MATH34042: Discrete Time Dynamical Systems David Broomhead (based on original notes by James Montaldi) School of Mathematics, University of Manchester 2008/9 Webpage: http://www.maths.manchester.ac.uk/∼dsb/Teaching/DynSys email: [email protected] Overview (2 lectures) Dynamical systems continuous/discrete; autonomous/non-autonomous. Iteration, orbit. Applica- tions (population dynamics, Newton’s method). A dynamical system is a map f : S S, and the set S is called the state space. Given an initial point x0 S, the orbit of x0 is the sequence x0,x1,x2,x3,... obtained by repeatedly applying f: ∈ → n x1 = f(x0), x2 = f(x1), x3 = f(x2),...,xn = f (x0), ... Basic question of dynamical systems: given x0, what is behaviour of orbit? Or, what is behaviour of most orbits? + + Example Let f : R R be given by f(x)= √x. Let x0 = 2. Then x1 = f(2)= √2 1.4142. And ≈ x = √2 1.1892 and x = √1.1892 1.0905. 2 p → 3 The orbit of≈2 therefore begins {2.0, 1.4142,≈ 1.1892, 1.0905, ...}. The orbit of 3 begins {3.0, 1.7321, 1.3161, 1.1472, ...}. The orbit of 0.5 begins {0.5, 0.7071, 0.8409, .9170, .9576, ...}. All are getting closer to 1. Regular dynamics fixed points, periodic points and orbits. Example f(x)= cos(x). Globally attracting fixed point at x = 0.739085 . ··· Chaotic dynamics Basic idea is unpredictability. There is no “typical’ behaviour. Example f(x)= 4x(1 − x), x [0,1]: Split the interval into two halves: L = [0, 1 ] and R = [ 1 ,1]. We will ∈ 2 2 prove that given any sequence of Ls and Rs, say LLRRLRRLLL... there is an initial point x0 such that x0 L, x1 L, x2 R,... (see ‘itineraries’ in §2) ∈ ∈ ∈ Fractals We see relation between fractal sets and dynamics, and what it means to say the dimen- sion of a set is (say) 1.5. Bifurcations How does dynamical system — or behaviour of orbits — depend on external param- eters. Eg fµ(x)= µx(1 − x) — how can dynamics change as µ changes? 0.1 §1 MATH34042: Discrete Time Dynamical Systems 1 Regular dynamics (mostly 1-dimensional) As usual, f : S S is a dynamical system. Almost always, we assume S R. Given any x0 S, ⊂ ∈ then its orbit is {x0,x1,x2,...} where xn+1 = f(xn). → 1.1 Fixed points A fixed point is one where f(p)= p. Proposition 1.1 If xn p and f is continuous then p is a fixed point. → PROOF A function f is continuous at x if, whenever xn x then f(xn) f(x). So we have lim f(xn)= f(p→) → xn p → but the sequence (xn)n=1 is exactly the same as (f(xn−1))n=1, so we have ∞ ∞ xn p f(xn−1) p and, therefore, f(p)= p as required. → ⇐⇒ → Theorem 1.2 (Fixed point theorem) Let I be a compact1 interval and suppose f : I I is contin- uous. Then f has a fixed point. → The basic truth of this can be seen with a convincing picture. Imagine drawing a graph of f inside a square whose sides are I. The graphis a continuouscurve, consisting of points (x,y) where y = f(x). Since f(x) is defined for all values of x I, the graph must intersect the diagonal of the square at least once. However, on the diagonal x =∈y, so at such an intersection we have simultaneously y = x and y = f(x) and hence a fixed point. Counter-example on an unbounded interval: f(x)= x + 1 (f : R R) has no fixed points! → PROOF Define g(x)= f(x)− x. This function is continuous, since f is continuous. Now let us write I = [xmin,xmax]. We notethat since we havea dynamicalsystem f(xmin) I and so g(xmin) 0. ∈ ≥ Similarly, g(xmax) 0. It follows from the intermediate value theorem that there exists at least one point p I where g≤(p)= 0 and hence that there is at least one fixed point of f. ∈ 1.2 Graphical analysis Useful in 1-dimensional systems (S R). The basic idea is that suggested by the ‘convincing picture’ approach to the above fixed point⊂ theorem. Examples including f(x)= x2 on [0,1], f(x)= 3.0x(1 − x) on [0,1], and f(x)= x2 + x on R. Fixed points occur at intersection of graph with diagonal. 1for us, compact = closed and bounded 2008/9 1.2 February 11, 2009 §1 MATH34042: Discrete Time Dynamical Systems 1 1 1 1 µ 4 b 0 0 0 p 0 1 0 1 0 1 f(x)= x2 f(x)=3.0x(1-x) f(x)= x2 + x 1.3 Stability of fixed points Let f : S S have fixed point p S. We say p is attracting or asymptotically stable if ε > 0 such that ∈ ∃ → dist(x0,p) < ε = xn p and dist(xn,p) < ε n > 0 ∀ R Rk Here ‘dist(x0,p)’ is the distance between⇒ x→0 and p. If S then dist(a,b)= |a − b|; if S then dist(a,b)= a − b = (a − b )2 + + (a − b⊂k)2. (For metric space fans, ‘dist’ is⊂ any p 1 1 k metric—we will ink practicek stick to R in this course.)··· A point p is repelling if ε > 0 s.t. dist(x0,p) < ε (x0 = p) implies there is an n > 0 for which ∃ 6 dist(xn,p) > ε. (If you are interested in reading more about the various concepts of stability Milnor has a nice, short article in Scholarpedia, see: http://www.scholarpedia.org/article/Attractor) Example if f : R R is linear, so f(x)= ax for some a, then 0 is a fixed point. It is attracting if |a| <1 and repelling if |a| >1. → n We can solve the linear difference equation explicitly: xn = a x0. Letting |a| <1, and choosing any n ε>0, then |x0| <ε = |xn| = |a ||x0| <ε and limn xn = 0. So x = 0 is an attracting fixed point in this case. A similar argument shows that x = 0 is repelling if |a| >1. ⇒ →∞ Given ε > 0, write Nε(p)= {x S ||x − p| < ε}. Called the ε-neighbourhood of p. ∈ Beware: If S = [0,1], then Nε(0) = [0,ε) (not (−ε,ε), as x < 0 is not in S). If f : I I (I R) is differentiable, then a fixed point is hyperbolic if |f (p)| = 1. ⊂ ′ 6 2 Example→ in f(x)= x + x, the origin is a fixed point which is not hyperbolic, as f′(0)= 1. Theorem 1.3 (Stability theorem) Suppose f : I I is differentiable and f′ is continuous (ie f is of class C1) and suppose p is a hyperbolic fixed point. Then → (a) |f′(p)| < 1 implies p is attracting, (b) |f′(p)| > 1 implies p is repelling. If f (p)= 1 one needs a finer analysis (using eg second or higher order Taylor series). ′ ± PROOF (Outline) (a) Let |f (p)| <r<1. Then ε > 0 s.t. whenever |x − p| < ε we have ′ ∃ |f(x)− f(p)| < r. |x − p| That is, |f(x)−f(p)| <r|x−p|. But f(p)= p, so this is the same as |f(x)−p| <r|x−p|. In particular, 2 if x Nε(p) then f(x) Nε(p) so we can keep applying f: |f(f(x)− p| <r|f(x)− p| <r |x − p|. ∈ ∈ 2008/9 1.3 February 11, 2009 §1 MATH34042: Discrete Time Dynamical Systems And similarly |f3(x)− p| <r3|x − p|. By induction fn(x)− p| <rn|x − p|. And as n so rn 0 so fn(x) p. → ∞ (b)→ Same argument→ turned on head: let 1<r< |f′(p)|. Then ε > 0 s.t. x Nε(p) |f(x)− p| > r|x − p|. Details left to reader! ∃ ∈ ⇒ Example f : R R, f(x)= cosx has a unique fixed point p (in fact p 0.739085). Now |f′(x)| = |sinx| 1, and since 0<p<π/2 it follows that |sinp| <1 so p is hyperbolic,≈ and attracting. ≤ → Basins of attraction If p is an attracting fixed points, it is natural to ask which initial conditions provide orbits that converge to p. The basin of attraction of p is the set of all these initial conditions: B(p)= {x0 S | xn p}. ∈ These basins of attraction can be very complicated in genera→l, but there is one general property: they are always open. Proposition 1.4 The basin of attraction of any attracting fixed point of a continuous dynamical system is open. PROOF Uses property that if f : S S is continuous and U S is open, then f−1(U) is open. ⊂ → 1.4 Stability of periodic orbits Dynamical system: f : S S. A periodic point of period k > 0 is a point p S for which the condition fk(p)= p is true2. That is, p is a fixed point of fk. Note: if p has period∈k, then it also has period 2k, 3k etc. The prime→ period (also, primitive or fundamental period) is the smallest positive period k. That is, it’s the smallest k such that xk = x0. If p is a periodic point of period k, then the sequence p,f(p),...,fk−1(p) is a periodic orbit. Example f(x)=−x3 has three periodic points: a (attracting) period 1 point (ie, fixed point) at x = 0 and a (repelling) period 2 point at x = 1; the corresponding periodic orbit is {1,−1}.