Economics Dynamics Problems 168

# Economics Dynamics Problems 168 - 152 Economic Dynamics the...

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Unformatted text preview: 152 Economic Dynamics the forces are directing the system away from the ﬁxed point. We can immediately conclude, therefore, that the ﬁxed point cannot be a stable point. Can we conclude that for any initial value of x and y, positioning the system in quadrants I or III, that the trajectory will tend over time to the ﬁxed point? No, we cannot make any such deduction! For instance, if the system began in quadrant I, and began to move towards the ﬁxed point, it could over time pass into quadrant IV, and once in quadrant IV would move away from the ﬁxed point. In fact, this is precisely the trajectory shown in ﬁgure 4.5. Although the trajectory shown in ﬁgure 4.5 moves from quadrant I into quadrant IV, this need not be true of all initial points beginning in quadrant I. Depending on the initial value for x and y, it is quite possible for the system to move from quadrant I into quadrant II, ﬁrst moving towards the ﬁxed point and then away from it once quadrant II has been entered. This would be the situation, for example, if the initial point was (x0 , y0 ) = (4, 2) (see exercise 2). This complex nature of the solution paths can be observed by considering the direction ﬁeld for the differential equation system. The direction ﬁeld, along with the equilibrium lines are shown in ﬁgure 4.7. Why the dynamic forces seem to operate in this way we shall investigate later in this chapter. Example 4.5 The following system of linear differential equations ˙ x = −3x + y ˙ y = x − 3y with initial condition x0 = 4 and y0 = 5 has solution equations x(t) = 9e2t − 1 2e4t and y(t) = 1 + 9e2t 2e4t which gives rise to a trajectory which approaches the ﬁxed point (x∗ , y∗ ) = (0, 0), as shown in ﬁgure 4.8. The path of x(t) and y(t), represented by the phase line, as t increases is shown by the direction of the arrows. Figure 4.7. ...
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