Chow T.L. Mathematical Methods for Physicists: A Concise Introduction
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Parseval's identity for Fourier integrals
We arrived earlier at Parseval's identity for Fourier series. An analogy exists for
Fourier integrals. If g and G are Fourier transforms of f x and Fx
respectively, we can show that
Z
1
ÿ1
f xF*xdx
1
2
Z
1
ÿ1
gG*d; 4:54
where F *x is the complex conjugate of Fx. In particular, if Fxf x and
hence Gg, then we have
Z
1
ÿ1
f x
jj
2
dx
Z
1
ÿ1
g
jj
d: 4:54
Equation (4.53), or the more general Eq. (4.54), is known as the Parseval's iden-
tity for Fourier integrals. Its proof is straightforward:
Z
1
ÿ1
f xF*xdx
Z
1
ÿ1
1
2
p
Z
1
ÿ1
ge
ÿix
d
1
2
p
Z
1
ÿ1
G*
0
e
i
0
x
d
0
dx
Z
1
ÿ1
d
Z
1
ÿ1
d
0
gG*
0
1
2
Z
1
ÿ1
e
ixÿ
0
dx
Z
1
ÿ1
dg
Z
1
ÿ1
d
0
G*
0
0
ÿ
Z
1
ÿ1
gG*d:
Parseval's identity is very useful in understanding the physical interpretation of
the transform function g when the physical signi®cance of f x is known. The
following example will show this.
Example 4.10
Consider the following function, as shown in Fig. 4.21, which might represent the
current in an antenna, or the electric ®eld in a radiated wave, or displacement of a
damped harmonic oscillator:
f t
0 t < 0
e
ÿt=T
sin !
0
tt> 0
:
186
FOURIER SERIES AND INTEGRALS
Its Fourier transform g! is
g!
1
2
p
Z
1
ÿ1
f te
ÿi!t
dt
1
2
p
Z
1
ÿ1
e
ÿt=T
e
ÿi!t
sin !
0
tdt
1
2
2
p
1
!!
0
ÿ i=T
ÿ
1
! ÿ !
0
ÿ i=T
:
If f t is a radiated electric ®eld, the radiated power is proportional to jf tj
2
and the total energy radiated is proportional to
R
1
0
f t
jj
2
dt. This is equal to
R
1
0
g!
jj
2
d! by Parseval's identity. Then jg!j
2
must be the energy radiated
per unit frequency interval.
Parseval's identity can be used to evaluate some de®nite integrals. As an exam-
ple, let us revisit Example 4.8, where the given function is
f x
1 x
jj
< a
0 x
jj
> a
(
and its Fourier transform is
g!
2
r
sin !a
!
:
By Parseval's identity, we have
Z
1
ÿ1
f x
fg
2
dx
Z
1
ÿ1
g!
fg
2
d!:
187
PARSEVAL'S IDENTITY FOR FOURIER INTEGRALS
Figure 4.21. A damped sine wave.
This is equivalent to
Z
a
ÿa
1
2
dx
Z
1
ÿ1
2
sin
2
!a
!
2
d!;
from which we ®nd
Z
1
0
2
sin
2
!a
!
2
d!
a
2
:
The convolution theorem for Fourier transforms
The convolution of the functions f x and Hx, denoted by f H, is de®ned by
f H
Z
1
ÿ1
f uHx ÿ udu: 4:55
If g! and G! are Fourier transforms of f x and Hx respectively, we can
show that
1
2
Z
1
ÿ1
g!G!e
i!x
d!
Z
1
ÿ1
f uHx ÿ udu: 4:56
This is known as the convolution theorem for Fourier transforms. It means that
the Fourier transform of the product g! G!, the left hand side of Eq. (55), is
the convolution of the original function.
The proof is not dicult. We have, by de®nition of the Fourier transform,
g!
1
2
p
Z
1
ÿ1
f xe
ÿi!x
dx; G!
1
2
p
Z
1
ÿ1
Hx
0
e
ÿi!x
0
dx
0
:
Then
g!G!
1
2
Z
1
ÿ1
Z
1
ÿ1
f xHx
0
e
ÿi!xx
0
dxdx
0
: 4:57
Let x x
0
u in the double integral of Eq. (4.57) and we wish to transform from
(x, x
0
)to(x; u). We thus have
dxdx
0
@x; x
0
@x; u
dudx;
where the Jacobian of the transformation is
@x; x
0
@x; u
@x
@x
@x
@u
@x
0
@x
@x
0
@u
þ
þ
þ
þ
þ
þ
þ
þ
þ
þ
þ
þ
þ
þ
þ
þ
þ
þ
10
01
þ
þ
þ
þ
þ
þ
þ
þ
1:
188
FOURIER SERIES AND INTEGRALS
Thus Eq. (4.57) becomes
g!G!
1
2
Z
1
ÿ1
Z
1
ÿ1
f xHu ÿ xe
ÿi!u
dxdu
1
2
Z
1
ÿ1
e
ÿi!u
Z
1
ÿ1
f xHu ÿ xdu
dx
F
Z
1
ÿ1
f xHu ÿ xdu
Ff H
fg
: 4:58
From this we have equivalently
f H F
ÿ1
g!G!fg1=2
Z
1
ÿ1
e
i!x
g!G!;
which is Eq. (4.56).
Equation (4.58) can be rewritten as
Ff
fg
FH
fg
Ff H
fg
g Ff
fg
; G FH
fg
;
which states that the Fourier transform of the convolution of f(x) and H(x) is equal
to the product of the Fourier transforms of f(x) and H(x). This statement is often
taken as the convolution theorem.
The convolution obeys the commutative, associative and distributive laws of
algebra that is, if we have functions f
1
; f
2
; f
3
then
f
1
f
2
f
2
f
1
commutative;
f
1
f
2
f
3
f
1
f
2
f
3
associative;
f
1
f
2
f
3
f
1
f
2
f
1
f
3
distributive:
9
>
>
=
>
>
;
4:59
It is not dicult to prove these relations. For example, to prove the commutative
law, we ®rst have
f
1
f
2
Z
1
ÿ1
f
1
uf
2
x ÿ udu:
Now let x ÿ u v, then
f
1
f
2
Z
1
ÿ1
f
1
uf
2
x ÿ u du
Z
1
ÿ1
f
1
x ÿ vf
2
vdv f
2
f
1
:
Example 4.11
Solve the integral equation yxf x
R
1
ÿ1
yurx ÿ udu, where f x and
rx are given, and the Four ier transforms of yx; f x and rx exist.
189
THE CONVOLUTION THEOREM FOR FOURIER TRANSFORMS
Solution: Let us denote the Fourier transforms of yx; f x and rx by
Y!; F!; and R! respectively. Taking the Fourier transform of both sides
of the given integral equation, we have by the convolution theorem
Y!F!Y!R! or Y!
F!
1 ÿ R!
:
Calculations of Fourier transforms
Fourier transforms can often be used to transform a diÿerential equatio n which is
dicult to solve into a simpler equation that can be solved relatively easy. In
order to use the transform methods to solve ®rst- and second-order diÿerential
equations, the transforms of ®rst- and second-order derivatives are needed. By
taking the Fourier transform with respect to the variable x, we can show that
a F
@u
@x
iF u;
b F
@
2
u
@x
2
ý!
ÿ
2
Fu;
c F
@u
@t
@
@t
Fu:
9
>
>
>
>
>
>
>
>
>
=
>
>
>
>
>
>
>
>
>
;
4:60
Proof: (a) By de®nition we have
F
@u
@x
Z
1
ÿ1
@u
@x
e
ÿix
dx;
where the factor 1=
2
p
has been dropped. Using integration by parts, we obtain
F
@u
@x
Z
1
ÿ1
@u
@x
e
ÿix
dx
ue
ÿix
þ
þ
þ
þ
1
ÿ1
i
Z
1
ÿ1
ue
ÿix
dx
iFu:
(b) Let u @v=@x in (a), then
F
@
2
v
@x
2
ý!
iF
@v
@x
i
2
Fv:
Now if we formally replace v by u we have
F
@
2
u
@x
2
ý!
ÿ
2
Fu;
190
FOURIER SERIES AND INTEGRALS
provided that u and @u=@x ! 0asx !1. In general, we can show that
F
@
n
u
@x
n
i
n
Fu
if u;@u=@x; ...;@
nÿ1
u=@x
nÿ1
!1as x !1.
(c) By de®nition
F
@u
@t
Z
1
ÿ1
@u
@t
e
ÿix
dx
@
@t
Z
1
ÿ1
ue
ÿix
dx
@
@t
Fu:
Example 4.12
Solve the inhomogeneous diÿerential equation
d
2
dx
2
p
d
dx
q
ý!
f xRx; ÿ1x 1;
where p and q are constants.
Solution: We transform both sides
F
d
2
f
dx
2
p
df
dx
qf
()
i
2
piqFfx
fg
FRx
fg
:
If we denote the Fourier transforms of f x and Rx by g and G, respec-
tively,
Ffx
fg
g; FRx
fg
G;
we have
ÿ
2
ip qgG; or gG=ÿ
2
ip q
and hence
f x
1
2
p
Z
1
ÿ1
e
ix
gd
1
2
p
Z
1
ÿ1
e
ix
G
ÿ
2
ip q
d:
We will not gain anything if we do not know how to evaluate this complex
integral. This is not a dicult problem in the theory of functions of comp lex
variables (see Chapter 7).
191
CALCULATIONS OF FOURIER TRANSFORMS
The delta function and the Green's function method
The Green's function method is a very useful technique in the solution of partial
diÿerential equations. It is usually used when boundary conditions, rather than
initial conditions, are speci®ed. To appreciate its usefulness, let us consider the
inhomogeneous diÿerential equation
Lxf xÿf xRx4:61
over a domain D, with L an arbitrary diÿerential ope rator, and a given con-
stant. Suppose we can expand f x and Rx in eigenfunctions u
n
of the operator
LLu
n
n
u
n
:
f x
X
n
c
n
u
n
x; Rx
X
n
d
n
u
n
x:
Substituting these into Eq. (4.61) we obtain
X
n
c
n
n
ÿ u
n
x
X
n
d
n
u
n
x:
Since the eigenfunctions u
n
x are linearly independent, we must have
c
n
n
ÿ d
n
or c
n
d
n
=
n
ÿ :
Moreover,
d
n
Z
D
u
n
*Rxdx:
Now we may write c
n
as
c
n
1
n
ÿ
Z
D
u
n
*Rxdx;
therefore
f x
X
n
u
n
n
ÿ
Z
D
u
n
*x
0
Rx
0
dx
0
:
This expression may be written in the form
f x
Z
D
Gx; x
0
Rx
0
dx
0
; 4:62
where Gx; x
0
is given by
Gx; x
0
X
n
u
n
xu
n
*x
0
n
ÿ
4:63
and is called the Green's function . Some authors prefer to write Gx; x
0
; to
emphasize the dependence of G on as well as on x and x
0
.
192
FOURIER SERIES AND INTEGRALS
What is the diÿerential equation obeyed by Gx; x
0
? Suppose f x
0
in Eq.
(4.62) is taken to be x
0
ÿ x
0
), then we obtain
f x
Z
D
Gx; x
0
x
0
ÿ x
0
dx Gx; x
0
:
Therefore Gx; x
0
is the solution of
LGx; x
0
ÿGx; x
0
x ÿ x
0
; 4:64
subject to the appropriate boundary conditions. Eq. (4.64) shows clearly that the
Green's function is the solution of the problem for a unit point `source'
Rxx ÿ x
0
.
Example 4.13
Find the solution to the diÿerential equation
d
2
u
dx
2
ÿ k
2
u f x4:65
on the interval 0 x l, with u0ul0, for a general function f x.
Solution: We ®rst solve the diÿerential equation which Gx; x
0
obeys:
d
2
Gx; x
0
dx
2
ÿ k
2
Gx; x
0
x ÿ x
0
: 4:66
For x equal to anything but x
0
(that is, for x < x
0
or x > x
0
), x ÿ x
0
0and
we have
d
2
G
<
x; x
0
dx
2
ÿ k
2
G
<
x; x
0
0 x < x
0
;
d
2
G
>
x; x
0
dx
2
ÿ k
2
G
>
x; x
0
0 x > x
0
:
Therefore, for x < x
0
G
<
Ae
kx
Be
ÿkx
:
By the boundary condition u00 we ®nd A B 0, and G
<
reduces to
G
<
Ae
kx
ÿ e
ÿkx
; 4:67a
similarly, for x > x
0
G
>
Ce
kx
De
ÿkx
:
193
THE DELTA FUNCTION AND THE GREEN'S FUNCTION METHOD
By the boundary condition ul0 we ®nd Ce
kl
De
ÿkl
0, and G
>
can be
rewritten as
G
>
C
0
e
kxÿl
ÿ e
ÿkxÿl
; 4:67b
where C
0
Ce
kl
.
How do we determine the constants A and C
0
? First, continuity of G at x x
0
gives
Ae
kx
ÿ e
ÿkx
C
0
e
kxÿl
ÿ e
ÿkxÿl
: 4:68
A second constraint is obtained by integ rating Eq. (4.61) from x
0
ÿ " to x
0
",
where " is in®nitesimal:
Z
x
0
"
x
0
ÿ"
d
2
G
dx
2
ÿ k
2
G
"#
dx
Z
x
0
"
x
0
ÿ"
x ÿ x
0
dx 1: 4:69
But
Z
x
0
"
x
0
ÿ"
k
2
Gdx k
2
G
>
ÿ G
<
0;
where the last step is required by the continuity of G. Accordingly, Eq. (4.64)
reduces to
Z
x
0
"
x
0
ÿ"
d
2
G
dx
2
dx
dG
>
dx
ÿ
dG
<
dx
1: 4:70
Now
dG
<
dx
xx
0
þ
þ
þ
þ
þ
Ake
kx
0
e
ÿkx
0
and
dG
>
dx
xx
0
þ
þ
þ
þ
þ
C
0
ke
kx
0
ÿl
e
ÿkx
0
ÿl
:
Substituting these into Eq. (4.70) yields
C
0
ke
kx
0
ÿl
e
ÿkx
0
ÿl
ÿAke
kx
0
e
ÿkx
0
1: 4:71
We can solve Eqs. (4.68) and (4.71) for the constants A and C
0
. After some
algebraic manipulation, the solution is
A
1
2k
sinh kx
0
ÿ l
sinh kl
; C
0
1
2k
sinh kx
0
sinh kl
194
FOURIER SERIES AND INTEGRALS
and the Green's function is
Gx; x
0
1
k
sinh kx sinh kx
0
ÿ l
sinh kl
; 4:72
which can be combined with f x to obtain ux:
ux
Z
l
0
Gx; x
0
f x
0
dx
0
:
Problems
4.1 (a) Find the period of the function f xcosx=3cosx=4.
(b) Show that, if the function f tcos !
1
t cos !
2
t is periodic with a
period T , then the ratio !
1
=!
2
must be a rational number.
4.2 Show that if f x Pf x, then
Z
aP=2
aÿP=2
f xdx
Z
P=2
ÿP=2
f xdx;
Z
Px
P
f xdx
Z
x
0
f xdx:
4.3 (a) Using the result of Example 4.2, prove that
1 ÿ
1
3
1
5
ÿ
1
7
ÿ
4
:
(b) Using the result of Example 4.3, prove that
1
1 3
ÿ
1
3 5
1
5 7
ÿ
ÿ 2
4
:
4.4 Find the Fourier series which represents the function f xjxj in the inter-
val ÿ x .
4.5 Find the Fourier series which represents the function f xx in the interval
ÿ x .
4.6 Find the Fourier series which represents the function f xx
2
in the inter-
val ÿ x .
4.7 Represent f xx; 0 < x < 2, as: (a) in a half-range sine series, (b) a half-
range cosine series.
4.8 Represent f xsin x,0< x <, as a Fourier cosine series.
4.9 (a) Show that the function f x of period 2 which is equal to x on ÿ1; 1
can be represented by the following Fourier series
ÿ
i
e
ix
ÿ e
ÿix
ÿ
1
2
e
2ix
1
2
e
ÿ2ix
1
3
e
3ix
ÿ
1
3
e
ÿ3ix
:
(b) Write Parseval's identity corresponding to the Fourier series of (a).
(c) Determine from (b) the sum S of the series 1
1
4
1
9
P
1
n1
1=n
2
.
195
PROBLEMS