倒立振り子の安定性に関する数値計算

1
2007MI226
2
[1]
l 0 y
y
η(t)
l
θ
0
1
x, y
θ y η(t)
(1)
dx
dt
d 2 x
dt 2
dy
dt
d 2 y
dt 2
x = l 0 sin θ, y = −l 0 cos θ + η(t) (1)
= l 0 cos θ dθ
dt
= −l 0 sin θ dθ
dt + l 0 cos θ d2 θ
dt 2
= l 0 sin θ dθ
dt + η′ (t)
= l 0 cos θ dθ
dt + l 0 sin θ d2 θ
dt 2 + η′′ (t)
m d2 x
dt 2 = 0, y
md2 = −mg (m
dt2 ) m
−l 0 sin θ dθ
dt + l 0 cos θ d2 θ
dt 2 = 0
l 0 cos θ dθ
dt + l 0 sin θ d2 θ
dt 2
x
= −g − η ′′ (t)
()× cos θ + ()× sin θ
d 2 θ
dt 2
= − g + η′′ (t)
l 0
sin θ (2)
g (g = 9.8 (m/s 2 )
3
[2]
dx
dt = f(t, x) (t 0 ≤ t ≤ t F ), x(t 0 ) = x 0 (3)
x(t) R d
f [t 0 , t F ] × R d R d ()
[t 0 , t F ]
t 0 < t 1 < · · · < t n < t n+1 < · · · < t N = t F
t n h n = t n+1 − t n
n
h = (t F − t 0 )/N
x(t) n
x(t n ) x n x n (n =
1, 2, ... )
(discrete variable method)
(Euler method)
x n+1 = x n + hf(t n , x n ) (4)
x 0 (4)
x 1 , x 2 , . . .
(Runge-Kutta method)
x n x n+1
k 1 = f(t n , x n )
(
k 2 = f t n + h 2 , x n + h )
2 k 1
(
k 3 = f t n + h 2 , x n + h )
2 k 2
)
= f
(t n + h, x n + hk 3
k 4
x n+1 = x n + h 6
(k 1 + 2k 2 + 2k 3 + k 4
)

4
l 0 = 9.8 (m)
η(t) = sin(ωt) (ω ) (5)
θ(0) = 3.04, θ ′ (0) = 0.0 (6)
(2) (
) ω
C ++
[3]
y
y
5
0
-5
-10
-10 -5 0 5 x
5 ω = 87.0 , h=1/20
y
5
5
0
0
-5
ω = 15
-5
-10
-10
-10 -5 0 5
x
-10 -5 0 5
x
6 ω = 87.0 , h=1/40
2 ω = 15.0 , h=1/20
y
5
0
-5
-10
ω = 16
-10 -5 0 5
3 ω = 16.0 , h=1/20
y
5
0
-5
-10
-10 -5 0 5 x
4 ω = 86.0 , h=1/20
x
5
5.1
ω 1.099.0
ω = 1.015.0
ω = 16.0
ω
ω = 87.0
ω = 99.0
ω = 87.0
ω
[1] 1994
[2]
2000
[3] Windows
2
2003.