18.303: Notes on the1d-Laplacian Green’s functionSteven G. JohnsonOctober 12, 2011In class, we solved for the Green’s function G(x,x0) of the 1d Poisson equation−d2dx2u = f where u(x) is a function on [0,L] with Dirichlet boundaries u(0) = u(L) = 0.We obtained:G(x,x0) =(−x0L(x − L) x > x01 −x0Lx x ≤ x0=(x01 −xLx > x0x1 −x0Lx ≤ x0,which can be derived in any number of ways. Once we understand delta “functions”and distributions, we can derive it from −∂2∂ x2G(x,x0) = δ (x − x0), where the δ (x − x0)yields a a slope discontinuity of 1 in G at x = x0. Instead, we first derived it the “long”way, by taking a width-∆x approximation for the delta function (= 1/∆x in [x0,x0+ ∆x]and = 0 otherwise) on the right-hand side, solving, and then taking the limit ∆x → 0on the solution (whereas the limit ∆x → 0 of the right-hand side does not exist as anordinary function).Regardless of how it is obtained, we get the G(x,x0) above. Clearly, this G satisfiesthe reciprocity property G(x,x0) = G(x0,x), which (as shown in class) follows from thefact that the Laplacian is real-symmetric with these boundary conditions. Physically,G(x,x0) could be thought of as a “plucked string:” a stretched string where you haveconcentrated a finite force at one “point” x0in the string.It is also good to check thatu(x) =ˆL0G(x,x0) f (x0)dx0satisfies −d2dx2u = f for any f (x). Note that u(0) = u(L) = 0 automatically, sinceG(0,x0) = G(L,x0) = 0 by construction. If we use delta functions, taking the secondderivative of G is easy. However, let’s do it the “hard way” instead, performing onederivative at a time and avoiding differentiating any discontinuous function. We can dothis by breaking the integral up into two pieces [0,x] and [x, L] based on the two piecesof G(x,x0):u(x) =1 −xLˆx0x0f (x0)dx0+ xˆLx1 −x0Lf (x0)dx0,1where we have pulled the x terms out of the x0integrals. The first derivative is then:u0(x) = −1Lˆx0x0f (x0)dx0+ˆLx1 −x0Lf (x0)dx0+1 −xLx f (x) −x1 −xLf (x),where the terms on the second line [which cancel each other] come from the derivativesof the integration limits. [Recall the fundamental theorem of calculus:ddx´xh(x0)dx0=h(x) andddx´xh(x0)dx0= −h(x) for any h.] Since x only appears in the integrationlimits in u0, the second derivative is then given by the fundamental theorem of calculusagain:u00(x) = −1Lx f (x) −1 − xLf (x) = − f (x),and thus −u00= f as desired.More abstractly, the key properties are that G(x,x0) is continuous: G(x,x−) =G(x,x+) [where x±denote the limits as x0→ x from the left and right], while G0(x,x0) =∂∂ xG(x,x0) has a jump discontinuity G0(x,x−) = G0(x,x+)−1, and G00(x,x0) =∂2∂ x2G(x,x0) =0 for x 6= x0. That means:u(x) =ˆx0G(x,x0) f (x0)dx0+ˆLxG(x,x0) f (x0)dx0,u0(x) =ˆx0G0(x,x0) f (x0)dx0+ˆLxG0(x,x0) f (x0)dx0+G(x,x−) f (x) −G(x,x+) f (x),u00(x) = G0(x,x−) f (x) − G0(x,x+) f (x) = − f (x)as desired (omitting the G00= 0 terms from the last line).So, all of this can be solved without delta functions or discontinuities, and peopledid this historically. But it is much easier if we don’t have to continually dodge discon-tinuities and can differentiate them naturally. Generalizing functions as distributions(or solving the “weak form” of the PDE), and setting up the rules for differentiatingdistributions, will allow us to do
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