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jim delta 1_11_11
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A short Word note from Phil to Jim, dated 1.11.11, restating the problem in Cartesian coordinates only. Question 1 asks for the 3D operator with L(1/|r-r'|) = delta; the answer given is -(1/4π) times the Laplacian. Question 2 asks for the analogous 2D operator acting on the same kind of kernel, which Phil says he cannot answer. He doubts the 2D Laplacian works, since it takes ln|r-r'| to a delta function. Some symbols in the equations were lost in extraction.
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Jim, let's start all over and everything is stated here only in Cartesian coordinates. 1.11.11
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Question #1: What Cartesian 3D differential operator Lxyz makes this equation be true:
Lxyz ( 1/ ) = δ(x-x')δ(y-y')δ(z-z')
Answer:
Lxyz = - (1/4π) (∂x2 + ∂y2 + ∂z2)
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Question #2: What Cartesian 2D differential operator Lxy makes this equation be true:
Lxy ( 1/ ) = δ(x-x')δ(y-y')
Answer:
Please put your Cartesian-coordinates answer in the box provided above.
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I don't know the answer. It might be that no operator Lxy exists which solves Problem #2. Or maybe it is some horrible thing like this, with horrible coefficient functions,
Lxy = a(x,y)∂x + a(y,x)∂y + b(x,y) ∂x2 + b(y,x) ∂y2 + c(x,y)∂x∂y + lots of higher terms.
I am pretty sure this is NOT the answer
Lxy = ∂x2 + ∂y2
The reason is that I know that
(∂x2 + ∂y2) [ ln(] = constant * δ(x-x')δ(y-y')
and it seems unlikely that two different functions would solve this equation.