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
- Temam, Roger. Navier-Stokes equations: theory and numerical analysis. Vol. 343. Oxford University Press, 2001.
- Majda, A. J., A. L. Bertozzi, and A. Ogawa. “Vorticity and Incompressible Flow. Cambridge Texts in Applied Mathematics.” Applied Mechanics Reviews 55 (2002): 77.
Table of Contents
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 $\div u=0$.
Stead fluid state
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
$$
\begin{cases}
u_t+u\cdot\div u+\div p=f+\mu\Delta u,\\
\div u=0\\
u|_{t=0}=u_0,\:\div u_0=0.
\end{cases}
$$
- For $n=2$, we know almost all the things, but for $n=3$, we almost know nothing!
- If $\mu=0$, then the fluid is called ideal fluid, and the NSE reduce to Euler equation:
$$
\begin{cases}
u_t+u\cdot \div u+\div p=f,\\
\div u=0.
\end{cases}
$$ - 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.
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