Mattias Blennow

Mathematical Methods for
Physics and Engineering:
Solutions manual


ii

2015–2017 Mattias Blennow

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Contents
Chapter 1 Solutions: Scalars and Vectors
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1.6
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1.14
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1.32
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1.37
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Chapter 2 Solutions: Tensors
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2.1
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2.15
2.16
2.17
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2.19
2.20
2.21
2.22
2.23
2.24
2.25
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2.27
2.28
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2.31
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Chapter 3 Solutions: PDEs and Modelling
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3.3
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3.6
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Chapter 4 Solutions: Symmetries and Group Theory
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4.1
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4.10
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4.36
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Chapter 5 Solutions: Function Spaces
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5.1
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5.25
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Chapter 6 Solutions: Eigenfunction Expansions
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6.1
6.2
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6.6
6.7
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6.40
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Chapter 7 Solutions: Green’s Functions
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7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
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7.10
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7.23
7.24
7.25
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7.35
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Chapter 8 Solutions: Variational Calculus
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8.1
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8.11
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8.13
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8.22
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8.24
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8.28
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8.32
8.33
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xiv

Contents

Chapter 9 Solutions: Calculus on Manifolds
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9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
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9.10
9.11
9.12
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9.23
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Chapter 10 Solutions: Classical Mechanics and Field Theory
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xv

10.1
10.2
10.3
10.4
10.5
10.6
10.7
10.8
10.9
10.10
10.11

10.12
10.13
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10.24
10.25
10.26
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10.31
10.32
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10.34
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CHAPTER

1

Solutions: Scalars and
Vectors

Solution 1.1 Using the relations v = v i ei , ei · ej = δij , and ei × ej = εijk ek , we find that
a) v · (k w + u) = v i ei · ej (kwj + uj ) = v i δij (kwj + uj ) = kv i wi + v i ui ,
b) [(kv) × w]( + m) = [kv i ei × wj ej ]( + m) = εijk kv i wj ( + m)ek ,
c) v × (w × u) = v i ei × (wj ej × uk ek ) = v i wj uk ei × εjk e = v i wj uk εi

m εjk

d) (v × w) × u = (v i ei × wj ej ) × uk ek = v i wj uk εij e × ek = v i wj uk εij ε

em ,


km em .

Solution 1.2 We start by writing the vectors v and w in terms of the vector basis e1 , e2 ,
and e3 and obtain
v × w = (v 1 e1 + v 2 e2 + v 3 e3 ) × (w1 e1 + w2 e2 + w3 e3 ).
Using the relation e1 × e2 = −e2 × e1 = e3 (and the corresponding relations for other
combinations), we find that
v × w = v 1 v 2 e3 − v 1 v 3 e2 − v 2 w1 e3 + v 2 w3 e1 + v 3 w1 e2 − v 3 w2 e1
= (v 2 w3 − v 3 w2 )e1 + (v 3 w1 − v 1 w3 )e2 + (v 1 w2 − v 2 w1 )e3 .

Solution 1.3 Writing the vector v = v i ei , we find that
ej · v = ej · v i ei = δji v i = v j .
The vector component v j can therefore be found by taking the scalar product of ej and v.
The corresponding relation for the primed system can be found by instead writing v = v i ei
and taking the scalar product with ej in the same fashion.
Solution 1.4 We use that the inner product of any two basis vectors is given by ei ·ej = δij
and the cross product relation of Eq. (1.7).

1

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2

Mathematical Methods for Physics and Engineering: Solutions manual
a) We find the magnitudes by using the relation |v|2 = v · v
|v1 |2 = (3e1 − e2 ) · (3e1 − e2 )
= 32 e1 · e1 + (−1)2 e2 · e2 + 2 · 3 · (−1)e1 · e2 = 10,
|v2 |2 = (2e2 ) · (2e2 ) = 22 e2 · e2 = 4,

|v3 |2 = (−e1 + e2 + 5e3 ) · (−e1 + e2 + 5e3 ) = (−1)2 + 12 + 52 = 27.
√
√
It follows that the magnitudes are |v1 | = 10, |v2 | = 2, and |v3 | = 3 3.
b) The inner products between the different pairs of the given vectors are given by
v1 · v2 = (3e1 − e2 ) · 2e2 = (−1)2 = −2,
v1 · v3 = (3e1 − e2 ) · (−e1 + e2 + 5e3 ) = 3(−1) + (−1)1 = −4,
v2 · v3 = 2e2 · (−e1 + e2 + 5e3 ) = 2 · 1 = 2.
c) The cross product between the different pairs of vectors are given by
v1 × v2 = (3e1 − e2 ) × 2e2 = 3e1 × 2e2 = 6e3 ,
v1 × v3 = (3e1 − e2 ) × (−e1 + e2 + 5e3 )
= 3e1 × e2 + 3e1 × 5e3 − e2 × (−e1 ) − e2 × 5e3
= −5e1 − 15e2 + 2e3 ,
v2 × v3 = 2e2 × (−e1 + e2 + 5e3 ) = 2e2 × (−e1 ) + 2e2 × 5e3 = 2e3 + 10e1 .
Note that the cross product of any vector with itself is equal to zero.
d) The angles between any two vectors can be found through Eq. (1.5). For our given
cases, we find that
α12 = arccos
α13 = arccos
α23 = arccos

v1 · v2
|v1 | |v2 |
v1 · v3
|v1 | |v3 |
v2 · v3
|v2 | |v3 |

= arccos
= arccos

= arccos

−2
√
2 10
−4
√
3 30
2
√
6 3

108◦ ,
104◦ ,
79◦ .

e) The volume of the parallelepiped is given by the triple vector product
V = v1 · (v2 × v3 ) = (3e1 − e2 ) · (2e3 + 10e1 ) = 3 · 10 − 1 · 0 + 0 · 2 = 30.
Note that the triple product is invariant under cyclic permutations of the vectors.
Solution 1.5 We start by considering v · (v × w). Inserting the result from Solution 1.2,
we find
v · (v × w) = (v 1 e1 + v 2 e2 + v 3 e3 ) · [(v 2 w3 − v 3 w2 )e1 + (v 3 w1 − v 1 w3 )e2
+ (v 1 w2 − v 2 w1 )e3 ]
= v 1 (v 2 w3 − v 3 w2 ) + v 2 (v 3 w1 − v 1 w3 ) + v 3 (v 1 w2 − v 2 w1 ).
These terms cancel pairwise and therefore v is orthogonal to v × w. That this is true also
for w follows by using the anti-symmetry of the cross product and renaming v ↔ w.

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Solutions: Scalars and Vectors

3

Solution 1.6 The inner products of the vectors are given by
e1 · e1 =
e2 · e2 =

e
1 3
√3 = + = 1,
2 2
2
e
e
e
1 1 1
√1 + √2 + √3 = + + = 1,
3 3 3
3
3
3

e
e
e
√1 − √3 · √1 −
2
2
2

e1
e2
e3
√ +√ +√
·
3
3
3
e3
2
e2 + √
3
6

e3
2
e2 + √
3
6

e
√1 −
6

e3 · e3 =

e
√1 −
6


e1 · e2 =

e
e
√1 − √3
2
2

·

e
e
e
√1 + √2 + √3
3
3
3

e1 · e3 =

e
e
√1 − √3
2
2

·

e
√1 −

6

·

e
e
e
e
√1 + √2 + √3 · √1 −
3
3
3
6
√
1
1
2
= √ −
+ √ = 0.
3
3 2
3 2

1 2 1
+ + = 1,
6 3 6

1
1
= √ + 0 − √ = 0,

6
6

2
e3
e2 + √
3
6

e2 · e3 =

=

1
1
= √ + 0 − √ = 0,
12
12

2
e3
e2 + √
3
6

From the above computations, we conclude that the given vectors constitute an orthonormal
vector basis. From the cross product
e1 × e2 =

e

e
√1 − √3
2
2

×

e
e
e
√1 + √2 + √3
3
3
3

1
= √ e1 −
6

1
2
e2 + √ e3 = e3
3
6

also follows that the set is right-handed.
Solution 1.7 Going from the Einstein summation convention to regular sum notation, we
have
N


δii =

N

δii =
i=1

1 = N.
i=1

In particular, for N = 3, we would have
δii = 1 + 1 + 1 = 3.

Solution 1.8 Writing out the non-zero terms of ei εijk v j wk , we find that
v × w = e1 (ε123 v 2 w3 + ε132 v 3 w2 ) + e2 (ε231 v 3 w1 + ε213 v 1 w3 )
+ e3 (ε312 v 1 w2 + ε321 v 2 w1 ).
Using the explicit expressions for the components of the permutation symbol now gives
v × w = e1 (v 2 w3 − v 3 w2 ) + e2 (v 3 w1 − v 1 w3 ) + e3 (v 1 w2 − v 2 w1 ).

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4

Mathematical Methods for Physics and Engineering: Solutions manual

Solution 1.9 Explicitly writing out the sum in the ε-δ-relation, we find that
εijk εk

m


= εij1 ε1

m

+ εij2 ε2

m

+ εij3 ε3

m.

If i and j are equal, then all of the terms are zero by virtue of the anti-symmetry in the
first permutation symbol. This is also true of the expression δi δjm − δim δj , which is also
anti-symmetric under the exchange of i and j. Assuming that i = 1 and j = 2, we find that
ε12k εk

m

= ε123 ε3

m

= ε3

m.

This expression is equal to one if = 1 = i and m = 2 = j, minus one if = 2 = j
and m = 1 = i, and zero otherwise. This is also true of the expression δ1 δ2m − δ1m δ2 .

The corresponding argumentation can be made for any other choice of different i and j. It
follows that the ε-δ-relation is valid for all choices of i and j.
Solution 1.10 The ε-δ-relation results in
εijk εjk = εijk εk

j

= δi δjj − δij δj = 3δi − δi = 2δi .

Solution 1.11 Explicit computation results in
a) v · w = v = (x2 e1 − x1 e2 ) · (x3 e1 − x1 e3 ) = x2 x3 ,
b) v × w = (x2 e1 − x1 e2 ) × (x3 e1 − x1 e3 ) = x1 x2 e2 + x1 x3 e3 + (x1 )2 e1 = x1 x,
c) v · (w × x) = x · (v × w) = (x1 )3 + x1 (x2 )2 + x1 (x3 )2 = x1 x 2 .
Solution 1.12 The vectors v + w and v − w are orthogonal if
0 = (v + w) · (v − w) = v 2 + w · v − v · w − w 2 .
Since v · w = w · v, the middle two terms cancel. Moving the last term to the left-hand side
results in
v 2 = w 2 ⇐⇒ |v| = |w|.
The diagonals are therefore orthogonal only if the vectors v and w have the same magnitude.

Solution 1.13 The cube diagonals can be taken to be
v1 = e1 + e2 + e3

and

v2 = e1 + e2 − e3 ,

which are both vectors √
from one of the cube corners to a diagonally opposite one. For both
of these, we find |vi | = 3. The inner product of the diagonals is given by

v1 · v2 = 1 + 1 − 1 = 1.
Using Eq. (1.5) to find the angle between the vectors, we find that
α = arccos

v1 · v2
vi2

= arccos

1
3

70.5◦ .

Note: Depending on the diagonals chosen, the angle may also be found to be 180◦ − 70.5◦ =
109.5◦ .

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Solutions: Scalars and Vectors

5

Solution 1.14 The squared magnitude of the vector v × w is given by
|v × w|2 = (v × w) · (v × w) = v · [w × (v × w)],
where we have used the cyclic property of the triple product a · (b × c). Applying the bac-cab
rule for the triple cross product, we find
|v × w|2 = v · [v(w · w) − w(v · w)] = v 2 w 2 − (v · w)2 .
Using Eq. (1.5) directly gives us the relation

|v × w|2 = |v|2 |w|2 [1 − cos2 (α)] = |v|2 |w|2 sin2 (α),
where α is the angle between v and w.
Solution 1.15 The vector d(t, s) is given by
d(t, s) = (t − 2s)e1 + (3 − t)e2 + se3 .
The squared distance between the points on the line is given by
d2 = d 2 = 2t2 − 4ts + 5s2 − 6t + 9.
The distance therefore has a minimum whenever the partial derivatives of this expression
with respect to both t and s are equal to zero, i.e.,
∂d2
∂d2
= 4(t − s) − 6 = 0,
= −4t + 10s = 0.
∂t
∂s
This system of equations has the solution t = 5/2, s = 1. The difference vector for these
values is
1
d = (e1 + e2 + 2e3 )
2
and the tangent vectors are given by v1 = e1 − e2 and v2 = 2e1 − e3 , respectively. We
therefore find the scalar products
1
1
(1 − 1) = 0 and v2 · d = (2 − 2) = 0,
2
2
implying that the tangent vectors are orthogonal to the separation vector between the
closest points.
v1 · d =


Solution 1.16
a) The distance from the origin is given by
d(t) = |x(t)| =
=

x(t)2 =

r02 cos2 (ωt) + r02 sin2 (ωt) + v02 t2

r02 + v02 t2 .

b) The velocity and acceleration vectors are given by the first and second time derivatives
of the position vector, respectively. We find that
˙
v(t) = x(t)
= −r0 ω sin(ωt)e1 + r0 ω cos(ωt)e2 + v0 e3
= r0 ω[− sin(t)e1 + cos(t)e2 ] + v0 e3 ,
ă = r0 ω 2 [cos(ωt)e1 + sin(ωt)e2 ].
a(t) = x(t)

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6

Mathematical Methods for Physics and Engineering: Solutions manual

Solution 1.17 The magnitude of the angular momentum vector is given by |L| =
and is constant if L 2 is constant. By taking its time derivative, we find that


L2

dL2
dL
= 2L ·
= 2L · (v × L) = 0,
dt
dt
since the cross product is orthogonal to both v and L. Similarly, the time derivative of the
inner product with v is given by
dL
d(v · L)
=v·
= v · (v × L) = 0
dt
dt
under the assumption that v is constant. Thus, both the magnitude of the angular momentum and its inner product with v are constant in time.
Solution 1.18 Applying the definitions of the divergence and curl gives
a)
∇ · x = ∂i xi = δii = 3,
∇ × x = ei εijk ∂j xk = ei εijk δjk = ei εijj = 0,
b)
∇ · v1 = ∂i εijk aj xk = εijk aj δik = εiji aj = 0,
∇ × v1 = ei εijk ∂j εk

ma

xm = ei (δi δjm − δim δj )a δjm = 2a,

c)

∇ · v2 = ∂1 x2 − ∂2 x1 = 0,
∇ × v2 = e1 ∂3 x1 + e2 ∂3 x2 + e3 [∂1 (−x1 ) − ∂2 x2 ] = −2e3 .
We have here used that ∂i xj = ∂xj /∂xi = δij as well as the ε-δ-relation.
Solution 1.19 Working in index notation and using the standard rules for partial derivatives, we find that
a)
∇ · (φx) = ∂i φxi = xi ∂i φ + φ∂i xi = xi ∂i φ + φδii = x · ∇φ + 3φ,
b)
∇ · (x × ∇φ) = ∂i (εijk xj ∂k φ) = εijk [(∂i xj )∂k φ + xj ∂i ∂k φ]
= εijk [δij ∂k φ + xj ∂i ∂k φ] = 0,
where we have used that the first term is symmetric under exchange of i and j and
the last under exchange of i and k.
c)
∇ · (φ∇φ) = ∂i (φ∂i φ) = φ∂i ∂i φ + (∂i φ)∂i φ = φ∇2 φ + (∇φ)2 ,

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Solutions: Scalars and Vectors

7

d)
∇ × (x × ∇φ) = ei εijk ∂j εk

mx

∂m φ

= ei (δi δjm − δim δj )(δj ∂m φ + x ∂j ∂m φ)
= ei (∂i φ + xi ∂j ∂j φ − 3∂i φ − xj ∂j ∂i φ)

= x∇2 φ − 2∇φ − (x · ∇)∇φ.

Solution 1.20 Using the commutativity of the partial derivatives, we find that
∇ × ∇φ = ei εijk ∂j ∂k φ
= {rename j and k} = ei εikj ∂j ∂k φ
= {εijk is anti-symmetric} = −ei εijk ∂j ∂k φ = −∇ × ∇φ.
Subtracting ∇ × ∇φ from both sides and dividing with two gives ∇ × ∇φ = 0. We also find
that
∇ · (∇ × v) = εijk ∂i ∂j v k = εijk ∂j ∂i v k
= {rename i and j} = εjik ∂i ∂j v k
= {εijk is anti-symmetric} = −εijk ∂i ∂j v k = −∇ · (∇ × v).
In the same manner it therefore follows that ∇ · (∇ × v) = 0.
Solution 1.21 We find that
∇ × (∇ × v) = ei εijk ∂j εk

m∂

v m = ei (δi δjm − δim δj )∂j ∂ v m

= ei (∂i ∂j v j − ∂j ∂j v i ) = ∇(∇ · v) − ∇2 v.

Solution 1.22 Taking the scalar product of S and v1 , it follows from Eq. (1.36) that
i

i

−1 j
−1 j
S · v1 = εji1 i2 ...iN −1 v1i1 v2i2 . . . vNN−1
v1 = −εi1 ji2 ...iN −1 v1i1 v2i2 . . . vNN−1

v1

i

−1 i1
= {rename i1 and j} = −εji1 i2 ...iN −1 v1j v2i2 . . . vNN−1
v1

i

−1 j
= −εji1 i2 ...iN −1 v1i1 v2i2 . . . vNN−1
v1 = −S · v1 .

It follows that S · v1 = 0. The same line of argumentation can be followed for any of the
vectors vk .
Solution 1.23 A point on the surface in either of the cases can be described by the vector
x(s, t) = se1 + te2 + fi (s, t)e3 , where the functions fi can be deduced by solving for x3 from
the functions φi that define the surfaces. The directed area element will then be given by
Eq. (1.59). We find that
a)
x(s, t) = se1 + te2 + (5 − t − s)e3 ,
dS = (e1 − e3 ) × (e2 − e3 )ds dt = (e1 + e2 + e3 )ds dt,
∇φ1 = ∇(x1 + x2 + x3 ) = e1 + e2 + e3 .

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8


Mathematical Methods for Physics and Engineering: Solutions manual
b)
x(s, t) = se1 + te2 + (s2 + t2 )e3 ,
dS = (e1 + 2se3 ) × (e2 + 2te3 )ds dt = (e3 − 2se1 − 2te2 )ds dt,
∇φ2 = ∇[(x1 )2 + (x2 )2 − x3 ] = 2x1 e1 + 2x2 e2 − e3 = 2se1 + 2te2 − e3 .
c)
x(s, t) = se1 + te2 + [r0 cos(ks) − 4]e3 ,
dS = [e1 − kr0 sin(ks)e3 ] × e2 = e3 + kr0 sin(ks)e1 ,
∇φ3 = ∇[x3 − r0 cos(kx1 )] = e3 + kr0 sin(kx1 )e1 = e3 + kr0 sin(ks)e1 .

In all cases, we find that ∇φi is parallel to dS.
Solution 1.24 Starting from the left-hand side, we obtain
[(x × ∇) × (x × ∇)]φ = ei εijk εj

mx

∂m εknp xn ∂p φ

= ei (δk δim − δkm δi )εknp x (δmn ∂p φ + xn ∂m ∂p φ)
= ei (εkip xk ∂p φ + εknp xk xn ∂p φ − εkkp xi ∂p φ
− εknp xi xn ∂k ∂p φ).
In this expression, the second term vanishes due to εknp being anti-symmetric under exchange of k and n whereas xk xn is symmetric, the third term vanishes due to εkkp = 0, and
the last term vanishes due to the anti-symmetry of εknp and the symmetry of ∂k ∂p φ. We
are left with
[(x × ∇) × (x × ∇)]φ = ei εkip xk ∂p φ = −ei εikp xk ∂p φ = −x × ∇φ.

Solution 1.25 The expression on the left-hand side of the relation can be written
∇ × (∇φ × a) = ei εijk ∂j εk

m∂


φam = ei (δi δjm − δim δj )am ∂j ∂ φ

= ei (aj ∂j ∂i φ − ai ∂j ∂j φ) = ∇(∇φ · a) − a∇2 φ.
In order for the quoted relation to hold, φ must therefore satisfy ∇2 φ = 0.
Solution 1.26 Looking at the vector field v(kx), we can take its derivative with respect
to k and obtain
d(k n v(x))
n
dv(kx)
=
= nk n−1 v(x) = v(kx),
dk
dk
k
where we have used the stated property v(kx) = k n v(x). We can also use the chain rule to
compute the derivative according to
dv(kx)
d(kxi ) ∂v(kx)
xi ∂v(kx)
=
=
.
dk
dk
∂kxi
k ∂xi
Equating the two expressions results in
nv(kx) = xi ∂i v(kx).


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Solutions: Scalars and Vectors

9

For k = 1 we therefore find the sought relation
nv(x) = xi ∂i v(x) = (x · ∇)v(x).
The second expression can be written as
∇ · (x(x · v)) = ∂i (xi xj v j ) = δii xj v j + δij xi v j + xi xj ∂i v j = 4x · v + x · (x · ∇)v.
Using the relation just derived, we find that
∇ · (x(x · v)) = (4 + n)x · v.

Solution 1.27 We start by finding the acceleration by differentiating v with respect to
time, keeping in mind that both ω and x generally depend on time
d
ω × (x − x0 ) = ω˙ × (x − x0 ) + ω × x˙ = α × (x − x0 ) + ω × [ω × (x − x0 )]
dt
= α × (x − x0 ) + ω[ω · (x − x0 )] − (x − x0 )ω 2 ,

a=

where we have introduced the angular acceleration α = ω˙ and identified x˙ with the velocity
v.
The divergence and curl of the velocity field can be found as
∇ · v = ∇ · [ω × (x − x0 )] = −ω · (∇ × x) = 0,
∇ × v = ∇ × [ω × (x − x0 )] = ω(∇ · x) − (ω∇)x = 2ω,
where we have used that x0 and ω are constant with respect to the spatial coordinates, that
∇ × x = 0, ∇ · x = 3, and (k · ∇)x = k.

The corresponding consideration for the acceleration field results in
∇ · a = ∇ · [α × (x − x0 ) + ω[ω · (x − x0 )] − (x − x0 )ω 2 ]
= ω 2 − 3ω 2 = −2ω 2 ,
∇ × a = ∇ × [α × (x − x0 ) + ω[ω · (x − x0 )] − (x − x0 )ω 2 ] = 2α.

Solution 1.28 a) The path can be parametrised as
x(t) = r0 cos(t)e1 + r0 sin(t)e2 ,
where 0 ≤ t ≤ π/2. This implies that
dx =

dx
dt = r0 [− sin(t)e1 + cos(t)e2 ]dt
dt

and we find that
π
2

F · dx =

kr02 [cos(t)e2 − sin(t)e1 ] · [− sin(t)e1 + cos(t)e2 ]dt

0

Γ

=

π
2


kr02

2

2

[cos (t) + sin (t)]dt =

π
2

kr02

0

dt =
0

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kr02 π
.
2