Micromechanics of composites/Proof 2

Tensor-vector identity 2
Let $$\mathbf{v}$$ be a vector field and let $$\boldsymbol{S}$$ be a second-order tensor field. Let $$\mathbf{a}$$ and $$\mathbf{b}$$ be two arbitrary vectors. Show that

\boldsymbol{\nabla} \bullet [(\mathbf{v}\cdot\mathbf{a})(\boldsymbol{S}\cdot\mathbf{b})] = \mathbf{a}\cdot[\{\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)\}\cdot\mathbf{b}] ~. $$

Proof:

Using the identity $$\boldsymbol{\nabla} \bullet (\varphi~\mathbf{u}) = \mathbf{u}\cdot\boldsymbol{\nabla} \varphi + \varphi~\boldsymbol{\nabla} \bullet \mathbf{u}$$ we have

\boldsymbol{\nabla} \bullet [(\mathbf{v}\cdot\mathbf{a})(\boldsymbol{S}\cdot\mathbf{b})] = (\boldsymbol{S}\cdot\mathbf{b})\cdot\boldsymbol{\nabla} (\mathbf{v}\cdot\mathbf{a}) + (\mathbf{v}\cdot\mathbf{a})~\boldsymbol{\nabla} \bullet (\boldsymbol{S}\cdot\mathbf{b}) ~. $$ From the identity $$\boldsymbol{\nabla} (\mathbf{u}\cdot\mathbf{v}) = \boldsymbol{\nabla}\mathbf{u}^T\cdot\boldsymbol{\nabla} \mathbf{v} + \boldsymbol{\nabla} \mathbf{v}^T\cdot\mathbf{u}$$, we have $$\boldsymbol{\nabla} (\mathbf{v}\cdot\mathbf{a}) = \boldsymbol{\nabla} \mathbf{v}^T\cdot\mathbf{a} + \boldsymbol{\nabla} \mathbf{a}^T\cdot\mathbf{v}$$.

Since $$\mathbf{a}$$ is constant, $$\boldsymbol{\nabla} \mathbf{a} = 0$$, and we have

(\boldsymbol{S}\cdot\mathbf{b})\cdot\boldsymbol{\nabla} (\mathbf{v}\cdot\mathbf{a})= (\boldsymbol{S}\cdot\mathbf{b})\cdot(\boldsymbol{\nabla} \mathbf{v}^T\cdot\mathbf{a})~. $$ From the relation $$ \mathbf{a}\cdot(\boldsymbol{A}^T\cdot\mathbf{b}) = \mathbf{b}\cdot(\boldsymbol{A}\cdot\mathbf{a})$$ we have

(\boldsymbol{S}\cdot\mathbf{b})\cdot(\boldsymbol{\nabla} \mathbf{v}^T\cdot\mathbf{a}) = \mathbf{a}\cdot[\boldsymbol{\nabla} \mathbf{v}\cdot(\boldsymbol{S}\cdot\mathbf{b})] ~. $$ Using the relation $$\boldsymbol{A}\cdot(\boldsymbol{B}\cdot\mathbf{b}) = (\boldsymbol{A}\cdot\boldsymbol{B})\cdot\mathbf{b}$$, we get

\boldsymbol{\nabla} \mathbf{v}\cdot(\boldsymbol{S}\cdot\mathbf{b}) = (\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S})\cdot\mathbf{b} ~. $$ Therefore, the final form of the first term is

(\boldsymbol{S}\cdot\mathbf{b})\cdot\boldsymbol{\nabla} (\mathbf{v}\cdot\mathbf{a})= \mathbf{a}\cdot[(\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S})\cdot\mathbf{b}] ~. $$ For the second term, from the identity $$\boldsymbol{\nabla} \bullet (\boldsymbol{S}^T\cdot\mathbf{v}) = \boldsymbol{S}:\boldsymbol{\nabla} \mathbf{v} + \mathbf{v}\cdot(\boldsymbol{\nabla} \bullet \boldsymbol{S})$$ we get, $$\boldsymbol{\nabla} \bullet (\boldsymbol{S}\cdot\mathbf{b}) = \boldsymbol{S}^T:\boldsymbol{\nabla} \mathbf{b} + \mathbf{b}\cdot(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)$$.

Since $$\mathbf{b}$$ is constant, $$\boldsymbol{\nabla} \mathbf{b} = 0$$, and we have

(\mathbf{v}\cdot\mathbf{a})~\boldsymbol{\nabla} \bullet (\boldsymbol{S}\cdot\mathbf{b}) = (\mathbf{v}\cdot\mathbf{a})~[\mathbf{b}\cdot(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)] = \mathbf{a}\cdot[\{\mathbf{b}\cdot(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)\}~\mathbf{v}] ~. $$ From the definition $$ (\mathbf{u}\otimes\mathbf{v})\cdot\mathbf{a} = (\mathbf{a}\cdot\mathbf{v})\mathbf{u} $$, we get

[\mathbf{b}\cdot(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)]~\mathbf{v} = [\mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)]\cdot\mathbf{b} ~. $$ Therefore, the final form of the second term is

(\mathbf{v}\cdot\mathbf{a})~\boldsymbol{\nabla} \bullet (\boldsymbol{S}\cdot\mathbf{b}) = \mathbf{a}\cdot[\mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)]\cdot\mathbf{b} ~. $$ Adding the two terms, we get

\boldsymbol{\nabla} \bullet [(\mathbf{v}\cdot\mathbf{a})(\boldsymbol{S}\cdot\mathbf{b})] = \mathbf{a}\cdot[(\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S})\cdot\mathbf{b}] + \mathbf{a}\cdot[\mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)]\cdot\mathbf{b} ~. $$ Therefore,

{ \boldsymbol{\nabla} \bullet [(\mathbf{v}\cdot\mathbf{a})(\boldsymbol{S}\cdot\mathbf{b})] = \mathbf{a}\cdot[\{\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)\}\cdot\mathbf{b}] } \qquad\qquad\qquad\square $$