Троян В.Н. Принципы решения обратных геофизических задач
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ˆ
I
m
g
m
(u
k
, I
m
, h
m
, ξ
m
)
g
m
(h
m
, ξ
m
) =
1
σ
2
P
k
y
km
ϕ
k
(h
m
, ξ
m
)
P
k
ϕ
2
k
(h
m
, ξ
m
)
.
h
m
ξ
m
m
g
m
h
m
, ξ
m
= max g
m
(u
k
,
ˆ
I
m
, h
m
, ξ
m
).
α
2
m
=
ˆ
I
2
m
σ
2
m
X
k
ϕ
2
k
(
ˆ
h
m
,
ˆ
ξ
m
) ≥ α
2
0P
e
,
M = m+1
α
2
m
< α
2
0P
e
,
ˆ
~
θ = {
ˆ
I
1
,
ˆ
h
1
,
ˆ
ξ
1
,
ˆ
I
2
,
ˆ
h
2
,
ˆ
ξ
2
, . . . ,
ˆ
I
m
,
ˆ
h
m
,
ˆ
ξ
m
}.
f(~r)
u
i
i
(s, r)
i
u
i
f(~r) L
i
u
i
=
Z
L
i
f(~r)dl.
u
i
=
Z
Ω
f(~r)δ(L
i
(~r))d~r,
Ω
δ(·) δ
˜
R
u
i
=
˜
R
i
[f(~r)],
˜
R
i
[·] ≡
Z
d~rδ(L
i
(~r))[·].
f(~r)
f(r, ϕ)
~r = (r, ϕ)
(n, θ)
(n, θ), u(n, θ) =
Z
L
f(
√
l
2
+ n
2
, θ − arctan(l/n))dl.
u(n, θ) =
˜
R[f(r, ϕ)].
f(r, ϕ) =
1
2π
2
∞
Z
−∞
π
Z
0
1
r cos(θ − ϕ) − n
∂u(n, θ)
∂n
dndθ
f(r, ϕ) =
˜
R
−1
[u(n, θ)].
˜
R
−1
= −
1
2π
P HD,
D : q(n, θ) = D[u(n, θ)] = lim
∆n→0
u(n + ∆n, θ) − u(n, θ)
∆n
H : h(n
′
, θ) = H[q(n
′
, θ)] = −
1
π
lim
ε→0
{
n
′
−ε
Z
−∞
q(n, θ)
n
′
− n
dn +
∞
Z
n
′
+ε
q(n, θ)
n
′
− n
dn}
P : p(r, ϕ) = P [h(n
′
, θ)] =
π
Z
0
h(r cos(θ − ϕ), θ)dθ
f(~r)
f(~r) =
M
X
j=1
θ
j
ψ
j
(~r).
u
i
u
i
=
M
X
j=1
C
ij
θ
j
,
C
ij
=
˜
R
i
[ψ
j
(~r)].
ψ
j
(~r) =
1 ~r ∈ Ω
j
,
0
~r 6∈ Ω
j
.
C
ij
C
ij
=
C L
i
T
Ω
j
6= ∅,
0
L
i
T
Ω
j
= ∅.
~u = C
~
θ + ~ε,
~ε ~ε ∈ N(0, R
ε
)
~
θ
~
θ ∈ N(h
~
θi, R
θ
)
~
θ
˜
~
θ = max
~
θ
p(
~
θ)p(~u/
~
θ).
~ε
~
θ
˜
~
θ = min
~
θ
[(~u − C
~
θ)
T
R
−1
ε
(~u − C
~
θ) + (
~
θ − h
~
θi)
T
R
−1
θ
(
~
θ − h
~
θi)].
1.
ˆ
~
θ = (C
T
R
−1
ε
C + R
−1
θ
)
−1
(C
T
R
−1
ε
~u + R
−1
θ
h
~
θi),
R(
ˆ
~
θ) = (C
T
R
−1
ε
C + R
−1
θ
)
−1
.
2.
˜
~
θ = h
~
θi+ R
θ
C
T
(CR
θ
C
T
+ R
ε
)
−1
(~u − Ch
~
θi),
R(
˜
~
θ) = R
θ
− R
θ
C
T
(CR
θ
C
T
+ R
ε
)
−1
CR
θ
.
R
ε
= σ
2
ε
I
R
θ
= σ
2
θ
I
1.
ˆ
~
θ = (C
T
C + αI)
−1
(C
T
~u + αh
~
θi),
R(
ˆ
~
θ) = σ
2
ε
(C
T
C + αI)
−1
,
α = σ
2
ε
/σ
2
θ
2.
˜
~
θ = h
~
θi+ C
T
(CC
T
+ αI)
−1
(~u − Ch
~
θi),
R(
˜
~
θ) = σ
2
θ
− σ
2
θ
C
T
(CC
T
+ αI)
−1
C.
R(
ˆ
~
θ)
R(
˜
~
θ)
R(
ˆ
~
θ)
R(
˜
~
θ)
C
~
θ
(0)
R
(0)
θ
~
θ
(0)
= h
~
θi
R
(0)
θ
= R
θ
.
~
θ
~
θ
(1)
=
~
θ
(0)
+ α
(1)
∆
~
θ
(1)
,
∆
~
θ
(1)
= R
(0)
θ
C
(1)T
, α
(1)
= e
1
/δ
2
1
,
e
1
= u
1
− C
(1)
~
θ
(1)
, δ
2
1
= σ
2
ε1
+ C
(1)
∆
~
θ
(1)
,
C
(1)
C
R
(1)
θ
= R
(0)
θ
− ∆
~
θ
(1)
∆
~
θ
(1)T
/δ
2
1
.
(i − 1)
~
θ
(i−1)
, R
(i−1)
θ
.
i
~
θ
~
θ
(i)
=
~
θ
(i−1)
+ α
(i)
∆
~
θ
(i)
,
α
(i)
= e
i
/δ
2
i
, e
i
= u
i
− C
(i)
~
θ
(i−1)
,
δ
2
i
= σ
2
εi
+ C
(i)
∆
~
θ
(i)
, ∆
~
θ
(i)
= R
(i−1)
θ
C
(i)T
,
C
(i)
i C i
R
(i)
θ
= R
(i−1)
θ
− ∆
~
θ
(i)
∆
~
θ
(i)T
/δ
2
i
.