Problem 12 - Differential Equations and Linear Algebra

Question

Characterize the equilibrium point for the system \(\mathbf{x}^{\prime}=A \mathbf{x}\) and sketch the phase portrait. $$A=\left[\begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array}\right]$$

Step-by-Step Solution

Answer

The equilibrium point for the given system is found to be at the origin. The eigenvalues are purely imaginary (\(\lambda_1 = i\), \(\lambda_2 = -i\)), indicating that the equilibrium is a center. Therefore, the system is stable but not asymptotically stable, with closed orbits around the equilibrium point. The eigenvectors \(\mathbf{x}_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\) and \(\mathbf{x}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\) represent directions of clockwise and counterclockwise rotation, respectively. This leads to a phase portrait displaying concentric circles around the origin, illustrating the stable oscillatory behavior of the system.

Step 1: Calculate the Eigenvalues of Matrix A

To find the eigenvalues, we need to compute the characteristic equation of matrix A, which is given by: \[\det(A - \lambda I) = 0\] where \(\lambda\) is the eigenvalue and \(I\) is the identity matrix. The characteristic equation for matrix A is: \[\begin{vmatrix}0 - \lambda & -1 \\1 & 0 - \lambda\end{vmatrix} = (0 - \lambda)(0 - \lambda) - (-1)(1) = \lambda^2 + 1\] Set the equation to 0 and solve for \(\lambda\): \[\lambda^2 + 1 = 0\] \[\lambda^2 = -1\] \[\lambda = \pm i\] The eigenvalues are \(\lambda_1 = i\) and \(\lambda_2 = -i\).

Step 2: Calculate the Eigenvectors of Matrix A

To compute the eigenvectors for each eigenvalue, we have to solve the equation \((A - \lambda I) \mathbf{x} = 0\), where \(\mathbf{x}\) is the eigenvector. 1. For \(\lambda_1 = i\), we have: \[\begin{bmatrix} -i & -1 \\ 1 & -i \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = 0\] We can convert this complex matrix equation into two real matrix equations: \[ \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = 0 \] We can solve these equations to get the eigenvector corresponding to eigenvalue \(\lambda_1 = i\): \[x_1+x_2 = 0 \implies x_2 = -x_1 \implies x_1 = x_1\] So, the eigenvector for \(\lambda_1 = i\) is \(\mathbf{x}_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\). 2. For \(\lambda_2 = -i\), we have: \[\begin{bmatrix} i & -1 \\ 1 & i \end{bmatrix} \begin{bmatrix} x_3 \\ x_4 \end{bmatrix} = 0\] We can convert this complex matrix equation into two real matrix equations: \[ \begin{bmatrix} -1 & -1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x_3 \\ x_4 \end{bmatrix} = 0 \] We can solve these equations to get the eigenvector corresponding to eigenvalue \(\lambda_2 = -i\): \[-x_3-x_4 = 0 \implies x_4 = -x_3 \implies x_3 = x_3\] So, the eigenvector for \(\lambda_2 = -i\) is \(\mathbf{x}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\).

Step 3: Determine Equilibrium Points

To find the equilibrium points for the system, we solve \(\mathbf{x'} = A \mathbf{x} = 0\). In this case, \[A \mathbf{x} = \begin{bmatrix}0 & -1 \\1 & 0 \end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \end{bmatrix} = 0\] This equation simplifies to: \[\begin{cases} -x_2 = 0 \\ x_1 = 0 \end{cases}\] Both equations imply that the only solution is \(\mathbf{x} = \begin{bmatrix}0 \\ 0\end{bmatrix}\). Therefore, the system has a single equilibrium point at the origin.

Step 4: Analyze and Sketch Phase Portrait

Since the eigenvalues are purely imaginary (\(\lambda_1 = i\) and \(\lambda_2 = -i\)), the equilibrium point at the origin is a center. This means that the trajectories around the equilibrium point are closed orbits and the system is stable but not asymptotically stable. To sketch the phase portrait, we can use the eigenvectors and eigenvalues to draw the orbits around the equilibrium point. For eigenvalue \(\lambda_1 = i\), the eigenvector \(\mathbf{x}_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\) represents a direction of clockwise rotation. For eigenvalue \(\lambda_2 = -i\), the eigenvector \(\mathbf{x}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\) represents a direction of counterclockwise rotation. Thus, the phase portrait will show concentric circles around the origin, depicting the stable oscillatory behavior of the system.

Key Concepts

Eigenvalues

Eigenvalues are fundamental to understanding linear transformations in linear algebra. They are scalar values that provide insight into the behavior and properties of a matrix. To find the eigenvalues of a matrix, you must solve its characteristic equation:

• This equation is formed using the determinant of the matrix subtracted by a scalar times the identity matrix, all set equal to zero.
• For the given matrix in this exercise, A, the characteristic equation is \[ \det(A - \lambda I) = (\lambda^2 + 1) = 0 \]

The solutions to this equation, \(\lambda = \pm i\), are the eigenvalues of matrix A. Here, they are purely imaginary, suggesting that any transformation represented by this matrix will result in rotations rather than contractions or stretches.

Eigenvectors

While eigenvalues indicate the magnitude of the transformation, eigenvectors point to directionally stable lines within the vector space. These vectors remain unchanged except for scaling when a matrix is applied.

• To find an eigenvector corresponding to each eigenvalue, solve the equation \[ (A - \lambda I) \mathbf{x} = 0 \]
• For \(\lambda = i\), the eigenvector is found to be \[ \mathbf{x}_1 = \begin{bmatrix} 1 \ -1 \end{bmatrix} \]
• Similarly, for \(\lambda = -i\), the eigenvector \[ \mathbf{x}_2 = \begin{bmatrix} 1 \ 1 \end{bmatrix} \] is obtained.

These eigenvectors define the axes of rotation for the system, giving insight into how the space is maintained under transformation.

Equilibrium Points

An equilibrium point occurs where a dynamic system reaches a state where its changes are nullified. In other words, a system is at equilibrium when its derivative is zero. For the system described by \[\mathbf{x'} = A \mathbf{x}\]setting \[A \mathbf{x} = 0\]yields the equilibrium point.

• By solving the matrix equation for zero, we find the equilibrium point at the origin, \[ \mathbf{x} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \]

Since there are no other points satisfying this condition, the origin is the sole equilibrium point. Its properties will determine the nature of the phase portrait.

Phase Portrait Analysis

Phase portrait analysis offers a visual snapshot of a system's behavior over time, grounded in the eigenvalues and eigenvectors found earlier. Here, with purely imaginary eigenvalues, the phase portrait highlights circular trajectories around the equilibrium point.

• The trajectories composed of revolving circles imply stable, non-asymptotic behavior, named a center in phase portrait terminology.
• Each trajectory corresponds to different initial conditions and reflects an oscillating nature, neither increasing nor decreasing as time progresses.

The counterclockwise and clockwise eigenvector directions illustrate how each trajectory develops, forming a harmonious and steady dynamic system with no net movement towards or away from the equilibrium point.