Homework 7 Solutions
Chapter 5, Exercise 15
Part a: Let g be the same function from Exercise 2. By the Poisson summation
formula,
∞∑
n=−∞
sin2(πn+ πα)
π2(n+ α)2
=
∞∑
−∞
ĝ(n+ α) =
∞∑
−∞
g(n)e2πinα = 1.
Since sin(πn+ πα) = (−1)n sin(πα), the desired identity follows immediately.
Part b: It suffices to prove the claim when α ∈ (0, 1), since the sum on the
left hand side only depends on the fractional part of α. So let 0 < α < 1.
First we note that for any N ∈ N
N−1∑
n=−N
1
n+ 1
2
= 0,
since the positive terms cancel the negative ones. Next note that for any α ∈
(0, 1),
1
n+ α
−
1
n+ 1
2
=
∫ 1/2
α
1
(n+ x)2
dx,
and thus by combining the previous two identities we get
N−1∑
n=−N
1
n+ α
=
∫ 1/2
α
N−1∑
n=−N
1
(n+ x)2
dx.
Letting N →∞ on both sides and using uniform convergence on [α, 1/2] of the
series inside the integral, we get
lim
N→∞
N−1∑
n=−N
1
n+ α
=
∫ 1/2
α
π2
sin2(πx)
dx =
π
tan(πα)
,
where the RHS is interpreted as 0 for α = 1/2, and we used the fact that the
antiderivative of csc2 u is − cotu.
1
Chapter 5, Problem 1
Let us write U(y, t) = u(e−y, t). Then we see by the chain rule that
Ut(y, t) = ut(e
−y, t), Uy(y, t) = −e−yux(e−y, t),
Uyy(y, t) = e
−2yuxx(e
−y, t) + e−yux(e
−y, t).
In particular, if ut = x
2uxx + axux, then
Ut(y, t) = ut(e
−y, t)
= e−2yuxx(e
−y, t) + ae−yux(e
−y, t)
= Uyy(y, t) + (1− a)Uy(y, t).
Thus U satisfies the constant-coefficient equation Ut = Uyy + (1− a)Uy. Taking
Fourier transforms, we see that Û satisfies the following ODE for fixed ξ:
∂tÛ(ξ, t) = −4π2ξ2Û(ξ, t) + 2πiξ(1− a)Û(ξ, t),
where Û(ξ, t) =
∫
R U(y, t)e
−2πiξydy is the Fourier transform in the y variable.
Solving the ODE for Û gives the solution
Û(ξ, t) = Û(ξ, 0)e(−4π
2ξ2+2(1−a)πiξ)t,
so by inverting the Fourier transform,
U(y, t) =
∫
R
Û(ξ, 0)e(−4π
2ξ2+2(1−a)πiξ)te2πiξydξ.
This last expression is simply the convolution of U(y, 0) = f(e−y) =: F (y) with
the kernel pt(y) :=
∫
R e
(−4π2ξ2+2(1−a)πiξ)te2πiξydξ. By completing the square
and then using the fact that
∫
R e
−(cξ+d)2dξ =
√
π/c2 for c ∈ R, we obtain that
pt(y) = e
−(y+(1−a)t)2/4t
∫
R
e
−
(
2πξt1/2− i
2t1/2
(
y+(1−a)t
))2
dξ =
1
√
4πt
e−(y+(1−a)t)
2/4t.
Thus U(y, t) = pt ∗ F (y) = 1√4πt
∫
R F (z)e
−(y−z+(1−a)t)2/4tdz, so that (trans-
forming variables back to the original function),
u(x, t) = U(− log x, t)
=
1
√
4πt
∫
R
F (z)e−(− log x−z+(1−a)t)
2/4tdz
=
1
√
4πt
∫ ∞
0
f(v)e−(log(v/x)+(1−a)t)
2 dv
v
,
where we substituted z = − log v in the final line, so that dz became −dv
v
,
F (z) = f(e−z) became f(v), and the limits changed from (−∞,∞) to (0,∞).
2
Chapter 5, Problem 5
Part a: Proceed by showing the contrapositive (i.e., prove that if u solves the
heat equation but violates the stated maximum principle, then we can find a
solution to the heat equation which violates the statement made in part a).
To this end, let u solve the heat equation, then let A := max∂′R u, and sup-
pose u has a strict maximum at (x0, t0) ∈ R\∂′R. Now consider the function
v := A− u, and check that this also solves the heat equation. Clearly, v ≥ 0 on
∂′R, but v(x0, t0) < 0, since u(x0, t0) > A.
Part b: Suppose that v (as defined in the book) has a minimum at (x1, t1) ∈ R,
where x1 /∈ {a, b} and t1 6= 0. Since v has a minimum at (x1, t1),
vt(x1, t1) ≤ 0 and vxx(x1, t1) ≥ 0.
Here we are using the fact that t1 > 0 which ensures that v is actually differ-
entiable at (x1, t1), and we are also using x1 /∈ {a, b} (since boundary minima
need not satisfy the second derivative test). The above expression implies that
vxx(x1, t1)−vt(x1, t1) ≥ 0. On the other hand, since u solves the heat equation,
vt = ut + � = uxx + � = vxx + �, and thus vxx(x1, t1) − vt(x1, t1) = −� < 0,
contradiction.
Part c: Assume (see part a) that u ≥ 0 on ∂′R. For � > 0, we know from
part b that v� := u + �t achieves its minimum at some point (x�1, t
�
1) ∈ ∂′R,
which means that
u(x, t) + �t = v�(x, t) ≥ v�(x�1, t
�
1) = u(x
�
1, t
�
1) + �t
�
1 ≥ �t
�
1,
where we used the fact that u ≥ 0 on ∂′R in the final inequality. Consequently,
we find that u(t, x) ≥ �(t�1−t) for every � > 0, as desired. The claim now follows
by letting �→ 0.
3
Chapter 6, Problem 8
Part a: We set β = 2π|x| on both sides of the given expression. In order to
prove the expression, we need to take the Fourier transform (in the x variable)
of both sides and check they are equal. First the left side:∫
R
e−2π|x|e−2πiξxdx =
∫ ∞
0
e−2πx(1+iξ)dx+
∫ 0
−∞
e2πx(1−iξ)dx
=
1
2π
[
1
1 + iξ
+
1
1− iξ
]
=
1
π(1 + ξ2)
.
Now the right side: we recall that
∫
R e
−cx2e−2πiξxdx =
√
π
c
e−π
2ξ2/c, and so by
changing the order of integration, we get∫
R
[ ∫ ∞
0
e−u
√
πu
e−4π
2×2/4udu
]
e−2πiξxdx =
∫ ∞
0
e−u
√
πu
[ ∫
R
e−π
2×2/ue−2πiξxdx
]
du
=
∫ ∞
0
e−u
√
πu
[√
u
π
e−uξ
2
]
du
=
1
π
∫ ∞
0
e−(1+ξ
2)udu
=
1
π(1 + ξ2)
,
thus proving the claim.
Part b: Using the subordination principle, we may write
e−2π|ξ|y =
∫ ∞
0
e−u
√
πu
e−π
2|ξ|2y2/udu,
where ξ ∈ Rd and |ξ| = (ξ21 + …+ ξ2d)
1/2. Thus we find that
(continued on next page)
4
P (d)y (x) =
∫
Rd
e2πix·ξe−2π|ξ|ydξ
=
∫
Rd
e2πix·ξ
[ ∫ ∞
0
e−u
√
πu
e−π
2|ξ|2y2/udu
]
dξ
=
∫ ∞
0
e−u
√
πu
[ ∫
Rd
e2πix·ξe−π
2|ξ|2y2/udξ
]
du
=
∫ ∞
0
e−u
√
πu
[
ud/2
πd/2yd
e
−u |x|
2
y2
]
du
= π−(d+1)/2y−d
∫ ∞
0
u(d−1)/2e
−u(1+ |x|
2
y2
)
du
=
π−(d+1)/2y−d
(1 +
|x|2
y2
)(d+1)/2
∫ ∞
0
v(d−1)/2e−vdv
=
Γ((d+ 1)/2) y
π(d+1)/2(y2 + |x|2)(d+1)/2
.
In the second-to-last equality, we made the substitution v = u(1 +
|x|2
y2
), and in
the final line we used the definition of the gamma function: Γ(z) =
∫∞
0
vz−1e−vdv,
and also multiplied the numerator and denominator by yd+1.
5