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I am looking for an identity for the following derivative:

$$\nabla[( \mathbf{a} \cdot \mathbf{b} )\mathbf{c}] \cdot \mathbf{d},$$

where $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, and $\mathbf{d}$ are column vectors.

My approach to expand the expression is as follows:

\begin{align*} \nabla[( \mathbf{a} \cdot \mathbf{b} )\mathbf{c}] \cdot \mathbf{d} &= [ ( \mathbf{a} \cdot \mathbf{b} ) D \mathbf{c} + \mathbf{c} \otimes\nabla(\mathbf{a} \cdot \mathbf{b}) ] \cdot \mathbf{d} \\ &= ( \mathbf{a} \cdot \mathbf{b} ) D \mathbf{c} \cdot \mathbf{d} + \{ \mathbf{c} \otimes [ (D\mathbf{a})^\top \mathbf{b} + (D\mathbf{b})^\top \mathbf{a}] \} \cdot \mathbf{d}\\ &= ( \mathbf{a} \cdot \mathbf{b} ) D \mathbf{c} \cdot \mathbf{d} + \{ \mathbf{c} \otimes [ (D\mathbf{a})^\top \mathbf{b} + (D\mathbf{b})^\top \mathbf{a}] \}^\top \mathbf{d}\\ &= ( \mathbf{a} \cdot \mathbf{b} ) D \mathbf{c} \cdot \mathbf{d} + \{ [ (D\mathbf{a})^\top \mathbf{b} + (D\mathbf{b})^\top \mathbf{a}] \otimes \mathbf{c} \} \mathbf{d}\\ &= ( \mathbf{a} \cdot \mathbf{b} ) D \mathbf{c} \cdot \mathbf{d} + (\mathbf{c} \cdot \mathbf{d}) [ (D\mathbf{a})^\top \mathbf{b} + (D\mathbf{b})^\top \mathbf{a}]. \end{align*}

Here, $D \equiv \nabla^\top$, i.e., $\nabla$ stands for the gradient operator and $D$ denotes the Jacobian.

My question is, is this expansion correct? Moreover, in my understanding, the resulting expression is again a column vector. Is it right?

Jullienne Franz
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    what is the meaning of the [] brackets? Is there no derivative on $\bf d$? – Calvin Khor Jan 14 '22 at 08:23
  • @CalvinKhor Thank you for your interest in my post. Here, I am taking the derivative of the product $\mathbf{(\mathbf{a} \cdot \mathbf{b})\mathbf{c}}$. So, the brackets simply means $\mathbf{(\mathbf{a} \cdot \mathbf{b})\mathbf{c}}$ are grouped together. – Jullienne Franz Jan 14 '22 at 08:26
  • @CalvinKhor Here, I am not taking the derivative of $\mathbf{d}$, and to my understanding the derivative is a directional derivative in the direction of the vector $\mathbf{d}$. – Jullienne Franz Jan 14 '22 at 08:42
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    The expression $\nabla [(a\cdot b)c]$ is then the gradient of a vector then (i.e. its Jacobian)? And what does $Dc\cdot d$ mean? This seems to be the dot product between a matrix and a vector. – Andreas Lenz Jan 14 '22 at 08:43
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    Well when I compute in index notation $${\let\del\partial} \del_i (a_j b_j c_k)=(\del_ia_j)b_jc_k+a_j (\del_ib_j) c_k+a_j b_j \del_ic_k$$ which has $c$ on the right? Since the Jacobian $\nabla f$ of $f$ is $\del_j f_i$, $\nabla f^Th = \del_i f_j h_j$, so this is $$ \nabla ((a\cdot b) c)= ((\nabla a)^Tb) \otimes c + (\nabla b^Ta)\otimes c + (a\cdot b)\nabla c $$ Now as mentioned above, this is a matrix, so I don't know how to take the dot product with a vector. – Calvin Khor Jan 14 '22 at 08:45
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    @AndreasLenz I see. Actually, I am trying to get the directional derivative of the product $(a \cdot b) c$ in the direction of the vector $d$. So, should I write $D((a\cdot b)c)d$ instead of the original expression? I'm sorry, so the dot product between a matrix and a vector does not make sense, right? – Jullienne Franz Jan 14 '22 at 08:58
  • @CalvinKhor I understand. About the identity you mentioned, is that the right order? I mean, how do I evaluate the product $((\nabla a)^\top b) \otimes c$? I am confused about these actually. – Jullienne Franz Jan 14 '22 at 09:01
  • If I'm honest, I cannot easily understand the matrix notation with $\otimes$ etc. It looks cleaner but its just not better. The version that I can say is 100% correct is the index notation verison written above. The directional derivative in direction $v$ is the operator $(v\cdot \nabla)=v_i\del_i$. – Calvin Khor Jan 14 '22 at 09:03
  • @JulienneFranz Yes, that expression would make more sense. Notice however that to the best of my knowledge the directional derivative is usually defined on scalars. And yes, the dot product is defined for two vectors of the same length. – Andreas Lenz Jan 14 '22 at 09:07
  • @CalvinKhor Thank you. So will it make sense to write $d \cdot \nabla((a\cdot b)c)$? but then, again, this notation is confusing as it is different from $D((a\cdot b)c)d$. – Jullienne Franz Jan 14 '22 at 09:16
  • @AndreasLenz Yes, I understand that for scalar function $f$, the directional derivative in the direction of a vector $d$ is expressed as $\nabla f \cdot d$. But how about for vector-valued function, will it also make sense to compute its directional derivative? I found this related post regarding this https://math.stackexchange.com/questions/2995762/directional-derivatives-for-vector-valued-functions – Jullienne Franz Jan 14 '22 at 09:19
  • Yes, but I would always write it with brackets like $(d\cdot \nabla)$. I can convert the above comment into an answer but it will a little while later. – Calvin Khor Jan 14 '22 at 09:43
  • @CalvinKhor Yes, please. Many thanks! – Jullienne Franz Jan 14 '22 at 10:03

2 Answers2

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It is said in comments that OP wants to compute the directional derivative in direction $d$. I'll write $v$ instead of $d$. The usual extension of a map defined for scalar-fields to vector fields is to apply the known map componentwise, at least in cartesian coordinates (see e.g. the vector Laplacian). That is, I would write $${\let\del\partial} \big((v\cdot \nabla) ((a\cdot b) c)\big)_k: = v_i\del_i ( a_j b_j c_k)$$ one has by the usual 1D Leibniz rule $$ v_i\del_i ( a_j b_j c_k) = (v_i\del_ia_j) b_j c_k + a_j (v_i\del_ib_j) c_k + a_j b_j v_i\del_ic_k$$ which can be also written $$ (v\cdot \nabla) ((a\cdot b) c) = (((v\cdot \nabla)a)\cdot b) c+(a\cdot ((v\cdot \nabla)b)) c+(a\cdot b) (v\cdot \nabla)c$$ I would write $(\nabla f)_{ij} = \del_j f_i$ and hence $ v_i\del_i f_j = (\nabla f^Tv)_j $. So you can rewrite the above only using matrix products, if you wanted:

$$((v\cdot \nabla)a)\cdot b = ((v\cdot \nabla)a)^Tb=(\nabla a^Tv)^Tb = v^T(\nabla a)b , \\ a\cdot ((v\cdot \nabla)b) =a^T\nabla b^Tv \\ (a\cdot b) (v\cdot \nabla)c = a^Tb \nabla c^T v$$ i.e. $$ (v\cdot \nabla) ((a\cdot b) c) = (v^T(\nabla a)b)c + (a^T\nabla b^Tv)c+ a^Tb \nabla c^T v$$

Calvin Khor
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I think I arrive at the same conclusion but in slightly different way. Using differentials $$ d\mathbf{f} = (d\mathbf{a})^T\mathbf{b} \mathbf{c}+ \mathbf{a}^T d\mathbf{b} \mathbf{c}+ \mathbf{a}^T\mathbf{b} d\mathbf{c} $$ Now using $d\mathbf{a}=\mathbf{J}_a d\mathbf{x}$... and rearranging terms, you will find that the Jacobian of $\mathbf{f}$ which is a matrix, is $$ \mathbf{J}_f = \mathbf{c} \left[ \mathbf{b}^T \mathbf{J}_a + \mathbf{a}^T \mathbf{J}_b \right] + \mathbf{a}^T\mathbf{b} \mathbf{J}_c $$ The directional derivative is $\mathbf{J}_f \mathbf{d}$.

Steph
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