I start with the following product that Wikipedia (https://en.wikipedia.org/wiki/Dyadics#Dyadic.2C_outer.2C_and_tensor_products) calls a dyadic product, or outer product, or tensor product. In 3D Cartesian coordinates, it looks like this:

\begin{equation} \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) \end{equation}

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

\begin{equation} \nabla \cdot \nabla \mathbf{v} \end{equation}

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

Is this correct?


1 Answer 1


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 $\partial_i\partial_jv_i$.

  • $\begingroup$ In the context in which I am using it, the divergence means the first of your two summations. I assume the first summation is for del^2 v and the second summation is del (del dot v)? Perhaps I should clarify in my question? $\endgroup$ Commented Sep 25, 2017 at 23:19
  • $\begingroup$ Yes, that is the case. $\nabla^2 \mathbf{v} = \partial_i\partial_iv_j$ and $\boldsymbol\nabla\boldsymbol\nabla \cdot\mathbf{v} = \partial_i\partial_jv_i$. $\endgroup$ Commented Sep 25, 2017 at 23:23

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