Using the Chain Rule for Vector Calculus: A Tutorial

[binbagsss]

TL;DR Summary

chain rule order of differentiation in the product

This is probably a stupid question, but I have never realised that there's an order things should be done in the chain rule , so for example

$\nabla(\bf{v}.\bf{v})=2\bf{v} (\nabla\cdot \bf{v})$

and not

$2 \bf{v} \cdot \nabla \bf{v}$

Is there an obvious way to see / think of this from the chain rule, say in 1-D, preferably through looking at the limit definition?
Thanks

 

[anuttarasammyak]

TL;DR Summary: chain rule order of differentiation in the product

and not

It contains gradient of vector which is a tough object.

 

[PeroK]

TL;DR Summary: chain rule order of differentiation in the product

This is probably a stupid question, but I have never realised that there's an order things should be done in the chain rule , so for example

$\nabla(\bf{v}.\bf{v})=2\bf{v} (\nabla\cdot \bf{v})$

and not

$2 \bf{v} \cdot \nabla \bf{v}$

Is there an obvious way to see / think of this from the chain rule, say in 1-D, preferably through looking at the limit definition?
Thanks

The gradient of a scalar function is a vector. All these identities follow from the definition.

 

[pasmith]

TL;DR Summary: chain rule order of differentiation in the product

This is probably a stupid question, but I have never realised that there's an order things should be done in the chain rule , so for example

$\nabla(\bf{v}.\bf{v})=2\bf{v} (\nabla\cdot \bf{v})$

and not

$2 \bf{v} \cdot \nabla \bf{v}$

Is there an obvious way to see / think of this from the chain rule, say in 1-D, preferably through looking at the limit definition?
Thanks

Using suffix notation, we can form five vectors from two copies of $\mathbf{v}$ and a single $\nabla$: $$\begin{array}{cc} \nabla (\mathbf{v} \cdot \mathbf{v}) & \partial_i(v_jv_j), \\ \nabla \cdot (\mathbf{v} \mathbf{v}) & \partial_j(v_iv_j) , \\ \mathbf{v} \cdot (\nabla \mathbf{v}) & v_j \partial_i v_j, \\ \mathbf{v} (\nabla \cdot \mathbf{v}) & v_i \partial_j v_j, \\ (\mathbf{v} \cdot \nabla) \mathbf{v} & v_j \partial_j v_i. \end{array}$$ Applying the product rule to the first two we have $$\begin{split} \nabla (\mathbf{v} \cdot \mathbf{v}) &= 2\mathbf{v} \cdot (\nabla \mathbf{v}) \\ \nabla \cdot (\mathbf{v} \mathbf{v}) &= (\mathbf{v} \cdot \nabla) \mathbf{v} + \mathbf{v}(\nabla \cdot \mathbf{v}).\end{split}$$ This is about the point at which suffix notation becomes clearer than vector notation.

EDIT: For completeness, we can also form these three using the cross product: $$\begin{array}{cc} \nabla \times (\mathbf{v} \times \mathbf{v}) & \epsilon_{ijk}\epsilon_{klm}\partial_j(v_lv_m) \\ \mathbf{v} \times (\nabla \times \mathbf{v}) & \epsilon_{ijk}\epsilon_{klm} v_j\partial_l v_m \\ (\mathbf{v} \times \nabla) \times \mathbf{v} & -\epsilon_{ijk}\epsilon_{klm} v_l\partial_mv_j \end{array}$$ These can, however, be expressed in terms of the previous vectors by use of the identity $$\epsilon_{ijk}\epsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}.$$

 

[Mark44]

@pasmith, what operation is implied in this product: $\mathbf{v} \mathbf{v}$? (The 2nd of your 5 examples)

 

[pasmith]

@pasmith, what operation is implied in this product: $\mathbf{v} \mathbf{v}$? (The 2nd of your 5 examples)

Tensor product: $(\mathbf{v}\mathbf{v})_{ij} = v_i v_j$.