Problem 53
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 complex conjugates. Analyze the stability of the equilibrium \((0,0)\), and classify the equilibrium according to whether it is a stable spiral, an unstable spiral, or a center. \(A=\left[\begin{array}{rr}0 & -1 \\ 1 & 0\end{array}\right]\)
Step-by-Step Solution
Verified Answer
The equilibrium \((0,0)\) is a center and neutrally stable.
1Step 1: Find the Eigenvalues
To determine the eigenvalues of matrix \(A\), use the characteristic equation: \( \det(A - \lambda I) = 0 \). Here, \( I \) is the identity matrix. Calculate the determinant: \[ \begin{vmatrix} -\lambda & -1 \ 1 & -\lambda \end{vmatrix} = (-\lambda)(-\lambda) - (-1)(1) = \lambda^2 + 1 \]. This results in the equation \( \lambda^2 + 1 = 0 \). Solving for \( \lambda \) yields the eigenvalues \( \lambda_1 = i \) and \( \lambda_2 = -i \).
2Step 2: Determine the Type of Equilibrium Point
The nature of the eigenvalues (purely imaginary \(i\) and \(-i\)) suggests that the system's equilibrium is characterized by a center. In such cases, trajectories in the phase plane are closed paths, indicating neutral stability but not asymptotic stability.
3Step 3: Analyze the Stability of the Equilibrium Point
To classify the stability, observe that the eigenvalues have no real part \( \Re(\lambda) = 0 \). Therefore, the equilibrium point \((0,0)\) is a center. Centers are neutrally stable as they are characterized by closed orbits around the equilibrium point without spiraling inwards or outwards.
Key Concepts
EigenvaluesStability AnalysisEquilibrium Points
Eigenvalues
Eigenvalues are critical in analyzing systems of differential equations. They are values of the scalar \( \lambda \) that serve as solutions to the characteristic equation derived from a square matrix \( A \). For our example, eigenvalues \( \lambda \) solve the determinant equation \( \det(A - \lambda I) = 0 \) where \( I \) is the identity matrix. The eigenvalues indicate how solutions to the differential equation evolve over time.
In the given matrix \( A = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix} \), we solve the characteristic equation \( \lambda^2 + 1 = 0 \), resulting in complex conjugate eigenvalues \( \lambda_1 = i \) and \( \lambda_2 = -i \).
These complex eigenvalues suggest a rotational or spiraling behavior due to their imaginary parts. The lack of a real component implies no exponential growth or decay, merely oscillatory movement.
In the given matrix \( A = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix} \), we solve the characteristic equation \( \lambda^2 + 1 = 0 \), resulting in complex conjugate eigenvalues \( \lambda_1 = i \) and \( \lambda_2 = -i \).
These complex eigenvalues suggest a rotational or spiraling behavior due to their imaginary parts. The lack of a real component implies no exponential growth or decay, merely oscillatory movement.
Stability Analysis
Stability analysis examines how small perturbations to the system’s equilibrium will react over time. For linear systems described by differential equations, the sign of the real part of eigenvalues determines stability.
When eigenvalues have negative real parts, the system's equilibrium point is asymptotically stable, indicating solutions eventually converge to the equilibrium. If they have positive real parts, the equilibrium is unstable, as solutions move away from the point over time.
In our system with eigenvalues \( \lambda_1 = i \) and \( \lambda_2 = -i \), the real parts \( \Re(\lambda) = 0 \) imply neither growth nor decay, just movement in closed paths. This leads to a classification as a neutrally stable system, where solutions neither converge nor diverge but perpetually orbit around the equilibrium point.
When eigenvalues have negative real parts, the system's equilibrium point is asymptotically stable, indicating solutions eventually converge to the equilibrium. If they have positive real parts, the equilibrium is unstable, as solutions move away from the point over time.
In our system with eigenvalues \( \lambda_1 = i \) and \( \lambda_2 = -i \), the real parts \( \Re(\lambda) = 0 \) imply neither growth nor decay, just movement in closed paths. This leads to a classification as a neutrally stable system, where solutions neither converge nor diverge but perpetually orbit around the equilibrium point.
Equilibrium Points
Equilibrium points, or steady states, in a differential equation system occur where changes over time become zero; i.e., \( \frac{d \mathbf{x}}{dt} = 0 \). This means the point remains constant when the system reaches this state.
For our system defined by \( \frac{d \mathbf{x}}{dt} = A \mathbf{x}(t) \), the equilibrium at \((0,0)\) is reached when the vector \( \mathbf{x} \) becomes zero. At this point, \( A \mathbf{x}(t) \) also equals zero, ceasing any temporal change.
The type of equilibrium is determined by the eigenvalues. Given that these eigenvalues are purely imaginary, the equilibrium point is classified as a "center." This indicates trajectories circle around but do not change their distance from it over time, exemplifying neutral stability.
For our system defined by \( \frac{d \mathbf{x}}{dt} = A \mathbf{x}(t) \), the equilibrium at \((0,0)\) is reached when the vector \( \mathbf{x} \) becomes zero. At this point, \( A \mathbf{x}(t) \) also equals zero, ceasing any temporal change.
The type of equilibrium is determined by the eigenvalues. Given that these eigenvalues are purely imaginary, the equilibrium point is classified as a "center." This indicates trajectories circle around but do not change their distance from it over time, exemplifying neutral stability.
Other exercises in this chapter
Problem 51
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
View solution Problem 52
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
View solution Problem 54
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
View solution Problem 55
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
View solution