Micromechanics of composites/Proof 3

Surface-volume integral relation 1
Let $$\Omega$$ be a body and let $$\partial{\Omega}$$ be its surface. Let $$\mathbf{n}$$ be the normal to the surface. Let $$\mathbf{v}$$ be a vector field on $$\Omega$$ and let $$\boldsymbol{S}$$ be a second-order tensor field on $$\Omega$$. Show that

\int_{\partial{\Omega}} \mathbf{v}\otimes(\boldsymbol{S}^T\cdot\mathbf{n})~\text{dA} = \int_{\Omega} [\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)]~\text{dV} ~. $$

Proof:

Recall the relation

\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}] ~. $$ Integrating over the volume, we have

\int_{\Omega} \boldsymbol{\nabla} \bullet [(\mathbf{v}\cdot\mathbf{a})(\boldsymbol{S}\cdot\mathbf{b})]~\text{dV} = \int_{\Omega} \mathbf{a}\cdot[\{\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)\}\cdot\mathbf{b}~\text{dV}~. $$ Since $$\mathbf{a}$$ and $$\mathbf{b}$$ are constant, we have

\int_{\Omega} \boldsymbol{\nabla} \bullet [(\mathbf{v}\cdot\mathbf{a})(\boldsymbol{S}\cdot\mathbf{b})]~\text{dV} = \mathbf{a}\cdot\left[\left\{\int_{\Omega} [\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)]~\text{dV}\right\}\cdot\mathbf{b}\right]~. $$ From the divergence theorem,

\int_{\Omega} \boldsymbol{\nabla} \bullet \mathbf{u}~\text{dV} = \int_{\partial{\Omega}} \mathbf{u}\cdot\mathbf{n}~\text{dA} $$ we get

\int_{\Omega} \boldsymbol{\nabla} \bullet [(\mathbf{v}\cdot\mathbf{a})(\boldsymbol{S}\cdot\mathbf{b})]~\text{dV} = \int_{\partial{\Omega}} [(\mathbf{v}\cdot\mathbf{a})(\boldsymbol{S}\cdot\mathbf{b})]\cdot\mathbf{n}~\text{dA} ~. $$ Using the relation

[(\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}] $$ we get

\int_{\Omega} \boldsymbol{\nabla} \bullet [(\mathbf{v}\cdot\mathbf{a})(\boldsymbol{S}\cdot\mathbf{b})]~\text{dV} = \int_{\partial{\Omega}} \mathbf{a}\cdot[\{\mathbf{v}\otimes(\boldsymbol{S}^T\bullet\mathbf{n})\}\cdot\mathbf{b}]~\text{dA} ~. $$ Since $$\mathbf{a}$$ and $$\mathbf{b}$$ are constant, we have

\int_{\Omega} \boldsymbol{\nabla} \bullet [(\mathbf{v}\cdot\mathbf{a})(\boldsymbol{S}\cdot\mathbf{b})]~\text{dV} = \mathbf{a}\cdot\left[\left\{\int_{\partial{\Omega}} \mathbf{v}\otimes(\boldsymbol{S}^T\bullet\mathbf{n})~\text{dA}\right\}\cdot\mathbf{b}\right] ~. $$ Therefore,

\mathbf{a}\cdot\left[\left\{\int_{\partial{\Omega}} \mathbf{v}\otimes(\boldsymbol{S}^T\bullet\mathbf{n})~\text{dA}\right\}\cdot\mathbf{b}\right]= \mathbf{a}\cdot\left[\left\{\int_{\Omega} [\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)]~\text{dV}\right\}\cdot\mathbf{b}\right]~. $$ Since $$\mathbf{a}$$ and $$\mathbf{b}$$ are arbitrary, we have

{ \int_{\partial{\Omega}} \mathbf{v}\otimes(\boldsymbol{S}^T\bullet\mathbf{n})~\text{dA} = \int_{\Omega} [\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)]~\text{dV} } \qquad\qquad\qquad\square $$