Alinhac S. Hyperbolic Partial Differential Equations
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6.4 General Multipliers 91
is called the energy of u at time t, and we have
∂D
t
edσ = E
u
(t) − E
u
(0) +
Λ
t
edσ.
i) In the special case where N =(0, 1) (the flat part of ∂D
t
), we obtain
e = X
0
[Σ(∂
i
u)
2
+(∂
t
u)
2
]+2(∂
t
u)A,
e = X
0
Σ[∂
i
u +
X
i
X
0
∂
t
u]
2
+ X
−1
0
(∂
t
u)
2
(X
2
0
− ΣX
2
i
).
Hence, if X is nonspacelike, the quadratic form e is nonnegative, and so
is the energy.
ii) On the other hand, if X = ∂
t
, on the lateral boundary Λ
t
of D
t
,
e = N
0
[Σ(∂
i
u)
2
+(∂
t
u)
2
] − 2(∂
t
u)B,
and we proved in Theorem 6.5 that it is nonnegative if N is nonspacelike.
The following theorem summarizes the interplay between the choice of X
and the geometry of ∂D.
Theorem 6.7. Assume that, on the upper part of ∂D, N and X are
nonspacelike. Then
D
t
(u)(Xu)dxds = E
u
(t) − E
u
(0) +
Λ
t
edσ +
D
t
Qdxds,
where E
u
(t) and e are nonnegative.
The proof of the general case is a tedious but straightforward computation
that we leave to the reader. One can also find in Exercise 13 a more
geometric proof.
This theorem leaves of course open the question about the sign of Q,which
is a delicate question in general. We saw in Section 6.2 a first example,
where X = e
b(r−t)
∂
t
generates a nonnegative Q. The next section is devoted
to a second example where X = ∂
r
, generating an “almost nonnegative ” Q,
and a third example will be seen in Section 6.7 where X =(r
2
+t
2
)∂
t
+2tr∂
r
also generates an “almost nonnegative ” Q. Other examples are given in
Exercise 4 and Exercises 7, 8, 10, 12 of Chapter 7.
92 Chapter 6 Energy Inequalities for the Wave Equation
6.5 Morawetz Inequality
In constrast with Sections 6.1-6.4 where we use timelike multipliers to
obtain energy inequalities, the computation below due to Morawetz uses
the spacelike multiplier X = ∂
r
. Of course, in this case, the energy on hor-
izontal slabs {t = T } will not be positive, but interesting quadratic terms
Q in ∂u will appear. This inequality can only be used when coupled with
an energy inequality giving a control of E
u
(t).
Theorem 6.8. Let u ∈ C
2
(R
3
x
× [0,T[) be sufficiently decaying when
|x|→∞and satisfy u =0. Then for all 0 ≤ t<T,
4π
t
0
u
2
(0,s)ds +
S
t
r
−1
[Σ(∂
i
u)
2
− (∂
r
u)
2
]dxds ≤ 4E
u
.
Note that Σ(∂
i
u)
2
− (∂
r
u)
2
= |(R/r)u|
2
.
We give a detailed proof of this theorem as an exercise in the technique
just described in Sections 6.1–6.4, following the same three Steps as before.
Proof:
Step 1. Establishing a differential identity. We write 2(u)(∂
r
u)asasum
of terms in divergence form plus a sum Q of quadratic terms in ∂
t
u, ∂
i
u,
just as in Sections 6.2 or 6.4. More precisely,
(∂
2
t
u)(∂
r
u)=(∂
2
t
u)(Σω
i
∂
i
u)
= ∂
t
[(∂
t
u)(∂
r
u)] −
1
2
Σ∂
i
[ω
i
(∂
t
u)
2
]+
1
2
(∂
t
u)
2
Σ∂
i
ω
i
,
(∂
2
j
u)(∂
r
u)=∂
j
[(∂
j
u)(∂
r
u)] −
1
2
Σ∂
i
[ω
i
(∂
j
u)
2
] − Σ
i
(∂
j
u)(∂
i
u)(∂
j
ω
i
)
+
1
2
(∂
j
u)
2
Σ∂
i
ω
i
.
Since
∂
j
(ω
i
)=r
−1
(δ
ij
− ω
i
ω
j
), Σ∂
i
ω
i
=
2
r
,
Σ(∂
i
u)(∂
j
u)(∂
j
ω
i
)=r
−1
[Σ(∂
i
u)
2
− (∂
r
u)
2
],
we obtain, gathering the terms,
2(u)(∂
r
u)=∂
t
[2(∂
t
u)(∂
r
u)] + Σ∂
i
{ω
i
[Σ(∂
j
u)
2
− (∂
t
u)
2
] − 2(∂
i
u)(∂
r
u)}+ Q,
Q =
2
r
[Σ(∂
i
u)
2
− (∂
r
u)
2
]+
2
r
[(∂
t
u)
2
− Σ(∂
i
u)
2
].
6.5 Morawetz Inequality 93
We note here a special feature of great importance: the sum Q of the
quadratic terms in ∂u contains two different terms. The first term is
nonnegative and gives a control of (R/r)u. The second term is an
expression with no special sign, but which is of the form
a[(∂
t
u)
2
− Σ(∂
i
u)
2
],
with a =2/r. The following lemma allows us to further transform this
expression.
Lemma 6.9. The following identity holds:
au(u)=∂
t
[au∂
t
u −
1
2
(∂
t
a)u
2
] − Σ∂
i
[au∂
i
u −
1
2
(∂
i
a)u
2
] − a[(∂
t
u)
2
− Σ(∂
i
u)
2
]+
u
2
2
a.
We skip the straightforward proof, which uses the by now familiar tech-
nique of step 1. This lemma suggests considering the product
2(u)
∂
r
u +
u
r
,
in which the bad quadratic terms cancel.
Step 2. Integration over the domain. Since we introduced the singular
term u/r,weconsidernowtheintegralof2(u)(∂
r
u +
u
r
)notinthefull
strip S
t
, but in the domain
{(x, s), 0 ≤ s ≤ t, |x|≥>0},
which is the exterior of a thin cylinder. The upper and lower boundary
terms are the integrals
|x|≥
dx
∂
r
u +
u
r
(∂
t
u)
taken at time t and 0, respectively. The lateral boundary term on the
cylinder |x| = with outer normal N =(−ω, 0) is
−1
{[(∂
t
u)
2
− Σ(∂
i
u)
2
]+2(∂
r
u)
2
+2u∂
r
u +
−1
u
2
}
2
dtdσ
S
2
,
where dσ
S
2
is the area element on the unit sphere of R
3
. Letting go to
zero, this last term goes to
4π
t
0
u
2
(0,s)ds.
94 Chapter 6 Energy Inequalities for the Wave Equation
Since u = 0, there is no step 3 here. Since the multiplier X = ∂
r
is
spacelike, the upper and lower boundary integrals have no special sign, and
we have to control them. We write
2
(∂
r
u +
u
r
)(∂
t
u)dx
≤
(∂
t
u)
2
+
∂
r
u +
u
r
2
dx,
and expand the last term. For the double product, we integrate by parts
2
(∂
r
u)
u
r
dx =2
(∂
r
u)urdrdω =
∂
r
(u
2
)rdrdω = −
u
2
r
2
dx,
hence finally
2
(∂
r
u +
u
r
)(∂
t
u)dx
≤
[(∂
t
u)
2
+(∂
r
u)
2
]dx ≤ 2E
u
,
which completes the proof.
6.6 KSS Inequality
The following inequality is named after its authors M. Keel, H. Smith and
C. Sogge (See Notes).
Theorem 6.10. There exists C such that, for all u ∈ C
2
(R
3
x
× [0,T[)
sufficiently decaying when |x|→+∞,andt<T,
[log(2 + t)]
−1/2
S
t
(1 + r)
−1
|∂u|
2
(x, s)dxds
1/2
≤ CE
1/2
u
(0) + C
t
0
||(u)(·,s)||
L
2
ds.
Proof: We only handle the case u = f with u(x, 0) = 0, (∂
t
u)(x, 0) = 0,
the homogeneous case being handled similarly. In contrast with the pre-
ceding inequalities, the proof here does not involve a multiplier method
and integrations by parts. It uses only the standard energy inequality,
combined with three new ideas, which are truncation, scaling, and
dyadic decomposition.
Step 1. The aim of this first step is to prove the inequality
t
0
|x|≤2
|∂u|
2
dxds
1/2
≤ C
t
0
||(u)(·,s)||
L
2
ds.
6.6 KSS Inequality 95
Note that this is a part of the inequality of Theorem 6.10 which is slightly
better than the general inequality, since the factor [log(2 + t)]
−1/2
is not
there. The idea is to use a clever truncation process jointly with the
strong Huygens principle. We define the disjoint regions R
k
in
the (x, t)spaceby
R
k
= {(x, t),k≤|x| + t<k+1,k=0, 1,...}
and define f
k
to be f in R
k
and zero otherwise, thus f =Σf
k
. Let then v
k
be the solution of the Cauchy problem
v
k
= f
k
,v
k
(x, 0) = 0, (∂
t
v
k
)(x, 0) = 0.
Obviously, u =Σv
k
. The function v
k
is zero for t + |x| <kbecause the
propagation speed is less than 1, and is also zero for t ≥ k +1+|x| by the
strong Huygens principle. Hence, in the cylinder
|x|≤2,l≤ t ≤ l +1,
v
k
vanishes identically unless |k − l|≤2. This implies, for some constant
C, the inequality
l+1
l
|x|≤2
|∂u|
2
dxdt =
l+1
l
|x|≤2
|Σ
|k−l|≤2
∂v
k
|
2
dxdt
≤ 5Σ
|k−l|≤2
l+1
l
|x|≤2
|∂v
k
|
2
dxdt.
For each k, the standard energy inequality applied to v
k
yields
max
t≤l+1
||∂v
k
(·,t)||
2
L
2
≤ 4
l+1
0
||f
k
(·,t)||
L
2
dt
2
.
Fixing N ≤ t and assuming l ≤ N − 1, we thus obtain
l+1
l
|x|≤2
|∂u|
2
dxdt ≤ CΣ
|k−l|≤2
N
0
||f
k
(·,t)||
L
2
dt
2
.
Summing on l and taking the square roots, this yields
N
0
|x|≤2
|∂u|
2
dxdt
1/2
≤ C
⎡
⎣
Σ
0≤k≤N +1
N
0
||f
k
(·,t)||
L
2
dt
2
⎤
⎦
1/2
.
96 Chapter 6 Energy Inequalities for the Wave Equation
In the righthand side of the above inequality, we recognize the standard
euclidean norm in R
N+2
of
N
0
A(t)dt, where the vector A(t)is
A(t)=(||f
0
(·,t)||
L
2
,...,||f
N+1
(·,t)||
L
2
).
Now
||A(t)|| =(Σ||f
k
(·,t)||
2
L
2
)
1/2
≤||f (·,t)||
L
2
,
since the f
k
have disjoint supports in x for fixed t. The norm of the integral
of A being less than the integral of the norm of A, we obtain finally
N
0
|x|≤2
|∂u|
2
dxdt
1/2
≤ C
N
0
||f(·,t||
L
2
dt.
We remark now that the above inequality holds in fact for any t instead
of N:
t
0
|x|≤2
|∂u|
2
dxds
1/2
≤ C
t
0
||f(·,s)||
L
2
ds.
To see this, let N be the integer part of t; splitting
t
0
|x|≤2
|∂u|
2
dxds =
N
0
+
t
N
≤ C
t
0
||f(·,s)||
L
2
ds
2
+max
N≤s≤t
|x|≤2
|∂u|
2
(x, s)dx
gives the result.
Step 2. The aim of this second step is to take advantage of the homo-
geneity properties of the wave equation to pass from the inequality of
step 1 (controlling the solution in the cylinder |x|≤2) to an inequality in
an annulus
t
0
R≤|x|≤2R
|x|
−1
|∂u|
2
dxds
1/2
≤ C
t
0
||(u)(·,s)||
L
2
ds.
Again, this is a part of the inequality of Theorem 6.10 with no logarithmic
factor. Note that the constant C does not depend on R! The idea is to use
a scaling argument, which is as follows: Let v be any C
2
function with
v = g and zero traces on {t =0}. Then, for any R>0, the function
w
R
(y,τ)=v(Ry, Rτ) is the solution of w
R
= g
R
with traces (0, 0) and
g
R
(y,τ)=R
2
g(Ry, Rτ).
6.6 KSS Inequality 97
We write, using the change of variables x = Ry, s = Rτ,
t
0
R≤|x|≤2R
|x|
−1
|∂v|
2
(x, s)dxds
=
t/R
0
1≤|y|≤2
R
3
|y|
−1
|∂v|
2
(Ry, Rτ)|
2
dydτ
≤ R
t/R
0
|y|≤2
|∂w
R
|
2
(y,τ)dydτ.
We now use for the function w
R
the inequality proved in step 1 and obtain
t/R
0
|y|≤2
|∂w
R
|
2
dydτ ≤ C
t/R
0
||g
R
(·,τ)||
L
2
dτ
2
.
From the definition of g
R
we get
||g
R
(·,τ)||
2
L
2
=
|g
R
(y,τ)|
2
dy = R
g(x, Rτ )
2
dx,
t/R
0
||g
R
(·,τ)||
L
2
dτ = R
−1/2
t
0
||g(·,s)||
L
2
ds.
Hence, finally,
t
0
R≤|x|≤2R
|x|
−1
|∂v|
2
(x, s)dxds ≤ C
t
0
||g(·,s)||
L
2
ds
2
.
Step 3. Dyadic decomposition. To finish the proof, we remark first that
t
0
|x|≥t
(1 + |x|)
−1
|∂u|
2
dxds ≤ C log(2 + t)max
0≤s≤t
E
u
(s),
so we need only prove
t
0
|x|≤t
(1 + |x|)
−1
|∂u|
2
dxds ≤ C log(2 + t)
t
0
||f(·,s)||
L
2
ds
2
.
Let N be the smallest integer such that t ≤ 2
N+1
. Wesplittheintegral
into a sum of integrals on dyadic shells
t
0
|x|≤t
(1 + |x|)
−1
|∂u|
2
dxds ≤
t
0
|x|≤2
N +1
(1 + |x|)
−1
|∂u|
2
dxds
98 Chapter 6 Energy Inequalities for the Wave Equation
=
t
0
|x|≤1
(1 + |x|)
−1
|∂u|
2
dxds
+Σ
0≤k≤N
t
0
2
k
≤|x|≤2
k+1
(1 + |x|)
−1
|∂u|
2
dxds.
Using N + 2 times the inequality of step 2, we have the same bound for
each term, and the result follows.
It is important to understand precisely what Theorem 6.10 says: We have,
for any α>0,
S
t
∩{|x|≥αt}
(1 + r)
−1
|∂u|
2
dxds ≤ C
t
0
ds
1+s
|∂u|
2
dx
≤ C(max
0≤s≤t
E
u
(s)) log(2 + t).
In other words, the standard energy inequality would imply Theorem 6.10
if the integral on S
t
on the lefthand side were replaced by an integral on
S
t
∩{|x|≥αt} only. Thus, Theorem 6.10 provides an improved behavior
of all derivatives of the solution far away inside the light cone (that is, in
regions |x|≤αt, α < 1), while Theorem 6.3 provides an improved behavior
of some special derivatives of the solution close to the boundary of the light
cone (that is, in regions |x|≥αt, α > 0).
6.7 Conformal Inequality
We turn now to an energy inequality obtained, in the spirit of Section 6.4,
using the nonspacelike multiplier
K
0
=(r
2
+ t
2
)∂
t
+2rt∂
r
.
Theorem 6.11. For n ≥ 3 there exists C such that, for all u ∈ C
2
(R
n
x
×
[0,T[), sufficiently decaying when |x|→+∞,andallt<T,
E
c
u
(t)
1/2
≤ CE
c
u
(0)
1/2
+ C
t
0
||g(·,s)||
L
2
ds.
Here, g =(r
2
+ t
2
)
1/2
u, and the conformal energy E
c
is defined by
E
c
(t)=E
c
u
(t)=
1
2
[(Su)
2
+ |Hu|
2
+ |Ru|
2
+ u
2
](x, t)dx.
6.7 Conformal Inequality 99
Recall from Chapter 5 the definitions of the Lorentz fields
S = t∂t + r∂
r
,H
i
= t∂
i
+ x
i
∂
t
,R= x ∧ ∂.
Proof: The first step of the proof is as in Sections 6.1–6.4.
Step 1. Establishing a differential identity. We write (u)(K
0
u)asthe
sum Q of quadratic terms in ∂u and terms in divergence form:
(r
2
+ t
2
)(∂
2
t
u)(∂
t
u)=
1
2
∂
t
[(r
2
+ t
2
)(∂
t
u)
2
] − t(∂
t
u)
2
,
(r
2
+ t
2
)(∂
2
i
u)(∂
t
u)=∂
i
[(r
2
+ t
2
)(∂
i
u)(∂
t
u)] −
1
2
∂
t
[(r
2
+ t
2
)(∂
i
u)
2
]
− 2x
i
(∂
t
u)(∂
i
u)+t(∂
i
u)
2
,
tx
i
(∂
2
t
u)(∂
i
u)=∂
t
[tx
i
(∂
t
u)(∂
i
u)] −
1
2
∂
i
[tx
i
(∂
t
u)
2
]
− x
i
(∂
t
u)(∂
i
u)+
t
2
(∂
t
u)
2
,
tx
i
(∂
2
j
u)(∂
i
u)=∂
j
[tx
i
(∂
i
u)(∂
j
u)] −
1
2
∂
i
[tx
i
(∂
j
u)
2
]
− tδ
ij
(∂
i
u)(∂
j
u)+
t
2
(∂
j
u)
2
.
Gathering the terms, we obtain finally
(u)(K
0
u)=
1
2
∂
t
[(r
2
+ t
2
)((∂
t
u)
2
+Σ(∂
i
u)
2
)+4rt(∂
t
u)(∂
r
u)]
+Σ∂
i
[tx
i
(−(∂
t
u)
2
+Σ(∂
j
u)
2
) − (r
2
+ t
2
)(∂
t
u)(∂
i
u)
− 2rt(∂
r
u)(∂
i
u)] + (n − 1)t[(∂
t
u)
2
− Σ(∂
i
u)
2
].
We see that Q reduces to a multiple of (∂
t
u)
2
− Σ(∂
i
u)
2
. Just as in the
proof of Morawetz inequality, we use Lemma 6.9 to tranform this term :
(n − 1)t[(∂
t
u)
2
− Σ(∂
i
u)
2
]=− (n − 1)tu(u)+∂
t
(n − 1)tu(∂
t
u)
−
(n − 1)
2
u
2
− Σ∂
i
[(n − 1)tu∂
i
u].
Step 2. Integration over the domain. Integrating over the strip S
t
,we
obtain the identity
S
t
(u)(K
0
u +(n − 1)tu)dxds =
˜
E
u
(t) −
˜
E
u
(0),
100 Chapter 6 Energy Inequalities for the Wave Equation
where the modified energy
˜
E is
˜
E
u
(t)=
1
2
{(r
2
+ t
2
)[(∂
t
u)
2
+Σ(∂
i
u)
2
]+4rt(∂
t
u)(∂
r
u)
+2(n − 1)tu(∂
t
u) − (n − 1)u
2
}dx.
This energy is the sum of two different contributions: The first two terms
come from using the multiplier K
0
and are nonnegative, since K
0
is non-
spacelike; this can be checked directly by writing
(r
2
+ t
2
)[(∂
t
u)
2
+(∂
r
u)
2
+ |
R
ra
u|
2
]+4rt(∂
t
u)(∂
r
u)
=(Su)
2
+(t∂
r
u + r∂
t
u)
2
+(r
2
+ t
2
)|
R
r
u|
2
.
The last two terms come from using Lemma 6.9 to get rid of the bad
quadratic terms Q, and it is not immediately clear that they do not destroy
the positivity of
˜
E. So we pause for a moment to discuss the sign of
˜
E.
Step 2’. Consider the simple identity Σ∂
i
(x
i
u
2
)=nu
2
+2ru∂
r
u.Writing
2(n − 1)ut∂
t
u =2(n − 1)u(Su − r∂
r
u)
and using the above identity, we obtain
[2(n − 1)tu∂
t
u − (n − 1)u
2
]dx =
[2(n − 1)uSu +(n − 1)
2
u
2
]dx,
which gives
˜
E =
1
2
(Su +(n − 1)u)
2
+(t∂
r
u + r∂
t
u)
2
+(r
2
+ t
2
)
R
r
u
2
dx.
Noticing that
|Hu|
2
=Σ(H
i
u)
2
=(t∂
r
u + r∂
t
u)
2
+ t
2
[Σ(∂
i
u)
2
− ∂
r
u)
2
]
=(t∂
r
u + r∂
t
u)
2
+
t
2
r
2
|Ru|
2
,
we finally obtain
˜
E =
1
2
{(Su +(n − 1)u)
2
+ |Hu|
2
+ |Ru|
2
}dx.
This computation can be considered a true miracle! It proves that the
modified energy is always nonnegative.