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