FUNCTION SPACES
Analysis of systems of linear algebraic equations leads naturally to the notion of a linear
space in which vectors (i.e., the unknowns in the problem) and formation of linear
combinations (i.e., the algebraic conditions imposed on the unkowns) are the principle
ingredients. Considering linear systems in this setting of a linear space results in efficiency
of expression and economy of effort. We are motivated to try to discover similar structures
for the analysis of partial differential equations. Here, of course, functions are going to play
the role played by vectors in linear algebra and the matrices will be replaced by partial
differential operators. An additional difference is that the spaces of functions will be
required to be closed under passage to the limit besides being closed under the operation
of forming linear combinations.
1. Spaces of Functions
A linear space X is a collection of objects x, y, z, . . . together with a set of scalars A, B, C. . .
and two operations:
i) Addition ∀x, y ∈ X, x + y ∈ X
ii) Scalar Multiplication ∀x ∈ X, Ax ∈ X for all scalars A
Examples of linear spaces are familiar from linear algebra and algebra. These spaces are
generally examples of finite dimensional linear spaces. For 1 ≤ p < ∞, let L p U denote the
linear space of functions which are defined and whose p-th power is integrable on an open
set U ⊂ R n . That is, the functions which are defined on U and for which
∫I | fx| p dx < ∞.
1. 1
The open set U can be bounded or unbounded and the functions clearly need not be
continuous in order for (1.1) to be satisfied. In fact the set of singularities permitted for
functions in L p U is quite large and, as a result, the integral used must be more general
than the Riemann integral. However, for what we propose to do, we will not require any
detailed knowledge of a more general integration theory.
The linear structure on L p U is defined by
∀f, g ∈ L p U, ∀A, B ∈ R, let
Af + Bgx = Afx + Bgx
∀x ∈ U
1. 2
That L p U is a linear space is the assertion of,
Proposition 1.1 ∀f, g ∈ L p U, ∀A, B ∈ R,
This result follows from the inequality
for each p, 1 ≤ p < ∞,
Af + Bg ∈ L p U.
|A + B| p ≤ 2 p−1 |A| p + |B| p . ∀A, B ∈ R.
Note that the scalars here are taken to be the real numbers.
A real valued function, N⋅, defined on a linear space X is a norm if it has the following
properties:
1.
2.
3.
NAx = |A| Nx
∀A ∈ R, ∀x ∈ X
Nx + y ≤ Nx + Ny ∀x, y ∈ X
Nx ≥ 0 ∀x ∈ X and Nx = 0 iff x = 0
1
A norm is defined on L p U, 1 ≤ p < ∞, by
|| f || p =  ∫U | fx| p dx 1/p ,
for f ∈ L p U.
Then L p U becomes a normed linear space. That Nf = || f || p satisfies 1 and 3 is clear but
2 requires proof. Using the inequality
p
q
a b ≤ ap + bq
∀a, b ∈ R, and for
1 + 1 = 1,
p
q
1. 3
leads to
Proposition 1.2 ( Holder and Minkowski inequalities)
(a) If
1 + 1 = 1, then
p
q
||f g|| 1 ≤ || f|| p || g|| q
(b) If 1 ≤ p < ∞, then ∀f ∈ L p U
∀f ∈ L p U, ∀g ∈ L q U
||f + g|| p ≤ || f|| p + || g|| p .
We should be careful to notice that || f − g|| p = 0 does not imply that fx = gx at all x ∈ U.
In fact, f and g can differ at infinitely many points in U so long as the set of points contains
no positive volume subsets (i.e., f and g can differ on ”sets of measure zero”). This means
that functions in L p U are defined only up to sets of measure zero and that altering a
function in L p U on such a set does not produce a different function in L p U. Then
functions in L p I are, in fact, equivalence classes of pointwise defined functions where the
equivalence relation is equality ”almost everywhere”. When we say ”let f be an arbitrary
function in L p U" , what we really mean is ”let f be an arbitrary representative of one of
the equivalence classes of functions in L p U" .
Convergence
To say a sequence f n x in L p U converges of to a limit f in L p U means
||f n − f|| p → 0 as n → ∞. This type of convergence is weaker than uniform convergence on U
and neither implies nor is implied by pointwise convergence on U. Every sequence f n x
in L p U that is convergent must be a Cauchy sequence; i.e., it must satisfy
||f n − f m || p → 0 as m, n → ∞. In fact, it is true in any normed linear space that every
convergent sequence must be a Cauchy sequence. On the other hand, in an arbitrary
normed linear space it is not necessarily the case that every Cauchy sequence converges
to a limit that belongs to the space. A space with the property that every Cauchy sequence
is convergent is said to be complete. The following result, known as the Riesz-Fischer
Theorem in analysis, asserts that L p U is complete. The proof requires techniques of an
integration theory more general than Riemann integration.
Proposition 1.3 (Riesz-Fischer) For 1 ≤ p < ∞, L p U is complete for the norm || f || p .
For each p, 1 ≤ p < ∞, L p U is a complete, normed linear space. Such spaces are called
Banach spaces. We shall be most interested at this point in the special cases of L 1 U, the
absolutely integrable functions and L 2 U, the square integrable functions . The space L 2 U
enjoys some properties not shared by L 1 U.
2
2. Inner Product Spaces
An inner product space is a linear space X, on which there is defined a mapping which
associates to every pair of elements x, y ∈ X, a scalar value which we denote by x, y X .
This mapping must have the following properties
1 x, y X = y, x X
∀x, y ∈ X
2 Ax + By, z X = Ax, z X + By, z X , ∀x, y, z ∈ X, ∀A, B ∈ R,
3 x, x X ≥ 0 ∀x ∈ X, and x, x X = 0 iff x = 0
2. 1
The mapping is called an inner product on X.
The most familiar example of an inner product space is the space R n of n-tuples
⃗
x = x 1 , . . . , x n , where the inner product is defined as
y  R n = ∑ i=1 x i y i .
x⃗, ⃗
n
2. 2
An inner product space has a norm, induced by the inner product. That is,
|| x|| X = x, x 1/2
X
for x ∈ X
2. 3
defines a norm on the linear space X. Recall that the norm defines a meaning for distance
in the linear space X. In any inner product space, the following results are valid.
Proposition 2.1 (Cauchy-Schwartz and Triangle inequalities )
a) |x, y X | ≤ ||x|| X ||y|| X
b) || x + y|| X ≤ ||x|| X + ||y|| X ,
∀x, y ∈ X
∀x, y ∈ X
A Cauchy sequence in the inner product space X is a sequence of elements x m  ⊂ X
with the property that || x m − x k || X → 0 as m, k → ∞. A sequence of elements x m  ⊂ X is
said to be convergent if there exists an element x ∈ X such that ||x m − x|| X → 0 as m → ∞.
The inner product space X is said to be complete if every Cauchy sequence in X is
convergent. It is well known that the inner product space R n is complete (this is just a
consequence of the fact that the real numbers have been constructed to be complete).
We now consider two additional examples of inner product spaces. The first example is not
a function space but will be important later.
The Space of Square Summable Sequences
Let ℓ 2 denote the linear space of infinite sequences of real numbers ⃗
x = x 1 , x 2 , . . . . This
space carries the same linear and inner product structure as R n if we define
Ax⃗ + By⃗ = Ax 1 + By 1 , Ax 2 + By 2 , . . . 
∞
y  2 = ∑ i=1 x i y i
x⃗, ⃗
and
∀x, y ∈ ℓ 2 ∀A, B ∈ R
x || 2 = x⃗, ⃗
x  1/2
|| ⃗
2
2. 4
2. 5
Of course the inner product and norm have to be restricted to those sequences for which
the infinite sums are finite. Such sequences are said to be square summable sequences
and we use the notation ℓ 2 to indicate the linear space of all such sequences. It is
straightforward to show that the dimension of the space ℓ 2 is infinite. In fact, by showing
there is a basis for ℓ 2 which is in one to one correspondence with the natural numbers, it
3
follows that the dimension of this space is equal to the cardinality of the natural numbers.
Proposition 2.2 Every Cauchy sequence in ℓ 2 is convergent.
Proof- Let x⃗n  denote a Cauchy sequence in ℓ 2 . Then
2
x n || 2 = ∑ i x m
− x n
||x⃗m − ⃗
i
i
2
→ 0 as m, n → ∞.
This implies that for every  > 0, ∃ N  such that
2
x m
− x n
i
i
<  for all m, n > N  .
Then for each i, x m
is a Cauchy sequence in R, and since R is complete
i
m
⃗ = X 1 , X 2 , . . . 
x i → X i as m → ∞, for some real number X i . It remains now to show that X
⃗ in ℓ 2 .
belongs to ℓ 2 and that x⃗n  converges to X
2
M
x m
− x n
∑ i=1
i
i
For m, n > N  ,
∞
+ ∑ i=M+1 x m
− x n
i
i
2
< , for arbitrary M
and since each sum is nonnegative,
2
M
x m
− x n
∑ i=1
i
i
< .
Now fix m and let n tend to infinity. This leads to
M
x m
− Xi
∑ i=1
i
2
< ,
∀m > N  and every M > 1.
Since M is arbitrary, let M tend to infinity to conclude that for every  > 0 there is an N  such
that
∞
x m
− Xi
∑ i=1
i
2
< ,
∀m > N  .
Then it follows that
⃗−X
⃗ m
X
2
→ 0 as m → ∞,
and
⃗
X
2
≤
⃗−X
⃗ m
X
2
+
⃗ m
X
2
≤ +
⃗ m
X
2
⃗ ∈ ℓ 2 ).■
< ∞ (so X
⃗ m in ℓ 2 where E m
It should be noted that the set of vectors E
= δ im , forms a basis for
i
ℓ 2 in the sense that for any ⃗
x ∈ ℓ 2 , the sequence of elements
m
⃗ i
⃗
x m = ∑ i=1 ⃗
x, E
2
⃗ i
E
is a Cauchy sequence in ℓ 2 converging to ⃗
x as m → ∞. There is, in general, no finite
⃗ ′ s which equals ⃗
combination of E
x.
The Function Space L 2
In the special case, p = q = 2, the Holder inequality looks like the Cauchy-Schwartz
inequality. In fact, in this special case, L p U = L 2 U is an inner product space. The inner
product on L 2 U is defined by
f, g 2 = ∫U fx gx dx,
and
|| f || 2 = f, f 1/2
2
2. 6
L 2 U is the only one of the L p U spaces that supports an inner product. Since it is an
L p U space, it has all the relevant properties such as completeness.
4
Two elements in an inner product space are said to be orthogonal if their inner product
is zero. In L 2 U this has no visualizable significance (e.g. it does not mean the graphs of
two orthogonal functions are orthogonal trajectories). However, orthogonality in L 2 U does
imply linear independence. Since it is possible to generate an infinite family of orthogonal
functions in L 2 U, it follows that L 2 U is infinite dimensional. An infinite dimensional inner
product space that is complete in the norm induced by the inner product is called a Hilbert
space. Note that if U is bounded, so that ∫ 1 dx = |U| < ∞, it follows that L 2 U is contained
U
in L 1 U To see this suppose f ∈ L 2 U and write
|| f|| 1 = ∫U | f| dx = ∫U 1 | f| dx ≤ ||1|| 2 || f|| 2
When U is not bounded, then neither space is contained in the other. For example consider
the functions
1/x if
fx =
0
x>1
and
if x < 1
gx =
0 < x < 100
1/ x if
0
otherwise
.
Then f belongs to L 2 R but does not belong to L 1 R, while g belongs to L 1 R but does
not belong to L 2 R. These two spaces have many functions that are in both spaces but
each contains some functions that are not in the other.
3. The Fourier Integral Transform
For functions which are defined on all of R n we can define an alternative representation for
the function. This alternative representation is called the Fourier transform of the function.
The Fourier transform of the everywhere defined function fx⃗ is defined as follows
⃗  = 2π −n ∫ n fx⃗ e −i ⃗x⋅α⃗ dx.
T F f = Fα
R
Here, we will restrict our attention to the one dimensional case where we have
T F f = Fα = 2π −1 ∫ fx e −i xα dx
R
3. 1
The notations T F f and Fα will be used interchangeably to denote the Fourier transform of
the function f(x). Note that the transform does not exist for any function f(x) for which the
improper integral (3.1) fails to converge. If the integral converges, it defines a possibly
complex valued function of the real variable α. Evidently, a sufficient condition for the
Fourier transform to exist is that f is absolutely integrable, i.e., f ∈ L 1 R.
Example 3.1 Some Fourier Transforms
(a) Consider
fx =
1 if
|x| < 1
0 if
|x| > 1
Then f ∈ L 1 R and
−i xα
1
α
Fα = 1 ∫ fx e −i xα dx = 1 ∫ e −i xα dx = e
| x=1 = sin
πα
2π R
2π −1
−i2πα x=−1
(b) Consider fx = e −|x| . ∈ L 1 R. Then
5
∞
0
∞
Fα = 1 ∫ fx e −i xα dx = 1 ∫ e −|x| e −i xα dx = 1 ∫ e x−i xα dx + 1 ∫ e −x−i xα dx
R
−∞
−∞
0
2π
2π
2π
2π
x1−iα
−x1+iα
1
1
= 1 e
| 0−∞ + e
|∞  = 1  1 + 1  = π
2π 1 − iα
1 + iα 0
2π 1 − iα
1 + iα
1 + α2
where we used the result
e −x1+iα = 0.
lim e x1−iα = lim
x→∞
x→−∞
(c) The Gaussian function fx = e −x belongs to L 1 R and
2
∞
2
Fα = 1 ∫ fx e −i xα dx = 1 ∫ e −x e −i xα dx
2π R
2π −∞
∞
∞
2
2
2
2
2
= 1 ∫ e −x +ixα−α /4+α /4 dx = e −α /4 1 ∫ e − x+iα dx
2π −∞
2π −∞
But
∞
∞
∫ −∞ e − x+iα 2 dx = ∫ −∞ e −z 2 dz = π
hence
Fα =
1 e −α 2 /4 = T F e −x 2 .
4π
Each of the functions in this example belongs to L 1 R and each transform is a continuous
function of the transform variable α. In addition, it appears that the transform tends to zero
as α → ±∞. In fact, this is true in general. We will use the notation F ∈ C 0 R to indicate
that F is continuous on R 1 and Fα tends to zero as |α| tends to infinity. Note that a function
in C 0 R is not necessarily absolutely integrable. The function pairs  fx, Fα listed
above are examples of Fourier transform pairs.
Theorem 3.1 If f ∈ L 1 R then Fα = T F fx exists and is a continuous function of
α ∈ R 1 . Moreover, F ∈ C 0 R.
The proof of this theorem requires the dominated convergence theorem for Lebesgue
integrals and it therefore omitted. We list now several useful properties of the Fourier
transform.
Theorem 3.2 If fx and gx have Fourier transforms Fα, Gα, respectively, then
1.
2.
T F Afx + Bgx = AFα + BGα
T F fbx = 1 F α
∀b ≠ 0
b
| b|
∀A, B ∈ R
A transformation with property 1 is said to be linear and one with property 2 is said to be
homogeneous.
Problem 1 Use the definition of the Fourier transform to prove theorem 3.2
Problem 2 Use theorem 3.22 to show that
(a) T F I A x =
sinαA
πα
where
I A x =
1 if
|x/A| < 1
0 if
|x/A| > 1
=
1 if
|x| < A
0 if
|x| > A
6
b
1
(b) T F e −b|x|  = π
for b > 0
b2 + α2
2
1 e −α 2 /4b
for b > 0
(c) T F e −bx =
4πb
Theorem 3.3 If fx has Fourier transform Fα, then for all real values c the following
transforms exist and are related to Fα as indicated
1.
2.
T F fx − c = e −icα Fα
T F e icx fx = Fα − c
Problem 3 Use the definition of the Fourier transform to prove theorem 3.3
Theorem 3.4 If both fx and f ′ x have Fourier transforms then the transform of the
derivative, f ′ x, is given in terms of Fα = T F  f by
T F  df/dx  = iαFα.
More generally, if fx and all its derivatives up to order m have Fourier transforms, then
T F  d k f/dx k  = iα k Fα for k = 1, 2, . . . , m
Theorem 3.5 If both fx and xf x have Fourier transforms then the transform of x f x, is
given in terms of Fα = T F  f by
T F  xfx  = i d/dαFα.
More generally, if fx and x k fx have Fourier transforms, for k = 1, 2, . . . , m then
T F  x k fx = i d/dα k Fα
for k = 1, 2, . . . , m
Problem 4 Use the definition of the Fourier transform to prove theorem 3.4
Problem 5 Use the definition of the Fourier transform to prove theorem 3.5
For functions fx and gx defined on R, we formally define the convolution product of f
and g as
f ∗ gx = ∫ fx − y gy dy
R
3. 2
Problem 6 Use the definition of the convolution product and the change of variable,
z = x − y, to show that f ∗ gx = g ∗ fx.
Theorem 3.6 If fx and gx belong to L 1 R, then f ∗ g ∈ L 1 R and
T F f ∗ g = 2πFα Gα.
The L 1 R Inversion Theorem
We have seen that for each f ∈ L 1 R, there is a Fourier transform F ∈ C 0 R. In some
7
sense, knowledge of one of these functions is equivalent to knowledge of the other; i.e.,
they are two different representations for the same information. To make this assertion
more precise, we have the next theorem.
Theorem 3.7 If fx belongs to L 1 R, and, in addition, the Fourier transform F is also in
L 1 R, then
φx = ∫ Fα e i xα dα ∈ C 0 R
R
and
|| f − φ|| L 1 = 0.
The assertion of this theorem is that if f and F both belong to L 1 R, then the inverse
Fourier transform of F is defined by
i xα
T −1
dα
F Fα = ∫ Fα e
R
3. 3
and T −1
F Fα = fx where the equality is equality in the sense of L 1 R. Then 3. 3 is
known as the Fourier inversion formula. By comparing (3.3) with (3.1), we arrive at the
following result which can be used to increase the number of transform pairs.
Theorem 3.8 If fx belongs to L 1 R, and, in addition, the Fourier transform F is also in
L 1 R, then
T F Fα = 1 f−α.
2π
Problem 7 Use theorem 3.8 and previous transform pairs to show that:
a T F
sinAx
πx
= 1 I A −x = 1 I A x
2π
2π
b T F
2b
b2 + x2
= e −b|α|
c T F e −x
2 /4b
=
b −bα 2
e
π
for b > 0
for b > 0
Symmetry
Symmetry in the graph of the absolutely integrable function fx implies certain related
symmetries for the Fourier transform, Fα. Since, in general, fx may be complex valued,
we will write it as the sum of real and imaginary parts, each of which is then the sum of an
odd and an even function,
fx = Re f E x + Re f O x + i Im f E x + Im f O x.
Now
so
e −ixα = cosxα + i sinxα
1 ∫ ∞  Re f E x cosxα + Im f O x. sinxα dx
Fα = π
0
∞
i
+ π ∫  Im f E x cosxα − Re f O x. sinxα dx.
0
Clearly Fα can be complex valued even when fx is real valued, but in any case there are
8
certain correspondences between the evenness or oddness of f and that of F:
∗∗∗ f∗∗∗
even
complex
even
odd
∗∗∗ F∗∗∗
complex
odd
real
even
imaginary
real
odd
imaginary
even
odd
real
imaginary
imaginary
real
4. The Fourier Transform in L 2 R
Note that the functions
f 1 x = I A x,
f 2 x = e −b|x| ,
and
f 3 x = e −bx
2
all belong to L 1 R and each has a Fourier transform in C 0 R by theorem 3.1. Note further
that these transforms
F 1 α =
sinαA
πα ,
b
1
F 2 α = π
b2 + α2
and
F 3 α =
1 e −α 2 /4b
4πb
all belong to L 2 R, but only F 2 and F 3 belong to L 1 R. Then only F 2 and F 3 can be
inverted using Theorem 3.7 to recover the functions f 2 and f 3 . So in what sense is it
possible to say that the inverse transform of F 1 is f 1 ?
If f ∈ L 1 R then its Fourier transform exists and belongs to C 0 R. For an arbitrary
function f ∈ L 2 R, it is not necessarily the case that f ∈ L 1 R and it is not clear then that
its Fourier transform exists. In order to extend the Fourier transform to L 2 R we will use an
idea that is pervasive in the study of partial differential equations. We first introduce a
special subspace of L 2 R where transforming and inverting works more smoothly. This
subspace has the additional property that the results which hold on the subspace can be
extended to the whole space, L 2 R, by passing to the limit. We define this subspace as
follows.
The Space of Test Functions A function φx defined on R is a test function if φ is
infinitely differentiable and vanishes on the complement of some closed bounded interval
a, b; i. e. , φ has compact support. We denote the linear space of test functions by C ∞c R. It
is obvious that C ∞c R ⊂ L 1 R and C ∞c R ⊂ L 2 R but a more surprising fact is true.
Theorem 4.1 For every f ∈ L p R, 1 ≤ p < ∞, there exists a sequence φ n  ⊂ C ∞c R
such that ||φ n − f|| p → 0 as n → ∞.
We describe this by saying that C ∞c R is a dense subspace of L p R. The proof of this
theorem as well as additional information about test functions will be given later when we
develop the theory of generalized functions. We now proceed to use the test functions to
9
extend the Fourier transform to L 2 R.
Theorem 4.2 For every φ ∈ C ∞c R, the Fourier transform, Φ, exists and, in addition, Φ
belongs to L 2 R. Moreover,
|| φ|| 22 = 2π|| Φ|| 22
4. 1
It follows now from theorems 4.1 and 4.2 that
1) for every f ∈ L 2 R, there exists a sequence φ n x ∈ C ∞c R such that
||φ n − f|| 2 → 0 as n → ∞.
(hence
||φ n − φ m || 2 → 0 as m, n → ∞ )
2) since φ n ∈ C ∞c R, the Fourier transforms, Φ n , exist and,
in addition, Φ n ∈ L 2 R. Moreover,
|| φ n − φ m || 22 = 2π|| Φ n − Φ m || 22 → 0 as m, n → ∞
3) since L 2 R is complete, and the sequence Φ n  is Cauchy, there exists
a unique v ∈ L 2 R such that
|| Φ n − v|| 2 → 0 as n → ∞
4) since φ n → f, Φ n → v in L 2 R, and Φ n = T F φ n , we define v = T F  f
Note that we can also invert the transformation as follows,
1) for every F ∈ L 2 R, there exists a sequence Ψ n x ∈ C ∞c R such that
||Ψ n − F|| 2 → 0 as n → ∞.
(hence
||Ψ n − Ψ m || 2 → 0 as m, n → ∞ )
2) since Ψ n ∈ C ∞c R, the inverse Fourier transforms,
ψ n x = ∫ Ψ n α e i xα dα
R
exist and, in addition, ψ n ∈ L 2 R. Moreover,
|| ψ n − ψ m || 22 = 2π|| Ψ n − Ψ m || 22 → 0 as m, n → ∞
3) since L 2 R is complete, and the sequence ψ n  is Cauchy, there exists
ψ ∈ L 2 R such that
|| ψ n − ψ|| 2 → 0 as n → ∞
−1
4) since Ψ n → F, ψ n → ψ in L 2 R, and ψ n = T −1
F Ψ n , we define ψ = T F  F
5) since T F f − ψ = F − F = 0, and || f − ψ|| 22 = 2π||T F f − ψ|| 22 = 0, we have f = ψ.
We summarize these observations in the following theorem.
Theorem 4.3 For every f ∈ L 2 R there exists a unique F ∈ L 2 R such that F = T F  f in
the sense that
if φ n x ∈ C ∞c R is such that ||φ n − f|| 2 → 0 as n → ∞,
10
then Φ n = T F φ n  ∈ L 2 R is such that ||Φ n − F|| 2 → 0 as n → ∞,
In addition T −1
F F = f, where the equality is in the sense of L 2 R.
Problem 8 Show that the definition of the L 2 R Fourier transform F does not depend on
the choice of the sequence φ n 
Each of the Fourier transform properties detailed in theorems 3.2 to 3.6 holds for the
L 2 R extension of the Fourier transform. In addition, we have
Theorem 4.4 For f, g ∈ L 2 R the Fourier transforms, F, G satisfy
f, g 2 = ∫R fx g ∗ x dx = 2π ∫R Fα G ∗ α dα = 2πF, G 2
4. 2
Here g ∗ , G ∗ are used to denote complex conjugates. Even though we are dealing
exclusively with real valued functions f, g the Fourier transforms may be complex valued and
we have therefore stated the result for the complex form of the inner product on L 2 R. Note
that when f = g (4.2) reduces to (4.1). The result (4.1) is known as the Parseval relation
and (4.2) is called the Plancherel relation. Together they assert that the Fourier transform is
a Hilbert space isometry, meaning that T F maps L 2 R onto itself in a one to one norm and
inner product preserving fashion.
5. Some Applications of the Fourier Transform
Consider the functions
gx =
1 − | x|
if
| x| < 1
0
if
| x| > 1
px =
and
| x|
0
if
if
| x| < 1
| x| > 1
Both g and p belong to L 2 R, in fact g is continuous with compact support while p has
compact support but is discontinuous at | x| = 1. Neither function is differentiable in the
classical sense. Since g and p belong to L 2 R, they have Fourier transforms in L 2 R,
given by
Gα = 22 1 − cos α,
α
Pα = 22 α sin α − 1 − cos α.
α
We observe that iα Gα ∈ L 2 R, but iα Pα ∉ L 2 R. Since iα Gα = T F g ′ x we
conclude that g has a derivative in the sense of L 2 R and this derivative is given by
fx = T −1
iα Gα
F
=
sgnx
if
| x| < 1
0
if
| x| > 1
.
On the other hand, iα Pα ∉ L 2 R implies that p does not have a derivative in L 2 R
(although we will see later that p has a derivative in the sense of generalized functions).
Problem 9 Find the Fourier transforms and use them to find the derivatives in the
L 2 −sense of
11
fx =
1 − x2
if
x2 < 1
0
if
x2 > 1
gx =
,
x2
0
if
if
x2 < 1
x2 > 1
Theorem 5.1 For f, g ∈ L 2 R, the following statements are equivalent
(a) f ′x = gx where the equality is in the L 2 R sense
(b) iα Fα = Gα
fx + h − fx
(c) ||D h f − g|| 2 → 0 as h → 0, where D h fx =
for h > 0
h
(d) there exists a sequence φ n x ∈ C ∞c R such that
and ||φ ′n − g|| 2 → 0 as n → ∞,
||φ n − f|| 2 → 0
Proof-a  b
If f and f ′ ∈ L 2 then T F f ′  = i αFα ∈ L 2 so clearly a  b
b  c : If f ∈ L 2 then for h > 0, theorem 3.3 implies,
T F D h fx = T F fx + h − fx/h =
=
Since
e ihα − 1
ihα
e ihα − 1 → 1 as h → 0
ihα
e ihα − 1Fα
h
iαFα =: G h α
(L’Hopital’s rule), it follows that
T F D h fx = G h α → iα Fα =: Gα
as
h → 0 .(convergence in L 2 )
Since the Fourier transform is an isometry on L 2 this implies
−1
′
D h fx = T −1
F G h α  T F Gα = g = f
as h  0.
c  a :Note that for any test function, φ
D h fx, φ 2 = h −1 fs, φs − he j  2 − fx, φx 2 = fx, −D h φx 2
Now c) implies that as h tends to zero, D h fx, φ 2  g, φ 2 and for any test function, φ
fx, −D h φx 2  f, −φ ′  2
hence we conclude g = f ′ ∈ L 2 .
d  a : Suppose, as n → ∞, φ n  f
For any test function ψ,
Then g = f ′ ∈ L 2 .
and
φ ′n  g in L 2
φ ′n , ψ 0 = φ n , −ψ ′  0 ∀n
↓
↓ as n  ∞
g, ψ 0 = −f, ψ ′  0
That a) implies d) can be proved using the mollifier theorem.■
The examples have suggested that havingf and αFα in L 2 implies that f ′ is in L 2 , which is
to say that f has a derivative in the L 2 sense. While having a derivative in the L 2 sense is not
the same as having a derivative in the classical point wise sense, it does imply some
12
additional regularity as the next proposition will show.
Proposition 5.2 Suppose that f and f ′ ∈ L 2 R. Then f is continuous on R and
f → 0 as | x| → ∞.
Proof- For arbitrary x, y ∈ R, x > y, we can use the Cauchy-Schwarz inequality to write
x
x
x
| fx − fy|= |∫ f ′ s ds |≤  ∫ 1 2 ds 1/2  ∫ f ′ s 2 ds 1/2 ≤ | x − y| 1/2 ||f ′ || 2 ,
y
y
y
b
from which it is clear the f is continuous at each point. Next, since ∫ fx ± f ′ x 2 dx ≥ 0, for
a
all a,b we have
b
b
b
∫ a fx 2 dx + ∫ a f ′ x 2 dx ≥ 2 ∫ a fxf ′ xdx ,
and since f is continuous,
b
2 ∫ fxf ′ xdx = fb 2 − fa 2 .
a
Now since f and f’ are both square integrable,
b
b
lim ∫ fx 2 dx = lim ∫ f ′ x 2 dx = 0
a,b→∞
a
a,b→∞
a
where a and b are allowed to tend independently to plus infinity. Combining these last three
results leads to the conclusion that fx tends to a constant as x → ∞. Since f is square
integrable, this constant must be zero. Similar reasoning implies f tends to zero as x tends
to minus infinity.■
The proposition asserts that having a derivative in the L 2 -sense does imply some additional
smoothness for the function. Precisely, it says the following:
i.e.,
f and αFα in L 2 implies that f ′ is in L 2
f and f ′ in L 2 implies that f is in C 0
We might expect this to generalize to,
f and α p Fα in L 2 for p ≤ M implies that f p is in L 2 for p ≤ M
f and f p in L 2 for p ≤ M implies that f p is in C 0 for p ≤ ?
This precise assertion is given in the next proposition.
Recall that if f ∈ L 1 R then F = T F f ∈ C 0 R. If, in addition, F ∈ L 1 R,
−1
−1
then T −1
F F ∈ C 0 R and ||T F F − f|| 1 = 0, which means T F Fx and fx belong to the
same equivalence class in L 1 R. More generally,
if α p Fα ∈ L 1 R then
and
p
T −1
F iα Fα ∈ C 0 R
p
p
|| 1 = 0;
||T −1
F iα Fα − f
i.e., f ∈ L 1 R has continuous derivatives up to order p if it is the case that
α p Fα ∈ L 1 R. Evidently, decay at infinity for the Fourier transform of a function in L 1 R
implies smoothness for the function. Similarly, theorem 3.5 implies that if x k fx ∈ L 1 R,
then the Fourier transform, F, must have continuous derivatives up to order k. We would
like to consider this connection between decay at infinity for f or F and smoothness for F
or f, in the symmetric setting of L 2 R.
13
Theorem 5.3 (Sobolev Embedding Theorem in R 1 ) Suppose f ∈ L 2 R
and α p Fα ∈ L 2 R for p ≤ M. Then f has derivatives in the L 2 R sense of all orders
≤ M, and these L 2 −derivatives, f j , belong to C 0 for j ≤ M − 1.
Proof- Recall that if f ∈ L 1 R then F = T F f ∈ C 0 R. If, in addition, F ∈ L 1 R,
−1
−1
then T −1
F F ∈ C 0 R and ||T F F − f|| 1 = 0, which means T F Fx and fx belong to the
same equivalence class in L 1 R. This will form the basis for our proof.
Since we assumed that α p Fα ∈ L 2 R for p ≤ M, it follows that
p
j
f = T −1
F iα Fα, for p ≤ M, where this equality is in the L 2 R sense. Note that
p
α Fα ∈ L 2 R for p ≤ M, implies
1 + α 2 
M/2
∫R 1 + α 2  M Fα 2 dα = C 2F < ∞
Fα ∈ L 2 R; i. e. ,
It remains to be discovered for which integers q it is true that α q Fα ∈ L 1 R, since for
these values of q it is then the case that
q
T −1
F iα Fα ∈ C 0 R
∫R | α| q |Fα| dα = ∫R
Write
≤
q
q
T −1
x
F iα Fα = f
and
| α| q
M/2
1 + α 2  |Fα| dα
2 M/2
1 + α 
2q
∫R | α| 2 M dα
1 + α 
≤ CF
for almost all x.
1/2
∫R 1 + α 2  M |Fα| 2 dα
2q
∫R | α| 2 M dα
1 + α 
1/2
1/2
It only remains to be seen for which integers q it is true that
IM, q =
∫R
| α| 2q
dα
M
1 + α 2 
1/2
is finite.
It is easy to check that if M is 1 then we get q = 0. If M is 2, then q ≤ 1, and if M is 3 then
q ≤ 2. Evidently, IM, q is finite if q ≤ M − 1. That is,
if
α p Fα ∈ L 2 R for p ≤ M,
then
f q x ∈ C 0 R for q ≤ M − 1. ■
We will now use this result to consider several example problems for PDE’s.
Example 5.1 Suppose ∇ 2 ux, y = 0,
ux, 0 = fx
x ∈ R, y > 0
x ∈ R,
where f ∈ L 2 R. If Uα, y = T F ux, y denotes the Fourier transform, in x, of u(x,y),
then
and
− α 2 Uα, y = T F ∂ xx ux, y
∂ yy Uα, y = T F ∂ yy ux, y,
− α 2 Uα, y + ∂ yy Uα, y = 0,
Uα, 0 = Fα.
y>0
14
Uα, y = e −y| α| Fα, y > 0.
It follows that
Now, two things are clear. First,
||Uα, y − Uα, 0|| 2 = || e −y| α| Fα − Fα|| 2 → 0, as y → 0 +
so, by theorem 4.4,
||ux, y − ux, 0|| 2 = || ux, y − fx|| 2 → 0, as y → 0 +.
i.e., ux, y tends (in the L 2 − sense) to fx as x, y tends to the boundary. Second, for all
positive integers, p and q
T F ∂ px ∂ qy ux, y = iα p ∂ qy Uα, y = iα p ∂ qy e −y| α| Fα = iα p −|α| q e −y| α| Fα.
Now, because of the presence of e −y| α| this expression belongs to L 2 R for all positive y
and all positive integers p,q. It follows that ux, y has L 2 −derivatives of all orders in both x
and y for all x and all positive y. Then by theorem 5.2, ux, y has continuous derivatives of
all orders as well. Then u is infinitely differentiable in x and y for all x and all positive y, given
only that f is square integrable.
To complete the solution, note that
−y| α|
T −1
=
F e
2y
= gx, y (see prob 7)
y + x2
2
−y| α|
Fα = 1 g⋅, y ∗ f
T −1
F e
2π
and
(Theorem 3.6).
Then
fz
y ∞
ux, y = 1 g⋅, y ∗ f = π ∫
dz.
2
−∞
2π
y + x − z 2
Problem 10 Show that in the special case fx = I A x this solution becomes
1 arctan x + A
ux, y = π
y
− arctan x −y A
.
What would be the solution if fx = I A x − 2A?
Example 5.2 Suppose
∂ t ux, t − ∂ xx ux, t = 0,
ux, 0 = fx,
x ∈ R, t > 0,
x ∈ R,
where f ∈ L 2 R. If Uα, t = T F ux, t denotes the Fourier transform, in x, of u(x,t), then
− α 2 Uα, t = T F ∂ xx ux, t
∂ t Uα, t = T F ∂ t ux, t,
and
It follows that
∂ t Uα, t − α 2 Uα, t = 0,
Uα, 0 = Fα.
t>0
Uα, t = e −t α Fα, t > 0.
2
As in the previous example, it is clear that
||Uα, t − Uα, 0|| 2 =
e −tα Fα − Fα
2
2
→ 0, as t → 0 +
15
so, by theorem 4.4,
|| ux, t − ux, 0|| 2 = || ux, t − fx|| 2 → 0, as t → 0 +.
i.e., ux, t tends (in the L 2 − sense) to fx as x, t tends to the initial line. Second, for all
positive integers, p and q
T F ∂ px ∂ qt ux, t = iα p ∂ qt Uα, t = iα p ∂ qt e −t α Fα = iα p −α 2  q e −t α Fα.
2
2
Now, because of the presence of e −t α this expression belongs to L 2 R for all positive t
and all positive integers p,q. It follows that ux, t has L 2 −derivatives of all orders in both x
and t for all x and all positive t. Then by theorem 5.2, ux, t has continuous derivatives of all
orders as well. Then u is infinitely differentiable in x and t for all x and all positive t, given
only that f is square integrable.
To complete the solution note that
2
T −1
F
and
T −1
F
x2
−
π
e
=
e 4t = hx, t (see prob 7)
t
2
e −t α Fα = 1 h⋅, t ∗ f
(Theorem 3.6).
2π
−t α 2
Then
ux, t = 1 h⋅, t ∗ f =
2π
∞
1 ∫
4πt −∞
x − y 2
4t
e
fydy
−
Problem 11 Show that in the special case fx = I A x this solution becomes
A
1 ∫
4πt −A
ux, t =
=
where
erfx =
Example 5.3 Suppose
1
2
erf
1
2
x − y 2
4t
e
dy
−
A+x
t
− erf
1
2
x−A
t
,
2 ∫ ∞ e −y 2 dy.
π 0
∂ tt ux, t − a 2 ∂ xx ux, t = 0,
ux, 0 = fx,
∂ t ux, 0 = gx,
x ∈ R, t > 0,
x ∈ R,
x ∈ R,
where f, g ∈ L 2 R. If Uα, t = T F ux, t denotes the Fourier transform, in x, of ux, t,
then
− α 2 Uα, t = T F ∂ xx ux, t
∂ tt Uα, t = T F ∂ tt ux, t,
and
∂ tt Uα, t + a 2 α 2 Uα, t = 0,
Uα, 0 = Fα
∂ t Uα, 0 = Gα.
t>0
It follows that
16
Gα
Uα, t = cosaαt Fα + sinaαt aα ,
t > 0.
Gα
Gα
= 1 e iaαt Fα + e −iaαt Fα + 1 e iaαt
− e −iaαt
iα
iα
2
2a
In this case we have
Gα
cosaαtFα − Fα + sinaαt aα
2
Gα
≤ || cosaαtFα − Fα|| 2 + sinaαt aα
→ 0, as t → 0 +,
||Uα, t − Uα, 0|| 2 =
2
implying that
|| ux, t − ux, 0|| 2 = || ux, t − fx|| 2 → 0, as t → 0 +.
Similarly,
||∂ t Uα, t − ∂ t Uα, 0|| 2 = || − aα sinaαtFα + cosaαt Gα − Gα|| 2
≤ || − aα sinaαtFα|| 2 + ||cosaαt Gα − Gα|| 2 → 0, as t → 0 +,
from which it follows, using theorem 4.4, that
||∂ t ux, t − ∂ t ux, 0|| 2 = || ∂ t ux, t − gx|| 2 → 0, as t → 0 +.
Then the initial conditions are satisfied if the limits are understood to mean L 2 −limits.
Note that by theorem 3.3,
± iaαt
T −1
Fα = fx ± at
F e
and
T −1
e ± iaαt
F
Gα
iα
= hx ± at
where
Gα = iα Hα = T F h ′ x; i. e.
h ′ x = gx.
Then
ux, t = 1 fx + at + fx − at +
2
= 1 fx + at + fx − at +
2
1 hx + at − hx − at
2a
1 ∫ x+at gs ds.
2a x−at
In contrast to the previous two examples, the transform of the solution in this case contains
no ”smoothing operator” in the form of a rapidly decreasing exponential function of α. Here
the modifier in the solution is an oscillatory function of αt and does not decrease at all as |α|
tends to inifinity. Here, in order to have
T F ∂ px ∂ qt ux, t = iα p ∂ qt Uα, t = iα p −iaα q Uα, t
belong to L 2 R, it would be necessary to place conditions on the data; e.g., if
α 3 Fα and α 2 Gα both belong to L 2 R, then this is sufficient to imply
iaα 3 Uα, t belongs to L 2 R, which implies then that u(x,t) has continuous derivatives of
up to order two with respect to both x and t. If the data is not this smooth, then the PDE is
not satisfied in the classical sense but only in some weaker sense. The smoothness of the
solution in this case depends on the smoothness of the data.
In each of these examples, the Fourier transform of the solution equals the Fourier
transform of the data multiplied by a ”modifier” that is a function of the transform variable, α,
17
and the nontransform variable, y or t. When the exponent in the modifier is real and
negative for all relevant values of the nontransform variable then the modifier decreases
exponentially as | α| tends to infinity, resulting in the extreme smoothness of the solutions to
the equation, independent of the smoothness of the data. This is the case for the Laplace
and heat equations. If the exponent in the modifier is imaginary but linear in α, then the
modifier acts as a ”translation operator” as prescribed in theorem 3.3. Then the solution is
only as smooth as the data but the data is transmitted without distortion. This is the case for
the transport and wave equations (in the case of the transport equation with a lower order
term, the lower order term introduces some distortion, see example 5.6 below). Thus, even
without obtaining explicit formulas for the solutions in these problems, we can deduce
qualitative properties of the solutions from looking at the transforms of the solutions. We will
now consider additional examples and see what we can determine about their solutions.
Example 5.4 Suppose ∇ 2 ux, y, z = 0,
ux, y, 0 = fx, y
x, y ∈ R 2 , z > 0
x, y ∈ R 2 ,
f ∈ L 2 R 2 .
where
1 2 ∫ ∫ e −ixα 1 −iyα 2 ux, y, z dx dy
R R
2π
= T F ux, y, z
Uα 1 , α 2 , z =
If
denotes the Fourier transform, in (x,y) of ux, y, z, then
⃗, z = T F ∂ xx u + ∂ yy u
− α 21 + α 22  Uα
⃗, z = T F ∂ zz ux, y, z,
∂ zz Uα
and
⃗, z + ∂ zz Uα
⃗, z = 0,
− α 21 + α 22  Uα
⃗, 0 = Fα
⃗ .
Uα
It follows that
⃗| =
⃗, z = e −z| α⃗ | Fα
⃗ , z > 0, where |α
Uα
z>0
α 21 + α 22 .
It is clear that the qualitative behavior of the solution to Laplace’s equation is essentially the
same in 3 dimensions as it was in 2 dimensions. The formula representing the solution in
terms of the data might be more complicated in the higher dimensional case but the
smoothness and limit at the boundary results are not changed. The heat equation shows a
similar lack of variation in qualitative behavior as the number of dimensions increases.
Example 5.5 Suppose
∂ tt ux, y, t − a 21 ∂ xx ux, y, t − a 22 ∂ xx ux, y, t = 0,
ux, y, 0 = fx, y,
∂ t ux, y, 0 = 0,
x, y ∈ R 2 , t > 0,
x, y ∈ R 2 ,
x, y ∈ R 2 ,
where f ∈ L 2 R 2 . If Uα 1 , α 2 , t = T F ux, y, t denotes the Fourier transform, in (x,y), of
ux, y, t, then
⃗, t = T F ∂ xx ux, y, t + ∂ yy ux, y, t
− α 21 + α 22  Uα
⃗, t = T F ∂ tt ux, y, t,
∂ tt Uα
and
18
⃗, t − a 21 α 21 + a 22 α 22 Uα
⃗, t = 0,
∂ tt Uα
⃗, 0 = Fα
⃗
Uα
⃗, 0 = 0.
∂ t Uα
It follows that
⃗, t = cos t a 1 α 1  2 + a 2 α 2  2
Uα
= 1 e it
2
a 1 α 1  2 +a 2 α 2  2
⃗ ,
Fα
⃗  + e −it
Fα
t>0
t > 0.
a 1 α 1  2 +a 2 α 2  2
⃗
Fα
It is apparent from this that while the solution to the wave equation in 2 dimensions
continues to have the same properties in terms of lack of smoothness and limiting behavior
as t tends to zero as we observed for the 1-dimensional solution, the modifier that multiplies
the transform of the data is no longer a simple translation operator. The solution depends
on the data in a much more complicated way than what we saw for the 1-dimensional wave
equation. This is consistent with the remark that solutions to the wave equation are very
sensitive to changes in the dimension.
Example 5.6 Suppose
∂ t ux, t − a ∂ x ux, t + b ux, t = 0,
ux, 0 = fx,
x ∈ R, t > 0,
x ∈ R,
where f ∈ L 2 R. If Uα, t = T F ux, t denotes the Fourier transform, in x, of ux, t, then
∂ t Uα, t − a iα − bUα, t = 0,
Uα, 0 = Fα
It follows that
and
t>0
Uα, t = e iaα−bt Fα = e −bt e iatα Fα,
t > 0.
ux, t = e −bt fx + at, ∀x, t > 0.
Clearly as in the case of the wave equation, there is L 2 −convergence to the initial condition
as t tends to zero and there is no smoothing of the data since the modifier is a simple
translation operator. The effect of the lower order term in the equation is simply to introduce
exponential time growth or decay of the solution depending on the sign of the coefficient b.
In more dimensions the solution becomes
⃗, t = e ia⃗⋅α⃗−bt Fα
⃗  = e −bt e ita⃗⋅α⃗ Fα
⃗
Uα
= e −bt e ita 1 α 1 . . . e ita n α n Fα 1 , . . . , α n 
and
⃗
a = a 1 , . . . , a n 
⃗
α = α 1 , . . . , α n ,
ux⃗, t = e −bt fx 1 + a 1 t, . . . , x n + a n t, ∀x⃗, t > 0.
Evidently the solutions to the transport equation do not have the same sensitivity to
dimension as the solutions to the wave equation.
Example 5.7 Suppose
∂ tt ux, t − a 2 ∂ xx ux, t + d 2 ux, t = 0,
ux, 0 = fx,
∂ t ux, 0 = 0,
x ∈ R, t > 0,
x ∈ R,
x ∈ R,
where f ∈ L 2 R. If Uα, t = T F ux, t denotes the Fourier transform, in x, of ux, t, then,
∂ tt Uα, t + a 2 α 2 + d 2 Uα, t = 0,
Uα, 0 = Fα
t>0
19
∂ t Uα, 0 = 0.
It follows that
Uα, t = costαQα Fα,
t > 0,
= 12 e itαQα + e −itαQα  Fα,
where
Qα =
a 2 + d/α 2 .
Note that if d = 0 then Q = a and the exponent in the modifier is linear in α. Then the
modifier is just a simple shifting operator leading to The d’Alembert solution of the wave
equation where the speed of propagation is equal to a = Q. On the other hand, if d is not
zero then the exponent in the modifier is not linear in α, and the modifier is not a shifting
operator. Instead, expitαQα acts like a shift operator in which the propagation speed,
Q, is a function of α. Propagation where waves with different frequencies propagate at
different speeds is sometimes referred to as ”dispersive” wave propagation. Each of the
following examples exhibits a similar type of modifier and, consequently describes some
kind of dispersive propagation.
∂ t ux, t − a 2 ∂ xxx ux, t = 0,
ux, 0 = fx,
i) Dispersion equation
x ∈ R, t > 0,
x ∈ R,
and
Uα, t = e −ita
2α3
Fα,
∂ t ux, t − i ∂ xx ux, t = 0,
ux, 0 = fx,
ii) Schrodinger equation
x ∈ R, t > 0,
x ∈ R,
and
Uα, t = e −itα Fα,
2
∂ tt ux, t + ∂ xxxx ux, t = 0,
ux, 0 = fx,
∂ t ux, 0 = gx,
iii) Vibrating beam equation
x ∈ R, t > 0,
x ∈ R,
x ∈ R,
and
Uα, t = Cosα 2 t Fα +
=
1
2
e itα + e −itα
2
2
sinα 2 t
Gα
α2
2
2
Fα + 1 2 e itα − e −itα Gα.
2α
In each of these cases, it is evident that the solutions will tend to the initial condition in the
L 2 − sense as t tends to zero and that there is no smoothing of the data. It is also evident that
the modifier acts as some version of the shifting operator that appears in the solution of the
simple wave equation but that the shifting action varies with the transform variable α.
Exactly how this will act on the data when the Fourier transform is inverted is not clear,
beyond the fact that the solution will involve a convolution integral operator of the form
ux, t = ∫ Kx − y, t fy dy.
R
Problem 12 Analyze the following problem as to assumption of the initial condition,
20
smoothness of the solution and whether the propagation is more wave-like or diffusion-like.
∂ t ux, t + ∂ xxxx ux, t = 0,
ux, 0 = fx,
x ∈ R, t > 0
x ∈ R.
Example 5.8 The following example illustrates some of the tricks that are often employed
in order to find the inverse transform using only the short table of transforms that can be
developed without contour integrations. Suppose
∇ 2 ux, y = 0
x ∈ R, 0 < y < 1,
ux, 0 = 0,
x ∈ R,
∂ y ux, 1 = gx ∈ L 2 R,
x ∈ R.
Apply Fourier transform in x:
− α 2 Uα, y + U ′′ α, y = 0,
Uα, 0 = 0,
U ′ α, 1 = Gα.
Then, solving the ode, we get
Uα, y =
Gα sinh |αy|
α cosh |α|
−|α|y
|α|y
sinh |αy|
1
= e |α| − e −|α| =
e −|α|1−y − e −|α|1+y 
cosh |α|
e +e
1 + e −2|α|
But
∞
= ∑e −|α|2n+1−y − e −|α|2n+1+y ,
n=0
so
Uα, y =
Now
Gα ∞ −|α|2n+1−y
− e −|α|2n+1+y 
∑e
|α| n=0
e −|α|b − e −|α|a = ∫ a e −|α|s ds
b
|α|
|α|
∞
so
2n+1+y
Uα, y = ∑ Gα ∫
e −|α|s ds
2n+1−y
n=0
−|α|s
T −1
=
F e
Recall
∞
−|α|s
T −1
gz dz
 = 1 ∫ −∞ 2 2s
F Gα e
2π
s + x − z 2
and
∞
so
2s
s2 + x2
2n+1+y
ux, y = ∑ ∫
2n+1−y
n=0
1 ∫∞
2s
gz dz ds
2π −∞ s 2 + x − z 2
∞
∞
2n+1+y
2s
= 1 ∑∫ ∫
ds gz dz
−∞
2n+1−y
2π n=0
s 2 + x − z 2
Then, since
2n+1+y
∫ 2n+1−y
2s
ds = log
s 2 + x − z 2
2n + 1 + y 2 + x − z 2
2n + 1 − y 2 + x − z 2
,
we get
21
∞
∞
ux, y = 1 ∑ ∫ log
2π n=0 −∞
=∫
∞
−∞
2n + 1 + y 2 + x − z 2
2n + 1 − y 2 + x − z 2
gz dz
gx − z, y gz dz
where
∞
gx − z, y = 1 ∑ log
2π n=0
2n + 1 + y 2 + x − z 2
2n + 1 − y 2 + x − z 2
.
6. Orthogonal Families and Generalized Fourier Series
Using a basis for representing arbitrary vectors in R n has numerous advantages in dealing
with problems in linear algebra. For most computational purposes, it is convenient if the
basis is an orthonormal basis, meaning that the vectors in the basis are mutually orthogonal
⃗ 1,..., ⃗
unit vector. If u
u n  is such an orthonormal basis then an arbitrary ⃗
v ∈ R n can be
uniquely expressed as
n
⃗
v = ∑ j=1 v⃗ ⋅ ⃗
uj ⃗
uj.
It is our aim now to develop such representations in infinite dimensional function spaces.
Let U ⊂ R n denote a bounded open and connected set in R n and consider functions
f, g ∈ L 2 U. The functions are said to be orthogonal if
f, g 2 = ∫U fx gx dx = 0.
A countable family g 1 x, g 2 x, . . . ,  of functions in L 2 U is said to be an orthogonal
family in L 2 U if g i , g j  2 = 0 if i ≠ j. The orthogonal family g j x is said to be an
orthonormal family if, in addition to be mutually orthogonal, the functions satisfy ||g j || 2 = 1
for every j; e.g., The family of functions G n x = sinnπx, n = 1, 2, . . . is an orthogonal
family in L 2 0, 1, and the family g n x = 2 sinnπx, n = 1, 2, . . . is an orthonormal family
in L 2 0, 1.
Suppose g 1 x, g 2 x, . . . ,  is an orthonormal family of functions in L 2 U, and for an
arbitrary f ∈ L 2 U, form the infinite sum
∞
∑ n=1
f, g n  2 g n x.
It is not clear that this sum is convergent in L 2 U, and even if it is convergent, it is not
evident that it is convergent to f. In any case, we refer to the sum as the generalized
Fourier series for f and we refer to the coefficients f n = f, g n  2 as the generalized
Fourier coefficients for f.
Proposition 6.1 Suppose g 1 x, g 2 x, . . . ,  is an orthonormal family of functions in
L 2 U, and for an arbitrary f ∈ L 2 U, let f n = f, g n  2 . Then for any integer N, and any
choice of constants a 1 , a 2 , . . . a N we have
i)
fx − ∑ n=1 a n g n x
ii)
fx − ∑ n=1 a n g n x
N
2
2
N
2
= || f|| 22 − ∑ n=1 f, g n  22 + ∑ n=1 f n − a n  2
N
≥
N
fx − ∑ n=1 f n g n x
N
2
Proof- For N a fixed positive integer, let
22
S N x = ∑ n=1 a n g n x.
N
Then
∫U fx − S N x 2 dx = ∫U fx 2 dx − 2 ∫U fxS N xdx + ∫U S N x 2 dx.
But
N
N
N
∫U fxS N xdx = ∫U fx ∑ n=1
a n g n x dx = ∑ n=1 a n ∫ fx g n xdx = ∑ n=1 a n f n
U
and
N
N
∫U S N x 2 dx = ∫U ∑ m=1
a m g m x ∑ n=1 a n g n x dx
= ∑ m=1 a m ∑ n=1 a n ∫ g n xg m xdx = ∑ n=1 a 2n
N
N
N
U
Then
since g n , g m  2 = δ mn
N
N
∫U fx − S N x 2 dx = ∫U fx 2 dx − 2 ∑ n=1
a n f n + ∑ n=1 a 2n
= ∫ fx 2 dx − ∑ n=1 f 2n + ∑ n=1 f 2n − 2 ∑ n=1 a n f n + ∑ n=1 a 2n
N
N
N
N
U
= ∫ fx 2 dx − ∑ n=1 f 2n + ∑ n=1 f n − a n  2 .
N
N
U
This is the result (i). Since ∑ n=1 f n − a n  2 ≥ 0, we get the result (ii). This last result asserts
that among all linear combinations of the functions g 1 , . . . , g N , the one that is closest to f in
the L 2 U − norm, is the combination with a n = f n = f, g n  2 , n = 1, . . . , N. ■
N
Theorem 6.2 Suppose g 1 x, g 2 x, . . . ,  is an orthonormal family of functions in L 2 U,
and for an arbitrary f ∈ L 2 U, let f n = f, g n  2 . Then
1 (Bessel’s inequality)
∞
f 2n ≤ || f || 22 = ∫ fx 2 dx.
∑ n=1
U
2 (Riemann-Lebesgue lemma)
f n → 0 as n → ∞
Proof-(1) Choosing a n = f n in the sum S N x, and using the results of the previous
proposition,
0 ≤ ∫ f − S N  2 dx = ∫ f 2 dx − ∑ n=1 f 2n ,
U
U
N
i.e.,
N
f 2n ≤ ∫ f 2 dx = || f || 22 .
∑ n=1
U
Since this result holds for all positive integers N, and the right side of the estimate does not
depend on N, we are entitled to let N tend to infinity to obtain (1). Then the n-th term test
implies the result (2).■
In the special case that the orthogonal family is 1, cos x, sin x, cos 2x, sin 2x, . . .  in L 2 0, 2π
then the (2) takes the form
2π
∫ 0 fx sin nx dx → 0,
2π
∫ 0 fx cos nx dx → 0,
as
n → ∞.
This is what is often referred to as the Riemann-Lebesgue lemma.
The Bessel’s inequality implies that for an arbitrary f ∈ L 2 U, the sequence
f n = f, g n  2 of generalized Fourier coefficients belongs to ℓ 2 . Then it is also evident that the
sequence of partial sums S N x, is a Cauchy sequence and therefore converges in L 2 U to
23
some limit, Sx. However, it is not necessarily the case that S = f. An orthonormal family
with the property that for every f ∈ L 2 U, the generalized Fourier series converges to
f in L 2 U is said to be a complete orthonormal family.
Theorem 6.3 Suppose g 1 x, g 2 x, . . . ,  is an orthonormal family of functions in L 2 U,
and for an arbitrary f ∈ L 2 U, let f n = f, g n  2 . Then the following assertions are all
equivalent:
1.
2.
3.
4.
5.
g 1 x, g 2 x, . . . ,  is a complete orthonormal family
∞
f 2n = || f || 22
∑ n=1
f n = 0 ∀n,
if and only if f = 0
∞
∀f, g ∈ L 2 U, f, g 2 = ∑ n=1 f n g n
|| f − S N || 2 → 0 as N → ∞
As a result of this theorem, it follows that L 2 U is isometrically isomorphic to ℓ 2 .
Theorem 6.4 Suppose g 1 x, g 2 x, . . . ,  is a complete orthonormal of functions in L 2 U.
Then for an every f ∈ L 2 U, f n = f, g n  2 belongs to ℓ 2 with ||f n || ℓ 2 = || f || 2 . Conversely,
N
for every f n  ∈ ℓ 2 , the sequence S N = ∑ n=1 f n g n , converges to a unique limit, f, in L 2 U
and ||f n || ℓ 2 = || f || 2 .
Just as the Fourier transform provided an alternative but equivalent representation for
functions in L 2 R n , the sequence of generalized Fourier coefficients provides an alternative
but equivalent representation for functions in L 2 U. The functions in the complete
orthonormal family are the elements of a countable basis for L 2 U, and every function in
L 2 U can be expressed uniquely as a (possibly infinite) linear combination of these
elements. While it is relatively easy to find examples of orthonormal families of functions in
L 2 U, it is not clear how to determine whether or not a family is complete. We shall see
now one source of complete orthonormal families.
Sturm-Liouville Systems
Consider the problem of finding all scalars λ for which there exist nontrivial solutions to the
following boundary value problem
−d/dxpx du/dx + qxux = λrxux,
a < x < b,
6. 1
C 1 ua + C 2 u ′ a = 0,
C 3 ub + C 4 u ′ b = 0,
Clearly the trivial solution satisfies 6. 1 for all choices of the parameter λ. Any value λ
which leads to a nontrivial solution will be called an eigenvalue of the problem and the
corresponding nontrivial solution will be called an eigenfunction of the problem,
corresponding to the eigenvalue λ. Note that if u = ux is an eigenfunction corresponding
to the eigenvalue λ, the for every nonzero constant k, the function kux is also an
eigenfunction for the same eigenvalue. A problem of the form (6.1) is called a
Sturm-Liouville problem.
Theorem 6.5 Suppose the coefficients in the Sturm-Liouville problem (6.1) satisfy
24
px, p ′ x, qx and rx are all continuous on [a,b]
px > 0, rx > 0 for all x ∈ a, b
C 21 + C 22 > 0, and C 23 + C 24 > 0,
Then
i) the Sturm-Liouville problem (6.1) has countably many real eigenvalues
λ 1 ≤ λ 2 ≤. . . ≤ λ n → +∞
ii) for each eigenvalue there is a single independent eigenfunction, and
eigenfunctions corresponding to distinct eigenvalues satisfy
b
∫ a ux, λ j  ux, λ k  rx dx = 0
j ≠ k.
iii) The family of normalized eigenfunctions ux, λ n  are a complete
orthonormal family in L 2 a, b for the weighted inner product
b
f, g 2 = ∫ a fx gx rx dx.
The proof of the Sturm-Liouville theorem is beyond the scope of this course. Instead we will
list several examples of S-L problems and the associated eigenvalues and eigenfunctions.
In each of the following examples we have px = rx = 1, qx = 0, on a, b = 0, 1.
Then since rx = 1, the weighted inner product of the theorem reduces to the usual L 2
inner product.
Example 6.1- Choose
Then
C 1 = C 3 = 1, C 2 = C 4 = 0,
−u" x = λux 0 < x < 1,
u0 = u1 = 0.
In this case, we find
λ n = n 2 π 2 , n = 1, 2, . . .
u n x = sinnπx
Example 6.2- Choose
Then
C 1 = C 3 = 0, C 2 = C 4 = 1,
−u" x = λux 0 < x < 1,
u ′ 0 = u ′ 1 = 0.
In this case, we find
λ n = n 2 π 2 , n = 0, 1, 2, . . .
u n x = cosnπx
Example 6.3- Choose
Then
C 1 = C 4 = 1, C 2 = C 3 = 0,
−u" x = λux 0 < x < 1,
u0 = u ′ 1 = 0.
In this case, we find
λ n = n −
1
2
 2 π 2 , n = 1, 2, . . .
25
u n x = sin n −
Example 6.4- Choose
Then
1
2
πx
C 1 = C 4 = 0, C 2 = C 3 = 1,
−u" x = λux 0 < x < 1,
u ′ 0 = u1 = 0.
In this case, we find
λ n = n − 12  2 π 2 , n = 1, 2, . . .
u n x = cos n − 12 πx
Problem 12 Show that if Lux = −d/dxpx du/dx + qxux,
then for smooth functions ux, vx
Lu, v 2 = u, Lv 2 − pxu ′ v − v ′ u| ba
Show that if ux, vx both belong to D 1 = u ∈ C 2 a, b : ua = ub = 0 then this
reduces to
Lu, v 2 = u, Lv 2 ;
i. e. , L is self adjoint.
Show the result continues to hold if D 1 is replaced by any one of the following domains for
L:
D 2 = u ∈ C 2 a, b : u ′ a = u ′ b = 0
D 3 = u ∈ C 2 a, b : ua = u ′ b = 0
D 4 = u ∈ C 2 a, b : u ′ a = ub = 0.
Problem 13 Show that if λ is an eigenvalue of problem (6.1), then λ is necessarily real.
Problem 14 Show that if λ, μ are distinct eigenvalues of problem (6.1), then associated
eigenfunctions must satisfy
b
∫ a rx ux, λ ux, μ dx = 0.
Example 6.5- Consider the problem,
−u" x = λux −1 < x < 1,
u−1 = u1
u ′ −1 = u ′ 1.
The boundary conditions in this problem are called periodic boundary conditions and are
not of the type indicated in problem (6.1). Nevertheless, in this slightly modified S-L problem
there are an infinite set of eigenvalues λ n = n 2 π 2 , n = 1, 2, . . . but now there are two
independent eigenfunctions for each eigenvalue, except the eigenvalue zero for which there
is just one eigenfunction. We have
λ0 = 0
λn = n2π2
u 0 x = 1
u n x = cosnπx,
v n x = sinnπx,
26
Alternatively, we can express the eigenfunctions as follows,
λ0 = 0
λn = n2π2
u 0 x = 1
u n x = e inπx ,
v n x = e −inπx ,
In the first case, the Fourier expansion would have the form
fx =
1
2
∞
a 0 + ∑ n=1 a n cosnπx + b n sinnπx
where
1 ∫ 1 fx cosnπx dx,
an = π
0
1 ∫ 1 fx sinnπx dx,
bn = π
0
and in the second case it has the form
∞
fx = ∑ n=−∞ c n e −inπx ,
1
c n = 1 ∫ fx e −inπx dx.
2π 0
In either case we have ||S N − f || 2 → 0, as N → ∞.
7. Applications to Partial Differential Equations
Consider the following problem for Laplace’s equation
−∇ 2 ux, y = Fx, y
0 < x, y < 1
ux, 0 = ux, 1 = 0
0 < x < 1,
u0, y = u1, y = 0,
0 < y < 1,
7. 1
and recall the eigenfunctions from example 6.1, u n x = sinnπx, n = 1, 2. . . We note that
the functions
U mn x, y = u m x u n y
satisfy
−∇ 2 U mn x, y = −u" m x u n y − u m x u" n y
= λ m u m x u n y + λ n u m x u n y = λ m + λ n  u m x u n y.
Evidently, the functions U mn x, y are eigenfunctions for the problem (7.1), with eigenvalues
μ mn = λ m + λ n = m 2 + n 2 π 2 ,
m, n = 1, 2, . . .
It can be shown that since u n x is a complete orthogonal family in L 2 0, 1 then it is
necessarily the case that the orthogonal family U mn x, y = u m x u n y is a complete
orthogonal family in L 2 0, 1 × 0, 1. We can use this family of eigenfunctions to construct
a Hilbert scale for L 2 0, 1 × 0, 1 = L 2 U. We define
H s U =
f ∈ L 2 U : 1 + μ mn  s/2 f mn ∈ ℓ 2 ,
s ≥ 0.
Now this scale can be used to asses the regularity of the solution to problem (7.1). First we
show how this solution can be constructed.
Since the family U mn x, y = u m x u n y is a complete orthogonal family in L 2 U, any
function in L 2 U can be expressed in terms of this family. In particular,
∞
∞
∞
∞
Fx, y = ∑ m,n>1 F mn U mn = ∑ m=1 ∑ n=1 F mn u m x u n y
and
ux, y = ∑ m,n>1 C mn U mn = ∑ m=1 ∑ n=1 C mn u m x u n y.
27
Since F is known, we can compute the coefficients F mn from
F, U mn  2 = ∫ ∫ Fξ, η U mn ξ, η dξ dη
U
∞
∞
= ∫ ∫ ∑ j=1 ∑ k=1 F jk u j ξ u k ηU mn ξ, η dξ dη
U
∞
∞
1
1
0
0
= ∑ m=1 ∑ n=1 F jk ∫ u j ξu m ξdξ ∫ u k η u n ηdη
∞
∞
= ∑ m=1 ∑ n=1 F jk u j , u m  2 u k , u n  2
= F mn || u m || 22 || u n || 22 = 4 F mn .
The coefficients, C mn , in the expansion for the solution, ux, y, are computed from the
equation by noting that
∞
∞
−∇ 2 ux, y = ux, y = −∇ 2 ∑ m=1 ∑ n=1 C mn U mn x, y
∞
∞
∞
∞
= ∑ m=1 ∑ n=1 − ∇ 2 U mn x, y C mn
= ∑ m=1 ∑ n=1 μ mn U mn x, y C mn ,
and
∞
∞
−∇ 2 ux, y − Fx, y = ∑ m=1 ∑ n=1  μ mn C mn − F mn U mn x, y = 0,
from which it follows that
mn
C mn = F
μ mn ,
m, n = 1, 2, . . .
i.e.,
∞
∞
mn
ux, y = ∑ m=1 ∑ n=1 F
μ mn u m x u n y.
Now observe that if F ∈ L 2 U = H 0 U, then F mn ∈ ℓ 2 and
mn
1 + μ mn  C mn = 1 + μ mn  F
μ mn ~ F mn ∈ ℓ 2 .
This implies that u ∈ H 2 U if F ∈ H 0 U. More generally, if
F ∈ H m U then 1 + μ mn  m/2 F mn ∈ ℓ 2 and
mn
1 + μ mn  m+2/2 C mn = 1 + μ mn  m/2 1 + μ mn  F
μ mn
~ 1 + μ mn  m/2 F mn ∈ ℓ 2 .
This implies that u ∈ H m+2 U if F ∈ H m U; i. e, the solution is ”2 steps up the regularity
scale” from the data.
Now consider the related problem
−∇ 2 ux, y = 0
0 < x, y < 1
ux, 0 = f 0 x, ux, 1 = f 1 x
0 < x < 1,
u0, y = g 0 y u1, y = g 1 y, 0 < y < 1.
7. 2
For convenience, suppose f 0 x = g 0 y = 0, and note then that ux, y = vx, y + wx, y
where
−∇ 2 vx, y = 0
0 < x, y < 1
28
vx, 0 = 0, vx, 1 = f 1 x
v0, y = 0, v1, y = 0,
0 < x < 1,
0 < y < 1,
− ∇ 2 wx, y = 0
wx, 0 = 0, wx, 1 = 0,
w0, y = 0, w1, y = g 1 y,
0 < x, y < 1
0 < x < 1,
0 < y < 1.
7. 3a
and
7. 3b
Instead of expanding v and w in terms of the eigenfunctions U mn (this is called the ”full
eigenfunction expansion”) we will expand v in terms of the eigenfunctions u m y and we will
expand w in terms of the eigenfunctions u n x. The reason for this will become clear in a
moment. We write
∞
∞
vx, y = ∑ m=1 v m y u m x = ∑ m=1 v m y sinmπx
∞
∞
wx, y = ∑ n=1 w n x u n y = ∑ n=1 w n x sinnπy.
Substituting these expressions into the equations 7. 3a, b for v and w leads to
∞
−v" m y + mπ 2 v m y sinmπx = 0
∑ m=1
and
∞
−w" n x + nπ 2 w n x sinnπy = 0.
∑ n=1
Since the eigenfunctions are orthogonal, they are certainly linearly independent and then it
follows from these last 2 equations that
−v" m y + mπ 2 v m y = 0,
−w" n x + nπ 2 w n x = 0,
m = 1, 2, . . .
n = 1, 2, . . .
7. 4a
7. 4b
Substituting the eigenfunction expansions into the boundary conditions, leads to
v m 0 = 0
w n 0 = 0
v m 1 = f 1 , u m  2 = f m
1 ,
w n 1 = g 1 , u n  2 = g n
1 ,
m = 1, 2, . . .
n = 1, 2, . . .
7. 5a
7. 5b
The general solutions for 7. 4a, b are given by
v m y = Ae mπy + Be −mπy ,
w n x = Ce nπx + De −nπx
and then the boundary conditions 7. 5a, b lead to the results
e mπy − e −mπy ,
v m y = f m
1
e mπ − e −mπ
e nπx − e −nπx ,
w n x = g n
1
e nπ − e −nπ
and
∞
e mπy − e −mπy sinmπx + ∑ ∞ g n e nπx − e −nπx sinnπy.
ux, y = ∑ m=1 f m
1
n=1 1
e mπ − e −mπ
e nπ − e −nπ
Note first that
e mπy − e −mπy ≈ e −mπ1−y for 0 < y < 1, m = 1, 2, . . .
e mπ − e −mπ
which means that
e mπy − e −mπy ≈ C e −mπy f m with y = 1 − y > 0 for 0 < y < 1.
v m y = f m
1
1
e mπ − e −mπ
29
But then the decaying exponential factor ensures that
mπ p+q v m y ∈ ℓ 2 for all positive integers p, q and all y, 0 < y < 1,
which implies that ∂ px ∂ qy vx, y ∈ L 2 U for all positive integers p, q. A similar argument
shows that ∂ px ∂ qy wx, y ∈ L 2 U for all positive integers p, q, and together, these results
show the smoothness we have come to expect of solutions to Laplace’s equation on the
interior of the domain. Note that this smoothness is independent of the data smoothness
since we assumed only L 2 regularity for the data.
Note finally that
e mπy − e −mπy → f m as y → 1
f m
1
1
e mπ − e −mπ
hence
v m y − f m
1
ℓ2
→ 0 as y → 1.
Then the isometric isomorphism between L 2 0, 1 and ℓ 2 ensures that
|| vx, y − f 1 x|| 2 → 0 as y → 1.
Similarly,
|| wx, y − g 1 y|| 2 → 0 as x → 1,
and thus
ux, y → f 1 x in L 2 0, 1 as y → 1
and
ux, y → g 1 y in L 2 0, 1 as x → 1
i.e., ux, y satisfies the boundary conditions in the L 2 − sense.
Expanding the solution in terms of just one or the other of the 2 families of
eigenfunctions in the problem was necessary in the case of inhomogeneous boundary data
since the 2-d eigenfunctions U mn x, y vanish at all points of the boundary. Expanding the
solution in terms of one of the 1-d families of eigenfunctions is referred to as a ”partial
eigenfunction expansion”.
Consider now the following initial boundary value problems
∂ t ux, t = K ∂ xx ux, t
ux, 0 = fx,
u0, t = u1, t = 0
0 < x < 1, t > 0,
0 < x < 1,
t > 0,
7. 6
and
∂ tt wx, t = a 2 ∂ xx wx, t
wx, 0 = fx,
∂ t wx, 0 = 0,
w0, t = w1, t = 0
0 < x < 1, t > 0,
0 < x < 1,
0 < x < 1,
t > 0.
7. 7
If we were to assume a solution of the form ux, t = Xx Tt for (7.6), then on substituting
this into the pde, we would find
T ′ t
X́ " x
=
.
KTt
Xx
Since the two sides of this equation depend on different variables, it follows that both sides
30
are equal to the same constant, −λ 2 . This fact, together with the boundary conditions imply
that Xx satisfies the S-L problem in example 6.1; i.e., X n x = sinnπx, n = 1, 2, . . . . We
conclude that the solution to (7.6) can be expanded in terms of this family of eigenfunctions.
Similarly, it is easy to follow the same reasoning to see that the solution of (7.7) can be
expanded in terms of the same family. Then
∞
ux, t = ∑ n=1 u n t sinnπx
and
∞
wx, t = ∑ n=1 w n t sinnπx.
7. 8
Substituting these expansions into the respective problems leads to
∞
u ′n t + Knπ 2 u n t sinnπx = 0,
∑ n=1
∞
u n 0 − f n  sinnπx = 0,
∑ n=1
and
∞
w n " t + a 2 nπ 2 w n t sinnπx = 0,
∑ n=1
∞
w n 0 − f n  sinnπx = 0,
∑ n=1
∞
w ′n 0  sinnπx = 0,
∑ n=1
where
fn =
f, sinnπx 2
1
= 2 ∫ fx sinnπx dx, n = 1, 2, . . .
0
sinnπx, sinnπx 2
so that
N
f n sinnπx → fx in L 2 0, 1 as N → ∞.
∑ n=1
7. 9
The independence of the eigenfunctions then implies that for n = 1, 2, . . .
u ′n t + Knπ 2 u n t,
u n 0 = f n ,
and
w n " t + a 2 nπ 2 w n t = 0,
u n t = f n e −Knπ
Then
2t
∞
and
w n 0 = f n ,
w ′n 0 = 0.
w n t = f n cosnπat,
hence
ux, t = ∑ n=1 f n e −Knπ t sinnπx
and
wx, t = ∑ n=1 f n cosnπat sinnπx,
2
∞
∞
∂ t wx, t = ∑ n=1 f n −nπa sinnπat sinnπx
f n e −Knπ t → f n
2
It is evident that
and
in ℓ 2 as t → 0
f n cosnπat → f n ,
f n −nπa sinnπat → 0
in ℓ 2 as t → 0,
and this implies that
ux, t → fx in L 2 0, 1 as t → 0
wx, t → fx,
∂ t wx, t → 0 in L 2 0, 1 as t → 0.
In addition,
∂ pt ∂ qx ux, t = ∑ n=1 C p,q n 2p+q f n e −Knπ t sinnπx
∞
2
31
C p,q n 2p+q f n e −Knπ t ∈ ℓ 2 .
2
and for each t > 0,
Then it follows that has ux, t has L 2 -derivatives of all orders with respect to both x and t,
for all x in [0,1] and all positive t. This is result is independent of the smoothness of the
data, so long as the data is at least in L 2 0, 1. There is no such smoothness in the solution
of the wave equation, since
|w n t| = | f n cosnπat| ≤ | f n |.
Then f ∈ L 2 0, 1  f n ∈ ℓ 2  w n t ∈ ℓ 2  wx, t ∈ L 2 0, 1 for each t, t ≥ 0. Additional
insight into this lack of smoothing is obtained if we use the identity
cosnπat sinnπx =
to write
∞
wx, t = ∑ n=1
1
2
1
2
sin nπx + at + sin nπx − at
f n sin nπx + at + sin nπx − at.
Then it follows from (7.9) that as N → ∞,
N
∑ n=1
1
2
f n sin nπx + at + sin nπx − at →
1
2
fx + at + fx − at in L 2 0, 1.
from which it is clear that the solution can be no smoother than the data. In fact, if the data
is assumed to be only in L 2 0, 1, there is some question about the meaning of the
differential equation. If we suppose f belongs to H 2 0, 1, then the differential equation is
satisfied in the L 2 − sense.
Notice that since each of the functions sinnπx is periodic with period equal to 2, the
series in (7.9) is convergent to a function which is also periodic with period equal to 2. So,
while the series does converge to fx on 0, 1, it converges to ̂fx, the periodic extension
of fx on the whole real line. More precisely, ̂fx, denotes the function which is the odd
extension to −1, 0 of fx, and is then the 2-periodic extension of this function. The
convergence properties of the series in (7.9) are then dictated by the smoothness of this
extension.
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