Homework 4 Solutions
Laplacian in Polar Coordinates
Suppose that we have a C2 function u : R2 → R, and we define a new function
v : [0,∞) × [0, 2π) → R by v(r, θ) = u(r cos θ, r sin θ). The question is then
asking us to prove the identity
(uxx + uyy)(r cos θ, r sin θ) = vrr(r, θ) +
1
r
vr(r, θ) +
1
r2
vθθ(r, θ).
To prove this, we compute using the chain rule:
vr(r, θ) = ux(r cos θ, r sin θ) cos θ + uy(r cos θ, r sin θ) sin θ,
vθ(r, θ) = −ux(r cos θ, r sin θ)r sin θ + uy(r cos θ, r sinθ)r cos θ.
Differentiating again using the chain rule, we find:
vrr(r, θ) = uxx(r cos θ, r sin θ) cos
2 θ + 2uxy(r cos θ, r sin θ) cos θ sin θ
+uyy(r cos θ, r sin θ) sin
2 θ,
vθθ(r, θ) = uxx(r sin θ, r sin θ)r
2 sin2 θ − ux(r cos θ, r sin θ)r cos θ
−2uxy(r cos θ, r sin θ)r2 sin θ cos θ
+uyy(r cos θ, r sin θ)r
2 cos2 θ − uy(r cos θ, r sin θ)r sin θ.
Now just add the expressions for vrr,
1
r
vr, and
1
r2
vθθ and check that it equals
(uxx + uyy)(r cos θ, r sin θ). The identity sin
2 θ + cos2 θ = 1 will be useful.
This proves the formula for the laplacian in polar coordinates. The separation
of variables is done quite explicitly on page 22 of the book.
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Chapter 2, Exercise 19
Separation of variables leads to the solutions u(x, y) = f(x)g(y), where f(x) =
c1 sin(ax) + c2 cos(ax) and g(y) = d1e
ay + d2e
−ay (the other solution where f
has an exponential form and g has a sinusoidal form is not possible unless f = 0
due to the boundary conditions). The boundary conditions say f(0) = f(1) = 0
which force c2 = f(0) = 0 and hence a ∈ πZ. Assuming without loss of gen-
erality that a > 0, we throw out the unbounded (“unreasonable”) part of the
solution for g, corresponding to d1e
ay. So we have showed that any bounded
solution u(x, y) = f(x)g(y) has the form ce−nπy sin(nπx).
Then the general solution is given by superposition:
u(x, y) =
∞∑
n=1
ane
−nπy sin(nπx).
If we define
Py(x) =
∞∑
n=−∞
e−|n|πyeiπx,
then we claim that
u(x, y) = (f ∗ Py)(x) =
1
2
∫ 1
−1
f(z)Py(x− z)dz,
where f is defined on [−1, 0] by odd reflection. Indeed, f has Fourier coefficients
f̂(n) =
1
2
∫ 1
−1
f(x)e−iπxdx = −i sign(n)a|n|/2,
while Py has Fourier coefficients
P̂y(n) =
1
2
∫ 1
−1
Py(x)e
−iπxdx = e−|n|πy,
and therefore f ∗Py has Fourier coefficients which are the product of those of the
individual functions, namely −i sign(n)a|n|e−|n|πy/2, which agree with those of
u(·, y).
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Chapter 3, Exercise 3
Let us enumerate a sequence of intervals {Ik}∞k=1 as
I1 = [0, 1/2], I2 = [1/2, 1],
I3 = [0, 1/3], I4 = [1/3, 2/3], I5 = [2/3, 1],
I6 = [0, 1/4], I7 = [1/4, 1/2], I8 = [1/2, 3/4], I9 = [3/4, 1], ….
Then let fk denote the indicator function of the interval Ik, i.e., fk(x) = 1 if
x ∈ Ik and fk(x) = 0 otherwise.
It is clear that any point x lies in Ik infinitely often, but x also lies in the
complement of Ik infinitely often. Hence fk(x) alternates between 0 and 1 in-
finitely often, for any x.
On the other hand, the lengths of the intervals Ik are clearly tending to 0
as k → ∞, in fact |Ik| ≤ 1/n as soon as k > n(n + 1)/2. Hence
∫
fk(x)
2dx =
|Ik| → 0.
Correction: Our example was constructed on the interval [0, 1]. The book
actually asks to construct an example on [0, 2π], which is very similar (just
consider fk(
x
2π
)).
3
Chapter 3, Exercise 5
One may solve the problem by using the hint to find an explicit antideriva-
tive for (log x)2. However, we use a different approach.
We claim that for any α > 0, there exists some constant C = C(α) such that
| log(1/x)| ≤ Cx−α for every x ∈ (0, 2π]. Indeed, L’Hopital’s rule shows that
lim
x→0+
log(x)
x−α
= lim
x→0+
1/x
−αx−α−1
= −α−1 lim
x→0+
xα = 0.
Using this bound, we get that | log(1/x)|2 ≤ C2x−2α = C ′x−2α. Letting α = 1/4
gives | log(1/x)|2 ≤ C ′x−1/2. Consequently, if 0 < a < b then∫ b
a
| log(1/x)|2dx ≤ C ′
∫ b
a
x−1/2dx ≤ 2C ′b1/2 = C ′′b1/2.
In particular, fn must be a Cauchy sequence since if n > m then∫ 2π
0
(fn(x)− fm(x))2dx =
∫ 1/m
1/n
| log(1/x)|2dx ≤ C ′′m−1/2,
which can clearly be made smaller than some predefined � if m is large enough.
Nevertheless, fn does not converge in R because the limit would be f which is
unbounded, hence not in R. Indeed, any unbounded function has infinite upper
Darboux sums, hence is not Riemann-integrable.
4
Chapter 3, Exercise 8
Recall Parseval’s identity which says
∑
n∈Z
|f̂(n)|2 =
1
2π
∫ π
−π
|f(x)|2dx.
Part a: Using Part b of Exercise 6 of Chapter 2, we see that f̂(0) = π/2,
f̂(n) = − 2
πn2
for n odd, and f̂(n) = 0 otherwise. Consequently, Parseval gives
π2
4
+
4
π2
∑
n∈Z, odd
n−4 =
1
2π
∫ π
−π
x2dx =
1
3
π2.
In particular,
8
π2
∞∑
n=0
(2n+ 1)−4 =
4
π2
∑
n∈Z, odd
n−4 =
[
1
3
−
1
4
]
π2 =
1
12
π2.
Hence
∑∞
0 (2n+ 1)
−4 = π4/96. Letting S =
∑∞
1 n
−4, we see that
S =
∑
n≥1, even
n−4 +
∑
n≥1, odd
n−4 =
S
24
+
π4
96
=⇒ S =
16
15
·
π4
96
=
π4
90
.
Part b: Using Exercise 4b of Chapter 2, the Fourier coefficients are f̂(n) =
4
iπn3
sign(n), if n is odd and 0 otherwise. Then applying Parseval,
32
π2
∞∑
n=0
(2n+ 1)−6 =
16
π2
∑
n∈Z, odd
n−6 =
1
π
∫ π
0
x2(π − x)2dx =
1
30
π4.
This gives the desired identity
∑∞
n=0(2n+1)
−6 = π6/960, from which the other
identity may be derived using the same odd-even decomposition as in part a
(after noting that 960 · 63
64
= 945).
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