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**Unformatted text preview: **Non-Linear Dynamics Homework Solutions Week 8 Chris Small March 10, 2007 Please email me at [email protected] with any questions or concerns reguarding these solutions. 8.1.1 For the following prototypical examples, plot the phase portrait as μ varies. a) ˙ x = μx − x 2 ˙ y = − y In order to do this we first find the nullclines. The x-nullclines are given by x = 0 and x = μ while the y-nullcline is given by y = 0. Our fixed points occur at intersection between x and y nullclines, giving us the origin and ( μ, 0) as our fixed points. Next, we plot the nullclines and then determine the directions of the flow along the nullclines by considering various values for each region. From this point we can get a pretty good idea of what kind of fixed points we have and how trajectories are going to behave. We could also find the Jacobian matrices to do much of this, by finding eigenvalues and stabilities at the fixed points. See Figure 1 for phase portraits for μ less than, equal to and greater than zero are given. b) ˙ x = μx − x 3 ˙ y = − y Figure 1: Phase Portraits for μ < 0, μ = 0 and μ > 0, from left to right 1 Figure 2: Phase Portraits for μ < 0, μ = 0 and μ > 0, from left to right As in part (a), we find our x-nullclines to be given by x = 0 and x = ± √ − μ and our y-nullcline to be y = 0. From this we conclude that our bifurcation occurs at μ = 0, so we show phase portraits for values of μ to either side of μ = 0 (see Figure 2). As above, to make these plots, first graph the nullclines, then determine the flow aong them, then fill in the rest of the portrait around them in, using the linearization of the fixed points to determine eigenvectors and stabilities if neccessary. 8.1.7 Find and classify all bifurcations of the system ˙ x = y − ax ˙ y = − by + x 1 + x As with most problems in this subject, we begin by finding our x-nullcline to be given by y = ax and our y-nullcline to be given by y = x/ ( b (1 + x )). The intersectios between these are the origin and the point (1 /ab − 1 , 1 /b − a ). Next, we compute the Jacobian to be J = parenleftbigg − a 1 1 (1+ x ) 2 − b parenrightbigg For the origin, T = − ( a + b ) and D = ab − 1 and we find that T = − ( a + b ) and D = ab − ( ab ) 2 for the other fixed point. Note that when either of a or b is zero, the origin becomes the only fixed point, but not through a collision, but through a process by which the second fixed point explodes to infinity in an ω- diamondsolid ski slope bifurcation (see an earlier homework assigment). To find our more traditional kinds of bifurcations, we think about these fixed points and what sort of trace determinant transitions we can have. Doing this we find that if we increase the value of ab from just below 1 to just above, the sign of D for the origin changes. Note also, that when ab = 1, our second fixed point merges with the origin. As it passes by the origin changes stability. Thus we conclude that along the curvechanges stability....

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