Kelly J.J. Graduate Mathematical Physics, With MATHEMATICA Supplements
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136 5 Integral Transforms
1 0 1 2 3
ΩΩ
0
0
5
10
15
20
f
Ω
0
T10
Ω
0
T4
Figure 5.2. Fourier transform of chopped sinusoid:
Ωt
<T t T
.
5.2.7.3 Step Function
Often it is useful to idealize a quantity, such as a force or voltage, that has a rapid onset
followed by a sustained level using a step function, but the corresponding Fourier trans-
form requires special handling as a generalized function because the step function is not
absolutely integrable. First we consider a rectangular pulse of finite duration, for which
f t<t<t Τ
˜
f Ω
ΩΤ
1
Ω
(5.70)
In the limit that Τ!, the contribution of
ΩΤ
varies rapidly in phase and would tend to
average to zero. This can be done in a more controlled fashion using a convergence factor,
gt lim
!0
<t
t
(5.71)
such that
˜gΩ lim
!0
0
t ExpΩ t
lim
!0
ExpΩt Expt
Ω
0
lim
!0
Ω
(5.72)
where we leave the limit unevaluated because we recognize that this result must be treated
as a generalized function. Hence, we find that the Fourier transform of <t is given by / Ω,
but this is not a rigorous proof and we must verify that the inverse Fourier transform does
provide the desired result. However, because the pole at the origin is on the integration path
5.2 Fourier Transform 137
Re#Z'
Im#Z'
t0
Re#Z'
Im#Z'
t!0
Figure 5.3. Contours used in Eq. (5.74) for Fourier representation of step function.
for the inverse transform, we must evaluate its contribution carefully. Thus, the inversion
integral
lim
!0
2Π
Ωt
Ω
Ω
L
M
M
N
M
M
O
0 t<0
1 t>0
(5.73)
with a pole in the lower half-plane just below the real axis for ! 0
, where closure is
made either up or down according to the sign of t, does indeed produce a step function
as sketched in Fig. 5.3. Therefore, the Fourier transform of the Heaviside step function
becomes
<t lim
!0
1
2Π
Ωt
Ω
Ω < lim
!0
Ω
Ω
(5.74)
where the notation Ω
implies addition to Ω of an infinitesimal imaginary part, , that has
the effect of shifting the pole slightly below the real axis with the limit ! 0
taken later.
Notice that this infinitesimal imaginary contribution is similar to the attenuation property
of Fourier transforms, such that represents the damping coefficient that suppresses infi-
nite times as if we had applied a convergence factor, Expt, to the step function.
Alternatively, consider the principal-value integral
2Π
Ωt
Ω
Ω
1
2
1 2<t
L
M
M
N
M
M
O
1
2
t<0
1
2
t>0
(5.75)
using the contours sketched in Fig. 5.4. Solving for
<t
1
2
2Π
Ωt
Ω
Ω (5.76)
and evaluating the Fourier transform of both sides, we obtain
Ωt
<tt (5.77)
1
2
Ωt
t
t
Ωt
2Π
Ω
'
t
Ω
'
Ω
'
(5.78)
138 5 Integral Transforms
Re#Z'
Im#Z'
t0
Re#Z'
Im#Z'
t!0
Figure 5.4. Contours used in Eq. (5.77) for Fourier representation of step function.
where we assume that the time integration can be performed first. This result can be
expressed in a compact operator representation as
<Π∆Ω
Ω
(5.79)
in which we understand that the delta function applies under a Fourier integral and that
the principal value is to be taken for the singularity at the origin. This representation is
equivalent to the former, except that now we imagine that the limit ! 0
is taken first
and that the pole pushes the contour upward, making a small semicircular indentation in
the portion of the contour that is on the real axis. Consequently, we interpret the “bump in
the road” as
< lim
!0
Ω
Π∆Ω
Ω
(5.80)
5.2.8 Fourier Transform of Derivatives
Many applications of Fourier transforms exploit the simple transformation properties of
derivatives to transform a differential equation into an algebraic equation for the Fourier
transform. Assuming that the Fourier transforms exist for f t and its derivatives, the sim-
plest method for deducing the Fourier transform of derivatives is to differentiate under the
integral
f t
Ω
2Π
Ωt
˜
f Ω f
'
t
Ω
2Π
Ωt
Ω
˜
f Ω (5.81)
and thus to identify
f
'
t
Ω
f t
(5.82)
Therefore, each derivative in a differential equation can be replaced by a power of (Ω)
according to
(
(t
BΩ (5.83)
(
n
(t
n
BΩ
n
(5.84)
5.3 Green Functions via Fourier Transform 139
5.2.9 Summary
A brief table of Fourier transforms and their properties is given in Table 5.2. The para-
meters Τ, Ω
0
, Γ, Σ, and Λ are real. We also assume that the required derivatives exist and
integrals converge. Be careful in applying theorems for derivatives and other operations to
functions with discontinuities.
Table 5.2. Brief table of Fourier transforms.
Type f t
˜
f Ω
definition f t
Ω
2Π
Ωt
˜
f Ω
˜
f Ω
t
Ωt
f t
step function <t lim
!0
Ω
Π∆Ω
Ω
delta function ∆t Τ
ΩΤ
chopped sinusoid
Ω
0
t
<Τ t 2
SinΩΩ
0
Τ
ΩΩ
0
shifting f t Τ
ΩΤ
˜
f Ω
attenuation
Γt
f t
˜
f Ω Γ
dilatation f ΛtΛ
1
˜
f Ω/ Λ
convolution f t
Τgt ΤhΤ
˜
f Ω ˜gΩ
˜
hΩ
derivatives f
n
tΩ
n
˜
f Ω
multiplication by powers tft
˜
f
'
Ω
Gaussian Exp
t
2
2Σ
2
2ΠΣ
2
ExpΩ
2
Σ
2
/ 2
exponential
Γt
Γt
<t
2Π∆Ω Γ
ΩΓ
trigonometric CosΩ
0
t
SinΩ
0
t
Π∆Ω Ω
0
∆ΩΩ
0
Π∆Ω Ω
0
∆ΩΩ
0
5.3 Green Functions via Fourier Transform
The Fourier transform can be very useful in analyzing the response of a linear system to
external stimuli. In this section we illustrate this technique using a few relatively simple
examples.
5.3.1 Example: Green Function for One-Dimensional Diffusion
Suppose that Ψx, t represents the temperature at position x and time t relative to the
background temperature for an effectively infinite homogeneous system. If the initial tem-
perature Ψx, 0 f x is elevated, heat will diffuse to distant regions according to the
140 5 Integral Transforms
heat diffusion equation
(
2
Ψ
(x
2
1
Κ
(Ψ
(t
(5.85)
where the diffusion constant Κ is proportional to the thermal conductivity of the material
and inversely proportional to its heat capacity per unit volume. Thus, we expect that the
temperature for neighboring regions will initially rise as heat passes through them but will
eventually return to ambient levels as the energy spreads out over an infinite volume. One
method for determining the evolution of temperature throughout this sample is to employ
a spatial Fourier transform to write
Ψx, t
k
2Π
kx
˜
Ψk,t (5.86)
˜
Ψk,t
x
kx
Ψx, t (5.87)
such that the differential equation becomes
k
2
˜
Ψk,t
1
Κ
(
˜
Ψk,t
(t
˜
Ψk,t
˜
Ψk, 0 ExpΚxk
2
(5.88)
The initial condition requires
˜
Ψk, 0
˜
f k
x
kx
f x (5.89)
The inverse Fourier transform becomes
Ψx, t
k
2Π
kx
˜
f k ExpΚtk
2
k
2Π
x
'
f x
'
Exp
Κtk
2
kx x
'
(5.90)
The integral with respect to k can be evaluated by completing the square in the exponential,
whereby
Ψx, t4ΠΚt
1/ 2
x
'
f x
'
Exp
x x
'
2
4Κt
(5.91)
takes the form of a convolution integral.
Suppose that the region of elevated temperature is initially restricted to a plane, such
that
f xΨ
0
∆xΨx, t
Ψ
0
4ΠΚt
1/ 2
Exp
x
2
4Κt
(5.92)
describes both the time and spatial temperature variation produced by a localized distur-
bance. This special solution is an example of a Green function representing the response
5.3 Green Functions via Fourier Transform 141
of a linear system to a localized perturbation. The Green function for a one-dimensional
diffusion equation with a localized initial condition satisfies the equation
(
2
(x
2
1
Κ
(
(t
Gx x
'
,t∆x x
'
Gx x
'
,t4ΠΚt
1/ 2
Exp
x x
'
2
4Κt
(5.93)
Notice that this Green function is a nascent delta function of Gaussian form with width
parameter Σ
2Κt, such that
lim
t!0
Gx x
'
,t∆x x
'
(5.94)
Solutions for more general initial conditions are then obtained using the convolution
Ψx, 0 f xΨx, t
Gx x
'
,t f x
'
x
'
(5.95)
Therefore, the Green function describes the essential dynamics of the problem and reduces
problems with specialized initial conditions to an integral which can be evaluated numeri-
cally, if not analytically. The Green function for this problem is sketched in Fig. 5.5. The
delta function at x 0 must be cut off for plotting. It is clear from the dimensions of the
original differential equation that the spatial scale must be proportional to
Κt. Thus, the
width of the Gaussian temperature profile spreads at a rate proportional to t
1/ 2
as the dis-
turbance propagates. The Green function is often described as a propagator – it governs
the propagation of a disturbance (cause) at one location or time to a different location or
later time (effect). The net effect of a distributed disturbance is obtained from the folding
or convolution of the propagator with the profile of the disturbance.
5.3.2 Example: Three-Dimensional Green Function for Diffusion Equations
In three dimensions the initial-value problem for diffusion within an infinite uniform
medium takes the form
1
Κ
(Ψ
r,t
(t
+
2
Ψ
r,t , Ψ
r, 0Ψ
0
r (5.96)
where Ψ
0
r is the initial spatial distribution and where we assume that Κ is a positive
constant. We assume that the initial distribution is localized and at least piecewise contin-
uous so that its Fourier transform exists in all three dimensions. Using a three-dimensional
Fourier transform
Ψ
r,t
˜
Ψ
k,t Exp
k ,
r
3
k
2Π
3
(5.97)
˜
Ψ
k,t
Ψ
r,t Exp
k ,
r
3
r (5.98)
142 5 Integral Transforms
Nt
x
G#x,t'
Nt
Figure 5.5. Dissipation of temperature disturbance (arb. units).
we find
1
Κ
(
(t
+
2
Ψ0
1
Κ
(
(t
k
2
˜
Ψ
k,t Exp
k ,
r
3
k
2Π
3
0 (5.99)
If this equation is to be satisfied for any
r, we must require the integrand to vanish. Hence,
we obtain
1
Κ
(
(t
k
2
˜
Ψ
k,t0
˜
Ψ
k,t
˜
Ψ
0
k ExpΚk
2
t (5.100)
where
˜
Ψ
0
k
Ψ
0
r Exp
k ,
r
3
r (5.101)
is the initial spectral distribution. To obtain the Green function, we substitute
Ψ
0
r∆
r
r
'
˜
Ψ
0
kExp
k ,
r
'
(5.102)
such that
G
r
r
'
,t
ExpΚk
2
t Exp
k ,
r
r
'
3
k
2Π
3
(5.103)
can be factored according to
G
r
r
'
,t
3
i1
ExpΚk
2
i
t Expk
i
,r
i
r
'
i
k
i
2Π
(5.104)
5.3 Green Functions via Fourier Transform 143
where the index i runs over the three Cartesian components. Each factor is a one-dimen-
sional Green function for the diffusion equation; hence, using our previous result, we find
that
G
r
r
'
,t
3
i1
4ΠΚt
1/ 2
Exp
r
i
r
'
i
2
4Κt
(5.105)
depends upon spatial coordinates through
r
r
'
. Therefore, we finally obtain
G
r
r
'
,t4ΠΚt
3/ 2
Exp
r
r
'
2
4Κt
(5.106)
such that
Ψ
r,t
G
r
r
'
,tΨ
0
r
'
3
r
'
(5.107)
is the general solution for the initial-value problem for the diffusion equation in a uniform
medium.
5.3.3 Example: Green Function for Damped Oscillator
The Fourier transform of the differential equation
m
(
2
(t
2
2mΓ
(
(t
k
xtmft (5.108)
for a driven harmonic oscillator with linear damping becomes an algebraic equation
mΩ
2
2ΩmΓk
˜xΩ m
˜
f Ω (5.109)
where
˜
f Ω is the Fourier transform of the driving force (per unit mass). Thus, we imme-
diately obtain a particular solution of the form
˜xΩ
˜
f Ω
Ω
2
0
Ω
2
2ΩΓ
(5.110)
where Ω
0
k/m is the natural frequency for free oscillation. A complete solution can
then be expressed in the form
xtx
0
t
Ω
2Π
Ωt
˜
f Ω
Ω
2
0
Ω
2
2ΩΓ
(5.111)
where x
0
t is a solution to the homogeneous equation
(
2
(t
2
2Γ
(
(t
Ω
2
0
x
0
t0 (5.112)
designed to satisfy the appropriate boundary conditions. Often one can assume that x
0
0
if f t is of finite duration and occurs after any earlier motions have decayed away due to
the damping term.
144 5 Integral Transforms
Alternatively, we recently demonstrated that a general solution to this equation can be
expressed as
xtx
0
t
Gt t
'
f t
'
t
'
(5.113)
where the Green function satisfies
(
2
(t
2
2Γ
(
(t
Ω
2
0
Gt∆t (5.114)
with
t<0 Gt0 ,
(Gt
(t
0 (5.115)
We can demonstrate equivalence between these representations by constructing the spec-
tral (frequency) representation of the Green function directly. Assuming that
f t∆t
˜
f Ω 1 (5.116)
is a unit impulse at time t
'
and that the mass is initially at rest, we obtain
˜
GΩ
1
Ω
2
0
Ω
2
2ΩΓ
(5.117)
such that
Gt
Ω
2Π
Ωt
Ω
2
0
Ω
2
2ΩΓ
(5.118)
is the corresponding temporal representation. Thus, we can use either of the representa-
tions
xtx
0
t
1
2Π
Ωt
˜
GΩ
˜
f Ω Ω (5.119)
xtx
0
t
Gt t
'
f t
'
t
'
(5.120)
Just to belabor the point, we substitute the spectral representation of Gt into the second
equation
xtx
0
t
Ω
2Π
Ωtt
'
Ω
2
0
Ω
2
2ΩΓ
f t
'
t
'
(5.121)
and perform the integration over t
'
to obtain
˜
f Ω
t
'
Ωt
'
f t
'
xtx
0
t
Ω
2Π
Ωt
Ω
2
0
Ω
2
2ΩΓ
˜
f Ω (5.122)
as before. Alternatively, we could have substituted
˜
f Ω into the first version and per-
formed the integration over Ω to obtain the second. (Try it!)
5.3 Green Functions via Fourier Transform 145
To complete the analysis, we must construct explicit expressions for Gt by inverting
the Fourier transform. First, we consider the underdamped case for which the integrand
has two poles
Γ < Ω
0
Ω
ΓΩ
R
with Ω
R
Ω
2
0
Γ
2
(5.123)
symmetrically placed about the imaginary axis in the lower half of the complex Ω-plane.
In order to keep the exponential factor finite, we must close the contour with great semi-
circles in the upper half-plane for t<t
'
or the lower half-plane for t>t
'
. Thus, we find
Gt,t
'
0fort<t
'
because there are no poles in the upper half-plane while for t>t
'
t>t
'
Gt,t
'
1
2Π
C
Ωt
Ω
2
0
Ω
2
2ΩΓ
Ω
R
Ω
tt
'
R
Ω
tt
'
(5.124)
where the negative sign accounts for the clockwise contour and where the residues are
Γ < Ω
0
R
ExpΩ
t t
'
2Ω
R
(5.125)
such that
Γ < Ω
0
Gt,t
'
Ω
1
R
ExpΓt t
'
SinΩ
R
t t
'
<t t
'
(5.126)
describes damped oscillations following a sharp blow. If the damping is small the oscil-
lation frequency is near the natural frequency and the oscillations persist for a long time
when the poles are near the real axis. Next, we consider the overdamped case
Γ > Ω
0
Ω
ΓΚ with Κ
Γ
2
Ω
2
0
(5.127)
where both poles are found on the negative imaginary axis with residues
Γ > Ω
0
R
ExpΓ Κt t
'
2Κ
(5.128)
such that
Γ > Ω
0
Gt,t
'
Κ
1
ExpΓt t
'
SinhΚt t
'
<t t
'
(5.129)
describes the damping of the initial velocity. (Note that Γ > Κ.) Finally, for the critically
damped case the two poles coalesce into a single double pole with residue
ΓΩ
0
R t t
'
ExpΓt t
'
(5.130)
such that
ΓΩ
0
Gt,t
'
t t
'
ExpΓt t
'
<t t
'
(5.131)
These results are summarized in Table 5.3. Notice that the underdamped and over-
damped solutions are related by the replacement Ω
R
!Κas Γ increases and that the crit-
ically damped solution is related to either of the other two by the limit Ω
R
! 0orΚ!0.