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

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Unformatted text preview: 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 SolidCircle Circle 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 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...
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This note was uploaded on 02/20/2012 for the course TFY 4305 taught by Professor Jenso.andersen during the Fall '11 term at Norwegian Univ. of Science & Technology.

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

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