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Chapter 7: Problem 7

In each of Problems 7 and 8 find the general solution of the given system of equations. Also draw a direction field and a few of the trajectories. In each of these problems the coefficient matrix has a zero eigenvalue. As a result, the pattern of trajectories is different from those in the examples in the text. $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{4} & {-3} \\ {8} & {-6}\end{array}\right) \mathbf{x} $$

Short Answer

Expert verified
In summary, for the given linear system of differential equations, we found that the eigenvalues are both 0, and there is only one linearly independent eigenvector. Therefore, the general solution has the form: $$ \mathbf{x}(t) = c_1\begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_2 t \begin{pmatrix} 1 \\ 2 \end{pmatrix}, $$ where \(c_1\) and \(c_2\) are constants. When plotting the trajectories, we observed that the system has a line of equilibria and the trajectories are parallel to this line. This pattern is different from the typical spiral pattern seen in textbooks for linear systems of differential equations.

Step by step solution

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01

Eigenvalues and Eigenvectors

To determine the eigenvalues, we solve the equation \(\det(A-\lambda I) = 0\) where A is the given coefficient matrix and I is the identity matrix: $$ \det\left(\begin{array}{cc} 4-\lambda & -3 \\ 8 & -6-\lambda \end{array}\right) = (4-\lambda)(-6-\lambda) - (-3)(8) = 0 $$ After solving this equation, we'll have the eigenvalues. Then, for each eigenvalue, plug them into the equation \((A-\lambda I)\mathbf{v} = \mathbf{0}\) to solve for the corresponding eigenvectors \(\mathbf{v}\).
02

General Solution

Once we've found the eigenvalues and eigenvectors, the general solution for a linear system of the form \(\mathbf{x}^{\prime} = A\mathbf{x}\) can be written as: $$ \mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v_1} + c_2 e^{\lambda_2 t} \mathbf{v_2} $$ where \(\lambda_1\) and \(\lambda_2\) are eigenvalues, \(\mathbf{v_1}\) and \(\mathbf{v_2}\) are the corresponding eigenvectors, and \(c_1\) and \(c_2\) are constants.
03

Trajectories

Plot a direction field for the given system, and then draw a few trajectories. Notice the pattern of these trajectories and how they are different from the typical patterns in the textbook, as a result of having a zero eigenvalue. Discuss the differences briefly. Solution:
04

Eigenvalues and Eigenvectors

Let us calculate the determinant of \((A-\lambda I)\), and we obtain: $$ (4-\lambda)(-6-\lambda) - (-3)(8) = \lambda^2 - 0\lambda = 0 $$ The eigenvalues are \(\lambda_1 = 0\) and \(\lambda_2 = 0\). Find eigenvectors for \(\lambda_1 = 0\) in \((A-\lambda_1 I)\mathbf{v} = \mathbf{0}\): $$ \left(\begin{array}{cc} 4 & -3 \\ 8 & -6 \end{array}\right) \begin{pmatrix} v_{11} \\ v_{12} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$ We can obtain \(\mathbf{v_1} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}\). Since there is only one linearly independent eigenvector, we can obtain a generalized eigenvector.
05

General Solution

Our general solution is given by: $$ \mathbf{x}(t) = c_1 e^{0 \cdot t}\begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_2 t e^{0 \cdot t}\begin{pmatrix} 1 \\ 2 \end{pmatrix} = c_1\begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_2 t \begin{pmatrix} 1 \\ 2 \end{pmatrix}, $$ where \(c_1\) and \(c_2\) are constants.
06

Trajectories

When plotting the direction field and some trajectories for this system, the patterns of trajectories are less standard. The system has a line of equilibria along the line \(x=cy\), where \(c\) is any constant. The trajectories are parallel to this line and don't either spiral in or out, which is different from the regular trajectory patterns we see in textbooks.

Key Concepts

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

Eigenvalues
A fundamental concept in linear algebra and differential equations is the idea of eigenvalues. These are special numbers that give us essential insights into the behavior of a system of equations. To find eigenvalues, we solve the characteristic equation, given by \( \det(A - \lambda I) = 0 \), where \( A \) is a matrix, and \( \lambda \) is the eigenvalue. Typically, this involves setting up a determinant based on the matrix and solving for \( \lambda \).

In the example provided, this process revealed a zero eigenvalue (\( \lambda_1 = 0 \) and \( \lambda_2 = 0 \)), which leads to interesting dynamics in the system, making the trajectories different from more common cases. Zero eigenvalues can indicate a line of equilibria, highlighting the importance of understanding the roots of your system.
Eigenvectors
Once eigenvalues are determined, the next step is finding eigenvectors, which provide a direction in the vector space. Eigenvectors associated with each eigenvalue solve the equation \( (A - \lambda I) \mathbf{v} = \mathbf{0} \). These vectors retain their direction upon the transformation described by the matrix \( A \).

In the solution, after identifying the eigenvalue as zero, we calculated the eigenvectors and obtained \( \mathbf{v_1} = \begin{pmatrix} 1 \ 2 \end{pmatrix} \). With only one linearly independent eigenvector, we used a generalized eigenvector to completely describe the system. This scenario is common in systems where multiple eigenvalues are identical, requiring both regular and generalized eigenvectors.
Direction Field
A directional field gives a visual representation of a differential equation system. It helps illustrate the tendencies or directions that trajectories will follow across the plane. By plotting small arrows on a grid, we can see how solutions evolve over time.

For our system, with constraints from zero eigenvalues, the direction field showed parallel lines that indicate a line of equilibria. Instead of spiraling or behaving chaotically, the trajectories align in a more orderly pattern. Understanding these directions can reveal crucial insights about the stability and long-term behavior of the system.
Trajectory Patterns
In differential equations, trajectory patterns describe how solutions behave over the long term. Regular patterns include spirals, circles, or straight lines, typically influenced by the nature of the eigenvalues.

However, in the case with zero eigenvalues, our system displayed a unique pattern with a line of equilibria—trajectories that didn't spiral in or out but remained parallel. This pattern stands out because it demonstrates stability and indicates that solutions don't diverge widely with small perturbations. Understanding these behavior patterns helps predict how systems might behave under different conditions.

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Most popular questions from this chapter

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

Consider the system $$ \mathbf{x}^{\prime}=\mathbf{A x}=\left(\begin{array}{rrr}{5} & {-3} & {-2} \\\ {8} & {-5} & {-2} \\ {-4} & {-5} & {-4} \\ {-4} & {3} & {3}\end{array}\right) \mathbf{x} $$ (a) Show that \(r=1\) is a triple eigenvalue of the coefficient matrix \(\mathbf{A},\) and that there are only two linearly independent eigenvectors, which we may take as $$ \xi^{(1)}=\left(\begin{array}{l}{1} \\ {0} \\ {2}\end{array}\right), \quad \xi^{(2)}=\left(\begin{array}{r}{0} \\ {2} \\ {-3}\end{array}\right) $$ Find two linearly independent solutions \(\mathbf{x}^{(1)}(t)\) and \(\mathbf{x}^{(2)}(t)\) of Eq. (i). (b) To find a third solution assume that \(\mathbf{x}=\xi t e^{t}+\mathbf{\eta} e^{\lambda} ;\) thow that \(\xi\) and \(\eta\) must satisfy $$ (\mathbf{A}-\mathbf{1}) \xi=0 $$ \((\mathbf{A}-\mathbf{I}) \mathbf{\eta}=\mathbf{\xi}\) (c) Show that \(\xi=c_{1} \xi^{(1)}+c_{2} \mathbf{\xi}^{(2)},\) where \(c_{1}\) and \(c_{2}\) are arbitrary constants, is the most general solution of Eq. (iii). Show that in order to solve Eq. (iv) it is necessary that \(c_{1}=c_{2}\) (d) It is convenient to choose \(c_{1}=c_{2}=2 .\) For this choice show that $$ \xi=\left(\begin{array}{r}{2} \\ {4} \\ {-2}\end{array}\right), \quad \mathbf{\eta}=\left(\begin{array}{r}{0} \\ {0} \\ {-1}\end{array}\right) $$ where we have dropped the multiples of \(\xi^{(1)}\) and \(\xi^{(2)}\) that appear in \(\eta\). Use the results given in Eqs. (v) to find a third linearly independent solution \(\mathbf{x}^{(3)}\) of Eq. (i). (e) Write down a fundamental matrix \(\Psi(t)\) for the system (i). (f) Form a matrix T with the cigenvector \(\xi^{(1)}\) in the first column and with the eigenvector \(\xi\) and the generalized eigenvector \(\eta\) from Eqs. (v) in the other two columns. Find \(\mathbf{T}^{-1}\) and form the product \(\mathbf{J}=\mathbf{T}^{-1} \mathbf{A} \mathbf{T}\). The matrix \(\mathbf{J}\) is the Jordan form of \(\mathbf{A} .\)

Let $$ \mathbf{J}=\left(\begin{array}{ccc}{\lambda} & {1} & {0} \\ {0} & {\lambda} & {1} \\ {0} & {0} & {\lambda}\end{array}\right) $$ where \(\lambda\) is an arbitrary real number. (a) Find \(\mathbf{J}^{2}, \mathbf{J}^{3},\) and \(\mathbf{J}^{4}\). (b) Use an inductive argument to show that $$ \mathbf{J}^{n}=\left(\begin{array}{ccc}{\lambda^{n}} & {n \lambda^{n-1}} & {[n(n-1) / 2] \lambda^{n-2}} \\ {0} & {\lambda^{n}} & {n \lambda^{n-1}} \\\ {0} & {0} & {\lambda^{n}}\end{array}\right) $$ (c) Determine exp(Jt). (d) Observe that if you choose \(\lambda=2\), then the matrix \(\mathbf{J}\) in this problem is the same as the matrix \(\mathbf{J}\) in Problem \(17(f)\). Using the matrix T from Problem \(17(f),\) form the product Texp(Jt) with \(\lambda=2\). Observe that the resulting matrix is the same as the fundamental matrix \(\Psi(t)\) in Problem \(17(e) .\)

The coefficient matrix contains a parameter \(\alpha\). In each of these problems: (a) Determine the eigervalues in terms of \(\alpha\). (b) Find the critical value or values of \(\alpha\) where the qualitative nature of the phase portrait for the system changes. (c) Draw a phase portrait for a value of \(\alpha\) slightly below, and for another value slightly above, each crititical value. $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{\alpha} & {10} \\ {-1} & {-4}\end{array}\right) \mathbf{x} $$

In each of Problems 23 and 24 ; (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. Also draw the trajectories in the \(x_{1} x_{1}-\) and \(x_{2} x_{3}-\) planes. (c) For the initial point in part (b) draw the corresponding trajectory in \(x_{1} x_{2} x_{3}\) -space. $$ \mathbf{x}^{\prime}=\left(\begin{array}{ccc}{-\frac{1}{4}} & {1} & {0} \\\ {-1} & {-\frac{1}{4}} & {0} \\ {0} & {0} & {-\frac{1}{4}}\end{array}\right) \mathbf{x} $$
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