# The Azimuth Project Blog - fluid flows and infinite dimensional manifolds (part 2)

## Or: inviscid incompressible fluids: dry water?

This page is a blog article in progress, written by Tim van Beek. To see discussions of this article while it was being written, visit the Azimuth Forum. For the final polished article, go to the Azimuth Blog.

Last time on fluid flows and infinite dimensional manifolds we set the stage by explaining infinite dimensional manifolds. Then we looked at a simple example: the inviscid Burgers equation. We saw this was the equation for geodesics in the diffeomorphism group of the circle.

Now let’s look at a more interesting example! It will still be a simplified model of fluid flow: it will describe an ideal fluid that is incompressible. I’ll start by explaining these concepts. We will then see how the equation of motion for ideal incompressible fluids can be interpreted as a geodesic equation.

En route I will also repeat some stuff from classical vector analysis, mostly for my own sake. The last time I seriously had to calculate with it was when I attended a class on “classical electrodynamics”, which was almost 15 years ago!

When we delve into differential geometry, it is always a good idea to look both at the “coordinate free” formulation using abstract concepts like differential forms, and also at the “classical vector analysis” part, that is best for calculating stuff once suitable coordinates have been chosen. Our fluid flows will take place in a smooth, orientable, compact, $n$-dimensional Riemannian manifold $M$, possibly with a smooth boundary $\partial M$.

I will frequently think of $M$ as an open set in $\mathbb{R}^2$ or $\mathbb{R}^3$, so I will use the globally defined coordinate chart of Euclidean coordinates on $\mathbb{R}^n$ denoted by $x, y$ (and $z$, if needed) without further warning.

Before we continue: Last time reader “nick” pointed out a blog post by Terence Tao about the same topic as ours, but—as could be expected—assuming a little bit more of a mathematical background: The Euler-Arnold equation. If you are into math, you might like to take a look at it.

So, let us start with the first important concept: the “ideal fluid”.

### What is an ideal fluid?

When you are a small parcel in a fluid flow, you will feel two kinds of forces:

external forces like gravity that are there whether or not your fellow fluid parcels surround you or are absent,

internal forces that come from your interaction with the other fluid parcels.

If there is friction between you and other fluid parcels, for example, then there will be a force slowing down faster parcels and speeding up slower parcels. This is called viscosity, which I already explained back in the post Eddy Who?. High viscosity means that there is a lot of friction: think of honey.

The presence of viscosity leads to shear stress whenever there are differences in the velocities of nearby fluid parcels. These lead to the formation of eddies and therefore to turbulence. This complicates matters considerably! For this reason, sometimes people like to simplify matters and to assume that the fluid flow that they consider has zero viscosity. This leads us to the physics definition of an ideal fluid:

An ideal fluid (as physicists say) is a fluid with zero viscosity.

As you can guess, I have also a mathematical definition in store for you:

An ideal fluid (as mathematicians say) is a fluid with the following property: For any motion of the fluid there is a (real valued) function $p(x, t)$ called the pressure such that if $S$ is a surface in the fluid with a chosen unit normal $n$, the force of stress exerted across the surface $S$ per unit area at $x \in S$ at time $t$ is $p(x,t) n$.

This implies that there is no force acting tangentially to the surface $S$: This picture is from

• Alexandre Chorin and Jerrold E. Marsden, A Mathematical Introduction to Fluid Mechanics, 3rd edition, Springer, New York 1993.

An ideal fluid cannot form eddies by itself without the help of external forces, nor can eddies vanish once they are present. So this simplification exclude a lot of very interesting phenomena, including everything that is usually associated with the term “turbulence”. But it is a necessary simplification for describing fluid flow using geodesic equations, because something moving along a geodesic doesn’t lose energy due to friction! So we will have to stick with it for now.

Historically, ideal fluids were almost exclusively studied during the 19th century, because the mathematics of viscous fluids seemed to be too hard—which it still is, although there has been a lot of progress. T his led to a schism of theoretical hydrodynamics and engineering hydrodynamics, because engineers had to handle effects like turbulence that ideal fluids cannot model. A very problematic aspect is that no body with a subsonic velocity feels any drag force in an ideal fluid. This is known as D’Alembert’s paradox. This means that one cannot find out anything about optimal design of ships or aircrafts or cars using ideal fluids as a model. This situation was overcome by the invention of “boundary layer techniques” by the physicist Ludwig Prandtl at the beginning of the 20th century.

John von Neumann is cited by Richard Feynman in his physics lectures as having said that “ideal water” is like “dry water”, because it is so unlike real water. This is what the subtitle of this post alludes to. I don’t think this is quite fair to say: Along these lines one could say that quantum mechanics is the theory of stagnant light, because it does not include relativistic effects like quantum field theory does. Of course every mathematical model is always just an approximation to a Gedankenexperiment. And ideal fluids still have their role to play.

Maybe I will tell you more about this in a follow-up post, but before this one gets too long, let us move on to our second topic: incompressible fluids and ‘volume preserving’ diffeomorphisms.

### What is an incompressible fluid flow?

If you are a parcel of an incompressible fluid, this means that your volume does not change over time. But your shape may, so if you start out as a sphere, after some time you may end up as an ellipsoid. Let’s make this mathematically precise.

But first note, that “incompressible” in the sense above means that the density of a given fluid parcel does not change over time. It does not mean that the density of the whole fluid is everywhere the same. A fluid like that is actually called homogeneous. So we have two different notions:

incompressible means that the volume of an infinitesimal fluid parcel does not change as it moves along the fluid flow,

homogeneous means that the density at a given time is everywhere the same, that is: constant in space.

This distinction is important, but for now we will study fluid flows that are both homogeneous and incompressible.

Let us see how we can make the notion of “incompressible” mathematically precise:

Remember from the last post: The flow of each fluid package is described by a path on $M$ parametrized by time, so that for every time $t \gt t_0$ there is a diffeomorphism

$g^t : M \to M$

defined by the requirement that it maps the initial position $x$ of each fluid package to its position $g^t(x)$ at time $t$: Now let’s assume our fluid flow is incompressible. What does that mean for the diffeomorphisms that describe the flow? Assuming that we have a fixed volume form $\mu$ on $M$, these diffeomorphisms must conserve it:

$\mathrm{SDiff}(M) := \{ f \in \mathrm{Diff}(M): f^* \mu = \mu$

For people who need a reminder of the concepts involved (which includes me), here it is:

Remember that $M$ is a smooth orientable Riemannian manifold of dimension $n$. A volume form $\mu$ is a $n$-form that vanishes nowhere. In $\mathbb{R}^3$ with Cartesian coordinates $x, y, z$ the canonical example would be

$\mu = d x \wedge d y \wedge d z$

The dual basis of $d x, d y, d z$ is denoted by $\partial_x, \partial_y, \partial_z$ in our example.

Given two manifolds $M, N$ and a differentiable map $f: M \to N$, we can pull back a differential form $\mu$ on $N$ to one on $M$ via

$f^{*} \mu_p (v_1, ..., v_n) := \mu_{f(p)} (d f(v_1), ..., d f(v_n))$

For the math nerds: Remember that we see the group of diffeomorpisms $\mathrm{Diff}(M)$ as a Fréchet Lie group modelled on the Fréchet space of vector fields on $M$, $\mathrm{Vec}(M)$. For those who would like to read more about this concept, here is a very readable paper:

• Karl-Hermann Neeb, Monastir Summer School: infinite-dimensional Lie groups.

$\mathrm{SDiff}(M)$ is clearly a subgroup of $\mathrm{Diff}(M)$. It is less obvious, but true, that it is a closed subgroup and therefore itself a Lie group. What about its Lie algebra? For a vector field to give a flow that’s volume preserving, it must have zero divergence. So, the vector fields that form the tangent space $T_{i d} \mathrm{SDiff}(M)$ consist of all smooth vector fields $X$ with zero divergence:

$div(X) = 0$

These vector fields form a vector space we denote by $\mathrm{SVec}(M)$. Remember $T_{i d}$ stands for the tangent space at the identity element of the group $\mathrm{SDiff}(M)$, which is the identity diffeomorphism $id$ of $M$. The tangent space at the identity of a Lie group is a Lie algebra, so $\mathrm{SVec}(M)$ is a Lie algebra.

I will need a little refresher about the definition of divergence. Then I will point you to a proof of the claim above, namely that zero-divergence vector fields form the Lie algebra of volume preserving diffeomorphism. This may seem obvious on an intuitive level, if you ever learned that the zero-divergence vector fields have “no sinks and no sources”, for example in a course on classical electromagnetism.

So, what is the divergence, again?

The divergence of a vector field $V$ with respect to a volume form $\mu$ is the unique scalar function $\mathrm{div}(V)$ such that:

$\mathrm{div}(V) = d (i_V \mu)$

Here, $i_X$ is the contraction with X. Contraction means that you feed the vector $X$ in the first slot of the differential form $\mu$, and therefore reduce the function $\mu$ of $n$ vector fields to one of $n-1$ vector fields.

When we use our standard example $M = \mathbb{R}^3$, we of course write a vector field as

$V = V_x \partial_x + V_y\partial_y + V_z \partial_z$

where $V_x, V_y$ and $V_z$ are smooth real-valued functions. The divergence of $X$ is then

$\mathrm{div}(V) = \partial_x V_x + \partial_y V_y + \partial_z V_z$

which we get if we plug in the expression for $V$ into the formula $d(i_V \mu)$.

So, how does one see that ‘zero divergence’ of a vector field is equivalent to ‘volume preserving’ for the flow it generates?

If we write $\phi(t) = (x(t), y(t), z(t))$ for the path of a fluid particle and $u$ for its velocity, then we have of course

$\frac{d \phi}{d t} = u$

For a scalar function $f(t, x(t), y(t), z(t))$ we get

$\frac{d f}{d t} = \frac{\partial f}{\partial t} + u \cdot \grad(f)$

Here $\cdot$ is the scalar product. The latter part is often written with the help of the nabla operator $\nabla$ as

$u \cdot \grad(f) = u \cdot \nabla \; f$

This is really just a handy short notation, there is no mystery behind it: it’s just like how we write the divergence as $div(X) = \nabla \cdot X$ and the curl as $curl(X) = \nabla \times X$.

The operator

$D_t = \partial_t + u \cdot \nabla$

appears so often that it has its own name: it is called the material derivative.

Why ‘material’? Because if we follow a little bit of material—what we’re calling a parcel of fluid—something about it can change with time for two different reasons. First, this quantity can explicitly depend on time: that’s what the first term, $\partial_t$, is about. Second, this quantity can depend on where you are, so it changes as the parcel moves: that’s what $u \cdot \nabla$ is about.

Now suppose we have a little parcel of fluid. We’ve been talking about it intuitively, but mathematically we can describe it at time zero as an open set $W_0$ in our manifold. After a time $t$, it will be mapped by the fluid flow $g^t$ to

$W_t := g^t (W_0)$

This describes how our parcel moves. We define the fluid to be ‘incompressible’ iff the volume of $W_t$ for all choices of $W_0$ is constant, that is:

$0 = \frac{d}{d t} \int_{W_t} d \mu$

If we write $J^t$ for the Jacobian determinant of $g^t$, then we have

$\displaystyle{ 0 = \frac{d}{d t} \int_{W_t} d \mu = \frac{d}{d t} \int_{W_0} J^t d \mu }$

So in a first step we get that a fluid flow is incompressible iff the Jacobian determinant $J$ is $1$ for all times, which is true iff $g^t$ is volume preserving.

It is not that hard to show by a direct calculation that

$\partial_t J = div(u) J$

If you don’t want to do it yourself, you can look it up in a book that I already mentioned:

• Alexandre Chorin and Jerrold E. Marsden, A Mathematical Introduction to Fluid Mechanics, 3rd edition, Springer-Verlag, New York 1993.

This is the connection between ‘volume preserving’ and ‘zero divergence’! Inserting this into our equation of incompressibility, we finally get:

$\begin{array}{ccl} 0 &=& \displaystyle{ \frac{d}{d t} \int_{W_t} d \mu } \\ &=& \displaystyle{\frac{d}{d t} \int_{W_0} J^t d \mu } \\ &=& \displaystyle{\int_{W_0} div(u) J d \mu } \end{array}$

which is true for all open sets $W_0$ iff $\mathrm{div}(u) = 0$. The equation of continuity for a fluid flow is:

$\frac{\partial \rho}{\partial t} + div(\rho u) = 0$

This says that mass is conserved. Written with the material derivative it is:

$\frac{D \rho}{D t} + \rho div(u) = 0$

So, since $\mathrm{div}(u) = 0$, we get

$\frac{D \rho}{D t} = 0$

which is what we intuitively expect, namely that the density is constant for a fluid parcel following the fluid flow.

### Euler's equation for the ideal incompressible fluid

The equation of motion for an ideal incompressible fluid is Euler’s equation:

$\partial_t u + (u \cdot \nabla) u = - \nabla p$

$p$ is the pressure function mentioned in the mathematical definition of an ideal fluid above. As I already mentioned, to be precise I should say that we also assume that the fluid is homogeneous. This means that the density $\rho$ is constant both in space and time and therefore can be cancelled from the equation of motion.

If $M$ has a nonempty (smooth) boundary $\partial M$, the equation is supplemented by the boundary condition that $u$ is tangential to $\partial M$.

How can we turn this equation into a geodesic equation on $\mathrm{SDiff}(M)$? Our strategy will be the same as last time when we handled the diffeomorphism group of the circle. We will define the necessary gadgets of differential geometry on $\mathrm{SDiff}(M)$ using the already existing ones on $M$. First we define them on $T_{id}\mathrm{SDiff}(M)$. Then, for any diffeomorphism $\phi \in \mathrm{SDiff}(M)$, we use right translation by $\phi$ to define them on $T_{\phi}\mathrm{SDiff}(M)$. After that, we can use the version of the abstract version of the geodesic equation for right invariant metrics to calculate the explicit differential equation behind it.

Let us start with defining right invariant vector fields on $\mathrm{SDiff}(M)$. A right invariant vector field $U$ is a vector field such that there is a $u \in \mathrm{SVec}(M)$ such that $U_{\phi} = u \circ \phi$. In the following, we restrict ourselves to right invariant vector fields only.

We define the usual $L^2$ scalar product of vector fields $u, v$ on $M$ just as last time:

$\displaystyle{ \langle u, v \rangle \coloneqq \int_M \langle u_x, v_x \rangle \; d \mu (x) }$

The scalar product used on the right is of course the one on $M$. For two right invariant vector fields $U, V$ with $U_{\phi} = u \circ \phi$ and $V_{\phi} = v \circ \phi$, we define the scalar product on $T_{\phi}\mathrm{SDiff}(M)$ by

$\langle U, V \rangle_{\phi} \coloneqq \langle u, v \rangle$

This definition induces a right invariant metric on $\mathrm{SDiff}(M)$. Note that it is right invariant because we have measure preserving diffeomorphisms only. It is not right invariant on the larger group of all diffeomorphims $\mathrm{Diff}(M)$!

For an incompressible ideal fluid without external fields the only kind of energy one has to consider is the kinetic energy. The scalar product that we use is actually proportional to the kinetic energy of the whole fluid flow at a fixed time. So geodesics with respect to the induced metric will correspond to Hamilton’s extremal principle. In fact it is possible to formulate all this in the language of Hamiltonian systems, but I will stop here and return to the quest of calculating the geodesic equation.

Last but not least, we define the following right invariant connection:

$\nabla_{U_{\phi}} V_{\phi} \coloneqq (\nabla_{u} v) \circ \phi$

Here $\nabla$ on the right is the connection on $M$—sorry, this is not quite the same as the $\nabla$ we’d been using earlier! But in $\mathbb{R}^3$ or any other dimension of Euclidean space, $\nabla_u v$ is just another name for $(u \cdot \nabla) v$, so don’t get scared.

Remember from last time that the geodesic equation says

$\nabla_u u = 0$

where $u$ is the velocity vector of our geodesic, say

$u(t) = \frac{d}{d t} \gamma(t)$

where $\gamma$ is the curve describing our geodesic. We saw that for a right-invariant metric on a Lie group, this equation says

$\partial_t u = ad^*_u u$

where the coadjoint operator $\mathrm{ad}^*$ is defined by

$\langle \mathrm{ad}^*_u v, w \rangle = \langle v, \mathrm{ad}_u w \rangle = \langle v, [u, w] \rangle$

For simplicity, let us specialize to $\mathbb{R}^3$, or an open set in there. What can we say about the right hand side of the above equation in this case? First, we have the vector identiy

$\nabla \times (u \times w) = - [u, w] + u \; \nabla \cdot w - w \; \nabla \cdot u$

Since we are talking about divergence-free vector fields, we actually have

$[u, w] = - \nabla \times (u \times w)$

Also note that for a scalar function $f$ and the divergence-free vector field $u$ we have

$\begin{array}{ccl} \langle u, \nabla f \rangle &=& \int_M \langle u(x), \nabla f(x) \rangle d \mu (x) \\ &=& \int_M \nabla \cdot (f(x) u(x)) d \mu (x) \\ &=& \int_{\partial M} f(x) \; \langle u, n \rangle \; d S (x) \\ &=& 0 \end{array}$

The last term is zero because of our boundary condition that says that the velocity field $u$ is tangent to $\partial M$.

So, now I am ready to formulate my claim that

$\mathrm{ad}^*_u v = - (\nabla \times v) \times u + \nabla f$

for some yet undetermined scalar function $f$. This can be verified by a direct calculation:

$\begin{array}{ccl} \langle ad^*_u v, w \rangle &=& \langle v, ad_u w \rangle \\ &=& \langle v, [u, w] \rangle = \int_M \langle v_x, [u_x, w_x] \rangle \;d \mu(x) \\ &=& - \int_M \langle v_x, \nabla \times (u_x \times w_x) \rangle \;d \mu(x) \end{array}$

What next? We can use the following 3 dimensional version of Green’s theorem for the curl operator:

$\int_M ( \langle \nabla \times a, b \rangle - \langle a, \nabla \times b \rangle ) d \mu = \int_{\partial M} \langle a \times b, n \rangle d S$

That is, the curl operator is symmetric when acting on vector fields that have no tangetial component to $\partial M$. Note that I deliberatly forget to talk about function spaces that every vector field needs to belong to and regularity assumptions on the domain $M$ and its boundary, because this is a blog post and not a math lecture . But the operators we use on vector fields obviously depend on such assumptions.

If you are interested in how to extend the symmetric curl operator to a self adjoint operator, for example, you could look it up here:

• R. Hiptmair, P.R. Kotiuga, S. Tordeux: Self-adjoint curl operators (arXiv).

Since our vector fields are supposed to be tangential to $\partial M$, we have that the boundary term in our case is

$\int_{\partial M} \langle u_x \times \w_x \times v_x, n \rangle \; dS = 0$

because $u_x \times \w_x$ is normal, and therefore $u_x \times \w_x \times v_x$ is tangential to $\partial M$, so that the scalar product with the normal vector $n$ is zero.

So we can shift the curl operator from right to left like this:

$\begin{array}{ccl} - \int_M \langle v_x, \nabla \times (u_x \times w_x) \rangle \;d \mu(x) &=& - \int_M \langle \nabla \times v_x, u_x \times w_x \rangle \;d \mu(x) \\ &=& - \int_M \langle (\nabla \times v_x) \times u_x, w_x \rangle \;d \mu(x) \end{array}$

In the last step we used the cyclicity of the relation of the vector product and the volume spanned by three vectors:

$\langle a \times b, c \rangle = \mu(a, b, c) = \mu (c, a, b) = \langle c \times a, b \rangle$

This verifies the claim, since the part $\nabla f$ does not contribute, as stated above.

And now, yet another vector identity comes to our rescue:

$(\nabla \times v) \times u = (u \cdot \nabla) v - u_k \nabla v_k$

So, we finally end up with

$\begin{array}{ccl} \mathrm{ad}^*_u u &=& - (u \cdot \nabla) u - u_k \nabla u_k + \nabla f \\ &=& - (u \cdot \nabla) u + \nabla g \end{array}$

for some function $g$. Why? Since the middle term $u_k \nabla u_k = \frac{1}{2} \nabla u^2$ is actually a gradient, we can absorb this summand and $\nabla f$ into one summand with a new function, $\nabla g$.

Thanks to this formula we derived, the abstract and elegant equation for a geodesic on any Lie group

$\partial_t u = ad^*_u u$

becomes, in this special case

$\partial_t u = - (u \cdot \nabla) u + \nabla g$

If we can convince ourselves that $-g$ is the pressure $p$ of our fluid, we get Euler’s equation:

$\partial_t u + (u \cdot \nabla) u = - \nabla p$

Wow! Starting with abstract stuff about infinite-dimensional Lie groups, we’ve almost managed to derive Euler’s equation as the geodesic equation on $\mathrm{SDiff}(M)$! We’re not quite done: we still need to talk about the role of the function $g$, and why it’s the pressure. But that will have to wait for another post.