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F(t i )

âˆ†

ti

t t

Figure 14.2: A continous function (left) and an approximation to this function that is

constant within nite intervals (right).

R

P

the summation over ti can be replaced by an integration: ti ( ) ( ) d . The

!

integration variable plays the role of the summation variable ti and the time-interval

is replaced d . The response of the oscillator to a continuous force F(t) is then given by

Zt 1

x(t) = )) F( )d :

sin (!0 (t (14.9)

;1 m!0

;

< t because in the summation (14.8)

The integration is only carried out over times

extends only over the times ti < t.

14.2. YOU HAVE SEEN GREEN'S FUNCTIONS BEFORE! 177

With a slight change in notation this result can be written as:

Z1

x(t) = G(t )F ( )d (14.10)

;1

(

with

for t <

0

G(t ) = (14.11)

1 for t >

m!0 sin (!0 (t ; ))

The function G(t ) in expression (14.11) is called the Green's function of the harmonic

oscillator. Note that equation (14.10) is very similar to the response of a linear lter that

was derived in equation (11.55) of section (11.7). This is not surprising, in both examples

the response of a linear system to impulsive input was determined, it will be no surprise

that the results are identical. In fact, the Green's function is de ned as the response of

a linear system to a delta-function input. Although Green's functions often are presented

in a rather abstract way, one should remember that:

The Green's function of a system is nothing but the impulse response of a

system, i.e. it is the response of the system to a delta-function excitation.

14.2 You have seen Green's functions before!

Although the concept of a Green's function may appear to be new to you, you have already

seen ample examples of Green's function, although the term \Green's function" might not

have been used in that context. One example is the electric eld generated by a point

charge q in the origin that was used in section (4.1):

r

E(r) = 4 q^r2 (4:2) again

"0

Since this is the electric eld generated by a delta-function charge in the origin, this eld

is very closely related to the Green's function for this problem. The eld equation (4.12)

of the electric eld is invariant for translations in space. This is a complex way of saying

that the electric eld depends only on the relative position between the point charge and

the point of observation.

Problem a: Show that this implies that the electric eld at location r due to a point

charge at location r0 is given by:

q (r r0 )

E(r) = 4 " (14.12)

;

r0 3

0 jr ; j

Now suppose that we don't have a single point charge, but a system of point charges qi

at locations ri . Since the eld equation is linear, the electric eld generated by a sum of

point charges is the sum of the elds generated by each point charge:

X

E(r) = 4 qi" (r rri)3 (14.13)

;

0 i

i jr ; j

CHAPTER 14. GREEN'S FUNCTIONS, PRINCIPLES

178

Problem b: To which expression of the previous section does this equation correspond?

Just as in the previous sections we now make the transition from a nite number of

discrete inputs (either pushes of the swing or point charges) to an input function that is

a continuous function (either the applied force to the oscillator as a function of time or a

continuous electric charge). Let the electric charge per unit volume be denoted by (r),

this means that the electric charge in a volume dV is given by (r)dV .

Problem c: Replacing the sum in (14.13) by an integration over volume and using that

the appropriate charge for each volume element dV show that the electric eld for a

continuous charge distribution is given by:

ZZZ (r0 ) (r r0 )

0

E(r) = 3 dV (14.14)

;

r0

4 "0 jr ; j

where the volume integration is over r0 .

Problem d: Show that this implies that the electric eld can be written as

ZZZ

G(r r0 ) (r0 )dV 0

E(r) = (14.15)

with the Green's function given by

0 ) = 1 (r r0)

G(r r 4 " (14.16)

;

r0 3

0 jr ; j

Note that this Green's function has the same form as the electric eld for a point charge

shown in (14.13).

Problem e: Show that the Green's function is only a function of the relative distance

r r0 . ExplainRRR the integral (14.15) can be seen as a three-dimensional convo-

why

lution: E(r) = G(r r0 ) (r0 )dV 0 .

;

;

The main purpose of this section was not to show you that you had seen an example of

a Green's function before. Instead, it provides an example that the Green's function is

not necessarily a function of time and that the Green's function is not necessarily a scalar

function the Green's function (14.16) depends only on the position and not on time and

it describes a vector eld rather than a scalar. The most important thing to remember is

that the Green's function is the impulse response of a linear system.

Problem f: You have seen another Green's function before if you have worked through

section (13.4) where the response of a particle in syrup was treated. Find the Green's

function in that section and spot the equivalent expressions of the equations (14.10)

and (14.11) of this section

14.3. THE GREEN'S FUNCTION AS IMPULSE RESPONSE 179

14.3 The Green's function as impulse response

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