Rigidity for totally integrable convex billiards

Transcription

1 Rigidity for totally integrable convex billiards ICMAT, Madrid, November 11-15, 13 Michael Bialy Tel Aviv University Rigidity for totally integrable convex billiards p. 1/3

2 Introduction 1.Birkhoff billiards. Definitions: Billiard ball map Caustics, Invariant curves, Phase portraits. Lazutkin, R.Douady KAM type result. Integrable billiards. Elliptic versus circular billiards Total integrability. 3.Birkhoff conjecture: The only integrable billiards are circles and Ellipses. Rigidity for totally integrable convex billiards p. /3

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5 Approaches: 1) Bolotin: Polynomial integrals-requires smoothness of certain complex algebraic curve. ) Delshams, Ramirez-Ros: Small perturbation of ellipse splitting of separatrices. 3) Kaloshin and Sorrentino: If an integrable billiard isc conjugate to an ellipse (resp. a circle) in a neighborhood of the boundary, then it is an ellipse (resp. a circle). 4) Treschev: Computer experiments on local integrability. 5) Baryshnikov,Zharnitsky and later Glutsyuk: Birkhoff distributions My approach is based on the so called E.Hopf rigidity phenomenon: If geodesics on the torus have no conjugate points then the metric is flat. No conjugate points condition turns out to be almost equivalent to total integrability. The result for case of K=, I proved many years ago. Later on M. Wojtkowski suggested the use of Mirror formula for an alternative proof. I shall show that Mirror formula works well for constant curvature surfaces and for magnetic billiards. Rigidity for totally integrable convex billiards p. 5/3

6 Results Theorem 1. Let γ be smooth simple closed curve on a constant curvature surface S having positive geodesic curvature k. If the billiard ball orbits have no conjugate points. Then γ is a circle. Theorem. Letγ be smooth simple closed curve as above ons. If the billiard ball map is totally integrable thenγ is a circle. Theorem 3. Theorems 1, can be generalized to magnetic billiards on constant curvature surfaces. Remark: Magnetic billiards in ellipses seem to be non-integrable: Robnik, M, Berry, M V, 1985, J.Phys.A 18, , Classical billiards in magnetic fields. In contrast with the D case: Theorem 4. Billiard orbits for D 3 always have conjugate points. Example. Billiard in a ball. North and South poles are conjugate points. Rigidity for totally integrable convex billiards p. 6/3

7 Mirror equation on surfaces 1. Mirror formula in the plane:.mirror formula on surface: Y 1 a + 1 b = k(x) sinφ. Y (a)+ Y Y (b) = k(x) sinφ, Y(t) = sinht(k = 1); Y(t) = t(k = ); Y(t) = sint(k = 1). 3. Mirror formula for magnetic field β. The Jacobi fields change according to the effective curvaturek +β : Y β = 1 K +β sin( K +β t), for K +β >, Y β = Y β = t, for K +β =, 1 (K +β ) sinh( (K +β )t), for K +β <. Rigidity for totally integrable convex billiards p. 7/3

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9 Magnetic mirror formula β Y β (a)+ Y (b) = (k(x) βcosφ) Y β Y β sinφ Theorem. If the billiard has no conjugate points, then there exists a measurable function on the phase cylinder a : Ω R such that < a(x, Φ) < L(x, Φ) which satisfies the mirror equation: Y Proof by the following picture: Y (a(x,φ))+ Y Y (L(x 1,Φ 1 ) a(x 1,Φ 1 )) = k(x) sinφ. Rigidity for totally integrable convex billiards p. 9/3

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11 Proof of rigidity for Hyperbolic plane Mirror equation reads: coth(a(x,φ))+coth(l(x 1,Φ 1 ) a(x 1,Φ 1 )) = k(x) sinφ Fort >,cotht is a convex function ontwithcotht > 1. Takeφ = π/ to get that k(x) > 1 for anyx. So the domain must be convex with respect to horocycles on the hyperbolic plane. Moreover, using convexity one has coth a(x,φ)+l(x 1,Φ 1 ) a(x 1,Φ 1 ) k(x) sinφ. This can be written in the equivalent form a(x,φ)+l(x 1,Φ 1 ) a(x 1,Φ 1 ) arctanh ( ) sinφ k(x) Rigidity for totally integrable convex billiards p. 11/3

12 Integrate the last inequality with respect to the invariant measure dµ to get Ldµ = 4 dx dx π/ π arctanh arctanh ( sinφ k(x) ) sinφdφ = ( ) sinφ sinφdφ. k(x) Here P is the perimeter of the boundary curve γ. For every x compute the inner integral on the right hand side integrating by parts π/ arctanh ( sinφ k(x) ) sinφdφ = k(x) π/ = π (k(x) k (x) 1). cos φ k (x) sin φ dφ = Rigidity for totally integrable convex billiards p. 1/3

13 Using Santalo formula Ldµ = πa (A is the area of the domain) we obtain the following inequality A Using Gauss-Bonnet we can write it in the form (k(x) k (x) 1)dx. A π +A k (x) 1dx and therefore k (x) 1dx π. However, it then follows from the next lemma stating the opposite inequality that the curve γ must be a circle. Rigidity for totally integrable convex billiards p. 13/3

14 Lemma for K=-1 Lemma. For any simple closed curve γ on the hyperbolic plane which is convex with respect to horocycles the following inequality holds true I = k (x) 1dx π, where the equality is possible only for circles. Proof. The integral can be estimated from above by Cauchy-Schwartz I ( )1 ( )1 1 (k(x) 1)dx (k(x)+1)dx = ((A+π P)(A+π+P)), where Gauss-Bonnet formula is applied. The last expression gives ((A+π P)(A+π +P)) 1 = (A +4πA P +4π ) 1 π, since by the isoperimetric inequality on the hyperbolic plane A +4πA P. Rigidity for totally integrable convex billiards p. 14/3

15 Proof of Rigidity for the Hemisphere Write the mirror equation for the hemisphere: Claim cot(a(x,φ))+cot(l(x 1,Φ 1 ) a(x 1,Φ 1 )) = k(x) sinφ ( ) a+b cot cota+cotb, for alla,b in the range(;π) satisfying a+b π/. Proof. This becomes obvious in the case when bothaandbbelong to(;π/] simply by the convexity ofcot on this interval. In the remaining case when one of the numbers, sayalies in (;π/) andbin(π/;π) and the average is π/ we can writea = π/ x δ and b = π/+x, wherex,δ are non-negative andx+δ < π/. We need to prove Indeed we have by the trigonometric formula tan δ tan(x+δ) tanx. tan(x+δ) tanx = (tanδ)(1+tanxtan(x+δ)) tanδ tan δ. Rigidity for totally integrable convex billiards p. 15/3

16 Apply the Claim to the followinga = a(x,φ), b = L(x 1,Φ 1 ) a(x 1,Φ 1 ) : cot a(x,φ)+l(x 1,Φ 1 ) a(x 1,Φ 1 ) k(x) sinφ. This can be written in the equivalent form: a(x,φ)+l(x 1,Φ 1 ) a(x 1,Φ 1 ) Ldµ = 4 dx dx π π/ arctan ( ) sinφ. k(x) ( ) sinφ arctan sinφdφ = k(x) ( ) sinφ arctan sinφdφ. k(x) Rigidity for totally integrable convex billiards p. 16/3

17 Compute the inner integral on the right hand side integrating by parts π/ arctan ( sinφ k(x) ) sinφdφ = k(x) π/ = π ( k (x)+1 k(x)). cos φ k (x)+sin φ dφ = Using Santalo formula Ldµ = πa again we obtain the following inequality A ( k (x)+1 k(x))dx. Using Gauss-Bonnet we write it in the form A k (x)+1dx (π A), which leads to k (x)+1dx π. However, the following lemma then implies that the curve γ must be a circle, thus completing the proof of the main theorem for the hemisphere. Rigidity for totally integrable convex billiards p. 17/3

18 Lemma for K=+1 case Lemma. For any simple closed curve on the hemisphere the following inequality holds I = k (x)+1dx π, where the equality happens only for circles. Proof ( k (x)+1+1)dx Then this can be rewritten as ( k (x)+1 1)dx ( k(x)dx) = (π A). (I P)(I +P) (π A), and hence I P +A 4πA+4π 4π. In the above inequality I used the isoperimetric inequality on the sphere: ThusI π. The proof is completed. P +A 4πA. Rigidity for totally integrable convex billiards p. 18/3

19 Remarks on the Lemmas for Horocyclicly convexγ H, I 1 = k (x) 1dx π, for convexγ S. I = k (x)+1dx π, Notice that the second inequality follows from Fenchel inequality for integral curvature of curves in R 3. In fact it was understood by S.Tabachnikov that the analog of Fenchel inequality (with oposit sign) exists also for spacelike curves in LorentzianR 1,. Rigidity for totally integrable convex billiards p. 19/3

20 Rigidity for magnetic billiards Magnetic Santalo formula: Ldµ = πa NoticeY β looks differently in the casesk +β is negative, zero or positive. So the mirror equation also depends. Let us stick tok = 1 andβ (,1) so thatk +β < (other cases are treated analogously). Mirror equation gives coth( 1 β a(x,φ))+coth( 1 β (L(x 1,φ 1 ) a(x 1,φ 1 )) = k(x) βcosφ 1 β sinφ. One can see that for givenx, the minimum of the right hand side equals k (x) β / 1 β, which is attained for some angleφ. Comparing this value with the left hand side, which is obviously strictly greater than 1, we get 1 β < k (x) β. Thus,k(x) 1 for allx, so that the curveγ must be convex with respect to horocycles. Moreover, by the convexity of coth we have: Rigidity for totally integrable convex billiards p. /3

21 1 β coth 1 β [a(x,φ)+l(x 1,φ 1 ) a(x 1,φ 1 )] k(x) βcosφ. sinφ Or equivalently a(x,φ)+l(x 1,φ 1 ) a(x 1,φ 1 ) arctanh 1 β sinφ 1 β k(x) βcosφ Integrating: Ldµ dx π arctanh 1 β sinφ 1 β k(x) βcosφ sinφdφ. Amazingly both LHS and RHS of the last do not depend onβ. So we end with the same inequality as before which is possible only for circles by the lemma. k (x) 1dx π, Rigidity for totally integrable convex billiards p. 1/3

22 Open questions Are there totally integrable outer billiards different from ellipses? Does this type rigidity extends to variable magnetic fields? How about billiards on curved surfaces? Does totally integrable curved billiard table necessarily admitss 1 symmetry group? Given a twist symplectic map of the cylinder. One can show that the condition of no conjugate points is in fact equivalent to existence of measurable field of non-vertical tangent lines which is invariant under the twist map. Does it follow that this field of lines is necessarily smooth? How about higher dimensional convex billiards? Is there a billiard characterization of spheres? Rigidity for totally integrable convex billiards p. /3

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