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5 Fourier series.
Just like Taylor expansions, Fourier series is a way of expressing a function as an infinite sum of other functions. In the case of the Taylor expansion, we write
f (x) =
∑ (^) f (n)(0) n! xn
which means that we expand f with respect to the basis { 1 , x, x^2 ,.. .}. In the case of Fourier cosine series we write f (x) = a 0 2
× 1 +
m=
am cos
( (^) mπx b
which means that we expand f with respect to the basis
1 , cos
( (^) πx b
, cos
( (^2) πx b
; in the case of Fourier sine series we write f (x) =
m=
am sin
( (^) mπx b
which means that we expand f with respect to the basis
sin
( (^) πx b
, sin
( (^2) πx b
; and in case of the full Fourier series we write f (x) = a 0 2
m=
am cos
( (^) mπx b
m=
bm sin
( (^) mπx b
which means that we expand f with respect to the basis
1 , cos
( (^) πx b
, cos
( (^2) πx b
,... , sin
( (^) πx b
, sin
( (^2) πx b
Let us now find out how to compute the coefficients a 0 , a 1 ,... and b 1 , b 2 ,.. .. The following is more or less from [7].
5.1 The coefficients.
The reason it is useful to expand in terms of cosines; sines; or cosines and sines is that they possess the following property:
DEFINITION 5.1 Let f and g be functions. Then we say that f and g are orthogonal in L^2 (Ω) if (f, g)L (^2) (Ω) = 0, where (·, ·)L (^2) (Ω) is the L^2 (Ω)-inner-product.
Let us now see that any two distinct elements in the collection
1 , cos
( (^) πx b
, cos
( (^2) πx b
and any two distinct elements in the collection
sin
( (^) πx b
, sin
( (^2) πx b
are orthogonal on [0, b].
LEMMA 5.2 For distinct m and n we have ∫
[0,b]
cos
( (^) mπx b
cos
( (^) nπx b
dx = 0
and we also have (^) ∫
[0,b]
cos^2
( (^) mπx b
dx = b 2
for m 6 = 0; and for m = 0 we have (^) ∫
[0,b]
dx = b.
Also, for distinct m and n we have ∫
[0,b]
sin
( (^) mπx b
sin
( (^) nπx b
dx = 0
and we also have (^) ∫
[0,b]
sin^2
( (^) mπx b
dx = b 2
PROOF: See homework.
Also, any two distinct elements in the collection
1 , cos
( (^) πx b
, cos
( (^2) πx b
,... , sin
( (^) πx b
, sin
( (^2) πx b
are or- thogonal on [−b, b]:
LEMMA 5.3 For all m and n we have ∫
[−b,b]
cos
( (^) mπx b
sin
( (^) nπx b
dx = 0;
for distinct m and n we have (^) ∫
[−b,b]
cos
( (^) mπx b
cos
( (^) nπx b
dx = 0
and (^) ∫
[−b,b]
sin
( (^) mπx b
sin
( (^) nπx b
dx = 0
and (^) ∫
[−b,b]
cos
( (^) mπx b
dx =
[−b,b]
sin
( (^) mπx b
dx = 0.
Also, (^) ∫
[−b,b]
cos^2
( (^) mπx b
dx =
[−b,b]
sin^2
( (^) mπx b
dx = b
and for m = 0, (^) ∫
[−b,b]
cos^2
( (^) mπx b
dx =
[−b,b]
dx = 2b.
PROOF: See homework.
Let us now see how lemma 5.2 will help us expand a function f defined on [0, b]. Suppose that we have already found an expansion in terms of cosines for f. That means that we can write
f (x) = a^0 2
m=
am cos
( (^) mπx b
(5.1) on [0, b].
Fix n and multiply (5.1) by cos
( (^) nπx b
and integrate over [0, b]. Then we have ∫
[0,b]
f (x) cos
( (^) nπx b
dx =
[0,b]
a 0 2 cos
( (^) nπx b
(5.2) dx
m=
[0,b]
am cos
( (^) mπx b
cos
( (^) nπx b
(5.3) dx.
The value on the right hand side of (5.2) depends on what n is: If n = 0, then by lemma 5.2 we have
2 b
[0,b]
f (x)dx =
b
[0,b]
a 0 2 (5.4) dx = a 0
which determines what a 0 must be. If n > 0 we have
2 b
[0,b]
f (x) cos
( (^) nπx b
dx =
b
m=
[0,b]
am cos
( (^) mπx b
cos
( (^) nπx b
(5.5) dx = an
and
an =
b
[0,b]
x cos
( (^) nπx b
(5.15) dx
[
2 x nπ sin
( (^) nπx b
)]
[0,b]
2 b n^2 π^2
[
cos
( (^) nπx b
)]
[0,b]
4 b (5.17) n (^2) π 2
Therefore, if the Fourier series converges, then we have
x = b 2 −^
n=
4 b n^2 π^2 cos
( (^) nπx b
on [0, b].
But just like the case of the Taylor expansion, there are issues with convergence − for the Taylor expansion there is a so-called radius of convergence, as you may recall. Let us first see what we mean by convergence.
DEFINITION 5.4 We say that a sequence of function f 1 , f 2 ,... converges to a function f in Lp(Ω), if ‖f − fm‖Lp(Ω) → 0.
EXAMPLE: Let us see if the partial sums of
n=1(1^ −^ x)xn−^1 converge. We have ∑^ N n=
(5.18) (1 − x)xn−^1 = (1 − x) + (1 − x)x + (1 − x)x^2 +... + (1 − x)xN^ −^1
= (1 − x)
[
1 + x +... + xN^ −^1
]
(5.20) = 1 − xN^.
For |x| < 1 we have xN^ → 0 as N → ∞; for x = −1 the partial sum oscillates; for x = 1 we have xN^ = 1; and for |x| > 1 the partial sum diverges. Let us therefore only consider x such that − 1 ≤ x ≤ 1. For 0 ≤ p < ∞ we have ∥∥ ∥∥ ∥^1 −
∑^ N
n=
(1 − x)xn−^1
p
Lp[− 1 ,1]
[− 1 ,1]
|xN p|dx =
N p + 1
which converges to 0 as N → ∞. For p = ∞ we have ∥∥ ∥∥ ∥^1 −
∑^ N
n=
(1 − x)xn−^1
L∞[− 1 ,1]
= max
|xN^ | : x ∈ [− 1 , 1]
This means that the sum converges in Lp[− 1 , 1] for p < ∞ but not in L∞[− 1 , 1].
In our case, the approximating functions f 1 , f 2 ,... are going to be the partial sums a 0 2
a 0 2
( (^) πx b
a 0 2
( (^) πx b
2 πx b
for example, and we are hoping that the partial sums will converge in Lp^ (for p = 2 or ∞) to the function we are expanding. For Fourier series we will see two different theorems regarding convergence, one for convergence in the L^2 - norm, and one for convergence in the L∞-norm. Let us begin with the convergence result for the L^2 -norm.
5.2 L^2 -convergence.
In this section, we will prove the following theorem for the sine series on [0, b]:
THEOREM 5.5 Let f be in L^2 [0, b]. Then the Fourier sine series converges to f in L^2 [0, b].
The proofs of the analogous results of the Fourier cosine series and the full Fourier series are very similar. We begin with a lemma which shows that the Fourier sine series of f is the closest thing to f which is in the form of a Fouries series.
LEMMA 5.6 (The least squares approximation) Let f be a function in L^2 [0, b]. Fix an integer N. From the collection of all possible c 1 ,... , cN , the Fourier coefficients b 1 ,... , bN minimize
EN (c 1 ,... , cN ) =
f −
∑^ N
m=
cm sin
( (^) mπx b
L^2 [0,b]
PROOF: We have
E N^2 =
f −
∑^ N
m=
cm sin
( (^) mπx b
2
L^2 [0,b]
[0,b]
|f (x)|^2 dx − 2
∑^ N
m=
cm
[0,b]
f (x) sin
( (^) mπx b
(5.24) dx
∑^ N
m=
∑^ N
n=
cmcn
[0,b]
sin
( (^) mπx b
sin
( (^) nπx b
(5.25) dx.
Because of lemma 5.2, the last term in (5.23) is equal to
∑^ N m=
c^2 m
[0,b]
sin^2
( (^) mπx b
dx = b 2
∑^ N
m=
c^2 m.
Thus we have
E N^2 =
[0,b]
|f (x)|^2 dx +
∑^ N
m=
b 2
[
cm −
b
[0,b]
f (x) sin
( (^) mπx b
dx
] 2
− b 2
∑^ N
m=
[0,b]
f (x) sin
( (^) mπx b
dx
by completing the square. The right hand side of (5.26) is minimized when
cm =^2 b
[0,b]
f (x) sin
( (^) mπx b
dx
that is when c 1 ,... , cN are the Fourier coefficients.
{ This means that the Fourier expansion is the closest thing to^ f^ in the subspace spanned by vectors sin
( (^) πx b
, sin
( (^2) πx b
. Now we prove that the Fourier sine series converges somewhere; we do not yet know what it converges to, but we hope that it is f.
LEMMA 5.7 Let f be a function in L^2 [0, b] and let
m=1 bm^ sin^
( (^) mπx b
∑ be the Fourier sine series of^ f^. Then ∞ m=1 bm^ sin^
( (^) mπx b
is in L^2 [0, b].
LEMMA 5.10 Let f be a function in L^2 [0, b] and let
m=1 bm^ sin^
( (^) mπx b
be the Fourier sine series of f. Then
f (x) =
∑^ ∞
m=
bm sin
( (^) mπx b
in L^2 [0, b].
PROOF: Fix n. Then ∫
[0,b]
[
f (x) −
∑^ ∞
m=
bm sin
( (^) mπx b
)]
sin
( (^) nπx b
(5.32) dx
[0,b]
f (x) sin
( (^) nπx b
dx −
∑^ ∞
m=
bm
[0,b]
sin
( (^) mπx b
sin
( (^) nπx b
(5.33) dx
= bnb 2 − bnb 2
This means that for all n, ( f (x) −
∑^ ∞
m=
bm sin
( (^) mπx b
, sin
( (^) nπx b
L^2 [0,b]
Since
sin
( (^) πx b
, sin
( (^2) πx b
is complete, we therefore know that
f (x) =
∑^ ∞
m=
bm sin
( (^) mπx b
Finally, from the orthogonality of the basis elements, we can deduce Parseval’s identity.
PROPOSITION 5.11 Let f be in L^2 [0, b] and let
m=1 bm^ sin^
( (^) mπx b
be the Fourier sine series of f. Then
‖f ‖^2 L (^2) [0,b] = b 2
∑^ ∞
m=
b^2 m.
Let us now work on some examples.
EXAMPLE: Consider the function f (x) = 1 on [0, b]. It is clear that f is in L^2 [0, b]. We can therefore find the Fourier sine expansion of f and expect it to be well-behaved. The coefficients (given by (5.7))
bm =
b
[0,b]
sin
( (^) mπx b
(5.36) dx
[
mπ cos
( (^) mπx b
)]b 0
mπ (5.38) [1 − cos(mπ)]
=
mπ (5.39) [1 − (−1)m].
This means that if m is even we have bm = 4/mπ and if m is odd, bm = 0.
5.3 L∞-convergence.
In this section we state a theorem concerned with L∞-convergence:
THEOREM 5.12 Let f , f ′^ and f ′′^ be continuous on [0, b]. The Fourier sine series converges in L∞[0, b] to f.
PROOF: This is a complicated theorem. We will omit the proof.
5.4 Existence for Laplace’s equation.
Now we have the technology to consider the existence proof in one special setting. In this section we will consider two problems:
- Find ϕ such that ∆ϕ = 0 on Ω with boundary condition ϕ = g on ∂Ω
- Find ϕ such that ∆ϕ = 0 on Ω with boundary condition ∇ϕ · N = g on ∂Ω.
The following discussion is from [2] and can also be found in [7] and [8]. We will consider these two on the very special domain Ω = {(x, y) ∈ R^2 : 0 ≤ x ≤ a and 0 ≤ y ≤ b}. We will consider a very special form of solution ϕ(x, y) = X(x)Y (y). That is, the solution will split into two parts one part which only depends on x and another which only depends on y. This technique is known as variables separation. Since we know that solutions are unique, we know that if we were to find a solution of this form, it is actually the only solution to this problem. Let us apply the two dimensional Laplacean to ϕ(x, y) = X(x)Y (y). The equation we wish to solve then is X′′(x)Y (y) + X(x)Y ′′(y) = 0 on Ω.
This means that we want to solve
X′′(x) X(x)
−Y ′′(y) Y (y)
Since the left hand side of (5.40) depends only on x and the right hand side of (5.40) depends only on y, both sides must be equal to a constant c. Thus we have
(5.41) X′′(x) = cX(x) and Y ′′(y) = −cY (y).
The ODE (5.41) is a differential eigenproblem and depending on the sign of the eigenvalue c, eigenfunction X (and Y ) changes:
- If c < 0, then we can write c = −λ^2 , with λ > 0 and from (5.41) we have
(5.42) X(x) = A sin(λx) + B cos(λx) and Y (y) = C sinh(λy) + D cosh(λy).
- If c = 0, then
(5.43) X(x) = A + Bx and Y (y) = C + Dy.
- If c > 0, then we can write c = λ^2 where λ > 0, and we have
(5.44) X(x) = A sinh(λx) + B cosh(λx) and Y (y) = C sin(λy) + D cos(λy).
To see which solution we want, we have to be more specific about the boundary values we want.
is also harmonic. To make this function fit the condition (5.48) we use what we know about Fourier series. We suppose that h is well-behaved enough that it has a Fourier sine expansion (see theorem 5.5 and theorem 5.12). To satisfy (5.48) we need to find a 1 , a 2 ,... such that
h(y) =
∑^ ∞
m=
am sinh
( (^) mπa b
sin
( (^) mπy b
By letting bm = am sinh
( (^) mπa b
we see that what we are looking for are Fourier coefficients.
5.4.2 A Neumann boundary condition.
The method of separation of variables is also amenable to the Neumann problem (problem 2 from above), and you will be asked to obtain such a solution in a homework.
References
[1] Schr¨odinger equation. Wikipedia, the free encyclopedia, http: // en. wikipedia. org/ wiki/ Schr% C3% B6dinger equation.
[2] William E. Boyce and Richard C. DiPrima. Elementary differential equations and boundary value problems. John Wiley & Sons Inc., New York, 1965.
[3] Lawrence C. Evans. Partial differential equations, volume 19 of Graduate Studies in Mathematics. American Mathematical Society, Providence, RI, 1998.
[4] David Gilbarg and Neil S. Trudinger. Elliptic partial differential equations of second order, volume 224 of Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences]. Springer-Verlag, Berlin, second edition, 1983.
[5] Elliott H. Lieb and Michael Loss. Analysis, volume 14 of Graduate Studies in Mathematics. American Mathematical Society, Providence, RI, second edition, 2001.
[6] Elias M. Stein. Singular integrals and differentiability properties of functions. Princeton Mathematical Series, No. 30. Princeton University Press, Princeton, N.J., 1970.
[7] Walter A. Strauss. Partial differential equations. John Wiley & Sons Ltd., Chichester, second edition, 2008. An introduction.
[8] E. C. Zachmanoglou and Dale W. Thoe. Introduction to partial differential equations with applications. Dover Publications Inc., New York, second edition, 1986.
