Fluid Mechanics
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
x1
x1
A
τ
τ
A
x2
x3
x2
x3
x1 , x2 , x3 - the fixed rectangular coordinate system. ~x = (x1 , x2 , x3 ) - The position of the
particle A.
Say A is a particle moving with fluid:
~ = (X1 , X2 , X3 )
• at t = 0 it occupies the postion X
• at time t it has moved to the position ~x = (x1 , x2 , x3 )
~ and t:
Then ~x is determined as a function of X
(1.1)
~ t)
~x = ϕ(X,
or
1
xi = ϕi (Xi , t)
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PROF. VLADIMIR VLADIMIROV
~ is fixed while t varies then (1.1) gives the path of the particle A intially at X.
~
If X
We assume, that initially distinct points remain distinct throughout the entire motion; it
means that the transformation (1.1) posesses the inverse
~ = Φ(~x, t)
X
(1.2)
or
Xi = Φi (xi , t)
It is also usually assumed that ϕi and Φi posses continuous derivatives up to the third
order 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
~ during the course of time:
or X
• ρ = ρ(~x, t), ~u(~x, t) - Eulerian description
~ t), ~u(X,
~ t) - Lagrangian description
• ρ = ρ(X,
• ~x - spatial variables, i.e. Eulerian variables
~ - material variables, i.e. Lagrangian variables
• X
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
~ t) and conversely. To indicate a particular set
also a funtion of Lagrangian variables (X,
of variables we write either
~ t)
F = F (X,
or
~ t), t)
F = F (~x(X,
Geometrically
~ t) is the value of F experienced at time t by the particle initially at X;
~
• F (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)
for the
•
•
•
•
~ t)
∂F
∂F (~x, t)
∂F (X,
dF
≡
≡
∂t
∂t
dt
∂t
two possible time derivatives of F , obviously they are quite different quantites:
dF
is called the material derivative of F ;
dt
d
is the differentiation following the motion of the fluid;
dt
dF
measures the rate of change of F following a particle;
dt
∂F
gives the rate of F apparent to a viewer stationed at the position ~x.
∂t
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3
1.3. Velocity and Acceleration
The velocity of a particle is given by the definition
(1.4)
~u ≡
d~x
dt
ui ≡
~ t)
dxi
∂ϕ(X,
≡
.
dt
∂t
As defined, ~u is a function of the material variables. However, one usually deals with the
spatial 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 be
able to proceed in the opposite direction; to detemine the motion (1.1) from ~u(~x, t). This
transistion is given by solving the system of ODE’s
d~x
= ~u(~x, t)
dt
~
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)
~ t)
dui
∂ui (X,
≡
dt
∂t
~
dui ~x(X, t), t
∂ui
∂ui dxk
dui
≡
=
+
dt
dt
∂t
∂xk dt
ai ≡
≡uk
or
~a =
∂~u
∂~u
+ ~u.∇(~u) ≡
+ (~u.∇)~u.
∂t
∂t
Correspondently, the general formula for differentiation following the motion of the fluid
is:
4
(1.6)
PROF. VLADIMIR VLADIMIROV
∂F
dF
=
+ (~u.∇)F
dt
∂t
~ t), t
F = F (~x, t) = F ~x(X,
~ t)
∂F
dF
∂F
∂F
∂F ∂F ∂xk (X,
∂F
=
=
=
≡
+
+ uk
+ (~u.∇)F
dt
∂t X~
∂t
∂xk
∂t
∂t
∂xk
∂t
≡uk
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 instantaneously at the position ~x.
1.4. Jacobian and Euler’s Formula
~ coordinates is given by
The basic transformation between Eulerian ~x and Lagrangian X
(1.1) and (1.2). The Jacobian of the transformation is:
∂(x1 , x2 , x3 )
J =
= det
∂(X1 , X2 , X3 )
∂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)
dx1 dx2 dx3 = J dX1 dX2 dX3
Euler’s formula for J
We shall use Euler’s formula:
(1.9)
dJ
= J ∇.~u
dt
Proof. Let Aik be the cofactor of the element
minant:
∂xi
∂Xk
in the expression of the Jacobian deter-
∂xi
Ajk = J δij
∂Xk
The derivative of J is,
TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS
d
dJ
=
dt
dt
=
∂xi
∂Xk
Aik =
5
∂ui
∂ui ∂xj
Aik =
Aik
∂Xk
∂xj ∂Xk
=J δij
∂ui
∂ui
δij J =
J = J ∇.~u
∂xj
∂xi
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
Z
F dτ.
τ (t)
Its derivative is given by the important formula
(1.11)
d
dt
Z
τ (t)
F dτ =
Z
τ (t)
dF
+ F ∇.~u dτ
dt
Proof. Consider the correspondence between the Eulerian coordinates and the Lagrangian
coordinates (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)
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PROF. VLADIMIR VLADIMIROV
Then,
Z
F dτ =
τ (t)
Z
~ t)J dτ0
F (X,
τ0
where dτ = J dτ0 ; dτ = dx1 dx2 dx3 , dτ0 = dX1 dX2 dX3 (see (1.8)). Differentiation yields
d
dt
Z
Z ∂J ∂F +F
dτ0
J
F dτ =
∂t X~
∂t X~
τ (t)
τ0
Z dF
=
+ F ∇.~u dτ
dt
τ (t)
where we have used (1.3) and (1.9).
Another form of (1.11) can be obtained after transformation
∂F
dF
+ F (∇.~u) =
+ ∇.(~uF )
dt
∂t
(1.12)
which gives
(1.12a)
d
dt
Z
F dτ =
τ (t)
Z
τ (t)
∂F
dτ +
∂t
I
F (~u.~n) dS
∂τ (t)
~n
τ (t)
∂τ (t)
It should be emphasised that (1.11) and (1.12) express a kinematical theorem, independant
of 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=
Z
ρ dτ
τ
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7
Physics
We postulate the principle of conservation of mass. “The mass of a fluid in material volume
τ does not change as τ (t) moves with the fluid”.
d
dt
(1.13)
Z
ρ dτ = 0
τ (t)
now from (1.13) and (1.11) it follows that
Z dρ
+ ρ(∇.~u) dτ = 0
dt
τ (t)
and since τ (t) is an arbitrary volume, this implies:
dρ
+ ρ(∇.~u) = 0.
dt
This is the Eulerian form of the equation of continuity. We can express this in the alternative form
(1.14)
∂ρ
+ ∇.(ρ~u) = 0.
∂t
Now, multiplying (1.14) by J and using (1.9) we derive the Lagrangian form of the equation
of continuity:
(1.15)
d
(ρJ ) = 0
ρJ = ρ0
dt
where ρ = ρ0 is the initial density distribution. For future use, let us also write that for
an arbitrary function F (~x, t)
(1.16)
(1.16a)
d
dt
Z
ρF dτ =
τ (t)
Z
τ (t)
ρ
dF
dτ
dt
(prove it!)
To prove this equality one should use (1.9) and (1.14).
1.7. The Rate of Strain Tensor
∂ui
, namely
Let us introduce a simple decomposition of the tensor ∂x
k
(1.17)
where
∂ui
= eik + Ωik
∂xk
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The tensors eik
PROF. VLADIMIR VLADIMIROV
1 ∂uk
∂uk
∂ui
1 ∂ui
Ωik =
+
−
eik =
2 ∂xk
∂xi
2 ∂xk ∂xk
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 is
given by the formula:
d
d
(dxi ) =
dt
dt
∂xi
dXk
∂Xk
=
∂ui
∂ui
dXk =
dxj
∂Xk
∂xj
or simply
d
(d~x) = (d~x.∇)~u.
dt
(1.18)
From (1.18) we get
d
(dℓ2 ) = 2eik dxi dxk
dt
where dℓ = |d~x|. Thus the tensor eik is a measure of the rate of change of the squared
element of arc following a fluid motion.
(1.19)
In a rigid motion dℓ ≡ const, whence a necessary condition that a motion be localy and
instantaneously rigid is that eik = 0 (see in homework). For this reason, eik is called
the deformation tensor (or the rate of strain tensor). The tensor eik − 31 ejj δik is also of
interest, for its vanishing is the necessary and sufficient condition that the motion, locally
and instantaneously preserves angles (prove it!).
If eik = 0 everwhere in the fluid, then the motion is rigid and
1
~u = ~ω × ~r + const.
2
ω
~
where 2 is the constant angular velocity of the motion. Equation (1.19a) can be derived
analytically as the integral of the system of first order partial differential equations eik = 0.
(1.19a)
1.8. Vorticity and the Potential of Velocity
Let us consider the velocity field in the neighbourhood of a fixed point P . Denoting the
evaluation of a quantity at the point P by a superscript, we have near P
ui =
uPi
+ rk
∂ui
∂rk
P
+ O(r 2 )
TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS
9
where ~r denotes the radius vector from P . Neglecting terms of order r 2 and using (1.17)
we get
(1.20)
ui = uPi + ePik rk + ΩPik rk
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 = ePik ri rk
then the second term can be written in the form
(1.21)
1
∇ e
2
or
1 ∂e
.
2 ∂xi
This term represents a veloctiy field normal at each point to the quadric surface e = const
which passes through that point. In this velocity field there are 3 mutually perpendicular
directions 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 in
these directions.
r2
~n
extension and rotation
r1
pure extension
The final term in (1.20) may be written as
(1.22)
where ~ω = ∇ × ~u is the vorticity vector.
1 P
~ω × ~r
2
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PROF. VLADIMIR VLADIMIROV
The form (1.22) shows clearly that the final term in (1.20) represents a rigid motion of
angular velocity 21 ~ω P .
By considering the above results, the identity (1.17) can be fully interpreted. For an
arbitrary motion, the velocity ~u in the neighbourhod of a fixed point P is given, up to
terms of order r 2 , by
P
(1.23)
~u = ~u + ∇
1
1
e + ~ω P × ~r
2
2
where e = eik ri rk is the rate of strain quadric and ~ω = ∇ × ~u is the vorticity vector. Thus
an 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 angular
velocity of rotation is 12 ~ω P . 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 instantaneously
a 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, then
the relative motion of any element of that position consists of a pure deformation, and is
called 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
∂2ϕ
∂ul
= εikl
≡0
ωi = εikl
∂xk
∂xk ∂xl
it is identically equal to zero, since it is the scalar product of a symmetric tensor
∂2ϕ
and antisymmetric tensor εikl . So
∂xk ∂xl
~u = ∇ϕ ⇒ ~ω = 0.
Step 2: Let ~ω = ∇ × ~u = 0. Then let
ϕ=
Z
~
x
~u.d~x
x~0
the definition of the potential ϕ(~x, t). Evidently ∇ϕ = ~u. However, we have to
prove that ϕ(~x, t) is a single valued function. It means that the integral,
TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS
Z
11
~
x
~u.d~x
x~0
does not depend on the path of integration. This statement is equivalent to the
equality
I
(1.24a)
~u.d~x = 0
γ
for any closed circuit γ. Using stokes theorem
I
γ
~u.d~x =
Z
ω
~ .~n dS
Sγ
for any surface Sγ ‘based’ on γ. If ω
~ = 0 then
valued function.
H
~u.d~x = 0 and ϕ(~x, t) is a single
γ
1.9. Streamlines, Vortex Lines and Trajectories
A curve, everywhere tangent to a given continuous vector field is called a vector-line. In
particular, the vecor-lines of the velocity field are called streamlines, and the vector-lines
of the vorticity are called vortex-lines.
Equations of Streamline
(1.25)
~x = ~x(s);
or
d~x
= ~u ~x(s), t ;
ds
t = const.
dy
dz
dx
=
=
;
u(x, y, z, t)
v(x, y, z, t)
w(x, y, z, t)
t = const.
Equation of a Vortex line
(1.26)
d~x
=ω
~ ~x(s), t ;
ds
t = const.
We define a trajectory to be the curve traced out by a particle as time progresses.
12
PROF. VLADIMIR VLADIMIROV
Equation of a Trajectory (or Pathline)
d~x
= ~u ~x(t), t
dt
(1.27)
or
dy
dz
dx
=
=
= dt
u(x, y, z, t)
v(x, y, z, t)
w(x, y, z, t)
u
=
0
then the streamlines and
with suitable initial conditions. If ~u is independent of t, ∂~
∂t
trajectories 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 whose
resultant and moment (torque) are equivalent to those of the actual forces exerted by the
material outside S upon that inside.”
• ~t - surface force (stress);
• ~n - the outward unit normal to S.
~t
~t
~n
~t
It is assumed that ~t = ~t(~x, t, ~n), for all ~x ∈ S. Now we can set the fundamental principle
of 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 the
resultant force on the volume
d
dt
(2.1)
Z
ρ~u dτ =
τ (t)
Z
I
ρf~ dτ +
τ (t)
~t dS
∂τ (t)
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
Z
Z
I
d~u
~
~t dS
ρ
dτ =
ρf dτ +
τ (t) dt
τ (t)
∂τ (t)
This equality is valid for any instant t and does not contain any differentiation of integrals
over variable volume τ (t). Therefore here the integration over a moving volume can be
replaced without loss of generality by integration over a fixed volume.
Z
d~u
ρ
dτ =
τ0 dt
(2.2)
Z
ρf~ dτ +
τ0
I
~t dS.
∂τ0
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 → 0
we obtain
(2.3)
lim ℓ
ℓ→0
−2
I
~t dS = 0
∂τ0
that is, the stress forces in local equilibrium!
x3
~ı
x1
Σ
~n
~k
~
x2
Consider the tetrahedron with vertex at an arbitrary point ~x and with three of its faces parallel 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 n1 Σ, n2 Σ, n3 Σ,
Σ. Now let us consider (2.3). Since ℓ2 ∼ Σ we
can 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 :
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PROF. VLADIMIR VLADIMIROV
(2.4)
ti = Tik nk ;
Tik = Tik (~x, t).
The matrix (tensor) Tik is called the stress tensor. Using (2.4) we transform (2.2) to the
form
Z
dui
ρ
dτ =
dt
τ0
and since τ0 is arbitrary
(2.5)
ρ
Z ∂Tik
ρfi +
dτ
∂xk
τ0
∂Tik
dui
.
= ρfi +
dt
∂xk
This is the equation of motion discovered by Cachy (1828). As the result we have obtained
four equations (2.5) and (1.14)
dui
∂Tik
= ρfi +
ρ
dt
∂xk
dρ
+ ρ∇ · ~u = 0
dt
(2.6)
They relate to each other four quantities ρ, ~u and contain still undefined Tik . Tensor Tik (the
stress tensor) can be expressed in terms of ρ, ~u by direct mechanical or thermodynamical
assumptions. 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 conservation 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
Z
ρ ~r × ~u dτ =
τ
Z
where τ is an arbitrary material volume.
Proof. Using (1.9), (1.3) we get
τ
ρ ~r × f~ dτ +
I
∂τ
~r × ~t dS
TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS
d
dt
(2.9)
15
d~u
ρ ~r × ~u dτ = ρ ~r ×
dτ
dt
τ
τ
Z
I
Z
= ρ ~r × f~ dτ +
~r × ~t dS − T~0 dτ
Z
Z
τ
∂τ
where T0i ≡ ǫikl Tkl (prove it!). By virtue of (2.7) we have T~0 = 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
d~u
ρ
= ρf~ − ∇ × p
dt
(2.12)
dp + ρ(∇.~u) = 0
dt
and usually are referred to as Euler’s equations. This system contains four equations for
five 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 the
∂ρ
∂p
necessary conditions are ∂p
6= 0, ∂ρ
6= 0. From physical reasons one can see, that
always
c2 ≡
∂p
>0
∂ρ
16
PROF. VLADIMIR VLADIMIROV
where c(ρ) is the Speed of Sound. (the increasing of pressure does correspond to
the increasing of density). Equations (2.12) can be written as
(2.13)
c2
1
∇p(ρ)
=
−∇w
=
−
∇ρ
~
u
+
(~
u
.∇)~
u
=
−
t
ρ
ρ
ρt + ∇.(ρ~u) = 0
s
∂p
c(ρ) ≡
∂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 of
equations.
ii) Incompressible Fluid: It corresponds to the case ρ(p) = const. = ρ0 . Equations
(2.12) can be written as
(2.14)
1
~
~ut + (~u.∇)~u = − ρ0 ∇p + f
∇.~u = 0
ρ = ρ0 = const.
This system of four equations also contains four unknown functions, so it is also
closed.
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
1
eik ≡
2
∂ui
∂uk
+
∂xk
∂xi
is the rate of deformation tensor, µ is a constant coefficient known as the coefficient of
dynamic viscosity; the combination ν ≡ µρ is 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 famous
Navier-Stokes equations for a viscous incompressible fluid:
TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS
17
1
~ut + (~u.∇)~u = − ∇p + ν△~u + f~
ρ0
∇.~u = 0
∂2
△
≡
∂xi ∂xj
(2.16)
This is also a closed system of equations which represent the most basic mathematical
model in the modern fluid dynamic.
Note. The main argument towards (2.14) is; the ‘viscous part’ of stresses Tik should be
proportional to the ‘amplitude’ of deformations, which can be measured by eik . In particular, the ‘viscous stresses’ must be zero in the absence of deformations eik ≡ 0 (which
~ +Ω
~ × ~r with constants U
~ and Ω,
~
corresponds to the motion of fluid as a solid body, ~u = U
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 basis
of 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 the
fluid can not ‘leak’ or ‘penetrate’ into the boundary. For the stationary (fixed) boundary
∂τ it means that
~u n ≡ ~u.~n = 0 on ∂τ
(2.17)
where ~n is the normal unit vector to ∂τ . If the boundary is moving with the velocity
~ =U
~ (t), then (2.17) should be modified as
U
(2.18)
~
~u − U
n
~ .~n = 0 or ~u.~n = U
~ · ~n
≡ ~u − U
on ∂τ
~ = 0; the conditions for U
~ =
In the remainder of this lecture we take U
6 0 can be restored
in a straight forward way.
It is proven that it all flow boundary ∂τ is solid, then for the perfect fluid (2.17) gives us
the full set of boundary data; any additional to (2.17) boundary condition will produce
a mathematically controversial problem. So, for the perfect fluid the system of governing
equations is usually written together with the boundary conditions as
18
PROF. VLADIMIR VLADIMIROV
1
d~u
= − p(ρ) + f~
ρ
dt
dp
+ ρ(∇.~u) = 0
dt
~u.~n = 0 on ∂τ
(2.19)
for the compressible baratropic fluid (2.13) and
(2.20)
d~u
1
= − ∇p + f~
dt
ρ0
∇.~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 a
viscous 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 unified
into one
(2.23)
~u = ~u n + ~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 between
the molecules of fluid and solid). Mathematically the doubled number of boundary conditions ((2.17) and (2.22) for viscous fluid in comparison with (2.17) only for an inviscid
fluid) 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 (for
ν = 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 written
as the set of the Navier-Stokes equations along with the non-slip boundary conditions:
TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS
(2.24)
19
1
d~u
~
in τ
dt = − ρ ∇p + ν△~u + f
0
∇.~u = 0
ρ0 = const.
~u = 0 on ∂τ
The systems of governing equations (2.19), (2.20) and (2.24) represent the main result of
this Lecture. In order to make the mathematical formulations complete, all these systems
must be complemented by the initial data for the velocity field
(2.25)
~u(~x, 0) = u~0 (~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 of
external forcing is described by Euler’s equations (2.14):
(2.26)
d~u = − 1 ∇p
dt
ρ0
∇.~u = 0
In Cartesian coordinates (x, y, z) the components of velocity are ~u = (u, v, w). Write the
component 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
v + uv + vv + wv
t
x
y
z
wt + uwx + vwy + wwz
ux + vy + wz
1
px
ρ0
1
= − py
ρ0
1
= − pz
ρ0
=0
=−
20
PROF. VLADIMIR VLADIMIROV
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)
1
ut + uux + vuy = − px
ρ0
1
vt + uvx + vvy = − py
ρ0
ux + vy = 0
Example 2.2. A perfect incompressible fluid of a constant density ρ = ρ0 in the absence
of external forces is described by Euler’s equations (2.15):
(2.31)
d~u = − 1 ∇p
dt
ρ0
∇.~u = 0
In the cylindrical coordinates (r, ϕ, z) the components of velocity are ~u = (u, v, w). Write
the 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)
d
In (2.31) operators dt
, ∇p and ∇.~u must be expressed in the cylindrical coordinates.
Euler’s equations take the form,
(2.33)
du
dt
z
}|
{ v2
1
= − pr
u
+
(~
u
.∇)u
−
t
r
ρ0
dv
dt
}|
{ uv
1
z
vt + (~u.∇)v +
pϕ
=−
r
rρ0
dw
dt
}|
{
z
1
w
+
(~
u
.∇)w
= − pz
t
ρ0
ur + u + 1 vϕ + wz = 0
r r
TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS
21
where
∇=
∂ 1 ∂ ∂
,
,
∂r r ∂ϕ ∂z
so, that
~u.∇ = u
∂
v ∂
∂
+
+w
∂r r ∂ϕ
∂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,
du
dt
z
}|
{ v2
1
u
+
uu
+
wu
= − pr
t
r
z −
r
ρ0
dv
dt
}|
{ uv
z
vt + uvr + wvz +
=0
r
dw
dt
}|
{
z
1
wt + uwr + wwz = − pz
ρ0
u
u r + + wz = 0
r
(2.35)
The second equation can be rewritten for γ = rv;
dγ
= γt + uγr + wγz = 0
dt
(2.36)
That equation represents a special case of the conservation of the circulation of
velocity, 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
22
PROF. VLADIMIR VLADIMIROV
du
dt
z
}|
{
1
u
+
uu
+
wu
pr
t
r
z = −
ρ0
dw
dt
(2.38)
}|
{
z
1
w
+
uw
+
ww
pz
t
r
z = −
ρ0
u
ur + + wz = 0.
r
The condition v ≡ 0 is natural from the view point of equation (2.36). Indeed if we
take 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 incompressible fluid of a constant density ρ0 in a gap between two concentric cylinders of radii
R2 > R1 . Obtain;
i) the general solution of this type: calculate its vorticity ω from Stokes’s theorem
and from the formula
1
1
ω = (rv)r − uϕ ,
r
r
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:
(2.38a)
y
v
u
r
ϕ
τ
x
R1
R2
TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS
23
i) A plane steady rotationally-symmetric solution in polar coordinates (r, ϕ) can be
obtained 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
v2
1
uu
−
= − pr
r
r
ρ0
uv
(2.40)
uvr +
=0
in τ
r
ur + u = 0
r
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
(ru)r = 0;
r
the only solution to this is u(r) = Qr with 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
1
v2
= pr or p = ρ0
(2.42)
r
ρ0
Therefore the required solution is
(2.43)
u≡0
v = V (r)
Z
r
V 2 (ξ)
dξ
ξ
p = P (r) = ρ0
Z
r
V 2 (ξ)
dξ
ξ
with an arbitrary function V (r). The vorticity of this flow is
(2.44)
(2.45)
V
1
Ω(r) = (rV )r = Vr + .
r
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:
Γ = Γ(r) = 2πrV (r)
24
PROF. VLADIMIR VLADIMIROV
ii) Solution with constant vorticity Ω0 according to (2.44) must satisfy the equation
1
(rV )r = Ω0
r
It follows that,
1
(rV )r = Ω0 r ⇒ rV = Ω0 r 2 + C
2
1
C
⇒ V = Ω0 r +
(2.46)
2
r
which contains two arbitrary constants Ω0 and C.
iii) The irrotational (potential) solution satisfies the equation
1
(rV )r = 0
r
and can be obtained from (2.46) after taking Ω0 = 0
(2.47)
Γ
2πr
=const. This definition of Γ is in agreement with
V =
where we have taken C =
(2.45).
Γ
2π
Example 2.4 (Lamb’s form of Euler’s equation). A perfect barotropic gas in the absence
of external forcing is described by Euler’s equations (2.13)
(2.48)
d~u = −∇w w = w(ρ) dw = dp
dt
ρ
ρt + ∇.(ρ~u) = 0
Show that the first equation in (2.48) can be written as
(2.49)
u2
~ut + ~ω × ~u = −∇ w +
2
where ~ω = ∇ × ~u. Write the counterpart of equation (2.49) for the perfect incompressible
fluid.
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
(~ω × ~u )i = −ǫikl ǫlmn uk
∂un
∂xm
= −(δim δkn − δin δkm )uk
= −uk
25
∂un
∂xm
∂ui
∂uk
+ uk
.
∂xi
∂xk
or in vector form:
~u 2
~ω × ~u = −∇
+ (~u.∇) u
2
and (2.50) is proven. Substitution of (2.50) into the first equation (2.48) gives the required
form (2.49). For the perfect incompressible fluid the transformation (2.49) is the same, so
the equation is
(2.51)
~ut + ~ω × ~u = −∇
~u 2
p
+
ρ0
2
Example 2.5 (Sound Waves. Linearization). Compressible fluid (gas) is described by
(2.13):
(2.52)
c2
~
u
+
(~
u
.∇)~
u
=
−
∇ρ
t
ρ
ρt + ∇.(ρ~u) = 0
s
∂p
c = c(ρ) =
∂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 + ρ˜0 (~x, t)
Substitution of (2.53) into (2.52) yields;
[c(ρ0 + ρ˜)]2
u˜t + u˜u˜x = −
ρ˜x
ρ0 + ρ˜
ρ˜t + (ρ0 + ρ˜)˜
u x=0
26
PROF. VLADIMIR VLADIMIROV
Here for the sake of simplicity we consider one-dimensional problem (general 3D problem
is given in the Homework). Now we use the lineraization procedure. It means that we keep
only the linear (with respect to ‘tilde’ variables) terms and neglect all their squares, cubes,
etc. The result is:
2
u˜ = − c0 ρ˜
t
x
ρ0
ρ˜t + ρ0 u˜x = 0
now we can derive equations for ρ˜ only or for u˜ only.
ρ˜tt − c20 ρ˜xx = 0
Both equations have the form
(2.54)
u˜tt − c20 u˜xx = 0
ftt − c20 fxx = 0
c20 = const.
which is well known as the wave equations. It can be solved by introducing of new independent variables
ξ ≡ x − c0 t
and
η ≡ x + c0 t.
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 − c0 t) + h(x + c0 t)
which represents the combination of ‘the wave travelling to the right’ and ’the wave travelling to the left’. The propagation of both waves takes place without any deformations of
their profiles. The absence of the deformations is vitally important in our life: it allows us
to hear the signal in the same form, as it has been generated!
Example 2.6 (Archimede’s Theorem). Now we consider an incompressible fluid, either
viscous (2.16) or inviscid (2.14). For the state of rest (~u ≡ 0) both these systems of
equations are reduced to
(2.55)
0=−
1
∇p + ~g
ρ0
where ~g is the gravity field, representing in this case the force per unit mass f~ = ~g . This
is 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 of
Partial Differential Equations for pressure:
TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS
27
px = 0
py = 0
p = −ρg
z
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 equilibrium. Let us consider a solid volume V submerged into the fluid.
x3
∂τ
~n
~g
z
τ
x2
V
x1
y
x
The force acting on the solid is
F~ = −
I
p~u dS
(justify this!)
∂τ0
The use of (2.21) and transformations yield:
−
I
pui dS = −
∂τ
I
(p0 − ρgx3 )ni dS = I1 + I2
∂τ
where we have;
I1 = −p0
I
∂τ
=0
ni dS
28
PROF. VLADIMIR VLADIMIROV
prove it! Also,
I2 = ρg
I
Z
x3 ni dS
∂τ
∂x3
dτ
τ ∂xi
0
= ρg 0 V
1
= −ρgi V
= ρg
So finally we have
F~ = −ρ~g V
(2.57)
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 flows
will 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
Z
ρui dτ =
Z
ρfi dτ +
τ0
τ0
I
(ti − ρui uk nk ) dS
(prove it!)
∂τ0
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
Z
F~ dτ =
τ (t)
now taking F~ ≡ ρ~u this gives
(3.1a)
d
dt
Z
τ (t)
ρ~u dτ =
Z
τ (t)
∂ F~
dτ +
∂t
I
Z
τ (t)
∂ρ~u
dτ +
∂t
I
∂τ (t)
∂τ (t)
F~ ~u.~n dS
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
(3.1b)
Z
τ (t)
∂ρ~u
dτ =
∂t
Z
ρf~ dτ +
τ (t)
I
∂τ (t)
29
h
i
~t − ρ~u ~u.~n dS
In this expression we moved the t-differentiation under the integral sign. It means that
equality is valid at any instant t separately (t in τ (t) plays a part of a parameter). Therefore in (3.1b) the integration over a moving volume can be replaced (without the loss of
generality) by integration over a fixed volume: (the fact that τ (t) is moving does not play
any role in (3.1b)!)
(3.1c)
Z
τ0
∂ρ~u
dτ =
∂t
Z
ρf~ dτ +
I
∂τ0
τ0
h
i
~t − ρ~u ~u.~n dS.
The partial derivative on the left hand side can be pulled out of the integral sign. The
result is the required expression (3.1)
I h
Z
Z
i
d
~t − ρ~u ~u.~n dS
ρf~ +
ρ~u dτ =
dt τ0
∂τ0
τ0
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 − ρui uk nk = (Tik − ρui uk )nk
is known as the flux of momentum through the fixed ‘test’ surface of unit area and with
the unit normal vector ~n. The first term in P~ represents the flux of momentum due to the
surface force ~t; the second term −ρui uk nk 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 external
force field is zero. Then if ∂τ0 denotes the surface of the obstacle and ∂Σ denotes a ‘control
surface’ enclosing ∂τ0 , we have the following formula for the force F acting on the obsticle:
(3.3)
F~ = −
I
~t dS =
∂τ0
I
∂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~ . The
values 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
and (3.1)-(3.3) become slightly simpler.
Pi = (−pδik − ρui uk )nk
30
PROF. VLADIMIR VLADIMIROV
~n
∂τ0
~n
f~ = 0
∂Σ
3.2. The energy transfer equation
Let K denote the kinetic energy of a volume τ
1
K≡
2
Z
ρui uk dτ.
τ
Then for an arbitrary material volume τ (t) we have
(3.5)
dK
=
dt
Z
τ (t)
ρfi ui dτ +
I
~t.~u dS −
∂τ (t)
Z
Tik eik dτ
τ (t)
where eik is the rate-of-strain tensor (1.17).
Proof. We show this by differentiating the kinetic energy,
(prove it!)
TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS
31
Z
d
dK
ρui ui dτ
=
dt
dt τ (t)
Z
dui
=
ρui
dτ
dt
τ (t)
Z
∂Tik
=
ui ρfi +
dτ
∂xk
τ (t)
Z
Z ∂ui
∂ui Tik
dτ
− Tik
=
ρui fi dτ +
∂xk
xk
τ (t)
τ (t)
Z
Z
I
∂ui
∂uk
1
Tik
+
dτ
=
ρui fi dτ +
ui Tik nk dS −
2 τ (t)
∂xk
∂xi
τ (t)
∂τ (t)
Z
Z
Z
=
ρui fi dτ +
ui ti dS −
Tik eik dτ
τ (t)
∂τ (t)
τ (t)
where
1
eik =
2
ti = Tik nk
∂ui
∂uk
+
∂xk
∂xi
It states that the rate of change of kinetic energy of a moving volume is equal to the
rate 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
Z
ρ f~.~u dτ −
τ (t)
Z
p ~u.~n dS +
∂τ (t)
Z
p∇.~u dτ
(prove it!)
τ (t)
The last term is the rate at which work is done by the pressure in changing the volume of
fluid elements.
A simplification of (3.5) and (3.6) may be obtained if f~ = −∇ × Φ, where Φ = Φ(~x) is a
time independent potential. In this case we can introduce the potential energy
Z
Π ≡ ρΦ dτ
τ
and (3.5) becomes
(3.7)
d
(K + Π) =
dt
I
~t.~u dS −
∂τ
Z
Tik eik dτ
τ
It is interesting to consider separately the term
(please prove it!)
32
PROF. VLADIMIR VLADIMIROV
D≡−
Z
Tik eik dτ
τ
which represents the rate of energy dissipation in the volume τ . For a perfect fluid Tik =
−pδik and the integrand is
∂ui
= −p∇ × ~u.
∂xi
One can see, that it gives the last term in (3.6), which is equal to zero for an inviscid
in-compresible fluid. For the viscous incompressible fluid
Tik eik = −peii = −p
(2.12)
Tik = −pδik + 2µeik
so
(3.8)
D = −2µ
Z
eik eik dτ.
τ
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. The
equations of motion, (2.12), are
ρ~a = −∇p − ∇Φ
dρ + ρ∇.~u = 0
dt
(3.9)
where acceleration is
d~u
∂~u
=
+ (~u.∇)~u.
dt
∂t
There is a well known vector identity,
~a =
1 2
(~u.∇)~u = ~ω × ~u + ∇
u
2
(3.10)
so ~a = ~ut + ~ω × ~u + ∇
1 2
u
2
, ~ω ≡ ∇ × ~u. Calculations of ∇ × ~a yields
∂~ω
+ ∇ × (~ω × ~u)
∂t
d~ω
(3.11)
− (~ω .∇)~u + ~ω (∇.~u).
=
dt
From (3.11) and the equation for ρ, (3.9), we obtain
∇ × ~a =
TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS
d
dt
33
~ω
~ω
1
=
.∇ ~u + ∇ × ~a.
ρ
ρ
ρ
Next, by virtue of ~a = −∇(w + Φ) we obtain
d
dt
(3.12)
~ω
~ω
=
.∇ ~u
ρ
ρ
This equation governs the convection of vorticity in a baratropic fluid and called Helmholtz’s
equation. It can be integrated explicitly using the Lagrangian coordinates. Let us introduce
a 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
=0
dt
or
~
~c = ~c(X).
Setting, in (3.13), t = 0 we get
ω~0 = ρ0~c
ω~0 ≡ ~ω (~x, 0)
ρ0 ≡ ρ(~x, 0)
that allow us to write the explicit expression for ~c
~c =
~
ω~0 (X)
~
ρ0 (X)
which finally leads to
(3.14)
ωi
ω0k ∂xi
=
.
ρ
ρ0 ∂Xk
There are three important consequences from equation (3.14)
3.4. The Lagrange-Cauchy theorem on the persistence of irrotationality
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 it
means that any flow that appears from the state of rest or from a uniform motion is
potential.
34
PROF. VLADIMIR VLADIMIROV
3.5. Vortex lines are material lines
set of particles which composes a vortex line at one instant will continue to form a vortex
line at later instants! The proof lies in the fact that a direction (material differential) d~x
once tangent to a vortex line is carried by the fluid so that it is always tangent to a vortex
line. Indeed if at t = 0
~ = ω~0 dS
d~x = dX
then at any other instant
dxi =
ρ0
∂xi
∂xi
dXk = dS ω0k
= ωi dS
∂Xk
∂Xk
ρ
=dX
k
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 = ω0 dS
and
dℓ =
ρ0
ωdS
ρ
where ω0 = |ω~0 |, ω = |~ω |. Hence, we can write
ρ
ρ0
dℓ = dℓ0 .
ω
ω0
(3.15)
This formula shows, that the stretching of vortex lines increases the vorticity (here is an
open road to turbulence!).
Plane flow
In the plane flow ~ω is normal to the flow plane:
(3.15a)
ωx = ωy = 0
ωz = ω =
∂v ∂u
−
= 0.
∂x ∂y
hence (~ω .∇)~u = 0 and (3.12) gives
d
dt
(3.16)
In other words
ω
ρ
ω
ρ
ω
rρ
ω
~ .
=f X
ρ
=const. following each particle. Similarly, in an axially-symmetrix flow
ωr = ωz = 0
Hence
or
ωθ = ω =
=const. following each particle (prove it!).
∂v ∂u
−
.
∂z
∂r
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35
3.6. Bernouliun theorems (1738)
A first integral of the equations of motion (either for compressible fluid (3.10) or incompressible fluid (3.11)) is commonly called a Bernoulli equation. There are various forms
which it can take, depending on the particular kinematical or dynamical assumptions made
about the motion, though in all cases the basic expression
(3.17)
1
H ≡ u2 +
2
Z
dp
+Φ
ρ
is present. For incompressible fluid, (3.11), it takes the form
(3.18)
1
p
H ≡ u2 +
+ Φ.
2
ρ0
By means of (3.10) the first equation (either from (3.10) or (3.11)) takes the form
∂~u
+ ~ω × ~u = −∇H.
∂t
u
If we assume that the flow is steady (i.e. ∂~
= 0), then we arrive to the following important
∂t
result.
(3.18a)
Bernoulli theorem
Consider the steady baratropic flow of a perfect fluid. Then if ~ω × ~u ≡ 0 in the flow region
we 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 constant
on 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 general
result. Equation (3.18a) takes the form
∂ϕ
∇
+H =0
∂t
which can be integrated to obtain
36
PROF. VLADIMIR VLADIMIROV
∂ϕ
+ H = f (t)
∂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 great
importance in fluid dynamics.
(3.21)
3.7. The Stream function
It is possible to introduce a stream function whenever the equation of continuity can be
written as the sum of two derivatives.
Plane flows of an incompressible fluid
In this case we have
u = u(x, y, t)
so the equation of continuity
v = v(x, y, t)
w≡0
ρ ≡ const.
ρt + ∇.(ρ~u) = 0
gives us
(3.22)
∇.~u = 0
or
ux + vy = 0.
R
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
∂ψ
∂ψ
v = −ψx = − .
∂y
∂x
It is evident, that the curves ψ = const. are the streamlines of the field (prove it!); ψ is
called the stream function.
From the definition of vorticity (3.15a) and (3.23) we obtain the following important
equation satisfied by ψ, namely
(3.23)
u = ψy =
∂2ψ ∂2ψ
(3.24)
△ψ = ψxx + ψyy =
+ 2 = −ω.
∂x2
∂y
This equation serves to determine ψ when the vorticity magnitude is known. Now for the
steady 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 into
account 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)
The equation of continuity takes the form:
w ≡ 0.
∂
∂
(rv) + (ru) = 0
(prove it!)
∂r
∂z
hence we can define a stream function ψ = ψ(r, z, t) such that
(3.28)
1
u = ψr
r
The equation relating ψ to the vorticity is:
(3.29)
1
v = − ψx .
r
∂ 2 ψ ∂ 2 ψ 1 ∂ψ
+ 2 −
= −rω
∂z 2
∂r
r ∂r
In steady flow H and ω are connected by
(3.30)
(3.31)
H = H(ψ)
ω = −rH ′ (ψ)
(prove it!).
(prove it!).
Thus any solution of the equation
1
ψzz + ψrr − ψr = rf (ψ)
r
provides an example of steady axially symmetric flow.
38
PROF. VLADIMIR VLADIMIROV
Example 3.2 (Hill’s spherical vortex). Show that,
1
ψ = Ar 2 (a2 − z 2 − r 2 )
2
ω = 5Ar
H = −5Aψ + const.
is an exact solution for an ideal incompressible steady axially symmetric flow. Draw the
picture 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 find
equations satisfied by ~u, ρ and p as functions of the Lagrangian variables Xα , t. Indeed,
the first equation in (2.12)
can be transformed using
(3.32)
d~
u
dt
=
d2 ~
x
dt2
d~u
= f~ − ∇(p)
dt
and multiplying both sides by
∂ 2 xi
− fi
∂t2
∂xi
.
∂Xα
The result is,
1 ∂p
∂xi
=−
∂Xα
ρ ∂Xα
~ X,
~ t), p = p(X,
~ t), f~ = f(
~ t). This equation together with (1.16)
where xi = xi (X,
(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
Γ≡
I
39
~u.d~x.
γ
If now γ moves with the fluid, i.e. γ = γ(t), we may calculate the rate of change of the
circulation around this moving circuit
I
d
dΓ
~u.d~x.
=
dt
dt γ(t)
Kelvin’s Theorem
The baratropic flow of a perfect fluid under the action of a conservative extraneous force
is circulation-preserving
d
Γ=0
dt
where (3.33) holds for any material circuit γ(t). In order to prove (3.33) consider the
transformation to Lagrangian coordinates
(3.33)
d
dt
(3.34)
I
I
d
~ t) ∂xi dXk
ui(X,
ui dxi =
dt γ0
∂Xk
γ(t)
I ∂ui
∂xi
+ ui
dXk
uit
=
∂Xk
∂Xk
γ0
I
dui
=
dxi
γ(t) dt
I
≡
ai dxi
γ(t)
where γ0 = γ(0) and we also used
2
I
I
∂ui
~u
ui
= 0.
d
dXk =
∂Xk
2
γ0
γ0
Now for barotropic flow we recall, from (2.13)-(2.14), that acceleration ~a is derivable from
a 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=
I
~u.d~x =
γ
I
(∇ϕ).d~x =
γ
I
γ
dϕ ≡ 0
40
PROF. VLADIMIR VLADIMIROV
if the potential function is single-valued.
It is always true for a simply connected domain of flow, so the potential flows with nonzero
circulation 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 different
circulations around different components of boundary.
γ2
Γ
Γ2
γ1
Γ3
Γ1
γ
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)
1 ∂Tik
dui
= fi +
dt
ρ0 ∂xk
∂ui
=0
in τ .
∂xi
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, f~
is the given external force (per unit mass).
The stress tensor has been specified in eq2.12, as
(4.2)
Tik = −pδik + 2µeik
1
eik ≡
2
∂ui
∂uk
+
∂xk
∂xi
where p is pressure, µ is the coefficient of dynamic viscosity, eik is the rate-of-strain (or rateof-deformation) tensor. The substitution of (4.2) into (4.1) produces the famous NavierStokes equations
TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS
d~u = f~ − 1 ∇p + ν△~u
dt
ρ0
∇.~u = 0
(4.3)
where ν ≡
µ
ρ0
41
is 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
~
~u = U
on ∂τ
for a fixed boundary, and
(4.4a)
~ For the boundary condition (4.4) the stress
for a boundary moving with a velocity U.
vector ~t applied from the boundary ∂τ to the fluid is generally non-zero
(4.5)
ti = Tik nk = −pni + 2µeik nk 6= 0
on ∂τ
where ~n is the external unit normal vector to ∂τ . The total force acting on the fluid (from
the wall) will appear as the surface integral
Fi =
Z
ti dS =
∂τ
Z
∂τ
Tik nk dS = −
|
Z
∂τ
pni dS + 2µ
|
{z
}
the force of pressure
Z
∂τ
eik nk dS .
{z
}
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 + 2µeik nk = 0
on ∂τ
which corresponds to the absence of surface forcing. In this situation ~u 6= 0 on ∂τ .
42
PROF. VLADIMIR VLADIMIROV
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(v + uv + vv + wv ) = −p + µ(v + v + v )
t
x
y
z
y
xx
yy
zz
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
u = u(y, z, t)
⇒ ρut = −px + µ(uyy + uzz )
0 = −pz ⇒
p = p(x, t)
ux = 0
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 direction
x > 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
G
uyy + uzz = − .
µ
(4.9)
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 + 2µeik nk .
The first term in the right hand side is the normal force of pressure. To analyse the second
term we introduce the local (at the wall) coordinates
where the plane (x, z) is tangent to the wall, ~n is antiparallel to the y-axis ~n = (0, −1, 0).
ux
uy + vx uz + wx
0 uy uz
∂uk
∂ui
vy
vz + wy = uy 0 0
+
= uy + vx
2eik =
∂xk
∂xi
u +w v +w
w
u 0 0
z
x
z
y
z
z
TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS
43
y
u(y, z)
~n
x
z
for the unidirectional flows, which implies
0 uy uz
0
−uy
2µeik nk = µ uy 0 0 −1 = µ 0 .
uz 0 0
0
0
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)
du ˜
t1 = µuy = µ
dy at the wall
Formulae (4.10) and (4.11) were originally proposed by Newton, which implies the viscosity
as 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 between
the pressures imposed at the two ends of the tube: Hagen (1839), Poisenille (1840). The
flow is axisymmetric u = u(r), see (2.38), then equation (4.9) implies
1 ∂
G
(rur ) = −
r ∂r
µ
(△ - in cylindrical coordinates, see Arfken) then the general solution is
△u =
u(r) =
G
(−r 2 + A log r + B)
4µ
44
PROF. VLADIMIR VLADIMIROV
p 0 > p1
r
p1
p0
0
ℓ
x
• no singularity at r = 0 ⇒ A = 0
• u = 0 at r = a (non-slip), which implies
(4.12)
u=
G 2
(a − r 2 )
4µ
• the tangential stress at the wall (see (4.11) with y → −r):
1
du = − Ga
t1 = µ dr r=a
2
So, the total frictional force on a length ℓ of the tube is,
1
Ga = πa2 (p0 − p1 ).
2πaℓ
2
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 any
section of the tube:
Z a
πGa4
πa4 (p0 − p1 )
Q=
u2πr dr =
=
.
8µ
8µℓ
0
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 plates
The flow is plane, which implies u = u(y). Now equation (4.9) implies
µuyy = −G
the general solution
TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS
z
45
~n
H
p0
p1
H
2
0
x
~n
G 2
y + ay + b
2µ
Use of the non-slip boundary condition yields:
u(y) = −
a, b - constant.
u = 0 at y = 0 ⇒ b = 0
GH
u = 0 aty = H ⇒ a =
2µ
(4.13)
"
2 #
H
G H2
− y−
⇒ u(y) =
2µ 4
2
It is the parabolic profile with
GH 2
H
at y = .
8µ
2
The tangential stress on the wall, equation (4.11) implies
umax =
du
GH
t˜1 = −µ
=
.
dy
2
So, the total force on both walls is GHℓ, where ℓ is the length of the channel. It is once
again 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 difference
with 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
46
PROF. VLADIMIR VLADIMIROV
y
U0
~n
H
u(y)
0
u=0
~n
• u = U0 at y = H ⇒ u = UH0 y.
So, the required profile is linear:
(4.14)
u(y) =
U0
y
H
The tangential stress on the walls:
µU0
µU0
t˜1 |y=0 =
.
t˜1 |y=H = −
H
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 sin α, p =const.
stress free surface
y
H
g sin α
g
p =co
nst.
t1 =
0
u (y )
α
x
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 sin α
µ
the general solution
ρg sin α 2
y + ay + b
with a, b constant.
µ
Using the boundary conditions we obtain the velocity profile
u(y) = −
(4.15)
u=
G
y(2H − y)
2µ
G ≡ ρg sin α.
Calculating the flux:
Q=
one can find the relation
Z
k=
k
u dy =
Gh3
3µ
3µQ
ρg sin α
31
0
4.9. Unsteady unidirectional flows
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
y
τ
u → 0 at y → ∞
x
u = U0 cos ωt
48
PROF. VLADIMIR VLADIMIROV
The equation we have to solve is
ut = νuyy
in τ.
The boundary conditions are u(y, t) = U0 cos ωt at y = 0. We are looking for a solution in
a complex form:
iωt
u(y, t) = ℜ{e
′′
F (y)} ⇒ iωF = νF ⇒ F (y) = A exp
− (1 + i)
this is finite as y → ∞
⇒ u(y, t) = U0 exp
ω 2ν
ω 12 ω 21 y cos ωt −
y
−
2ν
2ν
4.10. Dynamical Similarity and the Reynolds Number
The Navier Stokes equations give
∂p
∂ui
∂ 2 ui
∂ui
=−
+ uk
+µ
ρ
∂t
∂xk
∂xi
∂xk ∂xk
∂uk
=0
∂xk
L
′
~u
~x
tU
p
′
~x =
t′ =
p′ =
U
L
L
ρU 2
′
′
∂ui
1 ∂ 2 u′i
∂p′
′ ∂ui
+
+
u
=
−
∂t′
k
∂x′k
∂x′i Rℓ ∂x′k ∂x′k
⇒
∂u′k
=0
∂x′k
~u =
Then the Reynolds number is defined to be
ρLU
µ
Result: all problems with the same Rℓ are the same! (= dynamical similarity).
Rℓ =
TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS
The ‘drag’ force F~ ,
Z 1 ∂u′i
∂u′k
′
−p δik +
Fi = −
Tij nj dS = −ρU L
nk dS ′
+
′
′
R
∂x
∂x
ℓ
∂τ
i
k
~
F
⇒ 1 2 2 = C0 (Rℓ )
ρU L
2
I
2
2
where C0 is the drag coefficient. This a universal formula for all spheres etc.
C0
0.8
0.4
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