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MATHM242 Lecture Notes - first half
Robert Bowles
January 3, 2007
Pierre-Simon Laplace 1749-1827
Adrien-Marie Legendre 1752-1833
Jean-Baptiste Fourier 1768-1830
Freidrich Wilhelm Bessel 1784-1846
Jacques Charles Francois Sturm 1803-1855
Joseph Liouville 1809-1882
Hermann Ludwig Ferdinand von Helmholtz 1821-1894
2
Contents
A Separation of Variables 6
A1 A problem from M241 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
A2 Separation of Variables – The Laplace Equation . . . . . . . . . . . . . . . . . . . . . 10
A2.1 Separation of Variables for Laplace Equation in Cartesian Coordinates . . . . 10
A3 An example—The influence of boundary conditions . . . . . . . . . . . . . . . . . . . 12
A4 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
A4.1 Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
A4.2 An alternative approach to Orthogonality . . . . . . . . . . . . . . . . . . . . 14
A4.3 Fourier Coefficients from a more general viewpoint. . . . . . . . . . . . . . . . 14
A4.4 Sturm-Liouville Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
A5 An example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
A6 The heat or diffusion equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
A6.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
A6.2 Example—A radiating rod and generalised Fourier Series . . . . . . . . . . . 18
A7 Laplaces Equation in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 20
A7.1 Changing from Cartesian to Polar Coordinates in 2-D . . . . . . . . . . . . . 20
A7.2 Solution of Laplaces equation in polar coordinates. . . . . . . . . . . . . . . . 21
A8 The Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
A8.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
A8.2 The wave equation in 1-D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
A8.3 Determination of wave frequency . . . . . . . . . . . . . . . . . . . . . . . . . 24
A8.4 Initial Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
A9 Higher Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
A9.1 The wave equation in 2 and 3-D. Helmholtz Equation. . . . . . . . . . . . . . 25
A9.2 Properties of solutions of Helmholtz equation with homogenous boundaryconditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
B Series solution of differential equations 27
B1 Regular singular points. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
B1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
B1.2 First-order equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
B1.3 Ordinary points and regular singular points . . . . . . . . . . . . . . . . . . . 29
B2 The Gamma function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
B3 Legendre’s equation—solution about an ordinary point. . . . . . . . . . . . . . . . . 33
B3.1 Legendre’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
B3.2 Separation of variables for Laplaces Equation in spherical polar coordinates. . 33
B3.3 Series Solution of Legendre’s equation. . . . . . . . . . . . . . . . . . . . . . . 34
3
B3.4 Legendre Polynomials Pn and Laplace’s equation . . . . . . . . . . . . . . . . 36B4 Solution about a regular singular point . . . . . . . . . . . . . . . . . . . . . . . . . . 37
B4.1 An example—the indicial equation . . . . . . . . . . . . . . . . . . . . . . . . 38B4.2 The failure of the method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39B4.3 A general approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40B4.4 An example—the roots of the indicial equation are equal. . . . . . . . . . . . 43
B5 Bessel’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45B5.1 The wave equation and Helmholtz equation in polar coordinates . . . . . . . 45B5.2 Series solution of Bessel’s equation . . . . . . . . . . . . . . . . . . . . . . . . 46B5.3 The functions J0 and Y0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47B5.4 The functions Jn and Yn. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
C Properties of Legendre Polynomials 51
C1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51C2 Rodrigue’s Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
C2.1 Its a polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52C2.2 It takes the value 1 at 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52C2.3 It satisfies the equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52C2.4 And that’s it . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
C3 Orthogonality of Legendre Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . 53C4 What is
∫ 1−1 P2
n dx? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53C5 Generalised Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54C6 A Generating Function for Legendre Polynomials . . . . . . . . . . . . . . . . . . . . 55
C6.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55C6.2 Derivation of the generating function. . . . . . . . . . . . . . . . . . . . . . . 55C6.3 Applications of the Generating Function. . . . . . . . . . . . . . . . . . . . . 56C6.4 Solution of Laplace’s equation. . . . . . . . . . . . . . . . . . . . . . . . . . . 57
C7 Example: Temperatures in a Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
D Oscillation of a circular membrane 61
D1 The Problem and the initial steps in its solution . . . . . . . . . . . . . . . . . . . . 61D1.1 The problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61D1.2 Separating out the time dependence . . . . . . . . . . . . . . . . . . . . . . . 61D1.3 Separating out the θ dependence . . . . . . . . . . . . . . . . . . . . . . . . . 62D1.4 On Bessel Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63D1.5 Determination of λ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65D1.6 Incorporating the Initial Conditions . . . . . . . . . . . . . . . . . . . . . . . 66D1.7 Orthogonality of Bessel Functions . . . . . . . . . . . . . . . . . . . . . . . . 67D1.8 The solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
4
Chapter A
Separation of Variables
A1 A problem from M241
In this section we look at a problem from MATHM241 involving the method of separation ofvariables and we approach it ”from first principles”.
Solve the wave equation
utt = c2uxx, −L ≤ x ≤ L, t ≥ 0, (A.1)
for u(x, t) with the initial condition ut(x, 0) = L2 − x2 and the boundary conditions u(−L, t) =u(L, t) = 0.
We try a solution of the type u(x, t) = X(x)T (t) and substitute to yield XT ′′ = c2X ′′T or
c2X ′′
X=T ′′
T.
Both sides of this expression must be a constant, independent of both x and t. Call this constantµ. Then
T ′′ = µT, (A.2)
and
X ′′ = (µ/c2)X. (A.3)
This second equation is forced by the boundary conditions X(−L) = X(L) = 0. This is anhomogeneous system and the only obvious solution is X = 0. However we have seen in algebracourses that linear systems of the type Ax = 0 have solutions other than x = 0 if A is singular.Here this will correspond to nonzero solutions for (A.3) for special values of µ - an infinite numberof special values in fact.
If µ > 0 then the solutions are X = A exp(√µx/c)+B exp(−√
µx/c) where A and B are chosenso that X(L) = X(−L) = 0. This turns out to be impossible for non-zero values of A and B so wediscount µ < 0.
If µ = 0 the the solution is X = Ax+B and again the boundary conditions give A = B = 0.
If µ < 0 on the other hand then the solutions for X have a different character - they areoscilliatory. It turns out that in this case the bondary conditions can be satisfied. The solutionsare X = A cos(
√−µx/c) + B sin(√−µx/c). Putting x = ±L, using the properties of the odd and
even sine and cosine functions and writing√−µ/c as γ gives the equations
0 = A cos γL+B sin γL, 0 = A cos γL−B sin γL
5
or (cos γL sin γLcos γL − sin γL
)(AB
)
=
(00
)
.
This has solution, for non-zero A and B if the determinant of the matrix is zero. We have(cos γL)(− sin γL)−(sin γL)(cos γL) = −2 sin 2γL = 0, giving 2γL = nπ, n = · · · ,−2,−1, 0, 1, 2, · · · .We discount n = 0 as this corresponds to µ = 0 which we have already discounted. For the momentlet us consider only n > 0. The values of γ = γn = nπ/2L that we have found by this mecha-nism are known as eigenvalues of the problem and the corresponding solutions for X are known aseigenfunctions or normal modes for the oscillation.
What are these eigenfunctions? For a particular value of n we need solutions for A and B to
(cos(nπ/2) sin(nπ/2)cos(nπ/2) − sin(nπ/2)
)(AB
)
=
(00
)
.
If n = 2m+ 1 is odd (n = 1, 3, 5, · · · , m = 0, 1, 2, · · · ) then this becomes
(0 (−1)m
0 −(−1)m
)(AB
)
=
(00
)
,
giving B = 0, A is undetermined and X = A cos γnx. These happen to be even solutions of thespatial problem (the problem for X). If n = 2m is even (n = 2, 4, 6, · · · ,m = 1, 2, 3, · · · ) then
((−1)m 0(−1)m 0
)(AB
)
=
(00
)
,
with solution A = 0 and B undetermined and X = B sin γnx. These are odd solutions of the spatialproblem. We now see why we need only take n and so γ positive, as a change in the sign of n (γ)leaves the eigenfunction either unchanged (for the cosine, the even solution) or simply changes itssign (for the sine, the odd solution). This change of sign can be easily reproduced by a change insign of B and the negative values of n, therefore, intoduce no new solutions.
-3 -2 -1 1 2 3
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0.25
0.5
0.75
1
-3 -2 -1 1 2 3
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0.5
1
-3 -2 -1 1 2 3
-1
-0.5
0.5
1
· · ·Even, m = 0, Odd m = 1, Even, m = 1, · · ·
γ = π/2L, X = A cos γx γ = π/L, X = A sin(γx) γ = 3π/2L, X = A cos(γx)
The time dependence of these solutions (temporal properties) is given by the solution of thetemporal equation T ′′ = µT or T ′′ + c2γ2
mT = 0 with solution
T (t) = A cos(γmct) +B sin(γmct).
The frequency of vibration of each normal mode is therefore ωm = cγm. A pure normal mode ofvibration is u(x, t) = X(x)T (t) or
Even, u(x, t) = cos(γmx)(Am cos(γmct) +B sin(γmct)), γm = (m+ 12)π/L, m = 0, 1, 2, · · ·
Odd, u(x, t) = sin(γmx)(Am cos(γmct) +B sin(γmct)), γm = mπ/L, m = 1, 2, 3, · · ·
6
The general solution is an arbitrary linear combination of all of these. This satisfies the differ-ential equation and the boundary conditions at x = ±L.
u(x, t) =∞∑
m=0
cos[(m+ 1
2)πx/L] (Am cos
[(m+ 1
2)πct/L]+Bm sin
[(m+ 1
2)πct/L])
+
∞∑
m=1
sin [mπx/L] (Cm cos [mπct/L] +Dm sin [mπct/L]) (A.4)
The constants in this solution are chosen so that the initial conditions are satisfied. They arecoefficients in a Fourier Series. The first, u(x, 0) = 0, gives
0 =
∞∑
m=0
cos[(m+ 1
2)πx/L](Am) +
∞∑
m=1
sin [mπx/L] (Dm) ,
leading to Am = Dm = 0. The second, ut(x, 0) = L2 − x2, gives
L2 − x2 =
∞∑
m=0
cos[(m+ 1
2)πx/L] (
(m+ 12)πcBm/L
)+
∞∑
m=1
sin [mπx/L] (mπcCm/L) .
We might expect Am = Dm = 0 and the time dependence of the solution to be like sin t as thisfunction is zero but has non-zero derivavtive at t = 0. Now L2 − x2 is even so will have an evenFourier series. Therefore Cm = 0. We need to find Bm, therefore, so that
L2 − x2 =
∞∑
m=0
cos[(m+ 1
2)πx/L] (
(m+ 12)πcBm/L
).
Multiply through by cos[(n + 12 )πx/L] (n ≥ 0) and integrate from −L to L to give
∫ L
−L(L2 − x2) cos
[(n+ 1
2)πx/L]
dx =
∞∑
m=0
Bm[c(m+ 12)π/L]
∫ L−L cos
[(m+ 1
2 )πx/L]cos[(n+ 1
2)πx/L]
dx. (A.5)
Using the formula 2 cosα cos β = cos(α+ β) + cos(α − β), makes the terms in the right hand sidesum
Bm[c(m+ 1
2)π/L]
2
∫ L
−Lcos [(n+m+ 1)πx/L] + cos [(n−m)πx/L] dx = Bmc(m+ 1
2)πδnm
as sinNπ = 0 for integer N and δnm = 0 if n 6= m and 1 if n = m. Hence in the sum only oneterm remains - that with m = n and, using the fact that the integrand is even to change the rangeof integration to [0, L],
Bn =2
cπ(n + 12)
∫ L
0(L2 − x2) cos
[(n+ 1
2)πx/L]
dx
=2
cπ(n + 12)
[
L2 − x2
[(n + 12)π/L]
sin[(n+ 12)πx/L]−
2x
[(n+ 12 )π/L]2
cos[(n+ 12)πx/L] +
2
[(n + 12)π/L]3
sin[(n+ 12 )πx/L]
]L
0
=2
cπ(n + 12)
2(−1)n
[(n + 12)π/L]3
=64(−1)nL3
π4c(2n + 1)4.
7
So we have
u(x, t) =64L3
π4c
∞∑
n=0
cos[(n+ 1
2)πxL
]sin[(n+ 1
2)πctL
]
(2n + 1)4.
An animation of this solution obtained with the Mathematica commands
u[x_,t_]:=(64/Pi^4) Sum[Cos[(n + 0.5)Pi x]Sin[(n + 0.5)Pit]/(2n+1)^4,{n, 0, 100}];
string:=Table[Plot[u[x,t],{x, -1,1},PlotRange->{-.8,.8}],{t, 0, 4Pi, 4Pi/40}];
Export["stringloop.gif", string, ConversionOptions -> {"Loop" -> True}];
is available at http://www.ucl.ac.uk/~ucahdrb/MATHM242/stringloop.gif.
A2 Separation of Variables – The Laplace Equation
A2.1 Separation of Variables for Laplace Equation in Cartesian Coordinates
A2.1a In 3D
Consider ∇2φ = 0 in three dimensions
φxx + φyy + φzz = 0. (A.6)
We look for special solutions of the type φ(x, y, z) = X(x)Y (y)Z(z). Substitution gives
X ′′Y Z +XY ′′Z +XY Z ′′ = 0
and dividing by XY Z,X ′′
X︸︷︷︸
1
+Y ′′
Y+Z ′′
Z︸ ︷︷ ︸
2
= 0.
If we vary x 1 might vary, but 2 cannot. Hence to maintain equality 1 must be a constant, asay. Similarly varying y, we see Y ′′/Y = b say and hence Z ′′/Z = −(a+ b). Thus
X ′′ − aX = 0, Y ′′ − bY = 0, Z ′′ + (a+ b)Z = 0.
a and b are called separation constants. Note the signs. The solutions are
X(x) = exp(±√a x), Y (y) = exp(±
√b y), Z(z) = exp(±i
√a+ b z)
and
φab︸︷︷︸
solution for any a and any b
= exp(±√a x) exp(±
√b y) exp(±i
√a+ b z)
is a solution, where φab represents eight solutions in all, one for each choice of possible ± signs.The general solution is
φ(x, y, z) =∑
a,b
φab =∑
a,b
Ca,b exp(±√a x) exp(±
√b y) exp(±i
√a+ b z),
where again, each term in the sum represents eight contributions and each Ca,b represents eightcorresponding constants.
8
A2.1b In 2D
The 2-D problem can be solved by putting a = −b, for then φab is independent of z. In this caselet us write a = −b = m2 so that
φab = exp(±mx) exp(±imy) = φm
where φm represents four possible choices of signs. We can combine these exponentials to givealternative φm:
φm = Am coshmx cosmy+Bm coshmx sinmy+Cm sinhmx cosmy+Dm sinhmx sinmy. (A.7)
These are exponential in the x-direction and sinusoidal in the y-direction. Alternatively, choosinga = −m2, b = m2 or im instead of m gives a solution with these behaviours reversed:
φm = Am coshmy cosmx+Bm coshmy sinmx+Cm sinhmy cosmx+Dm sinhmy sinmx. (A.8)
A2.1c Zero separation constant
We have seen above what happens for positive and negative m2. The case m = 0 is also valid andwe must look at this also. We have a = b = 0 so that X satisfies X ′′ = 0 and Y satisfies a similarequation. So
φ0 = (Ax+B)(Cy +D).
This may also be obtained from A.8 by letting m → 0 and writing Am = BD, Bm = AD/m,Cm = CB/m and Dm = AC/m2
Exercise: Derive the 2-D solution using the argument in A2.1a
A3 An example—The influence of boundary conditions
y
x
l
h
1
2
3
4
Solve ∇2u = 0 in the rectangle 0 ≤ x ≤ L,0 ≤ y ≤ h, with boundary conditions 1 :
u(0, y) = 0, 2 : u(x, h) = K sin(πx/L), 3 :
u(L, y) = 0, 4 : u(x, 0) = K sin(πx/L).These boundary conditions enter into thechoice of allowable separation constants asfollows:
1. 2 and 4 suggest that the solution has oscilliatory behaviour in x so we take solutions oftype (A.8).
2. We must not forget the case m = 0 but solutions with the x-variation Ax+B are precludedby 1 and 3 .
3. 1 forces us to choose Am = Cm = 0.
4. We ensure that 3 is also satisfied by making the choice of m such that sinmL = 0 implyingm = nπ/L for integer n. The value of n must be non-zero as we have dealt with the casem = 0 already and we may as well make it postive since as sin is odd it corresponds merelyto a change in sign of Bm and Dm.
9
5. Note now that the boundary conditions in x have fixed the allowed separation constants andwe have
um = sin(πx/L)(Bm cosh(πy/L) +Dm sinh(πy/L)).
We go on to choose Bm and Dm to satisfy 2 and 4 .Note that in this simple example thesolution is only going to consist of terms with n = 1. In more complicated examples we woulduse Fourier series to represent the boundary conditions and would have to include all valuesof n.
6. We find B1 = K and B1 cosh(πh/L) +D1 sinh(πh/L) = K and finally
u(x, y) = K sin(πx/L) (cosh(πy/L) − tanh(πh/2L) sinh(πy/L)) .
Exercise: Show u(L/2, h/2) = K sech(πh/2L).
A4 Fourier Series
A4.1 Orthogonality
A4.1a Trigonometric Formulae
We have the following trigonometric formulae
cos a cos b = 12 (cos(a− b) + cos(a+ b)) ,
sin a sin b = 12 (cos(a− b) − cos(a+ b)) ,
sin a cos b = 12 (sin(a+ b) − sin(a− b)) .
A4.1b Integrals
We may use the formulae to show that∫ L
−Lcos(mπx/L) cos(nπx/L) dx =
∫ L
−Lsin(mπx/L) sin(nπx/L) dx = Lδmn,
∫ L
−Lsin(mπx/L) cos(nπx/L) dx = 0.
We say that the set of functions {sin(nπx/L), cos(nπx/L)} , n = 0, 1, 2, . . . are orthogonal with
respect to the inner product (f, g) =∫ L−L f(x)g(x) dx.
A4.1c Fourier Coefficients
The results above imply that if we can write
f(x) =a0
2+
∞∑
n=1
an cos(nπx/L) + bn sin(nπx/L)
then on multiplying through, for example, by cos(mπx/L) and integrating over the interval [−L,L],only one term survives on the right hand side and the result is Lam. Similarly bm can be foundand we have
an =1
L
∫ L
−Lf(x) cos(nπx/L) dx, bn =
1
L
∫ L
−Lf(x) sin(nπx/L) dx. (A.9)
It is in fact possible to show that you can in fact write f(x) in this way.
10
A4.1d Graphs
-3 -2 -1 1 2 3
-1
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0.5
1
-3 -2 -1 1 2 3
-1
-0.5
0.5
1
Graphs of sinnx cosnx for n = 0, 1, 2, 3, 4, 5. Note how one set is odd and the other even and howtheir rate of oscilation increases with n. Large values of n represent the rapidly changing componentsof f(x) in its Fourier series. Such components can be expected if f(x) has a discontinuity.
A4.2 An alternative approach to Orthogonality
A4.3 Fourier Coefficients from a more general viewpoint.
Consider the problemy′′ + λy = 0, y(−L) = y(L) = 0.
One solution is simply y = 0 but there are other non-zero solutions if n takes on certain values.These are known as eigenvalues and determining these is part of the problem. It is an eigenvalue
problem. We have y(x) = A sin(√λx) + B cos(
√λx). Applying the boundary conditions, using
the facts that sin, cos are odd and even respectively, we find 0 = A cos√λL − B sin
√λL, 0 =
A cos√λL + B sin
√λL. These imply that A cos
√λL = 0 and B sin
√λ = 0 and we cannot allow
both A and B to be zero as this would give the trivial zero solution.. If we choose A = 0 and B nonzero (the odd solution) then sin
√λL = 0 so that λ = n2π2/L2. If we choose B = 0 and A non zero
(the even solution), then λ must take values satisfying cos√λL = 0, so that λ = (n+ 1
2 )2π2/L2.Now denote the solution obtained with a choice of eigenvalue λ = λn as yn and ym a solution
obtained with λ = λm. Then
y′′n + λny = 0, yn(−L) = yn(L) = 0, (A.10)
y′′m + λym = 0, ym(−L) = ym(L) = 0. (A.11)
Take yn times (A.11) - ym times (A.10) to obtain
(yny′′
m − ymy′′
n) + (λm − λn)ynym = 0
and integrating and using integration by parts,
[yny
′
m − ymy′
n
]L
−L−∫ L
−Ly′ny
′
m − y′my′
n dx+ (λm − λn)
∫ L
−Lynym dx = 0
so that[yny
′
m − ymy′
n
]L
−L+ (λm − λn)
∫ L
−Lynym dx = 0.
The boundary conditions satisfied at x = ±L are such that this becomes
(λm − λn)
∫ L
−Lynym dx = 0,
11
so if the eigenvalues are distinct so that λm 6= λn we have to conclude that∫ L
−Lynym dx = 0
and that the eigenfunctions yn and ym are orthogonal with respect to the inner product defined by(f, g) =
∫ L−L f(x)g(x) dx.
A4.4 Sturm-Liouville Theory
The above approach can be generalised to obtain the eqivalent of the Spectral Theorem in linearalgebra. In this context it is referred to as Sturm-Liouville theory. Consider the differential equation
d
dx
[
r(x)dy
dx
]
+ [q(x) + λp(x)] y = 0 (A.12)
on [x1, x2] witha1y(x1) + b1y
′(x1) = 0, a2y(x2) + b2y′(x2) = 0, (A.13)
with a1,b1 nor a2,b2 not both zero. We will also add the constraints that p(x) and r(x) are continuousand postive on [x1, x2] and that q(x) is continuous on (x1, x2). If now λ1, λ2, · · · are distinct valuesof λ for which (A.12) and (A.13) have non-zero solutions y1, y2, · · · , then
1. {λn} is an infinite unbounded set.
2. As above
(λm − λn)
∫ x2
x1
p(x)ym(x)yn(x) dx =[r(x)(ymy
′
n − y′myn)]x2
x1,
and the boundary conditions (A.13) always ensure that
(λm − λn)
∫ x2
x1
p(x)ym(x)yn(x) dx = 0
so that if the eigenvalues are distinct then∫ x2
x1
p(x)ym(x)yn(x) dx = 0 (A.14)
and the eigenfunctions are orthogonal with respect to the weight function p(x) or, in otherwords, with the inner product defined as (f, g) =
∫ x2
x1p(x)f(x)g(x) dx.
3. In the cases where r(x) = 0 at an endpoint of the interval the boundary conditions there maybe relaxed.
4. The result will also be valid if the boundary conditions are replaced by the insistence that yis periodic outside [x1, x2] so that y(x1) = y(x2), y
′(x1) = y′(x2).
5. The set of functions {yn} form a basis for functions on [x1, x2]. In other words we can find ageneralised Fourier Series:
f(x) =
∞∑
1
anyn (A.15)
and it converges!
6. Since the functions yn are orthogonal the coefficients in this series are given by
an =
∫ x2
x1p(x)yn(x)f(x) dx∫ x2
x1p(x)y2
n(x) dx=
(f, yn)
(yn, yn). (A.16)
12
A5 An example
y
x
l
1
2
3
4 Solve ∇2u = 0 in the semi-infinite rectangle0 ≤ x ≤ L, y ≥ 0, with boundary condi-tions 1 : ux(0, y) = 0, 2 : u(L, y) = 0, 3 :
u(x, 0) = f(x), 4 : u(x, y) → 0 as y → ∞.We proceed by separation of variables andmake the following decisions
1. We want the exponential behaviour to be in y so that we can impose 4 . This also allows usto use Fourier Series to represent f(x). We discount the solutions exponentially growing in y.We also discount the solution arising from the zero separation constant which cannot satisfyboth 1 and 2 . So we consider
u(x, y) =∑
m
exp(−my)(Am cos(mx) +Bm sin(mx)). (A.17)
2. 1 suggests that we only consider the cosine dependence in x. The parts of (A.24) dependingon sin(mx) cannot satisfy this condition.
u(x, y) =∑
m
exp(−my)Am cos(mx). (A.18)
3. To satisfy 2 , we must have cos(mL) = 0 implying mL = (n + 12)π. As before we need only
consider m > 0. Therefore
u(x, y) =∞∑
n=0
An exp(−(n+ 12)πy/L) cos((n + 1
2)πx/L). (A.19)
4. To find the An we put y = 0 and hope to write
f(x) =∞∑
n=0
An cos((n + 12)πx/L) (A.20)
5. If the separated from of solution was X(x)Y (y) then it is important to realise that the problemof determining X(x) satisfying 1 and 2 is a Sturm-Liouville eigenvalue problem namely
X ′′ +m2X = 0, X ′(0) = X(L) = 0.
Hence its solutions, the cos((n + 12)πx/L), n = 0, 1, 2, · · · will form an orthogonal basis and
we can find An so that (A.20) is possible and converges. In particular we know
∫ L
0cos((n + 1
2)πx/L) cos((m+ 12)πx/L) dx = 0
if n 6= m1 and so also that
An =
∫ L0 f(x) cos((n+ 1
2)πx/L) dx∫ L0 cos2((n+ 1
2)πx/L) dx=
2
L
∫ L
0f(x) cos((n + 1
2)πx/L) dx. (A.21)
1to check use the trigonometric formulae, but we know its true!
13
Exercise: Show that if f(x) = 0 for ǫ < x ≤ L but f(x) = 1/ǫ for 0 ≤ x ≤ ǫ then
u(x, y) =
∞∑
n=0
sin rnrn
exp(−rny/ǫ) cos(rnx/ǫ), rn = ǫ(n+ 12)π/L.
A6 The heat or diffusion equation
A6.1 Definition
The heat or diffusion equation is
∇2u = a2ut (A.22)
where t represents time and a2 is a typically constant. In one dimension it reduces to
uxx = a2ut. (A.23)
A6.2 Example—A radiating rod and generalised Fourier Series
A thin rod of length l has its lateral edges insulated. Its left end is kept at a temperature ofzero whilst its right radiates heat freely into air of temperature zero. Initially the rod has a giventemperature distribution and we need to find the temperature distribution in the rod at any latertime.
This reduces to solving
uxx = a2ut, 0 ≤ x ≤ l, t ≥ 0 (A.24)
with boundary and initial conditions
1 : u(0, t) = 0, 2 : hu(l, t) + ux(l, t) = 0, 3 : u(x, 0) = f(x),
where h is a positive constant, the so-called surface conductivity.
1. We start by looking for a solution u(x, t) = X(x)T (t) and find
X ′′
X= a2T
′
T= µ,
where µ is the separation constant.
2. The equation for T is that of exponential growth or decay and as we expect the temperatureto decay rather than grow we must reject values with µ > 0.
3. The possibility of zero separation constant leads to X ′′ = 0 and T ′ = 0 so that X = Ax+B,T = C so that u = XT = Ax + B, redefining A and B. 1 implies B = 0 and then 2requires hAl +A = 0 and we deduce A = 0 as h and l are positive. We therefore discard thezero separation constant.
4. We deduce that µ is negative and write it as µ = −λ2 with λ > 0. Thus
T ′ = −λ2T/a2, =⇒ T = C exp(−λ2t/a2).
X ′′ + λ2X = 0, X(0) = 0, hX(l) +X ′(l) = 0. (A.25)
14
5. We see that all solutions decay in time at a rate proportional to λ2 but that the values of λare determined by the eigenvalues of the Sturm-Liouville problem (A.25). The solution is
X(x) = B sin(λx),
having used 1 to discount the cosine dependence and where 2 fixes λ to be solutions of
λ cos λl + h sin λl = 0. (A.26)
6.2 4 6 8 10 12
-10
-7.5
-5
-2.5
2.5
5
7.5
10
If we write λ = z/l and hl = 1/α, then the allowedvalues of z satisfy
tan z = −αz
which graphical considerations show has an infi-nite number of solutions, zn, say as expected. Wewrite λn = zn/l and deduce
u(x, t) =
∞∑
1
Bn exp(−λ2nt/a
2) sin λnx.
7. Finally we need to choose Bn so that 3 is satisfied and that
f(x) =
∞∑
1
Bn sinλnx.
This is not a Fourier series representation but rather a generalised Fourier series. Howeverwe know we are attempting to write f(x) in terms of an orthogonal basis so that
Bn =
∫ l0 f(x) sinλnx dx∫ l0 sin2 λnx dx
.
A7 Laplaces Equation in Polar Coordinates
A7.1 Changing from Cartesian to Polar Coordinates in 2-D
The ∇2 operator in Cartesian coordinates has the form
∇2 =∂2
∂x2+
∂2
∂y2
We ask what the from is in polar coordinates with x = r cos θ and y = r sin θ. To find out we usethe chain rule. This gives use the results
∂
∂x=∂r
∂x
∂
∂r+∂θ
∂x
∂
∂θ,
∂
∂y=∂r
∂r
∂
∂r+∂θ
∂y
∂
∂θ.
15
To work out these partial derivatives we need r = r(x, y) and θ = θ(x, y). These functions are
r =√
x2 + y2,
θ = tan−1(y/x)
so that∂r
∂x=
x√
x2 + y2= cos θ,
∂r
∂r=
y√
x2 + y2= sin θ,
∂θ
∂x= − y
x2 + y2= −sin θ
r,
∂θ
∂y=
x
x2 + y2=
cos θ
r.
These imply
∂2
∂x2+
∂2
∂y2=
(
cos θ∂
∂r− sin θ
r
∂
∂θ
)(
cos θ∂
∂r− sin θ
r
∂
∂θ
)
+
(
sin θ∂
∂r+
cos θ
r
∂
∂θ
)(
sin θ∂
∂r+
cos θ
r
∂
∂θ
)
= cos θ
{
cos θ∂2
∂r2+
sin θ
r2∂
∂θ− sin θ
r
∂2
∂r∂θ
}
−
sin θ
r
{
− sin θ∂
∂r+ cos θ
∂2
∂r∂θ− cos θ
r
∂
∂θ− sin θ
r
∂2
∂θ2
}
+
sin θ
{
sin θ∂2
∂r2− cos θ
r2∂
∂θ+
cos θ
r
∂2
∂r∂θ
}
+
cos θ
r
{
cos θ∂
∂r+ sin θ
∂2
∂r∂θ− sin θ
r
∂
∂θ+
cos θ
r
∂2
∂θ2
}
=(cos2 θ + sin2 θ)
(∂2
∂r2+
1
r
∂
∂r+
1
r2∂2
∂θ2
)
=∂2
∂r2+
1
r
∂
∂r+
1
r2∂2
∂θ2.
So Laplaces equation for u(r, θ) in polar coordinates reads
urr +1
rur +
1
r2uθθ = 0 (A.27)
A7.2 Solution of Laplaces equation in polar coordinates.
A7.2a Separation of Variables
We can solve (A.27) by separation of variables. Write
u(r, θ) = R(r)Θ(θ),
16
substitute in, multiply by r2 and divide by RΘ to get
r2R′′
R+ r
R′
R+
Θ′′
Θ= 0.
ThusΘ′′
Θ= constant = −λ2
say. The choice of this form of separation constant simplifies matters later but for the present λcan be any complex number. Thus
Θ′′ + λ2Θ = 0, (A.28)
r2R′′ + rR′ − λ2R = 0. (A.29)
We want solutions of (A.28) that are periodic in θ, ie Θ(θ + 2π) = Θ(θ). Thus
λ = n > 0,
Θ(θ) = An cosnθ +Bn sinnθ
for any constants An and Bn. Alternatively if λ = 0 we have the solution of (A.28)
Θ(θ) = A0 +B0θ
but periodicity in θ implies B0 = 0.
Note that from our knowledge of Sturm-Liouville systems we know that the set {1, cosnθ, sinnθ}form an orthogonal basis for the set of periodic functions on [0, 2π].
For the radial function R(r) satisfying (A.29) with λ = n try as a solution R = rα and substituteto find
α(α− 1) + α− n2 = 0
So that α = ±n and we have two solutions rn and r−n as long as n 6= 0. If n = 0 then (A.29)becomes
R′′ +1
rR′ = 0
with solution
R = C +D ln r
the most general solution will be a sum of the solutions found for all separation constants. So
u(r, θ) = A0 +D0 ln r︸ ︷︷ ︸
from n = 0 rewriting C and D
+
∞∑
n=1
(
rn +Dn
rn
)
(An cosnθ +Bn sinnθ)
︸ ︷︷ ︸
This is only one way of writing this
.
A7.2b Axisymmetric Solutions
Solutions to ∇2u = 0 which have uθ = 0 are known as axisymmetric. In polar coordinates thisis u = A ln r + B, arising from the zero separation constant, It may also be obtained by directintegration of urr + ur/r = 0 which may be rewritten as (rur)r = 0.
17
A7.2c An example
If we are interested only in the interior of the circle r = 1 with the solution regular at r = 0 andu(1, θ) = h(θ), then we must discount the ln r and r−n solutions. Therefore
u = A0 +
∞∑
n=1
rn (An cosnθ +Bn sinnθ)
and evaluating the solution at r = 1,
h(θ) = A0 +∞∑
n=1
(An cosnθ +Bn sinnθ)
so that
A0 =1
2π
∫ 2π
0h(φ) dφ
{An, Bn} =1
π
∫ 2π
0h(φ){cos nφ, sinnφ} dφ.
The 2π rather than π in the definition of A0 arises from the a0/2 in the result (A.9). Note that A0
is the mean part of h.
A8 The Wave Equation
A8.1 Definition
The wave equation isc2∇2ψ = ψtt (A.30)
for ψ(x, t), where c corresponds to a wave speed. We are often interested in solutions that areperiodic in time.
A8.2 The wave equation in 1-D
In one dimension (A.30) reduces toc2ψxx = ψtt. (A.31)
It may be solved using separation of variables. Writing ψ(x, t) = X(x)T (t) we find
X ′′
X=
1
c2T ′′
T= µ,
where µ is a separation constant. In many problems we can discount the possibility that µ > 0as this leads to solutions that are exponetially growing in both x and t and have no physicalsignificance. The case µ = 0 leads to the solution ψ = (Cx+D)(At+B) which does not describeperiodic motion so that often this can be discounted also. Finally we face the case µ < 0 and writeµ = −λ, λ > 0. The solutions are straightforward:
X(x) = A cos(√λ x) +B sin(
√λ x). (A.32)
T (t) = C cos(c√λ t) +D sin(c
√λ t). (A.33)
These are periodic in x and t. The wavelength of the solution is 2π/√λ while
√λ is referred to as
the wave number. The period of the solution is 2π/c√λ and c
√λ is the frequency. Note that the
wave speed is frequency divided by wave number or the wavelength divided by the period.
18
A8.3 Determination of wave frequency
Often the wave speed c is given and it is required to find the possible frequencies and wavelengthsof possible waves. For example the waves may be on a string stretched between x = 0 and x = l.This translates into boundary conditions ψ(0, t) = ψ(l, t) = 0. These boundary conditions are tobe applied to (A.32) which then becomes a Sturm-Liouville problem for the eigenvalues λn. Aknowledge of these gives the allowable frequencies with which the string can oscillate.
Exercise: Find the frequencies of oscillation of the system satisfying ψxx = ψtt, ψ(0, t) =ψ(1, t) = 0
A8.4 Initial Conditions
General initial conditions can be imposed by expressing them as Fourier Series. Again we areguaranteed that we can write arbitrary initial conditions in terms of the eigenfunctions that ariseas we find the eigenvalues corresponding to possible frequencies.
A9 Higher Dimensions
A9.1 The wave equation in 2 and 3-D. Helmholtz Equation.
We can treat (A.30) by the method of separation of variables, writing ψ(x, t) = u(x)T (t), substutingand dividing by c2XT to find
∇2u
u=
T ′′
c2T= −λ,
say, where as above we can restrict our attention here to the case λ > 0. The solution for thetemporal dependence of the solution is given by (A.33). The spatial dependence and also thevalues of λ are determined by
∇2u+ λu = 0 (A.34)
and the spatial boundary conditions. This is the Helmholtz Equation. It is the two- or three-dimensional eqivalent of u′′ + λu = 0. Its solution together with boundary conditions forms aneigenvalue problem for λ.
A9.2 Properties of solutions of Helmholtz equation with homogenous boundary
conditions
1. Generally in any particular geometry (within a rectangle or disk or sphere for relatively simpleexamples) and with homogenous boundary conditions, there will be non-zero solutions onlyfor given values of λ and as in the case of the Sturm-Liouville system, there will be an infinitenumber of these λn say.
2. Associated with each of these eigenvalues is an eigenfunction un say satisfying
∇2un + λnun = 0
and appropriate boundary conditions.
3. Eigenfunctions arising from distinct eigenvalues are orthogonal. Consider the case where(A.34) is to be solved within a region D (for example the interior of a sphere) with boundaryconditions on the boundary of D, denoted as ∂D of the type
an · ∇u+ bu = 0, on ∂D, (A.35)
19
relating u and its normal derivative on the boundary. Then if un and um are solutionscorresponding to distinct eigenvalues then
∫
D
un(x)um(x) dx = 0 (A.36)
where the integral is a multiple integral over D.
This is easily proved. Proceeding as in the one-dimensional case leads us to
∫
D
um∇2un − un∇2um dx + (λn − λm)
∫
D
umun dx = 0.
However the following formula may be used to rewrite the first integral.
∇ · (f(∇g) − g(∇f)) = f∇2g − g∇2f
and use of the divergence theorem
∫
D
∇ · F dx =
∫
∂DF · n dx,
leads to ∫
D
(um∇un − un∇um) · n dx + (λn − λm)
∫
D
umun dx = 0.
The boundary conditions (A.35) ensure that the first integral is in fact zero and we have ourresult.
Helmholtz equation also appears in the solution of the heat equation in 2 and 3D by separation ofvariables. In our example of the wave equation, once the eigensolutions are found we have
ψ(x, t) =∑
n
(An cos(c√λ t) +Bn sin(c
√λ t))un(x).
As in one dimension the initial conditions can be satisfied by expressing them in a generalisedFourier series. Consider the case where ψ(x, 0) = f(x), ψt(x, 0) = 0. The second of these equationsimplies that Bn = 0 whilst the first will be satisfied if we choose
An =
∫
Df(x)un(x)dx∫
Du2
n(x)dx.
20
Chapter B
Series solution of differential
equations
B1 Regular singular points.
B1.1 Introduction
We consider the second-order equation
d2w
dz2+ p(z)
dw
dz+ q(z)w = 0 (B.1)
for w(z) and ask when we can find solutions of the type
w(z) =
∞∑
k=0
akzk, (B.2)
or, more generally,
w(z) =
∞∑
k=0
akzk+σ, a0 6= 0, (B.3)
where σ is not necessarily an integer. The solution (B.2) assumes the solution is analytic, whilst(B.3) allows for poles in w(z) and, if σ is not an integer, branch cuts.
B1.2 First-order equations
We look to the first order equation for guidance. Consider
dw
dz+ p(z)w = 0,
where we assume p(z) has a Laurent series expansion
p(z) = · · · + c−1
z+ c0 + c1z + · · · .
The integrating factor method shows that
w(z) = A exp
(
−∫ z
p(ξ) dξ
)
,
21
so we need consider
−∫ z
p(ξ) dξ = · · · + c−2
z− c−1 ln z + C
︸︷︷︸
const of integration—set to 0
−c0z + · · · .
Consider the three cases:
1. If p(z) is analytic at z = 0, then ci = 0, i < 0 and then w(z) is analytic and solutions of type(B.2) can be found.
2. If p(z) has a simple pole at z = 0 then c−1 6= 0 but ci = 0, i < −1 and
w(z) = A exp(−c−1 ln z) exp(−c0z − · · · ) =1
zc−1× An analytic function.
The solution need not be analytic at z = 0 but we can look for a generalised solution of type(B.3).
3. If the singularity in p(z) is more severe then c−k 6= 0 for some k > 1 and w(z) has a factorexp(− c−k
1−kz1−k), (like exp(−1/z) for example) which has an essential singularity at z = 0 and
we cannot express the solution as a power series.
So for first-order equations we conclude
1. If the coefficient p(z) is analytic at a point then the solution is and a power series solution ispossible.
2. If p(z) is singular, but only as singular as a simple pole, then the solution is not analytic, buta generalised power series solution is possible.
3. If p(z) is any more singular at a point, then the differential equation has solutions that arenot anyalytic at that point.
B1.3 Ordinary points and regular singular points
A similar analysis in the case of second-order equations
d2w
dz2+ p(z)
dw
dz+ q(z)w = 0
leads us to make the following definitions
1. The point z = z0 is an ordinary point of the o.d.e. if p(z) and q(z) are analytic at z = z0.The solution is then analytic at z = z0 and a solution of type (B.2) can be found.
2. If either p(z)or q(z) are singular at z = z0, but (z − z0)p(z) and (z − z0)2q(z) are analytic at
z = z0 then z0 is a regular singular point of the o.d.e. and solutions of type (B.3) are possible.
22
B2 The Gamma function
1. The Gamma function, Γ(x) is defined in terms of the integral
Γ(x) =
∫∞
0tx−1 exp(−t) dt.
Note that this integral only converges, for x > 0 (strictly, Re(x) > 0). This does not meanthat the Gamma function exists only for Re(x) > 0, merely that it cannot be represented bythis integral for Re(x) > 0.
2. It satisfies the recursion relationΓ(x+ 1) = xΓ(x), (B.4)
since
Γ(x+ 1) =
∫∞
0tx exp(−t) dt = [−tx exp(−t)]∞0 +
∫∞
0xtx−1 exp(−t) dt
= xΓ(x), (x > 0).
3. In addition Γ(1) = 1 as∫∞
0 exp(−t) dt = 1 .This means that
Γ(2) = 1Γ(1) = 1.1 = 1
Γ(3) = 2Γ(2) = 2.1 = 2
Γ(4) = 3Γ(3) = 3.2 = 6
...
Γ(n+ 1) = nΓ(n) = n! Γ(n) = (n− 1)! (B.5)
The Gamma function extends the factorial function to non-integer arguments.
4. Rearranging, we have
Γ(x) =Γ(x+ 1)
x, (B.6)
so that, as x→ 0,
Γ(x) ∼ 1
x. (B.7)
It also allows us to extend the Gamma function, and hence the factorial function, to negativearguments - but there is a problem at zero and at negative integers. If a negative number isx = −n+ e, for n > 0, 0 < e < 1, then
Γ(x) = Γ(−n+ e) =Γ(−n+ 1 + e)
(−n+ e)
=Γ(−n+ 2 + e)
(−n+ e)(−n+ 1 + e)
=Γ(−n+ 3 + e)
(−n+ e)(−n+ 1 + e)(−n+ 2 + e)
...
=Γ(1 + e)
(−n+ e)(−n+ 1 + e)(−n+ 2 + e) · · · (e− 1)(e)︸ ︷︷ ︸
n of these
.
23
Thus Γ(−n+ e) can be calculated in terms of Γ(1+ e) for which the integral definition can beused. This is an example of ”analytic continuation” of a function defined in one region of thecomplex plane into a different region where the originbal definition does not apply. We shallsee others later. Note that if we let e→ 0 in the above, corresponding to attempt to evaluateΓ(−n), then the (e) in the denominator gives rise to a singularity. The Gamma function hasa simple pole at x = −n with residue (−1)n/n!.
5. A graph of the Gamma function for real argument. Note the singularities at x = 0 at at thenegative integers.
-4 -2 2 4
-20
-10
10
20
Plot[Gamma[x], {x, -4.5, 4.5}, PlotRange -> {-20, 25},PlotPoints -> 200]
6. One aspect of Γ in the complex plane.
-4-2
02
4ReHzL -5
-2.5
02.55
ImHzL-505
10ReHGHzLL
-4-2
02
4ReHzL
Plot3D[Re[Gamma[x + I y]], {x, -4, 4}, {y, -6, 6},PlotPoints -> {100,
100},PlotRange -> {-5, 10}, Lighting -> False, Mesh -> False,FaceGrids -> All,
AxesLabel -> {"Re(z)", "Im(z)", "Re(Gamma(z))"}]
7. What is (1/2)!? This corresponds to Γ(1 + 1/2) = 1/2Γ(1/2).
Γ(1/2) =
∫∞
0t−1/2 exp(−t) dt =
∫∞
0u−1 exp(−u2)2u du = 2(
√π/2)
=√π,
24
so (1/2)! =√π/2. Show that (−1/2)! = 2
√π.
8. The derivative of the Gamma function may be calculated, although we do not show how todo so here. It turns out that, for integers,
Γ′(n + 1)
Γ(n+ 1)= −γ +
n∑
s=1
1
s. (B.8)
Here γ is a constant, known as Eulers constant.
γ = limn→∞
[n∑
k=1
1
k− lnn
]
=
[n∑
k=1
1
k−∫
∞
1
1
kdk
]
≈ 0.577216. (B.9)
1 2 3 4 5
0.25
0.5
0.75
1
1.25
1.5
1.75
2
1 2 3 4 5
0.25
0.5
0.75
1
1.25
1.5
1.75
2
γ is the shaded area above the curve.
9. Recall that differentiating with respect to an exponent can be done using logarithms. To finddzλ/dλ, write y(λ) = zλ, so that
ln y = λ ln z
1
y
dy
dλ= ln z
dzλ
dλ= (ln z)zλ. (B.10)
10. Consider the product
(c+ 1)(c + 2)(c + 3) · · · (c+ k − 1)(c+ k)︸ ︷︷ ︸
k terms
. (B.11)
If c = 0, then this is k! = Γ(k + 1). We can use the Gamma function to write it for other calso. Consider
Γ(k + c+ 1) = (k + c)Γ(k + c)
= (k + c)(k + c− 1)Γ(k + c− 1)
...
= (k + c)(k + c− 1) · · · (c+ 2)(c + 1)︸ ︷︷ ︸
k terms
Γ(c+ 1), (B.12)
25
so that (B.11) is Γ(k+ c+ 1)/Γ(c+ 1) = (k+ c)!/c! as it is often written, even for nonintegerc.
B3 Legendre’s equation—solution about an ordinary point.
B3.1 Legendre’s Equation
φ
x0
xθ
r
x
y
zreplacemen
φ
θz = 1
z = 0
z = −1
Legendre’s equation arises in the solution of Laplace’s equation ∇2u = 0 in spherical polarcoordinates. We will use it here as an example of obtaining solutions about a regular point of anequation although we will need the results later in the course. The equation is
1
r2∂
∂r
(
r2 sin θ∂u
∂r
)
+1
r2∂
∂θ
(
sin θ∂u
∂θ
)
+1
r2∂
∂φ
(1
sin θ
∂u
∂φ
)
= 0. (B.13)
B3.2 Separation of variables for Laplaces Equation in spherical polar coordi-
nates.
In this course we consier only axisymmetric solutions with no variation on the azimuthal variableφ so we set the operator ∂/∂φ to zero. We look for solutions of the type u(r, θ) = rλΘ(θ). Thisis taking a bit of a shortcut - we could approach the problem looking for solutions of the typeu(r, θ) = R(r)Θ(θ) but I already know that the solution to the equation for the radial dependenceof the solution (R(r)) is in powers of r. I can see this from the equation (B.13) as each part willreduce the power in rλ by 2 so that each has the same radial dependence which could cancel outof the equation. Substitution gives
sin θλ(λ+ 1)rλ + rλ(sin θΘ′
)′= 0.
As expected the factor rλ can be cancelled through. The equation simplifies if we write z = cos θ,Θ(θ) = w(z). The chain rule then gives
d
dθ= − sin θ
d
dz,
so that
sin θλ(λ+ 1)w − sin θd
dz
(
− sin2 θdw
dz
)
= 0,
but sin2 θ = 1 − cos2 θ = 1 − z2, so that
d
dz
(
(1 − z2)dw
dz
)
+ λ(λ+ 1)w = 0.
26
This is Legendre’s equation, or
(1 − z2)d2w
dz2− 2z
dw
dz+ λ(λ+ 1)w = 0. (B.14)
Comparing this with our standard form for ordinary differential equations we see that
p(z) =−2z
1 − z2, q(z) =
λ(λ+ 1)
1 − z2,
so that p(z) and q(z) are regular at z = 0 (the equator) and we expect w(z) to have solutions thatare analytic there. However we might expect the solutions to be singular (in general) at z = ±1(the poles).
B3.3 Series Solution of Legendre’s equation.
We try therefore
w(z) =
∞∑
r=0
arzr,
and substitute into (B.14). Note that
w′ =
∞∑
r=0
rarzr−1, w′ =
∞∑
r=0
r(r − 1)arzr−2.
We obtain∞∑
0
r(r − 1)arzr−2
︸ ︷︷ ︸
from 1w′′
+
∞∑
0
[
−r(r − 1)︸ ︷︷ ︸
−z2w′′
−2r︸︷︷︸
−2zw′
+λ(λ+ 1)︸ ︷︷ ︸
λ(λ+1)w
]
arzr = 0,
or∞∑
0
r(r − 1)arzr−2 +
∞∑
0
[λ(λ+ 1) − r(r + 1)
]arz
r = 0.
This expression has to be true for all z and this can be achieved by setting the coefficients of eachpower of z to zero. The first sum gives zero for r = 0 and r = 1 and then a term in a2 and a3zfrom r = 2 and r = 3. Following this idea we see that the coefficient of zr obtained from the twosums is
(r + 1)(r + 2)ar+2 +[λ(λ+ 1) − r(r + 1)
]ar.
This can be made a bit more formal by writing the first sum as
∞∑
0
r(r − 1)arzr−2 =
∞∑
2
r(r − 1)arzr−2 =
∞∑
0
(r + 2)(r + 1)ar+2zr,
using a new r to take us through the sum with rold = rnew +2. The two sums can now be combinedto read
∞∑
0
[
(r + 2)(r + 1)ar+2 +[λ(λ+ 1) − r(r + 1)
]ar
]
zr = 0.
Each term of this must be zero, leading to a recursion relation between the coefficients ar
ar+2 =r(r + 1) − λ(λ+ 1)
(r + 1)(r + 2)ar. (B.15)
We can use this to find two linearly independent solutions of Legendre’s equation as it relatesevery second coefficient.
27
1. w1(z): Choose a0 = 1, a1 = 0 and use (B.15) to find a2, a4, · · · , a1 = a3 = a5 = · · · = 0. Thisis an even solution.
2. w2(z): Choose a0 = 0, a1 = 1 and use (B.15) to find a3, a5, · · · , a0 = a2 = a4 = · · · = 0. Thisis an odd solution.
Notice that, if instead of choosing a0 = 1, we had chosen it to be another non-zero number, 3 say,then a2 would be multiplied by 3 so that a4 will also. In fact all coefficients will be multiplied by3 and the solution gemnerated in this way would be 3w1(z). As it is linear, the general solution ofLegendre’s equation is an arbitrary linear combination of these two,
w(z) = Aw1(z) +Bw2(z).
If we had started the recursion relation with the choice a0 = 2, a1 = −5, we would generate thesolution w(z) = 2w1(z) − 5w2(z).
We can get recursion relations for the coefficients of the even and odd solutions separately asfollows. First the even solution. If we put r = 2m, a2m = αm, for integer m = 0, 1, 2, · · · , in (B.15)then
αm+1 = a2m+2 =2m(2m+ 1) − λ(λ+ 1)
(2m+ 1)(2m + 2)a2m =
2m(2m+ 1) − λ(λ+ 1)
(2m+ 1)(2m + 2)αm,
a recursion relation, with α0 = 1 for the coefficients of w1(z) =∑
∞
m=0 αmz2m.
The odd solution can be approaced by writing r = 2m+ 1, a2m+1 = βm, m = 0, 1, 2, · · · , giving
βm+1 = a2m+3 =(2m+ 1)(2m + 2) − λ(λ+ 1)
(2m+ 2)(2m + 3)a2m+1 =
(2m+ 1)(2m+ 2) − λ(λ+ 1)
(2m+ 2)(2m + 3)βm,
β0 = 1, as the coeffiecients for w2(z) =∑
∞
m=0 βmz2m+1 = z
∑∞
m=0 βmz2m.
B3.4 Legendre Polynomials Pn and Laplace’s equation
We expected the solutions we found above to be regular at the origin and indeed they were. Howeverit is simple to show that the radius of converegnce of the series is 1, so indicating that there islikely to be a singularity in the solution a distance 1 from the origin in the complex plane. Thisis entirely consistent with our observation that Legendre’s equation has regular singular point atz = ±1. This corresponds to the poles of our sperical coordinate system.
However there are solutions that are not singular at the poles. These are attained for specialvalues of λ. Note that if λ = N is an integer then the numerator in (B.15) is zero if r = N so thataN+2 and all subsequent coeffiecients are zero. The solutions are polynomials of degree N whichare regular everywhere. There is also a second value of λ for which λ(λ+ 1) = N(N + 1) which wewill return to below.
When these are suitably normalised - by picking an a0 or a1 other than 1 - are known as theLegendre Polynomials, PN (z). They are the only solution of (B.14) that are regular at z = ±1.PN (z) is a polynomial of degree N .
If we had come across Legendre’s equation in applying the method of separation of variables toLaplace’s equation in spherical polars then we would want our solutions to be regular at the poles,z = ±1. Recall that z = cos θ. This requires the solutions to be polynomials of degree N which, inturn, fixes the radial behaviour of the solutions. We have
λ(λ+ 1) = N(N + 1) =⇒ λ2 + λ = N2 +N =⇒ λ2 + λ+ 14 = N2 +N + 1
4
(λ+ 12)2 = (N + 1
2 )2 =⇒ (λ+ 12) = ±(N + 1
2) =⇒ λ = N, λ = −(N + 1). (B.16)
28
Thus the expression u(r, θ) = rN PN (cos θ) is a solution of Laplace’s equation, regular at the poles.So is u(r, θ) = r−N−1 PN (cos θ) and so is the general linear combination
u(r, θ) =
∞∑
n=0
(
Anrn +
Bn
rn+1
)
Pn(cos θ). (B.17)
We return to Legendre’s Polynomials later in Section C. Here we list the first few and show agraph. Note how they oscillate like the trigonometric functions, with increasingly rapid oscillationsas the order of the polynomial increases.
P0(x) = 1,
P1(x) = x,
P2(x) = (3x2 − 1)/2,
P3(x) = (5x3 − 3x)/2.
-1 -0.5 0.5 1
-1
-0.5
0.5
1
Plot[Evaluate[Table[LegendreP[n, x], {n, 0, 6}]], {x, -1, 1}]
B4 Solution about a regular singular point
First we do an example, then a general theory.
B4.1 An example—the indicial equation
Consider the equation2z2w′′ + (2z2 − z)w′ + w = 0, (B.18)
for w(z). Dividing through by 2z2 we see that the equation has a regular singular point at z = 0.With our previous notation, we can identify
p(z) = 1 − 1
2z, q(z) =
1
2z2,
so that p(z) and q(z) are singular at z = 0 although zp(z) = z − 12 and z2q(z) = 1
2 are regular. Wetherefore look for a solution
w(z) = zc∞∑
k=0
akzk =
∞∑
k=0
akzk+c, a0 6= 0.
29
Substitution into (B.18) gives
∞∑
k=0
[
(k + c)(k + c− 1)︸ ︷︷ ︸
from z2w′′
−12(k + c)
︸ ︷︷ ︸
−zw′/2
+12
︸︷︷︸
w/2
]
akzk+c +
∞∑
k=0
(k + c)︸ ︷︷ ︸
z2w′
akzk+c+1 = 0. (B.19)
Again we ensure this is true by setting each successive coefficient of powers of z equal to zero.The lowest power of z appearing is zc, obtained from the first bracket with k = 0. Setting thiscoeffiecient to zero gives
c(c− 1) − 12c+ 1
2 = 0, =⇒ c2 − 32c+ 1
2 = (c− 1)(c− 12 ) = 0.
This is the indicial equation which we shall write as F (c) = 0, defining F (c) = (c− 1)(c− 12). The
two possible roots of this give two possible values of c (c = 12 , c = 1). Note that the term in square
brackets in (B.19) is F (k + c) - we obtained F (c) by taking k = 0.The other coefficients of powers of z can be set to be zero if the ak satisfy a particular reccurence
relation. Setting the coefficient of zk+c, k > 0, to zero gives
akF (k + c) + (k + c− 1)ak−1 = 0,
or
ak = −(k + c− 1)
F (k + c)ak−1 = − (k + c− 1)
(k + c− 1)(k + c− 12)ak−1 =
−ak−1
(k + c− 12). (B.20)
So, using this formula repeatedly (k times), finishing with k = 1
ak =(−1)k
(k + c− 12)(k + c− 3
2 ) · · · (c+ 32)
︸ ︷︷ ︸
k=2
(c+ 12)
︸ ︷︷ ︸
k=1
a0.
We have so far shown that the two functions
w1,2(z) = zc∑
akzk, c = 1, c = 1
2 ,
where a0 is non-zero and subsequent ak given by the recurrence relation (B.20) satisfy the differentialequation (B.18). These are obviously independent solutions as they have different qualitativebehavious near the origin - one behaves like
√z, the other like z. We have solved the recurrence
relation but the solution can be written in a more convenient form using the Gamma function. Seesection B2, item 10.
Γ(k + c+ 12) = (k + c+ 1
2)Γ(k + c− 12) = (k + c+ 1
2)(k + c− 12) · · · (c+ 1
2)Γ(c+ 12),
so the denominator in our solution is, choosing a0 = 1, without loss of generality, Γ(k+c+ 12 )/Γ(c+ 1
2)and
ak =(−1)kΓ(c+ 1
2)
Γ(k + c+ 12)
=(−1)k(c− 1
2)!
(k + c− 12)!
, (B.21)
using the generalised factorial function. We have therefore w(z) = Aw1(z) +Bw2(z) where
w1(z) = z12
∞∑
k=0
(−1)kzk
k!= z
12 exp(−z), w2(z) = z
∞∑
k=0
(−1)kΓ(32)zk
Γ(k + 32)
. (B.22)
30
B4.2 The failure of the method
This method will fail to provide two soltions to the differential equation however if either
1. F (c) has repeated roots, as then we only get one solution.
2. F (c) has roots that differ by an integer. This is trickier to see. If the two roots are c1 andc2 with c2 > c1 then a solution will be given by c2 will be valid. However if c2 = c1 +m forinteger m then the recursion relation similar to (B.20) will usually fail to give a value for am
as it will involve division by F (c1 +m) = 0.
B4.3 A general approach
We will proceed with a general approach to see what can be done in the above cases. This will leadus to Frobenius’ method for solving differential equations. Consider the operator D(w) defined by
D(w) = w′′ + P (z)w′ +Q(z)w.
We want solutions, w, to D(w) = 0. Let
zP (z) =
∞∑
0
pnzn, z2Q(z) =
∞∑
0
qnzn, w(z, λ) = zλ
∞∑
n=0
anzn, a0 6= 0,
where we allow w to be a function both of z and of λ which we have not yet chosen. For themoment we suppress the dependence on λ for clarity of presentation. Direct substitution gives
D(w) =∞∑
k=0
(k + λ)(k + λ− 1)akzk+λ−2 +
∞∑
n=0
pnzn
∞∑
k=0
(k + λ)akzk+λ−2 +
∞∑
n=0
qnzn
∞∑
k=0
akzk+λ−2,
or
D(w) =[λ(λ− 1) + p0λ+ q0
]a0z
λ−2
︸ ︷︷ ︸
lowest power of z, from k = 0, n = 0,
+
∞∑
k=1
[(k + λ)(k + λ− 1) + p0(k + λ) + q0
]akz
k+λ−2
︸ ︷︷ ︸
from n = 0, k = 1, 2, 3, · · · ,
+
∞∑
k=1
k∑
r=1
[(λ+ k − r)pr + qr
]ak−rz
k+λ−2
︸ ︷︷ ︸
from the rest. A power of zk+λ−2 is obtained from (q1z)(ak−1zλ+k−3), (q2z2)(ak−2zλ+k−4),· · · ,(qkzk)(a0zλ−2), for example.
.
(B.23)
As before, we define F (λ) = λ(λ− 1) + p0λ+ q0. We can make the last two terms add to givezero if the ak satisfy the recursion relation
ak =−1
F (λ+ k)
k∑
r=1
[(λ+ k − r)pr + qr
]ak−r, a0 = 1. (B.24)
This is the equivalent to the usual recursion relation between the coefficients. However, we havenot yet chosen λ and we must consider ak = ak(λ). If, for a particular λ, the coefficients in w(z)are defined in this way, then we have
D(w) = a0zλ−2F (λ).
We have a solution if the right hand side of this expression is zero.
31
B4.3a The general solution and the indicial equation
If the indicial equation F (λ) = 0 has two distinct roots λ = λ1, λ2, then we have D(w) = 0 withthese choices of λ and we have two solutions of the equation
w1,2(z) = w1,2(z, λ1,2) = zλ1,2
∞∑
k=0
ak(λ1,2)zk,
with ak(λ1,2) defined by (B.24).
B4.3b When the roots of the indicial equation are equal
If F (λ) = 0 has a repeated root, λ = λ̃, say, then F (λ) = (λ− λ̃)2 and
D(w) = a0zλ−2(λ− λ̃)2. (B.25)
One solution is obtained by putting λ = λ̃, w(z) = w1(z, λ̃) =∑
∞
k=0 ak(λ̃)zk+λ̃. The other can beobtained by differentiating (B.25) with respect to λ. The operator D involves only differentiationwith respect to z so it will commute with differentiating with respect to λ. Hence
D(∂w
∂λ
)
= 2a0zλ−2(λ− λ̃) + a0(λ− λ̃)2zλ−2 ln z,
see B2 section 9. If we now put λ = λ̃, we see that
D(∂w
∂λ
∣∣∣∣λ=λ̃
)
= 0,
so that a second solution is w2(z) = ∂w(z,λ)∂λ |λ=λ̃. Now
w(z, λ) =∞∑
k=0
ak(λ)zk+λ,
so
w1(z)∞∑
k=0
ak(λ̃)zk+λ̃, (B.26)
w2(z) =∞∑
k=0
(∂ak
∂λ
∣∣∣∣λ=λ̃
zλ̃+k + ak(λ̃)zλ̃+k ln z
)
= w1(z) ln z +∞∑
k=1
(∂ak
∂λ
∣∣∣∣λ=λ̃
)
zλ̃+k, (B.27)
where the sum starts from k = 1 rather than k = 0 as ∂a0
∂λ = ∂1∂λ = 0. Finally
w(z) = Aw1(z) +Bw2(z).
B4.3c When the roots of the indicial equation differ by an integer
This topic is not part of the course but is covered here for completeness. Using the greater of thetwo roots gives a solution with no problem. We have seen that the problem in this case arises whenusing the lower of the two roots of the indicial equation λ1 say, to calculate am where the secondroot is λ2 = λ1 +m. We have in this case
D(w) = a0zλ−2(λ− λ1)(λ− λ2)
32
where, from the discussion above the coefficients ak(λ) generally have simple poles in λ at λ = λ1 fork ≥ m. These arise from the factor F (λ+k) = (λ+k−λ1)(λ+k−λ2) = (λ−λ1 +k)(λ−λ1 +k−m)in the denominator of (B.24). In fact this relation becomes
ak =(−1)k
(λ− λ1 + k)(λ− λ1 + k −m)
k∑
s=1
[(λ+ k − s)ps + qs
]ak−s, a0 = 1. (B.28)
When used with k = m to find am we see a factor of (λ−λ1) in the expression for am. This is thenpropagated into am+1, am+2, and so on through repeated use of the recursion relation. Note thata zero is only actually attained when we put λ = λ1 and that the coefficients, and so w(z, λ), arewell defined for other values of λ. Note also that it is possible that the numerator of the recursionrelation will also be zero at k = m and this problem might not arise. Having noted these propertiesof the coefficients we consider
w̄(z) = limλ→λ1
(λ− λ1)w(z, λ).
If the power series for w̄ is considered then the first m − 1 terms will be zero as ak is finite atλ = λ1 and the first non-zero term is limλ→λ1
(λ − λ1)am(λ)zm+λ1 = b0zλ2 , say for some finite
b0. We denote the higher coefficients by brzm+r so that w̄ =
∑∞
r=0 brzr+m+λ1=zλ2
∑∞
r=0 brzr. The
coefficients br will satisfy the expression (B.28) with k = r +m, am+r = br, i.e.
br =(−1)k
(λ− λ1 +m+ r)(λ− λ1 + r)
k∑
s=1
[(λ+m+ r − s)ps + qs
]br−s, (B.29)
and taking the limit λ→ λ1, but noting that λ1 +m = λ2 gives
br =(−1)k
(λ2 − λ1 + r)(r)
k∑
s=1
[(λ2 + r − s)ps + qs
]br−s. (B.30)
This is precisely the recursion relation obtained from (B.28) when we put λ = λ2. In other wordsw̄ is only a multiple of w(z, λ2) and not a second independent solution of the equation. Howeverconsider differentiating with respect to λ. We have
D(w̄) = a0(λ− λ1)2(λ− λ2)z
λ−2
so that
D(∂w̄
∂λ
)
= 2a0(λ− λ1)(λ− λ2)zλ−2 + a0(λ− λ1)
2(λ− λ2)zλ−2 ln z,
which does give zero as λ → λ1 and is independent of w(z, λ2) due to the logarithmic behaviour.Hence a second independent solution in the case where the roots of the indicial equation differ byan integer is
w1(z) = limλ→λ1
∂
∂λ
[(λ− λ1)w(z, λ)
].
B4.4 An example—the roots of the indicial equation are equal.
We solve
zw′′ + w′ + zw = 0. (B.31)
33
This has a regular singular point at z = 0 and we look for a solution w(z) = zλ∑
∞
r=0 arzr, a0 6= 0.
Substitution gives
∞∑
r=0
[(r + λ)(r + λ− 1)︸ ︷︷ ︸
zw′′
+ (r + λ)︸ ︷︷ ︸
w′
]arz
r+λ−1 +
∞∑
r=0
arzr+λ+1
︸ ︷︷ ︸
zw
= 0.
This is written
a0zλ−1[λ(λ−1)+λ
]+a1z
λ[(λ+1)λ+(λ+1)
]+
∞∑
r=2
{[(r+λ)(r+λ−1)+(r+λ)
]ar+ar−2
}
zr+λ−1 = 0.
The indicial equation arises from the first of these terms and is λ2 = 0 so we have the root λ = 0repeated. The next term, with the now forced choice λ = 0 gives a1[1] = 0 forcing a1 = 0. The lastterm gives the recurrence relation
ar = − ar−2
(r + λ)2.
The series for w(z) will be in even powers of z. As in section B3.3 we can put a2r = αr, and write
w(z) =∞∑
r=0
αrz2r, αr = − αr−1
(2r + λ)2= − αr−1
22(r + 12λ)2
, α0 = 1.
We can find explicit expressions for the coefficients. Note that the Frobenius method requires usto find the coefficients as functions of λ so we do not yet put λ = 0. We have
αr =(−1)r
22r
1
(r + 12λ)2(r − 1 + 1
2λ)2 · · · (2 + 12λ)2 (1 + 1
2λ)2︸ ︷︷ ︸
last use when r = 1
. (B.32)
We find one solution by putting λ = 0 giving
αr =(−1)r
22r(r!)2, w1(z) =
∞∑
r=0
(−1)r
(r!)2
(z
2
)2r. (B.33)
The second solution is
w2(z) = w1(z) ln z + zλ∞∑
r=1
∂αr
∂λz2r∣∣∣λ=0
.
We can find the derivative of α2r by using logarithmic differentiation. Taking the logarithm of(B.32) gives
lnαr = ln
((−1)2r
2r
)
− 2
r∑
n=1
ln(
n+ 12λ)
,
so, differentiating,
1
αr
∂αr
∂λ= −2
1
2
r∑
n=1
1
n+ 12λ,
and putting λ = 0 and substituting for αr when λ = 0, which we already know from above, we get
∂αr
∂λ
∣∣∣λ=0
= − (−1)r
2r(r!)2Sr, Sr =
r∑
n=1
1
n.
34
So
w2(z) = w1(z) ln(z) −∞∑
r=1
(−1)r
(r!)2
(z
2
)2rSr, (B.34)
andw(z) = Aw1(z) +Bw2(z).
B5 Bessel’s Equation
Bessel’s equation arises in the solution of the wave equation in polar coordinates and many otherplaces.
B5.1 The wave equation and Helmholtz equation in polar coordinates
The wave equation isc2∇2φ = φtt
and we have seen in section A9.1 that if we look for solutions of the type φ(x, t) = u(x)T (t), thenwe obtain Helmholtz equation for u,
∇2u+ λ2u = 0,
with λ2 = ω2/c2 with ω the frequency of the wave - see section A8.1 to A8.3. The operator ∇2 inpolar coordinates is given by (A.27) so we have
urr +1
rur +
1
r2uθθ + λ2u = 0.
If we look for solutions for u(r, θ) of the form u(r, θ) = R(r)Θ(θ) then we find the equations
Θ′′ + ν2Θ = 0,
R′′ +1
rR′ +
(
λ2 − ν2
r2
)
R = 0.
This is an example in a later homework sheet, where you will show that in many cases ν is aninteger. This is Bessel’s equation, but we can put it into standard form by writing z = λr andR(r) = w(z). Hence, as d
dr = λ ddz , we have
λ2wzz + λ2 1
zwz +
(
λ2 − λ2ν2
z2
)
w = 0,
orz2w′′ + zw′ + (z2 − ν2)w = 0. (B.35)
This is Bessel’s equation of order ν.
B5.2 Series solution of Bessel’s equation
Bessel’s equation has a regular singular point at z = 0 so we may search for solutions w(z) =zλ∑
∞
r=0 arzr. Substitution gives, as we have seen in many examples before,
∞∑
r=0
[
(λ+ r)(λ+ r − 1) + (λ+ r) − ν2]
arzr+λ +
∞∑
r=0
arzλ+r+2 = 0.
35
The indicial equation arises from the coefficent of zλ and reads
λ(λ− 1) + λ− ν2 = λ2 − ν2 = 0, =⇒ λ = ±ν.The coefficient of zλ+1 gives a1 = 0 and the coefficient of zλ+r+2 gives
ar =−ar−2
(r + λ)2 − ν2, r = 2, 3, 4 · · · . (B.36)
We now have three possibilities, relating to the roots of the indicial equation, λ = ±ν, and theirdifference.
1. 2ν is not an integer. In this case the two independent solutions are given by λ = ±ν.
2. ν = 0, i.e. equal roots
3. 2ν is an integer and the roots of the indicial equation differ by an integer. This turns outto be the most important case but we will not cover the method of Frobenius in this case ingreat detail, although we shall certainly need the results that arise.
We return to case 1. The two independent solutions are, noting that the series is actually in z2,
w1,2(z) = z±ν∞∑
r=0
brz2r
where, putting br = a2r in (B.36), gives
br =−br−1
(2r + λ)2 − ν2=
−br−1
4(r + 12(λ− ν))(r + 1
2(λ+ ν)),
giving
br = b0(−1)r
22r
1
(r + 12(λ− ν))(r − 1 + 1
2(λ− ν)) · · · (1 + 12(λ− ν))
︸ ︷︷ ︸
r=1
×
1
(r + 12 (λ+ ν))(r − 1 + 1
2(λ+ ν)) · · · (1 + 12 (λ+ ν))
︸ ︷︷ ︸
r=1
(B.37)
or
br = b0(−1)r
22r
Γ(1 + 12 (λ− ν))
Γ(r + 1 + 12(λ− ν))
Γ(1 + 12(λ+ ν))
Γ(r + 1 + 12(λ+ ν))
,
asΓ(r + a+ 1) = (r + a)Γ(r + a) = (r + a)(r + a− 1) · · · (a+ 1)Γ(a+ 1).
Putting λ = ±ν and writing Γ(1 + a) = a!, gives
w1(z) = b0zν
∞∑
r=0
(−1)r
r!
ν!
(r + ν)!
(z
2
)2r,
w2(z) = b01
zν
∞∑
r=0
(−1)r
r!
(−ν)!(r − ν)!
(z
2
)2r.
The Bessel functions that you will find tabulated do not take b0 = 1 as we usually have done.Instead they have b0 =
(12
)ν/ν! and are written as J±ν ,
J±ν(z) =
∞∑
r=0
(−1)r
r!
1
(r ± ν)!
(z
2
)2r±ν. (B.38)
36
B5.3 The functions J0 and Y0.
The case where the roots are equal, case 2, corresponds to the example we covered in section B4.4and we found there that there were two solutions w1 and w2 as given by (B.33) and (B.34). Thetwo solutions that are tabulated are w1, here denoted as J0, Bessel’s function of order zero, and alinear combination of w1 and w2, denoted Y0 and known as Bessel’s function of the second kind oforder zero. We have
J0(z) =
∞∑
r=0
(−1)r
(r!)2
(z
2
)2r, (B.39)
Y0(z) =2
πw2(z) +
2
π(γ − ln 2)w1(z) =
2
π
∞∑
r=0
(−1)r
(r!)2
(z
2
)2r[ln z − ln 2 + γ − Sr]
=2
π
∞∑
r=0
(−1)r
(r!)2
(z
2
)2r [
ln(z
2
)
− ψ(1 + r)]
(B.40)
where γ is Eulers constant as described in section B2, item 8 and ψ(1 + r) = Γ′(1 + r)/Γ(1 + r) =Sr − γ. The important thing that you need to remember about Y0 is that it has a logarithmicsingularity at the origin. J0 on the other hand is finite.
B5.4 The functions Jn and Yn.
Case 3 above is when 2ν is an integer. This occurs if ν is a half-integer as well as if ν is integer.However the solutions we have found in (B.38), are fine and independent when ν is a half integer,with the extension of the factorial function to non-integer arguments. In fact the solutions for halfintegers have a remarkably simple form, hidden by these series solutions, and which we return toin a homework question later.
If ν is an integer ν = n, say, then Jn, n > 0, defined by (B.38), is a valid solution. Note inpassing that Jn behaves like zn for small z. A problem arises in putting ν = −n in however. Thefactorial term (r− ν)! in the denominator has a pole when (r− ν) is a negative integer. See sectionB2, parts 4 and 5, recalling that a! = Γ(a + 1). Therefore the first n terms in the series approachzero as ν approaches n, as the terms (−n)!, (−n + 1)!, · · · , (−1)! arise as r takes the values 0, 1,· · · , n− 1. Hence the series starts at r = n and may be written as
J−n(z) =∞∑
r=n
(−1)r
r!
1
(r − n)!
(z
2
)2r−ν=
∞∑
r=0
(−1)(r + n)
(r + n)!
1
(r)!
(z
2
)2r+2n−n
= (−1)n∞∑
r=0
(−1)r
r!
1
(r + n)!
(z
2
)2r+n= (−1)n Jn(z). (B.41)
Hence the series is well behaved with ν = −n but gives a multiple of the solution we alreadyhave obtained with ν = n. We need a second solution. The derivation below is not on the syllabusand the detail is omitted. The second solution is
w2(z) = limλ→−n
∂
∂λ
[
(λ+ ν)
∞∑
r=0
b(λ)rz2r+λ
]
,
37
with br given by (B.37). This gives, after quite a bit of work,
w2(z) =1
2n
−1
(n− 1)!
{(z
2
)n∞∑
r=0
(z
2
)2r (−1)r
r!(r + n)![2 ln z −Q(r + n, n)] −
(z
2
)−nn−1∑
r=0
(z
2
)2r (n− 1 − r)!
r!
}
Q(r, n) =
n−1∑
k=1
1
k − n+
r∑
k=n+1
1
k − n+
r∑
k=1
1
k.
The actual tabulated solution is a linear combination of this and Jn, denoted Yn, the Bessel functionof order n of the second kind. The important thing to notice is that it is sungular at the origin,with a pole of order n.
Yn(z) = − 1
π
(z
2
)−nn−1∑
r=0
(n− 1 − r)!
r!
(z2
4
)r
+2
πln(z/2) Jn(z)−
1
π
(z
2
)n∞∑
r=0
ψ(1 + r) + ψ(1 + r + n)
r!(r + n)!
(−z2
4
)r
. (B.42)
In fact the second solution Yn is also given by
Yn(z) = limν→n
Jν(z) cos νπ − J−ν(z)
sin νπ,
which is very similar to differentiation with respect to ν and then putting ν = n, just as we obtainedsolutions by differentiation with respect to λ.
In summary:
2 4 6 8 10 12 14
-0.4
-0.2
0.2
0.4
0.6
0.8
1
Table[Plot[BesselJ[i, x], {x, 0, 15}, PlotPoints -> 50], {i, 0, 5}];Show[%]
2 4 6 8 10 12 14
-3
-2.5
-2
-1.5
-1
-0.5
0.5
1
38
Table[Plot[BesselY[i, x], {x, 0, 15}, PlotPoints -> 50], {i, 0, 5}];
Show[%,PlotRange -> {-3, 1}]
For small z
Jp(z) ≈ 1
Γ(p+ 1)
(z
2
)p(B.43)
Y0(z) ≈ (2/π) (ln(z/2) + γ + . . .) (B.44)
Yp(z) ≈ −Γ(p)
π
(2
z
)p
(B.45)
39
Chapter C
Properties of Legendre Polynomials
C1 Definitions
The Legendre Polynomials are the everywhere regular solutions of Legendre’s Equation,
(1 − x2)u′′ − 2xu′ +mu = [(1 − x2)u′]′ +mu = 0, (C.1)
which are possible only if
m = n(n+ 1), n = 0, 1, 2, · · · . (C.2)
We write the solution for a particular value of n as Pn(x). It is a polynomial of degree n. If n iseven/odd then the polynomial is even/odd. They are normalised such that Pn(1) = 1.
P0(x) = 1,
P1(x) = x,
P2(x) = (3x2 − 1)/2,
P0(x) = (5x3 − 3x)/2.
C2 Rodrigue’s Formula
They can also be represented using Rodrigue’s Formula
Pn(x) =1
2nn!
dn
dxn(x2 − 1)n. (C.3)
This can be demonstrated through the following observations
C2.1 Its a polynomial
The right hand side of (C.3) is a polynomial.
C2.2 It takes the value 1 at 1.
If
v(x) =1
2nn!
dn
dxn(x2 − 1)n,
40
then, treating (x2 − 1)n = (x− 1)n(x+ 1)n as a product and using Leibnitz’ rule to differentiate ntimes, we have
v(x) =1
2nn!(n!(x+ 1)n + terms with (x− 1) as a factor) ,
so that
v(1) =n!2n
2nn!= 1.
C2.3 It satisfies the equation
Finally
(1 − x2)v′′ − 2xv′ + n(n+ 1)v = 0,
since, if h(x) = (1 − x2)n, then h′ = −2nx(1 − x2)n−1, so that
(1 − x2)h′ + 2nxh = 0.
Now differentiate n+ 1 times, using Leibnitz, to get
(1 − x2)hn+2 − 2(n + 1)xhn+1 − 2(n + 1)n
2hn + 2nxhn+1 + 2n(n+ 1)hn = 0,
or
(1 − x2)hn+2 − 2xhn+1 + n(n+ 1)hn = 0.
As the equation is linear and v ∝ h, v satisfies the equation also.
C2.4 And that’s it
Thus v(x) is proportional to the regular solution of Legendre’s equation and v(1) = P(1) = 1, sov(x) = P(x).
C3 Orthogonality of Legendre Polynomials
The differential equation and boundary conditions satisfied by the Legendre Polynomials forms aSturm-Liouiville system (actually a ”generalised system” where the boundary condition amountsto insisting on regularity of the solutions at the boundaries). They should therefore satisfy theorthogonality relation
∫ 1
−1Pn(x) Pm(x) dx = 0, n 6= m. (C.4)
If Pn and Pm are solutions of Legendre’s equation then
[(1 − x2) P′
n]′ + n(n+ 1)Pn = 0, (C.5)
[(1 − x2) P′
m]′ +m(m+ 1)Pm = 0. (C.6)
Integrating the combination Pm(C.5)−Pn(C.6) gives
∫ 1
−1Pm[(1 − x2) P′
n]′ − Pn[(1 − x2) P′
m]′ dx+ [n(n+ 1) −m(m+ 1)]
∫ 1
−1Pn Pm dx = 0.
41
Using integration by parts gives, for the first integral,
[Pm(1 − x2) P′
n
]1
−1−[Pn(1 − x2) P′
m
]1
−1−∫ 1
−1P′
m(1 − x2) P′
n −P′
n(1 − x2) P′
m︸ ︷︷ ︸
=0
dx = 0,
as Pm,n and their derivatives are finite at x = ±1 (i.e. they are regular there). Hence, if n 6= m
∫ 1
−1Pn Pm dx = 0. (C.7)
C4 What is∫ 1
−1 P2n dx?
We can evaluate this integral using Rodrigue’s formula. We have
In =
∫ 1
−1P2
n dx =1
22n(n!)2
∫ 1
−1
dn
dxn(x2 − 1)n
dn
dxn(x2 − 1)n dx.
Integrating by parts gives
(22n(n!)2)In =
[dn−1
dxn−1(x2 − 1)n
]1
−1
−∫ 1
−1
dn−1
dxn−1(x2 − 1)n
dn+1
dxn+1(x2 − 1)n dx.
Note that differentiating (x2−1)n anything less than n times leaves an expression that has (x2−1)as a factor so that the first of these two terms vanishes. Similarly, integrating by parts n timesgives
In =(−1)n
22n(n!)2
∫ 1
−1(x2 − 1)n
d2n
dx2n(x2 − 1)n dx.
The (2n)th derivative of the polynomial (x2 − 1)n, which has degree 2n is (2n)!. Thus
In =(−1)n(2n)!
22n(n!)2
∫ 1
−1(x2 − 1)n dx.
The completion of this argument is left as an exercise. One way to proceed is to use the transfor-mation s = (x+ 1)/2 to transform the integral and then use a reduction formula to show that
∫ 1
0sn(1 − s)n ds =
(n!)2
(2n+ 1)!.
The final result is ∫ 1
−1P2
n dx =2
2n + 1. (C.8)
C5 Generalised Fourier Series
Sturm-Liouiville theory does more than guarantee the orthogonality of Legendre polynomials, italso shows that we can represent functions on [−1, 1] as a sum of Legendre Polynomials. Thus forsuitable f(x) on [−1, 1] we have the generalized Fourier series
f(x) =
∞∑
0
an Pn(x). (C.9)
42
To find the coefficients an, we multiply both sides of this expression by Pm(x) and integrate toobtain
∫ 1
−1Pm(x)f(x) dx =
∞∑
0
an
∫ 1
−1Pn(x) Pm(x) dx = am
2
2m+ 1,
so that
an =
(
n+1
2
)∫ 1
−1f(x) Pn(x) dx. (C.10)
C6 A Generating Function for Legendre Polynomials
C6.1 Definition
We consider a function of two variables G(x, t) such that
G(x, t) =∞∑
n=0
Pn(x)tn, (C.11)
so that the Legendre Polynomials are the coefficients in the Taylor series of G(x, t) about t = 0.Our first task is to identify what the function G(x, t) actually is.
C6.2 Derivation of the generating function.
We know that, in spherical polar coordinates, the function r−1 is harmonic, away from r = 0, i.e.
∇2 1
r= ∇2 1
|x| = 0.
It is a harmonic function independent of φ. Similarly 1/(|x − x0|) is harmonic away from x = x0.If x0 is a unit vector in the z-direction, then
|x− x0|2 = (x− x0).(x − x0)
= x.x− 2x.x0 + x0.x0
= r2 − 2r cos θ + 1. (C.12)
So the function1√
r2 − 2r cos θ + 1
is harmonic, regular at the origin and independent of φ. We should therefore be able to write it inthe form,
1√r2 − 2r cos θ + 1
=
∞∑
n=0
Anrn Pn(cos θ). (C.13)
43
φ
x0
x
x − x0
θ
r
x
y
z
If we can show that An = 1, then the replacement of cos θ by x and r by t gives the required result.To find An, we evaluate the function along the positive z-axis, putting cos θ = 1, noting that
1√r2 − 2r + 1
=1
√
(r − 1)2=
1
|r − 1| =1
1 − r= 1 + r + r2 + r3 + · · · , |r| < 1.
So∞∑
n=0
rn =
∞∑
n=0
Anrn Pn(1) =
∞∑
n=0
Anrn,
so that An = 1. Thus, with x = cos θ and t = r,
G(x, t) =1√
1 − 2xt+ t2=
∞∑
n=0
tn Pn(x). (C.14)
C6.3 Applications of the Generating Function.
Generating functions can be applied in many ingenious ways, somethimes best left for examinationquestions. As an example, we can differentiate G(x, t) with respect to t to show that
∂G
∂t=
x− t
(1 − 2xt+ t2)3/2=⇒ (1 − 2xt+ t2)
∂G
∂t= (x− t)G.
Now write G(x, t) as a sum of Legendre polynomials to get
(1 − 2xt+ t2)
∞∑
n=0
nPn(x)tn−1 = (x− t)
∞∑
n=0
Pn(x)tn.
Now comparing the coefficients of t0 gives
P1(x) = xP0(x),
so that, as P0 = 1, we have P1 = x, as expected. Comparing the general coefficient of tn, n > 0,gives
(n + 1)Pn+1 −2xnPn +(n− 1)Pn−1 = xPn −nPn−1,
44
or, rearranging(n+ 1)Pn+1 −(2n+ 1)xPn +nPn−1 = 0, (C.15)
a recursion relation for Legendre polynomials.Differentiating G(x, t) with respect to x, and proceeding in a similar way yields the result
P′
n+1 −2xP′
n + P′
n−1 = Pn, n ≥ 1. (C.16)
Combining (C.15) and (C.16), or obtaining a relationship between Gx and Gt, shows
P′
n+1 −P′
n−1 = (2n + 1)Pn . (C.17)
These need not be learnt.
C6.4 Solution of Laplace’s equation.
Remember where we first came across Lagrange polynomials. If ∇2u = 0 and u is regular at θ = 0, πin spherical polar coordinates and with ∂/∂φ = 0 then
u(r, θ) =∞∑
n=0
(
Anrn +
Bn
rn+1
)
Pn(cos θ). (C.18)
C7 Example: Temperatures in a Sphere
The steady temperature distribution u(x) inside the sphere r = a, in spherical polar coordinates,satisfies ∇2u = 0. If we heat the surface of the sphere so that u = f(θ) on r = a for some functionf(θ), what is the temperature distribution within the sphere?
The equation and boundary conditions do not depend on φ so we know that u is of the form(C.18). Further more we expect u to be finite at r = 0 so that Bn = 0. We find the coefficients An
by evaluating this on r = a. We require
f(θ) =
∞∑
n=0
Anan Pn(cos θ).
We can find An using the orthogonality of the polynomials (C.7). However in (C.7), the integrationis with respect to x and not cos θ. If x = cos θ, then dx = − sin θ dθ. The interval −1 ≤ x ≤ 1 isthe interval −π ≥ θ ≥ 0. Multiply through by − sin θPm(θ) and integrate in θ to obtain
∫ 0
π− sin θf(θ) Pm(cos θ) dθ =
∞∑
n=0
(anAn)
∫ 0
π− sin θPn(cos θ) Pm(cos θ) dθ
=
∞∑
n=0
(anAn)
∫ 1
−1Pn(x) Pm(x) dx =
2amAm
2m+ 1. (C.19)
So
u(r, θ) =
∞∑
n=0
(
n+1
2
)( r
a
)nPn(cos θ)
∫ π
0f(ν) Pn(cos ν) sin ν dν. (C.20)
Let us heat the northern hemisphere and leave the southern half cold, so that f(θ) = 1 for 0 ≤ θ ≤π/2 and f(θ) = 0 for π/2 < θ ≤ π. Then the integral in (C.20) is
∫ π/2
0sin ν Pn(cos ν) dν =
∫ 1
0Pn(x) dx.
45
An integration of (C.17) gives
(2n + 1)
∫ 1
xPn(q) dq = [Pn+1 −Pn−1]
1x = Pn−1(x) − Pn+1(x), n > 1.
We know that∫ 10 P0(q) dq = 1 so
u(r, θ) =1
2+
1
2
∞∑
n=1
(r
a
)nPn(cos θ)(Pn−1(0) − Pn+1(0)).
Note that the temperature at the centre of the hemisphere (r = 0) is 1/2, which might be expected.The Legendre polynomials of odd degree are odd and will be zero at the origin so that the coefficientsin the sum will be zero for even values of n. Hence
u(r, θ) =1
2+
1
2
(r
a
) ∞∑
m=0
(r
a
)2mP2m+1(cos θ)(P2m(0) − P2(m+1)(0)). (C.21)
We need the values at the origin of the polynomials of even degree. Putting x = 0 in (C.14) gives
∞∑
n=0
tn Pn(0) =1√
1 + t2(C.22)
= 1 − 1
2t2 +
1.3
2.2
t4
2!− 1.3.5
2.2.2
t6
3!+ · · ·
=
∞∑
m=0
(−1)m(2m− 1)(2m− 3) · · · 3.1
2mm!t2m
=
∞∑
m=0
(−1)m(2m)!
22m(m!)2t2m.
Therefore
P2m(0) − P2(m+1)(0) = (−1)m(2m)!
22m(m!)2− (−1)m+1 (2m+ 2)!
22m+2((m+ 1)!)2
= (−1)m(2m)!
22m(m!)2
(
1 +2m+ 1
2m+ 2
)
,
and
u(r, θ) =1
2+
1
2
(r
a
) ∞∑
m=0
( r
a
)2mP2m+1(cos θ)(−1)m
(2m)!
22m(m!)2
(
1 +2m+ 1
2m+ 2
)
.
We will evaluate this expression along the axis of the sphere. Here cos θ = ±1, depending if we arein the northern or southern hemisphere. The polynomials in appearing are odd so take the value±1 at ±1. This can be accounted for by allowing r to be negative so that it measures the directeddistance from the centre in a northerly direction.
u(r) =1
2+
1
2
(r
a
) ∞∑
m=0
( r
a
)2m(−1)m
(2m)!
22m(m!)2
(
1 +2m+ 1
2m+ 2
)
.
A graph of the solution obtained by summing this series to 1,4,7,10,13 terms is shown below
46
-1 -0.5 0.5 1
0.25
0.5
0.75
1
1.25
1.5
We can see that the convergence is not good near the poles. The line that actually attains thevalues 0 and 1 at the poles is the exact solution
u(r) =1
2+
(r/a)2 +√
1 + (r/a)2 − 1
2(r/a)√
1 + (r/a)2.
This is attained as follows. Equation (C.21) tells us
u(r) =1
2+
1
2
(r
a
) ∞∑
m=0
( r
a
)2m(P2m(0) − P2(m+1)(0)).
Identifying t with (r/a), and using (C.22), recognising that the sum only contains the even powersof t, gives
∞∑
m=0
(r/a)2m P2m(0) −∞∑
m=0
(r/a)2m−2 P2m(0) =1
√
1 + (r/a)2− 1
(r/a)21
√
1 + (r/a)2
Changing the index in the second sum, using P0 = 1, we find
∞∑
m=0
(r/a)2m (P2m(0) − P2m+2(0)) −1
(r/a)2=
1√
1 + (r/a)2− 1
(r/a)21
√
1 + (r/a)2.
And so
u(r) =1
2+
1
2
r
a
(
1√
1 + (r/a)2− 1
(r/a)2√
1 + (r/a)2+
1
(r/a)2
)
,
which simplifies to the result above.
47
Chapter D
Oscillation of a circular membrane
D1 The Problem and the initial steps in its solution
D1.1 The problem
If we have a circular drum, radius a, and hit it, we will set the drum vibrating. To study thisvibration, we need to solve the wave equation
c2∇2ψ = ψtt, 0 ≤ r ≤ a (D.1)
with c the speed of wave motion in the drum’s material and ψ the displacement of the drum’ssurface. We have the boundary condition
ψ(a, θ) = 0, 0 ≤ θ < 2π, (D.2)
corresponding to the drum being fixed at its circular edge. We also expect ψ to be finite at r = 0,the drum’s centre. It we ”hit” the drum, so that, at t = 0, ψ = 0, ψt(r, θ) = f(r), then we mustalso impose this initial condition.
D1.2 Separating out the time dependence
We look for oscilliatory solutions, writing
ψ(r, θ, t) = u(r, θ) exp(iωt). (D.3)
This is equivalent to looking for solutions of the type ψ = u(x)T (t) and knowing in advance thatthe equation for T will have the form T ′′ + ω2T = 0 where ω is related to the separation constant.This has solutions proportional to cosωt and sinωt and we choose to write these in exponentialform. The value of ω is the frequency of the disturbance. Doing so leads to
c2∇2u = −ω2u, or ∇2u+ λ2u = 0, (D.4)
(after dividing through by the exponential factor). Here ω = λc and we need to solve Helmholtz’equation for the spatial part of the solution. We should expand a little on the shorthand, exp(iωt),that we are using to describe the temporal part of the solution. The solution of the equation forT is T = Aω cos(ωt) + Bω sin(ωt). To write this in exponential form we can take the real part of(Aω − iBω)(cos ωt + i sinωt), i.e. of (Aω − iBω) exp(iωt). We therefore have a complex constantmultiplying (D.3) which we have omitted. We can write this complex constant in modulus-argumentform as Rω exp(iǫωt).
48
Note that, defining ω2 = λc2, we would have ended up with λ rather than λ2 in (D.4), but thatwould lead to lots of √’s in what follows. Note now that to find the frequency of oscillation weneed to solve the eigenvalue problem (D.4), with u(a, θ) = 0 (from (D.2) and u finite at r = 0.
urr +1
rur +
1
r2uθθ + λ2u = 0. (D.5)
D1.3 Separating out the θ dependence
We are looking for solutions that are 2π-periodic in θ as we have a circular drum. We know that ifwe look for solutions of the form u(r, θ) = R(r)Θ(θ) then the θ-dependence will satisfy Θ′′+p2Θ = 0,where to further ensure periodicity in θ, p2 ≥ 0 and finally to ensure 2π-periodicity p = n for integern, n = 0, 1, 2, 3, . . .. The case n = 0 corresponds to solutions with no θ-dependence. In general theθ-dependence of the solution is like sin(nθ) and cos(nθ). We choose to write both of these togetheras exp(inθ) (strictly Rn exp(inθ) exp(iǫn)) and look for solutions of the type u = R(r) exp(inθ).Subsitution into (D.5), dividing out the exponential factor and multiply by r2 gives
r2R′′ + rR′ + (λ2r2 − n2)R = 0, (D.6)
and if we write z = λr, R = w(z),
z2w′′ + zw′ + (z2 − n2)w = 0. (D.7)
We have obtained solutions of this equation for integer n through series and found that it has twoindependent solutions. We can then write
w(z) = An Jn(z) +Bn Yn(z). (D.8)
The point z = 0 corresponds to the centre of the membrane and we wish our solution to beanalytic here. This means that we must set Bn = 0 and we consider only solutions finite at z = 0,w(z) = An Jn(z).
At this stage our solution is of the form
ψ(r, θ, t) =∑
λ
∞∑
n=0
Anλ Jn(λr) exp(inθ) exp(icλt) (D.9)
where Anλ = Rnλ exp(iǫλ) exp(iǫn) are (complex) constants that we need to find so as to satisfy theinitial conditions. In relating this to our definitions above we have used Rnλ = RωRn. We returnto this solution later, but for the present we look more closely at the properties of the solutions ofBessel’s equation so as to firstly enable us to fix possible values of λ and secondly to express theinitial conditions as generalised Fourier Series.
D1.4 On Bessel Functions
D1.4a Expression as a series
For general p we have the series solution
Jp(x) =
∞∑
j=0
(−1)j
Γ(j + 1)Γ(j + p+ 1)
(x
2
)2j+p(D.10)
49
with Γ(z) the Gamma function, extending the factorial function to non-integer argument and withn! = Γ(n + 1). The second independent solution is J−p(x) if p is not an integer. However if p isan integer then as J−n = (−1)nJn, these solutions are linearly dependent. To cover this case alsoa second linearly independent solution Yp(x) is taken. (Note Yp is often written Np.) For small x
Jp(x) ≈ 1
Γ(p+ 1)
(x
2
)p(D.11)
Y0(x) ≈ (2/π) (ln(x/2) + γ + . . .) (D.12)
Yp(x) ≈ −Γ(p)
π
(2
x
)p
(D.13)
D1.4b Behaviour for large x.
For large x all these solutions behave like a damped sinusoidal function with y(x) ≈ A sin(x+ǫ)/√x.
This is easy to demonstrate, writing w(x) = f(x)y(x) and substituting into (D.7)
x2(f ′′y + 2f ′y′ + fy′′) + x(f ′y + fy′) + (x2 − n2)fy = 0
The coeficient of y′ can be made zero if 2x2f ′ + xf = 0 giving f ∝ x−1/2. Choosing f = x−1/2 anddividing by x3/2 gives
y′′ +
(
1 +1/4 − n2
x2
)
y = 0.
For large x, this is well approximated by y′′ + y = 0 so that y = A sin(x+ ǫ) and the result.
2 4 6 8 10 12 14
-0.4
-0.2
0.2
0.4
0.6
0.8
1
Table[Plot[BesselJ[i, x], {x, 0, 15}, PlotPoints -> 50], {i, 0, 5}];Show[%]
2 4 6 8 10 12 14
-3
-2.5
-2
-1.5
-1
-0.5
0.5
1
50
Table[Plot[BesselY[i, x], {x, 0, 15}, PlotPoints -> 50], {i, 0, 5}];
Show[%,PlotRange -> {-3, 1}]
All the solutions have an infinite number of zeros. The zeros of Jn are denoted jnm so thatJn(jnm) = 0 and 0 < jn1 < jn2 < jn3 < . . ..
jn1 jn2 jn3 jn4 jn5 · · ·n = 0 2.4048 5.5201 8.6537 11.7915 14.9309 · · ·n = 1 3.8317 7.0156 10.173 13.323 16.471 · · ·n = 2 5.1356 8.4172 11.6198 14.796 17.960 · · ·
......
......
......
D1.5 Determination of λ
We have the boundary condtion ψ(a, θ) = 0, i.e. w(λa) = 0. Thus λ is determined so that λa = jnm
or λ = jnm/a, m = 1, 2, 3, · · · . Thus the possible frequencies of the drum’s vibration are determinedas the doubly infinite family
ω = ωnm = cjnm/a (D.14)
For each frequency the vibration is described by
Jn(jnmr/a) exp(icjnmt/a) exp(inθ) (D.15)
The general solution is an arbitray linear combination of all these modes so that (D.9) becomes thedouble sum
ψ(r, θ, t) =
∞∑
n=0
∞∑
m=1
Anm Jn(jnmr/a) exp(inθ) exp(icjnmt/a). (D.16)
Here now Anm = Rnm exp(iǫn) exp(iǫm). The constants Rnm can be related to the amplitude of aparticular mode, ǫn to its orientation realtive to the line θ = 0 and ǫm to its phase (where in thesinusoidal temporal cycle it started).
n = 0, m = 1
51
n = 0, m = 2
n = 1, m = 2
n = 3, m = 4
<< NumericalMath‘BesselZeros‘;n = 1; m = 2;jnm = BesselJZeros[n, m][[m]];
radial[r_] = BesselJ[n, jnm r];azimuth[theta_] = Re[Exp[I n theta]];
wrap[f_, r_, theta_] = If[r < 1, f[r, theta], 0];polarr[x_, y_] = Sqrt[x^2 + y^2];
polartheta[x_, y_] = If[x > 0, ArcTan[y/x], If[y > 0, Pi/2 + ArcTan[-x/y],
-Pi/2 - ArcTan[x/y]]];mode[r_, theta_] = radial[r]azimuth[theta];
surf = Plot3D[wrap[mode, polarr[x, y], polartheta[x,y]], {x, -1, 1},{y, -1, 1},
52
PlotRange -> All, PlotPoints -> {100,100}, Lighting -> False, Mesh ->False,
Axes -> False, Boxed -> False];cont = ContourPlot[wrap[mode, polarr[x, y],
polartheta[x, y]], {x, -1,1}, {y, -1, 1}, PlotRange -> All, PlotPoints-> {100, 100},
FrameTicks->None];Show[GraphicsArray[{surf, cont}]]
D1.6 Incorporating the Initial Conditions
Our initital conditions are that, at t = 0, ψ = 0 and ψt = f(r). Putting t = 0 in (D.16) gives
0 =
∞∑
n=0
∞∑
m=1
Anm Jn(jnmr/a) exp(inθ).
Differentiating and putting t = 0 in (D.16)
f(r) =
∞∑
n=0
∞∑
m=1
(icjnm/a)Anm Jn(jnmr/a) exp(inθ)
where Anm are complex constants (as above) and it is understood that we take the real part of thesum. The right hand side has no θ-dependence and we can deduce that we need only the componentn = 0 from the first sum. Thus we must find A0m = Am = Rm exp(iǫm), say, such that, taking realparts,
0 =
∞∑
m=1
Rm exp(iǫm) Jn(j0mr/a). (D.17)
f(r) =∞∑
m=1
(icj0m/a)Rm exp(iǫm) Jn(j0mr/a). (D.18)
This can be achieved by setting exp(iǫm) = −i, remembering that Rm is real. The choice −i isdriven by a desire for neatness in (D.18).
Note that we have been a little overgeneral in our presentation here. Right at the start wecould have described the time-dependence by the solution sin(ωt). This is zero but has a non-zerotime derivative at t = 0 and is obviously that which is required for our particular initial conditions.This corresponds to our current choice of a purely imaginary value for exp(iǫm). Similarly we couldhave realised that, as the initial and boundary conditions had no θ-dependence, neither the finalsolution and included only the n = 0 mode from the start. If there was some θ-dependence inthe initial condition, then we would deal with each Fourier component (in θ) separately. We arelooking to represent f(r) as a generalised Fourier Series in r, otherwise known as a Fourier-BesselSeries. This is possible as the differential equation satisfied by the Bessel functions, together withthe boundary conditions - finiteness at r = 0 and 0 at r = a are a Sturm-Liouvuille problem.
D1.7 Orthogonality of Bessel Functions
The set of functions Jn(λr) in 0 ≤ r ≤ a) with λ = j0m/a are eigenfunctions of the Sturm-Liouivilleproblem
r2R′′ + rR′ + (λ2r2 − n2)R = 0, R(0) finite, R(a) = 0 (D.19)
We can rewite this as[rR′]′
+
(
λ2r − n2
r
)
R = 0,
53
with solution R = Jn(λr). Let λ = λi be such that Jn(λia) = 0 and λ be a different value. We have
[r Jn(λir)
′]′ +
(
λ2i r −
n2
r
)
Jn(λir) = 0, (D.20)
[r Jn(λr)′
]′ +
(
λ2r − n2
r
)
Jn(λr) = 0. (D.21)
If we multiply (D.20) by Jn(λr), (D.21) by Jn(λir), subtract and integrate we get∫ a
0Jn(λr)
[r Jn(λir)
′]′ − Jn(λir)
[r Jn(λr)′
]′dr + (λ2
i − λ2)
∫ a
0r Jn(λir) Jn(λr) dr
− n2
∫ a
0(Jn(λir) Jn(λr) − Jn(λr) Jn(λir))︸ ︷︷ ︸
=0
dr
r= 0. (D.22)
Use integration by parts on the first integral to get
[Jn(λr)
{r Jn(λir)
′}− Jn(λir)
{r Jn(λr)′
}]−∫ a
0Jn(λr)′r Jn(λir)
′ − Jn(λir)′r Jn(λr)′
︸ ︷︷ ︸
=0
dr
+(λ2i − λ2)
∫ a
0r Jn(λir) Jn(λr) dr = 0.
Now Jn(λr)′ = λ J′n(λr) so, using the fact that Jn(λia) = 0, we have the result
∫ a
0r Jn(λir) Jn(λr) dr = aλi J′n(λia)
Jn(λa)
λ2 − λ2i
. (D.23)
Thus if λ 6= λi and J(λa) = 0, i.e. λ is another, distinct, eigenvalue, then∫ a
0r Jn(λir) Jn(λr) dr = 0. (D.24)
To see what happens if λ = λi, let λ→ λi and use l’Hopital’s rule.
∫ a
0r Jn(λir)
2 dr = aλi J′
n(λia)
∂∂λ Jn(λa)
∣∣λ=λi
∂∂λ(λ2 − λ2
i )∣∣λ=λi
=a2
2
[J′n(λia)
]2. (D.25)
We can take λ and λi as different roots of J0, j0m. We shall see in the section on the generatingfunction that follows that
Jn±1(x) =n
xJn(x) ∓ J′n(x), =⇒ (J′0)
2 = (J1)2,
and we can easily evaluate the values of J1 at the zeros of J0.
D1.8 The solution
We can now consider (D.18), multiply by r J0(rj0n/a) and integrate between 0 and a. Only oneelement of the sum survives due to the orthogonality and we have
∫ a
0r J(rj0n/a)f(r) dr = Rn(cj0n/a)(a
2/2)[J1(j0n)]2,
54
giving Rn We make the substitution for A0m = Rm exp(iǫm) = −iRm in (D.16) and take the realpart of the result to yield
ψ(r, θ, t) =∞∑
m=1
2∫ a0 r J(rj0m/a)f(r) dr
cj0ma[J1(j0m)]2J0(rj0m/a) sin(cj0mt/a). (D.26)
The animation at http://www.ucl.ac.uk/∼ucahdrb/MATHM242/ illustrates this solution andused the following commands
<< NumericalMath‘BesselZeros‘;f[x_] = Sin[2Pi x]; stop = 10; c = 1;j0m = BesselJZeros[0,
stop]; coef = Map[2NIntegrate[x BesselJ[0, # x] f[x], {x, 0, 1}]/(c# BesselJ[1, #]^2) &,
j0m]; soln[r_, t_] := Tr[coef*Map[BesselJ[0, r#]&, j0m]*Map[Sin[c # t] &, j0m]];
res[t_] := Module[{},wrap[f_, r_]=If[r < 1, f[r, t], 0]; polarr[x_, y_] = Sqrt[x^2 + y^2];
surf = Plot3D[wrap[soln, polarr[x, y]], {x, -1, 1}, {y, -1, 1}, PlotRange -> {-.3, .3},
PlotPoints -> {20, 20},Lighting -> False, Mesh -> False, Axes -> False, Boxed -> False]];
Export["res.gif", Table[res[i], {i, 0, 6, .01}]]
D1.8a Generating Function
There is a generating function for the Bessel functions Jn. It turns out that
n=∞∑
n=−∞
Jn(x)tn = exp
(x
2
(
t− 1
t
))
. (D.27)
This can be derived by considering solutions of Helmholtz’ equation for G(x) in Cartesian and inpolar coordinates.
∇2G+G = 0, Gxx +Gyy +G = 0, r2Grr + rGr +Gθθ + r2G = 0.
One solution is G = exp(iy), corresponding to a plane wave. In polar coordinates, this is G =exp(ir sin θ). We have seen that the solution in polar coordinates can be found by separating outthe angular dependence exp(inθ) leaving the radial dependence satisfying Bessel’s equation. Theplane wave is regular at the origin so we do not need the Yn solutions. We know from section A9that we should be able to write G as follows
G = exp(ir sin θ) =
∞∑
n=−∞
An Jn(r) exp(inθ) (D.28)
for some constants An. If we write t = exp(iθ) so that sin θ = (t− t−1)/2, and replace r by x thenwe have
exp
(x
2
(
t− 1
t
))
=
n=∞∑
n=−∞
AnJn(x)tn. (D.29)
To show that the An are all equal to one we use our knowledge of the expansions of Jn(x) for smallx, obtained from the series solutions. From (D.11)
Jnn(0) = 2−n.
We start by isolating a particular Jm from the sum (D.28) by effectively using the orthogonalityproperties of cos(nθ) and sin(nθ) which lie behind the idea of Fourier Series. We multiply (D.28)by exp(−imθ) and integrate
∫ 2π
0exp(ir sin θ − imθ)dθ =
∞∑
n=−∞
An Jn(r)
∫ 2π
0exp(inθ − imθ)dθ = 2πAm Jm(r)
55
Now differentiate m times with respect to r and put r = 0 to find∫ 2π
0im(sinm θ) exp(ir sin θ) exp(−imθ)dθ = 2πAm Jm
m(r),
=⇒∫ 2π
0im sinm θ exp(−imθ)dθ = 2πAm/2
m.
Using the exponential form for sin θ gives
Am =2m
2π
∫ 2π
0
1
2m
(exp iθ−exp(−iθ)
)mexp(−imθ)dθ =
1
2π
∫ 2π
0(1−exp(−2imθ))mdθ =
1
2π2π = 1,
as the only non-zero contribution to the integral comes from integrating the first term in thebinomial expansion of the integrand.
D1.8b Use of the Generating Function
1. We see that
Gx =1
2
(
t− 1
t
)
G
so that∞∑
n=−∞
tn J′n(x) =1
2
∞∑
n=−∞
tn+1 Jn(x) − 1
2
∞∑
n=−∞
tn−1 Jn(x),
and comparing coefficients of powers of t,
J′n(x) =1
2(Jn−1(x) − Jn+1(x)) . (D.30)
2. If we make the replacement t→ −1/t, we see that
G(x,−1/t) = exp
(1
2
(
−1
t+ t
))
= exp
(1
2
(
t− 1
t
))
= G(x, t),
so that∞∑
n=−∞
Jn(x)(−1)n
tn=
∞∑
n=−∞
tn Jn(x).
However the left hand side is also∑
∞
n=−∞tn J−n(x)(−1)n, using −n, rather than n to take
us through the sum. Thus∞∑
n=−∞
tn J−n(x)(−1)n =∞∑
n=−∞
tn Jn(x),
and comparing coefficients of tn,
Jn(x) = (−1)n J−n(x), (D.31)
so thatJ−1(x) = − J1(x), J−2(x) = J2(x).
Thus from (D.30), with n = 0,J′0(x) = − J1(x). (D.32)
3. If we choose to differentiate G with respect to t, then we can show
2n
xJn(x) = Jn−1(x) + Jn+1(x). (D.33)
56
