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The reason for this is that in expression (14.3) the motion of the swing was determined

before the push (t < 0), during the push (0 < t < ) and after the push (t > ).

The requirement that the displacement x and the velocity x were continuous then led to

_

the system of equations (14.4) with four unknowns. However, in the end we took the

0 and never used the solution for time 0 < t < . This suggests that this

limit !

method of solution is unnecessarily complicated. This is indeed the case. In this section

an alternative derivation of the Green's function (14.11) is given that is based directly on

the idea that the Green's function G(t ) describes the motion of the oscillator due to a

delta-function force at time :

G(t ) + !2 G(t ) = 1 (t ) : (14.17)

0 m ;

Problem a: For t = the delta function vanishes and the right hand side of this ex-

6

pression is equal to zero. We are looking for the causal Green's function, this is the

solution where the cause (the force) precedes the e ect (the motion of the oscillator).

Show that these conditions imply that for t = the Green's function is given by:

6

(

G(t ) = 0 cos (! (t )) + B sin (! (t )) for t < (14.18)

A for t >

0 0

; ;

where A and B are unknown integration constants.

The integration constants follow from conditions at t = . Since we have two unknown

parameters we need to impose two conditions. The rst conditions is that the motion of

the oscillator is continuous at t = . If this would not be the case the velocity of the

oscillator would be in nite at that moment.

Problem b: Show that the requirement of continuity of the Green's function at t =

implies that A = 0.

The second condition requires more care. We will derive the second condition rst math-

ematically and then explore the physical meaning. The second condition follows by in-

tegrating expression (14.17) over t from " to + " and by taking the limit " 0.

; #

Integrating (14.17) in this way gives:

Z +" Z Z +"

2 +" G(t )dt = 1

G(t )dt + ! (t )dt : (14.19)

0 m ;"

;" ;"

;

Problem c: Show that the right-hand side is equal to 1=m, regardless of the value of ".

Problem d: Show that the absolute value of the middle term is smaller than 2"!0 max (G),

2

where max (G) is the maximum of G over the integration interval. Since the Green's

function is nite this means that the middle term vanishes in the limit " 0. #

_ _

Problem e: Show that the left term inh(14.19) iis equal to G(t = +" ); G(t = " ). ;

t= +"

_

This quantity will be denoted by G(t ) t= ;" .

CHAPTER 14. GREEN'S FUNCTIONS, PRINCIPLES

180

Problem f: Show that in the limit " 0 equation (14.19) gives:

#

h _ it= +" 1

G(t ) t= ;" = m : (14.20)

Problem g: Show that this condition together with the continuity of G implies that the

integration constants in expression (14.18) have the values A = 0 and B = 1=m!0 ,

i.e. that the Green's function is given by:

(

for t <

G(t ) = 0 1 sin (! (t (14.21)

for t >

))

0

m!0 ;

A comparison with (14.11) shows that the Green's function derived in this section is iden-

tical to the Green's function derived in section (14.1). Note that the solution was obtained

here without invoking the motion of the oscillator during the moment of excitation. This

also would have been very di cult because the duration of the excitation (a delta function)

is equal to zero, if it can be de ned at all.

There is however something strange about the derivation in this section. In section

(14.1) the solution was found by requiring that the displacement x and its rst derivative

x were continuous at all times. As used in problem b the rst condition is also met by

_

_

the solution (14.21). However, the derivative G is not continuous at t = .

Problem h: Which of the equations that you derived above states that the rst derivative

is not continuous?

Problem i: G(t ) denotes the displacement of the oscillator. Show that expression

(14.20) states that the velocity of the oscillator is changes discontinuously at t = .

Problem j: Give a physical reason why the velocity of the oscillator was continuous in

the rst part of section (14.1) and why the velocity is discontinuous for the Green's

function derived in this section. Hint: how large is the force needed produce a nite

jump in the velocity of a particle when the force is applied over a time interval of

length zero (the width of the delta-function excitation).

How can we reconcile this result with the solution obtained in section (14.1)?

Problem k: Show that the change in the velocity in the solution x(t) in equation (14.7)

is proportional to F(ti ) , i.e. that

_ ;" 1

x]ttii +" = m F(ti ) (14.22)

This means that the change in the velocity depends on the strength of the force times

the duration of the force. The physical reason for this is that the change in the velocity

depends on the integral of the force over time divided by the mass of the particle.

Problem l: Derive this last statement also directly from Newton's law (F = ma).

14.4. THE GREEN'S FUNCTION FOR A GENERAL PROBLEM 181

When the force is nite and when 0, the jump in the velocity is zero and the velocity

!

is continuous. However, when the force is in nite (as is the case for a delta function), the

jump in the velocity is nonzero and the velocity is discontinuous.

In many applications the Green's function is the solution of a di erential equation

with a delta function as excitation. This implies that some derivative, or combination of

derivatives, of the Green's function are equal to a delta function at the point (or time)

of excitation. This usually has the e ect that the Green's function or its derivative are

not continuous functions. The delta function in the di erential equation usually leads to

a singularity in the Green's function or its derivative.

14.4 The Green's function for a general problem

In this section, the theory of Green's functions are treated in a more abstract fashion.

Every linear di erential equation for a function u with a source term F can symbolically

be written as:

Lu = F : (14.23)

For example in equation (14.1) for the girl on the swing, u is the displacement x(t) while

L is a di erential operator given by

d2 + m!2

L = m dt2 (14.24)

0

where is understood that a di erential operator acts term by term on the function to the

right of the operator.

Problem a: Find the di erential operator L and the source term F for the electrical

eld treated in section (14.2) from the eld equation (4.12).

In the notation used in this section, the Green's function depends on the position vector

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