Problem 19
Question
Show that \(\begin{array}{ll}0 & \text { is an equilibrium of }\end{array}\) $$ \left[\begin{array}{l} x_{1}(t+1) \\ x_{2}(t+1) \end{array}\right]=\left[\begin{array}{ll} -1.4 & 0 \\ -0.5 & 0.1 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ and determine its stability.
Step-by-Step Solution
Verified Answer
The equilibrium at 0 is unstable because the eigenvalue \(-1.4\) has magnitude greater than 1.
1Step 1: Identify the System
The given system is a discrete time linear system of the form \[\begin{bmatrix} x_1(t+1) \ x_2(t+1) \end{bmatrix} = \begin{bmatrix} -1.4 & 0 \ -0.5 & 0.1 \end{bmatrix}\begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix}\] where \(A = \begin{bmatrix} -1.4 & 0 \ -0.5 & 0.1 \end{bmatrix}\) is the system matrix.
2Step 2: Determine the Equilibrium Point
To find the equilibrium point, set the next state equal to the current state: \[\begin{bmatrix} x_1(t+1) \ x_2(t+1) \end{bmatrix} = \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix}.\]At equilibrium, this reduces to \[\begin{bmatrix} -1.4 & 0 \ -0.5 & 0.1 \end{bmatrix}\begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\]Solving this, we see \((x_1, x_2) = (0, 0)\). Thus, \(0\) is the equilibrium.
3Step 3: Find Eigenvalues of A
To determine stability, find the eigenvalues of matrix \(A\). Set up the characteristic polynomial:\[\text{det}(A - \lambda I) = \begin{vmatrix} -1.4 - \lambda & 0 \ -0.5 & 0.1 - \lambda \end{vmatrix} = 0\]Solving this, we get the equation \((-1.4 - \lambda)(0.1 - \lambda) = 0\).This gives eigenvalues \(\lambda_1 = -1.4\) and \(\lambda_2 = 0.1\).
4Step 4: Assess Stability of the Equilibrium
An equilibrium of a discrete system is stable if all eigenvalues of the system matrix have magnitudes less than 1.For \(\lambda_1 = -1.4\), \(|-1.4| = 1.4\) which is greater than 1.Thus, the equilibrium \(0\) is unstable because one eigenvalue's magnitude exceeds 1.
Key Concepts
Understanding Discrete Time Linear SystemsGrasping the Basics of EigenvaluesExploring the System MatrixPerforming Stability Analysis
Understanding Discrete Time Linear Systems
In mathematics, a **discrete time linear system** is a type of dynamic model that describes the evolution of variables over discrete time steps using linear transformations. Such systems are frequently used in engineering, economics, and physics for modeling processes that are observed at regular intervals. A discrete time linear system can often be described using a matrix equation, where the next state of the system is a linear combination of its current states.
For example, consider a system where variables are represented by a state vector, and changes from one time step to the next are governed by a system matrix. These models are simple yet powerful, being able to depict complex behavior over time while remaining computationally manageable.
For example, consider a system where variables are represented by a state vector, and changes from one time step to the next are governed by a system matrix. These models are simple yet powerful, being able to depict complex behavior over time while remaining computationally manageable.
- State vectors capture the current status of the system.
- System matrices define how current states transition to future states.
- Time is indexed discretely, which means computations are performed at specified intervals.
Grasping the Basics of Eigenvalues
**Eigenvalues** are vital in understanding how a matrix behaves, particularly in systems of equations that model real-world processes. Eigenvalues are numbers that provide insights into the system's dynamics by revealing how transformations represented by the matrix affect vectors in its domain.
In a discrete time linear system, eigenvalues tell us how the system evolves over time. If you begin with a vector in the direction of an eigenvector, the action of the matrix is simply to scale this vector by its corresponding eigenvalue.
In a discrete time linear system, eigenvalues tell us how the system evolves over time. If you begin with a vector in the direction of an eigenvector, the action of the matrix is simply to scale this vector by its corresponding eigenvalue.
- They give insight into the "stretching" factors of linear transformations.
- Eigenvalues are crucial for stability analysis, determining whether solutions grow, shrink, or oscillate over time.
- For a system matrix, finding eigenvalues involves solving the characteristic polynomial derived from the matrix.
Exploring the System Matrix
The **system matrix** serves as a foundational element in describing discrete time linear systems. It defines how the system's state evolves from one time step to the next. This matrix is vital since it encapsulates all the parameters that govern the system's behavior.
A system matrix, often denoted as 'A,' holds the linear coefficients for a system of equations, acting on the state vector to yield a new state. In this context, the matrix essentially maps the system's dynamics:
A system matrix, often denoted as 'A,' holds the linear coefficients for a system of equations, acting on the state vector to yield a new state. In this context, the matrix essentially maps the system's dynamics:
- Each element of the matrix represents the influence of one state variable on another.
- Determining the eigenvalues and eigenvectors of the system matrix allows for analyzing the stability and behavior of the system.
- Stability conditions for the equilibrium can be directly derived by examining the system matrix.
Performing Stability Analysis
**Stability analysis** assesses the long-term behavior of equilibrium points within dynamic systems. In a discrete time linear system, stability concerns determine whether the system returns to an equilibrium state after a disturbance or diverges.
The key to stability lies in the eigenvalues of the system matrix. Stability analysis checks the magnitude of these eigenvalues.
The key to stability lies in the eigenvalues of the system matrix. Stability analysis checks the magnitude of these eigenvalues.
- If all eigenvalues have absolute values less than one, the system's equilibrium is stable.
- If any eigenvalue has an absolute value greater than or equal to one, the system's equilibrium is unstable, suggesting potential divergence or oscillation.
- This assessment helps in understanding and controlling how a system behaves over time, particularly after any external perturbations.
Other exercises in this chapter
Problem 18
Compute $$\lim _{(x, y) \rightarrow(0,0)} \frac{3 x y}{x^{2}+y^{3}} $$ along lines of the form \(y=m x\), for \(m \neq 0 .\) What can you conclude?
View solution Problem 19
Find the gradient of each function. $$ f(x, y)=\sqrt{x^{3}-3 x y} $$
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Find the linearization of \(f(x, y)\) at the indicated point \(\left(x_{0}, y_{0}\right).\) $$ f(x, y)=\sqrt{x}+2 y ;(1,0) $$
View solution Problem 19
In Problems 17-24, find the indicated partial derivatives. $$ g(x, y)=e^{x+3 y} ; g_{y}(2,1) $$
View solution