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Chapter 5: Problem 12

Find the equilibria of $$x_{t+1}=\frac{x_{t}}{0.3+x_{t}}, \quad t=0,1,2, \ldots$$ and use the stability criterion for an equilibrium point to determine whether they are stable or unstable.

Short Answer

Expert verified
Equilibrium points are \( x_e = 0 \) (unstable) and \( x_e = 0.7 \) (stable).

Step by step solution

01

Define Equilibrium

An equilibrium point is a value of \( x \) such that the sequence doesn't change, i.e., \( x_{t+1} = x_t = x_e \). In other words, we solve \( x_e = \frac{x_e}{0.3 + x_e} \) for \( x_e \).
02

Solve for Equilibrium

Substitute \( x_e \) into the given equation: \( x_e = \frac{x_e}{0.3 + x_e} \). Multiply both sides by \( 0.3 + x_e \) to get \( x_e (0.3 + x_e) = x_e \), which simplifies to \( 0.3 x_e + x_e^2 = x_e \). Subtract \( x_e \) from both sides to form the equation \( x_e^2 - 0.7x_e = 0 \).
03

Factor the Equation

Factor the equation \( x_e^2 - 0.7x_e = 0 \) as \( x_e(x_e - 0.7) = 0 \). This gives us two potential equilibria: \( x_e = 0 \) or \( x_e = 0.7 \).
04

Determine Stability Criterion

The stability of an equilibrium point \( x_e \) is determined by the derivative of the right-hand side of the difference equation, \( f(x) = \frac{x}{0.3 + x} \). Calculate \( f'(x) \) and evaluate at each equilibrium point. If \( |f'(x_e)| < 1 \), the equilibrium is stable; if \( |f'(x_e)| > 1 \), it is unstable.
05

Compute Derivative and Evaluate Stability at \( x_e = 0 \)

The derivative of \( f(x) \) is \( f'(x) = \frac{0.3}{(0.3 + x)^2} \). At \( x_e = 0 \), \( f'(0) = \frac{0.3}{0.3^2} = \frac{0.3}{0.09} = 3.33 \), indicating that the equilibrium \( x_e = 0 \) is unstable since \( |3.33| > 1 \).
06

Evaluate Stability at \( x_e = 0.7 \)

Evaluate \( f'(x) \) at \( x_e = 0.7 \): \( f'(0.7) = \frac{0.3}{(0.3 + 0.7)^2} = \frac{0.3}{1} = 0.3 \). Since \( |0.3| < 1 \), the equilibrium \( x_e = 0.7 \) is stable.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stability Analysis
Stability analysis is a fundamental tool used to determine the behavior of dynamic systems over time. In the context of difference equations, it helps us understand how small changes in initial conditions can affect long-term behavior.
Stability is often determined by examining the derivative of the system's function at equilibrium points. This involves evaluating the change or "slope" of the function, providing insight into whether small perturbations will return to equilibrium or diverge away.
  • If the absolute value of this derivative is less than one, the equilibrium is stable, indicating that the system will return to this point even if slightly disturbed.
  • If the absolute value is greater than one, the equilibrium is unstable, meaning that even small disturbances will cause the system to move away from this point.
Understanding stability is crucial for predicting the behavior of systems described by equations, like population models, economic forecasts, or physical phenomena.
Difference Equations
Difference equations are discrete analogs of differential equations, often used to model scenarios where changes occur at specific intervals. They provide a way to describe the evolution of quantities in a step-by-step manner, which is particularly useful in digital or time-stepped models.
In mathematical modeling, difference equations set the scene for the dynamics of a system. For example, given the difference equation \(x_{t+1} = \frac{x_{t}}{0.3 + x_{t}}\), we can analyze the behavior of the variable \(x\) as it progresses over time. Each "t" represents a specific point in time, such as days, months, or years.
  • Example fields where difference equations are used include population dynamics, where they predict how a population size will change based on factors like birth and death rates.
  • Financial models may also use them to forecast stock or investment values over time.
Equilibria Stability
Equilibria stability refers to the behavior of distinct points or states, known as equilibria, where a system tends to settle over time if undisturbed. In mathematical modeling, identifying and understanding these points is key to predicting long-term dynamics.
When dealing with the given function \(x_{t+1} = \frac{x_{t}}{0.3 + x_{t}}\), we find the equilibria by ensuring that the system remains constant, i.e., \(x_{t+1} = x_t = x_e\). Solving this equation reveals potential equilibrium values such as \(x_e = 0\) and \(x_e = 0.7\).
  • The system is stable at \(x_e = 0.7\) as small deviations will gravitate the system back to \(0.7\).
  • Conversely, for \(x_e = 0\), the system is unstable, meaning any small deviation will result in a move away from zero.
This analysis provides insights into the stability of populations, chemical concentrations, or mechanical systems.
Mathematical Modeling
Mathematical modeling is the process of creating equations and formulas to represent real-world systems. It transforms complex phenomena into understandable mathematical forms, making it easier to predict and analyze outcomes.
In the given example, \(x_{t+1} = \frac{x_{t}}{0.3 + x_{t}}\) models a scenario where current state influences the next state in a system. This could mirror real-life situations, such as the spread of diseases or ecological dynamics.
  • Effective modeling requires understanding the key factors of a system and translating them into mathematical terms.
  • It aids in making projections and testing interventions or changes in controlled environments before applying them in the real world.
Overall, mathematical modeling bridges abstract math with practical applications, providing essential tools for analyzing and solving complex problems across diverse fields.

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Most popular questions from this chapter

Hill's equation for the oxygen saturation of blood states that the level of oxygen saturation (fraction of hemoglobin molecules that are bound to oxygen) in blood can be represented by a function: $$ f(P)=\frac{P^{n}}{P^{n}+30^{n}} $$ where \(P\) is the oxygen concentration around the blood \((P \geq 0)\) and \(n\) is a parameter that varies between different species. (a) Assume that \(n=1\). Show that \(f(P)\) is an increasing function of \(P\) and that \(f(P) \rightarrow 1\) as \(P \rightarrow \infty\). (b) Assuming that \(n=1\) show that \(f(P)\) has no inflection points. Is it concave up or concave down everywhere? (c) Knowing that \(f(P)\) has no inflection points, could you deduce which way the curve bends (whether it is concave up or concave down) without calculating \(f^{\prime \prime}(P) ?\) (d) For most mammals \(n\) is close to 3. Assuming that \(n=3\) show that \(f(P)\) is an increasing function of \(P\) and that \(f(P) \rightarrow 1\) as \(P \rightarrow \infty\) (e) Assuming that \(n=3\), show that \(f(P)\) has an inflection point, and that it goes from concave up to concave down at this inflection point. (f) Using a graphing calculator plot \(f(P)\) for \(n=1\) and \(n=3\). How do the two curves look different?

Use the Newton-Raphson method to find a numerical approximation to the solution of $$ e^{x}=4 x $$ in the interval \((2,3)\) correct to six decimal places.

Find the general solution of the differential equation. $$ \frac{d y}{d x}=x(1+x), x>0 $$

Fish are indeterminate growers; that is, their length \(L(t)\) increases with age \(t\) throughout their lifetime. If we plot the growth rate \(d L / d t\) versus age \(t\) on semilog paper, a straight line with negative slope results, meaning that: $$ \frac{d L}{d t}=A e^{-k t} $$ where \(A>0\) and \(k>0\) are both coefficients that depend on the species of fish, and the habitat that it is growing in. (a) Find the solution for this differential equation (your solution will include \(A\) and \(k\) as unknown constants, as well as one additional unknown constant \(C\) from the antiderivative). (b) Find the values for the constants \(A, k, C\), that would fit the solution to the following data \(L(0)=5, L(1)=10\), and $$ \lim _{t \rightarrow \infty} L(t)=30. $$ (c) Graph the solution \(L(t)\) as a function of \(t\).

(a) Use the stability criterion to characterize the stability of the equilibria of $$x_{t+1}=\frac{5 x_{t}^{2}}{4+x_{t}^{2}}, \quad t=0,1,2, \ldots$$ (b) Use cobwebbing to find the limit that \(x_{t}\) converges to as \(t \rightarrow \infty\) if (i) \(x_{0}=0.5\) and (ii) \(x_{0}=2\).
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