Identity for divergence of vector product

Question

I'm reviewing this paper and they state the following "vector identity":

$ \nabla \cdot (\textbf{u}\textbf{v}) = \nabla \textbf{u} \cdot \textbf{v} + (\nabla \cdot \textbf{v})\textbf{u}$

I tried to find some material online about this identity but didn't have any luck. I'm pretty sure $\textbf{u}$ and $\textbf{v}$ are both vectors, as they're both bold face in the paper. Can someone help to provide more information on this identity? Is it valid? Can you give the proof?

EDIT

After some discussion, it seems that $\textbf{u}$ should be interpreted as a scalar (remove boldface), not a vector. I will clarify this with the original source and follow up if there is any further amendments. If it is a scalar, the answer below is correct.

Answer 1

This is, fundamentally, just another case of the product rule for derivatives.

For the notation to make sense ($\nabla u$ and $\nabla\cdot v$ defined), $u$ should be a scalar-valued function of a vector variable, and $v$ should be a vector-valued function of that variable. Then $\nabla u$ is the gradient of $u$, and $\nabla\cdot v$ is the divergence of $v$. The two terms on the right are both scalars - the first is the dot product of the vector-valued gradient of $u$ and the vector-valued function $v$, while the second is the product of the scalar-valued divergence of $v$ and the scalar-valued function $u$.

To prove it, we just go down to components. I'll assume three dimensions here, although it works in more generality (Let $A,B,C$ be the components of $v$; $v(x,y,z)=(A(x,y,z),B(x,y,z),C(x,y,z))$): \begin{align*}\nabla\cdot (uv)\quad &?\quad \nabla u\cdot v + (\nabla\cdot v) u\\ \frac{\partial (uA)}{\partial x}+\frac{\partial (uB)}{\partial y}+\frac{\partial (uC)}{\partial z}\quad &?\quad \left(\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial u}{\partial z}\right)\cdot (A,B,C) + \left(\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}\right)u\\ u\frac{\partial A}{\partial x}+A\frac{\partial u}{\partial x}+u\frac{\partial B}{\partial y}+B\frac{\partial u}{\partial y}+u\frac{\partial C}{\partial z}+C\frac{\partial u}{\partial z} &= A\frac{\partial u}{\partial x}+B\frac{\partial u}{\partial y}+C\frac{\partial u}{\partial z}+u\frac{\partial A}{\partial x}+u\frac{\partial B}{\partial y}+u\frac{\partial C}{\partial z} \end{align*} Not too bad there. It's just all the same terms rearranged a bit.

There's another way to look at this. If you go on with mathematics long enough, you'll encounter the language of manifolds and differential forms, which provides a unifying framework for all of these vector calculus notions. In particular, both the gradient and the divergence are examples of the exterior derivative, and the exterior derivative has a product rule. Here, working in $n$ dimensions, we have $u$ as a $0$-form and $v$ as an $(n-1)$-form. The gradient takes $0$-forms to $1$-forms, the divergence takes $(n-1)$-forms to $n$-forms, and the dot product between $1$-forms and $(n-1)$ forms is the wedge product, taking us to a $n$-form. Then we just apply the exterior derivative's product rule: $$\nabla\cdot (uv) = d(u\wedge v) = du\wedge v + (-1)^0 u\wedge dv = \nabla u\cdot v + u(\nabla \cdot v)$$

Answer 2

For this to make sense $u$ is a scalar function and $\mathbf{v} = (v_1, \ldots, v_n)$ is a vector function.

We can calculate the partial derivatives using the chain rule:

$$\frac{\partial}{\partial x_j}(u\mathbf{v})_i = \frac{\partial}{\partial x_j}(uv_i) = \frac{\partial u}{\partial x_j}v_i + u\frac{\partial v_i}{\partial x_j}$$

Therefore, the Jacobi matrix is given by $$\nabla(u\mathbf{v}) = \left[\frac{\partial}{\partial x_j}(u\mathbf{v})_i\right]_{1\le i,j\le n} = \left[\frac{\partial u}{\partial x_j}v_i + u\frac{\partial v_i}{\partial x_j}\right]_{1\le i,j\le n} = \mathbf{v}\otimes (\nabla u) + u (\nabla\mathbf{v})$$

The divergence is obtained by taking the trace:

$$\nabla \cdot (u\mathbf{v}) = \operatorname{Tr}\left(\nabla(u\mathbf{v})\right) = \operatorname{Tr}(\mathbf{v}\otimes (\nabla u)) + u \operatorname{Tr}(\nabla\mathbf{v}) = \mathbf{v}\cdot (\nabla u) + u(\nabla\cdot \mathbf{v})$$

Answer 3

Using Einstein notation, we can see it more easily: $$ \nabla \cdot (u \textbf{v}) = \partial_i (uv_i) = \partial_iu \cdot v_i + u\cdot\partial_iv_i = \nabla u\cdot \textbf{v} + u(\nabla \cdot \textbf{v}). $$