Problem 14 - Differential Equations and Linear Algebra

Question

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

Step-by-Step Solution

Answer

The equilibrium point for the system is an unstable node, with eigenvalues \(\lambda_{1}=1\) and \(\lambda_{2}=3\), and corresponding eigenvectors \(\mathbf{v}_{1}=\begin{pmatrix}1 \\ 1\end{pmatrix}\) and \(\mathbf{v}_{2}=\begin{pmatrix}1 \\ -1\end{pmatrix}\). The phase portrait has directions along the lines \(y=x\) and \(y=-x\), with trajectories moving away from the origin.

Step 1: Compute the eigenvalues

Let's find the eigenvalues of the matrix A. To do this, we must solve the following characteristic equation: \[\text{det}(A-\lambda I)=0\] where \(I\) is the identity matrix and \(\lambda\) represents the eigenvalues. The characteristic equation is given by: \[\text{det}\left(\begin{array}{rr} 2-\lambda & -1 \\ 1 & 2-\lambda \end{array}\right)=(2-\lambda)^{2}-1\cdot(-1)=0\] Solving for \(\lambda\), we get: \[\lambda^2-4\lambda+3=0\] The eigenvalues for this quadratic equation are: \[\lambda_{1}=1, \quad \lambda_{2}=3\]

Step 2: Compute the eigenvectors

For each eigenvalue, we need to compute the corresponding eigenvector by solving the following equation: \[(A-\lambda_{i} I)\mathbf{v}_{i}=\mathbf{0}\] For \(\lambda_{1}=1\), we have: \[\left(\begin{array}{rr} 1 & -1\\ 1 & 1 \end{array}\right)\left(\begin{array}{c}v_{1} \\ v_{2}\end{array}\right)=\mathbf{0}\] From the first row, we get \(v_{1}=v_{2}\). Therefore, the eigenvector corresponding to \(\lambda_{1}=1\) is: \[\mathbf{v}_{1}=\begin{pmatrix}1 \\ 1\end{pmatrix}\] For \(\lambda_{2}=3\), we have: \[\left(\begin{array}{rr} -1 & -1\\ 1 & -1 \end{array}\right)\left(\begin{array}{c}v_{1} \\ v_{2}\end{array}\right)=\mathbf{0}\] From the first row, we get \(v_{1}=-v_{2}\). Therefore, the eigenvector corresponding to \(\lambda_{2}=3\) is: \[\mathbf{v}_{2}=\begin{pmatrix}1 \\ -1\end{pmatrix}\]

Step 3: Characterize the equilibrium point and sketch the phase portrait

As both eigenvalues are real and positive, the equilibrium point is an unstable node. To sketch the phase portrait, we will plot the eigenvectors and draw curves that follow the directions of the eigenvectors. The eigenvector \(\mathbf{v}_{1}\) indicates that the phase portrait direction is along the line \(y=x\). The eigenvector \(\mathbf{v}_{2}\) indicates that the phase portrait direction is along the line \(y=-x\). Since the equilibrium point is unstable, trajectories will move away from the origin (the equilibrium point) along these lines. To summarize, the equilibrium point of the given system is an unstable node, with phase portrait directions specified by the eigenvectors \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\).

Key Concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is crucial when analyzing the behavior of dynamical systems. In essence, eigenvalues indicate the growth rate and direction of the system, while eigenvectors describe the line along which the stretching or compressing happens.

For a matrix A, eigenvalues are found by solving the characteristic equation, which involves the determinant of A - \(\lambda I\), where I is the identity matrix and \lambda represents the eigenvalues. Once the eigenvalues are determined, the corresponding eigenvectors are calculated by solving (A - \(\lambda_i I\))\textbf{v}_i = \textbf{0} for each \lambda_i.

For the system in question, with matrix A having eigenvalues 1 and 3, we have two distinct eigenvalues indicating two directions of movement, described by their corresponding eigenvectors. The eigenvectors provide the important geometric insight needed for sketching the phase portrait, as they outline the principal directions for the system's evolution.

Phase Portrait Sketching

The phase portrait sketching is a visual representation of the trajectory that solutions to a system of differential equations will follow over time. It is especially helpful in understanding the qualitative behavior of dynamical systems. After obtaining the eigenvalues and eigenvectors, one can begin sketching the phase portrait by drawing lines corresponding to the eigenvectors, known as the system's principal axes.

In this specific case, the eigenvectors suggest two principal directions: one along y = x and another along y = -x. The phase portrait will exhibit arrows indicating the direction of motion. Since the eigenvalues are positive, we expect the trajectories to move away from the equilibrium point, which is the origin in this case. This visualization facilitates comprehension of how the system behaves over time without needing to solve for every possible solution explicitly.

Stability of Dynamical Systems

The stability of dynamical systems can be inferred by analyzing the eigenvalues of the system's matrix. The signs and values of the eigenvalues determine the nature of the equilibrium point. For instance:

• If all eigenvalues have negative real parts, the system is stable, and the equilibrium point is attracting.
• If at least one eigenvalue has a positive real part, the system is unstable, and the equilibrium point is repelling.
• If any eigenvalues are strictly complex, the system can exhibit spiraling behavior, which can be stable or unstable depending on the real part of the eigenvalues.

In the given exercise, both eigenvalues are real and positive. Therefore, the equilibrium point is characterized as an unstable node. Trajectories will move away from the equilibrium, signifying that any small deviation from the equilibrium will cause the system to diverge further over time. This information about stability helps in predicting the long-term behavior of the system under small perturbations.