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?

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

2 Answers 2


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!


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


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 $\boldsymbol{u}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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