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q2 ;3=2 ) q 2x sin ;x

; ;3=2 )

x cos x + O(x 4 + O(x

1 4

x ; ;

Table 15.1: Leading asymptotic behaviour of the Bessel function and Neumann function

of order zero.

(1)

degree zero and is denoted by H0 (kr). In general the Hankel functions are simply linear

combinations of the Bessel function and the Neumann function:

(1)

Hm (x) Jm (x) + iNm (x) (15.36)

(2)

Hm (x) Jm (x) iNm (x)

;

(1) (kr) behaves for large values of r as exp(+i (kr);i =4)=q kr

Problem k: Show that H0

(2) (kr) behaves as exp(;i (kr) ; i =4)=q kr. Use this to

2

and that in this limit H0 2

argue that the Green's function is given by

(1)

G(r) = CH0 (kr) (15.37)

where the constant C still needs to be determined.

Problem l: This constant follows from the requirement (15.29) at the source. Use (15.36)

and the asymptotic value of the Bessel function and the Neumann function given

in table (15.1) to derive the asymptotic behavior of the Green's function near the

source and use this to show that C = ;i=4.

This result implies that in two dimensions the Green's function of the Helmholtz equation

is given by

(1)

G2D (r) = 4 H0 (kr) : (15.38)

;i

Summarizing these results and reverting to the more general case of a source at location

r0 it follows that the Green's functions of the Helmholtz equation is in one, two and three

dimensions given by:

G1D (x x0 ) = ;i eikjx;x0j

2k

2D (r r ) = ;i H (1) (k r0

G (15.39)

0 0

4 ikjr;r j jr ; j)

G3D (r r0 ) = ;1 e jr;r0 j

0

4

Note that in two and three dimensions the Green's function is singular at the source

r0 .

CHAPTER 15. GREEN'S FUNCTIONS, EXAMPLES

202

Problem m: Show that these singularity is integrable, i.e. show that when the Green's

function is integrated over a sphere with nite radius around the source the result

is nite.

There is a physical reason why the Green's function in two and three dimensions has an

integrable singularity. Suppose one has a source that is not a point source but that the

source is constant within a sphere R radius R centered around the origin. The response

with

p to this source is given by p(r) = r0 <R G(r r0)dV 0 where the integration over the variable

r0 is over a sphere with radius R. It follows from this expression that the response in the

origin is given by Z

p(r =0) = 0 G(r =0 r0 )dV 0 : (15.40)

r <R

Since the excitation of this eld is nite everywhere, the response p(r = 0) should be nite

as well. This implies that the integral (15.40) should be nite as well, which is a di erent

way of stating that the singularity of the Green's function must be integrable.

15.4 The wave equation in 1,2,3 dimensions

In this section we will consider the Green's function for the wave equation in 1,2 and 3

dimensions. This means that we consider solutions to the wave equation with an impulsive

source at location r0 at time t0 :

1 @ 2 G(r t r0 t0 ) = (r r ) (t t )

t r0 t0 ) c2

2

r G(r (15:22) again:

0 0

@t2

; ; ;

It was shown in the previous section that this Green's function depends only on the

r0 and the relative time t t0. For the case of a source in the origin

relative distancejr ; j ;

(r0 = 0) acting at time zero (t0 = 0) the time domain solution follows by applying a

Fourier transform to the solution G(r !) of the previous section. This Fourier transform

is simplest in three dimensions, hence we will start with this case.

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

use the relation k = !=c and the properties of the delta function to show that the

Green's function is in the time given by

G3D (r t) = 4 1 r t r : (15.41)

c

; ;

Problem b: Now consider the wave equation with a general source term S(r t):

1 @ 2 p(r t0 ) = S(r t) :

2

r p(r t) c2 @t2 (15.42)

;

Use the Green's function (15.41) to show that a solution of this equation is given by

Z S t jr;r0j

p(r t) = 41 0

c ;

0 dV : (15.43)

r

;

jr ; j

15.4. THE WAVE EQUATION IN 1,2,3 DIMENSIONS 203

r0 is always positive, the response p(r t) depends only on the source

Note that since jr ; j

function at earlier times. The solution therefore has a causal behavior and the Green's

function (15.41) is called the retarded Green's function. However, in several applications

one does not want to use a Green's function that depends on excitation on earlier times.

An example is re ection seismology. In that case one records the wave eld at the surface,

and from these observations one wants to reconstruct the wave eld at earlier times while it

was being re ected o layers inside the earth. (See the treatment in section 6.3 and paper

of Schneider 53].) A Green's function with waves that propagate towards the source and

are then annihilated by the source can be obtained by replacing the radiation condition

(15.30) by @G=@r = The only di erence is the minus sign in the right hand side,

;ikG.

this is equivalent to replacing k by ;k.

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