Problem 1
a) If O(εn ) denotes terms of order εn then
∂h1
∂h1
+ε
ẋ(t) + O(ε2 )
∂t
∂x
∂h1
+ O(ε2 )
= εf (x(t), t) + ε
∂t
ż(t) = ẋ(t) + ε
Introducing
δf (x, t) = f (x, t) − f¯(x), i.e.
Z
T
δf (x, t0 )dt0 = 0
0
we have
∂h1
¯
+ O(ε2 )
ż(t) = εf (x(t)) + ε δf (x(t), t) +
∂t
Thus
Z
t
h1 (x, t) = −
δf (x, t0 )dt0
0
satisfies the periodicity condition 0 = h1 (x, 0) = h1 (x, T ) and given that x = z + O(ε) we end up
with
ż(t) = εf¯(x(t)) + O(ε2 ) = εf¯(z(t)) + O(ε2 )
b) Using
sin(ωt)
ω
cos(ωt)
v = x sin(ωt) + ẋ
ω
u = x cos(ωt) − ẋ
we obtain
u̇ = ẋ cos(ωt) − ω sin(ωt)x − ẍ
sin(ωt)
− ẋ cos(ωt)
ω
sin(ωt)
ε(x2 − 1)ẋ + ω02 x − εh cos(ωt)
ω
cos(ωt)
v̇ = ẋ sin(ωt) + ω cos(ωt)x + ẍ
− ẋ sin(ωt)
ω
cos(ω)
= ω cos(ωt)x +
−ε(x2 − 1)ẋ − ω02 x + εh cos(ωt)
ω
= −ω sin(ωt)x +
Using the detuning ε∆ = ω 2 − ω02 the system results in
∆
sin(ωt)
2
u̇ = ε − sin(ωt)x +
(x − 1)ẋ − h cos(ωt)
ω
ω
∆
cos(ωt)
2
v̇ = ε
cos(ωt)x +
−(x − 1)ẋ + h cos(ωt)
ω
ω
where on the right hand side x and ẋ have to be replaced by
x = u cos(ωt) + v sin(ωt)
ẋ = −ω sin(ωt)u + ω cos(ωt)v
1
Thus u and v are slow variables and we map apply averaging. The linear and the constant
contribution are easy to deal with. Using
Z 2π
Z 2π
1
1
1
2
cos (x)dx =
sin2 (x)dx =
2π 0
2π 0
2
and the orthogonality between the different trigonometric functions we have (T = 2π/ω)
1
T
Z
T
∆
∆
sin(ωt)xdt = −
ω
2ω
∆
∆
cos(ωt)xdt =
ω
2ω
−
0
Z
1 T
T 0
Z T
1
sin(ωt)
(−h cos(ωt))dt = 0
T 0
ω
Z
1 T cos(ωt)
h
h cos(ωt)dt =
T 0
ω
2ω
We are thus left with the nonlinear contribution
(x2 − 1)ẋ = (u2 cos2 (ωt) + v 2 sin2 (ωt) + 2uv sin(ωt) cos(ωt) − 1)(−ω sin(ωt)u + ω cos(ωt)v)
If we observe that
Z 2π
1
1
cos2 (x) sin2 (x)dx = ,
2π 0
8
1
2π
Z
0
2π
1
cos (x)dx =
2π
4
Z
0
2π
sin4 (x)dx =
3
8
and that all the other integrals involving odd powers of trigonometric functions will vanish we
end up with
Z
1 T sin(ωt) 2
v u
u
3u
(x − 1)ẋdt = −u2 − v 2
+ 2uv +
T 0
ω
8
8
8 2
u
1
=
1 − (u2 + v 2 )
2
4
Z T
3v
1
cos(ωt) 2
v
u v
−
(x − 1)ẋdt = −u2 − v 2 + 2uv +
T 0
ω
8
8
8 2
v
1
1 − (u2 + v 2 )
=
2
4
If we combine all the results the averaged equation of motion reads
ε
∆
1 2
2
u̇ =
− v + u 1 − (u + v )
2
ω
4
ε ∆
1 2
h
2
v̇ =
u + v 1 − (u + v ) +
2 ω
4
ω
A simple linear rescaling yields the expression used in the lecture notes with σ = ∆/ω and
γ = −h/(2ω).
2