Problem 12
Question
Use the stability criterion to characterize the stability of the equilibria of $$ x_{t+1}=\frac{x_{t}}{0.3+x_{t}}, \quad t=0,1,2, \ldots $$
Step-by-Step Solution
Verified Answer
The equilibrium \( x_t = 0 \) is unstable, and \( x_t = 0.7 \) is stable.
1Step 1: Identify Equilibrium Points
Equilibrium points occur when the state doesn't change, i.e., when \( x_{t+1} = x_t \). Thus, set the equation \( \frac{x_t}{0.3 + x_t} = x_t \) and solve for \( x_t \).
2Step 2: Solve for x_t
Rearranging the equation \( \frac{x_t}{0.3 + x_t} = x_t \), we multiply both sides by \( 0.3 + x_t \) to eliminate the fraction: \( x_t = x_t(0.3 + x_t) \). Expanding gives \( x_t = 0.3x_t + x_t^2 \), leading to \( x_t^2 + 0.3x_t - x_t = 0 \), or \( x_t^2 - 0.7x_t = 0 \). Factoring out \( x_t \) results in \( x_t(x_t - 0.7) = 0 \). Therefore, the equilibrium points are \( x_t = 0 \) and \( x_t = 0.7 \).
3Step 3: Linearize the Equation
To assess stability, linearize the recurrence around the equilibria. Consider the function \( f(x) = \frac{x}{0.3 + x} \). The equilibrium stability depends on the derivative value at equilibrium points \( x_e \).
4Step 4: Take the Derivative
Calculate \( f'(x) \) by applying the quotient rule: \( f'(x) = \frac{(0.3 + x)1 - x(1)}{(0.3 + x)^2} = \frac{0.3}{(0.3 + x)^2} \).
5Step 5: Evaluate the Derivative at Equilibria
Evaluate \( f'(x) \) at each equilibrium point. - For \( x_e = 0 \): \( f'(0) = \frac{0.3}{(0.3 + 0)^2} = \frac{0.3}{0.09} = \frac{10}{3} \). - For \( x_e = 0.7 \): \( f'(0.7) = \frac{0.3}{(0.3 + 0.7)^2} = \frac{0.3}{1} = 0.3 \).
6Step 6: Determine Stability
According to stability criterion, an equilibrium is stable if \( |f'(x_e)| < 1 \). Considering:- For \( x_e = 0 \), \( \left| \frac{10}{3} \right| > 1 \) implies the equilibrium is unstable. - For \( x_e = 0.7 \), \( |0.3| < 1 \) implies the equilibrium is stable.
Key Concepts
Equilibrium PointsLinearizationStability CriterionDerivatives in Dynamics
Equilibrium Points
Equilibrium points are fundamental to understanding the dynamics of a system. They occur in scenarios where the system remains unchanged over time. When analyzing a discrete dynamic equation like \[ \frac{x_t}{0.3 + x_t} = x_t \]we find equilibrium points by setting the next state equal to the current state. This means the equation simplifies to \( x_{t+1} = x_t \). In this context, equilibrium points are the values of \( x_t \) that satisfy this condition over time.
Identifying these points involves solving the transformed equation, leading us to find that \( x_t = 0 \) and \( x_t = 0.7 \) are equilibria. These points reveal critical insights into the behavior of the system, serving as anchors where the dynamics could potentially stabilize or diverge from.
Identifying these points involves solving the transformed equation, leading us to find that \( x_t = 0 \) and \( x_t = 0.7 \) are equilibria. These points reveal critical insights into the behavior of the system, serving as anchors where the dynamics could potentially stabilize or diverge from.
Linearization
Linearization allows us to simplify complex nonlinear dynamic equations, making them easier to analyze. By focusing on small perturbations around equilibrium points, the nonlinear function can be approximated as a linear one. This approximation involves calculating the derivative of the function at the equilibrium and using it as a linear function's slope.
Take the function \( f(x) = \frac{x}{0.3+x} \) and find its derivative. In our example, by applying the quotient rule, we find that \[ f'(x) = \frac{0.3}{(0.3 + x)^2} \].
This derivative serves as a key ingredient in our linear approximation. By evaluating it at the equilibrium points, we can investigate how small deviations will evolve, providing us with insight into the stability behavior around those points.
Take the function \( f(x) = \frac{x}{0.3+x} \) and find its derivative. In our example, by applying the quotient rule, we find that \[ f'(x) = \frac{0.3}{(0.3 + x)^2} \].
This derivative serves as a key ingredient in our linear approximation. By evaluating it at the equilibrium points, we can investigate how small deviations will evolve, providing us with insight into the stability behavior around those points.
Stability Criterion
The stability criterion is a tool used to determine whether an equilibrium point will attract or repel nearby trajectories. The primary rule is simple: an equilibrium is stable if the magnitude of the derivative at that point is less than one. This is summarized mathematically as \[ |f'(x_e)| < 1 \].
In our system, we found two equilibria: \( x_e = 0 \) and \( x_e = 0.7 \). By evaluating the derivative - At \( x_e = 0 \), we found \( f'(0) = \frac{10}{3} \), which is greater than one. Thus, this point is unstable.
- At \( x_e = 0.7 \), with \( f'(0.7) = 0.3 \), which is less than one, suggesting that this equilibrium point is stable.
This rule allows us to predict the system's future behavior by only examining small perturbations around equilibria, simplifying the complexity of our original nonlinear system.
In our system, we found two equilibria: \( x_e = 0 \) and \( x_e = 0.7 \). By evaluating the derivative - At \( x_e = 0 \), we found \( f'(0) = \frac{10}{3} \), which is greater than one. Thus, this point is unstable.
- At \( x_e = 0.7 \), with \( f'(0.7) = 0.3 \), which is less than one, suggesting that this equilibrium point is stable.
This rule allows us to predict the system's future behavior by only examining small perturbations around equilibria, simplifying the complexity of our original nonlinear system.
Derivatives in Dynamics
Derivatives play a crucial role in understanding the dynamic behavior of systems by providing a glimpse into the rate of change. They tell us how a slight change in one variable affects another. In the context of stability analysis, derivatives help linearize a nonlinear system, making it easier to analyze by determining the sensitivity of the system at specific points.
In our example, the derivative \( f'(x) \) calculated was \[ \frac{0.3}{(0.3 + x)^2} \].
By evaluating this derivative at equilibrium points, we obtain crucial insights into the system's responsiveness. It enables us to employ the stability criterion to determine if an equilibrium is stable. This approach bridges the gap between intricate nonlinear systems and straightforward linear analysis, revealing underlying dynamics effectively.
In our example, the derivative \( f'(x) \) calculated was \[ \frac{0.3}{(0.3 + x)^2} \].
By evaluating this derivative at equilibrium points, we obtain crucial insights into the system's responsiveness. It enables us to employ the stability criterion to determine if an equilibrium is stable. This approach bridges the gap between intricate nonlinear systems and straightforward linear analysis, revealing underlying dynamics effectively.
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