Differential Equations on the two dimensional torus 1 Constant ...

November 1, 2010 9b-3The solution in R 2 has the form( ) px(t) = ω 1 t, y(t) = ω 1 tqLet π : R 2 → T 2 denote the canonical projection.When t = q/ω 1 , we have x(t) = q, y(t) = p, so the projection π(γ 0 ) = γinto T 2 is a closed curve (topological circle) in T 2 .Since all the solutions are vertical tranlates of γ 0 we have that allsolutions in T 2 are periodic of the same period.2. ω 2ω 1is irrational.This case is more interesting. We will show that every orbit in T 2 isdense in T 2 . Hence, all of T 2 is a minimal set.Let us show that the orbit through (0, 0) is dense in T 2 . The analogousstatement for other orbits is similar.Consider the circle S 1 = π({x = 0}).There is a first return map F : S 1 → S 1 (also called Poincaré map)obtained by taking a point (0, y) and taking the point (1, F (y)) wherethe orbit through (0, y) hits the line {x = 1}.Let us compute F .The orbit through (0, y) is {(ω 1 t, ω 2 t + y), t ∈ R}.We getHence,ω 1 t = 1ω 2 t + y = F (y).t = 1/ω 1 , F (y) = y + ω 2 /ω 1So, letting α = ω 2ω 1we get that F (y) = y + α (mod 1).To show that the orbits of X are dense, it suffices to show that theiterates {F n (y), n ∈ Z} are dense mod 1