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set1 - Copy (2) - TFY4305 solutions exercise set 1 2011...

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TFY4305 solutions exercise set 1 2011 Problem 2.2.3 The equation reads ˙ x = x (1 - x 2 ) , (1) The fixed points are the solutions to x (1 - x 2 ) = 0. This yields x = 0 , x = ± 1 . (2) Furthermore, f ( x ) = 1 - 3 x 2 Since f ( x = 0) = 1, x = 0 is an unstable fixed point . Since f ( x = ± 1) = - 1, x = ± 1 are stable fixed points . Greater Greater Less SolidCircle SolidCircle Circle Less Minus 1.5 Minus 1.0 Minus 0.5 0.5 1.0 1.5 x Minus 2 Minus 1 1 2 f LParen1 x RParen1 Figure 1: Vector field including the three fixed points. The exact solution can be found by separation of variables. dx x (1 - x 2 ) = dt , (3) The left-hand side can be rewritten using partial fractional decomposition. integraldisplay dx bracketleftbigg 1 x - 1 2 1 x - 1 - 1 2 1 x + 1 bracketrightbigg = integraldisplay dt . (4) 1
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2 Integration yields 1 2 ln vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle x 2 x 2 - 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = t + C , (5) where C is an integration constant. Exponentiating gives ± x 2 x 2 - 1 = Ke 2 t , (6) where K = e 2 C . This is quadratic equation for x and can be easily solved. For the upper sign, we obtain x ( t ) = ± Ke t Ke 2 t - 1 . (7) For the lower sign, we find x ( t ) = ± Ke t e 2 t + 1 . (8) Note that the solutions tend to x = ± 1 as t → ∞ . In the first case, we have x
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