x R d, λ R, f smooth enough. Goal: compute ( follow ) equilibrium solutions as λ varies, i.e. compute solutions (x, λ) to 0 = f(x, λ).

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2 Continuation of equilibria Problem Parameter-dependent ODE ẋ = f(x, λ), x R d, λ R, f smooth enough. Goal: compute ( follow ) equilibrium solutions as λ varies, i.e. compute solutions (x, λ) to 0 = f(x, λ). Oliver Junge, Institute for Mathematics, University of Paderborn 2

3 Continuation of equilibria Structure of the Solution Let u 0 = (x 0, λ 0 ) s.t. f(u 0 ) = 0 and rank(f (u 0 )) = N. Then locally f 1 (0) is a one-dimensional manifold in R d+1. x f 1 (0) u 0 λ Oliver Junge, Institute for Mathematics, University of Paderborn 3

4 Continuation of equilibria Tangent vector Let c : ( 1, 1) R d+1 be a local parametrization of f 1 (0). Then f(c(s)) = 0, i.e. In other words f (c(s)) c (s) = 0. T c(s) f 1 (0) = ker f (c(s)). f 1 (0) u 0 ker f (u 0 ) Oliver Junge, Institute for Mathematics, University of Paderborn 4

5 Continuation of equilibria Fixing the orientation We will use ker f (c(s)) to move along c. In order to fix a consistent direction, we require that c (s) = 1 and det f (c(s)) c (s) T > 0. Exercise 1 Show that f (c(s)) c (s) T is regular. Oliver Junge, Institute for Mathematics, University of Paderborn 5

6 Continuation of equilibria The induced tangent vector For a d d + 1-matrix A with rank(a) = d let t = t(a) R d+1 be the unique vector such that (i) At = 0; (ii) t = 1; (iii) det A t T > 0. We call t(a) the tangent vector induced by A. Exercise 2 Show that the set of all d (d + 1)-matrices with full rank is open. Oliver Junge, Institute for Mathematics, University of Paderborn 6

7 Continuation of equilibria The associated differential equation Consider the ODE c = t(f (c)). (1) Since d f(c(s)) = f (c(s)) c (s) = 0, f is constant along solutions of ds (1). Idea: in order to compute the path c, solve (1). Given data: vector field t(f ( )) and condition f(c(s)) = 0. predictor-corrector methods. Oliver Junge, Institute for Mathematics, University of Paderborn 7

8 Continuation of equilibria Predictor-corrector scheme Given u 0 with f(u 0 ) < ε, find sequence u 1, u 2,... of points such that f(u i ) < ε for some prescribed accuracy ε > 0. f 1 (0) u 1 u 2 u 0 (i) Predictor: solve the initial value problem u = t(f (u)), u(0) = u i, by an explicit scheme, e.g. one Euler-step: where h is the stepsize. ũ i+1 = u i + ht(f (u i )), Oliver Junge, Institute for Mathematics, University of Paderborn 8

9 Continuation of equilibria (ii) Corrector: compute u i+1 = argmin f(u)=0 u ũ i+1 by some iteration scheme, e.g. Newton s method. u i+1 t(f (u i+1 )) u i ũ i+1 A necessary condition for u i+1 is f(u i+1 ) = 0, t(f (u i+1 ))(u i+1 ũ i+1 ) = 0. Oliver Junge, Institute for Mathematics, University of Paderborn 9

10 Computation of bifurcation points Saddle-node bifurcation x x λ λ Example 1 ẋ = λ + x 2 λ 0 λ < 0 λ 0 λ = 0 0 λ > 0 Oliver Junge, Institute for Mathematics, University of Paderborn 10

11 Computation of bifurcation points Saddle-node bifurcation Condition: dim ker f x (x, λ ) = 1 (but still rank(f (x, λ )) = d). Detection: during path-following, check sign of det f x (x(s), λ(s)). Computation: Solve (Moore, Spence, 1980) f(x, λ) = 0 f x (x, λ)φ = 0 l T φ 1 = 0, l R d. Oliver Junge, Institute for Mathematics, University of Paderborn 11

12 Computation of bifurcation points Saddle-node bifurcation Let ψ f x (x, λ ) = 0. Theorem 1 If dim ker f (x, λ ) = d and dim ker f x (x, λ ) = 1, then (x, φ, λ ) is a regular zero of the extended system, if and only if ψ f xx (x, λ )φ φ 0. The point (x, λ ) is a quadratic turning point. Oliver Junge, Institute for Mathematics, University of Paderborn 12

13 Computation of bifurcation points Hopf bifurcation Consider the system ẋ = λx y x 3 ẏ = x for various λ R. Oliver Junge, Institute for Mathematics, University of Paderborn 13

14 Computation of bifurcation points Eigenvalue movement Linearization in 0, eigenvalues: Df(0) = λ 1 1 0, µ 1,2 = λ 2 ± λ µ Im µ Re Oliver Junge, Institute for Mathematics, University of Paderborn 14

15 Computation of bifurcation points The Theorem of Hopf (x 0, λ 0 ) equilibrium, iω 0 simple eigenvalue of f x (x 0, λ 0 ), ir σ(f x (x 0, λ 0 )) = {iω, iω}, C = {(x(λ), λ)} local path of equilibria, µ(λ) + iν(λ) eigenvalue of f x (x(λ), λ) with µ(λ 0 ) + iν(λ 0 ) = iω, If dµ dλ (λ 0) 0, then there exists a unique one-parameter family of periodic solutions (with positive period) in a neighborhood of (x 0, λ 0 ). Oliver Junge, Institute for Mathematics, University of Paderborn 15

16 Computation of bifurcation points Computing Hopf points Consider an eigenvector v + iw at iω of D := d x f(x 0, λ 0 ), i.e. D(v + iw) = iω(v + iw) Dv = ωw and Dw = ωv. Exercise 3 Show that v and w are linearly independent. Thus D 2 v = ω 2 v and D 2 w = ω 2 w. Oliver Junge, Institute for Mathematics, University of Paderborn 16

17 Computation of bifurcation points Computing Hopf points Consider the extended system F (x, ϕ, λ, ν) = f(x, λ) (d x f(x, λ) 2 + ν 2 I)ϕ (ϕ, ϕ) 1 l T ϕ = 0. If l is not orthogonal to span{v, w}, then there exists a unique (up to its sign) vector ϕ 0 span{v, w}, such that (x 0, ϕ 0, λ 0, ω) is a regular a solution of this system. a Roose,Hlavacek, 85 Oliver Junge, Institute for Mathematics, University of Paderborn 17

18 Goal: compute periodic solution of x = f(x), x R d, i.e. a solution x(t) with x(t) = x(t + T ) for some unknown period T = 2π/ω. The transformation y(t) = x(t/ω) yields the system y = 1 ω f(y), for which ȳ(t) = x(t/ω) is a 2π-periodic solution. Oliver Junge, Institute for Mathematics, University of Paderborn 18

19 Boundary value problem: y = 1 ω f(y) y(0) = y(2π). Let ϕ t (, ω) denote the flow of this ode. Shooting method: solve S(ξ, ω) = ϕ 2π (ξ, ω) ξ = 0. Because of the S 1 -symmetry of each periodic solution we need an additional phase condition. Simple choice: ξ 1 = ξ 1 Oliver Junge, Institute for Mathematics, University of Paderborn 19

20 The variational equation In order to solve S(ξ, ω) = 0, ξ 1 = ξ 1 by Newton s method, we need to compute DS. D ξ S(ξ, ω) = D ξ ϕ 2π (ξ, ω) I can be computed via the variational equation t D ξϕ t (ξ, ω) = 1 ω Df(ϕt (ξ, ω))d ξ ϕ t (ξ, ω), D ξ ϕ 0 (ξ, ω) = I. For S (ξ, ω) we compute ω ω ϕt = t ω 2 ẋ(t/ω) = t ω 2 f(x(t/ω)), Oliver Junge, Institute for Mathematics, University of Paderborn 20

21 thus S (ξ, ω) = ω ω ϕ2π (ξ, ω) = 2π ω 2 f(ϕ2π (ξ, ω)). Advantage of shooting: stability of the periodic solution is directly given by the eigenvalues of D ξ ϕ 2π (ξ, ω). Oliver Junge, Institute for Mathematics, University of Paderborn 21

22 Galerkin approach Let C r 2π = {u C r (R, R d ) : u(s + 2π) = u(s)}. Consider the operator F : C 1 2π R C 0 2π, F (y, ω) = dy dt 1 ω f(y). A zero of this operator is a periodic solution of the boundary value problem. Oliver Junge, Institute for Mathematics, University of Paderborn 22

23 Line of Reasoning (i) Choose a suitable, countable basis of the underlying space countable system of equations; (ii) using finitely many modes, numerically compute an approximate solution (Galerkin); (iii) construct a restricted domain for F that isolates the numerical zero; (iv) using topological arguments, show that there actually exists a zero in the restricted domain. Oliver Junge, Institute for Mathematics, University of Paderborn 23

24 Setting F (y, y,..., y (r) ) = 0, y(t) R, F : R r+1 R smooth. We look for functions y : R R, y C r, y(t) = y(t + 2π/ω) for some ω R such that F (y, y,..., y (r) ) = 0. Rescaling time, x(t) = y(t/ω), yields the map ˆF : R C r 1 C 0 1 ˆF : (ω, x) F (x, ωx,..., ω r x (r) ) Oliver Junge, Institute for Mathematics, University of Paderborn 24

25 ˆF induces a map on the Fourier coefficients (c k ) k Z of x. Note that c k = c k, k Z, so it suffices to consider { l 2 = (c k ) k=0 C N : c k 2 < }. k=0 Correspondingly l 2 r = { (c k ) k=0 l 2 u C r 1 : c k is the k-th Fourier coefficient of u }. Induced map F : R l 2 r l 2 0. Oliver Junge, Institute for Mathematics, University of Paderborn 25

26 Phase Condition Since the original ODE is autonomous: x C1 r solution x( + t) solution t [0, 1]. Numerically more favorable: regularize by phase condition ϕ(c) = 0. Full system: Φ : R l 2 r R l 2 0 Φ(ω, c) = (ϕ(c), F (ω, c)). Oliver Junge, Institute for Mathematics, University of Paderborn 26

27 We look for (ω, c) such that Φ(ω, c) = 0. Equivalently: look for fixed points of G = id + Φ. Define and consider the sets A k (ω, c) = c k P k (ω, c) = (ω, c 0,..., c k ) Q k (ω, c) = (c k+1, c k+2,...) Z k = {(ω, c) : P k G(ω, c) = P k (ω, c)} Z = k 0 Z k Oliver Junge, Institute for Mathematics, University of Paderborn 27

28 Proposition 1 If G is continuous, Z M is compact for some M and Z k for k M, then Z is nonempty and all points in Z are fixed points of G. Proof of Existence (i) Find compact restricted domain for G; (ii) Show that Z M is nonempty for some (small) M; (iii) Show that Z k is nonempty for k > M. Oliver Junge, Institute for Mathematics, University of Paderborn 28

29 (i) Construction of the Restricted Domain (i) Use Newton s method on a Galerkin projection of G in order to estimate a fixed point; (ii) F C r x C r c k = O(k r ) as k. Restricted domain: compact set Ω D R l 2 r, where D = D 0 D M 1 {c k C : c k r k }, k=m where D 0 D M 1 C M contains the numerically computed fixed point. Oliver Junge, Institute for Mathematics, University of Paderborn 29

30 (ii) Z M is nonempty Define the multivalued map G M : P M (Ω D) P M (R l 2 ) (ω, c) {P M G(ω, c) : P M (ω, c) = (ω, c) and (ω, c) Ω D}. We will show that every continuous selector of G M has a fixed point. Note that the set Q M (Ω D) determines the size of the images of G M. Oliver Junge, Institute for Mathematics, University of Paderborn 30

31 (iii) Z k is nonempty for k M Definition 1 A map f : Ω D C is linearly dominated on the ball B 0 (r) A k (Ω D) if f(ω, c) = L(ω)c k + g(ω, c), for all (ω, c) Ω D, c k r, where L : I C, Ω I, and g are continuous functions, such that sup Ω D g(ω, c) < r inf ω I L(ω) r. (2) Oliver Junge, Institute for Mathematics, University of Paderborn 31

32 r r R R f( Ω,D ) f( Ω,D ) (a) Linearly dominated. (b) Not linearly dominated. R = r inf ω I L(ω), D = D { c k = r}. Oliver Junge, Institute for Mathematics, University of Paderborn 32

33 Existence Theorem Theorem 2 Let Ω D be compact. Suppose that there exists a starshaped enclosure ĜM for G M such that (N, L) is an index pair for ĜM and Λ((N, L), ĜM) 0, where Λ is the Lefschetz number of the index pair (N, L). Assume furthermore that the maps A k G are linearly dominated for all k > M, then there exists a fixed point for the map G. Oliver Junge, Institute for Mathematics, University of Paderborn 33

34 Example Consider y + y + σy δy + y 2 = 0, σ = 2, δ = 3. Induced map on Fourier coefficients A k F : (ω, c) ( iω 3 k 3 ω 2 k 2 + σωik δ)c k + l Z c l c k l, k 0. Computations: real version. Phase condition ϕ(c) = imag(c 1 ) = 0. G(ω, c) = (ω, c) + (ϕ(c), F (ω, c)) as above. Oliver Junge, Institute for Mathematics, University of Paderborn 34

35 Numerical fixed point ω = 1.39, c 0 = 2.46, c 1 = 0.813, c 2 = i , c 3 = i , c 4 = i , c 5 = i , c 6 = i , c 7 = i , c 8 = i , c 9 = i Oliver Junge, Institute for Mathematics, University of Paderborn 35

36 Numerical fixed point: decay of coefficients c k β α k k Oliver Junge, Institute for Mathematics, University of Paderborn 36

37 Restricted domain Ω = [1.36, 1.44], D = D 0 D 1 D 2 B 0 (βα k ), with α = 0.1, β = 11.5, and D 0 = [2.43, 2.49], D 1 = [0.8, ], k=3 D 2 = [ 0.003, 0.04] [0.005, 0.06]i, imag(d 0 ) = 0, since the Fourier series is real valued. imag(d 1 ) = 0 due to the phase condition. Oliver Junge, Institute for Mathematics, University of Paderborn 37

38 Linear domiation for k > 2 Lemma: [ ] c l c k l β 2 α k 2 1 α 2 + k 1. l Z Lemma: The map A k G is linearly dominated on the ball B 0 (βα k ) for k > 2. Proof: By the estimate we need to verify that [ ] β 2 α k 2 1 α 2 + k 1 < βα k min ω Ω ( iω 3 k 3 ω 2 k 2 + σωik δ ). Rewrite the rhs, use Ω = [ω, ω], plug in ω and ω where appropriate, function k e(k, ω, ω, δ, σ) such that A k is linearly dominated, Oliver Junge, Institute for Mathematics, University of Paderborn 38

39 if e(k) > 0. Result: e(k) is increasing and e(3) > 0. Index pair for lower modes We need to numerically construct an index pair for Ĝ2 (with, hopefully, a non-zero Lefschetz number). 7d 3d by exploiting the structure of the map; Oliver Junge, Institute for Mathematics, University of Paderborn 39

40 Explicit multivalued map a 1 ( ω 2 δ + 1)a 1 + 2a 1 a 2 + 2ā 0 a 1 + c 2 I I 1 a 2 ( 4 ω 2 δ + 1)a 2 + (8 ω 3 2σ ω)b 2 + 2ā 0 a 2 + a ( c c 1 )I I 1 b 2 ( 8 ω 3 + 2σ ω)a 2 + ( 4 ω 2 δ + 1)b 2 + 2ā 0 b 2 + c I 0 + c 1 I I 1, where I 0 = [ 1, 1], I 1 = [ 1, 1], and ā 0 = ā 0 (a 1, a 2, b 2 ) and ω = ω(a 1, a 2, b 2 ) are intervals. Oliver Junge, Institute for Mathematics, University of Paderborn 40

41 Isolating neighborhood Oliver Junge, Institute for Mathematics, University of Paderborn 41

42 Index pair (N, L) := ( G M (I), G M (I)\I ), where G M : P P (P a partition of the search box ) such that for C P we have Ĝ2(C) G M (C). Induced homology map: f k = 1 k = 3, 0 else. Λ((N, L), Ĝ2) = 1. Oliver Junge, Institute for Mathematics, University of Paderborn 42

43 Theorem 3 The differential equation with parameters σ = 2 and δ = 3 possesses a periodic orbit y(t) = k Z c k e ikωt with c k = c k and ω [1.36, 1.44] c 0 [2.43, 2.49] c 1 [0.8, ] c 2 [ 0.003, 0.04] [0.005, 0.06]i c k {z C z β α k }, β = 11.5, α = 0.1, k > 2. Oliver Junge, Institute for Mathematics, University of Paderborn 43