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An Introduction to First-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 y t+1 = y t + ry t = y t (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 model’s 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 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)
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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)
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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
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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.
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a > 1, y 0 > b/(1-a) Increasing, monotonic, unbounded y(t) time y0y0
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a > 1, y 0 < b/(1-a) Decreasing, monotonic, unbounded y(t) time y0y0
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0 < a < 1, y 0 < b/(1-a) Increasing, monotonic, bounded, convergent y(t) time y0y0 y*
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0 b/(1-a) Decreasing, monotonic, bounded, convergent y(t) time y0y0 y*
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-1 < a < 0 bounded, oscillatory, convergent y(t) time y0y0 y*
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a < -1 unbounded, oscillatory, divergent y(t) time y0y0 y*
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a = 1, b = 0 constant y(t) time y* and y 0
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a = 1, b > 0 constant increasing y(t) time y0y0
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a = 1, b < 0 constant decreasing y(t) time y0y0
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a = -1 finite, bounded oscillatory y(t) time y0y0 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 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
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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 the algebra of the analytical solution to the equation.
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