prove that $\vec V ∇^2s=∇·((∇s)\vec V)-∇s.∇\vec V$

Question

I am given $\vec V ∇^2s=∇·((∇s)\vec V)-∇s.∇\vec V$ where V is any vector field and S is an arbitrary scalar and I am not sure how I can prove this.

Answer 1

When trying to prove identities it is often easier to start with the more complicated side and show that it reduces to the simpler side. Are you familiar with index notation? Repeated indices imply a summation, and $\partial_{i}=\frac{\partial}{\partial x_{i}}$.

We have that $(\nabla s)\vec{V}$ has the components;

$$\left[(\nabla s)\vec{V}\right]_{ij}=[\nabla s]_{j}[\vec{V}]_{i}=(\partial_{j}s)V_{i}$$

And that $\nabla s\cdot \nabla\vec{V}$ has the components;

$$\left[\nabla s\cdot \nabla\vec{V}\right]_{i}=[\nabla\vec{V}]_{ij}[\nabla s]_{j}=(\partial_{j}V_{i})(\partial_{j}s)$$

So that in index form the right hand side $\nabla\cdot((\nabla s)\vec{V})-\nabla s\cdot\nabla\vec{V}$ reads; \begin{align*} \left[\nabla\cdot((\nabla s)\vec{V})-\nabla s\cdot\nabla\vec{V}\right]_{i} &= \partial_{j}\left((\partial_{j}s)V_{i}\right)-(\partial_{j}V_{i})(\partial_{j}s) \\ &= \partial_{j}(\partial_{j}s)V_{i}+(\partial_{j}s)(\partial_{j}V_{i})-(\partial_{j}V_{i})(\partial_{j}s) \\ &= \nabla\cdot(\nabla s)V_{i} \\ \Rightarrow \left[\nabla\cdot((\nabla s)\vec{V})-\nabla s\cdot\nabla\vec{V}\right]_{i} &= V_{i}\nabla^{2}s \\ \end{align*}

This is true for each component so we have;

$$ \nabla^{2}V=\nabla\cdot((\nabla s)\vec{V})-\nabla s\cdot\nabla\vec{V} $$

As required.