Alinhac S. Hyperbolic Partial Differential Equations
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Chapter 7
Variable Coefficient Wave
Equations and Systems
7.1 What is a Wave Equation?
Consider, in R
n
x
× R
t
, a second order partial differential operator of the
form
L ≡ ∂
2
t
+2Σb
i
(x, t)∂
2
it
− Σa
ij
(x, t)∂
2
ij
+ L
1
,a
ij
= a
ji
,
where all coefficients are real and C
∞
,andL
1
is a first order operator. We
would like L to be an operator similar to the wave operator, and to enjoy
the same properties: Finite speed of propagation, energy inequalities, etc.
We saw in Chapter 2 that, for an operator in the plane, it is natural to
require that its principal part should be the principal part of a product
a real vector fields. Here, suppose first that L is homogeneous (that is,
L
1
≡ 0) with constant coefficients. For any ξ ∈ R
n
, we can construct an
operator L
ξ
by letting L act on functions of t and s = x · ξ only:
L(v(x · ξ, t)) = [∂
2
t
+2(b · ξ)∂
2
st
− (Σa
ij
ξ
i
ξ
j
)∂
2
s
]v(s, t)=[L
ξ
v](x · ξ, t).
The requirement that L
ξ
should be strictly hyperbolic as an operator in
the plane (s, t)means
δ =(b · ξ)
2
+Σa
ij
ξ
i
ξ
j
> 0,ξ=0.
This will motivate the following definitions.
S. Alinhac, Hyperbolic Partial Differential Equations, Universitext,
DOI 10.1007/978-0-387-87823-2_7, © Springer Science+Business Media, LLC 2009
112 Chapter 7 Variable Coefficient Wave Equations and Systems
Definition 7.1. We call L hyperbolic (with respect to t)inaregion
Ω ⊂ R
n
x
× R
t
if, for all (x, t) ∈ Ω and all ξ ∈ R
n
,
δ =(b(x, t) · ξ)
2
+Σa
ij
(x, t)ξ
i
ξ
j
≥ 0.
In other words, L is hyperbolic if the roots −λ
k
(x, t, ξ) (k =1, 2)inτ of
the characteristic equation
τ
2
+2Σb
i
(x, t)τξ
i
− Σa
ij
(x, t)ξ
i
ξ
j
=0
are real. The operator L is strictly hyperbolic if these roots are real and
distinct for ξ =0.Theλ
k
are the “characteristic speeds” of the equation.
Note that when no cross terms (that is, terms like b
i
∂
2
it
) are present, the
strict hyperbolicity of L is easily seen: this just means that the quadratic
form Σa
ij
ξ
i
ξ
j
is positive definite. In general, we can get rid (at least locally)
of cross terms by the following procedure: let us write (with new first order
terms L
1
)
L =(∂
t
+Σb
i
∂
i
)
2
− Σ[a
ij
+ b
i
b
j
]∂
2
ij
+ L
1
.
If, in the region under consideration, we can perform a change of variables
X
1
= φ
1
(x, t),..., X
n
= φ
n
(x, t),T= t
in such a way that the vector field ∂
t
+Σb
i
∂
i
becomes ∂
T
(see Chapter 1,
Section 1.8), then the operator L takes the form
¯
L = ∂
2
T
− Σ¯a
ij
∂
2
X
i
X
j
+
¯
L
1
,
for some new coefficients ¯a
ij
and lower order terms
¯
L
1
.
From now on, we always tacitly assume that L is strictly hyperbolic in the
region we consider.
7.2 Energy Inequality for the Wave Equation
We consider a domain D of the closed half-space R
n
x
× [0, ∞[exactlyas
in Chapter 6, Section 6.3. The following theorem gives an extension of
the standard energy inequality in the case of a variable coefficients wave
equation.
Theorem 7.2. Let L = ∂
2
t
− Σa
ij
(x, t)∂
2
ij
be a strictly hyperbolic wave
equation. Assume that there exists 0 <α
0
≤ 1 such that, for all (x, t) ∈ D,
ξ ∈ R
n
,
Σa
ij
(x, t)ξ
i
ξ
j
≥ α
0
|ξ|
2
.
7.2 Energy Inequality for the Wave Equation 113
Assume that the outward normal N =(N
1
,...,N
n
,N
0
) on the upper part
of ∂D satisfies
N
2
0
≥ Σa
ij
N
i
N
j
.
Then, for all u ∈ C
2
(
¯
D) sufficiently decaying as |x|→+∞,
max
0≤s≤t
E
u
(s)
1/2
≤ [E
u
(0)
1/2
+
√
2
t
0
||f(·,s)||
L
2
ds]exp
2
α
0
t
0
A(s)ds
.
Here, f = Lu, the energy E
u
(t) is defined as
E
u
(t)=
1
2
Σ
t
[(∂
t
u)
2
+Σa
ij
(∂
i
u)(∂
j
u)](x, t)dx,
and the amplification factor A is
A(t)=max
x
[Σ|∂
t
a
ij
| +Σ|∂
k
a
ij
|](x, t).
Proof: The steps of the proof are exactly the same as for the constant
coefficients case. Once again, we restrict ourselves to the case of a real
function u.
Step 1. Establishing a differential identity. We write the product (Lu)(∂
t
u)
as usual:
(∂
2
t
u)(∂
t
u)=
1
2
∂
t
[(∂
t
u)
2
],
(a
ij
∂
2
ij
u)(∂
t
u)=∂
i
[a
ij
(∂
j
u)(∂
t
u)] − a
ij
(∂
j
u)(∂
2
ti
u) − (∂
i
a
ij
)(∂
t
u)(∂
j
u).
To handle the second term above, we use the symmetry of a
ij
= a
ji
:
a
ij
(∂
j
u)(∂
2
ti
u)=∂
t
[a
ij
(∂
i
u)(∂
j
u)] − (∂
t
a
ij
)(∂
i
u)(∂
j
u) − a
ij
(∂
i
u)(∂
2
tj
u).
Summing over i, j gives us
2Σa
ij
(∂
j
u)(∂
2
ti
u)=∂
t
[Σa
ij
(∂
i
u)(∂
j
u)] − Σ(∂
t
a
ij
)(∂
i
u)(∂
j
u).
Gathering the terms, we obtain
2(Lu)(∂
t
u)=∂
t
[(∂
t
u)
2
+Σa
ij
(∂
i
u)(∂
j
u)] + Σ∂
i
[−2Σa
ij
(∂
j
u)(∂
t
u)] + Q,
Q = −Σ(∂
t
a
ij
)(∂
i
u)(∂
j
u)+2Σ(∂
i
a
ij
)(∂
j
u)(∂
t
u).
Step 2. Integration on the domain. We integrate in the truncated domain
D
t
using Stokes formula:
D
t
(Lu)(∂
t
u)dxds = E
u
(t) − E
u
(0) +
Λ
t
Idσ +
D
t
Qdxds.
114 Chapter 7 Variable Coefficient Wave Equations and Systems
The boundary integrand I is
2I = N
0
[(∂
t
u)
2
+Σa
ij
(∂
i
u)(∂
j
u)] − 2(∂
t
u)Σa
ij
N
i
∂
j
u.
Introducing the vector fields
¯
∂
i
= ∂
i
−
N
0
N
i
∂
t
,
which span the tangent plane to ∂D,wecanwrite
I = N
0
Σa
ij
¯
∂
i
u
¯
∂
j
u + N
−1
0
(∂
t
u)
2
(N
2
0
− Σa
ij
N
i
N
j
).
The assumptions of the theorem imply I ≥ 0.
Step 3. Handling the remainder terms. We handle the term
(Lu)(∂
t
u)dxds
exactly as before:
(Lu)(∂
t
u)dxds
≤
√
2
t
0
||f(·,s)||
L
2
E
u
(s)
1/2
ds.
It remains to deal with the “error term”
D
t
Qdxds.Wehave
|Q|≤2A[(∂
t
u)
2
+Σ(∂
i
u)
2
] ≤ 2
A
α
0
[(∂
t
u)
2
+Σa
ij
(∂
i
u)(∂
j
u)],
hence by integration
D
t
Qdxds
≤
2
α
0
t
0
A(s)E
u
(s)ds.
We want to apply the Gronwall lemma (see Chapter 2, Lemma 2.16) to
the inequality we have obtained so far:
E(t) ≤ E(0) +
√
2
t
0
||f(·,s)||
L
2
E(s)
1/2
ds +
2
α
0
t
0
A(s)E(s)ds.
We use the same trick as before: For t
≤ t<T,wewrite
E(t
) ≤ E(0) +
√
2
t
0
||f(·,s)||
L
2
E(s)
1/2
ds +
2
α
0
t
0
A(s)E(s)ds.
Applying now the lemma, we get
max
0≤t
≤t
E(t
) ≤ [E(0) +
√
2
t
0
||f(·,s)||
L
2
E(s)
1/2
ds]exp
2
α
0
t
0
A(s)ds
.
This implies immediately the conclusion.
7.3 Symmetric Systems 115
7.3 Symmetric Systems
7.3.1 Definitions and Examples
Just as in Chapter 2, it is useful to consider also first order N ×N systems
L = S(x, t)∂
t
+ΣA
i
(x, t)∂
i
+ B(x, t).
Here, S, A
i
,andB are C
∞
N × N matrices, and the operator acts on
functions
u : R
n
x
× R
t
⊃ Ω → C
N
.
We will always assume S to be invertible. To define hyperbolicity, we
can follow the same path as in Section 1: Assuming B = 0 and constant
coefficients, we let L act on functions of t and s = x ·ξ only; we thus define
asystemL
ξ
L
ξ
v(s, t)=S∂
t
v +(ΣA
i
ξ
i
)∂
s
v.
For this system to be hyperbolic in the (s, t)-plane, we require S
−1
ΣA
i
ξ
i
to have only real eigenvalues. This motivates the following definitions.
Definition 7.3. We call L hyperbolic (with respect to t) in the region
Ω ⊂ R
n
x
× R
t
if, for all (x, t) ∈ Ω and ξ ∈ R
n
,thematrixS
−1
ΣA
i
ξ
i
has
real eigenvalues. The operator L is strictly hyperbolic if these eigenvalues
are also distinct for ξ =0.
Definition 7.4. The system L is symmetric if the matrices S and A
i
are
hermitian. The operator is symmetric hyperbolic if, moreover, S is positive
definite.
We will see below why symmetry is an essential feature to obtain energy
inequalities. Let us comment about the concept of symmetric hyperbolic
system: If S = id, symmetry clearly implies hyperbolicity, since an her-
mitian matrix has real eigenvalues. More generally, let X =0suchthat
S
−1
AX = λX: This implies
t
¯
XAX = λ
t
¯
XSX.
Since
t
¯
XSX > 0, λ is the quotient of two real quantities, hence λ is real.
In other words, a symmetric hyperbolic operator is actually hyperbolic.
Note that an operator may be symmetric hyperbolic without being strictly
hyperbolic, a fact which is important in applications.
Example 7.5. Just as in Chapter 2, we can reduce a variable coefficients
wave equation
P = ∂
2
t
− Σa
ij
∂
2
ij
116 Chapter 7 Variable Coefficient Wave Equations and Systems
to a first order symmetric system. To do this, set u
0
= ∂
t
u, u
i
= ∂
i
u;then
the n +1 equations
∂
t
u
0
=Σa
ij
∂
j
u
i
, Σ
j
a
ij
∂
t
u
j
=Σ
j
a
ij
∂
j
u
0
form a symmetric system L for which
S
00
=1,S
0i
=0,S
ij
= a
ij
.
Note that the condition a>>0 (this notation meaning “positive defi-
nite”) of strict hyperbolicity of P is equivalent to the condition S>>0
which makes L symmetric hyperbolic.
Example 7.6. The nonlinear system of compressible isentropic Euler
equations is
∂
t
ρ + div(ρu)=0,∂
t
u
i
+(Σu
j
∂
j
)u
i
+ ρ
−1
∂
i
P =0,i=1,...,n.
Here, ρ>0 is the density of the fluid, u(x, t) ∈ R
n
is its velocity as
observed at time t at the point x,andP is the pressure, considered here
as a given function of ρ (depending on the particular physical fluid we are
considering), of the form, say,
P = P (ρ)=Aρ
γ
,A>0,γ>1.
If we introduce the new unknown function c instead of ρ,
c = αρ
(γ−1)/2
,
α(γ − 1)
2
=(Aγ)
1/2
,
we obtain a new system
D
t
c +
c(γ − 1)
2
div u =0,
D
t
u
i
+
c(γ − 1)
2
∂
i
c =0,i=1,...,n,
where D
t
≡ ∂
t
+Σu
i
∂
i
.Foreach¯c given, the linearized system
D
t
c +
¯c(γ − 1)
2
div u =0,D
t
u
i
+
¯c(γ − 1)
2
∂
i
c =0,i=1,...,n,
is symmetric hyperbolic, but not strictly hyperbolic as soon as n ≥ 3.
Example 7.7. The Maxwell system is
∂
t
E +curlB =0,∂
t
B − curlE =0,
where E and B are, respectively, the electric field and the magnetic field.
Considered as a 6 × 6 system with unknown u =(E,B), it is a symmetric
hyperbolic system.
7.3 Symmetric Systems 117
7.3.2 Energy Inequality
Let us consider a domain D ⊂ R
n
x
× [0, ∞[ as in Section 7.2, with outer
normal N =(N
1
,...,N
n
,N
0
). We have the following energy inequality.
Theorem 7.8. Let
L = S∂
t
+ΣA
i
∂
i
+ B
be a symmetric hyperbolic system in
¯
D, and assume, for some constant
α
0
> 0 and all (x, t) ∈ D, X ∈ C
N
,
t
¯
XS(x, t)X ≥ α
0
|X|
2
.
Define the energy E
u
(t) by
E
u
(t)=
1
2
Σ
t
[
t
uSu](x, t)dx.
Assume that, on the upper part of ∂D, the hermitian matrix N
0
S +ΣN
i
A
i
is nonnegative. Then, for all u ∈ C
1
(
¯
D) sufficiently decaying when |x|→
+∞,
max
0≤s≤t
E
u
(s)
1/2
≤
E
u
(0)
1/2
+
2
α
0
1/2
t
0
||f(·,s)||
L
2
ds
exp
2
α
0
t
0
A(s)ds
.
Here, the amplification factor A is
A(t)=max
x
||C(x, t)||,C=
B +
t
¯
B
2
−
1
2
(∂
t
S +Σ∂
i
A
i
).
Proof: The steps of the proof are again essentially the same as before.
For simplicity, we assume that S, A
i
, B,andu are real.
Step 1. Establishing a differential identity. We write
t
u(Lu)asasumof
terms in divergence form and quadratic terms in u; Using the symmetry
of S and A
i
we obtain
t
uS∂
t
u =
1
2
∂
t
[
t
uSu] −
1
2
t
u(∂
t
S)u,
t
uA
i
∂
i
u =
1
2
∂
i
[
t
uA
i
u] −
1
2
t
u(∂
i
A
i
)u.
Gathering the terms,
2
t
uLu = ∂
t
[
t
uSu]+Σ∂
i
[
t
uA
i
u]+2
t
uCu, C =
B +
t
B
2
−
1
2
(∂
t
S +Σ∂
i
A
i
).
118 Chapter 7 Variable Coefficient Wave Equations and Systems
Step 2. Integration over the domain. Integrating
t
uLu in the domain D
t
,
we obtain using Stokes formula
D
t
t
uLudxds = E
u
(t) − E
u
(0) +
Λ
t
Idσ +
D
t
t
uCudxds.
The lateral boundary terms integrand I is given by
2I =
t
u(N
0
S +ΣN
i
A
i
)u.
The assumption of the theorem implies I ≥ 0.
Step 3. Handling the remainder term. We handle now the two remaining
integrals on D
t
.Since
|
t
uCu|≤||C|||u|
2
≤
||C||
α
0
(
t
uSu),
D
t
t
uCudxds
≤
2
α
0
t
0
A(s)E(s)ds.
Also, as usual,
D
t
t
uLudxds
≤
2
α
0
1/2
t
0
||f(·,s)||
L
2
E(s)
1/2
ds.
We have obtained so far the inequality
E(t) ≤ E(0) +
2
α
0
1/2
t
0
||f(·,s)||
L
2
E(s)
1/2
ds +
2
α
0
t
0
A(s)E(s)ds.
The end of the proof is exactly similar to that for Theorem 7.2.
7.4 Finite Speed of Propagation
The two theorems of Sections 7.2, 7.3 have important corollaries, which
display the finite speed of propagation property for hyperbolic wave equa-
tions or symmetric hyperbolic systems. We give here the simplest form of
this corollary for the wave equation.
Theorem 7.9. Let us consider in R
n
x
×[0, ∞[ a hyperbolic wave equation
Lu = ∂
2
t
u − Σa
ij
∂
2
ij
u + a
0
∂
t
u +Σa
i
∂
i
u + bu.
7.4 Finite Speed of Propagation 119
Assume that the coefficients a
ij
are C
1
, and that, for some constant
λ
0
> 0 and all ξ ∈ R
n
,ξ=0, they satisfy everywhere
0 < Σa
ij
(x, t)ξ
i
ξ
j
≤ λ
2
0
|ξ|
2
.
Assume that the other coefficients a
0
, a
i
, b are bounded. Consider u ∈ C
2
(R
n
x
× [0, ∞[) with
Lu = f, u(x, 0) = u
0
(x), (∂
t
u)(x, 0) = u
1
(x).
Assume that u
0
, u
1
vanish for |x|≥M , and that f vanishes for
|x|≥M + λ
0
t.Thenu vanishes for |x|≥M + λ
0
t.
Proof: The proof is a typical application of the energy inequality of
Theorem 7.2: In this theorem, we assume L to be homogeneous of order
two. Here, we assume that lower order terms are present. Let R>M and
consider the (rotationally invariant) domain D
R
defined by
D
R
= {(x, t),t≥ 0,M + λ
0
t ≤ r = |x|≤2R − M − λ
0
t}.
An outward normal N to the upper boundary of D
R
is given by
N
i
= ±
x
i
|x|
,N
0
= λ
0
,
the sign depending on which point of ∂D we are considering. In all cases,
N satisfies the condition
N
2
0
= λ
2
0
ΣN
2
i
≥ Σa
ij
N
i
N
j
of Theorem 7.2. Moreover, since D
R
is compact, the assumptions of
Theorem 7.2 are satisfied with some α
0
> 0andsomeA<∞.When
applying Theorem 7.2 to the domain D
R
,wewrite
∂
2
t
u − Σa
ij
∂
2
ij
u = f − a
0
∂
t
u − Σa
i
∂
i
u − bu = g.
Hence we obtain, for some C, the inequality
φ(t) ≡ max
0≤s≤t
E
u
(s)
1/2
≤ C
t
0
||g(·,s)||
L
2
ds.
It is easy to handle the derivatives of u in g :
||a
0
∂
t
u +Σa
i
∂
i
u||
L
2
≤ CE
1/2
u
.
120 Chapter 7 Variable Coefficient Wave Equations and Systems
The control of bu is more delicate, since the energy inequality controls
only the derivatives of u. We simply write in D
R
, with various constants
C depending on R,
u(x, t)=
t
0
(∂
t
u)(x, s)ds, |u(x, t)|
2
≤ C
t
0
(∂
t
u)
2
(x, s)ds,
Σ
t
u
2
(x, t)dx ≤ C
t
0
ds
Σ
t
(∂
t
u)
2
(x, s)dx ≤ C
t
0
E(s)ds ≤ C max
0≤s≤t
E(s).
Finally, the inequality gives us, since f =0inD
R
,
φ(t) ≤ C
t
0
E(s)
1/2
+ C
t
0
φ(s)ds ≤ C
t
0
φ(s)ds.
Applying the Gronwall lemma yields finally u =0inD
R
.SinceR is
arbitrary, this completes the proof.
We leave it to the reader as Exercise 7 to establish an analogue of this
theorem for systems.
7.5 Klainerman’s Method
Klainerman’s method is an “energy inequality method” which allows one,
in global situations, to obtain pointwise estimates of solutions to vari-
able coefficients wave equations. The denomination “energy method” is
chosen to emphasize the fact that no attempt is made to represent the
solution by some formula, in contrast with the parametrix methods (see
section 7.7). Not only this method gives a good qualitative information on
the behavior of the solutions, but it is an essential tool in studying nonlinear
perturbations of the wave equation.
Consider a variable coefficients wave equation L in R
n
x
×R
t
. To emphasize
the fact that L is a perturbation of the wave equation, we write, with x
0
= t,
L = ∂
2
t
− Δ
x
+Σ
0≤i,j≤n
a
ij
(x, t)∂
2
ij
,
where the coefficients a
ij
are supposed to be “small” (see step 2 below).
There are three steps in the method:
Step 1. Commuting Lorentz fields. Let k be an integer at most equal to
(n +2)/2, and compute
Z
k
Lu =0=[Z
k
,L]u + L(Z
k
u).