24
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 Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at . http://www.jstor.org/page/info/about/policies/terms.jsp JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected]. Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access to The American Mathematical Monthly. http://www.jstor.org

Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

Embed Size (px)

Citation preview

Page 1: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

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 .Accessed: 15/03/2012 22:10

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].

Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access toThe American Mathematical Monthly.

http://www.jstor.org

Page 2: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

CAN ONE HEAR THE SHAPE OF A DRUM?

MARK KAC, The Rockefeller University, New York

To George Eugene Uhlenbeck on the occasion of his sixty-fifth birthday

"La Physique ne nous donne pas seulement l'occasion de resoudre des problemes . .. elle nous fait presentir la solution." H. POINCARE.

Before I explain the title and introduce the theme of the lecture I should like to state that my presentation will be more in the nature of a leisurely excursion than of an organized tour. It will not be my purpose to reach a specified des- tination at a scheduled time. Rather I should like to allow myself on many occasions the luxury of stopping and looking around. So much effort is being spent on streamlining mathematics and in rendering it more efficient, that a solitary transgression against the trend could perhaps be forgiven.

r

Q

FIG. 1

1. And now to the theme and the title. It has been known for well over a century that if a membrane 2, held fixed

along its boundary F (see Fig. 1), is set in motion its displacement (in the direc- tion perpendicular to 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 properties of the mem- brane and on the tension under which the membrane is held.

I shall choose units to make C2= =2

Page 3: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

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 harmonic in time, they represent the putre tones the membrane is capa- ble of producing. These special solutions are also known as normal modes.

To find the normal modes we substitute U(-)eiwt into the wave equation and see that U must satisfy the equation 2 V2U+co2U=O with the boundary condition U=O on the boundary F of Q, corresponding to the membrane being held fixed along its boundary.

The meaning of "U =0 on F" should be made clear; for sufficiently smooth boundaries it simply means that U(p)->O as p approaches a point of F (from the inside). To show that a membrane is capable of producing a discrete spectrum of pure tones i.e. that there is a discrete sequence of c's W<_ ?2 ?Co3_ ? * for which nontrivial solutions of

V2U + (2U =0 , U = 0 on ,

exist, was one of the great problems of 19th century mathematical physics. Poincare struggled with it and so did many others.

The solution was finally achieved in the early years of our century by the use of the theory of integral equations.

We now know and I shall ask you to believe me if you do not, that for regions Q bounded by a smooth curve F there is a sequence of numbers X1_X2_ * * -

called eigenvalues such that to each there corresponds a function 4'(-), called an eigenfunction, such that

2 172V, + Xn4n - 0

and 41'(-)->O as p->a point of F. It is customary to 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 following problem: Let ?2 and ?2 be two plane regions bounded 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 for each n the eigenvalue N,, for ?1 is equal to the ei-genvalue AP1

Page 4: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

CAN ONE HEAR THE SHAPE OF A DRUM? 3

for 02. Question: Are the regions i1 and 02 congruent in the sense of Euclidean geometry?

I first heard the problem posed this way some ten years ago from Professor Bochner. Much more recently, when I mentioned it to Professor Bers, he said, almost at once: "You mean, if you had perfect pitch could you find the 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 whether we can determine Q if we know all the eigenvalues of the eigenvalue problem

V2U+XU=0 in Q, U =0 o0n F.

3. Before I go any further let 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 from the knowledge of all the eigenvalues, and to impress upon you the multi- tude of connections between our problem and various parts of mathematics and physics.

It should perhaps be stated at this point that throughout the paper only asymptotic properties of large eigenvalues will be used. This may represent, of course, a serious loss of information and it may perhaps be argued that precise knowledge of all the eigenvalues may be sufficient to determine the shape of the membrane. It should also be pointed out, howTever, that quite recently John Milnor constructed two noncongruent sixteen 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 first pertinent result is that one can "hear" the area of Q. This is an old result with a fascinating history which I shall now relate briefly.

At the end of October 1910 the great Dutch physicist H. A. Lorentz was invited to G6ttingen to deliver the Wolfskehl lectures. 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 from the principal be used to invite eminent scientists to lecture at Gottingen.

Lorentz gave five lectures under the overall title "Alte und neue Fragen der Physik"-Old and new problems of physics-and at the end of the fourth lecture he spoke as follows (in free translation from the original Germ-Ian): "In conclu- sion there is a mathematical probleimi which perhaps will arouse the interest of mathematicians who are present. It originates in the radiation theory of Jeans.

"In an enclosure with a perfectly reflecting surface there can form standing

Page 5: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

4 PAPERS IN ANALYSIS

electromagnetic wAAaves analogous to tones of ain organ pipe; we shall confine our attention to very high overtones. Jeans asks for the energy in the frequency interval dv. To this end he calculates the numlber of overtones which lie between the frequencies v and v+dy and multiplies this number by the energy which be- longs to the frequenlcy v, and which according to a theorem of statistical mllechan- ics is the samne for all frequencies.

"It is here that there arises the mathematical problem to prove that the num- ber of sufficiently high overtones which lies between v and v+dv is independent of the shape of the enclosure and is simiiply proportional to its volumlle. For many simnple shapes for wAhich calculations canl be carried out, this theoreml has been verified in a Leiden dissertation. There is no doubt that it holds in general even for multiply connected regions. Sinilar theoremns for other vibrating structures like inembranes, air inasses, etc. should also hold."

If one expresses this conjecture of Lorentz in terms of our imembrane, it enmerges in the form:

V (X) < 2 7r

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 apocryphal report that Hilbert predicted that the theorem would not be proved in his life time. Well, he was w-rong by many, many years. For less than two years later Hermiian Weyl, 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 proof was a crow ning achievement of this beautiful theory. I\lany subsequent developnments in the theory of differential and integral equations (especially the wNork of Coturant and his school) can be traced directly to Weyl's memoir on the conjecture of Lorentz.

5. Let me now consider briefly a different physical problem which too is closely related to the problenm of the distribution of eigenvalues of the Laplacian.

It can be taken as a basic postulate of classical statistical mechanics that if a system of 1l particles confined to a volume Q is in equilibrium wTith a thermo- stat of temperature T the probability of finding specified particles at ri,, r2,

rm (tn lel nd r -

rMf (wvithin volume elements dr1, dr2, . , drAzl) is

Page 6: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

CAN ONE HEAR THE SHAPE OF A DRUM? 5

exp [- V(rly ..

** rml)0 dr-, ...

dr,- L kTdr

* J. exp V(rl ... rm) dr, ... drm

where V(r-1, ,rll) is the interactioni potential 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 corresponding assumiption in quantum statistical mechanics seems much more complicated. One starts with the Schr6dinger equation

2 V - V(ri, ..., r_>) Eq Q 2- , where h is the Planck constant) 2m 27

with the boundary condition limn 4(ri. , 1) =0, whenever at least one rk

approaches the boundary of Q. (This boundary condition has the effect of con- fining the particles to Q.) Let E1 < E2 ? E3 < be the eigenvalues and 4'11, 4, . the corresponding normalized eigenfunctions. Then the basic postu- late is that the probability of finding specified 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 finding specified 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

Page 7: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

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 expressible in termiis of the X's and corresponding 41(r)'s.

The formula for the probability of finding specified particles at -ri, rm turns out to be

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

H ~ L L _ _ _ _ _ _ _ _ _ _ - d r 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 immllediately leads to the conjecture that as

LmkT_

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

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

-e_ n X

2r) 2e-X,- 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 interior of Q. If we believe Weyl's result that (in the two-dimensional case)

IV (X) /_ -I I

1%), 2er

it follows immediately by an Abelian theorem that

1 0 1 __E e -XnT - T O,-- 0

and hence that

-X2 1 r2 -XT

E eXnr) -Ie dX. n=1 27rT 2ro

Setting A (X) = x <x 4'(r), wNve can record the last result as

Page 8: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

CAN ONE HEAR THE SHAPE OF A DRUM? 7

de-Xr A (X) e-Xr dX, r -> 0.

Since A (X) is nondecreasing we can apply the Hardy-Littlewood-Karamata Tauberian theorem and conclude what everyone would be tempted to conclude, namely that

2- N A (X)= + (r)2 , -->o,

f or every r in the interior of Q. Though this asymptotic formula is thus nearly "obvious" on "physical

grounds," it was not until 1934 that Carleman succeeded in supplying a rigorous proof.

In concluding this section it may be worthwhile to say a word about the "strategy" of our approach.

We are primarily interested, of course, in asymptotic properties of X,, for large n. This can be approached by the device of studying the Dirichlet series

00

E e-Xnt

n=1

for small t. This in turn is most conveniently approached through the series 00 {D00

Ze CXt,(P) = fe-tdA(x) n1=1

and thus we are led to the Abelian-Tauberian interplay described above.

6. It would seem that the physical intuition ought not only provide the mathematician with interesting and challenging conjectures, but also show him the way toward a proof and toward possible generalizations.

The context of the theory of black body radiation or that of quantum statis- tical mechanics, however, is too far removed from elementary intuition and too full of daring and complex physical extrapolations to be of much use even in seeking the kind of understanding that makes a mathematician comfortable, let alone in pointing toward a rigorous proof.

Fortunately, in a much more elementary context the problem of the dis- tribution of eigenvalues of the Laplacian becomes quite tractable. Proofs emerge as natural extensions of physical intuition and interesting generalizations come within reach.

7. The physical context in question is that of diffu sion theory, another branch of nineteenth century mathematical physics.

Imagine "stuff," initially concentrated at p( (xo, yo)), diffusing through a plane region Q bounded by F. Imagine furthermore that the stuff gets absorbed ( "eaten") at the boundary.

Page 9: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

8 PAPERS IN ANALYSIS

The concentration PQ(p| r; t) of matter at r-((x, y)) at time t obeys the differential equation of diffusion

aPe 1 (a) - =t 2

the boundary condition

(b) PQ(p I r; t) - 0 as r approaches a boundary point,

and the initial condition

(c) PQ(P |r; t) -*(r-p) as t ->0;

here (-r-p) is the Dirac "delta function," with "value" oo if -r= p and 0 if r p'. The boundary condition (b) expresses the 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 diffusion constant equal to 2.

As is well known the concentration PQ(P I -; t) can be expressed in terms of the eigenvalues Xz, and normalized eigenfunctions 4,(r-) of the problem

'72v +X4/=0 in Q, 0 on P.

In fact, PQ(p I r; t)= En-], 6-fl'4' (p)Abl(r). NOw, for small t, it appears intuitively clear that particles of the diffusing

stuff will not have had enough time to have felt the influence of the boundary Q. As particles begin to diffuse they may not be aware, so to speak, of the disaster that awaits them when they reach the boundary.

We may thus expect that in some approximate sense

PQ(p I r; t) - Po(p I r; t), as t O ,

where Po(p1 r; t) still satisfies the same diffusion equation

(a') P VPO

and the same initial condition

(c') Po(p ;t) = P-p), t0,

but is otherwise unrestricted. Actually there is a slight additional restriction without which the solution

is not unique (a remarkable fact discovered some years ago by D. V. Widder). The restriction is that Po > 0 (or more generally that Po be bounded from below).

A similar restriction for PQ is not needed since for diffusion in a bounded region it follows automatically.

Page 10: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

CAN ONE HEAR THE SHAPE OF A DRUM? 9

An explicit formula for Po is, of course, well known. It is

Po(pI r; t) exp - - ]

where ||- denotes the Euclidean distance between p and r. I can now state a little more precisely the principle of "not feeling the bound-

ary" explained a moment ago. The statement is that as t >0

00 1 F r ~P~2 1.o(

Pnp | r; t) = 1 -,1 e -A()nr exp- 2- np r; t), PQ(

n=1 2w1t 1 21-

where - stands here for "is approximately equal to." This is a bit vague but let it go at that for the moment.

If we can trust this formula even for p = r we get 00 1

x'-Ant 2 1 e-e n(r)

n=l 27rt

and if we display still more optimism we can integrate the above and, making use of the normalization condition

2

n a(r-) d-r'=: 1,

obtain

I:e-Xnt, Q. 1= 1 2 7rt

We recognize immediately the formulas discussed a while back in connection with the quantum-statistical-mechanical treatment of the ideal gas. If we apply the Hardy-Littlewood-Karamata theorem, alluded to before, we obtain as corollary the theorems of 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 from a rigor- ous treatnment as we were before? True, diffusion is more familiar than black body radiation or quantum statistics. But familiarity gives comfort, at best, and comfort may still be (and often is) miles away from the rigor demanded by mathematics.

Let us see then what we can do about tightening the loose talk. First let me dispose of a few minor items which may cause you worry. When I write i/ = 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.

Page 11: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

10 PAPERS IN ANALYSIS

Let me assume that r is sufficiently regular so that no ambiguity arises i.e.

P(p r; t) -O 0 as r -> a boundary point of Q,

means exactly what it says, while 41 = 0 on r means

4 0 as r -* a boundary point 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 containing P. Now, to more pertinent items. If the mathematical theory of diffusion corre-

sponds in any way to physical reality we should have the inequality

r_IIP - l11 exp [- 71? P(Q (P; t) 0 Po(p|I;t) ; L 2t

For surely less stuff will be found at r at time t if there is a possibility of 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 center at p totally contained in Q. Let its bound- ary act as an absorbing barrier and denote by PQ (P I r; t), r E Q, the corresponding concentration at r at time t.

In other words, PQ satisfies the differential equation

OPQ 1 (a ~ ~ ~ __ = - V72PQ (a") --=-2P

and the initial condition

(C") PQ(P r; t) - (r -p) as t O .

It also satisfies the boundary condition

(b"/) PQ(p I Y; 1) -O 0 as Y- a boundary point of Q.

Again it appears obvious that

PQ (P I|r; t) < PQ (P |r; t), rCQ,

for the diffusing stuff which reaches the boundary of Q is lost as far as PQ is con- cerned but need not be lost as a contribution to PQ.

Q has been chosen so simply because PQ(P| r'; t) is known explicitly, and, in particular

Page 12: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

CAN ONE HEAR THE SHAPE OF A DRUM? 11

P I(p r (m2 + n

2)w2 1

PQ(P p; t) = E exp __- , a2 m,n _ 2a

odd integers

where a is the side of the square. The combined inequalities

2w1 rlir - Pll

PQ(p | ;t) < PQ(p

| r; t) < 27rt

hold for all IrCQ and in particular for r=p. In this case we get

4 M2 2)72 0

x V2 -

a2 m,n exp [-)?t < E e t() < a 2 mmn 2a' n== ~'( 2,7rt

odd integers

and it is a simple matter to 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 integrates over Q the inequality

4 E exp[ (m2+ n2)w7r2 < ??

-Xntg 2

-2 ex 2a2 t E e 7?n P)

odd

one obtains

4 E exp L (m2--t)] _ e JJ i/(p) dp m,n 2a2 n= Q odd

We now cover Q wRrith a 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 = Ce-A Je J p) dp ? , ff P/(p) dp n=1 n=l Q n=1 Q (a)

_ 4N(a) E exp - ---)] mon L 2a odd

Page 13: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

12 PAPERS IN ANALYSIS

and, integrating the inequality PQ(P1 |; t) < 1/2wrt over Q we get e-Zt <_ &/2irt.

a fa

a

FIG. 2

Noting that N(a)a2= |Q(a) we record the fruits of our latest labor in the form of the inequality

4 (Mn2 + n2)w2] _ _

Q(a) I 2 E exp L ) j < e <nt < odd

From the fact (already noted above) that

4 (n2 + n2)7r2 1 lim 2rt- E exp - t =1 t+ a2 ma n L 2a2

odd

we conclude easily that 00 00

Q(a) ? < lim inf 2nrt Z e-nt < lim sup 27rt EI e->n <? l t0-~O n=1 t0 n=l

and since, by choosing a sufficiently small, we can make |Q(a) arbitrarily close to Q we must have limt o 27rt " , e-nt= | or, in other words,

E e xnt r t -- 0. n=1 27rt

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

2wt2 exp L- 2 paprt) -- 27rt

Page 14: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

CAN ONE HEAR THE SHAPE OF A DRUM? 13

PQ(P| r;) PQP r;t, r Q,

are utterly obvious on intuitive grounds 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 diffusion is but a macroscopic manifestation of microscopic Brownian motion.

Under suitable physical assumptions PQ(P| Y; t) can be interpreted as the probability density of finding a free Brownian particle at r at time t if it started on its erratic journey at t = 0 from p and if it gets absorbed when it comes to the boundary of Q.

If a large number N of independent free Brownian particles are 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 percentage error is of the order 1/V/N continuous diffusion theory is an excellent approximation when N is large.

A significant deepening of this point of view was achieved in the early twenties by Norbert Wiener. Instead of viewing the problem as a problem in statistics of particles he viewed it as a problem in statistics of paths. Without entering into details let me review briefly what is involved here.

Consider the set of all continuous curves r(T), 0 <r < cc, starting from some arbitrarily chosen origin 0. Let Q1, $22, , ?Q, be open sets and t1 <t2 < . . . <tn ordered instants of time. The Einstein-Smoluchowski theory required that (with suitable units)

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 - I. L xt2t W-iener has shoxvn that it is possible to construct a completely additive mea-

sure on the space of all continuous curves r(r) emanating from the origin such that the set of curves p+ r() which at times tl <t2 < <tn find themselves in open sets Q,, Q2, , Qn respectively, has measure given by the Einstein- Smoluchowski formula above.

The set of curves such that p+r(T)E-, 0<T<t, and p+r(t)EA (A an open set) turns out to be measurable and it can be shown, if Q has sufficiently

Page 15: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

14 PAPERS IN ANALYSIS

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 principle of not feeling the boundary.

In fact, as the reader no doubt sees, the inequalities in question are simply a consequence of the fact that if sets a, 6(, e are such that

UIc(cC e

then meas. @, < meas. (L < meas. C. One final remark before we go on. The set of curves for which

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) satisfies the diffusion equation 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 longer clear how to interpret the boundary condition that

Ps2(p r; t)- 0 when r- rI.

This difficulty forces the classical theory of diffusion to consider reasonably smooth boundaries. The probabilistic interpretation of P2(pj r; t) provides a natural definition of a generalized solttion of the boundary value problem under consideration.

10, We are now sure that we can hear the area of a drum and it mnay seem that Nwe spent a lot of effort to achieve so little.

Let me now show you that the approach we used can be extended to yield more, but to avoid certain purely geometrical complications I shall restrict my- self to convex drums.

We have achieved our first success by introducing the principle of not feeling the boundary. But if - is close to the boundary F of Q then the diffusing particles starting from p- will, to some extent, begin to be influenced by F.

Let q be the point on F closest to - and let 1(p) be the straight line perpendicu- lar to the line joining - and q. (See Fig. 3.) Then a diffusing particle starting from p will see for a short time the boundary F as the straight line 1(P).

One may say, using again somewhat picturesque language, that, for small t, the particle has not had time to feel the curvature of the boundary.

Page 16: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

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) satisfies again the diffusion equation

OP 1 - = - 72P At 2

with the initial condition P--(p- r) as t-*O, but with the boundary condition

Pl(-,)(p ; t) -* 0 as r approaches a point on 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 lP( P; t P

It is well known that

1e-26'/t Pi(P) cpI; t) = 2 -t

where b=||q-P|| =minimal distance from p to P. Thus (hopefully!)

P; t) d = E e-t e21 dp PQ(

P P n=1 27rt 27rt Q

d

Here |I2 12wxt is our old friend from before and it remains to calculate asymp- totically (as t-?0) the integral foe-281tdi. To do this consider the curve F(5) of points in Q whose "distance" from P is a. (See Fig. 4.)

Page 17: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

16 PAPERS IN ANALYSIS

For small enough &, F(s) is well defined (and even convex) aind the major contribution to our integral comes from small 8.

If L(6) denotes the length of F(5) we have

60 fe 282tdp - J e 252ItL(Q) d6 + something less than | Q 25o2/t

and hence, neglecting an exponentially small term (as well as terms of order t) 60 .o\/ ro L

Je-22 /t dp -, -Vt f e2x2L(xcvt) d V v-tL 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 Z eeX7t _ - ____ for tO 72=l 2,72t 4 V/2i7rt

and so we can also "hear" the length of the circumiiference of the drum! The last asymptotic formula was proved only a few years ago by the Swedish

mathematician Ake Pleijel [2] using an entirely different approach. It is worth remarking that we can now prove that if all the frequencies of a

drum are equal to those of a circular drum then the drum must itself be circular. This follows at once from the classical isoperimetric inequality which states that L2 ?4 Q| with equality occurring only for a circle.

By pitch alone one can thus determine whether a drum is circular or not!

l(p)

FIG. 5

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

1SP 1P )<P p|pt)=-2 2 _ -p I p 27rt 2w7t

Page 18: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

CAN ONE HEAR THE SHAPE OF A DRUM? 17

wvhich is simply a refinement of 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 for PQ(P |; t) and this is a little more difficult. We "inscribe" the rectangle R(p, t) as shown in Fig. 5, wJThere h(t), the height of the shaded segment, is to be determined a 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 from the picture that the y-axis bisects the sides of the rectangle which are parallel to the x-axis.

Now consider PR(P | P; t). This notation is perhaps confusing since it suggests that we are dealing with a boundary value problem in w Thich the boundary varies with time. This is not the case. What we have in mind is the followATing: fix t, find PR(t)(p-r; r) which is defined unambiguously, and finally set r=t. The result is PR(Pl |r; t). A convenient expression is

P pi ; t)=2{E ep[-n22-x --1 ) PR( X ) E (ex [_ _ 2] - exp [ep -

x {~exp -W n 2]expL -w (i)2])

where 3=a-h(t) = q-pI-h(t). Now let h(t) =EX\t and, assuming that l(p) is actually tangent to the curve (which for a convex curve will happen waTith at most a denumerable number of exceptional points w), we have

b(t) b(t) lim = rlln = t->o h(t) t- E/t

and consequently

/ _ 1 2b 12 E exp [-- n2 -exp L--(n +j-)_) = 1 + o(l).

This is not quite enough, however, and one needs the stronger estimate

E (exp [_ b n2] - exp [- b + 1 )]) = 1 +

This will surely be the case, for example, if the curvature exists at q, for this would imply that h(t)-b2(t) and the o(x/t) term above would then be an enormous overestimate. Very mild additional regularity conditions at nearly all points q would insure o(-/t). Without entering into 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 from below as t->O, we also have

2W2 exp 2

Z (exp [_ - exp [-- n +-

= 1-e-252It + exponentially small terms.

Page 19: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

18 PAPERS IN ANALYSIS

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

, _ u(p I P; t) dp >{V PR(P p, t) dp

2+ wt))f~(~v (1 - -2a2e t + exponentially small terms) dp. 2 Xt Q eV t )

FIG. 6

Except then for exponentially 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

where one neglects terms of order t and exponentially small 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 arbitrarily small, the asymptotic formula

2 | L 1 e-x , _ _ =

2rt 4 V/ 2rt follows.

12. If our overall strategy of attack on the problem is right we should be able to go on and for points very close to a smooth boundary replace the bound- ary locally by suitable circles of curvature.

Page 20: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

CAN ONE HEAR THE SHAPE OF A DRUM? 19

A result of Pleijel suggests strongly that for a simply connected drum with a smooth boundary (i.e. without corners and with curvature existing at 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 expression for Pu(pj |; t) if 2 is a circle.

Rather than yield to despair over this sad state of affairs let me devote the remainder of the lecture to polygonal drums, i.e. drums whose boundaries are polygons. This study will show beyond the shadow of a doubt that the constant term in our asymptotic expansion owes its existence to the overall curvature of the boundary.

13. Before I go on I need an expression for Ps(a,)(p ; t) where S(GO) is an infinite wedge of angle 6o. In other words Ps(oo) is the solution of

lop I

d V 2

subject to the usual initial condition Ps(o0J(pJ t)r - -) t--*0, and vanishing as r approaches a point on either side of the angle 0o.

This is a very old, very classical, problem and if 0O = rm, with m an integer, it can be solved by the familiar method of images. For m not an integer, Som- merfeld invented 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 elementary approach in which Ps(o) (p I; t) is represented by a suitable contour integral. Carslaw transforms the integral into an infinite series of Bessel functions but for our purposes it is best to resist the temptation of Bessel func- tions and to reduce the integral to a different form. I shall skip the details (though some are quite instructive) and simply reproduce the final result.

Set

vQxe)= (Ij2irt) E-a--2kG exp [- - 2rpcos(l -a - 2kGo)+ p2] 0-a-r< 2006 2t

<O-a+?r

r r2p2r rp exp [ . exp - cosh y

fsin-A L ~2t L t d ) i - - - - ' -- dy , 00} 44rftot fir iw ) 7

cosh i-y +-(G-a) - - cos-

where the summation E is extended over k's satisfying the inequality under the summation sign and p (p,a), r= (r, ).

Page 21: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

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 renmains in the resulting expression for v(a) -v(-a) is a collection of termns easily identifiable xvith those 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. In the 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:

IF 0 < a < 0- only k = O is allowed,

2

IF < a < 60, only k = 1 is allowed,

2

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 expressions

for Ps(oO)(pj p; t) in the three sectors. For 0 < (x < 0 - 7r/2

r2 exp - (1- cos 2a)

Ps(p | p; t) -2t 27rt

- r2- r2

expl 0 L - rX exp L--cosh y

- sin - g~ dy So/ 47rOot J r 7r2

cosh y - cos 0000

P ) exp [_ r exp o - cosh y

2wt 2w

++sin ' r o dr. 00S 47rOot J_OO a r( 7r 2

cosh -- y + 27r, > cos

For 7r/2 <ao < 00

1 exp (I- - (1cos 2(Oo - a) Pspl p;t)= -_

27rt 27r

+ the same two integrals as above

Page 22: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

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 <aoz< r/2

exp [--(I-cos 2a) exp [-- (1-cos 2(Oo-a)]

P'S(p p; t) = -_ 2 r 27rt 2wrt 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 from p to a side of the wedge.

FIG. 8

14. To simplify matters somewhat let me assume that the polygonal drum is convex and that every angle is obtuse.

At each vertex we draw perpendiculars to the sides of the polygon thus ob- taining N shaded sectors (where N is the number of sides or vertices of our polygon).

Page 23: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

22 PAPERS IN ANALYSIS

Now let p be a point in 2. Stuff diffusing from p will either "see" the boundary as a straight line or, if p is near a vertex, as an infinite wedge.

We may as well say that the boundary will appear to the diffusing particle 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 integration over Q this gives the principal term IQI/2 2t. Next, each Ps(b P; t) contains two compli- cated looking integrals which have to be integrated over the wedge.

Fortunately, the second of these integrates out to 0, while the first yields, upon integration over S(0O),

n 2 00 __ dy -?sin )f

(1 + cosh y) (cosh - 2cos 00 0

This is only the contribution of one wedge; to get the total contribution one must sum over all wedges.

Thus the total contribution is

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

Finally, if - is in the shaded sector of the wedge S(0O) we get, on integrating over the sector,

M exp [--(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 contribution from the shaded sectors is 1/27r Zo0 cot (Oo-7r/2). The remaining contribution is easily seen to be

-- 1J K(L-2a E cot (Ooy---) e-262/t d5

L 1 1 ( 7r Finally,+ -Z cot od- u

4 V27rt 2 2w Oo k 2/

Finally, for a polygonal drum

Page 24: Can One Hear the Shape of a Drum? - UC Davis Mathematicshunter/m207b/kac.pdf · CAN ONE HEAR THE SHAPE OF A DRUM? MARK KAC, The Rockefeller University, New York To George Eugene Uhlenbeck

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 + cosh y)(cos 7r/woy-cosw7r2/0)

with the understanding that each 0o satisfies the inequality wr/2 <Go <r. If the polygon has N sides, and if we let N o-> o in such a way that each 0o>7r, then the constant term approaches

2+ r 0f dy

8wJ (1 + cosh y)2

This should strengthen our belief that for simply connected smooth drums the constant is universal and equal to 6.

15. What happens for multiply connected drums? If the drum as well as the holes are polygonal the answer is easily obtained.

One only needs Ps(.) (p1 p; t) for 6o satisfying the inequality r <o <27r and this is easily gotten from the general formula quoted above.

Near the holes the diffusing particles will "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)!, where r is the number of holes. It is thus natural to conjec- ture that for a smooth drum with r smooth holes

Le\nt ISI _ L 1

n=1 21rt 4 V2wt

and that therefore one can "hear" the connectivity of the drum! One can, of course, speculate on whether in general one can hear the Euler-

Poincare characteristic and raise all sorts of other interesting questions. As our study of the polygonal drum shows, the structure of the constant term

is quite complex since it combines metric and topological features. Whether these can be properly disentangled remains to be seen.

This is an expanded version of a lecture which was filmed under 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 differential and integral equations, Proc. Second Berkeley Symposium on Mathematical Statistics and Probability, 1957, pp. 189-215.

2. A. Pleijel, A study of certain Green's functions with applications in the theory of vibrating membranes, Arkiv for Matematik, 2 (1954) 553-569.