6
$\begingroup$

I am studying the Navier-Stokes equations and I have the equation in the form: $$\rho \dfrac{\partial{\mathbf{u}}}{\partial{t}} + \rho (\mathbf{u}\cdot\nabla)\mathbf{u} - \mu\nabla^2\mathbf{u} + \nabla p = \rho f$$

Can someone explain me what does $ (\mathbf{u}\cdot\nabla)\mathbf{u}$ mathematically (generally) mean here?

$\endgroup$
2
  • $\begingroup$ I added the mark-up needed to properly display the MathJax. I apologize if the meaning changed. You got it almost correct! $\endgroup$
    – garyp
    Commented Dec 22, 2017 at 16:24
  • 1
    $\begingroup$ Related, if not dupe of, physics.stackexchange.com/q/160229/25301 $\endgroup$
    – Kyle Kanos
    Commented Dec 22, 2017 at 20:31

2 Answers 2

7
$\begingroup$

If you're just concerned with what $(\textbf{u}\cdot \nabla)\textbf{u}$ means mathematically then it isn't too complicated. $\textbf{u}$ and $\nabla$ are both vectors so, $$\textbf{u}\cdot \nabla=u_x \partial_x+u_y \partial_y+u_z \partial_z.$$ Now just apply this "directional" derivative operator to each component of the vector $\textbf{u}$ to get, $$(\textbf{u}\cdot \nabla)\textbf{u}=(u_x \partial_x+u_y \partial_y+u_z \partial_z)\textbf{u}=\left((u_x \partial_x+u_y \partial_y+u_z \partial_z)u_x,(u_x \partial_x+u_y \partial_y+u_z \partial_z)u_y,(u_x \partial_x+u_y \partial_y+u_z \partial_z)u_z\right).$$ This term is called the convection term in the context of the Navier-Stokes equation. For a quick discussion of the convection and diffusion terms in the Navier-Stokes equation see the following post: Convective and Diffusive terms in Navier Stokes Equations

Hope this helps!

$\endgroup$
4
$\begingroup$

It is customary to define the material derivative of a vector field $\boldsymbol{A}$ as

$$\frac{D\boldsymbol{A}}{Dt}\equiv\frac{\partial\boldsymbol{A}}{\partial t}+\left(\boldsymbol{u}\cdot\boldsymbol{\nabla}\right)\boldsymbol{A}$$

where $\boldsymbol{u}$ is the velocity field of the fluid, so the Navier-Stokes equations are written in the following manner

$$\rho\frac{D\boldsymbol{u}}{Dt}=\mu\nabla^{2}\boldsymbol{u}-\boldsymbol{\nabla}p+\rho\boldsymbol{f}$$

The best way to understand this type of derivative is to think of a particle tracing the stream of the fluid. Lets denote the position of this particle as $\boldsymbol{x}$, and the velocity field of the fluid as $\boldsymbol{u}$. Also, denote by $\boldsymbol{A}=\boldsymbol{A}\left(\boldsymbol{x}\left(t\right),t\right)$ some quantity related to the particle. How does this quantity change in time? You can easily see by differentiating that

$$\frac{{\rm d}\boldsymbol{A}}{{\rm d}t}=\frac{\partial\boldsymbol{A}}{\partial t}+\frac{\partial\boldsymbol{A}}{\partial x}\frac{{\rm d}x}{{\rm d}t}+\frac{\partial\boldsymbol{A}}{\partial y}\frac{{\rm d}y}{{\rm d}t}+\frac{\partial\boldsymbol{A}}{\partial z}\frac{{\rm d}z}{{\rm d}t} =\\{}\\=\frac{\partial\boldsymbol{A}}{\partial t}+\Bigg[\frac{{\rm d}x}{{\rm d}t}\frac{\partial}{\partial x}+\frac{{\rm d}y}{{\rm d}t}\frac{\partial}{\partial y}+\frac{{\rm d}z}{{\rm d}t}\frac{\partial}{\partial z}\Bigg]\boldsymbol{A} =\\{}\\=\frac{\partial\boldsymbol{A}}{\partial t}+\Bigg[u_{x}\frac{\partial}{\partial x}+u_{y}\frac{\partial}{\partial y}+u_{z}\frac{\partial}{\partial z}\Bigg]\boldsymbol{A} =\\{}\\=\frac{\partial\boldsymbol{A}}{\partial t}+\left(\boldsymbol{u}\cdot\boldsymbol{\nabla}\right)\boldsymbol{A}$$

according to the chain rule. What does this mean? You can divide the change in $\boldsymbol{A}$ into two effects

  • The change in the field $\boldsymbol{A}$ at a specific point, in time. This is described by the term $\frac{\partial\boldsymbol{A}}{\partial t}$.

  • The change in the field $\boldsymbol{A}$ due to the change of the evaluation point $\boldsymbol{x}$. This is due to the flow of the particle in question, and is represented by the term $\left(\boldsymbol{u}\cdot\boldsymbol{\nabla}\right)\boldsymbol{A}$. This is essentially a directional derivative in the direction of the particle's velocity $\boldsymbol{u}$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.