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How can I simplify

$\nabla \cdot \left(\vec{f}(x)\delta_S(x)\right)$ where $\nabla \cdot$ is 3D divergence operator and $\vec{f}$ is a 3D vector valued function. The delta function $\delta_S(x)$ is delta measure defined on a surface in the volume.

I tried

$=\left(\nabla \cdot \vec{f}(x)\right)\delta_S(x)+ \vec{f}(x)\cdot \nabla\delta_S(x)$

Is this true $\vec{f}(x)\cdot \nabla\delta_S(x) = -\nabla \cdot \vec{f}\delta_S(x)$?

How do I simplify this? Am I going in the right direction?

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  • $\begingroup$ You applied the product rule correctly. The distributional derivative (in one variable) is defined as $\int f(x) \partial \delta(x)/\partial x =\int f(x) \delta(x) =-f'(0)$. Applying this to your problem: $$\nabla \cdot (\mathbf{f}(x)\delta(x))= (\nabla \cdot \mathbf{f})\delta(x) + \mathbf{f} \cdot \nabla \delta(x) = (\nabla \cdot \mathbf{f})\delta(x) - (\nabla \cdot\mathbf{f}) \delta(x) =0 $$ $\endgroup$
    – Cyclone
    Jul 10, 2016 at 9:18
  • $\begingroup$ Thanks, I find this counterintuitive. I would have thought that you get surface divergence of $f$. Any intuition behind why this is 0 would be appreciated. $\endgroup$
    – AnandJ
    Jul 10, 2016 at 9:56
  • $\begingroup$ Warning: I am a physicist, so this is not a rigorous argument. However maybe it helps to think of the delta distribution $\delta(x)$ as an object which is 0 everywhere away from $x=0$. Then $\mathbf{f}(x) \delta(x)$ is also constant = 0 everywhere away from $x=0$ (hence derivative vanishes there). The reason why there is no non-zero contribution from $x=0$ is the definition of the distributional derivative via partial integration which leads to this additional minus sign. Can you explain why you expected the surface divergence of $\mathbf{f}$? $\endgroup$
    – Cyclone
    Jul 14, 2016 at 5:25
  • $\begingroup$ I expected derivative components along the surface to stay and get surface divergence or divergence in the tangent space of the surface due to delta function. This would happen for example in 2D case and delta function is along one of the axes. $\endgroup$
    – AnandJ
    Jul 17, 2016 at 5:02

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