Quantiation of unique ontinuation
for the wave equation with Dirihlet boundary ondition
Kim Dang PHUNG
1
Introdution and main result
The purpose of this note is to prove the following result:
Theorem .Let be a bounded onneted C 2 domain in Rn , n > 1, and ! be a non-empty
open subset of . Let us onsider the following wave equation in R, with initial data and homo-
geneous Dirihlet boundary ondition :
8
<
:
t2 u u = 0 in R
u = 0 on R
u (; 0) = u0 , t u (; 0) = u1 in
(1.1)
.
Then, there exist a onstant > 0 and a time T > 0 suh that for all initial data (u0 ; u1 ) 2 H01 (
) L2 (
), (u0 ; u1 ) 6= 0, the solution u of ( 1.1) satises the following estimate :
k(u0 ; u1 )k2L2 (
)H
1 (
)
k(u0 ;u1 )k2H 1 (
)L2 (
) Z Z
T
k(u ;u )k2 0
e 0 1 L2 (
)H 1 (
)
0 !
ju (x; t)j2 dxdt .
Remark that estimate (1.2) is equivalent to
k(u0 ; u1 )k2L2 (
)H 1 (
) k(u0 ;u1)kH1 (
)L2 (
) k(u0 ; u1 )k2H01 (
)L2 (
) ,
ln 2 + kukL2(!0(0;T ))
(1.2)
(1.3)
or equivalently,
k(u0 ; u1)k2L2 (
)H
1 (
)
e="
Z TZ
0
!
ju (x; t)j2 dxdt + " k(u0 ; u1)k2H01 (
)L2(
) , 8" > 0 ,
(1.4)
where the value of the onstant > 0 may hanged from line (1.2) to line (1.4) but not its dependene.
Similar kind of estimates already appears in [ R℄, [ LR℄ in the framework of boundary ontrol and
stabilization for hyperboli equations. Here, we use the ideas from [ R℄.
1
2
2.1
Proof of Theorem
step 1: Energy estimates
First, reall the onservation identity of the energy for the solution u of
8
t2 u u = 0 in R
<
u = 0 on R
(2.1)
:
u (; 0) = u0 , t u (; 0) = u1 in ,
whih is, for all t 2 R
Z (2.2)
k(u0; u1 )k2H01 (
)L2 (
) = jt u (x; t)j2 + jru (x; t)j2 dx .
Furthermore, we have the following inequality: there exists a onstant > 0 suh that for all T 1,
we get
Z TZ
T k(u0 ; u1 )k2L2 (
)H 1 (
) ju (x; t)j2 dx .
(2.3)
0 Indeed, let 2 C01 ((0; T )) suh that 0 1, 1 on T4 ; 34T and j0j T8 . Multiplying (2.1)
by u and integrating over (0; T ) by parts yields:
R R
kruk2L2 (
( T4 ; 34T )) 0T (t) jru (x; t)j2 dxdt
= R R0RT R
t2u (x; t) (t) u (x; t) dxdt R R
= R0T R
t u (x; t) 0 (t) u (x; t) dxdt + 0T (t) jtu (x; t)j2 dxdt (2.4)
R0T R
jt u (x; t)j2 dxdt + k0 t ukL2 (
(0;T )) kukL2 (
(0;T ))
0T jt u (x; t)j2 dxdt + T8 kt ukL2 (
(0;T )) kukL2 (
(0;T ))
Adding ktuk2L2(
( T4 ; 34T )), using Poinare inequality and onservation of energy (2.2), then there
exists a onstant > 0 suh that for all T 1,
p
(2.5)
T k(u0 ; u1 )k2H01 (
)L2 (
) T kt ukL2 (
(0;T )) k(u0 ; u1 )kH01 (
)L2 (
) ,
that is
Z TZ
T k(u0 ; u1 )k2H01 (
)L2 (
) jt u (x; t)j2 dx .
(2.6)
0
Now,
we dedue (2.3) from (2.6). Indeed, let v (x; t) = R0t u (x; `) d` 1 u1 (x), then v (x; t) =
Rt
0 u (x; `) d` + u1 (x) and t v (x; t) = u (x; t),
t2 v (x; t) = t u (x; t) R
= tu (x; 0)R+t 0t t2u (x; `) d`
(2.7)
= u1 (x) + 0 u (x; `) d`
= v (x; t) .
1
Also, v (x; 0) = u1 (x), tv (x; 0) = u0 (x), and the homogeneous Dirihlet boundary ondition is satised. Consequently,
k(u0; u1 )k2L2 (
)H 1 (
) = 1 u1 ; u0 2H01 (
)L2 (
)
= k(vR (x;R0) ; t v (x; 0))k2H01 (
)L2(
) T R0T R
jt v (x; t)j2 dxdt (2.8)
T 0T ju (x; t)j2 dxdt ,
that is (2.3).
2
2.2
step 2: FBI transform
Let F (z) = 21 RR eiz e
onsider
2 d
p
, then F (z) =
F (z ) = F (z ) =
We have,
)
(
1 jIm zj2 jRe zj2
i
e 2 (Im z Re z)
2 e 4
p
. Also, with > 0, let us
1 Z eiz e ( )2 d .
2 R
jF (z )j = 2 e 4 (jIm zj2
2
and, denoting Fb the Fourier transform of F ,
F ( ) = e
jRe zj2 )
(2.9)
,
(2.10)
( )2 .
(2.11)
Let s; `o 2 R, we introdue the following gaussian transformation whih is a partiular ase of the
Fourier-Bros-Iagolnitzer transformation in [ LR℄ :
W`o ; (x; s) =
Z
R
F (`o + is
`)(`)u(x; `)d`
where 2 C01(R). We remark that s2F (`o + is `) = `2F (`o + is `), so
R
s2 W`o ; (x; s) = RR `2 F (`o + is `)(`)u(x; `)d`
= R F(`o + is `) 00(`)u(x; `) + 20(`)t u(x; `) + (`)t2 u(x; `) d`.
As u is solution of the wave equation, W`o ; satises:
8
R
< s2 W`o ; (x; s) + W`o ; (x; s) = R F (`o + is `) [00 (`)u(x; `) + 20 (`)t u(x; `)℄ d`
W`o ; (x; s) = 0 for x 2 :
W`o ; (x; 0) = (F u(x; )) (`o ) for x 2 .
In another hand, we also have for some T > 0,
ku (x; )kL2 (( T2 1; T2 +1)) ku(x; ) F u(x; )kL2(( T2 1; T2 +1))
+ kF u(x; )kL2(( T2 1; T2 +1))
ku(x; ) F u(x; )kL2(R)
1=2
+ Rt2( T2 1; T2 +1) jWt; (x; 0)j2 dt .
But using Parseval equality and (2.11),
ku(x; )
F u(x; )kL2 (R)
= p12 u(x; )
1
p
(2.14)
(2.15)
( ) ( ) ( )
2 ( ) ( )
( ( )) ( )
p 1
2 kt (u(x; ))kL2 (R) ,
3
(2.13)
u(x; )L2 (R)
\
F
2 1=2
R 2 \
e
u
x;
d
R
2 1 =2
R
\
p12 R u
x;
d
2 1 =2
p
R p22 1 R t \
u x;
d
= p12
(2.12)
(2.16)
so,
Z
(
t2 2
T
1; T +1
2
)
j(t)u(x; t)j2 dt 4 12 kt (u(x; ))k2L2 (R) + 2
Z
(
t2 2
T
1; T +1
2
)
jWt; (x; 0)j2 dt (2.17)
Now, reall that from the Cauhy theorem we have:
Proposition .-
suppose that
Do =
then
Let f be a holomorphi funtion in a domain D C . Let a; b > 0, z 2 C . We
(x; y) 2 R2 ' C n jx Re(z)j a; jy Re(z)j b D ,
Z
1
f (z ) =
ab
Z
jx
( + iy) dxdy .
f x
Re(z) 2 + y Im(z) 2 1
a
b
j j
j
Choosing z = t 2 T2 1; T2 + 1 R and x + iy = `o + is, we dedue that
R
1 R
jW`o +is; (x; 0)j d`o ds
jWt; (x; 0)j ab
j
`
t
j
a
Rjsjb
1 R o
ab
j` tja jsjb jW`o ; (x; s)j dsd`o 1=2
Ro
R
R
2 dsd` 1=2 R
(2.18)
1
ab
dsd`
j
W
(
x;
s
)
j
o
`
;
o
o
j`o tja jsjb
j`o tja jsjb
R
1 =2
R
,
p2ab j`o tja jsjb jW`o ; (x; s)j2 dsd`o
and with a = b = 1,
R
R
R
2 dt R
2 dsd` dt .
j
W
(
x;
0)
j
j
W
(
x;
s
)
j
T
T
T
T
t;
`
;
o
t2
1; 2 +1) R j`o tj1 jsj1R o
t2( 2 1; 2 +1)
R (2
t2( T2 1; T2 +1) `o 2( T2 2; T2 +2) jsj1 jW`o ; (x; s)j2 dsd`odt (2.19)
R
R
2 `o 2( T2 2; T2 +2) jsj1 jW`o ; (x; s)j2 dsd`o .
Consequently, from (2.17), (2.19) and integrating over , we get
R R
2
2
1 R R
0
t2( T2 1; T2 +1) j(t)u(x; t)j dtdx 4 2 R j (t) u(x; t) + (t) t u(x; t)j dtdx +4 R`o2( T2 2; T2 +2) R
Rjsj1 jW`o;(x; s)j2 dsdx d`o .
(2.20)
Now reall the following quantiation result for unique ontinuation of ellipti equation with Dirihlet
boundary ondition:
Proposition .Let be a bounded onneted C 2 domain in Rn , n > 1, and ! be a non-empty
open subset of . Let f = f (x; s) 2 L2 (
( 2; 2)). Then there exists C > 0 suh that for all
w = w (x; s) 2 H 2 (
( 2; 2)) solution of
2
s w + x w = f in ( 2; 2)
w = 0 on ( 2; 2)
for all " > 0, we have :
R
R
R
R
2 dxds
2 dxds + R
2 dxds CeC=" R
j
f
(
x;
s
)
j
j
w
(
x;
s
)
j
j
w
(
x;
s
)
j
jsj2 jsj2 !
jsj1 +e 4=" Rjsj2 R
jw (x; s)j2 dxds .
4
Applying to W`0 ;, we dedue that for all " > 0,
R
R
2
jsj1 R
jW`R0 ; (x; s)j dxds2
jW` ; (x; s)j dxds
e 4=" jsj
R 2 R 0
2
C="
+Ce Rjsj2 R! jW
`0 ; (x; s)j dxds
R
+CeC=" jsj2 R F (`0 + is `) [00(`)u(x; `) + 20(`)tu(x; `)℄ d`2 dxds .
(2.21)
Consequently, from (2.20) and (2.21), there exists a onstant C > 0, suh that for all " > 0,
R R
2
tR2( TR2 1; T2 +1) j(t)u(x; t)j dtdx
4 12 RRj0 (t) u(x; t) +R (t) Rt u(x; t)j2 dtdx
+4e 4=" `o2( T2 2; T2 +2) jsj2 jW`0; (x; s)j2 dxds d`o
+4CeC=" R`o2( T2 2; T2 +2) Rjsj2 R! jW`0 ;(x; s)j2 dxds d`o
+4CeC=" R`o2( T2 2; T2 +2) Rjsj2 R
RR F(`0 + is `) [00(`)u(x; `) + 20(`)t u(x; `)℄ d`2 dxds d`o .
(2.22)
1
Let dene
T > 16, we hoose 2 C01 ((0; T )), 0 1, 1
2 C0 (R) more preisely
now:
let
3
T
T
T
3
T
supp(00 ) K . Let
on 4 ; 4 . Furthermore,
let K = 0; 4 [ 4 ; T suh that supp(0) = K and
5
T
T
3
T
T
T
K0 = 8 ; 8 suh that dist(K; Ko ) = 8 . We will hoose `0 2 2 2; 2 + 2 K0 .
Till the end of the proof, CT will denote a generi positive onstant independent of " and but
dependent on (
; T ), whose value may hange from line to line.
The rst term of the seond member of (2.22) beomes, using (2.2),
1 Z Z j0 (t) u(x; t) + (t) tu(x; t)j2 dtdx CT k(u0; u1)k2 1 2
(2.23)
2 R
2
H0 (
)L
(
)
The seond term of the seond member of (2.22) beomes, using (2.10),
R
R
R
e 4=" `o 2( T 2; T +2) jsj2 jW`0 ; (x; s)j2 dxds d`o
2
2
= e 4=" R`o2( T2 2; T2 +2) Rjsj2 R
RR F (`o + is `)(`)u(x; `)d`2 dxds d`o
2
2
R
R
R R p
e 4=" `o 2( T2 2; T2 +2) jsj2 0T 2 e 4 (jsj2 j`o `j2 ) ju(x; `)j d` dxds d`o
2
2
p
R
R
R R
2 e2 e 4=" `o 2( T2 2; T2 +2) jsj2 0T ju(x; `)j d` dxds d`o
2
p
R R
2 e2 e 4=" 4 4T 0T ju(x; `)j2 d`dx
CT 2 e22 e 4=" k(u0 ; u1 )k2H01 (
)L2 (
)
The third term of the seond member of (2.22) beomes, using (2.10),
R
R
R
eC=" `o 2( T 2; T +2) jsj2 ! jW`0 ; (x; s)j2 dxds d`o
2
2
= eC=" R`o 2( T2 2; T2 +2) Rjsj2 R! RR F (`o + is `)(`)u(x; `)d`2 dxds d`o 2
2
R
R
R R p
eC=" `o 2( T2 2; T2 +2) jsj2 ! 0T 2 e 4 (jsj2 j`o `j2 ) j(`)u(x; `)j d` dxds d`o
2
2
p
R
R
R R
2 e2 eC=" `o 2( T2 2; T2 +2) jsj2 ! 0T ju(x; `)j d` dxds d`o
R R
42 T e22 eC=" ! 0T ju(x; t)j2 dtdx
5
(2.24)
(2.25)
The fourth term of the seond member of (2.22) beomes, using (2.10) and the hoie of ,
R
2
R
R R
eC=" `o 2( T 2; T +2) jsj2 R F (`0 + is `) [00 (`)u(x; `) + 20 (`)t u(x; `)℄ d` dxds d`o
2
2
2
2
R
R R p
R
C="
e Ko jsj2 K 2 e 4 (jsj2 j`o `j2 ) j00 (`)u(x; `) + 20 (`)t u(x; `)j d` dxds d`o
2
2
p
2
R
R R
R
2 e2 eC=" CT Ko jsj2 K e 4 j`o `j2 (ju(x; `)j + jt u(x; `)j) d` dxds d`o
2
p
2
R R
CT 2 e2 e 4 dist(K;K0 )2 eC=" K (ju(x; `)j + jt u(x; `)j) d`2 dx
2
K;K0 )2 eC=" k(u ; u )k2 1
CT 2 e22e 2 dist(
0 1 H0 (
)L2 (
)
2 4 T 2
2
2
C="
2
2
8 e
CT e
k(u0 ; u1)kH01 (
)L2 (
)
(2.26)
We nally obtain from (2.23), (2.24), (2.25), (2.26) and (2.22), that
R R
2
2
CT
t2( T2 1; T2 +1) j(t)u(x; t)j dtdx 2 k(u0 ; u1 )kH01 (
)L2 (
)
+CT 2 e22 e 4=" Rk(uR0; u1)k2H01 (
)L2(
)
(2.27)
+42T e222 eC="2 ! 0T ju(x; t)j2 dtdx
T
+CT 2 e 2 4 82 eC=" k(u0; u1)k2H01 (
)L2(
) .
We begin to hoose 2 = 1" in order that
R R
2
2
t2( T2 1; T2 +1) j(t)u(x; t)j dtdx "CT k(u0 ; u1 )kH01 (
)L2 (
)
+e 1="CT k(uR0; uR1)k2H01 (
)L2(
)
+e(3+C)="
CT ! 0T ju(x; t)j2dtdx
+CT exp 3 21 T822 + C 1" k(u0; u1)k2H01 (
)L2(
) .
(2.28)
We nally hoose T > 16 large enough suh that 3 12 T822 + C 1 that is 82 (8 + 2C ) T 2, to
dedue
that
R R
R R
2
2
t2( T2 1; T2 +1) ju(x; t)j dtdx t2( T2 1; T2 +1) j(t)u(x; t)j dtdx
R R
"CT k(u0 ; u1)k2H01 (
)L2 (
) + e(3+C )=" CT ! 0T ju(x; t)j2 dtdx .
(2.29)
Now we onlude from (2.3), that there exist a onstant > 0 and a time T > 0 large enough suh
that for all " 2 (0; 1) small enough, we have
Z Z T
2
="
k(u0; u1 )kL2 (
)H 1 (
) e
ju(x; t)j2 dtdx + " k(u0; u1 )k2H01 (
)L2 (
)
(2.30)
!
0
Referenes
[ LR℄ G. Lebeau and L. Robbiano, Stabilisation de l'equation des ondes par le bord, Duke Math. J.
86 (1997) 465-491.
[ R℄ L. Robbiano, Fontion de o^ut et ontr^ole des solutions des equations hyperboliques, Asymptoti
Analysis 10 (1995) 95-115.
6