Let $\textbf{v}\in C^2$ be a velocity field and $D$ the deformation tensor defined as $$d_{ij} = \dfrac{1}{2}\left(\dfrac{\partial v_i}{\partial x_j}+\dfrac{\partial v_j}{\partial x_i}\right).$$

Then $2\nabla\cdot D = \nabla^2\textbf{v}+\nabla(\nabla\cdot\textbf{v})$. The proof goes as follows:

Let $1\leq i \leq 3$ and $d_i = (d_{i1}\ d_{i2} \ d_{i3})$. Then we calculate:

\begin{align} 2\nabla \cdot d_i &= \sum_{j=1}^3\partial_j(\partial_jv_i+\partial_i v_j) \\ &= \sum_{j=1}^3(\partial_j^2v_i + \partial_j\partial_iv_i) \\ &=\nabla^2v_i+ \partial_i\left(\sum_{j=1}^3\partial_jv_i\right)\\ &=\nabla^2v_i + \partial_i(\nabla\cdot v_i). \end{align}

Hence $2\nabla\cdot D = \nabla^2\textbf{v}+\nabla(\nabla\cdot\textbf{v})$.

Can someone verify if this is correct?

Question: how physically realistic is the condition $\textbf{v}\in C^2$?

  • $\begingroup$ Remark: this equality makes the derivation of the (conserved, Newtonian) Navier-Stokes equation rather straight-forward, using $T = -\bar{p}\textbf{1}_3 + 2\mu S$ and $S = D - 1/3(\nabla\cdot\textbf{v})\textbf{1}_3$, with $\bar{p}$ the mechanical pressure. $\endgroup$ Commented Jan 20, 2020 at 13:14
  • $\begingroup$ What line in the calculation don't you understand? Also, you would hope $\boldsymbol{v} \in C^{2}$ at least, else after taking Laplacians the equations won't make much sense. $\endgroup$ Commented Jan 20, 2020 at 13:24
  • $\begingroup$ @mattos I understand every line in the calculation, it is $\textit{my}$ calculation, not a book's. I just want someone to verify whether it's correct. And $\textbf{v}\in C^2$ is not necessary for the Laplacian - I would say. It is sufficient though. However, it is a necessary condition for the switching of the partial derivatives from the second to the third equality. $\endgroup$ Commented Jan 20, 2020 at 13:30
  • 1
    $\begingroup$ You might have wanted to explain that in your post then. And as far as I can tell, your computation is correct. $\endgroup$ Commented Jan 20, 2020 at 13:32

1 Answer 1


It's not quite correct, there's a small index error.

\begin{align} 2\nabla \cdot d_i &= \sum_{j=1}^3\partial_j(\partial_jv_i+\partial_i v_j) \\ &= \sum_{j=1}^3(\partial_j^2v_i + \partial_j\partial_iv_i) \quad\text{should have $\partial_j\partial_iv_j$} \\ &=\nabla^2v_i+ \partial_i\left(\sum_{j=1}^3\partial_jv_i\right) \quad\text{should be $\partial_jv_j$ which sums to $\nabla\cdot\mathbf v$}\\ &=\nabla^2v_i + \partial_i(\nabla\cdot v_i). \quad\text{should be $\partial_i(\nabla\cdot\mathbf v)$} \end{align}

This is a good example of a situation where Einstein notation can be cleaner. I use $f_{a,b}$ to represent the derivative of $f_a$ with respect to the basis variable $x_b$, and repeated indices are summed.

$$\begin{align}2\nabla\cdot d_i&=\left(v_{i,j}+v_{j,i}\right)_{,j} \\ &=v_{i,jj}+v_{j,ij} \\ &=\partial_j\partial_jv_i+(v_{j,j})_{,i} \\ &=\nabla^2v_i+(\nabla\cdot\mathbf v)_{,i} \\ &=(\nabla^2\mathbf v+\nabla(\nabla\cdot\mathbf v))_i\end{align}$$

  • $\begingroup$ Ah yes, thank you. $\endgroup$ Commented Jan 23, 2020 at 13:46

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