What does $\nabla \nabla$ mean? (nabla nabla, del del)

I see that used in many books but none of them defines what this actually means, since $$\nabla$$ is typically seen as a vector, but plain vector-vector multiplication does not exist.

For instance, in the identity:

$$\nabla \times \nabla \times A = \nabla \nabla \cdot A - \nabla ^ 2 A$$

$$\bar{G}(\mathbf{r},\mathbf{r}') = \left[\bar{I} + \frac{\nabla \nabla}{k_0^2}\right] \frac{e^{i k_0 |\mathbf{r} - \mathbf{r}'|}}{4 \pi |\mathbf{r} - \mathbf{r}'|}$$

• It has to be understood as $\nabla (\nabla A)$, i.e. you take the divergence of $A$, which yields $\nabla A$ and then you take the gradient of this scalar, i.e. $\nabla (\nabla A)$. Commented Feb 23, 2023 at 14:13
• Thank you, so in the second expression both the parenthesis and the dot are omitted but implied? Commented Feb 23, 2023 at 14:18
• Also, do you mean divergence $\nabla \cdot A$ ? Or do you mean gradient $\nabla A$ ? Commented Feb 23, 2023 at 14:20
• The dot should be included: $\nabla (\nabla \cdot A)$. Commented Feb 23, 2023 at 14:28
• $\nabla f$ for a scalar function $f$, $\nabla \cdot A$ for a vector valued function $A$ and $\nabla \times A$ all have a meaning. In each case it might be better thought of as 3 different operators $\nabla, \nabla \cdot$ and $\nabla \times$
– Paul
Commented Feb 23, 2023 at 14:33

If we take the double gradients, i.e., the gradient of the gradient, we obtain the so-called Hessian matrix, here donated by $$\nabla\!\nabla$$ in stead of $$\nabla^2$$. However, not only a few mathematicians like to use $$\nabla^2$$, which becomes to bewilder many engineers due to the fact that $$\nabla^2$$ stands for the Laplacian were already deeply ingrained in the education. Especially, when some writers do not differentiate the scalar notations with vector/matrix ones. So to avoid any confusion, the Laplacian is donated by $$\nabla^2=\nabla\cdot\nabla$$, the divergent of the gradient and is a scalar, while the Hessian Matrix is the gradient of the gradient, therefore explicitly expressed as $$\nabla\!\nabla$$. From the word-width point of view, $$\nabla^2$$ can save spaces for $$\nabla\cdot\nabla$$, but not for $$\nabla\!\nabla$$. Really, really wish that a dedicated campaign for unified notation can be populated to the students! Then life would be much easier.
For instance, if we take the gradient of \begin{align} \nabla^\mathrm{T}f(\mathbf x)=(\mathbf{A}^T+\mathbf{A})\mathbf{x}, \end{align} it then arrives at \begin{align} \nabla\!\nabla f(\mathbf x)&=\nabla\!\nabla\left(\mathbf x^T\mathbf{A}\mathbf x\right)\\ &=\nabla\left[(\mathbf{A}^T+\mathbf{A})\mathbf{x}\right]\\ &=\left(\mathbf{A}^T+\mathbf{A}\right)^T\\ &=\mathbf{A}+\mathbf{A}^T. \end{align}
• I don't think $\nabla^2$ to denote the Hessian matrix is common, even among mathematicians. When it is needed to discuss higher-order derivatives of multi-variable functions, most mathematicians use generalized $\mathrm D$ notation, as seen in Lawrence C. Evans's book "Partial Differential Equations". In this book, $\mathrm D^2$ is used to denote the Hessian matrix. Commented Mar 11 at 16:50