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). \(\frac{d \mathbf{x}}{d t}=\left(\begin{array}{ll}{3} & {-2} \\ {4} & {-1}\end{array}\right) \mathbf{x}\)

Short answer

In conclusion, we have found the eigenvalues of the matrix as \(\lambda_1 = 2 + i\) and \(\lambda_2 = 2 - i\). The eigenvectors corresponding to these eigenvalues are \(\mathbf{v}_1 = \begin{pmatrix} 1+i \\ 2 \end{pmatrix}\) and \(\mathbf{v}_2 = \begin{pmatrix} 1-i \\ 2 \end{pmatrix}\). The critical point (0,0) is classified as an unstable spiral since the real parts of both eigenvalues are positive. The trajectories in the phase plane form an outward spiral, while in the \(x_1\) vs \(t\) graph, they display an oscillatory behavior with an increasing amplitude. To create accurate plots, a computer software such as MATLAB or Mathematica would be needed.

Step-by-step solution

Find the matrix eigenvalues

To find the eigenvalues of the matrix, we have to solve the following equation for \(\lambda\): \(det(A - \lambda I) = 0\) Where \(A = \begin{pmatrix} 3 & -2 \\ 4 & -1 \end{pmatrix}\), and I is the identity matrix. So, we have to find the determinant of the following matrix: \(\begin{vmatrix} 3-\lambda & -2 \\ 4 & -1-\lambda\end{vmatrix}= (3-\lambda)(-1-\lambda) - (-2) \cdot 4\) Now, we solve for \(\lambda\): \((3-\lambda)(-1-\lambda) + 8 = 0\) \(-3 + 4\lambda - \lambda^2 + 8 = 0\) After simplifying the equation, we get the quadratic equation: \(\lambda^2 - 4\lambda + 5 = 0\)

Solve the quadratic equation for eigenvalues

We can solve the equation \(\lambda^2 - 4\lambda + 5 = 0\) with the quadratic formula: \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) \(\lambda = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(5)}}{2(1)}\) \(\lambda = \frac{4 \pm \sqrt{16 - 20}}{2}\) So, the eigenvalues are \(\lambda_1 = 2 + i\) and \(\lambda_2 = 2 - i\)

Find the eigenvectors

To find the eigenvectors, we plug the eigenvalues back into the equation \((A - \lambda I) \mathbf{x} = \mathbf{0}\). For the eigenvalue \(\lambda_1 = 2 + i\), we have: \(\left(\begin{array}{cc}{1-i} & {2} \\ {4} & {-1-i}\end{array}\right) \left(\begin{array}{c}{x_1}\\ {x_2}\end{array}\right) = \left(\begin{array}{c}{0}\\ {0}\end{array}\right)\) Using row reduction or any other method, we can find the eigenvector corresponding to the eigenvalue \(\lambda_1 = 2 + i\) as \(\mathbf{v}_1 = \begin{pmatrix} 1+i \\ 2 \end{pmatrix}\). Similarly, for the eigenvalue \(\lambda_2 = 2 - i\), we have: \(\left(\begin{array}{cc}{1+i} & {2} \\ {4} & {-1+i}\end{array}\right) \left(\begin{array}{c}{x_1}\\ {x_2}\end{array}\right) = \left(\begin{array}{c}{0}\\ {0}\end{array}\right)\) The eigenvector corresponding to the eigenvalue \(\lambda_2 = 2 - i\) is \(\mathbf{v}_2 = \begin{pmatrix} 1-i \\ 2 \end{pmatrix}\).

Analyzing the stability of the critical point (0,0)

Since the real parts of both eigenvalues are positive, the critical point (0,0) is an unstable spiral.

Sketching phase plane trajectories and \(x_1\) vs \(t\) graphs

Sketching phase plane trajectories and \(x_1\) vs \(t\) graphs requires understanding of how the system behaves with respect to time given the eigenvalues and eigenvectors. Since the eigenvalues are complex conjugates with positive real parts, the trajectories in the phase plane will form an outward spiral, representing an unstable spiral behavior. When sketching \(x_1\) vs \(t\) graphs, trajectories would have an oscillatory behavior, with the amplitude increasing over time due to the unstable nature of the critical point (0,0). Finally, please note that the accurate plots of the trajectories mentioned in part (d) of the problem require the use of a computer software such as MATLAB or Mathematica.

Key concepts

Eigenvalues

Eigenvalues are critical in understanding the behavior of linear systems. To find them, we start with a matrix, let's call it \( A \), and solve the characteristic equation \( \text{det}(A - \lambda I) = 0 \). Here, \( \lambda \) represents the eigenvalues, and \( I \) is the identity matrix. For our matrix \( A = \begin{pmatrix} 3 & -2 \ 4 & -1 \end{pmatrix} \), we derived the equation \( \lambda^2 - 4\lambda + 5 = 0 \). Solving this quadratic equation using the quadratic formula gives us the complex eigenvalues \( \lambda_1 = 2 + i \) and \( \lambda_2 = 2 - i \). Complex eigenvalues often indicate oscillatory behavior, which plays a significant role in analyzing the dynamics of the system.

Eigenvectors

Once we have the eigenvalues, the next step is to find the corresponding eigenvectors. Eigenvectors give direction to the system's behavior. We find them by plugging each eigenvalue back into the equation \((A - \lambda I) \mathbf{x} = \mathbf{0}\). For \( \lambda_1 = 2 + i \), the eigenvector \( \mathbf{v}_1 \) is \( \begin{pmatrix} 1+i \ 2 \end{pmatrix} \). Similarly, for \( \lambda_2 = 2 - i \), we find \( \mathbf{v}_2 = \begin{pmatrix} 1-i \ 2 \end{pmatrix} \). These vectors define the trajectory direction in the phase plane.
Understanding eigenvectors is essential as they reveal how perturbations evolve over time, showing both direction and scale of movements in our system.

Stability Analysis

Stability analysis helps determine the behavior of a system near a critical point, like \((0,0)\) in our exercise. We look at the real parts of eigenvalues:

• If both real parts are negative, the critical point is stable.
• If both are positive, it is unstable.
• Mixed signs indicate a saddle point.

Here, \( \lambda_1 = 2 + i \) and \( \lambda_2 = 2 - i \), both having positive real parts, suggest an unstable spiral.
This means any small disturbance will cause the system to spiral away from the critical point, increasing the distance over time. Such analysis is crucial for applications where system stability is a concern, such as control systems and mechanical structures.

Phase Plane Trajectories

Phase plane trajectories allow us to visualize the paths of the system in a two-dimensional space. They show how the system evolves over time. With eigenvalues \( \lambda_1 = 2 + i \) and \( \lambda_2 = 2 - i \), we expect spiral trajectories. These spirals indicate oscillatory but diverging behavior due to positivity in the real parts.
Drawing these trajectories helps in understanding system dynamics, providing a graphical representation of stability and behavior over time. For practical accuracy in sketching, software tools like MATLAB can plot these complex trajectories efficiently, confirming the theoretical expectations of system behavior.