18.303 Problem Set 6 Solutions
Problem 1: (5+10) Scalar Helmholtz equation
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Recall that usual scalar wave equation [e.g. for pressure waves u(x, t)] is equation Âu + f (x, t) = ∂∂t2u , where f (x, t) is
an external force density (pressure) on the string and  = ∇ · c∇ for c(x) > 0 (e.g. c ~ springiness, or “bulk modulus”).
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(a) Suppose that ũ solves Âũ + f˜(x, t) = ∂∂t2ũ and satisfies the boundary conditions, e.g. ũ = 0 on the boundaries
¯
ũ
for Dirichlet. Now, consider u = Re ũ = ũ+
2 . Clearly, u satisfies the same boundary conditions (e.g. Dirichlet
or Neumann). It also satisfies the PDE:
"
#
1 ∂ 2 ũ ∂ 2 ũ
∂2u
=
+ 2
∂t2
2 ∂t2
∂t
h
i
1
Âũ + f˜ + Âũ + f˜
=
2
= Âu + f,
˜ ¯
˜
f
˜
since f = f +
2 = Re f . The key factors that allowed us to do this are (i) linearity, and (ii) the real-ness of the
PDE (the PDE itself contains no i factors or other complex coefficients).
(b) In this case, we have ü = −ω 2 u = Âu + f . Multiplying both sides by eiωt , we obtain −ω 2 v = Âv + g, or Ĥv = g
where Ĥ = −Â − ω 2 .
Since  is self-adjoint for Dirichlet boundary conditions (from class), and −ω 2 is trivially self-adjoint (just
a real number), then Ĥ is self-adjoint. However, since −Â is positive-definite (from class) and −ω 2 is negativedefinite (obviously), Ĥ is in generally indefinite. The eigenvalues of Ĥ are the eigenvalues of −Â minus ω 2 , and
hence if ω 2 is large enough (larger than the smallest eigenvalue of −Â), the eigenvalues of Ĥ can have either
sign.
Problem 2: ((5+10)+2 points) Green’s functions
(a) We solve for g(r) in 3d as follows:
(i) For r > 0, −∇2 g − ω 2 g = 0 and hence −ω 2 g = ∇2 g = 1r (rg)00 =⇒ h00 = −ω 2 h where h(r) = rg(r). The
iωr
−iωr
solution to this is h(r) = ceiωr + de−iωr for some constants c and d, or g(r) = ce +de
r
It is a little more tricky to determine whether we should use the c or the d term than in class, since
both decay at the same rate. The ratio c/d will be determined by some kind of boundary condition at
infinity, but what might this be? It is acceptable for you to just punt on this here; since e±iωr are complex
conjugates of each other, your analysis will apply equally well to either one, and you can arbitrarily pick
one, say ceiωr /r, to analyze.
However, to see why there should be a sensible choice, recall that this operator arose in the previous
problem by assuming a time dependence e−iωt multiplying the solution, in which case we are looking at
iω(r−t)
+de−iω(r+t)
wave solutions ce
, where the c term describes waves moving out towards r → ∞, while
r
the d term describes waves moving in from infinity. In wave problems, we typically impose a boundary
condition of outgoing waves at infinity, in which case we must set d = 0. (However, the choice would have
been reversed if we picked the opposite sign convention, e+iωt , for the time dependence.)
(ii) Let’s focus on g(r) = ceiωr /r. As in class, the 1/r singularity is no problem in 3d (it is cancelled by the
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Jacobian factor r2 dr), so g is a regular distribution. Given an arbitrary test function q(x), we now evaluate
ˆ
(Âg){q} = g{Âq} = g Âq


ˆ π
ˆ 2π
ˆ ∞
( 
 2
g ∂2
(((
−ω
gq
−
sin θ dθ 
dφ
r2 dr
=
(rq) + (θ,(
φ(
derivatives
((( of q) 

2
(
{z
} 
|
r
∂r
0
0
0
integrate to 0, from class


 
¨
ˆ ∞ 
 
∂2


 
c −ω 2 rqeiωr − eiωr 2 [rq]  dr
sin θ dθ dφ 
∂r

 0
{z
} 
|
int. by parts


 
¨
ˆ ∞

 
∞
∂
∂

 2 iωr
 
=
sin θ dθ dφ  −ceiωr [rq] + c
−ω rqe + iωeiωr [rq] dr
∂r
∂r


0
0
{z
} 
|
int. by parts
¨
ˆ ∞
( ((
∞
(2 e(
iωr
2
iωr
iωr
(
(
=
+
c
dr
sin θ dθ dφ cq(0) − icωe
[rq]
−ω
rqe
+
ω
[rq]
(
0
(((
=
0
= 4πcq(0),
and hence c = 1/4π . Thus
0
G(x, x0 ) =
eiω|x−x |
4π|x − x0 |
assuming boundary conditions such that d = 0. More generally, since exactly the same result applies to the
de−iωr /r term, we obtain,
0
0
ceiω|x−x | + de−iω|x−x |
0
G(x, x ) =
|x − x0 |
for c + d = 1/4π , with the ratio c/d being set by the boundary conditions at ∞. The value at x = x0
being irrelevant in the distribution sense, e.g. we can assign it to zero, since this is a regular distribution
with a finite integral, similar to class.)
(b) The ω → 0 limit gives 1/4π|x − x0 | as in class, by inspection.
Problem 3: Born again (20 points)
The Green’s function where ω = ω2 is
0
G0 (x, x0 ) =
eiω2 |x−x |
4π|x − x0 |
from above. Write ω 2 = ω22 + c(x), where c = 0 in the ω2 region and c = ω12 − ω22 in the ω1 volume. Then we can write
the PDE Âu = f as
(−∇2 − ω22 )u = f + cu,
and hence we can write a Born-Dyson series
u = B̂f + B̂cu = B̂f + B̂cB̂f + · · ·
in class, where B̂ = (−∇2 − ω22 )−1 is integration against the Green’s function G0 . If we let u0 = B̂f as in class, then
we have the first Born approximation:
ˆ
u(x) ≈ u0 + B̂cu0 = u0 (x) + G0 (x, x0 )c(x0 )u0 (x0 ).
As in the notes, f = δ(x − x0 ), so u0 (x) = G0 (x, x0 ). The next term has an integrand that is only nonzero inside
V , giving:
ˆ
G0 (x, x0 )G0 (x0 , x0 ).
(ω12 − ω22 )
V
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However, as in the notes, if the volume V is small and and far from both the source point x0 and the observation point
x, then the integrand is nearly constant and we can approximately pull it out of the integral. In particular, if x1 is an
arbitrary point in V , we need the diameter of V to be much smaller than |x0 − x1 | and |x − x1 | to make the 1/r terms
in G0 approximately constant as in class, and we need the diameter of V to be much smaller than 1/ω2 to make the
oscillatory numerator of G0 approximately constant in the integral. Under these conditions, we obtain simply
u(x) ≈ G0 (x, x0 ) + (ω12 − ω22 )G0 (x, x1 )G0 (x1 , x0 ) V,
which is much simpler than the form of the Born approximation from class.
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