Apply Chain rule to vector function with chained dot and cross product?

Question

Okay, I got

$\mathbf{v}=(\mathbf{u}_{n-1}-\mathbf{u}_{n})\times(\mathbf{u}_{n}-\mathbf{u}_{n+1})$

and

${e}_{n}=\mathbf{v}\cdot\mathbf{v}=((\mathbf{u}_{n-1}-\mathbf{u}_{n})\times(\mathbf{u}_{n}-\mathbf{u}_{n+1}))^{2}$

with $\mathbf{u}_n=(u_{n,x},u_{n,y},u_{n,z})$

I would like to compute the partial differtial of $e_n$ with respect to $u_{n,x}$

What I tried:

$\frac{\partial e_{n}}{\partial u_{n,x}}=\frac{\partial(\mathbf{v}\cdot\mathbf{v})}{\partial u_{n,x}}$

$=\left(\frac{\partial\mathbf{v}}{\partial u_{n,x}}\cdot\mathbf{v}+\frac{\partial\mathbf{v}}{\partial u_{n,x}}\cdot\mathbf{v}\right)$

$=2\left(\frac{\partial\mathbf{v}}{\partial u_{n,x}}\cdot\mathbf{v}\right)$

$=2\cdot\mathbf{v}\cdot\left(\frac{\partial(u_{n-1}-u_{n})}{\partial u_{n,x}}\times(u_{n}-u_{n+1})+\frac{\partial(u_{n}-u_{n+1})}{\partial u_{n,x}}\times(u_{n-1}-u_{n})\right)$

$=2\cdot\mathbf{v}\cdot\left(\begin{pmatrix}-1\\ 0\\ 0 \end{pmatrix}\times(u_{n,x}-u_{n+1,x})+\begin{pmatrix}1\\ 0\\ 0 \end{pmatrix}\times(u_{n-1,x}-u_{n,x})\right)$

Can I do this this way, or do I have to use the Chain rule? If so, why?

Answer 1

As Christian Blatter has pointed, there are no composition of maps involved, so the chain rule does not apply. All you need is to use the product rule for derivatives. This applies in the usual way also for dot and cross products, as, at the end, they are just linear combinations of products of components.

Let's see: $$ \frac{\partial e_n}{\partial u_{n,x}}=\frac{\partial (\mathbf{v}\cdot \mathbf{v})}{\partial u_{n,x}}=\left(\frac{\partial \mathbf{v}}{\partial u_{n,x}}\right)\cdot \mathbf{v}+\mathbf{v}\cdot\left(\frac{\partial \mathbf{v}}{\partial u_{n,x}}\right)=2\mathbf{v}\cdot\left(\frac{\partial \mathbf{v}}{\partial u_{n,x}}\right) \tag{1} $$ which is the formula you have obtained and it's correct. Now, \begin{align} \frac{\partial \mathbf{v}}{\partial u_{n,x}}&=\frac{\partial}{\partial u_{n,x}}[(\mathbf{u}_{n-1}-\mathbf{u}_n)\times(\mathbf{u}_{n}-\mathbf{u}_{n+1})]\\ &=\left[\frac{\partial (\mathbf{u}_{n-1}-\mathbf{u}_n)}{\partial u_{n,x}}\right]\times(\mathbf{u}_{n}-\mathbf{u}_{n+1})+(\mathbf{u}_{n-1}-\mathbf{u}_n)\times \left[\frac{\partial (\mathbf{u}_{n}-\mathbf{u}_{n+1})}{\partial u_{n,x}}\right]\\ &=(-1,0,0)\times(\mathbf{u}_{n}-\mathbf{u}_{n+1})+(\mathbf{u}_{n-1}-\mathbf{u}_n)\times(1,0,0)\\ &=(-1,0,0)\times(\mathbf{u}_{n}-\mathbf{u}_{n+1})-(1,0,0)\times (\mathbf{u}_{n-1}-\mathbf{u}_n)\\ &=(1,0,0)\times(\mathbf{u}_{n+1}-\mathbf{u}_{n-1}) \end{align} where I have used standard relations for the cross product: $\textbf{a}\times\textbf{b}=-\textbf{b}\times\textbf{a}$, $(-\textbf{a})\times\textbf{b}=-\textbf{a}\times\textbf{b}$ and $\textbf{a}\times(\textbf{b}+\textbf{c})=\textbf{a}\times\textbf{b}+\textbf{a}\times\textbf{c}$.

By plugging this result in $(1)$ and having into account the formula $(\textbf{a}\times\textbf{b})\cdot(\textbf{c}\times\textbf{d})=(\textbf{a}\cdot\textbf{c})(\textbf{b}\cdot\textbf{d})-(\textbf{a}\cdot\textbf{d})(\textbf{b}\cdot\textbf{c})$ we obtain: $$ \frac{\partial e_n}{\partial u_{n,x}}=2\{(u_{n-1,x}-u_{n,x})[(\textbf{u}_{n}-\textbf{u}_{n+1})\cdot(\textbf{u}_{n+1}-\textbf{u}_{n-1})]+(u_{n+1,x}-u_{n,x})[(\textbf{u}_{n-1}-\textbf{u}_{n})\cdot(\textbf{u}_{n+1}-\textbf{u}_{n-1})]\}. $$ This can be written in components, but it does not seem to simplify too much.

In the special case that $\{\textbf{u}_n\}$ form an orthonormal sets of vectors $\textbf{u}_n\cdot \textbf{u}_m=\delta_{n,m}$ the result becomes more compact: $$ \frac{\partial e_n}{\partial u_{n,x}}=2(2u_{n,x}-u_{n-1,x}-u_{n+1,x}). $$

Answer 2

As there are no compositions of maps involved the chain rule need not be applied, just the rules of vector algebra and the product rule.

You have defined $$v:=(a-x)\times(x-b),\quad \psi:=v\cdot v\ ,$$ and want to know ${\partial\psi\over\partial x_1}$. You begin correctly with $${\partial \psi\over\partial x_1}=2v\cdot{\partial v\over\partial x_1}\ .\tag{1}$$ In order to compute the vector ${\partial v\over\partial x_1}$ we note that $${\partial\over\partial x_1}(a-x)=(-1,0,0)^T,\quad{\partial\over\partial x_1}(x-b)=(1,0,0)^T\ ,$$ which shines up in your calculation. Since the vector product is bilinear we now get $${\partial v\over\partial x_1}=(-1,0,0)^T\times(x-b)+(a-x)\times(1,0,0)^T=(1,0,0)^T\times(b-a)=(0,a_3-b_3,b_2-a_2)^T\ .$$Now plug this into $(1)$; maybe some simplification will result.