Алексеев Г.В. Классические методы математической физики. Часть 2
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Ω
y
x
Γ
2
δ
2δ
x
0
*
γ
γ
)
0
(x
S
δ
S )
0
2
(x
δ
0
0
Ω
u
g
Γ
a
u(x) =
(
1
2πa
R
Γ
a
(a
2
−ρ
2
)
|x−y|
2
g(y)ds
y
, x ∈ Ω,
g(x), x ∈ Γ
a
C
2
(Ω)∩C(Ω) g ∈ C(Γ
a
)
g ≡ 1 Γ
a
u ≡ 1 Ω
g = 1, u = 1
1 =
1
2π
2π
Z
0
(a
2
− ρ
2
)dψ
a
2
− 2aρcos(ϕ − ψ) + ρ
2
=
1
2πa
Z
Γ
a
a
2
− ρ
2
r
2
ds
y
, ρ = |x| ≤ a.
g(x
0
) x
0
∈ Γ
a
u(x) − g(x
0
) =
1
2πa
Z
Γ
a
[g(y) − g(x
0
)]
a
2
− ρ
2
r
2
ds
y
, ρ = |x|.
ε > 0 x
0
S
2δ
= S
2δ
(x
0
) 2δ
g δ
y Γ
1
Γ
a
S
2δ
|g(y) − g(x
0
)| <
ε
2
.
Γ
2
= Γ
a
\Γ
1
, I
k
(x, x
0
) =
1
2πa
Z
Γ
k
[g(y) −g(x
0
)]
a
2
− ρ
2
r
2
ds
y
, k = 1, 2.
|u(x) − g(x
0
)| ≤ |I
1
(x, x
0
)| + |I
2
(x, x
0
)| ∀x ∈ Ω.
|I
1
(x, x
0
)| <
ε
2
1
2πa
Z
Γ
1
a
2
− ρ
2
r
2
ds
y
<
ε
2
1
2πa
Z
Γ
a
a
2
− ρ
2
r
2
ds =
ε
2
∀x ∈ Ω.
|I
2
(x, x
0
)|
S
δ
= S
δ
(x
0
) x
0
δ
u(x) x → x
0
x
S
δ
y ∈ Γ
2
r ≡ |x − y| > δ
[0, 2π] g |g(y)| ≤ M
′
= const Γ
a
|I
2
(x, x
0
)| ≤
2M
′
(a
2
− ρ
2
)
2πaδ
2
Z
Γ
2
ds ≤
2M
′
(a
2
− ρ
2
)
δ
2
, ρ = |x|.
δ x → x
0
ρ = |x| → a
δ
1
> 0
|I
2
(x, x
0
)| ≤
ε
2
|x − x
0
| < δ
1
.
ε
|u(x) − g(x
0
)| → 0 |x − x
0
| → 0.
u ∈ C(Ω) u ∈ C
2
(Ω)
(i)
Ω = Ω
a
g
u(x) =
(
1
2πa
R
Γ
a
(ρ
2
−a
2
)
|x−y|
2
g(y)ds
y
, x ∈ Ω
e
,
g(x), x ∈ Γ
a
u ∈ C
2
(Ω
e
) ∩ C
0
(Ω
e
)
(i)
Γ
a
a x
0
ρ = |x − x
0
|
ρ = |x − x
0
| Ω
e
Γ
a
Ω u Ω
u(x) =
1
2πa
Z
Γ
a
a
2
− |x|
2
|x − y|
2
u(y)ds
y
, x ∈ Ω.
u
1
g = u|
Γ
a
∈ C(Γ
a
)
u
1
= u Ω
Ω
u Γ
a
x = 0
R
2
u Ω
R
2
\Ω
R
2
R
3
§6
§4
R
2
u : Ω
e
→ R Ω
e
Ω ⊂ R
2
R > 0 C = C
R
(u)
∂u(x)
∂x
≤
C
R
|x|
2
,
∂u(x)
∂y
≤
C
R
|x|
2
|x| ≥ R.
Ω
Γ
R
R Ω Γ
R
Ω
eR
= {x ∈ Ω
e
: |x| > R} u Ω
eR
Ω
eR
x = (x, y) Γ
R
u(x) =
1
2πR
Z
Γ
R
ρ
2
− R
2
r
2
u(y)ds
y
=
=
1
2πR
Z
|y|=R
|x|
2
− R
2
|x − y|
2
u(y)ds
y
, x ∈ Ω
eR
.
ρ = |x| =
p
x
2
+ y
2
y = (ξ, η) r = |x − y| =
p
(x − ξ)
2
+ (y − η)
2
x
∂u(x)
∂x
=
1
2πR
Z
Γ
R
∂
∂x
(
ρ
2
− R
2
r
2
)u(y)ds
y
.
∂u/∂x
∂
∂x
(
ρ
2
− R
2
r
2
) =
2xr
2
− 2(ρ
2
− R
2
)(x − ξ)
r
4
=
=
2
r
4
[xr(r − ρ) + ρx(r −ρ) + ξρ
2
+ R
2
(x − ξ)].
ρ ≥ 2R R ≤ ρ/2 r ≥ ρ − R ≥ ρ/2 ⇒ 1/r ≤ 2/ρ
|x| ≤ ρ |x − ξ|/r ≤ r |r − ρ| ≤ R |ξ| ≤ R
|
∂
∂x
(
ρ
2
− R
2
r
2
)| ≤
2|x||r − ρ|
r
3
+
2|x||r − ρ|ρ
r
4
+
2|ξ|ρ
2
r
4
+
2R
2
|x − ξ|
r
4
≤
88R
ρ
2
.
|
∂u(x)
∂x
| ≤
88R
ρ
2
1
2πR
Z
Γ
R
|u(y)|ds
y
≤
M
R
ρ
2
|x| ≥ 2R.
M
R
= 88M
′
R
M
′
R
= sup
|y|=2R
|u(y)|
|∂u/∂y|
C
R
= sup
|x|=2R
|u(x)|
O(1/|x|
2
)
∆u = 0, x ∈ Ω, u|
Γ
a
= g
Ω = {x = (x, y, z) ∈ R
3
: |x| < a}
∆u = 0, x ∈ Ω
e
, u|
Γ
a
= g, u(x) = o(1) |x| → ∞
Ω
e
= R
3
\Ω Ω Γ
a
= ∂Ω
k(x, y) =
a
2
− |x|
2
|x − y|
3
, x ∈ R
3
\ Γ
a
, y ∈ Γ
a
,
Ω
k
u(x) =
1
4πa
Z
Γ
a
a
2
− |x|
2
|x − y|
3
g(y)dσ
y
, x ∈ Ω.
k |x| = a
x = y
1
4πa
Z
|y|=a
a
2
− |x|
2
|x − y|
3
dσ
y
≡
1
|x| < a,
−
a
|x|
|x| > a.
g Γ
a
a
u
u(x) =
(
1
4πa
R
Γ
a
a
2
−|x|
2
|x−y|
3
g(y)dσ
y
, x ∈ Ω,
g(x), x ∈ Γ
a
.
u
u(x) =
(
1
4πa
R
Γ
a
|x|
2
−a
2
|x−y|
3
g(y)dσ
y
, x ∈ Ω
e
,
g(x), x ∈ Γ
a
.
u(x) =
1
4πa
Z
|y|=a
a
2
− |x|
2
|x − y|
3
u(y)dσ
y
.
Ω u ∈ C(Ω)
Γ
a
Ω x = 0
R
3
u Ω R
3
\Ω
x → ∞
u(x) = o(1) |x| → ∞.
u ∈ C
2
(Ω
e
)
Ω
e
= R
3
\Ω
Ω
R > 0 C = C
R
(u)
u(x) ≤
C
R
|x|
, |
u(x)| ≤
C
R
|x|
2
|x| ≥ R.
Ω
e
= R
3
\Ω
|x|
−1
|x|
−2
|x| → ∞
u
Ω
e
u
Ω
e
u
n = 3
Ω
e
u : Ω
e
→ R
Ω
e
≡ R
3
\Ω
Ω
e
R
3
Ω R
3
Γ ∈ C
1
Ω
e
f Ω Ω
e
g Γ
∆u = −f.
u ∈ C
2
(Ω)
Ω
x ∈ Ω
u = g Γ.
u ∈ C
2
(Ω
e
)
Ω
e
≡ Ω
e
∪Γ
x ∈ Ω
e
u(x) = o(1) |x| → ∞.
u ∈ C
2
(Ω) ∩
C
1
(
Ω) Ω
∂u
∂n
= g
Γ.
n Γ
Ω
u ∈ C
2
(Ω
e
) ∩
C
1
(
Ω
e
) Ω
e
u ∈
C
2
(Ω) ∩C
1
(Ω) Ω
∂u
∂n
+ au = g
Γ.
a : Γ → R Γ
u ∈ C
2
(Ω
e
)∩
C
1
(
Ω
e
) Ω
e
Γ
1
Γ
2
Γ
Γ = Γ
1
∪ Γ
2
,
Z
Γ
ϕdσ =
Z
Γ
1
ϕdσ +
Z
Γ
2
ϕdσ ∀ϕ ∈ C(Γ).
Γ
1
Γ
2
g
1
: Γ
1
→ R
g
2
: Γ
2
→ R
Ω
Ω
e
Γ
1
2
Γ
Γ
Ω
Ω
e
u ∈ C
2
(Ω) ∩ C
1
(Ω) Ω
u = g
1
Γ
1
,
∂u
∂n
+ au = g
2
Γ
2
.
u ∈
C
2
(Ω
e
)
∩C
1
(
Ω
e
) Ω
e
u ∈ C
2
(Ω) ∩C(Ω)
u ∈ C
2
(Ω
e
) ∩ C(Ω
e
)
u = u
2
− u
1
Ω Ω
e
Ω
Ω
e
Γ
u ˜u
C
2
(Ω) ∩ C(Ω)
u|
Γ
= g ˜u|
Γ
= ˜g
|g(x) − ˜g(x)| ≤ ε ∀x ∈ Γ.
|u(x) − ˜u(x)| ≤ ε Ω.
v = u − ˜u
Ω Ω Γ |v| ≤ ε
u ˜u
C
2
(Ω
e
) ∩ C(Ω
e
)
|u(x) − ˜u(x)| ≤ ε Ω
e
.
u ∈ C
2
(Ω)∩C
1
(Ω)
u
1
u
2
u = u
2
− u
1
Ω C
1
(Ω) ∂u/∂n = 0 Γ
v = u
Z
Ω
|∇u|
2
dx = −
Z
Ω
∆u udx +
Z
Γ
∂u
∂n
udσ.
∆u = 0 Ω ∂u/∂n = 0 Γ
Z
Ω
|∇u|
2
dx ≡
Z
Ω
"
∂u
∂x
2
+
∂u
∂y
2
+
∂u
∂z
2
#
dx = 0 ⇒
∇u = 0
Ω ⇒ u = .
u ∈ C
2
(Ω
e
) ∩ C
1
(Ω
e
)
Ω
e
u
1
u
2
B
R
R Γ
R
Ω
Ω
R
Ω
R
= Ω
e
∩ B
R
= B
R
\ Ω
Ω
R
u = u
2
− u
1
, v = u ∂Ω
R
= Γ
R
∪ Γ ∆ u = 0 Ω
R