Double logarithmic inequality with a sharp constant in four space dimensions

0.19 MB·English

by  Mohamed Majdoub

DOUBLE LOGARITHMIC INEQUALITY WITH A SHARP CONSTANT
IN FOUR SPACE DIMENSIONS
MOHAMED MAJDOUBAND TAREK SAANOUNI
Abstract. We prove a Log Log inequality with a sharp constant in four dimensions
for radially symmetricfunctions. Wealso show that theconstant in theLog estimate is
almost sharp.
3
1
0 1. Introduction and statement of the results
2
The Sobolev embeddings in four dimensions [1],
n
Ja W2,p ֒ L4−82p for 1 p < 2 and W2,p ֒ C2−p4 for 2< p <
→ ≤ → ∞

0 fails in the limiting case p = 2. In the setting of a bounded domain we have the injection
1 W2,2 Lq for any q < . The function log(1 log x ) is a conterexample if the domain
] is a su⊂bset of the unit ba∞ll. Moreover, H2 := W−2,2 fu|nc|tions are in a so-colled Orlicz space
P
[5], i.e. their exponential powers are integrable functions. Precisely, we have the following
A
Adams’ type inequality.
.
h
Theorem 1.1 ([13],Theorem2.2). For any α (0,32π2) there exists a constant C(α) > 0
t
a ∈
such that
m
[ (1.1) eαu(x)2 1 dx C(α) u 2 u W2,2(R4) with ∆u 1,
R4 | | − ≤ k kL2 ∀ ∈ k kL2 ≤
1 Z (cid:16) (cid:17)
v and this inequality is false for α > 32π2.
3
5 We stress that α = 32π2 becomes admissible if we require u 1 raher than
W2,2
3 k k ≤
∆u 1, where
2 k kL2 ≤
1. kuk2W2,2 = k∆uk2L2 +k∇uk2L2 +kuk2L2.
0
Theorem 1.2 ([16], Theorem 1.4). There exists a constant C > 0 such that for any
3
domain Ω R4
1
⊂
:
v (1.2) sup e32π2u(x)2 1 dx C
| |
i − ≤
X u∈H02(Ω), kukH2≤1 ZΩ(cid:16) (cid:17)
r and this inequality is sharp.
a
In this work, we prove that in the radial case we can control the L norm with H˙2
∞
norm and a stronger norm with Logarithmic growth or double logarithmic growth. The
Date: January 14, 2013.
1991 Mathematics Subject Classification. 49K20, 35L70.
Key words and phrases. Limiting Sobolev embedding,Moser-Trudinger inequality,best constants.
M. M. is grateful tothe Laboratory of PDEand Applications at theFaculty of Sciences of Tunis.
T. S. is grateful to theLaboratory of PDE and Applications at theFaculty of Sciences of Tunis.
1
2 M.Majdoub,T.Saanouni,.../Globalwell-posedness ofa4D...
inequality is sharp for the double logarithmic growth. Similar results proved in two di-
mensions in [10, 4] was applied in [11, 7] to prove global well-poseness of semilinear wave
and Schro¨dinger equations with nonlinearity growing exponentially.
For any α (0,1), we denote by Cα := Cα(R4) the space of α-H¨older continuous
∈
functions endowed with the norm
u(x) u(y)
kukCα = kukCα(R4) := kukL∞(R4)+sup| x −y α |.
x=y
6 | − |
Moreover, C˙α := C˙α(R4) denotes the homogenous space of α-H¨older continuous functions
endowed with the semi norm
u(x) u(y)
u = u := sup| − |.
k kC˙α k kC˙α(R4) x y α
x=y
6 | − |
WofeRa4lscoendteefirneedtahtetrhaetioorNigαin(uw)i:t=h krk∆audukCki˙uLα2s.rFaonrdanBy p:=osiBtiv.e Trehael nsupmacbeerHr2,(ΩB)r sistatnhdesbfaolrl
1 0
the completion in the Sobolev space H2 of smooth and compactly supported functions.
H2 (Ω) (respectively H2 (Ω)) is the space of radially symmetric functions of H2(Ω)
0,rad rad 0
(respectively H2(Ω)).
Our first result reads
Theorem 1.3. (Double Log estimate) Let α (0,1). A positive constant C exists
α
∈
such that for any function u (H2 C˙α)(B), we have
∈ 0,rad∩
1
(1.3) kuk2L∞ ≤ 8π2αk∆uk2L2log e3+CαNα(u) log(2e+Nα(u)) .
Moreover, the constant 1 in the above(cid:16)inequality is spharp. (cid:17)
8π2α
The second result of this paper is the following
Theorem 1.4. (Log estimate) Let α (0,1). For any λ > 1 there exists C > 0
∈ 8π2α λ
such that for any function u (H2 C˙α)(B), we have
∈ 0,rad∩
(1.4) kuk2L∞ ≤ λk∆uk2L2log Cλ+Nα(u) .
Moreover, the above inequality is false for λ = (cid:16)1 . (cid:17)
8π2α
We derive the following global estimate.
Corollary 1.5. (Global Log estimate) Let α (0,1). For any λ > 1 and any
∈ 8π2α
µ (0,1], there exists C > 0 such that for any function u (H2 Cα)(R4), we have
∈ λ ∈ rad∩
(1.5) kuk2L∞ ≤ λkuk2µlog Cλ+ 8αµ−uαkukCα .
µ
(cid:18) k k (cid:19)
Where u 2 := (1+3µ) ∆u 2 +3µ u 2 .
k kµ k kL2 k kH1
Remark 1.6. When we deal with higher order derivatives, we cannot reduce the problem
to the radial case as in dimension two for example. The reason is that, for a givenfunction
u W2,2, we do not know wether or not u♯ (the Schwarz symmetrization of u) still belongs
∈
to W2,2. Even if this is the case, no inequality of the form ∆u♯ ∆u is known to
L2 L2
k k ≤k k
hold. To overcome this difficulty, one can try to apply a suitable comparison principle as
M.Majdoub,T.Saanouni,.../Globalwell-posedness ofa4D... 3
in [16]. Since in our case we need to control H¨older norms also, this method fails to reduce
our problem to the radial case. This is why we restrict ourselves to the radial setting.
Finally, we mention that C will be used to denote a constant which may vary from line
to line. We also use A . B to denote an estimate of the form A CB for some absolute
≤
constant C and A B if A. B and B . A.
≈
2. A Littlewood-Paley proof
We prove that inequality (1.4) can be obtained with an unknown absolute constant
instead of any λ > 1 . To do so, we give a brief review of the Littlewood-Paley theory.
8π2α
We refer to [6] for more details. Denote by the annulus ring defined by
0
C
3 8
:= x R4 such that < x < ,
0
C { ∈ 4 | | 3}
and choose two nonnegative radial functions χ C (B ) and ϕ C ( ) such that
∈ 0∞ 43 ∈ 0∞ C0
χ+ ϕ(2 j.)= 1 on R4 and ϕ(2 j.) = 1 on R4 0 .
− −
−{ }
j N j Z
X∈ X∈
Define the frequency projectors by
(∆˙ u) := ϕ(2 j.) u for j Z,
j −
F F ∈
∆ u = 0 if j 2, (∆ u) = χ u and ∆ u := ∆˙ u for j 0.
j 1 j j
≤ − F − F ≥
Recall that
k∆uk2L2 ≈ (cid:16)Xj∈Z24jk∆˙juk2L2(cid:17) and kukC˙α ≈ sju∈pZ(cid:16)2jαk∆˙ jukL∞(cid:17).
We have the following result in the whole space
Proposition 2.1. Let α (0,1). There exists a positive constant C := C such that for
α
any function u ( α H∈2)(R4), one has
∈ C ∩
(2.6) kuk2L∞ ≤ Ckuk2L2 +Ck∆uk2L2log e+Nα(u) .
(cid:16) (cid:17)
Proof. We have
m 1
− ∞
u= ∆ u+ ∆ u= ∆ u+ ∆ u+ ∆ u
1 j 1 j j
− −
j N j=0 j=m
X∈ X X
where m is an integer to fix later. Using Bernstein inequality, we get
m 1
− ∞
u L∞ C ∆ 1u L2 +C 22j ∆ju L2 + 2−jα(2jα ∆ju L∞)
k k ≤ k − k k k k k
j=0 j=m
X X
m 1 1
− ∞
C u +C√m 24j ∆ u 2 2 +C 2 jα u
≤ k kL2 k j kL2 − k kC˙α
(cid:16) Xj=0 (cid:17) (cid:16)jX=m (cid:17)
2 mα
C u +√m ∆u + − u .
≤ k kL2 k kL2 1 2 αk kC˙α
−
(cid:16) − (cid:17)
So
2 2mα
kuk2L∞ ≤ C kuk2L2 +mk∆uk2L2 + (1 −2 α)2kuk2C˙α .
−
(cid:16) − (cid:17)
4 M.Majdoub,T.Saanouni,.../Globalwell-posedness ofa4D...
Taking for E(x) the integer part of any real x,
m := max 1,1+E(2log (N (u)2) ,
2 α
(cid:16) (cid:17)
the proof is achieved.
Clearly, if u is supported in the unit ball, then by Poincar´e inequality and Proposition
2.1, we get
kuk2L∞ ≤ Cαk∆uk2L2log C0+Nα(u)
for some constant C big enough. (cid:16) (cid:17)
0
3. Proof of Theorem 1.3
To prove (1.3) and the fact that the constant is sharp, it is sufficient to show that
∆u 2 log e3+C N (u) log(2e+N (u))
inf k kL2 0 α α =8π2α.
u∈(H02,rad∩C˙α)(B) h kuk2L∞p i
Let prove, first, the optimality of the constant 8π2α in the previous equality. Define for
ε > 0, the functions
vε(x) :=  q321π2 log(1ε)− q8π2|xε|l2oqg(21επ)21l+og(q1ε)8πlo21glo(g|(x11ε|)) iiff ε|x14|≤≤|εx14|,≤ 1,
 (−|x|+1)(|x|−2)2 if 1 < x < 2
q2π2log(1ε) | |
 0 if x 2.
 | | ≥

Clearly uε(x) :=vε(2x) ∈ H02(B). Moreover, for small ε > 0, we have
1 1 1
kuεkL∞(B) = kvεkL∞(B2) = r32π2 log(ε)+ 8π2log(1)
ε
q
and
2
u = 2 v .
k εkLip(B) k εkLip(B2) ≤ π 2ε12 log(1)
ε
Since u u 1 α u α , we get q
k εkC˙α ≤ k εkL−∞ k εkLip
u C log(1ε)12−α.
k εkC˙α ≤ α εα4
Using the fact that ∆u 2 = ∆v 2 = 1+O( 1 ), we have
k εkL2(B) k εkL2(B2) log(1ε)
N (u ) C log(1ε)12−α.
α ε ≤ α εα4
So, for C > 0,
0
∆u 2
lim k εkL2 log e3+C N (u ) log(2e+N (u )) 8π2α.
ε→0 kuεk2L∞ h 0 α ε p α ε i ≤
M.Majdoub,T.Saanouni,.../Globalwell-posedness ofa4D... 5
Finally
∆u 2 log e3+C N (u) log(2e+N (u))
inf k kL2 0 α α 8π2α.
u∈(H02,rad∩C˙α)(B) h kuk2L∞p i ≤
Letusprovetheoppositeinequality. Withoutlossofgenerality wecannormalize u L∞ =
k k
1. Moreover, using a translation argument we may assume that u(0) = 1. Since u vanishes
on the boundary, we deduce that
x
u u( ) u(0) = 1.
k kC˙α ≥ | x − |
| |
Moreover, if u = 1 then u(x) = 1 x α and the inequality is evident. In fact
k kC˙α − | |
1 u(x) = u(x) 1 x α thus u(x) 1 x α, moreover if u(x ) > 1 x α then
0 0
− | − | ≤ | | ≥ − | | − | |
|u(1x0)x−0uα(1)| > 1 and kukC˙α > 1, which is absurd. In the sequel we assume that kukC˙α > 1.
For−D| >| 1, we denote the space
(3.7) K := u H2 (B), u(r) 1 Drα, for any 0 <r 1 .
D { ∈ 0,rad ≥ − ≤ }
It is sufficient to prove that for some C > 0, we have
α
C D D
8π2α inf inf ∆u 2 log e3+ α log(2e+ ) .
≤ D 1u KDk kL2 ∆u L2s ∆u L2
≥ ∈ h k k k k i
Consider the minimizing problem
(3.8) I[u] := ∆u 2
k kL2(B)
among the functions belonging to the set K . This is a variational problem with obstacle.
D
It has a unique minimizer u which is variationally characterized by
∗
(3.9) ∆v∆u ∆v 2 , v K .
∗ ≥ k kL2(B) ∀ ∈ D
ZB
Moreover u W3, (B), (see [12]). Hence we have an open radially symmetric set
∗ ∞
∈
:= x B,u (x) > 1 D x α .
∗
O { ∈ − | | }
Now, foranyv C ( )andanyrealnumber τ smallenough,wehaveu +τv 1 Drα
∈ 0∞ O | | ∗ ≥ −
thus u +τv K . So by (3.9) we have
∗ D
∈
∆(u +τv)∆u ∆u 2 .
∗ ∗ ≥ k ∗kL2(B)
ZB
Taking τ positive then negative and v := ∆u , we have
∗ ∗
v ∆v = 0, v C ( ).
∗ ∀ ∈ 0∞ O
Z
O
Thus u is biharmonic on ,
∗
O
1 d dv
(r3 ∗) = ∆v = ∆2u = 0 and u C ( ).
∗ ∗ ∗ ∞
r3dr dr ∈ O
So, there exists two real numbers a and b such that
1 d du a
v (r)= (r3 ∗)= +b.
∗
r3dr dr r2
6 M.Majdoub,T.Saanouni,.../Globalwell-posedness ofa4D...
With a straightforward computation, and using the boundary condition, there exists a
real number c such that
c
u (r)= b c+ +alog(r)+br2.
∗
− − r2
Now, by the boundary condition du∗(1) = 0, we have
dr
c
u (r)= b c+ +2(c b)log(r)+br2.
∗
− − r2 −
Moreover, u cannot start to be biharmonic at r = 0 because of boundary condition. So
∗
there exists a real number r (0,1] such that
0
∈
1 Drα if 0 r r ,
u∗(r)= b(r2 1)+c( 1 −1)+2(c b)log(r) if r ≤< r≤ 10.
(cid:26) − r2 − − 0 ≤
Since u C2(B), we have
∗
∈
1 Drα = (r2 1 2log(r ))b+( 1 1+2log(r ))c,
− 0 0 − − 0 r2 − 0
0
αDrα 1 = 2(r 1 )b+ 2 (1 1 )c,
 − 0− 0 − r0 r0 − r02
 −α(α−1)Dr0α−2 = 2(1+ r12)b+ r22(r32 −1)c.
0 0 0
We consider the two last equations
−αDr0α−1 = 2(r0 − r10) r20(1− r102) b := A b .
α(α 1)Drα 2 2(1+ 1 ) 2 ( 3 1) c c
(cid:18) − − 0− (cid:19) r02 r02 r02 − !(cid:18) (cid:19) (cid:18) (cid:19)
Let x := r2. With a simple computation, we obtain
0
8
det(A) = (r2 1)2,
−r5 0 −
0
α
αD x2 2
b = [α+ ],
4 1 x 1 x
− −
αDxα2+1 2
c = [α 2+ ].
4 x 1 − 1 x
− −
Substituting in the first equation of the precedent system, we obtain
4(x 1)2
D(x) = − .
α
x2[4(x 1)2+α(2+α(1 x))(x 1 log(x))+α((α 2)(1 x) 2)(1 x+xlog(x))]
− − − − − − − −
Now, let us compute ∆u 2 . Since
k ∗kL2(B)
Dα(α+2)rα 2 if 0 r r ,
− 0
∆u (r) = − ≤ ≤
∗ 4(c b)
(cid:26) r−2 +8b if r0 ≤ r ≤ 1,
we obtain
r0 1 4(c b)
k∆u∗k2L2(B) = 2π2 (Dα(α+2))2 r2α−1dr+ r3( r−2 +8b)2dr
h Z0 Zr0 i
= (π(α+2))2αD2xα 16π2(c b)2log(x)+32π2b2(1 x2)+64π2b(c b)(1 x).
− − − − −
It follows that
(α+2 (α 2)x2)2 2α2
k∆u∗k2L2(B) =π2xαD2 α(α+2)2−α2logx (−1 x−)4 +(1 x)3(α+2−αx)((α−4)x2+2x−α−2) .
h − − i
M.Majdoub,T.Saanouni,.../Globalwell-posedness ofa4D... 7
We denote
(α+2 (α 2)x2)2 2α2
g(x): = π2xα α(α+2)2 α2logx − − + (α+2 αx)((α 4)x2+2x α 2) ,
− (1 x)4 (1 x)3 − − − −
h − − i
4(x 1)2
D(x) = − ,
xα2[4(x 1)2+α(2+α(1 x))(x 1 log(x))+α((α 2)(1 x) 2)(1 x+xlog(x))]
− − − − − − − −
FC(x) := k∆u∗k2L2(B)log e3+CNα(u∗) log(2e+Nα(u∗)) ,
h p i
log(2e+1/ g(x))
= D2(x)g(x)log e3+C .
s g(x)
p
h i
It is sufficient to prove that a constant C exists such that
α
(3.10) F 8π2α on (0,1].
Cα ≥
We have
1 4
g(x) ∽ π2α2(α+2)2xαlog( ) and D(x) ∽ ,
x α(2+α)xα2 log(1)
x
where ∽ is used to indicate that the ratio of the two sides goes to 1 as x goes to zero.
Thus
4 2 1 C
F (x) ∽ π2α2(α+2)2xαlog( )log(e3 + )
C α(2+α)xα2( log(x)) x g(x)
(cid:16) − (cid:17)
log e3+ C p
∽ 16π2 √g(x)
(cid:16)log(1) (cid:17)
x
∽ 8π2α.
Consequently, there exists x (0,1) such that
α
∈
F (x) 8π2α for all x [0,x ].
C α
≥ ∈
Now, to study the behaviour of D(x) for x 1, we denote
→
y := x 1, h(y) := 4y2+α(2 αy)(y log(y+1)) α((α 2)y+2)( y+(y+1)log(1+y)).
− − − − − −
An easy computation yields to
h(y) = 4y2+α(2 αy)(y log(y+1)) α((α 2)y+2)( y+(y+1)log(1+y))
− − − − −
y2 y2
= 4y2+α(2 αy)( +o(y2)) α((α 2)y+2)( +o(y2))
− 2 − − 2
= 4y2+o(y2), as y 0.
→
Hence,
D(1 )= 1 and inf D =D(x ) > 0.
− α
[xα,1]
Moreover, g 0 and g(x) = 0 for any x (0,1] because if g(x) = 0 then u is harmonic
∗
≥ 6 ∈
on B, which is absurd. Thus g y > 0, 1 y on [x ,1] and
≥ α g ≤ α α
F (x) D2(x )y log(e3 +C√y ).
C α α α
≥
8π2α
Taking C = 1+ eD2(xα)yα ,we have
α √yα
F (x) 8π2α for all x [0,1].
Cα ≥ ∈
8 M.Majdoub,T.Saanouni,.../Globalwell-posedness ofa4D...
4. Proof of Theorem 1.4
The proof of Theorem 1.4 is similar to that of Theorem 1.3.
Let λ > 1 , in order to prove (1.4) it is sufficient to prove that for some C > 0, we
8π2α λ
have
inf k∆uk2L2log(Cλ+Nα(u)) 1.
u∈(H02,rad∩C˙α)(B) kuk2L∞ ≥ λ
Arguing as previously, it is sufficient to prove that for some C > 0, we have
λ
1 D
inf inf ∆u 2 log C + ,
λ ≤ D 1u KDk kL2 λ ∆u L2
≥ ∈ (cid:16) k k (cid:17)
where the set K is already defined in (3.7). Since for all C > 1, the function
D
1
t t2log(C + )
7−→ t
is increasing, it is sufficient to minimize I[u] among the functions belonging to the set K .
D
Consider u a such minimizer. Recall that with previous computations, we have
∗
(α+2 (α 2)x2)2 2α2
k∆u∗k2L2(B) =π2xαD2 α(α+2)2−α2logx (−1 x−)4 + (1 x)3(α+2−αx)((α−4)x2+2x−α−2) ,
h − − i
1
H(x):=k∆u∗k2L2(B)log(C+Nα(u∗))=D2(x)g(x)log C+ g(x) ,
(cid:16) (cid:17)
(α+2 (α 2)x2)2 2α2
g(x):=π2xα α(α+2)2 α2logx − − + (α+2 pαx)((α 4)x2+2x α 2) .
− (1 x)4 (1 x)3 − − − −
h − − i
4(x 1)2
D(x)= − .
xα2[4(x 1)2+α(2+α(1 x))(x 1 log(x))+α((α 2)(1 x) 2)(1 x+xlog(x))]
− − − − − − − −
Recall also that
1 4
g(x) ∽ π2α2(α+2)2xαlog( ), D(x) ∽ and H(x) ∽ 8π2α.
x α(2+α)xα2 log(1)
x
Therefore, there exists x (0,1) such that
λ
∈
λH(x) 1 for all x [0,x ].
λ
≥ ∈
Moreover, via previous calculus
D(1 )= 1 and inf D = D(xλ) > 0.
−
[xλ,1]
Note also that g 0 and g(x) = 0, x (0,1] because if g(x) = 0 then u is harmonic on
∗
≥ 6 ∀ ∈
B which is absurd. Thus g y > 0 on [x ,1]. So
λ λ
≥
λH(x) λD2(xλ)y log(C ).
λ λ
≥
1
Taking Cλ = 1+eλD2(xλ)yλ, we have
λH(x) 1 for all x [0,1].
≥ ∈
Now, let us prove that (1.4) is false for λ = 1 which means that it is sharp. Precisely,
8π2α
we show that a sequence of functions u (H2 C˙α)(B) exists such that for n big
n ∈ 0,rad ∩
enough the following holds
kunk2L∞ > 8π12αk∆unk2L2log nα2 +Nα(un) .
(cid:16) (cid:17)
M.Majdoub,T.Saanouni,.../Globalwell-posedness ofa4D... 9
Take the sequence x := 1 := a2 and the sequence of functions
n n n
1 D rα if 0 r a ,
n n
un(r):= b (r2 1)+c ( 1− 1)+2(c b )log(r) if a≤< r≤ 1,
(cid:26) n − n r2 − n − n n ≤
where
4(xn 1)2
Dn = α − ,
xn2 4(xn 1)2+α(2+α(1 xn))(xn 1 log(xn))+α((α 2)(1 xn) 2)(1 xn+xnlog(xn))
− − − − − − − −
h α i
αDn xn2 2
bn = α+ ,
4 1 xn 1 xn
αDn xnα2+−1 h − 2 i
cn = α 2+ .
4 xn 1 − 1 xn
− h − i
Using previous computations it is sufficient to prove that
Hn := k∆unk2L2log nα2 +Nα(un)
(cid:16) D (cid:17)
= k∆unk2L2log nα2 + ∆un
n L2
(cid:16) k k (cid:17)
1
= Dn2gnlog nα2 + √gn < 8π2α,
(cid:16) (cid:17)
where g := g(x ) and
n n
(α+2 (α 2)x2)2 2α2
g(x) := π2xα α(α+2)2 α2logx − − + (α+2 αx)((α 4)x2+2x α 2) .
− (1 x)4 (1 x)3 − − − −
h − − i
We have, for some sequence of positive real numbers β vanishing at infinity,
n
1 4
g < π2α2(2+α)2xαlog( )(1+β ) and D ∽ .
n n xn n n α(2+α)xα2 log( 1 )
n xn
Where ∽ is used here to indicate that the ratio of the two sides goes to 1 when n goes to
infinity. Thus, for some sequence β vanishing at infinity,
n
1 1
Hn < Dn2π2α2(2+α)2xαnlog(xn)(1+βn)log nα2 + π2α2(2+α)2xαlog( 1 )(1+β )
(cid:16) n xn n (cid:17)
16π2 1 q
α
< log(x1n)(1+βn)log(cid:16)n2 + πα(2+α) xαnlog(x1n)(1+βn)(cid:17)
16π2 α 1 αqα 1
< log(x1n)(1+βn)h2 log(xn)+log(cid:16)n2xn2 + πα(2+α) log(x1n)(1+βn)(cid:17)i
q
To conclude, it is sufficient to take the limit as n goes to infinity.
5. Case of the whole space
Theorems 1.3 and 1.4 were stated in the unit ball. If the function u is supported in a
B , a simple scaling argument gives
R
1
u 2 ∆u 2 log e3+C RαN (u) log(2e+RαN (u)) .
k kL∞(BR) ≤ 8π2αk kL2(BR) 0 α α
Similarly, a simple scaling argument in Thheorem 1.4 yieldsp i
10 M.Majdoub,T.Saanouni,.../Globalwell-posedness ofa4D...
Corollary 5.1. (Log estimate) Let α (0,1). For any λ > 1 there exists C > 0
∈ 8π2α λ
such that for any R > 0 and any radial function u (H2 C˙α)(B ), we have
∈ 0 ∩ R
(5.11) kuk2L∞ ≤λk∆uk2L2log Cλ+RαNα(u) .
(cid:16) (cid:17)
Now, in the whole space we have the following result.
Corollary 5.2. (Global Log estimate) Let α (0,1). For any λ > 1 and any
∈ 8π2α
µ (0,1], there exists C > 0 such that for any radial function u (H2 Cα)(R4), we
λ
∈ ∈ ∩
have
(5.12) kuk2L∞ ≤ λkuk2µlog Cλ+ 8αµ−uαkukCα ,
µ
(cid:16) k k (cid:17)
where u 2 := (1+3µ) ∆u 2 +3µ u 2 .
k kµ k kL2 k kH1
Proof. Let α (0,1), λ > 1 , µ (0,1] and a radial function u (H2 Cα)(R4). Fix
∈ 8π2α ∈ ∈ ∩
a radially symmetric function φ C (B ) such that 0 φ 1, φ = 0 near zero and
φ 1, ∆φ 1. Let φ := φ(∈µ.) a0∞nd u4 := φ u. Assu≤me (≤without loss of generality)
|∇ | ≤ | | ≤ µ 2 µ µ
that u L∞ = u(0). Then,
k k | |
kuµkL∞ = kukL∞ and kuµkC˙α ≤ kukCα.
Applying Corollary 5.1, we obtain
kuk2L∞ ≤ λk∆uµk2L2log Cλ+ 8αµ∆−uαku2kCα .
(cid:16) k µkL2 (cid:17)
Now,
∆u 2 = ∆φ u 2 + ∆uφ 2 +4 φ u 2
k µkL2 k µ kL2 k µkL2 k∇ µ∇ kL2
+ 2 ∆φ uφ ∆u+4 ∆φ u φ u+4 φ ∆u φ u
µ µ µ µ µ µ
∇ ∇ ∇ ∇
Z Z Z
µ4
u 2 + ∆u 2 +µ2 u 2 .
≤ 16k kL2 k kL2 k∇ kL2
+ 2(I)+4(II)+4(III),
where
µ2
(I) = ∆φ uφ ∆u ( u 2 + ∆u 2 ),
µ µ ≤ 8 k kL2 k kL2
Z
µ3
(II) = ∆φ u φ u ( u 2 + u 2 ),
µ ∇ µ∇ ≤ 16 k kL2 k∇ kL2
Z
µ
(II) = φ ∆u φ u ( ∆u 2 + u 2 ).
µ ∇ µ∇ ≤ 4 k kL2 k∇ kL2
Z
The proof is achieved because x x2log(C + C), C > 0 is increasing.
→ λ x
We also have the following result
Corollary 5.3. Let α (0,1). For any λ > 1 , a constant C > 0 exists such that for
∈ 8π2α λ
any radial function u (H2 Cα)(R4), we have
∈ ∩
u Cα
u L∞ u L2 + ∆u L2 λlog e+Cλ k k .
k k ≤ k k k k s ∆u L2
(cid:16) k k (cid:17)