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Magazine Chapter 9: Problem 10
Find the equilibrium points and assess the stability of each. \(x^{\prime}=x-e^{y}, y^{\prime}=2 \ln x+y-6\)
Short Answer
Expert verified
The equilibrium point is \( (e^2, 2) \) and it is unstable.
Step by step solution
01
- Find the Equilibrium Points
To find the equilibrium points, we need to set the derivatives equal to zero: \( x' = x - e^y = 0 \) \( y' = 2 \ln x + y - 6 = 0 \).This results in the equations: \( x = e^y\) and \( y = 6 - 2 \ln x \). Substitute \( x = e^y \) into the second equation: \( y = 6 - 2 \ln(e^y) = 6 - 2y. \) Solve for \( y \): \( y + 2y = 6 \Rightarrow 3y = 6 \Rightarrow y = 2. \) Substitute \( y = 2 \) back into \( x = e^y \): \( x = e^2. \)
02
- Calculate the Jacobian Matrix
The Jacobian matrix, \( J, \) of the system is defined as:\[ J = \begin{bmatrix} \frac{\fracpartial x'}{\fracpartial x} & \frac{\fracpartial x'}{\fracpartial y} \ \frac{\fracpartial y'}{\fracpartial x} & \frac{\fracpartial y'}{\fracpartial y} \ \end{bmatrix} \].First, compute the partial derivatives:\begin{align*} \frac{\fracpartial x'}{\fracpartial x} &= 1, \ \frac{\fracpartial x'}{\fracpartial y} &= -e^y, \ \frac{\fracpartial y'}{\fracpartial x} &= \frac{2}{x}, \ \frac{\fracpartial y'}{\fracpartial y} &= 1. \end{align*}Hence, we have:\begin{bmatrix} 1 & -e^y \ \frac{2}{x} & 1 \end{bmatrix}.
03
- Evaluate the Jacobian at the Equilibrium Point
Substitute the equilibrium point \( (e^2, 2) \) into the Jacobian matrix:\begin{bmatrix} 1 & -e^2 \ \frac{2}{e^2} & 1 \end{bmatrix}.This results in:\begin{bmatrix} 1 & -e^2 \ \frac{2}{e^2} & 1 \end{bmatrix}.
04
- Find the Eigenvalues of the Jacobian Matrix
To find the stability of the equilibrium point, compute the eigenvalues of the Jacobian matrix:Find the determinant of \( J - \lambda I \):\begin{align*} \det(J - \lambda I) = \begin{vmatrix} 1 - \lambda & -e^2 \ \frac{2}{e^2} & 1 - \lambda \end{vmatrix} = (1 - \lambda)^2 - (-e^2) \frac{2}{e^2} = 1 - 2 \lambda + \lambda^2 + 2. \end{align*}Solve the characteristic equation \( \lambda^2 - 2\lambda + 3 = 0 \):\begin{align*} \lambda = \frac{2 \pm \sqrt{4 - 12}}{2} = 1 \pm i. \end{align*}The eigenvalues are complex numbers with a real part equal to 1.
05
- Assess the Stability
Because the real part of the eigenvalues (1 \pm i) is positive, the equilibrium point \( (e^2, 2) \) is unstable.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Jacobian Matrix
The Jacobian matrix is essential in studying nonlinear dynamical systems. It consists of all first-order partial derivatives of a vector-valued function. For a system of differential equations, the Jacobian matrix helps us understand how changes in the variables affect the system's behavior.
For an example system with variables \( x \) and \( y \), the Jacobian matrix \( J \) is given by:
\[ J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} \]
This matrix is used to assess the stability of equilibrium points by analyzing its eigenvalues. A Jacobian matrix provides insights into how small deviations from an equilibrium point will evolve over time, thus determining if the point is stable or not.
For an example system with variables \( x \) and \( y \), the Jacobian matrix \( J \) is given by:
\[ J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} \]
This matrix is used to assess the stability of equilibrium points by analyzing its eigenvalues. A Jacobian matrix provides insights into how small deviations from an equilibrium point will evolve over time, thus determining if the point is stable or not.
Equilibrium Points
Equilibrium points in a system of differential equations are where the system is in a state of balance. These points occur when the derivatives of all variables are zero.
To find equilibrium points, set the equations of the system to zero and solve for the variables. For instance:
\[ x' = x - e^y = 0 \]
\[ y' = 2 \ln x + y - 6 = 0 \]
After solving these equations, we get the values of \( x \) and \( y \) that make both derivatives zero. These values are the equilibrium points. Properly analyzing these points helps in predicting system behavior around these points.
To find equilibrium points, set the equations of the system to zero and solve for the variables. For instance:
\[ x' = x - e^y = 0 \]
\[ y' = 2 \ln x + y - 6 = 0 \]
After solving these equations, we get the values of \( x \) and \( y \) that make both derivatives zero. These values are the equilibrium points. Properly analyzing these points helps in predicting system behavior around these points.
Eigenvalues
Eigenvalues are crucial in determining the behavior of a dynamical system near an equilibrium point. They are derived from the Jacobian matrix by solving the characteristic equation.
For a matrix \( J \), the eigenvalues \( \lambda \) are found by solving:
\[ \det(J - \lambda I) = 0 \]
These eigenvalues determine the nature of the equilibrium points. If all eigenvalues have negative real parts, the equilibrium is stable. If any eigenvalue has a positive real part, the equilibrium is unstable.
For a matrix \( J \), the eigenvalues \( \lambda \) are found by solving:
\[ \det(J - \lambda I) = 0 \]
These eigenvalues determine the nature of the equilibrium points. If all eigenvalues have negative real parts, the equilibrium is stable. If any eigenvalue has a positive real part, the equilibrium is unstable.
Determinant
The determinant of a matrix is a scalar value that can indicate certain properties of the matrix. For a Jacobian matrix \( J \), the determinant helps in finding eigenvalues, which in turn are used to assess the stability of equilibrium points.
The determinant for a 2x2 matrix is calculated as:
\[ \det \begin{bmatrix} a & b \ c & d \end{bmatrix} = ad - bc \]
When evaluating the stability of equilibrium points, finding the determinant of the modified Jacobian matrix \( J - \lambda I \) is an essential step.
The determinant for a 2x2 matrix is calculated as:
\[ \det \begin{bmatrix} a & b \ c & d \end{bmatrix} = ad - bc \]
When evaluating the stability of equilibrium points, finding the determinant of the modified Jacobian matrix \( J - \lambda I \) is an essential step.
Differential Equations
Differential equations describe how variables change over time. They are fundamental in modeling various real-world phenomena like population growth, chemical reactions, and mechanical systems.
A general form of a differential equation is:
\[ y' = f(x, y) \]
Here, \( y' \) represents the rate of change of \( y \) with respect to \( x \).
To analyze these systems, we often look for equilibrium points and study the behavior around these points using methods like the Jacobian matrix and eigenvalues.
A general form of a differential equation is:
\[ y' = f(x, y) \]
Here, \( y' \) represents the rate of change of \( y \) with respect to \( x \).
To analyze these systems, we often look for equilibrium points and study the behavior around these points using methods like the Jacobian matrix and eigenvalues.
