18.303 Problem Set 7
Due Wednesday, 27 October 2010.
Problem 1: Reciprocity
In class, we showed that if  is self-adjoint ( = Â∗ ), then Â−1 is as well (if it exists). From the fact that −∇2 is
real-symmetric, we concluded that the Green’s function G0 (x, x0 ) = G0 (x0 , x), a property called “reciprocity.”
(a) In the notes, section 2.1, we considered the operator −c∇2 for a real function c(x) > 0 on some Ω with Dirichlet
boundaries, and showed that its Green’s function is simply G(x, x0 ) R= G0 (x, x0 )/c(x0 ), where G0 is the Green’s
function of −∇2 on Ω. If we define the integral operator Ĝu = Ω G(x, x0 )u(x0 )dn x0 , show explicitly from
G0 (x, x0 ) = G0 (x0 , x) = real (not as in class by using the fact that Ĝ is the inverse of a self-adjoint operator)
that Ĝ = Ĝ∗ for an appropriately chosen inner product hu, vi. [Hint: remember problem 2(c) of pset 2.]
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(b) In electrostatics,
R the potential V of a charge density ρ satisfies −∇ V = ρ/ε0 . The potential energy of a charge
distribution is ρV , since V is the potential energy per unit charge; this is the energy to bringR the charges
together from infinity. If ρ(x) = ρ1 (x) + ρ2 (x) and V = V1 + V2 with −∇2 V1,2 = ρ1,2 /ε0 , then R ρ1 V2 is the
energy required to bring the ρ1 charges in from infinity with the ρ2 charges already there, and ρ2 V1 is the
energy required
to bring
the ρ2 charges in from infinity with the ρ1 charges already there. Using reciprocity,
R
R
show that ρ1 V2 = ρ2 V1 , which means that the energy doesn’t depend on which charges you hold fixed and
which ones you bring in from infinity. (ρ1 and ρ2 are purely real, of course.)
Problem 2: Born approximations
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.
P∞
(a) Suppose we want to solve Âu = f . Show that we can write u = u0 + B̂u = k=1 B̂ k u0 for some integral operator
B̂ (similar to what we did in the notes for −∇ · c∇). Explicitly write B̂u as some integral of u and G0 .
(b) Similar to section 3 of the notes, suppose that c = 0 everywhere except in a small volume V (centered at x1 )
where c(x) = c1 . Suppose f (x) = δ(x − x0 ) and we want the solution at x, where x and x0 are far from x1
compared to the diameter of V . Suppose Ω = R3 so that G0 (x, x0 ) = 1/4π|x−x0 |. Apply the Born approximation
u ≈ u0 + B̂u0 , and explicitly write the (approximate) solution u(x) as a sum of G0 term from the source at x0
and some term from a “source” at x1 due the inhomogeneity, keeping only the lowest-order term with respect to
the small parameter (the diameter of V compared to |x0 − x1 | and |x − x1 |).
Problem 3: Born again
In this problem, you will compare a Born approximation to an exact numerical solution for a 1d problem, using the
d
d
c dx
u on
finite-difference approximation from the previous problem sets, considering the 1d operator Âu = − 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. Use an n × n
finite-difference approximation with n = 100.
(a) Give Matlab commands to form a finite-difference approximation A to Â, similar to problem 3 of pset 2 except
now the c(x) function is in between the derivatives (between the D matrices). Careful: remember that D is
(n + 1) × n, and produces the derivatives not at the grid points but halfway between the grid points. [Note: it
looks like there was a typo on pset 2: it computed DT D/∆x2 and said this was a discrete d2 /dx2 , but of course
that is actually a discrete −d2 /dx2 . I’ve fixed this online.]
(b) Using section 2.2.2 of the notes, write down the Born approximation for the solution u to Âu = δ(x − x0 ) [that is,
u = G(x, x0 ), the Green’s function of Â). You should be able to perform all the integrals explicitly (by breaking
it up into two cases, x < x0 and x > x0 , you should get a few easy integrals). Plot your approximate “scattered”
part u − u0 = B̂u0 , defined as in the notes, versus x for x0 = L/2 = 1/2.
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(
(∆x)−1 j = n/2
to find the finite(c) In Matlab, use a discrete “delta” function δ(x − L/2) vector b where bj =
0
otherwise
αn∆x/2
difference solution u = A−1 b (A \ b in Matlab). Also compute the finite-difference u0 solution A−1
,
0 b/e
d2
where A0 is the finite-difference approximation to − dx2 . Plot u−u0 , and compare with your Born approximation
from the previous part—plot them on the same plot so that you can compare quantitatively.
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