Rigidity for totally integrable convex billiards ICMAT, Madrid, - PowerPoint PPT Presentation

Rigidity for totally integrable convex billiards ICMAT, Madrid, November 11-15, 2013 Michael Bialy bialy@post.tau.ac.il Tel Aviv University Rigidity for totally integrable convex billiards – p. 1/23

Introduction 1.Birkhoff billiards 2. 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. 2/23

Rigidity for totally integrable convex billiards – p. 3/23

Rigidity for totally integrable convex billiards – p. 4/23

Approaches: 1) Bolotin: Polynomial integrals-requires smoothness of certain complex algebraic curve. 2) Delshams, Ramirez-Ros: Small perturbation of ellipse–splitting of separatrices. 3) Kaloshin and Sorrentino: If an integrable billiard is C 2 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=0, 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. 5/23

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 2. Let γ be smooth simple closed curve as above on S . If the billiard ball map is totally integrable then γ is a circle. Theorem 3. Theorems 1,2 can be generalized to magnetic billiards on constant curvature surface S . Remark: Magnetic billiards in ellipses seem to be non-integrable: Robnik, M, Berry, M V, 1985, J.Phys.A 18, 1361-1378, ’Classical billiards in magnetic fields’. In contrast with the 2D 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. 6/23

Mirror equation on surfaces 1. Mirror formula in the plane: 1 a + 1 b = 2 k ( x ) sin φ . 2.Mirror formula on surface: ′ ′ Y Y ( a ) + Y Y ( b ) = 2 k ( x ) sin φ , Y ( t ) = sinh t ( K = − 1); Y ( t ) = t ( K = 0); Y ( t ) = sin t ( K = 1) . 3. Mirror formula for magnetic field β . The Jacobi fields change according to the effective curvature K + β 2 : 1 K + β 2 t ) , for K + β 2 > 0 , � Y β = K + β 2 sin( � Y β = t, for K + β 2 = 0 , 1 − ( K + β 2 ) t ) , for K + β 2 < 0 . � Y β = sinh( � − ( K + β 2 ) Rigidity for totally integrable convex billiards – p. 7/23

Rigidity for totally integrable convex billiards – p. 8/23

Magnetic mirror formula ′ ′ Y ( a ) + Y ( b ) = 2( k ( x ) − β cos φ ) β β Y β Y β sin φ Theorem . If the billiard has no conjugate points, then there exists a measurable function on the phase cylinder a : Ω → R such that 0 < a ( x, Φ) < L ( x, Φ) which satisfies the mirror equation: ′ ′ Y Y ( a ( x, Φ)) + Y Y ( L ( x − 1 , Φ − 1 ) − a ( x − 1 , Φ − 1 )) = 2 k ( x ) sin φ . Proof by the following picture: Rigidity for totally integrable convex billiards – p. 9/23

Rigidity for totally integrable convex billiards – p. 10/23

Proof of rigidity for Hyperbolic plane Mirror equation reads: coth ( a ( x, Φ)) + coth ( L ( x − 1 , Φ − 1 ) − a ( x − 1 , Φ − 1 )) = 2 k ( x ) sin φ For t > 0 , coth t is a convex function on t with coth t > 1 . Take φ = π/ 2 to get that k ( x ) > 1 for any x . 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 ) sin φ. 2 This can be written in the equivalent form � sin φ � a ( x, Φ) + L ( x − 1 , Φ − 1 ) − a ( x − 1 , Φ − 1 ) ≥ arctanh 2 k(x) Rigidity for totally integrable convex billiards – p. 11/23

Integrate the last inequality with respect to the invariant measure dµ to get � P � π � sin φ � � L dµ ≥ 2 dx arctanh sin φ d φ = k(x) 0 0 � P � π/ 2 � sin φ � = 4 dx arctanh sin φ d φ. k(x) 0 0 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/ 2 � π/ 2 cos 2 φ � sin φ � arctanh sin φ d φ = k(x) k 2 (x) − sin 2 φ d φ = k(x) 0 0 = π � 2 ( k ( x ) − k 2 ( x ) − 1) . Rigidity for totally integrable convex billiards – p. 12/23

� L dµ = 2 πA ( A is the area of the domain) we obtain the following Using Santalo’ formula inequality � P � A ≥ ( k ( x ) − k 2 ( x ) − 1) dx. 0 Using Gauss-Bonnet we can write it in the form � P � A ≥ 2 π + A − k 2 ( x ) − 1 dx 0 and therefore � P � k 2 ( x ) − 1 dx ≥ 2 π. 0 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. 13/23

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 � P � I = k 2 ( x ) − 1 dx ≤ 2 π, 0 where the equality is possible only for circles. Proof. The integral can be estimated from above by Cauchy-Schwartz � 1 � 1 �� P 2 �� P 2 1 2 , I ≤ ( k ( x ) − 1) dx ( k ( x ) + 1) dx = (( A + 2 π − P )( A + 2 π + P )) 0 0 where Gauss-Bonnet formula is applied. The last expression gives 2 = ( A 2 + 4 πA − P 2 + 4 π 2 ) 1 1 2 ≤ 2 π, (( A + 2 π − P )( A + 2 π + P )) since by the isoperimetric inequality on the hyperbolic plane A 2 + 4 πA − P 2 ≤ 0 . Rigidity for totally integrable convex billiards – p. 14/23

Proof of Rigidity for the Hemisphere Write the mirror equation for the hemisphere: cot ( a ( x, Φ)) + cot ( L ( x − 1 , Φ − 1 ) − a ( x − 1 , Φ − 1 )) = 2 k ( x ) sin φ Claim � a + b � ≤ cot a + cot b cot , 2 2 for all a, b in the range (0; π ) satisfying a + b ≤ π/ 2 . 2 Proof. This becomes obvious in the case when both a and b belong to (0; π/ 2] simply by the convexity of cot on this interval. In the remaining case when one of the numbers, say a lies in (0; π/ 2) and b in ( π/ 2; π ) and the average is ≤ π/ 2 we can write a = π/ 2 − x − δ and b = π/ 2 + x , where x, δ are non-negative and x + δ < π/ 2 . We need to prove tan δ 2 ≤ tan( x + δ ) − tan x . 2 Indeed we have by the trigonometric formula tan( x + δ ) − tan x = (tan δ )(1 + tan x tan( x + δ )) ≥ tan δ ≥ tan δ 2 . � 2 2 2 Rigidity for totally integrable convex billiards – p. 15/23

Apply the Claim to the following a = 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 ) sin φ. 2 This can be written in the equivalent form: � sin φ � a ( x, Φ) + L ( x − 1 , Φ − 1 ) − a ( x − 1 , Φ − 1 ) ≥ arctan . 2 k ( x ) � P � π � sin φ � � L dµ ≥ 2 dx arctan sin φ dφ = k ( x ) 0 0 � P � π/ 2 � sin φ � = 4 dx arctan sin φ dφ. k ( x ) 0 0 Rigidity for totally integrable convex billiards – p. 16/23

Compute the inner integral on the right hand side integrating by parts � π/ 2 � π/ 2 cos 2 φ � sin φ � arctan sin φ dφ = k ( x ) k 2 ( x ) + sin 2 φ dφ = k ( x ) 0 0 = π � 2 ( k 2 ( x ) + 1 − k ( x )) . � L dµ = 2 πA again we obtain the following inequality Using Santalo’ formula � P � A ≥ ( k 2 ( x ) + 1 − k ( x )) dx. 0 Using Gauss-Bonnet we write it in the form � P � A ≥ k 2 ( x ) + 1 dx − (2 π − A ) , 0 which leads to � P � k 2 ( x ) + 1 dx ≤ 2 π. 0 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. 17/23

Lemma for K=+1 case Lemma. For any simple closed curve on the hemisphere the following inequality holds � P � I = k 2 ( x ) + 1 dx ≥ 2 π, 0 where the equality happens only for circles. Proof � P � P �� P � 2 � � = (2 π − A ) 2 . ( k 2 ( x ) + 1+1) dx · ( k 2 ( x ) + 1 − 1) dx ≥ k ( x ) dx 0 0 0 Then this can be rewritten as ( I − P )( I + P ) ≥ (2 π − A ) 2 , and hence I 2 ≥ P 2 + A 2 − 4 πA + 4 π 2 ≥ 4 π 2 . In the above inequality I used the isoperimetric inequality on the sphere: P 2 + A 2 − 4 πA ≥ 0 . Thus I ≥ 2 π . The proof is completed. Rigidity for totally integrable convex billiards – p. 18/23