Show $\left(\vec{A}\cdot\nabla\right)\vec{A} = \nabla\left( \frac{A^2}{2} \right) - \vec{A}\times\left(\nabla\times\vec{A}\right)$

Question

\begin{equation} \begin{aligned} \left(\vec{A}\cdot\nabla\right)\vec{A} &= \nabla\left( \frac{A^2}{2} \right) - \vec{A}\times\left(\nabla\times\vec{A}\right)\\ \left(\vec{A}\cdot\nabla\right)\vec{A} &= A_x\frac{\partial A_x}{\partial x} + A_y\frac{\partial A_y}{\partial y} + A_z\frac{\partial A_z}{\partial z}\\ \left(\vec{A}\cdot\nabla\right)\vec{A} &= \dots \end{aligned} \end{equation} I'm stuck. Any help would be greatly appreciated.

Answer 1

You obtain the above relation by putting $\vec A=\vec B $ in the following relation: $$\nabla(\vec A\cdot \vec B)=(\vec A\cdot \nabla)\vec B+(\vec B\cdot \nabla)\vec A+\vec B \times (\nabla \times \vec A)+\vec A \times (\nabla \times \vec B)$$

and then prove this, which is quite easy using Levi-Cevita notation.

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For an alternative detailed analysis, consider this:

If we let the components of the vector field $\vec A$ be $(f,g,h)$ and use subscripts for differentiation then, looking at the $x$-component \begin{align} ((\nabla\times \vec A) \times \vec A)_x &= ((\nabla\times \vec A)_yh-(\nabla\times \vec A)_zg \\ & =(f_z-h_x)h-(g_x-f_y)g \\ &= (f_yg+f_zh) - (g_xg+h_xh) \end{align} if we insert an extra term $f_xf$ in each parenthesis the value of the whole is not changed and we have:

\begin{align} (f_xf+f_yg+f_zh) - (f_xf+g_xg+h_xh) &= \vec A\cdot \nabla f -\frac12 \frac{\partial}{\partial x}(f^2+g^2+h^2) \\ &= \left(\vec A\cdot \nabla \vec A -\frac12\nabla |\vec A|^2 \right)_x \end{align}