special functions of mathematical (geo-)physics · preface these are the lecture notes of the...

Special Functions of Mathematical(Geo-)Physics

Dr. M. Gutting

Universitat Siegen – AG Geomathematik

July 14, 2015

Preface

These are the lecture notes of the lecture about special functions and their applicationsin mathematical (geo-)physics that took place in the summer term 2015 at the universityof Siegen. The contents can be summarized as follows.

The lecture gives an elementary approach to the theory of special functions in mathemat-ical physics with special emphasis on geophysically relevant aspects. The essential topicsof the lecture are in chronological order: the Gamma function, orthogonal polynomials,spherical polynomials (scalar, vectorial, and tensorial case), and Bessel functions. Allfields will be assisted by geophysically relevant applications.The lecture is a good preparation for further activities in the field of geomathematicssuch as “Constructive Approximation”, “Geomathematics”, “Potential Theory”, “InverseProblems”, etc.

3

Contents

1 Introduction 71.1 Example: Gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2 Example: Geomagnetics (Maxwell’s Equations) . . . . . . . . . . . . . . . 9

2 The Gamma Function 13

3 Orthogonal Polynomials 233.1 Properties of Orthogonal Polynomials . . . . . . . . . . . . . . . . . . . . . 303.2 Quadrature Rules and Orthogonal Polynomials . . . . . . . . . . . . . . . 383.3 The Jacobi Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423.4 Ultraspherical Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . 513.5 Application of the Legendre Polynomials in Electrostatics . . . . . . . . . . 593.6 Hermite Polynomials and Applications . . . . . . . . . . . . . . . . . . . . 633.7 Laguerre Polynomials and Applications . . . . . . . . . . . . . . . . . . . . 66

4 Spherical Harmonics 714.1 Basic Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 714.2 Orthogonal Invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 754.3 Polynomials on the Unit Sphere in R3 . . . . . . . . . . . . . . . . . . . . . 77

4.3.1 Homogeneous Polynomials . . . . . . . . . . . . . . . . . . . . . . . 774.3.2 Harmonic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . 834.3.3 Harmonic Polynomials on the Sphere . . . . . . . . . . . . . . . . . 88

4.4 Closure and Completeness of Spherical Harmonics . . . . . . . . . . . . . . 934.5 The Funk-Hecke Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1024.6 Green’s Function with Respect to the Beltrami Operator . . . . . . . . . . 1064.7 The Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

5 Vectorial Spherical Harmonics 1135.1 Notation for Spherical Vector Fields . . . . . . . . . . . . . . . . . . . . . . 1135.2 Definition of Vector Spherical Harmonics . . . . . . . . . . . . . . . . . . . 1185.3 The Helmholtz Decomposition Theorem . . . . . . . . . . . . . . . . . . . 1195.4 Closure and Completeness of Vector Spherical Harmonics . . . . . . . . . . 1225.5 Vectorial Addition Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 1285.6 Vectorial Funk–Hecke Formulas . . . . . . . . . . . . . . . . . . . . . . . . 1345.7 Homogeneous Harmonic Vector Polynomials . . . . . . . . . . . . . . . . . 1375.8 Vectorial Beltrami Operator . . . . . . . . . . . . . . . . . . . . . . . . . . 1415.9 Example: Fluid Flow (Navier–Stokes Equation) . . . . . . . . . . . . . . . 145

5

1 Introduction

The main topics of this lecture can be briefly summarized:

• Finding orthogonal/orthonormal basis systems in (weighted) Hilbert spaces. Theyare directly related to (partial) differential equations and many (geo-)physical ap-plications.

• Green’s functions and corresponding integral formulas which are also related tospecific (P)DEs.

We will consider 1D systems (and some generalizations to Rn) as well as systems on thesphere S2 ⊆ R3 (and some generalizations to Sn ⊆ Rn+1).

1.1 Example: Gravitation

Newton’s law about the mutual attraction of two masses informs us that the attractiveforce, called gravitation, is directed along the line connecting the two centers of the objectsand is proportional to both masses as well as to the squared inverse of the distance betweenthe two objects. In consequence,∫

G%(x)

a− x|a− x|3

dV (x) = ∇a

∫G

%(x)

|a− x|dV (x) (1.1)

is the gravitational field of the Earth (with the interior density distribution %),

V (a) =

∫G

%(x)

|x− a|dV (x) (1.2)

is the so-called gravitational potential. These formulas explain that the line, along whichNewton’s body fell, would be indeed a straight line, directed radially and going exactlythrough the Earth’s center of mass if the Earth had a perfectly spherical shape and themass inside the Earth were distributed homogeneously or rotationally symmetric. Thegravitational field obtained in this way would be perfectly symmetric. In reality, however,the situation is much more complicated. The topographic features of ∂G, mountains andvalleys, are very irregular, leading to a mathematical simplification at global scale. Theactual gravitational field is influenced by strong irregularities in the density % within theEarth G. Spherical signatures of the gravitational field in frequency domain show an ex-ponential smoothing effect in the exterior with increasing distance from the Earth’s body.

We ideally assume that all mass is contained within a sphere of radius R that modelsapproximately the Earth’s surface, i.e., ∂G = S2

R = {x ∈ R3 : |x| = R}. In other words,

7

8 Chapter 1. Introduction

the heterogenous mass is supposed to be contained in the ball B3R = {x ∈ R3 : |x| < R}

of radius R around the origin (where R can be taken as the mean Earth’s radius). Underthese circumstances, we investigate the gravitational field via the Newtonian gravitationalpotential

V (a) =

∫B3R

%(x)

|x− a|dV (x) (1.3)

with the density function % for non-terrestrial data a ∈ R3 \ B3R. From potential theory,

it is well-known that the potential and the density are related by the Poisson equationwhich holds in the ball B3

R, i.e.,

∆xV (x) =

(∂2

∂x21

+∂2

∂x22

+∂2

∂x23

)V (x) = −4π%(x) , x ∈ B3

R. (1.4)

Inserting this formally in (1.3) and using Green’s second theorem, we find that

V (a) =

∫B3R

−1

4π|x− a|∆xV (x)dV (x)

=−∫B3R

V (x)∆x1

4π|x− a|dV (x) (1.5)

−∫S2R

(1

4π|x− a|∂V

∂ν(x)− V (x)

∂

∂νx

1

4π|x− a|

)dS(x),

where ν denotes the surface unit normal pointing to the exterior, i.e., to R3\B3R. Obviously,

∆x1

4π|x−a| = 0 for all x ∈ B3R. Moreover, we can use the following bilinear expansion of

the single pole:

1

|x− a|= 1

R

∞∑n=0

(R|a|

)n+1( |x|R

)n 4π2n+1

n∑k=−n

Yn,k(x|x|

)Yn,k

(a|a|

), (1.6)

where |x| ≤ R < |a| and Yn,k denotes the spherical harmonic of degree n and order k(see chapter on scalar spherical harmonics). Roughly spoken, (1.6) expresses the singlepole as a series expansion in terms of multipoles. It is the key structure for modelingthe gravitational potential with respect to frequencies, i.e., degree n and order k. Theidentity (1.6) gives us the following representation of the potential

V (a) =

∫S2R

V (x) ∂∂νx

1R

∞∑n=0

12n+1

n∑k=−n

( |x|R

)nYn,k

(x|x|

)(R|a|

)n+1Yn,k

(a|a|

)(1.7)

− 1R

∞∑n=0

12n+1

n∑k=−n

( |x|R

)nYn,k

(x|x|

)(R|a|

)n+1Yn,k

(a|a|

)∂V∂ν

(x)dS(x).

Since the surface is a sphere of radius R, we can easily compute the normal derivative,i.e.,

∂∂νx

( |x|R

)nYn,k

(x|x|

)= n

R

( |x|R

)n−1Yn,k

(x|x|

), (1.8)

Chapter 1. Introduction 9

and on the surface S2R we have |x| = R. If we assume V to be sufficiently smooth, we

obtain from (1.7) with the help of (1.8) that

V (a) = 1R

∞∑n=0

n∑k=−n

12n+1

∫S2R

(nRV (x)− ∂V

∂ν(x))Yn,k

(x|x|

)dS(x)

(R|a|

)n+1Yn,k

(a|a|

). (1.9)

Even more, ∫S2R

∂V∂ν

(x)Yn,k(x|x|

)dS(x) = −n+1

R

∫S2R

V (x)Yn,k(x|x|

)dS(x). (1.10)

Finally, this leads us to the expansion

V (a) = 1R

∞∑n=0

n∑k=−n

∫S2R

V (x) 1RYn,k

(x|x|

)dS(x)

(R|a|

)n+1Yn,k

(a|a|

). (1.11)

This shows us how the gravitational potential can be expanded in a series of sphericalharmonics. Moreover, the spheriodization of the Earth implies a frequency dependentdescription of V outside and on the sphere S2

R.

Our work shows that spherical harmonics form a complete orthonormal basis for thesquare-integrable functions on the sphere. Therefore, this basis system plays the samerole in spherical Fourier analysis as the trigonometric polynomials in one dimension. Thisaspect is the essential prerequisite for constructive approximation on the sphere (see, e.g.,[8, 7, 15]). It is applied in many areas, for example, in physical geodesy, in geomagnetics(see below), even spaceborne vectorial or tensorial data can be handled in modern satellitegeodesy, such as satellite-to-satellite tracking or satellite-gravity-gradiometry. Furtherapplications can be found, e.g., in quantum mechanics, in many other geomathematicalproblems or in crystallography using also results of higher dimensions.

1.2 Example: Geomagnetics (Maxwell’s Equations)

The basis of all electromagnetic considerations is the full system of Maxwell’s equationsgiven by

∇x ∧ e(x, t) +∂

∂tb(x, t) = 0

∇x · d(x, t) = Ff (x, t)

∇x ∧ h(x, t)− ∂

∂td(x, t) = jf (x, t)

∇x · b(x, t) = 0 ,

where the unknowns are defined as follows (note that capital letters are used for scalars,lower-case letter for vectors in R3):

d electric displacemente electric fieldh magnetic displacementb magnetic fieldFf density of free chargesjf density of free currents.

10 Chapter 1. Introduction

All quantities are understood as averages over a unit volume in space. The electric andmagnetic displacement, d and h, can be written as

d(x, t) = ε0e(x, t) + p(x, t) (1.12)

h(x, t) =1

µob(x, t)−m(x, t) , (1.13)

where p is the averaged polarization, m is the (averaged) magnetization, ε0 is the permit-tivity of the vacuum and µ0 is the permeability of vacuum.The total charge and current density, respectively, can be written as the sum of the freecharges and currents and the bounded ones, i.e.

F (x, t) = Ff (x, t) + Fb(x, t), j(x, t) = jf (x, t) + jb(x, t)

where it is well-known that

∇x·︸︷︷︸=div

p(x, t) = −Fb(x, t), ∇x · jb(x, t) = − ∂

∂tFb(x, t)

and

∇x∧︸︷︷︸=curl

m(x, t) = − ∂

∂tp(x, t) + jb(x, t) .

We can now reformulate Maxwell’s equations.

∇x · e(x, t)(1.12)=

1

ε0

∇x · (d(x, t)− p(x, t)) =1

ε0

(Ff (x, t) + Fb(x, t))

⇒ ∇x · e(x, t) =1

ε0

F (x, t) (1.14)

∇x ∧ e(x, t) = − ∂

∂tb(x, t) (1.15)

∇x · b(x, t) = 0 (1.16)

∇x ∧ b(x, t)(1.13)= µ0 (∇x ∧ h(x, t) +∇x ∧m(x, t))

= µ0

(jf (x, t) +

∂

∂td(x, t) +∇x ∧m(x, t)

)⇒ ∇x ∧ b(x, t) = µ0

(jf (x, t) +∇x ∧m(x, t) + ε0

∂

∂te(x, t) +

∂

∂tp(x, t)

)(1.17)

In most geomathematical problems this system of equations is too detailed to describe theoccurring phenomena. They have to be reduced as follows. Let L be the typical lengthscale of the discussed geomathematical problem and T be the typical time scale. In mostproblems we have L = 102 km − 103 km and T = hours - days, such that we get for thetypical velocity of the system

L

T� c

Chapter 1. Introduction 11

where c is the speed of light (c = 299 792 458 m/s). Thus, it can be shown, that the termsε0

∂∂te(x, t)+ ∂

∂tp(x, t) can be neglected. Hence, Maxwell’s equations partially decouple and

the resulting equations for the magnetic field are

∇x · b(x, t) = 0

∇x ∧ b(x, t) = µ0 (jf (x, t) +∇x ∧m(x, t)) .

Since div curl = 0 we furthermore can conclude that (by applying ∇· to the secondequation)

∇x · (µ0 (jf (x, t) +∇x ∧m(x, t))) = 0 . (1.18)

For solving this system of equations, data of the magnetic field of the Earth are pri-marily available in the exterior of the Earth, i.e. at the Earth’s surface or at satellitealtitude. Thus, we can assume that the magnetization m of the surrounding medium canbe neglected. We, therefore, arrive at the Pre-Maxwell equations

∇x · b(x, t) = 0 (1.19)

∇x ∧ b(x, t) = µ0jf (x, t) . (1.20)

Furthermore, we have due to (1.18)

∇x · jf (x, t) = 0 .

In earlier concepts, geoscientist assumed that the current density j is also negligible inthe spherical shell S2

(σ1,σ2) = {x ∈ R3 |σ1 < |x| < σ2}, in which the magnetic field data ismeasured. This yields

∇x · b(x, t) = 0, ∇x ∧ b(x, t) = 0 .

Hence, the magnetic field b can be written as the gradient of a scalar potential U , i.e.

b(x, t) = ∇xU(x, t), x ∈ S2(σ1,σ2) ,

where U fulfills the Laplace equation

∆xU(x, t) = 0, x ∈ S2(σ1,σ2) .

This so-called Gauss-representation yields a spherical harmonic expansion of the scalarpotential U which is similar to the modeling of the gravitational field of the Earth.

Modern satellite missions like CHAMP, which is measuring the Earth’s magnetic field, arelocated in the ionosphere, a region where the assumption jf = 0 is not valid. Therefore,we have to deal with the Pre-Maxwell equations

∇x · b(x, t) = 0, ∇x ∧ b(x, t) = µ0jf (x, t), x ∈ S2(σ1,σ2).

A new concept of modeling this situation has to be applied which is the so-called Mie-representation. This also yields the need for basis systems for vector-valued functions onthe sphere, i.e. for vector spherical harmonics.

2 The Gamma Function

In what follows, we introduce the classical Gamma function. Its essential properties willbe explained (for a more detailed discussion the reader is referred, e.g., to [14].For real values x > 0 we consider the integrals

(α)∫ 1

0e−ttx−1dt and

(β)∫∞

1e−ttx−1dt.

In order to show the convergence of (α) we observe that 0 < e−ttx−1 ≤ tx−1 holds true forall t ∈ (0, 1]. Therefore, for ε > 0 sufficiently small, we have∫ 1

ε

e−ttx−1dt ≤∫ 1

ε

tx−1dt =tx

x

∣∣∣∣1ε

=1

x− εx

x.

Consequently, for all x > 0, the integral (α) is convergent.To assure the convergence of (β) we observe that

e−ttx−1 =1∑∞k=0

tk

k!

tx−1 ≤ 1tn

n!

tx−1 =n!

tn−x+1

for all n ∈ N and t ≥ 1. This shows us that∫ A

1

e−ttx−1dt ≤ n!

∫ A

1

1

tn−x+1dt = n!

t−n+x

x− n

∣∣∣∣A1

=n!

x− n

(1

An−x− 1

)provided that A is sufficiently large and n is chosen that n ≥ x + 1. Thus, the integral(β) is convergent.

Lemma 2.1For all x > 0, the integral ∫ ∞

0

e−ttx−1dt

is convergent.

Definition 2.2The function x 7→ Γ(x), x > 0, defined by

Γ(x) =

∫ ∞0

e−ttx−1dt

is called Gamma function.

13

14 Chapter 2. The Gamma Function

Obviously, we have the following properties:

1. Γ is positive for all x > 0,

2. Γ(1) =∫∞

0e−tdt = 1.

We can use integration by parts to obtain for x > 0:

Γ(x+ 1) =

∫ ∞0

e−ttxdt

= −e−ttx∣∣∞0−∫ ∞

0

(−e−t)xtx−1dt

= x

∫ ∞0

e−ttx−1dt

= x Γ(x).

Lemma 2.3The Gamma function Γ satisfies the functional equation

Γ(x+ 1) = xΓ(x), x > 0.

Moreover,

Γ(x+ n) = (x+ n− 1) · · · (x+ 1)xΓ(x) =n∏i=1

(x+ i− 1)Γ(x) for x > 0, n ∈ N

Γ(n+ 1) =n∏i=1

(i)Γ(1) = n! for n ∈ N0.

In other words, the Gamma function can be understood as an extension of factorials.

Lemma 2.4The Gamma function Γ is differentiable for all x > 0 and we have

Γ′(x) =

∫ ∞0

e−t(ln(t))tx−1dt.

Proof. See exercises.An analogous proof can be used to show that Γ is infinitely often differentiable for allx > 0 and

Γ(k)(x) =

∫ ∞0

e−t(ln(t))ktx−1dt, k ∈ N.

Lemma 2.5 (Gauß’ Expression of the Second Logarithmic Derivative)For x > 0

(Γ′(x))2 ≤ Γ(x)Γ′′(x).

Equivalently, we have (d

dx

)2

ln(Γ(x)) =Γ′′(x)

Γ(x)−(

Γ′(x)

Γ(x)

)2

> 0,

i.e. x 7→ ln(Γ(x)), x > 0, is a convex function or Γ is logarithmic convex.

Chapter 2. The Gamma Function 15

Proof.

(Γ′(x))2

=

(∫ ∞0

e−t(ln(t))tx−1dt

)2

=

(∫ ∞0

e−t2 t

x−12 (ln(t))e−

t2 t

x−12 dt

)2

.

The Cauchy-Schwarz inequality yields (note that equality cannot occur since the twofunctions are linearly independent):

(Γ′(x))2 ≤

∫ ∞0

(e−

t2 t

x−12

)2

dt

∫ ∞0

(e−

t2 t

x−12 (ln(t))

)2

dt

=

∫ ∞0

e−ttx−1dt

∫ ∞0

e−ttx−1(ln(t))2dt

= Γ(x) Γ′′(x).

Moreover,

d2

dx2ln(Γ(x)) =

d

dx

Γ′(x)

Γ(x)

=Γ′′(x)Γ(x)− (Γ′(x))2

(Γ(x))2︸︷︷︸>0

=Γ′′(x)

Γ(x)−(

Γ′(x)

Γ(x)

)2

> 0.

Note: ln(Γ(·)) is convex, i.e. for t ∈ (0, 1)

ln (Γ(tx+ (1− t)y)) ≤ t ln(Γ(x)) + (1− t) ln(Γ(y))

= ln(Γt(x)

)+ ln

(Γ1−t(y)

)= ln

(Γt(x) · Γ1−t(y)

)which is equivalent to Γ(tx+ (1− t)y) ≤ Γt(x) · Γ1−t(y) with x, y > 0.

Next we are interested in the behavior of the Gamma function Γ for large positive valuesx.

Theorem 2.6 (Stirling’s Formula)For x > 0 ∣∣∣∣ Γ(x)√

2πxx−1/2e−x− 1

∣∣∣∣ ≤√

2

πx.

Proof. Regard x as fixed and substitute

t = x(1 + s), −1 ≤ s <∞, dt

ds= x.


in the defining integral of the Gamma function. This gives us

Γ(x) =

∫ ∞0

e−ttx−1dt =

∫ ∞−1

e−x−xsxx−1(1 + s)x−1xds = xxe−x∫ ∞−1

(1 + s)x−1e−xsds︸︷︷︸=I(x)

.

Our aim is to verify that I(x) satisfies∣∣∣∣∣I(x)−√

2π

x

∣∣∣∣∣ ≤ 2

x.

Then we have ∣∣∣∣∣ Γ(x)

xxe−x−√

2π

x

∣∣∣∣∣ ≤ 2

x, i.e.

∣∣∣∣ Γ(x)

xx−1/2e−x√

2π− 1

∣∣∣∣ ≤√

2

xπ.

For that purpose we write

(1 + s)xe−xs = exp (−x(s− ln(1 + s))) = e−xu2(s)

where

u(s) =

{|s− ln(1 + s)| 12 , s ∈ [0,∞)

−|s− ln(1 + s)| 12 , s ∈ (−1, 0).

Taylor expansion of u2 for s ∈ (−1,∞) at 0:

u2(s) = u2(0) +d

dsu2(0)s+

d2

ds2u2(ϑs)

s2

2

= 0 +

(1− 1

1 + 0

)s+

1

(1 + ϑs)2

s2

2

Therefore,

u2(s) =s2

2

1

(1 + sϑ)2

with 0 < ϑ < 1. We interpret ϑ as a uniquely defined function of s, i.e., ϑ : s 7→ ϑ(s),such that

u(s)

s=

1√2

1

(1 + sϑ(s))

is a positive continuous function for s ∈ (−1,∞) with the property∣∣∣∣u(s)

s− 1√

2

∣∣∣∣ =

∣∣∣∣ 1√2

(1

1 + sϑ(s)− 1

)∣∣∣∣ =1√2

∣∣∣∣ sϑ(s)

1 + sϑ(s)

∣∣∣∣=

∣∣∣∣sϑ(s)u(s)

s

∣∣∣∣ = |ϑ(s)| · |u(s)| ≤ |u(s)|.

From u2(s) = s− ln(1 + s) follows that

2u du =s

1 + sds.


Obviously, s : u 7→ s(u), u ∈ R is of class C(1)(R) and thus,

I(x) =

∞∫−1

(1 + s)x−1e−xs ds = 2

+∞∫−∞

e−xu2 u

s(u)du.

We are able to deduce that∣∣∣∣I(x)− 2√

2

∫ ∞0

e−xu2

du

∣∣∣∣ =

∣∣∣∣∣∣2∞∫

−∞

e−xu2 u

s(u)du−

√2

+∞∫−∞

e−xu2

du

∣∣∣∣∣∣=

∣∣∣∣∣∣2∞∫

−∞

e−xu2

(u

s(u)− 1√

2

)du

∣∣∣∣∣∣≤ 2

∞∫−∞

e−xu2

∣∣∣∣ u

s(u)− 1√

2

∣∣∣∣ du≤ 2

∞∫−∞

e−xu2 |u| du

= 4

∞∫0

e−xu2

u du.

Note that we have the following integrals (for α, x > 0)∫ ∞0

e−tα

dtu=tα=

1

α

∫ ∞0

e−uu1/α−1du =1

αΓ( 1

α) = Γ(α+1

α),∫ ∞

0

tx−1e−tα

dtu=tα=

1

α

∫ ∞0

e−uux−1α u

1−αα du =

1

αΓ( x

α),∫ ∞

0

tx−1e−αt2

dtu=αt2

=1

2√α

∫ ∞0

e−u1√u

(uα

)x−12du =

1

2αx/2

∫ ∞0

e−uux/2−1du =1

2αx/2Γ(x

2).

Therefore, for x = 1 and α = x:∫ ∞0

e−xu2

du = 12x−1/2Γ

(12

)=

√π

2√x,

for x = 2 and α = x: ∫ ∞0

e−xu2

u du = 12x−1Γ(1) =

1

2x.

This yields: ∣∣∣∣I(x)−√

2 2 12

√π

x

∣∣∣∣ =

∣∣∣∣∣I(x)−√

2π

x

∣∣∣∣∣ ≤ 41

2x=

2

x.

This leads to the desired result.


Remark 2.7Stirling’s formula can be rewritten in the form

limx→∞

Γ(x)√2πxx−1/2e−x

= 1 .

An immediate application is the limit relation

limx→∞

Γ(x+ a)

xaΓ(x)= 1, a > 0. (2.1)

This can be seen from Stirling’s formula by

limx→∞

Γ(x+ a)√

2π(x+ a)x+a− 12 e−x−a

= 1 (2.2)

due to the relation

(x+ a)x+a− 12 = xx+a− 1

2 (1 +a

x)x+a− 1

2

and the limits

limx→∞

(1 + ax)x

ea= 1, lim

x→∞(1 +

a

x)a−

12 = 1.

Lemma 2.8 (Duplication Formula)For x > 0 we have

2x−1Γ(x

2

)Γ

(x+ 1

2

)=√π Γ(x).

Proof. We consider the function x 7→ Φ(x), x > 0, defined by

Φ(x) =2x−1Γ(x

2)Γ(x+1

2)

Γ(x)

for x > 0. Setting x + 1 instead of x we find with the following functional equation forthe numerator

2xΓ

(x+ 1

2

)Γ(x

2+ 1)

= 2x−1xΓ(x

2

)Γ

(x+ 1

2

),

such that the numerator satisfies the same functional equation as the denominator. Thismeans Φ(x+ 1) = Φ(x), x > 0. By repetition we get for all n ∈ N and x fixed Φ(x+n) =Φ(x). We let n tend toward∞. For the numerator of Φ(x+n) we then find by use of theresult in Remark 2.7, i.e. by using twice (2.1) that

limn→∞

2x+n−1Γ(x+n2

)Γ(x+n+12

)

2x+n−1(n2

)x2 (n

2)x+12

(Γ(n

2))2

= 1.


For the denominator we consider

limn→∞

Γ(x+ n)

2x+n−1(n2

)x2 (n

2)x+12

(Γ(n

2))2

= limn→∞

Γ(x+ n)

2x+n−1(n2

)x+ 12 (n

2)n−1e−n2π

(Γ(n2

))2

(n2

)n−1e−n2π

= limn→∞

Γ(x+ n)

2x+n(n2

)n+x− 12 e−nπ

( (Γ(n

2))2

(n2)n−1e−n2π

)−1

= limn→∞

Γ(x+ n)√

2πnn+x− 12 e−n√π

( (Γ(n

2))2

(n2)n−1e−n2π

)−1

=1√π

since Stirling’s formula yields that

limn→∞

Γ(n2)

√2π(n

2)n2−1

2 e−n2

= 1,

i.e.

limn→∞

(Γ(n

2))2

2π(n2)n−1e−n

= 1,

and by the same arguments as in Remark 2.7 (set a in (2.2) to x and x in (2.2) to n) wefind that

limn→∞

Γ(x+ n)√

2πnn+x− 12 e−n

= 1.

We therefore get for every x > 0 and all n ∈ N

Φ(x) = Φ(x+ n) = limn→∞

Φ(x+ n) =√π.

A periodic function with this property must be constant. This proves the lemma.

Thus far, the Gamma function Γ is defined for positive values, i.e., x ∈ R>0 (cf. Figure2.1). We are interested in an extension of Γ to the real line R (or even to the complexplane C) if possible.

Definition 2.9The so-called Pochhammer factorial (x)n with x ∈ R, n ∈ N is defined by

(x)n = x(x+ 1) . . . (x+ n− 1) =n∏i=1

(x+ i− 1).

For x > 0 it is clear that

(x)n =Γ(x+ n)

Γ(x)or

(x)nΓ(x+ n)

=1

Γ(x).


Figure 2.1: The Gamma function on the real line R.

The left hand side is defined for x > −n and gives the same value for all n ∈ N withn > −x. We may use this relation to define 1

Γ(x)for all x ∈ R, and we see that this

function vanishes for x = 0,−1,−2, . . . (cf. Figure 2.1).This leads us to the following conclusion: The Gamma integral is absolutely convergentfor x ∈ C with Re(x) > 0, and represents a holomorphic function for all x ∈ C withRe(x) > 0. Moreover, the Pochhammer factorial (x)n can be defined for all complex x.Therefore we have a definition of 1

Γ(x)for all x ∈ C.

Lemma 2.10The Γ-function is a meromorphic function that has simple poles in 0,−1,−2, . . .. Thereciprocal x 7→ 1

Γ(x), is an entire analytic function.

Lemma 2.11For x ∈ C,

1

Γ(x)= lim

n→∞n−xx

n−1∏k=1

(1 + x

k

).

Proof. The identity(x)nΓ(n)

Γ(n)

Γ(x+ n)=

1

Γ(x)

is valid for all x ∈ C and all n > −Re(x). Furthermore it is easy to see that

(x)nΓ(n)

= x(x+ 1)(x+ 2) . . . (x+ n− 1)

1 · 2 . . . (n− 1)= x

n−1∏k=1

(1 +

x

k

).

Stirling’s formula tells us that for positive x ∈ R

limn→∞

Γ(x+ n)

nxΓ(n)= 1.


and1

Γ(x)= lim

n→∞

Γ(x+ n)

nxΓ(n)Γ(x)= lim

n→∞n−xx

n−1∏k=1

(1 +

x

k

)which proves the lemma for x > 0.

To determine if this limit is also defined for x ≤ 0 we consider once again s − ln(1 + s)with −1 < s <∞:

0 ≤ s− ln(1 + s) =s2

2

1

(1 + ϑs)2, ϑ = ϑ(s) ∈ (0, 1).

Therefore, we can put s = 1k

and estimate the right hand side with its maximum (ϑ = 0):

0 ≤ 1k− ln

(1 + 1

k

)≤ 1

2k2.

This immediately proves that limn→∞

n∑k=1

(1k− ln

(1 + 1

k

))exists and is positive. Moreover,

limn→∞

n∑k=1

(1k− ln

(1 + 1

k

))= lim

n→∞

n∑k=1

(1k− ln(k + 1) + ln(k)

)= lim

n→∞

(n∑k=1

1k− ln(n+ 1)

)

= limn→∞

(n−1∑k=1

1k− ln(n)

)= γ

where γ denotes Euler’s constant

γ = limm→∞

(m−1∑k=1

1

k− lnm

)≈ 0.5772156649 . . . .

Assume now that x ∈ R. If k ≥ 2|x|, then

0 ≤ xk− ln

(1 + x

k

)< x2

k2(2.3)

and

n∏k=1

(1 +

x

k

)=

n∏k=1

(1 +

x

k

)exk e−

xk

=n∏j=1

exj

n∏k=1

(1 +

x

k

)e−

xk .

For k ≥ k0 = b2|x|c we obtain by multiplying (2.3) with −1 and applying the exponentialfunction to it:

1 ≥(1 + x

k

)e−

xk > e−

x2

k2


which shows that

limn→∞

n∏k=1

(1 + x

k

)e−

xk =

∞∏k=1

(1 + x

k

)e−

xk

exists for all x. Moreover,

n∏j=1

exj = exp

(x

n∑j=1

1j

)

= exp

(x

n∑j=1

1j− x ln(n) + x ln(n)

)

= nx exp

(x

n∑j=1

1j− x ln(n)

)= nxeγx.

Therefore, limn→∞

n−xxn∏k=1

(1 + x

k

)exists for all x ∈ R and it holds that

limn→∞

n−xxn∏k=1

(1 + x

k

)= eγxx

∞∏k=1

(1 + x

k

)e−

xk

where the infinite product is also convergent for all x ∈ R. By similar arguments theseresults can be extended for all x ∈ C.The proof of the previous lemma also shows us the following.

Lemma 2.12For x ∈ C,

1

Γ(x)= xeγx

∞∏k=1

(1 +

x

k

)e−

xk .

3 Orthogonal Polynomials

We consider weighted Hilbert spaces of square-integrable functions on intervals in R. Ingeneral, these spaces are denoted by L2

λ([a, b]), where a, b can also be −∞,∞, respectively.They possess the scalar product, for F,G ∈ L2

λ([a, b]),

〈F,G〉dλ =

∫ b

a

F (x)G(x) dλ(x). (3.1)

λ is a nondecreasing function on R which has finite limits as x→ ±∞ and whose inducedpositive measure dλ has finite moments of all orders, i.e.,

µr = µr(dλ) =

∫Rxr dλ(x) <∞ (3.2)

for r ∈ N0 with µ0 > 0. If λ is absolutely continuous, the scalar product becomes

〈F,G〉dλ =

∫ b

a

F (x)G(x)w(x) dx. (3.3)

Here w is a non-negative function which is Lebesgue-measurable and for which∫ baw(x) dx > 0. It is called a weight function.

Definition 3.1Two functions F,G ∈ L2

λ([a, b]) are orthogonal if 〈F,G〉dλ = 0. If additionally ‖F‖dλ =‖G‖dλ = 1, they are called orthonormal.

Example 3.2The trigonometric functions Fn(x) = cos(nx), n ∈ N0, form a system of orthogonalfunctions on the interval (0, π) with the weight function w(x) = 1 since∫ π

0

cos(mx) cos(nx) dx = 0 (3.4)

for n 6= m.

Definition 3.3Let {xk}k∈N0 be a given orthonormal set (either finite or infinite). To an arbitrary real-valued function F let there correspond the formal Fourier expansion

F (x) ∼∞∑k=0

Fkxk(x) (3.5)

23

24 Chapter 3. Orthogonal Polynomials

with the coefficients Fk defined by

Fk = 〈F, xk〉dλ =

∫ b

a

F (x)xk(x) dλ(x) , k ∈ N0. (3.6)

These coefficients are called Fourier coefficients of F with respect to the given orthonormalsystem.

Theorem 3.4Let xk, Fk be as before with F ∈ L2

λ([a, b]). Let l ≥ 0 be a fixed integer and ak ∈ R,k = 0, . . . , l. The integral ∫ b

a

(F (x)−

l∑k=0

akxk

)2

dλ(x) (3.7)

becomes minimal if and only if ak = Fk for k = 0, . . . , l.

Proof. We consider the integral in question with G =∑l

k=0 akxk:

‖F −G‖2dλ = 〈F, F 〉dλ − 2Re(〈F,G〉dλ) + 〈G,G〉dλ (3.8)

= ‖F‖2dλ − 2Re

( l∑k=0

ak〈F, xk〉dλ)

+l∑

k,j=0

akaj 〈xk, xj〉dλ︸︷︷︸δk,j

= ‖F‖2dλ +

l∑k=0

|ak|2 − 2Re

( l∑k=0

akFk

)

= ‖F‖2dλ −

l∑k=0

|Fk|2 +l∑

k=0

|ak − Fk|2

which is minimal if and only if ak = Fk for all k = 0, . . . , l. Therefore, the truncatedFourier series is the best approximating element.

Remark 3.5We also find Bessel’s equality for F , Fk, and xk as in Theorem 3.4:∥∥∥F − l∑

k=0

Fkxk

∥∥∥2

dλ= ‖F‖2

dλ −l∑

k=0

|Fk|2 (3.9)

and Bessel’s inequality (the left hand side in (3.9) is ≥ 0):

l∑k=0

|Fk|2 ≤ ‖F‖2dλ . (3.10)

Lemma 3.6For F , Fk, and xk as in Theorem 3.4 the Fourier series∑

k∈N0

Fkxk (3.11)

converges in the L2λ-norm (in the L2

λ-sense) to an L2λ-function S and (F − S)⊥xk for all

k ∈ N0.

Chapter 3. Orthogonal Polynomials 25

Proof. For n ≥ m holds:∥∥∥ n∑k=m

Fkxk

∥∥∥2

dλ=

n∑k=m

‖Fkxk‖2dλ =

n∑k=m

|Fk|2 (3.12)

due to the Theorem of Pythagoras and the orthonormality of the functions xk.Bessel’s inequality (3.10) gives us the convergence of the series

∑k∈N0|Fk|2. Thus, for all

ε > 0 there exists an m ∈ N, such that for n ≥ m

n∑k=m

|Fk|2 < ε. (3.13)

Therefore, the sequence (Sn)n with

Sn =n∑k=0

Fkxk (3.14)

is a Cauchy sequence with respect to the L2λ-norm. Since L2

λ([a, b]) is complete, thereexists S ∈ L2

λ([a, b]) which is the limit of this sequence, i.e.,

S = limn→∞

Sn =∑k∈N0

Fkxk (3.15)

in the sense of L2λ([a, b]).

It remains to show the orthogonality of S − F : Let j ∈ {0, . . . , l}.

〈S − F, xj〉dλ =⟨∑k∈N0

Fkxk, xj

⟩dλ− 〈F, xj〉dλ =

∑k∈N0

Fk 〈xk, xj〉dλ︸︷︷︸δk,j

−Fj = 0, (3.16)

since for Fn → F , Gn → G holds that 〈Fn, Gn〉dλ → 〈F,G〉dλ. Now, we only have toshow completeness of the system {xk}k. Then, F −S = 0 (in the L2

λ-sense), i.e., the limitof the Fourier series is the function F . Therefore, we consider properties of Fourier seriesin a general inner product space.

Theorem 3.7Let {xk}k∈N ⊂ X be a sequence of orthonormal elements of the inner product space X.Consider the following statements:

(A) The elements xk are closed in X, i.e., for any element x ∈ X and for all ε > 0 thereexist an n ∈ N and coefficients a1, . . . , an ∈ K such that∥∥∥x− n∑

k=1

akxk

∥∥∥X≤ ε . (3.17)

(B) The Fourier series of any y ∈ X converges to y (in the norm of X), i.e.,

limn→∞

∥∥∥y − n∑k=1

〈y, xk〉Xxk∥∥∥X

= 0 . (3.18)


(C) Parseval’s identity holds, i.e., for all y ∈ X,

‖y‖2X = 〈y, y〉X =

∞∑k=1

|〈y, xk〉X |2 . (3.19)

(D) The extended Parseval identity holds, i.e., for all x, y ∈ X,

〈x, y〉X =∞∑k=1

〈x, xk〉X〈xk, y〉X . (3.20)

(E) There is no strictly larger orthonormal system containing the set {xk}k∈N.

(F) {xk}k∈N is complete, i.e., if y ∈ X and 〈y, xk〉X = 0 for all k ∈ N, then y = 0.

(G) An element of X is determined uniquely by its Fourier coefficients, i.e., if 〈x, xk〉X =〈y, xk〉X for all k ∈ N, then x = y.

Then, it holds that

(A)⇐⇒ (B)⇐⇒ (C)⇐⇒ (D) =⇒ (E)⇐⇒ (F )⇐⇒ (G) . (3.21)

If X is a complete inner product space, then also (E) =⇒ (D) and all statements areequivalent.

Proof. Assume (A). Due to the minimizing property of the truncated Fourier series (seeTheorem 3.4 which holds for any orthonormal system {xk}k)∥∥∥y − n∑

k=1

〈y, xk〉Xxk∥∥∥X≤∥∥∥y − n∑

k=1

akxk

∥∥∥X≤ ε , (3.22)

where the last estimate is provided by (A). If on the other hand (B) holds, we canapproximate any y by its truncated Fourier series which shows closure. Thus, (A) ⇐⇒(B).By orthogonality,⟨

x−n∑k=1

〈x, xk〉Xxk, y −n∑k=1

〈y, xk〉Xxk⟩X

= 〈x, y〉X −n∑k=1

〈x, xk〉X〈xk, y〉X . (3.23)

Using the Schwarz inequality,∣∣∣〈x, y〉X − n∑k=1

〈x, xk〉X〈xk, y〉X∣∣∣ ≤ ∥∥∥x− n∑

k=1

〈x, xk〉Xxk∥∥∥X·∥∥∥y − n∑

k=1

〈y, xk〉Xxk∥∥∥X. (3.24)

If (B) holds, the right-hand members both tend to 0, hence (B) =⇒ (D).Selecting x = y in (D) shows (C), i.e., (D) =⇒ (C).Since

0 ≤∥∥∥y − n∑

k=1

〈y, xk〉Xxk∥∥∥2

X= ‖y‖2

X −n∑k=1

|〈y, xk〉X |2 , (3.25)


we see (C) =⇒ (B) and thus, (A)⇐⇒ (B)⇐⇒ (C)⇐⇒ (D).Now, assume (A) and suppose that {xk}k∈N∪{w}, w 6= xk, is also an orthonormal system.This system is also closed in X. Since (A) =⇒ (C)

‖w‖2X =

∞∑k=1

|〈w, xk〉X |2 + |〈w,w〉X |2 , and ‖w‖2X =

∞∑k=1

|〈w, xk〉X |2. (3.26)

Thus, 〈w,w〉X = 0 which contradicts ‖w‖X = 1. This means that (A) =⇒ (E).Suppose there is a 0 6= y ∈ X, such that 〈y, xk〉X = 0 for all k. Then, the set{xk}k∈N ∪ {y/ ‖y‖X} would be a strictly larger orthonormal system than{xk}k∈N. There-fore, (E)⇐⇒ (F ).Suppose that 〈w, xk〉X = 〈y, xk〉X , k ∈ N. Then, 〈w− y, xk〉X = 0, k ∈ N. Assuming (F ),w − y = 0 and (F ) =⇒ (G).If (F ) were false, we could find z 6= 0 with 〈z, xk〉X = 0, k ∈ N. For any y, 〈y, xk〉X =〈y+ z, xk〉X , k ∈ N. So y and y+ z would be two distinct elements with the same Fouriercoefficients. Thus, (G) would be false and we obtain (G) =⇒ (F ). This completes thechain of implications (3.21).Assume now that additionally X is complete. We want to show (G) =⇒ (B) which is themissing implication.Let w ∈ X and consider

Sn =n∑k=1

〈w, xk〉Xxk . (3.27)

For n > m, we find that

‖Sn − Sm‖2X =

n∑k=m+1

|〈w, xk〉X |2 . (3.28)

Bessel’s inequality gives us∞∑k=1

|〈w, xk〉X |2 <∞. (3.29)

Thus, for a given ε > 0, we can find an N(ε) such that the right hand side of (3.28) is lessthan ε for all m,n ≥ N(ε). Thus, (Sn)n is a Cauchy sequence and since X is complete, itconverges to an element S ∈ X.Let l be fixed and n ≥ l.

〈S − Sn, xl〉X = 〈S, xl〉X − 〈Sn, xl〉X = 〈S, xl〉X − 〈w, xl〉X (3.30)

and by the Schwarz inequality

|〈S, xl〉X − 〈w, xl〉X | = |〈S − Sn, xl〉X | ≤ ‖S − Sn‖X ‖xl‖X = ‖S − Sn‖X . (3.31)

Together with the convergence of Sn to S this gives us

〈S, xl〉X = 〈w, xl〉X for l ∈ N. (3.32)


By (G), this implies that S = w, such that the convergence reads as follows

limn→∞

∥∥∥w − n∑k=1

〈w, xk〉Xxk∥∥∥X

= 0 . (3.33)

This is precisely (B).Now we start to consider polynomials.

Definition 3.8The space of real polynomials up to degree n is denoted by Πn. The space of real poly-nomials of all degrees is Π, Πn ⊂ Π for all n ∈ N0. For P,Q ∈ Π we use the scalarproduct

〈P,Q〉dλ =

∫RP (x)Q(x)dλ(x) (3.34)

and the induced norm.

Note that by definition of λ (finite moments) these integrals exist.

Definition 3.9The scalar product (3.34) is called positive definite on Π if ‖P‖dλ > 0 for all P ∈ Π,P 6≡ 0. It is called positive definite on Πn if ‖P‖dλ > 0 for all P ∈ Πn, P 6≡ 0.

Theorem 3.10The scalar product (3.34) is positive definite on Π if and only if

det Mk =

∣∣∣∣∣∣∣∣∣µ0 µ1 · · · µk−1

µ1 µ2 · · · µk...

......

µk−1 µk · · · µ2k−2

∣∣∣∣∣∣∣∣∣ > 0, k ∈ N.

It is positive definite on Πn if and only if det Mk > 0 for k = 1, 2, . . . , n+ 1.

Proof. See tutorials.

Definition 3.11Polynomials whose leading coefficient is 1, i.e. Pk(x) = xk + . . . with k ∈ N0, are calledmonic polynomials. The set of monic polynomials of degree n is denoted by Π◦n.Monic real polynomials Pk, k ∈ N0, are called monic orthogonal polynomials w.r.t. themeasure dλ if

〈Pk, Pl〉dλ = 0 for k 6= l, k, l ∈ N0 and ‖Pk‖dλ > 0 for k ∈ N0.

Normalization yields the orthonormal polynomials Pk(x) = Pk(x)/ ‖Pk‖dλ.

Lemma 3.12Let {Pk}k=0,1,...,n be monic orthogonal polynomials. If Q ∈ Πn satisfies 〈Q,Pk〉 = 0 fork = 0, 1, . . . , n, then Q ≡ 0.


Proof. Write Q as Q(x) =n∑i=0

aixi. Then

0 = 〈Q,Pn〉 =n∑i=0

ai〈xi, Pn〉

and

〈xi, Pn〉 = 〈Pi + Ri−1︸︷︷︸∈Πi−1

, Pn〉 where Pi(x) = xi +Ri−1(x)

= 〈Pi, Pn〉+ 〈Ri−1, Pn〉= δi,n · 〈Pn, Pn〉+ ri−1〈xi−1 +Ri−2, Pn〉= δi,n · 〈Pn, Pn〉+ ri−1〈Pi−1 + Si−2, Pn〉= · · · = δi,n · 〈Pn, Pn〉

This gives us 0 = 〈Q,Pn〉 = an〈Pn, Pn〉. Since 〈Pn, Pn〉 > 0, we obtain an = 0. Similarly,we can show that an−1 = 0, an−2 = 0, . . . , a0 = 0.

Lemma 3.13A set P0, . . . , Pn of monic orthogonal polynomials is linearly independent. Moreover, any

polynomial P ∈ Πn can be uniquely represented in the form P =n∑k=0

ckPk for some real

coefficients ck, i.e. P0, . . . , Pn is a basis of Πn.

Proof. Ifn∑k=0

γkPk ≡ 0, taking the scalar product on both sides of the equation with Pj,

j = 0, . . . , n, yields that γj = 0. This gives us linear independence.

If we take the scalar product with Pj on both sides of P =n∑k=0

ckPk we find that

〈P, Pj〉 =n∑k=0

ck〈Pk, Pj〉 = cj〈Pj, Pj〉 , j = 0, . . . , n.

The difference P −∑n

k=0 ckPk is orthogonal to P0, . . . , Pn and by the previous lemma hasto be identically zero.

Theorem 3.14If the scalar product 〈·, ·〉dλ is positive definite on Π, there exists a unique infinite sequence{Pk}k∈N0 of monic orthogonal polynomials.

Proof. The polynomials Pk can be constructed by applying Gram-Schmidt orthogonal-ization to the sequence of powers {Ek}k∈N0 where Ek(x) = xk. Therefore, we chooseP0 = 1 and for k ∈ N we use the recursion

Pk = Ek −k−1∑l=0

clPl , cl =〈Ek, Pl〉〈Pl, Pl〉

. (3.35)


Since the scalar product is positive definite, 〈Pl, Pl〉 > 0. Thus, the monic polynomial Pkis uniquely defined and by construction orthogonal to all Pj, j < k.The prerequisites of this theorem are fulfilled if λ has many points of increase, i.e. pointst0 such that λ(t0 + ε) > λ(t0 − ε) for all ε > 0. The set of all points of increase of λ iscalled support of the measure dλ, its convex hull is the support interval of dλ.

Theorem 3.15If the scalar product 〈·, ·〉dλ is positive definite on Πn, but not on Πm for all m > n, thereexist only n+ 1 orthogonal polynomials P0, . . . , Pn.


Theorem 3.16If the moments of dλ exist only for r = 0, 1, . . . , r0, there exist only n + 1 orthogonalpolynomials P0, . . . , Pn, where n = br0/2c.

Proof. The Gram-Schmidt procedure can be carried out as long as the scalar productsin (3.35) including 〈Pk, Pk〉 exist. This is the case as long as 2k ≤ r0, i.e. k ≤ n = br0/2c.

3.1 Properties of Orthogonal Polynomials

For this section we assume that dλ is a positive measure on R with an infinite number ofpoints of increase and with finite moments of all orders.

Definition 3.17An absolutely continuous measure dλ(x) = w(x)dx is symmetric (with respect to theorigin) if its support interval is [−a, a] with 0 < a ≤ ∞ and w(−x) = w(x) for all x ∈ R.

Theorem 3.18If dλ is symmetric, then

Pk(−x) = (−1)kPk(x) , k ∈ N0.

Thus, Pk is either an even or an odd polynomial depending on the degree k.

Proof. Set Pk(x) = (−1)kPk(−x). We compute for k 6= l

〈Pk, Pl〉dλ = (−1)k+l

∫RPk(−x)Pl(−x)dλ(x)

= (−1)k+l

∫RPk(x)Pl(x)dλ(−x)

= (−1)k+l

∫RPk(x)Pl(x)dλ(x)

= (−1)k+l〈Pk, Pl〉dλ = 0.

Since all Pk are monic, Pk(x) = Pk(x) by the uniqueness of monic orthogonal polynomials.


Theorem 3.19Let dλ be symmetric on [−a, a], 0 < a ≤ ∞, and

P2k(x) = P+k (x2) P2k+1(x) = xP−k (x2).

Then {P±k } are the monic orthogonal polynomials with respect to the measure dλ±(x) =

x∓12w(x

12 )dx on [0, a2].

Proof. We only prove the assertion for P+k , the other case follows analogously.

Obviously, the polynomials P+k are monic. By symmetry holds that

0 = 〈P2k, P2l〉dλ = 2

∫ a

0

P2k(x)P2l(x)w(x)dx

for k 6= l. Therefore, we obtain for k 6= l

0 = 2

∫ a

0

P+k (x2)P+

l (x2)w(x)dx =

∫ a2

0

P+k (t)P+

l (t)t−12w(t

12 )dt.

Theorem 3.20All zeros of Pk, k ∈ N, are real, simple, and located in the interior of the support interval[a, b] of dλ.

Proof. Since∫R Pk(x)dλ(x) = 0 for k ≥ 1, there must exist at least one point in the

interior of [a, b] at which Pk changes sign. Let x1, x2, . . . , xn, n ≤ k, be all these points.If we had n < k, then due to orthogonality∫

RPk(x)

n∏j=1

(x− xj)dλ(x) = 0,

which is impossible since the integrand does no longer change sign. Therefore, we obtainthat n = k.

Theorem 3.21The zeros of Pk+1 alternate with those of Pk, i.e.

x(k+1)k+1 < x

(k)k < x

(k+1)k < x

(k)k−1 < · · · < x

(k)1 < x

(k+1)1 ,

where x(k+1)j , x

(k)i are the zeros in descending order of Pk+1 and Pk, respectively.

Proof. Later (uses the Christoffel-Darboux formula which we will get very soon).

Theorem 3.22In any open interval (c, d) in which dλ ≡ 0 there can be at most one zero of Pk.


Proof. We perform a proof by contradiction. Suppose there are two zeros x(k)i 6= x

(k)j in

(c, d). Let all the other zeros (within (c, d) or not) be denoted by x(k)n . Then holds∫

RPk(x)

∏n6=i,j

(x− x(k)n )dλ(x) =

∫R

∏n6=i,j

(x− x(k)n )2(x− x(k)

j )(x− x(k)i )dλ(x) > 0 ,

since the integrand is non-negative outside of (c, d). This is a contradiction to the orthog-

onality of Pk to polynomials of lower degree such as∏n6=i,j

(x− x(k)n ).

Theorem 3.23For any monic polynomial P ∈ Π◦n holds∫

RP 2(x)dλ(x) ≥

∫RP 2n(x)dλ(x)

where equality is only achieved for P = Pn. In other words, Pn minimizes the integral onthe left hand side above over all P ∈ Π◦n, i.e.

minP∈Π◦n

∫RP 2(x)dλ(x) =

∫RP 2n(x)dλ(x).

Proof. Due to Lemma 3.13 the polynomial P can be represented as follows

P (x) = Pn(x) +n−1∑k=0

ckPk(x).

Therefore, ∫RP 2(x)dλ(x) =

∫RP 2n(x)dλ(x) +

n−1∑k=0

c2k

∫RP 2k (x)dλ(x).

This shows the desired inequality. Equality holds if and only if c0 = c1 = . . . = cn−1 = 0,i.e. for P = Pn.

Remark 3.24If we consider the function

Φ(a0, a1, . . . , an−1) =

∫R

(xn +

n−1∑k=0

akxk

)2

dλ(x) ,

we can compute the partial derivatives and set them equal to zero which yields∫RP (x)xkdλ(x) = 0 , k = 0, 1, . . . , n− 1.

These are exactly the conditions of orthogonality that Pn has to satisfy.Furthermore, the Hessian of Φ is 2Mn of Theorem 3.10 which is positive definite. Thisconfirms that Pn gives us a minimum.


Theorem 3.25Let 1 < p <∞. Then, the extremal problem of determining

minP∈Π◦n

∫R|P (x)|pdλ(x)

possesses the unique solution Pn.

Proof. The search for the desired minimum is equivalent to the problem of best-approximation of xn by polynomials of degree ≤ n− 1 in the Lp-norm.The problem of best approximation in normed spaces is uniquely solvable if the space isstrictly convex, i.e. if from ‖x‖ = ‖y‖ = 1 and x 6= y we can conclude that ‖x+ y‖ < 2.A normed space X is strictly convex if and only if ‖x+ y‖ = ‖x‖+ ‖y‖ yields x = αy ory = αx with an α ≥ 0. The Minkowski inequality guarantees this for the Lp-norm.

For further details see e.g. [1, 5, 13, 18, 19, 23] or other books on functional analysis.

Orthogonal polynomials fulfill a three-term recurrence relation which can be used for:

• generating values of the polynomials and their derivatives,

• computation of the zeros as eigenvalues of a symmetric tridiagonal matrix via therecursion coefficients,

• normalization of the orthogonal polynomials,

• efficient evaluation of expansions in orthogonal polynomials.

The reason for the existence of these three-term recurrences is the shift property of thescalar product, i.e.

〈xU, V 〉dλ = 〈U, xV 〉dλ ∀ U, V ∈ Π.

Note that there are other scalar products that do not possess this property (even thoughthey are positive definite).

Theorem 3.26Let Pk, k ∈ N0, be the monic orthogonal polynomials w.r.t. the measure dλ (see Definition3.11). Then,

P−1(x) = 0, P0(x) = 1, Pk+1(x) = (x− αk)Pk(x)− βkPk−1(x), k ∈ N0, (3.36)

where

αk =〈xPk, Pk〉dλ〈Pk, Pk〉dλ

, k ∈ N0,

βk =〈Pk, Pk〉dλ〈Pk−1, Pk−1〉dλ

, k ∈ N.

The index range is infinite (k ≤ ∞) if the scalar product is positive definite on Π. Itis finite (k ≤ d − 1) if the scalar product is positive definite on Πd, but not on Πn withn > d.


Proof. Since Pk+1 − xPk is a polynomial of degree ≤ k, we can write

Pk+1(x)− xPk(x) = −αkPk(x)− βkPk−1(x) +k−2∑j=0

γk,jPj(x)

for certain constants αk, βk, γk,j, where P−1(x) = 0 and empty sums are also zero.We take the scalar product with Pk on both sides which gives us (using orthogonality):

〈Pk+1, Pk〉 − 〈xPk, Pk〉 = −αk〈Pk, Pk〉 − βk〈Pk−1, Pk〉+k−2∑j=0

γk,j〈Pj, Pk〉

=⇒ −〈xPk, Pk〉 = −αk〈Pk, Pk〉

=⇒ αk =〈xPk, Pk〉〈Pk, Pk〉

This proves the relation for αk. For βk we need to take the scalar product with Pk−1

where k ≥ 1:

〈Pk+1, Pk−1〉 − 〈xPk, Pk−1〉 = −αk〈Pk, Pk−1〉 − βk〈Pk−1, Pk−1〉+k−2∑j=0

γk,j〈Pj, Pk−1〉

=⇒ −〈xPk, Pk−1〉 = −βk〈Pk−1, Pk−1〉

Since 〈xPk, Pk−1〉 = 〈Pk, xPk−1〉 = 〈Pk, Pk + Rk−1〉 with Rk−1 ∈ Πk−1, we find that〈xPk, Pk−1〉 = 〈Pk, Pk〉 which provides us with

βk =〈Pk, Pk〉〈Pk−1, Pk−1〉

.

As a last step we take the scalar product on both sides with Pi, i < k − 1, and obtain

〈Pk+1, Pi〉 − 〈xPk, Pi〉 = −αk〈Pk, Pi〉 − βk〈Pk−1, Pi〉+k−2∑j=0

γk,j〈Pj, Pi〉

=⇒ −〈xPk, Pi〉 = γk,i〈Pi, Pi〉

Here we make use of the shift property of the scalar product, i.e. 〈xPk, Pi〉 = 〈Pk, xPi〉 = 0since xPi ∈ Πk−1. Therefore, γk,i = 0 for i < k − 1 which finally proves (3.36).

If the index range in Theorem 3.26 is finite (k ≤ d − 1), the relations for αd and βd stillmake sense, βd > 0, but the polynomial Pd+1 that results from (3.36) has norm ‖Pd+1‖ = 0(see also Theorem 3.15 and the exercises).

Remark 3.27Although β0 in (3.36) can be arbitrary since it is multiplied with P−1 ≡ 0, we define itfor later purposes as

β0 = 〈P0, P0〉 =

∫Rdλ(x) .

Note that βk > 0 for all k ∈ N0 and for n ∈ N0

‖Pn‖2 = βnβn−1 . . . β1β0.


Remark 3.28There is a converse result to Theorem 3.26 saying that any sequence of polynomials whichsatisfy a three-term recurrence relation of the form (3.36) with all βk positive is orthogonalw.r.t. a positive measure with infinite support.

Theorem 3.29Let the support interval [a, b] of dλ be finite. Then,

a ≤ αk ≤ b , k ∈ N0,

0 < βk ≤ max{a2, b2} , k ∈ N,

where the index range is k ≤ ∞ or k ≤ d (d as in Theorem 3.26), respectively.

Proof. Since for x in the support of dλ we know that a ≤ x ≤ b, the definition of αk inTheorem 3.26 immediately yields the desired estimates.By definition 0 < βk and it remains show the upper bound. We notice that

‖Pk‖2 = 〈Pk, Pk〉 = |〈xPk−1, Pk〉|

and apply the Cauchy-Schwarz inequality to obtain

‖Pk‖2 ≤ max{|a|, |b|} ‖Pk−1‖ ‖Pk‖ .

Therefore,

‖Pk‖‖Pk−1‖

≤ max{|a|, |b|} and βk =‖Pk‖2

‖Pk−1‖2 ≤ max{a2, b2}.

Definition 3.30If the index range in Theorem 3.26 is infinte, the Jacobi matrix associated with themeasure dλ is the infinite, symmetric, tridiagonal matrix

J∞ =

α0

√β1 0√

β1 α1

√β2√

β2 α2

√β3

. . . . . . . . .

0

.

Its leading principal minor matrix of size n× n is denoted by

Jn = [J∞][1:n,1:n] =

α0

√β1 0√

β1 α1

√β2√

β2 α2

√β3

. . . . . . . . .√βn−2 αn−2

√βn−1

0√βn−1 αn−1

.

If the index range in Theorem 3.26 is finite (k ≤ d − 1), then Jn is well-defined for0 ≤ n ≤ d.


Theorem 3.31The zeros x

(n)i of Pn (the orthonormal version Pn) are the eigenvalues of the Jacobi matrix

Jn of order n. The corresponding eigenvectors are given by p(x

(n)i

)where

p(x) =[P0(x), P1(x), . . . , Pn−1(x)

]T.

Proof. We know from the tutorials that the following system of equations holds

xp(x) = JnP(x) +√βnpn(x)en ,

where en = [0, 0, . . . , 1]T is the n-th unit vector in Rn.

If we put x = x(n)i in this equation, the second summand on the right hand side drops

out. Since the first component of the vector p(x(n)i ) is 1/

√β0, the vector cannot be 0 and

is indeed the eigenvector to the eigenvalue x(n)i .

Theorem 3.32 (Christoffel-Darboux Formula)Let Pk denote the orthonormal polynomials with respect to the measure dλ. Then,

n∑k=0

Pk(x)Pk(t) =√βn+1

Pn+1(x)Pn(t)− Pn(x)Pn+1(t)

x− t

andn∑k=0

(Pk(x)

)2

=√βn+1

(P ′n+1(x)Pn(x)− P ′n(x)Pn+1(x)

).

Proof. We use the recurrence relation for the orthonormal polynomials (see exercises),i.e. √

βk+1Pk+1(t) = (t− αk)Pk(t)−√βkPk−1(t) , k ∈ N0, P−1(t) = 0, P0(t) = 1√

β0,

and multiply this by Pk(x) to obtain:√βk+1Pk+1(t)Pk(x) = (t− αk)Pk(t)Pk(x)−

√βkPk−1(t)Pk(x).

Now we interchange the roles of t and x and subtract the first relation, i.e.√βk+1Pk+1(x)Pk(t)−

√βk+1Pk+1(t)Pk(x)

= (x− αk)Pk(x)Pk(t)−√βkPk−1(x)Pk(t)−

((t− αk)Pk(t)Pk(x)−

√βkPk−1(t)Pk(x)

)= (x− t)Pk(x)Pk(t)−

√βk

(Pk−1(x)Pk(t)− Pk−1(t)Pk(x)

).

Therefore,

(x− t)Pk(x)Pk(t) =√βk+1

(Pk+1(x)Pk(t)− Pk+1(t)Pk(x)

)−√βk

(Pk−1(t)Pk(x)− Pk−1(x)Pk(t)

).


Now we sum up both sides from k = 0 to k = n and make use of the teleskop sum on theright (note that P−1 = 0):

n∑k=0

Pk(x)Pk(t) =√βn+1


x− t.

For the second part we take the limit t→ x on both sides and on the right hand side wefind


x− t= Pn+1(x)

Pn(t)− Pn(x)

x− t+ Pn+1(x)

Pn(x)

x− t

− Pn(x)Pn+1(t)− Pn+1(x)

x− t− Pn(x)

Pn+1(x)

x− t

= Pn+1(x)Pn(t)− Pn(x)

x− t− Pn(x)

Pn+1(t)− Pn+1(x)

x− tt→x−−→ Pn+1(x)

(−P ′n(x)

)− Pn(x)

(−P ′n+1(x)

)= P ′n+1(x)Pn(x)− P ′n(x)Pn+1(x).

Corollary 3.33Let Pk denote the monic orthogonal polynomials with respect to the measure dλ. Then,

n∑k=0

(n∏

i=k+1

βi

)Pk(x)Pk(t) =

n∑k=0

βn · βn−1 · . . . · βk+1Pk(x)Pk(t)

=Pn+1(x)Pn(t)− Pn(x)Pn+1(t)

x− t.

Proof. Put Pk = Pk/ ‖Pk‖ in the first formula of Theorem 3.32 and use from Theorem3.26 that

√βn+1 = ‖Pn+1‖ / ‖Pn‖ together with

‖Pn‖2 = βnβn−1 . . . β1β0 =n∏i=0

βi.

This provides us with

n∑k=0

Pk(x)Pk(t) =n∑k=0

1

‖Pk‖2Pk(x)Pk(t)

=‖Pn+1‖‖Pn‖


‖Pn+1‖ ‖Pn‖ (x− t)n∑k=0

‖Pn‖2

‖Pk‖2Pk(x)Pk(t) =Pn+1(x)Pn(t)− Pn(x)Pn+1(t)

x− tn∑k=0

(n∏

i=k+1

βi

)Pk(x)Pk(t) =


x− t


Remark 3.34From the second part of Theorem 3.32 we obtain the inequality

P ′n+1(x)Pn(x)− P ′n(x)Pn+1(x) > 0.

This can be used to prove Theorem 3.21 the following way.Let τ and σ be consecutive zeros of Pn, such that P ′n(τ)P ′n(σ) < 0 (which holds since allzeros are simple). Then, the inequality above tells us that

−P ′n(τ)Pn+1(τ) > 0 and − P ′n(σ)Pn+1(σ) > 0.

This implies that Pn+1 has opposite signs at τ and σ. Therefore, there is at least one zeroof Pn+1 between τ and σ. In this way we find at least n− 1 zeros of Pn+1.

For the largest zero of Pn, i.e. for τ(n)1 , holds P ′n(τ

(n)1 ) > 0 and by the inequality above

Pn+1(τ(n)1 ) < 0. The polynomial Pn+1 has another zero at the right of τ

(n)1 since Pn+1(x) >

0 for x sufficiently large.A similar argument holds for the smallest zero of Pn, i.e. for τ

(n)n . This proves Theorem

3.21.

3.2 Quadrature Rules and Orthogonal Polynomials

Let in this section dλ be a measure with bounded or unbounded support, positive definiteor not. An n-point quadrature rule for dλ is a formula of the type∫

RF (x)dλ(x) =

n∑j=1

wjF (xj) +Rn(F ) (3.37)

The mutually distinct points xj are called nodes, the numbers wj are the weights of thequadrature rule. Rn(F ) is the remainder or error term.

Definition 3.35The quadrature rule (3.37) possesses degree of exactness d if Rn(P ) = 0 for all P ∈ Πd.It has precise degree of exactness d if it has degree of exactness d, but not d+ 1.A quadrature rule (3.37) with degree of exactness d = n− 1 is called interpolatory.

Regarding interpolatory quadrature rules

A quadrature rule is interpolatory if and only if it is obtained by integration from theLagrange interpolation, i.e. from

F (x) =n∑j=1

F (xj)Lj(x) + rn−1(F ;x)

where Lj(x) =n∏i=1i 6=j

x−xixj−xi . Thus, we obtain

wj =

∫RLj(x)dλ(x) , j = 1, 2, . . . , n and Rn(F ) =

∫Rrn−1(F ;x)dλ(x) .


It is well-known that for P ∈ Πn−1 holds that rn−1(P ;x) ≡ 0, i.e. Rn(P ) = 0 andd = n− 1.Given any n distinct nodes an interpolatory quadrature can always be constructed.

Theorem 3.36Let 0 ≤ k ≤ n be an integer. The quadrature rule (3.37) has degree of exactness d =n− 1 + k if and only if the following two conditions are both fulfilled:

(i) (3.37) is interpolatory,

(ii) The node polynomial ωn(x) =n∏j=1

(x− xj) satisfies

∫Rωn(x)P (x)dλ(x) = 0 for all P ∈ Πk−1.

Proof. First, let the degree of exactness be d = n − 1 + k. Obviously, the quadraturerule is interpolatory. For P ∈ Πk−1 we know that ωnP ∈ Πn+k−1. Therefore,∫

Rωn(x)P (x) dλ(x) =

n∑j=1

wjωn(xj)P (xj) = 0 (3.38)

by definition of ωn.Let now the conditions (i) and (ii) be fulfilled. Let P ∈ Πn+k−1. We have to prove thatthe remainder term Rn(P ) = 0. Polynomial long division by the node polynomial ωnyields that P = Qωn +R, where Q ∈ Πk−1 and R ∈ Πn−1. Then,∫

RP (x) dλ(x) =

∫RQ(x)ωn(x) dλ(x) +

∫RR(x) dλ(x) =

n∑j=1

wjR(xj) , (3.39)

since the first integral vanishes due to (ii) and the second integral is evaluated exactly bythe quadrature rule due to (i). However,

R(xj) = P (xj)−Q(xj)ωn(xj) = P (xj). (3.40)

Therefore, ∫RP (x) dλ(x) =

n∑j=1

wjP (xj) , (3.41)

i.e., Rn(P ) = 0.

Remark 3.37If dλ is positive definite, then k = n is optimal. k = n+ 1 requires that ωn is orthogonalto all elements of Πn, i.e. also to itself, which is not possible.


Gauss quadratures

Definition 3.38The quadrature rule (3.37) with k = n is called Gauss quadrature rule with respect to themeasure dλ. Its degree of exactness is d = 2n− 1.

Remark 3.39The second condition in Theorem 3.36 shows that for a Gauss quadrature (i.e. k = n) wehave ωn = Pn. Therefore, the nodes xGj are the zeros of the n-th orthogonal polynomialwith respect to dλ. The weights wGj can be found by interpolation.Note that for the Gauss quadratures we assume that dλ is positive definite.

Theorem 3.40All nodes xGj of the Gauss quadrature rule are mutually distinct and contained in theinterior of the support interval [a, b] of dλ. All weights wGj are positive.

Proof. Since the nodes xGj are the zeros of Pn, the first part follows from Theorem 3.20.Now we consider the weights:

0 <

∫RL2i (x)dλ(x) =

n∑j=1

wGj L2i (x

Gj ) = wGi

for i = 1, 2, . . . , n since L2i (·) ∈ Π2n−2 ⊂ Π2n−1.

Theorem 3.41Let wGj be the weights and xGj be the nodes of a Gauss quadrature rule. Then

n∑j=1

wGj F (xGj ) =

∫RP2n−1(F ; t)dλ(x)

where P2n−1(F ; ·) is the Hermite interpolation polynomial of degree 2n− 1 which satisfiesthe equations

P2n−1(F ;xGj ) = F (xGj ) , P ′2n−1(F ;xGj ) = F ′(xGj ) for j = 1, 2, . . . , n.

Proof. The Hermite interpolation polynomial is given by

P2n−1(F ;x) =n∑j=1

(Hj(x)F (xGj ) +Kj(x)F ′(xGj )

), (3.42)

where

Hj(x) =(1− 2(x− xGj )L′j(x

Gj ))L2j(x), (3.43)

Kj(x) =(x− xGj )L2j(x), (3.44)

Lj(x) =n∏i=1i 6=j

x− xGixGj − xGi

. (3.45)


Note that Hj(xGi ) = δj,i, H

′j(x

Gi ) = 0, as well as Kj(x

Gi ) = 0, K ′j(x

Gi ) = δj,i. Obviously,

Hj, Kj ∈ Π2n−1. Hence,∫RP2n−1(F ;x) dλ(x) =

n∑j=1

(∫RHj(x) dλ(x)F (xGj ) +

∫RKj(x) dλ(x)F ′(xGj )

). (3.46)

Moreover, ∫RHj(x) dλ(x) =

n∑i=1

wi(1− 2(xGi − xGj )L′j(x

Gj ))L2j(x

Gi )

=n∑i=1

wi(1− 2(xGi − xGj )L′j(x

Gj ))δj,i = wj, (3.47)

and ∫RKj(x) dλ(x) =

n∑i=1

wi(xGi − xGj )L2

j(xGi ) =

n∑i=1

wi(xGi − xGj )δj,i = 0 . (3.48)

This finally gives us the desired result.

Corollary 3.42If F ∈ C(2n)([a, b]) and [a, b] is the support interval of dλ, then the remainder term of theGauss quadrature formula can be expressed as

Rn(F ) =F (2n)(x)

(2n)!

∫R

(Pn(x))2 dλ(x)

with a x ∈ (a, b).

Proof. We know from the theory of interpolation (cf. e.g. [5]) that

F (x) = P2n−1(F ;x) + r2n−1(F ;x) = P2n−1(F ;x) +F (2n)(τ(x))

(2n)!

n∏j=1

(x− xj)2

with x ∈ (a, b). This yields for the remainder of the numerical integration that

Rn(F ) =

∫Rr2n−1(F ;x)dλ(x)

=

∫R

F (2n)(τ(x))

(2n)!

n∏j=1

(x− xj)2dλ(x).

The mean value theorem of integration completes the proof.

Theorem 3.43The first n orthogonal polynomials Pk, k = 0, . . . , n − 1, are discrete orthogonal in thesense

n∑j=1

wGj Pk(xGj )Pl(x

Gj ) = δk,l ‖Pk‖2 , k, l = 0, 1, . . . , n− 1,

where xGj , wGj are the nodes and weights of the n-point Gauss quadrature formula.


Proof. This follows directly since the degree of exactness of the n-point Gauss quadratureis 2n − 1 and the product of Pk and Pl with k, l ∈ {0, 1, . . . , n − 1} is a polynomial ofdegree ≤ 2n− 2.

Remark 3.44If a > −∞, b ≤ ∞, it can be desired to have x0 = a. To achieve this we replace n byn + 1 in (3.37) and write ωn+1(x) = (x − a)ωn(x). The optimal formula (see Theorem3.36 requires ωn to satisfy∫

Rωn(x)P (x)(x− a)dλ(x) = 0 , ∀P ∈ Πn−1.

Therefore, ωn(x) = Pn(x; dλa) with dλa(x) = (x− a)dλ(x).The remaining n nodes have to be zeros of Pn(·; dλa). The resulting rule is called Gauss-Radau rule.If also b < ∞ and a as well as b are both desired as nodes, we find the (n + 2)-pointGauss-Lobatto rule similarly with the measure dλa,b(x) = (x− a)(b− x)dλ(x).These rules have the degrees of exactness equal to 2n and 2n+ 1, respectively.

3.3 The Jacobi Polynomials

Definition 3.45Let dλ(x) = w(x)dx with w(x) = (1− x)α(1 + x)β, α, β > −1, with the support interval

[−1, 1]. The corresponding orthogonal polynomials are the Jacobi polynomials P(α,β)n

which are normalized by the condition

P (α,β)n (1) =

(n+ α

n

)=

Γ(n+ α + 1)

Γ(n+ 1)Γ(α + 1).

(Note that we generalize the notation of the binomial coefficients naturally via the Gammafunction.)

Remark 3.46Similar to Theorem 3.18 we find the identity

P (α,β)n (x) = (−1)nP (β,α)

n (−x)

and so we obtain

P (α,β)n (−1) = (−1)n

(n+ β

n

).


Theorem 3.47For α > −1 hold

P(α,α)2n (x) =

Γ(2n+ α + 1)Γ(n+ 1)

Γ(n+ α + 1)Γ(2n+ 1)P

(α,−12

)n (2x2 − 1)

= (−1)nΓ(2n+ α + 1)Γ(n+ 1)

Γ(n+ α + 1)Γ(2n+ 1)P

(−12,α)

n (1− 2x2) ,

P(α,α)2n+1 (x) =

Γ(2n+ α + 2)Γ(n+ 1)

Γ(n+ α + 1)Γ(2n+ 2)xP

(α,12

)n (2x2 − 1)

= (−1)nΓ(2n+ α + 2)Γ(n+ 1)

Γ(n+ α + 1)Γ(2n+ 2)xP

(12,α)

n (1− 2x2) .

Proof. We consider the first relation and use the notation dλ1(x) = (1−x)α(1+x)αdx =

(1− x2)αdx and dλ2(x) = (1− x)α(1 + x)−12dx. It suffices to prove that

I =

∫RP

(α,−12

)n (2x2 − 1)P (x)dλ1(x) =

1∫−1

P(α,−1

2)

n (2x2 − 1)P (x)(1− x2)αdx = 0,

where P ∈ Π2n−1.If P is an odd polynomial, this is fulfilled. Therefore, let P be even, i.e. P (x) = R(x2)with R ∈ Πn−1. Then

I =

1∫−1

P(α,−1

2)

n (2x2 − 1)R(x2)(1− x2)αdx

= 2

1∫0

P(α,−1

2)

n (2x2 − 1)R(x2)(1− x2)αdx

=

1∫0

P(α,−1

2)

n (2x− 1)R(x)(1− x)αx−12dx

= 2−α−12

1∫−1

P(α,−1

2)

n (x)R(x+12

)(1− x)α(1 + x)−12dx

= 2−α−12

∫R

P(α,−1

2)

n (x)R(x+12

)dλ2(x) = 0.

A similar argument can be used to prove the second relation of the theorem.

Remark 3.48 (Special Cases)(a) For α = β we are dealing with the special case of ultraspherical polynomials (or

Gegenbauer polynomials) which due to the previous theorem are even or odd poly-


nomials (depending on n being even or odd).

C(λ)n (x) =

Γ(α + 1)Γ(n+ 2α + 1)

Γ(2α + 1)Γ(n+ α + 1)P (α,α)n (x)

=Γ(λ+ 1

2)Γ(n+ 2λ)

Γ(2λ)Γ(n+ λ+ 12)P

(λ−12,λ−1

2)

n (x) ,

where α = β = λ− 12, λ > −1

2since α > −1. If α = −1

2(or λ = 0), the polynomial

C(0)n vanishes identically for n ≥ 1.

Another consequence of that theorem is that Jacobi polynomials with α = ±12

orβ = ±1

2can be expressed by ultraspherical polynomials.

(b) For α = β = −12

we obtain the Chebyshev polynomials of first kind Tn, i.e.

P(−1

2,−1

2)

n (x) =

∏ni=1(2i− 1)

2nn!Tn(x) =

∏ni=1(2i− 1)

2nn!cos(nϑ),

where x = cos(ϑ).

(c) For α = β = 12

we obtain the Chebyshev polynomials of second kind Un, i.e.

P(12,12

)n (x) =

∏ni=0(2i+ 1)

2n(n+ 1)!Un(x) =

∏ni=0(2i+ 1)

2n(n+ 1)!

sin((n+ 1)ϑ)

sin(ϑ),

where x = cos(ϑ).

(d) The mixed variants α = −12, β = 1

2and α = 1

2, β = −1

2yield the Chebyshev

polynomials of third kind Vn and of forth kind Wn, respectively, i.e.

P(−1

2,12

)n (x) =

∏ni=1(2i− 1)

2nn!Vn(x) =

∏ni=1(2i− 1)

2nn!

cos( (2n+1)ϑ2

)

cos(ϑ2)

,

P(12,−1

2)

n (x) =

∏ni=1(2i− 1)

2n(n+ 1)!Wn(x) =

∏ni=1(2i− 1)

2n(n+ 1)!

sin( (2n+1)ϑ2

)

sin(ϑ2)

,

where x = cos(ϑ).

(e) For α = β = 0 we find the Legendre polynomials Pn, i.e.

P (0,0)n (x) = C

(12

)n (x) = Pn(x).

Note that in this case the weight function is w(x) = 1, x ∈ [−1, 1].

Theorem 3.49The Jacobi polynomials y = P

(α,β)n (x) satisfy the following linear homogeneous differential

equation of the second order

(1− x2)y′′ + (β − α− (α + β + 2)x)y′ + n(n+ α + β + 1)y = 0

ord

dx

((1− x)α+1(1 + x)β+1y′

)+ n(n+ α + β + 1)(1− x)α(1 + x)βy = 0.


Proof. We note that since y ∈ Πn we have that

d

dx

((1− x)α+1(1 + x)β+1y′

)= (1− x)α(1 + x)βz

where z ∈ Πn. To show that z is a constant multiple of y we have to prove the orthogo-nality to any P ∈ Πn−1, i.e.∫ 1

−1

d

dx

((1− x)α+1(1 + x)β+1y′

)P (x)dx = 0.

We use integration by parts on the left hand side which yields (since α + 1 > 0 andβ + 1 > 0):∫ 1

−1

d

dx

((1− x)α+1(1 + x)β+1y′

)P (x)dx = −

∫ 1

−1

(1− x)α+1(1 + x)β+1y′P ′(x)dx

=

∫ 1

−1

yd

dx

((1− x)α+1(1 + x)β+1P ′(x)

)︸︷︷︸C(1−x)α(1+x)βR(x)

dx

where we performed another integration by parts and R ∈ Πn−1. Therefore, the integralvanishes. The constant factor can be calculated by comparing the highest terms, i.e.

y = knxn + . . . , y′ = nknx

n−1 + . . . , y′′ = n(n− 1)knxn−2 + . . . ,

and

0 =d

dx

((1− x)α+1(1 + x)β+1y′

)+ C(1− x)α(1 + x)βy

= −(α + 1)(1− x)α(1 + x)β+1y′ + (β + 1)(1− x)α+1(1 + x)βy′

+ (1− x)α+1(1 + x)β+1y′′ + C(1− x)α(1 + x)βy

= −(α + 1)(1− x)α(1 + x)β(1 + x)y′ − (β + 1)(1− x)α(1 + x)β(x− 1)y′

− (1− x)α(1 + x)β(x2 − 1)y′′ + C(1− x)α(1 + x)βy .

Thus,

C = −(−n(α + 1)− n(β + 1)− n(n− 1)) = n(n+ α + β + 1) .

Theorem 3.50Let α, β > −1. The differential equation

(1− x2)y′′ + (β − α− (α + β + 2)x)y′ + γy = 0,

where γ is a parameter, has a polynomial solution not identically zero if and only ifγ = n(n+α+β+ 1), n ∈ N0. This solution is C ·P (α,β)

n (x), C 6= 0, and no solution which

is linearly independent of P(α,β)n can be a polynomial.


Proof. Substitute y =∑∞

ν=0 aν(x− 1)ν in the differential equation. This gives us

0 = −(x+ 1)∞∑ν=2

ν(ν − 1)aν(x− 1)ν−1 − (2(α + 1) + (α + β + 2)(x− 1))∞∑ν=1

νaν(x− 1)ν−1

+ γ

∞∑ν=0

aν(x− 1)ν

= −(x− 1 + 2)∞∑ν=2

ν(ν − 1)aν(x− 1)ν−1 − 2(α + 1)∞∑ν=1

νaν(x− 1)ν−1

− (α + β + 2)∞∑ν=1

νaν(x− 1)ν + γ∞∑ν=0

aν(x− 1)ν

= −∞∑ν=2

ν(ν − 1)aν(x− 1)ν − 2∞∑ν=1

(ν + 1)νaν+1(x− 1)ν − 2(α + 1)∞∑ν=0

(ν + 1)aν+1(x− 1)ν

− (α + β + 2)∞∑ν=1

νaν(x− 1)ν + γ∞∑ν=0

aν(x− 1)ν

Thus, the coefficients have to fulfill the relation

(−ν(ν − 1)− (α + β + 2)ν + γ) aν − 2ν(ν + 1)aν+1 − 2(α + 1)(ν + 1)aν+1 =

(γ − ν(ν + α + β + 1)) aν − 2(ν + 1)(ν + α + 1)aν+1 = 0

for ν ∈ N0. If we assume that y is a polynomial, we can suppose that an denotes the lastnonzero coefficient, i.e. an+1 = 0. Therefore, the factor in front of an in the recurrencerelation above has to vanish, i.e.

γ = n(n+ α + β + 1).

On the other hand, if this condition for γ holds, we find that ai = 0 for i ≥ n + 1 sincethe factor of aν+1 6= 0.

Let γ = n(n+ α + β + 1) and let z be a second solution of the differential equation, i.e.

d

dx

((1− x)α+1(1 + x)β+1y′

)+ n(n+ α + β + 1)(1− x)α(1 + x)βy = 0,

d

dx

((1− x)α+1(1 + x)β+1z′

)+ n(n+ α + β + 1)(1− x)α(1 + x)βz = 0.

Multiply the first equation by z and the second by y and subtract them:

0 =d

dx

((1− x)α+1(1 + x)β+1y′

)z − d

dx

((1− x)α+1(1 + x)β+1z′

)y

=d

dx

((1− x)α+1(1 + x)β+1(y′z − z′y)

)Thus, for all x ∈ [−1, 1]

(1− x)α+1(1 + x)β+1(y′z − yz′) = c = const.,

If we let x→ ±1, we see that y and z cannot both be polynomials unless c = 0. Therefore,y′z = z′y for all x ∈ (−1, 1), i.e. y and z are linearly dependent, z(x) = cP

(α,β)n .


Remark 3.51The Jacobi polynomials can also be defined as the polynomial solutions of the correspond-ing differential equation that additionally take the value

P (α,β)n (1) =

(n+ α

n

).

Definition 3.52The hypergeometric function F (sometimes denoted by 2F1) is defined by

F (a, b; c;x) =∞∑k=0

(a)k(b)k(c)k

xk

k!

=Γ(c)

Γ(a)Γ(b)

∞∑k=0

Γ(a+ k)Γ(b+ k)

Γ(c+ k)

xk

k!

or its analytic continuation with a, b ∈ R, c ∈ R \ {−N0}, x ∈ (−1, 1).If a = −n or b = −n, n ∈ N0, the hypergeometric function reduces to a polynomial in xwhose degree is n.

Theorem 3.53For the Jacobi polynomials hold

P (α,β)n (x) =

(n+ α

n

)F(−n, n+ α + β + 1;α + 1; 1−x

2

)=∞∑k=0

Γ(n+ α + 1)

Γ(n+ 1)Γ(α + 1)

(−n)kΓ(k + 1)

Γ(n+ α + β + k + 1)Γ(α + 1)

Γ(n+ α + β + 1)Γ(α + 1 + k)(−1)k

(x− 1

2

)k=

Γ(n+ α + 1)

n!Γ(n+ α + β + 1)

n∑k=0

Γ(n+ α + β + 1 + k)

Γ(α + 1 + k)

(n

k

)(x− 1

2

)kThis can be reformulated into

P (α,β)n (x) =

1

2nn!

n∑k=0

(n

k

)Γ(n+ α + 1)

Γ(n+ α + 1− k)

Γ(n+ β + 1)

Γ(β + 1 + k)(x− 1)n−k(x+ 1)k

=1

2nn!

n∑k=0

(n

k

)(n− k + α + 1)k(k + β + 1)(n−k+1)(x− 1)n−k(x+ 1)k

=1

2n

n∑k=0

(n+ α

k

)(n+ β

n− k

)(x− 1)n−k(x+ 1)k

Proof. This can be shown via the properties of the hypergeometric function, in particularits differential equation (see literature).Another way uses the Rodriguez representation of the Jacobi polynomials which will followsoon.Note that

(−n)k(−1)k

k!=

n!

k!(n− k)!=

(n

k

).


Corollary 3.54The leading coefficient of the Jacobi polynomial P

(α,β)n of degree n is given by

k(α,β)n = 2−n

(2n+ α + β

n

).

Proof. Consider the representation

P (α,β)n (x) =

Γ(n+ α + 1)

n!Γ(n+ α + β + 1)

n∑k=0

Γ(n+ α + β + 1 + k)

Γ(α + 1 + k)

(n

k

)(x− 1

2

)kof Theorem 3.53 in the following limit:

k(α,β)n = lim

x→∞x−nP (α,β)

n (x)

=Γ(n+ α + 1)

n!Γ(n+ α + β + 1)

Γ(n+ α + β + 1 + n)

Γ(α + 1 + n)

(n

n

)1

2n

= 2−n(

2n+ α + β

n

).

Corollary 3.55For the derivative of the Jacobi polynomials holds

d

dxP (α,β)n (x) =

n+ α + β + 1

2P

(α+1,β+1)n−1 (x).

Proof. See exercises. (This follows immediately if both sides are expanded according toTheorem 3.53.)

Theorem 3.56 (Rodriguez’ Formula)Let α, β > −1. Then

(1− x)α(1 + x)βP (α,β)n (x) =

(−1)n

2nn!

(d

dx

)n ((1− x)n+α(1 + x)n+β

)Proof. We use Leibniz’ rule for the differentiation of products to find(

d

dx

)n ((1− x)n+α(1 + x)n+β

)=

n∑k=0

(n

k

)(d

dx

)k(1− x)n+α

(d

dx

)n−k(1 + x)n+β

=n∑k=0

(n

k

)(1− x)αRn−k(x)(1 + x)βSk(x)

= (1− x)α(1 + x)β%(x)


where Rj, Sj ∈ Πj, deg(Rj) = deg(Sj) = j, j = 0, . . . , n, and % ∈ Πn, deg(%) = n. Indetail (we will need that later):

(1− x)α(1− x)β%(x)

=n∑k=0

(n

k

)(k−1∏i=0

(n+ α− i)

)(−1)k(1− x)n+α−k

(n−k+1∏i=0

(n+ β − i)

)(1 + x)n+β−(n−k)

=n∑k=0

(n

k

)Γ(n+ α + 1)

Γ(n+ α− k + 1)

Γ(n+ β + 1)

Γ(β + k + 1)(−1)k(1− x)n+α−k(1 + x)β+k.

Therefore,

%(x) =n∑k=0

(n

k

)Γ(n+ α + 1)

Γ(n+ α− k + 1)

Γ(n+ β + 1)

Γ(β + k + 1)(−1)k(1− x)n−k(1 + x)k

We now have to show that % = C · P (α,β)n with a constant C. It suffices to show that

1∫−1

(1− x)α(1− x)β%(x)R(x)dx = 0

for all R ∈ Πn−1. We use partial integration:

1∫−1

(1− x)α(1− x)β%(x)R(x)dx =

1∫−1

(d

dx

)n ((1− x)n+α(1 + x)n+β

)R(x)dx

=

((d

dx

)n−1 ((1− x)n+α(1 + x)n+β

)R(x)

)∣∣∣∣∣1

−1︸︷︷︸=0

−1∫

−1

(d

dx

)n−1 ((1− x)n+α(1 + x)n+β

)R′(x)dx

= . . . = (−1)n1∫

−1

((1− x)n+α(1 + x)n+β

)R(n)(x)dx = 0

since deg(R) ≤ n − 1, i.e. R(n) ≡ 0. Now we just have to determine the constant C.Consider x = 1 in %, i.e. only the summand k = n remains, i.e.

%(1) =

(n

n

)Γ(n+ α + 1)

Γ(α + 1)

Γ(n+ β + 1)

Γ(n+ β + 1)(−1)n2n = (−1)n2nn!

(n+ α

n

)= (−1)n2nn! P (α,β)

n (1).

Thus, C = 2nn!(−1)n

.


Theorem 3.57Let α, β > −1. Then for n ∈ N

h(α,β)n =

∥∥P (α,β)n

∥∥2=

∫ 1

−1

(P (α,β)n (x)

)2(1− x)α(1 + x)βdx

=2α+β+1

2n+ α + β + 1

Γ(n+ α + 1)Γ(n+ β + 1)

Γ(n+ 1)Γ(n+ α + β + 1)

and

h(α,β)0 =

∥∥∥P (α,β)0

∥∥∥2

=

∫ 1

−1

(P

(α,β)0 (x)

)2

(1− x)α(1 + x)βdx

= 2α+β+1 Γ(α + 1)Γ(β + 1)

Γ(α + β + 2).

Proof. Let n ∈ N. We have from the proof of Theorem 3.56∫ 1

−1

(P (α,β)n (x)

)2(1− x)α(1 + x)βdx = k(α,β)

n

∫ 1

−1

P (α,β)n (x)xn(1− x)α(1 + x)βdx

=(−1)n

2nn!k(α,β)n

∫ 1

−1

(d

dx

)n ((1− x)n+α(1 + x)n+β

)xndx

= 2−nk(α,β)n

∫ 1

−1

(1− x)n+α(1 + x)n+βdx

= 2−2n

(2n+ α + β

n

)∫ 1

−1

(1− x)n+α(1 + x)n+βdx

where we used partial integration and Corollary 3.54.∫ 1

−1

(P (α,β)n (x)

)2(1− x)α(1 + x)βdx = 2−2n

(2n+ α + β

n

)∫ 1

−1

(1− x)n+α(1 + x)n+βdx

= 2α+β

(2n+ α + β

n

)∫ 1

−1

(1− x

2

)n+α(1 + x

2

)n+β

dx

= 2α+β+1

(2n+ α + β

n

)∫ 1

0

yn+β(1− y)n+αdy

= 2α+β+1 Γ(2n+ α + β + 1)

Γ(n+ 1)Γ(n+ α + β + 1)

Γ(n+ α + 1)Γ(n+ β + 1)

Γ(2n+ α + β + 2)

=2α+β+1

2n+ α + β + 1

Γ(n+ α + 1)Γ(n+ β + 1)

Γ(n+ 1)Γ(n+ α + β + 1).

The case n = 0 follows similarly.

Theorem 3.58The Jacobi polynomials fulfill the following three term recurrence relation:

2(n+ 1)(n+ α + β + 1)(2n+ α + β)P(α,β)n+1 (x) =

(2n+ α + β + 1)((2n+ α + β + 2)(2n+ α + β)x+ α2 − β2

)P (α,β)n (x)

− 2(n+ α)(n+ β)(2n+ α + β + 2)P(α,β)n−1 (x)


for n ∈ N with P(α,β)0 (x) = 1 and P

(α,β)1 (x) = 1

2(α + β + 2)x+ 1

2(α− β).

Proof. Consequence of Theorem 3.26, the previous constants and a lot of calculations.Some more details are given in the lecture.

3.4 Ultraspherical Polynomials

In Remark 3.48 we have already presented the ultraspherical (or Gegenbauer) polynomials

C(λ)n and given their connection to the general Jacobi polynomials, i.e.

C(λ)n (x) =

Γ(α + 1)Γ(n+ 2α + 1)

Γ(2α + 1)Γ(n+ α + 1)P (α,α)n (x)

=Γ(λ+ 1

2)Γ(n+ 2λ)

Γ(2λ)Γ(n+ λ+ 12)P

(λ−12,λ−1

2)

n (x) ,

where α = β = λ− 12, λ > −1

2since α > −1. This relation gives us the following properties

for λ 6= 0:

1. The value at 1 is C(λ)n (1) =

(n+2λ−1

n

), the symmetry relation is given by

C(λ)n (−x) = (−1)nC(λ)

n (x),

the polynomials are either even or odd (depending on n) due to Theorem 3.47.

2. The differential equation (whose solution is y = C(λ)n ) becomes

(1− x2)y′′ − (2λ+ 1)xy′ + n(n+ 2λ)y = 0. (3.49)

3. We know the explicit representation, i.e.

C(λ)n (x) =

(n+ 2λ− 1

n

)F (−n, n+ 2λ;λ+ 1

2; 1−x

2)

=Γ(λ+ 1

2)

Γ(2λ)n!

n∑k=0

(n

k

)Γ(n+ 2λ+ k)

Γ(λ+ 12

+ k)

(x− 1

2

)k.

4. The leading coefficient is

k(λ)n = lim

x→∞x−nC(λ)

n (x) = 2n(n+ λ− 1

n

).

5. The derivative is again an ultraspherical polynomial, i.e.

d

dxC(λ)n (x) = 2λC

(λ+1)n−1 (x). (3.50)


6. The Rodriguez representation for the Gegenbauer polynomials reads as follows

C(λ)n (x) =

(−1)nΓ(λ+ 12)Γ(n+ 2λ)

2nn! Γ(2λ)Γ(n+ λ+ 12)

(1− x2)12−λ(d

dx

)n((1− x2)n+λ−1

2

)=

(−2)nΓ(n+ λ)Γ(n+ 2λ)

n! Γ(λ)Γ(2n+ 2λ)(1− x2)

12−λ(d

dx

)n((1− x2)n+λ−1

2

).

7. Their norm is computed to be

h(λ)n =

∥∥C(λ)n

∥∥2=

21−2λΓ(n+ 2λ)

n!(n+ λ)(Γ(λ))2π .

8. The corresponding three term recurrence is given by

(n+ 1)C(λ)n+1(x) = 2(n+ λ)xC(λ)

n (x)− (n+ 2λ− 1)C(λ)n−1(x) (3.51)

for n ∈ N with C(λ)0 (x) = 1 and C

(λ)1 (x) = 2λx.

Lemma 3.59If n ≥ 1, C

(0)n ≡ 0, but

limλ→0

C(λ)n (x)

λ=

2

nTn(x)

with the Chebyshev polynomials of first kind Tn.

Proof. We remember that

Tn(x) =2nn!∏n

i=1(2i− 1)P

(−12,−1

2)

n (x).

Now we consider

C(λ)n (x)

λ=

Γ(λ+ 12)

λΓ(2λ)

Γ(n+ 2λ)

Γ(n+ λ+ 12)P

(λ−12,λ−1

2)

n (x)

=2Γ(λ+ 1

2)

Γ(2λ+ 1)

Γ(n+ 2λ)

Γ(n+ λ+ 12)P

(λ−12,λ−1

2)

n (x)

λ→0−−→2 Γ(1

2)

Γ(1)

Γ(n)

Γ(n+ 12)P

(−12,−1

2)

n (x)

=2 n!

n

Γ(12)

Γ(n+ 12)P

(−12,−1

2)

n (x)

=2 n!

n

1∏ni=1(i− 1/2)

P(−1

2,−1

2)

n (x)

=2

n

n! 2n∏ni=1(2i− 1)

P(−1

2,−1

2)

n (x) =2

nTn(x).

Note that C(λ)0 = 1, also as λ → 0. Combining Theorem 3.47 and Theorem 3.53 for the

ultraspherical polynomials we obtain the following representations.


Lemma 3.60For n ∈ N0 hold

C(λ)2n (x) =

(2n+ 2λ− 1

2n

)F (−n, n+ λ;λ+ 1

2; 1− x2) ,

C(λ)2n+1(x) =

(2n+ 2λ

2n+ 1

)xF (−n, n+ λ+ 1;λ+ 1

2; 1− x2) .

Proof. Consider the even case first. We start with the relation between Gegenbauer andJacobi polynomials and apply Theorem 3.47.

C(λ)2n (x) =

Γ(λ+ 12)Γ(2n+ 2λ)

Γ(2λ)Γ(2n+ λ+ 12)P

(λ−12,λ−1

2)

2n (x)

=Γ(λ+ 1

2)Γ(2n+ 2λ)

Γ(2λ)Γ(2n+ λ+ 12)

Γ(2n+ λ+ 12)Γ(n+ 1)

Γ(n+ λ+ 12)Γ(2n+ 1)

P(λ−1

2,−1

2)

n (2x2 − 1)

Now we use the representation of Theorem 3.53:

C(λ)2n (x) =

Γ(λ+ 12)Γ(2n+ 2λ)Γ(n+ 1)

Γ(2λ)Γ(n+ λ+ 12)Γ(2n+ 1)

(n+ λ− 1

2

n

)F (−n, n+ λ;λ+ 1

2; 1− x2)

=

(2n+ 2λ− 1

2n

)F (−n, n+ λ;λ+ 1

2; 1− x2).

The odd case can be shown analogously.

Lemma 3.61For n ∈ N0 hold

C(λ)2n (x) = (−1)n

(n+ λ− 1

n

)F (−n, n+ λ; 1

2;x2)

C(λ)2n+1(x) = (−1)n2λ

(n+ λ

n

)x F (−n, n+ λ+ 1; 3

2;x2)

Proof. Consider the even case. This time we apply the second variant of Theorem 3.47.

C(λ)2n (x) =

Γ(λ+ 12)Γ(2n+ 2λ)

Γ(2λ)Γ(2n+ λ+ 12)P

(λ−12,λ−1

2)

2n (x)

=Γ(λ+ 1

2)Γ(2n+ 2λ)

Γ(2λ)Γ(2n+ λ+ 12)(−1)n

Γ(2n+ λ+ 12)Γ(n+ 1)

Γ(n+ λ+ 12)Γ(2n+ 1)

P(−1

2,λ−1

2)

n (1− 2x2)

=Γ(λ+ 1

2)Γ(2n+ 2λ)Γ(n+ 1)

Γ(2λ)Γ(n+ λ+ 12)Γ(2n+ 1)

(−1)n(n− 1

2

n

)F (−n, n+ λ; 1

2; 1−(1−2x2)

2)

=Γ(λ+ 1

2)Γ(2n+ 2λ)Γ(n+ 1)

Γ(2λ)Γ(n+ λ+ 12)Γ(2n+ 1)

(−1)nΓ(n+ 1

2)

Γ(n+ 1)Γ(12)F (−n, n+ λ; 1

2;x2)


Due to the duplication formula of Lemma 2.8, we find that

Γ(λ+ 12)

Γ(2λ)=

Γ(12)

22λ−1Γ(λ), (3.52)

Γ(n+ λ+ 12)

Γ(2n+ 2λ)=

Γ(12)

22n+2λ−1Γ(n+ λ), (3.53)

Γ(n+ 12)

Γ(12)

=Γ(2n)

22n−1Γ(n), (3.54)

such thatΓ(λ+ 1

2)

Γ(n+ λ+ 12)

Γ(2n+ 2λ)

Γ(2λ)= 22nΓ(n+ λ)

Γ(λ).

Therefore,

C(λ)2n (x) =

Γ(12)

22λ−1Γ(λ)

22n+2λ−1Γ(n+ λ)

Γ(12)

Γ(2n)

22n−1Γ(n)

(−1)n

Γ(2n+ 1)F(−n, n+ λ; 1

2;x2)

=2 Γ(n+ λ)(−1)n

2 n Γ(λ)Γ(n)F(−n, n+ λ; 1

2;x2)

= (−1)n(n+ λ− 1

n

)F(−n, n+ λ; 1

2;x2). (3.55)

The odd case can be shown analogously.

Corollary 3.62For N ∈ N0 holds the explicit representation

C(λ)N (x) =

bN/2c∑m=0

(−1)mΓ(N −m+ λ)

Γ(λ)Γ(m+ 1)Γ(N − 2m+ 1)(2x)N−2m.

Proof. Let N = 2n, i.e. bN/2c = n. we use the definition of the hypergeometric functionF (Definition 3.52) in Lemma 3.61.

C(λ)N (x) = (−1)n

Γ(n+ λ)

Γ(λ)Γ(n+ 1)F (−n, n+ λ; 1

2;x2)

= (−1)nΓ(n+ λ)

Γ(λ)Γ(n+ 1)

n∑k=0

(−n)k(n+ λ)k(1

2)k

x2k

k!

We resolve the Pochhammer symbols as follows (note that the duplication formula ofLemma 2.8 is used in (3.58)):

(−n)k = (−1)kΓ(n+ 1)

Γ(n+ 1− k), (3.56)

(n+ λ)k =Γ(n+ λ+ k)

Γ(n+ λ), (3.57)

(12)k =

Γ(k + 12)

Γ(12)

=Γ(2k)

22k−1Γ(k). (3.58)


We obtain:

C(λ)N (x) =

n∑k=0

(−1)n+k Γ(n+ λ)

Γ(λ)Γ(n+ 1)

Γ(n+ 1)

Γ(n+ 1− k)

Γ(n+ λ+ k)

Γ(n+ λ)

Γ(12)

Γ(k + 12)

x2k

k!

=n∑k=0

(−1)n−kΓ(n+ λ+ k)

Γ(λ)Γ(n− k + 1)Γ(k + 1)

22k−1Γ(k)

Γ(2k)x2k

=n∑k=0

(−1)n−kΓ(n+ λ+ k)

Γ(λ)Γ(n− k + 1)

(2x)2k

Γ(2k + 1). (3.59)

Now, we shift the index using k = n−m or m = n− k:

C(λ)N (x) =

n∑m=0

(−1)mΓ(n+ n−m+ λ)

Γ(λ)Γ(m+ 1)

(2x)2n−2m

Γ(2n− 2m+ 1)

=

bN/2c∑m=0

(−1)mΓ(N −m+ λ)

Γ(λ)Γ(m+ 1)Γ(N − 2m+ 1)(2x)N−2m . (3.60)

For N = 2n+ 1, the proof can be performed analogously.

Theorem 3.63We have for λ > 0 that

max−1≤x≤1

∣∣C(λ)n (x)

∣∣ = C(λ)n (1) =

(n+ 2λ− 1

n

).

For λ < 0 holds

max−1≤x≤1

∣∣C(λ)n (x)

∣∣ =∣∣C(λ)

n (x′)∣∣

where x′ is one of the two maximum points nearest to 0 if n is odd; x′ = 0 if n is even.

Proof. Let n ≥ 1. We consider

n(n+ 2λ)F (x) = n(n+ 2λ)(C(λ)n (x)

)2+ (1− x2)

(d

dxC(λ)n (x)

)2

.

Then F (x) =(C

(λ)n (x)

)2

if ddxC

(λ)n (x) = 0 or x = ±1, i.e.

max−1≤x≤1

(C(λ)n (x)

)2 ≤ max−1≤x≤1

F (x).


Now we differentiate the equation above and use the differential equation (3.49):

n(n+ 2λ)F ′(x) = n(n+ 2λ)2C(λ)n (x)

(d

dxC(λ)n (x)

)− 2x

(d

dxC(λ)n (x)

)2

+ (1− x2)2

(d

dxC(λ)n (x)

)(d2

dx2C(λ)n (x)

)= 2

(d

dxC(λ)n (x)

)(n(n+ 2λ)C(λ)

n (x) + (1− x2)

(d2

dx2C(λ)n (x)

)−x(d

dxC(λ)n (x)

))= 2

(d

dxC(λ)n (x)

)((2λ+ 1)x

(d

dxC(λ)n (x)

)− x

(d

dxC(λ)n (x)

))= 4λx

(d

dxC(λ)n (x)

)2

Therefore, we find that if λ > 0, F is increasing in [0, 1]. If λ < 0, F is decreasing in[0, 1]. This (together with the symmetry relation) gives us the desired result.

Remark 3.64For the Legendre polynomials (λ = 1

2) holds that

|Pn(x)| ≤ Pn(1) = 1

and

|P ′n(x)| =∣∣∣C(3/2)

n−1 (x)∣∣∣ ≤ (n− 1 + 3− 1

n− 1

)=

(n+ 1

n− 1

)=n(n+ 1)

2

for x ∈ [−1, 1].

Theorem 3.65 (Generating Function)For λ > 0 and h ∈ (−1, 1) holds

∞∑n=0

hnC(λ)n (x) = (1− 2hx+ h2)−λ ,

where x ∈ [−1, 1].

Proof. First we check the convergence of the series.∣∣∣∣∣∞∑n=0

hnC(λ)n (x)

∣∣∣∣∣ ≤∞∑n=0

|h|n∣∣C(λ)

n (x)∣∣

≤∞∑n=0

|h|n(n+ 2λ− 1

n

)≤ C

∞∑n=0

n2λ|h|n <∞


since (n+ 2λ− 1

n

)=

Γ(n+ 2λ)

Γ(n+ 1)Γ(2λ)≤ Cn2λ .

Thus, we have absolute and uniform convergence of the series.

Now we consider the recurrence relation (3.51):

∞∑n=1

nC(λ)n (x)hn−1 = 2x

∞∑n=1

(n+ λ− 1)C(λ)n−1(x)hn−1 −

∞∑n=1

(n+ 2λ− 2)C(λ)n−2(x)hn−1

where C(λ)−1 ≡ 0. Denote for a fixed x ∈ [−1, 1]

Φ(h) =∞∑n=0

C(λ)n (x)hn.

Then we obtain (the differentiation is performed with respect to h)

Φ′(h) =∞∑n=1

nC(λ)n (x)hn−1

h1−λ (hλΦ(h))′

= λΦ(h) + hΦ′(h) =∞∑n=0

λC(λ)n (x)hn +

∞∑n=1

nC(λ)n (x)hn

=∞∑n=0

(n+ λ)C(λ)n (x)hn =

∞∑n=1

(n+ λ− 1)C(λ)n−1(x)hn−1

h2−2λ(h2λΦ(h)

)′= 2λhΦ(h) + h2Φ′(h) =

∞∑n=0

2λC(λ)n (x)hn+1 +

∞∑n=1

nC(λ)n (x)hn+1

=∞∑n=0

(n+ 2λ)C(λ)n (x)hn+1 =

∞∑n=1

(n+ 2λ− 2)C(λ)n−2(x)hn−1

where C(λ)−1 ≡ 0. Therefore,

Φ′(h) = 2xh1−λ (hλΦ(h))′ − h2−2λ

(h2λΦ(h)

)′= 2x

(λhλ−1h1−λΦ(h) + h1−λhλΦ′(h)

)−(2λh2λ−1h2−2λΦ(h) + h2−2λh2λΦ′(h)

)= 2x (λΦ(h) + hΦ′(h))−

(2λhΦ(h) + h2Φ′(h)

)This yields

(1− 2hx+ h2)Φ′(h) = (2xλ− 2hλ)Φ(h)

orΦ′(h)

Φ(h)= −λ 2h− 2x

1− 2hx+ h2.

We perform integration (with respect to the variable h).

ln(Φ(h)) = −λ ln(1− 2hx+ h2) + C = ln((1− 2hx+ h2)−λ

)+ C


and finallyΦ(h) = C(1− 2hx+ h2)−λ

where C = 1, since Φ(0) = 1.

Remark 3.66For λ = 0 we can find the generating function of the Chebyshev polynomials of first kind,i.e.

− ln(1− 2hx+ h2) =∞∑n=1

2

nTn(x)hn

=∞∑n=1

2

n

n∏k=1

2k

2k − 1P

(−12,−1

2)

n (x)hn .

Remark 3.67The generating function can also be used to define the corresponding orthogonal polyno-mials as coefficients of the series expansion.

Remark 3.68For q ∈ N, q ≥ 3, and t ∈ [−1, 1],

Pn(q; t) =1(

n+q−3n

)C(q−2

2)

n (t) =Γ(n+ 1)Γ(q − 2)

Γ(n+ q − 2)C

(q−2

2)

n (t)

denote the Legendre polynomials of dimension q (see e.g. [17]).

Theorem 3.69We can derive the following relations (n ∈ N0):

(1− x2)d

dxC(λ)n (x) =

1

2(n+ λ)

((n+ 2λ− 1)(n+ 2λ)C

(λ)n−1(x)− n(n+ 1)C

(λ)n+1(x)

)(3.61)

= −nxC(λ)n (x) + (n+ 2λ− 1)C

(λ)n−1(x) (3.62)

= (n+ 2λ)xC(λ)n (x)− (n+ 1)C

(λ)n+1(x) (3.63)

= 2λ(1− x2)C(λ+1)n−1 (x) , (3.64)

nC(λ)n (x) = x

d

dxC(λ)n (x)− d

dxC

(λ)n−1(x) , (3.65)

(n+ 2λ)C(λ)n (x) =

d

dxC

(λ)n+1(x)− x d

dxC(λ)n (x) (3.66)

as well as (for n ∈ N)

d

dx

(C

(λ)n+1(x)− C(λ)

n−1(x))

=2(n+ λ)C(λ)n (x) (3.67)

=2λ(C(λ+1)n (x)− C(λ+1)

n−2 (x)). (3.68)


Proof. Relation (3.61) is proved by induction. The case n = 0 is obvious. Taking n to

n+ 1 we then use the three-term recurrence (3.51) for C(λ)n+1 and obtain

(1− x2)d

dxC

(λ)n+1(x) =

2(n+ λ)

n+ 1C(λ)n (x)− 2(n+ λ)

n+ 1x2C(λ)

n (x) (3.69)

+2(n+ λ)

n+ 1x(1− x2)

d

dxC(λ)n (x)− n+ 2λ− 1

n+ 1(1− x2)

d

dxC

(λ)n−1(x) .

Now, the induction assumption is applied twice, for (1 − x2) ddxC

(λ)n (x) as well as for

(1 − x2) ddxC

(λ)n−1(x). Next, the three-term recurrence (3.51) helps to substitute xC

(λ)n+1,

xC(λ)n−1, and x2C

(λ)n (here we use it twice).

This leads to a formula that only contains the ultraspherical polynomials C(λ)n+2, C

(λ)n ,

and C(λ)n−2. Simplifying their coefficients concludes the induction and yields (3.61) (the

coefficient of C(λ)n−2 turns out to be 0).

The three-term recurrence (3.51) for (n+ 1)C(λ)n+1(x) in (3.61) gives us (3.63) and the step

between (3.63) and (3.62) is once again only (3.61).(3.64) is a direct consequence of (3.50) for the derivative of the Gegenbauer polynomials.To prove (3.65), apply the derivative to (3.62) and rearrange the terms such that

nC(λ)n (x)− x d

dxC(λ)n (x) +

d

dxC

(λ)n−1(x) = (3.70)

(n+ 2λ)d

dxC

(λ)n−1(x)− (n− 1)x

d

dxC(λ)n (x)− (1− x2)

d2

dx2C(λ)n (x) .

Using (3.50) for the derivative of the Gegenbauer polynomials on each term of the righthand side as well as (3.62), the right hand side becomes 0 as desired.(3.66) is shown analogously to (3.65), only now (3.63) takes the role of (3.62) before.Finally, (3.67) is obtained by adding (3.65) to (3.66) and (3.68) follows directly from(3.64) or from the formula for the derivative of the Gegenbauer polynomials (3.50).

3.5 Application of the Legendre Polynomials inElectrostatics

Let x ∈ R3 and %(x) be a charge distribution with total charge∫R3 %(x)dx. Then the

fundamental equations of electrostatics are the electrostatic pre-Maxwell equations. LetE denote the electric field, then for x ∈ R3

div E(x) = 4π%(x)

curl E(x) = 0 .

We introduce the electric potential Φ and solve the equations by E(x) = −∇Φ(x) whichgives us the following Poisson equation

∆Φ(x) = −4π%(x).


Example 3.70A point charge q at x0 ∈ R3 yields

%(x) = qδ(x− x0)

with the delta-distribution δ in R3 (equality holds in the weak sense). We obtain

∆Φ(x) = 0 for x ∈ R3 \ {x0}. (3.71)

The solution is given by Φ(x) = q|x−x0| for x ∈ R3 \ {x0}. The corresponding electric field

can be calculated to be

E(x) = qx− x0

|x− x0|3for x ∈ R3 \ {x0}.

Another approach to (3.71) introduces polar coordinates (r, ϕ, t) for x ∈ R3 \ {0} by

x1 = r√

1− t2 cosϕ , x2 = r√

1− t2 sinϕ , x3 = rt

where r = |x| > 0, ϕ ∈ [0, 2π), t ∈ [−1, 1]. This gives us the representation of the Laplaceequation in polar coordinates, i.e.[(

∂

∂r

)2

+2

r

∂

∂r+

1

r2

∂

∂t(1− t2)

∂

∂t+

1

r2

1

1− t2

(∂

∂ϕ

)2]

Φ(r, ϕ, t) = 0

for x ∈ R3 \ {0}. By separation of variables we set

Φ(r, ϕ, t) = U(r)V (ϕ)W (t)

and insert this in the equation above, i.e.

V (ϕ)W (t)

(∂

∂r

)2

U(r) + V (ϕ)W (t)2

r

∂

∂rU(r)

+1

r2U(r)V (ϕ)

∂

∂t(1− t2)

∂

∂tW (t) +

1

r2U(r)W (t)

1

1− t2

(∂

∂ϕ

)2

V (ϕ) = 0.

Next we multiply by r2(1− t2) and divide by U(r)V (ϕ)W (t):

r2(1− t2)U ′′(r)

U(r)+ 2r(1− t2)

U ′(r)

U(r)+ (1− t2)2W

′′(t)

W (t)− 2t(1− t2)

W ′(t)

W (t)+V ′′(ϕ)

V (ϕ)= 0

which can be rewritten as

−r2(1− t2)U ′′(r)

U(r)− 2r(1− t2)

U ′(r)

U(r)− (1− t2)2W

′′(t)

W (t)+ 2t(1− t2)

W ′(t)

W (t)=V ′′(ϕ)

V (ϕ).

The left hand side only depends on r and t, the righthand side only on ϕ. Thus, theequation can only be fulfilled if both sides are equal to a constant λ ∈ R. Therefore,

V ′′(ϕ)

V (ϕ)= λ

⇔ V ′′(ϕ)− λV (ϕ) = 0 , ϕ ∈ [0, 2π).


The non-trivial solutions of this differential equation are linear combinations of V1 =exp(i

√λϕ) and V2 = exp(−i

√λϕ). V1 and V2 have to be 2π-periodic. This leads to a

discretization of λ, i.e. λ = m2, m ∈ N0. The linear independent solutions are given by

Vm(ϕ) = exp(imϕ) , ϕ ∈ [0, 2π) , m ∈ Z.

Now we consider the left hand side for these values of λ:

− m2

(1− t2)= r2U

′′(r)

U(r)+ 2r

U ′(r)

U(r)+ (1− t2)

W ′′(t)

W (t)− 2t

W ′(t)

W (t)

⇔ − m2

(1− t2)− (1− t2)

W ′′(t)

W (t)+ 2t

W ′(t)

W (t)= r2U

′′(r)

U(r)+ 2r

U ′(r)

U(r).

By the same argument as before we find that this equation can only hold if both sides areequal to a constant λ, i.e.

− m2

(1− t2)− (1− t2)

W ′′(t)

W (t)+ 2t

W ′(t)

W (t)= λ

⇔ (1− t2)W ′′(t)− 2tW ′(t) +

(λ+

m2

1− t2

)W (t) = 0

for t ∈ [−1, 1] and

λ = r2U′′(r)

U(r)+ 2r

U ′(r)

U(r)

⇔ 0 = r2U ′′(r) + 2rU ′(r)− λU(r)

for r > 0.

Symmetric Problems

For the case of a point charge located at x0 we can assume (after a certain rotation of thecoordinate system) that x0 is on the ε(3)-axis, i.e. x0 = (0, 0, r0)T where r0 = |x0| > 0. Inthis case it is clear that the solution of (3.71) has a rotational symmetry, i.e. Φ does notdependent on ϕ. We can use

Φ(r, ϕ, t) = U(r)W (t)

with r > 0 and t ∈ [−1, 1] (this corresponds to m = 0 before). Therefore, we find theequation for W :

(1− t2)W ′′(t)− 2tW ′(t) + λW (t) = 0 , t ∈ [−1, 1].

This is the Legendre differential equation (or the Gegenbauer differential equation withλ = 1/2) which possesses a polynomial solution (in physical context: a solution with finiteenergy) if and only if

λ = n(n+ 1) , n ∈ N0.


The solutions are the Legendre polynomials Wn(t) = Pn(t), t ∈ [−1, 1], n ∈ N0.Now we consider the second equation for this values of λ:

r2U ′′(r) + 2rU ′(r)− n(n+ 1)U(r) = 0 , r > 0.

This ODE possesses two linearly independent solutions given by

U1,n(r) = rn , n ∈ N0,

U2,n(r) =1

rn+1, n ∈ N0.

Together we find the two linearly independent solutions of the Laplace equation in therotational symmetric case to be

Φ1,n(r) = rnPn(t),

Φ2,n(r) =1

rn+1Pn(t)

for r > 0, t ∈ [−1, 1], n ∈ N0. Every solution can be expressed as a linear combination ofthese two, i.e. there exist coefficients An, Bn such that

Φ(r, t) =∞∑n=0

(Anr

n +Bn1

rn+1

)Pn(t).

For the case of a point charge located in x0 = (0, 0, r0)T we know a solution, i.e.

Φ(x) =q

|x− x0|, x ∈ R3 \ {x0}.

We use this solution to compute the coefficients An and Bn in the expansion. Restrictingthe solutions to the ε(3)-axis, the point x possesses the polar coordinates r = |x|, t = 1,ϕ = 0, i.e.

Φ(x) =q

|r − r0|, r > 0, r 6= r0.

We use the well-known geometric series and obtain

Φ(x) =

q

r−r0 = qr

1

1− r0r

= qr

∞∑n=0

(r0r

)nfor r > r0 ,

qr0−r = q

r01

1− rr0

= qr0

∞∑n=0

(rr0

)nfor r < r0 .

Comparing this to the expansion for t = 1 (remember that Pn(1) = 1), i.e.

Φ(r, 1) =∞∑n=0

Anrn +

Bn

rn+1, r > 0,

results in the following coefficients:

An =

{0 if r > r0q

rn+10

if r < r0

Bn =

{qrn0 if r > r0

0 if r < r0


thus, we finally obtain

Φ(x) = Φ(r, t) =q

|x− x0|=

q∞∑n=0

rn0rn+1Pn(t) for |x| = r > r0 = |x0| ,

q∞∑n=0

rn

rn+10

Pn(t) for |x| = r < r0 = |x0| .

From a mathematical point of view there is another interesting aspect. Canceling thecharge q leads to (x 6= x0)

1

|x− x0|=

∞∑n=0

rn0rn+1Pn(t) for |x| = r > r0 = |x0| ,

∞∑n=0

rn

rn+10

Pn(t) for |x| = r < r0 = |x0| .

The value |x− x0| can be expressed using polar coordinates, i.e.

|x− x0|2 = |(r√

1− t2 cosϕ, r√

1− t2 sinϕ, rt− r0)T |2

= r2(1− t2) + (rt− r0)2

= r2

(1 +

r20

r2− 2

r0

rt

).

Take r > r0 and we get for r > r0

1

r

√1 +

r20r2− 2 r0

rt

=1

|x− x0|=∞∑n=0

rn0rn+1

Pn(t)

which is equivalent to

1√1 +

r20r2− 2 r0

rt

=∞∑n=0

(r0

r

)nPn(t).

Analogously, for r < r0

1√1 + r2

r20− 2 r

r0t

=∞∑n=0

(r

r0

)nPn(t).

Substituting h = r0/r, respectively h = r/r0, we obtain the result on the generatingfunction of the Legendre polynomials.

3.6 Hermite Polynomials and Applications

The Hermite polynomialsHn are the unique orthogonal polynomials in L2w(R) with w(x) =

exp(−x2), i.e.

1. Hn is a polynomial of degree n,


2.∫∞−∞Hn(x)Hm(x) exp(−x2)dx = 0 for m 6= n,

3. ‖Hn‖2 =√π2nn!, n ∈ N0.

By this definition we can calculate

H0(x) = 1

H1(x) = 2x

H2(x) = 4x2 − 2

H3(x) = 8x3 − 12x

Rodriguez’ representation and the explicit representation are given by

Hn(x) = (−1)n exp(x2)

(d

dx

)n (exp(−x2)

),

Hn(x) =

bn/2c∑k=0

(−1)kn!

k!(n− 2k)!(2x)n−2k .

The following recurrence relation holds with H0 ≡ 1, H−1 ≡ 0:

Hn(x) = 2xHn−1(x)− 2nHn−2(x) for n ∈ N, x ∈ R.

The derivative is given by H ′n(x) = 2nHn−1(x) for n ∈ N0, x ∈ R, and the followingdifferential equation is solved by the Hermite polynomials (n ∈ N0, x ∈ R):

H ′′n(x)− 2xH ′n(x) + 2nHn(x) = 0.

Furthermore, for un(x) = exp(−x2/2)Hn(x) we have (n ∈ N0, x ∈ R):

u′′n(x) + (2n+ 1− x2)un(x) = 0.

For further details and properties of Hermite polynomials we refer to the tutorials ande.g. to [14, 20].

Application in Quantum Mechanics

The starting point of any system is its Hamilton function

E = H(pi, qj, t)

where qj are (unified) coordinates, pi are the impulse coordinates, t is the time. H is thetotal energy of the system.The step from classical mechanics to quantum mechanics is performed by substitutionrules:

E → i~∂

∂tp→ −i~∇


The coordinates qj are substituted by the wave function, where |Ψ|2 is the probabilitydensity, i.e. ∫

R3

|Ψ(x, t)|2dx = 1 for all t.

Let e.g. Ψ be the wave function of a particle in a potential V , then the classical Hamiltonfunction is

E =p2

2m+ V (x).

This becomes by our substitution

i~∂

∂tΨ(x, t) = − ~2

2m∆xΨ(x, t) + V (x)Ψ(x, t)

or

i~Ψ = HΨ with H = − ~2

2m∆x + V (x).

Note that ~ = h2π≈ 1.05457 · 10−34Js is Plank’s constant.

If V and thus H are time-independent, we use the following approach

Ψ(x, t) = ϕ(x) exp(− iEt~ ) ,

such that for all t

i~∂

∂tΨ(x, t) = i~ϕ(x)−iE~ exp(− iEt

~ ) = Hϕ(x) exp(− iEt~ )

⇔ Hϕ(x) = Eϕ(x).

This is the time-independent Schrodinger equation. H is the time-independent Hamiltonoperator and E is the energy of the system, which is an unknown. Furthermore,

1 =

∫R3

|Ψ(x, t)|2dx =

∫R3

|ϕ(x) exp(− iEt~ )|2dx =

∫R3

|ϕ(x)|2dx,

i.e. the probability density is time-independent. Thus, the eigenvalue problem

Hϕ = Eϕ

has to be solved with∫R3 |ϕ(x)|2dx = 1.

One-dimensional Oscillation

Let m be a mass connected to a wall by a spring with spring constant f . The classicalHamilton function of this (1-D) system is

Hk,l(p, x) =p2

2m+mω2x2

2

with ω =√f/m the eigenfrequency of the system. By our substitution rules we obtain

the operator

H = − ~2

2m

d2

dx2+mω2x2

2.


This H-operator describes e.g. the oscillation of a molecule with two atoms and m is therelative mass of the atoms. Thus, the eigenvalue problem for the oscillation is (x ∈ R):(

− ~2

2m

d2

dx2+mω2x2

2

)ϕ(x) = Eϕ(x).

Defining a new coordinate by y = x/b where b =√

~mω

and setting ε = 2E~ω this becomes

d2

dy2ϕ(by)︸︷︷︸u(y)

+(ε− y2)ϕ(by)︸︷︷︸u(y)

= 0 .

This equation possesses a solution if and only if ε = 2n + 1 where n ∈ N0. The solutionis of the form

un(y) = const. · exp(−y2

2)Hn(y)

for n ∈ N0. The constant is determined by the condition∫ ∞−∞|ϕn(x)|2dx = 1

⇔∫ ∞−∞|un(y)|2dy =

1

b.

Thus, we can summarize

ϕn(x) = cnHn(xb) exp(− x2

2b2)

for n ∈ N0 and x ∈ R. These functions are the eigenfunctions of the 1-D oscillationcorresponding to the eigenvalues (energy)

En = (n+ 12)~ω , n ∈ N0.

This shows that the energy of a quantum-mechanical oscillator can only take discretevalues, i.e. the quantization of energy.

3.7 Laguerre Polynomials and Applications

For α > −1 the Laguerre polynomials L(α)n , n ∈ N0, are uniquely defined by

1. L(α)n is a polynomial of degree n defined on R≥0 = [0,∞),

2.∫∞

0L

(α)m (x)L

(α)n (x) exp(−x)xαdx = 0 for n 6= m,

3.∥∥∥L(α)

n

∥∥∥2

= Γ(α + 1)(n+αn

)= Γ(n+α+1)

Γ(n+1).


The Laguerre polynomials admit the following Rodriguez’ representation and explicitrepresentation (n ∈ N0, x ∈ R≥0):

L(α)n (x) = exp(x)

x−α

n!

(d

dx

)n (exp(−x)xn+α

)=

n∑k=0

Γ(n+ α + 1)

Γ(k + α + 1)

(−x)k

k!(n− k)!.

By these properties we get:

L(α)0 (x) = 1

L(α)1 (x) = 1 + α− x

L(α)2 (x) = 1

2

((1 + α)(2 + α)− 2(2 + α)x+ x2

)and we find the recursion formula (n ∈ N, n ≥ 2, x ∈ R≥0):

nL(α)n (x) = (2n+ α− 1− x)L

(α)n−1(x)− (n+ α− 1)L

(α)n−2(x)

and including the derivative (n ∈ N, x ∈ R≥0)

xd

dxL(α)n (x) = nL(α)

n (x)− (n+ α)L(α)n−1(x).

The differential equation (with α > −1, x ∈ R≥0)

xy′′ + (1 + α− x)y′ + γy = 0

possesses a polynomial solution if and only if γ = n ∈ N0. This solution is given byy(x) = const.L

(α)n (x), x ∈ R≥0.

For further details and properties of Laguerre polynomials we refer to the tutorials ande.g. to [14, 20].

Eigenoscillations of an n-fold Pendulum

Let l be the total length of the pendulum, each section has length a = l/n. The angles ofeach section are collected in a vector, i.e. ϕ = (ϕ1, . . . , ϕn)T .The dynamic of the pendulum is described by (linearized, near a stable equilibrium, i.e.ϕk = 0, k = 1, . . . , n):

Mϕ+ Cϕ = 0

where M is the mass matrix with entries

Mi,k = a2 min(i, k)

and C is the matrix of restitutional forces with Ci,i = iga, g = 9.81. Small vibrations aregiven by ϕ = ϕ sin(ωt).

Mϕ+ Cϕ = 0

⇔ −Mϕω2 sin(ωt) + Cϕ sin(ωt) = 0

⇔ (C − ω2M)ϕ = 0.


We decompose M by Cholesky decomposition:

M = UTU =⇒ U = a

1 · · · 1. . .

...0 1

and we set x = Uϕ where x = (x0, . . . , xn−1)T . This gives us(

C − ω2UTU)ϕ = 0

⇔(CU−1 − ω2UT

)x = 0

⇔((UT )−1CU−1 − ω2I

)x = 0

⇔(ag(UT )−1CU−1 − a

gω2I)x = 0

⇔ (A− λI)x = 0

This is an eigenvalue problem for the matrix

A =

1 −1 0 0−1 3 −2 0

0 −2 5 −3. . .

. . . . . . . . . . . . 0. . . . . . . . . −n+ 1

0 0 −n+ 1 2n− 1

Writing out the eigenvalue problem explicitly we obtain

0x−1 + 1x0 − 1x1 = λx0

−kxk−1 + (2k + 1)xk − (k + 1)xk+1 = λxk

−(n− 1)xn−2 + (2n− 1)xn−1 = λxn−1

where k = 1, . . . , n− 2. We have to append x−1 and xn such that this equation holds fork = 0, . . . , n− 1, i.e. x−1 can be chosen arbitrarily and xn = 0 to fulfill the last line.If we take a look at the recurrence relation for the classical Laguerre polynomials (α = 0):

−kLk−1(x) + (2k + 1)Lk(x)− (k + 1)Lk+1(x) = xLk(x).

We can identify x with λ and Lk(x) with xk. Hence, xk = Lk(λ) for k = 0, . . . , n satisfiesthe equations above for all λ. Since we have to fulfill xn = 0 we have

Ln(λ) = xn = 0.

The eigenfrequencies λ1, . . . , λn of the system

Ax = λx


are determined by this equation. Therefore, let λn,1, . . . , λn,n be the zeros of Ln in [0,∞).Then the eigenfrequencies of the system are

ωk =√

gaλn,k , k = 1, . . . , n.

The corresponding eigenmodes to the eigenvalue ωk can be calculated from

xj = Lj(λn,k) , j = 0, . . . , n− 1

and ϕ = U−1x. All interesting properties like energy etc. of the system can be expressedusing Laguerre polynomials.

4 Spherical Harmonics

4.1 Basic Notation

We start by introducing some basic spherical notation.

S2 = {ξ ∈ R3 | |ξ| = 1}

is the unit sphere in R3, we use Greek letters for elements of S2.Polar coordinate representation of x ∈ R3:

x(r, ϕ, t) =

r√

1− t2 cos(ϕ)

r√

1− t2 sin(ϕ)rt

where r = |x| ∈ R≥0 is the distance to the origin, ϕ ∈ [0, 2π) the longitude and t =cos(ϑ) ∈ [−1, 1] the polar distance and ϑ ∈ [0, π] the latitude. The canonical basis inR3 is denoted by ε1, ε2, ε3 and another orthonormal basis is given by the moving frameconsisting of the following three vectors (depending on the spherical coordinates ϕ andt):

εr(ϕ, t) =

√1− t2 cos(ϕ)√1− t2 sin(ϕ)

t

, εϕ(ϕ, t) =

− sin(ϕ)cos(ϕ)

0

, εt(ϕ, t) =

−t cos(ϕ)−t sin(ϕ)√

1− t2

.

εϕ and εt are tangential vectors. Note the vector product εr ∧ εϕ = εt.

The gradient ∇ in R3 can be composed into a radial and an angular part, i.e.

∇ = εr∂

∂r+

1

r∇∗ where ∇∗ = εϕ

1√1− t2

∂

∂ϕ+ εt√

1− t2 ∂∂t

and the tangential operator ∇∗ is called surface gradient. Another tangential operator isthe surface curl gradient L∗ which is defined by L∗ξF (ξ) = ξ ∧ ∇∗ξF (ξ) for F ∈ C(1)(S2),ξ ∈ S2, i.e. in local coordinates:

L∗ = −εϕ√

1− t2 ∂∂t

+ εt1√

1− t2∂

∂ϕ.

We canonically define the surface divergence

∇∗ξ · f(ξ) =3∑i=1

∇∗ξFi(ξ) · εi

71

72 Chapter 4. Spherical Harmonics

and the surface curl

L∗ξ · f(ξ) =3∑i=1

L∗ξFi(ξ) · εi,

where f = [F1, F2, F3]T ∈ c(1)(S2). Note that we use lower-case letters for vector fieldsand capital letters for scalar fields. The same convention applies to the correspondingfunction spaces such as c(1)(S2) for the space of continuously differentiable vector fieldsand C(1)(S2) for the space of continuously differentiable scalar fields on the sphere.The surface curl represents a scalar-valued function on the unit sphere:

L∗ξ · f(ξ) = ∇∗ξ · (f(ξ) ∧ ξ).

Finally, the Laplace operator ∆ can be decomposed into

∆ =∂2

∂r2+

2

r

∂

∂r+

1

r2∆∗ξ where ∆∗ =

∂

∂t(1− t2)

∂

∂t+

1

1− t2∂2

∂ϕ2

and ∆∗ denotes the Beltrami operator. It holds that ∇∗ · ∇∗ = L∗ · L∗ = ∆∗.

Definition 4.1 (Regular Region on the Sphere)A bounded region Γ ⊂ S2 is called regular, if its boundary ∂Γ is an orientable piecewisesmooth Lipschitzian manifold of dimension 1. An example is a spherical cap of radius raround ξ ∈ S2, i.e., C(ξ, r) = {η ∈ S2 : 1− r ≤ ξ · η ≤ 1}.

The following spherical versions of the theorems of Gauß and Stokes are well-known.

Theorem 4.2 (Surface Theorems of Gauß and Stokes)Suppose that Γ ⊂ S2 is a regular region with continuously differentiable boundary curve

∂Γ. Let f be a tangential vector field of class c(1)tan(Γ), i.e., f(ξ) · ξ = 0 for all ξ ∈ Γ. Then,∫

Γ

∇∗ξ · f(ξ)dS(ξ) =

∫∂Γ

νξ · f(ξ)dσ(ξ),∫Γ

L∗ξ · f(ξ)dS(ξ) =

∫∂Γ

τξ · f(ξ)dσ(ξ),

where dσ is the arc element.

τξ denotes the positively oriented unit tangential vector of the boundary curve ∂Γ atξ ∈ ∂Γ. The unit normal vector νξ points into the exterior of Γ and is perpendicular toτξ and ξ, i.e., νξ is perpendicular to ∂Γ in the point ξ ∈ ∂Γ, but tangential to S2.It is important to point out the assumption of the vector field f being tangential inTheorem 4.2. This causes an additional term in Green’s formulas involving ∇∗, but itdoes not affect those for L∗, which is due to the fact that ∇∗ · ξ = 2, but L∗ · ξ = 0 forξ ∈ S2. The normal derivative is given by

∂F∂ν

(ξ) = νξ · ∇∗ξF (ξ) = τξ · L∗ξF (ξ)

for F ∈ C(1)(Γ).

Chapter 4. Spherical Harmonics 73

Lemma 4.3Let Γ ⊂ S2 be a regular region with continuously differentiable boundary ∂Γ. Suppose

that F , G are of class C(1)(Γ). Then,∫Γ

G(η)∇∗ηF (η)dS(η) +

∫Γ

F (η)∇∗ηG(η)dS(η)

=

∫∂Γ

νη(F (η)G(η))dσ(η) + 2

∫∂Γ

η(F (η)G(η))dσ(η),∫Γ

G(η)L∗ηF (η)dS(η) +

∫Γ

F (η)L∗ηG(η)dS(η)

=

∫∂Γ

τη(F (η)G(η))dσ(η).

Theorem 4.4Let G ∈ C(2)(Γ), Γ ⊂ S2 be a regular region with a continuously differentiable boundary∂Γ and a unit outward normal vector field ν. Then, we have

1. Green’s first surface identity for F ∈ C(1)(Γ), i.e.,∫Γ

(∇∗ξG(ξ)

)·(∇∗ξF (ξ)

)dS(ξ) +

∫Γ

F (ξ)∆∗ξG(ξ)dS(ξ)

=

∫∂Γ

F (ξ)∂

∂νG(ξ)dσ(ξ) , (4.1)

2. Green’s second surface identity for F ∈ C(2)(Γ), i.e.,∫Γ

F (ξ)∆∗ξG(ξ)−G(ξ)∆∗ξF (ξ)dS(ξ)

=

∫∂Γ

F (ξ)∂

∂νG(ξ)−G(ξ)

∂

∂νF (ξ)dσ(ξ) . (4.2)

Proof. This is an immediate consequence of the surface Gauß theorem (Theorem 4.2).

The aforementioned statements (Lemma 4.3 and Theorem 4.4) hold as well for the entiresphere S2 instead of a subregion Γ, thereby observing that the occurring boundary inte-grals vanish. For functions F ∈ C(1)(S2) and tangential vector fields f ∈ c

(1)tan(S2), this

implies the following identities:∫S2f(η) · ∇∗ηF (η)dS(η) = −

∫S2F (η)∇∗η · f(η)dS(η),∫

S2f(η) · L∗ηF (η)dS(η) = −

∫S2F (η)L∗η · f(η)dS(η),∫

S2∇∗η · f(η)dS(η) =

∫S2L∗η · f(η)dS(η) = 0.

Definition 4.5Let ξ ∈ S2. A function of the form Gξ : S2 → R, η 7→ Gξ(η) = G(ξ·η) with G : [−1, 1]→ Ris called (ξ-)zonal function on S2.


Theorem 4.6Let G ∈ L2([−1, 1]). Then

∫S2

G(ξ · η)dS(η) = 2π

1∫−1

G(t)dt

for all ξ ∈ S2.

Proof. We know that∫S2Gξ(η)dS(η) =

∫S2G(ξ · η)dS(η). Consider first the case ξ = ε3.

Note that η = (√

1− t2 cos(ϕ),√

1− t2 sin(ϕ), t)T with −1 ≤ t ≤ 1, 0 ≤ ϕ < 2π. There-fore, ε3 · η = t and

∫S2

G(ε3 · η)dS(η) =

1∫−1

2π∫0

G(t)

∣∣∣∣ ∂η∂ϕ ∧ ∂η∂t∣∣∣∣ dϕdt .

We compute

∂η

∂ϕ=

√1− t2(− sin(ϕ))√1− t2 cos(ϕ)

0

,∂η

∂t=

−t√1−t2 cos(ϕ)−t√1−t2 sin(ϕ)

1

,

and find that ∂η∂ϕ· ∂η∂t

= 0,∣∣∣ ∂η∂ϕ ∣∣∣ =

√1− t2 and

∣∣∂η∂t

∣∣ = 1√1−t2 . Together with the rule

|x ∧ y|2 = |x|2|y|2 − (x · y)2 for x, y ∈ R3 this yields that∣∣∣∣ ∂η∂ϕ ∧ ∂η∂t∣∣∣∣ = 1 .

Thus, ∫S2

Gε3(η)dS(η) =

1∫−1

2π∫0

G(t)dϕdt = 2π

∫ 1

−1

G(t)dt .

Now let A ∈ SO(3) = {B ∈ R3×3| detB = 1} be a rotation with Aξ = ε3, i.e. ξ =A−1ε3 = AT ε3. Then:∫

S2

G(ξ · η)dS(η) =

∫S2

G((AT ε3) · η)dS(η) =

∫S2

G(ε3 · (Aη))dS(η)

=

∫S2

G(ε3 · α) detAT︸︷︷︸=1

dS(α) where η = ATα

= 2π

∫ 1

−1

G(t)dt .


4.2 Orthogonal Invariance

Systems of equations which maintain their form when the coordinate axes are subjected toan arbitrary rotation are said to be rotationally, or orthogonally, invariant. The orthogonalinvariance is, of course, closely related to the group O(3) of all orthogonal transformations,i.e., the group of all t ∈ R3×3 such that ttT = tT t = i, i = (δi,j)i,j=1,2,3 denotes theunit matrix of R3×3. The set of all rotations, i.e., SO(3) = {t ∈ O(3) : det t = 1} is asubgroup called the special orthogonal group.We briefly recapitulate some properties of these groups:

(i) Let ξ, η ∈ S2. Then, there exists an orthogonal transformation t ∈ O(3) with η = tξand an orthogonal transformation s ∈ SO(3) with η = sξ.

(ii) For every t ∈ O(3),tξ · tη = ξ · η, ξ, η ∈ S2.

(iii) Suppose that ξ ∈ S2. The set Oξ(3) = {t ∈ O(3) : tξ = ξ} is a subgroup of O(3).Analogously, the set SOξ(3) = {t ∈ SO(3) : tξ = ξ} is a subgroup of SO(3).

(iv) For every t ∈ O(3), we have det t = ±1. If det t = 1, t is called a rotation, whilefor det t = −1, t is called a reflection.

(v) Let t, t′ ∈ O(3) with det t = 1, det t′ = −1. Then,

tξ ∧ tη = t(ξ ∧ η), ξ, η ∈ S2,

t′ξ ∧ t′η = −t′(ξ ∧ η), ξ, η ∈ S2.

The following definitions will prove useful for our later considerations.

Definition 4.7Let F ∈ L2(S2), f ∈ l2(S2), and t ∈ O(3). For scalar and vector fields the operator Rt isdefined by

Rt : L2(S2)→ L2(S2), RtF (ξ) = F (tξ), (4.3)

Rt : l2(S2)→ l2(S2), Rtf(ξ) = tTf(tξ), (4.4)

respectively. RtF and Rtf are called the t-transformed fields.

Remark 4.8Note that l2(S2) denotes the Hilbert space of square-integrable vector fields on the unitsphere S2 with the canonical scalar product. The Banach space of continuous vector fieldsis given by c(S2) with the canonical norm.

Definition 4.9Let X be a subspace of L2(S2) or l2(S2). X is called invariant with respect to orthogonaltransformations or, equivalently, orthogonally invariant if, for all F ∈ X and for allorthogonal transformations t ∈ O(3), the function RtF is of class X.An orthogonally invariant space X is called reducible if there exists a proper subspaceX ′ ⊂ X which itself is invariant with respect to orthogonal transformations.


Note that the expressions invariant with respect to rotations and invariant with respectto reflections are understood in analogy to the aforementioned definition.A linear, orthogonally invariant space which is not reducible is called irreducible. It shouldbe noted that each orthogonally invariant space of dimension 1 is irreducible.

Lemma 4.10Let (X, 〈·, ·〉) be an orthogonally invariant Hilbert subspace of the space L2(S2). Let X1

be an orthogonally invariant subspace of X. Then, the orthogonal complement X⊥1 of X1

is orthogonally invariant as well.

Proof. For all F ∈ X1, F⊥ ∈ X⊥1 , and for all orthogonal transformations t ∈ O(3), wehave

〈F,RtF⊥〉 =

∫S2F (ξ)F⊥(tξ)dS(ξ) (4.5)

=

∫S2F (tTη)F⊥(η)dS(tTη)

= | det tT |∫S2F (tTη)F⊥(η)dS(η)

=

∫S2RtTF (η)F⊥(η)dS(η) = 0,

since RtTF ∈ X1. This implies that RtF⊥ ∈ X⊥1 and, therefore, X⊥1 is invariant with

respect to orthogonal transformations.Analogous results can be formulated for Hilbert spaces of square-integrable vector andtensor fields. Lemma 4.10 shows that each orthogonally invariant Hilbert space can becompletely decomposed into invariant parts.

In view of the last result, we are particularly interested in irreducible spaces, i.e., spacesthat provide us with elements that are invariant with respect to certain orthogonal trans-formations. The following results help us to analyze the structure of such rotationallyinvariant functions.

Lemma 4.11Let F be a function of class C(S2) with RtF (ξ) = F (ξ) for all transformations t ∈ SO(3)and all ξ ∈ S2. Then,

F = F (ε3) = C = const. (4.6)

Proof. For all ξ ∈ S2, there exists a rotation t ∈ SO(3) with tξ = ε3. Consequently, forevery ξ ∈ S2, we have F (ξ) = RtF (ξ) = F (tξ) = F (ε3) = C = const.

Theorem 4.12Let η ∈ S2 be fixed. Furthermore, suppose F is of class C(S2) with RtF (ξ) = F (ξ) for allt ∈ SOη(3) and for all ξ ∈ S2. Then, F can be represented in the form

F (ξ) = Φ(ξ · η), ξ ∈ S2, (4.7)

Φ being a function Φ : [−1, 1]→ R.


Proof. Without loss of generality, let η = ε3 (if this were not true, we could use thefunction G(ξ) = Rt′F (ξ), where t′ ∈ O(3) with t′ε3 = η). With ξ = tε3 +

√1− t2η′ we

have, by assumption, that

F (tε3 +√

1− t2η′) = F (tε3 +√

1− t2η′′), (4.8)

for all points η′, η′′ of the unit circle. Hence, F depends only on t = ξ ·ε3 and is, therefore,a function of t alone, as desired.

Lemma 4.13Let η ∈ S2 be fixed. Let F ∈ C(S2) with RtF (ξ) = (det t)F (ξ) for all t ∈ Oη(3) and allξ ∈ S2. Then, F = 0.

Proof. Suppose that ξ is an element of S2. There exists a reflection t ∈ Oη(3) withtξ = ξ, but then — by assumption — we have F (ξ) = RtF (ξ) = −F (ξ), hence, F (ξ) = 0.

Note that in Lemma 4.11 and Theorem 4.12, the rotations can as well be replaced by re-flections, i.e., in the scalar case, we need not distinguish between rotations and reflections.In the vectorial case, however, this is not true anymore (see literature).

4.3 Polynomials on the Unit Sphere in R3

4.3.1 Homogeneous Polynomials

Definition 4.14A polynomial P on Rn, n ∈ N, is called homogeneous of degree m ∈ N if P (λx) = λmP (x)for all λ ∈ R and all x ∈ Rn. The space of all homogeneous polynomials of degree m onRn is denoted by

Homm(Rn) =

P ∣∣∣ P (x) =∑|α|=m

Cαxα , x ∈ Rn , α ∈ Nn

0

,

(note that for the multi-index holds |α| =n∑i=1

αi).

The set of their restrictions to a set D ⊂ Rn is defined by

Homm(D) = {P |D : P ∈ Homm(Rn)} .

Example 4.15P (x) = x2

1 + x22 + x2

3 ∈ Hom2(R3), therefore P |S2 ∈ Hom2(S2) although P |S2 ≡ 1, i.e.deg(P |S2) = 0.The index m of Homm(S2) refers to the degree of the original polynomial on R3.

Theorem 4.16The set of functions {x 7→ xα}|α|=n is a basis of Homn(R3) and

M(3;n) = M(n) = dim Homn(R3) =(n+ 1)(n+ 2)

2.


Proof. Homogeneous polynomials of degree n in R3 have the form

P (x) =∑|α|=n

Cαxα =

∑α1+α2+α3=n

Cα1,α2,α3xα11 x

α22 x

α33

where α1, α2, α3 ∈ N0. Let P (x) = 0, then

n∑α1=0

(n−α1∑α2=0

Cα1,α2,n−α1−α2xα22 x

n−α1−α23

)xα1

1 = 0.

For x2, x3 ∈ R fixed we have a polynomial in R with respect to x1 which is zero for allx1 ∈ R. Therefore,

n−α1∑α2=0

Cα1,α2,n−α1−α2xα22 x

n−α1−α23 = 0.

Keep now just x3 fixed and we obtain a polynomial in x2 which is zero for all x2 ∈ R.Thus,

Cα1,α2,n−α1−α2xn−α1−α23 = 0⇒ Cα1,α2,n−α1−α2 = 0 for all |α| = n.

Therefore, the set {x 7→ xα}|α|=n is a basis of Homn(R3).The dimension is #{x 7→ xα}|α|=n. We have n+ 1 choices for α1 ∈ {0, . . . , n}, n+ 1− α1

choices for α2 ∈ {0, . . . , n− α1} and in the end 1 choice for α3 = n− α1 − α2. This givesus:

M(3;n) = M(n) = dim Homn(R3) =n∑

α1=0

#{0, . . . , n− α1}

=n∑

α1=0

(n+ 1− α1)

= (n+ 1)2 − n(n+ 1)

2

=(n+ 1)(n+ 2)

2.

Formally we can insert the gradient operator in R3 into a homogeneous polynomial Hn ∈Homn(R3), i.e.

Hn(∇) =∑|α|=n

Cα∂|α|

∂xα11 ∂x

α22 ∂x

α33

.

If Un ∈ Homn(R3) with Un(x) =∑|β|=n

Cβxβ, then holds:

Hn(∇)Un(x) =∑|α|=n

∑|β|=n

CαCβ

(∂

∂x1

)α1

xβ11

(∂

∂x2

)α2

xβ22

(∂

∂x3

)α3

xβ33 .


Now all summands are zero except the ones with α = β such that

Hn(∇)Un(x) =∑|α|=n

CαCα

3∏i=1

(∂

∂xi

)αixαii

=∑|α|=n

CαCα α1! α2! α3!︸︷︷︸α!

∈ R .

Lemma 4.17The mapping

〈· , · 〉Homn(R3) : Homn(R3)× Homn(R3)→ Rgiven by

〈Hn, Un〉Homn(R3) = Hn(∇x)Un(x)

defines an inner product on Homn(R3).The set of monomials {

x 7→ 1√α!xα∣∣∣α ∈ N3

0, |α| = n

}is an orthonormal system in the Hilbert space

(Homn(R3), 〈· , · 〉Homn(R3)

).

Proof. Let Hn, Un ∈ Homn(R3).

1. 〈Hn, Hn〉Homn(R3) = Hn(∇x)Hn(x) =∑|α|=n

C2αα! ≥ 0 and

〈Hn, Hn〉Homn(R3) = 0 ⇔∑|α|=n

C2αα! = 0

⇔ C2α = 0 ∀ α with |α| = n

⇔ Hn ≡ 0 .

2. 〈Hn, Un〉Homn(R3) =∑|α|=n

CαCαα! = 〈Un, Hn〉Homn(R3)

3. Let also Vn ∈ Homn(R3) and λ, µ ∈ R.

〈Hn, λUn + µVn〉Homn(R3) =∑|α|=n

Cα(λCα + µDα)α!

= λ∑|α|=n

CαCαα! + µ∑|α|=n

CαDαα!

= λ〈Hn, Un〉Homn(R3) + µ〈Hn, Vn〉Homn(R3) .

For the monomials M(n)α (x) = 1√

α!xα =

∑|α′|=n

1√α!δα,α′x

α′ we have:

〈M (n)α ,M (n)

α 〉Homn(R3) =∑|α′|=n

(1√α!δα,α′

)2

α′! = 1

〈M (n)α ,M

(n)α 〉Homn(R3) =

∑|α′|=n

1√α!δα,α′

1√α!δα′,αα

′! = 0.


This concludes our proof.

By the fundamental theorem of Fourier analysis an orthonormal expansion of Hn ∈Homn(R3) is given by

Hn(x) =∑|α|=n

〈Hn,M(n)α 〉Homn(R3)M

(n)α (x)

=∑|α|=n

Hn(∇y)1√α!yα

1√α!xα

=∑|α|=n

1

α!(Hn(∇y)y

α)xα

= Hn(∇y)1

n!

∑|α|=n

n!

α!yαxα

Thus, we have

Hx(x) = Hx(∇y)

(1

n!(x · y)n

)= 〈Hn,

1

n!(x ·2)n〉Homn(R3) .

Theorem 4.18The mapping

KHomn(R3) : R3 × R3 → R

(x, y) 7→ KHomn(R3)(x, y) =1

n!(x · y)n

is the reproducing kernel in(Homn(R3), 〈· , · 〉Homn(R3)

), i.e.

(i) KHomn(R3)(x, ·) ∈ Homn(R3) for all x ∈ R3 and KHomn(R3)(·, y) ∈ Homn(R3) for ally ∈ R3,

(ii) 〈Hn, KHomn(R3)(x, ·)〉Homn(R3) = Hn(x) for all x ∈ R3.

Proof. The second property is shown above. The first holds since (2 · y)n as well as(x · 2)n are homogeneous polynomials of degree n for any fixed x, y ∈ R3 and thus,KHomn(R3)(x,2), KHomn(R3)(2, y) ∈ Homn(R3) for any fixed x, y ∈ R3.

It should be noted that a reproducing kernel is always unique (see exercises).

Theorem 4.19Let {Hn,j}j=1,...,M(n), M(n) = (n+1)(n+2)

2, be an orthonormal system in Homn(R3), then

holds for all x, y ∈ R3 that

M(n)∑j=1

Hn,j(x)Hn,j(y) =1

n!(x · y)n = KHomn(R3)(x, y).


Proof. Since the reproducing kernel in a Hilbert space is unique, we have to show thatM(n)∑j=1

Hn,j(x)Hn,j(y) is a reproducing kernel in Homn(R3). Clearly,

M(n)∑j=1

Hn,j(x)Hn,j(2) ∈ Homn(R3),

M(n)∑j=1

Hn,j(2)Hn,j(y) ∈ Homn(R3),

for any fixed x, y ∈ R3. Let Hn ∈ Homn(R3), then

Hn(x) =

M(n)∑j=1

〈Hn, Hn,j〉Homn(R3)Hn,j(x)

and we have that

〈Hn,

M(n)∑j=1

Hn,j(x)Hn,j(2)〉Homn(R3) =

M(n)∑j=1

Hn,j(x)〈Hn, Hn,j〉Homn(R3) = Hn(x) .

Therefore,M(n)∑j=1

Hn,j(x)Hn,j(y) is the reproducing kernel in the Hilbert space Homn(R3)

with the inner product 〈· , · 〉Homn(R3).

Interpolation Problem

Given measurements (xj, bj) ∈ R3 × R, j = 1, . . . ,M(n) = 12(n + 1)(n + 2) determine

F ∈ Homn(R3) with the property

F (xj) = bj , j = 1, . . . ,M(n).

When does a solution exist? If a solution exists, how does it look like?

Definition 4.20A system of M(n) points {x1, . . . , xM(n)} ⊂ R3 is called a fundamental system relative toHomn(R3) if the matrix A with the entries Aj,k = Hn,j(xk), j, k = 1, . . . ,M(n), is regular,where {Hn,j} is an orthonormal system in Homn(R3).

Lemma 4.21For each n ∈ N0 there exists a fundamental system relative to Homn(R3).

Proof. Let n ∈ N0 be fixed. We construct the points {x1, . . . , xM(n)} inductively. Clearlythere exists a point x1 ∈ R3 with

Hn,1(x1) 6= 0.


Now assume there exist {x1, . . . , xm}, m < M(n), such that

(Hn,j(xk))j,k=1,...,m

is regular.Furthermore, assume that for every y ∈ R3 \ {x1, . . . , xm} the matrix

Hn,1(x1) . . . . . . Hn,1(xm) Hn,1(y)...

......

......

...Hn,m+1(x1) . . . . . . Hn,m+1(xm) Hn,m+1(y)

is singular. This means that there exists a row i ∈ {1, . . . ,m+ 1} such that this row canbe represented as a linear combination of the other rows

[Hn,j(x1), . . . , Hn,j(xm), Hn,j(y)] , j ∈ {1, . . . ,m+ 1} \ {i} .

In particular, Hn,i(y) is representable as linear combination of {Hn,j(y)}j∈{1,...,m+1}\{i} forall y ∈ R3. But his is a contradiction to the linear independence of {Hn,1, . . . , Hn,m+1}.This inductive construction can be continued until we reach m = M(n) = 1

2(n+1)(n+2).

If the points of the interpolation problem are a fundamental system relative to Homn(R3),the problem can be solved uniquely.

Theorem 4.22Let {Hn,j}j=1,...,M(n) be an orthonormal system in Homn(R3) and {xk}k=1,...,M(n) be afundamental system relative to Homn(R3). Then each F ∈ Homn(R3) is uniquely repre-sentable in the form (x ∈ R3)

F (x) =

M(n)∑j=1

ajKHomn(R3)(xj, x) =

M(n)∑j=1

aj(xj · x)n

n!.

Proof. It is clear that F can be written as

F =

M(n)∑k=1

bkHn,k

where bk = 〈F,Hn,k〉Homn(R3), k = 1, . . . ,M(n). Let {aj}j=1,...,M(n) be the solution of thesystem of linear equations

M(n)∑j=1

ajHn,k(xj) = bk , k = 1, . . . ,M(n).

The system is uniquely solvable since the points xk form a fundamental system (and thus,the matrix is regular). Then:

F =

M(n)∑k=1

M(n)∑j=1

ajHn,k(xj)Hn,k =

M(n)∑j=1

aj

M(n)∑k=1

Hn,k(xj)Hn,k =

M(n)∑j=1

ajKHomn(R3)(xj, ·).

Note that the system can be very ill-conditioned.


4.3.2 Harmonic Polynomials

Definition 4.23Let D ⊂ R3 be open and connected. F ∈ C(2)(D) is called harmonic if ∆xF (x) = 0 for

all x ∈ D. The set of all harmonic functions in C(2)(D) is denoted by Harm(D).

Definition 4.24The set of all homogeneous harmonic polynomials on R3 with degree m ∈ N0 is denotedby

Harmm(R3) ={P ∈ Homm(R3)|∆P = 0

}.

Moreover, we define for m ∈ N0

Harm0,...,m(R3) =m⊕i=0

Harmi(R3) ,

Harm0,...,∞(R3) =∞⋃i=0

Harm0,...,i(R3).

Let H ∈ Homn(R3). There is a representation

H(x) =n∑j=0

An−j(x1, x2)xj3

where An−j ∈ Homn−j(R2). Now let H ∈ Harmn(R3):

0 = ∆xH(x) =n∑j=0

(∂2

∂x21

+∂2

∂x22

)An−j(x1, x2)xj3 +

n∑j=0

∂2

∂x23

xj3An−j(x1, x2).

Note that (∂2

∂x21

+∂2

∂x22

)Ar(x1, x2) = 0 for r ∈ {0, 1} .

Therefore,

0 =n−2∑j=0

(∂2

∂x21

+∂2

∂x22

)An−j(x1, x2)xj3 +

n∑j=2

j(j − 1)xj−23 An−j(x1, x2)

=n−2∑j=0

∆x,2An−j(x1, x2)xj3 +n−2∑j=0

(j + 2)(j + 1)xj3An−j−2(x1, x2)

Theorem 4.25Let n ∈ N0. Let An and An−1 be homogeneous polynomials of degree n and n− 1 in R2.For j = 0, . . . , n− 2 we define recursively

An−j−2(x1, x2) = − 1

(j + 1)(j + 2)∆x,2An−j(x1, x2) .


Then H : R3 → R given by

H(x1, x2, x3) =n∑j=0

An−j(x1, x2)xj3

is in Harmn(R3). Moreover,

dim Harmn(R3) = 2n+ 1.

Proof. We already know that H ∈ Harmn(R3) from our considerations before. Thedegrees of freedom for H occur in the choices of An and An−1.

Aj ∈ Homj(R2) ⇒ Aj(x) =∑|α|=j

Cαxα

where α ∈ N20. Thus,

dim Homj(R2) =

j∑k=0

1 = j + 1 .

This yields that

dim Harmn(R3) = dim Homn(R2) + dim Homn−1(R2) = (n+ 1) + ((n− 1) + 1)) = 2n+ 1 .

Theorem 4.26For n ≥ 2 the space Homn(R3) can be decomposed into the orthogonal and direct sum

Homn(R3) = Harmn(R3)⊕(Harmn(R3)

)⊥and(Harmn(R3)

)⊥= |· |2Homn−2(R3) =

{Ln(x)

∣∣∣Ln(x) = |x|2Hn−2(x) , Hn−2 ∈ Homn−2(R3)}.

Proof. The first result is just the standard decomposition of functional analysis. For thesecond result let n ≥ 2, Hn−2 ∈ Homn−2(R3) and Kn ∈ Harmn(R3) be arbitrary, then

〈| · |2Hn−2, Kn〉Homn(R3) = 〈Hn−2| · |2, Kn〉Homn(R3)

= Hn(∇x)∆xKn(x) = 0

Therefore, | · |2Hn−2 ∈ (Harmn(R3))⊥

.

Since dim Homn(R3) = M(n) = (n+1)(n+2)2

, dim Harmn(R3) = 2n + 1, dim Homn−2(R3) =

M(n− 2) = (n−1)n2

and thus,

dim(Harmn(R3)

)⊥= dim Homn(R3)− dim Harmn(R3) = dim Homn−2(R3),

we find the desired equality of the spaces.


Corollary 4.27Analyzing this theorem iteratively we obtain the decomposition

Homn(R3) =

bn/2c⊕i=0

| · |2iHarmn−2i(R3).

Remark 4.28Let {Hn,j}j=−n,...,n be an orthonormal system in Harmn(R3). Then this system can becompleted to an orthonormal system in Homn(R3), i.e.

{Hn,j}j=−n,...,n ∪ {Un,j}j=1,...,M(n−2), M(n− 2) = 12n(n− 1),

where {Un,j}j=1,...,M(n) is an orthonormal system in (Harmn(R3))⊥

.

Lemma 4.29For Hn ∈ Homn(R3) holds

ProjHarmn(R3)Hn(x) =

bn/2c∑s=0

(−1)sn!(2n− 2s)!

(2n)!(n− s)!s!|x|2s∆s

Hn(x) .


Theorem 4.30 (Addition Theorem in Harmn(R3))Let {Hn,j}j=−n,...,n be an orthonormal system in Harmn(R3) with respect to 〈· , · 〉Homn(R3).Then for every x, y ∈ R3 with x = |x|ξ, y = |y|η (ξ, η ∈ S2) we obtain

n∑j=−n

Hn,j(x)Hn,j(y) =2nn!

(2n)!|x|n|y|nPn(ξ · η) ,

where Pn denotes the Legendre polynomial of degree n.

Proof. Let {Kn,j}j=1,...,M(n) with M(n) = 12(n+ 1)(n+ 2) be the orthonormal system in

Homn(R3) which completes the basis from {Hn,j}. Due to the uniqueness of the repro-ducing kernel in Homn(R3) we know that

M(n)∑j=1

Kn,j(x)Kn,j(y) =(x · y)n

n!

for x, y ∈ R3. We project both sides to Harmn(R3):

ProjHarmn(R3)

M(n)∑j=1

Kn,j(x)Kn,j(y)

=n∑

j=−n

Hn,j(x)Hn,j(y)


and using Lemma 4.29 together with the property that ∆x(x ·y)n = n(n−1)|y|2(x ·y)n−2:

ProjHarmn(R3)

((x · y)n

n!

)=

1

n!

bn/2c∑s=0

(−1)s(2n− 2s)!(n!)2

(n− 2s)!(n− s)!s!(2n)!|x|2s|y|2s(x · y)n−2s

=(2n+ 1)2nn!

(2n+ 1)!|x|n|y|n

bn/2c∑s=0

(−1)s(2n− 2s)!

2n(n− 2s)!(n− s)!s!(ξ · η)n−2s

=2nn!

(2n)!|x|n|y|n

bn/2c∑s=0

(−1)s(2n− 2s)!

2n(n− 2s)!(n− s)!s!(ξ · η)n−2s

︸︷︷︸=Pn(ξ·η)

due to the explicit representation of the Legendre polynomials. For details see Corollary3.62 with λ = 1

2and

2N−mΓ(N −m+ 1

2)

Γ(12)

=(2N − 2m)!

2N(N −m)!

resulting from the duplication formula of Lemma 2.8.

Theorem 4.31For Hm ∈ Harmm(R3), Kn ∈ Harmn(R3) we have that

〈Hm|S2 , Kn|S2〉L2(S2) =δn,mµn〈Hm, Kn〉Homn(R3)

where µn = (2n+1)!4π2nn!

.

Proof. Idea of the proof:By the fundamental theorem of potential theory holds for all x ∈ R3 with |x| < 1 that

Kn(x) =1

4π

∫S2

(1

|x− y|∂

∂ν(y)Kn(y)−Kn(y)

∂

∂ν(y)

1

|x− y|

)dS(y) .

Therefore,

Hm(∇x)Kn(x) =1

4π

∫S2

(Hm(∇x)

1

|x− y|∂


∂

∂ν(y)Hm(∇x)

1

|x− y|

)dS(y).

Now we have

Hm(∇x)1

|x− y|= (−1)m

(2m)!

2mm!

Hm(x− y)

|x− y|2m+1=

(2m)!

2mm!

Hm(y − x)

|x− y|2m+1.

At x = 0 holds:

Hm(∇x)Kn(x)|x=0 =

{0 for m 6= n

〈Hm, Kn〉Homn(R3) for n = m


Therefore,

1

4π

∫S2

(Hm(y)

|y|2m+1

∂


∂

∂ν(y)

Hm(y)

|y|2m+1

)dS(y)

=

{0 for m 6= n

2mm!(2m)!〈Hm, Kn〉Homn(R3) for n = m

Next we see that on the sphere S2

∂

∂ν(y)Kn(y) = ν(y)∇yKn(y) =

∂

∂rKn(rξ) =

∂

∂rrnKn(ξ) = nrn−1Kn(ξ)

⇒ ∂

∂ν(y)Kn(y)

∣∣∣∣S2

= nKn(ξ)

∂

∂ν(y)Hm(y)

∣∣∣∣S2

= mHm(ξ)

⇒ ∂

∂ν(y)

Hm(y)

|y|2m+1=

(∂

∂ν(y)Hm(y)

)1

|y|2m+1+Hm(y)

∂

∂ν(y)

1

|y|2m+1

= mHm(y)1

|y|2m+1+Hm(y)(−2m− 1)

1

|y|2m+2

⇒ ∂

∂ν(y)

Hm(y)

|y|2m+1

∣∣∣∣S2

= mHm(ξ)− (2m+ 1)Hm(ξ) = −(m+ 1)Hm(ξ)

Together this yields that

1

4π

∫S2

(Hm(y)

|y|2m+1

∂


∂

∂ν(y)

Hm(y)

|y|2m+1

)dS(y)

=1

4π

∫S2

(nHm(ξ)Kn(ξ) + (m+ 1)Hm(ξ)Kn(ξ)) dS(ξ)

=n+m+ 1

4π

∫S2Hm(ξ)Kn(ξ)dS(ξ).

Altogether we find the desired result.

We reformulate the addition theorem.

Corollary 4.32Let {Hn,j}j=−n,...,n be an orthonormal system in Harmn(R3) with respect to 〈· , · 〉Homn(R3).Then for every x, y ∈ R3 with x = |x|ξ, y = |y|η (ξ, η ∈ S2) we find that

n∑j=−n

√µnHn,j(x)

√µnHn,j(y) =

2n+ 1

4π|x|n|y|nPn(ξ · η) .


4.3.3 Harmonic Polynomials on the Sphere

Definition 4.33For m ∈ N0 and a set D ⊂ R3

Harmm(D) ={P |D : P ∈ Harmm(R3)

},

Harm0,...,m(D) ={P |D : P ∈ Harm0,...,m(R3)

},

Harm0,...,∞(D) ={P |D : P ∈ Harm0,...,∞(R3)

}.

The elements of the space Harmn(S2), n ∈ N0, are called (scalar) spherical harmonics ofdegree n.

From the comparison of 〈· , · 〉L2(S2) and 〈· , · 〉Homn(R3) (see Theorem 4.31) it immediatelyfollows that for Yn ∈ Harmn(S2) and Ym ∈ Harmm(S2)

〈Yn, Ym〉L2(S2) =

∫S2Yn(ξ)Ym(ξ)dS(ξ) = 0

if n 6= m. (dS(·) denotes the surface element of the sphere S2.)

Theorem 4.34Let n ∈ N0.

dim Harmn(R3) = dim Harmn(S2) = 2n+ 1.

Proof. It is clear that

dim Harmn(R3) ≥ dim Harmn(S2).

Assume that m = dim Harmn(S2) < dim Harmn(R3) = 2n+ 1.

Let {Hn} ⊂ Harmn(S2) be a linear independent system in Homn(R3). Let Yj = Hn,j|S2 ,i.e. Hn,j(x) = rnYj(ξ). Suppose that F ∈ C(S2) with

F (ξ) =n∑

j=−n

ajYj(ξ) = 0.

Consider the boundary value problem: find U ∈ C(2)(S2int) ∩ C(0)(S2

int) such that

∆U = 0 in S2int , U |S2 = F .

This boundary value problem is uniquely solvable by

F (ξ) =n∑

j=−n

ajYj(ξ) = 0 ⇒ F (ξ) = 0.

⇒ U(x) =n∑

j=−n

ajrnYj(ξ) ⇒ U(x) = 0,


where x = rξ. Thus,

⇒n∑

j=−n

ajrnYj(ξ) = 0 ∀ r < 1

⇒n∑

j=−n

ajHn,j(x) = 0 ∀ x ∈ S2int

The left hand side is a polynomial, i.e.

n∑j=−n

ajHn,j(x) = 0 ∀ x ∈ R3.

The polynomials Hn,j are linearly independent, therefore, aj = 0 for all j.

Lemma 4.35Any spherical harmonic Yn ∈ Harmn(S2), n ∈ N0, is an infinitely often differentiableeigenfunction of the Beltrami operator ∆∗ corresponding to the eigenvalue (∆∗)∧(n) =−n(n+ 1), i.e.

∆∗ξYn(ξ) = (∆∗)∧(n)Yn(ξ) ∀ξ ∈ S2 .

The sequence {(∆∗)∧(n)}n∈N0 is called (spherical) symbol of the Beltrami operator.

Proof. Yn ∈ Harmn(S2) is the restriction of a polynomial Hn ∈ Harmn(R3) to the sphere,i.e.

∆xHn(x) = 0 ∀x ∈ R3 =⇒

((∂

∂r

)2

+2

r

∂

∂r+

1

r2∆∗ξ

)Hn(rξ) = 0

Due to the homogeneity of Hn: Hn(x) = rnYn(ξ) with r > 0. Thus,

0 = ∆xHn(x) = n(n− 1)rn−2Yn(ξ) +2

rnrn−1Yn(ξ) +

1

r2rn∆∗ξYn(ξ)

= rn−2(n2 − n+ 2n+ ∆∗ξ

)Yn(ξ)

Since r > 0, it follows that

0 =(n2 − n+ 2n+ ∆∗ξ

)Yn(ξ)

∆∗ξYn(ξ) = −(n2 + n)Yn(ξ) = −n(n+ 1)Yn(ξ).

Definition 4.36We denote a complete orthonormal system in (Harmn(S2), 〈·, ·〉L2(S2)) by {Yn,j}j=−n,...,n,i.e.

1. 〈Yn,j, Yn,k〉L2(S2) = δj,k for all j, k ∈ {−n, . . . , n}.

2. If 〈F, Yn,j〉L2(S2) = 0 for all j ∈ {−n, . . . , n} with F ∈ Harmn(S2), then F = 0.

We call n the degree of the spherical harmonic Yn,j and j the order.


Remark 4.37The family {Yn,j}n∈N0,j=−n,...,n forms an L2(S2)-orthonormal system in Harm0,...,∞(S2).

Theorem 4.38 (Addition Theorem of Spherical Harmonics)If {Yn,j}j=−n,...,n is an L2(S2)-orthonormal system in Harmn(S2), n ∈ N0, then

n∑j=−n

Yn,j(ξ)Yn,j(η) =2n+ 1

4πPn(ξ · η)

for all (ξ, η) ∈ S22, where Pn is the Legendre polynomial of degree n. Moreover,

n∑j=−n

(Yn,j(ξ))2 =

2n+ 1

4πfor all ξ ∈ S2,

‖Yn,j‖C(S2) = supξ∈S2|Yn,j(ξ)| ≤

√2n+ 1

4πfor all j = −n, . . . , n .

Proof. This result follows immediately from the addition theorem in Harmn(R3), seeTheorem 4.30, respectively Corollary 4.32 together with Theorem 4.31.

Remark 4.39Using the addition theorem it is easy to show that the mapping

(ξ, η) 7→ KHarmn(S2)(ξ, η) =2n+ 1

4πPn(ξ · η)

is the reproducing kernel in Harmn(S2), i.e.

2n+ 1

4π

∫S2Yn(η)Pn(ξ · η)dS(η) = Yn(ξ).

Lemma 4.40For every Yn ∈ Harmn(S2) holds

supξ∈S2|Yn(ξ)| ≤

√2n+ 1

4π‖Yn‖L2(S2) .

Proof. Due to Fourier theory it is clear that by Parseval’s identity

‖Yn‖2L2(S2) =

n∑j=−n

|〈Yn, Yn,j〉L2(S2)|2

and we have the representation

Yn =n∑

j=−n

〈Yn, Yn,j〉L2(S2)Yn,j


on the sphere S2. Therefore,

|Yn(ξ)|2 ≤n∑

j=−n

〈Yn, Yn,j〉2L2(S2) (Yn,j(ξ))2

≤n∑

j=−n

〈Yn, Yn,j〉2L2(S2)

n∑j=−n

(Yn,j(ξ))2

=2n+ 1

4π‖Yn‖2

L2(S2) .

Example 4.41 (Complex-valued Spherical Harmonics)Let n ∈ N0, j ∈ Z with −n ≤ j ≤ n. The function

Y Cn,j : S2 −→ C

ξ 7→ Y Cn,j(ξ) = (−1)j

√2n+ 1

4π

(n− j)!(n+ j)!

Pn,j(cos(ϑ))eijϕ

is called (complex) spherical harmonic of degree n and order j where ϑ, ϕ are the sphericalcoordinates of ξ and

Pn,j : [−1, 1] −→ R

t 7→ Pn,j(t) = (1− t2)j2dj

dtjPn(t)

is the associated Legendre function of degree n and order 0 ≤ j ≤ n. For negative ordersholds the symmetry relation

Pn,−j(t) = (−1)j(n− j)!(n+ j)!

Pn,j(t).

These spherical harmonics are orthonormal with respect to the scalar product of thespace L2

C(S2) of complex-valued square-integrable functions on the unit sphere S2. Theiraddition theorem reads as follows

n∑j=−n

Y Cn,j(ξ)Y

Cn,j(η) =

2n+ 1

4πPn(ξ · η).

Example 4.42 (Real Fully Normalized Spherical Harmonics)Let n ∈ N0, j ∈ Z with −n ≤ j ≤ n. The function

Y Rn,j : S2 −→ R

ξ 7→ Y Rn,j(ξ) =

√2n+ 1

4π

(n− |j|)!(n+ |j|)!

Pn,|j|(cos(ϑ))

√

2 cos(jϕ) : j < 01 : j = 0√

2 sin(jϕ) : j > 0


is called (real) fully normalized spherical harmonic of degree n and order j where ϑ, ϕ arethe spherical coordinates of ξ and

Pn,j : [−1, 1] −→ R

t 7→ Pn,j(t) = (1− t2)j2dj

dtjPn(t)

is the associated Legendre function of degree n and order 0 ≤ j ≤ n.

The Y Rn,j are orthonormal with respect to the scalar product of L2

R(S2) and the additiontheorem holds as written in Theorem 4.38. The real spherical harmonics Y R

n,j and thecomplex spherical harmonics Y C

n,j (of Example 4.41) are related via

Y Rn,j(ξ) =

√

2−δ0,j2

(Y Cn,j(ξ) + Y C

n,j(ξ))

: j ≤ 0

(−1)j√

22i

(Y Cn,j(ξ)− Y C

n,j(ξ))

: j > 0

for all ξ ∈ S2, n ∈ N0 and j ∈ Z with −n ≤ j ≤ n.

Figure 4.1: Spherical harmonics Y3,2 (left) and Y7,0 (right).


Figure 4.2: Spherical harmonics Y7,−5 (left) and Y7,7 (right).

4.4 Closure and Completeness of Spherical Harmonics

Lemma 4.43For all t ∈ [−1, 1] and all h ∈ (−1, 1) hold

∞∑n=0

hnPn(t) =1√

1 + h2 − 2ht,

∞∑n=0

(2n+ 1)hnPn(t) =1− h2

(1 + h2 − 2ht)32

.

Proof. The first result is the generating function which we have in the more generalsetting of Gegenbauer polynomials. For the second equation see exercises.

Theorem 4.44 (Poisson Integral Formula)If F is continuous on S2, then

limh→1−

supξ∈S2

∣∣∣∣∣∣ 1

4π

∫S2

1− h2

(1 + h2 − 2h(ξ · η))32

F (η)dS(η)− F (ξ)

∣∣∣∣∣∣ = 0 .

Proof. Since |h| < 1 and ‖Pn‖C[−1,1] = 1 we use Lemma 4.43 and find

1∫−1

1− h2

(1 + h2 − 2ht)32

dt =∞∑n=0

(2n+ 1)hn1∫

−1

Pn(t) · 1 dt

︸︷︷︸=δn,0‖P0‖2L2

= 2.


Figure 4.3: Spherical harmonics Y10,2 (left) and Y10,9 (right).

Therefore, we use Theorem 4.6:

1

2

1

2π

∫S2

1− h2

(1 + h2 − 2h(ξ · η))32

dS(η) = 1, ξ ∈ S2.

Choose ξ ∈ S2 arbitrary, but fixed. Then

1

4π

∫S2

(1− h2)(F (η)− F (ξ))

(1 + h2 − 2h(ξ · η))32

dS(η) =1

4π

∫S2

(1− h2)F (η)

(1 + h2 − 2h(ξ · η))32

dS(η)− F (ξ) .

For h ∈ [12, 1) we split the left integral into two parts, i.e.∫

S2

. . . =

∫−1≤ξ·η≤1− 3√1−h

. . . +

∫1− 3√1−h≤ξ·η≤1

. . .

For t ∈ [−1, 1− 3√

1− h] we find

1 + h2 − 2ht = (1− h)2 + 2h(1− t) ≥ 2h(1− t) ≥ 2h3√

1− h .

This leads to

1− h2

(1 + h2 − 2ht)32

≤ 1− h2

(2h 3√

1− h)32

=1 + h

(2h)32

· 1− h√1− h

≤ 2√

1− h


Figure 4.4: Spherical harmonics Y10,−6 (left) and Y10,6 (right).

which gives us for the first integral of the splitting the following estimate:∣∣∣∣∣∣∣∫

−1≤ξ·η≤1− 3√1−h

(1− h2)(F (η)− F (ξ))

(1 + h2 − 2h(ξ · η))32

dS(η)

∣∣∣∣∣∣∣≤

∫−1≤ξ·η≤1− 3√1−h

(1− h2)(‖F‖C(S2) + ‖F‖C(S2))

(1 + h2 − 2h(ξ · η))32

dS(η)

= 2 ‖F‖C(S2) 2π

1− 3√1−h∫−1

1− h2

(1 + h2 − 2ht)32

dt

≤ 4π ‖F‖C(S2)

1− 3√1−h∫−1

2√

1− hdt

= 4π ‖F‖C(S2) 2√

1− h (1− 3√

1− h+ 1)︸︷︷︸≤2

≤ 16π ‖F‖C(S2)

√1− h h→1−−−−→ 0

uniformly w.r.t. ξ ∈ S2.

Since F ∈ C(S2) and S2 is compact, F is uniformly continuous, i.e. there exists a functionµ : h→ µ(h) with lim

h→1−µ(h) = 0 such that

|F (ξ)− F (η)| ≤ µ(h)


for all η ∈ S2 that satisfy 1− 3√

1− h ≤ ξ · η ≤ 1. µ(h) is independent of ξ.Now we consider the second part of the integral:∣∣∣∣∣∣∣

∫1− 3√1−h≤ξ·η≤1

(1− h2)(F (η)− F (ξ))

(1 + h2 − 2h(ξ · η))32

dS(η)

∣∣∣∣∣∣∣≤

∫1− 3√1−h≤ξ·η≤1

(1− h2)|F (η)− F (ξ)|(1 + h2 − 2h(ξ · η))

32

dS(η)

≤ µ(h)

∫S2

1− h2

(1 + h2 − 2h(ξ · η))32

dS(η) = µ(h)4πh→1−−−−→ 0 .

Altogether we find the desired convergence result.

Theorem 4.45Let F ∈ C(S2). Then the series

∞∑n=0

hnn∑

j=−nF∧(n, j)Yn,j(ξ) with the coefficients

F∧(n, j) = 〈F, Yn,j〉L2(S2) converges uniformly w.r.t. all ξ ∈ S2 for h ∈ (0, h0) withh0 < 1 and

limh→1−

∞∑n=0

hnn∑

j=−n

F∧(n, j)Yn,j(ξ) = F (ξ)

where this convergence is uniform.


First show uniform convergence of the series using the definition of the Fourier coefficientsand the addition theorem (Theorem 4.38). Estimate the integrand with the sup-norm andwith |Pn(t)| ≤ 1 (for t ∈ [−1, 1]).For the limit process use Lemma 4.43 to obtain a representation like in Theorem 4.44.The dominated convergence theorem allows to interchange the infinite series and integral.

Corollary 4.46The system {Yn,j}n∈N0,−n≤j≤n is closed in

(C(S2), ‖·‖C(S2)

), i.e. for all F ∈ C(S2) and for

all ε > 0 exists a finite linear combinationN∑n=0

n∑j=−n

dn,jYn,j such that

∥∥∥∥∥F −N∑n=0

n∑j=−n

dn,jYn,j

∥∥∥∥∥C(S2)

≤ ε .

In other words:

span(Yn,j, n ∈ N0,−n, . . . , n)‖·‖C(S2) = C(S2) .


Proof. Let F ∈ C(S2) and ε > 0 be arbitrary. According to Theorem 4.45 there existsh = h(ε) < 1 fixed such that∥∥∥∥∥

∞∑n=0

hnn∑

j=−n

F∧(n, j)Yn,j − F

∥∥∥∥∥C(S2)

≤ ε

2.

Moreover, the theorem tells us that there exists N = N(ε) such that∥∥∥∥∥∞∑n=0

hnn∑

j=−n

F∧(n, j)Yn,j −N∑n=0

hnn∑

j=−n

F∧(n, j)Yn,j

∥∥∥∥∥C(S2)

≤ ε

2

since the series converges uniformly for fixed h < 1 (here h = h(ε)). Together this givesus ∥∥∥∥∥∥∥F −

N∑n=0

n∑j=−n

hnF∧(n, j)︸︷︷︸=dn,j

Yn,j

∥∥∥∥∥∥∥C(S2)

≤ ε

2+ε

2= ε .

Bernstein Summability

Now, we consider Bernstein summability to obtain this result. The point of departure forBernstein summability is the so-called Bernstein kernel of degree n.

Definition 4.47The Bernstein kernel of degree n is given by

KB,n : [−1, 1]→ R

t 7→ KB,n(t) =n+ 1

4π

(1 + t

2

)n. (4.9)

Remark 4.48The name Bernstein is motivated by the fact that the kernel is proportional to the Bern-

stein polynomial of degree n with the parameter ν = n, i.e., Bν,n(t) =(nν

)tν(1 − t)n−ν

scaled to the interval [−1, 1] (t is the polar distance between ξ and η, i.e., t = ξ · η). Forfurther details we refer to [9] and the references therein.

First, we mention some important properties of the Bernstein kernel that can easily beverified (see also the tutorials).

Lemma 4.49The Bernstein kernel (4.9) possesses the following properties:

(i) For all t ∈ [−1, 1] and n = 0, 1, . . ., we have

K∧B,n(0) = 2π

∫ 1

−1

KB,n(t) dt = 1.


(ii) For all t ∈ [−1, 1],KB,n(t) ≥ 0.

(iii) For all t ∈ [−1, 1),limn→∞

KB,n(t) = 0.

(iv) For k = 0, . . . , n,

2π

∫ 1

−1

KB,n(t)Pk(t) dt = K∧B,n(k) =n!

(n− k)!

(n+ 1)!

(n+ k + 1)!=

(nk

)(n+k+1

k

) ,where Pk is the Legendre polynomial of degree k.

(v) For k ∈ N fixed,K∧B,n(k) < K∧B,n+1(k).

(vi) For k ∈ N0 fixed, K∧B,n(k)→ 1 as n→∞, i.e.,

limn→∞

K∧B,n(k) = 1.

Now, suppose that F is continuous on S2. We use the property (i) to guarantee that∫S2KB,n(ξ · η)F (η)dS(η) = F (ξ) +

∫S2KB,n(ξ · η) (F (η)− F (ξ)) dS(η), (4.10)

where ξ ∈ S2. We split S2 into two parts depending on a parameter γ ∈ (0, 1) , i.e., intothe spherical cap of radius γ around ξ ∈ S2:

C(ξ, γ) = {η ∈ S2 : 1− γ ≤ ξ · η ≤ 1}

and the remaining sphere S2\C(ξ, γ). The corresponding split of the integral (4.10) yields∫S2. . . =

∫S2\C(ξ,γ)

. . .+

∫C(ξ,γ)

. . . .

On the one hand, we find with (ii)∣∣∣∣∫S2\C(ξ,γ)

KB,n(ξ · η)F (η)dS(η)

∣∣∣∣ ≤ 4π ‖F‖C(S2)

∫ 1−γ

−1

KB,n(t) dt

≤ 2 ‖F‖C(S2)

(1− γ

2

)n+1

.

On the other hand, F is uniformly continuous on S2. Thus, there exists a positive functionµ : γ 7→ µ(γ) with

limγ→0+

µ(γ) = 0,

such that|F (ξ)− F (η)| ≤ µ(γ)


for all η ∈ S2 with 1− γ ≤ ξ · η ≤ 1. Thus, it follows that∣∣∣∣∫C(ξ,γ)

KB,n(ξ · η)(F (η)− F (ξ))dS(η)

∣∣∣∣ ≤ µ(γ)

∫S2KB,n(ξ · η)dS(η)

= µ(γ).

Summarizing our results, we obtain the following estimate

supξ∈S2

∣∣∣∣∫S2KB,n(ξ · η)F (η)dS(η)− F (ξ)

∣∣∣∣= sup

ξ∈S2

∣∣∣∣∫S2KB,n(ξ · η) (F (η)− F (ξ)) dS(η)

∣∣∣∣≤ sup

ξ∈S2

∣∣∣∣ ∫S2\C(ξ,γ)

KB,n(ξ · η)F (η)dS(η)− F (ξ)

∫S2\C(ξ,γ)

KB,n(ξ · η)dS(η)

∣∣∣∣+ sup

ξ∈S2

∣∣∣∣ ∫C(ξ,γ)

KB,n(ξ · η) (F (η)− F (ξ)) dS(η)

∣∣∣∣≤ sup

ξ∈S2

∣∣∣∣ ∫S2\C(ξ,γ)

KB,n(ξ · η)F (η)dS(η)

∣∣∣∣+‖F‖C(S2)

∫S2\C(ξ,γ)

KB,n(ξ · η)dS(η) + µ(γ)

≤ 2‖F‖C(S2)

(1− γ

2

)n+1+ µ(γ).

Now, we choose γ = n−1/2, such that γ → 0 for n→∞, i.e.,

µ(

1√n

)→ 0 for n→∞,(

1− 12√n

)n+1

→ 0 for n→∞.

This shows us that

limn→∞

supξ∈S2

∣∣∣∣∫S2KB,n(ξ · η)F (η)dS(η)− F (ξ)

∣∣∣∣ = 0

and we obtain the following result.

Theorem 4.50For F ∈ C(S2),

limn→∞

supξ∈S2

∣∣∣∣∫S2KB,n(ξ · η)F (η)dS(η)− F (ξ)

∣∣∣∣ = 0.

Now, it can be readily seen that the Legendre series of the Bernstein kernel is given by

KB,n(t) =n∑k=0

K∧B,n(k)2k + 1

4πPk(t)


for all t ∈ [−1, 1]. This shows us that∫S2KB,n(ξ · η)F (η)dS(η) =

n∑k=0

K∧B,n(k)2k + 1

4π

∫S2Pk(ξ · η)F (η)dS(η)

=n∑k=0

K∧B,n(k) ProjHarmk(F )(ξ).

Thus, we finally have the ‘Bernstein summability’ of a Fourier series expansion in termsof spherical harmonics.

Theorem 4.51For F ∈ C(S2),

limn→∞

supξ∈S2

∣∣∣∣ n∑k=0

K∧B,n(k)k∑

j=−k

F∧(k, j)Yk,j(ξ)− F (ξ)

∣∣∣∣ = 0.

Theorem 4.51 also enables us to prove the closure of the system of spherical harmonicsin the space C(S2).

Corollary 4.52The system {Yk,j}k∈N0, j=−k,...,k is closed in C(S2), i.e., for any given ε > 0 and eachF ∈ C(S2), there exists a linear combination

N∑k=0

k∑j=−k

dk,jYk,j

such that ∥∥∥F − N∑k=0

k∑j=−k

dk,jYk,j

∥∥∥C(S2)

≤ ε.

Proof. Given F ∈ C(S2). Then, for any given ε > 0, there exists an integer N = N(ε)such that

supξ∈S2

∣∣∣∣ N∑k=0

k∑j=−k

K∧B,N(k)F∧(k, j)︸︷︷︸=dk,j

Yk,j(ξ)− F (ξ)

∣∣∣∣ ≤ ε.

This proves Corollary 4.52.Since for all F ∈ C(S2) holds the inequality

‖F‖L2(S2) ≤√

4π ‖F‖C(S2) (4.11)

we immediately find that {Yn,j}n∈N0,−n≤j≤n is also closed in(C(S2), ‖·‖L2(S2)

).

Corollary 4.53The system {Yn,j}n∈N0,−n≤j≤n is closed in

(L2(S2), ‖·‖L2(S2)

).


Proof. We know thatC(S2)

‖·‖L2(S2) = L2(S2) .

Let F ∈ L2(S2) and ε > 0. Then there exists G ∈ C(S2) such that ‖F −G‖L2(S2) ≤ε2. Due to Corollary 4.46 and estimate (4.11) there exists a finite linear combinationN∑n=0

n∑j=−n

an,jYn,j corresponding to G such that

∥∥∥∥∥G−N∑n=0

n∑j=−n

an,jYn,j

∥∥∥∥∥L2(S2)

≤ ε

2

Combining these two results yields:∥∥∥∥∥F −N∑n=0

n∑j=−n

an,jYn,j

∥∥∥∥∥L2(S2)

≤ ε

2+ε

2= ε .

Now we can apply Theorem 3.7 (“closure is equivalent to completeness etc.”) to thesystem of spherical harmonics. This gives us L2-Fourier approximation on the sphere, i.e.

limN→∞

∥∥∥∥∥F −N∑n=0

n∑j=−n

〈F, Yn,j〉L2(S2)Yn,j

∥∥∥∥∥L2(S2)

= 0 ∀F ∈ L2(S2) .

Remark 4.54(a) Let X ∈ {C(S2), Lp(S2)}p∈[1,∞). If F,G ∈ X with

limN→∞

∥∥∥∥∥G−N∑n=0

n∑j=−n


∥∥∥∥∥X

= 0 ,

then F = G almost everywhere on S2.

(b) For X ∈ {C(S2), Lp(S2)}p∈[1, 43

]∪[4,∞) it is unknown whether the truncated Fourier

series converges for all F ∈ X, i.e. a truncated Fourier series with L2-Fouriercoefficients might not yield an approximation in the X-topology.

(c) If F : S2 → R is Lipschitz continuous, then

limN→∞

∥∥∥∥∥F −N∑n=0

n∑j=−n


∥∥∥∥∥C(S2)

= 0 ,

i.e. the Fourier series is uniformly convergent in this case.

Remark 4.55Using the completeness property of the spherical harmonics system we can prove that the

spherical harmonics Yn ∈ Harmn(S2) are the only C(∞)(S2)-eigenfunctions of the Beltramioperator. ∆∗ only has the eigenvalues −n(n+ 1) with n ∈ N0.


4.5 The Funk-Hecke Formula

In this section we present a central result in the theory of spherical harmonics, i.e., theFunk–Hecke formula, which is closely related to the irreducibility of the space of sphericalharmonics.

Lemma 4.56Let Hn : R3 → R be a homogeneous harmonic polynomial of degree n with the followingproperties:

1. Hn(tx) = Hn(x) for all orthogonal transformations t ∈ SOε3(3), i.e., all rotationsthat leave ε3 invariant,

2. Hn(ε3) = 1.

Then, Hn is uniquely determined and coincides with the so-called Legendre harmonic Lnof degree n, i.e.,

Hn(x) = Ln(x) = rnPn(t), x = r(tε3 +√

1− t2(cosϕε1 + sinϕε2)),

where Pn is the Legendre polynomial of degree n.

Proof. We already know from Theorem 4.25 that Hn as a homogeneous harmonic poly-nomial of degree n can be written in the form

Hn(x) =n∑k=0

An−k(x1, x2)xk3,

where the coefficients An−k(x1, x2) fulfill a recursion for k = 0, . . . , n − 2. Therefore, Hn

is uniquely determined by the homogeneous polynomials An : (x1, x2) 7→ An(x1, x2) andAn−1 : (x1, x2) 7→ An−1(x1, x2). Condition (i) implies that these polynomials depend onlyon x2

1 + x22. Thus, we find with a constant Cn−k

2, such that

An−k(x1, x2) =

{0 , n− k odd,

Cn−k2

(x21 + x2

2)n−k2 , n− k even.

(4.12)

For x = ε3 in (4.12), we get x21 + x2

2 = 0 and x3 = 1 such that C0 = 1. In order todetermine Cn−k

2for even integers n− k, we see that(∂2

∂x21

+∂2

∂x22

)(x2

1 + x22)

n−k2 = 4

(n− k

2

)2

(x21 + x2

2)n−k−2

2 .

In connection with the recursion of Theorem 4.25 we find the recursion relation

4Cn−k2

(n− k

2

)2

(x21 + x2

2)n−k−2

2 = −(k + 2)(k + 1)Cn−k−22

(x21 + x2

2)n−k−2

2

such that(n− k)2Cn−k

2+ (k + 2)(k + 1)Cn−k−2

2= 0,


k = 0, 2, . . . , n− 2. In other words, since C0 = 1,

Cn−k2

= 1 for k = n,

Cn−k2

= (−1)n−k2 2k−n

n!((n−k

2)!)2

k!for n− k even,

for k = 0, . . . , n− 1. This shows us that Hn is uniquely determined by the conditions (i)and (ii).It remains to prove that Ln(x) = rnPn(t) (with r = |x|, t = ε3 · ξ, and ξ = x

|x|) is a

homogeneous harmonic polynomial that fulfills (i) and (ii).Obviously, Ln ∈ Homn(R3) and by the addition theorem (Theorem 4.30) we find that

Ln(x) = |x|nPn(ε3 · ξ) =(2n)!

2nn!

n∑j=−n

Hn,j(ε3)Hn,j(x) ∈ Harmn(R3),

where Hn,j, j = −n, . . . , n, is an orthonormal basis of Harmn(R3). Condition (ii) is alsoobvious, since Pn(1) = 1. For condition (i), consider any t ∈ SOε3(3)

Ln(tx) = |x|nPn(ε3 · tξ) = |x|nPn(tT ε3 · ξ) = |x|nPn(ε3 · ξ) = Ln(x),

since SOε3(3) is a group and, therefore, tT also leaves ε3 invariant.

Corollary 4.57Let Hn ∈ Harmn(R3) with the following properties:

1. Hn(tx) = Hn(x) for all orthogonal transformations t ∈ SOη(3), i.e., all rotationsthat leave η ∈ S2 invariant,

2. Hn(η) = 1.

Then, Hn is uniquely determined by

Hn(x) = |x|nPn(ξ · η), x = |x|ξ, ξ ∈ S2,

where Pn is the Legendre polynomial of degree n.

Proof. This follows from Lemma 4.56 by simply rotating η to ε3.For any function F ∈ L2(S2), the transformation ξ 7→ tξ, ξ ∈ S2, produces a changein the functional values of F . Let F,G ∈ L2(S2), t ∈ O(3). Observing the change ofcoordinates ζ = tξ, we find that∫

S2F (tξ)G(ξ)dS(ξ) = | det t|

∫S2F (ζ)G(tT ζ)dS(ζ)

(observe that dS(tξ) = | det t|dS(ξ)). Thus, it follows that

〈RtF,G〉L2(S2) = 〈F,RtTG〉L2(S2).


Theorem 4.58The space Harmn(S2) of spherical harmonics of degree n is irreducible.

Proof. Assume that there exists an invariant subspace Y ⊂ Harmn(S2) of dimensiondim(Y ) < dim(Harmn(S2)) = 2n+1. Then, we would be able to show that the orthogonalcomplement Y ⊥ of Y in Harmn(S2) (with respect to 〈·, ·〉L2(S2)) is an invariant subspace (seeLemma 4.10). Now, because of the invariance of Y and Y ⊥, G and G⊥ being representedin terms of the L2(S2)-orthonormal system {Yn,j}, respectively,

G =

dim(Y )∑j=1

Yn,j(ε3)Yn,j ∈ Y,

G⊥ =

dim(Harmn(S2))∑j=dim(Y )+1

Yn,j(ε3)Yn,j ∈ Y ⊥

satisfy G(t·) = G, G⊥(t·) = G⊥ for all t ∈ SOε3(3). Moreover, G,G⊥ do not vanishidentically (note that there exist elements in Y , Y ⊥ different from zero at ε3). Thus, notall values of Yn,j(ε

3), j = 1, . . . , dim(Y ) or j = dim(Y ) + 1, . . . , dim(Harmn(S2)) are zero.By Lemma 4.56 there exists a constant a ∈ R \ {0} such that G = aG⊥, in contradictionto our assumption.The irreducibility of Harmn(S2) leads us to simple representations of spherical harmonics(see, e.g., [8]).

Lemma 4.59Let Zn be a member of Harmn(S2). There exist 2n + 1 orthogonal transformationst−n, . . . , tn such that any Yn ∈ Harmn(S2) can be represented with real numbers a−n, . . . , anin the form

Yn =n∑

j=−n

ajZn(tj·).

Proof. Consider the setX = {Zn(t·) : t ∈ O(3)}.

Clearly, there exist t−n, . . . , tn ∈ O(3) such that

X = span{Zn(t−n·), . . . , Zn(tn·)}.Therefore, Zn ∈ X implies Zn(t·) ∈ X. From the irreducibility of Harmn(S2) (see The-orem 4.58), it follows that X is an invariant non-void space of dimension dim(X) =dim(Harmn(S2)). Moreover, the 2n+ 1 linearly independent spherical harmonics

Zn(t−n·), . . . , Zn(tn·)form a basis.

Lemma 4.60There exist 2n+ 1 points η−n, . . . , ηn of the unit sphere S2 such that any Yn ∈ Harmn(S2)can be represented with real numbers a−n, . . . , an in the form

Yn(ξ) =n∑

j=−n

ajPn(ηj · ξ), ξ ∈ S2.


Proof. From Lemma 4.59 we know that there exist 2n + 1 orthogonal transformationst−n, . . . , tn such that any Yn ∈ Harmn(S2) admits the form

Yn(ξ) =n∑

j=−n

ajPn(ε3 · tjξ), ξ ∈ S2.

But this means that Lemma 4.60 follows with ηj = tTj ε3, j = −n, . . . , n.

Theorem 4.61 (Funk–Hecke Formula)Let G ∈ L1([−1, 1]), n ∈ N0. Then,∫

S2G(ξ · ζ)Pn(ζ · η)dS(ζ) = 2π

∫ 1

−1

G(t)Pn(t) dt︸︷︷︸=G∧(n)

Pn(ξ · η)

for all ξ, η ∈ S2. G∧(n) is called Legendre coefficient of G.

Proof. Let Vn(ξ, η) =∫S2 G(ξ ·α)Pn(η ·α)dS(α) for ξ, η ∈ S2 and t ∈ O(3) an orthogonal

transformation. Then,

Vn(tξ, tη) =

∫S2G(tξ · α)Pn(tη · α)dS(α)

=

∫S2G(tξ · tβ)Pn(tη · tβ)dS(tβ)

= | det t|︸︷︷︸=1

∫S2G(ξ · β)Pn(η · β)dS(β) = Vn(ξ, η).

Furthermore, Vn(ξ, ·) with fixed ξ ∈ S2 is a spherical harmonic by the addition theorem,i.e., for any η ∈ S2,

Vn(ξ, η) =

∫S2G(ξ · α)Pn(η · α)dS(α)

=

∫S2

n∑j=−n

G(ξ · α)4π

2n+ 1Yn,j(α)Yn,j(η)dS(α)

=n∑

j=−n

4π

2n+ 1Yn,j(η)G∧ξ (n, j) .

This means that Vn(ξ, ·) is a harmonic polynomial that is invariant under orthogonaltransformations that leave ξ invariant. Hence, by Lemma 4.56 or, more precisely, byCorollary 4.57

Vn(ξ, η) = αnPn(ξ · η).

Setting η = ξ we obtain (using Theorem 4.6)

αn = αnPn(ξ · ξ) =

∫S2G(ξ · α)Pn(ξ · α)dS(α)

= 2π

∫ 1

−1

G(t)Pn(t) dt = G∧(n)

which corresponds to the desired Legendre coefficient.


Corollary 4.62Let G ∈ L1([−1, 1]). For every Yn ∈ Harmn(S2), we have∫

S2G(ξ · η)Yn(η)dS(η) = G∧(n)Yn(ξ) , ξ ∈ S2,

where G∧(n) is the Legendre coefficient as defined in Theorem 4.61.

Proof. This is a direct consequence of the Funk–Hecke formula (Theorem 4.61) and thefact that the Legendre polynomial Pn is the reproducing kernel in Harmn(S2) (see Remark4.39).

4.6 Green’s Function with Respect to the BeltramiOperator

Definition 4.63The function G(∆∗; ·, ·) : (ξ, η) 7→ G(∆∗; ξ, η), −1 ≤ ξ · η < 1, is called Green’s functionon S2 with respect to the operator ∆∗, if it satisfies the following properties:

(i) for every ξ ∈ S2 the function η 7→ G(∆∗; ξ, η) is twice continuously differentiable onthe set S2 \ {ξ} such that

∆∗ηG(∆∗; ξ, η) = − 1

4π,−1 ≤ ξ · η < 1,

(ii) for every ξ ∈ S2 the function

η 7→ G(∆∗; ξ, η)− 1

4πln(1− ξ · η)

is continuously differentiable on S2.

(iii) for all orthogonal transformations A

G(∆∗; Aξ,Aη) = G(∆∗; ξ, η),

(iv) for every ξ ∈ S2

1

4π

∫S2G(∆∗; ξ, η)dS(η) = 0.

Note that for −1 ≤ ξ · η < 1

ln |ξ − η| = 1

2ln(2− 2ξ · η) =

1

2ln(1− ξ · η) +

1

2ln(2).

Lemma 4.64G(∆∗; ·, ·) is uniquely determined by its defining properties (i)-(iv).The explicit representation of the Green function for −1 ≤ ξ · η < 1 is

G(∆∗; ξ, η) =1

4πln(1− ξ · η) +

1

4π− 1

4πln(2).


Proof. See tutorials.By applying the second Green surface theorem we see that

−k(k + 1)

∫S2G(∆∗; ξ, η)Yk(η)dS(η) = (1− δ0,k)Yk(ξ).

Therefore, we can determine the spectral representation of the Green function as thefollowing bilinear expansion for −1 ≤ ξ · η < 1

G(∆∗; ξ, η) =∞∑n=1

n∑k=−n

1

−n(n+ 1)Yn,k(ξ)Yn,k(η)

=∞∑n=1

2n+ 1

4π

1

−n(n+ 1)Pn(ξ · η).

Theorem 4.65 (Third Green Surface Theorem for ∇∗)Let ξ be a fixed point of S2. Suppose that F is a continuously differentiable function onS2. Then

F (ξ) =1

4π

∫S2F (η)dS(η)−

∫S2

(∇∗ηG(∆∗; ξ · η)

)·(∇∗ηF (η)

)dS(η) .

Proof. Let ξ ∈ S2 be fixed. For each sufficiently small % > 0 Green’s first surface identity(Theorem 4.4) gives us∫

ξ·η<1−%

(F (η)∆∗ηG(∆∗; ξ · η) +∇∗ηF (η) · ∇∗ηG(∆∗; ξ · η)

)dS(η)

=

∫ξ·η=1−%

F (η)∂

∂νηG(∆∗; ξ · η)dσ(η)

where ν is the unit normal to the circle that consists of all points η ∈ S2 with ξ ·η = 1−%.ν points outward the set of points η ∈ S2 with ξ · η ≤ 1− %. Thus,

νη = − η√1− (ξ · η)2

∧ (η ∧ ξ).

We use the differential equation for Green’s function and obtain∫ξ·η<1−%

F (η)∆∗ηG(∆∗; ξ · η)dS(η) = − 1

4π

∫ξ·η<1−%

F (η)dS(η) .

Moreover,∫ξ·η=1−%

F (η)∂

∂νηG(∆∗; ξ · η)dσ(η)

=

∫ξ·η=1−%

F (η)ξ − (1− %)η√

1− (1− %)2·(− 1

4π%(ξ − (1− %)η)

)dσ(η)

= − 1

4π

∫ξ·η=1−%

F (η)

√1− (1− %)2

%dσ(η)

= − 1

4π

√1− (1− %)2

%2π√

1− (1− %)2F (η%) = −2− %2

F (η%)


where we used the mean value theorem with η% ∈ {η ∈ S2|1− ξ · η = %}. The continuityof F yields that F (η%)→ F (ξ) as η% → ξ for %→ 0 such that

lim%→0

∫1−ξ·η=%

F (η)∂

∂νηG(∆∗; ξ · η)dσ(η) = −F (ξ).

This gives us the desired result.

Corollary 4.66 (Third Green Surface Theorem for L∗)Under the assumptions of Theorem 4.65

F (ξ) =1

4π

∫S2F (η)dS(η)−

∫S2

(L∗ηG(∆∗; ξ · η)

)·(L∗ηF (η)

)dS(η) .

Proof. We use an analogous Green’s surface identity for L∗. Note that instead of thenormal vector the tangential vector

τη =ξ ∧ η√

1− (ξ · η)2

is required. The same reasoning as in Theorem 4.65 is used to prove this corollary.

Theorem 4.67 (Third Green Surface Theorem for ∆∗)Let ξ be a fixed point of S2. Suppose that F is a twice continuously differentiable functionon S2. Then

F (ξ) =1

4π

∫S2F (η)dS(η) +

∫S2G(∆∗; ξ · η)

(∆∗ηF (η)

)dS(η) .

Proof. From Green’s second surface identity (Theorem 4.4) we get for a sufficiently small% > 0 that∫

ξ·η≤1−%

(G(∆∗; ξ · η)∆∗ηF (η)− F (η)∆∗ηG(∆∗; ξ · η)

)dS(η)

=

∫ξ·η=1−%

(G(∆∗; ξ · η)

∂

∂ν(η)F (η)− F (η)

∂

∂ν(η)G(∆∗; ξ · η)

)dσ(η)

We use the defining properties of the Green function. The continuous differentiability ofF on S2 leads us to

lim%→0

∫1−ξ·η=%

G(∆∗; ξ · η)∂

∂νηF (η)dσ(η) = 0.

Together with the results of the proof of Theorem 4.65 we find the desired result.For more details see the lecture on potential theory.

Theorem 4.68 (Differential Equation for ∆∗ on S2)Let H ∈ C(S2) with

1

4π

∫S2H(ξ)dS(ξ) = 0.


Let F ∈ C(2)(S2) satisfy the Beltrami differential equation

∆∗F = H.

Then

F (ξ) =

∫S2G(∆∗; ξ · η)H(η)dS(η) , ξ ∈ S2

and F is uniquely determined with

1

4π

∫S2F (ξ)dS(ξ) = 0.

Other solutions take the form F = F + C with a constant C ∈ R.

Proof. Direct consequence of Theorem 4.67.

4.7 The Hydrogen Atom

The hydrogen atom is the simplest atom we can think of. It consists of an electron withthe charge −e and mass m in the Coulomb potential of a nucleus with charge e and massM where M � m.The starting point of the quantum-mechanical discussion of the hydrogen atom is theHamilton function of an electron with charge −e in the Coulomb potential of a nucleuswith charge e, i.e.

H(p, x) =p2

2µ− e2

4πε0|x|, x ≥ 0,

where x is the relative distance of the electron and the nucleus and µ is the reduced massgiven by

µ =mM

m+M≈ m,

since M � m. Applying the substitution rules from before (in this case we only needpj → −i~ ∂

∂xj, j = 1, 2, 3) we obtain the Hamilton operator by

H = − ~2

2m∆x −

e2

4πε0|x|.

Thus, the stationary Hamilton equation is (x ∈ R3)

Hψ(x) = Eψ(x) ⇔ − ~2

2m∆xψ(x)− e2

4πε0|x|ψ(x) = Eψ(x)

where ψ ∈ L2(R3) with ‖ψ‖L2(R3) = 1 and E < 0 such that we obtain bounded states.

By separation of variables ψ(x) = U(r)F (ξ), x = rξ, r = |x| > 0 and ξ ∈ S2. We use therepresentation of the Laplace operator in polar coordinates and get:

− ~2

2m

(∂2

∂r2+

2

r

∂

∂r+

1

r2∆∗ξ

)U(r)F (ξ)− e2

4πε0rU(r)F (ξ)− E U(r)F (ξ) = 0.


The radial derivatives only apply to U and the Beltrami operator applies to F such that

− ~2

2m

(∂2

∂r2+

2

r

∂

∂r

)U(r)F (ξ)− ~2

2m

U(r)

r2∆∗ξF (ξ)− e2

4πε0rU(r)F (ξ)− E U(r)F (ξ) = 0.

Dividing by U(r), F (ξ) and − ~22m

and multiplying by r2 this is equivalent to

r2U′′(r)

U(r)+ 2r

U ′(r)

U(r)+

2m

~2

(e2r

4πε0

+ Er2

)= −

∆∗ξF (ξ)

F (ξ)

for all r > 0, ξ ∈ S2. This can only be true if both sides are equal to a constant λ ∈ R.Thus, we obtain for the right hand side (ξ ∈ S2):

−∆∗ξF (ξ)

F (ξ)= λ ⇔ ∆∗ξF (ξ) = −λF (ξ).

This is the Beltrami differential equation which has a polynomial solution if and only ifλ = l(l + 1), l ∈ N0. In this case the solution is a multiple of a spherical harmonic ofdegree l, i.e. F (ξ) = C1Yl,m(ξ), m = −l, . . . , l.For the left hand side we get with λ = l(l + 1) and after dividing by r2 and multiplyingwith U(r):

U ′′(r) +2

rU ′(r) +

(2m e2

4πε0~2r+

2mE

~2− l(l + 1)

r2

)U(r) = 0 , r > 0.

To solve this equation we use

P (r) = rU(r)

P ′(r) = U(r) + rU ′(r)

P ′′(r) = U ′(r) + U ′(r) + rU ′′(r) = 2U ′(r) + rU ′′(r)

such that we have (r > 0)

U ′′(r) +2

rU ′(r) =

P ′′(r)

r.

Using the abbreviation aB = 4πε0~2m e2

(Bohr’s atomic radius) we find that

P ′′(r) +

(2

aBr+

2E4πε0

e2aB− l(l + 1)

r2

)P (r) = 0.

Next we introduce the Rydberg energy

ER =~2

2ma2B

=m e4

2~2(4πε0)2

and multiply the previous equation by a2B such that

a2BP′′(r) +

(2aBr

+E

ER− a2

Bl(l + 1)

r2

)P (r) = 0.


We introduce dimensionless coordinates by % = r/aB, P (%) = P (r) and η =√−E/ER

(observe that E < 0). This yields us that

d2

d%2P (%) +

(2

%− l(l + 1)

%2− η2

)P (%) = 0 , % > 0.

To solve this equation we set (% > 0):

P (%) = V (%)%l+1 exp(−η%).

We obtain an equation for V given by

d2

d%2V (%) + 2

d

d%V (%)

(l + 1

%− η)

+ V (%)2

%(1− η(l + 1)) = 0.

This equation is multiplied by % and we set y = 2η%, V (y) = V (%) such that

yd2

dy2V (y) + (2l + 2− y)

d

dyV (y) +

(1

η− l − 1

)V (y) = 0.

Comparing this equation to the differential equation of the Laguerre polynomials L(α)n

given by (y > 0)

yd2

dy2L(α)n (y) + (α + 1− y)

d

dyL(α)n (y) + nL(α)

n (y) = 0,

we see that α = 2l + 1 and that we have a polynomial solution if and only if

1

η− l − 1 = nr , nr ∈ N0 .

The solution is then given by

V (y) = C2L(2l+1)nr (y) = C2L

(2l+1)n−l−1(y)

with n = nr + l + 1 = 1/η, n ∈ N. This means that for 1/η only values in N are allowedwhich gives the discretization of energy by (n ∈ N)

1

η= n ⇔ En

ER= −η2 ⇔ En = − e2

aB4πε0

1

2n2= − m e4

2~2(4πε0)2n2.

Resubstituting V into P , then P into U and finally U into ψ:

U(r) =1

rP (r) =

1

rP (%) =

1

rV (%)%l+1 exp(−η%)

= C21

rL

(2l+1)n−l−1( 2r

naB)( raB

)l+1 exp(− rnaB

)

= C21aB

(n2)l︸︷︷︸

C2

L(2l+1)n−l−1( 2r

naB)( 2rnaB

)l exp(− rnaB

)


We get after normalizing ‖ψ‖L2(R3) = 1:

ψn,l,m(x) = CUn,l(r)Fl,m(ξ)

=1√a3B

√(n− l − 1)!

(n+ l)!

2

n2

(2r

naB

)lL

(2l+1)n−l−1

(2r

naB

)exp

(− r

naB

)Yl,m(ξ).

n ∈ N is called the main quantum number, l = 0, . . . , n−1 is called the angular momentumquantum number and m = −l, . . . , l (sometimes renumbered to m = 1, . . . , 2l+1) is calledthe magnetic quantum number.For n = 1 we get for the energy E1 the well-known ionization energy of hydrogen, i.e.

1

4πε0

E1 = −13.6eV .

Figure 4.5: Radial probability densities r2U2n,l(r) for n = 1, 2, 3 and corresponding l.

5 Vectorial Spherical Harmonics

5.1 Notation for Spherical Vector Fields

Vector fields on the unit sphere f : S2 → R3 can be written in the form

f(ξ) =3∑i=1

Fi(ξ)εi , ξ ∈ S2,

where the component functions Fi are given by Fi(ξ) = f(ξ) · εi (vector fields will bedenoted by lower case letters). First we introduce function spaces for vector fields on thesphere.l2(S2) denotes the space of all square-integrable vector fields on S2 which is a Hilbertspace in connection with the scalar product

〈f, g〉l2(S2) =

∫S2f(ξ) · g(ξ)dS(ξ) , f, g ∈ l2(S2).

The space c(p)(S2), 0 ≤ p ≤ ∞, consists of all p-times continuously differentiable vectorfields on the sphere. c(S2) = c(0)(S2) is a Banach space with respect to the norm

‖f‖c(S2) = supξ∈S2|f(ξ)| , f ∈ c(S2).

We know thatc(S2)

‖·‖l2(S2) = l2(S2) .

As in the scalar case there is the norm estimate

‖f‖l2(S2) ≤√

4π ‖f‖c(S2) .

Any vector field f ∈ c(S2) can be decomposed into its normal and its tangential part:

f(ξ) = fnor(ξ) + ftan(ξ)

where the normal and tangential field are given by

ξ 7→ pnorf(ξ) = fnor(ξ) = (f(ξ) · ξ)ξ , ξ ∈ S2,

ξ 7→ ptanf(ξ) = ftan(ξ) = f(ξ)− (f(ξ) · ξ)ξ , ξ ∈ S2,

with the projection operators pnor and ptan. For f, g ∈ c(S2), ξ ∈ S2, holds

f(ξ) · g(ξ) = fnor(ξ) · gnor(ξ) + ftan(ξ) · gtan(ξ) .

113

114 Chapter 5. Vectorial Spherical Harmonics

The space can also be decomposed accordingly

cnor(S2) = {f ∈ c(S2)|f = pnorf},ctan(S2) = {f ∈ c(S2)|f = ptanf},c(S2) = cnor(S2)⊕ ctan(S2),

as well as

l2nor(S2) = cnor(S2)‖·‖l2(S2) ,

l2tan(S2) = ctan(S2)‖·‖l2(S2) ,

l2(S2) = l2nor(S2)⊕ l2tan(S2),

We remember the surface gradient operator

∇∗ = εϕ1√

1− t2∂

∂ϕ+ εt√

1− t2 ∂∂t

which contains the tangential derivatives of the gradient and the surface curl gradientoperator (L∗ξF (ξ) = ξ ∧∇∗ξF (ξ) for F ∈ C(1)(S2))

L∗ = −εϕ√

1− t2 ∂∂t

+ εt1√

1− t2∂

∂ϕ

where L∗F is a tangential vector field perpendicular to ∇∗F , i.e. for ξ ∈ S2

∇∗ξF (ξ) · L∗ξF (ξ) = 0.

Moreover, we have also defined the surface divergence

∇∗ξ · f(ξ) =3∑i=1

∇∗ξFi(ξ) · εi

and the surface curl

L∗ξ · f(ξ) =3∑i=1

L∗ξFi(ξ) · εi

where f = (F1, F2, F3)T ∈ c(1)(S2). Note that the surface curl represents a scalar-valuedfunction on the unit sphere:

L∗ξ · f(ξ) = ∇∗ξ · (f(ξ) ∧ ξ).

We already know that for F ∈ C(2)(S2):

∇∗ξ · ∇∗ξF (ξ) = ∆∗ξF (ξ) and L∗ξ · L∗ξF (ξ) = ∆∗ξF (ξ) .

Moreover, from the tutorials we obtain that

L∗ξ · ∇∗ξF (ξ) = 0 and ∇∗ξ · L∗ξF (ξ) = 0 .

Chapter 5. Vectorial Spherical Harmonics 115

The application of these differential operators to zonal functions yields for F ∈ C(1)[−1, 1]that

∇∗ξF (ξ · η) = F ′(ξ · η)(η − (ξ · η)ξ),

L∗ξF (ξ · η) = F ′(ξ · η)(ξ ∧ η),

and for F ∈ C(2)[−1, 1]

∆∗ξF (ξ · η) = −2(ξ · η)F ′(ξ · η) + (1− (ξ · η)2)F ′′(ξ · η).

The following lemmas are mostly results from vector analysis.

Lemma 5.1The tangential field of f vanishes, i.e. ftan(ξ) = 0 for all ξ ∈ S2, if and only if f(ξ)·τ(ξ) = 0for every unit vector τ(ξ) that is perpendicular to ξ.


Lemma 5.2Let f ∈ c(S2) with ∫

Cτξ · f(ξ)dσ(ξ) = 0

for every curve C ⊂ S2. Thenftan(ξ) = 0

for all ξ ∈ S2.

Proof. Choose any point ξ0 ∈ S2. Let τξ0 be a unit vector with τξ0 · ξ0 = 0. Then, thereis a curve C ⊂ S2 which passes through ξ0 and has τξ0 as tangent vector at ξ0. Let Cξ0 beany subset of C containing ξ0. Then,∫

Cξ0

τξ · f(ξ)dσ(ξ) = 0.

Now we let the length of Cξ0 tend to zero and find that τξ0 ·f(ξ0) = 0. The previous lemmayields that then ftan(ξ0) = 0. Since ξ0 has been chosen arbitrarily, we have ftan = 0.

Lemma 5.3Let F ∈ C(1)(S2) with ∇∗ξF (ξ) = 0 for all ξ ∈ S2. Then F is constant and conversely.

Let F ∈ C(1)(S2) with L∗ξF (ξ) = 0 for all ξ ∈ S2. Then F is constant and conversely.


Lemma 5.4Let f ∈ c(S2) be a tangential vector field with∫



for every closed curve C ⊂ S2. Then, there is a scalar field P on S2 such that

f(ξ) = ∇∗ξP (ξ)

where P ∈ C(1)(S2) and is unique up to a constant.

Proof. Take an arbitrary, but fixed ξ0 ∈ S2. We set

P (ξ) =

∫ ξ

ξ0

τη · f(η)dσ(η)

with the integral along any curve that starts at ξ0 and ends at ξ ∈ S2. Then, for ξ0, ξ1 ∈ S2

and any curve C ⊂ S2 starting at ξ0 and ending at ξ1 holds

P (ξ1)− P (ξ0) =

∫ ξ1

ξ0

τη · f(η)dσ(η) and

P (ξ1)− P (ξ0) =

∫ ξ1

ξ0

τη · ∇∗ηP (η)dσ(η),

since the surface gradient acts like an ordinary gradient in R3 when integrating along lineson S2.Therefore, we combine the two equations to∫ ξ1

ξ0

τη · (f(η)−∇∗ηP (η))dσ(η) = 0

for any curve C ⊂ S2 connecting ξ0 and ξ1. By Lemma 5.2 we find that

f(ξ)−∇∗ξP (ξ) = 0 , ξ ∈ S2.

The proof that P is continuously differentiable can be found e.g. in [3] or in lectures onvector analysis. One basic idea is to take P constant on each straight line passing throughS2 in the normal direction.

For proving uniqueness we consider ∇∗ξP1(ξ) = ∇∗ξP2(ξ), ξ ∈ S2, which implies that∇∗ξ(P1 − P2)(ξ) = 0, i.e. due to Lemma 5.3 P1 − P2 = const.

Theorem 5.5Let f ∈ c(1)(S2) be a tangential vector field. Then, L∗ξ · f(ξ) = 0, ξ ∈ S2, if and only ifthere is a scalar field P such that

f(ξ) = ∇∗ξP (ξ) , ξ ∈ S2,

and P (called potential function for f) is unique up to an additive constant.Similarly, ∇∗ξ · f(ξ) = 0, ξ ∈ S2, if and only if there is a scalar field S such that

f(ξ) = L∗ξS(ξ) , ξ ∈ S2,

and S (called stream function for f) is unique up to an additive constant.


Proof. The condition f = ∇∗P implies that L∗ · f = L∗ · ∇∗P = 0 and f = L∗S impliesthat ∇∗ · f = ∇∗ · L∗S = 0.

Conversely, assume that L∗ξ · f(ξ) = 0, ξ ∈ S2. Then we can use the surface theorem ofStokes to obtain that ∫


for every closed curve C ⊂ S2. By Lemma 5.4 there exists a scalar field P such thatf = ∇∗P where P is unique up to a constant.

Suppose now that ∇∗ξ ·f(ξ) = 0, ξ ∈ S2. Then for ξ ∈ S2 (note that f = ftan is a tangentialfield):

L∗ξ · (ξ ∧ f(ξ)) = (ξ ∧∇∗ξ) · (ξ ∧ f(ξ))

= (ξ · ξ)(∇∗ξ · f(ξ))− (ξ · f(ξ))(∇∗ · ξ)= ∇∗ξ · f(ξ)− 2(ξ · f(ξ))

= ∇∗ξ · f(ξ)− 2(ξ · ftan(ξ))

= ∇∗ξ · f(ξ),

i.e. L∗ξ · (ξ ∧ f(ξ)) = 0, since ∇∗ξ · f(ξ) = 0. Thus, as before there exists a scalar field S(unique up to a constant) such that

−ξ ∧ f(ξ) = ∇∗ξS(ξ) , ξ ∈ S2.

This is equivalent to

−ξ ∧ (ξ ∧ f(ξ)) = (ξ ∧∇∗ξ)S(ξ) , ξ ∈ S2

or f = L∗S on S2, because

−ξ ∧ (ξ ∧ f(ξ)) = (−ξ · f(ξ))ξ − (−ξ · ξ)f(ξ)

= −(ξ · ftan(ξ))ξ + f(ξ)

= f(ξ).

Theorem 5.6Let f ∈ c(1)(S2) be a tangential vector field such that for all ξ ∈ S2

∇∗ξ · f(ξ) = 0 and L∗ξ · f(ξ) = 0.

Then f = 0 on S2.



5.2 Definition of Vector Spherical Harmonics

Definition 5.7For ξ ∈ S2 and F ∈ C(0i)(S2) we define the operators o(i) : C(0i)(S2) → c(S2), i = 1, 2, 3,by

o(1)ξ F (ξ) = ξF (ξ)

o(2)ξ F (ξ) = ∇∗ξF (ξ)

o(3)ξ F (ξ) = L∗ξF (ξ)

where we used the abbreviation

0i =

{0 if i = 1,1 if i = 2, 3.

It is clear that o(1)F ∈ cnor(S2) and o(2)F, o(3)F ∈ ctan(S2).

Remark 5.8Let Fη ∈ C(0i)(S2) be an η-zonal function, i.e. Fη(ξ) = F (ξ · η). Then

o(1)ξ Fη(ξ) = ξF (ξ · η),

o(2)ξ Fη(ξ) = ∇∗ξF (ξ · η) = F ′(ξ · η)(η − (ξ · η)ξ),

o(3)ξ Fη(ξ) = L∗ξF (ξ · η) = F ′(ξ · η)(ξ ∧ η).

By Green’s integral formulas we can introduce the adjoint operators.

Definition 5.9For f ∈ c(0i)(S2) and G ∈ C(0i)(S2) (i = 1, 2, 3) we have that

〈o(i)G, f〉l2(S2) = 〈G,O(i)f〉L2(S2).

Therefore, for f ∈ c(0i)(S2), ξ ∈ S2:

O(1)ξ f(ξ) = ξ · pnorf(ξ) ,

O(2)ξ f(ξ) = −∇∗ξ · ptanf(ξ) ,

O(3)ξ f(ξ) = −L∗ξ · ptanf(ξ) .

Lemma 5.10Let F ∈ C(2)(S2). Then

(i) If i 6= j, i, j ∈ {1, 2, 3}, then O(i)ξ o

(j)ξ F (ξ) = 0, ξ ∈ S2.

(ii) O(i)ξ o

(i)ξ F (ξ) =

{F (ξ) , i = 1,−∆∗ξF (ξ),i = 2, 3.

Proof. Known properties of the differential operators ∇∗, L∗.


Definition 5.11For any Yn ∈ Harmn(S2) the vector field

y(i)n = o(i)Yn , n ≥ 0i , i ∈ {1, 2, 3},

is called a vector spherical harmonic of degree n and type i.By harm(i)

n we denote the space of all vector spherical harmonic of degree n and type i.We also set the spaces

harm0 = harm(1)0 ,

harmn =3⊕i=1

harm(i)n , n ≥ 1.

Remark 5.12We have ξ ∧ y(1)

n = 0, ξ · y(2)n = 0 and ξ · y(3)

n = 0.

Theorem 5.13If {Yn,k}n∈N0,k=−n,...,n is an L2(S2)-orthonormal system of scalar spherical harmonics, thenthe system of

y(i)n,k(ξ) =

1√µ

(i)n

o(i)ξ Yn,k(ξ) , i = 1, 2, 3; n ≥ 0i; k = −n, . . . , n;

with

µ(i)n =

∥∥O(i)o(i)Yn,k∥∥L2(S2)

=

{1 , i = 1,

−(∆∗)∧(n) = n(n+ 1) , i = 2, 3,

forms an l2(S2)-orthonormal system of vector spherical harmonics, i.e.∫S2y

(i)n,k(ξ) · y

(j)m,l(ξ)dS(ξ) = δn,mδk,lδi,j .

Proof. This follows directly from the orthogonality of the scalar spherical harmonics andthe properties of the operators o(i), O(i) with i = 1, 2, 3.

5.3 The Helmholtz Decomposition Theorem

Theorem 5.14 (Helmholtz Decomposition Theorem)Let f ∈ c(1)(S2). Then there exist uniquely determined functions F1 ∈ C(1)(S2) and

F2, F3 ∈ C(2)(S2) satisfying ∫S2Fi(ξ)dS(ξ) = 0 , i = 2, 3,

such that

f(ξ) =3∑i=1

o(i)Fi(ξ) = F1(ξ)ξ +∇∗ξF2(ξ) + L∗ξF3(ξ) , ξ ∈ S2.


The functions Fi are given by

F1(ξ) = O(1)ξ f(ξ) , ξ ∈ S2,

F2(ξ) = −∫S2G(∆∗; ξ, η)O(2)

η f(η)dS(η) , ξ ∈ S2,

F3(ξ) = −∫S2G(∆∗; ξ, η)O(3)

η f(η)dS(η) , ξ ∈ S2.

Proof. Any vector field f ∈ c(1)(S2) can be written as

f = fnor + ftan ,

with

fnor = pnorf ∈ c(1)nor(S2) ,

ftan = pnorf ∈ c(1)tan(S2) .

Clearly, fnor(ξ) = o(1)F1(ξ) with F1(ξ) = O(1)ξ f(ξ) for ξ ∈ S2.

For the tangential field ftan the Stokes theorem for S2 (closed surface, no boundary) yieldsthat ∫

S2L∗η · ftan(η)dS(η) = 0.

Thus, L∗ · ftan is suitable as a right hand side of the Beltrami differential equation and byTheorem 4.68 we find F3 ∈ C(2)(S2) such that

∆∗F3 = L∗ · ftan and

∫S2F3(η)dS(η) = 0 .

In other words,L∗ · L∗F3 = L∗ · ftan or L∗ · (ftan − L∗F3) = 0.

Note that the difference ftan − L∗F3 is still a tangential vector field. Therefore, we cannow use Theorem 5.5 which gives us a scalar field F2 ∈ C(2)(S2) with

ftan − L∗F3 = ∇∗F2 or ftan = ∇∗F2 + L∗F3.

If∫S2 F2(η)dS(η) 6= 0, we replace F2 by F2 = F2 − 1

4π

∫S2 F2(η)dS(η). Then holds that

∇∗F2 = ∇∗F2 and

∫S2F2(η)dS(η) = 0 .

Assume there exists another triple Gi, i = 1, 2, 3, such that

f(ξ) = o(1)ξ F1(ξ) + o

(2)ξ F2(ξ) + o

(3)ξ F3(ξ) ,

f(ξ) = o(1)ξ G1(ξ) + o

(2)ξ G2(ξ) + o

(3)ξ G3(ξ) .


Then, it follows that F1 = O(1)f = G1, i.e. F1 is uniquely defined. Thus,

o(2)ξ F2(ξ) + o

(3)ξ F3(ξ) = o

(2)ξ G2(ξ) + o

(3)ξ G3(ξ) .

We now apply O(2) and O(3) which yield

∆∗F2 = ∆∗G2 ,

∆∗F3 = ∆∗G3 .

Hence, we find uniqueness up to a constant and the normalization conditions on F2 andF3 imply that F2 = G2 and F3 = G3.The specific representation of F2, F3 involving integrals with Green’s function G(∆∗; ·, ·)follows directly from Theorem 4.68.

Remark 5.15A vector field of the form

ξ 7→ F1(ξ)ξ +∇∗ξF2(ξ) , ξ ∈ S2,

is called spheroidal, one of the form

ξ 7→ L∗ξF3(ξ) , ξ ∈ S2,

is said to be toroidal. Thus, the Helmholtz decomposition theorem represents the decom-position of a c(1)-vector field into its spheroidal and its toroidal parts. It also implies anorthogonal decomposition of the space c(∞)(S2), i.e.

c(∞)(S2) = c(∞)(1) (S2)⊕ c(∞)

(2) (S2)⊕ c(∞)(3) (S2)

where

c(∞)(1) (S2) = c(∞)

nor (S2) ,

c(∞)(2) (S2) =

{f ∈ c(∞)(S2) | O(1)f = O(3)f = 0

},

c(∞)(3) (S2) =

{f ∈ c(∞)(S2) | O(1)f = O(2)f = 0

}.

These definitions can also be extended to c(k)(S2), k ∈ N, and to l2(S2), i.e.

l2(i)(S2) = {o(i)F | F ∈ C(∞)(S2)}‖·‖l2(S2) .

Therefore, we find the orthogonal decompositions

l2(S2) = l2nor(S2)⊕ l2tan(S2) ,

l2tan(S2) = l2(2)(S2)⊕ l2(3)(S2) .


5.4 Closure and Completeness of Vector SphericalHarmonics

Next, we prove the closure and completeness of vector spherical harmonics intrinsicallyon the sphere by vectorial variants of the scalar zonal Bernstein kernels. Essential toolsare the theory of the Green function with respect to the Beltrami operator of Section 4.6and the Helmholtz decomposition theorem (Theorem 5.14).We start our considerations by convolving Green’s function with respect to the Beltramioperator against the Bernstein kernel (4.9) of Definition 4.47.

BGn(ξ · η) = (G(∆∗; · ) ∗KB,n( ·η)) (ξ) =

∫S2G(∆∗; ξ · ζ)KB,n(ζ · η)dS(ζ).

Written in terms of a Legendre series, we find the finite sum

BGn(ξ · η) =n∑k=1

2k + 1

4π

K∧B,n(k)

−k(k + 1)Pk(ξ · η).

Note that the Bernstein kernel is a polynomial and the Green function is of class L1([−1, 1]),hence, the existence of the convolution integral as bandlimited Legendre expansion is ob-vious.Next, we are interested in the Bernstein summability of Fourier expansions in terms ofvector spherical harmonics. To this end, we need some preparatory material (more pre-cisely, Lemmas 5.16 and 5.17). Essential tool of our considerations is the Green functionwith respect to the Beltrami operator.

Lemma 5.16For i ∈ {1, 2, 3},

limn→∞

‖Fi − F (n)i ‖C(S2) = 0, (5.1)

where Fi, i = 1, 2, 3, are the functions occurring in the Helmholtz decomposition theorem(Theorem 5.14)

F1(ξ) = O(1)ξ f(ξ), (5.2)

Fi(ξ) = −∫S2G(∆∗; ξ · η)O(i)

η f(η)dS(η), i = 2, 3, (5.3)

and F(n)i , i = 1, 2, 3, are given by

F(n)1 (ξ) =

∫S2KB,n(ξ · η)O(1)

η f(η)dS(η), (5.4)

F(n)i (ξ) = −

∫S2BGn(ξ · η)O(i)

η f(η)dS(η), i = 2, 3. (5.5)

Proof. Clearly, the case i = 1 of Lemma 5.16 is easy to handle. It follows immediatelyfrom the scalar theory. Thus, it remains to study the cases i = 2, 3. We start from theconvolution integrals

F(n)i (ξ) = −

(BGn ∗O(i)f

)(ξ) = −


η f(η)dS(η), (5.6)


i = 2, 3. It is not difficult to see that

‖Fi − F (n)i ‖C(S2) = ‖G(∆∗; ·) ∗O(i)f −BGn ∗O(i)f‖C(S2)

≤ ‖O(i)f‖C(S2)‖G(∆∗; ·)−BGn‖L1([−1,1]). (5.7)

Since both kernels G(∆∗; ·) and BGn are of class L2([−1, 1]) and, for all k ∈ N0, theLegendre coefficients of the Bernstein kernelK∧B,n(k) converge to 1 for n tending to infinity,we are able to deduce that

limn→∞

‖G(∆∗; ·)−BGn‖L2([−1,1]) = 0. (5.8)

Obviously, this implies L1-convergence as well as ‖Fi − F (n)i ‖C(S2) → 0 for i = 1, 2, 3 and

n→∞, as required.

Considering the o(i)-derivatives, we have to verify the following lemma.

Lemma 5.17For i ∈ {1, 2, 3},

limn→∞

supξ∈S2

∣∣∣o(i)ξ Fi(ξ)− o

(i)ξ F

(n)i (ξ)

∣∣∣ = 0. (5.9)

Proof. The case i = 1 is obvious and it is not difficult to see that, for i ∈ {2, 3},

‖o(i)ξ Fi(ξ)− o

(i)ξ F

(n)i (ξ)‖c(S2) (5.10)

= supξ∈S2

∣∣∣∣o(i)ξ

∫S2G(∆∗; ξ · η)O(i)

η f(η)dS(η)−o(i)ξ


η f(η)dS(η)

∣∣∣∣= sup

ξ∈S2

∣∣∣∣ ∫S2o

(i)ξ G(∆∗; ξ · η)O(i)

η f(η)dS(η)−∫S2o

(i)ξ BGn(ξ · η)O(i)

η f(η)dS(η)

∣∣∣∣,where it is clear that the operator o(i) can be drawn inside the two integrals. This leadsus to the following estimate:

supξ∈S2

∣∣∣∣ ∫S2o

(i)ξ G(∆∗; ξ · η)O(i)

η f(η)dS(η)−∫S2o

(i)ξ BGn(ξ · η)O(i)

η f(η)dS(η)

∣∣∣∣≤ sup

ξ∈S2

∫S2

∣∣∣o(i)ξ G(∆∗; ξ · η)− o(i)

ξ BGn(ξ · η)∣∣∣ ∣∣O(i)

η f(η)∣∣ dS(η)

≤ ‖O(i)f‖C(S2) supξ∈S2

∫S2

∣∣∣o(i)ξ G(∆∗; ξ · η)− o(i)

ξ BGn(ξ · η)∣∣∣ dS(η). (5.11)

We have to study the convergence of the last integral. In more detail, we are interestedin proving that

limn→∞

supξ∈S2

∫S2

∣∣∣o(i)ξ G(∆∗; ξ · η)− o(i)

ξ BGn(ξ · η)∣∣∣ dS(η) = 0. (5.12)


For that purpose, we notice that the Bernstein kernels o(i)ξ BGn(ξ · η), i = 2, 3, admit the

following (Legendre) series expansions

o(2)ξ BGn(ξ · η) =

n∑k=1

2k + 1

4π

K∧B,n(k)

−k(k + 1)P ′k(ξ · η) (η − (ξ · η)ξ) , (5.13)

o(3)ξ BGn(ξ · η) =

n∑k=1

2k + 1

4π

K∧B,n(k)

−k(k + 1)P ′k(ξ · η) (ξ ∧ η) . (5.14)

Moreover, an easy calculation shows that the application of the o(i)-operators, i = 2, 3, tothe Green function with respect to the Beltrami operator leads us to the identities

o(2)ξ G(∆∗; ξ · η) = − 1

4π

η − (η · ξ)ξ1− η · ξ

, o(3)ξ G(∆∗; ξ · η) = − 1

4π

ξ ∧ η1− η · ξ

. (5.15)

Consequently, for i = 2, 3, our integral can be expressed in the form∫S2

∣∣∣o(i)ξ G(∆∗; ξ · η)− o(i)

ξ BGn(ξ · η)∣∣∣ dS(η) (5.16)

=

∫S2

∣∣∣∣−1

4π

o(i)ξ (ξ · η)

1− ξ · η− −1

4π

n∑k=1

2k + 1

k(k + 1)K∧B,n(k)P ′k(ξ · η)o

(i)ξ (ξ · η)

∣∣∣∣dS(η)

=1

4π

∫S2

∣∣∣o(i)ξ (ξ · η)

∣∣∣ ∣∣∣∣ 1

1− ξ · η−

n∑k=1

2k + 1

k(k + 1)K∧B,n(k)P ′k(ξ · η)

∣∣∣∣dS(η)

=1

2

∫ 1

−1

√1− t2

∣∣∣∣ 1

1− t−

n∑k=1

2k + 1

k(k + 1)K∧B,n(k)P ′k(t)

∣∣∣∣ dt.At this point, we use the recurrence relation (3.61) for the Legendre polynomials, i.e.,λ = 1

2in the version for ultraspherical polynomials. This gives us the identity∫

S2

∣∣∣o(i)ξ G(∆∗; ξ · η)− o(i)

ξ BGn(ξ · η)∣∣∣ dS(η) (5.17)

=1

2

∫ 1

−1

√1 + t

1− t

∣∣∣∣1− (1− t)n∑k=1

K∧B,n(k)Pk+1(t)− Pk−1(t)

t2 − 1

∣∣∣∣ dt=

1

2

∫ 1

−1

√1 + t

1− t

∣∣∣∣1 +1

1 + t

n∑k=1

K∧B,n(k) (Pk+1(t)− Pk−1(t))

∣∣∣∣ dt .For the occurring sum, it follows that

n∑k=1

K∧B,n(k) (Pk+1(t)− Pk−1(t))

=K∧B,n(n)Pn+1(t) +K∧B,n(n− 1)Pn(t)−K∧B,n(2)P1(t)−K∧B,n(1)P0(t)

+n−1∑k=2

(K∧B,n(k − 1)−K∧B,n(k + 1)

)Pk(t) , (5.18)


where a simple calculation shows that

K∧B,n(k − 1)−K∧B,n(k + 1) = K∧B,n+1(k)(2k + 1)2

(n+ 2). (5.19)

We put (5.19) into (5.18) getting the following result

n∑k=1

K∧B,n(k) (Pk+1(t)− Pk−1(t)) (5.20)

= K∧B,n(n)Pn+1(t) +K∧B,n(n− 1)Pn(t)−K∧B,n(2)P1(t)−K∧B,n(1)P0(t)

+2

n+ 2

n−1∑k=2

K∧B,n+1(k)(2k + 1)Pk(t)

=2

n+ 2

n+1∑k=0

K∧B,n+1(k)(2k + 1)Pk(t)− (1 + t)

=2

n+ 2(n+ 1)

(1 + t

2

)n+1

− (1 + t) .

Keeping this result in mind, we return to the integral (5.17). As a matter of fact, theidentity (5.17) can be rewritten in the form

1

2

∫ 1

−1

√1 + t

1− t

∣∣∣∣1 +1

1 + t

n∑k=1

K∧B,n(k) (Pk+1(t)− Pk−1(t))

∣∣∣∣ dt (5.21)

=1

2

∫ 1

−1

√1 + t

1− t

∣∣∣∣n+ 1

n+ 2

(1 + t

2

)n∣∣∣∣ dt.

Clearly, as the Bernstein kernel is non-negative, we are left with the integral expression∫S2

∣∣∣o(i)ξ G(∆∗; ξ · η)− o(i)

ξ BGn(ξ · η)∣∣∣ dS(η) =

1

2

n+ 1

n+ 2

∫ 1

−1

√1 + t

1− t

(1 + t

2

)ndt

=Γ(n+ 3

2)

Γ(12) Γ(n+ 2)

, (5.22)

which follows by induction. It is well-known that the value of our integral can be estimatedas follows:

1√2n+ 2

<Γ(n+ 3

2)

Γ(12) Γ(n+ 2)

<2√

2n+ 2. (5.23)

Therefore, we immediately obtain the convergence of our integral for n→∞. In addition,we get information about the speed of the convergence, i.e.,∫

S2

∣∣∣o(i)ξ G(∆∗; ξ · η)− o(i)

ξ BGn(ξ · η)∣∣∣ dS(η) = O(n−1/2). (5.24)

Moreover, the convergence obviously is uniform with respect to ξ.After these preparations, we are now in a position to establish the Bernstein summabilityof the Fourier series in terms of vector spherical harmonics.


Theorem 5.18For any vector field f of class c(1)(S2),

limn→∞

supξ∈S2

∣∣∣∣f(ξ)−3∑i=1

n∑k=0i

k∑j=−k

K∧B,n(k)(f (i))∧

(k, j)y(i)k,j(ξ)

∣∣∣∣ = 0, (5.25)

where, as usual, 01 = 0 and 0i = 1, i = 2, 3, and the Fourier coefficients of f are given by(f (i))∧

(k, j) = 〈f, y(i)k,j〉l2(S2) =

∫S2f(η) · y(i)

k,j(η)dS(η). (5.26)

Proof. From Lemma 5.17, we know that, for f ∈ c(1)(S2),

limn→∞

supξ∈S2

∣∣∣f(ξ)−3∑i=1

o(i)ξ F

(n)i (ξ)

∣∣∣ = limn→∞

supξ∈S2

∣∣∣ 3∑i=1

o(i)ξ Fi(ξ)−

3∑i=1

o(i)ξ F

(n)i (ξ)

∣∣∣≤

3∑i=1

limn→∞

supξ∈S2

∣∣∣o(i)ξ Fi(ξ)− o

(i)ξ F

(n)i (ξ)

∣∣∣ = 0. (5.27)

The expression o(1)ξ F

(n)1 (ξ) can be expressed in the form

o(1)ξ F

(n)1 (ξ) = o

(1)ξ

∫S2KB,n(ξ · η)O(1)

η f(η)dS(η)

=n∑k=0

K∧B,n(k)o(1)ξ

k∑j=−k

(O(1)f

)∧(k, j)Yk,j(ξ)

=n∑k=0

k∑j=−k

K∧B,n(k)(O(1)f

)∧(k, j)y

(1)k,j(ξ), (5.28)

where we have(O(1)f

)∧(k, j) =

∫S2O(1)η f(η)Yk,j(η)dS(η) =

∫S2f(η) · o(1)

η Yk,j(η)dS(η)

=

∫S2f(η) · y(1)

k,j(η)dS(η) =(f (1))∧

(k, j). (5.29)

Furthermore, for i = 2, 3, it is not difficult to see that

o(i)ξ F

(n)i (ξ) = −o(i)

ξ


η f(η)dS(η)

=n∑k=1

K∧B,n(k)

k(k + 1)o

(i)ξ

k∑j=−k

(O(i)f

)∧(k, j)Yk,j(ξ)

=n∑k=1

k∑j=−k

K∧B,n(k)√k(k + 1)

(O(i)f

)∧(k, j)y

(i)k,j(ξ). (5.30)


Taking a look at the coefficients(O(i)f

)∧(k, j), we find(

O(i)f)∧

(k, j) =

∫S2O(i)η f(η)Yk,j(η)dS(η) =

∫S2f(η) · o(i)

η Yk,j(η)dS(η)

=√k(k + 1)

∫S2f(η) · y(i)

k,j(η)dS(η) =√k(k + 1)

(f (i))∧

(k, j). (5.31)

The identities (5.28) and (5.29) as well as (5.30) and (5.31) allow us to conclude

o(i)ξ F

(n)i (ξ) =

n∑k=1

k∑j=−k

K∧B,n(k)(f (i))∧

(k, j)y(i)k,j(ξ), i = 1, 2, 3. (5.32)

In connection with (5.27), we obtain

limn→∞

supξ∈S2

∣∣∣∣f(ξ)−3∑i=1

o(i)ξ F

(n)i (ξ)

∣∣∣∣ (5.33)

= limn→∞

supξ∈S2

∣∣∣∣f(ξ)−3∑i=1

n∑k=0i

k∑j=1−k

K∧B,n(k)(f (i))∧

(k, j)y(i)k,j(ξ)

∣∣∣∣ = 0,

provided that f ∈ c(1)(S2). This is the desired result.Next, a well-known density argument enables us to verify the closure of the vector sphericalharmonics

{y

(i)k,j

}i,k,j

in the space c(S2).

Theorem 5.19For any given ε > 0 and each f ∈ c(S2), there exist N ∈ N and coefficients d

(i)k,j, i = 1, 2, 3,

k = 0, . . . , N , j = −k, . . . , k, such that∥∥∥∥f − 3∑i=1

N∑k=0i

k∑j=−k

d(i)k,jy

(i)k,j

∥∥∥∥c(S2)

≤ ε. (5.34)

Proof. Indeed, if we take any f ∈ c(S2) and any ε > 0, we find a field g ∈ c(1)(S2) suchthat supξ∈S2 |f(ξ)− g(ξ)| < ε

2. Due to Theorem 5.18, there also exists an integer N with

supξ∈S2

∣∣∣∣g(ξ)−3∑i=1

N∑k=0i

k∑j=−k

K∧B,N(k)(g(i))∧

(k, j)y(i)k,j(ξ)

∣∣∣∣ < ε

2. (5.35)

By combining both inequalities, we obtain

supξ∈S2

∣∣∣∣f(ξ)−3∑i=1

N∑k=0i

k∑j=−k

K∧B,N(k)(g(i))∧

(k, j)y(i)k,j(ξ)

∣∣∣∣ < ε, (5.36)

i.e., the coefficients are d(i)k,j = K∧B,N(k)

(g(i))∧

(k, j).By standard arguments, this immediately gives us the closure in c(S2) with respect to

‖ · ‖l2(S2) as well as in l2(S2) which in turn leads to completeness of the system{y

(i)k,j

}i,k,j

in l2(S2).Summarizing our results, we therefore obtain the following theorem.


Theorem 5.20Let {y(i)

n,j}i=1,2,3;n∈N0i;j=−n,...,n be defined by Definition 5.11. Then, the following statements

are valid:

(i) The system of vector spherical harmonics is closed in c(S2) with respect to ‖ · ‖c(S2).

(ii) The system is complete in l2(S2) with respect to 〈·, ·〉l2(S2).

Once more, part (i) of this theorem says that any continuous vector field on S2 can beapproximated arbitrarily close by finite linear combinations of vector spherical harmonics,while part (ii) is equivalent (see Theorem 3.7) to the fact that every vector field f ∈ l2(S2)can be represented by its Fourier (orthogonal) series in terms of the l2(S2)-orthogonal

system{y

(i)n,j

}, i.e.,

limN→∞

∥∥∥∥f − 3∑i=1

N∑n=0i

n∑j=−n

(f (i))∧(n, j)y(i)n,j

∥∥∥∥l2(S2)

= 0, (5.37)

where the Fourier coefficients are given by (5.26).

5.5 Vectorial Addition Theorem

Our interest is the formulation of a vectorial analog of the addition theorem for scalarspherical harmonics (Theorem 4.38). This vectorial addition theorem assures the exis-tence of a reproducing kernel of tensorial structure which is a basic tool, e.g., in thetheory of vectorial zonal functions (cf. [8, 11] and the references therein).

Let {y(i)n,j}i=1,2,3; j=−n,...,n be an l2(S2)-orthonormal basis of harmn(S2), n ∈ N0, as de-

fined by Definition 5.11, corresponding to an L2(S2)-orthonormal basis {Yn,j}j=−n,...,n of

Harmn(S2), i.e., y(i)n,j = (µ

(i)n )−1/2o(i)Yn,j, n ∈ N0i . Then, the announced vectorial analog

of the addition theorem deals with the question of determining the expression

2n+ 1

4πp(i,k)n (ξ, η) =

n∑j=−n

y(i)n,j(ξ)⊗ y

(k)n,j(η), ξ, η ∈ S2, (5.38)

where ⊗ denotes the dyadic tensor product of two vectors. Note that rank-2 tensor fieldsare denoted by lower case bold letters. Furthermore, we require the following bits ofnotation for tensor fields:

The identity tensor in R3×3, i.e.,

i =3∑i=1

εi ⊗ εi , (5.39)

is projected onto the tangential components at a point on the unit sphere. This definesthe surface identity tensor field

itan(ξ) = i− ξ ⊗ ξ , ξ ∈ S2. (5.40)


The surface rotation tensor field is given by

jtan(ξ) = ξ ∧ i =3∑i=1

(ξ ∧ εi)⊗ εi , ξ ∈ S2. (5.41)

Our purpose is to explain how a vectorial counterpart of tensorial nature for the Legen-dre polynomial comes into play. For that purpose, we first extend in canonical way thedefinition of the o(i)-operators defined of Definition 5.7 to vector fields.

Suppose that f : S2 → R3 is a sufficiently smooth vector field with the representation

f(ξ) =3∑

k=1

Fk(ξ)εk, Fk(ξ) = f(ξ) · εk, ξ ∈ S2.

Then, we set

o(i)ξ f(ξ) =

3∑k=1

(o(i)ξ Fk(ξ))⊗ ε

k =3∑

k=1

(o(i)(f(ξ) · εk)

)⊗ εk, (5.42)

where i = 1, 2, 3. Thus, o(i) maps scalar functions to vector fields and vector fields torank-2 tensor fields, respectively.In accordance with this nomenclature, (5.38) can be expressed as follows:

n∑j=−n

y(i)n,j(ξ)⊗ y

(k)n,j(η) = (µ(i)

n )−1/2(µ(k)n )−1/2

n∑j=−n

o(i)ξ Yn,j(ξ)⊗ o

(k)η Yn,j(η)

= (µ(i)n )−1/2(µ(k)

n )−1/2o(i)ξ o

(k)η

n∑j=−n

Yn,j(ξ)Yn,j(η)

= (µ(i)n )−1/2(µ(k)

n )−1/2 2n+ 1

4πo

(i)ξ o

(k)η Pn(ξ · η).

In other words, p(i,k)n : S2 × S2 → R3×3, n ∈ N, is given in terms of the one-dimensional

Legendre polynomial by

p(i,k)n (ξ, η) = (µ(i)

n )−1/2(µ(k)n )−1/2o

(i)ξ o

(k)η Pn(ξ · η), ξ, η ∈ S2, (5.43)

and p(1,1)0 (ξ, η) = o

(1)ξ o

(1)η P0(ξ ·η) = ξ⊗η. Altogether, this leads us to the following variant

of the addition theorem for vector spherical harmonics.

Theorem 5.21 (Addition Theorem for harmn(S2))Let {y(i)

n,j}i=1,2,3;j=−n,...,n be an l2(S2)-orthonormal basis of the space harmn(S2), n ∈ N0.Then,

n∑j=−n

y(i)n,j(ξ)⊗ y

(k)n,j(η) =

2n+ 1

4πp(i,k)n (ξ, η)

= (µ(i)n )−1/2(µ(k)

n )−1/2 2n+ 1

4πo

(i)ξ o

(k)η Pn(ξ · η)


holds for ξ, η ∈ S2, n ∈ N, and i, k ∈ {1, 2, 3}. For (i, k) = (1, 1), we have

n∑j=−n

y(1)n,j(ξ)⊗ y

(1)n,j(η) =

2n+ 1

4πp(1,1)n (ξ, η),

for n ∈ N0, i.e., also for n = 0.

Definition 5.22The kernel p

(i,k)n : S2 × S2 → R3×3, i, k ∈ {1, 2, 3}, given by

p(i,k)n (ξ, η) = (µ(i)

n )−1/2(µ(k)n )−1/2o

(i)ξ o

(k)η Pn(ξ · η) =

4π

2n+ 1

n∑j=−n

y(i)n,j(ξ)⊗ y

(k)n,j(η) (5.44)

is called the Legendre rank-2 tensor kernel of type (i, k) and degree n ∈ N0i,k with respect

to the dual system of operators o(i), O(i), i ∈ {1, 2, 3}. Note that we use the notation

N0i,k =

{N0 , (i, k) = (1, 1),

N , else.

The kernel

pn =3∑i=1

3∑k=1

p(i,k)n

is called Legendre rank-2 tensor kernel of degree n ∈ N with respect to the dual system ofoperators o(i), O(i), i = 1, 2, 3, and p0 = p

(1,1)0 .

The main problem to be solved is the evaluation of p(i,k)n as introduced by (5.43). As

auxiliary results we verify the following identities.

Lemma 5.23For ξ, η ∈ S2,

o(2)ξ (ξ − (ξ · η)η) = itan(ξ)− (η − (ξ · η)ξ)⊗ η,

o(3)ξ (ξ − (ξ · η)η) = jtan(ξ)− (ξ ∧ η)⊗ η,

o(2)ξ (η ∧ ξ) = −jtan(η)− ξ ⊗ η ∧ ξ,

o(3)ξ (η ∧ ξ) = (ξ · η)itan(ξ)− (η − (ξ · η)ξ)⊗ ξ.

Proof. We prove the first and third formulas. The second and fourth formulas follow bysimilar arguments, since we know that L∗ξ = ξ ∧∇∗ξ . First, we get

o(2)ξ (ξ − (ξ · η)η) = o

(2)ξ

3∑l=1

((ξ · εl)− (ξ · η)(η · εl))εl

=3∑l=1

(εl − (ξ · εl)ξ)⊗ εl − ((η · εl)(η − (ξ · η)ξ)⊗ εl)

= itan(ξ)− (η − (ξ · η)ξ)⊗ η.


Furthermore, we have

o(2)ξ (η ∧ ξ) = o

(2)ξ

3∑l=1

((η ∧ ξ) · εl)εl

= o(2)ξ

3∑l=1

((εl ∧ η) · ξ)εl

=3∑l=1

(εl ∧ η − ((εl ∧ η) · ξ)ξ)⊗ εl

=3∑l=1

(−η ∧ εl)⊗ εl − ((η ∧ ξ) · εl)ξ ⊗ εl

= −jtan(η)− ξ ⊗ η ∧ ξ.

which yields the third formula.If F : [−1, 1]→ R is sufficiently smooth, then, for ξ, η ∈ S2, we obtain from Lemma 5.23:

o(1)ξ o(1)

η F (ξ · η) = F (ξ · η)ξ ⊗ η,

o(1)ξ o(2)

η F (ξ · η) = F ′(ξ · η)ξ ⊗ (ξ − (ξ · η)η),

o(1)ξ o(3)

η F (ξ · η) = F ′(ξ · η)ξ ⊗ (η ∧ ξ).

Similar results hold for o(2)ξ o

(1)η F (ξ·η) and o

(3)ξ o

(1)η F (ξ·η). Treating the tangential operators,

we find, for ξ, η ∈ S2,

o(2)ξ o(2)

η F (ξ · η) = ∇∗ξ ⊗ (F ′(ξ · η)(ξ − (ξ · η)η))

= (∇∗ξF ′(ξ · η))⊗ (ξ−(ξ · η)η)+F ′(ξ · η)∇∗ξ ⊗ (ξ−(ξ · η)η)

= F ′′(ξ · η)(η − (ξ · η)ξ)⊗ (ξ − (ξ · η)η)

+F ′(ξ · η)(itan(ξ)− (η − (ξ · η)ξ)⊗ η)

and

o(2)ξ o(3)

η F (ξ · η) = ∇∗ξ ⊗ (F ′(ξ · η)η ∧ ξ)= (∇∗ξF ′(ξ · η))⊗ (η ∧ ξ) + F ′(ξ · η)∇∗ξ ⊗ (η ∧ ξ)= F ′′(ξ · η)(η − (ξ · η)ξ)⊗ (η ∧ ξ)

+F ′(ξ · η) (−jtan(η)− ξ ⊗ (η ∧ ξ)) .

Similar calculations yield the following formulas

o(3)ξ o(2)

η F (ξ · η) = F ′′(ξ · η)(ξ ∧ η)⊗ (ξ − (ξ · η)η)

+F ′(ξ · η)(jtan(ξ)− (ξ ∧ η)⊗ η)

and

o(3)ξ o(3)

η F (ξ · η) = F ′′(ξ · η)(ξ ∧ η)⊗ (η ∧ ξ)+F ′(ξ · η)((ξ · η)itan(ξ)− (η − (ξ · η)ξ)⊗ ξ).


Applying these results to the scalar Legendre polynomial Pn, we obtain the followingrepresentation for the Legendre rank-2 tensor kernel p

(i,k)n of type (i, k) and degree n ∈ N0i,k

with respect to the dual system of operators o(i), O(i), i, k ∈ {1, 2, 3}.

Theorem 5.24Suppose that n ∈ N0i,k , i, k ∈ {1, 2, 3}. Then, the identities

p(1,1)n (ξ, η) =Pn(ξ · η)ξ ⊗ η,

p(1,2)n (ξ, η) =

1√n(n+ 1)

P ′n(ξ · η)ξ ⊗ (ξ − (ξ · η)η),

p(1,3)n (ξ, η) =

1√n(n+ 1)

P ′n(ξ · η)ξ ⊗ η ∧ ξ,

p(2,1)n (ξ, η) =

1√n(n+ 1)

P ′n(ξ · η)(η − (ξ · η)ξ)⊗ η,

p(3,1)n (ξ, η) =

1√n(n+ 1)

P ′n(ξ · η)ξ ∧ η ⊗ η,

p(2,2)n (ξ, η) =

1

n(n+ 1)(P ′′n (ξ · η)(η − (ξ · η)ξ)⊗ (ξ − (ξ · η)η)

+ P ′n(ξ · η)(itan(ξ)− (η − (ξ · η)ξ)⊗ η)),

p(2,3)n (ξ, η) =

1

n(n+ 1)(P ′′n (ξ · η)(η − (ξ · η)ξ)⊗ η ∧ ξ

+ P ′n(ξ · η)(−jtan(η)− ξ ⊗ η ∧ ξ)),

p(3,2)n (ξ, η) =

1

n(n+ 1)(P ′′n (ξ · η)ξ ∧ η ⊗ (ξ − (ξ · η)η)

+ P ′n(ξ · η)(jtan(ξ)− ξ ∧ η ⊗ η)),

p(3,3)n (ξ, η) =

1

n(n+ 1)(P ′′n (ξ · η)ξ ∧ η ⊗ η ∧ ξ

+ P ′n(ξ · η)((ξ · η)itan(ξ)− (η − (ξ · η)ξ)⊗ ξ))

hold for all (ξ, η) ∈ S2 × S2.

The case ξ = η in the previous theorem is of particular interest. Observing that Pn(1) = 1,P ′n(1) = 1

2n(n+1), and P ′′n (1) = 1

8n(n+1)(n(n+1)−2), we obtain the following corollary.

Corollary 5.25For n ∈ N0i,k , i, k ∈ {1, 2, 3}, and all ξ ∈ S2,

p(1,1)n (ξ, ξ) = ξ ⊗ ξ,

p(1,i)n (ξ, ξ) = p(i,1)

n (ξ, ξ) = 0, i = 2, 3,

p(2,2)n (ξ, ξ) = p(3,3)

n (ξ, ξ) =1

2itan(ξ),

p(2,3)n (ξ, ξ) = −p(3,2)

n (ξ, ξ) = −1

2jtan(ξ).


It follows readily that

trace ((η − (ξ · η)ξ)⊗ (ξ − (ξ · η)η)) = −(ξ · η)(1− (ξ · η)2)

andtrace ((ξ ∧ η)⊗ (η ∧ ξ)) = −(1− (ξ · η)2).

Hence, we get the following identities from Theorem 5.24.

Lemma 5.26For n ∈ N0i,k , i, k ∈ {1, 2, 3}, and all ξ, η ∈ S2, we have

trace (p(1,1)n (ξ, η)) =Pn(ξ · η)(ξ · η),

trace (p(1,2)n (ξ, η)) =trace (p(2,1)

n (ξ, η))

=1√

n(n+ 1)P ′n(ξ · η)(1− (ξ · η)2),

trace (p(2,2)n (ξ · η)) =

1

n(n+ 1)(P ′′n (ξ · η)(1− (ξ · η)2)(ξ · η) + 2P ′n(ξ · η)),

trace (p(3,3)n (ξ, η)) =Pn(ξ · η),

trace (p(1,3)n (ξ, η)) =trace (p(3,1)

n (ξ, η)) = trace (p(2,3)n (ξ, η))

=trace (p(3,2)n (ξ, η)) = 0.

Taking ξ = η in Lemma 5.26, we get in connection with Theorem 5.21 the following result.

Lemma 5.27Let n ∈ N0i and i ∈ {1, 2, 3}. If y

(i)n,j, j = −n, . . . , n, forms an l2(S2)-orthonormal basis of

harm(i)n (S2), then

n∑j=−n

∣∣∣y(i)n,j(ξ)

∣∣∣2 = tracen∑

j=−n

y(i)n,j(ξ)⊗ y

(i)n,j(ξ) =

2n+ 1

4π.

Every vector spherical harmonic y(i)n ∈ harm(i)

n (S2) of degree n and type i can be repre-sented by its orthogonal expansion

y(i)n =

n∑j=−n

an,jy(i)n,j

withan,j = 〈y(i)

n , y(i)n,j〉l2(S2), j = −n, . . . , n.

The application of the Cauchy–Schwarz inequality in combination with Lemma 5.27 yieldsthe estimate

|y(i)n (ξ)|2 ≤

( n∑j=−n

a2n,j

)( n∑j=−n

|y(i)n,j(ξ)|2

)=

2n+ 1

4π

n∑j=−n

a2n,j =

2n+ 1

4π‖y(i)

n ‖2l2(S2) (5.45)

for all ξ ∈ S2. Thus, we finally obtain the following lemma.


Lemma 5.28Suppose y

(i)n is a member of harm(i)

n (S2), n ∈ N0i , i ∈ {1, 2, 3}. Then,

‖y(i)n ‖c(S2) ≤

√2n+ 1

4π‖y(i)

n ‖l2(S2).

In particular,

‖y(i)n,j‖c(S2) ≤

√2n+ 1

4π.

5.6 Vectorial Funk–Hecke Formulas

Next, we deal with generalizations of the Funk–Hecke formula (Theorem 4.61) to thevectorial case. It turns out that we find two variants:

1. Let g(·, η) : S2 → R3 be a vector field which is invariant with respect to all orthogonaltransformations t ∈ SO(3) leaving η ∈ S2 fixed. Determine the integral∫

S2g(ξ, η) · y(i)

n (ξ)dS(ξ), (5.46)

where y(i)n ∈ harm(i)

n (S2), n ∈ N0i .

2. Let G ∈ L1([−1, 1]), η ∈ S2 fixed. Determine the integral∫S2G(ξ · η)y(i)

n (ξ)dS(ξ), (5.47)

where y(i)n ∈ harm(i)

n (S2), n ∈ N0i .

Remark 5.29Note that the integral (5.46) is scalar-valued, while (5.47) is vector-valued. This differ-ence causes completely different ways of establishing the Funk–Hecke formulas. The firstvariant uses certain properties of invariant vector fields, while the second one is based onthe Cartesian representation of vector spherical harmonics. We focus on the first variant(see, e.g., [9] for the other).

We remember the representation theory needed for our studies on the Funk–Hecke formula(see Sections 4.2 and 4.5).Now, assume that f, g are of class l2(S2), t ∈ SO(3). Then, it follows that

〈Rtf, g〉l2(S2) = 〈f,RtT g〉l2(S2). (5.48)

But this means that the adjoint operator of Rt is given by RtT . Let F ∈ C(1)(S2) andt ∈ SO(3). Then, we find with (4.4) of Definition 4.7 that

o(1)ξ RtF (ξ) = ξF (tξ) = tT tξF (tξ) =

(Rto

(1)F)

(ξ) (5.49)


and by the chain rule

o(2)ξ RtF (ξ) = ∇∗ξF (tξ) = tT (∇∗F )(tξ) =

(Rto

(2)F)

(ξ). (5.50)

An analogous result holds for o(3). Together with (5.48), for F ∈ C(1)(S2) and anyt ∈ SO(3), we obtain that, for all i ∈ {1, 2, 3} and ξ ∈ S2,

o(i)ξ RtF (ξ) = Rto

(i)F (ξ) (5.51)

and for f ∈ c(1)(S2)

O(i)ξ Rtf(ξ) = RtO

(i)f(ξ). (5.52)

Therefore, we remember in connection with the results of Section 4.2 and Theorem 4.58that the space l2(i)(S2) is SO(3)-invariant for all i ∈ {1, 2, 3}.Another coordinate-free classification of vector spherical harmonics can be given by look-ing at the following system of partial differential equations:

ξ∆∗ξ(ξ · f(ξ)) + n(n+ 1)f(ξ) = 0, n ≥ 0, (5.53)

∇∗ξ(∇∗ξ · f(ξ)) + n(n+ 1)f(ξ) = 0, n ≥ 1, (5.54)

L∗ξ(∇∗ξ · (f(ξ) ∧ ξ)) + n(n+ 1)f(ξ) = 0, n ≥ 1. (5.55)

Solutions y(1)n of (5.53) fulfill ξ∧y(1)

n (ξ) = 0 and, consequently, there exists a scalar function

F such that y(1)n (ξ) = ξF (ξ). In connection with (5.53), this leads to

ξ(∆∗ξF (ξ) + n(n+ 1)F (ξ)) = 0, (5.56)

which means that F is a spherical harmonic of degree n, i.e., solutions of (5.53) are of the

form y(1)n (ξ) = ξYn(ξ).

For solutions y(2)n of (5.54), we get ξ · y(2)

n (ξ) = 0 and ∇∗ξ · (ξ∧ y(2)n (ξ)) = 0, such that there

exists a scalar function G with y(2)n (ξ) = ∇∗ξG(ξ). Together with (5.54), this leads to

∇∗ξ(∆∗ξG(ξ) + n(n+ 1)G(ξ)) = 0. (5.57)

Consequently, we have∆∗ξG(ξ) + n(n+ 1)G(ξ) = const. (5.58)

This means that, up to a constant, G is a spherical harmonic of degree n, i.e., solutionsof (5.54) are of the form y

(2)n (ξ) = ∇∗ξYn(ξ).

Analogously, solutions y(3)n of (5.55) fulfill both ξ ·y(3)

n (ξ) = 0 and ∇∗ξ ·y(3)n (ξ), which means

that there exists a scalar function H such that y(3)n (ξ) = L∗ξH(ξ). Consequently,

L∗ξ(∆∗ξH(ξ) + n(n+ 1)H(ξ)) = 0 (5.59)

and, therefore, y(3)n (ξ) = L∗ξYn(ξ).

As we have seen, the solutions of the differential equations (5.53)–(5.55) are the vec-tor spherical harmonics. This observation has immediate consequences for the spacesharm(i)

n (S2) of vector spherical harmonics. In fact, they can be regarded as “the smallest”orthogonally invariant spaces.


Theorem 5.30The spaces harm(i)

n (S2) of vector spherical harmonics are orthogonally invariant and irre-ducible.

Proof. The orthogonal invariance is a direct consequence of the invariant differentialoperators of (5.53)–(5.55). To be more concrete, suppose that there exists an orthogonallyinvariant subspace of harm(i)

n (S2).The application of the operators O(i) to the respective elements would — because ofDefinition 5.11 — generate an orthogonally invariant subspace in the space of scalarspherical harmonics. This, however, is a contradiction to the irreducibility of the spacesHarmn(S2).Theorem 5.30 shows that vector spherical harmonics have the same significance for spher-ical vector fields as the spherical harmonics in the theory of scalar spherical fields. Furtherconsequences for vector spherical harmonics can be found, e.g., in [11].Assume that F ∈ L2(S2) with RtF = F for all t ∈ SOη(3). Then, we already know thatthere exists a function F ∈ L2([−1, 1]) such that F (ξ) = F (ξ · η), ξ ∈ S2. Furthermore,we have shown in Lemma 4.56 that if F is, in addition, a spherical harmonic of degree n,i.e., F ∈ Harmn(S2), there exists a constant C ∈ R such that

F (ξ) = C Pn(ξ · η), ξ ∈ S2. (5.60)

A generalization of these results to the vectorial case can be written down as follows byusing (5.51) or (5.52), respectively:

1. If f ∈ c(1)(S2) satisfies Rtf = f for all t ∈ SOη(3), η fixed, then there exist functionsFi ∈ C([−1, 1]), i = 1, 2, 3, such that

O(i)ξ f(ξ) = Fi(ξ · η), ξ ∈ S2. (5.61)

2. Let i ∈ {1, 2, 3} and y(i)n ∈ harm(i)

n (S2), n ∈ N0i , with Rty(i)n = y

(i)n for all rotations

t ∈ SOη(3), η fixed. There exists a constant C ∈ R such that

y(i)n (ξ) = C o

(i)ξ Pn(ξ · η), ξ ∈ S2. (5.62)

Let η ∈ S2 be fixed. Assume that g(·, η) ∈ c(1)(S2) is a spherical vector field withRtg(ξ, η) = g(ξ, η), ξ ∈ S2, for all t ∈ SOη(3). It follows from the considerations above

that, for i ∈ {1, 2, 3}, the functions O(i)ξ g(ξ, η) = Gi(ξ · η) depend only on the inner

product ξ · η. Thus, we may define in analogy to Theorem 4.61

(O(i)g)∧(n) = 2π

∫ 1

−1

Gi(t)Pn(t) dt. (5.63)

It follows from Corollary 4.62 that y(i)n = o(i)Yn ∈ harm(i)

n (S2), n ∈ N0i , satisfies∫S2g(ξ, η) · y(i)

n (ξ)dS(ξ) =

∫S2O

(i)ξ g(ξ, η)Yn(ξ)dS(ξ) (5.64)

= (O(i)g)∧(n)Yn(η).

This leads us to the first variant of the vectorial Funk–Hecke formula.


Theorem 5.31 (Funk–Hecke Formula)Let η ∈ S2 be fixed. Assume that g(·, η) ∈ c(1)(S2) satisfies

Rtg(ξ, η) = g(ξ, η) (5.65)

for all t ∈ SOη(3) and all ξ ∈ S2. Then, for i ∈ {1, 2, 3} and y(i)n ∈ harm(i)

n (S2), n ≥ 0i,∫S2g(ξ, η) · y(i)

n (ξ)dS(ξ) = (µ(i)n )−1(O(i)g)∧(n)O(i)

η y(i)n (η), (5.66)

where (O(i)g)∧(n) is defined by (5.63).

By virtue of the addition theorem for vector spherical harmonics, we immediately obtainthe following consequence.

Corollary 5.32Let η ∈ S2 be fixed, g(·, η) ∈ c(1)(S2). Assume that

Rtg(ξ, η) = g(ξ, η) (5.67)

for all t ∈ SOη(3) and all ξ ∈ S2. Then, for all ζ ∈ S2, i ∈ {1, 2, 3}, and n ∈ N0i ,∫S2g(ξ, η)Tp(i,i)

n (ξ, ζ)dS(ξ) = (µ(i)n )−1(O(i)g)∧(n) o

(i)ζ Pn(ζ · η). (5.68)

5.7 Homogeneous Harmonic Vector Polynomials

We could identify scalar spherical harmonics as homogeneous harmonic polynomials re-stricted to the unit sphere. This gave us many results in scalar theory. In this sectionwe develop a relation between the vector spherical harmonics of Section 5.2 and homoge-neous harmonic vector polynomials. The separation into tangential/normal componentson the one hand and the Cartesian components on the other hand make these relationsmore difficult than in the scalar case. However, we are given an alternative system of vec-tor spherical harmonics and an alternative way to prove the completeness of the vectorspherical harmonics of Section 5.2.

Definition 5.33A vector field hn : R3 → R3 is called a homogeneous harmonic vector polynomial of degreen if hn · εi ∈ Harmn(R3) for every i ∈ {1, 2, 3}.

Definition 5.34Let n ∈ N0i , Hn ∈ Homn(R3), and i ∈ {1, 2, 3}. The operators o

(i)n are defined by

o(1)n Hn(x) =

((2n+ 1)x− |x|2∇x

)Hn(x) , (5.69)

o(2)n Hn(x) = ∇xHn(x) , (5.70)

o(3)n Hn(x) = x ∧∇xHn(x) . (5.71)


Lemma 5.35Let Hn ∈ Harmn(R3), n ∈ N0i . Then, o

(i)n Hn is a homogeneous harmonic vector polyno-

mial of degree

deg(i)(n) =

n+ 1 if i = 1,

n− 1 if i = 2,

n if i = 3.

(5.72)

Proof. The cases i = 2, 3 are straightforward, thus, we can restrict ourselves to the casei = 1. Observe that the components of o

(1)n Hn are homogeneous of degree n + 1. For

j = 1, 2, 3 and n ∈ N, we have that

∆x

((o(1)n Hn(x)

)· εj)

= ∆x

((2n+ 1)xjHn(x)− |x|2 ∂

∂xjHn(x)

)(5.73)

= 2(2n+ 1)εj · ∇xHn(x)−(∆x|x|2

) ∂

∂xjHn(x)−

(2∇x|x|2

)·(∇x

∂

∂xjHn(x)

)= 2(2n+ 1)

∂

∂xjHn(x)− 6

∂

∂xjHn(x)− 4x · ∇x

∂

∂xjHn(x)

= (4n− 4)∂

∂xjHn(x)− 4(n− 1)

∂

∂xjHn(x) = 0,

since

x · ∇x∂

∂xjHn(x) = (n− 1)

∂

∂xjHn(x) (5.74)

and ∂∂xjHn(x) is a homogeneous harmonic polynomial of degree n− 1. The case n = 0 is

trivial.

Lemma 5.36The space of functions⋃

n∈N0

3⊕i=1

Harmn(S2)εi ={Yn,kε

i : n ∈ N0, k = −n, . . . , n, i = 1, 2, 3}

(5.75)

is closed in l2(S2) with respect to ‖·‖l2(S2).

Proof. Let f ∈ l2(S2). There exist F1, F2, F3 ∈ L2(S2) such that

f = F1ε1 + F2ε

2 + F3ε3. (5.76)

Let ε > 0. Then, there exist coefficients a(i)n,k , i = 1, 2, 3, n = 0, . . . , Ni, k = −n, . . . , n,

such that ∥∥∥F1 −N1∑n=0

n∑k=−n

a(1)n,kYn,k

∥∥∥2

L2(S2)<ε

3, (5.77)

∥∥∥F2 −N2∑n=0

n∑k=−n

a(2)n,kYn,k

∥∥∥2

L2(S2)<ε

3, (5.78)

∥∥∥F3 −N3∑n=0

n∑k=−n

a(3)n,kYn,k

∥∥∥2

L2(S2)<ε

3. (5.79)


Set N = max{N1, N2, N3} and set

a(1)n,k = 0 for N1 < n ≤ N, −n ≤ k ≤ n, (5.80)

a(2)n,k = 0 for N2 < n ≤ N, −n ≤ k ≤ n, (5.81)

a(3)n,k = 0 for N3 < n ≤ N, −n ≤ k ≤ n. (5.82)

It is clear that

g =3∑i=1

N∑n=0

n∑k=−n

a(i)n,kYn,kε

i ∈⋃n∈N0

3⊕i=1

Harmn(S2)εi (5.83)

and

‖f − g‖2l2(S2) =

∫S2

3∑i=1

(Fi(ξ)−Gi(ξ))2 dS(ξ) (5.84)

=3∑i=1

‖Fi −Gi‖2L2(S2) <

ε

3+ε

3+ε

3= ε

which proves the closure in l2(S2) with respect to ‖·‖l2(S2). As in the scalar case, we

can use closure to find completeness in l2(S2). Using the representation of the gradient inpolar coordinates, i.e.,

∇x = εr∂

∂r+

1

r∇∗ξ = ξ

∂

∂r+

1

r∇∗ξ , (5.85)

we obtain

(2n+ 1)x− |x|2∇x = (2n+ 1)rξ − r2(ξ∂

∂r+

1

r∇∗ξ)

(5.86)

= ξ(

(2n+ 1)r − r2 ∂

∂r

)− r∇∗ξ .

Now, let Yn ∈ Harmn(S2), n ∈ N0i , and x = rξ, r > 0:

o(1)n (rnYn(ξ)) = (n+ 1)rn+1Yn(ξ)ξ − rn+1∇∗ξYn(ξ) , (5.87)

o(2)n (rnYn(ξ)) = nrn−1Yn(ξ)ξ + rn−1∇∗ξYn(ξ) , (5.88)

o(3)n (rnYn(ξ)) = rnL∗ξYn(ξ) . (5.89)

Using the definition of the o(i)-operators and restricting the functions to r = 1, we get(with Hn(x) = rnYn(ξ)):(

o(1)n Hn(x)

) ∣∣∣r=1

= (n+ 1)o(1)ξ Yn(ξ)− o(2)

ξ Yn(ξ) , (5.90)(o(2)n Hn(x)

) ∣∣∣r=1

= no(1)ξ Yn(ξ) + o

(2)ξ Yn(ξ) , (5.91)(

o(3)n Hn(x)

) ∣∣∣r=1

= o(3)ξ Yn(ξ) (5.92)


for ξ ∈ S2. Rearranging the equations (5.90) – (5.92), we find that

o(1)ξ Yn(ξ) =

1

2n+ 1

(o(1)n Hn(x)

) ∣∣∣r=1

+1

2n+ 1

(o(2)n Hn(x)

) ∣∣∣r=1

, (5.93)

o(2)ξ Yn(ξ) = − n

2n+ 1

(o(1)n Hn(x)

) ∣∣∣r=1

+n+ 1

2n+ 1

(o(2)n Hn(x)

) ∣∣∣r=1

, (5.94)

o(3)ξ Yn(ξ) =

(o(3)n Hn(x)

) ∣∣∣r=1

. (5.95)

Note that o(1)ξ Y0(ξ) = (o1

0H0(x)) |r=1 = (xH0(x)) |r=1. This shows the following lemma.

Lemma 5.37The spaces of vector spherical harmonics harm(i)

n (S2), i = 1, 2, 3, and the spaces of scalarspherical harmonics Harmm(S2) with m = n− 1, n, n+ 1 are related by

harm(i)n (S2) ⊂

3⊕j=1

Harmn−1(S2)εj ⊕3⊕j=1

Harmn+1(S2)εj , i = 1, 2, (5.96)

harm(3)n (S2) ⊂

3⊕j=1

Harmn(S2)εj. (5.97)

The space of vector spherical harmonics of all three types is related by

harmn(S2) ⊂n+1⊕

m=n−1

3⊕i=1

Harmm(S2)εi . (5.98)

Altogether, we obtain the following theorem as a consequence of Lemma 5.37.

Theorem 5.38The l2(S2)-orthonormal system {y(i)

n,k}i=1,2,3,n∈N0i,k=−n,...,n is closed in l2(S2) and it is com-

plete in l2(S2) with respect to ‖·‖l2(S2).

This means that, for every f ∈ l2(S2) and ε > 0, there exists an N ∈ N such that

∥∥∥f − 3∑i=1

N∑n=0i

n∑k=−n

〈f, y(i)n,k〉l2(S2)y

(i)n,k

∥∥∥l2(S2)

< ε . (5.99)

Definition 5.39A second system of vector spherical harmonics is given by

y(1)n,k =

√n+ 1

2n+ 1y

(1)n,k −

√n

2n+ 1y

(2)n,k , (5.100)

y(2)n,k =

√n

2n+ 1y

(1)n,k +

√n+ 1

2n+ 1y

(2)n,k , (5.101)

y(3)n,k = y

(3)n,k , (5.102)

where n ∈ N0i and k = −n, . . . , n.


Theorem 5.40The l2(S2)-orthonormal system {y(i)

n,k}i=1,2,3,n∈N0i,k=−n,...,n is

closed and complete in l2(S2), i.e., for all f ∈ l2(S2) and ε > 0 there exists N ∈ Nsuch that ∥∥∥f − 3∑

i=1

N∑n=0i

n∑k=−n

〈f, y(i)n,k〉l2(S2)y

(i)n,k

∥∥∥l2(S2)

< ε . (5.103)

Remark 5.41The system of vector spherical harmonics of Definition 5.39 does not strictly separate intoradial and tangential vector spherical harmonics. However, note that

∆∗ξ y(1)n−1,k(ξ) = −n(n+ 1)y

(1)n−1,k(ξ) , n ∈ N, −(n− 1) ≤ k ≤ n− 1, (5.104)

∆∗ξ y(2)n+1,k(ξ) = −n(n+ 1)y

(2)n+1,k(ξ) , n ∈ N0, −(n+ 1) ≤ k ≤ n+ 1, (5.105)

∆∗ξ y(3)n,k(ξ) = −n(n+ 1)y

(3)n,k(ξ) , n ∈ N, −n ≤ k ≤ n. (5.106)

Thus, this system is a set of eigenfunctions of the Beltrami operator which is appliedcomponentwise. Moreover, we can show that, for n ∈ N0i and k = −n, . . . , n,

y(1)n,k(ξ) =

1√(n+ 1)(2n+ 1)

(−∇x

(1

rn+1Yn,k(ξ)

)) ∣∣∣r=1

, (5.107)

y(2)n,k(ξ) =

1√n(2n+ 1)

(∇x (rnYn,k(ξ)))∣∣∣r=1

, (5.108)

y(3)n,k(ξ) =

1√n(n+ 1)

(x ∧∇x

(1

rn+1Yn,k

)) ∣∣∣r=1

=1√

n(n+ 1)(x ∧∇x (rnYn,k))

∣∣∣r=1

. (5.109)

5.8 Vectorial Beltrami Operator

Next, we develop a vectorial analog of the Beltrami operator ∆∗, denoted by ∆∗, suchthat the vector spherical harmonics of class harmn(S2) can be recognized as eigenfunctionsof the vectorial operator ∆∗. In particular, it turns out that the operator ∆∗ correspondsto the orthogonal decomposition with respect to the operators o(i), i = 1, 2, 3, that is tosay, ∆∗f ∈ c(i)(S2) for all i = 1, 2, 3, provided that f ∈ c

(2)(i) (S2).

Our construction is based on the componentwise application of the (scalar) Beltramioperator ∆∗. The point of departure is the following convention: If f ∈ c(2)(S2) is of theform

f(ξ) =3∑i=1

εiFi(ξ), ξ ∈ S2, (5.110)

then ∆∗f is understood to be

∆∗f(ξ) =3∑i=1

εi∆∗Fi(ξ), ξ ∈ S2. (5.111)

Observing this setting, we are able to deduce the following identities.


Lemma 5.42The following statements are true.

1. Let F : S2 → R be sufficiently smooth. Then,

∆∗o(1)F = o(1)(∆∗ − 2)F + 2o(2)F, (5.112)

∆∗o(2)F = −2o(1)∆∗F + o(2)∆∗F, (5.113)

∆∗o(3)F = o(3)∆∗F. (5.114)

2. Let f : S2 → R3 be a sufficiently smooth vector field. Then,

O(1)∆∗f = (∆∗ − 2)O(1)f + 2O(2)f, (5.115)

O(2)∆∗f = −2∆∗O(1)f + ∆∗O(2)f, (5.116)

O(3)∆∗f = ∆∗O(3)f. (5.117)

Proof. We verify only the first formula of part (i), since the other assertions follow bysimilar arguments. Assume that the function under consideration is a spherical harmonicYn of class Harmn(S2), n ∈ N0. Then, it follows from (5.93) that

∆∗ξo(1)ξ Yn(ξ) = ∆∗ξ

1

2n+ 1

(o(1)n rnYn(ξ)|r=1 + o(2)

n rnYn(ξ)|r=1

)(5.118)

= −(n+ 1)(n+ 2)

2n+ 1(o(1)n rnYn(ξ))|r=1 −

n(n− 1)

2n+ 1(o(2)n rnYn(ξ))|r=1

holds for all ξ ∈ S2, since the first summand is of degree n+ 1 and the second summandis of degree n− 1 (see Lemma 5.35). Using (5.90) and (5.91), we obtain

∆∗ξo(1)ξ Yn(ξ) = −(n+ 1)2(n+ 2)

2n+ 1o

(1)ξ Yn(ξ) +

(n+ 1)(n+ 2)

2n+ 1o

(2)ξ Yn(ξ)

−(n− 1)n2

2n+ 1o

(1)ξ Yn(ξ)− n(n− 1)

2n+ 1o

(2)ξ Yn(ξ)

= (−n(n+ 1)− 2)o(1)ξ Yn(ξ) + 2o

(2)ξ Yn(ξ)

= o(1)ξ (−n(n+ 1)− 2)Yn(ξ) + 2o

(2)ξ Yn(ξ)

= o(1)ξ (∆∗ξ − 2)Yn(ξ) + 2o

(2)ξ Yn(ξ). (5.119)

Thus, our formula is true for every spherical harmonic. The completeness of the systemof vector spherical harmonics in l2(S2) then implies the validity for all sufficiently smoothfunctions. Part (ii) follows from the adjointness of the operators o(i) and O(i) and theself-adjointness of ∆∗. Lemma 5.42 helps us to find a vectorial Beltrami operator byobserving the following identities

(∆∗ + 2)o(1)Yn = −n(n+ 1)o(1)Yn + 2o(2)Yn, (5.120)

∆∗o(2)Yn = 2n(n+ 1)o(1)Yn − n(n+ 1)o(2)Yn, (5.121)

∆∗o(3)Yn = −n(n+ 1)o(3)Yn. (5.122)


In consequence, we have

pnor(∆∗ + 2)o(1)Yn = −n(n+ 1)o(1)Yn, (5.123)

ptan∆∗o(2)Yn = −n(n+ 1)o(2)Yn, (5.124)

ptan∆∗o(3)Yn = −n(n+ 1)o(3)Yn. (5.125)

In other words, (5.123)–(5.125) motivate the introduction of the vectorial Beltrami oper-ator in the following way.

Definition 5.43The operator ∆∗ : c(2)(S2)→ c(S2) given by

∆∗ = pnor(∆∗ + 2)pnor + ptan∆∗ptan (5.126)

is called vectorial Beltrami operator, where the application of ∆∗ on vector fields is un-derstood in the sense of (5.111).

As an immediate consequence of Lemma 5.42, we obtain the following result.

Lemma 5.44If F : S2 → R and f : S2 → R3 are sufficiently smooth, then, for i ∈ {1, 2, 3},

∆∗o(i)F = o(i)∆∗F, (5.127)

O(i)∆∗f = ∆∗O(i)f. (5.128)

Lemma 5.44 motivates us to characterize the spectrum of the operator ∆∗.

Theorem 5.45Any vector spherical harmonic yn ∈ harmn(S2) of degree n ∈ N0 is an infinitely oftendifferentiable eigenfunction of the vectorial Beltrami operator ∆∗ with respect to theeigenvalue (∆∗)∧(n) = n(n+ 1). Conversely, any infinitely often differentiable eigenfunc-tion of ∆∗ is a vector spherical harmonic.

Remark 5.46The vector spherical harmonics can be seen to be the eigenfunctions of an operator ∆∗

that can be introduced (without projection operators) only by the use of differentiationprocesses in R3. More explicitly, the following theorem is valid.

Theorem 5.47Let ∆∗ be given by

∆∗ξf(ξ) = ∆∗ξf(ξ)− 2(ξ ∧∇ξ) ∧ f(ξ)− 2f(ξ), f ∈ c(2)(S2). (5.129)

Then, for n ∈ N0,

∆∗ξyn + n(n+ 1)yn = 0, yn ∈ harmn(S2). (5.130)


We now come to the second variant of the vectorial Funk–Hecke formula as announced in(5.47). The basic ideas to handle this problem are the representation of vector sphericalharmonics by means of restrictions of homogeneous harmonic vector polynomials and thecomponentwise application of the (scalar) Funk–Hecke formula.Let Yn ∈ Harmn(S2) be a spherical harmonic. Then, we know from Lemma 5.35 that theCartesian components of the spherical vector field

ξ 7→ o(i)n r

nYn(ξ)|r=1, ξ ∈ S2, i ∈ {1, 2, 3}, n ∈ N0i , (5.131)

are (scalar) spherical harmonics of degree deg(i)(n) (which is n− 1, n, or n+ 1 dependingon the type i, see Lemma 5.35). Thus, we get, for G ∈ L1([−1, 1]) and η ∈ S2,∫

S2G(ξ · η)o(i)

n rnYn(ξ)|r=1dS(ξ) = G∧(deg(i)(n))o(i)

n rnYn(η)|r=1. (5.132)

We know from the formulas (5.93)–(5.95) how a vector spherical harmonic is expressibleby restrictions of homogeneous harmonic vector polynomials. Combining these results,we get the second variant of the vectorial Funk–Hecke formula.

Theorem 5.48 (Funk–Hecke Formula)Let G be of class L1([−1, 1]). Assume that Yn ∈ Harmn(S2). Then, for all η ∈ S2 and forall n ∈ N,∫

S2G(ξ · η)o

(1)ξ Yn(ξ)dS(ξ) = G∧(1,1)(n)o(1)

η Yn(η) +G∧(1,2)(n)o(2)η Yn(η), (5.133)∫

S2G(ξ · η)o

(2)ξ Yn(ξ)dS(ξ) = G∧(2,1)(n)o(1)

η Yn(η) +G∧(2,2)(n)o(2)η Yn(η), (5.134)∫

S2G(ξ · η)o

(3)ξ Yn(ξ)dS(ξ) = G∧(3,3)(n)o(3)

η Yn(η), (5.135)

where the coefficients G∧(i,j)(n) are given by

G∧(1,1)(n) =1

2n+ 1((n+ 1)G∧(n+ 1) + nG∧(n− 1)) , (5.136)

G∧(1,2)(n) =1

2n+ 1(G∧(n− 1)−G∧(n+ 1)) , (5.137)

G∧(2,1)(n) =n(n+ 1)

2n+ 1(G∧(n− 1)−G∧(n+ 1)) , (5.138)

G∧(2,2)(n) =1

2n+ 1(nG∧(n+ 1) + (n+ 1)G∧(n− 1)) , (5.139)

G∧(3,3)(n) = G∧(n). (5.140)

Remark 5.49For n = 0, we know that o

(i)ξ Yn(ξ) = 0, i = 2, 3, but we can find the following analog to

Theorem 5.48: ∫S2G(ξ · η)o

(1)ξ Y0(ξ)dS(ξ) = G∧(1)o(1)

η Y0(η). (5.141)


In terms of the coefficients G∧(i,j)(n), we set

G∧(1,1)(0) = G∧(1), (5.142)

G∧(1,2)(0) = G∧(2,1)(0) = G∧(2,2)(0) = G∧(3,3)(0) = 0.

Note that the space l2(3)(S2) is invariant with respect to the defined integral operator,

while l2(1)(S2) and l2(2)(S2) are not. However, it is clear that l2(1)(S2)⊕ l2(2)(S2) is an invariant

subspace of l2(S2).

5.9 Example: Fluid Flow (Navier–Stokes Equation)

Our interest is to give a brief derivation of the equations of thermodynamics and fluiddynamics which are used to forecast an atmospheric state. For further details of thededuction of the fundamental equations the reader is referred to, e.g., [2, 21]. We considera meteorological field F = F(t, x) (scalar or vectorial) that depends both on time t andspace x and assume its differentiability with respect to both arguments. By the Taylorexpansion

F(t+ ∆t, x+ ∆x) = F(t, x) +∂F∂t

∆t+ (∆x · ∇x)F , (5.143)

where ∆t and ∆x are displacements in time and space, such that the flow velocity u ofthe air is given by ∆x

∆t= u, we derive

dFdt

= (u · ∇)F +∂F∂t

. (5.144)

The term on the left hand side is the Lagrangian time derivative of F (the rate of changefollowing a small parcel of air), the term ∂F

∂ton the right hand side is the Eulerian time

derivative of F (rate of change at a fixed point). For the purpose of weather forecasts,the interesting quantity is the local change, i.e.,

∂F∂t

=dFdt− (u · ∇)F . (5.145)

If F is conserved on particles of fluids, i.e., dFdt

= 0, the local value at a fixed point canvery well be changing because of the advection brought in by the term −(u · ∇)F . Now,we consider several choices for F and the corresponding physical laws.The first law of thermodynamics tells us the following relation between temperature and(specific) volume/density of a moving parcel of air

cvdT

dt+ p

dα

dt= T

ds

dt= Q, (5.146)

where we introduce the following quantities

T temperaturecv specific heat at constant volumep pressure

α (specific) volumes (specific) entropyQ total heating per unit mass


Combining (5.146) with the density % = 1α

, we obtain

cvdT

dt− p

%

d%

dt= Q. (5.147)

The law of mass conservation says that

∂%

∂t+∇ · (%u) = 0 (5.148)

such that, by use of (5.144),d%

dt+ %∇ · u = 0 . (5.149)

Now, we combine (5.147), (5.149), and (5.144) applied to T to get

cv∂T

∂t+ cv(u · ∇)T +

p

%∇ · u = Q. (5.150)

Moreover, we remember the ideal gas law, i.e.,

p = %cRT (5.151)

with cR being the gas constant per unit mass. With the help of Newton’s second lawof motion we can relate the inertial acceleration of an element of air and the net forcesacting on it, i.e., the pressure gradient, gravity, boundary-induced friction, and viscosity.Velocities and accelerations are measured relative to the rotating frame of the solid Earth.Therefore, we introduce Coriolis/centrifugal forces. We find that the Lagrangian rate ofchange of the velocity u relative to the rotating Earth is governed by the following equation(all forces are expressed per unit mass of air)

du

dt= −2ω ∧ u− 1

%∇p−∇Φ + ν∆u+ f, (5.152)

where we use the following notation

2ω ∧ u Coriolis forceω rotation vector of the Earth∇p pressure gradient∇Φ apparent gravity

ν∆u frictionν kinematic viscosityf other forces.

Note that the apparent gravity∇Φ is due to the gravitational potential and the centrifugalforce −ω ∧ (ω ∧ x) at the position x in a frame rotating with the Earth and having itsorigin at the Earth’s center. Equation (5.152) is the Navier–Stokes equation of motionand acceleration relative to the Earth. Applying (5.144) to the flow velocity u, we cansummarize (5.148) and (5.150)–(5.152) to the full set of forecasting equations:

∂u

∂t= −(u · ∇)u− 2ω ∧ u− 1

%∇p−∇Φ + ν∆u+ f , (5.153)

cv∂T

∂t= −cv(u · ∇)T − p

%∇ · u+Q , (5.154)


∂%

∂t= −(u · ∇)%− %∇ · u , (5.155)

p = %cRT . (5.156)

Note that in many climate simulation models and many weather prediction models prog-nostic equations for local water substance concentration are included. The distribution ofwater substance greatly effects the heating rate Q. We neither discuss water conservationequations nor the corresponding variations of the gas constant cR and the specific heatconstant cv.We now assume incompressibility of the fluid, i.e., ∇ · u = 0 or d%

dt= 0. This is justi-

fied if the flow is not strongly heated and the density % only slightly varies around thecharacteristic density %0, i.e., %

%0≈ 0.1. This leads us to the simplified equations

∂u

∂t= −(u · ∇)u− 2ω ∧ u− 1

%0

∇p−∇Φ + ν∆u+ f , (5.157)

∇ · u = 0 . (5.158)

Finally, we concentrate on tangential streams and assume the Earth to possess a sphericalshape, i.e., we discuss tangential, surface divergence free vector fields which we will latercall type 3 vector fields (see chapter on vector spherical harmonics). This gives us thetangential variants of (5.157) and (5.158):

∂u

∂t= −(u · ∇∗)u− 2ω ∧ u− 1

%0

∇∗p−∇∗Φ + ν∆∗u+ f , (5.159)

∇∗ · u = 0 . (5.160)

Note that we denote the tangential parts of differential operators with a star. Goingover to the weak formulation of these equations using type 3 test functions, e.g., vectorspherical harmonics of type 3, we find that both the pressure surface gradient ∇∗p andthe gravity term ∇∗Φ are of type 2 (see chapter on vector spherical harmonics). Thus,they both are orthogonal to our test functions. Therefore, we only have to deal with thefollowing Navier–Stokes equation of motion (in the weak sense)

∂u

∂t= −(u · ∇∗)u− 2ω ∧ u+ ν∆∗u+ f , (5.161)

∇∗ · u = 0 , (5.162)

where f is projected to the space of type 3 vector fields. Further details on the constructionand numerical implementation of the corresponding Galerkin scheme can be found in[6]. It should be noted that a Poisson equation for the pressure can be derived thatinvolves tensor spherical harmonics (see [11] and the references therein) and the surfacecurl/surface vorticity (see [6] and the references therein).

Bibliography

[1] Alt, H.W.: Lineare Funktionalanalysis. Springer Lehrbuch, Berlin, 2006.

[2] Ansorge, R., Sonar, T.: Mathematical Models of Fluid Dynamics. Wiley-VCH, 2009.

[3] Backus, G.E., Parker, R., Constable, C.: Foundations of Geomagnetism, CambrigdeUniversity Press, 1996.

[4] Butzer, P.L., Nessel, R.J.: Fourier Analysis and Approximation. Birkhauser, Basel,Stuttgart, 1971.

[5] Davis, P.J.: Interpolation and Approximation. Dover Publ., 1975.

[6] Fengler, M.J.: Vector Spherical Harmonic and Vector Wavelet Based Non-LinearGalerkin Schemes for Solving the Incompressible Navier–Stokes Equation on theSphere. PhD-Thesis, Geomathematics Group, TU Kaiserslautern, Shaker, 2005.

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[10] Freeden, W., Michel, V.: Multiscale Potential Theory. Birkhauser, 2004.

[11] Freeden, W., Schreiner, M.: Spherical Functions of Mathematical Geosciences.Springer, Berlin, 2009.

[12] Gautschi, W.: Orthogonal Polynomials, Computation and Approximation. OxfordUniversity Press, 2004.

[13] Heuser, H.: Funktionalanalysis. Teubner, Stuttgart, 2006.

[14] Lebedew, N.N.: Spezielle Funktionen und ihre Anwendungen. BI, 1973.

[15] Michel, V.: Lectures on Constructive Approximation—Fourier, Spline, and WaveletMethods on the Real Line, the Sphere, and the Ball. Birkhauser, 2012.

[16] Muller, C.: Spherical Harmonics, Lecture Notes in Mathematics, 17. Springer, 1966.

[17] Muller, C.: Analysis of Spherical Symmetrics in Euclidean Spaces. Springer, 1998.

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[18] Reed, M., Simon, B.: Functional Analysis I. Academic Press, New York, 1972.

[19] Rudin, W.: Functional Analysis. McGraw-Hill, Boston, 1991.

[20] Szego, G.P.: Orthogonal polynomials. American Math. Soc., 1967.

[21] Teman, R.: Navier–Stokes Equations and Non-Linear Functional Analysis. SIAM,1983.

[22] Watson, G.N.: A Treatise on the Theory of Bessel Functions, 2nd ed., CambridgeUniversity Press, Cambridge, 1966.

[23] Yoshida, K.: Functional Analysis. Springer, Berlin, 1980.

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