The Sound of Music - zlu/talks/2013-ucigraduateseminar/2013-uci... · The Sound of Music Symmetry Zhiqin Lu University of California, Irvine November 8, 2013

The Sound of Music

Zhiqin Lu

University of California, Irvine

November 8, 2013

The Sound of Music Symmetry

Zhiqin Lu

University of California, Irvine

November 8, 2013

In 1966, M. Kac asked the the following question:

Question

If two planar domains have the same spectrum, are theyidentical up to rigid motions of the plane?

(Can one hear the shape of a drum?)

In 1966, M. Kac asked the the following question:

Question

If two planar domains have the same spectrum, are theyidentical up to rigid motions of the plane?

(Can one hear the shape of a drum?)

For a domain Ω in R2, the above mentioned spectrum is theset of eigenvalues of the Laplace operator ∆ with Dirichletboundary condition. This is the set of all real numbers λ suchthat there exists a smooth function u on Ω satisfying theLaplace equation and boundary condition

∆u(x, y) :=∂2u

∂x2+∂2u

∂y2= −λu(x, y),

u = 0 on the boundary of Ω.

The eigenvalues of a bounded domain from a discrete set of(0,∞)

0 < λ1 < λ2 ≤ λ3 . . . .

The set of eigenvalues λk∞k=1 is in bijection with theresonant frequencies a drum would produce if Ω were itsdrumhead. With a perfect ear one could hear all thesefrequencies and therefore know the spectrum. In other words,if two drums sound identical to a perfect ear and thereforehave identical spectra, then are they the same shape?

Let’s take a look at the simplest case and assume that ourdrum is actually a string which is mathematically described bythe ordinary differential equation

f ′′(x) = −λf(x), f(0) = f(`) = 0.

We know from calculus that the solutions are

fk(x) = sin

(kπx

`

), with λ = λk =

k2π2

`2

for k = 1, 2, · · · .

Question

Can one hear the shape of a string?

The “shape” of the string is just its length, so we canformulate the question as: if we know the set of all λk∞k=1,then do we know the length of the string?

` =

√π2

λ1

.

This shows that we can hear the length of the string based onthe first eigenvalue.

The “shape” of the string is just its length, so we canformulate the question as: if we know the set of all λk∞k=1,then do we know the length of the string?

` =

√π2

λ1

.

This shows that we can hear the length of the string based onthe first eigenvalue.

Exercise

Prove that one can hear the shape of a rectangle.

1 Can you compute the eigenvalues of other domains?

2 disks, annuals, etc...

3 How to answer the question of Kac’s?

Mathematically, the question is equivalent to determiningwhether or not the following map

Λ :M→ R∞, Ω 7→ (λ1, λ2, · · · , λk, · · · ),

is injective, where M is the moduli space of all boundeddomains in R2.





Λ :M→ R∞, Ω 7→ (λ1, λ2, · · · , λk, · · · ),






Λ :M→ R∞, Ω 7→ (λ1, λ2, · · · , λk, · · · ),






Λ :M→ R∞, Ω 7→ (λ1, λ2, · · · , λk, · · · ),


Counterexample of Kac’s question

Gordon, Webb, and Wolpert gave a counterexample:

Figure : Identical sounding drums

1 Many open problems remain.

2 Motivated by these problems we shall investigate theinjectivity of Λ restricted to certain subsets of M.

3 A natural choice is the set of convex n-gons since this setcan be identified with a finite dimensional manifold withcorners.

4 Surprisingly, even within this subset Kac’s question is asubtle problem. Durso (1988) proved that one can hearthe shape of a triangle, so in fact Λ restricted to themoduli space of Euclidean triangles is injective. For n > 3the problem is widely open.













The spectrum not only determines the resonant frequencies ofvibration but also the so called heat trace.

∞∑k=1

e−λkt.

The above is convergent for any t > 0.In particular, λk →∞ and ∆ is an unbounded operator.


∞∑k=1

e−λkt.

The above is convergent for any t > 0.

In particular, λk →∞ and ∆ is an unbounded operator.


∞∑k=1

e−λkt.

The above is convergent for any t > 0.In particular, λk →∞ and ∆ is an unbounded operator.

Paul Erdos: A proof from THE BOOK

Theorem

There are infinitely manyprime numbers.

Proof.

∑p:prime number

1

p=∞.


Theorem


Proof.

∑p:prime number

1

p=∞.


Theorem


Proof.

∑p:prime number

1

p=∞.

How to prove that∞∑k=1

e−λkt

is finite for any t > 0?

The heat kernel is a smooth function H(x, y, t) such that∂tH(x, y, t) = ∆xH(x, y, t)H(x, y, t) = 0 for x on the boundaryH(x, y, 0) = δx(y) is the dirac function

Symbolically,

H = e∆t. ∂t(e∆t) = ∆(e∆t)

∞∑k=1

e−λkt = Tr (e∆t) =

∫Ω

H(x, x, t)dx.


Symbolically,

H = e∆t.

∂t(e∆t) = ∆(e∆t)

∞∑k=1


∫Ω

H(x, x, t)dx.


Symbolically,

H = e∆t. ∂t(e∆t)

= ∆(e∆t)

∞∑k=1


∫Ω

H(x, x, t)dx.


Symbolically,

H = e∆t. ∂t(e∆t) = ∆(e∆t)

∞∑k=1


∫Ω

H(x, x, t)dx.


Symbolically,

H = e∆t. ∂t(e∆t) = ∆(e∆t)

∞∑k=1


∫Ω

H(x, x, t)dx.

The heat kernel in R2 is

1

4πte−|x−y|2

4t .

Can we compute the heat kernel for any domain explicitly?Yes and No!

By our life experience, H(x, y, t) is very small when d(x, y) isbig or t is small!


1

4πte−|x−y|2

4t .

Can we compute the heat kernel for any domain explicitly?

Yes and No!



1

4πte−|x−y|2

4t .




1

4πte−|x−y|2

4t .


By our life experience,

H(x, y, t) is very small when d(x, y) isbig or t is small!


1

4πte−|x−y|2

4t .



The idea of Kac

Let Ω be the n-polygon.

If x is in the interior of Ω, then for y close enough, then

H(x, y, t) ≈ 1

4πte−|x−y|2

4t .

If x is near the edge (assuming that the edge is x2 = 0), theabove computation should have a correction term by theSchwarz reflexion principle:

H(x, y, t) ≈ 1

4πt

(e−|x−y|2

4t − e−|x′−y|2

4t

), x = (x1,−x2)

The idea of Kac

Let Ω be the n-polygon.If x is in the interior of Ω, then for y close enough, then

H(x, y, t) ≈ 1

4πte−|x−y|2

4t .


H(x, y, t) ≈ 1

4πt

(e−|x−y|2

4t − e−|x′−y|2

4t

), x = (x1,−x2)

The idea of Kac

Let Ω be the n-polygon.If x is in the interior of Ω, then for y close enough, then

H(x, y, t) ≈ 1

4πte−|x−y|2

4t .


H(x, y, t) ≈ 1

4πt

(e−|x−y|2

4t − e−|x′−y|2

4t

), x = (x1,−x2)

Thus we have

H(x, x, t) =1

4πtif x is in the interior

H(x, x, t) =1

4πt(1− e−

x22t ) if x is near the boundary

∑e−λkt ≈

∫Ω

1

4πtdx+

∫∂Ωε

1

4πte−

x22t dx1dx2

≈ |Ω|4πt− |∂Ω|

8√πt

as t→ 0.

A more accurate result

∞∑k=1

e−λkt ∼ |Ω|4πt− |∂Ω|

8√πt

+n∑i=1

π2 − α2i

24παi+O(e−1/t) t ↓ 0.

This shows that the area, the perimeter, and the sum over theangles of π2−α2

24παare all spectral invariants.

Let

k(x, y, t) =∞∑k=1

e−λkt −

(|Ω|4πt− |∂Ω|

8√πt

+n∑i=1

π2 − α2i

24παi

).

Conjecture (Lu)

There exists a number β > 0 such that for any ε > 0 we have

C(ε)e−β+εt ≤ k(x, y, t) ≤ C(ε)−1e−

β−εt

for a constant C(ε).

Let

k(x, y, t) =∞∑k=1

e−λkt −

(|Ω|4πt− |∂Ω|

8√πt

+n∑i=1

π2 − α2i

24παi

).

Conjecture (Lu)

There exists a number β > 0 such that for any ε > 0 we have

C(ε)e−β+εt ≤ k(x, y, t) ≤ C(ε)−1e−

β−εt

for a constant C(ε).

At the current technology, we can only make use of threeequations.

More spectrum invariants?

At the current technology, we can only make use of threeequations. More spectrum invariants?

The spectrum also determines the wave trace. Imagine aconvex n-gon is a billiard table. The set of closed geodesics isprecisely the set of all paths along which a billiard ball hit witha pool cue could roll, such that the ball returns to its startingpoint. The set of lengths of closed geodesics, which is knownas the length spectrum, is related to the (Laplace) spectrumby a deep result proven by Duistermaat and Guillemin in thelate 1970s.

The wave trace is a tempered distribution defined by

∞∑k=1

ei√λkt.

Duistermaat and Guillemin proved in that the singularities ofthe wave trace is contained in the length spectrum.

Question

Can we hear the shape of a trapezoid?

4 dimensional moduli space!

Answer: partially yes, by L-Rowlett!

Question




Question




Theorem (L-Rowlett-2013)

If two acute trapezoids have the same spectrum, then they areidentical up to rigid motions of the plane.

Figure : An acute trapezoid

Lemma

The length of the shortest closed geodesic of an acutetrapezoid is twice the height, that is twice the distance fromthe base to the opposite parallel side.

The four equations

h = h1

12(B + b)h = 1

2(B1 + b1)h1

`+ `′ +B + b = `1 + `′1 +B1 + b1

1α(π−α)

+ 1β(π−β)

= 1α1(π−α1)

+ 1β1(π−β1)

The deduction

h(cscα + csc β) = `+ `′ = L− 2A/h

where L is the perimeter and A is the area.

Therefore

cscα + csc β = cscα1 + csc β1.

The deduction

h(cscα + csc β) = `+ `′ = L− 2A/h

where L is the perimeter and A is the area. Therefore

cscα + csc β = cscα1 + csc β1.

We have cscα + csc β = cscα1 + csc β1

1α(π−α)

+ 1β(π−β)

= 1α1(π−α1)

+ 1β1(π−β1)

Lemma (A Putnam-type problem)

We haveα = α1, β = β1

if α ≤ β, α1 ≤ β1.

solved by L-Rowlett, 2013


1α(π−α)

+ 1β(π−β)

= 1α1(π−α1)

+ 1β1(π−β1)



if α ≤ β, α1 ≤ β1.



1α(π−α)

+ 1β(π−β)

= 1α1(π−α1)

+ 1β1(π−β1)



if α ≤ β, α1 ≤ β1.


Theorem

Let p, q be real numbers. Then the solution of the system ofequations

csc(α) + csc(β) = p;

(α(π − α))−1 + (β(π − β))−1 = q,

if it exists, must be unique for 0 < β ≤ α ≤ π/2.

First, let’s use the second equation to show that each αuniquely determine a β. Solving the second equation for β interms of α and q leads to a quadratic equation in β,

β2(1− qα(π − α)) + β(π(qα(π − α)− 1))− α(π − α) = 0.

Since the trapezoid is acute, β ≤ π2

, so indeed α and quniquely determine

β = β(α) =π

2−

√π2

4+

α(π − α)

1− qα(π − α).

We can prove the lemma if we prove that the function

g(α) := csc(α) + csc(β(α))

has unique solution α for any given p. Implicit differentiationgives

β′(α) = −((α(π − α))−1)′

((β(π − β))−1)′,

and so the derivative

g′(α) = − csc(α) cot(α)+csc(β(α)) cot(β(α))·((α(π − α))−1)′

((β(π − β))−1)′.

Let’s see if we can relate g′(α) to the following function

f(α) :=csc(α) cot(α)

((α(π − α))−1)′=α2(π − α)2 cosα

(2α− π) sin2 α.

Since

g′(α)

((α(π − α))−1)′= − csc(α) cot(α)

((α(π − α))−1)′+

csc(β(α)) cot(β(α))

((β(π − β))−1)′,

we see that

g′(α) =π − 2α

α2(π − α)2(f(α)− f(β))

It turns out that the logarithmic derivative of f is pleasantlysimple

f ′(α)

f(α)=

2

π − 2α+

2

α+

2

α− π− 2 cot(α)− tan(α).

Need to prove that the above expression is negative forα ∈ (0, π/2).

It turns out that the logarithmic derivative of f is pleasantlysimple

f ′(α)

f(α)=

2

π − 2α+

2

α+

2

α− π− 2 cot(α)− tan(α).

Need to prove that the above expression is negative forα ∈ (0, π/2).

The use of MatLab

Figure : Graph of (log f)′

Thank you!

Documents

The Sound of Music - zlu/talks/2013-ucigraduateseminar/2013-uci... · The Sound of Music Symmetry Zhiqin Lu University of California, Irvine November 8, 2013