Thesis

MEMBRANE HARMONICS
ASPEN DEVRIES
1. Abstract
In this paper we look at the question “can you hear the shape of a drum?”,
posed by Marc Kac in his famous 1966 paper [2]. We begin with the one-
dimensional analogue of the question and proceed to consider the special
case of rectangular drums. In both cases, “hearing” the drum (knowing
its spectrum) determines the geometry of the drum. We then compare the
lowest eigenvalue in the spectrum of a square drum to that of a circular
drum. We go on to present a proof of the Courant-Hilbert theorem which was
devised by Gladwell [6] and check that it applies to the first 9 eigenmodes of
the Helmholtz equation with Dirichlet boundary conditions on the rectangle.
Finally, we consider a set of drums constructed by Gordon et al [3] which
provide an answer to Kac’s question. We present a proof of isospectrality
constructed by Buser [7].
2. Introduction
For a given operator Ô : A ⊂ D → D we can characterize eigenvalues and
eigenfunctions generally as scalars λ and functions f(x) such that
Ôf(x) = λf(x).
Equations of this form are known as eigenvalue equations. The Helmholtz
equation is an example of an eigenvalue equation. We formulate it as in [1],
section 11.1.10: Define C2(Ω) to be the set of all u : Ω → R such that u and
its first and second derivatives are continuous (here Ω = Ω∪∂Ω is the union
of Ω with its boundary, called the closure of Ω) ([1], 447). Then define the
set
C2
D(Ω) = {u ∈ C2
(Ω) : x ∈ ∂Ω ⇒ u(x) = 0}.
In words, this is the set of twice differentiable functions that vanish on the
boundary of Ω. Gockenbach [1] also defines the operator
LD : C2
D(Ω) → C(Ω),
LDu = −∆u.
where in Rn, twe define the Laplacian ∆ = n
i=1
∂2
∂x2
i
.Then the Helmholtz
equation with eigenfunction f and associated eigenvalue E, is given by
LDf = Ef.
1

2 ASPEN DEVRIES
This equation has a wide range of applicability in mathematics and in
physics. For instance, if we assume the time variation of the Electric field
E and Magnetic field B is of the form e−iωt (i.e. the time variation is
simple harmonic), Maxwell’s equations reduce to the Helmholtz equations
∆E+k2E = 0, and ∆B+k2B = 0 where k is a constant known as the wave
number [13]. The Helmholtz equation also describes a particle in a box: if
we take the function f to be the wavefunction associated with the system
and E to be the energy of the particle, the Helmholtz equation takes the
form of the time independent Schroedinger equation [4].
In this paper, we’ll be concerned with problems more like the particle
in a box problem in which the function f is subject to Dirichlet boundary
conditions. More precisely, for a domain Ω ⊂ R2, we are interested in finding
functions f : Ω → R and numbers E ∈ R satisfying
∆f + Ef = 0
f = 0 on ∂Ω
where Ω denotes the closure of Ω, and ∂Ω denotes the boundary of Ω.
Poisson, Kirchhoff, Lamé, Mathieu, and Clensch have found analytic (as
opposed to numeric) expressions that describe the oscillation of elementary
shapes such as the rectangle, the triangle, the circle, and the ellipse [4]. We
will look at the solution to the above equation on a rectangle, but first we
find an exact solution for the analogous one dimensional problem.
Though the one dimensional case is rather elementary, it is a widely stud-
ied case in both physics. It will motivate our consideration of a two dimen-
sional drum and begins to give us a sense of the relation between solutions
of the Helmholtz equation with Dirichlet boundary conditions on a domain
and the choice of domain.
3. 1D Helmholtz equation on a line
In one dimension, the Helmholtz equation with dirichlet boundary condi-
tions becomes
∆f + Ef = 0
f = 0 on ∂Ω
where Ω ∈ R, E ∈ R and f : Ω → R. We let Ω = (0, a) for some a ∈ R. In
this case ∂Ω = {{0}, {a}} so that the boundary conditions amount to the
requirement that f(0) = f(a) = 0. Rewriting the Helmholtz equation in
terms of the 1-dimensional Laplacian operator, we have
d2
dx2
f + Ef = 0

MEMBRANE HARMONICS 3
so that
−
d2
dx2
f = Ef.
This is simply a second order linear differential equation with complex char-
acteristic roots r1 = −i
√
E, r2 = i
√
E so that the general solution to the
equation above is f(x) = c1cos(
√
Ex) + c2sin(−
√
Ex) for any constants
c1, c2 (see [1], p.86). To find the particular solution which takes boundary
conditions into account, we must use the constraints imposed on f:
f(0) = 0
so that
c1cos(0) + c2sin(0) = 0
and hence
c1 = 0.
Applying the second boundary condition, we get
f(a) = 0
so that
c1cos(
√
E · a) + c2sin(−
√
E · a) = 0.
Plugging in our result that c1 = 0, we have
0 · cos(
√
E) + c2sin(−
√
Ea) = 0.
Using a a property of the sine function we can bring out the negative sign
to get
c2sin(−
√
Ea) = −c2sin(
√
Ea) = 0.
Since c1 = 0 and we are not concerned with trivial solutions, it can-
not be the case that c2 = 0 (or f would be the zero function). So for
−c2sin(
√
Ea) = 0 to be satisfied, it must be the case that sin(
√
Ea) = 0
which is true exactly when
√
Ea = ±nπ, where n = 1, 2, 3, .... Solving for E,
we have E = n2π2
a2 . It follows that the particular solution to this boundary
value problem is given by
f(x) = Asin(
nπx
a
), E =
n2π2
a2
for n ∈ N and A ∈ R. Notice that for any choice of A, f(x) is a solution
to the problem with an eigenvalue that is independent of A. As a result,
for any given eigenvalue, there are multiple choices of f that would solve
the problem. However, if we were to limit ourselves to consider only eigen-
functions normalized with respect to the inner product norm (as defined in
equation 3.14 of [1]) then we see that a given eigenvalue uniquely specifies an
associated eigenfunction f(x), up to a change in sign. To be more explicit,
requiring that the eigenfunction f(x) be normalized would be requiring that
f =
a
0
|f(x)|2dx = 1

4 ASPEN DEVRIES
where |f(x)| denotes the absolute value of the function f(x). Continuing,
we see that
a
0
|Asin(
nπx
a
)|2dx =
a
0
|A|2sin2(
nπx
a
)dx = A2
a
0
sin2(
nπx
a
)dx
= A2
a
2
= 1.
Solving this equation for A, we have
±A2
=
2
a
so that
A = ±
2
a
.
Recall that a is some chosen real number so that the value of A (and as a
result, the values of the function f(x)) are determined up to a sign when we
require that f be normalized with respect to the inner product norm. Thus
the normalized eigenfunctions are
fn(x) =
2
a
sin(
nπx
a
)
with associated eigenvalues
En =
n2π2
a2
.
We refer to the set {En|n ∈ N} as the spectrum of the string. More
precisely, {En|n ∈ N} is the spectrum of the Laplacian on the string. More
generally, the set of eigenvalues that correspond to solutions of the Helmholtz
equation with Dirichlet boundary conditions on a domain is refered to as
the spectrum of the Laplacian on that domain.
We go on to derive the eigenvalues and normalized eigenfunctions for the
Dirichlet problem on a rectangle. As in this one dimensional case, we will
use separation of variables to ﬁnd a solution.
4. 2D Helmholtz equation on a rectangle
We hope to derive the solution of the 2D Helmholtz equation on a rec-
tangle with Dirichlet boundary conditions. Consider the set of equations
−∆f = Ef
f = 0 on ∂Ω
Where Ω = (0, a)×(0, b) is the inside of a rectangle with width a and height
b. We can solve this set of equations using the method of separation of
variables. This method looks for solutions to the equation that are “sep-
arated” (we will say what we mean by this shortly). We then justify that

the “separated” solution gives us all solutions to the set of equations using
arguments of Courant and Hilbert [8].
Suppose that the function of the form f(x, y) = X(x)Y (y) satisfies the
2D Helmholtz equation stated above. Functions of this form are often called
“separated” since they can be written as the product of two functions that
each depend on one independent variable. We substitute the function into
the Helmholtz equation and perform some manipulations:
−∆f(x, y) = −∆X(x)Y (y) = −[
d2
dx2
+
d2
dy2
]X(x)Y (y)
= −
d2
dx2
X(x)Y (y) −
d2
dy2
X(x)Y (y)
since the function X(x) is independent of the variable y by assumption,
we can write d2
dx2 X(x)Y (y) = Y (y) d2
dx2 X(x). Similarly, d2
dy2 X(x)Y (y) =
X(x) d2
dy2 Y (y). The equation above now becomes
= −Y (y)
d2
dx2
X(x) − X(x)
d2
dy2
Y (y)
so that the differential equation can be written as
= −Y (y)
d2
dx2
X(x) − X(x)
d2
dy2
Y (y) = EX(x)Y (y)
We are only interested in non-trivial solutions, so that neither X(x) nor
Y (y) is allowed to be the zero function. Dividing by X(x)Y (y), we have
−
1
X(x)
d2
dx2
X(x) −
1
Y (y)
d2
dy2
Y (y) = E
or equivalently,
−
1
X(x)
d2
dx2
X(x) = E +
1
Y (y)
d2
dy2
Y (y)
Notice that the left-hand side of the above equation depends only on the
variable x, while the right side depends only on y. As a result, differentiating
both sides with respect to x makes the right side vanish:
d
dx
[−
1
X(x)
d2
dx2
X(x)] = 0.
which implies that the function 1
X(x)
d2
dx2 X(x) is a constant with respect
to x. Since this function depends only on x, it must be a constant function.
If we take the derivative with respect to y of both sides, we see similarly
that the function E + 1
Y (y)
d2
dy2 Y (y), and hence 1
Y (y)
d2
dy2 Y (y), is constant. In
particular, we have

6 ASPEN DEVRIES
−
1
X(x)
d2
dx2
X(x) = E1
and
−
1
Y (y)
d2
dy2
Y (y) = E2
where E1, E2 ∈ R satisfy E = E1 + E2. We can rewrite these equations in
the familiar form of the 1D Helmholtz equation by multiplying through by
the appropriate functions.
d2
dx2
X(x) = E1X(x)
and
d2
dy2
Y (y) = E2Y (y).
We already derived the solution to the 1D Helmholtz equation with Dirichlet
boundary conditions in the previous section. Therefore we can conclude that
X(x), E1 are of the form
X(x) = A1sin(
nπx
a
), E1 =
n2π2
a2
where n ∈ N, while Y (y), E2 are given by
Y (y) = A2sin(
mπy
b
), E2 =
m2π2
b2
where m ∈ N. We can easily check that functions of the form f(x, y) =
A1A2sin(nπx
a )sin(mπy
b ) with associated eigenvalues Enm = n2π2
a2 + m2π2
b2 sat-
isfy the 2d Helmholtz equation with Dirichlet boundary conditions on a
rectangle by plugging the function into the diﬀerential equation and check-
ing that it’s zero on the rectangle. The fact that all non-trivial solutions are
of this form is a result of Courant and Hilbert in Methods of Mathematical
Physics Vol. 1 [8].
To ﬁnd the normalized eigenfunctions with respect to the inner product
norm, we must choose the constant A1A2 so that
f =
a
0
|f(x)|2dx = 1
=
b
0
a
0
(A1A2sin(
nπx
a
)sin(
mπy
b
))2 = (A1A2)2
ab
4
Solving for A1, A2 we have
A1A2 = ±
2
√
ab
.

In summary, we’ve found that for the Dirichlet problem on a rectangle, the
normalized eigenfunctions and corresponding eigenvalues of the Helmholtz
equation are
fnm(x, y) =
2
√
ab
sin(
nπx
a
)sin(
mπy
b
)
and
Enm =
n2π2
a2
+
m2π2
b2
.
The eigenvalues and eigenfunctions that we have obtained in this section
characterize the vibration modes of an idealized rectangular drum. Our
derivation of the solutions gives us a mathematical sense of how the choice
of domain constrains the frequencies of vibration of the drum. In the next
section, we consider the solutions of this problem on a different boundary
and make a comparison between the resulting spectra.
5. 2D Helmholtz equation on a disk
We do not go into the details of the solution to the Helmholtz equation on
a disk. For the derivation and statement of the solutions, we refer the reader
to pages 469-485 of [1]. Instead, we will discuss the solution qualitatively
and quote a result the strongly motivates the discussion of drum spectra.
Our discussion closely follows the discussion of [1].
The statement of the Helmholtz equation on a circular disk is as follows.
Define the region
Ωc = {(x1, x2) ∈ R2
: x2
1 + x2
2 < A},
where A > 0 is the radius of the disk. The Helmholtx equation with dirichlet
boundary conditions on the disk then takes the familiar form
−∆f(x) = λf(x), x ∈ Ωc
f(x) = 0, x ∈ ∂Ωc.
The only technique we have for finding eigenfunctions of a differential
operator in two dimensions is separation of variables [1]. However, imple-
menting separation of variables for Dirichlet boundary conditions on a circle
is problematic because the boundary condition does not separate when carte-
sian coordinates are used. It’s therefore necessary to re-express the problem
in polar coordinates. If we define
g(r, θ) = f(x1, x2)
as in [1], where (r, θ) are polar coordinates that coorespond to the cartesian
coordinates (x1, x2), the Dirichlet boundary condition simplifies to
v(A, θ) = 0, −π ≤ θ < π.

8 ASPEN DEVRIES
To see that these boundary conditions do indeed separate, apply separation
of variables with g(r, θ) = R(r)T(θ). Then the above equation becomes
R(A)T(θ) = 0, −π <≤ θ < π.
In particular, the dirichlet boundary condition reduces to the constraint
that either T(θ) = 0 for all θ or R(A) = 0 for all θ. T depends only on the
variable θ so if it were zero for all θ on the boundary, it would have to be
zero everywhere on the disk (T has no radial dependence). That is, T would
be the zero function. But then g would also be the zero function, which
doesn’t interest us. Thus the dirichlet boundary condition simplifies to the
condition that R(A) = 0.
The use of polar coordinates imposes additional constraints on the func-
tions T and R. In particular, we require that T and its first derivative are
singlevalued when θ is π and −π, namely that
T(−π) = T(π)
and
∂T(θ)
∂θ
(−π) =
∂T(θ)
∂θ
(π).
The analogous condition for R is that R(0) is finite.
Just as the boundary conditions take on a different form in polar coor-
dinates, so does the Laplacian. The Laplacian in polar coordinates is given
by
∆ =
∂2
∂r2
+
1
r2
∂2
∂θ2
+
1
r
∂
∂r
.
Now that we’ve re-expressed the Helmholtz equation in polar coordinates,
it is possible to apply separation of variables and find expressions that al-
low is to compute approximate solutions to the Helmholtz equation on the
disk. The resulting solutions are unlike the the solutions in the case of a
rectangular boundary in that they cannot be expressed in terms of elemen-
tary functions; to find a solution, one must use a power series expansion of
the solution which has infinitely many terms. Usable solutions have to be
computed by finding the terms in the power series and summing them up.
Since it is not feasable to do this for infinitely many terms, the solutions
that we use for computations are approximate. The associated eigenvalues
are also not given in explicit form: the eigenvalues must also be computed
numerically.
Given the solutions to the Helmholtz equation on a disk, it is possible
to compute a value for the lowest frequency in the spectrum for a circular
drum ([1], p.484). Its value, accurate to 3 decimal places is given in [1]:
F10 ≈ 5.783
1
A2
.

Note that the actual frequency of vibration of the drum is the square root
of this value divided by 2π. This result in combination with the results of
section 5 allow for a comparison between the spectrum of the rectangular
drum and that of the circular drum.
Using the expression for the eigenvalues for the Helmholtz equation on
the rectangle, we can compute the lowest frequency in the spectrum for a
square drum with sides of length A :
E11 =
12π2
(2A)2
+
12π2
(2A)2
E11 =
π2
2A2
= 0.853F10
so that the square drum with sides of length 2A has a lower ”lowest fre-
quency” than the circular drum. This means that if we could ”hear” the
spectra (knew the full set of eigenvalues corresponding to the respective
boundaries) of the circular drum of radius A and the square drum with side
lengths 2A we could tell that they are different drums, since we would be
able to hear the lowest frequency.
We can also compute the smallest eigenvalue in the spectrum for a square
drum with area πA2 (sides of length
√
πA):
E11 =
2π2
πA2
=
2π
A2
= 1.086F10
so that the square drum with area πA2 has a higher lowest frequency than
the circular drum of radius A. Again, this means that knowing the spectrum
of these two drums allows us to distinguish between them.
In the next section we will develop the notion of a vibrating drum more
precisely. Doing so gives a new framework in which to think of the Helmholtz
equation and its solutions. One fascinating question that arises from this
framework seeks to understand what constraints eigenvalues place on the
domain, if any.
6. The inverse spectral problem
Now that we’ve solved the eigenvalue problem (namely, the Helmholtz
equation with boundary conditions) on different domains, a natural ques-
tion to ask is whether we can distinguish between boundaries if we know the
set of functions f and numbers E that solve the eigenvalue problems on the
boundaries. More specifically, is a boundary uniquely determined by the set
of eigenvalues E corresponding to solutions of the eigenvalue problem?
This question is often expressed as ‘can you hear the shape of a drum?’.
Here, “drum” is meant to represent the domain Ω (often referred to as ’mem-
brane’) on which oscillations specified by f are occurring. It also represents

10 ASPEN DEVRIES
the fact that f is subject to Dirichlet conditions so that the membrane
doesn’t oscillate on its boundary, as is the case for a drumhead. The phrase
‘can you hear the shape of a membrane?’ is analogous to the phrase ‘does
the spectrum uniquely determine the shape of a membrane?’.
This question of determining domains from spectra is known more for-
mally as The Inverse Spectral Problem, which is one of the main problems
in spectral geometry. Since the question was ﬁrst formulated in 1966 by
Mark Kac, it has developed a rich history that is worth considering.
In his paper, Kac [2] formulated the inverse spectral problem as follows.
Let Ω1 and Ω2 be two plane regions bounded by curves Γ1 and Γ2 respec-
tively, and consider the eigenvalue problems:
1
2
∆U + λU = 0 in Ω1
with
U = 0 on Γ1
and
1
2
∆U + λU = 0 in Ω2
with
U = 0 on Γ2.
Assume that for each n the eigenvalue λn for Ω1 is equal to the eigenvalue
µn fo Ω2. Can we then say that Ω1 and Ω2 are congruent in the sense of
Euclidean geometry? [2]
We recognize the familiar form of the Helmholtz equation above. Es-
sentially what Kac wanted to understand was the nature of the relationship
between eigenvalues of the Helmholtz equation with Dirichlet boundary con-
ditions imposed on a boundary, and the chosen boundary. One can get a
stronger physical intuition for the problem by seeing how it arises from the
wave equation when using separation of variables.
The wave equation is given by
∂2f
∂t2
= c2
∆f
where c is a constant. The value of this constant depends on the type of wave
being considered. For electromagnetic waves propagating in free space, c is
the speed of light [13]. For waves on a drumhead, c depends on the tension in
the drumhead and its physical properties [2]. For a two-dimensional drum,
the function appearing in the wave equation is a function f = f(x, y, t)
that depends on two spacial dimensions and time. It gives us the vertical

displacement of the drum at each point on the drum head, and at each point
in time. If we suppose that f is separable in time so that we can write
f(x, y, t) = U(x, y)T(t)
we can substitute back into the wave equation and see that the problem
simpliﬁes:
∂2
∂t2
U(x, y)T(t) = c2
∆U(x, y)T(t)
⇒ U(x, y)
∂2
∂t2
T(t) = T(t)c2
∆U(x, y)
We use our trick of dividing by the separable function f(x, y, t) as before so
that
⇒
1
T(t)
∂2
∂t2
T(t) =
1
U(x, y)
c2
∆U(x, y)
The left side of the equation depends only on time while the right side
depends only on spatial variables x and y. Diﬀerentiating both sides with
respect to time gives
∂
∂t
[
1
T(t)
∂2
∂t2
T(t)] = 0
which implies that 1
T(t)
∂2
∂t2 T(t) is a constant function. In particular,
1
T(t)
∂2
∂t2
T(t) = λ
and
1
U(x, y)
c2
∆U(x, y) = λ
for some constant λ ∈ R. Rearranging these equations gives
∂2T(t)
∂t2
= λT(t)
which is just a one-dimensional Helmholtz equation in the variable t and
∆U(x, y) =
λ
c2
U(x, y)
which is a two-dimensional Helmholtz equation in the variables x and y.
Physically, U(x, y) gives the displacement of the drum at a snapshot in
time, while T(t) is the function that allows us to see how the displacement
of the drumhead evolves with time.
Just two years before Kac wrote his paper, John Milnor constructed two
non congruent 16-dimensional tori whose Laplace-Beltrami operators have
the same set of eigenvalues [4]. Still, it does not necessarily follow that mem-
branes in higher or lower dimensions aren’t determined by their spectra. For
instance, it is possible to “hear the shape” of a string (the 1D analogue of a

12 ASPEN DEVRIES
drum). We prove this using the results from section 3.
In physics, it is known that for a string of length L, the “normal modes”
of vibration– modes of vibration that correspond to musical notes– are de-
scribed by a function f(x, t) = U(x)eiωt [2] which is separable in time. In
fact, we can treat any solution to the Helmholtz equation with eigenvalue λ
as representing normal modes of vibration of the domain. This is possible
because given a solution g(x) to the Helmholtz equation with eigenvalue λ,
the function h(x, t) = g(x)ei
√
λt is a solution to the wave equation on the
domain. If we think of the eigenfunctions for the Helmholtz equation in this
light (as normal modes of vibration) then it seems natural to refer to the
spectrum of a drum as its “harmonics”.
As we saw previously, substituting the separable function f(x, t) = U(x)eiωt
into the wave equation leaves us with the Helmholtz equation
∆U = EU
where we let E = λ
c2 . In section 3, we found that if we fix the ends of the
string so that boundary conditions (U(0) = U(L) = 0, the allowed values of
E are
E =
n2π2
L2
, n = 1, 2, 3, ...
.
Suppose for contradiction that there are two strings with identical spectra
but different lengths. That is, suppose that the set {n2π2
L2
1
|n ∈ N} is equal
to the set {m2π2
L2
1
|m ∈ N}, and that L1 = L2. Since the sets are equal and
clearly have a smallest value, there smallest values have to be equal. In
particular, it must hold that
12π2
L2
1
=
12π2
L2
2
so that
L1 = ±L2.
Lengths are positive, which means L1 = L2 (a contradiction). Therefore,
if two strings have the same spectrum, their lengths (which fully specifies
shape in one dimension) must also be equivalent.
It has been proven that for each n ≥ 4, there are domains in the unit
sphere Sn−1 in Rn that are non-congruent and isospectral [4]. The proof
involves looking at reflection groups that act on the same Euclidean space
Rn for n ≥ 4 and applying some advanced group-theoretic arguments [4].
In 1992, Kac’s question was finally answered when Gordon and Webb 1992
[3] were able to apply a proof of Sunada’s Theorem (which gives a method

of constructing pairs of isospectral manifolds) by [14] to construct non-
congruent isospectral membranes that lie in a plane. Gordon and Webb’s
example has given rise to 17 classes of examples that demonstrate that one
cannot hear the shape of a drum in general [4].
Before focusing on the central question of this section, we will further
develop our understanding of the dynamics of vibrating drums by looking
at a particular trait of the eigenfunctions of the Helmholtz equation with
Dirichlet boundary conditions that is constrained by their respective eigen-
functions.
7. Constraints on Eigenvalues and Eigenfunctions
While the detailed relation between boundary conditions and eigenmodes
of operators (the set of eigenvalues and associated eigenfunctions) is com-
plicated, there have been a number of results that put concrete restrictions
on the eigenmodes. We proceed to discuss one such constraint.
Courant’s nodal line theorem relates to Dirichlet eigenfunctions of elliptic
operators. It says that if eigenvalues of an operator are put in increasing
order, the corresponding sequence of eigenfunctions will have a certain prop-
erty. If we look at the points at which an eigenfunction is zero and plot those
points within the boundary of interest, we will get a series of curves (nodal
lines) that divide the region within the boundary into a certain number of
pieces called nodal domains. Courant’s Theorem tells us that the number
of these nodal domains corresponding to the n-st term of the sequence of
eigenfunctions is at most n if the eigenvalues all have multiplicity 1.
We will give a statement and proof of Courant’s Nodal Line Theorem
(CNLT) based upon those in [6]. First, we revisit the k-dimensional Helmholtz
equation
(1) ∆u + λu = 0, x ∈ D ⊂ Rk
.
We note that in [6] a positive bounded function ρ(x) multiplies u in the above
equation. For consistency with our previous discussion, we take ρ(x) = 1.
For m = 1, choosing ρ(x) = 1 amounts to declaring that the ratio T
µ of the
tension in the string to linear mass density is constant [13]. In higher di-
mensions, this assumption amounts to constraining the physical properties
of the drum (or higher dimensional analogue thereof). In Kac’s formulation
of the question, he takes ρ(x) = 1/2 rather than 1. This just amounts to a
diﬀerent choice of units [2].
The nodal set of u(x) is deﬁned as the set of points x such that u(x) = 0
[6]. According to [6], for D ⊂ Rk, the nodal set of an eigenfunction of the
Helmholtz equation is locally composed of hypersurfaces (e.g. a line for
k = 2 and a plane for k = 3) of dimension k − 1. These hypersurfaces

14 ASPEN DEVRIES
cannot end in the interior of D, which implies that they are either closed,
or begin and end at the boundary [6]. Thus for k = 2, the nodal set of an
eigenfunction u(x) of the 2D Helmholtz equation with Dirichlet boundary
conditions is made up of continuous curves, called nodal lines, which are
either closed or begin and end at the boundary [6].
Courant’s Nodal Line Theorem
Before stating and proving CNLT, we establish some necessary back-
ground. We express the Helmholtz boundary value problem in its weak
form. Define
(u, v)D =
D
u · vdx, u, v D =
D
uvdx.
It is well known that the Helmholtz equation with Dirichlet boundary
conditions has infinitely many positive eigenvalues
0 < λ1 ≤ λ2 ≤ λ3 ≤ · · · .
According to [6], the corresponding eigenfunctions {ui}∞
1 , where the ui are
orthonormal, form a complete set of L2(D). We say that u ∈ H1
0 (D) is a
weak solution of (1) if
(2) −(u, v)D + λ u, v D = 0
for all v ∈ H1
0 (D) 1
The Rayleigh quotient for such a u is defined as
λR ≡
(u, u)D
u, u D
.
Theorem 7.1 (Rayleigh quotient Theorem [6]). If u, ui D = 0 for each
i ∈ {1, 2, ..., N−1}, then λR ≥ λN , with equality if and only if u(x) = uN(x).
Now that we have established the necessary background, we will prove
CNLT.
Theorem 7.2 (Courant Hilbert Theorem [6]). Suppose the eigenvalues λi
of (1) are ordered increasingly and un(x) is an eigenfunction corresponding
to λn. If λn has multiplicity r ≥ 1, so that
λn−1 < λn = λn+1 = · · · = λn+r−1 < λn+r,
then un(x) has at most n + r − 1 nodal domains.
Proof. We present a proof of the theorem formulated by [6]. Suppose that
un(x) has M sign domains Di such that ∪M
i=1Di = D. Define
1A description of the Sobolev space and the necessary background material can be
found on p. 579, 618-9 of [1].

wi(x) =
βiun(x), x ∈ Di
0, otherwise
.
This is a piecewise function whose j − th component takes the value of the
eigenfunction βjun(x) on the j −th sign domain Dj, and vanishes elsewhere.
It’s important to realize that since the Di are disjoint (definition of nodal
domain), the functions (wi(x))M
1 are orthogonal (namely, wi, wj D = 0
whenever i = j). We choose the βi so that the wi are orthonormal i.e.
wi, wi Di = 1. Also define
v(x) =
M
i=1
ciwi(x),
M
i=1
c2
i = 1.
We compute the Rayleigh quotient with respect to this function. With
the right choice of the coefficients ci, the Rayleigh Quotient Theorem will
yield the desired inequality. First we compute the numerator of λR.
(v, v)D =
D
v · vdx
=
D
[
M
i=1
ciwi(x)] · [
M
i=1
ciwi(x)]dx
Using linearity of the Laplacian, we can write
=
D
M
i=1
ci wi(x) ·
M
i=1
ci wi(x)dx
Here we have a Lebesgue integral involving the function wi(x), which
is not defined on ∂Di. Recall that since Di ⊂ Rl, its boundary is a subset
of the nodal set which has dimension k − 1 [6]. It is a well known theorem
in Lebesgue integration that any subspace of Rl with dimension less than l
has Lebesgue measure zero in Rl. Further, it is known that functions that
are equivalent almost everywhere on a set (that is, they only differ on a set
of Lebesgue measure zero) integrate to the same value on that set. Thus if
we define the function
ωi(x) =
βiλnun(x), x ∈ Di
0, otherwise
then D wi(x)dx = D ωi(x)dx (since the functions differ only on ∂Di
which has Lebesgue measure zero in D). Similarly, we could define a func-
tion ωhk that is equal to wh(x) · wk(x) almost everywhere on D. For
brevity, we’ll just accept that the properties of Lebesgue integration that

16 ASPEN DEVRIES
we’ve discussed allow us to treat the terms wi(x), wh(x) · wk(x) as
though they have value zero where they are undefined. Then we can define
∆ωi(x) =
∆(βiun(x)), x ∈ Di
0, otherwise
without affecting any of the computations that follow. Since βi is constant
for each i, linearity of the Laplacian gives ∆(βiun(x)) = βi∆un(x). However,
un was chosen to be an eigenfunction with eigenvalue λn so that ∆un(x) =
λnun(x) and so
∆ωi(x) =
λnβiun(x)), x ∈ Di
0, otherwise
Continuing with our manipulation of (v, v)D, we have
(v, v)D =
D
[c1 w1(x)+· · ·+cM wM (x)]·[c1 w1(x)+· · ·+cM wM (x)]dx
Expanding the integrand gives us M2 terms of the form chck wh(x)· wk(x).
We can then split the integral into M2 integrals over D (we can do this be-
cause it is a sum of finitely many terms that are normalized and therefore
bounded on the integration region). All terms with h = k vanish since the
wi are orthogonal and thus the wi(x), which are constant multiples of the
wi (since we choose the undefined points to be zeroes) must also be orthog-
onal (a constant times an integral that evaluates to zero is still zero).
We are left with the integral
=
D
c2
1 w1(x) · w1(x)dx + · · · +
D
c2
M wM (x) · wM (x)dx
=
D1
c2
1 w1(x) · w1(x)dx + · · · +
DM
c2
M wM (x) · wM (x)dx
= c2
1
D1
w1(x) · w1(x)dx + · · · + c2
M
DM
wM (x) · wM (x)dx
=
M
i=1
c2
i
Di
wi(x) · wi(x)dx
=
M
i=1
c2
i
Di
λnβiun(x) · λnβiun(x)dx
=
M
i=1
c2
i
Di
(λnβiun(x))2
dx
(3) =
M
i=1
c2
i λ2
n
Di
(βiun(x))2
dx

Recall that βi was chosen so that wi was normalized on Di with respect to
the inner product norm. In particular, wi
2 = Di
(βiun(x))2dx = 12 = 1.
Substituting this back into (3), we have
(3) =
M
i=1
c2
i λ2
n · 1
= λ2
n
M
i=1
c2
i
= λn
We now compute the denominator of λR:
v, v D =
D
(
M
i=1
ciwi(x))2
dx
=
D
[c1w1(x) + · · · + cM wM (x)][c1w1(x) + · · · + cM wM (x)]dx
As before the expanded integrand gives M2 terms. We separate those terms
into M2 integrals. Due to orthogonality of the wi’s, D wh(x)wk(x)dx = 0
when h = k. The integral then becomes
=
D
c2
1w1(x)w1(x)dx + · · · +
D
c2
M wM (x)wM (x)dx
=
D1
c2
1w1(x)w1(x)dx + · · · +
D2
c2
M wM (x)wM (x)dx
= c2
1
D1
w1(x)w1(x)dx + · · · + c2
M
D2
wM (x)wM (x)dx
M
i=1
c2
i wi, wi Di =
M
i=1
c2
i · 1 = 1.
We can now compute the Rayleigh quotient with respect to v:
λR =
(v, v)D
v, v D
=
λn
1
= λn
Thus
(4) λR = λn.
We just computed λR with respect to v and found that λR = λn. We
apply theorem 6.1 to get another constraint on λR, which will allows us to
construct the desired inequality.
In order to apply Theorem 6.1, we must choose coeﬃcients ci so that
v, ui = 0 for each i = 1, 2, ..., M − 1. The only constraint we have put
on the coeﬃcients (ci)M
1 is that their squares sum to one. We add M − 1

18 ASPEN DEVRIES
constraints, namely v, ui = 0 for i = 1, 2, ..., M − 1. There must be a
collection of coefficients satisfying these constraints since together they form
a system of M equations and M unknowns. Once we choose the coefficients
that make v, ui = 0 for each i = 1, 2, ..., M − 1 (which we know exist), we
can apply Theorem 6.1 with N = M. Doing so gives the inequality
(5) λR ≥ λM
We have that λR = λn. Additionally, we used Theorem 6.1 to demonstrate
that this same λR satisfies λR ≥ λM . Combining these expressions involving
λR gives
λn = λR ≥ λM .
The eigenvalues are arranged in increasing order so that λn ≥ λM implies
n ≥ M.
Since r − 1 ≥ 0 (each eigenvalue λn appears at least once so that r ≥ 1), we
have
n + r − 1 ≥ n.
We combine these two inequalities to get
M ≤ n ≤ n + r − 1
so that
M ≤ n + r − 1.
We can apply the Courant-Hilbert Theorem to the rectangular domain
considered in section 3. Recall that on a rectangular drum, the Helmholtz
equation with Dirichlet boundary conditions has normalized eigenfunctions
fij(x, y) =
2
√
ab
sin(
iπx
a
)sin(
jπy
b
)
where i, j ∈ {1, 2, 3, ...}, with associated eigenvalues
Eij =
i2π2
a2
+
j2π2
b2
.
Let a = b = 1. We compute the lowest 9 eigenvalues:
E11 =
12π2
12
+
12π2
12
= 2π2
E12 =
12π2
12
+
22π2
12
= 5π2
E21 = 5π2
E22 = 8π2
E13 = 10π2
E31 = 10π2
E23 = 13π2

E32 = 13π2
E33 = 18π2
We now arrange them in increasing order
E11 ≤ E12 ≤ E21 ≤ E22 ≤ E13 ≤ E31 ≤ E23 ≤ E32 ≤ E33
and define
λ1 = E11, λ2 = E12, λ3 = E21, λ4 = E22, λ5 = E13, λ6 = E31, λ7 = E23, λ8 = E32, λ9 = E33.
We graph the 9 eigenfunctions along with their contour plots. In Figure
1, the nodal sets are the points along the rectangles shown in the contour
plots. The regions within the rectangles are the nodal domains.
We see in the figure that the number of nodal domains of the vibrational
modes corresponding to λ1, λ2, λ3, λ4, λ5, λ6, λ7, λ8, λ9 are 1, 2, 2, 4, 3, 3, 6, 6, 9
(respectively). Observe that the number of nodal domains is less than the
index n on lambda for each n ∈ {1, ..., 9}, as predicted by CNLT.
It turns out that the statement of the Courant-Hilbert theorem can be
made stronger; the number of of nodal domains is in fact bounded above
by n. This is a theorem known as the Hermann-Pleijel theorem. A proof is
given in [6].
While the success of Courant’s Nodal Line theorem in bounding the nodal
domains for the first 9 vibrational modes of the square might make us hopeful
that Kac’s question could be answered positively, it turns out that it is
not the case. In the next section we consider a counterexample of Kac’s
question which shows that the harmonics of drums don’t always determine
their shape.
Proof of Isospectrality
Transplantation is a method of proving isospectrality of domains con-
structed by a particular algebraic process that involves Sunada triples (for
further discussion of the construction of these domains see [12, 7]). The
transplantation method was originally formulated by Berard [?]. [?] gives
a way of constructing a map called a Transplantation mapping between
eigenfunctions of the specially constructed domains. We will use Transplan-
tation to prove that a pair of domains constructed by Gordon et al. [3] are
isospectral, following the proof of [7]. Consider the domains in Figure 2.
The the proof of isospectrality by [7] will hold for the domains of Fig. 2,
which are shown to be constructed with equilateral triangles. These domains
are not very interesting as shown since a reflection about a line bisecting the
central triangle of either domain defines an isometry between the domains.
To make the proof of isospectrality more interesting and more fruitful in
the context of this paper, we replace the equilateral triangles in Fig. 1 by
acute-angled scalene triangles (see Fig. 3). In fact, we could replace the

20 ASPEN DEVRIES
Figure 1. Parts a − g of this figure plot the eigenfunctions
fij(x, y) = 2sin(iπx)sin(jπy) of the Helmholtz equation with
Dirichlet boundary conditions imposed on a unit square, with
eigenvalues Eij = (i2 + j2)π2 for i, j ∈ {1, 2, 3}. Below each
plot of an eigenfunction is its contour plot. For a given con-
tour plot, the set of points that lie on the rectangles, includ-
ing those on the unit square, make up the nodal set of the
square for the corresponding vibrational mode. The nodal
lines are the edges of the rectangles, while the nodal domains
are the regions within the rectangles. M is the number of
nodal domains for a given mode.
equilateral triangles by any building block possessing three edges around
which to reflect and still get a pair of isospectral domains [4]. The three
unequal side lengths of the scalene triangle are represented by the three
distinct types of line used in Fig. 3.
As can be seen in Fig. 3, any two triangles that meet along a line are
mirror images in that line. It’s also important to see that these domains
are not isometric (if they were, a proof of isospectrality would be superflu-
ous). Suppose for contradiction that the domains are isometric i.e. there is
an isometry that maps M1 to M2. View both domains as a central triangle

Figure 2. A pair of isospectral domains constructed by [3].
Each type of segment (dotted, solid, dashed) represents a dif-
ferent segment length. These domains are sometimes called
“propellers” due to their shape.
connecting three “propellers”. These propellers are constructed so that each
propeller has a distinctly shaped point. As a result, any isometry between
the domains would have to send the triangle labeled 3 in Figure 3 to the
triangle labeled 13. Since these triangles are scalene, there is only one map
that does this. That map cannot be an isometry because it maps all trian-
gles but those labeled 3 and 4 to points outside of M2. Thus there can be
no isometry between M1 and M2.
Let f be a function that satisﬁes ∆f = λf (λ ∈ R) for the Dirichlet prob-
lem corresponding to M1. Let f0, f1, ..., f6 denote the functions obtained by
restricting f to each of the 7 triangles of M1, as labeled in Figure 3. To be
explicit, we deﬁne the functions
fi(x) = f(x) for x ∈ Di
where Di is the interior of the triangle labeled with i ∈ {0, 1, ..., 6} in Fig.
2. Note that we do not require fi to vanish on ∂Di.

22 ASPEN DEVRIES
Figure 3. The isospectral domains constructed by [3],
shown with acute-angled scalene triangles. They are some-
times called “warped propellers”.
We will show that we can use f to define a piecewise function g such that
∆g = λg for the Dirichlet problem corresponding to M2. We could compute
a transplantation matrix using a somewhat complex expressionfrom [14]. In
this case, the domains are simple enough that one can essentially guess a
matrix and then prove that the matrix indeed gives a map between eigen-
functions of the domains with the same eigenvalue. Seeing as the background
necessary to discuss Sunada triples and the algebraic structures involved in
the Transplantation method (see [12, 14, 4] for further discussion) is beyond
the scope of this paper, we will use the latter method.
Essentially what we want to do is to define a piecewise function on M2 in
terms of the fi on M1. We will need to perform isometries on the fi in order
for them to “fit” into triangles on M2. However, because eigenfunctions of
the laplacian are known to remain eigenfunctions of the Laplacian when an
isometry is applied, this will not pose a problem. The function we define on
M2 needs to be smooth (i.e. have a continuous first and second derivative)
so that we can apply the Laplacian to it on all of M2. Then because its
peices are linear combinations of the fi (these will be composed with the
appropriate isometries), it will have eigenvalue λ as we hoped. We also need
the function to be zero on the boundary so that we are actually comparing

drums and not the harmonics of pizza dough.
Before we can construct the desired function, we need a reflection prin-
ciple for the Helmholtz equation with Dirichlet boundary conditions. In
particular, we need a way of extending an eigenfunction across a boundary
where it vanishes to a function that is smooth across the boundary. The
reflection principle gives us just that.
Theorem 7.3 (Reflection Principle for the Helmholtz equation). Let D be
a domain (i.e. an open, connected set) in R2, which is symmetric about a
line L, and d denote the (nonempty) intersection of L and D, while D+
and D− designate, respectively, the two symmetric parts into which D −d is
divided by L. Suppose u(x1, ..., xn) is a real, single-valued, twice continuously
differentiable solution of the Helmholtz equation in D+, such that
limx→xu(x) = 0,
for c = (x1, ..., xn) in D+ and x = (x1, ..., xn) in d. Then the function U
defined in D by
U(x) =



u(x), x ∈ D+
0, x ∈ d,
−u(x∗), x ∈ D−,
where x∗ is the mirror image of x with respect to the line L, is a solution of
the Helmholtz equation throughout D with continuous first and second order
derivatives.
A proof of this theorem for Rn can be found in [10].
We can describe how the reflection principle applies to our construction of
g as follows. Suppose that for some q ∈ {0, ..., 6}, the piece fq of f vanishes
along a line Lq which lies on the boundary of D+
q (analogous to D+). Now
consider the interior of the triangle obtained by flipping the region D+
q about
Lq (see figure above). We denote the resulting region by D−
q (analogous to
D−). The region Dq (analogous to D described in the theorem) is the union
of D+
q , D−
q , Lq, provided we take out the endpoints of Lq. The reflection
principle allows us to define a function
Uq(x) =



fq(x), x ∈ D+
q
0, x ∈ dq,
−fq(x∗), x ∈ D−
q ,
where dq is the intersection of D and Lq.
Before continuing we will discuss a subtlety that will arise in notation
when we begin to construct g. By construction we have Uq(x) = −fq(x∗)
on D−
q . In our construction for g, we often refer to the restriction of Uq

24 ASPEN DEVRIES
Figure 4. An example of domains
D, D−, D+ and line L that satisfy
the hypotheses of the reflection prin-
ciple.
to D−
q as the function −fq. This function takes on the same value of the
function −fq(x) (namely, the composition of the function s(y) = −y with
fq(x)), though calling this function −fq is slightly misleading since fq(x) is
not defined on D−
q , whereas the restriction of Uq to D−
q is defined on D−
q .
Since the region D−
q is typically not contained in M1, we end up thinking
of the restriction of Uq to D−
q as the function −fq(x) defined on D+
q which
makes the distinction between the functions even more subtle. We say this
another way. We define values of g on M2 in terms of values of f on M1.
The restriction of Uq to D−
q does not lie on M1. Thus, in order to use the
restriction of Uq to define a value of g, we must map values of the restriction
of Uq on D−
q to values of −fq(x) defined on D+
q , before mapping them to val-
ues of g on M2. It is much more convenient to work with −fq(x) rather than
the restricted function for the sake of defining g. The restriction function
isn’t of real importance until we are verifying the g we construct is smooth.
We’ll start by choosing any peice fp for p ∈ {0, ..., 6} and performing an
isometry on it so that it is defined over any triangle we choose on M2. This
choice can affect the final function one gets, but as long as it has the proper-
ties want, this fact is inconsequential. We’ll start by placing f2 in 7 (revisit
Figure 3 for labelings). Tentatively, we’re letting g7 = f2 ◦ τ7,2 though we
will add and subtract terms until function g is what we want it to be. Here,
for h ∈ {7, 8, ..., 13}, k ∈ {0, 1, ..., 6}, τh,k is the isometry (i.e. composition

of length-preserving reflections, rotations and/or translations) from the tri-
angle h on M2 to the triangle k on M1 (see a discussion of these isometries
at the end of the section). To be sure that g7 continues smoothly into the
surrounding triangles we use the fact that in M1, f2 continued smoothly
into f5 across the dotted line and f0 across the dashed line. f2 is zero at the
solid line, so we use the reflection principle to define the function −f2 that
f2 continues smoothly into.
Now our tentative g function has g7 = f2 ◦τ7,2, g9 = f0 ◦τ9,0, g8 = f5 ◦τ8,5,
and g11 = −f2 ◦ τ11,2. To check that all of these peices fit together the way
we want them to, we think of 2, 7 as reference triangles and recall the prop-
erty of both domains that any two triangles that meet on a line are mirrored
along that line. Thus, as long as f2 and f5 share the same edge in M1 as
f5 ◦ τ8,5 and f2 ◦ τ7,2 do on M2, and similarly for the other two edges, we
will maintain the reflectional symmetry of the domains hence get attached
pieces that “fit” in M2.
So far we’ve defined a function on the central triangle of M2 and extended
it to a smooth function defined on 7 and the three neighboring triangles 9,11
and 8. Every time we change the value of our function g on a triangle, we
guarantee smoothness on all neighboring triangles by doing what we did in
the first step with 7 and its neighbors. Additionally, when we change the
value of g(x) on a triangle that lies on the boundary of M2, we must add
the appropriate function to ensure that g vanishes on the boundary of M2.
Once we add that function, we again add components to neighboring
triangles to ensure smoothness, and add more peices as needed to ensure
the function vanishes at the boundary of M2. We continue this process
until we have defined a smooth function on all of M2 that vanishes on the
boundary of M2. This procedure is somewhat akin to applying makeup; you
put a dab on your face, smooth it out, clean up the edges, then dab some
more on and smooth that out until the whole area is neatly covered and
smooth. This process is more concisely demonstrated in an image than in
words. The construction of g is broken into steps in the Figure 5.

26 ASPEN DEVRIES
2-3-5
1+2+4
0+5-40+3-1
4-6-3
6+0-2
1-6-5
Figure5.Inthisfigure,weabusenotationsothatit’seasiertoseewherethepeices
offaremapped.Theindices0,...,6representthepeicesf0,...,f6off(respectively).
Ourchoiceofnotationneglectsthefactthatthepeicesoff(whicharedefinedonM1)
mustbecomposedwithisometriesoftheplaneinordertobeconsideredfunctionson
M2.Forinstance,a“0”onthecentraltriangleofM1wouldrepresentf0whilea“0”
onthebottom-mosttriangleofM2wouldbethefunctionf0◦τ10,0.Thenecessary
isometriesareincludedinthedefinitionofg.Thenumbersinthefigurearemeantto
conveywherepeicesoffarebeingmappedonM2,ratherthanthepreciseformof
theconstructedfunction.Weusedanumberofdevicestomakeiteasiertofollowthe
stepsoftheconstructionofg.Afterensuringthatagivenpeicecontinuedsmoothlyto
allneighboringtriangles,wehighlighteditwithpurple.Thishelpeduskeeptrackof
whichpeiceshadbeen“smoothedout”andwhichhadnot.Anytimeweperformeda
stepthatinvolvedsmoothing,weusedabluearrowbetweenthedomainimmediately
precedingandfollowingthestep.Similarly,redarrowsindicateastepthatinvolves
makinggvanishattheboundariesofM2.Theresultingarrangementofindicieson
M2specifiestheg.

The result of our method is the piecewise function
g(x) =



g7(x), x ∈ D7
g8(x), x ∈ D8
g9(x), x ∈ D9
g10(x), x ∈ D10
g11(x), x ∈ D11
g12(x), x ∈ D12
g13(x), x ∈ D13
=



f1 ◦ τ7,1(x) + f2 ◦ τ7,2(x) + f4 ◦ τ7,4(x), x ∈ D7
f0 ◦ τ8,0(x) + f5 ◦ τ8,5(x) − f4 ◦ τ8,4(x), x ∈ D8
f0 ◦ τ9,0(x) + f3 ◦ τ9,3(x) − f1 ◦ τ9,1(x), x ∈ D9
f1 ◦ τ10,1(x) − f6 ◦ τ10,6(x) − f5 ◦ τ10,5(x), x ∈ D10
f0 ◦ τ11,0(x) + f6 ◦ τ11,6(x) − f2 ◦ τ11,2(x), x ∈ D11
f4 ◦ τ12,4(x) − f3 ◦ τ12,3(x) − f6 ◦ τ12,6(x), x ∈ D12
f2 ◦ τ13,2(x) − f5 ◦ τ13,5(x) − f3 ◦ τ13,3(x), x ∈ D13
We have deﬁned a piecewise function g that is smooth at the bound-
aries by construction (and therefore smooth everywhere by smoothness of
f0, ..., f6). We now argue that g(x) satisﬁes ∆g = λg on M2.
We chose f so that ∆f = λf on M1. It follows that for i ∈ {0, ..., 6},
∆fi = ∆f = λf on Di so that fi is an eigenfunction of the Laplacian. It is
known that eigenfunctions of the Laplacian and their associated eigenvalues
are still eigenfunctions of the Laplacian with the same eigenvalues when
an isometry is performed. More precisely, the Laplacian “commutes” with
isometries (this is an exercise in many undergraduate texts) so that for each
fi, τj,i, we have
∆(fi ◦ τj,i) = (∆fi) ◦ τj,i.
To see that for i ∈ {0, ..., 6} and j ∈ {7, ..., 13} fi◦τj,i is also an eigenfunction
of the Helmholtz equation on Dj with eigenvalue λ, consider the following
argument. By linearity of the Laplacian we can write
∆(
i,j
fi ◦ τj,i) =
i,j
∆(fi ◦ τj,i).
Since the Laplacian commutes with isometries we have
=
i,j
∆(fi) ◦ τj,i
=
i,j
(λfi) ◦ τj,i = λ(
i,j
fi ◦ τj,i)
which shows that fi ◦ τj,i is indeed an eigenfunction of the Laplacian with
eigenvalue λ. Furthermore, All pieces gk of g can be expressed as a sum of the

28 ASPEN DEVRIES
form i,j λfi ◦τj,i so that indeed ∆gj = λgj for each k ∈ {7, ..., 13}. That is,
∆g(x) =



∆g7(x), x ∈ D7
∆g8(x), x ∈ D8
∆g9(x), x ∈ D9
∆g10(x), x ∈ D10
∆g11(x), x ∈ D11
∆g12(x), x ∈ D12
∆g13(x), x ∈ D13
=



λg7(x), x ∈ D7
λg8(x), x ∈ D8
λg9(x), x ∈ D9
λg10(x), x ∈ D10
λg11(x), x ∈ D11
λg12(x), x ∈ D12
λg13(x), x ∈ D13
= λg(x)
Note that we didn’t have to worry about ∆g being defined at boundaries
of triangles because g(x) is smooth on M2 (and so has a continuous second
derivative). We’ve verified that g(x) satisfies the Helmholtz equation on M2.
The last thing to check is that g vanishes at the boundary of M2.
We begin with the dashed boundary line of triangle 10 which we denote
B1 (see figure below) and work our way clockwise around the boundary of
M2 . We have g(x) = f1 ◦ τ7,1(x) + f2 ◦ τ7,2(x) + f4 ◦ τ7,4(x) on B1. Take any
point y on the boundary line B1. We have g(y) = f1 ◦ τ7,1(y) + f2 ◦ τ7,2(y) +
f4 ◦ τ7,4(y) = 0 + 0 + 0 = 0. Since g(x) is zero at every point on B1, g(x)
vanishes on B1 as hoped.
Each of the terms was computed by visualizing the isometries τa,b to see
where they map the point y ∈ M2 on M1 and then finding the values ofgj
using properties of the fi. The easiest way to do this is by noting that the
isometries between the triangular regions of M2 and M1 will match sides of
the triangles that have equal length. Since the triangles are scalene, this is
fairly straight forward. If we use the visualization of Fig. 2, this amounts
to saying that the isometry carries dashed lines to dashed lines; dotted lines
to dotted and solid lines to solid.
We continue to compute g(x) on B2. We abuse notation a bit to make
it easier to visualize our steps. We have g(x) = f0(dashed) + f6(dashed) −
f2(dashed) = f0(dashed) + 0 − f2(dashed). Since f is smooth and f0, f2
meet on the dashed line in M1, we must have f0(dashed) = f2(dashed) so
that g(x) = 0 on B2. Then on B3 we have g(x) = f4(dashed)−f6(dashed)−
f3(dashed) = We trust that the reader can verify as we did that g(x) van-
ishes on the remainder of the boundary of M2.
We chose an arbitrary eigenfunction f of the Helmholtz equation with
Dirichlet boundary conditions on M1. That eigenfunction had an associ-
ated eigenvalue λ ∈ R. We then gave a constructive proof that there exists

Figure 6. We label the boundary segments of M2 so that it
is easier to check that boundary conditions are verified. M1
is included for reference.
an eigenfunction g of the Helmholtz equation with Dirichlet boundary condi-
tions on M2 that also had eigenvalue λ. Since all eigenvalues for the Dirichlet
problem on M1 are also eigenvalues for the corresponding problem on M2, it
must be the case that the spectrum of M1 is contained within the spectrum
of M2. We use the same method to show that given an eigenfunction G(x)
of the Helmholtz equation with Dirichlet boundary conditions on M2 with
associated eigenvalue w, we can find an eigenfunction of the problem on
M1 with eigenvalue w. Since we’ve already covered the details of the proof
process, we will only show the critical steps for the construction of F(x).
We define the restriction Gi(x) of G(x) to the ith triangle in M2 by
Gi(x) = G(x) for x ∈ Di
where i ∈ {7, 8, ..., 13}. The steps for constructing a smooth function F
defined on M1 are given in Figure y.

30 ASPEN DEVRIES
9-13-12
7+12-8
8+11+9
7+13-11
8-13-10
7+10-911-10
-12
Figure7.ThepiecesG7,...,G13ofGarelooselyrepresentedbytheindices7,...,13.Weusedthe
sameorganizationalmethodsasinthepreviousﬁgure.Weusetheﬁnalarrangementofindiceson
M1toconstructafunctionFwhichwearguehasthedesiredproperties.

We now a have a function
F(x) =



F0(x), x ∈ D0
F1(x), x ∈ D1
F2(x), x ∈ D2
F3(x), x ∈ D3
F4(x), x ∈ D4
F5(x), x ∈ D5
F6(x), x ∈ D6
=



G8 ◦ τ0,8(x) + G9 ◦ τ0,9(x) + G11 ◦ τ0,11(x), x ∈ D0
G10 ◦ τ1,10(x) + G7 ◦ τ1,7(x) − G9 ◦ τ1,9(x), x ∈ D1
G13 ◦ τ2,13(x) + G7 ◦ τ2,7(x) − G11 ◦ τ2,11(x), x ∈ D2
G13 ◦ τ3,13(x) + G9 ◦ τ3,9(x) − G12 ◦ τ3,12(x), x ∈ D3
G7 ◦ τ4,7(x) + G12 ◦ τ4,12(x) − G8 ◦ τ4,8(x), x ∈ D4
G8 ◦ τ5,8(x) − G13 ◦ τ5,13(x) − G10 ◦ τ5,10(x), x ∈ D5
G11 ◦ τ6,11(x) − G10 ◦ τ6,10(x) − G12 ◦ τ6,12(x), x ∈ D6
constructed to be smooth which is a solution to the Helmholtz equation on
M1 by similarity with the case for g(x). All that is left to check is that F(x)
vanishes on the boundary of M1. To do this, we follow the same logic as
for the previously constructed g(x). We see that indeed this is the case. To
quickly check this, refer to the figure below, and carry out the procedure
described when we checked that g satisfied the boundary conditions.
We have given a constructive proof that the spectrum of M2 is contained
within the spectrum of M1. We’ve already shown, however, that the spec-
trum of M1 is contained within the spectrum of M2. Therefore equality
holds: M1 and M2 are isospectral.
We have proven that the domains are isospectral. In doing so we applied
isomorphisms to assemble peices of a function into a new function. We will
spell out some of these isomorphisms now. As defined in [9], an isometry of
R2 is a distance-preserving map τ : R2 → R2 such that for all u and v in R2,
|τ(u) − τ(v)| = |u − v|.
Reflections about a line, rotations about a point, translations by a vector
a and any composition thereof all define isometries of the plane [9]. The
easiest way of specifying what the isomorphisms τh,k are is by taking advan-
tage of the reflectional symmetry of M1 and M2. Because of this symmetry,
all of the isomorphisms τh,k between triangles of M2 and triangles of M1 can
be described in terms of reflections. We label the edges of the triangle used
to build domain M1, M2 as in the figure below.

32 ASPEN DEVRIES
Figure 8. We label the boundary segments of M1 so that it
is easier to check that boundary conditions are verified. M2
is included for reference.
Figure 9. The triangle used to generate the warped pro-
pellers. We label the sides to make it easier to specify isome-
tries between triangles in M1 and M2.
Let rz, z ∈ {a, b, c} denote the reflection of the triangle about the edge
z. It’s straightforward to express the τh,k in terms of these reflections. We
write out the first few:
τ7,0 = I
τ7,1 = rc
τ7,2 = rb

τ7,3 = ra ◦ rb
τ7,4 = ra
τ7,5 = rb ◦ rc
τ7,6 = rc ◦ ra
τ13,6 = ra ◦ rc ◦ rc ◦ rb
τ13,5 = rc ◦ rb ◦ rc ◦ rb
τ13,3 = rb ◦ ra ◦ rc ◦ rb.
8. Conclusion
We’ve seen that the harmonics of a drum do not specify its shape in
all cases. The pair of drums we examined are one of a class of 17 pairs
of drums that can be proven isospectral with the transplantation method.
Perhaps there is something about the method of constructing the drums
that makes the spectrum less informative about drum shape as it is in other
types of drums. Though we can’t hear the shapes of this class of drums,
our intuition as people who rely on sound to distinguish between musical
instruments or to sense the closeness of objects is that the spectrum of the
drum should encode a lot of information about the drum, whether or not
it can be used to determine the exact shape of the drum. Questions along
these lines are currently being explored. For instance, it has been found that
some drums that are isospectral and nonisometric can be distinguished by
counting nodal domains [16]. The work we did to motivate Kac’s question
and better understand the dynamics of drums is barely a glance through a
keyhole at the complexities and insights oﬀered by the pursuit to understand
sound and shape.

34 ASPEN DEVRIES
References
[1] S. Gockenbach. Partial Differential Equations. Siam. 2002. Defined the eigenvalue prob-
lem and provided necessary background; gave techniques for solving differential equa-
tions.
[2] Mark Kac. Can one hear the shape of a drum? Am. Math. Month. Vol 73, No. 4, Part
2: Papers in Analysis (Apr., 1966). Formalized the question of determining shape from
spectra.
[3] Gordon and Webb. You can’t hear the shape of a drum. American Scientist. 1992.
Provided a negative answer to Kac’s question.
[4] Giraud and Thas. Hearing Shapes of Drums. Arxiv (2010). Provided an extensive
overview of research related to Kac’s question.
[5] Grieser and Svenja. Hearing the Shape of a Triangle. Not. of the AMS (2013). Discussed
the eigenvalue problem on a triangle and argues that triangles can be distinguished
from each other based on their spectrum.
[6] G.M.L Gladwell. Courant’s Nodal Line Theorem and its Discrete Counterparts. De-
partment of Mathematics, University of Waterloo. 2001. Provided a proof of the
Courant-Hilbert Theorem.
[7] Peter Buser, John Conway, Peter Doyle, Klaus-Dieter Semmler. Some Planar isospec-
tral Domains. International mathematics Research notes, No. 9. 1994. Provided a proof
of isospectrality of the non isometric drums using transplantation.
[8] Courant and Hilbert. Methods of Mathematical physics Vol. 1. Interscience Publishers.
1953. Gave a justification that the separation of variables solution to the 2D helmholtz
equation was unique.
[9] Artin. Algebra (Second Edition). Prentice Hall (2011). Gave a discussion of isometries.
[10] Diaz and Ludford. Reflection principle for linear elliptic second order partial differ-
ential equations with constant coefficients. Annali di Matematica Pura ed Applicata
(1955). A statement and proof of the reflection principle.
[11] Peter Buser. Isospectral Riemann Surfaces. Annales de l’institut Fourier (1986). Gives
a precise description of the transplantation method.
[12] Gordon, Webb and Wolpert. Isospectral plane domains and surfaces via Riemannian
orbifolds(1992). Describes the algebraic structure of isospectral drums.
[13] Taylor. Classical Mechanics. University Science Books (2005). Provided a discussion
of the wave equatino in the context of physics.
[14] Bérard. Transplantation et isospectralité I (Transplantation and isospectrality I).
Séminaire de Théorie spectrale et géométrie (1990-1991). The first paper to develop
the transplantation method.
[15] Bérard. Domaines plans isospectraux á la Gordon-Webb-Wolpert: une preuve terre
á terre (Isospectral plane domains of Gordon-Webb-Wolpert: a down to earth proof).
Séminaire de Théorie spectrale et géométrie (1991-1992). Provides a proof of isospec-
trality of plane domains constructed by Gordon, Webb and Wolpert.
[16] Gnutzmann, Smilansky, Sondergaard. Resolving isospectral drums by counting nodal
domains. J. Phys. A. Math. Gen. 2005.
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