# Navier–Stokes equations

Lecture notes taken from C.Y. Wang’‘s short course.

Please note that the complete notes is now available at Changyou Wang’s homapage.

### Reference

1. Temam, Roger. Navier-Stokes equations: theory and numerical analysis. Vol. 343. Oxford University Press, 2001.
2. Majda, A. J., A. L. Bertozzi, and A. Ogawa. “Vorticity and Incompressible Flow. Cambridge Texts in Applied Mathematics.” Applied Mechanics Reviews 55 (2002): 77.

#### Flow map

Let $\Omega\subset \R^n$ be a fluid, $n=2,3$. $\rho$ is the density of fluid, $u:\omega\to\R^n$ is the velocity field of fluid. The flow map is defined as:
\begin{align*}
x:\Omega\times[0,+\infty)&\to\R^n\\
(\alpha,t)&\mapsto x(\alpha,t),
\end{align*}
where $\alpha$ is called the Lagrangian coordinates and $x(\alpha,t)$ is called the Eulerian coordinates. We must have the following equation:
$$\begin{cases} \frac{d x(\alpha,t)}{d t}=u(x(\alpha,t),t)\ x(\alpha,0)=\alpha. \end{cases}$$
What’s we concern is the acceleration
\begin{align*}
a&=\frac{d^2}{d t^2}x(\alpha,t)=\frac{d}{dt}u(x,t)=u_t+\frac{\partial u}{\partial x_i}\frac{d x^i}{d t}\\
&=u_t+\frac{\partial u}{\partial x_i}u^i=u_t+(u\cdot\nabla)u,
\end{align*}
$\frac{D}{D t}:=\frac{\partial }{\partial t}+(u\cdot\nabla)$ is called material derivative.

By the Newtonian second law, we have $Ma=F$, that is $\rho\frac{Du}{Dt}=F$, where $F=f+T$, where $f$ is the external body force and $T$ is the friction force.

thus
$$0=\frac{\rd \vol(O_t)}{\rd t}=\frac{\rd }{\rd t}\int_O\det \left(\frac{\pt x}{\pt \alpha}\right)\rd \alpha=\int_O\frac{\rd }{\rd t}\det\left(\frac{\pt x}{\pt \alpha}\right)\rd \alpha.$$
If we denote $A=(a_{ij})$, where $a_{ij}=\frac{\pt x^i(\alpha,t)}{\pt \alpha^j}$, and the cofactor of $a_{ij}$ is $A_{ij}$, then apply the ralation $A_{ij}a_{jk}=\delta_{ik}$, we obtain
\begin{align*}
\frac{\rd}{\rd t}\det\left(\frac{\pt x}{\pt \alpha}\right)
&=A_{ij}\frac{\rd }{\rd t}\left(\frac{\pt x^i}{\pt \alpha^j}\right)
=A_{ij}\frac{\pt}{\pt\alpha^j}\left(u^i(x,t)\right)\\
&=A_{ij}\frac{\pt u^i}{\pt x^k}\frac{\pt x^k}{\pt \alpha^j}
=\sum_i\frac{\pt u^i}{\pt x^i}.
\end{align*}
Therefore,
$$0=\int_O\sum_i\frac{\pt u^i}{\pt x^i}\rd\alpha,\quad\forall O\subset\Omega,$$
i.e., $x$ is incompressible iff $\div u=0$.

Since in this case the fluid is force balance, we have
$$\int_O f\rd v+\int_{\partial O}\tau\cdot \nu\rd \sigma=0,$$
where $f$ is the body force and $\tau$ is the cauchy stress tensor, it is a matrix of $n\times n$, and $\rd v$ is the volume elements, $\rd\sigma$ is the area elements, $\nu$ is the unit outer normal vector field.

By the divergence theorem (view $\tau$ as a vector with three components, each component is a row of $\tau$.) we have the vector styled equation:
$$f+\div \tau=0.$$
Usually, $\tau=-pI_n+\sigma$, where $p$ is the pressure, and $I_n$ is the identity matrix and $\sigma$ is the viscous stress.

#### Strain rate

The differential of $u:\Omega\to\R^3$ will be denoted as $Du$, set $Du=\dd (u)+\Omega(u)$, where $\dd(u)=\frac{1}{2}\left[Du+(Du)^T\right]$ is the rate-of-strain tensor and $\Omega(u)=\frac{1}{2}\left[Du-(Du)^T\right]$ is the rate of expansion of the flow. If $\sigma$ is linearly depended on $\dd(u)$, we say in this case the fluid is simple fluid, denoted as $\sigma=L(\dd(u))$. Note that, for the simple fluid, we have
$$L(Q^T\dd(u)Q)=Q^TL(\dd(u))Q,\quad \forall Q\in O(n),$$
Thus,
$$\sigma=2\mu\dd(u)+\lambda\tr(\dd(u))I_n =2\mu\dd(u)+\lambda(\div u)I_n,$$
where $\mu$ is called shear viscosity.

Now, $\tau=-pI_n+\sigma=-pI_n+2\mu\dd(u)+\lambda(\div u)I_n$, from which we obtain, in the steady case:
$$0=f+\div(-pI_n+2\mu\dd(u)+\lambda(\div u)I_n),$$
in general, we obtain the Navier-Stokes equation:
$$\rho\frac{D u}{D t}=f+\div(-pI_n+2\mu\dd(u)+\lambda(\div u)I_n)\tag{NSE}.$$

#### Conservation of Mass

By the conservation of mass, we have
$$\int_O\rho_t=\frac{\rd}{\rd t}\int_O\rho\rd x =\int_{\partial O}\rho u\cdot \nu\rd\sigma =-\int_O\div(\rho u),$$
that is
$$\rho_t+\div(\rho u)=0.$$
Thus, the fluid is incompressible iff $\rho_t+u\div \rho=0$.

#### Homogenous NSE

If $\rho\equiv1$, then the NSE becomes
$$\begin{cases} u_t+u\cdot\div u+\div p=f+\mu\Delta u,\\ \div u=0\\ u|_{t=0}=u_0,\:\div u_0=0. \end{cases}$$

1. For $n=2$, we know almost all the things, but for $n=3$, we almost know nothing!
2. If $\mu=0$, then the fluid is called ideal fluid, and the NSE reduce to Euler equation:
$$\begin{cases} u_t+u\cdot \div u+\div p=f,\\ \div u=0. \end{cases}$$
3. If, furthermore, we have $f=0$, then, by a simple calculation (recall that we have $\div u=0$),
$$-\Delta p=-\div((u\cdot\nabla)u)=\begin{cases}\tr[(\nabla u)^2]\\div^2(u\otimes u)\end{cases},$$
where $(u\otimes v)_{ij}=u^iv^i$, they are both useful in the study of fluid.