In my book the Navies-Stokes equations are stated as:

Let $\Omega$ be a domain in $\mathbb R^n$, $n\geq 2$ and let $T\in (0,\infty)$. The incompressible Navier-Stokes equations for velocity $u(x,t):\Omega \times \mathbb (0,T) \to \mathbb R^n$ and pressure $p(x,t):\Omega \times (0,T) \to \mathbb R$ are \begin{align} \partial_t u-\nu \Delta u &+(u\cdot\nabla) u + \nabla p = f \tag 1 \\ &\text{div }u = 0 \tag 2 \end{align} Here $f$ is a given body force and $\nu > 0$ is the viscosity constant. The equations are coupled with an initial condition $$ u(x,0)=u_0(x), \quad \text{div }u_0 =0 \tag 3 $$

I know this notation, so far so good!

However, at Clay Mathematics Institute they use a different notation. Are the following equations equivalent to the equations above? If so, can you explain the difference in notation? How can we go from $(4)-(5)$ to $(1)-(2)$?

These equations are to be solved for an unknown velocity vector $u(x,t)=(u_i(x,t))_{1\leq i \leq n}\in \mathbb R^n$ and pressure $p(x,t)\in \mathbb R$, defined for position $x\in \mathbb R^n$ and time $t \geq 0$. We restrict attention here to incompressible fluids filling all of $\mathbb R^n$. The Navier–Stokes equations are then given by \begin{align} \frac{\partial}{\partial t} u_i + \sum_{j=1}^{n} u_j \frac{\partial u_i}{\partial x_j}&=\nu \Delta u_i - \frac{\partial p}{\partial x_i} + f_i(x,t) \tag 4\\ \text{div }u &= \sum_{i=1}^{n}\frac{\partial u_i}{\partial x_i}=0 \tag 5 \end{align} with initial conditions $$ u(x,0)=u^{\circ}(x) \tag 6 $$ Here $u^{\circ}(x)$ is a given, $C^{\infty}$ divergence-free vector field on $\mathbb R^n$, $f_i(x,t)$ are the components of a given, externally applied force (e.g gavity), $\nu$ is a positive coefficient (the viscosity) and $\Delta = \sum_{i=1}^{n} \frac{\partial^2}{\partial x_i^2}$ is the Laplacian in the space variables.

Question 1:

I guess $(4)$ is written componentwise for each $u_i(x,t)$. However, the author use $i$ for both $u_i(x,t)$ and the summation in $(5)$. Isn't there a distinction between $i$ for $u_i(x,t)$ and $i$ in the summation? Say we use $k$ in $(5)$ instead $$ \sum_{k=1}^{n}\frac{\partial u_k}{\partial x_k}=0 \tag 7 $$

Question 2:

Is the following correct? Suppose $n=2$, $u(x,t)=(u_1(x,t),u_2(t))$.

So for $u_1(x,t)$ we have 2 equations: \begin{align} \frac{\partial}{\partial t} u_1(x,t) &+ u_1(x,t) \frac{\partial u_1 (x,t)}{\partial x_1} + u_2(x,t) \frac{\partial u_1(x,t)}{\partial x_2}\\ &=\nu \bigg (\frac{\partial^2 u_1(x,t)}{\partial x_1^2} + \frac{\partial^2 u_1(x,t)}{\partial x_2^2} \bigg ) - \frac{\partial p(x,t)}{\partial x_1} + f_1(x,t) \tag 8 \\ \quad \text{div }u &= \frac{\partial u_1(x,t)}{\partial x_1} + \frac{\partial u_2(x,t)}{\partial x_2}=0 \tag 9 \end{align} And for $u_2(x,t)$ we have these 2 equations: \begin{align} \frac{\partial}{\partial t} u_2(x,t) &+ u_1(x,t) \frac{\partial u_2 (x,t)}{\partial x_1} + u_2(x,t) \frac{\partial u_2 (x,t)}{\partial x_2}\\ &=\nu \bigg (\frac{\partial^2 u_2 (x,t)}{\partial x_1^2} + \frac{\partial^2 u_2 (x,t)}{\partial x_2^2} \bigg ) - \frac{\partial p(x,t)}{\partial x_2} + f_2(x,t) \tag{10} \\ \quad \text{div }u &= \frac{\partial u_1(x,t)}{\partial x_1} + \frac{\partial u_2(x,t)}{\partial x_2}=0 \tag{11} \end{align}

Question 3:

I haven't seen $u(x,t)=(u_i(x,t))_{1\leq i \leq n}\in \mathbb R^n$ before. Is it equivalent to the notation $u(x,t)=(u_1(x,t), \dots, u_n(x,t))$? Or does $u(x,t)=(u_i(x,t))_{1\leq i \leq n}\in \mathbb R^n$ mean: "pick $i$ distinct functions", i.e. if $n=2$ we have the two functions \begin{align} u(x,t)&=(u_1(x,t)) \tag{12}\\ u(x,t)&=(u_2(x,t)) \tag{13} \end{align}

  • 1
    $\begingroup$ It actually looks exactly the same to me. Can you help me understand what terms you think are different? Maybe the nonlinear term $u\cdot \nabla u$? $\endgroup$ Jun 26, 2020 at 7:18
  • $\begingroup$ Said another way, I thought (1), (2) was defined by (4), (5). So I don't know how to begin answering your question? $\endgroup$ Jun 26, 2020 at 7:25
  • $\begingroup$ Hi @CalvinKhor! (1), (2) are from a book. In the link from Clay only (4), (5) are given ((1), (2) are not given at all). $\endgroup$
    – JDoeDoe
    Jun 26, 2020 at 7:35
  • $\begingroup$ Hi :) I'm not interested in where you found the formulas, that doesn't answer my question. Whats the difference, to use a simpler example, between saying that $\partial_t u = g$ and $\partial_t u_i = g_i$? And how does this not fully answer how to go from $(4),(5)$ to $(1),(2)$? Do you know what $\Delta u$ means? And what $u\cdot \nabla$ is? (I'd assume yes, since you said you know this notation. Yet knowing this notation, to me, means that you can translate $(4),(5)$ to $(1),(2)$, which is clearly not the case...?) $\endgroup$ Jun 26, 2020 at 7:39
  • $\begingroup$ I guess you are getting stuck somewhere when trying to go from (1) to (4), what is the problematic step? $\endgroup$ Jun 26, 2020 at 7:45

1 Answer 1


Any book worth even half the wood that was needed to print it would have explained that $(1)$ is shorthand for $(4)$. First you should know $(1)$ is not one equation, but $n$ equations. And the same is true for $(4)$, which writes the $n$ equations explcitly in terms of the cartesian coordinates i.e. one equation for each $i=1,2,\dots,n$. Third, you should know that for vector fields $u$, $\partial_t u$ has components $\partial_t u_i$, i.e. $(\partial_t u)_i = \partial_t u_i$. In words, the time derivative is applied componentwise. Next, the Laplacian, a scalar operator, is also applied componentwise: $(\Delta u)_i := \Delta u_i$. Finally, the most likely confusing term is the nonlinear term. The operator $u\cdot \nabla $ is a scalar operator, defined as $u\cdot \nabla := \sum_j u_j \partial_j$, and it is applied componentwise: $$ (u\cdot \nabla u)_i := ((u\cdot \nabla) u)_i := (u\cdot \nabla) u_i = \sum_j u_j \partial_j u_i.$$ Note that something completely different, $u\cdot (\nabla u)$ would describe a dot product of a vector and a matrix. If you want matrix multiplication it would be $(u\cdot \nabla)u=(\nabla u)u$.

tl;dr Looking at each $i$th component in $(1)$ leads to $(4)$.

Response to edited-in questions: Firstly I didn't cover (2) and (5) because I thought knowing the notation would mean you knew all about the divergence operator, which is covered in every book about multivariable calculus. $(5)$ is literally nothing but the definition of $(2)$. If you are not sure about multivariable calculus, I would suggest you revise that before tackling the topics covered by Fefferman in that paper.

  1. $i$ is used in both cases to refer to components, and writing it this way is a tiny bit easier on the mind when you want to apply the divergence operator to (4). But don't place too much stock in $i$ c.f. dummy variable. Its absolutely correct (if annoying) to write\begin{align}\frac{\partial}{\partial t} u_{🥴}+\sum_{🦊=1}^{n} u_{🦊} \frac{\partial u_{🥴}}{\partial x_{🦊}} &=\nu \Delta u_{🥴}-\frac{\partial p}{\partial x_{i}}+f_{🥴}(x, t), \quad 🥴=1,2,\dots,n \tag{4'}\\ \operatorname{div} u &=\sum_{M\in\{2,4,6,\dots,2n\}} \frac{\partial u_{M/2}}{\partial x_{M/2}}=0 \tag{5'}\end{align} PS I'm sure Fefferman assumed familliarity with multivariable calc, which is maybe why he didn't bother writing something to the effect of $i=1,2,\dots,n$, but strictly speaking it should have been there (to emphasise that it was $n$ equations).

  2. Yes but no. The reason no is because you repeated an equation. The divergence equation is not an equation for $u_1$ or $u_2$ separately. Unlike the other operators it is not a scalar operator. It takes in a vector valued function and spits out a scalar, which is the opposite of the gradient operator. So while $(1)$ is $n$ equations, $(2)$ is only one equation. In total there are $n+1$ equations, but you have written $n+2$ equations.

  3. I would say its never the second one but some writers are not so careful, or have different conventions from me. So always look at context. At least, I cannot think of any scenario in multivariable calc where I have seen the second option. I personally would write $$u=(u_i)_i = (u_i)_{i=1,2,\dots,n} = (u_i)_{i=1}^n = (u_i)_{1\le i\le n}= (u_1,\dots,u_n)^T = \begin{pmatrix}u_1\\ \vdots \\ u_n\end{pmatrix}$$

  • $\begingroup$ I tried to cover everything you might need but please ask if you don't understand $\endgroup$ Jun 26, 2020 at 13:16
  • $\begingroup$ Hi! Thank you! I understand if my question was ambiguous and based on your answer I tried to specify my questions. Due to the limited size for comments I updated my original post. Thanks in advance if you have the time to answer the new questions. $\endgroup$
    – JDoeDoe
    Jun 26, 2020 at 19:35
  • $\begingroup$ @JDoeDoe I have updated my answer for your new questions $\endgroup$ Jun 27, 2020 at 1:11

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