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

10.2 Stiffness matrix method in Cartesian coordinates 153
which is the classical equation for Love waves. Once we have found the roots of this
equation, we can obtain the displacement amplitudes at the interfaces as eigenvectors of
the equation Ku = 0. Thereafter, displacements caused by Love waves within the layer
or the half-space can be obtained by means of the analytic continuation technique given
previously.
Example 10.3: SH line source at some depth in a layer over
an elastic half-space
The geometry is as in Example 10.2, but now the medium is subjected to an impulsive
line source at some elevation z ′ and in direction y, which corresponds to a body load
b y (x, z, t) = δ(t) δ(x) δ(z − z ′ ) (10.51)
Carrying out a Fourier transform in x and t, we obtain the source in the frequency–
wavenumber domain as
˜b y (k, z,ω) = δ(z − z ′ ) (10.52)
which is equivalent to a surface traction p y = 1 in a horizontal plane at elevation z = z ′ .If
the source is in the upper layer at a depth ζ = z 1 − z ′ with complement η = h − ζ = z ′ − z 2
and defining κ = s 2 µ 2 /s 1 µ 1 then
⎧ ⎫ ⎧ ⎫ ⎧ ⎫
⎨ coth ks 1 ζ −1/ sinh ks 1 ζ 0 ⎬ ⎨ ũ y1 ⎬ ⎨ 0 ⎬
ks 1 µ 1 −1/ sinh ks
⎩
1 ζ coth ks 1 ζ + coth ks 1 η −1/ sinh ks 1 η ũ
⎭ ⎩ y
⎭ = 1
⎩ ⎭
0 −1/ sinh ks 1 η coth ks 1 η + κ ũ y2 0
(10.53)
whereas if the source is in the half-space at ζ = z 2 − z ′ and defining again κ = s 2 µ 2 /s 1 µ 1 ,
then the system is
⎧ ⎫ ⎧ ⎫ ⎧ ⎫
⎨ coth ks 1 h −1/ sinh ks 1 h 0 ⎬ ⎨ ũ y1 ⎬ ⎨ 0 ⎬
ks 1 µ 1 −1/ sinh ks
⎩
1 h coth ks 1 h + κ coth ks 2 ζ −κ/sinh ks 2 ζ ũ
⎭ ⎩ y2
⎭ = 0
⎩ ⎭
0 −κ/sinh ks 2 ζ κ(coth ks 2 ζ + 1) ũ y 1
(10.54)
We added tildes to the components to remind us that that the displacements in these
expressions are cast in the frequency–wavenumber domain, i.e., ũ y = ũ y (k,ω). While the
above systems could be solved analytically and the result simplified with the aid of basic
trigonometric identities, in the normal use of the stiffness matrix method – especially for a
system with several layers – this is done numerically for each frequency and wavenumber,
as it is also in the propagator matrix method. After thus finding the response functions,
we obtain the displacements in space–time by carrying out numerically an inverse Fourier
transform over wavenumbers and, if needed, over frequencies, that is,
( ) 1 2 ∫ +∞ ∫ +∞
u y (x, t) =
ũ y e i(ωt−kx) dkdω (10.55)
2π −∞ −∞
and similar expressions for the other components. If the medium has attenuation, the system
has no singularities, so the numerical integrals are feasible. However, the kernels for
layered media are likely to be wavy, and they will exhibit sharp peaks and undulations that