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 .
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CAN ONE HEAR THE SHAPE OF A DRUM?
MARK KAC, The RockefellerUniversity,New YorkTo George Eugene Uhlenbeck on the occasion of his sixty-fifthbirthday"La Physique ne nous donne pas seulementl'occasion de resoudredes problemes . .. elle nous
fait presentirla solution." H. POINCARE.

BeforeI explain the titleand introducethe themeof the lectureI should liketo 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 specifieddestination at a scheduled time. Rather I should like to allow myselfon manyoccasions the luxury of stopping and looking around. So much effortis being
spent on streamliningmathematics and in renderingit more efficient,that asolitary 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 fixedalong its boundaryF (see Fig. 1), is set in motionits displacement(in the direction perpendicularto its originalplane)
F(x,y; t)-F (p; t)obeys the wave equation
aF
at2

=

c2 72F,

wherec is a certain constant dependingon the physical propertiesof the membrane and on the tension under which the membraneis held.
I shall choose units to make C2==2

2

IN ANALYSISPAPERSOf special interest (both to the mathematician and to the musician) aresolutions of the form
F(p-; t) =: U(p)eiwl,

for,being harmonicin time,theyrepresentthe putretones the membraneis capable of producing.These special solutions are also known as normal modes.
To find the normal modes we substitute U(-)eiwtinto the wave equationand 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.
smoothThe meaning of "U =0 on F" should be made clear; forsufficiently
boundaries it simplymeans that U(p)->O as p approaches a point of F (fromthe
inside). To show that a membraneis capable of producinga discretespectrum*
forof pure tones i.e. that there is a discretesequence of c's W<_?2 ?Co3_?
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 theuse 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 numbersX1_X2_ * * called eigenvalues such that to each therecorrespondsa function4'(-), called aneigenfunction,such that
+ Xn4n-2172V,0

and 41'(-)->O as p->a point of F.
It is customaryto normalize the 4t'sso that= 1.

J / (p)-dp

Note that I use dp to denote the element of integration(in Cartesian coordinates,e.g., dp-dxdy).
2. The focal point of my expositionis the followingproblem:
Let ?2 and ?2 be two plane regionsbounded by curves F, and F2 respectively,
and considerthe 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

for02. 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 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 stripped of picturesque language theproblemis whetherwe can determineQ if we know all the eigenvalues of theeigenvalue problem
V2U+XU=0

U =0

in Q,
o0n F.

3. Before I go any furtherlet me say that as far as I know the problem isstill unsolved. Personally,I believe that one cannot "hear" the shape of a tambourine 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 knowledgeof all the eigenvalues, and to impressupon you the multitude of connectionsbetweenour problemand various parts of mathematicsand
physics.
It should perhaps be stated at this point that throughoutthe paper only
asymptoticpropertiesof large eigenvalues will be used. This may represent,ofcourse, a serious loss of informationand it may perhaps be argued that precise
to determinethe shape of theknowledgeof all the eigenvalues may be sufficient
membrane. It should also be pointed out, howTever,that quite recentlyJohn
Milnor constructedtwo noncongruentsixteen dimensionaltori whose LaplaceBetrami 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 firstpertinentresult is that one can "hear" the area of Q. This is an
old result with a fascinatinghistorywhich 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 stipulated that in case the prize is not awarded the proceeds fromthe principal beused to invite eminentscientiststo lecture at Gottingen.
Lorentz gave fivelecturesunder 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 freetranslationfromthe original Germ-Ian):"In conclusion there is a mathematical probleimi
which perhaps will arouse the interestof
mathematicianswho are present. It originatesin the radiation theoryof Jeans.
"In an enclosurewith a perfectlyreflectingsurfacethere can formstanding

4

PAPERSIN 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 overtoneswhichlie between
the frequenciesv and v+dy and multipliesthis numberby the energywhich belongs to the frequenlcy
v,and whichaccordingto a theoremofstatistical mllechanics is the samneforall frequencies.
"It is herethat therearises the mathematicalproblemto prove that the number of sufficiently
high overtoneswhich lies between v and v+dv is independentof the shape of the enclosureand is simiiply
proportionalto its volumlle.
For many
simnpleshapes forwAhich
calculations canl be carried out, this theoremlhas beenverifiedin a Leiden dissertation.There is no doubt that it holds in general evenformultiplyconnected regions.Sinilar theoremns
forother vibratingstructures
like inembranes,air inasses, etc. should also hold."
If one expresses this conjecture of Lorentz in terms of our imembrane,itin the form:
enmerges
V (X)

27r

<

Here N(X) is the number of eigenvalues less than X, | I the area of Q andmeans that
N0 (X)
X

a

|Q|

27r

X

There is an apocryphalreportthat Hilbert predictedthat the theoremwould
not be proved in his life time. Well, he was w-rongby many, many years. Forless than two years later HermiianWeyl, who was present at the Lorentz' lecture and whose interestwas aroused by the problem, proved the theorem inquestion, i.e. that as X >oc

x.

N1(X)

Weyl used in a masterlyway the theory of integral equations, which histeacher Hilbert developed only a few,years before,and his proofwas a crowningachievement of this beautiful theory. I\lany subsequent developnmentsin theand integralequations (especially the wNork
theoryof differentialof Coturantand
his school) can be traced directlyto Weyl's memoiron the conjectureof Lorentz.
5. Let me now consider brieflya differentphysical problem which too isof the distributionof eigenvaluesof the Laplacian.
closelyrelated to the problenm
It can be taken as a basic postulate of classical statistical mechanics that if
a systemof 1l particlesconfinedto a volume Q is in equilibriumwTitha thermostat of temperatureT the probabilityof findingspecifiedparticlesat ri,,r2,
rm (tnlelndr

rMf(wvithinvolume elementsdr1,dr2,.

-, drAzl)isCAN ONE HEAR

exp [-

L

* J.

exp

THE

..

kTdr
V(rly

SHAPE

rml)0 dr-,**

V(rl ...

5

OF A DRUM?

rm)

...

dr,-

dr, ...

drm

where V(r-1,,rll) is the interactionipotential of the particles and k=R/Nwith R the "gas constant" and N the Avogadro number.
For identical particles each of mass m obeying the so called Boltzmainii
statisticsthe correspondingassumiptionin quantum statistical mechanicsseems
much more complicated. One starts with the Schr6dingerequation
2 V
2m

-

V(ri, ...,

r_>)

Q

2- , whereh is the Planck constant)27

Eqwith the boundary condition limn4(ri.
, 1) =0, whenever at least one rk
approaches the boundary of Q. (This boundary condition has the effectof conbe the eigenvalues andfining the particles to Q.) Let E1 < E2 ? E3 <.
thecorresponding
Then thebasic postunormalized
4,
eigenfunctions.
4'11,, ur(withinlate is that the probabilityof findingspecifiedparticlesat ri, r2,
dr1* **, d-rm)is

je
s=l

E,/kT

2-+-

A(rl

ri)9drl . . . dr31*

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, r1) _O.
definition,means that V( r-,ofClassically, the probability findingspecifiedparticles at r1,, r7 isclearlydr1

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

2V2A(r) =-V(r) -> 0

as

IN ANALYSISrthe boundary of 9,r, r) are easily expressiblein termiis
it is clear that the Es as well as theof the X's and corresponding41(r)'s.
The formulafor the probabilityof findingspecifiedparticles at -ri,rmturnsout to be

exp [-m--T]'fl(rk)

H ~ LLr0
k=1_ _ _ _ _ _ _ _ _ _ -

Ant

dri,

---]

E exp

Now, as 1->0 (or as T-*>occ)the quantum mechanical result should go overleads to the conjecture that as
into the classical one and this immllediately
LmkT_
EeAn(F)

i
n=1

?1=1

container Q I consider a twoIf instead of a realistic three-dimnensional
dimensional one, the result would still be the same
X
-e_n

0

2e-X,2r)n=1

n=1

except that now I Q| is the area of Q ratherthan 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)IV (X) /_ -I

I

2er

1%),

it followsimmediatelyby an Abelian theoremthat
1 0

e
__E
and hence that

n=1

Setting A (X) =

r2
E-X2
eXnr)

-XnT

-

1

TO,--

1 -XT
27rT

-Ie

2ro

0

dX.

can record the last result as
x <x4'(r), wNve

CAN ONE HEAR

THE

SHAPE

OF A DRUM?

e-XrdX,

de-XrA (X)7r -> 0.

Since A (X) is nondecreasingwe can apply the Hardy-Littlewood-Karamata
Tauberian theoremand conclude what everyonewould be temptedto conclude,
namely that

A(X)=

+

N

2-

(r)2

,

-->o,

for every r in the interiorof Q.
Though this asymptotic formula is thus nearly "obvious" on "physicalgrounds,"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,,forlarge n. This can be approached by the device of studyingthe Dirichlet series
00

E

e-Xnt

n=1

forsmall t. This in turn is most convenientlyapproached throughthe series
00

ZeCXt,(P)

n1=1

{D00

=

fe-tdA(x)and thus we are led to the Abelian-Tauberian interplaydescribed above.
6. It would seem that the physical intuition ought not only provide themathematicianwith interestingand challengingconjectures,but also show him
the way toward a proofand toward possible generalizations.
The contextof the theoryof black body radiation or that of quantum statistical mechanics,however,is too far removedfromelementaryintuitionand too
full of daring and complex physical extrapolationsto be of much use even inseeking 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 distributionof eigenvaluesof the Laplacian becomes quite tractable. Proofsemerge
as natural extensionsof physical intuitionand interestinggeneralizationscome
withinreach.
7. The physical context in question is that of diffusion theory,anotherbranch of nineteenthcenturymathematical physics.
Imagine "stuff,"initiallyconcentratedat p( (xo, yo)), diffusingthrougha
plane regionQ bounded by F. Imagine furthermorethat the stuffgets absorbed( "eaten") at the boundary.

8

IN ANALYSISPAPERS

The concentrationPQ(p| r; t) of matter at r-((x, y)) at time t obeys thedifferentialequation of diffusion
aPe

- =t

(a)

1

2

the boundary condition(b)

PQ(p I r; t) -0 as r approachesa boundarypoint,

and the initial condition

as t ->0;

PQ(P |r;t) -*(r-p)(c)

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 absorbing and the initial condition (c) the fact that initiallyall the "stuff"was concentrated at ip.
I have again chosen units so as to make the diffusionconstant equal to 2.
As is well knowntheconcentrationPQ(P I -; t) can be expressedin termsof theeigenvalues Xz, and normalizedeigenfunctions4,(r-) of the problem
'72v

in Q,

+X4/=00 on P.
In fact, PQ(p Ir; t)= En-],

6-fl'4'(p)Abl(r).

clear that particlesof the diffusing
NOw,forsmall t, it appearsintuitively
oftheboundaryQ.
stuffwillnothavehad enoughtimeto havefelttheinfluence
As particlesbeginto diffuse
theymaynotbe aware,so to speak,of the disaster
thatawaitsthemwhentheyreachthe boundary.
We maythusexpectthatin someapproximatesense
PQ(p I r; t) -Po(p I r; t),

as t O,equationwherePo(p1r; t) stillsatisfiesthesame diffusion

P

(a')

VPO

and the same initialcondition(c')

Po(p ;t) =

P-p),

t0,

but is otherwiseunrestricted.
withoutwhichthe solution
Actuallythereis a slightadditionalrestriction
is notunique (a remarkablefactdiscoveredsomeyearsago by D. V. Widder).
thatPo be boundedfrombelow).
is thatPo > 0 (ormoregenerally
The restriction
in a bounded
forPQ is not neededsincefordiffusion
A similarrestriction
regionit followsautomatically.

THE

CAN ONE HEARSHAPE

OF A DRUM?

9

An explicitformulaforPo is, of course, well known. It isexp -Po(pIr;t)-]denotes the Euclidean distance between p and r.
where ||I can now state a littlemorepreciselythe principleof "not feelingthe boundary" explained a momentago.
The statementis that as t >0F
100
r ~P~2|e
1.o(-A()nr=-,12npr; t),expPQ( r; t)
1
Pnp
1

2w1t

n=1

21-

where - stands here for "is approximatelyequal to." This is a bit vague but
let it go at that forthe moment.
If we can trust this formulaeven forp = r we get
00

x'-Ant

e-e
n=l

1

1

2

n(r)

27rt

and if we display still more optimismwe can integratethe above and, making
use of the normalizationcondition2n d-r'=: 1,a(r-)
obtain
I:e-Xnt,
1=1

Q.

27rt

We recognizeimmediatelythe formulasdiscussed a whileback in connection
with the quantum-statistical-mechanicaltreatment of the ideal gas. If weapply the Hardy-Littlewood-Karamata theorem,alluded to before, we obtainas corollarythe theoremsof Carleman and Weyl.
To do this,however,we mustbe allowed to interpret as meaning"asynmptotic to."8. Now, a little mathematicalsoul-searching.Aren't we as far froma rigorous treatnment
as 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 bymathematics.
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 thereis always a question of interpretation.

10

PAPERSIN ANALYSIS

Let me assume that r is sufficiently
regularso 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

40 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(pr; t)dr= 1

for every open set A containingP.
Now, to morepertinentitems. If the mathematicaltheoryof diffusioncorresponds in any way to physical realitywe should have the inequality
exp[P(Q(P;- l11
r_IIP71?

Lt) Po(p|I;t);02t

For surelyless stuffwill be foundat r at time t ifthereis a possibilityof matterbeing destroyed(on the boundary F of Q) than if there were no possibilityof such destruction.
Now let Q be a square withcenterat p totallycontained in Q. Let its boundary act as an absorbingbarrierand denote by PQ(P Ir; t), rE Q, the corresponding
concentrationat r at time t.
In other words,PQ satisfiesthe differentialequation

(a")
(a ~

~

~

OPQ

--=-2P
= __

1

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 boundary
pointofQ.

Again it appears obvious that
PQ(P I|r;t) < PQ(P |r; t),rCQ,forthe diffusingstuffwhichreaches the boundaryof Q is lost as faras PQ is concerned but need notbe lost as a contributionto PQ.
Q has been chosen so simplybecause PQ(P| r'; t) is known explicitly,and, inparticular

CAN

ONE

PQ(PP I(p p;

t) =

HEAR

SHAPE

THE

E

a2

exp

OF

r

n

+

(m2
__-

_

m,n

11

A DRUM?

2a

2)w2

1

,

odd integers

wherea is the side of the square.
The combined inequalities

rlir- Pll
|

PQ(p

<

;t)

PQ(p

|

2w1

r; t) <

27rt

hold forall IrCQ and in particularfor r=p. In this case we get

4
a

a2
2

exp [-)?t

m,n

mmn

M2

0

2)72

2a'

odd integers

x V2 -

< E e

t()

~'(

n==

<

2,7rt

and it is a simple matterto prove that as t-*O we have asymptotically

4
a2

-

exp

,

odd integers

Thus asymptotically for t->O

[

(m2 +

2a2

e-

'2(p)-1n?=1

1n2)7r2

L--------1L-

27rt

12t and Carleman's theorem

follows.
It is only a little harderto prove Weyl's theorem.
If one integratesover Q the inequality

4 E exp[(m2+n2)w7r2

ex

-2

2a2

??

<

-Xntg

E e

t

2

7?n P)

odd

one obtains

4

E exp Lm,n(m2--t)]

_

2a2

odd

e

n=

JJ
i/(p)

dp

Q

a net of squares of side a, as shown in Fig. 2, and keep
We now cover Q wRrith
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 =

n=1

_

n=l

Je
Ce-A

4N(a)

J
Q

p) dp ?

E exp mon

odd

L

,

n=1

---)]
2a

ff
Q (a)

P/(p) dp

12

PAPERSIN ANALYSISt) < 1/2wrtover Q we getand, integratingthe inequality PQ(P1|;
<_

e-Zt

&/2irt.

afaaFIG. 2

Noting that N(a)a2=
formof the inequality

Q(a) I2

we record the fruitsof our latest labor in the|Q(a)+ n2)w2](Mn2
E expL
) j <

4

_

_

e <nt<

odd

From the fact (already noted above) that4

lim 2rt- E exp
a2
ma n
t+
odd

L

1
(n2+ n2)7r2
2a2

-t =1

we conclude easily that
00

Q(a)

00

<? lim inf2nrt Z e-nt < lim sup 27rtEI e->n <?
t0-~O

n=1

t0

n=l

l

and since, by choosinga sufficiently
small, we can make |Q(a) arbitrarilyclose
to Q we must have limto 27rt " e-nt= |or, in other words,,

E e xnt r

n=1

27rt

t -- 0.

9. Are we now throughwith rigor?Not quite. For while the inequalities

expL- 2wt22
paprt)--

27rt

CAN ONE HEARTHE

PQ(P| r;)

SHAPE

13

OF A DRUM?

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 choosingit toexhibityet another physical context.
It has been known since the early days of this century,throughthe work of
Einstein and Smoluchowski,that diffusionis but a macroscopic manifestationof microscopicBrownian motion.
Under suitable physical assumptions PQ(P| Y; t) can be interpretedas theprobabilitydensityof findinga freeBrownian particle at r at time t if it started
on its erraticjourney at t = 0 fromp and ifit gets absorbed when it comes to theboundary of Q.
If a large numberN of independentfreeBrownian particlesare startedfrom, thenATf(fP(pr;t)dris the average numberof these particles which are found in A at time t. Since
the statisticalpercentageerroris of the order 1/V/N continuousdiffusiontheory
is an excellent approximationwhen N is large.
A significantdeepening of this point of view was achieved in the earlytwentiesby Norbert Wiener. Instead of viewing the problem as a problem instatisticsof 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 <r < cc, starting from some
, ?Q,be open sets and t1<t2 < . . . <tnarbitrarilychosen origin0. Let Q1,$22,
ordered instants of time. The Einstein-Smoluchowskitheory required that(with suitable units)
Prob. {p +=Q(tl) E Qp+ r(tj'*S Po(p ri; t1)Po(r.

Q2

p + r(tn) E

n}

t1) .

P0( rnr;

xt2t
L-

I.

r2;t2 -

tn-tn)

dr.

drn

n

where,as before,Po(p I r; 2) =-exp

W-ienerhas shoxvnthat it is possible to constructa completelyadditive measure on the space of all continuouscurves r(r) emanating fromthe originsuchthat the set of curves p+ r() which at times tl<t2 <<tn find themselves, Qn respectively,has measure given by the Einsteinin open sets Q,, Q2,
Smoluchowskiformulaabove.
The set of curves such that p+r(T)E-,and p+r(t)EA(A an
0<T<t,
open set) turns out to be measurable and it can be shown, if Q has sufficiently

14

PAPERSIN ANALYSIS

smooth boundaries, that this measure is equal tofPu(p|

dr-.

r;t)This is not a trivial statement and it should come as no surprise that it
triviallyimplies the inequalities we needed a while back to make precise theprincipleof 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(cCethen meas. @,< meas. (L < meas. C.
One finalremarkbeforewe go on. The set of curves forwhichp+r(T)eSSQ

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 V72Puas well as the initial condition

lim fPQ(p; r; 1) dr = 1,
forall open sets A such that 'EA.
It is, however,no longerclear how to interpretthe boundary conditionthat
Ps2(p r;t)-0 when r-rI.

This difficulty
forcesthe classical theoryof diffusionto considerreasonablysmooth boundaries. The probabilistic interpretationof P2(pj r; t) provides anatural definitionof a generalizedsolttionof the boundary value problemunderconsideration.
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 certainpurelygeometricalcomplicationsI shall restrictmyself to convex drums.
We have achieved our firstsuccess by introducingthe principleofnot feelingthe boundary.But if is close to the boundaryF of Q then the diffusingparticles
startingfromp-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 perpendicular to the line joining and q. (See Fig. 3.) Then a diffusing
particlestartingfrom
p will see fora shorttime the boundary F as the straightline 1(P).
One may say, using again somewhat picturesque language, that, forsmall t,
the particle has not had time to feel the curvatureof the boundary.

CAN ONE HEAR

THE

SHAPE

15

OF A DRUM?

If this principle is valid I should be allowed to approximate (for small t)r; t),

PR(p r; t) by Pl()(pwhere Pi(p)(p Jr;t) satisfiesagain the diffusionequation

OP - 1
=
72P
At 2

-

r) as t-*O,but with the boundarycondition

with the initial conditionP--(p-a pointon 1(p; t) -* 0 as r approaches

Pl(-,)(pqrFIG. 4

FIG. 3Carryingthis optimism as far as possible we would expect that to a good
approximation,P( (P;)dPI P;
lP(t PQ|

It is well known that
1e-26'/t

Pi(P) cpI;

where b=||q-P||

t) =

2 -t

=minimal distance fromp to P. Thus (hopefully!)

PQ(

P; P

e-t E

t) dP =
n=1

27rt

e21 ddp

27rtQ

is our old friendfrombeforeand it remainsto calculateasympHere |I2 12wxt
totically(as t-?0) the integralfoe-281tdi. To do thisconsiderthe curveF(5) of
pointsin Q whose"distance"fromP is a. (See Fig. 4.)

16

PAPERSIN 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 lengthof F(5) we havefe 282tdp

J

60

-

d6 + somethingless than | Q

e 252ItL(Q)25o2/tand hence, neglectingan exponentiallysmall term (as well as termsof order t)
Je-22/t dp-,

60.o\/

e2x2L(xcvt)

-Vt f

V
v-tL

d

ro

Je

2x2 dxL-LV2?rtis the lengthof F.
We are finally led to the formulawhereL=L(O)X0

ZeeX7t
72=l

_

I Q|

-

2,72t

L

1

4

____

V/2i7rtfor tOand so we can also "hear" the lengthof the circumiiferenceof the drum!
The last asymptoticformulawas proved only a fewyears ago by the Swedish
mathematicianAke Pleijel [2] using an entirelydifferentapproach.
It is worthremarkingthat we can now prove that if all the frequenciesof adrum are equal to those of a circulardrumthen the drum must itselfbe circular.
This follows at once from the classical isoperimetricinequality which statesthat L2 ?4
Q| with equality occurringonly fora circle.
By pitch alone one can thus determinewhethera drum is circularor not!

l(p)FIG. 5

1 1 Can the heuristicargumentagain be made rigorous?Indeed it can. First,
we use the inequality

1SP_
1P )<P-pp|pt)=-2
I p

27rt

2w7t

2

CAN ONE HEAR

THE

SHAPE

17

OF A DRUM?

is simply a refinementof the one used previously,namely, PQ(l |; t)
wvhich< Ij/2rt,and which can be proven the same way.
Next we need a preciselower estimateforPQ(P |; t) and this is a little more
We "inscribe"the rectangleR(p, t) as shown in Fig. 5, wJTheredifficult.
h(t), theheightof the shaded segment,is to be determineda littlelater.
Let the side of R along the base of the segment be b(t) and the other sidebe w(t). It should be clear fromthe picturethat the y-axisbisectsthe sides of therectanglewhich are parallel to the x-axis.
Now considerPR(P | P; t). This notationis perhaps confusingsince it suggeststhat we are dealing with a boundary value problem in wThichthe boundary
varies with time. This is not the case. What we have in mind is the followATing:
and finallyset r=t.
fix t, findPR(t)(p-r; r) whichis definedunambiguously,
The result is PR(Pl |r; t). A convenientexpressionis) X
PR(P pi ; t)=2{EE(ex ep[-n22-x[_ _2] - exp [ep)-w (i)2])-W n2]expL

x {~exp---1= q-pI-h(t).
Now let h(t) =EX\t and, assumingthat l(p) is
at mostactually tangentto the curve (whichfora convex curve will happen waTitha denumerable numberof exceptional points w),we havewhere3=a-h(t)limt->oh(t)and consequentlyE/

exp [--

b(t)

_

1

n2 -exp

b(t)

rlln

=

t-

E/t

=

12

2b

L--(n

= 1 + o(l).

+j-)_)

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, ifthe curvatureexists at q, forthis
would imply that h(t)-b2(t) and the o(x/t) term above would then be anenormousoverestimate.Very mild additional regularityconditionsat nearly all
points q would insure o(-/t). Without enteringinto a discussionof these conditions let us simply assume the boundary to be such as to guarantee at leasto(Vt).
Since w(t) remainsbounded frombelow as t->O, we also haveZ (exp [_

2W2

exp

- exp [--n +=2

1-e-252It+ exponentiallysmall terms.

18

PAPERSIN ANALYSIS

We are now almost through.We write (cf. Fig. 6),_

u(pI P;t) dp >{V2+wt))f~(~v
2 Xt

Q eV t )p, t) dp

PR(Psmallterms)dp.
(1 - -2a2e t + exponentially

FIG. 6

Except thenforexponentiallysmall termsand the factor I +o(\It) in front
we have the integralT(1e-2821t)dp

Q(EV t)which, as before,can be seen to be asymptotically| Q(EVt) |--+2tL4

whereone neglectstermsof ordert and exponentiallysmall terms.Since asympQ I -Lc-\t one can obtain the inequality
totically IQ(EVt)?21 _(L+E')

27rt

4

1
27rt

where e' is related in a simple way to E. Since E' can be made arbitrarilysmall,the asymptoticformulae-xfollows.

| ,2

2rt

_

L_

1

=

4 V/2rt12. If our overall strategyof attack on the problem is rightwe should be
able to go on and forpoints veryclose to a smoothboundary replace the boundary locally by suitable circlesof curvature.

CAN ONE HEAR

THE

SHAPE

19

OF A DRUM?

A result of Pleijel suggests stronglythat fora simplyconnected drum witha smooth boundary (i.e. without cornersand with curvature existingat every
point) one has
2j

1

L

Z?e~f~'-A 2irtn1

4 V2wt+_,6

UnfortunatelyI am unable to obtain this, for the exasperating reason
that I am unable to get a workable expressionforPu(pj|; t) if 2 is a circle.
Rather than yield to despair over this sad state of affairslet me devote theremainderof the lecture to polygonaldrums,i.e. drums whose boundaries arepolygons.This study will show beyond the shadow of a doubt that the constanttermin our asymptoticexpansion owes its existenceto the overall curvatureof
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

-) t--*0, and vanishingt)r subject to the usual initialconditionPs(o0J(pJ
as r approaches a point on eitherside of the angle 0o.
This is a very old, veryclassical, problemand if0O= rm, with m an integer,
it can be solved by the familiarmethod of images. For m not an integer,Sommerfeldinvented 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 suitablecontourintegral.Carslaw transformsthe integralinto an infiniteseriesof Bessel
functionsbut forour purposes it is best to resistthe temptationof Bessel functions and to reduce the integral to a differentform. I shall skip the details(though some are quite instructive)and simplyreproducethe finalresult.
SetvQxe)= (Ij2irt)

exp [E-a--2kG
0-a-r<
2006- 2rpcos(l-a2t<O-a+?rexp[

r r2p2r.

i
fsin-A
L44rftot
~2t
)
00}
-

-

-

-

'

- 2kGo)+ p2]

exp
--

fir

rp

- cosh yLiw tcosh i-y +-(G-a))

7-- cos-d ,dywherethe summation E is extendedover k's satisfyingthe inequality underthesummationsign and p (p,a), r= (r, ).

20

PAPERSIN ANALYSIS

Then
Ps(00)(p r; t) = v(a) - v(-a).
Note that if0 = 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
xviththose obtained by the method of images.
Let us now assume that w 2 <00 <w and see what Ps(p- p; t) is in this case.
theIn
expression for v(a) when xve set O=a the inequality under the E sign
becomes -w<2k0o<w
and only k=O is allowed. In v(-ax) the inequality is
2c - r<2kOo<2cx+ r and what k's to take depends on (x.
We see that:

0 < a <
IF

2

only k = O is allowed,

60, only k =

a <

<

2

IF

0-

1 is allowed,

but for Oo-w 2 <ca < v12 both k = 0 and k = 1 are allowed. (See Fig. 7.)
Let us now put r =p (so that p =r) and write doxvn in detail the expressionsfor Ps(oO)(pj p; t) in the three sectors. For 0 < (x < 0 - 7r/2

exp
Ps(p | p; t)-r2(1- cos 2a)
27rt

-2t

sin -

-

expl
0

So/

P
++sin

)

L

Pspl p;t)=r2

rX exp L--coshg~

47rOot J

exp[_

'

r

47rOot00SFor 7r/2<ao < 00

r2-

-

2wt

r

cosh

ro

- cos
y
0000

exp
o

J_OO

dr.

y

- coshr(cosh -- y + 27r,
2w

exp

1
-_

27rt

(I- (1cosdy

7r2

2(Oo - a)27r+ the same two integralsas above

a

y

>

cos

7r2

CAN

ONE

HEAR

THE

OF

SHAPE

21

A DRUM?

7r

2~~~~~~
I/
/ /

I
I
I

\
\

and fk=l2

I k=Oort,'

2

/

k
\

0- -

~~~~~~~~~~/~~~~~~~~/~~~~~~~/

2

2

I

/

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

exp[--(I-cos
P'S(p p;t) = 27rt
-_

2a)

2wrt

exp[-2(1-cos 2(Oo-a)]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 simplifymatterssomewhat 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 obtaining N shaded sectors (where N is the number of sides or vertices of ourpolygon).

22

PAPERSIN ANALYSIS

fromp will either"see" the boundary
Now let p be a point in 2. Stuffdiffusing
a
line
if
p
is
near
a
vertex,as
an infinitewedge.
straight
or,as
We may as well say that the boundary will appear to the diffusingparticlet) by
as the nearest wedge, and that consequently we may replace P Q(-;
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 afterintegrationoverQ thisgives the principal term IQI/22t. Next, each Ps(b P; t) contains two complicated looking integralswhich have to be integratedover the wedge.
Fortunately,the second of these integratesout to 0, while the firstyields,
upon integrationover S(0O),
-?sin

2

n

__
dy

00

)f

(1 + cosh y) (cosh

00

2cos

0

This is only the contributionof one wedge; to get the total contributionone
must sum over all wedges.
Thus the total contributionis-

Z(sin-)--d- y - cos-)(1 + coshy)(coshFinally, if is in the shaded sectorof the wedge S(0O) we get, on integratingover the sector,
cos 2(a)Mexp [--(10-7r_T2

JO t27rt

exp

~2

--(1-

+

(0-o a))1
cos 2(O
[rdr=

2wt

-

and the total contributionfromthe shaded sectorsis
The remainingcontributionis easily seen to be-- 1J K(L-2a

Finally,+E

cot (Ooy---)
L

1

odu

4 V27rtFinally, fora polygonal drum

1(

-Z cotk2)

1/27rZo0 cot (Oo-7r/2).

e-262/t d5

22w Oocot (0 -7r

2/

CAN ONE HEAR

X0

E e-Ant

n=

L

| S

THE

SHAPE

OF A DRUM?

23

I

~____
2 rt
at \/27rtw1

4

87r'o\

\[so

00

2

00/

~~~dy

oo (1 + coshy)(cos7r/woy-cosw7r2/0)

with the understandingthat each 0o satisfiesthe inequality wr/2<Go <r. If theo in such a way that each 0o>7r, thenpolygon has N sides, and if we let N o->termapproaches
the constant2+r 0fdy

8wJ(1 + coshy)2

This should strengthenour belief that forsimplyconnected smooth drums theconstant is universal and equal to 6.
15. What happens formultiplyconnected 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<o <27r and thisis easily gottenfromthe generalformulaquoted above.
Near the holes the diffusingparticleswill "see" concave wedges but nothing
will change in principle.
If we let all polygons approach smooth curves it turns out the constant
approaches (1 -r)!, wherer is the numberof holes. It is thus natural to conjecture that fora smoothdrum with r smoothholes
Le\nt

ISI

n=1

21rt

_

L

1

4 V2wtand that thereforeone can "hear" the connectivityof the drum!
One can, of course, speculate on whetherin general one can hear the EulerPoincare characteristicand raise all sortsof otherinterestingquestions.
As our study of the polygonaldrumshows,the structureof theconstantterm
is quite complex since it combines metric and topological features. Whether
these can be properlydisentangledremains to be seen.
This is an expanded version ofa lecturewhichwas filmedunder the auspices of the Committeeon 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.