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An Introduction toFirst-Order Linear
Difference Equations
With Constant Coefficients
Courtney Brown, Ph.D.
Emory University
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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
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Relation to Interest
yt+1 = yt + ryt = yt(1+r)
r = rate of interest
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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 models behavior
Determine the equilibrium values
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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
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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
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Analytic Solution: Part I
y1 = ay0 + b y2 = ay1 + b
y2 = a(ay0 + b) + b
y2 = a2
y0 + ab + by2 = a
2y0 + (a+1)b
y3 = ay2 + b
y3
= a(a2y0
+ ab + b) + b
y3 = a3y0 + a
2b + ab + b
y3 = a3y0 + b(a
2 + a + 1)
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Part II: Proof by Induction Proves That
yn = any0 + b(1 + a + a
2+ + an-1), then
To find the sum 1 + a + a2+ + an-1 write
S = 1 + a + a2+ + an-1
-aS = -(a + a2+ + an-1 + an)
S aS = 1 an
S(1 a) = 1 a
n
S = (1 an)/(1 a)
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Part III: Thus, the solution for yn is
yn = an
y0 + b(1 an
)/(1 a)or, more conveniently,
yn = b/(1-a) + [y0 b/(1-a)]an, for a1
We like to write it this way because b/(1-a) isthe equilibrium value, y*.
If a=1, then go back to
yn = an
y0 + b(1 + a + a2
+ + an-1
) to obtainyn = y0 + b(n), the solution when a=1
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Note: These solutions for the first-order linear
difference equation with constant coefficientsare used analytically to describe the time
paths with words.
Computing the time paths with a computer isdone using the original equation,
y(t+1) = ay(t) + b
using programming loops.
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a > 1, y0 > b/(1-a)
Increasing, monotonic, unbounded
y(t)
time
y0
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a > 1, y0 < b/(1-a)
Decreasing, monotonic, unbounded
y(t)
time
y0
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0 < a < 1, y0 < b/(1-a)
Increasing, monotonic, bounded, convergent
y(t)
time
y0
y*
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0 < a < 1, y0 > b/(1-a)
Decreasing, monotonic, bounded, convergent
y(t)
time
y0
y*
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-1 < a < 0
bounded, oscillatory, convergent
y(t)
time
y0
y*
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a < -1
unbounded, oscillatory, divergent
y(t)
time
y0
y*
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a = 1, b = 0
constant
y(t)
time
y* and y0
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a = 1, b > 0
constant increasing
y(t)
time
y0
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a = 1, b < 0
constant decreasing
y(t)
time
y0
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a = -1
finite, bounded oscillatory
y(t)
time
y0
y*
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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 oscillatoryAll of this can be deduced from the solution
yn = b/(1-a) + [y0 b/(1-a)]an, for a1
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Special Cases
a = 1, b = 0 constant
a = 1, b > 0 constant increasing
a = 1, b < 0 constant decreasing
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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 thealgebra of the analytical solution to the
equation.
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