# 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

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

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

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

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.

• 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 $F = \nabla u$. Oct 24, 2023 at 14:02
• @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 Oct 24, 2023 at 14:17

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.