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Is there a way to apply integration by parts or the divergence theorem to the integral:

$$ \int_{\Omega} \langle F'(x)n(x),n(x)\rangle dx $$

where

  • F: $\mathbb{R}^m\rightarrow\mathbb{R}^m$
  • n: $\mathbb{R}^m\rightarrow\mathbb{R}^m$
  • $F^\prime(x) = \begin{bmatrix}\nabla F_1(x)^T\\\vdots\\\nabla F_m(x)^T\end{bmatrix}$
  • $\langle x,y\rangle = x^Ty$

To write it another way, I'd like to rewrite the integral

$$ \int_{\Omega} n(x)^T F'(x)n(x) dx $$

as a surface integral on $\partial \Omega$ where we shed the derivative off of $F$. Please assume $\Omega$ to be as smooth as necessary and other properties on $F$ to get us as close as possible. Likely, $m$ will be 3, but I'd like a result for a general $m$ if possible. Thanks!

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1 Answer 1

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This can be done using index notation (and also vector notation). We have \begin{multline} \left<F'(x)n(x),n(x)\right> = \mathbf{n}\cdot(\mathbf{n}\cdot\boldsymbol\nabla\mathbf{F})= n_in_j\partial_iF_j = \partial_i(n_in_jF_j) - \partial_i(n_in_j)F_j \\= \boldsymbol\nabla \cdot [\mathbf{n}(\mathbf{n}\cdot \mathbf{F})] - [\boldsymbol \nabla\cdot(\mathbf{n}\mathbf{n})]\cdot\mathbf{F} \end{multline} Divergence theorem then gives $$ \int_{\Omega}\mathbf{n}\cdot(\mathbf{n}\cdot\boldsymbol\nabla\mathbf{F})d^mV = \int_{\partial\Omega} (\mathbf{n}\cdot\mathbf{F})(\mathbf{n}\cdot d^{m-1}\mathbf{A}) - \int_\Omega[\boldsymbol\nabla\cdot(\mathbf{n}\mathbf{n})]\cdot\mathbf{F}d^mV $$ where $d^m V$ is the $m$-dimensional volume element and $d^{m-1}\mathbf{A}$ is the directed $m-1$-dimensional area element of the $m-1$-dimensional surface $\partial\Omega$.

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