Divergence of gradient of vector field

$$\nabla \mathbf{v} = \nabla \otimes \mathbf{v} = \nabla \mathbf{v}^T = \left( \begin{matrix} \frac{\partial v_1}{\partial x_1} & \frac{\partial v_2}{\partial x_1} & \frac{\partial v_3}{\partial x_1} \\ \frac{\partial v_1}{\partial x_2} & \frac{\partial v_2}{\partial x_2} & \frac{\partial v_3}{\partial x_2} \\ \frac{\partial v_1}{\partial x_3} & \frac{\partial v_2}{\partial x_3} & \frac{\partial v_3}{\partial x_3} \\ \end{matrix} \right)$$

I am wondering how to reduce the divergence of this object to a simpler form:

$$\nabla \cdot \nabla \mathbf{v}$$

I believe that this reduces to the following: $$\nabla \cdot \nabla \mathbf{v} = \nabla^2 \mathbf{v}$$

Is this correct?

Most people would interpret that as $\nabla^2\mathbf{v}$. However, it is somewhat ambiguous, and could be interpreted as $\boldsymbol\nabla(\boldsymbol\nabla\cdot\mathbf{v})$ depending on your convention for tensor divergence.
Or, in summation notation, there is ambiguity as to whether $\boldsymbol\nabla\cdot\boldsymbol \nabla \mathbf{v}$ means $\partial_i\partial_i v_j$ or $\partial_i\partial_jv_i$.
• Yes, that is the case. $\nabla^2 \mathbf{v} = \partial_i\partial_iv_j$ and $\boldsymbol\nabla\boldsymbol\nabla \cdot\mathbf{v} = \partial_i\partial_jv_i$. Commented Sep 25, 2017 at 23:23