2013 REU Lecture notes

Sasha Kiselev

1. The passive scalar equation.

Let us start with the simplest partial differential equation (PDE) in the world. Let θ(x, t) be

function of time and space. Then the following simple PDE is called transport equation.

∂tθ(x, t) + a · ∇θ(x, t) = 0,

where a ∈ Rd is a vector. This an evolution equation that comes with initial data θ(x, 0) = θ0(x).

How to solve it? We can think of this equation as stating that a directional derivative of θ in direction

(1, a) vanishes. Hence θ should be constant along this direction. Indeed, set z(s) = θ(x+ as, t+ s).

Then z′(s) = a∇θ + ∂tθ = 0. Hence indeed, θ is constant along every such line. Now given some

point (x, t), take s = 0 and s = −t. This gives θ(x, t) = θ0(x− at), solving the equation at least if θ0is sufficiently smooth - has continuous derivatives in all directions justifying our manipulations. The

solution tells us that under evolution the initial data is just transported with speed a.

Now let u(x, t) be a vector field in Rd which may depend on both space and time. Consider the

equation

∂tθ(x, t) + u(x, t) · ∇θ(x, t) = 0. (1)

By analogy with our previous example, we can think of a scalar function θ transported by vector field

u(x, t) - except that now the velocity is not constant. Examples would include cream you add to your

coffee and then mix it, or transport of vapor in atmosphere, or plant pollen transported by wind.

The term ”passive” in the name of the equation stresses that u is given to us and does not depend

on the function θ. This equation may be used to describe evolution of the density of some compound

in ambient fluid flow described by u(x, t). Here are some simple examples of common model fluid

flows. Shear flow u(x, y) = (f(y), 0) models directed transport in one direction with varying velocity

profile. Water flow in pipes or channels look like this before turbulence takes over. Cellular flow

u(x, y) = (− sin y, sinx) describes rotational motion. A flow like that goes on in your pot when you

cook unless boiling starts and again turbulence takes over. These are sample flows defined on two

dimensions and do not depend on time (such flows are called stationary). Draw these flows!

To solve (1), consider a curve Φ(x, t) defined by

d

dtΦ(x, t) = u(Φ(x, t), t), Φ(x, 0) = x. (2)

In dimension d, the above equation is a system of d first order ordinary differential equations where

x is just a parameter.

Let us recall a particular version of the existence and uniqueness result for solutions of such systems.1

2

Theorem 1. Let y ∈ Rd, and suppose that a function f(y, t) is globally bounded, continuous and

Lipschitz in the first variable: |f(y, t)| ≤ M for every y, t and |f(y1, t) − f(y2, t)| ≤ L|y1 − y2| forevery y1, y2, t. Then the system of the first order differential equations given by

y′ = f(y, t), y(0) = y0 (3)

has a unique solution in C1((0,∞],Rd). The latter notation means that the solution takes values in

Rd and has first order derivative in time that is continuous and bounded.

Proof. I assume that you have seen a proof of this or similar result and so I will only sketch the

argument. Please feel in the gaps; you can also find more detailed presentations on internet. The

system (3) can be rewritten in the equivalent integral form

y(t) = y0 +

∫ t

0

f(y(s), s) ds. (4)

To show existence of solution, proceed with Picard iteration defining

yk(t) = y0 +

∫ t

0

f(yk−1(s), s) ds, k = 1, . . . (5)

Observe that

|y1(t)− y0| ≤Mt.

Inductively,

|yk(t)− yk−1(t)| ≤∫ t

0

(f(yk−1(s), s)− f(yk−2(s), s)) ds ≤ L

∫ t

0

|yk−1(s)− yk−2(s)| ds ≤MLk−1tk

k!.

It follows that the series

y0 +∞∑k=1

(yk − yk−1)

converges absolutely and uniformly on any finite time interval [0, T ]. Denote this limit Y (t). By

definition of iterates (5) and continuity of f, Y (t) solves the integral equation (4). It follows then

that Y (t) is C1 in time.

For uniqueness, suppose that we have two solutions Y1(t) and Y2(t) and consider their difference

W (t). Consider f(t) = |W (t)|2. Then f(0) = 0, and

f ′(t) = 2|⟨W (t),W ′(t)⟩| ≤ |⟨Y1(t)− Y2(t), f(Y1(t), t)− f(Y2(t), t)⟩| ≤ L|W (t)|2 = Lf(t).

Here ⟨·, ·⟩ denotes inner product in Rd. By Gronwall lemma which we state below, W (t) ≡ 0.

Proposition 2 (Gronwall Lemma). Let I = (t0, t1) be an interval. Suppose that g(t), h(t) are

continuous and f ∈ C1(I) satisfies

f ′(t) ≤ a(t)f(t) + b(t) for t ∈ I and f(t0) = f0.

Then for all t ∈ I,

f(t) ≤ f0e∫ tt0

a(s) ds+

∫ t

t0

e∫ ts a(r) drb(s) ds.

3

Proof. Exercise.

Coming back to passive scalar we see that if we require that u is globally bounded and sufficiently

smooth (Lipschitz would do), then solutions of the equation (2) for each given initial condition x exist

globally and are unique. These solutions are called characteristic curves or simply characteristics.

Now consider that

d

dtθ(Φ(x, t), t) = ∂tθ +∇θ · d

dtΦ = (∂tθ + u · ∇θ)|(Φ(x,t),t) = 0.

This implies that θ(Φ(x, t), t) = θ0(x) for all t ≥ 0. Observe that under our assumptions on u, the

map x 7→ Φ(x, t) is invertible for each t. Indeed, if we had Φ(x2, t) = Φ(x1, t) for some t and x1 = x2,

it would contradict the uniqueness of solutions to systems of ODE theorem, as we can look at our

characteristic equation back in time and find two distinct solutions corresponding to the same value

at time t. Therefore we can hope to solve the equation by using characteristics, and solution can be

explicit if we could explicitly invert the map Φ(x, t) for each t : then θ(x, t) = θ0(Φ−1(x, t)).

Exercise. Perform these computations for some simple examples of flows, such as shear flows.

We can say more by using the inverse function theorem.

Theorem 3 (Inverse Function Theorem). Let F : Rd 7→ Rd be a continuously differentiable mapping.

If the derivative of F is invertible at x = p, then the function F is invertible in some neighborhood

of the point F (p), and its inverse F−1 is a continuously differentiable function. If F possesses higher

regularity (say n continuous derivatives), then so does F−1.

Recall that the derivative DF of the mapping F is a matrix with elements (DF )jk(x) = ∂jFk(x),

and its invertibility is equivalent to the Jacobian J(x, t) = det(DF (x, t)) not being equal to zero.

Now let us compute the evolution of the Jacobian of the map Φ(x, t).

Proposition 4. Let J(t, x) = det(DΦ(x, t)). Then

J ′(t, x) = (∇ · u)(Φ(x, t), t)J(t, x). (6)

Proof. Let us compute

J ′(t, x) =

∣∣∣∣∣∣∂∂t

∂Φ1

∂x1. . . ∂Φn

∂x1

. . . . . . . . .∂∂t

∂Φ1

∂xn. . . ∂Φn

∂xn

∣∣∣∣∣∣+∣∣∣∣∣∣

∂Φ1

∂x1

∂∂t

∂Φ2

∂x1. . . ∂Φn

∂x1

. . . . . . . . . . . .∂Φ1

∂xn

∂∂t

∂Φ2

∂xn. . . ∂Φn

∂xn

∣∣∣∣∣∣+ · · ·+

∣∣∣∣∣∣∂Φ1

∂x1. . . ∂

∂t∂Φn

∂x1

. . . . . . . . .∂Φ1

∂xn. . . ∂

∂t∂Φn

∂xn

∣∣∣∣∣∣ . (7)

Observe that∂

∂t

∂Φ1

∂xj=

∂

∂xju1(Φ(x, t), t) = ∂ku1(Φ(x, t), t)

∂Φk

∂xj,

where using Einstein’s convention we sum in the repeated index k. So the the first column in the first

determinant on the right hand side of (7) is a sum of k = 1, . . . , d columns. Each column except the

one corresponding to k = 1 is a multiple of a another column of the determinant. Using properties

of determinants, we see that the first term is equal to ∂1u1(Φ(x, t), t)J(t, x). A similar argument can

be used for all other terms on the right hand side of (7), leading to the result of the proposition.

4

Since J(x, 0) = 1, we have that the Jacobian of Φ(x, t) never vanishes provided that the divergence

of u is bounded. Hence by inverse function theorem Φ−1(x, t) is continuously differentiable.

Summarizing the discussion, let us state

Theorem 5. Suppose that the vector field u(x, t) is C1(Rd,R). Then the passive scalar equation (1)

has a unique C1 solution θ(x, t) = θ0(Φ−1(x, t)). This solution can be shown to be more regular if the

initial data and the vector field u are more than C1 in spacial coordinates.

An important class of vector fields u is given by incompressible flows.

Definition 6. The vector field u is called incompressible if the map Φ(x, t) defined by characteristics

is volume preserving for all t : for every Ω ∈ Rd, we have vol(Φ(Ω, t)) = vol(Ω).

Many fluid flows around us satisfy this property. For example, water can be regarded as incom-

pressible for most phenomena in nature. Air can be regarded as incompressible for most atmospheric

processes, but not for modeling flows around airplanes.

Proposition 4 allows us to describe all incompressible flows in simple terms if we also recall the

change of variable formula in multidimensional integrals. Let Ω be domain in Rd, ϕ : Rd 7→ Rd a C1

mapping, and f a continuous function defined on ϕ(Ω). Then∫ϕ(Ω)

f(y) dy =

∫Ω

f(ϕ(x))| detDϕ(x)| dx. (8)

Given this reminder, we can now state

Theorem 7. The flow defined by u is incompressible if an only if u is divergence free for all times:

∇ · u(x, t) = 0 for all x, t.

Proof. In view of (6), if u is divergence free for all times, then J(x, t) ≡ 1. On the other hand,

apply the change of variable formula (8) to the case of ϕ(x) = Φ(x, t) and f ≡ 1. We get

vol(Φ(Ω, t)) =

∫Φ(Ω,t)

dy =

∫Ω

| det(DΦ(x, t)| dx =

∫Ω

dx = vol(Ω).

The converse is left as an exercise.

The solutions of passive scalar equation with incompressible u obey a number of special properties.

The initial data for passive scalar equation is often taken to be decaying or periodic. In the latter

case, we are effectively working not in Rd, but on the torus Td.

Corollary 8. Suppose that u is incompressible and θ(x, t) is the solution of passive scalar equation

∂tθ+ u · ∇θ = 0 with smooth decaying (or periodic) initial data θ0(x). Then for every p > 0, we have∫|θ(x, t)|p dx =

∫|θ0(x)|p dx

for all t. The integral is taken over all Rd in the case of decaying and over the period cell in the case

of the periodic initial data.

5

Proof. For the incompressible flow, the characteristic map Φ(x, t) conserves volume, and so

detDΦ(x, t) ≡ 1. Thus by change of variable formula and expression for the solution of passive

scalar,∫Φ(Rd,t)

|θ(y, t)|p dy =

∫Rd

|θ(Φ(x, t), t)|p| detDΦ(x, t)| dx =

∫Rd

|θ(Φ(x, t), t)|p dx =

∫Rd

|θ0(x)|p dx.

In fact more is true; we can replace |θ|p with any function f(θ). The way to think about it is that

dynamics just moves around the initial data, without distorting volumes. So all integral measures of

size of the initial data will be conserved by the solution. The maximum or minimum of the initial

data will also be conserved; they are just moved around by the flow.

2. Burgers equation: the simplest active scalar

The passive scalar equation is linear: if θ1 and θ2 are its solutions, so is θ1 + θ2 (with initial data

equal to the sum of the original ones). In applications one often encounters situations where the

function u(x, t) is not some given function, but depends on the solution θ itself. This makes the

equation nonlinear, and equations of this class are called active scalars. Here the scalar function θ

is not just moved around by a given vector field u, but it actively participates in forming this vector

field. The simplest equation of this class is Burgers equation, which appears in gas dynamics and as

a basic model of traffic flow.

Consider a highway, which we will model as just a line. Denote by ρ(x, t) the density of cars, and

by u(x, t) the speed of their motion. Consider the number of cars in some interval I = (a, b) at time

t - this quantity is given by∫Iρ(x, t) dx. Let us advance by a small time interval δt. The number of

cars that enters I from the left is approximately equal to ρ(a, t)u(a, t)δt; any correction to that is

of higher order δt2. Similarly, the number of cars exiting I at the right end point is approximately

ρ(b, t)u(b, t)δt. Therefore we can write that∫I

(ρ(x, t+ δt)− ρ(x, t)) dx = ρ(a, t)u(a, t)δt− ρ(b, t)u(b, t)δt+O((δt)2).

The last O correction is a standard notation for terms that are bounded from above by some constant

times (δt)2. Dividing by δt and passing to the limit δt→ 0 we get∫I

∂tρ(x, t) dx = ρ(a, t)u(a, t)− ρ(b, t)u(b, t).

Dividing by b− a and passing to the limit again, we obtain

∂tρ+ ∂x(uρ) = 0.

Of course you can object to these limiting procedures since cars are discrete and kind of big objects.

But it is a question of relative scale; to a Martian observing New Jersey highway from Mars cars

look truly infinitesimal. Moreover, we are just building a model. By virtue of magic (behind which

is more math actually), many discrete problems in nature are modeled well by continuous calculus.

Now to make the equation complete we need to somehow specify u. It is not unreasonable to assume

that u is determined by the density ρ. At low density, the speed is maximal. As freeway becomes

6

crowded it slows down. The simplest relationship to capture this effect is u = 1 − ρ, 0 ≤ ρ ≤ 1,

measuring speed in units of maximal speed. This gives us equation

∂tρt + (1− 2ρ)∂xρ = 0, ρ(x, 0) = ρ0(x).

Setting θ = 1− 2ρ, we arrive at a more common form of this equation called Burgers equation

∂tθ + θ∂xθ = 0, θ(x, 0) = θ0(x). (9)

This equation was first introduced in the early twentieth century, and still appears in current research

(well, with a few other terms added).

Let us try to solve this equation following the ideas we developed for passive scalar equation. The

characteristic curves now should be defined by ddtΦ(x, t) = θ(Φ(x, t), t), which does not seem very

useful since θ is itself unknown. However, observe that if θ is a solution of (9), it should be constant

on such characteristic curves:d

dtθ(Φ(x, t), t) = θt(Φ(x, t), t) + θx(Φ(x, t), t)

d

dtΦ(x, t) = θt(Φ(x, t), t) + θ(Φ(x, t), t)θx(Φ(x, t), t) = 0.

So characteristic curves should be just lines, and θ should be constant on these lines. The slope

of these lines is determined by initial data. Specifically, define characteristic line starting at x by

Φ(x, t) = x + θ0(x)t. Define solution θ implicity by θ(x + θ0(x)t, t) = θ0(x). We can now find the

solution θ(y, t) if we can solve for x from y = x+ θ0(x)t. The picture one can have in mind is finding

a characteristic line, and in particular its origin at time zero, which will arrive at point y at time

t. By the inverse function theorem, we can solve for x in terms of y if the derivative 1 + tθ′0(x)

of the right hand side never vanishes. This will happen for all sufficiently small t provided that

infRθ′0(x) = M > −∞. In fact, we can find the inverse for all t < 1/M if M is negative and for all t

if M ≥ 0. Then we set θ(y, t) = θ0(x(y, t)). Using the equation

y = x(y, t) + θ0(x(y, t))t (10)

one can then check by direct computation that such solution θ(y, t) indeed solves the Burgers equation

∂tθ + θ∂yθ = 0. Let us summarize the result.

Theorem 9. Let θ0 be smooth (basically, at least C1) initial data. If θ′0(x) ≥ 0, then for all t ≥ 0

there exists a smooth solution θ(y, t) of Burgers equation given by θ(y, t) = θ0(x(y, t)) where x(y, t) is

determined from (10). If θ′0(x) turns negative somewhere, denote M = |infθ′0(x)|. Then for t < 1/M,

there exists a unique smooth solution of the Burgers equation determined by the same formula as

before.

The only part of theorem that we need to comment on is uniqueness. But we noted that any smooth

solution of Burgers equation should stay constant on characteristic lines and so must coincide with

the solution defined in Theorem 9.

The constraint on time of existence of smooth solutions is a new feature compared to the passive

scalar equation, and it is a nonlinear phenomenon. Such constraint is not a technical artifact -

solutions of Burgers equation do become singular as the derivative of θ(x, t) becomes infinite at some

critical time. Indeed, characteristic lines have different slopes and will in general cross at some time.

At this time solution develops an infinite derivative or even a jump in value, also called shock. Such

7

phenomenon is called ”finite time blow up” of a solution. Generally, singularities of solutions to

different PDE are often of great interest. These singularities may be telling us about interesting

dramatic phenomena (such as traffic jam) or may be warning us about limitations of the model. For

the Burgers equation, we note that the estimate of existence time given in Theorem is in fact sharp,

and finite time blow up of solutions to Burgers equation does happen for wide classes of initial data.

In fact, one can continue these solutions past singular time, and make sense of solutions which are

not regular. In the process, interesting things happen, and in particular uniqueness of solutions may

be lost. But we will not pursue this line of thought here. Instead, let us state a result describing

finite time blow up for a useful and broad class of initial data θ0(x) - periodic data.

Theorem 10. Let θ0(x) be smooth and periodic. Then smooth solution of Burgers equation with

initial data θ0(x) exists for all time if and only if θ0(x) is constant. Otherwise the maximal time till

which smooth solution exists is given by tB = 1/M, where as before M = |minxθ′0(x)|.

Proof. Note that since θ0 is periodic then, unless it is constant, its derivative has to turn negative

somewhere. Hence according to our earlier considerations we should expect blow up. Indeed, suppose

that θ0(x) is not constant. Then we can find x1 < x2 such that θ(x1) > θ(x2). the characteristic lines

corresponding to x1 and x2 cross if

x1 + θ0(x1)t = x2 + θ(x2)t, or t =x2 − x1

θ(x1)− θ(x2).

But a smooth solution has to stay constant on characteristic line, which is impossible if lines cor-

responding to different values cross. Hence smooth solution cannot exist past such crossing time.

Therefore smooth solution can only persist for

tB ≤ infx2>x1,θ(x1)>θ(x2)x2 − x1

θ(x1)− θ(x2).

Now the expression on the right hand side is exactly 1/M, by definition of derivative and by mean

value theorem.

Let us outline an alternative method of proof of finite time blow up for some solutions of Burgers

equation. It is more crude and gives much less detail than the explicit method of characteristics. But

it has an advantage of being more easily extendable to more complicated equations, with appropriate

modifications. This method is often called Lyapunov functional method. The idea is to find a quantity

which should stay finite if solution is smooth, and then show that it in fact does not stay finite. This

is usually achieved by showing monotonicity of this quantity by deriving some sort of differential

inequality for it.

Consider smooth periodic and odd initial data θ0(x). Suppose the period is equal to 2, that θ0(x) <

0 for x ∈ (0, 1), and θ0(x) is also odd with respect to all integer x = n. That is, for every n ∈ Z we

have θ(n + x) = −θ(n − x). For example, sin πx satisfies this property. We first show that periodic

and odd symmetries are conserved by solutions of the equation (the periodic part we actually took

for granted so far).

Lemma 11. Suppose that the initial data θ0 is smooth, periodic and odd. Then the smooth solution

of Burgers equation corresponding to this initial data remains periodic and odd while it exists.

8

Proof. Consider the oddness. Define η(x, t) = −θ(−x, t). Then ηt(x, t) = −θt(−x, t), and ηx(x, t) =θx(−x, t). Thus η(x, t) satisfies the same Burgers equation as θ(x, t). Moreover, since θ0 is odd, initial

data for η and θ are the same. By uniqueness of smooth solutions, θ = η, and so θ is odd for all

times, at least while it remains smooth. Periodicity can be established in a similar way.

Now let us define our Lyapunov functional

L(t) = −∫ 1

0

θ(x)

xdx.

If θ is smooth and odd, the integral is well defined since θ(x)/x ≤ C. Also L(0) > 0 by our assumptions

on θ0. Suppose that θ(x, t) remains smooth for all times. Consider

L′(t) = −∫ 1

0

∂tθ(x, t)

xdx =

∫ 1

0

θ(x, t)θx(x, t)

xdx.

Let us integrate by parts, using the fact that θθx = ∂x(θ)2/2. We obtain

L′(t) =

∫ 1

0

θ(x, t)2

x2dx.

The off integral terms vanish if θ(x, t) is smooth and satisfies our oddness assumptions. But now

recall Cauchy-Schwartz inequality which states that∣∣∣∣∫ fg dx

∣∣∣∣ ≤ (∫f 2 dx

)1/2(∫g2 dx

)1/2

.

Applying this to L, we see that∣∣∣∣∫ 1

0

θ(x)

xdx

∣∣∣∣ ≤ (∫ 1

0

θ(x, t)2

x2dx

)1/2

.

The second factor that should appear on the right hand side in our case is just a square root of∫ 1

0dx

and is equal to one. It follows that

L′(t) =

∫ 1

0

θ(x, t)2

x2dx ≥ L(t)2.

This is a differential inequality which implies that L(t) becomes infinite in finite time, contradicting

smoothness of the solution. Indeed, we have that L′(t)/L(t)2 ≥ 1. Integrating this from s = 0 to

s = t we get ∫ t

0

L′(s)/L(s)2 ds = 1/L(0)− 1/L(t) ≥ t.

This leads to L(t) ≥ L(0)/(1− L(0)t), and hence finite time blow up.

9

3. Classical equations of fluid mechanics

Our first section discussed passive scalar equation, which was linear. For nice initial data, the

unique solution exists and remains smooth for all times. We reduced the solution to essentially solving

a system of ordinary differential equations, something that is reasonably easy to do numerically. In

the second section, we discussed Burgers equation, the simplest active scalar. This equation is

nonlinear, and we saw that even starting from smooth initial data, the solution can lead to shock

in finite time. Still, the equation can by analyzed in fair detail using the method of characteristics.

The classical equations of fluid mechanics add one more twist in the game: non-locality. These are

nonlinear and nonlocal equations and this makes analysis an order of magnitude harder. What does

nonlocal mean? For the Burgers equation, all terms are evaluated at the same spacial point x. For

most equations of fluid dynamics, equation at a given point has to take into account what happens

everywhere else in the domain. Such equations are much harder to compute and to understand.

The most classical equation of fluid mechanics is Euler’s equation describing motion of incompress-

ible, ideal fluid. Its discovery goes back to 1757 when it was derived by Leonard Euler. Consider

fluid particle moving with the flow, tracing a curve x(t). Then x′(t) = u(x(t), t) is the fluid flow

velocity. The acceleration of the particle is then given by a(t) = x′′(t) = ddtu(x(t), t). Using chin rule

we obtain

aj(t) = ∂tu(x(t), t) + ∂iuj(x(t), t) · x′i(t) = ut + ui∂iuj,

where by usual convention we are summing in repeated indexes. A common notation for the last

term on the right hand side is also (u · ∇)u.

By Newton’s second law, the acceleration should be proportional to force acting on the small

volume of fluid we trace. Here comes a big assumption: we call fluid ideal if there exists a scalar

function, called pressure p(x, t), with the following properties. Given any region Ω within the fluid,

the force exerted by the surrounding fluid at each point x ∈ ∂Ω is equal to −p(x, t)n, where n is

the unit outside normal to Ω at point x. In this way, the overall force F acting on Ω is equal to

F = −∫∂Ωp(x, t)n(x)dσ(x), where σ is the surface measure on ∂Ω. The assumption that fluid is ideal

means, in particular, that there is no friction or viscosity in the fluid.

Now recall divergence theorem, which is just a version of fundamental theorem of calculus in higher

dimensions. Given a vector field G(x) and a domain Ω, we have∫∂Ω

(G · n) dσ(x) =∫Ω

∇ ·Gdx.

To apply this to our formula for F , dot it with any fixed vector e. We get

F · e = −∫∂Ω

n(x) · (ep(x, t))dσ(x) = −∫Ω

∇ · (ep(x, t)) dx = −∫Ω

e · ∇p(x, t) dx.

Since e is arbitrary, we see that

F = −∫Ω

∇p(x, t) dx.

Incompressibility means that the density of fluid is constant in both space and time, and we can set

it equal to one without loss of generality. Then the second Newton’s law for our small volume of

10

fluid looks as follows: ∫Ω

(ut + (u · ∇)u) dx = −∫Ω

∇p(x, t) dx.

Given that the domain Ω is arbitrary, we arrive at Euler equation

ut + (u · ∇)u = −∇p, ∇ · u = 0, u(x, 0) = u0(x). (11)

The ∇ · u = 0 condition expresses incompressibility. The equation is valid in any dimension, but the

relevant cases are d = 2 and d = 3. The initial data is often taken decaying in Rd or periodic. It

should be noted that the assumption of ideal flow works well for flows with very small viscosity, like

air or water, at moderate speeds of motion. But Euler equation misses some important effects. A

paradox discovered by d’Alambert in 18th century is that ideal fluids cannot provide lift - airplanes

would not be able to fly in ideal fluid. A related classical equation of fluid mechanics is the Navier-

Stokes equation. It is almost a hundred years younger than Euler equation, and takes into account

effects of viscosity and dissipation. The equation differs from (11) by addition of ∆u term, similar

to the one you have seen in heat equation:

ut + (u · ∇)u− ν∆u = −∇p, ∇ · u = 0, u(x, 0) = u0(x). (12)

Euler and Navier-Stokes equations are really systems of d + 1 equations for d + 1 functions - d

components of velocity and pressure. This does look much more complicated than Burgers or passive

scalar, so how can we hope to even make some qualitative bridge here? And also, how does one study

solutions of such complex systems, how to even show solutions exist for short time? These questions

go beyond our scope here, but we will at least provide very rough big picture.

First of all, conservation laws and symmetries are important for understanding PDEs. The main

conservation law we have for Euler and Navier-Stokes equations is energy:∫|u|2 dx decays (for

Navier-Stokes) or stays constant (for Euler) for smooth solutions. This can be checked from the

equation by integration by parts.

∂t

∫|u|2 dx = 2

∫u · ut dx = −2

∫uj(ui∂i)uj dx+ ν

∫uj∂

2i uj =

−∫ui∂i(uj)

2 dx− ν

∫(∂iuj)

2 dx = −∫(∂iui)u

2j dx− ν

∫(∂iuj)

2 dx = −ν∫

|∇u|2 dx.

Here in the last step we used that u is divergence free. We also ignored boundary terms when

integrating by parts, which is justified in periodic or decaying case. There are other conserved

quantities for Euler equation, but energy is the most useful one for proving existence of solutions since

it controls ”the size” of the solution. You see that in the Navier-Stokes case the energy is decaying,

and this corresponds well to the intuition that friction and viscosity should dissipate energy. One

can also expect that it will be easier to control solutions in the Naver-Stokes case compared to Euler.

From here it turns out that the story differs in two and three dimensions. In two dimensions

one can find additional useful quantities that are conserved and one can bring Euler equation to a

form where it will look almost like equations we considered earlier. A useful quantity to introduce is

11

vorticity ω, defined by ω = curlu. Recall that

curlu =

(∂u3∂x2

− ∂u2∂x3

,∂u1∂x3

− ∂u3∂x1

,∂u2∂x1

− ∂u1∂x2

). (13)

You may also recall a formal determinant formula for curl :

curlu =

∣∣∣∣∣∣i j k

∂1 ∂2 ∂3u1 u2 u3

∣∣∣∣∣∣Another convenient way to express the curl operator is through third order anti-symmetric tensor

ϵijk. It is defined as follows. ϵ123 = 1; more generally, if i, j, k is some permutation of 1, 2, 3 then

ϵijk = (−1)σ, where σ is the order of the permutation. For example, ϵ321 = −1 since it takes

three neighbor switches to arrive to 123 from 321. If values of any two indexes are the same, the

tensor is zero. If we switch values of any nearby indexes, the tensor changes sign: for example

ϵijk = −ϵjik = ϵjki. Now one can check that (curlu)i = ϵijk∂xjuk. Also,

ϵijkajbk = a× b =

∣∣∣∣∣∣i j k

a1 a2 a3b1 b2 b3

∣∣∣∣∣∣is another expression for the cross product of vectors a and b.

Vorticity describes rotational motion in flows. To gain some intuition, let us pause for a little and

look at some examples. Draw all these flows!

12

Example 1. Strain and jet flows. The simplest two dimensional strain flow is given by u(x, t) =

(−γx1, γx2), where γ is a parameter. The flow is stationary (does not depend on time) and divergence

free. You can check that

(u · ∇)u = (−γx1∂x1 + γx2∂x2)(−γx1i+ γx2j) = γ2(x1i+ x2j),

and so Euler equation holds with pressure p(x) = γ2|x|2/2. Jet flow is a similar flow in three dimen-

sions, given by u(x, t) = (−γ1x1,−γ2x2, (γ1 + γ2)x3). Note that vorticity of the strain flow is zero:

ω = curlu = (∂1u2 − ∂2u1)k = 0. The same can be checked for jet flows.

Example 2. Vortex flow. Let us take u(x, t) = (−βx2, βx1). Note that the streamlines - curves

such that the flow is tangent to them at every point - are just circles. The vorticity ω for this flow

is equal to 2β.

To see connection between vorticity and rotational motion more clearly, consider an arbitrary point

(x0, t) and Taylor expansion of fluid velocity near this point:

u(x0 + h, t) = u(x0, t) +∇u(x0, t)h+O(h2), h ∈ R3,

where ∇u is a matrix ∂xjui. The matrix ∇u has a symmetric part D and anti-symmetric part Ω,

given by

D =1

2

(∇u+ (∇u)t

), Ω =

1

2

(∇u− (∇u)t

).

The symmetric matrix D is called deformation or rate of strain matrix. One can diagonalize it, and

so in the right basis the matrix D acts by stretching some directions and compressing others. One

can check that incompressibility of u is equivalent to the fact that sum of the eigenvalues of D is

equal to zero, and stretching in some directions is compensated by compression in others so that the

volume does not change.

The action of the anti-symmetric part can be linked with vorticity. Namely, we have that

Ωh =1

2ω × h, (14)

where × denotes the cross product. To see the relation (14) more clearly, let us write

(ω × h)i = ϵijkϵjlm∂xlumhk.

Here is a useful identity for summing the expressions involving ϵ : ϵijkϵilm = δjlδkm − δjmδlk. Here

δij = 1 if i = j and 0 otherwise. Please verify it - you can do it by explicit calculation, or by a

quicker argument taking advantage of properties of ϵ. Using this identity, we compute

ϵijkϵjlm∂xlumhk = −ϵjikϵjlm∂xl

umhk = (δimδkl − δilδkm)∂xlumhk = ∂xk

uihk − ∂xiukhk = 2Ωikhk.

To see how this describes rotational component of motion, choose for simplicity a basis in which

vorticity is equal to (0, 0, ω). Then

ω × (x− x0) =

0 −ω 0

ω 0 0

0 0 0

(x− x0).

13

The equations for the particle trajectories corresponding to this kind of motion are given by

x′1(t) = −ωx2(t), x′2(t) = ωx1(t), x1(0) = A, x2(0) = B.

Observing that in this case x′′1(t) + ω2x1 = 0, and solving the system, we get x1(t) = A cos(ωt) −B sin(ωt), x2(t) = B cos(ωt) + A sin(ωt). This is exactly rotational motion with angular velocity ω.

Thus, at each point vorticity describes infinitesemal rotational component of fluid motion near that

point.

Rotational motion is ubiquitous in fluids. Just think of swirling water going down the pipe in

bathroom, or think of hurricanes and Jupiter’s red spot. In virtually any flow that is not still and

especially if it is being stirred you can observe many eddies which form spontaneously. All this

makes vorticity a natural and useful quantity to look at. Let us now derive Euler and Navier-Stokes

equations in vorticity form, by taking curl of (11) and (12). Observe that curl∇p = ϵijk∂j∂kp = 0

due to anti-symmetry of the tensor. For the nonlinearity we have

ϵljk∂j(ui∂iuk) = ui∂iϵljk∂juk + ϵljk∂jui∂iuk.

The first term on the right hand side is equal to (u ·∇)ω. A computation shows that the second term

on the right hand side can be represented as −(ω · ∇)u. Hence Navier-Stokes equation in vorticity

form is given by

∂tω + (u · ∇)ω − ν∆ω = (ω · ∇)u; (15)

Euler equation corresponds to ν = 0 case. To make the system complete, we now need to prescribe

how u can be expressed in terms of ω. Let us compute

curlω = curlcurlu = ϵijk∂jϵklm∂lum = ϵkijϵklm∂j∂lum = (δilδjm−δimδjl)∂j∂lum = ∂i∂mum−∂2l ui = −∆u.

The last equality follows from the incompressibility of u and the usual notation ∆ = ∂21 + ∂22 + ∂23 .

Hence, at least formally, we can write u = −(∆)−1curlω. For the inversion of Laplacian one can use

Green’s function.

Proposition 12. Suppose that f(x) is a smooth compactly supported (or decaying sufficiently fast)

function. Then in two dimensions

f(x) =1

2π

∫R2

(log |y|)∆f(x− y) dy (16)

and in three dimensions

f(x) =1

4π

∫R3

|y|−1∆f(x− y) dy. (17)

Similar formula is available in higher dimensions; in dimension d > 3 the kernel is cd|x− y|2−d.

Proof. Let us carry out the computation in two dimensions. First, a direct computation shows

that ∆ log |y| = 0 except at y = 0 where it is undefined. Indeed, ∂j log |y| = yj/|y|2, and ∂2j log |y| =|y|−2 − 2y2j/|y|4. Summing over j = 1, 2 we get zero. To simplify (16), we therefore would like to

integrate by parts bringing Laplacian from f onto the logarithmic function. We cannot quite do

14

that, because of singularity at zero, so we cut a small circle of radius δ around zero, and perform

integration by parts outside it according to Green’s theorem. Thus, observe first that∫R2

(log |y|)∆f(x− y) dy = limδ→0

∫Bc

δ

(log |y|)∆f(x− y) dy,

where Bδ denotes a ball (circle) of radius δ centered at zero, and Bcδ its complement. Then∫

Bcδ

(log |y|)∆f(x− y) dy =

∫∂Bc

δ

∂f

∂n(x− y) log |y| dσ(y)−

∫Bc

δ

∇(log |y|)∇f(x− y) dy =∫∂Bc

δ

(∂

∂n(f(x− y)) log |y| − f(x− y)

∂ log |y|∂n

)dσ(y) +

∫Bc

δ

(∆ log |y|)f(x− y) dy.

The last term on the right hand side is equal to zero by our earlier computation. Let us consider the

integral over the boundary, which is just a circle of radius δ around the origin, and see what happens

as we take δ to zero. First, the normal on the boundary of Bcδ points in the radial direction towards

the origin, and so derivative in normal direction is the same as derivative with respect to −|y|. Sincef is smooth, ∂

∂n(f(x− y)) is a smooth bounded function. Then∣∣∣∣∣∫∂Bc

δ

∂

∂n(f(x− y)) log |y| dσ(y)

∣∣∣∣∣ ≤ C

∫ 2π

0

δ log δ dϕ ≤ C1δ log δ → 0

as δ → 0. What remains is

−∫∂Bc

δ

f(x− y)∂ log |y|∂n

dσ(y) =

∫|y|=δ

f(x− y)|y|−1 dσ(y) =

∫ 2π

0

(f(x) +O(δ)) dϕ→ 2πf(x)

as δ → 0. This is exactly what we need to establish (16).

Now given Proposition 12, we see that we can complement (15) by u = −curl(∆)−1ω(x), where

action of (∆)−1 is given by (16) or (17). One can check that in three dimensions this leads to explicit

formula

u(x) = − 1

4π

∫R3

y × ω(x− y)

|y|3dy,

which is called Bio-Savart law. In two dimensions, u = (u1(x1, x2), u2(x1, x2), 0), and the vorticity

vector has only one component ω = (0, 0, ω3(x1, x2)). Then the formula simplifies to

u(x) = − 1

2π

∫R2

y⊥

|y|2ω(x− y) dy,

where y⊥ = (−y2, y1) (and is orthogonal to y = (y1, y2), hence the notation).

The vorticity form of Euler and Navier-Stokes equations has several interesting and useful features.

First, pressure has been eliminated from the system. Second, a key difference between three and two

dimensional cases is much more apparent in this form. Indeed, in two dimensions vorticity has only

third nonzero component. Therefore, the term (ω · ∇)u on the right hand side of (15) is equal to

ω3∂3u = 0, as u does not depend on x3. Therefore, in two dimensions the Euler equation becomes

∂tω + (u · ∇)ω = 0, ω(x, 0) = ω0(x) u(x) =1

2π

∫R2

(x− y)⊥

|x− y|2ω(y) dy. (18)

15

Thus the Euler equation in two dimensions in vorticity form is an active scalar, similar to the

Burgers equation we considered earlier. It is also an active scalar with incompressible velocity u. An

immediate consequence is that, as we discussed earlier for passive scalar, we get a whole new set of

controlled quantities such as∫|ω(x)|p dx for all p > 0. In particular, we also see that the maximum

of ω cannot grow in two dimensions. This is very useful addition to conservation of energy, and is

helpful in analysing the solutions. But compared to the Burgers equation there is one major new

effect: the law defining u from the active scalar ω is nonlocal. This is a serious difference, making

analysis of 2D Euler equation much more delicate than that of the Burgers equation. Nevertheless,

the question of global existence of smooth solutions can be answered for the 2D Euler equation.

Theorem 13. Let ω0(x) R2 7→ R be smooth, quickly decaying (or periodic) function. Then the 2D

Euler equation (18) with initial data ω0 has a unique smooth global solution ω(x, t). The rate of

growth of derivatives of ω(x, t) can be bounded from above by double exponential in time:

∥∂αω∥∞ ≤ exp(Cα expCαt).

Here ∥f(x)∥∞ = supx|f(x)|. The proof of this theorem goes beyond the scope of these notes, but

let us outline one of the key differences with the Burgers equation (for which, as we know, finite time

blow up can happen). For the Burgers equation, u = θ, the active scalar itself. For the 2D Euler

equation, u = ∇⊥(−∆)−1ω. Notice that in this case the velocity u is ”one derivative more regular”.

This extra smoothness is what stands behind the proof of global regularity of solutions for 2D Euler

equation. A similar result is true for the Navier-Stokes equation in two dimensions which, intuitively,

should be better behaved than Euler equation due to presence of dissipation.

But there is another equation which looks very similar to (13) - called surface quasi-geostrophic

(SQG) equation. It looks exactly like (18), but u is less regular:

∂tθ + (u · ∇)θ = 0, θ(x, 0) = θ0(x) u(x) =1

2π

∫R2

(x− y)⊥

|x− y|3θ(y) dy. (19)

The only difference with (18) is higher power of singularity in the definition of u. The SQG equation

appears in atmospheric science. It describes, under some assumptions and approximations, evolution

of temperature near the surface of Earth. For this equation, global regularity vs finite time blow up

question is open. Numerical simulations show that solutions of SQG equation are certainly capable

of creating sharp gradients. One can think of these as modeling one of those days when temperature

changes by 30 degrees within hours (we have such days sometimes here in Madison!). See below a

picture of simulation by Constantin, Lai, Sharma, Tseng and Wu (’11)

In three dimensions, the question of global regularity vs finite time blow up is also open both

for Euler and Navier-Stokes equation. The extra term (ω · ∇)u present in (15) is called ”vorticity

stretching” term, since now the maximum of vorticity may grow. This makes control of solutions

a much subtler enterprise. Clay Institute’s seven Millenium problems outline major and central

challenges in mathematics, and offer million dollar prize for solution of each of these problems. The

question of global regularity vs finite time blow up of solutions to Navier-Stokes equation in three

dimensions with smooth initial data is one of these seven problems. To make progress here, one likely

has either to be lucky to find new well hidden conservation laws in the equation, or to discover some

16

Summary of Results on the SQGGeneralized SQG: inviscid case

Generalized SQG: dissipat ive caseGeneralized SQG with singular velocit ies

Logarithmically supercrit ical SQGNumerical results

Figure: Contours of from t= 0 to t= 16Jiahong Wu Generalized surface quasi-geostrophic equat ions

mechanisms of dynamical nonlinearity depletion, or to construct blow up! My impression though that

most physicists tend to think there is no blow up. Models which appear simpler, like SQG equation,

may be important stepping blocks in understanding 3D Euler and Navier-Stokes equations.

With this background, I can state the first proposed project. Burgers equation is understood fairly

well. It is nonlinear, but local - and this helps. Already SQG equation is very poorly understood,

and the main difference, apart from one dimension up, is non-locality. How about trying to work at

some intermediate model to gain experience and insight? Here is a couple of candidates, which I’d

all call ”nonlocal Burgers”.

∂tθ + (θ(x+ h)± θ(x− h))θx = 0, θ(x, 0) = θ0(x), x ∈ T1. (20)

For simplicity let us look at this equation on the circle (or, which is the same, consider periodic

initial data on the line). Here h is some fixed number. Thinking in the context of modeling traffic in

which we derived Burgers equation, one can imagine that a driver surveys not only speed at a single

point, but certain distance ahead and behind. If we implement such assumption, this will lead to

equation like (20). The questions would be to build a theory for such equations parallel to that of

Burgers equation.

Project I.

17

1. Prove local existence of smooth solutions. The method of characteristics is likely to help here,

but nonlocality will make it harder.

2. Can solutions of (20) blow up in finite time?

3. If the answer to 2. is no, it would be extremely interesting: a very simple sort of nonlocality

makes equation more regular. I think it more likely that the answer is yes. In this case the next

question is: for Burgers equation, any initial data other than constant lead to blow up. Is this still

true for (20)? Or solutions may stay globally smooth for some initial data?

4. Lastly, the hardest question - given initial data, can we give some qualitative picture of the

dynamics? Can we determine if it leads to blow up in finite time? Here numerical simulations may

be useful to gain insight and suggest some answers.

4. Mixing for dissipative passive scalar

In this section we are coming back to the passive scalar equation, but with a different kind of

questions. So far we discussed issues such as existence, regularity, and blow up of solutions. This is

really the first step in studying a PDE. With more difficult equations, even this first step may be

hard, deep and subtle. But of course we are also interested in more detailed qualitative properties

of solutions. One such question is mixing. Here is a particular context that I would like to consider.

The equation

∂tθ + A(u · ∇)θ −∆θ = 0, θ(x, 0) = θ0(x) (21)

is called a dissipative passive scalar if the vector field u us just given to us. Here A is a coupling

constant, and let us assume that u is divergence free. Let us consider equation (21) for simplicity on

a unit two dimensional torus T2 (or with period one periodic initial data on R2, which is the same).

Let us first think of the case where there is no flow: A = 0. Then all we have is heat equation. Heat

equation has a tendency to bring solution everywhere to its average value. The easiest way to see

that is through the Fourier series. Given a continuous function f(x) defined on the torus with values

in C, we define

f(k) =

∫T2

f(x)e−2πik·x dx.

Here k = (k1, k2) ∈ Z2, so that the complex exponential is periodic on unit torus. The most natural

space to work with Fourier series is actually not continuous functions, but square integrable ones,

f(x) that satisfy∫T2 |f(x)|2 x. Such space of square integrable functions is denoted L2(T2, dx). This

space can be supplied with inner product

⟨f, g⟩ =∫T2

f(x)g(x) dx,

and a norm

∥f∥2 = ⟨f, f⟩ =∫T2

|f(x)|2 dx.

The norm can play a role of distance in this space, with distance between two functions f and g is

defined as ∥f − g∥. L2 is a linear space, as if f, g ∈ L2, then so is a1f + a2g for any a1, a2 ∈ C. Sincewe defined inner product, we can also define orthogonality by saying that f and g are orthogonal if

18

⟨f, g⟩ = 0. Then functions e2πik·x, k ∈ Z2, that appear in the definition of Fourier transform can be

viewed as an orthonormal basis in L2. Indeed, one can check that

⟨e2πik·x, e2πim·x⟩ =∫T2

e2πi((k1−m1)x1+(k2−m2)x2) dx = δk1m1δk2m2

where as before δij = 1 if i = j and 0 otherwise. One can also prove that this basis is complete:

there are no functions which are orthogonal to all the exponents. Then Fourier coefficients f(k) are

just coefficients in expansion of f in this orthogonal basis. A little more work is needed to set it all

up rigorously, but we will not pursue it here. Given what has been said, we can recover f from its

Fourier coefficients by summing up the basis expansion

f(x) =∑k∈Z2

f(k)e2πik·x.

Also, one has the Parseval identity

∥f∥2 =∫T2

|f(x)|2 dx =∑k∈Z2

|f(k)|2. (22)

Now let us apply Fourier transform to the heat equation on the torus ((21) with A = 0). We

obtain for each k ∈ Z2

∂tθ(k, t) + 4π2|k|2θ(k, t) = 0, θ(k, 0) = θ0(k).

This follows from integrating by parts, as∫T2

∆f(x)e−2πik·x dx =

∫T2

∇f · ∇e−2πik·x dx =

∫T2

f(x)∆e−2πik·x dx =

−∫T2

f(x)4π2|k|2e−2πik·x dx = −4π2|k|2f(k).

This equation can be solved explicitly, and we get θ(k, t) = θ0(k)e−4π2|k|2t. The only mode that is not

decaying but remains constant in time is the mean, customarily denoted by

θ ≡ θ(0, t) =

∫T2

θ(x, t) dx.

Thus using (22) we obtain

∥θ(x, t)− θ∥2 =∑k =0

|θ(k, t)|2 =∑k =0

|θ0(k)|2e−8π2|k|2t ≤∑k =0

|θ0(k)|2e−8π2t = ∥θ0(x)− θ∥2e−8π2t.

Hence heat equation indeed drives θ(x, t) towards its average, and we have an estimate on the rate

of convergence.

Let us now come back to (21) with A = 0. It is our everyday experience that fluid flow should

be helping diffusion. When we add cream to a cup of coffee, very few of us are content to wait for

diffusion to make the mixture uniform. Coffee might well get cold by that time. Usually we make

a swirl, and the mixture is uniform right away. An interesting, and much studied question is how

to see this effect mathematically, and in particular how the structure and the geometry of the flow

affect its mixing properties. I will focus on one particular notion of mixing effectiveness. Usually, the

19

regime of large A is of interest, since in this case the influence of the flow is strong. Also, in many

situations in practice diffusion is very weak - and this is equivalent to strong flow by changing the

time scale in the equation.

Definition 14. A flow u is called relaxation enhancing (RE) if for every τ, δ > 0 there exists a

stirring intensity A(τ, δ) such that if A > A(τ, δ) then ∥θ(x, τ)−θ∥2 < δ∥θ0(x)−θ∥2, for every initial

data θ0.

In other words, a flow is RE if, stirred quickly enough, it can bring the solution as close as we

want to the mean, as soon as we want. Which flows are RE? It is easier to say which flows are

not. Let us for now consider flows that do not depend on time. A good example is cellular flow

that we discussed earlier, u(x1, x2) = (− sinx2, sin x1). Recall that u = ∇⊥(−∆)−1ω, where ω is

the vorticity of the flow. The function ψ = (−∆)−1ω is called the stream function of the flow, and

then u = ∇⊥ψ = (−∂2ψ, ∂1ψ). The name stream function is motivated by the fact that the flow is

tangential to its level sets. In our cellular flow example, ω = ∂1u2−∂2u1 = cos x1+cos x2. The stream

function ψ(x) = (−∆)−1ω(x) in our case happens to be equal to vorticity, as −∆cos x1,2 = cos x1,2. In

general, on the torus, inversion of Laplacian is best done by Fourier series: if f(x) =∑

k∈Z2 f(k)e2πik·x,

then

(−∆)−1f(x) =∑k∈Z2

f(k)

4π2|k|2e2πik·x.

Note, however, that this definition makes sense only if f(0) = 0. Otherwise, we are dividing a nonzero

number by zero and this is undefined. The vorticity, however, consists of derivatives of the velocity

and is automatically mean zero. Let us now state

Proposition 15. Given a smooth divergence free velocity field u(x) defined on T2, we can find a

smooth stream function ψ(x) defined on T2 such that u = ∇⊥ψ = (∂2ψ,−∂1ψ) if and only if u is

mean zero. In this case we also have (u · ∇)ψ = 0.

Proof. Given u, define ψ = (−∆)−1(∂1u2 − ∂2u1). Then observe that

∂2ψ = (−∆)−1(∂1∂2u2 − ∂22u1) = (−∆)−1(−∂21u1 − ∂22u1) = u1

if and only if u1 is mean zero; otherwise the non zero mean is not recovered in inversion. The second

component of u can be verified similarly.

For the second statement, note that (u · ∇)ψ = u1∂1ψ + u2∂2ψ = ∂2ψ∂1ψ − ∂1ψ∂2ψ = 0.

It turns out that in two dimensions, all mean zero smooth incompressible flows are not RE.

Intuitively, the flow u should not be of much help to diffusion in mixing up the initial data coinciding

with stream function ψ(x). Indeed, (u · ∇)ψ = 0, so initially the term corresponding to u in the

equation is just zero. This simple idea is behind the proof below.

Theorem 16. Suppose that u is divergence free, smooth, mean zero, defined on T2. Then u is not

RE.

Proof. Consider the equation

∂tθ + A(u · ∇)θ −∆θ = 0,

20

and set initial data θ0(x) = ψ(x), the stream function of the flow u. Note that by definition, the stream

function is mean zero as inversion of Laplacian gives us mean zero function. Hence to show that u

is not RE, we must find τ, δ > 0 such that no matter how large A is, we have ∥θ(x, τ)∥2 > δ∥ψ∥2.Observe first that when we multiply the equation by θ(x, t) and integrate in x, we get

∂t1

2

∫T2

θ2(x, t) dx = −A∫T2

(u · ∇)(θ(x, t)2/2) dx+

∫T2

∆θ(x, t)θ(x, t) dx =

A

2

∫T2

(∇ · u)θ(x, t)2 dx−∫T2

|∇θ(x, t)|2 dx.

The first integral on the right hand side is zero since u is divergence free, so we obtain

∂t

∫T2

θ2(x, t) dx = −2

∫T2

|∇θ(x, t)|2 dx.

Integrating this identity in time from t = 0 to t = T , we obtain that∫T2

θ2(x, T ) dx−∫T2

θ2(x, 0) dx = −2

∫ T

0

∫T2

|∇θ(x, t)|2 dxdt.

It follows that that ∫ ∞

0

∫T2

|∇θ(x, t)|2 dxdt ≤ 1

2

∫T2

θ20(x) dx. (23)

Next take inner product of the equation with ψ :

∂t

∫T2

θ(x, t)ψ(x) dx = −A∫T2

(u · ∇)θ(x, t)ψ(x) dx+

∫T2

∆θ(x, t)ψ(x) dx.

Integrating by parts we get∫T2

(u · ∇)θ(x, t)ψ(x) dx = −∫T2

θ(x, t)∇ · (u(x)ψ(x)) dx =

−∫T2

θ(x, t)(∇ · u)ψ(x) dx−∫T2

θ(x, t)(u · ∇)ψ(x) dx = 0

since u is divergence free and (u · ∇)ψ = 0. Also∫T2

∆θ(x, t)ψ(x) dx = −∫T2

∇θ(x, t) · ∇ψ(x) dx,

and ∫T2

∇θ(x, t) · ∇ψ(x) dx ≤ 1

2

∫T2

(|∇θ(x, t)|2 + |∇ψ(x)|2

)dx.

Therefore we obtain∣∣∣∣∂t ∫T2

θ(x, t)ψ(x) dx

∣∣∣∣ ≤ 1

2

∫T2

(|∇θ(x, t)|2 + |∇ψ(x) dx|2

)dx. (24)

But at time t = 0 the left hand side is equal to ∥ψ∥2. Also, according to (23),

1

2

∫ ∞

0

∫T2

|∇θ(x, t)|2 dxdt ≤ ∥θ0∥2/4 = ∥ψ∥2/4.

21

For the left hand side of (24) to decay below the value of, say, 12∥ψ∥2 by time τ, we would need

1

2

∫ τ

0

∫T2

|∇θ(x, t)|2 dx+ 1

2

∫ τ

0

∫T2

|∇ψ(x)|2 dx ≥ ∥ψ∥2

2.

But the first summand does not exceed ∥ψ∥2/4. So the needed inequality can hold only if

1

2τ∥∇ψ∥2 ≥ ∥ψ∥2

4, or τ ≥ ∥ψ∥2

2∥∇ψ∥2. (25)

Hence by definition, no matter what A is,∫T2 θ(x, t)ψ(x) dx will stay above ∥ψ∥2/2 until time τ

defined on the right hand side of (25) Since∫T2

θ(x, t)ψ(x) dx ≤ ∥θ(x, t)∥∥psi(x)∥,

this implies a similar lower bound on ∥θ(x, t)∥2 for t ≤ τ and so u is not RE.

The examples of RE flows are harder to find. Such flows do exist, and can be constructed by a

fairly intricate construction, first done by Kolmogorov in 1950s. Description of this construction is

beyond the scope of these notes. The construction does not give a good idea of how common RE

flows are. What I propose as a project in this direction is a mostly computational study of the mixing

properties of various classes of flows.

Project II.

1. Implement a numerical scheme for the equation (21) with a cellular flow. Measure the decay of

deviation from the mean and its dependence on the stirring strength A. We know what to expect

here.

2. Now take a random flow, defined as follows.

u(x) =∑k∈Z2

ak(ω)e2πik·x.

Here ω is not vorticity, but is a standard notation to indicate randomness. Take ak(ω) to be equal

to some decaying factor times uniform random variable in (−1, 1). Every mathematics software

package, including Python, contains a random number generator, so such ak can be easily generated.

Compute the decay of deviation from the mean for these flows. Random flows are often used to

simulate turbulence, and turbulent flows are ubiquitous in nature. So it is very interesting to get an

insight whether such random flows might be RE.

3. Now to be honest, the turbulent flows in nature are also time dependent. So the next step would

be to carry out computation for time dependent random flows. One would again expect improvement

in mixing properties.

4. If there is interest, I could tell more about Kolmogorov’s construction of RE flows, and we can

play with those. There may be some analytical results to prove in this direction, though it is not clear

if we’ll have enough time. Numerical simulations may actually provide some intuition for proofs.

22

5. Chemotaxis and fluid flow

So far, we talked about equations modeling diffusion and fluid flow. Another important effect that

we did not talk about so far is chemotaxis. This effect is prominent in many biological processes.

Chemotaxis describes motion of a species that is able to detect some favorable chemical compound

and move towards higher concentration of this compound. Bees rely on chemotaxis to find flowers,

and basically any situation where animals rely on a sense of smell is an instance of chemotaxis as well.

The model we are going to discuss is called Keller-Segel equation, and it has been derived in late

1960s in the context of modeling behavior of certain kinds of bacteria and mold. Slime mold, when

starving, uses the following mechanism to escape unfavorable location. In normal conditions, slime

mold cells move and feed independently. Under stress, they start to exude chemical that serves as

attractant for other cells, and mold uses it to achieve high concentrations. Then part of the mold cells

form a stem, with other mold cells becoming spores sitting on top of it. From the height of the stem

the mold disperses over wider area, hopefully escaping to better locations. There are also species of

bacteria that use similar mechanisms to aggregate in certain situations. Let us denote ρ(x, t) the

density of bacteria/mold, and c(x, t) the density of attractive chemical. Then the Keller-Segel system

reads

∂tρ+ χ∇(ρ∇c)−∆ρ = 0 (26)

∂tc−∆c = ρ

The second equation is just a heat equation for the chemical c(x, t) with the source term ρ, since it is

bacteria that exude the chemical. In the first equation, only the second, chemotactic, term requires

explanation, the rest is just heat equation. In the chemotactic term, χ is just a coupling constant

describing the strength of chemical attraction. The ∇(ρ∇c) expression describes the tendency of

density ρ to move towards higher concentration of the chemical c. Indeed, take an arbitrary region

Ω, and look at ∫Ω

∇(ρ∇c) dx =

∫∂Ω

ρ∂c

∂ndσ(x),

where the equality follows from divergence theorem. Looking at the equation (26), and ignoring the

diffusion term ∆ρ for now, we see that

∂t

∫Ω

ρ dx = −∫∂Ω

ρ∂c

∂ndσ(x). (27)

The normal derivative in the divergence theorem is in the direction of outside normal. So if the

concentration of c inside Ω is higher than outside, we have ∂c/∂n < 0. This will make the right hand

side in (27) positive, and hence bacteria will be moving into Ω increasing total amount of bacteria

in Ω (which is equal to∫Ωρ(x, t) dx).

There is a common simplification of (26). On can assume that chemical diffuses very quickly,

and its concentration promptly adjusts to the stationary state satisfying −∆c = ρ, so that one can

remove the ∂tc term in the second equation of (26). This leads to c = (−∆)−1ρ. We will consider

the equation either in whole plane or in periodic setting, and then inversion of the Laplacian can be

23

carried out as we discussed in the earlier sections. This leads to a single equation

∂tρ+ χ∇(ρ∇(−∆)−1ρ)−∆ρ = 0, ρ(x, 0) = ρ0(x) ≥ 0. (28)

Let us consider this equation in the plane R2 with smooth and decaying initial data ρ0. One can

show local existence of smooth, decaying solutions to (28) by setting up fixed point argument in

an appropriate function space. One can also show that nonnegative solution remains nonnegative,

which is natural since we think of ρ as density. An interesting feature of (28) is that its solutions can

blow up in finite time, by creating infinite concentration of ρ - similarly to the phenomenon observed

for mold. Let us sketch the argument that shows that finite time blow up must happen if total mass

of mold is sufficiently large.

Theorem 17. Suppose that the initial data ρ0(x) to (28) is nonnegative, smooth, decaying, and

satisfies∫R2 ρ0(x) dx > 8πχ−1. Then solution to (28) blows up in finite time.

Proof. First, by integrating (28), observe that∫R2 ρ(x, t) dx remains constant as far as solution

stays smooth, since all other terms in the equation are derivatives and vanish when integrated over

the entire R2. This is reasonable, since we expect the total mass of bacteria/mold to stay constant.

To prove finite time blow up, we will consider an appropriate Lyapunov functional. We expect

solution to essentially collapse into a single point. Let us consider the second moment of solution,

M2(t) =∫R2 |x|2ρ(x, t) dx. The choice of such Lyapunov functional is not obvious - it is usually found

by trial and error, and may appear magical afterwards. Then we have

M ′2(t) =

∫R2

|x|2∂tρ dx = −χ∫R2

|x|2∇(ρ∇(−∆)−1ρ) dx+

∫R2

|x|2∆ρ dx.

Observe that, by Green’s formula∫R2

|x|2∆ρ dx = −∫R2

∇|x|2∇ρ dx =

∫R2

∆(|x|2)ρ dx = 4

∫R2

ρ dx.

Let us recall the formula for (−∆)−1ρ in two dimensions:

(−∆)−1ρ(x) = − 1

2π

∫R2

log |x− y|ρ(y) dy.

Then ∫R2

|x|2∇(ρ∇(−∆)−1ρ) dx = − 1

2π

∫R2

∇|x|2ρ(x)∫R2

x− y

|x− y|2ρ(y) dy dx.

Note that ∇|x|2 = 2x. Now to compute the integral on the right hand side, we apply a useful trick.

If we relabel x by y and y by x, the value of the integral does not change of course. We then replace

the integral by 1/2 of the integral as it is and 1/2 of the relabeled version. This is often called

24

”symmetrization”. We get∫R2

∫R2

ρ(x)x · x− y

|x− y|2ρ(y) dy dx =

1

2

∫R2

∫R2

x · (x− y)

|x− y|2ρ(x)ρ(y) dy dx+

1

2

∫R2

∫R2

y · (y − x)

|y − x|2ρ(y)ρ(x) dx dy =

1

2

∫R2

∫R2

x · (x− y) + y(y − x)

|x− y|2ρ(x)ρ(y) dy dx =

1

2

∫R2

∫R2

|x− y|2

|x− y|2ρ(x)ρ(y) dy dx =

1

2

(∫R2

ρ(x) dx

)2

.

Combining all these computation together, we obtain

M ′2(t) = 4

∫R2

ρ dx− χ

2π

(∫R2

ρ(x) dx

)2

. (29)

Now recall that∫R2 ρ(x, t) dx =

∫R2 ρ0(x) dx for all times. Then if

∫R2 ρ0(x) dx >

8πχ

the right hand

side of (29) is a fixed negative constant. Thus in finite time M2(t) must become negative. But

this is not possible as the integrand defining M2 is nonnegative. This is a contradiction. The only

possibility is that our solution becomes singular - which makes integrations by parts in the proof not

justified. The value of mass 8π/χ is called ”critical mass”. Any initial data with mass larger than

critical leads to finite time blow up. It has also been shown that any smaller mass leads to globally

regular solutions.

Keller-Segel equation has been extensively studied. I outlined the simplest proof of blow up, but

there are other proofs providing more information on how blow up happens and showing convergence

to δ function - in the limit finite mass gets concentrated in one point. There is one natural question

that has not been fully addressed though. In many instances in nature chemotaxis happens in

ambient fluid or gas which may be in motion. Then the question is how fluid flow may affect the

process.

Project III.

1. Consider the Keller-Segel model with additional passive advection term:

∂tρ+ (u · ∇)ρ+ χ∇(ρ∇(−∆)−1ρ)−∆ρ = 0, (30)

where u is smooth and divergence free. Build a numerical scheme simulating (30) (we can help you

with that).

2. Experiment with different examples of u, such as cellular or shear flows. The question is whether

the presence of flow can suppress blow up in certain situations. Intuitively, it seems likely that fluid

flow can disperse high concentration of ρ and delay if not prevent the blow up.

3. In some cases, try to prove analytically that presence of fluid flow prevents blow up for some

initial data with mass higher than critical. To attempt this argument you would need to learn a bit

more material than these notes provide, but not an unreasonable amount.