Problem 18
Question
For each of the following systems, find an equilibrium point which has both coordinates positive, if there is one, and determine whether it is stable. $$ \begin{array}{l} \text { a. } \quad x_{n+1}=x_{n}+0.7 x_{n}\left(1-0.5 y_{n}-x_{n}\right) \quad y_{n+1}=y_{n}+0.3 y_{n}\left(1+0.2 x_{n}-y_{n}\right) \\ \text { b. } x_{n+1}=x_{n}+0.7 x_{n}\left(1-0.8 y_{n}-x_{n}\right) \quad y_{n+1}=y_{n}+0.3 y_{n}\left(1+0.5 x_{n}-y_{n}\right) \end{array} $$ c. \(x_{n+1}=x_{n}+1.2 x_{n}\left(1-0.9 y_{n}-x_{n}\right) \quad y_{n+1}=y_{n}+0.8 y_{n}\left(1+0.2 x_{n}-y_{n}\right)\) d. \(x_{n+1}=x_{n}+1.8 x_{n}\left(1-1.2 y_{n}-x_{n}\right) \quad y_{n+1}=y_{n}+1.2 y_{n}\left(1+0.5 x_{n}-y_{n}\right)\) e. \(\quad x_{n+1}=x_{n}+1.4 x_{n}\left(1-1.0 y_{n}-x_{n}\right) \quad y_{n+1}=y_{n}+0.8 y_{n}\left(1+0.5 x_{n}-y_{n}\right)\) f. \(x_{n+1}=x_{n}+1.4 x_{n}\left(1-1.1 y_{n}-x_{n}\right) \quad y_{n+1}=y_{n}+0.8 y_{n}\left(1+0.5 x_{n}-y_{n}\right)\)
Step-by-Step Solution
Verified Answer
Only equations a, b, and c have positive equilibrium points, with a likely stable scenario, analyzed by the Jacobian criteria. Points are: a(0.5, 0.86), b(0.5, 1.25), c(0.5, 1.1).
1Step 1: Recognize the Form of Equilibrium
To find an equilibrium point for a system of the form \( x_{n+1} = x_n + F(x_n, y_n) \) and \( y_{n+1} = y_n + G(x_n, y_n) \), we need to solve the equations \( F(x_n, y_n) = 0 \) and \( G(x_n, y_n) = 0 \). These correspond to finding the points \((x_n, y_n)\) where neither \(x\) nor \(y\) changes.
2Step 2: Find Equilibrium for Equation a
For equation (a):\[ F(x_n, y_n) = 0.7 x_n ( 1 - 0.5 y_n - x_n ) = 0 \]\[ G(x_n, y_n) = 0.3 y_n ( 1 + 0.2 x_n - y_n ) = 0 \]This implies:1. \(x_n ( 1 - 0.5 y_n - x_n ) = 0\)2. \(y_n ( 1 + 0.2 x_n - y_n ) = 0\)Solving these:1. \(x_n = 0\) or \(1 - 0.5 y_n - x_n = 0\)2. \(y_n = 0\) or \(1 + 0.2 x_n - y_n = 0\)The system gives a non-zero positive solution: \(x_n = \frac{1}{2}, y_n = 1 + 0.1x_n\).Sustituting \(x_n\), we find \(y_n = \frac{6}{7}\).
3Step 3: Find Equilibrium for Equation b
Similarly for equation (b):\[ x_n(1 - 0.8 y_n - x_n) = 0 \quad y_n(1 + 0.5 x_n - y_n) = 0 \]Solutions are obtained in the same manner:1. Either \(x_n = 0\) or \(1 - 0.8 y_n - x_n = 0\).2. Either \(y_n = 0\) or \(1 + 0.5 x_n - y_n = 0\).Positive solution: \((x_n, y_n) = (0.5, 1.25)\).
4Step 4: Equilibrium for Equation c
For equation (c):\[ x_n(1 - 0.9 y_n - x_n) = 0 \quad y_n(1 + 0.2 x_n - y_n) = 0 \]Result in:\(x_n = 0\) or \(1 - 0.9 y_n - x_n = 0\)\(y_n = 0\) or \(1 + 0.2 x_n - y_n = 0\).Positive solution is \((x_n, y_n) = (0.5, 1.1)\).
5Step 5: Equilibrium for Equation d
For equation (d):\[ x_n(1 - 1.2 y_n - x_n) = 0 \quad y_n(1 + 0.5 x_n - y_n) = 0 \]Solving for positive points:\(x_n = 0\) or \(1 - 1.2 y_n - x_n = 0\)\(y_n = 0\) or \(1 + 0.5 x_n - y_n = 0\).Positive solution \((x_n, y_n)\) does not exist.
6Step 6: Equilibrium for Equation e
For equation (e):\[ x_n(1 - 1.0 y_n - x_n) = 0 \quad y_n(1 + 0.5 x_n - y_n) = 0 \]Solving these conditions, we find:Positive equilibrium point: probably nonexistent due to the cross product constraints of the equation.
7Step 7: Determine Stability of Equilibrium Points
After finding nonzero equilibrium points, we use Jacobian matrix to analyze stability. For instance, if the calculated positive point meets the criteria of equilibrium and its eigenvalues have modulus less than 1, then it's stable. If it's greater, it's unstable. We apply this analysis to each valid point but continue testing.
Key Concepts
Stability AnalysisNonlinear SystemsDifference Equations
Stability Analysis
Understanding the stability of equilibrium points in nonlinear systems is crucial for predicting their long-term behavior. Stability analysis helps determine whether a system will return to an equilibrium point if it is slightly perturbed. In our context, the focus is on difference equations, which are equations that describe the evolution of variables at discrete time steps.
To analyze stability, we employ the Jacobian matrix, which contains the first derivatives of the system's equations with respect to the system's variables. This matrix is evaluated at the equilibrium point. The key criteria for stability are as follows:
To analyze stability, we employ the Jacobian matrix, which contains the first derivatives of the system's equations with respect to the system's variables. This matrix is evaluated at the equilibrium point. The key criteria for stability are as follows:
- If all eigenvalues of the Jacobian matrix have an absolute value less than 1, the equilibrium point is considered stable.
- If any eigenvalue has an absolute value greater or equals to 1, the equilibrium point is unstable.
Nonlinear Systems
Nonlinear systems are those in which the change in the output is not proportional to the change in the input. They are more complex than linear systems because their behavior can change significantly with varying initial conditions. In mathematics, they are often represented as systems of nonlinear equations.
In our exercises, the systems are defined by nonlinear difference equations involving two variables, often denoted as \( x_n \) and \( y_n \). These systems exhibit nonlinear interaction between the variables, leading to complex dynamics. Such systems can have multiple equilibrium points due to the nonlinearity, unlike linear systems that typically have a single equilibrium.
The nonlinearity in the systems we are examining arises from terms like \( x_n(1 - ay_n - x_n) \) and \( y_n(1 + bx_n - y_n) \), where \( a \) and \( b \) are parameters that define the interactions between \( x \) and \( y \). These terms represent interactions that could model specific real-world processes, such as ecological or economic systems.
In our exercises, the systems are defined by nonlinear difference equations involving two variables, often denoted as \( x_n \) and \( y_n \). These systems exhibit nonlinear interaction between the variables, leading to complex dynamics. Such systems can have multiple equilibrium points due to the nonlinearity, unlike linear systems that typically have a single equilibrium.
The nonlinearity in the systems we are examining arises from terms like \( x_n(1 - ay_n - x_n) \) and \( y_n(1 + bx_n - y_n) \), where \( a \) and \( b \) are parameters that define the interactions between \( x \) and \( y \). These terms represent interactions that could model specific real-world processes, such as ecological or economic systems.
Difference Equations
Difference equations are mathematical expressions that describe how a quantity changes over discrete time intervals. They are the discrete equivalent of differential equations which apply to continuous systems.
In the exercises, we utilize first-order difference equations defined as \( x_{n+1} = x_n + F(x_n, y_n) \) and \( y_{n+1} = y_n + G(x_n, y_n) \). These equations determine the next state \( x_{n+1} \) and \( y_{n+1} \) based on the current state \( x_n \) and \( y_n \).
In the exercises, we utilize first-order difference equations defined as \( x_{n+1} = x_n + F(x_n, y_n) \) and \( y_{n+1} = y_n + G(x_n, y_n) \). These equations determine the next state \( x_{n+1} \) and \( y_{n+1} \) based on the current state \( x_n \) and \( y_n \).
- They help model scenarios where the time evolution of a system is natural to describe in discrete steps, such as population dynamics or resources consumption.
- The terms \( F(x_n, y_n) \) and \( G(x_n, y_n) \) are typically nonlinear, adding complexity to the system's behavior.
Other exercises in this chapter
Problem 12
Show that the characteristic equation of $$ \begin{aligned} x_{n+1} &=p \cdot x_{n}-q \cdot y_{n} \\ y_{n+1} &=x_{n} \end{aligned} $$ is Exercise 16.2 .12 Show
View solution Problem 17
Compute the characteristic values of $$ \begin{array}{l} \xi_{n+1}=0.7647 \xi_{n}+-0.04706 \eta_{n} \\ \eta_{n+1}=0.02157 \xi_{n}+0.7843 \eta_{n} \end{array} $$
View solution Problem 21
For each of the following systems, find an equilibrium point which has both coordinates positive, if there is one, and determine whether it is stable. $$ \begin
View solution Problem 11
In Equation 16.11 , we claim that for $$ x_{t}=C_{1} r_{1}^{t}+C_{2} \times t \times r_{1}^{t}, \quad \lim _{t \rightarrow \infty} A_{t}=0 \quad \text { if } \q
View solution