hw2sol - Solutions for Homework 2 1. 3.1.1 The graph of f...

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Solutions for Homework 2 1. 3.1.1 The graph of f ( x ) = 1 + rx + x 2 is a parabola, opening upwards. The minimum occurs where f ( x ) = 0, that is, at x 0 = r 2 . At this point, the value of f is f ( x 0 ) = 1 r 2 2 + r 2 4 = 1 r 2 4 . If | r | < 2, then f ( x 0 ) > 0, therefore f ( x ) > 0 for all x , and there are no fixed points. If | r | > 2, then f ( x 0 ) < 0, and therefore there are two fixed points. They are easy to compute: f ( x ) = 0 x = r ± r 2 4 2 . We will write x ± = r ± r 2 4 2 . (1) Since the graph of f is a parabola opening upwards, it must be true that f ( x ) < 0 and f ( x + ) > 0, therefore x is stable and x + is unstable. Let us plot the bifurcation diagram. This requires plotting x and x + as functions of r . We could do that, but life becomes easier if we solve the equation 1 + rx ± + x 2 ± = 0 for r : r = x ± 1 x ± . The function r ( x ) = x 1 x is really easy to plot: 1
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-4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4 x r Now the graph of x as a function of r is obtained simply by reversing x and r in the above plot. Here is what you get when you do that, remembering that the smaller of the two fixed points is stable, and the larger is unstable: -4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4 r x This is the bifurcation diagram. There are two saddle-node bifurcations, one at r = 2 and the other at r = 2. I’ll leave it to you to plot the qualitatively different vector fields. 3.1.3 Let’s first think about the fixed point:
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hw2sol - Solutions for Homework 2 1. 3.1.1 The graph of f...

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