As a realistic application of Fourier series we now derive first integral equations satisfied by Mathieu functions, from which subsequently their Fourier series are obtained.
Integral Equations and Fourier Series for Mathieu Functions
Our first goal is to establish Whittaker’s integral equations that Mathieu functions satisfy, from which we then obtain their Fourier series representations.
We start from an integral representation V (r)=
π
−π
f (z+ixcosθ+iysinθ, θ ) dθ (14.97) of a solutionV of Laplace’s equation with a twice differentiable functionf (v, θ ).Apply- ing∇2toV we verify that it obeys Laplace’s PDE. Separating variables in Laplace’s PDE suggests choosing the product formf (v, θ )=ekvφ (θ ).Substituting the elliptical variables of Eq. (13.163) we rewriteV as
R(ξ )(η)ekz= π
−π
φ (θ )ek(z+iccoshξcosηcosθ+icsinhξsinηsinθ )dθ (14.98) with normalizationR(0)=1.Sinceξ andηare independent variables we may setξ=0, which leads to Whittaker’s integral representation
(η)= π
−π
φ (θ )exp(ickcosθcosη) dθ, (14.99)
whereck=2√qfrom Eq. (13.180). Clearly,is even in the variableηand periodic with periodπ.In order to prove thatφ∼we check howφ (θ )is constrained when(η)is taken to obey the angular Mathieu ODE
d2
dη2 +(λ−2qcos 2η)(η)
= π
−π
φ (θ )exp(ickcosθcosη)
ã
λ−2qcos 2η+(ickcosθsinη)2−ickcosθcosη dθ. (14.100) Here we integrate the last term on the right-hand side by parts, obtaining
d2
dη2 +(λ−2qcos 2η)(η)
=φ (θ )(−ickcosηsinθ )exp(ickcosθcosη) π
θ=−π
+ π
−π
φ (θ )exp(ickcosθcosη)[λ−2qcos 2η−ickcosθcosη]dθ +
π
−π
−φ′(θ )(−ickcosηsinθ )+φ (θ )ickcosηcosθ exp(ickcosθcosη) dθ
= π
−π
expickcosθcosη
φ (θ )(λ−2qcos 2η)+φ′(θ )ickcosηsinθ dθ, (14.101) where the integrated term vanishes ifφ (−π )=φ (π ),which we assume to be the case.
Integrating once more by parts yields d2
dη2 +(λ−2qcos 2η)(η)= −φ′(θ )expickcosθcosη π
θ=−π
+ π
−π
exp(ickcosθcosη)
φ (θ )(λ−2qcos 2η)+φ′′(θ ) dθ, (14.102) where the integrated term vanishes if φ′ is periodic with periodπ, which we assume is the case. Therefore, ifφ (θ )obeys the angular Mathieu ODE, so does the integral(η), in Eq. (14.99). As a consequence,φ (θ )∼(θ ), where the constant may be a function of the parameterq.
Thus, we have the main result that a solution(η)of Mathieu’s ODE that is even in the variableηsatisfies the integral equation
(η)=n(q) π
−π
e2i√qcosθcosη(θ ) dθ. (14.103) When these Mathieu functions are expanded in a Fourier cosine series and normalized so that the leading term is cosnη, they are denoted by cen(η, q).
Similarly, solutions of Mathieu’s ODE that are odd inηwith leading term sinnη in a Fourier series are denoted by sen(η, q), and they can similarly be shown to obey the integral equation
sen(η, q)=sn(q) π
−π
sin
2i√qsinηsinθ
sen(θ, q) dθ. (14.104) We now come to the Fourier expansion for the angular Mathieu functions and start with
se1(η, q)=sinη+ ∞
ν=1
βν(q)sin(2ν+1)η, βν(q)= ∞
à=ν
βà(ν)qà (14.105) as a paradigm for the systematic construction of Mathieu functions of odd parity. Notice the key point that the coefficientβν of sin(2ν+1)η in the Fourier series depends on the parameterq and is expanded in a power series. Moreover, se1is normalized so that the coefficient of the leading term, sinη,is unity, that is, independent ofq.This feature will become important when se1is substituted into the angular Mathieu ODE to determine the eigenvalueλ(q).
The fact that theβν power series inqstarts with exponentνcan be proved by a simpler but similar series for se1(η, q):
se1(η, q)= ∞
ν=0
γν(q)sin2ν+1η, γν(q)= ∞
à=0
γà(ν)qà, (14.106) which is useful for this demonstration alone. However, since we need to expand
sin2ν+1η= ν m=0
Bνmsin(2m+1)η (14.107)
with Fourier coefficients Bνm= 1
π π
−π
sin2ν+1ηsin(2m+1)η dη=(−1)m 22ν
2ν+1 ν−m
(14.108) that we can look up in a table of integrals (see Gradshteyn and Ryzhik in the Additional Readings of Chapter 13), this proof gives us an opportunity to introduce theBνmthat are nonzero only ifm≤νand are important ingredients of the recursion relations for the lead- ing terms of se1(and all other Mathieu functions of odd parity). Substituting Eq. (14.107) into Eq. (14.106) we obtain
se1(η, q)= ∞
ν=0
ν m=0
Bνmγν(q)sin(2m+1)η. (14.109) Comparing this expression for se1with Eq. (14.105) we find
βν(q)= ∞ m=ν
Bmνγm(q). (14.110)
Here, the sum starts withm=νbecauseBmν=0 form < ν.
Next we substitute Eq. (14.106) into the integral Eq. (14.104) forn=1, where we insert the power series for sin(2i√qsinηsinθ ). This yields
se1(η, q) 2π s1(q) = 1
2π π
−π
sin(2i√qsinηsinθ )se1(θ, q) dθ
= 1 2π s1(q)
∞
m=0
γm(q)sin2m+1η
=i√q ∞
m,ν=0
qmγν(q)sin2m+1η 22m+1 (2m+1)!
1 2π
π
−π
sin2ν+2m+2θ dθ, (14.111) from which we obtain the recursion relations
γm(q)= 22m+1
(2m+1)!qmi√qs1(q) ∞
ν=0
γν(q) π
−π
sin2ν+2m+2θ dθ, (14.112) upon comparing coefficients of sin2m+1η. This shows that the power series forγm(q)starts withqm.Using Eq. (14.110) proves that the power series forβm(q)also starts withqm, and this confirms Eq. (14.105). The integral in Eq. (14.112) can be evaluated analytically and expressed via the beta function (Chapter 8) in terms of ratios of factorials, but we do not need this formula here.
Our next goal is to establish a recursion relation for the leading term βν(ν) of se1, in Eq. (14.105). We substitute Eq. (14.105) into the integral Eq. (14.104) forn=1, where we insert the power series for sin(2i√qsinηsinθ )again, along with the expansion
1 2π s1(q)=i
∞
m=0
αmqm+1/2. (14.113)
Here, the extra factor,i√q, cancels the corresponding factor from the sine in the integral equation. This yields
∞
ν=0
βà(λ)qà+νsin2ν+1η 22ν+1 (2ν+1)!
1 2π
π
−π
sin2ν+2λ+2θ dθ
= ∞
m,à=0
αmqm+àβà(ν)sin(2ν+1)η. (14.114) Here, we replace sin2ν+1ηby sin(2m+1)ηusing Eq. (14.107). Upon comparing the co- efficients ofqNsin(2ν+1)ηforN=à+νwe obtain the recursion relation
N ν=n
N−ν λ=0
βN(λ)−ν 22ν
(2ν+1)!BνnBνλ=
N−n m=0
αmβN(n)−m. (14.115)
Now we substitute Eq. (14.108) to obtain the main recursion relation for the leading coefficientsβν(ν)of se1:
N ν=n
N−ν λ=0
βN(λ)−ν 22ν(2ν+1)!
2ν+1 ν−n
2ν+1 ν−λ
=
N−n m=0
αmβN(n)−m. (14.116)
Example 14.7.1 LEADINGCOEFFICIENTS OFse1
We evaluate Eq. (14.116) starting withN=0, n=0.For this case we findβ0(0)=α0β0(0), orα0=1 because the coefficient of sinηin se1, β0(0)=1,by normalization. ForN=1, n=0 Eq. (14.116) yields
α0β1(0)+α1β0(0)=β1(0)+ 1 4ã3!
3 1
β0(0)
3 1
+β0(1) 3
0
, (14.117)
whereβ0(1)=0 andβ1(0)drops out, a general feature. Of course,β1(0)=0 because sinηin se1has coefficient unity. This yieldsα1=3/8.
The caseN=1,n=1 yields
−1 4ã3!
3 0
β0(0) 3
1
=α0β1(1), (14.118)
orβ1(1)= −1/8.The leading term is obtained from the general casen=N, (−1)N
22N(2N+1)!
2N+1 0
β0(0)
2N+1 N
=α0βN(N ), (14.119) as
βN(N )= (−1)N 22N(2N+1)!
2N+1 N
, (14.120)
which was first derived by Mathieu. ForN=1 this formula reproduces our earlier result,
β1(1)= −1/8.
In order to determine the first nonleading term βN(N )+1 of se1, Eq. (14.105), and the eigenvalueλ1(q)we substitute se1 into the angular Mathieu ODE, Eq. (13.181), using the trigonometric identities
2 cos 2ηsin(2ν+1)η=sin(2ν+3)η+sin(2ν−1)η and
d2sin(2ν+1)η
dη2 = −(2ν+1)2sin(2ν+1)η.
This yields 0=d2se1
dη2 +(λ1−2qcos 2η)se1=q(sinη−sin 3η)+λ1sinη−sinη +
∞
ν=1
λ1−(2ν+1)2
(−1)νqν
22νν!(ν+1)!+βν+1(ν) qν+1+ ã ã ã
sin(2ν+1)η
−q ∞
ν=1
(−q)ν
22νν!(ν+1)!+βν(ν)+1qν+1+ ã ã ã
sin(2ν+3)η+sin(2ν−1)η
=
λ1−1+q−q
− q
222!+β2(1)q2+ ã ã ã
sinη +sin 3η
−q−q q2
242!3!+β3(2)q3
+
λ1−32
− q
222!+β2(1)q2
+sin(2ν+1)η
λ1−(2ν+1)2
(−q)ν
22νν!(ν+1)!+βν(ν)+1qν+1
−qsin(2ν+1)η
(−q)ν+1
22(ν+1)(ν+1)!(ν+2)!+βν+2(ν+1)qν+2
−qsin(2ν+1)η
(−q)ν−1
22(ν−1)(ν−1)!ν!+βν(ν−1)qν
+ ã ã ã. (14.121) In this series the coefficient of each power ofq within different sine terms must vanish;
that of sinηbeing zero yields the eigenvalue λ1(q)=1−q−1
8q2+β2(1)q3+ ã ã ã, (14.122) withβ2(1)=1/26coming from the vanishing coefficient ofq2in sin 3η.Setting the coeffi- cient of(−q)ν in sin(2ν+1)ηequal to zero yields the identity
1−(2ν+1)2 1
22νν!(ν+1)!+ 1
22(ν−1)(ν−1)!ν!=0, (14.123) which verifies the correct determination of the leading terms βν(ν) in Eq. (14.120). The vanishing coefficient ofqν+1in sin(2ν+1)ηyields
(−1)ν+1 22νν!(ν+1)!+
1−(2ν+1)2 βν(ν)+1−βν(ν−1)=0, (14.124) which implies the main recursion relation for nonleading coefficients,
4ν(ν+1)βν+1(ν) = −βν(ν−1)+ (−1)ν+1
22νν!(ν+1)!, (14.125) for the first nonleading terms. We verify that
βν+1(ν) = (−1)ν+1ν
22ν+2(ν+1)!2 (14.126)
satisfies this recursion relation. Higher nonleading terms may be obtained by setting to zero the coefficient ofqν+2, etc. Altogether we have derived the Fourier series for
se1(η, q)=sinη+ ∞
ν=1
(−q)ν
22νν!(ν+1)!+ (−q)ν+1ν 22ν+2(ν+1)!2+ ã ã ã
ãsin(2ν+1)η.
(14.127) A similar treatment yields the Fourier series for se2n+1(η, q)and se2n(η, q).An invariance of Mathieu’s ODE leads to the symmetry relation
ce2n+1(η, q)=(−1)nse2n+1(η+π/2,−q), (14.128) which allows us to determine the ce2n+1 of period 2π from se2n+1. Similarly, ce2n(η+π/2,−q)=se2n(η, q)relates these Mathieu functions of periodπ to each other.
Finally, we briefly outline a derivation of the Fourier series for ce0(η, q)=1+
∞
n=1
βn(q)cos 2nη, βn(q)= ∞
m=n
βm(n)qm, (14.129) as a paradigm for the Mathieu functions of periodπ.Note that this normalization agrees with Whittaker and Watson and with Hochstadt in the Additional Readings of Chapter 13, whereas in AMS-55 (for the full reference see footnote 4 in Chapter 5) ce0differs by a factor of 1/√
2.The symmetry relation from the Mathieu ODE, ce0
π
2 −η,−q
=ce0(η, q), (14.130)
implies
βn(−q)=(−1)nβn(q); (14.131) that is,β2ncontains only even powers ofq andβ2n+1only odd powers.
The fact that the power series forβn(q)in Eq. (14.129) starts withqncan be proved by the similar expansion
ce0(η, q)= ∞
n=0
γn(q)cos2nη, γn(q)= ∞
à=0
γà(n)qà, (14.132) as for se1in Eqs. (14.105) to (14.112). Substituting Eq. (14.132) into the integral equation
ce0(η, q)=c0(q) π
−π
e2i√qcosθcosηce(θ, q) dθ, (14.133) inserting the power series for the exponential function (odd powers cos2m+1θ drop out) and equating the coefficients of cos2mηyields
γm(q)=c1(q)(−q)m 22m (2m)!
∞
à=0
γà(q) π
−π
cos2m+2àθ dθ. (14.134)
This recursion relation shows that the power series forγm(q)starts withqm.We expand cos2nη=
n m=0
Anmcos 2mη (14.135)
with Fourier coefficients Anm= 2
π π/2
−π/2
cos2nηcos 2mη dη= 1 22n−1
2n n−m
, (14.136)
which are nonzero only whenm≤n. Using this result to replace the cosine powers in Eq. (14.132) by cos 2mηwe obtain
βn(q)= ∞
m=n
Amnγm(q), (14.137)
confirming Eq. (14.129).
Proceeding as for se1 in Eqs. (14.113) to (14.120) we substitute Eq. (14.129) into the integral Eq. (14.133) and obtain
∞
m,à,ν,λ=0
(−1)m 22m (2m)!qà+m
m ν=0
Amνcos(2νη)Amλ
= ∞
m,à,ν=0
αmβà(ν)qm+àcos(2νη), (14.138) with
1 2π c1(q)=
∞
m=0
αmqm. (14.139)
Upon comparing the coefficient ofqNcos(2νη)withN=m+à,we extract the recursion relation for leading coefficientsβn(n)of ce0
N m=ν
(−1)m 22m (2m)!Amν
m λ=0
βN−m(λ) Amλ= N m=0
αmβN−m(ν) , (14.140) withAnmin Eq. (14.136).
Example 14.7.2 LEADINGCOEFFICIENTS FORce0 The caseN=0, ν=0 of Eq. (14.140) yields
A200β0(0)=α0β0(0), (14.141) withA00=1 andβ0(0)=1 from normalizing the leading term of ce0to unity so thatα0=1 results.
The caseN=1, ν=0 yields A00β1(0)A00−2A10
β0(0)A10+β0(1)A11 =α0β1(0)+α1β0(0), (14.142)
withβ0(1)=0 by Eq. (14.129). This simplifies to β1(0)−1
2β0(0)=α1β0(0)+β1(0), (14.143) whereβ1(0) drops out. We know already thatβ1(0)=0 from the leading term unity of ce0. Therefore,α1= −1/2.
For the caseN=1, ν=1 we obtain
−2A11β0(0)A10=α0β1(1), (14.144) withA10=1/2=A11,from whichβ1(1)= −1/2 follows. For the caseN=2, ν=2 we find
24
4!23β0(0)A20=α0β2(2), (14.145) withA20=3/8,from whichβ2(2)=2−4follows. The general caseN, ν=Nyields
(−1)N 22N
(2N )!AN Nβ0(0)AN0=α0βN(N ), (14.146) withAN N=2−2N+1, AN0=22N1−1
2N
N
, from which the leading term
βN(N )= (−1)N 22N−1(2N )!
2N N
= (−1)N
22N−1N!2 (14.147)
follows.
The nonleading termsβN(N )+1of ce0are best determined from the angular Mathieu ODE by substitution of Eq. (14.129), in analogy with se1,Eqs. (14.121) to (14.127). Using the identities
2 cos(2nη)cos 2η=cos(2n+2)η+cos(2n−2)η, d2
dη2cos(2nη)= −(2n)2cos(2nη), (14.148) we obtain
d2ce0
dη2 +
λ0(q)−2qcos 2η ce0=0
=λ0(q)−q
−q 2+ 7
27q3
+ ã ã ã +
∞
n=1
λ0−4n2
cos(2nη)
(−q)n
22n−1n!2+βn+2(n) qn+2
−q ∞
n=1
(−q)n
22n−1n!2+βn(n)+2qn+2
×
cos(2n+2)η+cos(2n−2)η . (14.149)
Setting the coefficient of cos(2nη)forn=0 to zero yields the eigenvalue λ0= −1
2q2+ 7
27q4+ ã ã ã. (14.150) The coefficient of cos(2nη)qnyields an identity,
(−1)n+1 4n2
22n−1n!2+ (−1)n
22n−3(n−1)!2=0, (14.151) which shows that the leading term in Eq. (14.147) was correctly determined. The coeffi- cient ofqn+2cos(2nη)yields the recursion relation
−4n2βn(n)+2−βn(n+−11)+(−1)n+1
22nn!2 + (−1)n
22n+1(n+1)!2 =0. (14.152) It is straightforward to check that
βn+2(n) =(−1)n+1 n(3n+4)
22n+3(n+1)!2 (14.153) satisfies this recursion relation. Altogether we have derived the formula
ce0(η, q)=1+cos 2η
−1 2q2+ 7
27q3+ ã ã ã
+cos 4η q2
25 + ã ã ã
+cos 6η
− q3 2732 + ã ã ã
=1+ ∞
n=1
cos(2nη)
(−q)n
22n−1n!2+(−1)n+1n(3n+4)qn+2 22n+3(n+1)!2 + ã ã ã
. (14.154) Similarly one can derive
ce1(η, q)=cosη+ ∞
n=1
cos(2n+1)η
(−q)n
22nn!(n+1)!− (−q)n+1n
22n+2(n+1)12+ ã ã ã
, (14.155) whose eigenvalue is given by the power series
λ1(q)=1+q−1 8q2− 1
26q3+ ã ã ã. (14.156)
FIGURE14.12 Angular Mathieu functions. (From Gutiérrez-Vega et al., Am. J. Phys. 71: 233 (2003).)
Exercises
14.7.1 Determine the nonleading coefficientsβn(n)+2for se1. Derive a suitable recursion relation.
14.7.2 Determine the nonleading coefficientsβn+4(n) for ce0. Derive the corresponding recursion relation.
14.7.3 Derive the formula for ce1, Eq. (14.155), and its eigenvalue, Eq. (14.156).
Additional Readings
Carslaw, H. S., Introduction to the Theory of Fourier’s Series and Integrals, 2nd ed. London: Macmillan (1921);
3rd ed., paperback, New York: Dover (1952). This is a detailed and classic work; includes a considerable discussion of Gibbs phenomenon in Chapter IX.
Hamming, R. W., Numerical Methods for Scientists and Engineers, 2nd ed. New York: McGraw-Hill (1973), reprinted Dover (1987). Chapter 33 provides an excellent description of the fast Fourier transform.
Jeffreys, H., and B. S. Jeffreys, Methods of Mathematical Physics, 3rd ed. Cambridge, UK: Cambridge University Press (1972).
Kufner, A., and J. Kadlec, Fourier Series. London: Iliffe (1971). This book is a clear account of Fourier series in the context of Hilbert space.
Lanczos, C., Applied Analysis, Englewood Cliffs, NJ: Prentice-Hall (1956), reprinted Dover (1988). The book gives a well-written presentation of the Lanczos convergence technique (which suppresses the Gibbs phenom- enon oscillations). This and several other topics are presented from the point of view of a mathematician who wants useful numerical results and not just abstract existence theorems.
Oberhettinger, F., Fourier Expansions, A Collection of Formulas. New York, Academic Press (1973).
Zygmund, A., Trigonometric Series. Cambridge, UK: Cambridge University Press (1988). The volume contains an extremely complete exposition, including relatively recent results in the realm of pure mathematics.
I NTEGRAL T RANSFORMS