Question

In each of Problems 24 through 27 the eigenvalues and eigenvectors of a matrix \(\mathrm{A}\) are given. Consider the corresponding system \(\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}\). $$ \begin{array}{l}{\text { (a) Sketch a phase portrait of the system. }} \\\ {\text { (b) Sketch the trajectory passing through the initial point }(2,3) \text { . }} \\ {\text { (c) For the trajectory in part (b) sketch the graphs of } x_{1} \text { versus } t \text { and of } x_{2} \text { versus } t \text { on the }} \\ {\text { same set of axes. }}\end{array} $$ $$ r_{1}=1, \quad \xi^{(1)}=\left(\begin{array}{r}{-1} \\ {2}\end{array}\right) ; \quad r_{2}=-2, \quad \xi^{(2)}=\left(\begin{array}{c}{1} \\\ {2}\end{array}\right) $$

Short answer

Question: Sketch the phase portrait, trajectory, and graphs of the given system with eigenvalues and eigenvectors and the initial point (2, 3). Answer: To sketch the phase portrait, trajectory, and graphs of the given system, follow these steps: 1. Find the general solution using eigenvalues and eigenvectors. 2. Find the specific solution using the initial conditions. 3. Use the specific solution to calculate the trajectory and sketch the phase portrait, trajectory, and graphs of \(x_1\) vs \(t\) and \(x_2\) vs \(t\). For this given system, the specific solution is: $$\mathbf{x}(t) = \frac{1}{3} e^{t} \begin{pmatrix}{-1} \\ {2} \end{pmatrix} + \frac{5}{3} e^{-2t} \begin{pmatrix}{1} \\ {2} \end{pmatrix}$$ The trajectory is: $$x_1(t)=-\frac{1}{3} e^t+\frac{5}{3} e^{-2t}$$ $$x_2(t)=\frac{2}{3} e^t+\frac{10}{3} e^{-2t}$$ Sketch the phase portrait, trajectory, and graphs of \(x_1\) vs \(t\) and \(x_2\) vs \(t\) based on the above solutions.

Step-by-step solution

Find the general solution using eigenvalues and eigenvectors

Since we have two distinct eigenvalues \(r_1\) and \(r_2\), and their respective eigenvectors \(\xi^{(1)}\) and \(\xi^{(2)}\), we can find the general solution of the given system as follows: $$\mathbf{x}(t)=c_{1} e^{r_{1} t} \xi^{(1)}+c_{2} e^{r_{2} t} \xi^{(2)}$$ Considering the given eigenvalues and eigenvectors, the general solution becomes: $$\mathbf{x}(t) = c_1 e^{t} \begin{pmatrix}{-1} \\ {2} \end{pmatrix} + c_2 e^{-2t} \begin{pmatrix}{1} \\ {2} \end{pmatrix}$$

Find the specific solution using the initial conditions

We are given the initial condition \((2,3)\), so we have \(x_1(0)=2\) and \(x_2(0)=3\). Now, let's substitute these initial conditions into the general solution to find the constants \(c_1\) and \(c_2\). $$\begin{cases} - c_1 + c_2 = 2 \\ 2 c_1 + 2 c_2 = 3 \end{cases}$$ Solving this system of equations, we get \(c_1 = \frac{1}{3}\) and \(c_2 = \frac{5}{3}\). So, the specific solution is: $$\mathbf{x}(t) = \frac{1}{3} e^{t} \begin{pmatrix}{-1} \\ {2} \end{pmatrix} + \frac{5}{3} e^{-2t} \begin{pmatrix}{1} \\ {2} \end{pmatrix}$$

Obtain the trajectory and create sketches

With the specific solution, calculate the trajectory: $$x_1(t)=-\frac{1}{3} e^t+\frac{5}{3} e^{-2t}$$ $$x_2(t)=\frac{2}{3} e^t+\frac{10}{3} e^{-2t}$$ Now, we can sketch the phase portrait, trajectory, and graphs of \(x_1\) vs \(t\) and \(x_2\) vs \(t\). This cannot be shown in text, but the steps to create the sketches are as follows: 1. Draw the eigenvector directions corresponding to the eigenvalues on the phase portrait. 2. Sketch the trajectory passing through the initial point (2, 3) by considering the behavior of the system along the eigenvectors and eigenvalues. 3. Plot the functions \(x_1(t)\) and \(x_2(t)\) on the same set of axes. Observing the behavior of these functions, we can notice that \(x_1(t)\) goes to infinity as \(t\) goes to infinity, while \(x_2(t)\) converges to 0 as \(t\) goes to infinity. By following these steps, you will have the desired phase portrait, trajectory, and required graphs.

Key concepts

Phase Portrait

A phase portrait is a graphical representation that shows all the possible trajectories of a dynamic system in the phase plane. Each trajectory corresponds to a set of initial conditions, forming a distinct curve that represents the evolution of the system over time.

To sketch a phase portrait for a given system of differential equations, first identify the eigenvalues and eigenvectors. These will provide you with critical information on the behavior of trajectories near the equilibrium points. In our exercise, the eigenvalues are 1 and -2, which suggests the presence of a source (an unstable point where trajectories move away from) and a sink (a stable point where trajectories are attracted to), respectively.

The portraits consist of various trajectories showing how the state of a system (given by the vector \(\mathbf{x}\)) changes over time. In systems with two variables, these trajectories are curves in the two-dimensional plane. Trajectories do not intersect in a phase portrait, except possibly at equilibrium points, where trajectories might start or end.

Differential Equations

Differential equations are equations that relate a function with one or more of its derivatives. In the context of our exercise, the differential equation \(\mathbf{x}'=\mathbf{A}\mathbf{x}\) describes how the vector \(\mathbf{x}\) changes over time. The solution to a system of differential equations yields the trajectory of the system which plots all possible states (or positions) \(\mathbf{x}\) as time \(t\) progresses.

The role of the eigenvalues and their associated eigenvectors cannot be overstated. They determine the fundamental behavior of the system. Positive eigenvalues indicate growth in the direction of the eigenvector whereas negative eigenvalues show decay. When eigenvalues are real and distinct, as they are in this exercise, the system exhibits clear exponential growth or decay along the directions defined by the eigenvectors.

Trajectory Sketching

Trajectory sketching involves plotting the path that a system follows in the phase plane as it progresses over time from a specific initial condition. To sketch the trajectory for our system of differential equations, you must use the specific solution obtained by applying the initial conditions to the general solution of the system.

Starting from the initial point (2, 3), the trajectory will follow the combination of behaviors prescribed by the eigenvalues and eigenvectors. In this case, we'll sketch the trajectory using the direction fields provided by the eigenvectors and watch the solution curve as it moves following the tendency towards growth given by the positive eigenvalue for \(x_1\) and decay given by the negative eigenvalue for \(x_2\).

Typically, the trajectory will be inclined towards the eigenvector associated with the dominant (largest absolute value) eigenvalue, which can be seen by the coefficients that scale the exponential terms in the solution vector.

Initial Value Problem

An initial value problem in the context of differential equations is a problem where the solution to the differential equation is required to satisfy a specific starting condition, referred to as the 'initial value'. In our exercise, the initial condition is the point (2, 3), which is the state of the system at time \(t=0\).

The initial condition allows us to solve for unique scalar coefficients (in our case, \(c_1\) and \(c_2\)) that make the general solution a specific solution tailored to the given starting point of the system. Once the initial value problem is solved and the specific solution is found, it shows us not just how the system moves, but exactly where it will be at any point in time starting from \(t=0\) onwards. This particular solution is vital for accurately sketching the trajectory and understanding the system’s behavior over time.

By solving initial value problems, we gain a deep insight into the specifics of how dynamic systems evolve, and this is crucial for applications ranging from physics and engineering to economics and biology.