Problem 7
Question
In Problems \(7-12\), find all equilibria of each system of differential equations and use the analytical approach to determine the stability of each equilibrium. \(\frac{d x_{1}}{d t}=-x_{1}+2 x_{1}\left(1-x_{1}\right)\) \(\frac{d x_{2}}{d t}=-x_{2}+5 x_{2}\left(1-x_{1}-x_{2}\right)\)
Step-by-Step Solution
Verified Answer
Equilibria are (0,0) and \((\frac{2}{3}, \frac{2}{3})\).
1Step 1: Setting the Differential Equations to Zero
To find the equilibrium points, we need to set the right-hand side of each differential equation to zero. Thus, we have:- For \( \frac{dx_1}{dt} \), set \( -x_1 + 2x_1(1-x_1) = 0 \).- For \( \frac{dx_2}{dt} \), set \( -x_2 + 5x_2(1-x_1-x_2) = 0 \).
2Step 2: Solving for Equilibrium in \(x_1\) Equation
First, solve \( -x_1 + 2x_1 (1-x_1) = 0 \). Rewriting, we have \( x_1 (2 - 3x_1) = 0 \).This gives \( x_1 = 0 \) or \( 2 - 3x_1 = 0 \) which solves to \( x_1 = \frac{2}{3} \).
3Step 3: Solving for Equilibrium in \(x_2\) Equation
Using the equation \( -x_2 + 5x_2(1-x_1-x_2) = 0 \), rewrite it as \( x_2[5(1-x_1-x_2) - 1] = 0 \).So, \( x_2 = 0 \) or \( 5 - 5x_1 - 5x_2 - 1 = 0 \), which simplifies to \( 5x_1 + 5x_2 = 4 \).
4Step 4: Finding Equilibrium Points
From previous steps, combine conditions:1. \( (x_1, x_2) = (0,0) \),2. \( x_1 = \frac{2}{3} \) and plugging into \( 5(\frac{2}{3}) + 5x_2 = 4 \) leading to \( 5x_2 = \frac{2}{3} \) and \( x_2 = \frac{2}{3} \), leading to \( (x_1, x_2) = (\frac{2}{3}, \frac{2}{3}) \).
Key Concepts
Equilibrium PointsStability AnalysisAnalytical Methods
Equilibrium Points
Equilibrium points in a differential equation system occur where the rate of change is zero; that is, where the system is at rest or unchanging. To find these points, we need to solve for when the derivatives with respect to time equal zero. Let's consider the equations:
\(-x_1 + 2x_1(1-x_1) = 0 \) and \(-x_2 + 5x_2(1-x_1-x_2) = 0 \).
By solving these equations, we discern potential equilibrium points where the behaviors of \( x_1 \) and \( x_2 \) are constant. Thus, the system's state is identified by finding the intersections of these solutions, such as \((x_1, x_2) = (0, 0)\) and \((x_1, x_2) = \left(\frac{2}{3}, \frac{2}{3}\right)\).
This understanding forms the base for further analyzing how the system behaves near these points.
- For \( x_1 \): \( \frac{dx_1}{dt} = -x_1 + 2x_1(1-x_1) \)
- For \( x_2 \): \( \frac{dx_2}{dt} = -x_2 + 5x_2(1-x_1-x_2) \)
\(-x_1 + 2x_1(1-x_1) = 0 \) and \(-x_2 + 5x_2(1-x_1-x_2) = 0 \).
By solving these equations, we discern potential equilibrium points where the behaviors of \( x_1 \) and \( x_2 \) are constant. Thus, the system's state is identified by finding the intersections of these solutions, such as \((x_1, x_2) = (0, 0)\) and \((x_1, x_2) = \left(\frac{2}{3}, \frac{2}{3}\right)\).
This understanding forms the base for further analyzing how the system behaves near these points.
Stability Analysis
To decide the nature of equilibrium points, we perform a stability analysis. This analysis helps us determine if a system will remain at equilibrium or diverge when slightly disturbed. The stability of a point informs us whether it is "stable" (the system returns to equilibrium), "unstable" (the system deviates further from equilibrium) or "neutral" (the system stays at new close states).
To perform stability analysis:
If the real parts of all eigenvalues are negative, the equilibrium is stable; if positive, it is unstable. For example, a Jacobian matrix derived from our system indicates the nature of stability of \((x_1, x_2) = (0, 0)\) in contrast to \((x_1, x_2) = \left(\frac{2}{3}, \frac{2}{3}\right)\), aiding in predicting long-term system behaviors.
To perform stability analysis:
- Compute the Jacobian matrix, which consists of partial derivatives of the right-hand side of each differential equation with respect to each variable.
- Evaluate the Jacobian at each equilibrium point.
- Analyze the eigenvalues obtained from the Jacobian.
If the real parts of all eigenvalues are negative, the equilibrium is stable; if positive, it is unstable. For example, a Jacobian matrix derived from our system indicates the nature of stability of \((x_1, x_2) = (0, 0)\) in contrast to \((x_1, x_2) = \left(\frac{2}{3}, \frac{2}{3}\right)\), aiding in predicting long-term system behaviors.
Analytical Methods
Analytical methods offer precise ways to solve and understand differential equations without resorting to numerical approximations. These methods, crucial for finding exact solutions, simplify the process of determining behaviors in complex systems like our given problem.
For this differential equation system, solutions came by systematically algebraically manipulating:
For this differential equation system, solutions came by systematically algebraically manipulating:
- Setting the derivatives to zero to identify equilibrium points, where changes cease.
- Simplifying equations to solve for individual \(x_1\) and \(x_2\) to deduce rest states.
- Analyzing the system’s behavior around these equilibria through linearization and examining the Jacobian matrix.
Other exercises in this chapter
Problem 7
Consider $$ \begin{array}{l} \frac{d x_{1}}{d t}=x_{1}+3 x_{2} \\ \frac{d x_{2}}{d t}=-x_{1}+2 x_{2} \end{array} $$ Determine the direction vectors associated w
View solution Problem 7
Use the eigenvalue approach to analyze all equilibria of the given Lotka- Volterra models of interspecific competition. \(\frac{d N_{1}}{d t}=3 N_{1}\left(1-\fr
View solution Problem 8
Consider $$ \begin{array}{l} \frac{d x_{1}}{d t}=-x_{2} \\ \frac{d x_{2}}{d t}=x_{1}+x_{2} \end{array} $$ Determine the direction vectors associated with the fo
View solution Problem 8
Use the eigenvalue approach to analyze all equilibria of the given Lotka- Volterra models of interspecific competition. \(\frac{d N_{1}}{d t}=4 N_{1}\left(1-\fr
View solution