FUNCTIONS ON THE CIRCLE
Chapter
Q
(FOURIER ANALYSIS)
chapter we shall study periodic functions of a real variable. The
importance of such functions derives from the fact that many natural and
physical phenomena are oscillatory, or recurrent. In the early 19th century,
J. B. J. Fourier laid down the foundations of the study of periodic functions
in his treatise Analytic Theory of Heat. There remained a few gaps and
difficulties in Fourier's theory and much mathematical energy during the
The invention of
19th century was expended in the study of these problems.
Lebesgue's theory of integration in the early 20th century finally laid the
foundations to this theory. Our exposition will not follow this chrono
logical pattern; but rather will try to develop the way of thinking about
Fourier series which emerged during the late 19th century.
A periodic function is one whose behavior is recurrent.
That is, there
is a certain number L, called the period of the function, such that the function
repeats itself over every interval of length L,
In this
f(x
From
our
functions
+
L)= f(x)
for all
point of view (which
begins by discarding
x 6
R
is very much a posteriori) the study of periodic
the notion of periodicity in favor of a change
in the geometry of the domain. That is, to
a fixed period, we make the
functions with
We shall think of the real line
functions
are
just
the functions
452
as
on
study the collection of all periodic
underlying space periodic instead.
wound around a circle, and our periodic
the circle.
6.1
To fix the
complex
mapping
we
9
->
9 + / sin 9
cos
continuously
In)
+
=
In the past few
fix)
chapters
on
on
R which
are
a
Y which is
points
go onto
want: It winds the
function of eiB which
periodic
of
on
period
Y
are
2n
pre
:
forallxeP
we
have been
studying
of view of differentiation.
from the
Y is
onto
end
Thus the continuous functions
the continuous functions
fix
eie of the real numbers
A continuous function
with 9.
453
particular circle in mind: the set Y of
We have already seen that there is a
length 2n, except that both
This mapping does precisely what we
real line around Y.
cisely
=
a
one.
interval of
on an
point.
same
varies
shall have
numbers of modulus
one-to-one
the
ideas,
Approximation by Trigonometric Polynomials
the behavior of functions
point
Taylor
an expansion into polynomials, and we have related the co
efficients to the subsequent derivatives of the function. Since the simplest
periodic functions are the trigonometric polynomials, we attempt to expand
a given periodic function in a series of trigonometric polynomials.
This
is the so-called Fourier series of the function. The interesting fact here
is that the relevant coefficients are found by integration. In fact, as we shall
see, the Fourier series of a function is a sort of an expansion in terms of an
We have studied the
expansion,
orthonormal basis in the vector space of continuous functions
with the inner product
<f,g>
=
on
the circle
^zfjiOM8)d9
Finally, as the circle is the set of complex numbers of modulus one, it is the
boundary of the unit disk in C and we can study the relation between Taylor
expansions in the disk and the Fourier expansions on the circle for suitable
It will turn out that for such functions the
functions.
can
6.1
also be obtained
by integration
on
Taylor coefficients
the circle.
Approximation by Trigonometric Polynomials
We shall
functions
simplest
begin
on
with the attitude that
the circle.
According
we
are
studying complex-valued
view, the function e'e is the
This attitude is really just a con
to this
and the most basic function.
venience; the point of view of strictly real-valued functions would consign us
to consider cos 9, sin 9 as the elementary building blocks of our theory.
454
6
Functions
But, since eie
more
cos
=
the Circle
on
(Fourier Analysis)
9 + i sin 9, there is little difference, and
we
select the
comfortable notation.
Our purpose is to describe a given function on the circle in terms of the
More precisely, if the series
powers of e', both positive and negative.
CO
aj""
71=
(6.1)
00
We ask the
converges for all 9, it defines a function on the circle.
question: Can we express any periodic function as such a series?
converse
If
only
many of the {an} in (6.1) are nonzero, there is no problem of con
vergence, and the sum defines a function, called a trigonometric polynomial.
finitely
This
we
the
subject gets off the ground once
given function, and that leads us
to our first
Proposition
Let
1.
P(9)
=
know how to compute the
^=_N anel"e
be
{an}
from
proposition.
trigonometric polynomial.
a
Then
d9
2n J-n
for
all
m.
Proof.
1
"
2
c"
27T,n=-N
a
J
e-<> dd
-
1
=
2tt
Now, given
a
series of
a
am
2tt + 0
continuous function
on
trigonometric polynomials,
=
am
the circle, if it has an expansion into
could expect that the coefficients
we
of this series will be related to the function in the
same
way.
Thus
we
form
this definition.
Definition 1.
Let
/
be
a
continuous function
on
the circle.
The nth
Fourier coefficient of /is
A
\in)=l-f JiSin*dcp
(6.2)
6.1
The Fourier series
Approximation by Trigonometric Polynomials
of/ is
455
the series
CO
fin)eM
n
=
(6.3)
oo
Examples
1.
sin
0
Let/(0)
=
sin 9.
Since
=
2i
its Fourier series is
2i
2i
From
(6.2)
f"
deduce
we can
sin 9 eie d9
=
2nJ-n
2. Since
3.
/()
cos
Let/(0)
=
1 f"
=
=
cos2
=
=
{0
n
easily computed):
sin 9 e~ie d9
i(e'm9
+
=i
2i
2nJ-n
e~im6),
the Fourier series of
cos
md is
Then
0e-'"* <ty
=
-/- f (1
47t J
+
cos
2^)e-'"* d<p
-n
-2,2
=
U
- f"
^
2i
cos2 9.
2% J -n
a
()
m0
is also
(as
o
#
-
2, 0, 2
Thus the Fourier series of cos2 0 is
e-i29
+
(Notice
i
+
iei2e
that cos2 9
=
1/2(1
trigonometric polynomial.)
+ cos
29)
=
1/2(1
+
l/2(ew
+
e~iB))
is
a
456
6
Functions
4.
Let/(0)
/(")
=
77-2
the Circle
n2
^f i*2
=
/(0)
on
iFourier Analysis)
02.
-
<P2)e-in* dcp
~
1
71
9tt^
n
^L#~^J_/^
=
/(n)=-^-f 4>V^rf</>
=
2nJ-K
by
"
T"
(-!)"
4
=
0
n*0
n
Thus the Fourier series of
integrations by parts.
two
=
n
9
^ 2^>
(6.4)
+
3
n
n*o
Notice that by the
is
comparison test,
this series does converge to
a
continuous function of e,e :
2n2
(ew)n
}
^
+
3
2(-l)^
n#o
n#0
n
given function 7t2
In order to conclude that this is the
need
more
theoretical
5. It is not necessary for
expansion. It need
a
Fourier
(6.2), (6.3)
m~
to be
0
=
Let
computable.
us
=
shall
-^-e
inQ
2mn
0
o
n
n
n in
27rm
even,
odd
the
a
expressions
compute the Fourier series of
0<O
^-fe"'U<j>
2n Jo
we
function to be continuous to have
only be integrable for
m=hf0d^\
!in)
02,
investigations.
n
le~im
^ 0
-
1]
6.1
Approximation by Trigonometric Polynomials
Thus the Fourier series
1
457
of/ is
einB
1
2+niJ^
n
Recapturing the
Notice that
^
odd
Function from Its Fourier Series
claim of convergence in Definition 1 is made.
In particular,
not
to
for
the
test
does not
appears
converge,
comparison
no
the series
(6.5)
apply. However, we cannot conclude that convergence fails; only that the
question can be exceedingly difficult. We ask instead what appears to be a
simpler question: Does the Fourier series identify the given function, and if
We now try to investigate the recapture of a function by its
so, in what way ?
Fourier series, deliberately leaving aside all questions of convergence.
Let / be a given integrable function on the circle and consider the
"function"
AnV
9i0)=
n=
oo
By definition of f(n),
1
oo
0(0)=
oo
n=
Now
we
t\
*-ft J
interchange
9iO)
Well, it is
^-\
=
L%
<j
n
fit)**-''" d*
ti
and
J",
e'"Wd<l>
fit)
%
obtaining
r\
co
too bad it turned out this way because
problem,
convergence
like it
or
not.
In
fact,
we are
still up
the situation is
against
worse:
a
it is
untenable because
.'(-*)
(6.6)
e
seemingly insurmountable obstacle can
be overcome, so long as we are not solely interested in pointwise convergence,
by a subtle mathematical technique: that of inserting convergence factors.
converges for
no
values of (b.
This
458
If
6
we
Functions
replace
on
the series
the Circle
(Fourier Analysis)
the series
(6.6) by
2 /-|"lein(('-*)
n
=
(6.7)
oo
this series converges beautifully for r < 1 and the series (6.6) is in some ideal
sense the limit of (6.7) as r tends to 1.
Stepping backward two steps, this
causes us to now consider the series
2 fin)rweM
9ir,9)=
n=
and the limit lim
(6.8)
oo
of
gir, 9) (hoping
course
that it is
/(0)).
Notice that the
r->i
series (6.8) does converge since the Fourier coefficients {/()} are bounded
(Problem 1) and the comparison test applies. Now, proceeding as above
but this time with g(r, 9), we obtain
1
9ir, 0)
and here
a sum
rn
rf
Z7l J
we can
uniformly.
it is
=
rl-l^'-W d<b
K
71
o=
and
interchange
The
in the above
sum
of two
geometric
1
1 + r2
The function
integral
-
+
")
e
P(r, t) is called
converges
nicer form since
(6-9)
^ f M*
2nJ-n
1 +
r2
-
-
r
2r
cos t
(named after its French
fishy), and the association of/
Poisson's kernel
technique
to g is called the Poisson transform.
=
a
-
1
not because its whole
iPflir, 0)
question
(re")"
+
r2
r(e"
be put in
T-^-T,
re"
1
-
the series in
can
00
(re-")"
=
1 +
I -re-"
discoverer,
j" because
CO
rMe'"<
=
1
series.
CO
P(r, t)
r <
00
Thus, the Poisson transform
1
11 j.
+
r
is
2
,Tcos(0
2r
m
-
,,
q>)
deb
=
2 /(H)r""<r**
=-,
(6.10)
6.1
Approximation by Trigonometric Polynomials
459
takes continuous functions on the circle into continuous functions of r, 9
for |r| < 1 ; that is, into continuous functions on the open unit disk. We shall
later
see
partial
the
importance of the Poisson
equations.
transform from the
point of view
of
differential
Examples (Some Poisson Transforms)
6. We
can
find the Poisson transform of functions
=
irVi29
Thinking of r, 9
(using z re'9
=
Pfiz)
=
+
i
+
ir2ei2e
polar
as
x
+
iy,
i(z2
+
z2)]
=
i[l
+
z
=
[1
i(re~i&)2
+
coordinates in the disk,
=
=
the circle
quite
Equation (6.10).
Using Example 3 we have
=
Pfir, 9)
on
notation and
explicitly, using some complex
example, consider f(9) cos2 9.
re',e
=
iireiB)2]
we can
rewrite this
iy):
x
+ Re
i[l
+
For
z2]
=
i(l
+
x2
-
y2)
Clearly,
lim
Pfir, 9)
lim
=
Pfiz)
=
-
2
x2+}>2->1
r^l
(1
+
x2
(1
-
x2))
-
=
x2
=
cos2 0
7. The Poisson transform of
1
0>O
=\0
0<O
is
given by
1
1
Pf{r>e)
2
1
Pf(z)
Now,
series.
=
+
=
-
,_,
r|n|e'"9
niL
/
2
+ -Im
we can use
y
2
=
nL2i[-n
r"e-inB\
^J
n>0
-
z2"
2
1
z"\
-
+
/r"einB
1
2
1
=
+
-
Im
Taylor expansions
to
2
:
+ 1
TT
obtain
a
closed form for this
460
6
Functions
on
the Circle
(Fourier Analysis)
Now
Jr^-j(JnW)-Mf^)
(We
once
have used real-variable techniques to find this closed
it is found it is valid for all z, \z\ <
1.) Thus
P/(Z)12
As
now
|z|
form, but
iimm(ii)zj
\1
it
-
1, Pf(z) has
-*
a
limit except for
show that except for these two
z -> 1, z ->
1. We shall
values, limP/(r, 0) =/(0).
-
r-l
lim
Pf^, 9)
=
P/(l, 0)
+
ei9)(l
ei9)(l
=
1 + I Im
ln(if!)
(6.11)
Now
l+eie
(1
(1
_
~
1
-
eie
-
-
-
e-'9)
e-'9)
1 +
eie
-
e~iB
l
ew
-
e~m
-
1
_
~
_
+ 1
j sin 0
=
<612)
(l-cos0)
Since In
Since
z
=
In
|z|
+ / arg z, Im In
is pure
(6.12)
imaginary,
(n
Imlnr^
,
Thus, referring back
Pf(r, 9)
=
\
+
2
r->l
_
1
7T
1
=
z
0<O
to
(6.11)
fc)
\2/
1 /
+
arg
have
n
2
lim
=
0>O
-
12
1 + e"
we
z
=
1
if0>O
7t\
_(__j
=
0
if,<0
for any
complex
number.
6.1
We
still
are
Approximation by Trigonometric Polynomials
hoping that it is true for all/that lim Pf(r, 9) =f(9).
461
Of course,
r->l
this turns out to be true.
To
Poisson's kernel.
rewrite the Poisson kernel
p(r< 0
First
we
(1
n,
(n)
(iii)
l'l><5,
,2
rf
-
>
uniformly
the
some
properties
as r
for all values of r, t,
-*
-
\
r)
*-::
2r(l
r)2
r <
1
>ooasr->l
-2=-
(1
following properties:
l+r
-
+
r
^
t
0, P(r, ()->0
x-ri
^
as
r->l.
If
-
~
-
cos
d)
2r(l
-
cos
S)
1.
goes up and the
(iv)
of
as
fixed value of r, the
peak
This
verify
t)
conclude the
easily
we
1_r2
=
cos
-
On the other hand, for values of
(1
a
2r(l
+
0
x
P(r,0)
Pir,=
For
have to
1 -r2
P(r, 0
/\
we
=
From this reformulation
(i)
this
see
graph of P(r, r) is drawn in Figure 6.1.
valleys get larger and deeper. Finally,
^fp(r,t)dt
=
As
r
-*
1
,
l
computed directly; however it is easier to use Equation (6.10)
particular case where / is the function which is identically one (see
Problem 2).
can
be
in the
Theorem 6.1.
Iff is
a
continuous function
on
the circle,
limP/(/-,0)=/(0)
r-l
Proof.
Using property (iv) above
(Pf)(r, 8)
-
f(8)
=
i-
f
2tt J_
we can
[f(<j>)
-
write
Pf(r, 8)
f(9)]P(r, 8-<j,)d<t>
f(8)
as
an
integral,
462
6
Functions
on
the Circle (Fourier
Analysis)
nP(r,l)
Figure 6.1
For any S > 0
we
(Pf)(r, 8)
-
break up the
f(6)
=
integral into
1f
Z7TJ|<,_e|.s{
+
[f(<f>)
-
two
pieces :
f(8)]P(r,
8
-
<j>) d<j>
T-fJ\i),-e\Z6 Uift-fWmr.d-fidt
Z.TT
Now, by (iii) the integrand in the lower integral tends to zero as r -> 1 and, by
continuity, \f(<j>) f(8)\ is small for all <j> near enough to 8 so that we can make
the first integral small by taking S small.
More
precisely,
let
e
> 0 be
\f(<t>)-m\<l
Given that
{
8, by (iii),
Jl-6|a
given.
Let S be such that
if|0-0|<8
there is
an
-q > 0 such that for
P(,^_0)^<-^
Z.
||/ I co
(6.13)
\r
1 1 < -q,
6.1
Then for \r
Approximation by Trigonometric Polynomials
1 1 < ij,
-
\Pfir, 6) -f(r, 0)1
f
<;
+
|/(0) -f(8)\P(r,
^/
i77
f
Jl-l2f
ll/IU
We
seem
<j,) d<f>
-
P(r,6-<j>)d4>
+ =-2 11/11.
~2'
8
\f(<f>)-f(0)\P(r,8-<f,)d<l>
-t'rl
have not
463
+
*
P(r, 8
-
<f>) d<f>
TS
2||/|U"e
to have come
a long way away from our
original quest, but we
The content of Theorem 6.1 is this: Let /be a continuous
the circle. Its Fourier series
really.
function
on
fin)eM
n=
~
oo
is too hard to
regards convergence, but it does represent /in some
almost converges
to /; that is, if we put in factors to
the convergence and consider instead
relevant
ensure
study
It
sense.
as
"
"
00
Pfir, 0)
finy^e
=
71=
then for
r
00
very close to
make
For
~
important
example,
Collorary
oo, then f is
1.
the
1, this function is
assertions based
on
very close
any information
Iff is a continuous function
of its Fourier series,
on
to/.
on
This allows
us
the Fourier series
the circle,
to
of/.
andY^=-x |/()|
<
sum
00
fin)e'"e
/(#)=
n=
oo
Proof. The condition allows us to conclude on the basis of the comparison test
that the Fourier series converges; the essential content here is that it converges to/.
464
In
6
Functions
the Circle
fact, by the comparison test,
(Pf)(r, 8)
a
(Fourier Analysis)
we can
conclude that
2 f(n)r"e""
=
n=
is
on
-
oo
continuous function
the closed unit disk : all
on
r
< 1
Then for any
.
8, by
Theorem 6.1.
f(9)
=
lim
Pf(r,8)= lim
r-+l
In
r~*
1
f
n=
f(n)rMeM
f
=
oo
f(n)e'"e
oo
n=
particular, if/()
vanishes for all but finitely many n, then /is a trigono
Thus the trigonometric polynomials are precisely the class
metric polynomial.
of continuous functions
mined
by
on
the circle with
only finitely
basic consequence is that
its Fourier series.
coefficients.
A
Iff and g
2.
Collorary
thenf=g.
more
are
continuous
on
a
many
nonzero
function is
the circle
andf(n)
Fourier
uniquely deter
=
c}(n)for alln,
Proof, fg is continuous on V, and (fg)'(n)=f(n)
g(n)=0 for all n.
Applying the first corollary to / g we see that it is the sum of its Fourier series,
which is identically zero. Thus fg=0, so / g.
=
Conditions
on
the Fourier coefficients of
a
function, such
as
that in
Corollary 1 are not hard to come by. For example, suppose / is a twice
continuously differentiable periodic function. Then by integrating by parts
,
we
have
Since
/"
is continuous
these bounds
Thus
\f(ri)\
on
on
the circle, it is bounded, say
the Fourier coefficients
< oo.
of/:
by
M.
We obtain
6.1
Corollary 3.
Approximation by Trigonometric Polynomials
Iff is
a
C2 function
the circle, it is the
on
sum
of its
465
Fourier
series.
We shall have
6.1 does allow
As
series.
to
one
approximate
mate it
by
better result in Section 6.4.
to make deductions
last
it tells
application,
Nevertheless, Theorem
the convergence of the Fourier
that although we may not be able
on
us
by its Fourier series, we can nevertheless
sequence of trigonometric polynomials.
function
a
some
A continuous function
Corollary 4.
metric
an even
us
on
the circle is
approximable by trigono
polynomials.
Using the notion of uniform continuity,
Proof.
we can
Theorem 6.1, that the 8 chosen so that (6.13) is true is
in the rest of the argument we find an r < 1 such that
2=-> f(.n)rMe'"0
Now, the series
there is
an
N such that the
partial
8).
Thus
where within
e
of Pf(r,
12(0) -f(9)\
0,
as
<
be sure, in the proof of
of 8. Thus,
independent
for all 6
\Pf(r, 8)-f(9)\<e
for all
approxi
12(0)
-
uniformly to Pf(r, 6), if r < 1. Thus
Q of the terms between TV and N is every
converges
sum
Pfir, 0)1
+
IP/0", 0) -/(0)l <e+e=2e
desired.
EXERCISES
1. Find the Fourier series of the
(a)
(c)
f(8)
/(0)
=
82
=
e'"
following functions
(b) /(0)=cos50
p, > 0, not necessarily
(d)
-7T<0<
0
-2<e<
/(0)
=
77
77
-0 +
0
(e)
/(0)=|sin0|
2
O<0<^
-
<0<77
_
2
_
an
on
integer.
the circle.
466
6
Functions
(f)
f(8)
the Circle
on
=
(g)
sin 0 +
cos
(Fourier Analysis)
0
/
/(0)
=
0
-7T^0<
1
-2*0*0
0
O<0<-
{
-
77
2
77
-<0<7T
1
~
2~
(h) f(8)=e
(i) f(8)=e
2. Find the Poisson transforms of the
(a)
(b)
(c)
(d)
(e)
(f)
following functions
on
the circle:
cos3 0 + sin3 0
(1
+ cos2
0)-1
Exercise
1(c).
Exercise 1(d).
Exercise 1(g).
(l+e')2
PROBLEMS
1
.
Show that the Fourier coefficients
defined
on
the circle
are
f(n)
of
a
continuous function /
bounded :
|/()|< H/ll =max{|/(0)|: -77<0<77}
2. Show that
r~
\
2rr J_
P(r, t) dt=\
by computing the Poisson transform of the function 1.
3. Show that if /is a real-valued function on the circle, f(n) =f(n)~.
4. (a)
Show that the Poisson transform of /can be written
Pfir, 0) =/(0)
+
2 (f(-n)zf(n)
+
j\n)z")
0, n > 0, then Pf is the
vergent power series in the unit disk.
(b)
Show that if
=
sum
of
a
con
6.2
F(e">)
=
where F
Pf(z)
467
Show that if /can be written in the form
(c)
f(8)
Laplace's Equation
=
can
be written
as
a
convergent series in powers of
z, z,
then
F(z)
5. What is the Poisson transform of these functions?
(a)
(b)
(c)
expte'8)
(1+2Z)"1
(z+z)"
6. We
following
can
(l+cos20)-'
(d)
(e)
(f)
ln(5 + z)
exp(cos0)
the approximation theorem
use
(Corollary 4)
to prove the
fact.
(Weierstrass Approximation Theorem). Iff is a continuous function on the
s > 0, there is a polynomial
P(x) a0 + aYx +
+anx"
interval [0, 1] and
such that
\f(x)
=
P(x)|
-
<
1
for all
x e
[0, 1]
Prove it
according to this idea: First extend /as a continuous function on
[ n, 7t] so that/( 71) fin). Now view the extended function
function on the circle and, by Corollary 4, approximate it by a trigono-
the interval
as a
=
N
metric
polynomial
of the form
2
n=
{einB :
functions
6.2
-N <n
<
N}
an e'"e-
Now
use
the fact that the 2N
-N
can
be
approximated by polynomials
in 0.
Laplace's Equation
techniques described in the previous section came out of Poisson's
the theory of heat flow.
Suppose D is a domain in the plane
(representing a homogeneous metallic plate); we wish to study the tempera
Let
ture distribution on this plate subject to certain sources of heat energy.
u(x, t) be the temperature at the point x at time /. We shall see in Chapter 8
The
work
that,
on
as
a
function
w
consequence of the law of energy conservation, the temperature
behaves according to this partial differential equation (appropri
ately called the
heat
du
ld2u
1
_
dt~a1\dx^
equation) :
d2u\
+
'dy2)
(6.14)
468
6
Functions
on
the Circle
(Fourier Analysis)
Now, suppose our sources of heat maintain the temperature at the boundary
of D, and there is no other source or loss of heat. Then, as t -* oo the
temperature distribution will tend toward equilibrium: that state at which
du/dt
=
0.
therefore
This equilibrium (or steady-state) temperature distribution
satisfy Laplace's equation:
d2u
dx-2
must
d2u
+
^
=
d?
This is sometimes denoted Am
=
0.
Solutions of Laplace's
equation are called
harmonic functions.
The Poisson transform has to do with the solution of this
problem when
D is the unit disk.
perature distribution f(9)
function
on
defined for
u(r, 9)
Suppose then,
the unit circle ;
r <
order to attack this
circle
r
=
constant
we
1 such that Aw
problem, we assume that
by its Fourier series :
that
we are
wish to find
=
steady-state
given a tem
a
continuous
0 and
u can
be
w(l, 0) =/(0). In
represented, on each
2 ,(>)<"
u(r,9)=
n=
(6.16)
oo
Our conditions become
an(l)=:f(n)
(i)
,-n
(We
d 1
*
Au
(11)
=
^fj(9)e-i"Bd9
=
du\
d2u
rdrH)+W2
(6.17)
n
<6-18>
=
have rewritten Au in terms of polar coordinates
Fourier series.
Laplacian.)
Now, computing (ii)
by
term
term in the series
apply it to
polar form of
so we can
We leave it to the reader to derive the
(6.16),
we
the
the
obtain
OO
ir2<
0=
Since the
zero
+
ra'n
-
function is
n2an)einB
=
0
represented only by
the
zero
Fourier series
we
deduce that
r2a'^
+
ra'n-n2an
=
0
(6.19)
6.2
for all
This
n.
1,
r",
We have
ordinary differential equation
n
r~"
n=0
only
we must have a
=
=
condition
boundary
one
r
0,
=
/(n)r|n|,
easily
solved
:
so
(6.17),
log
the solutions
however
~
r
r,
we
do want the
'"' are excluded.
Thus
and the solution must have the form
f(n)r^einB
u(r,9)=
oo
n
Hence, if the problem is solvable, the solution
which is Poisson's transform.
must be
is
0
logr
functions continuous at
469
Laplace's Equation
given by
Poisson's transform.
Conversely,
the
following
is
a
solu
tion.
Theorem 6.2.
be
only verify
integral sign
entiate under the
A"
=
continuous
J/*1"1'"*
-
We need
Proof.
a
function
on
the circle.
There is
a
values
fi
boundary
transform off:
is the Poisson
*. D
f
u, harmonic in the disk and assuming the
unique function
u
Let
-
that
5
u
/>> r.-'2"^-) *
i +
is indeed harmonic.
Since
we can
differ
t J_. /WA l+r2-27cos(0-0) *
need only show that the Poisson kernel is harmonic. That can be done
direct computation, or by referring back to Equation (6.9). There we have
we
by
^e)=rr^-1+r^ TJ^--1 rz
=
=
Now, (1
-
z)"1
-1 +2 Re
is
a
G-J
seen that
complex differentiable function, and we have already
see Problem 5.7.2
a function satisfies Laplace's equation,
the real part of such
Thus AP
0.
=
+
470
is
6
Functions
on
the Circle
(Fourier Analysis)
To recapitulate, Laplace's equation for the disk with given boundary values
easily solved by Fourier methods. If/is the boundary temperature distri
bution, the solution is
OO
00
/(/0r"V'"9=/(0)+ (/(-n)z"+/(n)z")
00=
n=
(since r|n|e'"9
n=
oo
z" for
=
0, rweinB
n >
=
1
z" for
n <
0).
Examples
the solution of Laplace's equation with boundary values
cos3 9 + 3 sin 30. This is easy to do, for we can easily
/(0)
recognize the function as the boundary value of the real part of a com
plex differentiable function. Since
8. Find
=
0
cos
=
-
2
(z
on
the unit
for
z
=
\z\
<
z,
+
circle,
e,B.
sin 30
z)
=
2i
we
(z3
-
z3)
have
Thus the solution is
1 since it is
given by
clearly harmonic.
the
same
9. Solve
Laplace's equation with boundary
Since |0| is not a trigonometric polynomial, we
Fourier expansion.
fin)
=
-f
2n J
=
t-
\9\e-'"B
d0
=
-\ 9e~inB
2n J
-K
Co(eM + e-'"6) d9
2% Jo
1
rn
=
nn
f(o)
=
-
71
Ce
Jo
=
nn
=
2
cos
must
fee
cos
nB d9
Jo
n
r
-
values
2n Jo
-
1
sin n0 d9
-fed9
^
Jo
n
=
d9 +
expression
nB
=
o
zl
nn2
( odd
# 0
\n
/(0)
for all
=
|0|.
compute the
'dd
6.2
471
Laplace's Equation
Finally,
=
z" + z"
2
n
"/ (Z)
=
o
2
2
~
n
The
2
_
o
2
n
7t B=i
z2"+1+z2n+1
n
=
172
/")
(zn
=i
,
+
1\2
1)
odd
to the above in the
problem analogous
case
of
a
general domain is
problem is to
Dirichlet's
Dirichlet's problem. More precisely,
given domain D and function /defined on D, a function harmonic
in D and taking the given boundary values. In 1931, O. Perron gave an
elementary, but extremely clever argument which proved the existence of a
solution to Dirichlet's problem. Poisson's method plays a strategic role in
known
as
find for
a
Perron's arguments, which we shall not go into here.
However, we shall
harmonic function
be
at
most
one
:
there
can
is
that
the
solution
unique
verify
with given boundary values. This follows from the mean value property of
harmonic functions.
Proposition 2. Suppose
A(z0 R) cz D, then
u
is
a
harmonic
function
in the domain D.
If
,
(z0)
that is,
=
+
We
can
expand
ReW)
dB
of its values around any
is the average
u(z0)
Proof.
2^f_ "Oo
u
in
a
circle in D contained in D.
Fourier series around any circle \z
z0
1
=
r,
r^R:
2 an(r)eine
u(z)=
n=
Since A
=
0,
where
r<R
z
=
z0
+ re'
oo
we must
have
a(r)
=
/(>"", where /(0))
=
u(z0 + Re"), already seen.
Thus
u(z)
=
2 fin)
\z
n
1
u(z0)
=
/(0)
=
ZTT
z0|""e'" <"*<= -*o>
-
f
J
_
1
.
f(8)dd=
Z7T
c11
J
u(z0 + Re">) d8
_
472
6
Functions
the Circle
on
Corollary 1. Suppose
Ifu>0on dD, then u >
Let
Proof.
us
u
0
is harmonic
throughout
on
the closed and bounded domain D.
D.
suppose that the conclusion is false.
We shall derive
< 0.
which
takes its minimum value.
bounded, and it is interior
That at
some
point z0 inside
contradiction.
D, u(z0)
u
iFourier Analysis)
a
We may take for z0 a point at
There is such a point since D is closed and
0 on dD.
Let A(z0 R) be the largest disk
boundary of A(z0 R) must touch dD (see
Figure 6.2), for if not we could find a larger disk centered at z0 and contained in D.
Thus there are points on the circle \z
R at which u > 0.
Since (z0) is the
z0\
average value of u on this circle, and u(z0) < 0, there must be points on this circle
at which u < u(z0) in order to compensate.
But u(z0) is the minimum value of u,
so we have a contradiction.
More precisely, since u(z) > u(z0) for all z e D,
u(z0 + Re'e)
u(z0) > 0 for all 0. On the other hand, by the mean value property
centered at
z0
to D since
contained in D.
u
>
,
The
,
=
f
((z0 + Rew)
'-71
When the
identically
-
u(z0)) dd
=
0
integral of a continuous nonnegative function is
zero.
Thus,
w(z0 + Re">)
=
w(z0)
for all 8
This contradicts the fact that for
some
0, u(za
Figure 6.2
+
Reie)
> 0.
zero, that function is
6.2
473
Laplace's Equation
Corollary 2. A function harmonic on a closed and bounded domain
uniquely determined by its boundary values.
Suppose that
Proof.
Then
u
+ 6, v
in D, thus
v
positive
u>v
Since
v
>
u
u
u,
+
both harmonic in D, but
v are
e are
both positive
dD.
u
=
v on
Let
dD.
e
By Corollary 2, they
are
We conclude that
>
>
0.
both
in D
v>ue
s
on
D is
e is arbitrary, we may now let it tend
v in D.
throughout D. Thus u
to
zero.
u
v
and
=
Another
problem
distribution
boundary,
on
and
of heat transfer is this : find the
the unit disk
other
no
flow, denoted q, is
a
assuming
source or
vector field
a
given
loss of heat.
on
steady-state temperature
through the
Now the velocity of heat
rate of heat flow
the domain and it is
a
law of thermo
dynamics that this field is proportional to the temperature gradient, but
oppositely directed. Thus, in this problem, our given data are the rate of heat
flow perpendicular to the boundary of the unit disk, which is proportional to
du/dr on the boundary. By the law of conservation of energy, since we are
assuming a steady state, the total energy change is zero, thus we must impose
this condition:
$*_K du/dr(e,B) d9
0.
=
Thus, the mathematical formulation
of this problem (known as Neumann's problem) is this : Find a function u
harmonic in the unit disk such that du/dr(e,B) assumes given boundary values
We
#(0).
impose
the condition
this condition in order to obtain
will
see
in Problem
We
8.)
JZ #(0) d9
a
0.
=
(It
is necessary to
solution, for mathematical
again
solve this
problem by
reasons,
impose
as
you
Fourier methods.
Find
00
u(reie)
=
2 anir)einB
00
(i) idujdr)(eie) gi9), Am 0. Again, this leads to the ordinary
differential equation (6.19) with the boundary condition a'n(\)
gin). The
solution continuous at the origin is \n\~1g(n)rM. Thus the solution must be
given by
so
that
=
=
=
u(reiB)
=
V
-,
We will omit the
^ r^eine
(6.20)
n
proof that
this function does solve Neumann's
the argument is much like that in Theorem 6. 1
.
problem;
collapse
We can, of course,
474
(6.20)
6
Functions
into
integral
an
u(reie)
on
=
the Circle
formula :
f IV"
f
\_zn
ai<P)e~""P dcp r\n\e'"8
-
-
oo
1
(Fourier Analysis)
J
n
-
*
rne-in(fi-<l>)
r*
n
.= i
i
/",(*)
Re
nJ-K
=
-
f"
nJ-n
=
#
n
i
(ye'(9_*))ni
oo
i /"
=
r"ein(e-<l>y
oo
+
^Ln=i
"
<?(</>) Re[ln(l
-
re^9"*')]
deb
Now
|1 -reu\2
=
l+r2-2cost
so
Re
ln(l
re1')
-
=
i ln|l
relt\2
-
Thus the solution to Neumann's
=
\ ln(l
problem
+
r2
-
2r
cos
takes the form
t)
(6.20)
or
*
u(reiB)
=
f g(<b) ln[l
^2nJ-K
+
r2
-
2r
cos(0
-
</>)]
deb
EXERCISES
3. Solve Dirichlet's
(b)
(C)
(d)
4.
problem in
f(8) sin2 0
/(0)=772-02
=
-
the disk with these
boundary conditions:
cos2 0
/as is given in Exercise 1(c).
(e) /(0) as is given in Exercise 2(f).
Solve Neumann's problem with these boundary conditions:
sin 0 + 2 cos 20
(a) f(8)
=
/w-i
(c)
i:
Si!
/as is given in Exercise 3(a).
6.2
Laplace's Equation
475
PROBLEMS
7. Show that the
8
AU
=
8.
Laplacian
( du\
is
given in polar coordinates by
d2u
r7rV^)+W2
it is necessary that
Verify that
R
1
f
2tt
g(8)d8
=
Q
J-
for there to be
a
function
u
harmonic in the disk such that
du
limr-,i
(r, 8) =g(8)
or
9.
Verify by direct computation that P(r, 8) is harmonic.
/is a complex differentiable function (it satisfies the
Cauchy-Riemann equations), then /is harmonic.
11. We can prove, using the Poisson transform, this remarkable fact
about complex differentiable functions :
10. Show that if
Theorem.
disk.
Suppose that f is a complex differentiable function on the unit
sum of a convergent power series centered at the
origin.
Then f is the
The
goes like this: Let
#(0) =f(e[0). Since /is harmonic in the
10), it solves Dirichlet's problem with the boundary values g.
Pg(re">). Now prove this fact.
(a) If the Poisson transform Pg is complex differentiable, then
g(n) 0 for n < 0. (Hint: Apply d/dx + i d/dy to the expression
proof
disk (Problem
Thus f(re">)
=
=
Pg(re')
=
+
2 ig(-n)z" + g(n)z")
Deduce from
(b)
fire")
<?(0)
(a) that
2 gin)z"
=
n
=
0
12. Under what conditions
13.
(a)
mean
f(z0)
Show that
if/ is
a
on/
g is
P(fg)
=
continous function
P(f)P(g)l
on
the domain D with the
value property :
If"
=
^J
/(z0
+
Re") d8
for every A(z0
,
R)
<=
D
476
6
Functions
the Circle
on
then /satisfies
maximum
a
(Fourier Analysis)
principle : f(z0)
<
max{/(z):
z e
dD}, for
every
z0e D.
(b) Conclude that
function
a
having
the
mean
value property
is harmonic.
14. Prove: A bounded function defined
harmonic,
6.3
on
the entire
plane which
is
must be constant.
Fourier Sine and Cosine Series
There
are
expansion
hand.
notationally
many
of
different ways of expressing the Fourier
on the dictates of the problem at
function, depending mostly
a
We shall devote this section to the
development
of these various
expressions.
First of all, since the main physical study is that of real-valued functions
should introduce the purely real notation.
We merely convert the
we
Fourier
expansion
e'"e
=
cos
2/(")e'"9 via tne
nB + i sin n9
Thus the Fourier
expansion
expressions
inB
e~~
=
cos
0
-
i sin n9
n >
0
will take the form
00
do
An
+
n
=
cos
n9 +
B sin 0
(6.21)
0
where the ,4's and B's
are
found from the Fourier coefficients
C
=/()
follows :
oo
00
C einB
n=
C(cos
=
n=
oo
=
n9 + i sin
n9)
oo
C0
+
2 [(C.
+
C_)cos n9
+
i(C
-
C_)
sin
n=l
Thus
,40
=
Notice that
C0
if/ is
/f
=
Cn
+
C_
B
=
i{Cn-C.}
n>0
real valued
1
c-n=2n!
fi<t>y'"Pd<t>=
yJ
f^e~itt*d$
=
c
0]
as
Thus
we
477
Fourier Sine and Cosine Series
6.3
have
A0
C0
=
f_Ji<b)d(b
=
1 i-n
A
2 Re
=
C
=
-
n
J
fi<b)
cos
neb deb
n>0
(6.22)
fieb)
sin
n< d</>
n >
0
(6.23)
-n
1 c"
B
2 Im
=
C
=
-
Furthermore,
C
K^
=
C_
iBn)
+
i(/l
=
-
iBn)
n>0
Examples
Express the Fourier series of n2
Example 4, we have
10.
3
-
92 in the form (6.21).
From
n
n*o
Thus
(-1)"
2
^0=-tt2
3
and
ni
-
we
An
=
4^-
obtain this Fourier
B2
=
%-
+ 4
3
Notice that
2
>o
equality
v
nk
(-1)"
n2
fact
y
12
0
cos
n0
n
justified by Corollary 1
Evaluating
the Fourier series does converge.
interesting
=
expansion :
^
is
P
to Theorem 6.1
at
0,
we
since
obtain this
478
6
Functions
e-inB
2
2~nh
Thus
+
Reading
eine
n2
have the real Fourier series
we
cos0
4
7T
0
(Fourier Analysis)
Express the Fourier series of |0| in the form (6.21).
Example 9, we have the Fourier series for |0| :
11.
from
7t
the Circle
on
2
=
~
Evaluating
2~~
at 0
=
0,
obtain
we
~
k
n2
8
12. As
usual, trigonometric polynomials
can
be handled
directly,
without computation of integrals :
e
e-;V
+
1
=
2
=
16
(2
3
cos4 9
=
-
16
/
cos
40 + 8
1
+
(**"
cos
+
20 +
4e2M
+ 6 +
4e-2i9
+
e'4iB)
6)
1
cos
-
2
8
77
20 +
-
cos
40
8
Even and Odd Functions
A function of a real variable is called
x, and it
is
an
odd function if f(x)
=
an even
-f(x).
function
odd functions is even, and the product of an odd and
If/is an odd function on the interval [ A, A~\, then
f
We
can
f(t)
dt
=
f
conclude that
Fourier series is
f(t)
if/
dt +
is
\Af(t)
an even
dt=-
function
cosine series.
purely
for all n, so the integrals (6.23) all vanish.
Fourier series is purely a sine series.
a
if/(x) =/(
Notice that the
\Af(t)
on
even
dt +
product
case
Similarly, if/is
f/(0
f(eb)
an
of two
function is odd.
the interval
For in this
x) for all
[
dt
n,
=
0
n\, its
sin neb is odd
odd function its
479
Fourier Sine and Cosine Series
6.3
Example
13. The Fourier series of 0 is of the form
B sin n9
since 0 is
an
lf"
,
-1
7t
=
odd function.
0sinn0d0
Here
10
--cos0
=
7t n
J-n
1 r*
-n
Thus 0 has the Fourier series
2
+-|
n J
cos
2
</0=--(-l)n
n
n
-n
( l)"/n
n0
sin n9.
n=l
have been done for periodic functions of
arising in physics do not usually have such a
convenient period, yet they are subject to Fourier methods merely by a
normalization. Suppose that / is a periodic function of period L. Then
g(B) f(L9/2n) is periodic of period 27r. For
Now, all
period
our
computations
Periodic functions
2n.
=
giO
Now, if g
in)
+
can
3(0)
=
be
A0
=/(|(0 2,)) =/(! L) =/(f?)
+
+
expanded
(A
+
in
cos
=
,(0)
Fourier series :
a
w0 +
Bn sin n0)
n=l
then
we can
fix)
write
/27rx\
=
g^f
"
=a0+
where (as is easy to compute
1
Ao
=
B
=
cos(-r--)
by
the
rL/2
2
An
fix)dx
fix) sin
T\
LJ-L/2
27tnx
=
Bn
(2nnx\
-
rL'2
nAS
[-r-)
=
(6-24)
2L~~ix)
2nnx
/(x)cos-
dx
(6.25)
,
dx
L
.
+
change of coordinates eb
,L/2
t\
2
(2nnx\
2A
(6.26)
480
6
Functions
on
the Circle
With these formulas the Fourier
made
(Fourier Analysis)
analysis of functions periodic of period
L is
possible.
Fourier Cosine Series
There
two more variations which are, as
yet
are
the
study of partial differential equations.
with period L and define
Let/be
we
a
shall see, of value in
function
given periodic
O<0<7T
(6.27)
-7T<0<O
Then g is an even function on the interval
a Fourier series involving only cosines :
[
n,
n\,
so
it
can
be
expressed by
OO
g(9)
2
A0+
=
An
cos
n0
=i
where
A0
=
zn
f
J
g(6)
d9
A
=
-\J
n
-n
g(9)
n0 d9
cos
-n
*
=
fg(9) d9
Jo
-
n
=
-
n
f g(0) cos
n9 d9
Jo
Now, making the substitution g(9) =f(L9/n) in the interval 0
00
fix)
=
A0+
0
< n,
these
nnx
E4,cos
=
T
\ /(*)
L, Jo
dx
(6.28)
-.
n=l
Ao
<
become
expressions
a
=
fix)
t\
Li Jo
cos
~r dx
L
We pause to remind the reader that the use of equality in
(6.28) is not literal, it holds only if the series converge
and
continuously differentiable).
(6.24), (6.28) are valid, where
(6-29)
Equations (6.24)
(say if g is twice
The point is that in such cases the expansions
the coefficients are defined by (6.25), (6.26), or
6.3
Fourier Sine and Cosine Series
481
(6.29), respectively. The choice of these expansions is free it is usually
dependent on the demands of the particular problem at hand. Equation
(6.28) is called the Fourier cosine series for the function /. Of course, if we
define # as an odd function, instead of the expression (6.28) we can obtain the
Fourier sine series for/:
nnx
fix)=
Bsin
L,
(6.30)
k=l
where
2
Bn
=
rL
nnx
-j fix) sin
dx
(6.31)
We leave the verification of this
possibility
to the readers as a
problem.
EXERCISES
5. Find the Fourier
(a)
(b)
(c)
(d)
(e)
expansions into sines
and cosines for these functions
sink 0
k
/as given
/as given
/as given
in Exercise
positive integer
3(a).
in Exercise 1(g).
in Exercise 1(b).
6. Find the function whose Fourier expansion is
a
2^=
-
e'"e/in.
1. Find the Fourier sine and Fourier cosine series for these
.
=
=
=
0<x<l/2
1/2<x<1
1
(d) /(*)
(e)
f(x)
=
=
...
(f)
/(*)
(g) f(x)
=
=
{
sin(77x)
jx
an
0<x<l/2
l/2<x<l
{!_*
sin(7rx)
8. Show that any
function and
+
cos(77x)
periodic
function
on
the circle is the
sum
of
an even
odd function.
9. What is the Fourier
/(0)?
periodic
period 1
l,allx
/(x)
sin(2rrx)
f(x)
/(*) cos(2rrx)
functions of
(a)
(b)
(c)
:
cos8 0
expansion of f(8) +f(n
8)
in terms of that for
482
6
6.4
Functions
on
the Circle
(Fourier Analysis)
The One-Dimensional Wave and Heat
Equations
physics, Fourier analysis begins with the study of wave motions. Sup
we have a homogeneous string of density p and length L lying on the
horizontal axis in the plane which is kept extended by equal and opposite
forces of magnitude k at the end points. If we pluck the string, it will follow
a motion which is (classically) determined by Newton's laws.
We shall
derive the differential equation governing the motion. At some time t the
string has a shape somewhat like that pictured in Figure 6.3. We shall refer
to a point on the string according to the distance s, measured along the string
from the left end point. The position in the plane of the point at distance s1
at time / will be denoted by z(s, t).
This is the function that fully describes
In
pose
the motion.
Now, if
argue as if the string were a collection of points, we will get
For the only forces acting on the string are those obtained by
we
nowhere.
transferring the equal, but opposite
along the string. Thus, at any point
there
is
can
be
Now
motion.
no
inadequate
and
we
forces at the end
the
sum
points tangentially
acting is zero, so
model of the string
of the forces
As that is contrary to fact, this
select another.
must
consider the
string as a large finite collection of segments and
equation of motion from Newton's laws. Having
done that, we can idealize by letting the number of segments become infinite
(as their lengths tend to zero) and obtain a differential equation. Let s0
and s0 + As be the end points of such a segment (see Figure 6.4).
The mass
of this segment is pAs and the forces acting on it are opposed tangential
forces of magnitude k acting at the end points.
Letting T(s) be the tangent
vector at the point s, these forces are thus
kT(s0), kT(s0 + As), respectively.
If A is the acceleration of this segment, we have by Newton's laws
we
again try
to
pAsA
Now, T(s)
=
deduce the
=
k[T(s0
+
dz(s, t)/ds
As)
-
T(j0)]
and lim A
=
dz/dt(s0 t).
,
Thus
As->0
k
A
A
=
(dzlds)(s0
+
As, t)
now
letting
d2zt
(dzlds)(s0 t)
,
As
p
and
-
As
=
->
0
we
kd2zt
obtain the equation of motion:
Wis0,t) -^-2is0,t)
6.4
The One- Dimensional Wave and Heat
Equations
483
Figure 6.3
This
equation,
called the one-dimensional
d2Z
VOL
ds2
c1 dt2
wave
equation,
is
usually written
(632)
legitimate since both k, p are positive).
We now make the (physically plausible) assumption that the horizontal
motion is negligible (for we are interested only in almost horizontal wave
motions with small fluctuations). This assumption allows the replacement
of s by the horizontal coordinate x, and the positive vector z by only the
vertical coordinate y. Thus (6.32) becomes simply
(where
the substitution
c2
=
kjp
is
d2l L8ll
dx2
(6.33)
c2 dt2
The motion of the string is completely governed by this
equation and the initial displacement and velocity:
yis,0)=fis)
d_y is, 0)
dt
=
partial
(6.34)
g(s)
-kT(s,,)
So
+ AS
Figure 6.4
differential
kT(s + as)
484
6
Functions
on
the Circle
iFourier Analysis)
technique for solving this differential equation with boundary conditions
in the theory of ordinary differential equations.
We find an
set
the
and
of
solutions
of
independent
general equations
hypothesize that the
solution we seek is a linear combination of these. We then identify the
coefficients by substituting the initial conditions. However, the situation is
The space of solutions of
more complicated than in the one-variable theory.
is
infinite
so
the
solution
cannot be picked out
dimensional,
particular
(6.32)
of the general solution by means of simple linear algebra. This difficulty
will be overcome, as we shall see, because the form of the general solution
will be that of a Fourier expansion and so the initial data will give us the
coefficients by Fourier methods.
Let us now solve the differential equation
The
is the
same as
1 d2y
d2y
,--22
dx^'STt
(6-35)
_
for
a
function y defined
on
the interval
[0, L] and where these conditions
must
be satisfied
y(0, 0
=
0
y(L,t)
=
8^(x,0)
y(x,0)=f(x)
all
0
=
t
(6.36)
g(x)
(6.37)
given functions/ g. First, we put aside the initial data (6.37) and find all
Equation (6.35) subject to (6.36). Since we have no tech
we have to make a guess at the form of the solution, and
available,
niques
hope that our guess is general enough (of course, in the end it will turn out to
be so). The guess that works is
for
solutions of
y(x, t)
and
(6.35)
=
F(x)G(t)
becomes
F"(x)G(0
or, what is the
F"(x)
Fix)
=
^P(x)G"(0
same
1
(since
G"(t)
=
c2 G(t)
we
exclude the
zero
solution),
6.4
The One- Dimensional Wave and Heat Equations
The left-hand side is
Since
they
independent
the same,
are
they
are
of t, and the
both constant.
485
right is independent
Thus, there
of
must be
a
x.
A
such that
^
^
c2
A
=
F
=
x
G
Now, incorporating the conditions (6.36),
boundary
F"
value
XF
-
F(0)
We
can
that the
=
for
0
0
F(L)
=
some
=
I
(6.38)
0
(6.39)
First of all,
problem.
cx
exp(x/Ax) + c2 exp( ~Jkx)
the
boundary conditions (6.39),
=
F(0)
from
we see
(6.38)
form of F is
general
Substituting
0
=
find all solutions of this
Fix)
arrive at this one-variable
we
problem :
-
0
c, + c2
=
In order for there to be
a
=
F(L)
=
ct
solution for both
have
we
exp(N/lL) +
equations
we
c2
exp(-N/l)
must have c1
=
c2
and
exp(N/iL)
Thus
the
we
=
must have
only possible
F(x)
-JXL)
exp(
2^JXL
=
expl
Corresponding
Ix
-
expl
to the solution
n
some n >
(6.38), (6.39)
1
[nni\
=
2nni for
solutions of
exp(2N/AL)
or
nni\
P(x)
or
1
yJX
=
Therefore,
nni/L.
are
Inn \
2\ sinl
xl
.
Jx
=
0,
=
=
sin(7tn/L)x,
we now
all
n >
0
solve for G:
n
G"--1}?G
The solutions
are
Spanned by G(t)
=
cos(nn/Lc)t, sininn/Lc)t.
Thus, all
486
6
Functions
solutions of
(6.35)
of the form
[nnx\
.
the Circle
on
[nn \
We
F(x)Git)
are
these:
Inn
[nnx\
.
sinl^r)C0Sfcv
(Fourier Analysis)
\
sinhr)smM
(6.40)
initial conditions
(6.37) and hope to find a
(6.40) which has those initial conditions.
Of course, the linear combination will satisfy (6.37) since it is a linear differ
ential equation. (However, we must caution the reader that ours will be an
infinite linear combination so questions of convergence are inevitable. If
the initial data are well behaved, these problems disperse as you shall see in
Problem 15.) Thus we seek
now
particular
return to our
linear combination of the functions
yix, t)
satisfying
A.
=
(nn \
cos
the conditions
yix, 0)
=
dy
/
,(x)
But
t
sm|
x
Inn
2 A sinl
nn
^
\
x
I
(nn
.
\
-(x,0)=2^^sin(-x)
=
we can
=
5|sin
(6.37):
00
fix)
nn
+
solve these
equations,
into Fourier sine series.
for these
are
just
the
expansions of/ and g
following
We collect this discussion into the
proposition.
Proposition 3. If the functions f g defined on the interval [0, L]
(say at least twice differentiable), then the wave equation
are
well
behaved
dx2~~?dT
with the
boundary datayifJ, t) 0, y(L, t)
(dy/dt)(x, 0) g(x) has a solution. The
=
=
yix, t)
(nnt \
=
n
=
l
r.
=
0 and the initial data y(x,
solution is given
tnnt\\
by
(ttnx\
A"C0S[Tc )+Bsin(-)jsin( )
0) =/(x),
The One- Dimensional Wave and Heat Equations
6.4
487
where
rL
2
=
"
(nnx\
.
^^
IJ
2c rL
,
B"
Sm\~L~)
~n
(nnx\
.
_
^
J
=
,
Sinl~L~)
Examples
14. Solve the
equation
wave
d2y 1 d2x
ch?~4li2
on
the interval
yix, 0)
Now
so
A
B_
=
sin 2x
=
c
=
=
4
[0, 7r] with initial data
2, L
(dy/dt)(x, 0)
c"
n
,
.
.
nnJo
=
0
if
n
is
x
,
=
4
,
dx
y(x, 0)
is
just
sin 2x,
Now
1.
(" 1
cos
-
2x
.
2
nnJo
t
N
sin(nx)
=
,
dx
even
We concentrate
Bn
2, A2
=
sin(nx)
x
sin
sin2
The Fourier sine series for
7t.
=
0 unless
=
now on
2
2
2
nn
n
nn
rn
=
Jo
the
case
where
cos(2x) sin(nx)
n
is odd :
dx
(6.41)
Now
f*
'o
cos(2x) sin(nx)
dx
cos(2x) cos(nx)
71
2
0n
sin(2x) sin(nx)
n
4
+
r"
("
r
n
Jo
cos(2x) sin(x)
dx
488
6
Functions
the Circle
on
iFourier Analysis)
Thus
/
4\ r"
A
1
\
/ Jo
n
2
1
dx
cos(2x) sin(nx)
=
(1
-
n
cos
nn)
=
(n odd)
-
n
Now, putting the result of this computation into (6.41):
-16
2n
2
B
=
4
n
n
nn
Thus the solution is
n\n2
given by
"
16
v(x, t)
=
cos f sin
4)
-
2x
7r
,,
i
odd
=
n
=
cos t sin
n
15. Solve the
y(x, 0)
The
2x
nz(n2 -4)
sin
=
x
expressions
can
transfer.
this
=
cos
-
+ 2m
-
3)
with initial data
dy
(x, 0)
are
=
0
the Fourier sine series
read off the solution:
5t
sin
x
sin 5x + 2
+ cos
2
cos
3r sin 6x
2
Transfer
Another
which
l)2(4m2
for the initial conditions
we can
sin 2mx
1
TTiZj
dt
t
Heat
sin(mf)
+
equation
+ sin 5x + 2 sin 6x
for those functions; thus
y(x, t)
2
m^i 7;
(2m
same wave
.
sin nx
,
16
y(x, t)
nt\2
sin
)
physical problem
be solved in
which
gives
rise to
a
partial
differential
equation
similar way is the problem of one-dimensional heat
We shall derive this equation here (the derivation in Chapter 8 of
a
higher dimensions shall be seen to be completely analogous).
Suppose
given a thin homogeneous rod of length L lying on the
horizontal axis. Let u(x, t) be the temperature at x at time t. We assume
that there is no heat loss, and the temperatures at the end points are maintained
constant.
Now the basic physical law here is that the flow of heat is pro
the temperature gradient, but points in the opposite direction.
to
portional
equation
we
in
are
The One- Dimensional Wave and Heat Equations
6.4
Thus, during
small interval of time At the heat
a
(energy) passing
489
from left to
If we select
x0 is proportional to (du/dx)(x0) At.
a segment of the rod with end points
x0 and x0 + Ax the increase in energy
in that segment of the rod is proportional to
right through
point
a
du
/
\
du
(
\
-(--(x0 Ax)At) (--(x0)A,)
+
+
(6.42)
On the other hand, the increase in energy is proportional to the product of
mass and the change in temperature.
Thus (6.42) is proportional to
the
An
Letting k2 be the
Ax.
period
constant of
proportionality
we
have, for the
of time At:
Am
Ax
k2
=
ox
Dividing by
At and
Ax
(x0
+
Ax)
letting
-
dx
(x0)
At
both tend to zero,
we
obtain the heat equation :
d2"
1 5"
(6-43)
i?o-rdi?
We
now
propose to solve
M(0, 0
=
0
(6.43) given
u(L, 0
=
the
boundary conditions
0
(6.44)
and the initial temperature distribution
(6.45)
u(x, 0) =/(x)
The
technique is the same as that for the wave equation.
u(x, 0 F(x)G(t). (6.43) becomes
of the form
We try
a
solution
=
1
G'(t)F(x)
=
T1F"(x)G(t)
k2
Dividing by F(x)G(t),
_
~F=X
again
find that there must be
a
A such that
X
G'
F"
we
_
G~k2
equation, subject to the initial conditions (6.44) again has only the
nni/L. The
solutions sin(7inx/Z,), n > 0, corresponding to the choices yJ~X
The first
=
490
6
second
Functions
equation
n
,
the Circle
on
(Fourier Analysis)
becomes
n
G=-L^G
which has the solutions
T22
G(0
For
=
exp^)t
convenience, let
us
write C
00
=
n/Lk.
We
now
try
to fit the series
/nnx\
24,exp(-C2n20sin()
to
the initial conditions.
Evaluating
of/(x).
(6.46)
at t
=
0
we
find that the
{An}
must be
the Fourier sine coefficients
Proposition 4.
(say
at
least twice
If the function f defined on the interval [0, L] is well-behaved
continuously differentiable), then the heat equation
d2u
1 du
k2~dt==~dx2
with the
f(x)
has
boundary
a
solution.
2
A
data
=
j-Lrr
y(0, 0
=
The solution
.
0
y(L, t) and the initial condition u(x, 0)
is given by (6.46) where C
n/Lk and
=
=
=
(nnx\
Zjo/(x)sin^ jdx
equations readily and conveniently led us to the
analysis. Actually this could have been (and in
fact was) anticipated on physical grounds, for we should expect periodic
behavior in these circumstances. Other partial differential equations arising
out of physics can be solved by similar techniques, but we do not necessarily
end up with a sequence of solutions of the general equation which are made
Thus the Fourier analysis does not apply,
up of trigonometric functions.
Now, the
wave
and heat
considerations of Fourier
whereas the fundamental ideas may carry
this: a partial differential operator P is given
a
solution /of
Pif)
=
0
over.
on a
The
typical situation
certain domain D;
we
is
seek
6.4
subject
to certain
The One- Dimensional Wave and Heat Equations
boundary
conditions "5" and initial data
491
/(x, 0) g(x).
First,
boundary conditions B,
subject
without regard to initial conditions. \f {Su ...,Sn, ...} are these solutions,
then we try to find a linear combination
2 a Sn which fits our initial data:
we
find all solutions of P(f)
=
0
to
=
the
2ZanSn(x,0)=g(x)
In
our
typical
situation the
convenient inner
are
product
readily computable :
o
The
=
S(x, 0)
orthonormal in the
are
the space of all initial data.
sense
In this
of
case
some
the an
<Sn g}
,
cases
cases
on
of the heat and
of this method.
wave
There
are
equations
many
described above
are
of such
examples
more
just special
orthogonal
expansions; discussions of them can be found in most texts of mathematical
physics.
Finally, we cannot really expect to be able to follow through such a
program for every partial differential equation, thus the general theory does
In one approach, local solu
not follow such an explicit line of reasoning.
tions are sought through examination of Taylor expansions (everything
involved is assumed analytic). This is the Cauchy-Kowalewski theory. A
more
recent attack has its roots in the above
ideas,
well
as
as
the Picard
The vector space of differentiable functions is provided with a
tion of distance and length which is suited to the given problem so that one
theorem.
no
can
questions of existence and uniqueness (as in the Picard theorem) and
provide usable approximations with estimates derived from the initial data.
This study is one of the most active branches of modern mathematics.
resolve
EXERCISES
10. Solve the
d2y
wave
equation
d2x
Jx2~'dt2
the interval (0, 1) with the boundary data >'(0, 0
following initial data
on
(a)
Kx,0)
(b)
y(x, 0)
=
=
sinx
cos3
jf
dy
77X
cos nx
(x, 0)
(x, 0)
=0
=
0
=
sin
=^(1, /),
nx
and the
492
6
Functions
on
the Circle
(c)
Xx,0)=x(x-1)
(d)
y(x, 0)
(Fourier Analysis)
^(x,0)=0
dy
=
cos 77X
at
(x, 0)
=
sin
=
sin
77X
*
(e)
11
=
dy
0
at
Solve the heat
.
3tt
(x, 0)
it
x
2
+ sin
-
x
2
equation
d2u
du
=
~dt
on
yix, 0)
A~c~x2
the interval (0, L) with the
(0, 0
=
and the
0
=
boundary data
u(L, t)
following initial
(a)
u(x, 0)
(b)
u(x, 0)
(c)
u(x, 0)
(d)
u(x, 0)
=
sin
=
cos
data
x
77 X
=
x(x
L)
-
577X
77X
sin
+ 3 sin
(a) Show that the function uix, t) =ax + b solves the heat equation
(0, L), with boundary data
12.
on
=
the interval
u(0,t)
=
b,u(L,t)=aL + b
(b) Show that if
u(0,t)
=
t,
u,
u(L,t)
v
=
solve the heat equation with boundary data
t2
w(0, 0=0
+ solves the heat equation with the
(c) Solve the heat equation
then
du
d2u
~t ~~dx~2
v(L, 0
same
=
0
boundary
data
as u.
6.4
The One- Dimensional Wave and Heat
the interval
493
Equations
(0, 1) with boundary data (0, 0
1, (1, 0
u(x, 0) e".
initial data given in the problem of heat flow may be the
on
=
=
initial data
e' and
=
13. The
rate of
flow of heat energy; or what is the same, the gradient of the temperature.
Show that the solution of the heat equation
1
d2u
du
_
k~2~dt~'dx2
the interval (0, L) with
on
data
(dujdx)(x, 0) =/(x)
is
boundary
given by
.
77HX
n=l
J-,
data
u(L, 0)
=
0
=
u(L, t)
and initial
2 Aexp( C2n2t)cos
where C is
Z
2
A=
nn
a
constant, and
77X
(
/(x)cos
Jn
dx
L
14. Solve the heat
equation given
dujSx(x, 0) cos nx/L,
du/dx(x, 0) sin 77X/Z,.
Solve the differential equation
(a)
(b)
15.
=
d2u
d2u
_
dX2
on
in Exercise 11 with this initial data:
=
~~d~y~2
+
"
the interval (0,77) with boundary data
initial data
u(0, t)
=0
=
(1, 0
and the
u(x, 0) =/(x).
16. Do the
same
where the differential
equation is
d2u du
d2u du
_
~dx~2~dt'~~dt2l)x
PROBLEMS
defining the function y(x, t) in Proposition 2
uniformly and absolutely under the stated conditions. Does this
15. Show that the series
converges
observation suffice to deduce the conclusion of Proposition 2 ?
16. We may be given, in the heat problem, the gradient of the tempera
Show that the general solution of the heat equation
ture as boundary data.
with boundary data
du
8
-(o,o=o=-a,o
494
6
Functions
can
be written
the Circle (Fourier
on
as a
Analysis)
Fourier cosine series.
Solve the
equation
d2u
du
~dt~~dT2
on
the interval (0,
du
t-
dx
77) with the boundary conditions
du
(0,/)=0
=
dx
(t,/)
and the initial conditions
u(x, 0)
(a)
=
sin
x
du
(b)
dx
(x, 0)
=
sin x
17. Solve the differential
equation
t'2u
du
~
dt
dx2
with the boundary data
0.
tions u(x, 0)
du/dx(0, t) =0, du/dx(L, t)
=
h and initial condi
=
18. Solve
on
Laplace's equation
d2u
d2u
dx
dy2-
the infinite
rectangle
0 <y <L,0 <x
(see Figure 6.4) with the boundary
values
u(x, 0)
=
0
=
w(x, L)
"(0, y) =f(y)
du
7r(0,y)=g(y)
dx
Show that the
assumption
that
u
is bounded
implies
that the third condition
is unnecessary: the solution is uniquely determined by its boundary values.
19. Find the bounded solution of the differential equation Am + u
0
=
in the infinite rectangle (Figure 6.5) with the boundary conditions
w(x, 0)
=
0
=
"(0, y) =f(y)
(x, L)
The Geometry
6.5
495
of Fourier Expansions
Figure 6.5
The
6.5
We
of Fourier
Geometry
return to the
now
functions of
Fourier series of
a
of functions
study
us
the
consider the real Fourier series of
a
that
circle;
function converges to that function ;
we
is, periodic
in which the
sense
have
only Corol
uniform convergence.
continuous real-valued function/:
laries 1 and 3 of Theorem 6.1 which deal with
Let
on
We still have not studied the
2n.
period
Fxpansions
pointwise
00
A0
[A
+
cos nx
+
B sin nx]
(6.47)
n=l
A0
If*
fix)
=
2n J
dx
An
=
-n
1 f"
fix) cos
-\
nJ-n
nx
B
Since the Fourier series of
applying
these definitions to
cos nx
sin
mx
dx
a
trigonometric
cos nx,
=
I
sin
_
nx
sin
mx
dx
=
n
J
fix)
sin
nx
dx
-n
we
find, by
(6.48)
m
#
m
n
n
=
m
# 0
2n
n
=
m
=
I
[n
-
function is itself,
n
10
J
1 c"
=
sin nx, that
all n,
0
dx
n
=
m
(6.49)
0
(6.50)
496
6
Functions
on
the Circle
(Fourier Analysis)
geometric way of interpreting these equations which sheds light
subject. We consider C(Y) as a vector space endowed with the inner
product
There is
on
a
the
</> 0>
This inner
f
=
J
dx
fix)gix)
-IT
product, of
course, defines
a
notion of distance
(recall Section
lid
1/2
Il/-ffll2
which is
f
=
J
l/(x)
-
gix)\2
dx
distinct from the uniform,
quite
(6.51)
ir
or
supremum distance
||/- ^|| =max{|/(x)-<7(x)|: -7r<x<7t}
We shall call the distance
of convergence in this
(6.51) the
sense as
mean
mean
square distance, and we shall speak
More precisely,
square convergence.
0 as n
oo.
/ -?/(mean square) if ||/ -f\\2
Now the importance of the equations above is that they imply that the
functions cos nx, sin nx are mutually orthogonal in the vector space C(Y)
with this inner product. Thus, we can interpret (6.47) as an orthogonal
expansion. Let us make these new definitions:
->
1
^
,
Coto
=
^
,
i2n)
cos nx
x
Cix)
7TTT72
->
=
n
_
sjn
Then the collection
orthonormal set.
A
Ao
nx
=
j=-
Jn
Cn,Sn is, according to Equations (6.48)-(6.50),
If/ is any function on the circle,
1_
~
sin
_
S(X)
7
(2*)1'2
"
f
J-
ff
,
1
,
dX
(27T)1'2
An='ff(x)CSipdx
JnJ-n
Jn
</, Cq>
(2k)1'2
=
<f'C">
yjn
1
b.-C mdx
Jn
_
~
=
<*$>
an
6.5
so
the Fourier
The
expansion (6.47)
</, c0>c0
can
[</, cnycn
+
Geometry of Fourier Expansions
be rewritten
+
497
as
</, s>s]
n=l
and is thus the infinite-dimensional
element in
an
interpretation
to
of the
space in terms of
has
consequences for
Theorem 6.3.
(6.47) be
analog
inner
product
important
Let
f
be
continuous
a
orthogonal expansion of an
orthonormal basis.
an
This
us.
function
on
the unit circle, and let
its Fourier series.
all trigonometric
(i) Among
f is
polynomials of degree
N, the closest
at most
N
A0
iAn
+
+
cos nx
Bn sin nx)
(6.52)
n=l
(ii) (Bessel's inequality)
i-
f
Zn J
\f(x)\2
dx
>
V
1
+
z=\
-K
Proof. In order to verify
expansions (Theorem 1.8).
these
(A2
facts,
+
we use
The functions
B2)
(6.53)
the basic theorem
C0,Ci,
.
..,CN
,
Si,
.
on
orthogonal
SN form
..,
an
orthonormal basis for the space SN of trigonometric polynomials of degree at most
The orthogonal projection of /into this space is
TV.
/o
=
</, Co>c0 +
2 </.
n=
c>c. + </, sys
1
same as (6.52). Thus, according to Theorem
(0 ll/ll22=!l/ol[22+ll/-/oll22
(ii) foranyweS*, IIZ^/V < 11/- wY22
(ii) directly implies Theorem 6.3(i). According to (i),
which is the
il/iu2
>
11/0II22 =/, c2
2 i'f c2
+
+
1.8
/, sn/)2
n=l
ll/ll22>277^02
+
77
2
A2 + B2
n=l
Since this is true for all N,
ing Bessel's inequality.
we can
take the limit
on
the
right
as
N
x
,
thus obtain
498
6
Functions
on
the Circle (Fourier
Analysis)
Now, it is clear that for trigonometric polynomials, Bessel's inequality is
actually equality. For if/ is such a trigonometric polynomial, there is an
N such that/e Sw so/
Thus, by (i) above ||/||2= ||/0||2, and ||/0||2 is
/0
just the right-hand side of Bessel's inequality. Since any function can be
uniformly approximated by trigonometric polynomials (although not
necessarily by its Fourier series), we should expect Bessel's inequality to be
always equality. This is the case.
=
.
,
Corollary. (Parseval's Equality) If f is
(6.47), then
a
continuous
function
on
the unit
circle and has the Fourier series
1
rn
f
2ji
J
\f(x)\2
dx
=
A02+l-f (A2
+ B
2
Z =1
-71
Proof. We continue the notation of Theorem 6.3. Let e > 0 be given. By
Corollary 4 of Theorem 6.1, there is a trigonometric polynomial w such that
\'w
/|| < e. Then
||vf
-f',22=j
-
\w~f\2dx<\v-f\'2
j
dx<2ne2
Now, since w is a trigonometric polynomial, there is an
/o be the projection of /into S,v. Then by (i) and (ii),
ll/"22
=
This becomes,
"/o V + !!/-/o'!22
as
l/(x)|2
dx < 2nA02 +
1
r-
2
77
n
sum
\
Zn J
to
<
j./o V + 2772
A,2 + B2 + 2t72
i
=
infinity only increases the right-hand side,
l/(x)|2 dx
_
H/ol 22 + |'/- W\\22
Let
in the above argument,
'-it
Since the
<
TV such that weSn.
<
A02 +
-
2
( A2 +
z=i
B2) +
f'
Now, since e was arbitrary we may let it tend to zero. The resulting inequality,
together with Bessel's inequality, gives Parseval's equality.
Finally,
we
note that Parseval's
equality
can
be
expressed in
terms of the
into
expansion
series of
a
of Fourier Expansions
The Geometry
6.5
f(n)e'"B.
complex exponentials:
n=
/(0)
A02
so we
f(n)
^0
=
A2
|./(0)|2
=
\(An
=
+
+
B2
fi-n)
iB)
=
2(|/()|2
=
499
Since
oo
HAn-iBn)
\fi-n)\2)
+
have
z-f
\fid)\2d9=
Zn
J
jr
\f(n)\2
n=
oo
Examples
16. Since cos2 6
1r
=
02
-
\
77 + 7 + 77
ie'29,
+
=
2Tt2/3
=
n
8
16
4
16
27T J-n
tt2
+
1113
COS4 0d0
17.
\e~i2B
=
(-1)V"V
+ 2
71*0
47T4
167t5
,
+
J
.9
*-^
-n
j_
,ion4
We conclude that
i
n4
The
90
partial
sum
to
2_2
F3(9)
degree
3 of the Fourier series of
0 +
cos
The square of the
mean
cos
-
28
-
-
cos
?44
71*
square distance between 92
1
1
16
81
1
1
8
~~
"81
10
1
1
9
16
81
~
90
_3_
16" 80
9Z is
30
is
1
-
2
i
2
=
n2
-
n2 and this
sum
500
6
Functions
the Circle
on
(Fourier Analysis)
Figure 6.6
In
Figure
18.
\9\
6.6 the
has the Fourier
2~rcA00(2n
+
From Parseval's
tt2
tt^ +
_
~
3
4
The third
r,
n
,
F3(0)
The
-
92 and F3(9)
are
drawn.
expansion
ei(2"+1)9
2
n
graphs of 7t2
=
l)2
equality,
4
find
1
n2
partial
2
/
sum
r
^o (2n + l)4
of the Fourier series of
cos
+
tc4
1
i0 (2n + l)4
---(cos(,
mean
we
_
~
|0|
96
is
30\
)
square distance between
|0| and this trigonometric poly-
6.5
The
Geometry of Fourier Expansions
501
nomial is
1
1
*
,
4
*
t'2(2n
l)4
+
1
3
"
96
81
^
96
81
96
(see Figure 6.7).
Mean square
approximation is interesting from the physical point of view.
wave equation
(suitably normalized)
Consider the solution of the
u(x, 0
(A
=
n
The
(kinetic)
/
=
cos nt
+
sin
B
nt) sin
(6.54)
nx
0
energy of the
wave at
time
t
is
proportional
to
du
dx
dt
Now, by Parseval's equality that
Fourier sine coefficients of du/dt:
can
easily
be
computed in
terms
of the
du
=
ejt
2 niB
=
cos nt
-
A
sin
nt) sin
nx
o
du
dx
const
dt
=
(In2(B
(Because of our normalizations,
cos nt
-
A sin nt)2)
the constant is not
Figure 6.7
relevant; it might
as
wel
502
6
be 1 .)
Now the maximum value of the
Functions
on
the Circle
(Fourier Analysis)
right-hand
side is
CO
n\A2
B2)
+
71=1
(see Problem 20) so this is the maximum kinetic energy of the wave. Now,
according to our geometric considerations above, the Mh partial sum of
(6.54) provides the best approximation to the solution wave in the sense of
energy.
wave
Furthermore, the difference in
and this
approximation
n2A2
is
energy levels between the solution
readily computable,
it is
B2)
+
n>N
Since energy is the
square
important concept in the study of
approximation is well suited for this study.
waves, this
mean
EXERCISES
17.
Compute these integrals by Fourier methods:
(a)
J
cos8 3d dd
71
fid dd
sin2
(b)
71
(c)
an
integer
l-r2
L
I + r2
TC
(d)
/j, not
r
-
2r
cos(8
-
<j>)
2d<j>
l-r2
,
1
L LCS<7,l+r2-2rcos(e-oi)J d<f>
Tt
18.
(e)
j
(f)
f.'
82d8
4d6
Approximate
within 10-3 in
(a)
\8\8
cos
(b)
(c)
(d)
the
mean.
nd
^(W^
sin3 8
ecos
cos
8
given
function
by
a
trigonometric polynomial
to
6.6
Differential Equations
on
the Circle
503
PROBLEMS
20. Show that the maximum of
A sin nt)2 is B2 + A2
(B cos nt
{/} be a sequence of continuous functions on the circle. Show
that if / ->/ uniformly, then / ->/ in mean. Show by example that the
21. Let
converse
statement is false.
22. Prove:
1
=-
f
We
now
g
are
integrable
real-valued functions
on
the circle
f(8)g(8)d8= 2 fin)g(n)
Differential
6.6
if/,
Equations
turn to a
on
the Circle
slightly different problem involving ordinary
differential
equations. We propose to find all periodic solutions of a linear constant
coefficient equation. The particular theory which results is not in itself of
vital
importance,
but it is worthwhile to
study because of the symmetry of the
general theory of
results and because it presents the simplest example of the
differential operators on compact manifolds.
As
seen, it is valuable in the theory of ordinary differential
complex-valued functions. We return then to our
of the Fourier expansion of a function/: 2/(")^'"e- Our first
concerns the computation of the Fourier coefficients of the derivative
we
have
already
equations to
original form
result
of
a
allow
function.
Proposition
Let f be
5.
The Fourier series
fin)
=
The
f.
off
a
continuously differentiable function on the circle.
by term by term differentiation. That is,
is obtainable
infn)
(6.55)
proof is by integration by parts.
Tt
Tt
fin)
=
^/
f'(8)e>" d8
=
=
^
in
f(8)e-'
+
-,
f
Tn J
f(8)e-' >d6
infin)
Thus, if the differentiable function / has the Fourier series 2 A e'"9, then
/' is 2 "A e'"6- I* follows from the fundamental
the Fourier series of
theorem of calculus that
so
long
as
it has
no
we can
also
integrate
Fourier series term
constant term: if /has the Fourier series
by term,
2 AemB,
then
504
6
Jo /has
the Fourier series
Functions
on
the Circle
(Fourier Analysis)
2 iin)~1AeinB.
A useful consequence of
osition 5. in conjunction with Bessel's inequality is that
differentiable function is the sum of its Fourier series.
6.
Proposition
^L-a)fin)eM
is
If f
Prop
continuously
a
continuously differentiable function, then f(9)
a
=
holds for all 0.
By Bessel's inequality
Proof.
2 |/'()l2<>
co
n=
Using the
above
2
proposition
l/()i
=
<
Thus
2 l/(")l
1/(0)1
+
a
fin)
by Schwarz's inequality
i
1/(0)1
1 of Theorem 6.1
continuous function
*=o aiX',
we
< oo
applies.
Given
the circle.
on
want to
fin)
2
+
( 2 Vf2( 2 \f\n)\2\12
Corollary
Now, suppose g is
F(X) Xk +
=
then obtain
l/(0)i+2
< o, so
nomial
we
find
a
periodic
a
poly
f
function
such that
fm+
i
The fact that
zV=ff
=
(6.56)
0
we are
interested in
periodic
functions is
local results, such as Picard's theorem, are hardly
consider the simplest differential equation :
a new
applicable.
f'=9
=
0.
we
f m)deb
J
example,
+
know that /must be
c
-n
However, / will be
must
For
(6-57)
By local considerations
fiO)=
twist and the
c: for this we
only if f(n) =f( n)
0.
Thus (6.57) has a solution if and only if $(0)
have J_: g(eb) deb
We have already recognized this condition in the above discussion of
a
periodic
=
function
=
6.6
integration of
g(n) for all n.
must have
Now
Fourier series.
(0)
we
Differential Equations
For
by (6.55), if/'
This necessary condition shows up
0.
=
return to the
we
must have
again by taking
n
505
inf(n)
=
0:
=
we
general
case
If
(6.56).
F(in)f(n)
=
we
g(n).
look at the Fourier
Thus
we
must have
Otherwise, the equation does not have a
On the other hand, if this condition is satisfied, then the equation
0 whenever
solution.
is
g
the Circle
=
series of both sides this becomes
(n)
=
on
F(in)
easily solved since
=
0.
must
we
have/()
=
F(i/i)-1^(n).
The solution is the
function whose Fourier series is
f iWgfa*
(6.58)
n=^oo F(in)
Theorem 6.4.
F(iri)
=
0.
Let
Let
LF
F(X) 2*=o ctXl, and
differential operator
=
let nu...,na be the roots
of
be the
Lf(/)=ci/(i)
i
=
0
(i) The space of periodic solutions of LF(f) 0 is spanned by exp(/10),
...,exp(ina6).
(ii) Given any periodic function g, the equation LFf= g has a solution if and
only if(nt) 0, 1 < i < a. The solution is uniquely determined by specifying
the Fourier coefficients] (n^, 1 < i <o.
=
=
{F(in)f(n)).
Now if LF (/)
0, we
0.
necessarily except when F(in)
Since nu ..., are the roots of this equation, (i) is proven.
g(n). Thus
If g is a periodic function and LT(f) =g, we must have F(in)f(ri)
now that
for
this
condition
<
ct
is
a
<
Suppose
1
/
equation.
necessary
g(ni) =o,
this condition is satisfied. Then if /is a solution we must have
Proof.
must have
The Fourier coefficients of LF(f)
F(in)f(n)
=
0 for all n,
so
/()
are
=
=
0
=
=
/()=|^r
(6-59)
#!,...,.
and the f(nt), 1 < i < a can be freely chosen. Upon specification of these coeffi
cients the Fourier series of /is uniquely determined. The only question is this:
function? The answer is yes
are the numbers (6.59) the Fourier coefficients of a
then
For
one.
at
least
\F(iri)\ >C\n\ for some constant C
when F is of degree
and all sufficiently large
gin)
F(in)
<lc
n
(Problem 24),
gin)
n
<
and thus
c(2^Y2(2l^m1/2<
(6-60)
506
6
Functions
the Circle
on
(Fourier Analysis)
for the tail end of the series, and thus the
the Fourier series
=_
We
is
uniformly
can
get
a
to
looking form for the solution,
For then
second degree).
least
ei"9
=
(6-62)
(Problem 24)
1
ti
-TT-T-^
n= -oo
-
F(in)
eb)
We
in terms of
integral formula.
can now
Let
of F(in)
nomial F.
Let
=
F(X)
Let
0.
deb
write the conclusions of Theorem 6.4
using (6.61).
Theorem 6.5.
(6.61) is given by
e'n(8-<t>)
9i4>)
^f 9i<b)FiO
solutions
and the solution
n
oo
T-f
2nJ-n
an
of F
F(in)2nJ-n
=_,
1
degree
F(in)
continuous function
oo
if the
TT",
=-,
=
The theorem is thus proven.
continuous function.
a
P(W=
a
Hence
(6-61)
much better
large enough (at
defines
of the whole series is finite.
J^'""
F(m)
/(*)= I
converges
sum
=
LF
2), and let nu...,na be the
differential operator defined by the poly
jL0 ctXl (k
be the
explicitly
>
emB
m
=
F(in)
=-oo
77^771,
",
tin
Then the equation LT(f)
g has
the
are
All solutions
form
of
=
fiO)
=
^f
znJ-n
9i4>)H0
a
solution
if and only if g(n ,)
-eb)deb+i Cj exp(inJ- 0)
j=\
=
0,1
<i<o.
(6.63)
6.6
Thus
a
Differential Equations
constant coefficient differential
of the form
(6.63) (defined
on
its
operator
range),
called
on
an
on
the Circle
the circle has
an
507
inverse
integral operator.
Examples
19. Find
gi9)
=
cos
periodic
a
29
=
i-ie120
The characteristic
has
no
solution of /"
cos
29.
Now
e-'2B).
+
is
polynomial
Thus there exists
roots.
/=
1 and Fiin)
n2
1
X2
FiX)
a unique solution and it is given by
=
=
(6.58):
=
g(n)e"">
/"
-
3/'
2f=n2
+
(X2 -3X+2)
F(X)
no integral
=
-i26\
e~'2B\
(e'2B
29
cos
-
20. Solve:
is
11 /"29
f"8
-
has
roots.
-92.
The characteristic
(X- l)(X
=
Since n2
-
-
92 has
polynomial
0
2) so that again F(in)
(by Example 4) the Fourier
=
series
einB
2tt2
^- 2S(-D-^r
+
j
n
ti#o
the solution is
/(0)
=
21.
(by 6.58)
^-22(-D" n2in2
f(k)
solution is
=
g.
This has
a
einB
+ 3m
-
2)
solution if
0(0)
=
0.
In this
case
the
given by
einB
fie)=C+2Zgin)k
n#o
(in)
=
22. Find all solutions of /" +4/= 0. Here, the roots of X2 + 4
0 are 2i, therefore, all solutions are periodic of period 27t: e2'9,
e~2'9 span the space of solutions.
Notice, however, that there
solutions off" + 5/= 0 which
periodic.
are
are no
508
6
Functions
on
the Circle
(Fourier Analysis)
EXERCISES
19. Find all
periodic solutions of these differential equations:
y" + 2iy'+l5y=0
0
v(4) + 10/ 3)
10/' + 9/
9y
/5)
/4) + 2/' + l=0
20. Find periodic solutions of these differential equations :
sin 58 + cos 50
(a) v<4> + 2y" + v
(b) y" + 6y' + 9y n<-82
expfcos 8)
(c) v<5) + v
(a)
(b)
(c)
-
-
=
-
=
=
=
PROBLEMS
23.
F is
Suppose
C> 0, and
an
polynomial
a
integer
24. Show that if F is
Fie)-.
n= -oo
t
is
a
not
root
a
If/
of
degree
Show that there is
n
at least
a
> N.
2, then
.FO'H)
are
^
polynomial
a
continuous function
25.
of degree at least k.
\F(iri)\ >C\n\k for
TV" such that
on
the circle.
two continuous functions
convolution of /and g,
on
the circle,
define/*^, the
by
1
(f*g)iO)=^\
f(<j>)g(6-<j>)d<j>
(a) Show that/* g =g */.
(b) Show that the Fourier coefficients off*g are f(n)g(n).
(c) Show that the differential equation Lr(f) g, where LF is the
constant coefficient operator associated to the polynomial F is solved by
f g* F, where F has the Fourier coefficients F'iri)'1.
=
26. Let
P1(r,t)
=
\
P(r,r)dr
(a) Show that
lim
Pi(r, t)
is 0 if
t <
0, and 1 if
/ >
0.
r-l
(b) Show that for
*(0)
=
lim
r-*l
any C1 function g
on
the circle
f g'(4,)Pi(r,8-^)d<t,
J
_
00
(////: lim Pi(r, 0) "has the Fourier series"
2 e'^/in.)
6.7
Taylor Series
6.7
Taylor
Series and Fourier Series
509
and Fourier Series
boundary of the unit
disk we discover connections between Fourier series and Taylor series which
These con
are of enormous significance in complex function theory.
the
next
the
fact
nections cannot be fully exploited until we learn
(in
chapter)
that complex differentiable functions can be expanded in a power series. In
this section we shall explore the relationship between the Fourier and Taylor
expansions of such a function defined on the unit disk, assuming its Taylor
If
take the attitude that the unit circle is the
we now
series converges
Let/(z)
on
the disk.
=0 anz"
=
on
tne unit disk.
In
polar coordinates this becomes
CO
f(rew)
(6.64)
Y,anrHeine
=
77
=
0
which is, for each r, a Fourier series. Using the Fourier theoretic material
at hand we get a most remarkable collection of integral formulas for func
tions which are sums of convergent power series.
00
Theorem 6.6.
Letf(z)
*0 anz"
=
71
Then
(i)
we
For each
r <
f
2nJ-n
(ii) M
=
convergent power series in {\z\ <, 1}.
1,
C f(rete)e-inB d9
z
a
have these equations.
=
2nJ-n
for
be
=
=
reie,
fireie)einB
d9
=
0
an
=
^/(n)(0)
n >
0
n\
n >
0
kfj{ei*\ r2Xrcos(9-eb)d+
+
r <
1.
By Equation (6.64), for fixed r, ar" is the nth Fourier coefficient of
f(rew) for n > 0, and for negative n the Fourier coefficient vanishes. This is just
Proof.
510
6
Functions
the Circle
on
(Fourier Analysis)
what part (i) says explicitly. Equation (6.64) also says that f(re'e) is the Poisson
integral of f(ei0) and thus we obtain (ii). Then (iii) follows from resumming the
series, using the fact that the negative coefficients vanish.
|
f(z)=
71
=
2
az=
0
71
=
have
f(<t>)e-1"* d<j>
_
r"
e'*
2n\j^7^zd*
=
the last
we
/ z\"
If"
1
ff
0 Z77 J
Explicitly,
change being accomplished by summing the geometric series.
There
are
several
the above theorem.
more or
less immediate conclusions
First of all, the
sum
one can
draw from
of a convergent power series
on
the
unit disk is
completely determined by its boundary values, by Equation (iii),
known as Cauchy' s formula. This of course follows from the maximum
principle verified in the last chapter for analytic functions. The Cauchy
formula itself implies the maximum principle (see Problem 27). A more
important implication is that the sum of a convergent power series in the disk
is analytic; that is, it can be expanded in a power series centered at any
point.
Corollary. Let f(z) 2tT=o anz" m the disk {\z\ < 1}.
\zo\ < 1,/cfln be expanded in a power series centered at z0
=
Proof.
Then
for
any z0,
By Cauchy's formula
1
e"
r*
Now
1
1
/
1
Z-Zp
_
e'*
In the disk
e">
z
e'*
-z0-(z- z0)
[ze C: \z z0\
< 1
e'*
-
zoj
z0
\
e'*
-
z0
\z0\, the last factor is the
series :
\
-
fo \e'*
-
zo)
T
sum
of
a
geometric
6.7
which
may substitute in the
we
/(Z)
the series
we
=
Taylor
Series and Fourier Series
511
We obtain
integral.
1[^/1/(^(^^^](Z-Z0)"
being convergent for all z such that \z
more integral formulas :
z0
\
<
1
|z0 1
.
As
a
consequence
have still
f^-fjji^j^-^^
for any
\z0\
z0,
We shall
<
1.
in the next
chapter that these integral formulas can be ex
(basically the fundamental theorem of calculus)
and are just very special cases of general formulas. We conclude now with
an approximation theorem which should be contrasted with the Weierstrass
approximation theorem (Problem 7) for a real variable.
see
in yet another way
plained
Theorem 6.7.
mable by
Let
f be
polynomials
C
27E J-n
in
z
fid)eM d9
a
continuous
function
on
the circle,
f is approxi-
if and only if
0
=
n >
0
(6.65)
/ is approximable by polynomials, there is a sequence {/} of poly
ft -s> f uniformly. Since (6.65) is readily verified for any poly
nomial, it thus holds also for / by continuity of the integral.
Conversely, if (6.65) is verified, then the Poisson transform of /is the sum of a
If
Proof.
nomials such that
convergent power series :
2z"
P/(z)=
71
Since
on
(6-66)
I*I<1
0
Pf(re")^-f(9)
\Pf(re,e) -f(8)\
Now
=
< e
as
r-*l, then given e>0, there is
the circle \z\
=
r, the
series
(6.66)
such that
Pfiz)-
2
an
r<\ such that
for all 8.
az"
<e
\z\=r
converges
uniformly,
so
there is
an
TV"
512
Then,
6
Functions
the unit
circle,
f(z)- 2
ar"z'
on
the Circle
on
(Fourier Analysis)
f(e">)- 2
71
=
a-r-e'"1
0
<\f(e">)-Pf(rel)\
+
Pfire")
-
2 aire")"
<2e
71=0
independently
of 8.
EXERCISES
21.
Integrate:
*
^3ie
(a)
/
(b)
L&
j_ Aiie
ie
Ae2te + e"
2e<"
,
dd
1
-
-dd
1/4)"
k,
n
positive integers
PROBLEMS
27. Deduce the maximum
principle for convergent
power series
on
the
disk from Cauchy's formula.
28. Using the results of Problem 11, verify that these assertions for
function defined on the disk are equivalent :
(a)
a
f(z) =2az"
71=1
f is
(b)
complex differentiable.
(c) /is uniformly approximable by polynomials.
0 for n > 0.
(d) /is harmonic and /( n)
=
6.8
Summary
The function
exp(2nit/L) wraps the real line around the circle so
that every interval of length L covers the circle once.
The collection of
periodic functions of period L may be viewed as the collection of functions
on
f(t)
=
the circle.
If/ is a piecewise continuous
efficient is
f(n)
=
^fj(eb)ei"Ueb
function
on
the
circle, its nth Fourier
co
6.8
513
Summary
The Fourier series off is the series
fin)eiB
=
n
oo
The Poisson transform
Pfir, 0)
=
f
f fi
2nJ-n
Theorem.
J// is
defined by
the disk
gir, 9)
off is
1
<
1 +
M
r
2
~
2r
continuous function
a
Pfir, 9)
=
the function
r
the unit disk
on
given by
T
n
m
(
cos({?
<p)
d(b
fin)r(n)eM
=
=-
of the circle andg
is the function
on
1
<
g(\,9)=fi9)
then g is continuous
inside :
d2g
s2g
+ -4
-|
dx2
dy2
n
for
0
=
r <
isatisfies Laplace's equation)
1
Iff is continuously differentiable
Theorem.
of
the disk and harmonic
on
on
the circle, then it is the
sum
its Fourier series:
Rn)eine
fiO)=
77=
Suppose
00
is harmonic in
u
a
closed and bounded domain D in the
plane.
Then
(i)
if
A(a, R)^D
1
u(a)
(ii)
if
u
r"
=
<
2n J
M
on
u(a
+
d9
(mean value property)
M inside D
(maximum principle)
RelB)
-Tt
dD,
u
<
A function harmonic on a closed and bounded domain is uniquely deter
mined by its boundary values.
Ce'"e, we can rewrite
If the real-valued function/has the Fourier series
514
it
6
Functions
on
the Circle
(Fourier Analysis)
as
00
A0
A
+
n9 +
cos
sin n9
B
71=1
where
^\[feb)dtb
Ao
=
An
=
2Re
C
B
=
2 Im
C
Co
=
*
If /is
C
=
=
+
C_
j(C,,
-
=
f f(eb)
-
C_)
-
=
-
n
n0 rf<
f /((/>) sin neb deb
J-*
C1 periodic function of period L, it
a
cos
can
be
expanded
in
a
Fourier
cosine series
fix)
S,
=
nnx
A0+ 2ZAn
COS
Ao=j
I fix) dx
L, Jq
or a
-
L
n=l
A
=
-
L,
f /(x) cos
Jo
-
dx
L
Fourier sine series
.S,
7tnx
/(x)= 2^nsin
L,
-
7t
2
B"
=
rL
l
,
s
.
nnx
limSm-L-
the wave
the
=
equation.
Given the C2
equation
~
dx2
with the
c2 a/2
boundary data
y(0, 0
=
0
y(L, t)
=
0
periodic functions /
g of
period
L
6.8
Summary
515
and the initial data
^(x,0)
yix, 0)= fix)
has
a
The solution is
solution.
=
5(x)
given by
nnt
nnt
yix, 0
An
=
cos
I-
-
Lc
B
nnx
sin
sin
-
Lc
where
^
ZJo/(x)sin()rfx B.-7Joff(x)sm()
=
the heat
equation.
Given the C2
periodic
function
equation
82u
1 du
l?dt~dP
with the
boundary
m(0, 0
0
=
data
=
u(L, t)
and the initial condition
u(x,0)=/(x)
has
a
solution
given by
oo
u(x,t)=
TJftX
Aexp(-CVOsin
where C
=
7T/Z.A: and
2
An
-
U
71=1
rL
LJ
Consider C(Y)
</, g>
=
nnx
,
fWsin~rdx
=
as
f"
J
it
endowed with the inner
/<?(*)
dx
product
/
of
dx
period
L the
516
6
Functions
the Circle
on
iFourier Analysis)
Let
1
CoW
is
{Cn S'n)
,
rewritten
=
=
orthonormal set.
an
,
7=-
sin
,
S(x)
_
c=^-
T^m
{2njT2
nx
=
The Fourier series of
a
function
/ can
be
as
</, cyc
71
cos nx
Cn(x)
=
</, s>s
+
0
17=1
parseval's equality
1 f" |/(x)|2 dx
ZnJ-n
=
V
+
Z=l
equations on
differential
1
G42
+
.
.
.
,
na be the
0.
a
Let
polynomial
LF
and let
be the differential
=
F(in)
=_oo
71^711
Tio-
equation LF(f)
All solutions
fie)
=
eM
He)
Then the
Let F be
the circle.
integer solutions of F(in)
operator defined by the polynomial F. Let
nx,
B2)
=
are
g has
=
a
solution if and
only
if g(nt)
=
0, 1
<
i
< a.
of the form
f f 0-)^(0 -eb)deb+t cj cxp(inj9)
OO
Le?
Theorem.
71
disk.
(i)
(ii)
anz" be
/(z) =2
=
Then these equations
are
^- f fireiB)e~inB d9
2n J
f* firew)einB d9
1
/*
1
.*
..
convergent power series
=
valid:
an
-.f\0)
n!
=
-n
271 J-
a
0
=
0,
n >
1
,,..
g'*
0
-
r <
r2
n >
1
0,
r <
1
on
the unit
6.8
If /is
differentiable in
complex
power series in
a
Summary
domain D, it can be
point in D.
expanded
517
in
a
disk centered at any
some
FURTHER READING
theory of Fourier series is exposed in these texts :
Seeley, An Introduction to Fourier Series and Transforms, W. A. Benjamin,
The
R.
Inc., New York, 1966.
Kreider, Kuller, Ostberg, Perkins, An Introduction to Linear Analysis,
Addison-Wesley, Reading, Mass., 1966.
Hardy and Rogosinski, Fourier Series, Oxford University Press, New
York, 1956.
applications to physics and the development of other partial
differential equations can be pursued in
E. Butkov, Mathematical Physics, Addison-Wesley, Reading, Mass., 1968.
O. D. Kellogg, Foundations of Potential Theory, Dover, New York, 1953.
Further
MISCELLANEOUS PROBLEMS
29. Let
2 /()
Show that
1/2 (see Example 7).
=
What is
2 (-!)"/()?
go
n=
30. Let
has
a
/ be
a
piecewise continuous function
at 0; that is, the limits
on
the circle.
Suppose /
jump discontinuity
lim f(x)
=
lim f(x)
a
=
b
*-0
*>0
*->0
x<0
both exist, but
are
different.
Show that
1
limPf(r,0)
7-.1
=
-(a + b)
^
(Hint: Follow the proof of Theorem
using the substitution
2
i*
"
i
i.
=
(" P(r, _# d<b
Jn
=
|
P(r, -$) d<j>)
J-n
6.1 for
0>O, 0<O independently,
6
Functions
the Circle
on
31. Show that
/is
an
(Fourier Analysis)
infinitely
differentiable function
on
the circle if
and only if this condition on the Fourier coefficients is satisfied: for every
k > 0 there is an M > 0 such that
M
for all
\fin)\<k
32.
f(z)'
JK
=
Suppose
P(z)
-^-JQiz)
where P is
an
n
integer
analytic on {|z| < 1} and Q is
complex numbers a0
TV and
,
aNf(k-N)+--(Hint: Let Qiz)
+ a0
=
aN
f(k)
z* +
=
a
.
.
polynomial.
,
Show that there is
such that
aN
for all k
0
+
.
<
0
State and prove the
a0.)
converse asser
tion.
Suppose that /is complex differentiable in the annulus [r < \z\
Using the polar form of the Cauchy-Riemann equations, show that
33.
2
f(z)=
n
=
-
where the
<
R}.
a.z"
x
a can
be computed from the Fourier coefficients
fin)
of f(pew)
for any p between r and R.
34. Show that if / is analytic in the punctured disk {0 < |z| <R) and
bounded, then /extends analytically to the entire disk.
35. Show that if
u
is harmonic in the disk
{\z
a\ <R], then for
every
r<R,
=
R2~r*
/"
1
uia + re'*)
uia + Re'*)
T1?\
R2 + r2
2nRJ_K
2
36. (Harnack's principle)
{|z-a| <R}, then
R-r
R +
r
R
37. Suppose {} is
=
u
-^cosi8
<f>)
d<f>
is harmonic and nonnegative in the disk
R+r
<uia + re<)<-
disk {|z
u(z)
If
2Rr
a\
<
R}.
a
r
sequence of
nonnegative harmonic functions
2 uia) < oo. Then
Suppose also that
2 u(z)
converges for all
z
in that disk, and
u
is harmonic.
on
the
6.8
519
Summary
38. Show that
(-1)" _7r
f
.fi
2n + 1
39.
4
the
Verify
trigonometric identity
sin(2TV+l)0
1
-
2
+
>
+i
cos
71
40.
SN(9)
sum
=
0
to be evaluated is
Using the identity of Problem 39, obtain Dirichlet's integral for the
sums of the Fourier series off:
=
2
AQ +
1
iA
cos
r" sin(TV
+)</>
sin
(/>
sin **
2t7 Jn
Using
"0 + # sin "e>
1
2tt J_
.
sin
1
n=
41
-
.
.
2 (e2'8)")
+ 2Re
Z
partial
=
2
The
(Hint:
\
2/70
the Dirichlet
tinuous function
on
integral (Problem 40), verify that for / a
point 0O
the circle which is differentiable at the
,
con
then
oo
/(0O)
=
A0 +
2 iA" cos "e<> + B" sin n6)
71=1
(Hint:
SN(80)
=
The
-
if"
f(80)
sin
=
TV^
expressions
1\
^ J sin^TV j ^
+
,
M>
r cos <j>f(8o + 4>)-f(8o)
;
2
in brackets
sin
are
+
-
</>
0 -/(<.)
i^
,,
sin
+
cos
Nftf (80
continuous functions.)
+
+)-/ (80)] d<[>
520
6
Functions
on
the Circle
42. Solve the differential
v(0)=0
(Fourier Analysis)
equation
y(L)=0
x.
by Fourier methods, where (a) g is constant, (b) g(x) L
43. Suppose we want to study the problem of heat transfer through a
homogeneous rod with insulated ends : heat does not flow through the ends.
=
If the rod is assumed to lie
the interval 0 <L
on
x
< L
this amounts to the
boundary conditions
du
du
-(0,t)=0=
dx
dx
(L,t)
The initial condition may be given either as the temperature distribution
(u(x, 0)) or the initial heat flow [(du/dt)(x, 0)]. Show that with these
boundary conditions and either kind of initial condition the heat equation
(6.43) can be uniquely solved.
problem with (a) constant initial
cos(x/L), (c) initial heat flow x(x
44. Solve the insulated end heat
perature, (b) initial temperature
45.
Suppose that
Show that
a0
+
2=i
u
is
a
=
=
tem
L).
real-valued function harmonic in the unit disk.
is the real part of an analytic function.
(Hint: Write u(z)
and
add
a
+
anz"),
pure imaginary-valued harmonic
(a-z~"
=
u
negative Fourier coefficients.)
on the unit disk, there is a unique
harmonic real- valued function v such that v(0) =0, u + iv is analytic.
Using the relation between the Fourier expansions of v and u (Problem 45)
find an integral form for v in terms of the boundary values of u.
47. If u and v are as in Problem 45, show that the families of curves
{u constant}, {v
constant} are orthogonal.
48. (The convolution transform) Let # be a continuous function on the
circle and define the transformation G : C(T) -* C(T) by
function with the
46. If
u
same
is harmonic and real-valued
=
=
n
i
G(f)(8)
=
j
f(<p)g(8
-
<)>)
deb
(a) the eigenvalues of G are the Fourier coefficients g(n) of g.
(b) The nonzero eigenvalues of g form a sequence converging to zero.
(c) The eigenspaces associated to the nonzero eigenvalues are finite
Show that
dimensional.
(d) The Fourier series of G(f) is
2d(n)f(n)e"
6.8
(e) G(u) =/has
a
Summary
521
solution if and only iff is orthogonal to the kernel
of G.
49. Under what conditions is the convolution transform
(Problem 48)
symmetric; skew-symmetric; one-to-one?
The
Laplace Transform
50. The
defined
on
Laplace transform is useful in the study of differential equations
the positive real axis, R+.
If /is a bounded function on R+,
define
L(f)(s)
=
j
Show that
e-fit)dt
L(f) is defined for
51. A function /defined
is
a
bounded function.
all
on
s
> 0.
R* is
of exponential order s0 if exp( s0t)f(t)
a function, L(f) is defined for
Show that for such
s>s0.
52.
Compute these Laplace transforms :
(a)
L(1)(j)=7
(b)
L(t")
(c)
L(e<)
n\
53.
=
(s-D
(d) L(e-')=?
Verify these properties of the Laplace transform:
(a) L is linear.
/un
(b)
v/vi
if/a(x)
=
[
x<0
Q
L(fa)
(c)
x>a,fora>0
//(*-)
=
e-asL(/)
L(/')=^(/)-/(0)
54. Notice that by the above problems we see that the Laplace transform
of a polynomial is a polynomial in I Is. More generally, we might expect
- oo.
Is this true?
at least that if /is of exponential type, L(f)(s) -+0 as s
(c) in Problem 53, Laplace transformation trans
differential equation into an algebraic problem. For example,
If / is a
1 >-(0)
0, y'(0) =0 on R+
suppose we want to solve y" + y
55. Because of property
forms
a
=
=
,
.
6
Functions
solution
Uf)
Analysis)
have
must
we
the Circle (Fourier
on
sLif)-fi<0)
=
Lif) =sLif)
/'(0)
-
=
s2Lif)
-
5/(0) -/'(0)
Thus, using the differential equation and the initial conditions
L(f"+f)=LH)
s2Lif)
L(/)
Lif)=-
+
s
~
~
~
sis2 + 1)
2
s
\s
+
i)~ 2 \s ij
-
Reading Problem 52(a)-(d) backward,
f(t)
=
1
i(e" + e~")
-
=
l-
we
obtain the solution
cos t
equations by Laplace transformation :
1
l
K0)=0 /(0)
e'
l
y(0)
y'(0)=0
y"-y
I
y(0)
y'-2y=e1
v(0)
y' + 3y' + 2y=e~' + t
y'(0) =0
Solve these systems by Laplace transformation:
(a) yi+y2=e-2'
yi(0) =0=y2(0)
1
y'i + 2^i
1
>'i(0)=0
/,(0)
(b) yl-y'1=yl
1
0.
y"i + yi
2y2
y2(0)
yiiO)
Solve these
(a)
(b)
(c)
(d)
56.
y+v
=
=
=
=
=
=
=
.
=
=
=
=
57. Define this convolution for continuous functions
if* 0X0=
f fir)git-r)dr
>o
Show that
Lif*g)=Lif)Lig).
equation
58. Solve the differential
v" + 2/ + j;
where
=
/(O
X0)
=
0
/(0)
=
1
/is of exponential type (recall Problem 51).
a complex variable
59. Show that the function of
Lif)iz)
=
f
Jo
*/(0
dt
on
R+
6.8
523
Summary
is
complex differentiable (if / is continuous and bounded) in the domain
{zeC: Rez>0}.
The Fourier
Transform
60. We
consider the collection of continuous functions defined
now
all of R.
We shall discuss the Fourier transform, which is the
Fourier series:
/"periodic
fin)
=
\
ir-
2nJ-
/defined
f(8)e~
'">
dd
/(
f fin)e"">
Fourier series of /=
Of course, since
we are
working
on an
| f
=
F(/)(x)
71= -a
on
i2n)"2J-<a
=
-
on
analog of
R
fix)e~
'*
d$
f /(&>'<* d
(277)"2 J_
infinite interval,
we
want to be assured
The most appropriate class of
out of these observations :
that the Fourier transform is defined.
functions will
(i)
(ii)
(iii)
(iv)
come
If/ is integrable on R, /is a bounded continuous function.
The Fourier transform/^ /is linear.
ii&f
if'Y
=
/'
=
iixfY
Since Fourier transformation interchanges the operations of differentiation
and multiplication, we select the class of functions such that the effect of all
such operations produces an integrable function. This is the Schwartz
class SiR) of test functions : / is a Schwartz function, fe SiR), if and only if
/"is C"\ and for all positive integers n and k the function
dk
{x"f)
T"
dx
"
fe SiR) implies fe SiR)._ (Hint: (x"/)(t)
integrable" implies "(i$)'"fM bounded" implies "f-2/<"> integrable.")
61. For/ g e S(R) define the convolution by
is
integrable
if*gix)=
Show that
f
on
R.
Show that
fiy)g(x-y)dy
if*g)A =fg.
524
6
Functions
the Circle
on
62.
Borrowing again
that/(x)=F(/)(x):
(Fourier Analysis)
from the theory of Fourier series,
we
should expect
^)=^!^*
(6-67)
If we try to verify this directly we enter difficulties similar to that in Theorem
6.1 : the integral on the right is
(2^f-J-/^(X-dt^
we cannot apply Fubini's theorem since
J e'*-" d does not exist.
difficulty is overcome by introducing the convergence factor e~rM, then
letting y->0. More precisely, define the Poisson transform off, a function
defined on the upper half plane by
and
The
Pfix
+ iy)
=
p^lTi j Jifcxplfa
-
y
If |] #
Prove these assertions :
(i) if/ is integrable
Pfix + iy)
=
-
f
nJ-m
on
/(/)
R,
-
7j
(x~t)2+y2
.
dt
and
limP/(x
+
/=/(x)
y-0
f
(Hint: Integrate
co
J ^exp[('f(jc
The second statement follows
Poisson kernel
y[(x
-
t)
as
y
|f |] dt;
over
R+, R~, independently.
does the result of Theorem 6. 1
t)2 + y2]'1 has the
behavior
,
since this
that for the
disk.)
(ii) Pf is harmonic in the upper half plane and thus solves Dirichlet's
problem there with the boundary values /.
(iii) if fe SiR),
lim
Pf(x
+
same
iy)=F(f)(x)
V-.0
so
the inversion formula
(6.67) holds
on
S(R).
as