# Continuum mechanics/Tensor-vector identities
## Tensor-vector identity - 1

    $[(\mathbf {v} \bullet \mathbf {a} )({\boldsymbol {S}}\bullet \mathbf {b} )]\cdot \mathbf {n} =\mathbf {a} \cdot [\{\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )\}\cdot \mathbf {b} ]~.$
Proof:
Using the identity $\mathbf {a} \cdot ({\boldsymbol {A}}^{T}\cdot \mathbf {b} )=\mathbf {b} \cdot ({\boldsymbol {A}}\cdot \mathbf {a} )$ we have
    $\mathbf {n} \cdot [(\mathbf {v} \bullet \mathbf {a} )({\boldsymbol {S}}\bullet \mathbf {b} )]=\mathbf {b} \cdot [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}^{T}\cdot \mathbf {n} )]~.$
Also, using the definition $(\mathbf {u} \otimes \mathbf {v} )\cdot \mathbf {a} =(\mathbf {a} \cdot \mathbf {v} )\mathbf {u}$ we have
    $(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}^{T}\cdot \mathbf {n} )=[({\boldsymbol {S}}^{T}\cdot \mathbf {n} )\otimes \mathbf {v} ]\cdot \mathbf {a} ~.$
Therefore,
    $\mathbf {n} \cdot [(\mathbf {v} \bullet \mathbf {a} )({\boldsymbol {S}}\bullet \mathbf {b} )]=\mathbf {b} \cdot [\{({\boldsymbol {S}}^{T}\cdot \mathbf {n} )\otimes \mathbf {v} \}\cdot \mathbf {a} ]~.$
Using the identity $\mathbf {a} \cdot ({\boldsymbol {A}}^{T}\cdot \mathbf {b} )=\mathbf {b} \cdot ({\boldsymbol {A}}\cdot \mathbf {a} )$ we have
    $\mathbf {b} \cdot [\{({\boldsymbol {S}}^{T}\cdot \mathbf {n} )\otimes \mathbf {v} \}\cdot \mathbf {a} ]=\mathbf {a} \cdot [\{({\boldsymbol {S}}^{T}\cdot \mathbf {n} )\otimes \mathbf {v} \}^{T}\cdot \mathbf {b} ]~.$
Finally, using the relation $(\mathbf {u} \otimes \mathbf {v} )^{T}=\mathbf {v} \otimes \mathbf {u}$, we get
    $\mathbf {a} \cdot [\{({\boldsymbol {S}}^{T}\cdot \mathbf {n} )\otimes \mathbf {v} \}^{T}\cdot \mathbf {b} ]=\mathbf {a} \cdot [\{\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\cdot \mathbf {n} )\}\cdot \mathbf {b} ]~.$
Hence,
    ${[(\mathbf {v} \bullet \mathbf {a} )({\boldsymbol {S}}\bullet \mathbf {b} )]\cdot \mathbf {n} =\mathbf {a} \cdot [\{\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )\}\cdot \mathbf {b} ]}\qquad \qquad \qquad \square$

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