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Which order do you take derivative of delta function?

  1. Jul 21, 2012 #1
    If you have I=∫∫dxdy [∇xy δ(x-y)] f(x)g(y)

    where ∇x is the derivative with respect to x (and similarly for y), then doesn't it matter which order you take the derivatives? For example:

    I=∫∫dxdy f(x) ∇x [∇y δ(x-y)] g(y)
    =∫dx f(x) ∇x[-g'(x)]=∫dx f(x) [-g''(x)]

    whereas if you take the other order:

    I=∫∫dxdy ∇y [∇x δ(x-y)] f(x)g(y)
    =∫∫dxdy g(y)∇y [∇x δ(x-y)] f(x)
    =∫dx g(x) [-f''(x)]
  2. jcsd
  3. Jul 21, 2012 #2


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    I don't follow your calculation. Assuming notation isn't misleading, and writing things mathematics-style rather than physics-style, the calculation would go

    \int_{\mathbb{R}} \int_{\mathbb{R}} (\nabla_x \nabla_y \delta(x-y)) f(x) g(y) \, dx \, dy
    &= - \int_{\mathbb{R}} \int_{\mathbb{R}} (\nabla_y \delta(x-y)) \nabla_x(f(x) g(y)) \, dx \, dy
    \\& = - \int_{\mathbb{R}} \int_{\mathbb{R}} (\nabla_y \delta(x-y)) \nabla_x(f(x) g(y)) \, dy \, dx
    \\& = \int_{\mathbb{R}} \int_{\mathbb{R}} \delta(x-y) \nabla_y \nabla_x(f(x) g(y)) \, dy \, dx
    \\& = \cdots

    Of course, I think the end result is the same -- just continue using integration by parts to switch between the result you get from this and the two results you got.

    EDIT: I see how your calculation works now: if the inner integral is in terms of y, you are factoring the [itex]\nabla_x[/itex] outside of the inner integral.
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