Троян В.Н. Принципы решения обратных геофизических задач
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1
(2π)
3
Z
∞
Z
−∞
Z
exp[−i(
~
k
′
, ~x
′
−~x)]
k
2
0
− k
′
2
dk
′
= −
1
(4π)
exp[i|
~
k
0
||~x −~x
′
|]
|~x −~x
′
|
.
G(~x, ~x
′
) = exp[−i(
~
k
0
, ~x −~x
′
)](
−1
4π
exp[i|
~
k
0
||~x −~x
′
|]
|~x −~x
′
|
) =
=
−1
4π
1
|~x −~x
′
|
exp[i|
~
k
0
||~x −~x
′
|(1 − cos θ)],
θ
~
k
0
~x − ~x
′
u
1
(~x, ω) =
Z
∞
Z
−∞
Z
G(~x, ~x
′
)
˜
β(~x
′
)d~x
′
=
= −
1
4π
Z
∞
Z
−∞
Z
˜
β(~x
′
, ω) exp[i|
~
k
0
||~x −~x
′
|(1 − cos θ)]
|~x −~x
′
|
d~x
′
.
u
B
(~x, ω) = u
0
(~x, ω)(1 + εu
1
(~x, ω)) =
= a
0
exp[i(
~
k
0
, ~x)] + a
0
Q(~x, ω,
~
k
0
),
Q(~x, ω,
~
k
0
) = −
exp[i(
~
k
0
, ~x)]
4π
Z
∞
Z
−∞
Z
β(~x
′
, ω)
|~x −~x
′
|
×
× exp[i|
~
k
0
||~x −~x
′
|(1 − cos θ)]d~x
′
.
β(x, ω)
β
0
(x) = 1−v
2
0
/v
2
(x)
δv
v
d
λ
≪ 1,
δv/v
d λ
˜u(t, ~x
r
, ~n) = u
mB
(t, ~x
r
, ~n, β
0
(~x)) + ξ(t, ~x
r
, ~n).
u
mB
u
mB
(t, ~x
r
, ~n, β
0
(~x)) = h(t) ∗ F
−1
ω
[u
B
(~x, ω, β
0
(~x))],
h(t)
F
−1
ω
u
B
(~x, ω) = a
0
exp[i(
~
k
0
, ~x)] + a
0
Q(~x
r
, ω,
~
k
0
).
Q(~x
r
, ω,
~
k
0
)
Q(~x
r
, ω,
~
k
0
) = −
1
4π
Z
∞
Z
−∞
Z
β
0
(~x
′
) ω
2
v
2
0
|~x
r
−~x
′
|
exp[i|
~
k
0
||~x
r
−~x
′
|] ×
×exp[i(
~
k
0
, ~x
r
)] exp[−i(
~
k
0
, ~x
r
−~x
′
)]d~x
′
=
=
−1
4πv
2
0
Z
∞
Z
−∞
Z
β
0
(~x
′
) ω
2
|~x
r
−~x
′
|
exp{i[|
~
k
0
||~x
r
−~x
′
| + (
~
k
0
, ~x
′
)]}d~x
′
=
=
1
4πv
2
0
Z
∞
Z
−∞
Z
(iω)
2
|~x
r
−~x
′
|
β
0
(~x
′
) exp{iω[
|~x
r
−~x
′
|
v
0
+
(~n, ~x
′
)
v
0
]}d~x
′
.
β
0
(~x)
β
0
(~x) = (1 −
v
2
0
v
2
(~x)
) =
M
X
j=1
α
j
ϕ
j
(~x),
ϕ
j
(~x) =
1
~x ∈ ∆Ω
j
,
0
~x 6∈ ∆Ω
j
.
β
0
(~x)
Q(~x
r
, ω,
~
k
0
) = (iω)
2
M
X
j=1
α
j
A
j
exp[iωτ
j
],
τ
j
=
|~x
r
−~x
j
|
v
0
+
(~n, ~x
j
)
v
0
, A
j
=
∆Ω
j
4πv
2
0
|~x
r
−~x
j
|
.
Q(~x
r
, ω, k
0
)
u
mB
u
mB
= a
0
h(t) ∗ F
−1
ω
h
exp[i(
~
k
0
, ~x
r
)]
i
+
+ a
0
h(t) ∗ F
−1
ω
(iω)
2
M
X
j=1
exp[iωτ
j
]A
j
α
j
=
= a
0
h(t) ∗ δ
t −
(~n, ~x
r
)
v
0
+ a
0
h(t) ∗
M
X
j=1
δ
(2)
(t − τ
j
)α
j
A
j
=
= a
0
h
t −
(~n, ~x
r
)
v
0
+ a
0
M
X
j=1
α
j
A
j
h
′′
(t − τ
j
).
˜u(t, ~x
r
, ~n)
u
mB
(t, ~x
r
, ~n, β
0
(~x))
u(t, ~x
r
, ~n) =
M
X
j=1
α
j
C
j
(t, ~x
r
, ~n) + ε(t, ~x
r
, ~n),
u(t, ~x
r
, ~n) = ˜u(t, ~x
r
, ~n) − a
0
h(t − (~n, ~x
r
)/v
0
),
C
j
(t, ~x
r
, ~n) = A
j
h(t − τ
j
).
x
ri
(i = 1, 2, . . . , n)
~u = C~α + ~ε,
~u = ku(t, x
ri
, ~n)k
n
i=1
C = kC
j
(t, x
ri
, ~n)k
n M
i=1,j=1
~α
λ(~x) µ(~x) ρ(~x)
p s θ(~x)
Φ(R
3
)
θ(~x) ∈ Φ(~x ∈ R
3
) ⇒ (λ(~x), µ(~x), ρ(~x)).
n
(R
n
)
Φ(~x ∈ R
3
)
P
⇒ R
n
{u(x
i
, t
i
), i = 1 ÷
n} ∈ R
n
, x ∈ R
3
n
θ(~x)
u
i
= P
i
(θ) + ε
i
, ε
i
∈ N( 0, K
ε
),
P
i
(θ)
Φ(~x ∈ R
3
)
R
n
ε
i
K
ε
0
f
L
θ
L
θ
ϕ = f,
L
θ
: L
θ
(αϕ + βψ) = αL
θ
ϕ + βL
θ
ψ.
H : H
i
ϕ↔
Z
T
Z
Ω
Z
V
d~xdτdΩh
i
(~n, ~n
′
, t−τ)ϕ(~x, ~n, τ)δ(~x −~x
i
),
Ω ~n
′
u
i
= H
i
L
−1
θ
~
f + ε
i
.
P
i
(θ) = H
i
L
−1
θ
~
f.
θ
0
ϕ
0
= L
−1
0
f.
θ
θ
0
θ = θ
0
+ δθ,
δθ ≪ θ
0
ϕ
ϕ = ϕ
0
+ L
−1
0
δL
θ
ϕ,
δL
θ
= L
0
−L
θ
u
i
= H
i
[ϕ
0
+ L
−1
0
δL
θ
ϕ] + ε
i
.
δθ
kH
i
L
−1
0
δL
θ
(ϕ − ϕ
0
)k
2
E(ε
2
i
)
≪ 1
ϕ ϕ
0
u
i
= H
i
[ϕ
0
+ L
−1
0
δL
θ
ϕ
0
] + ˜ε
i
,
˜ε
i
ε
i
˜u
i
= hh
i
|L
−1
0
δL
θ
ϕ
0
i
V,T,Ω
+ ˜ε
i
hξ|ηi
T,V,Ω
=
Z
Ω
Z
V
ξ(~x, t, Ω) ∗ η(~x, t, Ω)d~xdΩ
∗
˜u
i
= u
i
− hh
i
|ϕ
0
i
T,V,Ω
u
i
θ
0
(~x)
δL
ν(δθ)
˜u
i
= hL
−1∗
0
h
i
|δL
θ
ϕ
0
i
V T Ω
+ ˜ε
i
=
= hhG
∗
h
i
|L
′
ν
ϕ
0
i
T Ω
|ν(δθ)i
V
+ ˜ε
i
,
G = L
−1
0
G
∗
L
′
ν
L
′
ν
:
δu
δν
= hG
∗
h |L
′
ν
ϕ
0
i
T Ω
.
ν(δθ)
p
i
= hϕ
out
|S|ϕ
in
i
T Ω
,
ϕ
in
= ϕ
0
= Gf
θ
0
ϕ
out
= G
∗
h
i
S = L
′
ν
ϕ
in
ϕ
out