Троян В.Н. Принципы решения обратных геофизических задач
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˜u = P ν + ˜ε,
˜u = k˜u
1
, ˜u
2
, . . . , ˜u
n
k
T
˜ε = k˜ε
1
, ˜ε
2
, . . . , ˜ε
n
k
T
ν =
|ν
1
i
. . .
|ν
M
i
P =
hp
11
| . . . hp
1µ
| . . . hp
1M
|
. . . . . . . . . . . .
hp
n1
| . . . hp
nµ
| . . . hp
nM
|
p
i
u
i
u
i
= hp
i
|ν(δθ)i
V
.
p
i
= δ(L
i
(~x))
ϕ
in
ϕ
out
L
0
= −∆ +
1
v
2
0
(x)
∂
2
∂t
2
.
L
θ
= −∆ +
1
v
2
(x)
∂
2
∂t
2
,
θ
0
= v
0
θ = v(x) x ∈ R
3
ν =
1
v
2
0
1 −
v
2
0
v
2
(x)
S = −
∂
2
∂t
2
p = hϕ
out
|
∂
2
∂t
2
|ϕ
in
i
T
ϕ
in
ϕ
out
L
0
~ϕ = ρ
0
∂
2
∂t
2
~ϕ − [(λ
0
+ µ
0
)
~
∇
~
∇ · ~ϕ + µ∆~ϕ +
~
∇λ
0
~
∇ · ~ϕ +
+
~
∇µ
0
×
~
∇ × ~ϕ + 2(
~
∇µ
0
·
~
∇)~ϕ].
L~ϕ = ρ
∂
2
∂t
2
~ϕ − [(λ + µ)
~
∇
~
∇ · ~ϕ + µ∆~ϕ +
~
∇λ
~
∇ · ~ϕ +
+
~
∇µ ×
~
∇ × ~ϕ + 2(
~
∇µ ·
~
∇)~ϕ].
θ
0
θ
θ
0
=
|λ
0
(x)i
|µ
0
(x)i
|ρ
0
(x)i
θ =
|λ(x)i
|µ(x)i
|ρ(x)i
λ(x) µ(x)
ρ(x)
p = khp
λ
|hp
µ
|hp
ρ
|k S = kS
λ
|S
µ
|S
ρ
k.
δL
θ
= L
0
− L = δL
λ
+ δL
µ
+ δL
ρ
.
λ(x) = λ
0
(x) + δλ(x), µ(x) = µ
0
(x) + δµ(x),
ρ(x) = ρ
0
(x) + δρ(x).
δL
λ
δL
λ
: δL
λ
~ϕ = δλ
~
∇
~
∇ · ~ϕ +
~
∇δλ
~
∇ · ~ϕ =
~
∇(δλ
~
∇ · ~ϕ),
~
∇(φψ) = ψ
~
∇φ +
φ
~
∇ψ
δL
µ
δL
µ
: δL
µ
~ϕ = δµ
~
∇
~
∇ · ~ϕ + δµ∆~ϕ +
~
∇δµ ×
~
∇ × ~ϕ +
+2(
~
∇δµ ·
~
∇)~ϕ = 2
~
∇ · (δµ
~
∇~ϕ) +
~
∇ × (δµ
~
∇ × ~ϕ).
δL
ρ
δL
ρ
: δL
ρ
~ϕ = −δρ∂
2
~ϕ/∂t
2
.
~u
i
= u
i
−u
i0
= hp
iλ
|δλi
V
+hp
iµ
|δµi
V
+hp
iρ
|δρi
V
+ε
i
= hp
iθ
|δθi
V
+ε
i
,
hp
iθ
|δθi
V
=
Z
V
~ϕ
out
⊙ δL
θ
~ϕ
in
d~x = hh~ϕ
out
|S
θ
|~ϕ
in
i
T
|δθi
V
.
⊙
∗ (·, ·)
~
ξ ⊙~η ≡ (
~
ξ(t) ∗ ~η(t)).
p
λ
S
λ
: p
λ
= h~ϕ
out
|S
λ
|~ϕ
in
i
T
= −
~
∇ · ~ϕ
out
∗
~
∇ · ~ϕ
in
p
µ
S
µ
S
µ
: p
µ
= h~ϕ
out
|S
µ
|~ϕ
in
i
T
=
= (
~
∇ × ~ϕ
out
) ⊙ (
~
∇ ×
~ϕ
in
) − 2
~
∇~ϕ
out
⊕
~
∇~ϕ
in
~
∇~ϕ ⊕
~
∇
~
ψ =
3
X
i=1
3
X
j=1
∇
i
ϕ
j
∗ ∇
i
ψ
j
.
p
ρ
S
ρ
S
ρ
: p
ρ
= h~ϕ
out
|S
ρ
|~ϕ
in
i
T
=
∂
∂t
~ϕ
out
⊙
∂
∂t
~ϕ
in
~ϕ
out
t
i
p
s ~ϕ
in
x
~ϕ
out
~ϕ
in
~ϕ
out
~ϕ
in
~ϕ
in
~ϕ
out
~ϕ
in
p
s
~ϕ
in
~ϕ
out
~ϕ
in
s
u = P ν + ε, u
i
=
M
X
µ=1
hp
µi
|ν
µ
(δθ)i
V
+ ε
i
.
l(ν)
l(ν) =
n
X
i=1
α
i
u
i
= (~α ·~u),
l(ν) ≡ hl|νi
V
l
X
(ν) = (
4
3
πr
3
)
−1
H(r − |x − X|),
l
X
(ν) =
1
abc
H(a/2 − |x
1
− X
1
|)H(b/2 − |x
2
− X
2
|) ×
×H(c/2 − |x
3
− X
3
|),
H(x) =
1 x > 0,
0
x < 0.
H(x)
H(x)
η
X
µ
= l(ν
µ
) −
M
X
µ
′
=1
(~α · h~p
µ
′
|ν
µ
′
i
V
) − (~α ·~ε) =