Given that the question is purely mathematical and has nothing to do with electromagnetism, I still provide some context. Studying on the book 'Antenna theory and design' by R.S. Elliott, pag. 22, I am faced with this expression:
$$ \frac{\rho}{\epsilon_0}\nabla\psi-j\omega\mu_0\psi \mathbf{J} $$
where $ \rho $ is the charge density, $ \mathbf{J}$ is the current density, $ \omega $ is the angular frequency, $\epsilon_0,\mu_0 $ are the electromagnetic constanst of vacuum and $\psi = \frac{e^{-jk\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}}}{\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}} $ ($ (x_0,y_0,z_0) $ fixed point). That said, charges and currents are related by the continuity equation: $\nabla\cdot \mathbf{J}=-j\omega\rho $, that my book claims to use to move from the above expression to the following:
$$\frac{1}{j\omega\epsilon_0}[(\mathbf{J}\cdot\nabla)\nabla\psi+k^2\psi \mathbf{J}] $$
where $ k^2 =\omega^2\mu_0\epsilon_0 $.
I am trying to prove this but I cannot. In particular, the two expressions are identical if it happens that: $-(\nabla\cdot\mathbf{J})\nabla\psi=(\mathbf{J}\cdot\nabla)\nabla\psi $. If I write mathematical steps I obtain:
$$-(\nabla\cdot\mathbf{J})\nabla\psi = -\sum_{j=1,2,3}\partial_jJ_j \sum_{i=1,2,3,}\mathbf{e}_i\partial_i\psi = -\sum_{i=1,2,3}\mathbf{e}_i\sum_{j=1,2,3}[\partial_j(J_j\partial_i\psi)-J_j\partial_i\partial_j\psi] = $$ $$=-\sum_{i=1,2,3}\mathbf{e}_i\sum_{j=1,2,3}\partial_j(J_j\partial_i\psi) +\sum_{i=1,2,3}\mathbf{e}_i\sum_{j=1,2,3}J_j\partial_i\partial_j\psi = -\sum_{i=1,2,3}\mathbf{e}_i\sum_{j=1,2,3}\partial_j(J_j\partial_i\psi) + (\mathbf{J}\cdot\nabla)\nabla\psi$$
therefore at this point the first addendum at the end should be null, but I don't understand why. Does anyone see the reason?