Micromechanics of composites/Proof 11

Identity 1
Let $$\boldsymbol{A}$$ and $$\boldsymbol{B}$$ be two second order tensors. Show that

\boldsymbol{A}:\boldsymbol{B} = (\boldsymbol{A}^T\cdot\boldsymbol{B}):\boldsymbol{\mathit{1}}~. $$

Proof:

Using index notation,

\boldsymbol{A}:\boldsymbol{B} = A_{ij}~B_{ij} = A^T_{ji}~B_{ij} = A^T_{ji}~B_{ik}~\delta_{jk} = [\boldsymbol{A}^T\cdot\boldsymbol{B}]_{jk}~\delta_{jk} = (\boldsymbol{A}^T\cdot\boldsymbol{B}):\boldsymbol{\mathit{1}} ~. $$ Hence,

{ \boldsymbol{A}:\boldsymbol{B} = (\boldsymbol{A}^T\cdot\boldsymbol{B}):\boldsymbol{\mathit{1}} \qquad \square } $$

Identity 2
Let $$\boldsymbol{A}$$ be a second order tensor and let $$\mathbf{a}$$ and $$\mathbf{b}$$ be two vectors. Show that

\boldsymbol{A}:(\mathbf{a}\otimes\mathbf{b}) = (\boldsymbol{A}\cdot\mathbf{b})\cdot\mathbf{a} ~. $$

 Proof:

It is convenient to use index notation for this. We have

\boldsymbol{A}:(\mathbf{a}\otimes\mathbf{b}) = A_{ij}~a_i~b_j = (A_{ij}~b_j)~a_i = (\boldsymbol{A}\cdot\mathbf{b})\cdot\mathbf{a} ~. $$ Hence,

{ \boldsymbol{A}:(\mathbf{a}\otimes\mathbf{b}) = (\boldsymbol{A}\cdot\mathbf{b})\cdot\mathbf{a} \qquad \square } $$

Identity 3
Let $$\boldsymbol{A}$$ and $$\boldsymbol{B}$$ be two second order tensors and let $$\mathbf{a}$$ and $$\mathbf{b}$$ be two vectors. Show that

(\boldsymbol{A}\cdot\mathbf{a})\cdot(\boldsymbol{B}\cdot\mathbf{b}) = (\boldsymbol{A}^T\cdot\boldsymbol{B}):(\mathbf{a}\otimes\mathbf{b}) ~. $$

Proof:

Using index notation,

(\boldsymbol{A}\cdot\mathbf{a})\cdot(\boldsymbol{B}\cdot\mathbf{b}) = (A_{ij}~a_j)(B_{ik}~b_k) = (A_{ij}~B_{ik})(a_j~b_k) = (A^T_{ji}~B_{ik})(a_j~b_k) = (\boldsymbol{A}^T\cdot\boldsymbol{B}):(\mathbf{a}\otimes\mathbf{b}) ~. $$ Hence,

{ (\boldsymbol{A}\cdot\mathbf{a})\cdot(\boldsymbol{B}\cdot\mathbf{b}) = (\boldsymbol{A}^T\cdot\boldsymbol{B}):(\mathbf{a}\otimes\mathbf{b}) \qquad \square } $$

Identity 4
Let $$\boldsymbol{A}$$ be a second order tensors and let $$\mathbf{a}$$ and $$\mathbf{b}$$ be two vectors. Show that

(\boldsymbol{A}\cdot\mathbf{a})\otimes\mathbf{b} = \boldsymbol{A}\cdot(\mathbf{a}\otimes\mathbf{b}) \qquad \text{and} \qquad \mathbf{a}\otimes(\boldsymbol{A}\cdot\mathbf{b}) = [\boldsymbol{A}\cdot(\mathbf{b}\otimes\mathbf{a})]^T = (\mathbf{a}\otimes\mathbf{b})\cdot\boldsymbol{A}^T ~. $$

Proof:

For the first identity, using index notation, we have

[(\boldsymbol{A}\cdot\mathbf{a})\otimes\mathbf{b}]_{ik} = (A_{ij}~a_j)~b_k = A_{ij}~(a_j~b_k) = A_{ij}~[\mathbf{a}\otimes\mathbf{b}]_{jk} = \boldsymbol{A}\cdot(\mathbf{a}\otimes\mathbf{b}) ~. $$ Hence,

{ (\boldsymbol{A}\cdot\mathbf{a})\otimes\mathbf{b} = \boldsymbol{A}\cdot(\mathbf{a}\otimes\mathbf{b}) \qquad \square } $$

For the second identity, we have

[\mathbf{a}\otimes(\boldsymbol{A}\cdot\mathbf{b})]_{ij} = a_i~(A_{jk}~b_k) = (a_i~b_k)~A_{jk} = (a_i~b_k)~A^T_{kj} = [(\mathbf{a}\otimes\mathbf{b})\cdot\boldsymbol{A}^T]_{ij} ~. $$ Therefore,

\mathbf{a}\otimes(\boldsymbol{A}\cdot\mathbf{b}) = (\mathbf{a}\otimes\mathbf{b})\cdot\boldsymbol{A}^T ~. $$ Now, $$\mathbf{a}\otimes\mathbf{b} = [\mathbf{b}\otimes\mathbf{a}]^T$$ and $$(\boldsymbol{A}\cdot\boldsymbol{B})^T = \boldsymbol{B}^T\cdot\boldsymbol{A}^T$$. Hence,

(\mathbf{a}\otimes\mathbf{b})\cdot\boldsymbol{A}^T = (\mathbf{b}\otimes\mathbf{a})^T\cdot\boldsymbol{A}^T = [\boldsymbol{A}\cdot(\mathbf{b}\otimes\mathbf{a})]^T ~. $$ Therefore,

{ \mathbf{a}\otimes(\boldsymbol{A}\cdot\mathbf{b}) = [\boldsymbol{A}\cdot(\mathbf{b}\otimes\mathbf{a})]^T = (\mathbf{a}\otimes\mathbf{b})\cdot\boldsymbol{A}^T \qquad \square } $$