Navier Stokes: $(u⋅∇)u$ vs $u⋅∇u$

Question

I can find this term stated both ways in different literature.
Are they equivalent?
It's weird because the dot is a dot product in (u⋅∇), but ∇u being a gradient of a vector field, would (presumably) produce a (jacobian?) matrix which would turn that dot-product dot into a regular vector-matrix multiply where I have to swap the matrix and the vector (because as it's written, the vector will be on the left side).

Jacques Langner · Accepted Answer

Here $vec{u}$ is a vector field $vec{u} = (u_{x},u_{y}, u_{z})^T$ where $u_{x}, u_{y}$ and $u_{z}$ are scalar fields. So writing things out for clarity we have:
begin{align}
(vec{u}cdotnabla)vec{u} &= (u_xfrac{partial}{partial x} +u_yfrac{partial}{partial y}+u_zfrac{partial}{partial z})vec{u} 
&= (vec{u}cdotnabla u_{x},vec{u}cdotnabla u_{y},vec{u}cdotnabla u_{z})^T
end{align}
Furthermore:
begin{align}
vec{u}cdotnablavec{u} &= vec{u}cdot(nabla u_x,nabla u_y,nabla u_z)^T 
&= (vec{u}cdotnabla u_{x},vec{u}cdotnabla u_{y},vec{u}cdotnabla u_{z})^T
end{align}
The second expression is indeed equivalent to the first. The first way of writing it is definitely more explicit and clear though. See this answer.

AnotherShruggingPhysicist · Answer

The two expressions are equivalent.  For the second form, you left-multiply by the vector at the end to return to a vector result.
The first form is simpler, because it avoids the confusion you highlighted, and makes the physical meaning of the term (looking for the change along the direction of $u$, which is what $ucdotnabla$ projects out).

Emilio Pisanty · Answer

In general, I find $nablavec u$ to be slightly ambiguous notation, which should be either avoided or explained when it is first introduced (which can be done).
Usually, however, it refers to a matrix whose entries are
$$
(nablavec u)_{ij} = frac{partial u_j}{partial x_i},
$$
and the notation $vec u cdot nabla vec u$ is usually the left-hand vector-matrix product, giving you the vector
$$
left( vec u cdot nabla vec u right)_j = sum_i u_i frac{partial u_j}{partial x_i}.
$$
Of course, this coincides with
$$
left( (vec u cdot nabla) vec u right)_j = left[sum_i u_i frac{partial}{partial x_i}right] u_j,
$$
i.e., the two notations are equivalent.
In the terms that you describe,

It's weird because the dot is a dot product in (u⋅∇), but ∇u being a gradient of a vector field, would (presumably) produce a (jacobian?) matrix which would turn that dot-product dot into a regular vector-matrix multiply where I have to swap the matrix and the vector (because as it's written, the vector will be on the left side).

this isn't particularly weird at all: dot products and vector-matrix products are both special cases of tensor contractions, i.e., taking two objects with indices and summing over one paired index. The only differences between the two are superficial and cosmetic.

Navier Stokes: $(u⋅∇)u$ vs $u⋅∇u$

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