Алексеев Г.В. Классические методы математической физики. Часть 2
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t=t
0
t=t
1
t=t
2
S
R
3R
R
2R
S
S
R
1
R
R
C
2
R
n
R
3
R
n
l
∂
2
u
∂t
2
= a
2
∂
2
u
∂x
2
Q
T
= (0, l) × (0, T ] 0 < T < ∞
u|
x=0
= 0, u|
x=l
= 0 (0, T ]
u|
t=0
= ϕ
0
(x),
∂u
∂t
t=o
= ϕ
1
(x)
(0, l).
a > 0
ϕ
0
ϕ
1
u
u(x, t) = T (t)X(x).
T (t) X(x)
T
′′
(t)X(x) = a
2
T (t)X
′′
(x).
a
2
XT
T
′′
(t)
a
2
T (t)
=
X
′′
(x)
X(x)
.
t x
x t
λ
T
′′
(t) + a
2
λT (t) = 0
T
X
′′
(x) + λX(x) = 0
X
X(0) = 0, X(l) = 0.
λ
λ
λ < 0, λ = 0 λ > 0
λ < 0
X(x) = C
1
e
√
−λx
+ C
2
e
−
√
−λx
,
C
1
C
2
C
1
+ C
2
= 0, C
1
e
√
−λl
+ C
2
e
−
√
−λl
= 0.
C
1
= 0, C
2
= 0 X(x) ≡ 0
λ = 0
X(x) = C
1
+ C
2
x.
C
1
+ C
2
· 0 = 0 C
1
+ C
2
l = 0
C
1
= 0, C
2
= 0 X(x) ≡ 0
λ > 0 X(x) = C
1
cos
√
λx+
+C
2
sin
√
λx C
1
·1 + C
2
·
0 = 0 C
1
cos
√
λl + C
2
sin
√
λl = 0 C
1
= 0
C
2
sin
√
λl = 0 C
2
6= 0
X(x) ≡ 0 sin
√
λl = 0
√
λ = (kπ/l) k
λ = λ
k
λ
k
=
kπ
l
2
, k = 1, 2, 3, ... .
X
k
(x) = sin
kπx
l
,
k
λ
−k
= λ
k
k
λ
k
X
k
λ λ
k
T
k
(t) = a
k
cos
kπat
l
+ b
k
sin
kπat
l
,
a
k
b
k
u
k
(x, t) = T
k
(t)X
k
(x) =
a
k
cos
kπat
l
+ b
k
sin
kπat
l
sin
kπx
l
, k = 1, 2, ... .
u
k
k a
k
b
k
u(x, t) =
∞
X
k=1
a
k
cos
kπat
l
+ b
k
sin
kπat
l
sin
kπx
l
t
Q
T
= [0, l] × [0, T ]
x t Q
T
a
k
b
k
t
∂u
∂t
=
∞
X
k=1
kπa
l
−a
k
sin
kπat
l
+ b
k
cos
kπat
l
sin
kπx
l
.
ϕ
0
(x) =
∞
X
k=1
a
k
sin
kπx
l
, ϕ
1
(x) =
∞
X
k=1
kπa
l
b
k
sin
kπx
l
.
ϕ
0
ϕ
1
(0, l)
C[0, l]
a
k
b
k
ϕ
0
ϕ
1
a
k
=
2
l
l
Z
0
ϕ
0
(x)sin
kπx
l
dx, b
k
=
2
kπa
l
Z
0
ϕ
1
(x)sin
kπx
l
dx.
a
k
b
k
Q
T
∂u/∂t
x t Q
T
(x, t) ∈ Q
T
u
k
T
k
X
k
t x
a
k
b
k
Q
T
∂u/∂t
x t Q
T
ϕ
0
[0, l]
ϕ
0
(0) = ϕ
0
(l) = 0, ϕ
′′
0
(0) = ϕ
′′
0
(l) = 0,
ϕ
1
[0, l]
u
ϕ
1
(0) = ϕ
1
(l) = 0,
u
Q
T
(x, t) ∈ Q
T
t ∈ [0, T ] x ∈ [0, l]
x t
Q
T
a
k
= −
l
π
3
b
(3)
k
k
3
, b
k
= −
l
π
3
a
(2)
k
ak
3
,
b
(3)
k
=
2
l
l
Z
0
ϕ
′′′
0
(x)cos
kπx
l
dx, a
(2)
k
=
2
al
l
Z
0
ϕ
′′
1
(x)sin
kπx
l
dx.
∞
X
k=1
|a
(2)
k
|
k
,
∞
X
k=1
|b
(3)
k
|
k