Powers D.L. Boundary Value Problems and Partial Differential Equations
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108 Chapter 1 Fourier Series and Integrals
Fourier Integral Representation Theorem. Let f (x) be sectionally smooth on
every finite interval, and let
∞
−∞
|f (x)|dx be finite. Then at every point x,
∞
0
A(λ) cos(λx) +B(λ) sin(λx)
dλ =
1
2
f (x+) +f (x−)
,
−∞< x < ∞, (8)
where
A(λ) =
1
π
∞
−∞
f (x ) cos(λx) dx, B(λ) =
1
π
∞
−∞
f (x ) sin(λx) dx. (9)
Equation (8) is called the Fourier integral representation of f (x); A(λ) and B(λ)
in Eq. (9) are the Fourier integral coefficient functions of f (x). The right-hand
sideofEq.(8)wasalsoseenintheFourierseriesconvergencetheorem.Since
f (x ) is sectionally smooth, the expression in Eq. (8) is the same as f (x) almost
everywhere, so we often write Eq. (7) instead of Eq. (8).
Example 2.
The function
f (x) =
1, |x| < 1,
0, |x| > 1
has the Fourier integral coefficient functions
A(λ) =
1
π
∞
−∞
f (x) cos(λx) dx =
1
π
1
−1
cos(λx) dx =
2sin(λ)
πλ
,
B(λ) = 0.
Since f (x) is sectionally smooth, the Fourier integral representation is legit-
imate, and we write
f (x) =
∞
0
2sin(λ)
πλ
cos(λx) dλ.
(Actually the integral equals
1
2
at x =±1, so equality is not strictly correct at
these two points.)
Example 3.
Find the Fourier integral representation of f (x) = exp(−|x|).
Solution: Direct integration gives
A(λ) =
1
π
∞
−∞
exp
−|x|
cos(λx) dx, (10)
1.9 Fourier Integral 109
A(λ) =
2
π
∞
0
e
−x
cos(λx) dx, (11)
A(λ) =
2
π
e
−x
(−cos(λx) + λ sin(λx))
1 +λ
2
∞
0
=
2
π
1
1 +λ
2
. (12)
B(λ) = 0, because exp(−|x|) is even. Since exp(−|x|) is continuous and sec-
tionally smooth, we may write
exp(−|x|) =
2
π
∞
0
cos(λx)
1 +λ
2
dλ, −∞ < x < ∞.
These two examples illustrate the fact that, in general, one cannot evaluate
the integral in the Fourier integral representation. It is the theorem stated in
the preceding that allows us to write the equality between a suitable function
and its Fourier integral.
If f (x) is defined only in the interval 0 < x < ∞, one can construct an even
or odd extension whose Fourier integral contains only cos(λx) or sin(λx).
These are called the Fourier cosine and sine integral representations of f ,re-
spectively.
Let f (x ) be defined and sectionally smooth for 0 < x < ∞,andlet
∞
0
|f (x)|dx < ∞.Thenwewrite:
Fourier cosine integral representation
f (x) =
∞
0
A(λ) cos(λx ) dλ, 0 < x < ∞
with A(λ) =
2
π
∞
0
f (x) cos(λx) dx,
Fourier sine integral representation
f (x) =
∞
0
B(λ) sin(λx) dλ, 0 < x < ∞
with B(λ) =
2
π
∞
0
f (x ) sin(λx) dx.
Example 4.
Find the Fourier sine and cosine integral representations of f (x) given for 0 < x
by
f (x) =
sin(x), 0 < x <π,
0,π<x.
110 Chapter 1 Fourier Series and Integrals
Since f (x) = 0 for x >π, the integral for B(λ) reduces to one over the interval
0 < x <π:
B(λ) =
2
π
∞
0
f (x) sin(λx) dx =
2
π
∞
0
sin(x) sin(λx) dx
=
2
π
sin((λ −1)x)
2(λ −1)
−
sin((λ +1)x)
2(λ +1)
π
0
=
2
π
sin((λ −1)π )
2(λ −1)
−
sin((λ +1)π )
2(λ +1)
.
This expression can be simplified by using the fact that
sin
(λ ±1)π
=sin(λπ ±π)=−sin(λπ).
Then, creating a common denominator, we obtain
B(λ) =
−2sin(λπ )
π(λ
2
−1)
.
Hence the Fourier sine integral representation of f (x) is
f (x) =
∞
0
−2sin(λπ )
π(λ
2
−1)
sin(λx) dλ, 0 < x.
Since f (x) is continuous for 0 < x, the equality holds at every point.
Similarly, we can compute the cosine coefficient function
A(λ) =
−2(1 +cos(λπ ))
π(λ
2
−1)
,
and the cosine integral representation of f (x) is
f (x) =
∞
0
−2(1 +cos(λπ ))
π(λ
2
−1)
cos(λx) dλ, 0 < x.
Note that both A(λ) and B(λ) have removable discontinuities at λ = 1.
It seems to be a rule of thumb that if the Fourier coefficient functions A(λ)
and B(λ) can be found in closed form for some function f (x), then the inte-
gral in the Fourier integral representation cannot be carried out by elementary
means, and vice versa. (See Exercise 3.)
Rules for operations on Fourier integrals generally follow the lines men-
tioned in Section 1.5 for Fourier series. In particular: If f (x) is continuous and
if both f (x) and f
(x) have Fourier integral representations, then
f (x ) =
∞
0
A(λ) cos(λx) +B(λ) sin(λx)
dλ,
1.9 Fourier Integral 111
f
(x) =
∞
0
−λA(λ) sin(λx) +λB(λ) cos(λx)
dλ.
Example 5.
Let f (x) = exp(−|x|) as in Example 3. Then its derivative is
f
(x) =
−e
−x
, 0 < x,
e
x
, x < 0.
Clearly, f (x) is continuous, and both f (x) and f
(x) have Fourier integral rep-
resentations. The one for f (x) is in Example 3. Thus we have
f
(x) =
∞
0
2
π
−λ
1 +λ
2
sin(λx) dλ.
EXERCISES
1. Sketch the even and odd extensions of each of the following functions, and
find the Fourier cosine and sine integrals for f .Eachfunctionisgivenin
the interval 0 < x < ∞.
a. f (x) = e
−x
;
b. f (x) =
1, 0 < x < 1,
0, 1 < x;
c. f (x) =
π −x, 0 < x <π,
0,π<x.
2. Find the Fourier integral representation of the following function f
t
(x).
This is sometimes called a “window” because it is “open” for t − h < x <
t +h.
f
t
(x) =
1, |x −t|< h,
0, |x −t|> h.
3. Find the Fourier integral representation of each of the following functions.
a. f (x) =
1
1 +x
2
;
b. f (x) =
sin(x)
x
.
(Hint: To evaluate the Fourier integral coefficient functions, consult the
Fourier integral representations found in the examples.)
4. In Exercise 3b, the integral
∞
−∞
|f (x)|dx is not finite. Nevertheless, A(λ)
and B(λ) do exist (B(λ) = 0). Find a rationale in the convergence theo-
rem for saying that this function can be represented by its Fourier integral.
(Hint: See Example 1.)
5. Find the Fourier integral representation of each of these functions:
112 Chapter 1 Fourier Series and Integrals
a.
f (x ) =
sin(x), −π<x <π,
0, |x|>π;
b. f (x) =
sin(x), 0 < x <π,
0, otherwise;
c. f (x) =
|sin(x)|, −π<x <π,
0, otherwise.
6. Show that if k and K are positive, then the following are true:
a.
∞
0
e
−kx
sin(x) dx =
1
1 +k
2
;
b.
∞
0
1 −e
−Kx
x
sin(x) dx = tan
−1
(K);
c.
∞
0
sin(x)
x
dx =
π
2
.
(Part (a) by direct integration, (b) by integration of (a) with respect to k
over the interval 0 to K,(c)bylimitof(b)asK →∞.)
7. Starting from Exercise 6c, show that
∞
0
sin(λz)
λ
dλ =
π/2, 0 < z,
0, z = 0,
−π/2, z < 0.
Is this the Fourier integral of some function?
8. Change the variable of integration in the formulas for A and B, and justify
each step of the following string of equalities. (Do not worry about chang-
ing order of integration.)
f (x) =
1
π
∞
0
∞
−∞
f (t)
cos(λt) cos(λx) +sin(λt) sin(λx)
dt dλ
=
1
π
∞
−∞
f (t)
∞
0
cos
λ(t −x)
dλ dt
=
1
π
∞
−∞
f (t)
lim
ω→∞
sin(ω(t −x))
t −x
dt
= lim
ω→∞
1
π
∞
−∞
f (t)
sin(ω(t −x))
t −x
dt.
The last integral is called Fourier’s single integral.Sketchthefunction
sin(ωv)/v as a function of v for several values of ω. What happens near
1.10 Complex Methods 113
v = 0? Sometimes notation is compressed and, instead of the last line, we
write
f (x ) =
∞
−∞
f (t)δ(t −x ) dt.
Although δ is not, strictly speaking, a function, it is called Dirac’s delta func-
tion.
1.10 Complex Methods
Fourier series
Suppose that a function f (x) equals its Fourier series
f (x ) = a
0
+
∞
n=1
a
n
cos(nx) + b
n
sin(nx).
(We use period 2π for simplicity only.) A famous formula of Euler states that
e
iθ
=cos(θ) + i sin(θ), where i
2
=−1.
Some simple algebra then gives the exponential definitions of the sine and
cosine:
cos(θ) =
1
2
e
iθ
+e
−iθ
, sin(θ) =
1
2i
e
iθ
−e
−iθ
.
By substituting the exponential forms into the Fourier series of f we arrive at
the alternate form
f (x ) = a
0
+
1
2
∞
n=1
a
n
e
inx
+e
−inx
−ib
n
e
inx
−e
−inx
= a
0
+
1
2
∞
n=1
(a
n
−ib
n
)e
inx
+(a
n
+ib
n
)e
−inx
.
We a re n ow l ed t o de fine complex Fourier coefficients for f :
c
0
=a
0
, c
n
=
1
2
(a
n
−ib
n
), c
−n
=
1
2
(a
n
+ib
n
), n =1, 2, 3,....
In terms of these two coefficients, we have
f (x) = c
0
+
∞
n=1
c
n
e
inx
+c
−n
e
−inx
=
∞
−∞
c
n
e
inx
. (1)
114 Chapter 1 Fourier Series and Integrals
This is the complex form of the Fourier series for f .Itiseasytoderivethe
universal formula
c
n
=
1
2π
π
−π
f (x)e
−inx
dx, (2)
which is valid for all integers n, positive, negative, or zero. The complex form is
used especially in physics and electrical engineering. Sometimes the function
corresponding to a Fourier series can be recognized by use of the complex
form.
Example.
The series
∞
n=1
(−1)
n+1
n
cos(nx)
may be considered the real part of
∞
n=1
(−1)
n+1
n
e
inx
=
∞
n=1
(−1)
n+1
n
e
ix
n
(3)
because the real part of e
iθ
is cos(θ). The series on the right in Eq. (3) is recog-
nized as a Taylor series,
∞
n=1
(−1)
n+1
n
e
ix
n
=ln
1 +e
ix
.
Some manipulations yield
1 +e
ix
= e
ix/2
e
ix/2
+e
−ix/2
=2e
ix/2
cos
x
2
,
ln
1 +e
ix
=
ix
2
+ln
2cos
x
2
.
The real part of ln(1+e
ix
) is ln(2cos(x/2)) when −π<x <π.Thus,wederive
the relation
ln
2cos
x
2
∼
∞
n=1
(−1)
n+1
n
cos(nx), −π<x <π. (4)
(The series actually converges except at x =±π, ±3π,....)
1.10 Complex Methods 115
Fourier integral
The Fourier integral of a function f (x) defined in the entire interval −∞ <
x < ∞ can also be cast in complex form:
f (x) =
∞
−∞
C(λ)e
iλx
dλ. (5)
The complex Fourier integral coefficient function is given by
C(λ) =
1
2π
∞
−∞
f (x )e
−iλx
dx. (6)
It is simple to show that
C(λ) =
1
2
A(λ) −iB(λ)
, (7)
where A and B are the usual Fourier integral coefficients. The complex Fourier
integral coefficient is often called the Fourier transform of the function f (x).
Example.
Find the complex Fourier integral representation of
f (x ) =
1, −a < x < a,
0, x < |a|.
The coefficient function (or transform) of f is
C(λ) =
1
2π
a
−a
e
−iλx
dx =
1
2π
e
−iλx
−iλ
a
−a
=
1
2π
e
iλa
−e
−iλa
iλ
=
sin(λa)
πλ
.
The representation of f is
f (x) =
∞
−∞
sin(λa)
πλ
e
iλx
dλ, −∞ < x < ∞.
Of course, at x =±a,theintegralconvergesto1/2.
This example brings out a fact about symmetry: If f (x) is even, C(λ) is real;
if f (x) is odd, C(λ) is imaginary.
The Fourier integral or transform may be used to solve differential equations
on the interval −∞ < x < ∞, in much the same way that Laplace transform
is used.
116 Chapter 1 Fourier Series and Integrals
EXERCISES
1. Use the complex form
a
n
−ib
n
=
1
π
π
−π
f (x )e
−inx
dx, n = 0,
to find the Fourier series of the function
f (x) = e
αx
, −π<x <π.
2. Find the complex Fourier series for the “square wave” with period 2π :
f (x ) =
1, 0 < x <π,
−1, −π<x < 0.
3. Find the complex Fourier integral representation of the following func-
tions:
a. f (x) =
e
−x
, x > 0,
0, x < 0;
b. f (x) =
sin(x), 0 < x <π,
0, elsewhere.
4. Find the complex Fourier integral for
a. f (x) =
xe
−x
, 0 < x,
0, x < 0;
b. f (x) = e
−α|x|
sin(x).
5. Relate the functions and series that follow by using complex form and Tay-
lor series.
a. 1 +
∞
n=1
r
n
cos(nx) =
1 −r cos(x)
1 −2r cos(x) + r
2
, 0 ≤r < 1;
b.
∞
n=1
sin(nx)
n!
=e
cos(x)
sin
sin(x)
.
6. Show by integrating that
π
−π
e
inx
e
−imx
dx =
0, n = m,
2π, n = m,
and develop the formula for the complex Fourier coefficients using this idea
of orthogonality.
1.11 Applications of Fourier Series and Integrals 117
7. Find the function f (x) whose complex Fourier coefficient function is
given.
a. C(λ) =
1, −1 <λ<1,
0, otherwise;
b. C(λ) = e
−|λ|
.
8. Show that the complex Fourier coefficient of f (x) = e
−x
2
is
C(λ) =
e
−λ
2
/4
2
√
π
.
Use a change of variable in the exponent. You need to know that
∞
−∞
e
−z
2
dz =
√
π.
1.11 Applications of Fourier Series and Integrals
Fourier series and integrals are among the most basic tools of applied mathe-
matics. In what follows, we give just a few applications that do not fall within
the scope of the rest of this book.
A. Nonhomogeneous Differential Equation
Many mechanical and electrical systems may be described by the differential
equation
¨y +α˙y + βy = f (t).
The function f (t) is called the “forcing function,” βy the “restoring term,” and
α˙y the “damping term.” It is known (see Section 0.2) that: (1) a sine or cosine
in f (t) will cause functions of the same period in y(t);(2)iff (t) is broken
down as a sum of simpler functions, y(t) canbebrokendowninthesameway.
Suppose that f (t) is periodic with period 2π , and let its Fourier series be
f (t) =a
0
+
∞
n=1
a
n
cos(nt) + b
n
sin(nt).
Then a particular solution y(t) will be periodic with period 2π; it and its deriv-
atives have Fourier series
y(t) = A
0
+
∞
n=1
A
n
cos(nt) + B
n
sin(nt),