Problem 44
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. \(A=\left[\begin{array}{rr}-1 & -5 \\ 4 & -3\end{array}\right]\)
Step-by-Step Solution
Verified Answer
The equilibrium \((0,0)\) is a stable spiral, as the eigenvalues \(-2 \pm 3i\) have a negative real part.
1Step 1: Find the Eigenvalues
Compute the eigenvalues of matrix \( A \). The eigenvalues are solutions to the characteristic equation \( \det(A - \lambda I) = 0 \). First, determine the characteristic polynomial: \[ \det \left( \begin{bmatrix} -1 & -5 \ 4 & -3 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \ 0 & \lambda \end{bmatrix} \right) = \det \begin{bmatrix} -1 - \lambda & -5 \ 4 & -3 - \lambda \end{bmatrix} \] This results in the equation: \[ (-1-\lambda)(-3-\lambda) - (-20) = \lambda^2 + 4\lambda + 13 = 0 \] Solve this using the quadratic formula: \[ \lambda = \frac{-4 \pm \sqrt{16 - 52}}{2} = -2 \pm i\sqrt{9} \] Therefore, the eigenvalues are \( -2 + 3i \) and \( -2 - 3i \).
2Step 2: Analyze the Real Part of Eigenvalues
Check the real part of the complex eigenvalues to determine stability. The real part of both eigenvalues is \(-2\). If the real part is negative, the equilibrium point \((0,0)\) is stable. If the real part were positive, it would be unstable.
3Step 3: Classify the Type of Equilibrium
Since the eigenvalues are complex conjugates \(-2 \pm 3i\) with a negative real part, the equilibrium point is a stable spiral. Complex eigenvalues generally indicate spiral behavior, while a negative real part means it spirals inward.
Key Concepts
EigenvaluesStability AnalysisEquilibrium ClassificationCharacteristic Equation
Eigenvalues
Eigenvalues are fundamental in understanding differential equations, especially when dealing with systems of equations. They provide insights into the behavior of a system by analyzing the matrix involved in the equations. In the exercise, we consider the matrix \( A \), from which we derive eigenvalues to understand the system's behavior.
To find eigenvalues, we solve the characteristic equation \( \det(A - \lambda I) = 0 \). This involves determining values of \( \lambda \) that make the determinant of this matrix zero. These values help us predict how solutions to the differential equations behave over time.
In our case, the eigenvalues are complex numbers \(-2 \pm 3i\). Complex eigenvalues typically suggest oscillatory behavior, indicative of spirals in the system's phase plane. The real part (\(-2\) here) significantly influences stability, making it a crucial component of our analysis.
To find eigenvalues, we solve the characteristic equation \( \det(A - \lambda I) = 0 \). This involves determining values of \( \lambda \) that make the determinant of this matrix zero. These values help us predict how solutions to the differential equations behave over time.
In our case, the eigenvalues are complex numbers \(-2 \pm 3i\). Complex eigenvalues typically suggest oscillatory behavior, indicative of spirals in the system's phase plane. The real part (\(-2\) here) significantly influences stability, making it a crucial component of our analysis.
Stability Analysis
Stability analysis determines whether solutions to a system of differential equations remain bounded over time. When analyzing stability, the real part of the eigenvalues plays an essential role.
If the real part of an eigenvalue is negative, the equilibrium point is stable, meaning solutions will gravitate towards it over time. Conversely, a positive real part indicates instability, as solutions diverge away.
In the given exercise, both eigenvalues derived from \( A \) have a real part of \(-2\). This negative value ensures that the equilibrium at the origin \((0,0)\) is stable. The system naturally spirals inward, closer to the equilibrium.
If the real part of an eigenvalue is negative, the equilibrium point is stable, meaning solutions will gravitate towards it over time. Conversely, a positive real part indicates instability, as solutions diverge away.
In the given exercise, both eigenvalues derived from \( A \) have a real part of \(-2\). This negative value ensures that the equilibrium at the origin \((0,0)\) is stable. The system naturally spirals inward, closer to the equilibrium.
Equilibrium Classification
Equilibrium classification provides further information about the nature of a point's stability and behavior in a system modeled by differential equations.
Equilibrium points can be categorized as stable spirals, unstable spirals, or centers, often determined by the nature of their eigenvalues. In our specific case, the complex conjugate eigenvalues \(-2 \pm 3i\) guide this classification.
These eigenvalues place the system in the stable spiral category. The complex part, \(3i\), induces oscillations around the equilibrium, while the negative real part, \(-2\), ensures that these oscillations steadily decrease in amplitude, drawing the state towards \((0,0)\).
Thus, the equilibrium is not just stable, but it also exhibits a spiraling inward trajectory, which is typical of stable spirals.
Equilibrium points can be categorized as stable spirals, unstable spirals, or centers, often determined by the nature of their eigenvalues. In our specific case, the complex conjugate eigenvalues \(-2 \pm 3i\) guide this classification.
These eigenvalues place the system in the stable spiral category. The complex part, \(3i\), induces oscillations around the equilibrium, while the negative real part, \(-2\), ensures that these oscillations steadily decrease in amplitude, drawing the state towards \((0,0)\).
Thus, the equilibrium is not just stable, but it also exhibits a spiraling inward trajectory, which is typical of stable spirals.
Characteristic Equation
The characteristic equation is central in determining the eigenvalues of a matrix, which in turn provides crucial insights into the system's behavior.
To compute this equation, we consider the matrix \( A \) and subtract \( \lambda I \) from it. The next step involves finding the determinant of this resulting matrix and setting it equal to zero: \( \det(A - \lambda I) = 0 \).
This equation is usually quadratic for two-dimensional systems, resulting in a polynomial like the one in our problem: \( \lambda^2 + 4\lambda + 13 = 0 \). Solving this using the quadratic formula yields the eigenvalues \(-2 \pm 3i\).
Thus, the characteristic equation serves as the gateway to understanding the dynamical behavior of systems modeled by differential equations.
To compute this equation, we consider the matrix \( A \) and subtract \( \lambda I \) from it. The next step involves finding the determinant of this resulting matrix and setting it equal to zero: \( \det(A - \lambda I) = 0 \).
This equation is usually quadratic for two-dimensional systems, resulting in a polynomial like the one in our problem: \( \lambda^2 + 4\lambda + 13 = 0 \). Solving this using the quadratic formula yields the eigenvalues \(-2 \pm 3i\).
Thus, the characteristic equation serves as the gateway to understanding the dynamical behavior of systems modeled by differential equations.
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