Can One Hear the Shape of a Drum? Author(s): Mark Kac Reviewed work(s): Source: The American Mathematical Monthly, Vol. 73, No. 4, Part 2: Papers in Analysis (Apr., 1966), pp. 1-23 Published by: Mathematical Association of America Stable URL: http://www.jstor.org/stable/2313748 . Accessed: 15/03/2012 22:10
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http://www.jstor.org CAN ONE HEAR THE SHAPE OF A DRUM?
MARK KAC, The RockefellerUniversity, New York To George Eugene Uhlenbeck on the occasion of his sixty-fifthbirthday "La Physique ne nous donne pas seulement l'occasion de resoudre des problemes . .. elle nous fait presentirla solution." H. POINCARE. Before I explain the title and introducethe theme of the lecture I should like to state that my presentationwill be more in the nature of a leisurelyexcursion than of an organized tour. It will not be my purpose to reach a specifieddes- tination at a scheduled time. Rather I should like to allow myselfon many occasions the luxury of stopping and looking around. So much effortis being spent on streamliningmathematics and in renderingit more efficient,that a solitary transgressionagainst the trend could perhaps be forgiven.
r
Q
FIG. 1
1. And now to the theme and the title. It has been known forwell over a centurythat if a membrane 2, held fixed along its boundary F (see Fig. 1), is set in motionits displacement(in the direc- tion perpendicularto its original plane)
F(x,y; t)-F (p; t) obeys the wave equation aF = c2 72F, at2 where c is a certain constant depending on the physical propertiesof the mem- brane and on the tension under which the membraneis held. I shall choose units to make C2==2 2 PAPERS IN ANALYSIS
Of special interest (both to the mathematician and to the musician) are solutions of the form
F(p-; t) =: U(p)eiwl, for,being harmonicin time,they representthe putretones the membraneis capa- ble of producing.These special solutions are also known as normal modes. To find the normal modes we substitute U(-)eiwtinto the wave equation and see that U must satisfythe equation 2 V2U+co2U=O with the boundary condition U=O on the boundary F of Q, correspondingto the membranebeing held fixedalong its boundary. The meaning of "U =0 on F" should be made clear; for sufficientlysmooth boundaries it simplymeans that U(p)->O as p approaches a point of F (fromthe inside). To show that a membraneis capable of producinga discrete spectrum of pure tones i.e. that there is a discrete sequence of c's W<_?2 ?Co3_? * for which nontrivialsolutions of V2U + (2U =0 , U = 0 on , exist, was one of the great problems of 19th century mathematical physics. Poincare struggledwith it and so did many others. The solution was finallyachieved in the early years of our centuryby the use of the theoryof integralequations. We now know and I shall ask you to believe me ifyou do not, that forregions Q bounded by a smooth curve F there is a sequence of numbers X1_X2_ * * - called eigenvalues such that to each therecorresponds a function4'(-), called an eigenfunction,such that
- 2172V, + Xn4n 0 and 41'(-)->O as p->a point of F. It is customaryto normalize the 4t's so that
J / (p)-dp= 1.
Note that I use dp to denote the element of integration(in Cartesian co- ordinates,e.g., dp-dxdy). 2. The focal point of my exposition is the followingproblem: Let ?2 and ?2 be two plane regionsbounded by curves F, and F2 respectively, and consider the eigenvalue problems:
V2U + XU 0 in Q 22V + AV =0 in Q2 with with U= on F, V= on F2.
Assume that foreach n the eigenvalue N,,for ?1 is equal to the ei-genvalueAP1 CAN ONE HEAR THE SHAPE OF A DRUM? 3
for 02. Question: Are the regions i1 and 02 congruentin the sense of Euclidean geometry? I firstheard the problem posed this way some ten years ago fromProfessor Bochner. Much more recently,when I mentionedit to ProfessorBers, he said, almost at once: "You mean, if you had perfectpitch could you findthe shape of a drum.'" You can now see that the "drum" of my title is more like a tambourine (which really is a membrane) and that stripped of picturesque language the problem is whetherwe can determineQ if we know all the eigenvalues of the eigenvalue problem V2U+XU=0 in Q, U =0 o0n F. 3. Before I go any furtherlet me say that as far as I know the problem is still unsolved. Personally, I believe that one cannot "hear" the shape of a tam- bourine but I may well be wrong and I am not prepared to bet large sums either way. What I propose to do is to see how much about the shape can be inferred fromthe knowledge of all the eigenvalues, and to impress upon you the multi- tude of connectionsbetween our problemand various parts of mathematicsand physics. It should perhaps be stated at this point that throughoutthe paper only asymptoticproperties of large eigenvalues will be used. This may represent,of course, a serious loss of informationand it may perhaps be argued that precise knowledgeof all the eigenvalues may be sufficientto determinethe shape of the membrane. It should also be pointed out, howTever,that quite recentlyJohn Milnor constructedtwo noncongruentsixteen dimensional tori whose Laplace- Betrami operators have exactly the same eigenvalues (see his one page note "Eigenvalues of the Laplace operator on certain mlanifolds"Proc. Nat. Acad. Sc., 51(1964) 542). 4. The firstpertinent result is that one can "hear" the area of Q. This is an old result with a fascinatinghistory which I shall now relate briefly. At the end of October 1910 the great Dutch physicist H. A. Lorentz was invited to G6ttingento deliver the Wolfskehllectures. Wolfskehl,by the way, endowed a prize for proving, or disproving,Fermat's last theorem and stipu- lated that in case the prize is not awarded the proceeds fromthe principal be used to invite eminentscientists to lecture at Gottingen. Lorentz gave five lecturesunder the overall title "Alte und neue Fragen der Physik"-Old and new problemsof physics-and at the end of the fourthlecture he spoke as follows (in freetranslation from the original Germ-Ian):"In conclu- sion there is a mathematical probleimiwhich perhaps will arouse the interestof mathematicianswho are present. It originatesin the radiation theoryof Jeans. "In an enclosure with a perfectlyreflecting surface there can formstanding 4 PAPERS IN ANALYSIS
electromagneticwAAaves analogous to tones of ainorgan pipe; we shall confineour attention to very high overtones. Jeans asks for the energy in the frequency intervaldv. To this end he calculates the numlberof overtoneswhich lie between the frequenciesv and v+dy and multipliesthis numberby the energywhich be- longs to the frequenlcyv, and whichaccording to a theoremof statistical mllechan- ics is the samnefor all frequencies. "It is here that therearises the mathematicalproblem to prove that the num- ber of sufficientlyhigh overtoneswhich lies between v and v+dv is independent of the shape of the enclosureand is simiiplyproportional to its volumlle.For many simnpleshapes for wAhichcalculations canl be carried out, this theoremlhas been verifiedin a Leiden dissertation.There is no doubt that it holds in general even for multiplyconnected regions.Sinilar theoremnsfor other vibratingstructures like inembranes,air inasses, etc. should also hold." If one expresses this conjecture of Lorentz in terms of our imembrane,it enmergesin the form:
V (X) < 27r Here N(X) is the number of eigenvalues less than X, | I the area of Q and means that
N0 (X) |Q|
X a X 27r There is an apocryphalreport that Hilbert predictedthat the theoremwould not be proved in his life time. Well, he was w-rongby many, many years. For less than two years later HermiianWeyl, who was present at the Lorentz' lec- ture and whose interest was aroused by the problem, proved the theorem in question, i.e. that as X >oc
N1(X) x.
Weyl used in a masterly way the theory of integral equations, which his teacher Hilbert developed only a few,-years before,and his proofwas a crowning achievement of this beautiful theory. I\lany subsequent developnmentsin the theoryof differentialand integralequations (especially the wNorkof Coturantand his school) can be traced directlyto Weyl's memoiron the conjectureof Lorentz. 5. Let me now consider brieflya differentphysical problem which too is closelyrelated to the problenmof the distributionof eigenvalues of the Laplacian. It can be taken as a basic postulate of classical statistical mechanics that if a systemof 1l particles confinedto a volume Q is in equilibriumwTith a thermo- stat of temperatureT the probabilityof findingspecified particles at ri,,r2, rm (tn lel nd r - rMf(wvithin volume elementsdr1, dr2, . , drAzl)is CAN ONE HEAR THE SHAPE OF A DRUM? 5
.. ... exp LkTdrV(rly rml)0 dr-, dr,- [- **
* J. exp V(rl ... rm) dr, ... drm where V(r-1, ,rll) is the interactionipotential of the particles and k=R/N with R the "gas constant" and N the Avogadro number. For identical particles each of mass m obeying the so called Boltzmainii statistics the correspondingassumiption in quantum statistical mechanicsseems much more complicated. One starts with the Schr6dingerequation
2 V - V(ri, ..., r_>) Eq 2- , whereh is the Planck constant) 2m Q27 with the boundary condition limn4(ri. , 1) =0, whenever at least one rk approaches the boundary of Q. (This boundary condition has the effectof con- fining the particles to Q.) Let E1 < E2 ? E3 < be the eigenvalues and 4'11,4, . thecorresponding normalized eigenfunctions. Then thebasic postu- late is that the probabilityof findingspecified particles at ri, r2, , ur(within dr1* **, d-rm)is
j E,/kT 2-+- e A(rl * ri)9drl . . . dr31 s=l
Z e-EsIkT s=1
There are actually no known particles obeying the Boltzmann statistics. But don't let this worry you for our purposes this regrettable fact is immilaterial. Now, let us specialize our discussion to the case of an ideal gas w7hich,by definition,means that V( r-, , r 1) _O. Classically, the probability of findingspecified particles at r1, , r7 is clearly
dr1 dr3
where I Q | is now the volume of Q. Quantum mechanically the answer is not nearly so explicit. The Schr6dinger equation for an ideal gas is ft2
2m 9 - and the equation is obviously separable. If I now consider the three-dimensional (rather than the 3MH-dimenlsional) eigenvalue problem 6 PAPERS IN ANALYSIS
2V2A(r) = - r V(r) -> 0 as r the boundary of 9, it is clear that the Es as well as the , r) are easily expressiblein termiis of the X's and corresponding41(r)'s. The formulafor the probabilityof findingspecified particles at -ri, rm turns out to be
exp [-m-- T]'fl(rk)
H ~ LL ______- dr i, k=1 r0 Ant E exp ---]
Now, as 1->0 (or as T-*> occ) the quantum mechanical result should go over into the classical one and this immllediatelyleads to the conjecture that as
LmkT_
EeAn(F) i ?1=1 n=1
If instead of a realistic three-dimnensionalcontainer Q I consider a two- dimensional one, the result would still be the same
X n -e_ 2e-X,-2r) 0 n=1 n=1 except that now I Q| is the area of Q rather than the volume. Clearly the result is expected to hold only for r in the interiorof Q. If we believe Weyl's result that (in the two-dimensionalcase)
I IV (X) /_ -I 1%), 2er it followsimmediately by an Abelian theoremthat 1 0 1 __Ee-XnT - T O,-- 0 and hence that 1 r2 -XT E-X2eXnr) -Ie dX. n=1 27rT 2ro
Setting A (X) = x de-XrA (X) e-XrdX, r -> 0. Since A (X) is nondecreasingwe can apply the Hardy-Littlewood-Karamata Tauberian theoremand conclude what everyonewould be temptedto conclude, namely that 2- N A (X)= + (r)2 , -->o, for every r in the interiorof Q. Though this asymptotic formula is thus nearly "obvious" on "physical grounds,"it was not until 1934 that Carleman succeeded in supplyinga rigorous proof. In concluding this section it may be worthwhileto say a word about the "strategy" of our approach. We are primarilyinterested, of course, in asymptotic propertiesof X,,for large n. This can be approached by the device of studyingthe Dirichlet series 00 E e-Xnt n=1 for small t. This in turn is most convenientlyapproached throughthe series 00 {D00 ZeCXt,(P) = fe-tdA(x) n1=1 and thus we are led to the Abelian-Tauberian interplaydescribed above. 6. It would seem that the physical intuition ought not only provide the mathematicianwith interestingand challengingconjectures, but also show him the way toward a proof and toward possible generalizations. The contextof the theoryof black body radiation or that of quantum statis- tical mechanics,however, is too far removed fromelementary intuition and too full of daring and complex physical extrapolations to be of much use even in seeking the kind of understandingthat makes a mathematician comfortable, let alone in pointingtoward a rigorousproof. Fortunately, in a much more elementary context the problem of the dis- tributionof eigenvalues of the Laplacian becomes quite tractable. Proofsemerge as natural extensionsof physical intuitionand interestinggeneralizations come withinreach. 7. The physical context in question is that of diffusion theory,another branch of nineteenthcentury mathematical physics. Imagine "stuff,"initially concentrated at p( (xo, yo)), diffusingthrough a plane region Q bounded by F. Imagine furthermorethat the stuffgets absorbed ( "eaten") at the boundary. 8 PAPERS IN ANALYSIS The concentrationPQ(p| r; t) of matter at r-((x, y)) at time t obeys the differentialequation of diffusion aPe 1 (a) - =t 2 the boundary condition (b) PQ(p I r; t) - 0 as r approachesa boundarypoint, and the initial condition (c) PQ(P |r;t) -*(r-p) as t ->0; here (-r-p) is the Dirac "delta function,"with "value" ooif -r= p and 0 if r p'. The boundary condition (b) expressesthe fact that the boundary is 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 diffusionconstant equal to 2. As is well knownthe concentrationPQ(P I -; t) can be expressedin termsof the eigenvalues Xz, and normalized eigenfunctions4,(r-) of the problem '72v +X4/=0 in Q, 0 on P. In fact, PQ(p Ir; t)= En-], 6-fl'4' (p)Abl(r). NOw,for small t, it appearsintuitively clear that particlesof the diffusing stuffwill not have had enoughtime to have feltthe influence of the boundary Q. As particlesbegin to diffusethey may not be aware,so to speak,of the disaster that awaits themwhen they reach the boundary. We may thusexpect that in someapproximate sense PQ(p I r; t) - Po(p I r; t), as t O , wherePo(p1 r; t) stillsatisfies the same diffusionequation (a') P VPO and the same initialcondition (c') Po(p ;t) = P-p), t0, but is otherwiseunrestricted. Actuallythere is a slightadditional restriction without which the solution is not unique (a remarkablefact discovered some years ago by D. V. Widder). The restrictionis thatPo > 0 (or moregenerally that Po be boundedfrom below). A similarrestriction for PQ is not neededsince fordiffusion in a bounded regionit followsautomatically. CAN ONE HEAR THE SHAPE OF A DRUM? 9 An explicit formulafor Po is, of course, well known. It is Po(pIr; t) exp - - ] where ||- denotes the Euclidean distance between p and r. I can now state a little moreprecisely the principleof "not feelingthe bound- ary" explained a momentago. The statementis that as t >0 1 F 00 r ~P~2 | = -A()nr -,1e 1.o( PnpPQ( r; t) 1 exp- 2- npr; t), n=1 2w1t 1 21- where - stands here for "is approximatelyequal to." This is a bit vague but let it go at that for the moment. If we can trust this formulaeven forp = r we get 00 1 x'-Ant 2 1 e-e n(r) n=l 27rt and if we display still more optimismwe can integratethe above and, making use of the normalizationcondition 2 a(r-)n d-r'=: 1, obtain I:e-Xnt, Q. 1=1 27rt We recognizeimmediately the formulasdiscussed a while back in connection with the quantum-statistical-mechanicaltreatment of the ideal gas. If we apply the Hardy-Littlewood-Karamata theorem,alluded to before, we obtain as corollarythe theoremsof Carleman and Weyl. To do this, however,we must be allowed to interpret as meaning "asynmp- totic to." 8. Now, a little mathematical soul-searching.Aren't we as far froma rigor- ous treatnmentas we were before? True, diffusionis more familiar than black body radiation or quantum statistics. But familiaritygives comfort,at best, and comfortmay still be (and oftenis) miles away fromthe rigordemanded by mathematics. Let us see then what we can do about tighteningthe loose talk. First let me dispose of a few minoritems which may cause you worry. When I writei/ = 0 on F or P(p I r; t)-O as r approaches a boundary point of Q there is always a question of interpretation. 10 PAPERS IN ANALYSIS Let me assume that r is sufficientlyregular so that no ambiguityarises i.e. P(p r; t) -O 0 as r -> a boundarypoint of Q, means exactly what it says, while 41= 0 on r means 4 0 as r -* a boundarypoint of Q. Likewise, P(p| r; t)->5(r-p) as t-*0, has the obvious interpretation,i.e. lim fJP(p r; t)dr= 1 for every open set A containingP. Now, to morepertinent items. If the mathematical theoryof diffusioncorre- sponds in any way to physical reality we should have the inequality r_IIP- l11 exp [- 71? P(Q(P; t) Po(p|I;t)0 ; L 2t For surelyless stuffwill be found at r at time t if thereis a possibilityof mat- ter being destroyed (on the boundary F of Q) than if there were no possibility of such destruction. Now let Q be a square with centerat p totally contained in Q. Let its bound- ary act as an absorbingbarrier and denote by PQ(P Ir; t), rE Q, the corresponding concentrationat r at time t. In other words, PQ satisfiesthe differentialequation OPQ 1 (a")(a ~ ~ ~ __--=-2P = - V72PQ and the initial condition (C") PQ(P r; t) - (r -p) as Ot . It also satisfiesthe boundary condition (b"/) PQ(p I Y;1) -O 0 as Y- a boundarypoint of Q. Again it appears obvious that PQ(P I|r;t) < PQ(P |r; t), rCQ, forthe diffusingstuff which reaches the boundary of Q is lost as far as PQ is con- cerned but need not be lost as a contributionto PQ. Q has been chosen so simply because PQ(P| r'; t) is known explicitly,and, in particular CAN ONE HEAR THE SHAPE OF A DRUM? 11 r (m2 + n 1 PQ(PP I(p p; t) = E exp __- 2)w2 , a2 m,n _ 2a odd integers where a is the side of the square. The combined inequalities rlir- Pll | | 2w1 PQ(p ;t) < r; t) < PQ(p 27rt hold for all IrCQ and in particular for r=p. In this case we get 0 M2 4 2)72 x V2 - a2 m,n exp [-)?t < E e t() < a2 mmn 2a' n== ~'( 2,7rt odd integers and it is a simple matterto prove that as t-*O we have asymptotically 4 [ (m2 + n2)7r2 1 - exp ------1L- a2 , L 2a2 27rt odd integers Thus asymptotically for t->O n?=1 e- '2(p)-1 12t and Carleman's theorem follows. It is only a little harder to prove Weyl's theorem. If one integratesover Q the inequality 4 ?? 2 E exp[ (m2+ n2)w7r2 < -Xntg e P) -2 ex 2a2 t E 7?n odd one obtains 4 E exp L (m2--t)] _ e JJi/(p) dp m,n 2a2 n= Q odd We now cover Q wRritha net of squares of side a, as shown in Fig. 2, and keep only those contained in Q. Let N(a) be the number of these squares and let Q(a) be the union of all these squares. We have Z e-Xn = JeCe-A J p) dp ? , ff P/(p) dp n=1 n=l Q n=1 Q (a) _ 4N(a) E exp - ---)] mon L 2a odd 12 PAPERS IN ANALYSIS and, integratingthe inequality PQ(P1|; t) < 1/2wrtover Q we get e-Zt <_ &/2irt. a fa a FIG. 2 Noting that N(a)a2= |Q(a) we record the fruitsof our latest labor in the formof the inequality 4 (Mn2+ n2)w2] _ _ Q(a) I 2 E expL ) j < e 00 00 Q(a) n -- E e xnt r t 0. n=1 27rt 9. Are we now throughwith rigor?Not quite. For while the inequalities expL- 2wt22 paprt) -- 27rt CAN ONE HEAR THE SHAPE OF A DRUM? 13 PQ(P| r;) PQP r;t, r Q, are utterlyobvious on intuitivegrounds they must be proved. Let me indicate a way of doing it which is probably by far not the simplest. I am choosing it to exhibit yet another physical context. It has been known since the early days of this century,through the work of Einstein and Smoluchowski,that diffusionis but a macroscopic manifestation of microscopicBrownian motion. Under suitable physical assumptions PQ(P| Y; t) can be interpretedas the probabilitydensity of findinga freeBrownian particle at r at time t if it started on its erraticjourney at t = 0 fromp and if it gets absorbed when it comes to the boundary of Q. If a large number N of independentfree Brownian particlesare started from , then ATf(fP(p r;t)dr is the average number of these particles which are found in A at time t. Since the statistical percentageerror is of the order 1/V/N continuousdiffusion theory is an excellent approximationwhen N is large. A significantdeepening of this point of view was achieved in the early twentiesby Norbert Wiener. Instead of viewing the problem as a problem in statisticsof particleshe viewed it as a problem in statisticsof paths. Without enteringinto details let me review brieflywhat is involved here. Consider the set of all continuous curves r(T), 0 Prob. {p + (tl) E Qp + r(t Q2 . p + r(tn) E n} = j'*S Po(p ri; t1)Po(r r2;t2 - t1) . P0( rnr; tn-tn) dr. drn Q n where,as before, Po(p I r; 2) =-exp xt2tL- I. W-ienerhas shoxvnthat it is possible to constructa completelyadditive mea- sure on the space of all continuous curves r(r) emanating fromthe origin such that the set of curves p+ r() which at times tl smooth boundaries, that this measure is equal to fPu(p| r;t) dr-. This is not a trivial statement and it should come as no surprise that it trivially implies the inequalities we needed a while back to make precise the principleof not feelingthe boundary. In fact, as the reader no doubt sees, the inequalities in question are simplya consequence of the fact that if sets a, 6(, e are such that UIc(cC e then meas. @,< meas. (L < meas. C. One finalremark before we go on. The set of curves forwhich p + r(T)eS SQ O _ 7 < t and p+ r(t) E A is measurable even if the boundary of Q is quite Nwild.The measure can still be written as JfAP3(p I r; t)d r and it can be shown that in the interior of Q, PQ(P I r; t) satisfiesthe diffusionequation aP20t= 2 V72Pu as well as the initial condition lim fPQ(p; r; 1) dr = 1, for all open sets A such that 'EA. It is, however,no longerclear how to interpretthe boundary condition that Ps2(p r; t)- 0 when r-rI. This difficultyforces the classical theoryof diffusionto considerreasonably smooth boundaries. The probabilistic interpretationof P2(pj r; t) provides a natural definitionof a generalizedsolttion of the boundary value problem under consideration. 10, We are now sure that we can hear the area of a drum and it mnayseem that Nwespent a lot of effortto achieve so little. Let me now show you that the approach we used can be extended to yield more,but to avoid certain purelygeometrical complications I shall restrictmy- self to convex drums. We have achieved our firstsuccess by introducingthe principleof not feeling - the boundary. But if is close to the boundary F of Q then the diffusingparticles startingfrom p- will, to some extent,begin to be influencedby F. - Let q be the point on F closest to and let 1(p) be the straightline perpendicu- - lar to the line joining and q. (See Fig. 3.) Then a diffusingparticle starting from p will see fora short time the boundary F as the straightline 1(P). One may say, using again somewhat picturesque language, that, for small t, the particle has not had time to feel the curvatureof the boundary. CAN ONE HEAR THE SHAPE OF A DRUM? 15 If this principle is valid I should be allowed to approximate (for small t) PR(p r; t) by Pl()(p r; t), where Pi(p)(p Jr;t) satisfiesagain the diffusionequation OP 1 - = - 72P At 2 with the initial condition P--(p- r) as t-*O,but with the boundarycondition Pl(-,)(p ; t) -* 0 as r approachesa pointon 1(p q r FIG. 3 FIG. 4 Carrying this optimism as far as possible we would expect that to a good approximation , P Q| ( (P;)dP I P;lP( t P It is well known that 1e-26'/t Pi(P) cpI; t) = 2 -t where b=||q-P|| =minimal distance fromp to P. Thus (hopefully!) P; P t) dP = e-t E e21 ddp PQ( n=1 27rt 27rt Q Here |I2 12wxtis our old friendfrom before and it remainsto calculateasymp- totically(as t-?0) the integralfoe-281tdi. To do thisconsider the curveF(5) of pointsin Q whose"distance" from P is a. (See Fig. 4.) 16 PAPERS IN ANALYSIS For small enough &, F(s) is well defined (and even convex) aind the major contributionto our integralcomes fromsmall 8. If L(6) denotes the length of F(5) we have 60 fe 282tdp - J e 252ItL(Q) d6 + somethingless than | Q 25o2/t and hence, neglectingan exponentiallysmall term (as well as termsof order t) 60.o\/ ro L Je-22 /t dp-, -Vt f e2x2L(xcvt) d v-tLV Je 2x2 dx -LV2?rt where L=L(O) is the length of F. We are finally led to the formula X0 I Q| L 1 _ - for ZeeX7t ____ tO 72=l 2,72t 4 V/2i7rt and so we can also "hear" the lengthof the circumiiferenceof the drum! The last asymptoticformula was proved only a fewyears ago by the Swedish mathematicianAke Pleijel [2] using an entirelydifferent approach. It is worthremarking that we can now prove that if all the frequenciesof a drum are equal to those of a circulardrum then the drum must itselfbe circular. This follows at once from the classical isoperimetricinequality which states that L2 ?4 Q| with equality occurringonly for a circle. By pitch alone one can thus determinewhether a drum is circular or not! l(p) FIG. 5 1 1 Can the heuristicargument again be made rigorous?Indeed it can. First, we use the inequality 1SP 1P ) wvhichis simply a refinementof the one used previously, namely, PQ(l |; t) < Ij/2rt,and which can be proven the same way. Next we need a precise lower estimate forPQ(P |; t) and this is a little more difficult.We "inscribe"the rectangleR(p, t) as shown in Fig. 5, wJThereh(t), the heightof the shaded segment,is to be determineda little later. Let the side of R along the base of the segment be b(t) and the other side be w(t). It should be clear fromthe picturethat the y-axisbisects the sides of the rectanglewhich are parallel to the x-axis. Now considerPR(P | P; t). This notation is perhaps confusingsince it suggests that we are dealing with a boundary value problem in wThich the boundary varies with time. This is not the case. What we have in mind is the followATing: fix t, findPR(t)(p-r; r) whichis definedunambiguously, and finallyset r=t. The result is PR(Pl |r; t). A convenientexpression is ) X E (ex _ 2] - - PR(P pi ; t)=2{E ep[-n22-x[_ exp [ep --1 ) x {~exp -W n2]expL -w (i)2]) where3=a-h(t) = q-pI-h(t). Now let h(t) =EX\t and, assumingthat l(p) is actually tangentto the curve (which fora convex curve will happen waTithat most a denumerable numberof exceptional points w),we have b(t) b(t) lim = rlln = t->oh(t) t- E/t and consequently 12 / _ 1 2b E exp [-- n2 -exp L--(n +j-)_) = 1 + o(l). This is not quite enough, however,and one needs the strongerestimate E (exp [_ b n2] - exp [- b + 1 )]) = 1 + This will surelybe the case, forexample, if the curvatureexists at q, forthis would imply that h(t)-b2(t) and the o(x/t) term above would then be an enormousoverestimate. Very mild additional regularityconditions at nearly all points q would insure o(-/t). Without enteringinto a discussion of these condi- tions let us simply assume the boundary to be such as to guarantee at least o(Vt). Since w(t) remains bounded frombelow as t->O, we also have 2W2 exp 2 Z (exp [_ - exp [-- n +- = 1-e-252It + exponentiallysmall terms. 18 PAPERS IN ANALYSIS We are now almost through.We write (cf. Fig. 6) , _ PR(P t) u(pI P;t) dp >{V p, dp 2+wt))f~(~v (1 - -2a2e t + exponentiallysmall terms) dp. 2 Xt Q eV t ) FIG. 6 Except then forexponentially small terms and the factor I +o(\It) in front we have the integral T (1 e-2821t)dp Q(EV t) which, as before,can be seen to be asymptotically L | Q(EVt) |--+2t 4 whereone neglectsterms of ordert and exponentiallysmall terms.Since asymp- totically IQ(EVt) Q I -Lc-\t one can obtain the inequality ?21 _ (L+E') 1 27rt 4 27rt where e' is related in a simple way to E. Since E' can be made arbitrarilysmall, the asymptoticformula | 2 L 1 e-x , _ _ = 2rt 4 V/2rt follows. 12. If our overall strategyof attack on the problem is right we should be able to go on and forpoints very close to a smoothboundary replace the bound- ary locally by suitable circlesof curvature. CAN ONE HEAR THE SHAPE OF A DRUM? 19 A result of Pleijel suggests stronglythat for a simply connected drum with a smooth boundary (i.e. without cornersand with curvature existingat every point) one has 2j L 1 1 Z?e~f~'-A n + _, 2irt 4 V2wt 6 Unfortunately I am unable to obtain this, for the exasperating reason that I am unable to get a workable expressionfor Pu(pj |; t) if 2 is a circle. Rather than yield to despair over this sad state of affairslet me devote the remainderof the lecture to polygonaldrums, i.e. drums whose boundaries are polygons.This study will show beyond the shadow of a doubt that the constant termin our asymptoticexpansion owes its existenceto the overall curvature of the boundary. 13. Before I go on I need an expressionfor Ps(a,)(p ; t) where S(GO) is an infinitewedge of angle 6o. In other words Ps(oo) is the solution of lop I d 2 V subject to the usual initial conditionPs(o0J(pJ t)r - -) t--*0, and vanishing as r approaches a point on eitherside of the angle 0o. This is a very old, very classical, problemand if0O = rm, with m an integer, it can be solved by the familiarmethod of images. For m not an integer,Som- merfeldinvented a method which, so to speak, extends the method of images to a Riemann surface. A little later, in 1899 to be precise, H. S. Carslaw gave a more elementaryapproach in which Ps(o) (p I; t) is representedby a suitable contourintegral. Carslaw transformsthe integralinto an infiniteseries of Bessel functionsbut forour purposes it is best to resist the temptationof Bessel func- tions and to reduce the integral to a differentform. I shall skip the details (though some are quite instructive)and simply reproducethe finalresult. Set vQxe)= (Ij2irt) E-a--2kG exp [- - 2rpcos(l-a - 2kGo)+ p2] 0-a-r<2006 2t r r2p2r rp exp[ . exp - cosh y i t fsin-A) L - - ~2t- - ' -- L dyd , 00} 44rftot fir iw ) 7 cosh i-y +-(G-a) - cos- wherethe summation E is extendedover k's satisfyingthe inequality underthe summation sign and p (p,a), r= (r, ). 20 PAPERS IN ANALYSIS Then Ps(00)(p r; t) = v(a) - v(-a). Note that if 0 = vlm, with m an integer, the complicated integral is out, since the factor in front of it, to wit sin w2 00 =sin wim,is zero; what renmainsin the resulting expression for v(a) -v(-a) is a collection of termnseasily identifiable xvith those obtained by the method of images. Let us now assume that w 2 <00 IF 0 < a < 0- only k = O is allowed, 2 IF < a < 60, only k = 1 is allowed, 2 but for Oo -w 2 r2 exp - (1- cos 2a) Ps(p | p; t) -2t 27rt - r2- r2 - expl0 L rX exp L--cosh y - sin - g~ dy So/ 47rOot J r 7r2 cosh y0000 - cos P ) exp[_ r expo - cosh y ++sin ' r o dr. 00S 47rOot J_OO r( a 7r2 cosh -- y + 27r, > cos For 7r/2 1 exp (I- (1cos- 2(Oo - a) Pspl p;t)= -_ 27rt 27r + the same two integralsas above CAN ONE HEAR THE SHAPE OF A DRUM? 21 7r 2~~~~~~ I/ 0- - / / 2 / k \ I ~~~~~~~~~~/ \ I ~~~~~~~~/ \ I ~~~~~~~/ and fk=l2 I k=Oort,' 2 2 I / FIG. 7 and finally,for Oo-wr/2 exp[--(I-cos 2a) exp[-- (1-cos 2(Oo-a)] P'S(p p; t) = 27rt-_ 2wrt 2 r 2 + again the same two integrals. We should recognize r2(1-cos 2a) (and r2(1-cos 2(0o-av))) as being 252 where 3 is the distance fromp to a side of the wedge. FIG. 8 14. To simplifymatters somewhat let me assume that the polygonal drum is convex and that every angle is obtuse. At each vertex we draw perpendicularsto the sides of the polygon thus ob- taining N shaded sectors (where N is the number of sides or vertices of our polygon). 22 PAPERS IN ANALYSIS Now let p be a point in 2. Stuffdiffusing from p will either"see" the boundary as a straightline or, if p is near a vertex,as an infinitewedge. We may as well say that the boundary will appear to the diffusingparticle as the nearest wedge, and that consequently we may replace P Q(-; t) by PS(o0)(C3p; t), where S(00)) is the wedge nearest to p5. Now, each PS(P |; t) has 1/2wt as a term and after integrationover Q this gives the principal term IQI/22t. Next, each Ps(b P; t) contains two compli- cated looking integralswhich have to be integratedover the wedge. Fortunately,the second of these integratesout to 0, while the firstyields, upon integrationover S(0O), 2 00 n dy__ -?sin )f (1 + cosh y) (cosh - 2cos 00 0 This is only the contributionof one wedge; to get the total contributionone must sum over all wedges. Thus the total contributionis d -- Z(sin-) - - (1 + coshy)(cosh y - cos-) - Finally, if is in the shaded sector of the wedge S(0O) we get, on integrating over the sector, Mexp [--(1- cos 2(a) 0-7r_T2 JO t27rt ~ 2 exp --(1- cos 2(O(0-o a))1 + 2wt [rdr= - cot (0 - 2) and the total contributionfrom the shaded sectorsis 1/27rZo0 cot (Oo-7r/2). The remainingcontribution is easily seen to be -- 1J K(L-2a E cot (Ooy---) e-262/t d5 L 1 1 ( 7r Finally,+ od-u 2 -Z cot 4 V27rt 2w Oo k 2/ Finally, for a polygonal drum CAN ONE HEAR THE SHAPE OF A DRUM? 23 X0 | S L I E e-Ant ~____ n= 2 rt at \/27rt w1 4 \[so 2 00 ~~~dy 87r'o\ 00/ oo (1 + coshy)(cos 7r/woy-cosw7r2/0) with the understandingthat each 0o satisfiesthe inequality wr/2 2+r 0f dy 8wJ (1 + coshy)2 This should strengthenour belief that for simply connected smooth drums the constant is universal and equal to 6. 15. What happens for multiplyconnected drums? If the drum as well as the holes are polygonal the answer is easily obtained. One only needs Ps(.) (p1 p; t) for6o satisfyingthe inequality r Le\nt ISI _ L 1 n=1 21rt 4 V2wt and that thereforeone can "hear" the connectivityof the drum! One can, of course, speculate on whetherin general one can hear the Euler- Poincare characteristicand raise all sortsof other interestingquestions. As our study of the polygonaldrum shows,the structureof the constant term is quite complex since it combines metric and topological features. Whether these can be properlydisentangled remains to be seen. This is an expanded version of a lecturewhich was filmedunder the auspices of the Committee on Educational Media of the Mathematical Association of America. References 1. M. Kac, On some connections between probability theory and differentialand integral equations, Proc. Second Berkeley Symposium on Mathematical Statistics and Probability, 1957, pp. 189-215. 2. A. Pleijel, A study of certain Green's functionswith applications in the theoryof vibrating membranes,Arkiv for Matematik, 2 (1954) 553-569.