Continuum mechanics/Tensor-vector identities

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[edit] 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

[edit] 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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