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## Math 505, Mathematical Fluid Mechanics: Notes 1

This Spring ’16 semester, I am teaching a graduate Math 505 course, whose goal is to introduce the basic concepts and the fundamental mathematical problems in Fluid Mechanics for students both in math and engineering. The difficulty is to assume no background in both fluids and analysis of PDEs from the students. That’s it!

Anyhow, materials for my course are based on various books and lecture notes, one of which is the great lecture notes by V. Sverak (selected topics on fluid mechanics, 2011). More information, some pdf notes, and so on can be found from my course webpage!

1. Continuum assumption

Fluids are made up of many many discrete molecules that interact with one another. Throughout the course, we shall assume that fluid molecules are small enough to be infinitesimally close to one another (and so, of course, the number of molecules is infinite). That is, we shall work with the continuum models of fluids. The mathematical justification of the continuum dynamics of fluids (macroscopic description) from the deterministic Hamiltonian dynamics of discrete molecules (microscopic description) remains an outstanding unsolved problem (see, however, Quastel-Yau ’98 for stochastic particles). This is the Hilbert’s 6th problem. One could also formally derive the continuum models through the mesoscopic description as suggested by Boltzmann. Again, a rigorous derivation remains incomplete (cf., L. Saint-Raymond).

2. Eulerian and Lagrangian description of fluid motion

Let ${\Omega\subset \mathbb{R}^d}$ be the fluid domain, ${d\ge 1}$. By the continuum assumption, each point ${x\in \Omega}$ is viewed as a fluid “particle”. The motion of fluids is described by the velocity vector field

$\displaystyle u(x,t) \in \mathbb{R}^d$

at each particle ${x \in \Omega}$ and at a time ${t \in \mathbb{R}_+}$. This is the so-called Eulerian description of fluids, introduced by Euler in 1755. There is another common way to describe fluid motion, the Lagrangian description, which keeps track of the trajectory of particles. For each “initial” particle ${x\in \Omega}$, denote by ${X(t;x)}$ the new position of the particle ${x}$ at the time ${t}$, which is defined by the ODEs

$\displaystyle \frac{d}{dt}X(t;x) = u(X(t;x), t) , \qquad X(0,x) = x.$

That is, the map ${t \mapsto X(t;x)}$, as ${t}$ runs in ${\mathbb{R}_+}$, keeps track of the trajectory of the initial particle ${x}$, whereas the Lagrangian map ${x\mapsto X(t;x)}$ gives the new position of the particle ${x}$ when time evolves. By a view of the standard ODE theory (e.g., the standard Picard’s iteration), for each ${x}$, a trajectory ${X(t;x)}$ exists locally in time ${t}$ near zero, if ${u(x,t)}$ is continuous in ${x}$ and ${t}$. In addition, if ${u(x,t)}$ is Lipschitz in ${x}$, the local trajectory is unique.

In the case when ${\Omega}$ has a boundary, it is natural to assume that the fluids do not cross the boundary. That is,

$\displaystyle u(x,t) \cdot n(x)_{\vert_{\partial \Omega}} = 0$

in which ${n(x)}$ denotes the outer normal unit vector at ${x\in \partial \Omega}$. With the above boundary condition, it is easy to see that the trajectories ${X(t;x)}$, when exist, either remain on ${\partial \Omega}$ or never cross it. On the other words, fluids in the interior of ${\Omega}$ remain in the interior, and those on the boundary ${\partial \Omega}$ remain on the boundary.

2.1. Continuity equation

Let ${\rho(x,t)}$ be the density distribution of fluids. That is, the mass of fluids in the infinitesimal volume ${dx}$ is equal to ${\rho(x,t)dx}$, and the total of mass in an arbitrary domain ${O\subset \Omega}$ is defined by

$\displaystyle \int_O \rho(x,t) \; dx .$

Let ${O_t}$ be the image of ${O}$ under the map ${x\mapsto X(t;x)}$. The conservation of mass reads

$\displaystyle \int_{O_t} \rho(X, t)\; dX = \int_O \rho(x,0)\; dx .\ \ \ \ \ (1)$

Using the change of variables ${X = X(t;x)}$ for ${x \in O }$ and denoting the Jacobian determinant ${J(x,t) = \det \nabla_x X(t;x)}$, we have

$\displaystyle \int_{O_t} \rho(X, t)\; dX = \int_O \rho(X(t;x),t) J(x,t)\; dx .$

Since ${O}$ was arbitrary, the conservation of mass implies

$\displaystyle \rho(X(t;x),t) J(x,t) = \rho(x,0), \qquad \forall ~ x\in \Omega, \quad \forall ~t\in \mathbb{R}_+.\ \ \ \ \ (2)$

This is the conservation of mass in the Lagrangian coordinate. To write the conservation in the Eulerian coordinate, we take the time derivative of (2). A direct computation yields

Lemma 1 There holds ${\frac{d}{dt} J(x,t) = J(x,t) (\nabla_x \cdot u) (X(t;x),t).}$

Proof: Exercise. $\Box$

Here, ${\nabla_x}$ denotes the usual gradient vector. In addition, for any quantity ${f(x,t)}$, the rate of change of quantity ${f}$ along each particle trajectory is computed by

$\displaystyle \frac{d}{dt} f(X(t;x), t) = \partial_t f + \frac{d}{dt}X(t;x) \cdot \nabla_x f = (\partial_t + u (X(t;x),t)\cdot \nabla_x) f (X(t;x),t) .$

The derivative ${D_t: = \partial_t + u\cdot \nabla_x}$ is often referred to as the material derivative.

Hence, (2) yields

$\displaystyle (\partial_t + u(X(t;x),t) \cdot \nabla_x) \rho (X(t;x),t)+ \rho(X(t;x),t) \nabla_x \cdot u(X(t;x),t) = 0,$

for all ${x\in \Omega}$ and ${t \in \mathbb{R}_+}$. Here, assuming sufficient regularity of ${u(x,t)}$, the map ${x \rightarrow X(t;x)}$ is a diffeomorphism from ${\Omega}$ to itself. This shows that for all points ${(x,t)}$, there is a unique ${y}$ so that ${X(t;y) = x}$. That is, the above equation yields

Lemma 2 The density ${\rho(x,t)}$ satisfies the continuity equation:

$\displaystyle \rho_t + \mathrm{div}_x (\rho u) =0. \ \ \ \ \ (3)$

For an arbitrary fluid subdomain ${O \subset \Omega}$, using the continuity equation and the divergence theorem, we compute

$\displaystyle \frac{d}{dt} \int_O \rho(x,t)\; dx = - \int_O \mathrm{div}_x (\rho u) (x,t)\; dx = - \int_{\partial O} \rho u \cdot n \; d\sigma(x)$

which asserts that the rate of change of the total mass in ${O}$ is equal to the total density flux, ${j = \rho u}$, of the fluid through the boundary ${\partial O}$. The minus sign is due to the fact that ${n}$ is the outer normal unit vector. Of course, this is the alternate way to derive the continuity equation (3).

2.2. Transport theorem

It is straightforward to check that

Lemma 3 (Transport theorem) Let ${u(x,t)}$ be a velocity vector field, with ${u \cdot n =0}$ on ${\partial \Omega}$, and let ${D_t = \partial_t + u \cdot \nabla_x}$ be the corresponding material derivative. For ${\rho}$ solving the continuity equation (3), there holds

$\displaystyle \frac{d}{dt} \int_\Omega \rho f \; dx = \int_\Omega \rho D_t f\; dx$

for any smooth function ${f(x,t)}$.

Proof: Exercise. $\Box$ The lemma shows that the integral ${\int_\Omega \rho f\; dx}$ is conserved in time, provided ${f}$ solving the transport equation ${D_t f =0}$, or equivalently

$\displaystyle (\partial_t + u \cdot \nabla_x ) f(x,t) =0.$

That is, ${f}$ is constant along the particle trajectory ${X(t;x)}$, associated with the velocity field ${u(x,t)}$. In particular, ${f =1}$ solves the transport equation, and thus the transport theorem yields the conservation of the total mass in ${\Omega}$. In fact, if we let ${f = \chi_{O_t}}$ be the characteristic function on ${O_t = X(t;O)}$, then ${f}$ solves the transport equation (in the weak sense), and the transport theorem reassures the conservation of mass; see (1).

2.3. Incompressibility

We say the fluid flow is incompressible if the Lagrangian map ${x\mapsto X(t;x)}$ is a volume-preserving map, for all time ${t\in \mathbb{R}_+}$. Precisely, there holds

$\displaystyle \mathrm{Vol} (O_t) : = \int_{O_t}\; dX = \mathrm{Vol}(O) := \int_O \; dx$

for all subdomains ${O \subset \Omega}$. Here, ${O_t}$ denotes the image of ${O}$ under the map ${x\mapsto X(t;x)}$. Exactly as done in the previous section, this is equivalent to

$\displaystyle \mathrm{div}_x u =0.\ \ \ \ \ (4)$

For incompressible flows, the density is constant along the particle trajectories ${D_t \rho =0}$, or equivalently

$\displaystyle \rho(X(t;x),t) = \rho_0(x)$

for all ${x,t}$. In particular, homogeneity (i.e., constant density) of incompressible fluids propagates in time. Gas or air are compressible flows, whereas water is modeled by the incompressible flow.

For the incompressible flows, it is easy to check that the quantity

$\displaystyle \int_\Omega \varphi (\rho ) \; dx$

is conserved in time, for arbitrary smooth function ${\varphi(\cdot)}$ so that the integral is well-defined.

\newpage

3. Momentum equations

Certainly, the continuity equation does not constitute a complete set of equations to describe fluids, since the velocity field ${u(x,t)}$ itself is an unknown. In this section, we derive the momentum equations. The derivation uses the continuum version of the Newton’s second law:

$\displaystyle \rho a(x,t) = F(x,t), \ \ \ \ \ (5)$

in which ${\rho(x,t)}$ is the density, ${a(x,t)}$ the acceleration or the rate of change of the fluid velocity, and ${F(x,t)}$ the total force acting on the fluid. Here, in (5), the forces are understood as the net force acting on fluid parcels.

3.1. Acceleration

By definition, the acceleration is defined by

$\displaystyle a(X(t;x),t) := \frac{d^2}{dt^2} X(t;x) = \frac{d}{dt} u(X(t;x),t) = (\partial_t + u (X(t;x),t) \cdot \nabla_x) u (X(t;x),t).$

The above holds for all ${x}$ and ${t}$, and so for all points ${(X(t;x),t)}$. That is, the acceleration of fluid motion at each ${(x,t)}$ is

$\displaystyle a(x,t) = (\partial_t + u(x,t) \cdot \nabla_x) u (x,t).\ \ \ \ \ (6)$

For free particles, that is, for fluids that experience neither internal nor external forces ${F(x,t) =0}$, the velocity field satisfies

$\displaystyle (\partial_t + u \cdot \nabla_x) u = 0$

which is the inviscid Burgers equation. In the Lagrangian coordinates, this shows that the velocity field is constant along the particle trajectories ${ u(X(t;x), t) = u_0(x)}$ and so the trajectories are simply straight lines

$\displaystyle X(t;x) = x + t u_0(x).$

It is easy to construct smooth initial data ${u_0(x)}$ so that two trajectories with different initial velocity meet in a finite time, which results in the discontinuity of the velocity field.

In addition, using the transport theorem, Lemma 3, with ${f = 1, u, \frac12|u|^2}$, one has for free particles the conservation of mass, momentum, and energy

$\displaystyle \int_\Omega \rho\; dx, \qquad \int_\Omega \rho u\; dx , \qquad \frac12 \int_\Omega \rho |u|^2 \; dx .$

3.2. Examples of forces

An example of forces includes gravity, Coriolis, or electromagnetic forces that acts on the fluid. For instance, the gravity force is often taken to be

$\displaystyle - \rho(x,t) g e_z$

where ${e_z}$ denotes the upward vertical direction. In what follows, we shall ignore these forces.

3.3. Cauchy stress tensor

Cauchy stress tensor ${\sigma}$, a ${2}$-tensor ${\sigma_{ij}}$, accounts for the force acting on the boundary of fluid parcels. That is, for any fluid subdomain ${O \subset \Omega}$, the net force produced by the stress tensor is defined by

$\displaystyle \int_{\partial O} \sigma \cdot n \; d\sigma(x) = \int_{\partial O} [\sum_j \sigma_{ij} n_j]_i\; d\sigma(x),$

which yields the net force (due to the Cauchy stress)

$\displaystyle F: = \nabla \cdot \sigma$

in the (arbitrary) fluid domain ${O}$, by a view of the divergence theorem. Here, by convention, the ${i^{\mathrm{th}}}$-component of the vector ${\nabla \cdot \sigma}$ is ${\sum_j \partial_{x_j}\sigma_{ij}}$.

3.4. Ideal fluids

By definition, ideal fluid is defined by ideally setting the Cauchy stress tensor to be of the form

$\displaystyle \sigma = - p I_{\mathrm{d}}$

in which ${p}$ is the so-called the pressure of the fluid and ${I_\mathrm{d}}$ denotes the ${d\times d}$ identity matrix. By (5) and (6), the momentum equations for ideal fluids then read

$\displaystyle \rho (\partial_t + u \cdot \nabla_x) u + \nabla p = 0 . \ \ \ \ \ (7)$

The equations, together with the continuity equations, are referred to as the Euler equations.

3.5. Viscous fluids

To account for friction, one needs to take into account of the additional viscous stress tensor ${\sigma_{vis}}$. For Newtonian fluids, the viscous stress is assumed to be proportional to the gradient of velocity field:

$\displaystyle \sigma_{vis} = \lambda \mathrm{div} u I_{\mathrm{d}} + \nu (\nabla u + (\nabla u)^t)$

in which ${\lambda, \nu\ge 0}$ denote the LamÃ© viscosity coefficients. A direct computation yields the net viscous force

\displaystyle \begin{aligned} \nabla \cdot \sigma_{vis} &= \lambda \nabla \mathrm{div} u +\nu \Big(\sum_j \partial_{x_j} ( \partial_{x_i} u_j + \partial_{x_j} u_i)\Big)_i \\ &= (\lambda + \nu) \nabla \mathrm{div} u +\nu \Delta u. \end{aligned}

3.6. Compressible flows

Combining, the conservation of mass and momentum yields the compressible Euler (when no viscosity) and Navier-Stokes equations

\displaystyle \left \{ \begin{aligned} \rho_t + \mathrm{div} (\rho u) & =0 \\ \rho (\partial_t + u \cdot \nabla ) u + \nabla p &= \nu \Delta u + (\lambda + \nu) \nabla \mathrm{div} u \end{aligned}\right. \ \ \ \ \ (8)

For this set of equations to be complete, a pressure law is needed. For instance, a barotropic gas is the fluid flow where the pressure is an (invertible) function of density:

$\displaystyle p = p(\rho).$

In the literature, the full set of compressible flows takes into account of the conservation of energy as well.

Let us compute the rate of change of the total energy. The kinetic energy satisfies

\displaystyle \begin{aligned} \frac{d}{dt} \int_\Omega \rho \frac{|u|^2}{2}\; dx &= \int_\Omega \rho D_t \frac{|u|^2}{2} \; dx = \int_\Omega \rho D_t u \cdot u \; dx \\ &= \int_\Omega \Big[ \nu \Delta u + (\lambda + \nu) \nabla \mathrm{div} u -\nabla p \Big]\cdot u \; dx \\ &= - \nu \int_\Omega |\nabla u|^2 \; dx - (\lambda + \nu) \int_\Omega|\mathrm{div} u|^2 \; dx + \int_\Omega p(\rho) \mathrm{div} u \; dx, \end{aligned}

whereas the potential energy satisfies

$\displaystyle \frac{d}{dt}\int_\Omega P(\rho) \; dx = - \int_\Omega P'(\rho) \mathrm{div} (\rho u)\; dx = - \int_\Omega \Big[ P'(\rho) \rho - P(\rho) \Big] \mathrm{div} u\; dx.$

Comparing, we take ${P(\rho)}$ so that

$\displaystyle P'(\rho) \rho - P(\rho) = p(\rho),\ \ \ \ \ (9)$

or equivalently, ${P(\rho) = \rho \int_1^\rho z^{-2} p(z)\; dz}$. For instance, in the case of the ${\gamma}$– law pressure ${ p(\rho) = \rho^\gamma}$, we take ${P(\rho) = \frac{1}{\gamma-1} \rho^\gamma}$. The total energy satisfies

\displaystyle \begin{aligned} \frac{d}{dt} \int_\Omega \Big[ \rho \frac{|u|^2}{2} + P(\rho) \Big]\; dx = - \nu \int_\Omega |\nabla u|^2 \; dx - (\lambda + \nu) \int_\Omega|\mathrm{div} u|^2 \; dx \end{aligned}

with ${P(\rho)}$ defined as in (9). In particular, the total energy is decreasing in time. For ideal fluids, the total energy is constant in time (for smooth solutions).

3.7. Incompressible flows

When the flow is assumed to be incompressible, the Euler and Navier-Stokes equations are

\displaystyle \left \{ \begin{aligned} \rho_t + u \cdot \nabla \rho & =0 \\ \rho (\partial_t + u \cdot \nabla ) u + \nabla p &= \nu \Delta u \\ \nabla \cdot u &=0. \end{aligned}\right. \ \ \ \ \ (10)

Unlike in the compressible case, this set of equations is complete and the pressure itself is an unknown function. The most popular model is when fluid is incompressible and homogenous (${\rho =1}$), and is often referred to simply as the Euler (${\nu =0}$) and Navier-Stokes equations.

We have the (kinetic) energy equality

\displaystyle \begin{aligned} \frac{d}{dt} \int_\Omega \rho \frac{|u|^2}{2} \; dx = - \nu \int_\Omega |\nabla u|^2 \; dx . \end{aligned}

Here, we note that since ${\rho}$ satisfies the transport equation, with the incompressible velocity vector field, the potential energy is conserved in time; see Section 2.

3.8. Reynolds number

Consider the incompressible homogenous Navier-Stokes equations. Let ${T}$ be the time unit, ${L}$ the length unit, and ${U}$ the velocity unit, with ${L = UT}$. Let us introduce the change of variables

$\displaystyle t \mapsto \frac tT , \qquad x\mapsto \frac xL, \qquad u\mapsto \frac uU.$

We then arrive at the non-dimensional Navier-Stokes equations:

\displaystyle \left \{ \begin{aligned} u_t + u \cdot \nabla u+ \nabla p &= \frac{1}{Re} \Delta u \\ \nabla \cdot u &=0 \end{aligned}\right. \ \ \ \ \ (11)

with ${Re = \frac{UL}{\nu}}$ being called the physical Reynolds number. The problem of small viscosity limit or high Reynolds number has a very long story. Indeed, it is one of the most classical subjects in fluid dynamics. It interests most prominent physicists such as Lord Rayleigh, W. Orr, A. Sommerfeld, Heisenberg, W. Tollmien, H. Schlichting, among many others. It was already noted by Reynolds himself in his seminal experiment (1883) that the Reynolds number governs the transition from laminar to turbulent flows. The studies became active around 1930, motivated by the study of the boundary layer around wings. In airplanes design, it is crucial to study the boundary layer around the wing, and more precisely the transition between the laminar and turbulent regimes, and even more crucial to predict the point where boundary layer splits from the boundary. A large number of papers has been devoted to the estimation of the critical Reynolds number of classical shear flows (Blasius profile, exponential suction/blowing profile, etc…). It was Heisenberg in 1924 who first estimated the critical Reynolds number of parallel shear flows. C. C Lin and then Tollmien around 1940s completed the picture with lower and upper stability branches, respectively for parallel flows and boundary layers. Most of the physical literature, together with many mathematical insights, on the subject is well documented by Drazin and Reid in their famous book on hydrodynamics instability. I will be sure to come back to this topic near the end of the course.

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