Mathematics , Fall,
August ,
Contents
Hilbert space and orthonormal bases
. Norms, inner products and Schwarz inequality . . . . . . . . .
. Bessels inequality and orthonormal bases . . . . . . . . . . . .
. Problems on Hilbert space . . . . . . . . . . . . . . . . .
Second order linear ordinary dierential equations
. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. The initial value problem. . . . . . . . . . . . . . . . . . . . .
. The boundary value problem and Green functions . . . . . . .
. Eigenfunction expansion of the Green function . . . . . . . . .
. Problems on ordinary dierential equations . . . . . . .
Generalized functions. functions and all that.
. Dual spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. Problems on Linear Functionals . . . . . . . . . . . . . . . . .
. Generalized functions . . . . . . . . . . . . . . . . . . . . . . .
. Derivatives of generalized functions . . . . . . . . . . . . . . .
. Problems on derivatives and convergence of distribu
tions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. Distributions over R
n
.......................
. Poissons Equation . . . . . . . . . . . . . . . . . . . . . . . .
The Fourier Transform
. The Fourier Inversion formula on SR. . . . . . . . . . . . . .
. The Fourier transform over R
n
..................
. Tempered Distributions . . . . . . . . . . . . . . . . . . . . . .
. Problems on the Fourier Transform . . . . . . . . . . . .
Applications of the Fourier Transform
. The Heat Equation by Fourier transform. . . . . . . . . . . . .
. Poisssons equation by Fourier transform . . . . . . . . . . . .
. Advanced, Retarded and Feynman Propagators for the forced
harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . .
. The wave equation by Fourier transform . . . . . . . . . . . .
.. Problems . . . . . . . . . . . . . . . . . . . . . . . . .
Laplace Transform
. The Inversion Formula . . . . . . . . . . . . . . . . . . . . . .
. Application of Uniqueness . . . . . . . . . . . . . . . . . . . .
. Problems on the Laplace Transform . . . . . . . . . . . .
Asymptotic Expansions
. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. The Method of Integration by Parts. . . . . . . . . . . . . . .
. Laplaces Method. . . . . . . . . . . . . . . . . . . . . . . . . .
.. Watsons Lemma . . . . . . . . . . . . . . . . . . . . .
.. Stirlings formula . . . . . . . . . . . . . . . . . . . . .
. The Method of Stationary Phase . . . . . . . . . . . . . . . .
. The Method of Steepest Descent . . . . . . . . . . . . . . . . .
.. Constancy of phase and steepness of descent . . . . . .
. Problems on asymptotic expansions . . . . . . . . . . . .
Old Prelims
Appendix Review of Real numbers, Sequences, Limits, Lim
Sup, Set notation
. Set Theory Notation . . . . . . . . . . . . . . . . . . . . . . .
. How to communicate mathematics . . . . . . . . . . . . . . .
. Cute Homework Problems . . . . . . . . . . . . . . . . . . . .
Hilbert space and orthonormal bases
. Norms, inner products and Schwarz inequality
Denition. A norm on a real or complex vector space, V is a real
valued function, on V such that
. a. x for all x in V .
b. x only if x O.
. x x for all scalars and all x in V .
. x y x y.
Note that the converse of .b follows from . because O O
O.
Examples
. V R, x x.
. V C, x x.
.VR
n
with x
n
x
j
when x x
,...,x
n
.
.VC
n
with x
n
x
j
when x x
,...,x
n
.
. V C, . These are the continuous, complex valued functions on
,.
Here are two norms on V .
f
ftdt . H.
f
supft t , . H.
Exercise . Prove that the expression in
H.
. is a norm.
Here is an innite family of other norms on R
n
. Let p lt . Dene
x
p
n
j
x
j
p
/p
. H.
FACT These are all norms. And here is yet one more. Dene
x
max
j,...,n
x
j
. H.
Exercise . Prove that x
and x
are norms. Sadly, the proof that
x
p
is a norm is not that easy. So dont even try. Well prove the p
case later.
The unit ball of a norm on a vector space V is
BxVx.
A picture of the unit ball of a norm gives some geometric insight into the
nature of the norm. Its especially illuminating to compare norms by compar
ing their unit balls. Lets consider the family of norms dened in
H.
. and
H.
. on R
. The boundary of each unit ball is the curve x and clearly
goes through the points , and , . Its not hard to compute that the
unit ball of
is a square with sides parallel to the coordinate axes. The
unit ball of
is a disk contained in this square, and unit ball of
is
a diamond shape thing contained in this disk. Less obvious, but reasonable,
is the fact that the other unit balls lie inside the square and contain the
diamond. Draw pictures.
Denition An inner product on a real or complex vector space V
is a function on V V to the scalars such that
. x, x for all x V and x, x only if x .
. ax by, z ax, z by, z for all scalars a, b and all vectors x, y.
. x, y y, x.
Examples
.VR
n
with x, y
n
j
x
j
y
j
.
.VC
n
with x, y
n
x
j
y
j
. V C, with f, g
ftgtdt. It would be good for you
to verify yourself that this really is an inner product.
.Vl
, which is the standard notation for the set of sequences x
a
,a
, . . . such that
k
a
k
lt . Dene
x, y
k
a
k
b
k
when y b
,b
, . . . . The series entering into this denition converges
absolutely because of the identity ab a
b
/. It would be very
good for you to verify that this really denes an inner product on l
.
Theorem . The Schwarz inequality. In any inner product space
x, y x, x
/
y, y
/
.H
Proof If x or y is zero then both sides are . So we can assume x
and y . In case the inner product space is complex choose C such
that and x, y is real. If the inner product space is real just take
. In either case let pt x ty
for all t R. Then
pt x ty, x ty x
t
y
tx, y
because x, y y, x Rex, y x, y. So, by the quadratic
formula, the discriminant, b
ac . That is , x, y
x
y
.
Thusx, y
x
y
. Now use . QED.
We are going to show next that in an inner product space one can always
produce a norm from the inner product by means of the denition
x
x, x. . H
Corollary . Dene by
H
.. Then
x y x y.
Proof Using
H
. we nd
xy
x y, x y x
y
Rex, y
x
xy y
xy
.
QED.
Corollary . is a norm.
Proof Using the previous corollary its easy to verify the three properties in
the denition of a norm. QED
. Bessels inequality and orthonormal bases
Denition An orthonormal sequence in an inner product space is a set
e
,e
, . . . which we allow to be nite or innite such that
e
j
,e
k
if j k and ifj k
Lemma . Bessels inequality Let e
,e
, . . . be an orthonormal set
in a real or complex inner product space V . Then for any x V
x
j
x, e
j
Proof Let a
k
x, e
k
. Then, for any integer n,
x
n
k
a
k
e
k
,x
n
k
a
k
e
k
x
n
k
a
k
e
k
, x x, e
k
a
k
j,k
a
j
a
k
e
j
,e
k
x
n
k
a
k
n
k
a
k
n
k
a
k
x
n
k
a
k
So
n
k
a
k
x
for all n. Now take the limit as n . QED
Denition In a vector space with a given norm we dene
lim
n
x
n
x
to mean
lim
n
x
n
x.
Were all familiar with the concept of an orthonormal basis in nite di
mensions e
,...,e
n
is an orthonormal basis of a nite dimensional inner
product space V if
a the set e
,...,e
n
is orthonormal and
b every vector x V is a sum x
n
j
a
j
e
j
.
If V is innite dimensional then we should expect that the concept of
orthonormal basis should, similarly, be given by the requirement a as
before and b every vector x V is a sum
x
j
a
j
e
j
..H
And this is right. Of course writing an innite sum means, as usual, a limit
of nite sums lim
n
x
n
j
a
j
e
j
. There is an unfortunate and
downright annoying aspect to the equation
H
., however. We see from
Bessels inequality that
H
. implies that
k
a
k
lt .
Suppose that we are given the orthonormal sequence e
,e
, . . . and a
sequence of real or complex numbers a
j
such that
k
a
k
lt . Does
the series
k
a
k
e
k
converge to some vector in V If not can we really say
then that we have coordinatized V if we dont even know which coordinate
sequences a
,a
, . . . actually correspond to vectors in V by the formula
H
. Here is an example of how easily convergence of the series
k
a
j
e
j
can fail, even when
k
a
k
lt .
Let F be the subspace of l
consisting of nitely nonzero sequences. Thus
x F if x a
,...,a
n
, , , , . . . . F is clearly a vector space and the inner
product on l
restricts to an inner product on F. Let e
,,,,...,e
, , , , , . . . , etc. The sequence e
,e
, . . . is orthonormal. Let a
j
j
.
Then
j
a
j
lt . Now the sequence of partial sums, x
n
n
a
j
e
j
converges to the vector x
a
j
e
j
in l
you verify this. But x is not
in F. So there is no vector in F whose coodinates are the nice sequence
j
. Of course we caused this trouble by making holes in l
. These
circumstances are analogous to the holes in the set Q of rational numbers.
For example the sequence s
n
n
k
/k is a sequence of rational numbers
whose limit is e. But e is not rational. So there is a hole in Q at e. Question
Can you immagine how intolerably complicated calculus would be if we had
to worry about these holes in Q E.g. f
x gives the maximum of F
on Q provided x is rational The same nuisance would arise if we allowed
holes when dealing with ON bases. We are going to eliminate holes
Denition. A sequence x
,x
, . . . in a normed vector space is a Cauchy
sequence if
lim
n,m
x
n
x
m
.
That is, for any gt there is an integer N such that x
n
x
m
lt whenever
n and m N.
Remark. If x
n
converges to a vector x in any normed space then the
sequence is a Cauchy sequence. Proof Same as proof of Proposition
propR
. in
the Appendix. Just replace by .
Denition. A normed vector space is complete if every Cauchy sequence
in V has a limit in V . A Banach space is a normed vector space which is
complete.
Examples. The spaces V in Examples . to . are complete.
Youre supposed to know this from rst year calculus.
In Example . the space C, is complete in the norm f
but not
in the norm f
. See if you can prove both of these statements.
Denition. A Hilbert space is an inner product space which is complete
in the associated norm
H
..
Examples. The Examples ., ., . are Hilbert spaces. But Example
. is not complete. The proof that Example . is complete will be given
only on popular demand. It is extremely unfortunate that the Example .
is not complete. In order to get a complete space one must throw in with
the continuous functions all the functions whose square is integrable. This is
such an important example that it gets its own notation.
Notation.
L
, is the set of functions f , C such that
ft
dt lt
For these functions we dene
f, g
ftgtdt
This is an inner product on L
, easy to verify and the associated norm
is
f
ft
dt
Of course if we wish to consider square integrable functions on some other
set, such as R we would denote it by L
R.
Just as the real numbers ll in the holes in the rational numbers so
also one may view L
, as lling in the holes in C, . Here is the
denition that makes this notion precise.
Denition. A subset A of a Hilbert space H is called dense if for any
x H and any gt there is a vector y in A such that x y lt . In
words you can get arbitrarily close to any vector in H with vectors in A. As
we know, the rational numbers are dense in the real numbers. For example
the rational number . is pretty close to . Similarly, if you
cut o the decimal expansion of a real number at the twentieth digit after
the decimal point then you have a rational number which is very close to the
given real number.
GOOD NEWS C, is dense in L
, . You may use this fact when
ever you nd it convenient. Its often best to prove some formula for an easy
to handle dense set rst, and then show that it automatically extends to the
whole Hilbert space. Well see this later in the context of Fourier transforms.
Here is the rst important consequence of completeness.
lem. Lemma . Suppose that H is a Hilbert space and that e
,e
, . . . is any
ON sequence. Let c
k
be any sequence of scalars such that
k
c
k
lt .
Then the series
k
c
k
e
k
converges to some unique vector x in H.
Proof Let s
n
n
k
c
k
e
k
. We must show that the sequence con
verges to a vector in H. Since we dont know in advance that there is a
limit, x, to which the sequence converges we will show instead that the se
quence is a Cauchy sequence. Suppose that n gt m. Then s
n
s
m
n
km
c
k
e
k
n
km
c
k
as m, n because the series
k
c
k
converges. Now, because H is assumed to be complete, we know
that there exists a vector x in H such that lim
n
s
n
x. Since limits are
unique, x is unique. You prove this on the way to your next class. Of
course we will write
x
k
c
k
e
k
,.
keeping in mind that this means that x is the limit of the nite sums. QED.
Lemma . In any inner product space the function x x, y is continu
ous for each xed element y.
Proof. If x
n
x then x
n
, y x, y x
n
x, y x
n
xy .
QED
Of course x, y is also a continuous function of y for each xed x. One
can either repeat the preceding proof or just use y, x x, y.
thmH. Theorem . Let e
,e
, . . . be an orthonormal sequence in a real or com
plex Hilbert space H. Then the following are equivalent.
a. e
,e
, . . . is a maximal ON set. That is, it is not properly contained
in any other ON set.
b. For every vector x H we have
x
k
a
k
e
k
where a
k
x, e
k
c. For every pair of vectors x and y in H we have
x, y
k
a
k
b
k
where a
k
x, e
k
and b
k
y, e
k
d. For every vector x in H we have
x
k
a
k
Proof We will show that a. b. c. d. a.
Assume that a. holds. Let x H. By Bessel we have
a
k
lt . By
Lemma
lem.
., y
a
k
e
k
exists. But xy, e
j
a
j
lim
n
n
k
a
k
e
k
,e
j
a
j
a
j
for all j. If x y then let h x y/x y. One can now
adjoin h to the original set and obtain a larger ON set. So we must have
x y . This proves that b. holds.
Assume now that b. holds. Then
x, y lim
n
n
j
a
j
e
j
, y lim
n
lim
m
n
j
a
j
e
j
,
m
k
b
k
e
k
lim
n
lim
m
min n,m
j
a
j
b
j
j
a
j
b
j
.
So c. holds.
Next, assume that c. holds. Put y x to derive that d. holds.
Finally, assume that d. holds. If e
,e
, . . . is not a maximal ON set
then there exists a vector x such that x, e
k
for all k. So the
coordinates, a
k
x, e
k
are all zero. But from d. we see that x
k
a
k
. So x . Contradiction. QED.
Now were ready for the denition of ON basis.
Denition. An ON sequence e
,e
, . . . in a Hilbert space H is an ON
basis of H if condition b. in Theorem
thmH.
. holds.
Of course we could have used any of the other conditions in Theorem
thmH.
.
for the denition of ON basis because theyre equivalent. So why did I use
condition b. for the denition Because surveys of your predecessors show
that its the most popular.
. Problems on Hilbert space
. Let f
,f
,...,f
be an orthonormal set in L
, . Assume that
A
xf
xdx and
B
xf
xdx
What can you say about the value of
xf
xdx
Give reasons.
. Let
u
x/
,x
and
u
x
x, x .
Suppose that f and g are in L
, and f g . Let a
j
u
j
xfxdx, j , and b
j
u
j
xgxdx. Show that
j
a
j
b
j
. Cite any theorem you use.
. Suppose that f , R satises
fx
dx
and
fxdx .
What can you say about the size of
xfxdx
. Suppose that u
,u
are O.N. vectors in an inner product space H. Let
f H and assume that
f
a
a
where a
j
f, u
j
for j , . Show that f a
u
a
u
.
. The Hermite polynomials are the sequence of polynomials H
n
x
uniquely determined by the properties
aH
n
x is a real polynomial of degree n, n , , , . . . with positive
leading coecient.
b The functions
u
n
xH
n
xe
x
/
form an O.N. sequence in L
R.
Fact that you may use If f is in L
R and
fxu
n
xdx for
n , , , . . . then f .
Let c
n
e
x
u
n
xdx.
Evaluate
n
c
n
.
. Let f
,f
, . . . be an O.N. set in a Hilbert space H. Prove that it is
an O.N. basis if and only if the nite linear combinations
n
j
a
j
f
j
n nite but arbitrary
are dense in H.
. Let f
,f
, . . . be an O.N. set in a Hilbert space H. Prove that it is an
O.N. basis if and only if Parsevals equality
g
j
f
j
,g
holds for a dense set of g in H.
. Suppose that f
,f
, . . . is an O.N. basis of a Hilbert space H and
that g
,g
, . . . is an O.N. sequence in H. Suppose further that
j
g
j
f
j
lt .
Prove that g
,g
, . . . is also an O.N. basis.
Hint Use Theorem
thmH.
. by supposing that there exists h such that
h, g
j
for all j.
Second order linear ordinary dierential equa
tions
. Introduction
Consider the dierential equation
u
x ux fx . O
on the interval x . Here f is a given function on this interval.
As you know, the solution to a second order ordinary dierential equation
requires specication of some additional information in order for the ODE
to pick out a solution uniquely. For example one might specify u and
u
to pick out a unique solution to
O
.. This is the initial value
problem both pieces of information are specied at the same endpoint. We
are going to be primarily concerned with the boundary value problem the
two pieces of information will be specied at opposite endpoints. Lets focus
on the boundary value problem for the dierential equation
O
. given by
the boundary data
u,u..O
Our goal is to show that the solution to
O
. and
O
. can be represented
in the form
ux
Gx, yfydy . O
for some function Gx, y. G is called the Green function for the boundary
value problem
O
.,
O
.. Here is the method for constructing the Green
function in this simple example.
STEP . Consider rst the homogeneous version of
O
., namely
u
u.O
Since this equation has constant coecients its easy to write down the general
solution. It is
ux Asinh x Bcosh x
We will need rst to nd the solutions which satises the LEFT HAND
boundary condition u . These are clearly given by choosing B .
We will see later that our procedure doesnt care what A is. So lets just
choose A . Denote the solution that weve now got by U. So
Ux sinh x
STEP . Do the same thing all over again, but using the RIGHT HAND
boundary condition. To this end we may write the general solution to
O
.
in the form
ux C sinhx Dcoshx
A solution that satises the right hand boundary condition is
V x sinhx
and this is the only one, up to a scalar multiple.
STEP . Dene
Wx det
Ux U
x
VxV
x
Inserting our particular U and V we can compute that
Wx sinh x . O
STEP . Dene
Gx, y
UxV y
Wy
, for x y
UyV x
Wy
, for y x .
.O
In our case this expression reduces to
Gx, y
sinh x sinhy
sinh
, for x y
sinh y sinhx
sinh
, for y x .
Notice that on the overlapping portion of these denitions, namely on the di
agonal x y, the two halves of these formulas agree. So the function Gx, y
is well dened even on the diagonal of the square , , . Moreover,
for each x, Gx, y is a continuous function of y and for each y Gx, y is a
continuous function of x because the limits from the left and right or up to
down agree on the diagonal.
NOW let
ux
Gx, yfydy . O
for ANY continuous function f.
I claim that u is a solution to the boundary value problem
O
.,
O
..
The easy part of this is seeing that the function u in
O
. satises the two
boundary conditions
O
.. Indeed the denition of G shows that G, y
for all y and G, y for all y. This is why we chose U and V as we
did. So u is zero at the two boundaries. To prove that u satises
O
. we
must dierentiate the right side of
O
. a couple of times. The naive reader
not you might think that we can just put d
/dx
under the integral sign.
But if you do this you will nd d
/dx
Gx, y Gx, y on each half of
the square because both Ux and V x satises this equation. This would
then give u
u , which is not what we want. So here is what we must
do. There is trouble with Gx, y as x passes through y. Although Gx, y is
continuous in x for xed y its rst derivative, G
x
x, y, actually has a jump
as x passes y, as we will see in a moment. So G
xx
x, y doesnt even exist
at x y. To deal with this we just put the trouble spot into the limits of
integration thus
ux
x
Gx, yfydy
x
Gx, yfydy.
In each integral the integrand is a nice function of x, provided y is where
the limits of integration allow it to go. There is never a crossing over the
diagonal. Here is the computation. As you know there is a contribution to
the derivative of u from the x that occurs in the limits of integration.
u
x Gx, x
fx Gx, x
fx .
x
G
x
x, yfydy
x
G
x
x, yfydy .
In the rst line the expression x
indicates that we are letting y approach
x from below, as it should, while x
indicates that y is approaching x from
above. But now recall that you admitted a while back that the function
y Gx, y is continuous. Therefore the rst line is zero. So
u
x
x
G
x
x, yfydy
x
G
x
x, yfydy, . O
which is just what we would have gotten if we had simply put d/dx under
the integral in
O
.. But the next time we wont be so lucky. Were going to
repeat this procedure for the second derivative. But well see that the rst
line is not zero this time. We nd from
O
.
u
xG
x
x, x
fx G
x
x, x
fx
x
G
xx
x, yfydy
x
G
xx
x, yfydy . O
This time the rst line does not give zero because G
x
x, y has a jump as y
passes by x. How much is the jump Just look at the denition of G and
the denition of W G
x
x, y UyV
x/Wy if y lt x. So G
x
x, x
UxV
x/Wx. Similarly G
x
x, x
U
xV x/Wx. So
G
x
x, x
G
x
x, x
UxV
xU
xV x
Wx
..O
Ha So the rst line in
O
. is just fx. But we have already observed that
G
xx
x, y Gx, y at least when y x, and this is where y is ranging in the
two integrals. Putting this together we now nd that u
x fx ux,
which is exactly the equation
O
.. Thus ux, as given by
O
., satises
both
O
. and
O
.. So Gx, y is the Green function for the boundary value
problem
O
.,
O
..
Our aim in this section is to carry out this procedure for a general second
order ordinary dierential equation. To this end we must be able to construct
the functions U and V and understand the function W. All of these are issues
pertaining to the initial value problem, which is much easier to deal with than
the boundary value problem and whose basic theory we will review in the
next section.
But before leaving this successful example lets rewrite informally the
preceeding computations. We actually showed that
d
/dx
Gx, yfydy fx . O
If we innocently rewrite this as
d
/dx
Gx, yfydy fx
then we could claim that we have shown that
d
/dx
Gx, y x y. . O
And what does
O
. mean It means exactly that
O
. holds, which we
have proved. So if you are fond of functions and who isnt then all is well.
In truth, this neat approach to Green functions, as embodied in the fun
damental formula
O
., can break down for a general second order dierential
equation. We need to see how this can happen before proceeding to a general
second order ODE. Here is a simple example of
BREAK DOWN
Lets modify the equation
O
. slightly. Consider the equation
u
x
ux fx x . . O
The general solution to the corresponding homogeneous equation u
u
is ux Asinx Bcosx. Since cos and cos the
desired function U is Ux sinx and the desired function V is ALSO
V x sinx. Consequently Wx for all x. The formula
O
. is
now meaningless. Thats life. So is there a Green function Answer No.
It will be important for us to understand exactly why and when such failure
occurs.
. The initial value problem.
We consider a closed bounded interval a, b. We wish to study several prob
lems associated with the dierential operator
Lux p
xu
xp
xu
xp
xux. . .
We will assume throughout that the coecients p
j
are continuous on a, b
and that p
x gt on the entire closed interval.
thmO Theorem . Existence and uniqueness for the initial value problem
Let f be a continuous function on a, b and let c and c
be two real num
bers. Then there exists a unique function u C
a, b satisfying
Lux fx for all x a, b and ua c and u
ac
...
Proof Use Picards method. See e.g. Kreyszig Advanced Engineering
Mathematics
defWronsk Denition . Wronskian Let f and g be two functions in C
a, b. The
Wronskian of f and g is the function
Wf, gx det
fx f
x
gx g
x
..
Recall that f and g are said to be linearly dependent on a, b if there are
constants and such that
fx gx
In this case we may dierentiate this equation and nd that
f
xg
x
when f and g are dierentiable. This shows that the rows of the matrix in
.
. are linearly dependent for every x a, b. Hence Wf, g . This
proves
lemO Lemma . If f and g are in C
a, b and are linearly dependent on the
interval a, b then
Wf, gx for all x a, b.
The partial converse of this is a little bit more subtle.
Lemma . Let u and v be two solutions of Lu . Then u and v are
linearly independent on a, b if and only if Wu, v is nowhere zero.
Proof Assume Lu Lv . If, for some point x
one has Wu, vx
then the points ux
,u
x
and vx
,v
x
in the plane are linearly
dependent vectors in R
. Therefore there exist nonzero constants , such
that
ux
vx
and u
x
v
x
.
Let gx ux vx. Then Lg . But gx
and g
x
.
Therefore gx for all x a, b by the uniqueness portion of Theorem
thmO
.. Hence u and v are linearly dependent on a, b. Conversely, if u and v are
linearly dependent on a, b then Lemma
lemO
. shows that Wu, v is identically
zero even if u and v are not solutions to Lu .
. The boundary value problem and Green functions
thmO Theorem . Let
Lu p
xu
p
xu
p
xu .
Au ua
u
a.
Bu ub
u
b.
for u C
a, b, where p
,p
,p
are in Ca, b, ,
,,
are real
numbers, and p
gt on a, b. Then the inhomogenous system
Lw f Ca, b
Aw
Bw
has a solution for all f,
,
if and only if the homogeneous system
Lu
Au
Bu
has no nonzero solution.
Proof Dene functions U, V , and F with the aid of the basic existence
and uniqueness theorem, Theorem
thmO
., so as to satisfy
LU , Ua
,U
a.
LV , V b
,V
b.
LF f, Fa F
a.
Then clearly AU and BV .
Thus for Dirichlet boundary conditions ,
,,
,U
and V look like this
Let
w cU dV F.
Then
Lw f .
Aw dAV .
Bw cBU .
Now assume that the system has no nonzero solutions. Since LU
AU it then follows that BU . Similarly AV . Hence for any
,
we may choose c and d so that w satises .
Conversely assume that has a solution for all f
,
,
and that has
a solution u not identically zero. Then there exists a function v such that
Lv .
Av .
Bv .
But ua, u
aR
is orthogonal to ,
as is also va, v
a and
neither is zero by uniqueness theorem. Hence one is a multiple of the
other. Say cua, u
a va, v
a. Then the function cu v has zero
initial data at a and satises Lcu v . cu v on a, b. But
Bu . Bv , contradiction.
Corollary . If the system has no non zero solution then
a the functions U and V constructed in the previous proof are linearly
independent and
b their Wronskian W is nowhere zero.
Proof It suces to show U and V are linearly independent. If they arent
then there exists a constant c such that U cV . But then BU cBV
. So U satises . But U is not identically zero on a, b.
thmO Theorem . Green function. Assume that the system has no non
zero solution. Then the system
Lu f
Au
Bu . .
has the unique solution
ux
b
a
Gx, yfydy . .
where
Gx, y
UxV y/p
yWy a x y b
UyV x/p
yWy a y x b
..
and
W UV
VU
...
Proof. As in the introduction to this section we are going to have to break
up the integral into two parts to take into account the jump in the derivative
of G. Thus we will write
ux
b
a
Gx, yfydy .
x
a
Gx, yfydy
b
x
Gx, yfydy. . .
The derivative of each integral with respect to x will have two terms, corre
sponding to the two appearances of x. Dierentiating rst with respect to
the x that appears in the limits of the integrals the fundamental theorm of
calculus gives
u
x Gx, x
fx Gx, x
fx .
x
a
G
x
x, yfydy
b
x
G
x
x, yfydy. .
Thus the contribution to u
x from the limits cancel because Gx, y is con
tinuous in y at y x. We therefore nd
u
x
x
a
G
x
x, yfydy
b
x
G
x
x, yfydy. . .
We need now the key identity that describes the jump of the rst derivative
of G. It is
G
x
x, x
G
x
x, x
p
x
. ..
This may be derived from
.
.. Just imitate the derivation of
O
.. Now
dierentiate
.
. to nd
u
xG
x
x, x
fx G
x
x, x
fx .
x
a
G
xx
x, yfydy
b
x
G
xx
x, yyfydy. .
fx/p
x
x
a
G
xx
x, yfydy
b
x
G
xx
x, yyfydy. .
Hence multiplying by p
x we nd
p
xu
x fx .
x
a
p
xG
xx
x, yfydy
b
x
p
xG
xx
x, yfydy. . .
Now multiply
.
. by p
x and multiply
.
. by p
x and add them to
.
. to nd
Lux fx
x
a
L
x
Gx, yfydy
b
x
L
x
Gx, yfydy .
fx .
because L
x
Gx, y o the diagonal. Finally, just as in the simple example
in the introduction, the reader may verify that, for each y a, b, the
function x Gx, y satises the boundary conditions in
.
.. Hence so
does u, by
.
. and
.
..
Denition . Eigenvalue. A complex number is called an eigenvalue
of the dierential operator L with boundary conditions given by A and B if
there exists a function u in C
a, b such that
Lu u .
Au .
Bu . .
The function u is called an eigenfunction with eigenvalue .
Denition . The dierential operator L is called symmetric on a, b if
Lu, v u, Lv
for all u and v in C
a, b which are zero in a neighborhood of each endpoint.
Fact If L is symmetric then L must have the form given in the next
theorem wherein we change notational and sign conventions, taking p
x
px lt . Give a little try to nd a proof of this fact by yourself.
thmO Theorem . Let p and q be continuous functions on a, b with px gt
for all x a, b. Let
Lu
d
dx
pxu
x qxux
.
Au ua
u
a.
Bu ub
u
b. .
Put
u, v
b
a
uxvxdx.
Then Lu, v u, Lv if u and v C
a, b and Au Bu Av
Bv .
Moreover there exists a strictly increasing sequence
n
of real eigenvalues
for the operator L with boundary conditions Au , Bu . In fact one
has lim
n
. The corresponding eigenfunctions u
n
may be chosen real.
Assume they are normalized
b
a
u
n
x
dx .
Then the sequence u
,u
, . . . forms an O. N. basis of L
a, b.
. Eigenfunction expansion of the Green function
We are going to use the notation
Kux
b
a
Kx, yuydy
whenever Kx, y is a function on the square a, b a, b. The operator
u Ku is called an integral operator. You can see from the repeated use of
such expressions in the previous sections that this is a useful notation.
Let
Lu
d
dx
px
du
dx
qxux .
Au and Bu as usual. .
If is not an eigenvalue of L for the boundary conditions A , B then
L
exists and is given by a Green function G
x,
Lufu
b
a
G
x, fd.
Let
,
, . . . , be the eigenvalues of L and u
,u
, . . . the corresponding nor
malized real eigenfunction. Then
thmO Theorem .
G
x, y
j
j
u
j
xu
j
y.
Informal Proof We assume all series converge and all interchanges are
legal.
Let
K
x, y
j
j
u
j
xu
j
y.
Now
b
a
K
x, yu
n
ydy
j
j
b
a
ujxu
j
yu
n
ydy .
n
u
n
x. .
because all the other terms are zero. We may write this as
K
u
n
n
u
n
Therefore
LK
u
n
u
n
But L G
Identity. So
LK
u
n
u
n
LG
u
n
.
Therefore K
u
n
G
u
n
for all n because L has trivial nullspace. Thus,
for each x
K
x, , u
n
G
x, , u
n
n
Therefore
K
x, y G
x, y.
This proves Theorem
thmO
.. QED
. Problems on ordinary dierential equations
. Apply Picards iteration method to the following initial value problem.
Find y
x, y
x, y
x.
y
xy, y
Ref Kreyszig, Advanced Engineering Mathematics sec.. in rd edi
tion
. Determine which of the following operators are symmetric on the interval
,
a Lu
d
u
dx
du
dx
u.
b Lu
d
u
dx
u.
c Lu
d
u
dx
u.
. Determine which of the operators in Problem have inverses for the
boundary conditions u
u
and for these operators nd the
Green functions.
. Let
Lu x
u
xu
on , .
a Show that L is symmetric on this interval.
b Find the Green function for L under Dirichlet boundary conditions
uu.
c For the same boundary conditions nd the eigenfunctions and eigen
values of L.
Hint Equations of the form ax
u
bxu
cu tend to have two linearly
independent solutions of the form u x
, complex. When only one can
be found in this form then x
ln x will give another one.
. Prove the Fact preceding Theorem
thmO
..
Generalized functions. functions and all
that.
. Dual spaces
The concept of a dual space arises naturally in dierential geometry, mechan
ics and general relativity. And we will need it later to understand generalized
functions.
Denition . Let V be a real or complex vector space. A function L
V scalars is called a linear functional if
Lx y Lx Ly
for all x and y in V and for all scalars and . In other words a linear
functional is a linear transformation from V into R or C if V is a complex
vector space. The dual space to V is the set, denoted V
, of all linear
functionals on V .
For example the function which is identically is a linear functional.
Check this against the denition. Moreover if L
and L
are linear func
tionals then so is the function aL
bL
for any scalars a and b. Check this
against the denition now Therefore V
is itself a vector space. Its a new
vector space constructed from the old one.
Example .. Denote by P
the vector space consisiting of polynomials
of degree less or equal to . This is a four dimensional vector space because
, t, t
,t
constitutes a basis. Here are some linear functionals on this space.
.pL
p p.
.pL
p
ptdt.
.pL
p
pt sin tdt.
It would be best if you, personally, verify that each of these functions on
P
are linear functionals.
The thing to take away from these examples is that there is no resem
blance between V and V
you cannot really identify any of these three
linear functionals on P
with elements of P
itself. RIGHT V
is really a
dierent vector space from V itself. We have constucted a new vector space
from the given one. This being the case, you have to regard the following
theorem as remarkable.
Theorem . If V is an ndimensional vector space then so is V
.
Proof. Let e
, ...e
n
be any basis of V . Then any vector x V can be
uniquely written
x
n
j
a
j
e
j
..
Uniqueness means that each a
j
is a function of x. Dene
L
j
xa
j
, j , ..., n.
Its straightforward to check that each function L
j
is a linear functional. We
will show that they form a basis of V
.
. They are linearly independent. Proof Suppose that M
n
j
c
j
L
j
. Then Me
k
n
j
c
j
L
j
e
k
n
j
c
j
jk
c
k
. So all the coe
cients c
k
are zero. Hence the functionals L
j
are linearly independent.
. They span V
. Proof Let L be any linear functional. Dene c
k
Le
k
. Claim Then L
n
k
c
k
L
k
. You can check this yourself by showing
that both sides of this equation agree on each e
j
and therefore on all linear
combinations of the e
j
. Thus they agree on all of V .
So we have now produced a basis of V
consisting of n elements. Hence
dimV
n. QED.
Terminology. The basis L
j
described in the preceding proof is called
the dual basis to the basis e
, ..., e
n
. It has the nice property that
L
j
e
k
jk
Philosophic considerations .. Having chosen the basis e
, ..., e
n
of
V we see that we automatically get a basis L
, ..., L
n
of V
. Since any vector
x in V can be written uniquely in the form . we can now dene a vector
L
x
in V
by the formula
L
x
n
j
a
j
L
j
In this way we get a map x L
x
from V onto V
. You can check easily
that this map is a linear, b onetoone and c onto V
. This is, as some
people would say, an isomorphism from V onto V
. With the help of this
map we could, if we wished, identify V and V
and even go so far as to say
that V and V
are the same space. But there is a catch A choice of basis
has been made in constructing this isomorphism. If Jim goes into one room
and chooses a basis e
, ..., e
n
to construct this isomorphism and Jane goes
into another room and chooses a basis the chances are that they will choose
dierent bases. Then they will arrive at dierent isomorphisms. So each
one will identify V with V
in dierent ways. Jim will say that the vector
x V corresponds to a certain linear functional L and Jane will say, no, it
corresponds to a dierent linear functional, M.
When an isomorphism between two vector spaces depends on someones
choice of a basis we say that the isomorphism is not natural. If you should
nevertheless decide to think of these two vector spaces as the same i.e.
identify them then sooner or later you will run into conceptual and even
computational trouble.
But there is an important circumstance in which one really can justify
identifying V and V
. Some readers might recognize the next theorem as
raising and lowering indices.
Theorem .. Suppose that V is a real nite dimensional vector space
and , is a given inner product on V . Then for any linear functional L on
V there is a unique vector y in V such that
Lx x, y for all x V.
Denote by L
y
the linear functional determined by y in this way. That is,
L
y
x x, y for all x V. . .
Then the map
yL
y
is a onetoone linear map of V onto V
. I.e. it is an isomorphism.
Proof. The map y L
y
is clearly linear. You better check this. It will
be good practice in dealing with these structures. Moreover this map is one
toone because if L
y
then in particular L
y
y . That is, y, y .
So y . Therefore the map y L
y
is onetoone. Hence, by the rank
theorem, the range of this map has the same dimension as the domain. But
if dim V n then by Theorem . dim V
n also. Hence the range is all
of V
. QED.
Moral .. We know that there are many inner products on any nite
dimensional vector space. But given a particular inner product on a real
nite dimensional vector space V , the preceding theorem provides a natural
way to identify V with V
without making any ad hoc choices of basis. Here
is a consequence of this identication that we live with every day.
Derivative versus gradient. Suppose that V is a nite dimensional
real vector space and f is a real valued function on V . For any point x in V
and any vector v V dene
v
fx
dfx tv
dt
t
This is the derivative of f in the direction v. For example if we choose any
basis e
, ..., e
n
of V we may write x
n
j
x
j
e
j
and then f is just a function
of n real variables, x
, ..., x
n
. The chain rule then gives
v
fx
n
j
v
j
f/x
j
x
where of course v
n
j
v
j
e
j
. This sum is clearly linear in v. In other
words the map v
v
fx is, for each x, a linear functional on V . One
often writes f
x for this linear functional. That is, f
xv
v
fx. So
f
x is in V
for each x V . Therefore the derivative , f
, is a function
from V into V
. If there is no natural way to identify V
with V then this
map to V
is the only object around that captures the notion of derivative
that youre familiar with. But if V has a given inner product, , , then we
can identify the linear functional f
x for each x with an element of V .
This is the gradient of f. That is,
fx f
x
identied to an element of V by Theorem .. Thus
f
xv
v
fx fx, v.
. Problems on Linear Functionals
Denition. A linear functional on a real or complex vector space V is a scalar
valued function f on V such that
i fx fx for all scalars and all x V .
and ii fx y fx fy for all x and y in V .
Which of the following expressions dene linear functionals on the given
vector space
.VR
,xx
,x
,x
. Explain why not if you think not.
a fx x
x
b fx x
c fx x
x
d fx
e fx
f fx sin x
g fx x
x
h fx x u where u is a xed vector and x u
j
x
j
u
j
. V C, real valued continuous functions on , .
aF
tdt for C,
bF/
cF
dF
eF
fF
t sin tdt
gF
t
dt
hF
t
tdt
iF
t/
tdt
jF
t/tdt
kF
sin tdt
Explain why not if you think not.
. Let V be the vector space of all nitely nonzero real sequences. A
sequence x x
,x
, . . . is called nitely nonzero if N x
k
for all
k N. If a a
,a
,a
, . . . is an arbitrary sequence of real numbers let
f
a
x
j
a
j
x
j
for x in V. .
a Show that the series converges for each x in V .
b Show that f
a
x is a linear function of x.
c Show that every linear functional on V has the form . That is, show
that if f is a linear functional on V then there exists a sequence, a, such that
fx f
a
xxV.
d Show that the sequence a, in part c is unique.
. Denote by P
the space of real valued polynomials of degree less or equal
to . This is a real vector space of dimension three. Right Dene an inner
product on P
by
p, q
ptqtdt.
You have already admitted that the function on P
dened by Lp
pt sin tdt is a linear functional. But we also know that every linear
functional is given uniquely by an element of the space P
with the help of
the inner product. Thus there is a unique polynomial f of degree at most
such that
Lp p, f for all p P
.
Find f.
. Generalized functions
If you want to measure the electric eld near some point in space you could
put a small charged piece of cork there and measure the force exerted by the
eld on the cork. In this way you have converted the problem of making
an electrical measurement to that of making a mechanical measurement. Of
course the force on the cork is a constant multiple of an average of the forces
at each point of the cork. If Ex is the eld strength at x and is the charge
distribution on the cork then the net force on the cork is
R
Exxd
x,
in appropriate units. In practice and even in some theories it is only these
averages that you can measure. For example if you really want to measure
the eld at a point x by the preceding method you would have to place a
point charge at x. But classical theory shows that the total electric energy of
the eld produced by a point charge is innite. The notion of a point charge
is therefore at best an idealization. Of course if you know in advance that the
electric eld is continuous then you can get better and better approximations
to the value of Ex by using a sequence of smaller and smaller corks. In
the classical theory of electomagnetic elds the electric eld tends to be
continuous and therefore it makes sense to talk about its value at a point.
The quantum theory of eletromagnetic elds, however, recognizes that there
are tremendous uctuations in the eld at very small scales and only the
averaged eld has a meaning.
The need for talking only about averages shows up also in measurement of
temperature. A thermometer is clearly measuring the average temperature
over the volume of the little bulb at the bottom. If the temerature varies
from point to point and is a continuous function of position then you can, in
principle, measure the temperature at a point by using a sequence of smaller
and smaller bulbs. Recall however that a typical small bulb will contain
on the order of
molecules. In accordance with statistical mechanics
the temperature at a point of a system has a really questionable meaning
because temperature is a measure of average kinetic energy of a large bunch of
molecules. So at an atomic size of scale temperature at a point is meaningless.
A pairing such as
R
Exxd
x, between an extensive quantity such
as charge extensive means that in a larger volume you have more of the
stu, such as charge, mass, etc. and an intensive quantity, such as the
electric eld in a larger volume you dont have more eld occurs often in
physics. The integral is linear in and denes a linear functional on the
vector space of test charges. The integral is also linear in E for xed .
We say that the integral is a bilinear pairing between the extensive quantity
and the intensive quantity E. Such bilinear pairings between two vector
spaces is common. Usually the elements in these dual vector spaces have a
physical interpretation, extensive for one and intensive for the other. Julian
Schwinger, in his book Sources and Fields emphasizes the duality between
sources and elds.
We are going to develop and use this notion of duality between very
smooth functions test functions and very rough functions e.g. delta
functions. It has proven to be a great simplifying machinery for under
standing partial dierential equations as well as Fourier transforms. We are
going to apply it to both.
The rst step is to understand really smooth functions.
Test functions
Lemma . Let
fx
e
/x
for x gt
for x
Then f is innitely dierentiable on the entire real line.
Proof First, recall that e
t
grows faster than any polynomial as t .
That is, lim
t
pte
t
for any polynomial. Second, you can see by
induction that for x gt the nth derivative d
n
e
/x
dx
n
p
n
/xe
/x
for
some polynomial p
n
. Convince yourself with the cases n , , . Third,
you can see easily that all of the derivatives of f exist at any point other
than x , and is zero to the left of . The only question then is whats
happening at x Sadly, one must go back to the denition of derivative
to answer this. But its not so hard. If we know that the rst n derivatives
exist at x and are zero there then the rst and second comments above
show that the n st right hand derivative is
lim
h
p
n
/he
/h
h
Of course the left hand derivative is clearly zero. So the two sided derivative
exists and is zero. This is the basis for an induction proof. Carry out the
case n yourself. QED.
Lemma Let
x fxf x
where f is the function constructed in Lemma . Then is an innitely
dierentiable function on R with support contained in the interval , .
Proof. is clearly innitely dierentiable by the repeated application
of the product rule for derivatives. To see that is zero outside the interval
, draw a picture. QED
Notation. C
c
is the standard notation for the set of inntiely dier
entiable functions with support in a nite interval. One says that these
functions have compact support.
We have now constructed one not identically zero function in C
c
. From
this function its easy to construct lots more. For example the function x
xx
is also in C
c
. By scaling and translating the argument
of and taking powers one clearly gets lots of such functions. In FACT there
are so many such functions that they are dense in L
R. Remember the
concept of density form the chapter on Hilbert space One of the homework
problems sketches how to show that any continuous function with compact
support is a limit of functions in C
c
.
Notation The space C
c
arises so often that it customarily is given a
special notation
DC
c
R.
The dual space of D is denoted, as usual, D
.
Terminology An element in D
is called a distribution or a generalized
function according to taste.
Examples . Suppose that f is a continuous function on R. Dene
L
f
fxxdx for D. . .
Then L
f
is a linear functional on D. Notice that even if f increases near
e.g. fx e
x
the integral makes sense because its really an integral over
some and in fact any interval that supports . So L
f
D
.
. Let
L
.
Then L
is also a linear functional on D. Clear So L
D
. But this
example is substantially dierent from the rst one. There is no continuous
function f, or even discontinuous function f, such that L
fxxdx.
. Its still OK if the function f in Example is not continuous but has
only some mild singularities. For example if fx /x
/
in
.
. then
the integral still exists for any test function . All we need of f is that
b
a
fxdx lt for any nite interval a, b. In particular the function
fx /x wont work in
.
.. The integral doesnt make sense for an
arbitrary test function . One says that f is locally integrable if
b
a
fxdx lt
for any nite interval a, b.
The lesson to be drawn from these examples is this any continuous
function f on R produces a generalized function L
f
, i.e. an element of
D
by means of the formula
.
.. But not every element of D
comes from
a continuous function in this way or even from a discontinuous function,
as we see in Example . Thats why we call the elements of D
generalized
functions. Neat terminology, huh
. There is an important instance that violates the wisdom of Example
. Its based strongly on cancellation of singularity.
Lemma Let D. Then
P
x
lim
a
xgta
x
x
dx . .
exists and is equal to
xgt
x
x
dx
x
x
dx . .
Proof If lt a lt b then
xgta
x
x
dx
xgtb
x
x
dx
altxb
x
x
dx . .
because
altxb
x
dx . The integrand in the last term in
.
. is bounded
on R by the mean value theorem. So we can let a and get a limit. This
also proves the validity of the representation
.
. of this limit. QED
The generalized function P
x
is called the Principal Part of /x. It
arises in many contexts. We will see it coming up later in the Feynman
propagator.
. Derivatives of generalized functions
Denition Let T D
. The derivative of T is the element T
of D
given
by
T
T
for all D. . D
What does this denition have to do with our well known notion of deriva
tive of a function First lets observe that at least T
is indeed a well dened
linear functional on D. The reason is that for any D the function
is
again in D so the right side of
D
. makes sense. The linearity of T
is clear.
Right To understand why this denition is justied consider the example
TL
f
spelled out in
.
.. Suppose that f is actually dierentiable in the
classical sense. i.e. in the sense that you grew up with. Then
L
f
f
xxdx
b
a
f
xxdx
where a and b are chosen so that o a, b
fxx
b
a
b
a
fx
xdx
b
a
fx
xdx
fx
xdx
L
f
So
L
f
L
f
..D
Stare at
D
. and
D
.. Do you see now why
D
. is a justiable denition
of derivative of a generalized function T Thats right. It agrees with the
classical notion of derivative when T L
f
and f is itself dierentiable
But
D
. has a well dened meaning even when its not of the form L
f
for
some dierentiable function f. Lets see what the denition
D
. gives when
TL
H
and H is the nondierentiable function given by
Hx
,x
, x lt
.
In this case we have, USING
D
.,
T
T
.
L
H
.
xdx .
.
by the fundamental theorem of calculus. So we have
L
H
L
..D
You could say ippantly that we have now dierentiated a nondierentiable
function, H and found H
. In truth the perfectly meaningful equation
D
. is often written as H
. But what does this equation mean It
means
D
.. For the next two weeks we will regard it as immoral to write
the equation H
.
Example L
We now have a notion of derivative of a generalized function. Lets end
with one more denition.
Denition A sequence T
n
of generalized functions converges to a gen
eralized function T if the sequence of numbers T
n
converges to T for
each D.
. Problems on derivatives and convergence of distri
butions.
. Dene
L
xxdx for C
c
R.
Using the denition
T
T
,
compute the rst four derivatives of L that is, compute T
, ..., T
.
Hint . Use the denition of derivative of a distribution.
Hint . Use the denition four times to compute the four derivatives.
. Suppose that f R R is continuous but not necessarily dierentiable.
Let
ux, t fx ct.
Show that u is a solution to the wave equation
u/t
c
u/x
in the distribution sense sometimes called the weak sense.
Nota Bene Since f is not necessarily dierentiable you cannot use f
in
the classical sense.
. Does
n
x n converge in the distribution sense
. Let p
t
xt
/
e
x
/t
. Find the folowing limits if they exist.
a. the pointwise limit as t .
b. the L
R limit.
c. the limit in D
.
You may use the following FACT that will be proved later
p
t
xdx t gt .
. Let be the function constructed in Lemma , except multiply it by a
positive constant such that
R
xdx
Such a constant can be found because the original is nonnegative and has
a strictly positive integral. Then let
n
x nnx
Our goal is to show that
n
converges in some sense to a delta function.
a. Show that
R
n
xdx for all positive integers n.
b. Suppose that g is a continuous function on R. Show that
lim
n
R
n
xgxdx g.
c. Use part b. and the denition of convergence of distributions to show
that
L
n
converges in the weak sense to L
.
. Prove that, for all D,
lim
x
xi
dx P
x
i..
This is often written as
lim
xi
P
x
i weak sense . .
Hint Review the proof of the lemma at the end of Section ..
. Distributions over R
n
The extension of the one dimensional notion of distribution to an n di
mensional distribution is straightforward. Denote by D the vector space
of innitely dierentiable functions on R
n
which are zero outside of a large
cube that can depend on the function. This space is sometimes denoted
C
c
R
n
. There are plenty of such functions. For example if
,...,
n
are each in C
c
R then the function x
,...,x
n
x
x
n
is in
C
c
R
n
. What cube could you use So is any nite linear combination of
these products. A distribution over R
n
is dened as a linear functional on
D. Of course we can make up examples of such ndimensional distributions
similar to the ones we already know in one dimension Let
L
f
R
n
fxxd
n
x.
As long as f has a nite integral over every cube this expression makes
sense and denes a linear functional on D, just as in one dimension. We also
have the ndimensional function dened by
L
..
The dierence from one dimension shows up when we consider dierentiation.
We now have partial derivatives. Here is the denition of partial derivative
as if you couldnt guess.
T
x
k
T/x
k
..
When n every distribution has a very intuitive interpretation as an
arrangement of charges, dipoles, quadrupoles, etc. Im going to explain this
interpretation in class in more detail. It gives physical meaning to every
element of D
. Poissons Equation
We will write as usual r x in R
.
Theorem .
r
..P
in the distribution sense. That is,
L
/r
L
We will break the proof up into several small steps.
Lemma . At r
/r
Proof. /r/x x/r
and
/r/x
/r
x
r
. So
/r /r
x
y
z
r
/r
/r
.
QED.
In view of this lemma you can see that we have only to deal now with
the singularity at r . Our notion of weak derivative is just right for doing
this.
The trick is to avoid the singularity until after one does some clever
integration by parts in the form of the divergence theorem. In case you
forgot your vector calculus identities a self contained review is at the end of
this section. I want to warn you that this is not the kind of proof that you
are likely to invent yourself. But the techniques are so frequently occurring
that there is some virtue in following it through at least once in ones life.
Lemma . Let D. Then
R
/rxdx lim
r
/rdx
Proof The dierence between the left and the right sides before taking the
limit is at most use spherical coordinates in the next step
r
/rd
x max
xR
x
r
/rd
x max
xR
x
QED.
Before really getting down to business lets apply the denitions.
T
/r
j
/x
j
T
/r
.
j
/x
j
T
/r
/x
j
.
T
/r
.
R
/rxd
x.
lim
r
/rxd
x. .
So what we really need to do is show that this limit is . To this end
we are going to apply some standard integration by parts identities in the
OK region r .
C
r
/rxd
x.
r
r
r
d
x by identity
P
..
r
r
n
r
ndA by the divergence theorem .
where n is the unit normal pointing toward the origin. The other boundary
term in this integration by parts identity is zero because we can take it over
a sphere so large that is zero on and outside it.
Now
r
/r ndA
r
ndA .
max
.
.
as . This gets rid of one of the terms in C
in the limit. For the other
one just note that
r
n /r/r /r
. So
r
r
ndA
r
xdA .
r
dA
r
x dA .
r
x dA .
Only one more term to get rid of
r
x dA max
x
x
because is continuous at x . This proves
P
..
Vector calculus identities.
If f is a real valued function and G is a vector eld, both dened on some
region in R
then
fG f Gf G . P
Application . Take f /r and G . Then we get
r
r
r
wherever r . . P
Application . Take f and G
r
. Then we get
r
r
r
wherever r . P
But
r
wherever r . So subtracting
P
. from
P
. we nd
r
r
r
wherever r . . P
This is the identity we need in the proof of
P
..
The Fourier Transform
The Fourier transform of a complex valued function on the line is
f
e
ix
fxdx . F
Here runs over R. The most useful aspect of this transform of functions is
that it interchanges dierentiation and multiplication. Thus if you dieren
tiate under the integral sign you get
d
d
f
e
ix
ixfxdx. . F
So
Fourier transform ofixfx
d
d
f. .
And an integration by parts never mind the boundary terms clearly gives
i
f
e
ix
f
xdx. . F
So
f
i
f.
We will see later that these formulas allow one to solve some partial dieren
tial equations. Moreover in quantum mechanics these two formulas amount
to the statement that the Fourier transform interchanges P and Q momen
tum and position operators.
But the usefulness of these formulas depends crucially on the fact that
one can also transform back and recover f from
f. To this end there is an
inversion formula that does the job. Our goal is to establish the most useful
properties of the Fourier transform and in particular to derive the inversion
formula and show how to use it to solve PDEs.
To begin with we must understand how to give honest meaning to the
formula
F
.. Since the integral is over an innite interval there is a conver
gence question right away. Suppose that
fxdx lt . . F
We will write f L
if
F
. holds. If f L
then there is no problem with
the existence of the integral in
F
. because lim
a
a
a
e
ix
fxdx exists.
Proof
a
a
b
b
e
ix
fxdx
axb
fxdx as a b . Use
the Cauchy convergence criterion now.
Of course even if f L
it can happen that f
is not in L
and/or that
xfx is not in L
. This is a nuisance in dealing with the identities
F
. and
F
.. We are going to restrict our attention for a while to a class of functions
that will make these issues easy to deal with.
Denition. A function f on R is said to be rapidly decreasing if
x
n
fx M
n
, n , , , ... . F
for some real numbers M
n
. In words x
n
fx is bounded on R for each n.
Examples . e
x
is rapidly decreasing.
.
x
is not rapidly decreasing because
F
. only holds for n , ,
but not for n or more.
. If f is rapidly decreasing then so is x
fx because x
n
x
fx
x
n
fx which is bounded in accordance with
F
.. Just replace n by n
in
F
.. Since any nite linear combination of rapidly decreasing functions
is also rapidly decreasing we see that pxfx is rapidly decreasing for any
polynomial p if f is rapidly decreasing.
. So by examples . and . we see that pxe
x
is rapidly decreasing
for any polynomial p.
. Any function in C
c
R is rapidly decreasing.
. Summary The space of rapidly decreasing functions is a vector space
and is closed under multiplication by any polynomial. And besides, there are
lots of these functions.
Lemma Any rapidly decreasing function is in L
.
Proof Apply
F
. for n and n to conclude that
x
fx M
for some real number M. So
fxdx M
x
dx lt .
QED
Using only rapidly decreasing functions in
F
. will allow us not to have
to worry about whether f and xfx are both in L
. Neat, huh
But to use
F
. we still need to worry about whether f
is in L
. So
we are going to restrict our attention for a while, even further, to a class of
functions that makes both
F
. and
F
. easy to deal with.
Notation S will denote the set of C
functions on R such that f and
each of its derivatives is rapidly decreasing.
Examples
.e
x
is innitely dierentiable and each derivative is just a polynomial
times e
x
. Weve already seen that these functions are rapidly decreasing.
So the function e
x
is in S.
.
x
is innitely dierentiable but is not rapidly decreasing. So this
function is not in S.
. Now here is a real nice thing about this space S. If f S then any
polynomial, p, times any derivative of f is again in S. CHECK THIS against
the denitions I know this may seem too good to believe. But we do know
that there are lots of functions in S. All of C
c
is contained in S. And
besides there are more, as we saw in Example .
STATUS If f S then f and all of its derivatives are in L
. So both
formulas
F
. and
F
. make sense. Moreover they are both correct because
the boundary terms that we ignored in deriving
F
. are indeed zero, since
these functions go to zero so quickly at . Check this at your leisure
In truth, here is the reason that the space S is so great.
Invariance Theorem. If f S then
f S.
Proof First notice that for any function f in L
we have the bound
f
e
ix
fxdx .
fxdx .
f
..
Since the right side doesnt depend on
f is bounded.
Now suppose that f S. Then so is f
,f
, etc. So f
,f
etc. are all
in L
. By
F
. it now follows that
n
f is bounded for each n , , ....
So
f is rapidly decreasing. But d
f/d is the Fourier transform of ixfx
which we have seen is also in S. So d
f/d is also rapidly decreasing. And
so on. QED.
Here is what the inversion formula will look like
fy
e
iy
fd. . F
Since we already know that
f is in S we know that the right hand side of
F
. makes sense. Contrast this with the example that you worked out in
the homework Take fx if x and f otherwise. Then f L
so
f makes sense. But
f itself decreases so slowly at that its not in L
. So
F
. doesnt make sense. Arent you glad that we are focusing on functions
in S
Notation
gy
e
iy
gd . F
Inversion Theorem. For f S equation
F
. holds.
In other words
f f.
We are going to spend the next few pages proving this formula. Not
because I fear that you might not trust me, but because the proof derives
some very useful identities along the way.
. The Fourier Inversion formula on SR.
To begin, here are some explicit computations of some important Fourier
transforms.
Gaussian identities
Let
p
t
x
t
e
x
/t
.F
Then we have the following three identities.
p
t
xdx . F
p
t
e
t
/
.F
e
t
x
t
e
x
/t
p
t
x. . F
Proof of
F
. Sneaky use of polar coordinates.
p
t
xdx
R
p
t
xp
t
ydxdy .
t
R
e
x
y
t
dxdy .
t
e
r
/t
rdrd .
.
Use the substitution s r
/t in the last step.
Proof of
F
.
To compute p
t
we need rst to multiply p
t
x by e
ix
before integration.
The exponent is a quadratic function of x for which we can complete the
square thus
x
t
ix
t
x it
t
/
Hence
p
t
e
t
/
t
e
t
xit
dx . F
The coecient of e
t
/
looks just like
p
t
xdx which is one. This
would prove
F
.. But not so fast. x has been translated by the imaginary
number it. You cant just translate the argument to get rid of it. Here
is how to get rid of it. Let a t. and suppose a gt just so that I
can draw pictures verbally. Take a large positive number R and consider
the rectangular contour in the complex plane which starts at R on the
real axis, moves along the real axis to R and then moves vertically up to
the point R ia. From there move horizontally, west to R ia and then
south back to the point R. Since e
z
/t
is an entire function, the integral
of this function around this rectangular contour is zero. Otherwise put,
R
R
e
xia
/t
R
R
e
x
/t
dx plus a little contribution from the vertical
sides of the rectangle. On the right end the integral is at most a times
the maximum value of the integrand on that segment. But e
Riy
/t
e
R
y
/t
, which goes to zero quite quickly for y a as R . So
the integrals along the end segments go to zero as R . So it is indeed
true that the coecient of e
t
in
F
. equals
p
t
xdx, which is one.
QED
Proof of
F
.e
t
/
t
p
/t
by
F
.. Since p
/t
is even we have
e
t
/
/
t
p
/t
x
t
e
x
/t
by
F
.. QED
Denition . The convolution of two functions f and g on R is given by
f gx
R
fx ygydy . F
lemF Lemma . Let g be a bounded continuous function on R. Then for each x
lim
t
p
t
gx gx.
Proof Since
p
t
xdx we have
p
t
gx gx g p
t
x gx .
gx yp
t
ydy gx .
gx y gxp
t
ydy. .
Given gt , gt gx y gx lt if y lt .
p
t
gx gx
gx y gxp
t
ydy .
ygt
gx y gxp
t
ydy .
p
t
ydy g
ygt
p
t
ydy .
g
ygt
p
t
ydy. .
lim
t
p
t
gx gx g
lim
t
y
p
t
ydy.
But
y
p
t
ydy
t
e
y
/t
dy
t
y
e
y
/t
dy .
t
te
y
/t
te
/t
as t . .
Therefore
lim
t
p
t
gx gx gt .
Hence
lim
t
p
t
gx gx .
Q.E.D.
Lemma . If f and g are in SR then
f gx
n
fe
ix
g.
Proof
f gx
R
fx ygydy .
n
R
R
fe
ixy
gyddy .
n
R
R
fe
ix
e
iy
gydyd .
n
R
fe
ix
gd. .
Q.E.D.
Theorem . If g is in SR then
g
x gx.
Proof. In the preceding lemma put f e
t
/
. Then, by
F
.,
fx
p
t
x. Thus
p
t
gx
R
e
t
e
ix
gd.
Let t . Use Lemma
lemF
. on the left and Dom. Conv. theorem on the right
to get
gx
R
e
ix
gd.
The asymmetric way in which the factor occurs in the inversion formula
is sometimes a confusing nuisance. It is useful to distribute this factor among
the forward and backward transforms. Dene
Fg
/
g.
The factor in front of g has clearly no eect on the onetoone or ontoness
property of the map g g. That is, F is a onetoone map of SR onto
SR. However the factor makes for a nice identity
corFPl Corollary . Plancherel formula for S If f and g are in SR then
Ff, Fg f, g .
Explicitly, this says
R
FfFgd
R
fxgxdx .
Proof. Phrasing this identity directly in terms of
f and g, it asserts that
R
f gd
R
fxgxdx
But g
gxe
ix
dx
gxe
ix
dx. Hence
R
f gd
R
R
fgxe
ix
dxd .
R
R
fe
ix
gxd
dx .
R
fxgxdx. .
Q.E.D.
. The Fourier transform over R
n
The denitions, key formulas, theorems and proofs for the Fourier transform
over R
n
are nearly identical to those over R. In this section we are going to
summarize the results weve obtained so far and formulate them over R
n
.
The Fourier transform of a complex valued function on R
n
is
f
R
n
e
ix
fxdx . F
Here runs over R
n
. Just as in one dimension, one must pay some attention
to the meaningfulness of this integral. But the ideas are similar.
As in one dimension, the Fourier transform over R
n
interchanges multi
plication and dierentiation. The analog of
F
. is
j
f
R
n
e
ix
ix
j
fxdx. . F
So
Fourier transform ofix
j
fx
j
f. .
An integration by parts never mind the boundary terms clearly gives the
analog of
F
.
i
j
f
R
n
e
ix
f/x
j
xdx. . F
So
f/x
j
i
j
f.
We will see later that these formulas allow one to solve some partial dieren
tial equations. Moreover in quantum mechanics these two formulas amount
to the statement that the Fourier transform interchanges P
j
and Q
j
mo
mentum and position operators.
We say that a function f is in L
R
n
if
R
n
fxd
n
x lt .
The formula
F
. makes perfectly good sense if f L
R
n
. But in
the end we are going to give meaning to
F
. for a much larger class of
functions and generalized functions, including e.g. delta functions and their
derivatives. To this end we need the ndimensional analog of S.
Denition . A function f on R
n
is said to be rapidly decreasing if
x
k
fx M
k
, k , , , ... . F
for some real numbers M
k
. In words x
k
fx is bounded on R
n
for each k.
In Exercise you will have the opportunity to show that the condition
F
. is equivalent to the statement that for any polynomial px
,...,x
n
in n real variables, the product pxfx is bounded.
lemF Lemma . Any rapidly decreasing function on R
n
is in L
R
n
.
Denition . SR
n
is the set of functions f in C
R
n
such that f and
each of its partial derivatives are rapidly decreasing.
Theorem . a. If f is in SR
n
then
f is also in SR
n
.
b. The map f
f is a onetoone linear map of SR
n
onto SR
n
.
c. The inverse is given by the inversion formula
fy
n
R
n
e
iy
fd
n
..F
Denition . The convolution of two functions f and g on R
n
is given
by
f gx
R
n
fx ygyd
n
y.F
The important identity in the next theorem is the basis for the application
of the Fourier transform to solution of partial dierential equations.
Theorem .
fg
fg.F
Proof. The proof consists of the following straight forward computation.
A reader who is concerned about the validity of any of these steps should
simply assume that f and g are in SR
n
, although the identity is valid quite
a bit more generally.
fg
R
n
f gxe
ix
dx .
R
n
R
n
fx ygydye
ix
dx .
R
n
R
n
fx ye
ixy
dxgye
iy
dy .
R
n
fgye
iy
dy .
fg.
Theorem . Plancherel formula Dene F
n/
. Then
F
L
R
n
L
R
n
.F
This is the Plancherel formula. As a consequence F is a unitary operator on
L
R
n
This is a minor restatement of Corollary
corFPl
. and extension to R
n
.
Finally, here are the ndimensional analogs of the important Gaussian
identities
F
.
F
..
Gaussian identities over R
n
.
Let
p
t
x
t
n
e
x
/t
xR
n
..F
Then
R
n
p
t
xdx . F
p
t
e
t
/
.F
e
t
x
t
n
e
x
/t
p
t
x. . F
. Tempered Distributions
Denition . A tempered distribution on R
n
is a linear functional on
SR
n
.
Example . n Suppose that f R C satises
fx const. x
n
for some n . F
In this case we say that f has polynomial growth. E.g. x
sin x has polynomial
growth. Take n in
F
.. Of course n will also do. If f has
polynomial growth then the integral
L
f
R
fxxdx . F
exists when S because fxx const. x
n
x
n
, which
goes to zero at like x
and so is integrable. Clearly L
f
is linear. Thus
any function of polynomial growth determines in this natural way a linear
functional on S. Remember that we did not need polynomial growth of f
when we made L
f
into a linear functional on D. So we have a smaller dual
space now. But the delta distribution and its derivatives are in this smaller
dual space anyway because
L
is a meaningful linear functional on S. And similarly for
k
.
Example . We cant allow the function fx e
x
in
F
. because
the function x e
x
is in S and so
F
. wouldnt make sense.
Denition . The Fourier transform of an element L S
is dened by
LL
for S. . F
Now you can see the virtue of using S as our new test function space the
right hand side of
F
. makes sense precisely because
is back in S when
is in S. After all, L is only dened on S. Had we attempted to use D this
denition would have failed because
is never in D when D if .
Example .
L
L
. In the outside world this is usually written
.
Proof. Let S. Then
L
L
by def.
F
..
by the def. of L
.
R
xdx by the def. of Fourier trans. at .
L
by yet another denition .
One thing to take away from this proof is that every step is just the
application of some denition. There is no mysterious computation. Let this
be a lesson to all of us
Example .
L
L
.F
After you have achieved a suitable state of sophistication you can write this
as
.
Or even as
e
ixy
dx y
if you dont feel too uncomfortable with the seeming nonsense of the left
side. But you better get used to it. Thats how
F
. is written in the
outside world.
Proof. of
F
.
L
L
by def.
F
..
R
d by def.of L
.
by the Fourier inversion formula .
Notice that this time the last step uses something deep.
Consistency of the two denitions of the Fourier transform. If f L
then we already have a meaning for
f. So is
L
f
L
f
Yes. Here is a denition chasing proof.
L
f
L
f
.
R
f
d.
R
f
R
e
ix
xdxd .
R
R
fe
ix
d
xdx .
R
fxxdx .
L
f
.
QED
. Problems on the Fourier Transform
. Find the Fourier transforms of the following functions
ae
x
be
x
/
c xe
x
/
dx
e
x
/
e fx
if a x b
otherwise.
f
x
. For in SR
dene
T
xx, dx.
Note that this is an integral over a line not over R
.
a Show that T is in S
R
.
b Find the Fourier transform,
T, of T explicitly explicitly enough to do
part c without nding
.
c Evaluate
T where
,
e
d Let L a
b
c
. Find all values of the real parameters
a, b, c such that
L
T.
. Compute
a.
b.
c.
L
x
d.
L
x
Hints . Use the denitions. . Use the denitions. . Use the
denitions.
. Show that a function f on R
n
is rapidly decreasing if and only if, for every
polynomial px
,...,x
n
, the function pxfx is bounded.
. Derive the identities
F
.
F
. from the onedimensional identities
F
.
F
..
Applications of the Fourier Transform
. The Heat Equation by Fourier transform.
The three fundamaental types of PDE, parabolic, elliptic and hyperbolic are
amenable to study by using the Fourier transform in slightly diering ways
for each. The prototype of these three kinds of PDEs are the heat equaiton,
Poisson equation and wave equation.
The heat equation for a function u ut, x in n spatial dimensions is
u/t /u t . .
with initial condition
u, x fx. . .
Here f is a given function on R
n
. As you probably know very well, there
is strong physical motivation for seeking a solution to
.
. for t only
bodies cool o as time moves forward. If you move backward in time temper
atures can increase in such a way as to produce singularities in the solution.
We will see on purely mathematical grounds why solving the equation
.
.
backwards in time can produce singularities.
Solution of
.
.,
.
. Apply the Fourier transform to u in the spatial
variable only. We obtain a function ut, satisfying
ut, /t /
u..
which we would like to satisfy the intial condition
u,
f..
Now for each R
n
the equation
.
. an ORDINARY dierential equation,
whose solution we know from elementary calculus. It is
ut, e
t/
f. . .
In order to nd ut, x itself we still have to do a reverse Fourier transform
on
.
.. But Hey We already know of a function whose Fourier transform
is the rst factor on the right in
.
.. It is the function
p
t
xt
n/
e
x
/t
..
whose Fourier transform we already computed. See
F
. and thereafter.
Use the ndimensional version of that function.
We also know that convolution goes over to multiplication under Fourier
transformation. Hence
p
t
f
p
t
f.
e
t/
f. .
By the uniqueness property of the Fourier transform we therefore nd
ut, x p
t
fx .
t
n
R
n
e
xy
/t
fyd
n
y.
. Poisssons equation by Fourier transform
We wish now to apply the preceding method to the equation
u...
Although the method works in n dimensions with n , we will carry
this out only in dimension n , where we can compare with our previous
solution.
Apply the Fourier transform to
.
. in ALL of the variables. Of course
we only have spatial variables this time. We nd
u..
This time the partial dierential equation has been reduced to an algebraic
equation there are no derivatives left at all. So you think that the solution
is
u/
..
Well, if that were right then we again have a solution for the Fourier transform
in the form of a product. Of course this will give u itself as a convolution.
In fact it is true that
r
/
And this would give
u
r
,..
which we already know from our previous work is actually correct. We saw
back in those days that could be e.g any charge distribution in D
with
compact support. Such a distribution is clearly in S
also and the compu
tations
.
.
.
. are actually correct.
But we are losing some solutions in the passage from
.
. to
.
. For
example we have the identity
...
This means that we could add onto the right side of
.
. and get
another solution. Back in x space this just means that we can add a constant
onto any solution u and get another solution. In fact if v as any harmonic
function on R
i.e. v we could add v onto any solution of
.
.
and get another solution. For example x y z is harmonic. So is
x
y
z
. There are, in fact, lots of harmonic polynomials on R
. Among
all the solutions to
.
. the solutions
.
. are the only ones that go to
zero as x . The solution
.
. is called the potential of . It is true
that all the solutions of
.
. dier from
.
. only in such a way. Here is
a problem that illuminates this.
ProblemPoisson by Fourier
Suppose that px, y, z is a harmonic polynomial on R
. That is, p .
Write
j
/
j
and let L be the distribution
L pi
,i
,i
.
Show that
L.
Recall that this means L
for all SR
.
. Advanced, Retarded and Feynman Propagators for
the forced harmonic oscillator
Let be a strictly positive constant. The harmonic oscillator equation with
frequency actually frequency / is
d
ut
dt
ut . . .
the general solution to . is
ut Ae
it
Be
it
...
where A and B are complex constants. Of course one can choose A and B so
that the solution is real and is expressible as a linear combination of sint
and cos t. But we are going to stick to the complex form
.
. because it
will be more revealing for our purposes.
The forced harmonic oscillator equation is
d
ut
dt
ut ft. . .
It will be adequate for our purposes just to assume that f C
c
R
and then not have to worry about any technical details. We can think of a
weight hanging from a spring which is attached to the ceiling. ut is the
displacement of the weight from its neutral position. For a few seconds we
push and pull on the weight up and down in accordance with the force ft
and then let go altogether at the upper bound of the support set of f. What
happens to the weight during and after the disturbance thats us acts via f
Of course that depends on the initial state of the weight at time t , say.
Was the weight in its neutral position u or was it elsewhere Was it
moving or stationary If we specify the intial position u and the initial
velocity u
then we must solve the Initial Value Problem for the equation
.
. with the specied f and specied initial data. Our goal however is to
address a dierent problem. We are going to study the equation
.
. from
the point of view of the Boundary Value Problem, somewhat in the spirit of
the section on Green functions. The novelty of the present section lies in the
fact that we will be interested in boundary conditions at . The function
f is sometimes called the source generating the motion u. For example in
the context of Maxwells equations the sources for the electric and magnetic
elds are charges and currents. As in
.
. they make the homogeneous
Maxwell equations inhomogeneous.
We saw in the section on Green functions that on an interval a, b one
can choose, at a, Dirichlet or Neumann conditions or a linear combination
of them. Similarly at b. In this sense we have four degrees of freedom in
the nature of our choice of boundary conditions. But you can only impose
two of them on a solution. Here are three dierent Green functions for the
equation
.
.. We will discuss their physical interpretations afterward. Let
G
r
t
sin t
t gt
t
..
G
a
t
t
sin t
t lt
..
G
F
t
i
e
it
t
e
it
t lt
.
Problem . Suppose that f C
c
R. Let
uGf
with G G
a
or G
r
or G
F
.
a. Show that in each case u is a solution to
.
.. Hint. Imitate the
proof of Theorem
thmO
.. Note that each of these three functions is continuous
and has a jump of one in its rst derivative at t .
b. Show that the solution G
r
f is zero in the distant past, i.e. for
suciently large negative time, depending on f.
c. Show that G
a
f is zero in the distant future.
Terminology. The Green function G
r
is called the retarded Green func
tion, and also sometimes called the retarded propagator. It propagates the
disturbance f into the future and depends on the the disturbance at some
earlier time. The etymology of the word retarded can be understood in the
context of electromagnetic theory, where the potential,at x t, of a changing
charge distribution depends not on the charge distribution at time t but at
an earlier time. The eld produced by the charge distribution travels only
with the speed of light, not instantly. The resulting potential is called the
retarded potential.
Similary the propagator G
a
is called the advanced propagator. As you
saw in part c. of Problem , u
a
t can be nonzero before the disturbance
f even begins Doesnt sound very causal, does it One says that G
a
propagates the disturbance into the past.
The third propagator, G
F
, is particularly important in understanding the
behavior of electrons and positrons in combination. G
F
is called the Feyn
man propagator for the single frequency . It propagates the disturbance
f into the future as a positive frequency wave and into the past as a neg
ative frequency wave. When this discussion is boosted up to three space
dimensions from the present zero space dimensions Feynmans propagator
has the interpretation of propagating electron wave functions forward in time
as positive energy wave functions and propagating positron wave functions
backward in time as negative energy wave functions. Some authors say, for
short, that positrons are negative energy electrons moving backward in time.
The Green functions above have been constructed directly in terms of
the homogeneous solutions in accordance with the method in the section on
Green functions. Here is how the Fourier transform method can be applied.
In some contexts its the more useful way to go.
Take the Fourier transform of
.
. in the t variable which is the only
variable around, this time. We will use s for the variable conjugate to t.
That is, we dene us
e
ist
utdt. We nd
s
us
fs. . .
For the homogeneous equation we need to put f . The solutions to
.
.
then include the functions at the two roots of
s
. Thus
us As Bs . .
gives solutions to the homogeneous equation on the Fourier transform side.
It is a FACT that these are the only solutions to
.
. when f .
Problem . Show that the Fourier transform of the solutions
.
. are
given by
.
. up to a constant. Thus we have recovered the solutions to
the homogeneous equation by the method of Fourier transforms.
Next we address the inhomogeneous equation. Note rst the identity
s
s
s
..
Dividing
.
. by the coecient of us we nd
us
fs
s
fs
s
..
TROUBLE We already know that
s
has a nonintegrable singularity and
therefore so does the right hand side of
.
.. As it stands the right hand
side therefore has no meaning. But we saw earlier that the function /s
has an interpretation as a distribution called the Principle Part of /s. It is
dened as
P
s
lt gt lim
c
sgtc
s
s
..
Lemma . Let
t /sgnt . .
Then
s iP
s
..
Proof
.
lim
a
a
a
t
tdt .
lim
a
a
a
te
itx
xdxdt .
lim
a
a
a
te
itx
dtxdx .
lim
a
cos ax
ix
xdx .
lim
a
xgt
cos ax
ix
xdx .
lim
a
x
cos ax
ix
x dx .
because
cos ax
ix
is an odd function. The next to the last line converges to
xgt
x
ix
dx by the RiemannLebesgue Lemma. The very last integral
may be rewritten
x
cos ax
x
ix
dx. So this integral converges to
x
ix
dx by the RiemannLebesgue Lemma also. Q.E.D.
Problem . Using Lemma show that
te
it
s iP
s
..
for any real .
Problem Let
F
t te
it
for gt . .
In view of
.
. and
.
. we should expect that all the solutions to
d
ut
dt
ut t. . .
should be expressible as a linear combination of the functions F
and the
solutions to the homogeneous equation
.
.. For example one expects
G
r
t AF
t BF
t Ce
it
De
it
.
for some constants A, B, C, D.
Find explicitly this representation of the three Green functions G
r
,G
a
and
G
F
in terms of the functions F
and the solutions,
.
., of the homogeneous
equation.
. The wave equation by Fourier transform
The physically appropriate problem is the Cauchy problem. This is the initial
value problem.
ut, x
t
ut, x . .
u, x fx .
u, x/t gx . .
As in the heat equation we will take the Fourier transform in the space
variables only. We nd
ut,
t
ut, . .
Just as in the case of the heat equation we have now an ordinary dierential
equation. But this time its second order in t. The general solution to
.
.
for each is
ut, A cos t B sin t . .
Upon Fourier transforming the initial conditions
.
. we see that we must
have A
f and B g. Hence the Fourier transform of u is
given by
ut,
f cos t g
sin t
...
That was the easy part. Now we have to Fourier transform back to nd u
itself. Notice rst that cos t is exactly the t derivative of sin t/. It
will suce then to nd a distribution
t
on R
whose Fourier transform, for
each t, is sin t/ . The solution to
.
.,
.
. is then clearly
ut, x
t
gx /t
t
fx. . .
Lemma . Let
t
be surface area element on the sphere x t in R
.
Then
R
e
ix
d
t
xt
sin t
..
Proof. The integral is just an integral over a sphere of radius r t. Use
spherical polar coordinates , . We can make use of the rotation invariance
of the integral by choosing the positive z axis in the direction of . Let
a . Then x ra cos . So the integral is
r
e
ira cos
sin dd r
sin ra
a
which you get by substituting s cos . This proves
.
. once one
observes that even if t lt
.
. is correct because sin is odd. QED
Now dene a distribution
t
on R
by the formula
t
t
xt
xd
t
x for SR
..
for t and dene
t
to be zero for t .
.. Problems
Problem . Prove that
t
sin t
for all t . .
Hints . Use the denition of Fourier transform of a distribution.
. Use the denition of
t
.
. Use Lemma .
Problem Explain why the solution
.
., in combination with the explicit
form
.
. of
t
, says that light travels with exactly speed one in our units.
Finally, lets consider the inhomogeneous wave equation
ut, x/t
ut, x t, x. . .
Assume, for ease of mind, that is in C
c
R
. If we rst Fourier transform
in the spatial variables only we nd
ut, /t
ut, t, . .
Notice that for each this equation looks just like the harmonic oscillator
equation
.
. with
. We can, and will, take over what we
learned from the forced harmonic oscillator. It will be suciently illuminat
ing to focus on just one of the three propagators that we studied for the
harmonic oscillator.
Put in
.
. and do the usual reverse Fourier transform to nd.
ut, t
e
it
t
e
it
t,
For each t and t
we see that we have the usual product of Fourier transforms
on the right in the spatial variables. Moreover we have the formula
.
.
at our disposal. Thus we nd
ut, x
tt
tt
t
, xdt
..
Not only does the integral represent a convolution but there is also a convolu
tion right in the middle of the integral. All in all this represents a convolution
over R
. We may write this as
uG..
where
Gt
t
..
is a distribution on R
for each t and all together is a distribution on R
.
This is one of the propagators for the inhomogeneous wave equation ..
One might wish to write
.
. more suggestively to emphasize its char
acter as a distribution over R
as Gt, x t
t
x. But one should not
lose sight of the fact that the second factor is not actually a function.
Remark to ponder. For the wave equation, as for the forced harmonic
oscilator, there are essentially three propagators for the inhomogeneous
equation
.
.. We arrived at one of them by the above procedure. Its
one half the advanced plus retarded. Where does the Feynman propagator
come from Hint Consider the identity lim
xi
P/x i which
you proved long ago in
.
..
Laplace Transform
Denition. The Laplace transform of a function f , C is the
function L
f
dened by
L
f
s
fte
st
dt. . L
The domain of L
f
is the set DL
f
consisting of those s R for which
fte
st
dt lt .
Examples. . ft e
t
, DL
f
. If f L
, then DL
f
, . E.g., if ft / t
then
DL
f
,.
. ft e
t
. Then L
f
s
e
st
dt lt if s gt . So DL
f
,.
. ft e
t
. Then DL
f
, . Note. If s
DL
f
then
s DL
f
for any s gt s
because
fte
st
fte
s
t
..
L
,
e
ss
t
..
bdd
Consequently DL
f
is always an interval. As we see from the above
examples it may be the empty set or of the form s
, , or s
, , or
,.
Remark. If s
DL
f
then L
f
has an analytic extension to the half
space Re z gt s
.
Proof Put z s i with s gt s
. Then
e
zt
ft e
st
ft L
,.
Hence the function
Lz
fte
zt
dt . L
exists for Re z gt s
.
By dierentiating under the integral sign easily justied we see that
Lz is analytic with
d
Lz/dz
tfte
zt
dt.
Example. ft e
t
, DL
f
,
Lz
e
zt
dt
z
which is analytic in Rez gt . But note that
L has an analytic extension to
the entire complex plane with the exception of a pole at z . This is typical
of Laplace transforms that arise in practice. They often have meromorphic
extensions to the entire plane. But the integral representation may be only
valid for Re z gt s
.
. The Inversion Formula
Suppose s
DL
f
and s
gt s
. Let
g
Ls
i.
fte
s
it
dt .
fte
s
t
..
L
,
e
it
dt. .
Put
ht
lt t lt
fte
s
t
t lt .
Then
g
hte
it
dt Fourier transform of h.
Hence
ht
ge
it
d.
This integral may converge pointwise in t or in the L
sense of convergence,
depending on the quality of g. But we ignore these questions.
Thus for t lt
fte
s
t
ge
it
d
by the Fourier inversion formula. So
ft
e
s
t
Ls
ie
it
d for t . . L
Corollary . Uniqueness Theorem The Laplace transform L
f
s of a
function f on , determines f uniquely up to values on a set of measure
zero if DL
f
is not empty.
. Application of Uniqueness
Problem. Find the function ft whose Laplace Transform is
Lz
z
z
z
z
.
Solution ft e
t
e
t
cos t.
Reason.
z/z
zi
zi
.
Note. In this problem Lz had simple poles. Higher order poles can be
dealt with using
d
Lz
dz
te
zt
ftdt.
But note.
L may have no singularities in the nite plane even though f is
not the zero function.
Example. Let f be in L
R with support in , a. Then
Lz
fte
zt
dt
a
fte
zt
dt
exists for all z C and is entire.
Nevertheless the case where
L has only poles and goes to zero at is
important. For this situation it is useful to rewrite the inversion formula
thus
ft
Ls
ie
s
it
d.
Putting z s
i, dz id we get
ft
i
s
i
s
i
Lze
zt
dz.
Sometimes the straight line integration contour can be deformed to surround
the poles of
L. See the homework problem.
. Problems on the Laplace Transform
. We have seen that if
Lz
fte
zt
dt
exists in the open half space Re z gt s
then for any real number s
gt s
f
may be expressed by
fz
e
s
t
Ls
ie
i
d.L
with z s i. This may clearly be rewritten
ft
i
C
Lze
zt
dz . L
where C is the straight line contour from s
i to s
i.
For some functions
Lz it may be possible to deform the contour to C
so as to include all the singular points of
L inside. If these singular points
are isolated poles then one could nd fz by the method of residues.
Justify this procedure for the function
Lz
z
z
z
z
z
and use it to nd ft for all t .
Asymptotic Expansions
Reference C.M. Bender and A.S. Orszag, Advanced Mathematical Methods
for Scientists and Engineers, Chapter . QA B
. Introduction
Denition If f and g are two real or complex valued functions on a set S
then one writes
f Og on S
if there is a real constant C such that
fx Cgx for all x S.
Example. x x
/
Ox
on , but not on ,
Denition If f and g are two functions on an interval a, then one writes
f Og as x
if f Og on M, for some M.
Example x x
/
Ox
as x .
Denition fx ogx as x means
lim
x
fx/gx .
Examples. i If fx /x
and gx /x then fx ogx as
x.
ii e
x
oe
x
as x .
More Notation fx xox as x means that fxx
ox as x .
Example.
x
x
o/x
as x .
We are going to study the four basic methods for determining the
Asymptotic behavior of an integral.
Four Methods
. Integration by parts page of NB
. Laplaces method p. of NB
. Stationary phase p. of NB
. Steepest descent p. of NB
. The Method of Integration by Parts.
PROBLEM Let
fx
x
e
t
/tdt, x gt .
Find the asymptotic behavior of fx as x .
SOLUTION Integrate by parts to nd
fx
x
/tde
t
.
e
t
t
x
x
e
t
t
dt .
e
x
x
e
x
e
t
t
dt .
Now it happens that the last term the integral is rather small compared
to the rst term for large x. This needs to be proved. But once it is proved
we can see that fx behaves like e
x
/x for large x, the precise meaning of
which can be stated in terms of the preceeding notation in the form
fx
e
x
x
o
e
x
x
as x .
We will do better than this. But rst we need the following tricky lemma.
Lemma . For any integer n we have
x
e
t
t
n
dt O
e
x
x
n
as x .
Proof
x
e
t
t
n
dt
x/
e
t
t
n
dt
x
x/
e
t
t
n
dt .
x/
e
t
dt
x
x/
e
t
x/
n
dt .
e
x/
e /x
n
e
x
e
x/
,.
which proves the lemma.
Corollary .
x
e
t
t
n
dt o
e
x
x
n
as x .
One can go a step further now in determining the asymptotic behavior of
f. Integrating by parts again
fx
e
t
t
x
e
t
t
x
x
e
t
t
dt e
x
x
x
O
e
x
x
as before.
Clearly we may continue this and get
fx e
x
x
x
x
x
n
x
n
o
x
n
.
This motivates the following denition.
Denition. Let
,
,
, . . . be a sequence of functions on a, such
that
n
xo
n
x. If f is a function on a, one says that the series
j
a
j
j
xa
j
are constants is an asymptotic expansion of f if for each
n,,,...
fx
n
j
a
j
j
xo
n
xx.
Thus the innite series
j
a
j
j
x need not converge for any x. In the
preceding example the series
n
n
x
n
is easily seen by the ratio test to
converge for no x.
. Laplaces Method.
.. Watsons Lemma
Theorem . Watsons Lemma Assume
A.
fte
ct
dt lt for some c gt .
B. For some gt and gt , f has the asymptotic expansion
ft t
k
a
k
t
k
as t . . As
Let
Ix
fte
xt
dt x gt c. . As
Then Ix has the asymptotic expansion
Ix
k
a
k
k
x
k
x . . As
where
x
u
x
e
u
du for x gt .
Proof First, note that for a simple power we have
t
e
xt
dt x
s
e
s
ds by putting s xt .
x
, gt . . As
Second, note that we may replace
fte
x
by
fte
x
and incur an
exponentially small error as in the following lemma.
Lemma . For any gt and f satisfying A
fte
xt
dt ox
as x for all gt . . As
Proof of Lemma For x gt c we have
fte
xt
dt
fte
ct
e
xct
dt .
e
xc
fte
ct
dt ox
x gt .
. As
Q.E.D.
Now in order to prove the theorem we choose N and let N .
We must show that
Ix
N
k
a
k
k
x
k
ox
x . As
since x
is the last retained term in the sum.
By the assumption B
ft
N
k
a
k
t
k
ot
N
t . . As
Hence there exists a constant K
and gt
ft
N
k
a
k
t
k
K
t
N
t . . As
Actually
As
. is weaker than
As
.. Hence
ft
N
k
a
k
t
k
Kt
N
t . As
for K K
a
N
. Thus
Ix
fte
xt
dt E
x where E
x ox
, x by Lemma
.
N
k
a
k
t
k
e
xt
dt
ft
N
k
a
k
t
k
e
xt
dt
..
E
x
E
x.
.
But by
As
.
ft
N
k
a
k
t
k
e
xt
dt
K
t
N
e
xt
dt .
K
t
N
e
xt
dt .
Kx
N
N
.
const. x
.
ox
x..
Thus
Ix
N
k
a
k
t
k
e
xt
dt E
xE
x where E
x ox
,x
.
N
k
a
k
t
k
e
xt
dt E
xE
xE
x where .
E
x ox
, x by Lemma .
N
k
a
k
x
k
kE
xE
xE
x. .
This proves
As
.. Q.E.D.
corAS Corollary . Assume A.
gte
ct
dt lt for some c gt .
B. g has an asymptotic expansion
gt
k
a
k
t
k
as t . . As
Let
Ix
gte
xt
dt. . As
Then
Ix
k
a
k
k/
x
k/
as x . As
is an asymptotic expansion of Ix.
Proof By the same argument as in the theorem, i.e., Lemma , we commit
an exponentially small error in replacing the integral by
gte
xt
dt. We
will write the remainder of the proof in an informal way because the details
of justication are exactly the same as in the proof of Watsons Lemma. In
fact we will write equality of series even though one must interpret all steps
in terms of nite sums and behavior as x or t . In other words we
will concentrate just on the algebra. Thus we have
E
x exponentially small error .
Oe
x
..
Ix
gte
xt
dt E
x.
k
a
k
t
k
e
xt
dt E
x.
k
a
k
t
k
e
xt
dt E
x. .
At this point we could replace the integrals by
k
t
k
e
xt
dt which can
be done explicitly by integration by parts, and this would give the answer
As
.. But instead lets derive it from Watsons Lemma.
Put s t
. Then dt ds/
s. So
Ix
k
a
k
s
k
e
xs
ds
s
E
x.
k
a
k
s
k/
e
xs
ds E
x.
k
a
k
k/
x
k/
Ex by Watsons Lemma with
,.
.
Recall that these equations are not identities but asymptotic expansions.
Q.E.D.
Using the same ideas, here is another method. Let
ft
e
xht
dt
where h need not be quadratic but looks like this
We assume for simplicity that h but if not this can be subtracted from
ht changing our result by an overall factor of e
th
. As in the quadratic
case the main contribution to ft will come, for large x, from a neighborhood
of t say t , the error being
e
tx
x
n
dx t
n/
n
n
n
/
exponentially small, e
xhs
. We are going to get power decay of fx. Now
since h is a minimum at t we have h
t . Hence for small t
ht
h
t
Ot
as t .
Assume h
gt . Our objective is to get the leading term in the asymptotic
expansion of fx if there is such an expansion. So we ignore the Ot
terms to get
fx
e
x
h
t
dt .
h
x
..
So
fx
xh
.
Denition. The function is
x
u
x
e
u
du, x gt . . As
Note
n n . As
Proof
e
u
du . So
As
. holds for n . Induction
Note that for x gt ,
x
u
x
e
u
du . As
u
x
e
u
e
u
du
x
du
du
e
u
xu
x
du xx.
So
x xx for x gt . . As
Thus if n n then n n n n . Q.E.D.
Some values i /
.
ii k
k
kk,k.
Proof
t
/
e
t
dt
e
s
ds
. The identity ii
follows now by induction.
corAS Corollary . Suppose that a lt lt b a or b or both could be innite and
that ht for t a, b. Assume that h and ht gt for all other
t in a, b. Assume that h is in C
a, b and that h
. Finally assume
that
A.
b
a
e
cht
dt lt for some c gt
AND
A.
t
e
xht
dt ox
/
as x gt .
Let
Ix
b
a
e
xht
dt x gt c.
Then
Ix
xh
ox
/
.
Proof By Taylors theorem with remainder we may write
ht
h
t
gt
for t near zero where g is smooth and g . Choose so small that
gt gt / for t lt .
Let s t
gt on , . then
ds/dt
gt tg
t/
gt.
At t this is one. So we may choose even smaller if necessary so that
ds/dt gt on , . Since s is a strictly monotone function of t on ,
we may solve for t in terms of s and we may expand t in terms of s to any
nite order. The expansion begins
t
dt
ds
s
s
d
t
ds
s
s
.
But dt/ds
s
/ds/dt
t
. Hence
t s Os
as s .
Now
Ix
e
xht
dt
alttltb,t
e
xht
dt.
By Assumption A
the second term is ox
/
. On the interval , make
the change of variables s t
gt. Then
e
xht
dt
e
x
h
s
dt
ds
ds lt , gt .
e
x
h
s
Osds .
/
xh
/
/
Ox
/
as x by Corollary
corAS
..
.
QED
.. Stirlings formula
Corollary .
lim
x
x
e
x
x
x/
Proof Make the substitution u x tx in Equation
As
. to nd
x
u
x
e
u
du .
xt
x
e
xt
xdt .
x
x
e
x
e
xt
t
x
dt .
x
x
e
x
e
xtlogt
dt .
So take ht t log t. Then h
t / t. Therefore h has a
minimum at t . Moreover h
. Hence, by Corollary
corAS
.,
e
xht
dt
/x ox
/
as x .
This proves Stirlings formula.
The most frequently used form of Stirlings formula is for integer values
of x. This takes the form
STIRLINGS FORMULA
lim
n
n
e
n
n
n/
. The Method of Stationary Phase
Ref. Sec.. of Bender and Orszag
Theorem . Let f and be dened on a, b with f C
a, b and real
with
t on a, b but
a . Assume that is in C
p
a, b and
p
a , while
n
a , n , , . . . , p and fa .
Let
Ix
b
a
fte
ixt
dt.
Then
Ix fae
ixai/p
p
x
p
a
/p
/p
p
Ox
as x
where I sgn
p
a.
Example. Ix
/
e
ixcos t
dt. Here t is a stationary point t
cos t, ft , ,
,
. So p . Therefore
Ix e
ixi/
x
/
..
Ox
as x .
Proof of Theorem Write for small
Ix
a
a
fte
ixt
dt
b
a
fte
ixt
dt.
The second integral is Ox
as x by the lemma. So we may
ignore it for any xed gt . Now the idea is to choose so small that
rst integral can be adequately approximated by using the rst terms of the
expansion of f and at a. Thus we use the approximation ft
.
fa and
t
.
a
p
a
p
ta
p
and we omit justication of this approximation
as do Bender and Orszag. Thus
Ix fa
a
a
e
ixa
p
ata
p
/p
dt O
x
.
fae
ixa
e
ixs
p
ds O
x
.
where
p
a
p
.
Case gt . Put s e
i/p
u
x
/p
. Then
s
p
e
i/
u
x
i
u
x
.
Therefore
Ix fa
e
ixai/
x
x
p
e
i/p
e
u
u
p
p
du.
Case lt yields the other sign and C is which must be rotated down.
The relevant substitution is s e
i/p
u
x
/p
.
Case gt . For p , , , . . . put
se
i/p
/x
/p
u
/p
on the lower half plane Imu . . As
This is a well dened continuous function of u in the shaded region if one
chooses the positive pth root on the positive x axis. Then
ds
e
i/p
x
/p
p
u
p
du . As
and also
s
p
e
i/
/xu iu/x. . As
So u ixs
p
. Hence as s traverses the interval , on the positive x
axis u traverses the contour C
on the negative y axis. Thus
e
ixs
p
ds e
i/p
/px
/p
C
e
u
u
p
du . As
But
C
e
u
u
p
du
x
p
e
u
u
p
du
C
e
u
u
p
du . As
by analyticity of e
u
u
p
in the open doted region. The singularity at the
origin is not too bad. The rst term on the right of
As
. converges to
/p as x with exponentially small error. Moreover on C
we may
write u re
i
with r x
p
and / . Thus
C
e
u
u
/p
du
/
e
re
i
r
p
e
i
p
p
ird
.
r
/p
/
e
r cos
d.
r
/p
/
e
r/
dt r
/p
O
r
.
x
/p
O
x
. . As
Now combine ,
As
.,
As
.,
As
and
As
.. Q.E.D.
. The Method of Steepest Descent
Objective To nd asymptotic behavior of
Ix
b
a
fte
xt
dt
where f and are analytic, a and b are complex or innite, and the integral
is along some contour in the complex plane.
Under these circumstances the contour can be deformed without changing
the value of the integral. We take advantage of this by deforming to a contour
on which the imaginary part of is essentially constant. This will eliminate
the oscillation problem.
Example . From Bender and Orszag, p. Example
Ix
ln t e
ixt
dt.
Note that the method of stationary phase fails t t has no stationary
point. Moreover ft ln t is singular at so integration by parts fails.
Step . Deform path from to . Now along C
,e
ixt
e
xT
e
ixu
,u,
where we have put t uiT. So for xed x gt ,
C
as T . Hence
Ix
i
ln t e
ixt
dt
i
ln t e
ixt
dt.
On these paths there is no oscillation of the exponential factor because it has
constant imaginary part on each path. To see this put t is, s lt on
the rst path and t is on the second path to get
Ix i
lnise
xs
ds ie
ix
ln ise
xs
ds.
These are both Laplace integrals and so the problem is now in principle
solved We have gotten rid of the oscillatory integrals by deforming the
path. It happens that the rst integral can be done explicitly Put u sx.
Then
i
lnise
xs
ds
i
x
lniu/xe
u
du .
i/x
lni/x
ln ue
u
du
..
Eulers constant
.
i ln x i/ i
x
..
The second term has an asymptotic expansion via Watsons Lemma
ln ise
xs
ds
n
is
n
n
e
xs
ds .
n
i
n
nx
n
u
n
e
u
du .
n
i
n
n
x
n
..
Note that this is asymptotic only and not convergent because the series for
ln is only converges in a neighborhood of s . This is all that is
required for aymptotic expansion. Thus
Ix
i ln n i
x
ie
ix
n
i
n
n
x
n
as x .
.. Constancy of phase and steepness of descent
Consider an analytic function t t it where t u iv is a com
plex variable. We have seen that the asymptotic properties of
b
a
fte
xt
dt
depends on the peaking property of . It is fortuitous that these peaking
properties are optimal along paths of constant as we shall now see.
Let C be a curve of constant and let t C.
Case a.
t . The Cauchy Riemann equations read
u
v
,
n
u
. So
u
u
v
v
u
v
v
u
.
So is and therefore is parallel to C. Hence, among all directions
starting from t, varies most rapidly in the tangent direction of C. Thus
ascends most rapidly in one direction along C and descends most rapidly in
the other direction along C. The level curves of are orthogonal to C. So
the curves of constant phase are also steepest curves for
e
xt
e
xt
.
In Example the endpoint t of the constant phase path C
is a point
with
t . Fortunately, with t u iv, t Re it v s in
that example reaches a maximum along C
at the end point t . This was
the source of our success in the asymptotic expansion of
C
.
Case b.
t . Typical behavior of near such a point is illustrated
by the case t t
n
near zero, n , , . . . since the expansion of around
a point t
is t a
tt
n
c Ot t
if
k
t
,k,...,n
and
n
t
.
Use polar coodinates t re
i
. Then
t
n
r
n
e
in
.
r
n
cos n ir
n
sin n .
i. .
The directions of most rapid variation of are the directions in which
cos n steepest ascent and cos n steepest descent. As one
increases one alternately reaches curves of steepest ascent and steepest de
scent. In either case sin n , so these are precisely the curves of constant
phase emanating from the origin.
In case n the graph of t is clearly saddle shaped
A point t
t is called a saddle point in this business. See Bender
and Orszag for more pictures.
In Example the maximum of along the curves C
and C
of constant
phase occurred at the endpoint of each path where
. If a maximm of
occurs at an interior point of a curve of constant phase it will necessarily
be a saddle point. This is a consequence of CauchyRiemann equations.
The following example illustrates this case.
Example . Bender and Orszag p. Asymptotic behavior of Bessel
function J
x.
Def. J
x
/
/
cosx cos d/.
Step . Get this into standard form. Thus
J
x Re
/
/
e
ixcos
d.
To coincide with others notation, we also rotate the plane by putting
t i. Now cos
e
i
e
i
e
t
e
t
cosh t. so
J
x Re
i
i
i
e
ixcosht
dt.
Step . Change to contour of constant phase.
Now the phase is the same at the two endspoint i cosh
i
i cos
and i cosh
i
also. Nevertheless they do not actually lie on a common
contour of constant phase.
To see what the constant phase contours are like write t u iv and
t i coshu iv icosh ucosh iv sinh usinh iv icosh ucos v
sinh ui sin v. So Imt cosh ucos v t and Ret sinh usin v
t. Therefore the horizontal lines v / and v / are curves of
constant phase . The curves coshucos v are also curves of constant
phase . One of them goes through . Thus
t t it t sinh usin v t cosh ucos v
and
J
x Re
i
i/
i/
e
xt
dt t u iv
Since Ret sinh usin v, e
xt
is very rapidly decreasing on the distant
parts of contours C
,C
, C. Hence we may change from the original contour
C
to the three contours C
,C
, C, of constant phase.
Step . Evaluation. On C
we have Imt , so
C
e
xt
dt is real.
Hence it contributes nothing to J
x. Similarly for
C
. Consequently we
have
J
x Re
i
C
e
ixcosh t
dt. . As
As we saw, t Rei cosh t sinh usin v goes to as we got to in
either direction on C. The asymptotic behavior is therefore determined by
any peaks in t on C. These are saddle points. To nd them set
t.
Thus i sinh t . t is clearly a saddle point. It is the only one on C.
Exercise.
Since the asymptotic behavior is determined by a neighborhood of on
C we can approximate C locally by a straight line whose slope is determined
thus
cosh ucos v .
For small u and v this is
u
v
. So
u
v
to second
order. Therefore
u
v
or u v.
The curves of constant phase look like this.
This pattern repeats itself as one goes up the y axis n units. Thus if we
parametrize locally on C by
t is lt s lt
we can expect some kind of approximation to J
x as x . Thus, to
leading order
J
x Re
i
s
s
e
ixt
dt .
Re
e
ix
i
e
xs
ids .
Re ie
ix
x
.
Ree
ix
nx
.
So
J
x
x
Ree
ix
Therefore
J
x
x
cos
x
x.
Note This agrees with the Example on p. if one compares ReIx.
To get the full asymptotic expansion of J
x we must go back to
As
.
and evaluate it more carefully. On C, t i t where t is real. We
parametrize the curve by r t which goes from to on both halves
of C. Now i cosh t i r so i sinh tdt dr. Therefore
dt
dr
i sinh t
dr
i
ir
dr
i
r
i
r
J
x Re
t
e
ix
e
xr
i
r
dr
it
Re ie
ix
e
xr
dr
r
So
J
x Re
e
ix
r
ir
e
xr
dr.
To get the full asymptotic expansion one now expands
ir
in a power
series valid for small r and informally integrates term by term getting a valid
asymptotic expansion. The nal form according to Bender and Orszag,
who get
dr
instead of my
ir
, is
J
x
x
x cos
x
x sin
x
where
x
k
k
k
kx
k
x
and
x
k
k
k
kx
k
x.
Reference for x and x, Bender and Orszag, p..
. Problems on asymptotic expansions
Last homework of
. Watsons Lemma Suppose that gt is a polynomially bounded function
on , . That is
gt C t
n
t lt
for some constant C and some integer n . Suppose further that g has an
asymptotic expansion as t
gt
n
a
n
t
n
,t.
Let
Ix
gte
xt
dt x gt .
Prove that Ix has the asymptotic expansion
Ix
n
na
n
x
n
x.
. Find the asymptotic behavior as x of Ix where
Ix
ln ite
xt
dt.
. Find the asymptotic behavior of J
x as x up to and including
terms of order x
i.e., x
/
.
Old Prelims
Math Prelim in class Friday, October ,
N.B. Use a writing style which leaves no ambiguity as to what your ar
gument is. Your writing style is a part of this test.
. Suppose that f , R satises
fx
dx
and
fxdx .
What can you say about
xfxdx
.Which of the following dierential operators are symmetric on the in
terval ,
a Lux x
u
x xu
x sin xux.
b. Lux u
x xu
xx
ux.
. Write the Green function for the boundary value problem
u
u f on ,
u
u
in two ways.
a. in closed form i.e. no series.
b. as an eigenfunction expansion.
. Which of the following are linear functionals on the given vector space
V
a. V R
, Fx
,x
x
x
b. V C, , F
xx
dx
Appendix Review of Real numbers, Se
quences, Limits, Lim Sup, Set notation
These notes are intended to be a fast review of elementary facts concerning
convergence of sequences of real numbers. The goal is to give a self contained
exposition of the notion of completeness of the real number system. Com
pleteness is responsibe for the existence of solutions to ordinary dierential
equations, partial dierential equations, eigenvalue problems, and almost ev
erything else that guides our intuition in analysis. The main ideas in these
notes are usually taught in a freshman calculus course. Some of these ideas
are developed over and over again in later courses. Chances are that this is
at least the third time around for most of you ... but it may have been a
while... .
Recall that a rational numbber is a number of the form m/n where m
and n are integers i.e. whole numbers. If x m/n is a rational number
it is always possible to choose m and n to have no positive integer factors in
common. We say that x is then represented in lowest terms.
Example of an irrational number. The suare root of is not rational.Proof
Assume that x m/n is represented in lowest terms and that x
. Then
m
n
. So m
is even i.e. is a multiple of . If m itself were odd then
m
would also be odd because k
k
k. So m must be even.
Say m j. Therefore j
n
. So j
n
. It now follows that n is
also even. This contradicts our assumption that m and n have no common
factors. So there is no rational number x satisfying x
. Q.E.D.
Denote by Q the set of rational numbers and by R the set of real numbers.
We will be using the symbols and in this course quite a bit. But it
is a universally accepted convention not to call them real numbers. So they
are not in R. Some authors like to adjoin them to R and then call the result
the extended real number system R . We will do this also.
Notation , R, , R , etc. But R still means the
set of real numbers and excludes .
Addition, subtraction, multplication and division but not by zero make
sense within Q and also within R. Both are elds. Moreover Q and R
both have an order relation on them. I.e. a lt b has a well dened meaning,
which is in fact nicely related to the algebraic operations. E.g. a lt b implies
ac lt bc, etc. The big dierence between Qand R is that Qis full of holes.
For example the sequence of rational numbers , /, /, /,
... would like to converge to
/
. But
/
isnt there in Q. There is a
hole there. The next few pages are devoted to making this anthropomorphic
discussion precise.The key concept is that of a Cauchy sequence. Theorem
on page says that there are no holes in R.
Denition. A set S R real numbers is bounded if there is a real number
M such that x M for all x in S.
S is bounded above if there is a real number M such that x M for all x
in S.
S is bounded below if for some M in R, x M for all x in S.
Denition. If M is a number such that x M for all x in S then M is called
an upper bound for S.
Denition. If L is an upper bound for S such that L M for all upper
bounds M of S then L is called the least upper bound for S. We write
L lubS.
In order to prove anytning about the real numbers one needs , of course,
a denition of them. But no axiomatic treatment of the real number system
will be given here. See Rudin, Principles of Analysis, Chapter , if you
really want one. Instead I am just going to list the key property order
completeness which any denition of the real numbers must include, either
in the following form or some equivalent form.
R is order complete that is, if S is a nonempty set of real numbers which
is bounded above then S has a least upper bound in R.
Remark. The set Q of rational numbers is not order complete. For example
if S x in Q x
lt then S does not have a least upper bound in Q.
But S does have a least upper bound in R. What is it
Denition. A sequence is a function from the set Z , , , . . . to R.
Notation si s
i
.
Denition. If s
n
is a sequence then t
k
is a subsequence if there is a
sequence integers n
k
such that n
lt n
lt n
. . . and t
k
s
n
k
.
Denition. A sequence s
n
is said to converge to a number L in R if for
each gt there is an integer N such that
s
n
L lt whenever n N.
Notation. L lim
n
s
n
or s
n
L as n .
Proposition . If s
n
is an increasing sequence of real numbers bounded
above then lim
n
s
n
exists and equals lubs
n
n , , . . ..
Proof Let L lubs
n
n
. Then s
n
L for all n. Let gt . Then
L is not an upper bound for s
n
n
. Thus there exists N such that
s
N
gt L . Since s
n
s
N
for all n gt N we have s
n
gt L for all n gt N.
Hence s
n
L gt . But s
n
L . Hence s
n
L lt if n N.
Remarks. If S is bounded below, then S is bounded above. By S we
mean x x S.
Greatest lower bound is dened analogously to least upper bound.
Notation. glb S greatest lower bound of S.
Notation. If S is a nonempty set of real numbers which is not bounded
above we write
lub S .
If S is nonempty and not bounded below we write glb S .
Exercise. Write out, but dont hand in Prove glbS lubS for
any nonempty set S.
Denition. For any sequence s
n
n
we write lim
n
s
n
if for each real
number M there is an integer N such that
s
n
M for all n N.
Similar denition for lim
n
s
n
.
Now let s
n
n
be an arbitrary sequence of real numbers. Set
a
k
lubs
n
n k.
Since s
n
nks
n
n k we have a
k
a
k
. Thus the sequence
a
k
k
is a decreasing sequence.
Using the preceding proposition and denitions, we see that either
a For all k, a
k
or
b For all k, a
k
is nite i.e., real and the set a
k
k
is bounded below
or
c For all k, a
k
is nite anda
k
k , , . . . is not bounded below.
In case b lim
k
a
k
exists and is real and we write
lim
n
sup s
n
lim
k
a
k
.
In case a we write
lim
n
sup s
n
.
In case c we write
lim
n
sup s
n
.
Remarks. sup stands for supremum and means the same thing as lub. inf
stands for innum and means the same as glb.
The preceding denition of limsup s
n
can be written succinctly in all
cases as
lim
n
sup s
n
lim
k
sups
n
n k.
Similarly we dene
lim
n
inf s
n
lim
k
infs
n
n k.
Proposition . If lim
n
s
n
L exists and is real i.e., nite then
limsup s
n
L liminf s
n
.
Conversely if limsup s
n
liminf s
n
and both are nite then lims
n
exists and
is equal to their common value.
Proof Assume lims
n
exists and denote it by L. Let gt . N depending
on such that s
n
n N L , L . Thus for all k N, a
k
sups
n
n k L . Hence lim
k
a
k
L . Since the inequality holds
for all gt and the left side is independent of we have lim
k
a
k
L. Thus
lim
n
sup s
n
L. Similarly lim
n
inf s
n
L. But note that
infs
n
n k sups
n
n k.
Hence one always has for any sequence
lim
n
inf s
n
lim
n
sup s
n
.
But we saw that
lim
n
sup s
n
L lim
n
inf s
n
.
Hence we have equality. Q.E.D.
Proof of converse. Assume
lim
n
sup s
n
lim
n
inf s
n
L
L is assumed nite, i.e., real.
Let
a
k
infs
n
nk
and
b
k
sups
n
n k.
Then lima
k
limb
k
L. Let gt . There is an integer N
depending
on , as usual such that a
k
L lt whenever k N
. Similarly N
b
k
L lt whenever k N
. Let N maxN
,N
. Then a
k
L lt
and b
k
L lt if k N. Thus L lt a
k
b
k
lt L if k N. In
particular L lt a
N
b
N
lt L .
Thus
L lt a
N
s
k
b
N
lt L k N.
So
s
k
L lt k N.
Q.E.D.
Denition. A sequence s
n
n
is CAUCHY if for each gt N
s
n
s
m
lt n, m N.
propR Proposition . Any convergent sequence s
n
n
of real numbers is Cauchy.
Proof Given gt N
s
n
L lt / n N.
Then
s
n
s
m
s
n
LLs
m
.
s
n
LLs
m
.
lt / / if n, m N. .
Q.E.D.
Theorem . Every Cauchy sequence of real numbers converges to a real
number.
Remarks. The property stated in Theorem is usually referred to as the
completeness of the set R of real numbers. The set of rational numbers is
not complete. I.e., there is a Cauchy sequence s
n
in Q which does not have
a limit in Q. E.g. take s
,s
., s
., s
., . . ., so that
s
n
, which is not in Q.
Proof of Theorem Let s
n
n
be a Cauchy sequence.
Step . The set s
n
n , , . . . is bounded because if we choose
in the denition of Cauchy sequence then we know that there is an integer
Ns
n
s
N
lt if n N. Hence s
n
s
n
s
N
s
N
s
n
s
N
s
N
s
N
if n N. Thus if we put K maxs
,s
,...,s
N
,s
N
then s
n
K for all n.
Step . We now know that lim
n
sup s
n
is a real number. Let L
lim
n
sup s
n
.
Claim lims
n
L. For let gt . Choose N such that s
n
s
m
lt
n, m N. Then
s
N
lt s
n
lt s
N
n N.
Thus if a
k
sups
n
n k then
s
N
a
k
s
N
k N.
But by denition L lim
k
a
k
. Thus taking the limit as k we get
s
N
Ls
N
. So s
N
L . Hence if n N then
s
n
Ls
n
s
N
s
N
L n N.
Q.E.D.
. Set Theory Notation
Notation. If A and B are subsets of a set S then the intersection of A and
B, A B, is dened by
A B x S x A and x B.
The union of A and B is dened by
A B x S x A or x B or both.
The complement of A is
A
c
x S x / A.
The dierence is
A B x A x / B.
If f S T is a function and B T then f
B is dened by
f
B x S fx B.
. How to communicate mathematics
Before starting the following homework here are some useful tips on how to
write for a reader other than yourself. The three most important rules are
Rule . Remove ambiguity.
Rule . Remove ambiguity.
Rule . Remove ambiguity.
Here are some examples of ambiguity.
Example . Consider the statement
..C
Is this statement true Well, it depends on what you mean by the symbol
.
Meaning . implies that. If this is what you mean by the symbol
then the statement
C
. says
implies that . . C
In this case statement
C
. is TRUE. Its equivalent to the standard if then
statement If then . All you have to do is add to both sides
of the equation to deduce correctly that .
Meaning . This implies that. If this is what you mean by the symbol
then the statement
C
.says, when translated into english,
. This implies that . . C
The statement
C
. now consists of two sentences, one of which is false. So
statement
C
. is false and therefore statement
C
. is FALSE.
So what to do In principle you could dene a symbol to mean anything
you want but only one meaning. But the symbol is quite universally
used to mean implies that. Even the TeX command for is back
slashimplies, not backslashThis implies. It would be easiest on readers to
use the symbol always to mean implies or implies that which has
the same logical content. Moreover there is already a standard symbol for
This imples that. Its the symbol .
This symbol can be translated into English in a number of logically equiv
alent ways. Here are some of them.
can be translated as
This implies that, Therefore, Hence, So, Consequently.
When writing for a journal you have to use the words, not the symbols.
Moreover good prose form is better if you dont use the same one of these ve
phrases at the beginning of several sentences in a row. Technically you could
be correct in starting ve sentences in a row with the word Therefore. But
some readers might question the quality of the highschool you went to.
Often linked with the evils of the misuse of the symbol is the fail
ure to punctuate sentences. The evolution of grammar over the last
years has been aimed at precision of communication. Its still far from per
fect in normal social, legal and political communication. We have to do
better. Keep in mind, for example, that the statement x . really is
an english sentence. It has a subject, x, a verb, equals, and an object, ,
AND A PERIOD at the end of the sentence. Yes, even a period can help
clarity in communication of mathematics. COMMUNICATION IN MATHE
MATICS CONSISTS OF A BUNCH OF GRAMMATICALLY COMPLETE
SENTENCES.
If your TA cant gure out what you mean, even before he tries to gure
out if youre right, expect trouble from him and me.
A nal word. Dont think that context is a reasonable basis for a reader
to decide how to interpret an ambiguous statement. More often than not this
kind of thinking just means that you expect the reader to interpret correctly
what you wrote because he/she already knows what the argument should be.
All of this was well understood quite some time ago. Here is some ancient
wisdom.
A good scientist knows how to say exactly what he means.
A good politician knows how to say exactly the opposite of what he
means.
A good philosopher knows how to say things with no exact meaning.
Lezar el Gralidin,
. Cute Homework Problems
. Find the lim sup and lim inf for the following sequences
a,,,,,,,,,...
b sinn/
n
c /n cos n
n
d /n
n
n
. If the lim sup of the sequence s
n
n
is equal to M, prove that the lim
sup of any subsequence is M.
. If s
n
n
is a bounded sequence of real numbers and lim
n
inf s
n
m,
prove that there is a subsequence of s
n
n
which converges to m. Also
prove that no subsequence of s
m
m
can converge to a limit less than m.
. Write the set of all rational numbers in , as r
,r
, . . .. Calculate
lim
n
sup r
n
and lim
n
inf r
n
.
. If s
n
, prove that lim
n
sup s
n
lim
n
inf s
n
.
. Prove that A B C A B A C.
. If f X Y is a function, and A and B are subsets of Y , prove
af
ABf
Af
B
bf
ABf
Af
B
cf
A
c
f
A
c
. Show by example that fA B fA fB is not always true.
Show by example that fA
c
fA
c
is not always true.
. Find at least seven errors in mathematical communication in the following
six statements and explain whats wrong in each case.
a. Let fx x for all real x.
b. gx x
c. The function g, which is increasing, is zero at x
d. The function g which is increasing is greater than zero when x gt .
e. The function which is increasing is less than zero when x lt
f. fx gx. x
g. Rewrite item f. so that it is grammatically AND mathematically
correct AND unambiguous.
..
Tempered Distributions . . . . . . . . . . . . . . . . . . . . . . Problems on the Fourier Transform . . . . .
............
Applications of the Fourier Transform . The Heat Equation by Fourier transform. . . . . . . . . . .
. . Poisssons equation by Fourier transform . . . . . . . . . . . . Advanced, Retarded and
Feynman Propagators for the forced harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . The
wave equation by Fourier transform . . . . . . . . . . . .. Problems . . . . . . . . . . . . . . . . . . . . . . . .
Laplace Transform . The Inversion Formula . . . . . . . . . . . . . . . . . . . . . . . Application of
Uniqueness . . . . . . . . . . . . . . . . . . . . . Problems on the Laplace Transform . . . . . . . . . . . .
Asymptotic Expansions . Introduction . . . . . . . . . . . . . . . . . The Method of Integration by
Parts. . . . Laplaces Method. . . . . . . . . . . . . . .. Watsons Lemma . . . . . . . . . .. Stirlings
formula . . . . . . . . . . The Method of Stationary Phase . . . . . The Method of Steepest Descent
. . . . . .. Constancy of phase and steepness . Problems on asymptotic expansions Old
Prelims . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . of descent . . . . . . . . . . . . . . . . .
...........................................
Appendix Review of Real numbers, Sequences, Limits, Lim Sup, Set notation . Set Theory
Notation . . . . . . . . . . . . . . . . . . . . . . . . How to communicate mathematics . . . . . . . . . . . . . . .
. Cute Homework Problems . . . . . . . . . . . . . . . . . . . .
.
Hilbert space and orthonormal bases
Norms, inner products and Schwarz inequality
Denition. A norm on a real or complex vector space, V is a real valued function, on V such
that . a. x for all x in V . b. x only if x O. . x x for all scalars and all x in V . . x y x y . Note that
the converse of .b follows from . because O O O . Examples . V R, x x. . V C, x x. . V Rn with
x
n
. V C with x x x , . . . , xn . . V C, . These are the continuous, complex valued functions on , .
Here are two norms on V .
n xj when x n xj when
x , . . . , xn .
ff
f tdt
H.
..
H. H.
supf t t ,
Exercise . Prove that the expression in . is a norm. Here is an innite family of other norms on
Rn . Let p lt . Dene
n
x
p
j
xj p
/p
.
H.
FACT These are all norms. And here is yet one more. Dene x
max xj
j,...,n
.
H.
j . . The unit ball of a norm on a vector space V is B x V x . Examples . and unit ball of is a
diamond shape thing contained in this disk. y n xj yj . . Less obvious. Its especially
illuminating to compare norms by H. such that ak lt . It would be good for you to verify
yourself that this really is an inner product. b . The unit ball of is a disk contained in this
square. z for all scalars a. . the proof that x p is a norm is not that easy. g f tgtdt. but
reasonable. Dene k x. y k ak b k when y b . ax by. b and all vectors x. V C. Denition An inner
product on a real or complex vector space V is a function on V V to the scalars such that . .
Sadly. x. V Cn with x. . y y.Exercise . A picture of the unit ball of a norm gives some
geometric insight into the nature of the norm. Well prove the p case later. . z ax. x. . and .
Prove that x and x are norms. . So dont even try. on R . which is the standard notation for the
set of sequences x a . comparing their unit balls. is the fact that the other unit balls lie inside
the square and contain the diamond. x. . x for all x V and x. . Draw pictures. with f. The series
entering into this denition converges absolutely because of the identity ab a b /. a . . y n xj yj .
. The boundary of each unit ball is the curve x and clearly goes through the points . x only if x
. V l . It would be very good for you to verify that this really denes an inner product on l . Lets
consider the family of norms dened in . z by. and H. . V Rn with x. Its not hard to compute
that the unit ball of is a square with sides parallel to the coordinate axes. y.
Proof Using the previous corollary its easy to verify the three properties in the denition of a
norm. Now use . y x y . we nd xy H x y. Proof Using . y. Corollary . x. In any inner product
space x. x Rex. x/ y. y x. is a norm. QED. b ac . H Proof If x or y is zero then both sides are .
So. Dene H x. y because x. If the inner product space is real just take . y y. y x x y y x y .. x ty
x t y tx. by the quadratic formula. The Schwarz inequality. H by . y is real. QED . x y x y Rex.
So we can assume x and y . y/ . That is . Thusx. We are going to show next that in an inner
product space one can always produce a norm from the inner product by means of the
denition x Corollary .Theorem . y x y . y x. x. In either case let pt x ty for all t R. the
discriminant. QED. . Then pt x ty. In case the inner product space is complex choose C such
that and x. Then xy x y .
QED Denition In a vector space with a given norm we dene n lim xn x to mean n lim xn x . .
Then for any x V x j x.. for any integer n. Now take the limit as n . . n n x x x x So n k k ak ek .
Bessels inequality Let e . . be an orthonormal set in a real or complex inner product space V .
. en is orthonormal and . . ek k n ak ak k ak k ak k ak x for all n. . en is an orthonormal basis
of a nite dimensional inner product space V if a the set e .k n n aj ak ej . . . which we allow to
be nite or innite such that ej . x x. . e . . ek ak k n j. . . . ek if j k and if j k Lemma . Were all
familiar with the concept of an orthonormal basis in nite dimensions e . ej Proof Let ak x. .
Bessels inequality and orthonormal bases Denition An orthonormal sequence in an inner
product space is a set e . ek . e . Then. x n ak ek k ak ek .
m . e . . . a limit of nite sums limn x n aj ej . . e . We see from H Bessels inequality that . n.
Question Can you immagine how intolerably complicated calculus would be if we had to
worry about these holes in Q E. Now the sequence of partial sums.g. . be given by the
requirement a as before and b every vector x V is a sum x j aj ej .b every vector x V is a sum
x n aj ej . . . H There is an unfortunate and j downright annoying aspect to the equation . . an
. as usual. . Of course we caused this trouble by making holes in l . But e is not rational. . and
a sequence of real or complex numbers aj such that ak lt . . We are going to eliminate holes
Denition. H And this is right.. A sequence x . But x is not in F . . in a normed vector space is a
Cauchy sequence if lim xn xm . k Suppose that we are given the orthonormal sequence e . .
So there is no vector in F whose coodinates are the nice sequence j . is orthonormal. . . Does
k the series ak ek converge to some vector in V If not can we really say k then that we have
coordinatized V if we dont even know which coordinate sequences a . Of course writing an
innite sum means. k Let F be the subspace of l consisting of nitely nonzero sequences. Let aj
j . j If V is innite dimensional then we should expect that the concept of orthonormal basis
should. . . The sequence e . . Let e . . a . xn j aj ej converges to the vector x aj ej in l you
verify this. . . . . . however. Thus x F if x a . . . . So there is a hole in Q at e. . . similarly.
implies that ak lt . These circumstances are analogous to the holes in the set Q of rational
numbers. actually correspond to vectors in V by the formula H . etc. Here is an example of
how easily convergence of the series aj ej k can fail. . x . . . e . . even when ak lt . . . F is
clearly a vector space and the inner product on l restricts to an inner product on F . . f x gives
the maximum of F on Q provided x is rational The same nuisance would arise if we allowed
holes when dealing with ON bases. . . . . n Then aj lt . For example the sequence sn n /k is a
sequence of rational numbers k whose limit is e.
Just as the real numbers ll in the holes in the rational numbers so also one may view L .
Notation. Denition. Proof Same as proof of Proposition . . This is such an important example
that it gets its own notation. the space C. Here is the denition that makes this notion precise.
for any gt there is an integer N such that xn xm lt whenever n and m N . In Example . . are
Hilbert spaces. But Example . Denition. Examples. In order to get a complete space one
must throw in with the continuous functions all the functions whose square is integrable. . A
Hilbert space is an inner product space which is complete H in the associated norm . in the
Appendix. A Banach space is a normed vector space which is complete. See if you can
prove both of these statements. The proof that Example . easy to verify and the associated
norm is f f t dt Of course if we wish to consider square integrable functions on some other
set. L .. is complete in the norm f but not in the norm f .. A normed vector space is complete if
every Cauchy sequence in V has a limit in V . are complete. Just replace by . C such that f t
dt lt For these functions we dene f.. Examples. Remark.That is. g f tgtdt This is an inner
product on L . is the set of functions f . is not complete. Youre supposed to know this from rst
year calculus. . It is extremely unfortunate that the Example . is not complete. as lling in the
holes in C. If xn converges to a vector x in any normed space then the propR sequence is a
Cauchy sequence. The spaces V in Examples . The Examples . such as R we would denote
it by L R. is complete will be given only on popular demand. to .
Similarly. A subset A of a Hilbert space H is called dense if for any x H and any gt there is a
vector y in A such that x y lt . . x. n Proof Let sn k ck ek . We must show that the sequence
converges to a vector in H. In words you can get arbitrarily close to any vector in H with
vectors in A. y is continuous for each xed element y. For example the rational number . In
any inner product space the function x x. the rational numbers are dense in the real numbers.
Since limits are unique. Of course we will write x ck ek . if you cut o the decimal expansion of
a real number at the twentieth digit after the decimal point then you have a rational number
which is very close to the given real number. . As we know. k keeping in mind that this
means that x is the limit of the nite sums. k Then the series ck ek k converges to some
unique vector x in H. . Lemma . e . Now. Suppose that n gt m. to which the sequence
converges we will show instead that the sequence is a Cauchy sequence. is dense in L . Let
ck be any sequence of scalars such that ck lt . Well see this later in the context of Fourier
transforms. . is any ON sequence. we know that there exists a vector x in H such that limn sn
x. . Here is the rst important consequence of completeness. Suppose that H is a Hilbert
space and that e . Then sn sm n n km ck ek km ck as m.Denition. You may use this fact
whenever you nd it convenient. is pretty close to . because H is assumed to be complete.
Since we dont know in advance that there is a limit. GOOD NEWS C. x is unique. n because
the series k ck converges. You prove this on the way to your next class. Lemma . lem. .
QED. Its often best to prove some formula for an easy to handle dense set rst. and then
show that it automatically extends to the whole Hilbert space.
x x. y is also a continuous function of y for each xed x. One can either repeat the preceding
proof or just use y. y xn x. . y. y x.Proof. If xn x then xn . QED Of course x. y xn x y .
Assume that a. Let x H. d. Lemma . . holds. ej aj limn n ak ek . y ak ek exists. . assume that
c. QED. b. holds. c. ek for all k. For every vector x in H we have x k ak Proof We will show
that a. Contradiction. we see that x k ak . Then n n m x. ek are all zero. e . . is a maximal ON
set. Next. So x . . ek and bk y. holds. If e . holds. By Bessel we have ak lt . y k ak bk where
ak x. . If x y then let h x y/ x y . e . ak x. But xy. ej k aj aj for all j. But from d. holds.thmH. Put
y x to derive that d. One can now adjoin h to the original set and obtain a larger ON set. be
an orthonormal sequence in a real or complex Hilbert space H. . Theorem . y lim lim n m j j
min n. . e . This proves that b. b. a. So we must have x y . k bk ek lim lim n m aj b j j aj b j .
So c. a. So the coordinates. For every pair of vectors x and y in H we have x. . Finally. ek d. .
it is not properly contained in any other ON set. is not a maximal ON set then there exists a
vector x such that x. holds. Assume now that b.. For every vector x H we have x k ak ek
where ak x. By lem.m aj ej . ek c. Let e . Then the following are equivalent. . y lim n j aj ej .
holds. That is. e . assume that d.
Of course we could have used any of the other conditions in Theorem . . An ON sequence e
. thmH.Now were ready for the denition of ON basis. holds. for the denition Because surveys
of your predecessors show that its the most popular. . So why did I use condition b. for the
denition of ON basis because theyre equivalent. in a Hilbert space H is an ON thmH.
Denition. e . in Theorem . . basis of H if condition b. .
j . Let f H and assume that f a a where aj f. Assume that A xf xdx and xf xdx B What can you
say about the value of xf xdx Give reasons.N. . Let f . Suppose that u . uj for j .. Show that aj .
f be an orthonormal set in L . Let and u x u x / . . Show that f a u a u . . . Problems on Hilbert
space . and bj bj f g . Let aj j uj xgxdx. . . f . . What can you say about the size of xf xdx . u
are O. and uj xf xdx. . Cite any theorem you use. f x dx and f xdx . vectors in an inner product
space H. x . Suppose that f and g are in L . Suppose that f . x x. R satises .
. sequence in L R. .N. basis of a Hilbert space H and that g . set in a Hilbert space H.N. . n .
. n . . Suppose that f . g .N.N.N. . . . be an O. set in a Hilbert space H. .N. then f . . gj for all j. .
. basis. . The Hermite polynomials are the sequence of polynomials Hn x uniquely
determined by the properties a Hn x is a real polynomial of degree n. is also an O. . c . g gj fj
lt . .N. thmH. Prove that it is an O. Fact that you may use If f is in L R and n . . Hint Use
Theorem . f . basis if and only if the nite linear combinations n aj fj n nite but arbitrary j are
dense in H. . sequence in H. with positive leading coecient. . . . b The functions un x Hn xex
form an O.N. by supposing that there exists h such that h. f . . . is an O. Prove that g . f . is an
O. Let f . Let cn Evaluate n / f xun xdx for ex un xdx. . basis if and only if Parsevals equality g
holds for a dense set of g in H. . . be an O. Prove that it is an O. Suppose further that j j fj . ..
Let f . . . g .
Here f is a given function on this interval. As you know.. O STEP . For example one might
specify u and O u to pick out a unique solution to . This is the initial value problem both
pieces of information are specied at the same endpoint. can be represented in the form O O
O ux Gx. O on the interval x . .. the solution to a second order ordinary dierential equation
requires specication of some additional information in order for the ODE to pick out a solution
uniquely. O for some function Gx. . We are going to be primarily concerned with the
boundary value problem the two pieces of information will be specied at opposite endpoints.
Our goal is to show that the solution to . given by the boundary data u . O value problem .
Lets focus O on the boundary value problem for the dierential equation .O G is called the
Green function for the boundary y. namely u u . O Since this equation has constant coecients
its easy to write down the general solution. u . Here is the method for constructing the Green
function in this simple example. . Consider rst the homogeneous version of . and . . These
are clearly given by choosing B ... yf ydy . It is ux A sinh x B cosh x We will need rst to nd the
solutions which satises the LEFT HAND boundary condition u . Second order linear ordinary
dierential equations Introduction u x ux f x Consider the dierential equation .
STEP . y sinh x sinhy .We will see later that our procedure doesnt care what A is. NOW let
ux Gx. in the form ux C sinhx D coshx A solution that satises the right hand boundary
condition is V x sinhx and this is the only one. namely on the diagonal x y. sinh for x y for y x
. So the function Gx. for each x. . Dene Gx. To this end we may write the general solution to .
O for x y for y x . . O . but using the RIGHT HAND O boundary condition. . Notice that on the
overlapping portion of these denitions. y is a continuous function of x because the limits from
the left and right or up to down agree on the diagonal. So U x sinh x STEP . Gx. sinh sinh y
sinhx . W y U yV x . y U xV y . y is well dened even on the diagonal of the square . the two
halves of these formulas agree. yf ydy . y is a continuous function of y and for each y Gx. up
to a scalar multiple. Dene U x U x W x det V x V x Inserting our particular U and V we can
compute that W x sinh x STEP . O In our case this expression reduces to Gx. Do the same
thing all over again. W y . Moreover. Denote the solution that weve now got by U . So lets
just choose A .
y on each half of the square because both U x and V x satises this equation. which is not
what we want. u x Gx. The naive reader not you might think that we can just put d /dx under
the integral sign. O which is just what we would have gotten if we had simply put d/dx under
O the integral in . There is never a crossing over the diagonal. But if you do this you will nd d
/dx Gx. This is why we chose U and VOas we did. . x f x Gx. y.for ANY continuous function f .
y is continuous. O O I claim that u is a solution to the boundary value O problem . Gx x.
There is trouble with Gx... a couple of times. Indeed the denition of G shows that G. So x u x
Gx x. But now recall that you admitted a while back that the function y Gx. The easy part of
this O seeing that the function u in . x f x x . yf ydy. y as x passes through y. yf ydy x Gx x.. .
To deal with this we just put the trouble spot into the limits of integration thus x ux Gx. while x
indicates that y is approaching x from above. Gx x. This would then give u u . as it should.
satises the two is boundary conditions . But the next time we wont be so lucky. yf ydy x Gx x.
y for all y. we O must dierentiate the right side of . actually has a jump as x passes y. As you
know there is a contribution to the derivative of u from the x that occurs in the limits of
integration. So Gxx x. yf ydy. Were going to . Although Gx. So here is what we must do. In
each integral the integrand is a nice function of x. as we will see in a moment. y doesnt even
exist at x y.. provided y is where the limits of integration allow it to go. Here is the
computation. y for all y and G. . yf ydy In the rst line the expression x indicates that we are
letting y approach x from below. Therefore the rst line is zero. y is continuous in x for xed y
its rst derivative. y Gx. To prove that u satises . So u is zero at the two boundaries. yf ydy x
Gx.
yf ydy f x . But well see that the rst O line is not zero this time. y has a jump as y passes by
x. as given by . So Gx. x O U xV x U xV x W x .repeat this procedure for the second
derivative. x f x x Gxx x. x U xV x/W x. satises is the both . So Gx x. x U xV x/W x. O Ha So
the rst line in . y is the Green function for the boundary value and O problem .. f O which O
exactlyO equation . . But before leaving this successful example lets rewrite informally the
preceeding computations. We nd from . u x Gx x. Similarly Gx x... yf ydy x Gxx x. O This time
the rst line does not give zero because Gx x. We actually showed that d /dx If we innocently
rewrite this as Gx. Our aim in this section is to carry out this procedure for a general second
order ordinary dierential equation. which is much easier to deal with than the boundary value
problem and whose basic theory we will review in the next section.. y Gx. So Gx x. All of
these are issues pertaining to the initial value problem. yf ydy f x .. yf ydy . Putting this
together we now nd that u x Ox ux. Thus ux. is just f x. y U yV x/W y if y lt x. But we have
already observed that Gxx x. x f x Gx x. and this is where y is ranging in the two integrals. To
this end we must be able to construct the functions U and V and understand the function W .
y at least when y x. O d /dx Gx.O .. x Gx x. How much is the jump Just look at the denition of
G and the denition of W Gx x.
Consider the equation u x ux f x x . b and let c and c be two real numbers. . The formula .
Since cos and cos the desired function U is U x sinx and the desired function V is O ALSO V
x sinx. . Then there exists a unique function u C a. O O The general solution to the
corresponding homogeneous equation u u is ux A sinx B cosx. . which we have proved.then
we could claim that we have shown that O d /dx Gx. We wish to study several problems
associated with the dierential operator Lux p xu x p xu x p xux. b and ua c and u a c . We will
assume throughout that the coecients pj are continuous on a. . We consider a closed
bounded interval a. Consequently W x for all x. Thats life. It will be important for us to
understand exactly why and when such failure occurs. can break down for a general second
order dierential equation. In truth. The initial value problem.. b satisfying Lux f x for all x a.
thmO Theorem . So if you are fond of functions and who isnt then all is well. b and that p x gt
on the entire closed interval. y x y. Here is a simple example of BREAK DOWN Lets modify
the equation . . O O And what does . this neat approach to Green functions. is now
meaningless. . We need to see how this can happen before proceeding to a general second
order ODE. Existence and uniqueness for the initial value problem Let f be a continuous
function on a. . b. as embodied in the funO damental formula . mean It means exactly that .
holds. . slightly. So is there a Green function Answer No.
v is nowhere zero. Let gx ux vx. Therefore gx for all x a. are linearly dependent for every x a.
v x in the plane are linearly dependent vectors in R . v is identically zero even if u and v are
not solutions to Lu . This proves lemO Lemma . b then Lemma . if u and v are lemO linearly
dependent on a. If. Recall that f and g are said to be linearly dependent on a. . such that ux
vx and u x v x . gx for all x a. The Wronskian of f and g is the function W f. b then W f. Let u
and v be two solutions of Lu . But gx and g x . Wronskian Let f and g be two functions in C a.
Then u and v are linearly independent on a. vx then the points ux . b if there are constants
and such that f x gx In this case we may dierentiate this equation and nd that f x g x when f
and g are dierentiable. b. Kreyszig Advanced Engineering Mathematics defWronsk Denition .
This shows that the rows of the matrix in . Therefore there exist nonzero constants . Proof
Assume Lu Lv . Hence W f. Conversely. Hence u and v are linearly dependent on a. for
some point x one has W u. shows that W u. . The partial converse of this is a little bit more
subtle. See e. If f and g are in C a. b. g . b. Lemma . Then Lg . b by the uniqueness portion of
Theorem thmO . ..Proof Use Picards method. b and are linearly dependent on the interval a.
b.g. gx det f x f x gx g x . u x and vx . b if and only if W u.
. Thus for Dirichlet boundary conditions . Theorem . Proof Dene functions U . V . and F with
the aid of the basic existence thmO and uniqueness theorem.. so as to satisfy LU . for u C a.
b. . Then the inhomogenous system Lw f Ca. Let Lu p xu p xu p xu Au ua u a Bu ub u b . V b
F a F a . . LF f. are real numbers. p are in Ca. LV . . b Aw Bw has a solution for all f . if and
only if the homogeneous system Lu Au Bu has no nonzero solution. thmO The boundary
value problem and Green functions Theorem . p . U and V look like this Let w cU dV F. and p
gt on a.. . where p . . . . . b. . . b. Then clearly AU and BV . U a . . U a V b .
as is also va. . v a. cu v on a. But U is not identically zero on a. Then the function cu v has
zero initial data at a and satises Lcu v . we may choose c and d so that w satises . If the
system has no non zero solution then a the functions U and V constructed in the previous
proof are linearly independent and b their Wronskian W is nowhere zero. Hence one is a
multiple of the other. Bv . Similarly AV . Then there exists a function v such that Lv Av Bv .
Say cua. contradiction. Hence for any . b. u a va. . b. Green function. and that has a solution
u not identically zero. thmO Theorem . But Bu . Now assume that the system has no nonzero
solutions. So U satises . u a R is orthogonal to . Assume that the system has no non zero
solution. v a and neither is zero by uniqueness theorem. . Proof It suces to show U and V are
linearly independent. . But then BU cBV . .Then Lw f Aw dAV Bw cBU . . Then the system Lu
f Au Bu . . Since LU AU it then follows that BU . But ua. Conversely assume that has a
solution for all f . If they arent then there exists a constant c such that U cV . Corollary .
yf ydy x Gx x.has the unique solution b ux a Gx. We therefore nd x b u x a Gx x. yf ydy x Gx
x.. We need now the key identity that describes the jump of the rst derivative of G. .
Dierentiating rst with respect to the x that appears in the limits of the integrals the
fundamental theorm of calculus gives u x Gx. Proof. Thus we will write b ux a x Gx. . . Gx.
The derivative of each integral with respect to x will have two terms. x Gx x. x f x x b . yf ydy.
. . x f x Gx. U xV y/p yW y U yV x/p yW y axyb ayxb . yf ydy. where Gx. yf ydy b . p x . x . y is
continuous in y at y x. yf ydy . It is Gx x. x a Gx. . yf ydy. y and W UV V U . Thus the
contribution to u x from the limits cancel because Gx. . . a Gx x. yf ydy . As in the introduction
to this section we are going to have to break up the integral into two parts to take into
account the jump in the derivative of G. . corresponding to the two appearances of x.
. to nd x b . yf ydy x Gxx x. Hence multiplying by p x we nd p xu x f x x b . y o the diagonal. .
yf ydy .. Hence so . v u. O . just as in the simple example in the introduction. x . yyf ydy. and .
yf ydy x x Gxx x. The function u is called an eigenfunction with eigenvalue . the y function x
Gx. dierentiate . . f x because Lx Gx. y satises the boundary conditions in . to nd u x Gx x. . a
Gxx x. . b. p xGxx x. b if Lu. the reader may verify that. . x f x Gx x. Finally. yf ydy x Lx Gx. . .
Denition . yf ydy. by .. a p xGxx x. The dierential operator L is called symmetric on a. b f x/p x
a Gxx x. by p x and multiply . does u. Lv for all u and v in C a. . Now . Lux f x a Lx Gx. b
which are zero in a neighborhood of each endpoint.This may be derived from .. b such that
Lu u Au Bu . by p x and add them to . Eigenvalue. .. yf ydy . a. Just imitate the derivation of .
Denition . A complex number is called an eigenvalue of the dierential operator L with
boundary conditions given by A and B if there exists a function u in C a. Now multiply . x f x x
b . for each . yyf ydy.
Then the sequence u . Assume they are normalized b a un x dx . Lu Put u.Fact If L is
symmetric then L must have the form given in the next theorem wherein we change
notational and sign conventions. In fact one has lim n . forms an O. You can see from the
repeated use of such expressions in the previous sections that this is a useful notation. . y is
a function on the square a. b. b. b with px gt for all x a. Then Lu. v a b . u . thmO Theorem . .
v u. The corresponding eigenfunctions un may be chosen real. . taking p x px lt . Let d pxu x
qxux dx Au ua u a Bu ub u b. yuydy whenever Kx. b and Au Bu Av Bv . Eigenfunction
expansion of the Green function We are going to use the notation b Kux a Kx. . b a. . uxvxdx.
Moreover there exists a strictly increasing sequence n of real eigenvalues for the operator L
with boundary conditions Au . Give a little try to nd a proof of this fact by yourself. N. b. Lv if u
and v C a. basis of L a. Let p and q be continuous functions on a. . . The operator u Ku is
called an integral operator. Bu .
We may write this as K u n Therefore L K un un . because all the other terms are zero. be
the eigenvalues of L and u . . j j Now b j K x. f d. yun ydy a un x. . u . Then thmO Theorem . .
G x. n un n j b ujxuj yun ydy a . y uj xuj y. . . a Let . . .Let d du px qxux dx dx Au and Bu as
usual. the corresponding normalized real eigenfunction. . y j uj xuj y. j Informal Proof We
assume all series converge and all interchanges are legal. If is not an eigenvalue of L for the
boundary conditions A . . Let K x. . b L u f u G x. Lu . B then L exists and is given by a Green
function G x.
Thus. un n Therefore K x. QED thmO . .But L G Identity.. Therefore K un G un for all n
because L has trivial nullspace. This proves Theorem . So L K un un L G un . . un G x. y. y G
x. for each x K x.
Let Lu x u xu on . Apply Picards iteration method to the following initial value problem. . y
Ref Kreyszig. dx d u c Lu u.. a Show that L is symmetric on this interval. dx dx du b Lu u.
complex. y x. b Find the Green function for L under Dirichlet boundary conditions u u .
Determine which of the following operators are symmetric on the interval . Find y x. c For the
same boundary conditions nd the eigenfunctions and eigenvalues of L. When only one can
be found in this form then x ln x will give another one. d u du u. Problems on ordinary
dierential equations . dx .. Hint Equations of the form ax u bxu cu tend to have two linearly
independent solutions of the form u x . Determine which of the operators in Problem have
inverses for the boundary conditions u u and for these operators nd the Green functions.
Advanced Engineering Mathematics sec. y x.. y xy. in rd edition . Prove the Fact preceding
Theorem . . thmO . a Lu .
RIGHT V is really a dierent vector space from V itself. functions and all that. p L p p. denoted
V . mechanics and general relativity. The dual space to V is the set. . . This is a four
dimensional vector space because .. It would be best if you. Moreover if L and L are linear
functionals then so is the function aL bL for any scalars a and b. p L p pt sin tdt. verify that
each of these functions on P are linear functionals. p L p ptdt. The thing to take away from
these examples is that there is no resemblance between V and V you cannot really identify
any of these three linear functionals on P with elements of P itself. t . And we will need it later
to understand generalized functions. t. Denote by P the vector space consisiting of
polynomials of degree less or equal to . Check this against the denition. Denition . A function
L V scalars is called a linear functional if Lx y Lx Ly for all x and y in V and for all scalars and
. . We have constucted a new vector space from the given one. This being the case. Here
are some linear functionals on this space. of all linear functionals on V . Its a new vector
space constructed from the old one. personally. Check this against the denition now
Therefore V is itself a vector space. In other words a linear functional is a linear
transformation from V into R or C if V is a complex vector space. Let V be a real or complex
vector space. Generalized functions. For example the function which is identically is a linear
functional. Dual spaces The concept of a dual space arises naturally in dierential geometry. .
t constitutes a basis. Example . . you have to regard the following theorem as remarkable.
. Claim Then L n ck Lk .. if we wished. They are linearly independent... Then any vector x V
can be uniquely written n x j aj ej . The basis Lj described in the preceding proof is called the
dual basis to the basis e .. Uniqueness means that each aj is a function of x. It has the nice
property that Lj ek jk Philosophic considerations . . With the help of this map we could. Its
straightforward to check that each function Lj is a linear functional. . Dene Lj x aj . This is. an
isomorphism from V onto V .. But there is a catch A choice of basis has been made in
constructing this isomorphism. .. You can check this yourself by showing k that both sides of
this equation agree on each ej and therefore on all linear combinations of the ej . . .. If V is an
ndimensional vector space then so is V .. Terminology. Let e . If Jim goes into one room ...
Then M ek n cj Lj ek n cj jk ck . They span V . Proof... Since any vector x in V can be written
uniquely in the form . Dene ck Lek . as some people would say. . Proof Let L be any linear
functional...Theorem . QED. n. Ln of V . Hence the functionals Lj are linearly independent. .
en . j . Hence dimV n. We will show that they form a basis of V . en of V we see that we
automatically get a basis L . b onetoone and c onto V . we can now dene a vector Lx in V by
the formula n Lx j aj L j In this way we get a map x Lx from V onto V . So we have now
produced a basis of V consisting of n elements. You can check easily that this map is a
linear. Proof Suppose that M n cj Lj j . Thus they agree on all of V . So all the coej j cients ck
are zero. . identify V and V and even go so far as to say that V and V are the same space.en
be any basis of V . Having chosen the basis e .
Then for any linear functional L on V there is a unique vector y in V such that Lx x. We know
that there are many inner products on any nite dimensional vector space. Ly x x. by the rank
theorem. Therefore the map y Ly is onetoone. y for all x V. But given a particular inner
product on a real nite dimensional vector space V . Here is a consequence of this
identication that we live with every day. QED. But if dim V n then by Theorem . Then the map
is a onetoone linear map of V onto V . the preceding theorem provides a natural way to
identify V with V without making any ad hoc choices of basis. it is an isomorphism. So each
one will identify V with V in dierent ways.. That is. So y . y. . Some readers might recognize
the next theorem as raising and lowering indices. Hence. is a given inner product on V . The
map y Ly is clearly linear. When an isomorphism between two vector spaces depends on
someones choice of a basis we say that the isomorphism is not natural.e. it corresponds to a
dierent linear functional. Denote by Ly the linear functional determined by y in this way. no. y
.. But there is an important circumstance in which one really can justify identifying V and V .
dim V n also. Proof. y Ly . That is. Hence the range is all of V .e. If you should nevertheless
decide to think of these two vector spaces as the same i. Theorem . the range of this map
has the same dimension as the domain. You better check this. Then they will arrive at dierent
isomorphisms. Jim will say that the vector x V corresponds to a certain linear functional L and
Jane will say. .and chooses a basis e . M . Suppose that V is a real nite dimensional vector
space and . y for all x V. I. Moreover this map is onetoone because if Ly then in particular Ly
y .. It will be good practice in dealing with these structures... Moral . identify them then sooner
or later you will run into conceptual and even computational trouble. . en to construct this
isomorphism and Jane goes into another room and chooses a basis the chances are that
they will choose dierent bases.
That is.. f . If there is no natural way to identify V with V then this map to V is the only object
around that captures the notion of derivative that youre familiar with. . One often writes f x for
this linear functional. Thus f xv v f x f x. is a function from V into V . That is. In other words
the map v v f x is. For example if we choose any basis e .. v. a linear functional on V .
Therefore the derivative . . xn . x . en of V we may write x n xj ej and then f is just a function j
of n real variables. The chain rule then gives n v f x j vj f /xj x n where of course v j vj ej . . for
each x. This sum is clearly linear in v. . then we can identify the linear functional f x for each x
with an element of V . f xv v f x. But if V has a given inner product.. .. f x f x identied to an
element of V by Theorem .. Suppose that V is a nite dimensional real vector space and f is a
real valued function on V .Derivative versus gradient. For any point x in V and any vector v V
dene v f x df x tv t dt This is the derivative of f in the direction v. This is the gradient of f . So f
x is in V for each x V ...
V C. and ii f x y f x f y for all x and y in V . real valued continuous functions on . Problems on
Linear Functionals Denition. A linear functional on a real or complex vector space V is a
scalar valued function f on V such that i f x f x for all scalars and all x V . a f x x x b f x x c f x
x x d f x e f x f f x sin x g f x x x h f x x u where u is a xed vector and x u xj uj j . x . V R . a F
tdt for C. Explain why not if you think not. Which of the following expressions dene linear
functionals on the given vector space . b F / c F d F e F f F t sin tdt t dt g F . x .. . x x .
a . admitted that the function on P dened by Lp linear functional. is called nitely nonzero if N
xk for all k N . If a a . This is a real vector space of dimension three. . j a Show that the series
converges for each x in V . . f for all p P . That is. A sequence x x . such that f x fa x x V . a .
ptqtdt. But we also know that every linear uniquely by an element of the space P with the
help of Thus there is a unique polynomial f of degree at most Lp p. q You have already pt sin
tdt is a functional is given the inner product. . show that if f is a linear functional on V then
there exists a sequence. is an arbitrary sequence of real numbers let fa x aj x j for x in V. .
Right Dene an inner product on P by p. b Show that fa x is a linear function of x. . Denote by
P the space of real valued polynomials of degree less or equal to . . . Let V be the vector
space of all nitely nonzero real sequences. . d Show that the sequence a. a. c Show that
every linear functional on V has the form . in part c is unique. . such that Find f . . x . h F t tdt
t/ tdt t/tdt i F j F k F sin tdt Explain why not if you think not.
The notion of a point charge is therefore at best an idealization. Of course if you know in
advance that the electric eld is continuous then you can get better and better approximations
to the value of Ex by using a sequence of smaller and smaller corks. For example if you
really want to measure the eld at a point x by the preceding method you would have to place
a point charge at x. In the classical theory of electomagnetic elds the electric eld tends to be
continuous and therefore it makes sense to talk about its value at a point. In accordance with
statistical mechanics the temperature at a point of a system has a really questionable
meaning because temperature is a measure of average kinetic energy of a large bunch of
molecules. If the temerature varies from point to point and is a continuous function of position
then you can. Recall however that a typical small bulb will contain on the order of molecules..
A pairing such as R Exxd x. measure the temperature at a point by using a sequence of
smaller and smaller bulbs. The need for talking only about averages shows up also in
measurement of temperature. Generalized functions If you want to measure the electric eld
near some point in space you could put a small charged piece of cork there and measure the
force exerted by the eld on the cork. Of course the force on the cork is a constant multiple of
an average of the forces at each point of the cork. In practice and even in some theories it is
only these averages that you can measure. The integral is linear in and denes a linear
functional on the . In this way you have converted the problem of making an electrical
measurement to that of making a mechanical measurement. such as charge. recognizes that
there are tremendous uctuations in the eld at very small scales and only the averaged eld
has a meaning. mass. between an extensive quantity such as charge extensive means that
in a larger volume you have more of the stu. If Ex is the eld strength at x and is the charge
distribution on the cork then the net force on the cork is R Exxd x. in appropriate units. A
thermometer is clearly measuring the average temperature over the volume of the little bulb
at the bottom. So at an atomic size of scale temperature at a point is meaningless. and an
intensive quantity. The quantum theory of eletromagnetic elds. etc. in principle. such as the
electric eld in a larger volume you dont have more eld occurs often in physics. however. But
classical theory shows that the total electric energy of the eld produced by a point charge is
innite.
That is. The only question then is whats happening at x Sadly. QED. Third. you can see by
induction that for x gt the nth derivative dn e/x dxn pn /xe/x for some polynomial pn . Carry
out the case n yourself. But its not so hard. Test functions Lemma . one must go back to the
denition of derivative to answer this. We are going to develop and use this notion of duality
between very smooth functions test functions and very rough functions e. If we know that the
rst n derivatives exist at x and are zero there then the rst and second comments above show
that the n st right hand derivative is lim h Of course the left hand derivative is clearly zero.
and is zero to the left of . Proof First. We say that the integral is a bilinear pairing between
the extensive quantity and the intensive quantity E. Lemma Let x f xf x pn /he/h h . limt ptet
for any polynomial. Convince yourself with the cases n . Let f x e/x for x gt for x Then f is
innitely dierentiable on the entire real line. The integral is also linear in E for xed . Usually the
elements in these dual vector spaces have a physical interpretation. you can see easily that
all of the derivatives of f exist at any point other than x . Julian Schwinger. It has proven to be
a great simplifying machinery for understanding partial dierential equations as well as Fourier
transforms. extensive for one and intensive for the other. Such bilinear pairings between two
vector spaces is common. . delta functions. in his book Sources and Fields emphasizes the
duality between sources and elds. . So the two sided derivative exists and is zero. This is the
basis for an induction proof. recall that et grows faster than any polynomial as t .vector space
of test charges. We are going to apply it to both.g. Second. The rst step is to understand
really smooth functions.
Then Lf is a linear functional on D. From this function its easy to construct lots more.
Remember the concept of density form the chapter on Hilbert space One of the homework
problems sketches how to show that any continuous function with compact support is a limit
of functions in Cc . Terminology An element in D is called a distribution or a generalized
function according to taste. is clearly innitely dierentiable by the repeated application of the
product rule for derivatives. But this example is substantially dierent from the rst one. .
Suppose that f is a continuous function on R. The dual space of D is denoted. Proof. Then is
an innitely dierentiable function on R with support contained in the interval . . or even
discontinuous function f . as usual. Notation The space Cc arises so often that it customarily
is given a special notation D Cc R. There is no continuous function f .g. Notice that even if f
increases near e. Clear So L D . such that L f xxdx. f x ex the integral makes sense because
its really an integral over some and in fact any interval that supports . By scaling and
translating the argument of and taking powers one clearly gets lots of such functions. D
.where f is the function constructed in Lemma . draw a picture. Let L . One says that these
functions have compact support. In FACT there are so many such functions that they are
dense in L R. Then L is also a linear functional on D. . Dene Lf f xxdx for D. For example the
function x x x is also in Cc . . . Cc is the standard notation for the set of inntiely dierentiable
functions with support in a nite interval. To see that is zero outside the interval . We have
now constructed one not identically zero function in Cc . So Lf D . QED Notation. Examples .
as we see in Example . One says that f is locally integrable if a f xdx lt for any nite interval a.
f x /x wont work in . . has but / only some mild singularities. For example if f x /x in ..e. The
lesson to be drawn from these examples is this any continuous function f on R produces a
generalized function Lf . . then the integral still exists for any test function . This . huh . D by
means of the formula . . QED The generalized function P x is called the Principal Part of /x.
also proves the validity of the representation . Proof If lt a lt b then x dx x x dx x x dx x . xgta
xgtb altxb because altxb x dx . It arises in many contexts. . We will see it coming up later in
the Feynman propagator.. is bounded on R by the mean value theorem.. Neat terminology.
All we need of f is that b f xdx lt for any nite interval a. Thats why we call the elements of D
generalized functions. of this limit. xgt x dx x . i. b. In particular the function a . The integral
doesnt make sense for an b arbitrary test function . So we can let a and get a limit. . Its
based strongly on cancellation of singularity. The integrand in the last term in . Its still OK if
the function f in Example is not continuous. But not every element of D comes from a
continuous function in this way or even from a discontinuous function. There is an important
instance that violates the wisdom of Example . Lemma Let D. b. Then P lim a x exists and is
equal to x dx x xgta x dx x . an element of .
Do you see now why . has a well dened meaning even when its not of the form Lf for D
some dierentiable function f . Lets see what the denition . Lf Lf . Suppose that f is actually
dierentiable in the classical sense.e. is a justiable denition of derivative of a generalized
function T Thats right. The linearity of T is clear. Derivatives of generalized functions Denition
Let T D . gives when T LH and H is the nondierentiable function given by Hx . in the sense
that you grew up with. . D . The reason is that for any D the function is D again in D so the
right side of . x xlt . The derivative of T is the element T of D given by T T for all D.. Right To
understand why this denition is justied consider the example . T Lf spelled out in . D What
does this denition have to do with our well known notion of derivative of a function First lets
observe that at least T is indeed a well dened linear functional on D. b f xxb a b f x xdx a So f
x xdx a f x xdx Lf D D Stare at . i. . D . makes sense... and . Then Lf a b f xxdx f xxdx b where
a and b are chosen so that o a. It agrees with the classical notion of derivative when T Lf and
f is itself dierentiable D But .
H and found H . . USING . . Lets end with one more denition. D You could say ippantly that
we have now dierentiated a nondierentiable function. In truth the perfectly meaningful
equation D . . . But what does this equation mean It means . xdx by the fundamental theorem
of calculus. Example L We now have a notion of derivative of a generalized function. T T LH
D . isD often written as H .. . Denition A sequence Tn of generalized functions converges to a
generalized function T if the sequence of numbers Tn converges to T for each D.In this case
we have. So we have LH L . For the next two weeks we will regard it as immoral to write the
equation H ..
.
Problems on derivatives and convergence of distributions.
xxdx for Cc R.
. Dene L Using the denition
compute the rst four derivatives of L that is, compute T , ..., T . Hint . Use the denition of
derivative of a distribution. Hint . Use the denition four times to compute the four derivatives. .
Suppose that f R R is continuous but not necessarily dierentiable. Let ux, t f x ct. Show that u
is a solution to the wave equation u/t c u/x in the distribution sense sometimes called the
weak sense. Nota Bene Since f is not necessarily dierentiable you cannot use f in the
classical sense. . Does
n
TT,
x n converge in the distribution sense
. Let pt x t/ ex /t . Find the folowing limits if they exist. a. the pointwise limit as t . b. the L R
limit. c. the limit in D . You may use the following FACT that will be proved later
pt xdx t gt .
. Let be the function constructed in Lemma , except multiply it by a positive constant such
that
R
xdx
Such a constant can be found because the original is nonnegative and has a strictly positive
integral. Then let n x nnx Our goal is to show that n converges in some sense to a delta
function. a. Show that R n xdx for all positive integers n. b. Suppose that g is a continuous
function on R. Show that
n
lim
R
n xgxdx g.
c. Use part b. and the denition of convergence of distributions to show that Ln converges in
the weak sense to L . . Prove that, for all D, lim
x dx P i xi x
.
.
This is often written as lim
P i weak sense xi x
.
.
Hint Review the proof of the lemma at the end of Section ..
.
Distributions over Rn
The extension of the one dimensional notion of distribution to an n dimensional distribution is
straightforward. Denote by D the vector space of innitely dierentiable functions on Rn which
are zero outside of a large cube that can depend on the function. This space is sometimes
denoted Cc Rn . There are plenty of such functions. For example if , . . . , n are each in Cc R
then the function x , . . . , xn x xn is in Cc Rn . What cube could you use So is any nite linear
combination of these products. A distribution over Rn is dened as a linear functional on D. Of
course we can make up examples of such ndimensional distributions similar to the ones we
already know in one dimension Let Lf f xxdn x .
Rn
As long as f has a nite integral over every cube this expression makes sense and denes a
linear functional on D, just as in one dimension. We also have the ndimensional function
dened by L . .
The dierence from one dimension shows up when we consider dierentiation. We now have
partial derivatives. Here is the denition of partial derivative as if you couldnt guess. T T /xk .
xk .
When n every distribution has a very intuitive interpretation as an arrangement of charges,
dipoles, quadrupoles, etc. Im going to explain this interpretation in class in more detail. It
gives physical meaning to every element of D
Theorem . In view of this lemma you can see that we have only to deal now with the
singularity at r . Our notion of weak derivative is just right for doing this. In case you forgot
your vector calculus identities a self contained review is at the end of this section. But the
techniques are so frequently occurring that there is some virtue in following it through at least
once in ones life. r . L/r L We will break the proof up into several small steps. So r /r /r x y z /r
/r . Lemma . r in the distribution sense.. P QED. Poissons Equation We will write as usual r x
in R . Let D. That is. At r /r Proof. Lemma . /r/x x/r and /r/x /r x . The trick is to avoid the
singularity until after one does some clever integration by parts in the form of the divergence
theorem. . Then R /rxdx lim r /rdx . I want to warn you that this is not the kind of proof that
you are likely to invent yourself.
. To this end we are going to apply some standard integration by parts identities in the OK
region r . r lim So what we really need to do is show that this limit is . . T/r j /x T/r j . r r n ndA
by the divergence theorem r r .Proof The dierence between the left and the right sides before
taking the limit is at most use spherical coordinates in the next step /rd x max x xR r r /rd x
max x xR QED. . The other boundary term in this integration by parts identity is zero because
we can take it over a sphere so large that is zero on and outside it. C r /rxd x P d x by identity
. r r where n is the unit normal pointing toward the origin. . Before really getting down to
business lets apply the denitions. . . . /xj T/r /xj j T/r R /rxd x /rxd x.
Application . Take f and G r . r r r P P But wherever r . Then we get . This proves . . r r dA r x
dA . Take f /r and G . Then we get wherever r . Vector calculus identities. . For the other one
just note that n /r/r /r . P P . max as . . we nd r wherever r . r r r Application . So r r ndA r xdA
r .Now r /r ndA r ndA . r r r P This is the identity we need in the proof of . This gets rid of one
of the terms in C in the limit. If f is a real valued function and G is a vector eld. P P P
wherever r . So subtracting . from . . both dened on some region in R then f G f G f G .. Only
one more term to get rid of x dA r x dA max x x because is continuous at x ..
To begin with we must understand how to give honest meaning to the F formula . The
Fourier Transform The Fourier transform of a complex valued function on the line is f eix f
xdx . But the usefulness of these formulas depends crucially on the fact that one can also
transform back and recover f from f . We will see later that these formulas allow one to solve
some partial dierential equations. Thus if you dierentiate under the integral sign you get d f d
So Fourier transform ofixf x eix ixf xdx. F f i f . .. Moreover in quantum mechanics these two
formulas amount to the statement that the Fourier transform interchanges P and Q
momentum and position operators. Suppose that f xdx lt . To this end there is an inversion
formula that does the job. d f . Since the integral is over an innite interval there is a
convergence question right away. Our goal is to establish the most useful properties of the
Fourier transform and in particular to derive the inversion formula and show how to use it to
solve PDEs. And an integration by parts never mind the boundary terms clearly gives i f So
eix f xdx. d . . F Here runs over R. The most useful aspect of this transform of functions is
that it interchanges dierentiation and multiplication. F . F .
n . only holds for n . . F for some real numbers Mn . will allow us not to have to worry about
whether f and xf x are both in L . huh . So by examples . ex is rapidly decreasing. . and F .
Summary The space of rapidly decreasing functions is a vector space and is closed under
multiplication by any polynomial. F . . Just replace n by n f in . In words xn f x is bounded on
R for each n. Examples . Since any nite linear combination of rapidly decreasing functions is
also rapidly decreasing we see that pxf x is rapidly decreasing for any polynomial p if f is
rapidly decreasing. If f is rapidly decreasing then so is xF x because xn x f x f n x F x which is
bounded in accordance with . holds. F Proof Apply . because lima a eix f xdx exists. So f xdx
M x dx lt . but not for n or more. Any function in Cc R is rapidly decreasing. and .We will write
f L if . there are lots of these functions. Denition. Use the Cauchy convergence criterion now.
This is a nuisance in dealing with the identities . QED F Using only rapidly decreasing
functions in . we see that pxex is rapidly decreasing for any polynomial p. ... Lemma Any
rapidly decreasing function is in L . a b F Proof a b eix f xdx axb f xdx as a b . And besides. If
f L then there is no problem with F a the existence of the integral in . Of course even if f L it
can happen that f is not in L and/or that F xf x is not in L .. .. .. . A function f on R is said to be
rapidly decreasing if xn f x Mn . for n and n to conclude that x f x M for some real number M .
x is not rapidly decreasing because . . . Neat. We are going to restrict our attention for a
while to a class of functions that will make these issues easy to deal with.
So we are going to restrict our F attention F a while. By . f . Weve already seen that these
functions are rapidly decreasing. QED. And besides there are more. f etc.But to use . So f is
rapidly decreasing. . Proof First notice that for any function f in L we have the bound f F eix f
xdx . But df /d is the Fourier transform of ixf x which we have seen is also in S. times any
derivative of f is again in S. here is the reason that the space S is so great. So f .. .
Invariance Theorem. and . x is innitely dierentiable but is not rapidly decreasing. Then so is f
. as we saw in Example . to a class of for functions that makes both . So both S F formulas . .
CHECK THIS against the denitions I know this may seem too good to believe. etc. So df /d is
also rapidly decreasing. Now here is a real nice thing about this space S. p. So this function
is not in S. Now suppose that f S. we still need to worry about whether f is in L . make sense.
easy to deal with. And so on. even further. . STATUS If f F then f and all of its derivatives are
in L . f xdx f . it now follows that n f is bounded for each n . Examples . But we do know that
there are lots of functions in S. So the function ex is in S. If f S then f S. All of Cc is contained
in S. are indeed zero. since these functions go to zero so quickly at . . are all F in L ..
Notation S will denote the set of C functions on R such that f and each of its derivatives is
rapidly decreasing. ex is innitely dierentiable and each derivative is just a polynomial times
ex . Since the right side doesnt depend on f is bounded. If f S then any polynomial. Moreover
they are both correct because F the boundary terms that we ignored in deriving . . . and ..
Check this at your leisure In truth.
Arent you glad that we are focusing on functions in S Notation g y eiy gd . . holds. So F .
Inversion Theorem. F Since we already know that f is in S we know that the right hand side
of F . But f itself decreases so slowly at that its not in L .Here is what the inversion formula
will look like f y eiy f d. but because the proof derives some very useful identities along the
way. makes sense. In other words f f . For f S equation . doesnt make sense. Contrast this
with the example that you worked out in the homework Take f x if x and f otherwise. We are
going to spend the next few pages proving this formula. Not because I fear that you might not
trust me. Then f L so f makes sense. F F .
The exponent is a quadratic function of x for which we can complete the square thus x ix x it
t / t t Hence pt et / e t xit dx . Sneaky use of polar coordinates. To compute pt we need rst to
multiply pt x by eix before integration. t F F . F pt xdx . . Proof of . Gaussian identities Let ex
/t t Then we have the following three identities. . F F F pt et / t e x ex /t pt x. . here are some
explicit computations of some important Fourier transforms. t F Proof of . The Fourier
Inversion formula on SR. pt x .. pt xdx R pt xpt ydxdy e x y t . . To begin. t t R dxdy /t er rdrd
Use the substitution s r /t in the last step. .
This F would prove . the integral of this function around this rectangular contour is zero. R R
exia /t R ex /t dx plus a little contribution from the vertical R sides of the rectangle. Since p/t
is even we have by . x has been translated by the imaginary number it. Let a t. You cant just
translate the argument to get rid of it. et t / / e F / p t /t by .. pt gx gx g pt x gx gx ypt ydy gx gx
y gxpt ydy. Since ez /t is an entire function. which goes to zero quite quickly for y a as R .
Take a large positive number R and consider the rectangular contour in the complex plane
which starts at R on the real axis. But not so fast.. . and suppose a gt just so that I can draw
pictures verbally. Here is how to get rid of it. So the integrals along the end segmentsF to
zero as R . . Let g be a bounded continuous function on R.The coecient of et / looks just like
pt xdx which is one. The convolution of two functions f and g on R is given by f gx lemF R f x
ygydy . But eRiy /t eR y /t . Otherwise put. QED F F /t x p t ex /t t Denition . Then for each x
limpt gx gx. . So it is indeed go t true that the coecient of e in .. F Lemma . t Proof Since pt
xdx we have . equals pt xdx. which is one. moves along the real axis to R and then moves
vertically up to the point R ia. On the right end the integral is at most a times the maximum
value of the integrand on that segment. QED Proof of . west to R ia and then south back to
the point R. From there move horizontally.
If f and g are in SR then f gx n f eix g. t .D. pt gx gx gx y gxpt ydy gx y gxpt ydy pt ydy g ygt .
. t Hence limpt gx gx . y limpt gx gx g lim t t But y y /t y /t pt ydy e dy e dy t t y /t tey /t te as t .
gt gx y gx lt if y lt . . . . t Q. pt ydy. . ygt pt ydy ygt g pt ydy.Given gt . Lemma . Therefore limpt
gx gx gt .E.
on the left and Dom. However the factor makes for a nice identity corFPl Corollary .E..Proof f
gx f x ygydy f eixy gyddy R R . Plancherel formula for S If f and g are in SR then Ff. theorem
on the right to get gx eix gd. . Use Lemma . Then. Conv. The factor in front of g has clearly
no eect on the onetoone or ontoness property of the map g g . . g . by . R The asymmetric
way in which the factor occurs in the inversion formula is sometimes a confusing nuisance. .
If g is in SR then gx gx. R n n n f eix eiy gydyd R R f eix gd. F is a onetoone map of SR onto
SR. That is. In the preceding lemma put f et / . Fg f. Q. . f x pt x. R lemF Let t . Dene Fg / g .
F Proof. It is useful to distribute this factor among the forward and backward transforms.
Thus pt gx et eix gd. R Theorem .D.
Hence f gxeix dxd R R f gd R . it asserts that f gd R R f xgxdx But g gxeix dx gxeix dx. .D.
Proof. Phrasing this identity directly in terms of f and g . R R f eix gxd dx f xgxdx.Explicitly. .
R Q. this says Ff Fgd R R f xgxdx .E. .
j . To this end we need the ndimensional analog of S. Just as in one dimension. But the
ideas are similar. delta functions and their derivatives. An integration by parts never mind the
boundary terms clearly gives the F analog of . . Moreover in quantum mechanics these two
formulas amount to the statement that the Fourier transform interchanges Pj and Qj
momentum and position operators. We say that a function f is in L Rn if f xdn x lt . F Here
runs over Rn . The Fourier transform of a complex valued function on Rn is f Rn eix f xdx . is
f j So Fourier transform ofixj f x eix ixj f xdx.g. The Fourier transform over Rn The denitions.
one must pay some attention to the meaningfulness of this integral. for a much larger class of
functions and generalized functions. f /xj ij f . As in one dimension. f . theorems and proofs for
the Fourier transform over Rn are nearly identical to those over R. including e. The analog of
.. Rn The formula . key formulas. F . ij f eix f /xj xdx. makes perfectly good sense if f L Rn . F
Rn . the Fourier transform over Rn interchanges multiF plication and dierentiation. Rn F So
We will see later that these formulas allow one to solve some partial dierential equations. But
in F the end we are going to give meaning to . In this section we are going to summarize the
results weve obtained so far and formulate them over Rn .
The inverse is given by the inversion formula f y n Rn eiy f dn . is equivalent to the statement
that for any polynomial px . .. . a. Denition . Theorem . Any rapidly decreasing function on Rn
is in L Rn .Denition .. The convolution of two functions f and g on Rn is given by f gx Rn f x
ygydn y . The map f f c. . . . k . . is a onetoone linear map of SRn onto SRn . F The important
identity in the next theorem is the basis for the application of the Fourier transform to solution
of partial dierential equations. F Denition . If f is in SRn then f is also in SRn . A function f on
Rn is said to be rapidly decreasing if xk f x Mk . . In words xk f x is bounded on Rn for each k.
In Exercise you will have the opportunity to show that the condition . b. F lemF Lemma . F
Proof. The proof consists of the following straight forward computation. xn in n real variables.
. g f g f . the product pxf x is bounded. F for some real numbers Mk . A reader who is
concerned about the validity of any of these steps should . Theorem . SRn is the set of
functions f in C Rn such that f and each of its partial derivatives are rapidly decreasing. .
although the identity is valid quite a bit more generally. . . Plancherel formula Dene F n/ . .
As a consequence F is a unitary operator on L Rn This is a minor restatement of Corollary . F
. Then F L Rn L Rn . . .simply assume that f and g are in SRn . . /t F F F pt et t e x / tn ex pt
x. Finally.. . f x ygydyeix dx f x yeixy dxgyeiy dy f gyeiy dy g f Theorem . . and extension to
Rn . . Let pt x ex /t tn Then pt xdx Rn corFPl x Rn .. f g f gxeix dx Rn Rn Rn Rn Rn Rn .F are
the ndimensional analogs of the important Gaussian here F identities . Gaussian identities
over Rn . F This is the Plancherel formula.
F F Now you can seeF virtue of using S as our new test function space the the right hand
side of .. Example . . After all. But the delta distribution and its derivatives are in this smaller
dual space anyway because L is a meaningful linear functional on S. Clearly Lf is linear. We
cant allow the function f x ex in . So we have a smaller dual space now. xn for some n .
wouldnt make sense. Denition . n Suppose that f R C satises f x const. makes sense
precisely because is back in S when is in S. Of course n will also do. F R exists when S
because f xx const. Had we attempted to use D this denition would have failed because is
never in D when D if .. x sin x has polynomial F growth. The Fourier transform of an element
L S is dened by L L for S. Take n in . A tempered distribution on Rn is a linear functional on
SRn .g. Thus any function of polynomial growth determines in this natural way a linear
functional on S. And similarly for k . Remember that we did not need polynomial growth of f
when we made Lf into a linear functional on D. If f has polynomial growth then the integral Lf
f xxdx . F In this case we say that f has polynomial growth. Example . Tempered Distributions
Denition . which goes to zero at like x and so is integrable. xn xn . L is only dened on S.
because F the function x ex is in S and so . . E.
. Or even as eixy dx y if you dont feel too uncomfortable with the seeming nonsense of the
left F side. R xdx by the def. Thats how . is written in the outside world. of Fourier trans. at L
by yet another denition One thing to take away from this proof is that every step is just the
application of some denition. Proof. F L L by def. . . L L . by the def. of L . But you better get
used to it. There is no mysterious computation. . . Example . F .of L by the Fourier inversion
formula Notice that this time the last step uses something deep. Then F L L by def. . Let S. .
L L . F After you have achieved a suitable state of sophistication you can write this as . R d
by def. In the outside world this is usually written . Let this be a lesson to all of us Example .
Proof. . of .
So is Lf Lf Yes. . . . . . Here is a denition chasing proof. Lf Lf R R R R .Consistency of the
two denitions of the Fourier transform. If f L then we already have a meaning for f . f d f eix
xdxd R R f eix d xdx f xxdx Lf QED .
Lx . Find the Fourier transforms of the following functions a ex b ex / / / c xex d x ex e f x f x
if a x b otherwise. of T explicitly explicitly enough to do part c without nding . c. Compute a..
b. a Show that T is in S R . Find all values of the real parameters LT . . c Evaluate T where .
c such that b c . T . e d Let L a a. . Problems on the Fourier Transform . dx. Note that this is
an integral over a line not over R . For in SR dene T xx. b Find the Fourier transform. b.
Use the denitions. . d. Use the denitions. .. from the onedimensional identities F F .. Lx Hints
. . Show that a function f on Rn is rapidly decreasing if and only if. . Derive the identities . the
function pxf x is bounded. Use the denitions. . . .. F F . . for every polynomial px . . . xn .
. x in n spatial dimensions is u/t /u t with initial condition u. . is the rst factor on the right in . .
But Hey We already know of a function whose Fourier transform . The heat equation for a
function u ut. whose solution we know from elementary calculus. We will see on purely
mathematical grounds why solving the equation . for t only bodies cool o as time moves
forward. . We obtain a function ut. Now for each Rn the equation . f . Poisson equation and
wave equation. In order to nd ut. It is ut. backwards in time can . produce singularities.
parabolic. satisfying ut. . It is the function pt x tn/ ex /t .. . . . elliptic and hyperbolic are
amenable to study by using the Fourier transform in slightly diering ways for each. As you
probably know very well. Apply the Fourier transform to u in the spatial variable only. . If you
move backward in time temperatures can increase in such a way as to produce singularities
in the solution. . there . on . Applications of the Fourier Transform The Heat Equation by
Fourier transform. . Solution of . x itself we still have to do a reverse Fourier transform . The
three fundamaental types of PDE. . Here f is a given function on Rn . et/ f . .. The prototype
of these three kinds of PDEs are the heat equaiton. /t / u which we would like to satisfy the
intial condition u. .. . an ORDINARY dierential equation. . x f x. is strong physical motivation
for seeking a solution to .
Poisssons equation by Fourier transform We wish now to apply the preceding method to the
equation u . We also know that convolution goes over to multiplication under Fourier
transformation. F t/ By the uniqueness property of the Fourier transform we therefore nd ut.
See . . . Of course we only have spatial variables this time. Apply the Fourier transform to . . .
.whose Fourier transform we already computed. . In fact it is true that / r . f . So you think that
the solution is u / . x pt f x exy /t f ydn y n t Rn . We nd u . This time the partial dierential
equation has been reduced to an algebraic equation there are no derivatives left at all. Well.
Although the method works in n dimensions with n . Use the ndimensional version of that
function. Hence pt f pt f e . we will carry this out only in dimension n . . if that were right then
we again have a solution for the Fourier transform in the form of a product. where we can
compare with our previous solution. . and thereafter. Of course this will give u itself as a
convolution. in ALL of the variables.
i . There are. support. Such a distribution is clearly in S also and the computations . that all
the solutions of . . v we could add v onto any solution of . lots of harmonic polynomials on R .
.. in fact. .And this would give . Show that Recall that this means L for all SR . are the only
ones that go to . The solution . . function on R i. . . We saw back in those days that could be
e. only in such a way. z is a harmonic polynomial on R . r which we already know from our
previous work is actually correct. . and get another solution. to . Write j /j and let L be the
distribution L pi . For example we have the identity u . zero as x . So is x y z . y. are actually
correct. .e. i . In fact if v as any harmonic . dier from . For example x y z is harmonic. L . This
means that we could add onto the right side of . But we are losing some solutions in the
passage from . . all the solutions to . Here is a problem that illuminates this. the solutions . p .
It is true .g any charge distribution in D with compact . . Back in x space this just means that
we can add a constant onto any solution u and get another solution. is called the potential of
. That is. and get another solution. ProblemPoisson by Fourier Suppose that px. Among . .
say. For example in . Was the weight in its neutral position u or was it elsewhere Was it
moving or stationary If we specify the intial position u and the initial velocity u then we must
solve the Initial Value Problem for the equation . and cos t. because it will be more revealing
for our purposes. The harmonic oscillator equation with frequency actually frequency / is d ut
ut . The forced harmonic oscillator equation is d ut ut f t. . What happens to the weight during
and after the disturbance thats us acts via f Of course that depends on the initial state of the
weight at time t .. . from the point of view of the Boundary Value Problem. where A and B are
complex constants. It will be adequate for our purposes just to assume that f Cc R and then
not have to worry about any technical details. . Our goal however is to . Of course one can
choose A and B so that the solution is real and is expressible as a linear combination of sin t
. The novelty of the present section lies in the fact that we will be interested in boundary
conditions at . is ut Aeit Beit . ut is the displacement of the weight from its neutral position.
Advanced. dt the general solution to . We are going to study the equation . . For a few
seconds we push and pull on the weight up and down in accordance with the force f t and
then let go altogether at the upper bound of the support set of f . The function f is sometimes
called the source generating the motion u. address a dierent problem. dt . with the specied f
and specied initial data. But we are going to stick to the complex form . We can think of a
weight hanging from a spring which is attached to the ceiling. somewhat in the spirit of the
section on Green functions. . . Retarded and Feynman Propagators for the forced harmonic
oscillator Let be a strictly positive constant.
But you can only impose two of them on a solution. Ga t . depending on f . Show that Ga f is
zero in the distant future. Dirichlet or Neumann conditions or a linear combination of them.
.the context of Maxwells equations the sources for the electric and magnetic .at x t. Imitate
the thmO proof of Theorem . b one can choose. Note that each of these three functions is
continuous and has a jump of one in its rst derivative at t .e. and also sometimes called the
retarded propagator. GF t i eit eit . It propagates the disturbance f into the future and
depends on the the disturbance at some earlier time. Let Gr t sin t sin t tgt t t tlt t tlt ..
equation . We will discuss their physical interpretations afterward. a. elds are charges and
currents. The etymology of the word retarded can be understood in the context of
electromagnetic theory. b. at a. c. In this sense we have four degrees of freedom in the
nature of our choice of boundary conditions. Here are three dierent Green functions for the ..
Problem . As in . Show that the solution Gr f is zero in the distant past. Similarly at b. of a
changing charge distribution depends not on the charge distribution at time t but at an earlier
time. Suppose that f Cc R. Let uGf with G Ga or Gr or GF . Show that in each case u is a
solution to . The Green function Gr is called the retarded Green function. .. Hint. We saw in
the section on Green functions that on an interval a. i. where the potential. The eld produced
by the charge distribution travels only . for suciently large negative time. . they make the
homogeneous Maxwell equations inhomogeneous. Terminology.
ua t can be nonzero before the disturbance f even begins Doesnt sound very causal. The
solutions to . . It is a FACT that these are the only solutions to . Problem . The resulting
potential is called the retarded potential. Thus we have recovered the solutions to the
homogeneous equation by the method of Fourier transforms. . The Green functions above
have been constructed directly in terms of the homogeneous solutions in accordance with
the method in the section on Green functions. . gives solutions to the homogeneous equation
on the Fourier transform side. s s s . in the t variable which is the only variable around. Thus
us As Bs . We will use s for the variable conjugate to t. For the homogeneous equation we
need to put f . then include the functions at the two roots of s . In some contexts its the more
useful way to go. That is. is particularly important in understanding the behavior of electrons
and positrons in combination. The third propagator. GF . Similary the propagator Ga is called
the advanced propagator. It propagates the disturbance f into the future as a positive
frequency wave and into the past as a negative frequency wave. . u .with the speed of light.
Show that the Fourier transform of the solutions . up to a constant. of Problem . Next we
address the inhomogeneous equation. Some authors say. . not instantly. . this time. that
positrons are negative energy electrons moving backward in time. when f . we dene us eist
utdt. are given by . Note rst the identity . . When this discussion is boosted up to three space
dimensions from the present zero space dimensions Feynmans propagator has the
interpretation of propagating electron wave functions forward in time as positive energy wave
functions and propagating positron wave functions backward in time as negative energy
wave functions. for short. . does it One says that Ga propagates the disturbance into the
past. Take the Fourier transform of . As you saw in part c. GF is called the Feynman
propagator for the single frequency . We nd s s f s. Here is how the Fourier transform method
can be applied.
TROUBLE We already know that s has a nonintegrable singularity and . Lemma . therefore
so does the right hand side of . . . ttdt a a lim lim a .. The next to the last line converges to x
dx by the RiemannLebesgue Lemma. . . It is dened as s P lt gt lim . . . c sgtc s s . . But we
saw earlier that the function /s has an interpretation as a distribution called the Principle Part
of /s. by the coecient of us we nd us f s f s s s . a a a teitx xdxdt teitx dtxdx lim lim a a cos ax
xdx a ix cos ax lim xdx a xgt ix cos ax lim x dx a x ix ax because cos ix is an odd function.
.Dividing . . As it stands the right hand side therefore has no meaning. s iP s . The very last
integral ix xgt . Let t /sgnt Then Proof a . .
GF in terms of the functions F and the solutions. . . . Problem .. s .. should be expressible as
a linear combination of the functions F and the . So this integral converges to ix Q. we should
expect that all the solutions to d ut ut t. C. .E. In view of . Problem Let F t teit for gt . for some
constants A. . of the homogeneous equation.may be rewritten x dx ix by the
RiemannLebesgue Lemma also. dt . Find explicitly this representation of the three Green
functions Gr . and . . . Using Lemma show that teit iP s for any real . solutions to the
homogeneous equation . D. Ga and . For example one expects Gr t AF t BF t Ceit Deit . B. x
cos ax x dx.D.
The wave equation by Fourier transform The physically appropriate problem is the Cauchy
problem. That was the easy part. . for each is ut. . The general solution to . . Notice rst that
cos t is exactly the t derivative of sin t/. Use spherical polar coordinates . . each t. . t . Then
sin t eix dt x t . We can make use of the rotation invariance . is sin t/ . x/t gx . It will suce then
to nd a distribution t on R whose Fourier transform. is then clearly ut. equation. The solution
to . Upon Fourier transforming the initial conditions . we see that we must have A f and B g .
As in the heat equation we will take the Fourier transform in the space variables only. But this
time its second order in t. f cos t g . This is the initial value problem. . x . . R . Proof. A cos t B
sin t . . Lemma . Let t be surface area element on the sphere x t in R . t u. . x f x u. x t gx /tt f
x. ut. . . We nd ut.. ut. Now we have to Fourier transform back to nd u itself. for .. x ut. Just as
in the case of the heat equation we have now an ordinary dierential . . Hence the Fourier
transform of u is given by sin t ut. The integral is just an integral over a sphere of radius r t. .
for ease of mind. . /t ut.of the integral by choosing the positive z axis in the direction of .
observes that even if t lt . Problem Explain why the solution . which you get by substituting s
cos . QED Now dene a distribution t on R by the formula t t xdt x for SR . . . Notice that for
each this equation looks just like the harmonic oscillator . If we rst Fourier transform in the
spatial variables only we nd ut. says that light travels with exactly speed one in our units.
Hints . x/t ut. xt for t and dene t to be zero for t . . . form . Prove that t sin t for all t . is correct
because sin is odd. . . that is in Cc R . and will. x t. t. Use the denition of t . x. in combination
with the explicit . Then x ra cos . So the integral is r eira cos sin dd r sin ra a . We can.
Assume. lets consider the inhomogeneous wave equation ut. equation . . with . of t . Let a .
Use the denition of Fourier transform of a distribution.. take over what we .. Finally. Problems
Problem . Use Lemma . . . once one . This proves .
Put in . But one should not lose sight of the fact that the second factor is not actually a
function. xdt . Thus we nd ut. For the wave equation. Not only does the integral represent a
convolution but there is also a convolution right in the middle of the integral. x t t tt t . the
inhomogeneous wave equation . and do the usual reverse Fourier transform to nd. All in all
this represents a convolution over R . We arrived at one of them by the above procedure.. as
for the forced harmonic oscilator. ut. . .learned from the forced harmonic oscillator. more
suggestively to emphasize its character as a distribution over R as Gt. For each t and t we
see that we have the usual product of Fourier transforms . . This is one of the propagators..
Its one half the advanced plus retarded. .. is a distribution on R for each t and all together is a
distribution on R . t eit eit t t. Remark to ponder. on the right in the spatial variables. x tt x. We
may write this as uG where G tt . It will be suciently illuminating to focus on just one of the
three propagators that we studied for the harmonic oscillator. there are essentially three
propagators for the inhomogeneous . you proved long ago in .. equation . at our disposal.
Where does the Feynman propagator come from Hint Consider the identity lim xi P /x i which
. . for One might wish to write . Moreover we have the formula .
If s DLf then Lf has an analytic extension to the half space Re z gt s .g. . f t et . If s DLf then f
test f tes t ess t L . . The Laplace transform of a function f . L The domain of Lf is the set DLf
consisting of those s R for which f test dt lt . f t et . or s . As we see from the above examples
it may be the empty set or of the form s . . . Then DLf . . L exists for Re z gt s .. Then Lf s .
Remark. Then ezt f t est f t L . . E. . So DLf s DLf for any s gt s because . . if f t / t then DLf . .
C is the function Lf dened by Lf s f test dt. . Hence the function Lz f tezt dt . bdd
Consequently DLf is always an interval. or . . If f L . f t et . then DLf . Proof Put z s i with s gt s
. . Note. Laplace Transform Denition. . Examples. DLf est dt lt if s gt .
L . But note that L has an analytic extension to the entire complex plane with the exception
of a pole at z . This is typical of Laplace transforms that arise in practice. Let g Ls i . They
often have meromorphic extensions to the entire plane. f t et . Example. Put ht Then g Hence
f tes t lt t lt t lt . . . . f tes i t dt f tes t ei t dt. But the integral representation may be only valid
for Re z gt s .By dierentiating under the integral sign easily justied we see that Lz is analytic
with dLz/dz tf tezt dt. . The Inversion Formula Suppose s DLf and s gt s . Lz ezt dt z which is
analytic in Rez gt . DLf . ht g ei t d. htei t dt Fourier transform of h.
Then Lz f te zt a dt f tezt dt . . Application of Uniqueness Problem. So f t es t Ls i ei t d for t .
Thus for t lt f tes t g ei t d by the Fourier inversion formula. . dz But note.This integral may
converge pointwise in t or in the L sense of convergence. z/z . But we ignore these
questions. a. z i z i Note. z z z Solution f t et et cos t. L Corollary . In this problem Lz had
simple poles. Reason. Example. Find the function f t whose Laplace Transform is Lz z .
depending on the quality of g. Uniqueness Theorem The Laplace transform Lf s of a function
f on . Let f be in L R with support in . Higher order poles can be dealt with using dLz tezt f tdt.
determines f uniquely up to values on a set of measure zero if DLf is not empty. L may have
no singularities in the nite plane even though f is not the zero function.
Nevertheless the case where L has only poles and goes to zero at is important. . dz id we
get f t i s i s i Lzezt dz. Putting z s i . For this situation it is useful to rewrite the inversion
formula thus f t Ls i es i t d. Sometimes the straight line integration contour can be deformed
to surround the poles of L.exists for all z C and is entire. See the homework problem.
This may clearly be rewritten f t i Lzezt dz C . We have seen that if Lz f tezt dt exists in the
open half space Re z gt s then for any real number s gt s f may be expressed by f z es t Ls i
ei d . .. L with z s i . If these singular points are isolated poles then one could nd f z by the
method of residues. Problems on the Laplace Transform . Justify this procedure for the
function Lz z z z z z and use it to nd f t for all t . L where C is the straight line contour from s i
to s i. For some functions Lz it may be possible to deform the contour to C so as to include all
the singular points of L inside.
but not on .S. Advanced Mathematical Methods for Scientists and Engineers. More Notation
f x x ox as x means that f x x ox as x . o/x as x . Examples.M. then one writes f Og as x if f
Og on M. Chapter . for some M . x x / Ox on . x x . Orszag. i If f x /x and gx /x then f x ogx as
x . Denition f x ogx as x means x lim f x/gx . Introduction Denition If f and g are two real or
complex valued functions on a set S then one writes f Og on S if there is a real constant C
such that f x Cgx for all x S. Denition If f and g are two functions on an interval a. Example x
x / Ox as x . QA B . Asymptotic Expansions Reference C. Example. ii ex oex as x . Example.
Bender and A.
tn x . Steepest descent p. Lemma . . of NB . x PROBLEM Let f x et /tdt. But once it is proved
we can see that f x behaves like ex /x for large x. This needs to be proved. Integration by
parts page of NB . of NB . Four Methods . But rst we need the following tricky lemma. Find
the asymptotic behavior of f x as x . x x We will do better than this. x gt . Laplaces method p.
.We are going to study the four basic methods for determining the Asymptotic behavior of an
integral. For any integer n we have x et ex dt O n as x . The Method of Integration by Parts.
SOLUTION Integrate by parts to nd x f x t /tdet . the precise meaning of which can be stated
in terms of the preceeding notation in the form f x ex ex o as x . Stationary phase p. of NB . x
t e x e dt t t x t ex e e dt x t Now it happens that the last term the integral is rather small
compared to the rst term for large x.
Proof x et dt tn x/ x/ et dt tn et dt x x/ x x/ et dt tn et dt x/n . . x ex dt o n as x . . tn x et One
can go a step further now in determining the asymptotic behavior of f . one says that the
series f x ex aj j x aj are constants is an asymptotic expansion of f if for each j n . . Let . such
that n x on x. . Denition. n f x j j aj j x on x x . n x x x x x x This motivates the following
denition. n is easily seen by the ratio test to xn . . ex/ e /xn ex ex/ . Corollary . . which proves
the lemma. Integrating by parts again f x et t x et t x x et ex dt ex O t x x x as before. If f is a
function on a. . . . Clearly we may continue this and get n o n . aj j x need not converge for
any x. In the n Thus the innite series preceding example the series converge for no x. . be a
sequence of functions on a. . .
Watsons Lemma Theorem . As f text dt x gt c. . As . . As ux eu du for x gt . Proof First. f has
the asymptotic expansion f t t Let Ix k f tect dt lt for some c gt . B. Lemma . . Laplaces
Method. . note that we may replace f tex by f tex and incur an exponentially small error as in
the following lemma. note that for a simple power we have t ext dt x s es ds by putting s xt . ..
Watsons Lemma Assume A. gt . As x . ak tk as t .. For any gt and f satisfying A f text dt ox as
x for all gt . For some gt and gt . . Second. As Then Ix has the asymptotic expansion Ix where
x k ak k xk x .
.. Q. As . . By the assumption B N f t k ak tk otN t . Hence N f t k ak tk KtN t . As since x is
the last retained term in the sum.Proof of Lemma For x gt c we have f text dt e f tect exct dt .
xc f tect dt ox x gt .D.E. is weaker than . As Hence there exists a constant K and gt N f t As k
ak tk K tN As t . We must show that N Ix k ak k ox x xk . As Actually . As Now in order to
prove the theorem we choose N and let N . .
Thus N Ix k N ak tk ext dt E x E x where E x ox . k N k ak xk k E x E x E x. N As f t ak t k k e
xt dt K K Kx tN ext dt tN ext dt . . . As This proves . x by Lemma . .E. . Q. N k ak tk ext dt N f t
ak tk ext dt E x. k E x . Thus Ix f text dt E x where E x ox . const.D. . .. x by Lemma .for K K
aN . ak tk ext dt E x E x E x where E x ox . N N . x ox x . But by . x .
. gtect dt lt for some c gt . k B. In other words we will concentrate just on the algebra. As
gtext dt. .e. Assume A.corAS Corollary . Lemma . We will write the remainder of the proof in
an informal way because the details of justication are exactly the same as in the proof of
Watsons Lemma.. i. . In fact we will write equality of series even though one must interpret all
steps in terms of nite sums and behavior as x or t . As is an asymptotic expansion of Ix. . As
ak k / as x xk/ . Ix gtext dt E x ak tk ext dt E x ak tk ext dt E x. . Proof By the same argument
as in the theorem. . . k k . we commit an exponentially small error in replacing the integral by
gtext dt. Thus we have E x exponentially small error Oe x . g has an asymptotic expansion gt
Let Ix Then Ix k ak tk as t .
exhs . for large x. Hence for small t ht h t Ot as t .. and this would give the answer As .E.At
this point we could replace the integrals by k tk ext dt which can be done explicitly by
integration by parts. Then dt ds/ s. Recall that these equations are not identities but
asymptotic expansions. from a neighborhood of t say t . . As in the quadratic case the main
contribution to f t will come. Q. Now since h is a minimum at t we have h t . the error being
etx xn dx tn/ n / nn exponentially small. k k ak k / Ex by Watsons Lemma with . . xk/ . But
instead lets derive from Watsons Lemma. . So Ix ds ak sk exs E x s k ak sk/ exs ds E x .
Using the same ideas. here is another method. it Put s t .D. Let f t exht dt where h need not
be quadratic but looks like this We assume for simplicity that h but if not this can be
subtracted from ht changing our result by an overall factor of eth . We are going to get power
decay of f x.
Our objective is to get the leading term in the asymptotic expansion of f x if there is such an
expansion. holds for n . x As ux eu du . As ux eu eu dux du du eu xux du xx.Assume h gt .
So we ignore the Ot terms to get f x So f x Denition. . Q. So . The identity ii follows now by
induction. h x . Some values i / . xh ux eu du. The function is x ex h t dt . As Proof eu du . ii k
k k k . . As Thus if n n then n n n n . Proof t/ et dt es ds . Induction Note that for x gt . k . So x
xx for x gt . x gt . As Note n n . .E.D. . .
b and that h . . Then Ix ox/ . . we may solve for t in terms of s and we may expand t in terms
of s to any nite order. At t this is one. Choose so small that gt gt / for t lt .t exht dt. So we may
choose even smaller if necessary so that ds/dt gt on . Let b Ix a exht dt x gt c. The expansion
begins t dt ds s d t ds s . then ds/dt gt tg t/ gt. Assume that h and ht gt for all other t in a.
Finally assume that b A. Now Ix exht dt alttltb. Let s t gt on . xh Proof By Taylors theorem
with remainder we may write ht h t gt for t near zero where g is smooth and g . Assume that h
is in C a.corAS Corollary . . Suppose that a lt lt b a or b or both could be innite and that ht for
t a. t exht dt ox/ as x gt . a echt dt lt for some c gt AND A. s s But dt/dss /ds/dtt . b. b. Since s
is a strictly monotone function of t on . Hence t s Os as s .
make the change of variables s t gt.. h s corAS / Ox/ as x by Corollary . Stirlings formula x ex
xx/ As Corollary .By Assumption A the second term is ox/ . . . The most frequently used form
of Stirlings formula is for integer values of x. to nd x ux eu du x tx ext xdt e .. exht dt /x ox/ as
x .. by Corollary . gt ds Osds . . This takes the form . xh // . Then h t / t. Moreover h . . x lim
Proof Make the substitution u x tx in Equation . On the interval . Then exht dt ex ex h s dt ds
lt . This proves Stirlings formula. x x x ext tx dt extlogt dt xx ex So take ht t log t. QED .
Hence. Therefore h has a corAS minimum at t .
STIRLINGS FORMULA lim n en nn/ n .
f aeixa eixs ds O x . . Sec. . Assume that is in C p a. Ix eix cos t dt. . The second integral is
Ox as x by the lemma. Example. Thus a Ix f a a eixa p p atap /p dt O x . So p . expansion of f
and at a. . while n a . b and real with t on a. p and f a . b with f C a. Here t is a stationary
point t cos t. Therefore Ix eixi/ x / / p x p a /p /p p Ox as x Ox as x . Now the idea is to choose
so small that rst integral can be adequately approximated by using the rst terms of the . n . .
Let f and be dened on a. . b and p a . Proof of Theorem Write for small a b Ix a f teixt dt a f
teixt dt. . b but a . f t . t a pa t ap and we omit justication of this approximation as do Bender
and Orszag. .. . Thus we use the approximation f t f a and p . Then Ix f aeixai/p where I sgn p
a. Let b Ix a f teixt dt.. The Method of Stationary Phase Ref. So we may ignore it for any xed
gt . of Bender and Orszag Theorem .
. Case gt . Then ei/p p u du /p p x ds and also . Case lt yields the other sign and C is which
must be rotated down. . As by analyticity of eu u p in the open doted region. x x xp ei/p e u u
p p du. . p u /p . As But e C u u p du eu u p du C eu u p du . put s ei/p /x/p u/p on the lower
half plane Imu . . As sp ei/ /xu iu/x. . x Case gt . converges to .where p a . . Hence as s
traverses the interval . As So u ixsp . The rst term on the right of . u /p The relevant
substitution is s ei/p x . on the positive x axis u traverses the contour C on the negative y
axis. TheAs singularity at the origin is not too bad. Put s ei/p Then sp ei/ Therefore eixai/ Ix f
a x u u i . Thus eixs ds ei/p /px/p C xp p eu u p du . . For p . As This is a well dened
continuous function of u in the shaded region if one chooses the positive pth root on the
positive x axis.
.. p. Step . u . and . Moreover f t ln t is singular at so integration by parts fails.. We take
advantage of this by deforming to a contour on which the imaginary part of is essentially
constant. . where we have put t u iT . Example Ix ln t eixt dt. Now along C . Thus eu u/p du C
/ ere r p ei i p p ird . r/p r/p As As As er cos d er/ dt r/p O r / / . . and the integral is along some
contour in the complex plane. As . . . Q. C as T . Moreover on C we may write u rei with r xp
and / .D. So for xed x gt . The Method of Steepest Descent b Objective To nd asymptotic
behavior of Ix a f text dt where f and are analytic. Example . eixt exT eixu . This will eliminate
the oscillation problem. a and b are complex or innite./p as x with exponentially small error.
Note that the method of stationary phase fails t t has no stationary point. Deform path from to
.E. x Now combine . Under these circumstances the contour can be deformed without
changing the value of the integral. From Bender and Orszag. Hence i i Ix ln t eixt dt ln t eixt
dt. As x/p O . .
To see this put t is. . Then i i lnisexs ds lniu/xeu du . It happens that the rst integral can be
done explicitly Put u sx. xn Note that this is asymptotic only and not convergent because the
series for ln is only converges in a neighborhood of s . s lt on the rst path and t is on the
second path to get Ix i lnisexs ds ieix ln isexs ds. i ln x i/ i . . This is all that is required for
aymptotic expansion. Thus i ln n i n Ix ieix in n as x . x i/x lni/x Eulers constant ln ueu du .
These are both Laplace integrals and so the problem is now in principle solved We have
gotten rid of the oscillatory integrals by deforming the path. in nxn un eu du in n . . Constancy
of phase and steepness of descent Consider an analytic function t t it where t u iv is a comb
plex variable. x x n .. x The second term has an asymptotic expansion via Watsons Lemma
ln ise xs ds n n n isn xs e ds n . We have seen that the asymptotic properties of a f text dt
.On these paths there is no oscillation of the exponential factor because it has constant
imaginary part on each path.
Case a. t . The Cauchy Riemann equations read u v , n u . So u u v v u v v u .
depends on the peaking property of . It is fortuitous that these peaking properties are optimal
along paths of constant as we shall now see. Let C be a curve of constant and let t C.
So is and therefore is parallel to C. Hence, among all directions starting from t, varies most
rapidly in the tangent direction of C. Thus ascends most rapidly in one direction along C and
descends most rapidly in the other direction along C. The level curves of are orthogonal to C.
So the curves of constant phase are also steepest curves for ext ext . In Example the
endpoint t of the constant phase path C is a point with t . Fortunately, with t u iv, t Re it v s in
that example reaches a maximum along C at the end point t . This was the source of our
success in the asymptotic expansion of C . Case b. t . Typical behavior of near such a point
is illustrated by the case t tn near zero, n , , . . . since the expansion of around a point t is t a t
t n c Ot t if k t , k , . . . , n and n t . Use polar coodinates t rei . Then tn rn ein rn cos n irn sin n
i. . . .
The directions of most rapid variation of are the directions in which cos n steepest ascent
and cos n steepest descent. As one increases one alternately reaches curves of steepest
ascent and steepest descent. In either case sin n , so these are precisely the curves of
constant phase emanating from the origin. In case n the graph of t is clearly saddle shaped
A point t t is called a saddle point in this business. See Bender and Orszag for more pictures.
In Example the maximum of along the curves C and C of constant phase occurred at the
endpoint of each path where . If a maximm of occurs at an interior point of a curve of
constant phase it will necessarily be a saddle point. This is a consequence of
CauchyRiemann equations. The following example illustrates this case. Example . Bender
and Orszag p. Asymptotic behavior of Bessel function J x. / Def. J x / cosx cos d/. Step . Get
this into standard form. Thus
/
J x Re
/
eix cos d.
To coincide with others notation, we also rotate the plane by putting i i t t t i. Now cos e e e e
cosh t. so J x Re i
ii
eix cosh t dt.
Step . Change to contour of constant phase. Now the phase is the same at the two endspoint
i cosh i i cos and i cosh i also. Nevertheless they do not actually lie on a common contour of
constant phase. To see what the constant phase contours are like write t u iv and t i coshu iv
icosh u cosh iv sinh u sinh iv icosh u cos v sinh ui sin v. So Imt cosh u cos v t and Ret sinh u
sin v t. Therefore the horizontal lines v / and v / are curves of constant phase . The curves
cosh u cos v are also curves of constant phase . One of them goes through . Thus t t it and J
x Re i t sinh u sin v
i/
t cosh u cos v
ext dt
i/
t u iv
Since Ret sinh u sin v, ext is very rapidly decreasing on the distant parts of contours C , C ,
C. Hence we may change from the original contour C to the three contours C , C , C, of
constant phase. Step . Evaluation. On C we have Imt , so C ext dt is real. Hence it
contributes nothing to J x. Similarly for C . Consequently we have J x Re eix cosh t dt. . i C
As we saw, t Rei cosh t sinh u sin v goes to as we got to in either direction on C. The
asymptotic behavior is therefore determined by any peaks in t on C. These are saddle points.
To nd them set t . Thus i sinh t . t is clearly a saddle point. It is the only one on C. Exercise.
Since the asymptotic behavior is determined by a neighborhood of on C we can approximate
C locally by a straight line whose slope is determined thus cosh u cos v . For small u and v
this is u v . So order. Therefore u v or u v. The curves of constant phase look like this.
As
uv
to second
This pattern repeats itself as one goes up the y axis n units. Thus if we parametrize locally on
C by t is ltslt
we can expect some kind of approximation to J x as x . Thus, to
in a power To get the full asymptotic expansion one now expands series valid for small r and
informally integrates term by term getting a valid asymptotic expansion. On C. . . Therefore J
x dt dr i sinh t i t i r ir dr it dr dr i r J x Re So i eix exr r Re ieix r ir dr exr r J x Re eix exr dr. As
To get the full asymptotic expansion of J x we must go back to . if one compares ReIx. So J x
Therefore cos x x . Now i cosh t i r so i sinh tdt dr. t i t where t is real. The nal form according
to Bender and Orszag.leading order J x Re s ixt e dt i s eix xs e Re ids i Re ieix x Reeix nx
Reeix x . and evaluate it more carefully. ir . x Note This agrees with the Example on p. . We
parametrize the curve by r t which goes from to on both halves of C.
.. Bender and Orszag. Reference for x and x. is J x where x cos x x sin x x k k x and x k k
kxk x k k k xk x . p.who get dr instead of my ir .
That is gt C tn tlt for some constant C and some integer n .. Find the asymptotic behavior as
x of Ix where Ix ln itext dt. Suppose further that g has an asymptotic expansion as t gt Let Ix n
an tn . Watsons Lemma Suppose that gt is a polynomially bounded function on . Prove that
Ix has the asymptotic expansion Ix n nan xn x .e. gtext dt x gt . x/ .. Problems on asymptotic
expansions Last homework of . Find the asymptotic behavior of J x as x up to and including
terms of order x i. . . . t . .
a. Math N. b. b. October . no series. Lux u x xu x x ux. Old Prelims Prelim in class Friday. u
u in two ways. Your writing style is a part of this test. Suppose that f .e. Write the Green
function for the boundary value problem u u f on . V C.B. as an eigenfunction expansion. F xx
dx . a Lux x u x xu x sin xux. . . . What can you say about xf xdx . . x x x b. R satises f x dx
and f xdx . F x . Use a writing style which leaves no ambiguity as to what your argument is.
Which of the following are linear functionals on the given vector space V a.Which of the
following dierential operators are symmetric on the interval . in closed form i. V R .
If m itself were odd then m would also be odd because k k k . Appendix Review of Real
numbers. Set notation These notes are intended to be a fast review of elementary facts
concerning convergence of sequences of real numbers. So there is no rational number x
satisfying x . But it is a universally accepted convention not to call them real numbers. Both
are elds. subtraction.e. So m is even i.. etc. . which is in fact nicely related to the algebraic
operations. /. Lim Sup. partial dierential equations.g. etc. The big dierence between Q and R
is that Q is full of holes. . a lt b implies ac lt bc. Addition. Denote by Q the set of rational
numbers and by R the set of real numbers. Therefore j n . We say that x is then represented
in lowest terms. We will be using the symbols and in this course quite a bit. R.E. Notation .
Sequences. We will do this also. Moreover Q and R both have an order relation on them.
Limits. I... Example of an irrational number.e. but it may have been a while. The main ideas
in these notes are usually taught in a freshman calculus course.D. It now follows that n is
also even. So m must be even. The goal is to give a self contained exposition of the notion of
completeness of the real number system. So they are not in R. Some of these ideas are
developed over and over again in later courses. For example the sequence of rational
numbers . This contradicts our assumption that m and n have no common factors. Q. is a
multiple of . eigenvalue problems. multplication and division but not by zero make sense
within Q and also within R. E. Chances are that this is at least the third time around for most
of you . Say m j.. The suare root of is not rational. . So j n . If x m/n is a rational number it is
always possible to choose m and n to have no positive integer factors in common.e. a lt b
has a well dened meaning. and almost everything else that guides our intuition in analysis.
Then m n . /. /.Proof Assume that x m/n is represented in lowest terms and that x . Recall that
a rational numbber is a number of the form m/n where m and n are integers i. Completeness
is responsibe for the existence of solutions to ordinary dierential equations. Some authors
like to adjoin them to R and then call the result the extended real number system R . But R
still means the set of real numbers and excludes . whole numbers. R .
if S is a nonempty set of real numbers which is bounded above then S has a least upper
bound in R. x M for all x in S. If sn is a sequence then tk is a subsequence if there is a
sequence integers nk such that n lt n lt n . We write L lubS. A sequence is a function from the
set Z . either in the following form or some equivalent form. If L is an upper bound for S such
that L M for all upper bounds M of S then L is called the least upper bound for S. would like
to converge to / . .which any denition of the real numbers must include.order completeness.
There is a hole there. . What is it Denition.. But S does have a least upper bound in R.
Chapter . If M is a number such that x M for all x in S then M is called an upper bound for S. .
of course. if you really want one. The next few pages are devoted to making this
anthropomorphic discussion precise. The set Q of rational numbers is not order complete.
Denition. a denition of them. Principles of Analysis. But no axiomatic treatment of the real
number system will be given here. . and tk snk . to R. In order to prove anytning about the
real numbers one needs . Denition. n . Remark. L lim sn or sn L as n . Notation. . S is
bounded above if there is a real number M such that x M for all x in S. Instead I am just going
to list the key property . Theorem on page says that there are no holes in R. For example if S
x in Q x lt then S does not have a least upper bound in Q. Denition. . But / isnt there in Q.
Notation si si . Denition. .. S is bounded below if for some M in R. R is order complete that is.
See Rudin.The key concept is that of a Cauchy sequence.. A set S R real numbers is
bounded if there is a real number M such that x M for all x in S. A sequence sn is said to
converge to a number L in R if for each gt there is an integer N such that sn L lt whenever n
N. Denition.
Thus the sequence ak is a decreasing sequence. . Hence sn L lt if n N . ak or b For all k.
Since sn n k sn n k we have ak ak . is not bounded below. k Using the preceding proposition
and denitions. n Proof Let L lubsn . Then sn L for all n. Write out. . But sn L . .
Notation.Proposition . Remarks. ak is nite andak k . Notation. Thus there exists N such that n
sN gt L . .. . If S is a nonempty set of real numbers which is not bounded above we write lub
S . Hence sn L gt . Exercise. For any sequence sn we write lim sn if for each real n n number
M there is an integer N such that sn M for all n N. n Now let sn be an arbitrary sequence of
real numbers. then S is bounded above. we see that either a For all k. but dont hand in Prove
glbS lubS for any nonempty set S. Let gt . Set n ak lubsn n k. If S is bounded below. . .e.
Greatest lower bound is dened analogously to least upper bound. If sn is an increasing
sequence of real numbers bounded above then lim sn exists and equals lubsn n . Similar
denition for lim sn . Since sn sN for all n gt N we have sn gt L for all n gt N .. ak is nite i.
Denition. glb S greatest lower bound of S. If S is nonempty and not bounded below we write
glb S . real and the set ak is bounded below k or c For all k. By S we mean x x S. . Then n L
is not an upper bound for sn . .
e. But note that n k infsn n k supsn n k. If lim sn L exists and is real i. sup stands for
supremum and means the same thing as lub. ak supsn n k L . k Proposition . Since the
inequality holds for all gt and the left side is independent of we have lim ak L. Conversely if
lim sup sn lim inf sn and both are nite then lim sn exists and is equal to their common value..
Hence lim ak L . In case c we write n lim sup sn . Proof Assume lim sn exists and denote it
by L. L . Thus n k lim sup sn L. k In case a we write n lim sup sn . Remarks. Thus for all k N .
inf stands for innum and means the same as glb. N depending on such that sn n N L . Let gt .
n k Similarly we dene n lim inf sn lim infsn n k. The preceding denition of lim sup sn can be
written succinctly in all cases as lim sup sn lim supsn n k. Similarly lim inf sn L. n . nite then n
lim sup sn L lim inf sn .In case b lim ak exists and is real and we write k n lim sup sn lim ak .
Hence one always has for any sequence n lim inf sn lim sup sn .
Q. So sk L lt k N. lim sup sn lim inf sn L n L is assumed nite. i. sn sm lt n. m N.e. m N.
Similarly N bk L lt whenever k N . Let gt . N .E. In particular L lt aN bN lt L . A sequence sn is
CAUCHY if for each gt N n propR Q.E. . as usual such that ak L lt whenever k N . n Proof
Given gt N sn L lt / n N.D. Then ak L lt and bk L lt if k N . Proof of converse.. Denition. . real.
Thus L lt aN sk bN lt L k N. Any convergent sequence sn of real numbers is Cauchy. There is
an integer N depending on . .But we saw that n lim sup sn L lim inf sn . Then sn sm sn L L
sm sn L L sm lt / / if n.D. n Hence we have equality. Let ak infsn n k and bk supsn n k.
Proposition . Then lim ak lim bk L. Thus L lt ak bk lt L if k N . Assume n Q. .E.D. Let N maxN .
. The set of rational numbers is not complete.E. . . The property stated in Theorem is usually
referred to as the completeness of the set R of real numbers. E.Theorem .. sN . n Proof of
Theorem Let sn be a Cauchy sequence.g. Hence if n N then sn L sn sN sN L n N. I. m N .
Thus if ak supsn n k then sN ak sN k N. Set Theory Notation Notation. s . Choose N such
that sn sm lt n. Every Cauchy sequence of real numbers converges to a real number. s . The
set sn n . Thus if we put K maxs .. . So sN L . . If A and B are subsets of a set S then the
intersection of A and B. s . Then sN lt sn lt sN n N. n Step . . . A B.. s . But by denition L limk
ak .e. . there is a Cauchy sequence sn in Q which does not have a limit in Q. is bounded
because if we choose in the denition of Cauchy sequence then we know that there is an
integer N sn sN lt if n N . take s . We now know that lim sup sn is a real number. so that sn ..
. . . sN then sn K for all n. For let gt . . Thus taking the limit as k we get sN L sN . Hence sn
sn sN sN sn sN sN sN if n N . Remarks. is dened by A B x S x A and x B. . Step . Let L Claim
lim sn L. which is not in Q. Q.. n lim sup sn .D.
. / The dierence is If f S T is a function and B T then f B is dened by f B x S f x B.The union
of A and B is dened by A B x S x A or x B The complement of A is Ac x S x A. A B x A x B. /
or both.
. it depends on what you mean by the symbol .says. Remove ambiguity. Moreover there is
already a standard symbol for This imples that. not backslashThis implies. says implies that
.then statement If then . All you have to do is add to both sides of the equation to deduce
correctly that . Rule . Its equivalent to the standard if . Rule . Remove ambiguity. C . now
consists of two sentences. So what to do In principle you could dene a symbol to mean
anything you want but only one meaning. Here are some examples of ambiguity. when
translated into english. Remove ambiguity. So C C statement . Even the TeX command for is
backslashimplies. This implies that. If this is what you mean by the symbol C then the
statement . Meaning . implies that. C . It would be easiest on readers to use the symbol
always to mean implies or implies that which has the same logical content. If this is what you
mean by the symbol C then the statement . Meaning . Consider the statement .. Example . C
Is this statement true Well. . is FALSE. Its the symbol . C The statement . is TRUE. . The
three most important rules are Rule . But the symbol is quite universally used to mean
implies that. is false and therefore statement . This implies that . C In this case statement .
How to communicate mathematics Before starting the following homework here are some
useful tips on how to write for a reader other than yourself. one of which is false.
Keep in mind. expect trouble from him and me. Consequently. . Hence. A good scientist
knows how to say exactly what he means. Here is some ancient wisdom. A nal word.
COMMUNICATION IN MATHEMATICS CONSISTS OF A BUNCH OF GRAMMATICALLY
COMPLETE SENTENCES. . All of this was well understood quite some time ago. equals.
Often linked with the evils of the misuse of the symbol is the failure to punctuate sentences.
More often than not this kind of thinking just means that you expect the reader to interpret
correctly what you wrote because he/she already knows what the argument should be. Lezar
el Gralidin. and an object. Yes. really is an english sentence. . We have to do better. But
some readers might question the quality of the highschool you went to. So. If your TA cant
gure out what you mean. even a period can help clarity in communication of mathematics.
AND A PERIOD at the end of the sentence. not the symbols. can be translated as This
implies that. Here are some of them. Dont think that context is a reasonable basis for a
reader to decide how to interpret an ambiguous statement.This symbol can be translated into
English in a number of logically equivalent ways. A good philosopher knows how to say
things with no exact meaning. legal and political communication. When writing for a journal
you have to use the words. for example. A good politician knows how to say exactly the
opposite of what he means. It has a subject. even before he tries to gure out if youre right.
Therefore. a verb. Its still far from perfect in normal social. Moreover good prose form is
better if you dont use the same one of these ve phrases at the beginning of several
sentences in a row. that the statement x . The evolution of grammar over the last years has
been aimed at precision of communication. Technically you could be correct in starting ve
sentences in a row with the word Therefore. x.
The function g. . Find the lim sup and lim inf for the following sequences a . If the lim sup of
the sequence sn is equal to M . . prove that lim sup sn lim inf sn . Write the set of all rational
numbers in . Find at least seven errors in mathematical communication in the following six
statements and explain whats wrong in each case. . r . b. . n n . is zero at x . Also n prove
that no subsequence of sm can converge to a limit less than m. . . . prove a f A B f A f B c f
Ac f Ac b f A B f A f B . Show by example that f A B f A f B is not always true. which is
increasing. . . . Calculate lim sup rn and lim inf rn . . Cute Homework Problems . If sn . Prove
that A B C A B A C. b sinn/ n c /n cos n n d /nn n . . prove that there is a subsequence of sn
which converges to m. a. m . . gx x c. . as r . Show by example that f Ac f Ac is not always
true. If sn is a bounded sequence of real numbers and lim inf sn m. prove that the lim n sup
of any subsequence is M . n n . n n . . and A and B are subsets of Y . If f X Y is a function. ..
Let f x x for all real x..
d. so that it is grammatically AND mathematically correct AND unambiguous. e. The function
which is increasing is less than zero when x lt f. Rewrite item f. . x g. f x gx. The function g
which is increasing is greater than zero when x gt .