Problem 16 - Differential Equations and Linear Algebra

Question

Characterize the equilibrium point for the system \(\mathbf{x}^{\prime}=A \mathbf{x}\) and sketch the phase portrait. $$A=\left[\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right]$$

Step-by-Step Solution

Answer

The equilibrium point for the system \(\mathbf{x}^{\prime} = A\mathbf{x}\) with the given matrix \(A = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}\) is an unstable node at the origin. The eigenvalues are \(\lambda_1 = 1\) and \(\lambda_2 = 5\), and the corresponding eigenvectors are \(\mathbf{v}_1 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}\) and \(\mathbf{v}_2 = \begin{bmatrix} 1 \\ 3 \end{bmatrix}\). The phase portrait displays trajectories moving away from the origin along the directions of these eigenvectors.

Step 1: Calculate the eigenvalues of the matrix A

Let us calculate the eigenvalues of matrix \(A\) by solving the characteristic equation, which is given as: \(\det(A - \lambda I) = 0\) where \(\lambda\) denotes the eigenvalues and \(I\) is the identity matrix. For the given matrix \(A\), the characteristic equation becomes: \(\det\left(\begin{bmatrix} 2-\lambda & 1\\ 3 & 4-\lambda\end{bmatrix}\right) = (2-\lambda)(4-\lambda) - (3)(1) = \lambda^2 - 6\lambda + 5 = (\lambda-5)(\lambda-1) = 0\) Hence, the eigenvalues are \(\lambda_1 = 1\) and \(\lambda_2 = 5\).

Step 2: Calculate the eigenvectors of the matrix A

Now, let's calculate the eigenvectors associated with each eigenvalue by solving the equation: \((A - \lambda I) \mathbf{v} = 0\), where \(\mathbf{v}\) is the eigenvector. For the eigenvalue \(\lambda_1 = 1\): \(\begin{bmatrix} 1 & 1\\ 3 & 3 \end{bmatrix}\mathbf{v}_1 = \mathbf{0}\) We can see that there is one linearly independent eigenvector for this system: \(\mathbf{v}_1 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}\) For the eigenvalue \(\lambda_2 = 5\): \(\begin{bmatrix} -3 & 1\\ 3 & -1 \end{bmatrix}\mathbf{v}_2 = \mathbf{0}\) We can also find one linearly independent eigenvector for this system: \(\mathbf{v}_2 = \begin{bmatrix} 1 \\ 3 \end{bmatrix}\)

Step 3: Characterize the equilibrium point

Since both eigenvalues \(1\) and \(5\) are real and positive, the equilibrium point at the origin \(\mathbf{x} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\) is an unstable node. The trajectories will move away from the origin along the eigenvectors \(\mathbf{v}_1\) and \(\mathbf{v}_2\).

Step 4: Sketch the phase portrait

To sketch the phase portrait, start by plotting the eigenvectors \(\mathbf{v}_1\) and \(\mathbf{v}_2\) and then draw the trajectories following the direction of the eigenvectors. Since both eigenvalues are positive, the trajectories will move away from the origin (unstable node). The direction of the trajectories will be determined by the eigenvalues: those associated with \(\lambda_1\) will move slower, while those associated with \(\lambda_2\) will move faster. In summary, the equilibrium point for the given system is an unstable node at the origin with eigenvectors \(\mathbf{v}_1 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}\) and \(\mathbf{v}_2 = \begin{bmatrix} 1 \\ 3 \end{bmatrix}\), and the phase portrait should be sketched to display the trajectories moving away from the origin along these eigenvectors directions.

Key Concepts

Eigenvalues

Eigenvalues are fundamental in analyzing the behavior of linear systems. They tell us whether the equilibrium point of a system is stable or unstable. To calculate them, we use the characteristic equation \[\det(A - \lambda I) = 0.\]Here, \(\lambda\) represents the eigenvalues and \(I\) is the identity matrix.

• By solving this equation for the given matrix \(A\), we find the eigenvalues \(\lambda_1 = 1\) and \(\lambda_2 = 5\).
• These eigenvalues are both real numbers and positive, indicating that the equilibrium point is an unstable node.

Understanding eigenvalues helps determine the stability and type of equilibrium of the given system. In our situation, both eigenvalues are positive, confirming instability.

Eigenvectors

Eigenvectors are vectors that give us the direction of phase trajectories of a system. These are found by solving the equation \[(A - \lambda I) \mathbf{v} = 0,\]for each eigenvalue. In our example:

• For \(\lambda_1 = 1\), the eigenvector \(\mathbf{v}_1 = \begin{bmatrix} -1 \ 1 \end{bmatrix}\) tells us the direction associated with this eigenvalue.


• For \(\lambda_2 = 5\), the eigenvector \(\mathbf{v}_2 = \begin{bmatrix} 1 \ 3 \end{bmatrix}\) provides another trajectory direction.

In a phase portrait, the trajectories follow the direction of these eigenvectors. Each eigenvector corresponds to a straight-line trajectory through the system's origin.
Understanding eigenvectors is crucial as they determine how solutions will evolve over time through the state space.

Unstable Node

An unstable node is a type of equilibrium point where solutions tend to move away over time. This occurs when all the eigenvalues of a system are real and positive.

• In our example, both eigenvalues \(\lambda_1 = 1\) and \(\lambda_2 = 5\) are positive, thus confirming the point is an unstable node.
• The eigenvectors \(\mathbf{v}_1\) and \(\mathbf{v}_2\) determine the directions in which the trajectories diverge from the origin.

Sketching the phase portrait, you would plot these eigenvectors starting at the origin. The trajectories will radiate outwards, indicating the system's tendency to move away from equilibrium.
This understanding allows us to visualize how disturbances to the system will grow over time, often a critical insight in control systems and stability analysis.