Micromechanics of composites/Proof 2

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

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

Proof:

Using the identity $\mathrm{\nabla }\bullet (\phi 𝐮)=𝐮\cdot \mathrm{\nabla }\phi +\phi \mathrm{\nabla }\bullet 𝐮$ we have

$\mathrm{\nabla }\bullet [(𝐯\cdot 𝐚)(𝑺\cdot 𝐛)]=(𝑺\cdot 𝐛)\cdot \mathrm{\nabla }(𝐯\cdot 𝐚)+(𝐯\cdot 𝐚)\mathrm{\nabla }\bullet (𝑺\cdot 𝐛).$

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

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

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

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

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

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

$\mathrm{\nabla }𝐯\cdot (𝑺\cdot 𝐛)=(\mathrm{\nabla }𝐯\cdot 𝑺)\cdot 𝐛.$

Therefore, the final form of the first term is

$(𝑺\cdot 𝐛)\cdot \mathrm{\nabla }(𝐯\cdot 𝐚)=𝐚\cdot [(\mathrm{\nabla }𝐯\cdot 𝑺)\cdot 𝐛].$

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

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

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

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

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

Therefore, the final form of the second term is

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

Adding the two terms, we get

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

Therefore,

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