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27-dynsys - CONTINUOUS DYNAMICAL SYSTEMS I Homework Section...

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CONTINUOUS DYNAMICAL SYSTEMS I Math 21b, O. Knill Homework: Section 9.1: 4,8,10,26,32,24*,46* until Tuesday CONTINUOUS DYNAMICAL SYSTMES. A differential equation d dt ~x = f ( ~x ) defines a dynamical system. The solu- tions is a curve ~x ( t ) which has the velocity vector f ( ~x ( t )) for all t . One often writes ˙ x instead of d dt x . So, we have the problem that we know a formula for the tangent at each point. The aim is to find a curve ~x ( t ) which starts at a given point ~v = ~x (0). IN ONE DIMENSION. A system ˙ x = g ( x, t ) is the general differential equation in one dimensions. Examples: If ˙ x = g ( t ) , then x ( t ) = R t 0 g ( t ) dt . Example: ˙ x = sin( t ) , x (0) = 0 has the solution x ( t ) = cos( t ) - 1. If ˙ x = h ( x ) , then dx/h ( x ) = dt and so t = R x 0 dx/h ( x ) = H ( x ) so that x ( t ) = H - 1 ( t ). Example: ˙ x = 1 cos( x ) with x (0) = 0 gives dx cos( x ) = dt and after integration sin( x ) = t + C so that x ( t ) = arcsin( t + C ). From x (0) = 0 we get C = π/ 2. If ˙ x = g ( t ) /h ( x ) , then H ( x ) = R x 0 h ( x ) dx = R t 0 g ( t ) dt = G ( t ) so that x ( t ) = H - 1 ( G ( t )). Example: ˙ x = sin( t ) /x 2 , x (0) = 0 gives dxx 2 = sin( t ) dt and after integration x 3 / 3 = - cos( t ) + C so that x ( t ) = (3 C - 3 cos( t )) 1 / 3 . From x (0) = 0 we obtain C = 1. Remarks: 1) In general, we have no closed form solutions in terms of known functions. The solution x ( t ) = R t 0 e - t 2 dt of ˙ x = e - t 2 for example can not be expressed in terms of functions exp , sin , log , · etc but it can be solved using Taylor series: because e - t 2 = 1 - t 2 + t 4 / 2! - t 6 / 3!+ . . . taking coefficient wise the anti-derivatives gives: x ( t ) = t - t 3 / 3+ t 4 / (32!) - t 7 / (73!)+ . . . .
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