� � � � � � 18.303 Problem Set 7 Due Wednesday, 27 October 2010. Problem 1: Reciprocity In class, we showed that if Aˆis self-adjoint ( Aˆ= Aˆ∗), then Aˆ−1 is as well (if it exists). From the fact that −�2 is real-symmetric, we concluded that the Green’s function G0(x, x�) = G0(x�, 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, x�) = G0(x, x�)/c(x�), where G0 is the Green’s function of −�2 on Ω. Gu = G(x, x�)u(x�)dnIf we define the integral operator ˆ� Ω x�, show explicitly from G0(x, x�) = G0(x�, x) = real (not as in class by using the fact that Gˆis the inverse of a self-adjoint operator) that Gˆ= Gˆ∗ for an appropriately chosen inner product �u, v�. [Hint: remember problem 2(c) of pset 2.] (b) In electrostatics, the potential V of a charge density ρ satisfies −�2V = 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 bring the charges together from infinity. If ρ(x) = ρ1(x) + ρ2(x) and V = V1 + V2 with −�2V1,2 = ρ1,2/ε0, then ρ1V2 is the energy required to bring the ρ1 charges in from infinity with the ρ2 charges already there, and ρ2V1 is the energy required to bring the ρ2 charges in from infinity with the ρ1 charges already there. Using reciprocity, show that ρ1V2 = ρ2V1, 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 Aˆ= −�2 + c(x) for some real function c(x) ≥ 0, on a domain Ω with Dirichlet boundary condi-tions. Let the c = 0 Green’s function be G0(x, x�), satisfying −�2G0(x, x�) = δ(x − x�). Let u0(x) = G0(x, x�)f(x�) be the solution to −�2u0 = f, as in the notes. (a) Suppose we want to solve Auˆ= f. Show that we can write u = u0 +ˆ�∞k=1 BˆkBu = u0 for some integral operator Bˆ(similar to what we did in the notes for −� · c�). Explicitly write ˆBu 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, x�) = 1/4π|x−x�. Apply the Born approximation u ≈ u0 +ˆ|term from the source at x0Bu0, and explicitly write the (approximate) solution u(x) as a sum of G0 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 dfinite-difference approximation from the previous problem sets, considering the 1d operator Auˆ= − dx � c dx u � on x ∈ [0, L], for c(x) > 0 and Dirichlet boundaries u(0) = u(L) = 0. From class, we already know the Green’s function G0(x, x�) for − d2 dx2 : � G0(x, x�) = x�(1 − x/L) x > x� . x(1 − x�/L) x < x� 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 Aˆ, 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/Δx2 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 ˆAu = δ(x− x0) [that is, u = G(x, x0), the Green’s function of Aˆ). 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” ˆpart u − u0 = Bu0, defined as in the notes, versus x for x0 = L/2 = 1/2. 1� (c) In Matlab, use a discrete “delta” function δ(x − L/2) vector b where bj = (Δx)−1 j = n/2 to find the finite-0 otherwise difference solution u = A−1b (A \ b in Matlab). Also compute the finite-difference u0 solution A−01b/eαnΔx/2 , d2where 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. 2MIT OpenCourseWare http://ocw.mit.edu 18.303 Linear Partial Differential Equations: Analysis and Numerics Fall 2010 For information about citing these materials or our Terms of Use, visit: