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