Question

(a) Find the eigenvalues and eigenvectors. (b) Classify the critical point \((0,0)\) as to type and determine whether it is stable, asymptotically stable, or unstable. (c) Sketch several trajectories in the phase plane and also sketch some typical graphs of \(x_{1}\) versus \(t .\) (d) Use a computer to plot accurately the curves requested in part (c). \(\frac{d \mathbf{x}}{d t}=\left(\begin{array}{rr}{-1} & {-1} \\ {0} & {-0.25}\end{array}\right) \mathbf{x}\)

Short answer

Question: Determine the eigenvalues, eigenvectors, type, and stability of the critical point (0,0) for the given matrix $A = \begin{pmatrix} -1 & -1 \\ 0 & -0.25 \end{pmatrix}$. Provide a rough sketch of the trajectories in the phase plane and a rough sketch of the graph of \(x_{1}\) versus \(t\).

Step-by-step solution

Determine the characteristic equation

To find the eigenvalues, first, we need to find the determinant of \((A - \lambda I)\) where \(A\) is the given matrix, and \(\lambda\) is the eigenvalue. $A = \begin{pmatrix} -1 & -1 \\ 0 & -0.25 \end{pmatrix}$, $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$, and $(A-\lambda I) = \begin{pmatrix} -1-\lambda & -1 \\ 0 & -0.25-\lambda \end{pmatrix}\(. The characteristic equation is formed by setting the determinant of \)(A - \lambda I)$ to zero.

Compute the eigenvalues

For the characteristic equation, we have \(|A - \lambda I|= (-1-\lambda)(-0.25-\lambda) - (0)(-1) = 0\). Solving for \(\lambda\), we get \(\lambda_{1}=-1\) and \(\lambda_{2}=-0.25\).

Compute the eigenvectors

To find the eigenvectors corresponding to each eigenvalue, we need to solve the following equation with each eigenvalue: \((A - \lambda_{i} I) v_{i}=0\). 1. For \(\lambda_{1}=-1\), we have $(A - \lambda_{1} I) v_{1} = \begin{pmatrix} 0 & -1 \\ 0 & 0.75 \end{pmatrix} v_{1}=0\(, which gives us \)v_{1}= \begin{pmatrix} 1 \\ 0 \end{pmatrix}$. 2. For \(\lambda_{2}=-0.25\), we have $(A - \lambda_{2} I) v_{2} = \begin{pmatrix} -0.75 & -1 \\ 0 & 0 \end{pmatrix} v_{2}=0\(, which gives us \)v_{2}= \begin{pmatrix} 1 \\ 0.75 \end{pmatrix}$. So, the eigenvalues are \(\lambda_{1} = -1\) and \(\lambda_{2} = -0.25\), and their respective eigenvectors are $v_{1} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\( and \)v_{2} = \begin{pmatrix} 1 \\ 0.75 \end{pmatrix}$. #b: Classifying the critical point#

Determine the type of critical point

The critical point is classified based on its eigenvalues. Since both eigenvalues are real and have the same sign, the critical point is a node. Since they are both negative, it is a stable node.

Determine the stability of the critical point

We can determine the stability of the critical point based on the value of the eigenvalues. Since both eigenvalues are negative, the critical point \((0,0)\) is asymptotically stable. #c: Sketching trajectories and graph of \(x_{1}\) vs. \(t\)

Sketching the trajectories

The two eigenvectors are $v_{1} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\( and \)v_{2} = \begin{pmatrix} 1 \\ 0.75 \end{pmatrix}\(. \)v_{1}\( gives the horizontal trajectories, while \)v_{2}$ gives the slanted trajectories. The trajectories will converge towards the critical point (0,0) as it is an asymptotically stable node. These trajectories will give us a rough sketch of the phase plane.

Sketching the graph of \(x_{1}\) vs \(t\)

The graph of \(x_{1}\) versus \(t\) will be exponentially decaying since the eigenvalues are negative. The rate of decay is determined by the eigenvalues, so we will have two curves that decay at different rates - one with a rate of \(-1\) (the curve from \(\lambda_{1}=-1\)) and another with a rate of \(-0.25\) (the curve from \(\lambda_{2}=-0.25\)). For part (d), use a suitable software like MATLAB, Python, or Wolfram Alpha to plot the curves mentioned in part (c) for an accurate plot.

Key concepts

Critical Points

Critical points play a crucial role when analyzing dynamic systems like the one described by the given matrix equation. The critical point we're focusing on is **(0,0)**, a point where the system settles when there are no external inputs. Critical points are sometimes referred to as equilibrium points. These points represent states where the system doesn't evolve further unless disturbed. To classify this critical point, we mainly rely on the eigenvalues and eigenvectors derived from the system's matrix. Eigenvalues give us a clear picture of the system's behavior around the critical point. Here,

• Eigenvalue \( \lambda_1 = -1 \)
• Eigenvalue \( \lambda_2 = -0.25 \)

Both eigenvalues are real and negative, indicating that any motion starting nearby will converge to the critical point. Each eigenvalue corresponds to a direction in which the system moves—indicating either convergence or divergence relative to the critical point.

Stability Analysis

Stability analysis helps determine how a system responds over time when it's slightly disturbed from its equilibrium state. For systems like the one we're handling, eigenvalues provide significant insights into stability. The type of eigenvalues indicates if the critical point is stable, unstable, or asymptotically stable. For the critical point (0,0), both eigenvalues \( \lambda_1 = -1 \) and \( \lambda_2 = -0.25 \) are negative. This tells us:

• The system is asymptotically stable. This means any trajectory starting from a nearby point in the phase plane will eventually return to the critical point over time.
• The negativity of the eigenvalues implies the trajectories decay over time without oscillating, leading them to the equilibrium point smoothly.

In practical terms, if this system represented, say, a population or temperature model, it would return to equilibrium without wild fluctuations, indicating robustness against small disturbances.

Phase Plane Trajectories

Visualizing the behavior of a system can be made easier through phase plane trajectories. The phase plane is essentially a graphical representation of the system's possible states plotted with respect to time. For our system:

• The eigenvector associated with \( \lambda_1 = -1 \) is \( \begin{pmatrix} 1 \ 0 \end{pmatrix} \), leading to horizontal trajectories approaching the critical point. These directions highlight the system's fast-decay path.
• The eigenvector for \( \lambda_2 = -0.25 \) is \( \begin{pmatrix} 1 \ 0.75 \end{pmatrix} \), showing slanted paths that approach more gradually due to the slower decay rate.

Both sets of trajectories converge to the critical point, illustrating the system's stability clearly. In this visualization, you will see trajectories of different curvature and speeds of convergence, all directed towards (0,0). By observing these trajectories and how they behave over time, we fully appreciate the system's built-in damping characteristics.