Math 54: Extra Topics on Differential Equations
The Phenomena of Wave and Vibration
Long Jin
August 14th, 2013
The phenomena of wave and vibration
North Pacific storm waves as seen from the NOAA M/V Noble Star,
Winter 1989.
Mechanical vibration (zero dimension)
The motion of a mechanical vibration is given by the second order
linear differential equation:
my 00 + by 0 + ky = F .
Here m > 0 is the mass or inertia, b > 0 is the friction or damping
term, k > 0 is the stiffness of the system, F = F (t) is the external
force.
Mass-Spring system.
Free vibration (F = 0): undamped b = 0
The equation becomes
my 00 + ky = 0.
This is a simple harmonic motion with frequency ω =
period 2π/ω:
p
k/m and
y = C1 cos(ωt) + C2 sin(ωt) = A sin(ωt + φ).
q
A = C12 + C22 is the amplitude, φ = arctan(C1 /C2 ) is the initial
phase.
Free vibration (F = 0): Different kind of damping
Now we have b > 0:
my 00 + by 0 + ky = 0.
There are three different cases: All have exponential decay but
with different behaviors.
Free vibration (F = 0): Different kind of damping
Now we have b > 0:
my 00 + by 0 + ky = 0.
There are three different cases: All have exponential decay but
with different behaviors.
Underdamped b 2 < 4mk
y = e αt (C1 cos βt + C2 sin βt) = Ae αt sin(βt + φ).
Free vibration (F = 0): Different kind of damping
Now we have b > 0:
my 00 + by 0 + ky = 0.
There are three different cases: All have exponential decay but
with different behaviors.
Underdamped b 2 < 4mk
y = e αt (C1 cos βt + C2 sin βt) = Ae αt sin(βt + φ).
Overdamped b 2 > 4mk
y = C1 e r1 t + C2 e r2 t .
Free vibration (F = 0): Different kind of damping
Now we have b > 0:
my 00 + by 0 + ky = 0.
There are three different cases: All have exponential decay but
with different behaviors.
Underdamped b 2 < 4mk
y = e αt (C1 cos βt + C2 sin βt) = Ae αt sin(βt + φ).
Overdamped b 2 > 4mk
y = C1 e r1 t + C2 e r2 t .
Critically damped b 2 = 4mk
y = (C1 + C2 t)e −rt .
Free vibration: Underdamped
When b 2 < 4mk, the solution to
my 00 + by 0 + ky = 0,
is
y = e αt (C1 cos βt + C2 sin βt) = Ae αt sin(βt + φ),
√
2
b
< 0, β = 4mk−b
. The solution still oscillates
where α = − 2m
2m
but also exponentially decays with “quasi-frequency” β/2π. In this
case the damping is too weak to prevent the oscillation.
Free vibration: Overdamped
When b 2 > 4mk, the solution to
my 00 + by 0 + ky = 0,
is
y = C1 e r1 t + C2 e r2 t
√
2
where r1,2 = − b± b2m−4mk < 0. In this case y has at most one
critical point which is either a local maximum or minimum. So the
damping is strong enough to prevent oscillation.
Free vibration: Critically damped
When b 2 = 4mk, the solution to
my 00 + by 0 + ky = 0,
is
y = (C1 + C2 t)e rt .
where r = −b/2m < 0. The behavior of y is similar to the
overdamped case: it has at most one critical point. However, the
decay rate is slower than the overdamped case. Moreover, this
situation is not stable: if we perturb b, it either falls into the
underdamped case or overdamped case.
Free vibration: Different situations
Time dependence of the system behavior on the values of the damping
rate.
Forced vibration
We consider the case of a sinusoidal external force:
my 00 + by 0 + ky = F0 cos γt.
The external force is also periodic.
Forced vibration: underdamped case
When 0 < b <
√
4mk, the solution of
my 00 + by 0 + ky = F0 cos γt.
is
√
y = Ae
−bt/2m
sin
4mk − b 2
t +φ
2m
!
F0
+p
sin(γt + θ).
(k − mγ 2 )2 + b 2 γ 2
The first term comes from the homogeneous equation, the second
term comes from the nonhomogeneous term, i.e. the external
force. We can think of the second term as a “steady state”.
Forced vibration: underdamped case
√
y = Ae
−bt/2m
sin
4mk − b 2
t +φ
2m
!
F0
sin(γt + θ).
+p
(k − mγ 2 )2 + b 2 γ 2
The second term has an extra factor (called gain factor) besides
the amplitude of the external force
1
M(γ) = p
.
(k − mγ 2 )2 + b 2 γ 2
q
k
b2
The critical point of M is given by γ = 0 or γ = γr = m
− 2m
2,
where the later one gives the maximum. Therefore if the external
force is at this frequency γr , the system gains most from the
oscillation and this is called phenomena resonance.
Forced vibration: resonance
When b = 0, the equation
my 00 + ky = F0 cos γt
has solution
y = A sin(ωt + φ) +
if γ 6= ω =
F0
sin(γt + θ)
k − mγ 2
p
k/m. The resonances appears when γ = ω:
y = A sin(ωt + φ) +
F0 t
sin ωt.
2mω
Here we see the solution is no longer bounded as t → ∞.
Resonances in real life
Resonance is a type of physical phenomena that the amplitude at
certain frequency is greater than others. It appears in all types of
waves: mechanical resonance, acoustic resonance, electromagnetic
resonance, nuclear magnetic resonance, electron spin resonance
and resonance of quantum wave functions.
Resonance. http://xkcd.com/228/
Mechanical Resonances
Tesla’s earthquake machine.
Acoustic Resonances
Shattering a wine glass with sound. Photo: stevenduong on Flickr
video on youtube
Schumann Resonance
Schumann resonance is the resonance of the Earth’s electromagnetic
field. Picture by Finnbogi Pétursson
Vibrating String: wave equation on an interval
Here is the wave equation on an interval:
2
∂2u
2∂ u
=
ω
, 0 < x < L, t > 0.
∂t 2
∂x 2
Vibrating String: separation of variables
The method of separation of variables can be applied to the wave
equation:
2
∂2u
2∂ u
=
ω
, 0 < x < π, t > 0.
∂t 2
∂x 2
u(0, t) = u(π, t) = 0, t > 0.
Let u(x, t) = X (t)T (t), we get an eigenvalue problem in the
x-direction
X 00 (x) + λX (x) = 0, X (0) = X (π) = 0
which gives λ = n2 , eλ = sin nx. The corresponding solutions are
X (x) = cos nωt sin nx or sin nωt sin nx.
Standing waves
Each eigenfunction (called modes in physics) corresponds to a
standing wave.
Some links for more pictures: standingwave1 standingwave2
Damped wave equation on an interval
We can also consider the damped wave equation: a > 0 is the
damping constant.
2
∂2u
∂u
2∂ u
=
ω
+
a
, 0 < x < π, t > 0.
∂t 2
∂t
∂x 2
u(0, t) = u(π, t) = 0, t > 0.
Again, we can use separation of variables: u(x, t) = X (x)T (t).
The eigenvalue problem X is as before and the equation for T is
T 00 (t) + aT 0 (t) + λω 2 T (t) = 0.
Here λ = n2 , n = 1, 2, . . .. Therefore some of the eigenfunctions
will behave as the underdamped case (a < 2nω); some as the
overdamped case (a > 2nω); some as the critically damped case
(a = 2nω).
Free waves: wave equation on the real line
Now we turn to the wave equation on the whole real line:
2
∂2u
2∂ u
=
ω
, x, t ∈ R.
∂t 2
∂x 2
Or equivalently
2
∂2u
2∂ u
−
ω
= 0, x, t ∈ R.
∂t 2
∂x 2
Notice that the left side ∂t2 − ω 2 ∂x2 = (∂t − ω∂x )(∂t + ω∂x ).
Transport equation
Let us start with the first order partial differential equation
∂u
∂u
= a , x, t ∈ R.
∂t
∂x
This is called the transport equation. The “data” are transported
along the characteristic curves: x(t) = x0 + at. Let
v (t) = u(x(t), t), then
v 0 (t) = ut (x(t), t) +
dx(t)
ux (x(t), t) = (ut + aux )(x(t), t) = 0.
dt
Therefore for initial value u(x, 0) = f (x), the solution is
u(x, t) = u(x − at, 0) = f (x − at).
Traveling waves
The solution to the transport equation corresponds to a traveling
wave:
Some links for more pictures: travelingwave waterwave1
waterwave2
General solution to the wave equation
The general solution to
2
∂2u
2∂ u
=
ω
, x, t ∈ R
∂t 2
∂x 2
is
u(x, t) = F (x + ωt) + G (x − ωt).
Here F and G are two functions. F represents a wave traveling to
the left, G a wave to the right. This can be seen by a change of
variable ξ = x + ωt, η = x − ωt, then the equation becomes
∂2u
∂ξ∂η = 0.
D’Alembert formula
For the initial value problem
2
∂2u
2∂ u
=
ω
, x, t ∈ R
∂t 2
∂x 2
∂u
(x, 0) = g (x), x ∈ R,
u(x, 0) = f (x),
∂t
the solution is given by the D’Alembert formula
Z
1
1 x+ωt
u(x, t) = [f (x + ωt) + f (x − ωt)] +
g (s)ds.
2
2 x−ωt
Wave equation in higher dimension
∂2u
(x, t) = ∆x u(x, t), t ∈ R, x ∈ Ω ⊂ Rn .
∂t 2
∂2
∂2
Here ∆x = ∂x
2 + · · · + ∂x 2 .
1
n
Maxwell’s equation shows that electromagnetic fields behave as waves, in
particular, light is a special kind of electromagnetic wave.
Eigenvalues and eigenfunctions
We can also use separation of variables for the boundary value
problem:
∂2u
(x, t) = ∆x u(x, t), t ∈ R, x ∈ Ω ⊂ Rn .
∂t 2
u(x, t) = 0, x ∈ ∂Ω, t ∈ R
The question is to solve the eigenvalue problem:
−∆v = λv , x ∈ Ω ⊂ Rn .
v (x) = 0, x ∈ ∂Ω.
The nontrivial solutions are called eigenfunctions for the (Dirichlet)
Laplacian on Ω. λ = ω 2 > 0 is the eigenvalue and ω corresponds
to the frequency of the solution to wave equation.
Scattering resonances and resonance states
When Ω is an unbounded domain in Rn , e.g., Ω = Rn \ O where O
is a bounded obstacle.
−∆v = λv , x ∈ Ω ⊂ Rn .
v (x) = 0, x ∈ ∂Ω.
We have to add certain condition at infinity. Then we are looking
for “outgoing” solutions and here λ is a complex number. The
convention is to write λ = ζ 2 where Im ζ < 0 and ζ is called a
(scattering) resonance of the (Dirichlet) Laplacian. Every resonant
wave has a certain decay since the wave might travel to “infinity”.
The real part of ζ gives the frequency and the imaginary part gives
the decay rate.
Laplace equation in polar coordinates
For special kind of domain, we can use separation of variables to
find the solution to Laplace equation or eigenvalue problem: For
example on the disk B = B(0, 1):
∂2u ∂2u
+
= 0, x 2 + y 2 < 1,
∂x 2 ∂y 2
we can use polar coordinates (r , θ) to change the domain to a
rectangle domain:
∂ 2 u 1 ∂u
1 ∂2u
+
+
= 0, 0 < r < 1, 0 < θ < 2π.
∂r 2
r ∂r
r 2 ∂θ2
Two dimensional disk: Laplace equation
For the boundary value problem, u|∂B = f , we can write
u = u(r , θ) satisfying
∂ 2 u 1 ∂u
1 ∂2u
+ 2 2 = 0, 0 < r < 1, 0 < θ < 2π
+
2
∂r
r ∂r
r ∂θ
∂u
with u(r , 0) = u(r , 2π), ∂u
∂r (r , 0) = ∂r (r , 2π) and u(1, θ) = f (θ).
We also have u(0, θ) should be a constant in θ. This can be solved
by separation of variables. In θ direction, we get periodic boundary
value problem:
Θ00 (θ) + λΘ(θ) = 0, Θ(0) = Θ(2π), Θ0 (0) = Θ0 (2π).
Again, we have λ = 0, Θ = 1; or λ = k 2 , Θ = cos kθ or sin kθ.
Two dimensional disk: Laplace equation
In r direction, we get Cauchy-Euler equation
r 2 R 00 (r ) + rR 0 (r ) − k 2 R(r ) = 0, 0 < r < 1
whose general solutions are (solving by change of variable
s = log r )
R = c1 + c2 s = c1 + c2 log r , if k = 0;
(1)
R = c1 e ks + c2 e −ks = c1 r k + c2 r −k , if k > 0.
(2)
for k > 0,
By the boundary condition in r , we should make c2 = 0 in both
case and we get the following solution:
∞
u(r , θ) ∼
a0 X k
+
r (ak cos kθ + bk sin kθ).
2
k=1
(3)
Two dimensional disk: Eigenvalues and resonances
For the eigenvalue problem on the disk:
∂2u ∂2u
+
= −λu, x 2 + y 2 < 1,
∂x 2 ∂y 2
with Dirichlet boundary condition, we can also use separation of
variables as before with the same answer in Θ. In r -direction, the
solutions are described by Bessel functions. The same method
applies to resonance problem
∂2u ∂2u
+
= −ζ 2 u, x 2 + y 2 > 1,
∂x 2 ∂y 2
again with Dirichlet boundary condition. Now we get Hankel
functions. Similarly we can treat higher dimension disks using
Bessel/Hankel functions.
Two dimensional sphere: Resonances
The Resonances for S 2 with Dirichlet boundary condition concentrated
along cubic curves, from Stefanov(2006).
Eigenvalue or resonances for general domain
For a general domain, it is unlikely that we can get the precise
values of eigenvalues or resonances. However, the distribution of
eigenvalues on the (positive) real line or the resonances in the
(lower half) complex plane can be studied asymptotically. This
asymptotic distribution is closely related to the geometry of the
domain. These are important questions in spectral theory and
scattering theory.
Weyl Law for eigenvalues
In 1912, Hermann Weyl proves that the number N(λ) of
eigenvalues of Dirichlet Laplacian on a bounded domain Ω ⊂ Rn
(counting multiplicities) less than λ has the following property:
N(λ) = (2π)−d ωd vol(Ω)λd/2 (1 + o(1)), λ → ∞.
Here ωd is the volume of the unit ball in Rd , o(1) is a term tends
to zero as λ → ∞. In other words N(λ) ∼ (2π)−d ωd vol(Ω)λd/2 .
This was improved to O(λ−1 ) later and in fact, for a large family
of domains Ω, the second term in this asymptotic expansion only
depends on the area of the boundary ∂Ω.
Can one hear the shape of the drum?
In 1966, Mark Kac published a popular article named “Can one
hear the shape of the drum?”. In the article, Kac asks whether the
eigenvalues of the (Dirichlet) Laplacian of a domain uniquely
determines the shape of the domain. The phrasing of the title is
due to Lipman Bers’s formulation: “if you had perfect pitch could
you find the shape of a drum.” (The eigenvalues are the square of
the frequencies of the wave.)
I
In general, the question has a negative answer. Gordon,
Webb, and Wolpert constructed two non-convex polygons
which have the same spectrum in 1992.
I
However, if we put more restrictions on the shape of the
domain, (say the boundary must be smooth), then the
question is still open. Zelditch gave an affirmative answer
under a very strong restriction.
I
Moreover, by Weyl law and isoperimetric inequality (or some
comparison theorem).
The distribution of resonances
I
The resonance-free region is a neighborhood of the real axis
that contains no resonances. Understanding of the
resonance-free region is related to the decay properties and
long time behavior of waves.
I
Keller shows that in the shadow area, the strength of the
wave decays faster than any negative power of the frequency
ω. This corresponds to the logarithm resonance-free region.
I
Melrose, Ivrii, Sjöstrand, Taylor: Logarithm resonance-free
region Im ζ > −M log |ζ| + CM for non-trapping obstacles.
I
Hargé-Lebeau (1994), Sjöstrand-Zworski (1995): Cubic
resonance-free region for smooth strictly convex obstacles
1
which shows that the decay is of O(e −cω 3 ).
I
Lots of open questions.
Geometry of eigenfunctions and resonance states
I
Another interesting question is to understand the behavior of
eigenfunctions or resonance states.
I
For a finite interval, the eigenfunctions are sine or cosine
functions and we can see the number of zeroes (or critical
points) of the eigenfunction is proportional to the eigenvalue.
I
It is natural to ask the same question for the
(n − 1)-dimensional measure of the eigenfunction in
n-dimensional domain. This is Yau’s conjecture on nodal set
(zero set) and critical set.
I
Donnelly-Fefferman first give an affirmative result for nodal
set of eigenfunctions in a large family of domain (analytic
manifold). Further results given by Sogge-Zelditch,
Colding-Minicozzi etc.
I
Lots of open questions.
Further reading on differential equations
I
Ordinary differential equations: Hurewicz,
Coddington-Levinson.
I
Partial differential equations: Evans, Taylor(three volumes),
John, etc.
Thanks for your attention.