18.303 Problem Set 7 Solutions
Problem 1: 10+10
R 1
ū(x)v(x)dn x, so that c∇2 is Hermitian
(a) Just as in pset 2, we should define the inner product to be hu, vi = Ω c(x)
R
R
R
(since hu, c∇2 i = ū∇2 v = v∇2 ū = 1c vc∇2 ū = hc∇2 u, vi ) and hence we expect that Ĝ = (c∇2 )−1 will be
Hermitian. We can check this explicitly by just writing out the integrals (omitting the complex conjugations
since everything here is real):
Z
Z
0
1
n 0 G0 (x, x )
0
n
u(x)
d x
v(x )
hu, Ĝvi =
d x
c(x)
c(x0 )
Ω
Ω
Z
Z
1
G0 (x0 , x)
=
dn x0
u(x)
v(x0 )
dn x
0)
c(x)
c(x
Ω
Ω
Z
Z
0
1
0
n
n 0 G0 (x, x )
u(x )
v(x)
=
d x
d x
0)
c(x
c(x)
Ω
Ω
= hĜu, vi,
where in the second line we have interchanged the order of the integrals and used reciprocity G0 (x, x0 ) = G0 (x0 , x),
and in the third line we have renamed the integration variables x ↔ x0 .
R
R
(b) We want to show that ρ1 V2 = ρ2 V1 , i.e. that hρ1 , V2 i = hV1 , ρ2 i (ρ and V being real) where we define
R
hu, vi = ūv as usual [with −∇2 and the resulting Green’s function operator Ĝ = (−∇2 )−1 being Hermitian
under this inner product, as shown in class]. There are two ways to do this: either express V in terms of ρ as
V1,2 = Ĝρ1,2 , or express ρ in terms of V as ρ1,2 = −∇2 V1,2 . In either case, the result then follows trivially from
the Hermitian property of Ĝ and −∇2 :
either hρ1 , V2 i = hρ1 , Ĝρ2 i = hĜρ1 , ρ2 i
= hV1 , ρ2 i
or hρ1 , V2 i = h−∇2 V1 , V2 i = hV1 , −∇2 V2 i
= hV1 , ρ2 i.
Problem 2: 10+10
Consider the operator  = −∇2 + c(x) for some real function c(x) ≥ 0, on a domain Ω with Dirichlet boundary
condiR
tions. Let the c = 0 Green’s function be G0 (x, x0 ), satisfying −∇2 G0 (x, x0 ) = δ(x − x0 ). Let u0 (x) = G0 (x, x0 )f (x0 )
be the solution to −∇2 u0 = f , as in the notes.
(a) Âu = f = −∇2 u + cu, which means −∇2 u = f − cu, which is the same as the c = 0 case with a new righthand-side f − cu instead of f . Hence we can quote the given solution from the c = 0 case to obtain the integral
equation:
Z
u(x) =
G0 (x, x0 ) [f (x0 ) − c(x0 )u(x0 )] dn x0
Ω
= u0 (x) + B̂u
where u0 is the c = 0 solution and
Z
B̂u = −
G0 (x, x0 )c(x0 )u(x0 )dn x0 .
Ω
Hence the Born–Dyson series u =
P∞
k=0
B̂ k u0 follows as in the notes.
(b) In this case, u0 (x) = G0 (x, x0 ) by definition of G0 . In the Born approximation, we must compute
Z
B̂u0 = −
G0 (x, x0 )c(x0 )u0 (x0 )dn x0
Ω
Z
dn x0
= −c1
2
0
0
V 16π |x − x | |x − x0 |
1
but as in the notes the integrand is almost constant. Taylor expanding to first order around x0 = x1 , we find
1
1
(x0 − x1 ) · (x1 − x)
,
≈
1−
|x0 − x|
|x1 − x|
|x1 − x|2
where the second term is of order (diameter V )/|x1 −x| compared to the first term (since |x0 −x1 | ≤ diameter V )
and hence we will neglect it; similarly for the x0 − x0 term. Hence, to lowest order we can set x0 = x1 in the
integrand, making the integrand a constant that integrates to the volume of V :
B̂u0 ≈ −c1 G0 (x, x1 )G0 (x1 , x0 ) (volume of V ).
Thus
u(x) ≈ G0 (x, x0 ) − c1 G0 (x, x1 )G0 (x1 , x0 ) (volume of V ),
which is the sum of a source at x0 and an effective/induced source at x1 (proportional to c1 , u0 at x1 , and the
volume of V ).
Problem 3: 10+10+5
In this problem, you will compare a Born approximation to an exact numerical solution for a 1d problem, using the
d
d
u on
finite-difference approximation from the previous problem sets, considering the 1d operator Âu = − dx
c dx
x ∈ [0, L], for c(x) > 0 and Dirichlet boundaries u(0) = u(L) = 0. From class, we already know the Green’s function
d2
G0 (x, x0 ) for − dx
2:
(
G0 (x, x0 ) =
x0 (1 − x/L) x > x0
.
x(1 − x0 /L) x < x0
We will consider L = 1 and a function c(x) = eαx that is nearly constant if α is small; take α = 0.01.
(a) Similar to pset 2, we will approximate  by:
L = 1; n = 100; dx = L / (n + 1);
xhalf= [0:n]’ * dx + dx/2;
alpha = 0.01;
D = diff1(n);
A = D’ * diag(exp(alpha * xhalf)) * D / dx^2;
The main difference from pset 2 is that now our diagonal c(x) matrix is in between the two D matrices. However, this means it is evaluated at different points than before: Du takes a vector of n values um at m∆x for
m = 1, 2, . . . n and produces n + 1 approximate derivatives u0m+1/2 at points halfway in between: (m + 0.5)∆x
for m = 0, 1, 2, . . . , n. Since c(x) multiplies u0 and not u, we must evaluate it at these n + 1 halfway points,
represented by the xhalf array above.
(b) The Born approximation is u = u0 + B̂u0 , where in this case u0 (x) = G0 (x, x0 )/c(x0 ) and B̂u0 is the integral
operator, from the notes:
Z L
d
d
0
0
0
B̂u0 =
G0 (x, x )
ln c(x )
u0 (x ) dx0
dx0
dx0
0
"(
#
Z L
0
−x
/L
x
>
x
1
0
0
= αx0
G0 (x, x0 ) [α]
dx0
e
(1 − x0 /L) x0 < x0
0
"
#
Z
Z
α x0 x0
x0 L
0
0
0
0
= αx0
G0 (x, x )dx −
G0 (x, x )dx .
1−
e
L
L x0
0
There are two cases: x < x0 and x > x0 . In the x < x0 case,
"
#
Z
Z x0 Z L
α x0 x x 0 0 x0 x0
x0
x0
0
0
B̂u0 = αx0
1−
1−
x dx + 1 −
x
1−
dx − x
1−
dx
e
L
L 0
L
L
L
L
x
x0
α x0 x x2 x0 x2 − x2
x0
L2 − x20
= αx0
1−
1−
+ 1−
x x0 − x − 0
− x L − x0 −
e
L
L 2
L
2L
L
2L
α x(L − x0 )(x0 − x)
= αx0
> 0.
e
2L2
2
−4
2
x 10
Born approximation
finite difference
1.5
1
u−u
0
0.5
0
−0.5
−1
−1.5
−2
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x
Figure 1: Scattered solution u − u0 in the Born approximation (= B̂u0 ) (red line) and from a 100-point finite-difference
approximation (blue circles).
In the x > x0 case:
#
Z L
Z
Z
x0 x0
x x0 0 0 x0 x x 0 0 x0
0
B̂u0 = αx0
1−
1−
dx
1−
x dx −
1−
x dx − x
e
L
L 0
L
L x0
L
L
x
x0 x x20
x0 x x2 − x20
x0
L2 − x2
α 1−
1−
−
1−
− x L−x−
= αx0
e
L
L 2
L
L
2
L
2L
α x0 (L − x)(x − x0 )
= − αx0
< 0.
e
2L2
α
"
In short:
α(x − x0 )
u(x) − u0 (x) = B̂u0 = −
2L2 eαx0
(
x(L − x0 ) x < x0
.
x0 (L − x) x > x0
(It’s okay if you don’t simplify things this much, however. Being lazy, I actually used Maple to simplify the final
expression.) The solution u−u0 is plotted as a solid red line in figure 1, where B̂u0 is computed with the command:
x = linspace(0,L, 1000); x0 = 0.5;
Bu0 = -alpha/(2*L^2*exp(alpha*x0)) * (x - x0) .* ...
(x .* (L - x0) .* (x < x0) + x0 .* (L - x) .* (x >= x0));
(c) In Matlab, b is formed by b = zeros(n,1); b(n/2) = 1/dx and A0 is just A0 = D’ * D/dx^2. Plotting u − u0
as blue dots in figure 1 by plot([1:n]’*dx, A \ b - A0 \ b / exp(alpha * x0), ’bo’), we find that it
agrees very well with the analytical Born approximation.
3