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\(13-14\) Find the equilibria of the difference equation. Determine the values of \(c\) for which each equilibrium is stable. \(x_{t+1}=\frac{c x_{t}}{1+x_{t}}\)

Short Answer

Expert verified
Equilibria are at \(x^* = 0\) and \(x^* = c - 1\). Stable for \(|c|<1\) and \(|c|>1\), respectively.

Step by step solution

01

Define Equilibrium

An equilibrium of a difference equation is a value \(x^*\) such that \(x_{t+1} = x_t = x^*\). For the given equation, this means solving \(x^* = \frac{c x^*}{1 + x^*}\).
02

Solve for Equilibrium

Rearrange the equation \(x^* = \frac{c x^*}{1 + x^*}\) to \(x^*(1 + x^*) = c x^*\). Assuming \(x^* eq 0\), divide both sides by \(x^*\) to get \(1 + x^* = c\). Then solve for \(x^*\) to find \(x^* = c - 1\). Also consider the case \(x^* = 0\), which trivially satisfies the equilibrium condition.
03

Determine Stability for \(x^* = 0\)

Stability is determined by evaluating the derivative \(f'(x) = \frac{d}{dx}\left(\frac{c x}{1 + x}\right)\). For \(x^* = 0\), compute \(f'(0) = \frac{c}{(1+0)^2} = c\). The equilibrium is stable if \(|f'(x^*)| < 1\). Hence, \(|c| < 1\) means \(x^* = 0\) is stable.
04

Determine Stability for \(x^* = c - 1\)

First, find \(f'(x)\) evaluated at \(x^* = c - 1\): \(f'(x) = \frac{c}{(1 + x)^2}\), thus \(f'(c - 1) = \frac{c}{c^2}\). Simplify to get \(f'(c - 1) = \frac{1}{c}\). The equilibrium is stable if \(|f'(x^*)| < 1\), so \(|1/c| < 1\) or equivalently \(|c| > 1\).

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Key Concepts

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

Stability Analysis
Stability analysis in mathematical models, like difference equations, is crucial to understanding how solutions behave over time. An equilibrium point is considered stable if small perturbations or deviations from this point shrink back to the equilibrium, rather than growing away from it.

To determine stability, we often look at the derivative of the function defining the difference equation. If the absolute value of this derivative at the equilibrium point is less than one, then the equilibrium is stable. This indicates that nearby points will converge to the equilibrium over time.

For example, consider the difference equation from the original problem, where the equilibrium is found by setting the derivative at the equilibrium point and checking if its absolute value is less than one. This approach helps us determine ranges of parameters (here, values of \(c\)) where the system remains stable.
Difference Equation
A difference equation is a mathematical formula that relates a sequence of numbers to its preceding elements. It is a tool used to understand how a quantity changes over discrete steps, often in time.

In our example, the difference equation is given as:
\[ x_{t+1} = \frac{c x_{t}}{1 + x_{t}}. \]
Here, the term \(x_{t+1}\) represents the next step or value in the sequence, expressed as a function of the current value \(x_{t}\).

Difference equations are widely used to model real-world phenomena where changes occur in steps, such as population dynamics or financial markets. By analyzing these equations, we can predict future behavior based on current or past states.
Equilibrium Points
Equilibrium points in the context of difference equations are values that remain constant over time once reached. This means that if the system reaches this value, it stays put in subsequent steps.

To find such equilibria, we set \(x_{t+1} = x_t = x^*\) and solve for \(x^*\). In the given problem, we find equilibria by solving:
\[ x^* = \frac{c x^*}{1 + x^*}. \]
This yields solutions like \(x^* = 0\) and \(x^* = c - 1\), where the choice depends on the constraints or conditions given by the equation

Equilibria help us understand long-term behavior and potential steady states of the system. If a system reaches equilibrium, it will not leave it unless subjected to external forces or changes in the parameters.

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