+ v0 cos(ωt) = v(t)

Problem 1
a)
ẋ(t) = −ωx0 sin(ωt) + v0 cos(ωt) = v(t)
v̇(t) = −ω 2 x0 cos(ωt) − ωv0 sin(ωt) = −ω 2 x(t)
b)
Φs (Φt (x0 , v0 )) =
=
=
=
v(t)
sin(ωs), −ωx(t) sin(ωs) + v(t) cos(ωs)
x(t) cos(ωs) +
ω
n
o
v0
x0 cos(ωt) + sin(ωt) cos(ωs) + {−ωx0 sin(ωt) + v0 cos(ωt)} cos(ωs),
ω
n
o
v0
−ω x0 cos(ωt) + sin(ωt) sin(ωs) + {−ωx0 sin(ωt) + v0 cos(ωt)} cos(ωs)
ω
v0
x0 {cos(ωt) cos(ωs) − sin(ωt) sin(ωs)} + {sin(ωt) cos(ωs) + cos(ωt) sin(ωs)} ,
ω
−ωx0 {cos(ωt) sin(ωs) + sin(ωt) cos(ωs)} + v0 {cos(ωt) cos(ωs) − sin(ωt) sin(ωs)})
v0
x0 cos(ω(t + s)) + sin(ω(t + s)), −ωx0 sin(ω(t + s)) + v0 cos(ω(t + s))
ω
Φt+s (x0 , v0 )
Problem 2
a) For instance: r2 = x2 + y 2 , x = r cos ϕ, y = r sin ϕ
rṙ = xẋ + v v̇ = x(v − x) + v(−x − x) = −r2
r2 ϕ̇ = −ẋv + v̇x = −(v − x)v + (−x − v)x = −r2
⇒
⇒
ṙ = −r
⇒
ϕ̇ = −1
r(t) = r(0)e−t
⇒
ϕ(t) = ϕ(0) − t
Thus
x(t) =
=
=
v(t) =
=
=
r(0)e−t cos(ϕ(0) − t)
r(0) cos(ϕ(0))e−t cos(t) + r(0) sin(ϕ(0))e−t sin(t)
x0 e−t cos(t) + v0 e−t sin(t)
r(0)e−t sin(ϕ(0) − t)
r(0) sin(ϕ(0))e−t cos(t) − r(0) cos(ϕ(0))e−t sin(t)
v0 e−t cos(t) − x0 e−t sin(t)
Φt (x0 , v0 ) = x0 e−t cos(t) + v0 e−t sin(t), v0 e−t cos(t) − x0 e−t sin(t)
b)
t = 0 : x0 = 0, v0 ≥ 0
x(t) = v0 e−t sin(t), v(t) = v0 e−t cos(t)
t = T : x(T ) = v0 e−T sin(T ) = 0, v(T ) = v0 e−T cos(T ) ≥ 0
v1 = v(T ) = v0 e−T
Thus
P : Σ → Σ,
vn 7→ vn+1 = vn e−2π
⇒
T = 2π