Geology Reference
In-Depth Information
1.6.1 Example 1-11. Propagation of an initially static disturbance.
Let us consider the classic initial value problem when the initial
displacement f(x) is given with a vanishing speed, namely,
2 u(x,t)/ t 2 - c 2
2 u/ x 2 = 0
(1.68)
u(x,0) = f(x)
(1.69)
u(x,0)/ t = 0
(1.70)
u,
u(x,t)/ x , u(x,t)/ t
0, as
(x 2 + t 2 )
(1.71)
Next, introduce the Fourier transform
+
U(
,t) =
e i
x u(x,t) dx
(1.72)
-
where t is regarded as a parameter. If we multiply Equation 1.68 by e i
x ,
integrate the result with respect to x over (-
,+
) by parts and apply the
“regularity condition” in Equation 1.71, we obtain
U tt ( ,t) + c 2 2 U( ,t) = 0 (1.73)
which is an ordinary differential equation in t. Equation 1.73 has the solution
U( ,t) = A( ) sin ct + B( ) cos ct (1.74)
Equations 1.69 and 1.70 can be similarly transformed, with the result that
U( ,0) = F( ) (1.75)
U t ( ,0) = 0 (1.76)
where F( ) is the Fourier transform of f(x). It follows that the solution to
Equation 1-74 satisfying Equations 1.75 and 1.76 is
U(
,t) = F(
) cos
ct
(1.77)
Following Equation 1.67, we write the solution u(x,t) as
+
u(x,t) = {1/(2
)}
e -i
x F(
) cos
ct d
(1.78)
-
which is the exact integral. A more enlightening interpretation is possible if we
observe that
F(
) cos
ct = 1/2 F(
) exp (i
ct) + 1/2 F(
) exp (-i
ct)
(1.79)
Thus, Equation 1.78 can be written as
+
u(x,t) = {1/(2
)}
e -i
x {1/2 F(
)e ic
t + 1/2 F(
)e -ic
t }d
(1.80)
-
which involves products of transforms. We turn to the Fourier Convolution
Th e o r e m , which relates k(x) and u(x) to their transforms K(
) and F(
), i.e.,
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