� � � 18.303 Problem Set 6 Due Wednesday, 20 October 2010. Problem 1: Regularization and Green’s functions (a) Let f(x) = 1for x > 0, and f(x) = 0 for x ≤ 0.√x (i) Explain why f defines a regular distribution, even though f(x) blows up as x 0+ .→ (ii) Let g(x) = − 1 1 for x > 0, and g(x) = 0 for x ≤ 0: g(x) matches the ordinary derivative f�(x)2 x3/2 everywhere f�(x) is defined (i.e. everywhere but x = 0). Explain why g(x) does not correspond to any regular distribution. (iii) Viewed as a distibution, f must have a derivative. Give an explicit formula for f�{φ} in terms of an integral of φ(x) − φ(0) (not φ�). Hint: f{φ} = lim� 0 ∞ φ(x) dx (why does this limit exist?), and integrate by parts � √x→using φ�(x) = d [φ(x) − φ(0)]. How is this different from trying to define a distribution directly from g(x)?dx (b) Consider the −�2 operator in two dimensions, with domain Ω = R2 (the whole space). [In Cartesian and 2 ∂2 ∂2 1 ∂ � ∂ � 1 ∂2cylindrical coordinates, � = ∂x2 + ∂y2 = r ∂r r ∂r + r2 ∂θ2 .] This has translational symmetry and rotational symmetry, so as in class the Green’s function −�2G(x, x�) = δ(x − x�) is of the form G(x, x�) = G(x − x�, 0) = g(|x − x�|) = g(r), with −�2g(r) = δ(x). Solve for g(r) in three steps, as in class: (i) For r > 0, −�2g(r) = 0; show that this gives a solution proportional to ln(r) with some unknown propor-tionality constant c [determined in (iii) below], not including additive constants (which are in the nullspace of −�2 and hence make the solution nonunique, even if we impose rotational symmetry). [The fact that ln(r) blows up at r → ∞ makes 2d more tricky than 3d.] (ii) Explain why your g(r) defines a regular distribution (i.e. why its integral against any (smooth, localized) test function φ is defined, even though the logarithm blows up as r 0 and r → ∞. Suggestion: write g{φ}� 2π � → as lim�→00 dθ � ∞ g(r)φ(r, θ)r dr, where the limit avoids the need to define the value of g(r) at r = 0. (iii) Using the distributional derivative (−�2g){φ} = g(r)[−�2φ] for a smooth localized test function φ(x), show that −�2g = δ(x) as desired [i.e. show (−�2g){φ} = = φ(0)] for an appropriate choice of c.�δ{φ} � 2π � Suggestion: write φ in cylindrical coordinates φ(r, θ), where = lim� 00 dθ � ∞ r dr, as in the previous →part. Problem 2: A simple integral-equation solver In this problem, you will implement a simple integral-equation solver in Matlab to solve for the Green’s function −�2G(x, x�) = δ(x − x�) in two dimensions, in the domain Ω given by the exterior of radius-1 circle at the origin, with Dirichlet boundaries (G vanishes at the surface dΩ of the circle). This problem is diagrammed in figure 1(left). A physical interpretation in electrostatics (for example) would be that δ(x − x�) corresponds to a line charge at x�, and G(x, x�) is the resulting potential (“voltage”) at x [in units where ε0 = 1], where the circle is a metal cylinder that is grounded (voltage = 0). If the circle weren’t there (“empty space”), the Green’s function would be c ln |x − x�| from the previous problem, for some c that you found; for simplicity, in this problem just suppose c = 1. With the circle there, the solution is changed—it not only has the ln |x − x�| term coming directly from the δ(x − x�), but there are also terms coming from the cylinder—compared to the empty-space solution, there are additional source terms on dΩ so that the total G satisfies the boundary condition. The (first-kind) integral-equation formulation of this (as discussed in class in 3d) is to put unknown sources with density σ(x) all around dΩ, so that the total solution: G(x, x�) = ln |x − x�| + dΩ σ(x��) ln |x − x��|dx�� satisfies the boundary condition G(x, x�) = 0 for x ∈ dΩ. We then try to find σ(x��), discretized/approximated in some way, to make this happen. (In the electrostatic interpretation, σ is an induced charge density on the cylinder surface.) Here, we will approximate the continuous distribution σ(x��) by N point sources σnδ(x�� − xn) for N equally-spaced points xn around the circle, with unknown amplitudes σn, and we will enforce the Dirichlet boundary condition approximately, only at another set of N points ym that are halfway in between the xn points. (Caveat: there are much 1� � R = 1domain Ωboundary dΩx'x'σ1σ2σ3σ4σNσNx1x2x3x3xNxNy1y2y3y4Figure 1: Schematic of 2d Green’s function problem. Left : for a domain Ω that is the exterior of a radius-1 cylinder, we want to solve for the Green’s function −�2G(x, x�) = δ(x, x�) with Dirichlet (zero) boundary conditions at dΩ. Right: approximate integral-equation problem for the Green’s function, in which we replace the surface by a set of unknown point sources σnδ(x − xn) at N points xn. We will then enforce the Dirichlet boundary conditions at N points yn on the cylinder, halfway in between the xn points. better ways to set up this approximation in a “serious” calculation; these are generally called Nyström methods.) This is shown schematically in figure 1(right). That is, we will solve: NG(ym, x�) = 0 ≈ ln |ym − x�| + σn ln |ym − xn|n=1 for m = 1, 2, . . . , N, giving N equations for the N unknowns σn. The key quantities will be the vector b with components bm = ln |ym − x�| and the matrix A with Amn = ln |ym − xn|. These are created in Matlab by the following commands, for N = 10011 and x� = (2, 0): N = 1001; dtheta = 2*pi / N; x_theta = [0:N-1]’ * dtheta; y_theta = x_theta + dtheta/2; x_x = cos(x_theta); x_y = sin(x_theta); y_x = cos(y_theta); y_y = sin(y_theta); b = log(sqrt((y_x - 2).^2 + y_y.^2)); o = ones(1, N); A = log(sqrt((y_x * o - o’ * x_x’).^2 + (y_y * o - o’ * x_y’).^2)); (a) Why didn’t we just choose yn = xn? (Caveat: as mentioned above, there are much better ways to handle this difficulty than what we are doing here.) (b) If we want to solve for the vector s = (σ1, σ2, . . .)T , what equation should s solve to enforce the boundary conditions above? (Hint: not quite As = b.) (c) Plot your solution s versus θ (x_theta) [it will be easier to read if you convert θ to degrees