4Partial differential equations

IB Methods

4.8 The wave equation

Consider a string x ∈ [0, L] undergoing small oscillations described by φ(x, t).

φ(x, t)

A

B

θ

A

θ

B

Consider two points

A

,

B

separated by a small distance

δx

. Let

T

A

(

T

B

) be

the outward pointing tension tangent to the string at

A

(

B

). Since there is no

sideways (x) motion, there is no net horizontal force. So

T

A

cos θ

A

= T

B

cos θ

B

= T. (∗)

If the string has mass per unit length

µ

, then in the vertical direction, Newton’s

second law gives

µδx

∂

2

φ

∂t

2

= T

B

sin θ

B

− T

A

sin θ

A

.

We now divide everything by T , noting the relation (∗), and get

µ

δx

T

∂

2

φ

∂t

2

=

T

B

sin θ

B

T

B

cos θ

B

−

T

A

sin θ

A

T

A

cos θ

A

= tan θ

B

− tan θ

A

=

∂φ

∂x

B

−

∂φ

∂x

A

≈ δx

∂

2

φ

∂x

2

.

Taking the limit

δx →

0 and setting

c

2

=

T/µ

, we have that

φ

(

x, t

) obeys the

wave equation

1

c

2

∂

2

φ

∂t

2

=

∂

2

φ

∂x

2

.

From IA Differential Equations, we’ve learnt that the solution can be written as

f

(

x − ct

) +

g

(

x

+

ct

). However, this does not extend to higher dimensions, but

separation of variables does. So let’s do that.

Assume that the string is fixed at both ends. Then

φ

(0

, t

) =

φ

(

L, t

) = 0 for

all

t

. The we can perform separation of variables, and the general solution can

then be written

φ(x, t) =

∞

X

n=1

sin

nπx

L

A

n

cos

nπct

L

+ B

n

sin

nπct

L

.

The coefficients

A

n

are fixed by the initial profile

φ

(

x,

0) of the string, while

the coefficients

B

n

are fixed by the initial string velocity

∂φ

∂t

(

x,

0). Note that we

need two sets of initial conditions, since the wave equation is second-order in

time.

Energetics and uniqueness

An oscillating string contains has some sort of energy. The kinetic energy of a

small element δx of the string is

1

2

µδx

∂φ

∂t

2

.

The total kinetic energy of the string is hence the integral

K(t) =

µ

2

Z

L

0

∂φ

∂t

2

dx.

The string also has potential energy due to tension. The potential energy of a

small element δx of the string is

T × extension = T (δs − δx)

= T (

p

δx

2

+ δφ

2

− δx)

≈ T δx

1 +

1

2

δφ

δx

2

+ ···

!

− T δx

=

T

2

δx

δφ

δx

2

.

Hence the total potential energy of the string is

V (t) =

µ

2

Z

L

0

c

2

∂φ

∂x

2

dx,

using the definition of c

2

.

Using our series expansion for φ, we get

K(t) =

µπ

2

c

2

4L

∞

X

n=1

n

2

A

n

sin

nπct

L

− B

n

cos

nπct

L

2

V (t) =

µπ

2

c

2

4L

∞

X

n=1

n

2

A

n

cos

nπct

L

+ B

n

sin

nπct

L

2

The total energy is

E(t) =

nπ

2

c

2

4L

∞

X

n=1

n

2

(A

2

n

+ B

2

n

).

What can we see here? Our solution is essentially an (infinite) sum of independent

harmonic oscillators, one for each

n

. The period of the fundamental mode (

n

= 1)

is

2π

ω

= 2

π ·

L

πc

=

2L

c

. Thus, averaging over a period, the average kinetic energy

is

¯

K =

c

2L

Z

2L/c

0

K(t) dt =

¯

V =

c

2L

Z

2L/c

0

V (t) dt =

E

2

.

Hence we have an equipartition of the energy between the kinetic and potential

energy.

The energy also allows us to prove a uniqueness theorem. First we show that

energy is conserved in general.

Proposition.

Suppose

φ

: Ω

×

[0

, ∞

)

→ R

obeys the wave equation

∂

2

φ

∂t

2

=

c

2

∇

2

φ

inside Ω × (0, ∞), and is fixed at the boundary. Then E is constant.

Proof. By definition, we have

dE

dt

=

Z

Ω

∂

2

ψ

∂t

2

∂ψ

∂t

+ c

2

∇

∂φ

∂t

· ∇φ dV.

We integrate by parts in the second term to obtain

dE

dt

=

Z

Ω

dφ

dt

∂

2

φ

∂t

2

− c

2

∇

2

φ

dV + c

2

Z

∂Ω

∂φ

∂t

∇φ · dS.

Since

∂

2

φ

∂t

2

=

c

2

∇

2

φ

by the wave equation, and

φ

is constant on

∂

Ω, we know

that

dE

φ

dt

= 0.

Proposition.

Suppose

φ

: Ω

×

[0

, ∞

)

→ R

obeys the wave equation

∂

2

φ

∂t

2

=

c

2

∇

2

φ

inside Ω × (0, ∞), and obeys, for some f, g, h,

(i) φ(x, 0) = f(x);

(ii)

∂φ

∂t

(x, 0) = g(x); and

(iii) φ|

∂Ω×[0,∞)

= h(x).

Then φ is unique.

Proof.

Suppose

φ

1

and

φ

2

are two such solutions. Then

ψ

=

φ

1

− φ

2

obeys the

wave equation

∂

2

ψ

∂t

2

= c

2

∇

2

ψ,

and

ψ|

∂Ω×[0,∞)

= ψ|

Ω×{0}

=

∂ψ

∂t

Ω×{0}

= 0.

We let

E

ψ

(t) =

1

2

Z

Ω

"

∂ψ

∂t

2

+ c

2

∇ψ · ∇ψ

#

dV.

Then since

ψ

obeys the wave equation with fixed boundary conditions, we know

E

ψ

is constant.

Initially, at

t

= 0, we know that

ψ

=

∂ψ

∂t

= 0. So

E

ψ

(0) = 0. At time

t

, we

have

E

ψ

=

1

2

Z

Ω

∂ψ

∂t

2

+ c

2

(∇ψ) · (∇ψ) dV = 0.

Hence we must have

∂ψ

∂t

= 0. So

ψ

is constant. Since it is 0 at the beginning, it

is always 0.

Example. Consider Ω = {(x, y) ∈ R

2

, x

2

+ y

2

≤ 1}, and let φ(r, θ, t) solve

1

c

2

∂

2

φ

∂t

2

= ∇

2

φ =

1

r

r

∂φ

∂r

+

1

r

2

∂

2

φ

∂θ

2

,

with the boundary condition

φ|

∂Ω

= 0. We can imagine this as a drum, where

the membrane can freely oscillate with the boundary fixed.

Separating variables with φ(r, θ, t) = T (t)R(r)Θ(θ), we get

T

00

= −c

2

λT, Θ

00

= −µΘ, r(R

0

)

0

+ (r

2

λ − µ)R = 0.

Then as before,

T

and Θ are both sine and cosine waves. Since we are in polars

coordinates, we need

φ

(

t, r, θ

+ 2

π

) =

φ

(

t, r, θ

). So we must have

µ

=

m

2

for

some m ∈ N. Then the radial equation becomes

r(rR

0

)

0

+ (r

2

λ − m

2

)R = 0,

which is Bessel’s equation of order m. So we have

R(r) = a

m

J

m

(

√

λr) + b

m

Y

m

(

√

λr).

Since we want regularity at

r

= 0, we need

b

m

= 0 for all

m

. To satisfy the

boundary condition

φ|

∂Ω

= 0, we must choose

√

λ

=

k

mi

, where

k

mi

is the

i

th

root of J

m

.

Hence the general solution is

φ(t, r, θ) =

∞

X

i=0

[A

0i

sin(k

0i

ct) + B

0i

cos(k

0i

ct)]J

0

(k

0i

r)

+

∞

X

m=1

∞

X

i=0

[A

mi

cos(mθ) + B

mi

sin(mθ)] sin k

mi

ctJ

m

(k

mi

r)

+

∞

X

m=1

∞

X

i=0

[C

mi

cos(mθ) + D

mi

sin(mθ)] cos k

mi

ctJ

m

(k

mi

r)

For example, suppose we have the initial conditions

φ

(0

, r, θ

) = 0,

∂

t

φ

(0

, r, θ

) =

g

(

r

). So we start with a flat surface and suddenly hit it with some force. By

symmetry, we must have

A

mi

, B

mi

, C

mi

, D

mi

= 0 for

m 6

= 0. If this does not

seem obvious, we can perform some integrals with the orthogonality relations to

prove this.

At t = 0, we need φ = 0. So we must have B

0i

= 0. So we are left with

φ =

∞

X

i=0

A

0i

sin(k

0i

ct)J

0

(k

0j

r).

We can differentiate this, multiply with J

0

(k

0j

r)r to obtain

Z

1

0

∞

X

i=0

k

0i

cA

0i

J

0

(k

0i

r)J

0

(k

0j

r)r dr =

Z

1

0

g(r)J

0

(k

0j

r)r dr.

Using the orthogonality relations for J

0

from the example sheet, we get

A

0i

=

2

ck

0i

1

[J

0

0

(k

0i

)]

2

Z

1

0

g(r)J

0

(k

0j

r)r dr.

Note that the frequencies come from the roots of the Bessel’s function, and are

not evenly spaced. This is different from, say, string instruments, where the

frequencies are evenly spaced. So drums sound differently from strings.

So far, we have used separation of variables to solve our differential equations.

This is quite good, as it worked in our examples, but there are a few issues with

it. Of course, we have the problem of whether it converges or not, but there is a

more fundamental problem.

To perform separation of variables, we need to pick a good coordinate system,

such that the boundary conditions come in a nice form. We were able to

perform separation of variables because we can find some coordinates that fit our

boundary conditions well. However, in real life, most things are not nice. Our

domain might have a weird shape, and we cannot easily find good coordinates

for it.

Hence, instead of solving things directly, we want to ask some more general

questions about them. In particular, Kac asked the question “can you hear the

shape of a drum?” — suppose we know all the frequencies of the modes of

oscillation on some domain Ω. Can we know what Ω is like?

The answer is no, and we can explicitly construct two distinct drums that

sound the same. Fortunately, we get an affirmative answer if we restrict ourselves

a bit. If we require Ω to be convex, and has a real analytic boundary, then yes!

In fact, we have the following result: let

N

(

λ

0

) be the number of eigenvalues

less than λ

0

. Then we can show that

4π

2

lim

λ

0

→∞

N(λ

0

)

λ

0

= Area(Ω).