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In my finite element method book, there is a notation which is confusing me. Given $v:R^2\rightarrow R^2$, I'm supposed to evaluate

$\sigma\cdot \nabla v^T$
where $\sigma$ is a smooth tensor valued function. What is confusing me is the notation $\nabla v^T$. How do I interpret the gradient of a vector? Is it a matrix or a vector? Also, should I interpret the equation as $\nabla (v^T)$ or $\nabla^T v$?

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The gradient of a vector field is

$$ (\nabla v)_{ij} = \frac{\partial v_i}{\partial x_j} $$

and so

$$ \sigma\cdot\nabla v^T=\sigma\cdot(\nabla v)^T=\sum_{i,j=1}^3\sigma_{ij}\frac{\partial v_i}{\partial x_j} $$

(given that $A\cdot B=\sum_{ij}A_{ij}B_{ji}$ and so $A\cdot B^T=\sum_{ij}A_{ij}B_{ij}$)

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  • $\begingroup$ So the gradient of a vector is a matrix then, right? $\endgroup$ – Paul Sep 15 '12 at 16:50
  • $\begingroup$ Yes, it is a matrix. The gradient raises by one the order of the object to which it is applied: a gradient of a scalar field is a vector field; the gradient of a vector field is a double tensor field, and so on. $\endgroup$ – enzotib Sep 15 '12 at 16:51
  • $\begingroup$ According to the book, the final result $\sigma\cdot\nabla v^T$ is supposed to be a scalar quantity. Your final result seems to be a vector, if I'm not mistaken... $\endgroup$ – Paul Sep 15 '12 at 16:53
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    $\begingroup$ No, you sum on both indexes, so the resulting dot product is a scalar. $\endgroup$ – enzotib Sep 15 '12 at 16:55
  • $\begingroup$ Awesome! Thank you, @enzotib! :) $\endgroup$ – Paul Sep 15 '12 at 17:01

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