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Let $f$ a scalar function and $\vec{v}$ a vector function. We can write a product as:

$f \vec{v} = \vec{F}$

If we differentiate the LHS of above equation by the chain rule we can obtain

\begin{align} \nabla \cdot (f \vec{v}) = f \, \nabla \cdot \vec{v} + \vec{v} \, \nabla f \end{align}

Integrating we have:

\begin{align} \int \nabla \cdot (f \vec{v}) = \int f \, \nabla \cdot \vec{v} + \int \vec{v} \, \nabla f \end{align}

By inner product definition

\begin{align} \int \mathbf{D} (f \vec{v}) = \langle f, \mathbf{D} \vec{v} \rangle + \langle \vec{v} , \mathbf{G} f \rangle \end{align}

where $\mathbf{D}$ and $\mathbf{G}$ are the divergent and gradient operators. That lead us to

\begin{align} \int \mathbf{D} \vec{F} = \langle f, \mathbf{D} \vec{v} \rangle + \langle \vec{v} , \mathbf{G} f \rangle \end{align}

And by Fundamental Theorem of Calculus

\begin{align} \left. \vec{F} \right|_{\partial\Omega} = \langle f, \mathbf{D} \vec{v} \rangle + \langle \vec{v} , \mathbf{G} f \rangle \end{align}

where ${\partial\Omega}$ is the region boundary.

Is that correct?

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  • $\begingroup$ What inner product definition is that? $\endgroup$ – Jackozee Hakkiuz Jul 12 '18 at 18:56

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