arXiv:1107.0847v1 [math.AP] 5 Jul 2011
THE GLASSEY CONJECTURE WITH RADIALLY SYMMETRIC
DATA
KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA
Abstract. In this paper, we verify the Glassey conjecture in the radial case
for all spatial dimensions. Moreover, we are able to prove the existence results
with low regularity assumption on the initial data and extend the solutions to
the sharp lifespan. The main idea is to exploit the trace estimates and KSS
type estimates.
Contents
1. Introduction
2. Sobolev type estimates
3. Space-time L2 estimates
4. Glassey conjecture when n ≥ 3
4.1. Glassey conjecture when p > pc and n ≥ 3
4.2. Glassey conjecture when p = pc and n ≥ 3
5. Glassey conjecture when p < pc and n ≥ 2
6. Glassey conjecture when n = 2, p > pc
References
1
5
10
15
16
18
21
25
27
1. Introduction
Let n ≥ 2, p > 1, = ∂t2 − ∆, and a, b be constants. Consider the following
nonlinear wave equations
u = a|∂t u|p + b|∇x u|p , (t, x) ∈ R × Rn
(1.1)
2
1
u(0, x) = u0 (x) ∈ Hrad
(Rn ), ∂t u(0, x) = u1 (x) ∈ Hrad
(Rn ) .
m
Here Hrad
stands for the space of spherically symmetric functions lying in the usual
Sobolev space H m .
In the 1980’s, Glassey made the conjecture that the critical exponent for the
problem to admit global small solutions is
2
pc = 1 +
n−1
in [4] (see also Schaeffer [16] and Rammaha [14]). The conjecture has been verified
in space dimension n = 2, 3 for general data (Hidano and Tsutaya [5] and Tzvetkov
Key words and phrases. Glassey conjecture, semilinear wave equations, Morawetz estimates,
KSS estimates.
The first author was partly supported by the Grant-in-Aid for Scientific Research (C)
(No.20540165 and 23540198), Japan Society for the Promotion of Science.
The second author was supported in part by NSFC 10871175 and 10911120383.
1
2
KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA
[24] independently) as well as radial data (Sideris [17] for n = 3). For higher
dimension n ≥ 4, there are only negative results available (blow up with upper
bound on expected sharp lifespan for p ≤ pc ) in Zhou [25].
The purpose of this paper is to verify this conjecture in the radial case for all
spatial dimensions, by proving global existence for p > pc . Moreover, we are able
to prove the results with low regularity assumption on the initial data and extend
the solutions to the sharp lifespan (for all 1 < p < 1 + 2/(n − 2)).
Before presenting our main results, let us first give a brief review of the history.
The problem is scale invariant in the Sobolev space Ḣ sc with
1
n
.
sc = + 1 −
2
p−1
For local well-posedness of the problem, it has been intensively studied at least
for p ∈ N, when the general result requires the initial data lie in H s × H s−1 for
s > max(sc , (n + 5)/4) (see Ponce and Sideris [13], Tataru [23], Fang and Wang
[1] and references therein). If p ≥ 3 or p = 2 with n ≥ 4, the problem is locally
well-posed in H s × H s−1 for s > sc , when the initial data have radial symmetry
or certain amount of angular regularity (see Fang and Wang [3] and references
therein).
For the long time existence of the solutions with C0∞ small data of size ǫ > 0,
it is well known also for the case of p ∈ N (even for the problem of quasilinear
equations). When p > pc , we have global existence. For p = pc , we have almost
global existence with lifespan Tǫ which satisfies
log(Tǫ ) ∼ ǫ1−p .
Instead, if p < pc , we have long time existence with lifespan
p−1
Tǫ ∼ ǫ− 1−(n−1)(p−1)/2 ,
see John and Klainerman [9], Klainerman [10], Sogge [20] and references therein.
Moreover, the estimate on the lifespan Tǫ is sharp for the problem with nonlinearity
|∂t u|p (see Rammaha [15] for p = 2 and n = 2, 3, Zhou [25] for p ∈ R and 1 < p ≤
pc ).
There is not much work on the long time existence with low-regularity small data.
2
1
In [8], Hidano and Yokoyama proved almost global existence for small Hrad
× Hrad
data when p = 2 and n = 3. It was generalized to the quasilinear problem in our
recent work [6]. If p ≥ 3 or p = 2 with n ≥ 4, we have global (almost global for
p = 3 and n = 2) in H s with s > sc and certain angular regularity (Sterbenz [22]
and Fang and Wang [3]).
We will use Λi to denote the norm of the initial data,
Λi := ku0 kḢ i (Rn ) + ku1 kḢ i−1 (Rn ) , i = 1, 2 .
n−1
,
Let ∂ = (∂√
x , ∂t ) with ∂x = (∂x1 , ∂x2 , . . . , ∂xn ), x = rω with r = |x| and ω ∈ S
2
and hri = 1 + r . Now we are ready to state our main results. The first result is
the global existence theorem for p > pc and n ≥ 3.
Theorem 1.1. Let n ≥ 3 and 1 + 2/(n − 1) < p < 1 + 2/(n − 2). Consider the
nonlinear wave equation (1.1). For any choice of s1 , s2 such that 1/2 ≤ s1 <
n/2 − 1/(p − 1) < s2 ≤ 1, there exist constants C, ǫ0 > 0, such that if
1
2
Λ1−s
Λs21 + Λ1−s
Λs22 ≤ ǫ0 ,
1
1
GLASSEY CONJECTURE
3
then we have a unique global solution u to (1.1) satisfying
2
1
u ∈ C([0, ∞); Hrad
(Rn )) ∩ C 1 ([0, ∞); Hrad
(Rn )) ,
′
k∂ukL∞ ([0,∞);L2 (Rn )) + kr−δ hri−1/2+δ ∂ukL2 ([0,∞)×Rn ) ≤ CΛ1 ,
′
where
k∂∂x ukL∞ ([0,∞);L2 (Rn )) + kr−δ hri−1/2+δ ∂∂x ukL2 ([0,∞)×Rn ) ≤ CΛ2 ,
1 − (s2 − s1 )(p − 1)
n − 2s2
(p − 1) , δ ′ =
.
4
2
In contrast, when p = pc , we have the almost global existence.
(1.2)
δ=
Theorem 1.2. Let n ≥ 3 and p = 1 + 2/(n − 1). Consider the nonlinear wave
equation (1.1). For any choice of s such that 1/2 < s ≤ 1, there exist constants
C, c, ǫ0 > 0, such that if
1/2
1/2
ǫ := Λ1 Λ2
s
+ Λ1−s
1 Λ2 ≤ ǫ0 ,
then we have a unique almost global solution u to (1.1) satisfying
2
1
u ∈ C([0, T∗ ]; Hrad
(Rn )) ∩ C 1 ([0, T∗ ]; Hrad
(Rn )) ,
k∂ukL∞([0,T∗ ];L2 (Rn )) + ǫ(p−1)/2 kr−δ hri−1/2+δ ∂ukL2 ([0,T∗ ]×Rn ) ≤ CΛ1 ,
k∂∂x ukL∞ ([0,T∗ ];L2 (Rn )) + ǫ(p−1)/2 kr−δ hri−1/2+δ ∂∂x ukL2 ([0,T∗ ]×Rn ) ≤ CΛ2 ,
where
n − 2s
δ=
(p − 1), T∗ = exp(cǫ1−p ) .
4
For the case 1 < p < pc , we expect a long time existence of the solution.
Theorem 1.3. Let n ≥ 2 and 1 < p < 1 + 2/(n − 1). Consider the nonlinear wave
equation (1.1). There exist constants C, c > 0, such that we have a unique solution
u to (1.1) satisfying
2
1
u ∈ C([0, T∗ ]; Hrad
(Rn )) ∩ C 1 ([0, T∗ ]; Hrad
(Rn )) ,
δ−1/2
k∂ukL∞ ([0,T∗ ];L2 (Rn )) + T∗
δ−1/2
where
k∂∂x ukL∞ ([0,T∗ ];L2 (Rn )) + T∗
1/2
kr−δ ∂ukL2 ([0,T∗ ]×Rn ) ≤ CΛ1 ,
kr−δ ∂∂x ukL2 ([0,T∗ ]×Rn ) ≤ CΛ2 ,
1/2
2(p−1)
T∗ = c(Λ1 Λ2 )− 2−(n−1)(p−1) ,
δ=
(
(n−1)(p−1)
,
2
(n−1)(p−1)
,
4
1
1 < p < 1 + n−1
1
1 + n−1
≤p<1+
2
n−1
= pc .
As can be observed from the statement, for p < pc , we do not require the
smallness of the initial data, in contrast to p ≥ pc .
Remark 1.1. In our Theorems, the assumptions posed on the initial data are
of “multiplicative form”, which is considered as one of the main innovations in
1/2 1/2
this paper. For example, in Theorem 1.3, the quantity Λ1 Λ2 is in fact scaleinvariant, and it scales like the homogeneous Sobolev space Ḣ 3/2 . The assumptions
in the other two Theorems are almost critical, which scale like Ḣ 3/2 ∩ Ḣ 3/2+ǫ for
p = pc and Ḣ sc −ǫ ∩ Ḣ sc +ǫ for p > pc , with the critical scaling regularity sc =
n/2 + 1 − 1/(p − 1). One of the advantages of using the “multiplicative form” is
that, for p ≥ pc , even if Λ1 is not so small, we still have (almost) global solutions
when Λ2 is sufficiently small.
4
KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA
Here, we would like to point out an interesting similarity between the Glassey
conjecture and the Strauss conjecture. Recall that for the Strauss conjecture, where
the nonlinearity is |u|p , we find similar phenomena. Besides the critical regularity
sc = n/2−2/(p−1), there is one more Sobolev regularity, namely sd = 1/2−1/p (see
Sogge [20] Section IV.4), as far as the radially symmetric functions are concerned.
The critical exponent p = p0 for this problem to have global small solutions is given
by the positive root of the equation
(n − 1)p2 − (n + 1)p − 2 = 0 .
There is an interesting relation between these two facts: if p > 1, we see that
sc > sd if and only if p > p0 ,
and the sharp lifespan for 1 < p < p0 has the order ǫ1/(sc −sd ) .
Interestingly enough, for the Glassey conjecture, the index 3/2 plays the same
role as sd . We have sc > 3/2 if and only if p > pc for p > 1, and the sharp
lifespan T∗ has also the order ǫ1/(sc −3/2) for p < pc . These observations strongly
suggest that, for the equation (1.1), by adding certain amount of angular regularity
if necessary, the minimal regularity for the problem to be well-posed is
3
, sc .
max
2
When n = 2 and p ≥ pc = 3, it seems to us that the methods to prove the
preceding theorems are not sufficient to give satisfactory results. In spite of that,
we can use the generalized Strichartz estimates of Smith, Sogge and Wang [19] to
prove the following global result for p > pc .
Theorem 1.4. Let n = 2 and p > 3. Consider the nonlinear wave equation (1.1).
There exist constants C, ǫ0 > 0, such that if
1/(p−1)
ǫ := Λ1
1−1/(p−1)
Λ2
≤ ǫ0 ,
then we have a unique global solution u to (1.1) satisfying
2
1
2 ≤ CΛ1 , k∂∂x ukL∞ L2 ≤ CΛ2 , k∂uk p−1 ∞ ≤ Cǫ .
u ∈ Ct Hrad
∩Ct1 Hrad
, k∂ukL∞
L
L
t Lx
t
x
t
x
Remark 1.2. For p = pc and n = 2, it has been proved in Fang and Wang [3] that
the problem has a unique almost global solution with almost critical regularity for
small data, which is not necessarily radial. A similar result for p > 3 and p ∈ N
has also been obtained there.
This paper is organized as follows. At the end of this section, we list our basic
notation. In the next section, we give several Sobolev type estimates related with
the trace estimates. In Section 3, we prove some space-time L2 estimates, which
are variants of the Morawetz-KSS estimates. In Sections 4 and 5, we give the proof
of the (almost) global results for n ≥ 3 (Theorems 1.1 and 1.2) and the scalesupercritical result for n ≥ 2 (Theorems 1.3), based on the results from Sections 2
and 3. In the last section, a simple proof for p > pc and n = 2 (Theorem 1.4) is
provided, by using the generalized Strichartz estimates√of [19].
Notation. Let δ ∈ (0, 1/2), δ ′ < δ. We denote D = −∆ and the homogeneous
Sobolev norm
kukḢ s = kDs ukL2 (Rn ) .
GLASSEY CONJECTURE
5
The homogeneous Sobolev space Ḣ s with s < n/2 is defined as the completion of
C0∞ with respect to the semi-norm k · kḢ s .
For fixed T > 0, we will use the following notation. We use k · kEi (i = 1, 2) to
denote the energy norm of order i,
kukE = kukE1 = k∂ukL∞([0,T ];L2 (Rn )) ,
kukE2 = k∂x ∂ukL∞ ([0,T ];L2 (Rn )) .
We will use k · kLE to denote the local energy norm,
kukLE = kukLE1 =
′
kr−δ hri−1/2+δ ∂ukL2 ([0,T ]×Rn )
′
+kr−1−δ hri−1/2+δ ukL2 ([0,T ]×Rn )
+(log(2 + T ))−1/2 r−δ hri−1/2+δ |∂u| +
+T δ−1/2 r−δ |∂u| + |u|
.
r
2
n
|u|
r
L2 ([0,T ]×Rn )
L ([0,T ]×R )
Here, when n ≤ 2, we will assume that there are only terms about ∂u. On the basis
of the space LE, we can define kukLE2 = k∂x ukLE and
LE ∗ = r−δ hriδ
′
−1/2
L2t,x + (log(2 + T ))−1/2 r−δ hriδ−1/2 L2t,x + T δ−1/2 r−δ L2t,x ,
where h ∈ f L2t,x means that h = f g for some g ∈ L2t,x . When T = ∞, by LE norm,
we mean
kukLE =
′
kukLE1 = kr−δ hri−1/2+δ ∂ukL2 ([0,∞)×Rn )
′
+kr−1−δ hri−1/2+δ ukL2 ([0,∞)×Rn )
+ supT >0 (log(2 + T ))−1/2 r−δ hri−1/2+δ |∂u| + |u|
r
|u|
δ−1/2
−δ
.
|∂u| + r
+ supT >0 T
r
L2 ([0,T ]×Rn )
L2 ([0,T ]×Rn )
2. Sobolev type estimates
In this section, we give several Sobolev type estimates related with the trace
estimates.
First, we state a variant of the Hardy inequality.
Lemma 2.1 (Hardy’s inequality). Let n ≥ 2 and 0 ≤ s ≤ 1 (s < 1 for n = 2).
Then we have
(2.1)
for any u ∈ H 1 .
s
kr−s ukL2x ≤ Ckuk1−s
L2 k∂r ukL2
6
KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA
Proof. We only need to give the proof for u ∈ C0∞ . First, we prove (2.1) for
s ≥ 1/2. Since s < n/2 and s ≤ 1, we have
r−s u
2
L2x
=
Z
=
≤
≤
≤
≤
∞
r−2s |u(rω)|2 rn−1 drdω
0
Z
Z ∞
1
|u(rω)|2 ∂r rn−2s drdω
n − 2s Sn−1 0
Z
Z ∞
1
−
∂r (|u(rω)|2 )rn−2s drdω
n − 2s Sn−1 0
Z
Z ∞
2
r1−2s |u||∂r u|rn−1 drdω
n − 2s Sn−1 0
2
kr1−2s ukL2x k∂r ukL2x
n − 2s
2
k(r−s |u|)(2s−1)/s |u|(1−s)/s kL2x k∂r ukL2x
n − 2s
2
(2s−1)/s
(1−s)/s
kr−s ukL2
kukL2
k∂r ukL2x ,
x
x
n − 2s
Sn−1
=
Z
where we have applied the Hölder inequality in the last step. This gives us the
required estimates with C = (2/(n − 2s))s and s ≥ 1/2. The case s = 0 is trivial.
For s ∈ (0, 1/2), we can use the result for s = 1/2 to prove the estimate as follows
r−s u
L2x
2s
=
r−1/2 |u|
≤
r−1/2 |u|
=
≤
2s
2s
|u|1−2s
1/s
Lx
L2x
|u|1−2s
2/(1−2s)
Lx
1−2s
r−1/2 u 2 kukL2x
L
s x
2
1−s
s
kukL2x k∂r ukL2x .
n−1
With the help of the Hardy inequality, it will be easy to prove trace estimates.
Lemma 2.2 (Trace estimates). Let n ≥ 2. If 1/2 ≤ s ≤ 1 (and s < 1 for n = 2),
then
(2.2)
1−s
s
2 ≤ Ckuk 2 k∂r uk 2
krn/2−s ukL∞
Lx
L
r Lω
x
for any u ∈ H 1 (Rn ). In particular, if s = 1/2, we have
(2.3)
1/2
1/2
x
x
2 ≤ Ckuk 2 k∂r uk 2 .
kr(n−1)/2 ukL∞
L
L
r Lω
GLASSEY CONJECTURE
7
Proof. We only need to give the proof for u ∈ C0∞ . The assumptions on s tell us
that n − 2s > 0, 0 ≤ 2s − 1 ≤ 1 and 2s − 1 < n/2. Then by using (2.1), we see that
Z
Z ∞
Rn−2s ku(Rω)k2L2ω = −Rn−2s
∂r |u(rω)|2 drdω
Sn−1 R
Z
Z ∞
rn−2s |u||∂r u|drdω
≤ 2
Sn−1 0
Z
Z ∞
r1−2s |u||∂r u|rn−1 drdω
= 2
Sn−1 0
1−2s
ukL2x k∂r ukL2x
2−2s
CkukL2 k∂r uk2s
L2x ,
x
≤ 2kr
≤
with C independent of R > 0. This completes the proof.
We will also need to use the following variant of the trace estimates for the proof
of Theorem 1.3 in the case of n = 2 and 2 ≤ p < 3.
Lemma 2.3. Let n ≥ 2. If s ≥ 0, then
√ s−(n−1)/2 1/2 s−(n−1)/2
1/2
2 ≤
2kr
ukL2 kr
∂x ukL2 ,
(2.4)
krs ukL∞
r Lω
x
x
for any u such that the right hand side is finite.
Proof. If u ∈ C0∞ , this inequality follows from a simple application of integration
by parts and the Cauchy-Schwarz inequality,
Z
s
2
2s
kr ukL2ω = r
|u(rω)|2 dω
Sn−1
Z
Z ∞
2s
= −r
∂R |u(Rω)|2 dRdω
Sn−1 r
Z
Z ∞
≤ 2
R2s |u(Rω)||∂R u(Rω)|dRdω
n−1
ZS
Zr ∞
≤ 2
R2s−(n−1) |u(Rω)||∂R u(Rω)|Rn−1 dRdω
≤
Sn−1 0
s−(n−1)/2
2kr
ukL2x krs−(n−1)/2 ∂r ukL2x .
Here the condition s ≥ 0 is used to control r2s by R2s .
In general, if u ∈ r(n−1)/2−s L2x and ∂x u ∈ r(n−1)/2−s L2x , we only need to construct a C0∞ sequence which is convergent to u in the corresponding norm. Define
ul,m (x) := ψl (x)(ρm ∗ u)(x) ,
R
where ψl (x) = ψ(x/l), ρm (x) = mn ρ(mx), ψ, ρ ∈ C0∞ , ρ ≥ 0, Rn ρ(x)dx = 1 and
ψ(x) ≡ 1 for |x| < 1. We recall the n-dimensional version of (4.2) of Lemma 4.2 in
our previous paper [6],
Z
ρm (y)
dy ≤ C|x|−α , α < n ,
(2.5)
|x
− y|α
n
R
where the constant C is independent of m ≥ 1.
We claim that there exists a function m = m(l) such that ul,m(l) → u in
r(n−1)/2−s L2x as l → ∞. If it is true, then we also have (∂x u)l,m(l) → ∂x u in
8
KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA
r(n−1)/2−s L2x . Notice that
∂x ul,m =
x
1
(ρm ∗ u)(x) + (∂x u)l,m (x) .
(∂x ψ)
l
l
For the first term, we see that
x
1
(ρm ∗ u)(x)kL2x
krs−(n−1)/2 (∂x ψ)
l
l
Z
x
1
=
ρm (x − y)u(y)dy
|x|s−(n−1)/2 (∂x ψ)
l
l
Rn
≤
≤
≤
L2x
C
1/2
k|x|s−(n−1)/2 ρ1/2
m (x − y)u(y)kL2y kρm (x − y)kL2y
l
C
k|x|s−(n−1)/2 ρ1/2
m (x − y)u(y)kL2y L2x
l
C
k|y|s−(n−1)/2 u(y)kL2y → 0 ,
l
L2x
as l → ∞, where we have used the inequality (2.5) with α = n − 1 − 2s and the
fact that s > −1/2. This gives us the convergence of ∂x ul,m(l) to ∂x u.
To complete the proof, it remains to prove the claim. Observe that
ul,m − u = ψl (x) ((ρm ∗ u)(x) − u(x)) + (ψl (x) − 1)u(x) .
For the second term, since
rs−(n−1)/2 (ψl (x) − 1)u(x) → 0 a.e. x ∈ Rn
as l → ∞, and
|rs−(n−1)/2 (ψl (x) − 1)u(x)|2 ≤ C|rs−(n−1)/2 u(x)|2 ∈ L1 ,
we see that, by Lebesgue’s dominated convergence theorem, (ψl (x) − 1)u(x) → 0
in r(n−1)/2−s L2x as l → ∞.
We only need to control the first term ψl (x) ((ρm ∗ u)(x) − u(x)). Since rs−(n−1)/2 u ∈
2
Lx , for any ǫ > 0, there exists a continuous function g such that
supp g ⊂ {x ∈ Rn : R1 ≤ |x| ≤ R2 }
for some 0 < R1 < R2 < ∞, and
krs−(n−1)/2 u − gkL2 ≤ ǫ .
To deal with the term ψl (x) ((ρm ∗ u)(x) − u(x)), we rewrite it as follows
ψl (x) ((ρm ∗ u)(x) − u(x)) = ψl (x) (ρm ∗ (u − G) + (ρm ∗ G − G) + G − u) ,
where G(x) := r−s+(n−1)/2 g. We easily see that
krs−(n−1)/2 ψl (x)(G − u)kL2 ≤ Ckg − rs−(n−1)/2 ukL2 ≤ Cǫ .
GLASSEY CONJECTURE
9
For the term involving ρm ∗ (u − G), we obtain
≤
krs−(n−1)/2 ψl (x)ρm ∗ (u − G)(x)kL2x
Z
C
|x|s−(n−1)/2 ρm (x − y)(u − G)(y)dy
Rn
L2x
≤
1/2
C k|x|s−(n−1)/2 ρ1/2
m (x − y)(u − G)(y)kL2y kρm (x − y)kL2y
≤
Ck|x|s−(n−1)/2 ρ1/2
m (x
≤
L2x
− y)(u − G)(y)kL2y L2x
Ck|y|s−(n−1)/2 (u − G)(y)kL2y ≤ Cǫ ,
where we have used the inequality (2.5) with α = n − 1 − 2s and the fact that
s > −1/2.
Finally, we consider the term involving (ρm ∗ G − G). Note that G is a uniformly
continuous function,
Z
|(ρm ∗ G)(x) − G(x)| =
ρm (x − y)(G(y) − G(x))dy
Rn
y
≤
sup
|y−x|<C/m,x,y∈suppG
|G(y) − G(x)| → 0
as m → ∞. Since supp ψl ⊂ {x ∈ Rn : |x| < Cl},
krs−(n−1)/2 ψl (x)((ρm ∗ G)(x) − G(x))kL2
≤ Ckrs−(n−1)/2 ((ρm ∗ G)(x) − G(x))kL2 (|x|<Cl)
≤ Cls+1/2
sup
|y−x|<C/m,x,y∈suppG
|G(y) − G(x)| → 0
as m → ∞, for any fixed l. This completes the proof.
As we may observe, all these estimates hold for general functions. Typically, we
will apply these estimates to ∂u, which is not radial, even if u is radial. This is the
main reason for us to state all the estimates above involving the L2ω norm. In this
way, as we can see in the following lemma, we can easily control ∂x u and ∂r u.
Lemma 2.4. Let u = u(x) be a radially symmetric function. Then
−1/2
(2.6)
|∂x u| = |∂r u| = An−1 k∂x ukL2ω
with An−1 = |Sn−1 |.
The proof is just a simple calculation. Since u is radial, we see ∂r u is radial.
Further,
x
x
∂x u = ∂r u, |∂x u| = | ||∂r u| = |∂r u| ,
r
r
and
1/2
k∂r ukL2ω = An−1 |∂r u| .
Thus,
−1/2
−1/2
|∂x u| = |∂r u| = An−1 k∂r ukL2ω = An−1 k∂x ukL2ω .
10
KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA
3. Space-time L2 estimates
In this section, we prove the space-time L2 estimates, which are variants of the
Morawetz-KSS estimates.
Consider the wave equation
u = F, (t, x) ∈ R × Rn
(3.1)
u(0, x) = u0 (x), ∂t u(0, x) = u1 (x) .
Lemma 3.1 (KSS type estimates). Let n ≥ 1, 0 ≤ δ < 1/2 and δ ′ < δ. For any
solution u = u(t, x) to the wave equation (3.1), we have the following inequality
(3.2)
kukE + kukLE ≤ C(k∂x u0 kL2x + ku1 kL2x + kF kL1t L2x ) ,
where C is independent of T > 0 and the functions u0 ∈ H 1 , u1 ∈ L2 and F ∈ L1t L2x .
This is a standard estimate now. The estimates of this type together with the
application to nonlinear wave equations originate from the work of Keel, Smith and
Sogge [11]. The variants with LE norm including the homogeneous weight r−δ are
due to Hidano and Yokoyama [7]. Here, for completeness, we give a proof.
Proof. To begin the proof, let us recall the classical local energy estimates of
Smith-Sogge (Lemma 2.2 in [18])
(3.3)
kβ(x)eitD f kL2 (R×Rn ) ≤ Cn,γ,β kf kḢ γ ,
for β ∈ C0∞ and 2γ ≤ n − 1. The inequality (3.2) follows from this inequality with
γ = 0 (and γ = 1 for n ≥ 3), together with the energy estimate.
First, owing to the Duhamel principle and a standard scaling argument, it is
enough to prove the following six inequalities
(3.4)
(3.5)
kr−δ eitD f kL2 ([0,1]×Rn ) ≤ Ckf kL2x , 0 ≤ δ < 1/2 ,
(log(2 + T ))−1/2 kr−δ hri−1/2+δ eitD f kL2 ([0,T ]×Rn ) ≤ Ckf kL2x , δ < 1/2 ,
′
(3.6)
kr−δ hri−1/2+δ eitD f kL2 (R×Rn ) ≤ Ckf kL2x , δ ′ < δ < 1/2 ,
(3.7)
kr−1−δ eitD f kL2 ([0,1]×Rn ) ≤ Ckf kḢx1 , 0 ≤ δ < 1/2 , n ≥ 3 ,
(3.8)
(log(2 + T ))−1/2 kr−1−δ hri−1/2+δ eitD f kL2 ([0,T ]×Rn ) ≤ Ckf kḢ 1 , δ < 1/2 , n ≥ 3
x
(3.9)
′
kr−1−δ hri−1/2+δ eitD f kL2 (R×Rn ) ≤ Ckf kḢ 1 , δ ′ < δ < 1/2 , n ≥ 3 .
x
We begin by the proof of the first three inequalities for r ≤ 1 and (3.6) for r > 1.
From (3.3) with γ = 0 and n ≥ 1, we see that
keitD f kL2 (R×{r≤1}) ≤ Ckf kL2 .
A standard scaling argument leads us to
(3.10)
sup 2−j/2 keitD f kL2 (R×{r≤2j }) ≤ Ckf kL2 ,
j∈Z
GLASSEY CONJECTURE
11
and so for any δ < 1/2,
kr−δ eitD f kL2 (R×{r≤1})
C 2(1/2−δ)j 2−j/2 keitD f kL2 (R×{2j−1 <r≤2j }) 2
≤
lj:j≤0
−j/2
≤
C sup 2
≤
Ckf kL2 .
j≤0
ke
itD
f kL2 (R×{2j−1 <r≤2j })
Similarly, for any δ ′ < δ < 1/2, since r ≤ hri, we obtain
kr−δ hriδ
≤
≤
′
−1/2 itD
e
′
δ −δ−1/2 itD
f kL2 (R×{r≥1})
kr
e f kL2 (R×{r≥1})
′
C 2(δ −δ)j 2−j/2 keitD f kL2 (R×{2j−1 ≤r≤2j }) 2
lj:j≥1
−j/2
≤
C sup 2
≤
Ckf kL2 ,
j≥1
ke
itD
f kL2 (R×{2j−1 ≤r≤2j })
which is (3.6) for r > 1.
It remains to prove (3.4) and (3.5) for r > 1. For (3.4), because of the assumption
δ ≥ 0, we can easily get by the energy estimates
kr−δ eitD f kL2 ([0,1]×{r>1})
≤ keitD f kL2 ([0,1]×Rn )
≤ keitD f kL∞ ([0,1];L2 (Rn ))
≤ Ckf kL2x .
For (3.5) with r > 1, we consider 1 ≤ r ≤ T and r ≥ T separately. For r ≥ T ,
since δ − 1/2 < 0 and r ≤ hri, we obtain
kr−δ hriδ−1/2 eitD f kL2 ([0,T ]×{r≥T })
≤
kr−1/2 eitD f kL2 ([0,T ]×{r≥T })
≤
keitD f kL∞ ([0,T ];L2 (Rn ))
≤
≤
T −1/2 keitD f kL2 ([0,T ]×Rn )
Ckf kL2x .
Now we give the estimate of (3.5) for 1 ≤ r ≤ T . By (3.10) and the elementary
inequality 2[10 log(2+T )] ≥ T (where [M ] denotes the greatest integer not greater
than M ), we have
≤
≤
kr−δ hriδ−1/2 eitD f kL2 (R×{1≤r≤T })
Ckr−1/2 eitD f kL2 (R×{1≤r≤T })
C 2−j/2 keitD f kL2 (R×{2j−1 ≤r≤2j }) 2
lj:1≤j≤10 log(2+T )
1/2
sup 2
−j/2
≤
C(log(2 + T ))
≤
C(log(2 + T ))1/2 kf kL2x .
j
ke
itD
f kL2 (R×{2j−1 ≤r≤2j })
This completes the proof of the first three inequalities.
The inequalities (3.7)-(3.9) follow from basically the same proof, by using (3.3)
with γ = 1 and Hardy’s inequality. For example, to prove (3.7) for n ≥ 3, we use
12
KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA
(3.3) with γ = 1 (since 1 ≤ (n − 1)/2 for n ≥ 3), which tells us that
keitD f kL2 (R×{r≤1}) ≤ Ckf kḢ 1 .
A standard scaling argument leads us to
(3.11)
sup 2−3j/2 keitD f kL2 (R×{r≤2j }) ≤ Ckf kḢ 1 ,
j∈Z
and so for any δ < 1/2,
kr−δ−1 eitD f kL2 (R×{r≤1})
≤
C 2(1/2−δ)j 2−3j/2 keitD f kL2 (R×{2j−1 <r≤2j }) 2
lj:j≤0
−3j/2
≤
C sup 2
≤
Ckf kḢ 1 .
j≤0
ke
itD
f kL2 (R×{2j−1 <r≤2j })
For (3.7) with r > 1, since δ ≥ 0, we can easily get by the energy estimates and
Hardy’s inequality (2.1) with s = 1,
kr−1−δ eitD f kL2 ([0,1]×{r>1})
≤ kr−1 eitD f kL2 ([0,1]×Rn )
≤ kDeitD f kL∞ ([0,1];L2 (Rn ))
≤ Ckf kḢ 1 ,
x
which completes the proof of (3.7).
When n ≥ 3, we can prove the following inhomogeneous KSS type estimates
with LE ∗ norm on F .
Lemma 3.2 (Inhomogeneous KSS type estimates). Let n ≥ 3, 0 < δ < 1/2,
and δ ′ < δ. For any solution u = u(t, x) to the wave equation (3.1), we have the
following inequality
(3.12)
kukE + kukLE ≤ C(k∂x u0 kL2x + ku1 kL2x + kF kLE ∗ ) ,
where C is independent of T > 0 and the functions u0 ∈ Ḣ 1 , u1 ∈ L2 and F ∈ LE ∗ .
Proof.
i) Let us first consider smooth solutions. For such a case, we have
the space-time L2 estimates even for certain small perturbations of the Minkowski
metric (see [21], [12] and our previous work [6]). Recall that using Lemma 2.3 and
(2.30) of [6], we can get
Z TZ
|∂u|2
u2
(3.13)
+ 2+2δ dxdt
T 2δ−1
2δ
r
0 {x∈Rn ; 1<r<T } r
Z TZ
|∂u|2
−1
u2
+ log(2 + T )
+ 3 dxdt
r
r
0 {x∈Rn ; 1<r<T }
Z TZ
|∂u|2
2
u
+
+ 3+2δ−2δ′ dxdt
1+2δ−2δ ′
r
r
n
0 {x∈R ; 1<r<∞}
≤ C(k∇u0 k2L2 (Rn ) + ku1 k2L2 (Rn ) )
Z TZ
|u||F |
|∂u||F | +
+C
dxdt ,
hri
0
Rn
GLASSEY CONJECTURE
13
for any smooth solution u to the wave equations (3.1), T > 1, δ ′ < δ and 0 < δ <
1/2. We will also need a slight variant of Lemma 2.2 of [6]. Observe that if we
choose the function
k
r
f (r) =
r+λ
with k = 1 − 2δ ∈ (0, 1) and λ > 0, then the same argument as in the proof of
Lemma 2.2 of [6] will tell us that
λ2δ−1
(3.14)
Z TZ
{x∈Rn ; r<λ}
0
≤
|∂u|2
r2δ
+
u2
r2+2δ
dxdt
C(k∇u0 k2L2 (Rn ) + ku1 k2L2 (Rn ) )
Z TZ
|u||F |
|∂u||F | + 2δ
+C
dxdt,
r (r + λ)1−2δ
0
Rn
where the constant C is independent of λ > 0. We only need to check the new
relations (instead of (2.15) and (2.16) there)
k(1 − k)λ2
f
rk−1
f
′
≤
−
,
∆
,
− f (r) ≥ (1 − k)
r
(λ + r)k
r
r3−k (λ + r)2+k
and substitute these new relations to (2.12) and (2.17) there.
On the basis of (3.13) and (3.14), together with the standard energy estimate
!
Z
Z Z
T
(3.15)
sup
t∈[0,T ]
Rn
|∂u|2 dx ≤ C
k∂u(0)k2L2(Rn ) +
0
Rn
|∂u||F |dxdt
,
it will be easy to prove the required estimates for the smooth solutions. Suppose
T > 1 first. By applying (3.15) to the integrals over {r > T }, we see that
≤
≤
≤
≤
≤
kuk2E + kuk2LE(r>T )
Z
Z TZ
2
2δ−1
sup
|∂u| dx + T
|∂u|2
u2
dxdt
r2δ
r2+2δ
Z Z
|∂u|2
−1 T
u2
+ log(2 + T )
+ 3 dxdt
r
r
0 {x∈Rn ; r>T }
Z TZ
|∂u|2
u2
+
dxdt
′ +
′
1+2δ−2δ
3+2δ−2δ
r
0 {x∈Rn ; r>T } r
Z
Z TZ
u2
sup
|∂u|2 + 2 dxdt
|∂u|2 dx + 3T −1
r
t∈[0,T ] Rn
0 {x∈Rn ; r>T }
Z
2
u
C sup
|∂u|2 + 2 dx
r
n
t∈[0,T ] R
Z
C sup
|∂u|2 dx
t∈[0,T ] Rn
0
{x∈Rn ; r>T }
+
t∈[0,T ] Rn
C(k∇u0 k2L2 (Rn )
+
ku1 k2L2 (Rn )
+
Z
0
T
Z
Rn
|∂u||F |dxdt),
14
KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA
where we have applied the Hardy inequality (2.1) with s = 1. For the integral over
{r < T }, we use (3.13) and (3.14) with λ = 1 to get
Z TZ
|u||F |
kuk2LE(r<T ) ≤ C(k∇u0 k2L2 (Rn ) + ku1 k2L2 (Rn ) ) + C
|∂u||F | +
dxdt.
r
0
Rn
Then an application of the Cauchy-Schwarz inequality yields the required estimate
(3.12) for T > 1.
To prove the general result for any T > 0, we only need to control the term
|u|
A[u] = T δ−1/2 r−δ |∂u| +
r
L2 ([0,T ]×Rn )
for T ∈ (0, 1]. To control this, we only need to apply (3.14) with λ = T and (3.15)
as follows:
2
|u|
+ kuk2E
A[u]2 + kuk2E = T 2δ−1 r−δ |∂u| +
r
2
n
L ([0,T ]×R )
2
|u|
≤ CT 2δ−1 r−δ |∂u| +
r
L2 ([0,T ]×Rn :r<T )
+CT −1 |∂u| +
|u|
r
2
L2 ([0,T ]×Rn :r>T )
|u|
≤ CT 2δ−1 r−δ |∂u| +
r
+C |∂u| +
≤
|u|
r
+ kuk2E
2
L2 ([0,T ]×Rn :r<T )
2
2
n
L∞
t ([0,T ];L (R ))
+ ku1 k2L2 (Rn ) )
|u||F |
|∂u||F | +
dxdt .
r
C(k∇u0 k2L2 (Rn )
Z TZ
+C
0
Rn
Once again, an application of the Cauchy-Schwarz inequality gives us the required
estimate (3.12) for T ≤ 1.
ii) We next consider the case where u is not smooth. By Lemma 3.1, we only need
to prove for the case u0 = u1 = 0. Fix T ∈ (0, ∞). Observe that for 0 < δ < 1/2,
we have the Hardy inequality
kr−δ ukL2 ≤ CkukḢ δ ≤ CkukH δ ,
which means r−δ L2x ⊂ H −δ , and so LE ∗ ⊂ L1t H −δ ([0, T ] × Rn ) if T < ∞. Thus by
the standard existence and uniqueness result of the linear wave equation, we have
u ∈ Ct Hx1−δ ∩ Ct1 Hx−δ ([0, T ] × Rn ).
We claim that there exists a sequence of smooth functions Fk such that Fk → F
in LE ∗ . If it is true, then uk are Cauchy sequence in E1 ∩ LE1 , and uk → u
in Ct Hx1−δ ∩ Ct1 Hx−δ ([0, T ] × Rn ). This tells us that {uk }∞
k=1 converges to u in
E1 ∩ LE1 , and so
kukE1 ∩LE1 = lim kuk kE1 ∩LE1 ≤ C lim kFk kLE ∗ = CkF kLE ∗ ,
k→∞
which implies (3.12).
k→∞
GLASSEY CONJECTURE
15
To complete the proof, it remains to prove the claim.
Proof of the claim. Without loss of generality, we give the proof for F ∈ r−δ L2t,x .
Let F̃ (t, x) be the zero extension of rδ F ∈ L2t,x ([0, TR] × Rn ) in R × Rn . Let φ(x) ∈
C0∞ (Rn ) be a function with the properties φ ≥ 0, Rn φ(x)dx = 1, φ = 1 near 0.
We will also choose its one-dimensional counterpart ψ(t) ∈ C0∞ (R). Define φk (x) =
2kn φ(2k x) and ψk (t) = 2k ψ(2k t). Then the standard results of approximations of
the identity give us
F̃k = (φk (x)ψk (t)) ∗t,x F̃ → F̃ in L2t,x ([0, T ] × Rn ) ,
that is,
Fk1 := r−δ F̃k → F in r−δ L2t,x ([0, T ] × Rn ) .
Notice that Fk1 is smooth except at x = 0. It suffices to set Fk (t, x) = (1 −
φ(2k x))Fk1 (t, x), which is smooth for any t, x. Indeed, by Lebesgue’s dominated
convergence theorem, we see
krδ (Fk − F )kL2t,x ([0,T ]×Rn )
= kF̃k (1 − φ(2k x)) − F̃ kL2t,x ([0,T ]×Rn )
≤ k(F̃k − F̃ )(1 − φ(2k x))kL2t,x ([0,T ]×Rn ) + kφ(2k x)F̃ kL2t,x ([0,T ]×Rn )
≤ kF̃k − F̃ kL2t,x ([0,T ]×Rn ) + kφ(2k x)F̃ kL2t,x ([0,T ]×Rn ) → 0 as k → ∞ .
This completes the proof of the claim and hence that of Lemma 3.2.
4. Glassey conjecture when n ≥ 3
Now we are ready to present our proof of the Glassey conjecture for radial initial
data.
Let us first formulate the setup of the proof for existence and uniqueness. Define
(4.1)
XT : = {
1
2
u ∈ C([0, T ]; Hrad
(Rn )) ∩ L∞ ([0, T ]; Hrad
(Rn )) :
1
∂t u ∈ C([0, T ]; L2rad (Rn )) ∩ L∞ ([0, T ]; Hrad
(Rn )),
kukE1 ∩E2 + kukLE1∩LE2 < ∞} .
For R1 > 0 and R2 > 0, we next define
X(R1 , R2 ; T ) := {u ∈ XT : kukEi + kukLEi ≤ Ri , i = 1, 2} .
Endowed with
(4.2)
ρ(u, v) := ku − vkE1 + ku − vkLE1 ,
it is easy to check that X(R1 , R2 ; T ) is complete with the metric ρ(u, v).
2
1
For fixed (u0 , u1 ) ∈ Hrad
× Hrad
, we define the iteration map
(4.3)
Φ[u](t) := u(0) (t) + I[N [u]] ,
where u(0) (t) = cos(tD)u0 + D−1 sin(tD)u1 is the solution of the linear Cauchy
problem,
(4.4)
N [u] := a|∂t u|p + b|∇x u|p ,
and
(4.5)
I[F ] :=
Z
0
t
sin((t − s)D)
F (s)ds .
D
16
KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA
For the nonlinearity N [u], we have the properties
|∂xα N [u]| ≤ C|∂u|p−1 |∂xα ∂u| , |α| ≤ 1,
(4.6)
and
|N [u] − N [v]| ≤ C(|∂u|p−1 + |∂v|p−1 )|∂(u − v)| .
(4.7)
Notice that v = Φ[u] is defined as the solution to the following equation
v = N [u], (t, x) ∈ R × Rn
(4.8)
v(0, x) = u0 (x), ∂t v(0, x) = u1 (x) .
We aim at showing that Φ is a contraction mapping of X(R1 , R2 ; T ), if we choose
R1 , R2 and T suitably.
4.1. Glassey conjecture when p > pc and n ≥ 3. Consider the nonlinear wave
equation (1.1) for p > pc and n ≥ 3.
Let us begin with the estimate of the homogeneous solution, u(0) , which follows
directly from the application of Lemma 3.1 to u and ∂x u with F = 0.
Proposition 4.1. Let n ≥ 2. There is a positive constant C1 , independent of
T > 0, such that the following estimates hold
(4.9)
ku(0) kE1 + ku(0) kLE1 ≤ C1 (k∂x u0 kL2 + ku1 kL2 ) ,
(4.10)
ku(0) kE2 + ku(0) kLE2 ≤ C1 (k∂x2 u0 kL2 + k∂x u1 kL2 ) .
Next, we give the estimate for the inhomogeneous part.
Proposition 4.2. Let pc < p < 1 + 2/(n − 2) and n ≥ 3, u ∈ X∞ and s1 , s2 such
that 1/2 ≤ s1 < n/2 − 1/(p − 1) < s2 ≤ 1. Set δ and δ ′ as in (1.2). Then there is
a positive constant C2 , such that the following estimates hold
s1
1−s2
s2 p−1
1
(4.11) kI[N [u]]kEi ∩LEi ≤ C2 (kuk1−s
kukLEi , i = 1, 2 .
E1 kukE2 + kukE1 kukE2 )
Moreover, if u, v ∈ X∞ , we have
(4.12) kΦ[u] − Φ[v]kE1 ∩LE1
≤ C3 ku − vkLE1
s1
1−s2
s2
1−s1
s1
1−s2
s2 p−1
1
×(kuk1−s
,
E1 kukE2 + kukE1 kukE2 + kvkE1 kvkE2 + kvkE1 kvkE2 )
for some C3 .
Proof. First, by Lemma 2.2, we have for any s ∈ [1/2, 1],
1−s
s
2 ≤ Ckuk 2 k∂r uk 2 .
krn/2−s ukL∞
Lx
L
r Lω
x
Fix s1 , s2 such that 1/2 ≤ s1 < n/2 − 1/(p − 1) < s2 ≤ 1. Then we have
(4.13)
s1
1−s2
s2
1
ku(rω)kL2ω ≤ Crs2 −n/2 hris1 −s2 (kuk1−s
L2 k∂r ukL2 + kukL2 k∂r ukL2 ) .
x
x
x
By Lemma 2.4, we have for u ∈ X∞ ,
(4.14)
s1
1−s2
s2
1
|∂u| ≤ Crs2 −n/2 hris1 −s2 (kuk1−s
E1 kukE2 + kukE1 kukE2 ) .
x
GLASSEY CONJECTURE
17
From (4.14), (4.6) and (1.2), it is clear that, for i = 1, 2,
X
′
krδ hri1/2−δ ∂xα N [u]kL2x
|α|=i−1
≤ C
′
X
|α|=i−1
krδ hri1/2−δ |∂u|p−1 ∂xα ∂ukL2x
s2 p−1
1
2
≤ C(kuk1−s
kuksE12 + kuk1−s
E
E1 kukE2 )
X1
′
krδ+(s2 −n/2)(p−1) hri1/2−δ +(s1 −s2 )(p−1) ∂xα ∂ukL2x
×
|α|=i−1
s1
1−s2
s2 p−1
1
= C(kuk1−s
E1 kukE2 + kukE1 kukE2 )
X
|α|=i−1
′
kr−δ hri−1/2+δ ∂xα ∂ukL2x .
It is easy to check that δ and δ ′ satisfy 0 < δ < 1/2 and δ ′ < δ. Now applying
Lemma 3.2 to ∂xα u with |α| ≤ 1 and u0 = u1 = 0, we have for i = 1, 2,
kI[N [u]]kEi ∩LEi
X
′
krδ hri1/2−δ ∂xα N [u]kL2 ([0,∞);L2x )
≤ C
|α|=i−1
1−s2
s2 p−1
s1
1
≤ C(kuk1−s
E1 kukE2 + kukE1 kukE2 )
≤
s1
1
C(kuk1−s
E1 kukE2
+
X
|α|=i−1
s2 p−1
2
kuk1−s
kukLEi
E1 kukE2 )
′
kr−δ hri−1/2+δ ∂xα ∂ukL2 ([0,∞);L2x )
.
This proves (4.11). A similar argument with (4.7) instead of (4.6) will yield (4.12).
With these two Propositions 4.1 and 4.2 in hand, it will be easy to show Theorem
1.1. Setting
Λi := ku0 kḢ i (Rn ) + ku1 kḢ i−1 (Rn ) , i = 1, 2 ,
we find by Propositions 4.1 and 4.2 that the mapping Φ, defined by (4.3), is a
contraction mapping from X(2C1 Λ1 , 2C1 Λ2 ; T ) into itself, for any T > 0 provided
that
(4.15)
and
(4.16)
1
2
C2 (2C1 )p−1 (Λ1−s
Λs21 + Λ1−s
Λs22 )p−1 ≤ 1/2 ,
1
1
1
2
C3 (4C1 )p−1 (Λ1−s
Λs21 + Λ1−s
Λs22 )p−1 ≤ 1/2 .
1
1
Define a positive constant C0 by
−(p−1)
C0
= max(2C3 (4C1 )p−1 , 2C2 (2C1 )p−1 ) .
Then we see that when
(4.17)
1
2
Λ1−s
Λs21 + Λ1−s
Λs22 ≤ C0 ,
1
1
the map Φ is a contraction mapping of X(2C1 Λ1 , 2C1 Λ2 ; T ) for any T > 0, the
global in time unique fixed point u ∈ X(2C1 Λ1 , 2C1 Λ2 ; ∞) is the solution which we
seek.
To complete the proof of Theroem 1.1, we also need to establish the regularity
of u, i.e.,
(4.18)
∂ti u ∈ C([0, ∞); H 2−i (Rn )), i = 0, 1,
18
KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA
and the uniqueness of the solution u.
First, for the problem of regularity, it suffices to show
∂ti u ∈ C([0, ∞); Ḣ 2−i (Rn )), i = 0, 1 .
In fact, using the inequalities (4.11), (4.15) and the fact that u ∈ LE2 , we have
k∂∂x (u(T ) − u(0))kL2x
= k∂∂x (Φ[u](T ) − u(0))kL2x
≤ k∂∂x I[N [u]](T )kL2x + k∂∂x (u(0) (T ) − u(0))kL2x
s1
1−s2
s2 p−1
1
≤ C2 (kuk1−s
E1 kukE2 + kukE1 kukE2 )
′
×kr−δ hri−1/2+δ ∂∂x ukL2 ([0,T ];L2x ) + o(1)
′
≤ kr−δ hri−1/2+δ ∂∂x ukL2 ([0,T ];L2x ) + o(1) = o(1)
as T → 0+. This proves the continuity at t = 0. Recall that our solution satisfies
u = Φ[u], which tells us that we can also view u as the solution to the linear wave
equation v = N [u](t0 + t) with initial data (u(t0 ), ∂t u(t0 )) at any other time
t0 ∈ (0, ∞). Then a similar argument will give us the continuity at any t ∈ [0, ∞).
Now, we turn to the proof of uniqueness. Assume there exists another solution
v ∈ X∞ ∩ Ct H 2 ∩ Ct1 H 1 , with the same initial data. Recall that u, v ∈ Ct H 2 ∩
Ct1 H 1 . If we restrict these solutions to small enough time interval [0, T ], owing to
∂∂xα (u − v)(0) = 0, we have
X
X
(k∂∂xα (u − v)kC([0,T ];L2x) + k∂∂xα ukC([0,T ];L2x) )
k∂∂xα vkC([0,T ];L2x) ≤
|α|=i−1
|α|=i−1
≤
o(1) + 2C1 Λi ,
as T → 0+. Using the inequality (4.12), we see that
′
kr−δ hri−1/2+δ ∂(u − v)kL2 ([0,T ];L2x )
=
≤
≤
′
kr−δ hri−1/2+δ ∂(Φ[u] − Φ[v])kL2 ([0,T ];L2x )
′
C3 kr−δ hri−1/2+δ ∂(u − v)kL2 ([0,T ];L2x )
s1
1−s2
s2
1−s1
s1
1−s2
s2 p−1
1
×(kuk1−s
E1 kukE2 + kukE1 kukE2 + kvkE1 kvkE2 + kvkE1 kvkE2 )
3 −δ −1/2+δ′
kr hri
∂(u − v)kL2 ([0,T ];L2x ) ,
4
provided T > 0 is small enough, where we have used (4.17) and (4.16). By this we
arrive at the conclusion that u = v for t ∈ [0, T ], which shows the uniqueness. This
completes the proof of Theorem 1.1.
4.2. Glassey conjecture when p = pc and n ≥ 3. Consider (1.1) for p = pc and
n ≥ 3.
The estimate of the homogeneous solution, u(0) , is given by Proposition 4.1. We
only need to give the estimate for the inhomogeneous part.
Proposition 4.3. Let p = pc and n ≥ 3, u ∈ XT and s ∈ (1/2, 1]. Define
(4.19)
δ=
n − 2s
(p − 1) .
4
GLASSEY CONJECTURE
19
Then there is a positive constant C4 , independent of T > 0, such that the following
estimates hold
(4.20)
1/2
1/2
s p−1
kukLEi , i = 1, 2 .
kI[N [u]]kEi∩LEi ≤ C4 log(2 + T )(kukE1 kukE2 + kuk1−s
E1 kukE2 )
Moreover, we have
(4.21)
kΦ[u] − Φ[v]kE1 ∩LE1
≤ C5 log(2 + T )ku − vkLE1
1/2
1/2
1/2
1/2
1−s
s p−1
s
,
×(kukE1 kukE2 + kuk1−s
E1 kukE2 + kvkE1 kvkE2 + kvkE1 kvkE2 )
for some C5 .
Proof. First, by (4.13) with s1 = 1/2 and s2 = s, we have
(4.22)
1/2
1/2
x
x
s
ku(rω)kL2ω ≤ Crs−n/2 hri1/2−s (kukL2 k∂r ukL2 + kuk1−s
L2 k∂r ukL2x ) .
x
By Lemma 2.4, we have for u ∈ XT ,
1/2
1/2
s
|∂u| ≤ Crs−n/2 hri1/2−s (kukE1 kukE2 + kuk1−s
E1 kukE2 ) .
(4.23)
From (4.23) and (4.6), it is clear that, for i = 1, 2,
X
krδ hri1/2−δ ∂xα N [u]kL2x
|α|=i−1
≤ C
X
|α|=i−1
1/2
krδ hri1/2−δ |∂u|p−1 ∂xα ∂ukL2x
1/2
s p−1
≤ C(kukE1 kukE2 + kuk1−s
E1 kukE2 )
X
krδ+(s−n/2)(p−1) hri1/2−δ+(1/2−s)(p−1) ∂xα ∂ukL2x
×
|α|=i−1
1/2
1/2
s p−1
= C(kukE1 kukE2 + kuk1−s
E1 kukE2 )
X
|α|=i−1
kr−δ hri−1/2+δ ∂xα ∂ukL2x .
Now applying Lemma 3.2 to ∂xα u with |α| ≤ 1 and u0 = u1 = 0, we have for i = 1, 2,
kI[N [u]]kEi ∩LEi
≤ C(log(2 + T ))1/2
X
|α|=i−1
krδ hri1/2−δ ∂xα N [u]kL2([0,T ];L2x )
1/2
1/2
s p−1
≤ C(log(2 + T )) (kukE1 kukE2 + kuk1−s
E1 kukE2 )
X
kr−δ hri−1/2+δ ∂xα ∂ukL2 ([0,T ];L2x )
×
1/2
|α|=i−1
1/2
1/2
s p−1
≤ C log(2 + T )(kukE1 kukE2 + kuk1−s
kukLEi .
E1 kukE2 )
This proves (4.20). A similar argument with (4.7) instead of (4.6) will yield (4.21).
Here, for later use, we record the following inequality which is a direct consequence
20
KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA
of the last one,
sup k∂∂x I[N [u]](t)kL2x (Rn )
(4.24)
t∈[0,T ]
≤
C4 (log(2 + T ))1/2 kr−δ hri−1/2+δ ∂x ∂ukL2 ([0,T ];L2x )
1/2
1/2
s p−1
.
×(kukE1 kukE2 + kuk1−s
E1 kukE2 )
With these two Propositions 4.1 and 4.3 in hand, the proof of Theorem 1.2
proceeds similarly to that of Theorem 1.1. With
Λi := ku0 kḢ i (Rn ) + ku1 kḢ i−1 (Rn ) , i = 1, 2 ,
we find by Propositions 4.1 and 4.3 that the mapping Φ, defined by (4.3), is a
contraction mapping from X(2C1 Λ1 , 2C1 Λ2 ; T ) into itself, for any T > 0 provided
that
(4.25)
1/2
1/2
s p−1
+ Λ1−s
≤ 1/2 ,
1 Λ2 )
1/2
1/2
s p−1
+ Λ1−s
≤ 1/2 .
1 Λ2 )
C4 log(2 + T )(2C1 )p−1 (Λ1 Λ2
and
(4.26)
C5 log(2 + T )(4C1 )p−1 (Λ1 Λ2
Define a positive constant C6 by
C6−1 = max(2C5 (4C1 )p−1 , 2C4 (2C1 )p−1 ) ,
and set T∗ according to
1/2
1/2
log(2 + T∗ )(Λ1 Λ2
s p−1
+ Λ1−s
= C6 ,
1 Λ2 )
which is possible in general only if
1/2
1/2
ǫ = Λ1 Λ2
s
+ Λ1−s
1 Λ2 ≪ 1 .
That is
T∗ = exp(C6 ǫ1−p ) − 2, ǫ ≪ 1 .
(4.27)
Since Φ is a contraction mapping in X(2C1 Λ1 , 2C1 Λ2 ; T∗ ), the unique fixed point
u ∈ X(2C1 Λ1 , 2C1 Λ2 ; T∗ ) is the solution which we seek.
To complete the proof of Theroem 1.2, we need also to establish the regularity
of u, i.e.,
(4.28)
∂ti u ∈ C([0, T∗ ]; H 2−i (Rn )), i = 0, 1,
and the uniqueness of the solution u.
First, for the problem of regularity, it suffices to show
∂ti u ∈ C([0, T∗ ]; Ḣ 2−i (Rn )), i = 0, 1 .
Indeed, since u ∈ LE2 (T∗ ), we know that
kr−δ hri−1/2+δ ∂x ∂ukL2([0,T∗ ];L2x ) < ∞ ,
and so
lim kr−δ hri−1/2+δ ∂x ∂ukL2 ([0,T ];L2x ) = 0 .
T →0+
GLASSEY CONJECTURE
21
Using the inequality (4.24) and (4.25), we have
=
≤
≤
≤
k∂∂x (u(T ) − u(0))kL2x
k∂∂x (Φ[u](T ) − u(0))kL2x
k∂∂x I[N [u]](T )kL2x + k∂∂x (u(0) (T ) − u(0))kL2x
C4 (log(2 + T ))1/2 kr−δ hri−1/2+δ ∂x ∂ukL2 ([0,T ];L2x )
1/2
1/2
s p−1
+ o(1)
×(kukE1 kukE2 + kuk1−s
E1 kukE2 )
kr−δ hri−1/2+δ ∂x ∂ukL2 ([0,T ];L2x ) + o(1) = o(1)
as T → 0+. This proves the continuity at t = 0. A similar argument will give us
the continuity at any t ∈ [0, T∗ ].
Now, we turn to the proof of uniqueness. Assume that there exists another
solution v ∈ XT∗ ∩ Ct H 2 ∩ Ct1 H 1 , with the same initial data. Recall that u, v ∈
Ct H 2 ∩ Ct1 H 1 . If we restrict these solutions to small enough time interval [0, T ],
owing to ∂∂xα (u − v)(0) = 0, we have
X
X
(k∂∂xα (u − v)kC([0,T ];L2x) + k∂∂xα ukC([0,T ];L2x) )
k∂∂xα vkC([0,T ];L2x) ≤
|α|=i−1
|α|=i−1
≤
o(1) + 2C1 Λi ,
as T → 0+. Using the inequality (4.12), we see that
kr−δ hri−1/2+δ ∂(u − v)kL2 ([0,T ];L2x )
=
≤
kr−δ hri−1/2+δ ∂(Φ[u] − Φ[v])kL2 ([0,T ];L2x )
C5 log(2 + T )kr−δ hri−1/2+δ ∂(u − v)kL2 ([0,T ];L2x )
1/2
1/2
1/2
1/2
1−s
s p−1
s
×(kukE1 kukE2 + kuk1−s
E1 kukE2 + kvkE1 kvkE2 + kvkE1 kvkE2 )
3 −δ −1/2+δ
≤
kr hri
∂(u − v)kL2 ([0,T ];L2x) ,
4
provided T > 0 is small enough, where we have used (4.26). By this, we conclude
that u = v for t ∈ [0, T ], which shows the uniqueness. This completes the proof of
Theorem 1.2.
5. Glassey conjecture when p < pc and n ≥ 2
In this section, we aim at giving the proof of Theorem 1.3 for n ≥ 2. As we will
see, the argument in the previous section can be adapted to the scale-supercritical
case p < pc , for n ≥ 3. The argument in the previous section does not apply when
n = 2, owing to the fact that current techniques do not yield the inhomogeneous
KSS type estimates (3.12) for n = 2. Alternatively, applying the homogeneous
estimates in Lemma 3.1 gives us the proof.
In this section, by δ in LE norm, we mean
(
(n−1)(p−1)
1
, 1 < p < 1 + n−1
2
(5.1)
δ=
(n−1)(p−1)
1
2
, 1 + n−1 ≤ p < 1 + n−1
= pc .
4
Note that 0 < δ < 1/2.
We aim at showing that Φ is a contraction mapping of X(R1 , R2 ; T ), if we choose
R1 , R2 and T suitably. As before, the estimate of the homogeneous solution, u(0) ,
22
KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA
is given by Proposition 4.1. We only need to obtain a similar estimate for the
inhomogeneous part.
Proposition 5.1. Let 1 < p < pc , u ∈ XT and δ as in (5.1). Then there is a
positive constant C7 , independent of T > 0, such that the following estimates hold
(5.2)
kI[N [u]]kEi + kI[N [u]]kLEi
≤ C7 T 1−(n−1)(p−1)/2 (kukE1 ∩LE1 kukE2∩LE2 )(p−1)/2 kukLEi , i = 1, 2 .
Moreover, we have
(5.3)
≤
kΦ[u] − Φ[v]kE1 ∩LE1
C8 T 1−(n−1)(p−1)/2 (kukE1 ∩LE1 kukE2 ∩LE2 + kvkE1 ∩LE1 kvkE2 ∩LE2 )(p−1)/2
×ku − vkLE1
for some C8 , independent of T > 0.
Remark 5.1. From the proof of (5.3), we can extract the following estimates. If
1 < p < min(pc , 2), then
kr−δ ∂(Φ[u] − Φ[v])kL2 ([0,T ];L2x )
(5.4)
≤
C8 T 1−(n−1)(p−1)/2 (kukE1 kukE2 + kvkE1 kvkE2 )(p−1)/2
×kr−δ ∂(u − v)kL2 ([0,T ];L2x ) .
If 2 ≤ p < 3 and n = 2,
kr−δ (Φ[u] − Φ[v])kL2 ([0,T ];L2x )
(5.5)
≤ C8 T (3−p)/4 (kukE1 kukE2 + kvkE1 kvkE2 )(p−2)/2
× kr−δ ∂ukL2 ([0,T ];L2x ) kr−δ ∂∂x ukL2 ([0,T ];L2x )
1/2
+ kr−δ ∂vkL2 ([0,T ];L2x ) kr−δ ∂∂x vkL2 ([0,T ];L2x )
×kr−δ ∂(u − v)kL2 ([0,T ];L2x ) .
Remark 5.2. From the proof of (5.2) with i = 2, we can extract the following
estimates. If 1 < p < min(pc , 2), then
(5.6)
k∂∂x (Φ[u](T ) − u(0))kL2x
≤ k∂∂x I[N [u]](T )kL2x + k∂∂x (u(0) (T ) − u(0))kL2x
≤ C7 T 1−(n−1)(p−1)/2 (kukE1 kukE2 )(p−1)/2 T δ−1/2 kr−δ ∂∂x ukL2 ([0,T ];L2x ) + o(1)
as T → 0+. If 2 ≤ p < 3 and n = 2,
(5.7)
k∂∂x (Φ[u](T ) − u(0))kL2x
≤ k∂∂x I[N [u]](T )kL2x + k∂∂x (u(0) (T ) − u(0))kL2x
1/2
≤ C7 (kukE1 kukE2 )(p−2)/2 kr−(p−1)/4 ∂ukL2 ([0,T ];L2 )
x
3/2
×kr−(p−1)/4 ∂∂x ukL2 ([0,T ];L2 ) + o(1)
x
as T → 0+.
Proof. We will deal with three different cases: 1 < p < 1 + 1/(n − 1) when n ≥ 2;
2 ≤ p < 3 when n = 2; and 1 + 1/(n − 1) ≤ p < pc when n ≥ 3.
GLASSEY CONJECTURE
23
Case i) 1 < p < 1 + 1/(n − 1) with δ = (n − 1)(p − 1)/2. First, by (2.3) and
Lemma 2.4, we have for u ∈ XT ,
|∂u| ≤ Cr−(n−1)/2 (kukE1 kukE2 )1/2 .
(5.8)
By (5.8) and (4.6), it is clear that, for i = 1, 2,
X
X
k|∂u|p−1 ∂xα ∂ukL2x
k∂xα N [u]kL2x ≤ C
|α|=i−1
|α|=i−1
≤
C(kukE1 kukE2 )(p−1)/2
=
(p−1)/2
C(kukE1 kukE2 )
X
|α|=i−1
X
|α|=i−1
kr−(n−1)(p−1)/2 ∂xα ∂ukL2x
kr−δ ∂xα ∂ukL2x .
Now applying Lemma 3.1 to ∂xα u with |α| ≤ 1 and u0 = u1 = 0, we have for i = 1, 2,
≤
kI[N [u]]kEi∩LEi
X
k∂xα N [u]kL1 ([0,T ];L2x )
C
≤
CT 1/2
≤
CT
1/2
≤
CT 1−δ (kukE1 kukE2 )(p−1)/2 T δ−1/2
≤
|α|=i−1
X
|α|=i−1
CT
k∂xα N [u]kL2 ([0,T ];L2x)
(kukE1 kukE2 )(p−1)/2
1−δ
(p−1)/2
(kukE1 kukE2 )
X
|α|=i−1
kr−δ ∂xα ∂ukL2 ([0,T ];L2x )
X
|α|=i−1
kr−δ ∂xα ∂ukL2 ([0,T ];L2x )
kukLEi .
This proves (5.2). A similar argument with (4.7) instead of (4.6) will yield (5.3).
Case ii) 1 + 1/(n − 1) ≤ p < pc , n ≥ 3 and δ = (n − 1)(p − 1)/4. In this case,
we may use Lemma 3.2 instead. Applying it to ∂xα u with |α| ≤ 1 and u0 = u1 = 0,
we have for i = 1, 2,
≤
≤
≤
≤
kI[N [u]]kEi + kI[N [u]]kLEi
X
krδ ∂xα N [u]kL2 ([0,T ];L2x )
CT 1/2−δ
|α|=i−1
CT
1/2−δ
CT
1−2δ
CT
1−2δ
(kukE1 kukE2 )(p−1)/2
(p−1)/2
(kukE1 kukE2 )
(p−1)/2
(kukE1 kukE2 )
X
|α|=i−1
kr−(n−1)(p−1)/2 rδ ∂xα ∂ukL2 ([0,T ];L2x )
T δ−1/2
X
|α|=i−1
kr−δ ∂xα ∂ukL2 ([0,T ];L2x)
kukLEi ,
where we have used (5.8) and (4.6). Using (4.7) instead of (4.6), (5.3) follows
similarly.
Case iii) 2 ≤ p < 3, n = 2 and δ = (p − 1)/4. Notice that Lemma 2.3 with
s = (3 − p)/4 > 0 gives us
√ −(p−1)/4 1/2
1/2
2 ≤
kr−(p−3)/4 ukL∞
2kr
ukL2 (R2 ) kr−(p−1)/4 ∂x ukL2 (R2 ) .
L
r
ω
x
x
24
KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA
Then for i = 1, 2, we obtain by using Lemma 3.1
≤
kI[N [u]]kEi + kI[N [u]]kLEi
X
k∂xα N [u]kL1 ([0,T ];L2x (R2 ))
C
≤
C(kukE1 kukE2 )(p−2)/2
≤
(p−2)/2
|α|=i−1
X
|α|=i−1
kr−(p−2)/2 ∂u∂xα ∂ukL1 ([0,T ];L2x (R2 ))
kr−(p−3)/4 ∂ukL2 ([0,T ];L∞
C(kukE1 kukE2 )
x )
X
−(p−1)/4 α
kr
∂x ∂ukL2 ([0,T ];L2x )
×
|α|=i−1
≤
2
C(kukE1 kukE2 )(p−2)/2 kr−(p−3)/4 ∂ukL2 ([0,T ];L∞
r Lω )
X
kr−(p−1)/4 ∂xα ∂ukL2 ([0,T ];L2x )
×
|α|=i−1
≤
CT 1−2δ (kukE1 kukE2 )(p−2)/2 (kukLE1 kukLE2 )1/2 kukLEi ,
where we have used (4.6) and Lemma 2.4.
The estimate (5.3) follows from the similar arguments by using (4.7) instead of
(4.6). This completes the proof.
With these two Propositions 4.1 and 5.1 in hand, it will be easy to show Theorem
1.3. Setting
Λi := ku0 kḢ i (Rn ) + ku1 kḢ i−1 (Rn ) , i = 1, 2 ,
we find by Propositions 4.1 and 5.1 that the mapping Φ, defined by (4.3), is a
contraction mapping from X(2C1 Λ1 , 2C1 Λ2 ; T ) into itself, provided that
C7 T 1−(n−1)(p−1)/2 (2C1 )p−1 (Λ1 Λ2 )(p−1)/2 ≤ 1/2
and
C8 T 1−(n−1)(p−1)/2 (4C1 )p−1 (Λ1 Λ2 )(p−1)/2 ≤ 1/2 .
Define a positive constant C9 by
−(1−(n−1)(p−1)/2)
C9
= max(2C8 (4C1 )p−1 , 2C7 (2C1 )p−1 ) ,
and set T∗ according to
−(1−(n−1)(p−1)/2)
C9
1−(n−1)(p−1)/2
T∗
(Λ1 Λ2 )(p−1)/2 = 1 ,
that is
(5.9)
p−1
T∗ = C9 (Λ1 Λ2 )− 2−(n−1)(p−1) .
Since Φ is a contraction mapping of X(2C1 Λ1 , 2C1 Λ2 ; T∗ ), the unique fixed point
u ∈ X(2C1 Λ1 , 2C1 Λ2 ; T∗ ) is the solution which we seek.
To complete the proof of Theroem 1.3, we also need to establish the uniqueness
of u in XT∗ , and the regularity of u, i.e.,
(5.10)
∂ti u ∈ C([0, T∗ ]; H 2−i (Rn )), i = 0, 1.
First, for the proof of uniqueness, assume that there exists another solution v ∈ XT∗ ,
with the same initial data. Recall the estimates (5.4) and (5.5). If we restrict these
GLASSEY CONJECTURE
25
solutions to small enough 0 < T < T∗ , we have
kr−δ ∂(u − v)kL2 ([0,T ];L2x )
= kr−δ ∂(Φ[u] − Φ[v])kL2 ([0,T ];L2x )
1 −δ
kr ∂(u − v)kL2 ([0,T ];L2x ) .
≤
2
Combining this with the fact that u and v share the same initial data, we conclude
that u = v for t ∈ [0, T ], which shows the uniqueness.
By Remark 5.2, 1 − (n − 1)(p − 1)/2 + δ − 1/2 > 0, and the fact that r−δ ∂∂x u ∈
2
L ([0, T∗ ]; L2x ) (and so
kr−δ ∂∂x ukL2 ([0,T ];L2x ) = o(1)
as T → 0+), we see that ∂∂x u(T ) converges to ∂∂x u(0) in L2x . This tells us that
the continuity at t = 0. A similar argument will give us the continuity at any
t ∈ [0, T∗ ]. This completes the proof of Theorem 1.3.
6. Glassey conjecture when n = 2, p > pc
For this case, it seems not enough for us to prove global results by applying the
KSS type estimates, mainly because we do not have the favorable inhomogeneous
estimates as (3.12) in Lemma 3.2.
Instead, we want to present a proof based on the recent generalized Strichartz
estimates of Smith, Sogge and Wang [19] (with the previous radial estimates in
Fang and Wang [2]).
Lemma 6.1 (Generalized Strichartz estimates). Let n = 2 and q ∈ (2, ∞). For
any solution u = u(t, x) to the wave equation (3.1), we have the following inequality
with s = 1 − 1/q,
(6.1)
2
2
k∂ukLq ([0,∞);L∞
≤ Cq (k∂x u0 kḢ s + ku1 kḢ s + kF kL1 Ḣ s ) ,
r Lω (R ))
x
x
t
x
where C is independent of the functions u0 , u1 and F .
With these estimates, we are able to present a simple proof of Theorem 1.4. Let
Λi := ku0 kḢ i (Rn ) + ku1 kḢ i−1 (Rn ) , i = 1, 2 .
By using (4.6) and the energy estimates, we have
(6.2)
2
k∂Φ[u]kL∞
t Lx
≤
CΛ1 + CkN [u]kL1t L2x
≤
2 ,
k∂ukL∞
CΛ1 + Ck∂ukp−1
t Lx
Lp−1 L∞
≤
CΛ2 + Ck∂x N [u]kL1t L2x
t
x
and
(6.3)
2
k∂∂x Φ[u]kL∞
t Lx
≤
2 .
CΛ2 + Ck∂ukp−1
k∂∂x ukL∞
t Lx
Lp−1 L∞
t
x
Recall the convex inequality
kf kḢ 1−θ ≤ kf kθL2 kf k1−θ
, θ ∈ [0, 1],
Ḣ 1
26
KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA
together with (4.6), Lemma 6.1 and Lemma 2.4. We see that for p > 3,
(6.4)
k∂Φ[u]kLp−1L∞
t
≤
≤
≤
x
k∂Φ[u]kLp−1L∞ L2
t
r
ω
Ck∂u(0)kḢ 1−1/(p−1) + CkN [u]kL1 Ḣ 1−1/(p−1)
t
1/(p−1) 1−1/(p−1)
CΛ1
Λ2
+
1−1/(p−1)
1/(p−1)
Ck∂ukp−1
k∂ukL∞L2 k∂∂x ukL∞ L2
Lp−1
L∞
x
x
t
t
t
x
.
Moreover, we have
(6.5)
≤
≤
2
k∂(Φ[u] − Φ[v])kL∞
t Lx
CkN [u] − N [v]kL1t L2x
2 .
C(k∂ukLp−1L∞ + k∂vkLp−1 L∞ )p−1 k∂(u − v)kL∞
t Lx
t
t
x
x
Let ǫ0 > 0 be the number such that
C(4Cǫ0 )p−1 = 1/2 .
If
1/(p−1)
ǫ = Λ1
1−1/(p−1)
Λ2
≤ ǫ0 ,
then we see that Φ is a contraction mapping in Y (2CΛ1 , 2CΛ2 , 2Cǫ). Here the
complete space Y (R1 , R2 , R3 ) is defined as
Y (R1 , R2 , R3 ) =
1
2 ≤ R1 ,
{u ∈ Ct Hrad
∩ Ct1 L2rad ; k∂ukL∞
t Lx
2 ≤ R2 , k∂uk p−1 ∞ ≤ R3 }
k∂∂x ukL∞
L
L
t Lx
t
x
2.
with the metric ρ(u, v) = k∂(u − v)kL∞
t Lx
To prove the regularity, we only need to show the continuity at t = 0. For that,
since ∂u ∈ Lp−1 ([0, ∞); L∞
x ), we have
k∂∂x (u(t) − u(0))kL2x
≤
k∂∂x I[N [u]](t)kL2x + k∂∂x (u(0) (t) − u(0))kL2x
≤
∞ 2
Ck∂ukp−1
Lp−1 ([0,t];L∞ ) k∂∂x ukLt Lx + o(1) = o(1)
≤
k∂x N [u]kL1 ([0,t];L2x ) + o(1)
x
as t → 0+. This tells us that u ∈ Ct Ḣ 2 ∩ Ct1 Ḣ 1 .
For uniqueness, suppose that there exists another solution v ∈ Y ∩Ct H 2 ∩Ct1 H 1 ,
with the same initial data. Using the inequality (6.5), we see that
=
≤
≤
k∂(u − v)kCt ([0,T ];L2x)
k∂(Φ[u] − Φ[v])kCt ([0,T ];L2x )
C(k∂ukLp−1([0,T ];L∞ ) + k∂vkLp−1 ([0,T ];L∞ ) )p−1 k∂(u − v)kCt ([0,T ];L2x )
t
x
t
x
o(1)k∂(u − v)kCt ([0,T ];L2x )
as T → 0+. Thus by choosing T > 0 small enough, we conclude that u = v for
t ∈ [0, T ], which shows the uniqueness. This completes the proof of Theorem 1.4.
GLASSEY CONJECTURE
27
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28
KUNIO HIDANO, CHENGBO WANG, AND KAZUYOSHI YOKOYAMA
Department of Mathematics, Faculty of Education, Mie University, 1577 Kurimamachiya-cho, Tsu, Mie 514-8507, Japan
E-mail address: hidano@edu.mie-u.ac.jp
Department of Mathematics, Zhejiang University, Hangzhou 310027, China
E-mail address: wangcbo@gmail.com
Hokkaido Institute of Technology, 7-15-4-1 Maeda, Teine-ku, Sapporo, Hokkaido
006-8585, Japan
E-mail address: yokoyama@hit.ac.jp