Continuum mechanics/Tensor-vector identities

Tensor-vector identity - 1

$[(𝐯\bullet 𝐚)(𝑺\bullet 𝐛)]\cdot 𝐧=𝐚\cdot [\{𝐯\otimes ({𝑺}^T\bullet 𝐧)\}\cdot 𝐛].$

Proof:

Using the identity $𝐚\cdot ({𝑨}^T\cdot 𝐛)=𝐛\cdot (𝑨\cdot 𝐚)$ we have

$𝐧\cdot [(𝐯\bullet 𝐚)(𝑺\bullet 𝐛)]=𝐛\cdot [(𝐯\cdot 𝐚)({𝑺}^T\cdot 𝐧)].$

Also, using the definition $(𝐮\otimes 𝐯)\cdot 𝐚=(𝐚\cdot 𝐯)𝐮$ we have

$(𝐯\cdot 𝐚)({𝑺}^T\cdot 𝐧)=[({𝑺}^T\cdot 𝐧)\otimes 𝐯]\cdot 𝐚.$

Therefore,

$𝐧\cdot [(𝐯\bullet 𝐚)(𝑺\bullet 𝐛)]=𝐛\cdot [\{({𝑺}^T\cdot 𝐧)\otimes 𝐯\}\cdot 𝐚].$

Using the identity $𝐚\cdot ({𝑨}^T\cdot 𝐛)=𝐛\cdot (𝑨\cdot 𝐚)$ we have

$𝐛\cdot [\{({𝑺}^T\cdot 𝐧)\otimes 𝐯\}\cdot 𝐚]=𝐚\cdot [\{({𝑺}^T\cdot 𝐧)\otimes 𝐯{\}}^T\cdot 𝐛].$

Finally, using the relation $(𝐮\otimes 𝐯{)}^T=𝐯\otimes 𝐮$, we get

$𝐚\cdot [\{({𝑺}^T\cdot 𝐧)\otimes 𝐯{\}}^T\cdot 𝐛]=𝐚\cdot [\{𝐯\otimes ({𝑺}^T\cdot 𝐧)\}\cdot 𝐛].$

Hence,

$[(𝐯\bullet 𝐚)(𝑺\bullet 𝐛)]\cdot 𝐧=𝐚\cdot [\{𝐯\otimes ({𝑺}^T\bullet 𝐧)\}\cdot 𝐛]◻$

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Tensor-vector identity 2

Let $𝐯$ be a vector field and let $𝑺$ be a second-order tensor field. Let $𝐚$ and $𝐛$ be two arbitrary vectors. Show that

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

Proof:

Using the identity $𝜵\bullet (\phi 𝐮)=𝐮\cdot 𝜵\phi +\phi 𝜵\bullet 𝐮$ we have

$𝜵\bullet [(𝐯\cdot 𝐚)(𝑺\cdot 𝐛)]=(𝑺\cdot 𝐛)\cdot 𝜵(𝐯\cdot 𝐚)+(𝐯\cdot 𝐚)𝜵\bullet (𝑺\cdot 𝐛).$

From the identity $𝜵(𝐮\cdot 𝐯)=𝜵{𝐮}^T\cdot 𝜵𝐯+𝜵{𝐯}^T\cdot 𝐮$, we have $𝜵(𝐯\cdot 𝐚)=𝜵{𝐯}^T\cdot 𝐚+𝜵{𝐚}^T\cdot 𝐯$.

Since $𝐚$ is constant, $𝜵𝐚=0$, and we have

$(𝑺\cdot 𝐛)\cdot 𝜵(𝐯\cdot 𝐚)=(𝑺\cdot 𝐛)\cdot (𝜵{𝐯}^T\cdot 𝐚).$

From the relation $𝐚\cdot ({𝑨}^T\cdot 𝐛)=𝐛\cdot (𝑨\cdot 𝐚)$ we have

$(𝑺\cdot 𝐛)\cdot (𝜵{𝐯}^T\cdot 𝐚)=𝐚\cdot [𝜵𝐯\cdot (𝑺\cdot 𝐛)].$

Using the relation $𝑨\cdot (𝑩\cdot 𝐛)=(𝑨\cdot 𝑩)\cdot 𝐛$, we get

$𝜵𝐯\cdot (𝑺\cdot 𝐛)=(𝜵𝐯\cdot 𝑺)\cdot 𝐛.$

Therefore, the final form of the first term is

$(𝑺\cdot 𝐛)\cdot 𝜵(𝐯\cdot 𝐚)=𝐚\cdot [(𝜵𝐯\cdot 𝑺)\cdot 𝐛].$

For the second term, from the identity $𝜵\bullet ({𝑺}^T\cdot 𝐯)=𝑺:𝜵𝐯+𝐯\cdot (𝜵\bullet 𝑺)$ we get, $𝜵\bullet (𝑺\cdot 𝐛)={𝑺}^T:𝜵𝐛+𝐛\cdot (𝜵\bullet {𝑺}^T)$.

Since $𝐛$ is constant, $𝜵𝐛=0$, and we have

$(𝐯\cdot 𝐚)𝜵\bullet (𝑺\cdot 𝐛)=(𝐯\cdot 𝐚)[𝐛\cdot (𝜵\bullet {𝑺}^T)]=𝐚\cdot [\{𝐛\cdot (𝜵\bullet {𝑺}^T)\}𝐯].$

From the definition $(𝐮\otimes 𝐯)\cdot 𝐚=(𝐚\cdot 𝐯)𝐮$, we get

$[𝐛\cdot (𝜵\bullet {𝑺}^T)]𝐯=[𝐯\otimes (𝜵\bullet {𝑺}^T)]\cdot 𝐛.$

Therefore, the final form of the second term is

$(𝐯\cdot 𝐚)𝜵\bullet (𝑺\cdot 𝐛)=𝐚\cdot [𝐯\otimes (𝜵\bullet {𝑺}^T)]\cdot 𝐛.$

Adding the two terms, we get

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

Therefore,

$𝜵\bullet [(𝐯\cdot 𝐚)(𝑺\cdot 𝐛)]=𝐚\cdot [\{𝜵𝐯\cdot 𝑺+𝐯\otimes (𝜵\bullet {𝑺}^T)\}\cdot 𝐛]◻$

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