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TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS PROF. VLADIMIR VLADIMIROV 1. The Kinematics of Continuous Medium 1.1. Lagrangian and Eulerian Coordinates General Idea Fluid ﬂow is represented mathematically by a continuous transformation of 3D Euclidean space into itself. Time, t is the parameter, describing the transformation x 1 x 1 x 2 x 2 x 3 x 3 A A τ τ x 1 ,x 2 ,x 3 - the ﬁxed rectangular coordinate system. x =(x 1 ,x 2 ,x 3 ) - The position of the particle A. Say A is a particle moving with ﬂuid: at t = 0 it occupies the postion X =(X 1 ,X 2 ,X 3 ) at time t it has moved to the position x =(x 1 ,x 2 ,x 3 ) Then x is determined as a function of X and t: (1.1) x = ϕ( X,t) or x i = ϕ i (X i ,t) 1

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TOPICS IN MATHEMATICALPHYSICS: FLUID DYNAMICS

1. The Kinematics of Continuous Medium

1.1. Lagrangian and Eulerian Coordinates

General Idea

Fluid flow is represented mathematically by a continuous transformation of 3D Euclideanspace into itself. Time, t is the parameter, describing the transformation

x1x1

x2x2

x3 x3

A

A

ττ

x1, x2, x3 - the fixed rectangular coordinate system. ~x = (x1, x2, x3) - The position of theparticle A.

Say A is a particle moving with fluid:

• at t = 0 it occupies the postion ~X = (X1, X2, X3)• at time t it has moved to the position ~x = (x1, x2, x3)

Then ~x is determined as a function of ~X and t:

(1.1) ~x = ϕ( ~X, t) or xi = ϕi(Xi, t)1

If ~X is fixed while t varies then (1.1) gives the path of the particle A intially at ~X.

We assume, that initially distinct points remain distinct throughout the entire motion; itmeans that the transformation (1.1) posesses the inverse

(1.2) ~X = Φ(~x, t) or Xi = Φi(xi, t)

It is also usually assumed that ϕi and Φi posses continuous derivatives up to the thirdorder in all variables. The flow is completely determined by the transformation (1.1).

Below we will consider the state of motion (say velocity ~u or density ρ) at given points ~x

or ~X during the course of time:

• ρ = ρ(~x, t), ~u(~x, t) - Eulerian description

• ρ = ρ( ~X, t), ~u( ~X, t) - Lagrangian description• ~x - spatial variables, i.e. Eulerian variables• ~X - material variables, i.e. Lagrangian variables

1.2. Partial and Material Derivatives

By means of (1.1), (1.2) any quantity F , which is a function of Eulerian variables (~x, t) is

also a funtion of Lagrangian variables ( ~X, t) and conversely. To indicate a particular setof variables we write either

F = F ( ~X, t) or F = F (~x( ~X, t), t)

Geometrically

• F ( ~X, t) is the value of F experienced at time t by the particle initially at ~X;• F (~x, t) is the value of F felt by the particle instantaneously at the position ~x.

We shall use the symbols

(1.3)∂F

∂t≡∂F (~x, t)

∂t

dF

dt≡∂F ( ~X, t)

∂t

for the two possible time derivatives of F , obviously they are quite different quantites:

•dF

dtis called the material derivative of F ;

•d

dtis the differentiation following the motion of the fluid;

•dF

dtmeasures the rate of change of F following a particle;

•∂F

∂tgives the rate of F apparent to a viewer stationed at the position ~x.

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 3

1.3. Velocity and Acceleration

The velocity of a particle is given by the definition

(1.4) ~u ≡d~x

dtui ≡

dxi

dt≡∂ϕ( ~X, t)

∂t.

As defined, ~u is a function of the material variables. However, one usually deals with thespatial form:

~u = ~u(~x, t).

In most problems it is sufficient to know ~u(~x, t) rather than the actual motion (1.1).

We have introduced the velocity field in terms of the motion (1.1). It is important to beable to proceed in the opposite direction; to detemine the motion (1.1) from ~u(~x, t). Thistransistion is given by solving the system of ODE’s

d~x

dt= ~u(~x, t)

with the intial condition ~x(0) = ~X.

The acceleration is the rate of change of velocity experienced by a moving particle:

(1.5) ~a ≡d~u

dt.

So, the acceleration ~a(~x, t) can be computed directly in terms of the velocity field ~u(~x, t)

ai ≡dui

dt≡∂ui( ~X, t)

∂t

dui

dt≡

dui

(~x( ~X, t), t

)

dt=∂ui

∂t+∂ui

∂xk

dxk

dt≡uk

or

~a =∂~u

∂t+ ~u.∇(~u) ≡

∂~u

∂t+ (~u.∇)~u.

Correspondently, the general formula for differentiation following the motion of the fluidis:

dF

dt=∂F

∂t+ (~u.∇)F

F = F (~x, t) = F(~x( ~X, t), t

)

dF

dt≡∂F

∂t

∣∣∣∣

~X

=∂F

∂t+∂F

∂xk

∂xk( ~X, t)

∂t≡uk

=∂F

∂t+ uk

∂F

∂xk

=∂F

∂t+ (~u.∇)F(1.6)

Equation (1.3) may be interpreted as expressing, for an arbitrary quantity F = F (~x, t),the time rate of cahnge of F apparent to a viewer situated on the moving particle instan-taneously at the position ~x.

1.4. Jacobian and Euler’s Formula

The basic transformation between Eulerian ~x and Lagrangian ~X coordinates is given by(1.1) and (1.2). The Jacobian of the transformation is:

J =∂(x1, x2, x3)

∂(X1, X2, X3)= det

(∂xi

∂Xk

)

From the assumption, that (1.1) possesses a differentiable inverse, (1.2), it follows that

(1.7) 0 < J <∞ (for invertibility J 6= 0).

Geometrically, J represents the dilation of an infintesimal volume as it follows the motion:

(1.8) dx1dx2dx3 = J dX1dX2dX3

Euler’s formula for J

We shall use Euler’s formula:

(1.9)dJ

dt= J∇.~u

Proof. Let Aik be the cofactor of the element ∂xi

∂Xkin the expression of the Jacobian deter-

minant:

∂xi

∂Xk

Ajk = J δij

The derivative of J is,

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 5

dJ

dt=

d

dt

(∂xi

∂Xk

)

Aik =∂ui

∂Xk

Aik =∂ui

∂xj

∂xj

∂Xk

Aik

=J δij

=∂ui

∂xj

δijJ =∂ui

∂xi

J = J∇.~u

1.5. The Transport Theorem

If a fluid is assumed to be incompressible, that is to move without a change in volume(1.8), we get J ≡ 1 in (1.8) and (1.9) yields

(1.10) ∇.~u = 0

The Transport Theorem

Let τ = τ(t) denote an arbitrary volume, which is moving with a fulid.

Also let F (~x, t) be a scalar or vector function of position. Consider the volume integral

τ(t)

F dτ.

Its derivative is given by the important formula

(1.11)d

dt

τ(t)

F dτ =

τ(t)

(dF

dt+ F∇.~u

)

Proof. Consider the correspondence between the Eulerian coordinates and the Lagrangiancoordinates (1.1), (1.2).

• In x1, x2, x3 - a moving volume τ(t)• In X1, X2, X3 - a fixed volume τ0• τ(t) ↔ τ0 under transformations (1.1),(1.2)

Then,

τ(t)

F dτ =

τ0

F ( ~X, t)J dτ0

where dτ = J dτ0; dτ = dx1dx2dx3, dτ0 = dX1dX2dX3 (see (1.8)). Differentiation yields

d

dt

τ(t)

F dτ =

τ0

(

J∂F

∂t

∣∣∣∣

~X

+ F∂J

∂t

∣∣∣∣

~X

)

dτ0

=

τ(t)

(dF

dt+ F ∇.~u

)

where we have used (1.3) and (1.9).

Another form of (1.11) can be obtained after transformation

(1.12)dF

dt+ F (∇.~u) =

∂F

∂t+ ∇.(~uF )

which gives

(1.12a)d

dt

τ(t)

F dτ =

τ(t)

∂F

∂tdτ +

∂τ(t)

F (~u.~n) dS

∂τ(t)

~n

τ(t)

It should be emphasised that (1.11) and (1.12) express a kinematical theorem, independantof any meaning attached to F .

1.6. Equation of Continuity

The density function ρ = ρ(~x, t) > 0. The mass of the fluid occupying a region τ is:

M =

τ

ρ dτ

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 7

Physics

We postulate the principle of conservation of mass. “The mass of a fluid in material volumeτ does not change as τ(t) moves with the fluid”.

(1.13)d

dt

τ(t)

ρ dτ = 0

now from (1.13) and (1.11) it follows that

τ(t)

(dρ

dt+ ρ(∇.~u)

)

dτ = 0

and since τ(t) is an arbitrary volume, this implies:

(1.14)dρ

dt+ ρ(∇.~u) = 0.

This is the Eulerian form of the equation of continuity. We can express this in the alter-native form

(1.15)∂ρ

∂t+ ∇.(ρ~u) = 0.

Now, multiplying (1.14) by J and using (1.9) we derive the Lagrangian form of the equation

of continuity :

(1.16)d

dt(ρJ ) = 0 ρJ = ρ0

where ρ = ρ0 is the initial density distribution. For future use, let us also write that foran arbitrary function F (~x, t)

(1.16a)d

dt

τ(t)

ρF dτ =

τ(t)

ρdF

dtdτ (prove it!)

To prove this equality one should use (1.9) and (1.14).

1.7. The Rate of Strain Tensor

Let us introduce a simple decomposition of the tensor ∂ui

∂xk, namely

(1.17)∂ui

∂xk

= eik + Ωik

where

eik =1

2

(∂ui

∂xk

+∂uk

∂xi

)

Ωik =1

2

(∂uk

∂xk

−∂ui

∂xk

)

The tensors eik and Ωik are respectively the symmetric and skew-symmetric part of ∂ui

∂xk.

The deformation (rate of strain) tensor

Let d~x denote a material element of arc. It’s rate of change during the fluid motion isgiven by the formula:

d

dt(dxi) =

d

dt

(∂xi

∂Xk

dXk

)

=∂ui

∂Xk

dXk =∂ui

∂xj

dxj

or simply

d

dt(d~x) = (d~x.∇)~u.(1.18)

From (1.18) we get

(1.19)d

dt(dℓ2) = 2eikdxidxk

where dℓ = |d~x|. Thus the tensor eik is a measure of the rate of change of the squaredelement of arc following a fluid motion.

In a rigid motion dℓ ≡ const, whence a necessary condition that a motion be localy andinstantaneously rigid is that eik = 0 (see in homework). For this reason, eik is calledthe deformation tensor (or the rate of strain tensor). The tensor eik − 1

3ejjδik is also of

interest, for its vanishing is the necessary and sufficient condition that the motion, locallyand instantaneously preserves angles (prove it!).

If eik = 0 everwhere in the fluid, then the motion is rigid and

(1.19a) ~u =1

2~ω × ~r + const.

where ~ω2

is the constant angular velocity of the motion. Equation (1.19a) can be derivedanalytically as the integral of the system of first order partial differential equations eik = 0.

1.8. Vorticity and the Potential of Velocity

Let us consider the velocity field in the neighbourhood of a fixed point P . Denoting theevaluation of a quantity at the point P by a superscript, we have near P

ui = uPi + rk

(∂ui

∂rk

)P

+ O(r2)

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 9

where ~r denotes the radius vector from P . Neglecting terms of order r2 and using (1.17)we get

(1.20) ui = uPi + eP

ikrk + ΩPikrk

We must now interpret the various terms in this formula.

The first term on the right represents a uniform translation of velocity ~uP . If we set

e = ePikrirk

then the second term can be written in the form

(1.21) ∇1

2e or

1

2

∂e

∂xi

.

This term represents a veloctiy field normal at each point to the quadric surface e = constwhich passes through that point. In this velocity field there are 3 mutually perpendiculardirections which are suffering no instantaneous rotation (the axes of strian). The principal(or eigen-) values of e measure the rates of extension per unit length of fluid elements inthese directions.

extension and rotation

pure extension

r2~n

r1

The final term in (1.20) may be written as

(1.22)1

2~ωP × ~r

where ~ω = ∇× ~u is the vorticity vector.

The form (1.22) shows clearly that the final term in (1.20) represents a rigid motion ofangular velocity 1

2~ωP .

By considering the above results, the identity (1.17) can be fully interpreted. For anarbitrary motion, the velocity ~u in the neighbourhod of a fixed point P is given, up toterms of order r2, by

(1.23) ~u = ~uP + ∇

(1

2e

)

+1

2~ωP × ~r

where e = eikrirk is the rate of strain quadric and ~ω = ∇× ~u is the vorticity vector. Thusan arbitrary instantaneous state of continuous motion is at each translation, a dilation

along three mutually perpendicular axes, and the rigid motion of these axes. The angularvelocity of rotation is 1

2~ωP . This result amply esablished that ~ω represents the local and

instantaneous rate of rotation of the fluid.

If eik = 0 at a point, it is apparent from (1.23) that the motion is locally and instantaneouslya rotation, while if eik = kδik the motion is a combination of pure expansion and rotation.On the other hand, if throughout a finite portion of fluid we have ~ω = 0, Ωik = 0, thenthe relative motion of any element of that position consists of a pure deformation, and iscalled irrotational ! In this case ~a is derivable from a potential ϕ(~x, t):

(1.24) ~u = ∇ϕ (prove it!).

Problem

Prove, that the motions of irrotaional flows and potential flow coincide.

~ω = ∇× ~u = 0 ⇔ ~u = ∇ϕ

Step 1: Let ~u = ∇ϕ. Then

ωi = εikl

∂ul

∂xk

= εikl

∂2ϕ

∂xk∂xl

≡ 0

it is identically equal to zero, since it is the scalar product of a symmetric tensor∂2ϕ

∂xk∂xland antisymmetric tensor εikl. So

~u = ∇ϕ⇒ ~ω = 0.

Step 2: Let ~ω = ∇× ~u = 0. Then let

ϕ =

∫ ~x

~x0

~u.d~x

the definition of the potential ϕ(~x, t). Evidently ∇ϕ = ~u. However, we have toprove that ϕ(~x, t) is a single valued function. It means that the integral,

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 11

∫ ~x

~x0

~u.d~x

does not depend on the path of integration. This statement is equivalent to theequality

(1.24a)

γ

~u.d~x = 0

for any closed circuit γ. Using stokes theorem

γ

~u.d~x =

~ω.~n dS

for any surface Sγ ‘based’ on γ. If ~ω = 0 then∮

γ~u.d~x = 0 and ϕ(~x, t) is a single

valued function.

1.9. Streamlines, Vortex Lines and Trajectories

A curve, everywhere tangent to a given continuous vector field is called a vector-line. Inparticular, the vecor-lines of the velocity field are called streamlines, and the vector-linesof the vorticity are called vortex-lines.

Equations of Streamline

(1.25) ~x = ~x(s);d~x

ds= ~u

(~x(s), t

); t = const.

or

dx

u(x, y, z, t)=

dy

v(x, y, z, t)=

dz

w(x, y, z, t); t = const.

Equation of a Vortex line

(1.26)d~x

ds= ~ω

(~x(s), t

); t = const.

We define a trajectory to be the curve traced out by a particle as time progresses.

Equation of a Trajectory (or Pathline)

(1.27)d~x

dt= ~u

(~x(t), t

)

or

dx

u(x, y, z, t)=

dy

v(x, y, z, t)=

dz

w(x, y, z, t)= dt

with suitable initial conditions. If ~u is independent of t,(

∂~u∂t

= 0)

then the streamlines andtrajectories coincide (prove it!).

In this case the flow is called steady or stationary.

2. The Equations of Motion

Our aim is to derive the equations, which describe the action of forces.

2.1. The Stress Principle (Cauchy, 1827)

“Upon any imagined closed surface S there exists a distribution of stress vector ~t whoseresultant and moment (torque) are equivalent to those of the actual forces exerted by thematerial outside S upon that inside.”

• ~t - surface force (stress);• ~n - the outward unit normal to S.

~t

~t

~t ~n

It is assumed that ~t = ~t(~x, t, ~n), for all ~x ∈ S. Now we can set the fundamental principleof the dynamics of fluid motion - the principle of conservation of linear momentum:

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 13

The rate of change of linear momentum of a material volume τ(t) equals theresultant force on the volume

(2.1)d

dt

τ(t)

ρ~u dτ =

τ(t)

ρ~f dτ +

∂τ(t)

~t dS

Here ~f(~x, t) is the external force per unit mass. By means of our transport theorem, (1.11),and conservation of mass, (1.13), we can transform (2.1) to

τ(t)

ρd~u

dtdτ =

τ(t)

ρ~f dτ +

∂τ(t)

~t dS

This equality is valid for any instant t and does not contain any differentiation of integralsover variable volume τ(t). Therefore here the integration over a moving volume can bereplaced without loss of generality by integration over a fixed volume.

(2.2)

τ0

ρd~u

dtdτ =

τ0

ρ~f dτ +

∂τ0

~t dS.

A result of great importance follows from (2.2). Let ℓ be the ‘typical’ or ‘characteristic’size of τ0; then ℓ3 will be the volume of τ0. Dividing both sides of (2.2) by ℓ2, letting ℓ→ 0we obtain

(2.3) limℓ→0

ℓ−2

∂τ0

~t dS = 0

that is, the stress forces in local equilibrium!

x1

x2

x3

~ı~

~k

Σ~n

Consider the tetrahedron with vertex at an ar-bitrary point ~x and with three of its faces par-allel to the coordinate planes. The normals are:

−~ı,−~. − ~k, ~n with ~n = (n1, n2, n3) and with~ı = (1, 0, 0), etc. The areas are n1Σ, n2Σ, n3Σ,Σ. Now let us consider (2.3). Since ℓ2 ∼ Σ wecan write

~t(~n

)+ n1~t

(~ı)

+ n2~t(~)

+ n3~t(~k

)= 0

where ~t(~n

)≡ ~t

(~x, t, ~n

). Also by Newton’s second law we have ~t

(~ı)

= −~t(−~ı

), ~t

(~)

=

−~t(− ~

), ~t

(~k

)= −~t

(− ~k

). Now we obtain,

~t(~n

)= n1

~t(~ı)

+ n2~t(~)

+ n3~t(~k

)

Therefore ~t may be expressed as a linear function of the components ni:

(2.4) ti = Tiknk; Tik = Tik(~x, t).

The matrix (tensor) Tik is called the stress tensor. Using (2.4) we transform (2.2) to theform

τ0

ρdui

dtdτ =

τ0

(

ρfi +∂Tik

∂xk

)

and since τ0 is arbitrary

(2.5) ρdui

dt= ρfi +

∂Tik

∂xk

.

This is the equation of motion discovered by Cachy (1828). As the result we have obtainedfour equations (2.5) and (1.14)

(2.6)

ρdui

dt= ρfi +

∂Tik

∂xk

dt+ ρ∇ · ~u = 0

They relate to each other four quantities ρ, ~u and contain still undefined Tik. Tensor Tik (thestress tensor) can be expressed in terms of ρ, ~u by direct mechanical or thermodynamicalassumptions. The various possibilities for this form the physical core of fluid mechanics.

2.2. Conservation of angular momentum

Tensor Tik in (2.6) has nine independent components. However to guarantee the conserva-tion of angular momentum, we have to accept that it is symmetric

(2.7) Tik = Tki

which is known as the Boltzmann postulate.

Theorem 1 (Boltzmann postulate). For an arbitrary continuous medium satisfying (2.6),(2.7) we have

(2.8)d

dt

τ

ρ(~r × ~u

)dτ =

τ

ρ(~r × ~f

)dτ +

∂τ

~r × ~t dS

where τ is an arbitrary material volume.

Proof. Using (1.9), (1.3) we get

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 15

d

dt

τ

ρ(~r × ~u

)dτ =

τ

ρ

(

~r ×d~u

dt

)

=

τ

ρ(~r × ~f

)dτ +

∂τ

~r × ~t dS −

~T0 dτ(2.9)

where T0i ≡ ǫiklTkl (prove it!). By virtue of (2.7) we have ~T0 = 0 and (2.8) is proven.Conversely, if (2.8) holds for arbitrary volumes then Tik must be symmetric.

2.3. Perfect Fluid

A perfect fluid is by definition a material for which

(2.10) ~t = −p~n

p is called the pressure: when p > 0 the vectors ~t acting on a closed surface tend to

compress the fluid inside. Comparing ~t(~n

)= n1

~t(~ı)+n2

~t(~)+n3

~t(~k

)and ~t

(n

)= −p~n

we find p(~n

)= p

(~ı)

= p(~)

= p(~k

). That is, p is independent on ~n

p = p(x, t)

Also we can obtain

(2.11) Tik = −pδik

The equations of motion (2.6) take Euler’s (1755) form

(2.12)

ρd~u

dt= ρ~f −∇× p

dp

dt+ ρ(∇.~u) = 0

and usually are referred to as Euler’s equations. This system contains four equations forfive variables ρ, ~u, p, so it is not closed and it is not solvable. In order to make it closed,one should introduce one more equation, based on physical reasoning.

2.4. Barotropic Fluid

A perfect fluid with ρ = ρ(p) is called the baratropic fluid. We will consider two basic cases

i) Compressible Fluid (gas): with an invertible function ρ(p) ⇔ p(ρ). Here thenecessary conditions are ∂ρ

∂p6= 0, ∂p

∂ρ6= 0. From physical reasons one can see, that

always

c2 ≡∂p

∂ρ> 0

where c(ρ) is the Speed of Sound. (the increasing of pressure does correspond tothe increasing of density). Equations (2.12) can be written as

(2.13)

~ut + (~u.∇)~u = −1

ρ∇p(ρ) = −∇w = −

c2

ρ∇ρ

ρt + ∇.(ρ~u) = 0

c(ρ) ≡

∂p

∂ρ

Here w = w(ρ) is a thermodynamic function, called the enthalpy. System (2.13)contains four equations for four unknown functions, so it is a closed system ofequations.

ii) Incompressible Fluid: It corresponds to the case ρ(p) = const. = ρ0. Equations(2.12) can be written as

(2.14)

~ut + (~u.∇)~u = −1

ρ0∇p+ ~f

∇.~u = 0

ρ = ρ0 = const.

This system of four equations also contains four unknown functions, so it is alsoclosed.

2.5. Viscous Incompressible Fluid

If the internal friction (viscosity) in a fluid is present, the expression for the stress tensor(2.11) should be generalized as

(2.15) Tik = −pδik + 2µeik

where

eik ≡1

2

(∂ui

∂xk

+∂uk

∂xi

)

is the rate of deformation tensor, µ is a constant coefficient known as the coefficient ofdynamic viscosity; the combination ν ≡ µ

ρis called the coefficient of kinematic viscosity.

The form of ‘viscous term’ (2.15) is also called the Newtonian viscosity.

Substitution of (2.14) into (2.6) (in the case of incompressible fluid) gives the famousNavier-Stokes equations for a viscous incompressible fluid:

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 17

(2.16)

~ut + (~u.∇)~u = −1

ρ0∇p+ ν~u+ ~f

∇.~u = 0

≡∂2

∂xi∂xj

This is also a closed system of equations which represent the most basic mathematicalmodel in the modern fluid dynamic.

Note. The main argument towards (2.14) is; the ‘viscous part’ of stresses Tik should beproportional to the ‘amplitude’ of deformations, which can be measured by eik. In par-ticular, the ‘viscous stresses’ must be zero in the absence of deformations eik ≡ 0 (which

corresponds to the motion of fluid as a solid body, ~u = ~U + ~Ω×~r with constants ~U and ~Ω,see Homework).

2.6. Boundary Conditions

For the formulation of any particular problem of fluid motion the governing equations (2.6),(2.13), (2.14), or (2.16) must be complemented by the appropriate boundary conditions.There are many versions of such conditions, the particular one must be chosen on the basisof the physical and mathematical analysis of the problem.

The non-leak boundary conditions

If a fluid flow is bounded by solid walls, then the most evident condition requires that thefluid can not ‘leak’ or ‘penetrate’ into the boundary. For the stationary (fixed) boundary∂τ it means that

(2.17) ~un ≡ ~u.~n = 0 on ∂τ

where ~n is the normal unit vector to ∂τ . If the boundary is moving with the velocity~U = ~U(t), then (2.17) should be modified as

(2.18)(~u− ~U

)n≡

(~u− ~U

).~n = 0 or ~u.~n = ~U · ~n on ∂τ

In the remainder of this lecture we take ~U = 0; the conditions for ~U 6= 0 can be restoredin a straight forward way.

It is proven that it all flow boundary ∂τ is solid, then for the perfect fluid (2.17) gives usthe full set of boundary data; any additional to (2.17) boundary condition will producea mathematically controversial problem. So, for the perfect fluid the system of governingequations is usually written together with the boundary conditions as

(2.19)

d~u

dt= −

1

ρp(ρ) + ~f

dp

dt+ ρ(∇.~u) = 0

~u.~n = 0 on ∂τ

for the compressible baratropic fluid (2.13) and

(2.20)

d~u

dt= −

1

ρ0

∇p+ ~f

∇.~u = 0 ρ0 = const.

~u.~n = 0 on ∂τ

for the inviscid incompressible fluid (2.14).

The non-slip boundary condition

The non-leak boundary conditions (2.17), (2.18) leaves the tangential velocity component

(2.21) ~u t = ~u− ~n(~n.~u) on ∂τ

undetermined; ~u t can be found as a result of solving (2.19) or (2.20). In contrary, for aviscous fluid (2.16) the vector ~u t also must be zero.

(2.22) ~u t = 0 on ∂τ

So, for viscous fluid both (2.17) and (2.18) must hold. Both these conditions can be unifiedinto one

(2.23) ~u = ~un + ~u t = 0 on ∂τ

which is well known as the non-slip boundary condition.

The physical reason for (2.22) is the friction on the molecular level (the attraction betweenthe molecules of fluid and solid). Mathematically the doubled number of boundary con-ditions ((2.17) and (2.22) for viscous fluid in comparison with (2.17) only for an inviscidfluid) is required following to the doubling of spatial order of governing PDEs. Indeed,the Laplacian operator in the R.H.S. of (2.16) is of the second order, while without it (forν = 0), the highest order of spatial derivative correspond to ∇ in the L.H.S. of (2.16).

As the result, the governing equations for a viscous incompressible fluid are often writtenas the set of the Navier-Stokes equations along with the non-slip boundary conditions:

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 19

(2.24)

d~u

dt= −

1

ρ0∇p+ ν~u+ ~f in τ

∇.~u = 0 ρ0 = const.

~u = 0 on ∂τ

The systems of governing equations (2.19), (2.20) and (2.24) represent the main result ofthis Lecture. In order to make the mathematical formulations complete, all these systemsmust be complemented by the initial data for the velocity field

(2.25) ~u(~x, 0) = ~u0(~x) in τ

2.7. Plane rotationally-symmetric and axisymmetric motions

Example 2.1. A perfect incompressible fluid of constant density ρ = ρ0 in the absence ofexternal forcing is described by Euler’s equations (2.14):

(2.26)

d~u

dt= −

1

ρ0

∇p

∇.~u = 0

In Cartesian coordinates (x, y, z) the components of velocity are ~u = (u, v, w). Write thecomponent form of Euler’s equation for:

i) general three-dimensional (3D) motions;ii) plane (2D) motions.

Solution: Taking the above cases separately we get

i) In general 3D motions we describe u, v, w, p as

(2.27) u = u(x, y, z, t) v = v(x, y, z, t) w = w(x, y, z, t) p = p(x, y, z, t)

The component form of equations is

(2.28)

ut + uuk + vuy + wuz = −1

ρ0px

vt + uvx + vvy + wvz = −1

ρ0py

wt + uwx + vwy + wwz = −1

ρ0

pz

ux + vy + wz = 0

ii) Plane (2D) motions:

(2.29) u = u(x, y, t) v = v(x, y, t) w ≡ 0 p = p(x, y, t)

These equations reduce down to;

(2.30)

ut + uux + vuy = −1

ρ0px

vt + uvx + vvy = −1

ρ0

py

ux + vy = 0

Example 2.2. A perfect incompressible fluid of a constant density ρ = ρ0 in the absenceof external forces is described by Euler’s equations (2.15):

(2.31)

d~u

dt= −

1

ρ0

∇p

∇.~u = 0

In the cylindrical coordinates (r, ϕ, z) the components of velocity are ~u = (u, v, w). Writethe component form of Euler’s equation for

i) general three-dimensional (3D) motions;ii) rotationally-symmetric motions;iii) axisymmetric motions.

Solution: Again breaking the cases down separately we get.

i) For general 3D motions

(2.32) u = u(r, ϕ, z, t) v = v(r, ϕ, z, t) w = w(r, ϕ, z, t) p = p(r, ϕ, z, t)

In (2.31) operators ddt

, ∇p and ∇.~u must be expressed in the cylindrical coordinates.Euler’s equations take the form,

(2.33)

dudt

︷ ︸︸ ︷

ut + (~u.∇)u−v2

r= −

1

ρ0pr

dvdt

︷ ︸︸ ︷

vt + (~u.∇)v +uv

r= −

1

rρ0

dwdt

︷ ︸︸ ︷

wt + (~u.∇)w = −1

ρ0pz

ur +u

r+

1

rvϕ + wz = 0

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 21

where

∇ =

(∂

∂r,1

r

∂ϕ,∂

∂z

)

so, that

~u.∇ = u∂

∂r+v

r

∂ϕ+ w

∂z

ii) Rotationaly-symmetric motions: Here all functions do not depend on ϕ

(2.34) u = u(r, z, t) v = v(r, z, t) w = w(r, z, t) p = p(r, z, t)

so in (2.33) we can put ∂∂ϕ

= 0 that leads to equations,

(2.35)

dudt

︷ ︸︸ ︷ut + uur + wuz −

v2

r= −

1

ρ0pr

dvdt

︷ ︸︸ ︷vt + uvr + wvz +

uv

r= 0

dwdt

︷ ︸︸ ︷wt + uwr + wwz = −

1

ρ0pz

ur +u

r+ wz = 0

The second equation can be rewritten for γ = rv;

(2.36)dγ

dt= γt + uγr + wγz = 0

That equation represents a special case of the conservation of the circulation ofvelocity, which we will consider later.

iii) Axisymmetric motions: Here in addition to (2.34) also v ≡ 0,

(2.37) v ≡ 0 u = u(r, z, t) w = w(r, z, t) p = p(r, z, t)

so (2.35) reduces to

(2.38)

dudt

︷ ︸︸ ︷ut + uur + wuz = −

1

ρ0pr

dwdt

︷ ︸︸ ︷wt + uwr + wwz = −

1

ρ0pz

ur +u

r+ wz = 0.

The condition v ≡ 0 is natural from the view point of equation (2.36). Indeed if wetake γ|t=0 = 0, it follows that γ = 0 at any t.

Example 2.3. Consider a plane steady rotationally-symmetric flow of an inviscid incom-pressible fluid of a constant density ρ0 in a gap between two concentric cylinders of radiiR2 > R1. Obtain;

i) the general solution of this type: calculate its vorticity ω from Stokes’s theoremand from the formula

(2.38a) ω =1

r(rv)r −

1

ruϕ,

where u, v are r, ϕ-components of velocity.ii) the solution with constant vorticity ω = Ω0: does this solution contain any other

arbitrary parameters?iii) the irrotational potential solution.

Solution:

y

x

u

v

r

τ

R1

R2

ϕ

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 23

i) A plane steady rotationally-symmetric solution in polar coordinates (r, ϕ) can beobtained from (2.34) after the elimination of the dependence on z and t;

(2.39) u = u(r) v = v(r) w = 0 p = p(r)

Those functions must satisfy to the equations

(2.40)

uur −v2

r= −

1

ρ0

pr

uvr +uv

r= 0 in τ

ur +u

r= 0

and the non-leak boundary conditions

(2.41) u(r) = 0 at r = R1 and r = R2.

The third equation in (2.40) is

1

r(ru)r = 0;

the only solution to this is u(r) = Q

rwith an arbitrary constant Q. Substitution to

the boundary condition (2.41) shows that the only possibility is Q = 0 and u ≡ 0.

Taking it (u ≡ 0) into account, we see that the only equation which follows from(2.40) is

(2.42)v2

r=

1

ρ0pr or p = ρ0

∫ r V 2(ξ)

ξdξ

Therefore the required solution is

(2.43) u ≡ 0 v = V (r) p = P (r) = ρ0

∫ r V 2(ξ)

ξdξ

with an arbitrary function V (r). The vorticity of this flow is

(2.44) Ω(r) =1

r(rV )r = Vr +

V

r.

It can be obtained both from (2.38a) and from Stokes’s theorem (see Homework).We can also introduce the circulation of velocity as:

(2.45) Γ = Γ(r) = 2πrV (r)

ii) Solution with constant vorticity Ω0 according to (2.44) must satisfy the equation

1

r(rV )r = Ω0

It follows that,

(rV )r = Ω0r ⇒ rV =1

2Ω0r

2 + C

⇒ V =1

2Ω0r +

C

r(2.46)

which contains two arbitrary constants Ω0 and C.iii) The irrotational (potential) solution satisfies the equation

1

r(rV )r = 0

and can be obtained from (2.46) after taking Ω0 = 0

(2.47) V =Γ

2πr

where we have taken C = Γ2π

=const. This definition of Γ is in agreement with(2.45).

Example 2.4 (Lamb’s form of Euler’s equation). A perfect barotropic gas in the absenceof external forcing is described by Euler’s equations (2.13)

(2.48)

d~u

dt= −∇w w = w(ρ) dw =

dp

ρ

ρt + ∇.(ρ~u) = 0

Show that the first equation in (2.48) can be written as

(2.49) ~ut + ~ω × ~u = −∇

(

w +u2

2

)

where ~ω = ∇× ~u. Write the counterpart of equation (2.49) for the perfect incompressiblefluid.

Solution: The transformation is based on the use of the identity

(2.50) (~u.∇)~u = ~ω × ~u+ ∇

(u2

2

)

ω ≡ ∇u.

In order to prove it let us write ~ω × ~u = −~u × ~ω = −~u× (∇× ~u). So, in components

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 25

(~ω × ~u )i = −ǫiklǫlmnuk

∂un

∂xm

= −(δimδkn − δinδkm)uk

∂un

∂xm

= −uk

∂uk

∂xi

+ uk

∂ui

∂xk

.

or in vector form:

~ω × ~u = −∇

(~u 2

2

)

+ (~u.∇)u

and (2.50) is proven. Substitution of (2.50) into the first equation (2.48) gives the requiredform (2.49). For the perfect incompressible fluid the transformation (2.49) is the same, sothe equation is

(2.51) ~ut + ~ω × ~u = −∇

(p

ρ0+~u 2

2

)

Example 2.5 (Sound Waves. Linearization). Compressible fluid (gas) is described by(2.13):

(2.52)

~ut + (~u.∇)~u = −c2

ρ∇ρ

ρt + ∇.(ρ~u) = 0

c = c(ρ) =

∂p

∂ρ

A particular solution of this system is the state of rest:

~u ≡ 0 ρ = ρ0 = const. c0 = c(ρ0)

Let us consider small deviations (perturbations) of the state of rest, which we denote with‘tildes’,

(2.53) ~u = 0 + ~u(~x, t) ρ = ρ0 + ρ0(~x, t)

Substitution of (2.53) into (2.52) yields;

ut + uux = −[c(ρ0 + ρ)]2

ρ0 + ρρx

ρt +[(ρ0 + ρ)u

]

x= 0

Here for the sake of simplicity we consider one-dimensional problem (general 3D problemis given in the Homework). Now we use the lineraization procedure. It means that we keeponly the linear (with respect to ‘tilde’ variables) terms and neglect all their squares, cubes,etc. The result is:

ut = −c20ρ0ρx

ρt + ρ0ux = 0

now we can derive equations for ρ only or for u only.

ρtt − c20ρxx = 0 utt − c20uxx = 0

Both equations have the form

(2.54) ftt − c20fxx = 0 c20 = const.

which is well known as the wave equations. It can be solved by introducing of new inde-pendent variables

ξ ≡ x− c0t and η ≡ x+ c0t.

In these variables (2.54) takes the form fξη = 0 (prove it!). This has the general solution,

f(ξ, η) = g(ξ) + h(η)

with arbitrary funtions g(ξ) and h(η). In the original variables x, t this solution is

f(x, t) = g(x− c0t) + h(x+ c0t)

which represents the combination of ‘the wave travelling to the right’ and ’the wave trav-elling to the left’. The propagation of both waves takes place without any deformations oftheir profiles. The absence of the deformations is vitally important in our life: it allows usto hear the signal in the same form, as it has been generated!

Example 2.6 (Archimede’s Theorem). Now we consider an incompressible fluid, eitherviscous (2.16) or inviscid (2.14). For the state of rest (~u ≡ 0) both these systems ofequations are reduced to

(2.55) 0 = −1

ρ0∇p+ ~g

where ~g is the gravity field, representing in this case the force per unit mass ~f = ~g. Thisis the equations for hydrostatic equilibrium. In full, they can be written as

∇p = ρ0~g ~u = 0

In Cartesian coordinates (x, y, z), such that ~g = (0, 0,−g) with g > 0 we get the system ofPartial Differential Equations for pressure:

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 27

px = 0

py = 0

pz = −ρg

Solution of this system is

(2.56) p = p(z) = ρ0 − ρgz

where p0 is a constant. The represents the distribution of pressure in hydrostatic equilib-rium. Let us consider a solid volume V submerged into the fluid.

y

x

z

x1

x2

x3

~n

τ

V

∂τ~g

The force acting on the solid is

~F = −

∂τ0

p~u dS (justify this!)

The use of (2.21) and transformations yield:

∂τ

pui dS = −

∂τ

(p0 − ρgx3)ni dS = I1 + I2

where we have;

I1 = −p0

∂τ

ni dS

= 0

prove it! Also,

I2 = ρg

∂τ

x3ni dS

= ρg

τ

∂x3

∂xi

= ρg

001

V

= −ρgiV

So finally we have

(2.57) ~F = −ρ~gV

which gives us Archimedes’s theorem: “The buoyancy force is equal to the weight of the

fluid in the submerged volume of the solid.”

Further exercises on this subject can be found in the homework! Examples of viscous flowswill be given later.

3. Further elements of kinematics and dynamics

3.1. The momentum transfer equation

The principle of conservation of linear momentum (2.1) may be transformed by (1.12a),the transport theorem, into the form

(3.1)d

dt

τ0

ρui dτ =

τ0

ρfi dτ +

∂τ0

(ti − ρuiuknk) dS (prove it!)

expressed the rate of change of momentum of a fixed volume τ0.

Proof. We can prove this by using the transport theorem, (1.11). We transform equation(1.11) to give

d

dt

τ(t)

~F dτ =

τ(t)

∂ ~F

∂tdτ +

∂τ(t)

~F(~u.~n

)dS

now taking ~F ≡ ρ~u this gives

(3.1a)d

dt

τ(t)

ρ~u dτ =

τ(t)

∂ρ~u

∂tdτ +

∂τ(t)

ρ~u(~u.~n

)dS

Now combining (3.1a) with the law for changing of momentum, (2.1), gives

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 29

(3.1b)

τ(t)

∂ρ~u

∂tdτ =

τ(t)

ρ~f dτ +

∂τ(t)

[

~t− ρ~u(~u.~n

)]

dS

In this expression we moved the t-differentiation under the integral sign. It means thatequality is valid at any instant t separately (t in τ(t) plays a part of a parameter). There-fore in (3.1b) the integration over a moving volume can be replaced (without the loss ofgenerality) by integration over a fixed volume: (the fact that τ(t) is moving does not playany role in (3.1b)!)

(3.1c)

τ0

∂ρ~u

∂tdτ =

τ0

ρ~f dτ +

∂τ0

[

~t− ρ~u(~u.~n

)]

dS.

The partial derivative on the left hand side can be pulled out of the integral sign. Theresult is the required expression (3.1)

d

dt

τ0

ρ~u dτ =

τ0

ρ~f +

∂τ0

[

~t− ρ~u(~u.~n

)]

dS

Because of the physical interpretation of the final term, (3.1) is known as the momentum

transfer equation. The vector Pi

(3.2) Pi ≡ ti − ρuiuknk = (Tik − ρuiuk)nk

is known as the flux of momentum through the fixed ‘test’ surface of unit area and withthe unit normal vector ~n. The first term in ~P represents the flux of momentum due to thesurface force ~t; the second term −ρuiuknk is known as the ‘convective’ flux of momentum(the minus sign appears since the positive influx of momentum into τ0 takes place when~u.~n < 0).

Example 3.1. Let us determine the force on an obstacle immersed in a steady flow.Suppose that the fluid occupies the entire exterior of some obstacle, and that the externalforce field is zero. Then if ∂τ0 denotes the surface of the obstacle and ∂Σ denotes a ‘controlsurface’ enclosing ∂τ0, we have the following formula for the force F acting on the obsticle:

(3.3) ~F = −

∂τ0

~t dS =

∂Σ

[~t− ρ~u(~u.~n)

]dS

If we can prescribe ~t, ~u and ρ at the ‘remote surface’ Σ ⇒ then we can calculate ~F . Thevalues of functions at the ‘remote surface’ or ‘at infinity’ is often known.

For the perfect fluid (2.10) we have

(3.4) ti = −pni Tik = −pδik Pi = (−pδik − ρuiuk)nk

and (3.1)-(3.3) become slightly simpler.

~n

~n

∂τ0

~f = 0

∂Σ

3.2. The energy transfer equation

Let K denote the kinetic energy of a volume τ

K ≡1

2

τ

ρuiuk dτ.

Then for an arbitrary material volume τ(t) we have

(3.5)dK

dt=

τ(t)

ρfiui dτ +

∂τ(t)

~t.~u dS −

τ(t)

Tikeik dτ (prove it!)

where eik is the rate-of-strain tensor (1.17).

Proof. We show this by differentiating the kinetic energy,

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 31

dK

dt=

d

dt

τ(t)

ρuiui dτ

=

τ(t)

ρui

dui

dtdτ

=

τ(t)

ui

(

ρfi +∂Tik

∂xk

)

=

τ(t)

ρuifi dτ +

τ(t)

(∂uiTik

∂xk

− Tik

∂ui

xk

)

=

τ(t)

ρuifi dτ +

∂τ(t)

uiTiknk dS −1

2

τ(t)

Tik

(∂ui

∂xk

+∂uk

∂xi

)

=

τ(t)

ρuifi dτ +

∂τ(t)

uiti dS −

τ(t)

Tikeik dτ

where

ti = Tiknk eik =1

2

(∂ui

∂xk

+∂uk

∂xi

)

It states that the rate of change of kinetic energy of a moving volume is equal to therate at which the work is begin done on the volume by external forces, diminished by a‘dissipation’ term, involving the interaction of stress and deformation. For a perfect fluid(3.5) simplifies to

(3.6)dK

dt=

τ(t)

ρ(~f.~u

)dτ −

∂τ(t)

p(~u.~n

)dS +

τ(t)

p∇.~u dτ (prove it!)

The last term is the rate at which work is done by the pressure in changing the volume offluid elements.

A simplification of (3.5) and (3.6) may be obtained if ~f = −∇ × Φ, where Φ = Φ(~x) is atime independent potential. In this case we can introduce the potential energy

Π ≡

τ

ρΦ dτ

and (3.5) becomes

(3.7)d

dt(K + Π) =

∂τ

~t.~u dS −

τ

It is interesting to consider separately the term

D ≡ −

τ

Tikeik dτ

which represents the rate of energy dissipation in the volume τ . For a perfect fluid Tik =−pδik and the integrand is

Tikeik = −peii = −p∂ui

∂xi

= −p∇× ~u.

One can see, that it gives the last term in (3.6), which is equal to zero for an inviscidin-compresible fluid. For the viscous incompressible fluid

(2.12) Tik = −pδik + 2µeik

so

(3.8) D = −2µ

τ

eikeik dτ.

One can see that the ‘viscous dissipation’ (3.8) is always negative, D < 0.

3.3. Convection of vorticity

Consider perfect barotropic fluid, which can be either compressible or incompressible. Theequations of motion, (2.12), are

(3.9)

ρ~a = −∇p−∇Φ

dt+ ρ∇.~u = 0

where acceleration is

~a =d~u

dt=∂~u

∂t+ (~u.∇)~u.

There is a well known vector identity,

(3.10) (~u.∇)~u = ~ω × ~u+ ∇

(1

2u2

)

so ~a = ~ut + ~ω × ~u+ ∇(

12u2

), ~ω ≡ ∇× ~u. Calculations of ∇× ~a yields

∇× ~a =∂~ω

∂t+ ∇× (~ω × ~u)

=d~ω

dt− (~ω.∇)~u+ ~ω(∇.~u).(3.11)

From (3.11) and the equation for ρ, (3.9), we obtain

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 33

d

dt

(~ω

ρ

)

=

(~ω

ρ.∇

)

~u+1

ρ∇× ~a.

Next, by virtue of ~a = −∇(w + Φ) we obtain

(3.12)d

dt

(~ω

ρ

)

=

(~ω

ρ.∇

)

~u

This equation governs the convection of vorticity in a baratropic fluid and called Helmholtz’sequation. It can be integrated explicitly using the Lagrangian coordinates. Let us introducea vector field ~c(~x, t) according to the definition

(3.13) ωi ≡ ρck∂xi

∂Xk

this being possible since J 6= 0. Substitution of (3.13) into (3.12) yields

d~c

dt= 0 or ~c = ~c( ~X).

Setting, in (3.13), t = 0 we get

~ω0 = ρ0~c ~ω0 ≡ ~ω(~x, 0) ρ0 ≡ ρ(~x, 0)

that allow us to write the explicit expression for ~c

~c =~ω0( ~X)

ρ0( ~X)

(3.14)ωi

ρ=ω0k

ρ0

∂xi

∂Xk

.

There are three important consequences from equation (3.14)

3.4. The Lagrange-Cauchy theorem on the persistence of irrota-tionality

If a fluid particle or a portion of fluid (material volume) is initially in irrotational motion(~ω = 0), then it will retain this property throughout its entire history. In particular itmeans that any flow that appears from the state of rest or from a uniform motion ispotential.

3.5. Vortex lines are material lines

set of particles which composes a vortex line at one instant will continue to form a vortexline at later instants! The proof lies in the fact that a direction (material differential) d~xonce tangent to a vortex line is carried by the fluid so that it is always tangent to a vortexline. Indeed if at t = 0

d~x = d ~X = ~ω0dS

then at any other instant

dxi =∂xi

∂Xk

dXk = dS ω0k

=dXk

∂xi

∂Xk

=ρ0

ρωidS

and d~x is tangent to a vortex-line. In other words, if d~x is parrallel to ~ω at any instant,this property will be valid forever.

Let dℓ be a material arc along a vortex line, then

dℓ0 = ω0dS and dℓ =ρ0

ρωdS

where ω0 = | ~ω0|, ω = |~ω|. Hence, we can write

(3.15)ρ

ωdℓ =

ρ0

ω0dℓ0.

This formula shows, that the stretching of vortex lines increases the vorticity (here is anopen road to turbulence!).

Plane flow

In the plane flow ~ω is normal to the flow plane:

(3.15a) ωx = ωy = 0 ωz = ω =∂v

∂x−∂u

∂y= 0.

hence (~ω.∇)~u = 0 and (3.12) gives

(3.16)d

dt

ρ

)

orω

ρ= f

(~X

).

In other words ωρ

=const. following each particle. Similarly, in an axially-symmetrix flow

ωr = ωz = 0 ωθ = ω =∂v

∂z−∂u

∂r.

Hence ωrρ

=const. following each particle (prove it!).

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 35

3.6. Bernouliun theorems (1738)

A first integral of the equations of motion (either for compressible fluid (3.10) or incom-pressible fluid (3.11)) is commonly called a Bernoulli equation. There are various formswhich it can take, depending on the particular kinematical or dynamical assumptions madeabout the motion, though in all cases the basic expression

(3.17) H ≡1

2u2 +

∫dp

ρ+ Φ

is present. For incompressible fluid, (3.11), it takes the form

(3.18) H ≡1

2u2 +

p

ρ0

+ Φ.

By means of (3.10) the first equation (either from (3.10) or (3.11)) takes the form

(3.18a)∂~u

∂t+ ~ω × ~u = −∇H.

If we assume that the flow is steady (i.e. ∂~u∂t

= 0), then we arrive to the following importantresult.

Bernoulli theorem

Consider the steady baratropic flow of a perfect fluid. Then if ~ω×~u ≡ 0 in the flow regionwe have

(3.19) H ≡ const.

On the other hand, if ~ω × ~u 6= 0 then there exists a set of surfaces

(3.19a) H = const.

each one covered by a network of vortex-lines and steamlines. In particular, H is constanton streamlines. The case (3.19) includes very important irrotational (or potential) flows,in which

(3.20) ~ω ≡ 0 ~u = ∇ϕ

with a velocity potential ϕ = ϕ(~x, t). In the case (3.20) one can derive a more generalresult. Equation (3.18a) takes the form

(∂ϕ

∂t+H

)

= 0

which can be integrated to obtain

(3.21)∂ϕ

∂t+H = f(t)

where f(t) is an arbitrary function of time. Equation (3.21) is known as the Bernoulli

theorem for irrotational flow ; for steady flow it reduces to (3.19). Equations (3.19), (3.19a)and (3.20) constitute complete integrals of the equations of motion; they are of greatimportance in fluid dynamics.

3.7. The Stream function

It is possible to introduce a stream function whenever the equation of continuity can bewritten as the sum of two derivatives.

Plane flows of an incompressible fluid

In this case we have

u = u(x, y, t) v = v(x, y, t) w ≡ 0 ρ ≡ const.

so the equation of continuity

ρt + ∇.(ρ~u) = 0

gives us

(3.22) ∇.~u = 0 or ux + vy = 0.

It follows that the line integral∫u dy− v dx from some fixed point (x0, y0) to the variable

point (x, y) defines a (possibly mutiple-valued) function ψ = ψ(x, y, t) such that

(3.23) u = ψy =∂ψ

∂yv = −ψx = −

∂ψ

∂x.

It is evident, that the curves ψ = const. are the streamlines of the field (prove it!); ψ iscalled the stream function.

From the definition of vorticity((3.15a) and (3.23)

)we obtain the following important

equation satisfied by ψ, namely

(3.24) ψ = ψxx + ψyy =∂2ψ

∂x2+∂2ψ

∂y2= −ω.

This equation serves to determine ψ when the vorticity magnitude is known. Now for thesteady plane flow the equation of motion (3.18a) is equivalent to

H = H(ψ) ω = −∂H

∂ψ(prove it!).

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 37

Thus any solution of the equation

(3.25) ψ = f(ψ)

provides an example of steady two-dimensional flow; of course one must also take intoaccount boundary conditions on ψ.

In irrotational flow a velocity potential ϕ exists and

(3.26) u = ϕx = ψy v = ϕy = −ψx.

The complex function

(3.27) w = w(z, t) ≡ ϕ+ iψ z ≡ x+ iy

is therefore analytic (prove it); this fact is very useful for solving particular problems.

Axially symmetric flows

In the cylindrical coordinates (z, r, θ) velocity components

u = u(z, r, t) v = v(z, r, t) w ≡ 0.

The equation of continuity takes the form:

(3.28)∂

∂r(rv) +

∂z(ru) = 0 (prove it!)

hence we can define a stream function ψ = ψ(r, z, t) such that

(3.29) u =1

rψr v = −

1

rψx.

The equation relating ψ to the vorticity is:

(3.30)∂2ψ

∂z2+∂2ψ

∂r2−

1

r

∂ψ

∂r= −rω (prove it!).

In steady flow H and ω are connected by

(3.31) H = H(ψ) ω = −rH ′(ψ) (prove it!).

Thus any solution of the equation

ψzz + ψrr −1

rψr = rf(ψ)

provides an example of steady axially symmetric flow.

Example 3.2 (Hill’s spherical vortex). Show that,

ψ =1

2Ar2(a2 − z2 − r2)

ω = 5Ar

H = −5Aψ + const.

is an exact solution for an ideal incompressible steady axially symmetric flow. Draw thepicture of streamlines (qualitatively).

3.8. Lagrangian form of the equations of motion

For the case of a perfect fluid (see section 2, equations (2.10)-(2.12)); it is simple to findequations satisfied by ~u, ρ and p as functions of the Lagrangian variables Xα, t. Indeed,the first equation in (2.12)

d~u

dt= ~f −∇(p)

can be transformed using d~udt

= d2~xdt2

and multiplying both sides by ∂xi

∂Xα. The result is,

(3.32)

(∂2xi

∂t2− fi

)∂xi

∂Xα

= −1

ρ

∂p

∂Xα

where xi = xi( ~X, t), p = p( ~X, t), ~f = ~f( ~X, t). This equation together with (1.16)

(1.16) ρJ = ρ0

form the system of governing equations written in Lagrangian coordinates.

3.9. Kelvin’s circulation theorem (1869)

The circulation around any closed curve (circuit) in the fluid is defined by the integral

dx

γ

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 39

Γ ≡

γ

~u.d~x.

If now γ moves with the fluid, i.e. γ = γ(t), we may calculate the rate of change of thecirculation around this moving circuit

dt=

d

dt

γ(t)

~u.d~x.

Kelvin’s Theorem

The baratropic flow of a perfect fluid under the action of a conservative extraneous forceis circulation-preserving

(3.33)d

dtΓ = 0

where (3.33) holds for any material circuit γ(t). In order to prove (3.33) consider thetransformation to Lagrangian coordinates

d

dt

γ(t)

uidxi =d

dt

γ0

ui( ~X, t)∂xi

∂Xk

dXk

=

γ0

(

uit

∂xi

∂Xk

+ ui

∂ui

∂Xk

)

dXk

=

γ(t)

dui

dtdxi

γ(t)

aidxi(3.34)

where γ0 = γ(0) and we also used

γ0

ui

∂ui

∂Xk

dXk =

γ0

d

(~u 2

2

)

= 0.

Now for barotropic flow we recall, from (2.13)-(2.14), that acceleration ~a is derivable froma potential, therefore the last integral in (3.34) vanishes and (3.33) is proven.

One can see that for a potential flow

~u = ∇ϕ

we have the circulation to be

Γ =

γ

~u.d~x =

γ

(∇ϕ).d~x =

γ

dϕ ≡ 0

if the potential function is single-valued.

It is always true for a simply connected domain of flow, so the potential flows with nonzerocirculation in this are forbidden. However it is not true for a multi-connected domain,where we can have a multi-valued potential ϕ(~x, t) and, consequently, several differentcirculations around different components of boundary.

γ

γ1

γ2

γ3

Γ

Γ1

Γ2

Γ3

4. Viscous Fluid

4.1. The Navier-Stokes equations

The general equations for an incompressible continuous medium have been derived as (see(2.6))

(4.1)

dui

dt= fi +

1

ρ0

∂Tik

∂xk

∂ui

∂xi

= 0 in τ .

Here ~u = ~u(~x, t) = (u, v, w) = (u1, u2, u3) is the velocity field, ~x = (x, y, z) = (x1, x2, x3) -

Cartesian coordinates, ρ = ρ0 = const. - the density of fluid, Tik(~x, t) - the stress tensor, ~fis the given external force (per unit mass).

The stress tensor has been specified in eq2.12, as

(4.2) Tik = −pδik + 2µeik eik ≡1

2

(∂ui

∂xk

+∂uk

∂xi

)

where p is pressure, µ is the coefficient of dynamic viscosity, eik is the rate-of-strain (or rate-of-deformation) tensor. The substitution of (4.2) into (4.1) produces the famous Navier-Stokes equations

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 41

(4.3)

d~u

dt= ~f −

1

ρ0∇p+ ν~u

∇.~u = 0

where ν ≡ µ

ρ0is the coefficient of kinematic viscosity, τ is the region occupied by the fluid.

4.2. The Boundary Conditions

For a solid boundary of flow we have introduced the non-slip boundary conditions (see(2.21) and (2.22)):

(4.4) ~u = 0 on ∂τ

for a fixed boundary, and

(4.4a) ~u = ~U on ∂τ

for a boundary moving with a velocity ~U . For the boundary condition (4.4) the stressvector ~t applied from the boundary ∂τ to the fluid is generally non-zero

(4.5) ti = Tiknk = −pni + 2µeiknk 6= 0 on ∂τ

where ~n is the external unit normal vector to ∂τ . The total force acting on the fluid (fromthe wall) will appear as the surface integral

Fi =

∂τ

ti dS =

∂τ

Tiknk dS = −

∂τ

pni dS

︸ ︷︷ ︸

the force of pressure

+ 2µ

∂τ

eiknk dS

︸ ︷︷ ︸

viscous force

.

Let us introduce also the free-surface (or the stress-free) boundary condition

(4.6) ~t = 0 on ∂τ

or

(4.6a) −pni + 2µeiknk = 0 on ∂τ

which corresponds to the absence of surface forcing. In this situation ~u 6= 0 on ∂τ .

4.3. Unidirectional Flows

~u = (u, 0, 0) in Cartesian coordinates, (x, y, z). The full Navier-Stokes equations are:

(4.7)

ρ(ut + uux + vuy + wuz) = −px + µ(uxx + uyy + uzz)

ρ(vt + uvx + vvy + wvz) = −py + µ(vxx + vyy + vzz)

ρ(wt + uwx + vwy + wwz) = −pz + µ(wxx + wyy + wzz)

ux + vy + wz = 0

for unidirectional flow they become

ρ(ut + uux) = −px + ν(uxx + uyy + uzz)

0 = −py

0 = −pz

ux = 0

⇒u = u(y, z, t)

p = p(x, t)

⇒ ρut = −px + µ(uyy + uzz)

since neither the first nor the last term depends on x we obtain

px = −G(t)

when function G > 0, the pressure gradient represents a uniform body force in the directionx > 0. Finally, the main equation for unidirectional flows is

(4.8) ρut = G(t) + µ(uyy + uzz).

In the case of steady flows ut = 0 and G = const, which implies

(4.9) uyy + uzz = −G

µ.

Equations (4.8), (4.9) are to be solved subject to the boundary conditions.

4.4. Tangential Stress on a Wall

The Cauchy stress vector at the wall is

ti = (−pδik + 2µeik)|wall uk = −pni + 2µeiknk.

The first term in the right hand side is the normal force of pressure. To analyse the secondterm we introduce the local (at the wall) coordinateswhere the plane (x, z) is tangent to the wall, ~n is antiparallel to the y-axis ~n = (0,−1, 0).

2eik =∂ui

∂xk

+∂uk

∂xi

=

ux uy + vx uz + wx

uy + vx vy vz + wy

uz + wx vz + wy wz

=

0 uy uz

uy 0 0uz 0 0

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 43

~n

u(y, z)

y

x

z

for the unidirectional flows, which implies

2µeiknk = µ

0 uy uz

uy 0 0uz 0 0

0−10

= µ

−uy

00

.

So, the tangential stress (applied to fluid) is directed against the flow

(4.10) t1 = −µuy.

If we wish to obtain the force acting on the wall we have to change sign:

(4.11) t1 = µuy = µ

(du

dy

)∣∣∣∣at the wall

Formulae (4.10) and (4.11) were originally proposed by Newton, which implies the viscosityas it appears in the Navier-Stokes equations is called the ‘Newtonian viscosity’.

4.5. Poisenille Flow in a Tube

Consider a flow in a long tube of circular section under the action of a difference betweenthe pressures imposed at the two ends of the tube: Hagen (1839), Poisenille (1840). Theflow is axisymmetric u = u(r), see (2.38), then equation (4.9) implies

u =1

r

∂r(rur) = −

G

µ

( - in cylindrical coordinates, see Arfken) then the general solution is

u(r) =G

4µ(−r2 + A log r +B)

p0 > p1

p1p0

0 ℓ x

r

• no singularity at r = 0 ⇒ A = 0• u = 0 at r = a (non-slip), which implies

(4.12) u =G

4µ(a2 − r2)

• the tangential stress at the wall (see (4.11) with y → −r):

t1 = µdu

dr

∣∣∣∣r=a

= −1

2Ga

So, the total frictional force on a length ℓ of the tube is,

2πaℓ

(1

2Ga

)

= πa2(p0 − p1).

Such an expression was to be expected from the conservation of momentum (steady flow⇒ total force = 0). A quantity of practical importance is the flux of volume past anysection of the tube:

Q =

∫ a

0

u2πr dr =πGa4

8µ=πa4(p0 − p1)

8µℓ.

The law Qa−4 ∼ constant was established experimentally by Hagen and Poisneille.

4.6. Plane Poisenille Flow

Consider the unidirectional flow in a gap between two parallel platesThe flow is plane, which implies u = u(y). Now equation (4.9) implies

µuyy = −G

the general solution

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 45

x~n

~n

p1p0

0

z

H

H2

u(y) = −G

2µy2 + ay + b a, b - constant.

Use of the non-slip boundary condition yields:

u = 0 at y = 0 ⇒ b = 0

u = 0 aty = H ⇒ a =GH

(4.13) ⇒ u(y) =G

[

H2

4−

(

y −H

2

)2]

It is the parabolic profile with

umax =GH2

8µat y =

H

2.

The tangential stress on the wall, equation (4.11) implies

t1 = −µdu

dy=GH

2.

So, the total force on both walls is GHℓ, where ℓ is the length of the channel. It is onceagain equal to the total force due to pressure.

4.7. Plane Couette Flow

Consider a plane unidirectional flow in a channel between two parallel plates. The differencewith the previous case: G ≡ 0 (no pressure gradient), but the top plate is moving.(4.9) combined with G = 0 gives uyy = 0. Then the general solution is u = ay + b, with a

and b constant. Now considering boundary conditions we obtain

• u = 0 at y = 0 ⇒ b = 0

u = 0

U0

u(y)

~n

~n

0

y

H

• u = U0 at y = H ⇒ u = U0

Hy.

So, the required profile is linear:

(4.14) u(y) =U0

Hy

The tangential stress on the walls:

t1|y=0 =µU0

Ht1|y=H = −

µU0

H.

The opposite signs are natural because the upper plate provides the driving!

4.8. Flow in a layer on an inclined plane due to gravity

In this case, elementary consideration (see Figure) shows that (4.9) still works, G ⇒ρg sinα, p =const.

x

y

α

p =const.

t1 = 0

stress free surface

H

u(y)

g sinα

g

The boundary conditions

• u = 0 at y = 0

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 47

• t1 = 0 at y = H .

The flow is plane, u = u(y), and equation (4.9) implies

uyy = −ρg sinα

µ

the general solution

u(y) = −ρg sinα

µy2 + ay + b with a, b constant.

Using the boundary conditions we obtain the velocity profile

(4.15) u =G

2µy(2H − y) G ≡ ρg sinα.

Calculating the flux:

Q =

∫ k

0

u dy =Gh3

3µone can find the relation

k =

(3µQ

ρg sinα

) 1

3

If ut 6= 0 ⇒ (4.8) and G = 0 implies

(4.16) ut = ν(uyy + uzz).

It is the famous ‘diffusion equation’. It is interesting to consider a flow caused by an

oscillating plane boundary

τ

x

y

u = U0 cosωt

u→ 0 at y → ∞

The equation we have to solve is

ut = νuyy in τ.

The boundary conditions are u(y, t) = U0 cosωt at y = 0. We are looking for a solution ina complex form:

u(y, t) = ℜeiωtF (y) ⇒ iωF = νF ′′ ⇒ F (y) = A exp

− (1 + i)( ω

)

this is finite as y → ∞

⇒ u(y, t) = U0 exp

−( ω

) 1

2

y

cos

ωt−( ω

) 1

2

y

4.10. Dynamical Similarity and the Reynolds Number

The Navier Stokes equations give

ρ

(∂ui

∂t+ uk

∂ui

∂xk

)

= −∂p

∂xi

+ µ∂2ui

∂xk∂xk

∂uk

∂xk

= 0

L

~u′

=~u

U~x

=~x

Lt′ =

tU

Lp′ =

p

ρU2

∂u′i∂t′

+ u′k∂u′i∂x′k

= −∂p′

∂x′i+

1

Rℓ

∂2u′i∂x′k∂x

′k

∂u′k∂x′k

= 0

Then the Reynolds number is defined to be

Rℓ =ρLU

µ

Result: all problems with the same Rℓ are the same! (= dynamical similarity).

TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 49

The ‘drag’ force ~F ,

Fi = −

∂τ

Tijnj dS = −ρU2L2

−p′δik +1

Rℓ

(∂u′i∂x′k

+∂u′k∂x′i

)

nk dS ′

⇒~F

12ρU2L2

= C0(Rℓ)

where C0 is the drag coefficient. This a universal formula for all spheres etc.

0.4

0.8

C0