By Tabachnikov S.
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2 that three consecutive points are related by the equation: ∂H(t−1 , t0 ) ∂H(t0 , t1 ) + = 0. ∂t0 ∂t0 Considering t0 as a function of t−1 and t1 , implicit differentiation yields: ∂t0 ∂ 2 H(t−1 , t0 ) =− ∂t−1 ∂t−1 ∂t0 ∂ 2 H(t−1 , t0 ) ∂ 2 H(t0 , t1 ) + , ∂t0 2 ∂t0 2 45 ∂ 2 H(t0 , t1 ) ∂t0 =− ∂t1 ∂t0 ∂t1 ∂ 2 H(t−1 , t0 ) ∂ 2 H(t0 , t1 ) + ∂t0 2 ∂t0 2 (since the invariant circle is the graph of a Lipshitz function the derivatives exist almost everywhere). In view of the above inequality, the denominators are positive whenever (t −1 , t1 ) is sufficiently close to (t¯−1 , t¯1 ).
Call a billiard trajectory glancing if, for some bounce, the angle of reflection is smaller than . One distinguishes positive and negative -glancing, according to whether the angle with the positive or negative direction of the billiard curve is taken into account. Corollary 5. If the curvature of a smooth convex billiard curve vanishes at some point then, for every > 0, there exists a trajectory that is both positively and negatively -glancing. fig. 41 Proof. Let V be the phase cylinder S 1 × [−π, π] of the billiard transformation T .
Mather on noexistence of invariant curves of a billiard transformation ([Ma 1]). These results are in sharp contrast with those 44 of Lazutkin. First, we formulate the twist property of the billiard map analytically. As before, H(t, t ) denotes the Euclidean distance between two points on the billiard curve whose length parameters are t and t . The following statement simply means that the curve is convex. Lemma 3. ∂ 2 H(t, t ) >0 ∂t ∂t Proof. Consider the segment tt of a billiard trajectory, and let α and α be the corresponding angle variables.