Geology Reference
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6.7.4 Difficulties with simple difference formulation.
6.7.4.1 Two space dimensions.
There are subtleties and problems in the basic approach. To understand
these them, let us consider the details that appear in the formulation, and take as
our starting point, Equation 6.99. For simplicity, we define the “slowness”
function s(x,y,z) by
s(x,y,z) = 1/c(x,y,z) (6.100)
and consider two space dimensions first. In the coordinate space (x,z), Equation
6.99 becomes
(
/ x) 2 + (
/ z) 2 = s 2 (x,z)
(6.101)
Now, we introduce the notation
u(x,z) =
x (x,z)
(6.102)
so that Equation 6.101 becomes
u 2 + (
/ z) 2 = s 2 (x,z)
(6.103)
Equation 6.103 can therefore be solved to yield
z (x,z) = {s 2 -u 2 } F(u) (6.104)
Next, we differentiate Equation 6.102 with respect to z, and combine the result
with the x derivative of Equation 6.104. This successively gives u(x,z)/ z =
zx (x,z) = xz (x,z) = F( u) / x, and hence, the first-order, nonlinear, wave-like
governing differential equation
u/ z - F( u) / x = 0 (6.105)
Why is Equation 6.105 physically significant? Let us consider the total
differential for the phase function (x,z),
d =
x dx +
z dz
(6.106)
If we use Equations 6.102 and 6.104, we can rewrite Equation 6.106 as
d = u dx + F(u) dz (6.107)
Thus, if the function u(x,z) is known, numerically say, so that F(u) is likewise
known, then the travel-time between any two points A and B can be performed
by the extremely simple integration
B B B
| = u dx + F(u) dz
(6.108)
A A
A
6.7.4.2 Three space dimensions.
Now, with a clearer motivation in mind, we reconsider the three-
dimensional eikonal equation
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