Eduardo Kausel-Fundamental solutions in elastodynamics_ a compendium-Cambridge University Press (2006)

46 Two-dimensional problems in full, homogeneous spaces
0.4
0.2
µ rg r
0
-0.2
-0.4
0 10
ωr
20 30
α
Figure 3.6a: Cylindrical cavity in full space with ν = 0.3: frequency
response at r/r 0 = 5 for pressure pπr0 2 = 1. Solid line, real part; dashed
line, imaginary part.
with t ′ = t − t 0 , where t 0 = (r − r 0 )/α is the time of arrival of P waves at the receiver station.
From this asymptotic structure and the properties of Fourier transforms, 5 we infer that the
magnitude of the response immediately behind the wave front is (see also Fig. 3.6b)
u(r, r 0 , t + 0 ) = lim
ω→∞
[
iω gr (ω) e iωt0] = α a2
µ
√
r0
r = 1
ρα
√
r0
r
(3.56)
with t + 0
the instant immediately after passage of the wave front, and ρ the mass density.
A strategy to evaluate the impulse response would be to work in delayed time t ′ = t −
t 0 ≥ 0, use the asymptotic expansion for the Hankel functions beginning at an appropriately
large frequency ω 0 (say, 0 > 2π, that is, ω 0 = 2πα/r 0 ), and express the contribution of
the two tails in terms of the sine integral, for which accurate and efficient polynomial
approximations exist:
u r (r, r 0 , t) = r 0
2πµ
∫ +ω0
−ω0
[
]
H (2)
1
( P ) e iωt0
e iω(t−t0) dω
2H (2)
1
( 0 ) − a −2 0 H (2)
0
( 0 )
+ 1 √ [
r0
1 − 2 ]
2ρα r π Si (ω 0(t − t 0 ))
(3.57)
Observe that the kernel of the above integral tends to a constant value at zero frequency.
Also, the denominator is never zero for any frequency or Poisson’s ratio, which means that
the integrand has no real poles. Figures 3.6a,b show the results for an impulsive pressure
of strength p πr 2 0 = δ(t) at a dimensionless distance r/r 0 = 5 in a full space with Poisson’s
ratio ν = 0.30. This pressure is consistent with that of a line of pressure in Section 3.6,
which facilitates comparisons.
5 Papoulis, A., 1962, The Fourier integral and its applications, McGraw-Hill.