Алексеев Г.В. Классические методы математической физики. Часть 2
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Z
Ω
R
|∇u|
2
dx =
Z
Γ
∂u
∂n
udσ +
Z
Γ
R
∂u
∂n
udσ.
n ∂Ω
R
Ω
R
Ω
Γ
Γ
n
B
n
R
R
Ω
R
R
3
|∇u|
2
|x| → ∞
|x|
−4
Ω
R
R = |x|
R
3
R → ∞
R
Ω
e
|∇u|
2
dx
Ω
e
R
Γ
R
(∂u/∂n)udσ
R → ∞
(u∂u/∂n)|
Γ
R
R → ∞
O(R
−3
) Γ
R
4πR
2
/3
R
2
R → ∞
∂u/∂n = 0 Γ
R
Ω
e
|∇u|
2
dx = 0
Ω
e
u = u
0
= const
u
0
= 0 ⇒ u
1
= u
2
a
(i) a ∈ C(Γ), a ≥ 0 Γ,
R
Γ
adσ > 0
(i) u ∈ C
2
(Ω) ∩C
1
(Ω)
u
1
u
2
u = u
2
−u
1
∆u = 0 Ω ∂u/∂n = −au Γ
Z
Ω
|∇u|
2
dx +
Z
Γ
au
2
dσ = 0.
a ≥ 0 |∇u| = 0 Ω ⇒ u = u
0
=
const
u = u
0
u
2
0
Z
Γ
adσ = 0.
u
0
= 0 =⇒ u
1
= u
2
a
a ∈ C(Γ) a ≥ 0 Γ u ∈ C
2
(Ω
e
)∩
C
1
(
Ω
e
)
Ω
e
meas Γ
1
> 0, a ∈ C(Γ
2
), a ≥ 0 Γ
2
meas Γ
1
≥ 0, a ∈ C(Γ
2
), a ≥ 0 Γ
2
,
R
Γ
2
adσ > 0
measΓ
2
> 0
(ii) (iii) u ∈ C
2
(Ω) ∩ C
1
(Ω)
u
1
u
2
u = u
2
−u
1
∆u = 0 Ω, u = 0 Γ
1
,
∂u
∂n
= −au
Γ
2
Z
Ω
|∇u|
2
dx +
Z
Γ
2
au
2
dσ = 0.
a ≥ 0 Γ
2
|∇u| = 0 Ω ⇒ u =
u
0
= const
Ω
u
2
0
Z
Γ
2
adσ = 0.
u = 0 Γ
1
u
0
= 0 u
1
= u
2
u
0
= 0 ⇒ u
1
= u
2
measΓ
2
> 0 a ∈ C(Γ
2
) a ≥ 0 Γ
2
u ∈ C
2
(Ω
e
) ∩ C
1
(Ω
e
)
Ω
e
u
1
u
2
Ω
R
= Ω
e
∩ B
R
u = u
2
− u
1
R → ∞
R
Ω
e
|∇u|
2
dx −
R
Γ
(∂u/∂n)udσ = 0
u = 0 Γ
1
∂u/∂n = −au Γ
2
Z
Ω
e
|∇u|
2
dx +
Z
Γ
2
au
2
dσ = 0.
a ≥ 0 Γ
2
|∇u| = 0 Ω
e
⇒ u =
u
0
= const
Ω
e
u
0
= 0 ⇒ u
1
= u
2
Γ = Γ
2
Γ
2
> 0
Γ
2
≥ 0
Γ
2
= 0
Γ = Γ
1
u(x) = O(1) |x| → ∞.
R
3
R
3
R
2
measΓ
2
≥ 0
u ∈ C
2
(Ω
e
) ∩ C
1
(Ω
e
)
R
2
u ∈ C
2
(Ω) ∩ C
1
(Ω)
Z
Ω
∆udx = −
Z
Ω
fdx.
u v ≡ 1
R
Ω
fdx = −
R
Γ
(∂u/∂n)dσ
∂u/∂n = g Γ
Z
Ω
fdx +
Z
Γ
gdσ = 0.
u ∈ C
2
(Ω) ∩C
1
(Ω)
f = 0
Z
Γ
gdσ = 0.
Ω
u
−
R
Γ
∂u/∂ndσ
Γ Ω
R
Ω
fdx
Ω
Ω x ∈ Ω
f = 0
§4 K
K
Ω
R
2
Ω u
(x −
x
0
)(y − y
0
) x
0
= (x
0
, y
0
) ∈ Ω
K R x
0
Ω
x
0
= (0, 0)
R = 1
K u
u(x) =
a
0
2
+
∞
X
k=1
ρ
k
(a
k
cos kϕ + b
k
sin kϕ),
a
k
b
k
u
u K a
k
b
k
|a
k
| ≤ M, |b
k
| ≤ M k = 1, 2, ...
x = ρ cos ϕ, y = ρ sin ϕ i =
√
−1
(x+iy)
k
= [ρ(cos ϕ+i sin ϕ)]
k
= (ρ exp
iϕ
)
k
= ρ
k
exp
ikϕ
= ρ
k
(cos kϕ+i sin kϕ).
[(a
k
−ib
k
)(x+iy)
k
] =
ρ
k
(a
k
cos kϕ + b
k
sin kϕ) a
a
u(x) =
a
0
2
+
∞
X
k=1
[(a
k
− ib
k
)(x + iy)
k
].
a
0
2
+
∞
X
k=0
∞
X
l=0
[(a
k+1
− ib
k+1
)C
l
k+l
i
l
x
k
y
l
].
C
l
k+l
k
2
+ l
2
6= 0 C
l
k+l
<
2
k+l
a
k
b
k
M
|x| < 1/2 |y| < 1 /2
2M
∞
X
k=0
∞
X
l=0
2
k+l
|x|
k
|y|
l
.
(0, 0)
u
u
(0, 0) u
x y u (0, 0)
Ω
u
1
, u
2
, ..., u
n
, ...
Ω
Ω Γ Ω
Ω u
Ω
Γ ε > 0
N m, n > N Γ |u
n
(x) −u
m
(x)| < ε
m, n |u
n
(x)−u
m
(x)| <
ε
Ω
u ∈ C(Ω) Ω
u = lim
n→∞
u
n
Ω
x
0
= (x
0
, y
0
) ∈ Ω
K x
0
a Ω u
n
Ω
K
u
n
(x) =
1
2πa
Z
∂K
a
2
− ρ
2
r
2
u
n
(y)ds
y
.
Ω
u(x) =
1
2πa
Z
∂K
a
2
− ρ
2
r
2
u(y)ds
y
.
u|
∂K
∈ C(Γ) u
K
Ω
x
0
= (x
0
, y
0
) ∈ Ω n
u
n+1
(x) ≥ u
n
(x)
Ω
Ω u
Ω
K
0
R x
0
K
0
Ω u
n+p
− u
n
= v
n,p
p
v
n,p
≥ 0 K
0
K
∗
K
0
R + ε
Ω v
n,p
K
∗
v
n,p
(ρ, ϕ) =
1
2π
2π
Z
0
v
n,p
(R + ε, ψ)
(R + ε)
2
− ρ
2
(R + ε)
2
+ ρ
2
− 2(R + ε)ρcos( ϕ − ψ)
dψ.
−1 ≤ cos(ϕ − ψ) ≤ 1
R + ε − ρ
R + ε + ρ
≤
(R + ε)
2
− ρ
2
(R + ε)
2
+ ρ
2
− 2(R + ε)ρcos(ϕ − ψ)
≤
R + ε + ρ
R + ε − ρ
.
v
n,p
(R + ε, ψ) ≥ 0
1
2π
R + ε − ρ
R + ε + ρ
2π
Z
0
v
n,p
(R+ε, ψ)dψ ≤ v
n,p
(ρ, ϕ) ≤
1
2π
R + ε + ρ
R + ε − ρ
2π
Z
0
v
n,p
(R+ε, ψ)dψ.
R
2
(1/2π)
R
2π
0
v
n,p
(R + ε, ψ)dψ = v
n,p
(x
0
)
R + ε − ρ
R + ε + ρ
v
n,p
(x
0
) ≤ v
n,p
(x) ≡ v
n,p
(ρ, ϕ) ≤
R + ε + ρ
R + ε − ρ
v
n,p
(x
0
) ∀x ∈
K
0
.
v
n,p
(x
0
) → 0 n, p → ∞
u
n
K
0
u K
0
y
Ω x
0
l
Ω
l x
0
y
Ω
δ
K
0
l x
1
K
1
δ/2
u
n
K
0
x
1
u
n
K
1
K
2
δ/2 K
2
l
K
1
K
i
(i = 0, 1 , ..., N)
l y K
N
l y
u
n
K
i
K
N
y
u
n
Ω
0
Ω Ω
0
K
1
, ..., K
N
Ω u
n
K
i
Ω
0
Ω
Ω
0
Ω
M
γ > 0
Ω
0
Ω δ = γ/2
x
0
= (x
0
, y
0
) Ω
0
Ω
∂u
∂x
(x
0
) =
1
πδ
2
Z
K
∂u
∂x
dxdy =
4
πγ
2
Z
∂K
u cos(n, x)ds.
u n
∂K
∂u
∂x
(x
0
)
≤
4
πγ
2
M · πγ =
4M
γ
.
x
0
u
Ω
0
x
Ω
0
y
Ω
Ω
0
Ω
Ω
0
u ∈ C
2
(R
2
)
R
2
u u(x) ≥
m = co nst
x ∈ R
2
u
m ≥ 0 u
x = (x, y) u (0, 0)
u = const K
(0, 0) R x
K u
u(x) =
1
2π
Z
2π
0
u(R, ψ)
R
2
− ρ
2
R
2
+ ρ
2
− 2Rρ cos(ϕ − ψ)
dψ, ρ = |x|.
R − ρ
R + ρ
u(0, 0) ≤ u(x) ≤
R + ρ
R − ρ
u(0, 0).
R → ∞ u(0, 0) ≤ u(x) ≤ u(0, 0) ⇒ u(x) = u(0, 0)
u
R
2
u
x
0
x
0
u u
x
0
u
x
0
x
0
x
0
K R x
0
x
0
u
1
K u K
u − u
1
≡ v v K
x
0
∂K v
x
0
∈ K x
0
v(x
0
) ≡ 0
u(x
0
) = u
1
(x
0
)
v
ε
(x, x
0
) =
M ln(|x − x
0
|/R)
ln(ε/R)
.
M = sup
x∈K
|v(x)| ε x
0
K
ε
= {x ∈ K : ε < |x−x
0
|} v
ε
K
ε
K
ε
K
ε
ρ = R M ρ = ε v
∂K
ε
K
ε
|v(x)| ≤ v
ε
(x, x
0
).
v
ε
v
∂K
ε
K
ε
x
0
|v(x
0
)| ≤ v
ε
(x
0
, x
0
) = M
ln(|x
0
− x
0
|/R)
ln(ε/R)
.