The Mathematical Theory of Maxwell s Equations

Transcription

1 The Mathematical Theory of Maxwell s Equations Andreas Kirsch and Frank Hettlich epartment of Mathematics Karlsruhe Institute of Technology (KIT) Karlsruhe, Germany c January 4, 203

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3 Contents Introduction 7. Maxwell s Equations The Constitutive Equations Special Cases Boundary and Radiation Conditions Exercises Expansion into Wave Functions 2 2. Separation in Spherical Coordinates Legendre Polynomials Spherical Harmonics The Boundary Value Problem for the Laplace Equation in a Ball Bessel Functions The Boundary Value Problems for the Helmholtz Equation for a Ball Expansion of Electromagnetic Waves Exercises Scattering From a Perfect Conductor A Scattering Problem for the Helmholtz Equation Formulation of the Problem Representation Theorems Volume and Surface Potentials Boundary Integral Operators Uniqueness and Existence A Scattering Problem for the Maxwell System Formulation of the Problem

4 4 CONTENTS Representation Theorems Vector Potentials and Boundary Integral Operators Uniqueness and Existence Exercises The Variational Approach to the Cavity Problem Formulation of the Problem Sobolev Spaces Basic Properties of Sobolev Spaces of Scalar Functions Basic Properties of Sobolev Spaces of Vector Valued Functions The Helmholtz ecomposition and Further Results on Sobolev Spaces The Cavity Problem The Variational Formulation and Existence Uniqueness and Unique Continuation The Time ependent Cavity Problem Exercises Boundary Integral Equation Methods for Lipschitz omains Sobolev Spaces on Boundaries Boundary-Sobolev Spaces of Scalar Functions Boundary-Sobolev Spaces of Vector-Valued Functions The Case of a Sphere Surface Potentials Boundary Integral Equation Methods Exercises Appendix on Vector Calculus Table of ifferential Operators and Their Properties Elementary Facts from ifferential Geometry Integral Identities Surface Gradient and Surface ivergence Bibliography 243

5 Preface This book arose from a lecture on Maxwell s equations given by the authors between?? and The emphasis is put on three topics which are clearly structured into Chapters 2, 3, and 4. In each of these chapters we study first the simpler scalar case where we replace the Maxwell system by the scalar Helmholtz equation. Then we investigate the time harmonic Maxwell s equations. In Chapter we start from the (time dependent) Maxwell system in integral form and derive... 5

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7 Chapter Introduction. Maxwell s Equations Electromagnetic wave propagation is described by particular equations relating five vector fields E,, H, B, J and the scalar field ρ, where E and denote the electric field (in V/m) and electric displacement (in As/m 2 ) respectively, while H and B denote the magnetic field (in A/m) and magnetic flux density (in V s/m 2 = T =Tesla). Likewise, J and ρ denote the current density (in A/m 2 ) and charge density (in As/m 3 ) of the medium. Here and throughout the lecture we use the rationalized MKS-system, i.e. V, A, m and s. All fields will be assumed to depend both on the space variable x R 3 and on the time variable t R. The actual equations that govern the behavior of the electromagnetic field, first completely formulated by Maxwell, may be expressed easily in integral form. Such a formulation has the advantage of being closely connected to the physical situation. The more familiar differential form of Maxwell s equations can be derived very easily from the integral relations as we will see below. In order to write these integral relations, we begin by letting S be a connected smooth surface with boundary S in the interior of a region Ω 0 where electromagnetic waves propagate. In particular, we require that the unit normal vector ν(x) for x S be continuous and directed always into one side of S, which we call the positive side of S. By τ(x) we denote the unit vector tangent to the boundary of S at x S. This vector, lying in the tangent plane of S together with a vector n(x), x S, normal to S is oriented so as to form a mathematically positive system (i.e. τ is directed counterclockwise when we sit on the positive side of S, and n(x) is directed to the outside of S). Furthermore, let Ω R 3 be an open set with boundary Ω and outer unit normal vector ν(x) at x Ω. Then Maxwell s equations in integral form 7

8 8 CHAPTER. INTROUCTION state: H τ dl = d dt ν ds + J ν ds (Ampère s Law) (.a) S S E τ dl = d dt S B ν ds (Law of Induction) (.b) S ν ds = S ρ dx (Gauss Electric Law) (.c) Ω Ω B ν ds = 0 (Gauss Magnetic Law) (.d) Ω To derive the Maxwell s equations in differential form we consider a region Ω 0 where µ and ε are constant (homogeneous medium) or at least continuous. In regions where the vector fields are smooth functions we can apply the Stokes and Gauss theorems for surfaces S and solids Ω lying completely in Ω 0 : curl F ν ds = F τ dl (Stokes), (.2) S div F dx = S F ν ds (Gauss), (.3) Ω Ω where F denotes one of the fields H, E, B or. We recall the differential operators (in cartesian coordinates): div F(x) = curl F(x) = 3 j= F j x j (x) F 3 (x) F 2 (x) x 2 x 3 F (x) F 3 (x) x 3 x F 2 (x) F (x) x x 2 (divergenz, ivergenz ) (curl, Rotation ). With these formulas we can eliminate the boundary integrals in (.a-.d). We then use the fact that we can vary the surface S and the solid Ω in arbitrarily. By equating the integrands we are led to Maxwell s equations in differential form so that Ampère s Law, the Law of Induction and Gauss Electric and Magnetic Laws, respectively, become:

9 .. MAXWELL S EQUATIONS 9 B t t + curl x E = 0 (Faraday s Law of Induction) curl x H = J (Ampere s Law) div x = ρ (Gauss Electric Law) div x B = 0 (Gauss Magnetic Law) We note that the differential operators are always taken w.r.t. the spacial variable x (not w.r.t. time t!). Therefore, in the following we often drop the index x. Physical remarks: The law of induction describes how a time-varying magnetic field effects the electric field. Ampere s Law describes the effect of the current (external and induced) on the magnetic field. Gauss Electric Law describes the sources of the electric displacement. The forth law states that there are no magnetic currents. Maxwell s equations imply the existence of electromagnetic waves (as ligh, X-rays, etc) in vacuum and explain many electromagnetic phenomena. Literature wrt physics: J.. Jackson, Klassische Elektrodynamik, de Gruyter Verlag Historical Remark: ates: André Marie Ampère ( ), Charles Augustin de Coulomb ( ), Michael Faraday (79 867), James Clerk Maxwell (83 879) It was the ingeneous idea of Maxwell to modify Ampere s Law which was known up to that time in the form curl H = J for stationary currents. Furthermore, he collected the four equations as a consistent theory to describe the electromagnetic fields. (James Clerk Maxwell, Treatise on Electricity and Magnetism, 873). Conclusion. Gauss Electric Law and Ampere s Law imply the equation of continuity ρ t = div t = div ( curl H J ) = div J because div curl = 0.

10 0 CHAPTER. INTROUCTION.2 The Constitutive Equations In this general setting the equations are not yet consistent (there are more unknowns than equations). The Constitutive Equations couple them: = (E, H) and B = B(E, H) The electric properties of the material are complicated. In general, they not only depend on the molecular character but also on macroscopic quantities as density and temperature of the material. Also, there are time-dependent dependencies as, e.g., the hysteresis effect, i.e. the fields at time t depend also on the past. As a first approximation one starts with representations of the form = E + 4πP and B = H 4πM where P denotes the electric polarization vector and M the magnetization of the material. These can be interpreted as mean values of microscopic effects in the material. Analogously, ρ and J are macroscopic mean values of the free charge and current densities in the medium. If we ignore ferro-electric and ferro-magnetic media and if the fields are small one can model the dependencies by linear equations of the form = εe and B = µh with matrix-valued functions ε : R 3 R 3 3 (dielectric tensor), and µ : R 3 R 3 3 (permeability tensor). In this case we call the media inhomogenous and anisotropic. The special case of an isotropic medium means that polarization and magnetization do not depend on the directions. In this case they are just real valued functions, and we have with functions ε, µ : R 3 R. = εe and B = µh In the simplest case these functions ε and µ are constant. This is the case, e.g., in vacuum. We indicated already that also ρ and J can depend on the material and the fields. Therefore, we need a further equation. In conducting media the electric field induces a current. In a linear approximation this is described by Ohm s Law: J = σe + J e where J e is the external current density. For isotropic media the function σ : R 3 R is called the conductivity. Remark: If σ = 0 then the material is called dielectric. In vacuum we have σ = 0, ε = ε AS/V m, µ = µ 0 = 4π 0 7 V s/am. In anisotropic media, also the function σ is matrix valued.

11 .3. SPECIAL CASES.3 Special Cases Vacuum In regions of vacuum with no charge distributions and (external) currents (that is, ρ = 0 and J e = 0) the law of induction takes the form µ 0 H t + curl E = 0. ifferentiation wrt time t and use of Ampere s Law yields that is, µ 0 2 H t 2 + ε 0 curl curl H = 0 ; ε 0 µ 0 2 H t 2 + curl curl H = 0. / ε 0 µ 0 has the dimension of velocity and is called the speed of light: c 0 = / ε 0 µ 0. From curl curl = div it follows that the components of H are solutions of the linear wave equation c 2 2 H 0 H = 0. t 2 Analogously, one derives the same equation for the electric field: c 2 2 E 0 E = 0. t 2 Remark: Heinrich Rudolf Hertz ( ) showed also experimentally the existence of electromagenetic waves about 20 years after Maxwell s paper (in Karlsruhe!). Electrostatics If E is in some region Ω constant wrt time t (tha is, in the static case) the law of induction reduces to curl E = 0 in Ω. Therefore, if Ω is simply connected there exists a potential u : Ω R with E = u in Ω. Gauss Electric Law yields in homogeneous media the Poisson equation ρ = div = div(ε 0 E) = ε 0 u for the potential u. The electrostatics is described by this basic elliptic partial differential equation u = ρ/ε 0. Mathematically, this is the subject of potential theory.

12 2 CHAPTER. INTROUCTION Magnetostatics The same technique does not work in magnetostatics because, in general, curl H 0. However, because div B = 0 we conclude the existence of a vector potential A : R 3 R 3 with B = curl A in. Substituting this into Ampere s Law yields (for homogeneous media Ω) after multiplication with µ 0 µ 0 J = curl curl A = div A A. Since curl = 0 we can add gradients u to A without changing B. We will see later that we can choose u such that the resulting potential A satisfies div A = 0. This choice of normalization is called Coulomb gauge. With this normalization we also get in the magnetostatic case the Poisson equation A = µ 0 J. We note that in this case the Laplacian has to be taken componentwise. Time Harmonic Fields Under the assumptions that the fields allow a Fourier transformation w.r.t. time we set E(x; ω) = (F t E)(x; ω) = H(x; ω) = (F t H)(x; ω) = R R E(x, t) e iωt dt, H(x, t) e iωt dt, etc. We note that the fields E, H etc are now complex valued, i.e, E( ; ω), H( ; ω) : R 3 C 3 (and also the other fields). Although they are vector fields we denote them by capital Latin letters only. Maxwell s equations transform into (because F t (u ) = iωf t u) the time harmonic Maxwell s equations iωb + curl E = 0, iω + curl H = σe + J e, div = ρ, div B = 0. Remark: The time harmonic Maxwell s equation can also be derived from the assumption that all fields behave periodically w.r.t. time with the same frequency ω. Then the forms E(x, t) = e iωt E(x), H(x, t) = e iωt H(x), etc satisfy the time harmonic Maxwell s equations.

13 .3. SPECIAL CASES 3 With the constitutive equations = εe and B = µh we arrive at iωµh + curl E = 0, (.4a) iωεe + curl H = σe + J e, (.4b) div(εe) = ρ, (.4c) div(µh) = 0. (.4d) Eliminating H or E, respectively, from (.4a) and (.4b) yields ( ) curl iωµ curl E + (iωε σ) E = J e. (.5) and ( ) curl iωε σ curl H ( ) + iωµ H = curl iωε σ J e, (.6) respectively. Usually, one writes these equations in a slightly different way by introducing the constant values ε 0 > 0 and µ 0 > 0 in vacuum and relative values (dimensionless!) µ r, ε r R and ε c C, defined by µ r = µ µ 0, ε r = ε ε 0, ε c = ε r + i σ ωε 0. Then equations (.5) and (.6) take the form ( ) curl curl E k 2 ε c E = iωµ 0 J e, (.7) µ r ( ) ( ) curl curl H k 2 µ r H = curl J e, (.8) ε c ε c with the wave number k = ω ε 0 µ 0. We conclude from this again that div(µ r H) = 0 because div curl vanishes. For the electric field we conclude that div(ε c E) = iωµ 0 div J k 2 e = i ωε 0 div J e. In vacuum we have ε c =, µ r = and thus curl curl E k 2 E = iωµ 0 J e, (.9) curl curl H k 2 H = curl J e. (.0) For J e = 0 this reduces the vector Helmholtz equations E+k 2 E = 0 and H+k 2 H = 0. ###. Aequivalenz: ivergenz frei und vector Helmholtz als Lemma, Also the following observation will be useful. Lemma.2 Is E a divergence free solution of the vector Helmholtz equation in a domain then x E is a solution of the scalar Helmholtz equation.

14 4 CHAPTER. INTROUCTION Proof: By the formulae?? and?? we obtain δ(x E) = div (x E) = div ( (E ) x + (x )E ) = div ( (E ) x ) + div E = 3 i= j = 3 2 E j x x 2 j + 2 div(e) i = A x = k 2 x E. Example.3 In the case J e = 0 and in vacuum the fields E(x) = p e ik d x ik d x and H(x) = (p d) e ### are solutions of the time harmonic Maxwell s equations (.9), (.0) provided d is a unit vector in R 3 and p C 3 with p d = 0. Such fields are called plane time harmonic fields with polarization vector p C 3 and direction d. We make the assumption ε c =, µ r = for the rest of Section.3. Taking the divergence of these equations yields div H = 0 and k 2 div E = iωµ 0 div J e ; that is, div E = (i/ωε 0 ) div J e. Comparing this to (.4c) yields the time harmonic version of the equation of continuity div J e = iωρ. With the vector identity curl curl = + div equations (.0) and (.9) can be written as E + k 2 E = iωµ 0 J e + ε 0 ρ, (.) H + k 2 H = curl J e. (.2) We consider the magnetic field first and introduce magnetic Hertz potentials: div H = 0 implies the existence of a vector potential A with H = curl A. Thus (.2) takes the form and thus curl( A + k 2 A) = curl J e A + k 2 A = J e + ϕ (.3) for some scalar field ϕ. On the other hand, if A and ϕ satisfy (.3) then H = curl A and E = iωε 0 (curl H J e ) = iωµ 0 A iωε 0 (div A ϕ) We set p d = 3 j= p jd j even for p C 3

15 .4. BOUNARY AN RAIATION CONITIONS 5 satisfies the Maxwell system (.4a) (.4d). Analogously, we can introduce electric Hertz potentials if J e = 0. Since then div E = 0 there exists a vector potential A with E = curl A. Substituting this into (.) yields curl( A + k 2 A) = 0 and thus A + k 2 A = ϕ (.4) for some scalar field ϕ. On the other hand, if A and ϕ satisfy (.4) then E = curl A and H = iωµ 0 curl E = iωε 0 A + iωµ 0 (div A ϕ) satifies the Maxwell system (.4a) (.4d). As a particular example we take the magnetic Hertz vector A of the form A(x) = u(x) ẑ with a scalar function u and the uni vector ẑ = (0, 0, ) R 3. Then H = curl(uẑ) = E = iωµ 0 ẑ + ( u, u ), 0, x 2 x iωε 0 ( u/ x 3 ). ### TE und TM ausfuehrlicher und efinition formulieren (s. altes Skript, Jackson) If u is independent of x 3 then E has only a x 3 component and this mode is called E-mode or TM-mode (for transverse-magnetic). Analogously, the H-mode or TE-mode is defined. In any case, the vector Helmholtz equation for A reduced to the scalar Helmholtz equation for u. ###.4 Boundary and Radiation Conditions Maxwell s equations hold only in regions with smooth parameter functions ε r, µ r and σ. If we consider a situation in which a surface S separates two homogeneous media from each other, the constitutive parameters ε, µ and σ are no longer continuous but piecewise continuous with finite jumps on S. While on both sides of S Maxwell s equations (.4a) (.4d) hold, the presence of these jumps implies that the fields satisfy certain conditions on the surface. To derive the mathematical form of this behaviour (the boundary conditions) we apply the law of induction (.b) to a narrow rectangle-like surface R, containing the normal n to the surface S and whose long sides C + and C are parallel to S and are on the opposite sides of it, see the following figure.

16 6 CHAPTER. INTROUCTION When we let the height of the narrow sides, AA and BB, approach zero then C + and C approach a curve C on S, the surface integral B ν ds will vanish in the limit because t R the field remains finite (note, that the normal ν is the normal to R lying in the tangential plane of S). Hence, the line integrals E C + τ dl and E C τ dl must be equal. Since the curve C is arbitrary the integrands E + τ and E τ coincide on every arc C; that is, n E + n E = 0 on S. (.5) A similar argument holds for the magnetic field in (.a) if the current distribution J = σe + J e remains finite. In this case, the same arguments lead to the boundary condition n H + n H = 0 on S. (.6) If, however, the external current distribution is a surface current; that is, if J e is of the form J e (x + τn(x)) = J s (x)δ(τ) for small τ and x S and with tangential surface field J s and σ is finite, then the surface integral R J e ν ds will tend to C J s ν dl, and so the boundary condition is n H + n H = J s on S. (.7) We will call (.5) and (.6) or (.7) the transmission boundary conditions. A special and very important case is that of a perfectly conducting medium with boundary S. Such a medium is characterized by the fact that the electric field vanishes inside this medium, and (.5) reduces to n E = 0 on S (.8) Another important case is the impedance- or Leontovich boundary condition n H = λ n (E n) on S (.9)

17 .4. BOUNARY AN RAIATION CONITIONS 7 which, under appropriate conditions, may be used as an approximation of the transmission conditions. The same kind of boundary occur also in the time harmonic case (where we denote the fields by capital Latin letters). Finally, we specify the boundary conditions to the E- and H-modes derived above. ### Wo??? ### We assume that the surface S is an infinite cylinder in x 3 direction with constant cross section. Furthermore, we assume that the volume current density J vanishes near the boundary S and that the surface current densities take the form J s = j s ẑ for the E-mode and J s = j s ( ν ẑ ) for the H-mode. We use the notation v] := v + v for the jump of the function v at the boundary. Also, we abbreviate (only for this table) σ = σ iωε. We list the boundary conditions in the following table. Boundary condition E-mode H-mode ] transmission u] = 0 on S, µ u ν = 0 on S, ] σ u ν = js on S, u] = j s on S, impedance λ k 2 u + σ u ν = j s on S, k 2 u λ iωµ u ν = j s on S, perfect conductor u = 0 on S, u ν = 0 on S. The situation is different for the normal components. We consider Gauss Electric and Magnetic Laws and choose Ω to be a box which is separated by S into two parts Ω and Ω 2. We apply (.c) first to all of Ω and then to Ω and Ω 2 separately. The addition of the last two formulas and the comparison with the first yields that the normal component n has to be continuous as well as (application of (.d)) B n. With the constitutive equations one gets n (ε r, E ε r,2 E 2 ) = 0 on S and n (µ r, H µ r,2 H 2 ) = 0 on S. Conclusion.4 The normal components of E and/or H are not continuous at interfaces where ε c and/or µ r have jumps. The Silver-Müller radiation condition ### Physik, Sommerfeld, Silver-Müller ### Reference Problems uring our course we will consider two classical boundary value problems.

18 8 CHAPTER. INTROUCTION (Cavity with an ideal conductor as boundary) Let R 3 be a bounded domain with sufficiently smooth boundary and exterior unit normal vector ν(x) at x. Let J e : C 3 be a vector field. etermine a solution (E, H) of the time harmonic Maxwell system curl E iωµh = 0 in, (.20a) curl H + (iωε σ)e = J e in, (.20b) ν E = 0 on. (.20c) ε c, µ r ν E = 0 (Scattering by an ideal conductor) Given a bounded region and some solution E i and H i of the unperturbed time harmonic Maxwell system curl E i iµ 0 H i = 0 in R 3, curl H i + iε 0 E i = 0 in R 3, determine E, H of the Maxwell system curl E iµ 0 H = 0 in R 3 \, curl H + iε 0 E = 0 in R 3 \, such that E satisfies the boundary condition ν E = 0 on, and E and H have the decompositions into E = E s + E i and H = H s + H i in R 3 \ with some scattered field E s, H s which satisfy the Silver-Müller radiation condition lim x x lim x x ( H s (x) x x ( E s (x) x x uniformly with respect to all directions x/ x. + ) ε0 E s (x) µ 0 ) µ0 H s (x) ε 0 = 0 = 0 Remark: For general µ r, ε c L (R 3 ) we have to give a correct interpretation of the differential equations ( variational or weak formulation ) and transmission conditions ( trace theorems ).

19 .5. EXERCISES 9 E i, H i ν E = 0 E s, H s.5 Exercises

20 20 CHAPTER. INTROUCTION

21 Chapter 2 Expansion into Wave Functions 2. Separation in Spherical Coordinates The starting point of the investigations is the consideration of the boundary value problems inside or outside the simple geometry of a ball in case of a homogeneous medium. According to boundary conditions on the sphere we are interested in solutions of the time-harmonic Maxwell equations with components u of E and H, which can be separated by u(x) = v(r) K(ˆx) with a radial depending factor v : R >0 R and a spherical part K : S 2 R, where r = x = x 2 + x x 2 3 and ˆx = x S 2 in the unit sphere S 2 R 3. Using spherical x coordinates for x R 3, r cos ϕ sin θ x = r sin ϕ sin θ with r R >0, ϕ 0, 2π), ψ 0, π], r cos θ we have ˆx = (cos ϕ sin θ, sin ϕ sin θ, cos θ). We already saw that the essential differential operator in the modelling of electromangentic waves is the Laplace operator. It occurs directly in the stationary situation (see page???), in case of the usual polarizations as principal part of the Helmholtz equation and also in the full Maxwell system, where the components turned out to be solutions of the vector Helmholtz equation in particular (see page???). Thus, a first step must be a representation of the Laplace operator in spherical coordinates. Lemma 2. The Laplace operator in R 3 for spherical coordinates (r, ϕ, θ) R >0 0, 2π) 0, π] is given by = ( r 2 ) ( 2 + r 2 r r r 2 sin 2 + sin θ ) θ ϕ 2 r 2 sin θ θ θ = 2 r r r + r 2 sin 2 θ 2 2 ϕ 2 + r 2 sin θ θ ( sin θ θ ).

22 22 CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS Proof: Let the unit basis vectors of the spherical coordinate system be denoted by e r = x g r r, e ϕ = x g ϕ ϕ, and e θ = x g θ θ where g r = x =, g r ϕ = x = r sin θ and g ϕ θ = x = r. With the jacobian of the θ transformation, det x x = g r g ϕ g θ = r 2 sin θ, and the divergence theorem we obtain r0 +ε r ϕ0 +ε ϕ θ0 +ε θ r 0 ϕ 0 θ 0 u g r g ϕ g θ d(r, θ, ϕ) = u dx = W ε ( u) ν ds W ε in a domain W ε = {x(r, θ, ϕ) R 3 : (r, θ, ϕ) (r 0, r 0 + ε r ) (θ 0, θ 0 + ε θ ) (ϕ 0, ϕ 0 + ε ϕ ). Substitution of the unit normal vectors and an application of Fubinis theorem lead to r0 +ε r θ0 +ε θ ϕ0 +ε ϕ r 0 θ 0 θ0 +ε θ ϕ0 +ε ϕ = = θ 0 + ϕ 0 u g r g ϕ g θ d(r, θ, ϕ) ( u) e ϕ 0 }{{} = u r r0 +ε r ϕ0 +ε ϕ r 0 r0 +ε r θ0 +ε θ ϕ0 +ε ϕ r 0 θ 0 ϕ 0 g ϕ g θ r 0 +ε r r 0 dϕ dθ + ( u) θ 0 +ε θ e 3 g ϕ 0 }{{} r g ϕ dϕ dr θ 0 = u g θ θ r0 +ε r θ0 +ε θ r (r2 sin θ u r ) + 2 u sin θ ϕ + 2 θ r 0 ( u) ϕ 0 +ε ϕ e 2 g θ 0 }{{} r g θ dθ dr ϕ 0 = gϕ u ϕ u (sin θ ) dϕ dθ dr θ Since the identity holds for any ε r, ε θ, ε ϕ > 0 and the integrand is a continuous function we conclude the given representation of the Laplace operator. efinition 2.2 Functions u : Ω R which satisfy the Laplace equation u = 0 in Ω are called harmonic functions in Ω. Complex valued functions are harmonic if real and imaginary parts are harmonic. The Laplace operator separates into a radial part, which is r 2 r (r2 ) = 2 + 2, and the r r 2 r r spherical part which is called the Laplace-Beltrami operator, a differential operator on the unit sphere. efinition 2.3 The differential operator S : C 2 (S 2 ) C(S 2 ) with representation S = sin 2 θ 2 ϕ + 2 sin θ ( sin θ ). θ θ in spherical coordinates is called Laplace-Beltrami operator on S 2.

23 2.. SEPARATION IN SPHERICAL COORINATES 23 With the notation of the Beltrami operator and r = r we obtain = r r r + r 2 S = r ( ) 2 r r 2 r + r 2 S. ### Remark: analoge Zerlegung im R n, Aufgabe: im R 2 (Polarkoordinaten) berechnen ### A substitution of the guess u(x) = u(rˆx) = v(r)k(ˆx) into the Laplace/Helmholtz equation with k C leads to 0 = u(rˆx) + k 2 u(rˆx) = v (r) K(ˆx) + 2 r v (r) K(ˆx) + r 2 v(r) SK(ˆx) + k 2 v(r) K(ˆx). Thus on sets with v(r) 0 and K(ˆx) 0 we obtain v (r) + 2 r v (r) v(r) + r 2 S K(ˆx) K(ˆx) + k 2 = 0. If there is a nontrivial solution of the supposed form, it follows the existence of a constant λ such that v satisfies the differential equation r 2 v (r) + 2r v (r) + ( k 2 r 2 + λ ) v(r) = 0 (2.) and K the equation S K(ˆx) = λk(ˆx). (2.2) The last equation can be read in a functional analytic contents. The problem of determine functions K can be formulated as looking for eigenfunctions K and eigenvalues λ of the Laplace-Beltrami operator S. In view of irichlet boundary conditions like u = f on S 2 the idea is to determine a complete set of eigenfunctions of the Laplace-Beltrami operator. Then we can hope to find a solution for any given f on S 2 by an expansion in terms of eigenfunctions. We observe that the spherical equation (2.2) is independent of k. Thus, results will be useful in any case, the Laplace as well as the Helmholtz equation. If we consider the expilcit representation of the Laplace-Beltrami operator in spherical coordinates the eigenvalue equation (2.2) becomes 2 ϕ K + sin θ ( sin θ ) K = λ sin 2 θ K. 2 θ θ Assuming eigenfunctions of the form K(ˆx) = y (ϕ) y 2 (θ) leads to y (ϕ) y (ϕ) + sin θ ( sin θ y 2 (θ) ) y 2 (θ) λ sin 2 θ = 0 (2.3) provided y (ϕ) 0 and y 2 (θ) 0. If the decomposition is valid there exists a constant µ C such that y = µy. Since we are interested in differentiable solutions u, the function y must

24 24 CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS be 2π periodic. By the fundamental system of the linear ordinary differential equation we conclude that µ = m 2 with m Z and the general solution is y (ϕ) = c e imϕ + c 2 e imϕ = c cos(mϕ) + c 2 sin(mϕ) with arbitrary constants c, c 2 C and c, c 2 C, respectively. We assume that the reader is familar with the classical Fourier theory. In particular with the fact that every function f L 2 ( π, π) allows an expansion in the form f(t) = 2π m Z f m e imt with Fourier coefficients f m = 2π π f(s)e ims ds, m Z. The convergence of the series must be understood in the L 2 -sense, that is, π π f(t) 2π M m= N π f m e imt 2 dt 0, N, M. Thus by the previous result we conclude, that the possible functions y for m Z constitute a complete orthonormal system in L 2 ( π, π). Next we consider the dependency on θ. Using the result µ = m 2 we get from (2.3) the differential equation sin θ ( sin θ y 2(θ) ) ( m 2 + λ sin 2 θ ) y 2 (θ) = 0 for y 2. With the substitution z = cos θ (, ] and w(z) = y 2 (θ) we conclude an associated Legendre differential equation ( ( z 2 )w (z) ) ) (λ + m2 w(z) = 0. (2.4) z 2 An investigation of this equation will be the main task of the next section leading to the so called Legendre polynomials and to a complete orthonormal system of spherical surface harmonics. Obviously the radial part of separated solutions given by the ordinary differential equation (2.) is depending on the wave number k. If k = 0 we obtain the Euler equation r 2 v (r) + 2r v (r) + λv(r) = 0. (2.5) By the guess v(r) = r µ we compute µ(µ + ) = λ, which leads to a fundamental set of solutions v (r) = r λ und v 2 (r) = r 2 4 λ.

25 2.2. LEGENRE POLYNOMIALS 25 Later we will see that λ = n(n + ) for n N 0 are the eigenvalues of the Laplace-Beltrami operator. Thus v (r) = r n and v 2 (r) = r (n+). According to non singular solutions of the Laplace equation inside a ball we have to choose v(r) = r n. In the exterior of a ball v 2 will lead to solutions with asymptotic behavior u(x) 0 for x. In case of k 0 we rewrite the differential equation (2.) by the substitution z = kr and hatv(kr) = kr v(r) into z 2ˆv (z) + zˆv (z) + ( z 2 + λ ) ˆv(z) = 0 4 and with the already mentioned eigenvalues λ = n(n + ) for n N we obtain the Bessel differential equation ( z 2ˆv ( (z) + zˆv (z) + z 2 n + ) ) 2 ˆv(z) = 0. (2.6) 2 In Section 2.5 we will discuss the Bessel functions, which solve this differential equation. Combining spherical surface harmonics and the corresponding Bessel functions will lead to solutions of the Helmholtz equation and further on also of Maxwells equations. ### Aufgabe: Separation in kartesischen Koordinaten ### 2.2 Legendre Polynomials Let us consider the case of harmonic functions u, that is u = 0. We already saw that a separation in spherical coordinates leads to functions of the form u(x) = r µ (c e imϕ + c 2 e i m ϕ )y 2 (θ), where y 2 (θ) = w(cos θ) is determined by the associated Legendre differential equation (2.4). We discuss this ordinary differential equation (2.4) first for the special case m = 0; that is, d ( z 2 ) w (z) ] λ w(z) = 0, < z <. (2.7) dz This is a differential equation of Legendre type. The coefficients vanish at z = ±. We want to determine solutions which are continuous up to the boundary, thus w C 2 (, +) C, +]. Theorem 2.4 (a) For λ = n(n + ), n N {0}, there exists exactly one solution w C 2 (, +) C, +] with w() =. We set P n = w. The function P n is a polynomial of degree n and is called Legendre polynomial of degree n. It satisfies the differential equation d ( x 2 ) P dx n(x) ] + n(n + ) P n (x) = 0, x.

26 26 CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS (b) If there is no n N {0} with λ = n(n + ) then there is no non-trivial solution of (2.7) in C 2 (, +) C, +]. Proof: We make a first only formal ansatz for a solution in the form of a power series: w(x) = a j x j j=0 and substitute this into the differential equation (2.7). A simple calculation shows that the coefficients have to satisfy the recursion a j+2 = j(j + ) + λ (j + )(j + 2) a j, j = 0,, 2,.... (2.8) The ratio test, applied to w (x) = a 2k x 2k and w 2 (x) = k=0 a 2k+ x 2k+ k=0 separately yields that the radius of convergence is one. Therefore, for any a 0, a R the function w is an analytic solution of (2.7) in (, ). Case : There is no n N {0} with a j = 0 for all j n; that is, the series does not reduce to a finite sum. We study the behaviour of w (x) and w 2 (x) as x tends to ±. First we consider w and split w in the form w (x) = k 0 k=0 a 2k x 2k + k=k 0 a 2k x 2k where k 0 is chosen such that 2k 0 (2k 0 +)+γ > 0 and (2k 0 +)+γ/2 ] / (2k 0 +)(k 0 +) ] /2. Then a 2k does not change its sign anymore for k k 0. Taking the logarithm of (2.8) for j = 2k yields ln a 2(k+) = ln ] 2k(2k + ) + λ (2k + )(2k + 2) + ln a 2k = ln ] (2k + ) + λ/2 (2k + )(k + ) + ln a 2k. Now we use the elementary estimate ln( u) u 2u 2 for 0 u /2 which yields (2k + ) + λ/2 ln a 2(k+) (2k + )(k + ) k + c k 2 + ln a 2k ] 2 (2k + ) + λ/2 + ln a 2k (2k + )(k + ) for some c > 0. Therefore, we arrive at the following estimate for ln a 2k : ln a 2k k j=k 0 k j + c j=k 0 j 2 + ln a 2k 0 for k k 0.

27 2.2. LEGENRE POLYNOMIALS 27 Using k j + j=k 0 k j=k 0 j+ j dt t = k k 0 dt t = ln k ln k 0 yields ] ln a 2k ln k + ln k 0 c j + ln a 2 2k 0 j=k 0 }{{} = ĉ and thus a 2k exp ĉ for all k k 0. k We note that a 2k /a 2k0 is positive for all k k 0 and thus w (x) a 2k0 a 0 a 2k0 + k 0 k= a2k a 2k0 exp ĉ ] x 2k k a 2k0 + exp ĉ a 2k0 k= k x2k = c exp ĉ a 2k0 ln( x2 ). From this we observe that w (x) + as x ± or w (x) as x ± (depending on the sign of a 2k0 ). By the same arguments one shows for positive a 2k0 + that w 2 (x) + as x + and w 2 (x) as x. For negative a 2k0 + the roles of + and have to be interchanged. In any case, the sum w(x) = w (x) + w 2 (x) is not bounded on, ] which contradicts our requirement on the solution of (2.7). Therefore, this case can not happen. Case 2: There is m N with a j = 0 for all j m; that is, the series reduces to a finite sum. Let m be the smallest number with this property. From the recursion formula we conclude that λ = n(n + ) for n = m 2 and, furthermore, that a 0 = 0 if n is odd and a = 0 if n is even. In particular, w is a polynomial of degree n. We can normalize w by w() = because w() 0. Indeed, if w() = 0 the differential equation ( x 2 ) w (x) 2x w (x) + n(n + ) w(x) = 0 would imply w () = 0. ifferentiating the differential equation would yield w (k) () = 0 for all k N, a contradiction to w 0. In the proof of this theorem we have proven more. We collect the results as a corollary. Corollary 2.5 The Legendre polynomials P n (x) = n j=0 a jx j have the properties (a) P n is even for even n and odd for odd n. (b) a j+2 = (c) + j(j + ) n(n + ) (j + )(j + 2) P n (x)p m (x) dx = 0 for n m. a j, j = 0,,..., n 2.

28 28 CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS Proof: Only part (c) has to be shown. We multiply the differential equation (2.7) for P n by P m (x), the differential equation (2.7) for P m by P n (x), take the difference and integrate. This yields 0 = + { P m (x) d dx ( x 2 ) P n(x) ] P n (x) d ( x 2 ) P dx m(x) ]} dx + n(n + ) m(m + ) ] + P n (x) P m (x) dx. The first integral vanishes by partial integration. This proves part (c). Before we return to the Laplace equation we prove some more results for the Legendre polynomials. Lemma 2.6 max P n(x) = for all n = 0,, 2,.... x Proof: For fixed n N we define the function Φ(x) := P n (x) 2 + x2 n(n + ) P n(x) 2 for all x, +]. We differentiate Φ and have Φ (x) = 2 P n(x) P n (x) + x2 n(n + ) P n (x) ] x n(n + ) P n(x) = 2 P n(x) n(n + ) P n (x) + d n(n + ) dx {( x2 ) P n(x)} + x P n(x) } {{ } = 0 by the differential equation ] = 2 x P n(x) 2 n(n + ). Therefore Φ > 0 on (0, ] and Φ < 0 on, 0); that is, Φ is monotonously increasing on (0, ] and monotonously decreasing on, 0). Therefore, 0 P n (x) 2 Φ(x) max{φ(), Φ( )}. From Φ() = Φ( ) = we conclude that P n (x) for all x, +]. The lemma is proven by noting that P n () =.

29 2.2. LEGENRE POLYNOMIALS Theorem 2.7 (Rodriguez) For all n N 0 we have P n (x) = 2 n n! d n dx n (x2 ) n, x R. Proof: First we prove that the right hand side solves the Legendre differential equation; that is, we show that ] d ( x 2 ) dn+ dx dx n+ (x2 ) n + n(n + ) dn dx n (x2 ) n = 0. (2.9) We observe that both parts are polynomials of degree n. We multiply the first part by x j for some j {0,..., n}, integrate and use partial integration two times. This yields for j A j := x j ] d ( x 2 ) dn+ dx dx n+ (x2 ) n dx = j x j ( x 2 ) dn+ dx n+ (x2 ) n dx = j (j ) x j 2 (j + ) x j] d n dx n (x2 ) n dx

30 30 CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS We note that no boundary contributions occur and, furthermore, that A 0 = 0. Now we use partial integration n times again. No boundary contributions occur either since there is always at least one factor (x 2 ) left. For j n the integral vanishes because the n-th derivative of x j and x j 2 vanish for j < n. For j = n we have A n = n(n + ) ( ) n n! (x 2 ) n dx. Analogously, we multiply the second part of (2.9) by x j, integrate, and apply partial integration n-times. This yields B j = n(n + ) x j d n dx n (x2 ) n dx = n(n + ) ( ) n n! and zero if j < n. This proves that the polynomial ] d ( x 2 ) dn+ dx dx n+ (x2 ) n + n(n + ) dn dx n (x2 ) n (x 2 ) n dx if j = n of degree n is orthogonal in L 2 (, ) to all polynomials of degree at most n and, therefore, has to vanish. Furthermore, d n dx n (x2 ) n = dn (x + ) n (x ) n] x= dx n From dn k dx = n ( ) n d k (x + )n k dxk k=0 (x )n n k = 0 for k we conclude x= d n dx n (x2 ) n = x= x= x= d n k (x )n dxn k. x= ( ) n 2 n dn (x )n 0 dxn = 2 n n! x= which proves the theorem. As a first application of the formula of Rodriguez we show: Theorem (a) P n (x) 2 dx = 2 2n + for all n = 0,, 2, (b) x P n (x) P n+ (x) dx = 2(n + ) (2n + )(2n + 3) for all n = 0,, 2,....

31 2.2. LEGENRE POLYNOMIALS 3 Proof: (a) We use the representation of P n by Rodriguez and n partial integrations, + d n dx n (x2 ) n dn dx n (x2 ) n dx = ( ) n + d 2n dx 2n (x 2 ) n] (x 2 ) n dx + + = ( ) n (2n)! (x 2 ) n dx = (2n)! ( x 2 ) n dx. + It remains to compute I n := ( x 2 ) n dx. We claim: I n = 2 2n(2n 2) 2 (2n + )(2n ) = 2 4n (n!) 2 (2n + )! for all n N. The assertion is true for n =. Let it be true for n, n 2. Then I n = I n + x 2 ( x 2 ) n dx = I n + = I n 2n I n and thus I n = 2n 2n + I n. + x d dx ] 2n ( x2 ) n dx This proves the representation of I n. We arrive at + P n (x) 2 dx = (2 n n!) 2 (2n)! 2 4n (n!) 2 (2n + )! = 2 2n +.

32 32 CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS (b) This is proven quite similarily: + x dn dx n (x2 ) n dxn+ dx n+ (x2 ) n+ dx = ( ) n+ + (x 2 n+ dn+ ) dx n+ + = ( ) n+ (x 2 ) n+ x d2n+ ] x dn dx n (x2 ) n dx dx 2n+ (x2 ) n }{{} =0 + (n + ) d2n dx 2n (x2 ) n dx }{{} =(2n)! = (n + ) ( ) n+ (2n)! I n+ = 2(n + )(2n)! (2n + 2)(2n)(2n 2) 2 (2n + 3)(2n + )(2n ) = (n + ) 2(2n n!)(2 n+ (n + )!) (2n + )(2n + 3) which yields the assertion. The formula of Rodriguez is also useful for proving recursion formulas for the Legendre polynomials. We prove only some of these formulas. Theorem 2.9 For all x R and n N, n 0, we have (a) P n+(x) P n (x) = (2n + ) P n (x), (b) (n + )P n+ (x) = (2n + )x P n (x) np n (x), (c) P n(x) = np n (x) + xp n (x), (d) xp n(x) = np n (x) + P n (x), (e) ( x 2 )P n(x) = (n + ) x P n (x) P n+ (x) ] = n x P n (x) P n (x) ], (f) P n+(x) + P n (x) = P n (x) + 2x P n(x), (g) (2n + )( x 2 )P n(x) = n(n + ) P n (x) P n+ (x) ],

33 2.2. LEGENRE POLYNOMIALS 33 (h) n P n+(x) (2n + )x P n(x) + (n + )P n (x) = 0. In these formulas we have set P = 0. Proof: (a) The formula is obvious for n = 0. Let now n. We calculate, using the formula of Rodriguez, ( P n+(x) d n+2 = 2 n+ (n + )! dx n+2 (x2 ) n+ d n = 2(n + ) d x (x 2 ) n]) 2 n+ (n + )! dx n dx = 2 n n! = P n (x) + d n dx n (x 2 ) n + 2n x 2 (x 2 ) n ] 2 n (n )! = P n (x) + 2n P n (x) + P n (x) which proves formula (a). d n dx n (x 2 ) n + (x 2 ) n ] (b) The orthogonality of the system {P n : n = 0,, 2,...} implies its linear independence. Therefore, {P 0,..., P n } forms a basis of the space P n of all polynomials of degree n. This yields existence of α n, β n R and q n 3 P n 3 such that P n+ (x) = α n xp n (x) + β n P n (x) + q n 3. The orthogonality condition implies that + + q n 3 P n+ dx = 0, q n 3 P n dx = 0, + x q n 3 (x) P n (x) dx = 0, thus + q n 3 (x) 2 dx = 0, and therefore q n 3 0. From = P n+ () = α n P n () + β n P n () = α n + β n we conclude that P n+ (x) = α n x P n (x) + ( α n )P n (x). We determine α n by Theorem 2.8: 0 = + = α n P n (x) P n+ (x) dx + x P n (x) P n (x) dx + ( α n ) + P n (x) 2 dx ] 2n = α n (2n )(2n + ) 2 2n + 2 2n,

34 34 CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS thus α n = 2n +. This proves part (b). n + (c) The definition of P n yields P n(x) = = = 2 n n! d n+ dx n+ (x2 ) n 2 n (n )! x 2 n (n )! d n dx n x(x2 ) n ] d n dx n (x2 ) n + n 2 n (n )! d n dx n (x2 ) n = x P n (x) + n P n (x). (d) ifferentiation of (b) and subtraction of (d) for n + instead of n yields (n + ) P n+(x) = (2n + )x P n(x) + (2n + )P n (x) np n (x), (n + ) P n+(x) = (n + ) 2 P n (x) + (n + )x P n(x), thus that is, 0 = (2n + ) (n + ) 2] P n (x) + (2n + ) (n + ) ] x P n(x) n P n (x) ; 0 = n 2 P n (x) + nx P n(x) n P n (x). Now we go back to the associated Legendre differential equation (2.4) and determine solutions in C 2 (, +) C, +] for m 0. Theorem 2.0 The functions P m n (x) = ( x 2 ) m/2 dm dx m P n(x), < x <, 0 m n, (2.0) are solutions of the differential equation (2.4) for λ = n(n + ); that is, d ( x 2 ) d ] dx dx P n m (x) + (n(n + ) m2 x 2 ) P m n (x) = 0, < x <, (2.) for all 0 m n. The functions P m n are called associated Legendre functions.

35 2.3. SPHERICAL HARMONICS 35 Proof: We compute ( x 2 ) d dx P n m (x) = m x ( x 2 ) ] d ( x 2 ) d dx dx P n m (x) m/2 dm dx m P n(x) + ( x 2 ) m/2+ = m Pn m (x) + m 2 x 2 ( x 2 m/2 dm ) dx P n(x) m m x ( x 2 ) m/2 dm+ dx m+ P n(x) dm+ (m + 2) x ( x 2 ) m/2 dx P n(x) + P m+2 m+ n (x) = m P m n (x) + m2 x 2 ifferentiating equation (2.7) m times yields m+ k=0 ( ) m + d k k The sum reduces to three terms only, thus x 2 P m n (x) (m + ) 2x x 2 dm+ dx m+ P n(x), P m+ n (x) + P m+2 n (x). dx k ( x2 ) dm+2 k dx m+2 k P n(x) + n(n + ) dm dx m P n(x) = 0. (2.2) ( x 2 ) dm+2 dx m+2 P n(x) (m + ) 2x dm+ dx m+ P n(x) n(n + ) m(m + ) ] d m dx m P n(x) = 0. We multiply the identity by ( x 2 ) m/2 and arrive at Pn m+2 (x) (m + ) 2x x 2 P m+ n (x) + n(n + ) m(m + ) ] P m n (x) = 0. Combining this with the previous equation by eliminating the terms involving Pn m+ (x) yields P m+2 n d dx ( x 2 ) d dx P m n (x) ] (x) and = m Pn m (x) + m2 x 2 x P m 2 n (x) n(n + ) m(m + ) ] Pn m (x) ( ) m 2 = n(n + ) P m x2 n (x) which proves the theorem. 2.3 Spherical Harmonics Now we return to the Laplace equation u = 0 and collect our arguments. We have shown that for n N and 0 m n the functions h m n (r, θ, ϕ) = c r n P m n (cos θ) e ±i m ϕ

36 36 CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS are harmonic functions in all of R 3 for any constant c C. Analogously, the functions v(r, θ, ϕ) = c r n P m n (cos θ) e ±imϕ are harmonic in R 3 \ {0}. They decay at infinity. Another observation on u is that these functions are polynomials. Theorem 2. The functions h m n (r, θ, ϕ) = r n P m n (cos θ) e ±imϕ with 0 m n are polynomials of degree n. Proof: First we consider the case m = 0, that is the functions h 0 n(r, ϕ, θ) = r n P n (cos θ). Let n be even, that is, n = 2l for some m. Then P n (x) = l a 2j x 2j. j=0 From r = x and x 3 = r cos θ we write u in the form ( ) h 0 n(x) = c x n x3 P n x = c x 2l l j=0 a 2j x 2j 3 x 2j = c l j=0 a 2j x 2j 3 x 2(m j), and this is obviously a polynomial of degree 2l = n. The same arguments hold for odd values of n. Now we show that also the associated functions; that is, for m > 0, are polynomials of degree n. We write h m n (r, θ, ϕ) = r n Pn m (cos θ) e imϕ = r n sin m θ dm dx P n(x) m e imϕ. x=cos θ and set Q = dm dx m P n. It holds as well as Therefore, cos ϕ = r cos θ = x 3 r, h m n (x) = r n m (x 2 + x 2 2) m/2 Q sin θ = r x 2 + x 2 2, x sin θ = x, sin ϕ = x 2 + x 2 2 ( x3 r ) (x + ix 2 ) m (x 2 + x 2 2) m/2 = r n m Q x 2. x 2 + x 2 2 ( x3 r ) (x + ix 2 ) m. The assertion follows because Q is a homogeneous polynomial of degree n m. Furthermore, the functions h m n (x) = c r n Pn m (cos θ) e ±i m ϕ are obviously homogeneous. polynomial of degree n N is called homogeneous, if A u(µx) = µ n u(x) for all x R n and µ R.

37 2.3. SPHERICAL HARMONICS 37 efinition 2.2 Let n N {0}. (a) Homogeneous harmonic polynomials of degree n are called spherical harmonics of order n. (b) The functions K n : S 2 C, defined as restrictions of spherical harmonics of degree n to the unit sphere are called spherical surface harmonics of order n. Therefore, the functions h m n (r, θ, ϕ) = r n P n m (cos θ)e imϕ are spherical harmonics of order n for n m n. Spherical harmonics which do not depend on ϕ are called zonal. Thus, in case of m = 0 the functions h 0 n are zonal spherical harmonics. For any surface spherical harmonic K n of order n the function ( ) x H n (x) = x n K n, x R 3, x is a homogeneous harmonic polynomial; that is, a spherical harmonic. We immediately have Lemma 2.3 (a) K n ( ˆx) = ( ) n K n (ˆx) for all ˆx S 2. (b) S 2 K n (ˆx) K m (ˆx) ds(ˆx) = 0 for all n m. Proof: Part (a) follows immediately since H n is homogeneous (set µ = ). (b) With Green s second formula in the region {x R 3 : x < } we have ( ) H n ds = (H n H m H m H n ) dx = 0. S 2 r H m H m r H n x Setting f(r) = H n (rˆx) = r n H n (ˆx) for fixed ˆx S 2 we have that f () = H r n(ˆx) = n H n (ˆx); that is, 0 = (m n) H n H m ds = (m n) K n K m ds. S 2 S 2 Now we determine the dimension of the space of spherical harmonics for fixed order n. Theorem 2.4 The set of spherical harmonics of order n is a vector space of dimension 2n +. In particular, there exists a system {Kn m : n m n} of spherical harmonics of order n such that { for m = l, Kn m Kn l ds = δ m,l = 0 for m l ; S 2 that is, {K m n : n m n} is an orthonormal basis of this vector space.

38 38 CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS Proof: Every homogeneous polynomial of degree n is necessarily of the form n H n (x) = A n j (x, x 2 ) x j 3 x R 3, (2.3) j=0 where A n j are homogeneous polynomials with respect to (x, x 2 ) of degree n j. Since H n is harmonic it follows that n n 0 = H n (x) = x j 3 2 A n j (x, x 2 ) + j(j ) x j 2 3 A n j (x, x 2 ), where 2 = 2 x 2 we conclude that + 2 x = j=0 j=2 denote the two-dimensional Laplace operator. From 2 A 0 = 2 A = 0 n 2 2 A n j (x, x 2 ) + (j + )(j + 2) A n j 2 (x, x 2 )] x j 3, j=0 and thus by comparing the coefficients A n j 2 (x, x 2 ) = (j + )(j + 2) 2A n j (x, x 2 ) for all (x, x 2 ) R 2, j = 0,..., n 2; that is (replace n j by j) A j 2 = (n j + )(n j + 2) 2A j for j = n, n,..., 2. (2.4) Also, one can reverse the arguments: If A n and A n are homogeneous polynomials of degree n and n, respectively, then all of the functions A j defined by (2.4) are homogeneous polynomials of degree j, and H n is a homogeneous harmonic polynomial of order n. Therefore, the space of all spherical harmonics of order n is isomorphic to the space { (An, A n ) : A n, A n are homogeneous polynomials of degree n and n, resp. }. From the representation A n (x, x 2 ) = n i=0 a i x i x n i 2 we note that the dimension of the space of all homogeneous polynomials of degree n is just n +. Therefore, the dimension of the space of all spherical harmonics of order n is (n + ) + n = 2n +. Finally, it is well known that any basis of this finite dimensional Euclidian space can be orthogonalized by the method of Schmidt. Remark: The set {K m n : n m n} is not uniquely determined. Indeed, for any orthogonal matrix A R 3 3 ; that is, A A = I, also the set {K m n (Ax) : n m n} is an orthonormal system. Indeed, the substitution x = Ay yields S 2 K m n (x) K m n (x) ds(x) = S 2 K m n (Ay) K m n (Ay) ds(y) = δ n,n δ m,m. Now we can state that the spherical harmonics Hn m = cr n P n m (cos θ)e imϕ with n m n, which we have determined by separation, constitute an orthogonal basis of the space of spherical harmonics of order n.

39 2.3. SPHERICAL HARMONICS 39 Theorem 2.5 The functions Hn m (r, θ, ϕ) = r n P n m (cos θ)e imϕ, n m n are spherical harmonics of order n. They are mutually orthogonal and, therefore, form an orthogonal basis of the (2n + )-dimensional space of all spherical harmonics of order n. Proof: We already know that H m n are spherical harmonics of order n N. Thus the theorem follows from x < H m n (x) H l n(x) dx = π 2π r 2n P n m (cos θ) P n l (cos θ) e i(m l)ϕ r 2 sin θ dϕ dθ dr = 0 for m l. We want to normalize these functions. First we consider the associated Legendre functions. Theorem P m n (x) 2 dx = 2 (n + m)!, m = 0,..., n, n N {0}. 2n + (n m)! Proof: The case m = 0 has been proven in Theorem 2.8 already. For m partial integration yields + P m n (x) 2 dx = + = ( x 2 ) m dm dx m P n(x) dm dx m P n(x) dx + d dx ] ( x 2 ) m dm d m dx P n(x) m dx P n(x) dx. (2.5) m Now we differentiate the Legendre differential equation (2.7) (m )-times (see (2.2) ### check! ### for m replaced by m ); that is, ( x 2 ) dm+ dx m+ P n(x) 2m x dm dx m P n(x) + n(n + ) m(m ) ] d m dx m P n(x) = 0, thus, after multiplication by ( x 2 ) m, ] d ( x 2 ) m dm dx dx P n(x) + n(n + ) m(m ) ] ( x 2 m dm ) m dx P n(x) = 0. m Substituting this into (2.2) ### check! ### yields + Pn m (x) 2 dx = n(n+) m(m ) ] + P m n (x) 2 dx = (n+m) (n m+) + P m n (x) 2 dx.

40 40 CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS This is a recursion formula for using the formula for m = 0. Now we define the normalized spherical surface harmonics Y m n Y m n (θ, ϕ) := + P m n (x) 2 dx with respect to m and yields the assertion by by (2n + ) (n m )! P n m (cos θ) e imϕ, n m n, n = 0,,... (2.6) 4π (n + m )! They form an orthonormal system in L 2 (S 2 ). x = (sin θ cos ϕ, sin θ sin ϕ, cos θ) S 2. We will identify Y m n (x) with Y m n (θ, ϕ) for From (2.2) for λ = n(n + ) we remember that Y m n satisfies the differential equation S Yn m (θ, ϕ) + n(n + )Yn m (θ, ϕ) sin θ Y ] m = sin θ θ n (θ, ϕ) θ + sin 2 θ 2 Y m n (θ, ϕ) ϕ 2 + n(n + ) Y m n (θ, ϕ) = 0, with the Laplace-Beltrami operator S (see page??). Thus, we now see that λ = n(n + ) for n N are the eigenvalues of S with eigenfunctions Y m n, m n. The Laplace-Beltrami operator S is symmetric; that is, (f, S g) L 2 (S 2 ) = ( S f, g) L 2 (S 2 ) for all f, g C 2 (S 2 ). (2.7) Indeed, by partial integration we see (f, S g) L 2 (S 2 ) = = π 2π 0 0 2π π 0 = 0 2π π 0 { f(θ, ϕ) sin θ f(θ, ϕ) θ 0 f(θ, ϕ) θ sin θ sin θ sin θ θ ] g(θ, ϕ) θ + ] g(θ, ϕ) dθ dϕ + θ g(θ, ϕ) θ dθ dϕ π 0 sin 2 θ π 0 sin θ sin θ } 2 g(θ, ϕ) sin θ dϕ dθ ϕ 2 2π 0 2π 0 f(θ, ϕ) 2 g(θ, ϕ) dϕ dθ ϕ 2 f(θ, ϕ) ϕ g(θ, ϕ) ϕ and this is symmetric with respect to f and g. Setting f = g we observe that S is also non-negative; that is, dϕ dθ (f, S f) L 2 (S 2 ) 0 for all f C 2 (S 2 ). (2.8) In the setting of abstract functional analysis we note that is a (densely defined and unbounded) operator from L 2 (S 2 ) into itself which is selfadjoint. This observation allows the use of general functional analytic tools to prove, e.g., that the eigenfunctions { Yn m : m n, n = 0,,... } of form a complete orthonormal system of L 2 (S 2 ). We want to prove this fact directly. First we show the following result.