0521842077pre

CB1005/Chen

0 521 84207 7

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January 29, 2006

14:8


stne o C

e fac Pr

page

ix

1

Preliminary algebra

1

2

Preliminary calculus

17

3

Complex numbers and hyperbolic functions

39

4

Series and limits

55

5

Partial differentiation

71

6

Multiple integrals

90

7

Vector algebra


104

8

Matrices and vector spaces

119

9

Normal modes

145

10

Vector calculus

156

11

Line, surface and volume integrals

176

v



CONTENTS

12

Fourier series

193

13

Integral transforms

211

14

First-order ODEs

228

15

Higher-order ODEs

246

16

Series solutions of ODEs


269

17

Eigenfunction methods for ODEs

283

18

Special functions

296

19

Quantum operators

313

20

PDEs: general and particular solutions

319

21

PDEs: separation of variables and other methods


335

22

Calculus of variations

353

23

Integral equations

374

24

Complex variables

386

25

Applications of complex variables

400

26

Tensors


420

27

Numerical methods

440

28

Group theory

461

29

Representation theory

480

vi


CONTENTS

30

Probability

494


31

Statistics

519

vii



e caf rP

The second edition of alicthemM ethodsM for Physic and Engier
carried
more than twice as many exercises, based on its various chapters, as did the first.
In the Preface we discussed the general question of how such exercises should
be treated but, in the end, decided to provide hints and outline answers to all
problems, as in the first edition. This decision was an uneasy one as, on the one
hand, it did not allow the exercises to be set as totally unaided homework that
could be used for assessment purposes, but, on the other, it did not give a full
explanation of how to tackle a problem when a student needed explicit guidance
or a model answer.
In order to allow both of these educationally desirable goals to be achieved, we
have, in the third edition, completely changed the way this matter is handled.
All of the exercises from the second edition, plus a number of additional ones
testing the newly added material, have been included in penultimate subsections
of the appropriate, sometimes reorganised, chapters. Hints and outline answers
are given, as previously, in the final subsections, tub y lno o t eht edrbmun-o
cise r x

. This leaves all even-numbered exercises free to be set as unaided
homework, as described below.
For the four hundred plus edrbmun-o
exercises, complete solutions are available, to both students and their teachers, in the form of siht
manual; these are in
addition to the hints and outline answers given in the main text. For each exercise,
the original question is reproduced and then followed by a fully worked solution.
For those original exercises that make internal reference to the text or to other
(even-numbered) exercises not included in this solutions manual, the questions
have been reworded, usually by including additional information, so that the
questions can stand alone. Some further minor rewording has been included to
improve the page layout.
In many cases the solution given is even fuller than one that might be expected
ix


PREFACE

of a good student who has understood the material. This is because we have
aimed to make the solutions instructional as well as utilitarian. To this end, we
have included comments that are intended to show how the plan for the solution
is formulated and have provided the justifications for particular intermediate
steps (something not always done, even by the best of students). We have also
tried to write each individual substituted formula in the form that best indicates
how it was obtained, before simplifying it at the next or a subsequent stage.
Where several lines of algebraic manipulation or calculus are needed to obtain a
final result, they are normally included in full; this should enable the student to
determine whether an incorrect answer is due to a misunderstanding of principles
or to a technical error.
The remaining four hundred or so edvn-umbr

exercises have no hints or
answers (outlined or detailed) available for general access. They can therefore
be used by instructors as a basis for setting unaided homework. Full solutions
to these exercises, in the same general format as those appearing in this manual
(though they may contain references to the main text or to other exercises), are
available without charge to accredited teachers as downloadable pdf files on the
password-protected website Teachers
wishing to have access to the website should contact
for registration details.
As noted above, the original questions are reproduced in full, or in a suitably
modified stand-alone form, at the start of each exercise. Reference to the main
text is not needed provided that standard formulae are known (and a set of tables
is available for a few of the statistical and numerical exercises). This means that,
although it is not its prime purpose, this manual could be used as a test or quiz
book by a student who has learned, or thinks that he or she has learned, the
material covered in the main text.
In all new publications, errors and typographical mistakes are virtually unavoidable, and we would be grateful to any reader who brings instances to our
attention. Finally, we are extremely grateful to Dave Green for his considerable
and continuing advice concerning typesetting in LATEX.
Ken Riley, Michael Hobson,
Cambridge, 2006

x


1

y anim le rP a rbegl

laimony P snoi e tauq

1.1 tI nac eb nwsoh tah eht laimony p
g(x) = 4x3 + 3x2 − 6x − 1
s a h g n i r u t s t n i o p ta
an iongatves fo tis iesoprt sa ws.olf

x = −1 dna

x=

1
2

dna e rht lae r sto r .reht go la euni t o C

(a) akeM a able t fo values fo
g(x) for egrint values of
se U it and the ionfrmat given above o t ward a aphgr and so ermindt
e h t s to r f o
g(x) = 0 as yeltacur as sible.po
(b) indF one etacur otr of
g(x) = 0 y b ionspect and ehnc ermindt ecispr
values for he t other wo t ots. r
(c) wSho tha
f(x) = 4x3 + 3x2 − 6x − k = 0 has y onl one ealr otr unles
7
−5 ≤ k ≤ 4 .

x we n b t

−2 dna


2.

(a) Straightforward evaluation of g(x) at integer values of x gives the following
table:
x
g(x)

−2
−9

−1
4

0
−1

1
0

2
31

(b) It is apparent from the table alone that x = 1 is an exact root of g(x) = 0 and
so g(x) can be factorised as g(x) = (x − 1)h(x) = (x − 1)(b2 x2 + b1 x + b0 ). Equating
the coefficients of x3 , x2 , x and the constant term gives 4 = b2 , b1 − b2 = 3,
b0 − b1 = −6 and −b0 = −1, respectively, which are consistent if b1 = 7. To find
the two remaining roots we set h(x) = 0:
4x2 + 7x + 1 = 0.
1



PRELIMINARY ALGEBRA

The roots of this quadratic equation are given by the standard formula as
√
−7 ± 49 − 16
α1,2 =
.
8
(c) When k = 1 (i.e. the original equation) the values of g(x) at its turning points,
x = −1 and x = 12 , are 4 and − 11
4 , respectively. Thus g(x) can have up to 4
subtracted from it or up to 11
added
to it and still satisfy the condition for three
4
(or, at the limit, two) distinct roots of g(x) = 0. It follows that for k outside the
range −5 ≤ k ≤ 74 , f(x) [= g(x) + 1 − k] has only one real root.

1.3 eigatInvs het iesoprt fo het ynomialp ionequat
f(x) = x7 + 5x6 + x4 − x3 + x2 − 2 = 0,
y b e d i n g c o p r sa w s . o l f
(a) yB
ni
ivestpo
(b) yB
ive n gat
s to r


f (x)

gnitrw eht ergd-htf laimony p aeprgni in eht noisexrp rof
eht mro f
7x5 + 30x4 + a(x − b)2 + c, wsho hat ethr is in actf yonl one
ot r of
f(x) = 0.
ing evalu t
f(1), f(0) dna
f(−1), and y b ingspect the ormf fo
v a l u e s fo
x, ermintd what you nac about the ionstp of the ealr
fo
f(x) = 0.

f(x) rof

(a) We start by finding the derivative of f(x) and note that, because f contains no
linear term, f can be written as the product of x and a fifth-degree polynomial:
f(x) = x7 + 5x6 + x4 − x3 + x2 − 2 = 0,
f (x) = x(7x5 + 30x4 + 4x2 − 3x + 2)
= x[ 7x5 + 30x4 + 4(x − 38 )2 − 4( 38 )2 + 2 ]

= x[ 7x5 + 30x4 + 4(x − 38 )2 +

23
16

].


Since, for positive x, every term in this last expression is necessarily positive, it
follows that f (x) can have no zeros in the range 0 < x < ∞. Consequently, f(x)
can have no turning points in that range and f(x) = 0 can have at most one root
in the same range. However, f(+∞) = +∞ and f(0) = −2 < 0 and so f(x) = 0
has at least one root in 0 < x < ∞. Consequently it has exactly one root in the
range.
(b) f(1) = 5, f(0) = −2 and f(−1) = 5, and so there is at least one root in each
of the ranges 0 < x < 1 and −1 < x < 0.
There is no simple systematic way to examine the form of a general polynomial
function for the purpose of determining where its zeros lie, but it is sometimes
2


PRELIMINARY ALGEBRA

helpful to group terms in the polynomial and determine how the sign of each
group depends upon the range in which x lies. Here grouping successive pairs of
terms yields some information as follows:
x7 + 5x6 is positive for x > −5,
x4 − x3 is positive for x > 1 and x < 0,
√
√
x2 − 2 is positive for x > 2 and x < − 2.
Thus, all three
√ terms are positive in the range(s) common to these, namely
−5 < x < − 2 and x > 1. It follows that f(x) is positive definite in these ranges
and there can be no roots of f(x) = 0 within them. However, since f(x) is negative
for large negative x, there must be at least one root α with α < −5.
1.5 onstrucC het icatqudr ionsequat tha have het wingfol pairs of ots:r
(a) −6, −3; (b) 0, 4; (c) 2, 2; (d) 3 + 2i, 3 − 2i, ewhr


i2 = −1.

Starting in each case from the ‘product of factors’ form of the quadratic equation,
(x − α1 )(x − α2 ) = 0, we obtain:
(a)

(x + 6)(x + 3) = x2 + 9x + 18 = 0;

(b)

(x − 0)(x − 4) = x2 − 4x = 0;

(c)

(x − 2)(x − 2) = x2 − 4x + 4 = 0;

(d) (x − 3 − 2i)(x − 3 + 2i) = x2 + x(−3 − 2i − 3 + 2i)
+ (9 − 6i + 6i − 4i2 )
= x2 − 6x + 13 = 0.

Trigon met c ies t d n
1.7 ovePr hat
π
cos
=
12

√


3+1
√
2 2

y b onsider g c
(a) eht smu fo eht seni fo
(b) eht sein fo eht smu fo

π/3 dna
π/3 dna

π/6,
π/4.

(a) Using
sin A + sin B = 2 sin

A+B
2
3

cos

A−B
2

,




SERIES SOLUTIONS OF ODES

y + z −3 y = 0, wsho hat the roign omesbc a egular

16.13 ro F het ionequat
singular point if the indept varible is hangedc morf
n d a s e r i i o n s l u t fo h e t f o r m
and xpandig e the ingesultr esionxpr for
si a dnosce niotuls htiw cito pmysa mrof

an z −n . yB ginste
du/dz in werspo of

y2 (z) = c z + ln z −
e wh r

x = 1/z. ecnH
y2 (z) = u(z)y1 (z)
z −1 , wsoh tah
y2 (z)

z ot

∞
0

y1 (z) =

1
2


+O

ln z
z

,

c si na y a r tib .tna s o c

With the equation in its original form, it is clear that, since z 2 /z 3 → ∞ as
z → 0, the origin is an irregular singular point. However, if we set 1/z = ξ and
y(z) = Y (ξ), with
1
dξ
= − 2 = −ξ 2
dz
z

d
d
= −ξ 2 ,
dz
dξ

⇒

then
−ξ 2


d
dξ

ξ2

−ξ 2

dY
dξ

+ ξ 3 Y = 0,

d2 Y
dY
+ ξY = 0,
+ 2ξ
dξ 2
dξ
2
1
Y + Y + Y = 0.
ξ
ξ

By inspection, ξ = 0 is a regular singular point of this equation, and its indicial
equation is
σ(σ − 1) + 2σ = 0

⇒


σ = 0, −1.

We start with the larger root, σ = 0, as this is ‘guaranteed’ to give a valid series
n
solution and assume a solution of the form Y (ξ) = ∞
n=0 an ξ , leading to
∞

∞

n(n − 1)an ξ n−1 + 2
n=0

n=0

Equating to zero the coefficient of ξ
am =

∞

nan ξ n−1 +

−am−1
m(m + 1)

m−1

⇒

an ξ n = 0.

n=0

gives the recurrence relation
am =

(−1)m
a0
(m + 1) (m!)2

and the series solution in inverse powers of z,
∞

y1 (z) = a0
n=0

(−1)n
.
(n + 1) (n!)2 z n

To find the second solution we set y2 (z) = f(z)y1 (z). As usual (and as intended),
278


SERIES SOLUTIONS OF ODES

all terms with f undifferentiated vanish when this is substituted in the original
equation. What is left is
0 = f (z)y1 (z) + 2f (z)y1 (z),
which on rearrangement yields
f

2y
= − 1.
f
y1
This equation, although it contains a second derivative, is in fact only a first-order
equation (for f ). It can be integrated directly to give
ln f = −2 ln y1 + c.
After exponentiation, this equation can be written as
df
A
1
A
1
= 2
+
= 2 1−
−
2
dz
2×1 z
3 × 22 z 2
y1 (z)
a0
1
A
1
= 2 1+ +O
,
z
z2

a0
where A = ec .
Hence, on integrating a second time, one obtains
f(z) =

A
a20

z + ln z + O

1
z

,

which in turn implies
1
1
A
1
+
−
z + ln z + O
a0 1 −
2
z
2z
12z 2
a0
ln z

1
= c z + ln z − + O
.
2
z

y2 (z) =

This establishes the asymptotic form of the second solution.
279

−2


SERIES SOLUTIONS OF ODES

16.15 The roign is na y idnaro point of the syhev bC ion,equat
(1 − z 2 )y − zy + m2 y = 0,
zσ

hwic eforht has seri ionslut of the ormf
(a) indF het encur ionshpelatr for het
erht tisx e laimony p snoitul

∞
0

σ = 0 dna

an z n rof


σ = 1.

an in the wot saec and wsho tha
Tm (z):

(i) rof
e ;smr t
(ii) rof
e .smr t

σ = 0, wehn

m is an ev n e g r, int the ynomial p having

1
2 (m + 2)

σ = 1, wehn

m si na d o e ,r g tni eht laimony p gnivah

1
2 (m + 1)

(b) Tm (z) is normalised so as o t have
rof
m = 0, 1, 2, 3.
(c) wohS tah eht egnidopsrc gnietamr-o seir snoitul
rieht tsr w ef smre t


Tm (1) = 1. indF xplict e forms orf
Sm (z) have sa
1 3
z +
3!
1
1 − z2 −
2!
3
z − z3 −
3!
9
1 − z2 +
2!

S0 (z) = a0 z +
S1 (z) = a0
S2 (z) = a0
S3 (z) = a0

(a)(i) If, for σ = 0, y(z) =
of z r in
(1 − z 2 )

Tm (z)

∞
n=0


9 5
z +
5!
3 4
z −
4!
15 5
z −
5!
45 4
z +
4!

,
,
,
.

an z n with a0 = 0, the condition for the coefficient

∞

∞

n(n − 1)an z n−2 − z
n=0

n=0

nan z n−1 + m2


∞

an z n
n=0

to be zero is that
(r + 2)(r + 1)ar+2 − r(r − 1)ar − rar + m2 ar = 0,
r 2 − m2
ar .
⇒ ar+2 =
(r + 2)(r + 1)
This relation relates ar+2 to ar and so to a0 if r is even. For ar+2 to vanish,
in this case, requires that r = m, which must therefore be an even integer. The
non-vanishing coefficients will be a0 , a2 , . . . , am , i.e. 12 (m + 2) of them in all.
(ii) If, for σ = 1, y(z) =

∞
n=0

an z n+1 with a0 = 0, the condition for the coefficient
280


SERIES SOLUTIONS OF ODES

of z r+1 in
∞

(1 − z 2 )


∞

(n + 1)nan z n−1 − z
n=0

∞

(n + 1)an z n + m2

n=0

an z n+1
n=0

to be zero is that
(r + 3)(r + 2)ar+2 − (r + 1)rar − (r + 1)ar + m2 ar = 0,
(r + 1)2 − m2
⇒ ar+2 =
ar .
(r + 3)(r + 2)
This relation relates ar+2 to ar and so to a0 if r is even. For ar+2 to vanish, in
this case, requires that r + 1 = m, which must therefore be an odd integer. The
non-vanishing coefficients will be, as before, a0 , a2 , . . . , am−1 , i.e. 12 (m + 1) of them
in all.
(b) For m = 0, T0 (z) = a0 . With the given normalisation, a0 = 1 and T0 (z) = 1.
For m = 1, T1 (z) = a0 z. The required normalisation implies that a0 = 1 and so
T0 (z) = z.
For m = 2, we need the recurrence relation in (a)(i). This shows that
a2 =


02 − 22
a0 = −2a0
(2)(1)

⇒

T2 (z) = a0 (1 − 2z 2 ).

With the given normalisation, a0 = −1 and T2 (z) = 2z 2 − 1.
For m = 3, we use the recurrence relation in (a)(ii) and obtain
a2 =

12 − 32
4
a0 = − a0
(3)(2)
3

⇒

T3 (z) = a0 (z −

4z 3
).
3

For the required normalisation, we must have a0 = − 31 and consequently that
T3 (z) = 4z 3 − 3z.
(c) The non-terminating series solutions Sm (z) arise when σ = 0 but m is an odd

integer and when σ = 1 with m an even integer. We take each in turn and apply
the appropriate recurrence relation to generate the coefficients.
(i) σ = 0, m = 1, using the (a)(i) recurrence relation:
a2 =

0−1
1
a0 = − a0 ,
(2)(1)
2!

a4 =

4−1
3
a2 = − a0 .
(4)(3)
4!

Hence,
S1 (z) = a0 1 −

3
1 2
z − z4 −
2!
4!

.


(ii) σ = 0, m = 3, using the (a)(i) recurrence relation:
a2 =

0−9
9
a0 = − a0 ,
(2)(1)
2!
281

a4 =

4−9
45
a2 =
a0 .
(4)(3)
4!


SERIES SOLUTIONS OF ODES

Hence,
S3 (z) = a0 1 −

9 2 45 4
z + z +
2!
4!


.

(iii) σ = 1, m = 0, using the (a)(ii) recurrence relation:
a2 =

1−0
1
a0 =
a0 ,
(3)(2)
3!

a4 =

9−0
9
a2 =
a0 .
(5)(4)
5!

Hence,
S0 (z) = a0 z +

9
1 3
z + z5 +
3!
5!


.

(iv) σ = 1, m = 2, using the (a)(ii) recurrence relation:
a2 =

1−4
3
a0 = − a0 ,
(3)(2)
3!

a4 =

9−4
15
a2 = − a0 .
(5)(4)
5!

Hence,
S2 (z) = a0 z −

3 3 15 5
z − z −
3!
5!

282

.



71

niotcufegE sdohtem rof
id lai tn e r snoi tauq e

17.1 yB onsidergc
sniotcuf

h = f + λg htiw

h|h , ewhr
f dna

λ e ,la r evo rp ,tah ro f w o t

g,
f|f g|g ≥ 14 [ f|g + g|f ]2 .

The oinuctf
is de n y b

y(x) is ealr and ivestpo for la
y˜c (k) =

and it is given tha

y˜c (k)


x. Ist reiuo F senioc mrofsna t
∞
−∞

y(x) cos(kx) dx,

y˜c (0) = 1. voePr tha
y˜c (2k) ≥ 2[˜
yc (k)]2 − 1.

For any |h we have that h|h ≥ 0, with equality only if |h = |0 . Hence, noting
that λ is real, we have
0 ≤ h|h = f + λg|f + λg = f|f + λ g|f + λ f|g + λ2 g|g .
This equation, considered as a quadratic inequality in λ, states that the corresponding quadratic equation has no real roots. The condition for this (‘b2 < 4ac’)
is given by
[ g|f + f|g ]2 ≤ 4 f|f g|g ,

(∗)

from which the stated result follows immediately. Note that g|f + f|g is real
and its square is therefore non-negative.
The given datum is equivalent to
1 = y˜c (0) =

∞
−∞

y(x) cos(0x) dx =
283


∞

y(x) dx.
−∞


EIGENFUNCTION METHODS FOR ODES

Now consider
∞

y˜c (2k) =
=2
⇒

y˜c (2k) + 1 = 2

y(x) cos(2kx) dx

−∞
∞

−∞
∞
−∞

y(x) cos2 kx −

∞


y(x) dx,
−∞

y(x) cos2 kx.

In order to use (∗), we need to choose for f(x) and g(x) functions whose
product will form the integrand defining y˜c (k). With this in mind, we take
f(x) = y 1/2 (x) cos kx and g(x) = y 1/2 (x); we may do this since y(x) > 0 for all x.
Making these choices gives
∞
−∞

∞

y cos kx dx +

−∞
∞
−∞

y cos kx dx

2y cos kx dx

2

2

4˜
yc2 (k)

Thus,
y˜c (2k) + 1 = 2

∞
−∞

≤4
≤4
≤4

∞
−∞
∞
−∞
∞
−∞

y cos2 kx dx

∞

y dx,
−∞

y cos2 kx dx × 1,
y cos2 kx dx.

y(x) cos2 kx ≥ 2[˜
yc (k)]2


and hence the stated result.

yn (x) of a SturmLiovle ionequat

17.3 onsiderC the ealr ionsegfuct

(py ) + qy + λρy = 0,
in h w ic
es do not hange c sign in
y e s a n c r y b hangi c the s ign of al e ig nvalu s. or F
eht y ti ned

p(x), q(x) dna

ρ(x) aer yinuosltc di iablentr ealr ionsuctf and
a ≤ x ≤ b. ekaT

p(x)
p(x) as ivestpo oughtr the erval,int if
a ≤ x1 ≤ x2 ≤ b, ablishet

(λn − λm )

x2
x1

e cud D tah fi
fo ym (x).

a ≤ x ≤ b,


λn > λm then

[ The eadr yam nd it elhpfu o t etailusr this esultr y b inhgcsket het rst wfe
ionsegfuct fo the emyst
slaimony p
Pn (z) rof

ρyn ym dx = yn p ym − ym p yn

x2
x1

.

yn (x) must hangec sign wenbt wot esivuc sozer

y + λy = 0, hwti
n = 2, 3, 4, 5. ]

284

y(0) = y(π) = 0, and the eLgndr


EIGENFUNCTION METHODS FOR ODES

The function p (x) does not change sign in the interval a ≤ x ≤ b; we take it as
positive, multiplying the equation all through by −1 if necessary. This means that
the weight function ρ can still be taken as positive, but that we must consider all

possible functions for q(x) and eigenvalues λ of either sign.
We start with the eigenvalue equation for yn (x), multiply it through by ym (x) and
then integrate from x1 to x2 . From this result we subtract the same equation with
the roles of n and m reversed, as follows. The integration limits are omitted until
the explicit integration by parts is carried through:
ym (p yn ) dx +

ym q yn dx +

ym λn ρyn dx = 0,

yn (p ym ) dx +

yn q ym dx +

yn λm ρym dx = 0,

ym (p yn ) − yn (p ym )
ym p yn

x2
x1

−

dx + (λn − λm )

ym p yn dx − yn p ym
+


ym ρyn dx = 0,

x2
x1

yn p ym dx + (λn − λm )

ym ρyn dx = 0.

Hence
(λn − λm )

ym ρyn dx = yn p ym − ym p yn

x2
x1

.

(∗)

Now, in this general result, take x1 and x2 as successive zeros of ym (x), where m is
determined by λn > λm (after the signs have been changed, if that was necessary).
Clearly the sign of ym (x) does not change in this interval; let it be α. It follows
that the sign of ym (x1 ) is also α, whilst that of ym (x2 ) is −α. In addition, the
second term on the RHS of (∗) vanishes at both limits, as ym (x1 ) = ym (x2 ) = 0.
Let us now seupo
that the sign of yn (x) does not change in this same interval and
is always β. Then the sign of the expression on the LHS of (∗) is (+1)(α)(+1)β =
αβ. The first (+1) appears because λn > λm .

The signs of the upper- and lower-limit contributions of the remaining term on
the RHS of (∗) are β(+1)(−α) and (−1)β(+1)α, respectively, the additional factor
of (−1) in the second product arising from the fact that the contribution comes
from a lower limit. The contributions at both limits have the same sign, −αβ, and
so the sign of the total RHS must also be −αβ.
This contradicts, however, the sign of +αβ found for the LHS. It follows that it
was wrong to suppose that the sign of yn (x) does not change in the interval; in
other words, a zero of yn (x) does appear between every pair of zeros of ym (x).
285


EIGENFUNCTION METHODS FOR ODES

17.5 se U het iesoprt fo eLgndr ynomialsp o t y acr out the wingolf x- e
cise . r
(1 − x2 )y − 2xy + by = f(x) tah si dilav ni eht egnar
x = 0, in ermst of eLgndr ynomials. p
b = 14 dna
f(x) = 5x3 . yerif V it y b ectidr

(a) dinF eht noitusl fo
−1 ≤ x ≤ 1 and enit ta
(b) indF the xplict e ionslut if
.niotusb

[ x tic lp E smrof ro f eht e L rdn g slaimony p na c eb dnuo f ni y na .ko x bt e nI
Mathematical Methods for Physics and Engineering, 3rd edition, y het aer given
in ion Subsect 18.
]
(a) The LHS of the given equation is the same as that of Legendre’s equation and

2
so we substitute y(x) = ∞
n=0 an Pn (x) and use the fact that (1 − x )Pn − 2xPn =
−n(n + 1)Pn . This results in
∞

an [ b − n(n + 1) ]Pn = f(x).
n=0

Now, using the mutual orthogonality and normalisation of the Pn (x), we multiply
both sides by Pm (x) and integrate over x:
∞

an [ b − n(n + 1) ] δmn
n=0

⇒

am =

2
=
2m + 1

2m + 1
2[ b − m(m + 1) ]

1
−1
1

−1

f(z)Pm (z) dz,

f(z)Pm (z) dz.

This gives the coefficients in the solution y(x).
(b) We now express f(x) in terms of Legendre polynomials,
f(x) = 5x3 = 2[ 12 (5x3 − 3x) ] + 3[ x ] = 2P3 (x) + 3P1 (x),
and conclude that, because of the mutual orthogonality of the Legendre polynomials, only a3 and a1 in the series solution will be non-zero. To find them we
need to evaluate
1
4
2
= ;
f(z)P3 (z) dz = 2
2(3) + 1
7
−1
similarly,

1
−1

f(z)P1 (z) dz = 3 × (2/3) = 2.

Inserting these values gives
a3 =

4

1
7
3
= 1 and a1 =
2= .
2(14 − 12) 7
2(14 − 2)
4
286


EIGENFUNCTION METHODS FOR ODES

Thus the solution is
y(x) =
k Chec

1
1
5(2x3 − x)
1
P1 (x) + P3 (x) = x + (5x3 − 3x) =
.
4
4
2
4

:
30x2 − 5 140x3 − 70x

60x
− 2x
+
= 5x3 ,
4
4
4
60x − 60x3 − 60x3 + 10x + 140x3 − 70x = 20x3 ,

(1 − x2 )
⇒
which is satisfied.

17.7 onsiderC the set of ions,fuct
e rval int
−∞ < x < ∞, tah
unit weight ion,fuct ermindt whetr heac of the wingolf linear ros tape is
ianermtH when ingcta upon
d
+ x;
dx

(a)

{f(x)}, fo het ealr varible
→ 0 at least sa y lk quic as

x de n in the
x → ±∞. ro F
x , sa

−1

{f(x)}:
(b) − i

d
+ x2 ;
dx

(c) ix

d
;
dx

(d) i

d3
.
dx3

For an operator to be Hermitian over the given range with respect to a unit
weight function, the equation
∞
−∞

∞

f ∗ (x)[ g(x) ] dx =


−∞

g ∗ (x)[ f(x) ] dx

∗

(∗)

must be satisfied for general functions f and g.
d
(a) For =
+ x, the LHS of (∗) is
dx
∞
−∞

f ∗ (x)

dg
+ xg
dx

dx = f ∗ g
=0−

∞
−∞
∞
−∞


−

∞
−∞

df ∗
g dx +
dx
∞

df ∗
g dx +
dx

∞

f ∗ xg dx

−∞

f ∗ xg dx.

−∞

The RHS of (∗) is
∞
−∞

g ∗ (x)


df
+ xf
dx

∗

dx

=

∞
−∞
∞

g∗

df
dx
dx

df ∗
dx +
=
g
dx
−∞

∗

+

∞

∞

g ∗ xf dx

∗

−∞

gxf ∗ dx.

−∞

Since the sign of the first term differs in the two expressions, the LHS = RHS
and is ton
Hermitian. It will also be apparent that purely multiplicative terms
in the operator, such as x or x2 , will always be Hermitian; thus we can ignore
the x2 term in part (b).
287


EIGENFUNCTION METHODS FOR ODES

(b) As explained above, we need only consider
∞
−∞

f ∗ (x) −i


dg
dx

∞
−∞

dx = −if ∗ g
∞

=0+i

∞

+i

−∞

df ∗
g dx
dx

df ∗
g dx
dx

−∞

and
∞


g ∗ (x) −i

−∞

These are equal, and so
(c) For

= ix
∞
−∞

= −i

∗

df
dx

∞

=i

dx

g
−∞

df ∗
dx.
dx


d
is Hermitian, as is
dx

= −i

d
+ x2 .
dx

d
, the LHS of (∗) is
dx

f ∗ (x) ix

dg
dx

dx = ixf ∗ g

∞
−∞

∞

−i

x

−∞

∞

df ∗
g dx − i
dx

df ∗
g dx − i
x
=0−i
dx
−∞

∞

∞

f ∗ g dx

−∞

∗

f g dx.
−∞

The RHS of (∗) is given by
∞

−∞

Since, in general, −i
not Hermitian.

g ∗ (x)ix
∞
−∞

∗

df
dx

∞

= −i

dx

gx
−∞

df ∗
dx.
dx

∗

fg dx = 0, the two sides are not equal; therefore


is

d
d3
is the cube of the operator −i , which was shown in part (b)
3
dx
dx
to be Hermitian, it is expected that is Hermitian. This can be verified directly
as follows.
(d) Since

=i

The LHS of (∗) is given by
∞

i
−∞

f∗

∞

2
d3 g
∗d g
dx
=

if
dx3
dx2

−∞

−i

df ∗ dg
=0−i
dx dx
=0+i
=0+
Thus

2 ∗

d f
g
dx2
∞

∞

∞
−∞

−∞
∞
−∞


+i

−i

d3 f
ig ∗ 3 dx
dx
−∞

is confirmed as Hermitian.
288

df ∗ d2 g
dx
dx dx2

∗

∞

d2 f ∗ dg
dx
dx2 dx

−∞
∞ 3 ∗
−∞

df

g dx
dx3

= RHS of (∗).


EIGENFUNCTION METHODS FOR ODES

17.9 dinF na niotcufeg xnoisap e rof eht noistul whti y adnuobr sniotdc
y(0) = y(π) = 0 of the inhomgeus ionequat
d2 y
+ κy = f(x),
dx2
e wh r

κ is a ant osc and
0 ≤ x ≤ π/2,
π/2 < x ≤ π.

x
π−x

f(x) =

The eigenfunctions of the operator

=

d2
+ κ are obviously

dx2

yn (x) = An sin nx + Bn cos nx,
with corresponding eigenvalues λn = n2 − κ.
The boundary conditions, y(0) = y(π) = 0, require that n is a positive integer and
that Bn = 0, i.e.
2
sin nx,
π

yn (x) = An sin nx =

where An (for n ≥ 1) has been chosen so that the eigenfunctions are normalised
over the interval x = 0 to x = π. Since
is Hermitian on the range 0 ≤ x ≤
π, the eigenfunctions are also mutually orthogonal, and so the yn (x) form an
orthonormal set.
If the required solution is y(x) =
result
∞

n

an yn (x), then direct substitution yields the

(κ − n2 )an yn (x) = f(x).

n=1

Following the usual procedure for analysis using sets of orthonormal functions,

this implies that
am =

1
κ − m2

π
0

f(z)ym (z) dz

and, consequently, that
∞

y(x) =
n=1

2 sin nx
π κ − n2
289

2
π

π
0

f(z) sin(nz) dz.



EIGENFUNCTION METHODS FOR ODES

It only remains to evaluate
In =
=
=

π
0

sin(nx)f(x) dx

π/2
0

π

x sin nx dx +

(π − x) sin nx dx

π/2

π/2
−x cos nx π/2
cos nx
dx
+
n
n

0
0
π
−(π − x) cos nx π
(−1) cos nx
+
dx
+
n
n
π/2
π/2

=−

π/2

sin nx
π cos(nπ/2)
(1 − 1) +
2
n
n2

=0+

(n−1)/2

(−1)
n2


0

−

sin nx
n2

π
π/2

(1 + 1) for odd n and = 0 for even n.

Thus,
y(x) =

4
π

n odd

(−1)(n−1)/2
sin nx
n2 (κ − n2 )

is the required solution.

17.11 The id ialentr ro tape

is de n y b

y=−

d
dx

ex

dy
dx

− 14 ex y.

λn of the blemorp

e rmin D t he t eig nvalu s

yn = λn ex yn

0 < x < 1,

htiw y adnuobr sniotdc
dy
+ 1y = 0
dx 2

y(0) = 0,
(a) indF the espondigrc unormalised
e sp ct r o t h w ic he t
rof eh t
yn .

(b) yB making an ionegfuct xpansio, e solve het ionequat

ta

x = 1.

yn , and also a weight ionfuct
yn aer rothgnal. e,ncH selct a able suit ionrmalst

y = −ex/2 ,
subject o t the same y boundar ionsdtc as .yeviouslpr

290

0 < x < 1,

ρ(x) htiw