Lecture 18 Stable Manifold Theorem

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1 Lecture 18 Stable Manifold Theorem March 4, 2008 Theorem 0.1 (Stable Manifold Theorem) Let f diff k (M) and Λ be a hyperbolic set for f with hyperbolic constants λ (0, 1) and C 1. Then there exists an r > 0 such that for each p Λ there exists two embedded disks Wr s (p, f) and Wr u (p, f) which are tangent to E s p and E u p, respectively. There is an identification of these disks to E s p(r) and E u p(r) and Wr s (p, f) is a graph of a C k function σp s : E s p(r) E u p(r) with σp(0 s p ) = 0 p and D(σp) s 0 = 0 (Wr s (p, f) = {(σp(y), s y) : y E s p(r)}). The function σp s and its first k derivatives vary continuously with p. Similar statements hold for the unstable manifold. Moreover, for λ < λ < 1 and r > 0 sufficiently small W s r (p, f) = {q E u p(r) E s p(r) f j (q) E u f j (p) (r) Es f j (p)(r) and d(f j (q), f j (p)) C(λ ) j d(q, p) for all j 0}. The sets W s r (p, f) and W u r (p, f) are called the local stable and local unstable manifolds of p, respectively. We define the stable and unstable manifolds, respectively, as W s (p, f) = {q M d(f j (q), f j (p)) 0 as j } = n 0 f n (W s r (f n (p), f)) W s (p, f) = {q M d(f j (q), f j (p)) 0 as j } = n 0 f n (W s r (f n (p), f)). Remark 0.2 The sets W s r (p, f) and W u r (p, f) are not necessarily unique, but the sets W s (p, f) and W u (p, f) are unique. 1

2 Proposition 0.3 Let Λ be a hyperbolic set for f. Then for each p Λ, W s (p, f) is an immersed copy of E s p and W u (p, f) is an immersed copy of E u p. Proof. We know that Wr s (p, f) is an embedded copy of E s p(r) be a map. Since f is a diffeomorphism we know that f j (Wr s (p, f)) is an embedded image of E s p(r). Thus W s (p, f) is the union of embedded copies of a disk and W s (p, f) is an immersed copy. The proof for W u (p, f) is similar. We remark that the immersion can be very complicated (i.e. a homoclinic tangle). We will prove the Stable Manifold Theorem for a hyperbolic fixed point. The case of a periodic point is easily generalized. For a more general hyperbolic set see the proof in Katok and Hasselblatt or Robinson. As a reminder a fixed point is hyperbolic if no eigenvalues of Df p lie on the unit circle. Then there exists a splitting R n = E s E u such that there exists a λ > 1 and µ (0, 1) where each eigenvalue α whose generalized eigenspace lies in E u satisfies α > λ and each eigenvalue α whose generalized eigenspace lies in E u satisfies α < µ. Furthermore, there exists a C > 0 such that Dfp n E s < Cµ n and Dfp n E u < Cλ n for all n > 0. By suing an adapted metric we may suppose that C = 1. 1 Proof of the Stable Manifold Theorem The Graph Transform (or Hadamard) Method for proof uses the following idea: Take a trial function σ : E s p(r) E u p(r) that might give the stable manifold. Then define an appropriate operator Γ on the set of all such trial functions. We show that Γ is a contraction mapping and the unique fixed point gives the desired function. Remark 1.1 The other common method is the Perron Method. This uses integral equations similar to how we showed the existence and uniqueness of solutions to ODE s. Outline: We first show that Wr s (0, f) is the graph of a Lipschhitz function with Lipschitz constant less thatn some fixed α. To do this we take B N = N 1 j=0 f j (E u (r) E s (r)) and show B N+1 = f(b N ) B 0 and B = j=0 B j is the graph of a Lipschitz function. We then show the funciton is C 1 and prove a fact that will show it is C k. 2

3 1.1 Cone fields A useful tool is the use of cone fields. For α > 0 we define the stable and unstable cones by C s (α) = {(v s, v u ) E s E u v u α v s } C u (α) = {(v s, v u ) E s E u v s α v u } Normally, α > 0 is chosen sufficiently small so the cones are thin. A cone field for a set Λ with a splitting E s x E u x for each x Λ is then the set of cones at each point x Λ and α is a uniform constant (i.e. does not depend on x). As a reminder the minimum (or conorm) of a linear map A : R k R k is Av m(a) = inf v 0 v, This measures the minimum expansion of A. If A is invertible, then m(a) = A 1 1. For a hyperbolic fixed point with Df n E u < Cλ n it follows that m(df n E u) > Cλ n for n > 0. Let π s : R n E s and π u : R E[ u. We may ] Ass 0 assume that the point p is identified with the origin and Df = 0 A uu where A ss = π s (Df E s {0}) and A uu = π u (Df {0} E u). Since 0 is a hyperbolic fixed point there exists 0 < µ < 1 < λ such that m(a uu ) > λ > 1 and A ss < µ < 1. We now define the norm on R n to be the maximum component norms in E s and E u. Let B(r) = E s (r) E u (r). We now want to specify how thin to make the cone fields. These estimates are needed so that in the set B(r) the cones map into the interiors in a uniform way. Fix α > 0 sufficiently small and ɛ > 0 sufficiently small such that µ + ɛ α + ɛα + ɛ < 1, λ α 2ɛ ɛα > 1, and λα 2ɛ ɛ α > 1. Given such α and ɛ there exists an r > 0 such that for all q B(r) the matrix [ ] Ass (q) A Df q = su (q) A us (q) A uu (q) with A ss (q) < µ, m(a uu ) > λ, A su (q) < ɛ, A us (q) < ɛ, and Df q Df 0 < ɛ. 3

4 The next three lemmas give estimates on the α stable and unstable cone fields in B(r) under the action f. Lemma 1.2 Let q B(r). Then Df q C u (α) C u (α). Proof. Let v = (v s, v u ) C u (α) {0} and q B(r). Then π s Df q v π u Df q v = A su(q)v u + A ss (q)v s A uu (q)v u + A us (q)v s A su(q) v u + A ss (q) v s m(a uu (q)) v u A us (q) v s = A su(q) + A ss (q) v s /v u m(a uu (q)) A us (q) v s /v u ɛ + µα λ αɛ = α(µ + ɛα 1 λα 1 ɛ ) < α. Lemma 1.3 Let q, q B(r) and q {q}+c u (α). Then the following hold: 1. π s f(q ) π s f(q) α(µ + ɛα 1 ) π u (q q), 2. π u f(q ) π u f(q) (λ ɛα) π u (q q), and 3. f(q ) {f(q)} + C u (α). Proof. Let ψ(t) = q + t(q q) for t [0, 1]. Using the MVT we have π s f(q ) π s f(q) = π s f ψ(1) π s f ψ(0) sup 0 t 1 d dt (π s f ψ(t)) sup 0 t 1 A su (ψ(t)) pi u (q q) + sup 0 t 1 A ss (ψ(t)) π s (q q) (ɛ + µα) π u (q q) = α(µ + ɛα 1 ) π u (q q). For the second part let q = (q s, q u ), q = (q s, q u), and z = (q s, q u). So q q = (q z) + (q q), z q {0} E u, and q z E s {0}. 4

5 For y E u (r) and x E s (r) define h(y) = π u f(q s, y) and g(x) = π u f(x, 0). Then g(q s ) = π u f(q s, 0) = h(0). Also, Dg x = A us (x, 0) so Dg x ɛ. By the MVT we then have ɛr ɛ g s g(q s ) = h(0). Then Dh y = A u u(q s, +y) which is an isomorphism with minimum norm greater than λ and Dh y A uu (0, 0) < ɛ. So h E u (r) is onto E u ((λ ɛ)r h(0) ) E u ((λ 2ɛ)r) E u (r). So h has a unique fixed point. Also, by the Inverse Function Theorem the derivative of h 1 is given by D(h 1 ) h(y) = (Dh y ) 1 and has norm less than λ 1. Therefore, the MVT says that h 1 (w 2 ) h 1 (w 1 ) λ 1 w 2 w 1. Let w 2 = h(q u) = π u f(z) and w 1 = h(q u ) = π u f(q). So h 1 (w 2 ) = q u and h 1 (w 1 ) = q u. Then we have Also, we have λ π u (q q) π u f(z) π u f(q). π f(q ) π u f(z) sup A us π s (q q) ɛ π s (q q) ɛα π u (q q). Combining these we have π u f(q ) π u f(q) π u f(z) π u f(q) π u f(q ) π u f(z) (λ ɛα) π u (q q). For (3) we see that π s (f(q ) f(q) π u (f(q ) f(q) α(µ + ɛα 1 λ ɛα < α. Lemma 1.4 Assume that f(q 1 ) = q, f(q 1) = q, q {q} + C s (α), and q, q, q 1, q 1 B(r). Then 1. q 1 {q 1 } + C s (α), and 2. π s (q 1 q 1 ) (µ + ɛα) 1 π s (q q). 5

6 Proof. The proof of (1) follows similarly to the previous lemma. For (2) we let ψ(t) = q 1 + t(q 1 q 1 ) where t [0, 1]. Then π s (q q) = π s f ψ(t) π s f ψ(0) sup 0 t 1 d dt π s f ψ(t) sup A ss (ψ(t)) π s (q 1 q 1 ) + sup A su (ψ(t)) π i (q 1 q 1 ) (µ + ɛα) π s (q 1 q 1 ). Definition 1.5 An α-unstable disk is a graph of a C 1 function ϕ : E u (r) E s (r) such that Dϕ y α for all y E u (r). The next lemma is the key ingredient to the proof of the Stable Manifold Theorem. Lemma 1.6 Let D u 0 be an α-unstable disk over E u (r). Then 1. D u 1 = f(d u 0) B(r) is an α-unstable disk over E u (r) and diam[π u (f 1 (D u 1) D u 0 )] (λ ɛα) 1 2r. 2. Furthermore, let D u n = f(d u n 1) B(r). Then D u n is an α-unstable disk for n 1 and diam[π u ( n j=1 f j (D u j ))] (λ ɛα) n 2r. Proof. Since D0 u is the graph of a C 1 function ϕ 0 : E u (r) E s (r) with (Dϕ 0 ) y α. Let σ 0 : E u (r) B(r) be defined as σ 0 (y) = (ϕ 0 (y), y). Define h = π u f σ 0. Then Dh y A uu (0) = D(π u f σ 0 ) y A uu (0) = A uu (σ 0 (y)) + A us (σ 0 (y))d(ϕ 0 ) y A uu (0) A uu (σ 0 (y)) A uu (0) + A us (σ 0 (y)) D(ϕ 0 ) y ɛ + ɛα. So h is a C 1 local diffeomorphism and has an inverse and h(e u (r)) {h(0)} + E u ((λ ɛɛα)r) E u ((λ ɛ ɛα)r h(0) ). Notice that h(0) = π u f(ϕ 0 (0), 0) with ϕ 0 (0) r and π u f(0, 0) = 0. By the MVT applied to π u f at the points (0, 0) and (ϕ 0 (0), 0) we have h(0) sup( A us ) ϕ 0 (0) 0 ɛr. 6

7 Then h(e u (r)) E u ((λ ɛα ɛ)r) E(r) so h 1 (E u (r)) E u (r). Define the map σ 1 = f σ 0 h 1 E u (r). Hence, π u σ 1 = π u f σ 0 h 1 = h h 1 = id E u (r) and this implies that σ 1 is a graph over E u (r). For v u E u we have D(π u f σ 0 ) y v u = A uu (σ 0 (y))v u + A us (σ 0 (y))d(ϕ 0 (y)d(ϕ 0 ) y v u A uu (σ 0 (y))v u A us (σ 0 (y))d(ϕ 0 ) y v u m(a uu (σ 0 (y))) v u A us (σ 0 (y)) D(ϕ 0 ) y v u (λ ɛα) v u. Hence, m(dh y ) λ ɛα and D(h 1 ) y (λ ɛα) 1. and h 1 (y 2 ) h 1 (y 1 ) (λ ɛα) 1 y 2 y 1. Let ϕ i (y) = π s σ 1 (y) = π s f(ϕ 0 (h 1 (y)), h 1 (y)). Then D(ϕ 1 ) y A su (σ 0 h 1 (y))d(h 1 ) y + A ss (σ 0 h 1 (y))d(ϕ 0 ) h 1 (y)d(h 1 y ɛ(λ ɛα) 1 + µα(λ ɛα) 1 α µ+ɛα 1 < α. λ ɛα So now let D u 1 = σ 1 (E u (r)) and we have part (1). Part (2) follows inductively. From the above lemma we have a graph transform. Proposition 1.7 W s r (0) = j=0 f j (B(r)) is the graph of an α-lipschitz function ϕ s : E s (r) E u (r) with ϕ s (0) = 0. Proof. Let S n = n j=0 f j (B(r)) so W s r (0) = n=0 S n. Fix x E s (r) and D u 0(x) = {x} E u (r). By Lemma 1.6 we have D u n(x) is an unstable disk and [{x} E u (r)] S n = n f j (Dj u (x)) D0(x) u j=0 is a nested sequence of closed complete sets and the diameter is bounded above by (λ ɛα) n 2r. So intersection of [{x} E u (r)] W s r (0) is a unique point. Hence, W s r (0) is a graph over E s. We now show the graph is Lipschitz. Let q W s r (0) and assume that q W s r (0) [{q} + C u (α 1 )]. Then by Lemma 1.2 we get f j (q) {f j (q )} + 7

8 C u (α 1 )] and π u (f j (q) f j (q )) (λ ɛα 1 ) j π u (q q; ). Since (λ ɛα 1 ) j we get f j (q ) or j j (q) leaves B(r), but both are in the stable manifold, a contradiction. So Wr s (0) ({q} + C u (α 1 )) = {q}. Now for x y, x, y E u (r) since σ u (x) = (x, ϕ s (x)) + (y, ϕ s (y)) = σ u (y) it follows that σ u (y) is not contained in {σ u (x)} + C u (α 1 ), so σ u (y) {σ u (x)} + C s (α). Hence, Lip(π u σ s ) α. Proposition 1.8 If q W s r (0), then f j (q) α(µ + ɛα 1 ) j q so f j (q) 0 exponentially fast. Proof. Since W s r (0) is α-lipschitz we get π u (f j (q) 0)) α π s (f j (q) 0). So f j (q) {0} + C s (α). Applying Lemma 1.3 to f j (q) and f j (0) = 0 we get (µ + ɛα 1 ) f j 1 (q) (µ + ɛα 1 ) π s f j 1 (q) (µ + ɛα 1 ) π s (f j 1 (q) f j 1 (0) π s (f j (q) f j (0)) π s f j (q) Hence, (µ + ɛα 1 ) j q π s f j (q) and π u f j (q) α π s f j (q) α(µ + ɛα 1 ) j q. So f j (q) (µ + αɛ) j q max{1, α}. Proposition 1.9 W s r (0) is C 1 and tangent to E s at 0 To show this we need one more lemma. Let C s,0 (q) = C s (α) and C s,n (q) = (Df n q ) 1 C s (α). Lemma 1.10 Let q W s r (0). Then C s,n (q) C s,n 1 (q)... C s,0 (q) and there exists a positive angle between vectors in C s,n+1 (q) and the complement of C s,n (q). Furthermore, n 0 Cs,n (q) is a plane P q that is a graph of L q : E s E u that depends continuously on q. Proof. from Lemma 1.2 we have Df jj (q)c u (α 1 ) C u (α 1 ) so C s,1 (f j (q)) = [Df f j (q)] 1 C s (α) C s (α) = C s,0 (f j (q)). 8

9 Inductively, we have C s,n (q)... C s,0 (q). Applying Proposition 1.7 to the sequence of linear maps we get that C s,n (q) converges to a set that is a graph of a Lipschitz function and is a plane P q. From estimates in Lemma 1.2 we can show the existence of the positive angle. This implies that for q near q we have P q C,n+1 (q ) C s,n (q). So P q varies continuously with q. Proof of Proposition 1.9. We need to show that ϕ s is differentiable at x 0 = π s q with Dϕ s x 0 = L q and P q = T q Wr s (0). Then ϕ is C 1 since P q varies continuously. Also, P 0 = E s {0} so Wr s (0) is tangent to E s at 0. Let C s,n (f n+1 (q), δ) = C s,n (f n+1 (q)) πs 1 (E s (δ)). Sine f n 1 is differentiable there exists a δ > 0 such that f n 1 ({f n+2 (q)} + C s,0 (f n=1 (q), δ)) {q} + C s,n (q). Since σ s and f n+1 are continuous and ϕ s is α-lipschitz there exists an η > 0 such that if x x 0 η, then f n+1 (σ s (x)) {f n+1 (σ s (x 0 ))} + C s,0 (f n+1 (σ s (s 0 ), δ)). So σ s (x) {σ s (x 0 )} + C s,n (σ s (x 0 )). Hence, given n there exists an η > 0 such that x x 0 < η implies taht σ s (x) σ s (x 0 ) C s,n (σ s (x 0 )) so σ s is differentiable at x 0. Proposition 1.11 W s r (0) is C k. Proof. We have shown that this holds for k = 1. Assume that k 2. Define F (p, v) = (f(p), Df p v). This F is defined for (p, v) with p B(r) and v T p R n. So F is C k 1 with F (0, 0) = (0, 0 and [ ] DF (p,v) (p, v Df ) = p p D 2 f p (v, p ) + Df p v Then [ ] [ ] DF (0,0) (p, v Df0 0 p ) = 0 Df 0 v. So (0, 0) is a hyperbolic fixed point. By induction F has a C k 1 stable manifold at (0, 0). Thus, T W s r (0, f) is C k 1 and W s r (0, 0) is C k. 9