ME 680 Spring Geometrical Analysis of 1D Dynamical Systems


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1 ME 680 Spring 2014 Geometrical Analysis of 1D Dynamical Systems 1
2 Geometrical Analysis of 1D Dynamical Systems Logistic equation: n = rn(1 n) velocity function Equilibria or fied points : initial conditions n where you start and stay without evolving for all time. They correspond to zeros of the velocity function: f(n) Phase diagram n =0 n =1 n The length of the arrows magnitude of the velocity (function) at that point. 2
3 α limit set of a point (initial condition) n 0 : It is defined as the set of limit points of the trajectory started at n 0, for t . Thus, α(n 0 ) = n lim φ (n 0, t) = n t ω limit set of a point n 0 is the set ω(n 0 ) = n lim φ (n 0, t) = n t + Eistence of a potential function: Consider n = f(n) (Gradient Dynamical System) Let there be a function V(n) such that f(n) = V n Eample: for the Logistic equation f(n) = r n (1 n), i.e., V(n) = r n r n 3 3 3
4 Then, note that the equilibrium points for the system (a Gradient Dynamical System) are at the local etrema of the potential function. This is where the similarity with mechanical systems with potential energy functions ends!! Considering the Logistic equation: V(n) = rn2 2 + rn3 3 the plot of the potential function, and the equilibrium points are as follows: V(n) n=0 n=1 n 4
5 Observations: Oscillatory behavior is not possible in 1D autonomous Systems Trajectories approach the equilibrium point n =1, but never reach it in finite time. Invariant subspaces are regions I R in phase space where if n 0 I, then φ t, n 0 I for all negative and positive flow times ( < t < ). For the Logistic Equation, the invariant subspaces are: I 1 =, 0, I 2 = 0, I 3 = 0, 1, I 4 = 1, I 5 = {1, + } Thus, the state space is decomposed into: R = I 1 I 2 I 3 I 4 I 5 α and ω limit sets of any initial condition n 0 : 5
6 If n 0 I, the α limit set is 0 ω limit set is If n 0 I 2, then α and ω limit sets are the same If n 0 I 3, then α (n 0 ) = {0}, and ω(n 0 ) = {1} so on. Stability of Equilibria/Fied Points An equilibrium point of = f(), say =, is stable if ε > 0, δ (ε) > 0 such that for any initial condition 0, with 0 < δ, φ (t, 0 ) < ε for all t, 0 < t < +. Otherwise, it is unstable. 6
7 δ 0 This definition of stability is very difficult to use directly to deduce stability of an equilibrium point. One needs to a priori know the solution for every given initial condition starting inside the region of size δ. Thus, one really needs to find other criteria that can be used to characterize stability without solving the differential equation. If in addition, φ(t, 0 ) 0 as t, then is an asymptotically stable equilibrium. 7
8 Eamples: 1. Logistic Equation f(n)=rn(1 n) n =0 (unstable) n =1 n (asymptotically stable) 2. Quadratic System f(n)=an 2 3. Cubic System f(n) = an 3 n =0 (unstable) n n =0 (stable) n Observe: Any isolated stable equilibrium in 1D autonomous systems has to be asymptotically stable. 8
9 Linearization about equilibrium points Let be a fied point of = f(), i.e. f( ) = 0 To linearize about =, introduce a perturbation: Let = Then + = f( + ) df or f( ) d 0 0 (Taylor series epansion for small X ) df d 0 This is the linearized equation about = 9
10 Eample: Logistic Equation: n = r n(1 n) The equilibrium points are n = 1 and n = 0. Let us linearize the system about n = 1 Then n = n + n = 1 + n and n = r(1 + n) ( n) = nr rn 2 df r n r n dn n 0 This is the linearized system near n = 1. Note that n = 1 is linearly stable. We can make the connection between linear stability (i.e. stability of equilibrium for the linearized system) and nonlinear stability if (only if) df dn n0 0 (HartmanGrobman theorem) 10
11 Closing Remarks on Linearization = f() ; f( ) = 0 is an equilibrium There are a few ways to linearize the system. (i): df f( ) ( ) d (Taylor series epansion) df f( ) ( ) d Let. Then df d linearized system around an Equilibrium 11
12 (ii): Let = +. Then f( ) f() ˆ dfˆ f(0) ˆ d 0 dfˆ or d 0 General solution of dfˆ d dfˆ (0)ep( t) d eigenvalue if eigenvalue < 0, = is asymptotically stable if eigenvalue > 0, = is unstable is 12
13 If eigenvalue is 0, the equilibrium is called hyperbolic. Otherwise, it is called nonhyperbolic. According to the HartmanGrobman theorem, if is a hyperbolic equilibrium, stability conclusions drawn from linearized equation (linear stability) hold also for the nonlinear model (nonlinear stability) if df d 0,, then we have to look at higher order terms in the Taylor series to judge stability. 13
14 Bifurcations of equilibria in 1D Interesting dynamics can occur as system (or control) parameters vary: Equilibria can suddenly change in number or stability type. E: Consider the eample of a cantilever beam with a mass on top, with the mass being a control parameter: For mg < P cr (1 equilibrium) For mg > P cr (3 equilibriums) mg g g 14
15 Saddlenode bifurcation (fold, or turning point, blue sky bifurcation) A prototypical eample: = r + 2 here r is some control parameter The velocity functions for three distinct cases are as follows: r r < 0 two equil r = 0 one equil r > 0 no equil 15
16 We can present these results in a diagram of equilibrium solutions as a function of the parameter r. X unstable r =0 r stable This is a bifurcation diagram. (r = 0, = 0) is the bifurcation point. This is called a subcritical saddle  node bifurcation. 16
17 Supercritical saddle node bifurcation: Consider the system: 2 r f() X r =0 stable unstable r Linear Stability Analysis r > 0 : the equilibrium points are df The function derivative is d r For df r, we get 2 r ( asympt stable) d 17
18 df For r, we get 2 r (unstable) d df Note that both equilibria are hyperbolic 0 d At r = 0, however, theorem fails! df d 0 Consider the velocity function at r = 0: 0 i.e., HartmanGrobman The equilibrium at =0 is actually unstable! r = 0 one equilibrium 18
19 Another eample Consider a system governed by : r e f() We can determine the equilibria and find their stability via linearization: r < r r = r r > r 1 e  1 e  1 e  r (r) r (r) (r) What is the critical value of r? At critical value, and r must satisfy f( ) = 0 r   e  = 0 as well as r df d 0 1 e 0 0 r 1 19
20 Brief Introduction to Normal Forms In a sense, f() = r ± 2 are prototypical of all 1D systems undergoing a saddlenode bifurcation. Consider the system just studied: f (,r) r e with r 1, 0. Near the critical point, r r r 1 r and for small r r and, write 2 2 higher order 1 r (1 / 2 ) r / 2 terms f(,r) increasing r for r=r same form as that of supercritical saddle node bifurcation 20
21 f() = a + b 2 is the normal form of saddle  node bifurcation, i.e., all systems in 1D undergoing this bifurcation must locally possess this form. Transcritical Bifurcation The normal form for this bifurcation is = r 2 (similar to n = r n (1 n), the logistic equation). Consider the velocity function for different parameter values: r < 0 two equil r = 0 one equil r > 0 two equil 21
22 We can display the results in the form of a bifurcation diagram: =r =0 r=0 r This is called a transcritical bifurcation Eample: Lasers. See notes. 22
23 Eample: Laser threshold pump partially reflecting mirror Active material laser light At low energy levels each atom oscillates acting as a little antenna, but all atoms oscillate independently and emit randomly phased photons. At a threshold pumping level, all the atoms oscillate in phase producing laser! This is due to selforganization out of cooperative interaction of atoms. (Ref: Haken 1983, Strogatz s book) 23
24 Let n(t)  no. of photons Then, n = gain loss = G n N k n (escape or leakage thru endface) gain coeff > 0 no. of ecited atoms Note that k > 0, a rate constant Here τ = 1 k = typical life time of a photon in the laser Note however that N(t) = N o α n (because atoms after radiation of a photon, are not in an ecited state), i.e., n Gn(N n) kn or 2 o n (GNo k)n Gn 24
25 The corresponding bifurcation diagram is: lamp n laser =r n =0 N 0 =k/g N 0 No physical meaning 25
26 Pitchfork bifurcation Eamples: We have already seen the eample of buckling of a column as a function of the aial load: g mg g fluid Another eample is that of the onset of convection in a toroidal thermosyphan heating coil 26
27 The normal form for pitchfork bifurcation is: = r 3 = f( The behavior can be understood in terms of the velocity functions as follows: f() f() f() r < 0 r = 0 X r > 0 stable The bifurcation diagram is then: Supercritical pitchfork stable r =0 unstable stable r 27
28 Linear stability analysis Consider = r 3 = f() The equilibria are at 0, r df d 0 0 r 3(0) r is stable when r < 0 is unstable r > 0 what about when r = 0? The linear analysis fails!! For the nonzero equilibria: df d eigenvalue is negative if r > 0 2 i.e., these bifurcating equilbiria are asymp. stable. r 2 r 3( r ) 2r 28
29 Subcritical pitchfork The normal form is r, with equilibria 0, r The resulting bifurcation diagram is: 3 unstable X stable r =0 unstable r unstable 29
30 Usually, the unstable behavior is stabilized by higher order nonlinear terms, e.g., = r The resulting bifurcation diagram can be shown to be: unstable X stable r =0 unstable r unstable subcritical pitchfork, r = r P supercritical saddlenode, r = r S 30
31 Connection between simple bifurcations and the implicit function theorem Consider the system f(,r) Let f( 0, r 0 ) = 0 i. e., ( 0, r 0 ) be an equilibrium. Let f be continuously differentiable w.r.t. and r in some open region in the (, r) plane containing ( 0, r 0 ). Then if df 0 in a small neighborhood of ( 0, r 0 ), d (,r ) 0 0 we must have: f(, r) = 0 has a unique solution =(r) such that f((r),r)=0 furthermore, (r) is also continuously differentiable. No bifurcations arise so long as df 0 d (,r )
32 The figure below illustrates the idea through two points along a solution curve. At ( 1,r 1 ), the derivative df/d does not vanish, where as at ( 2,r 2 ), the derivative df/d vanishes. df/d0 ( 2,r 2 ) ( 1,r 1 ) df/d=0 r 32
33 Imperfect Bifurcations & Catastrophes Consider the buckling eample. If the load does not coincide with the ais of the column, what happens? g mg g mg symmetric loading asymmetric loading Real physical systems have imperfections and mathematical imposition of reflection symmetry is an idealization. Do the bifurcation diagrams change significantly if imperfections or perturbations are added to the model (velocity function)? This is related to the concept of structural stability or robustness of models. 33
34 Consider the twoparameter normal form: h r 3 we now have the two parameters, h and r. Note that it is a perturbation of the normal form for pitchfork bifurcation (h=0) y h y r 3 y h, h h (r) h c h (r) c y r r 0, only one equilibrium possible for any h 3 h h (r) c 3 equilibria in t his region h h (r) c r > 0, one or three equilibria possible 34
35 h increasing Imperfect bifurcations and catastrophes (cont d). Consider = h + r 3 = f( (h = 0 normal form of pitchfork) y() = r 3 with h We look for intersections of y h 3 y r y h, h h c(r) h h (r) c y r r 0, only one equilibrium possible for any h 3 r C / 3 h r > 0, one or three equilibria possible h (r) c 35
36 dy 2 r For critical point 0 r c c 3c 0 c d 3 Furthermore, 3 2r c r h c c rc c c 0 hc 3 3 h h=0 only one equil soln cusp r=0 r C equil solns equil solns r 36
37 1. 0 r C r 2. r C 0 r 3. 0 r C r 37
38 4. 0 h=0 h Alternately, in 3D we can visualize the solutions set as follows: h h=0 catastrophic surface r=0 r 38
39 1D system on a circle: [overdamped pendulum (acted by a constant torque )] The equation of motion is: O θ l g 2 ml b mglsin m e Let ml 2 θ e R be negligible (imagine pendulum in a vat of molasses) The resulting equation is b mglsin b or mgl sin mgl 39
40 mgl Let t ; b mgl (ratio of appl. torque to ma. gravitational torque) d Then sin where d We say that [0, 2 ] i.e., the phase space is a circle θ = θ θ =0 Consider the system: sin 40
41 if γ > 1, pendulum goes around the circle albeit nonuniformly If γ = 1, θ = π 2 is an equilibrium θ =/2 θ = θ =0 1, If 1, there are two equilibria : 1 and 2 which have opposite stability characteristics θ = θ 2 Clearly, there is a saddlenode bifurcation at γ = 1. θ 1 θ =0 41
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