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2 F(r) x2 @ 2 F x2 @F

@ = r2 @r2 + 1 ; r3 @r

j j (15.25)

@x2 r

j

P

and that the Laplacian N @ 2 F=@x2 is given by

j=1 j

@ 2 F + N 1 @F = 1 @ rN;1 @F :

2

r F(r) = (15.26)

;

@r2 r @r rN;1 @r @r

Using this expression the di erential equation for the Green's function in N-dimension is

given by

1 @ rN;1 @G + k2 G(r !) = (r) : (15.27)

rN;1 @r @r

This di erential equation is not di cult to solve for 1, 2 or 3 space dimensions for locations

away from the source (r = 0). However, we need to consider carefully how the source (r)

6

should be coupled to the solution of the di erential equation. For the case of one dimension

this can be achieved using the theory of section (14.3). The derivation of that section needs

to be generalized to more space dimensions.

This can be achieved by integrating (15.24) over a sphere with radius R centered at

the source and letting the radius go to zero.

15.3. THE HELMHOLTZ EQUATION IN 1,2,3 DIMENSIONS 199

Problem d: Integrate (15.24) over this volume, use Gauss' law and let the radius R go

to zero to show that the Green's function satis es

I @G

@r dS = 1 (15.28)

SR

where the surface integral is over a sphere SR with radius R in the limit R 0. Show

#

that this can also be written as

lim Sr @G = 1 (15.29)

r#0 @r

where Sr is the surface of a sphere in N dimensions with radius r.

Note that the surface of the sphere in general goes to zero as r 0 (except in one dimen-

#

sion), this implies that @G=@r must be in nite in the limit r 0 in more than one space

#

dimension.

The di erential equation (15.27) is a second order di erential equation. Such an equa-

tion must be supplemented with two boundary conditions. The rst boundary condition

is given by (15.29), this condition speci es how the solution is coupled to the source at

r =0. The second boundary condition that we will use re ects the fact that the waves

generated by the source will move away from the source. The solutions that we will nd

will behave for large distance as exp( ikr), but it is not clear whether we should use the

upper sign (+) or the lower sign (;).

Problem e: Use the Fourier transform (11.42) and the relation k = !=c to show;that the

r.

integrand in the Fourier transform to the time domain is proportional to exp c

;i!(t

Show that the waves only move away from the source for the upper sign. This means

that the second boundary condition dictates that the solution behave in the limit

r as exp(+ikr).

!1

The derivative of function exp(+ikr) is given by ik exp(+ikr), i.e. the derivative is ik times

the original function. When the Green's function behaves for large r as exp(+ikr), then

the derivative of the Green's must satisfy the same relation as the derivative of exp(+ikr).

This means that the Green's function satis es for large distance r:

@G = ikG : (15.30)

@r

This relation speci es that the energy radiates away from the source. For this reason

expression (15.30) is called the radiation boundary condition.

Now we are at the point where we can actually construct the solution for each dimen-

sion. Let us rst determine the solution in one space dimension.

Problem f: Show that for one dimension (N = 1) the di erential equation (15.27) has

away from the source the general form G = C exp( ikr), where r is the distance to

Use the result of problem e to show that the plus sign should

the origin: r = jxj.

be used in the exponent and equation (15.28) to derive that the constant C is given

by C = (Hint, what is the surface of a one-dimensional \volume"?) Show

;i=2k.

that this implies that the Green's function in one dimension is given by

G1D (x) = 2k eikjxj : (15.31)

;i

CHAPTER 15. GREEN'S FUNCTIONS, EXAMPLES

200

Before we go to two dimensions we will rst solve the Green's function in three dimensions.

Problem g: Make for three dimensions (N = 3) the substitution G(r) = f(r)=r and

show that (15.27) implies that away from the source the function f(r) satis es

@ 2 f + k2 f = 0 : (15.32)

@r2

This equation has the solution C exp( ikr). According to problem e the upper

sign should be used and the Green's function is given by G(r) = Ceikr =r. Show

that the condition (15.29) dictates that C = , so that in three dimensions the

;1=4

Green's function is given by:

eikr :

G3D (r) = (15.33)

;1

4r

The problem is actually most di cult in two dimensions because in that case the

Green's function cannot be expressed in the simplest elementary functions.

Problem h: Show that in two dimensions (N = 2) the di erential equation of the Green's

function is away from the source given by

@ 2 G + 1 @G + k2 G(r) = 0 r=0: (15.34)

@r2 r @r 6

Problem i: This equation cannot be solved in terms of elementary functions. However

there is a close relation between equation (15.34) and the Bessel equation

d2 F + 1 dF + (1 m2 )F = 0 : (15.35)

dx2 x dx x2

;

Show that the G(kr) satis es the Bessel equation for order m = 0.

This implies that the Green's function is given by the solution of the zeroth-order Bessel

equation with argument kr. The Bessel equation is a second order di erential equation,

there are therefore two independent solutions. The solution that is nite everywhere is

denoted by Jm (x), it is the called the regular Bessel function. The second solution is

singular at the point x = 0 and is called the Neumann function denoted by Nm (x). The

Green's function obviously is a linear combination of J0 (kr) and N0 (kr). In order to

determine how this linear combination is constructed it is crucial to consider the behavior

of these functions at the source (i.e. for x = 0) and at in nity (i.e. for x 1). The

required asymptotic behavior can be found in textbooks such as Butkov 14] and Arfken 3]

and is summarized in table (15.1).

Problem j: Show that neither J0 (kr) nor N0 (kr) behave for large values of r as exp (+ikr).

Show that the linear combination J0 (kr) + iN0 (kr) does behave as exp (+ikr).

The Green's function thus is a linear combination of the regular Bessel function and the

Neumann function. This particular combination is called the rst Hankel function of

15.3. THE HELMHOLTZ EQUATION IN 1,2,3 DIMENSIONS 201

J0 (x) N0 (x)

1 1 x2 + O(x4 ) 2 ln (x) + O(1)

x 0 4

! ;

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