Divergence of a displacement vector field multiplied by delta function I'm trying to work out why
$$ \boldsymbol{\nabla\cdot[u}\,\delta^3(\mathbf{r})]=0,
$$
where $\boldsymbol{u}$ is the displacement field of a source of stress, $\boldsymbol{\nabla\cdot u}\ne 0$, and $\delta^3(\mathbf{r})$ is the 3d delta function. I suspect that it could have something to do with
$$
     \delta^3(\mathbf{r})=-\frac{1}{4\pi}\nabla^2\frac{1}{r},
$$
but I could be wrong. Any suggestions would be appreciated!
 A: The distribution $D(\vec{r}) = \vec{\nabla} \cdot \left[\vec{u} \delta(\vec{r}) \right]$ is not equal to zero.  Two distributions are equal if and only if you get the same result when you multiply them by any test function $f$ and integrate the result over their domains.  This means that if $D(\vec{r}) = 0$, then the integral $$\int D(\vec{r}) f(\vec{r}) d^3\vec{r}$$
would have to vanish for any test function $f(\vec{r})$.  But integrating this expression by parts, we obtain
\begin{align}
\int D(\vec{r}) f(\vec{r}) \, d^3\vec{r} &= \int \vec{\nabla} \cdot \left[\vec{u} \delta(\vec{r})\right]f(\vec{r}) d^3\vec{r} \\
&= \int \vec{\nabla} \cdot \left[\vec{u} \delta(\vec{r}) f(\vec{r}) \right] d^3\vec{r} - \int \delta(\vec{r}) \vec{u} \cdot \vec{\nabla} f(\vec{r}) d^3\vec{r} \\
&= \oint \left[ \vec{u} \delta(\vec{r}) f(\vec{r}) \right] \cdot d\vec{a} - \left[\vec{u} \cdot \vec{\nabla} f \right]_{\vec{r} = 0}.
\end{align}
The first term does vanish, since the delta function is zero everywhere on the boundary.  The second term, however, does not vanish in general.  
The second term does automatically vanish when $\vec{u}(0) = 0$, though. In that case, we can say that $D(\vec{r}) = 0$.  Perhaps this is an implicit assumption concerning $\vec{u}$ in stress fields?  (It's been a long time since I've looked at continuum mechanics in detail.)
