Lecture 6: Limit Cycles and Index Theory • In this lecture we consider limit cycles in detail. • We exclude separatrix cycles (in particular homoclinic and heteroclinic connections connecting ! to " points – going from an ! to an " point takes an infinite time – not exactly periodic!) • Definition: A solution of $̇ = & $ through $' is said to be periodic if there exists a ( > 0 such that + ,, $' = + , + (, $' for all , ∈ ℝ. The minimum such ( is called the period of the periodic orbit. Dynamical Systems: Lecture 6 1 Proving a periodic orbit does not exist We may wish to prove that a particular second order dynamical system does not posses periodic orbits. Bendixson’s criterion: Let "̇ = $ ", & and &̇ = ' ", & . () (, If + is not identically zero and does not change sign within a simply (* (- connected region . of the phase plane, then the system has no closed orbits in .. Dynamical Systems: Lecture 6 2 Outline of proof • Assume a closed orbit Γ ⊂ # exists. Then the orbit is a parametric %& ) curve in $ such that = . %' * • So on a closed orbit Γ we get +,- = .,/ ⇒ ∮2 (.,/ − +,-) = 0 • Using Stoke’s theorem we then have 9+ 9. 7 + ,/ ,- = 0 8 9/ 9- where ; ⊂ # is the region enclosed by Γ <* <) • But if + is never zero, then it must have the same sign all over # <' <& and so the integral cannot be zero. Dynamical Systems: Lecture 6 3 Dulac’s criterion We consider the same differential equations but now allow both ! and " to be multiplied by another function #. Dulac’s criterion: Let #(%, ') be a continuously differentiable function, defined on a region ) ⊂ ℝ,. -(./) -(.2) If + is not identically zero and does not change sign in ), then -0 -3 the system %̇ = ! %, ' , '̇ = " %, ' has no closed orbits in ). Dynamical Systems: Lecture 6 4 Example using Bendixson’s criterion Let "̇ = $ ≝ & ", $ $̇ = " − ") − *$ ≝ + ", $ , * ≥ 0 Then .& .+ + = −* ." .$ Thus for * ≠ 0 there are no closed orbits. 12 42 45 For * = 0, there is an energy function − + and the system is 3 3 6 Hamiltonian, so we can study the trajectories using level sets (see Lec. 4). Dynamical Systems: Lecture 6 5 Second example Let "̇ = $ ≝ & ", $ $̇ = " − ") − *$ + ",$ ≝ - ", $ , * ≥ 0 then 0& 0- + = −* + ",. 0" 0$ Linearisation shows that (0,0) is a saddle, and (±1,0) are stable nodes for * > 1 and unstable nodes for 0 ≤ * < 1. There can be no closed orbits within regions where " is very large or very small compared to *, but orbits may pass through these regions – we are thus undecided using Bendixson. Dynamical Systems: Lecture 6 6 Gradient Systems • Equations are such that "̇ = −∇& • Such systems cannot have closed orbits since &̇ = ∇&. "̇ = −∇&. ∇& = − "̇ ( implies that, if "(*) is on a closed orbit of period ,, then 0 & " , − & " 0 = − . "̇ (1* ≠ 0 / But on a closed orbit we must have & " , = & " 0 implying "̇ = 0 so " 0 must be a fixed point, not an orbit Dynamical Systems: Lecture 6 7 Index Theory For two-dimensional systems the equations "' "̇# = % "#, "' "̇' = ( "#, "' ) define a vector field or flow. "# These vectors make an angle to the "#-axis: -# ( "#, "' ) "#, "' = tan % "#, "' Dynamical Systems: Lecture 6 8 Index Theory The index of a (non-intersecting, simple) curve Γ, "(Γ) is defined by 1 " Γ = ) +, -., -0 2( * Γ Γ Index = 0 Index = 1 Follow the curve Γ around anti-clockwise and measure how many times the vectors on the curve rotate anti-clockwise during one rotation. Dynamical Systems: Lecture 6 9 Properties of Indices • The index is an integer (you must rotate by a multiple of 2" to get back to the start). • If there are no equilibria inside Γ, then the index is zero ($ Γ = 0). • If Γ coincides with a closed orbit, then the index is 1. • If Γ encloses an isolated saddle equilibrium point, then the index is −1. If Γ encloses any other isolated equilibrium point then the index is 1. • The index of a curve Γ enclosing multiple isolated equilibrium points is the sum of the indices of the individual equilibrium points enclosed. Dynamical Systems: Lecture 6 10 Some observations • Given a closed curve that does not enclose any equilibrium points – can this be a trajectory? No: if this curve is a closed orbit then its index is 1, but as there are no equilibrium points the curve has index 0. • Can there be a closed trajectory surrounding a single saddle node? No: the index of a saddle is −1 and that of a closed orbit is 1. Dynamical Systems: Lecture 6 11 Return to Bendixson example Earlier problem: "̇ = $ ≝ & ", $ $̇ = " − ") − *$ + ",$ ≝ - ", $ , * ≥ 0 0& 0- + = −* + ", Index of curve = 1-1+1=1 0" 0$ this is the only possible closed orbit Index = 1 Index = 1 " = −√* Index = -1 " = √* Dynamical Systems: Lecture 6 12 Matlab simulation ! = 2 8 6 Unstable limit cycle 4 2 0 y -2 -4 -6 -8 -3 -2 -1 0 1 2 3 x Dynamical Systems: Lecture 6 13 Another example of reasoning Consider the system "̇# = "# 3 − "# − 2"( "̇( = "( 2 − "# − "( 3 − 2" − 2" −2" with Jacobian: # ( ( −"# 2 − "# − 2"( Equilibrium points: )∗ = (0,0), Jacobian: 3 0 , positive eigenvalues (unstable node) so /()∗) = 1 # 0 2 # )∗ = (0,2), Jacobian: −1 0 , negative eigenvalues (stable node) so /()∗) = 1 ( −2 −2 ( )∗ = (3,0), Jacobian: −3 −6 , negative eigenvalues (stable node) so /()∗) = 1 1 0 −1 1 )∗ = (1,1), Jacobian: −1 −2 , eigenvalues 0.414, -2.414 (saddle) so /()∗) = −1 3 −1 −1 3 Dynamical Systems: Lecture 6 14 Reasoning (continued) • For closed trajectory we need an index of 1. • Trajectories cannot cross (curves only meet at equilibrium points) and there are no equilibrium points in the 2nd, 3rd, or 4th quadrants. • There are trajectories on the "#- and "$-axes, so no trajectory can cross into the 2nd, 3rd, or 4th quadrants. Hence these quadrants are free of any part of a closed trajectory. Thus (0,0), (0,2) and (3,0) cannot lie inside a closed trajectory. • The point (1,1) is a saddle with index −1, so it cannot lie inside a closed trajectory. • Conclusion: there are no closed trajectories. Dynamical Systems: Lecture 6 15 Graphical illustration of reasoning !# Index = 0: not a trajectory No trajectory can cross here Index = -1: not a trajectory !" Index = 0: not a No trajectory can cross here trajectory Dynamical Systems: Lecture 6 16 Limit Cycles • Limit cycles are isolated periodic orbits that can be stable or unstable (a cycle around a linear centre is not isolated and thus not a limit cycle). • In the plane a limit cycle is the ! or " limit set of some trajectory other than itself. • Definition: A periodic orbit Γ is said to be stable if for every $ > 0 there is a neighbourhood ' of Γ such that for ( ∈ ' the distance between *(,, () and Γ is less than $. Γ is called asymptotically stable if it is stable and, for all points ( ∈ ', this distance tends to zero as , tends to infinity. Dynamical Systems: Lecture 6 17 Condition for stability Let "̇ = $ " have a periodic solution " = % & , 0 ≤ & ≤ *, then the periodic orbit Γ lies on %(&). • The periodic orbit is asymptotically stable only if 0 . ∇. $ % & 3& ≤ 0 / • For planar systems, if Γ is the 4 limit set of all trajectories in the neighbourhood of Γ then it is a stable limit cycle. • If it is the 5 limit set, then it is an unstable limit cycle. • If it is the 4 limit set for one trajectory and the 5 limit set for some other trajectory it is called a semi-stable limit cycle. Dynamical Systems: Lecture 6 18 Limit cycle example 1.5 Consider 1 "̇ = −% + " 1 − "( − %( ( ( ( ( %̇ = " + % 1 − " − % 0.5 In polar co-ords: 0 y )̇ = ) 1 − )( ( *̇ = 1 -0.5 For ) ≠ 1, )̇ > 0 and we spiral out. -1 For ) = 1, )̇ = 0 ⇒ a limit cycle, which -1.5 -1.5 -1 -0.5 0 0.5 1 1.5 must be semi-stable. x Dynamical Systems: Lecture 6 19 The Poincaré Map • An extremely important tool for the analysis of dynamical systems, sometimes called the ‘return map’. • Based on a hyperplane perpendicular to a periodic orbit. • Consider points close to !" on the orbit and where those points arrive back in the hyperplane after traversing the orbit. This defines a map ! ↦ $ ! • As the map is iterated, the intersection point ! moves in the hyperplane. • If we are on a periodic orbit then the return point is the original point. Dynamical Systems: Lecture 6 20 The Poincaré Map Hyperplane Σ #(!) ! !" Γ Dynamical Systems: Lecture 6 21 Example "̇ = −% + " 1 − "( − %( %̇ = " + % 1 − "( − %( In polar co-ords: )̇ = ) 1 − )( *̇ = 1 This has stable limit cycle, which is an attractor for ℝ( − 0 (see Lec. 5) Solving the equations: .) 1 - ( = - ./ ⇒ ) = ) 1 − ) 1 3(4 1 − ( − 1 2 )1 * = / + *1 Dynamical Systems: Lecture 6 22 Poincaré Map The hyperplane Σ is the ray " = "$ through the origin that is crossed every 2& seconds. 9 Thus ( 1 '((0) 0 ' ($ = 1 -./ 1 − + − 1 , ($ "0 and '(1) = 1 (a fixed point). 8 Also 3' 2 = ,-./ < 1 3( 456 so the limit cycle is stable. Dynamical Systems: Lecture 6 23.