Problem 55
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. $$ A=\left[\begin{array}{rr} 1 & 2 \\ -5 & -3 \end{array}\right] $$
Step-by-Step Solution
Verified Answer
The equilibrium \((0,0)\) is a stable spiral.
1Step 1: Identify Eigenvalues
First, determine the eigenvalues of matrix \( A \). For a 2x2 matrix \( A = \begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix} \), solve the characteristic equation: \( \lambda^2 - (a_{11} + a_{22}) \lambda + (a_{11}a_{22} - a_{12}a_{21}) = 0 \). In this case, with \( A = \begin{bmatrix} 1 & 2 \ -5 & -3 \end{bmatrix} \), we have \( \lambda^2 - (-2) \lambda - 11 = 0 \).
2Step 2: Solve Characteristic Equation
Solve the quadratic equation \( \lambda^2 + 2\lambda - 11 = 0 \) using the quadratic formula: \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1, b = 2, c = -11 \). Calculate the discriminant: \( b^2 - 4ac = 4 + 44 = 48 \), hence \( \lambda = \frac{-2 \pm \sqrt{-40}}{2} = -1 \pm i \sqrt{10} \). The eigenvalues are \( \lambda_1 = -1 + i\sqrt{10} \) and \( \lambda_2 = -1 - i\sqrt{10} \).
3Step 3: Analyze Real Part of Eigenvalues
Check the real part of the eigenvalues to determine stability. The real part of both eigenvalues is \(-1\), which is negative.
4Step 4: Classify Equilibrium Point
The eigenvalues are complex conjugates with a negative real part. This configuration indicates that the equilibrium point \((0,0)\) is a "stable spiral."
Key Concepts
EigenvaluesStability AnalysisEquilibrium Classification
Eigenvalues
Eigenvalues are a crucial concept when working with differential equations, especially for analyzing the behavior of solutions over time. For a system described by a matrix, like in our exercise, eigenvalues provide insights into the system's dynamics.
The matrix \( A \) in our problem is given as \( \left[\begin{array}{cc} 1 & 2 \ -5 & -3 \end{array}\right] \). To find its eigenvalues, we use the characteristic equation: \[ \lambda^2 - (a_{11} + a_{22}) \lambda + (a_{11}a_{22} - a_{12}a_{21}) = 0 \] where \( a_{11} = 1 \), \( a_{12} = 2 \), \( a_{21} = -5 \), and \( a_{22} = -3 \).
Solving this, we obtain the eigenvalues \( \lambda_1 = -1 + i\sqrt{10} \) and \( \lambda_2 = -1 - i\sqrt{10} \), which are complex conjugates. This means that the solutions to our differential equations will involve oscillatory components, influenced by these complex eigenvalues.
The matrix \( A \) in our problem is given as \( \left[\begin{array}{cc} 1 & 2 \ -5 & -3 \end{array}\right] \). To find its eigenvalues, we use the characteristic equation: \[ \lambda^2 - (a_{11} + a_{22}) \lambda + (a_{11}a_{22} - a_{12}a_{21}) = 0 \] where \( a_{11} = 1 \), \( a_{12} = 2 \), \( a_{21} = -5 \), and \( a_{22} = -3 \).
Solving this, we obtain the eigenvalues \( \lambda_1 = -1 + i\sqrt{10} \) and \( \lambda_2 = -1 - i\sqrt{10} \), which are complex conjugates. This means that the solutions to our differential equations will involve oscillatory components, influenced by these complex eigenvalues.
Stability Analysis
Stability analysis involves examining whether small perturbations or changes in the system will grow or decay over time. In our exercise, we use the real parts of the eigenvalues to determine stability.
When the real part of an eigenvalue is negative, the corresponding solution component decays exponentially, suggesting stability. Conversely, a positive real part suggests instability, as the solution component would grow over time.
For our eigenvalues \( \lambda_1 = -1 + i\sqrt{10} \) and \( \lambda_2 = -1 - i\sqrt{10} \), both share a negative real part of \(-1\). This indicates that both solution components will experience exponential decay, leading us to conclude that the system at equilibrium is stable.
Remember, stability in differential equations ensures that the system's behavior remains predictable and controlled when subject to minor disturbances.
When the real part of an eigenvalue is negative, the corresponding solution component decays exponentially, suggesting stability. Conversely, a positive real part suggests instability, as the solution component would grow over time.
For our eigenvalues \( \lambda_1 = -1 + i\sqrt{10} \) and \( \lambda_2 = -1 - i\sqrt{10} \), both share a negative real part of \(-1\). This indicates that both solution components will experience exponential decay, leading us to conclude that the system at equilibrium is stable.
Remember, stability in differential equations ensures that the system's behavior remains predictable and controlled when subject to minor disturbances.
Equilibrium Classification
Equilibrium classification provides a deeper understanding of how a system behaves near an equilibrium point like \((0,0)\). Equilibria can be classified based on the nature of their eigenvalues.
In our exercise, the equilibrium point \((0,0)\) is classified according to its eigenvalues, \( \lambda_1 = -1 + i\sqrt{10} \) and \( \lambda_2 = -1 - i\sqrt{10} \):
In our exercise, the equilibrium point \((0,0)\) is classified according to its eigenvalues, \( \lambda_1 = -1 + i\sqrt{10} \) and \( \lambda_2 = -1 - i\sqrt{10} \):
- Since both eigenvalues are complex and have negative real parts, the equilibrium at \((0,0)\) is classified as a **stable spiral**.
- A stable spiral implies that trajectories surrounding the equilibrium point spiral inwards over time, leading to stabilization at the equilibrium point.
Other exercises in this chapter
Problem 54
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