An Introduction to First-Order Linear Difference Equations With Constant Coefficients Courtney Brown, Ph.D. Emory University

Definition of a Difference Equation y(t+1) = ay(t) + b y(t+1) = Some constant of proportionality times y(t) plus some constant. Some interesting cases are  y(t+1) = ay(t) … exponential growth  y(t+1) = b … a horizontal line  y(t+1) = y(t) + b … a straight sloping line

Relation to Interest y t+1 = y t + ry t = y t (1+r) r = rate of interest

“To Do” List for Difference Equations Plot the equation over time Get an analytical solution for the difference equation if it is available Describe the model’s behavior Determine the equilibrium values

Equilibrium Values At equilibrium, y(t+1) = y(t) = y* y* = ay* + b y* - ay* = b y*(1-a) = b y* = b/(1-a) = the equilibrium value

Stability Criteria y(t) will be stable if |a| < 1 y(t) will be unstable if |a| > 1 y(t) will oscillate if a < 0 y(t) will change monotonically if a > 0 y(t) will converge to a stable equilibrium value if |a| < 1

Analytic Solution: Part I y 1 = ay 0 + b y 2 = ay 1 + b y 2 = a(ay 0 + b) + b y 2 = a 2 y 0 + ab + b y 2 = a 2 y 0 + (a+1)b y 3 = ay 2 + b y 3 = a(a 2 y 0 + ab + b) + b y 3 = a 3 y 0 + a 2 b + ab + b y 3 = a 3 y 0 + b(a 2 + a + 1)

Part II: Proof by Induction Proves That y n = a n y 0 + b(1 + a + a 2 + … + a n-1 ), then … To find the sum 1 + a + a 2 + … + a n-1 write S = 1 + a + a 2 + … + a n-1 -aS = -(a + a 2 + … + a n-1 + a n ) S – aS = 1 – a n S(1 – a) = 1 – a n S = (1 – a n )/(1 – a)

Part III: Thus, the solution for y n is y n = a n y 0 + b(1 – a n )/(1 – a) or, more conveniently, y n = b/(1-a) + [y 0 – b/(1-a)]a n, for a≠1 We like to write it this way because b/(1-a) is the equilibrium value, y*. If a=1, then go back to y n = a n y 0 + b(1 + a + a 2 + … + a n-1 ) to obtain y n = y 0 + b(n), the solution when a=1

Note: These solutions for the first-order linear difference equation with constant coefficients are used analytically to describe the time paths with words. Computing the time paths with a computer is done using the original equation, y(t+1) = ay(t) + b using programming loops.

a > 1, y 0 > b/(1-a) Increasing, monotonic, unbounded y(t) time y0y0

a > 1, y 0 < b/(1-a) Decreasing, monotonic, unbounded y(t) time y0y0

0 < a < 1, y 0 < b/(1-a) Increasing, monotonic, bounded, convergent y(t) time y0y0 y*

0 b/(1-a) Decreasing, monotonic, bounded, convergent y(t) time y0y0 y*

-1 < a < 0 bounded, oscillatory, convergent y(t) time y0y0 y*

a < -1 unbounded, oscillatory, divergent y(t) time y0y0 y*

a = 1, b = 0 constant y(t) time y* and y 0

a = 1, b > 0 constant increasing y(t) time y0y0

a = 1, b < 0 constant decreasing y(t) time y0y0

a = -1 finite, bounded oscillatory y(t) time y0y0 y*

Rules of Interpretation |a| > 1 unbounded [repelled from line b/(1-a)] |a| < 1 bounded [attracted or convergent to [b/(1-a)] a < 0 oscillatory a > 0 monotonic a = -1 bounded oscillatory All of this can be deduced from the solution y n = b/(1-a) + [y 0 – b/(1-a)]a n, for a≠1

Special Cases a = 1, b = 0 constant a = 1, b > 0 constant increasing a = 1, b < 0 constant decreasing

Words Are Important The words used to describe the behaviors of the first-order linear difference equation with constant coefficients are used in your text. These behaviors can all be deduced from the algebra of the analytical solution to the equation.