Essential Mathematical Methods for the Physical Sciences (Instructor Solution Manual, Solutions)

467 Pages·2011·2.407 MB·English

by  K. F. Riley, M. P. Hobson

Essential
Mathematical Methods
for the
Physical Sciences
Instructor Solution Manual
K F RILEY and M P HOBSON
———- OOO ———-
Contents
1 Matrices and vector spaces 1
2 Vector calculus 53
3 Line, surface and volume integrals 79
4 Fourier series 108
5 Integral transforms 137
6 Higher-order ODEs 168
7 Series solutions of ODEs 206
8 Eigenfunction methods for ODEs 226
9 Special functions 247
10 Partial differential equations 267
11 Solution methods for PDEs 289
12 Calculus of variations 323
13 Integral equations 354
14 Complex variables 374
15 Applications of complex variables 389
16 Probability 412
17 Statistics 444
ii
Preface
ForreasonsthatareexplainedintheprefacetoEssential Mathematical Methods
for the Physical Sciences the text of the third edition of Mathematical Methods
forPhysicsandEngineering(MMPE)(Cambridge: CambridgeUniversityPress,
2006) by Riley, Hobson and Bence, after a number of additions and omissions,
has been republished as two slightly overlapping texts. Essential Mathematical
MethodsforthePhysicalSciences(EMMPS)containsmostofthemoreadvanced
material, and specifically develops mathematical methods that can be applied
throughoutthephysicalsciences;anaugmentedversionofthemoreintroductory
material, principally concerned with mathematical tools rather than methods,
is available as Foundation Mathematics for the Physical Sciences. The full text
of MMPE, including all of the more specialized and advanced topics, is still
available under its original title.
Asinthe thirdeditionofMMPE, the penultimate subsectionofeachchap-
ter of EMMPS consists of a significant number of problems, nearly all of which
are based on topics drawn from several sections of that chapter. Also as in the
third edition, hints and outline answers are given in the final subsection, but
only to the odd-numbered problems, leaving all even-numbered problems free
to be set as unaided homework.
A separate manual containing complete solutions to the two hundred and
thirtyplusodd-numberedproblemsinEMMPSisavailabletostudents. Thisweb
site file contains fully-worked solutions for all the problems in EMMPS, both
the odd-numbered and the even-numbered; it is available only to registered
instructors. For each problem, the originalquestion is reproducedand followed
by a fully-worked solution. In some cases where the original problem makes
internal reference to the main text, the questions have been reworded, usually
by including additional information; the solutions themselves do contain some
numbered references to topics and results in the main text.
Inmany casesthe solutiongivenhere is evenfuller thanone that mightbe
expected of a good student who has understood the material. This is because
we have, where possible, aimed to make the solutions both utilitarian and the
basis for further instruction. To this end, we have included comments that are
intended toshowhowthe planforthe solutionis formulatedandhaveprovided
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
simplifyingitatthenextorasubsequentstage. Whereseverallinesofalgebraic
iii
iv CONTENTS
manipulation or calculus are needed to obtain a final result, they are normally
includedinfull;thisshouldhelptheinstructortodeterminewhetherastudent’s
incorrect answer is due to a misunderstanding of principles or to a technical
error.
Ken Riley, Michael Hobson,
Cambridge, 2009
Chapter 1
Matrices and vector spaces
1.1 Which of the following statements about linear vector spaces are true?
Where a statement is false, give a counter-example to demonstrate this.
(a) Non-singular N N matrices form a vector space of dimension N2.
×
(b) Singular N N matrices form a vector space of dimension N2.
×
(c) Complex numbers form a vector space of dimension 2.
(d) Polynomial functions of x form an infinite-dimensional vector space.
(e) Series a ,a ,a ,...,a for which N a 2 = 1 form an N-
{ 0 1 2 N} n=0| n|
dimensional vector space.
P
(f) Absolutely convergentseries form an infinite-dimensional vector space.
(g) Convergent series with terms of alternating sign form an infinite-
dimensional vector space.
We first remind ourselves that for a set of entities to form a vector space, they
must pass five tests: (i) closure under commutative and associative addition;
(ii) closure under multiplication by a scalar; (iii) the existence of a null vector
in the set; (iv) multiplication by unity leaves any vector unchanged; (v) each
vector has a corresponding negative vector.
(a) False. The matrix 0 , the N N null matrix, required by (iii) is not
N
×
non-singular and is therefore not in the set.
1 0 0 0
(b) Consider the sum of and . The sum is the unit matrix
0 0 0 1
(cid:18) (cid:19) (cid:18) (cid:19)
which is not singular and so the set is not closed; this violates requirement (i).
The statement is false.
(c) The space is closed under addition and multiplication by a scalar; multi-
plication by unity leaves a complex number unchanged; there is a null vector
(= 0+i0) and a negative complex number for each vector. All the necessary
conditions are satisfied and the statement is true.
1
2 CHAPTER 1. MATRICES AND VECTOR SPACES
(d) As in the previouscase, allthe conditions aresatisfied andthe statementis
true.
(e) This statement is false. To see why, consider b = a + a for which
n n n
N b 2 = 4 = 1, i.e. the set is not closed (violating (i)), or note that
n=0| n| 6
there is no zero vector with unit norm (violating (iii)).
P
(f) True. Note that an absolutely convergent series remains absolutely conver-
gent when the signs of all of its terms are reversed.
(g) False. Consider the two series defined by
a = 1, a =2( 1)n forn 1; b = ( 1)n forn 0.
0 2 n −2 ≥ n − −2 ≥
Theseriesthatisthe sumof a and b doesnothavealternatingsignsand
n n
{ } { }
so closure (required by (i)) does not hold.
1.2 Consider the matrices
0 i i √3 √2 √3
− 1 − −
(a) B= i 0 i , (b) C= 1 √6 1 .
 −  √8 − 
i i 0 2 0 2
−
   
Are they (i) real, (ii) diagonal, (iii) symmetric, (iv) antisymmetric, (v) sin-
gular, (vi) orthogonal, (vii) Hermitian, (viii) anti-Hermitian, (ix) unitary, (x)
normal?
(a) For matrix B:
Clearly, (i)-(iii) are not true whilst (iv) is.
(v) B = i(i2 0)+i(i2)=0 and so B is singular.
| | − −
(vi)From(v)itfollowsthatBhasnoinverse. Inparticular,itstransposecannot
be its inverse, i.e. B is not orthogonal.
(vii)
T
0 i i 0 i i
− −
(B∗)T = i 0 i = i 0 i =B,
 −   − 
i i 0 i i 0
− −
   
i.e. B is Hermitian.
(viii) In view of (vii), B cannot be anti-Hermitian.
(ix) As in (vi), B cannot be unitary.
(x) Since B is Hermitian, it commuteswith its Hermitianconjugate (itself) and
is therefore normal.
(b) For matrix C:
C is clearly real, i.e. satisfies (i), and, equally clearly, satisfies none of (ii)-
(iv).
(v)
3 √3 √2 √3 3 √3 √2 0
1 − − 1 −
C = 1 √6 1 = 1 √6 0
| | √8 (cid:12) − (cid:12) √8 (cid:12) (cid:12)
(cid:18) (cid:19) (cid:12) 2 0 2 (cid:12) (cid:18) (cid:19) (cid:12) 2 0 4 (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
3
1
= (√18+√2)
√32
= 1(3+1)=1=0.
4 6
Thus C is not singular.
(vi) Consider CTC, which is given by
√3 1 2 √3 √2 √3 1 0 0
1 − −
√2 √6 0 1 √6 1 = 0 1 0 =I .
3
8 −   −   
√3 1 2 2 0 2 0 0 1
− −
     
Thus C is orthogonal.
(vii) & (viii) In view of (i), (iii) and (iv), C cannot be either Hermitian or anti-
Hermitian.
(ix) In view of (i) and (vi), C is unitary.
(x) In view of (i) and (vi), C†C=CTC=I=CCT =CC†. Hence C is normal.
1.3 By considering the matrices
1 0 0 0
A= , B= ,
0 0 3 4
(cid:18) (cid:19) (cid:18) (cid:19)
showthatAB=0doesnotimplythateitherAorBisthezeromatrixbutthat
it does imply that at least one of them is singular.
We have
1 0 0 0 0 0
AB= = .
0 0 3 4 0 0
(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
Thus AB is the zero matrix 0 without either A=0 or B=0.
However,AB=0 A B = 0 =0 and thereforeeither A =0 or B =0
⇒| || | | | | | | |
(or both).
1.4 Evaluate the determinants
1 0 2 3
a h g
0 1 2 1
(a) h b f , (b) (cid:12) − (cid:12)
(cid:12) (cid:12) (cid:12) 3 3 4 2 (cid:12)
(cid:12) g f c (cid:12) (cid:12) − − (cid:12)
(cid:12) (cid:12) (cid:12) 2 1 2 1 (cid:12)
(cid:12) (cid:12) (cid:12) − − (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
and (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12)
gc ge a+ge gb+ge
0 b b b
(c) (cid:12) (cid:12).
(cid:12) c e e b+e (cid:12)
(cid:12) (cid:12)
(cid:12) a b b+f b+d (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(a)UsingtheelementsandcofactorsofthefirstrowinastraightforwardLaplace
expansion, we have
a h g
h b f = a(bc f)+h(fg hc)+g(hf gb)
(cid:12) (cid:12) − − −
(cid:12) g f c (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
4 CHAPTER 1. MATRICES AND VECTOR SPACES
= abc+2fgh af2 bg2 ch2.
− − −
(b) At each stage we subtract a suitable multiple of the first column from each
other column so as to make the first entry in each of the other columns zero;
then we use a Laplace expansion with a single term. Here this reduction is
carried out three times.
1 0 2 3 1 0 0 0
0 1 2 1 0 1 2 1
(cid:12) − (cid:12) = (cid:12) − (cid:12)
(cid:12) 3 3 4 2 (cid:12) (cid:12) 3 3 2 11 (cid:12)
(cid:12) − − (cid:12) (cid:12) − − − (cid:12)
(cid:12) 2 1 2 1 (cid:12) (cid:12) 2 1 2 7 (cid:12)
(cid:12) − − (cid:12) (cid:12) − (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12)(cid:12) 1 2 1 (cid:12)(cid:12) (cid:12)(cid:12) 1 0 0 (cid:12)(cid:12)
−
=1 3 2 11 = 3 8 8
(cid:12) − − − (cid:12) (cid:12) − − − (cid:12)
(cid:12) 1 2 7 (cid:12) (cid:12) 1 4 6 (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) 8 8 (cid:12) (cid:12) 8 0 (cid:12)
=1 − − = 1 − = 82 = 16.
4 6 4 2 − | | −
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(c)Inmakingthis red(cid:12)uctionwe ((cid:12)i)subtra(cid:12)ctg times(cid:12)the thirdrowfromthe first
row,(ii)subtractthesecondrowfromthefourth,(iii)usetheLaplaceexpansion,
(iv)subtractthesecondcolumnfromthethird,and(v)useaLaplaceexpansion
followed by direct evaluation.
gc ge a+ge gb+ge 0 0 a 0
0 b b b 0 b b b
(cid:12) (cid:12) = (cid:12) (cid:12)
(cid:12) c e e b+e (cid:12) (cid:12) c e e b+e (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) a b b+f b+d (cid:12) (cid:12) a b b+f b+d (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) 0 0 a 0 (cid:12) (cid:12) (cid:12)
0 b b
0 b b b
(cid:12) (cid:12) = a c e b+e
(cid:12) c e e b+e (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) a 0 d (cid:12)
(cid:12) a 0 f d (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) 0 b 0 (cid:12)
c b
=a c e b = ab
(cid:12) (cid:12) − a d
(cid:12) a 0 d (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) = ab(ab(cid:12) cd).(cid:12)
(cid:12) (cid:12)
−
1.5 Using the properties of determinants, solvewith a minimum of calculation
the following equations for x:
x a a 1
x+2 x+4 x 3
a x b 1 −
(a) (cid:12) (cid:12)=0, (b) x+3 x x+5 =0.
(cid:12) a b x 1 (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) x 2 x 1 x+1 (cid:12)
(cid:12) a b c 1 (cid:12) (cid:12) − − (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12)
(a) In view of the similarities between some rows and some columns, the prop-
ertymostlikelytobe usefulhereisthatifadeterminanthastworows/columns
5
equal (or multiples of each other) then its value is zero.
(i) We note that setting x=a makes the first and fourth columns multiples of
each other and hence makes the value of the determinant 0; thus x = a is one
solution to the equation.
(ii) Setting x=b makes the second and third rows equal, and again the deter-
minant vanishes; thus b is another root of the equation.
(iii) Setting x = c makes the third and fourth rows equal, and yet again the
determinant vanishes; thus c is also a root of the equation.
Since the determinant contains no x in its final column, it is a cubic poly-
nomial in x and there will be exactly three roots to the equation. We have
already found all three!
(b) Here, the presence of x multiplied by unity in every entry means that sub-
tracting rows/columns will lead to a simplification. After (i) subtracting the
firstcolumnfromeachoftheothers,andthen(ii)subtractingthefirstrowfrom
each of the others, the determinant becomes
x+2 2 5 x+2 2 5
− −
x+3 3 2 = 1 5 7
(cid:12) − (cid:12) (cid:12) − (cid:12)
(cid:12) x 2 1 3 (cid:12) (cid:12) 4 1 8 (cid:12)
(cid:12) − (cid:12) (cid:12) − − (cid:12)
(cid:12) (cid:12) = (cid:12)(x+2)( 40+7)+(cid:12)2( 28 8) 5( 1 20)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) − (cid:12) − − − − −
= 33(x+2) 72+105
− −
= 33x 33.
− −
Thus x= 1 is the only solution to the original (linear!) equation.
−
1.6 This problem considersa crystalwhose unit cell has base vectorsthat are
not necessarily mutually orthogonal.
(a) The basis vectors of the unit cell of a crystal, with the origin O at one
corner,aredenoted by e , e , e . The matrix G has elements G , where
1 2 3 ij
Gij =ei ej andHij arethe elements ofthe matrixH G−1. Showthat
· ≡
thevectorsf = H e arethereciprocalvectorsandthatH =f f .
i j ij j ij i· j
(b) If the vectors uPand v are given by
u= u e , v= v f ,
i i i i
i i
X X
obtain expressions for u, v, and u v.
| | | | ·
(c) If the basis vectors are each of length a and the angle between each pair
is π/3, write down G and hence obtain H.
(d) Calculate (i) the length of the normal from O onto the plane containing
the points p 1e , q 1e , r 1e , and (ii) the angle between this normal
− 1 − 2 − 3
and e .
1
(a) With f defined by f = H e , consider
i i j ij j
f e = H e Pe = G 1 G = G 1G =δ .
i· k ij j · k − ij jk − ik ik
j j
X X(cid:0) (cid:1) (cid:0) (cid:1)
6 CHAPTER 1. MATRICES AND VECTOR SPACES
Thus the f are the reciprocal vectors of the cell’s base vectors.
i
Now consider
f f = H e H e = H H G
i j ik k jm m ik jm km
·
k m k,m
X X X
= H H G = H δ =H =H .
jm ik km jm im ji ij
m k m
X X X
(b) With u= u e ,
i i i
P
1/2
u2 = u e u e = u G u u = u G u .
i i j j i ij j i ij j
| | ⇒| |  
i j i,j i,j
X X X X
 
Similarly,
1/2
v = v H v .
i ij j
| |  
i,j
X
 
For the scalar product of u and v,
u v= u e v f = u v δ = u v .
i i j j i j ij i i
·
i j i,j i
X X X X
(c) For i=j, e e =a2 whilst, for i=j, e e =a2cos(π/3)= 1a2. Thus
i· j 6 i· j 2
2 1 1
G= 1a2 1 2 1 .
2  
1 1 2
 
Thematrix2G/a2 hasdeterminant4andallofits co-factorsareeither3or 1.
±
The matrix H, computed using this data, is found to be
3 1 1
H G−1 = 1 1 −3 −1 .
≡ 2a2  − − 
1 1 3
− −
 
(d)(i) The normal to the plane is in the direction
(q 1e p 1e ) (r 1e p 1e ) (qr) 1f +(pr) 1f +(qp) 1f
− 2 − 1 − 3 − 1 − 1 − 2 − 3
− × − ∝
(cid:2) (cid:3)
i.e. in the direction f =pf +qf +rf .
1 2 3
A unit vector in this direction is
pf +qf +rf
nˆ = 1 2 3
(f f)1/2
·
and the distance from the origin to the plane is the scalar product of this unit
vector and the position vector of any one of the three points (necessarily, they