Computer Simulation and Applications
9 September 2008 today’s menu: extremely simple models
China Central Television, Beijing There are also models to study effects of earth quakes on this building Models to study effects of winds/storms
Computer Simulation and Applications 9 September 2008 todays menu: - - PowerPoint PPT Presentation
Computer Simulation and Applications 9 September 2008 todays menu: extremely simple models Models to study effects of winds/storms China Central Television, Beijing There are also models to study effects of earth quakes on this
9 September 2008 today’s menu: extremely simple models
China Central Television, Beijing There are also models to study effects of earth quakes on this building Models to study effects of winds/storms
Poincaré in ‘Science and Method’ (1903): “If we knew exactly the laws of nature and the situation of the universe at the initial moment, we could predict exactly the situation of that same universe at a succeeding moment. But even if it were the case that the natural laws had no longer any secret for us, we would still only know the initial situation approximately. If that enabled us to predict the succeeding situation with the same approximation, that is all we require, and we should say that the phenomenon had been predicted, that it is governed by laws. But it is not always so; it may happen that small differences in the initial conditions produce very great ones in the final phenomena. A small error in the former will produce an enormous error in the latter. Prediction becomes impossible, and we have the fortuitous phenomenon. “
a) Discretization of differential equations b) Poincaré map c) Functions seen as models for describing population levels which change at discrete times (as in our example of bacteria growth) d) Newton-Raphson method for solving equations f(x) =0 e) And many more ....
k times Define: fk (x0) := f ( f (f (x0))), k times Orbit of x0
Steady state
k f(x)=2x f(x)=2x(1-x) f(x)=2x(1-x) 0,01 0,01 0,9 1 0,02 0,0198 0,18 2 0,04 0,03881592 0,2952 3 0,08 0,074618489 0,41611392 4 0,16 0,13810114 0,485926251 5 0,32 0,23805843 0,499603859 6 0,64 0,362773228 0,499999686 7 1,28 0,462337626 0,5 8 2,56 0,497163091 0,5 9 5,12 0,499983904 0,5 10 10,24 0,499999999 0,5 11 20,48 0,5 0,5 12 40,96 0,5 0,5
x0 = 0.375 The line y=x, the diagonal, is an auxiliary “hulplijn” etcetera WHY do you get orbit? In words: Go from xo to the graph of the function, from the graph to the diagonal, from the diagonal to the graph, etc., etc. f(x) = 2x
x0 = 0.375 (xo, x1)
x0 = 0.375 (xo, x1) (?, x1)
x0 = 0.375 (xo, x1) (?, x1) = (x1, x1)
x0 = 0.375 (xo, x1) (x1, x1)
x0 = 0.375 (xo, x1) (x1, x1) (?, 0)=(x1, 0)
x0 = 0.375 (xo, x1) (x1, x1) (x1, x2)
x0 = 0.375 (xo, x1) (x1, x1) (x1, x2) (?, x2)
x0 = 0.375 (xo, x1) (x1, x1) (x1, x2) (?, x2) =(x2, x2)
x0 = 0.375 (xo, x1) (x1, x1) (x1, x2) (x2, x2)
x0 = 0.375 (xo, x1) (x1, x1) (x1, x2) (x2, x2) (x2, 0)
x0 = 0.375 (xo, x1) (x1, x1) (x1, x2) (x2, x2) (x2, x3)
x0 = 0.077 f(x) = 2x(1-x)
xo x1 x1 x2 x2 x3 x4
a) Fixed points 0 and ½ b) f’(0) = 2 and f’(1/2)=0 c) Thus (see theorem of previous slide): 0 is a repellor and ½ is an attractor d) Basin of attraction of ½ is (0,1) (do this with a) cobweb and b) algebraically) e) (-∞,0) U (1,∞): each initial value in this set has an orbit that tends toward -∞