Problem 31
Question
We consider differential equations of the form $$ \frac{d \mathbf{x}}{d t}=A \mathbf{x}(t) $$ where $$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] $$ The eigenvalues of A will be real, distinct, and nonzero. Analyze the stability of the equilibrium \((0,0)\), and classify the equilibrium according to whether it is a sink, a source, or a saddle point. $$ A=\left[\begin{array}{rr} -2 & 2 \\ 2 & 1 \end{array}\right] $$
Step-by-Step Solution
Verified Answer
The equilibrium \((0, 0)\) is a saddle point due to having eigenvalues of opposite signs.
1Step 1: Calculate the Eigenvalues
To determine the stability, we first find the eigenvalues of matrix \( A \). The eigenvalues \( \lambda \) are found by solving the characteristic equation: \( \text{det}(A - \lambda I) = 0 \). For our matrix: \[A = \begin{bmatrix} -2 & 2 \ 2 & 1 \end{bmatrix}\]The characteristic equation is: \[\begin{vmatrix} -2 - \lambda & 2 \ 2 & 1 - \lambda \end{vmatrix} = 0\]This simplifies to: \[(-2 - \lambda)(1 - \lambda) - (2)(2) = 0\]\[\lambda^2 + \lambda - 6 = 0\]Solving the quadratic equation, we find \( \lambda = 2 \) or \( \lambda = -3 \).
2Step 2: Analyze the Stability Based on Eigenvalues
With eigenvalues \( \lambda_1 = 2 \) and \( \lambda_2 = -3 \), we analyze the stability:- \( \lambda_1 = 2 \) is positive, indicating an unstable component.- \( \lambda_2 = -3 \) is negative, indicating a stable component.A system is a saddle point if there are real distinct eigenvalues of opposite signs. Since one eigenvalue is positive and the other is negative, the equilibrium point \((0,0)\) is a saddle point.
Key Concepts
EigenvaluesStability AnalysisSaddle Point
Eigenvalues
In the context of differential equations, eigenvalues offer significant insights into the system's behavior, especially its stability. To find the eigenvalues of a matrix, we solve the characteristic equation, which involves finding values for which the determinant of \(A - \lambda I = 0\), where \(I\) is the identity matrix and \(\lambda\) represents the eigenvalues. These eigenvalues help determine whether the system will evolve towards or away from equilibrium points over time.
For the matrix \( A = \begin{bmatrix} -2 & 2 \ 2 & 1 \end{bmatrix} \), we computed the characteristic equation as \( \lambda^2 + \lambda - 6 = 0 \). Solving this gives us two distinct eigenvalues, \( \lambda_1 = 2 \) and \( \lambda_2 = -3 \). These eigenvalues are real and distinct, crucial for determining the type of equilibrium, which we'll discuss next.
For the matrix \( A = \begin{bmatrix} -2 & 2 \ 2 & 1 \end{bmatrix} \), we computed the characteristic equation as \( \lambda^2 + \lambda - 6 = 0 \). Solving this gives us two distinct eigenvalues, \( \lambda_1 = 2 \) and \( \lambda_2 = -3 \). These eigenvalues are real and distinct, crucial for determining the type of equilibrium, which we'll discuss next.
Stability Analysis
Stability analysis helps understand how a system behaves in the vicinity of an equilibrium point, often by examining the eigenvalues of the system's matrix. In our case, the equilibrium is at \((0,0)\).
To conduct the analysis, we look at the signs of the eigenvalues:
To conduct the analysis, we look at the signs of the eigenvalues:
- If both eigenvalues are negative, the system is stable, meaning it will return to equilibrium after small disturbances.
- If both are positive, the system is unstable, which implies it will move away from the equilibrium after perturbations.
- If the eigenvalues have opposite signs, the system is classified as a saddle point.
Saddle Point
A saddle point for a system refers to an equilibrium state that is neither entirely stable nor unstable—it has characteristics of both. This occurs when the system has real eigenvalues of opposite signs.
At a saddle point, you can visualize the system's trajectory as being pushed away in one direction (related to the positive eigenvalue) while being pulled back in another direction (associated with the negative eigenvalue). This means that although part of the system's pattern wants to diverge, another desires to converge to the equilibrium.
In our matrix example \((0,0)\), with eigenvalues \( \lambda_1 = 2 \) and \( \lambda_2 = -3 \), one can predict instability in certain directions and stability in others. This makes transitioning through a saddle point a complex yet fascinating interplay of forces, often applicable in various real-world scenarios, like predicting population dynamics in ecology or analyzing financial systems.
At a saddle point, you can visualize the system's trajectory as being pushed away in one direction (related to the positive eigenvalue) while being pulled back in another direction (associated with the negative eigenvalue). This means that although part of the system's pattern wants to diverge, another desires to converge to the equilibrium.
In our matrix example \((0,0)\), with eigenvalues \( \lambda_1 = 2 \) and \( \lambda_2 = -3 \), one can predict instability in certain directions and stability in others. This makes transitioning through a saddle point a complex yet fascinating interplay of forces, often applicable in various real-world scenarios, like predicting population dynamics in ecology or analyzing financial systems.
Other exercises in this chapter
Problem 29
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