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Unformatted text preview: 642:527 SOLUTIONS: ASSIGNMENT 8 FALL 2007 Section 7.4: 7 (a) By introducing y = x we can reformulate the given equation as x = y, y = 2 P ml x bracketleftbigg 1 parenleftBig x l parenrightBig 2 bracketrightbigg 1 / 2 k m x. The right hand sides vanish when x = y = 0 so the origin is indeed a critical point. To linearize there we simply drop the factor [1 ( x/l ) 2 ] 1 / 2 in the second equation to obtain z = A z , z = bracketleftbigg x y bracketrightbigg , A = bracketleftbigg 1 2 P/ ( ml ) k/m bracketrightbigg . This system has eigenvalues = radicalbig 2 P/ ( ml ) k/m , so it is a center if P < k/ 2 l and a saddle point if P > k/ 2 l . For P < k/ 2 l the mass will oscillate around its equilibrium poition at the origin, but the saddle point is unstable so for P > k/ 2 l the mass will leave the neighborhood of the origin  this is buckling. FURTHER DISCUSSION (not asked for in problem): In the buckling regime, that is, when P > k/ 2 l , there are two additional singular points, located at x = x c , where x c = l radicalbig 1 ( kl/ 2 P ) 2 . One easily finds that these are centers, so that after the buckling the system can oscillate around these new singular points. There is also a figure 8 trajectory through the saddle point at the origin and trajectories which go around all three critical points. (b) The value k/lm is a bifurcation point for P : when P passes through this value the nature of the phase plane changes. (c) Omitted. Section 7.5: 4 (a) We write x = r cos , y = r sin , and remember that r and are functions of t . Then x = r cos r sin and y = r sin + r cos , and the equations become r cos r sin = r cos + r sin r 3...
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This note was uploaded on 09/29/2009 for the course 642 527 taught by Professor Speer during the Fall '07 term at Rutgers.
 Fall '07
 SPEER

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