Topic: Introduction to Green s functions

Transcription

1 Lecture notes on Variational and Approximate Methods in Applied Mathematics - A Peirce UBC 1 Topic: Introduction to Green s functions (Compiled 2 September 212) In this lecture we provide a brief introduction to Green s Functions Key Concepts: Green s Functions, Linear Self-Adjoint Differential Operators, 9 Introduction/Overview 91 Green s Function Example: A Loaded String Consider the forced boundary value problem Figure 1 Model of a loaded string Lu = u (x) = φ(x) u() = = u(1) Physical Interpretation: u(x) is the static deflection of a string stretched under unit tension between fixed endpoints and subject to a force distribution φ(x) Newtons per unit length shown in figure 1 Question: Since this is a linear equation can we invert the differential operator L = d2 dx 2 the solution in the form: u(x) = T (x) φ to obtain an expression for

2 2 911 Method 1: Variation of Parameters Homogeneous eqn: u = has solution u(x) = c 1 x + c 2 Part Solution: u(x) = xv 1 (x) + v 2 (x) u = v 1 (x) + {xv 1 + v 2} u = v 1 + { } Therefore Lu = v 1 = φ(x) Require { } = since we need another constraint to determine v 1 > v 2 uniquely [ ] [ ] [ ] x 1 v 1 1 v 2 = φ(x) Therefore u(x) = v 1 = φ(x)/( 1) x v 1 = φ(s)ds Therefore u p (x) = u c = c 1 x + c 2 + φds x v 2 = xφ(x)( 1) x v 2 = sφ(s)ds φ(s)ds x u() = c 2 =, u(1) = c 1 + u(x) = = φds + x(s 1)φ(s)ds + sφds x + 1 G(s, x)φ(s)ds where G(s, x) = sφ(s)ds φds 1 φds x s(x 1)φ(s)ds sφ(s)ds x(s 1) s > x sφ(s)ds = sφ(s)ds

3 Green s Functions for two-point Boundary Value Problems 3 Physical Interpretation: G(s, x) is the deflection at s due to a unit point load at x Figure 2 Displacement of a string due to a point loading G(s, x) = x(s 1) s > x Physical Interpretation of reciprocity: G(s, x) = G(x, s) Therefore deflection at s due to a unit point load at x = deflection at x due to a unit point load at s Figure 3 Physical interpretation of reciprocity

4 4 912 Method 2: The Adjoint Operator Lu = u = φ u() = = u(1) vluds = vu ds = vu ] 1 v u ds = vu ] 1 v u] 1 + Note: Since L = L, we say that L is formerly self-adjoint uv ds 1 = [vu uv ] 1 + ul vds vφds = v(1)u (1) v()u () u(1) v (1) + u() v () = v(1)u (1) v()u () + + u d2 vds (91) ds2 u d2 vds (92) ds2 Up till now other than being sufficiently differentiable, v has been arbitrary How can we choose v so that we obtain an expression of the form: u(x) = If v satisfies the following boundary value problem v(s, x)φ(s)ds (93) L v = d2 ds 2 v(s, x) = δ(s x) v() = = v(1) then (92) reduces to (93) How do we solve (94)? Method A: direct integration v ss = δ(s x) Recall H (x) = s(x) v s = H(s x) + A v(s, x) = H(s x)ds + As = H(χ)dχ + As = χh(χ) + As + B = (s x)h(s x) + As + B s x = χ (94)

5 Green s Functions for two-point Boundary Value Problems 5 = v(, x) = B = v(1, x) = (1 x)h(1 x) + A Therefore A = (x 1)H(1 x) = (x 1) Therefore v(s, x) = (s x)h(s x) + s(x 1) = (s x) + sx s = x(s 1) s > x Method B: Stitching in the region s < x and s > x v ss = thus: { A s + B v(s, x) = = v s < x A + s + B + = v + s > x We have 4 constants and only two boundary conditions so we need some additional conditions to determine v Continuity at x v(x, x) = v(x +, x) Jump Condition at x Therefore x+ε x ε A x + B = A +? + B + (95) v ss = δ(s x) x+ε v ss ds = δ(s x)dx = 1 x ε [v s ] x+ε x ε = 1 A + A = 1 (96) = v(, x) = B (97) = v(1, x) = A + + B + (98) Therefore [ x (1 x) 1 1 ] [ A A + ] = [ 1 ] A = (1 x)/1 = (x 1), A + = x Therefore v(s, x) = x(s 1) s > x

6 6 92 Summary Given a linear differential operator lu = f + BC we will be looking for a Green s Function satisfying L G = δ(ε x) + appropriate BC such that we can express the inverse operator for L in the form: u(x) = G(ε, x)f(ε)ds Ω 93 Applications (1) Boundary integral methods Heat Transfer, Fluid Flow, Elasticity, Electram? (2) Tomography Note: What is the analogue of the Green s Function in a discrete problem? Consider a linear operator A : R N R N eg the matrix problem Au = f Suppose we solve each of the problems a u a 1n a n1 a nn u 1 u n A T v k = e k = [ 1 k ] T Now define the matrix V whose columns comprise the v k so that V = v 1 v 2 v n f 1 f n v T k Au = u T A T v k = u T e k = u k since A T v k = e k u k = vk T Au = vk T f since Au = f v 1 v 2 u = f = V T f = A 1 f v n