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8

Spectral Theory

Introduction

This chapter is devoted to the spectral theory of self-adjoint, differentialoperators. We cover a number of different topics, beginning in §1 witha proof of the spectral theorem. It was an arbitrary choice to put thatmaterial here, rather than in Appendix A, on functional analysis. The mainmotivation for putting it here is to begin a line of reasoning that will becontinued in subsequent sections, using the great power of studying unitarygroups as a tool in spectral theory. After we show how easily this studyleads to a proof of the spectral theorem in §1, in later sections we use it invarious ways: as a tool to establish self-adjointness, as a tool for obtainingspecific formulas, including basic identities among special functions, and inother capacities.

Sections 2 and 3 deal with some general questions in spectral theory, suchas when does a differential operator define a self-adjoint operator, whendoes it have a compact resolvent, and what asymptotic properties does itsspectrum have? We tackle the latter question, for the Laplace operator ∆,by examining the asymptotic behavior of the trace of the solution operatoret∆ for the heat equation, showing that

(0.1) Tr et∆ = (4πt)−n/2 vol Ω + o(t−n/2), t ց 0,

when Ω is either a compact Riemannian manifold or a bounded domain inR

n (and has the Dirichlet boundary condition). Using techniques developedin §13 of Chapter 7, we could extend (0.1) to general compact Riemannianmanifolds with smooth boundary and to other boundary conditions, suchas the Neumann boundary condition. We use instead a different methodhere in §3, one that works without any regularity hypotheses on ∂Ω. Insuch generality, (0.1) does not necessarily hold for the Neumann boundaryproblem.

The study of (0.1) and refinements got a big push from [Kac]. As pursuedin [MS], it led to developments that we will discuss in Chapter 10. Theproblem of to what extent a Riemannian manifold is determined by the

2 8. Spectral Theory

spectrum of its Laplace operator has led to much work, which we do notinclude here. Some is discussed in [Ber], [Br], [BGM], and [Cha]. Wemention particularly some distinct regions in R

2 whose Laplace operatorshave the same spectra, given in [GWW].

We have not included general results on the spectral behavior of ∆ ob-tained via geometrical optics and its refinement, the theory of Fourier in-tegral operators. Results of this nature can be found in Volume 3 of [Ho],in [Shu], and in Chapter 12 of [T1].

Sections 4 through 7 are devoted to specific examples. In §4 we study theLaplace operator on the unit spheres Sn. We specify precisely the spectrumof ∆ and discuss explicit formulas for certain functions of ∆, particularly

(0.2) A−1 sin tA, A =(−∆ +

K

4(n − 1)2

)1/2

.

with K = 1, the sectional curvature of Sn. In §5 we obtain an explicitformula for (0.2), with K = −1, on hyperbolic space. In §6 we study thespectral theory of the harmonic oscillator

(0.3) H = −∆ + |x|2.

We obtain an explicit formula for e−tH , an analogue of which will be usefulin Chapter 10. In §7 we study the operator

(0.4) H = −∆ − K|x|−1

on R3, obtaining in particular all the eigenvalues of this operator. This

operator arises in the simplest quantum mechanical model of the hydrogenatom. In §8 we study the Laplace operator on a cone. Studies done in thesesections bring in a number of special functions, including Legendre func-tions, Bessel functions, and hypergeometric functions. We have includedtwo auxiliary problem sets, one on confluent hypergeometric functions andone on hypergeometric functions.

1. The spectral theorem

Appendix A contains a proof of the spectral theorem for a compact, self-adjoint operator A on a Hilbert space H. In that case, H has an orthonor-mal basis uj such that Auj = λjuj , λj being real numbers having only0 as an accumulation point. The vectors uj are eigenvectors.

A general bounded, self-adjoint operator A may not have any eigenvec-tors, and the statement of the spectral theorem is somewhat more subtle.The following is a useful version.

Theorem 1.1. If A is a bounded, self-adjoint operator on a separableHilbert space H, then there is a σ-compact space Ω, a Borel measure µ, a

1. The spectral theorem 3

unitary map

(1.1) W : L2(Ω, dµ) −→ H,

and a real-valued function a ∈ L∞(Ω, dµ) such that

(1.2) W−1AWf(x) = a(x)f(x), f ∈ L2(Ω, dµ).

Note that when A is compact, the eigenvector decomposition above yields(1.1) and (1.2) with (Ω, µ) a purely atomic measure space. Later in thissection we will extend Theorem 1.1 to the case of unbounded, self-adjointoperators.

In order to prove Theorem 1.1, we will work with the operators

(1.3) U(t) = eitA,

defined by the power-series expansion

(1.4) eitA =∞∑

n=0

(it)n

n!An.

This is a special case of a construction made in §4 of Chapter 1. U(t) isuniquely characterized as the solution to the differential equation

(1.5)d

dtU(t) = iAU(t), U(0) = I.

We have the group property

(1.6) U(s + t) = U(s)U(t),

which follows since both sides satisfy the ODE (d/ds)Z(s) = iAZ(s), Z(0)= U(t). If A = A∗, then applying the adjoint to (1.4) gives

(1.7) U(t)∗ = U(−t),

which is the inverse of U(t) in view of (1.6). Thus U(t) : t ∈ R is a groupof unitary operators.

For a given v ∈ H, let Hv be the closed linear span of U(t)v : t ∈ R;we say Hv is the cyclic space generated by v. We say v is a cyclic vectorfor H if H = Hv. If Hv is not all of H, note that H⊥

v is invariant underU(t), that is, U(t)H⊥

v ⊂ H⊥v for all t, since for a linear subspace V of H,

generally

(1.8) U(t)V ⊂ V =⇒ U(t)∗V ⊥ ⊂ V ⊥.

Using this observation, we can prove the next result.

Lemma 1.2. If U(t) is a unitary group on a separable Hilbert space H,then H is an orthogonal direct sum of cyclic subspaces.

Proof. Let wj be a countable, dense subset of H. Take v1 = w1 andH1 = Hv1

. If H1 6= H, let v2 be the first nonzero element P1wj , j ≥ 2,


where P1 is the orthogonal projection of H onto H⊥1 , and let H2 = Hv2

.Continue.

In view of this, Theorem 1.1 is a consequence of the following:

Proposition 1.3. If U(t) is a strongly continuous, unitary group on H,having a cyclic vector v, then we can take Ω = R, and there exists a positiveBorel measure µ on R and a unitary map W : L2(R, dµ) → H such that

(1.9) W−1U(t)W f(x) = eitxf(x), f ∈ L2(R, dµ).

The measure µ on R will be the Fourier transform

(1.10) µ = ζ,

where

(1.11) ζ(t) = (2π)−1/2 (eitAv, v).

It is not clear a priori that (1.10) defines a measure; since ζ ∈ L∞(R), wesee that µ is a tempered distribution. We will show that µ is indeed apositive measure during the course of our argument. As for the map W ,we first define

(1.12) W : S(R) −→ H,

where S(R) is the Schwartz space of rapidly decreasing functions, by

(1.13) W (f) = f(A)v,

where we define the operator f(A) by the formula

(1.14) f(A) = (2π)−1/2

∫ ∞

−∞f(t)eitA dt.

The reason for this notation will become apparent shortly; see (1.20). Mak-ing use of (1.10), we have

(1.15)

(f(A)v, g(A)v

)= (2π)−1

(∫f(s)eisAv ds,

∫g(t)eitAv dt

)

= (2π)−1

∫∫f(s)g(t)

(ei(s−t)Av, v

)ds dt

= (2π)−1/2

∫∫f(s)g(σ − s)ζ(σ) ds dσ

= 〈(fg), ζ〉= 〈fg, µ〉.

Now, if g = f , the left side of (1.15) is ‖f(A)v‖2, which is ≥ 0. Hence

(1.16)⟨|f |2, µ

⟩≥ 0, for all f ∈ S(R).


With this, we can establish:

Lemma 1.4. The tempered distribution µ, defined by (1.10)–(1.11), is apositive measure on R.

Proof. Apply (1.16) with f =√

Fs,σ, where

Fs,σ(τ) = (4πs)−1/2e−(τ−σ)2/4s, s > 0, σ ∈ R.

Note that this is a fundamental solution to the heat equation. For eachs > 0, Fs,0 ∗ µ is a positive function. We saw in Chapter 3 that Fs,0 ∗ µconverges to µ in S ′(R) as s → 0, so this implies that µ is a positivemeasure.

Now we can finish the proof of Proposition 1.3. From (1.15) we see thatW has a unique continuous extension

(1.17) W : L2(R, dµ) −→ H,

and W is an isometry. Since v is assumed to be cyclic, the range of Wmust be dense in H, so W must be unitary. From (1.14) it follows that iff ∈ S(R), then

(1.18) eisAf(A) = fs(A), with fs(τ) = eisτf(τ).

Hence, for f ∈ S(R),

(1.19) W−1eisAW f = W−1fs(A)v = eisτf(τ).

Since S(R) is dense in L2(R, dµ), this gives (1.9). Thus the spectral theoremfor bounded, self-adjoint operators is proved.

Given (1.9), we have from (1.14) that

(1.20) W−1f(A)W g(x) = f(x)g(x), f ∈ S(R), g ∈ L2(R, dµ),

which justifies the notation f(A) in (1.14).Note that (1.9) implies

(1.21) W−1AW f(x) = x f(x), f ∈ L2(R, dµ),

since (d/dt)U(t) = iAU(t). The essential supremum of x on (R, µ) is equalto ‖A‖. Thus µ has compact support in R if A is bounded. If a self-adjointoperator A has the representation (1.21), one says A has simple spectrum.It follows from Proposition 1.3 that A has simple spectrum if and only ifit has a cyclic vector.

One can generalize the results above to a k-tuple of commuting, bounded,self-adjoint operators A = (A1, . . . , Ak). In that case, for t = (t1, . . . , tk) ∈R

k, set

(1.22) U(t) = eit·A, t · A = t1A1 + · · · + tkAk.


The hypothesis that the Aj all commute implies U(t) = U1(t1) · · ·Uk(tk),where Uj(s) = eisAj . U(t) in (1.22) continues to satisfy the properties(1.6) and (1.7); we have a k-parameter unitary group. As above, for v ∈ H,we set Hv equal to the closed linear span of U(t)v : t ∈ R

k, and we say vis a cyclic vector provided Hv = H. Lemma 1.2 goes through in this case.Furthermore, for f ∈ S(Rk), we can define

(1.23) f(A) = (2π)−k/2

∫f(t)eit·A dt,

and if H has a cyclic vector v, the proof of Proposition 1.3 generalizes,giving a unitary map W : L2(Rk, dµ) → H such that

(1.24) W−1U(t)Wf(x) = eit·xf(x), f ∈ L2(Rk, dµ), t ∈ Rk.

Therefore, Theorem 1.1 has the following extension.

Proposition 1.5. If A = (A1, . . . , Ak) is a k-tuple of commuting, bounded,self-adjoint operators on H, there is a measure space (Ω, µ), a unitary mapW : L2(Ω, dµ) → H, and real-valued aj ∈ L∞(Ω, dµ) such that

(1.25) W−1AjWf(x) = aj(x)f(x), f ∈ L2(Ω, dµ), 1 ≤ j ≤ k.

A bounded operator B ∈ L(H) is said to be normal provided B and B∗

commute. Equivalently, if we set

(1.26) A1 =1

2

(B + B∗), A2 =

1

2i

(B − B∗),

then B = A1 + iA2, and (A1, A2) is a 2-tuple of commuting, self-adjointoperators. Applying Proposition 1.5 and setting b(x) = a1(x) + ia2(x), wehave:

Corollary 1.6. If B ∈ L(H) is a normal operator, there is a unitary mapW : L2(Ω, dµ) → H and a (complex-valued) b ∈ L∞(Ω, dµ) such that

(1.27) W−1BWf(x) = b(x)f(x), f ∈ L2(Ω, dµ).

In particular, Corollary 1.6 holds when B = U is unitary. We nextextend the spectral theorem to an unbounded, self-adjoint operator A ona Hilbert space H, whose domain D(A) is a dense linear subspace of H.This extension, due to von Neumann, uses von Neumann’s unitary trick,described in (8.18)–(8.19) of Appendix A. We recall that, for such A, thefollowing three properties hold:

(1.28)

A ± i : D(A) −→ H bijectively,

U = (A − i)(A + i)−1 is unitary on H,

A = i(I + U)(I − U)−1,


where the range of I − U = 2i(A + i)−1 is D(A). Applying Corollary 1.6to B = U , we have the following theorem:

Theorem 1.7. If A is an unbounded, self-adjoint operator on a sepa-rable Hilbert space H, there is a measure space (Ω, µ), a unitary mapW : L2(Ω, dµ) → H, and a real-valued measurable function a on Ω suchthat

(1.29) W−1AWf(x) = a(x)f(x), Wf ∈ D(A).

In this situation, given f ∈ L2(Ω, dµ), Wf belongs to D(A) if and only ifthe right side of (1.29) belongs to L2(Ω, dµ).

The formula (1.29) is called the “spectral representation” of a self-adjointoperator A. Using it, we can extend the functional calculus defined by(1.14) as follows. For a Borel function f : R → C, define f(A) by

(1.30) W−1f(A)Wg(x) = f(a(x))g(x).

If f is a bounded Borel function, this is defined for all g ∈ L2(Ω, dµ) andprovides a bounded operator f(A) on H. More generally,

(1.31) D(f(A)

)=

Wg ∈ H : g ∈ L2(Ω, dµ) and f(a(x))g ∈ L2(Ω, dµ)

.

In particular, we can define eitA, for unbounded, self-adjoint A, by

W−1eitAWg = eita(x)g(x).

Then eitA is a strongly continuous unitary group, and we have the followingresult, known as Stone’s theorem (stated as Proposition 9.5 in AppendixA):

Proposition 1.8. If A is self-adjoint, then iA generates a strongly con-tinuous, unitary group, U(t) = eitA.

Note that Lemma 1.2 and Proposition 1.3 are proved for a strongly con-tinuous, unitary group U(t) = eitA, without the hypothesis that A bebounded. This yields the following analogue of (1.2):

(1.32) W−1U(t)Wf(x) = eita(x)f(x), f ∈ L2(Ω, dµ),

for this more general class of unitary groups. Sometimes a direct construc-tion, such as by PDE methods, of U(t) is fairly easy. In such a case, theuse of U(t) can be a more convenient tool than the unitary trick involving(1.28).

We say a self-adjoint operator A is positive, A ≥ 0, provided (Au, u) ≥ 0,for all u ∈ D(A). In terms of the spectral representation, this says we have(1.29) with a(x) ≥ 0 on Ω. In such a case, e−tA is bounded for t ≥ 0,even for complex t with Re t ≥ 0, and also defines a strongly continuoussemigroup. This proves Proposition 9.4 of Appendix A.


Given a self-adjoint operator A and a Borel set S ⊂ R, define P (S) =χS(A), that is, using (1.29),

(1.33) W−1P (S)Wg = χS(a(x))g(x), g ∈ L2(Ω, dµ),

where χS is the characteristic function of S. Then each P (S) is an orthog-onal projection. Also, if S =

⋃j≥1 Sj is a countable union of disjoint Borel

sets Sj , then, for each u ∈ H,

(1.34) limn→∞

n∑

j=1

P (Sj)u = P (S)u,

with convergence in the H-norm. This is equivalent to the statement that

n∑

j=1

χSj(a(x))g → χS(a(x))g in L2-norm, for each g ∈ L2(Ω, dµ),

which in turn follows from Lebesgue’s dominated convergence theorem. By(1.34), P (·) is a strongly countably additive, projection-valued measure.Then (1.30) yields

(1.35) f(A) =

∫f(λ) P (dλ).

P (·) is called the spectral measure of A.One useful formula for the spectral measure is given in terms of the jump

of the resolvent Rλ = (λ−A)−1, across the real axis. We have the following.

Proposition 1.9. For bounded, continuous f : R → C,

(1.36) f(A)u = limεց0

1

2πi

∫ ∞

−∞f(λ)

[(λ− iε−A)−1 − (λ + iε−A)−1

]u dλ.

Proof. Since W−1f(A)W is multiplication by f(a(x)), (1.36) follows fromthe fact that

(1.37)1

π

∫ ∞

−∞

εf(λ)

(λ − a(x))2 + ε2dλ −→ f(a(x)),

pointwise and boundedly, as ε ց 0.

An important class of operators f(A) are the fractional powers f(A) =Aα, α ∈ (0,∞), defined by (1.30)–(1.31), with f(λ) = λα, provided A ≥ 0.Note that if g ∈ C([0,∞)) satisfies g(0) = 1, g(λ) = O(λ−α) as λ → ∞,then, for u ∈ H,

(1.38) u ∈ D(Aα) ⇐⇒ ‖Aαg(εA)u‖H is bounded, for ε ∈ (0, 1],

as follows easily from the characterization (1.31) and Fatou’s lemma. Wenote that Proposition 2.2 of Chapter 4 applies to D(Aα), describing it asan interpolation space.

2. Self-adjoint differential operators 9

We particularly want to identify D(A1/2), when A is a positive, self-adjoint operator on a Hilbert space H constructed by the Friedrichs method,as described in Proposition 8.7 of Appendix A. Recall that this arises asfollows. One has a Hilbert space H1, a continuous injection J : H1 → Hwith dense range, and one defines A by

(1.39)(A(Ju), Jv

)H

= (u, v)H1,

with

(1.40)D(A) =

Ju ∈ JH1 ⊂ H : v 7→ (u, v)H1

is

continuous in Jv, in the H-norm.

We establish the following.

Proposition 1.10. If A is obtained by the Friedrichs extension method(1.39)–(1.40), then

(1.41) D(A1/2) = J(H1) ⊂ H.

Proof. D(A1/2) consists of elements of H that are limits of sequences inD(A), in the norm ‖A1/2u‖H +‖u‖H . As shown in the proof of Proposition8.7 in Appendix A, D(A) = R(JJ∗). Now

(1.42) ‖A1/2JJ∗f‖2H = (AJJ∗f, JJ∗f)H = ‖J∗f‖2

H1.

Thus a sequence (JJ∗fn) converges in the D(A1/2)-norm (to an element g)if and only if (J∗fn) converges in the H1-norm (to an element u), in whichcase g = Ju. Since J∗ : H → H1 has dense range, precisely all u ∈ H1

arise as limits of such (J∗fn), so the proposition is proved.

Exercises

1. The definition (1.33) of the spectral measure P (·) of a self-adjoint operator Adepends a priori on a choice of the spectral representation of A. Show thatany two spectral representations of A yield the same spectral measure.(Hint: For f ∈ S(R), f(A) is well defined by (1.14), or alternatively by (1.36).)

2. Self-adjoint differential operators

In this section we present some examples of differential operators on amanifold Ω which, with appropriately specified domains, give unbounded,self-adjoint operators on L2(Ω, dV ), dV typically being the volume elementdetermined by a Riemannian metric on Ω.

We begin with self-adjoint operators arising from the Laplacian, makinguse of material developed in Chapter 5. Let Ω be a smooth, compact


Riemannian manifold with boundary, or more generally the closure of anopen subset Ω of a compact manifold M without boundary. Then, as shownin Chapter 5,

(2.1) I − ∆ : H10 (Ω) −→ H1

0 (Ω)∗

is bijective, with inverse we denote T ; if we restrict T to L2(Ω),

(2.2) T : L2(Ω) −→ L2(Ω) is compact and self-adjoint.

Denote by R(T ) the image of L2(Ω) under T . We can apply Proposition8.2 of Appendix A to deduce the following.

Proposition 2.1. If Ω is a region in a compact Riemannian manifold M ,then ∆ is self-adjoint on L2(Ω), with domain D(∆) = R(T ) ⊂ H1

0 (Ω)described above.

For a further description of D(∆), note that

(2.3) D(∆) = u ∈ H10 (Ω) : ∆u ∈ L2(Ω).

If ∂Ω is smooth, we can apply the regularity theory of Chapter 5 to obtain

(2.4) D(∆) = H10 (Ω) ∩ H2(Ω).

Instead of relying on Proposition 8.2, we could use the Friedrichs con-struction, given in Proposition 8.7 of Appendix A. This construction can beapplied more generally. Let Ω be any Riemannian manifold, with Laplaceoperator ∆. We can define H1

0 (Ω) to be the closure of C∞0 (Ω) in the space

u ∈ L2(Ω) : du ∈ L2(Ω,Λ1). The inner product on H10 (Ω) is

(2.5) (u, v)1 = (u, v)L2 + (du, dv)L2 .

We have a natural inclusion H10 (Ω) → L2(Ω), and the Friedrichs method

gives a self-adjoint operator A on L2(Ω) such that

(2.6) (Au, v)L2 = (u, v)1, for u ∈ D(A), v ∈ H10 (Ω),

with

(2.7)D(A) =

u ∈ H1

0 (Ω) : v 7→ (u, v)1 extends from H10 (Ω) → C to a

continuous linear functional L2(Ω) → C,

that is,

(2.8)D(A) =

u ∈ H1

0 (Ω) : ∃f ∈ L2(Ω) such that

(u, v)1 = (f, v)L2 ,∀v ∈ H10 (Ω)

.

Integrating (2.5) by parts for v ∈ C∞0 (Ω), we see that A = I −∆ on D(A),

so we have a self-adjoint extension of ∆ in this general setting, with domainagain described by (2.3).


The process above gives one self-adjoint extension of ∆, initially definedon C∞

0 (Ω). It is not always the only self-adjoint extension. For example,suppose Ω is compact with smooth boundary; consider H1(Ω), with innerproduct (2.5), and apply the Friedrichs extension procedure. Again wehave a self-adjoint operator A, extending I − ∆, with (2.8) replaced by

(2.9)D(A) =

u ∈ H1(Ω) : ∃f ∈ L2(Ω) such that

(u, v)1 = (f, v)L2 ,∀v ∈ H1(Ω).

In this case, Proposition 7.2 of Chapter 5 yields the following.

Proposition 2.2. If Ω is a smooth, compact manifold with boundary and∆ the self-adjoint extension just described, then

(2.10) D(∆) = u ∈ H2(Ω) : ∂νu = 0 on ∂Ω.

In case (2.10), we say D(∆) is given by the Neumann boundary condi-tion, while in case (2.4) we say D(∆) is given by the Dirichlet boundarycondition.

In both cases covered by Propositions 2.1 and 2.2, (−∆)1/2 is defined asa self-adjoint operator. We can specify its domain using Proposition 1.10,obtaining the next result:

Proposition 2.3. In case (2.3), D((−∆)1/2) = H10 (Ω); in case (2.10),

D((−∆)1/2) = H1(Ω).

Though ∆ on C∞0 (Ω) has several self-adjoint extensions when Ω has a

boundary, it has only one when Ω is a complete Riemannian manifold.This is a classical result, due to Roelcke; we present an elegant proof dueto Chernoff [Chn]. When an unbounded operator A0 on a Hilbert space H,with domain D0, has exactly one self-adjoint extension, namely the closureof A0, we say A0 is essentially self-adjoint on D0.

Proposition 2.4. If Ω is a complete Riemannian manifold, then ∆ isessentially self-adjoint on C∞

0 (Ω). Thus the self-adjoint extension withdomain given by (2.3) is the closure of ∆ on C∞

0 (Ω).

Proof. We will obtain this as a consequence of Proposition 9.6 of AppendixA, which states the following. Let U(t) = eitA be a unitary group on aHilbert space H which leaves invariant a dense linear space D; U(t)D ⊂ D.If A is an extension of A0 and A0 : D → D, then A0 and all its powers areessentially self-adjoint on D.


In this case, U(t) will be the solution operator for a wave equation, andwe will exploit finite propagation speed. Set

(2.11) iA0 =

(0 I

∆ − I 0

), D(A0) = C∞

0 (Ω) ⊕ C∞0 (Ω).

The group U(t) will be the solution operator for the wave equation

(2.12) U(t)

(fg

)=

(u(t)ut(t)

),

where u(t, x) is determined by

∂2u

∂t2− (∆ − 1)u = 0; u(0, x) = f, ut(0, x) = g.

It was shown in §2 of Chapter 6 that U(t) is a unitary group on H =H1

0 (Ω)⊕L2(Ω); its generator is an extension of (2.11), and finite propaga-tion speed implies that U(t) preserves C∞

0 (Ω) ⊕ C∞0 (Ω) for all t, provided

Ω is complete. Thus each Ak0 is essentially self-adjoint on this space. Since

(2.13) −A20 =

(∆ − I 0

0 ∆ − I

),

we have the proof of Proposition 2.3. Considering A2k0 , we deduce further-

more that each power ∆k is essentially self-adjoint on C∞0 (Ω), when Ω is

complete.

Though ∆ is not essentially self-adjoint on C∞0 (Ω) when Ω is compact,

we do have such results as the following:

Proposition 2.5. If Ω is a smooth, compact manifold with boundary,then ∆ is essentially self-adjoint on

(2.14) u ∈ C∞(Ω) : u = 0 on ∂Ω,its closure having domain described by (2.3). Also, ∆ is essentially self-adjoint on

(2.15) u ∈ C∞(Ω) : ∂νu = 0 on ∂Ω,its closure having domain described by (2.10).

Proof. It suffices to note the simple facts that the closure of (2.14) inH2(Ω) is (2.3) and the closure of (2.15) in H2(Ω) is (2.10).

We note that when Ω is a smooth, compact Riemannian manifold withboundary, and D(∆) is given by the Dirichlet boundary condition, then

(2.16)

∞⋂

j=1

D(∆j) = u ∈ C∞(Ω) : ∆ku = 0 on ∂Ω, k = 0, 1, 2, . . . ,


and when D(∆) is given by the Neumann boundary condition, then

(2.17)

∞⋂

j=1

D(∆j) = u ∈ C∞(Ω) : ∂ν(∆ku) = 0 on ∂Ω, k ≥ 0.

We now derive a result that to some degree amalgamates Propositions2.4 and 2.5. Let Ω be a smooth Riemannian manifold with boundary, andset

(2.18) C∞c (Ω) = u ∈ C∞(Ω) : supp u is compact in Ω;

we do not require elements of this space to vanish on ∂Ω. We say that Ωis complete if it is complete as a metric space.

Proposition 2.6. If Ω is a smooth Riemannian manifold with boundarywhich is complete, then ∆ is essentially self-adjoint on

(2.19) u ∈ C∞c (Ω) : u = 0 on ∂Ω.

In this case, the closure has domain given by (2.3).

Proof. Consider the following linear subspace of (2.19):

(2.20) D0 = u ∈ C∞c (Ω) : ∆ju = 0 on ∂Ω for j = 0, 1, 2, . . . .

Let U(t) be the unitary group on H10 (Ω)⊕L2(Ω) defined as in (2.12), with

u also satisfying the Dirichlet boundary condition, u(t, x) = 0 for x ∈ ∂Ω.Then, by finite propagaton speed, U(t) preserves D0 ⊕ D0, provided Ωis complete, so as in the proof of Proposition 2.4, we deduce that ∆ isessentially self-adjoint on D0; a fortiori it is essentially self-adjoint on thespace (2.19).

By similar reasoning, we can show that if Ω is complete, then ∆ is es-sentially self-adjoint on

(2.21) u ∈ C∞c (Ω) : ∂νu = 0 on ∂Ω.

The results of this section so far have involved only the Laplace operator∆. It is also of interest to look at Schrodinger operators, of the form−∆ + V , where the “potential” V (x) is a real-valued function. In thissection we will restrict attention to the case V ∈ C∞(Ω) and we will alsosuppose that V is bounded from below. By adding a constant to −∆ + V ,we may as well suppose

(2.22) V (x) ≥ 1 on Ω.

We can define a Hilbert space H1V 0(Ω) to be the closure of C∞

0 (Ω) in thespace

(2.23) H1V (Ω) = u ∈ L2(Ω) : du ∈ L2(Ω,Λ1), V 1/2u ∈ L2(Ω),


with inner product

(2.24) (u, v)1,V = (du, dv)L2 + (V u, v)L2 .

Then there is a natural injection H1V 0(Ω) → L2(Ω), and the Friedrichs

extension method provides a self-adjoint operator A. Integration by partsin (2.24), with v ∈ C∞

0 (Ω), shows that such A is an extension of −∆ + V .For this self-adjoint extension, we have

(2.25) D(A1/2) = H1V 0(Ω).

In case Ω is a smooth, compact Riemannian manifold with boundary andV ∈ C∞(Ω), one clearly has H1

V 0(Ω) = H10 (Ω). In such a case, we have

an immediate extension of Proposition 2.1, including the characterization(2.4) of D(−∆+V ). One can also easily extend Proposition 2.2 to −∆+Vin this case. It is of substantial interest that Proposition 2.4 also extends,as follows:

Proposition 2.7. If Ω is a complete Riemannian manifold and the func-tion V ∈ C∞(Ω) satisfies V ≥ 1, then −∆ + V is essentially self-adjoint onC∞

0 (Ω).

Proof. We can modify the proof of Proposition 2.4; replace ∆−1 by ∆−Vin (2.11) and (2.12). Then U(t) gives a unitary group on H1

V 0(Ω)⊕L2(Ω),and the finite propagation speed argument given there goes through. Asbefore, all powers of −∆ + V are essentially self-adjoint on C∞

0 (Ω).

Some important classes of potentials V have singularities and are notbounded below. In §7 we return to this, in a study of the quantum me-chanical Coulomb problem.

We record here an important compactness property when V ∈ C∞(Ω)tends to +∞ at infinity in Ω.

Proposition 2.8. If the Friedrichs extension method described above isused to construct the self-adjoint operator −∆ + V for smooth V ≥ 1, asabove, and if V → +∞ at infinity (i.e., for each N < ∞, ΩN = x ∈ Ω :V (x) ≤ N is compact), then −∆ + V has compact resolvent.

Proof. Given (2.25), it suffices to prove that the injection H1V 0(Ω) →

L2(Ω) is compact, under the current hypotheses on V . Indeed, if un isbounded in H1

V 0(Ω), with inner product (2.24), then dun and V 1/2unare bounded in L2(Ω). By Rellich’s theorem and a diagonal argument, onehas a subsequence unk

whose restriction to each ΩN converges in L2(ΩN )-norm. The boundedness of V 1/2un in L2(Ω) then gives convergence ofthis subsequence in L2(Ω)-norm, proving the proposition.

The following result extends Proposition 2.4 of Chapter 5.

Exercises 15

Proposition 2.9. Assume that Ω is connected and that either Ω is com-pact or V → +∞ at infinity. Denote by λ0 the first eigenvalue of −∆ + V .Then a λ0-eigenfunction of −∆ + V is nowhere vanishing on Ω. Conse-quently, the λ0-eigenspace is one-dimensional.

Proof. Let u be a λ0-eigenfunction of −∆ + V . As in the proof of Propo-sition 2.4 of Chapter 5, we can write u = u+ +u−, where u+(x) = u(x) foru(x) > 0 and u−(x) = u(x) for u(x) ≤ 0, and the variational characteriza-tion of the λ0-eigenspace implies that u± are eigenfunctions (if nonzero).Hence it suffices to prove that if u is a λ0-eigenfunction and u(x) ≥ 0 onΩ, then u(x) > 0 on Ω. To this end, write

u(x) = et(∆−V +λ0)u(x) =

∫

Ω

pt(x, y)u(y) dV (y).

We see that this forces pt(x, y) = 0 for all t > 0, when

x ∈ Σ = x : u(x) = 0, y ∈ O, O = x : u(x) > 0,since pt(x, y) is smooth and ≥ 0. The strong maximum principle (seeExercise 3 in §1 of Chapter 6) forces Σ = ∅.

Exercises

1. Let H1V (Ω) be the space (2.23). If V ≥ 1 belongs to C∞(Ω), show that the

Friedrichs extension also defines a self-adjoint operator A1, equal to −∆ + Von C∞

0 (Ω), such that D(A1/21 ) = H1

V (Ω). If Ω is complete, show that thisoperator coincides with the extension A defined in (2.25). Conclude that, inthis case, H1

V (Ω) = H1V 0(Ω).

2. Let Ω be complete, V ≥ 1 smooth. Show that if A is the self-adjoint extensionof −∆ + V described in Proposition 2.7, then

(2.26) D(A) = u ∈ L2(Ω) : −∆u + V u ∈ L2(Ω),

where a priori we regard −∆u + V u as an element of D′(Ω).3. Define T : L2(Ω) → L2(Ω, Λ1)⊕L2(Ω) by D(T ) = H1

V 0(Ω), Tu = (du, V 1/2u).Show that

(2.27) D(T ∗) = (v1, v2) ∈ L2(Ω, Λ1) ⊕ L2(Ω) : δv1 ∈ L2(Ω), V 1/2v2 ∈ L2(Ω).

Show that T ∗T is equal to the self-adjoint extension A of −∆ + V defined bythe Friedrichs extension, as in (2.25).

4. If Ω is complete, show that the self-adjoint extension A of −∆ + V in Propo-sition 2.7 satisfies

(2.28) D(A) = u ∈ L2(Ω) : ∆u ∈ L2(Ω), V u ∈ L2(Ω).

(Hint: Denote the right side by W. Use Exercise 3 and A = T ∗T to show thatD(A) ⊂ W. Use Exercise 2 to show that W ⊂ D(A).)


5. Let D = −i d/dx on C∞(R), and let B(x) ∈ C∞(R) be real-valued. Definethe unbounded operator L on L2(R) by

(2.29) D(L) = u ∈ L2(R) : Du ∈ L2(R), Bu ∈ L2(R), Lu = Du + iB(x)u.

Show that L∗ = D − iB, with

D(L∗) = u ∈ L2(R) : Du − iBu ∈ L2(R).Deduce that A0 = L∗L is given by A0u = D2u + B2u + B′(x)u on

D(A0) = u ∈ L2(R) : Du ∈ L2(R), Bu ∈ L2(R), D2u+B2u+B′(x)u ∈ L2(R).6. Suppose that |B′(x)| ≤ ϑB(x)2 + C, for some ϑ < 1, C < ∞. Show that

D(A0) = u ∈ L2(R) : D2u + (B2 + B′)u ∈ L2(R).(Hint: Apply Exercise 2 to D2 + (B2 + B′) = A, and show that D(A1/2) isgiven by D(L), defined in (2.29).)

7. In the setting of Exercise 6, show that the operator L of Exercise 5 is closed.(Hint: L∗L = A is a self-adjoint extension of D2 + (B2 + B′). Show thatD(A

1/21 ) = D(L) and also = D(L).) Also show that D(L∗) = D(L) in this

case.

3. Heat asymptotics and eigenvalue asymptotics

In this section we will study the asymptotic behavior of the eigenvalues ofthe Laplace operator on a compact Riemannian manifold, with or withoutboundary.

We begin with the boundaryless case. Let M be a compact Riemannianmanifold without boundary, of dimension n. In §13 of Chapter 7 we haveconstructed a parametrix for the solution operator et∆ of the heat equation

(3.1)( ∂

∂t− ∆

)u = 0 on R

+ × M, u(0, x) = f(x)

and deduced that

(3.2) Tr et∆ ∼ t−n/2(a0 + a1t + a2t

2 + · · · ), t ց 0,

for certain constants aj . In particular,

(3.3) a0 = (4π)−n/2 vol M.

This is related to the behavior of the eigenvalues of ∆ as follows. Letthe eigenvalues of −∆ be 0 = λ0 ≤ λ1 ≤ λ2 ≤ · · · ր ∞. Then (3.2) isequivalent to

(3.4)

∞∑

j=0

e−tλj ∼ t−n/2(a0 + a1t + a2t

2 + · · · ), t ց 0.

We will relate this to the counting function

(3.5) N(λ) ∼ #λj : λj ≤ λ,

3. Heat asymptotics and eigenvalue asymptotics 17

establishing the following:

Theorem 3.1. The eigenvalues λj of −∆ on the compact Riemannianmanifold M have the behavior

(3.6) N(λ) ∼ C(M)λn/2, λ → +∞,

with

(3.7) C(M) =a0

Γ(n2 + 1)

=vol M

Γ(n2 + 1)(4π)n/2

.

That (3.6) follows from (3.4) is a special case of a result known as Kara-

mata’s Tauberian theorem. The following neat proof follows one in [Si3].Let µ be a positive (locally finite) Borel measure on [0,∞); in the exampleabove, µ

([0, λ]

)= N(λ).

Proposition 3.2. If µ is a positive measure on [0,∞), α ∈ (0,∞), then

(3.8)

∫ ∞

0

e−tλ dµ(λ) ∼ at−α, t ց 0,

implies

(3.9)

∫ x

0

dµ(λ) ∼ bxα, x ր ∞,

with

(3.10) b =a

Γ(α + 1).

Proof. Let dµt be the measure given by µt(A) = tαµ(t−1A), and letdν(λ) = αλα−1dλ; then νt = ν. The hypothesis (3.8) becomes

(3.11) limt→0

∫e−λ dµt(λ) = b

∫e−λ dν(λ),

with b given by (3.10), and the desired conclusion becomes

(3.12) limt→0

∫χ(λ) dµt(λ) = b

∫χ(λ) dν(λ)

when χ is the characteristic function of [0, 1]. It would suffice to show that(3.12) holds for all continuous χ(λ) with compact support in [0,∞).

From (3.11) we deduce that the measures e−λdµt are uniformly bounded,for t ∈ (0, 1]. Thus (3.12) follows if we can establish

(3.13) limt→0

∫g(λ)e−λ dµt(λ) = b

∫g(λ)e−λ dν(λ),

for g in a dense subspace of C0(R+), the space of continuous functions on

[0,∞) that vanish at infinity. Indeed, the hypothesis implies that (3.13)


holds for all g in A, the space of finite, linear combinations of functions ofλ ∈ [0,∞) of the form ϕs(λ) = e−sλ, s ∈ (0,∞), as can be seen by dilatingthe variables in (3.11). By the Stone-Weierstrass theorem, A is dense inCo(R

+), so the proof is complete.

We next want to establish similar results on N(λ) for the Laplace oper-ator ∆ on a compact manifold Ω with boundary, with Dirichlet boundarycondition. At the end of §13 in Chapter 7 we sketched a construction ofa parametrix for et∆ in this case which, when carried out, would yield anexpansion

(3.14) Tr et∆ ∼ t−n/2(a0 + a1/2t

1/2 + a1t + · · ·), t ց 0,

extending (3.2). However, we will be able to verify the hypothesis of Propo-sition 3.2 with less effort than it would take to carry out the details of thisconstruction, and for a much larger class of domains.

For simplicity, we will restrict attention to bounded domains in Rn and

to the flat Laplacian, though more general cases can be handled similarly.Now, let Ω be an arbitrary bounded, open subset of R

n, with closure Ω.The Laplace operator on Ω, with Dirichlet boundary condition, was studiedin §5 of Chapter 5.

Lemma 3.3. For any bounded, open Ω ⊂ Rn, ∆ with Dirichlet boundary

condition, et∆ is trace class for all t > 0.

Proof. Let Ω ⊂ B, a large open ball. Then the variational characterizationof eigenvalues shows that the eigenvalues λj(Ω) of −∆ on Ω and λj(B) ofL = −∆ on B, both arranged in increasing order, have the relation

(3.15) λj(Ω) ≥ λj(B).

But we know that e−tL has integral kernel in C∞(B × B) for each t > 0,hence is trace class. Since e−tλj(Ω) ≤ e−tλj(B), this implies that the positiveself-adjoint operator et∆ is also trace class.

Limiting arguments, which we leave to the reader, allow one to showthat, even in this generality, if H(t, x, y) ∈ C∞(Ω × Ω) is, for fixed t > 0,the integral kernel of et∆ on L2(Ω), then

(3.16) Tr et∆ =

∫

Ω

H(t, x, x) dx.

See Exercises 1–5 at the end of this section.

Proposition 3.4. If Ω is a bounded, open subset of Rn and ∆ has the

Dirichlet boundary condition, then

(3.17) Tr et∆ ∼ (4πt)−n/2 vol Ω, t ց 0.

3. Heat asymptotics and eigenvalue asymptotics 19

Proof. We will compare H(t, x, y) with H0(t, x, y) = (4πt)−n/2e|x−y|2/4t,the free-space heat kernel. Let E(t, x, y) = H0(t, x, y) − H(t, x, y). Then,for fixed y ∈ Ω,

(3.18)∂E

∂t− ∆xE = 0 on R

+ × Ω, E(0, x, y) = 0,

and

(3.19) E(t, x, y) = H0(t, x, y), for x ∈ ∂Ω.

To make simple sense out of (3.19), one might assume that every point of∂Ω is a regular boundary point, though a further limiting argument canbe made to lift such a restriction. The maximum principle for solutions tothe heat equation implies

(3.20) 0 ≤ E(t, x, y) ≤ sup0≤s≤t,z∈Ω

H0(s, z, y) ≤ sup0≤s≤t

(4πs)−n/2 e−δ(y)2/4s,

where δ(y) = dist(y, ∂Ω). Now the function

ψδ(s) = (4πs)−n/2e−δ2/4s

on (0,∞) vanishes at 0 and ∞ and has a unique maximum at s = δ2/2n;we have ψδ(δ

2/2n) = Cnδ−n. Thus

(3.21) 0 ≤ E(t, x, y) ≤ max((4πt)−n/2e−δ(y)2/4t, Cnδ(y)−n

).

Of course, E(t, x, y) ≤ H0(t, x, y) also.Now, let O ⊂⊂ Ω be such that vol(Ω \ O) < ε. For t small enough,

namely for t ≤ δ21/2n where δ1 = dist(O, ∂Ω), we have

(3.22) 0 ≤ E(t, x, x) ≤ (4πt)−n/2e−δ(x)2/4t, x ∈ O,

while of course 0 ≤ E(t, x, x) ≤ (4πt)−n/2, for x ∈ Ω \ O. Therefore,

(3.23) lim supt→0

(4πt)n/2

∫

Ω

E(t, x, x) dx ≤ ε,

so

(3.24)

vol Ω − ε ≤ lim inft→0

(4πt)n/2

∫

Ω

H(t, x, x) dx

≤ lim supt→0

(4πt)n/2

∫

Ω

H(t, x, x) dx ≤ vol Ω.

As ε can be taken arbitrarily small, we have a proof of (3.17).

Corollary 3.5. If Ω is a bounded, open subset of Rn, N(λ) the counting

function of the eigenvalues of −∆, with Dirichlet boundary condition, then(3.6) holds.


Note that if Oε is the set of points in Ω of distance ≥ ε from ∂Ω and wedefine v(ε) = vol(Ω \ Oε), then the estimate (3.24) can be given the moreprecise reformulation

(3.25) 0 ≤ vol Ω − (4πt)n/2 Tr et∆ ≤ ω(√

2nt),

where

(3.26) ω(ε) = v(ε) +

∫ ∞

ε

e−ns2/2ε2

dv(s).

The fact that such a crude argument works, and works so generally,is a special property of the Dirichlet problem. If one uses the Neumannboundary condition, then for bounded Ω ⊂ R

n with nasty boundary, ∆need not even have compact resolvent. However, Theorem 3.1 does extendto the Neumann boundary condition provided ∂Ω is smooth. One can dothis via the sort of parametrix for boundary problems sketched in §13 ofChapter 7.

We now look at the heat kernel H(t, x, y) on the complement of a smooth,bounded region K ⊂ R

n. We impose the Dirichlet boundary condition on∂K. As before, 0 ≤ H(t, x, y) ≤ H0(t, x, y), where H0(t, x, y) is the free-space heat kernel. We can extend H(t, x, y) to be Lipschitz continuous on(0,∞)×R

n×Rn by setting H(t, x, y) = 0 when either x ∈ K or y ∈ K. We

now estimate E(t, x, y) = H0(t, x, y) − H(t, x, y). Suppose K is containedin the open ball of radius R centered at the origin.

Lemma 3.6. For |x − y| ≤ |y| − R, we have

(3.27) E(t, x, y) ≤ Ct−1/2e−(|y|−R)2/4t.

Proof. With y ∈ Ω = Rn \ K, write

(3.28) H(t, x, y) = (4πt)−1/2

∫ ∞

−∞e−s2/4t cos sΛ ds,

where Λ =√−∆ and ∆ is the Laplace operator on Ω, with the Dirichlet

boundary condition. We have a similar formula for H0(t, x, y), using insteadΛ0 =

√−∆0, with ∆0 the free-space Laplacian. Now, by finite propagationspeed,

cos sΛ δy(x) = cos sΛ0 δy(x),

provided

|s| ≤ d = dist(y, ∂K), and |x − y| ≤ d.

Thus, as long as |x − y| ≤ d, we have

(3.29) E(t, x, y) = (4πt)−1/2

∫

|s|≥d

e−s2/4t[cos sΛ0 δy(x)−cos sΛ δy(x)

]ds.

Exercises 21

Then the estimate (3.27) follows easily, along the same lines as estimateson heat kernels discussed in Chapter 6, §2.

When we combine (3.27) with the obvious inequality

(3.30) 0 ≤ E(t, x, y) ≤ H0(t, x, y) = (4πt)−n/2e−|x−y|2/4t,

we see that, for each t > 0, E(t, x, y) is rapidly decreasing as |x|+ |y| → ∞.Using this and appropriate estimates on derivatives, we can show thatE(t, x, y) is the integral kernel of a trace class operator on L2(Rn). We canwrite

(3.31) Tr(et∆0 − et∆P

)=

∫

Rn

E(t, x, x) dx,

where P is the projection of L2(Rn) onto L2(Ω) defined by restriction toΩ. Now, as t ց 0, (4πt)n/2E(t, x, x) approaches 1 on K and 0 on R

n \K.Together with the estimates (3.27) and (3.30), this implies

(3.32) (4πt)n/2

∫

Rn

E(t, x, x) dx −→ vol K,

as t ց 0. This establishes the following:

Proposition 3.7. If K is a closed, bounded set in Rn, ∆ is the Laplacian

on L2(Rn \K), with Dirichlet boundary condition, and ∆0 is the Laplacianon L2(Rn), then et∆0 − et∆P is trace class for each t > 0 and

(3.33) Tr(et∆0 − et∆P

)∼ (4πt)−n/2 vol K,

as t ց 0.

This result will be of use in the study of scattering by an obstacle K, inChapter 9. It is also valid for the Neumann boundary condition if ∂K issmooth.

Exercises

In Exercises 1–4, let Ω ⊂ Rn be a bounded, open set and let Oj be open with

smooth boundary such that

O1 ⊂⊂ O2 ⊂⊂ · · · ⊂⊂ Oj ⊂⊂ · · · ր Ω.

Let Lj be −∆ on Lj , with Dirichlet boundary condition; the correspondingoperator on Ω is simply denoted −∆.

1. Using material developed in §5 of Chapter 5, show that, for any t > 0, f ∈L2(Ω),

e−tLj Pjf −→ et∆f strongly in L2(Ω),


as j → ∞, where Pj is multiplication by the characteristic function of Oj .Don’t peek at Lemma 3.4 in Chapter 11!

2. If λν(Oj) are the eigenvalues of Lj , arranged in increasing order for each j,show that, for each ν,

λν(Oj) ց λν(Ω), as j → ∞.

3. Show that, for each t > 0,

Tr e−tLj ր Tr et∆.

4. Let Hj(t, x, y) be the heat kernel on R+×Oj×Oj . Extend Hj to R

+×Ω×Ω soas to vanish if x or y belongs to Ω\Oj . Show that, for each x ∈ Ω, y ∈ Ω, t > 0,

Hj(t, x, y) ր H(t, x, y), as j → ∞.

Deduce that, for each t > 0,Z

Oj

Hj(t, x, x) dx րZ

Ω

H(t, x, x) dx.

5. Using Exercises 1–4, give a detailed proof of (3.16) for general bounded Ω ⊂R

n.6. Give an example of a bounded, open, connected set Ω ⊂ R

2 (with roughboundary) such that ∆, with Neumann boundary condition, does not havecompact resolvent.

4. The Laplace operator on Sn

A key tool in the analysis of the Laplace operator ∆S on Sn is the formulafor the Laplace operator on R

n+1 in polar coordinates:

(4.1) ∆ =∂2

∂r2+

n

r

∂

∂r+

1

r2∆S .

In fact, this formula is simultaneously the main source of interest in ∆S

and the best source of information about it.To begin, we consider the Dirichlet problem for the unit ball in Euclidean

space, B = x ∈ Rn+1 : |x| < 1:

(4.2) ∆u = 0 in B, u = f on Sn = ∂B,

given f ∈ D′(Sn). In Chapter 5 we obtained the Poisson integral formulafor the solution:

(4.3) u(x) =1 − |x|2

An

∫

Sn

f(y)

|x − y|n+1dS(y),

where An is the volume of Sn. Equivalently, if we set x = rω with r =|x|, ω ∈ Sn,

(4.4) u(rω) =1 − r2

An

∫

Sn

f(ω′)

(1 − 2rω · ω′ + r2)(n+1)/2dS(ω′).

4. The Laplace operator on Sn 23

Now we can derive an alternative formula for the solution of (4.2) if weuse (4.1) and regard ∆u = 0 as an operator-valued ODE in r; it is an Eulerequation, with solution

(4.5) u(rω) = rA−(n−1)/2f(ω), r ≤ 1,

where A is an operator on D′(Sn), defined by

(4.6) A =(−∆S +

(n − 1)2

4

)1/2

.

If we set r = e−t and compare (4.5) and (4.4), we obtain a formula for thesemigroup e−tA as follows. Let θ(ω, ω′) denote the geodesic distance on Sn

from ω to ω′, so cos θ(ω, ω′) = ω · ω′. We can rewrite (4.4) as

(4.7)

u(rω) =2

Ansinh(log r−1) r−(n−1)/2

·∫

Sn

f(ω′)[2 cosh(log r−1) − 2 cos θ(ω, ω′)

]−(n+1)/2dS(ω′).

In other words, by (4.5),

(4.8) e−tAf(ω) =2

Ansinh t

∫

Sn

f(ω′)(2 cosh t − 2 cos θ(ω, ω′)

)(n+1)/2dS(ω′).

Identifying an operator on D′(Sn) with its Schwartz kernel in D′(Sn×Sn),we write

(4.9) e−tA =2

An

sinh t

(2 cosh t − 2 cos θ)(n+1)/2, t > 0.

Note that the integration of (4.9) from t to ∞ produces the formula

(4.10) A−1e−tA = 2Cn(2 cosh t − 2 cos θ)−(n−1)/2, t > 0,

provided n ≥ 2, where

Cn =1

(n − 1)An=

1

4π−(n+1)/2Γ

(n − 1

2

).

With the exact formula (4.9) for the semigroup e−tA, we can proceed togive formulas for fundamental solutions to various important PDE, partic-ularly

(4.11)∂2u

∂t2− Lu = 0 (wave equation)

and

(4.12)∂u

∂t− Lu = 0 (heat equation),

where

(4.13) L = ∆S − (n − 1)2

4= −A2.


If we prescribe Cauchy data u(0) = f, ut(0) = g for (4.11), the solution is

(4.14) u(t) = (cos tA)f + A−1(sin tA)g.

Assume n ≥ 2. We obtain formulas for these terms by analytic continuationof the formulas (4.9) and (4.10) to Re t > 0 and then passing to the limitt ∈ iR. This is parallel to the derivation of the fundamental solution to thewave equation on Euclidean space in §5 of Chapter 3. We have

(4.15)A−1e(it−ε)A = −2Cn

[2 cosh(it − ε) − 2 cos θ

]−(n−1)/2,

e(it−ε)A =2

Ansinh(it − ε)

[2 cosh(it − ε) − 2 cos θ

]−(n+1)/2.

Letting ε ց 0, we have

(4.16)A−1 sin tA =

limεց0

−2Cn Im (2 cosh ε cos t − 2i sinh ε sin t − 2 cos θ)−(n−1)/2

and

(4.17)

cos tA =

limεց0

−2

AnIm(sin t)(2 cosh ε cos t − 2i sinh ε sin t − 2 cos θ)−(n+1)/2.

For example, on S2 we have, for 0 ≤ t ≤ π,

(4.18)A−1 sin tA = −2C2(2 cos θ − 2 cos t)−1/2, θ < |t|,

0, θ > |t|,with an analogous expression for general t, determined by the identity

(4.19) A−1 sin(t + 2π)A = −A−1 sin tA on D′(S2k),

plus the fact that sin tA is odd in t. The last line on the right in (4.18)reflects the well-known finite propagation speed for solutions to the hyper-bolic equation (4.11).

To understand how the sign is determined in (4.19), note that, in (4.15),with ε > 0, for t = 0 we have a real kernel, produced by taking the−(n − 1)/2 = −k + 1/2 power of a positive quantity. As t runs from 0to 2π, the quantity 2 cosh(it − ε) = 2 cosh ε cos t − 2i sinh ε sin t movesonce clockwise around a circle of radius 2(cosh2 ε + sinh2 ε)1/2, centered at0, so 2 cosh ε cos t− 2i sinh ε sin t− 2 cos θ describes a curve winding onceclockwise about the origin in C. Thus taking a half-integral power of thisgives one the negative sign in (4.14).

On the other hand, when n is odd, the exponents on the right side of(4.15)–(4.17) are integers. Thus

(4.20) A−1 sin(t + 2π)A = A−1 sin tA on D′(S2k+1).


Also, in this case, the distributional kernel for A−1 sin tA must vanish for|t| 6= θ. In other words, the kernel is supported on the shell θ = |t|. This isthe generalization to spheres of the strict Huygens principle.

In case n = 2k + 1 is odd, we obtain from (4.16) and (4.17) that

(4.21) A−1 sin tA f(x) =1

(2k − 1)!!

( 1

sin s

∂

∂s

)k−1(sin2k−1 s f(x, s)

)s=t

and

(4.22) cos tA f(x) =1

(2k − 1)!!sin s

( 1

sin s

∂

∂s

)k(sin2k−1 s f(x, s)

)s=t

,

where, as in (5.66) of Chapter 3, (2k − 1)!! = 3 · 5 · · · (2k − 1) and

(4.23) f(x, s) = mean value of f on Σs(x) = y ∈ Sn : θ(x, y) = |s|.We can examine general functions of the operator A by the functional

calculus

(4.24) g(A) = (2π)−1/2

∫ ∞

−∞g(t)eitA dt = (2π)−1/2

∫ ∞

−∞g(t) cos tA dt,

where the last identity holds provided g is an even function. We can rewritethis, using the fact that cos tA has period 2π in t on D′(Sn) for n odd,period 4π for n even. In concert with (4.22), we have the following formulafor the Schwartz kernel of g(A) on D′(S2k+1), for g even:

(4.25) g(A) = (2π)−1/2(− 1

2π

1

sin θ

∂

∂θ

)k ∞∑

k=−∞g(θ + 2kπ).

As an example, we compute the heat kernel on odd-dimensional spheres.Take g(λ) = e−tλ2

. Then g(s) = (2t)−1/2e−s2/4t and

(4.26) (2π)−1/2∑

k

g(s + 2kπ) = (4πt)−1/2∑

k

e−(s+2kπ)2/4t = ϑ(s, t),

where ϑ(s, t) is a “theta function.” Thus the kernel of e−tA2

on S2k+1 isgiven by

(4.27) e−tA2

=(− 1

2π

1

sin θ

∂

∂θ

)k

ϑ(θ, t).

A similar analysis on S2k gives an integral, with the theta function appear-ing in the integrand.

The operator A has a compact resolvent on L2(Sn), and hence a discreteset of eigenvalues, corresponding to an orthonormal basis of eigenfunctions.Indeed, the spectrum of A has the following description.

Proposition 4.1. The spectrum of the self-adjoint operator A on L2(Sn)is

(4.28) spec A =1

2(n − 1) + k : k = 0, 1, 2, . . .

.


Proof. Since 0 is the smallest eigenvalue of −∆S , the definition (4.6) showsthat (n − 1)/2 is the smallest eigenvalue of A. Also, (4.20) shows thatall eigenvalues of A are integers if n is odd, while (4.19) implies that alleigenvalues of A are (nonintegral) half-integers if n is even. Thus spec A iscertainly contained in the right side of (4.28).

Another way to see this containment is to note that since the functionu(x) given by (4.5) must be smooth at x = 0, the exponent of r in thatformula can take only integer values.

Let Vk denote the eigenspace of A with eigenvalue νk = (n−1)/2+k. Wewant to show that Vk 6= 0 for k = 0, 1, 2, . . . . Moreover, we want to iden-tify Vk. Now if f ∈ Vk, it follows that u(x) = u(rω) = rA−(n−1)/2f(ω) =rkf(ω) is a harmonic function defined on all of R

n+1, which, being homoge-neous and smooth at x = 0, must be a harmonic polynomial, homogeneousof degree k in x. If Hk denotes the space of harmonic polynomials, homo-geneous of degree k, restriction to Sn ⊂ R

n+1 produces an isomorphism:

(4.29) ρ : Hk≈−→ Vk.

To show that each Vk 6= 0, it suffices to show that each Hk 6= 0.Indeed, for c = (c1, . . . , cn+1) ∈ C

n+1, consider

pc(x) = (c1x1 + · · · + cn+1xn+1)k.

A computation gives

∆pc(x) = k(k − 1)〈c, c〉(c1x1 + · · · + ckxk)k−2,

〈c, c〉 = c21 + · · · + c2

k.

Hence ∆pc = 0 whenever 〈c, c〉 = 0, so the proposition is proved.

We now want to specify the orthogonal projections Ek of L2(Sn) on Vk.We can attack this via (4.10), which implies

(4.30)∞∑

k=0

ν−1k e−tνkEk(x, y) = 2Cn(2 cosh t − 2 cos θ)−(n−1)/2,

where θ = θ(x, y) is the geodesic distance from x to y in Sn. If we setr = e−t and use νk = (n−1)/2+k, we get the generating function identity

(4.31)

∞∑

k=0

rkν−1k Ek(x, y) = 2Cn(1 − 2r cos θ + r2)−(n−1)/2

=

∞∑

k=0

rkpk(cos θ);

in particular,

(4.32) Ek(x, y) = νk pk(cos θ).


These functions are polynomials in cos θ. To see this, set t = cos θ andwrite

(4.33) (1 − 2tr + r2)−α =∞∑

k=0

Cαk (t) rk,

thus defining coefficients Cαk (t). To compute these, use

(1 − z)−α =

∞∑

j=0

(j + α − 1

j

)zj ,

with z = r(2t − r), to write the left side of (4.33) as

∞∑

j=0

(α

j

)rj(2t − r)j =

∞∑

j=0

j∑

ℓ=0

(j + α − 1

j

)(j

ℓ

)(−1)ℓrj+ℓ(2t)j−ℓ

=

∞∑

k=0

[k/2]∑

ℓ=0

(−1)ℓ

(k − ℓ + α − 1

k − ℓ

)(k − ℓ

ℓ

)(2t)k−2ℓrk.

Hence

(4.34) Cαk (t) =

[k/2]∑

ℓ=0

(−1)ℓ

(k − ℓ + α − 1

k − ℓ

)(k − ℓ

ℓ

)(2t)k−2ℓ.

These are called Gegenbauer polynomials. Therefore, we have the following:

Proposition 4.2. The orthogonal projection of L2(Sn) onto Vk has kernel

(4.35) Ek(x, y) = 2Cnνk Cαk (cos θ), α =

1

2(n − 1),

with Cn as in (4.10).

In the special case n = 2, we have C2 = 1/4π, and νk = k + 1/2; hence

(4.36) Ek(x, y) =2k + 1

4πC

1/2k (cos θ) =

2k + 1

4πPk(cos θ),

where C1/2k (t) = Pk(t) are the Legendre polynomials.

The trace of Ek is easily obtained by integrating (4.35) over the diagonal,to yield

(4.37) Tr Ek = 2CnAnνk C(n−1)/2k (1) =

2νk

n − 1C

(n−1)/2k (1).

Setting t = 1 in (4.33), so (1 − 2r + r2)−α = (1 − r)−2α, we obtain

(4.38) Cαk (1) =

(k + 2α − 1

k

), e.g., Pk(1) = 1.

Thus we have the dimensions of the eigenspaces Vk:


Corollary 4.3. The eigenspace Vk of −∆S on Sn, with eigenvalue

λk = ν2k − 1

4(n − 1)2 = k2 + (n − 1)k,

satisfies

(4.39) dim Vk =2k + n − 1

n − 1

(k + n − 2

k

)=

(k + n − 2

k − 1

)+

(k + n − 1

k

).

In particular, on S2 we have dim Vk = 2k + 1.

Another natural approach to Ek is via the wave equation. We have

(4.40)

Ek =1

2T

∫ T

−T

e−iνkteitA dt

=1

2T

∫ T

−T

cos t(A − νk) dt,

where T = π or 2π depending on whether n is odd or even. (In eithercase, one can take T = 2π.) In the special case of S2, when (4.18) is used,comparison of (4.36) with the formula produced by this method producesthe identity

(4.41) Pk(cos θ) =1

π

∫ θ

−θ

cos(k + 12 )t

(2 cos t − 2 cos θ)1/2dt,

for the Legendre polynomials, known as the Mehler-Dirichlet formula.

Exercises

Exercises 1–5 deal with results that follow from symmetries of the sphere.The group SO(n + 1) acts as a group of isometries of Sn ⊂ R

n+1, hence asa group of unitary operators on L2(Sn). Each eigenspace Vk of the Laplaceoperator is preserved by this action. Fix p = (0, . . . , 0, 1) ∈ Sn, regarded asthe “north pole.” The subgroup of SO(n + 1) fixing p is a copy of SO(n).

1. Show that each eigenspace Vk has an element u such that u(p) 6= 0. Concludeby forming

Z

SO(n)

u(gx) dg

that each eigenspace Vk of ∆S has an element zk 6= 0 such that zk(x) =zk(gx), for all g ∈ SO(n). Such a function is called a spherical function.

2. Suppose Vk has a proper subspace W invariant under SO(n + 1). (HenceW⊥ ⊂ Vk is also invariant.) Show that W must contain a nonzero sphericalfunction.

3. Suppose zk and yk are two nonzero spherical functions in Vk. Show that theymust be multiples of each other. Hence the unique spherical functions (up

Exercises 29

to constant multiples) are given by (4.35), with y = p. (Hint: zk and yk areeigenfunctions of −∆S , with eigenvalue λk = k2 + (n− 1)k. Pick a sequenceof surfaces

Σj = x ∈ Sn : θ(x, p) = εj ⊂ Sn,

with εj → 0, on which zk = αj 6= 0. With βj = yk|Σj , it follows thatβjzk − αjyk is an eigenfunction of −∆S that vanishes on Σj . Show that, forj large, this forces βjzk − αjyk to be identically zero.)

4. Using Exercises 2 and 3, show that the action of SO(n+1) on each eigenspaceVk is irreducible, that is, Vk has no proper invariant subspaces.

5. Show that each Vk is equal to the linear span of the set of polynomials of theform pc(x) = (c1x1 + · · · + cn+1xn+1)

k, with 〈c, c〉 = 0.(Hint: Show that this linear span is invariant under SO(n + 1).)

6. Using (4.9), show that

(4.42) Tr e−tA =2 sinh t

(2 cosh t − 2)(n+1)/2.

Find the asymptotic behavior as t ց 0. Use Karamata’s Tauberian theoremto determine the asymptotic behavior of the eigenvalues of A, hence of −∆S .Compare this with the general results of §3 and also with the explicit resultsof Corollary 4.3.

7. Using (4.27), show that, for A on Sn with n = 2k + 1,

(4.43)Tr e−tA2

=A2k+1√

4πt

„

− 1

2π

1

sin θ

∂

∂θ

«k

e−θ2/4t

˛

˛

˛

˛

θ=0

+ O(t∞)

= (4πt)−n/2 A2k+1 + O“

t−n/2+1”

,

as t ց 0. Compare the general results of §3.8. Show that

(4.44) e−πi(A−(n−1)/2)f(ω) = f(−ω), f ∈ L2(Sn).

(Hint: Check it for f ∈ Vk, the restriction to Sn of a homogeneous harmonicpolynomial of degree k.)

Exercises 9–13 deal with analysis on Sn when n = 2. When doing them, lookfor generalizations to other values of n.

9. If Ξ(A) has integral kernel KΞ(x, y), show that when n = 2,

(4.45) KΞ(x, y) =1

4π

∞X

ℓ=0

(2ℓ + 1)Ξ

„

ℓ +1

2

«

Pℓ(cos θ),

where cos θ = x · y and Pℓ(t) are the Legendre polynomials.10. Demonstrate the Rodrigues formula for the Legendre polynomials:

(4.46) Pk(t) =1

2kk!

„

d

dt

«k“

t2 − 1”k

.


(Hint: Use Cauchy’s formula to get

Pk(t) =1

2πi

Z

γ

(1 − 2zt + z2)−1/2z−k−1 dz

from (4.33); then use the change of variable 1−uz = (1− 2tz + z2)1/2. Thenappeal to Cauchy’s formula again, to analyze the resulting integral.)

11. If f ∈ L2(S2) has the form f(x) = g(x · y) =P

ϕℓPℓ(x · y), for some y ∈ S2,show that

(4.47) ϕℓ =2ℓ + 1

4π

Z

S2

f(z)Pℓ(y · z) dS(z) =

„

ℓ +1

2

«Z 1

−1

g(t)Pℓ(t) dt.

(Hint: UseR

S2 Ek(x, z)Eℓ(z, y) dS(z) = δkℓ Eℓ(x, y).) Conclude that g(x · y)is the integral kernel of ψ(A − 1/2), where

(4.48) ψ(ℓ) =4π

2ℓ + 1ϕℓ = 2π

Z 1

−1

g(t)Pℓ(t) dt.

This result is known as the Funk-Hecke theorem.12. Show that, for x, y ∈ S2,

(4.49) eikx·y =

∞X

ℓ=0

(2ℓ + 1) iℓ jℓ(k) Pℓ(x · y),

where

(4.50) jℓ(z) =

„

π

2z

«1/2

Jℓ+1/2(z) =1

2

1

ℓ!

„

z

2

«ℓ Z 1

−1

(1 − t2)ℓ eizt dt.

(Hint: Take g(t) = eikt in Exercise 11, apply the Rodrigues formula, andintegrate by parts.) Thus eikx·y is the integral kernel of the operator

4π e(1/2)πi(A−1/2) jA−1/2(k).

For another approach, see Exercises 10 and 11 in §9 of Chapter 9.13. Demonstrate the identities

(4.51)

»

(1 − t2)d

dt+ ℓt

–

Pℓ(t) = ℓPℓ−1(t)

and

(4.52)d

dt

»

(1 − t2)d

dtPℓ(t)

–

+ ℓ(ℓ + 1)Pℓ(t) = 0.

Relate (4.52) to the statement that, for fixed y ∈ S2, ϕ(x) = Pℓ(x ·y) belongsto the ℓ(ℓ + 1)-eigenspace of −∆S .

Exercises 14–19 deal with formulas for an orthogonal basis of Vk (for S2).We will make use of the structure of irreducible representations of SO(3),obtained in §9 of Appendix B, Manifolds, Vector Bundles, and Lie Groups.

14. Show that the representation of SO(3) on Vk is equivalent to the representa-tion Dk, for each k = 0, 1, 2, . . . .

5. The Laplace operator on hyperbolic space 31

15. Show that if we use coordinates (θ, ψ) on S2, where θ is the geodesic distancefrom (1, 0, 0) and ψ is the angular coordinate about the x1-axis in R

3, then

(4.53) L1 =∂

∂ψ, L± = i e±iψ

»

± ∂

∂θ+ i cot θ

∂

∂ψ

–

.

16. Set

(4.54) wk(x) = (x2 + ix3)k = sink θ eikψ.

Show that wk ∈ Vk and that it is the highest-weight vector for the represen-tation, so

L1wk = ik wk.

17. Show that an orthogonal basis of Vk is given by

wk, L−wk, . . . , L2k− wk.

18. Show that the functions ζkj = Lk−j− wk, j ∈ −k,−k + 1, . . . , k − 1, k, listed

in Exercise 17 coincide, up to nonzero constant factors, with zkj , given by

zk0 = zk,

the spherical function considered in Exercises 1–3, and, for 1 ≤ j ≤ k,

zk,−j = Lj−zk, zkj = Lj

+zk.

19. Show that the functions zkj coincide, up to nonzero constant factors, with

(4.55) eijψ P jk (cos θ), −k ≤ j ≤ k,

where P jk (t), called associated Legendre functions, are defined by

(4.56) P jk (t) = (−1)j(1 − t2)|j|/2

„

d

dt

«|j|

Pk(t).

5. The Laplace operator on hyperbolic space

The hyperbolic space Hn shares with the sphere Sn the property of havingconstant sectional curvature, but for Hn it is −1. One way to describe Hn

is as a set of vectors with square length 1 in Rn+1, not for a Euclidean

metric, but rather for a Lorentz metric

(5.1) 〈v, v〉 = −v21 − · · · − v2

n + v2n+1,

namely,

(5.2) Hn = v ∈ Rn+1 : 〈v, v〉 = 1, vn+1 > 0,

with metric tensor induced from (5.1). The connected component G of theidentity of the group O(n, 1) of linear transformations preserving the qua-dratic form (5.1) acts transitively on Hn, as a group of isometries. In fact,SO(n), acting on R

n ⊂ Rn+1, leaves invariant p = (0, . . . , 0, 1) ∈ Hn and

acts transitively on the unit sphere in TpHn. Also, if A(u1, . . . , un, un+1)t


= (u1, . . . , un+1, un)t, then etA is a one-parameter subgroup of SO(n, 1)taking p to the curve

γ = (0, . . . , 0, xn, xn+1) : x2n+1 − x2

n = 1, xn+1 > 0.

Together these facts imply that Hn is a homogeneous space.There is a map of Hn onto the unit ball in R

n, defined in a fashion similarto the stereographic projection of Sn. The map

(5.3) s : Hn −→ Bn = x ∈ Rn : |x| < 1

is defined by

(5.4) s(x, xn+1) = (1 + xn+1)−1x.

The metric on Hn defined above then yields the following metric tensor onBn:

(5.5) ds2 = 4(1 − |x|2

)−2n∑

j=1

dx2j .

Another useful representation of hyperbolic space is as the upper halfspace R

n+ = x ∈ R

n : xn > 0, with a metric we will specify shortly. Infact, with en = (0, . . . , 0, 1),

(5.6) τ(x) = |x + en|−2(x + en) − 1

2en

defines a map of the unit ball Bn onto Rn+, taking the metric (5.5) to

(5.7) ds2 = x−2n

n∑

j=1

dx2j .

The Laplace operator for the metric (5.7) has the form

(5.8)

∆u =n∑

j=1

xnn ∂j

(x2−n

n ∂ju)

= x2n

n∑

j=1

∂2j u + (2 − n)xn ∂nu.

which is convenient for a number of computations, such as (5.9) in thefollowing:

Proposition 5.1. If ∆ is the Laplace operator on Hn, then ∆ is essen-tially self-adjoint on C∞

0 (Hn), and its natural self-adjoint extension hasthe property

(5.9) spec(−∆) ⊂[1

4(n − 1)2,∞

).

5. The Laplace operator on hyperbolic space 33

Proof. Since Hn is a complete Riemannian manifold, the essential self-adjointness on C∞

0 (Hn) follows from Proposition 2.4. To establish (5.9), itsuffices to show that

(−∆u, u)L2(Hn) ≥(n − 1)2

4‖u‖2

L2(Hn),

for all u ∈ C∞0 (Hn). Now the volume element on Hn, identified with the

upper half-space with the metric (5.7), is x−nn dx1 · · · dxn, so for such u we

have

(5.10)

((−∆ − 1

4(n − 1)2

)u, u

)L2

=

∫ [(∂nu)2 −

( (n − 1)u

2xn

)2]x2−n

n dx1 · · · dxn

+n−1∑

j=1

∫(∂ju)2x2−n

n dx1 · · · dxn.

Now, by an integration by parts, the first integral on the right is equal to

(5.11)

∫

Rn+

[∂n

(x−(n−1)/2

n u)]2

xn dx1 · · · dxn.

Thus the expression (5.10) is ≥ 0, and (5.9) is proved.

We next describe how to obtain the fundamental solution to the waveequation on Hn. This will be obtained from the formula for Sn, via ananalytic continuation in the metric tensor. Let p be a fixed point (e.g., thenorth pole) in Sn, taken to be the origin in geodesic normal coordinates.Consider the one-parameter family of metrics given by dilating the sphere,which has constant curvature K = 1. Spheres dilated to have radius > 1have constant curvature K ∈ (0, 1). On such a space, the fundamentalkernel A−1 sin tA δp(x), with

(5.12) A =(−∆ +

K

4(n − 1)2

)1/2

,

can be obtained explicitly from that on the unit sphere by a change ofscale. The explicit representation so obtained continues analytically to allreal values of K and at K = −1 gives a formula for the wave kernel,

(5.13) A−1 sin tA δp(x) = R(t, p, x), A =(−∆ − 1

4(n − 1)2

)1/2

.

We have

(5.14) R(t, p, x) = limεց0

−2Cn Im[2 cos(it − ε) − 2 cosh r

]−(n−1)/2,

where r = r(p, x) is the geodesic distance from p to x. Here, as in (4.10),Cn = 1/(n − 1)An. This exhibits several properties similar to those in


the case of Sn discussed in §4. Of course, for r > |t|, the limit vanishes,exhibiting the finite propagation speed phenomenon. Also, if n is odd, theexponent (n− 1)/2 is an integer, which implies that (5.14) is supported onthe shell r = |t|.

In analogy with (4.25), we have the following formula for g(A)δp(x), forg ∈ S(R), when acting on L2(Hn), with n = 2k + 1:

(5.15) g(A) = (2π)−1/2(− 1

2π

1

sinh r

∂

∂r

)k

g(r).

If n = 2k, we have

(5.16)

g(A) =

1

π1/2

∫ ∞

r

(− 1

2π

1

sinh s

∂

∂s

)k

g(s)(cosh s − cosh r

)−1/2sinh s ds.

Exercises

1. If n = 2k + 1, show that the Schwartz kernel of“

−∆− (n− 1)2/4− z2”−1

on

Hn, for z ∈ C \ [0,∞), is

Gz(x, y) = − 1

2iz

„

− 1

2π

1

sinh r

∂

∂r

«k

eizr,

where r = r(x, y) is geodesic distance, and the integral kernel of et(∆+(n−1)2/4),for t > 0, is

Ht(x, y) =1√4πt

„

− 1

2π

1

sinh r

∂

∂r

«k

e−r2/4t.

6. The harmonic oscillator

We consider the differential operator H = −∆+ |x|2 on L2(Rn). By Propo-sition 2.7, H is essentially self-adjoint on C∞

0 (Rn). Furthermore, as a spe-cial case of Proposition 2.8, we know that H has compact resolvent, soL2(Rn) has an orthonormal basis of eigenfunctions of H. To work outthe spectrum, it suffices to work with the case n = 1, so we considerH = D2 + x2, where D = −i d/dx.

The spectral analysis follows by some simple algebraic relations, involv-ing the operators

(6.1)a = D − ix =

1

i

( d

dx+ x

),

a+ = D + ix =1

i

( d

dx− x

).

6. The harmonic oscillator 35

Note that on D′(R),

(6.2) H = aa+ − I = a+a + I,

and

(6.3) [H, a] = −2a, [H, a+] = 2a+.

Suppose that uj ∈ C∞(R) is an eigenfunction of H, that is,

(6.4) uj ∈ D(H), Huj = λjuj .

Now, by material developed in §2,

(6.5)D(H1/2) = u ∈ L2(R) : Du ∈ L2(R), xu ∈ L2(R),

D(H) = u ∈ L2(R) : D2u + x2u ∈ L2(R).Since certainly each uj belongs to D(H1/2), it follows that auj and a+uj

belong to L2(R). By (6.3), we have

(6.6) H(auj) = (λj − 2)auj , H(a+uj) = (λj + 2)a+uj .

It follows that auj and a+uj belong to D(H) and are eigenfunctions. Hence,if

(6.7) Eigen(λ,H) = u ∈ D(H) : Hu = λu,we have, for all λ ∈ R,

(6.8)a+ : Eigen(λ,H) → Eigen(λ + 2,H),

a : Eigen(λ + 2,H) → Eigen(λ,H).

From (6.2) it follows that (Hu, u) ≥ ‖u‖2L2 , for all u ∈ C∞

0 (R); hence, inview of essential self-adjointness,

(6.9) spec H ⊂ [1,∞), for n = 1.

Now each space Eigen(λ,H) is a finite-dimensional subspace of C∞(R),and, by (6.2), we conclude that, in (6.8), a+ is an isomorphism of Eigen(λj ,H)onto Eigen(λj + 2,H), for each λj ∈ spec H. Also, a is an isomorphismof Eigen(λj ,H) onto Eigen(λj − 2,H), for all λj > 1. On the other hand,a must annihilate Eigen(λ0,H) when λ0 is the smallest element of spec H,so

(6.10)u0 ∈ Eigen(λ0,H) =⇒ u′

0(x) = −xu0(x)

=⇒ u0(x) = K e−x2/2.

Thus

(6.11) λ0 = 1, Eigen(1,H) = span(e−x2/2

).

Since e−x2/2 spans the null space of a, acting on C∞(R), and since eachnonzero space Eigen(λj ,H) is mapped by some power of a to this null


space, it follows that, for n = 1,

(6.12) spec H = 2k + 1 : k = 0, 1, 2, . . . and

(6.13) Eigen(2k + 1,H) = span(( ∂

∂x− x

)ke−x2/2

).

One also writes

(6.14)( ∂

∂x− x

)k

e−x2/2 = Hk(x) e−x2/2,

where Hk(x) are the Hermite polynomials, given by

(6.15)

Hk(x) = (−1)kex2( d

dx

)k

e−x2

=

[k/2]∑

j=0

(−1)j k!

j!(k − 2j)!(2x)k−2j .

We define eigenfunctions of H:

(6.16) hk(x) = ck

( ∂

∂x− x

)k

e−x2/2 = ckHk(x)e−x2/2,

where ck is the unique positive number such that ‖hk‖L2(R) = 1. To eval-uate ck, note that

(6.17) ‖a+hk‖2L2 = (aa+hk, hk)L2 = 2(k + 1)‖hk‖2

L2 .

Thus, if ‖hk‖L2 = 1, in order for hk+1 = γka+hk to have unit norm, weneed γk = (2k + 2)−1/2. Hence

(6.18) ck =[π1/22k(k!)

]−1/2.

Of course, given the analysis above of H on L2(R), then for H = −∆ +|x|2 on L2(Rn), we have

(6.19) spec H = 2k + n : k = 0, 1, 2, . . . .In this case, an orthonormal basis of Eigen(2k + n,H) is given by

(6.20) ck1· · · ckn

Hk1(x1) · · ·Hkn

(xn)e−|x|2/2, k1 + · · · + kn = k,

where kν ∈ 0, . . . , k, the Hkν(xν) are the Hermite polynomials, and the

ckνare given by (6.18). The dimension of this eigenspace is the same as

the dimension of the space of homogeneous polynomials of degree k in nvariables.

We now want to derive a formula for the semigroup e−tH , t > 0, calledthe Hermite semigroup. Again it suffices to treat the case n = 1. To somedegree paralleling the analysis of the eigenfunctions above, we can producethis formula via some commutator identities, involving the operators

(6.21) X = D2 = −∂2x, Y = x2, Z = x∂x + ∂xx = 2x ∂x + 1.


Note that H = X + Y . The commutator identities are

(6.22) [X,Y ] = −2Z, [X,Z] = 4X, [Y,Z] = −4Y.

Thus, X,Y , and Z span a three-dimensional, real Lie algebra. This isisomorphic to sl(2, R), the Lie algebra consisting of 2 × 2 real matrices oftrace zero, spanned by

(6.23) n+ =

(0 10 0

), n− =

(0 01 0

), α =

(1 00 −1

).

We have

(6.24) [n+, n−] = α, [n+, α] = −2n+, [n−, α] = 2n−.

The isomorphism is implemented by

(6.25) X ↔ 2n+, Y ↔ 2n−, Z ↔ −2α.

Now we will be able to write

(6.26) e−t(2n++2n−) = e−2σ1(t)n+ e−2σ3(t)α e−2σ2(t)n− ,

as we will see shortly, and, once this is accomplished, we will be motivatedto suspect that also

(6.27) e−tH = e−σ1(t)X eσ3(t)Z e−σ2(t)Y .

To achieve (6.26), write

(6.28)

e−2σ1n+ =

(1 −2σ1

0 1

)=

(1 x0 1

),

e−2σ3α =

(e−2σ3 0

0 e2σ3

)=

(y 00 1/y

),

e−2σ2n− =

(1 0

−2σ2 1

)=

(1 0z 1

),

and

(6.29) e−2t(n++n−) =

(cosh 2t − sinh 2t− sinh 2t cosh 2t

)=

(u vv u

).

Then (6.26) holds if and only if

(6.30) y =1

u=

1

cosh 2t, x = z =

v

u= − tanh 2t,

so the quantities σj(t) are given by

(6.31) σ1(t) = σ2(t) =1

2tanh 2t, e2σ3(t) = cosh 2t.


Now we can compute the right side of (6.27). Note that

(6.32)

e−σ1Xu(x) = (4πσ1)−1/2

∫e−(x−y)2/4σ1u(y) dy,

e−σ2Y u(x) = e−σ2x2

u(x),

eσ3Zu(x) = eσ3 u(e2σ3x).

Upon composing these operators we find that, for n = 1,

(6.33) e−tHu(x) =

∫Kt(x, y)u(y) dy,

with

(6.34) Kt(x, y) =exp

[− 1

2 (cosh 2t)(x2 + y2) + xy]/

sinh 2t

(2π sinh 2t

)1/2.

This is known as Mehler’s formula for the Hermite semigroup. Clearly, forgeneral n, we have

(6.35) e−tHu(x) =

∫Kn(t, x, y)u(y) dy,

with

(6.36) Kn(t, x, y) = Kt(x1, y1) · · ·Kt(xn, yn).

The idea behind passing from (6.26) to (6.27) is that the Lie algebrahomomorphism defined by (6.25) should give rise to a Lie group homo-morphism from (perhaps a covering group G of) SL(2, R) into a group ofoperators. Since this involves an infinite-dimensional representation of G(not necessarily by bounded operators here, since e−tH is bounded only fort ≥ 0), there are analytical problems that must be overcome to justify thisreasoning. Rather than take the space to develop such analysis here, wewill instead just give a direct justification of (6.33)–(6.34).

Indeed, let v(t, x) denote the right side of (6.33), with u ∈ L2(R) given.The rapid decrease of Kt(x, y) as |x| + |y| → ∞, for t > 0, makes it easyto show that

(6.37) u ∈ L2(R) =⇒ v ∈ C∞((0,∞),S(R)

).

Also, it is routine to verify that

(6.38)∂v

∂t= −Hv.

Simple estimates yielding uniqueness then imply that, for each s > 0,

(6.39) v(t + s, ·) = e−tHv(s, ·).


Indeed, if w(t, ·) denotes the difference between the two sides of (6.39), thenwe have w(0) = 0, w ∈ C(R+,D(H)), ∂w/∂t ∈ C(R+, L2(R)), and

d

dt‖w(t)‖2

L2 = −2(Hw,w) ≤ 0,

so w(t) = 0, for all t ≥ 0.Finally, as t ց 0, we see from (6.31) that each σj(t) ց 0. Since v(t, x)

is also given by the right side of (6.27), we conclude that

(6.40) v(t, ·) → u in L2(R), as t ց 0.

Thus we can let s ց 0 in (6.39), obtaining a complete proof that e−tHu isgiven by (6.33) when n = 1.

It is useful to write down the formula for e−tH using the Weyl calculus,introduced in §14 of Chapter 7. We recall that it associates to a(x, ξ) theoperator

(6.41)

a(X,D)u = (2π)−n

∫a(q, p)ei(q·X+p·D)u(x) dq dp

= (2π)−n

∫a(x + y

2, ξ

)ei(x−y)·ξu(y) dy dξ.

In other words, the operator a(X,D) has integral kernel Ka(x, y), for which

a(X,D)u(x) =

∫Ka(x, y)u(y) dy,

given by

Ka(x, y) = (2π)−n

∫a(x + y

2, ξ

)ei(x−y)·ξ dξ.

Recovery of a(x, ξ) from Ka(x, y) is an exercise in Fourier analysis. Whenit is applied to the formulas (6.33)–(6.36), this exercise involves computinga Gaussian integral, and we obtain the formula

(6.42) e−tH = ht(X,D)

on L2(Rn), with

(6.43) ht(x, ξ) = (cosh t)−n e−(tanh t)(|x|2+|ξ|2).

It is interesting that this formula, while equivalent to (6.33)–(6.36), has asimpler and more symmetrical appearance.

In fact, the formula (6.43) was derived in §15 of Chapter 7, by a differentmethod, which we briefly recall here. For reasons of symmetry, involvingthe identity (14.19), one can write

(6.44) ht(x, ξ) = g(t,Q), Q(x, ξ) = |x|2 + |ξ|2.Note that (6.42) gives ∂t ht(X,D) = −Hht(X,D). Now the compositionformula for the Weyl calculus implies that ht(x, ξ) satisfies the following


evolution equation:

(6.45)

∂

∂tht(x, ξ) = −(Q ht)(x, ξ)

= −Q(x, ξ)ht(x, ξ) − 1

2Q,ht2(x, ξ)

= −(|x|2 + |ξ|2)ht(x, ξ) +1

4

∑

k

(∂2

xk+ ∂2

ξk

)ht(x, ξ).

Given (6.44), we have for g(t,Q) the equation

(6.46)∂g

∂t= −Qg + Q

∂2g

∂Q2+ n

∂g

∂Q.

It is easy to verify that (6.43) solves this evolution equation, with h0(x, ξ) =1.

We can obtain a formula for

(6.47) e−tQ(X,D) = hQt (X,D),

for a general positive-definite quadratic form Q(x, ξ). First, in the case

(6.48) Q(x, ξ) =

n∑

j=1

µj(x2j + ξ2

j ), µj > 0,

it follows easily from (6.43) and multiplicativity, as in (6.36), that

(6.49) hQt (x, ξ) =

n∏

j=1

(cosh tµj

)−1 · exp−

n∑

j=1

(tanh tµj)(x2j + ξ2

j )

.

Now any positive quadratic form Q(x, ξ) can be put in the form (6.48) via alinear symplectic transformation, so to get the general formula we need onlyrewrite (6.49) in a symplectically invariant fashion. This is accomplishedusing the “Hamilton map” FQ, a skew-symmetric transformation on R

2n

defined by

(6.50) Q(u, v) = σ(u, FQv), u, v ∈ R2n,

where Q(u, v) is the bilinear form polarizing Q, and σ is the symplecticform on R

2n; σ(u, v) = x · ξ′ − x′ · ξ if u = (x, ξ), v = (x′, ξ′). When Q has

the form (6.48), FQ is a sum of 2 × 2 blocks

(0 µj

−µj 0

), and we have

(6.51)n∏

j=1

(cosh tµj

)−1=

(det cosh itFQ

)−1/2

.

Passing from FQ to

(6.52) AQ =(−F 2

Q

)1/2,

Exercises 41

the unique positive-definite square root, means passing to blocks(

µj 00 µj

),

and when Q has the form (6.48), then

(6.53)

n∑

j=1

(tanh tµj)(x2j + ξ2

j ) = tQ(ϑ(tAQ)ζ, ζ

),

where ζ = (x, ξ) and

(6.54) ϑ(t) =tanh t

t.

Thus the general formula for (6.47) is

(6.55) hQt (x, ξ) =

(cosh tAQ

)−1/2

e−tQ(ϑ(tAQ)ζ,ζ).

Exercises

1. Define an unbounded operator A on L2(R) by

D(A) = u ∈ L2(R) : Du ∈ L2(R), xu ∈ L2(R), Au = Du − ixu.

Show that A is closed and that the self-adjoint operator H satisfies

H = A∗A + I = AA∗ − I.

(Hint: Note Exercises 5–7 of §2.)2. If Hk(x) are the Hermite polynomials, show that there is the generating func-

tion identity

∞X

k=0

1

k!Hk(x)sk = e2xs−s2

.

(Hint: Use the first identity in (6.15).)3. Show that Mehler’s formula (6.34) is equivalent to the identity

∞X

j=0

hj(x)hj(y)sj =

π−1/2(1 − s2)−1/2 exp

(1 − s2)−1h

2xys − (x2 + y2)s2i

ff

· e−(x2+y2)/2,

for 0 ≤ s < 1. Deduce that

∞X

j=0

Hj(x)2sj

2jj!= (1 − s2)−1/2e2sx2/(1+s), |s| < 1.

4. Using

H−s =1

Γ(s)

Z ∞

0

e−tH ts−1 dt, Re s > 0,


find the integral kernel As(x, y) such that

H−su(x) =

Z

As(x, y)u(y) dy.

Writing Tr H−s =R

As(x, x) dx, Re s > 1, n = 1, show that

ζ(s) =1

Γ(s)

Z ∞

0

ys−1

ey − 1dy.

See [Ing], pp. 41–44, for a derivation of the functional equation for the Rie-mann zeta function, using this formula.

5. Let Hω = −d2/dx2 + ω2x2. Show that e−tHω has integral kernel

Kωt (x, y) = (4πt)−1/2 γ(2ωt)1/2 e−γ(2ωt)[(cosh 2ωt)(x2+y2)−2xy]/4t,

where

γ(z) =z

sinh z.

6. Consider the operator

Q(X, D) = −„

∂

∂x1− iωx2

«2

−„

∂

∂x2+ iωx1

«2

= −∆ + ω2|x|2 + 2iω

„

x2∂

∂x1− x1

∂

∂x2

«

.

Note that Q(x, ξ) is nonnegative, but not definite. Study the integral kernelKQ

t (x, y) of e−tQ(X,D). Show that

KQt (x, 0) = (4πt)−1 γ(2ωt) e−τ(2ωt)|x|2/4t,

where

τ(z) = z coth z.

7. Let (ωjk) be an invertible, n × n, skew-symmetric matrix of real numbers (son must be even). Suppose

L = −n

X

j=1

„

∂

∂xj− i

X

k

ωjkxk

«2

.

Evaluate the integral kernel KLt (x, y), particularly at y = 0.

8. In terms of the operators a, a+ given by (6.1) and the basis of L2(R) given by(6.16)–(6.18), show that

a+hk =√

2k + 2 hk+1, ahk =√

2k hk−1.

7. The quantum Coulomb problem

In this section we examine the operator

(7.1) Hu = −∆u − K|x|−1u,

acting on functions on R3. Here, K is a positive constant.

7. The quantum Coulomb problem 43

This provides a quantum mechanical description of the Coulomb forcebetween two charged particles. It is the first step toward a quantum me-chanical description of the hydrogen atom, and it provides a decent approx-imation to the observed behavior of such an atom, though it leaves out anumber of features. The most important omitted feature is the spin of theelectron (and of the nucleus). Giving rise to further small corrections arethe nonzero size of the proton, and relativistic effects, which confront onewith great subtleties since relativity forces one to treat the electromagneticfield quantum mechanically. We refer to texts on quantum physics, such as[Mes], [Ser], [BLP], and [IZ], for work on these more sophisticated modelsof the hydrogen atom.

We want to define a self-adjoint operator via the Friedrichs method.Thus we want to work with a Hilbert space

(7.2) H =u ∈ L2(R3) : ∇u ∈ L2(R3),

∫|x|−1|u(x)|2 dx < ∞

,

with inner product

(7.3) (u, v)H = (∇u,∇v)L2 + A(u, v)L2 − K

∫|x|−1u(x)v(x) dx,

where A is a sufficiently large, positive constant. We must first show thatA can be picked to make this inner product positive-definite. In fact, wehave the following:

Lemma 7.1. For all ε ∈ (0, 1], there exists C(ε) < ∞ such that

(7.4)

∫|x|−1|u(x)|2 dx ≤ ε‖∇u‖2

L2 + C(ε)‖u‖2L2 ,

for all u ∈ H1(R3).

Proof. Here and below we will use the inclusion

(7.5) Hs(Rn) ⊂ Lp(Rn), ∀ p ∈[2,

2n

n − 2s

), 0 ≤ s <

n

2,

from (2.42) of Chapter 4. In Chapter 13 we will establish the sharper resultthat Hs(Rn) ⊂ L2n/(n−2s)(Rn); for example, H1(R3) ⊂ L6(R3). We willalso cite this stronger result in some arguments below, though that couldbe avoided.

We also use the fact that (if B = |x| < 1 and χB(x) is its characteristicfunction),

χBV ∈ Lq(R3), for all q < 3.

Here and below we will use V (x) = |x|−1. Thus the left side of (7.4) isbounded by

(7.6) ‖χBV ‖Lq · ‖u‖2L2q′ + ‖u‖2

L2 ≤ C‖u‖2Hσ(R3) + ‖u‖2

L2(R3),


where we can take any q′ > 3/2; take q′ ∈ (3/2, 3). Then (7.6) holdsfor some σ < 1, for which L2q′

(R3) ⊃ Hσ(R3). From this, (7.4) followsimmediately.

Thus the Hilbert space H in (7.2) is simply H1(R3), and we see thatindeed, for some A > 0, (7.3) defines an inner product equivalent to thestandard one on H1(R3). The Friedrichs method then defines a positive,self-adjoint operator H + AI, for which

(7.7) D((H + AI)1/2

)= H1(R3).

Then

(7.8) D(H) = u ∈ H1(R3) : −∆u − K|x|−1u ∈ L2(R3),where −∆u − K|x|−1u is a priori regarded as an element of H−1(R3) ifu ∈ H1(R3). Since H2(R3) ⊂ L∞(R3), we have

(7.9) u ∈ H2(R3) =⇒ |x|−1u ∈ L2(R3),

so

(7.10) D(H) ⊃ H2(R3).

Indeed, we have:

Proposition 7.2. For the self-adjoint extension H of −∆−K|x|−1 definedabove,

(7.11) D(H) = H2(R3).

Proof. Pick λ in the resolvent set of H; for instance, λ ∈ C \ R. Ifu ∈ D(H) and (H − λ)u = f ∈ L2(R3), we have

(7.12) u − KRλV u = Rλf = gλ,

where V (x) = |x|−1 and Rλ = (−∆ − λ)−1. Now the operator of multipli-cation by V (x) = |x|−1 has the property

(7.13) MV : H1(R3) −→ L2−ε(R3),

for all ε > 0, since H1(R3) ⊂ L6(R3) ∩ L2(R3) and V ∈ L3−ε on |x| < 1.Hence

MV : H1(R3) −→ H−ε(R3),

for all ε > 0. Let us apply this to (7.12). We know that u ∈ D(H) ⊂D(H1/2) = H1(R3), so KRλV u ∈ H2−ε(R3). Thus u ∈ H2−ε(R3), for allε > 0. But, for ε > 0 small enough,

(7.14) MV : H2−ε(R3) −→ L2(R3),


so then u = KRλ(V u)+Rλf ∈ H2(R3). This proves that D(H) ⊂ H2(R3)and gives (7.11).

Since H is self-adjoint, its spectrum is a subset of the real axis, (−∞,∞).We next show that there is only point spectrum in (−∞, 0).

Proposition 7.3. The part of spec H lying in C \ [0,∞) is a bounded,discrete subset of (−∞, 0), consisting of eigenvalues of finite multiplicityand having at most 0 as an accumulation point.

Proof. Consider the equation (H − λ)u = f ∈ L2(R3), that is,

(7.15) (−∆ − λ)u − KV u = f,

with V (x) = |x|−1 as before. Applying Rλ = (−∆−λ)−1 to both sides, weagain obtain (7.12):

(7.16) (I − KRλMV )u = gλ = Rλf.

Note that Rλ is a holomorphic function of λ ∈ C \ [0,∞), with values inL(L2(R3),H2(R3)). A key result in the analysis of (7.16) is the following:

Lemma 7.4. For λ ∈ C \ [0,∞),

(7.17) RλMV ∈ K(L2(R3)),

where K is the space of compact operators.

We will establish this via the following basic tool. For λ ∈ C\[0,∞), ϕ ∈C0(R

3), the space of continuous functions vanishing at infinity, we have

(7.18) MϕRλ ∈ K(L2) and RλMϕ ∈ K(L2).

To see this, note that, for ϕ ∈ C∞0 (R3), the first inclusion in (7.18) follows

from Rellich’s theorem. Then this inclusion holds for uniform limits of suchϕ, hence for ϕ ∈ C0(R

3). Taking adjoints yields the rest of (7.18).Now, to establish (7.17), write

(7.19) V = V1 + V2,

where V1 = ψV, ψ ∈ C∞0 (R3), ψ(x) = 1 for |x| ≤ 1. Then V2 ∈ C0(R

3), soRλMV2

∈ K. We have V1 ∈ Lq(R3), for all q ∈ [1, 3), so, taking q = 2, wehave

(7.20) MV1: L2(R3) −→ L1(R3) ⊂ H−3/2−ε(R3),

for all ε > 0, hence

(7.21) RλMV1: L2(R3) −→ H1/2−ε(R3) ⊂ L2(R3).

Given V1 supported on a ball BR, the operator norm in (7.21) is boundedby a constant times ‖V1‖L2 . You can approximate V1 in L2-norm by a


sequence wj ∈ C∞0 (R3). It follows that RλMV1

is a norm limit of a se-quence of compact operators on L2(R3), so it is also compact, and (7.17)is established.

The proof of Proposition 7.4 is finished by the following result, whichcan be found as Proposition 7.4 in Chapter 9.

Proposition 7.5. Let O be a connected, open set in C. Suppose C(λ)is a compact-operator-valued holomorphic function of λ ∈ O. If I − C(λ)is invertible at one point p ∈ O, then it is invertible except at most on adiscrete set in O, and (I − C(λ))−1 is meromorphic on O.

This applies to our situation, with C(λ) = KRλMV ; we know thatI − C(λ) is invertible for all λ ∈ C \ R in this case.

One approach to analyzing the negative eigenvalues of H is to use polarcoordinates. If −K|x|−1 is replaced by any radial potential V(|x|), theeigenvalue equation Hu = −Eu becomes

(7.22)∂2u

∂r+

2

r

∂u

∂r+

1

r2∆Su − V(r)u = Eu.

We can use separation of variables, writing u(rθ) = v(r)ϕ(θ), where ϕ isan eigenfunction of ∆S , the Laplace operator on S2,

(7.23) ∆Sϕ = −λϕ, λ =(k +

1

2

)2 − 1

4= k2 + k.

Then we obtain for v(r) the ODE

(7.24) v′′(r) +2

rv′(r) + f(r)v(r) = 0, f(r) = −E − λ

r2− V(r).

One can eliminate the term involving v′ by setting

(7.25) w(r) = rv(r).

Then

(7.26) w′′(r) + f(r)w(r) = 0.

For the Coulomb problem, this becomes

(7.27) w′′(r) +[−E +

K

r− λ

r2

]w(r) = 0.

If we set W (r) = w(βr), β = 1/2√

E, we get a form of Whittaker’s ODE:

(7.28) W ′′(z) +[−1

4+

κ

z+

14 − µ2

z2

]W (z) = 0,

with

(7.29) κ =K

2√

E, µ2 = λ +

1

4=

(k +

1

2

)2.


This in turn can be converted to the confluent hypergeometric equation

(7.30) zψ′′(z) + (b − z)ψ′(z) − aψ(z) = 0

upon setting

(7.31) W (z) = zµ+1/2 e−z/2 ψ(z),

with

(7.32)a = µ − κ +

1

2= k + 1 − K

2√

E,

b = 2µ + 1 = 2k + 2.

Note that ψ and v are related by

(7.33) v(r) = (2√

E)k+1 rke−2√

Erψ(2√

Er).

Looking at (7.28), we see that there are two independent solutions, onebehaving roughly like e−z/2 and the other like ez/2, as z → +∞. Equiv-alently, (7.30) has two linearly independent solutions, a “good” one grow-ing more slowly than exponentially and a “bad” one growing like ez, asz → +∞. Of course, for a solution to give rise to an eigenfunction, weneed v ∈ L2(R+, r2 dr), that is, w ∈ L2(R+, dr). We need to have simulta-neously w(z) ∼ ce−z/2 (roughly) as z → +∞ and w square integrable nearz = 0. In view of (7.8), we also need v′ ∈ L2(R+, r2 dr).

To examine the behavior near z = 0, note that the Euler equation asso-ciated with (7.28) is

(7.34) z2W ′′(z) +(1

4− µ2

)W (z) = 0,

with solutions z1/2+µ and z1/2−µ, i.e., zk+1 and z−k, k = 0, 1, 2, . . . . Ifk = 0, both are square integrable near 0, but for k ≥ 1 only one is. Goingto the confluent hypergeometric equation (7.30), we see that two linearlyindependent solutions behave respectively like z0 and z−2µ = z−2k−1 asz → 0.

As a further comment on the case k = 0, note that a solution W behavinglike z0 at z = 0 gives rise to v(r) ∼ C/r as r → 0, with c 6= 0, hencev′(r) ∼ −C/r2. This is not square integrable near r = 0, with respect tor2 dr, so also this case does not produce an eigenfunction of H.

If b /∈ 0,−1,−2, . . . , which certainly holds here, the solution to (7.30)that is “good” near z = 0 is given by the confluent hypergeometric function

(7.35) 1F1(a; b; z) =

∞∑

n=0

(a)n

(b)n

zn

n!,


an entire function of z. Here, (a)n = a(a + 1) · · · (a + n − 1); (a)0 = 1. Ifalso a /∈ 0,−1,−2, . . . , it can be shown that

(7.36) 1F1(a; b; z) ∼ Γ(b)

Γ(a)ez z−(b−a), z → +∞.

See the exercises below for a proof of this. Thus the “good” solution nearz = 0 is “bad” as z → +∞, unless a is a nonpositive integer, say a = −j.In that case, as is clear from (7.35), 1F1(−j; b; z) is a polynomial in z, thus“good” as z → +∞. Thus the negative eigenvalues of H are given by −E,with

(7.37)K

2√

E= j + k + 1 = n,

that is, by

(7.38) E =K2

4n2, n = 1, 2, 3 . . . .

Note that, for each value of n, one can write n = j+k+1 using n choicesof k ∈ 0, 1, 2, . . . , n − 1. For each such k, the (k2 + k)-eigenspace of ∆S

has dimension 2k + 1, as established in Corollary 4.3. Thus the eigenvalue−E = −K2/4n2 of H has multiplicity

(7.39)

n−1∑

k=0

(2k + 1) = n2.

Let us denote by Vn the n2-dimensional eigenspace of H, associated to theeigenvalue λn = −K2/4n2.

The rotation group SO(3) acts on each Vn, via

ρ(g)f(x) = f(g−1x), g ∈ SO(3), x ∈ R3.

By the analysis leading to (7.39), this action on Vn is not irreducible,but rather has n irreducible components. This suggests that there is anextra symmetry, and indeed, as W. Pauli discovered early in the historyof quantum mechanics, there is one, arising via the Lenz vector (brieflyintroduced in §16 of Chapter 1), which we proceed to define.

The angular momentum vector L = x×p, with p replaced by the vectoroperator (∂/∂x1, ∂/∂x2, ∂/∂x3), commutes with H as a consequence of therotational invariance of H. The components of L are

(7.40) Lℓ = xj∂

∂xk− xk

∂

∂xj,

where (j, k, ℓ) is a cyclic permutation of (1, 2, 3). Then the Lenz vector isdefined by

(7.41) B =1

K

(L × p − p × L

)− x

r,


with components Bj , 1 ≤ j ≤ 3, each of which is a second-order differentialoperator, given explicitly by

(7.42) Bj =1

K(Lk∂ℓ + ∂ℓLk − Lℓ∂k − ∂kLℓ) −

xj

r,

where (j, k, ℓ) is a cyclic permutation of (1, 2, 3). A calculation gives

(7.43) [H,Bj ] = 0,

in the sense that these operators commute on C∞(R3 \ 0).It follows that if u ∈ Vn, then Bju is annihilated by H − λn, on R

3 \ 0.Now, we have just gone through an argument designed to glean from allfunctions that are so annihilated, those that are actually eigenfunctions ofH. In view of that, it is important to establish the next lemma.

Lemma 7.6. We have

(7.44) Bj : Vn −→ Vn.

Proof. Let u ∈ Vn. We know that u ∈ D(H) = H2(R3). Also, fromthe analysis of the ODE (7.28), we know that u(x) decays as |x| → ∞,roughly like e−|λn|1/2|x|. It follows from (7.42) that Bju ∈ L2(R3). It willbe useful to obtain a bit more regularity, using Vn ⊂ D(H2) together withthe following.

Proposition 7.7. If u ∈ D(H2), then, for all ε > 0,

(7.45) u ∈ H5/2−ε(R3).

Furthermore,

(7.46) g ∈ S(R3), g(0) = 0 =⇒ gu ∈ H7/2−ε(R3).

Proof. We proceed along the lines of the proof of Proposition 7.2, using(7.12), i.e.,

(7.47) u = KRλV u + Rλf,

where f = (H − λ)u, with λ chosen in C \R. We know that f = (H − λ)ubelongs to D(H), so Rλf ∈ H4(R3). We know that u ∈ H2(R3). Parallelto (7.13), we can show that, for all ε > 0,

(7.48) MV : H2(R3) −→ H1/2−ε(R3),

so KRλV u ∈ H5/2−ε(R3). This gives (7.45).Now, multiply (7.47) by g and write

(7.49) gu = KRλgV u + K[Mg, Rλ]V u + gRλf.


This time we have

MgV : H2(R3) −→ H3/2−ε(R3),

so RλgV u ∈ H7/2−ε(R3). Furthermore,

(7.50) [Mg, Rλ] = Rλ [∆,Mg] Rλ : Hs(R3) −→ Hs+3(R3),

so [Mg, Rλ]V u ∈ H7/2−ε(R3). This establishes (7.46).

We can now finish the proof of Lemma 7.6. Note that the second-orderderivatives in Bj have a coefficient vanishing at 0. Keep in mind theknown exponential decay of u ∈ Vn. Also note that Mxj/r : H2(R3) →H3/2−ε(R3). Therefore,

(7.51) u ∈ Vn =⇒ Bju ∈ H3/2−ε(R3).

Consequently,

(7.52) ∆(Bju) ∈ H−1/2−ε(R3), and V (Bju) ∈ L1(R3) + L2(R3).

Thus (H − λn)(Bju), which we know vanishes on R3 \ 0, must vanish

completely, since (7.52) does not allow for a nonzero quantity supportedon 0. Using (7.8), we conclude that Bju ∈ D(H), and the lemma isproved.

With Lemma 7.6 established, we can proceed to study the action of Bj

and Lj on Vn. When (j, k, ℓ) is a cyclic permutation of (1, 2, 3), we have

(7.53) [Lj , Lk] = Lℓ,

and, after a computation,

(7.54) [Lj , Bk] = Bℓ, [Bj , Bk] = − 4

KHLℓ.

Of course, (7.52) is the statement that Lj span the Lie algebra so(3) ofSO(3). The identities (7.54), when Lj and Bj act on Vn, can be rewrittenas

(7.55) [Lj , Ak] = Aℓ, [Aj , Ak] = Aℓ, Aj =K

2√−λn

Bj .

If we set

(7.56) M =1

2(L + A), N =

1

2(L − A),

we get, for cyclic permutations (j, k, ℓ) of (1, 2, 3),

(7.57) [Mj ,Mk] = Mℓ, [Nj , Nk] = Nℓ, [Mj , Nj′ ] = 0,

which is clearly the set of commutation relations for the Lie algebra so(3)⊕so(3). We next aim to show that this produces an irreducible representationof SO(4) on Vn, and to identify this representation. A priori, of course, onecertainly has a representation of SU(2) × SU(2) on Vn.


We now examine the behavior on Vn of the Casimir operators M2 =M2

1 +M22 +M2

3 and N2. A calculation using the definitions gives B ·L = 0,hence A · L = 0, so, on Vn,

(7.58)

M2 = N2 =1

4(A2 + L2)

=1

4

(L2 − K2

4λnB2

).

We also have the following key identity:

(7.59) K2(B2 − I) = 4H(L2 + I),

which follows from the definitions by a straightforward computation. If wecompare (7.58) and (7.59) on Vn, where H = λn, we get

(7.60) 4M2 = 4N2 = −(1 +

K2

4λn

)I on Vn.

Now the representation σn we get of SU(2)×SU(2) on Vn is a direct sum(possibly with only one summand) of representations Dj/2 ⊗ Dj/2, whereDj/2 is the standard irreducible representation of SU(2) on C

j+1, definedin §9 of Appendix B. The computation (7.60) implies that all the copies inthis sum are isomorphic, that is, for some j = j(n),

(7.61) σn =

µ⊕

ℓ=1

Dj(n)/2 ⊗ Dj(n)/2.

A dimension count gives µ(j(n) + 1

)2= n2. Note that on Dj/2 ⊗ Dj/2,

we have M2 = N2 = (j/2)(j/2 + 1). Thus (7.60) implies j(j + 2) =−1 + K2/4λn, or

(7.62) λn = − K2

4(j + 1)2, j = j(n).

Comparing (7.38), we have (j + 1)2 = n2, that is,

(7.63) j(n) = n − 1.

Since we know that dim Vn = n2, this implies that there is just one sum-mand in (7.61), so

(7.64) σn = D(n−1)/2 ⊗ D(n−1)/2.

This is an irreducible representation of SU(2) × SU(2), which is a doublecover of SO(4),

κ : SU(2) × SU(2) −→ SO(4).

It is clear that σn is the identity operator on both elements in ker κ, andso σn actually produces an irreducible representation of SO(4).


Let ρn denote the restriction to Vn of the representation ρ of SO(3)on L3(R3), described above. If we regard this as a representation ofSU(2), it is clear that ρn is the composition of σn with the diagonal mapSU(2) → SU(2) × SU(2). Results established in §9 of Appendix B implythat such a tensor-product representation of SU(2) has the decompositioninto irreducible representations:

(7.65) ρn ≈n−1⊕

k=0

Dk.

This is also precisely the description of ρn given by the analysis leading to(7.39).

There are a number of other group-theoretic perspectives on the quantumCoulomb problem, which can be found in [Eng] and [GS2]. See also [Ad]and [Cor], Vol. 2.

Exercises

1. For H = −∆ − K|x|−1 with domain given by (7.8), show that

(7.66) D(H) = u ∈ L2(R3) : −∆u − K|x|−1u ∈ L2(R3),where a priori, if u ∈ L2(R3), then ∆u ∈ H−2(R3) and |x|−1u ∈ L1(R3) +L2(R3) ⊂ H−2(R3).(Hint: Parallel the proof of Proposition 7.2. If u belongs to the right side of(7.66), and if you pick λ ∈ C \ R, then, as in (7.12),

(7.67) u − KRλV u = Rλf ∈ H2(R3).)

Complement (7.13) with

(7.68)

MV : L2(R3) −→\

ε>0

H−3/2−ε(R3),

MV :\

ε>0

H1/2−ε(R3) −→\

δ>0

H−3/4−δ(R3).

(Indeed, sharper results can be obtained.) Then deduce from (7.67) first thatu ∈ H1/2−ε(R3) and then that u ∈ H5/4−δ(R3) ⊂ H1(R3).)

2. As a variant of (7.4), show that, for u ∈ H1(R3),

(7.69)

Z

|x|−2|u(x)|2 dx ≤ 4

Z

|∇u(x)|2 dx.

Show that 4 is the best possible constant on the right. (Hint: Use the Mellintransform to show that the spectrum of r d/dr−1/2 on L2(R+, r−1dr) (whichcoincides with the spectrum of r d/dr on L2(R+, dr)) is is − 1/2 : s ∈ R,hence

(7.70)

Z ∞

0

|u(r)|2dr ≤ 4

Z ∞

0

|u′(r)|2r2 dr.

This is sometimes called an “uncertainty principle” estimate. Why mightthat be? (Cf. [RS], Vol. 2, p. 169.)

Exercises 53

3. Show that H = −∆−K/|x| has no non-negative eigenvalues, i.e., only contin-uous spectrum in [0,∞). (Hint: Study the behavior as r → +∞ of solutionsto the ODE (7.28), when −E is replaced by +E ∈ [0,∞). Consult [Olv] fortechniques. See also [RS], Vol. 4, for general results.)

4. Generalize the propositions of this section, with modifications as needed, toother classes of potentials V (x), such as

V ∈ L2 + εL∞,

the set of functions V such that, for each ε > 0, one can write V = V1 +V2, V1 ∈ L2, ‖V2‖L∞ ≤ ε. Consult [RS], Vols. 2-4, for further generalizations.

Exercises on the confluent hypergeometric function

1. Taking (7.35) as the definition of 1F1(a; b; z), show that(7.71)

1F1(a; b; z) =Γ(b)

Γ(a)Γ(b − a)

Z 1

0

eztta−1(1 − t)b−a−1 dt, Re b > Re a > 0.

(Hint: Use the beta function identity, (A.23)–(A.24) of Chapter 3.) Showthat (7.71) implies the asymptotic behavior (7.36), provided Re b > Re a > 0,but that this is insufficient for making the deduction (7.37).

Exercises 2–5 deal with the analytic continuation of (7.71) in a and b, and acomplete justification of (7.36). To begin, write

(7.72) 1F1(a; b; z) =Γ(b)

Γ(b − a)Aψ(a,−z) +

Γ(b)

Γ(a)Aϕ(b − a, z)ez,

where, for Re c > 0, ψ ∈ C∞“

[0, 1/2]”

, we set

(7.73) Aψ(c, z) =1

Γ(c)

Z 1/2

0

e−ztψ(t)tc−1 dt,

and, in (7.72),

ψ(t) = (1 − t)b−a−1, ϕ(t) = (1 − t)a−1.

2. Given Re c > 0, show that

(7.74) Aψ(c, z) ∼ ψ(0)z−c, z → +∞,

and

(7.75) Aψ(c,−z) ∼ ψ( 12)

Γ(c)z−1ez/2, z → +∞.

3. For j = 0, 1, 2, . . . , set

(7.76) Aj(c, t) =1

Γ(c)

Z 1/2

0

e−zttjtc−1 dt,


so Aj(c, z) = Aψ(c, z), with ψ(t) = tj . Show that

Aj(c, z) =Γ(c + j)

Γ(c)z−c−j − 1

Γ(c)

Z ∞

1/2

e−zttc+j−1 dt,

for Re z > 0. Deduce that Aj(c, t) is an entire function of c, for Re z > 0,and that

Aj(c, z) ∼ Γ(c + j)

Γ(c)z−c−j , z → +∞,

if c /∈ 0,−1,−2, . . . .4. Given k = 1, 2, 3, . . . , write

ψ(t) = a0 + a1t + · · · + ak−1tk−1 + ψk(t)tk, ψk ∈ C∞

„»

0,1

2

–«

.

Thus

(7.77) Aψ(c, z) =

k−1X

j=0

ajAj(c, z) +1

Γ(c)

Z 1/2

0

e−ztψk(t)tk+c−1 dt.

Deduce that Aψ(c, z) can be analytically continued to Re c > −k when Rez > 0 and that (7.74) continues to hold if c /∈ 0,−1,−2, . . . , a0 6= 0.

5. Using tc−1 = c−1(d/dt)tc and integrating by parts, show that

(7.78) A0(c, z) = zA0(c + 1, z) − 1

2cΓ(c + 1)e−z/2,

for Re c > 0, all z ∈ C. Show that this provides an entire analytic continua-tion of A0(c, z) and that (7.74)–(7.75) hold, for ψ(t) = 1. Using

Aj(c, z) =Γ(c + j)

Γ(c)A0(c + j, z)

and (7.77), verify (7.75) for all ψ ∈ C∞“

[0, 1/2]”

. (Also again verify (7.74).)

Hence, verify the asymptotic expansion (7.36).

The approach given above to (7.36) is one the author learned from conversa-tions with A. N. Varchenko. In Exercises 6–15 below, we introduce anothersolution to the confluent hypergeometric equation and follow a path to theexpansion (7.36) similar to one described in [Leb] and in [Olv].

6. Show that a solution to the ODE (7.30) is also given by

z1−b1F1(1 + a − b; 2 − b; z),

in addition to 1F1(a; b; z), defined by (7.35). Assume b 6= 0,−1,−2, . . . . Set

(7.79) Ψ(a; b; z) =Γ(1 − b)

Γ(1 + a − b)1F1(a; b; z)+

Γ(b − 1)

Γ(a)z1−b

1F1(1+a−b; 2−b; z).

Show that the Wronskian is given by

W“

1F1(a; b; z), Ψ(a; b; z)”

= −Γ(b)

Γ(a)z−bez.

Exercises 55

7. Show that

(7.80) 1F1(a; b; z) = ez1F1(b − a; b;−z), b /∈ 0,−1,−2, . . . .

(Hint: Use the integral in Exercise 1, and set s = 1 − t, for the case Re b >Re a > 0.)

8. Show that

(7.81) Ψ(a; b; z) =1

Γ(a)

Z ∞

0

e−ztta−1(1 + t)b−a−1 dt, Re a > 0, Re z > 0.

(Hint: First show that the right side solves (7.30). Then check the behavioras z → 0.)

9. Show that

(7.82) Ψ(a; b; z) = zΨ(a + 1; b + 1; z) + (1 − a − b)Ψ(a + 1; b; z).

(Hint: To get this when Re a > 0, use the integral expression (7.81) forΨ(a + 1; b + 1; z), write ze−zt = −(d/dt)e−zt, and integrate by parts.)

10. Show that

(7.83) 1F1(a; b; z) =Γ(b)

Γ(b − a)e±πaiΨ(a; b; z) +

Γ(b)

Γ(a)e±π(a−b)i ez Ψ(b− a; b;−z),

where −z = e∓πiz, b 6= 0,−1,−2, . . . . (Hint: Make use of (7.80) as well as(7.79).)

11. Using the integral representation (7.81), show that under the hypothesesδ > 0, b /∈ 0,−1,−2, . . . , and Re a > 0, we have

(7.84) Ψ(a; b; z) ∼ z−α, |z| → ∞,

in the sector

(7.85) |Arg z| ≤ π

2− δ.

12. Extend (7.84) to the sector |Arg z| ≤ π − δ. (Hint: Replace (7.81) by anintegral along the ray γ = eiαs : 0 ≤ s < ∞, given |α| < π/2.)

13. Further extend (7.84) to the case where no restriction is placed on Re a.(Hint: Use (7.82).)

14. Extend (7.84) still further, to be valid for

(7.86) |Arg z| ≤ 3π

2− δ.

(Hint: See Theorem 2.2 on p. 235 of [Olv], and its application to this problemon p. 256 of [Olv].)

15. Use (7.83)–(7.86) to prove (7.36), that is,

(7.87) 1F1(a; b; z) ∼ Γ(b)

Γ(a)ez z−(b−a), z → +∞,

provided a, b /∈ 0,−1,−2, . . . .Remark: For the analysis of Ψ(b−a; b;−z) as z → +∞, the result of Exercise14 suffices, but the result of Exercise 13 does not. This point appears to havebeen neglected in the discussion of (7.87) on p. 271 of [Leb].


8. The Laplace operator on cones

Generally, if N is any compact Riemannian manifold of dimension m, pos-sibly with boundary, the cone over N, C(N), is the space R

+×N togetherwith the Riemannian metric

(8.1) dr2 + r2g,

where g is the metric tensor on N . In particular, a cone with vertex at theorigin in R

m+1 can be described as the cone over a subdomain Ω of theunit sphere Sm in R

m+1. Our purpose is to understand the behavior of theLaplace operator ∆, a negative, self-adjoint operator, on C(N). If ∂N 6= ∅,we impose Dirichlet boundary conditions on ∂C(N), though many otherboundary conditions could be equally easily treated. The analysis herefollows [CT].

The initial step is to use the method of separation of variables, writing∆ on C(N) in the form

(8.2) ∆ =∂2

∂r2+

m

r

∂

∂r+

1

r2∆N ,

where ∆N is the Laplace operator on the base N . Let µj , ϕj(x) denote theeigenvalues and eigenfunctions of −∆N (with Dirichlet boundary conditionon ∂N if ∂N 6= ∅), and set

(8.3) νj = (µj + α2)1/2, α = −m − 1

2.

If

g(r, x) =∑

j

gj(r)ϕj(x),

with gj(r) well behaved, and if we define the second-order operator Lµ by

(8.4) Lµg(r) =( ∂2

∂r2+

m

r

∂

∂r− µ

r2

)g(r),

then we have

(8.5) ∆g(r, x) =∑

j

Lµjgj(r)ϕj(x).

In particular,

(8.6) ∆(gjϕj) = −λ2gjϕj

provided

(8.7) gj(r) = r−(m−1)/2Jνj(λr).

8. The Laplace operator on cones 57

Here Jν(z) is the Bessel function, introduced in §6 of Chapter 3; there in(6.6) it is defined to be

(8.8) Jν(z) =(z/2)ν

Γ( 12 )Γ(ν + 1

2 )

∫ 1

−1

(1 − t2)ν−1/2eizt dt,

for Re ν > −1/2; in (6.11) we establish Bessel’s equation

(8.9)

[d2

dz2+

1

z

d

dz+

(1 − ν2

z2

)]Jν(z) = 0,

which justifies (8.6); and in (6.19) we produced the formula

(8.10) Jν(z) =(z

2

)ν ∞∑

k=0

(−1)k

k!Γ(k + ν + 1)

(z

2

)2k

.

We also recall, from (6.56) of Chapter 3, the asymptotic behavior

(8.11) Jν(r) ∼( 2

πr

)1/2

cos(r − πν

2− π

4

)+ O(r−3/2), r → +∞.

This suggests making use of the Hankel transform, defined for ν ∈ R+

by

(8.12) Hν(g)(λ) =

∫ ∞

0

g(r)Jν(λr)r dr.

Clearly, Hν : C∞0

((0,∞)

)→ L∞(R+). We will establish the following:

Proposition 8.1. For ν ≥ 0, Hν extends uniquely from C∞0

((0,∞)

)to

(8.13) Hν : L2(R+, r dr) −→ L2(R+, λ dλ), unitary.

Furthermore, for each g ∈ L2(R+, r dr),

(8.14) Hν Hνg = g.

To prove this, it is convenient to consider first

(8.15) Hνf(λ) =

∫ ∞

0

f(r)Jν(λr)

(λr)νr2ν+1 dr,

since, by (8.10), (λr)−νJν(λr) is a smooth function of λr. Set

(8.16) S(R+) = f |R+ : f ∈ S(R) is even.

Lemma 8.2. If ν ≥ −1/2, then

(8.17) Hν : S(R+) −→ S(R+).


Proof. By (8.10), Jν(λr)/(λr)ν is a smooth function of λr. The formula(8.8) yields

(8.18)∣∣∣Jν(λr)

(λr)ν

∣∣∣ ≤ Cν < ∞,

for λr ∈ [0,∞), ν > −1/2, a result that, by the identity

(8.19) J−1/2(z) =( 2

πz

)1/2

cos z,

established in (6.35) of Chapter 3, also holds for ν = −1/2. This readilyyields

(8.20) Hν : S(R+) −→ L∞(R+),

whenever ν ≥ −1/2. Now consider the differential operator Lν , given by

(8.21)Lνf(r) = −r−2ν−1 ∂

∂r

(r2ν+1 ∂f

∂r

)

= −∂2f

∂r2− 2ν + 1

r

∂f

∂r.

Using Bessel’s equation (8.9), we have

(8.22) Lν

(Jν(λr)

(λr)ν

)= λ2 Jν(λr)

(λr)ν,

and, for f ∈ S(R+),

(8.23)Hν(Lνf)(λ) = λ2Hνf(λ),

Hν(r2f)(λ) = LνHνf(λ).

Since f ∈ L∞(R+) belongs to S(R+) if and only if arbitrary iterated ap-plications of Lν and multiplication by r2 to f yield elements of L∞(R+),the result (8.17) follows. We also have that this map is continuous withrespect to the natural Frechet space structure on S(R+).

Lemma 8.3. Consider the elements Eb ∈ S(R+), given for b > 0 by

(8.24) Eb(r) = e−br2

.

We have

(8.25) HνE1/2(λ) = E1/2(λ),

and more generally

(8.26) HνEb(λ) = (2b)−ν−1E1/4b(λ).


Proof. To establish (8.25), plug the power series (8.10) for Jν(z) into(8.15) and integrate term by term, to get

(8.27) HνE1/2(λ) =

∞∑

k=0

(−1)k2−ν−2k

k!Γ(k + ν + 1)λ2k

∫ ∞

0

r2k+2ν+1e−r2/2 dr.

This last integral is seen to equal 2k+νΓ(k + ν + 1), so we have

(8.28) HνE1/2(λ) =

∞∑

k=0

1

k!

(−λ2

2

)k

= e−λ2/2 = E1/2(λ).

Having (8.25), we get (8.26) by an easy change of variable argument.In more detail, set r2/2 = bs2, or s = r/

√2b. Then set µ =

√2bλ, so

λr = µs. Then (8.28), which we can write as

(8.29)

∫ ∞

0

e−r2/2Jν(λr)rν+1 dr = λνe−λ2/2,

translates to

(8.30)

∫ ∞

0

e−bs2

Jν(µs)(2b)(ν+1)/2sν+1(2b)1/2 ds = (2b)−ν/2µνe−µ2/4b,

or, changing notation back,

(8.31)

∫ ∞

0

e−bs2

Jν(λs)sν+1 ds = (2b)−ν−1λνe−λ2/4b,

which gives (8.26).

From (8.26) we have, for each b > 0,

(8.32) HνHνEb = (2b)−ν−1HνE1/4b = Eb,

which verifies our stated Hankel inversion formula for f = Eb, b > 0. Toget the inversion formula for general f ∈ S(R+), it suffices to establish thefollowing.

Lemma 8.4. The space

(8.33) V = Span Eb : b > 0is dense in S(R+).

Proof. Let V denote the closure of V in S(R+). From

(8.34)1

ε

(e−br2 − e−(b+ε)r2) → r2e−br2

,

we deduce that r2e−br2 ∈ V, and inductively, we get

(8.35) r2je−br2 ∈ V, ∀ j ∈ Z+.


From here, one has

(8.36) (cos ξr)e−r2 ∈ V, ∀ ξ ∈ R.

Now each even ω ∈ S ′(R) annihilating (8.36) for all ξ ∈ R has the propertythat e−r2

ω has Fourier transform zero, which implies ω = 0. The assertion(8.33) then follows by the Hahn-Banach theorem.

Putting the results of Lemmas 8.2–8.4 together, we have

Proposition 8.5. Given ν ≥ −1/2, we have

(8.37) HνHνf = f,

for all f ∈ S(R+).

We promote this to

Proposition 8.6. If ν ≥ −1/2, we have a unique extension of Hν fromS(R+) to

(8.38) Hν : L2(R+, r2ν+1 dr) −→ L2(R+, λ2ν+1 dλ),

as a unitary operator, and (8.37) holds for all f ∈ L2(R+, r2ν+1 dr).

Proof. Take f, g ∈ S(R+), and use the inner product

(8.39) (f, g) =

∫ ∞

0

f(r)g(r)r2ν+1 dr.

Using Fubini’s theorem and the fact that Jν(λr)/(λr)ν is real valued andsymmetric in (λ, r), we get the first identity in

(8.40) (Hνf, Hνg) = (HνHνf, g) = (f, g),

the second identity following by Proposition 8.5. From here, given that thelinear space S(R+) ⊂ L2(R+, r2ν+1 dr) is dense, the assertions of Proposi-tion 8.6 are apparent.

We return to the Hankel transform (8.12). Note that

(8.41) Hν(rνf)(λ) = λνHνf(λ),

and that Mνf(r) = rνf(r) has the property that

(8.42) Mν : L2(R+, r2ν+1 dr) −→ L2(R+, r dr) is unitary.

Thus Proposition 8.6 yields Proposition 8.1.Another proof is sketched in the exercises. An elaboration of Hankel’s

original proof is given on pp. 456–464 of [Wat].


In view of (8.23) and (8.41), we have

(8.43)

Hν(r−αLµg) =

∫ ∞

0

Lµ(rαJν(λr))g rm dr

= −λ2

∫ ∞

0

grαJν(λr)rm dr

= −λ2Hν(r−αg).

Now from (8.5)–(8.13), it follows that the map H given by

(8.44) Hg =(Hν0

(r−αg0),Hν1(r−αg1), . . .

)

provides an isometry of L2(C(N)) onto L2(R+, λ dλ, ℓ2), such that ∆ iscarried into multiplication by −λ2. Thus (8.44) provides a spectral repre-sentation of ∆. Consequently, for well-behaved functions f , we have

(8.45)

f(−∆)g(r, x)

= rα∑

j

∫ ∞

0

f(λ2)Jνj(λr)λ

∫ ∞

0

s1−αJνj(λs)gj(s) ds dλ ϕj(x).

Now we can interpret (8.45) in the following fashion. Define the operator

ν on N by

(8.46) ν =(−∆N + α2

)1/2.

Thus νϕj = νjϕj . Identifying operators with their distributional kernels,we can describe the kernel of f(−∆) as a function on R

+×R+ taking values

in operators on N , by the formula

(8.47)f(−∆) = (r1r2)

α

∫ ∞

0

f(λ2)Jν(λr1)Jν(λr2)λ dλ

= K(r1, r2, ν),

since the volume element on C(N) is rm dr dS(x) if the m-dimensionalarea element of N is dS(x).

At this point it is convenient to have in hand some calculations of Han-kel transforms, including some examples of the form (8.47). We establishsome here; many more can be found in [Wat]. Generalizing (8.31), we cancompute

∫ ∞0

e−br2

Jν(λr)rµ+1 dr in a similar fashion, replacing the integralin (8.27) by

(8.48)

∫ ∞

0

r2k+µ+ν+1e−br2

dr =1

2b−k−µ/2−ν/2−1Γ

(µ

2+

ν

2+ k + 1

).


We get

(8.49)

∫ ∞

0

e−br2

Jν(λr)rµ+1 dr

= λν2−ν−1b−µ/2−ν/2−1∞∑

k=0

Γ(µ2 + ν

2 + k + 1)

k!Γ(k + ν + 1)

(−λ2

4b

)k

.

We can express the infinite series in terms of the confluent hypergeometricfunction, introduced in §7. A formula equivalent to (7.35) is

(8.50) 1F1(a; b; z) =Γ(b)

Γ(a)

∞∑

k=0

Γ(a + k)

Γ(b + k)

zk

k!,

since (a)k = a(a + 1) · · · (a + k − 1) = Γ(a + k)/Γ(a). We obtain, for Reb > 0, Re(µ + ν) > −2,(8.51)∫ ∞

0

e−br2

Jν(λr)rµ+1 dr

= λν2−ν−1b−µ/2−ν/2−1 Γ(µ2 + ν

2 + 1)

Γ(ν + 1)1F1

(µ

2+

ν

2+ 1; ν + 1;−λ2

4b

).

We can apply a similar attack when e−br2

is replaced by e−br, obtaining

(8.52)

∫ ∞

0

e−brJν(λr)rµ−1 dr

=(λ

2

)ν

b−µ−ν∞∑

k=0

Γ(µ + ν + 2k)

k!Γ(ν + k + 1)

(− λ2

2b2

)k

,

at least provided Re b > |λ|, ν ≥ 0, and µ + ν > 0; here we use

(8.53)

∫ ∞

0

e−brr2k+µ+ν−1 dr = b−2k−µ−νΓ(µ + ν + 2k).

The duplication formula for the gamma function (see (A.22) of Chapter 3)implies

(8.54) Γ(2k + µ + ν) = π−1/222k+µ+ν−1Γ(µ

2+

ν

2+ k

)Γ(µ

2+

ν

2+ k +

1

2

),

so the right side of (8.52) can be rewritten as

(8.55) π−1/2λν2µ−1b−µ−ν∞∑

k=0

Γ(µ2 + ν

2 + k)Γ(µ2 + ν

2 + 12 + k)

k!Γ(ν + 1 + k)

(−λ2

b2

)k

.


This infinite series can be expressed in terms of the hypergeometric func-tion, defined by

(8.56)

2F1(a1, a2; b; z) =∞∑

k=0

(a1)k(a2)k

(b)k

zk

k!

=Γ(b)

Γ(a1)Γ(a2)

∞∑

k=0

Γ(a1 + k)Γ(a2 + k)

Γ(b + k)

zk

k!,

for a1, a2 /∈ 0,−1,−2, . . . , |z| < 1. If we put the sum in (8.55) into thisform, and use the duplication formula, to write

Γ(a1)Γ(a2) = Γ(µ

2+

ν

2

)Γ(µ

2+

ν

2+

1

2

)= π1/22−µ−ν+1Γ(µ + ν),

we obtain

(8.57)

∫ ∞

0

e−brJν(λr)rµ−1 dr

=(λ

2

)ν

b−µ−ν Γ(µ + ν)

Γ(ν + 1)· 2F1

(µ

2+

ν

2,µ

2+

ν

2+

1

2; ν + 1;−λ2

b2

).

This identity, established so far for |λ| < Re b (and ν ≥ 0, µ + ν > 0),continues analytically to λ in a complex neighborhood of (0,∞).

To evaluate the integral (8.47) with f(λ2) = e−tλ2

, we can use the powerseries (8.10) for Jν(λr1) and for Jν(λr2) and integrate the resulting doubleseries term by term using (8.48). We get(8.58)∫ ∞

0

e−tλ2

Jν(r1λ)Jν(r2λ)λ dλ

=1

2t

(r1r2

4t

)ν ∑

j,k≥0

Γ(ν + j + k + 1)

Γ(ν + j + 1)Γ(ν + k + 1)

1

j!k!

(−r2

1

4t

)j(−r2

2

4t

)k

,

for any t, r1, r2 > 0, ν ≥ 0. This can be written in terms of the modifiedBessel function Iν(z), given by

(8.59) Iν(z) =(z

2

)ν ∞∑

k=0

1

k!Γ(ν + k + 1)

(z

2

)2k

.

One obtains the following, known as the Weber identity.

Proposition 8.7. For t, r1, r2 > 0,

(8.60)

∫ ∞

0

e−tλ2

Jν(r1λ)Jν(r2λ)λ dλ =1

2te−(r2

1+r22)/4t Iν

(r1r2

2t

).


Proof. The left side of (8.60) is given by (8.58). Meanwhile, by (8.59),the right side of (8.60) is equal to (1/2t)(r1r2/4t)ν times

(8.61)∑

ℓ,m≥0

1

ℓ!m!

(−r2

1

4t

)ℓ(−r2

2

4t

)m ∞∑

n=0

1

n!Γ(ν + n + 1)

(r1r2

4t

)2n

.

If we set yj = −r2j /4t, we see that the asserted identity (8.60) is equivalent

to the identity

(8.62)

∑

j,k≥0

Γ(ν + j + k + 1)

Γ(ν + j + 1)Γ(ν + k + 1)

1

j!k!yj1y

k2

=∑

ℓ,m,n≥0

1

ℓ!m!

1

n!Γ(ν + n + 1)yℓ+n1 ym+n

2 .

We compare coefficients of yj1y

k2 in (8.62). Since both sides of 8.62) are

symmetric in (y1, y2), it suffices to treat the case

(8.63) j ≤ k,

which we assume henceforth. Then we take ℓ + n = j, m + n = k and sumover n ∈ 0, . . . , j, to see that (8.62) is equivalent to the validity of(8.64)

j∑

n=0

1

(j − n)!(k − n)!n!Γ(ν + n + 1)=

Γ(ν + j + k + 1)

Γ(ν + j + 1)Γ(ν + k + 1)

1

j!k!,

whenever 0 ≤ j ≤ k. Using the identity

Γ(ν + j + 1) = (ν + j) · · · (ν + n + 1)Γ(ν + n + 1)

and its analogues for the other Γ-factors in (8.64), we see that (8.64) isequivalent to the validity of

(8.65)

j∑

n=0

j!k!

(j − n)!(k − n)!n!(ν+j) · · · (ν+n+1) = (ν+j+k) · · · (ν+k+1),

for 0 ≤ j ≤ k. Note that the right side of (8.65) is a polynomial of degreej in ν, and the general term on the left side of (8.65) is a polynomial ofdegree j − n in ν.

In order to establish (8.65), it is convenient to set

(8.66) µ = ν + j

and consider the associated polynomial identity in µ. With

(8.67)p0(µ) = 1, p1(µ) = µ, p2(µ) = µ(µ − 1), . . .

pj(µ) = µ(µ − 1) · · · (µ − j + 1),


we see that p0, p1, . . . , pj is a basis of the space Pj of polynomials ofdegree j in µ, and our task is to write

(8.68) pj(µ + k) = (µ + k)(µ + k − 1) · · · (µ + k − j + 1)

as a linear combination of p0, . . . , pj . To this end, define

(8.69) T : Pj −→ Pj , Tp(µ) = p(µ + 1).

By explicit calculation,

(8.70)p1(µ + 1) = p1(µ) + p0(µ),

p2(µ + 1) = (µ + 1)µ = µ(µ − 1) + 2µ = p2(µ) + 2p1(µ),

and an inductive argument gives

(8.71) Tpi = pi + ipi−1.

By convention we set pi = 0 for i < 0. Our goal is to compute T kpj . Notethat

(8.72) T = I + N, Npi = ipi−1,

and

(8.73) T k =

j∑

n=0

(k

n

)Nn,

if j ≤ k. By (8.72),

(8.74) Nnpi = i(i − 1) · · · (i − n + 1)pi−n,

so we have

(8.75)

T kpj =

j∑

n=0

(k

n

)j(j − 1) · · · (j − n + 1)pj−n

=

j∑

n=0

k!

(k − n)!n!

j!

(j − n)!pj−n.

This verifies (8.65) and completes the proof of (8.60).

Similarly we can evaluate (8.47) with f(λ2) = e−tλ/λ, as an infiniteseries, using (8.53) to integrate each term of the double series. We get(8.76)∫ ∞

0

e−tλJν(r1λ)Jν(r2λ) dλ

=1

t

(r1r2

t2

)ν ∑

j,k≥0

Γ(2ν + 2j + 2k + 1)

Γ(ν + j + 1)Γ(ν + k + 1)

1

j!k!

(− r2

1

4t2

)j(− r2

2

4t2

)k

,

provided t > rj > 0. It is possible to express this integral in terms of theLegendre function Qν−1/2(z).


Proposition 8.8. One has, for all y, r1, r2 > 0, ν ≥ 0,

(8.77)

∫ ∞

0

e−yλJν(r1λ)Jν(r2λ) dλ =1

π(r1r2)

−1/2Qν−1/2

(r21 + r2

2 + y2

2r1r2

).

The Legendre functions Pν−1/2(z) and Qν−1/2(z) are solutions to

(8.78)d

dz

[(1 − z2)

d

dzu(z)

]+

(ν2 − 1

4

)u(z) = 0;

Compare with (4.52). Extending (4.41), we can set

(8.79) Pν−1/2(cos θ) =2

π

∫ θ

0

(2 cos s − 2 cos θ

)−1/2cos νs ds,

and Qν−1/2(z) can be defined by the integral formula

(8.80) Qν−1/2(cosh η) =

∫ ∞

η

(2 cosh s − 2 cosh η

)−1/2e−sν ds.

The identity (8.77) is known as the Lipschitz-Hankel integral formula.

Proof of Proposition 8.8. We derive (8.77) from the Weber identity(8.60). Recall

(8.81) Iν(y) = e−πiν/2 Jν(iy), y > 0.

To work with (8.60), we use the subordination identity

(8.82) e−yλ =λ√π

∫ ∞

0

e−y2/4te−tλ2

t−1/2 dt;

cf. Chapter 3, (5.31) for a proof. Plugging this into the left side of (8.77),and using (8.60), we see that the left side of (8.77) is equal to

(8.83)1

2√

π

∫ ∞

0

e−(r21+r2

2+y2)/4t Iν

(r1r2

2t

)t−3/2 dt.

The change of variable s = r1r2/2t gives

(8.84)

√1

2π(r1r2)

−1/2

∫ ∞

0

e−s(r21+r2

2+y2)/2r1r2 Iν(s)s−1/2 ds.

Thus the asserted identity (8.77) follows from the identity

(8.85)

∫ ∞

0

e−szIν(s)s−1/2 ds =

√2

πQν−1/2(z), z > 0.


As for the validity of (8.85), we mention two identities. Recall from(8.57) that

(8.86)

∫ ∞

0

e−szJν(λs)sµ−1 ds =(λ

2

)ν

z−µ−ν Γ(µ + ν)

Γ(ν + 1)

· 2F1

(µ

2+

ν

2+

1

2,µ

2+

ν

2; ν + 1;−λ2

z2

).

Next, there is the classical representation of the Legendre function Qν−1/2(z)as a hypergeometric function:(8.87)

Qν−1/2(z) =Γ( 1

2 )Γ(ν + 12 )

Γ(ν + 1)(2z)−ν−1/2

2F1

(ν

2+

3

4,ν

2+

1

4; ν + 1;

1

z2

);

cf. [Leb], (7.3.7). If we apply (8.86) with λ = i, µ = 1/2 (keeping (8.81) inmind), then (8.85) follows.

Remark. Formulas (8.77) and (8.60) are proven in the opposite order in[W].

By analytic continuation, we can treat f(λ2) = e−ελλ−1 sin λt for anyε > 0. We apply this to (8.47). Letting ε ց 0, we get for the fundamentalsolution to the wave equation:

(8.88)

(−∆)−1/2 sin t(−∆)1/2

= − limεց0

(r1r2)α Im

∫ ∞

0

e−(ε+it)λ Jν(λr1)Jν(λr2) dλ

= − 1

π(r1r2)

α−1/2 limεց0

Im Qν−1/2

(r21 + r2

2 + (ε + it)2

2r1r2

).

Using the integral formula (8.80), where the path of integration is a suit-able path from η to +∞ in the complex plane, one obtains the followingalternative integral representation of (−∆)−1/2 sin t(−∆)1/2. The Schwartzkernel is equal to

0, if t < |r1 − r2|,(8.89)

1

π(r1r2)

α

∫ β1

0

[t2 − (r2

1 + r22 − 2r1r2 cos s)

]−1/2cos νs ds,(8.90)

if |r1 − r2| < t < r1 + r2, and

(8.91)1

π(r1r2)

α cos πν

∫ ∞

β2

[r21 + r2

2 + 2r1r2 cosh s − t2]−1/2

e−sν ds,

if t > r1 + r2, where

(8.92) β1 = cos−1(r2

1 + r22 − t2

2r1r2

), β2 = cosh−1

( t2 − r21 − r2

2

2r1r2

).


Recall that α = −(m − 1)/2, where m = dim N .We next show how formulas (8.89)–(8.91) lead to an analysis of the clas-

sical problem of diffraction of waves by a slit along the positive x-axis inthe plane R

2. In fact, if waves propagate in R2 with this ray removed, on

which Dirichlet boundary conditions are placed, we can regard the space asthe cone over an interval of length 2π, with Dirichlet boundary conditionsat the endpoints. By the method of images, it suffices to analyze the caseof the cone over a circle of circumference 4π (twice the circumference ofthe standard unit circle). Thus C(N) is a double cover of R

2 \ 0 in thiscase. We divide up the spacetime into regions I, II, and III, respectively,as described by (8.89), (8.90), and (8.91). Region I contains only pointson C(N) too far away from the source point to be influenced by time t;that the fundamental solution is 0 here is consistent with finite propagationspeed.

Since the circle has dimension m = 1, we see that

(8.93) ν = (−∆N )1/2 =(− d2

dθ2

)1/2

in this case if θ ∈ R/(4πZ) is the parameter on the circle of circumference4π. On the line, we have

(8.94) cos sν δθ1(θ2) =

1

2

[δ(θ1 − θ2 + s) + δ(θ1 − θ2 − s)

].

To get cos sν on R/(4πZ), we simply make (8.94) periodic by the method ofimages. Consequently, from (8.90), the wave kernel (−∆)−1/2 sin t(−∆)1/2

is equal to

(8.95)(2π)−1

[t2 − r2

1 − r22 + 2r1r2 cos(θ1 − θ2)

]−1/2if |θ1 − θ2| ≤ π,

0 if |θ1 − θ2| > π,

in region II. Of course, for |θ1 − θ2| < π this coincides with the free spacefundamental solution, so (8.95) also follows by finite propagation speed.

We turn now to an analysis of region III. In order to make this analysis,it is convenent to make simultaneous use both of (8.91) and of anotherformula for the wave kernel in this region, obtained by choosing anotherpath from η to ∞ in the integral representation (8.80). The formula (8.91)is obtained by taking a horizontal line segment; see Fig. 8.1.

If instead we take the path indicated in Fig. 8.2, we obtain the followingformula for (−∆)−1/2 sin t(−∆)1/2 in region III:

(8.96)

π−1(r1r2)−(m−1)/2

∫ π

0

(t2 − r2

1 − r22 + 2r1r2 cos s

)−1/2cos sν ds

− sin πν

∫ β2

0

(t2 − r2

1 − r22 − 2r1r2 cosh s

)−1/2e−sν ds

.


Figure 8.1

Figure 8.2

The operator ν on R/(4πZ) given by (8.93) has spectrum consisting of

(8.97) Spec ν =

0,1

2, 1,

3

2, 2, . . .

,

all the eigenvalues except for 0 occurring with multiplicity 2. The formula(8.91) shows the contribution coming from the half-integers in Spec ν van-ishes, since cos 1

2πn = 0 if n is an odd integer. Thus we can use formula(8.96) and compose with the projection onto the sum of the eigenspaces ofν with integer spectrum. This projection is given by

(8.98) P = cos2 πν

on R/(4πZ). Since sinπn = 0, in the case N = R/(4πZ) we can rewrite(8.96) as

(8.99) π−1(r1r2)−(m−1)/2

∫ π

0

(t2 − r2

1 − r22 + 2r1r2 cos s

)−1/2P cos sν ds.


In view of the formulas (8.94) and (8.96), we have

(8.100)

P cos sν δθ1(θ2)

=1

4

[δ(θ1 − θ2 + s) + δ(θ1 − θ2 − s)

+ δ(θ1 − θ2 + 2π + s) + δ(θ1 − θ2 + 2π − s)]

mod 4π.

Thus, in region III, we have for the wave kernel (−∆)−1/2 sin t(−∆)1/2 theformula

(8.101) (4π)−1(t2 − r2

1 − r22 + 2r1r2 cos(θ1 − θ2)

)−1/2.

Thus, in region III, the value of the wave kernel at points (r1, θ1), (r2, θ2)of the double cover of R

2 \ 0 is given by half the value of the wave kernelon R

2 at the image points. The jump in behavior from (8.95) to (8.101)gives rise to a diffracted wave.

Figure 8.3

We depict the singularities of the fundamental solution to the wave equa-tion for R

2 minus a slit in Figs. 8.3 and 8.4. In Fig. 8.3 we have the situ-ation |t| < r1, where no diffraction has occurred, and region III is empty.In Fig. 8.4 we have a typical situation for |t| > r1, with the diffracted wavelabeled by a “D.”

This diffraction problem was first treated by Sommerfeld [Som] andwas the first diffraction problem to be rigorously analyzed. For other ap-proaches to the diffraction problem presented above, see [BSU] and [Stk].

Generally, the solution (8.89)–(8.91) contains a diffracted wave on theboundary between regions II and III. In Fig. 8.5 we illustrate the diffractionof a wave by a wedge; here N is an interval of length ℓ < 2π. We now wantto provide, for general N , a description of the behavior of the distributionv = (−∆)−1/2 sin t(−∆)1/2 δ(r2,x2) near this diffracted wave, that is, astudy of the limiting behavior as r1 ց t − r2 and as r1 ր t − r2.

We begin with region II. From (8.90), we have v equal to

(8.102)1

2(r1r2)

α−1/2Pν−1/2(cos β1) δx2in region II,


Figure 8.4

Figure 8.5

where Pν−1/2 is the Legendre function defined by (8.79) and β1 is given by(8.92). Note that as r1 ց t − r2, β1 ր π.

To analyze (8.102), replace s by π − s in (8.79), and, with δ1 = π − β1,write

(8.103)

π

2Pν−1/2(cos β1) = cos πν

∫ π

δ1

(2 cos δ1 − 2 cos s

)−1/2cos sν ds

+ sin πν

∫ π

δ1

(2 cos δ1 − 2 cos s

)−1/2sin sν ds.


As δ1 ց 0, the second term on the right tends in the limit to

(8.104) sinπν

∫ π

0

sin sν

sin 12s

ds.

Write the first term on the right side of (8.103) as

(8.105)

cos πν

∫ π

δ1

(2 cos δ1 − 2 cos s)−1/2(cos sν − 1) ds

+ cos πν

∫ π

δ1

(2 cos δ1 − 2 cos s)−1/2 ds.

As δ1 ց 0, the first term here tends in the limit to

(8.106) cos πν

∫ π

0

cos sν − 1

sin 12s

ds.

The second integral in (8.105) is a scalar, independent of ν, and it is easilyseen to have a logarithmic singularity. More precisely,

(8.107)

∫ π

δ1

(2 cos δ1 − 2 cos s)−12 ds

∼(log

2

δ1

) ∞∑

j=0

Ajδj1 +

∞∑

j=1

Bjδj1, A0 = 1.

Consequently, one derives the following.

Proposition 8.9. Fix (r2, x2) and t. Then, as r1 ց t − r2,

(8.108)

(−∆)−1/2 sin t(−∆)1/2 δ(r2,x2)

=1

π(r1r2)

α−1/2

log2

δ1cos πν δx2

+

∫ π

0

cos sν − cos πν

2 cos 12s

ds δx2+ R1 δx2

,

where, for s > (m + 1)/2,

(8.109) ‖R1 δx2‖D−s−1 ≤ Cδ1 log

1

δ1, as δ1 ց 0.

The following result analyzes the second term on the right in (8.108).

Proposition 8.10. We have

(8.110)

∫ π

0

(2 cos

1

2s)−1

(cos sν − cos πν) ds

= cos πν− log ν +

K∑

j=0

ajν−2j

+

π

2sin πν + SK(ν),


where SK(ν) : Ds → Ds+2K , for all s.

The spaces Ds are spaces of generalized functions on N , introduced inChapter 5, Appendix A.

We turn to the analysis of v in region III. Using (8.91), we can write vas

(8.111)1

π(r1r2)

α−1/2 cos πν Qν−1/2(cosh β2) δx2, in region III,

where Qν−1/2 is the Legendre function given by (8.80) and β2 is given by(8.92). It is more convenient to use (8.96) instead; this yields for v theformula

(8.112)

1

π(r1r2)

α−1/2 ∫ π

0

(2 cosh β2 + 2 cos s)−1/2 cos sν ds

− sin πν

∫ β2

0

(2 cosh β2 − 2 cosh s)−1/2e−sν ds

.

Note that as r1 ր t − r2, β2 ց 0.The first integral in (8.112) has an analysis similar to that arising in

(8.103); first replace s by π − s to rewrite the integral as

(8.113)

cos πν

∫ π

0

(2 cosh β2 − 2 cos s)−1/2 cos sν ds

+ sinπν

∫ π

0

(2 cosh β2 − 2 cos s)−1/2 sin sν ds.

As β2 ց 0, the second term in (8.113) tends to the limit (8.104), andthe first term also has an analysis similar to (8.105)–(8.107), with (8.107)replaced by

(8.114)

∫ π

0

(2 cosh β2 − 2 cos s)−1/2 ds

∼(log

2

β2

)∑

j≥0

A′jβ

j2 +

∑

j≥1

B′jβ

j2, A′

0 = 1.

It is the second term in (8.112) that leads to the jump across r1 = t − r2,hence to the diffracted wave. We have

(8.115)

∫ β2

0

(2 cosh β2 − 2 cosh s)−1/2e−sν ds ∼∫ β2

0

ds√β2

2 − s2=

π

2.

Thus we have the following:


Proposition 8.11. For r1 ր t − r2,

(8.116)

(−∆)−1/2 sin t(−∆)1/2 δ(r2,x2)

=1

π(r1r2)

α−1/2

log2

β2cos πν δx2

+

∫ π

0

cos sν − cos πν

2 cos 12s

ds δx2− π

2sinπν δx2

+ R1δx2

,

where, for s > (m + 1)/2,

(8.117) ‖R1δx2‖D−s−1 ≤ Cβ2 log

1

β2, as β2 ց 0.

Note that (8.116) differs from (8.108) by the term π−1(r1r2)α−1/2 times

(8.118) −π

2sin πν δx2

.

This contribution represents a jump in the fundamental solution across thediffracted wave D. There is also the logarithmic singularity, (r1r2)

α−1/2

times

(8.119)1

πlog

2

δcos πν δx2

,

where δ = δ1 in (8.108) and δ = β2 in (8.116). In the special case whereN is an interval [0, L], so dim C(N) = 2, cos πν δx2

is a sum of two deltafunctions. Thus its manifestation in such a case is subtle.

We also remark that if N is a subdomain of the unit sphere S2k (of even

dimension), then cos πν δx2vanishes on the set N \ N0, where

(8.120) N0 = x1 ∈ N : for some y ∈ ∂N, dist(x2, y) + dist(y, x1) ≤ π.Thus the log blow-up disappears on N \N0. This follows from the fact thatcos πν0 = 0, where ν0 is the operator (8.46) on S2k, together with a finitepropagation speed argument.

While Propositions 8.9–8.11 contain substantial information about thenature of the diffracted wave, this information can be sharpened in a num-ber of respects. A much more detailed analysis is given in [CT].

Exercises

1. Using (7.36) and (7.80), work out the asymptotic behavior of 1F1(a; b;−z) asz → +∞, given b, b−a /∈ 0,−1,−2, . . . . Deduce from (8.51) that wheneverν ≥ 0, s ∈ R,

(8.121) limbց0

Z ∞

0

e−br2

Jν(r)r−is dr = 2−isΓ

“

12(ν + 1 − is)

”

Γ“

12(ν + 1 + is)

” .

Exercises 75

2. Define operators

(8.122) Mrf(r) = rf(r), J f(r) = f(r−1).

Show that(8.123)Mr : L2(R+, r dr) −→ L2(R+, r−1 dr), J : L2(R+, r−1 dr) −→ L2(R+, r−1 dr)

are unitary. Show that

(8.124) H#ν = JMrHνM−1

r

is given by

(8.125) H#ν f(λ) = (f ⋆ ℓν)(λ),

where ⋆ denotes the natural convolution on R+, with Haar measure r−1dr:

(8.126) (f ⋆ g)(λ) =

Z ∞

0

f(r)g(r−1λ)r−1 dr,

and

(8.127) ℓν(r) = r−1Jν(r−1).

3. Consider the Mellin transform:

(8.128) M#f(s) =

Z ∞

0

f(r)ris−1 dr.

As shown in (A.17)–(A.20) of Chapter 3, we have

(8.129) (2π)−1/2M# : L2(R+, r−1 dr) −→ L2(R, ds), unitary.

Show that

(8.130) M#(f ⋆ g)(s) = M#f(s) · M#g(s),

and deduce that

(8.131) M#H#ν f(s) = Ψ(s)M#f(s),

where(8.132)

Ψ(s) =

Z ∞

0

Jν(r−1)ris−2 dr =

Z ∞

0

Jν(r)r−is dr = 2−isΓ

“

12(ν + 1 − is)

”

Γ“

12(ν + 1 + is)

” .

4. From (8.126)–(8.132), give another proof of the unitarity (8.13) of Hν . Usingsymmetry, deduce that spec Hν = −1, 1, and hence deduce again theinversion formula (8.14).

5. Verify the asymptotic expansion (8.107). (Hint: Write 2 cos δ−2 cos s = (s2−δ2)F (s, δ) with F smooth and positive, F (0, 0) = 1. Then, with G(s, δ) =F (s, δ)−1/2,

(8.133)

Z π

δ

(2 cos δ − 2 cos s)−1/2 ds =

Z π

δ

G(s, δ)ds√

s2 − δ2.


Write G(s, δ) = g(s) + δH(s, δ), g(0) = 1, and verify that (8.133) is equal toA1 + A2, where

A1 =

Z π

δ

G(s, δ)ds

s= g(0) log

1

δ+ O

„

δ log1

δ

«

,

A2 =

Z π

δ

g(s)

»

1√s2 − δ2

− 1

s

–

ds + O(δ) = B2 + O(δ).

Show that

B2 = g(0)

Z π/δ

1

»

1√t2 − 1

− 1

t

–

dt + O(δ) = C2 + O(δ),

with

C2 =

Z ∞

1

»

1√t2 − 1

− 1

t

–

dt.

Use the substitution t = cosh u to do this integral and get C2 = log 2.)Next, verify the expansion (8.114).

Exercises on the hypergeometric function

1. Show that 2F1(a1, a2; b; z), defined by (8.56), satisfies

(8.134) 2F1(a1, a2; b; z) =Γ(b)

Γ(a2)Γ(b − a2)

Z 1

0

ta2−1(1 − t)b−a2−1(1 − tz)−a1 dt,

for Re b > Re a2 > 0, |z| < 1. (Hint: Use the beta function identity,(A.23)–(A.24) of Chapter 3, to write

(a2)k

(b)k=

Γ(b)

Γ(a2)Γ(b − a2)

Z 1

0

ta2−1+k(1 − t)b−a2−1 dt, k = 0, 1, 2, . . . ,

and substitute this into (8.39). Then use

∞X

k=0

(a1)k

k!(zt)k = (1 − tz)−a1 , 0 ≤ t ≤ 1, |z| < 1.)

2. Show that, given Re b > Re a2 > 0, (8.134) analytically continues in z toz ∈ C \ [1,∞).

3. Show that the function (8.134) satisfies the ODE

z(1 − z)d2u

dz2+

n

b − (a1 + a2 + 1)zodu

dz− a1a2u = 0.

Note that u(0) = 1, u′(0) = a1a2/b, but zero is a singular point for thisODE. Show that another solution is

u(z) = z1−b2F1(a1 − b + 1, a2 − b + 1; 2 − b; z).

4. Show that

2F1(a1, a2; b; z) = (1 − z)−a12F1

“

a1, b − a2; b; (z − 1)−1z”

.

References 77

(Hint: Make a change of variable s = 1 − t in (8.134).)For many other important transformation formulas, see [Leb] or [WW].

5. Show that

1F1(a; b; z) = limcր∞

2F1(a, c; b; c−1z).

We mention the generalized hypergeometric function, defined by

pFq(a; b; z) =

∞X

k=0

(a)k

(b)k

zk

k!,

where p ≤ q + 1, a = (a1, . . . , ap), b = (b1, . . . , bq), bj ∈ C \ 0,−1,−2, . . . ,|z| < 1, and

(a)k = (a1)k · · · (ap)k, (b)k = (b1)k · · · (bq)k,

and where, as before, for c ∈ C, (c)k = c(c + 1) · · · (c + k − 1). For more onthis class of functions, see [Bai].

6. The Legendre function Qν−1/2(z) satisfies the identity (8.87), for ν ≥ 0, |z| >1, and |Arg z| < π; cf. (7.3.7) of [Leb]. Take z = (r2

1 + r22 + t2)/2r1r2, and

compare the resulting power series for the right side of (8.77) with the powerseries in (8.76).

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