# I. FOURIER SERIES, FOURIER TRANSFORM

Document Sample

```					             Computational Methods and advanced Statistics Tools

I. FOURIER SERIES, FOURIER TRANSFORM

1. Fourier series

A function f is said to be periodic if it is deﬁned for all real x and if
there is a positive number T such that
(1)                            f (x + T ) = f (x), ∀x
According to this deﬁnition, a constant function is periodic with arbi-
trary period. Trigonometric functions, such as sin x, sin 2x, ..., sin nx,
or cos x, ..., cos nx, are periodic with period 2π. If T is a period for f
then f (x + nT ) = f (x), ∀n ∈ N, so that f is periodic with periods
2T, 3T, ..., too. A trigonometric polynomial is a ﬁnite sum of the form
k
a◦
(2)                         +     (an cos nx + bn sin nx)
2    n=1

Any trigonometric polynomial is periodic with period 2π.
Let us assume that f (x) is a periodic function with period 2π that can
be represented by a trigonometric series,
∞
a◦
(3)               f (x) =         +     (an cos nx + bn sin nx)
2    n=1

that is, we assume that this series converges and has f (x) as its sum.
Given such a function f (x), we want to determine the coeﬃcients an
and bn of the corresponding series (3). We ﬁrst determine a◦ . Inte-
grating on both sides of (3) from −π to π, and assuming that term-by
term integration of the series is allowed 1, we have
π                        π          ∞             π                       π
a◦
f (x) dx =               dx +         an           cos nx dx + bn         sin nx dx .
−π                 2     −π          n=1           −π                     −π

1This   is justiﬁed, for example, in the case of uniform convergence:
n
∀ǫ > 0, ∃¯ (ǫ) > 0 : ∀x ∈ R, |
n                                                        ¯
gk (x) − f (x)| < ǫ, ∀n > n(ǫ)
k=1

It is not guaranteed by weaker assumptions like, for example, point convergence:
n
∀x ∈ R, ∀ǫ > 0, ∃¯ (ǫ, x) > 0 : |
n                                                  ¯
gk (x) − f (x)| < ǫ, ∀n > n(ǫ, x)
k=1

1
2

The ﬁrst term on the right equals πa◦ . All the other integrals on the
right are zero, as can be readily seen by performing the integrations.
Hence our ﬁrst result is
π
a◦     1
(4)                       =         f (x) dx.
2    2π −π
We now determine a1 , a2 , ... by a similar procedure. We multiply (3) by
cos mx, where m is any ﬁxed positive integer, and integrate from −π
π
to π term by term, we see that −π f (x) cos mx dx becomes:
π                     ∞              π                               π
a◦
mx
cos(5) dx +            an            cos nx cos mx dx + bn           sin nx cos mx dx .
2        −π                 n=1            −π                              −π

The ﬁrst integral is zero. Now by the trigonometric identities
1                                   1
cos α cos β = [cos(α+β)+cos(α−β)] sin α cos β = [sin(α+β)+sin(α−β)]
2                                   2
we obtain
π
1 π                   1 π
cos nx cos mx dx =        cos(n + m)x dx +        cos(n − m)x dx,
−π                      2 −π                  2 −π
π
1 π                        1 π
sin nx cos mx dx =         sin(n + m)x dx +         sin(n − m)x dx.
−π                     2 −π                       2 −π
Integration shows that the four terms on the right are zero, except for
the last term in the ﬁrst line, which equals π when n = m. Since in (5)
this term is multiplied by am , the whole of (5) is equal to am π, and our
second result is
1 π
(6)           am =        f (x) cos mx dx, m = 1, 2, ...
π −π
We ﬁnally determine b1 , b2 , ... in (3). If we multiply (3) by sin mx,
where m ∈ N, and then integrate from −π to π, we have
π                              π               ∞
a◦
(7) sin mx dx =
f (x)                              +           (an cos nx + bn sin nx) sin mx dx
−π                              −π    2          1
Integrating term by term, we see that the right hand side becomes
π                   ∞                π                                π
a◦
sin mx dx+              an         cos nx sin mx dx + bn           sin nx sin mx dx
−π   2                  1             −π                             −π

The ﬁrst integral is zero. The next integral is of the type considered
before, and we know that it is zero for all n ∈ N. For the last integral
we need the identity
1
sin α sin β = − [cos(α + β) − cos(α − β)]
2
we obtain
π
1 π                    1 π
sin nx sin mx dx =         cos(n − m)x dx −       cos(n + m)x dx,
−π                       2 −π                   2 −π
3

The last term is zero. The ﬁrst term on the right is zero when n = m
and is π when n = m. Since in (7) this term is multiplied by bm , the
right hand side in (7) is equal to bm π, and our last result is
π
1
(8)            bm =                          f (x) sin mx dx,     m ∈ N.
π           −π

The above calculation can be performed for a periodic function with
any period T > 0, (f (x) = f (x + T ), ∀x ∈ R). Besides, since the
integrals (4,6, 8) exist under wide conditions (e.g. if f is piecewise
continuous), we give the following deﬁnition:

Def. 1 (Fourier series) If f (x) is a piecewse continuous, periodic func-
tion with period 2π, the Fourier series associated with f (x) is the (for-
mal) series
∞
a◦
(9)          f (x) ∼    +     (an cos nx + bn sin nx)
2    n=1

where the Fourier coeﬃcients an , bn are given by:
1 π
(10)        an =      f (x) cos nx dx, n = 0, 1, 2, ...
π −π
π
1
(11)          bn =                          f (x) sin nx dx,    n = 1, 2, ...
π       −π

More generally, for a piecewise continuous, periodic function with pe-
riod T , the associated Fourier series is
∞
a◦              2π            2π
(12)    f (x) ∼        +     an cos( nx) + bn sin( nx)
2    n=1
T             T
with Fourier coeﬃcients:
T /2
2                                    2π
(13)        an =                            f (x) cos(      nx) dx, n ∈ N◦
T       −T /2                        T
T /2
2                                    2π
(14)         bn =                            f (x) sin(      nx) dx, n ∈ N
T          −T /2                     T
△

Remarks 1.A . a) An equivalent choice is to express the coeﬃcients
by integrals from 0 to T , by virtue of periodicity of the function f :
T
2                               2π
(15)         an =                       f (x) cos(      nx) dx, n ∈ N◦
T          0                    T
T
2                                2π
(16)          bn =                          f (x) sin(      nx) dx, n ∈ N
T           0                    T
4

Actually the same numbers are obtained by integrating on any interval
[c, c + T ], with ﬁxed c ∈ R.
b) The symbol ∼ reminds that the trigonometric expansion is still
formal: by that symbol nothing is said about convergence, and, if the
series is convergent, the sum is not necessarily f (x).
1  T
c) The constant term in (12), i.e. a◦ /2 = T 0 f (x) dx, is the mean
value of f in one period.
d) Finally, an equivalent exponential form of the Fourier series holds,
by use of the Euler formulas
1                       1
cos α = (eiα + e−iα ), sin α = (eiα − e−iα ).
2                       2i
Indeed setting
a−n = an , b−n = −bn ,            per n ∈ N, e b◦ = 0
the Fourier series of f becomes:
N
a◦                                  2π                       2π
f (x) ∼    + lim               an cos          nx + bn sin              nx
2   N →∞
n=1
T                        T
N
a◦                    an − ibn i 2π nx an + ibn −i 2π nx
=    + lim                      e T +            e T
2   N →∞
n=1
2                2
N                                         N
an − ibn i 2π nx                2π                               2π
(17) lim
=                            e T = lim          cn ei T nx =                   cn ei T nx
N →∞
n=−N
2             N →∞
n=−N                          n∈Z

where the coeﬃcients cn are given by
T
an − ibn   1                        2π
(18)                   cn =            =                  f (x)e−i T nx
2       T        0
2
¯
In particular, if f (x) is a real function, then cn = c−n .

2. Theorems on Fouries series

Theorem 1.1 (A priori bound on Fourier coeﬃcients)
If the periodic function f admits continuous r−th derivative in [0, T ]
with r ≥ 1, then the coeﬃcients of the Fourier exponential series satisfy
|cn | ≤ c|n|−r , ∀n ∈ Z,
where the constant c is independent of n. Similarly, |an | ≤ cn−r , |bn | ≤
˜
cn−r , if an , bn are the coeﬃcients of the trigonometric series.
˜
2    ¯
c denotes the complex conjugate of c, i.e. ℜc − i · ℑc.
5

PROOF. Let f ∈ C r ([0, T ]). The derivative of a periodic function is
still periodic, so integrating by parts r times the expression (18 )
T           T
1                   T      2π                                                 T      2π
cn =        f (x)              e−i T nx                      −           f ′ (x)         e−i T nx dx
T                 −i2πn                           0       0                 −i2πn
T                                                              r        T
1 T                 ′        −i 2π nx          1                      T                               2π
=+                     f (x)e       T       dx = +                                             f (r) (x)e−i T nx dx.
T i2πn     0                                   T                    i2πn           0
Therefore
r           T
1      T                                                 2π
|cn | ≤                                          |f (r) (x)||e−i T nx | dx
T     2π|n|                0
r           T
1    T                                                           c
≤          r 2π
|f (r) (x)| dx ≤
T |n|                       0                                      |n|r
where                                           r
T
c=           max |f (r) (x)|.
2π    x∈[0,T ]

Since an − ibn = 2cn , choosing tildec = 2c the proof is complete. △

Def. A function f (x) is said piecewise continuous in (a, b) ⊂ R if
the function is continuous except in a ﬁnite number of points xj ∈ (a, b),
j = 1, ..., N, where there exist the ﬁnite limits:
f (xj + 0) = limx→xj + (x), f (xj − 0) = limx→xj − f (x).

Such type of regularity is part of the Dirichlet condition, assumed in
the most usual convergence result, that we state without proof:

Theorem 1.2 (Dirichlet’s Criterion) Let the function f (x) be peri-
odic (with period T ). Let both f (x) and f ′ (x) be piecewise continuous.
Then the trigonometric series associated with f is convergent to f (x)
in each point where f is continuous; while in each point x◦ where f is
discontinuous, the series is convergent to the half-sum
f (x◦ + 0) + f (x◦ − 0)
.
2

Example 1.B (Fourier expansion of a periodic step function) - Let
f (x) = 1, x ∈ [2nπ, (2n + 1)π]0, elsewhere
which satisﬁes the Dirichlet conditions, with period T = 2π.                                                   The
Fourier coeﬃcients a◦ , an , bn are directly calculated:
1                      1 π
a◦ =           πf (x) dx =        1 dx = 1,
π −π                   π 0
6

1.0
0.8
0.6
0.4
0.2
0.0

−5                     0                     5

x

Figure 1. Graph of the Fourier series calculated by the
ﬁrst 25 harmonics, i.e. for n from 0 to 12.

π                                   π
1                                 1                                 1 sin(nx)
an =               πf (x) cos(nx) dx =                      cos(nx) dx =                         =0
π          −π                     π            0                    π    n             0
π                                        π
1                                      1                             1   cos(nx)
bn =                 πf (x) sin(nx) dx =                    sin(nx) dx =        −                    =
π      −π                              π   0                         π      n           0
1
=      [1 − (−1)n ].
nπ
2
Thus bn vanishes for even n, it equals nπ for odd n. The Fourier series
is:
∞                                   ∞
1                                   1            2
+   [an cos(nx) + bn sin(nx)] = +                 sin [(2n + 1)x] .
2 n=1                               2 n=0 (2n + 1)π

Remarks 1.C (On the computation of Fourier coeﬃcients)
a) In order to compute the coeﬃcients of the Fourier series, several
remarks are useful.
A function f (x) is said to be even if f (−x) = f (x), ∀x; it is odd if
f (−x) = −f (x), ∀x. If the function f (x) is even, then all the Fourier
coeﬃcients bn vanish: indeed, under a substitution y = −x,
T                                    T /2
2                          2π        2                                2π
bn =                  f (x) sin( nx) dx =                     f (x) sin(      nx) dx =
T        0                 T         T           −T /2                T
T /2                                                 T /2
2                                 2π            2                             2π
f (−y) sin(−             ny) dy = −                  f (y) sin(      ny) dy = −bn
T   −T /2                         T             T        −T /2                T
7

Thus 2bn = 0, so that bn = 0. Similarly, it turns out that a◦ = an = 0
if the function is odd.
b) Therefore an even periodic function admits a series expansion with
only cosine terms; while an odd periodic function admits a series
expansion with only sine terms. The next remark regards an arbitrary
function f (x), deﬁned in some interval, for example in (0, π). We can
extend it both to an even function and to an odd function. Indeed,
we can deﬁne it in the interval (−pi, 0) so that it becomes even:
f◦ (x) = f (x), x ∈ (0, π) lim f (x), x = 0f (−x), x ∈ (−π, 0)
x→0+

Another choice is to deﬁne an odd extension of f (x):
f1 (x) = f (x), x ∈ (0, π)0, x = 0 − f (−x), x ∈ (−π, 0)
Finally, one can extend the functions f◦ (x) and f1 (x) to the whole of
R, so that the resulting function is periodic with period 2π. Assuming
the Dirichlet conditions to hold, it follows that
∞
1
(19)                  f◦ (x) = a◦ +     an cos(nx)
2     n=1

where
π                                   π
1                               2
an =            f◦ (x) cos(nx) dx =                f (x) cos(nx) dx
π   −π                          π   0

and
∞
(20)                       f1 (x) =         bn sin(nx)
n=1

where
π                                  π
1                               2
bn =            f◦ (x) sin(nx) dx =                f (x) sin(nx) dx
π   −π                          π   0

Now under a restriction of (19), (20) to the interval (0, π), it follows
that a function with domain (0, π) can be exapnded in a cosine-series
as well as in a sine-series.

3. Application of the Fourier series to the harmonic
oscillator

Let us compute a particular solution of the damped harmonic oscillator
with a forcing term
(21)                        x + µx + ω 2 x = f (t)
¨    ˙
8

when f (t) is a real valued periodic function with period T . Although
not real-valued, it is convenient ﬁrst to consider
f (t) = ρ · eiΩt
where Ω = 2π is the pulsation of the forcing term. In this case a
T
particular solution in the form const. · eiΩt is easily found:
ρeiΩt
x∗ (t) =
ω 2 − Ω2 + iΩµ
As a second step let us consider f (t) given by a ﬁnite Fourier expansion
in exponential form:
N
f (t) =               cn eiΩnt
n=−N

¯
where cn = c−n since f (t) is real valued. A particular solution with
period T is given by
N
∗                                                  cn eiΩnt
x (t) =             x∗ (t),
n         x∗ (t) =
n
n=−N
ω 2 − n2 Ω2 + inΩµ
Notice that x∗ is real since the addends of the sum with opposite indices
n, −n are complex conjugate. Moreover x∗ (t) is a particular solution
n
of the diﬀerential equation
x + µx + ω 2x = cn eiΩnt .
¨    ˙
Thus we can immediately verify:
N                                           N
∗    ∗        2 ∗
x + µx + ω x =
¨    ˙                        (¨∗
xn   +   µx∗
˙n     +   ω 2x∗ )
n      ==          cn eiΩnt = f (t).
n=−N                                           n=−N

Finally, let f (t) admit a Fourier expansion with inﬁnitely many nonzero
coeﬃcients:
∞
f (t) =               cn eiΩnt
n=−∞
A particular solution is looked for in the form:
∞
∗                               cn
(22)              x (t) =                                           eiΩnt
n=−∞
ω2    −   n2 Ω2    + inΩµ

To this end we have to check the convergence of (22); then, provided
its convergence, to check whether is it a solution of the diﬀerential
equation. Now a suﬃcient condition to derive k times term by term
is the convergence of the series of the k−th order derivatives:
∞
cn (iΩn)k
(23)                                            eiΩnt
n=−∞
ω 2 − n2 Ω2 + inΩµ
9

uniformly with respect to t. On the other hand we know that f ∈ C r
implies the bound on the Fourier coeﬃcients |cn | ≤ cn−r . Therefore
the general term of (26) is bounded by
nk
cΩk                                     ≤ Cnk−r−2
nr   (ω 2 − n2 Ω2 )2 + n2 µ2 Ω2
for some constant C > 0 independent of n. So the series (26) is uni-
formly convergent with respect to t if
r + 2 − k > 1;
In particular, the series (22) is a solution of the diﬀerential equation if
r + 2 − 2 > 1 (k = 2), i.e. if the function f (t) is at least in C 2 .
Notice: for µ small enough, the particular solution x∗ (t) ampliﬁes the
harmonics n = ±[ω/Ω], (where [·] denotes the integral part), of the
function f (t) for which ω 2 − n2 Ω2 attains its minimum, while the other
harmonics are damped.

4. Application of the Fourier series to partial
differential equations

The partial diﬀerential equations here considered are of the following
types:
The one dimensional wave equation

∂2u      ∂2u
= a2 2
∂t2      ∂x
A string is stretched between two ﬁxed points a distance L apart. The
string has an equilibrium position in which it lies on the line segment
joining the points. The unknown function u(x, t) is the displacement
of a generic point x of the string at the instant t. The constant a2
is equal to τ /ρ where the tension τ and the density ρ are supposed
constant. No external forces are present and vibrations are uniquely
due to helasticity. Boundary conditions
u(0, t) = 0, u(L, t) = 0
and initial conditions
∂u
u(x, 0) = f (x),   (x, 0) = g(x)
∂t
are associated with the diﬀerential equations (the extreme points are
ﬁxed; positions and velocities are given at t = 0).
The heat equation:
∂u
= k∆u,
∂t
10

where ∆ is the Laplace operator
∂2u ∂2u ∂2u
∆u =      +     + 2.
∂x2 ∂y 2  ∂z
The temperature in the point (x, y, z) of a body at time t is repre-
sented by u(x, y, z, t), while the diﬀusion constant k depends on ther-
mal conductivity, speciﬁc heat and density, all assumed to be constant.
Moreover, no internal heat sources are assumed. As above, boundary
conditions (for all times) and initial conditions (in the points of the
body) have to be associated with the heat equation.

The Laplace’s equation

∆u = 0.
The Laplace’s equation appears in many ﬁelds of Physics. In the frame-
work of thermal conduction, u can represent the temperature in sta-
tionary conditions i.e. the temperature after a suﬃciently long time (it
is obtained from the heat equation if t−independent solutions, ∂u = 0,
∂t
are looked for).

Example 4.A Let us consider the one-dimensional heat equation with
boundary and initial conditions:
∂u      ∂2u
(24)                                                      =2 2
∂t      ∂x
(25)                                u(0, t) = u(3, t) = 0, ∀t ≥ 0
u(x, 0) = 5 sin(4πx) − 3 sin(8πx) + 2 sin(10πx), x ∈ [0, 3]
Besides, the solution is required to remain bounded ∀(x, t) ∈ [0, 3]×R+ .
(26)
Let us solve the equation by separation of variables. By such a method,
we look for a solution (if possible) of the form
u(x, t) = f (x)g(t)
. Then the equation reads f (x)g(t) = 2f ′′ (x)g(t). After separating
˙
variables,
f ′′ (x)        ˙
g(t)
=       .
f (x)           g(t)
Since the left hand side depends only on x, the right hand side depends
only on t, and the equation holds for all (x, t) where x, t are independent
variables,
f ′′ (x)        ˙
g(t)
∃λ ∈ R :               =λ=         .
f (x)           g(t)
Therefore we obtain two ordinary diﬀerential equations depending on
an arbitrary parameter λ:
f ′′ = λf,   ˙
g = 2λg
11

with general solutions
√                  √
λx
(27)    f (x) = c1 e            + c2 e−       λx
,        g(t) = ce2λt   if λ = 0

f (x) = c1 x + c2 ,                   g(t) = c       if λ = 0.
+
Now, the boundedness requirement ∀t ∈ R implies λ < 0. Therefore,
setting λ = −ω 2 and c1 = 1 (a + ib = c2 , the general solutions (27) can
2
¯
be written:
2
f (x) = a cos(ωx) + b sin(ωx),                        g(t) = ce−2ω t .
Hence a solution of the heat equation turns out to be
2
u(x, t) = f (x)g(t) = ce−2ω t [a cos(ωx) + b sin(ωx)] =
2
= e−2ω t [A cos(ωx) + B sin(ωx)]
with constants A, B, ω to be determined by using the boundary condi-
tions in (26):
2
u(0, t) = 0 =⇒ A = 0, i.e. u(x, t) = e−ω t B sin(ωx)

u(3, t) = 0 =⇒ B sin(3ω) = 0
This last equality, in turn, implies
either B = 0, arbitrary ω,                           or arbitrary B, ω = nπ/3, n ∈ Z.
Now B = 0 is immediately rejected since it gives an identically vanish-
ing solution. Therefore a nontrivial solution of the diﬀusion equation
in (26) with zero boundary condition has the form:
2 2        nπ
(28)          u(x, t) = Be−2n π t/9 sin( x), n ∈ N
3
By the superposition principle (which holds for linear diﬀerential equa-
tions), the sum of solutions of a homogeneous diﬀerential equation is
still a solution; so for any choice of n1 , n2 , n3 ∈ N the function
2 2        n1 π           2 2          n2 π           2 2          n3 π
B1 e−2n1 π t/9 sin(      x)+B2 e−2n2 π t/9 sin(      x)+B3 e−2n3 π t/9 sin(      x)
3                           3                           3
is still a solution of the equation in (26). By comparing this type of
solution with the initial condition in (26) we ﬁnd:
B1 = 5, n1 π/3 = 4π, B2 = −3, n2 π/3 = 8π, B3 = 2, n3 π/3 = 10π
i.e. n1 = 12, n2 = 24, n3 = 30. Thus the unique (bounded) solution
of the diﬀusion problem (26) is
2                                         2                         2
u(x, t) = 5e−32π t sin(4πx) − 3e−128π t sin(8πx) + 2e−200π t sin(10πx)
12

5
u(x,t)

0
−5

0.0                0.5           1.0         1.5            2.0           2.5         3.0

x

Figure 2. Graph of the function u(x, t) for some values of
t: t = 0 (continuous line), t = 0.001 and t = 0.01

5
u(x,t)

0

−5                                                              0.010
0.008

0.006
0.0
0.5
0.004
t

1.0
1.5
x     2.0
0.002

2.5
3.00.000

Figure 3. Graph of the function u(x, t), for (x, t) ∈ (0, 3) × (0, 0.01)

Example 4.B We solve the same diﬀusion equation with a diﬀerent
initial value function:
∂u      ∂2u
(29)                                         =2 2
∂t      ∂x
(30)                   u(0, t) = u(3, t) = 0, ∀t ≥ 0
u(x, 0) = 25x(3 − x), x ∈ [0, 3]
Again
(31) by separation of variables we obtain a solution of the heat equa-
tion (28), but it is no more suﬃcient to make a ﬁnite superposition of
13

such functions to solve the problem (31). We try by a superposition
of inﬁnitely many functions:
∞
2 π 2 t/9        π
u(x, t) =         Bn e−2n               sin(n x),
n=1
3
where we require u(x, 0) = 25x(3 − x), that is
∞
π
25x(3 − x) =            Bn sin(n x), x ∈ (0, 3).
n=1
3
This amounts to expand in a sine Fourier series the function 25x(3−x).
According to Remarks 2 (b), this is possible if we regard the function
as a restriction of an odd periodic function deﬁned for all x ∈ R, with
period 6:
f2 (x) = 25x(3 − x), x ∈ (0, 3)25x(3 + x), x ∈ (−3, 0)... and so on.
Then we have the usual Fourier coeﬃcients for an odd function of
period T :
T /2                                             T /2
2                           2π       4                                            2π
Bn =                 f2 (x) sin(n x) dx =                               f2 (x) sin(n      x) dx =
T         −T /2             T        T                  0                         T
3
2                                                       900
=                25x(3 − x) sin(nπx/3) dx =                        [1 − (−1)n ]
3    0                                                  n3 π 3
Finally
∞
1800                        2 2
u(x, t)           (2m + 1)−3 e−2(2m+1) π t/9 sin((2m + 1)πx/3).
π 3 m=0
The graph of u(x, t) is ﬁrst represented for some values of t (Fig. 4),
then for (x, t) ∈ (0, 3) × (0, 1) (Fig. 5).

5. Fourier transform

Def. (Direct Fourier transform) Let the function f (t) be given for any
t ∈ R. If the integral
∞
f (t)e−iωt dt
0
ˆ
exists ∀ω ∈ R, then f (t) is said to admit the Fourier transform f (ω) :
+∞
(32)                                ˆ
F (f ) = f (ω) =             f (t)e−iωt dt.
−∞
14

50
40
30
u(x,t)

20
10
0

0.0   0.5        1.0       1.5       2.0      2.5       3.0

x

Figure 4. For initial shape u(x, 0) = 25x(3 − x), graph of
the solution u(x, t) when t = 0, t = 0.01, t = 0.1
u(x,t)

x

t

Figure 5. For initial shape u(x, 0) = 25x(3 − x), graph of
the solution u(x, t), (x, t) ∈ (0, 3) × (0, 1)

ˆ
The operator F , which transforms f (t) into f (ω), is said ”Fourier
transform”.

Usually the variable t is a time variable, in which case the variable ω is
ˆ
a frequency. The Fourier transform f (ω) is a complex valued function
that can be written in exponential notation as
ˆ
f (ω) = A(ω)eiφ(ω)
15

indicatore di [−a,a]

1.0
0.8
0.6
p_a(x)

0.4
0.2
0.0

−a                              a

Figure 6. Graph of pa (t).

where the real function A(ω) is said Fourier spectrum, A2 (ω) is the
power spectrum, and φ(ω) is the phase angle.
ˆ
Remark 5.A The complex conjugate of f(ω) is equal to the Fourier
transform of the conjugate of f calculated in −ω:
+∞                        +∞
¯
ˆ                                               ¯              ˆ
¯
f (ω) = (            f (t)e−iωt dt)¯ =          f (t)eiωt dt = f (−ω).
−∞                            −∞

In particular, if f (t) is a real valued function, then
¯
ˆ        ˆ
¯        ˆ
f (ω) = f (−ω) = f (−ω)
and it turns out that A(ω) is an even function and φ(ω) is an odd
function.

Example 5.B We want to determine the Fourier transform of the
indicator function of the interval [−a, a],
(33)                                  pa (t) = 1,    t ∈ [−a, a]
0, |t| > a
where
(34) a > 0 is ﬁxed. In this case the Fourier transform (32) reduces
to the integral
a                        a
−iωt        e−iωt              eiωa − e−iωa    sin ωa
ˆ
pa (ω) =              e      dt =               =                 =2
−a                  −iω      −a             iω            ω
ˆ
for ω = 0 (while pa (0) = 2a).                      Fig. 6a and Fig. 6b show pa (t) and
ˆ
pa (ω), respectively.
16

Trasf. di Fourier di p_a(t)

2a

p(ω)
^

Figure 7. Graph of pa (ω).
ˆ

Remark 5.C No suﬃcient condition for the existence of the Fourier
transform has yet been stated. The expression (32) easily suggests
that if the function f (t) is absolutely integrable on R, that is
∞
(35)                                       |f (t)| dt < +∞,
−∞

ˆ
then f (ω) exists for any ω ∈ R. In other words, under this assumption
the integral (32) exists and is ﬁnite. Such a condition is suﬃcient but
not necessary. Indeed there exist functions, such as sin(ω◦ t)/ω, for
which the Fourier transform exists, but the property (35) does not
hold.

Lemma 5.1 (Riemann’s lemma) Let f (t) be an absolutely integrable
function in the (bounded or unbounded) interval (a, b). Then
b
lim            f (t)e−iωt dt = 0.
ω→±∞     a
PROOF. For the sake of simplicity, we prove the Riemann’s lemma
by assuming that f ∈ C 1 (a, b) and the interval (a, b) is bounded. The
statement still holds under the only assumption of absolute integrabil-
ity. If f (t) is of class C 1 (a, b) then, integrating by parts, we obtain:
b                                      b               b
e−iωt           i
f (t)e−iωt dt = f (t)                  −               f ′ (t)e−iωt dt.
a                                  −iω     a       ω   a

Since (a, b) is a bounded interval,
b
1
|           f (t)e−iωt dt| ≤         C, C = |f (b)| + |f (a)| + (b − a) · max |f ′ (t)|
a                            |ω|                                      t∈[a,b]
17

and the statement is proved.             △
ˆ
Remark 5.D Let f (ω) be the Fourier transform of an absolutely
integrable function f (t) on R. Then, by Riemann’s Lemma,
ˆ
lim f(ω) = 0.
ω→±∞

Thus the Riemann’s Lemma implies a condition about the behaviour
at inﬁnity of the Fourier transform.

The following is a regularity property.
Theorem 5.2 If f (t) is absolutely integrable, then its Fourier trans-
ˆ
form f (ω) is continuous on R.
PROOF. Let ω◦ ∈ R be ﬁxed, and let
+∞
ˆ ˆ         ˆ
∆f = f (ω) − f(ω◦ ) =                 f (t)[e−iωt − e−iω◦ t ] dt =
−∞
+∞
=            f (t)e−iω◦ t [e−i(ω−ω◦ )t − 1] dt.
−∞
We need to prove that
ˆ
∀ǫ > 0, ∃δ(ǫ) > 0 : |ω − ω◦ | < δ =⇒ |∆f | < ǫ.
Since f (t) is absolutely integrable, there is R = R(ǫ) > 0 such that
−R                    +∞
1
|f (t)|dt +           |f (t)|dt ≤ ǫ.
−∞                    R                 4
Besides, for such R = R(ǫ) we choose δ(R(ǫ)) such that
+∞               −1
−iδt         1
max ·|e           − 1| ≤ ǫ              |f (t)|dt
t∈[−R,R]                   2         −∞
Now using
|e−i(ω−ω◦ )t − 1| ≤ |e−i(ω−ω◦ )t | + 1 = 2
and
|e−i(ω−ω◦ )t − 1| ≤ |e−iδt − 1|, for |ω − ω◦ | ≤ δ, δt ≤ π/2,
it follows that
−R                     +∞                  R
ˆ                                                                                    1 1
|∆f | ≤ 2           |f (t)|dt +            |f (t)|dt +      |f (t)||e−i(ω−ω◦ )t −1|dt ≤ ǫ+ ǫ = ǫ
−∞                     R                   −R                            2 2
whenever |ω − ω◦ | ≤ δ.           △

Now let us consider the inversion problem, assuming f (t) absolutely in-
ˆ
tegrable on R. If f (ω) = F (f ), the problem is how to reconstruct f (t)
ˆ
starting from f(ω). The following inversion theorem is stated without
proof, but a heuristic argument for it will be given in the following
section.
18

Theorem 5.3 (Inverse Fourier transform) - Let the function f (t) be
absolutely integrable on R and let f (t), f ′ (t) be piecewise continuous
in any bounded interval. Then the inversion formula
+∞                                    +R
1              ˆ              1                      ˆ
(36) f (t) =      P            f(ω)eiωt dω =     lim                 f (ω)eiωt dω
2π     −∞                     2π R→+∞          −R

holds in all points t where f (t) is continuous, while the formula
+∞
1                              1                    ˆ
(37)        [f (t◦ + 0) + f (t◦ − 0)] =    P                  f (ω)eiωt dω
2                             2π            −∞

ˆ
holds in all discontinuity points t◦ . The operator F −1 trasforming f (ω)
into f (t) (at least where f is continuous) is said the inverse Fourier
transform. △

Remark 5.E The integral in (36)is computed as a Cauchy princi-
pal value. The integral is convergent as a Cauchy Principal Value if
+R
the limit limR→+∞ −R (...) exists and is ﬁnite. Recall that the usual
generalized integral of a function g(t) is convergent if the double limit
R
lim          g(t) dt
R,S→+∞   −S

exists and is ﬁnite with R, S diverging to +∞ independently of each
other. Of course if the integral is convergent in the generalized sense,
then it is convergent in the sense of the Cauchy principal value. But
the opposite implication is not true in general. An example is the
function g(t) = t, which is integrable in the Cauchy principal value
sense:
+∞                    +R
P             t dt = lim              t dt = 0.
−∞              R→+∞     −R

The same function is not integrable in the generalized sense, since many
diﬀerent values are attained by the double limit prescription: for exam-
R                       2R
ple −2R tdt → −∞, while −R tdt → +∞. A similar situation regards
any odd, piecewise continuous function.

Remark 5.F In some textbooks the direct ad the inverse Fourier
in
transforms are deﬁned√ a slightly diﬀerent way, attributing to both
the same coeﬃcient 1/ 2π (instead of 1 and 1/2π, respectively). Ac-
tually any other choice is good, provided that the product of the two
coeﬃcients is 1/2π.

ˆ
Remark 5.G (Symmetry property) If f (ω) is the Fourier transform
of f (t) (and if both satisfy the hypotheses of Theorem 5.3) then the
19

ˆ
Fourier transform of f (t) is 2πf (−ω), at least in the points in which f
is continuous. Indeed
+∞                         +∞
ˆ         ˆ                  1        ˆ
F (f) =      f (t)e−iωt dt = 2π          f(t)e−iωt dt = 2πf (−ω).
−∞                     2π −∞

Example 5.H Compute the Fourier transform of
sin(ω◦ t)
f (t) =            .
t
By Example 5.B, the indicator function pa (t) of the interval (−a, a),
sin(ωa)
ˆ
pa (ω) = 2            .
ω
Hence the given function f (t) can be seen as a Fourier transform:
sin(ω◦ t)     1
f (t) =             = pω◦ (t),
ˆ
t         2
where pω◦ is the indicator function of (−ω◦ , ω◦ ). Finally, by the sym-
metry property
ˆ         1                     1
f(ω) ≡ F [ˆω◦ ](ω) = 2π pω◦ (−ω)
p
2                     2
at least in the points where pω◦ is continuous. Explicitly the result is:
sin(ω◦ t)                  ˆ
(38) if f (t) =              , then           f (ω) = π, |ω| < ω◦
t
1
(39)                                                    π, ω = ±ω◦
2
0, |ω| > ω◦
(40)
Example 5.I Using (40), compute the integral
+∞
dt
I(x, a) =               sin(at) cos(tx)      .
0                           t
sin(at)
First, the Fourier transform of f (t) =             t
is known by (40):
+∞
ˆ           sin(at) −iωt                π
f(ω) =             e     dt = π, |ω| < a , |ω| = a0, |ω| > a.
−∞      t                        2
On the other hand
+∞                    +∞
ˆ             sin(at) −iωt          sin(at) iωt
f (ω) =               e    dt =            e dt
−∞      t              −∞     t
0                    +∞                        +∞
sin(at) iωt             sin(at) iωt           sin(at) iωt
=                 e dt +                  e dt =                [e + e−iωt ]dt
−∞        t             0         t             0        t
+∞
sin(at)
=2                    cos(tω)dt = 2 · I(ω; a)
0           t
20

Hence
+∞
dt     1ˆ     π         π
I(x, a) ≡            sin(at) cos(tx)
= f (x) = , |x| < a , |x| = a0, |x| > a.
0                    t     2      2         4
As a by-product, by choosing x = 0 and a = 1, the computation of the
integral
∞
sin t      π
(41)                                dt =
0      t       2
is obtained. Notice that (41) is an important example of a generalized
integral which is convergent without being absolutely convergent.
+∞      2
Example 5.L Compute: a) the Gauss integral −∞ e−αt dt ; b) the
2
Fourier transform of the Gauss function e−αt (α > 0).
+∞     2
a) Let I −∞ e−αx dx. By Fubini’s theorem
+∞                 2         +∞                        +∞
2               −αx2                                    2                 2
I =                e        dx       =             e−αx dx                e−αy dy
−∞                           −∞                         −∞
+∞           +∞
2 +y 2 )
=                      e−α(x               dxdy
−∞       −∞
Under a change of variables (x, y) → (ρ, θ) from cartesian coordinates
to polar coordinates, the integral becomes
2π       +∞
2                 π
I2 =                      e−αρ ρdρdθ =             .
0        0                               α
b) Let us derive
+∞
ˆ
f (ω) = F (f ) =
2
e−αt e−iωt dt
−∞

with respect to ω. The derivation, with respect to ω, commutes with
integration, with respect to t, because the integral is uniformly conver-
2
gent with respect to ω; then we can integrate by parts taking −2αte−αt
as a derived factor:
ˆ
df       +∞
2
=        −ite−αt e−iωt dt
dω      −∞
+∞
ω −αt2 −iωt       ω ˆ
=           −      e  e     dt = − f (ω).
−∞        2α                2α
ˆ
So f (ω) is a solution of the ﬁrst order diﬀerential equation
ˆ
df    ω ˆ
= − f (ω).
dω    2α
whose general solution is
ω          2
ˆ       ˆ
f (ω) = f (0)e− 4α
21

ˆ
As for f (0), it is noting else than the Gauss integral previously com-
puted:
+∞
ˆ               2       π
f (0) =      e−αt dt =     .
−∞              α
Hence the remarkable result that the Fourier transform of the Gauss
function is a Gauss function (up to a coeﬃcient and a parameter):
π      ω2
F [exp(−αt2 )] =             exp(− ).
α      4α
In particular, choosing α = 1 ,
2
2 /2        √           2 /2
F (e−t      )=       2πe−ω          .
[ To sum up the above given examples:
(42)                      pa (t) = 1 t ∈ (−a, a)
/
0, t ∈ (−a, a)
pa (ω) = 2 sin(ωa) (43)
ˆ             ω

sin(ω◦ t)                ˆ
(44)     f (t) =                            f (t) = π/2, |ω| < ω◦
t
(45)                                                π/4, |ω| = ω◦
0, |ω| > ω◦
(46)
2
ˆ                 π −ω2 /(4α)
(47)     f (t) = e−αt                   f (ω) =             e         .]
α

6. Properties of the Fourier transform

Thm. 6.1 (Linearity property) Let the functions f1 , f2 admit Fourier
ˆ                    ˆ
transform F (f1 ) = f1 and F (f2 ) = f2 , respectively. Then for arbitrary
coeﬃcients c1 , c2 ∈ C there exists the Fourier transform of c1 f1 (t) +
c2 f2 (t) and
ˆ       ˆ
F (c1 f1 + c2 f2 ) = c1 f1 + c2 f2 .
ˆ ˆ
Vice-versa, let the functions f1 , f2 admit inverse Fourier transform
ˆ               ˆ
F −1 (f1 ) = f1 , F −1 (f2 ) = f2 , respectively. Then the same holds for
any linear combintation:
ˆ       ˆ
F −1 (c1 f1 + c2 f2 ) = c1 f1 + c2 f2 .               △
PROOF. A straightforward consequence of the linearity property of
the integral. F and F −1 are linear operators, too. △.
22

Trasf. di Fourier di p_a(t)cos(omega0 t)

Figure 8. Graph of the Fourier transform of the modulated
indicator function pa (t) cos(ω◦ t), with ω◦ = 7, a = 5.

Thm. 6.2 (Frequency shifting) Let the function f (t) admit Fourier
ˆ
transform f (ω). For any constant ω◦ ∈ R,
ˆ
F [eiω◦ t f (t)] = f (ω − ω◦ ),          ˆ
F −1 [f (ω − ω◦ )] = eiω◦ t f (t).
PROOF.
+∞
F [eiω◦ t f ] =                              ˆ
f (t)e−i(ω−ω◦ )t dt = f(ω − ω◦ )     △
−∞

Ex. 6A Compute the Fourier transform of the modulated indicator
function
f (t) = pa (t) cos(ω◦ t),
where pa (t) is the indicator of (−a, a) and ω◦ is real and ﬁxed. By Eu-
ler’s formula, linearity, frequency shifting and by the above examples,
ˆ                              1
f (ω) = F [pa (t) cos(ω◦ t)] = [F (pa (t)eiω◦ t ) + F (pa (t)e−iω◦ t )]
2
1
p                ˆ
= [ˆa (ω − ω◦ ) + pa (ω + ω◦ )]
2
sin[a(ω − ω◦ )] sin[a(ω + ω◦ )]
=                   +                  .
ω − ω◦              ω + ω◦

Thm. 6.3 (Time shifting) Let the function f (t) admit Fourier trans-
ˆ
form f (ω). Then
ˆ
F [f (t − t◦ )] = e−iωt◦ f(ω),                     ˆ
F −1 [e−iωt◦ f(ω)] = f (t − t◦ ).
23

Trasf. di Fourier di p_a(t+2a)+p_a(t−2a)

4a
4
3
2
1
0
−1
−2

−15      −10        −5        0          5         10     15

Figure 9. Graph of the Fourier transform of pa (t + 2a) +
pa (t − 2a), sum of two indicators centered at −2a, 2a, with
a=1

In particular it follows that
ˆ
F [f (t − t◦ )] = e−iωt◦ f (ω) = A(ω)ei(φ(ω)−ωt◦ ) .
PROOF.
+∞
F [f (t − t◦ )] =           f (t − t◦ )e−iωt dt
−∞
+∞
= e−iωt◦          f (t − t◦ )e−iω(t−t◦ ) dt
−∞
+∞
= e−iωt◦                                 ˆ
f (t)e−iωt dt = e−iωt◦ f(ω). △
−∞

Ex. 6B Compute the Fourier transform of the following functions:
f1 (t) = pa (t + 2a) + pa (t − 2a),          f2 (t) = pa (t + 2a) − pa (t − 2a).
Using the linearity property, the time shifting theorem and (43) it
follows that
ˆ                                    4 sin(ωa) cos(2ωa)
f1 (ω) = pa (ω)ei2aω + pa (ω)e−i2aω =
ˆ             ˆ
ω
and
ˆ                                     4i sin(ωa) sin(2ωa)
f2 (ω) = pa (ω)ei2aω − pa (ω)e−i2aω =
ˆ             ˆ                                   .
ω
24

Thm. 6.4 (Time scaling) Let the function f (t) admit Fourier trans-
ˆ
form f (ω) and let a ∈ R − {0} be ﬁxed. Then
1 ˆω
F [f (at)] =          f ( ).
|a| a
PROOF. If a < 0 the change of variable τ = at changes the extreme
value −∞ into +∞ and vice-versa, so that
+∞                                  +∞                                         +∞
F [f (at)] =          f (at)e−iωt dt =                    f (τ )e−iωτ /a dτ /a, a > 0−              f (τ )e−iωτ /a dτ /a, a < 0
−∞                                  −∞                                          −∞
+∞
1                                                 1 ˆω
=                    f (τ )e−iωτ /a dτ =                 f ( ). △
|a|        −∞                                     |a| a

Let us prove that the Fourier operator transforms derivatives in the
time representation into multiplication by corresponding monomials in
the frequency representation, and vice-versa. This property makes the
Fourier transform able to solve diﬀerential equations.

Thm. 6.5 (Transformation of derivatives into multiplications and
vice-versa) Let f ∈ C n (R) and let f (r) (t) be absolutely integrable on R
for each r ≤ n. Then
ˆ
F [f (n) ] = (iω)n f (ω)               and                 ˆ
F −1 [(iω)n f (ω)] = f (n) .
Similarly, assuming that tn f (t) is absolutely integrable,
ˆ
dn f (ω)                 ˆ
dn f(ω)
F [(−it)n f (t)] =   and F −1 [          ] = (−it)n f (t).
dω n                 dω n
PROOF. The proof is based on integration by parts and the fact that
(48)                lim f (r) (t) = 0,              r = 0, 1, ..., n − 1
t→±∞

since f (r) (t), r = 0, 1, . . . , n is absolutely integrable on R. We can
realize this fact by writing
t
(r)             (r)
f         (t) = f         (0) +            f (r+1) (τ )dτ.
0

Since f (r+1) (τ ) is absolutely integrable, then the following limit exists
and is ﬁnite:
+∞
L = lim f (r) (t) = f (r) (0) +                               f (r+1) (τ )dτ.
t→+∞                                          0

Now, either L = 0 or L = 0 : we prove that the second case is impossi-
ble. Ab absurdo, let L = 0. Then there is an interval [T, +∞) in which
25

|f (r) (t)| > L/2, so that f (r) (t) cannot be absolutely integrable. Thus
(48) is proved. Consider
+∞
(n)
F (f         )(ω) =            f (n) (t)e−iωt dt
−∞
+∞
+∞
= f (n−1) (t)e−iωt       −∞
+ iω               f (n−1) (t)e−iωt dt
−∞
(n−1)
= iω F (f     )(ω)
(n−1)
since limt→±∞ f      (t) = 0. By iterating such relation the statement
follows. Finally, by direct inspection the trasform of (−it)f (t) is equal
ˆ
to df ()/dω, and the iteration on n concludes the proof. △
ˆ
Remark 6C Let the function f (t) amit a Fourier transform f (ω) and
let us consider the integral function
t
ϕ(t) =           f (τ )dτ.
−∞

ˆ
Let us suppose that ϕ admits, in turn, the Fourier transform ϕ(ω) =
˙
F (ϕ). Then, recalling ϕ(t) = f (t), Thm. 6.5 applies:
ˆ                   ˙      ˆ
f(ω) = F (f ) = F (ϕ) = iω ϕ(ω).
Therefore
t                     ˆ
f (ω)
(49)                 F               f (τ )dτ     =         .
−∞                     iω
Notice that a condition for the existence of the Fourier trasform is ϕ to
be absolutely integrable, whence in particular limt→+∞ ϕ(t) = 0, that
is
+∞
f (t) dt = 0.
−∞
If this last condition does not hold, then (49) has to be modiﬁed.

Ex. 6D Compute the Fourier transform of the triangular function
1                         1
qa (t) = (x+a), −a ≤ x ≤ 0− (x−a), 0 ≤ x ≤ a0, elsewhere (a > 0).
a                         a
One can easily verify that the triangular function is given by the inte-
gral
1 t
qa (t) =        [pa/2 (τ + a/2) − pa/2 (τ − a/2)]dτ.
a −∞
Hence, from (49) and from Ex.6B, we get
1 1
F (qa ) =      F [pa/2 (t + a/2) − pa/2 (t − a/2)]
iω a
1 −4i sin(ωa/2) sin(2ωa/4)        4 sin2 (ωa/2)
=                                   =                .
iωa               ω                      aω 2
26

Ex. 6E Using the symmetry property and Ex. 6D, compute the
Fourier transform of the function
sin2 at
f (t) =         .
πat2
In fact by Ex. 6D, changing a/2 into a,
sin2 at    4 sin2 at 2                                 2
=           ·      ˆ
= g , where g(t) = q2a (t) ·
πat2        2at2    4π                              4π
g
By symmetry F [ˆ] = 2πg(−ω), so that
sin2 at                     2
F[   2
] = 2πg(−ω) = 2π q2a (−ω) = q2a (ω).
πat                      4π
As a by-product of this example the value of the following integral
follows:
+∞
sin2 (at)
dt = π q2a (0) = π.
−∞      at2

7. Convolution product, filtering and the heat equation
on a line

Def. (Convolution product) The convolution product of two functions
f1 (t), f2 (t) deﬁned for t ∈ R, is the generalized integral (provided it
exists)
+∞
(50)             (f1 ∗ f2 )(t) =         f1 (x)f2 (t − x)dx.
−∞

Remark 7A For the convolution product the usual associative and
distributive (with respect to sum) properties hold. Moreover, the
commutative property can be easily proved: by a change of variable
t − x = y,
+∞                          +∞
f1 ∗ f2 (t) =         f1 (x)f2 (t − x)dx =        f1 (t − y)f2 (y)dy = f2 ∗ f1 (t).
−∞                           −∞

Remark 7B By the convolution product functions are regularized.
More precisely, let f (x) be a function coming from experimental mea-
surments, but cotaining irregular details due to undesired eﬀects (for
example noise in reception of an electrical signal, or atmospherics in
27

u(x,t) per t=0.01,t=1,t=10

1.0
t=0.01       t=1
0.8

t=10
0.6
u(x,t)

0.4
0.2
0.0

−10     −5            0               5         10       15   20

x

Figure 10. Graph of the solution of the heat equation,
starting from an initial function u(x, 0) with compact sup-
port, the indicator of (0, 10). Evidence of its ”diﬀusion” is
given at times t = 0.01, t = 1, t = 10.

misurazione soggetta a disturbo
1.0
0.8
0.6
0.4
0.2
0.0

0           2             4                6        8        10

x

Figure 11. Graph of the function f (x), resulting from a
(hypothetic) measurement performed in the interval [0, 10]
and perturbed by some noise

an astronomic measurement, etc.). f can have a graph like in Fig.
9. If we perform the convolution product of f with a regular ﬁltering
function R(x), then we could obtain a regular version of the function
f , free from undesidered eﬀects. Indeed
∞
(f ∗ R)(t) =                  f (x)R(t − x)dx
−∞
28

f*R con sigma=0.01

1.0
0.8
0.6
0.4
0.2
0.0

0          2          4           6          8          10

t

Figure 12. Graph of the function f ∗ R, for σ = 0.01,
where the convolution product is performed in [0, 10]; notice
that the eﬀects of noise are still reproduced.

f*R con sigma=0.5
1.0
0.8
0.6
0.4
0.2

0          2          4           6          8          10

t

Figure 13. Graph of the function f ∗R, for σ = 0.5. In this
case the eﬀects of noise are remarkably reduced, although not
completely eliminated. Notice that for larger values of σ, e.g.
σ = 5, the convolution product f ∗ R is no longer suitable to
represent f (x).

and now the dependence on t is through the egular function R. Assum-
ing that the operations of derivative and integral commute, (f ∗ R)(t)
is diﬀerentiable as many times as R(x)! By normalization reasons, it
29
∞
is appropriate to choose R(x) ≥ 0 such that             −∞
R(x)dx = 1, e.g.
1            x2
R(x) = √ exp − 2 .
σ 2π          2σ
Notice that, if σ is very small, then the function f (x) is reproduced
by f ∗ R even in details (Figure 11, where σ = 0.01) while for larger
(but not too large) values of σ the function f (x) is reproduced without
relevant perturbations (for example, in Figure 12 σ = 0.5 was chosen).

Thm. 7.1 (Convolution theorem in time domain, convolution theorem
in frequency domain)
If the functions f1 , f2 are suﬃciently regular 3 then
(51)                                   ˆ        ˆ
F (f1 ∗ f2 )(ω) = f1 (ω) · f2 (ω)

(52)                         ˆ ˆ
F −1 (f1 ∗ f2 )(t) = (f1 ∗ f2 )(t).
Scheme of proof.
Let us consider the fourier transfor of a convolution product:
+∞              +∞
F (f1 ∗ f2 )(ω) =           e−iωt dt        f1 (x)f2 (t − x)dx
−∞                −∞
+∞     +∞
=                 e−iωt f1 (x)f2 (t − x)dt dx
−∞     −∞
By the change of variables (t, x) → (y, x), where x = x and y = t − x,
we obtain
+∞      +∞
F (f1 ∗ f2 )(ω) =                  e−iω(x+y) f1 (x)f2 (y)dy dx
−∞      −∞
+∞                        +∞
=          e−iωx f1 (x) dx                             ˆ        ˆ
e−iωy f2 (y) dy = f1 (ω) · f2 (ω)
−∞                       −∞
and vice-versa
ˆ ˆ
F −1 f1 · f2 (t) = (f1 ∗ f2 ) (t) △
Remark 7C (The Fourier transform applied to the diﬀusion equation

on a line )
Consider the heat equation with initial condidion f (x) for ∈ R:
∂u    ∂2u
= k 2 u(x, 0) = f (x), −∞ < x < +∞|u(x, t)| < M, −∞ < x < +∞, t > 0
∂t    ∂x
3A  generic assumption of regularity so as to ensure the existence of the Fourier
transforms, the change in the order of integration, etc.
30

In this problem it is useful to consider the Fourier transform of both
sides with respect to x, so that the second order derivative (with respect
to x) is transformed into a multiplication by the square of the conjugate
variable (Thm. 6.5). Denoting p the Fourier variable conjugate with
x, we obtain:
dF (u)
= −kp2 · F (u)
dt
where the Fourier transform of u is denoted by F (u). Here we deal with
a ﬁrst order ordinary diﬀerential equation, since the unknown function
F (u) no more depends on x. The general solution of such ordinary
equation is:
2t
(53)                     F [u(x, t)] = Ce−kp
Setting t = 0 in (54) we see that
F [u(x, 0)] = F [f (x)] = C
so that
2
F [u] = F [f ] · e−kp t .
Now we can apply the convolution theorem (Thm. 7.1):
2
u(x, t) = f (x) ∗ F −1 [e−kp t ]
From (47) we know that
2     π −p2 /4α            1        2               2
F [e−αx ] =   e       , or F [ √     · e−x /(4kt) ] = e−kp t .
α                   4πkt
Then the solution is the convolution product
+∞
1        2                       1          2
u(x, t) = f (x) ∗   √      e−x /4kt =          f (z) √      e−(x−z) /4kt dz.
4πkt               −∞            4πkt
Notice that in this formula the well known ”heat kernel” appears:
1      2
G(z; t) = √      e−z /4kt
4πkt
which coincides with a Gaussian probability density with mean zero
and variance 2kt. So we have three facts: a) the diﬀusion equation on
the line with initial condition u(x, 0) = f (x) is solvable with a known
explicit kernel:
+∞
u(x, t) =        f (z)G(x − z; t)dz.
−∞

b) The physical interpretation of u(x, t) as the temperature at time t
suggests that a choice of a compact support initial temperature f (x)
gives rise to a temperature which is at once everywhere nonzero (!):
u(x, t) = 0 ∀x ∈ R, ∀t > 0. An instantaneous diﬀusion to all points
of space takes place. c) Setting σ 2 = 2kt, the convolution of any
1    2   2
function f (x) with the probability density σ√2π e−z /2σ is convergent
31

u(x,t) per t=0.01,t=1,t=10

1.0
t=0.01   t=1
0.8

t=10
0.6
u(x,t)

0.4
0.2
0.0

−10       −5         0           5           10     15       20

x

Figure 14. Graph of the solution of the heat equation,
starting from an initial function u(x, 0) with compact sup-
port, the indicator of (0, 10). Its diﬀusion is represented at
times t = 0.01, t = 1, t = 10.

to f as σ → 0. This has a probabilistic interpretation: if f (x) is
normalized so as to be the density of a random variable X, and if X,
1     2  2
Zσ are supposed independent, then f ∗ σ√2π e−z /2σ is the density of
the sum X + Zσ , where Zσ ∼ N(0; σ 2 ). X + Zσ is just a disturbed
version of the random variable X, and converges in distribution to X
as σ → 0, just as Zσ converges in distribution to the constant 0.

8. The Impulse δ(t), or Dirac’s Delta
The so called impulse ”function”, or ”Dirac’s Delta” is an important

tool in Fourier analysis. Such so called ”function” is usually introduced
as a limit (in a sense to be explained) of a sequence fn (t) of functions
such that
∞
(54) fn (t) ≥ 0,              fn (t)dt = 1 and          lim fn (t) = 0, ∀t = 0.
−∞                             n→∞

For example the indicator functions in smaller and smaller intervals
can be chosen:
n           n        1 1
fn (t) = p1/n (t) = , t ∈ [− , ]0, elsewhere
2          2        n n
Gaussian density functions
√
fn (t) = n2π exp −nt2 /2
32
√
with mean µ = 0 and standard deviation σ = 1/ n can be chosen, too.
Now, if the ”limit” of the fn ’s was understood in the usual sense, the
limit ”function” would be vanishing ∀t = 0 and such that f (0) = +∞.
Therefore a diﬀerent meaning is needed by the limit underlying (54).

Def. I.8A - (Dirac’s Delta) Let the sequence of functions {fn (t)}n∈N
satisfy the properties (54). Let ϕ(t) be any continuous function in t = 0
(sometimes called ”test function”). Dirac’s Delta is the functional
acting on any test function ϕ by the correspondence:
+∞
(55)                ϕ → lim            fn (t)ϕ(t) dt
n→∞    −∞

Dirac’s Delta is represented by the symbol δ(t) inserted in a (only for-
mal) integral notation:
+∞                           +∞
(56)               δ(t)ϕ(t)dt := lim            fn (t)ϕ(t) dt
−∞                    n→∞    −∞

Remark I.8B
1) It can be shown that the deﬁnition (56) is well-posed, i.e. the limit
exists and, moreover, it is independent of the sequence fn , provided it
satisﬁes (55). Hence, for example, we can perform computations by use
of the simple sequence n p1/n t) of indicator functions.
2
2) According this deﬁnition, Dirac’s Delta is a functional acting on
test functions ϕ. It is clear that the symbol δ(t) is not a function, but
has the merit to keep an integral-type notation, which recalls the ap-
proximation rule (56): for example, it is usual to say that the functions
fn ’s, when satisfying (55), are ”approximants of Dirac’s delta”.
Moreover, such a notation reminds the main properties of integrals,
above all linearity: Dirac’s Delta belongs to a wide class of objects
called ”linear functionals”, which properly extends the class of integral
operators of the type ϕ → f (t)ϕ(t)dt, where f is a ﬁxed function.

Thm. I.8.1 Let ϕ be a continuous function in the origin. Then
∞
(57)                         δ(t)ϕ(t) dt = ϕ(0)
−∞

PROOF
Let us prove the assertion for two types of probability densities, the
uniform and the normal one: similar ideas can be adapted to any other
sequence of densities satisfying (54). Let fn (t) = n p1/n (t). From Def.
2
8A it follows that
1
+∞                           +n
n
δ(t)ϕ(t) dt = lim            ϕ(t)dt = ϕ(0)
−∞                  n→∞ 2     1
−n
33

Indeed
1
+n
n
min1 ϕ(t) = mn ≤                     ϕ(t)dt ≤ Mn = max1 ϕ(t)
1
[− n ,+ n ]        2             1
−n
1
[− n ,+ n ]

where, by continuity in 0,
lim mn = ϕ(0),           lim Mn = ϕ(0),
n→∞                       n→+∞

whence the formula in the ﬁrst case. In fn is the Gaussian probability
2
density with mean zero ad variance σn = 1/n, for any ǫ > 0 one can
restrict the integral to the interval [−σn /ǫ, +σn /ǫ], where
+σn /ǫ
mn · α(ǫ) ≡ mn                    fn (t)dt
−σn /ǫ
+σn /ǫ                              +σn /ǫ
≤             fn (t)ϕ(t)dt ≤ Mn                   fn (t)dt ≡ Mn · α(ǫ)
−σn /ǫ                              −σn /ǫ
As above, mn , Mn are convergent to ϕ(0) as n → ∞ by continuity of ϕ.
On the other hand the integral of fn between extremes proportional to
the standard deviation σn is independent of n, so that α only depends
on ǫ. Moreover α(ǫ) tends to 1, the total probability, as ǫ → 0. Hence
ϕ(0) is the limit as n → ∞. △
Remark I.8C The translated Dirac’s Delta δa (t) ≡ δ(t − a) can be
deﬁned as the functional
+∞                                 +∞
δa (t)ϕ(t)dt = lim                 fn (t − a)ϕ(t) dt
−∞                          n→∞     −∞
By Thm. I.8.1 and a substitution t − a = τ, the action of δa on a test
function ϕ is
+∞
δa (t)ϕ(t)dt = ϕ(a)
−∞
if ϕ is continuous in a.
Remark I.8D Not only probability densities satisfying (54), but some
further classes of functions are good approximants of δ(t), for example
the sequence of functions
sin(nt)                    n
fn (t) =         , ∀t = 0, fn (0) =
πt                      π
even if they don’t satisfy fn ≥ 0, and limn→∞ fn (t) = 0 for any t = 0.
∞
Indeed one can prove that −∞ sin(nt) dt = 1 and that
πt
∞
sin(nt)
lim                  ϕ(t)dt = ϕ(0).
n→∞     −∞      πt
Another approximant sequence of δ(t) is
e−int
(58)                    fn (t) = i         ,    as n → ∞.
πt
34

Remark I.8E Dirac’s Delta is not a function, but a linear functional
deﬁned as the limit of integrals (56). By this reason Dirac’s Delta
belongs to a wide class of objects called linear functionals or distribu-
tions, whose theory was an outstanding achievement of Mathematical
Analysis and Physics in the 20-th century. In particular, δ is the
distribution assuming the value

δ(t)ϕ(t)dt = ϕ(0)
R
on any test function ϕ continuous in 0. With an abuse of language, it
is often said that δ is a generalized function such that δ(t) = 0, ∀t = 0,
δ(0) = +∞, and R δ(t)dt = 1 : a way to say that δ(t) it is an
idealization, or model, of any impulse , any physical or engineering
phaenomenon with relevant intensity in a very small time interval.

To conclude, let us see some Fourier transforms involvng Dirac’s Delta.
Aa an immediate cosequence of the deﬁnition
F (δ) = 1,        F −1 1) = δ.
By virtue of the symmetry property
F (1) = 2πδ(ω),           F −1 [2πδ(ω)] = 1.
From such results and from the time shifting property another conse-
quence is
F (δa ) = e−iωa and F −1 (e−iωa = δa (t) = δ(t − a).

Remark I.8F Let us compute the Fourier transorm of the function
cos(ω◦ t). From Euler’s formulas, from linearity property, and from the
above transforms we obtain
1             1
F [cos(ω◦ t)] = F (eiω◦ t ) + F (e−iω◦ t )
2             2
= π[δ(ω − ω◦ ) + δ(ω + ω◦ )].

Remark I.8G Compute the Fourier transform of the Heaviside func-
tion H(t)
H(t) = I[0,∞) (t) = 0, t < 0 + 1 t ≥ 0.
Notice that
+∞                                     +R
F (H) =          H(t)e−iωt dt =         lim             H(t)e−iωt dt
−∞                          R→+∞     −R
R                                  R
−iωt                   e−iωt
= lim               e      dt = lim
R→+∞     0                  R→+∞        −iω     0
1
=    + lim fR (ω)
iω R→+∞
35

where fR (ω) = ω e−iωR . By (58), since i e πt converges to δ(t) as n → ∞
i                          −int

in the distributional sense,

1
F (H) = πδ(ω) +           .
iω

Remark I.8H As an application of the above example, given the
ˆ
Fourier transform f(ω) of f (t), let us compute the Fourier transform
t
ˆ
g (ω) of g(t) = −∞ f (τ )dτ. The answer was found

ˆ
f(ω)
ˆ
g (ω) =
iω

+∞
under the assumption that −∞ f (τ )dτ = 0. Now we look for a more
general expression. We notice that

t                +∞
g(t) = g(t) =        f (τ )dτ =         f (τ )H(t − τ )dτ = (f ∗ H)(t)
−∞                −∞

Hence, by the convolution product theorem

ˆ       ˆ      ˆ               1
ˆ
g (ω) = f (ω) · H(ω) = f(ω) · πδ(ω) +
iω

ˆ
f (ω)
ˆ
= f (ω)πδ(ω) +
iω

ˆ          ˆ
where f(ω)δ(ω) = f (0)δ(ω) and

+∞
ˆ
f(0) =          f (t)dt.
−∞
36

f (t)                                                  ˆ
f(ω)

c1 f1 (t) + c2 f2 (t)                                     ˆ           ˆ
c1 f1 (ω) + c2 f2 (ω)

f ′ (t)                                                   ˆ
iω f(ω)

f (n) (t)                                                    ˆ
(iω)n f (ω)
n   ˆ
tn f (t)                                                     f
in d dω(ω)
n

f (t − t◦ )                                                   ˆ
e−iωt◦ f(ω)
eiω◦ t f (t)                                           ˆ
f(ω − ω◦ )
(f1 ∗ f2 )(t)                                          ˆ        ˆ
f1 (ω) · f2 (ω)
1 ˆ          ˆ
f1 · f2                                                  (f
2π 1
∗ f2 )(ω)
pa (t) = I(−a,a) (t)                                   2 sin(aω)/ω
sin(ω◦ t)/t                                            ˆ
f(ω) = I(−ω◦ ,ω◦ ) (ω)
2                                                  π −ω 2 /(4α)
e−αt                                                       α
e
δ(t)                                                   1
1                                                      2πδ(ω)
δ(t − a)                                               e−iωa
cos(ω◦ t)                                              π[δ(ω − ω◦ ) + δ(ω + ω◦ )]
H(t) = I(0,∞) (t)                                      πδ(t) + 1/(iω)
pa (t + 2a) + pa (t − 2a)                              4 sin(ωa) cos(2ωa)/ω
pa (t + 2a) − pa (t − 2a)                              4i sin(ωa) sin(2ωa)/ω
1 ˆ ω
f (at)                                                 |a|
f(a )
1                         1                         2
qa (t) = a (x + a) · I(−a,0) (t) − a (x − a) · I(0,a) (t) 4 sin (ωa/2)/(aω 2)
sin2 at/(πat2 )                                        q2a (ω)
t                                                     ˆ           ˆ
−∞
f (τ )dτ                                     f(0)πδ(ω) + f(ω)/(iω)

```
DOCUMENT INFO
Shared By:
Categories:
Stats:
 views: 65 posted: 5/30/2010 language: English pages: 36