# In fluid-dynamics: $2\nabla\cdot D = \nabla^2\textbf{v}+\nabla(\nabla\cdot\textbf{v})$, where $D$ is the deformation tensor.

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 $$\textbf{v}\in C^2$$?

• 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. Jan 20 '20 at 13:14
• 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. Jan 20 '20 at 13:24
• @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. Jan 20 '20 at 13:30
• You might have wanted to explain that in your post then. And as far as I can tell, your computation is correct. Jan 20 '20 at 13:32

\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}