Троян В.Н. Принципы решения обратных геофизических задач
Подождите немного. Документ загружается.
= hl −(~α · ~p
µ
)|ν
µ
i
V
−
X
µ
′
6=µ
h(~α · ~p
µ
′
)|ν
µ
′
i
V
− (~α ·~ε).
ˆ
l(ν) = (~α ·~u),
E(η
X
µ
)
2
= (
~
Λ − Q~α)
∗
K(
~
Λ − Q~α),
Q = kP |Ik I
α
T
Q
∗
= kα
1
. . . α
n
k×
×
hp
11
(x)| . . . hp
1µ
(x)| . . . hp
1M
(x)| 1 . . . 0
. . . . . . . . . . . . . . . . . . . . . . . .
hp
i1
(x)| . . . hp
iµ
(x)| . . . hp
iM
(x)| 0 . . . 0
. . . . . . . . . . . . . . . . . . . . . . . .
hp
n1
(x)| . . . hp
nµ
(x)| . . . hp
nM
(x)| 0 . . . 1
,
Λ
∗
= k h0 . . . hl(x) . . . h0 | 0 . . . 0 k
1 µ M M + 1 M + n
.
K
K =
K
ν
1
ν
1
. . . K
ν
1
ν
M
K
ν
1
ε
. . . . . . . . . . . .
K
ν
M
ν
1
. . . K
ν
M
ν
M
K
ν
M
ε
K
εν
1
. . . K
εν
M
K
εε
,
K : hϕ|K
ν
µ
ν
µ
′
|ψi = Ehϕ|ν
µ
ihν
µ
′
|ψi,
K
εε
= E(~ε · ~ε
T
), K
εν
µ
: E(εhν
µ
|ψi) = K
εν
µ
|ψi.
~α
ˆ
~α = (QKQ
∗
)
−1
QK
~
Λ.
ν
µ
ν
µ
′
ν
µ
ε
ˆ
~α
µ
= (P
µ
K
ν
µ
ν
µ
P
∗
µ
+
X
µ
′
6=µ
P
µ
′
K
ν
µ
′
ν
µ
′
P
∗
µ
′
+ K
εε
)
−1
hp
µ
K
ν
µ
ν
µ
|li,
P
µ
K
ν
µ
ν
µ
P
∗
µ
=
hp
µ1
|
. . .
hp
µn
|
K
ν
µ
ν
µ
k|p
µ1
i. . . |p
µn
ik =
=
hp
µ1
(x)|K
ν
µ
ν
µ
(x, x
′
)|p
µ1
(x
′
)i . . . hp
µ1
(x)|K
ν
µ
ν
µ
(x, x
′
)|p
µn
(x
′
)i
. . . . . . . . .
hp
µn
(x)|K
ν
µ
ν
µ
(x, x
′
)|p
µ1
(x
′
)i . . . hp
µn
(x)|K
ν
µ
ν
µ
(x, x
′
)|p
µn
(x
′
)i
µ
(
X
µ
′
6=µ
P
µ
′
K
ν
µ
′
ν
µ
′
P
µ
′
+ K
εε
)
E(η
µ
)
2
= hl|K
ν
µ
ν
µ
− K
ν
µ
ν
µ
P
∗
µ
(P
µ
K
ν
µ
ν
µ
P
∗
µ
+
+
X
µ
′
6=µ
P
µ
′
K
ν
µ
′
ν
µ
′
P
∗
µ
′
+ K
εε
)
−1
P
µ
K
ν
µ
ν
µ
|li
E(η
µ
)
2
= hl|F
−1
µ
|li,
F
µ
= [K
ν
µ
ν
µ
− K
ν
µ
ν
µ
P
∗
µ
(P
µ
K
ν
µ
ν
µ
P
∗
µ
+
+
X
µ
′
6=µ
P
µ
′
K
ν
µ
′
ν
µ
′
P
∗
µ
′
+ K
εε
)
−1
P
µ
K
ν
µ
ν
µ
]
−1
.
µ
µ
L
θ
ϕ = s, L
θ
= ρ(x)
∂
2
∂t
2
−
∂
∂x
[E(x)
∂
∂x
]
E(x) E(x) = λ(x) + 2µ(x)
P
λ(x), µ(x) E(x) = µ(x) S
L
θ(i)
= ρ
i
∂
2
∂t
2
− E
i
∂
2
∂x
2
ρ
i
∂
2
ϕ
i
∂t
2
− E
i
∂
2
ϕ
i
∂x
2
= 0
E
i
= const; ρ
i
= const;
ϕ
i
= f
i
(t ± x/v
i
);
v
i
=
p
E
i
/ρ
i
v
ip
=
p
(λ
i
+ 2µ
i
)/ρ
i
; v
is
=
p
µ
i
/ρ
i
∆t
i
= ∆x
i
/v
i
= ∆t = const
ϕ(t) =
X
n
ϕ
n
δ(t − 2n∆t);
φ(ω) =
Z
∞
−∞
ϕ(t)e
iωt
dt =
X
n
ϕ
n
e
2inω∆t
;
z = e
2iω∆t
; φ(z) =
P
n
ϕ
n
z
n
ϕ
i
= ϕ
i+1
E
i
∂ϕ
i
∂x
= E
i+1
∂ϕ
i+1
∂x
1 + r
i
= t
i
;
¯
t
i
= 1 + ¯r
i
; P
a
= (1/2)ρvω
2
a
2
; a =
t,
r,
1
.
P
1
= P
r
+ P
t
⇒ ρ
i
v
i
= ρ
i
v
i
r
2
i
+ ρ
i+1
v
i+1
t
2
i
ρ
i
v
i
(1 − r
2
i
) − ρ
i+1
v
i+1
(1 + r
i
)
2
= 0
ρ
i
v
i
(1 − r
i
) − ρ
i+1
v
i+1
(1 + r
i
) = 0
r
i
=
ρ
i
v
i
− ρ
i+1
v
i+1
ρ
i
v
i
+ ρ
i+1
v
i+1
= −¯r
i
r
1
t = 2∆t; t
1
= 1 + r
1
¯r
1
= −r
1
;
¯
t
1
= 1 + ¯r
1
r
2
t = 4∆t
t
1
r
2
¯
t
1
+ r
1
¯r
0
r
1
r
1
, r
2
, . . . , r
n
v
1
, ρ
1
v
i
ρ
i
(i > 1); v
i+1
ρ
i+1
=
1 − r
i
1 + r
i
v
i
ρ
i
(2 ≤ i ≤ n + 1)
D
i
exp(−iω(t − x/v
i
)), u
i
exp(−iω(t + x/v
i
))
1
v
2
∂
2
ϕ
∂t
2
−
∂
2
ϕ
∂x
2
= 0
ϕ(x, t) ↔ ϕ(x, ω) =
∞
Z
−∞
ϕ(x, t)e
iωt
dt
∂
2
ϕ(ω, x)
∂x
2
+
ω
2
v
2
ϕ(ω, x) = 0
ϕ(ω, x) = D(ω) exp(−iω(t−x/v
i
))+u(ω) exp(−iω(t+x/v
i
))
D
i+1
= t
i
D
′
i
+ ¯r
i
u
i+1
; −t
i
D
′
i
= ¯r
i
u
i+1
− D
i+1
u
′
i
= r
i
D
′
i
+
¯
t
i
u
i+1
u
′
i
− r
i
D
′
i
=
¯
t
i
u
i+1
0 −t
i
1 −r
i
u
′
i
D
′
i
=
¯r
i
−1
¯
t
i
0
u
i+1
D
i+1
0 −t
i
1 −r
i
−1
=
1
t
i
−r
i
t
i
−1 0
;
1
t
i
−r
i
t
i
−1 0
¯r
i
−1
¯
t
i
0
=
=
−r
i
¯r
i
/t
i
+
¯
t
i
r
i
/t
i
−¯r
i
/t
i
1/t
i
=
1
t
i
1 r
i
r
i
1
r
i
= −¯r
i
;
¯
t
i
− r
i
¯r
i
/t
i
= 1/t
i
u
′
i
D
′
i
=
1
t
i
1 r
i
r
i
1
u
i+1
D
i+1
u
i
= u
′
i
exp(iω∆t)
D
i
= D
′
i
exp(−iω∆t)
u
′
i
D
′
i
=
exp(−iω∆t) 0
0 exp(iω∆t)
u
i
D
i
u
i
D
i
=
1
t
i
exp(iω∆t) 0
0 exp(−iω∆t)
1 r
i
r
i
1
u
i+1
D
i+1
=
=
1
t
i
exp(iω∆t) r
i
exp(iω∆t)
r
i
exp(−iω∆t) exp(−iω∆t)
u
i+1
D
i+1
z = exp(2iω∆t); w = exp(iω∆t)
u
D
i
=
1
wt
i
z r
i
z
r
i
1
u
D
i+1
u
D
1
=
1
wt
1
z r
1
z
r
1
1
1
wt
2
z r
2
z
r
2
1
u
D
3
=
=
1
w
2
t
1
t
2
z
2
+ zr
1
r
2
z
2
r
2
+ zr
1
zr
1
+ r
2
zr
1
r
2
+ 1
u
D
3
=
=
1
w
2
t
1
t
2
z
2
(1 + r
1
r
2
/z) z
2
(r
2
+ r
1
/z)
zr
1
+ r
2
zr
1
r
2
+ 1
u
D
3
=
=
1
w
2
z
2
F
2
(1/z) z
2
G
2
(1/z)
G
2
(z) F
2
(z)
u
D
3
F
2
(z) =
1
t
1
t
2
(1 + r
1
r
2
z); G
2
(z) =
1
t
1
t
2
(r
2
+ r
1
z)