AMath 567, Autumn 2011
Sample Solutions for Assignment 8.
Reading: Sec. 7.1-7.4, Sec. 8.1-8.2.
1. Apply Parseval’s relation to the function f (x) = x on (−π, π) to evaluate
The Fourier coefficients of f (x) = x are: fˆ(0) =
n 6= 0,
√1
2π
Rπ
−π
P∞
1
n=1 n2 .
x dx = 0, and for
1 Z π −inx
fˆ(n) = √
xe
dx
2π −π
"
#
Z π
i −inx π
i −inx
1
x e
e
dx (using integration by parts)
= √
−
n
−π n
2π
−π
1
i
= √
π (e−inπ + einπ ) + (e−inπ − einπ )
2π n
√ i
√ i
=
2π cos(nπ) = ± 2π .
n
n
By Parseval’s identity, the sum of squares of absolute values of the Fourier
coefficients is equal to the square of the L2 [−π, π]-norm of x, namely,
Z π
x2 dx =
−π
2π 3
.
3
Since fˆ(0) = 0 and fˆ(−n) = −fˆ(n), the sum of squares of the Fourier
coefficients is twice the sum of squares of those corresponding to positive n;
hence
∞
∞
X
X
2π 3
1
π2
1
=
,
or,
=
.
4π
2
2
3
6
n=1 n
n=1 n
2. Exercise 7.3 on p. 183.
First note that the functions
1
1
1
√ , √ cos(kx), √ sin(kx), k = 1, 2, . . . ,
π
π
2π
form an orthonormal basis for L2 [−π, π].
Since
the functions cos(kx)
, k = 0, 1, 2, . . . are even functions, we have
Rπ
Rπ
−π cos(kx) cos(jx) dx = 2 0 cos(kx) cos(jx) dx, and so the set
√
1
2
√ , √ cos(kx), k = 1, 2, . . . ,
π
π
1
is an orthonormal set in L2 [0, π]. RSimilarly, since the functions
sin(kx), k =
Rπ
π
1, 2, . . . are odd functions, we have −π sin(kx) sin(jx) dx = 2 0 sin(kx) sin(jx) dx,
and so the set
√
2
√ sin(kx), k = 1, 2, . . . ,
π
is an orthonormal set in L2 [0, π].
Let f ∈ L2 [0, π]. Then f can be extended to [−π, π] in such a way that the
extended function is in L2 [−π, π] and is either even or odd:
(
fe (x) =
f (−x) −π ≤ x < 0
, fo (x) =
f (x)
0≤x≤π
(
−f (−x) −π ≤ x < 0
.
f (x)
0≤x≤π
In either case, let g denote the extended function. Then g can be expanded
in a Fourier series consisting of sin’s and cos’s:
g(x) =
where
∞
a0 X
(ak cos(kx) + bk sin(kx)),
+
2
k=1
1Zπ
g(x) cos(kx) dx, k = 0, 1, 2, . . . ,
ak =
π −π
1Zπ
bk =
g(x) sin(kx) dx, k = 1, 2, . . . .
π −π
Since fe (x) is an even function and sin(kx) is odd and cos(kx) is even, we
have
Z π
−π
fe (x) sin(kx) dx = 0,
Z π
−π
fe (x) cos(kx) dx = 2
Z π
0
fe (x) cos(kx) dx.
Thus f (x) can be expanded on [0, π] as
f (x) =
∞
α0 X
+
αk cos(kx),
2
k=1
where
2Zπ
f (x) cos(kx) dx.
π 0
P
Equivalently, f (x) = ∞
α̂ f (x), where the fk ’s are the functions defined
k=0
Rπ k k
in the problem and α̂k = 0 f (x)fk (x) dx. Since this series converges to fe in
L2 [−π, π], it converges to f in L2 [0, π], and so the functions fk , k = 0, 1, 2, . . .,
form a complete orthonormal set, i.e., an orthonormal basis for L2 [0, π].
Similarly, since fo (x) is an odd function, we have
αk =
Z π
−π
fo (x) cos(kx) dx = 0,
Z π
−π
fo (x) sin(kx) dx = 2
2
Z π
0
fo (x) sin(kx) dx.
Thus f (x) can be expanded on [0, π] as
∞
X
f (x) =
βk sin(kx),
k=1
where
2Zπ
βk =
f (x) sin(kx) dx.
π 0
P
β̂ e (x), where the ek ’s are the functions defined
Equivalently, f (x) = ∞
k=1
Rπ k k
in the problem and β̂k = 0 f (x)ek (x) dx. Since this series converges to fo in
L2 [−π, π], it converges to f in L2 [0, π], and so the functions ek , k = 1, 2, . . .,
form a complete orthonormal set, i.e., an orthonormal basis for L2 [0, π].
3. Exercise 7.7 on p. 184.
Let y = (π/L)x and define v(y, t) ≡ u(x, t) = u((L/π)y, t). Then vt (y, t) =
ut (x, t) and vyy (y, t) = (L/π)2 uxx (x, t). Thus v(y, t) satisfies the initialboundary value problem:
vt = (π/L)2 vyy ,
v(0, t) = v(π, t) = 0, t > 0,
v(y, 0) = f ((L/π)y), 0 ≤ y ≤ π.
q
Since the set of sin functions { 2/π sin(ny) : n = 1, 2, . . .} form an orthonormal basis for L2 [0, π], f ((L/π)y) can be expanded in terms of these
functions:
∞
X
2Zπ
fn sin(ny), fn =
f ((L/π)s) sin(ns) ds.
f ((L/π)y) =
π 0
n=1
Look for a solution v(y, t) of the form v(y, t) =
tiating term-by-term, we find
vt (y, t) =
∞
X
P∞
vn0 (t) sin(ny), vyy (y, t) = −
n=1
∞
X
vn (t) sin(ny). Differenvn (t)n2 sin(ny).
n=1
n=1
2
Equating Fourier coefficients of vt and (π/L) vyy gives
vn0 (t) = −(π/L)2 n2 vn (t), n = 1, 2, . . . .
Solving for vn and applying the initial condition:
2 n2 t
vn (t) = e−(π/L)
vn (0) = e−(π/L)
2 n2 t
fn .
Substituting this into the expression for v(y, t) gives
∞
X
v(y, t) =
2 n2 t
e−(π/L)
fn sin(ny).
n=1
Finally, since v(y, t) = u(x, t),
u(x, t) =
∞
X
2 n2 t
e−(π/L)
n=1
3
fn sin(n(π/L)x).
4. Exercise 7.10 on p. 184.
If u(x, t) = f (x + ct) + g(x − ct), then, differentiating using the chain rule, we
find ut (x, t) = cf 0 (x+ct)−cg 0 (x−ct), so that utt (x, t) = c2 f 00 (x+ct)+c2 g 00 (x−
ct); ux (x, t) = f 0 (x+ct)+g 0 (x−ct), so that uxx (x, t) = f 00 (x+ct)+g 00 (x+ct).
Thus, utt = c2 uxx .
5. Exercise 8.8 on p. 213.
Let ϕx (z) = hx, zi and ϕy (z) = hy, zi be two linear functionals in H ∗ and
define an inner product on H ∗ by
hϕx , ϕy i = hy, xi.
First check that this is an inner product:
(a) hϕx , ϕx i = hx, xi ≥ 0 with equality if and only if x = 0, which is the
case if and only if ϕx = 0;
(b) hϕy , ϕx i = hx, yi = hy, xi = hϕx , ϕy i;
(c) and for any scalars α and β, hϕx , αϕy + βϕw i = hϕx , ϕᾱy+β̄w i = hᾱy +
β̄w, xi = αhy, xi + βhw, xi = αhϕx , ϕy i + βhϕx , ϕw i.
Now show that H ∗ is complete with the norm defined by this inner product.
Let (ϕxn ) be a Cauchy sequence in H ∗ . Then given > 0 there exists N
such that n, m > N implies
kϕxn − ϕxm k = hϕxn − ϕxm , ϕxn − ϕxm i1/2 < .
But since
hϕxn − ϕxm , ϕxn − ϕxm i = hxn − xm , xn − xm i,
this means that (xn ) is a Cauchy sequence in H, hence has a limit x ∈ H.
Claim: ϕxn converges to ϕx in H ∗ . To see this, note that
kϕxn − ϕx k = hϕxn − ϕx , ϕxn − ϕx i1/2 = hxn − x, xn − xi1/2 = kxn − xk → 0.
Thus any Cauchy sequence (ϕxn ) in H ∗ has a limit ϕx in H ∗ , where x =
limn→∞ xn , so H ∗ is a Hilbert space with this inner product.
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