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I have a problem in the derivation to get to the final expression on this page. I am given the following:

$ \sigma_{i j}=-P \delta_{i j}+2\mu \epsilon_{i j}+\left(\kappa-\frac{2}{3}\mu \right) \epsilon_{k k} \delta_{i j}$

where $\sigma$ represents the stress tensor and $\epsilon$ represents the strain rate tensor. The Kronecker's delta is defined as:

$\delta_{ij} = \begin{cases} 1, & \text{if } i=j,\\ 0, & \text{if } i\neq j. \end{cases}$

$\epsilon_{i j}$ is defined as:

$\epsilon_{i j}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right)$

$\epsilon_{kk}$ is defined as:

$\epsilon_{k k}=\bigtriangledown \cdot \vec{u}=\frac{\partial u_i}{x_i}+\frac{\partial u_j}{x_j}+\frac{\partial u_k}{x_k} $

$P$ is pressure, $u$ is velocity, while $\mu$ and $\kappa$ are just different constants.

The exercises is to solve the following expression $\bigtriangledown \cdot\vec{\sigma}$.

Then the answer is given to be:

$\bigtriangledown \cdot \vec\sigma=-\bigtriangledown P+\mu \bigtriangledown^2 \vec{u}+\left(\kappa+\frac{\mu}{3}\right)\bigtriangledown \left(\bigtriangledown \cdot \vec{u}\right)$

I don't understand how they managed to get to the last expression. Some questions I have is:

  • What happened to the Kroneker's delta?
  • What is the difference between $\bigtriangledown^2 \vec{u}$ and $\bigtriangledown(\bigtriangledown \cdot \vec{u})$?
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    indeed that expression is not the gradient of $\sigma_{ij}$ but the gradient of the diagonal components of the tensor, so $i=j$, this explains why the k-delta disappears.also laplacian of u and grad of the divergence of u are very different you can verify that explicitly – 2017-01-20
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    Thank you for your comment. I still does not manage to get to the final answer though. – 2017-01-20

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