Justifying Vector Calculus Identities: $(u \times \nabla) \times v$

Question

The following relates to a conversation I had in the comments of an earlier post.

When considering an expression like $(u \times \nabla) \times v$, it is supposedly possible to apply the vector triple product identity in an unjustified (though "formally correct") way to get a correct simplification such as $(u\times \nabla) \times v = \nabla(u\cdot v) - (v\cdot \nabla) u$. In particular, the vector triple product formula states that for vectors $a,b,c \in \Bbb R^3$, we have $$ (a \times b) \times c = (a \cdot c)b - (b \cdot c)a = b(a \cdot c) - (c \cdot b)a. $$ If we treat $\nabla$ as a vector, then "plugging in" to the above yields $(u\times \nabla) \times v = \nabla(u\cdot v) - (v\cdot \nabla) u$.

Is there a rigorous approach to simplifying $(u \times \nabla) \times v$ (and other such expressions) that makes use of vector identities in a similar fashion?

---

Edit: Having derived a correct simplification in my answer below, it seems that the supposed identity I give above is incorrect (or at least, it seems to be since I don't see how the two expressions would be equal; I have not gone through the trouble of finding a $u,v$ that disprove the identity, though).

In some sense, the solution $(u \times \nabla) \times v = u \cdot (\nabla v) - u(\nabla \cdot v)$ fits the form of $(a \times b) \times c = (a \cdot c)b - a(b \cdot c)$, though it is perhaps counterintuitive that the first term should be $u \cdot (\nabla v)$ which, if the triple product rule can be directly applied in this context, should be our interpretation of $(u \cdot v)\nabla$.

Interestingly, we find that $$ u \times (\nabla \times v) = u \cdot (\nabla v) - (u \cdot \nabla)v $$ The associated vector identity is $a \times (b \times c) = (a \cdot c)b - (a \cdot b) c$. Again, the second term seems to "behave as expected".

Answer 1

Take $f: \Bbb R^3 \to \Bbb R$, $u,v:\Bbb R^3 \to \Bbb R^3$, write $v = (v_1,v_2,v_3)$ and take $e_1,e_2,e_3$ to be the canonical basis vectors (that is, $i,j,k$). We can write $v = \sum_{j=1}^3 v_j e_j$, and $$ \nabla \cdot v = \sum_{i=1}^3 e_i \cdot \frac{\partial v}{\partial x_i}, \qquad \nabla \times v = \sum_{i=1}^3 e_i \times \frac{\partial v}{\partial x_i}, \\ (v \cdot \nabla)f = \sum_{i=1}^3 (v \cdot e_i) \frac{\partial f}{\partial x_i}, \qquad (v \times \nabla)f = \sum_{i=1}^3 (v \times e_i) \frac{\partial f}{\partial x_i},\\ (u \times \nabla) \times v = \sum_{i=1}^3 \sum_{j=1}^3 [(u \times e_i) \times e_j] \frac{\partial v_j}{\partial x_i}. $$ With that stated, we can now apply the triple vector product identity to get $$ \begin{align} \sum_{i=1}^3 \sum_{j=1}^3 [(u \times e_i) \times e_j] \frac{\partial v_j}{\partial x_i} &= \sum_{i=1}^3 \sum_{j=1}^3 [(u \cdot e_j)e_i - (e_i \cdot e_j)u] \frac{\partial v_j}{\partial x_i} \\ &= \sum_{i=1}^3 \sum_{j=1}^3 u_j\frac{\partial v_j}{\partial x_i} e_i - \sum_{i=1}^3 \sum_{j=1}^3 \delta_{ij} u\frac{\partial v_j}{\partial x_i} \\ &= \sum_{j=1}^3 u_j \sum_{i=1}^3 \frac{\partial v_j}{\partial x_i} e_i - u\sum_{i=1}^3 \frac{\partial v_i}{\partial x_i} \\ & = \sum_{j=1}^3 u_j \nabla v - u\sum_{i=1}^3 \frac{\partial v_i}{\partial x_i} = u \cdot (\nabla v) - u(\nabla \cdot v). \end{align} $$

Answer 2

For what it's worth, the theory of geometric algebra can give you this calculation (and many others) as a one-liner.

$$ u\times (\nabla \times v) = i\left(u\wedge i(\nabla \wedge v)\right) = - u\cdot (\nabla \wedge v) = u\cdot(\nabla v) - (u\cdot\nabla)v, $$ where $\wedge$ denotes the exterior product, $i = e_1\wedge e_2\wedge e_3$ denotes the unit pseudoscalar of $\mathbb{R}^3$, and we used that $a\times b = i(a\wedge b)=a\cdot(ib)$ along with the "cab-bac" identity.

Answer 3

According to this page you seem to miss a term $\boldsymbol{u}\times(\nabla\times\boldsymbol{v})$ on RHS. The deduction is exactly the same.

I mean, the equality should be $$(\boldsymbol{u}\times\nabla)\times\boldsymbol{v}=\boldsymbol{u}\times(\nabla\times\boldsymbol{v})+\boldsymbol{u}\cdot\nabla\boldsymbol{v}-\boldsymbol{u}(\nabla\cdot\boldsymbol{v})$$