< Previous | Next | Contents >

Numerical Example of Affine Difference Equation

We have difference equation y(k+1) = 2 y(k) 5, what is the solution for initial value y(0) = 1 and what is the equilibrium value? How is the stability of this equilibrium value?

Solution

k = 0: y(0) = 1 initial value

k = 1: y(1) = 2 y(0) 5 = -3

k = 2: y(2) = 2 y(1) 5 = -11

= 2 ( 2 y(0)) 5) 5 = 4 y(0) -15

k = 3: y(3) = 2 y(2) 5 = -27

= 2 (4 y(0) -15) 5 = 8 y(0) -35

and so on. For big k value, it is easier to use the

Closed-form Solution

Rather than recursive calculation:

a = 2, b = - 5, k = 10 then Difference Equation Tutorial: Numerical Example of Affine DE

Equilibrium Value

Difference Equation Tutorial: Numerical Example of Affine DE or Difference Equation Tutorial: Numerical Example of Affine DE

If Difference Equation Tutorial: Numerical Example of Affine DE = 5 then y(k) = 2*5 - 5 = 5 forever. It remain constant at 5 regardless whatever the value of k.

Difference equation y(k+1) = 2 y(k) 5 has solution of Difference Equation Tutorial: Numerical Example of Affine DE and equilibrium value of Difference Equation Tutorial: Numerical Example of Affine DE .

Stability:

Since a = 2 > 1, the equilibrium is unstable. If the initial value y(0) is smaller than the equilibrium value, the graph of y(k) versus k is exponentially decreasing without bound. If the initial value of y(0) is bigger than the equilibrium value, the graph is exponentially increasing without bound.

Difference Equation Tutorial: Numerical Example of Affine DE Difference Equation Tutorial: Numerical Example of Affine DE


< Previous | Next | Contents >

Preferable reference for this tutorial is

Teknomo, Kardi (2015) Difference Equation Tutorial. http:\\people.revoledu.com\kardi\ tutorial\DifferenceEquation\