Doubling property for the Laplacian
and its applications (Course Chengdu 2007)
K.-D. PHUNG
1
The original approach of N. Garofalo and F.H. Lin
For simplicity, we reproduce the proof of N. Garofalo and F.H. Lin in the simplest case of the Laplacian.
©
ª
Let Byo ,R = y ∈ RN +1 , |y − yo | < R , with yo ∈ RN +1 and R > 0. The unit sphere is defined by
S N = ∂B0,1 .
Theorem 4.1 .Let D ⊂ RN +1 , N ≥ 1, be a connected bounded open set such that B0,1 ⊆ D.
2
If v = v (y) ∈ H (D) is a solution of ∆y v = 0 in D, then for any 0 < M ≤ 1/2, we have
R
Ã
!
2
Z
|∇v (y)| dy Z
B0,1
2
2
N +1
|v (y)| dy ≤ 2
exp (ln 4) R
|v (y)| dy .
2
|v
(y)|
dσ
B0,2M
B
0,M
∂B0,1
Proof .- We will use the following three formulas. Let R > 0,
Z
Z RZ
f (y) dy =
f (rs) rN drdσ (s) ,
B0,R
d
dR
Z
Z
Z
f (Rs) RN dσ (s) =
f (y) dy =
B0,R
SN
Z
B0,R
(4.1)
SN
0
∂f
(y) dy =
∂yi
f (y) dσ (y) ,
(4.2)
∂B0,R
Z
SN
f (Rs) si RN dσ (s) .
(4.3)
The identity (4.1) is the formula of change of variable in spherical coordinate. (4.2) comes from (4.1)
when f is a continuous function. The identity (4.3), available when ∇f is integrable, traduces the
Green formula. Indeed,
R
R
∂f
(y) dy = ∂B0,R f (y) νi (y) dσ (y)
B0,R ∂yi
R
x
.
= S N f (Rs) si RN dσ (s) because on ∂B0,R , ν (x) = R
Now, we will introduce the following quantities. Let r > 0,
R
2
H (r) = ∂B0,r |v (y)| dσ (y) ,
R
2
D (r) =
|∇v (y)| dy ,
B0,r
1
(4.4)
and
rD (r)
.
(4.5)
H (r)
The goal is to show that N is a non-decreasing function for 0 < r ≤ 1. To this end, we will compute
the derivatives of H and D.
R
2
The computation of the derivative of H (r) = S N |v (rs)| rN dσ (s) gives
R
R
2
H0 (r) = N rN −1 S N |v (rs)| dσ (s) + S N 2v (rs) (∇v)|rs · srN dσ (s)
¡
¢
R
= N 1r H (r) + S N ∇v 2 |rs · srN dσ (s)
¡ 2¢ N
RrR
= N 1r H (r) + R0 S N div
¡
¢∇v r drdσ (s)
= N 1r H (r) + B0,r ∆ v 2 dy ,
N (r) =
¡ ¢
2
but ∆ v 2 = 2v∆v + 2 |∇v| . Since ∆y v = 0, one finds
R
2
|∇v (y)| dy
H0 (r)
1
B
= N + 2 0,r
.
H (r)
r
H (r)
Next, one remarks that
Z
Z
2
|∇v (y)| dy =
v (y) ∇v (y) ·
B0,r
indeed,
R
B0,r
∂yi v∂yi vdy
∂B0,r
(4.6)
y
dσ (y) ,
|y|
R
R
= B0,r ∂yi [v∂yi v] dy − B0,r v∂y2i vdy
R
yi
= ∂B0,r [v∂yi v] |y|
dσ (y) because ∆y v = 0 and on ∂B0,R , ν (y) =
(4.7)
y
|y|
.
Consequently, (4.6) becomes from (4.7)
H0 (r)
1
=N +2
H (r)
r
On another hand, the derivative of D (r) =
R
∂B0,r
RrR
0
SN
y
|y| dσ (y)
v (y) ∇v (y) ·
.
H (r)
¯
¯2
¯
¯
¯(∇v)|ρs ¯ ρN dρdσ (s) is
(4.8)
¯
¯2
¯
¯
¯(∇v)|rs ¯ rN dσ (s)
¯
¯2
R ¯
¯
= 1r S N ¯(∇v)|rs ¯ rs · srN dσ (s)
µ¯
¯2 ¶
RrR
¯
¯
= 1r 0 S N div ¯(∇v)|rs ¯ rs rN drdσ (s)
³
´
R
2
= 1r B0,r div |∇v| y dy ,
D0 (r) =
R
SN
³
´
³
´
³
´
R
2
2
2
2
but div |∇v| y = |∇v| divy+∇ |∇v| ·y, with divy = N +1. It remains to compute B0,r ∇ |∇v| ·
ydy, when v = v (y) ∈ H 2 (D). One have
³¡
¢2 ´
R
∂yi ∂yj v
· yi dy
B
0,r
R
2
= 2 B0,r ∂yj v∂yi yj vyi dy
£
¤
R
R
R
= 2 B0,r ∂yj ∂yj v∂yi vyi dy − 2 B0,r ∂y2j v∂yi vyi dy − 2 B0,r ∂yj v∂yi v∂yj yi dy
´
³
´
³
R
R
y
2
yi
∂yi v |y|
dσ (y) − 2 B0,r |∂yi v| dy
= 2r ∂B0,r ∂yj v |y|j
y
y
because ∆y v = 0 and on ∂B0,R , ν (y) = |y| = R .
Consequently, D0 (r) =
N −1
r D (r)
+2
¯
¯
(y) ·
∂B0,r ¯∇v
R
D (r)
1
= (N − 1) + 2
D (r)
r
dσ (y), that is
¯
¯
∇v (y) ·
∂B0,r ¯
R
0
¯2
y ¯
|y| ¯
D (r)
2
¯2
y ¯
|y| ¯
dσ (y)
.
(4.9)
Finally, the computation of the derivative of N (r) = rD(r)
H(r) gives
·
¸
1 D0 (r) H0 (r)
N 0 (r) = N (r)
+
−
,
r
D (r)
H (r)
(4.10)
and one conclude from (4.4), (4.8), (4.9) and (4.10) that
 µR
¯
¯2 ¶ ³ R
´ ³R
¯
2
y ¯
∇v
·
dy
|v|
dy
− ∂B0,r v∇v ·
¯
¯
|y|
∂B0,r
∂B0,r

N 0 (r) = N (r) 
2

D (r) H (r)
y
|y| dy
´2 

 .

Thanks to the Cauchy-Schwarz inequality, one deduce that N 0 (r) ≥ 0 i.e., N is non-decreasing on
D(r)
]0, 1]. Therefore, for r ≤ 1, N (r) ≤ N (1) that is H(r)
≤ 1r N (1). Thus, from (4.6) and (4.5), one have
1
1
H0 (r)
− N ≤ 2N (1) .
H (r)
r
r
Consequently, ∀r ≤ 1,
d
dr
µ µ
¶¶
µ
¶
H (r)
d
1
ln
≤
2N
(1)
ln
.
rN
dr
r
(4.11)
By integrating (4.11) between R > 0 and 2R ≥ 1, one finds
¶
µ
H (2R) 1
≤ 2N (1) (ln 2) ,
ln
H (R) 2N
that is ∀0 < R ≤ 1/2,
Z
Z
2
N
2
|v (2Rs)| (2R) dσ (s) ≤ 2N eN (1) ln 4
SN
|v (Rs)| RN dσ (s) .
SN
One conclude that for any M ≤ 1/2,
R
R 2M R
2
2 N
|v (y)| dy = 0
drdσ (s)
N |v (rs)| r
B0,2M
R M RS
2
N
= 2 0 S N |v (2Rs)| (2R) dRdσ (s)
R
R
M
2
≤ 2N +1 eN (1) ln 4 0 S N |v (Rs)| RN dRdσ (s)
R
2
≤ 2N +1 eN (1) ln 4 B0,M |v (y)| dy .
Comment .- The above computations can be generalized to an elliptic operator of second order (see
[ GaL], [ Ku2]).
2
The improved approach of I. Kukavica
It seems more natural to consider the monotonicity properties of the frequency function defined by
³
´
R
R
2
2
2
2
|∇v
(y)|
r
−
|y|
dy
r B0,r |∇v (y)| dy
B0,r
instead of R
.
R
2
2
|v (y)| dy
|v (y)| dσ (y)
B0,r
∂B0,r
Following the ideas of Kukavica ([ Ku], [ KN], see also [ AE]), one obtains the following three lemmas.
3
2.1
Monotonicity formula
We present the following lemmas.
Lemma A .Let D ⊂ RN +1 , N ≥ 1, be a connected bounded open set such that Byo ,Ro ⊂ D
with yo ∈ D and Ro > 0. If v = v (y) ∈ H 2 (D) is a solution of ∆y v = 0 in D, then
³
´
R
2
2
|∇v (y)| r2 − |y − yo | dy
Byo ,r
Φ (r) =
is non-decreasing on 0 < r < Ro ,
R
2
|v (y)| dy
By ,r
o
and
d
ln
dr
Z
2
|v (y)| dy =
Byo ,r
1
(N + 1 + Φ (r)) .
r
Lemma B .Let D ⊂ RN +1 , N ≥ 1, be a connected bounded open set such that Byo ,Ro ⊂ D
with yo ∈ D and Ro > 0. Let r1 , r2 , r3 be three real numbers such that 0 < r1 < r2 < r3 < Ro . If
v = v (y) ∈ H 2 (D) is a solution of ∆y v = 0 in D, then
ÃZ
Z
!α ÃZ
2
2
|v (y)| dy ≤
|v (y)| dy
Byo ,r2
1
r
ln r2
,
Byo ,r3
¶−1
1
r
ln r2
1
|v (y)| dy
Byo ,r1
µ
where α =
!1−α
2
+
1
1
r
ln r3
∈ ]0, 1[.
2
The above two results are still available when we are closed to a part Γ of the boundary ∂Ω under the
homogeneous Dirichlet boundary condition on Γ.
Lemma C .Let D ⊂ RN +1 , N ≥ 1, be a connected bounded open set with boundary ∂D. Let
Γ be a non-empty Lipschitz open subset of ∂D. Let ro , r1 , r2 , r3 , Ro be five real numbers such that
0 < r1 < ro < r2 < r3 < Ro . Suppose that yo ∈ D satisfies the following three conditions:
i). Byo ,r ∩ D is star-shaped with respect to yo ∀r ∈ ]0, Ro [ ,
ii). Byo ,r ⊂ D ∀r ∈ ]0, ro [ ,
iii). Byo ,r ∩ ∂D ⊂ Γ ∀r ∈ [ro , Ro [ .
If v = v (y) ∈ H 2 (D) is a solution of ∆y v = 0 in D and v = 0 on Γ, then
ÃZ
Z
!α ÃZ
2
2
|v (y)| dy ≤
Byo ,r2 ∩D
where α =
1
2.2
|v (y)| dy
|v (y)| dy
Byo ,r1
µ
1
r
ln r2
Byo ,r3 ∩D
¶−1
1
r
ln r2
1
+
1
r
ln r3
∈ ]0, 1[.
2
Proof of Lemma B
Let
Z
2
H (r) =
|v (y)| dy .
Byo ,r
By applying Lemma A, we know that
d
1
lnH (r) = (N + 1 + Φ (r)) .
dr
r
4
!1−α
2
,
Next, from the monotonicity property of Φ, one deduces that the following two inequalities
³
´
Rr
2)
ln H(r
= r12 N +1+Φ(r)
dr
H(r1 )
r
≤ (N + 1 + Φ (r2 )) ln rr21 ,
³
´
Rr
3)
ln H(r
= r23 N +1+Φ(r)
dr
H(r2 )
r
≥ (N + 1 + Φ (r2 )) ln rr32 .
Consequently,
³
H(r2 )
H(r1 )
ln rr21
ln
´
³
≤ (N + 1) + Φ (r2 ) ≤
ln
H(r3 )
H(r2 )
ln rr32
´
,
and therefore the desired estimate holds
α
H (r2 ) ≤ (H (r1 )) (H (r3 ))
µ
where α =
1
r
ln r2
2.3
,
¶−1
1
r
ln r2
1
1−α
+
1
1
r
ln r3
.
2
Proof of Lemma A
We introduce the following two functions H and D for 0 < r < Ro :
R
2
H (r) = By ,r |v (y)| dy ,
³
´
R o
2
2
D (r) =
|∇v (y)| r2 − |y − yo | dy .
By ,r
o
First, the derivative of H (r) =
recall the Green formula
Z
Z
2
|v| ∂ν Gdσ (y) −
∂Byo ,r
RrR
0
2
SN
|v (ρs + yo )| ρN dρdσ (s) is H 0 (r) =
Z
³
´
2
∂ν |v| Gdσ (y) =
∂Byo ,r
R
2
∂Byo ,r
|v (y)| dσ (y). Next,
Z
³
´
2
∆ |v| Gdy .
2
|v| ∆Gdy −
Byo ,r
Byo ,r
2
We apply it with G (y) = r2 − |y − yo | where G|∂Byo ,r = 0, ∂ν G|∂Byo ,r = −2r, and ∆G = −2 (N + 1).
It gives
³
´³
´
R
R
2
2
2
1
2
H 0 (r) = 1r By ,r (N + 1) |v| dy + 2r
∆
|v|
r
−
|y
−
y
|
dy
o
By³o ,r
o
´
R
2
1
N +1
2
= r H (r) + r By ,r div (v∇v) r − |y − yo | dy
´³
´
R o ³
2
2
= N r+1 H (r) + 1r By ,r |∇v| + v∆v r2 − |y − yo | dy .
o
Consequently, when ∆y v = 0,
H 0 (r) =
that is
H 0 (r)
H(r)
=
N +1
r
+
1 D(r)
r H(r)
µ
d
dr
= 2r
= 2r
r2
RrR
0
R
(A.1)
the second equality in Lemma A.
Now, we compute the derivative of D (r) =
D0 (r) =
1
N +1
H (r) + D (r) ,
r
r
RrR
0
SN
¯
¯2 ¡
¢
¯
¯
¯(∇v)|ρs+yo ¯ r2 − ρ2 ρN dρdσ (s):
¶
¯
¯2
¯
¯2
R
¯
¯
¯
¯
¯(∇v)|ρs+yo ¯ ρN dρdσ (s) − S N r2 ¯(∇v)|rs+yo ¯ rN dσ (s)
¯
¯2
¯
¯
¯(∇v)|ρs+yo ¯ ρN dρdσ (s)
RrR
0
SN
Byo ,r
SN
2
|∇v| dy .
5
(A.2)
Here, we remark that
R
2r By
2
o ,r
|∇v| dy
=
R
2
N +1
4
) · ∇v| dy
r D (r) + r Byo ,r |(y − y
³o
´
R
2
− 1r By ,r ∇v · (y − yo ) ∆v r2 − |y − yo | dy
o
(A.3)
,
indeed,
´
³
R
2
2
(N + 1) By ,r |∇v| r2 − |y − yo | dy
³o
³
´
´
³
³
´´
R
R
2
2
2
2
= By ,r div |∇v| r2 − |y − yo | (y − yo ) dy − By ,r ∇ |∇v| r2 − |y − yo |
· (y − yo ) dy
o
³
³
´´
Ro
2
2
= − By ,r ∂yi |∇v| r2 − |y − yo |
(yi − yoi ) dy because on ∂Byo ,r , r = |y − yo |
³
´
R o
R
2
2
= − By ,r 2∇v∂yi ∇v r2 − |y − yo | (yi − yoi ) dy − By ,r |∇v| (−2 (yi − yoi )) (yi − yoi ) dy
o
o
³
´
R
R
2
2
2
= − By ,r 2∇v∂yi ∇v r2 − |y − yo | (yi − yoi ) dy + 2 By ,r |∇v| |y − yo | dy ,
o
o
´
2
2
2
∂
v∂
v
r
−
|y
−
y
|
(yi − yoi ) dy
y
o
y
y
j
By ,r
³ i j
³
´´
R o
2
2
= − By ,r ∂yj (yi − yoi ) ∂yj v∂yi v r − |y − yo |
dy
³
´
R o
2
+ By ,r ∂yj (yi − yoi ) ∂yj v∂yi v r2 − |y − yo | dy
³
´
R o
2
+ By ,r (yi − yoi ) ∂y2j v∂yi v r2 − |y − yo | dy
³
´
R o
2
+ By ,r (yi − yoi ) ∂yj v∂yi v∂yj r2 − |y − yo | dy
o
= 0 because on³∂Byo ,r , r = |y´− yo |
R
2
2
+ By ,r |∇v| r2 − |y − yo | dy
o
³
´
R
2
+ By ,r (y − yo ) · ∇v∆v r2 − |y − yo | dy
R o
2
−2 By ,r |(y − yo ) · ∇v| dy .
and
−
³
R
o
Therefore,
(N + 1)
R
Byo ,r
|∇v|
2
³
´
2
r2 − |y − yo | dy
= 2r2
+2
R
R
Byo
2
|∇v| dy − 4
R
2
|(y − yo ) · ∇v| dy
´
2
2
(y
−
y
)
·
∇v∆v
r
−
|y
−
y
|
dy ,
o
o
,r
Byo ,r
By³
o ,r
and this is the desired estimate (A.3).
Consequently, from (A.2) and (A.3), we obtain, when ∆y v = 0, the following formula
Z
N +1
4
2
D0 (r) =
D (r) +
|(y − yo ) · ∇v| dy .
r
r Byo ,r
The computation of the derivative of Φ (r) =
Φ0 (r) =
D(r)
H(r)
(A.4)
gives
1
[D0 (r) H (r) − D (r) H 0 (r)] ,
(r)
H2
which implies using (A.1) and (A.4) that
i
h
£
¤
R
2
H 2 (r) Φ0 (r) = N r+1 D (r) + 4r By ,r |(y − yo ) · ∇v| dy H (r) − N r+1 H (r) + 1r D (r) D (r)
o
³ R
´
2
= 1r 4 By ,r |(y − yo ) · ∇v| dyH (r) − D2 (r) ≥ 0 ,
o
indeed, thanks to an integration by parts and using Cauchy-Schwarz inequality, we have
³R
´2
D2 (r) = 4 By ,r v∇v · (y − yo ) dy
³R o
´ ³R
´
2
2
≤ 4 By ,r |(y − yo ) · ∇v| dy
|v|
dy
³R o
´ Byo ,r
2
≤ 4 By ,r |(y − yo ) · ∇v| dy H (r) .
o
6
Therefore, we have proved the desired monotonicity for Φ and this completes the proof of Lemma A.
2.4
Proof of Lemma C
Under the assumption Byo ,r ∩ ∂D ⊂ Γ for any r ∈ [ro , Ro ), we extend v by zero in Byo ,Ro \D and
denote by v its extension. Since v = 0 on Γ, we have

in Byo ,Ro ,
 v = v1D
v=0
on Byo ,Ro ∩ ∂D ,

∇v = ∇v1D in Byo ,Ro .
Now, we denote Ωr = Byo ,r ∩ D, when 0 < r < Ro . In particular, Ωr = Byo ,r , when 0 < r < ro . We
introduce the following three functions:
R
2
H (r) = Ωr |v (y)| dy³,
´
R
2
2
D (r) =
|∇v (y)| r2 − |y − yo | dy ,
Ωr
and
Φ (r) =
D (r)
≥0.
H (r)
Our goal is to show that Φ is a non-decreasing function. Indeed, we will prove that the following
equality holds
d
d
1
lnH (r) = (N + 1) lnr + Φ (r) .
(C.1)
dr
dr
r
Therefore, from the monotonicity of Φ, we will deduce that (in a similar way than in the proof of
Lemma A) that
³
´
³
´
2)
3)
ln H(r
ln H(r
H(r1 )
H(r2 )
≤ (N + 1) + Φ (r2 ) ≤
,
ln rr21
ln rr32
and this will imply the desired estimate
ÃZ
Z
!α ÃZ
2
|v (y)| dy ≤
Ωr2
1
r
ln r2
1
2
|v (y)| dy
|v (y)| dy
Byo ,r1
µ
where α =
!1−α
2
,
Ωr3
¶−1
1
r
ln r2
1
+
1
r
ln r3
.
2
First, we compute the derivative of H (r) =
H 0 (r) =
=
=
=
=
R
Byo ,r
2
|v (y)| dy =
RrR
0
SN
2
|v (ρs + yo )| ρN dρdσ (s).
R
2
|v (rs + yo )| rN dσ (s)
SRN
2
1
+ yo )| rs · srN dσ
r S N |v (rs ³
´ (s)
R
2
1
|v (y)| (y − yo ) dy
r Byo ,r div
³
´
R
2
2
1
r Byo ,r (N + 1) |v (y)| + ∇ |v (y)| · (y − yo ) dy
R
N +1
2
r H (r) + r Ωr v (y) ∇v (y) · (y − yo ) dy .
Next, when ∆y v = 0 in D and v|Γ = 0, we remark that
Z
D (r) = 2
v (y) ∇v (y) · (y − yo ) dy ,
Ωr
7
(C.2)
(C.3)
indeed,
³
´
2
2
|∇v| r2 − |y − yo | dy
³
³
´´
³
³
´´
R
R
2
2
= Ωr div v∇v r2 − |y − yo |
dy − Ωr vdiv ∇v r2 − |y − yo |
dy
³
´
³
´
R
R
2
2
= − Ωr v∆v r2 − |y − yo | dy − Ωr v∇v · ∇ r2 − |y − yo | dy
because
on ∂Byo ,r , r = |y − yo | and v|Γ = 0
R
= 2 Ωr v∇v · (y − yo ) dy because ∆y v = 0 in D .
R
Ωr
Consequently, from (C.2) and (C.3), we obtain
H 0 (r) =
1
N +1
H (r) + D (r) ,
r
r
(C.4)
and this is (C.1).
On another hand, the derivative of D (r) =
D0 (r) = 2r
= 2r
RrR
0
RrR
R0
SN
SN
¯
¯2 ¡
¢
¯
¯
¯(∇v)|ρs+yo ¯ r2 − ρ2 ρN dρdσ (s) is
¯
¯2
¯
¯
¯(∇v)|ρs+yo ¯ ρN dρdσ (s)
(C.5)
2
|∇v (y)| dy .
Ωr
Here, when ∆y v = 0 in D and v|Γ = 0, we will remark that
2r
R
Ωr
2
|∇v (y)| dy
=
R
2
4
N +1
|(y − yo ) · ∇v (y)| dy
r D (r) + r Byo³
,r
´
R
2
2
+ 1r Γ∩By ,r |∂ν v| r2 − |y − yo | (y − yo )
o
· νdσ (y) ,
(C.6)
indeed,
³
´
R
2
2
(N + 1) Ωr |∇v| r2 − |y − yo | dy
³
³
´
´
³
³
´´
R
R
2
2
2
2
= Ωr div |∇v| r2 − |y − yo | (y − yo ) dy − Ωr ∇ |∇v| r2 − |y − yo |
· (y − yo ) dy
³
´
³
³
´´
R
R
2
2
2
2
= Γ∩By ,r |∇v| r2 − |y − yo | (y − yo ) · νdσ (y) − Ωr ∂yi |∇v| r2 − |y − yo |
(yi − yoi ) dy
o
³
´
R
2
2
2
= Γ∩By ,r |∇v| r − |y − yo | (y − yo ) · νdσ (y)
³
´
R o
R
2
2
2
− Ωr 2∇v∂yi ∇v r2 − |y − yo | (yi − yoi ) dy + 2 Ωr |∇v| |y − yo | dy ,
and
³
´
2
∂yj v∂y2i yj v r2 − |y − yo | (yi − yoi ) dy
³
³
´´
R
2
= − Ωr ∂yj (yi − yoi ) ∂yj v∂yi v r2 − |y − yo |
dy
³
´
R
2
2
+ Ωr ∂yj (yi − yoi ) ∂yj v∂yi v r − |y − yo | dy
³
´
R
2
+ Ωr (yi − yoi ) ∂y2j v∂yi v r2 − |y − yo | dy
³
´
R
2
+ Ωr (yi − yoi ) ∂yj v∂yi v∂yj r2 − |y − yo | dy
³
³
´´
R
2
= − Γ∩By ,r νj (yi − yoi ) ∂yj v∂yi v r2 − |y − yo |
dσ (y)
o
³
´
R
2
2
2
+ Ωr |∇v| r − |y − yo | dy
+0 because ∆y v = 0 in D
R
2
− Ωr 2 |(y − yo ) · ∇v| dy .
−
R
Ωr
Therefore, when ∆y v = 0 in D, we have
³
´
³
´
R
R
2
2
2
2
(N + 1) Ωr |∇v| r2 − |y − yo | dy = Γ∩By ,r |∇v| r2 − |y − yo | (y − yo ) · νdσ (y)
³
´
R o
2
−2 Γ∩By ,r ∂yj vνj ((yi − yoi ) ∂yi v) r2 − |y − yo | dσ (y)
R o
R
2
2
+2r2 Ωr |∇u| dy − 4 Ωr |(y − yo ) · ∇v| dy .
8
By using the fact that v|Γ = 0, we get ∇v = (∇v · ν) ν on Γ and we deduce that
³
´
³
´
R
R
2
2
2
2
(N + 1) Ωr |∇v| r2 − |y − yo | dy = − Γ∩By ,r |∂ν v| r2 − |y − yo | (y − yo ) · νdσ (y)
R o
R
2
2
+2r2 Ωr |∇v| dy − 4 Ωr |(y − yo ) · ∇v| dy ,
and this is (C.6).
Consequently, from (C.5) and (C.6), when ∆y v = 0 in D and v|Γ = 0, we have
Z
Z
³
´
4
N +1
1
2
2
2
0
D (r)+
D (r) =
|(y − yo ) · ∇v (y)| dy+
|∂ν v| r2 − |y − yo | (y − yo )·νdσ (y) .
r
r Ωr
r Γ∩Byo ,r
(C.7)
The computation of the derivative of Φ (r) =
Φ0 (r) =
D(r)
H(r)
gives
1
[D0 (r) H (r) − D (r) H 0 (r)] ,
H 2 (r)
which implies from (C.4) and (C.7), that
´
³ R
2
H 2 (r) Φ0 (r) = 1r 4 Ωr |(y − yo ) · ∇v (y)| dy H (r) − D2 (r)
³
´
R
2
2
+ 1r Γ∩By ,r |∂ν v| r2 − |y − yo | (y − yo ) · νdσ (y) H (r) .
o
Thanks to (C.3) and Cauchy-Schwarz inequality, we obtain that
Z
2
0≤4
|(y − yo ) · ∇v (y)| dy H (r) − D2 (r) .
Ωr
the inequality 0 ≤ (y − yo ) · ν on Γ hols when Byo ,r ∩ D is star-shaped with respect to yo for any
r ∈ ]0, Ro [. Therefore, we get the desired monotonicity for Φ which completes the proof of Lemma C.
3
Quantitative unique continuation property for the Laplacian
Let D ⊂ RN +1 , N ≥ 1, be a connected bounded open set with boundary ∂D. Let Γ be a non-empty
Lipschitz open part of ∂D. We consider the Laplacian in D, with a homogeneous Dirichlet boundary
condition on Γ ⊂ ∂Ω:

in D ,
 ∆y v = 0
v=0
on Γ ,
(4.12)

v = v (y) ∈ H 2 (D) .
The goal of this section is to describe interpolation inequalities associated to solutions v of (4.12).
Theorem 4.2 .Let ω be a non-empty open subset of D. Then, for any D1 ⊂ D such that
∂D1 ∩ ∂D b Γ and D1 \(Γ ∩ ∂D1 ) ⊂ D, there exist C > 0 and µ ∈ ]0, 1[ such that for any v solution
of (4.12), we have
µZ
Z
2
2
|v (y)| dy ≤ C
D1
¶µ µZ
|v (y)| dy
ω
D
Or in a equivalent way,
9
2
|v (y)| dy
¶1−µ
.
Theorem 4.3 .Let ω be a non-empty open subset of D. Then, for any D1 ⊂ D such that
∂D1 ∩ ∂D b Γ and D1 \(Γ ∩ ∂D1 ) ⊂ D, there exist C > 0 and µ ∈ ]0, 1[ such that for any v solution
of (4.12), we have
Z
µ ¶ 1−µ
Z
Z
1 µ
2
2
|v (y)| dy ≤ C
|v (y)| dy + ε
|v (y)| dy
ε
D1
ω
D
2
∀ε > 0 .
Proof of Theorem 4.2 .- We divide the proof into two steps.
Step 1 .- We apply Lemma B, and use a standard argument (voir e.g., [ Ro]) which consists to
construct a sequence of balls chained along a curve. More precisely, we claim that for any non-empty
compact sets in D, K1 and K2 , such that meas(K1 ) > 0, there exists µ ∈ ]0, 1[ such that for any
v = v (y) ∈ H 2 (D), solution of ∆y v = 0 in D, we have
µZ
Z
2
2
|v (y)| dy ≤
¶µ µZ
2
|v (y)| dy
K2
¶1−µ
.
|v (y)| dy
K1
(4.13)
D
©
ª
Indeed, let δ > 0 and qj ∈ RN for j = 0, 1, · · · , m, one can construct a sequence of balls Bqj ,δ j=0,..,m ,
such that the following inclusion hold

K1 ⊃ Bq0 ,δ



K2 ⊂ Bqm ,δo for some δo > 0
Bqj+1 ,δ ⊂ Bqj ,2δ ∀j = 0, .., m − 1



Bqj ,3δ ⊂ D ∀j = 0, .., m .
Then, thanks to Lemma B, there exist α, α1 , µ ∈ ]0, 1[, such that
R
R
2
2
|v (y)| dy ≤ Bq ,δ |v (y)| dy
K2
m
o
´1−α
´α ³R
³R
2
2
|v (y)| dy
≤ Bq ,δ |v (y)| dy
B
q
,3δ
´1−α
´α ³Rm
³R m
2
2
|v (y)| dy
|v
(y)|
dy
≤ Bq
D
,2δ
µ³ m−1
´1−α
´1−α1 ¶α ³R
´α1 ³R
R
2
2
2
|v
(y)|
dy
|v
(y)|
dy
|v
(y)|
dy
≤
D
D
Bq
,δ
m−1
≤···
³R
≤ Bq
0
2
|v (y)| dy
,δ
´µ ³R
2
D
|v (y)| dy
´1−µ
,
which implies the desired inequality (4.13).
Step 2 .- We apply Lemma C, and choose yo in a neighborhood of the part Γ such that the conditions
i, ii, iii, hold. Next, by an adequate partition of D, we deduce from (4.13) that for any D1 ⊂ D such
that ∂D1 ∩ ∂D b Γ and D1 \(Γ ∩ ∂D1 ) ⊂ D, there exist C > 0 and µ ∈ ]0, 1[ such that for any
v = v (y) ∈ H 2 (D) such that ∆y v = 0 on D and v = 0 on Γ, we have
µZ
Z
2
2
|v (y)| dy ≤ C
D1
¶µ µZ
|v (y)| dy
ω
2
¶1−µ
|v (y)| dy
.
D
This completes the proof.
Remark .- From standard minimization technique, the above inequality implies
Z
µ ¶ 1−µ
Z
Z
1 µ
2
2
|v (y)| dy ≤ C
|v (y)| dy + ε
|v (y)| dy
ε
D1
ω
D
2
1
µ
10
∀ε > 0 .
R
R
R
2
2
2
Indeed, we denote A = D1 |v (y)| dy 6= 0, B = ω |v (y)| dy and E = D |v (y)| dy. We know that
µ 1−µ
there exist C > 0 and µ ∈ ]0, 1[ such that A ≤ CB E
. Therefore,
µ
1
µ
A≤C B
E
A
¶ 1−µ
µ
.
¡ 1 ¢ 1−µ
1
1
1
µ
µ
Now, if E
. And, if E
A ≤ ε , then A ≤ C B ε
A > ε , then A ≤ εE. Consequently, one obtain the
desired interpolation inequality. Conversely, suppose that there exist C > 0 and µ ∈ ]0, 1[ such that
Z
µ ¶ 1−µ
Z
Z
1 µ
2
2
|v (y)| dy ≤ C
|v (y)| dy + ε
|v (y)| dy
ε
D1
ω
D
2
then, we choose ε =
1A
2E
1
µ
∀ε > 0 ,
in order to get A ≤ 2CB µ E 1−µ .
Comment .- The above computations can be generalized to solutions of the following elliptic system

∆y u = f
in D ,


 u=0
on Γ ,
u = u (y) ∈ H 2 (D) ,



−1
(−∆y ) f ∈ H 2 ∩ H01 (D) ,
in order to get the following estimate
Z
D1
4
µ°
°2
°
2
−1 °
|u (y)| dy ≤ C °(−∆y ) f °
2
Z
L (D)
2
+
¶µ µZ
2
|u (y)| dy
¶1−µ
|u (y)| dy
ω
.
D
Quantitative unique continuation property for the elliptic
operator ∂t2 + ∆
In this section, we present the following result (to be compared to [ LeR]).
Theorem 4.4 .Let Ω be a Lipschitz connected bounded open set of RN , N ≥ 1. We choose
T > 0 and δ ∈ ]0, T /2[. We consider the elliptic operator of second order in Ω × ]0, T [ with a
homogeneous Dirichlet boundary condition on ∂Ω × (0, T ),
 2
on Ω × ]0, T [ ,
 ∂t u + ∆u = 0
u=0
in ∂Ω × ]0, T [ ,
(4.14)

u = u (x, t) ∈ H 2 (Ω × ]0, T [) .
Then, for any ϕ ∈ C0∞ (Ω × (0, T )), ϕ 6= 0, there exist C > 0 and µ ∈ ]0, 1[ such that for any u
solution of ( 4.14), we have
Z
T −δ
ÃZ
Z
T
2
Ω
T
2
|u (x, t)| dxdt ≤ C
δ
!1−µ ÃZ
Z
2
|u (x, t)| dxdt
0
Ω
!µ
Z
|ϕu (x, t)| dxdt
0
.
Ω
Proof .- We apply Theorem 4.2 with D = Ω × ]0, T [, Ω × ]δ, T − δ[ ⊂ D1 , y = (x, t), ∆y = ∂t2 + ∆.
11
5
Quantitative unique continuation property for the sum of
eigenfunctions
The goal of this section is to obtain the following results (to be compared to [ LZ] or [ JL]).
Theorem 4.5 .Let Ω be a bounded open set in RN , N ≥ 1, either convex or C 2 and connected.
Let ω be a non-empty open subset in Ω. Then, there exists C > 0 such that for any sequence {aj }j≥1
of real numbers and any integer M > 1, we have
¯
¯2
¯
Z ¯ X
√
X
¯
¯
¯
¯ dx ,
a2j ≤ CeC λM
a
e
(x)
j j
¯
¯
ω ¯1≤j≤M
¯
1≤j≤M
where {λj }j≥1 and {ej }j≥1 are the eigenvalues and eigenfunctions of −∆ in H01 (Ω), constituting an
orthonormal basis in L2 (Ω).
Or in an equivalent way,
Theorem 4.6 .Let Ω be a bounded open set in RN , N ≥ 1, either convex or C 2 and connected.
Let ω be a non-empty open subset in Ω. Then, there exists C > 0 such that for any sequence {aj }j≥1
of real numbers and any R > λ1 , we have
¯
¯2
¯
Z ¯ X
√
X
¯
¯
C R
2
¯
aj ≤ Ce
aj ej (x)¯¯ dx ,
¯
ω ¯{j;λ ≤R}
¯
{j;λj ≤R}
j
where {λj }j≥1 and {ej }j≥1 are the eigenvalues and eigenfunctions of −∆ in H01 (Ω), constituting an
orthonormal basis in L2 (Ω).
Proof of Theorem 4.5 .- We divide the proof into three steps.
Step 1 .- For any aj ∈ R, we introduce the solution
³p ´
X
X
w (x, t) =
aj ej (x) ch
λj t − χ (x)
aj ej (x) ,
j≤M
j≤M
where χ ∈ C0∞ (ω), χ = 1 in ω
e b ω. Recall that cht = (et + e−t ) /2. Therefore, w solves
 2
∂ w + ∆w = ∆f
in Ω × ]0, T [ ,


 t
w=0
on ∂Ω × ]0, T [ ,
w = ∂t w = 0
on ω
e × {0} ,



w = w (x, t) ∈ H 2 (Ω × ]0, T [) ,
P
for any T > 0, where f = −χ
aj ej ∈ H02 (Ω). We denote by w the extension of w by zero in
j≤M
ω
e × ]−T, 0[. Therefore, w solves
 2
 ∂t w + ∆w = ∆f 1Ω×(0,T )
w=0

w=0
in Ω × ]0, T [ ∪ ω
e × ]−T, 0[ ,
in ω
e × ]−T, 0[ ,
on ∂Ω × ]0, T [ .
At present, we define D, a connected open set in RN +1 , satisfying the following six conditions:
i). Ω × ]δ, T − δ[ ⊂ D for some δ ∈ ]0, T /2[ ;
ii). ∂Ω × ]δ, T − δ[ ⊂ ∂D ;
iii). D ⊂ Ω × ]0, T [ ∪ ω
e × ]−T, 0[ ;
12
iv). there exists a non-empty open subset ωo b D ∩ ω
e × ]−To , 0[ for some To ∈ ]0, T [ ;
v). D ∈ C 2 if Ω is C 2 and connected ;
vi). D is convex with an adequate choice of (δ, To ) if Ω is convex .
In particular, w ∈ H 2 (D).
Step 2 .- We claim that there exists g ∈ H 2 (Ω × ]−T, T [) ∩ H01 (Ω × ]−T, T [) ⊂ H 2 (D) such that
½ 2
∂t g + ∆g = ∆f 1Ω×(0,T ) in Ω × ]−T, T [ ,
g=0
on ∂ (Ω × ]−T, T [) ,
and
kgkL2 (D) ≤ kf kL2 (Ω×]0,T [) .
(4.15)
Indeed, we will proceed with six substeps when Ω is C 2 and connected (the case where Ω is convex is
well-known since then Ω × ]−T, T [ is convex). We denote h = ∆f 1Ω×(0,T ) ∈ L2 (Ω × ]−T, T [).
Substep 1: one recall that h ∈ L2 (Ω × ]−T, T [) implies the existence of g ∈ H01 (Ω × ]−T, T [).
Substep 2: thanks to the interior regularity for elliptic systems, for any D0 b Ω × ]−T, T [, g ∈
H 2 (D0 ).
Substep 3: thank to the boundary regularity for elliptic systems, but not closed to the boundary
Ω × {−T, T }, g is also locally in H 2 because Ω is C 2 .
Substep 4: we extend the solution at t = T as follows.
Let h (x, t) = h (x, t) for (x, t) ∈ Ω × ]−T, T [ and h (x, t) = −h (x, 2T − t) for (x, t) ∈ Ω × ]T, 3T [. Thus
h ∈ L2 (Ω × ]−T, 3T [). Let g (x, t) = g (x, t) for (x, t) ∈ Ω × [−T, T [ and g (x, t) = −g (x, 2T − t) for
(x, t) ∈ Ω × [T, 3T ]. Thus, g solves
½ 2
∂t g + ∆g = h in Ω × ]−T, 3T [ ,
g=0
on ∂ (Ω × ]−T, 3T [) .
By applying the boundary regularity as in substep 3, one obtain that g ∈ H 2 (Ω × ]0, 2T [). In particular, g ∈ H 2 (Ω × ]0, T [).
Substep 5: we extend in a similar way at t = −T in order to conclude that g ∈ H 2 (Ω × ]−T, 0[).
−1
Substep 6: finally, we multiply ∂t2 g + ∆g = h by (−∆) g and integrate by parts over Ω × ]−T, T [,
in order to obtain
RT
RT R
2
2
−1
k∂t gkH −1 (Ω) dt + kgkL2 (Ω×]−T,T [) = 0 Ω −∆f (x) (−∆) g (x, t) dxdt
−T
RT R
−1
= 0 Ω f (x) (−∆) (−∆) g (x, t) dxdt because f ∈ H02 (Ω)
≤ kf kL2 (Ω×]0,T [) kgkL2 (Ω×]−T,T [) from Cauchy-Schwarz .
which gives the desired inequality (4.15).
Step 3 .- Finally, we apply Theorem 4.3 with ∆y = ∂t2 + ∆, v = w − g in D with Γ = ∂Ω × ]0, T [
and Ω × ]δ, T − δ[ ⊂ D1 ⊂ D such that ∂D1 ∩ ∂D b Γ and D1 \(Γ ∩ ∂D1 ) ⊂ D in order that
µ ¶ 1−µ
Z
Z
1 µ
2
2
|w − g| dy ≤ C
|w − g| dy + ε
|w − g| dy
ε
D1
ωo
D
Z
2
∀ε > 0 ,
which implies that
Z
2
|w| dy ≤ C
D1
µ ¶ 1−µ
Z
Z
1 µ
2
2
|g| dy + ε
|w| dy
ε
D
D
∀ε ∈ ]0, 1[ ,
where we have used that w = 0 in ωo . From (4.15), we conclude that there exist C > 0 and µ ∈ ]0, 1[
such that
Z
δ
T −δ
µ ¶ 1−µ
Z TZ
Z TZ
1 µ
2
2
|w (x, t)| dxdt ≤ C
|f (x)| dxdt + ε
|w (x, t)| dxdt
ε
Ω
0
Ω
0
Ω
Z
2
13
∀ε > 0 .
On another hand, we have the following inequalities
RT R
0
RT R
0
Z
δ
T −δ
2
Ω
Ω
2
|f (x)| dxdt
|w (x, t)| dxdt
¯
¯2
¯
R T R ¯¯
P
¯
= 0 Ω ¯χ (x)
aj ej (x)¯ dxdt
¯
¯
j≤M
¯
¯2
¯
R ¯¯
P
¯
≤ T ω ¯χ (x)
aj ej (x)¯ dx ,
¯
¯
j≤M
¯
¯2
¯
¡p ¢
R T R ¯¯ P
P
¯
= 0 Ω¯
aj ej (x) ch
λj t − χ (x)
aj ej (x)¯ dxdt
¯j≤M
¯
j≤M
¯2
¯
¯
√
R ¯¯
P
P 2
¯
aj ej (x)¯ dx ,
aj + 2T ω ¯χ (x)
≤ 2T e2 λM T
¯
¯
j≤M
j≤M
¯2
¯
¯2
¯
¯
Z T −δ Z
Z ¯
Z ¯X
³p ´¯¯
X
¯
¯
¯
2
¯
¯
¯ dx .
¯
dxdt
≤
2
|w
(x,
t)|
dxdt
+
2T
χ
(x)
a
e
(x)
a
e
(x)
ch
λ
t
j j
j j
j ¯
¯
¯
¯
δ
Ω
ω¯
Ω ¯j≤M
¯
¯
j≤M
Consequently, from the last four inequalities, we deduce that for any ε > 0,
¯
¯2
¡p ¢¯¯
R T −δ R ¯¯ P
P 2
aj ≤ δ
(T − 2δ)
λj t ¯ dxdt
aj ej (x) ch
Ω ¯¯
¯
j≤M
j≤M
¯
¯2
¯
¯
¡ 1 ¢ 1−µ
R ¯
P
¯
µ
T ω ¯χ (x)
aj ej (x)¯ dx
≤ 2C ε
¯
¯
j≤M

¯
¯2 
¯
¯
√
R
P
P
¯
¯
a2j + T ω ¯χ (x)
aj ej (x)¯ dx
+4ε T e2 λM T
¯
¯
j≤M
j≤M
¯
¯2
¯
R ¯¯
P
¯
+2T ω ¯χ (x)
aj ej (x)¯ dx .
¯
¯
j≤M
Choosing ε =
−2δ)
1 (T√
8 T e2 λM T
, we obtain the existence of C > 0 such that
X
j≤M
6
a2j ≤ Ce
√
C λM
¯
¯2
¯
Z ¯X
¯
¯
¯
aj ej (x)¯¯ dx .
¯
ω ¯j≤M
¯
Application to the wave equation
From the idea of L. Robbiano which consists to use an interpolation inequality of Hölder type for the
elliptic operator ∂t2 + ∆ and the FBI transform introduced by G. Lebeau et L. Robbiano, we obtain
the following estimate of logarithmic type.
Theorem 4.9 .Let Ω be a bounded open set in RN , N ≥ 1, either convex or C 2 and connected.
Let ω be a non-empty open subset in Ω. Then, for any β ∈ ]0, 1[, there exist C > 0 and T > 0 such
that for any solution u of
 2
in Ω × ]0, T [ ,
 ∂t u − ∆u = 0
u=0
on ∂Ω × ]0, T [ ,

(u, ∂t u) (·, 0) = (u0 , u1 ) ,
14
with non-identically zero initial data (u0 , u1 ) ∈ H01 (Ω) × L2 (Ω), we have
k(u0 ,u1 )k
C k(u
1 (Ω)×L2 (Ω)
H0
!1/β
0 ,u1 )kL2 (Ω)×H −1 (Ω)
k(u0 , u1 )kH 1 (Ω)×L2 (Ω) ≤ e
kukL2 (ω×]0,T [) .
0
7
Application to the heat equation
In this section, we propose the following result.
Theorem 4.7 .Let Ω be a bounded open set in RN , N ≥ 1, either convex or C 2 and connected.
Let ω be a non-empty open subset in Ω. Then, for any T > 0, there exists C > 0 such that for any u
solution of

 ∂t u − ∆u = 0 in Ω × ]0, T [ ,
u=0
on ∂Ω × ]0, T [ ,
(2.3)

u (·, 0) = uo ,
with non-identically zero initial data uo ∈ H01 (Ω), and for any to ∈ ]0, T [, we have
0
C@
kuo kL2 (Ω) ≤ Ce
1
to
kuo k2 1
H0 (Ω)
+to
kuo k2 2
L (Ω)
1
to
+to
1
A
ku (·, to )kL2 (ω) .
Comment .- We also have that
C
kuo kH −1 (Ω) ≤ Ce
kuo k2 2
L (Ω)
kuo k2
H −1 (Ω)
!
ku (·, to )kL2 (ω) .
Remark .- The quantitative unique continuation property from ω × {to } for parabolic operator with
space-time coefficients was established by L. Escauriaza, F.J. Fernandez and S. Vessella [ EFV]).
Proof of Theorem 4.7 .- We decompose the proof into two steps. First, in step 1, we will prove that
the solution u of (2.3) satisfies the following estimate
Ã
!
2
kuo kH 1 (Ω)
2
2
0
kuo kL2 (Ω) ≤ exp 2to
ku (·, to )kL2 (Ω) .
2
kuo kL2 (Ω)
Next, in step 2, we will prove that the solution u of (2.3) satisfies the following estimate
Z
Ω
¶1−µ µZ
¶µ
µ
Z
C
2
2
2
|u (x, 0)| dx
|u (x, to )| dx
.
|u (x, to )| dx ≤ C e to
Ω
ω
Finally, the above inequalities imply the existence of C > 0 such that
!
Ã
2
kuo kH 1 (Ω) Z
C/µ
1
1
2
2
0
|u (x, to )| dx .
kuo kL2 (Ω) ≤ C µ e to exp 2to
µ kuo k2L2 (Ω)
ω
Proof of step 1 .- Let us introduce for almost t ∈ [0, T ] such that the solution of (2.3) satisfies
u (·, t) 6= 0, the following quantity.
2
ku (x, t)kH 1 (Ω)
0
Φ (t) =
.
2
ku (x, t)kL2 (Ω)
15
We begin to check that Φ is a non-increasing function on [0, T ]. This monotonicity property holds
d
because for any initial data in a dense set of H01 (Ω), we have that dt
Φ (t) ≥ 0. Indeed, from the
following two equalities
(
2
2
1 d
2 dt ku (x, t)kL2 (Ω) + ku (x, t)kHo1 (Ω) = 0 ,
2
2
1 d
2 dt ku (x, t)kH 1 (Ω) + k∆u (x, t)kL2 (Ω) = 0 ,
0
we can deduce that
h
i
d
2
2
2
4
Φ (t) =
− k∆u (x, t)kL2 (Ω) ku (x, t)kL2 (Ω) + ku (x, t)kH 1 (Ω) .
4
0
dt
ku (x, t)kL2 (Ω)
Therefore, we get by classical density argument and Cauchy-Schwarz inequality that for any solution
u of (2.3), u (·, t) 6= 0 a.e., and any t ∈ [0, T ], Φ (t) ≤ Φ (0). On another hand, we also have that
1 d
2
2
ku (x, t)kL2 (Ω) + Φ (t) ku (x, t)kL2 (Ω) = 0 ,
2 dt
which implies that
1 d
2
2
ku (x, t)kL2 (Ω) + Φ (0) ku (x, t)kL2 (Ω) .
2 dt
Therefore, by Gronwall Lemma, we get the desired estimate
0≤
Proof of step 2 .- Let λ1 ,λ2 ,· · · and e1 ,e2 ,· · · be the eigenvalues and eigenfunctions
of −∆ in
P
αj ej in L2 (Ω) where
H01 (Ω), constituting an orthonormal basis in L2 (Ω). For any uo = u (·, 0) =
j≥1
R
P
αj ej (x) e−λj t . Let to ∈ ]0, T [. We
αj = Ω uo ej dx, the solution u of (2.3), can be written u (x, t) =
j≥1
introduce (see [ Lin] or [ CRV]) the solution
³p ´
X
X
w (x, t) =
αj ej (x) e−λj to ch
λj t − χ (x)
αj ej (x) e−λj to ,
j≥1
j≥1
e b ω. Therefore, w solves
where χ ∈ C0∞ (ω), χ = 1 in ω
 2
∂ w + ∆w = ∆f
in Ω × ]0, T [ ,


 t
w=0
on ∂Ω × ]0, T [ ,
w = ∂t w = 0
on ω
e × {0} ,



w = w (x, t) ∈ H 2 (Ω × ]0, T [) ,
where f = −χu (·, to ) ∈ H02 (Ω). Consequently, in a similar way than in the proof of Theorem 4.5, for
any δ ∈ ]0, T /2[, there exist C > 0 and µ ∈ ]0, 1[ such that we have
Z
T −δ
δ
Z
µ ¶ 1−µ
Z TZ
Z TZ
1 µ
2
2
|w (x, t)| dxdt ≤ C
|f (x)| dxdt + ε
|w (x, t)| dxdt
ε
Ω
0
Ω
0
Ω
2
∀ε > 0
On another hand, the following inequalities hold.
Z TZ
Z
2
2
|f (x)| dxdt ≤ T
|χ (x) u (x, to )| dx ,
0
Z
T
Ω
ω
Z
2
|w (x, t)| dxdt ≤ 2T
0
P
j≥1
Ω
√
αj2 e−2(λj to − λj T )
X
√
αj2 e−2(λj to −
λj T )
Z
j≥1
=
{j≥1;
≤e
2
2 Tto
P
√
λj ≤ tTo
P
j≥1
αj2
}
√
αj2 e−2(λj to − λj T ) +
,
16
2
+ 2T
|χ (x) u (x, to )| dx ,
ω
P
√
{j≥1;
λj > tTo
}
αj2 e−2(λj to −
√
λj T )
Z
δ
T −δ
¯
¯2
Z ¯X
Z T −δ Z
Z
³p ´¯¯
¯
2
2
−λj to
¯
¯ dxdt ≤ 2
α
e
(x)
e
ch
λ
t
|w
(x,
t)|
dxdt+2T
|χ (x) u (x, to )| dx ,
j
j
j
¯
¯
Ω ¯ j≥1
δ
Ω
ω
¯
Consequently, from the fast five inequalities, we deduce that for any ε > 0,
P 2 −2λj to
P 2 −2(λj to −√λj δ)
(T − 2δ)
αj e
≤ (T − 2δ)
αj e
j≥1
¯j≥1
¯2
¯
¡p
¢
R T −δ R ¯¯ P
¯
αj ej (x) e−λj to ch
≤ δ
λj t ¯ dxdt
Ω ¯¯
¯
j≥1
¡ 1 ¢ 1−µ
R
2
µ
|χ (x) u (x, to )| dx
≤ 2T C Ã
ε
ω
!
2
R
2
2 Tto P 2
+4T ε e
αj + ω |χ (x) u (x, to )| dx
j≥1
R
2
+2T ω |χ (x) u (x, to )| dx .
Finally, there exists C > 0 such that for any to > 0,
R
P 2 −2λj to
2
|u (x, to )| dx =
αj e
Ω
j≥1
≤C
¡ 1 ¢ 1−µ
R
µ
ε
ω
2
C
|u (x, to )| dx + εe to
R
Ω
2
|u (x, 0)| dx
∀ε ∈ ]0, 1[ ,
which implies the desired estimate ,
µZ
¶1−µ µZ
¶µ
Z
C(1−µ)
2
2
2
|u (x, 0)| dx
|u (x, to )| dxdt
.
|u (x, to )| dx ≤ C µ e to
Ω
8
Ω
ω
Notes on the papers in reference
... to be completed. (some reference to papers of Alessandrini have to be add too...)
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18