Problem 49
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. $$ A=\left[\begin{array}{rr} 4 & 5 \\ -3 & -3 \end{array}\right] $$
Step-by-Step Solution
Verified Answer
The equilibrium (0,0) is an unstable spiral because the real part of the eigenvalues is positive.
1Step 1: Determine the eigenvalues of matrix A
To find the eigenvalues of the matrix \(A\), we start by solving the characteristic equation \(\det(A - \lambda I) = 0\), where \(I\) is the identity matrix. For matrix \(A\), this gives us the equation: \[ \begin{vmatrix} 4-\lambda & 5 \ -3 & -3-\lambda \end{vmatrix} = 0 \]. Calculate the determinant: \((4-\lambda)(-3-\lambda) - (-3)(5) = 0\). This simplifies to \(\lambda^2 - \lambda + 27 = 0\). Solving for \(\lambda\), find the roots using the quadratic formula: \(\lambda = \frac{1 \pm \sqrt{1 - 4 \times 27}}{2} = \frac{1 \pm \sqrt{-107}}{2}\). The eigenvalues are \(\lambda = \frac{1}{2} \pm i\frac{\sqrt{107}}{2}\).
2Step 2: Analyze the real part of the eigenvalues
The eigenvalues are \(\lambda_1 = \frac{1}{2} + i \frac{\sqrt{107}}{2}\) and \(\lambda_2 = \frac{1}{2} - i \frac{\sqrt{107}}{2}\). Since the real part of both eigenvalues is positive (\(\frac{1}{2}\)), this indicates that any perturbation from the equilibrium point \((0,0)\) will grow exponentially over time.
3Step 3: Classify the type of equilibrium based on eigenvalues
The equilibrium point \((0,0)\) is classified based on the eigenvalues. With a positive real part for the eigenvalues and because they are complex conjugates, \((0,0)\) is classified as an unstable spiral. In this scenario, solutions spiral away from the equilibrium.
Key Concepts
Stability AnalysisComplex EigenvaluesUnstable Spiral
Stability Analysis
When we analyze the stability of a differential equation system, we're interested in how the system behaves over time, particularly near equilibrium points. In this case, the system being examined is defined by a matrix equation. The matrix \( A \) determines the evolution of the system. To assess stability, we explore the matrix's eigenvalues.
The equilibrium point in question is \((0,0)\). We use the characteristic equation \( \det(A - \lambda I) = 0 \) to find eigenvalues. The eigenvalues reveal important information because they indicate how perturbations or deviations from equilibrium evolve.
The equilibrium point in question is \((0,0)\). We use the characteristic equation \( \det(A - \lambda I) = 0 \) to find eigenvalues. The eigenvalues reveal important information because they indicate how perturbations or deviations from equilibrium evolve.
- If all eigenvalues have negative real parts, the equilibrium is stable (attracting spiral or fixed point).
- If there's a positive real part, the equilibrium is unstable (solutions grow away).
- Zero real parts might indicate a center (neutral stability), depending on additional conditions.
Complex Eigenvalues
Eigenvalues can be real or complex. In our differential equation exercise, the eigenvalues of matrix \( A \) are complex conjugates: \( \lambda_1 = \frac{1}{2} + i \frac{\sqrt{107}}{2} \) and \( \lambda_2 = \frac{1}{2} - i \frac{\sqrt{107}}{2} \).
Complex eigenvalues imply that the system's solutions involve oscillatory behavior — a mix of rotations and scaling. This is because the presence of the imaginary part induces a sinusoidal component in solutions. The real part of these eigenvalues determines whether these oscillations grow or decay over time.
Complex eigenvalues imply that the system's solutions involve oscillatory behavior — a mix of rotations and scaling. This is because the presence of the imaginary part induces a sinusoidal component in solutions. The real part of these eigenvalues determines whether these oscillations grow or decay over time.
- If positive, the spiral grows (unstable spiral).
- If negative, it shrinks (stable spiral).
- With zero, it remains constant in the amplitude but oscillatory (center).
Unstable Spiral
An unstable spiral describes the dynamics where solutions move away from an equilibrium point, tracing out a spiral path. For the differential equation considered, the computed eigenvalues \( \lambda_1 = \frac{1}{2} + i \frac{\sqrt{107}}{2} \) indicate an unstable scenario due to the positive real part \( \frac{1}{2} \).
This means any small deviation or disturbance from the equilibrium \((0,0)\) will lead to solutions spiraling outward exponentially. The trajectory in the phase space keeps growing in amplitude away from the equilibrium, illustrating the instability.
This means any small deviation or disturbance from the equilibrium \((0,0)\) will lead to solutions spiraling outward exponentially. The trajectory in the phase space keeps growing in amplitude away from the equilibrium, illustrating the instability.
- This behavior contrasts with a stable spiral (solutions spiral inwards).
- It reflects instability in the system, often met in systems with feedback loops that reinforce deviations.
Other exercises in this chapter
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