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Ch.0. Group Work Units Continuum Mechanics Course (MMC) - ETSECCPB - UPC
1
Prove the following expression holds true: ε ε = 6ijk ijk
Unit 1
Unit 1 - Solution 2
ε ε ε ε ε
ε ε
εε ε ε ε εε ε
εε ε
+ + ++ + + ++ + + +
111 111 112 112 113 113
121 121 122
132 1
122
1 32
123
31 131 1 3
23
33
1
13
ε ε =ijk ijk
( ) ( ) ( ) ( ) ( ) ( )= + − − + − − + + + − − =
= + + + + + =
1·1 1 · 1 1 · 1 1·1 1·1 1 · 11 1 1 1 1 1 6
1
23
ε
ε
ε =
=
= −1
0
1ijk
k
ijk
ij
ε = 1ijk
ε = −1ijk
ε ε ε εε ε ε ε ε εε ε ε ε
ε
ε
ε
ε
+ + + ++ + + ++ + + +
211 211 212 212
221 221 222 222 223 223
232 23
213
2 23
21
3231 231 233
3
εε ε ε εε ε ε ε
ε ε ε ε ε
εε
εε
+ + + +
+ + + +
+ + +
311 311 313 313
322 322 323 323
331 33
312 3
1 332 33
321 3
2 33
1
21
3
2
333
3
Prove the following property of the tensor product: ( ) ( )⊗ = ⊗· ·u v w u v w
Unit 2
Unit 2 - Solution 4
( ) ( )⊗ = · · · u v w u v w
Vector Matrix Scalar Vector
Vector Vector
Vector (1st order tensor)
Matrix (2nd order tensor)
( ) [ ] ( ) ( ) = ⊗ = ⊗ = = ·k i i k i i kik ikc u v w u v wu v w u v w
(vector)c
k-component of vector c
( ) [ ] = ⊗ = ⊗ = ·k i i i i kkkc u v u v wu v w w k-component of vector c
( ) ( )⊗ = ⊗· ·u v w u v w
6
Prove the following properties of the scalar product:
=) · ·a u v v u
=) · 0b u 0
( ) ( ) ( )α β α β+ = +) · · ·c u v w u v u w
> ↔ ≠) · 0 d uu u 0
= ↔ =) · 0 e uu u 0
) 0 , , f = ≠ ≠ ↔ ⊥u·v u 0 v 0 u v
Unit 3 5
Unit 3 - Solution
7
= = =) · ·i i i ia u v v uu v v u
= =) · 0 0ib uu 0
( ) ( ) ( ) ( )α β α β α β α β+ = + = + = +) · · ·i i i i i i ic u v u v u wu v w u v u w
= ↔ = == ≥ → > ↔ ≠
0 0 , ) · 0
0 0 i i i
i ii i
u u ud u u
u uu 0
uuu
· cosθ=u·v u v
=· ·u v v u
=· 0u 0
( ) ( ) ( )α β α β+ = +·u v w u·v u·w
> ↔ ≠· 0 uu u 0
6
Unit 3 - Solution
) As it is developed in section d: 0 0i i ie u u u= ↔ =
( ) ( )
( ) ( )πθ θ θ
= = ≠ ≠ ≠ → → ≠ ≠ = = ≠
= → = → = → ⊥
1 12 2
1 12 2
· 0) , 0 , 0
· 0
Then · · cos cos 0 2
i i
i i
u uf
v v
u uuu 0 v 0 u v
v v v
u v u v u v
= ↔ =· 0 uu u 0
= ≠ ≠ ↔ ⊥· 0 , , u v u 0 v 0 u v
7
Unit 4
8
When does the relation hold true? =· ·n T T n
Unit 4 - Solution 9
· ·n T = T n
Vector Matrix
Vector
(vector)c
= → =
= → = =*
·
· k i ik
k ki i i ki
c nT
c T n nT*
c n Tc T n
* if Tk k ik kic c T= =
Unit 4 - Solution 10
=In compact notation: · ·n T T n
{ }= ∈In index notation: k 1,2,3i ik ki inT T n
[ ] [ ] [ ][ ] = In index notation: TTn T T n
ccTc
[ ] =
11 12 13 11 12 13 1
1 2 3 21 22 23 21 22 23 2
31 32 33 31 32 33 3
T T T T T T cc c c T T T T T T c
T T T T T T c[ ]
=
1
1 2 3 2
3
cc c c c
c
Unit 5 10
Prove the following equalities:
( )( )
=
=
) : ·
) ·· ·
Ta Tr
b Tr
A B A B
A B A B
Unit 5 - Solution 11
( ) [ ] = = = = = ) · ·T T Tik ki ij ijikkk ki
a c Tr A B A BA B A B A B
( ) [ ] [ ] [ ]= = = = = =) · · ki ik ij jikk ki ikb c Tr A B A BA B A B A B
k j→
i jk i→→
( )=: ·TTrA B A B
( )=·· ·TrA B A B
Unit 6
14
Prove the following properties of the open product:
( ) ( )⊗ ≠ ⊗) a u v v u
( ) ( ) ( ) ( )⊗ = ⊗ = =) · · · ·b u v w u v w u v w v w u
( )α β α β⊗ + = ⊗ + ⊗) c u v w u v u w
( ) ( ) ( )⊗ = ⊗ =) · · ·d u v w u v w w u v
12
Unit 6 - Solution
15
[ ][ ]
⊗ = = =⊗ =
) only if
i jij
i j i ji jij
a u vu v v u i j
v u v u
u v ( ) ( )⊗ ≠ ⊗u v v u
[ ] ( ) [ ] ( ) ( )[ ] ( ) ( )
→ ⊗ = ⊗⊗ = = = → ⊗ = ⊗
· · ·)
· · ·i
j i j j i j jiji
ub w u v w u v w
uv w u v w u v w
u vv w v w u u v w
( ) ( ) ( ) ( )⊗ = ⊗ = =· · · ·u v w u v w u v w v w u
( ) ( ) [ ] [ ]α β α β α β α β ⊗ + = + = + = ⊗ + ⊗ ) i j j i j i j ij ijijc u v w u v u wu v w u v u w
( )α β α β⊗ + = ⊗ + ⊗u v w u v u w
( ) ( ) ( ) ( ) ( ) ⊗ = = = = ⊗ = = ⊗ ) · · ·i i j i i j i i j j i ij j jd u v w u v w u v w w u vu v w u v w w u v
( ) ( ) ( )⊗ = ⊗ =· · ·u v w u v w w u v
13
Unit 7
16
Prove the following properties of the dot product:
14
= =) · ·a 1A A A 1
( )+ = +) · · ·b A B C A B A C
( ) ( )= =) · · · · · ·c A B C A B C A B C
≠) · ·d A B B A
Unit 7 - Solution
17
[ ] [ ]
[ ] [ ]
δ δ δ δ = = = = = = = = =
= =
) · · ·
· ·
Tik kj ij kj ik kj ki jk kiij ij ji ji
T
ji ij
a A A A A A T T T1A A A 1 A 1
A 1 A 1 = =· ·1A A A 1
( ) [ ] [ ] [ ] + = + = + = + = + ) · · ·ik ik kj kj ik kj ik kjkj ij ijijb A A B C A B A CA B C B C A B A C
( )+ = +· · ·A B C A B A C
( ) ( ) ( ) ( ) [ ] = = = = = ) · · · · · ·ik kl lj ik kl lj ik kl lj ijij ijc A B C A B C A B CA B C A B C A B C
( ) ( )= =· · · · · ·A B C A B C A B C
[ ][ ] [ ]
= = ≠
→ = → = → ==
) · If A sym A and B sym B · · ·
ik kj ik kj kj ik ik kjijT
ij ji ij jiik kjij
d A B A B B A B A
A BB A
A B
A B B AB A
15
18
Prove the following properties:
( ) ( ) ( ) ( )= = = = =) : · · · · :T T T Ta Tr Tr Tr TrA B A B B A A B B A B A
( )= =) : :b Tr1 A A A 1
( ) ( ) ( )= =) : · · : · :T Tc A B C B A C A C B
( ) ( )⊗ =) : · ·d A u v u A v
( ) ( ) ( )( )⊗ ⊗ =) : · ·e u v w x u w v x
Unit 8 16
Unit 8 - Solution
19
( )
( )
( )
( )
=
=
=
=
= = = = =
= = = = = =
= = = = =
= = = =
· ·
· ·
· ·
· ·
T T T Tik kj ik ki ki ki ij ijii j i
T T T Tik kj ik ki ki ki ij ij ij ijii j i
T T T Tik kj ik ki ik ik ij ijii j i
T T T Tik kj ik ki iii j i
Tr A B A B A B A B
Tr B A B A B A B A A B
Tr A B A B A B A B
Tr B A B A B
A B A B
B A B A
A B A B
B A B A = =k ik ij ij ij ijA B A A B
=) : ij ija A BA B
= = =: :ij ij ij ijB A A BB A A B
( ) ( ) ( ) ( )= = = = =: · · · · :T T T TTr Tr Tr TrA B A B B A A B B A B A
, k i i j→ →
, k i i j→ →
k j→
k j→
17
Unit 8 - Solution
20
( )= =: :Tr1 A A A 1
1
0 ij
ij
if i jif i j
δδ
= = = ≠
( ) ( ) ( )= =: · · : · :T TA B C B A C A C B
( )δ= = =) : ij ij iib A A Tr1 A A
( )δ= = =: ij ij iiA A TrA 1 A
[ ] ( ) ( ) ( ) ( ) = = = = = c) A · · · :T T Tij ij ik kj ik ij kj ki ij kj kjij kj
A B C B A C B A C CB C B A B A C
( ) ( ) = = = = = · : · · :T T T T T Tki ij kj ij kj ki ij jk ik ikik
B A C A C B A C B BB A C A C A C B
18
Unit 8 - Solution
[ ] ( ) [ ] ( )⊗ = = = = =) · · ·ij ij i j i ij j i ij j iij id A A u v u A v u A v u vu v A u A v
( ) ( )⊗ =: · ·A u v u A v
[ ] [ ] ( )( ) ( )( )⊗ ⊗ = = = =) · ·i j i j i i j j i i j jij ije u v w x u w v x u w v xu v w x u w v x
( ) ( ) ( )( )⊗ ⊗ =: · ·u v w x u w v x
19
Prove the following equality: ε= 1 2 3det ijk i j kA A A A
Unit 9 20
Unit 9 - Solution 21
ε ε εε εε ε
ε
ε
εεε ε
ε+ + +
+ + + ++ + + +
+ + + ++ +
11 21 31 11 21 32 11 21 33
11 22 31 11 22 32 11 22 33
11 23 31 11 23 32 11 23 33
12 21 31
111 112 113
121 122
131 133
211 212
22
1
12 21 32 12 21 33
12 22 31 12 2
1
32
21
2
1 22 2 32
3
3
A A A A A A A A AA A A A A A A A AA A A A A A A A A
A A A A A A A A AA A A A A A ε
ε ε
ε εε ε
ε
ε
ε
ε
εε
+ ++ + + +
+ + + +
+ + + +
+ + + =
2 12 22 33
12 23 31 12 23 32 12 23 33
13 21 31 13 21 32 13 21 33
13 22 31 13 22 32 13
223
232 233
311 313
322 3 22 33
23
13 23 31 13 23 3
23
331 332
321
1
312
2 13 2 333 33 3
A A AA A A A A A A A A
A A A A A A A A AA A A A A A A A AA A A A A A A A A
ε =1 2 3ijk i j jA A A
2
ε
ε
ε =
=
= −1
0
1ijk
k
ijk
ij
1ijkε =
1ijkε = −
3
1
First, can be computed as: ε 1 2 3ijk i j jA A A
Unit 9 - Solution
11 22 33 12 23 31 13 21 32 13 22 31 12 21 33 11 23 32A A A A A A A A A A A A A A A A A A= + + − − −
Then: 11 12 13
11 22 33 12 23 31 13 21 3221 22 23
13 22 31 12 21 33 11 23 3231 32 33
det det A A A
A A A A A A A A AA A A
A A A A A A A A AA A A
+ + = = − − −
A
ε= 1 2 3det ijk i j kA A A A
22
Prove the following equality: = × = − ×c a b b a
Unit 10 23
Unit 10 - Solution
[ ] [ ]× = = = − = − ×ijk j k ijk k j ikj k ji ie a b e b a e b aa b b a
= × = − ×c a b b a
24
Given the vector determine:
( )= = + +1 2 3 1 1 2 2 1 3ˆ ˆ ˆx x x x x xv v x e e e
∇
∇×
∇⊗ = ∇
) Divergence: ·
) Rotation:
) Gradient:
a
b
c
v
v
v v
Unit 11 25
Unit 11 - Solution
[ ]
= + + → =
1 2 3
1 2 3 1 1 2 2 1 3 1 2
1
ˆ ˆ ˆ x x x
x x x x x x x xx
v e e e v
∂∂ ∂ ∂∇ = = + + =
∂ ∂ ∂ ∂31 2
1 2 3
In index notation: · i
i
vv v vx x x x
v
2 3 1x x x+
[ ] [ ]∇ = ∇ =
∂ ∂ ∂ ∂ ∂ ∂ = + + = ∂ ∂ ∂ ∂ ∂ ∂
1 2 3
1 2 1 2 3 1 2 11 2 3 1 2 3
1
·
, ,
T
x x xx x x x x x x x
x x x x x xx
v v
a) Divergence:
In matrix notation:
2 3 1x x x+
1 1×3 1×1 3×
26
Unit 11 - Solution
3 32 1 1 212 13 21 23 31 32
1 1 2 2 3 3
In index notation:
kijk i i i i i i
j
v vv v v v ve e e e e e ex x x x x x x
∂ ∂∂ ∂ ∂ ∂ ∂∇× = = + + + + +
∂ ∂ ∂ ∂ ∂ ∂ ∂v
b) Rotation:
[ ]
∂ ∂+ ∂ ∂
∂ ∂ ∇× = + = − ∂ ∂ − ∂ ∂+
∂ ∂
3 2123 132
2 3
3 1213 231 1 2
1 32 1 3
2 1312 321
1 2
0 In matrix notation: 1
v ve ex xv ve e x xx x
x x xv ve ex x
v
In compact notation:
( ) ( )∇× = − + −1 2 2 2 1 3 3ˆ ˆ1x x x x xv e e
27
Unit 11 - Solution
∂ ∂ ∂ ∂ ∂ ∂ ∇× = × = = ∂ ∂ ∂ ∂ ∂ ∂
∂ ∂∂ ∂ ∂ ∂= − + − + − ∂ ∂ ∂ ∂ ∂
1 1 2 31
22 1 2 3
31 2 3
3
3 32 1 21 2
2 3 3 1 1
Calculated directly in matrix notation:
ˆ ˆ ˆ
det
ˆ ˆ
xvv
x x x xv
v v vx
v vv v v vx x x x x
e e e
v
e e ( ) ( ) = − + − ∂
13 1 2 2 2 1 3 3
2
ˆ ˆ ˆ1x x x x xx
e e e
28
Unit 11 - Solution
[ ] [ ] [ ][ ] [ ]
∂ ∂ ∂ ∇ ⊗ = ∇ = ∇ = = ∂ ∂ ∂
12 3 2
1 2 3 1 2 1 1 3 12
1 2
3
10
0 0
T
xx x x
x x x x x x x x xx
x x
x
v v v
c) Gradient:
In matrix notation: In compact notation:
[ ] [ ] ∂∇ ⊗ = ∇ =
∂j
ij iji
vx
v v
∇⊗ = ∇ =
⊗ + ⊗ + ⊗ + ⊗ + ⊗ + ⊗2 3 1 1 2 1 2 1 3 1 3 2 1 1 2 2 1 2 3 1ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆx x x x x x x xv ve e e e e e e e e e e e
3 1×
1 3×3 3×
29
Establish the following identities involving a smooth scalar field and a
smooth vector field v :
φ
( )φ φ φ∇ = ∇ + ∇) · · ·a v v v
( )φ φ φ∇ = ∇ ⊗ + ∇⊗) b v v v
Unit 12 30
Unit 12 - Solution
( ) [ ]φ φ φφ φ φ φ∂ ∂∂ ∂
∇ = = + + = ∇ + ∇∂ ∂ ∂ ∂
) · · ·i ii i
i i i i
va v vx x x xv
v v v
( )φ φ φ∇ = ∇ + ∇· · ·v v v
( ) ( )[ ]
[ ] [ ]φ φφ φ φ φ φ
∂ ∂∂ ∇ = ∇ = = + = ∇ ⊗ + ∇⊗ = ∂ ∂ ∂)
j jj ij ijij
i i i
vb v
x x x
vv v v v
[ ]φ φ= ∇ ⊗ + ∇⊗ ij
v v ( )φ φ φ∇ = ∇ ⊗ + ∇⊗v v v
31
Establish the following identities involving the smooth scalar fields and ,
smooth vector fields u and v , and a smooth second order tensor field A :
φ ψ
( ) ( ) ( )) a φψ φ ψ φ ψ∇⊗ = ∇⊗ + ∇⊗
( )φ φ φ∇ = ∇ +∇) · · ·b A A A
( ) ( )∇ = ∇ + ∇) · · · · :c A v A v A v
( ) ( ) ( )∇ ⊗ = ∇ ⊗ + ∇⊗) · · ·d u v u v u v
( ) ( )∇ × = ∇× − ∇×) · · ·e u v u v u v
Unit 13 32
Unit 13- Solution
( ) ( ) [ ] [ ] ( ) ( )) i ii i
i i i
ax x xφψ φ ψφψ ψ φ φ ψ φ ψ φ ψ φ ψ
∂ ∂ ∂ ∇ = = + = ∇ + ∇ = ∇ + ∇ ∂ ∂ ∂
( ) ( ) ( )φψ φ ψ φ ψ∇⊗ = ∇⊗ + ∇⊗
( )[ ]
[ ] [ ] [ ]
[ ]
φ φ φφ φ φ φ φ
φ φ
∂ ∂ ∂∂ ∂ ∇ = = + = + = ∇ + ∇ = ∂ ∂ ∂ ∂ ∂
= ∇ +∇
) · ·
· ·
ij ij ijij ij j i ijj
i i i i i
j
A Ab A A
x x x x x
AA A A
A A
( ) ( )∇ = ∇ + ∇· · · · :A v A v A v
( )φ φ φ∇ = ∇ +∇· · ·A A A
( ) ( ) [ ] [ ] [ ] [ ] ( )∂ ∂ ∂
∇ = = + = ∇ + ∇ = ∇ + ∇∂ ∂ ∂
) · · · · · :ij j ij jj ij j j ij ij
i i i
A v A vc v A A
x x xA v A v v A v A v
33
Unit 13- Solution
( ) ( ) ( )[ ] [ ] [ ] ( )∂ ∂∂
∇ ⊗ = = + = ∇ + ∇ = ∇ + ∇ ∂ ∂ ∂) · · · ·i j ji
j i j i ijj ji i i
u v vud v ux x x
u v u v u v u v u v
( ) ( ) ( )∇ ⊗ = ∇ ⊗ + ∇⊗· · ·u v u v u v
( ) ( ) ( )
[ ] [ ] [ ] [ ] ( )
εε ε ε ε ε
∂ ∂ ∂ ∂∂ ∂∇ × = = = + = − =
∂ ∂ ∂ ∂ ∂ ∂
= ∇× − ∇× = ∇× − ∇×
) ·
· ·
ijk j k j k j jk kijk ijk k ijk j k kij j jik
i i i i i i
k k j j
u v u v u uv ve v u v ux x x x x x
u v
v u u v u v u v
( ) ( )∇ × = ∇× − ∇×· · ·u v u v u v
34
Establish the following identities involving the smooth scalar field and the
smooth vector fields u and v :
φ
( ) ( ) ( )∇ = ∇ + ∇) · · ·a u v u v v u
( ) ( ) ( )φ φ φ∇× = ∇ × + ∇×) b v v v
( ) ( ) ( ) ( ) ( )∇× × = ∇ − ∇ + ∇ − ∇) · · · ·c u v u v v u u v u v
Unit 14 35
Unit 14 - Solution
( ) ( ) [ ] [ ] [ ] [ ] ( ) ( )∂ ∂ ∂
∇ = = + = ∇ + ∇ = ∇ + ∇ ∂ ∂ ∂) · · ·j j j j
j j ij j j iji ii i i
u v u va v u
x x xu v u v u v u v v u
( ) ( ) ( )∇ = ∇ + ∇· · ·u v u v v u
( ) ( ) [ ]
( ) ( )
φ φφ ε ε φε ε φ φε
φ φ
∂ ∂ ∂∂ ∇× = = + = ∇ + = ∂ ∂ ∂ ∂
= ∇ × + ∇×
)
k k kijk ijk k ijk ijk k ijkji
j j j j
i
v v vb v vx x x x
v
v v ( ) ( ) ( )φ φ φ∇× = ∇ × + ∇×v v v
( ) [ ] ( )
( ) ( )
εε ε ε ε ε ε
δ δ δ δ δ δ δ δ
∂ × ∂ ∂ ∂ ∇× × = = = + = ∂ ∂ ∂ ∂
∂ ∂= − + − =
∂ ∂
)
...
klm l mk l mijk ijk ijk lmk m ijk lmk li
j j j j
l mil jm im jl m il jm im jl l
j j
u v u vc v ux x x x
u vv ux x
u vu v
ijk pqk ip jq iq jpε ε δ δ δ δ= −
36
Unit 14- Solution
... j ji ij i i j
j j j j
u vu vv v u ux x x x
∂ ∂∂ ∂= − + − =∂ ∂ ∂ ∂
[ ] [ ] ( )[ ] [ ] ( ) [ ] [ ]= ∇ − ∇ + ∇ + ∇ =· ·ji j i i j ji
u v u v u v u v
[ ] [ ] ( )[ ] [ ] ( ) [ ] [ ]= ∇ − ∇ + ∇ + ∇ =· ·ij j i i j ji
u v u v u v u v
( ) ( ) ( ) ( ) = ∇ − ∇ + ∇ − ∇ = · · · ·i i i i
u v v u u v u v
( ) ( ) ( ) ( ) = ∇ − ∇ + ∇ − ∇ · · · ·i
u v v u u v u v
( ) ( ) ( ) ( ) ( )∇× × = ∇ − ∇ + ∇ − ∇· · · ·u v u v v u u v u v
37
Use the Generalized Divergence Theorem to show that
where is the position vector of i j ij
S
x n dS Vδ=∫ix .jn
V v
dS dV∂
∗ = ∗∇∫ ∫A n A
Generalized Divergence Theorem: 3x
2x
1x
V∂
V
n
n
Unit 15 38
Unit 15 - Solution
Applying the Generalized Divergence Theorem:
Applying the definition of gradient of a vector: The Generalized Divergence Theorem in index notation leads to:
= ⊗∫ ∫i jS S
x n dS dSx n
∂
⊗ = ⊗∇∫ ∫V V
dS dVx n x
[ ] [ ]∂ ∂∇ = → ∇ =
∂ ∂ j i
ij iji j
x xx x
x x
ii j ij ij
jS V V
xx n dS dV dV Vx
δ δ∂= = =
∂∫ ∫ ∫ i j ijS
x n dS Vδ=∫
39