Алексеев Г.В. Классические методы математической физики. Часть 2
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(λ
k
− λ
m
)ρ(x)X
k
(x)X
m
(x) +
d
dx
{p(x)[X
m
(x)X
′
k
(x) − X
k
(x)X
′
m
(x)]} = 0.
x l
(λ
m
− λ
k
)
l
Z
0
ρ(x)X
k
(x)X
m
(x)dx = p(x)[X
m
(x)X
′
k
(x) − X
k
(x)X
′
m
(x)] |
x=l
x=0
.
(λ
k
− λ
m
)
l
Z
0
ρ(x)X
k
(x)X
m
(x)dx = 0.
λ
m
6= λ
k
λ
1
< λ
2
< ...
X
1
, X
2
, ...
ρ
[p(x)X
′
k
(x)]
′
− q(x)X
k
(x) = −λ
k
ρ(x)X
k
(x).
X
k
λ
k
= −
l
Z
0
{[p(x)X
′
k
(x)]
′
− q(x)X
k
(x)}X
k
(x)dx.
λ
k
=
l
Z
0
[p(x)(X
′
k
(x))
2
+ q(x)X
2
k
(x)]dx − [p(x)X
k
(x)X
′
k
(x)] |
x=l
x=0
.
[p(x)X
k
(x)X
′
k
(x)] |
x=l
x=0
≤ 0.
p(x) ≥ p
0
> 0 q(x) ≥ 0
λ
k
≥ 0 k = 1, 2, ...
0 ≤ λ
1
< λ
2
< ....
1)X(0) = X(l) = 0; 2)X
′
(0) = X
′
(l) = 0;
3)X
′
(0) − h
1
X(0) = 0, X
′
(l) + h
2
X(l) = 0.
h
1
h
2
C
1
[0, l]
v ∈ C
1
[0, l]
v(x) =
∞
X
k=1
a
k
X
k
(x),
v
l
Z
0
|v(x) −
N
X
k=1
a
k
X
k
(x)|
2
dx → 0
N → ∞.
a
k
a
k
=
l
Z
0
ρ(x)v(x)X
k
(x)dx.
λ
k
ρ(x) (0, l)
{X
k
}
∞
k=1
C
1
[0, l]
λ
1
= 0 λ
1
X
1
q(x) ≡ 0
ρ = 1 p = 1 q = 0
a = 1
X
′′
+ λX = 0 (0, l).
α = γ = 1 β = 0 δ = 0
X(0) = X(l) = 0.
λ
k
=
kπ
l
2
, X
k
(x) = sin
kπ
l
x k = 1, 2, ... .
α = γ = 0 β = −1 δ = 1
X
′
(0) = X
′
(l) = 0.
λ
k
X
k
λ
k
=
kπ
l
2
, X
k
(x) = cos
kπ
l
x, k = 0 , 1, 2, ... .
λ
0
= 0, X
0
(x) =
α = δ = 1 β = γ = 0
X(0) = 0, X
′
(l) = 0.
λ
k
=
(2k + 1 )π
2l
2
, X
k
(x) = sin
(2k + 1)π
2l
x, k = 1 , 2, ... .
′
α = δ = 0 β = −1 γ = 1
X
′
(0) = 0, X(l) = 0.
λ
k
=
(2k + 1)π
2l
2
, X
k
(x) = cos
(2k + 1)π
2l
x, k = 1 , 2, ... .
α = 1 β = 0 δ = 1
X(0) = 0, X
′
(l) + γX(l) = 0.
λ
tg
√
λl = −
√
λ
γ
.
λ
k
, k = 1, 2, ...
X
k
(x) = sin
p
λ
k
x, k = 1 , 2, ... .
λ = λ
k
T
k
(t) = a
k
cos
p
λ
k
t + b
k
sin
p
λ
k
t,
a
k
b
k
u
k
(x, t) = X
k
(x)T
k
(t) = (a
k
cos
p
λ
k
t + b
k
sin
p
λ
k
t)X
k
(x)
k = 1, 2, ...
u(x, t) =
∞
X
k=1
(a
k
cos
p
λ
k
t + b
k
sin
p
λ
k
t)X
k
(x).
x t
u |
t=0
= ϕ
0
(x) =
∞
X
k=1
a
k
X
k
(x),
∂u
∂t
|
t=0
= ϕ
1
(x) =
∞
X
k=1
b
k
p
λ
k
X
k
(x).
ϕ
0
ϕ
1
{X
k
}
[0, l]
a
k
b
k
ρX
k
x
l λ
k
> 0
a
k
=
l
Z
0
ρ(x)ϕ
0
(x)X
k
(x)dx, b
k
=
1
√
λ
k
l
Z
0
ρ(x)ϕ
1
(x)X
k
(x)dx, k = 1, 2, ... .
a
k
b
k
u
ϕ
0
ϕ
1
ρ(x)
∂
2
u
∂t
2
=
∂
∂x
p(x)
∂u
∂x
− q(x)u + f(x, t)
f
X
k
u|
x=0
= g
1
(t), u|
x=l
= g
2
(t), t ∈ (0, T ],
C
2
(Q
T
)
C
2
(Q
T
)
u
1
u
2
C
2
(Q
T
) u =
u
1
− u
2
ρ(x)
∂
2
u
∂t
2
=
∂
∂x
p(x)
∂u
∂x
− q(x)u
(x, t) ∈ Q
T
u|
t=0
= 0,
∂u
∂t
t=0
= 0
(0, l)
u|
x=0
= 0, u|
x=l
= 0 (0, T ].
u(x, t) ≡ 0 Q
T
E(t) = E
u
(t) =
1
2
l
Z
0
"
ρ(x)
∂u
∂t
2
+ p(x)
∂u
∂x
2
+ q(x)u
2
#
dx.
t E t
dE(t)
dt
=
l
Z
0
ρ(x)
∂u
∂t
∂
2
u
∂t
2
+ p(x)
∂u
∂x
∂
2
u
∂x∂t
+ q(x)u
∂u
∂t
dx.
u Q
T
dE(t)
dt
=
l
Z
0
ρ(x)
∂
2
u
∂t
2
−
∂
∂x
p(x)
∂u
∂x
+ q(x)u
∂u
∂t
dx + p(x)
∂u
∂x
∂u
∂t
x=l
x=0
.
u
Q
T
∂u
∂t
x=0
= 0
∂u
∂t
x=l
= 0
dE(t)
dt
= 0,
E(t) = const [0, T ].
E(0) =
1
2
l
Z
0
"
ρ(x)
∂u
∂t
2
+ p(x)
∂u
∂x
2
+ q(x)u
2
#
t=0
dx = 0.
E(t) = 0
u(x, t) ≡ 0 Q
T
u
1
= u
2
∂u
∂x
− h
1
u
x=0
= g
1
(t),
∂u
∂x
+ h
2
u
x=l
= g
2
(t)
(0, T ].
h
1
h
2
u
u
1
u
2
u
i
|
t=0
= ϕ
i
(x),
∂u
i
∂t
t=0
= ψ
i
(x)
(0, l), i = 1, 2.
u = u
1
− u
2
Q
T
ϕ(x) = ϕ
1
(x)−ϕ
2
(x), ϕ
′
(x) = ϕ
′
1
(x)−ϕ
′
2
(x)
ψ(x) = ψ
1
(x)−ψ
2
(x) [0, l].
u ∈ C
2
(Q
T
)
u|
t=0
= ϕ(x),
∂u
∂t
t=0
= ψ(x)
(0, l).
ε > 0 δ = δ(ε)
max
kϕk
C[0,l]
, kϕ
′
k
C[0,l]
, kψk
C[0,l]
< δ, (kϕk
C[0,l]
= max
x∈[0,l]
|ϕ(x)|)
kuk
C(
Q
T
)
≡ max
(x,t)∈
Q
T
|u(x, t)| < ε.
E
u
u ∈ C
2
(Q
T
)
Q
T
E(t) ≡ E(0)
E(t) ≡ E(0) =
1
2
l
Z
0
ρ(x)ψ
2
(x) + p(x)[ϕ
′
(x)]
2
+ q(x)ϕ
2
(x)
dx.
M = max{max
x∈[0,l]
ρ(x), m ax
x∈[0,l]
p(x), max
x∈[0,l]
q(x)}.
E(t) ≡
1
2
l
Z
0
[ρ(x)(u
t
)
2
+ p(x)(u
x
)
2
+ q(x)u
2
]dx <
3
2
Mlδ
2
.
l
Z
0
p(x)( u
x
)
2
dx < 3Mlδ
2
.
t ∈ [0, T ]
u [0, l] δ
u C(Q
T
)
u|
x=0
= 0
u(x, t) =
x
Z
0
u
x
dx.
|u(x, t)| ≤
x
Z
0
|u
x
|dx =
x
Z
0
1
p
p(x)
·
p
p(x)|u
x
|dx.
b
Z
a
f(x)g(x)dx
≤
b
Z
a
f
2
(x)dx
1/2
·
b
Z
a
g
2
(x)dx
1/2
.
|u(x, t)| ≤
x
Z
0
dx
p(x)
1/2
·
x
Z
0
p(x)( u
x
)
2
dx
1/2
.
p(x) ≥ p
0
> 0
|u(x, t)| < M
1
δ ∀(x, t) ∈ Q
T
.
M
1
= l
p
3M/p
0
δ = ε/M
1
ϕ, ϕ
′
ψ
C[0, l]
l
Z
0
ϕ
2
(x)dx,
l
Z
0
[ϕ
′
(x)]
2
dx
l
Z
0
ψ
2
(x)dx.
u(0, t) = 0
|u(0, t)| |u(l, t)|
f C(Q
T
)
Ω
x = (x
1
, x
2
, ..., x
n
) R
n
Γ Q
T
= Ω ×(0, T ] Q
T
∂
2
u
∂t
2
= Lu.
L
Lu =
n
X
i,j=1
∂
∂x
i
a
ij
(x)
∂u
∂x
j
− a(x)u,
a
ij
a Ω
a
ij
= a
ji
∈ C
1
(Ω), a ∈ C( Ω),
n
X
i,j=1
a
ij
(x)ξ
i
ξ
j
≥ β
n
X
i=1
ξ
2
i
∀x ∈ Ω, β = const > 0, a(x) ≥ 0.
n
X
i,j=1
a
ij
(x)ξ
i
ξ
j
,
L
Ω Ω
Q
T
u|
Γ
= 0 (0, T ]
u|
t=0
= ϕ
0
(x),
∂u
∂t
t=0
= ϕ
1
(x)
Ω.
u(x, t) = v( x)T (t).