Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 2.1. For three spatial dimensions, rewrite the following expressions in index notation
and evaluate or simplify them using the values or parameters given, and the definitions of
δ
ij and
ε
ijk wherever possible. In b) through e), x is the position vector, with components xi.
a)
b⋅c
where b = (1, 4, 17) and c = (–4, –3, 1)
b)
u⋅ ∇
( )
x
where u a vector with components ui.
c)
∇
φ
, where
φ
=h⋅x
and h is a constant vector with components hi.
d)
∇ × u
, where u = Ω × x and Ω is a constant vector with components Ωi.
e)
C⋅x
, where
C=
1 2 3
0 1 2
0 0 1
“
#
$
%
$
&
‘
$
(
$
Solution 2.1. a)
b)
u⋅ ∇
( )
x=uj
∂
∂
xj
xi=u1
∂
∂
x1
%
&
‘
(
)
* +u2
∂
∂
x2
%
&
‘
(
)
* +u3
∂
∂
x3
%
&
‘
(
)
*
+
,
–
.
/
0
x1
x2
x3
+
,
–
–
–
.
/
0
0
0
=
u1
∂
x1
∂
x1
#
$
%
&
‘
( +u2
∂
x1
∂
x2
#
$
%
&
‘
( +u3
∂
x1
∂
x3
#
$
%
&
‘
(
u1
∂
x2
#
%
&
( +u2
∂
x2
#
%
&
( +u3
∂
x2
#
%
&
(
u1
∂
x3
∂
x1
#
$
%
&
‘
( +u2
∂
x3
∂
x2
#
$
%
&
‘
( +u3
∂
x3
∂
x3
#
$
%
&
‘
(
)
*
+
+
+
+
+
+
+
,
–
.
.
.
.
.
.
.
=
u1⋅1+u2⋅0+u3⋅0
u1⋅0+u2⋅1+u3⋅0
u1⋅0+u2⋅0+u3⋅1
)
*
+
+
+
,
–
.
.
.
=uj
δ
ij =
u1
u2
u3
)
*
+
+
+
,
–
.
.
.
=ui
c)
∇
φ
=
∂φ
∂
xj
=
∂
∂
xj
hixi
( )
=hi
∂
xi
∂
xj
=hi
δ
ij =hj=h
d)
∇ × u=∇ × Ω×x
( )
=
ε
ijk
∂
∂
xj
ε
klmΩlxm
( )
=
ε
ijk
ε
klmΩl
δ
jm =
δ
il
δ
jm −
δ
im
δ
jl
( )
Ωl
δ
jm =
δ
il
δ
jj −
δ
ij
δ
jl
( )
Ωl
0 0 1
&
)
x3
&
)
x3
&
)
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 2.2. Starting from (2.1) and (2.3), prove (2.7).
Solution 2.2. The two representations for the position vector are:
x=x1e1+x2e2+x3e3
, or
x=“
x
1“
e
1+“
x
2“
e
2+“
x
3“
e
3
.
Develop the dot product of x with e1 from each representation,
e1⋅x=e1⋅x1e1+x2e2+x3e3
( )
=x1e1⋅e1+x2e1⋅e2+x3e1⋅e3=x1⋅1+x2⋅0+x3⋅0=x1
,
and
e1⋅x=e1⋅#
x
1#
e
1+#
x
2#
e
2+#
x
3#
e
3
( )
=#
x
1e1⋅#
e
1+#
x
2e1⋅#
e
2+#
x
3e1⋅#
e
3=#
x
iC1i
,
set these equal to find:
x1=“
x
iC1i
,
where
Cij =ei⋅#
e j
is a 3 × 3 matrix of direction cosines. In an entirely parallel fashion, forming
the dot product of x with e2, and x with e2 produces:
x2=“
x
iC2i
and
x3=“
x
iC3i
.
Thus, for any component xj, where j = 1, 2, or 3, we have:
xj=“
x
iCji
,
which is (2.7).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 2.3. For two three-dimensional vectors with Cartesian components ai and bi, prove the
Cauchy-Schwartz inequality: (aibi)2 ≤ (ai)2(bi)2.
Solution 2.3. Expand the left side term,
(aibi)2=(a1b
1+a2b2+a3b3)2=a1
2b
1
2+a2
2b2
2+a3
2b3
2+2a1b
1a2b2+2a1b
1a3b3+2a2b2a3b3
,
then expand the right side term,
(ai)2(bi)2=(a1
2+a2
2+a3
2)(b
1
2+b2
2+b3
2)
=a1
2b
1
2+a2
2b2
2+a3
2b3
2+(a1
2b2
2+a2
2b
1
2)+(a1
2b3
2+a3
2b
1
2)+(a3
2b2
2+a2
2b3
2).
Subtract the left side term from the right side term to find:
(ai)2(bi)2−(aibi)2
=(a1
2b2
2−2a1b
1a2b2+a2
2b
1
2)+(a1
2b3
2−2a1b
1a3b3+a3
2b
1
2)+(a3
2b2
2−2a2b2a3b3+a2
2b3
2)
=(a1b2−a2b
1)2+(a1b3−a3b
1)2+(a3b2−a2b3)2=a×b2.
Thus, the difference
(ai)2(bi)2−(aibi)2
is greater than zero unless a = (const.)b then the
difference is zero.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 2.4. For two three-dimensional vectors with Cartesian components ai and bi, prove the
triangle inequality:
a+b≥a+b
.
Solution 2.4. To avoid square roots, square both side of the equation; this operation does not
change the equation’s meaning. The left side becomes:
a+b
( )
2=a2+2ab+b2
,
and the right side becomes:
a+b2=(a+b)⋅(a+b)=a⋅a+2a⋅b+b⋅b=a2+2a⋅b+b2
.
So,
a+b
( )
2−a+b2=2a b −2a⋅b
.
Thus, to prove the triangle equality, the right side of this last equation must be greater than or
equal to zero. This requires:
a b ≥a⋅b
or using index notation:
ai
2bi
2≥aibi
,
which can be squared to find:
ai
2bi
2≥(aibi)2
,
and this is the Cauchy-Schwartz inequality proved in Exercise 2.3. Thus, the triangle equality is
proved.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 2.5. Using Cartesian coordinates where the position vector is x = (x1, x2, x3) and the
fluid velocity is u = (u1, u2, u3), write out the three components of the vector:
u⋅ ∇
( )
u=ui
∂
uj
∂
xi
( )
.
Solution 2.5.
a)
u⋅ ∇
( )
u=ui
∂
uj
%
‘
(
* =u1
∂
uj
%
‘
(
* +u2
∂
uj
%
‘
(
* +u3
∂
uj
%
‘
(
* =
u1
∂
u1
∂
x1
%
&
‘
(
)
* +u2
∂
u1
∂
x2
%
&
‘
(
)
* +u3
∂
u1
∂
x3
%
&
‘
(
)
*
u1
∂
u2
%
‘
(
* +u2
∂
u2
%
‘
(
* +u3
∂
u2
%
‘
(
*
+
,
–
–
–
/
0
–
–
–
( )
∂
∂
xi
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 2.6. Convert
∇ × ∇
ρ
to indicial notation and show that it is zero in Cartesian
coordinates for any twice-differentiable scalar function
ρ
.
Solution 2.6. Start with the definitions of the cross product and the gradient.
∇ × ∇
ρ
( )
=
ε
ijk
∂
∂
xj
∇
ρ
( )
k=
ε
ijk
∂
2
ρ
∂
xj
∂
xk
Write out the vector component by component recalling that
ε
ijk = 0 if any two indices are equal.
Here the “i” index is the free index.
ε
ijk
∂
2
ρ
∂
xj
∂
xk
=
ε
123
∂
2
ρ
∂
x2
∂
x3
+
ε
132
∂
2
ρ
∂
x3
∂
x2
ε
213
∂
2
ρ
∂
x1
∂
x3
+
ε
231
∂
2
ρ
∂
x3
∂
x1
ε
312
∂
2
ρ
∂
x1
∂
x2
+
ε
321
∂
2
ρ
∂
x2
∂
x1
!
“
#
#
#
#
$
#
#
#
#
%
&
#
#
#
#
‘
#
#
#
#
=
∂
2
ρ
∂
x2
∂
x3
–
∂
2
ρ
∂
x3
∂
x2
–
∂
2
ρ
∂
x1
∂
x3
+
∂
2
ρ
∂
x3
∂
x1
∂
2
ρ
∂
x1
∂
x2
−
∂
2
ρ
∂
x2
∂
x1
!
“
#
#
#
#
$
#
#
#
#
%
&
#
#
#
#
‘
#
#
#
#
=0
,
where the middle equality follows from the definition of
ε
ijk (2.18), and the final equality follows
when
ρ
is twice differentiable so that
∂
2
ρ
∂
xj
∂
xk
=
∂
2
ρ
∂
xk
∂
xj
.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 2.7. Using indicial notation, show that a × (b × c) = (a ⋅ c)b − (a ⋅ b)c. [Hint: Call d ≡ b
× c. Then (a × d)m =
ε
pqmapdq =
ε
pqmap
ε
ijqbicj. Using (2.19), show that (a × d)m = (a ⋅ c)bm − (a ⋅
b)cm.]
Solution 2.7. Using the hint and the definition of
ε
ijk produces:
(a × d)m =
ε
pqmapdq =
ε
pqmap
ε
ijqbicj =
ε
pqm
ε
ijq bicjap = –
ε
ijq
ε
qpm bicjap.
Now use the identity (2.19) for the product of epsilons:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 2.8. Show that the condition for the vectors a, b, and c to be coplanar is
ε
ijkaibjck = 0.
Solution 2.8. The vector b × c is perpendicular to b and c. Thus, a will be coplanar with b and c
if it too is perpendicular to b × c. The condition for a to be perpendicular with b × c is:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 2.9. Prove the following relationships:
δ
ij
δ
ij = 3,
ε
pqr
ε
pqr = 6, and
ε
pqi
ε
pqj = 2
δ
ij.
Solution 2.9. (i)
δ
ij
δ
ij =
δ
ii =
δ
11 +
δ
22 +
δ
33 = 1 + 1 + 1 = 3.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 2.10. Show that C⋅CT = CT⋅C = δ, where C is the direction cosine matrix and δ is the
matrix of the Kronecker delta. Any matrix obeying such a relationship is called an orthogonal
matrix because it represents transformation of one set of orthogonal axes into another.
Solution 2.10. To show that C⋅CT = CT⋅C = δ, where C is the direction cosine matrix and δ is
the matrix of the Kronecker delta. Start from (2.5) and (2.7), which are
“
x j=xiCij
and
xj=“
x
iCji
,
respectively, and change the index “i” into “m” on (2.5):
“
x j=xmCmj
. Substitute this into (2.7) to
find:
xj=“
x
iCji =xmCmi
( )
Cji =CmiCji xm
.
However, we also have xj =
δ
jmxm, so
δ
jm xm=CmiCji xm→
δ
jm =CmiCji
,
which can be written:
δ
jm =CmiCij
T=
C⋅CT,
and taking the transpose of the this produces:
δ
jm
( )
T=
δ
mj =CmiCij
T
( )
T=Cmi
TCij =
CT⋅C.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 2.11. Show that for a second-order tensor A, the following quantities are invariant
under the rotation of axes:
I1 = Aii ,
I2=
A
11 A
12
A21 A22
+
A22 A23
A32 A33
+
A
11 A
13
A31 A33
, and I3 = det(Aij).
[Hint: Use the result of Exercise 2.8 and the transformation rule (2.12) to show that I‘1 = A‘ii =
Aii.= I1. Then show that AijAji and AijAjkAki are also invariants. In fact, all contracted scalars of the
form AijAjk ⋅⋅⋅ Ami are invariants. Finally, verify that
I2=1
2I1
2−Aij Aji
“
#$
%
,
I3=1
3Aij Ajk Aki −I1Aij Aji +I2Aii
“
#$
%
. Because the right-hand sides are invariant, so are I2 and I3.]
Solution 2.11. First prove I1 is invariant by using the second order tensor transformation rule
(2.12):
“
A
mn =CimCjn Aij
.
Replace Cjn by
Cnj
T
and set n = m,
“
A
mn =CimCnj
TAij →“
A
mm =CimCmj
TAij
.
Use the result of Exercise 2.8,
δ
ij =CimCmj
T=
, to find:
I1=“
A
mm =
δ
ij Aij =Aii
.
Thus, the first invariant is does not depend on a rotation of the coordinate axes.
Now consider whether or not AmnAnm is invariant under a rotation of the coordinate axes.
Start with a double application of (2.12):
“
A
mn “
A
nm =CimCjn Aij
( )
CpnCqm Apq
( )
=CjnCnp
T
( )
CimCmq
T
( )
Aij Apq
.
From the result of Exercise 2.8, the factors in parentheses in the last equality are Kronecker delta
functions, so
“
A
mn “
A
nm =
δ
jp
δ
iq Aij Apq =Aij Aji
.
Thus, the matrix contraction AmnAnm does not depend on a rotation of the coordinate axes.
The manipulations for AmnAnpApm are a straightforward extension of the prior efforts for
Aii and AijAji.
“
A
mn “
A
np “
A
pm =CimCjn Aij
( )
CqnCrp Aqr
( )
CspCtm Ast
( )
=CjnCnq
T
( )
CrpCps
T
( )
CimCmt
T
( )
Aij Aqr Ast
.
Again, the factors in parentheses are Kronecker delta functions, so
“
A
mn “
A
np “
A
pm =
δ
jq
δ
rs
δ
it Aij Aqr Ast =Aiq Aqs Asi
,
which implies that the matrix contraction AijAjkAki does not depend on a rotation of the coordinate
axes.
Now, for the second invariant, verify the given identity, starting from the given definition
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Now work from the given identity. The triple matrix product AijAjkAki has twenty-seven terms:
A
11
3+A
11A
12 A21 +A
11A
13 A31 +A
12 A21 A
11 +A
12 A22 A21 +A
12 A23 A31 +A
13 A31A
11 +A
13 A32 A21 +A
13 A33 A31 +
A21 A
11A
12 +A21 A
12 A22 +A21A
13 A32 +A22 A21A
12 +A22
3+A22 A23 A32 +A23 A31 A
12 +A23 A32 A22 +A23 A33 A32 +
A31A
11A
13 +A31A
12 A23 +A31 A
13 A33 +A32 A21A
13 +A32 A22 A23 +A32 A23 A33 +A33 A31 A
13 +A33 A32 A23 +A33
3
These can be grouped as follows:
Aij Ajk Aki =3(A12 A23 A31 +A13 A32 A21)+A11(A11
2+3A12 A21 +3A13 A31)+
A22 (3A21A
12 +A22
2+3A23 A32 )+A33 (3A31A
13 +3A32 A23 +A33
2)
(b)
The remaining terms of the given identity are:
−I1Aij Aji +I2Aii =I1(I2–Aij Aji )=I1(I2+2I2−I1
2)=3I1I2–I1
3
,
where the result for I2 has been used. Evaluating the first of these two terms leads to:
3I1I2=3(A
11 +A22 +A33 )(A
11A22 −A
12 A21 +A22 A33 −A23 A32 +A
11A33 −A
13 A31)
=3(A
11 +A22 +A33 )(A
11A22 +A22 A33 +A
11A33 )−3(A
11 +A22 +A33 )(A
12 A21 +A23 A32 +A
13 A31)
.
Adding this to (b) produces:
Aij Ajk Aki +3I1I2=3(A12 A23 A31 +A13 A32 A21)+3(A11 +A22 +A33 )(A11A22 +A22 A33 +A11A33 )+
A
11(A
11
2−3A23 A32 )+A22 (A22
2−3A
13 A31)+A33 (A33
2−3A
12 A21)
=3(A
12 A23 A31 +A
13 A32 A21 −A
11A23 A32 −A22 A
13 A31 −A33 A
12 A21)+
3(A
11 +A22 +A33 )(A
11A22 +A22 A33 +A
11A33 )+A
11
3+A22
3+A33
3
(c)
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 2.12. If u and v are vectors, show that the products ui
υ
j obey the transformation rule
(2.12), and therefore represent a second-order tensor.
Solution 2.12. Start by applying the vector transformation rule (2.5 or 2.6) to the components of
u and v separately,