Problem 17
Question
Show that \(\left.\right|_{0} ^{0}\) is an equilibrium of $$ \left[\begin{array}{l} x_{1}(t+1) \\ x_{2}(t+1) \end{array}\right]=\left[\begin{array}{ll} -0.6 & 0 \\ -0.3 & 0.3 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ and determine its stability.
Step-by-Step Solution
Verified Answer
The origin is an equilibrium and is stable since all eigenvalues have magnitudes less than one.
1Step 1: Define Equilibrium
An equilibrium point \( x_1, x_2 \) is a point where the state does not change over time. Mathematically, this means \( \mathbf{x}(t+1) = \mathbf{x}(t) \). Hence, for \\[\begin{bmatrix}x_{1}(t+1) \\x_{2}(t+1)\end{bmatrix}\ = \begin{bmatrix}x_{1}(t) \\x_{2}(t)\end{bmatrix}\]\to hold, we need:\[\begin{bmatrix}-0.6 & 0 \-0.3 & 0.3\end{bmatrix}\begin{bmatrix}x_{1}(t) \x_{2}(t)\end{bmatrix} = \begin{bmatrix}x_{1}(t) \x_{2}(t)\end{bmatrix}\]
Key Concepts
Stability AnalysisEquilibrium PointDynamical Systems
Stability Analysis
Stability analysis in dynamical systems is a crucial aspect that determines how a system behaves over time. It helps us understand whether a system will return to equilibrium after a disturbance. If slight perturbations cause the system to deviate further from equilibrium, the system is considered unstable.
There are various methods to perform stability analysis, but one common approach is to examine the eigenvalues of the system's matrix. A dynamical system defined by a linear matrix \[A = \begin{bmatrix} -0.6 & 0 \-0.3 & 0.3 \end{bmatrix}\]is analyzed for stability by determining the eigenvalues of matrix \( A \). If all eigenvalues have magnitudes less than 1, the equilibrium point is stable.
In our example, calculating the eigenvalues involves solving the characteristic equation \( \det(A - \lambda I) = 0 \), resulting in eigenvalues \( \lambda_1 \) and \( \lambda_2 \). If both \( | \lambda_1| < 1\) and \( |\lambda_2| < 1\), stability is confirmed.
There are various methods to perform stability analysis, but one common approach is to examine the eigenvalues of the system's matrix. A dynamical system defined by a linear matrix \[A = \begin{bmatrix} -0.6 & 0 \-0.3 & 0.3 \end{bmatrix}\]is analyzed for stability by determining the eigenvalues of matrix \( A \). If all eigenvalues have magnitudes less than 1, the equilibrium point is stable.
In our example, calculating the eigenvalues involves solving the characteristic equation \( \det(A - \lambda I) = 0 \), resulting in eigenvalues \( \lambda_1 \) and \( \lambda_2 \). If both \( | \lambda_1| < 1\) and \( |\lambda_2| < 1\), stability is confirmed.
Equilibrium Point
An equilibrium point in a dynamical system is where the system can "rest," meaning that if the system starts in this state, it will remain there indefinitely as long as there are no external influences. For example, if we say that \[\mathbf{x}(t+1) = \mathbf{x}(t)\]this implies that the system does not evolve over time from this point, providing us an essential criterion for system behavior analysis.
To find an equilibrium point, we equate the future state to the current state with the given rule:\[\begin{bmatrix}x_{1}(t+1) \x_{2}(t+1)\end{bmatrix} = \begin{bmatrix}x_{1}(t) \x_{2}(t)\end{bmatrix}.\]This leads to solving the equation:
To find an equilibrium point, we equate the future state to the current state with the given rule:\[\begin{bmatrix}x_{1}(t+1) \x_{2}(t+1)\end{bmatrix} = \begin{bmatrix}x_{1}(t) \x_{2}(t)\end{bmatrix}.\]This leads to solving the equation:
- \[-0.6x_1 + 0x_2 = x_1\]
- \[-0.3x_1 + 0.3x_2 = x_2\]
Dynamical Systems
Dynamical systems form the core framework for modeling systems evolving over time. It is often expressed in terms of difference or differential equations that dictate how the state of the system changes.
In linear dynamical systems, like our example, the state at the next time point is derived from the current state by a matrix operation. The system can be written in the form:\[\mathbf{x}(t+1) = A \mathbf{x}(t)\]
Where \( A \) is the transformation matrix that applies to the state vector \( \mathbf{x}(t) \). In our exercise, \[A = \begin{bmatrix} -0.6 & 0 \-0.3 & 0.3 \end{bmatrix},\]which captures how both variables \(x_1\) and \(x_2\) evolve independently or influence each other. Understanding such systems helps predict outcomes, such as stability and equilibrium, making them crucial in fields like engineering, economics, and biology.
In linear dynamical systems, like our example, the state at the next time point is derived from the current state by a matrix operation. The system can be written in the form:\[\mathbf{x}(t+1) = A \mathbf{x}(t)\]
Where \( A \) is the transformation matrix that applies to the state vector \( \mathbf{x}(t) \). In our exercise, \[A = \begin{bmatrix} -0.6 & 0 \-0.3 & 0.3 \end{bmatrix},\]which captures how both variables \(x_1\) and \(x_2\) evolve independently or influence each other. Understanding such systems helps predict outcomes, such as stability and equilibrium, making them crucial in fields like engineering, economics, and biology.
Other exercises in this chapter
Problem 17
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