pdfs.semanticscholar.org · DUKE MATHEMATICAL JOURNAL Vol. 114, No. 2, c 2002 WAVE KERNELS RELATED TO SECOND-ORDER OPERATORS PETER C. GREINER, DAVID HOLCMAN, and YAKAR KANNAI Abstract

DUKE MATHEMATICAL JOURNALVol. 114, No. 2,c© 2002

WAVE KERNELS RELATED TO SECOND-ORDEROPERATORS

PETER C. GREINER, DAVID HOLCMAN, and YAKAR KANNAI

AbstractThe wave kernel for a class of second-order subelliptic operators is explicitly com-puted. This class contains degenerate elliptic and hypo-elliptic operators (such asthe Heisenberg Laplacian and the Grusin operator). Three approaches are used tocompute the kernels and to determine their behavior near the singular set. The for-mulas are applied to study propagation of the singularities. The results are expressedin terms of the real values of a complex function extending the Carnot-Caratheodorydistance, and the geodesics of the associated sub-Riemannian geometry play a crucialrole in the analysis.

Contents1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .330

1.1. Sub-Riemannian (Carnot-Caratheodory) metrics. . . . . . . . . . 3331.2. A curve in the complex plane and the complex action. . . . . . . . 3371.3. Lorentz-Carnot-Caratheodory metric. . . . . . . . . . . . . . . . 3401.4. Separation of variables. . . . . . . . . . . . . . . . . . . . . . . 341

2. The wave kernel for the Grusin operator . . . . . . . . . . . . . . . . . 3432.1. The boundary of the forbidden set. . . . . . . . . . . . . . . . . 3462.2. The zero of the phase and the integration path. . . . . . . . . . . 3482.3. Explicit formula for the integral . . . . . . . . . . . . . . . . . . 3492.4. Analysis of the wave kernel near the singularities. . . . . . . . . . 3532.5. The operator satisfies the wave equation. . . . . . . . . . . . . . 3552.6. Source not at the origin. . . . . . . . . . . . . . . . . . . . . . 357

3. Wave kernels in one dimension. . . . . . . . . . . . . . . . . . . . . . 3583.1. Wave kernel for the harmonic oscillator. . . . . . . . . . . . . . 3583.2. Wave kernel for the Klein-Gordon operator. . . . . . . . . . . . . 360

4. The Heisenberg wave kernel via the heat kernel. . . . . . . . . . . . . . 3614.1. Deforming the path of integration. . . . . . . . . . . . . . . . . 363

DUKE MATHEMATICAL JOURNALVol. 114, No. 2, c© 2002Received 10 April 2001. Revision received 7 September 2001.2000Mathematics Subject Classification. Primary 35L80, 53C17; Secondary 35H20.Holcman and Kannai’s work supported by the Minerva Foundation, Germany.

329

330 GREINER, HOLCMAN, and KANNAI

4.2. Computation of the wave kernel. . . . . . . . . . . . . . . . . . 3655. Wave kernels via the continuation method. . . . . . . . . . . . . . . . 371

5.1. The Heisenberg Laplacian. . . . . . . . . . . . . . . . . . . . . 3735.2. Degenerate elliptic operators. . . . . . . . . . . . . . . . . . . . 376

6. Directions for further studies. . . . . . . . . . . . . . . . . . . . . . . 385References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .386

1. IntroductionWe study in this paper properties of fundamental solutions of wave equations associ-ated with several subelliptic second-order self-adjoint operatorsL. We give an explicitexpression for the Grusin operator, the Heisenberg Laplacian, and the harmonic os-cillator.

Recall that the general solution of the wave equation

∂2wL

∂t2= −LwL ,

wL(0) = f,

∂wL(0)

∂t= g, (1.1)

has the formal expression

wL(t) =sin(t L1/2)

L1/2g + cos(t L1/2) f. (1.2)

It suffices to compute the fundamental solutionwL that satisfies equation (1.1) andwL(0) = 0, ∂wL(0)/∂t = δ0, whereδ0 denotes the Dirac distribution at the originzero, that is,

wL(t) =sin(L1/2t)

L1/2, (1.3)

or to compute∂wL/∂t = cos(t L1/2)—the solution of (1.1) wherewL(0) = δ0,∂wL(0)/∂t = 0.

We consider the wave kernel for a number of second-order operators. We obtainin some cases an explicit representation for the kernel and some information aboutpropagation of singularities.

The wave kernel for the standard wave equation in the generaln-dimensionalEuclidean spaceRn was computed first by J. Hadamard (see R. Courant and D. Hilbert[5]), who considered more generally the case whereL is elliptic. The kernel is of theform δ(p)(t2

− |x|2) for the usual (Euclidean) Laplacian. In the general elliptic case,

the solution is represented as a sum of terms, starting from the less regular to the moreregular (see [5]). Once again the leading term is proportional toδ(p)(t2

−|x|2), where

WAVE KERNELS RELATED TO SECOND-ORDER OPERATORS 331

now |x| is the distance betweenx and the origin in the Riemannian metric associatedwith the second-order elliptic operator.

The wave equation for a subellipticL has been considered in [10]. The finitespeed of propagation was established in [12]. The first systematic analysis of thewave kernel for the Heisenberg Laplacian was performed in [13], where the “lightcone” was described and propagation of singularities discussed. The computations aresomewhat complicated. In [15] the broad features of the propagation of singularitiesfor the Heisenberg Laplacian are also described using a different method based ondeforming the path of integration in the complex plane. The geometry of the lightcone is rather different from the standard Euclidean case.

It turns out that finite speed of propagation is associated with the so-calledCarnot-Caratheodory distance defined as the sub-Riemannian length of a minimiz-ing geodesic (see [4], [14]); the formula for the wave kernel, and the full light cone,involve all geodesics. Sub-Riemannian geometry differs substantially (even locally)from usual Riemannian geometry. On the other hand, a complex-valued functionf ,appearing in the integral representation for the heat kernel on Heisenberg group (see[6]), was shown in [1] to satisfy a Hamilton-Jacobi equation with the symbol of theHeisenberg Laplacian as Hamiltonian. Critical points and critical values of this func-tion f (extended analytically to the complex plane) correspond to sub-Riemanniangeodesics and their lengths, respectively. A curve on which the functionf is realis constructed in [15] and in [1]. Our formulas involve integration along this curve.Moreover, adding a time-dependent term tof , we obtain a complex phase satisfyinga Hamilton-Jacobi equation with the symbol of the wave operator as Hamiltonian.

Observe that in both [13] and [15] propagation of singularities is studied withoutactually computing the wave kernel. In [13] the kernel is given as a limit of expres-sions containing integrals (or an infinite series); one could presumably get a closedform with extra effort. No attempt at calculating the kernel is made in [15]; the ap-pearance of fractional powers in [15, (8.7)] makes explicit computations difficult. Oneof the main purposes of the present paper is to obtain a more explicit formula for theHeisenberg wave kernel. Known properties of singularities (such as propagation) arethen easily obtained. Moreover, the relationship between the sub-Riemannian geom-etry and complex integration formulas (such as in [6] and [1]) is put into context.

Explicit formulas for model operators (such as the wave kernels for the Heisen-berg Laplacian or the Grusin operator), while interesting in their own right, may alsooffer new insights into the problem and may serve as principal terms in approxima-tions for more general cases.

Three methods are applied in this paper for explicit computation of the wavekernels. The first involves separation of variables, summation of series containingHermite polynomials, and deformation of integration path in the complex plane. This


approach is utilized in Section 2 for solving (1.1) if −L/2 is the Grusin operator inR2,

−L

2=

1

2

( ∂2

∂x2+ x2 ∂

2

∂y2

), (1.4)

and in Section 3 for the case where−L/2 is the Harmonic oscillator inR1,

−L

2=

1

2

( ∂2

∂x2− x2

). (1.5)

As an additional illustration, we solve in Section 3 the Klein-Gordon equation. Theintegral representing the wave kernel for the Grusin operator may be evaluated explic-itly by the residue theorem (Proposition4), and all properties (geometry of the lightcone, behavior near this cone, band structure) may be read off the resulting (ratherexplicit) formula. On the other hand, the formula for the harmonic oscillator (Propo-sition 7) involves an integral over a path where the phase is purely imaginary, and itseems that the integral may not be easily evaluated. To the best of our knowledge,wave kernels for the Grusin operator and for the harmonic oscillator were not calcu-lated before, nor was the propagation of singularities for the Grusin operator studiedin detail.

The second method involves inversion of the transmutation formula (Proposition9; see [8]) and deforming the integration path used in the integral formula for the heatkernel from the real axis to a path in the complex plane where the exponent is real.This approach is described in Section 4 and applied for the case where−L/2 is theHeisenberg Laplacian1H defined onR2n+1

= {(x1, . . . , x2n, x0)} by

1H =1

2

( n∑j =1

( ∂

∂x2 j −1+2α j x2 j

∂

∂x0

)2+

1

2

n∑j =1

( ∂

∂x2 j−2α j x2 j −1

∂

∂x0

)2), (1.6)

whereα1, . . . , αn are positive constants. Throughout most of the paper we considerthe isotropic case in which all theα j ’s are equal to a constantα (see, e.g., (4.11),(5.15)). (We comment on the extension to the general anisotropic case at variouspoints in the paper.) The expression for the wave kernel using a complex contour isgiven in Theorem2 (formula (4.14)). One may rewrite the formula using integrationover real intervals (Theorem3). The role of the geodesics emerges clearly, as theintegration is performed over intervals of the form(d2

2k−1,d22k), whered j is the length

of the j th geodesic. One may obtain a closed form (not involving integrations) ifx = (x1, . . . , x2n) = 0 (Theorem4). The leading singularities of the wave kernel arecalculated (and are compatible with the results of [13]). Observe that in Section 4 wedeal with the kernel of cos(t L1/2), unlike the rest of the paper where sin(t L1/2)/L1/2

is treated.


The third method is based on an analytic continuation of the Green function ofthe operator−L + ∂2/∂y2 and uses an idea due to M. Taylor [15]. This method is de-scribed in Section 5 and applied to the Heisenberg Laplacian, as well to the case where−L is a degenerate elliptic operator of the type studied in [2]. This class contains theGrusin operator and the Baouendi-Goulaouic operator as special cases. While the re-sults for the Heisenberg Laplacian parallel those of Section 4, new phenomena occurfor degenerate operators—if certain dimensions are odd (as is the case for the Grusinoperator), the integral representing the kernel is computable (by the residue theoremor otherwise) and yields a simple expression for the kernel. The results coincide withthose of Section 2.

There is of course a certain amount of redundancy in rederiving the same resultsby different methods. We feel, however, that each method has its advantages. Thusseparation of variables is directly applicable to the harmonic oscillator; using the heatkernel, we may compute cos(t L1/2) directly; and analytic continuation of the Greenfunction enables a straightforward calculation for degenerate elliptic operators with-out prior computation of the associated heat kernel. Moreover, one should not forgetthat separation of variables underlies the computation of heat kernels in [15], as wellas that of the Green kernels in [3].

An entirely different method for computing wave kernels for certain second-orderoperators was suggested in [9]. The method is based on transmutation formulas andon the Trotter product formula. Some kind of a “Feynman integral representation” isobtained, and the expression for a wave kernel involves differentiating to a high ordera very high-dimensional integral. In [9] expressions were obtained for the wave kernelcos(t L1/2) where−L/2 is the harmonic oscillator and when−L/2 is the HeisenbergLaplacian. A direct proof of the identity of the expressions from [9] with the expres-sion obtained here appears to be nontrivial.

In the remainder of this section we collect some preliminary material concerningsub-Riemannian geometry, complex action, and separation of variables.

1.1. Sub-Riemannian (Carnot-Caratheodory) metricsRecall the definition of sub-Riemannian (Carnot-Caratheodory, also known as C-C)metrics (see [4, pp. 4 – 7]): LetX1, . . . , Xm be smooth vector fields on a manifoldM .For x ∈ M andv ∈ Tx M ,

‖v‖2x = inf

{u2

1 + · · · + u2m s.t.u1X1(x)+ · · · + umXm(x) = v

}. (1.7)

In particular,‖v‖2x = ∞ if v is not contained in sp(X1, . . . , Xm).

The lengthl (c) of an absolutely continuous curvec(t) (a ≤ t ≤ b) contained inM (absolute continuityis well defined in terms of local charts) is given by the integral∫ ba ‖c(t)‖c(t) dt, and the energy ofc is equal to

∫ ba ‖c(t)‖2

c(t) dt.


The distance between two points is defined byd(p,q) = inf l (c), where theinfimum is taken over all absolutely continuous curves joiningp andq.

In this paper we always assume that the vector fieldsX1, . . . , Xm and their brack-ets[Xi , X j ], 1 ≤ i, j ≤ m, span the tangent spaceTx M at every point ofM , andM isconnected. By Chow’s theorem (see [4, p. 15]), any two points inM can be joined byan absolutely continuous curve with finite length. Henced(p,q) < ∞ for any pointsp,q ∈ M . (Note that we consider here only the so-called step-two case.)

We can define the Hamiltonian associated with the sub-Riemannian metric by

H(x, ξ) =1

2

m∑i =1

〈Xi , ξ〉2. (1.8)

Note that we do not have a finite metric defined on the tangent bundle; we areforced to study the cotangent bundle. It is well known that any two pointsp,q maybe joined by a curve whose length equalsd(p,q). Thus the distance betweenp andq is attained as the length of a minimizing geodesic joiningp andq. Moreover, thegeodesic curves are projections ontoM of bicharacteristics of HamiltonianH (see[14], [1]). Observe that if we normalized the “time” to be equal to 1, thend2(p,q) isequal to the energy of the minimizing geodesic joiningp with q and is also equal totwice the actionScomputed along the corresponding bicharacteristic.

Perhaps the simplest example of a sub-Riemannian metric is the metric associatedwith the Grusin operator. In the Grusin plane,R2, the sub-Riemannian metric is givenby the vectors

X1 =

(10

), X2 =

(0x

).

The vector fields span the tangent plane everywhere, except along the linex = 0. Butsince

[X1, X2] =

(01

),

Chow’s conditions are satisfied and it follows that the sub-Riemannian distance be-tween any two points is finite (see [4, p. 24]).

In the complement of the linex = 0, the sub-Riemannian metric is Riemannian,G = (R2,ds), whereds2

= dx2+ dy2/x2. The Hamiltonian is given by

H(x, y, η, ξ) =1

2(ξ2

+ x2η2) (1.9)

and is equal to the symbol of the Grusin operator.The distance between two points is

d(P, Q) = infc(t)∈C1([0,1],G), c(0)=P, c(1)=Q

∫ 1

0‖c(t)‖c(t) dt. (1.10)


A simple computation yields the Euler-Lagrange equations

y

x2= b,

x +y2

x3= 0, (1.11)

whereb is a constant. All geodesics may be computed explicitly from (1.11). In par-ticular, the geodesics starting at the origin are given by

x(t) =c

bsin(bt),

y(t) =c2

b

( t

2−

sin(2bt)

4b

), (1.12)

whereb andc are arbitrary real parameters. It is easy to see that these geodesics areprojections of certain bicharacteristics—the solutions of the system

x =∂H(x, y, η, ξ)

∂ξ= ξ,

ξ = −∂H(x, y, η, ξ)

∂x= −xη2,

y =∂H(x, y, η, ξ)

∂ξ= x2η,

η = −∂H(x, y, η, ξ)

∂y= 0,

(1.13)

with the initial conditionsx(0) = y(0) = 0, ξ(0) = c, η(0) = b. A similar systemwas studied in [6] and in [13] for the Heisenberg group.

Observe that

y

x2=

b

2

( t

sin2(bt)−

cos(bt)

bsin(bt)

)=

1

2µ(θ), (1.14)

where

µ(θ) =θ

sin2 θ− cot(θ), (1.15)

andθ = bt.It follows that if x 6= 0, then for every solution of the equation

2y

x2= µ(θ) (1.16)

there corresponds a geodesic joining the origin with the point(x, y). The graph of thefunctionµ is portrayed in Figure 1. The Hamiltonian is constant along any bichar-


0 2 4 6 8 10 12 14x

2

4

6

8

10

12

14

16

18

Figure 1. φ → φ/sin2 φ − cotφ

acteristic. For a geodesic starting at the origin,H is equal to(1/2)ξ(0)2 = (1/2)c2.Hence the energy is equal to

2S(x, y) =

∫ t

02H(t)dt =

∫ t

0c2 dt = c2t =

c2θ

b, (1.17)

and a similar computation shows that the length of the geodesic is equal tocθ/b. (Notethat onceθ is found from (1.16), c/b is obtained from the first equation in (1.12).) Ifx 6= 0, then the number of geodesics joining(x, y) with the origin is finite and growsfrom 1 to∞ asy/x2 varies from 0 to∞. If x = 0, thenµ(θ) = ∞ so thatθ = kπfor any integerk > 0. Correspondingly, there exist infinitely many geodesics joiningthe origin to(0, y) with lengths satisfyingd2

= 2πk|y|. (Herec/b is calculated fromthe second equation of (1.12).)

Another example is the Heisenberg groupHn. The Carnot-Caratheodory metricassociated with the left-invariant vector fields has been discussed in detail (see [13],[1], [4]). In particular, the Hamiltonian is the symbol of the Heisenberg Laplacian. Ifx = (x1, . . ., x2n) 6= 0, then there exist (in the isotropic case) finitely many geodesicsjoining the origin with(x1, . . ., x2n, x0) parametrized by the solutions of the equationµ(θ) = 2x0/r 2, wherer 2

= |x|2

=∑2n

i =1 x2i . If x = 0, then there exist infinitely


many geodesics joining the origin with(0, x0) parametrized by the Cartesian productof S2n−1 with the set of nonzero integers. Otherwise, the computations are similar tothe case of the Grusin plane.

Degenerate elliptic operators of the kind studied in Section 5.2 form a general-ization of the Grusin operator. A subclass consists of operators of the form

L = −

(( ∂

∂x1

)2+

( ∂

∂x2

)2+ |x1|

2 ∂2

∂y2

), (1.18)

wherexi ∈ Vi (i = 1,2), V1,V2 are real Euclidean vector spaces, and∂/∂xi denotesthe gradient inVi (i = 1,2). In order to cover both the Heisenberg Laplacian and thedegenerate elliptic operators, let us replacey by x0/a, wherea is a positive constant.In the case of operators described by (1.18), we setr 2

= |x1|2. The Hamiltonian is

the symbol of the operator. Once again we treat here only geodesics starting at theorigin. The following propositions hold in all cases.

PROPOSITION1There is a finite number of geodesics joining the origin with(x, x0) if and only if r 6=

0. They are parametrized by the solutionsθ of equation (1.16). Their lengths increasewith θ . Withφ1 denoting the first critical value ofµ, there is only one geodesic if andonly if 2|x0|/r 2 < µ(φ1). The number of geodesics increases to∞ with |x0|/r 2.

The C-C distance dc(x, x0) between the origin and the point(x, x0) is given bythe length of the shortest geodesic joining these points and d2

c (x, x0) = 2S(x, |x0|),where S is the action along the shortest geodesic.

PROPOSITION2There is an infinite number of geodesics that join the origin to the point(0, x0) oflength

d2k = 2πk

|x0|

a, k = 1,2, . . . . (1.19)

The Carnot-Caratheodory distance from the origin to(0, x0) is given by dc(0, x0) =√

2π(|x0|/a).

These propositions are proved in [1] for the Heisenberg case; the degenerate ellipticcase is very similar.

1.2. A curve in the complex plane and the complex actionWe continue to use the notation introduced in Section 1.1. Thus letx denote either thevector(x1, . . ., x2n) (the Heisenberg group case) or the vector(x1, x2) ∈ V1,V2 (thegeneralized Grusin case), and letx0 ∈ R1, r = |x| (Heisenberg) orr = |x1| (Grusin),


z ∈ C, anda ∈ R+. Consider the function

f (x, x0, z; a) =a

2r 2zcoth(az)− i x0z. (1.20)

Thenf ′(x, x0, z; a) =

a

2r 2

(coth(az)−

az

sinh2(az)

)− i x0 (1.21)

and

f ′′(x, x0, z; a) =a2r 2

sinh2(az)

(− 1 +

az

tanh(az)

), (1.22)

where′ denotes differentiation with respect toz. The function f (x, x0, z; a), whichappears in the heat kernel of the Heisenberg Laplacian (wherea = 2α) and other de-generate operators (for the Grusin operatora = 1), has been studied in, for example,[15] and [1], and it may be regarded as a complex action, associated with complexHamiltonian mechanics and extending the C-C metric to the complex plane. In fact,f satisfies the following analog of the Hamilton-Jacobi equation (cf., e.g., [1]):

H(x, x0,∇ f )+ z∂ f

∂z= f. (1.23)

In the sequel we sometimes suppress the parametera. Let 00,x,x0 denote the set(besides the imaginary axis) in the complex plane wheref is real; that is, let

00,x,x0 ={z ∈ C, Im f (x, x0, z) = 0, Rez 6= 0

}. (1.24)

We recall the main properties of the curve00,x,x0 (see [1], [15]). Here we assume,without loss of generality, thatx0 ≥ 0 and that geodesics are understood with respectto a Carnot-Caratheodory metric associated withH (see Fig. 2).

PROPOSITION3(1) If f (x, x0, z) is real and∂ f (x, x0, z)/∂z = 0, then z is purely imaginary.(2) If r > 0, then the number N of purely imaginary solutions of f′(x, x0, z) =

0 is finite depending onλ = 2x0/r 2. More precisely, the set Zf ′ of purelyimaginary zeros of f′ is given by

Z f ′ =

{iθ

a, θ ∈ R − πZ s.t.

θ

sin2 θ− cotθ = λ

}(1.25)

(see Fig. 1). Let pk denote the k th positive root of the equationtanθ = θ ; theelements(i θ j ) j =1,...,N of Z f ′ are such thatθ1 < π/a < θ2 ≤ p1/a < θ3 <

2π/a < · · · < θ2K ≤ pK /a ≤ θ2K+1 < (K + 1)π/a, where K =[N/2]. (Itmay happen thatθ2K = θ2K+1. Then the three curves intersecting at iθN formangles ofπ/3 radians with each other, one of them being the imaginary axis;see Fig. 3 and paragraph (5) below.)


Figure 2. The curve0 in the complex plane fora = 2

(3) 00,x,x0 is symmetric with respect to the imaginary axis. If x0 = 0, then00,x,x0

coincides with the real axis. If r> 0, then the curve branches off to∞ in bothdirections from iθN and the branches are asymptotic to the linesλ| Rez| =

Im z. Between iθ2k−1 and iθ2k, 00,x,x0 encircles the pole ik(π/a) of f .(4) Let 0x,x0 denote the union of00,x,x0 and [0, i θ1] ∪ [i θ2, i θ3] ∪ · · · ∪

[i θ2K , i θ2K+1]. The real function f is strictly increasing along the path0x,x0

assuming all values between ar2/2 and∞.(5) Let iθ be a zero of f′(x, x0, z). It is also a zero of f′′(x, x0, z) if and only ifθ

is equal to one of the real numbers pk, and in this case, f(3)(x, x0, i pk) 6= 0and f(i θk) = (a/2)r 2(θ2

k /(sinθk)2).

(6) If r > 0, then there exist N geodesic curves joining(x, x0) with the origin.The length of the j th geodesic is given by

d j (x, x0) =

√2 f (x, x0, i θ j ), 1 ≤ j ≤ N;

and d1(x, x0) ≤ d2(x, x0) ≤ · · · ≤ dN(x, x0).

Observe that the equation definingZ f ′ , (1.25), is equivalent to (1.16).The caser = 0 is degenerate. In that case, the functionf (0, x0, z) is real if and

only if z is purely imaginary.0 coincides with the ray Im(τ ) > 0, traversed twice in


Figure 3. 0 near a double root off

opposite directions. The numbersθk tend toπk/a asr → 0 (λ → ∞) (cf. also Taylor[15, pp. 82–83]).

We thus see from Proposition3 that√

2 f can be interpreted as a “distance” along0x,x0.

In several applications (nonisotropic Heisenberg Laplacian, certain degenerate el-liptic operators), we have to use a more general form of the functionf . Leta1, . . . ,am

be positive numbers, and letx ∈ Rm. Set

f (x, x0, z) =

m∑j =1

a j x2j

2zcoth(a j z)− i x0z. (1.26)

Assume, without loss of generality, thata1 ≤ a2 ≤ · · · ≤ ap < ap+1 = · · · = am.Setx′′

= (xp+1, . . . , xm), r = |x′′|. It is well known (see [1]) that Proposition3 is

valid, mutatis mutandis, in this case as well.

1.3. Lorentz-Carnot-Caratheodory metricIn analogy to the standard case, we introduce a Lorentz-Carnot-Caratheodory Hamil-tonian onM × R defined at a point(m, ξ, t, τ ) ∈ T∗(M × R) by Q(m, ξ, τ ) =

τ2/2 − H(m, ξ), whereH(m, ξ) is the Hamiltonian defined in formula (1.8). (Q is


independent oft .) For the model cases discussed in Sections 1.1 and 1.2,m = (x, x0).In these cases, consider the functions

F(x, x0, t, z,a) = i x0z +t2

2−

a

2r 2zcoth(az) (1.27)

and

φ(x, x0, t, z) = −a

2r 2 coth(az)+ i x0 +

t2

2z. (1.28)

We need these functions to analyze the wave kernel for all operators discussed in thesequel. In our applications, the functionφ satisfies the equation

∂φ(x, x0, t, z)

∂z+ Q

(∇mφ,

∂φ

∂t

)= 0. (1.29)

As an example, note that in the particular case of the Grusin plane we have

Qx,y,t (ξ, η, τ ) =τ2

− ξ2− x2η2

2, (1.30)

and we can check by computation thatφ is a solution of

∂φ(x, y, t, z)

∂z+ Q

(∂φ∂x,∂φ

∂y,∂φ

∂t

)= 0. (1.31)

In the general case, it follows from Proposition3 that f (x, x0, z) = t2/2 −

F(x, x0, t, z) = zφ(x, x0, t, z) − t2/2 is a complexification of the action com-puted along the bicharacteristics ofH , and for z = i θ1, 2 f (x, x0, i θ1) is exactlythe square of Carnot-Caratheodory distance from zero to the point(x, x0). Hence2F(x, x0, t, i θ1) is the square of the associated Lorentz indefinite metric. Note thatequation (1.29) may have other solutions, not of the type (1.28).

1.4. Separation of variablesThe general solution of the wave equation for the Grusin operator may be found usingseparation of variables. Writingu(x, t) = eikth(x)g(y), we obtain two families ofsolutions:

ua,n(x, y, t) = eiat√

2n+1 cos(a2y)Hn(ax)e−a2x2/2

and

va,n(x, y, t) = eiat√

2n+1 sin(a2y)Hn(ax)e−a2x2/2, (1.32)

wherea is a real parameter,n is a nonnegative integer, andHn is thenth Hermitepolynomial,

Hn(x) = (−1)nex2 dn

dxne−x2

, (1.33)


so thatz(x) = Hn(x)e−x2/2 is a solution of the ordinary differential equation (ODE)

z′′+ (2n + 1 − x2)z = 0. (1.34)

The Hermite polynomialsHn are orthogonal, and∫R

Hn(x)Hm(x)e−x2

dx = δnmπ1/22nn!. (1.35)

Recall the Mehler formula (see W. Magnus, F. Oberhettinger, and R. Soni [11,p. 252]), according to which for all realx, y and complexz (|z| < 1),

+∞∑n=0

Hn(x)Hn(y)zn

2nn!=

1√

1 − z2ey2

−(y−zx)2/(1−z2). (1.36)

We want to express the Dirac distributionδ((x, y); (0,0)) using the familyua,n. Re-call that in the distribution sense

δy =1

2π

∫R

e−iwy dw =2

π

∫+∞

0a cos(a2y)da; (1.37)

the last equality in (1.37) follows from a change of variablew = a2. Using the baseinduced by the Hermite polynomials, we have in the distribution sense

δ(x1 − x2) =

∞∑0

Hn(x1)e−x21/2Hn(x2)e−x2

2/2

‖Hn‖2

, (1.38)

where‖Hn‖2

=√πn!2n. Replacingx1, x2 by ax1,ax2, we get

δ(x1 − x2) =

∞∑0

Hn(ax1)e−a2x21/2Hn(ax2)e−a2x2

2/2

‖Hn‖2a

, (1.39)

where‖Hn‖2a =

√πn!2n/a. Hence in two-dimensional space, the Dirac distribution

at (0,0) has the form

δ((x, y); (0,0)

)=

∞∑0

2

π

∫+∞

0

a2

√πn!2n

cos(a2y)Hn(ax)e−a2x2/2Hn(0)da.

(1.40)Applying (1.40) to the functionφ(x, y) = f (x)g(y) ∈ D(R2), we get

g(0) =1

2π

∫R

g(w)e−iw0 dw =2

π

∫+∞

0g(a2)a da (1.41)

and

f (0) =

∞∑0

Hn(0)( f, hn,a)

‖Hn‖a, (1.42)


wherehn,a(x) = Hn(ax)e−(ax)2/2 and( f, hn,a) =∫

R f (x)Hm(ax)e−a2x2/2 dx.In performing the computations connected with the separation of variables, we

also use the following formula (see [11, p. 83]) for the Bessel functionJν :

2iπ Jν(αz) = zν∫ 0+

−∞

e(α/2)(t−z2/t)t−ν−1 dt, (1.43)

where Re(α) > 0 and | arg(t)| ≤ π , and the integral is extended over a contourstarting at∞, going clockwise around 0, going back to∞, and never cutting thesemiaxisx < 0. This contour can be deformed so that it becomes parallel to thex-axis and

2iπ Jν(αz) = zν∫ c+i ∞

c−i ∞e(α/2)(t−z2/t)t−ν−1 dt, (1.44)

wherec, α > 0 and Reν > 0. Also, it is well known that

J1/2(z) =sinz

√πz/2

. (1.45)

2. The wave kernel for the Grusin operatorIn this chapter, we study the properties of the fundamental solution of the wave equa-tion associated to the Grusin operatorL/2 = −(1/2)(∂2/∂x2

+ x2(∂2/∂x20)) us-

ing separation of variables. We are interested in computing the fundamental solutionthat satisfies the initial conditionu(x, x0,0) = 0 andut (x, x0,0) = δ(0,0)), whereδ(0,0)) denotes the Dirac distribution at(0,0) for the variable(x, x0). The kernel canbe expressed as

Kw(x, x0, t) =sin(t L1/2)

L1/2δ(0,0). (2.1)

Applying formula (1.40), we obtain

Kw(x, x0, t) =

∞∑0

2

π

∫+∞

0

sin(√

2n + 1at)

a√

2n + 1

a2

√πn!2n

× cos(a2x0)Hn(ax)e−a2x2/2Hn(0)da. (2.2)

We wish to sum the series (2.2) so as to obtain a more manageable form for the kernel.The situation is summed up in the following theorem.

THEOREM 1The wave kernel Kw(x, x0, t) defined by Kw(x, x0, t) = (sin(t L1/2)/L1/2)δ(0,0) isgiven by

Kw(x, x0, t) =K (x, x0, t)+ K (x,−x0, t)

2. (2.3)


Here

K (x, x0, t) = i K ′

0

∫C

√1

u sinh(u)

1

8(x, x0, t,u)du, (2.4)

where K′

0 = 1/(2(π)2) is a constant,8(x, x0, t, z) is the phase given by the expres-sion

8(x, x0, t, z) = i x0 +t2

2z−

x2 coth(z)

2, (2.5)

and C (the precise description is given in the following subsections) is a closed con-tour of integration lying outside the set whereRe8 > 0 and avoiding the zeros ofsinh(z).

The phase8 satisfies the Hamilton-Jacobi equation

2∂φ

∂z=

(∂φ∂t

)2−

(∂φ∂x

)2− x2

( ∂φ∂x0

)2. (2.6)

The wave kernel satisfies the finite speed property: it vanishes identically for t2 <

d2c (x, x0). Equivalently, the kernel is zero before the first geodesic of the C-C metric

arrives at the point(x, x0). Moreover, the kernel vanishes when the time satisfies theconditions

2 f (i θ2k) < t2 < 2 f (i θ2k+1), (2.7)

where the pointsθk are introduced in Proposition3.

Remark.The singularities of the wave kernel are computed using the zeros of thefunction8(x, x0, t, z) defined by (1.28) with a = 1 (or by (2.5)). When time in-creases to the valuet =

√2 f (i θ2k+1), this means that a new geodesic hits the point

(x, x0) and then the kernel becomes singular.

ProofAll the computations in this paragraph are to be understood in the distribution sense.From the identityJ1/2(z) = sinz/

√πz/2 and from formula (1.44), we obtain for

arbitraryc′ > 0,

sinαz

αz=

√π

2

1

2iπ

∫ c′+i ∞

c′−i ∞e(α/2)(u−z2/u)u−3/2 du. (2.8)

Let us chooseα = 1 andz = at√

2n + 1; thenc′= a2t2c, wherec > 0. Then

sinat√

2n + 1

at√

2n + 1=

√π

2

1

2iπ

∫ a2t2c+i ∞

a2t2c−i ∞e(1/2)(u−(2n+1)(at)2/u)u−3/2 du, (2.9)


and using the change of variableu = (at)2v for at > 0,

sinat√

2n + 1

a√

2n + 1=

√π

2

1

2iπa

∫ c+i ∞

c−i ∞e(1/2)(a

2t2v−(2n+1)/v)v−3/2 dv, (2.10)

we see that the kernel takes the form

Kw(x, x0, t) = −i K0

∫∞

0

∞∑0

∫ c+i ∞

c−i ∞e(1/2)(a

2t2v−(2n+1)/v)v−3/2 dva

n!2ncos(a2x0)

· Hn(ax)e−a2x2/2Hn(0)da (2.11)

with

K0 =

√1

2

1

2π

2

π=

1

π2√

2. (2.12)

To sum the series, we use the Mehler formula (1.36) (see [11, p. 252]). Setz =

e−1/v so that|z| < 1 since Re(−1/v) < 0 to get

e−1/2v+∞∑n=0

Hn(ax)Hn(0)e−n/v

2nn!=

e−1/2v

√1 − e2/v

e−(ax)2e−2/v/(1−e−2/v). (2.13)

Usinge−1/2v/√

1 − e2/v = 1/√

2 sinh(1/v) and cos(a2x0) = (1/2)(eia2x0+e−ia2x0),the kernel can be expressed asKw(x, x0, t) = (1/2)(K (x, x0, t) + K (x,−x0, t)),where

K (x, x0, t) = −i K0

∫+∞

0

∫ c+i ∞

c−i ∞v−3/2 a

√2 sinh(1/v)

· e−(ax)2e−2/v/(1−e−2/v)−a2x2/2+ia2x0+(1/2)(a2t2v) dv da.

Changing to a new variablez = 1/v, dv/v = −dz/z, and√v = 1/

√z, the contour

Rev = c is transformed to the circleC(1/2c,1/2c), centered at(1/2c,0) with radius1/2c. The previous integral becomes

K (x, x0, t) = i K0

∫C

∫+∞

0

az−1√z√

2 sinh(z)

· e−(ax)2e−2z/(1−e−2z)−a2x2/2+ia2x0+(1/2)(a2t2/z) dz da, (2.14)

and the term in the exponential can be rewritten as

(ax)2e−2z

1 − e−2z−

a2x2

2+ ia2x0 +

1

2

(a2t2

z

)= a2

(−

x2 coth(z)

2+ i x0 +

t2

2z

).

Recall that the phase is given by

8(x, x0, t, z) = −x2 coth(z)

2+ i x0 +

t2

2z. (2.15)


An elementary computation proves that Hamilton-Jacobi equation (2.6) is satisfied.Now for Re8 = Re(−x2 coth(z)/2 + t2/(2z)) < 0,∫

∞

0aea28 da =

−1

28, (2.16)

so that the kernel is

K (x, x0, t) =i K0

2

∫C

dz

√1

2zsinh(z)

1

8(x, x0, t, z), (2.17)

which is equivalent to

K (x, x0, t) =i K0

2

∫C

dz

√z

2 sinh(z)

1

F(x, x0, t, z), (2.18)

where we recall

F(x, x0, t, z) = i x0z +t2

2−

x2

2zcoth(z).

It follows from here that ift < |x|, then the kernel is zero (unlessx = x0 =

0). Indeed, Re8 < 0 if Rez > 0. Now since8 is a continuous function of thecomplex variablez, 8(x, x0, t,0) 6= 0, there exists a neighborhoodV of zero suchthat8(x, x0, t, z) does not vanish forz ∈ V . Hence the constantc can be chosenlarge enough so that the circleC(1/2c,1/2c) is small enough and contained inV .Finally, a simple application of the residue theorem implies thatK (x, x0, t) = 0.

We complete the proof in the following subsections, where we also describe thecontour of integration.

2.1. The boundary of the forbidden setIn this paragraph we describe the contours of integration which are used in the proof ofTheorem1. From the discussion leading to (2.17) and (2.18), we see that the contourshave to be contained in the region Re(8) < 0.

By analytic continuation, we may enlarge the region to Re(8) ≤ 0.Let us denoteλ = x2/t2. Then the set of points2x,t , where Re(8) = 0, is

defined by

2x,t =

{z = u + i v ∈ C,

u

u2 + v2= λ

sinhu coshu

sin2 v + sinh2 u

}. (2.19)

The curve2x,t depends on the time and can be described as follows. It is symmetricwith respect tov → −v. Thev-axis is always contained in2x,t sinceu = 0 is asolution.

There are the following two cases.


(1) If λ ≥ 1, the curve has no intersection with the axisv = 0 except at the origin.In this case, there exists no bifurcation point on thev-axis.

(2) If λ < 1, then there are several branches that intersect the imaginary axis (seeFig. 4). First by continuity, there exists a curve starting at the point(uλ,0),where tanh(uλ)/uλ = λ with a vertical tangent, connecting to a point(0, vλ),wherevλ is a solution of the following equation:(sinv

v

)2= λ < 1. (2.20)

Equation (2.20) has a finite number of solutions depending on the relative sizesof (sin pk/pk)

2 andλ, wherepk is thek-root of tanx = x. Note first that fort, x fixed the non-purely-imaginary part of the curve2x,t is bounded. Thisfollows from the fact thatu cannot go to∞ in the expression of2x,t .

Now Reφ > 0 on the real interval[0,uλ], and for v large enoughRe(8) < 0 uniformly in u ∈ [0,uλ]. By continuity, we deduce that thereexists a continuous curve that joins the point(0, vλ) to the point(uλ,0). Bythe inverse function theorem, we have a unique tangent at a neighborhood ofthe point(uλ,0). Hence the curve starting at this point is unique.

More precisely, consider first the case where

1

1 + p21

< λ < 1, (2.21)

so that equation (2.20) has only one solution. Hence there exists one curve connecting(uλ,0) to (0, vλ), and the tangent at(0, vλ) is parallel to the real axis.

In general, at the point(0, vλ), the curve2x,t has a horizontal tangent exceptwhenvλ = (pk) for a certaink. At these points, a bifurcation appears. This resultsfrom a simple perturbation analysis: writingv0 = vλ + δ near the point(0, vλ) in(2.19) gives foru small

2vλ(1 −

vλ

tan(vλ)

)δ =

(1

λ− 1

)u2. (2.22)

Now, except at pointsvλ = pk, the perturbationδ is of second order and only onehorizontal branch can start from this point. At the bifurcation pointspk, we obtain,after some computations,(

1 −cos(2vλ)

λ

)δ2

=

(1

λ− 1

)u2. (2.23)

Thusδ is linear inu, and the curve has two tangents that are not horizontal.For 1/(1 + p2

n+1) < λ < 1/(1 + p2n), we obtain 2n + 1 solutions

v1(t), . . . , v2n+1(t), and whenλ = 1/(1 + p2n), a double solution appears.


−25 −20 −15 −10 −5 0 5 10 15 20 25

u

−15

−10

−5

0v

5

10

15

u/(u2+ v2)− 1/25 sinh(u) cosh(u)/(sin(v)2 + sinh(u)2) = 0

Figure 4. Curve2x,t for λ = 1/25

At the first time whenλ = 1/(1 + p21), the curve2x,t jumps to reach the upper

point (0, v2), whereπ < v2 < 2π . Indeed, otherwise the branch of the curve startingat (0, v2) would return to itself and could not be connected to the rest of the curve.

Thus(uλ,0) is connected to(0, v2), and then an arc joins(0, v2(t)) to (0, v1(t))in the regionu > 0. After some time all tangents of the curve near the imaginaryaxis become horizontal until we reach the second bifurcation point 1/(1 + p2

2) = λ.As time goes on, the part of the curve near the imaginary axis turns around the point(0, kπ), and the two pointsv2p+1(t) < (p + 1)π < v2p+2(t) converge to(p + 1)πfor p ∈ N whent converges to∞.

2.2. The zero of the phase and the integration pathWe see in this subsection that the relevant pole of the function 1/8 is located at theintersection of the curves0 and2.

It has been proved in Section 2.1 that the curve2 has a unique branch startingon the positiveu-axis and intersecting the positivev-axis. This curve cuts the curve0


only once. This intersection point is exactly the pole.Recall that by Proposition3, the function f is strictly increasing along the curve

0. Moreover,

8(x, x0, t, z) = −x2

2coth(z)+ i x0 +

t2

2z= 0

is equivalent to

t2=

zx2

tanh(z)− 2i x0z = 2 f (x, x0, z), (2.24)

so that ifz is a zero of8, then Im f (z) = 0. If t2 > x2, thent2 is bigger than theminimum of 2f along0. This proves that there is precisely one solutionz on 0 ofequation (2.24). This point is at the intersection of the two curves0 (Im f = 0) and2 (Re8 = 0).

The converse holds as well. Ift2/z − x2/tanh(z) is purely imaginary (equalsiα,say), then multiplying byz, we get 28 = i (α + 2x0)z, so that the equation Im8 = 0implies that if Imz > 0, thenα + 2x0 = 0 and8 = 0. If z is purely imaginary,the statement remains correct due to the strictly increasing property off along0.Indeed, no jump occurs as Im(z) tends to zero due to the continuity of the zero withrespect to the arguments. The other purely imaginary zeros of8 do not contribute tothe integral.

Hence we may deform the path of integrationC and choose it starting in theregion where Re8 < 0 (taking into account the singularity), going along the imag-inary axis, and avoiding the polesikπ for k ∈ N. This completes the statement ofTheorem1 and the proof of formula (2.4).

We finish the proof of the theorem in the next subsection, after obtaining an ex-plicit formula.

2.3. Explicit formula for the integralWe obtain an explicit expression for the wave kernel when we perform the integrationalong the contour described in the previous paragraph and apply the residue theorem.

PROPOSITION4The wave kernel Kw for the Grusin operator is given at a point(x, x0), where x0 6= 0,


by the expression

Kw(x, x0, t)

=1

2π

(√z(x, x0, t)

2 sinh(z(x, x0, t))

1

−t2/(z(x, x0, t))+ x2z(x, x0, t)/sinh2(z(x, x0, t))

+

√z(x, x0, t)

2 sinh(z(x, x0, t))

1


),

(2.25)

where z(x, x0, t) is the unique solution on0 of F(x, x0, t, z) = 0, andF ′(x, x0, t, z) 6= 0; that is,

t2= x2 z(x, x0, t)

tanhz(x, x0, t)− 2i x0z(x, x0, t), (2.26)

and the denominator in formula (2.25) does not vanish. The kernel is analytic exceptat points where F′(x, x0, t, z(x, x0, t)) = 0. Put differently,

Kw(x, x0, t) = −1

4π t

(√z(x, x0, t)

2 sinh(z(x, x0, t))

∂z(x, x0, t)

∂t

+

√z(x, x0, t)

2 sinh(z(x, x0, t))

∂ z(x, x0, t)

∂t

),

where an elementary computation yields that

∂z(x, x0, t)

∂t=

2t

t2/(z(x, x0, t))− x2z(x, x0, t)/ sinh2(z(x, x0, t)). (2.27)

When x= 0 and x0 6= 0, the wave kernel is given for t> 0 by

Kw(0, x0, t) =(−1) j

2π |x0|

t√4x0 sin(−t2/(2|x0|))

χ{sin(t2/(2|x0|))<0}, (2.28)

whereχI is the characteristic function of the interval I and j∈ N is such that(2 j −

1)π < t2/(2x0) < 2 jπ . Moreover, the kernel is singular for t2= 2k|x0|π , where

k ∈ Z.

ProofSince the path avoids the zeros of sinh(z) = 0 and stays in the right half of thecomplex planeu ≥ 0, we can apply the residue theorem. Our path of integrationCcontains only singularities generated by the zeros of the function8(x, x0, t, z). Letus recall the expression of the kernelK ,

K (x, x0, t) = i K0/2∫

C

1

8(x, x0, t, z)w(z)dz, (2.29)


where

w(z) =

√z

2 sinh(z), (2.30)

and C has been described earlier. We have seen that the relevant solution of8(x, x0, t, z) = 0 is unique and is located exactly at the intersection of the curves0x,x0 and2x,t .

In the case wherex0 6= 0, we compute the residue at points where8′(x, x0, t, z) 6= 0, wherez satisfies the equation

t2= x2 z(x, x0, t)

tanhz(x, x0, t)− 2i x0z(x, x0, t). (2.31)

In this case, the function(x, x0, t) → z(x, x0, t) is an analytic function of(x, x0, t).We discuss below the set of points(x, x0, t) in R3 where the derivative is zero (seealso A. Nachman [13]).

Applying the residue theorem, we get∫C

1

8(x, x0, t, z)w(z)dz = 2iπ

√z(x, x0, t)

2 sinhz(x, x0, t)

1

8′(x, x0, t, z(x, x0, t)). (2.32)

Hence

K (x, x0, t) = 2πK0w(z(x, x0, t))

8′(x, x0, t, z(x, x0, t))(2.33)

and

8′(x, x0, t, z(x, x0, t)

)= −

t2

2z+

x2z

2 sinh2(z), (2.34)

so that finally

K (x, x0, t) =1

2π

·

(√z(x, x0, t)

2 sinh(z(x, x0, t))

2


).

(2.35)

Now to obtainKw, we need to computeK (x,−x0, t). But by symmetry, the solutionη ∈ C of 8(x,−x0, t, η) = 0 is exactlyz, wherez is the solution of8(x, x0, t, z) =

0. The pole in the integrand occurring in the expression forK (x,−x0, t) is exactlyz,and

K (x,−x0, t) = 2πK0w(z(x, x0, t))

8′(x, x0, t, z(x, x0, t)). (2.36)


We conclude that the wave kernel is given by

Kw(x, x0, t)

=1

2π

(√z(x, x0, t)

2 sinh(z(x, x0, t))

1


+

√z(x, x0, t)

2 sinh(z(x, x0, t))

1


).

(2.37)

Kw(x, x0, t) is thus analytic except at the critical points of the phase8.To finish the proof, consider the case wherex = 0. Since the expression of the

kernel is analytic in its arguments, whenx = 0 the expression is still valid and thepole is now located atz0 = i t 2/(2x0). Using expression (2.37) for z = z0, we obtain

Kw(0, x0, t) =1

πRe

(√z0

2 sinh(z0)

1

−t2/z0

). (2.38)

If sin(t2/(2x0)) ≥ 0 is positive, we obtain a purely imaginary number andKw(0, x0, t) vanishes. This proves that cancellation phenomena exist for an infinitenumber of time intervals. Now suppose that sin(t2/(2x0)) < 0. Then by evaluating(2.38), we obtain forx0 > 0,

Kw(0, x0, t) =t

2x0π

√1

−4x0 sin(t2/(2x0)). (2.39)

If x0 < 0, then

Kw(0, x0, t) = −t

2x0π

√1

−4x0 sin(t2/(2x0)). (2.40)

In formulas (2.39) and (2.40), the sign of the square root alternates.It follows from this expression that the singularities are located at the pointst2

=

2k|x0|π .

Remark.We could have derived the formula (2.28) of Proposition4 from formula(2.4) of Theorem1. For in this case the integration contour is contained in the imag-inary axis where the polez0 = i t 2/(2x0) is located. More precisely, the contour0starts at the origin, continues along the imaginary axis encircling the polez0, andreturns to the origin, oriented clockwise. Whenx tends to zero, the curve0x,x0 con-verges pointwise to the imaginary axis andF(x, x0, z, t) converges toF(0, x0, t, z),


except at the pointsiπk for k ∈ Z. Since Kw(x, x0, t) = (1/2)(K (x, x0, t) +

K (x,−x0, t)), we need to evaluate the integral

K (0, x0, t) =i

π2

∫0

√u

sinh(u)

1

2uix0 + t2du, (2.41)

where0 is the part of the contour contained in the imaginary axis. The pointsiπkdo not really contribute a singularity to the integral. In fact, the computation showsthat the integral along the integration contour vanishes and only the singularity at thepoint z0 is relevant.

We may now prove the statements of Theorem1 about the finite speed of propagation,using the expression of the wave kernel obtained above. Note that it is possible todeform the path of integration below theu-axis. Using this remark, it is possible toprove that the kernel is vanishing before it reachesd2

c (x, x0).Indeed, forx2 < t2 < d2

c (x, x0), the zero of the function8(x, x0, t, z) is exactlysituated on the imaginary axis between 0 andi θ1 with θ1 < π . Deforming the contourof integration in order to include the pole of 1/8, we obtain a certain residue. Thepole where the residue is computed is of the formiα, whereα > 0. Using exactly thesame argument, we see that the residue for the kernelK (x,−x0, t) is computed at thepole−iα. Since the kernelKw is an odd function ofz, the two residues cancel.

This result is valid each time thatf (x, x0, t, z) = 0 has a solutioniα in theimaginary axis such thatα ∈ [2kπ, (2k+1)π ], k ∈ N. Applying the description of thecurve0x,x0, given in Proposition3, this situation appears exactly when 2f (i θ2k) <

t2 < 2 f (i θ2k+1), whereθk are defined in the same proposition and, as above, thetermsK (x, x0, t) andK (x,−x0, t) cancel.

2.4. Analysis of the wave kernel near the singularitiesWe describe here the wave kernel near the singularities. We have found in the lastparagraph an explicit expression for the wave kernel. We are going to focus on the set

Z(x, x0, t) ={z ∈ C s.t. F(x, x0, t, z) = 0 andF ′(x, x0, t, z) = 0

}. (2.42)

The characteristic set may be parametrized by a real parameterθ . After someelementary computations, we find that

S =

{(x, x0, t) ∈ R3 s.t.

x2

t2=

sin2 θ

θ2and

2x0

x2=θ − cosθ sinθ

sin2 θ, θ ∈ R − πZ

},

S = S− ∪ S+, (2.43)


where

S± =

{(x, x0, t) ∈ R3 s.t.x = ±t

∣∣∣sinθ

θ

∣∣∣and

2x0

x2=θ − cosθ sinθ

sin2 θ, θ ∈ R − πZ

}.

The main result of this subsection is the following.

PROPOSITION5The singular set of the wave kernel is S. This set is the disjoint union of the sets S+

and S−. Moreover, near the singular set, the main singularity of the kernel is given bythe following expression:

Kw(x, x0, t) ∼1

2π

(√θk

2 sin(θk)

)1

(x2/sin2 θk)(−1 + θk/tanθk)

·

(√H(x, x0, t, i θk)

t2 − d2k (x, x0)

+

√H(x,−x0, t,−i θk)

t2 − d2k (x,−x0)

),

where dk is the length of the k th geodesic joining(x, x0) with the origin, k =

1, . . . , N, H is an analytic function, nonzero on S andθk is a solution of the equationF ′(x, x0, t, i θ) = 0 with F real, andθk is not a root oftanθ = θ . Whenθk is one ofthe points pk, Kw(x, x0, t) grows like1/(t2

− d2k (x, x0))

3/2.

ProofThe singular setsZ andS are essentially the same as the singular sets analyzed andshown graphically in [13]. Here we study the behavior ofKw near the setS anddetermine the singularity there.

In the set Z(x, x0, t), the points z = i pk = i tanpk are isolated andF ′′(x, x0, t, i pk) = 0, but the third derivative is not zero. At points ofScorrespondingto the latter, the singularity of the wave kernel is of higher order.

For z in a neighborhood ofZ(x, x0, t) − {i pk , k ∈ N}, F(x, x0, t, z) has thefollowing expansion:

F(x, x0, t, z) = F(x, x0, t, i θk)+ F ′(x, x0, t, i θk)(z − i θk)

+ (z − i θk)2H(x, x0, t, z)

= (z − i θk)2H(x, x0, t, z), (2.44)

whereH(x, x0, t, z) does not vanish in a neighborhood ofi θk, and by Taylor expan-sion we have

−t2

z(x, x0, t)+

x2z(x, x0, t)

sinh2(z(x, x0, t))=

x2

sinθk

(− 1 +

θk

tanθk

)(z − i θk)+ o(z − i θk)


and

Kw(x, x0, t)

∼1

2π

(√θk

2 sin(θk)

)1


( 1

(z − i θk)+

1

(z + i θk)

).

(2.45)

In order to express (2.45) in terms of the distance to the setS, note that

F(x, x0, t, z) =t2

2− f (x, x0, z) = (z − i θk)

2H(x, x0, t, z), (2.46)

and using the length of thek th geodesic,f (x, x0, z) ∼ d2k (x, x0)/2, we see that

Kw(x, x0, t) ∼1

2π

(√θk

2 sin(θk)

)1


·

(√H(x, x0, t, z)

t2 − d2k (x, x0)

+

√H(x,−x0, t, z)

t2 − d2k (x,−x0)

),

whereH does not vanish. Whenθk is one of the pointspk, the same type of analysisshows thatK grows like 1/(t2

− d2k (x, x0))

3/2.

2.5. The operator satisfies the wave equationUsing integration by parts, we prove that the wave kernelKw satisfies the wave equa-tion. Starting with the fact thatKw(x, x0, t) = (K (x, x0, t)+ K (x,−x0, t))/2, weonly need to show thatK satisfies the wave equation. Recall that

K (x, x0, t) = K0

∫C

1

8(x, x0, z, t)w(z)dz, (2.47)

wherew(z) =√

z/(2 sinh(z)) and8(x, x0, z, t) = i x0z + t2/2 − x2zcoth(z)/2,K0 is constant, andC is a closed contour that may enclose singularities but not passthrough them. We have already proved thatKw is zero fort small enough. Note thatw satisfies

w′(z) =w(z)

2

(1

z−

1

tanh(z)

), (2.48)

and recall the formula for the derivatives of8:

8′(z) = i x0 −x2

2

( 1

tanhz−

z

sinh2(z)

). (2.49)


By direct computations,

∂2K (x, x0, t)

∂t2= K0

∫C

(−

1

82(x, x0, z, t)+

2t2

83(x, x0, z, t)

)w(z)dz, (2.50)

∂2K (x, x0, t)

∂x2= K0

∫C

( z

tanh(z)82(x, x0, z, t)

+2x2z2

tanh2(z)83(x, x0, z, t)

)w(z)dz, (2.51)

x2∂2K (x, x0, t)

∂y2= −2K0

∫C

( z2x2

83(x, x0, z, t)

)w(z)dz, (2.52)

and

∂2K (x, x0, t)

∂x2+ x2∂

2K (x, x0, t)

∂y2

= K0

∫C

( z

tanh(z)82(x, x0, z, t)

+2x2z2

sinh2(z)83(x, x0, z, t)

)w(z)dz.

Consider the first term of the right-hand side in the last expression. Using theproperties of the functionw and integrating by parts, we obtain∫

C

z

tanh(z)82(x, x0, z, t)w(z)dz =

∫C

w(z)− 2zw′(z)

82(x, x0, z, t)dz

=

∫C

( w(z)

82(x, x0, z, t)− 2

[ wz

82(x, x0, z, t)

])dz

+ 2∫

C

( z

82(x, x0, z, t)

)′

w(z)dz

=

∫C

w(z)

82(x, x0, z, t)dz+ 2

∫C

w(z)

82(x, x0, z, t)dz

− 4∫

C

zw(z)8′(x, x0, z, t)

83(x, x0, z, t)dz

= 3∫

C

w(z)

82(x, x0, z, t)dz

− 4∫

C

(i x0z −

x2

2

( z

tanh(z)−

z2

sinh2(z)

))w(z)

83(x, x0, z, t)dz. (2.53)


Finally,

∂2K (x, x0, t)

∂x2+ x2∂

2K (x, x0, t)

∂y2

= 3∫

C

w(z)

82(x, x0, z, t)dz− 4

∫C

i x0 zw(z)

83(x, x0, z, t)dz

+ 2∫

C

x2w(z)z

tanh(z)83(x, x0, z, t)dz

= −

∫C

w(z)

82(x, x0, z, t)dz+ 4

∫C

8(x, x0, z, t)− i x0z

83(x, x0, z, t)w(z)dz

+ 2∫

C

x2w(z)z

tanh(z)83(x, x0, z, t)dz

= −

∫C

w(z)

82(x, x0, z, t)dz+ 2

∫C

t2

83(x, x0, z, t)w(z)dz

=∂2K (x, x0, t)

∂t2. (2.54)

Observe thatw plays the role of the transport term in wave theory.

2.6. Source not at the originIn this subsection we make several remarks concerning the case where the Dirac dis-tribution is given at a point(y, y0) 6= (0,0) and the observer is fixed at the point(x, x0). An analog of Theorem1 is valid in this case as well. The following is true.

PROPOSITION6The wave kernel Kw(x, x0, t) defined by K(x, x0, t) = (sin(t L1/2)/L1/2)δ(y, y0),where−L/2 is the Grusin operator, is given by

Kw(y, y0, x, x0, t, z) =K (y, y0, x, x0, t, z)+ K (y, y0, x, x0, t, z)

2, (2.55)

where

K (x, x0, t) = K ′

0

∫0

√1

u sinh(u)

1

8(x, x0, t,u)du, (2.56)

and K′

0 = 1/(2π)2,0 is an appropriate contour, and8(y, y0, x, x0, t, z) is the phasegiven by the expression

8(y, y0, x, x0, t, z) = −x2

+ y2

2 tanhz+

xy

sinhz+ i (y0 − x0)+

t2

2z. (2.57)

Set f(y, y0, x, x0, t, z) = z8(y, y0, x, x0, t, z). The singularities of the integrand arelocated at the zeros of F(y, y0, x, x0, t, z) = 0. More precisely, the singularity is lo-


cated at the intersection ofRe8(y, y0, x, x0, t, z) = 0 andIm f (y, y0, x, x0, t, z) =

0.

ProofThe proof is similar to the proof of Theorem1, and using the same arguments, weobtain the following expression of the phase:

8(y, y0, x, x0, t, z) = −x2

+ y2

2 tanhz+

xy

sinhz+ i (y0 − x0)+

t2

2z. (2.58)

The contour starts in the region where Re8(y, y0, x, x0, t, z) < 0, but it turns out tobe more complex.

The analog of Proposition4 is more complex and will be discussed in a future work.

3. Wave kernels in one dimensionIn this section we study the wave kernel for two operators. The first operator is theharmonic oscillator∂xx − x2, where we are able to give an integral representationformula, and the second is the Klein Gordon operator∂xx − a2 (a is a constant),where we recover well-known results.

3.1. Wave kernel for the harmonic oscillatorThe purpose of this subsection is to discuss the wave kernel for the equation and initialconditions

∂t tu = ∂xxu − x2u,

∂tu(x,0) = δx,

u(x,0) = 0.

We obtain the following result.

PROPOSITION7The wave kernel for the operator∂xx − x2 can be expressed in the form

K (x, t) =i

4π

∫C

√1

2zsinh(z)et2/2z−x2 coth(z)/2 dz, (3.1)

where C is a contour symmetric with respect to the x-axis going through the origin,obtained by a smooth deformation of the circle C(1/2c,1/2c). The singularity atz = 0 is essential. For t< |x| the kernel vanishes. The phaseφ(x, t, z) = t2/(2z)−

x2 coth(z)/2 satisfies the Hamilton-Jacobi equation

2∂φ

∂z= −

(∂φ∂t

)2+

(∂φ∂x

)2− x2. (3.2)


ProofUsing expression (1.38) for the Dirac operator and separation of variables, we canexpress the kernel in the form

K (x, t) =

+∞∑n=0

1

n!2n√π

sin(√

2n + 1t)√

2n + 1Hn(0)Hn(x)e

−x2/2. (3.3)

We now use integral formula (2.10) to express the time dependence:

sint√

2n + 1√

2n + 1=

√π

2

1

2iπ

∫ c+i ∞

c−i ∞e(1/2)(t

2v−(2n+1)/v)v−3/2 dv. (3.4)

Then the kernel assumes the form

Kw(x, x0, t)

=1

2iπ√

2

∞∑0

∫ c+i ∞

c−i ∞e(1/2)(t

2v−(2n+1)/v)v−3/2 dv1

n!2nHn(x)e

−x2/2Hn(0).

By Mehler formula (1.36) for Rev > 0, we have

e−1/2v+∞∑n=0

Hn(x)Hn(0)e−n/v

2nn!=

e−1/2v

√1 − e2/v

e−x2e−2/v/(1−e−2/v). (3.5)

We can simplify as in (2.14), and we find that in the new variablez = 1/v,dv/v = −dz/z, and

√v = 1/

√z, the contourx = c is transformed to a circle

C = C(1/2c,1/2c), centered at(1/2c,0) of radius 1/2c, obtaining finally

K (x, t) =i

4π

∫C

√1

2zsinh(z)et2/(2z)−x2 coth(z)/2 dz. (3.6)

The details of the computation are given in Section 2 for the Grusin operator.We can deformC to a closed contour symmetric with respect to theu-axis (u =

Re(z)) as follows. The contour starts outside the region defined by Re(t2/(2z) −

x2 coth(z)/2) > 0 and Rez ≥ 0 (see Fig. 4 for the curve2x,t ). At the point where2x,t meets the imaginary axis for the first time, we continue along this axis until thenext pointvλ (whereλ = x2/t2), defined in Section 2.1. Then the contour is continuedby being allowed to come back to Rez > 0, avoiding the singularity at a multiple ofiπ . This construction is continued each time a singularity has to be avoided. Thecontour reaches the origin along the imaginary axis. Then the path is symmetrizedwith respect to theu-axis. Thus the contour avoids the singularities sinh(z) = 0 forz 6= 0 and stays in the regionu ≥ 0.


To prove that the operator satisfies finite speed propagation and vanishes whent < x, we may use the residue theorem, and since the singularity atz = 0 is remov-able,

K (x, t) = 0 whent < x. (3.7)

We may verify by an elementary computation that the phaseφ satisfies Hamilton-Jacobi equation (3.2).

Remark.Whent > x, the point zero is an essential singularity and yields the maincontribution. Indeed, the integral on the imaginary axis can be expressed as

Ka(x, t)

=i

4π

∫ a

−a

√1

2y sin(y)

(cos

( t2

2y−

x2 coth(y)

2

)− i sin

( t2

2y−

x2 coth(y)

2

))dy,

which reduces by symmetry of the first term (understood as principal value) to

Ka(x, t) =1

4π

∫ a

−a

√1

2y sin(y)sin

( t2

2y−

x2 coth(y)

2

)dy, (3.8)

so that

Ka(x, t) =1

4π

∫ a

−a

√1

2y sin(y)sin

( t2− x2

2y

)dy + O(1). (3.9)

Ka depends (to a first approximation) only ont2− x2, and we have fora close to

zero,

Ka(x, t) ∼1

2√

2π

∫ a

0

1

ysin

( t2− x2

2y

)dy + O(1). (3.10)

The integral in (3.10) is convergent, as can be seen by using the following change ofvariablez = 1/y.

3.2. Wave kernel for the Klein-Gordon operatorIn this paragraph we show how it is possible to recover the well-known result concern-ing the wave kernel for the translation-invariant Klein-Gordon operator. The result isgiven in term of the Bessel functionJ0. We prove the following.

PROPOSITION8The wave kernel for the equation

∂2

∂t2w = ∂xxw − a2w,

w(x,0) = 0,

wt (x,0) = δ0(x), (3.11)


is given by

K (x, t) =1

2J0(a

√t2 − x2)H(t2

− x2), (3.12)

where H is the Heaviside function and J0 is the Bessel function.

ProofThe general solution of (3.11) is given by the family of functionseikx sin(

√k2 + a2t)

andeikx cos(√

k2 + a2t). Integrating the family (recall (1.9)), we obtain

K (x, t) =1

2π

∫R

cos(kx)sin(

√k2 + a2t)

√k2 + a2

dk. (3.13)

Using the relation

sin(√

k2 + a2t)√

k2 + a2=

√π/2

2iπ

∫ 0+

−∞

e(1/2)(u−(k2+a2)t2/u)u−3/2 du (3.14)

(see (1.43)), we get

K (x, t) =1

2π

√π/2

2iπ

∫R

∫ 0+

−∞

e(1/2)(u−(k2+a2)t2/u)u−3/2 ducos(kx)dk. (3.15)

But ∫R

e−k2t2/2u cos(kx)dk =

√2πu

te−x2u/(2t2) (3.16)

for u ∈ C − R− (usual cut). Hence

K (x, t) =1

2

1

2iπ

∫ 0+

−∞

e((1−x2/t2)/2)(u−t2a2/(1−x2/t2))u−1 du, (3.17)

and applying (1.43) again, we see that fort2− x2 > 0,

K (x, t) =1

2J0(a

√t2 − x2). (3.18)

It is well known thatK is zero fort2 < x2, and if H denotes the Heaviside function,we obtain (3.12).

4. The Heisenberg wave kernel via the heat kernelIn this section we construct a representation formula for the Heisenberg wave kernel(the kernel of cos(

√−21H t)) by inverting the so-called transmutation formulas. We

also have to deform an integration path in the complex plane. The results are expressedin Theorems2 and3 of this section. In the corollaries, analyticity results are given,and the behavior of the kernel near its singular support is described.


PROPOSITION9Let L be a nonnegative self-adjoint operator, u> 0. Then

e−Lu=

1√

4πu

∫∞

−∞

e−t2/(4u) cos(√

L t)dt =1

√πu

∫∞

0e−t2/(4u) cos(

√Lt)dt.

(4.1)

Proposition9 is a well-known transmutation formula. A proof may be found in [8].

PROPOSITION10For every a6= 0, u > 0, and nonnegative integer n,

e−a2/(2u)

un+1=

4n+1/2

√u

∫∞

0e−t2/(4u) ∂

∂tδ(n−1/2)(t2

− 2a2)dt. (4.2)

ProofWe have∫

∞

0e−t2/(4u) ∂

∂tδ(n−1/2)(t2

−2a2)dt = 1/(4u)∫

∞

0e−t2/(4u)δ(n−1/2)(t2

−2a2)2t dt.

Setting t2− 2a2

= y, we see (recalling the formula for fractional differentiation,cf. [5, pp. 739 – 740]) that the right-hand side is, by definition, equal to

1

4u

∫∞

0e−(2a2

+y)/(4u)δ(n−1/2)(y)dy =e−a2/(2u)

4u

∫∞

0

e−y/(4u)

0(1/2)

∂n

∂yn

( 1√

y

)dy

=e−a2/(2u)

0(1/2

)(4u)n+1

∫∞

0 e−y/(4u)/√

y dy

=

√π

√π (4u)n+1/2

e−a2/(2u)=

e−a2/u

4un+1/2,

and the proposition follows.

Note that one may derive (4.2) from (4.1) and the well-known expressions forcos(

√−1 t).

We set

P(x, x0; u) =1

(2πu)n+1

∫∞

−∞

e− f (x,x0;τ)/uV(τ )dτ, (4.3)

wheref (x, x0; τ) = ατ coth(2ατ)|x|

2− i x0τ = f (x, x0, τ ;α) (4.4)


and

V(τ ) =

( 2ατ

sinh(2ατ)

)n. (4.5)

Hereα is a positive constant andf (x, x0, τ ;α) is the function introduced in Sec-tion 1.2.

PROPOSITION11Let

1H =1

2

[ n∑j =1

( ∂

∂x2 j −1+ 2αx2 j

∂

∂x0

)2

+1

2

n∑j =1

( ∂

∂x2 j− 2αx2 j −1

∂

∂x0

)2]. (4.6)

ThenP(x, x0; t) = et1H (x, x0; 0,0). (4.7)

A proof may be found in [1]. Observe that the kernelP(x, x0; t) is the fundamentalsolution of the heat equation

∂P

∂t= 1H P (4.8)

with the singularity at the origin.

4.1. Deforming the path of integrationWe deform the path of integration used in formula (4.3) from the real axis to the curve0 discussed in Section 1.2. This path has been introduced for studying singularities ofthe Heisenberg wave kernel by Taylor [15, pp. 80 – 86]. We consider the case wherex0 ≥ 0; the other case is similar.

PROPOSITION12If x 6= 0, x0 6= 0, then

P(x, x0; u) =1

(2πu)n+1

∫0x,x0

e− f (x,x0,τ )/uV(τ )dτ. (4.9)

ProofLet N = N(x0/|x|

2) denote the number of purely imaginary zeros of∂ f/∂τ , andfor R → ∞ consider the closed contour0R (see Fig. 5) formed by the interval{−R ≤ τ1 ≤ R, τ2 = 0}, the vertical segmentsτ1 = ±R joining (−R,0) and(R,0) with the unbounded branches of0x,x0, and the portion of0x,x0 between thoseintersection points.


Figure 5. The integration path0R

By Cauchy’s integral theorem,∫0R

e− f (x,x0;τ)/uV(τ )dτ = 0.

Recall thatV(R + i τ2) ∼ 2(R + i τ2)n asR → ∞. Moreover,

f (x, x0; R + i τ2) = (R + i τ2) coth(R + i τ2)− i Rx0 + x0τ2,

so that for everyε > 0,

|e− f (x,x0;R+i τ)/u| ≤ exp

[−(|R| + τ2)(1 − ε)+ x0τ2

u

]if R is large enough. The length of the vertical interval is proportional (asymptoti-cally) to R (see Prop. 3). Hence the integral on the vertical lineτ1 = R, and similarlyon τ1 = −R, tends to zero asR → ∞. The proposition follows from (4.3).

Remark.Recall that00,x,x0 denotes the closure of the set of non-purely-imaginarypoints of0x,x0. The analyticity ofV(τ ) and of f (x, x0; τ) on (i θ2k, i θ2k+1) impliesthat the two integrals over(i θ2k, i θ2k+1), performed in opposite directions, cancel out,and we may rewrite (4.9) as

P(x, x0; u) =1

(2πu)n+1

∫00,x,x0

e− f (x,x0,τ )/uV(τ )dτ. (4.10)


4.2. Computation of the wave kernelWe analyze the wave kernel for the isotropic Heisenberg Laplacian, that is, the solu-tionw(x, x0; t) of the partial differential equation

∂2w

∂t2=

[ n∑j =1

( ∂

∂x2 j −1+ 2αx2 j

∂

∂x0

)2

+

n∑j =1

( ∂

∂x2 j− 2αx2 j −1

∂

∂x0

)2]w(x, x0, t) (4.11)

with the initial conditionsw(x, x0; 0) = δ(x, x0; 0,0) and(∂w/∂t)w(x, x0; 0) = 0.This is the kernel of cos(

√−21H t), and we calculate it by combining Propositions9,

10, and12and by settingL = −21H .

PROPOSITION13If x 6= 0, x0 6= 0, then

P(x, x0; 2u) =1

2πn+1√

u

∫0x,x0

V(τ )∫

∞

0e−t2/(4u)

×∂

∂tδ(n−1/2)(t2

− 2 f (x, x0; τ))

dt dτ. (4.12)

ProofRecall thatf (x, x0; τ) is positive on0x,x0. By (4.2) (and by settingf = a2),

1

(4πu)n+1e− f (x,x0,τ )/(2u)

=1

2πn+1

1√

u

∫∞

0e−t2/(4u) ∂

∂tδ(n−1/2)(t2

− 2 f (x, x0; τ))

dt.

Substituting in (4.9), we get (4.12).

Note also that

P(x, x0; 2u) =1

2πn+1√

u

∫00,x,x0

V(τ )∫

∞

0e−t2/(4u)

×∂

∂tδ(n−1/2)

(t2

− 2 f (x, x0; τ))

dt dτ. (4.13)


THEOREM 2If x 6= 0, x0 6= 0, then

w(x, x0; t) = cos(√

−21H t)(x, x0)

=1

πn+1/2

∫0x,x0

∂

∂tδ(n−1/2)(t2

− 2 f (x, x0; τ))V(τ )dτ. (4.14)

ProofSetL = −21H , and apply Proposition9 to (4.12), getting the relation

1√

4πu

∫∞

−∞

e−t2/(4u)w(x, x0; t)dt

=1

2πn+1√

u

∫0x,x0

V(τ )∫

∞

0e−t2/(4u) ∂

∂tδ(n−1/2)(t2

− 2 f (x, x0; τ))

ds dτ.

(4.15)

Recall that on the unbounded branch of0x,x0, coth(2τ) → 1 asτ → ∞, so that

f (x, x0; τ) = Re f (x, x0; τ) ∼ |x|2τ1 + x0τ2 ≥ δ|τ |

for a certainδ = δ(|x|, x0) > 0. Proceeding as in the proof of Proposition10 andsettingt2

− 2 f (x, x0; τ) = y, we see that the inner integral on the right-hand side of(4.15) is equal to

(e− f (x,x0;τ)/(2u)/(

√π un+1)

) ∫∞

0 e−y/(4u)/√

y dy, so that for fixedu, the right-hand side of (4.15) is proportional to∫

0x,x0

V(τ )e− f (x,x0;τ)/(2u)∫

∞

0

e−y/(4u)

√y

dy dτ

≤ C∫0x,x0

V(τ )e−δ|τ | dτ∫

∞

0

e−y/(4u)

√y

dy dc.

Hence we may interchange the order of integration in (4.15) to obtain the equation∫∞

−∞

e−t2/(4u)w(x, x0; t)dt = 2∫

∞

0e−t2/(4u)w(x, x0; t)dt

=1

πn+1/2

∫∞

0e−t2/(4u)

[ ∫0x,x0

V(τ )∂

∂tδ(n−1/2)(t2

− 2 f (x, x0; τ))dτ]

ds.

(4.16)

Relation (4.16) holds for allu > 0; hence for all 1/u > 0 and by uniqueness for theLaplace transform, we get (4.14).

Note. Formula (4.14) remains valid if the integration is extended over00,x,x0.


COROLLARY 1If x 6= 0, x0 6= 0 and t 6= d j for all i ≤ j ≤ N, thenw(x, x0; t) is a real analyticfunction of all its arguments.

Note. Recall the formulad j (x, x0) =√

2 f (x, x0; i θ j ) for i ≤ j ≤ N. Note also thatw is a solution of the Heisenberg wave equation.

ProofLet t0 be different fromd j for all 1 ≤ j ≤ N. Note thatw(x, x0; t) ≡ 0 for t < d1

(finite speed; see [13] or [15]). We may assume thatt0 ∈ (d2k+1,d2k+2) for a certaink(or dN < t0). The only possible singularities ofw may arise from the contributions ofthe parts of0x,x0 betweeni θ2k+1 andi θ2k+2 (or i θN and∞). There exists a positiveε such that the interval (t0 − ε, t0 + ε) contains nod j . Recall that∂ f/∂τ 6= 0 betweeni θ2k−1 and i θ2k+2 (or betweeni θN and∞). The reality of f on 0x,x0 along withthe Cauchy-Riemann equations imply that there exists an analytic functionh( f ) (theinverse function off ) such thath(d2

2k+1/2) = i θ2k+1, h(d22k+2/2) = i θ2k+2. Let

ψ = C∞

0 (t0 − ε, t0 + ε) be such thatψ(t) ≡ 1 for t in a neighborhood oft0. To proveCorollary1, it suffices to prove the regularity of the distributionv(x, x0; t) given by

v(x, x0; t) =

∫0x,x0

δ(n−1/2)(t2− 2 f (x, x0; τ)

)V(τ )ψ

(√2 f (x, x0; τ)

)dτ.

Introduce a new real variableσ = 2 f (x, x0, τ ). Then the functionh(σ/2) is welldefined in(

√t0 − ε,

√t0 + ε ) ⊃ suppψ . Hence

v(x, x0; t) =

∫δ(n−1/2)(t2

− σ)V

(h(σ

2

))ψ(σ)

h′(σ/2)

2dσ.

SetV(h(σ/2))ψ(σ)(h′/2)(σ/2) = g(σ ). Thusg(σ ) is real analytic neart20 , and so is

v(x, x0; t) =

∫g(t2

− σ)δ(n−1/2)(σ )dσ

neart0.

It is possible to use the same change of variables on all of00,x,x0 in order to ob-tain a representation ofw(x, x0; t). Recall thatf (x, x0; τ) maps the branch of0x,x0

which joinsi θ2k−1 andi θ2k and for whichτ > 0 in a differentially invertible manneronto(d2

2k−1/2,d2k/2). Lethk(σ ) denote the inverse function off (x, x0; τ) defined on

(d22k−1/2,d

2k/2). Setwk(σ ) = V(hk(σ ))+ V(−hk(σ )). Similarly, lethN(σ ) denote

the inverse off (x, x0; τ), defined on(d2N/2,∞) and parametrizing the right branch

of 0N joining i θN and∞. The note after the proof of Theorem2 yields the following.


THEOREM 3If x 6= 0, x0 6= 0, then

w(x, x0; t) = cos(√

−21H t)(x, x0)

=1

πn+1

{ [N/2]∑k=1

∫ d2k

d22k−1

∂

∂tδ(n−1/2)(t2

− σ)Wk(σ )h′

k(σ/2)

2dσ

+

∫∞

d2N

∂

∂tδ(n−1/2)(t2

− σ)Wk(σ )h′

k

2

(σ2

)dσ

}. (4.17)

One may read off (4.17) the nature of the singularities ofw(x, x0; t) whent is neard j . An alternative determination of the singularities may be found in Nachman [13,p. 713]. We restrict ourselves to the leading singularity.

COROLLARY 2Let x 6= 0, x0 6= 0, and let all theθ j be distinct. Then there exist Cj (x, x0) such that

w(x, x0; t) ∼ C j (x, x0)δ(n)(t − d j ) if j is odd,

w(x, x0, t) ∼ C j (x, x0)(t − d j )−n−1 if j is even, (4.18)

for t near dj (x, x0).

ProofConsider first the case wherej is odd, j = 2k − 1. By Theorem3, the only con-tribution to the singularities ofw(x, x0; t) for t neard2k−1 or dN arises from in-tegration near the local minimum off at i θ2k−1 (or σ = d2

2k−1). By assumption,(∂2 f/∂τ2)(x, x0; i θ2k−1) 6= 0, so thath′

k(σ/2) ∼ C j /(σ − d22k−1)

1/2. (We denotedifferent constants byC j .) Hence the main singularity ofw(x, x0; t) is given by

C j

∫d2

2k−1

∂

∂tδ(n−1/2)(t2

− σ)dσ

(σ − d22k−1)

1/2

= C j

∫0

∂

∂t

δ(n−1/2)(t2− d2

2k−1 − σ)√σ

dσ

= C j

∫0

∂

∂t

(1

2

∂

∂t

)(n)δ(−1/2)(t2

− d22k−1 − σ)

dσ√σ

= C j

∫0

∂

∂t

(1

2

∂

∂t

)(n) 1√t2 − d2

2k−1 − σ

dσ√σ,

where the integrations and the differentiations are to be understood in the distributionsense, and only the left endpoint of theσ -interval is indicated. Applying this distribu-tion to a test functionϕ(t), we see that the leading part ofw(x, x0; t) applied toϕ(t)


is

C j

∫ ∫0

∂

∂t

(1

t

∂

∂t

)(n) 1√t2 − d2

2k−1 − σ

dσ√σϕ(t)dt

= C j

∫ ∫ t2−d2

2k−1

0

1√t2 − d2

2k−1 − σ

dσ√σ

[(1

t

∂

∂t

)n ∂

∂tϕ](t)dt

= C j

∫∞

d2k−1

[(1

t

∂

∂t

)n ∂

∂tϕ](t)dt = C jϕ

(n)(2d22k−1)+ · · ·

(recall that∫ a

0 dσ/√(a − σ) σ = π if a > 0), yielding the first relation in (4.18).

Similarly, if j = 2k is even, thenf has a local maximum ati θ2k (σ = d2k ). Then

the main singularity ofw(x, x0; t) is given by

C j

∫ d22k ∂

∂tδ(n−1/2)(t2

− σ)dσ

(d22k − σ)1/2

= C j

∫0

∂

∂tδ(n−1/2)(t2

− d22k + σ)

dσ√σ

= C j

∫0

∂

∂t

(1

2

∂

∂t

)(n) 1√t2 − d2

k + σ

dσ√σ.

Applying this distribution to a test functionϕ(t), we get the leading term

C j

∫ ∫0

(1

t

∂

∂t

)(n) 1√t2 − d2

2k + σ

dσ√σϕ(t)dt

= C j

∫ ( ∫0

dσ√(t2 − d2

k + σ)σ

)[(1

t

∂

∂t

)n ∂

∂tϕ](t)dt.

Note that 1/√(a + σ)σ makes sense fora < 0 as well if σ > −a and∫ 1

max(−a,0) dσ/√(a + σ)σ ∼a→0 C ln |a|. Hence the leading term is equal to

C j

∫ln |t2

− d2k |

[[1

t

∂

∂t

]n ∂

∂tϕ

](t)dt.

Sett − dk = ρ. Then ln|t2− d2

k | = ln |ρ| + ln(2dk + ρ) ∼ ln |ρ| asρ → 0 and

C j

∫ln |ρ|

[[ 1

ρ + dk

∂

∂ρ

]n ∂

∂ρϕ

](dk + ρ)dρ ∼ C j

∫ϕ(dk + ρ)

ρn+1dρ,

where the last integral is understood in the distribution sense.


Remark.The preceding results continue to hold in the anisotropic case (described atthe end of Sec. 1.2) if the conditionx 6= 0 is replaced throughout byx′′

6= 0.

The wave kernel may be computed ifx = 0. We discuss the isotropic case first. Recallthatd2

k (0, x0) = kπx0/α.

THEOREM 4For every positive integer n andα, x0 > 0, there exist constants aj,k, k =

1,2, . . . , 1 ≤ j ≤ n − 1, such that

w(0, x0, t) =

∞∑k=0

n−1∑j =0

a j,k∂

∂tδ(n+ j −1/2)(t2

− d2k ). (4.19)

If n = 1, then

w(0, x0, t) =

√π

2α

∞∑k=1

(−1)k+1k∂

∂tδ(1/2)(t2

− d2k ). (4.20)

ProofBy (4.3),

P(0, x0; 2u) =1

(4πu)n+1

∫∞

−∞

ei x0τ/(2u)V(τ )dτ.

There exists a positiveε such that for everyk,n there exists a functionWk,n(τ )

holomorphic in |τ − πki/(2α)| < 2ε, Wk,n(πki/(2α)) 6= 0, and V(τ ) =

Wk,n(τ )/(τ − πki/(2α))n. Using simple estimates (cf. [1]), it follows that

P(0, x0; 2u) =1

(4πu)n+1

∞∑k=0

∫|τ−πki/(2α)|=ε

ei x0τ/(2u) Wk,n(τ )

(τ − πki/(2α))ndτ.

(4.21)Applying the residue theorem, we see that

P(0, x0; 2u) =1

un+1

∞∑k=1

e−πkx0/(4αu)n−1∑j =0

b j,k

u j=

∞∑k=0

e−d2k/(4u)

n−1∑j =0

b j,k

un+1+ j.

(4.22)Application of (4.2) with a2

= d2k/2 yields

P(0, x0; 2u) =

∞∑k=1

1√πu

∫∞

0

n−1∑j =0

a j,ke−t2/(4u) ∂

∂tδ(n+ j −1/2)(t2

− d2k )dt, (4.23)

and (4.19) follows from Proposition9 (P(0, x0; 2u) = e−2u1H (0, x0)).


If n = 1, then (see [1, p. 654])

P(0, x0; 2u) =1

16αu2

∞∑k=1

(−1)k+1ke−d2k/(4u).

Applying (4.2) once again witha2= d2

k/2, we get (4.20).

In the anisotropic case, the functionV(τ ) has poles at the pointsπki/(2α j ), k =

1,2, . . . , 1 ≤ j ≤ n. We leave the formulation and proof of the anisotropic analogof (4.19) to the diligent reader.

5. Wave kernels via the continuation methodRecall thatL denotes a second-order positive semidefinite self-adjoint operator. Set

WL(t) =sin(L1/2t)

L1/2, (5.1)

so thatWL(t) is the (operator-valued) solution of the wave equation

∂2WL

∂t2= −LWL (5.2)

with the initial conditions

WL(0) = 0, W′

L(0) = I . (5.3)

We find an explicit representation for the kernel ofWL(t) when−L/2 is the Heisen-berg Laplacian and when−L is a degenerate elliptic operator of the type studied in[2]. Our method involves analytic continuation of the Green function ofL − ∂2/∂y2,and it is applicable wheneverL is positive definite or zero is in the continuous spec-trum of L (so thatL−1/2 is well defined, at least as a closed operator with a densedomain). The Green function(−L + ∂2/∂y2)−1 is defined as that (operator-valued)solutionG(y) of the equation(−L + ∂2/∂y2)G(y) = I · δ(y) which tends to zero as|y| → ∞. The main tool is the following proposition, essentially due to Taylor [15].

PROPOSITION14(i) Let L be a positive semidefinite operator. Then

1

2

(− L +

∂2

∂y2

)(L−1/2e−|y|L1/2

) = −I , (5.4)

1

2L−1/2e−|y|L−1/2

= −

(− L +

∂2

∂y2

)−1, (5.5)

WL(t) = limε→0

Im

(2(

− L +∂2

∂y2

)−1(·, i t + ε)

). (5.6)


(ii) Let m be a positive integer. Then for every a> 0,

limε→0+

Im1

(a2 + (i t + ε)2)m= c′

m

(1

t

∂

∂t

)m−1δ(t − a)

t, (5.7)

limε→0+

Im1

(a2 + (i t + ε)2)m−1/2= c′′

m

(1

t

∂

∂t

)m−1 H(t − a)√

t2 − a2, (5.8)

where the limits are understood in the distribution sense (i.e., in D′m(R+) of

t), and c′m, c′′m are negative constants.

ProofNote that

∂

∂ye−|y|L1/2

= −L1/2sign(y)e−|y|L1/2, (5.9)

∂2

∂y2e−|y|L1/2

= Lsign2(y)e−|y|L1/2− 2L1/2δ(y)e−|y|L1/2

, (5.10)

implying (5.4). The limits of L−(1/2)e−|y|L1/2and of (L − ∂2/∂y2)−1 as |y| →

+∞ vanish. The solution of the second-order ordinary differential equation (iny)(−L +∂2/∂y2)u = −I is uniquely determined by the limits limy=±∞ u(y), and (5.5)

follows. The operator-valued functionL−(1/2)e−yL1/2is holomorphic in the half-plane

Rey > 0, its boundary values aty = i t satisfy the wave equation (5.2), its imagi-nary part is uniformly bounded in compact subsets of the half-plane Rey ≥ 0, andthe functionL−(1/2)sin(L1/2t) = − Im(L−(1/2)e−i t L 1/2

) also satisfies the initial con-ditions (5.3), proving (5.6). To prove (5.7), set L = −12m+1. Then the kernel of(−L + ∂2/∂y2)−1 is given by(

12m+1 +∂2

∂y2

)−1(x, y; 0,0) =

cm

(|x|2 + y2)m(cm < 0) (5.11)

(for x ∈ R2m+1). On the other hand, it is well known that fort > 0,

W−12m+1(t)(x,0) = dm

(1

t

∂

∂t

)m−1δ(t − |x|)

t(dm > 0). (5.12)

Formula (5.7) follows from (5.11), (5.12), and (5.6). Similarly, to prove (5.8), setL = −12m, and recall the formulas (forx ∈ R2m)(

12m +∂2

∂y2

)(x, y; 0,0) =

cm

(x2 + y2)(2m−1)/2, (5.13)

and (fort > 0)

W−1m(t)(x,0) = dm

(1

t

∂

∂t

)m−1 H(t − |x|)√

t2 − x2. (5.14)


Recall (see [3], [2]) that the Green kernels of operators such as the Heisenberg Lapla-cian, the Heisenberg Laplacian+∂2/∂y2, and certain degenerate elliptic second-orderoperators are known to be of the form∫

R

V(z)

f (z)qdz,

whereV is an analytic function ofz (only), whereasf is a complex-valued action (ofthe type introduced and discussed in Sec. 1.2) associated with the symbol ofL (orL − ∂2/∂y2). The integration path may be deformed to a contour on whichf is realand (5.7), (5.8) may be applied.

5.1. The Heisenberg LaplacianHere we consider for simplicity only the isotropic case. Thus, given a positive integern and a positive constantα, the Heisenberg Laplacian1H is defined onR2n+1

=

{(x1, . . . , x2n, t)} by

1H =1

2

n∑j =1

( ∂

∂x2 j −1+ 2αx2 j

∂

∂y

)2+

1

2

n∑j =1

( ∂

∂x2 j− 2αx2 j −1

∂

∂x0

)2. (5.15)

R. Beals and P. Greiner [3] computed the Green kernel of 21H + ∂2/∂y2 with apole at the origin (ofR2n+2):

−

(21H +

∂2

∂y2

)−1(x1, . . . , x2n, x0, y)

= cn

∫∞

−∞

( 2ατ

sinh(2ατ)

)n dτ(|x|

2ατcoth(2ατ)+ y2/2 − i x0τ)n+1/2

, (5.16)

where|x|2

=∑2n

j =1 x2j andcn is a positive constant. We utilize this computation in

order to derive formulas for the wave kernelWH (t)(x, x0; 0,0), whereH denotes theoperator−21H . Setting

V(τ ) =

( 2ατ

sinh(2ατ)

)n, f (x, x0; τ) = |x|

2ατcoth(2ατ)− i x0τ, (5.17)

we may rewrite (5.16) with a differentcn as

−

(21H +

∂2

∂y2

)−1(x1, . . . , xn, x0, y) = cn

∫∞

−∞

V(τ )

[2 f (τ )+ y2]n+1/2. (5.18)

Note that f (τ ) = f (x, x0; τ) = f (|x|, x0, τ ; 2α). Hence the properties of the curve0x,x0 (the set wheref is real) are as described in Proposition3.

Remark.If x0 = 0, then0x,x0 coincides with the real axis.


It is well known that one may perform the integrations in (5.16) or (5.18) on the de-formed contour0x,x0. (Note that Ref > 0 betweenR and0x,x0, so that the fractionalpower is single valued.) The following is a consequence of Proposition14.

THEOREM 5If x 6= 0, then for t> 0,

WH (t)(x, x0; 0,0) = cn

∫0x,x0

(1

t

∂

∂t

)n H(

t −√

2 f (x, x0; τ))V(τ )dτ√

t2 − 2 f (x, x0; τ). (5.19)

ProofNote thatV(τ ) andτ coth(2ατ) are both even functions and are real ifτ ∈ R. HenceV(−τ) = V(τ ) and f (−τ) = f (τ ) (x0 is real!). Moreover, the mapτ → −τ maps0x,x0 onto itself, with the orientation reversed. Hence

Im∫0

V(τ )dτ

(2 f + (i t + ε)2)n+1/2

=1

2i

( ∫0

V(τ )dτ

(2 f (τ )+ (i t + ε)2)n+1/2−

∫0

V(τ )dτ

(2 f (τ )+ (−i t + ε)2)n+1/2

)=

1

2i

( ∫0

V(τ )dτ

(2 f (τ )+ (i t + ε)2)n+1/2+

∫0

V(−τ)d(−τ)

(2 f (−τ)+ (−i t + ε)2)n+1/2

)(we use the fact thatf (τ ) = f (τ ) on 0)

=1

2i

( ∫0

V(τ )dτ

(2 f (τ )+ (i t + ε)2)n+1/2−

∫0

V(τ )dτ

(2 f (τ )+ (−i t + ε)2)n+1/2

)=

∫0

V(τ ) Im[ 1

(2 f (τ )+ (i t + ε)2)n+1/2

]dτ. (5.20)

Application of (5.8) (with m = n + 1) to (5.18) proves (5.19).

Remark.As in Section 4, (5.19) may be written as involving only integrals of a realvariable.

If x = 0, x0 6= 0, then we may assumex0 > 0. The integrals in (5.16) and(5.18)may be computed using the residue theorem: the functionV(τ ) has a pole of ordernat τ = kπ i /(2α), k = ±1,±2, . . . , and we may deform the path of integration sothat only the poles with Im(τ ) > 0 matter. Fory 6= 0 there is no singularity atτ = 0.For everyε > 0 sufficiently small, we have

−

(21H +

∂2

∂y2

)−1(0, x0, y) = cn

∞∑k=1

∫|τ−kπ i /(2α)|=ε

V(τ )dτ

(−4i x0τ + y2)n+1/2. (5.21)


It is well known that there exist infinitely many geodesics joining(0, x0) to theorigin with lengths

dk =

√kπx0

α, k = 1,2, . . . . (5.22)

THEOREM 6If x = 0, x0 > 0, then for t> 0,

WH (t)(0, x0; 0,0) =

∞∑k=1

n−1∑j =0

cn,k, j

(1

t

∂

∂t

)n+ j H(t −√

kπx0/α)√t2 − kπx0/α

, (5.23)

where cn,k, j are functions of x0.

(Note that for fixedt the sum in (5.22) is finite.)

ProofBy (5.17), we have for eachk = 1,2, . . . that

V(τ ) =Wk(τ )

(τ − kπ i /(2α))n,

whereWk(τ ) is regular nearτ = kπ i /(2α). Hence∫|τ−kπ i /(2α)|=ε

V(τ )dτ

(−2i x0τ + y2)n+1/2

=2π i

(n − 1)!

∂n−1

∂τn−1

( (Wk(τ )

(−2i x0τ + y2)n+1/2

)τ=kπ i /2

=2π i

(n − 1)!

n−1∑j =0

(n − 1

j

)W(n−1− j )

k

·

(kπ i

2α

) (n + 1/2) · · · (n + 1/2 + j − 1)(2i x0)j

[kπ/α + y2]n+1/2+ j. (5.24)

Inserting (5.24) in (5.21) and using (5.8), we get (5.23).

Remark.Note that for every nonnegative integerm and a positive numberc,

δm−1(t2− c2) =

( 1

2t

∂

∂t

)m H(t − c)√

t2 − c2. (5.25)

Moreover, (5.1) implies that(∂/∂t)WH (t) = cos(√

−21H t). Hence Theorems5 and6 are equivalent to Theorems2 and4, respectively.


5.2. Degenerate elliptic operatorsIn this subsection we computeWL whenL is a degenerate elliptic operator of the typeconsidered in [2]. For simplicity we consider here a subclass, consisting of operatorsof the form

L = −1

2

(( ∂

∂x1

)2+

( ∂

∂x2

)2+ (Bx1, Bx1)

∂2

∂x20

), (5.26)

wherexi ∈ Vi (i = 1,2), V1,V2 are real Euclidean vector spaces, and∂/∂xi denotesthe gradient inVi (i = 1,2), ( , ) denotes the inner product inV1, andB is a positivedefinite matrix onV1. It was proved in [2] that−L has a fundamental solution of theform

G(x1, x2; x0, x01, x0

2, x00) = −c

∫∞

−∞

V(τ )dτ

f (x1, x01, x2 − x0

2, x0 − x00, τ )

q, (5.27)

where

V(τ ) = det( Bτ

sinh(Bτ)

)1/2, (5.28)

f (x1, x01, x2, x0, τ ) = −i x0τ +

τ

2

(B coth(Bτ)(x1 − x0

1), x1 − x01

)+ τ

(B tanh

( Bτ

2

)x1, x0

1

)+

|x2 − x02|

2

2, (5.29)

q =dim V1 + dim V2

2, (5.30)

andc is a positive constant. We consider here only the case wherex01 = 0 (without

loss of generality, we may assumex02 = t0

= 0), leaving the case wherex01 6= 0 to

the future.

Examples.Special cases of (5.26) are the Grusin operator∂2/∂x2+ x2(∂2/∂x2

0) andthe Baouendi-Goulaouic operator∂2/∂x2

1 + ∂2/∂x22 + (x1(∂/∂x0))

2.

SettingV2 = V2 × R, we see from (5.27) that a fundamental solution of 2L − ∂2/∂y2

with a pole at the origin is given by the following (special case) of (5.27):

G(x1, x2, x0, y; 0,0,0,0) = c∫

∞

−∞

V(τ )dτ

( f (x1,0, x2, x0, τ )+ y2/2)q+1/2. (5.31)

Set n = dim V1, let 0 ≤ a1 ≤ a2 ≤ · · · ≤ an denote the eigenvalues ofB(repeated according to their multiplicities), let(x1) j denote the component ofx1 in thej th eigendirection, and letx′′

1 denote the projection ofx onto the eigenspace belonging


to the largest eigenvaluean. Then

f (x1,0, x2, x0, τ ) = −i x0τ +τ

2

n∑j =1

a j (x1)2j coth(a j τ)+

|x2|2

2. (5.32)

Thus f (x1,0, x2, x0, τ ) is the sum of a positive number and a function of the form(1.26).

Note thatG is arbitrarily small if|(x1, x2, x0, y)| is sufficiently large. Hence theright-hand side of (5.31) represents the kernel of(−L + ∂2/∂y2)−1. Observe thatProposition3 holds if x′′

1 6= 0, x0 ≥ 0 (andL is the (degenerate) Laplace-Beltramioperator associated with the distanced on V1 × V2). Hence

−

(− L +

∂2

∂y2

)−1(x1, x2, x0, y; 0,0,0,0)

= c∫0x1,t

V(τ )dτ

( f (x1,0, x2, x0, τ )+ y2/2)q+1/2. (5.33)

The next theorem follows from (5.33) and Proposition14 in the same manneras in the proof of Theorem5. Note, however, the distinction between the case wheredim V1+dim V2 is even and where it is odd. For fixedx1, x2, andx0, let hk(σ ) denote

the inverse of the function (ofτ )√

2 f (x1, x2, x0, τ ) restricted to that part of0x1,x0 (0does not depend onx2) lying betweeni22k−1 andi22k and in the right half-plane.

THEOREM 7Let x′′

1 6= 0.(i) If dim V1 + dim V2 = 2p, where p is a positive integer, then

WL(t)(x1, x2, x0; 0,0,0)

= c′

∫0x,x0

(1

t

∂

∂t

)p H(

t −

√2 f (x1, x2,0, x0, τ )

)√t2 − 2 f (x1, x2,0, x0, τ )

V(τ )dτ. (5.34)

(ii) If dim V1 + dim V2 = 2p + 1, where p is a nonnegative integer, then

WL(t)(x1, x2, x0; 0,0,0) = c′′1

t

(−

1

t

∂

∂t

)pRe

V(hk(t))

h′

k(t). (5.35)

(Here d2k−1 < t ≤ d2k or dN < t .)

Proof


We apply Proposition14 to

(− L +

∂2

∂y2

)−1(x1, x2, x0, y; 0,0,0,0)

= −c∫0x,x0

V(τ )dτ

(2 f (x1,0, x2, x0, τ )+ y2)q+1/2(c > 0). (5.36)

If dim V1 + dim V2 = 2p, wherep is a positive integer, thenq = p and (5.8)is applicable, withm = p + 1, and (5.34) follows as in the proof of Theorem5. Ifdim V1 + dim V2 = 2p + 1, wherep is a nonnegative integer, thenq = p + 1/2, sothatq + 1/2 = p+ 1 and (5.7) is applicable to the integrand, withm = p+ 1. Hence

WL(t)(x1, x2, x0; 0,0,0) = c∫0x,x0

V(τ )(1

t

∂

∂t

)p δ(t −

√2 f (·, τ )

)t

dτ. (5.37)

Now let t ∈ (d2k−1,d2k) or t > dN . Thenh′

k(σ ) is nonzero in an open interval(d2k−1,d2k) (or dN,∞), and we may change the variable of integration fromτ to σ ,integrating over an interval rather than over an arc of0x,x0. Recall that we also have totake into account the branch of0x,x0 lying in the left half-plane, whereτ = −hk(σ )

and the orientation is reversed. Hence

WL(t)(x1, x2, x0; 0,0,0) = c∫

V(hk(σ ))

h′

k(σ )

(1

t

∂

∂t

)p δ(t − σ)

tdσ

+ c∫

V(hk(σ ))

h′

k(σ )

(1

t

∂

∂t

)p δ(t − σ)

tdσ, (5.38)

the integration is performed over an interval containingt (if t < d1, thenWL(t)(x1, x2, x0; 0,0,0) = 0), and (5.35) follows.

Remark.For both the Grusin operator and the Baouendi-Goulaouic operator,n = 1andx′′

1 = x1. More generally, ifB is a scalar operator, then the same equalityx′′

1 = x1

holds. The analysis of the case wherex1 6= 0, x′′

1 = 0 is very complicated (cf. [1])and is not attempted here. The case wherex1 = 0, x0 = 0 is relatively simple, as thefollowing theorem shows.

THEOREM 8Let x1 = 0, x0 = 0. If dim V1 + dim V2 = 2p, where p is a positive integer, then

WL(t)(0, x2,0; 0,0,0) = c(1

t

∂

∂t

)p H(t − |x2|)√t2 − x2

2

. (5.39)


If dim V1 + dim V2 = 2p + 1, where p is a nonnegative integer, then

WL(t)(0, x2,0; 0,0,0) = c(1

t

∂

∂t

)p δ(t − |x2|)

t. (5.40)


ProofIt follows from (5.29) and (5.33) that

−

(− L +

∂2

∂y2

)−1(0, x2,0, y; 0,0,0,0) = c

∫∞

−∞

V(τ )dτ

((x22 + y2)/2)q+1/2

=c1

(x22 + y2)q+1/2

. (5.41)

If dim V1+dim V2 = 2p, wherep is a positive integer, thenq = p and we may apply(5.8) to (5.41) (with m = p + 1), obtaining (5.39). If dim V1 + dim V2 = 2p + 1,wherep is a nonnegative integer, thenq = p + 1/2 and we may apply (5.7) to (5.41)(with m = p + 1), obtaining (5.40).

The case wherex1 = 0, x0 6= 0 is considerably more complicated. The results dependon the parity of the multiplicities of the eigenvalues ofB. We treat here only the casewhere B is scalar, and we distinguish between even and oddn. The general case,where each eigenvalue has either even or odd multiplicity, is essentially a combinationof the scalar cases.

THEOREM 9Let x1 = 0, x0 > 0. Assume that B= aI and dim V1 = n = 2n′, where n′ is apositive integer. Ifdim V1 + dim V2 = 2p, where p is a positive integer, then

WL(t)(0, x2, x0; 0,0,0)

=

∞∑k=1

n′−1∑

j =0

ck, j ·

(1

t

∂

∂t

)p+ j H(

t −

√2kπx0/a + x2

2

)√t2 − 2kπx0/a − x2

2

. (5.42)


WL(t)(0, x2, x0; 0,0,0) =

∞∑k=1

n′−1∑

j =0

c′

k, j ·

(1

t

∂

∂t

)p+ j δ(t −

√2kπx0/a + x2

2

)t

.

(5.43)The constants ck, j and c′k, j in (5.42) and (5.43) also depend on a and on n.

Note that as in Theorem6, the sums in (5.42) and in (5.43) are finite fort fixed.

ProofLet x0 > 0 (the case wherex0 < 0 is treated similarly). Thenf (0,0, x2, x0, τ ) =

−i x0τ + x22/2 is real forτ in the imaginary axis and is positive if Imτ > 0. The


functionV(τ ) defined in (5.28) is given by

V(τ ) =

( aτ

sinh(aτ)

)n′

,

and it has poles of ordern′ at τ = kπ i /a, k = ±1,±2, . . . . As in the discussionleading to(5.18), we get from (5.36) the representation (valid fory 6= 0 andε > 0sufficiently small)

−

(− L +

∂2

∂y2

)−1(0, x2, x0, y; 0,0,0,0)

= c∞∑

k=1

∫|τ−kπ i /a|=ε

V(τ )dτ

[−2i x0τ + x22 + y2]q+1/2

= c∞∑

k=1


Vk(τ )

(τ − kπ i /a)n′

dτ

[−2i x0τ + x22 + y2]q+1/2

,

(5.44)

whereVk is regular nearτ = kπ i /a. But


Vk(τ )

(τ − kπ i /a)n′

dτ

[−2i x0τ + x22 + y2]q+1/2

=2π i

(n′ − 1)!

n′−1∑

j =1

(n′− 1

j

)· V (n′

− j −1)k

(kπ i

a

) (q + 1/2)(q + 1/2 + 1) · · · (q + 1/2 + j − 1)(2i x0)j

|2kπx0/a + x22 + y2|

q+1/2+ j. (5.45)

If dim V1+dim V2 = 2p, wherep is a positive integer, thenq+1/2+ j = p+ j +1/2,and we may apply (5.8) (m = p+ j +1) and substitute (5.45) in (5.44) to obtain (5.42).If dim V1 + dim V2 = 2p + 1, wherep is a nonnegative integer, thenq + 1/2 + j =

p + j + 1; we may apply (5.7) (m = p + j + 1), and, substituting (5.45) in (5.44),we get (5.43).

If n is odd, thenV(τ ) is no longer single-valued. First we have to determine theboundary values ofV(τ ) on the imaginary axis.

PROPOSITION15Let n be an odd integer, n= 2n′

+ 1, and let a denote a positive number. Then thereexist temperate (one-dimensional) distributions E+, E− such that

E+(s) = limε→0+

( a(is + ε)

sinh(a(is + ε))

)n/2,

E−(s) = limε→0+

( a(is − ε)

sinh(a(is − ε))

)n/2. (5.46)


Moreover,singsupp(E+) = singsupp(E−) = {kπ/a, k = ±1,±2, . . .}. Outside ofthe singular support, the functions E+(s) and E−(s) coincide with each other andare real if |s| < π/a or 2k(π/a) < |s| < (2k + 1)(π/a), k = ±1,±2, . . . , and arepurely imaginary with E+(s) = −E−(s) if (2k − 1)(π/a) < |s| < 2k(π/a), k =

±1,±2, . . . , and in this case,

E+(s) = (−1)k+1i(

|as|

| sin(as)|

)n/2. (5.47)

Both E+ and E− are n′th derivatives of a locally integrable function whose singular-ities are of the form1/(s − k(π/a))1/2.

ProofThe function sinh(z)/z vanishes if and only ifz = kπ i, k = ±1,±2, . . . . Moreover,sinh(z)/z is positive ifz is real. Hence the function(aiτ/sinh(aiτ))n/2 is well definedin the half-planes Im(τ ) > 0 and Im(τ ) < 0 with (aiτ/sinh(aiτ))n/2 positive if τ isreal (except at the poles). The zeros of sinh(i z) at z = kπ/a are simple, so that( aiz

sinh(aiz)

)n/2= O

( 1

|z − kπ/a|)n/2

)(5.48)

for Im(z) 6= o, z nearkπ/a. Hence the limits in (5.46) exist as distributions (see[7, p. 63]). If s is real, then the function sin(as)/(as) is positive for|s| < π/a or2k(π/a) < |s| < (2k + 1)(π/a), k = ±1,±2, . . . , and negative elsewhere. Hencethe functions(aiz/sinh(aiz))1/2 and (aiz/sinh(aiz))n/2 have branch points atz =

kπ/a, k = ±1,±2, . . . , and cuts in the intervals where sin(as)/(as) is negative. Thepositive choice of the square root nearz = 0 (in particular, forz = is with |s| < π/a)implies thatE+(s) andE−(s) must be negative for 2π < s < 3π and then positivefor 3π < s < 5π , and so on. Hence the alternating signs for the imaginary case(5.47). Recall that

z

sin(z)= (−1)k

( kπ

z − kπ

)+ hk(z), (5.49)

wherehk(z) is holomorphic nearkπ . Hence the function (having a branch point atkπ ) ( aiτ

sinh(aiτ)

)n/2=

( aiτ

sinh(aiτ)

)n′

·

( aiτ

sinh(aiτ)

)1/2

can be expanded in an algebraic Laurent series in powers of((τ − kπ)/a)1/2, startingwith −n = −(2n′

+ 1). This expansion may be integratedn′ times.

Set nowE = E+ − E−. ThenE(s) = 0 if |s| < π/a or 2k(π/a) < |s| < (2k +

1)(π/a), k = ±1,±2, . . . , andE(s) is purely imaginary if(2k − 1)(π/a) < |s| <


2k(π/a), k = ±1,±2, . . . ; then

E(s) = (−1)k+12i(

|as|

| sin(as)|

)n/2. (5.50)

THEOREM 10Let x1 = 0, x0 > 0. Assume that B= aI anddim V1 = n = 2n′

+ 1, where n′ is anonnegative integer. Ifdim V1 + dim V2 = 2p, where p is a positive integer, then

WL(t)(0, x2, x0; 0,0,0) = c∫

∞

0E(s)

(1

t

∂

∂t

)p H(

t −

√2sx0 + x2

2

)√t2 − 2sx0 + x2

2

ds. (5.51)


WL(t)(0, x2, x0; 0,0,0) = c∫

∞

0E(s)

(1

t

∂

∂t

)p δ(

t −

√2sx0 + x2

2

)t

ds, (5.52)

and there exist real constants ck such that the leading singularity of WL(t)(0, x2,

x0; 0,0,0) is given by

∞∑k=0

c2k+1H(

t −√(2k + 1)π/(ax0)− π/(2ax0)

)(t −

√(2k + 1)π/(ax0)

)n/2+p+

+

∞∑k=1

c2k H(

t −√(2k)π/(ax0)− π/(2ax0)

)(t −

√(2k)π/(ax0)

)n/2+p−

. (5.53)

(The distributions 1/sn/2+p+ and 1/sn/2+p

− are defined, e.g., in [7, pp. 68 – 71].) Notethat the integrals in (5.51) and in (5.52) are actually performed over compact intervals,and note that the sum in (5.53) is finite if t is fixed.

ProofLet y 6= 0. Recall that by (5.33),

−

(− L +

∂2

∂y2

)−1(x1, x2, x0, y; 0,0,0,0) = c

∫0x1,t

(aτ/sinh(aτ))n/2 dτ

(−2i x0τ + x22 + y2)q+1/2

,

(5.54)wherec is a positive constant,V(τ ) = (aτ/sinh(aτ))n/2, andq = p + (1/2). Anestimate similar to the one used in the proof of Proposition12 implies that for everyε > 0, R > 0, the integral in (5.54) may be extended over the contour composedof {−∞ < τ1 ≤ −ε, τ2 = R}, {τ1 = −ε,0 ≤ τ2 ≤ R}, {−ε ≤ τ ≤ ε, τ2 = 0},{τ1 = ε,0 ≤ τ2 ≤ R}, and{ε ≤ τ1 < ∞, τ2 = R}. For every positive integerj , set


Rj = ( j + 1/2)π . Then∫

∞

ε|V(i R j + σ)| dσ ,

∫−∞

−ε|V(i R j + σ)| dσ areO(Rn/2

j )

while (−2i x0τ + x22 + y2)−q−1/2

= O(R−q−1/2j ) for Im τ = Rj , j → ∞. Hence

−(−L+∂2

∂y2)−1(x1, x2, x0, y; 0,0,0,0)

= c lim j →∞

( ∫ 0

Rj

V(−ε + is)d(is)

[−2i x0(−ε + is)+ x22 + y2]q+1/2

+

∫ ε

−ε

V(−ε + is)d(is)

[−2i x0(−ε + is)+ x22 + y2]q+1/2

+

∫ Rj

0

V(ε + is)d(is)

[−2i x0(ε + is)+ x22 + y2]q+1/2

= c( ∫ 0

∞

V(−ε + is)d(is)

[−2i x0(−ε + is)+ x22 + y2]q+1/2

+

∫ ε

−ε

V(−ε + is)d(is)

[−2i x0(−ε + is)+ x22 + y2]q+1/2

+

∫∞

0

V(ε + is)d(is)

[−2i x0(ε + is)+ x22 + y2]q+1/2

)).

(As a side remark, note that the integrals do not converge absolutely if dimV2 ≤ 1.The following argument is valid without absolute convergence; one could also intro-duce additional artificialx2-variables and apply a method of descent.) The regularityof the integrands atτ = 0 (note thaty 6= 0) implies that we may letε tend to zeroand obtain the representation

−

(−L+

∂2

∂y2

)−1(x1, x2, x0, y; 0,0,0,0) = ci

∫∞

0

E(s)ds

[y2 + 2x0s + x22]q+1/2

. (5.55)

Recall thatE(s) is purely imaginary. Hence

Im(

− L +∂2

∂y2

)−1(x1, x2, x0, y; 0,0,0,0)

= ci∫

∞

0E(s) Im

( 1

[y2 + 2x0s + x22]q+1/2

)ds. (5.56)

Puttingy = i t + ε, wheret > 0, and lettingε tend to zero from the right, we get (atleast formally) (5.51) and (5.52) from (5.56), (5.7), and (5.8). The intersection of thewave front setW F(E(s)) with the setW F′(K )R1, whereK is one of the right-handsides of (5.7) or (5.8), is empty. By [7, Th. 8.2.13], the “integrals” in (5.51) and (5.52)make sense and are the limits of the integrals in (5.56).


Formula (5.53) follows from the observation that near the branch pointskπ/a, the leading singularity ofE+(E−) is given by (1/(s − kπ/a − i 0)n/2)(1/(s − kπ/a + i 0)n/2).

For the Grusin operatorn = 1,dim V2 = 0, so thatp = 0, and we get from (5.52) thesimple expression

WL(t)(0, x0; 0,0) = c′

∞∑k=1

(−1)k∫ 2kπ

(2k−1)π

(|s|

| sin(s)|

)1/2δ(

t −√

2x0s)

tds

= (−1) j c′

( t2/(2x0)

− sin(t2/(2x0))

)1/2, (5.57)

where j is a positive integer determined by(2 j − 1)π < t2/(2x0) < 2 jπ , andWL(t) = 0 if no such j exists.

6. Directions for further studiesWe suggest here a certain number of open problems connected to this paper. Thefirst question is how to extend the methods to find the wave kernel on a Riemannianmanifold (Mn, g) for a degenerate operatorL =

∑m1 Łi Ł∗

i , where Łi arem vectorfields such that their Lie brackets generate the tangent space at each point of themanifold. In particular, we are interested in the role played by the geometry.

Another possible extension is to consider the wave kernel for

∂2

∂t2w = 1w − V(x)w,

w(P,0) = 0,

wt (P,0) = δP, (6.1)

where V is a double well potential. A related question in dimension 2, is to study

∂2

∂t2w = ∂xxw + V(x)∂x0x0w,

w(x, x0,0) = 0,

wt (x, x0,0) = δP. (6.2)

We can choose, for example,V(x) = x2(1− x)2. The pointP can be(0,0) or (1,0).What is the picture of interferences?

Finally, we suggest a problem in the direction of the nonlinear wave equation. Inparticular, we are interested in the global existence or possible blow up in finite time


for the following wave equation:

∂2

∂t2w = 1Hw + |w|

p−1w,

w(P,0) = g,

wt (P,0) = f, (6.3)

where p ≤ pc = 1 + 2/n is the critical exponent in the Stein-Sobolev inequalityon the Heisenberg group and where1H is the Heisenberg Laplacian. The initial dataare smooth enough. We expected some anisotropic phenomena due to the interactionbetween the nonlinearities and the propagation along the bicharacteristics.

Acknowledgment.We are very much indebted to the referee for a careful reading ofthe manuscripts and for comments that greatly improved the exposition.

References

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GreinerDepartment of Mathematics, University of Toronto, Toronto, Ontario M5S 3G3, Canada;[email protected]

HolcmanDepartment of Mathematics, Weizmann Institute of Science, Rehovot 76100, Israel;[email protected]; current: Department of Physiology, University of Californiaat San Francisco, Keck Center, 513 Parnassus Ave., San Francisco, California 94143-0444, USA

KannaiDepartment of Mathematics, Weizmann Institute of Science, Rehovot 76100, Israel;[email protected]


http://www.ams.org/mathscinet-getitem?mr=89h:35177

http://www.ams.org/mathscinet-getitem?mr=84e:58074

http://www.ams.org/mathscinet-getitem?mr=88b:53055

http://www.ams.org/mathscinet-getitem?mr=90f:53081

http://www.ams.org/mathscinet-getitem?mr=88a:22021

mailto:[email protected]



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pdfs.semanticscholar.org · DUKE MATHEMATICAL JOURNAL Vol. 114, No. 2, c 2002 WAVE KERNELS RELATED TO SECOND-ORDER OPERATORS PETER C. GREINER, DAVID HOLCMAN, and YAKAR KANNAI Abstract