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TL:DR

Is the following relation true where $\vec{F}$ is a vector field and $v$ is a scalar field $v : \mathbb{R}^n \rightarrow \mathbb{R}$?

$$ v(\nabla \cdot \vec{F}) = (\nabla \cdot v \vec{F}) $$

Context and Attempt

I am trying to show that

$$ \begin{equation} \int_{\Omega} v (\nabla \cdot \nabla u)\ d\Omega = \oint_{\partial \Omega = \Gamma} v \nabla u\ \cdot \vec{n}\ d\Gamma - \int_{\Omega} \nabla v \cdot \nabla u\ d\Omega \tag{1} \end{equation} $$

which is a part of a step in the derivation of the weak form of Poisson's equation (see also this MathSE post and this video.

To do this, I use the Divergence theorem,

$$ \int_{\Omega} \nabla \cdot \vec{F} = \oint_{\partial \Omega = \Gamma} \vec{n} \cdot \vec{F}\ d\Gamma $$

and set the vector field $\vec{F} = \nabla u$. This way, equation (1) becomes

$$ \begin{equation} \int_{\Omega} v(\nabla \cdot \vec{F})\ d\Omega \tag{2} \end{equation} $$

but to take advantage of the product rule to get equation (2) into the form of equation (1), my current derivation would require the below relation:

$$ v(\nabla \cdot \vec{F}) = (\nabla \cdot v \vec{F}) $$

This seems to be true intuitively by simply claiming that $\nabla$ is a vector $\vec{\nabla}$ with components $\nabla_i = \frac{\partial}{\partial x_i}$ and $\vec{F}$ is of course just a vector with components $F_i$ where both have $\forall i \in [1...n]$ where $n$ is the dimension of the domain $\Omega$. Note that $v$ is a scalar field so it is just a function $v: \mathbb{R}^n \rightarrow \mathbb{R}$. If $\vec{\nabla} \cdot \vec{F}$ is just the dot product of two vectors, then from the definition of the dot product and including the multiplication of the dot product by the scalar field $v$ as shown below.

$$ v \sum_{i=1}^{n} \nabla_i F_i = \sum_{i=1}^{n} \nabla_i v F_i = \vec{\nabla} \cdot v \vec{F} $$

If the above is true, then I can show equation (1) from the following steps:

\begin{aligned} \int_{\Omega} v (\nabla \cdot \nabla u)\ d\Omega &= \int_{\Omega} \nabla \cdot v \nabla u\ d\Omega, && v(\nabla \cdot \vec{F}) = \nabla \cdot v \vec{F} && \text{(s1)} \\\\ &= \int_{\Omega} \nabla \cdot v \vec{F}\ d\Omega, && \vec{F} = \nabla u && \text{(s2)} \\\\ &= \int_{\partial \Omega = \Gamma} v \vec{F}\cdot \vec{n}\ d\Gamma, && \text{Divergence theorem} && \text{(s3)} \\\\ &= \int_{\Omega} v(\nabla \cdot \vec{F})\ d\Omega + \int_{\Omega} \nabla v \cdot \vec{F}\ d\Omega && \text{Product rule on (s2)} && \text{(s4)} \\\\ &= \int_{\Omega} v(\nabla \cdot \nabla u)\ d\Omega + \int_{\Omega} \nabla v \cdot \nabla u\ d\Omega && \text{Substitute (s4) } \vec{F} = \nabla u && \text{(s5)} \\\\ &= \int_{\partial \Omega = \Gamma} v \nabla u \cdot \vec{n}\ d\Gamma, && \text{Substitute (s3) } \vec{F} = \nabla u && \text{(s6)} \end{aligned}

Then using (s4) and (s5),

$$ \int_{\Omega} v (\nabla \cdot \nabla u)\ d\Omega = \int_{\partial \Omega = \Gamma} v \nabla u \cdot \vec{n}\ d\Gamma - \int_{\Omega} \nabla v \cdot \nabla u\ d\Omega $$

which matches equation (1). I am a bit suspicious about my derivation here, though, because the taking the lefthand side of (s1) and the right hand side of (s5),

$$ \int_{\Omega} v (\nabla \cdot \nabla u)\ d\Omega = \int_{\Omega} v (\nabla \cdot \nabla u)\ d\Omega + \int_{\Omega} \nabla v \cdot \nabla u\ d\Omega $$

which would imply that

$$ \int_{\Omega} \nabla v \cdot \nabla u\ d\Omega = 0 $$

but this doesn't seem right.

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  • $\begingroup$ To obtain your desired identity, use the product rule $\nabla\cdot (vF) =v \nabla \cdot F + \nabla v \cdot F$ and the divergence theorem applied to $vF$, with $F = \nabla u$. $\endgroup$ Oct 24, 2023 at 14:02
  • $\begingroup$ @kieransquared yep, I just figured this out and answered a related question I asked accordingly with steps. See the MathSE question understanding vector calculus identity in deriving weak form of heat equation $\endgroup$ Oct 24, 2023 at 14:17

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The relation fails already for one-variable functions. So if you take $F(x,y,z)=(f(x),0,0)$ and $v(x,y,z)=g(x)$, you are asking whether $$ gf'=(gf)', $$ which is clearly not the case in general.

A true form of the Leibniz product rule in vector calculus is the following: $$ \nabla\cdot (vF)=\nabla v \cdot F +v\nabla\cdot F.$$ This is not far from your formula, but you are missing the $v\nabla\cdot F$ term.

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