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MIT 18 303 - Green's functions and inverse operators

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Lecture 12: 4 October 2010Lecture 12: 4 October 2010 New topic: Green's functions and inverse operators. For an invertible matrix A, Ax=b is solved by x=A-1b, where A-1A=AA-1=I. For a general invertible operator A on functions, with Au=f, it is natural to ask whether we can define an operator A-1 such that u=A-1f. We can, and this leads to the topics of Green's functions and delta functions, which will take us some time to cover. Approach #1: fundamental solution. In the matrix case, x=A-1b is given by xi=Σi'(A-1)i,i'bi'. For Au=f, the analogue of xi is u(x) for some point x, the analogue of bi' is f(x'), and the analogue of the sum is an integral: u(x) = ∫Ω G(x,x') f(x') dx' = A-1f, where G(x,x') is a Green's function or fundamental solution. That is, if A is a differential operator, we should generally expect A-1 to be an integral operator. This is a perfectly good definition of G(x,x'): whatever function makes that integral give you the solution u(x) to Au=f. However, if we apply that definition to actually find G(x,x'), we run into difficulty. Since A is an operator that only acts on x, not x', evidently we must have: Au(x) = ∫Ω [A G(x,x')] f(x') dx' = f(x) . What does this mean about the function [A G(x,x')]? It must only have contributions from f(x), so it must be zero for x'≠x. But around x'=x it must have integral 1, so that it integrates to f(x) when multiplied by f(x'). But how can a function have a nonzero integral if it is only nonzero at one point? Evidently, it must be "infinite" at that point, and we are led informally to define a delta function δ(x-x'): this is zero for x≠x', infinite for x=x', and has integral 1. Then G is "defined" by AG(x,x')=δ(x-x'). This is a perfectly good equation, except for one thing: δ(x-x') is not a function in the usual sense, and our definition of it is almost nonsense. Approach #2: superposition Another, equivalent way to think about A-1 is that it is the solution to AA-1=I: that is, each column gm of A-1 solves Agm=em, where em is the unit vector in the m-th direction (the m-th column of I). Then, we write any b as a sum of the components of b times em, and hence Ax=b is solved by the superposition: the sum of the components of b times gm. For a Green's function, in principle we want to do something similar: break up f(x) into localized "pieces", solve Ag for each piece, and then add them up. The trick is what the "pieces" look like. For the 1d problem -u''=f(x) for u(0)=u(L)=0, we broke up f(x) approximately into a sum of piecewise-constant pieces, of width Δx each. At the end, we will try to take the Δx→0 limit to recover f(x) and the exact G(x,x'). We write each piece in terms of a "box" function s(x)=1 for x in [0,Δx] and s(x)=0 otherwise. We write f(x) as approximatelyΣmf(mΔx)[s(x-mΔx)/Δx]Δx, multiplying and dividing by Δx to make it look more like an integral in the limit Δx→0. We then solve: g''x'(x) = s(x-x')/Δx for gx'(x), so that the solution to Au=f will approximately be Σmf(mΔx)gmΔx(x)Δx by superposition. In the limit Δx→0 we will get G(x,x') from g. Note that the right-hand side s(x-x')/Δx does not have a well-defined limit as Δx→0, at least not in the ordinary definition of a "function", but note that it seems to behave like our (still non-rigorous) δ(x-x') in that it is zero for x≠x' in this limit, while its area is always 1. To solve for gx'(x), we play the trick of breaking up the problem into three regions with constant right-hand sides, solving in each region, and then piecing together the solutions afterwards. In particular, the three regions here are: • x<x': g''x'(x)=0, g(0)=0. Thus, gx'=αx for some α. • x>x'+Δx: g''x'(x)=0, g(L)=0. Thus, gx'=β(x-L) for some β. • x in [x',x'+Δx]: -g''x'(x)=1/Δx. Thus, gx'=-x2/2Δx + γx + κ for some γ and κ To determine the four unknown constants α, β, γ, and κ, we must put all the solutions together by enforcing the appropriate continuity conditions. From the PDE, g''x' is finite everywhere, gx' and g'x' must be continuous (making gx' piecewise twice-differentiable). Imposing these continuity conditions at x' and x'+Δx gives us four equations which we can solve for the four unknowns. We obtain β=-(x'+Δx/2)/L, α=1+β, γ=β+1+x'/Δx, and κ=-(x')2/2Δx. (This is the "hard way" to get a Green's function; we will learn an easier way once we can more carefully define delta functions.) Further reading: Strang book, section 1.4.MIT OpenCourseWarehttp://ocw.mit.edu 18.303 Linear Partial Differential Equations: Analysis and NumericsFall 2010 For information about citing these materials or our Terms of Use, visit:


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