Computation of $(\textbf{u}\cdot \nabla)\textbf{u}$ in the Navier stokes equation: inner or outer product?

Question

The Navier Stokes equation for an incompressible flow of a homogenenous fluid reads \begin{align*} \nabla\cdot \textbf{u} &= 0\\ \rho \frac{\partial }{\partial t}\textbf{u} + \rho(\textbf{u}\cdot \nabla)\textbf{u} &= -\nabla p + \mu \nabla ^2 \textbf{u} \end{align*} Regarding, $$(\textbf{u}\cdot \nabla)\textbf{u} $$ the two components of this term are: $$ \textbf{u} = \begin{bmatrix} u \\ v \end{bmatrix} \quad \text{and} \quad \nabla = \begin{bmatrix} \frac{\partial }{\partial x} \\ \frac{\partial }{\partial y} \end{bmatrix} $$ and the sign $\cdot$ being the dot product. However in the example that I was given, the computation has been done as follows: \begin{align*} (\textbf{u}\cdot \nabla)\textbf{u} &= \Big(\begin{bmatrix} u \\ v \end{bmatrix} \begin{bmatrix} \frac{\partial }{\partial x} & \frac{\partial }{\partial y} \end{bmatrix}\Big)\begin{bmatrix} u \\ v \end{bmatrix} \end{align*} where $\begin{bmatrix} u \\ v \end{bmatrix} \begin{bmatrix} \frac{\partial }{\partial x} & \frac{\partial }{\partial y} \end{bmatrix}$ is an outer product (Notation: $\otimes$ and not a dot product $\cdot$) and hence a matrix. So for me, the notation $\textbf{u}\cdot \nabla$ is misleading. What am I getting wrong?

Example Compute the acceleration of a fluid at $(x,y) = (1,2)$ at time $t = 2$. with velocity $(u,v) = (1, x^2t)$

Fluid acceleration $$ \frac{D\textbf{u}}{Dt} = \frac{\partial \textbf{u}}{\partial t} +(\textbf{u} \cdot \nabla) \textbf{u} $$

solution (given to me): \begin{align*} \frac{D\textbf{u}}{Dt} &=\frac{\partial \textbf{u}}{\partial t} +(\textbf{u} \cdot \nabla) \textbf{u}\\ &=\frac{\partial }{\partial t}\begin{bmatrix} 1\\x^2t \end{bmatrix} + \Big( \begin{bmatrix} 1\\x^2t \end{bmatrix} \cdot \begin{bmatrix} \frac{\partial}{\partial x}&\frac{\partial}{\partial y} \end{bmatrix}\Big) \begin{bmatrix} 1\\x^2t \end{bmatrix}\\ &=\begin{bmatrix} 0\\x^2 \end{bmatrix} + \begin{bmatrix} \frac{\partial 1}{\partial x} & \frac{\partial 1}{\partial y}\\ \frac{\partial x^2t}{\partial x} & \frac{\partial x^2t}{\partial y} \end{bmatrix}\cdot \begin{bmatrix} 1\\x^2t \end{bmatrix}\\ &=\begin{bmatrix} 0\\x^2 \end{bmatrix} + \begin{bmatrix} 0 & 0\\ 2xt & 0 \end{bmatrix}\cdot \begin{bmatrix} 1\\x^2t \end{bmatrix}\\ &=\begin{bmatrix} 0\\x^2 \end{bmatrix} + \begin{bmatrix} 0\\2xt \end{bmatrix}\\ &= \begin{bmatrix} 0\\x^2 + 2xt \end{bmatrix}\\ \end{align*}

Now if I try to compute firs the dot product inside the parenthesis (instead of the outer product, as it was proposed in the answers), I get \begin{align*} &=\frac{\partial }{\partial t}\begin{bmatrix} 1\\x^2t \end{bmatrix} + \Big( \begin{bmatrix} 1&x^2t \end{bmatrix} \cdot \begin{bmatrix} \frac{\partial}{\partial x}\\\frac{\partial}{\partial y} \end{bmatrix}\Big) \begin{bmatrix} 1\\x^2t \end{bmatrix}\\ &=\frac{\partial }{\partial t}\begin{bmatrix} 1\\x^2t \end{bmatrix} + \Big( \frac{\partial 1}{\partial x} + \frac{\partial x^2t}{\partial y}\Big) \begin{bmatrix} 1\\x^2t \end{bmatrix}\\ &=\frac{\partial }{\partial t}\begin{bmatrix} 1\\x^2t \end{bmatrix} + \Big( 0 + 0 \Big) \begin{bmatrix} 1\\x^2t \end{bmatrix}\\ &=\begin{bmatrix} 0 \\x^2 \end{bmatrix} + \Big( 0 \Big) \begin{bmatrix} 1\\x^2t \end{bmatrix}\\ &=\begin{bmatrix} 0\\x^2 \end{bmatrix} \end{align*}

Answer 1

Let $$\eqalign{ x &= \pmatrix{x\\y},\quad u = \pmatrix{u\\v}=\pmatrix{1\\tx^2} \\ }$$ Then the matrix-valued gradient $\,G=(\nabla u)^T$ has components $$\eqalign{ G_{ij} &= \frac{\partial u_i}{\partial x_j} \;=\; \pmatrix{ \frac{\partial u}{\partial x}&\frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x}&\frac{\partial v}{\partial y} } = \pmatrix{0&0\\2xt&0} \\ }$$ and can be used to write the material derivative as $$\eqalign{ \frac{Du}{Dt} &= \frac{\partial u}{\partial t} + Gu \\ &= \pmatrix{0\\x^2} + \pmatrix{0&0\\2xt&0}\pmatrix{1\\tx^2} \\ &= \pmatrix{0\\x^2} + \pmatrix{0\\2xt} \\ &= \pmatrix{0\\x^2+2xt} \\ }$$ which can be evaluated at $\;(t,x,y)=(2,1,2)$ $$\eqalign{ \frac{Du}{Dt} &= \pmatrix{0\\1^2+2(1)(2)} = \pmatrix{0\\5} \\ }$$

Answer 2

$\mathbf{u} \cdot \nabla$ is the dot product of a vector $\mathbf{u}$ and a "vector" $\nabla$. Thus it should be interpreted as a "scalar". Therefore $(\mathbf{u} \cdot \nabla) \mathbf{u}$ is a scalar multiplication of the "scalar" $\mathbf{u} \cdot \nabla$ and the vector $\mathbf{u}$. In other words,

\begin{align} ((\mathbf{u} \cdot \nabla) \mathbf{u})_i &= (\mathbf{u} \cdot \nabla) \mathbf{u}_i \\ &= \left( \sum_j \mathbf{u}_j \nabla_j \right) \mathbf{u}_i \\ &= \sum_j \mathbf{u}_j \nabla_j \mathbf{u}_i \end{align}

Notice that $(\mathbf{u} \cdot \nabla) \mathbf{u} = \mathbf{u} \cdot (\nabla \mathbf{u})$, where $\nabla \mathbf{u}$ is an outer product between the "vector" $\nabla$ and the vector $\mathbf{u}$:

\begin{align} ((\mathbf{u} \cdot \nabla) \mathbf{u})_i &= (\mathbf{u} \cdot (\nabla \mathbf{u}))_i \\ &= \left(\sum_j \mathbf{u}_j (\nabla \mathbf{u})_j\right)_i \\ &= \left(\sum_j \mathbf{u}_j (\nabla_j \mathbf{u})\right)_i \\ &= \sum_j (\mathbf{u}_j (\nabla_j \mathbf{u}))_i \\ &= \sum_j \mathbf{u}_j \nabla_j \mathbf{u}_i \end{align}

Answer 3

Either the computation has been done incorrectly or you have misunderstood it. It should be $$ \begin{align*} (\textbf{u}\cdot \nabla)\textbf{u} &= \Big(\begin{bmatrix} u & v \end{bmatrix} \begin{bmatrix} \frac{\partial }{\partial x} \\ \frac{\partial }{\partial y} \end{bmatrix} \Big) \begin{bmatrix} u \\ v \end{bmatrix} \\ &= \Big( u \frac{\partial }{\partial x} + v \frac{\partial }{\partial y} \Big) \begin{bmatrix} u \\ v \end{bmatrix} \\ &= \begin{bmatrix} u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} \\ u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y} \end{bmatrix} \end{align*} $$