jim delta 1_13_11

Jim, 1. I think you proposed by phone that a solution to Lxy(1/r) = 2π δ(2)(r) might be this Lxy([1/r]) = (1/ (∂x2+∂y2)( [1/r] ) = 2π δ(2)(r) = δ(r)/r But dimensionally this says L-3 = L-2, so I think what you intended was: Lxy([1/r]) = (∂x2+∂y2)( ([1/r]) = 2π δ(2)(r) = δ(r)/r 2. Assuming this is what you meant, you are then claiming this (perhaps apart from a constant) 22D(r[1/r]) = 2π δ(2)(r) = δ(r)/r where 22D = ∂x2+∂y2 is the 2D Laplacian. In cylindrical coordinates we know that 3D2 = r-1∂r(r ∂r ) + r-2∂φ2 + ∂z2 So for polar coordinates r and φ it must be true that 2D2 = ∂x2+∂y2 = r-1∂r(r ∂r ) + r-2∂φ2 Therefore 22DF(r) = r-1∂r(r ∂r F(r) ) = r-1( r ∂r2F(r) + ∂rF(r)) = ∂r2F(r) + r-1 ∂rF(r) Let's take F(r) = r[1/r] so this then says 22D(r[1/r]) = ∂r2(r[1/r]) + (1/r) ∂r(r[1/r]) Thus, a new term has been added to your original proposal which was just ∂r2(r[1/r]). 3. Using my Part 2 methods of the last doc, I was able to show these interesting facts: limε→0 ∂r2(r[1 /]) = – 2π δ(2)(r) limε→0 {(1/r)∂r(r [1/]) } = + 2π δ(2)(r) After doing this, I realized that it was obvious from the divergence theorem applied to ψ, ∫V dV 2ψ = ∫S dS ∂nψ which applied to a 2D disk of radius a says !Syntax Error, Irdr !Syntax Error, Idθ 2ψ = a!Syntax Error, Idθ ∂rψ Setting ψ = r [1/r] we get !Syntax Error, Irdr !Syntax Error, Idθ 2(r [1/r]) = a!Syntax Error, Idθ ∂r(r [1/r])|r=a = a!Syntax Error, Idθ ∂r(1) = 0 So that 2(r [1/r]) = 0 4. So the bad news is that your two terms exactly cancel so that Lxy (1/r) = 22D (r[1/r]) = – 2π δ(2)(r) + 2π δ(2)(r) = 0 The good news is that we now have two different solutions to the problem which are these - ∂r2(r [1/r]) = 2π δ(2)(r) = δ(r)/r +(1/r)∂r(r[1/r]) = 2π δ(2)(r) = δ(r)/r which correspond to (Maple) // this stuff is wrong, see Question #2 doc review Lxy(1) = - (1/[ x2 ∂x2 + y2∂y2 +2xy∂x∂y + 2x∂x + 2y∂y ] Lxy(2) = +(1/)[ x2(∂x2+∂y2) + y2(∂x2+∂y2) + 2x∂x + 2y∂y] and any linear combination must also be viable. If we add these two we get, for example, Lxy(3) = +(1/)[ x2∂y2 + y2∂x2 – 2xy∂x∂y] = (1/) (x∂y- y∂x)2 But since ∂φ = (∂φy)∂y + (∂φx)∂x = (x∂y -y∂x), we get Lxy(3) = (1/r) ∂φ2 and this is just the kind of term we know we can add in since Lxy(3 F(r) = 0.