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Chapter 10: Problem 52

Solve each of the following linear systems. (a) \(\mathbf{X}^{\prime}=\left(\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right) \mathbf{X}\) (b) \(\mathbf{X}^{\prime}=\left(\begin{array}{rr}1 & 1 \\ -1 & -1\end{array}\right) \mathbf{X}\) Find a phase portrait of each system. What is the geometric significance of the line \(y=-x\) in each portrait?

Short Answer

Expert verified
(a) Node along y = x and y = -x. (b) Shear along y = -x. Line y = -x is an invariant line in each.

Step by step solution

01

Understand the System

For a system of the form \(\mathbf{X}^{\prime}=A\mathbf{X}\), we are given two matrices, \(A\), one for each part (a and b). We need to analyze these matrices to find solutions and phase portraits.
02

Step 1a: Analyze Matrix (a)

For matrix (a), \(A=\left(\begin{array}{cc}1 & 1 \ 1 & 1\end{array}\right)\). The characteristic equation is derived from \(\det(A - \lambda I) = 0\). Calculate \(\det\begin{pmatrix}1 - \lambda & 1 \ 1 & 1 - \lambda\end{pmatrix} = (1-\lambda)^2 - 1\). Set this equal to zero to find eigenvalues.
03

Step 2a: Compute Eigenvalues of (a)

Solving \((1-\lambda)^2 - 1 = 0\) gives \(\lambda^2 - 2\lambda = 0\), resulting in \(\lambda_1 = 2\), \(\lambda_2 = 0\).
04

Step 3a: Find Eigenvectors for (a)

For \(\lambda_1 = 2\), solve \((A - 2I)\mathbf{v} = 0\). For \(\lambda_2 = 0\), solve \(A\mathbf{v} = 0\). These give eigenvectors for \(\lambda_1\) as \((1, 1)\) and for \(\lambda_2\) as \((1, -1)\).
05

Step 4a: Sketch Phase Portrait for (a)

The eigenvectors provide direction fields. \( (1, 1)\) indicates a line along \(y = x\), and \((1, -1)\) the line \(y = -x\). The system behaves as a node, with trajectories moving away from the origin along these directions.
06

Step 1b: Analyze Matrix (b)

For matrix (b), \(A=\left(\begin{array}{cc}1 & 1 \ -1 & -1\end{array}\right)\). Again, find the characteristic equation \(\det(A - \lambda I) = 0\). Calculate \(\det\begin{pmatrix}1 - \lambda & 1 \ -1 & -1 - \lambda\end{pmatrix} = (1-\lambda)(-1-\lambda) + 1\). Set this equal to zero.
07

Step 2b: Compute Eigenvalues of (b)

Solve \((1-\lambda)(-1-\lambda) + 1 = 0\); it results in \(\lambda^2 = 0\), giving a repeated eigenvalue \(\lambda = 0\).
08

Step 3b: Find Eigenvectors for (b)

Solve \(A\mathbf{v} = 0\) for eigenvectors with \(\lambda = 0\). The eigenvector is \((1, -1)\). Find a generalized eigenvector by solving \((A - 0I)\mathbf{v}=\mathbf{v}_1\), which typically gives another solution that differentiates trajectories.
09

Step 4b: Sketch Phase Portrait for (b)

With a repeated eigenvalue and one eigenvector \((1, -1)\), trajectories lie along \(y = -x\), indicating a shear with no real motion radiating from the origin. It represents a line of constant solutions.
10

Geometric Significance of y=-x

For both systems, the line \(y = -x\) is significant as it represents a direction of specific eigenvectors. In system (a), solutions diverge along this line, while in (b), it is the primary direction of shear. The line serves as a key trajectory or invariant direction.

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Key Concepts

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

Eigenvalues and Eigenvectors
In the study of linear systems of differential equations, eigenvalues and eigenvectors play a crucial role in finding solutions.
  • Eigenvalues (\( \lambda \)) are special numbers derived from a matrix equation known as the characteristic equation. They represent the scale factor for eigenvectors during the system's transformation.
  • Eigenvectors are non-zero vectors that change only in magnitude, not direction, when the matrix is applied. They define critical directions in the system's phase space.
For each given system, we first determine eigenvalues by solving the characteristic equation: \[\det(A - \lambda I) = 0\]. Upon finding eigenvalues, we substitute them back into the transformed equation \((A - \lambda I)\mathbf{v} = 0\) to solve for the corresponding eigenvectors. The eigenvectors provide insight into the trajectory and behavior of solutions over time, as they reveal the system's inherent structure.
Phase Portraits
Phase portraits are graphical representations that illustrate the trajectories of a system of differential equations in the phase space. They visualize how solutions evolve over time.
  • A typical phase portrait for a 2D linear system involves plotting trajectories of solutions as curves, which emanate from or converge to critical points, such as equilibrium points.
  • For example, in system (a), the phase portrait shows trajectories moving away from the origin, known as a node-like behavior, guided by eigenvectors \((1, 1)\) and \((1, -1)\).
  • In system (b), trajectories typically create a shear pattern due to repeated eigenvalues, mainly representing motion along the line \(y = -x\).
Phase portraits thus serve as a crucial tool in understanding the qualitative behavior of dynamic systems and predicting long-term outcomes.
Matrix Analysis
Matrix analysis is fundamental in solving linear systems of equations. It involves manipulating matrices to derive solutions or understand system dynamics.
  • The given problem represents linear systems as \(\mathbf{X}^{\prime} = A\mathbf{X}\), transforming the vector \(\mathbf{X}\) by multiplying it with matrix \(A\).
  • Understanding the properties of matrix \(A\), such as its determinant, eigenvalues, and eigenvectors, allows us to dissect the system's behavior.
  • Calculating the characteristic equation through the determinant helps us find eigenvalues, which are instrumental in predicting system dynamics.
Through matrix analysis, we systematically reduce complex systems to simpler forms that unveil the nature of solutions. This simplifies the understanding of how changes in components affect outcomes.
Characteristic Equation
The characteristic equation is a foundational step in analyzing matrices within linear systems. It is derived from the equation \[\det(A - \lambda I) = 0\], where \(A\) is the system matrix and \(\lambda\) is a scalar.
  • This equation allows us to find the eigenvalues of the matrix, which indicate specific scaling factors during transformations.
  • The roots of the characteristic equation correspond to the eigenvalues that help define the behavior of solutions, such as stability and oscillations.
  • In the provided exercises, solving these equations led to identifying important eigenvalues: \(\lambda = 2, 0\) (for system a) and \(\lambda = 0\) (for system b), which pointed out the system's inherent dynamics.
Understanding the characteristic equation is crucial for delving deeper into eigenvalues, which ultimately guides the interpretation and visualization of the system's phase portrait.

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Most popular questions from this chapter

Find the general solution of the given system. $$ \mathbf{X}^{\prime}=\left(\begin{array}{rrr} 1 & -1 & 2 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right) \mathbf{x} $$

Use variation of parameters to solve the given system. \(\mathbf{X}^{\prime}=\left(\begin{array}{lll}1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 3\end{array}\right) \mathbf{X}+\left(\begin{array}{c}e^{t} \\ e^{2 t} \\ t e^{3 t}\end{array}\right)\)

Verify that the vector \(\mathbf{X}_{p}\) is a particular solution of the given system. $$ \mathbf{X}^{\prime}=\left(\begin{array}{rr} 2 & 1 \\ 1 & -1 \end{array}\right) \mathbf{X}+\left(\begin{array}{r} -5 \\ 2 \end{array}\right) ; \quad \mathbf{X}_{p}=\left(\begin{array}{l} 1 \\ 3 \end{array}\right) $$

In Problems 9 and 10 , solve the given initial-value problem. \(\mathbf{X}^{\prime}=\left(\begin{array}{rr}-1 & -2 \\ 3 & 4\end{array}\right) \mathbf{X}+\left(\begin{array}{l}3 \\\ 3\end{array}\right), \quad \mathbf{X}(0)=\left(\begin{array}{r}-4 \\\ 5\end{array}\right)\)

Verify that the vector \(\mathbf{X}\) is a solution of the given system. $$ \begin{aligned} &\frac{d x}{d t}=-2 x+5 y \\ &\frac{d y}{d t}=-2 x+4 y ; \quad \mathbf{X}=\left(\begin{array}{c} 5 \cos t \\ 3 \cos t-\sin t \end{array}\right) e^{t} \end{aligned} $$
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