Observability for heat equations
Kim Dang PHUNG
Yangtze Center of Mathematics, Sichuan University,
Chengdu 610064, China.
Université d’Orléans, Laboratoire de
Mathématiques - Analyse, Probabilités, Modélisation - Orléans,
CNRS FR CNRS 2964, 45067 Orléans cedex 2, France.
E-mail: kim_dang_phung@yahoo.fr
Abstract
This talk describes di¤erent approaches to get the observability for
heat equations without the use of Carleman inequalities.
Contents
1 The heat equation and observability
2
2 Our motivation
3
3 Our strategy
3.1 Proof of "Hölder continuous dependence from one
) "Sum of Laplacian eigenfunctions" . . . . . .
3.2 Proof of "Hölder continuous dependence from one
) "Observability" . . . . . . . . . . . . . . . . .
3.3 Proof of "Hölder continuous dependence from one
) "Re…ned Observability" . . . . . . . . . . . .
4
point in time"
. . . . . . . . .
point in time"
. . . . . . . . .
point in time"
. . . . . . . . .
7
4 What I hope
4.1 Logarithmic convexity method . . . . . . . . . . . . . . . . . . .
4.2 Weighted logarithmic convexity method . . . . . . . . . . . . . .
9
9
10
4
4
5 What I can do
12
5.1 The frequency function . . . . . . . . . . . . . . . . . . . . . . . . 12
5.2 The frequency function with weight . . . . . . . . . . . . . . . . . 14
5.3 The heat equation with space-time potential . . . . . . . . . . . . 19
This talk was done when the author visited School of Mathematics & Statistics, Northeast
Normal University, Changchun, China, (July 4-21, 2011).
1
6 What already exists
21
6.1 Monotonicity formula . . . . . . . . . . . . . . . . . . . . . . . . 21
6.1.1 Proof of Lemma B . . . . . . . . . . . . . . . . . . . . . . 22
6.1.2 Proof of Lemma A . . . . . . . . . . . . . . . . . . . . . . 23
6.1.3 Proof of Lemma C . . . . . . . . . . . . . . . . . . . . . . 24
6.2 Quantitative unique continuation property for the Laplacian . . . 28
6.3 Quantitative unique continuation property for the elliptic operator @t2 + . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
6.4 The heat equation and the Hölder continuous dependence from
one point in time . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
1
The heat equation and observability
We consider the heat equation in the solution u = u(x; t)
8
u=0
in
(0; +1) ,
< @t u
u=0
on @
(0; +1) ,
:
u ( ; 0) 2 L2 ( ) ,
(1.1)
living in a bounded open set in Rn , n 1, either convex or C 2 and connected,
with boundary @ . It is well-known that the above problem is well-posed and
have a unique solution u 2 C [0; T ] ; L2 ( ) \ L2 0; T ; H01 ( ) for all T > 0.
The observability problem consists in proving the following estimate
Z
2
ju (x; T )j dx
C
Z
0
T
Z
!
2
ju (x; t)j dxdt
for some constant C > 0 independent on the initial data. Here, T > 0 and ! is
a non-empty open subset in .
2
In the literature, two ways allow to prove such observability estimate. One
is due to the work of Fursikov and Imanuvilov based on global Carleman inequalities (see [FI]). The other proof is established by Lebeau and Robbiano
(see [LR]. See [Le] for an english version). We resume the Lebeau-Robbiano
strategy as follows.
Z TZ
2
2
ku ( ; T )kL2 ( ) C
ju (x; t)j dxdt
0
!
*
controllability in …nite and in…nite dimension
*
X
2
j=1;::;N
for any faj g where (ej ;
ary conditions (0 < 1
jaj j
j)
C
Ce
p
N
Z
X
2
aj ej (x) dx
! j=1;::;N
solves the eigenvalue problem with Dirichlet boundbeing the corresponding eigenvalues).
2
*
For any > 0 and any non-trivial ' 2 C01 (
(0; T )), there are C > 0 and
2 (0; 1), such that for any w 2 H 2 (
(0; T )) with @t2 +
w = f and
wj@ = 0, it holds
1
kwkH 1 (
( ;T
))
C kwkH 1 (
kf kL2 (
(0;T ))
(0;T ))
+ k'wkL2 (
(0;T ))
The above interpolation inequality is proved using Carleman inequalities.
Recently, a shortcut of the Lebeau-Robbiano strategy is given in [M].
2
Our motivation
In application to bang-bang control (see [W]), we need the following re…ned
observability estimate from measure set in time.
Z Z
ku ( ; T )kL2 ( ) C
ju (x; t)j dxdt
E
!
for some constant C > 0 independent on the initial data. Here, E (0; T ) is a
measurable set of positive measure and ! is a non-empty open subset in .
Further, we want to be able to extend the proof to heat equations with
space-time potentials.
The approach describes in this talk is linked to parabolic quantitative unique
continuation (see [BT], [Li], [L], [P], [EFV], [K], [KT] and references therein).
3
3
Our strategy
We are able to prove that
Hölder continuous dependence from one point in time
=) Observability from a measure set in time.
More precisely,
ku ( ; t)kL2 (
C ku ( ; 0)kL2 (
)
ku ( ; T )kL2 (
3.1
bj e
jt
P
j=1;::;N
2
0
@C
j=1;::;N
bj e
X
j=1;::;N
p
which implies when t = 1=
P
j=1;::;N
+
Z Z
C
ju (x; t)j dxdt
)
E
bj e
j=
p
2
N
jt
!
ej (x). Then it holds for any t > 0
1 0
Z
2A B
C=t
jbj j
@Ce
11
2
X
jt
bj e
! j=1;::;N
C
ej (x) dxA
N
Ce2
p
CeC
N
P
j=1;::;N
p
This gives the desired estimate with bj e
3.2
8t > 0
Proof of "Hölder continuous dependence from one point
in time" ) "Sum of Laplacian eigenfunctions"
Choose u (x; t) =
X
1
eC=t ku ( ; t)kL2 (!)
)
N
R
!
j=
p
bj e
P
j=
p
j=1;::;N bj e
N
2
N
j=
p
2
N
1
ej (x) dx
= aj .
Proof of "Hölder continuous dependence from one point
in time" ) "Observability"
Let ` 2 [0; T ) and `1 2 (`; T ]. Let z > 1. Introduce the decreasing sequence
f`m gm 1 , which converges to ` given by
`m+1 = ` +
1
(`1
zm
4
`) .
.
Then
1
(z 1) (`1 `) > 0 .
zm
We start with the following interpolation estimate. Let ! be a nonempty open
subset of . For any 0 t1 < t2 T ,
`m
ku ( ; t2 )kL2 (
C1 t C2t
e 2 1 ku ( ; t2 )kL2 (!) + " ku ( ; t1 )kL2 (
"
)
Let 0 < `m+2 < `m+1
ku ( ; t)kL2 (
`m+1 =
t < `m < T . We get
C1 t
e
"
)
8" > 0 .
)
C2
`m+2
ku ( ; t)kL2 (!) + " ku ( ; `m+2 )kL2 (
8" > 0 .
)
Recall that
ku ( ; `m )kL2 (
ku ( ; t)kL2 (
)
)
.
Therefore,
ku ( ; `m )kL2 (
C1 t
e
"
)
C2
`m+2
ku ( ; t)kL2 (!) + " ku ( ; `m+2 )kL2 (
)
8" > 0 .
Integrating it over t 2 (`m+1 ; `m ), it gives
(`m
`m+1 ) ku ( ; `m )kL2 (
R `m
C2
`m+2
C1 `m+1
" e
)
ku ( ; t)kL2 (!) dt
`m+1
`m+1 ) ku ( ; `m+2 )kL2 ( ) 8" > 0 .
+" (`m
That is
ku ( ; `m )kL2 (
h
)
1
`1
zm
`z 1
i
C1 C2
" e
h
1
z m+1
`1 ` z 1
+" ku ( ; `m+2 )kL2 (
)
i
R `m
`m+1
8" > 0 .
ku ( ; t)kL2 (!) dt
Then
ku ( ; `m )kL2 (
Take d = 2C2
h
dz m+2
1 C1
z C2
R `m
1
z m+1
`1 ` z 1
+" ku ( ; `m+2 )kL2 (
i
)
R `m
ku ( ; t)kL2 (!) dt
8" > 0 .
`m+1
. It gives
ku ( ; `m )kL2 (
"1+ e
)
dz m+2
8" > 0 .
ku ( ; `m+2 )kL2 (
)
, then
( +1)dz m+2
1 C1
z C2
i
h
ku ( ; t)kL2 (!) dt
`m+1
dz m+2
e
)
1
1
`1 ` z(z 1)
" e
Take " = e
1 1 C1 2C2
" z C2 e
R `m
`m+1
ku ( ; `m )kL2 (
)
ku ( ; t)kL2 (!) dt .
5
e
(2+ )dz m+2
ku ( ; `m+2 )kL2 (
)
Take z =
q
+2
+1 ,
then
e (2+
q
)dz m
+1 C1
+2 C2
ku ( ; `m )kL2 (
R `m
(2+ )dz m+2
e
)
ku ( ; t)kL2 (!) dt .
`m+1
ku ( ; `m+2 )kL2 (
)
Change m to 2m0 and sum the above from m0 = 1 to in…nity give the desired
result with E = (0; T ). When E = (0; T ), we take ` = 0 and `1 = T , and get
2
(2+ )2C2 4 T1
e
e (2+
P
)dz 2
e
q
m0 1
P
0
qm
Finally,
r
r
+2
+1
+2
+1
1
3
5
ku ( ; T )kL2 (
ku ( ; `2 )kL2 ( )
(2+ )dz 2m
0
)
ku ( ; `2m0 )kL2 (
R `2m0
+1 C1
1
+2 C2 `2m0 +1 ku ( ; t)kL2 (!)
R
+1 C1 `2
+2 C2 0 ku ( ; t)kL2 (!) dt .
ku ( ; T )kL2 (
r
)
+ 1 C1 2C2 T1 (2+
e
+ 2 C2
)
p
e
)
(2+ )dz 2m
0 +2
ku ( ; `2m0 +2 )kL2 (
)
dt
p
p
+2[
+2+
+1]
Z
T
ku ( ; t)kL2 (!) dt .
0
In particular,
ku ( ; 1)kL2 (
)
r
+ 1 C1 2C2 (2+
e
+ 2 C2
r
+ 1 C1 2C2 (2+
e
+ 2 C2
)
p
p
p
+2[
+2+
+1]
Z
1
ku ( ; t)kL2 (!) dt
0
and
ku ( ; m)kL2 (
for any m
)
)
p
p
+2[
+2+
+1]
p
Z
m
m 1
1. Now, take T such that M < T
ku ( ; t)kL2 (!) dt
M + 1 for some M
1,
T
2
ku ( ; T )kL2 ( )
M ku ( ; M )kL2 ( )
PM
ku ( ; m)kL2 (
q m=1
p
+1 C1 2C2 (2+ )
+2 C2 e
q
p
+1 C1 2C2 (2+ )
+2 C2 e
q
p
+1 C1 2C2 (2+ )
+2 C2 e
)
p
p
+2[
+2+
+1]
Rm
m=1 m 1 ku ( ; t)kL2 (!)
p
p
R
+2[
+2+
+1] M
ku ( ; t)kL2 (!) dt
0
p
p
R
T
+2[
+2+
+1]
ku ( ; t)kL2 (!) dt .
0
We conclude that for any T 1,
r
1
+ 1 C1 2C2 T1 (2+
ku ( ; T )kL2 ( )
e
T
+ 2 C2
6
)
p
PM
p
p
+2[
+2+
+1]
Z
0
dt
T
ku ( ; t)kL2 (!) dt ,
and for any T > 1
ku ( ; T )kL2 (
2
T
)
r
+ 1 C1 2C2 (2+
e
+ 2 C2
Combining the case T
ku ( ; T )kL2 (
)
p
p
p
+2[
+2+
+1]
Z
T
ku ( ; t)kL2 (!) dt .
0
1 and the case T > 1, we get the following observability.
r
Z
3
+ 1 C1 4C2 (1+ T1 )(2+ )2 T
ku ( ; t)kL2 (!) dt
e
T
+ 2 C2
0
)
for any T > 0.
3.3
Proof of "Hölder continuous dependence from one point
in time" ) "Re…ned Observability"
In this proof, C denotes a positive constant which may change of value from
line to line. Let ` 2 [0; T ) and `1 2 (`; T ]. Let z > 1. Introduce the decreasing
sequence f`m gm 1 , which converges to ` given by
`m+1 = ` +
1
(`1
zm
`) .
Then
1
(z 1) (`1 `) > 0 .
zm
We start with the following interpolation estimate. Introduce !
eb!
any 0 t1 < t2 T ,
`m
ku ( ; t2 )kL2 (
)
`m+1 =
C1 t C2t
e 2 1 ku ( ; t2 )kL2 (e!) + " ku ( ; t1 )kL2 (
"
)
8" > 0 .
By Nash and Poincare inequality,
ku ( ; t2 )kL2 (e!)
C3
ku ( ; t2 )kL1 (!) + " kru ( ; t2 )kL2 (
"n=2
8" > 0 .
)
By an energy method,
kru ( ; t2 )kL2 (
C4
)
1=2
(t2
t1 )
ku ( ; t1 )kL2 (
)
Therefore, from the above three estimate, denoting
=
C5 = C1 C3
C1 C4
p
C2
1+
n
n
+ ,
2
2
n=2
, C6 = (n + 1) C2 ,
7
.
. For
we get
ku ( ; t2 )kL2 (
C5 t C6t
e 2 1 ku ( ; t2 )kL1 (!) + " ku ( ; t1 )kL2 (
"
)
Let 0 < `m+2 < `m+1
ku ( ; t)kL2 (
t < `m < T . We get
C5 t
e
"
)
8" > 0 .
)
C6
`m+2
ku ( ; t)kL1 (!) + " ku ( ; `m+2 )kL2 (
8" > 0 .
)
Recall that
ku ( ; `m )kL2 (
ku ( ; t)kL2 (
)
.
)
Therefore,
ku ( ; `m )kL2 (
C5 t
e
"
)
C6
`m+2
ku ( ; t)kL1 (!) + " ku ( ; `m+2 )kL2 (
)
8" > 0 .
Integrating it over t 2 (`m+1 ; `m ), it gives
(`m
`m+1 ) ku ( ; `m )kL2 (
R `m
C6
`m+2
C5 `m+1
" e
)
ku ( ; t)kL1 (!) dt
`m+1 ) ku ( ; `m+2 )kL2 ( ) 8" > 0 .
+" (`m
`m+1
That is
ku ( ; `m )kL2 (
h
)
1
`1
zm
`z 1
i
C5 C6
" e
h
1
z m+1
`1 ` z 1
+" ku ( ; `m+2 )kL2 (
)
i
R `m
`m+1
8" > 0 .
ku ( ; t)kL1 (!) dt
Then
ku ( ; `m )kL2 (
Take d = 2C6
h
dz m+2
1 C5
z C6
q
e
+" ku ( ; `m+2 )kL2 (
i
)
R `m
ku ( ; t)kL1 (!) dt
8" > 0 .
`m+1
. It gives
"1+ e
)
dz m+2
8" > 0 .
ku ( ; `m )kL2 (
)
e
ku ( ; `m+2 )kL2 (
(2+ )dz m+2
ku ( ; t)kL1 (!) dt .
`m+1
)
ku ( ; `m+2 )kL2 (
)
then
(2+ )dz m
1 C5
z C6
1
z m+1
`1 ` z 1
, then
R `m
+2
+1 ,
h
ku ( ; t)kL1 (!) dt
`m+1
( +1)dz m+2
1 C5
z C6
i
ku ( ; `m )kL2 (
R `m
dz m+2
e
Take z =
)
1
1
`1 ` z(z 1)
" e
Take " = e
1 1 C5 2C6
" z C6 e
R `m
`m+1
ku ( ; `m )kL2 (
e
)
ku ( ; t)kL1 (!) dt .
8
(2+ )dz m+2
ku ( ; `m+2 )kL2 (
)
Change m to 2m0 and sum the above from m0 = 1 to in…nity give the desired
result with E = (0; T ). By using a property of density points, we are able
to replace (0; T ) by a measurable set of positive measure (see [PW2]). For
simplicity, when E = (0; T ), we take ` = 0 and `1 = T , and get
2
(2+ )2C6 4 T1
r
r
+2
+1
+2
1
3
5
+1
ku ( ; T )kL2 ( )
e
(2+ )dz 2
e
ku ( ; `2 )kL2 ( )
h
0
P
(2+ )dz 2m
ku ( ; `2m0 )kL2 (
m0 1 e
R
P
1 C5 `2m0
m0 1 z C6 `2m0 +1 ku ( ; t)kL1 (!) dt
R
1 C5 T
z C6 0 ku ( ; t)kL1 (!) dt .
)
e
(2+ )dz 2m
0 +2
ku ( ; `2m0 +2 )kL2 (
That is
ku ( ; T )kL2 (
4
)
r
+ 2 C5 2C6 T1 (2+
e
+ 1 C6
)
p
p
p
+2[
+2+
+1]
Z
T
0
Z
!
ju (x; t)j dxdt .
What I hope
Recall the following two energy identities. For t 2 (0; T ],
Z
Z
1 d
2
2
ju (x; t)j dx +
jru (x; t)j dx = 0 ,
2 dt
Z
Z
1 d
2
2
jru (x; t)j dx +
j u (x; t)j dx = 0 .
2 dt
4.1
Logarithmic convexity method
Recall that if t 7!logf is a convex function, then for any t1 < t2 and any
2 (0; 1),
logf ( t2 + (1
) t1 )
logf (t2 ) + (1
) logf (t1 )
which implies
f ( t2 + (1
1
) t1 )
[f (t2 )] [f (t1 )]
.
00
2
Further, when f 2 C , logf is a convex function if and only if (logf )
2
f 00 f (f 0 )
0.
2
f
9
=
)
i
Now take t = T , t1 = 0, t2 = T and f (t) =
00
(logf (t))
=
1
f2
4
R
2
j u (x; t)j dx
R
R
2
ju (x; t)j dx. Therefore,
2
ju (x; t)j dx
2
0 by Cauchy-Schwarz inequality,
R
2
2
jru (x; t)j dx
which implies
Z
Z
2
ju (x; t)j dx
t=T
2
ju (x; T )j dx
Z
1 t=T
2
.
ju (x; 0)j dx
This estimate has similar form than a quantitative Hölder continuous dependence. We refer to [FEF] for more advanced computation with this technique.
4.2
Weighted logarithmic convexity method
We would like reproduce similar computation than the previous subsection but
with
Z
2
f (t) =
ju (x; t)j G (x; t) dx ,
where G is a suitable weighted function.
Step 1 .- Make appear Br
.- Let Br be the ball of radius r and center
2
x0 , and contained in . Denote m0 = max jx x0 j . Let > 0. Take G (x; t) =
x2
e
jx
T
x0 j2
t+
R
. If logf is a convex function, then
2
R
ju (x; t)j e
2
ju (x; T )j e
1
"(1
)=
1
"(1
jx
T
)=
R
R
x0 j2
t+
jx
dx
x0 j2
t=T
dx
jx
2
ju (x; T )j e
x0 j2
2
jx
x0 j2
2
jx
x0 j2
ju (x; T )j e
R
2
ju (x; 0)j e
dx
dx + "
On the other hand,
=
R
R
R
R
ju (x; T )j e
2
Br
ju (x; T )j e
jx
2
Br
ju (x; T )j dx + e
2
dx
x0 j2
ju (x; T )j dx + e
Br
dx +
r2 R
r2
"
R
R
nBr
e
m0
T t
10
R
R
jx x0 j2
T+
1 t=T
dx
2
ju (x; 0)j e
jx x0 j2
T+
2
ju (x; 0)j e
2
nBr
jx x0 j2
T+
ju (x; T )j e
jx
1
dx
dx
x0 j2
8" > 0 .
dx
2
ju (x; T )j dx
2
ju (x; t)j e
jx
T
x0 j2
t+
dx .
Take t = T =2 i.e.,
R
= 1=2 and
2
ju (x; T =2)j e
jx x0 j2
T =2+
dx
> 0 such that
1
"
R
+ 12
+"
It implies
R
2
ju (x; T =2)j dx
for any " > 0. Take " =
R
2
ju (x; T )j dx
1
2
1
"(1
e
)=
r2
m0
e T =2 = 12 . Then
2
Br
R
ju (x; T )j dx
2
ju (x; T =2)j e
R
2
ju (x; 0)j e
jx x0 j2
T =2+
jx x0 j2
T+
dx
dx
8" > 0 .
m0 R
jx x0 j2
2
e T =2
ju (x; T =2)j e T =2+ dx
m0
m0
R
R
2
2
2e T =2 1" Br ju (x; T )j dx + 2e T =2 " ju (x; 0)j dx ,
R
hR
2
ju (x; T =2)j dx
2
ih
m0
2e T =2
i
2
ju (x; 0)j dx
R
1
. Then
ju (x; T =2)j dx
i1=2 hR
i1=2
hR
m0
2
2
4e T =2 Br ju (x; T )j dx
ju (x; 0)j dx
.
Step 2 .- Get good sign for a derivative. The choice of the weighted function
G can be seen as follows. The quantity
Z
Z
2
2
(@t
) ju (x; t)j G (x; t) dx +
ju (x; t)j (@t + ) G (x; t) dx
where G 2 C 1 , has two expressions
Z
Z
d
2
2
2
ju (x; t)j G (x; t) dx
@ ju (x; t)j G (x; t) ju (x; t)j @ G (x; t) d
dt
@
and
2
Z
2
jru (x; t)j G (x; t) dx +
Z
2
ju (x; t)j (@t +
) G (x; t) dx .
When (@t +
) G (x; t) = 0 and u = 0 on @ ,
Z
Z
1 d
2
2
ju (x; t)j G (x; t) dx +
jru (x; t)j G (x; t) dx = 0 .
2 dt
This can be compared with the …rst energy identity. The natural choice for G
is given by
jx x0 j2
1
G (x; t) =
e 4(T t+ ) ,
n=2
(T t + )
which satis…es (@t +
) G (x; t) = 0 and G (x; T ) =
Of course, we will need to compute
d
dt
11
R
1
n=2
2
e
jx
x0 j2
4
.
jru (x; t)j G (x; t) dx.
Conclusion .- There are two parts: "Make appear Br
" and "Get good
sign for a derivative". With the weighted logarithmic convexity method, "Make
appear Br
" was easy but "Get good sign for a derivative" is di¢ cult and
actually unsolved.
Extension .- The above computation can be extended to the heat equation
with second member. We consider the heat equation in the solution U = U (x; t)
with a suitable second member F = F (x; t)
8
U =F
in
(0; T ) ,
< @t U
U =0
on @
(0; T ) ,
:
U ( ; 0) 2 L2 ( ) .
The two energy identities become as follows. For t 2 (0; T ],
Z
Z
Z
1 d
2
2
jU (x; t)j dx +
jrU (x; t)j dx =
F (x; t) U (x; t) dx ,
2 dt
Z
Z
Z
1 d
2
2
jrU (x; t)j dx +
j U (x; t)j dx =
F (x; t) U (x; t) dx .
2 dt
Further,
R
R
2
2
1 d
jU (x; t)j G (x; t) dx + jrU (x; t)j G (x; t) dx
R2 dt
=
U (x; t) F (x; t) G (x; t) dx .
5
What I can do
5.1
The frequency function
De…ne for t 2 (0; T ],
R
N (t) = R
2
jru (x; t)j dx
2
ju (x; t)j dx
, whenever
Z
2
ju (x; t)j dx 6= 0 .
Then the …rst energy identity becomes as follows.
Z
Z
1 d
2
2
ju (x; t)j dx + N (t)
ju (x; t)j dx = 0 .
2 dt
On the other hand, we study the sign N 0 (t).
d
N (t) = 2
dt
R
2
j u (x; t)j dx
R
2
R
ju (x; t)j dx +
2
ju (x; t)j dx
12
2
R
2
jru (x; t)j dx
2
0
by Cauchy-Schwarz inequality. Therefore, for t 2 [ ; T ],
N (T )
N (t)
N( ) .
Backward uniqueness uses this technique when u ( ; 0) 2 H01 ( ) with N (t)
N (0). We have to solve
Z
Z
1 d
2
2
ju (x; t)j dx + N (0)
0
ju (x; t)j dx .
2 dt
That is
d 2tN (0)
e
dt
0
Integrating over (0; t),
Z
2
ju (x; 0)j dx
Z
2
ju (x; t)j dx
e2tN (0)
Z
.
2
ju (x; t)j dx .
In our problem, we use N (T ) N (t). We have to solve
Z
Z
1 d
2
2
ju (x; t)j dx + N (T )
ju (x; t)j dx 0 .
2 dt
That is
d 2tN (T )
e
dt
Z
2
ju (x; t)j dx
0.
Integrating over (0; T ),
2T N (T )
e
Z
2
ju (x; T )j dx
That is
2T N (T )
R
log R
Z
2
ju (x; 0)j dx .
2
ju (x; 0)j dx
2
ju (x; T )j dx
.
Extension .- The above computation can be extended to the heat equation
with second member. We consider the heat equation in the solution U = U (x; t)
with a suitable second member F = F (x; t)
8
U =F
in
(0; T ) ,
< @t U
U =0
on @
(0; T ) ,
:
U ( ; 0) 2 L2 ( ) .
The following inequality holds (see [BT]).
R
2
jF (x; t)j dx
d
N (t) R
.
2
dt
jU (x; t)j dx
13
Indeed,
d
dt N
(t) = 2
[
R
But
j U j2
R
R
F
2
U dx]
R
R
R
2 R
jrU j2 dx
F U dx
jU j2 dx+( jrU j2 dx)
R
2
( jU j2 dx)
R
2
jrU j dx
F U dx
R
.
2
jrU j dx
2
R
R
R
2
2
1
=
jrU j dx 12
F U dx
F U dx
4
R
R
R
2
2
1
=
U U dx 12
F U dx
F U dx
4
R
R
2
2
1
=
U + 12 F U dx
F U dx
4
R
R
R
2
2
2
1
U + 2 F dx jU j dx 14
F U dx
by Cauchy-Schwarz inequality. Finally,
=
5.2
Let
R
R
R
2
2
jrU j dx
R
F U (x; t) dx
R
2
jrU j dx
R
2
2
U + 21 F dx jU j dx
R
R
R
R
2
2
2
j U j dx jU j dx +
U F dx jU j dx +
2
1
2F
dx
R
2
jU j dx .
The frequency function with weight
> 0. Recall that for t 2 [0; T ]
G (x; t) =
1
(T
e
n=2
t+ )
jx x0 j2
4(T t+ )
.
De…ne for t 2 (0; T ],
R
Z
2
jru (x; t)j G (x; t) dx
2
N (t) = R
, whenever
ju (x; t)j dx 6= 0 .
2
ju (x; t)j G (x; t) dx
Step 1 .- Get good sign for a derivative. We claim that if
star-shaped w.r.t. x0 , then
d
N (t)
dt
That is
d
[(T
dt
1
N (t) .
t+
T
t + ) N (t)]
0.
Integrating over (t; T ),
N (T )
(T + ) N (t) .
14
is convex or
Recall that
1 d
2 dt
Z
2
ju (x; t)j G (x; t) dx +
Z
2
jru (x; t)j G (x; t) dx = 0 .
That is
1 d
2 dt
Z
2
ju (x; t)j G (x; t) dx + N (t)
Z
2
ju (x; t)j G (x; t) dx = 0 .
Therefore,
Z
Z
d
2 N (T )
2
2
ju (x; t)j G (x; t) dx +
ju (x; t)j G (x; t) dx
dt
T+
That is
2t N (T )
d
e T+
dt
Z
2
ju (x; t)j G (x; t) dx
Integrating over (0; T =2),
Z
T N (T )
2
T
+
ju (x; T =2)j G (x; T =2) dx
e
But
R
R
2
ju (x; T )j dx
0.
2
ju (x; 0)j G (x; 0) dx .
2
ju (x; T =2)j dx
m0 R
2
ju (x; T =2)j e
e 2T
2
where m0 = max jx
Z
0.
jx x0 j2
4(T =2+ )
dx .
x0 j . Therefore,
x2
T
e
N (T )
T+
R
R
R
ju(x;0)j2 G (x;0)dx
2
ju(x;T
=2)dx
R =2)j G (x;T
ju(x;0)j2 dx
jx
x0 j2
)
ju(x;T =2)j2 e 4(T =2+
m0 R
e 2T
ju(x;0)j2 dx
R
.
ju(x;T )j2 dx
That is
m0
N (T )
1+
On the other hand,
n
4
n
4
m0
1+
T
Finally,
n
+ N (T )
4
T
e 2T
log R
n
+1
4
e1+ 2T
log R
R
2
ju (x; 0)j dx
T
15
.
2
ju (x; T )j dx
R
2
ju (x; 0)j dx
2
ju (x; T )j dx
m0
1+
dx
e1+ 2T
log R
R
.
2
ju (x; 0)j dx
2
ju (x; T )j dx
.
Now, we prove the claim saying that if is convex or star-shaped w.r.t. x0 ,
then
d
1
N (t)
N (t) .
dt
T t+
x0
First, by an integration by parts and by using rG (x; t) = 2(Tx t+
) G (x; t),
we have
R
R
R
2
jruj G dx =
( uuG + ruurG ) dx + @ @ uuG d
R
x0
=
@t u 2(Tx t+
) ru uG dx .
Next, we compute
R
d
dt
2
d
dt
R
jruj G dx =
2
jruj G dx.
R
2
2ru@t ruG + jruj @t G dx
R
R
2
= 2 ( u@t uG + ru@t urG ) dx
jruj G dx
R
R
2
x x0
= 2 j@t uj G dx + @t u T t+ ruG dx
R
R
2
2
x0
+ r jruj rG dx + @ jruj 2(Tx t+
G d .
)
But
R
R
2
2
r jruj rG dx =
@i j@j uj @i G dx
R
2
= 2 R@j u@ij
u@i G dx
2
2
= 2 R @j u@i u@i G + @j u@i u@ij
G dx
+2 @ @j u@i u j @i G d
i
h
R
(xi x0i )(xj x0j )
@j (xi x0i )
+
= 2
G dx
ururG + @j u@i u
2
2(T t+ )
4(T t+ )
R
2 x x0
j@ uj (T t+ ) G d
@
R
R
R
2
x x0
x0
ruG dx + (T 1t+ )
jruj G dx 2
=
@t u Tx t+
2(T t+ ) ru
R
2
x0
j@ uj (Tx t+
G d .
)
@
Therefore,
R
d
dt
=
G dx
2
jruj G dx
R
2
x0
j@t uj G dx + 2 @t u Tx t+
ruG dx
R
2
1
+ (T t+ )
jruj G dx
R
R
2 x x0
2
j@ uj (T t+ ) G d + @ jruj 2(Tx
@
2
R
2
2
R
x0
t+ )
x x0
2(T t+ )
G d .
That is
d
dt
R
2
jruj G dx =
2
R
+ (T
R
@
@t u
R
1
t+ )
x x0
2(T t+ )
2
2
ru G dx
jruj G dx
x0
j@ uj 2(Tx t+
G d .
)
16
2
2
ru
G dx
Finally, we compute
d
dt N
(t)
d
dt
=
=
+
+
+
R
jruj2 G dx
R
1
2
juj G dx)
R
1
22
juj2 G dx)
(
R
R
d
jruj2 G dx dt
juj2 G dx
juj2 G dx
R
2
2
( juj G dx)
2
R
R
x0
@t u 2(Tx t+
2
) ru G dx
R
(T
2
1
2
juj G dx)
R
(
2
(
(
R
R
1
2
juj G dx)
R
2
+ (T 1t+ ) N
+
(t)
1
2
juj2 G dx)
R
(
=
d
dt N
2
2
jruj G dx
2
x x0
2(T t+ )
2
2
R
jruj G dx
2
@
(t)
R
j@ uj
@
R
1
22
( juj G dx)
1
(T t+ ) N (t)
R
1
t+ )
R
R
x x0
2(T t+ )
@t u
x
x
0
j@ uj2 2(T t+
G d
)
R
2
juj G dx
x x0
2(T t+ )
@t u
2
juj G dx
R
2
G d
juj G dx
2
ru G dx
R
R
R
2
ju (x; T )j e
2
Br
ju (x; T )j e
jx
x0 j2
4
jx
ju (x; T )j dx +
On the other hand,
=
R
R
jx
2
2
x0 j ju (x; T )j e
2
2
juj G dx
2
x0 )
0 for convex
.
2
Br
R
ru uG dx
by Cauchy-Schwarz inequality and the fact that (x
domain .
Step 2 .- Make appear Br
2
juj G dx
jx
dx
R
2
dx + nBr ju (x; T )j e
R
2
2
jx x0 j ju (x; T )j e
x0 j2
4
1
r2
x0 j
4
jx
jx
x0 j2
4
x0 j2
4
dx
dx .
2
dx
jx
x0 j2
4
(x x0 ) ju (x; T )j ( 2 ) re
dx
R
R
jx x0 j2
jx x0 j2
2
2
4
4
= 2 @ ((x x0 ) ) ju (x; T )j e
d + 2 n ju (x; T )j e
dx
R
jx x0 j2
4
+4
(x x0 ) u (x; T ) ru (x; T ) e
dx by integration by parts
R
jx x0 j2
2
4
2 n ju (x; T )j e
dx
R
R
jx x0 j2
jx x0 j2
2
2
2
2
1
4
4
16 jru (x; T )j e
dx + 21
jx x0 j ju (x; T )j e
dx ,
+2
by Cauchy-Schwarz inequality. Therefore,
R
jx x0 j2
2
4
ju (x; T )j e
dx
R
2
ju (x; T )j dx
Br
R
R
jx x0 j2
jx x0 j2
2
2
1
4
4
+ r2 4 n ju (x; T )j e
dx + 16 2 jru (x; T )j e
dx
2
R
R
jx x0 j
2
2
n
4
ju (x; T )j dx + 16
ju (x; T )j e
dx .
r2
4 + N (T )
Br
17
By step 1, if
is convex,
R
jx x0 j2
2
4
ju (x; T )j e
dx
R
2
ju
(x;
T
)j
dx
Br
+ 16
r2
n
4
+1
Take
T
m0 R
e1+ 2T
ju(x;0)j2 dx
R
ju(x;T )j2 dx
2
r
6
T + 4T 2 + 4
16
in order that
Then
log
2
=
16
r2
1+
"
n
+1
4
R
ju (x; T )j dx
=
m0
4
m0
4
T
2
e
1+
R
r
(n+4)m0
R
ju(x;0)j2 dx
R
ju(x;T )j2 dx
2
ju (x; 0)j dx
2
ju (x; T )j dx
2
2
m0
2T
#
jx
x0 j
4
2
dx .
31=2
7
5
=
1
.
2
m0 R
jx x0 j2
2
4
e4
dx
ju (x; T )j e
m0 R
2
4
ju (x; T )j dx .
2e
Br
B
r
@ T + 4T 2 + 4 16
1+
+ 1 log
e1+ 2T
log R
But
m0
4
n
4
2
2
0
2
ju (x; T )j e
T
m0
1+
R
T
2(
n
4 +1
1=2
)log
2
r2
+1
m
1+ 0 R
2T
e
ju(x;0)j2 dx
R
ju(x;T )j2 dx
m0
e1+ 2T
(n + 4) log
R
31=2 1
5 C
A
1
R
ju(x;0)j2 dx
ju(x;T )j2 dx
.
Finally,
R
2
2
ju (x; T )j dx
m0 R
e1+ 2T
ju(x;0)j2 dx
R
ju(x;T )j2 dx
That is
Z
2
ju (x; T )j dx
m
1+ 2T0
e
Z
1=2
2
( n4 +1)
r
1+ (n+4)m
0
2
ju (x; 0)j dx
2m0
r2
+1
2
Z
Br
R
2
Br
ju (x; T )j dx .
1
2
ju (x; T )j dx
where
n
4
+1
1+
r2
(n+4)m0
1=2
=
1+
n
4
+1
1+
r2
(n+4)m0
18
+1
2m0
r2
.
1=2
+1
2m0
r2
,
It implies
R
2
ju (x; T )j dx
m0
R
R
2
2
1
ju (x; T )j dx
" ju (x; 0)j dx
2e(1+ 2T ) 1
Br
" =(1 )
m0
R
R
2
2
1
2e(1+ 2T )
ju (x; T )j dx + " ju (x; 0)j dx 8" > 0 ,
Br
"
where
1
=
n
=
+1
4
1
r2
1+
(n + 4) m0
!
1=2
+1
2m0
.
r2
Conclusion .- There are two parts: "Make appear Br
" and "Get good
sign for a derivative". With the frequency function with weight, "Make appear
Br
" is unusual "Get good sign for a derivative" is more usual.
Now, recall that
ku ( ; T )kL2 (
C1 C2
e T ku ( ; T )kL2 (!) + " ku ( ; 0)kL2 (
"
)
)
8" > 0 ,
implies
ku ( ; T )kL2 (
Take ! = Br ,
3
T
)
r
then using the fact that r2
whenever
5.3
)2
Z
T
ku ( ; t)kL2 (!) dt .
0
= ,
m0
C1 = 2e and C2 =
2
ku ( ; T )kL2 (
+ 1 C1 4C2 (1+ T1 )(2+
e
+ 2 C2
)
C(n)
n
=
+1
4
r2
1+
(n + 4) m0
m0 = max jx
x0 j ,
x2
1=2
!
+1
m20
r2
2
1 1 C(n) m0 (1+ T1 )(1+ m20 )3
r
e
T m0
Z
0
T
ku ( ; t)kL2 (Br ) dt ,
is convex.
The heat equation with space-time potential
The above computation can be extended to the heat equation with second member. We consider the heat equation in the solution U = U (x; t) with a suitable
second member F = F (x; t)
8
U =F
in
(0; T ) ,
< @t U
U =0
on @
(0; T ) ,
:
U ( ; 0) 2 L2 ( ) .
19
The following inequality holds (see [PW]). If
d
N (t)
dt
T
is convex,
R
2
jF (x; t)j G (x; t) dx
1
.
N (t) + R
2
t+
jU (x; t)j G (x; t) dx
Consider the heat equation with potentials a 2 L1 (
n
L1 (
(0; T )) .
8
< @t v
v + av + b rv = 0
v=0
v ( ; 0) 2 L2 ( )
:
in
on @
(0; T )) and b 2
(0; T ) ,
(0; T ) ,
.
Take F = av + b rv. Then
d
N (t)
dt
T
1
N (t) + kakL1 (
t+
(0;T ))
+ kbkL1 (
(0;T ))
N (t) ,
whenever is convex. We are able to reproduce the above treatment to get a
re…ned observability estimate (see [PW2]).
In order to drop the convexity hypothesis, a local study is necessary (see
next section). Take w = u where 2 C01 (BR ) with 0
1 and = 1 in
BR=2 . Then
8
< @t w
Take F =
w=
:
d
N (t)
dt
2r ru
T
2r ru
u
w=0
w ( ; 0) 2 L2 (BR )
in BR (0; T ) ,
on @BR (0; T ) ,
.
u. Then
1
N (t) + 2
t+
We need to bound
R
R
R
2
2
j2r ruj G dx
j
uj G dx
+
2
.
R
R
2
2
j uj G dx
j uj G dx
R
2
2
j2r ruj G dx
j
uj G dx
+ R
R
2
2
j uj G dx
j uj G dx
by an easy function on time in order to reproduce the above treatment. Here,
we can choose t small to do so. Actually, it is not yet done but ideas already
may be set up in [EFV].
20
6
What already exists
Here, we recall most of the material from the works of I. Kukavica [Ku2] and L.
Escauriaza [E] for the elliptic equation and its application to the heat equation.
In the original paper dealing with doubling property and frequency function,
N. Garofalo and F.H. Lin [GaL] study the monotonicity property of the following
quantity
R
2
r B0;r jrv (y)j dy
.
R
2
jv (y)j d (y)
@B0;r
However, it seems more natural in our context to consider the monotonicity
properties of the frequency function (see [Ze]) de…ned by
R
6.1
2
B0;r
2
jrv (y)j r2 jyj
R
2
jv (y)j dy
B0;r
dy
.
Monotonicity formula
Following the ideas of I. Kukavica ([Ku2], [Ku], [KN], see also [E], [AE]), one
obtains the following three lemmas. Detailed proofs are given in [Ph3].
Lemma A .Let D RN +1 , N
1, be a connected bounded open set
D with yo 2 D and Ro > 0. If v = v (y) 2 H 2 (D) is a
such that Byo ;Ro
solution of y v = 0 in D, then
(r) =
R
2
Byo ;r
jrv (y)j
R
Byo ;r
r2
2
jy
yo j
2
dy
jv (y)j dy
is non-decreasing on 0 < r < Ro , and
Z
d
1
2
ln
jv (y)j dy = (N + 1 +
dr
r
Byo ;r
(r)) .
Lemma B .Let D RN +1 , N
1, be a connected bounded open set
such that Byo ;Ro
D with yo 2 D and Ro > 0. Let r1 , r2 , r3 be three real
numbers such that 0 < r1 < r2 < r3 < Ro . If v = v (y) 2 H 2 (D) is a solution
of y v = 0 in D, then
Z
Byo ;r2
2
jv (y)j dy
Z
Byo ;r1
2
jv (y)j dy
21
!
Z
Byo ;r3
2
jv (y)j dy
!1
,
1
where
=
1
r2
r1
ln
1
ln
r2
r1
+
1
ln
2 (0; 1).
r3
r2
The above two results are still available when we are closed to a part of
the boundary @ under the homogeneous Dirichlet boundary condition on ,
as follows.
Lemma C .Let D RN +1 , N
1, be a connected bounded open set
with boundary @D. Let
be a non-empty Lipschitz open subset of @D. Let
ro , r1 , r2 , r3 , Ro be …ve real numbers such that 0 < r1 < ro < r2 < r3 < Ro .
Suppose that yo 2 D satis…es the following three conditions:
i). Byo ;r \ D is star-shaped with respect to yo 8r 2 (0; Ro ) ,
ii). Byo ;r D 8r 2 (0; ro ) ,
iii). Byo ;r \ @D
8r 2 [ro ; Ro ) .
If v = v (y) 2 H 2 (D) is a solution of y v = 0 in D and v = 0 on , then
Z
Z
2
Byo ;r2 \D
jv (y)j dy
2
Byo ;r1
jv (y)j dy
!
Z
2
Byo ;r3 \D
jv (y)j dy
!1
1
where
6.1.1
=
1
1
ln
r2
r1
ln
r2
r1
+
1
ln
r3
r2
2 (0; 1).
Proof of Lemma B
Let
H (r) =
Z
2
Byo ;r
jv (y)j dy .
By applying Lemma A, we know that
d
1
lnH (r) = (N + 1 +
dr
r
(r)) .
Next, from the monotonicity property of , one deduces the following two inequalities
Rr
(r)
2)
ln H(r
= r12 N +1+
dr
H(r1 )
r
(N + 1 + (r2 )) ln rr21 ,
Rr
(r)
3)
ln H(r
= r23 N +1+
dr
H(r2 )
r
(N + 1 + (r2 )) ln rr32 .
Consequently,
ln
H(r2 )
H(r1 )
ln rr12
ln
(N + 1) +
22
(r2 )
H(r3 )
H(r2 )
ln rr32
,
and therefore the desired estimate holds
1
H (r2 )
(H (r1 )) (H (r3 ))
,
1
1
1
1
where
=
6.1.2
Proof of Lemma A
ln
r2
r1
ln
+
r2
r1
.
r3
r2
ln
We introduce the following two functions H and D for 0 < r < Ro :
R
2
H (r) = By ;r jv (y)j dy ,
o
R
2
2
D (r) =
jrv (y)j r2 jy yo j dy .
By ;r
o
RrR
2
First, the derivative of H (r) = 0 S N jv ( s + yo )j N d d (s) is given by
R
2
H 0 (r) = @By ;r jv (y)j d (y). Next, recall the Green formula
o
R
=
R
jvj @ Gd (y)
2
Byo ;r
jvj
Gdy
@
1
r
Byo ;r
o
yv
H 0 (r)
H(r)
=
N +1
r
+
jvj
Gdy .
2
r2
jvj
jy
r
2
2
jy
yo j
dy
jy
yo j
2
2
yo j
2r,
dy
dy .
= 0,
H 0 (r) =
that is
Gd (y)
yo j where Gj@Byo ;r = 0, @ Gj@Byo ;r =
Byo ;r
Consequently, when
2
jvj
2
R
2
1
(N + 1) jvj dy + 2r
Byo ;r
R
N +1
1
= r H (r) + r By ;r div (vrv) r2
R o
2
= N r+1 H (r) + 1r By ;r jrvj + v v
=
R
@Byo ;r
R
2
We apply it with G (y) = r2 jy
and G = 2 (N + 1). It gives
H 0 (r)
R
2
@Byo ;r
1 D(r)
r H(r)
N +1
1
H (r) + D (r) ,
r
r
(A.1)
the second equality in Lemma A.
Now, we compute the derivative of D (r).
D0 (r)
=
d
dr
R
r2
RrR
0
2
SN
(rv)j
s+yo
N
d d (s)
2
r2 (rv)jrs+yo rN d (s)
2
RrR
N
= 2r 0 S N (rv)j s+yo
d d (s)
R
2
= 2r By ;r jrvj dy .
SN
o
23
(A.2)
On the other hand, we have by integrations by parts that
R
R
2
2
2r By ;r jrvj dy = N r+1 D (r) + 4r By ;r j(y yo ) rvj dy
o
o
R
2
1
yo ) v r2 jy yo j dy .
r By ;r rv (y
(A.3)
o
Therefore,
R
2
2
(N + 1) By ;r jrvj r2 jy yo j dy
o
R
R
2
2
= 2r2 By ;r jrvj dy 4 By ;r j(y yo ) rvj dy
o
R o
2
+2 By ;r (y yo ) rv v r2 jy yo j dy ,
o
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) +
j(y yo ) rvj dy .
(A.4)
r
r Byo ;r
The computation of the derivative of
0
(r) =
(r) =
1
[D0 (r) H (r)
H 2 (r)
D(r)
H(r)
gives
D (r) H 0 (r)] ,
which implies using (A.1) and (A.4) that
Z
1
2
H 2 (r) 0 (r) =
4
j(y yo ) rvj dyH (r)
r
Byo ;r
2
!
D (r)
0,
indeed, thanks to an integration by parts and using Cauchy-Schwarz inequality,
we have
2
R
D2 (r) = 4 By ;r vrv (y yo ) dy
R o
R
2
2
4 By ;r j(y yo ) rvj dy
jvj dy
Byo ;r
o
R
2
4 By ;r j(y yo ) rvj dy H (r) .
o
Therefore, we have proved the desired monotonicity for
the proof of Lemma A.
6.1.3
and this completes
Proof of Lemma C
Under the assumption Byo ;r \ @D
for any r 2 [ro ; Ro ), we extend v by zero
in Byo ;Ro nD and denote by v its extension. Since v = 0 on , we have
8
in Byo ;Ro ,
< v = v1D
v=0
on Byo ;Ro \ @D ,
:
rv = rv1D in Byo ;Ro .
24
Now, we denote r = Byo ;r \D, when 0 < r < Ro . In particular,
when 0 < r < ro . We introduce the following three functions:
R
2
H (r) =
jv (y)j dy ,
r
R
2
2
D (r) =
jrv (y)j r2 jy yo j dy ,
r
= Byo ;r ,
r
and
D (r)
0.
H (r)
Our goal is to show that is a non-decreasing function. Indeed, we will prove
that the following equality holds
d
1
d
lnH (r) = (N + 1) lnr +
(r) .
(C.1)
dr
dr
r
Therefore, from the monotonicity of , we will deduce (in a similar way than in
the proof of Lemma A) that
(r) =
ln
H(r2 )
H(r1 )
ln rr21
ln
(N + 1) +
(r2 )
and this will imply the desired estimate
!
Z
Z
2
2
jv (y)j dy
jv (y)j dy
Byo ;r1
r2
Z
H(r3 )
H(r2 )
ln rr32
,
2
r3
jv (y)j dy
!1
,
1
where
=
1
ln
1
r2
r1
ln
r2
r1
+
1
ln
.
r3
r2
R
2
First, we compute the derivative of H (r) = By ;r jv (y)j dy.
o
R
2
H 0 (r) = S N jv (rs + yo )j rN d (s)
R
2
1
= r S N jv (rs + yo )j rs srN d (s)
R
2
= 1r By ;r div jv (y)j (y yo ) dy
o
R
2
2
= 1r By ;r (N + 1) jv (y)j + r jv (y)j (y yo ) dy
o
R
= N r+1 H (r) + 2r r v (y) rv (y) (y yo ) dy .
Next, when
yv
(C.2)
= 0 in D and vj = 0, we remark that
Z
D (r) = 2
v (y) rv (y) (y yo ) dy ,
(C.3)
r
indeed,
R
2
2
jrvj r2 jy yo j dy
h
i
h
R
R
2
= r div vrv r2 jy yo j
dy
vdiv
rv r2
r
R
R
2
=
v v r2 jy yo j dy
vrv r r2 jy
r
r
because
on @Byo ;r , r = jy yo j and vj = 0
R
= 2 r vrv (y yo ) dy because y v = 0 in D .
r
25
jy
2
2
yo j
yo j
dy
i
dy
Consequently, from (C.2) and (C.3), we obtain
H 0 (r) =
N +1
1
H (r) + D (r) ,
r
r
(C.4)
and this is (C.1).
On another hand, the derivative of D (r) is
D0 (r)
= 2r
= 2r
Here, when
2r
R
yv
R0
2
SN
(rv)j
N
d d (s)
s+yo
(C.5)
2
jrv (y)j dy .
r
= 0 in D and vj = 0, we will remark that
2
r
RrR
R
N +1
4
r D (r) + r Byo ;r
R
2
+ 1r \By ;r j@ vj r2
o
jrv (y)j dy
=
indeed,
2
j(y
yo ) rv (y)j dy
jy
2
yo j
(y
yo )
d (y)
(C.6)
R
2
2
(N + 1) r jrvj r2 jy yo j dy
R
2
2
= r div jrvj r2 jy yo j (y yo ) dy
R
2
2
r jrvj r2 jy yo j
(y yo ) dy
r
R
2
2
= \By ;r jrvj r2 jy yo j (y yo ) d (y)
R o
2
2
@yi jrvj r2 jy yo j
(yi yoi ) dy
r
R
2
2
= \By ;r jrvj r2 jy yo j (y yo ) d (y)
R o
2
2rv@yi rv r2 jy yo j (yi yoi ) dy
Rr
2
2
+2 r jrvj jy yo j dy ,
R
and
=
+
+
+
=
R
R
R
R
R
R
r
@yj v@y2i yj v r2
r
@yj (yi
r
@yj (yi
r
r
2
jy
yo j
yoi ) @yj v@yi v r
yoi ) @yj v@yi v r2
(yi
yoi ) @y2j v@yi v
(yi
yoi ) @yj v@yi v@yj r2
\Byo ;r
j
(yi
2
(yi
r
2
jy
2
2
26
2
jy
jy
yo j
2
jy
2
yo j
2
yo j
yoi ) @yj v@yi v r
+ r jrvj r2 jy yo j dy
+0 because y v = 0 in D
R
2
2 j(y yo ) rvj dy .
r
yoi ) dy
dy
dy
dy
2
yo j
dy
jy
yo j
2
d (y)
Therefore, when
=
yv
(N + 1)
R
\Byo ;r
2
R
= 0 in D, we have
R
2
r2
jrvj
r
2
r2
jrvj
2
jy
jy
yo j
2
yo j
@yj v j ((yi
R\Byo ;r 2
R
2
+2r
jruj dy 4
r
dy
(y
yo )
yoi ) @yi v) r
r
2
j(y
=
+2r
R
2
r
2
\Byo ;r
R
2
r
r2
jrvj
j@ vj
r
2
jrvj dy
4
R
jy
r
)
2
jy
2
2
jy
yo ) rvj dy .
By using the fact that vj = 0, we get rv = (rv
(N + 1)
R
d (y)
2
yo j
on
and we deduce that
(y
yo )
d (y)
2
j(y
yo ) rvj dy ,
and this is (C.6). Consequently, from (C.5) and (C.6), when
vj = 0, we have
D0 (r)
=
N +1
4
r D (r) + r
R
+ 1r \By ;r j@
o
R
r
2
vj
j(y
r2
(r) =
yv
= 0 in D and
2
yo ) rv (y)j dy
2
jy
The computation of the derivative of
0
d (y)
dy
2
yo j
yo j
yo j
(y
yo )
(r) =
D(r)
H(r)
gives
1
[D0 (r) H (r)
H 2 (r)
d (y) .
(C.7)
D (r) H 0 (r)] ,
which implies from (C.4) and (C.7), that
H 2 (r)
0
(r)
1
r
=
4
R
+ H(r)
r
r
R
j(y
2
yo ) rv (y)j dy H (r)
\Byo ;r
2
j@ vj
r2
jy
2
yo j
D2 (r)
(y
yo )
d (y)
Thanks to (C.3) and Cauchy-Schwarz inequality, we obtain that
0
4
Z
r
j(y
2
yo ) rv (y)j dy H (r)
D2 (r) .
The inequality 0 (y yo )
on holds when Byo ;r \ D is star-shaped with
respect to yo for any r 2 (0; Ro ). Therefore, we get the desired monotonicity
for which completes the proof of Lemma C.
27
6.2
Quantitative unique continuation property for the Laplacian
Let D RN +1 , N
1, be a connected bounded open set with boundary @D.
Let be a non-empty Lipschitz open part of @D. We consider the Laplacian in
D, with a homogeneous Dirichlet boundary condition on
@ :
8
in D ,
<
yv = 0
v=0
on ,
(D.1)
:
v = v (y) 2 H 2 (D) .
The goal of this section is to describe interpolation inequalities associated to
solutions v of (D.1).
Theorem D .Let ! be a non-empty open subset of D. Then, for any
D, there exist C > 0
D1 D such that @D1 \ @D b and D1 n( \ @D1 )
and 2 (0; 1) such that for any v solution of (D.1), we have
Z
2
D1
jv (y)j dy
Z
C
!
2
jv (y)j dy
Z
1
2
jv (y)j dy
D
.
Or in a equivalent way,
Theorem D’.Let ! be a non-empty open subset of D. Then, for any
D1 D such that @D1 \ @D b and D1 n( \ @D1 )
D, there exist C > 0
and 2 (0; 1) such that for any v solution of (D.1), we have
Z
D1
Z
1
2
jv (y)j dy
C
1
"
!
2
jv (y)j dy + "
Z
D
2
jv (y)j dy
8" > 0 .
Proof of Theorem D .- We divide the proof into two steps.
Step 1 .- We apply Lemma B, and use a standard argument (see 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 2 (0; 1) such that for any v = v (y) 2 H 2 (D),
solution of y v = 0 in D, we have
Z
K2
2
jv (y)j dy
Z
K1
2
jv (y)j dy
Z
D
1
2
jv (y)j dy
.
(D.2)
Step 2 .- We apply Lemma C, and choose yo in a neighborhood of the part
such that the conditions i, ii, iii, hold. Next, by an adequate partition of
D, we deduce from (D.2) that for any D1
D such that @D1 \ @D b
28
and D1 n( \ @D1 )
D, there exist C > 0 and 2 (0; 1) such that for any
v = v (y) 2 H 2 (D) such that y v = 0 on D and v = 0 on , we have
Z
D1
2
jv (y)j dy
C
Z
Z
2
!
jv (y)j dy
1
2
D
jv (y)j dy
.
This completes the proof.
6.3
Quantitative unique continuation property for the elliptic operator @t2 +
In this section, we present the following result.
Theorem E .Let be a bounded open set in Rn , n 1, either convex
2
or C and connected. We choose T2 > T1 and
2 (0; (T2 T1 ) =2). Let
f 2 L2 (
(T1 ; T2 )). We consider the elliptic operator of second order in
(T1 ; T2 ) with a homogeneous Dirichlet boundary condition on @
(T1 ; T2 ),
8 2
in
(T1 ; T2 ) ,
< @t w + w = f
w=0
on @
(T1 ; T2 ) ,
(E.1)
:
w = w (x; t) 2 H 2 (
(T1 ; T2 )) .
Then, for any ' 2 C01 (
(T1 ; T2 )), ' 6= 0, there exist C > 0 and
such that for any w solution of ( E.1), we have
R T2
T1 +
C
R
2
R T2 R
T1
R T2
T1
2 (0; 1)
R
jw (x; t)j dxdt
1
2
jw (x; t)j dxdt
2
j'w (x; t)j dxdt +
R T2 R
T1
2
jf (x; t)j dxdt
.
Proof .- First, by a di¤erence quotient technique and a standard extension at
fT1 ; T2 g, we check the existence of a solution u 2 H 2 (
(T1 ; T2 )) solving
@t2 u +
u=0
u=f
in
on @
(T1 ; T2 ) ,
(T1 ; T2 ) [
fT1 ; T2 g ,
such that
kukH 2 (
(T1 ;T2 ))
c kf kL2 (
(T1 ;T2 ))
,
for some c > 0 only depending on ( ; T1 ; T2 ). Next, we apply Theorem D with
D =
(T1 ; T2 ),
(T1 + ; T2
)
D1 , y = (x; t), y = @t2 + , and
v = w u.
29
6.4
The heat equation and the Hölder continuous dependence from one point in time
In this section, we presents the following result.
Theorem F .or C 2 and connected.
there are C > 0 and
8
<
Let be a bounded open set in Rn , n 1, either convex
Let ! be a nonempty open subset of , and T > 0. Then
2 (0; 1) such that any solution to
@t u
u=0
in
u=0
on @
:
u ( ; 0) 2 L2 ( ) ,
satis…es, for any to 2 (0; T ),
Z
2
ju (x; to )j dx
C
C e to
Z
(0; T ) ,
(0; T ) ,
Z
1
2
ju (x; 0)j dx
!
(F.1)
2
ju (x; to )j dx
.
Proof .- We divide the proof into three steps.
Step 1 .- Let 1 , 2 ,
and e1 ,e2 ,
be the eigenvalues and eigenfunctions
1
of the Laplacian
in H
(
),
constituting
an orthonormal
basis in L2 ( ).
0
R
P
2
For any uo = u ( ; 0) =
uo ej dx, the solution
j ej in L ( ) with
j =
j 1
P
jt
u to (F.1), can be written as u (x; t) =
. Let to 2 (0; T ). We
j ej (x) e
j 1
introduce (see [L] or [CRV]) the function
X
p
j to
w (x; t) =
ch
j ej (x) e
jt
(x)
j 1
where
solves
X
j ej
(x) e
j to
,
j 1
2 C01 (!), = 1 in !
e b !. Recall that cht = (et + e t ) =2. Then, w
8 2
in
(0; T ) ,
@ w+ w= f
>
>
< t
w=0
on @
(0; T ) ,
w
=
@
w
=
0
in
!
e
f0g
,
>
t
>
:
w = w (x; t) 2 H 2 (
(0; T )) ,
where f =
u ( ; to ) 2 H02 ( ).
We denote by w the extension of w
8 2
< @t w + w = f 1 (0;T )
w=0
:
w=0
to zero in !
e
in
in !
e
on @
( T; 0). Then, w solves
(0; T ) [ !
e
( T; 0) ,
(0; T ) .
( T; 0) ,
Now, we construct D, an open connected set in RN +1 , satisfying the following
six conditions:
30
i).
( ;T
) D with 2 (0; T =2) ;
ii). @
( ;T
) @D ;
iii). D
(0; T ) [ !
e ( T; 0) ;
iv). there is an nonempty open ! o b D \ !
e ( To ; 0) with To 2 (0; T ) ;
v). D 2 C 2 when is of class C 2 and connected ;
vi). D is convex with an adequate choice of the pair ( ; To ) when is convex
.
In particular, w 2 H 2 (D).
Step 2 .- We claim that there is g 2 H 2 (
H (D) such that
2
@t2 g +
g=0
g=
f1
(0;T )
( T; T )) \ H01 (
in
( T; T ) ,
on @ (
( T; T )) = @
( T; T ))
( T; T ) [
f Tg ,
and
kgkL2 (D)
kf kL2 (
(0;T ))
.
(F.2)
Indeed, we will proceed in six substep when is of class C 2 and connected (the
case where is convex is well-known because
( T; T ) is then convex). We
de…ne h = f 1 (0;T ) 2 L2 (
( T; T )).
Substep 1: recall that h 2 L2 (
( T; T )) implies the existence of such
g 2 H01 (
( T; T )).
Substep 2: thanks to the interior regularity theorem of elliptic systems, for
any D0 b
( T; T ), g 2 H 2 (D0 ).
Substep 3: thanks to the boundary regularity theorem of elliptic systems,
but far for
f T; T g, g is locally in H 2 because is of class C 2 .
Substep 4: we extend the solution at t = T as follows.
Let h (x; t) = h (x; t) when (x; t) 2
( T; T ) and h (x; t) = h (x; 2T t)
when (x; t) 2
(T; 3T ). Then h 2 L2 (
( T; 3T )). Let g (x; t) = g (x; t)
when (x; t) 2
[ T; T ) and g (x; t) = g (x; 2T t) when (x; t) 2
[T; 3T ].
Then, g solves
@t2 g +
g=0
g=h
in
( T; 3T ) ,
on @ (
( T; 3T )) = @
( T; 3T ) [
f T; 3T g .
By applying the boundary regularity theorem of elliptic systems as in substep
3, we get g 2 H 2 (
(0; 2T )). In particular, g 2 H 2 (
(0; T )).
Substep 5: we extend in a similar way the solution at t = T in order to
conclude that g 2 H 2 (
( T; 0)).
1
Substep 6: Finally, we multiply @t2 g + g = h by (
) g and integrate by
parts over
( T; T ), to get
RT
2
2
k@t gkH 1 ( ) dt + kgkL2 ( ( T;T ))
R TTR
1
= 0
f (x) (
) g (x; t) dxdt
RT R
1
= 0
f (x) (
)(
) g (x; t) dxdt because f 2 H02 ( )
kf kL2 ( (0;T )) kgkL2 ( ( T;T )) by Cauchy-Schwarz .
31
This gives the desired inequality (F.2).
Step 3 .- Finally, we apply Theorem D with y = @t2 + , v = w
where = @
( ;T
) and
( =2 + T =4; 3T =4
=2)
D1
that @D1 \ @D b and D1 n( \ @D1 ) D in order that
Z
Z
1
D1
2
jw
gj dy
1
"
C
!o
2
jw
gj dy + "
Z
D
2
gj dy
jw
g in D
D such
8" > 0 ,
which implies
Z
Z
1
2
D1
jwj dy
C
1
"
2
D
jgj dy + "
Z
2
D
jwj dy
8" 2 (0; 1) ,
where we used the fact that w = 0 in ! o . By (F.2), we conclude that there exist
C > 0 and 2 (0; 1) such that
R 3T =4
=2
=2+T =4
R
2
jw (x; t)j dxdt
Consequently, there exist
RT
R
1
1
"
C
RT R
+"
0
RT R
0
2
jf (x)j dxdt
2
jw (x; t)j dxdt
2 (0; T =2), C > 0 and
8" > 0 .
2 (0; 1) such that
1
RT R
2
jf (x)j dxdt
C 1"
RT R 0
2
+" 0
jw (x; t)j dxdt 8" > 0 .
2
jw (x; t)j dxdt
On the other hand, the following four inequalities hold.
Z
T
0
Z
T
0
Z
Z
2
jw (x; t)j dxdt
P
j 1
2(
2
je
2
jf (x)j dxdt
j to
2T
X
T
Z
!
2(
2
je
j to
2
j (x) u (x; to )j dx ,
p
jT
) + 2T
jT
)
2
!
j 1
p
Z
=
j (x) u (x; to )j dx ,
p
P
2( j to
2
jT )
e
j
p
T
fj 1; j to g
p
P
2( j to
2
jT )
+
e
j
p
T
fj 1; j > to g
2
2 Tto P 2
e
j ,
j 1
RT
R
P
j 1
j ej
(x) e
j to
ch
p
2
jt
dxdt
2
RT
+2T
32
R
!
R
2
jw (x; t)j dxdt
2
j (x) u (x; to )j dx ,
We deduce from the above …ve inequalities that for any " > 0,
P 2 2 j to
P 2 2( j to p j )
(T 2 )
e
(T 2 )
e
j 1
j
j
j 1
RT
2T C
R
1
"
P
j ej
j 1
1
T2
R
(x) e
j to
ch
2
p
2
jt
dxdt
j (x) u (x; to )j dx
!
P 2 R
2
j + ! j (x) u (x; to )j dx
!
+4T " e2 to
j 1
R
2
+2T ! j (x) u (x; to )j dx .
Finally, there is C > 0 such that for any to > 0,
R
2
ju (x; to )j dx =
P
2
2
je
j 1
C
1
"
j to
R
1
C
2
!
ju (x; to )j dx + "e to
which implies the desired estimate of Theorem F,
Z
2
ju (x; to )j dx
C e
C(1
to
)
Z
1
2
ju (x; 0)j dx
R
2
ju (x; 0)j dx 8" 2 (0; 1) ,
Z
!
2
ju (x; to )j dxdt
.
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