pdf (it, 2.898,97 KB, 25/01/13)

µ 2 + 2µ + n 2 = 0 ∆ = 4(1 − n 2 )
n ∈ N
n = 0 ∆ > 0 µ1 = 0 µ2 = −2 T (t) = d1+d2e −2t
d1, d2 ∈ R T (0) = 0 d1 = −d2
T0(t) = d1(1 − e −2t )
n = 1 ∆ = 0 µ = −1 T (t) =
d1e −t + d2te −t d1, d2 ∈ R T (0) = 0 d1 = 0
T1(t) = d2te t
n > 1 ∆ < 0 µ1 = −1 + i √ n 2 − 1 µ2 = −1 −
i √ n 2 − 1 T (t) = d1e −t cos( √ n 2 − 1 t) + d2e −t sin( √ n 2 − 1 t)
T (0) = 0 d1 = 0 Tn(t) = dne −t sin( √ n 2 − 1 t)
un(x, t) = Tn(t)Xn(x)
u0(x, t) = d0(1 − e −2t )c0 = a0(1 − e −2t )
u1(x, t) = d1te −t c1 cos x = a1te −t cos x
un(x, t) = dne −t sin( n 2 − 1 t) cn cos(nx) = ane −t sin( n 2 − 1 t) cos(nx).
t 0
u(x, t) =
x = ∂tu(x, 0) =
n=0
∂tu0(x, 0) = 2a0
∂tu1(x, 0) = a1 cos x
∂tun(x, 0) = an n2 − 1 cos(nx).
∞
un(x, t) t t = 0
n=0
∞
∞
∂tun(x, 0) = 2a0 + a1 cos x + an n2 − 1 cos(nx)
f(x) = x [0, π]
[−π, π] 2π R n > 1
1
2π
π
0
x dx = π
2
π 2
x cos(nx) dx =
π 0
2
π
|x| ≤ π
x = π 2
−
2 π
∞
n=1
(1 − (−1) n )
n 2
n=2
x sin(nx)
π −
n 0
2
π
sin(nx) dx = −
nπ 0
2(1 − (−1)n )
πn2 x = 2a0 + a1 cos x +
cos(nx) = π 4 2
− cos x −
2 π π
∞
n=2
k=1
∞ 1 − (−1) n
n2 cos(nx),
n=1
an n2 − 1 cos(nx).
a0 = π
4 a1 = − 4
π a2k = 0 a2k+1 = − 4 1
k ∈ N k ≤ 1
π 4k(1 + k)(2k + 1) 2
u(x, t) = π
4 (1 − e−2t ) − 4
π te−t cos x − 4
∞ e
π
−t sin( 4k(1 + k) t)
cos((2k + 1)x).
4k(1 + k)(2k + 1) 2