3 Solved Problems on Vector and Tensor Analysis - Assignment 2 | CEE 451, Assignments of Fluid Mechanics

Material Type: Assignment; Professor: Parker; Class: Environmental Fluid Mechanics; Subject: Civil and Environ Engineering; University: University of Illinois - Urbana-Champaign; Term: Unknown 1989;

Typology: Assignments

Pre 2010

Uploaded on 03/10/2009

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CEE 451G
Homework Assignment 2
This assignment involves the establishment of certain identities for use in vector
and tensor analysis. The assignment begins with a vector identity that is
established in any standard text on vector analysis:
w)vu(v)wu()wxv(xu
(1)
1. Use the above identity to establish the identity
imjljmilklmijk
(2)
In order to do this you need to remember the cyclic nature of ijk; the tensor is
positive when i, j and k follow in clockwise order (1, 2, 3; 2, 3, 1; 3, 1, 2), negative
when they follow in counterclockwise order (1, 3, 2; 3, 2, 1; 2, 1, 3) and otherwise
is equal to zero. Thus for example ijk = jki = kij = - ikj. Also recall that ij = 1 for i
= j and otherwise = 0, so that ij = ji.
2. Show that the following vector identity holds: for any scalar
)x(
,
0
xx
or0)(x
kj
2
ijk
(3)
3. Use the results of part 1 to establish the following vector identity:
)ux(xu)u(
2
1
u)u(
2
(4a)
l
m
jklmijkjj
ij
i
j
x
u
u)uu(
x2
1
x
u
uor
(4b)

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CEE 451G

Homework Assignment 2 This assignment involves the establishment of certain identities for use in vector and tensor analysis. The assignment begins with a vector identity that is established in any standard text on vector analysis: ux (vxw) (u w)v (u v) w              (1)

  1. Use the above identity to establish the identity ijk klmiljm jl im (2) In order to do this you need to remember the cyclic nature of ijk; the tensor is positive when i, j and k follow in clockwise order (1, 2, 3; 2, 3, 1; 3, 1, 2), negative when they follow in counterclockwise order (1, 3, 2; 3, 2, 1; 2, 1, 3) and otherwise is equal to zero. Thus for example ijk = jki = kij = - ikj. Also recall that ij = 1 for i = j and otherwise = 0, so that ij = ji.
  2. Show that the following vector identity holds: for any scalar (x)     , 0 x x x( ) 0 or j k 2 ijk            (3)
  3. Use the results of part 1 to establish the following vector identity: (u ) ux( xu ) 2 1 (u ) u     2         (4a) l m j j ijk klm j j i i j x u (uu) u 2 x

x u or u 

(4b)