Problem 57
Question
In Problems 57-66, we consider differential equations of the form \(\frac{d x}{d t}=A x(t)\) where \(A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]\) Analyze the stability of the equilibrium \(\mathbf{( 0 , 0 ) , \text { and classify the }}\) equilibrium. \(A=\left[\begin{array}{rr}-1 & -2 \\ 1 & 3\end{array}\right]\)
Step-by-Step Solution
Verified Answer
The equilibrium \((0,0)\) is an unstable proper or improper node.
1Step 1: Determine the Eigenvalues
To analyze the stability of the equilibrium at \((0,0)\), we need to find the eigenvalues of the matrix \(A\). These are found by solving the characteristic equation: \( \det(A - \lambda I) = 0 \). For the given matrix: \[ A = \begin{pmatrix} -1 & -2 \ 1 & 3 \end{pmatrix} \] The characteristic equation becomes: \[ \det\begin{pmatrix} -1 - \lambda & -2 \ 1 & 3 - \lambda \end{pmatrix} = (-1 - \lambda)(3 - \lambda) - (-2)(1) = \lambda^2 - 2\lambda + 1 = 0 \] \( \lambda^2 - 2\lambda + 1 = (\lambda - 1)^2 = 0 \). Solving, we find that \(\lambda = 1\) is a repeated eigenvalue.
2Step 2: Analyze the Stability
To determine the nature of the equilibrium, consider the sign and multiplicity of the eigenvalue. Since \(\lambda = 1\) is a positive repeated eigenvalue, it implies that the origin \((0,0)\) is an unstable equilibrium because trajectories will move away from the equilibrium point over time.
3Step 3: Confirm Classification
With eigenvalue analysis, the system can be classified. For the repeated eigenvalue \(\lambda = 1\), the corresponding eigenspace may not span \(\mathbb{R}^2\) completely, indicating the presence of a defective matrix. Thus, the system represents a non-diagonalizable matrix with one line of solutions in its phase plane, which suggests a non-stable node or improper node type.
Key Concepts
Stability AnalysisEigenvaluesEquilibrium Classification
Stability Analysis
When examining a differential equation system like \(\frac{d x}{d t}=A x(t)\), analyzing the stability is crucial. Stability analysis helps us understand how solutions behave over time, especially around equilibrium points like \((0,0)\). Typically, if solutions remain close or converge to an equilibrium point as time progresses, the equilibrium is considered stable. Conversely, if solutions diverge, the equilibrium is unstable.
In our given problem, we began by determining the eigenvalues of matrix \(A\). Knowing the eigenvalues allows us to deduce stability. In this specific case, the eigenvalue \(\lambda = 1\) is repeated and positive. This positive eigenvalue indicates instability.
To visualize, consider a simple physics analogy: it's like a ball placed on the crest of a hill rather than a valley. Any slight push moves the ball away, analogous to solutions diverging from the equilibrium point. This step in stability analysis gives us a clear indication of how the system behaves near equilibrium.
In our given problem, we began by determining the eigenvalues of matrix \(A\). Knowing the eigenvalues allows us to deduce stability. In this specific case, the eigenvalue \(\lambda = 1\) is repeated and positive. This positive eigenvalue indicates instability.
To visualize, consider a simple physics analogy: it's like a ball placed on the crest of a hill rather than a valley. Any slight push moves the ball away, analogous to solutions diverging from the equilibrium point. This step in stability analysis gives us a clear indication of how the system behaves near equilibrium.
Eigenvalues
Eigenvalues play an important role when analyzing the behavior of differential equations. They provide insights into the dynamics of the system represented by the matrix \(A\). To find them, we solve the characteristic equation obtained by \(\det(A - \lambda I) = 0\). Here, \(I\) represents the identity matrix.
For our matrix \(A\), solving the equation \[\lambda^2 - 2\lambda + 1 = 0\] yielded the eigenvalue \(\lambda = 1\) as a repeated value. The nature and sign of the eigenvalues are crucial. Positive eigenvalues suggest instability, whereas negative ones or those with negative real parts usually imply stability. A repeated eigenvalue indicates the system might not illuminate different solution paths in the phase plane.
Understanding the calculation helps clarify the behavior of complex systems and aids in classifying equilibriums. As we saw, our positive repeat in \(\lambda\) points towards instability, leading us to suggest that solutions will not neatly line up along predictable paths in the analysis phase.
For our matrix \(A\), solving the equation \[\lambda^2 - 2\lambda + 1 = 0\] yielded the eigenvalue \(\lambda = 1\) as a repeated value. The nature and sign of the eigenvalues are crucial. Positive eigenvalues suggest instability, whereas negative ones or those with negative real parts usually imply stability. A repeated eigenvalue indicates the system might not illuminate different solution paths in the phase plane.
Understanding the calculation helps clarify the behavior of complex systems and aids in classifying equilibriums. As we saw, our positive repeat in \(\lambda\) points towards instability, leading us to suggest that solutions will not neatly line up along predictable paths in the analysis phase.
Equilibrium Classification
Equilibriums in differential equations describe the states where the system remains constant, i.e., where there’s no change in the state over time. Proper classification of these equilibrium points is essential to predicting and understanding system behavior. Our case considered the equilibrium at \((0,0)\).
Equilibriums are broadly categorized by analyzing eigenvalues. For our system, repeated eigenvalues \(\lambda = 1\) indicates a more complicated classification: a non-stable node or an improper node. Here's why:
Equilibriums are broadly categorized by analyzing eigenvalues. For our system, repeated eigenvalues \(\lambda = 1\) indicates a more complicated classification: a non-stable node or an improper node. Here's why:
- A node type tells us that solutions might not simply converge to the point, given the eigenvalues.
- The system is in a state where solutions spread out, not congregating at equilibrium, denoting instability.
Other exercises in this chapter
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 Problem 56
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 58
We consider differential equations of the form \(\frac{d x}{d t}=A x(t)\) where \(A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]\
View solution Problem 59
We consider differential equations of the form \(\frac{d x}{d t}=A x(t)\) where \(A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]\
View solution