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Chapter 4: Problem 79

Consider the dynamical system \(\mathbf{x}_{k+1}=A \mathbf{x}_{k}\) (a) Compute and plot \(\mathbf{x}_{0}, \mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\) for \(\mathbf{x}_{0}=\left[\begin{array}{l}1 \\\ 1\end{array}\right]\) (b) Compute and plot \(\mathbf{x}_{0}, \mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\) for \(\mathbf{x}_{0}=\left[\begin{array}{l}1 \\\ 0\end{array}\right]\) (c) Using eigenvalues and eigenvectors, classify the origin as an attractor, repeller, saddle point, or none of these. (d) Sketch several typical trajectories of the system. $$A=\left[\begin{array}{rr} 2 & -1 \\ -1 & 2 \end{array}\right]$$

Short Answer

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(a) - (c) Reasoning鈥擳he origin is a repeller as eigenvalues (>1) show system diverges.

Step by step solution

01

Compute Successive Vectors for (a)

Start by computing \(\mathbf{x}_{1}\) using the matrix \(A\) and initial vector \(\mathbf{x}_{0} = \begin{bmatrix} 1 \ 1 \end{bmatrix}\): \[ \mathbf{x}_{1} = A \mathbf{x}_{0} = \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix} \begin{bmatrix} 1 \ 1 \end{bmatrix} = \begin{bmatrix} 1 \ 1 \end{bmatrix} \]Next, compute \(\mathbf{x}_{2}\): \[ \mathbf{x}_{2} = A \mathbf{x}_{1} = \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix} \begin{bmatrix} 1 \ 1 \end{bmatrix} = \begin{bmatrix} 1 \ 1 \end{bmatrix} \]Finally, compute \(\mathbf{x}_{3}\): \[ \mathbf{x}_{3} = A \mathbf{x}_{2} = \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix} \begin{bmatrix} 1 \ 1 \end{bmatrix} = \begin{bmatrix} 1 \ 1 \end{bmatrix} \]The vectors \(\mathbf{x}_{0}, \mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\) are all \(\begin{bmatrix} 1 \ 1 \end{bmatrix}\).
02

Compute Successive Vectors for (b)

Using vector \(\mathbf{x}_{0} = \begin{bmatrix} 1 \ 0 \end{bmatrix}\), calculate \(\mathbf{x}_{1}\): \[ \mathbf{x}_{1} = A \mathbf{x}_{0} = \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix} \begin{bmatrix} 1 \ 0 \end{bmatrix} = \begin{bmatrix} 2 \ -1 \end{bmatrix} \]Next, calculate \(\mathbf{x}_{2}\): \[ \mathbf{x}_{2} = A \mathbf{x}_{1} = \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 \ -1 \end{bmatrix} = \begin{bmatrix} 5 \ -4 \end{bmatrix} \]Finally, calculate \(\mathbf{x}_{3}\): \[ \mathbf{x}_{3} = A \mathbf{x}_{2} = \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix} \begin{bmatrix} 5 \ -4 \end{bmatrix} = \begin{bmatrix} 14 \ -13 \end{bmatrix} \]
03

Analyze Eigenvalues and Eigenvectors

Calculate the eigenvalues \(\lambda\) of matrix \(A\): Find the characteristic polynomial:\[\det(A - \lambda I) = \det \begin{bmatrix} 2-\lambda & -1 \ -1 & 2-\lambda \end{bmatrix} = (2-\lambda)^2 - 1 = 0\]Solve the quadratic equation:\[(\lambda - 3)(\lambda - 1) = 0\]Eigenvalues are \(\lambda = 3\) and \(\lambda = 1\).For \(\lambda = 3\), the eigenvector is found by solving:\[\begin{bmatrix} -1 & -1 \ -1 & -1 \end{bmatrix} \mathbf{v} = 0 \Rightarrow \mathbf{v} = c \begin{bmatrix} 1 \ 1 \end{bmatrix}\]For \(\lambda = 1\), the eigenvector is:\[\begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix} \mathbf{v} = 0 \Rightarrow \mathbf{v} = c \begin{bmatrix} 1 \ 1 \end{bmatrix}\]
04

Classify the Origin Using Eigenvalues

The eigenvalues are 3 and 1, both positive and greater than 1. This indicates that all trajectories move away from the origin, classifying it as a repeller.
05

Sketch Trajectories

To sketch trajectories, note that:- Initial points along direction \(\begin{bmatrix} 1 \ 1 \end{bmatrix}\) do not change, leading to straight and constant lines.- Initial points like \(\begin{bmatrix} 1 \ 0 \end{bmatrix}\) grow outwardly, reflecting linear scaling through trajectory maps paralleling eigenvectors.

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Key Concepts

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

Eigenvalues
Eigenvalues in dynamical systems help us understand the system's behavior over time. When you analyze a matrix, like the one given in your dynamical system, the eigenvalues tell you how trajectories evolve.
For your system with matrix \(A = \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix}\), the eigenvalues are found by solving the characteristic equation:\[\det(A - \lambda I) = (2-\lambda)^2 - 1 = 0\]
The solutions, or roots, of this equation are the eigenvalues. Here, they are \(\lambda = 3\) and \(\lambda = 1\).
  • If an eigenvalue is greater than one, it suggests trajectories moving away, indicating growth or repulsion.
  • If less than one, it implies shrinking, pointing towards attraction.
  • When it is exactly one, the size of the system remains constant.
Both eigenvalues here are greater than one, meaning trajectories move outward, classifying the system's origin as a repeller.
Eigenvectors
Eigenvectors are essential companions to eigenvalues in describing a matrix's transformation characteristics. For matrix \(A\), after finding each eigenvalue, eigenvectors are determined by plugging these eigenvalues into \(A - \lambda I\) and solving the resulting equations.
For \(\lambda = 3\), the associated eigenvector, \(\mathbf{v}\), can be found by solving:\[\begin{bmatrix} -1 & -1 \ -1 & -1 \end{bmatrix} \mathbf{v} = 0\]
This simplifies to \(\mathbf{v} = c \begin{bmatrix} 1 \ 1 \end{bmatrix}\). Similarly, for \(\lambda = 1\), you solve:\[\begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix} \mathbf{v} = 0\]
Leading to the same eigenvector \(\mathbf{v} = c \begin{bmatrix} 1 \ 1 \end{bmatrix}\).
  • Eigenvectors outline directions, in which the matrix transformation acts as simple scaling by the eigenvalue.
  • They provide a kind of roadmap, illustrating the "stable" or "natural" directions in the space influenced by the matrix.
Trajectories
In dynamical systems, trajectories depict the path followed as the system evolves over iterations or time. By observing trajectories, you can visualize how initial vectors change when multiplied by the transformation matrix.
With your dynamical system, you calculated trajectories for initial vectors such as \(\mathbf{x}_0 = \begin{bmatrix} 1 \ 1 \end{bmatrix}\) and \(\mathbf{x}_0 = \begin{bmatrix} 1 \ 0 \end{bmatrix}\).
  • The first vector \(\begin{bmatrix} 1 \ 1 \end{bmatrix}\) remained constant over successive steps. This indicates a stable trajectory.
  • Meanwhile, starting from \(\begin{bmatrix} 1 \ 0 \end{bmatrix}\), the vectors grew, leading to \(\mathbf{x}_3 = \begin{bmatrix} 14 \ -13 \end{bmatrix}\), showing outward movement in space.
Trajectories are critical for understanding both the speed and direction of movement within the system.
Matrix Multiplication
Matrix multiplication plays a central role in analyzing dynamical systems. Given a transformation matrix \(A\) and an initial vector \(\mathbf{x}_0\), multiplying them reveals the next state \(\mathbf{x}_1 = A \mathbf{x}_0\).
This process of multiplying the vector by the matrix is repeated to observe how the vector evolves over iterations:\(\mathbf{x}_{k+1} = A \mathbf{x}_k\).
  • This iterative multiplication allows understanding of how transformations by \(A\) affect the vector space.
  • In the exercise, performing this on \(\mathbf{x}_0 = \begin{bmatrix} 1 \ 1 \end{bmatrix}\) retained the same vector, signifying an equilibrium or consistent trajectory.
  • Matrix multiplication benefits from properties like distributivity and associativity, ensuring efficient calculations of successive states.
Thus, matrix multiplication is essential for projecting and understanding future states in any dynamical system.

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Most popular questions from this chapter

Prove that if \(A\) is diagonalizable, so is \(A^{T}\)

Find the general solution to the given system of differential equations. Then find the specific solution that satisfies the initial conditions. (Consider all functions to be functions of \(t .\) $$\begin{array}{l} x^{\prime}=x+\quad 3 z, \quad x(0)=2 \\ y^{\prime}=x-2 y+z, \quad y(0)=3 \\ z^{\prime}=3 x+\quad z, \quad z(0)=4 \end{array}$$

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Find the Perron root and the corresponding Perron eigenvector of \(A\) $$A=\left[\begin{array}{ll} 1 & 3 \\ 2 & 0 \end{array}\right]$$

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