10

d’Alembert’s Mystery and Bernoulli’s Law

Thus, I do not see, I admit, how one can satisfactorily explain by theory the
resistance of fluids. On the contrary, it seems to me that the theory, in all
rigor, gives in many cases zero resistance; a singular paradox which I leave
to future Geometers for elucidation. (d’Alembert)

In classical hydrodynamics the motion of non-viscous fluids is chiefly dis-
cussed. For the motion of viscous fluids, we have the differential equation
(NS) whose evaluation has been well confirmed by physical observations.
As for solutions of this differential equation, we have, aside from unidimen-
sional problems like those given by Lord Rayleigh, only the ones in which
inertia of the fluid is disregarded (Stokes equations) or plays no important
role. The bi-dimensional and dimensional problems, taking viscosity and
inertia into account (NS), still await solution. (Prandtl in Motion of fluids
with very little viscosity 1904)

10.1 Introduction

We recall the Euler equations for incompressible inviscid flow:

˙u + (u

· ∇)u + ∇p = 0

in Ω

× I,

∇ · u = 0

in Ω

× I,

u

· n = g

on Γ

× I,

u(

·, 0) = u

0

in Ω.

(10.1)

We here set the volume force f = 0, and assume the non-homogeneous slip
boundary condition u

· n = g, with n the outward unit normal to Γ , and g a

given function. We assume that



Γ

g ds = 0, so that the boundary condition

is compatible with the incompressibility condition

∇ · u = 0 in Ω, in view of

the divergence theorem stating that



Ω

∇ · u dx =



Γ

u

· n ds.

74

10 d’Alembert’s Mystery and Bernoulli’s Law

It may seem natural to expect that the Euler equations model the flow of

a fluid with very small viscosity (or very high Reynold’s number). We shall
below see that this statement is largely true, if interpreted in the right way.
However, we shall now show that this idea has a serious flaw, if interpreted in
the wrong way, which in particular was done during the initial mathematical
studies of fluid mechanics by even the great masters of the 18th century.

10.2 Bernoulli, Euler, Ideal Fluids and Potential
Solutions

The Euler equations were formulated by Euler in 1755 [38] building on
Daniel Bernoulli’s Hydrodynamik from 1738 and Johann Bernoulli’s (father of
Daniel) Hydraulica (Fig. 10.1), and thus predated the NS equations by almost
100 years.

In particular, Euler derived Bernoulli’s Law for stationary incompressible

inviscid irrotational flow from the Euler equations. We shall present Euler’s
derivation below.

The initial studies by the Bernoulli’s and Euler thus concerned ideal fluids,

with the terminology suggesting that these studies would be fundamental: In
the ideal world there would be ideal fluids behaving in an ideal way. Of course,
the hope was that the newly invented Calculus would be the ideal tool to
uncover the secrets of this ideal World.

In particular, the interest focused on stationary potential solutions with

the velocity u =

∇φ, where the potential φ is a harmonic function satisfying

Laplace’s equation ∆φ = 0 in the domain of the fluid together with the
Neumann boundary condition

∂φ
∂n

= u

· n = g. Obviously, such a velocity is

both irrotational (

∇ × u = 0) and divergence free (∇ · u = 0). We show below

that if u =

∇φ with ∆φ = 0, then there is a pressure p such that (u, p)

solves the stationary Euler equations. By solving Laplace’s equation, one can
thus construct stationary irrotational solutions to the Euler equations, and
thus fluid mechanics seems to be open for exploration by Calculus, with all
its capabilities of producing harmonic functions. Unfortunately for Calculus
(but fortunately for science, since potential solutions are pretty boring), these
hopes were almost instantly ruined by the discovery that this type of ideal
fluid theoretical predictions almost always were in complete disagreement with
observations and thus had no scientific value.

10.3 d’Alembert’s Mystery

This was pointed out by d’Alembert in 1752 in his famous paradox [29], com-
paring the Calculus prediction of zero drag/lift of an inviscid potential solu-
tion, with the undeniable observations of non-zero drag/lift in nearly inviscid
fluids such as air and water. We shall present d’Alembert’s Mystery below. We
shall also propose a (new) solution to the mystery: Briefly speaking, we will

10.5 Bernoulli’s Law

75

Fig. 10.1. Daniel Bernoulli (1700–1782), his father Johann Bernoulli (1667–1748),
and Joseph-Louis Lagrange (1736–1813).

show that the zero drag/lift inviscid potential solution is not stable and instead
a turbulent approximate solution develops, which has non-zero drag/lift. Our
resolution of the mystery is different from the standard solution by Prandtl
from 1904, which claims that even the slightest viscosity changes the flow
completely due to the presence of no-slip boundary layers. We will present
more material further below to open for the reader to judge which solution
of the mystery may be closest to the truth. Of course, we do not claim that
boundary layers never influence the flow and drag/lift, but we give evidence
that for very small viscosity (very large Reynold’s number), Prandtl’s no-slip
boundary layers may not be the true reason why non-zero drag/lift develops.

10.4 A Vector Calculus Identity

In our study of potential solutions we shall use the following identity, which
may be verified by direct computation: If u = (u

1

, u

2

, u

3

) is differentiable,

then

1
2

∇|u|

2

= (u

· ∇)u + u × (∇ × u).

(10.2)

10.5 Bernoulli’s Law

Let now u =

∇φ with ∆φ = 0 in the domain Ω of the fluid together with the

Neumann boundary condition

∂φ
∂n

= u

· n = g on Γ . Let us next define the

pressure p by the equation

1
2

|u|

2

+ p = C

in Ω,

(10.3)

where C is a constant. Taking the gradient of both sides of (10.3) and using
(10.2) recalling that

∇ × u = 0, we find that (u, p) satisfies

76

10 d’Alembert’s Mystery and Bernoulli’s Law

(u

· ∇)u + ∇p = 0, ∇ · u = 0 in Ω,

(10.4)

and we thus refer to (u, p) as a stationary potential solution to the Euler
equations with f = 0, noting that by construction u

· n = g on Γ . We have

already remarked that a potential velocity is irrotational. We note further that
the pressure in the Euler equations (with velocity boundary conditions) is only
determined up to a constant, which explains the presence of the (arbitrary)
constant C in Bernoulli’s Law.

Conversely, we see that a stationary irrotational solution (u, p) of the Euler

equations (10.1), also satisfies (10.3), which we refer to as Bernoulli’s Law. We
thus see that for stationary potential solutions, the stationary Euler equations
and Bernoulli’s Law boil down to the same thing, all according to Euler.
The same conclusion was reached by Lagrange (Fig. 10.1), who showed that
Bernoulli’s Law is the exact differential of the Euler equations.

From Bernoulli’s Law follows that for a potential solution (u, p), the pres-

sure p is large where the speed

|u| is small, and vice versa.

10.6 Potential Flow around a Circular Cylinder

Consider an infinitely long cylinder of diameter 1 oriented along the x

3

axis

and immersed in an inviscid fluid filling

R

3

with velocity (1, 0, 0) at infinity,

see Fig. 10.2. We now construct a corresponding potential solution u =

∇φ by

using Calculus, where we assume that u

3

= 0, and seek a function φ(x

1

, x

2

)

which is harmonic outside the disc x

2

1

+ x

2

2

≤ 1 occupied by the cylinder, such

that

∂φ
∂n

= 0 on the boundary of the disc.

We find that φ is equal to the real part of the analytic function w(z) = z+

1
z

with z = x

1

+ ix

2

and i the imaginary unit, that is,

φ(x

1

, x

2

) = (r +

1
r

) cos(β),

(10.5)

where (x

1

, x

2

) = (r cos(β), r sin(β)) is expressed in polar coordinates (r, β).

We verify readily that φ is harmonic outside the disc, because φ is the real part
of an analytic function, and that

∂φ
∂n

= 0 on the boundary of the disc. Further,

one can show by a little more work that lim

x

2
1

+x

2
2

→∞

∇φ(x

1

, x

2

) = (1, 0, 0),

so that the potential solution satisfies the uniform flow condition far away
from the cylinder. In Fig 10.2 we plot the streamlines of u, which are the level
curves of the imaginary part (r

−

1
r

) sin(β) of w(z).

We have now constructed a potential flow solution around the cylinder,

which intuitively looks quite convincing: the flow opens up to go around the
cylinder and then closes, pretty much as one may naively expect. Right?

10.7 Zero Drag/Lift of Potential Flow

Let us now compute the drag of the potential flow around the cylinder, that is
the net force on the cylinder in the x

1

direction from the pressure acting on the

10.7 Zero Drag/Lift of Potential Flow

77

Fig. 10.2. Potential flow around a cylinder

boundary of the cylinder. To this end we note that the flow speed

|u| evidently

is symmetric with respect to both the x

1

and the x

2

-axis, because the flow

after/below the cylinder is evidently a mirror image of the flow before/above
the cylinder. From Bernoulli’s Law we then deduce that also the fluid pressure
is symmetric before/after and above/below. This means that the pressure
around the cylinder balances out to zero, that is, there is no net force from
the flow on the cylinder: The drag is zero and evidently also the lift which
is the net force in the x

2

-direction. We see in particular that the pressure is

high at the stagnation point A with zero speed opposing the flow, and that
this high pressure is balanced by an equally large pressure in the opposite
direction at the stagnation point B at the back of the cylinder. And so it goes
for all symmetrically placed points on the boundary. One can easily generalize
this result to a body of arbitrary (non-symmetric) shape, subject to potential
flow which is uniform at infinity, we give the proof in the last section of this
chapter.

This is d‘Alembert’s Mystery: Potential flow uniform at infinity has zero

drag/lift. But massive experimental evidence indicates substantial non-zero
drag even if the flow is very slightly viscous such as air and water. The first
such experiments were performed by d’Alembert himself in order to present his
paradox. Further, with zero lift, flying would be impossible, again at variance
with everybody’s experience.

So how is science/mathematics to deal with d’Alembert’s Mystery? Evi-

dently, something must be wrong with the potential solution. But what? Is
it the assumption about inviscid flow, which is essentially Prandtl’s explana-
tion? Or is there a different explanation? To find this out let us solve the Euler
equations computationally using G2, instead of using analytical mathematics,
and see what we get. Before we plunge into this adventure, let us give another

78

10 d’Alembert’s Mystery and Bernoulli’s Law

highlight of Calculus applied to the Euler equations, which is as misleading
as the zero drag/lift of a potential solution.

10.8 Ideal Fluids and Vorticity

Taking the vorticity of the momentum equation in (10.1), we obtain with a
direct computation using e.g. (10.2), the following equation for the vorticity
ω

≡ ∇ × u:

˙

ω + (u

· ∇)ω − (ω · ∇)u = 0

in Ω

× I.

(10.6)

This may be viewed as a linear convection-reaction equation for the propaga-
tion of the vorticity ω with the fluid velocity u being given. Multiplying by ω
and integrating by parts with respect to x, we obtain

d

dt



Ω

ω

2

(x, t) dx +

1
2



Γ

u

· n ω

2

(s, t) ds

≤ G



Ω

ω

2

(x, t) dx

if the gradient of the velocity

∇u is bounded by the constant G, where Γ is

the boundary of Ω. It follows that if ω vanishes in Ω at initial time t = 0,
and no vorticity is convected into Ω through the boundary Γ where u

· n < 0,

then ω(t) vanishes in Ω for all t > 0. We are thus led to the conclusion that
in an ideal fluid with bounded velocity gradient, vorticity cannot be created.
In the computational example with an ideal fluid which we present below,
nevertheless vorticity seems to be generated. The only way out of this dead-
lock is that the assumption of a bounded velocity gradient is not verified, that
is, that the underlying Euler solution (u, p) is not a pointwise solution.

10.9 d’Alembert’s Computation of Zero Drag/Lift

We recall d’Alemberts (erroneous) computation of zero drag: Suppose there
is a stationary pointwise solution (u, p) to the Euler equations of inviscid
incompressible flow around a bluff body in a horizontal channel oriented in
the x

1

-direction, with the velocity u irrotational, i.e.,

∇ × u = 0. Integrating

the momentum equation over the domain, we obtain by partial integration,
considering the first component

0 =



Γ

b

pn

1

ds +



Γ

in

(u

· nu

1

+ pn

1

) ds +



Γ

out

(u

· nu

1

+ pn

1

) ds

where Γ

in

and Γ

out

denote the inflow and outflow boundaries of the channel,

and Γ

b

denotes the boundary of the immersed body. Assuming now that the

velocity is equal on in and outflow, which is natural if the channel is long, by
Bernoulli’s law (stating that

|u|

2

/2 + p is constant), the pressure will be as

well, and thus the inflow and outflow terms will cancel and therefore the drag



Γ

b

pn

1

ds will be zero.

10.10 A Reformulation of the Momentum Equation

79

Obviously, zero drag of a bluff body contradicts experience: All bluff bodies

show substantial drag with the major contribution coming from the pressure
distribution around the body with high pressure up front and low pressure in
the back, and not from viscosity. In particular, we can attribute only a small
part of the drag to viscosity and thus experience clearly indicates substantial
drag for inviscid flow. But d’Alemberts computation shows zero drag.

We shall now in a concrete example see in Chapters 11–12 that the trou-

ble with Alembert’s computation of zero drag is that the pointwise laminar
solution simply does not exist as a stable solution, which makes the computa-
tion meaningless. Instead a turbulent approximate solution develops and this
solution has a substantial drag close to that for a solution of the NS equations
with large Reynolds number.

10.10 A Reformulation of the Momentum Equation

Using the identity (10.2) we can rewrite the Euler momentum equation as
follows, assuming f = 0

˙u

− u × ω + ∇(

1
2

|u|

2

+ p) = 0,

(10.7)

a formulation which is sometimes used. In particular, we derive from this
equation Bernoulli’s Law for stationary irrotational flow.