Problem 51
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. $$ A=\left[\begin{array}{ll} -1 & 1 \\ -3 & 1 \end{array}\right] $$
Step-by-Step Solution
Verified Answer
The equilibrium \((0,0)\) is an unstable spiral due to complex eigenvalues with positive real parts.
1Step 1: Determine the eigenvalues
The eigenvalues of the matrix \( A \) can be found by solving the characteristic equation: \[ \det(A - \lambda I) = 0 \].Substitute the given matrix \( A = \begin{bmatrix} -1 & 1 \ -3 & 1 \end{bmatrix} \) in the equation:\[\begin{vmatrix} -1-\lambda & 1 \ -3 & 1-\lambda \end{vmatrix} = (-1-\lambda)(1-\lambda) - (-3)(1) \]Simplify: \[ \lambda^2 - \lambda + 3 = 0 \]
2Step 2: Solve for eigenvalues
The characteristic equation \( \lambda^2 - \lambda + 3 = 0 \) is a quadratic equation. Use the quadratic formula to find the solutions:\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1 \), \( b = -1 \), and \( c = 3 \).\[ \lambda = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(3)}}{2(1)} = \frac{1 \pm \sqrt{1 - 12}}{2} = \frac{1 \pm \sqrt{-11}}{2} \]Thus, the eigenvalues are \( \lambda = \frac{1}{2} + \frac{i \sqrt{11}}{2} \) and \( \lambda = \frac{1}{2} - \frac{i \sqrt{11}}{2} \).
3Step 3: Analyze the stability based on eigenvalues
Since the eigenvalues are complex conjugates of the form \( \frac{1}{2} \pm i \frac{\sqrt{11}}{2} \) with a positive real part, \( \frac{1}{2} > 0 \), this indicates that the equilibrium point \((0,0)\) is unstable. Complex eigenvalues with a positive real part indicate an unstable spiral.
Key Concepts
EigenvaluesStability AnalysisEquilibrium ClassificationComplex Conjugates
Eigenvalues
Eigenvalues are an essential concept in the analysis of differential equations, especially those involving systems like the one in this exercise. They are scalars that offer key insights into the characteristics of a linear transformation represented by a matrix. For a matrix like \( A \), the eigenvalues are found by solving the characteristic equation, which is derived by setting the determinant of \( A - \lambda I \) to zero, where \( I \) is the identity matrix.
In this case, the characteristic equation becomes \( \lambda^2 - \lambda + 3 = 0 \). Solving this with the quadratic formula, we find that the eigenvalues are \( \lambda = \frac{1}{2} + \frac{i \sqrt{11}}{2} \) and \( \lambda = \frac{1}{2} - \frac{i \sqrt{11}}{2} \). These eigenvalues include an imaginary component, indicating the potential for oscillatory behavior in the system.
In this case, the characteristic equation becomes \( \lambda^2 - \lambda + 3 = 0 \). Solving this with the quadratic formula, we find that the eigenvalues are \( \lambda = \frac{1}{2} + \frac{i \sqrt{11}}{2} \) and \( \lambda = \frac{1}{2} - \frac{i \sqrt{11}}{2} \). These eigenvalues include an imaginary component, indicating the potential for oscillatory behavior in the system.
Stability Analysis
Stability analysis is crucial for understanding the behavior of equilibria in a dynamical system. By examining the sign of the real parts of the eigenvalues, we can determine whether a system is stable or unstable.
For eigenvalues with complex conjugate form, like \( \frac{1}{2} \pm \frac{i \sqrt{11}}{2} \), the stability is largely determined by the real part. Here, the real part is \( \frac{1}{2} \), which is positive. This indicates that the system is fundamentally unstable. In other words, trajectories of the system will spiral outwards, away from the equilibrium point, as time progresses.
For eigenvalues with complex conjugate form, like \( \frac{1}{2} \pm \frac{i \sqrt{11}}{2} \), the stability is largely determined by the real part. Here, the real part is \( \frac{1}{2} \), which is positive. This indicates that the system is fundamentally unstable. In other words, trajectories of the system will spiral outwards, away from the equilibrium point, as time progresses.
Equilibrium Classification
In differential equations, equilibria can be classified based on the nature of the eigenvalues. There are three primary types of classification that are commonly used: stable spirals, unstable spirals, and centers.
- **Stable Spiral**: Occurs when the eigenvalues are complex with a negative real part. This causes trajectories to spiral in towards the equilibrium.
- **Unstable Spiral**: Occurs when the eigenvalues are complex with a positive real part, like in our exercise. This results in trajectories that spiral away from the equilibrium, indicating an unstable behavior.
- **Center**: Occurs when the real part of the eigenvalues is zero, causing purely cyclic behavior around the equilibrium with no spiraling effect.
- **Stable Spiral**: Occurs when the eigenvalues are complex with a negative real part. This causes trajectories to spiral in towards the equilibrium.
- **Unstable Spiral**: Occurs when the eigenvalues are complex with a positive real part, like in our exercise. This results in trajectories that spiral away from the equilibrium, indicating an unstable behavior.
- **Center**: Occurs when the real part of the eigenvalues is zero, causing purely cyclic behavior around the equilibrium with no spiraling effect.
Complex Conjugates
Complex conjugates play a pivotal role in shaping the dynamics around an equilibrium point. When a system's matrix has complex conjugate eigenvalues, the corresponding solutions to the differential equation involve oscillations.
In the current analysis, the eigenvalues are \( \lambda = \frac{1}{2} \pm \frac{i \sqrt{11}}{2} \). The imaginary part \( \pm \frac{i \sqrt{11}}{2} \) suggests the presence of a rotational, or oscillatory, component. The behavior of solutions in the phase plane often takes the form of spirals.
Complex conjugates lead to interesting trajectory paths, often seen as ellipses or spirals, depending on the real part's sign. This mix of real and imaginary parts allows a richer understanding of both the local and global behavior of the system around that equilibrium point.
In the current analysis, the eigenvalues are \( \lambda = \frac{1}{2} \pm \frac{i \sqrt{11}}{2} \). The imaginary part \( \pm \frac{i \sqrt{11}}{2} \) suggests the presence of a rotational, or oscillatory, component. The behavior of solutions in the phase plane often takes the form of spirals.
Complex conjugates lead to interesting trajectory paths, often seen as ellipses or spirals, depending on the real part's sign. This mix of real and imaginary parts allows a richer understanding of both the local and global behavior of the system around that equilibrium point.
Other exercises in this chapter
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