ECH259Final04Soln

Then the ode that vt satisfies 70 v ååååååå t

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Then the oDE that v(t) satisfies (70) v ÅÅÅÅÅÅÅ t = F H R + v L = F H R L + J F ÅÅÅÅÅÅÅ r N R v For this problem (71) F H r, m L = - r H r 2 - b r + m L Hence we get (72) J F ÅÅÅÅÅÅÅ r N R = - 3 R 2 + 2 b R - m Consider the case R=0. Then ECH259Final04Soln.nb 9

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(73) J F ÅÅÅÅÅÅÅ r N R = 0 = -m Hence (74) v H t L = V H 0 L Exp @ -m t D Thus we see that the solution with zero amplitude H U 1 = U 2 = 0 L is stable if m > 0 , and unstable if m < 0 . Let us next consider the stability of the second family (75) R = b ÅÅÅÅ 2 \$%%%%%%%%%%%%%%% b 2 ÅÅÅÅÅÅ 4 - m We can solve for m to get (76) m = b R - R 2 We can use this result to eliminate m from (71) so that (77) J F ÅÅÅÅÅÅÅ r N R = - 3 R 2 + 2 b R - b R + R 2 = - R H 2 R - b L Since the stability of this steady solution is determined by the sign of F/ R, we can conclude that the solution is stable when 2 R - b > 0 , or R > b ê 2 . Thus the steady solution is unstable when R < b ê 2 . From the bifurcation diagram we can conclude that as the student decreased m the point is reached when m = m c = 0 at which point a Hopf bifurcation occurs. This is a periodic solution with amplitude R = b . With further decrease in m the ampli- tude decreases. Now if m <0 and we start to then increase m we reach a point at which m = b 2 ê 4 the periodic solution vanishes and we get R=0 . Thus the behavior with respect to m is a hysteresis loop ECH259: Final1/2001 Page 10
• Fall '10
• BrianHiggin
• Trigraph, fn HxL, ÅÅÅÅÅÅÅÅÅ, tr HJL

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