Roy K.K. Potential theory in applied geophysics
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8.1 Layered Earth Problem in a Direct Current Domain 219
B
ii
(u) = 1, D
ii
(u) = O,
B
i(i+1)
(u) = B
ii
(u) + k
1
u
h
1
D
ii
u
−1
D
i(i+1)
(u) = D
ii
(u) + k
1
u
h
i
B
ii
u
−1
(8.63)
and
B
iN
(u) = B
i(N−1)
(u) + k
N−1
u
h
N−1
D
i(N−1)
u
−1
D
iN
(u) = D
i(N−1)
(u) + k
N−1
u
h
N−1
B
i(N−1)
u
−1
. (8.64)
(iv) The Kernel Function A
N
fortheNthLayerisgivenby
A
N
=q
⎡
⎢
⎢
⎣
N−1
1
s=o
(1 + k
S
)
B
NN
(u)
H
N
(u) − P
N
(u)
− 1
⎤
⎥
⎥
⎦
(8.65)
where B
NN
(u) = 1, it should be noted that B
N(N−1)
(u) = 1. For the kernels
B
iN
and D
iN
where i stands for the ith layer and N stands for the total number
of layers.
8.1.5 Kernels in Different Layers for a Five Layered Earth
In this section we are presenting the values of the kernels P
N
(u), H
N
(u), B
iN
(u) and D
iN
(u) for a five layered earth model. One can just write down the
expressions of the surface and subsurface kernels for any number of layers.
Let ρ
1
, ρ
2
, ρ
3
, ρ
4
, ρ
5
are the restivities of the five layers and h
1
,h
2
,h
3
,
h
4
and h
5
are the th icknesses of the five layer where the last layer thickness is
infinitely large. From the general expressions for the kernels for an N-layered
earth, we can write the expressions for th ese surface and subsurface kernels
as follows:
A. Surface Kernels
(i) For the first layer (i = 1)
(a)
P
2
(u) = k
1
u
h
1
H
2
(u) = 1 (8.66)
(b)
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
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
(8.67)
220 8 Direct Current Field Related Potential Problems
(c)
P
4
(u)=P
3
(u)+k
3
u
h
3
H
3
u
−1
= k
1
u
h
1
+ k
2
u
h
2
+ k
3
u
h
3
+ k
1
k
2
k
3
u
h
3
−h
2
+h
1
H
4
(u) = H
3
(u) + k
3
u
h
3
P
3
u
−1
=1+k
1
k
2
u
h
2
−h
1
+k
1
k
2
u
h
3
−h
1
+k
2
k
3
u
h
3
−h
2
(8.68)
(d)
P
5
(u)=P
4
(u)+k
4
u
h
4
H
4
u
−1
=k
1
u
h
1
+k
2
u
h
2
+k
3
u
h
3
+k
4
u
h
4
+k
1
k
2
u
h
3
−h
2
+h
1
+k
1
k
2
k
4
u
h
4
−h
2
+h
1
+k
1
k
3
k
4
u
h
4
−h
3
+h
1
+k
2
k
3
k
4
u
h
4
−h
3
+h
2
(8.69)
H
5
(u) =H
4
(u) + k
4
u
h
4
P
4
u
−1
=1+k
1
k
2
u
h
2
−h
1
+k
1
k
3
u
h
3
−h
1
+k
1
k
4
u
h
4
−h
1
+k
2
k
3
u
h
3
−h
2
+k
2
k
4
u
h
4
−h
2
+k
3
k
4
u
h
4
−h
3
+k
1
k
2
k
3
k
4
u
h
4
−h
3
+h
2
−h
1
(8.70)
B. Subsurface Kernels
(i) Kernels for the second layer (i = 2) are
(a)
B
22
(u) = 1
D
22
(u) = 0, (8.71)
(b)
B
23
(u) = B
22
(u) + k
2
u
h
2
D
22
u
−1
=1
D
23
(u) = D
22
(u) + k
2
u
h
2
B
22
u
−1
=k
2
u
h
2
. (8.72)
(c)
B
24
(u) = B
23
(u) + k
3
u
h
3
D
23
u
−1
=1+k
2
k
3
u
h
3
−h
2
.
D
24
(u) = D
23
(u) + k
3
u
h
3
B
23
u
−1
=k
2
u
h
2
+k
3
u
h
3
. (8.73)
8.1 Layered Earth Problem in a Direct Current Domain 221
(d)
B
25
(u) = B
24
(u) + k
4
u
4
D
24
u
−1
=1+k
2
k
3
u
h
3
−h
2
+k
2
k
4
u
h
4
−h
2
+k
3
k
4
u
h
4
−h
3
.
D
25
(u) = B
24
(u) + k
4
u
h
4
B
24
u
−1
=k
2
u
h
2
+k
3
u
h
3
+k
4
u
h
4
+k
2
k
3
k
4
u
h
4
−h
3
+h
2
. (8.74)
(ii) Kernels for the third layer (i = 3) are
(a)
B
33
(u) = 1
D
33
(u) = 0. (8.75)
(b)
B
34
(u) = B
33
(u) + k
3
u
h
3
B
33
u
−1
=1
D
34
(u) = D
33
(u) + k
3
u
h
3
B
33
u
−1
=k
3
u
h
3
. (8.76)
(c)
B
35
(u) = B
34
(u) + k
4
u
h
4
D
34
u
−1
=1+k
3
k
4
u
h
4
−h
3
D
35
(u) = B
34
(u) + k
4
u
h
4
B
34
u
−1
=k
3
u
h
3
+k
4
u
h
4
. (8.77)
(iii) Kernels for the fourth layer (i = 4)
(a)
B
44
(u) = 1 (8.78)
D
44
(u) = O,
(b)
B
45
(u) = B
44
(u) + k
4
u
h
4
D
44
u
−1
=1
D
45
(u) = D
44
(u) + k
4
u
h
4
B
44
u
−1
=k
4
u
h
4
.. (8.79)
(v) Kernels for the fifth layer (i = 5)
(a)
B
55
(u) = 1
D
55
(u) = O. (8.80)
8.1.6 Potentials in Different Media
Substituting the values of surface and subsurface kernels, we can write the
expressions for the potentials in different media as
222 8 Direct Current Field Related Potential Problems
φ
1
=q
∞
0
e
−mz
J
o
(mr) dm + q
∞
0
P
N
(u)
H
N
(u) −P
N
(u)
e
−mz
+e
mz
J
o
(mr) dm (8.81)
φ
2
=q
∞
0
e
−mz
J
o
(mr) dm + q
∞
0
(1 + k
1
)B
2N
H
N
(u) −ρ
N
(u)
− 1
e
−mz
J
o
(mr) dm + q
∞
0
(1 + k
1
)D
2N
(u)
H
N
(u) −P
N
(u)
e
mz
J
o
(mr) dm
q
∞
0
e
−mz
J
o
(mr)dm+q(1+k
1
)
∞
0
'
E
2
(u)e
−mz
+F
2
(u)e
mz
(
J
o
(mr) dm
(8.82)
φ
i
=q
∞
0
e
−mz
J
o
(mr) dm + q
i
Π
s=1
(1 + k
s
)
∞
0
'
E
i
(u) e
−mz
+F
i
(u) e
mz
(
J
o
(mr) dm (8.83)
φ
N
=q
∞
0
e
−mz
J
o
(mr) dm + q
N−1
Π
s=1
(1 + k
s
)
∞
0
E
NN
e
−mz
J
o
(mr) dm (8.84)
where
E
i
(u) =
B
iN
(u)
H
N
(u) − P
N
(u)
F
i
(u) =
D
iN
(u)
H
N
(u) − P
N
(u)
(8.85)
and
E
NN
=
B
NN
(u)
H
N
(u) −P
N
(u)
.
Kernel A
1
is reg ularly used by the geophysicists to compute the apparent
resistivity for an one-dimensional N-layered earth problem. Apparent resis-
tivity is defined as the resistivity of a fictitious homoge n eous medium, which
generates the same potential in the potential electrodes as one gets for an
inhomogeneous earth. Apparent resistivity is expressed as ρ
a
=K
∆φ
I
,where
K is the geometric factor, I is the current and ∆φ is the potential difference
8.2 Potential due to a P oint Source in a Borehole 223
between the two potential electrodes created by the direct current flow field.
For two electrode system, i.e. for on e current and one potential electrodes
∆φ = φ, because the other potential electrode is kept away by about 10 time
the distance between current and potential electrodes. Once the expression for
potential for an N layered earth for two-electrode system is obtained, one can
compute potential for any other electrode configuration. Potential at a point
due to different sources and sinks can be added or subtracted as the case may
be (mentioned in Chap. 7). One can get deeper and deeper information about
the subsurface by gradual increase in electrode separation. Further details
about D.C. resistivity sounding, electrode configuration, concept of apparent
resistivity is outside th e scope of this book. The readers may consult Keller
and Frisehknect (1966), Bhattacharya and Patra (1968), Koefoed(1979) and
Zhdanov and Keller (1994).
The constants in potential functions, which are determined applying
boundary conditions, are kernel functions because they contain information
about all the layer parameters. The expressions of the kernels are shown in
(8.66) to (8.85). There may b e some applications of the subsurface kernels
in surface to borehole measurements. Researchers in Potential theory may be
interested to know in detail about the behaviours of these subsurface kernels
and their dependence on resistivities at the boundar ies a nd layer thicknesses.
Both potentials and normal components of current densities will be contin-
uous across the boundaries. However refraction, reflection and transmission
of current lines may be studied with greater details for academic interest.
One gets potentials at all the subsurface nodes in finite element and finite
difference formulation of 2D/3D structures (Chap. 15). Therefore a compari-
son can be made between the subsurface potentials obtai ned analytically and
numerically. The researchers generally calibrate their finite element and finite
difference source codes against the responses obtained from an analytical solu-
tion. In this sector, the subsurface kernels may have some use. Behaviours of
the surface kernels can also be studied for electromagnetics and electromag-
netic tra n sients.
Deriving recurrence relation of kernel functions for an N-layered earth in
one form proposed by Mooney et al (1960) is demonstrated here. There is
one mor e appro ach for deriving the kernel function in the recurrence form as
available in Zhdanov and Keller (1994).
8.2 Potential due to a Point Source in a Borehole
with Cylindrical Coaxial Boundaries
Borehole of r adius ‘r’ contains drilling mud of resistivity ρ
1
. This cylindrical
mud column in the borehole (Fig. 8.2) is surrounded by a flushed zone of
resistivity ρ
2
. I t is termed as the flushed zone because most o f the pore fluids
are flushed out by the mud filtrate due to high pressure maintained in the
borehole. Radius of the outer boundary of the flushed zone is r
2
. The flushed
224 8 Direct Current Field Related Potential Problems
zone is surrounded by uncontaminated zone of resistivity ρ
3
. Radii of the inner
and outer boundaries are respectively given by r
2
and infinity. Thickness of the
bed is also assumed to be infinitely high so that the effects of shoulder beds are
negligibly small. Effect of the transitional invaded zone in between the flushed
zone and the uncontaminated zone is discussed in the next section where non-
laplacian equation will be involved in generating potential functions. Here
the potential φ is a function of r and z. In a source free region the potential
satisfies Laplace equation,
∇
2
φ = 0 (8.86)
i.e.,
∂
2
φ
∂r
2
+
1
r
∂φ
∂r
+
∂
2
φ
∂z
2
=0
because the potential will have the axial symmetry.
Applying the method of separation of variables i.e. φ = R(r)Z(z), we get
∂φ
∂r
=
dR
dr
.Z
∂
2
φ
∂r
2
=Z
d
2
R
dr
2
(8.87)
d
2
φ
∂z
2
=R
d
2
Z
dz
2
and (8.87) changes to the form
Z
d
2
R
dr
2
+
1
r
dR
dr
.Z + R
d
2
z
dz
2
= 0 (8.88)
1
R
d
2
R
dr
2
+
1
Rr
dR
dr
+
1
Z
d
2
Z
dz
2
= 0 (8.89)
1
R
d
2
R
dr
2
+
1
Rr
dR
dr
= −
1
Z
d
2
Z
dz
=m
2
. (8.90)
We get,
d
2
Z
dz
2
+ m
2
Z =0. (8.91)
The solutions of this equation are sin mz and cos mz.
The solution of the second equation is
I
0
(mr) and K
0
(mr) (8.92)
where I
0
(mr) and K
0
(mr) are respectively the modified Bessel’s function of
first and second kind and of zero order. Since inside a borehole, the potential
will be independent of sign of Z, i.e. with respect to the source point, the
potential above or below the source point will be same provided the distance
from the source remains the same. Therefore cos mz will be the proper poten-
tial function and not sin mz. The general solution of the potential function is
8.2 Potential due to a P oint Source in a Borehole 225
Fig. 8.2. Bore hole D.C. resistivity model with a current and a potential electrode
inside a borehole and with cylindrical coaxial boundaries
φ =
∞
0
A(m)J
0
(mr) cos mz dm +
∞
0
B(m)K
0
(m) cos mz dm (8.93)
where A(m) and B(m) are arbitrary kernel functions.
For this problem, the Weber Lipschitz’s integral can be written as
1
√
r
2
+z
2
=
2
π
∞
0
K
0
(mr) cos mz dm. (8.94)
Therefore, the potential d ue to a point source is
φ
0
=
ρ
1
I
4π
.
1
R
=
ρ
1
I
4π
1
√
r
2
+z
2
=
ρ
1
I
4π
.
2
π
∞
0
K
0
(mr) cos mz dm
=
ρ
1
I
2π
2
∞
0
K
0
(mz) cos mz dm. (8.95)
Since I
0
(mr) will be the better potential function within the borehole or
medium 1. Therefore the potential in medium 1 is
φ
1
= φ
0
+ φ
/
where φ
0
is the source potential and φ
/
is the perturbatio n potential.
Here
⇒ φ
1
=
ρ
1
I
2π
2
⎡
⎣
∞
0
K
0
(mz) cos mz dm +
∞
0
C
0
(m)I
0
(mr) cos mz dm
⎤
⎦
(8.96)
226 8 Direct Current Field Related Potential Problems
where
C
0
(m) =
2π
2
ρ
2
I
A
0
(m) .
φ
2
=
ρ
2
I
2π
2
.
2π
2
ρ
2
I
⎡
⎣
∞
0
A
2
(m) I
0
(mr) cos mz dm +
∞
0
B
2
(m) K
0
(mr) cos mz dm
⇒ φ
2
=
ρ
2
I
2π
2
∞
0
C
2
(m)I
0
(mr) cos mz dm
+
∞
0
D
2
(m) K
0
(mr) cos mz dm (8.97)
where
C
2
(m) =
2π
2
ρ
2
I
A
2
(m)
D
2
(m) =
2π
2
ρ
2
I
B
2
(m) .
φ
3
=
ρ
3
I
2π
2
.
2π
2
ρ
3
I
⎡
⎣
∞
0
B
3
(m) K
0
(mr )cosmz dm
⎤
⎦
⇒ φ
3
=
ρ
3
I
2π
.
∞
0
D
3
(m) K
0
(mr) cos mz dm (8.98)
where
D
3
(m) =
2π
2
ρ
3
I
B
3
(m) .
∂φ
2
∂r
=
ρ
2
I
2π
2
⎡
⎣
∞
0
C
2
(m) I
/
0
(mr) m cos mz dm
+
∞
0
D
2
(m) K
/
0
(mr) m cos mz dm
⇒
∂φ
2
∂r
=
ρ
2
I
2π
2
⎡
⎣
∞
0
C
2
(m) I
1
(mr) −K
1
(mr) D
2
(m)
⎤
⎦
mcos mz dm. (8.99)
8.2 Potential due to a P oint Source in a Borehole 227
φ
3
=
ρ
3
I
2π
2
∞
0
D
3
(m) K
0
(mr) cos mz dm. (8.100)
∂φ
3
∂r
=
ρ
3
I
2π
2
∞
0
D
3
(m) K
/
0
(mr) m cos mz dm
= −
ρ
3
I
2π
2
∞
0
D
3
(m) K
1
(mr) m cos mz dm. (8.101)
The boundary conditions are:
(i)
φ
1
|
r=1
= φ
2
|
r=1
φ
2
|
r=r
1
= φ
3
|
r=r
2
.
The normal component of the current density should be continuous across the
boundaries. i. e.,
1
ρ
1
∂φ
1
∂r
r=r
1
=
1
ρ
2
∂φ
1
∂r
r=r
1
1
ρ
2
∂φ
∂r
r=r
2
=
1
ρ
3
∂φ
∂r
r=r
2
. (8.102)
Assuming the radius of the cylinder to be unity, we put φ
1
= φ
2
at r = 1 and
get
ρ
1
I
2π
2
∞
0
K
0
(mr) cos mz dm +
∞
0
C
1
(m) I
0
(mr) cos mz dm
=
ρ
2
I
2π
2
⎡
⎣
∞
0
C
2
(m) I
0
(mr) cos mz dm +
∞
0
D
2
(m) K
0
(mr) cos mz dm
⇒
∞
0
{ρ
1
[C
1
(m) I
0
(mr) + K
0
(mr)] −ρ
2
[C
2
(m) I
0
(mr)
+D
2
(m) K
0
(mr)]}cos mz dm = 0. (8.103)
Similarly at
r=r
2
, φ
2
= φ
3
.
228 8 Direct Current Field Related Potential Problems
Therefore
ρ
2
I
2π
2
⎡
⎣
∞
0
C
2
(m)I
0
(mr
2
)cosmz dm+
∞
0
D
2
(m)K
0
(mr
2
)cosmz dm
⎤
⎦
=
ρ
3
I
2π
2
⎡
⎣
∞
0
D
3
(m)K
0
(mr
2
)cosmz dm
⎤
⎦
⇒
∞
0
{ρ
2
[C
2
(m) I
0
(mr
2
)+D
2
(m) K
0
(mr
2
)]
= ρ
3
D
3
(m) K
0
(mr
2
)}cos mz dm. (8.104)
From (8.103) and (8.104), the expressions within the bracket are zero, i.e.,
ρ
2
C
2
(m) I
0
(mr
2
)+ρ
2
D
2
(m) K
0
(mr
2
) − ρ
3
D
3
(m) K
0
(mr
2
)=0.
To apply the other boundary condition, i.e., the normal components of the
current densities are equal at the boundaries i.e.
J
1
=J
2
at r = r
1
and J
2
=J
3
at r = r
2
∂φ
1
∂r
,
∂φ
2
∂r
and
∂φ
3
∂r
are to be determined. Since
J=−σ∇φ,
∂φ
1
∂r
=
ρ
1
I
2π
2
⎡
⎣
∞
0
C
1
(m) I
/
0
(mr) cos mz m dm +
∞
0
K
/
0
(mr) cos mz m dm.
Since
I
/
0
(x) = I
1
(x)
K
/
0
(x) = −K
1
(x)
therefore,
∂φ
1
∂r
=
ρ
1
I
2π
2
⎡
⎣
∞
0
{C
1
(m) I
1
(mr) − K
1
(mr)}mcosmz dm
⎤
⎦
(8.105)
∂φ
2
∂r
=
ρ
2
I
2π
2
⎡
⎣
∞
0
C
2
(m) I
/
0
(mr) m cos mz dm
+
∞
0
D
2
(m) K
/
0
(mr) m cos mz dm
⎤
⎦
⇒
∂φ
2
∂r
=
ρ
2
I
2π
2
⎡
⎣
∞
0
{C
2
(m) I
1
(mr) − K
1
(mr) D
2
(m) m cos mz dm}
⎤
⎦
.
(8.106)