Let $f $ and $g$ be two smooth scalar valued functions. Compute
div$(∇f × ∇g).$
My attempt : div$(∇f × ∇g)=$ div$( \frac{\partial}{\partial x } (f.g) i + \frac{\partial}{\partial y } (f.g)j +\frac{\partial}{\partial z } (f.g)k ) $
after that im not able to proceed further
Best Answer
Let $\boldsymbol{F} = \left(F_1, F_2, F_3\right)$ and $\boldsymbol{G} = \left(G_1, G_2, G_3\right)$ be two vector fields.
Then, their vector product is defined as
$$ \boldsymbol{F}\times \boldsymbol{G} = (F_2G_3-F_3G_2, F_3G_1-F_1G_3, F_1G_2-F_2G_1) \Rightarrow. $$
$$ \begin{aligned} \text{div}\boldsymbol{F}\times \boldsymbol{G} &= \frac{\partial}{\partial x}(F_2G_3-F_3G_2) + \frac{\partial}{\partial y}(F_3G_1-F_1G_3) + \frac{\partial}{\partial z}(F_1G_2-F_2G_1) = \\ &= G_3\frac{\partial}{\partial x}F_2 + F_2\frac{\partial}{\partial x}G_3 - G_2\frac{\partial}{\partial x}F_3 - F_3\frac{\partial}{\partial x}G_2 + \\ &+ G_1\frac{\partial}{\partial y}F_3 + F_3\frac{\partial}{\partial y}G_1 - G_3\frac{\partial}{\partial y}F_1 - F_1\frac{\partial}{\partial y}G_3 + \\ &+ G_2\frac{\partial}{\partial z}F_1 + F_1\frac{\partial}{\partial z}G_2 - G_1\frac{\partial}{\partial z}F_2 - F_2\frac{\partial}{\partial z}G_1 = \\ &= G_1\left(\frac{\partial}{\partial y}F_3-\frac{\partial}{\partial z}F_2\right)+ G_2\left(\frac{\partial}{\partial z}F_1-\frac{\partial}{\partial x}F_3\right)+ G_3\left(\frac{\partial}{\partial x}F_2-\frac{\partial}{\partial y}F_1\right)- \\ &-F_1\left(\frac{\partial}{\partial y}G_3-\frac{\partial}{\partial z}G_2\right)+ F_2\left(\frac{\partial}{\partial z}G_1-\frac{\partial}{\partial x}G_3\right)+ F_3\left(\frac{\partial}{\partial x}G_2-\frac{\partial}{\partial y}G_1\right) = \\ &= \boldsymbol{G}\cdot\text{curl}\boldsymbol{F}-\boldsymbol{F}\cdot\text{curl}\boldsymbol{G}, \end{aligned} $$
where $\text{curl}\boldsymbol{F}$ is the the curl of the vector field $\boldsymbol{F}$, and it is defined as
$$ \text{curl}\boldsymbol{F} = \left(\frac{\partial}{\partial y}F_3-\frac{\partial}{\partial z}F_2, \frac{\partial}{\partial z}F_1-\frac{\partial}{\partial x}F_3, \frac{\partial}{\partial x}F_2-\frac{\partial}{\partial y}F_1\right). $$
Now, we have
$$ \text{div}\nabla{f}\times\nabla{g} = \nabla{g}\cdot\text{curl}(\nabla{f})-\nabla{f}\cdot\text{curl}(\nabla{g}). $$
Further, for any scalar function $f$, by definition of the $\nabla$ and $\text{curl}$,
$$ \begin{aligned} \text{curl}(\nabla{f}) &= \left(\frac{\partial}{\partial y}\left(\frac{\partial}{\partial z}f\right)-\frac{\partial}{\partial z}\left(\frac{\partial}{\partial y}f\right), \frac{\partial}{\partial z}\left(\frac{\partial}{\partial x}f\right)-\frac{\partial}{\partial x}\left(\frac{\partial}{\partial z}f\right), \frac{\partial}{\partial x}\left(\frac{\partial}{\partial y}f\right)-\frac{\partial}{\partial y}\left(\frac{\partial}{\partial x}f\right)\right) = (0, 0, 0) = \boldsymbol{0}. \end{aligned} $$
So, it is a zero vector.
Finally,
$$ \text{div}\nabla{f}\times\nabla{g} = \nabla{g}\cdot\text{curl}(\nabla{f})-\nabla{f}\cdot\text{curl}(\nabla{g}) = \nabla{g}\cdot\boldsymbol{0} - \nabla{f}\cdot\boldsymbol{0} = 0. $$