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

10.4 Stiffness matrix method for layered spheres 181
such that
⎧
⎨ ξ
˜F m = Q m F m =
⎩
⎧
⎨ ξ
˜H m = Q m H m =
⎩
1
1
⎫ ⎧
⎫
⎬ ⎨ f 11 f 12 0 ⎬
f
⎭ ⎩ 21 f 22 0
⎭
1 0 0 f 33
⎫ ⎧
⎫
⎬ ⎨ h 11 h 12 0 ⎬
h
⎭ ⎩ 21 h 22 0
⎭
1 0 0 h 33
(10.158)
(10.159)
which implies
ū = T n L n m Q−1 m Q m H m c
= T n L n m Q−1 ˜H m m c
and
¯p = T n L n m Q−1 m Q m F m c
= T n L n m Q−1 ˜F m m c
⇒
⇒
ū = T n L n m Q−1 m ũ (10.160)
¯p = T n L n m Q−1 m ˜p
= T n L n m Q−1 ˜K m ũ
(10.161)
For example, for an exterior region, the scaled, symmetric impedance matrix would be
˜K =−R 2 ˜F m
(
˜H m
) −1
⇒ ˜K = Q m KQ −1
m (10.162)
which relates the scaled force and displacement amplitudes vectors at the interface of the
outer region. More generally, in the case of a system of layers, we obtain a symmetric
global system matrix. However, to avoid proliferation of symbols, we continue labeling
the scaled vectors for the system as we did for any individual layer, and write the system
equation simply as
˜p = ˜K ũ ⇒ ũ = ˜K −1 ˜p (10.163)
In this equation, every third (SH) degree of freedom is uncoupled from the preceding
two (SV-P) degrees of freedom, so they should be solved separately. The advantage of
using a symmetric stiffness matrix lies in the saving in computational effort, which is a
factor of two. For notational transparency, however, we present the ensuing examples in
the standard, non-symmetric form.
10.4.3 Expansion of source and displacements into spherical harmonics
The stiffness matrix method described herein is based on a formulation in the frequency–
spheroidal-wavenumber domain. Thus, the sources p = p(R,φ,θ,ω), if any, must be
expressed in terms of spheroidal harmonics of order m,n, i.e., ˜p ≡ ˜p mn (R,ω)
Let p be the 3 × 1 vector of external tractions per solid angle at a given frequency, which
we assume to be defined over a spherical surface of constant radius R. Expansion into
spheroidal harmonics yields
∞∑ m∑
p = T n L n m ˜p mn (10.164)
m=0 n=0