $$A=\left[\begin{array}{rr} -2 & 3 \\ -3 & -2 \end{array}\right]$$

Step-by-Step Solution

Answer

The equilibrium point for the given system is a saddle point with eigenvalues \(\lambda_1 = -5\) and \(\lambda_2 = 1\), and eigenvectors \(\mathbf{v_1} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\) and \(\mathbf{v_2} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\). The phase portrait will exhibit saddle behavior, with the stable eigendirection along \(\mathbf{v_2}\) and the unstable eigendirection along \(\mathbf{v_1}\), and the equilibrium point at the origin (0,0).

Step 1: Calculating the Eigenvalues of matrix A

Firstly, we want to find the eigenvalues of matrix A. We can do this by solving the following equation: \(det(A - \lambda I) = 0\) Where λ represents the eigenvalues we need to find, and I is the identity matrix. Now, let's calculate the determinant: \(det(A - \lambda I) = det\begin{bmatrix} -2 - \lambda & 3\\ -3 & -2 - \lambda \end{bmatrix}\) Expanding the determinant, we get the characteristic equation \((-2 - \lambda)((-2) - \lambda) - (-3)(3) = 0\)

Step 2: Solving the Characteristic Equation

Now, let's solve this quadratic equation to find the eigenvalues: \((\lambda + 2)^{2} - 9 = 0 \) \(\lambda^{2} + 4\lambda - 5 = 0\) Solving the quadratic equation, we get two eigenvalues: \(\lambda_1 = -5, \lambda_2 = 1\)

Step 3: Finding the Eigenvectors

Next, we need to find the eigenvectors corresponding to the eigenvalues. Let's start with the first eigenvalue \(\lambda_1\). \(A\mathbf{v_1} = \lambda_1 \mathbf{v_1}\) \(\begin{bmatrix} -2 & 3\\ -3 & -2 \end{bmatrix}\mathbf{v_1} = -5\mathbf{v_1}\) Now, solving this equation, we get the eigenvector, \(\mathbf{v_1} = \begin{bmatrix} 1 \\ -1\end{bmatrix}\) Next, let's find the eigenvector for \(\lambda_2\), using the same method: \(\begin{bmatrix} -2 & 3\\ -3 & -2 \end{bmatrix}\mathbf{v_2} = \mathbf{v_2}\) Solving this equation, we get the eigenvector, \(\mathbf{v_2} = \begin{bmatrix} 1 \\ 1\end{bmatrix}\)

Step 4: Characterize the Equilibrium Point and Phase Portrait

Since we have a positive and a negative eigenvalue, we can conclude that the equilibrium point is a saddle point. To sketch the phase portrait, we need to plot the eigenvectors on the plane and draw the trajectory of the system based on those vectors. The phase portrait will exhibit saddle behavior, with the stable eigendirection along the eigenvector \(\mathbf{v_2}\) and the unstable eigendirection along the eigenvector \(\mathbf{v_1}\). The equilibrium point will be at the origin (0,0). In conclusion, the equilibrium point for the given system is a saddle point with eigenvectors \(\begin{bmatrix} 1 \\ -1 \end{bmatrix}\) and \(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\).

Key Concepts

Eigenvalues and Eigenvectors

Understanding the eigenvalues and eigenvectors of a matrix is a cornerstone in the study of systems of differential equations. They provide critical information about the behavior of the system, especially near equilibrium points. An eigenvalue is essentially a scalar that indicates whether and by how much an eigenvector is scaled during a transformation.

For a given square matrix A, an eigenvalue λ and its corresponding eigenvector v satisfy the equation Av = λv. To find them, one must calculate the determinant of the matrix subtracted by λ times the identity matrix, I, setting the resulting characteristic equation to zero. The solutions to this equation are the eigenvalues, and plugging these back into the equation provides the eigenvectors. These vectors indicate the direction of the invariant axes around the equilibrium points and reveal much about the system's local stability.

Phase Portrait

A phase portrait is a geometric representation of the trajectories of a dynamical system in the phase plane. Each point in this plane represents the state of the system at a given time, with the entire plane spanning all possible states. By plotting the eigenvectors associated with their eigenvalues, the phase portrait visualizes how the state of the system changes over time.

The arrows and curves on a phase portrait represent the system's trajectories, showing how it evolves from various starting points. For instance, in a two-dimensional system, the portrait provides a straightforward way to observe the stability or instability of equilibrium points. It also aids in the identification of patterns such as saddle points, nodes, or spirals, and is a powerful tool in understanding the overall dynamics within the system.

Characteristic Equation

The characteristic equation is the equation obtained when finding the eigenvalues of a matrix. By setting the determinant of A - λI to zero, where A is a given matrix, λ the eigenvalue, and I the identity matrix, one writes out the characteristic equation.

For a 2x2 matrix, this will usually yield a quadratic equation whose roots are the eigenvalues of the matrix. These roots are pivotal in determining the nature of equilibrium points and help predict how a system will behave. It is through this equation that one can discern the stability characteristics of the system and proceed to understand more complex behaviors exhibited by the system under analysis.

Saddle Point

A saddle point in the context of dynamical systems is an equilibrium point that is neither completely stable nor completely unstable. It is characterized by having eigenvalues with opposite signs, which implies that the system behaves differently along different directions - stable in some and unstable in others.

A saddle point is visually identified in a phase portrait as the point where trajectories approach the equilibrium along the stable direction and move away along the unstable direction. This mixed behavior makes saddle points particularly interesting, as they create a scenario where the stability of the system is highly dependent on the initial conditions and the direction from which the point is approached.