Quantiation of unique ontinuation for
an ellipti equation with Dirihlet boundary ondition
Kim Dang PHUNG
1
Introdution and main result
The purpose of this note is to prove the following result:
Let be a bounded onneted C 2 domain in Rn , n > 1. We hoose T > 0 and
Æ 2 (0; T=2). Let us onsider the following ellipti equation of seond order in (0; T ) with the
Dirihlet boundary ondition :
Theorem .-
8
<
:
t2 v + x v = 0 in (0; T )
v = 0 on (0; T )
v = v (x; t) 2 H 2 (
(0; T )) .
(1.1)
Then, for all ' 2 C01 (
(0; T )), ' 6= 0, there exist C > 0 and 2 (0; 1) suh that for all v solution
of ( 1.1), we have :
Z T ÆZ
Æ
jv (x; t)j2 dxdt C
Z TZ
0
!
jv (x; t)j2 dxdt
Z TZ
0
j'v (x; t)j2 dxdt
!1 .
(1.2)
Similar kind of estimates already appears in [ LR℄ from tehniques of Carleman inequality (see also
[ FI℄). Here, we will use the method desribed by [ E℄ (see also [ AE℄, [ GL℄).
2
Proof of Theorem, the results of [ E℄
Let be an open, bounded, onneted subset of Rn , n > 1, with a C 2 boundary . Let be a
non-empty subset of . Let us onsider the following ellipti equation of seond order in Rt , with
the Dirihlet boundary ondition :
8
<
:
t2 u + x u = 0 in R
u = 0 on R
u = u (x; t) 2 H 2 (
R) .
(2.1)
The goal of this Setion is to derive interpolation inequalities assoiated with the solutions u of (2.1).
Let Br denote the open ball of enter (xo ; to ) 2 R+ with radius r.
We propose to prove the two following lemmas : (the two following lemmas are stated in only one in
[ E℄)
1
Lemma 1 .Let r1 , r2 , r3 , Ro be four real numbers suh that 0 < r1 < r2 < r3 < Ro . Let
us onsider (xo ; to ) 2 R+ suh that BRo R, then there exists 2 (0; 1) suh that for all u
solution of ( 2.1), we have :
Z
Z
ju (x; t)j dxdt 2
Br2
ju (x; t)j dxdt
2
Br1
!
Z
Br3
ju (x; t)j dxdt
2
!1 .
(2.2)
Lemma 2 .Let ro , r1 , r2 , r3 , Ro be ve real numbers suh that 0 < r1 < ro < r2 < r3 < Ro .
Let us onsider (xo ; to ) 2 R+ satisfying the three following onditions :
i. Br \ (
R) is star-shaped with enter (xo ; to ) 8r 2 (0; Ro ) ,
ii. Br (
R) 8r 2 (0; ro ) ,
iii. Br \ (
R) 6= ; and Br \ ( R) ( R) 8r 2 [ro ; Ro ) .
Then there exists 2 (0; 1) suh that for all u solution of ( 2.1), we have :
Z
Br2 \(
R)
ju (x; t)j dxdt 2
Z
ju (x; t)j dxdt
2
Br1
!
Z
Br3 \(
R)
!1 ju (x; t)j dxdt
2
.
(2.3)
The proof of Theorem is dedued from Lemma 1 and Lemma 2 by an adequate partition of unity.
To apply Lemma 2, we will hoose xo in a neighborhood of the boundary suh that the onditions i,
ii, iii, hold.
3
Proof of Lemma 1
Let us denote y = (x; t) 2 Rn+1 , yo = (xo ; to ) and
funtions H , D, N , for 0 < r < Ro :
H (r) =
D (r) =
r = (rx ; t ).
We introdue the three following
R
2
BRr ju (y )j dy 2 2
1
2 Br jru (y )j r
jy yoj2 dy
(3.1)
and
N (r) =
D (r)
H (r)
0.
(3.2)
Our goal is to prove that N = N (r) is a non-dereasing funtion for r 2 (0; Ro ). Indeed, we will prove
that the following equality holds :
d
2
d
ln H (r) = (n + 1) ln r + N (r) .
dr
dr
r
So, from the non-dereasing property of N , we dedue that :
(r2 )
ln H
H (r1 )
H (r3 )
ln H
(r2 )
R
= (n + 1) ln rr12 + 2 rr12 Nr(r) dr
(n + 1) ln rr12 + 2N (r2 ) ln rr21 ,
R
= (n + 1) ln rr23 + 2 rr23 Nr(r) dr
(n + 1) ln rr23 + 2N (r2 ) ln rr32 .
2
(3.3)
(3.4)
(3.5)
Consequently,
(r2 )
ln H
H (r1 )
ln rr21
(n + 1) + 2N (r2 ) and we get what we wanted :
Z
Br2
with =
1r
ln r12
1r
ln r21
ju (y)j dy 2
+ ln
1r
1
3
r2
Z
ju (y)j dy
2
Br1
!
H (r3 )
ln H
(r2 )
,
r
ln r23
Z
ju (y)j dy
2
Br3
(3.6)
!1 ,
(3.7)
.
R R
First, we ompute the derivative of H (r) = 0r Sn ju (s + yo )j2 n dd (s) :
R
H 0 (r) = SRn ju (rs + yo )j2 rn d (s)
2
n d (s)
= 1r Sn ju (rs
+ yo )j rs sr R
2
= r1 Br div ju (y)j (y yo) dy
R = 1r Br (n + 1) ju (y)j2 + r ju (y)j2 (y yo) dy
2R
= n+1
r H (r) + r Br uru (y yo ) dy .
Next we have to remark that
D (r) =
indeed,
R
Br uru (y yo ) dy =
=
=
=
Z
Br
(3.8)
uru (y yo) dy ,
(3.9)
1 R uru r r2 jy y j2 dy
o
2 Br
1 R div r2 jy y j2 uru dy + 1 R
r 2 jy
o
2 Br
2
B
r
1R
2 jy y j2 div (uru) dy beause on B , r
r
o
2 Br r
R
2
2 2
1
jy yo j dy beause y u = 0 .
2 Br jru (y )j r
yoj2 div (uru) dy
= jy yo j
0
2 D(r)
Consequently, from (3.8) and (3.9), we prove : HH ((rr)) = n+1
r + r H (r) and this is (3.3).
Now, we ompute the derivative of D (r) =
2
1 R r R (ru)
js+yo 2 0 Sn
2
R R
D0 (r) = 21 drd r2 0r Sn (ru)js+yo n dd (s)
2
R R
= r 0r Sn (ru)js+yo n dd (s)
R
= r Br jru (y)j2 dy .
r2
2 n dd (s):
2
n
1R
2
2 S n r (ru)jrs+yo r d (s)
(3.10)
The omputation of the derivative of N (r) = HD((rr)) gives :
1
N 0 (r) = 2 [D0 (r) H (r) D (r) H 0 (r)℄ .
H (r)
As
monotoniity of N depends on the positivity of D0 (r) H (r) D (r) H 0 (r) =
h the desired non-dereasing
0 (r) i
H
0
D (r) D (r) H (r) H (r), we are redued from (3.3) to prove that
n+1
D (r) H (r) .
r
2 2
D (r) D0 (r)
r
3
(3.11)
By Cauhy-Shwarz inequality, we have :
2
R
uru (y yo ) dy
RBr
R
Br j(y yo ) ruj2 dy Br juj2 dy
R
Br j(y yo ) ruj2 dy H (r) .
D 2 (r ) =
(3.12)
Consequently, from (3.11) and (3.12), N is a non-dereasing funtion if
2
r
Z
Br
j(y yo ) ruj2 dy D0 (r) n +r 1 D (r)
.
(3.13)
Our goal is now redued to prove that for all u solution of (2.1), if 0 < r < Ro , then
Z
2
j(y yo ) ruj2 dy D0 (r) n +r 1 D (r) .
r Br
We begin to reall the following Rellih-Neas identity with vetor eld (y
(2.1) :
2div (((y
yo ) ru) ru) = div (y yo) jruj2
(3.14)
yo) for all u solution of
(n 1) jruj2 ,
(3.15)
indeed,
div (((y yo ) ru) ru) = ((y yo ) ru) u + r ((y yo ) ru) ru
= yi (y yo)j yj u yi u
= yi (y yo )jyj uyi u + (y yo )j y2i yj uyi u
= jruj2 + 12 r jruj2 (y yo )
div (y yo) jruj2 = jruj2 div (y yo ) + r jruj2 (y yo )
= (n + 1) jruj2 + 2yj uy2iyj u (y yo )i .
Consequently,
2
Z
Br
div (((y yo ) ru) ru) dy =
Z
Br
div (y yo) jruj2 dy (n 1)
Z
Br
jruj2 dy .
(3.16)
By multiplying (3.16) by r , and from (3.10) we dedue that
2r
Z
Sn
1
rs (ru)jrs+yo (ru)jrs+yo srn d (s) = r
By integrating (3.17) on (0; r) , we have
2
Z
Br
j(y yo) ruj2 dy =
Z
Br
R
R
2
rs (ru)jrs+yo srn d (s) (n 1) D0 (r) .
Sn 1
(3.17)
jy yo j2 jruj2 dy
Finally, from (3.1) and (3.18), we obtain :
2 Br j(y
Z
R
(n 1) D (r) .
yo ) ruj2 dy = r2 Br jruj2 dy Br r2 jy yo j2 jruj2 dy (n 1) D (r)
R
= r2 Br jruj2 dy + [ 2 (n 1)℄ D (r)
= rD0 (r) (n + 1) D (r) ,
and this is (3.14). The proof of Lemma 1 is now omplete.
4
(3.18)
(3.19)
4
Proof of Lemma 2
Let us denote y = (x; t) 2 Rn+1 , yo = (xo ; to) and r = (rx ; t ). As Br \ (
R) 6= ; and
Br \ ( R) ( R) for all r 2 [ro ; Ro[, we extend u to be zero outside R. As u = 0 on R,
we dedue that
8
u = uj
in BRo
>
>
<
u = 0 on BRo \ R
(4.1)
>
u
= ruj
in BRo
r
>
:
u = 0 in R .
Let us denote r = Br \ (
R), when 0 < r < Ro . We introdue the three following funtions :
R
2
Rr ju (y )j dy 2 2
1
2 r jru (y )j r
H (r ) =
D (r) =
(4.2)
jy yo j2 dy
and
D (r)
0.
(4.3)
H (r)
Our goal is to prove that N is a non-dereasing funtion. Indeed, we will prove that the following
equality holds :
d
2
d
ln H (r) = (n + 1) ln r + N (r) .
(4.4)
dr
dr
r
So, from the non-dereasing property of N , we dedue that :
N (r) =
R
(r2 )
ln H
= (n + 1) ln rr12 + 2 rr12 Nr(r) dr
H (r1 )
(n + 1) ln rr12 + 2N (r2 ) ln rr21 ,
H (r3 )
ln H
(r2 )
Consequently,
(r2 )
ln H
H (r1 )
ln rr21
R
= (n + 1) ln rr23 + 2 rr23 Nr(r) dr
(n + 1) ln rr23 + 2N (r2 ) ln rr32 .
(n + 1) + 2N (r2 ) and we get the desired result :
Z
r2
with =
1r
ln r12
1r
ln r21
ju (y)j dy 2
+ ln1r3
r2
1
(4.5)
Z
Br1
ju (y)j dy
2
!
(4.6)
H (r3 )
ln H
(r2 )
,
ln rr23
Z
ju (y)j dy
2
r3
(4.7)
!1 ,
(4.8)
.
R
R R
First, we ompute the derivative of H (r) = Br ju (y)j2 dy = 0r Sn ju (s + yo)j2 n dd (s) :
R
H 0 (r) = SRn ju (rs + yo )j2 rn d (s)
2
n d (s)
= 1r Sn ju (rs
+ yo )j rs sr R
2
= 1r Br div ju (y)j (y yo) dy
R = 1r Br (n + 1) ju (y)j2 + r ju (y)j2 (y yo) dy
2R
= n+1
r H (r) + r r uru (y yo ) dy .
5
(4.9)
Next we have to remark that
D (r) =
indeed,
R
r uru (y
yo ) dy =
=
=
=
=
Z
r
uru (y yo ) dy,
(4.10)
1R
r r2 jy yo j2 dy
2 Br uru
1 R div r2 jy y j2 uru dy + 1 R
2 jy y j2 div (uru) dy
r
o
o
2 Br
2 Br
2
1R
2
(uru) dy beause on Br , r = jy yoj
jy yoj div
2 Br r
2
2
1R
2
r
j
y
y
j
u
u
+
jr
u
j
dy
o
y
2 Br
R
2
2
1
2 jy y j dy beause u = 0 in R and u = 0.
y
o
j
2 r jruj r
0
2 D(r)
Consequently, from (4.9) and (4.10), we prove : HH ((rr)) = n+1
r + r H (r) and this is (4.4).
Now, we ompute the derivative of D (r) =
2
1 R r R (ru)
n
js+yo 2 0 S
2
R R
D0 (r) = 21 drd r2 0r Sn (ru)js+yo n dd (s)
2
R R
= r 0r Sn (ru)js+yo n dd (s)
R
= r r jru (y)j2 dy .
r2
2 n dd (s):
2
n
1 R r2 (ru)
n
jrs+yo r d (s)
2 S
(4.11)
D(r) gives :
The omputation of the derivative of N (r) = H
(r )
1
N 0 (r) = 2 [D0 (r) H (r) D (r) H 0 (r)℄ .
H (r)
As the desired non-dereasing monotoniity of N depends on the positivity of D0 (r) H (r) D (r) H 0 (r),
we are redued from (4.4) to prove that
n+1
D (r) H (r) .
r
2 2
D (r) D0 (r)
r
(4.12)
By Cauhy-Shwarz inequality, we have :
R
2
uru (y yo ) dy
r
R
R
r j(y yo ) ruj2 dy r juj2 dy
R
r j(y yo ) ruj2 dy H (r) .
D2 (r) =
(4.13)
Consequently, from (4.12) and (4.13), N is a non-dereasing funtion if
2
r
Z
r
j(y yo) ruj2 dy D0 (r) n +r 1 D (r)
Our goal is now redued to prove that for all u solution of (2.1), if ro
i,ii,iii, of Lemma 2 hold, then
2
r
Z
r
(4.14)
r < Ro and the hypothesis
j(y yo) ruj2 dy D0 (r) n +r 1 D (r)
6
.
.
(4.15)
We note that the ase 0 < r < ro is already studied in the proof of Lemma 1.
We begin to reall the following Rellih-Neas identity with vetor eld (y yo) for all u solution of
(2.1) :
yo ) ru) ru) = div (y yo) jruj2
2div (((y
Consequently,
2
Z
r
div (((y yo ) ru) ru) dy =
Z
r
ru (y)
Z
r
jruj2 dy .
= ( u (y)) on Br
R
(4.16)
(4.17)
\ R
with
R
yo) ru) ru) dy = Br \(
R) r1 ((y yo ) ru)2 d + Br \( R) ((y yo ) ru) u
d
R
R
= Br \(
R) r1 j(y yo) ruj2 d + Br \( R) ((x xo ) x ) j uj2 d ,
(4.18)
r div (((y
R
(n 1) jruj2 .
div (y yo ) jruj dy (n 1)
2
As Br \ ( R) ( R) and uj = 0, we have
= (x ; 0) 2 Rn R on . We obtain :
R
R
R
2
2
2
r div (y yo ) jruj dy = RBr \(
R) r jruj d + RBr \( R) (y yo ) jruj d
= Br \(
R) r jruj2 d + Br \( R) ((x xo ) x ) j uj2 d .
(4.19)
Consequently, from (4.11), (4.17), (4.18) and (4.19), we have
R
R
2 Br \(
R) 1r j(y yo ) ruj2 d + Br \( R) ((x xo ) x ) j uj2 d
R
= Br \(
R) r jruj2 d r1 (n 1) D0 (r) .
If ((x xo ) x ) 0 on , then
2
Z
Br \(
R)
j(y yo ) ruj2 d Z
Br \(
R)
r2 jruj2 d (n 1) D0 (r) .
(4.20)
(4.21)
So, by integrating (4.21), we dedue that
2
Z
r
j(y yo) ruj2 dy Z
r
jy yo j2 jruj2 dy
(n 1) D (r) .
(4.22)
From (4.2), (4.11) and (4.22), we obtain
R
R
R
2
2
2 jy y j2 jruj2 dy (n 1) D (r)
2
o
r
j(y yo ) ruj dy r R
jruj dy
2
2
(4.23)
r jruj dy + ( 2 (n 1)) D (r)
0
rD (r) (n + 1) D (r) ,
and this is (4.14). The hypothesis ((x xo ) x ) 0 on is of ourse true when Br \ (
R) is
star-shaped with enter (xo ; to ) for all r 2 (0; Ro). The proof of Lemma 2 is now omplete.
2
r
r
r
r
7
Referenes
[ AE℄ V. Adolfsson and L. Esauriaza, C 1; domains and unique ontinuation at the boundary, Comm.
Pure Appl. Math. 50, 3 (1997) 935-969.
[ E℄ L. Esauriaza, Communiation to E. Zuazua.
[ FI℄ A.V. Fursikov and O.Yu. Imanuvilov, Controllability of evolution equations, Leture Notes
Series,nÆ34, Seoul National University, Korea, 1996.
[ GL℄ N. Garofalo and F H. Lin, Monotoniity properties of variational integrals, Ap -weights and
unique ontinuation, Indiana Univ. Math. J. 35, 2 (1986) 245-268.
[ LR℄ G. Lebeau and L. Robbiano, Contr^ole exate de l'equation de la haleur, Comm. Part. Di. Eq.
20 (1995) 335-356.
8