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Unformatted text preview: Math 404  Homework 6 Example Student Solutions February 15, 2011 S: 6.1.8 Plot Phase Portraits for the following system (the van der Pol oscilla tor). ˙ x = y ˙ y = x + y (1 x 2 ) Answer First, note that we will have fixed points when ˙ x = 0 and ˙ y = 0. This will happen when: 0 = y 0 = x + 0 x = 0 Therefore, there is only one fixed point and it occurs at (0 , 0). We will thus focus our phase portrait graph around this point. Here we see that there is a closed orbit around the fixed point, and that within that closed orbit the fixed point acts like an unstable spiral. 1 Figure 1: Phase Portrait 2 S: 6.1.9 Plot Phase Portraits for the following system (the dipole fixed point). ˙ x = 2 xy ˙ y = y 2 x 2 Answer First note that we will have fixed points when ˙ x = 0 and ˙ y = 0. This will happen when: 0 = 2 xy and 0 = y 2 x 2 The second equation tells us that y = ± x and thus the first one implies that this only happens at (0 , 0). Therefore, there is only one fixed point, and we will focus our phase portrait around this point. Figure 2: Phase Portrait 3 S: 6.1.10 Plot Phase Portraits for the following system (the twoeyed mon ster). ˙ x = y + y 2 ˙ y = 1 2 x + 1 5 y xy + 6 5 y 2 Answer First note that we will have fixed points when ˙ x = 0 and ˙ y = 0. 0 = y + y 2 factros to give us ( y + 1) y = 0 and finally y = 1 and y = 0 When y = 0 and ˙ y = 0, we know that x = 0. Therefore, there is a fixed point at (0 , 0). When y = 1 and ˙ y = 0: 0 = 1 2 x 1 5 + x + 6 5 1 = 1 2 x x = 2 Therefore, there is another fixed point at ( 2 , 1). We will thus focus the phase portrait around these two points....
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 Spring '08
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 Math, Algebra, Numerical Analysis, phase portrait, Runge–Kutta methods, Numerical ordinary differential equations

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