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

Short answer

Based on the given system of two linear differential equations, we found the eigenvalues and eigenvectors, classified the stability of the critical point (0,0) as unstable, and sketched the trajectories in the phase plane. The eigenvalues are λ1=2 and λ2=-1, and the eigenvectors are v1= (2,1) and v2= (1,2). The critical point (0,0) is a saddle point, and the trajectories are stretched along the eigenvector corresponding to the positive eigenvalue, and compressed along the eigenvector corresponding to the negative eigenvalue.

Step-by-step solution

Finding eigenvalues of the matrix

To find the eigenvalues, we need to find the roots of the characteristic equation given by the determinant: \(\det(A - \lambda I) = 0\), where \(A = \begin{pmatrix} 3 & -2 \\ 2 & -2 \end{pmatrix}\) and \(I\) is the identity matrix. The characteristic equation is : \[ \begin{vmatrix} 3-\lambda & -2 \\ 2 & -2-\lambda \end{vmatrix} = (3-\lambda)(-2-\lambda) - (-2)(2) = \lambda^2-\lambda-2 \] Now, we find the roots (eigenvalues) of the characteristic equation: \[ \lambda^2-\lambda-2 = 0 \]

Finding eigenvectors of the matrix

The eigenvalues of the characteristic equation are \(\lambda_1=2\) and \(\lambda_2=-1\). Now, we find the eigenvectors corresponding to each eigenvalue by solving the system \((A-\lambda I)v=0\). For \(\lambda_1=2\) : \[ (A - \lambda_1 I)v_1 = \begin{pmatrix} 1 & -2 \\ 2 & -4 \end{pmatrix} \begin{pmatrix} v_{11} \\ v_{12} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \] We can see that the second row is a scalar multiple of the first row, so we only need to focus on one row: \[ v_{11} - 2 v_{12} = 0 \] We can choose \(v_{12} = 1\), which gives us \(v_{11} = 2\). So, \(v_1=\begin{pmatrix} 2 \\ 1 \end{pmatrix}\). For \(\lambda_2=-1\) : \[ (A - \lambda_2 I)v_2 = \begin{pmatrix} 4 & -2 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} v_{21} \\ v_{22} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \] Again, the second row is a scalar multiple of the first row, so we only need to focus on one row: \[ 4 v_{21} - 2 v_{22} = 0 \] We can simplify this to \(2v_{21} - v_{22} = 0\). Let \(v_{21}=1\). Then \(v_{22}=2\). So, \(v_2=\begin{pmatrix} 1 \\ 2 \end{pmatrix}\).

Classify the critical point and determine stability

The stability of the critical point \((0,0)\) can be determined using the eigenvalues found earlier. We have \(\lambda_1=2\) and \(\lambda_2=-1\). Since one eigenvalue is positive and the other is negative, the critical point \((0,0)\) is a saddle point and is unstable.

Sketching trajectories in the phase plane

To sketch the trajectories, we need to find the general solution of the system and simplify the expression. The general solution is given by: \[ \mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 \] Substituting the eigenvalues and eigenvectors, we get: \[ \mathbf{x}(t) = c_1 e^{2t} \begin{pmatrix} 2 \\ 1 \end{pmatrix} + c_2 e^{-t} \begin{pmatrix} 1 \\ 2 \end{pmatrix} \] Since the system is unstable, the trajectories will move away from the origin. For the given system, the trajectories will be stretched along the eigenvector corresponding to the positive eigenvalue (\(\begin{pmatrix} 2 \\ 1 \end{pmatrix}\)), and compressed along the eigenvector corresponding to the negative eigenvalue (\(\begin{pmatrix} 1 \\ 2 \end{pmatrix}\)).

Key concepts

Characteristic Equation

The characteristic equation is a crucial tool in solving systems of differential equations. It starts with finding the "eigenvalues" of a given matrix. In our scenario, the matrix is \( A = \begin{pmatrix} 3 & -2 \ 2 & -2 \end{pmatrix} \),\ which describes the system's dynamics. We use the formula \( \det(A - \lambda I) = 0 \) to derive the characteristic equation. Here, \( \lambda \) represents the eigenvalues, and \( I \) is the identity matrix. This gives us a quadratic equation:

• \( (3-\lambda)(-2-\lambda) - (-2)(2) = \lambda^2 - \lambda - 2 \)

The roots of this polynomial equation are the sought eigenvalues. Solving \( \lambda^2 - \lambda - 2 = 0 \), you find \( \lambda_1 = 2 \) and \( \lambda_2 = -1 \). These values tell us much about the system, including its stability and phase behavior.

Stability Analysis

Understanding stability requires scrutinizing the eigenvalues obtained from the characteristic equation. Stability analysis determines how solutions to a system of differential equations behave as time passes. A critical point, such as \((0,0)\) in our problem, is classified based on its eigenvalues:

• If all eigenvalues have negative real parts, the critical point is stable and eventually every trajectory leads to it.
• If any eigenvalue has a positive real part, the point is unstable, meaning trajectories will drift away.

In our case, \( \lambda_1 = 2 \) is positive, and \( \lambda_2 = -1 \) is negative, indicating a saddle point. This means that the system is unstable. Trajectories will diverge away from \((0,0)\), signaling instability in the equilibrium, and thus, the system does not tend to return to this point once perturbed.

Phase Plane Trajectories

Phase plane trajectories provide a visual representation of the system's behavior over time. We derive them from solving the system’s differential equation using its general solution:

• \( \mathbf{x}(t) = c_1 e^{2t} \begin{pmatrix} 2 \ 1 \end{pmatrix} + c_2 e^{-t} \begin{pmatrix} 1 \ 2 \end{pmatrix} \)

This expression combines the eigenvectors \( \begin{pmatrix} 2 \ 1 \end{pmatrix} \) and \( \begin{pmatrix} 1 \ 2 \end{pmatrix} \), along with the exponential growth or decay modulated by the eigenvalues \( \lambda_1 \) and \( \lambda_2 \).
The phase portrait sketches the trajectories based on these. You’ll notice that trajectories stretch along the direction of the eigenvector corresponding to \( \lambda_1 = 2 \) and compress along \( \lambda_2 = -1 \).
For the given unstable system, expect the paths to move away from the origin. The graphical representation of these trajectories showcases the directional tendencies of the system’s solutions, graphically depicting stability and instability regions.