5Hyperbolic equations
III Analysis of Partial Differential Equations
5 Hyperbolic equations
So far, we have been looking at elliptic PDEs. Since the operator is elliptic,
there is no preferred “time direction”. For example, Laplace’s equation models
static electric fields. Thus, it is natural to consider boundary value problems in
these cases.
Hyperbolic equations single out a time direction, and these model quantities
that evolve in time. In this case, we are often interested in initial value problems
instead. Let’s first define what it means for an equation to by hyperbolic
Definition
(Hyperbolic PDE)
.
A second-order linear hyperbolic PDE is a PDE
of the form
n+1
X
i,j=1
(a
ij
(y)u
y
j
)
y
i
+
n+1
X
i=1
b
i
(y)u
y
i
+ c(y)u = f
with y ∈ R
n+1
, a
ij
= a
ji
, b
i
, c ∈ C
∞
(R
n+1
), such that the principal symbol
Q(ξ) =
n+1
X
i,j=1
a
ij
(y)ξ
i
ξ
j
has signature (+
, −, −, . . .
) for all
y
. That is to say, after perhaps changing basis,
at each point we can write
q(ξ) = λ
2
n+1
ξ
2
n+1
−
n
X
i=1
λ
2
i
ξ
2
i
with λ
i
> 0.
It turns out not to be too helpful to treat this equation at this generality.
We would like to pick out a direction that corresponds to the positive eigenvalue.
By a coordinate transformation, we can locally put our equation in the form
u
tt
=
n
X
i,j=1
(a
ij
(x, t)u
x
i
)
x
j
+
n
X
i=1
b
i
(x, t)u
x
i
+ c(x, t)u.
Note that we did not write down a
u
t
term. It doesn’t make much difference,
and it is notationally convenient to leave it out.
In this form, hyperbolicity is equivalent to the statement that the operator
on the right is elliptic for each
t
(or rather, the negative of the right hand side).
We observe that
t
= 0 is a non-characteristic surface. So we can hope to
solve the Cauchy problem. In other words, we shall specify
u|
t=0
and
u
t
|
t=0
.
Actually, we’ll look at an initial boundary value problem. Consider a region of
the form R × U, where U ⊆ R
n
is open bounded with C
1
boundary.
t = 0
t = T
U
We define
U
t
= (0, t) × U
Σ
t
= {t} × U
∂
∗
U
t
= [0, t] × ∂U.
Then
∂U
T
= Σ
0
t Σ
T
t ∂
∗
U
T
.
The general initial boundary value problem (IVBP) is as follows: Let
L
be a
(time-dependent) uniformly elliptic operator. We want to solve
u
tt
+ Lu = f on U
T
u = ψ on Σ
0
u
t
= ψ
0
on Σ
0
u = 0 on ∂
∗
U
T
.
In the case of elliptic PDEs, we saw that Laplace’s equation was a canonical,
motivating example. In this case, if we take
L
=
−
∆, then we obtain the wave
equation. Let’s see what we can do with it.
Example.
Start with the equation
u
tt
−
∆
u
= 0. Multiply by
u
t
and integrate
over U
t
to obtain
0 =
Z
U
t
u
tt
u
t
− u
t
∆u
dx dt
=
Z
U
t
1
2
∂
∂t
u
2
t
− ∇ · (u
t
Du) + Du
t
· Du
dx dt
=
Z
U
t
1
2
∂
∂t
u
2
t
+ |Du|
2
− ∇ · (u
t
Du)
dx dt
=
1
2
Z
Σ
t
−Σ
0
u
2
t
+ |Du|
2
dx −
Z
∂
∗
U
t
u
t
∂u
∂ν
dS.
But
u
vanishing on
∂
∗
U
T
implies
u
t
vanishes as well. So the second term vanishes,
and we obtain
Z
Σ
t
u
2
t
+ |Du|
2
dx =
Z
Σ
0
u
2
t
+ |Du|
2
dx.
This is the conservation of energy! Thus, if a solution exists, we control
kuk
H
1
(Σ
t
)
in terms of
kψk
H
1
(Σ
0
)
and
kψ
0
k
L
2
(Σ
0
)
. We also see that the solution is uniquely
determined by
ψ
and
ψ
0
, since if
ψ
=
ψ
0
= 0, then
u
t
= D
u
= 0 and
u
is zero at
the boundary.
Estimates like this that control a solution without needing to construct it are
known as a priori estimates. These are often crucial to establish the existence
of solutions (cf. G˚arding).
We shall first find a weak formulation of this problem that only requires
u ∈ H
1
(
U
T
). Note that when we do so, we have to understand carefully what
we mean by
u
t
=
ψ
0
. We shall see how we will deal with that in the derivation
of the weak formulation.
Assume that
u ∈ C
2
(
¯
U
T
) is a classical solution. Multiply the equation by
v ∈ C
2
(
¯
U
T
) which satisfies v = 0 on ∂
∗
U
T
∪ Σ
T
. Then we have
Z
U
T
dx dt (f v) =
Z
U
T
dx dt (u
tt
v + Luv)
=
Z
U
T
dx dt
−u
t
v
t
+
X
a
ij
u
x
i
v
x
j
+
X
b
i
u
x
i
v
+ cu
+
Z
U
u
t
v dx
T
0
−
Z
T
0
dt
Z
∂U
X
a
ij
u
x
j
v dS
dt.
Using the boundary conditions, we find that
Z
U
T
fv dx dt =
Z
U
T
−u
t
v
t
+
X
a
ij
u
x
i
v
x
j
+
X
b
i
u
x
i
v + cuv
dx dt
−
Z
Σ
0
ψ
0
v dx. (†)
Conversely, suppose
u ∈ C
2
(
¯
U
T
) satisfies (
†
) for all such
v
, and
u|
Σ
0
=
ψ
and
u|
∂
∗
U
T
= 0. Then by first testing on
v ∈ C
∞
c
(
U
T
), reversing the integration by
parts tells us
0 =
Z
U
T
(u
tt
+ Lu − f)v dx,
since there is no boundary term. Hence we get
u
tt
+ Lu = f
on
U
T
. To check the boundary conditions, if
v ∈ C
∞
(
¯
U
T
) vanishes on
∂
∗
U
T
∪
Σ
T
,
then again reversing the integration by parts shows that
Z
U
T
(u
tt
+ Lu − f)v dx dt =
Z
Σ
0
(ψ
0
− u
t
)v dx.
Since we know that the LHS vanishes, it follows that ψ
0
= u
t
on Σ
0
. So we see
that our weak formulation can encapsulate the boundary condition on Σ
0
.
Definition
(Weak solution)
.
Suppose
f ∈ L
2
(
U
T
),
ψ ∈ H
1
0
(Σ
0
) and
ψ
0
∈
L
2
(Σ
0
). We say u ∈ H
1
(U
t
) is a weak solution to the hyperbolic PDE if
(i) u|
Σ
0
= ψ in the trace sense;
(ii) u|
∂
∗
U
T
= 0 in the trace sense; and
(iii) (†) holds for all v ∈ H
1
(U
T
) with v = 0 on ∂
∗
U
T
∪ Σ
T
in a trace sense.
Theorem (Uniqueness of weak solution). A weak solution, if exists, is unique.
Proof.
It suffices to consider the case
f
=
ψ
=
ψ
0
= 0, and show any solution
must be zero. Let
v(x, t) =
Z
T
t
e
−λs
u(x, s) ds,
where
λ
is a real number we will pick later. The point of introducing this
e
−λt
is that in general, we do not expect conservation of energy. There could be some
exponential growth in the energy, so want to suppress this.
Then this function belongs to H
1
(U
T
), v = 0 on Σ
T
∪ ∂
∗
U
T
, and
v
t
= −e
−λt
u.
Using the fact that u is a weak solution, we have
Z
U
T
u
t
ue
−λt
−
X
v
tx
j
v
x
i
e
λt
+
X
i
b
i
u
x
i
v + (c −1)uv −vv
t
e
λt
!
dx dt = 0.
Integrating by parts, we can write this as A = B, where
A =
Z
U
T
d
dt
1
2
u
2
e
−λt
−
X
a
ij
v
x
i
v
x
j
e
λt
−
1
2
v
2
e
λt
+
λ
2
u
2
e
−λt
+
X
a
ij
v
x
i
v
x
j
e
λt
+ v
2
e
λt
dx dt
B = −
Z
U
T
e
λt
X
a
ij
v
x
i
v
x
j
−
X
b
i
x
i
uv −
X
b
i
v
x
i
u + (c − 1)uv
dx dt.
Here
A
is the nice bit, which we can control, and
B
is the junk bit, which we
will show that we can absorb elsewhere.
Integrating the time derivative in
A
, using
v
= 0 on Σ
T
and
u
= 0 on Σ
0
, we
have
A = e
λT
Z
Σ
T
1
2
u
2
dx +
Z
Σ
0
X
a
ij
v
x
i
v
x
j
+ v
2
dx
λ
2
Z
U
T
u
2
e
−λt
+
X
a
ij
v
x
i
v
x
j
e
λt
+ v
2
e
λt
dx dt.
Using the uniform ellipticity condition (and the observation that the first line is
always non-negative), we can bound
A ≥
λ
2
Z
U
T
u
2
e
−λt
+ θ|Dv|
2
e
λt
+ v
2
e
λt
dx dt.
Doing some integration by parts, we can also bound
B ≤
c
2
Z
U
T
u
2
e
−λt
+ θ|Dv|
2
e
λt
+ v
2
e
λt
dx dt,
where the constant c does not depend on λ. Taking this together, we have
λ − c
2
Z
U
T
u
2
e
−λt
+ θ|Dv|
2
e
λt
+ v
2
e
λt
dx dt ≤ 0.
Taking
λ > c
, this tells us the integral must vanish. In particular, the integral of
u
2
e
λt
= 0. So u = 0.
We now want to prove the existence of weak solutions. While we didn’t
need to assume much regularity in the uniqueness result, since we are going
to subtract the boundary conditions off anyway, we expect that we need more
regularity to prove existence.
Theorem
(Existence of weak solution)
.
Given
ψ ∈ H
1
0
(
U
) and
ψ
0
∈ L
2
(
U
),
f ∈ L
2
(U
T
), there exists a (unique) weak solution with
kuk
H
1
(U
T
)
≤ C(kψk
H
1
(U)
+ kψ
0
k
L
2
(U)
+ kfk
L
2
(U
T
)
). (†)
Proof.
We use Galerkin’s method. The way we write our equations suggests we
should think of our hyperbolic PDE as a second-order ODE taking values in the
infinite-dimensional space
H
1
0
(
U
). To apply the ODE theorems we know, we
project our equation onto a finite-dimensional subspace, and then take the limit.
First note that by density arguments, we may assume
ψ, ψ
0
∈ C
∞
c
(
U
) and
f ∈ C
∞
c
(U
T
), as long as we prove the estimate (†). So let us do so.
Let
{ϕ
k
}
∞
k=1
be an orthonormal basis for
L
2
(
U
), with
ϕ
k
∈ H
1
0
(
U
). For
example, we can take
ϕ
k
to be eigenfunctions of
−
∆ with Dirichlet boundary
conditions.
We shall consider “solutions” of the form
u
N
(x, t) =
N
X
k=1
u
k
(t)ϕ
k
(x).
We want this to be a solution after projecting to the subspace spanned by
ϕ
1
, . . . , ϕ
N
. Thus, we want (
u
tt
+
Lu − f, ϕ
k
)
L
2
(Σ
t
)
= 0 for all
k
= 1
, . . . , N
.
After some integration by parts, we see that we want
¨u
N
, ϕ
k
L
2
(U)
+
Z
Σ
t
X
a
ij
u
N
x
i
(ϕ
k
)
x
j
+ b
i
u
N
x
i
ϕ
k
+ cu
N
ϕ
k
dx = (f, ϕ
k
)
L
2
(U)
.
(∗)
We also require
u
k
(0) = (ψ, ϕ
k
)
L
2
(U)
˙u
k
(0) = (ψ
0
, ϕ
k
)
L
2
(U)
.
Notice that if we have a genuine solution
u
that can be written as a finite sum
of the ϕ
k
(x), then these must be satisfied.
This is a system of ODEs for the functions
u
k
(
t
), and the RHS is uniformly
C
1
in
t
and linear in the
u
k
’s. By Picard–Lindel¨of, a solution exists for
t ∈
[0
, T
].
So for each
N
, we have an approximate solution that solves the equation
when projected onto
hϕ
1
, . . . , ϕ
N
i
. What we need to do is to extract from this
solution a genuine weak solution. To do so, we need some estimates to show
that the functions u
N
converge.
We multiply (
∗
) by
e
−λt
˙u
k
(
t
), sum over
k
= 1
, . . . , N
, and integrate from 0
to τ ∈ (0, T ), and end up with
Z
τ
0
dt
Z
U
dx
¨u
N
˙u
N
e
−λt
+
X
a
ij
u
N
x
i
˙u
N
x
j
+
X
b
i
u
N
x
i
˙u
N
+ cu
N
˙u
N
e
−λt
=
Z
τ
0
dt
Z
U
du(f ˙u
N
e
−λt
).
As before, we can rearrange this to get A = B, where
A =
Z
U
τ
dt dx
d
dt
1
2
( ˙u
N
)
2
+
1
2
X
a
ij
u
N
x
i
u
N
x
j
+
1
2
(u
N
)
2
e
−λt
+
λ
2
( ˙u
N
)
2
+
X
a
ij
u
N
x
i
u
N
x
j
+ (u
N
)
2
e
−λt
and
B =
Z
U
τ
dt dx
1
2
X
˙a
ij
u
N
x
i
u
N
x
j
−
X
b
i
u
N
x
i
˙u
N
+ (1 − c)u
N
˙u
N
+ f ˙u
N
e
−λt
.
Integrating in time, and estimating as before, for λ sufficiently large, we get
1
2
Z
Σ
τ
( ˙u
N
)
2
+ |Du
N
|
2
dx +
Z
U
τ
( ˙u
N
)
2
+ |Du
N
|
2
+ (u
N
)
2
dx dt
≤ C(kψk
2
H
1
(U)
+ kψ
0
k
2
L
2
(U)
+ kfk
2
U
T
).
This, in particular, tells us u
N
is bounded in H
1
(U
T
),
Since
u
N
(0) =
P
N
n=1
(
ψ, ϕ
k
)
ϕ
k
, we know this tends to
ψ
in
H
1
(
U
). So for
N large enough, we have
ku
N
k
H
1
(Σ
0
)
≤ 2kψk
H
1
(U)
.
Similarly, k˙u
N
k
L
2
(Σ
0
)
≤ 2kψ
0
k
L
2
(U)
.
Thus, we can extract a convergent subsequence
u
N
m
* u
in
H
1
(
U
) for some
u ∈ H
1
(U) such that
kuk
H
1
(U
T
)
≤ C(kψk
H
1
(U)
+ kψk
L
2
(U)
+ kfk
L
2
(U
T
)
).
For convenience, we may relabel the sequence so that in fact u
N
* u.
To check that
u
is a solution, suppose
v
=
P
M
k=1
v
k
(
t
)
ϕ
k
for some
v
k
∈
H
1
((0, T )) with v
k
(T ) = 0. By definition of u
N
, we have
(¨u
N
, v)
L
2
(U)
+
Z
Σ
t
X
i,j
a
ij
u
N
x
i
v
x
j
+
X
i
b
i
u
N
x
i
v + cuv dx = (f, v)
L
2
(U)
.
Integrating
R
T
0
dt using v(T ) = 0, we have
Z
U
T
−u
N
t
v
t
+
X
x
i
N
v
x
j
+
X
b
i
u
N
x
i
v + cuv
dx dt −
Z
Σ
0
u
N
t
v dx
=
Z
U
T
fv dx dt.
Now note that if
N > M
, then
R
Σ
0
u
N
t
v
d
x
=
R
Σ
0
ψ
0
v
d
x
. Now, passing to the
weak limit, we have
Z
U
T
−u
t
v
t
+
X
a
ij
u
x
i
v
x
j
+
X
b
i
u
x
i
v + cuv
dx dt −
Z
Σ
0
ψ
0
v dx
=
Z
U
T
fv dx dt.
So u
t
satisfies the identity required for u to be a weak solution.
Now for
k
= 1
, . . . , M
, the map
w ∈ H
1
(
U
T
)
7→
R
Σ
0
wϕ
k
d
x
is a bounded
linear map, since the trace is bounded in L
2
. So we conclude that
Z
Σ
0
uϕ
k
dx = lim
N→∞
Z
Σ
0
u
N
ϕ
k
dx = (ψ, ϕ
k
)
L
2
(H)
.
Since this is true for all
ϕ
k
, it follows that
u|
Σ
0
=
ψ
, and
v
of the form considered
are dense in H
1
(U
T
) with v = 0 on ∂
∗
U
T
∪ Σ
T
. So we are done.
In fact, we have
ess sup
t∈(0,T )
(k˙uk
L
2
(Σ
t
)
+ kuk
H
1
(Σ
t
)
) ≤ C · (data).
So we can say u ∈ L
∞
((0, T ), H
1
(U)) and ˙u ∈ L
∞
((0, T ), L
2
(U)).
We would like to improve the regularity of the solution. To motivate how we
are going to do that, let’s go back to the wave equation for a bit.
Suppose that in fact
u ∈ C
∞
(
U
T
) is a smooth solution to the wave equation
with initial conditions (
ψ, ψ
0
). We want a quantitative estimate for
u ∈ H
2
(Σ
t
).
The idea is to differentiate the equation with respect to
t
. Writing
w
=
u
t
, we
get
w
tt
− ∆w = 0
w|
Σ
0
= ψ
0
w
t
|
Σ
0
= ∆ψ
w|
∂
∗
U
T
= 0.
By the energy estimate we have for the wave equation, we get
kw
t
k
L
2
(Σ
t
)
+ kwk
H
1
(Σ
t
)
≤ C(kψ
0
k
H
1
(U)
+ k∆ψk
L
2
(U)
)
≤ C(kψ
0
k
H
1
(U)
+ kψk
H
2
(U)
).
So we now have control of
u
tt
and
u
tx
i
in
L
2
(Σ
t
). But once we know that
u
tt
is controlled in
L
2
, then we can use the elliptic estimate to gain control on the
second-order spatial derivatives of u. So
kuk
H
2
(Σ
t
)
≤ C(k∆uk
L
2
(Σ
t
)
) = Cku
tt
k
L
2
(Σ
t
)
.
So we control all second-derivatives of u in terms of the data.
Theorem.
If
a
ij
, b
i
, c ∈ C
2
(
U
T
) and
∂U ∈ C
2
, then for
ψ ∈ H
2
(
U
) and
ψ
0
∈ H
1
0
(U), and f, f
t
∈ L
2
(U
T
), we have
u ∈ H
2
(U
T
) ∩ L
∞
((0, T ); H
2
(U))
u
t
∈ L
∞
((0, T ), H
1
0
(U))
u
tt
∈ L
∞
((0, T ); L
2
(U))
Proof.
We return to the Galerkin approximation. Now by assumption, we have
a linear system with
C
2
coefficients. So
u
k
∈ C
3
((0
, T
)). Differentiating with
respect to t (assuming as we can f, f
t
∈ C
0
(
¯
U
T
)), we have
(∂
3
t
u
N
, ϕ
k
)
L
2
(U)
+
Z
Σ
t
X
a
ij
˙u
N
x
i
(ϕ
k
)
x
j
+
X
b
i
˙u
N
x
i
ϕ
k
+ c ˙u
N
ϕ
k
dx
= (
˙
f, ϕ
k
)
L
2
(U)
−
Z
Σ
t
X
˙a
ij
u
N
x
i
(ϕ
k
)
x
j
+
X
˙
b
i
u
N
x
i
ϕ
k
+ ˙cuϕ
k
dx.
Multiplying by
¨u
k
e
−λt
, summing
k
= 1
, . . . , N
, integrating
R
τ
0
d
t
, and recalling
we already control u ∈ H
1
(U
T
), we get
sup
t∈(0,T )
(ku
N
t
k
H
1
(Σ
t
)
+ ku
N
tt
k
L
2
(Σ
t
)
+ ku
N
t
k
H
2
(U
T
)
)
≤ C
ku
N
t
k
H
1
(Σ
0
)
+ ku
N
tt
k
L
2
(Σ
0
)
+ kψk
H
1
(Σ
0
)
+ kψ
0
k
L
2
(Σ
0
)
+ kfk
L
2
(U
T
)
+ kf
t
k
L
2
(U
T
)
.
We know
u
N
t
|
t=0
=
N
X
k=1
(ψ
0
, ϕ
k
)
L
2
(U)
ϕ
k
.
Since ϕ
k
are a basis for H
1
, we have
ku
N
t
k
H
1
(Σ
0
)
≤ kψ
0
k
H
1
(Σ
0
)
.
To control
u
N
tt
, let us assume for convenience that in fact
ϕ
k
are the eigenfunctions
−∆. From the fact that
(¨u
N
, ϕ
k
)
L
2
(U)
+
Z
Σ
t
X
i,j
a
ij
u
N
x
i
(ϕ
k
)
x
j
+
X
i
b
i
u
N
x
i
ϕ
k
+cu
N
ϕ
k
dx dt = (f, ϕ
k
)
L
2
(U)
,
integrate the first term in the integral by parts, multiply by
¨u
n
, and sum to get
ku
N
tt
k
Σ
0
≤ C(ku
N
k
H
2
(Σ
0
)
+ kfk
L
2
(U
T
)
+ kf
t
k
L
2
(U
T
)
).
We need to control
ku
N
k
H
2
(Σ
0
)
by
kψk
H
2
(Σ
0
)
. Then, using that ∆
ϕ
k
|
∂U
= 0
and u
N
is a finite sum of these ϕ
k
’s,
(∆u
N
, ∆u
N
)
L
2
(Σ
0
)
= (u
N
, ∆
2
u
N
)
L
2
(Σ
0
)
= (ψ, ∆
2
u
N
)
L
2
(Σ
0
)
= (∆ψ, ∆u
N
)
L
2
(Σ
0
)
.
So
ku
N
k
H
2
(Σ
0
)
≤ k∆u
N
k
L
2
(Σ
0
)
≤ Ckψk
2
H
(U).
Passing to the weak limit, we conclude that
u
t
∈ H
1
(U
T
)
u
t
∈ L
∞
((0, T ), H
1
0
(U))
u
tt
∈ L
∞
((0, T ), L
2
(U)).
Since
u
tt
+
Lu
=
f
, by an elliptic estimate on (almost) every constant
t
, we
obtain u ∈ L
∞
((0, T ), H
2
(U)).
We can now understand the equation as holding pointwise almost everywhere
by undoing the integration by parts that gave us the definition of the weak
solution. The initial conditions can also be understood in a trace sense.
Returning to the case
ψ ∈ H
1
0
(
U
) and
ψ
0
∈ L
2
(
U
), by approximating in
H
2
(
U
), by approximating in
H
2
(
U
),
H
1
0
(
U
) respectively, we can show that a
weak solution can be constructed as a strong limit in
H
1
(
U
T
). This implies the
energy identity, so that in fact weak solutions satisfy
u ∈ C
0
((0, T ); H
1
0
(U))
u
t
∈ C
0
((0, T ); L
2
(U))
This requires slightly stronger regularity assumptions on
a
ij
,
b
i
and
c
. Such
solutions are said to be in the energy class.
Finally, note that we can iterate the argument to get higher regularity.
Theorem. If a
ij
, b
i
, c ∈ C
k+1
(
¯
U
T
) and ∂U is C
k+1
, and
∂
i
t
u|
Σ
0
∈ H
1
0
(U) i = 0, . . . , k
∂
k+1
t
u|
Σ
0
∈ L
2
(U)
∂
i
t
f ∈ L
2
((0, T ); H
k−i
(U)) i = 0, . . . , k
then u ∈ H
k+1
(U) and
∂
i
t
u ∈ L
∞
((0, T ); H
k+1−i
(U))
for i = 0, . . . , k + 1.
In particular, if everything is smooth, then we get a smooth solution.
The first two conditions should be understood as conditions on
ψ
and
ψ
0
,
using the fact that the equation allows us to express higher time derivatives
of
u
in terms of lower time derivatives and spatial derivatives. One can check
that these condition imply
ψ ∈ H
k+1
(
U
) and
ψ
0
∈ H
k
(
U
), but the condition we
wrote down also encodes some compatibility conditions, since we know
u
ought
to vanish at the boundary, hence all time derivatives should.
Those were the standard existence and regularity theorems for hyperbolic
PDEs. However, there are more things to say about hyperbolic equations. The
“physicist’s version” of the wave equation involves a constant c, and says
¨u − c
2
∆x = 0.
This constant
c
is the speed of propagation. This tells us in the wave equation,
information propagates at a speed of at most
c
. We can see this very concretely
in the 1-dimensional wave equation, where d’Alembert wrote down an explicit
solution to the wave equation given by
u(x, t) =
1
2
(ψ(x −ct) + ψ(x + ct)) +
1
2c
Z
x+ct
x−ct
ψ
0
(y) dy.
Thus, we see that the value of
φ
at any point (
x, t
) is completely determined by
the values of ψ and ψ
0
in the interval [x − ct, x + ct].
(x, t)
t = 0
This is true for a general hyperbolic PDE. In this case, the speed of propa-
gation should be measured by the principal symbol
Q
(
ξ
) =
P
a
ij
(
y
)
ξ
i
ξ
j
. The
correct way to formulate this result is as follows:
Let
S
0
⊆ U
be an open set with (say) smooth boundary. Let
τ
:
S
0
→
[0
, T
]
be a smooth function vanishing on ∂S
0
, and define
D = {(t, x) ∈ U
T
: x ∈ S
0
, 0 < t < τ(x)}
S
0
= {(τ(x), x) : x ∈ S
0
}.
We say S
0
is spacelike if
n
X
i,j=1
a
ij
τ
x
i
τ
x
j
< 1
for all x ∈ S
0
.
Theorem.
If
u
is a weak solution of the usual thing, and
S
0
is spacelike, then
u|
D
depends only on ψ|
S
0
, ψ
0
|
S
0
and f|
D
.
The proof is rather similar to the proof of uniqueness of solutions.
Proof. Returning to the definition of a weak solution, we have
Z
U
T
−u
t
v
t
+
n
X
i,j=1
a
ij
u
x
j
v
x
i
+
n
X
i=1
b
i
u
x
i
+ cuv dx dt −
Z
Σ
0
ψ
0
v dx =
Z
U
T
fv dx dt.
By linearity it suffices to show that if
u|
Σ
0
= 0 if
ψ|
S
0
=
ψ
0
|
S
0
= 0 and
f|
D
= 0.
We take as test function
v(t, x) =
(
R
τ(x)
t
e
−λs
u(s, x) ds (t, x) ∈ D
0 (t, x) 6∈ D
.
One checks that this is in H
1
(U
T
), and v = 0 on Σ
T
∪ ∂
∗
U
T
with
v
x
i
= τ
x
i
e
−λτ
u(x, τ) +
Z
τ(x)
t
e
−λs
u
x
i
(x, s) ds
v
t
= −e
−λt
u(x, t).
Plugging these into the definition of a weak solution, we argue as in the previous
uniqueness proof. Then
Z
D
d
dt
1
2
u
2
e
−λt
−
1
2
X
a
ij
v
x
i
v
x
j
e
λt
−
1
2
v
2
e
λt
+
λ
2
u
2
e
−λt
+
X
a
ij
v
x
i
v
x
j
e
λt
+ v
2
e
λt
dx dt
=
Z
D
1
2
X
a
ij
v
x
i
v
x
j
e
λt
−
X
b
i
v
x
i
v −(c −1)uv
dx dt
Noting that
R
D
d
x
d
t
=
R
S
0
d
x
R
τ(x)
0
d
t
, we can perform the
t
integral of the
d
dt
term, and we get contribution from S
0
which is given by
I
S
0
=
Z
S
0
1
2
u
2
(τ(x), x)e
−λτ(x)
−
1
2
X
i,j
a
ij
τ
x
i
τ
x
j
u
2
e
−λτ
dx
We have used
v
= 0 on
S
0
and
v
x
i
=
τ
x
i
ue
−λτ
. Using the definition of a spacelike
surface, we have
I
S
0
>
0. The rest of the argument of the uniqueness of solutions
goes through to conclude that u = 0 on D.
This implies no signal can travel faster than a certain speed. In particular, if
X
i,j
a
ij
ξ
i
ξ
j
≤ µ|ξ|
k
for some
µ
, then no signal can travel faster than
√
µ
. This allows us to solve
hyperbolic equations on unbounded domains by restricting to bounded domains.