Unformatted text preview: Systems of ﬁrstorder 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 xline we have
y < 3x or −3x + y < 0 implying ˙
x < 0 so x is falling While to the left of the xline we have
y > 3x or −3x + y > 0 implying ˙
x > 0 so x is rising Similarly, to the right of the yline we have
y< x
or 0 < −3y + x
3 implying ˙
y > 0 so y is rising While to the left of the yline 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.
 Fall '11
 Dr.Gwartney
 Economics

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