# Different notation for equivalent equations? Navies-Stokes equations

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}

• It actually looks exactly the same to me. Can you help me understand what terms you think are different? Maybe the nonlinear term $u\cdot \nabla u$? Jun 26, 2020 at 7:18
• Said another way, I thought (1), (2) was defined by (4), (5). So I don't know how to begin answering your question? Jun 26, 2020 at 7:25
• 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). Jun 26, 2020 at 7:35
• 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...?) Jun 26, 2020 at 7:39
• I guess you are getting stuck somewhere when trying to go from (1) to (4), what is the problematic step? Jun 26, 2020 at 7:45

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 $$u=(u_i)_i = (u_i)_{i=1,2,\dots,n} = (u_i)_{i=1}^n = (u_i)_{1\le i\le n}= (u_1,\dots,u_n)^T = \begin{pmatrix}u_1\\ \vdots \\ u_n\end{pmatrix}$$

• I tried to cover everything you might need but please ask if you don't understand Jun 26, 2020 at 13:16
• 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. Jun 26, 2020 at 19:35
• @JDoeDoe I have updated my answer for your new questions Jun 27, 2020 at 1:11