Алексеев Г.В. Классические методы математической физики. Часть 2
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ε → 0
u ∈ C
2
(Ω)
Ω
(n)
⊂ Ω Ω
n → ∞ Ω
(n)
Ω
(n)
→ Ω
x
0
Ω x
0
Ω v ∈ C
2
(Ω) ∆v = 0 Ω
u v = 1/r
Z
Γ
1
|x − x
0
|
∂u(x)
∂n
x
− u(x)
∂
∂n
x
1
|x − x
0
|
σ
x
−
Z
Ω
∆u(x)
|x − x
0
|
x = 0.
x
0
∈ Γ
x
0
Γ Γ ∈ C
1
Γ
S
ε
(x
0
) ε > 0
x
0
Γ S
1ε
S
2ε
S
1ε
Ω S
2ε
Ω S
ε
Γ Γ
2ε
S
ε
Γ
1ε
Ω
ε
Ω Γ
2ε
S
1ε
Ω
u v = 1/r Ω
ε
Z
Ω
ε
∆u
r
x =
Z
Γ
2ε
1
r
∂u
∂n
x
− u
∂
∂n
x
1
r
σ
x
+
Z
S
1ε
1
r
∂u
∂n
x
− u
∂
∂n
x
1
r
σ
x
.
ε → 0 Ω
ε
ε → 0
Ω Γ
2ε
ε → 0
Γ
−2πu(x
0
)
ε → 0
4π 2π
Cu(x
0
) =
1
4π
Z
Γ
1
|x − x
0
|
∂u(x)
∂n
x
− u(x)
∂
∂n
x
1
|x − x
0
|
σ
x
−
1
4π
Z
Ω
∆u(x)
|x − x
0
|
x.
C = C(x
0
) =
1, x
0
∈ Ω,
c(x
0
), x
0
∈ ∂Ω,
0, x
0
6∈
Ω,
c(x
0
) ∈ [0, 1] x
0
x
0
c(x
0
) =
1/2
x
0
Γ x
0
Γ x
0
Γ x
0
c(x
0
) = α/4π α
Γ x
0
Ω u
Cu(x
0
) =
1
4π
Z
Γ
1
|x − x
0
|
∂u(x)
∂n
x
− u(x)
∂
∂n
x
1
|x − x
0
|
σ
x
.
x
0
∈ Ω
u(x
0
) =
1
4π
Z
Γ
1
|x − x
0
|
∂u(x)
∂n
x
− u(x)
∂
∂n
x
1
|x − x
0
|
σ
x
.
R
2
R
n
v(x) = ln(1/|x − x
0
|)
|x − x
0
| ≡
p
(x − x
0
)
2
+ (y − y
0
)
2
x x
0
v(x) = 1/(ω
n
|x − x
0
|
n−2
) |x − x
0
| =
p
(x
1
− x
0
1
)
2
+ (x
2
− x
0
2
)
2
+ . . . + (x
n
− x
0
n
)
2
n ≥ 3
ω
n
R
n
Cu(x
0
) =
1
2π
Z
Γ
ln
1
|x − x
0
|
∂u(x)
∂n
x
− u(x)
∂
∂n
x
ln
1
|x − x
0
|
σ
x
−
−
1
2π
Z
Ω
∆u(x) ln
1
|x − x
0
|
x
R
2
Cu(x
0
) =
1
ω
n
Z
Γ
1
|x − x
0
|
n−2
∂u(x)
∂n
x
− u(x)
∂
∂n
x
1
|x − x
0
|
n−2
σ
x
−
−
1
ω
n
Z
Ω
∆u(x)
1
|x − x
0
|
n−2
x
R
n
C
c(x
0
)
c(x
0
) = 1/2 x
0
Γ u Ω
Cu(x
0
) =
1
2π
Z
Γ
ln
1
|x − x
0
|
∂u(x)
∂n
x
− u(x)
∂
∂n
x
ln
1
|x − x
0
|
σ
x
.
x
0
∈ Ω
u(x
0
) =
1
2π
Z
Γ
ln
1
|x − x
0
|
∂u(x)
∂n
x
− u(x)
∂
∂n
x
ln
1
|x − x
0
|
σ
x
.
(n = 1) u
′′
= 0
u(x) = C
1
x + C
2
x
(a, b) u ∈ C
2
(a, b)
u
′′
= 0
x
n
[a, b] x b a
n
du(a)
dn
+
du(b)
dn
=
du(a)
d(−x)
+
du(b)
dx
= 0.
u (a, b)
(a, b)
(a, b) u ( α + β)/2
[α, β] (a, b)
u [α, β]
u [α, β]
u
α + β
2
=
1
β − α
Z
β
α
u(x)
x =
u(α) + u(β)
2
.
(a, b) u
(a, b)
(a, b)
(a, b)
u(x) ≡ const
(a, b) [a, b]
x = a x = b
R
n
R
n
R
n
n Ω ⊂ R
n
Γ = ∂Ω Ω
Ω Ω
Ω
Ω Ω
e
= R
n
\Ω Ω
Ω
e
W
W
e
W
e
0
¥
Ω
R
3
R
2
Ω
e
Ω
0
e
Ω
∞
e
Ω
e
Ω
∞
e
Ω
e
Ω
e
Ω
e
= Ω
e
∪ Γ
C
k
(Ω
e
) Ω
e
k Ω ∪ Γ
e
u ∈ C
k
(Ω
e
)
u(x) |x| → ∞
Ω R
3
Γ u ∈ C
1
(Ω) Ω
Z
Γ
∂u
∂n
σ = 0.
u v ≡ 1
u Ω
Ω
Ω
x
0
Ω
Γ
′
⊂ Ω u Ω u
Γ
′
u
Γ
′
Γ
′
u(x
0
) =
1
4π
Z
Γ
′
1
|x − x
0
|
∂u(x)
∂n
x
− u(x)
∂
∂n
x
1
|x − x
0
|
σ
x
.
x
0
6∈ Γ
′
1/|x − x
0
|
x
0
, y
0
, z
0
x
0
x
0
, y
0
, z
0
u
u
Ω x
0
∈ Ω
u(x
0
) =
1
4πa
2
Z
Γ
a
u(x)
σ,
Γ
a
a x
0
Ω
Ω x
0
∈ Ω
Γ
a
a x
0
Γ
a
u
B
a
≡ B
a
(x
0
)
a x
0
∈ Ω Γ
a
u(x
0
) =
1
4π
Z
Γ
a
1
|x − x
0
|
∂u(x)
∂n
x
− u(x)
∂
∂n
x
1
|x − x
0
|
σ.
|x − x
0
| = a Γ
a
Z
Γ
a
∂u
∂n
σ = 0,
∂
∂n
1
|x − x
0
|
=
∂
∂r
1
r
= −
1
r
2
= −
1
a
2
Γ
a
,
a = ρ
4πρ
2
u(x
0
) =
Z
Γ
ρ
u(x)
σ
ρ 0 a
4
3
πa
3
u(x
0
) =
a
Z
0
Z
Γ
ρ
u(x)
σ
ρ =
Z
B
a
u(x)
x
u(x
0
) =
1
V
a
Z
B
a
u(x)
x, V
a
=
4π
3
a
3
.
u
u
Ω
Ω u ≡
u
u
0
x
0
Ω
u
0
= max
x∈Ω
u(x) = u(x
0
) ≥ u(x) ∀x ∈ Ω.
x
0
Γ
a
a Ω
u
u(x
0
) =
1
4πa
2
Z
Γ
a
u(x)
σ ≤
1
4πa
2
Z
Γ
a
u
0
σ = u
0
.
x Γ
a
u(x) < u(x
0
) u Γ
a
≤ <
u
0
= u(x
0
)
u(x) = u(x
0
) = u
0
∀x ∈ Γ
a
.
a
u u
0
x
0
Ω
u(x) = u
0
Ω y ∈ Ω
x
0
y l Ω
Ω d
l Γ Ω
u(x) u
0
x
0
d/2 x
1
l
u(x
1
) = u
0
u(x) = u
0
x
1
d/2 x
2
l
u(x) = u
0
x
2
d/2 l
y
u(y) = u
0
u
Ω Ω
u
x
y
0
Ω
Ω
Ω
u
Γ
Ω
Ω = Ω
1
∪ Ω
2
Ω
1
Ω
2
u
Ω
1
Ω
2
u Ω
Ω
1
Ω
2
Ω
H(Ω)
Ω
u ∈ H(Ω) ∩ C(Ω) u : Ω → R
Ω Ω
u ∈ H( Ω)∩C(Ω)
Γ
m ≡ min
x∈Γ
u(x) < u(x) < max
x∈Γ
u(x) ≡ M ∀x ∈ Ω.
Ω u
M m
u Ω u
M m Ω
M m Γ Ω
u ∈ H( Ω) ∩C(Ω) Γ
u(x) ≡ 0 Ω
u Ω
u = 0 Ω
u ∈ H(Ω) ∩ C(Ω)
Γ u Ω
u, v ∈ H(Ω) ∩ C(Ω)
u ≤ v Γ u ≤ v Ω
v −u Ω Ω
Γ v − u ≥ 0 Ω
v ≥ 0 u, v ∈ H(Ω) ∩ C(Ω)
|u| ≤ v Γ.
|u| ≤ v Ω
−v ≤ u ≤ v
Γ −v ≤ u ≤ v Ω
|u| ≤ v Ω
v
Ω u Γ
u
u
u ∈ H(Ω) ∩ C(Ω)
|u(x)| ≤ max
x∈Γ
|u(x)| ∀x ∈ Ω.
M = max
x∈Γ
|u(x)|
u Γ Ω
Ω
R
n
n ≥ 2
u(x
0
) =
1
2πa
Z
Γ
a
u(x)ds, u(x
0
) =
1
πa
2
Z
K
a
u(x)dxdy.
Γ
a
K
a
a
x
0
ds Γ
n
X
i,j=1
a
ij
(x)
∂
2
u
∂x
i
∂x
j
+
n
X
i=1
b
i
(x)
∂u
∂x
i
+ c(x)u = f
a
ij
b
i
c
∆u = 0.
Ω u C
2
(Ω) ∩C(Ω)
x ∈ Ω
u = g Γ.