![{\displaystyle [(\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} ]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31a344970a0b7f91380ea49a642a8e1aa83d98c9)
Proof:
Using the identity
we have
![{\displaystyle \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} )]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a931704ed14f04f96186aa4ba841eef2af7a2ec1)
Also, using the definition
we have
![{\displaystyle (\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}^{T}\cdot \mathbf {n} )=[({\boldsymbol {S}}^{T}\cdot \mathbf {n} )\otimes \mathbf {v} ]\cdot \mathbf {a} ~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e65e3102db7772453a9ccaac8a945abfcaf08bd)
Therefore,
![{\displaystyle \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} ]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8dcf13178977ca8923a13c89badf31e252a29070)
Using the identity
we have
![{\displaystyle \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} ]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/26578ac487fb2b6520938aa9e2e3b74095fdcbed)
Finally, using the relation
, we get
![{\displaystyle \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} ]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9997ab1523c6dd6b53d23974b8ea7b7a226a4953)
Hence,
![{\displaystyle {[(\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ac235fbe48eff267f6e027fbea1dd91ea5fd1ab)
Let
be a vector field and let
be a second-order tensor field. Let
and
be two arbitrary vectors. Show that
![{\displaystyle {\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} ]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d08ec551dee6717ea19ddc9673fa3e2b12753f1)
Proof:
Using the identity
we have
![{\displaystyle {\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} )~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe8041dc2d467e6660bbae066c61384ef0b5981f)
From the identity
,
we have
.
Since
is constant,
, and we have
![{\displaystyle ({\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} )~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ca93a3629d0effa5ff8b091bd0d9fd3a728a00f)
From the relation
we have
![{\displaystyle ({\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} )]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a44e4b408226b3cb577d3199125d6eae2db7ff7)
Using the relation
, we
get
![{\displaystyle {\boldsymbol {\nabla }}\mathbf {v} \cdot ({\boldsymbol {S}}\cdot \mathbf {b} )=({\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}})\cdot \mathbf {b} ~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/478b7afdb08a4602e82d073bd90cd77347fe1658)
Therefore, the final form of the first term is
![{\displaystyle ({\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} ]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c99b14395e99482d694b6e938f41f57d013dec2c)
For the second term, from the identity
we get,
.
Since
is constant,
, and we have
![{\displaystyle (\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} ]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec683171beeb00e1b558fe3e092e3bd5f2804c95)
From the definition
, we get
![{\displaystyle [\mathbf {b} \cdot ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})]~\mathbf {v} =[\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})]\cdot \mathbf {b} ~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aebdf83a255bec52e82d681f983beb962abfbb25)
Therefore, the final form of the second term is
![{\displaystyle (\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} ~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e263dc37ea8e3edc44fa2c21a1009fb71ab58597)
Adding the two terms, we get
![{\displaystyle {\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} ~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50120190e3fffc149170abaa25a4d1e27a8b4845)
Therefore,
![{\displaystyle {{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/6631dc7cac9b8d11fa81cd0396e4bb3b47763edc)