Geology Reference
In-Depth Information
As in Chapter 1, we attempt separable solutions, but because the problem
is three-dimensional, the procedure must be applied twice in sequence. We
therefore take
P(x,y) = X(x)Y(y)
(3.36)
Substitution in Equation 3.35 leads to
- X"(x)/X(x) = Y"(y)/Y(y) + (
2
/c
2
- k
z
2
) (3.37)
Since the left side is a function of x (independent of y), and the right side is a
function of y (independent of x), each side must separately equal the same
positive constant k
x
2
, that is,
- X"(x)/X(x) = Y"(y)/Y(y) + (
2
/c
2
- k
z
2
) = k
x
2
(3.38)
This implies two equations, namely,
X"(x) + k
x
2
X(x) = 0
(3.39)
Y"(y)/Y(y) + (
2
/c
2
- k
z
2
) = k
x
2
(3.40)
The solution to Equation 3.39 is X(x) = A sin k
x
x + B cos k
x
x. Infinitely
rigid, impermeable walls are assumed at x = 0 and x = L
x
. Now, from Equation
3.32, the requirement U = 0 translates into one for the derivative P
x
(x,y) and
therefore X'(x). The condition X'(0) = 0 implies that A = 0, while the second
leads to X'(L
x
) = - k
x
B sin k
x
L
x
= 0. If nontrivial solutions are sought, we must
set k
x
L
x
=
l
, where
l
= 0,1,2,3, ... and so on. This leads to
X
l
(x) = B
l
cos
l
x/L
x
(3.41)
Next, let us consider Equation 3.40, and rewrite it in the form
Y"(y)/Y(y) = k
x
2
- (
2
/c
2
- k
z
2
) = - k
y
2
(3.42)
or
Y"(y) + k
y
2
Y(y) = 0 (3.43)
where k
y
2
is positive. This has the solution Y(y) = C sin k
y
y + D cos k
y
y.
Infinitely rigid, impermeable walls are also assumed at y = 0 and y = L
y
. We
similarly find that V = 0 implies a vanishing of P
y
(x,0). Hence Y'(0) = 0, so that
C = 0. Then, Y'(L
y
) = - D k
y
sin k
y
L
y
= 0 forces us to choose k
y
L
y
=
m
,
m
=
0,1,2,3, ... in order to obtain the nontrivial solutions
Y
m
(y) = D
m
cos
m
y/L
y
(3.44)
Thus, Equations 3.36, 3.41 and 3.44 together imply that p(x,y,z,t) can be
constructed from superpositions of the component solutions
p
l,m
(x,y,z,t) = P
l,m
cos
l
x/L
x
cos
m
y/L
y
exp i(
t-k
z
z)
(3.45)
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