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,

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

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Continuum mechanics

Curl of a gradient of a vector →

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,

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

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Curl of a gradient of a vector →