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Prove $$\operatorname{ div} (\phi A) = (\operatorname{grad} \phi) \cdot A + \phi \operatorname{ div} A$$ where $A$ is differentiable vector function and $\phi $ is differentiable scalar function.

Now i write $A =(A_1. A_2. A_3)$ and so i write as $ div (\phi A) = (f_x,f_y,f_z) \cdot (\phi A_1,\phi A_3,\phi A_3)$ which equals to $f_x(\phi A_1) + f_y(\phi A_2) + f_z(\phi A_3) $. how do i proceed?

thanks

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  • $\begingroup$ what is $f_x,f_y,f_z$? $\endgroup$ – Calvin Khor Jun 29 '19 at 10:51
  • $\begingroup$ partial derivatives w.r.t x $\endgroup$ – Gathdi Jun 29 '19 at 11:15
  • $\begingroup$ Where have you seen them written this way? Why not $\partial_x$ or $\partial/\partial x$? As for how to proceed, just use the 1D product rule $\endgroup$ – Calvin Khor Jun 29 '19 at 11:16
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    $\begingroup$ Yes thank you i was able to prove it $\endgroup$ – Gathdi Jun 29 '19 at 11:19
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    $\begingroup$ @Gathdi I think the the RHS should be $\text{grad}(\phi )\cdot \mathbf{A} +\phi\text{div}( \mathbf{A})$ $\endgroup$ – Robert Z Jun 29 '19 at 11:42
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Note that, by the product rule, for $i=1,2,3$, $$\frac{\partial (\phi A_i)}{\partial x_i}=\frac{\partial \phi}{\partial x_i} A_i+\phi \frac{\partial A_i}{\partial x_i}.$$ Then after summing over $i$ we get $$\text{div}(\phi \mathbf{A})=\sum_{i=1}^3\frac{\partial (\phi A_i)}{\partial x_i}=\sum_{i=1}^3\frac{\partial \phi}{\partial x_i}\cdot A_i+\phi\cdot\sum_{i=1}^3\frac{\partial A_i}{\partial x_i}=\text{grad}(\phi )\cdot \mathbf{A} +\phi\,\text{div}( \mathbf{A})$$ and we are done.

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We have $$\frac{\partial(\phi A_i)}{\partial x_j}= \frac{\partial\phi}{\partial x_j}A_i+ \phi\frac{\partial A_i}{\partial x_j}$$

so $$\nabla(\phi A) = \phi(\nabla A) + A \otimes (\nabla\phi)$$ where $[\vec{a}\otimes \vec{b}]_{ij} = a_ib_j$.

Therefore

$$\operatorname{div}(\phi A) = \operatorname{Tr} \nabla(\phi A) = \phi(\operatorname{Tr} (\nabla A)) + \operatorname{Tr}(A \otimes (\nabla\phi)) = \phi \operatorname{div} A + \langle \nabla \phi,A\rangle$$

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