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28

If certain conditions are satised, it is possible to identify an orthonormal basis {n1,n2,n3} such that = 1 n1n1 + 2 n2n2 + 3 n3n3 , (1.103)

which means that the matrix form of in the coordinate system dened by the new basis is diagonal:

=

1 0 00 2 0

0 0 3

. (1.104)

The orthogonal vectors n1, n2 and n3 that diagonalize are called the principal directions, and 1,2 and 3 are called the principal values of . From Eq. (1.105), one observes that the vector uxesthrough the surface of unit normal ni, i=1,2,3, satisfy the relation

fi = ni = ni = ini , i = 1, 2, 3 . (1.105)What the above equation says is that the vector ux through the surface with unit normal ni iscollinear with ni, i.e., ni is normal to that surface and its tangential component is zero. FromEq. (1.107) one gets:

( iI) ni = 0 , (1.106)where I is the unit tensor.

In mathematical terminology, Eq. (1.108) denes an eigenvalue problem. The principal directions andvalues of are thus also called the eigenvectors and eigenvalues of , respectively. The eigenvaluesare determined by solving the characteristic equation,

det( I) = 0 (1.107)or

11 12 1321 22 2331 32 33

= 0 , (1.108)which guarantees nonzero solutions to the homogeneous system (1.108). The characteristic equationis a cubic equation and, therefore, it has three roots, i, i=1,2,3. After determining an eigenvaluei, one can determine the eigenvectors, ni, associated with i by solving the characteristic system(1.108). When the tensor (or matrix) is symmetric, all eigenvalues and the associated eigenvectorsare real. This is the case with most tensors arising in uid mechanics.

Example 1.5.5. Principal values and directions

(a) Find the principal values of the tensor

=

x 0 z0 2y 0

z 0 x

.

(b) Determine the principal directions n1,n2,n3 at the point (0,1,1).

(c) Verify that the vector ux through a surface normal to a principal direction ni is collinear withni.

(d) What is the matrix form of the tensor in the coordinate system dened by {n1,n2,n3}?Solution:(a) The characteristic equation of is

0 = det( I) =x 0 z

0 2y 0z 0 x

= (2y ) x zz x

=

+ & 29

(2y ) (x z) (x + z) = 0.The eigenvalues of are 1=2y, 2=x z and 3=x+ z.(b) At the point (0, 1, 1),

=

0 0 10 2 0

1 0 0

= ik + 2jj + ki ,

and 1=2, 2=1 and 3=1. The associated eigenvectors are determined by solving the correspondingcharacteristic system:

( iI) ni = 0 , i = 1, 2, 3 .For 1=2, one gets

0 2 0 10 2 2 01 0 0 2

nx1ny1

nz1

=

00

0

= 2nx1 + nz1 = 00 = 0

nx1 2nz1 = 0

=

nx1 = nz1 = 0 .

Therefore, the eigenvectors associated with 1 are of the form (0, a, 0), where a is an arbitrary nonzeroconstant. For a=1, the eigenvector is normalized, i.e. it is of unit magnitude. We set

n1 = (0, 1, 0) = j .

Similarly, solving the characteristic systems 0 + 1 0 10 2 + 1 0

1 0 0 + 1

nx2ny2

nz2

=

00

0

of 2=1, and 0 1 0 10 2 1 0

1 0 0 1

nx3ny3

nz3

=

00

0

of 3=1, we nd the normalized eigenvectors

n2 =12(1, 0,1) = 1

2(i k)

andn3 =

12(1, 0, 1) =

12(i+ k) .

We observe that the three eigenvectors, n1 n2 and n3 are orthogonal:2

n1 n2 = n2 n3 = n3 n1 = 0 .(c) The vector uxes through the three surfaces normal to n1 n2 and n3 are:

n1 = j (ik + 2jj + ki) = 2j = 2 n1 ,n2 = 1

2(i k) (ik + 2jj + ki) = 1

2(k i) = n2 ,

n3 = 12(i + k) (ik + 2jj + ki) = 1

2(k+ i) = n3 .

2A well known result of linear algebra is that the eigenvectors associated with distinct eigenvalues ofa symmetric matrix are orthogonal. If two eigenvalues are the same, then the two linearly independenteigenvectors determined by solving the corresponding characteristic system may not be orthogonal. Fromthese two eigenvectors, however, a pair of orthogonal eigenvectors can be obtained using the Gram-Schmidtorthogonalization process; see, for example, [3].

30

(d) The matrix form of in the coordinate system dened by {n1,n2,n3} is

= 2n1n1 n2n2 + n3n3 = 2 0 00 1 0

0 0 1

.

The trace, tr , of a tensor is dened by

tr 3

i=1

ii = 11 + 22 + 33 . (1.109)

An interesting observation for the tensor of Example 1.5.5 is that its trace is the same (equal to 2)in both coordinate systems dened by {i, j,k} and {n1,n2,n3}. Actually, it can be shown that thetrace of a tensor is independent of the coordinate system to which its components are referred. Suchquantities are called invariants of a tensor.3 There are three independent invariants of a second-ordertensor :

I tr =3

i=1

ii , (1.110)

II tr 2 =3

i=1

3j=1

ijji , (1.111)

III tr 3 =3

i=1

3j=1

3k=1

ijjkki , (1.112)

where 2= and 3= 2. Other invariants can be formed by simply taking combinations of I,II and III. Another common set of independent invariants is the following:

I1 = I = tr , (1.113)

I2 =12(I2 II) = 1

2[(tr )2 tr 2] , (1.114)

I3 =16(I3 3I II + 2III) = det . (1.115)

I1, I2 and I3 are called basic invariants of . The characteristic equation of can be written as4

3 I12 + I2 I3 = 0 . (1.116)If 1, 2 and 3 are the eigenvalues of , the following identities hold:

I1 = 1 + 2 + 3 = tr , (1.117)

I2 = 12 + 23 + 31 =12[(tr )2 tr 2] , (1.118)

I3 = 123 = det . (1.119)

The theorem of Cayley-Hamilton states that a square matrix (or a tensor) is a root of its characteristicequation, i.e.,

3 I1 2 + I2 I3 I = O . (1.120)

3From a vector v, only one independent invariant can be constructed. This is the magnitude v=v v of

v.4The component matrices of a tensor in two dierent coordinate systems are similar. An important property

of similar matrices is that they have the same characteristic polynomial; hence, the coecients I1, I2 and I3and the eigenvalues 1, 2 and 3 are invariant under a change of coordinate system.

+ & 31

Note that in the last equation, the boldface quantities I and O are the unit and zero tensors, re-spectively. As implied by its name, the zero tensor is the tensor whose all components are zero.

Example 1.5.6. The first invariantConsider the tensor

=

0 0 10 2 0

1 0 0

= ik + 2jj + ki ,

encountered in Example 1.5.5. Its rst invariant is

I tr = 0 + 2 + 0 = 2 .

Verify that the value of I is the same in cylindrical coordinates.

Solution:Using the relations of Table 1.1, we have

= ik + 2jj + ki= (cos er sin e) ez + 2 (sin er + cos e) (sin er + cos e)

+ ez (cos er sin e)= 2 sin2 erer + 2 sin cos ere + cos erez +

2 sin cos eer + 2 cos2 ee sin eez +cos ezer sin eze + 0 ezez .

Therefore, the component matrix of in cylindrical coordinates {er, e, ez} is

=

2 sin2 2 sin cos cos 2 sin cos 2 cos2 sin

cos sin 0

.

Notice that remains symmetric. Its rst invariant is

I = tr = 2(sin2 + cos2

)+ 0 = 2 ,

as it should be.

' " # &

So far, we have used three dierent ways for representing tensors and vectors:(a) the compact symbolic notation, e.g., u for a vector and for a tensor;(b) the so-called Gibbs notation, e.g.,

3i=1

ui ei and3

i=1

3j=1

ij eiej

for u and , respectively; and(c) the matrix notation, e.g.,

=

11 12 1321 22 23

31 32 33

for .

32

Very frequently, in the literature, use is made of the index notation and the so-called Einsteinssummation convention, in order to simplify expressions involving vector and tensor operations byomitting the summation symbols.

In index notation, a vector v is represented as

vi 3

i=1

vi ei = v . (1.121)

A tensor is represented as

ij 3

i=1

3j=1

ij ei ej = . (1.122)

The nabla operator, for example, is represented as

xi

3i=1

xiei =

xi +

yj +

zk = , (1.123)

where xi is the general Cartesian coordinate taking on the values of x, y and z. The unit tensor I isrepresented by Kroneckers delta:

ij 3

i=1

3j=1

ij ei ej = I . (1.124)

It is evident that an explicit statement must be made when the tensor ij is to be distinguished fromits (i, j) element.

With Einsteins summation convention, if an index appears twice in an expression, then summation isimplied with respect to the repeated index, over the range of that index. The number of the free indices,i.e., the indices that appear only once, is the number of directions associated with an expression; itthus determines whether an expression is a scalar, a vector or a tensor. In the following expressions,there are no free indices, and thus these are scalars:

uivi 3

i=1

uivi = u v , (1.125)

ii 3

i=1

ii = tr , (1.126)

uixi

3

i=1

uixi

=uxx

+uyy

+uzz

= u , (1.127)

2f

xixior

2f

x2i

3i=1

2f

x2i=

2f

x2+

2f

y2+

2f

z2= 2f , (1.128)

where 2 is the Laplacian operator to be discussed in more detail in Section 1.4. In the followingexpression, there are two sets of double indices, and summation must be performed over both sets:

ijji 3

i=1

3j=1

ijji = : . (1.129)

+ &

The following expressions, with one free index, are vectors:

ijkuivj 3

k=1

3

i=1

3j=1

ijkuivj

ek = u v , (1.130)

f

xi

3i=1

f

xiei =

f

xi +

f

yj +

f

zk = f , (1.131)

ijvj 3

i=1

3

j=1

ijvj

ei = v (1.132)

uit

=

t

(i

ui ei

)=

ut

(1.133)

jixj

=

i

j

jixj

ei = (1.134)

ujuixj

=

i

j

ujuixj

ei = u u . (1.135)

Finally, the following quantities, having two free indices, are tensors:

uivj 3

i=1

3j=1

uivj eiej = uv , (1.136)

ikkj 3

i=1

3j=1

(3

k=1

ikkj

)eiej = , (1.137)

ujxi

3

i=1

3j=1

ujxi

eiej = u . (1.138)

Note that u in the last equation is a dyadic tensor.5

It is easy to show that the continuity and momentum equations,

t+ ( u) = 0 (1.139)

and

(ut

+ u u)

= p + + g , (1.140)

in index notation become

t+

( ui)xi

= 0 (1.141)

and

(uit

+ ujuixj

)= p

xi+

jixj

+ gi . (1.142)

5Some authors use even simpler expressions for the nabla operator. For example, u is also representedas iui or ui,i, with a comma to indicate the derivative, and the dyadic u is represented as iuj or ui,j .

' 2 - 5' (+ -$(' - ' ( ':

'+ Stokes '+ # -.# 6divergence theorem7 '+ Gauss

( & Stokes

=+ $' 84

'

- !

' S -$ 4+ ' S 1 3 !

, -)# .#)!

" (8 ' F + ( +' C '(:C

F dr = 0 .

!

-$' ' ' (

:+ ;( :+ # -.# $#

!

, -)# .#)!

,'+ '(5

0 =

(FG) dV =

[G F F G] dV =

F G dV =

G F dV .

;* 2 30% $,#

=+ ( - + @ ( / - 5 + ( '

dm

dt=

u n dS

m ' 52 ' ' u ( ' 3+' " -$' ' '

3:

dm

dt=

u n dS =

u dV . 6)7

( 3 ' (8' ' 8'

3 1+

' 3 - 6('( ' +72 '5 - !"

dm

dt= 0 =

u dV = 0 =

u dV = 0

6+ ' -7

;9

=

I =

S

x dydz + y dxdz + 2z dxdy

S ' 5 #(

x2 + y2 + z = 1 , 0 < z < 1

( z9) 6# ( 37 x

y

z

1 z

z1 x2 y2

+ S

1 + Ostrogradsky2 3 I:

I =

V

(1 + 1 + 2) dxdydz = 4

V

dxdydz.

,'+

I = 4 10

dz

dxdy = 4

10

2 dz = 4 10

(1 z) dz = 4[z z

2

2

]10

= 2

;;

=

I =

S

x2 dydz + y2 dxdz + z2 dxdy

' 8 C' ' x2 + y2 + z2 = 2 " + Ostrogradsky2 3:

I =

V

(2x + 2y + 2z) dxdydz

V = {(x, y, z) R3 | x2 + y2 + z2 2} " + (+ I '(

2 - '' ' = +

6r2 2 7 >5

x = r sin cos , y = r sin sin , z = r cos

0 r , 0 2 0

> 3$(' dV 3

dV = dxdydz =(x, y, z)(r, , )

drdd = r2 sin drdd .

, -)# .#)! !

$ :

I = 0

0

20

2r(sin cos + sin sin + cos ) r2 sin drdd

= 2 20

d

0

(sin2 cos+ sin2 sin + sin cos ) d 0

r3 dr

=4

2

20

d

[(12 1

4sin 2

)(sin + cos) +

sin2 2

]0

=4

2

20

[2(sin + cos) + 0

]d =

4

4[ cos + sin]20 =

4

4(1 + 1 + 0) = 0

.# Green!

"

"$ ' 6%7 ' 6.72 ' ! . Green ' ' . '+ 6symmetrical theorem7:

V

(2 2) dV =

S

( ) n dS . 67

' (8' - 3 3 (+ $ V S + - - ' ' '$ ' ' 67

@(3- -

F = 6 )7 + $ 6''7

;>

" ' '' +' ' 8' Laplace 3 V ' S2 :

S

ndS = 0 6 7

-.3

= = 1 ' $' ' Green V

(0 + 12) dV =

S

1 n dS

V

2 dV =

S

n dS

!/ 2 = 0 n = /n2 S

ndS = 0.

;?

" 2 3 ' (+' ' Green '(2

S

( ) n dS = 0

, -)# .#)!

3'5 + 8 (

" ( 5 ' C

dr = (P2) (P1) 6 7

#- ( C ' +' P1 P2 " ' P1 P2 '((2 ' 8' 6 7 - # '+ -' !

! " Reynolds

:+ 7# Reynolds 6Reynolds transport theorem7 ( 3-$' ' (8' .% "# 6conservation laws72 ( 84$ ('' ' 5 ' -$' -'- ' '

%! . Leibniz 6Leibniz integral rule7 ' 3 (' 4( 3$ = 8 $ Leibniz ( 3 (8'

:+ > . # Leibniz C f : R R R 3 f : R2 R (3 ' 3 #' 3

& " a b ( ' - + 67 :

d

dt

ba

f(x, t) dx = b

a

f

t(x, t) dx .

/ + ' $' '

(' ' -

* " ' f ( '' 32 f9f(x)2 ' ( Leibniz:

d

dt

b(t)a(t)

f(x) dx = f(b(t))db

dt f(a(t)) da

dt. 6&7

(8' ' 6&7 /' D " F ' ' f 2 ('( F (x)9f(x)

b(t)a(t)

f(x) dx = [F (x)]b(t)a(t) = F (b(t)) F (a(t)) =

d

dt

b(t)a(t)

f(x) dx =d

dt[F (b(t)) F (a(t))] = F (b(t))db

dt F (a(t))da

dt=

d

dt

b(t)a(t)

f(x) dx = f(b(t))db

dt f(a(t)) da

dt.

' '2 (' ' I(t)9[a(t), b(t)] ' #' 3

/ & .0)- ) Reynolds

x

a0

b0t9)

x

a1

b1t9t1

x

a(t)

b(t)t

.

.

.

.

.

- ( ( ' '' - 83+ ( + '

' ' ' # ' 3 ' ' '+ Euler 6Euler approach75 $ # Euler 6Eulerian coordinates7: x t !

/ & .0)- ) Reynolds

f/t + ' 8'' f 32 $ ' u f '' 6convective derivative7 ' '' f + 1 '

' ' ! (

V0

V (t)

X

x = (X, t)

t#"

t

" ,% %

.# .# $,

! . $, 6material control volume7 ' ' (8'

('' ' H( =+ ' + 3$ '2

( 3 V0 3 t9) (( + ( ( + 5 ' 3 V (t) # - 3 $ 3 ) 3 V (t) 5 ' 3 ( ( + - 3 !

!

G# 3'+ 3$( + dV (t)

3 V0 >@ ' J ' " 6dilation or dilatation or expansion7> V (t) 3+:

V (t) =

V (t)

dV =

V0

J dV0 ,

3' 3'+ +# V (t) 3 3 3

, ($ =$' 1 ( $ 3' ' '

' '.# )"# d/dt - 7$ - . . $, 05 " -% D/Dt -. ' '.# )"# d/dt - 7$/$ '. . $, 05 " -% /t

. >&

" u(x, t) ( 3+' J ' G# 3'+ 4 Euler Lagrange2

dJ

dt= u J . 6%7

-.3

" S3 + - 9(1, 2, 3) $ {1, 2, 3}2 + ' G# '

J =S3

sgn(1, 2, 3)x1X1

x2X2

x3X3

.

,5 3

dJ

dt=

S3

sgn(1, 2, 3)[d

dt

(x1X1

)x2X2

x3X3

+

x1X1

d

dt

(x2X2

)x3X3

+x1X1

x2X2

d

dt

(x3X3

)].

> ' $' 3 (8 3:

d

dt

(x1X1

)=

X1

(dx1dt

)=

u1X1

=u1x1

x1X1

+u1x2

x2X1

+u1x3

x3X1

63' ' (7 ' ' (8' ''

6%7 ' '

d(lnJ)dt

= u .

, ' ' ' ' ' 3+'

/ & .0)- ) Reynolds

:+ >( :+ 7# Reynolds

1 - Gauss2V (t)

(fu) dV =

S(t)

f u n dS .

+ ' 6 7 =$' 1 35 ' # ' ' f 3 V (t) ' # f 3 ' # f -+ (+( u ( ' S(t) V (t) ' 67 ' 6 7 ' 6*7 ' 6 )7 # Leibniz ( ' ' #- = 1 ,- $ ' " 7" -

:+ >* :+ 7# Reynolds 8 " 7"

1 F- = .*2 3+

d

dt

V (t)

f(x, t) u(x, t) dV =

V (t)

[(fu)t

+ u (fu) + (fu) u]

dV . 6 *7

-.3

" C ' 3+' '

u =3

i=1

ui ei

3

d

dt

V (t)

f(x, t) u(x, t) dV =3

i=1

eid

dt

V (t)

fui dV .

' ' (8' '' ( -+ ( - # '$ -$ @

3 ' -+' 5 ' :

67 '.# )"# . $,2

dV

dt=

d

dt

V (t)

dV .

" '(5 ' ' 6 ' 5

3 -7

6#7 '.# )"# # 5# @ 3 8

+ '(:

dm

dt=

d

dt

V (t)

dV = 0 . 6 7

67 '.# )"# # "# "# J + 3 @ 3 V (t)2 + $ ' ' F

( + V (t):

dJdt

=d

dt

V (t)

u dV = F . 6 7

= -+ # '

3 - (8 8$

('' ' 5 ' , % -+

/ & .0)- ) Reynolds !

V $% &% V #V (t) '$ $% '%u#u(t) u#u(x, y, z, t)

()% V#V dV

*+ m#V dV , #const.- $ m#V

)& J#V u dV # m u J#

V u dV

./$

%'

dVdt

= 0 dVdt

= ddt

V dV

$%

./$

%'

dmdt

= ddt

V dV #"

+

./$

%'

dJdt

= ddt

(mu) = m dudt

= m a dJdt

= ddt

V udV

%&

&%

$% % F = dJdt

= m a F = dJdt

= ddt

V udV

Newton

- $' *

.rigid body translation/ '

"

3 ' ' 7# ! +# 6rigid body translation72 u9u(t) ( + + + ' I

>

= f9 ' #- = 1 3:

dV (t)dt

=d

dt

V (t)

dV =

V (t)

u dV . 6 7

3' - # ' '

' 3+' u (3 8 ' ' ($ ' ' (+( " ' u -2 + C

1V (t)

dV (t)dt

= u . 6 .7

' ' 69const.72 ' ('' ' 5 +

0 = dV (t)dt

=

V (t)

u dV

/ & .0)- ) Reynolds

"-$ ' 6 7 :

dJdt

=

V (t)

[ut

+ u

t+ u u + uu + u u

]dV

=

V (t)

[

(ut

+ u u)

+ u(

t+ u + u

)]dV

=

V (t)

[

DuDt

+ u(

t+ (u)

)]dV

B'$ ' 8'

3

dJdt

=d

dt

V (t)

u dV =

V (t)

DuDt

dV .

# $%

, ( (

u = (x2 + y2) i + xy j + k

( J

& 8 ' ' ( :

a (b c) = b(a c) c(a b)0%6 a2 b c ( (

* "

u = x(2y 1) i y2 j + (z + 1) k ( ' 3+':

67 0 ).

1 )2!

6#7 ( (r, , z):

1r

r(rur) +

1r

u

+uzz

= 0 6. 7

67 (r, , ):

1r2

r(r2ur) +

1r sin

(sin u) +

1r sin

u

= 0 6..7

% 0 -0#

8 ' p2 p ' '2 - 67

(x, y, z):

p = px

i +p

yj +

p

zk 6.%7

6#7 ( (r, , z):

p = pr

er +1r

p

e +

p

zez 6.7

67 (r, , ):

p = pr

er +1r

p

e +

1r sin

p

e 6%)7

@" 687 (x, y) (('

4

5

ux =

ykai uy =

x.

=$ uz9)2 (8 ' 8'

3 '

) @" (r, z) 5

ur = 1r

zkai uz =

1r

r

3' ' $ 8 ( ,

-' 3+ ' $ u ' 3+' '8'

3 ' J

. -0 # -, #

( ( ' 3' 5 8:

a =dudt

6%7

u ( ( ' 3+'

u =drdt

6%&7

r = xi + yj + zk 6%*7

'$#

E (3-+ 8:

67

(x, y, z):

a =d2x

dt2i +

d2y

dt2j +

d2z

dt2k 6%7

6#7 ( (r, , z):

a =

[d2r

dt2 r

(d

dt

)2]er +

(rd2

dt2+ 2

dr

dt

d

dt

)e +

d2z

dt2ez 6%7

67 (r, , ):

a =

[d2r

dt2 r

(d

dt

)2 r sin2

(d

dt

)2]er

+

[2dr

dt

d

dt+ r

d2

dt2 r sin cos

(d

dt

)2]e

+(2 sin

dr

dt

d

dt+ 2r cos

d

dt

d

dt+ r sin

d2

dt2

)e 6% 7

& "# u 8 u 3+ 8:67

(x, y, z):

u = ux x

+ uy

y+ uz

z6%.7

6#7 ( (r, , z):

u = ur r

+ur

+ uz

z6%%7

67 (r, , ):

u = ur r

+ur

+

ur sin

6%7

* . -0 (u )u8 ( ( (u )u 3+ 8:67

(x, y, z):

(u )u =(ux

uxx

+ uyuxy

+ uzuxz

)i

+(ux

uyx

+ uyuyy

+ uzuyz

)j

+(ux

uzx

+ uyuzy

+ uzuzz

)k 6)7

6#7 ( (r, , z):

(u )u =(ur

urr

+ur

ur

u2

r+ uz

urz

)er

+(ur

ur

+ur

u

+urur

+ uzuz

)e

+(ur

uzr

+ur

uz

+ uzuzz

)ez 67

1 )2!

67 (r, , ):

(u )u =(ur

urr

+ur

ur

+u

r sin ur

u2 + u

2

r

)er

+

(ur

ur

+ur

u

+u

r sin u

+urur

u2 cot

r

)e

+(ur

ur

+ur

u

+u

r sin u

+urur

+uur

cot )e

6&7

"# "# -+

8 $2

D

Dt=

t+ u , 6*7

u ( ( 3+' 3+ 8:67

(x, y, z):

D

Dt=

t+ ux

x+ uy

y+ uz

z67

6#7 ( (r, , z):

D

Dt=

t+ ur

r+

ur

+ uz

z67

67 (r, , ):

D

Dt=

t+ ur

r+

ur

+

ur sin

6 7

" ' '2 #- ' '

D

Dt=

t+ u 6.7

" u ' 3+'2 #- ' '

DuDt

=ut

+ u u 6%7

0%6 "(+

(u )u = u (u) 67 u "# 0% # ,!# 6#' 7

. 30% "# # "# ' :

DuDt

= p + g 6&))7

g ( 3' ' #+'

>C $ ' 6&))7

% 0 $ " 0% # ,!#2 u2 4

3 # $ " '+ -.7%#2

D =12[u + (u)T ] 6&)7

" )!2

=12[u (u)T ] , 6&)&7

6 7

1 )2!

6#7 I : (ab)9a b67 I I9I6(7 I : I9*67 I9I 9 67 : ab 9 ( a) b

&* E (3-+ 8 ':

67 I : u = u6#7 (sI) = s2 s #- (67 (s ) = s + s 2 s #- (

& " (8

: u = ( u) u ( ) 6&)%7

& 8' -0 + ! :

= 2 D , 6&)7

D =12[u + (u)T ] 6&)7

0 $ 4

& 8 -0 + . 3+:

= 2u . 6&7

1 #' ' 8' ('' ' ' '

30% Navier-Stokes:

DuDt

= p + 2u + g 6&&70 ' $ ' 6&&7

&. E (3- ' ,' &:

" u(x, t) ( 3+' J ' G# 3'+ 4 Euler Lagrange2

dJ

dt= u J .

&% " J ' G# 3'+ Euler La-grange2 (8 . " 6steady ow7 3+:

d2J

dt2= J {( u) u} . 6&*7

-.36 d/dt 5 ($ ' D /Dt& B' ' #- = 1 Reynolds (8

; Leibniz

*) "(8 ' ( = 1 Reynolds:

d

dt

V (t)

f(x, t) u(x, t) dV =

V (t)

[(fu)t

+ u (fu) + (fu) u]

dV .

* 8

u (u) = u u + uu . 6&7

*& 67 !

1 )2! !

*% =+ ' (' ( :

u = ur(r, ) er + u(r, ) e

67 0 ' ' 3+ ' (+' ' IID -$ 4'

6#7 0 ' ' IID ' -2

u = u(r) e,

- -$ '

=

2IID

* !

"

& '%

1. T. Papanastasiou, G. Georgiou and A. Alexandrou, Viscous Fluid Flow, CRC Press, Boca Raton,1999.

& J. Marsden A. Tromba2 % & 61': " >

72 ,4'