$\nabla f \cdot \nabla |\nabla f|^2$

Question

I am trying to find (expand?) the following $\nabla f \cdot \nabla |\nabla f|^2$. Assuming rectangular coordinates I tried to expand it as such \begin{align} \nabla f \cdot \nabla |\nabla f|^2 & = f_i \partial_i(f_j^2) \\ & = 2 f_i f_{ij} f_j \\ & = 2 \nabla f^T \nabla(\nabla f) \nabla f, \end{align} where $i,j$ are derivatives. The last line I assumed from noting the similarity with $x^T A y = x_i A_{ij} y_j$ from matrix notation. Is this correct? is their a better or more formal way to write this?

Answer 1

Not a full answer, but maybe this will help somewhat.

Maybe consider the third identity under Differentiation.

$$\nabla(\vec{A}\cdot\vec{B} )= (\vec{A}\cdot\nabla)\vec{B} + (\vec{B}\cdot\nabla)\vec{A} + \vec{A}\times(\nabla\times\vec{B}) + \vec{B}\times(\nabla\times\vec{A})$$

where you have $\vec{A}=\vec{B}=\nabla f$. Then,

$$\nabla\left|\nabla f\right|^2=2(\nabla f \cdot\nabla)\nabla f = 2\sum_{i}\frac{\partial f}{\partial x^i}\frac{\partial^2 f}{\partial x^i\partial x^j}$$

$$\nabla f\cdot\nabla\left|\nabla f\right|^2=2\sum_{ij}\frac{\partial f}{\partial x^j}\frac{\partial f}{\partial x^i}\frac{\partial^2 f}{\partial x^i\partial x^j}=2\sum_{ij}\frac{\partial f}{\partial x^i}\frac{\partial^2 f}{\partial x^i\partial x^j}\frac{\partial f}{\partial x^j} = 2\sum_{ij}\frac{\partial f}{\partial x^i}H_{ij}\frac{\partial f}{\partial x^j} $$

where $H_{ij}$ is the Hessian matrix.