Prove a particular case of product rule for divergence

I am trying to verify the vector calculus identity in appendix B.2 in "Finite Elements Methods: A Practical Guide" (Whiteley 2017)

\begin{aligned} \nabla \cdot (vp\nabla u) &= v\nabla \cdot p\nabla u + p\nabla v \cdot \nabla u \end{aligned}

where $$v = v(x,y), p = p(x,y), u = u(x,y)$$ and "$$u$$ possesses all first and second partial derivatives while $$v$$ and $$p$$ possess all first partial derivatives".

To prove the above identity, I basically follow proof for product rule of divergence but substitute $$U = vp$$ and $$\mathbf{A} = \nabla u$$ but must be going wrong somewhere, here is my attempt (with labeled steps "(s#)" on far right of each line):

\begin{aligned} \nabla \cdot (vp\mathbf{A}) &= \sum_{k=1}^n \frac{\partial (vpA_k)}{\partial x_k} && \text{Definition divergence} && \text{(s1)} \\\\ &= \sum_{k=1}^n vp\frac{\partial A_k}{\partial x_k} + \sum_{k=1}^n v \frac{\partial p}{\partial x_k}A_k + \sum_{k=1}^n \frac{\partial v}{\partial x_k} p A_k && \text{Product rule} && \text{(s2)} \\\\ &= vp \sum_{k=1}^n \frac{\partial A_k}{\partial x_k} + v \sum_{k=1}^n \frac{\partial p}{\partial x_k}A_k + p \sum_{k=1}^n \frac{\partial v}{\partial x_k} A_k && \text{Factor non-subscripted terms} && \text{(s3)} \\\\ &= vp\nabla \cdot \mathbf{A} + v \sum_{k=1}^n \frac{\partial p}{\partial x_k}A_k + p \sum_{k=1}^n \frac{\partial v}{\partial x_k} A_k && \text{Definition divergence} && \text{(s4)} \\\\ &= vp\nabla \cdot \mathbf{A} + v \left[ \sum_{k=1}^n \frac{\partial p}{\partial x_k}\mathbf{e_k} \cdot \sum_{k=1}^n A_k \mathbf{e_k}\right] + p\sum_{k=1}^n \frac{\partial v}{\partial x_k} A_k && \text{Definition dot product, std basis } \mathbf{e_k} && \text{(s5)} \\\\ &= vp\nabla \cdot \mathbf{A} + v \nabla p \cdot \mathbf{A} + p\sum_{k=1}^n \frac{\partial v}{\partial x_k} A_k && \text{Definition gradient} && \text{(s6)} \\\\ &= vp\nabla \cdot \mathbf{A} + v \nabla p \cdot \mathbf{A} + p \nabla v \cdot \mathbf{A} && \text{(s5) and (s6) on last term} && \text{(s7)} \\\\ &= vp\nabla \cdot \nabla u + v \nabla p \cdot \nabla u + p \nabla v \cdot \nabla u && \mathbf{A} = \nabla u && \text{(s8)} \end{aligned}

Edit: The above proof could also be demonstrated by just using vector calculus identities

\begin{aligned} \nabla(\psi \phi) &= \phi \nabla \psi + \psi \nabla \phi && \text{Product rule scalar fields}\\\\ \nabla \cdot (U \mathbf{A}) &= U \nabla \cdot \mathbf{A} + (\nabla U) \cdot \mathbf{A} && \text{Product rule scalar field, vector field} \\\\ &= \psi\phi \nabla \cdot \nabla u + [\phi\nabla\psi + \psi\nabla\phi] \cdot \nabla u , && U = \psi\phi, \mathbf{A} = \nabla u \\\\ &= \psi\phi\nabla \cdot \nabla u + \phi\nabla\psi \cdot \nabla u + \psi\nabla\phi \cdot \nabla u \\\\ &= vp\nabla \cdot \nabla u + p\nabla v \cdot \nabla u + v\nabla p \cdot \nabla u , && v = \psi, p = \phi \end{aligned}

• Is it because the statement in the book is div of (p grad u) so it really should have had parentheses. Thus can be obtained with a single application of the product rule.
– dm63
Commented Oct 25, 2023 at 10:59

To get the desired identity take $$f = v$$ and $$\mathbf{g} = p \nabla u$$ and then use the product rule for divergence to get $$\nabla \cdot (f \mathbf{g}) = \nabla f \cdot \mathbf{g} + f \nabla \cdot \mathbf{g} = \nabla v \cdot p\nabla u + v \nabla \cdot (p\nabla u) = v \nabla \cdot (p\nabla u) + p \nabla v \cdot \nabla u.$$
Alternatively, you can get the result by using the identity you derived and noticing that $$v p \nabla \cdot \nabla u + v \nabla p \cdot \nabla u = v (p \nabla \cdot \nabla u + \nabla p \cdot \nabla u) = v \nabla \cdot (p \nabla u).$$