Unformatted text preview: Discrete dynamic systems
∗
Take ﬁrst x1 = 1, then
∗
f (x1 ) = 1
∗
∗
f (x1 ) = 2(4x1 − 3)−1/2 = 2 Hence
xt = 1 + 2(xt−1 − 1) = −1 + 2xt−1
∗
which is unstable since f (x1 ) = 2 > 1.
∗
Next consider x2 = 3
∗
f (x2 ) = 3
∗
∗
f (x2 ) = 2(4x2 − 3)−1/2 = 2
3 Hence
xt = 3 + 2
(xt−1 − 3)
3 2
= 1 + xt−1
3
∗
which is stable since f (x2 ) = 2/3 < 1. Example 3.19
yt+1 = f ( yt ) = 3.2yt − 0.8y2
t
Letting yt = y∗ for all t we can readily establish two equilibria: y∗ = 0 and y∗ =
1
2
2.75. Considering the nonzero equilibrium, then
f ( y∗ ) = 2.75
2
f ( y∗ ) = 3.2 − 1.6y∗ = −1.2
2
2
Hence, the linear approximation is
yt+1 = 2.75 − 1.2( yt − 2.75)
= 6.05 − 1.2yt
The situation is shown in ﬁgure 3.20. The solution to this model is
yt+1 = 2.75 + (−1.2)t ( y0 − 2.75)
which is oscillatory and explosive.
Although the linear approximation leads to an explosive oscillatory equilibrium,
the system in its nonlinear form exhibits a twocycle with values 2.0522 and
3.1978.12 What the linear approximation reveals is the movement away from y∗ =
2.75. What it cannot show is that it will converge on a twocycle. This example,
therefore, illustrates the care required in interpreting the stability of nonlinear
difference equations using their linear approximations.
12 This can be established quite readily with a spreadsheet or as explained in appendices 3.1 and 3.2. 129 ...
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 Fall '11
 Dr.Gwartney
 Economics, Equilibrium, Numerical Analysis, Approximation, Linear Approximation, DISCRETE DYNAMIC SYSTEMS, twocycle

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