Problem 16
Question
A Jacobian matrix and two equlibria are given. Determine if each is locally stable, unstable, or if the analysis is inconclusive. $$\begin{array}{l}{J=\left[ \begin{array}{cc}{2 x_{1}} & {-\sin x_{2}} \\\ {\cos x_{1}} & {0}\end{array}\right]} \\ {\text { (i) } \hat{x}_{1}=1, \hat{x}_{2}=-\pi} \\ {\text { (ii) } \hat{x}_{1}=1, \hat{x}_{2}=\pi}\end{array}$$
Step-by-Step Solution
Verified Answer
Both equilibria are unstable.
1Step 1: Evaluate Jacobian at the first equilibrium point
Substitute the values \(\hat{x}_{1} = 1\) and \(\hat{x}_{2} = -\pi\) into the Jacobian matrix \(J\).\[ J_1 = \begin{bmatrix} 2(1) & -\sin(-\pi) \ \cos(1) & 0 \end{bmatrix} = \begin{bmatrix} 2 & 0 \ \cos(1) & 0 \end{bmatrix} \]
2Step 2: Determine eigenvalues for the first equilibrium point
Calculate the eigenvalues of \(J_1\) using the determinant \(\det(J_1 - \lambda I) = 0\).\[ \det \left(\begin{bmatrix} 2 - \lambda & 0 \ \cos(1) & -\lambda \end{bmatrix}\right) = (2-\lambda)(-\lambda) = -\lambda(\lambda - 2) = 0\]The eigenvalues are \(\lambda = 0\) and \(\lambda = 2\).
3Step 3: Assess stability at the first equilibrium point
If any eigenvalue has a positive real part, the equilibrium is unstable. Since \(\lambda = 2\) is positive, the first equilibrium is unstable.
4Step 4: Evaluate Jacobian at the second equilibrium point
Substitute the values \(\hat{x}_{1} = 1\) and \(\hat{x}_{2} = \pi\) into the Jacobian matrix \(J\).\[ J_2 = \begin{bmatrix} 2(1) & -\sin(\pi) \ \cos(1) & 0 \end{bmatrix} = \begin{bmatrix} 2 & 0 \ \cos(1) & 0 \end{bmatrix} \]
5Step 5: Determine eigenvalues for the second equilibrium point
Calculate the eigenvalues of \(J_2\) in a similar manner to \(J_1\).\[ \det \left(\begin{bmatrix} 2 - \lambda & 0 \ \cos(1) & -\lambda \end{bmatrix}\right) = -\lambda(\lambda - 2) = 0\]The eigenvalues are \(\lambda = 0\) and \(\lambda = 2\).
6Step 6: Assess stability at the second equilibrium point
Similar to the first equilibrium, if any eigenvalue has a positive real part, it indicates instability. Since \(\lambda = 2\) is positive, the second equilibrium is unstable.
Key Concepts
EigenvaluesStability AnalysisEquilibrium Points
Eigenvalues
Eigenvalues are a fundamental concept in linear algebra and have key applications in stability analysis and differential equations. When examining the system's behavior near an equilibrium point, eigenvalues of the Jacobian matrix provide crucial insights. These are essentially numbers that describe how a dynamic system behaves in the neighborhood of equilibrium.
The process of determining eigenvalues typically involves solving the characteristic equation given by \( \ ext{det}(J - \lambda I) = 0 \), where \( J \) is the Jacobian matrix and \( I \) is the identity matrix of the same size. The solutions \( \lambda \) are the eigenvalues of the matrix.
The process of determining eigenvalues typically involves solving the characteristic equation given by \( \ ext{det}(J - \lambda I) = 0 \), where \( J \) is the Jacobian matrix and \( I \) is the identity matrix of the same size. The solutions \( \lambda \) are the eigenvalues of the matrix.
- Real parts of eigenvalues: Indicate whether perturbations grow or diminish over time.
- Positive real parts: Suggest that perturbations will grow, pointing towards system instability.
- Negative real parts: Suggest that perturbations will diminish, pointing toward system stability.
Stability Analysis
Stability analysis is crucial when assessing the behavior of dynamical systems near equilibrium points. It determines whether a system will remain close to an equilibrium when subjected to small disturbances, or if it will stray away significantly. This form of analysis often relies on evaluating the eigenvalues of the Jacobian matrix.
In the context of equilibrium analysis:
In the context of equilibrium analysis:
- An equilibrium point is said to be stable if all eigenvalues of the Jacobian matrix have negative real parts.
- If even one eigenvalue has a positive real part, the equilibrium is considered unstable, indicating that disturbances could cause the system to deviate far from equilibrium.
- A zero real part in an eigenvalue could mean the analysis is inconclusive, requiring further investigation or alternate approaches.
Equilibrium Points
Equilibrium points of a dynamical system are key positions where the system doesn't change, which means it is in a state of balance. However, just because a system is at equilibrium doesn’t mean it is stable.
Determining whether these are stable or unstable involves examining the Jacobian matrix evaluated at these points. This determines how the system evolves when started close to an equilibrium point.
Determining whether these are stable or unstable involves examining the Jacobian matrix evaluated at these points. This determines how the system evolves when started close to an equilibrium point.
- Equilibrium can be static (all directions remain unchanged) or dynamic (system returns to equilibrium after a disturbance).
- Stability assessment at equilibrium points needs identification of Jacobian matrix eigenvalues with their real parts.
- Depending on eigenvalues, the equilibrium can be stable, unstable, or require further research for more complex behaviors.
Other exercises in this chapter
Problem 15
Sketch several solution curves in the phase plane of the system of differential equations \(d \mathbf{x} / d t=A \mathbf{x}\) using the given eigenvalues and ei
View solution Problem 15
Given the system of differential equations \(d \mathbf{x} / d t=A \mathbf{x}\) , construct the phase plane, including the nullclines. Does the equilibrium look
View solution Problem 16
Sketch several solution curves in the phase plane of the system of differential equations \(d \mathbf{x} / d t=A \mathbf{x}\) using the given eigenvalues and ei
View solution Problem 16
Given the system of differential equations \(d \mathbf{x} / d t=A \mathbf{x}\) , construct the phase plane, including the nullclines. Does the equilibrium look
View solution