How to simplify expressions with del (or nabla) in them?

Question

I always find it difficult to simplify expressions or open brackets in expressions that have a 'Del' (or 'Nabla') in them.

For example, how would one go about simplifying this expression?:

$$\nabla\boldsymbol{\cdot}(\phi\nabla\psi)$$

($\phi$ and $\psi$ are both scalar fields)

I need it to become: $$[\phi\nabla^2\psi + (\nabla\phi)\boldsymbol{\cdot}(\nabla\psi)]$$

I would also love to know how to simplify those standard equations mentioned in Griffiths (for example - the expansion of the 'curl of the curl' of a vector field)

The only method I know is to

1. find out every single term in the expression (in terms of $a_x$, $a_y$ etc.)
2. and then cancel out the terms
3. and then find patterns and regroup the terms in the remaining expression

Is there a faster way to approach these 'simplify' (or 'expand') problems? Maybe there are some tricks or formulas that I am unaware of (maybe something analogous to the uv-rule for differentiating the product of two functions in simple calculus) $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$$

I understand that the uv-rule seems to work on my original expression. But I would still love some sort of formalization. The problem I have is that, in simple calculus, multiplying two functions does not have two meanings.

With Nabla however, I have two choices - Dot product and Cross product.

And I also have three choices for differentiation - Gradient, Divergence and Curl

To explain my concern better, try answering what would have been the simplification if the original expression was - $$\nabla \times (\phi\nabla\psi)$$

or maybe $$\nabla(v\boldsymbol{\cdot}\nabla\psi)$$ where $v$ is a vector-field

For the analogy, these three questions become the same question -

"Differentiation of something multiplied by the differential of something else"

Answer 1

You are computing the divergence of the vector field $\left(\phi \frac{\partial \psi}{\partial x_i}\right)_{i=1,\cdots,n}$, so you just get $$ \sum_{i=1}^n \frac{\partial}{\partial x_i} \left(\phi \frac{\partial \psi}{\partial x_i}\right) $$

using the product rule you simply get $$ \sum_{i=1}^n \left(\frac{\partial \phi}{\partial x_i} \frac{\partial \psi}{\partial x_i} + \phi \frac{\partial^2 \psi}{\partial x_i^2}\right) = \nabla \phi \cdot \nabla \psi + \phi \nabla^2\psi $$

once you know the result, you can "build" some mnemonic related to the product rule, but you still need to know what first and second order operators you must use.

Answer 2

There are many identities in vector calculus that can be referred to simplify such expressions.

Using $\nabla\cdot(\phi\mathbf A)=\phi\nabla\cdot\mathbf A+(\nabla\phi)\cdot\mathbf A$, which looks analogous to the product rule of differentiation, you get $$\nabla.(\phi\vec\nabla\psi)=\phi\nabla^2\psi + (\vec\nabla\phi).(\vec\nabla\psi)$$

EDIT:

Consider ${\color{red}{\mathbf C}}\times(\mathbf A\times\mathbf B)=\mathbf A(\mathbf {\color{red}{\mathbf C}}\cdot\mathbf B)-\mathbf B(\mathbf {\color{red}{\mathbf C}}\cdot\mathbf A)$ and ${\color{red}{\mathbf\nabla}} \times (\mathbf{A} \times \mathbf{B}) \ =\ \mathbf{A}({\color{red}{\mathbf\nabla}} {\cdot} \mathbf{B}) \,-\, \mathbf{B}({\color{red}{\mathbf\nabla}} {\cdot} \mathbf{A}) \,+\, (\mathbf{B} {\cdot} \nabla) \mathbf{A} \,-\, (\mathbf{A} {\cdot} \nabla) \mathbf{B}$ .

Where is the analogy? I think after deriving a few formulae listed in the link attached, you could tell apart where the analogy works and where it doesn't.

Answer 3

I'll expand what I've commented above here.

Using suffix notation and summation convention (since we are working with $\mathbb{R}^n$ there is no need to distinguish upstairs and downstairs indices, so just write everything downstairs), you can get, for example, \begin{align*}\require{color} [\nabla\times(\mathbf{A}\times\mathbf{B})]_i &=\epsilon_{ijk}\partial_j(\mathbf{A}\times\mathbf{B})_k\\ &=\epsilon_{ijk}\epsilon_{k\ell m}\partial_j A_\ell B_m\\ &=(\delta_{i\ell}\delta_{jm}-\delta_{im}\delta_{j\ell})\partial_j A_\ell B_m\\ &={\color{red}\partial_j A_i B_j}-{\color{blue}\partial_j A_j B_i}\\ \end{align*} Note that, by convention, $\partial$ acts on everything to its right. Whereas in the case of $\mathbf{C}\times(\mathbf{A}\times\mathbf{B})$ we would just "take out" the $A_i$ and $B_i$ from the red and blue term respectively, here we can't move them in front of the differential operator without paying for the noncommutativity: \begin{align*} {\color{brown}\partial_j A_i} B_j&={\color{brown}A_i\partial_j}B_j+({\color{brown}\partial_jA_i})B_j\\ &=A_i\partial_jB_j+B_j\partial_jA_i\\ \partial_j A_j B_i&=B_i\partial_jA_j+A_j\partial_jB_i. \end{align*} So $$ \nabla\times(\mathbf{A}\times\mathbf{B})=\mathbf{A}(\nabla\cdot\mathbf{B})-\mathbf{B}(\nabla\cdot\mathbf{A})+(\mathbf{B}\cdot\nabla)\mathbf{A}-(\mathbf{A}\cdot\nabla)\mathbf{B} $$ and you can see the additional terms are precisely what we get from moving something behind a $\nabla$ to in front of that $\nabla$.

Now we have done the calculation, you might reasonably ask the question: Can one immediately get from $$ \mathbf{C}\times(\mathbf{A}\times\mathbf{B})=\mathbf{A}(\mathbf{C}\cdot\mathbf{B})-\mathbf{B}(\mathbf{C}\cdot\mathbf{A}) $$ to a formula for $\nabla\times(\mathbf{A}\times\mathbf{B})$? To start with, we check the naive substitution still gives terms that make sense (i.e., it doesn't leave dangling $\nabla$). Then we see the formula involves pushing $\mathbf{A}$ (or $\mathbf{B}$ in the second term) in front of $\mathbf{C}$, so we need to compensate that by having something from $\nabla\mathbf{A}$ (or $\nabla\mathbf{B}$). So the formula has to read something like $$ \mathbf{A}(\nabla\cdot\mathbf{B})-\mathbf{B}(\nabla\cdot\mathbf{A})+(\nabla\mathbf{A})\ast\mathbf{B}-(\nabla\mathbf{B})\ast\mathbf{A} $$ where $\ast$ does some contraction between the rank-2 tensor and the vector. Now it is not hard to see in $(\nabla\mathbf{A})\ast\mathbf{B}$ the $\mathbf{B}$ has to contract with the $\nabla$ rather than $\mathbf{A}$ (because the term we are correcting has that), hence we obtain $$ \mathbf{A}(\nabla\cdot\mathbf{B})-\mathbf{B}(\nabla\cdot\mathbf{A})+(\mathbf{B}\cdot\nabla)\mathbf{A}-(\mathbf{A}\cdot\nabla)\mathbf{B}. $$

Curl of a curl: Similarly, $$ \mathbf{C}\times(\mathbf{A}\times\mathbf{B})=\mathbf{A}(\mathbf{C}\cdot\mathbf{B})-(\mathbf{C}\cdot\mathbf{A})\mathbf{B} $$ The right hand side still make sense when $\mathbf{A}=\mathbf{C}=\nabla$. $$ \nabla\times(\nabla\times\mathbf{B})=\nabla(\nabla\cdot\mathbf{B})-(\nabla\cdot\nabla)\mathbf{B}+\color{red}\text{correction} $$ We note that we didn't push any vector field pass a $\nabla$, so there are no correction terms. (We did change the order of $\mathbf{C}$ and $\mathbf{A}$ in $\mathbf{A}(\mathbf{C}\cdot\mathbf{B})$ but they are both the differential operator $\nabla$ so the symmetry of partial derivatives means there are no correction term).

However, I'd seriously advice against doing this eyeballing for anything more complicated. To see why, think about $(\mathbf{A}\times\nabla)\times\mathbf{B}=\mathbf{A}\cdot\nabla\mathbf{B}-\mathbf{A}(\nabla\cdot\mathbf{B})=\mathbf{A}\times(\nabla\times\mathbf{B})+(\mathbf{A}\cdot\nabla)\mathbf{B}-\mathbf{A}(\nabla\cdot\mathbf{B})$.