Can One Hear Kac

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Can One Hear the Shape of a Drum?Author(s): Mark KacReviewed work(s):Source: The American Mathematical Monthly, Vol. 73, No. 4, Part 2: Papers in Analysis (Apr.,1966), pp. 1-23Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/2313748.

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CAN ONE HEAR THE SHAPE OF A DRUM?MARK KAC, The RockefellerUniversity,New York

To George Eugene Uhlenbeck on the occasion of his sixty-fifthirthday"La Physique ne nous donne pas seulementl'occasion de resoudre des problemes .. elle nousfaitpresentir a solution." H. POINCARE.

Before explainthe title and introduce hethemeofthelecture shouldliketo state thatmy presentationwillbe more n the nature ofa leisurely xcursionthan of an organizedtour. It will not be my purpose to reach a specifieddes-tinationat a scheduled time. Rather I should like to allow myselfon manyoccasions the luxuryof stoppingand lookingaround. So much efforts beingspent on streamliningmathematicsand in rendering t more efficient,hat asolitarytransgressiongainst the trend could perhapsbe forgiven.r

Q

FIG. 11. And now to the themeand the title.It has been knownforwell over a century hat if a membrane 2,held fixedalongitsboundaryF (see Fig. 1), is set inmotion ts displacement in the direc-tionperpendicular o its originalplane)

F(x, ; t)-F (p; t)obeys thewave equation

aF = c2 72F,at2wherec is a certainconstantdependingon thephysicalproperties fthe mem-brane and on the tensionunderwhich themembrane s held.I shall choose units to make C2=2


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2 PAPERS IN ANALYSISOf special interest both to the mathematicianand to the musician) aresolutionsof theform

F(p-; t) =: U(p)eiwl,for,beingharmonic n time,they represent heputre onesthemembrane s capa-ble ofproducing.These special solutionsare also known as normalmodes.To findthe normalmodes we substitute U(-)eiwt nto the wave equationand see that U must satisfythe equation 2 V2U+co2U=O with the boundarycondition U=O on theboundaryF ofQ, correspondingo the membranebeingheld fixed long its boundary.The meaningof "U =0 on F" shouldbe made clear; for ufficientlymoothboundaries t simplymeansthat U(p)->O as p approachesa pointof F (from heinside). To show that a membrane s capable ofproducing discretespectrumofpure tones i.e. that there s a discretesequence ofc's WO as p->a pointof F.It is customaryto normalizethe4t'sso that/ (p)-dp= 1.

Note that I use dp to denote the element of integrationin Cartesian co-ordinates,e.g., dp-dxdy).2. The focal pointofmyexposition s the following roblem:Let ?2and ?2 be twoplane regionsboundedbycurvesF, and F2respectively,and considerthe eigenvalueproblems:

V2U + XU 0 in Q 22V + AV =0 in Q2with withU= on F, V= on F2.

Assumethatforeach n theeigenvalueN,, or?1 is equal to theei-genvalueP1


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CAN ONE HEAR THE SHAPE OF A DRUM? 3for 2. Question: Are the regions i1 and 02 congruent n the sense of Euclideangeometry?I first eard theproblem posed this way some ten years ago fromProfessorBochner. Much more recently,when I mentioned t to ProfessorBers, he said,almost at once: "You mean, if you had perfectpitch could you findthe shapeof a drum.'"You can now see that the "drum" of my title is more like a tambourine(which really is a membrane)and that strippedof picturesque language theproblem s whetherwe can determineQ if we know all the eigenvalues of theeigenvalue problem

V2U+XU=0 in Q,U =0 o0n F.3. Before go any further et me say that as faras I know the problem sstill unsolved. Personally, believe that one cannot "hear" the shape ofa tam-bourinebut I may well be wrongand I am not prepared to bet large sumseitherway.What I proposeto do is to see how much about the shape can be inferredfrom heknowledgeof all the eigenvalues, and to impress upon you the multi-tude ofconnectionsbetween our problemand variouspartsofmathematics ndphysics.It should perhaps be stated at this point that throughout he paper onlyasymptotic roperties f large eigenvalues will be used. This may represent, fcourse,a serious oss of information nd it may perhapsbe argued that preciseknowledge f all theeigenvalues may be sufficiento determine he shape ofthemembrane. It should also be pointed out, howTever,hat quite recentlyJohnMilnor constructed wo noncongruent ixteen dimensional tori whose Laplace-Betrami operatorshave exactly the same eigenvalues (see his one page note"Eigenvalues of the Laplace operatoron certain mlanifolds" roc. Nat. Acad.Sc., 51(1964) 542).4. The first ertinent esult s that one can "hear" the area ofQ. This is anold result witha fascinatinghistorywhich shall now relate briefly.At the end of October 1910 the great Dutch physicistH. A. Lorentz wasinvited to G6ttingento deliver the Wolfskehl ectures. Wolfskehl,by the way,endowed a prize for proving, or disproving,Fermat's last theoremand stipu-lated that in case the prize is not awarded the proceeds fromthe principal beused to invite eminent cientists o lecture at Gottingen.Lorentzgave five ecturesunder the overall title"Alte und neue FragenderPhysik"-Old and new problems f physics-and at the end of the fourth ecturehe spoke as follows in freetranslationfrom he original Germ-Ian): In conclu-sion there s a mathematical probleimi hich perhapswill arouse the interest fmathematicianswho are present. t originates n the radiation theoryof Jeans."In an enclosure with a perfectly eflectingurfacethere can form tanding


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4 PAPERS IN ANALYSISelectromagneticwAAavesnalogous to tones ofainorganpipe; we shall confine urattention to very highovertones. Jeans asks forthe energy n the frequencyintervaldv. To this end he calculates thenumlberfovertoneswhich ie betweenthefrequencies and v+dy and multiplies hisnumberby theenergywhich be-longs to thefrequenlcy,and which ccording o a theorem fstatistical mllechan-ics is the samne or all frequencies."It is herethatthere rises themathematicalproblem oprove thatthe num-berofsufficiently igh overtoneswhich ies between vand v+dv is independentofthe shape of theenclosure nd is simiiplyroportional o itsvolumlle. or manysimnplehapes forwAhichalculationscanl be carried out, thistheoreml as beenverified n a Leiden dissertation.There is no doubt that itholds ingeneralevenformultiply onnectedregions.Sinilar theoremnsorothervibrating tructureslike inembranes, ir inasses, etc. should also hold."If one expressesthis conjectureof Lorentz in terms of our imembrane, tenmergesn theform:

V (X) < 27rHere N(X) is the numberof eigenvalues less thanX, | I the area of Q andmeans that

N0 (X) |Q|X a X 27r

There is an apocryphalreport hat Hilbertpredicted hatthe theoremwouldnot be proved in his lifetime.Well, he was w-rongy many,manyyears. Forless than two years later HermiianWeyl, who was present at the Lorentz' lec-ture and whose interestwas aroused by the problem,proved the theorem nquestion, i.e. that as X >ocN1(X) x.

Weyl used in a masterlyway the theoryof integralequations, whichhisteacher Hilbertdeveloped onlya few,-earsbefore, nd hisproofwas a crowningachievementof this beautiful theory.I\lanysubsequent developnmentsn thetheory fdifferentialnd integral quations (especiallythe wNorkfCoturantndhisschool) can be traceddirectly oWeyl'smemoir n theconjectureof Lorentz.5. Let me now considerbriefly different hysical problem which too iscloselyrelated to theproblenmfthedistribution feigenvaluesof theLaplacian.It can be taken as a basic postulateofclassical statistical mechanicsthat ifa systemof 1l particlesconfined o a volumeQ is in equilibriumwTith thermo-statoftemperatureT theprobability ffinding pecified articles t ri,, 2,rm (tn lel nd r -rMf wvithinvolume elementsdr1, r2, , drAzl)s


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CAN ONE HEAR THE SHAPE OF A DRUM? 5exp [- V(rly .. ** rml)0 dr-, ... dr,-Tdr* J. exp V(rl ... rm) dr, ... drm

whereV(r-1, ,rll) is the interactioniotentialof the particlesand k=R/NwithR the "gas constant" and N the Avogadro number.For identical particles each of mass m obeying the so called Boltzmainiistatisticsthe correspondingssumiptionn quantumstatisticalmechanics eemsmuchmorecomplicated.One startswith theSchr6dinger quation2 V - V(ri, .., r_>) Eq Q - , whereh is thePlanckconstant)2m 27with the boundarycondition imn(ri. , 1)=0, whenever at least one rkapproaches the boundaryof Q. (This boundaryconditionhas the effect f con-finingthe particles to Q.) Let E1 < E2? E3< be the eigenvalues and4'11,, . thecorrespondingormalizedigenfunctions.henthebasicpostu-late is that theprobability f findingpecified articles t ri, r2, , ur withindr1* **, d-rm) s j ,/kT 2-+-e A(rl * ri)9drl . . . dr31

s=lZ e-EsIkTs=1

There are actually no known particles obeying the Boltzmann statistics. Butdon't let this worryyou forour purposes this regrettable fact is immilaterial.Now, let us specialize our discussion to the case of an ideal gas w7hich,bydefinition,means that V( r-, , r1) _O.Classically, the probabilityof finding pecifiedparticles at r1, , r7 isclearly

dr1 dr3where IQ is now the volume of Q.Quantum mechanically the answer is not nearly so explicit. The Schr6dingerequation for an ideal gas is

ft22m 9 -and the equation is obviously separable.If I now consider the three-dimensional (rather than the 3MH-dimenlsional)eigenvalue problem


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6 PAPERS IN ANALYSIS2V2A(r) = - rV(r) -> 0 as r the boundary of 9,

it is clear thattheEs as wellas the r) are easilyexpressiblentermiisof the X's and corresponding 1(r)'s.The formulaforthe probability f finding pecified articles at -ri, rmturns out to beexp [-m-- T]'fl(rk)H ~ LL _ _ _ _ _ _ _ _ _ _ - dr ,k=1 r0 AntE exp -- ]

Now, as 1->0 (or as T-*>occ)the quantum mechanicalresult should go overinto the classical one and this immllediatelyeads to the conjecturethat asLmkT_

E An(F) i?1=1 n=1

If instead of a realistic three-dimnensionalontainerQ I consider a two-dimensional one, the resultwould stillbe the same-e_n 2r)e-X,- 0

n=1 n=1except that now IQ| is the area ofQ rather than thevolume.Clearly the resultis expected to hold onlyforr in the interior fQ.If we believe Weyl's result that (in the two-dimensional ase)

IV (X) /_ -I I 1%),2erit follows mmediatelyby an Abelian theorem hat

1 0 1__EXnT - TO,-- 0and hencethat

-X2 1r2 -XTEeXnr) -Ie dX.n=1 27rT 2roSettingA (X)= x


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CAN ONE HEAR THE SHAPE OF A DRUM? 7de-XrA (X) e-Xr X, r -> 0.

Since A (X) is nondecreasingwe can apply theHardy-Littlewood-KaramataTauberian theorem nd conclude whateveryonewouldbe temptedto conclude,namely that2- NA X)= + (r)2 , -->o,

f r everyr in the interior fQ.Though this asymptoticformula is thus nearly "obvious" on "physicalgrounds," t was not until1934thatCarlemansucceeded insupplying rigorousproof.In concludingthis section it may be worthwhile o say a word about the"strategy"of our approach.We are primarily nterested, f course, in asymptotic propertiesofX,,forlargen. This can be approached by the device of studyingthe Dirichletseries

00E e-Xntn=1

for small t. This in turn s mostconveniently pproachedthrough he series00 {D00ZeXt,(P) = fe-tdA(x)n1=1

and thuswe are led to theAbelian-Tauberian nterplaydescribed above.6. It would seem that the physical intuitionought not only provide themathematicianwith nterestingnd challenging onjectures,but also showhimthe way toward a proofand towardpossible generalizations.The contextof thetheory f black bodyradiationorthat ofquantumstatis-

tical mechanics,however, s too farremovedfrom lementaryntuition nd toofull of daring and complexphysicalextrapolationsto be of much use even inseekingthe kind of understanding hat makes a mathematiciancomfortable,let alone in pointing owarda rigorousproof.Fortunately, n a much more elementarycontext the problemof the dis-tribution feigenvaluesof theLaplacian becomesquite tractable.Proofs mergeas natural extensionsofphysical ntuition nd interesting eneralizations omewithinreach.7. The physical context in question is that of diffuion theory, notherbranchofnineteenth enturymathematicalphysics.Imagine "stuff," nitiallyconcentrated t p( (xo,yo)), diffusing hroughplane region Q bounded by F. Imaginefurthermorehat the stuff ets absorbed( "eaten") at theboundary.


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8 PAPERS IN ANALYSISThe concentration Q(p| r; t) of matterat r-((x, y)) at time t obeys thedifferentialquation of diffusion

aPe 1(a) - =t 2the boundarycondition(b) PQ(p I r; t) 0 as r approaches boundarypoint,and the initialcondition(c) PQ(P |r; ) -*(r-p) as t ->0;here (-r-p) is the Dirac "delta function,"with"value" oo f -r= p and 0 ifr p'.The boundarycondition b) expressesthe factthat the boundary s absorb-ing and the initial condition(c) the fact that initially all the "stuff"was con-centrated at ip.I have again chosen units so as to make the diffusion onstant equal to 2.As iswellknown heconcentration Q(PI -; t) can be expressed n terms f theeigenvaluesXz,and normalizedeigenfunctions,(r-) of the problem

'72v +X4/=0 in Q,0 onP.In fact,PQ(p I ; t)= n-], 6-fl'4' (p)Abl(r).NOw, or mallt, t appears ntuitivelylear thatparticles f thediffusingstuff illnothave hadenough ime ohavefelt he nfluenceftheboundary .Asparticles egin odiffuseheymaynotbe aware, o to speak,ofthe disasterthat awaitsthemwhen hey each heboundary.We maythus xpect hat n someapproximateense

PQ(p r;t) - Po(p I r; t), as t O ,where o(p1r; t) still atisfieshesame diffusionquation(a') P VPOand the same nitial ondition(c') Po(p ;t) = P-p), t0,but s otherwisenrestricted.Actually here s a slight dditional estrictionithoutwhich hesolutionis notunique a remarkableactdiscoveredomeyears go by D. V. Widder).Therestrictions thatPo> 0 (ormore enerallyhatPobe bounded romelow).A similar estrictionorPQ is notneeded incefordiffusionn a boundedregiont follows utomatically.


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CAN ONE HEAR THE SHAPE OF A DRUM? 9An explicit formulaforPo is, of course,well known. It is

Po(pIr;t) exp - - ]where |- denotesthe Euclidean distance betweenp and r.I can now state a littlemoreprecisely he principle f "not feeling he bound-ary" explaineda momentago.The statement s that as t >0

00 1 F r ~P~2 1.o(Pnp r; t) = 1 -,1A()nr exp- 2- npr; t),Q( n=1 2w1t 1 21-where - stands here for "is approximately qual to." This is a bit vague butlet it go at that forthe moment.If we can trust this formula ven forp= r we get

00 1x'-Ant 2 1e-e n(r)n=l 27rtand ifwe display still moreoptimismwe can integratethe above and, makinguse of the normalization ondition

2n(r-)d-r'=: 1,obtain

I:e-Xnt, Q.1=1 27rtWe recognize mmediately heformulas iscusseda while back inconnectionwith the quantum-statistical-mechanical reatment of the ideal gas. If we

apply the Hardy-Littlewood-Karamata theorem, lluded to before,we obtainas corollarythe theoremsofCarlemanand Weyl.To do this, however,we mustbe allowed to interpret as meaning "asynmp-totic to."8. Now, a littlemathematicalsoul-searching.Aren't we as far from rigor-ous treatnments we were before?True, diffusion s more familiarthan blackbody radiation or quantum statistics. But familiarity ives comfort, t best,and comfortmay stillbe (and often s) milesaway from herigordemanded bymathematics.Let us see then what we can do about tightening he loose talk.First let medispose ofa fewminor temswhichmay cause you worry.WhenI write /= 0 on F orP(p Ir; t)-O as r approachesa boundary pointofQ there s always a question ofinterpretation.


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10 PAPERS IN ANALYSISLet me assume that r is sufficientlyegular o that no ambiguity rises i.e.

P(p r; t) -O 0 as r -> a boundarypointofQ,means exactly what it says, while41 0 on r means4 0 as r * a boundarypointofQ.Likewise,P(p| r; t)->5(r-p) as t-*0,has the obvious interpretation,.e.

lim fJP(p r; t)dr= 1forevery open set A containingP.

Now, to morepertinenttems. f themathematicaltheoryofdiffusionorre-sponds in any way to physical realitywe should have the inequalityr_IIP l11exp - 71?

P(QP; t) 0o(p|I;t) L 2tFor surely ess stuffwillbe foundat rat timet fthere sa possibility fmat-terbeingdestroyed on the boundaryF ofQ) than iftherewere no possibility

of such destruction.Now letQ be a square with center t p totallycontained nQ. Let itsbound-aryact as an absorbingbarrier nd denoteby PQ P I ; t), r Q,thecorrespondingconcentration t r at timet.In otherwords,PQ satisfies hedifferentialquationOPQ 1(a ~ ~ ~ __ = - V72PQa") --=-2P

and the initialcondition(C") PQ(P r;t)- (r -p) as t .It also satisfies he boundary condition(b"/) PQ(p Y; ) O 0 as Y- a boundaryoint fQ.

Again it appears obvious thatPQ P I|r; ) < PQP |r; t), rCQ,

for hediffusingtuffwhichreaches theboundaryofQ is lost as far as PQ is con-cernedbut neednot be lost as a contribution o PQ.Q has been chosenso simply because PQ(P| r'; t) is knownexplicitly, nd, inparticular


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CAN ONE HEAR THE SHAPE OF A DRUM? 11

P I(p r (m2 + n2)w2 1PQ(P p; t) = E exp __- ,a2 m,n _ 2aodd integerswherea is the side of the square.The combined nequalities

2w1rlir PllPQ(p | ;t) < PQ(p | r; t) < 27rthold forall IrCQ and in particular forr=p. In thiscase we get

4 M2 2)72 0 x V2 -a2 m,n exp[-)?t < E e t() O n?=1 e- '2(p)-1 12t and Carleman's theoremfollows.It is only a little harder to proveWeyl's theorem.If one integrates ver Q the inequality4 E exp[ (m2+ n2)w7r2 < ?? -Xntg 2-2 ex 2a2 t E e 7?nP)odd

one obtains4E xpL m2--t)] _ e JJ/(p) pm,n 2a2 n= Qodd

We nowcoverQ wRrithnet ofsquares ofside a, as shown n Fig. 2, and keeponly those contained in Q. Let N(a) be the numberof these squares and letQ(a) be the union of all these squares. We haveZ e-Xn = Ce-Ae J p) dp ? , ff /(p) dpn=1 n=l Q n=1 Q a)_ 4N(a) E exp - ---)]

mon L 2aodd


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12 PAPERS IN ANALYSISand, integratingthe inequality PQ(P1; t) < 1/2wrt ver Q we get e-Zt


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CAN ONE HEAR THE SHAPE OF A DRUM? 13PQ(P| r;) PQP r;t, r Q,

are utterly bvious on intuitivegroundstheymust be proved. Let me indicatea way ofdoingitwhich s probably byfarnot thesimplest. am choosing t toexhibit yet anotherphysicalcontext.It has been knownsincetheearlydays of thiscentury, hrough hework ofEinstein and Smoluchowski,that diffusion s but a macroscopicmanifestationofmicroscopicBrownianmotion.Under suitable physical assumptionsPQ(P| Y; t) can be interpreted s theprobabilitydensityoffinding freeBrownianparticleat r at timet if t startedon its erratic ourneyat t= 0 from and if t getsabsorbed when tcomes to theboundaryofQ.If a largenumberN of ndependentfreeBrownianparticles re startedfrom, thenATf(fP(p r;t)dr

is the average numberof theseparticleswhichare found n A at time t. Sincethe statisticalpercentage rror s ofthe order1/V/N continuousdiffusionheoryis an excellentapproximationwhen N is large.A significant eepeningof this point of view was achieved in the earlytwentiesby NorbertWiener. Instead ofviewingthe problemas a probleminstatistics f particleshe viewed it as a problemin statistics f paths.Withoutenteringnto details let me reviewbrieflywhat is involvedhere.Consider the set of all continuous curves r(T), 0


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14 PAPERS IN ANALYSISsmoothboundaries,that thismeasure s equal to

fPu(p| r;t) dr-.

This is not a trivial statementand it should come as no surprisethat ittrivially mpliesthe inequalitieswe needed a while back to make precisetheprincipleofnot feeling heboundary.In fact,as the reader no doubt sees, theinequalities n question are simplyconsequenceof the fact that if sets a, 6(, e are such thatUIc(cC e

thenmeas. @, meas. (L< meas. C.One finalremarkbeforewe go on. The set ofcurvesforwhichp + r(T)eS SQ O_ 7 < t and p+ r(t) E A

is measurable even if theboundaryofQ is quite Nwild.he measure can stillbewritten as JfAP3(p I r; t)dr and it can be shown that in the interiorof Q,PQ(P I r; t)satisfiesthe diffusionquation aP20t= 2V72Puas well as the initial conditionlim fPQ(p; r 1)dr = 1,

forall open sets A such that 'EA.It is, however,no longerclearhowto interpret heboundaryconditionthatPs2(p r; t)- 0 when r-rI.

This difficultyorces he classical theoryofdiffusion o considerreasonablysmoothboundaries. The probabilistic nterpretation f P2(pj r; t) provides anaturaldefinition f a generalizedolttionof theboundaryvalue problem underconsideration.10, We are now sure thatwe can hear the area ofa drum and it mnayeemthat Nwepent a lot ofeffort o achieve so little.Let me now showyou that the approach we used can be extendedto yieldmore,but to avoid certainpurelygeometrical omplications shall restrictmy-self to convex drums.We have achieved ourfirstuccessby introducing heprinciple f notfeelingtheboundary.But if -is close to theboundaryF ofQ thenthediffusingarticlesstartingfromp-will,to someextent,beginto be influenced y F.Let q be thepointonF closest to -and let1(p)be thestraightineperpendicu-lar to the ine oining -and q. (See Fig. 3.) Then a diffusing article tarting romp will see for shorttimetheboundaryF as thestraight ine1(P).One may say,using again somewhatpicturesque anguage,that,for small t,theparticlehas not had time to feelthecurvatureoftheboundary.


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CAN ONE HEAR THE SHAPE OF A DRUM? 15If this principle s valid I should be allowed to approximate (for small t)

PR(p r; t) by Pl()(p r; t),wherePi(p)(p Jr; ) satisfies gain the diffusionquationOP 1- = - 72PAt 2withthe initialconditionP--(p- r) as t-*O,but withtheboundarycondition

Pl(-,)(p ; t) * 0 as rapproaches point n1(p

q r

FIG. 3 FIG. 4

Carryingthis optimismas far as possiblewe would expect that to a goodapproximation, P Q ( (P;) P lP(; tP

It is well knownthat1e-26'/tPi(P) cpI; t) = 2 -t

whereb=||q-P|| =minimal distance from toP. Thus (hopefully )P; t) d = E-t e21 dpPQ( P P n=1 27rt 27rtQ

dHere |I2 12wxts our oldfriendrom eforend itremains o calculate symp-totically as t-?0) the ntegraloe-281tdi.To do this onsiderhecurveF(5) ofpointsnQwhose distance" rom is a. (See Fig. 4.)


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16 PAPERS IN ANALYSISFor small enough&,F(s) is well defined and even convex) aindthe majorcontribution o our integralcomes from mall 8.If L(6) denotesthe length of F(5) we have

60fe 282tdp - J 252ItL(Q) d6 + something ess than | Q 25o2/tand hence,neglecting n exponentially mall term as well as termsof order t)

60o\/ ro LJe-22 /t dp , -Vt f e2x2L(xcvt) d V-tL Je 2x2 dx -LV2?rt

whereL=L(O) is the lengthofF.We are finally led to the formulaX0 I Q| L 1Z eX7t _ - ____ for tO72=l 2,72t 4 V/2i7rt

and so we can also "hear" the lengthofthecircumiiferencef thedrumThe last asymptotic ormulawas provedonlya fewyears ago by the SwedishmathematicianAke Pleijel [2] usingan entirelydifferentpproach.It is worthremarkinghat we can nowprove that ifall thefrequencies f adrumare equal to those of a circulardrumthen thedrummust tself e circular.This follows at once from the classical isoperimetric nequality which statesthatL2 ?4 Q| withequality occurring nly fora circle.By pitchalone one can thus determinewhether drum s circularor notl(p)

FIG. 511 Can theheuristic rgument gain bemade rigorous? ndeed itcan. First,we use the inequality

1SP 1P )


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CAN ONE HEAR THE SHAPE OF A DRUM? 17wvhichs simplya refinement f the one used previously,namely,PQ(l |; t)< Ij/2rt, nd which can be proventhe same way.Next we need a precise ower estimatefor Q(P |; t) and this s a little moredifficult.We "inscribe"therectangleR(p, t) as shown in Fig. 5, wJThere(t), theheightofthe shaded segment, s to be determined little ater.Let the side of R along the base of the segmentbe b(t) and the othersidebe w(t). It shouldbe clear from hepicturethat they-axisbisects he sidesoftherectanglewhichare parallel to thex-axis.Now considerPR(P |P; t). This notation s perhapsconfusing ince t suggeststhat we are dealing with a boundary value problem in w hich he boundaryvarieswith time. This is not the case. What we have in mind s the followATing:fixt,find R(t)(p-r; r) which s defined nambiguously,nd finallyet r=t.The result s PR(Pl r; t). A convenient xpression s

P pi ; t)=2{E ep[-n22-x --1 )R( X E (ex [_ _ 2] - exp [ep -x {~exp -W n2]expL -w i)2])

where3=a-h(t) = q-pI-h(t). Nowleth(t)=EX\t and,assuminghat (p) isactuallytangento thecurve (whichfor convex curvewillhappenwaTitht mosta denumerablenumberofexceptionalpoints w), e haveb(t) b(t)lim = rlln =t->o (t) t- E/tand consequently

/ _ 1 2b 12E exp -- n2-exp L--(n +j-)_) = 1+ o(l).This is not quite enough,however, nd one needs the stronger stimate

E (exp [_ b n2] - exp - b + 1 )]) = 1+This willsurelybe thecase, for xample, f thecurvatureexistsat q,for hiswould imply that h(t)-b2(t) and the o(x/t) term above would then be anenormousoverestimate.Verymild additionalregularity onditions t nearlyallpointsq would insureo(-/t). Withoutenteringnto a discussionof thesecondi-tions let us simplyassume the boundary to be such as to guarantee at leasto(Vt).Since w(t) remainsboundedfrombelow as t->O,we also have2W2 exp 2Z (exp [_ -exp [-- n+-

= 1-e-252It + exponentiallymallterms.


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18 PAPERS IN ANALYSISWe are now almost through.We write cf. Fig. 6)

, _ u(pP; ) dp >{V PR(P p, t) dp2+wt))f~(~v (1 - -2a2e t + exponentiallymallterms) p.2Xt Q eV t

FIG. 6

Except thenforexponentially mall terms and the factor +o(\It) in frontwe have the integral T 1 e-2821t) pQ(EV t)which, as before, an be seen to be asymptoticallyL| Q(EVt) |--+2t 4

whereone neglects ermsofordert and exponentiallymall terms.Since asymp-totically IQ(EVt) Q I -Lc-\t one can obtain the inequality?21 _ (L+E') 1

27rt 4 27rtwheree' is related in a simple way to E. Since E' can be made arbitrarily mall,the asymptoticformula 2 L 1e-x , _ _ =2rt 4 V/ rtfollows.12. If our overall strategyof attack on the problem s rightwe should beable to go on and forpointsveryclose to a smooth oundaryreplacethebound-ary locallyby suitable circlesof curvature.


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CAN ONE HEAR THE SHAPE OF A DRUM? 19A resultof Pleijel suggests strongly hat for a simplyconnected drumwitha smoothboundary (i.e. without corners nd withcurvatureexisting t everypoint) one has

2j L 1 1Z?~f~'-A n + _,2irt 4 V2wt 6Unfortunately am unable to obtain this, for the exasperatingreasonthat I am unable to get a workableexpression orPu(pj ; t) if 2 is a circle.Ratherthanyield to despair over this sad state of affairset me devote theremainderof the lectureto polygonaldrums, .e. drumswhose boundariesarepolygons.This studywill showbeyondthe shadow of a doubt thattheconstantterm n our asymptotic xpansionowes its existenceto theoverall curvatureoftheboundary.13. BeforeI go on I need an expressionforPs(a,)(p ; t) whereS(GO) s aninfinitewedgeofangle6o. n other wordsPs(oo) s the solution of

lop Id V

subject to the usual initialconditionPs(o0J(pJ)r - -) t--*0, and vanishingas r approaches a pointon either ide of the angle0o.This is a veryold,veryclassical,problem nd if0O= rm, withman integer,it can be solved by the familiarmethodof images. For m not an integer, om-merfeld nventeda methodwhich,so to speak, extendsthe method of imagesto a Riemann surface.A little ater,in 1899 to be precise,H. S. Carslaw gavea moreelementary pproach in whichPs(o) (p ; t) is represented y a suitablecontour ntegral.Carslaw transformsheintegral nto an infiniteeriesofBesselfunctions ut forour purposes t is best to resistthetemptationof Bessel func-tions and to reduce the integralto a different orm. I shall skip the details(thoughsome are quite instructive) nd simply reproducethefinalresult.SetvQxe)= (Ij2irt) E-a--2kG exp[- - 2rpcos(l a - 2kGo)+ p2]0-a-r


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20 PAPERS IN ANALYSISThen

Ps(00)(p r; t) = v(a) - v(-a).Note that if0 = vlm, with m an integer, the complicated integral is out, sincethe factor in frontof it, to wit sin w2 00 =sin wim, s zero; what renmains n theresulting expression for v(a) -v(-a) is a collection of termns asily identifiablexvith those obtained by the method of images.Let us now assume that w 2 cosFor 7r/2ao < 00

1 exp (I- -1cos 2(Oo- a)Pspl p;t)= -_27rt 27r+ thesame two ntegrals s above


22/24

CAN ONE HEAR THE SHAPE OF A DRUM? 217r

2~~~~~~I 0- -/ / 2

k /

\ I ~~~~~~~~~~/\ I ~~~~~~~~/\ I ~~~~~~~/and fk=l2 I k=Oort,'2 2

I /

FIG. 7and finally, orOo-wr/2aoz< r/2

exp --(I-cos 2a) exp -- (1-cos 2(Oo-a)]P'S(p p; ) = -_ 2 r7rt 2wrt 2

+ again thesame two ntegrals.We should recognize r2(1-cos 2a) (and r2(1-cos 2(0o-av))) as being 252where 3is the distancefromp to a side of thewedge.

FIG. 814. To simplifymatters omewhat et me assume that the polygonaldrumis convex and that everyangle is obtuse.At each vertex we drawperpendiculars o the sides of thepolygonthus ob-tainingN shaded sectors where N is the numberof sides or verticesof ourpolygon).


23/24

22 PAPERS IN ANALYSISNow let p be a point n 2. Stuff iffusingrom will either see" theboundaryas a straight ineor,if p is near a vertex, s an infinitewedge.We may as well say that the boundary will appear to the diffusing article

as the nearest wedge, and that consequentlywe may replace PQ -; t) byPS(o0)(C3p; t), whereS(00)) s the wedgenearest to p5.Now, each PS(P |; t) has 1/2wtas a term and after ntegration verQ thisgives the principal term IQI/2 t. Next, each Ps(b P; t) contains two compli-cated looking ntegralswhich have to be integrated ver thewedge.Fortunately,the second of these integratesout to 0, while the firstyields,upon integration ver S(0O),n 2 00 __y-?sin )f

(1 + cosh y) cosh - 2cos00 0This is onlythecontribution fone wedge; to get the total contribution nemust sum over all wedges.Thus the total contributions

-- Z(sin-) d-(1 + coshy)(cosh - y - cos-)

Finally,if - is in the shaded sector of the wedgeS(0O)we get, on integratingover the sector,Mexp --(1- cos 2(a)

0-7r_T2 JO t27rt ~ 2exp --(1- cos 2(O0-o a))1+ 2wt [rdr= - cot (0 - 2)

and thetotal contribution rom he shaded sectors s 1/27r o0 cot (Oo-7r/2).The remaining ontribution s easily seen to be-- 1J K(L-2a E cot Ooy---) e-262/td5

L 1 1 ( 7rFinally,+ -Z cotd-4 V27rt 22w Oo k 2/Finally,for a polygonaldrum


24/24

CAN ONE HEAR THE SHAPE OF A DRUM? 23X0 | S L IE e-Ant ~____n= 2rt at \/ 7rtw1 4 \[so2 00 ~~~dy87ro\ 00/ oo (1 + coshy)(cos r/woy-cosw7r2/0)

withthe understanding hat each 0osatisfies he inequalitywr/2Go in such a way that each 0o>7r,thenthe constanttermapproaches2+ 0f dy8wJ (1 + cosh )2

This should strengthen ur belief thatforsimplyconnected smoothdrums theconstant s universaland equal to 6.15. What happensformultiply onnecteddrums?If thedrumas well as the holesare polygonalthe answer s easilyobtained.One onlyneedsPs(.) (p1p; t) for6osatisfyinghe inequality r o

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