How does one go about proving that the dot product of the gradient functions $\nabla{u} \cdot \nabla{v}$, where $u$ and $v$ are both scalar functions is equal to : $\frac{1}{2}[ \nabla^{2}{(uv)} – u\nabla^{2}{v} – v\nabla^{2}{u}] $.
I tried indexing this using Kronecker Delta however I couldn't arrive at a solution. Any thoughts would be much appreciated.
Best Answer
I would use Einstein notation (also called summation notation) to prove this. I would write:
$$ \begin{aligned} \nabla^2 (uv) & = \frac{\partial }{\partial x_i} \left( \frac{\partial (uv)}{\partial x_i}\right) \\ & = \frac{\partial }{\partial x_i} \left( u\frac{\partial v}{\partial x_i} + v\frac{\partial u}{\partial x_i}\right) \\ & = 2\frac{\partial u}{\partial x_i}\frac{\partial v}{\partial x_i} + u \frac{\partial^2 v}{\partial x_i \partial x_i} + v \frac{\partial^2 u}{\partial x_i \partial x_i} \end{aligned} $$
Since: $$ u \nabla^2 v = u \frac{\partial^2 v}{\partial x_i \partial x_i} $$ and $$ v \nabla^2 u = v \frac{\partial^2 u}{\partial x_i \partial x_i} $$
We have shown that:
$$ \frac{1}{2} \left[ \nabla^2 (uv) - u \nabla^2v - v\nabla^2u \right] = \frac{\partial u}{\partial x_i}\frac{\partial v}{\partial x_i} = \nabla u \cdot \nabla v $$