Economics Dynamics Problems 169

Economics Dynamics Problems 169 - Systems of first-order...

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Unformatted text preview: Systems of first-order differential equations 153 Figure 4.8. No matter what the initial point, it will be found that each trajectory approaches the fixed point (x∗ , y∗ ) = (0, 0). In other words, the fixed point (the equilibrium point) is globally stable. Considering the vector of forces for this system captures this feature. The equilibrium solution lines are ˙ y = 3x for x = 0 x ˙ y = for y = 0 3 with a fixed point at the origin. To the right of the x-line we have y < 3x or −3x + y < 0 implying ˙ x < 0 so x is falling While to the left of the x-line we have y > 3x or −3x + y > 0 implying ˙ x > 0 so x is rising Similarly, to the right of the y-line we have y< x or 0 < −3y + x 3 implying ˙ y > 0 so y is rising While to the left of the y-line we have y> x or 0 > −3y + x 3 implying ˙ y < 0 so y is falling All this information, including the vectors of force implied by the above results, is illustrated in figure 4.9. It is clear that no matter in which of the four quadrants the system begins, all the forces push the system towards the fixed point. This means that even if the trajectory crosses from one quadrant into another, it is still being directed towards the fixed point. The fixed point must be globally stable. If the initial point is (x0 , y0 ) = (4, 5), then the trajectory remains in quadrant I and tends to the fixed point (x∗ , y∗ ) = (0, 0) over time. However, figure 4.9 reveals much more. If the system should pass from one quadrant into an adjacent quadrant, then the trajectory is still being directed towards the fixed point, but the movement ...
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This note was uploaded on 12/14/2011 for the course ECO 3023 taught by Professor Dr.gwartney during the Fall '11 term at FSU.

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