Vector Identity in fem strong to weak form

Question

Hi all, Can anyone please help me understand (step by step) how the first term in the last relation appears? (these steps was to convert the first relation from the strong form to the weak form in the last relation) Thanks all

Answer 1

\begin{align} \nabla \cdot \boldsymbol{\Gamma} & = f \\ \overset{\text{Multiply with test-function } v}{\Rightarrow} v \nabla \cdot \boldsymbol{\Gamma} & = vf \\ \overset{\int_\Omega}{\Rightarrow} \int_\Omega v \nabla \cdot \boldsymbol{\Gamma} \mathrm d V & = \int_\Omega vf \mathrm d V \\ \overset{\text{Product rule of differentiation / Integration by parts}}{\Rightarrow} \int_\Omega \nabla \cdot \Big( v \boldsymbol{\Gamma}\Big) - \nabla v \cdot \boldsymbol{\Gamma} \mathrm d V & = \int_\Omega vf \mathrm d V \\ \overset{\text{Split integrals, apply Gauss/Divergence theorem on first term}}{\Rightarrow} - \int_\Omega \nabla v \cdot \boldsymbol{\Gamma} \mathrm d V + \int_{\partial \Omega} v \boldsymbol{\Gamma} \cdot \boldsymbol{n} \mathrm dS & = \int_\Omega vf \mathrm d V \end{align}