18.303 Problem Set 5 Solutions
Problem 1: (5+10+5)
(a) If an inverse Â−1 exists, then for any u, v we have hu, vi = hu, ÂÂ−1 vi = hÂ∗ u, Â−1 vi = h(Â−1 )∗ Â∗ u, vi, which
means that (Â−1 )∗ Â∗ = 1 and hence (Â−1 )∗ = (Â∗ )−1 . The adjoint of the inverse is the inverse of the adjoint!
If Â∗ = Â, it immediately follows that (Â−1 )∗ = Â−1 , i.e. the inverse is Hermitian.
´
(b) If Â−1 u(x) = Ω G(x, x0 )f (x0 )dn x0 and Â−1 is Hermitian under the usual inner product, then we have
¨
−1
u(x)G(x, x0 )v(x0 ) dn x dn x0
hu, Â vi =
Ω
¨
= hÂ−1 u, vi =
G(x, x0 )u(x0 )v(x) dn x dn x0
Ω
¨
0
G(x , x)u(x)v(x0 ) dn x dn x0
=
Ω
for all u, v. By comparing the two integrals, it follows that we have
G(x, x0 ) = G(x0 , x)
for all (or technically, almost all) x, x0 . This property is sometimes called reciprocity.
(c) The Green’s function G(x, x0 ) is precisely the diffusion from a source at x0 to the point x [i.e., it is the steadystate solution u(x) for f (x) = δ(x − x0 )]. If this is nonzero, then by reciprocity (above) it follows that G(x, x0 )
is also nonzero, so we must also have diffusion from a source at x to the point x0 [which is precisely G(x, x0 )].
So, “one-way” diffusion is impossible.
Problem 2: (5+10+5 points)
(a) g(x) is not locally integrable for intervals containing the origin. For example
ˆ b
ˆ b
dx
b
|g(x)|dx = lim+
= lim+ ln x| = ∞.
x
→0
→0
0
Therefore, f 0 {φ} = f {−φ0 } is a singular distribution.
(b) We write f 0 {φ} = lim→0+ f0 {φ}, and integrate by parts in f0 {φ} = f {−φ0 }:


ˆ −
ˆ ∞
f 0 {φ} = − lim+ 
ln(−x) φ0 (x)dx +
ln(x) φ0 (x)dx
| {z } | {z }
→0
−∞ | {z } | {z }
u
dv
−
u
dv
ˆ ∞
−1
1
φ(x)dx − ln()φ() −
φ(x)dx
x
→0
−∞ x
ˆ −
ˆ ∞
1
1
(((
((
= lim+
φ(x)dx +
φ(x)dx + lim+ [ln()
(φ()
−(
φ(−))].
(
(
(
x
→0
→0
((
−∞ x
(
´
In the last line, the first limit is precisely the Cauchy Principal Value of g(x)φ(x)dx (CPV = remove a ball
of radius around the singularity, do the integral, and then take the → 0+ limit). The second term vanishes
because, since φ(x) is continuous and infinitely differentiable, φ() − φ(−) vanishes at least as fast as as → 0,
so its product with ln vanishes in the limit as in part (i).1
ˆ
= − lim+ ln()φ(−) −
(c) Use f 0 {φ} = f {−φ0 } = f {−[φ − φ(0)]0 } = f 0 {φ − φ(0)}. Then substitute φ − φ(0) into the previous part, and
is now finite as x → 0 [since φ(x) − φ(0) goes to zero as x → 0, at
note that since the integrand φ(x)−φ(0)
x
least
proportionally
to
x
or
faster
as
in
the previous part], we can now just take the limit to write f 0 {φ} =
´∞
g(x)[φ(x) − φ(0)]dx without using the CPV.
−∞
1 This last fact is a little subtle. Naively, we could just write the Taylor expansion φ(x) = φ(0) + φ0 (0) + O(2 ) to obtain φ() − φ(−) =
2φ0 (0)+O(2 ). However, it turns out that test functions φ(x) are not analytic (do not have a convergent Taylor series) at all x. Nevertheless,
because φ is differentiable with bounded φ0 , it follows that φ(x) is “Lipshitz continuous:” |φ(x) − φ(y)| < K|x − y| for all x, y and for some
constant K > 0 (e.g. you can prove this from the mean-value theorem of analysis), so φ() − φ(−) < 2K and the result follows. But I
don’t expect you to provide this level of proof; a Taylor argument is acceptable.
1