Roy K.K. Potential theory in applied geophysics
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8.1 Layered Earth Problem in a Direct Current Domain 209
φ
i
=
Iρ
1
2π
.
1
(r
2
+z
2
)
1/2
+
∞
0
A
i
(m) e
−mz
+B
i
(m) e
mz
J
o
(mr)dm (8.5)
where A
i
and B
i
are the arbitrary constants determined from the boundary
conditions. At the air earth bou n d ary, the current density vector
J
= σ
E
cannot cut across the air earth bou nd ary because resistivity of air is infinitely
high. Therefore,
1
ρ
1
∂φ
1
∂z
=0atz=0whereρ
1
and φ
1
are respectively the
resistivity and potential of the first medium.
∂φ
∂z
z=o
=
⎧
⎨
⎩
Iρ
1
z
2π (r
2
+ z
2
)
1/2
+
∞
o
−A
1
e
−mz
+ B
mz
1
J
o
(mr) mdm
⎫
⎬
⎭
z=o
=
∞
0
(B
1
− A
1
)J
o
(mr) mdm = 0 (8.6)
Since the relation is true for any value of r,
B
1
=A
1
. (8.7)
Therefore, expression for potential in the first medium is
φ
1
=
Iρ
1
2π
.
1
(r
2
+z
2
)
1/2
+
∞
0
A
1
(m)
e
mz
+e
−mz
J
o
(mr) dm. (8.8)
Last layer thickness h
N
is infinitely high.
In this layer the perturbation potential decays down with depth. There-
fore, e
−mz
will be the appropriate p otential function and the expression for
potential in the last layer is
φ
N
=
Iρ
1
2π
.
1
(r
2
+ z
2
)
1/2
+
∞
0
e
−mz
J
0
(mr ) dm. (8.9)
To get the expressions for the unknown functions A
1
(m), A
2
(m) ...A
N
(m), it
is necessary to apply the boundary conditions (i) φ
i
= φ
i+1
and (ii) J
n
i
=J
n
i+1
at each boundary. Before applying the boundary conditions, it is necessary to
bring the source and perturbation potential in the same format. This is an
important task in these boundary value problems. This point is highlighted for
all the problems discussed in this volume. In this problem, we use the Weber
Lipschitz identity to bring the source potential and perturbation potential in
the same format, i.e.,
1
r
=
∞
0
e
−mz
J
o
(mr) dm. (8.10)
Substituting q =
Iρ
2π
,
210 8 Direct Current Field Related Potential Problems
we can write the expression for potentials in different media as,
φ
1
=q
∞
0
e
−mz
J
o
(mr) dm +
∞
0
A
1
e
−mz
+e
mz
J
o
(mr) dm
φ
2
=q
∞
0
e
−mz
J
o
(mr) dm +
α
0
A
2
e
−mz
+B
2
e
mz
J
o
(mr) dm
........................................................................
φ
i
=q
∞
0
e
−mz
J
o
(mr) dm +
α
0
A
i
e
−mz
+B
i
e
mz
J
o
(mr) dm
........................................................................
φ
N
=q
∞
0
e
−mz
J
o
(mr) dm +
α
0
A
N
e
−mz
J
o
(mr) dm (8.11)
Applying the boundary conditions at each boundary i.e.,
φ
i
= φ
i+1
and
1
ρ
i
∂φ
i
∂z
=
1
ρ
i+1
∂φ
i+1
∂z
at z = hi, (8.12)
one gets
A
1
e
−mh
1
+e
mh
1
− A
2
e
−mh
1
− B
2
e
mh
1
=0
A
1
ρ
2
e
mh
1
− e
−mh
1
− A
2
ρ
1
e
−mh
1
+ B
2
ρ
1
e
mh
1
= q (ρ
1
− ρ
2
) e
−mh
1
A
i
e
−mh
i
+B
i
e
mh
i
− A
i+1
e
−mh
i
− B
i+1
e
mhi
=0
ρ
i+1
−A
i
e
−mh
i
+B
i
e
mh
i
+ ρ
i
A
i+1
e
−mh
i
− B
i+1
e
mh
i
=q
ρ
i+1
− ρ
i
e
−mh
i
A
N−1
e
−mh
N−1
+B
N−1
e
−mh
N−1
− A
N
e
−mh
N−1
=0
− A
N−1
ρ
N
e
−mh
N−1
+B
N−1
ρ
N
e
mh
N−1
+A
N
ρ
N−1
e
−mh
N−1
=q
ρ
N
− ρ
N−1
e
mh
N−1
(8.13)
From this system of equations, it is possible to find out potential at a point
in any medium for an N-layered earth due to a point source. The factors A
i
s
and B
i
s are to be determined from the boundary conditions. They are termed
as the kernel functions because they carry i nformation about all the 2N-1
layer parameters. Here ρ
1
to ρ
N
are the layer resistivities and h
1
to h
N−1
the
thicknesses. Thickness of the Nth layer is infinitely high.
8.1 Layered Earth Problem in a Direct Current Domain 211
8.1.1 Cramer’s Rule
Cramer’s rule for evaluation of determinants states that if we have a set of
linear equations
a
11
x
1
+a
12
x
2
+a
13
x
3
+ ........................+a
1n
x
n
=k
1
a
21
x
1
+a
22
x
2
+a
23
x
3
+ ........................+a
2n
x
n
=k
2
a
n1
x
1
+a
n2
x
2
+a
n3
x
3
+ ........................+a
nn
x
n
=k
n
(8.14)
and if determinant of the coefficients is not equal to zero, i.e., if
|a| =
a
11
a
12
........ a
1n
a
21
a
22
........ a
2n
...... ..... ....... ......
a
n1
a
n2
...... a
nn
= 0 (8.15)
then
x
1
=
|D
1
|
|a|
,x
2
=
|D
2
|
|a|
..............x
n
=
|D
n
|
|a|
(8.16)
where D
r
is the determinant formed by replacing the elements a
1r
,a
2r
,.........
a
nr
of the column of |a| by k
1
, k
2
, k
3
.........k
n
.
8.1.2 Two Layered Earth Model
For a two layered earth problem, resistivity and thickness of the first layer
are ρ
1
and h
1
. Resistivity of the second layer is ρ
2
.h
2
, the thickness o f the
second layer is infinitely high. Equations for potentials in the two media can
easily b e obtained from (8.11) and (8.13). Only the kernels A
1
and A
2
are to
be evaluated and th ey can be expressed as the ratio of the two determinants.
Let the denominator of the quotient be D
2
,thus
D
2
=
x
h
1
+x
−h
1
−x
h
1
ρ
2
−x
h
1
+x
−h
1
ρ
1
x
h
1
(8.17)
obtained from (8.13) where u = x
2
=e
−2m
.
Let
D
2
=P
2
(x) + H
2
(x)
Then
P
2
(x)=
x
h
1
−x
h
1
−ρ
2
x
h
1
ρ
1
x
h
1
=(ρ
1
− ρ
2
)x
h
1
= −(ρ
2
− ρ
1
)u
h
1
=(ρ
2
+ ρ
1
)k
1
u
h
1
= −(ρ
2
+ ρ
1
)P
2
(u) (8.18)
where P
2
(u) = k
1
u
h
1
and
212 8 Direct Current Field Related Potential Problems
k
1
=
ρ
2
− ρ
1
ρ
2
+ ρ
1
.
k
1
is termed as the reflection factor. Similarly
H
2
(x) =
x
−h
1
−x
h
1
ρ
2
x
−h
1
ρ
1
x
h
1
=(ρ
2
+ ρ
1
)=(ρ
2
+ ρ
1
)H
2
(u) (8.19)
where H
2
(u) = 1.H
2
(u) is introduced to generate the Kernel function in the
form of a recurrence formula for a n N-layered earth. For a two-layer earth D
2
becomes
D
2
=(ρ
2
+ ρ
1
)[H
2
(u) −P
2
(u)] . (8.20)
Surface Kernel A
1
For a two-layer earth, the numerator of the quotient of the (8.13) obtained
from (8.11) is given by
N
21
=
0 −x
h
1
(ρ
2
− ρ
1
)x
h
1
ρ
1
x
h
1
. (8.21)
By adding the second column of the determinant with the first column, we
get
N
21
= −P
2
(x) = (ρ
2
+ ρ
1
)P
2
(u) (8.22)
Therefore the Kernel A
1
for a two layer earth is
N
21
D
2
=q
P
2
(u)
H
2
(u) −P
2
(u)
. (8.23)
This is a surface kernel because for surface geophysics where the measurements
are taken on the surface of the earth, we are mostly interested in A
1
.
Subsurface Kernel A
2
For the Kernel A
2
,thenumeratorN
22
from (8.13) is given by
N
22
=
x
h
1
+x
−h
1
0
ρ
2
−x
h
1
+x
−h
1
(ρ
2
− ρ
1
)x
h
1
. (8.24)
Let
N
22
=(N
22
)
1
+(N
22
)
2
where
(N
22
)
1
=
x
h
1
0
−ρ
2
x
h
1
(ρ
2
− ρ
1
)x
h
1
and
8.1 Layered Earth Problem in a Direct Current Domain 213
(N
22
)
2
=
x
−h
1
0
−ρ
2
x
−h
1
(ρ
2
− ρ
1
)x
h
1
. (8.25)
By adding the first column of (N
22
), to its second column it can be shown
that (N
22
)
1
= −P
2
(x) = (ρ
2
+ ρ
1
)P
2
(u) and
(N
22
)
2
=(ρ
2
+ ρ
1
)k
1
=(ρ
2
+ ρ
1
)[1+k
1
− H
2
(u)]
where H
2
(u) = 1. Therefore, we can write down the expression for subsurface
Kernel A
2
as
A
2
=
N
22
D
2
=q
1+k
1
H
2
(u) −ρ
2
(u)
− 1
. (8.26)
8.1.3 Three Layered Earth Mo del
For a three layered earth, the kernels to be determined from the (8.13) are
A
1
,A
2
,B
2
and A
3
.OnlyA
1
is the surface kernel. The denominator D
3
of
(8.13) can be written as
D
3
=
⎡
⎢
⎢
⎣
x
h
1
+x
−h
1
−x
h
1
−x
−h
1
0
ρ
2
−x
h
1
+x
−h
1
ρ
1
x
h
1
−ρ
1
x
−h
1
0
0x
h
2
x
−h
2
−x
h
2
0 −ρ
3
x
h
2
ρ
3
x
−h
2
ρ
2
x
h
2
⎤
⎥
⎥
⎦
. (8.27)
Let D
3
=P
3
(x) + H
3
(x)
where
P
3
(x) =
⎡
⎢
⎢
⎣
x
h
1
−x
h
1
−x
−h
1
0
−ρ
2
x
h
1
+ρ
1
x
h
1
−ρ
1
x
−h
1
0
0x
h
2
x
−h
2
−x
h
2
0 −ρ
3
x
h
2
ρ
3
x
−h
2
ρ
2
x
h
1
⎤
⎥
⎥
⎦
(8.28)
and
H
3
(x) =
⎡
⎢
⎢
⎣
x
−h
1
−x
h
1
−x
−h
1
0
ρ
2
x
−h
1
ρ
1
x
h
1
−ρ
1
x
−h
1
0
0x
h
2
x
−h
2
−x
h
2
0 −ρ
3
x
h
2
ρ
3
x
−h
2
ρ
2
x
h
2
⎤
⎥
⎥
⎦
. (8.29)
Let us now consider the determinant P
3
(x). Multiplying the third row by ρ
2
and adding it to the fourth row, we get
P
3
(x)=
⎡
⎢
⎢
⎣
P
2
(x) −x
−h
1
0
−ρ
1
x
−h
1
0
0 x
h
2
x
−h
2
−x
h
2
0 −(ρ
3
− ρ
2
) x
h
2
(ρ
3
+ ρ
2
) x
−h
2
0
⎤
⎥
⎥
⎦
. (8.30)
Developing it with respect to the last column, we get
P
3
(x)=−x
h
2
−x
−h
1
P
2
(x) ρ
1
x
−h
1
0(ρ
3
− ρ
2
) x
h
2
(ρ
3
+ ρ
2
) x
−h
2
. (8.31)
214 8 Direct Current Field Related Potential Problems
Developing next with respect to the last row, we get
P
3
(x) = (ρ
3
+ ρ
2
)P
2
(x) + (ρ
3
− ρ
2
)x
2h
2
x
h
1
−x
−h
1
−ρ
2
x
h
1
−ρ
1
x
−h
1
=(ρ
3
+ ρ
2
)
'
P
2
(x) − k
2
x
2h
2
H
2
x
−1
(
(8.32)
where
H
2
x
−1
=
x
h
1
−x
−h
1
ρ
2
x
h
1
ρ
1
x
−h
1
. (8.33)
It differs from the determinant H
2
(x) only in the power of x. Here k
2
,the
reflection factor at the boundary between the second and third layer is equal
to
k
2
=(ρ
3
− ρ
2
)/(ρ
3
+ ρ
2
).
Therefore,
P
3
(x) = −(ρ
2
+ ρ
1
)(ρ
3
+ ρ
2
)
'
P
2
(u) + k
2
u
h
2
H
2
u
−1
(
(8.34)
= −(ρ
2
+ ρ
1
)(ρ
3
+ ρ
2
)P
3
(u)
where
P
3
(u) =P
2
(u) + k
2
u
h
2
H
2
u
−1
=k
1
u
h
1
+k
2
u
h
2
. (8.35)
Similarly
H
3
(x)=x
h
2
−x
−h
1
H
2
(x) −ρ
1
x
−h
1
0 −(ρ
3
− ρ
2
) x
h
2
(ρ
3
+ ρ
2
) x
−h
2
=(ρ
3
+ ρ
2
) H
2
(x)+(ρ
3
− ρ
2
) x
2h
2
x
−h
1
−x
−h
1
ρ
2
x
−h
1
−ρ
1
x
−h
1
=(ρ
3
+ ρ
2
)
'
H
2
(x) − k
2
x
2h
2
P
2
x
−1
(
(8.36)
where
P
2
x
−1
=
x
−h
1
−x
−h
1
−ρ
2
x
−h
1
ρ
1
x
−h
1
=(ρ
2
+ ρ
1
)(ρ
3
+ ρ
2
)H
3
(u). (8.37)
Here
H
3
(u) = H
2
(u) + k
2
u
h
2
P
2
u
−1
=1+k
1
k
2
u
h
2
−h
1
(8.38)
and
D
3
=(ρ
2
+ ρ
1
)(ρ
3
+ ρ
2
)[H
3
(u) −P
3
(u)] . (8.39)
8.1 Layered Earth Problem in a Direct Current Domain 215
Surface Kernel A
1
For the Kernel A
1
, in a three layered earth model the numerator of the quo-
tient from (8.13) is given by
N
31
=
0 −x
−h
1
−x
−h
1
0
(ρ
2
− ρ
1
)x
h
1
ρ
1
x
h
1
−ρ
1
x
−h
1
0
0x
h
2
x
−h
2
−x
h
2
(ρ
3
− ρ
2
)x
h
2
−ρ
3
x
h
2
ρ
3
x
−h
2
ρ
2
x
h
2
(8.40)
The value of this determinant is
N
31
= −P
3
(x) = (ρ
2
+ ρ
1
)(ρ
3
+ ρ
2
)P
3
(u)
Therefore
A
1
=
N
31
D
3
=q
P
3
(u)
H
3
(u) −P
3
(u)
. (8.41)
Subsurface Kernel A
2
The numerator for the kernel A
2
may be written from (8.13) as
N
32
=
x
h
1
+x
−h
1
0 −x
−h
1
0
ρ
2
−x
h
1
+x
−h
1
(ρ
2
− ρ
2
)x
h
1
−ρ
1
x
−h
1
0
00x
−h
2
−x
h
2
0(ρ
3
− ρ
2
)x
h
2
ρ
3
x
−h
1
ρ
2
x
h
1
(8.42)
=(N
32
)
1
+(N
32
)
2
where
(N
32
)
1
=
x
h
1
0 −x
−h
1
0
−ρ
2
x
h
1
(ρ
2
− ρ
1
) x
h
1
−ρ
1
x
−h
1
0
00x
−h
2
−x
h
2
0(ρ
3
− ρ
2
) x
h
2
ρ
3
x
−h
2
ρ
2
x
h
1
(8.43)
and
(N
32
)
2
=
x
−h
1
0 −x
−h
1
0
ρ
2
x
−h
1
(ρ
2
− ρ
1
) x
h
1
−ρ
1
x
−h
1
0
0 −x
h
2
x
−h
2
−x
h
2
0 ρ
3
x
h
2
ρ
3
x
−h
2
ρ
3
x
h
2
. (8.44)
(N
32
)
1
can be shown to be equal to
= −P
3
(x) = (ρ
2
+ ρ
1
)(ρ
3
+ ρ
2
)P
3
(u). (8.45)
216 8 Direct Current Field Related Potential Problems
(N
32
)
2
= x
h
2
−x
−h
1
(N
22
)
2
−ρ
1
x
−h
1
0(P
3
− ρ
2
) x
h
2
(ρ
3
+ ρ
2
) x
−h
2
=(ρ
3
+ ρ
2
)(N
22
)
2
− (ρ
3
− ρ
2
)x
2h
2
x
−h
1
−x
−h
1
ρ
2
x
−h
1
−ρ
1
x
−h
1
=(ρ
3
+ ρ
2
)
'
(N
22
)
2
+k
2
x
2h
2
P
2
x
−1
(
=(ρ
2
+ ρ
1
)(ρ
3
+ ρ
2
)
'
1+k
1
− H
2
(u) −k
2
u
h
2
− P
2
u
−1
(
=(ρ
2
+ ρ
1
)(ρ
3
+ ρ
2
)[1+k
1
− H
3
(u)]
=(ρ
2
+ ρ
1
)(ρ
3
+ ρ
2
)[(1+k
1
)B
23
(u) −H
3
(u)] (8.46)
where
B
23
(u) = 1.
Therefore
N
32
=(ρ
2
+ ρ
1
)(ρ
3
+ ρ
2
)[(1+k
1
)B
23
(u) − H
3
(u) + P
3
(u)] (8.47)
and the kernel A
2
can be written as
A
2
=q
(1 + k
1
)B
23
(u)
H
3
(u) −P
3
(u)
− 1
. (8.48)
Subsurface Kernel B
2
The determinant N
33
for the numerator of the Kernel B
2
is
N
33
=
x
h
1
+x
−h
1
−x
h
1
00
ρ
2
−x
h
1
+x
−h
1
ρ
1
x
h
1
(ρ
2
− ρ
1
)x
h
1
0
0x
h
2
0 −x
h
2
0 −ρ
3
x
h
2
(ρ
3
− ρ
2
)x
h
2
ρ
2
x
h
2
(8.49)
=(N
33
)
1
+(N
33
)
2
where
(N
33
)
1
=
x
h
1
−x
h
1
00
−ρ
2
x
h
1
ρ
1
x
h
1
(ρ
2
− ρ
1
)x
h
1
0
0x
h
2
0 −x
h
2
0 −ρ
3
x
h
2
(ρ
3
− ρ
2
)x
h
2
ρ
2
x
h
2
= 0 (8.50)
(N
33
)
2
=
x
−h
1
−x
h
1
−x
h
1
0
ρ
2
x
−h
1
ρ
1
x
h
1
ρ
3
x
h
1
0
0x
h
2
0 −x
h
2
0 −ρ
3
x
h
2
0 ρ
2
x
h
2
=(ρ
2
+ ρ
1
)(ρ
3
+ ρ
2
)(1+k
1
)k
2
u
h
2
(8.51)
where
2ρ
2
ρ
2
+ρ
1
=1+k
1
.Ifwetake
8.1 Layered Earth Problem in a Direct Current Domain 217
D
23
(u) = k
2
u
h
2
,
we can write
N
33
=(ρ
2
+ ρ
1
)(ρ
3
+ ρ
2
)(1 + k
1
)D
23
(u)
and
B
2
=q(1+k
1
)
D
23
(u)
H
3
(u) −P
3
(u)
. (8.52)
Kernel A
3
The numerator of this kernel N
34
is
N
34
=
x
h
1
+x
−h
1
−x
h
1
−x
−h
1
0
ρ
2
−x
h
1
+x
−h
1
ρ
1
x
h
1
−ρ
1
x
−h
1
(ρ
3
− ρ
1
)x
h
1
0x
h
2
x
−h
2
0
0 −ρ
3
x
h
2
ρ
3
x
−h
2
(ρ
3
− ρ
2
)x
h
2
=(ρ
2
+ ρ
1
)(ρ
3
+ ρ
2
)[(1+k
1
)(1+k
2
) −H
3
(u) + P
3
(u)] (8.53)
Hence
A
3
=
q(1+k
1
)(1+k
2
)
H
3
(u) − ρ
3
(u)
− 1. (8.54)
8.1.4 General Expressions for the Surface and Subsurface Kernels
for an N-Layered Earth
After deriving the expressions for the surface and subsurface kernels upto a
three layered earth, we write down the recurrence formulae for the kernels
for an N-layered earth and write down the expressions for the potentials in
different media.
(i) Surface Kernel A
1
for an N -layered earth can be Written as
A
1
=q
P
N
(u)
H
N
(u) −P
N
(u)
(8.55)
where
P
N
(u) = P
N−1
(u) + k
N−1
u
h
N−1
H
N−1
u
−1
and
H
N
(u) = H
N−1
(u) + k
N−1
u
h
N−1
P
N−1
u
−1
(8.56)
where
P
2
(u) = k
1
u
h1
and
H
2
(u) = 1.
218 8 Direct Current Field Related Potential Problems
(ii) Kernels A
2
and B
2
of the Second Layer
The expressions for the Kernels A
2
and B
2
are respectively given by
A
2
= q
(1 + k
1
) B
2N
(u)
H
N
(u) −P
N
(u)
− 1
(8.57)
and
B
2
=q
(1 + k
1
)D
2N
(u)
H
N
(u) −P
N
(u)
(8.58)
where
B
22
(u) = 1 D
22
(u) = O
B
23
(u) = 1 D
23
(u) = k
2
u
h2
B
24
(u) = B
23
(u) + k
3
u
h3
D
23
(u
−1
)
D
24
(u) = B
23
(u) + k
3
u
h3
B
23
(u
−1
) (8.59)
B
2N
(u) = B
2(N−1)
(u) + k
N−1
u
h
N−1
D
2(N−1)
u
−1
D
2N
(u) = D
2(N−1)
(u) + k
N−1
u
h
N−1
B
2(N−1)
u
−1
. (8.60)
(iii) The Expressions for the Kernels A
i
and B
i
for ith Lay er can
be Written as
A
i
=q
⎡
⎢
⎢
⎣
i−1
1
s=1
(1 + k
S
)
B
iN
H
N
(u) − P
N
(u)
− 1
⎤
⎥
⎥
⎦
(8.61)
and
B
i
=q
⎡
⎢
⎢
⎢
⎣
i−1
1
s−1
(1 + k
S
)
D
iN
H
N
(u) −P
N
(u)
⎤
⎥
⎥
⎥
⎦
(8.62)
where
i−1
2
S=1
(1 + k
s
)=(1+k
1
)(1+k
2
)(1+k
3
) ......(1 + k
i−1
)
and
k
s
=
ρ
s+1
− ρ
s
ρ
s+1
+ ρ
s
.
B
iN
(u) and D
iN
(u) are related through the recurrence relations given as