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Chapter 7: Problem 12

(a) Find the eigenvalues of the given system. (b) Choose an initial point (other than the origin) and draw the corresponding trajectory in the \(x_{1} x_{2}\) plane. (c) For your trajectory in part (b) draw the graphs of \(x_{1}\) versus \(t\) and of \(x_{2}\) versus \(t .\) (d) For your trajectory in part (b) draw the corresponding graph in three- dimensional \(t x_{1} x_{2}\) space. $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{-\frac{4}{5}} & {2} \\ {-1} & {\frac{6}{5}}\end{array}\right) \mathbf{x} $$

Short Answer

Expert verified
Answer: To find the trajectory and graph the behaviors over time for a linear system of differential equations, follow these steps: 1) Find the eigenvalues of the matrix, 2) Choose an initial point, 3) Find the corresponding trajectory in the \(x_{1}x_{2}\) plane, 4) Graph \(x_{1}\) versus \(t\) and \(x_{2}\) versus \(t\), and 5) Graph the trajectory in three-dimensional \(t x_{1} x_{2}\) space.

Step by step solution

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01

Find the eigenvalues of the matrix

To find the eigenvalues, we need to first find the determinant of the matrix subtracted by the scalar, \(\lambda\): $$ \begin{vmatrix} {-\frac{4}{5}-\lambda} & {2}\\ {-1} & {\frac{6}{5}-\lambda} \end{vmatrix} $$ Now, expand this determinant and set it equal to zero: $$ (-\frac{4}{5}-\lambda)\left(\frac{6}{5}-\lambda\right)-(-1)(2)=0 $$ Solve for the eigenvalues, denoted by \(\lambda\).
02

Choose an initial point

Choose an initial point that is convenient and simple, for example: $$ x_{1}(0)=1, x_{2}(0)=1 $$
03

Find the corresponding trajectory in the \(x_{1}x_{2}\) plane

Once we find the eigenvalues, we can compute the eigenvectors. After that, we can represent the solution as a linear combination of the eigenvectors. With this representation, we can then find the trajectory using the initial point chosen in step 2. Be sure to sketch the trajectory in the \(x_{1} x_{2}\) plane.
04

Graph \(x_{1}\) versus \(t\) and \(x_{2}\) versus \(t\)

Using the solution found in step 3, plot the graphs of \(x_{1}(t)\) versus time \(t\) and \(x_{2}(t)\) versus time \(t\) on separate graphs.
05

Graph the trajectory in three-dimensional \(t x_{1} x_{2}\) space

Using the same solution found in step 3, plot the trajectory in the three-dimensional space represented by the \(t x_{1} x_{2}\) coordinates. This will give us a visual representation of how the system behaves over time in terms of the independent variables \(x_{1}\) and \(x_{2}\).

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Eigenvalues
Eigenvalues are special numbers associated with a square matrix. In the context of differential equations, they play a crucial role in determining the behavior of the system's solutions. - To find eigenvalues, we often set up and solve the characteristic equation. This equation derives from setting the determinant of the matrix subtracted by \(\lambda\) (a scalar) equal to zero. - The characteristic equation always takes the form of a polynomial, and solving it helps us find these important scalars. These scalars can reveal whether a system is stable, unstable, or oscillatory. For example, when dealing with the matrix from the problem, the calculated eigenvalues give us insight into the system's dynamics. They help predict whether solutions approach a steady state or move away indefinitely.
Eigenvectors
After finding eigenvalues, the next step is to determine eigenvectors. Eigenvectors correspond to each eigenvalue and help gain deeper insight into the system's orientation in space. - An eigenvector can be thought of as the direction in which a transformation (or matrix operation) stretches or compresses the plane. - To find an eigenvector, substitute each eigenvalue back into the equation \(A\mathbf{v} = \lambda\mathbf{v}\), where \mathbf{v}\ is the eigenvector and solve. Often, this computation involves matrix operations and leads to infinitely many solutions, represented in scalar multiples of the eigenvector. In our differential equations scenario, eigenvectors enable us to express solutions as linear combinations, thereby mapping trajectories respectful of orientation derived from the initial conditions of our system.
Linear Algebra
Linear algebra is the branch of mathematics that deals with vectors and matrices, providing the groundwork for analyzing systems of linear equations. - A key concept of linear algebra is the transformation of space, which directly affects the solutions of differential equations. - Through matrix manipulation, we can explore system solution paths and trajectories. The given matrix in the exercise is an example of how linear algebra underpins the solution process: from calculating determinants to finding eigenvalues and eigenvectors, it all relies on principles of linear algebra. Mastering these concepts ensures a solid foundation for understanding and predicting system dynamics effectively.
Phase Plane Analysis
Phase plane analysis is a powerful graphical method used to study differential equations. It provides a visual representation of trajectories of dynamical systems.- The phase plane is a two-dimensional plotting of variables, typically \(x_1\) vs. \(x_2\), representing the state of the system at any given time. - Each path on this plane, or trajectory, illustrates possible system transitions given initial conditions. By analyzing these paths, we infer a system's stability and predict future states.In the context of the exercise, after determining eigenvalues and eigenvectors, the system's behavior was visualized on the \(x_1 x_2\) plane. This analysis shows the path solutions take over time, helping to understand overall system dynamics, essential for predicting long-term behavior.

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Most popular questions from this chapter

Prove that \(\lambda=0\) is an eigenvalue of \(\mathbf{A}\) if and only if \(\mathbf{A}\) is singular.

The clectric circuit shown in Figure 7.9 .1 is described by the system of differential equations $$ \frac{d \mathbf{x}}{d t}=\left(\begin{array}{cc}{-\frac{1}{2}} & {-\frac{1}{8}} \\ {2} & {-\frac{1}{2}}\end{array}\right) \mathbf{x}+\left(\begin{array}{c}{\frac{1}{2}} \\ {0}\end{array}\right) I(t) $$ where \(x_{1}\) is the current through the inductor, \(x_{2}\) is the voltage drop across the capacitor, and \(I(t)\) is the current supplied by the external source. (a) Determine a fundamental matrix \(\Psi(t)\) for the homogeneous system corresponding to Eq. (i). Refer to Problem 25 of Section \(7.6 .\) (b) If \(I(t)=e^{-t / 2}\), determine the solution of the system (i) that also satisfies the initial conditions \(\mathbf{x}(0)=0\).

Find all eigenvalues and eigenvectors of the given matrix. $$ \left(\begin{array}{ccc}{1} & {0} & {0} \\ {2} & {1} & {-2} \\ {3} & {2} & {1}\end{array}\right) $$

Find the general solution of the given system of equations. $$ \mathbf{x}^{\prime}=\left(\begin{array}{lll}{3} & {2} & {4} \\ {2} & {0} & {2} \\ {4} & {2} & {3}\end{array}\right) \mathbf{x} $$

In this problem we indicate how to show that \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\), as given by Eqs. (9), are linearly independent. Let \(r_{1}=\lambda+i \mu\) and \(\bar{r}_{1}=\lambda-i \mu\) be a pair of conjugate eigenvalues of the coefficient matrix \(\mathbf{A}\) of \(\mathrm{Fq}(1)\); let \(\xi^{(1)}=\mathbf{a}+i \mathbf{b}\) and \(\bar{\xi}^{(1)}=\mathbf{a}-i \mathbf{b}\) be the corresponding eigenvectors. Recall that it was stated in Section 7.3 that if \(r_{1} \neq \bar{r}_{1},\) then \(\boldsymbol{\xi}^{(1)}\) and \(\bar{\xi}^{(1)}\) are linearly independent. (a) First we show that a and b are linearly independent. Consider the equation \(c_{1} \mathrm{a}+\) \(c_{2} \mathrm{b}=0 .\) Express a and \(\mathrm{b}\) in terms of \(\xi^{(1)}\) and \(\bar{\xi}^{(1)},\) and then show that \(\left(c_{1}-i c_{2}\right) \xi^{(1)}+\) \(\left(c_{1}+i c_{2}\right) \bar{\xi}^{(1)}=0\) (b) Show that \(c_{1}-i c_{2}=0\) and \(c_{1}+i c_{2}=0\) and then that \(c_{1}=0\) and \(c_{2}=0 .\) Consequently, a and b are linearly independent. (c) To show that \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\) are linearly independent consider the equation \(c_{1} \mathbf{u}\left(t_{0}\right)+\) \(c_{2} \mathbf{v}\left(t_{0}\right)=\mathbf{0}\), where \(t_{0}\) is an arbitrary point. Rewrite this equation in terms of a and \(\mathbf{b}\), and then proceed as in part (b) to show that \(c_{1}=0\) and \(c_{2}=0 .\) Hence \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\) are linearly independent at the arbitrary point \(t_{0}\). Therefore they are linearly independent at every point and on every interval.
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