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Problem c: Apply the Fourier transform (11.42) to the 3D Green's function (15.33) with

k replaced by and show that the resulting Green's function is in the time given

;k

by

t+ r

t) = 4 1 r

G3D advanced (r (15.44)

c

;

and that the following function is a solution of the wave equation (15.42):

Z S t + jr;r0j

p(r t) = 41 0

c

0 dV : (15.45)

r

;

jr ; j

Note that in this representation the wave eld is expressed in the source function at later

times. For this reason the Green's function (15.44) is called the advanced Green's function.

The fact that the wave equation has both a retarded and an advanced solution is that the

wave equation (15.42) is invariant for time-reversal, i.e. when one replaces t by the;t

equation does not change. In practice one works most often with the retarded Green's

function, but keep in mind that in some application such as exploration seismology the

advanced Green's functions are crucial. In the remaining part of this section we will focus

exclusively of the retarded Green's functions that represent causal solutions.

In order to obtain the Green's function for two dimensions in the time domain one

should apply a Fourier transform to the solution (15.38). This involves taking the Fourier

transform of a Hankel function, and it is not obvious how this Fourier integral should be

solved (although it can be solved). Here we will follow an alternative route by recognizing

that the Green's function in two dimensions is identical to the solution of the wave equation

in three dimensions when the source is not a point source but a cylinder-source. In other

words, we obtain the 2D Green's function by considering the wave eld in three dimensions

that is generated by a source that is distributed homogeneously along the z-axis. In order

to separate the distance to the origin from the distance to the z-axis the variables r and

are used, see gure (15.2).

Problem d: Show that Z1

G2D ( G3D (r t)dz :

t) = (15.46)

;1

CHAPTER 15. GREEN'S FUNCTIONS, EXAMPLES

204

z-axis

2

ПЃ +z2

r=

r

ПЃ y-axis

x-axis

Figure 15.2: De nition of the variables r and .

p

Problem e: Use the Green's function (15.41) and the relation r = 2 + z 2 to show that

p

+z2

1 Z1 t

2

c

;

p

G2D ( t) = 2 dz : (15.47)

2 + z2

;

0

Note that the integration interval has been changed from (;1 to (0 show

1) 1),

how this can be achieved.

The distance r in three dimensions does not appear in this expression anymore. Without

loss of generality we variable can therefore be replaced by r.

Problem f: The integral (15.47) (with replaced by r) can be solved by introducing

the new integration variable u r2 + z2 instead of the old integration variable z.

p

Show that the integral (15.47) can with this new variable be written as

;

1 Z 1 t u du

G2D (r t) = 2 c (15.48)

;

u 2 r2 p

;

r ;

pay attention to the limits of integration!

Problem g: Use the property (ax) = (x)= to rewrite this integral and evaluate the

jaj

resulting integral separately for t < r=c and t > r=c to show that:

(0 for t < r=c

G2D (r t) = (15.49)

1 1 for t > r=c

;r =c

2 p

;

t2 2 2

15.4. THE WAVE EQUATION IN 1,2,3 DIMENSIONS 205

One dimension

Two dimensions

Three dimensions

t=r/c

Figure 15.3: The Green's function of the wave equation in 1, 2 and 3 dimensios as a

function of time.

This Green's function as well as the Green's function for the three-dimensional case is

shown in gure (15.3). There is a fundamental di erence between the Green's function for

two dimensions and the Green's function (15.41) for three dimensions. In three dimensions

the Green's function is a delta function (t;r=c) modulated by the geometrical spreading

r. This means that the response to a delta function source has the same shape

;1=4

as the input function (t) that excites the wave eld. An impulsive input leads to an

impulsive output with a time delay given by r=c and the solution is only nonzero at the

wave front t = r=c. However, expressions (15.49) shows that an impulsive input in two

dimensions leads to a response that is not impulsive. The response has an in nite duration

p

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