How can I simplify $\nabla (X\cdot \nabla u)$?

Question

What is $\nabla (X\cdot \nabla u)$ where $X:U\subseteq\mathbb{R}^2\to\mathbb{R}^2$ is a vector field and $u:U\subseteq\mathbb{R}^2\to\mathbb{R}$ a scalar field?

Answer 1

To figure out what to do in two dimensions I will first go through all steps in

Three Dimensions

For any two vector fields $X,Y$ in three dimensions $$\tag{1} X\times(\nabla\times Y)=\varepsilon_{ghi}\,X_h\,\varepsilon_{ijk}\,\partial_j\,Y_k $$ where $\varepsilon_{abc}$ is the Levi-Civita symbol and all indices appearing twice are summed over $1,2,3\,.$ These are here all indices, except $g\,.$

Using the property $$\tag{2} \varepsilon_{ghi}\,\varepsilon_{ijk}=\varepsilon_{igh}\,\varepsilon_{ijk}=\delta_{gj}\,\delta_{hk}-\delta_{gk}\,\delta_{hj}\, $$ we get \begin{align} \varepsilon_{ghi}\,\varepsilon_{ijk}\,\partial_j\,Y_k&= \delta_{gj}\,\delta_{hk}\,\partial_j\,Y_k- \delta_{gk}\,\delta_{hj}\,\partial_j\,Y_k\\ &=\partial_g\,Y_h-\partial_h\,Y_g\,.\tag{3} \end{align} Therefore the right hand side of (1) is $$\tag{4} X_h\,\partial_g\,Y_h-X_h\,\partial_h\,Y_g\,. $$ Clearly, we can exchange $X$ and $Y$ to get $$\tag{5} Y\times(\nabla\times X)=Y_h\,\partial_g\,X_h-Y_h\,\partial_h\,X_g\,. $$ Therefore, \begin{align} X\times(\nabla\times Y)+Y\times(\nabla\times X)&= X_h\,\partial_g\,Y_h+Y_h\,\partial_g\,X_h-X_h\,\partial_h\,Y_g-Y_h\,\partial_h\,X_g\\ &=\partial_g(X_hY_h)-X_h\,\partial_h\,Y_g-Y_h\,\partial_h\,X_g\\ &=\nabla(X\cdot Y)-(X\cdot\nabla)Y-(Y\cdot\nabla)Y\,.\tag{6} \end{align} In short $$\tag{7} \boxed{\phantom{\Bigg|}\quad X\times(\nabla\times Y)+Y\times(\nabla\times X)=\nabla(X\cdot Y)-(X\cdot\nabla)Y-(Y\cdot\nabla)X\,. \quad} $$ With $Y=\nabla u$ where $u$ is a scalar function on $\mathbb R^3$ and using the well-known $$\tag{8} (\nabla\times (\nabla u))=\operatorname{rot grad}u=0 $$ the equation (7) gives $$\tag{9} \boxed{\phantom{\Bigg|}\quad (\nabla u)\times(\nabla\times X)=\nabla(X\cdot \nabla u)-(X\cdot\nabla)\nabla u-(\nabla u\cdot\nabla)X\, \quad} $$ from which you can easily back out $\nabla(X\cdot \nabla u)\,.$

Two Dimensions

The two-dimensional analogue of (1) is (4) where $g$ and $h$ only run through $\{1,2\}\,.$ With this in mind the middle part of (6) still holds: \begin{align} &X_h\,\partial_g\,Y_h+Y_h\,\partial_g\,X_h-X_h\,\partial_h\,Y_g-Y_h\,\partial_h\,X_g\\ &=\partial_g(X_hY_h)-X_h\,\partial_h\,Y_g-Y_h\,\partial_h\,X_g\,.\tag{10} \end{align} Again with $Y=\nabla u$ ($u$ now on $\mathbb R^2$) we get \begin{align}\require{cancel} &\cancel{X_h\,\partial_g\,\partial_h\,u}+(\partial_h\,u)\,(\partial_g\,X_h)-\cancel{X_h\,\partial_h\,\partial_g\,u}-(\partial_h\,u)\,(\partial_h\,X_g)\\ &=\partial_g(X_h\,\partial_h\,u)-X_h\,\partial_h\,\partial_g\,u-(\partial_h\,u)\,(\partial_h\,X_g)\tag{11} \end{align} which can be written in mixed notation as $$\boxed{\phantom{\Bigg|}\quad (\nabla u)^\top(\partial_g\,X_h-\partial_h\,X_g)=\nabla(X\cdot\nabla u) -(X\cdot \nabla)\,\nabla\,u-(\nabla u\cdot\nabla)\,X\,.\quad}\tag{12} $$ You can again back out $\nabla(X\cdot\nabla u)\,.$ You should compare (12) and (9).

Note that $(\partial_g\,X_h-\partial_h\,X_g)$ is an anti-symmetric $2\times 2$-matrix which we multiply from the left with the row vector $(\nabla u)^\top\,.$

Remarks

• There is nothing special about two dimensions. Formulas (10) to (12) obviously hold in every dimension. So do the following formulas:

• From a direct application of the product rule we obtain $$ \nabla(X\cdot\nabla u)=\partial_g\,(X_h\,\partial_h\,u)=X_h\,\partial_g\,\partial_h\,u+(\partial_g\,X_h)(\partial_h\,u)\,.\tag{13} $$ Clearly, $J_{gh}=\partial_g\,\partial_h\,u$ is the Jacobi matrix of $u$ so that the first term on the right hand side could be written as $X^\top J\,.$

• In (11) we could have cancelled another two terms: \begin{align}\require{cancel} &(\partial_h\,u)\,(\partial_g\,X_h)-\bcancel{(\partial_h\,u)\,(\partial_h\,X_g)}\\ &=\partial_g(X_h\,\partial_h\,u)-X_h\,\partial_h\,\partial_g\,u-\bcancel{(\partial_h\,u)\,(\partial_h\,X_g)}\tag{14} \end{align} which is nothing else than (13) as it must.

Answer 2

We know from the gradient of a dot product of two vector fields that:

$$ \nabla (\vec{x}\cdot \vec{v}) =\vec{x} \times (\nabla \times \vec{v}) + \vec{v} \times (\nabla \times \vec{x}) + (\vec{x} \cdot \nabla) \vec{v} + (\vec{v} \cdot \nabla) \vec{x}$$

In your case, we have $\vec{v} = \nabla u$, replacing this in the expression above, we have:

$$ \nabla (\vec{x}\cdot \vec{v}) =\vec{x} \times (\nabla \times \nabla u) + \nabla u \times (\nabla \times \vec{x}) + (\nabla \cdot \vec{x}) \nabla u + (\nabla u \cdot \nabla) \vec{x}$$

We then utilize the fact that the curl of a gradient is $0$. Hence, the first term vanishes. We are left with:

$$ \nabla (\vec{x}\cdot \vec{v}) =\nabla u \times (\nabla \times \vec{x}) + (\nabla \cdot \vec{x}) \nabla u + (\nabla u \cdot \nabla) \vec{x}$$

I can't really see any more way to simplify this that would be beneficial, but maybe I'm forgetting something.