Chapter 1.6, Problem 34E

To determine

To calculate: The value of the equilibria of discrete time dynamical system $x_{t+1}=\frac{x_t}{x_t-K}$ , also the values of parameters when there is no equilibrium, or equilibrium is negative or more than one equilibrium exists.

Answer to Problem 34E

The value of the equilibria of discrete time dynamical system $x_{t+1}=\frac{x_t}{x_t-K}$ are $x^{\ast }=0$ and $x^{\ast }=K+1$ . When $K=-1$ , thenboth the equilibrium points are same and when $K<-1$ , thesecond equilibrium point is negative.

Explanation of Solution

Given information:

The discrete time dynamical system, $x_{t+1}=\frac{x_t}{x_t-K}$ .

Formula used:

The steps to evaluate the equilibrium of discrete time dynamical system are as follows:

1. For discrete time dynamical system, $x_{t+1}=\frac{x_t}{x_t-K}$ the equilibrium is when $x_{t+1}\text{ and }{x}_t$ are replaced by $x^{\ast }$ , here K is a parameter.

2. With help of subtraction transfer all the term on left hand side and keep 0 on right hand side.

3. Factorize the expression so obtained.

4. Equate each factor to 0.

5. Interpret the results obtained.

Calculation:

Consider the provided discrete time dynamical system,

  $x_{t+1}=\frac{x_t}{x_t-K}$

Recall the steps to evaluate the equilibrium of discrete time dynamical system are as follows:

1. For discrete time dynamical system, $x_{t+1}=\frac{x_t}{x_t-K}$ the equilibrium is when $x_{t+1}\text{ and }{x}_t$ are replaced by $x^{\ast }$ , here K is a parameter.

2. With help of subtraction transfer all the term on left hand side and keep 0 on right hand side.

3. Factorize the expression so obtained.

4. Equate each factor to 0.

5. Interpret the results obtained.

Replace $x_{t+1}\text{ and }{x}_t$ by $x^{\ast }$ in the expression $x_{t+1}=\frac{x_t}{x_t-K}$ and solve.

  $x^{\ast }=\frac{x^{\ast }}{x^{\ast }-K}$

Multiply both sides by the denominator $\left[x^{\ast }-K\right]$ ,

  $\begin{array}{c}\left[x^{\ast }-K\right]x^{\ast }=\frac{x^{\ast }}{x^{\ast }-K}\times \left[x^{\ast }-K\right]\\ \left[x^{\ast }-K\right]x^{\ast }=x^{\ast }\end{array}$

Subtract the term $x^{\ast }$ from both the sides,

  $\begin{array}{c}\left[x^{\ast }-K\right]x^{\ast }-x^{\ast }=x^{\ast }-x^{\ast }\\ \left[x^{\ast }-K\right]x^{\ast }-x^{\ast }=0\end{array}$

Factor out the term $x^{\ast }$ ,

  $\begin{array}{c}\left[x^{\ast }-K\right]x^{\ast }-x^{\ast }=x^{\ast }-x^{\ast }\\ \left[x^{\ast }-K\right]x^{\ast }-x^{\ast }=0\\ x^{\ast }\left[x^{\ast }-K-1\right]=0\end{array}$

Equate each term to zero.

Either $x^{\ast }=0$ or $\left[x^{\ast }-K-1\right]=0$ .

Simplify the terms obtained above.

  $x^{\ast }-K-1=0$

Add K on both the sides also add 1 on both the sides,

  $\begin{array}{c}{x}^{\ast }-K-1+K+1=K+1\\ x^{\ast }=K+1\end{array}$

Therefore, either $x^{\ast }=0$ or $x^{\ast }=K+1$ .

Therefore, equilibrium points are $x^{\ast }=0$ and $x^{\ast }=K+1$ .

If $K=-1$ , thenboth the equilibrium points are same.

If $K<-1$ , the equilibrium point is negative.

Thus, the value of the equilibria of discrete time dynamical system $x_{t+1}=\frac{x_t}{x_t-K}$ are $x^{\ast }=0$ and $x^{\ast }=K+1$ . When $K=-1$ , thenboth the equilibrium points are same and when $K<-1$ , thesecond equilibrium point is negative.