Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 4: Problem 1
Determine the location and type of all critical points by linearization. $$\begin{array}{l}y_{1}^{\prime}=y_{2}+y_{2}^{2} \\\y_{2}^{\prime}=3 y_{1}\end{array}$$
Short Answer
Expert verified
Equilibrium points: (0,0) is a saddle point; (0,-1) is a center.
Step by step solution
01
Find the Equilibrium Points
To find the equilibrium points, set both equations to zero: 1. \( y_1' = y_2 + y_2^2 = 0 \) 2. \( y_2' = 3y_1 = 0 \). Solving \( y_2 + y_2^2 = 0 \), we factor to get \( y_2(y_2 + 1) = 0 \), giving \( y_2 = 0 \) or \( y_2 = -1 \). For \( y_2' = 3y_1 = 0 \), we have \( y_1 = 0 \). Thus, the equilibrium points are \((0, 0)\) and \((0, -1)\).
02
Linearize the System at Each Equilibrium Point
The linearization of the system requires the Jacobian matrix. For this system:The Jacobian matrix \( J \) is:\[J = \begin{pmatrix}\frac{\partial y_1'}{\partial y_1} & \frac{\partial y_1'}{\partial y_2} \\frac{\partial y_2'}{\partial y_1} & \frac{\partial y_2'}{\partial y_2}\end{pmatrix} = \begin{pmatrix}0 & 1 + 2y_2 \3 & 0\end{pmatrix}\]For \( (0, 0) \):\( J = \begin{pmatrix}0 & 1 \3 & 0\end{pmatrix} \)For \( (0, -1) \):\( J = \begin{pmatrix}0 & -1 \3 & 0\end{pmatrix} \).
03
Determine the Eigenvalues
Find the eigenvalues of the Jacobian matrix for each equilibrium point.For \( (0,0) \), solve the characteristic equation:\( \det(J - \lambda I) = \begin{vmatrix} -\lambda & 1 \ 3 & -\lambda \end{vmatrix} = \lambda^2 - 3 = 0 \). The eigenvalues are \( \lambda = \sqrt{3} \) and \( \lambda = -\sqrt{3} \).For \( (0,-1) \), solve:\( \det(J - \lambda I) = \lambda^2 + 3 = 0 \).The eigenvalues are \( \lambda = i\sqrt{3} \) and \( \lambda = -i\sqrt{3} \).
04
Classify Each Critical Point
Using the eigenvalues:- For \( (0, 0) \): Since one eigenvalue is positive and one is negative (\( \sqrt{3} \) and \( -\sqrt{3} \)), the origin \((0, 0)\) is a saddle point.- For \( (0, -1) \): The eigenvalues are purely imaginary (\( i\sqrt{3} \) and \( -i\sqrt{3} \)), which indicates that \((0, -1)\) is a center.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equilibrium Points
Equilibrium points of a system denote its stable states, where no movement occurs. To find them, set the derivative equations to zero. For our system:
- \( y_1' = y_2 + y_2^2 = 0 \)
- \( y_2' = 3y_1 = 0 \)
- \( y_2(y_2 + 1) = 0 \) gives \( y_2 = 0 \) or \( y_2 = -1 \).
- \( y_1 = 0 \)
Linearization
Linearization helps simplify nonlinear systems to predict their behavior around equilibrium points. By approximating a nonlinear system with its tangent line near a certain point, we gain insights into that point's nature.
In our system's case:
In our system's case:
- Jacobians are computed to provide linear approximations at equilibrium points.
- This involves calculating the partial derivatives of the system's equations.
Jacobian Matrix
The Jacobian matrix is a key mathematical tool when linearizing a system. It consists of partial derivatives used to approximate how the system behaves linearly around equilibrium points.
For our system:\[J = \begin{pmatrix}\frac{\partial y_1'}{\partial y_1} & \frac{\partial y_1'}{\partial y_2} \\frac{\partial y_2'}{\partial y_1} & \frac{\partial y_2'}{\partial y_2}\end{pmatrix} = \begin{pmatrix}0 & 1 + 2y_2 \3 & 0\end{pmatrix}\]Calculating this at each equilibrium point helps us determine respective linear approximations:
For our system:\[J = \begin{pmatrix}\frac{\partial y_1'}{\partial y_1} & \frac{\partial y_1'}{\partial y_2} \\frac{\partial y_2'}{\partial y_1} & \frac{\partial y_2'}{\partial y_2}\end{pmatrix} = \begin{pmatrix}0 & 1 + 2y_2 \3 & 0\end{pmatrix}\]Calculating this at each equilibrium point helps us determine respective linear approximations:
- At \((0, 0)\), the Jacobian is \( \begin{pmatrix}0 & 1 \3 & 0\end{pmatrix} \)
- At \((0, -1)\), the Jacobian is \( \begin{pmatrix}0 & -1 \3 & 0\end{pmatrix} \)
Eigenvalues
Eigenvalues are essential to understanding a system's stability. When you solve for them, they reveal the intrinsic behavior.
For our Jacobian matrices at each equilibrium point, they tell us how disturbances evolve:
For our Jacobian matrices at each equilibrium point, they tell us how disturbances evolve:
- \( J = \begin{pmatrix}0 & 1 \3 & 0\end{pmatrix} \) for \((0,0)\) gives eigenvalues \( \sqrt{3} \) and \( -\sqrt{3} \).
- \( J = \begin{pmatrix}0 & -1 \3 & 0\end{pmatrix} \) for \((0,-1)\) gives eigenvalues \( i\sqrt{3} \) and \( -i\sqrt{3} \).
Critical Points Classification
Critical points are classified based on the nature of their eigenvalues.
For our system:
For our system:
- At \((0,0)\), having one positive and one negative eigenvalue indicates a saddle point. This suggests an unstable equilibrium where trajectories diverge.
- At \((0,-1)\), purely imaginary eigenvalues correspond to a center. This implies stable, oscillatory motion around this point.
