The Navier-Stokes Equation

David A. Randall

Department of Atmospheric Science Colorado State University, Fort Collins, Colorado 80523

25.1 Analysis of the relative motion near a point

Suppose that the velocity of the ﬂuid at position r0x y z and time t is Vx y z t , and that the simultaneous velocity at a neighboring position r0 + r is V + V . We can write

u u u ----- ---------- x y z u x v v v v = ----- ----- ----- y . (25.1) x z z w z w w w ------------------ x y z

In vector form, this can be written as

V = D r , (25.2)

where D is called the deformation tensor because it describes how the ﬂuid element is being deformed by the non-uniform motion ﬁeld. Using Cartesian tensor notation, we can alternatively express (25.1) as

vi vi = ------xj , where i = 1 2 3 . (25.3) xj

We can decompose vi  xj into two parts which are symmetric and antisymmetric in the sufﬁces i and j . They are

1 v v = --------i + ------j , (25.4) ij 2 xj xi

and

Selected Nuggets 193

194 The Navier-Stokes Equation

1 v v = --------i – ------j , (25.5) ij 2 xj xi

respectively. The deformation D can now be expressed by

D = + , (25.6)

where we deﬁne two “parts” of D :

e e e 0 xx xy xz xy xz e e e , and 0 . (25.7) yx yy yz yx yz ezx ezy ezz zx zy 0

We show below that is related to the divergence, and is related to the vorticity. Correspondingly, V can be divided into two parts, i.e.

s a s a V = V + V , where Vi = eijXj , and Vi = ijXj . (25.8)

Here the superscripts s and a denote “symmetric” and “antisymmetric,” respectively.

We refer to as the rate of strain tensor, or simply as the strain tensor. The u v w diagonal elements of , i.e. e = ----- e = ----- e = ------, are called the normal strains, xx x yy y zz z and represent the rate of volume expansion, as illustrated in Fig. 25.1 a. The sum of the normal strains, i.e., the trace of the strain tensor, is the divergence of V , i.e.

exx + eyy + ezz = eii = V . (25.9)

The off-diagonal elements of , namely exy , eyz , and ezx , are called the shearing strains, and express the rate of shearing deformation, as illustrated in Fig. 25.1 b. They can be written as

1 u v e = e = ------- + ----- , xy yx 2y x 1 v w e = e = ------- + ------ , yz zy 2z y 1w u ezx = exz = ------+ ----- . 2 x z (25.10)

Selected Notes of David A. Randall 25.1 Analysis of the relative motion near a point 195

y a

dv u v ----- + ----- x y Dy

x 0 Dx du

y b

u v ----- + ----- y x Dy

d v x 0 Dx

Figure 25.1: a) The normal strains, representing the rate of volume expansion. b) The shearing strains, which represent the rate of shearing deformation.

The elements of the anti-symmetric tensor, , are

1 w v 1 u w v u = – = --------– ----- = – = ------- – ------ = – = ----- – ----- . (25.11) zy yz 2 y z xz zx 2 z x yx xy x y

Introducing the notation x = –2yz y = –2zx z = –2zx , we see that x y z is the vorticity vector

= V . (25.12)

Selected Notes of David A. Randall 196 The Navier-Stokes Equation

In tensor notation, (25.12) can be written as

vk i = ijk------, (25.13) xj

where ijk = 1 when i, j, k are in even permutation, and ijk = –1 when i, j, k are in odd

v2 v1 v u permutation. For example, z = 3 = 321------+ 312------= ----- – ----- , which gives x1 x2 x y twice the angular velocity about the z-axis, as illustrated in Fig. 25.2. Thus

-du

Dy v u ˙ ----- – ----- = ˙ + ˙ x y

dv ˙ Dx

Figure 25.2: The component of the vorticity in the z direction.

a u 0 –z –y x 1 a = -- 0 – y , (25.14) v 2 z x a z w –y x 0

a 1 V = -- r . (25.15) 2

25.2 Invariants of the strain tensor Consider the scalar product

Selected Notes of David A. Randall 25.2 Invariants of the strain tensor 197

r r = r r

2 2 2 = exxx + eyyy + ezzz + 2exyxy + eyzyz + ezxzx

= fx y z. (25.16)

By putting fx y z = constant , we obtain a surface of second order, the strain quadratic, also called the ellipsoid of strain.

We refer the strain quadratic to its principal axes X1 X2 X3 . Then (25.16) takes the form

2 2 2 e1X1 + e2X2 + e3X3 = FX1 X2 X3 = constant . (25.17)

Here e1 , e2 , and e3 are the principal values of the strain tensor, and are called the principal normal strains. Using (25.16), we see that

s 1 f u = ------= e x + e y + e z 2 x xx xy xz s 1 f v = ------= eyxx + eyyy + eyzz . (25.18) 2y s 1 f w = ------= ezxx + ezyy + ezzz 2z

In terms of the principal axes X1 X2 X3 ,

s 1 F s s v1 = ------= e1X1 v2 = e2X2 v3 = e3X3 . (25.19) 2X1

s s s s v1 v2 v3 are the projections of v upon the principal axes. Therefore, the symmetric part of V is composed of three expansions (or contractions) in the mutually orthogonal directions of the principal axes of the strain quadratic.

We next consider the sum of the diagonal elements of e, i.e.

exx + eyy + ezz = eii = V . (25.20)

This quantity is an invariant under the rotation of the axes of reference. Consider the

Selected Notes of David A. Randall 198 The Navier-Stokes Equation

transformation by which the surface fx y z = the constant of (25.16) is transformed into FX1 X2 X3 = the constant of (25.17). We can write the transformation between the two Cartesian coordinates in the schematic, tabular form shown in Table 25.1

x y z

X1 a1 b1 g1 X2 a2 b2 g2 X3 a3 b3 g3 Table 25.1: The transformation between two Cartesian coordinate systems. Here ai, bi, gi, are the direction cosines between the axes.

Here ai, bi, gi, are the direction cosines between the axes. We ﬁnd that

3 3 2 i = 1 ii = 0 , (25.21) i = 1 i = 1 ij + ij + i j = ij

where ij is Kronecker’s delta ij = 1 when i = j ij = 0 when i j . Substituting the expression

Xi = ix + iy + iz where i = 1 2 3 (25.22)

into (25.17), we obtain

3 3 2 i = 1 ii = 0 . (25.23) i = 1 i = 1 ij + ij + i j = ij

This expression must be identical with (25.16), so that

2 2 2 exx = 1e1 + 2e2 + 3e3 exy = 11e1 + 22e2 + 33e3 , 2 2 2 . (25.24) eyy = 1e1 + 2e2 + 3e3 eyz = 1 1e1 + 2 2e2 + 3 3e3 , e = 2e + 2e + 2e e = e + e + e , zz 1 1 2 2 3 3 zy 11 1 22 2 33 3

Selected Notes of David A. Randall 25.3 Stress 199

It follows that

3 2 2 2 = exx + eyy + ezz = i + i + i ei = e1 + e2 + e3 (25.25) i = 1

i.e., the sum exx + eyy + ezz = V is independent of the choice of reference. Of course, this quantity is called the divergence, and we “knew already” that it is independent of the choice of reference. Another invariant of the strain tensor is

2 2 2 exxeyy + eyyezz + ezzexx – exy – eyz – ezx = constant . (25.26)

The proof is left as an exercise.

25.3 Stress In a moving viscous ﬂuid, forces act not only normal to a surface but also

tangential to it. In Fig. 25.3, x , y and z denote respectively the forces per unit area acting upon the surfaces normal to the positive x, y, and z directions, respectively. In z + ------zz z z

–x = –x

+ ------yy y y 0 y

–y = –y

x + ------xx = – x x –z z

Figure 25.3: Sketch illustrating the normal and tangential forces acting on an element of fluid.

Selected Notes of David A. Randall 200 The Navier-Stokes Equation

general, each force is a vector with 3 components, so that

x = xxi + yxj + zxk , y = xyi + yyj + zyk , . (25.27) z = xzi + yzj + zzk .

Symbolically we can write an element of the stresses as iji = 1 2 3;j = 1 2 3 . Here ij is the i-component of the force per unit area exerted across a plane surface element normal to the j-direction. The stress is thus a tensor with nine elements, and can be represented by the matrix

xx xy xz T = (25.28) yx yy yz zx zy zz

For an arbitrarily chosen plane surface, whose normal vector is n , the force acting per unit area is n . See Fig. 25.4. The equilibrium condition is

–y = –y (25.29) n – xx – yy – zz = 0 ,

where is the area of the oblique surface, x , y , z , are the areas of the faces lying in the planes x = 0 y = 0 and z = 0 . Denoting the direction cosines of n by , and , we have ------x = , ------y = , ------z = . Therefore (25.29) can be written as

n = x + y + z , (25.30)

xn xx xy xz = , (25.31) yn yx yy yz zn zx zy zz

n = T n . (25.32)

Selected Notes of David A. Randall 25.3 Stress 201

z n n

C y B

–x

–y 0

A x –z

Figure 25.4: Sketch illustrating the normal force acting on a plane surface.

Equation (25.32) shows that matrix T transforms a vector n into another vector n , so that T is a tensor, called the stress tensor. See Fig. 25.5. The three diagonal elements, xx , yy , and zz , are normal stresses. The six off-diagonal elements are tangential stresses.

An important property of T is its symmetry, i.e. ij = ji . This symmetry can be deduced from the condition of moment equilibrium. Consider for example a moment about the z-axis due to yx and xy :

1 1 1 yz--x – xz--y = -- – xyz . (25.33) yx 2 xy 2 2 yx xy

The shear stresses acting on the negative x- and y-surfaces contribute also, so that the total moment about the z-axis is

yx – xyxyz . (25.34)

Euler’s equation (for a rotating rigid body) states that

Selected Notes of David A. Randall 202 The Navier-Stokes Equation

yx = –yx

xy xy = –xy 0 y

yx x

x Figure 25.5: Sketch illustrating the components of the stress tensor.

d I ------z = – xyz , (25.35) z dt yx xy

2 2 x + y 2 2 where I = ------= -----xyzx + y is the moment of inertia of z 12 12 the ﬂuid element xyz , and z is the angular velocity about the positive z-axis. Then (25.35) reduces to

2 2 dz -----x + y ------= – . (25.36) 12 dt yx xy

d In order that ------z remains ﬁnite, we need – 0 when x y 0 . It follows dt yx xy that yx = xy . Similar considerations establish that the stress tensor is symmetric, i.e.,

xy = yx yz = zy zx = xz . (25.37)

Selected Notes of David A. Randall 25.4 Invariants of the stress tensor 203

25.4 Invariants of the stress tensor As in the case of the strain tensor, we have a scalar product

2 2 2 r T r = 1X1 + 2X2 + 3X3 = constant , (25.38)

where 1 , 2 , and 3 are the principal stresses acting on the surface elements normal to the principal axes (X1, X2, X3) of the ellipsoid of stress. It can be shown that r T r is an invariant under the rotation of the axes of reference. The sum of the normal stresses, i.e., the trace of the stress tensor,

xx + yy + zz = ii = 1 + 2 + 3 = –3p , (25.39)

is also an invariant. For a moving fluid, (25.39) is the definition of pressure p , which reduces to the static pressure when the fluid is at rest.

25.5 The resultant forces due to the spatial variation of the stresses

The resultant force acting in the x-direction on a volume element of ﬂuid xyz is

+ ------x---xx yz – yz + + ------x---yy xz xx x xx xy y

xz xx xy xz –xyyz + xz + ------z xy – xzxy = ------+ ------+ ------xyz . z x y z (25.40)

Therefore the three components of the resultant force (vector) per unit volume are

X = ------x---x + ------x---y + ------x---z = ------x--j x y z xj yx yy yz yj Y = ------+ ------+ ------= ------ . (25.41) x y z xj Z = ------z---x + ------z---y + ------z---z = ------z--j x y z x j

Equation (25.41) can be compactly written as

Selected Notes of David A. Randall 204 The Navier-Stokes Equation

DivT = ------x + ------y + ------z x y z

= i------x--j + j------y--j + k------z--j . xj yj zj (25.42)

DivT is a tensor divergence and, therefore, is a vector.

25.6 Physical laws connecting the stress and strain tensors The relation between the stress tensor and the strain tensor may be established by the following assumptions. i)

ij = –pij + Fij , (25.43)

where Fij is the frictional stress tensor which depends on the space derivatives of the velocity, i.e. the strain tensor.

ij ii) The relationship between Fij and e is linear. iii) The relationship does not depend on the frame of reference. We have already shown that, with reference to the respective principal axes,

0 0 e 0 0 1 1 T = 0 0 and = 0 e 0 . (25.44) 2 2 0 0 3 0 0 e3

1 2 3 Because 1 , 2 , and 3 are normal stresses and e , e , e are normal strains, it is natural to assume that the principal axes of both tensors are the same. We therefore assume that

1 = – p + V + 2e1 2 = – p + V + 2e2 , (25.45) 3 = – p + V + 2e3

where p satisﬁes (25.39).

Writing direction cosines between the principal axes (X1, X2, X3) and axes of a Cartesian coordinate system (x, y, z) as shown in Table 25.2,

Selected Notes of David A. Randall 25.6 Physical laws connecting the stress and strain tensors 205

x y z

X1 a1 b1 g1 X2 a2 b2 g2 X3 a3 b3 g3 Table 25.2: Direction cosines between principle axes. we ﬁnd that

2 2 2 = + + xx 1 1 2 2 3 3 , (25.46) xy = 111 + 222 + 333

etc. Substituting (25.45) into (25.46), and using (25.24), we obtain

2 2 = – p + V + 2e + – p + V + 2e xx 1 1 2 2 2 + – p + V + 2 e – p + + 2 e , 33 xx (25.47) xy = – p + V + 2e111 + – p + V + 2e222 + – p + V + 2e333 = 2exy ,

etc. In general we have

ij = – pij + Vij + 2eij . (25.48)

This expression gives

ii = – 3p + V + 2 V

= – 3p + 3 + 2 V . (25.49)

Because we have deﬁned ij = –3p , we conclude that

3 + 2 + 0 . (25.50)

Thus we have the ﬁnal expression

Selected Notes of David A. Randall 206 The Navier-Stokes Equation

1 = – p + 2e – -- V ij ij ij 3 ij

= – p + F . ij ij (25.51)

25.7 The Navier-Stokes equation Substitution of (25.51) into (25.42) gives the resultant forces per unit volume, as expressed by the divergence of the stress tensor:

DivT = –p + DivF (25.52)

The equation of motion in the inertial frame (with subscript a omitted) is

DV ------= – – p + DivF (25.53) Dt

or, in Cartesian tensor notation,

Dvi vi vi p 1 ------= ------+ vj------= – ------– ------+ ------2 eij – -- Vij Dt t xj xi xi xj 3

p vi vj 2 vk = – ------– ------+ ------------+ ------ – --------. xi xi xj xj xi 3 xk (25.54)

This is called the Navier-Stokes equation.

When the ﬂuid is incompressible V = 0 , and is spatially constant, (25.54) reduces to

2 Dvi p vi ------= – ----- – ------+ ------ , (25.55) Dt x x 2 i xj

or, in vector form,

DV 2 ------= – – p + V . (25.56) Dt

Here we see the Laplacian of the vector V .

Selected Notes of David A. Randall 25.8 A proof that the dissipation is non-negative 207

1 v v It should be noted that e = --------i + ------j vanishes when V = r (rigid ij 2 xj xi rotation), and that r = 0 . This means that we do not have to worry about the distinction between Va = V + r and V in the expression F = DivF in the equation of relative motion. The frictional force is the same in the rotating frame as in the inertial frame.

25.8 A proof that the dissipation is non-negative The dissipation rate is given by

vi F V = Fij------ xj u u u = F11----- + F12----- + F13----- x y z (25.57) v v v + F ----- + F ----- + F ----- 21x 22y 23 z w w w + F ------ + F ------ + F ------ . 31 x 32 y 33 z

Since Fij = Fji , we have

vi vi vj = Fij------= Fji------= Fij------, (25.58) xj xj xi

so we can write

v v 2 = F ------i + ------j . (25.59) ij xj xi

But

1 v v 1vi F = 2 --------i + ------j – ------ ij ij 2 xj xi 3xi

1 v v 1 v v 1vi = 2 --------i + ------j 1 – + --------i + ------j – ------ , ij ij 2 xj xi 2 xj xi 3xi (25.60)

Selected Notes of David A. Randall 208 The Navier-Stokes Equation

1 v v 2v F = 2 --------i + ------j 1 – + ------i . (25.61) ij ij ij 2 xj xi 3xj

Substituting (25.61) into (25.59), we ﬁnd that

1 v v 2vi v v = --------i + ------j 1 – + ------ ------i + ------j ij ij 2 xj xi 3xj xj xi

4 2 = 2eijeij1 – ij + -- V ij . 3 (25.62)

This expression is obviously non-negative, since 1 – ij 0 and ij 0 . For a nondivergent ﬂow, the dissipation can be expanded, using Cartesian coordinates, as

= 2eijeij

1 u v 2 u w 2 = -- ----- + ----- + ----- + ------ 2 y x z x

v u 2 v w 2 + ----- + ----- + ----- + ------ x y z y

w u 2 w v 2 + ------ + ----- + ------ + ----- x z z z

u v 2 u w 2 w v 2 = ----- + ----- + ----- + ------ + ------+ ----- . y x z x y z (25.63)

This shows that is the sum of squares; again, it cannot be negative.

Selected Notes of David A. Randall