Cross product in distribution analysis

Question

So I have \begin{equation} \bar{J_t}(\bar{r},t)=\int d\bar{r}' G(\bar{r},\bar{r}')\left(-\nabla' \times( \nabla' \times \bar{J}(\bar{r}',t))\right) \end{equation} and in a text it says using IBP we can show: \begin{equation} \bar{J_t}(\bar{r},t)=- \nabla \times \left( \nabla \times \int d\bar{r}' G(\bar{r},\bar{r}')\bar{J}(\bar{r}',t) \right) \end{equation} provided $\bar{J}$ has compact support. I've done a class in PDE's so I'm familiar with distribution analysis, but we only did such manipulations with the Laplacian, so this is new to me. I was thinking of using the identity, \begin{equation} \nabla \times \left( \nabla \times \bar{A} \right) = \nabla \left( \nabla \cdot \bar{A} \right) - \nabla^2\bar{A} \end{equation} but I'm not sure. Any advice?

Answer 1

Observe \begin{align} -\nabla\times\left(\nabla \times \int dr'\ G(r, r')J(r', t)\right) =&\ -\int dr'\ \nabla\times(\nabla\times[G(r, r')J(r', t)])\\ =&\ -\int dr'\ \nabla\times(\nabla G(r, r')\times J(r', t)) \\ =&\ \int dr'\ J(r', t) \Delta G(r, r') -J(r', t)\cdot \nabla^2G(r, r'). \end{align} Since \begin{align} G(r, r') = \frac{-1}{4\pi |r-r'|} \end{align} then we see that \begin{align} \Delta G(r, r') = \delta(r-r') = \Delta' G(r, r') \end{align} and \begin{align} \nabla^2 G(r, r') = \delta(r-r') \mathbf{1}= (\nabla')^2G(r, r'). \end{align} Hence it follows \begin{align} \int& dr'\ J(r', t) \Delta G(r, r') -J(r', t)\cdot \nabla^2G(r, r')\\ =& \int dr'\ J(r', t) \Delta' G(r, r') -J(r', t)\cdot (\nabla')^2G(r, r')\\ =&\ \int dr'\ \nabla'\cdot J(r', t) \nabla' G(r, r')-\nabla' J(r', t) \cdot \nabla'G(r, r')\\ =&\ \int dr'\ -\nabla'(\nabla'\cdot J(r', t))G(r, r')+(\nabla')^2J(r', t) G(r, r')\\ =&\ \int dr'\ G(r, r')(-\nabla'\times(\nabla' \times J(r', t))). \end{align}