A local well-posedness result for the quasilinear wave
equation in R2+1
Dan-Andrei Geba
Department of Mathematics, Evans Hall
University of California at Berkeley, Berkeley, CA 94720-3840
Abstract
We investigate the local well-posedness for quasilinear wave equations in R2+1 .
Our work extends the geometric methods pioneered by Klainerman and KlainermanRodnianski for similar problems in R3+1 . The main new ingredient of the argument is
the use of two new vectorfields, the scaling vectorfield S and the angular momentum
vectorfield Ω, which complement the decay information provided by the Morawetz
vectorfield K.
1
Introduction
In this article we consider the following Cauchy problem:
∂t2 φ − g ij (φ)∂i ∂j φ = N (φ, ∂φ),
φ(0, x) = φ0 (x), ∂t φ(0, x) = φ1 (x), (t, x) ∈ R1+n .
(1)
where the metric g(u) = (gij (u))i,j=1,...,n is a smooth, uniformly positive definite matrix
and the nonlinearity N (φ, ∂φ) is quadratic in ∂φ. Assuming that the initial data satisfies
(φ0 , φ1 ) ∈ H s (Rn ) × H s−1 (Rn ), we will be interested in the local well-posedness of the initial
value problem (1), e.g. for what values of s there exists a unique local solution
φ ∈ C([0, T ], H s (Rn )) ∩ C 1 ([0, T ], H s−1 (Rn )).
The expected range for s is
n n+5
s > max{ ,
}
(2)
2
4
where the first exponent comes from scaling, while the second one, according to Linblad [9],
is connected with the concentration of null rays. Recently, Smith and Tataru [15] proved
1
that this is the range in the case of the dimensions n = 2, 3 for general systems of quasilinear
wave equations, while Klainerman and Rodnianski [8] showed it in the particular case of the
Einstein vacuum equations (n = 3) . For φ a smooth solution of equation (1), we have the
following H s -energy estimate, true for any s > 0 :
Z t
dτ )
(3)
k∂φ(t)kH s−1 . k∂φ(0)kH s−1 · exp( k∂φ(τ )kL∞
x
0
In order to obtain a local well-posedness result, one needs to control the integral term in (3).
The ’classical’ approach is by using the Sobolev inequality :
Z t
(4)
k∂φ(τ )kL∞
dτ ≤ t · sup k∂φ(τ )kHxs−1
x
0≤τ ≤t
0
where s needs to satisfy s > n2 + 1. Combining (3) and (4), together with a Picard iteration
argument, we obtain local well-posedness for s in the range s > n2 + 1. In the form presented above, the result appears in a paper by Hughes-Kato-Marsden [3]. Another way of
Rt
dτ , locally in time, is through Strichartz estimates
estimating the integral term 0 k∂φ(τ )kL∞
x
for k∂φkL4t L∞
(n = 2) and k∂φkL2t L∞
(n ≥ 3). Using such estimates, Ponce and Sideris [11]
x
x
n
3
proved that the semilinear wave equation φ = |∂φ|2 is well-posed in H 2 + 4 + (Rn ) for n = 2
n
1
and H 2 + 2 + (Rn ) when n ≥ 3. The scheme to keep in mind when attempting to prove a
local well-posedness result for the equation (1), based on Strichartz estimates (and of course
energy estimates), is captured in the following:
Heuristic argument In view of the energy estimate (3), one can make the bootstrap assumptions φ ∈ C([0, T ], H s (Rn )) and ∂φ ∈ L1t L∞
x . This implies that the metric g = g(φ)
satisfies the same conditions. So under these assumptions for g, one needs to prove Strichartz
estimates for the wave operator g = ∂t2 − g ij (t, x)∂i ∂j , with s ≥ s0 . This, in turn, will
imply local well-posedness in the space H s0 .
The first results were proved by Kapitanski [4] and Mockenhaupt-Seeger-Sooge [10] for
smooth metrics. A significant improvement came later, when Smith [12] proved, for n = 2, 3,
the full Strichartz estimates for metrics with C 2 coefficients. This is the sharp result, as it
was shown later by a counterexample in Smith-Sogge [13]. Smith relied on a technique based
on the approximation of the solution using a wave packet decomposition. This construction
turned out to be very important and was also used in the recent paper of Smith and Tataru
[15]. Observe that according to this scheme, in order to apply Smith’s result, one needs
H s -regularity of the coefficients, with s > n2 + 2, and so is not able to recover even the
classical result corresponding to s > n2 + 1. This is why one needs Strichartz-type estimates
for metrics with very rough coefficients. An important breakthrough took place when it was
first realized, independently by Bahouri-Chemin [1], [2] and Tataru [16], that in order to
go below this C 2 assumption on the coefficients, one needs to allow losses in the Strichartz
2
estimates which, though will not give the optimal expected result (2) for n = 2, 3, it will
nevertheless improve the classical result (s > n2 + 1). More precisely, Bahouri-Chemin [1],
[2] and Tataru [16] were able to show that for metrics with coefficients rougher than C 2 the
following Strichartz estimates hold
k∂ϕkL2t L∞
. kϕ(0)kH n2 + 12 + 14 + kϕ(0)kH n2 − 12 + 14
x
n ≥ 3,
(5)
k∂ϕkL4t L∞
. kϕ(0)kH n2 + 34 + 18 + kϕ(0)kH n2 − 14 + 18
x
n = 2.
(6)
Both papers are based on a parametrix construction for the local linearized problem on
small time intervals, followed by a summation of the estimates obtained on each one of these
intervals. Bahouri and Chemin used Fourier integral operators, while Tataru’s proof relied on
the FBI transform to localize precisely in space and frequency. Using similar ideas, he further
improved in [17] the local well-posedness theory for equation (1) up to s > n2 + 56 , when
n = 2, and s > n2 + 32 , for n ≥ 3. A later paper by Smith and Tataru [14] showed that based
on the heuristic argument previously presented, one cannot improve this result of Tataru.
One needs to stress the idea that all these results were based on regularity assumptions on the
coefficients of the metric, without taking advantage of the special structure of the nonlinear
equation. An important next step was taken by Klainerman in [6] and by Klainerman and
Rodnianski in [7], leading to further improvements in the local well-posedness theory. The
main new observation is that, due to the quasilinear structure of the equation, the coefficients
g ij (φ) of the equation (1) verify themselves an equation of the type
g g ij = N 0
(7)
where N 0 depends only on φ and ∂φ. The term g g appears crucially in the structure of the
Raychaudhuri equation for one of the Ricci coefficients and so it allows us to prove that this
Ricci coefficient is smoother than it was previously believed. Based on elliptic estimates,
one can then show that all Ricci coefficients are smoother. Both papers, dealing with the
case n = 3, are based on a geometric approach, using the foliation of the spacetime by null
cones and commutation of the linearized local equation with the modified version of Morawetz
vectorfield in order to prove that the conformal energy associated with the linearized equation
is bounded. The first paper by Klainerman [6] reproved Tataru’s result [17], while the
√
second one [7] by Klainerman and Rodnianski improved it up to s > n2 + 12 + 2−2 3 in
the case when n = 3. The recent papers who proved the sharp result, [8] and [15], used a
variation of the same idea explained above, the former one combining it with a wave packet
decomposition along the lines of [12]. Our work, which is in the spirit of Klainerman and
Rodnianski program, deals with the case n = 2. In this situation using only the Morawetz
vectorfield does not longer suffice to derive the needed decay information as in the case of
dimension n = 3. Instead, and this is the main novelty of our argument, we rely on two
3
other vectorfields, the scaling and the angular momentum vectorfield, which complement
the information provided by the Morawetz vectorfield. Using this method we improve the
previous result of Tataru [17], but we fail to match the sharp one s > 74 . Our main result is
the following:
Theorem 1.1 The Cauchy problem (1) with the metric g ij satisfying the assumptions below,
√
is locally well posed in H s for any s > s0 = 74 + 5−4 22 . Moreover, φ satisfies the following
Strichartz type estimate1
k∂φkL4[0,T ] L∞
≤ c T s−s0 kφ[0]kH s .
(8)
x
Conditions satisfied by the coefficients g ij : The metric g ij = g ij (z) is smooth and
uniformly positive definite with respect to bounded values of the parameter z ∈ R. Namely,
there exist positive constants M0 , A0 such that for a sufficiently large integer k
l
d
sup |
g ij | ≤ M0 ,
dz
|z|≤A0
∀0≤l≤k
M0−1 |ξ|2 ≤ g ij (z)ξi ξj ≤ M0 |ξ|2 , ∀|z| ≤ A0 ,
l
X
d
αβ
N αβ (z)| ≤ M0 ,
N (φ, ∂φ) =
N (φ)∂α φ∂β φ, sup |
dz
|z|≤A0
α,β
(9)
∀ 0 ≤ l ≤ k.
First we will reduce the problem to the proof of the boundedness of the conformal energy
for the solution of the reduced linear equation. This reduction is by now classical, see [6],
[7] and [8]. It is based on the parradiferential calculus in which one localizes the solution to
a certain dyadic frequency λ and the coefficients at frequency at most λa , for an a ∈ (0, 1).
This is followed by a T T ∗ -type argument which reduces the Strichartz estimate to a dispersive inequality. Finally, using Sobolev and trace theorems specially adapted to our situation,
we showed how to reduce our dispersive inequality to a bound for the conformal energy
associated with the linearized local problem. Secondly we will turn our attention to the geometry of the null cones. Here the important tools are: the null frame {L, L, A}, the optical
function u, the lapse function b, the affine parameter s, the second fundamental form κ and
the Ricci coefficients χ, χ, η, η, ξ. The main result is the proof of asymptotic properties for
the Ricci coefficients and for the lapse function. The last and the most important part of
the proof deals with the commutation between the wave operator g and the special vectorfields: the Morawetz vectorfield K, the scaling vectorfield S and the angular momentum
vectorfield Ω. When commuting we obtain error terms which are expressed relative to the
deformation tensor of these vectorfields. The deformation tensor is then estimated in terms
1
We denote the initial data for the equation (1) by φ[0] and say that φ[0] ∈ H s if (φ0 , φ1 ) ∈ H s × H s−1 .
4
of the asymptotics for the Ricci coefficients, deduced in the previous step. Boundedness for
the conformal energy is finally obtained by careful integration of the error terms.
2
The reduction of theorem 1.1 to a dispersive inequality
As mentioned in the introduction, this reduction is essentially the same with the one
presented in [6], [7] and [8]. This is why we will only sketch its main steps and refer the
reader to [7] for more details. We start first with:
2.1
Bootstrap argument
In order to prove Theorem 1.1 we have to obtain the following inequality2
k∂φkL1[0,T ] L∞
. k∂φ(0)kHxs−1 ,
x
s > s0
(10)
In view of the trivial Cauchy-Schwartz estimate
3
k∂φkL1[0,T ] L∞
≤ T 4 k∂φkL4[0,T ] L∞
x
x
and of the energy estimate (3), the proof of (10) is reduced to the following bootstrap
argument:
Theorem 2.1 If φ is a solution of (1), φ ∈ C([0, T ], H s ) ∩ C 1 ([0, T ], H s−1 ), with T ≤ 1 and
satisfying
k∂φkL4[0,T ] L∞
+ k∂φkL∞ Hxs−1 ≤ R
(11)
x
[0,T ]
then it will also verify
k∂φkL4[0,T ] L∞
≤ C(R)T s−s0 k∂φkL∞
x
[0,T ]
2.2
Hxs−1
(12)
Paradifferential approximation and linearization of the problem
In order to prove (12), we will use a Littlewood-Paley decomposition. In case of the
low frequencies the required estimate follows trivially applying the Sobolev inequality. The
delicate problem is in the case of the high frequencies λ ≥ Λ, where apart from the truncation
2
Throughout this paper we will use the notation A . B for A ≤ C · B where C is a universal constant.
5
of the coefficients caused by the commutation of the wave operator g(φ) with the LittlewoodPaley operator Pλ , we will further truncate them up to frequency λa , a ∈ (0, 1), using the
operator
X
Sλa =
Pµ .
µ<λa
We introduce the smooth metric
ij
ij
g≤λ
a = Sλa g (Sλa φ).
(13)
The inhomogeneous equation satisfied by the dyadic piece φλ is then the following one:
¯ g a φλ = −∂t2 φλ + g ij a ∂i ∂j φλ = Rλa ,
≤λ
≤λ
φλ |t=0 = φλ0 ,
∂t φλ |t=0 = φλ1 .
(14)
The right-hand side Rλa has the Fourier support contained in the set
{ξ : λ ≤ |ξ| ≤ 4λ}
and satisfies
kRλa (t)kḢ s . C(R)λ1−a cλ k∂φ(t)kL∞
kφ(t)kḢ s
(15)
x
P 2
for any s > 1 and t ∈ [0, T ], the constants cλ verifying
λ cλ ≤ 1. Theorem 2.1 is then an
immediate consequence of the following
Proposition 2.2 If φ satisfies the assumptions of Theorem 2.1, then for each λ ≥ Λ, where
Λ is a fixed large parameter, the following Strichartz estimate is true:
k∂φλ kL4[0,T ] L∞
≤ C(R) cλ T s−s0 k∂φkL∞
x
[0,T ]
for constants cλ verifying
2
λ cλ
P
Hxs−1
(16)
≤ 1.
For more details we refer the reader to Proposition 1.2 and Theorem 1.3 in [7].
Remark From now on, we can regard (14) as a linear equation with smooth coefficients,
depending on the parameter λ.
2.3
Restriction to frequency dependent time intervals
In this part we will reduce the proof of Proposition 2.2 to the proof of precise Strichartz
estimates (without losses) on small time intervals. Using the bootstrap assumption (11)
k∂φkL4[0,T ] L∞
≤ R,
x
we will partition the time interval [0, T ] into smaller intervals I such that they satisfy the
following three conditions:
6
- their total number is comparable to λ1−a ;
- the size of each I is bounded by T λ−(1−a) ;
- for each I
1−a
1−a
≤ Rλ− 4 .
k∂φkL4I L∞
≤ λ− 4 k∂φkL4[0,T ] L∞
x
x
(17)
Estimate (16) will then follow as a result of summing the exact Strichartz estimates over
the intervals I. We will use Duhamel formula to work from now on with an homogeneous
equation. We choose a to satisfy3
0< a=
√
8 − 4s
< 22 − 4 < 1.
1 − 4s + 4s0
The precise Strichartz estimate to be proved is:
kPλ ∂ψkL4I L∞
≤ C(R) |I| kψ[0]kḢ 47 + ,
x
(18)
where ψ is a solution of the linear wave equation
¯g aψ = 0
≤λ
with initial data ψ[0] such that
1
supp ψ̂[0] ∈ { λ ≤ |ξ| ≤ 2λ}
2
and
= s − s0 .
The metric g≤λa verifies for all nonnegative integers m and for all subintervals I:
k∂ 1+m g≤λa kL1I L∞
≤ λ−(1−a)+am R̄,
x
(19)
k∂ 1+m g≤λa kL2I L∞
≤λ
x
− 1−a
+am
2
R̄,
(20)
k∂ 1+m g≤λa kL4I L∞
≤λ
x
− 1−a
+am
4
R̄,
(21)
a2
+am
∞ ≤ λ 4
k∂ 1+m g≤λa kL∞
R̄,
I Lx
a2
+am
2 ≤ λ 4
k∂xm (∂ 2 g≤λa )kL∞
R̄,
I Lx
¯ g a g≤λa kL1 L∞ ≤ λ−(1−a)+am R̄
k∂ m ≤λ
x
I
(22)
(23)
(24)
R̄ depending only on the constants M0 and R.
Remarks 1. The estimates (19)-(21) follow immediately as a result of applying Hölder
inequality to the estimate (17).
2. The estimates (22)-(23) are deduced using Sobolev inequalities, while (24) appears in view
3
The particular choice for
√
22 − 4 will become later explicit.
7
of the fact that g g ij verifies (7). It is very easy to check that our linear wave equation
¯ g a ψ = 0 is invariant under rescaling. Introducing
≤λ
gλ (t, x) = g≤λa (λ−1 t, λ−1 x),
t x
ϕ(t, x) = ψ
,
,
λ λ
(25)
the estimate (18) is then further reduced to the proof of
Proposition 2.3 If ϕ is a solution of the homogeneous equation:
¯ g ϕ = −∂t2 ϕ + g ij ∂i ∂j ϕ = 0,
λ
λ
ϕ|t=0 = ϕ0 ,
(26)
∂t ϕ|t=0 = ϕ1
on the time interval I = [0, t∗ ] with t∗ ≤ λa and the metric gλ verifies the scaled versions of
(19)-(24):
k∂ 1+m gλ kL1I L∞
. λ−(1−a)(m+1) ,
x
k∂ 1+m gλ kL2I L∞
.λ
x
(27)
−(1−a)m
− 2−a
2
,
(28)
4−a
−(1−a)m
4
,
(29)
k∂ 1+m gλ kL4I L∞
. λ−
x
−ā−(1−a)m
∞ . λ
k∂ 1+m gλ kL∞
,
I Lx
(30)
−ā−(1−a)m
2 . λ
k∂ 2+m gλ kL∞
,
I Lx
¯ g gλ kL1 L∞ . λ−(2−a)−(1−a)m ,
k∂ m (31)
λ
I
x
(32)
√
2
where ā = 1 − a4 and the parameter a such that a < 22 − 4, then the following Strichartz
estimate holds true
kP ∂ϕkL4I L∞
. |t∗ | kϕ[0]kL2x ,
(33)
x
P is the operator of projection on the set {ξ :
inequality holds with a constant independent of λ.
1
2
≤ |ξ| ≤ 2} in Fourier space and the
Remark: Compared to the case n = 3, we notice the presence of inequality (29). Note also
that (30)-(31) are different from the corresponding inequalities in [7], due to the different
scaling in dimension n = 2.
2.4
Reduction of the Strichartz estimate to the dispersive inequality
This part contains three steps and follows identically the corresponding structure in [7]:
1. Equation (26) can be replaced with the geometric wave equation
8
p
p
1
1
(34)
∂t ( det gλ ∂t φ) + √
∂i (gλij det gλ ∂j φ) = 0
det gλ
det gλ
and this is due to the fact that the two wave operators differ only by lower order terms in
so far as the Strichartz estimates are concerned.
2. Using a modified version of the standard T T ∗ argument as in [6],[7], we can show that
the proof of the Strichartz estimate (33) is reduced to the proof of the following dispersive
inequality:
gλ φ = − √
Theorem 2.4 Under the hypothesis of Proposition 2.3, if φ is a solution of the linear wave
equation
gλ φ = 0,
(35)
φ|t=0 = φ0 ,
∂t φ|t=0 = φ1
with initial data φ[0] supported in the set {ξ :
t ≤ t∗ and a fixed arbitrary small > 0
kP ∂φ(t)kL∞
.
x
1
2
≤ |ξ| ≤ 2} in Fourier space, then for all
1
1
(1 + |t|) 2 −
kφ[0]kL1x .
(36)
3. Considering the same partition of unity as in [7], together with the additivity of the L1
norm and the standard Sobolev inequality, we will decompose the initial data φ[0] into a
sum of functions with almost disjoint supports contained in balls of radius 12 and therefore
reduce the dispersive inequality (36) to the following L2 − L∞ decay estimate:
Theorem 2.5 Under the hypothesis of Theorem 2.4, but with the initial data φ[t0 ] supported
in the ball B 1 (0), for all t0 ≤ t ≤ t∗ , an arbitrary small > 0, and a sufficiently large integer
2
m > 0,
m
X
1
k∂ k φ[t0 ]kL2x .
(37)
kP ∂φ(t)kL∞
.
1
x
(1 + |t − t0 |) 2 − k=1
Remark We will postpone the last step of the reduction until we have all the geometrical
ingredients needed in its proof.
3
3.1
The geometrical background of the problem
Basic geometric tools
From now on, in order to simplify notation, we will denote our underlying metric gλαβ = g αβ .
Therefore we will work with the space-time Lorentz metric
gαβ dxα dxβ = − dt2 + gij dxi dxj .
9
An immediate consequence of this is the fact that T = ∂t is geodesic (DT T = 0). We will
begin with the following
Definition 3.1 1.The spacelike hypersurfaces Σt are defined as the level hypersurfaces of
the time function t. The time axis Γ is the integral curve of the unit vectorfield T = ∂t ,
initiating at the origin.
2. The solution u of the eikonal equation:
g αβ ∂α u∂β u = 0
u(Γ ∩ Σt ) = t
(38)
is called the optical function. In connection with u, we will also define
u = 2t − u
and
s = t − u.
Cu denotes the level surfaces of function u, which are null cones with vertices on Γ. We
define by
L0 = −g αβ ∂α u∂β
the normal to the hypersurface Cu .
3. The function b given by
b−1 = − < L0 , ∂t > = ∂t u
(39)
is called the lapse function.
4. The collection of the following vectorfields
L = bL0
L = 2T − L ,
(40)
together with an unit vectorfield A, tangent to St,u = Σt ∩ Cu , is called a null frame.
Regarding St,u as embedded in Σt , we define the unit outward normal N .
5. The conformal energy for a function φ = φ(t, x) in the interior region and the exterior
region is given respectively by
Z
Eint (φ)(t) =
t2 · |∂φ|2 + |φ|2 · (1 − ω)
ZΣt
(41)
2
2
2
2
2
2
Eext (φ)(t) =
t · (|Lφ| + (|Aφ| ) + u · |Lφ| + |φ| · ω
Σt
where ω is a cut-off function equal to 1 in the region {u ≤ 2t }, whose derivative satisfies
|∂ω| . t−1 .
The full conformal energy is given by
E(φ)(t) = Eint (φ)(t) + Eext (φ)(t)
10
(42)
6. We introduce the following special vectorfields:
Ω = s · A angular momentum vectorf ield
1
S = (uL + uL) scalar vectorf ield
2
1
K = (u2 L + u2 L) M orawetz vectorf ield
2
(43)
(44)
(45)
The following facts are simple consequences of the previous definition:
Proposition 3.2 1. < L, L > = < L, L > = < L, A > = < L, A > = 0,
< L, L > = −2, < A, A > = 1.
2. The function s defined by s = t − u is the affine parameter of the vectorfield L.
3. Lu = 0, Lt = Ls = 1, Lu = 2.
4. Lu = 2b−1 , Lt = 1, Ls = 1 − 2b−1 , L u = 2 − 2b−1 .
5. Au = At = As = Au = 0.
6. L = T + N, L = T − N.
We recall the formula for the Ricci tensor
1 αβ 2
2
2
2
g ∂µβ gαν + ∂αν
gµβ − ∂µν
gαβ − ∂αβ
gµν +
2
+ g αβ gγδ Γγµβ Γδαν − Γγµν Γδαβ
Rµν = g αβ Rαµβν =
(46)
Relying on the previous proposition, in the particular case of RLL = RLALA , we have the
following remarkable decomposition
Proposition 3.3 ([7])
RLL = L(v) −
1 α β
L L g gαβ + E 0 ,
2
the terms
v = Lν g αβ ∂β gαν −
1 αβ
g L(gαβ )
2
and E 0 satisfy the following bounds:
|v| . |∂g| ,
|E 0 | . |∂g|2 .
An immediate consequence of this proposition is the following
Corollary 3.4 The quantities v and
1
E = − Lα Lβ g gαβ + E 0
2
11
(47)
(48)
satisfy the estimates:
|v| . λ−ā
Z s
sup |E| dρ . λa−2
(49)
(50)
0 Sρ+u,u
Proof The first estimate for v is straightforward due to the trivial bound |v| . |∂g| and
the condition (30). For E we have the bound
|E| . |∂g|2 + |g g|.
Therefore using (28) and (32), we obtain:
Z s
sup |E| dρ . k∂gk2L2
[0,t∗ ]
0 Sρ+u,u
L∞
x
+ kg gkL1[0,t
∗]
L∞
x
. λa−2 .
(51)
We will define now the Ricci coefficients and use them to describe the Levi-Civitta connection
defined by g.
Theorem 3.5 We define the following tensors on St,u :
χ = < DA L, A >,
χ = < DA L, A >,
1
1
< DL L, A >, η =
< DL L, A >,
2
2
1
ξ =
< DL L, A > .
2
Based on these, the connection D satisfies the following equations:
η =
DL L = − κN N L,
DL L = 2ηA + κN N L,
DL L = 2ηA + κN N L,
DA L = χA − κAN L,
DL A = ηL.
DL L = 2ξA − κN N L,
DA L = χA + κAN L,
DL A = ηL + ξL.
1
1
DA A = χL + χL.
2
2
(52)
(53)
(54)
(55)
(56)
(57)
Moreover:
χ = − χ − 2κAA ,
η = b−1 ∇
/ (b) + κAN ,
η = − κAN ,
ξ = κAN − η.
12
(58)
(59)
(60)
Remarks 1) The proof of this theorem follows exactly the same lines as the corresponding
result in [7], being just a simple application of the orthogonalities of the null frame {L, L, A}
and of the eikonal equation (38) together with the basic properties of the Levi-Civitta connection.
2) We notice that we can express all the other Ricci coefficients in terms of just χ, η, the
lapse function b and the second fundamental form κ.
3) We notice also that due to the fact St,u is a 1-dimensional surface, all our Ricci coefficients
can be treated as scalars. Therefore, if we denote by ∇
/ the induced covariant derivative on
St,u , the A-derivative of a Ricci coefficient will coincide with its ∇
/ -derivative.
Corollary 3.6 We have the following commutation properties:
[L, L] = 2(η − η)A + κN N L − κN N L,
[L, A] = − χA,
(61)
[L, A] = − χA + (η − κAN )L + ξL.
Among the tools that play an important role in the proof of the asymptotic properties are
the transport and the elliptic equations satisfied by the different Ricci coefficients. The ones
that will be used extensively are:
Lemma 3.7 The following transport and elliptic equations hold:
L(χ) = − (χ)2 − κN N χ − RLL ,
1
L(η) = − χη + χη + RALLL ,
2
L(b) = − bκN N ,
(62)
L(χ) − (χ)2 − (κN N + 2κAA )χ = 2∇
/ (η) + 2η 2 + RALLA .
(65)
(63)
(64)
Proof In this proof we use the equations for connection D and basic formulae for the
curvature tensor. In what concerns the transport equations, we present the proof for (63),
the other two equations being deduced similarly. Using (53) and taking an L-derivative we
can write
L(η) =
1
1
1
< DL DL L, A > + < DL L, DL A > = < DL DL L, A > +
2
2
2
1
1
1
+ < D[L,L] L, A > + < RLL L, A > + < DL L, DL A >
2
2
2
If we plug in the frame equations (55), (56), (57) and the commutation formula (61), we
obtain (63).
13
The deduction of the elliptic estimate (65) follows like this:
L(χ) = < DL DA L, A > + < DA L, DL A > = RALLA + < DA DL L, A > +
+ < D[L,A] L, A > + < DA L, DL A > = RALLA + < DA (2ηA + κN N L), A >
+ < D−χA + (η − κAN )L + ξL L, A > + < χA − κAN L, ηL + ξL > =
= RALLA + 2∇
/ (η) + κN N χ − χχ + 2(η − κAN )η + 2ηκAN .
Corollary 3.8 If we write the elliptic equation (65) in the form
∇
/ (η) + η 2 = H
(66)
k∇
/ ηkL2 (St,u ) + kηk2L4 (St,u ) . kHkL2 (St,u )
(67)
we have the following estimate
Proof We square equation (66) and integrate the resulting expression over St,u . Due to
R
/ (η)η 2 = 0, we obtain the identity
the fact that St,u ∇
Z
Z
2
4
|∇
/ η| + |η| =
H2
St,u
St,u
which implies (67).
3.2
Construction of the optical function u and the continuation
argument
We review here the main steps in the construction of u discussed in [7]. It will be, of
course, enough to describe its level hypersurfaces Cu , which are the union of null geodesics
x = x(s), starting on the time axis Γ from the vertex (u, 0) and having velocity in the
direction of the vector (1, $), with $ ∈ S 1 . Relative to our Lorentz metric g, the equations
for such a geodesic is:
j
k
d2 xi
i dx dx
+
Γ
= 0 (i, j, k = 0, . . . , 2),
jk
ds2
ds ds
dx
(0) = (1, $) .
ds
(68)
Obviously, these equations can then be rewritten in the form of a first order system for
6 dependent variables, and so, due to the basic existence theorem for ordinary differential
equations, we obtain a local solution. We are interested up to what value of s such a
local solution can be extended to. Heuristically, such a system can be approximated by the
following ODE :
dy
+ Γ(s) y 2 = 0 .
ds
14
A local solution can then be continued up to the value of the parameter s = smax , as long
as, for example,
smax |Γ(s)| 1 .
Due to the bound that we have for |Γijk |,
|Γijk | . |∂g| . λ−ā
and the fact that
t∗ λ−ā . λa−ā 1
we can argue that such a geodesic can be extended up to the value of the affine parameter
s = t∗ − u. It can be shown that the transport equation (64) for the lapse function b implies
that our geodesic intersect each time slice Σt , t ≤ t∗ . From the definitions, initial condition
of the optical function u and the geometry around the time axis Γ, we can check the following
Initial values For all t ∈ [0, t∗ ], there exists a constant R0 > 0 such that:
1
lim sup |χ − | + |η(s)| < R0 ,
s
s−→0
1
1
2
lim sup s kD(χ − )kL2 (St,u ) + k∇
/ (η)kL2 (St,u ) < R0 ,
s
s−→0
(69)
lim |b(s) − 1| + |A(St,u ) − 2πs| = 0
s→0
1
/ η(s)| = 0.
lim s3 |D(χ − )| + s2 |∇
s→0
s
We will rely on a continuation argument, used also in [7], whose essence we explain in the next
lines. We denote by s(t) the maximum value of s = t − u, for which the above initial values
can be extended in the neighborhood of Γ with the additional condition s(t) ≤ min{εR0−1 , t}
where ε 1. So in this region that we denote by ∆ ⊂ [0, t∗ ] × R2 , the following are true:
Bootstrap assumptions
1
|χ − | + |η(s)| ≤ R0 ,
(70)
s
1
1
/ (η)kL2 (St,u ) ≤ R0 ,
s 2 kD(χ − )kL2 (St,u ) + k∇
(71)
s
s(t) ≤ min{εR0−1 , t}.
(72)
We will show that under these assumptions, R0 can be chosen such that
min{εR0−1 , t} = t
and so from the maximality of ∆ we will conclude that the assumptions will then hold in
the whole region
{(s, t)|s ∈ [0, t], t ∈ [0, t∗ ]} .
Throughout this section and the following one, we will prove estimates in the region ∆.
15
3.3
Other geometric results
Based on the bootstrap assumptions (70)-(72), just as in [7], we can prove the following
result:
Proposition 3.9 A. Sobolev inequality For any smooth function f ,
f : St,u → R, 1 < p < ∞, we have the following Sobolev estimate:
1
1
/ f kLp (St,u ) + s− p kf kLp (St,u ) .
sup |f | . s1− p k∇
(73)
St,u
1
B. Trace estimates For any function h, h : Σt → R, h ∈ H 2 + (R2 ) and
∆s = ∪ 1 s≤ρ≤s St,t−ρ , we have the following inequalities:
4
1
1
kf kL2 (St,u ) . k∂ 2 − f kL2 (Σt ) + k∂ 2 + f kL2 (Σt ) .
(74)
1
kf k2L2 (St,u ) . kN (f )kL2 (∆s ) kf kL2 (∆s ) + kf k2L2 (∆s ) .
s
(75)
Remark These two results, in a more general form, can be found for the case n = 3 in
[7]. They will also survive in the general form in the case n = 2, but not to create too much
confusion, we presented them in the form perfectly suitable to our situation. We remark
also that the proof that we have right now is almost identical with the original one, the only
difference with respect to the case n = 3 being the fact that we are no longer able to recover
the isoperimetric inequality. We hope to provide in the future a different proof for (73)-(75),
which would rely more on the special structure of St,u .4
The next result will be extensively used in the proof of the asymptotic properties for the
Ricci coefficients.
Transport lemma If ΞA is an St,u -tangent covariant tensor satisfying the transport equation
L(ΞA ) + σ χ ΞA = FA ,
(76)
with the initial condition sσ ΞA (s) −→ 0 as s −→ 0 and (t, x) = (t, s, ω) ∈ ∆, the following
estimates hold:
If σ > − 12 then
Z
4 s σ
|Ξ(t, x)| ≤ σ
ρ |F | dρ
(77)
s 0
1
Additionally, if sσ− 2 kΞkL2 (St,u ) −→ 0 as s −→ 0, we have
Z s
1
4
ρσ− 2 kF kL2 (Su+ρ,u ) dρ
kΞkL2 (St,u ) ≤ σ− 1
s 2 0
4
(78)
This proof would take advantage of the fact that St,u is the intersection of the level sets for the functions
t and u.
16
If σ ≥ 0 and F defined in the whole time slab [0, t∗ ] × R2 , we have also
|Ξ(t, x)| ≤ 4kF kL1[0,t
.
L∞
x
(79)
4
sup ρ · |F |
σ ρ≤s
(80)
∗]
If σ > 0, it follows that
|Ξ(t, x)| ≤
Proof The transport equation (76) implies:
1 d
σ
1
|Ξ|2 + |Ξ|2 = −σ(χ − )|Ξ|2 + F · Ξ,
2 ds
s
s
which further yields
d 2σ
1
(s · |Ξ|2 ) = −2σs2σ (χ − )|Ξ|2 + 2s2σ F · Ξ.
ds
s
Integrating this equation with respect to s and taking into account the initial condition, we
obtain:
Z
Z s
2
2|σ| s 2σ
1
2
2
|Ξ(t, x)| ≤ 2σ
ρ |χ − ||Ξ| dρ + 2σ
ρ2σ |F · Ξ| dρ.
(81)
s
s
s
0
0
Using now the fact that |χ − 1s | ≤ R0 , contained in the assumptions, together with the
hypothesis, σ > − 21 , we can write
Z s
2|σ|
2
2
2
sup |Ξ| ≤
R0 s sup |Ξ| + 2σ
ρ2σ |F · Ξ| dρ.
2σ
+
1
s
ρ≤s
ρ≤s
0
Due to the fact that s ≤ s(t) and R0 s(t) 1, by the assumption (72), we can conclude that
Z s
4
2
ρ2σ |F · Ξ| dρ.
sup |Ξ| ≤ 2σ
s
ρ≤s
0
Finally using Gronwall inequality, we end up with
Z
4 s σ
|Ξ| ≤ σ
ρ |F | dρ.
s 0
This implies of course (77) and (80). (79) follows also immediately. To deduce (78) we
will continue from (81) by integrating it over the fixed surface St,u . Due to the fact that
R
R 2π
f ≈ s · 0 f (s, ω) dω, we obtain
St,u
Z s
2σ−1
2
ρ2σ−1 kΞk2L2 (Sρ+u,u ) dρ +
s
kΞkL2 (St,u ) ≤ 4|σ|
Z s0
+2
ρ2σ−1 kF kL2 (Sρ+u,u ) kΞkL2 (Sρ+u,u ) dρ.
0
Here applying twice Gronwall inequality, we come up with the desired result. Finally we
present integration results which will be useful in the sections which deal with commutation
of the D’Alembertian with the special vectorfields:
17
Proposition 3.10 (Integration results)
Σt , then the following estimate holds:
1. If f and h are smooth functions defined on
1
1
kf · h · ωkL2 (Σt ) . t− 2 sup khkL2 (St,u ) · E 2 (f )(t)
(82)
0≤u≤ 2t
2. In the context of our reduced linear problem, we can write the following integration by
parts:
Z
Z
V · g ψ · ψ · ω + O(λ− ) sup E(ψ)(τ )
(83)
V · Lψ · Lψ · ω =
τ ∈[t0 ,t]
[t0 ,t]×R2
[t0 ,t]×R2
provided the following conditions hold
kV kL∞ (Sτ,u ) . τ · λ−
Z t
3
τ − 2 sup kLV kL2 (Sτ,u ) + k∇
/ V kL2 (Sτ,u ) dτ . λ−
0≤u≤ τ2
t0
Also we have the estimate:
Z
V · ψ · Lψ · ω . λ− sup E(ψ)(τ )
(84)
τ ∈[t0 ,t]
[t0 ,t]×R2
if the previous conditions hold, relaxed by the fact that we can drop the term k∇
/ V kL2 (Sτ,u ) .
Remark The proof of these integration formulae is straightforward, as in the case of their
corresponding ones for n = 3, presented in [7].
Now we are ready to prove
4
Asymptotics properties of the Ricci coefficients
The main result of this section is:
Theorem 4.1 For sufficiently large values of λ and for the parameter a verifying
√
a < 22−4, the Ricci coefficients χ, η, any component R of the curvature, the lapse function
b and the second fundamental form κ satisfy the following estimates:
1
sup |χ − | + sup |κ| + sup |L(b)| . λ−ā
s
St,u
St,u
St,u
1
sup |η| + sup |∇
/ (b)| . s 2 λ−ā−
St,u
1−a
+
2
+ λ−ā
St,u
1
sup |η| . min{sλ−ā−(1−a) + λ−ā , s 2 λ−ā−
St,u
sup |b − 1| . min{sλ−ā , λ−(1−a) }
St,u
sup |L(b)| . sλ−ā−(1−a)
St,u
18
1−a
+
2
+ λ−ā }
kRkL2 (St,u ) + k∇κkL2 (St,u ) . λ−ā−
1−a
+
2
k |L(κA· )| + |L(κN · )| kL2 (St,u ) + k |L(κA· )| + |L(κN · )| kL2 (St,u ) . λ−ā−
1−a
1−a
+
2
1
k |∇
/ (κA· )| + |∇
/ (κN · )| kL2 (St,u ) . λ−ā− 2 + + s− 2 λ−ā
1−a
1
1
/ (η)kL2 (St,u ) + kL(η)kL2 (St,u ) . λ−ā− 2 + + s− 2 λ−ā
kD(χ − )kL2 (St,u ) + k∇
s
1
kL(b)kL2 (St,u ) . s 2 λ−ā
kL(b)kL2 (St,u ) . sλ−ā−
(85)
1−a
+
2
k∇
/ (b)kL2 (St,u ) . sλ−ā−
1−a
+
2
1
+ s 2 λ−ā
First, we will obtain the estimates for the second fundamental form κ and for the components
R of the curvature, using the bounds
|κ| . |∂g|
|R| . |∂ 2 g| + |∂g|2 ,
and the trace estimate (74). Next, we investigate the Ricci coefficient χ. To derive the
estimates, we use first the transport equation (62) together with the decomposition (47) of
the term RLL in order to infer that:
L(χ −
1
1
+ v) + 2χ(χ − + v) ≈ E ,
s
s
where |v| . |∂g| and |E| . |∂g|2 are the terms that appear in the decomposition of RLL .
Integrating this equation, we will obtain the bound
1
sup |χ − | . λ−ā
s
St,u
and, as an immediate consequence, the estimate:
1−a
1
1
kL(χ − )kL2 (St,u ) . λ−ā− 2 + + s− 2 λ−ā .
s
Using again the special structure of RLL , we will then write the transport equation for the
quantity ∇
/ (χ − 1s + v) and apply the Transport lemma in order to obtain the estimate for
k∇
/ (χ − 1s )kL2 (St,u ) . The estimate of the L derivative of χ − 1s will be obtained in connection
with the estimates for η. The third part of the proof deals with Ricci coefficient η. If we try
to investigate η using its transport equation (63), we will obtain the estimate supSt,u |η| .
. λ2a−ā−1 , which is worse than the one claimed in the theorem. This is due to the fact that
the curvature component present in the transport equation for η does not have a special
decomposition. There is though a better way to estimate η and this is with the help of the
elliptic equation (65), which we can write in the form:
∇
/ (η) + η 2 = H
19
where H will be evaluated using its transport equation. We will take advantage of the elliptic
estimate (67)
k∇
/ (η)kL2 (St,u ) + kηk2L4 (St,u ) . kHkL2 (St,u )
and the bound that we already have for kL(χ − 1s )kL2 (St,u ) . Thus, we obtain:
1
sup |η| . s 2 λ−ā−
1−a
+
2
+ λ−ā .
St,u
An immediate application of this estimate is the bound for kLηkL2 (St,u ) . We make here two
important remarks. The first one is the fact that the estimate for η is worse than the one
for χ − 1s and so this estimate is the one which dictates the range for a. Secondly, as in the
case n = 3, we do not have a good estimate for Lη. Finally, we turn our attention to the
lapse function b. We notice that estimates for all the quantities depending on b, with the
exception of Lb, are immediate consequences of previously obtained bounds . For Lb, we
write its transport equation and deduce the estimate using once again the Transport lemma.
4.1
Proof of the estimates for the second fundamental form κ and
for the components R of the curvature
The obvious estimate |κ| . |∂g| implies
sup |κ| . k∂gkL∞ (Σt )
St,u
which coupled with (30) gives us the desired estimate. Any component R of the curvature
satisfies
|R| . |∂ 2 g| + |∂g|2 ,
so using the trace estimate (74), we conclude that:
1
kRkL2 (St,u ) . k∂ 2 gkL2 (St,u ) + (A(St,u )) 2 k∂gk2L∞ (Σt )
1
1
1
. k∂ 2+ 2 + gkL2 (Σt ) + k∂ 2+ 2 − gkL2 (Σt ) + (A(St,u )) 2 k∂gk2L∞ (Σt )
. λ−ā−
1−a
+
2
(86)
1
+ s 2 λ−2ā .
Using the bound for a:
1
s 2 λ−2ā . λ−ā−
1−a
+
2
,
+
−ā− 1−a
2
1−a
+
2
. This implies also k∇κkL2 (St,u ) . λ−ā−
and so, kRkL2 (St,u ) . λ
following formulae for the derivatives of the second fundamental form κ:
. We have the
L(κN N ) = (DL κ)(N, N ) + 2κ(DL N, N ) = (DL κ)(N, N ) + 2κ2AN
L(κN N ) = (DL κ)(N, N ) + 2κ(DL N, N ) = (DL κ)(N, N ) + 2(2η − κAN )κAN
∇
/ (κN N ) = (DA κ)(N, N ) + 2κ(DA N, N ) = (DA κ)(N, N ) + 2(χ + κAA )κAN
20
L(κAA ) = (DL κ)(A, A) + 2κ(DL A, A) = (DL κ)(A, A) − 2κ2AN
L(κAA ) = (DL κ)(A, A) + 2κ(DL A, A) = (DL κ)(A, A) + 2(κAN − η)κAN
∇
/ (κAA ) = (DA κ)(A, A) + 2κ(DA A, A) = (DA κ)(A, A) − 2(χ + κAA )κAN
L(κAN ) = (DL κ)(A, N ) + κ(DL A, N ) + κ(DL N, A)
= (DL κ)(A, N ) + κAN (κAA − κN N )
L(κAN ) = (DL κ)(A, N ) + κ(DL A, N ) + κ(DL N, A)
= (DL κ)(A, N ) − (κAA − κN N )(κAN − 2η)
∇
/ (κAN ) = (DA κ)(A, N ) + κ(DA A, N ) + κ(DA N, A)
= (DA κ)(A, N ) + (κAA − κN N )(χ + κAA ).
We will show the proof for the quantities depending on κN N , the ones depending on κAA
and κAN following in a similar manner. Using the bootstrap assumptions (70), (72) and the
inequality for supSt,u |κ|, by taking the L2 (St,u ) norm in the above formulae we obtain:
kL(κN N )kL2 (St,u ) . λ−ā−
1−a
+
2
1
+ s 2 λ−2ā . λ−ā−
1−a
+
2
1−a
1
1
+ s 2 λ−ā (R0 + λ−ā ) . λ−ā− 2 + + s− 2 λ−ā
1−a
1−a
1
1
1
. λ−ā− 2 + + s 2 λ−ā (R0 + + λ−ā ) . λ−ā− 2 + + s− 2 λ−ā .
s
kL(κN N )kL2 (St,u ) . λ−ā−
k∇
/ (κN N )kL2 (St,u )
1−a
+
2
We notice that for the L derivatives the estimates are not the ones that we claimed, but as
soon as we have the estimate for η, we will deduce them immediately.
4.2
Proof of the estimates for χ −
1
s
and its derivatives
Let us remember first the transport equation for χ:
L(χ) = −χ2 − κN N χ − RLL
where RLL has the form
RLL = L(v) + E.
Due to Proposition 3.3, v satisfies |v| . |∂g| and the error term E has the property
|E| . |∂g|2 + |g g|. Using this decomposition we can write the transport equation in the
form:
1
1
1
1
L(χ − + v) + 2χ(χ − + v) = (χ − )2 + 2χ(v − κN N ) − E
(87)
s
s
s
2
Due to the bootstrap assumption |χ − 1s | ≤ R0 :
s2 |χ −
1
+ v| → 0 as s → 0
s
We are now in a position to apply (77) with σ = 2. Therefore we can infer that:
21
1
4
sup |χ − + v| . 2
s
s
St,u
s
Z
0
1
1
ρ2 |(χ − )2 + 2χ(v − κN N ) − E| dρ.
ρ
2
(88)
Using again the above bootstrap assumption and the fact that sR0 1, together with the
estimates for v and E, obtained in Corollary 3.4, we conclude that:
1
1
R0
sup |χ − | ≤ 4sR02 + 8s(R0 + )λ−ā + λa−2 ≤
+ λ−ā ,
s
s
2
St,u
(89)
which implies the desired estimate for supSt,u |χ − 1s |.
Remark From now on, throughout the rest of this section, this argument will be considered
implicit and so we will ignore in the deduction of further estimates the integral terms which
are part of the bootstrap argument. For more details see section 5.7 in [7].
An immediate consequence of this result is the estimate for kL(χ − 1s )kL2 (St,u ) . Writing once
again the transport equation in the form
1
1
L(χ − ) = −χ2 + 2 − κN N χ − RLL
s
s
we obtain by taking the L2 (St,u ) norm:
1−a
1−a
1
1
1
2
kL(χ − )kL2 (St,u ) . s 2 λ−ā (λ−ā + ) + λ−ā− 2 + . λ−ā− 2 + + s− 2 λ−ā
s
s
For the other two derivatives of χ −
1
s
(90)
we need the following
Lemma 4.2 The term v and the error E satisfy
1−a
1
kDvkL2 (St,u ) . λ−ā− 2 + + s− 2 λ−ā
Z s
1
1
ρ 2 kDEkL2 (Sρ+u,u ) dρ . λ2a−2ā− 2 +
(91)
0
Proof: v contains terms of the type Lν g αβ ∂β gαν , so its derivatives would depend on:
|Lν D(g αβ )∂β gαν | . |∂g|2
|Lν g αβ D∂β gαν | . |∂ 2 g|
|D(Lν )g αβ ∂β gαν | . |D(Lν )||∂g|.
For the first two terms the estimates are immediately due to the bounds that we have for κ
and R:
1
kLν D(g αβ )∂β gαν kL2 (St,u ) . s 2 λ−2ā
(92)
1−a
kLν g αβ D∂β gαν kL2 (St,u ) . λ−ā− 2 +
For the third term, we show how to obtain the estimate in the case when the derivative is
∇
/ , for L, respectively L, the approach is similar.
22
∇
/ (Lν ) = ∇
/ (g νµ < L, ∂µ >) =
=∇
/ (g νµ )Lµ + g νµ < DA L, ∂µ > +g νµ < L, DA ∂µ >
=∇
/ (g νµ )Lµ + g νµ Aζ Lδ Γδζµ + g νµ Aµ χ − g νµ Lµ κAN
Using the assumptions and the estimates proved so far we conclude :
1
|D(Lν )g αβ ∂β gαν | . λ−ā (λ−ā + ),
s
which yields
1
kD(Lν )g αβ ∂β gαν kL2 (St,u ) . s− 2 λ−ā .
(93)
Putting together (92) and (93) we obtain the estimate for kDvkL2 (St,u ) . The error term E
has the expression
1
E = − Lµ Lν g gµν + Lµ Lν g αβ gγδ (Γγµβ Γδαν − Γγµν Γδαβ )−
2
− L(Lν )g αβ ∂β gαν − Lν L(g αβ )∂β gαν ,
so for its derivatives we have to deal with the following types of terms:
1
|D(Lµ )Lν g gµν | . |g g| =⇒
s
Z s
1 1
=⇒
ρ 2 k |g g|kL2 (Sρ+u,u ) dρ . kg gkL1[0,t ] L∞
. λa−2
x
∗
ρ
0
|Lµ Lν Dg gµν | . |∂g g| =⇒
Z s
1
=⇒
ρ 2 k∂g g|kL2 (Sρ+u,u ) dρ . sk∂g gkL1[0,t
∗]
0
L∞
x
. sλ2a−3
1
|D(Lµ )Lν g αβ gγδ Γγµβ Γδαν | . |∂g|2 =⇒
s
Z s
1 1
. λa−2
=⇒
ρ 2 k (∂g)2 kL2 (Sρ+u,u ) dρ . k∂gk2L2 L∞
x
[0,t
]
∗
ρ
0
|Lµ Lν D(g αβ )gγδ Γγµβ Γδαν | . |∂g|3 =⇒
Z s
1
ρ 2 k(∂g)3 kL2 (Sρ+u,u ) dρ . sλa−2−ā
=⇒
0
ν αβ
|L L g gγδ DΓγµβ Γδαν | . |∂g||∂ 2 g| =⇒
Z s
3(1−a)
1
1
=⇒
ρ 2 k|∂g||∂ 2 g|kL2 (Sρ+u,u ) dρ . s 2 λ−ā− 2 +
µ
0
ν
|L(L )D(g αβ )∂β gαν | . |(∂g)3 | =⇒
Z s
1
=⇒
ρ 2 k(∂g)3 kL2 (Sρ+u,u ) dρ . sλa−2−ā
0
ν
|L(L )g αβ D(∂β gαν )| . |∂g||∂ 2 g| =⇒
Z s
3(1−a)
1
1
=⇒
ρ 2 k|∂g||∂ 2 g|kL2 (Sρ+u,u ) dρ . s 2 λ−ā− 2 +
0
23
1
|DL(Lν )g αβ ∂β gαν | . |∂g|(|∂ 2 g| + |∂g|2 + |∂g| + |∂(κN N )|) =⇒
s
Z s
1
1
ρ 2 k|∂g|(|∂ 2 g| + |∂g|2 + |∂g| + |∂(κN N )|)kL2 (Sρ+u,u ) dρ .
=⇒
s
0
1
. λa−2 + sλa−2−ā + s 2 λ−ā−
3(1−a)
+
2
3
+ sλ−2ā + s 2 λ−2ā−
(1−a)
+
2
Using the bound s ≤ λa , we obtain the desired result. Having obtained the estimates for
v and E, we can proceed to the investigation of the other two derivatives of χ − 1s . The
L derivative will be discussed later in connection with η. We deal now with the angular
derivative ∇
/ . The estimate is obtained using the transport equation for this quantity. To
simplify the notation we denote χ − 1s by w. Let us remember first the transport equation
for this quantity:
2
L(w + v) + 2χ(w + v) = (w + v)2 + v − v 2 − κN N χ − E.
s
Taking the angular derivative of this expression we obtain:
A(L(w + v)) + 2A(w)(w + v) + 2χA(w + v) = 2A(w + v)(w + v)+
2
+ A( v − v 2 − κN N χ − E)
s
Using the commutator [L, A] = −χA, we can write the previous equation in the form:
2
L(∇
/ (w + v)) + 3χ∇
/ (w + v) = 2∇
/ (v)(w + v) + ∇
/ ( v − v 2 − κN N χ − E)
s
(94)
We will apply (78) with σ = 3. Therefore, the only thing to be verified is that
5
/ (w + v)kL2 (St,u ) −→ 0 as s −→ 0
s 2 k∇
But this is true, because due to the bootstrap assumption (71) and the estimates for the
derivatives of v we have:
5
s 2 k∇
/ (w)kL2 (St,u ) ≤ R0 s2
5
5
s 2 k∇
/ (v)kL2 (St,u ) . s 2 λ−ā−
1−a
+
2
+ s2 λ−ā
So applying the above mentioned estimate, we infer that:
Z s
5
5
2
2
s k∇
/ (w + v)kL2 (St,u ) .
ρ 2 k2∇
/ (v)(w + v) + ∇
/ ( v − v 2 − κN N χ − E)kL2 (Sρ+u,u ) dρ
s
0
Using the estimates for w = χ − 1s , v and derivatives of v and E, we conclude that:
k∇
/ (w)kL2 (St,u ) . λ−ā−
1−a
+
2
1
+ s− 2 λ−ā
Finally we investigate L(χ − 1s ). We take advantage of the elliptic equation (65)
L(χ) − χ2 − (κN N + 2κAA )χ = 2∇
/ (η) + 2η 2 + RALLA
24
(95)
Let us denote L(χ) − χ2 by θ and the whole left side of the above equality by Θ. First we
can write:
L(θ) + 2χθ = 2(η − η)∇
/ (χ) + 3κN N χ2 + ((κN N )2 + 2RLL − L(κN N ))χ + κN N RLL − L(RLL )
This equation implies then
L(Θ) + 2χΘ = 2(η − η)∇
/ (χ) + 2(κN N − κAA )χ2 + 2 − T (κN N ) − L(κAA )+
+ (κN N )2 + κN N κAA + RLL χ + 2(κN N + κAA )RLL − L(RLL )
We will denote by F the right hand side of the above equation. Taking advantage of the
special structure for L(RLL ):
−L(RLL ) = −L(L(v) + E) = −L(L(v)) − L(E) + [L, L](v) =
= −L(L(v)) − L(E) + 2(η − η)∇
/ (v) + κN N (L(v) − L(v))
we can conclude that
/ (v) + (κN N + 2χ)L(v) − κN N L(v) = W.
L(Θ + L(v)) + 2χ(Θ + L(v)) = F 0 − L(E) + 2(η − η)∇
(96)
where we denoted by F 0 , all the terms in F but the last one. We rely once again on (78)
with σ = 2. The bootstrap assumption (71) and the estimate (91) imply
3
s 2 kΘ + L(v)kL2 (St,u ) −→ 0 as s −→ 0
Applying the result mentioned above, we deduce:
Z s
3
3
2
s kΘ + L(v)kL2 (St,u ) .
ρ 2 kW kL2 (Sρ+u,u ) dρ
0
Evaluating all the terms in F 0 we obtain:
Z s
1−a
1
3
1
/ (χ)kL2 (Sρ+u,u ) dρ . λ−ā− 2 + + s− 2 λ−ā
ρ 2 k(η − η)∇
3
s 2 Z0
s
3
1
1
ρ 2 kκ(χ)2 kL2 (Sρ+u,u ) dρ . s− 2 λ−ā
3
s 2 Z0
s
1−a
3
1
1
ρ 2 k(∂κ + (κ)2 )χkL2 (Sρ+u,u ) dρ . λ−ā− 2 + + s− 2 λ−ā
3
s 2 Z0
s
1−a
3
1
ρ 2 k(|κ| + |χ|)|R|kL2 (Sρ+u,u ) dρ . λ−ā− 2 +
3
s2 0
For all the other terms in W , we use the estimates (91) to argue that:
Z s
1
3
1
1
ρ 2 kL(E)kL2 (Sρ+u,u ) dρ . s− 2 λ−2ā+2a− 2 +
3
s 2 Z0
s
1−a
3
1
1
ρ 2 k(η − η)∇
/ (v)kL2 (Sρ+u,u ) dρ . λ−ā− 2 + + s− 2 λ−ā
3
s 2 Z0
s
1−a
3
1
1
ρ 2 k(|κ| + |χ|)|∂v|kL2 (Sρ+u,u ) dρ . λ−ā− 2 + + s− 2 λ−ā
3
s2 0
25
(97)
This allows to conclude:
kΘ + L(v)kL2 (St,u ) . λ−ā−
1−a
+
2
1
+ s− 2 λ−ā
which in turn implies
kΘkL2 (St,u ) . λ−ā−
1−a
+
2
1
+ s− 2 λ−ā
(98)
based on the estimates that we have for v. We will deal now with b, because the estimate
for |b − 1| will help us finish the proof for L(χ − 1s ). The transport equation for b is
L(b) = −bκN N
We can apply (77) with σ = 0, because the initial values (69) imply
|b − 1| −→ 0 as s −→ 0
Therefore, we obtain:
Z
|b − 1| .
s
|κ| dρ . min{sλ−ā , λ−(1−a) }
(99)
0
Rewriting Θ as
b−1 − 1
1
1
+ ( 2 − χ2 ) − (κN N + 2κAA )χ
Θ = L(χ − ) + 2
2
s
s
s
we can infer that
1
1
1
kL(χ − )kL2 (St,u ) . kΘkL2 (St,u ) + 2 kb − 1kL2 (St,u ) + k 2 − χ2 kL2 (St,u ) + kκχkL2 (St,u )
s
s
s
1
+
−
−ā
−ā− 1−a
2
+ s 2λ
.λ
(100)
which concludes our discussion.
4.3
Proof of the estimates for η and its derivatives
We remember what we have denoted by Θ
Θ = 2∇
/ (η) + 2η 2 + RALLA
We can write then
∇
/ (η) + η 2 = H
where H = 21 (Θ − RALLA ) satisfies
kHkL2 (St,u ) . λ−ā−
26
1−a
+
2
1
+ s− 2 λ−ā
Using (67), we can conclude that
k∇
/ (η)kL2 (St,u ) . kHkL2 (St,u ) . λ−ā−
kηk2L4 (St,u ) . kHkL2 (St,u ) . λ−ā−
1−a
+
2
1−a
+
2
1
+ s− 2 λ−ā
(101)
1
+ s− 2 λ−ā
We use the transport equation satisfied by η
1
L(η) + χη = −χκAN + RALLL
2
with the initial conditions
sη −→ 0
1
s 2 kηkL2 (St,u ) −→ 0
as s −→ 0, satisfied due to the assumption (70). Applying (80) with σ = 1, we conclude:
sup |η| . sup ρ(|κ||χ| + |R|) . sλ−ā−(1−a) + λ−ā
ρ≤s
St,u
Z s
1−a
1
1
1
ρ 2 k|κ||χ| + |R|kL2 (Sρ+u,u ) dρ . sλ−ā− 2 + + s 2 λ−ā
kηkL2 (St,u ) . 1
s2 0
At this point we take advantage of the Sobolev inequality (73) to infer that:
1
1
1
sup |η| . s 2 k∇
/ (η)kL2 (St,u ) + s− 2 kηkL2 (St,u ) . s 2 λ−ā−
1−a
+
2
+ λ−ā
(102)
St,u
Due to this result and the transport equation for η we obtain that L(η) satisfies the same
estimate as ∇
/ (η). Here we remark that these estimates for η imply the desired inequalities
for the L derivatives of the second fundamental form κ. As in [17], for the case n = 3,
we do not have a good estimate for L(η). This finishes the discussion concerning η and its
derivatives.
4.4
Proof of the estimates for b and its derivatives
We notice that already in the course of the proof for χ − 1s we obtained
Z s
|b − 1| .
|κ| dρ . min{sλ−ā , λ−(1−a) }
0
Due to the transport equation L(b) = −bκN N we infer immediately that
sup |L(b)| . λ−ā
St,u
(103)
1
2
kL(b)kL2 (St,u ) . s λ
−ā
The formula ∇
/ (b) = b(η − κAN ), together with the inequalities satisfied by η and κ:
27
sup |κ| . λ−ā ,
St,u
1
1−a
+
2
+ λ−ā ,
1
1−a
+
2
+ λ−ā
sup |η| . s 2 λ−ā−
St,u
provide us immediately with the estimates
sup |∇
/ (b)| . s 2 λ−ā−
St,u
(104)
k∇
/ (b)kL2 (St,u ) . sλ
−ā− 1−a
+
2
1
2
+s λ
−ā
For L(b) we write its transport equation
L(L(b)) = L(L(b)) + [L, L](b) = b ((κN N )2 + 2(κAN )2 − L(κN N ) − 2η 2 )
(105)
Due to the initial values (69), we have
L(b) −→ 0
1
s− 2 kL(b)kL2 (St,u ) −→ 0
as s −→ 0. We apply (77) and (78) with σ = 0 to deduce that:
sup |L(b)| . sλ−ā−(1−a)
St,u
(106)
kL(b)kL2 (St,u ) . sλ−ā−
1−a
+
2
This concludes the proof of the asymptotic estimates.
5
The reduction of Theorem 2.5 to conformal energy
estimates
We recall first, that according to Theorem 2.5, the estimate to be proved is:
kP ∂φ(t)kL∞
.
x
m
X
1
(1 + |t − t0 |)
1
−
2
k∂ k φ[t0 ]kL2x
k=1
for a sufficiently large integer m, where φ solves the equation g φ = 0, the metric g satisfies
(27)-(32), P is the operator of projection on the set {ξ : 21 ≤ |ξ| ≤ 2} in Fourier space and
the initial data φ[t0 ] is supported in the ball B 1 (0). The argument here follows to some
2
extent the same lines as the one in Section 7, [8]. For the interior region, due to the fact
that P is an operator acting on the scale of size 1 and 1 − ω is a cut-off function with the
scale of size t ≥ 1, we can basically write
P ∂φ · (1 − ω) ≈ P (∂φ · (1 − ω))
28
Therefore, using Bernstein inequality and the fact that |∂ω| . t−1 , we have the following
chain of estimates:
kP ∂φ · (1 − ω)kL∞ (Σt ) . k∂P (∂φ · (1 − ω)kL2 (Σt )
. k∂ 2 φ · (1 − ω)kL2 (Σt ) +
1
k∂φ · (1 − ω)kL2 (Σt )
t
(107)
1 1
E 2 (∂φ)(t)
t
.
Remark In the interior region the decay is t−1 , better than the overall decay estimate of
1
t− 2 . This consideration will turn out to be crucial in further steps of the proof.
For the exterior region, let us first remark that, due to 2t . s . t, we can consider s ≈ t.
Using the Sobolev inequality (73), we can infer that
1
kP ∂φkL∞ (St,u ) . t 2 k∇
/ P ∂φkL2 (St,u ) +
1
1
t2
kP ∂φkL2 (St,u )
(108)
For the second term, applying the trace estimate (75) together with the fact that t ≥ 1, we
conclude that:
1
t
1
2
kP ∂φkL2 (St,u ) .
.
1
1
2
t
1
1
t2
1
1
kN P ∂φkL2 2 (Σt,ext ) · kP ∂φkL2 2 (Σt,ext ) +
1
kP ∂φkL2 (Σt,ext ) .
t
(109)
1
2
E (∂φ)(t)
In the case of the first term in (108), again due to (75), it follows that:
1
1
1
1
t 2 k∇
/ P ∂φkL2 (St,u ) . t 2 kN ∇
/ P ∂φkL2 2 (Σt,ext ) · k∇
/ P ∂φkL2 2 (Σt,ext ) + k∇
/ P ∂φkL2 (Σt,ext )
(110)
1
For kN ∇
/ P ∂φkL2 2 (Σt,ext ) we use the trivial estimate
1
1
1
kN ∇
/ P ∂φkL2 2 (Σt,ext ) . k∂φkL2 2 (Σt,ext ) . E 4 (∂φ)(t),
while for k∇
/ P ∂φkL2 (Σt,ext ) it will be enough to prove that it decays like
k∇
/ P ∂φkL2 (Σt,ext ) .
(111)
1
t
, or more precisely:
1 1
E 2 (∂φ)(t).
t
(112)
Commuting P with ∇
/ directly as in Lemma 7.1, [8], together with the null-frame equations
contained in Theorem 3.5, we obtain the following chain of estimates:
k∇
/ P ∂φkL2 (Σt,ext ) . sup k∂α eβA kL∞ (Σt,ext ) · k∂φkL2 (Σt ) + kP ∇
/ ∂φkL2 (Σt,ext ) .
α,β
. max{kχkL∞ (Σt,ext ) , kηkL∞ (Σt,ext ) , kκkL∞ (Σt,ext ) } · k∂φkL2 (Σt ) + k∇
/ ∂φkL2 (Σt,ext )
1−a
1
1
1
1 1
1 1
. ( + λ−ā + t 2 λ−ā− 2 + )E 2 (∂φ)(t) + E 2 (∂φ)(t) . E 2 (∂φ)(t).
t
t
t
29
(113)
The inequality used here:
1−a
1
1
t 2 λ−ā− 2 + .
(114)
t
√
is the one that gives the range a < 22 − 4. Putting together (108)-(112), we infer that:
1
1
kP ∂φkL∞ (St,u ) . t− 2 E 2 (∂φ)(t)
(115)
Thus, due to (107) and (115), Theorem 2.5 will be an immediate consequence of the following:
Theorem 5.1 If φ is a solution of the linear wave equation
g φ = 0,
φ|t=0 = φ0 ,
(116)
∂t φ|t=0 = φ1
with the metric g verifying the condition (27)-(32), a <
supported in the ball B 1 (0), then:
√
22 − 4 and the initial data φ[t0 ]
2
sup E(∂φ)(τ ) . E(∂φ)(t0 )
(117)
τ ∈[t0 ,t∗ ]
6
Outline of the proof of Theorem 5.1
In the case of dimension n = 3, Klainerman and Rodnianski proved the corresponding decay
estimate using only the Morawetz vectorfield K. They relied on the following generalized
energy estimate:
Z
Q(K, ∂t )(φ) & E(φ)(t),
(118)
Σt
where
Q(K, ∂t )(φ) =
n−1 2
1 2
u |Lφ|2 + u2 |Lφ|2 + (u2 + u2 )|Aφ|2 + (n − 1)tφ∂t φ −
φ.
4
2
As it was first remarked by Klainerman in [5], this inequality is no longer true in the case of
dimension n = 2. It survives however in the form:
Z
Q(K, ∂t )(φ) + kφk2L2 (Σt ) & E(φ)(t).
(119)
Σt
Due to the fact that we do not control kφkL2 (Σt ) , we will have to use (119) with φ → ∂t φ, the
quantity k∂φkL2 (Σt ) being bounded through the classical energy estimate. This modification
will help us recover the desired estimate (117) only in the interior region. The novelty of our
argument comes in the deduction of the required estimate in the exterior region. We will
30
use two other vectorfields, S and Ω. We will commute them with the D’Alembertian g ,
according to the formula:
g (Xφ) − X(g φ) =
X
π αβ · Dα Dβ φ + Dα X παλ · Dλ φ −
1 λ X
D tr π · Dλ φ = Z,
2
(120)
where
X
παβ = Dα Xβ + Dβ Xα
is the deformation tensor of the vectorfield X relative to the metric g. For the equation
g (Xφ) = Z we will use the classical energy estimate (see also (125)):
Z
k∂Xφ(t)kL2 (Σt ) . k∂Xφ(t0 )kL2 (Σt0 ) +
t
C R t k∂ g(τ,·)kL∞ (Σ ) dτ
τ
kZ(τ, ·)kL2 (Στ ) dτ · e t0 t
. (121)
t0
As it can be seen from (117), the goal will be to prove that:
Z t
1
kZ(τ, ·)kL2 (Στ ) dτ . λ− sup E 2 (∂φ)(τ ).
(122)
τ ∈[t0 ,t]
t0
We will express Z in equation (120) relative to the null frame {L, L, A}. Accordingly, the
deformation tensor X π appears in terms of the Ricci coefficients χ, η, the lapse function b
and the second fundamental form κ. Therefore, in order to estimate Z in (122), we will
need the bounds provided by the Asymptotics Theorem 4.1 in section 4. A typical term that
appears in (122), which comes from the combination
Ω
πLL DL DL φ , is
Z
t
Z
t
kη · s · L∂φkL2 (Στ ) dτ .
t0
kηkL∞ (Στ ) · ks · L∂φkL2 (Στ ) dτ .
t
Z 0t
1
kηkL∞ (Στ ) dτ sup E 2 (∂φ)(τ )
.
τ ∈[t0 ,t]
t0
The estimate provided by Theorem 4.1 for η is
1
kηkL∞ (Στ ) . λa−ā− 2 +
Hence, to obtain the estimate (122), we need the condition
Z t
1
1
λa−ā− 2 dτ . λ2a−ā− 2 + . λ− ,
t0
√
which gives the range a < 22 − 4. The last section will gather the estimates obtained
in the next two sections for the vectorfields K, Ω and respectively S, and so conclude the
argument.
31
7
Energy estimates involving the Morawetz vectorfield
K
The central result of this section will be:
Theorem 7.1 Under the assumptions of Theorem 5.1 :
E(∂t φ)(t) . E(∂t φ)(t0 ) + λ− sup E(∂φ)(τ )
(123)
τ ∈[t0 ,t∗ ]
In order to prove this theorem we start first with
Lemma 7.2 ([7]) If w is a solution of the equation g w = f , Qαβ is the energy momentum
tensor associated to this equation
Qαβ = ∂α w∂β w −
1
gαβ (g µν ∂µ w∂ν w) ,
2
X is a vectorfield with the deformation tensor
X
O is a scalar function,
X
π̃ =
X
παβ = Dα Xβ + Dβ Xα ,
π − Og and
Q̄(X, Y ) = Q(X, Y ) +
n−1
n−1 2
O·w·Yw −
w Y (O) ,
4
8
then the following identity holds:
Z
Z
Z
Z
1
n−1
αβ X
Q̄(X, ∂t ) +
Q̄(X, ∂t ) =
Q
π̃αβ −
w2 g O +
2
8
n
n
Σ
Σt
[t0 ,t]×R
[t0 ,t]×R
Z t0
n−1
(Xw +
+
Ow)f.
4
[t0 ,t]×Rn
Proof Differentiating directly in the formula for Qαβ , we obtain:
Dβ Qαβ = Dα w f.
If we consider the 1-form Pα = Qαβ X β , previous equation leads to
Dα Pα = Xw f +
1
Qαβ X π αβ ,
2
which we can detail as:
1 X αβ
n−1
n−1 2
π̃ Qαβ + (Xw +
Ow)f −
w g O −
2
4
8
n−1 µ
1
−
D (O · w · Dµ w − Dµ O · w2 ).
4
2
D α Pα =
32
(124)
Denoting
P̄α = Pα +
n−1
n−1 2
O · w · Dα w −
w Dα O ,
4
8
it follows that:
Dα P̄α =
n−1
n−1 2
1 αβ X
Q
π̃αβ + (Xw +
O w)f −
w g O
2
4
8
Integrating this equation over the domain [t0 , t] × Rn , we obtain (124).
Corollary 7.3 For w a solution of the equation g w = f , the following estimate holds:
Z t
C R t k∂ g(τ,·)kL∞ (Σ ) dτ
τ
(125)
kf (τ, ·)kL2 (Στ ) dτ · e t0 t
k∂w(t)kL2 (Σt ) . k∂w(t0 )kL2 (Σt0 ) +
t0
Proof Taking in the previous lemma, X = ∂t and O = 0, we can rewrite (124) in the
form:
Z
Z
1
1
2
ij
(|∂t w| + g ∂i w · ∂j w) =
(|∂t w|2 + g ij ∂i w · ∂j w) +
2 Σt
2 Σt0
Z
Z
(126)
1
αβ
Q · ∂t gαβ +
∂t w · f
+
2 [t0 ,t]×Rn
[t0 ,t]×Rn
Using the fact that the metric g ij is uniformly elliptic and bounded, we obtain
Z t
2
2
k∂w(t)kL2 (Σt ) . k∂w(t0 )kL2 (Σt0 ) +
k∂t g(τ )kL∞ (Στ ) · k∂w(τ )k2L2 (Στ ) +
t0
(127)
+ kf (τ )kL2 (Στ ) · k∂w(τ )kL2 (Στ ) dτ
Applying Gronwall inequality to this equation concludes the proof.
Proposition 7.4 Under the assumptions of Theorem 5.1 :
E(∂t φ)(t) . E(∂t φ)(t0 ) + λ− sup E(∂φ)(τ ) +
τ ∈[t0 ,t∗ ]
Z
1 αβ K
+|
Q
π̃αβ + (K∂t φ + τ ∂t φ)g (∂t φ)|
[t0 ,t]×R2 2
(128)
Proof In the Lemma 7.2, we will choose w = ∂t φ, n = 2, X = K and O = 4t, this
last choice being motivated by the fact that in Minkowski space tr K π = 4t and that we are
looking for an O which would make K π̄ small. (124) then becomes:
Z
Z
Z
Z
1
1
αβ K
Q
π̃αβ −
(∂t φ)2 g (τ ) +
Q̄(K, ∂t ) =
Q̄(K, ∂t ) +
2
2
2
2
[t0 ,t]×R
[t0 ,t]×R
Σt
Σ
Z t0
(129)
+
(K∂t φ + τ ∂t φ)g (∂t φ).
[t0 ,t]×R2
We take apart the term Q̄(K, ∂t ) and write it in the more convenient form:
33
1
(∂t φ)2
2
1 2
1
1 2
2
2
= u |L∂t φ| + u |L∂t φ| + (u2 + u2 )|A∂t φ|2 + t · ∂t φ · ∂tt φ −
4
4
4
1
2
− (∂t φ)
2
Q̄(K, ∂t ) = Q(K, ∂t ) + t · ∂t φ · ∂tt φ −
We will be interested in the particular term:
Z
Z
t · ∂t φ · ∂tt φ =
∂t φ · (S∂t φ − (t − u)N ∂t φ).
Σt
Σt
We separate and integrate by parts the term involving N ∂t φ:
Z
Z
1
−
(t − u)N ∂t φ · ∂t φ = −
(t − u) N ((∂t φ)2 ) =
2
Σt
ZΣt
Z
1
1 −1
2
=
(N (t − u) + (χ + κAA )(t − u))(∂t φ) =
(b + sχ + sκAA )(∂t φ)2
2
2
Σt
Σt
Using the asymptotic estimates:
|b − 1| . λ− ,
1
|χ − | + |κ| . λ−ā ,
s
we conclude that:
Z
Z
−
(t − u)N ∂t φ · ∂t φ =
Σt
(1 + O(λ− ))(∂t φ)2
Σt
Hence
Z
Z
Q̄(K, ∂t ) =
Σt
Σt
1 2
1
1
u |L∂t φ|2 + u2 |L∂t φ|2 + (u2 + u2 )|A∂t φ|2 +
4
4
4
1
+ S∂t φ · ∂t φ + ( + O(λ− ))(∂t φ)2
2
(130)
Using the expression for S, it follows naturally that:
Z
Z
2
2
2
2
2
2
2
Q̄(K, ∂t ) + (∂t φ)2 ,
u |L∂t φ| + u |L∂t φ| + (u + u )|A∂t φ| .
Σt
Σt
which of course implies
Z
E(∂t φ)(t) .
Q̄(K, ∂t ) + k∂t φk2L2 (Σt ) .
(131)
Σt
Applying the energy estimate (125), for our reduced linear problem g φ = 0, we can infer
that:
R
C t k∂ g(τ,·)kL∞ (Στ ) dτ
k∂φ(t)kL2 (Σt ) . k∂φ(t0 )kL2 (Σt0 ) · e t0 t
,
34
which due to
k∂gkL1t L∞
. λ−(1−a) ,
x
yields
k∂φ(t)kL2 (Σt ) . k∂φ(t0 )kL2 (Σt0 )
This estimate allows us to write (131) in the form:
Z
Q̄(K, ∂t ) + E(∂t φ)(t0 )
E(∂t φ)(t) .
(132)
(133)
Σt
R
Checking easily that g (t) = g ij κij , the integral term − 21 [t0 ,t]×R2 (∂t φ)2 g (τ ) can be estimated as:
Z
Z t
1
2
|−
(∂t φ) g (τ )| .
k∂t φ(t)k2L2 (Στ ) kg ij κij kL∞ (Στ ) dτ
2 [t0 ,t]×R2
t0
(134)
. λ− sup E(∂t φ)(τ )
τ ∈[t0 ,t∗ ]
Putting together (133) and (134) with the obvious inequality
Z
Q̄(K, ∂t ) . E(∂t φ)(t0 ) ,
Σt0
we obtain (128).
Proposition 7.5 The deformation tensor
K =
K
π of the Morawetz vectorfield
1 2
(u L + u2 L) ,
2
satisfies
K
πLL = −2u2 κN N
K
πLL = 2u(4(b−1 − 1) − uκN N )
K
πAA = u2 χ + u2 χ
K
πLL = −4(u + b−1 u) + (u2 + u2 )κN N
K
πLA = −2u2 κAN
K
πLA = u2 ξ + u2 (η + κAN )
Proof Using the formula for K, we can write:
K
1 2
1
u (< Dα L, ∂β > + < Dβ L, ∂α >) + u2 (< Dα L, ∂β > + < Dβ L, ∂α >)+
2
2
1
1
1
1
+ ∂α u2 < L, ∂β > + ∂β u2 < L, ∂α > + ∂α u2 < L, ∂β > + ∂β u2 < L, ∂α >
2
2
2
2
παβ =
35
This implies
K
πLL = u2 < DL L, L > + u2 < DL L, L > + L(u2 ) < L, L > + L(u2 ) < L, L >=
= −2u2 κN N
K
πLL = u2 < DL L, L > + u2 < DL L, L > + L(u2 ) < L, L > + L(u2 ) < L, L >=
= u2 (−2κN N ) + (−4u)(2 − 2b−1 ) = 2u(4(b−1 − 1) − uκN N )
K
πAA = u2 < DA L, A > + u2 < DA L, A > + A(u2 ) < L, A > + A(u2 ) < L, A >=
= u2 χ + u 2 χ
1
1
K
πLA = u2 (< DL L, A > + < DA L, L >) + u2 (< DL L, A > + < DA L, L >)+
2
2
1
1
1
1
+ L(u2 ) < L, A > + A(u2 ) < L, L > + L(u2 ) < L, A > + A(u2 ) < L, L >=
2
2
2
2
1 2
= u (2η − 2κAN ) = −2u2 κAN
2
1
1
K
πLA = u2 (< DL L, A > + < DA L, L >) + u2 (< DL L, A > + < DA L, L >)+
2
2
1
1
1
1
+ L(u2 ) < L, A > + A(u2 ) < L, L > + L(u2 ) < L, A > + A(u2 ) < L, L >=
2
2
2
2
2
2
= u ξ + u (η + κAN )
1 2
1
u (< DL L, L > + < DL L, L >) + u2 (< DL L, L > + < DL L, L >)+
2
2
1
1
1
1
+ L(u2 ) < L, L > + L(u2 ) < L, L > + L(u2 ) < L, L > + L(u2 ) < L, L >=
2
2
2
2
2
2
−1
= u κN N + u κN N − 4u b − 4u
K
πLL =
This concludes the proof of this proposition. Next, we will analyze the integral term in (128)
which involves the deformation tensor K π. In this direction, we can prove the next
Lemma 7.6 Under the assumptions of Theorem 5.1 :
Z
1 2
1 αβ K
1
|
Q
π̃αβ − u (χ − )∂t φ · g (∂t φ) · ω | . λ− sup E(∂φ)(τ )
2
4
s
2
τ ∈[t0 ,t∗ ]
[t0 ,t]×R
(135)
Proof Detailing the term Qαβ K π̃αβ , we obtain:
1 K
1
( π̃LL |L∂t φ|2 + K π̃LL |L∂t φ|2 ) + (K π̃AA (L∂t φ · L∂t φ + |A∂t φ|2 ) +
4
2
K
2
K
+ π̃LL |A∂t φ| ) − π̃LA A∂t φ · L∂t φ − K π̃LA A∂t φ · L∂t φ
Qαβ K π̃αβ =
Using the expressions for the deformation tensor of K provided by Proposition 7.5, we have
the following set of estimates:
|K π̃LL |L∂t φ|2 | . λ−ā |uL∂t φ|2
|K π̃LL |L∂t φ|2 | . λ−ā |uL∂t φ|2
|K π̃AA |A∂t φ|2 | . λ−ā |uA∂t φ|2
36
|K π̃LL |A∂t φ|2 | . λ−ā |uA∂t φ|2
|K π̃LA A∂t φ · L∂t φ| . λ−ā |uL∂t φ| · |uA∂t φ|
1
|K π̃LA A∂t φ · L∂t φ| . (λ−ā + τ 2 λ−ā−
1−a
+
2
)|uL∂t φ| · |uA∂t φ|
These estimates imply directly:
Z
1 αβ K
1K
|
Q
π̃αβ −
π̃AA L∂t φ · L∂t φ | . λ− sup E(∂φ)(τ ).
2
4
τ ∈[t0 ,t∗ ]
[t0 ,t]×R2
Due to the fact that
K
and because
(136)
1
1
π̃AA = u2 (χ + ) + u2 (χ − )
s
s
1
|u2 (χ + ) L∂t φ · L∂t φ| . λ−ā |uL∂t φ| · |uL∂t φ|
s
K
| π̃AA L∂t φ · L∂t φ · (1 − ω)| . λ−ā |τ ∂∂t φ · (1 − ω)|2
we can rewrite (136) in the form:
Z
1 αβ K
1 2
1
|
Q
π̃αβ − u (χ − ) L∂t φ · L∂t φ · ω | . λ− sup E(∂φ)(τ ).
2
4
s
2
τ ∈[t0 ,t∗ ]
[t0 ,t]×R
(137)
We are in a position to apply one of our integration results, (83), for the second term in the
above integral. It will follow that:
Z
Z
1 2
1 2
1
1
u (χ − ) L∂t φ · L∂t φ · ω =
u (χ − ) ∂t φ · g (∂t φ) · ω +
s
s
[t0 ,t]×R2 4
[t0 ,t]×R2 4
(138)
+ O(λ− ) sup E(∂φ)(τ )
τ ∈[t0 ,t∗ ]
provided the conditions set in Proposition 3.10 are fulfilled. Let us verify them using the
asymptotic estimates:
Z
t
t0
Z
1
1
| u2 (χ − )| . τ 2 · λ−ā . τ · λ−
4
s
Z t
1−a
1
1
1
1 2
1
/ ( u (χ − ))kL2 (Sτ,u ) dτ .
τ 2 (λ−ā− 2 + + τ − 2 · λ−ā ) dτ
3 · sup k∇
4
s
τ 2 0≤u≤ 2t
t0
t
t0
. λ−
Z t
1−a
1
1
1 2
1
1
u (χ − ))kL2 (Sτ,u ) dτ .
τ 2 (λ−ā− 2 + + τ − 2 · λ−ā +
3 · sup kL(
4
s
τ 2 0≤u≤ 2t
t0
1
+ τ 2 · λ−2ā ) dτ . λ−
This concludes, of course, the proof of our lemma. Having in mind (128) and (135), in order
to obtain (123), we are left to prove:
Z
1
1
2
. λ− sup E(∂φ)(τ ). (139)
K∂
φ
+
τ
∂
φ
+
u
(χ
−
)
∂
φ
·
ω
(∂
φ)
t
t
t
g
t
4
s
τ ∈[t0 ,t∗ ]
[t0 ,t]×R2
37
Let us concentrate for a moment on g (∂t φ). We have the formula:
g (∂t φ) = −∂t g ij · ∂i ∂j φ − ∂t (tr κ) · ∂t φ − ∂t (Γjii ) · ∂j φ.
(140)
Due to this formula we can split g (∂t φ) into two parts: A which contains only second order
derivatives for φ and B which gathers the first order derivatives of φ. Accordingly, we will
prove (139) by considering first the part related to B and then the part related to A. We do
it like this because some of the information obtained by proving the ”B-part” will be used
in proving the ”A-part”. We will further divide our argument into the one concerning the
interior part of the integral and the one concerning its exterior part.
7.1
Estimate of the B-part
Due to the formula (140):
B = − ∂t (tr κ) · ∂t φ − ∂t (Γjii ) · ∂j φ = (|∂ 2 g| + |∂g|2 ) · ∂φ
(141)
First we prove the interior estimate:
Proposition 7.7 Under the assumptions of Theorem 5.1 :
Z
K∂t φ + τ ∂t φ) · (1 − ω) B . λ− sup E(∂φ)(τ ).
2
(142)
τ ∈[t0 ,t∗ ]
[t0 ,t]×R
Proof Estimating directly we obtain
Z
Z
2
2
(u L(∂t φ) + u L(∂t φ) + τ ∂t φ) · (1 − ω) · B K∂t φ + τ ∂t φ)(1 − ω B = [t0 ,t]×R2
[t0 ,t]×R2
Z t
1
.
τ · E 2 (∂φ)(τ ) · k(|∂ 2 g| + |∂g|2 ) · ∂φ · (1 − ω)kL2 (Στ ) dτ .
t
Z 0t
Z t
1
1
2
.
τ E 2 (∂φ)(τ )k∂ g∂φ(1 − ω)kL2 (Στ ) dτ +
τ E 2 (∂φ)(τ )k|∂g|2 · ∂φ(1 − ω)kL2 (Στ ) dτ
t0
t0
The second term is easily estimated as follows:
Z t
1
τ · E 2 (∂φ)(τ ) · k|∂g|2 · ∂φ(1 − ω)kL2 (Στ ) dτ .
t0
Z t
Z t
1
2
.
τ · E 2 (∂φ)(τ ) · k|∂g| kL∞ (Σt ) · k∂φ(1 − ω)kL2 (Στ ) dτ .
τ · λ−2ā · E(∂φ)(τ ) dτ .
t0
. λ
−
t0
sup E(∂φ)(τ ) ,
τ ∈[t0 ,t∗ ]
38
while for the first one we have to be more subtle in our approach, because we do not control
k∂φ(1 − ω)kL∞ (Σt ) . We proceed like this:
Z t
1
τ ·E 2 (∂φ)(τ ) · k∂ 2 g · ∂φ · (1 − ω)kL2 (Στ ) dτ .
t0
Z t
1
τ · E 2 (∂φ)(τ ) · k∂ 2 gkLp (Στ ) · k∂φ(1 − ω)kLq (Στ ) dτ ,
.
t0
where
1
1
1
+
= .
p
q
2
Due to Bernstein inequality we have the estimate:
1
1
k∂ 2 gkLp (Στ ) . λ 2 − p k∂ 2 gkL2 (Στ ) .
Using Sobolev inequality we obtain:
2
k∂φ · (1 − ω)kLq (Στ ) . k∂ 2− q φ · (1 − ω)kL2 (Στ ) .
Estimating the last term by interpolation, we infer that:
1− 2
2
2
q
k∂ 2− q φ · (1 − ω)kL2 (Στ ) . k∂ 2 φ · (1 − ω)kL2 (Σ
· k∂φ · (1 − ω)kLq 2 (Στ ) .
τ)
2
1
. τ −1+ q · E 2 (∂φ)(τ ).
Putting altogether the last four estimates, we obtain:
Z t
Z t
2
1
1
2
2
τ · E (∂φ)(τ ) · k∂ g · ∂φ · (1 − ω)kL2 (Στ ) dτ .
τ q · λ q −ā · E(∂φ)(τ ) dτ .
t0
t0
. λ− sup E(∂φ)(τ )
τ ∈[t0 ,t∗ ]
for q large enough, which ends the proof of this result. Therefore, we are left to prove that:
Proposition 7.8 Under the assumptions of Theorem 5.1 :
Z
1 2
1 . λ− sup E(∂φ)(τ ).
K∂
φ
+
τ
+
u
(χ
−
)
∂
φ
·
ω
·
B
t
t
4
s
2
τ ∈[t0 ,t∗ ]
[t0 ,t]×R
(143)
Proof Using the integral estimate (82) corresponding to the exterior region, we obtain:
Z
1 2
1 K∂t φ + τ + u (χ − ) ∂t φ · ω · B .
4
s
[t0 ,t]×R2
Z t
1
.
τ · E 2 (∂φ)(τ ) · k(|∂ 2 g| + |∂g|2 ) · ∂φ · ωkL2 (Στ ) dτ .
t
Z 0t
1
.
τ 2 · sup k|∂ 2 g| + |∂g|2 kL2 (Sτ,u ) · E(∂φ)(τ ) dτ .
t0
0≤u≤ τ2
39
Z
t
.
1
τ 2 · (λ−ā−
1−a
+
2
1
+ τ 2 λ−2ā ) · E(∂φ)(τ ) dτ .
t0
. λ− sup E(∂φ)(τ ).
τ ∈[t0 ,t∗ ]
This concludes the discussion concerning B.
7.2
Estimate of the A-part
According to formula (140)
A = −∂t g ij · ∂i ∂j φ = ∂g · ∂ 2 φ.
(144)
We will prove that:
Proposition 7.9 Under the assumptions of Theorem 5.1 :
Z
K∂t φ + τ ∂t φ) · (1 − ω) A . λ− sup E(∂φ)(τ ).
2
(145)
τ ∈[t0 ,t∗ ]
[t0 ,t]×R
Proof Estimating directly, as in Proposition 7.7, we obtain
Z
K∂t φ + τ ∂t φ) · (1 − ω) A .
2
[t0 ,t]×R
Z t
1
.
τ · E 2 (∂φ)(τ ) · k|∂g| · |∂ 2 φ| · (1 − ω)kL2 (Στ ) dτ .
t
Z 0t
.
k∂gkL∞ (Στ ) · E(∂φ)(τ ) dτ . λ− sup E(∂φ)(τ )
τ ∈[t0 ,t∗ ]
t0
We investigate now the exterior part of the integral. Due to the fact that in the exterior
region, in the energy expression, all the terms but the L-derivative come with a weight
comparable to τ , we can infer that:
Z
1 2
1 0
(χ
−
K∂
φ
+
τ
+
u
)
∂
φ
·
ω
·
A
t
t
.
4
s
[t0 ,t]×R2
Z t
1
.
τ · E 2 (∂φ)(τ ) · kA0 · ωkL2 (Στ ) dτ .
(146)
t0
Z t
.
k∂gkL∞ (Στ ) · E(∂φ)(τ ) dτ . λ− sup E(∂φ)(τ ) ,
τ ∈[t0 ,t∗ ]
t0
where A0 differs from A just through the term which involves two L-derivatives on φ. We can
further simplify our investigation by replacing DL DL φ = DL D2T −L φ with L(∂t φ), because
the difference between these two terms is of A0 -type. Therefore, the expression to estimate
is:
Z
1 2
1 K∂t φ + τ + u (χ − ) ∂t φ · ω · AL · L(∂t φ) ,
4
s
[t0 ,t]×R2
40
where AL = T πLL = −2κN N . This is the coefficient of DL DL φ in the expression of A, as
it can easily by commuting the wave operator g with the vectorfield T = ∂t . Using the
trivial estimate |AL | . |∂g|, we immediately obtain:
Z
|
u2 · L(∂t φ) · AL · L(∂t φ)| .
[t0 ,t]×R2
Z
(147)
t
k∂gkL∞ (Στ ) · E(∂φ)(τ ) dτ . λ
.
−
sup E(∂φ)(τ ).
τ ∈[t0 ,t∗ ]
t0
We will investigate now the term
Z
1
1 τ + u2 (χ − ) ∂t φ · ω · AL · L(∂t φ)
4
s
[t0 ,t]×R2
Again, we are in a position to apply one of our integration estimates, (84), provided
τ + 14 u2 (χ − 1s ) · AL verifies the conditions imposed in Proposition 3.10. We check them
one by one:
1 1
| τ + u2 (χ − ) · AL | . (τ + τ 2 λ−ā ) · |∂g| . λ−
4
s
Z t
1
1 1
τ + u2 (χ − ) · AL kL2 (Sτ,u ) dτ .
1 · sup kL
4
s
t0 τ 2 0≤u≤ τ2
Z t 1
1 2
1 −ā
2 −ā
.
λ
sup kL τ + u (χ − ) kL2 (Sτ,u ) + (τ + τ λ ) sup kL(AL )kL2 (Sτ,u ) dτ
1
4
s
0≤u≤ τ2
0≤u≤ τ2
t0 τ 2
Z t
1−a
1−a
1
3
1
−ā
2
+ τ λ−ā− 2 + + τ 2 λ−2ā + τ 2 λ−2ā− 2 + ) dτ
.
1 · (τ λ
t0 τ 2
. λ− sup E(∂φ)(τ )
τ ∈[t0 ,t∗ ]
This allows to conclude:
Z
1 2
1 τ + u (χ − ) ∂t φ · ω · AL · L(∂t φ) . λ− sup E(∂φ)(τ )
4
s
τ ∈[t0 ,t∗ ]
[t0 ,t]×R2
Remark Already at this stage with the estimates obtained so far, we have:
Z
1
1
2
. λ− sup E(∂φ)(τ )
τ
+
(χ
−
u
)
·
ω
·
∂
φ
·
(∂
φ)
t
g
t
4
s
2
τ ∈[t0 ,t∗ ]
[t0 ,t]×R
The last term to consider is
Z
[t0
u2 · L(∂t φ) · AL · L(∂t φ) · ω
,t]×R2
We will use once again (83). The conditions to be satisfied are:
|u2 AL | . τ 2 · λ−ā . τ · λ−
41
(148)
(149)
Z
t
t0
Z
t
t0
1
τ
3
2
1
τ
3
2
Z
2
t
0≤u≤ 2t
1
1−a
+
2
+ λ−ā ) dτ . λ−
1
1−a
+
2
+ λ−ā ) dτ . λ−
(τ 2 λ−ā−
/ (u AL )kL2 (Sτ,u ) dτ .
· sup k∇
t0
· sup kL(u2 AL )kL2 (Sτ,u ) dτ .
0≤u≤ 2t
Z
t
(τ 2 λ−ā−
t0
Therefore, we can write:
Z
u2 · L(∂t φ) · AL · L(∂t φ) · ω =
2
[t0 ,t]×R
Z
=
u2 · AL · ω · ∂t φ · g (∂t φ) + O(λ− ) sup E(∂φ)(τ )
(150)
τ ∈[t0 ,t∗ ]
[t0 ,t]×R2
Taking into account the previous Remark and the fact that u2 AL is ”dominated” by
τ + 14 u2 (χ − 1s ), we claim that we will also obtain
Z
(151)
u2 · AL · ω · ∂t φ · g (∂t φ) . λ− sup E(∂φ)(τ ).
τ ∈[t0 ,t∗ ]
[t0 ,t]×R2
This concludes the argument for this section.
8
Energy estimates involving the scaling vectorfield S
and the angular momentum vectorfield Ω
The main result of this section is:
Theorem 8.1 Under the assumptions of Theorem 5.1 :
1
k∂SφkL2 (Σt ) . k∂SφkL2 (Σt0 ) + λ− sup E 2 (∂φ)(τ )
(152)
τ ∈[t0 ,t∗ ]
1
k∂ΩφkL2 (Σt ) . k∂ΩφkL2 (Σt0 ) + λ− sup E 2 (∂φ)(τ )
(153)
τ ∈[t0 ,t∗ ]
First we prove some preliminary results:
8.1
Commutation results
Here we will record the commutation formula between a vectorfield X and the D’Alembertian
g . This result will be used crucially in this section. We start first with
Lemma 8.2 ([6]) For a vectorfield X with the deformation tensor X π and a 1-form V on
a space-time manifold with metric g and corresponding connection D, we have
Dσ (LX Vα ) = LX (Dσ Vα ) +
42
X
Γασλ V λ
(154)
where
X
Γασλ =
1
(Dα X πσλ + Dσ X παλ − Dλ X πασ ).
2
Proof The formula for the Lie derivative gives us :
LX Vα = X γ · Dγ Vα + Vγ · Dα X γ
This implies:
Dσ (LX Vα ) = X γ · Dσ Dγ Vα + Dσ X γ · Dγ Vα + Dσ Vγ · Dα X γ + Vγ · Dσ Dα X γ .
On the other hand:
LX (Dσ Vα ) = X γ · Dγ Dσ Vα + Dσ X γ · Dγ Vα + Dσ Vγ · Dα X γ .
Subtracting term by term these relations, we end up with:
Dσ (LX Vα ) − LX (Dσ Vα ) = X γ (Dσ Dγ Vα − Dγ Dσ Vα ) + Vγ · Dσ Dα X γ
= Rαµσλ V µ X λ + Vγ · Dσ Dα X γ
We investigate the
X
X
Γασλ term, which we can detail as
1
Γασλ = (Dα Dσ Xλ + Dα Dλ Xσ + Dσ Dα Xλ + Dσ Dλ Xα − Dλ Dα Xσ − Dλ Dσ Xα )
2
1
= (Dα Dσ Xλ + Dσ Dα Xλ + Rσµαλ X µ + Rαµσλ X µ ) = Dσ Dα Xλ + Rαλσµ X µ
2
This, of course, concludes the proof of the lemma. We are now ready to state and prove our
main commutation result:
Proposition 8.3 ([6]) For an arbitrary vectorfield X with the deformation tensor X π, trace
tr X π = g αβ X παβ
and traceless part
X
π̂αβ =
X
παβ −
1
tr X π gαβ ,
n+1
we have the following commutation formula:
1 λ X
D tr π · Dλ φ
2
1
1
1
=X π̂ αβ · Dα Dβ φ + Dα X π̂αλ · Dλ φ + (
− )Dλ tr X π · Dλ φ +
tr X π · g φ
n+1
2
n+1
(155)
g (Xφ) − X(g φ) =
X
π αβ · Dα Dβ φ + Dα X παλ · Dλ φ −
43
Proof Using the decomposition g = g αβ Dα Dβ , we can infer that:
g (Xφ) − X(g φ) = g αβ Dα Dβ (DX φ) − LX (g αβ Dα Dβ φ) = g αβ Dα (DX Dβ φ +
+ Dβ X γ · Dγ φ) − LX (g αβ )Dα Dβ φ − g αβ LX (Dα Dβ φ) = g αβ (Dα LX Dβ φ−
− LX Dα Dβ φ) + (g αγ · Dγ X β + g βγ · Dγ X α )Dα Dβ φ
Here we take advantage of (154) to conclude that:
1
g (Xφ) − X(g φ) = g αβ · (Dα X πβλ + Dβ X παλ − Dλ X παβ )Dλ φ +
2
1
+ X π αβ · Dα Dβ φ = Dα X παλ · Dλ φ − Dλ tr X π · Dλ φ + X π αβ · Dα Dβ φ,
2
which ends the proof. An immediate consequence of this result is the following:
Lemma 8.4 Under the assumptions of Theorem 5.1:
1 λ Ω
D tr π · Dλ φ
2
1
g (Sφ) = S π̂ αβ · Dα Dβ φ + Dα S π̂αλ · Dλ φ − Dλ tr S π · Dλ φ
6
g (Ωφ) = Ω π αβ · Dα Dβ φ + Dα Ω παλ · Dλ φ −
8.2
(156)
(157)
Proof of estimate (153)
We start first with:
Proposition 8.5 The deformation tensor Ω π of the angular momentum vectorfield
Ω = s·A
satisfies
Ω
πLL = 0
Ω
πLL = 4s(η − κAN )
Ω
πAA = 0
Ω
πLL = −2s(η − κAN )
Ω
πLA = −sχ + 1
Ω
πLA = −sχ + 1 − 2b−1
tr Ω π = −Ω πLL + Ω πAA = 2s(η − κAN )
Proof Using the expression for Ω, we can infer that:
Ω
παβ = s (< Dα A, ∂β > + < Dβ A, ∂α >) + ∂α s < A, ∂β > + ∂β s < A, ∂α >
44
(158)
This implies:
Ω
πLL = 2s < DL A, L > + 2L(s) < A, L > = 0
Ω
πLL = 2s < DL A, L > + 2L(s) < A, L > = 2s (−2ξ) = 4s (η − κAN )
Ω
πAA = 2s < DA A, A > + 2A(s) < A, A > = 0
Ω
πLA = s (< DL A, A > + < DA A, L >) + L(s) < A, A > + A(s) < A, L > =
= − sχ + 1
Ω
πLA = s(< DL A, A > + < DA A, L >) + L(s) < A, A > + A(s) < A, L > =
= − sχ + 1 − 2b−1
Ω
πLL = s(< DL A, L > + < DL A, L >) + L(s) < A, L > + L(s) < A, L > =
= s(−2η − 2η) = − 2s(η − κAN ).
Using the energy estimate (125) applied to equation (156), we obtain:
k∂ΩφkL2 (Σt ) . k∂ΩφkL2 (Σt0 ) +
Z t
1
+
kΩ π αβ · Dα Dβ φ + Dα Ω παλ · Dλ φ − Dλ tr Ω π · Dλ φkL2 (Στ ) dτ
2
t0
Detailing the first integral term, we infer that:
Z t
Z t
Ω αβ
k π Dα Dβ φkL2 (Στ ) dτ .
kΩ πLL DL DL φkL2 (Στ ) + kΩ πLL DL DL φkL2 (Στ ) +
t0
t0
Ω
+ k πAA DA DA φkL2 (Στ ) + kΩ πLA DA DL φkL2 (Στ ) + kΩ πLA DL DA φkL2 (Στ ) +
Ω
+ k πLL DL DL φkL2 (Στ ) dτ
The first and the third terms of the right hand side vanishes due to the formulae in Proposition 8.5. Investigating one by one the remaining terms we obtain:
Z t
Z t
Ω
ks(η − κAN )(L(Lφ) + κN N Lφ)kL2 (Στ ) dτ .
k πLL DL DL φkL2 (Στ ) dτ .
t0
t0
Z t
1−a
1
1
.
(1 + λ−ā τ )(τ 2 λ−ā− 2 + + λ−ā )E 2 (∂φ)(τ ) dτ .
t0
1
. λ− sup E 2 (∂φ)(τ )
τ ∈[t0 ,t∗ ]
Z
t
kΩ πLA DA DL φkL2 (Στ ) dτ .
t0
Z
t
t
Z 0t
.
1
ks(χ − )(A(Lφ) − χAφ − κAN Lφ)kL2 (Στ ) dτ
s
1
(τ λ−2ā + λ−ā )E 2 (∂φ)(τ ) dτ
t0
1
. λ− sup E 2 (∂φ)(τ )
τ ∈[t0 ,t∗ ]
45
t
Z
Z
Ω
t
k(−sχ + 1 − 2b−1 )(L(Aφ) + κAN Lφ)kL2 (Στ ) dτ
k πLA DL DA φkL2 (Στ ) dτ .
t
t0
Z 0t
.
1
(τ λ−2ā + λ−ā )E 2 (∂φ)(τ ) dτ
t0
1
. λ− sup E 2 (∂φ)(τ )
τ ∈[t0 ,t∗ ]
Z
t
kΩ πLL DL DL φkL2 (Στ ) dτ .
t0
Z
t
ks(η − κAN )(L(Lφ) − κN N Lφ + 2κAN Aφ)kL2 (Στ ) dτ
t
Z 0t
.
1
(1 + λ−ā τ )(τ 2 λ−ā−
1−a
+
2
1
+ λ−ā )E 2 (∂φ)(τ ) dτ
t0
1
. λ− sup E 2 (∂φ)(τ )
τ ∈[t0 ,t∗ ]
The other two terms that we encounter in the commutation formula (156) will be considered
together. Lowering the upper indices we obtain:
1
Dα Ω παβ · Dβ φ − Dα (tr Ω π) · Dα φ =
2
1
1
1
1
= ( DL Ω πLL + DL Ω πLL − DA Ω πAL + L(tr Ω π)) · Lφ+
4
4
2
4
1
1
1
1
+ ( DL Ω πLL + DL Ω πLL − DA Ω πAL + L(tr Ω π)) · Lφ+
4
4
2
4
1
1
1
+ (− DL Ω πLA − DL Ω πLA + DA Ω πAA − A(tr Ω π)) · Aφ
2
2
2
(159)
Detailing each parenthesis, we can infer that:
1
1
1
1
1
DL Ω πLL + DL Ω πLL − DA Ω πAL + L(tr Ω π) = (L(Ω πLL ) − 2η Ω πAL −
4
4
2
4
4
1
− κN N Ω πLL − 2ξ Ω πAL + κN N Ω πLL ) + (L(Ω πLL ) + 4κAN Ω πAL − 2κN N Ω πLL )−
4
1
1
1
1
− (A(Ω πAL ) − χ Ω πLL − χ Ω πLL − χ Ω πAA − κAN Ω πAL ) + L(−Ω πLL )
2
2
2
4
Remark We notice here the cancellation of 14 L(Ω πLL ) with the last term 41 L(−Ω πLL ). These
terms contain Lη for which, as mentioned before, we do not have good estimates. We wouldn’t
have been able to prove our result, unless this cancellation took place.
1
1
1
1
1
DL Ω πLL + DL Ω πLL − DA Ω πAL + L(tr Ω π) = (L(Ω πLL ) − 4η Ω πAL −
4
4
2
4
4
1
1
− 2κN N Ω πLL ) + (L(Ω πLL ) + 2κAN Ω πAL − κN N Ω πLL + κN N Ω πLL ) − (A(Ω πAL )−
4
2
1 Ω
1 Ω
1
− χ πLL − χ πLL − χ Ω πAA + κAN Ω πAL ) + L(−Ω πLL )
2
2
4
46
1
1
1
1
− DL Ω πLA − DL Ω πLA + DA Ω πAA − A(tr Ω π) = − (L(Ω πLA ) − 2η Ω πAA −
2
2
2
2
1
− κN N Ω πAL − η Ω πLL − ξ Ω πLL ) − (L(Ω πLA ) + 2κAN Ω πAA − κN N Ω πAL +
2
1
+ κAN Ω πLL ) + (A(Ω πAA ) − χ Ω πLA − χ Ω πLA ) + A(Ω πLL )
2
Taking into account the explicit formulae for the different components of the deformation
tensor provided in Proposition 8.5, we can reduce the above expressions to
1
1
1
1
DL Ω πLL + DL Ω πLL − DA Ω πAL + L(tr Ω π) = L(s(η − κAN ))−
4
4
2
4
1
1
− A(−s χ + 1 − 2b−1 ) − (η + κAN )(−s χ + 1 − 2b−1 )+
2
2
1
+ (η − κAN )( + s χ + s κAA − s κN N )
2
(160)
1
1
1
1
1
DL Ω πLL + DL Ω πLL − DA Ω πAL + L(tr Ω π) = − A(−s χ + 1)−
4
4
2
4
2
(161)
1
− η(−s χ + 1) − s χ(η − κAN )
2
1
1
1
1
− DL Ω πLA − DL Ω πLA + DA Ω πAA − A(tr Ω π) = − L(−s χ + 1)−
2
2
2
2
1
1
(162)
− L(−s χ + 1 − 2b−1 ) − A(s (η − κAN )) + ( κN N − χ)(−s χ + 1 − 2b−1 )+
2
2
1
+ ( κN N − χ)(−s χ + 1) − s(η − κAN )2
2
All the terms in these expressions which do not have a derivative with respect to the null
frame {L, L, A} and which we denote generically by X, can be estimated in the L∞ -norm as
follows
1−a
1
|X| . (1 + λ−ā τ )(τ 2 λ−ā− 2 + + λ−ā )
(163)
Then, the integral expressions corresponding to these terms satisfy:
Z t
Z t
1−a
1
1
kX ∂φkL2 (Στ ) dτ .
(1 + λ−ā τ )(τ 2 λ−ā− 2 + + λ−ā )E 2 (∂φ)(τ ) dτ
t0
t0
(164)
1
. λ− sup E 2 (∂φ)(τ )
τ ∈[t0 ,t∗ ]
We denote the terms which involve derivatives by Y . For their corresponding integrals we
split the discussion into the interior part, where we will use a similar argument as in the
case of Morawetz vectorfield K, and the exterior part, where we will take advantage of the
integral estimate (82):
Z t
Z t
Z t
kY ∂φkL2 (Στ ) dτ .
kY ∂φ ωkL2 (Στ ) dτ +
kY ∂φ (1 − ω)kL2 (Στ ) dτ .
t0
t0
t0
47
Z
t
τ
.
− 21
t
τ
.
t0
t
− 21
1
τ −1 kY (1 − ω)kL2 (Στ ) E 2 (∂φ)(τ ) dτ
sup kY kL2 (Sτ,u ) E (∂φ)(τ ) dτ +
0≤u≤ τ2
t0
Z
Z
1
2
t0
Z
1
2
t
0≤u≤ τ2
1
1
τ − 2 sup kY kL2 (Sτ,u ) E 2 (∂φ)(τ ) dτ
sup kY kL2 (Sτ,u ) E (∂φ)(τ ) dτ +
τ
≤u≤τ
2
t0
A typical ”Y ”-term is, for example, L(s(η − κAN )). Using the asymptotic estimates (85), we
can evaluate it as follows:
kL(s(η − κAN ))kL2 (Sτ,u ) . s kL(η − κAN )kL2 (Sτ,u ) + kη − κAN kL2 (Sτ,u )
. s (λ−ā−
. s λ−ā−
1−a
+
2
1−a
+
2
1
1
1
+ s− 2 λ−ā ) + s 2 (s 2 λ−ā−
1−a
+
2
+ λ−ā )
1
+ s 2 λ−ā
Hence, the general estimate for Y is:
sup kY kL2 (Sτ,u ) . τ λ−ā−
1−a
+
2
1
+ τ 2 λ−ā
(165)
0≤u≤τ
Therefore the integral above can be evaluated by:
Z t
Z t
1−a
1
1
kY ∂φkL2 (Στ ) dτ .
(τ 2 λ−ā− 2 + + λ−ā )E 2 (∂φ)(τ ) dτ
t0
t0
. λ
1
2
−
sup E (∂φ)(τ )
τ ∈[t0 ,t∗ ]
This concludes the proof of estimate (153).
8.3
Proof of estimate (152)
As in the previous section, we start with:
Proposition 8.6 The deformation tensor
S =
S
π of the scaling vectorfield
1
(uL + uL)
2
satisfies
S
πLL = −2uκN N
S
πLL = 4(b−1 − 1) − 2uκN N
S
πAA = uχ + uχ
S
πLL = −2(1 + b−1 ) + (u + u)κN N
S
πLA = −2uκAN
S
πLA = uξ + u(η + κAN )
tr S π = −S πLL + S πAA = uχ + uχ − (u + u)κN N + 2(1 + b−1 )
48
(166)
Proof Using the above formula for S, we infer that:
S
1
1
u(< Dα L, ∂β > + < Dβ L, ∂α >) + u(< Dα L, ∂β > + < Dβ L, ∂α >)+
2
2
1
1
1
1
+ ∂α u < L, ∂β > + ∂β u < L, ∂α > + ∂α u < L, ∂β > + ∂β u < L, ∂α >
2
2
2
2
παβ =
This implies
S
πLL = u < DL L, L > + u < DL L, L > + L(u) < L, L > + L(u) < L, L >=
= −2uκN N
S
πLL = u < DL L, L > + u < DL L, L > + L(u) < L, L > + L(u) < L, L >=
= u(−2κN N ) + (−2)(2 − 2b−1 ) = 4(b−1 − 1) − 2uκN N
S
πAA = u < DA L, A > + u < DA L, A > + A(u) < L, A > + A(u) < L, A >=
= uχ + uχ
1
1
S
πLA = u(< DL L, A > + < DA L, L >) + u(< DL L, A > + < DA L, L >)+
2
2
1
1
1
1
+ L(u) < L, A > + A(u) < L, L > + L(u) < L, A > + A(u) < L, L >=
2
2
2
2
1
= u (2η − 2κAN ) = −2u κAN
2
1
1
S
πLA = u(< DL L, A > + < DA L, L >) + u(< DL L, A > + < DA L, L >)+
2
2
1
1
1
1
+ L(u) < L, A > + A(u) < L, L > + L(u) < L, A > + A(u) < L, L >=
2
2
2
2
= uξ + u(η + κAN )
1
1
S
πLL = u(< DL L, L > + < DL L, L >) + u(< DL L, L > + < DL L, L >)+
2
2
1
1
1
1
+ L(u) < L, L > + L(u) < L, L > + L(u) < L, L > + L(u) < L, L >=
2
2
2
2
−1
= u κN N + u κN N − 2b − 2
As in the case of Ω, we will use the energy estimate (125) for the equation (157). This
implies:
k∂SφkL2 (Σt ) . k∂SφkL2 (Σt0 ) +
Z t
1
+
kS π̂ αβ · Dα Dβ φ + Dα S π̂αλ · Dλ φ − Dλ tr S π · Dλ φkL2 (Στ ) dτ
6
t0
Using π̂ instead of π will affect only two terms:
π̂LL = S πLL − 2 gLL = 2t κN N + 2(1 − b−1 )
1
1
S
π̂AA = S πAA − 2 gAA = u(χ + ) + u(χ − )
s
s
S
49
As before, we will first deal with the terms which do not involve derivatives of the deformation
tensor:
Z t
Z t
S
ku κN N (L(Lφ) − 2ξAφ + κN N Lφ)kL2 (Στ ) dτ
k π̂LL DL DL φkL2 (Στ ) dτ .
t0
t0
Z t
1−a
1
λ−ā ku L(∂φ)kL2 (Στ ) + τ λ−ā (τ 2 λ−ā− 2 + + λ−ā )k∂φkL2 (Στ ) dτ
.
t0
1
. λ− sup E 2 (∂φ)(τ )
τ ∈[t0 ,t∗ ]
Z
t
Z
t
Z
t
k(4(b−1 − 1) − 2uκN N )(L(Lφ) + κN N Lφ)kL2 (Στ ) dτ
k π̂LL DL DL φkL2 (Στ ) dτ .
t0
t0
Z t
.
λ−ā kτ L(∂φ)kL2 (Στ ) + τ λ−2ā k∂φkL2 (Στ ) dτ .
S
t0
1
. λ− sup E 2 (∂φ)(τ )
τ ∈[t0 ,t∗ ]
Z
t
S
k π̂LA DA DL φkL2 (Στ ) dτ .
k2u κAN (L(Aφ) − ξLφ − ηLφ)kL2 (Στ ) dτ
t0
t0
Z t
1−a
1
.
λ−ā ku L(∂φ)kL2 (Στ ) + τ λ−ā (τ 2 λ−ā− 2 + + λ−ā )k∂φkL2 (Στ ) dτ
t0
1
. λ− sup E 2 (∂φ)(τ )
τ ∈[t0 ,t∗ ]
Z
t
Z
S
t
k π̂LA DL DA φkL2 (Στ ) dτ .
k(u ξ + u(η + κAN ))(L(Aφ) + κAN Lφ)kL2 (Στ ) dτ
t0
Z t
1−a
1−a
1
1
.
(τ 2 λ−ā− 2 + + λ−ā )kτ L(∂φ)kL2 (Στ ) + τ λ−ā (τ 2 λ−ā− 2 + + λ−ā )k∂φkL2 (Στ ) dτ
t0
t0
1
. λ− sup E 2 (∂φ)(τ )
τ ∈[t0 ,t∗ ]
Z
t
Z
t
k π̂LL DL DL φkL2 (Στ ) dτ .
k(2t κN N + 2(1 − b−1 ))(L(Lφ) − κN N Lφ +
t0
t0
Z t
+ 2κAN Aφ)kL2 (Στ ) dτ .
λ−ā kτ L(∂φ)kL2 (Στ ) + τ λ−2ā k∂φkL2 (Στ ) dτ
S
t0
.λ
−
1
2
sup E (∂φ)(τ )
τ ∈[t0 ,t∗ ]
t
1
k π̂AA DA DA φkL2 (Στ ) dτ .
k(2s(χ − ) − u κAA )(A(Aφ) − χLφ−
s
t0
t0
Z t
− tr χLφ)kL2 (Στ ) dτ .
λ−ā kτ A(∂φ)kL2 (Στ ) + τ λ−2ā k∂φkL2 (Στ ) dτ +
t0
Z t
Z t
1
1
1
−
+
k κAA u∂φkL2 (Στ ) dτ . λ
sup E 2 (∂φ)(τ ) +
k κAA u∂φ(1 − η̄)kL2 (Στ ) dτ
τ ∈[t0 ,t∗ ]
t0 s
t0 s
Z
t
S
Z
To conclude the analysis of these terms, we have to deal with the last integral.
50
Z t
1
1 1
1
k κAA u (1 − η̄)kL2 (Στ ) E 2 (∂φ)(τ ) dτ .
k κAA u ∂φ(1 − η̄)kL2 (Στ ) dτ .
t0 τ s
t0 s
Z t
1
1
k κAA (1 − η̄)kL2 (Στ ) dτ sup E 2 (∂φ)(τ )
.
τ ∈[t0 ,t∗ ]
t0 s
Z
t
Due to the fact that
sup |κ| . min{sλ−ā−(1−a) , λ−ā }
St,u
2
the last L -norm can be estimated as follows :
Z 1 Z
Z τ
Z
2 1
1
1
2
2
k κAA (1 − η̄)kL2 (Στ ) .
|κ| dσ ds +
|κ|2 dσ ds .
2
2
s
0 s
1 s
Sτ,u
Sτ,u
Z τ
2 1
1
2 −2ā−2(1−a)
.
s
s
λ
ds
+
s λ−2ā ds . λ−2ā−2(1−a) + λ−2ā ln τ
2
2
s
s
0
1
This result obviously implies:
Z t
1
k κAA (1 − η̄)kL2 (Στ ) dτ . λ−
t0 s
1
Z
(167)
Next we investigate the terms which involve derivatives of the tr S π. Before we start, let us
first write tr S π in the form
1
tr S π = 4 + 2b−1 + 2s(χ − ) − 2u κAA − 2t κN N
s
Rt
The expression to evaluate is t0 kDα (tr S π)Dα φkL2 (Στ ) dτ . All the terms that appear from
the differentiation of tr S π, with the exception of u ∇
/ (κAA ) and t ∇
/ (κN N ), can be estimated in
a similar manner like the ”Y ”-term from the previous section. All these estimates yield the
1
desired upper bound λ− supτ ∈[t0 ,t∗ ] E 2 (∂φ)(τ ). We show the proof for the term involving
κAA . As before, we split the argument in the interior and the exterior part and use the
corresponding integral estimates:
Z t
Z t
1
1
ku ∇
/ (κAA )∂φkL2 (Στ ) dτ .
τ 2 sup k∇
/ (κAA )kL2 (Sτ,u ) E 2 (∂φ)(τ ) dτ +
t0
t0
Z
t
0≤u≤ τ2
1
2
k∇
/ (κAA )(1 − ω)kL2 (Στ ) E (∂φ)(τ ) dτ . λ
+
t0
−
Z
t
+
k∇
/ (κAA )(1 − ω)kL2 (Στ ) dτ ·
t0
1
2
· sup E (∂φ)(τ )
τ ∈[t0 ,t∗ ]
We are left to investigate the norm under the last integral, for which we need the following
weak estimate:
sup |∇
/ (κAA )| . s λ−ā−2(1−a)
Sτ,u
51
This is obtained immediately from the trivial bound
|
d
(∇
/ (κAA ))| . |∂ 3 g| + |∂g| · |∂ 2 g|
ds
and the following estimate (30) satisfied by the metric g:
−ā−(1−a)m
∞ . λ
k∂ 1+m gλ kL∞
,
t Lx
true for all integers 0 ≤ m. This implies then:
Z 1Z
Z
2
2
k∇
/ (κAA )(1 − ω)kL2 (Στ ) .
|∇
/ (κAA )| dσ ds +
0
Z
Sτ,u
s s2 λ−2ā−4(1−a) ds +
0
−2ā−4(1−a)
.λ
Z
1
1
.
τ
2
τ
2
Z
|∇
/ (κAA )|2 dσ ds .
Sτ,u
(λ−2ā−(1−a) + s−1 λ−2ā ) ds .
1
+ τλ
−2ā−(1−a)
+ λ−2ā ln τ
This estimate implies then
Z
t
k∇
/ (κAA )(1 − ω)kL2 (Στ ) dτ . λ−
(168)
t0
which is, of course, the desired bound. In order to conclude our argument we have to estimate
the term
Z t
kDα S π̂αβ · Dβ φkL2 (Στ ) dτ .
t0
As in the discussion for the angular momentum vectorfield Ω, in the decomposition of the
expression above we encounter two type of terms:
- terms which contain derivatives of the deformation tensor, which we denote generically by
Y 0;
- terms which do not contain derivatives of the deformation tensor, which we denote generically by X 0 . Through a very easy inspection, we have the estimates:
|Y 0 | . |Y | + |∇
/ (κ)|
1
|X 0 | . |X| + | τ κ|
s
(169)
As it can obviously be seen, these terms were covered previously, so this concludes our
discussion concerning the scaling vectorfield S.
Remark Comparing with the investigation of Ω, the discussion regarding S had difficulties
in evaluating the interior part of the integral terms, namely singularities appeared due to
negative powers of s. This is determined by the fact that, for the interior region, in the
case of S, u appears as minimal weight, while for the discussion regarding Ω, due also to
Ω = s · A, s is the weight that appeared more often.
52
9
The conclusion of the argument
At the end of section 5, we reduced the local well-posedness result up to the proof of Theorem
5.1, which we recall here in the form:
Theorem 9.1 Under the assumptions of Theorem 5.1 :
sup E(∂φ)(τ ) . E(∂φ)(t0 )
(170)
τ ∈[t0 ,t∗ ]
With the estimates obtained in the course of the previous two chapters, we can prove now
this theorem. We will proceed as follows:
9.1
Proof of the estimate (170) in the exterior region {s & 2t }
First we prove the following
Lemma 9.2 Under the assumptions of Theorem 5.1 we have the following commutator estimates:
k[∂, Ω]φkL2 (Σt ) . λ− k∂φkL2 (Σt )
(171)
k[∂, S]φkL2 (Σt ) . k∂φkL2 (Σt )
Proof It is obvious that the lemma will follow if we verify the commutator estimates for
∂ replaced with either L, L or A. For the angular momentum vectorfield Ω we have
[L, Ω] = L(s · A) − s · AL = (1 − s χ)A ,
[L, Ω] = L(s · A) − s · AL = (1 − 2 b−1 − s χ)A + 2s(κAN − η)N ,
[A, Ω] = A(s · A) − s · A2 = 0 ,
while for the scaling vectorfield S, the calculations go like this:
1
1
1
1
L(u)L + u[L, L] + L(u)L + u[L, L] = (−u(κAN + η))A +
2
2
2
2
1
1
+ (1 − uκN N )L + uκN N L
2
2
1
1
1
1
[L, S] = L(u)L + u[L, L] + L(u)L + u[L, L] = (−u(κAN + η))A +
2
2
2
2
1
1
+ (1 − b−1 + uκN N )L + (b−1 − uκN N )L
2
2
1
1
1
1
1
1
[A, S] = A(u)L + u[A, L] + A(u)L + u[A, L] = ( utr χ + uχ)A +
2
2
2
2
2
2
+ u(η − κAN )T
[L, S] =
These equations together with the asymptotic estimates yield the desired outcome.
53
Lemma 9.3 Under the assumptions of Theorem 5.1, the following estimates hold:
ku · A∂φ · η̄kL2 (Σt ) . kΩ∂φkL2 (Σt )
ku · L∂φ · η̄kL2 (Σt ) . kΩ∂φkL2 (Σt ) + kS∂φkL2 (Σt ) + k∂φkL2 (Σt )
(172)
ku · L∂φ · η̄kL2 (Σt ) . kΩ∂φkL2 (Σt ) + kS∂φkL2 (Σt ) + k∂φkL2 (Σt )
Proof The first estimate is almost trivial due to the fact that in the exterior region u ≈
t ≈ s. We can conclude
ku · A∂φ · η̄kL2 (Σt ) . ks · A∂φkL2 (Σt ) = kΩ∂φkL2 (Σt )
Let us observe that it is enough to prove only the estimates for L, the ones for L following
immediately from
u L = 2S − uL
In order to prove the estimates for L we use the equation:
g φ = − DL DL φ + DA DA φ =
= − L(Lφ) + A(Aφ) − (κN N +
1
1
χ)Lφ − χ Lφ = 0
2
2
(173)
This yields of course
|s · L(Lφ)| . |s · A(Aφ)| + |Lφ| + |Lφ|
which in turn implies:
ku · L(Lφ) · η̄kL2 (Σt ) . ks · L(Lφ)kL2 (Σt ) . kΩ∂φkL2 (Σt ) + k∂φkL2 (Σt )
(174)
Next we investigate |u · L(Lφ)|. We proceed like this:
|u · L(Lφ)| = |2S(Lφ) − u · L(Lφ)| . |S(Lφ)| + |u · L(Lφ)| + |u · [L, L]φ| .
. |S(Lφ)| + |u · L(Lφ)| + |u(η + κAN )Aφ| + |uκN N N φ|
This estimate together with the previous one yield:
ku · L(Lφ) · η̄kL2 (Σt ) . kΩ∂φkL2 (Σt ) + kS∂φkL2 (Σt ) + k∂φkL2 (Σt )
(175)
Finally:
|s · L(Aφ)| . |s · A(Lφ)| + |s · [L, A]φ| . |Ω(Lφ)| + |s · χ Aφ|
which enables us to infer that
ku · L(Aφ) · η̄kL2 (Σt ) . ks · L(Aφ)kL2 (Σt ) . kΩ∂φkL2 (Σt ) + k∂φkL2 (Σt )
(176)
Putting altogether the previous results, we finally conclude:
1
2
Eext
(∂φ)(t) . ku · A∂φ · η̄kL2 (Σt ) + ku · L∂φ · η̄kL2 (Σt ) + ku · L∂φ · η̄kL2 (Σt ) +
+ k∂φkL2 (Σt ) . kΩ∂φkL2 (Σt ) + kS∂φkL2 (Σt ) + k∂φkL2 (Σt ) .
1
2
. E (∂φ)(t0 ) + λ
−
1
2
sup E (∂φ)(τ )
τ ∈[t0 ,t∗ ]
54
(177)
9.2
Proof of the estimate (170) in the interior region {s . 2t }
Due to the result obtained in Theorem 7.1, for our purposes it will be enough to consider
Eint (∇φ), where ∇ designates spatial derivatives. We will prove the following result:
Theorem 9.4 Under the assumptions of Theorem 5.1 , we have the following estimate:
Eint (∇φ)(t) . Eint (∂t φ)(t) + E(∂φ)(t0 )
(178)
Proof The approach is direct, using integration by parts. When there is no ambiguity we
will write the terms symbolically.
Z
Z
2
ij
Eint (∇φ)(t) = t
g · ∂∂i φ · ∂∂j φ · (1 − ω) +
|∇φ|2 · (1 − ω)
Σt
Σt
Due to the energy estimate (132), the second term is bounded easily by:
Z
|∇φ|2 · (1 − ω) . k∂φ(t0 )kL2 (Σt0 ) . E(∂φ)(t0 )
(179)
Σt
For the first integral we proceed as follows:
Z
Z
2
ij
2
t
g · ∂∂i φ · ∂∂j φ · (1 − ω) = t
g ij · ∂t ∂i φ · ∂t ∂j φ · (1 − ω) +
Σt
Z Σt
g ij · g kl ∂k ∂i φ · ∂l ∂j φ · (1 − ω)
+ t2
Σt
Obviously, the first term has the upper bound:
Z
2
t
g ij · ∂t ∂i φ · ∂t ∂j φ · (1 − ω) . Eint (∂t φ)(t) ,
(180)
Σt
while for the second one we will use, as mentioned above, integration by parts
Z
2
t
g ij · g kl · ∂k ∂i φ · ∂l ∂j φ · (1 − ω) =
Σt
Z
Z
2
ij
kl
2
= −t
g · g · ∂i φ · ∂k ∂l ∂j φ · (1 − ω) + t
g · ∂g · ∂φ · ∂ 2 φ · (1 − ω) +
Σt
Z Σt
Z
+ t2
g · g · ∂φ · ∂ 2 φ · ∂ω = t2
|∆φ|2 · (1 − ω) +
Σt
ZΣt
Z
2
2
2
+t
g · ∂g · ∂φ · ∂ φ · (1 − ω) + t
g · g · ∂φ · ∂ 2 φ · ∂ω
Σt
Σt
Using the equation satisfied by φ, we can infer that:
∆φ = ∂tt φ + ∂g · ∂φ
55
(181)
If we plug in this expression for ∆φ in the above computation and take into account the
estimates for k∂gkL∞
, we obtain the following estimate:
x
Z
Z
2
ij
kl
2
t
g · g ∂k ∂i φ · ∂l ∂j φ · (1 − ω) . t
|∂tt φ|2 + |∂g|2 · |∂φ|2 +
Σt
Σt
2
+ |g| · |∂g| · |∂φ| · |∂ φ| + |g| · |g| · |∂φ| · |∂ 2 φ| · |∂ω| · (1 − ω) .
−
−
1
2
1
2
(182)
. Eint (∂t φ)(t) + λ E(∂φ)(t0 ) + λ E (∂φ)(t0 ) · Eint (∂φ)(t) +
1
1
2
+ E 2 (∂φ)(t0 ) · Eint
(∂φ)(t)
Putting together (179),(180),(182), we conclude that
1
1
2
Eint (∇φ)(t) . Eint (∂t φ)(t) + E(∂φ)(t0 ) + E 2 (∂φ)(t0 ) · Eint
(∂φ)(t)
which clearly implies the result claimed.
Remark The conclusion of this section, (170), will then be the cumulative result of (123),(177)
and (178).
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57