c39 - 18.03 Class 39, May 17 Dangers of linearization;...

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18.03 Class 39, May 17 Dangers of linearization; limit cycles; chaos; next steps in mathematics [1] The method we sketched on Monday works well "generically," i.e. almost all the time. Things get tricky if the trace and determinant of the Jacobian matrix lie on one of the dividing curves in the (tr,det) plane. If tr = 0 and det > 0 , the linearization gives centers, ellipses, periodic trajectories. But the actual trajectories of the nonlinear system may not be periodic. They may be densely packed spirals. These spirals will rotate in the same direction as the predicted ellipse, but they may be either stable or unstable. It's as if the nonlinearity jostles the linear phase portrait off onto one of the regions bounded by the tr = 0 line. Similarly, if det = (tr/2)^2 > 0 , the linear phase plane is a star or a defective node. The actual phase plane near the equilibrium may be more like a spiral, or more like a node. None of this is too bad, but there is one case where the actual phase portrait may not resemble the linear one at all. This is when the Jacobian matrix is degenerate, i.e. has determinant zero, i.e. has at least one eigenvalue which is zero. For example, x' y' = = 4xy x^2 - y^2 x' = 0 if either x = 0 or y = 0 . y' = 0 So the only critical point is at (0,0). if x = +- y . J(x,y) = [ f_x , f_y ; g_x , g_y ] = [ 4y , 4x ; 2x ; -2y ] so J(0,0) = [ 0 0 ; 0 0 ] What is the phase portrait of this linear system? It just says u' = 0 so every point is a constant solution. The eigenvalues are both zero, every vector is an eigenvector, it is a degenerate system. On the other hand, I showed a Matlab plot of the phase portrait: it has
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This note was uploaded on 01/31/2011 for the course MAT 17A taught by Professor Staff during the Winter '08 term at UC Davis.

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c39 - 18.03 Class 39, May 17 Dangers of linearization;...

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