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

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

This article is issued from Wikiversity - version of the Tuesday, September 11, 2007. The text is available under the Creative Commons Attribution/Share Alike but additional terms may apply for the media files.