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!n !2

2

2n ;

When the mode is a vector, one should take the transpose of the mode u (n) (r0 ). Note

the similarity between this expression for the Green's function of a continuous medium

with the Green's function (16.60) for a discrete system. In this sense, the Earth behaves

in the same way as a tri-atomic molecule. For both systems, the dyadic representation of

the Green's function provides a very compact way for accounting for the response of the

system to external forces.

Note that the response is strongest when the frequency ! of the external force is close

to one of the eigen-frequencies !n of the system. This implies for example for the Earth

that the modes with a frequency close to the frequency of the external forcing are most

strongly excited. If we jump up and down with a frequency of 1Hz, we excite the gravest

normal mode of the Earth with a period of about 1 hour only very weakly. In addition, a

mode is most e ectively excited when the inner product of the forcing F(r0 ) in (16.70) is

maximal. This means that a mode is most strongly excited when the spatial distribution

of the force equals the displacement u(n) (r0 ) of the mode.

Problem d: Show that a mode is not excited when the force acts only at one of the nodal

lines of that mode.

As a next step we consider the Green's function in the time domain. This function

follows by applying the Fourier transform (11.42) to the Green's function (16.71).

16.7. NORMAL MODES AND THE GREEN'S FUNCTION 229

Problem e: Show that this gives:

Z

0 t) = 1 X u(n) (r)u (n) (r0 ) 1 e;i!t d! :

G(r r (16.72)

;1 !2 !2

2 n;

n

The integrand is singular at the frequencies ! = !n of the normal modes. These singu-

larities are located on the integration path, as shown in the left panel of gure 16.4. At

the singularity at ! = !n the integrand behaves as 1= (2!R (! !n)). The contribution of

n ;

these singularities is poorly de ned because the integral 1= (! !n ) d! is not de ned.

;

? ?

x x

x x

Figure 16.4: The location of the poles and the integration path in the complex !-plane.

The poles are indicared with a cross. Left panel, the original situation where the poles are

located on the integration path at location !n. Right panel, the location of the poles

when a slight anelastic damping is present.

This situation is comparable to the treatment in section 13.4 of the response of a

particle in syrup to an external forcing. When this particle was subjected to a damping ,

the integrand in the Fourier transform to the time domain had a singularity in the lower

half plane. This gave a causal response as shown in equation (13.35) the response was

only di erent from zero at times later than the time at which the forcing was applied.

This suggests that we can obtain a well-de ned causal response of the Green's function

(16.72) when we introduce a slight damping. This damping breaks the invariance of the

problem for time-reversal, and is responsible for a causal response. At the end of the

calculation we can let the damping parameter go to zero. Damping can be introduced by

giving the eigen-frequencies of the normal modes a small negative imaginary component:

!n !n i , where is a small positive number.

! ;

Problem f: The time-dependence of the oscillation of a normal mode is given by e;i!nt .

Show that with this replacement the modes decay with a decay time that is given

by = 1= .

This last property means that when we ultimately set = 0 that the decay time because

in nite, in other words, the modes are not attenuated in that limit.

With the replacement !n !n i the poles that are associated with the normal

! ;

modes are located in the lower !-plane, this situation is shown in gure 16.4. Now that

the singularities are moved from the integration path the theory of complex integration

can be used to evaluate the resulting integral.

CHAPTER 16. NORMAL MODES

230

Problem g: Use the theory of contour integration as treated in chapter 13 to derive that

the Green's function is in the time domain given by:

8

> for t < 0

0

>

<

G(r r0 t) = > P u(n) (r)u (n) (r0) (16.73)

>

:n sin !nt for t > 0

! n

Hint: use the same steps as in the derivation of the function (13.35) of section 13.4

and let the damping parameter go to zero at the end of the integration.

This result gives a causal response because the Green's function is only nonzero at

times t > 0, which is later than the time t = 0 when the delta function forcing is nonzero.

The total response is given as a sum over all the modes. Each mode leads to a time signal

sin !nt in the modal sum, this is a periodic oscillation with the frequency !n of the mode.

The singularities in the integrand of the Green's function (16.72) at the pole positions

! = !n are thus associated in the time domain with a harmonic oscillation with angular

frequency !n. Note that the Green's function is continuous at the time t = 0 of excitation.

Problem h: Use the Green's function (16.73) to derive that the response of the system

to a force F(r t) is given by:

X 1 (n) Z Z t (n) 0 ;

u (r ) sin !n t t0 F(r0 t0 ) dt0 dV 0:

u(r t) = ! u (r) (16.74)

;1

;

nn

Justify the integration limit in the t0 -integration.

The results of this section imply that the total Green's function of a system is known once

the normal modes are known. The total response can then be obtained by summing the

contribution of each normal mode to the total response. This technique is called normal-

mode summation, it is often use to obtain the low-frequency response of the Earth to an

excitation 19]. However, in the seismological literature one usually treats a source signal

that is given by a step function at t = 0 rather than a delta function because this is a

more accurate description of the slip on a fault during an earthquake 2]. This leads to

a time-dependence (1 cos !nt) rather than the time-dependence sin !n t in the response

;

(16.73) to an delta function excitation.

16.8 Guided waves in a low velocity channel

In this section we will consider a system that strictly speaking does not have normal

modes, but that can support solutions that behave like travelling waves in one direction

and as modes in another direction. The waves in such a system propagate as guided

waves. Consider a system in two dimensions (x and z) where the velocity depends only

on the z-coordinate. We assume that the wave- eld satis es in the frequency domain the

Helmholtz equation (16.1):

2

2u + ! u = 0 : (16.75)

c2 (z)

r

16.8. GUIDED WAVES IN A LOW VELOCITY CHANNEL 231

In this section we consider a simple model of a layer with thickness H that extends from

z = 0 to z = H where the velocity is given by c1 . This layer is embedded in a medium with

a constant velocity c0 . The geometry of the problem is shown in gure 16.5. Since the

system is invariant in the x-direction, the problem can be simpli ed by a Fourier transform

Z1

over the x-coordinate:

U(k z)eikx dk :

u(x z) = (16.76)

;1

Problem a: Show that U(k z) satis es the following ordinary di erential equation:

!

2 2

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