observability for heat equations · 2018-12-10 · this talk describes di⁄erent approaches to get...

Observability for heat equations�

Kim Dang PHUNGYangtze 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: [email protected]

Abstract

This talk describes di¤erent approaches to get the observability forheat equations without the use of Carleman inequalities.

Contents

1 The heat equation and observability 2

2 Our motivation 3

3 Our strategy 43.1 Proof of "Hölder continuous dependence from one point in time"

) "Sum of Laplacian eigenfunctions" . . . . . . . . . . . . . . . 43.2 Proof of "Hölder continuous dependence from one point in time"

) "Observability" . . . . . . . . . . . . . . . . . . . . . . . . . . 43.3 Proof of "Hölder continuous dependence from one point in time"

) "Re�ned Observability" . . . . . . . . . . . . . . . . . . . . . 7

4 What I hope 94.1 Logarithmic convexity method . . . . . . . . . . . . . . . . . . . 94.2 Weighted logarithmic convexity method . . . . . . . . . . . . . . 10

5 What I can do 125.1 The frequency function . . . . . . . . . . . . . . . . . . . . . . . . 125.2 The frequency function with weight . . . . . . . . . . . . . . . . . 145.3 The heat equation with space-time potential . . . . . . . . . . . . 19

�This talk was done when the author visited School of Mathematics & Statistics, NortheastNormal University, Changchun, China, (July 4-21, 2011).

1

6 What already exists 216.1 Monotonicity formula . . . . . . . . . . . . . . . . . . . . . . . . 21

6.1.1 Proof of Lemma B . . . . . . . . . . . . . . . . . . . . . . 226.1.2 Proof of Lemma A . . . . . . . . . . . . . . . . . . . . . . 236.1.3 Proof of Lemma C . . . . . . . . . . . . . . . . . . . . . . 24

6.2 Quantitative unique continuation property for the Laplacian . . . 286.3 Quantitative unique continuation property for the elliptic opera-

tor @2t +� . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296.4 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<: @tu��u = 0 in � (0;+1) ,u = 0 on @� (0;+1) ,

u (�; 0) 2 L2() ,(1.1)

living in a bounded open set in Rn, n � 1, either convex or C2 and connected,with boundary @. It is well-known that the above problem is well-posed andhave a unique solution u 2 C

�[0; T ] ;L2 ()

�\ L2

�0; T ;H1

0 ()�for all T > 0.

The observability problem consists in proving the following estimateZ

ju (x; T )j2 dx � CZ T

0

Z!

ju (x; t)j2 dxdt

for some constant C > 0 independent on the initial data. Here, T > 0 and ! isa non-empty open subset in .

2

In the literature, two ways allow to prove such observability estimate. Oneis due to the work of Fursikov and Imanuvilov based on global Carleman in-equalities (see [FI]). The other proof is established by Lebeau and Robbiano(see [LR]. See [Le] for an english version). We resume the Lebeau-Robbianostrategy as follows.

ku (�; T )k2L2() � CZ T

0

Z!

ju (x; t)j2 dxdt

*controllability in �nite and in�nite dimension

*

Xj=1;::;N

jaj j2 � CeCp�N

Z!

��X

j=1;::;N

ajej (x)

��2

dx

for any fajg where (ej ; �j) solves the eigenvalue problem with Dirichlet bound-ary conditions (0 < �1 � �2 � � � � being the corresponding eigenvalues).

*

For any > 0 and any non-trivial ' 2 C10 (� (0; T )), there are C > 0 and� 2 (0; 1), such that for any w 2 H2 (� (0; T )) with

�@2t +�

�w = f and

wj@ = 0, it holds

kwkH1(�( ;T� )) � C kwk�H1(�(0;T ))

�kfkL2(�(0;T )) + k'wkL2(�(0;T ))

�1��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�nedobservability estimate from measure set in time.

ku (�; T )kL2() � CZE

Z!

ju (x; t)j dxdt

for some constant C > 0 independent on the initial data. Here, E � (0; T ) is ameasurable 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 withspace-time potentials.

The approach describes in this talk is linked to parabolic quantitative uniquecontinuation (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)k�L2()

�eC=t ku (�; t)kL2(!)

�1��8t > 0

+

ku (�; T )kL2() � CZE

Z!

ju (x; t)j dxdt

3.1 Proof of "Hölder continuous dependence from one pointin time" ) "Sum of Laplacian eigenfunctions"

Choose u (x; t) =P

j=1;::;N bje��jtej (x). Then it holds for any t > 0

Xj=1;::;N

��bje��jt��2 �0@C X

j=1;::;N

jbj j21A�

0B@CeC=t Z!

��X

j=1;::;N

bje��jtej (x)

��2

dx

1CA1��

which implies when t = 1=p�N

Pj=1;::;N

��bje��j=p�N ��2 ��Ce2

p�NP

j=1;::;N

��bje��j=p�N ��2��CeC

p�NR!

��Pj=1;::;N bje��j=

p�N ej (x)

��2 dx�1�� .

This gives the desired estimate with bje��j=p�N = aj .

3.2 Proof of "Hölder continuous dependence from one pointin time" ) "Observability"

Let ` 2 [0; T ) and `1 2 (`; T ]. Let z > 1. Introduce the decreasing sequencef`mgm�1, which converges to ` given by

`m+1 = `+1

zm(`1 � `) .

4

Then

`m � `m+1 =1

zm(z � 1) (`1 � `) > 0 .

We start with the following interpolation estimate. Let ! be a nonempty opensubset of . For any 0 � t1 < t2 � T ,

ku (�; t2)kL2() �C1" e

C2t2�t1 ku (�; t2)kL2(!) + " ku (�; t1)kL2() 8" > 0 .

Let 0 < `m+2 < `m+1 � t < `m < T . We get

ku (�; t)kL2() �C1" e

C2t�`m+2 ku (�; t)kL2(!) + " ku (�; `m+2)kL2() 8" > 0 .

Recall thatku (�; `m)kL2() � ku (�; t)kL2() .

Therefore,

ku (�; `m)kL2() �C1" e

C2t�`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() � C1" e

C2`m+1�`m+2

R `m`m+1

ku (�; t)kL2(!) dt+" (`m � `m+1) ku (�; `m+2)kL2() 8" > 0 .

That is

ku (�; `m)kL2() �h

1`1�`

zm

z�1

iC1" e

C2h

1`1�`

zm+1

z�1

i R `m`m+1

ku (�; t)kL2(!) dt+" ku (�; `m+2)kL2() 8" > 0 .

Then

ku (�; `m)kL2() � 1"

1zC1C2e2C2

h1

`1�`zm+1

z�1

i R `m`m+1

ku (�; t)kL2(!) dt+" ku (�; `m+2)kL2() 8" > 0 .

Take d = 2C2h

1`1�`

1z(z�1)

i. It gives

" e�dzm+2 ku (�; `m)kL2() � "1+ e�dz

m+2 ku (�; `m+2)kL2()� 1

zC1C2

R `m`m+1

ku (�; t)kL2(!) dt 8" > 0 .

Take " = e�dzm+2

, then

e�( +1)dzm+2 ku (�; `m)kL2() � e�(2+ )dz

m+2 ku (�; `m+2)kL2()� 1

zC1C2

R `m`m+1

ku (�; t)kL2(!) dt .

5

Take z =q

+2 +1 , then

e�(2+ )dzm ku (�; `m)kL2() � e�(2+ )dz

m+2 ku (�; `m+2)kL2()�q

+1 +2

C1C2

R `m`m+1

ku (�; t)kL2(!) dt .

Change m to 2m0 and sum the above from m0 = 1 to in�nity give the desiredresult with E = (0; T ). When E = (0; T ), we take ` = 0 and `1 = T , and get

e�(2+ )2C2

24 1T

r +2 +1r

+2 +1

�1

35ku (�; T )kL2()

� e�(2+ )dz2 ku (�; `2)kL2()�P

m0�1

�e�(2+ )dz

2m0ku (�; `2m0)kL2() � e�(2+ )dz

2m0+2 ku (�; `2m0+2)kL2()�

�P

m0�1

q +1 +2

C1C2

R `2m0`2m0+1

ku (�; t)kL2(!) dt

�q

+1 +2

C1C2

R `20ku (�; t)kL2(!) dt .

Finally,

ku (�; T )kL2() �r + 1

+ 2

C1C2e2C2

1T (2+ )

p +2[

p +2+

p +1]

Z T

0

ku (�; t)kL2(!) dt .

In particular,

ku (�; 1)kL2() �r + 1

+ 2

C1C2e2C2(2+ )

p +2[

p +2+

p +1]

Z 1

0

ku (�; t)kL2(!) dt

and

ku (�;m)kL2() �r + 1

+ 2

C1C2e2C2(2+ )

p +2[

p +2+

p +1]

Z m

m�1ku (�; t)kL2(!) dt

for any m � 1. Now, take T such that M < T �M + 1 for some M � 1,T2 ku (�; T )kL2()

�M ku (�;M)kL2()�PM

m=1 ku (�;m)kL2()�q

+1 +2

C1C2e2C2(2+ )

p +2[

p +2+

p +1]PM

m=1

Rmm�1 ku (�; t)kL2(!) dt

�q

+1 +2

C1C2e2C2(2+ )

p +2[

p +2+

p +1] RM

0ku (�; t)kL2(!) dt

�q

+1 +2

C1C2e2C2(2+ )

p +2[

p +2+

p +1] R T

0ku (�; t)kL2(!) dt .

We conclude that for any T � 1,

ku (�; T )kL2() �1

T

r + 1

+ 2

C1C2e2C2

1T (2+ )

p +2[

p +2+

p +1]

Z T

0

ku (�; t)kL2(!) dt ,

6

and for any T > 1

ku (�; T )kL2() �2

T

r + 1

+ 2

C1C2e2C2(2+ )

p +2[

p +2+

p +1]

Z T

0

ku (�; t)kL2(!) dt .

Combining the case T � 1 and the case T > 1, we get the following observability.

ku (�; T )kL2() �3

T

r + 1

+ 2

C1C2e4C2(1+

1T )(2+ )

2

Z T

0

ku (�; t)kL2(!) dt

for any T > 0.

3.3 Proof of "Hölder continuous dependence from one pointin time" ) "Re�ned Observability"

In this proof, C denotes a positive constant which may change of value fromline to line. Let ` 2 [0; T ) and `1 2 (`; T ]. Let z > 1. Introduce the decreasingsequence f`mgm�1, which converges to ` given by

`m+1 = `+1

zm(`1 � `) .

Then`m � `m+1 =

1

zm(z � 1) (`1 � `) > 0 .

We start with the following interpolation estimate. Introduce e! b ! � . Forany 0 � t1 < t2 � T ,

ku (�; t2)kL2() �C1"�e

C2t2�t1 ku (�; t2)kL2(e!) + " ku (�; t1)kL2() 8" > 0 .

By Nash and Poincare inequality,

ku (�; t2)kL2(e!) � C3"n=2

ku (�; t2)kL1(!) + " kru (�; t2)kL2() 8" > 0 .

By an energy method,

kru (�; t2)kL2() �C4

(t2 � t1)1=2ku (�; t1)kL2() .

Therefore, from the above three estimate, denoting

= ��1 +

n

2

�+n

2,

C5 = C1C3

�C1C4pC2

�n=2, C6 = (n+ 1)C2 ,

7

we get

ku (�; t2)kL2() �C5" e

C6t2�t1 ku (�; t2)kL1(!) + " ku (�; t1)kL2() 8" > 0 .

Let 0 < `m+2 < `m+1 � t < `m < T . We get

ku (�; t)kL2() �C5" e

C6t�`m+2 ku (�; t)kL1(!) + " ku (�; `m+2)kL2() 8" > 0 .

Recall thatku (�; `m)kL2() � ku (�; t)kL2() .

Therefore,

ku (�; `m)kL2() �C5" e

C6t�`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() � C5" e

C6`m+1�`m+2

R `m`m+1

ku (�; t)kL1(!) dt+" (`m � `m+1) ku (�; `m+2)kL2() 8" > 0 .

That is

ku (�; `m)kL2() �h

1`1�`

zm

z�1

iC5" e

C6h

1`1�`

zm+1

z�1

i R `m`m+1

ku (�; t)kL1(!) dt+" ku (�; `m+2)kL2() 8" > 0 .

Then

ku (�; `m)kL2() � 1"

1zC5C6e2C6

h1

`1�`zm+1

z�1

i R `m`m+1

ku (�; t)kL1(!) dt+" ku (�; `m+2)kL2() 8" > 0 .

Take d = 2C6h

1`1�`

1z(z�1)

i. It gives

" e�dzm+2 ku (�; `m)kL2() � "1+ e�dz

m+2 ku (�; `m+2)kL2()� 1

zC5C6

R `m`m+1

ku (�; t)kL1(!) dt 8" > 0 .

Take " = e�dzm+2

, then

e�( +1)dzm+2 ku (�; `m)kL2() � e�(2+ )dz

m+2 ku (�; `m+2)kL2()� 1

zC5C6

R `m`m+1

ku (�; t)kL1(!) dt .

Take z =q

+2 +1 , then

e�(2+ )dzm ku (�; `m)kL2() � e�(2+ )dz

m+2 ku (�; `m+2)kL2()� 1

zC5C6

R `m`m+1

ku (�; t)kL1(!) dt .

8

Change m to 2m0 and sum the above from m0 = 1 to in�nity give the desiredresult with E = (0; T ). By using a property of density points, we are ableto replace (0; T ) by a measurable set of positive measure (see [PW2]). Forsimplicity, when E = (0; T ), we take ` = 0 and `1 = T , and get

e�(2+ )2C6

24 1T

r +2 +1r

+2 +1

�1

35ku (�; T )kL2()

� e�(2+ )dz2 ku (�; `2)kL2()�P

m0�1

he�(2+ )dz

2m0ku (�; `2m0)kL2() � e�(2+ )dz

2m0+2 ku (�; `2m0+2)kL2()i

�P

m0�11zC5C6

R `2m0`2m0+1

ku (�; t)kL1(!) dt� 1

zC5C6

R T0ku (�; t)kL1(!) dt .

That is

ku (�; T )kL2() �r + 2

+ 1

C5C6e2C6

1T (2+ )

p +2[

p +2+

p +1]

Z T

0

Z!

ju (x; t)j dxdt .

4 What I hope

Recall the following two energy identities. For t 2 (0; T ],

1

2

d

dt

Z

ju (x; t)j2 dx+Z

jru (x; t)j2 dx = 0 ,

1

2

d

dt

Z

jru (x; t)j2 dx+Z

j�u (x; t)j2 dx = 0 .

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 impliesf (�t2 + (1� �) t1) � [f (t2)]� [f (t1)]1�� .

Further, when f 2 C2, logf is a convex function if and only if (logf)00 =f 00f�(f 0)

2

f2 � 0.

9

Now take t = �T , t1 = 0, t2 = T and f (t) =Rju (x; t)j2 dx. Therefore,

(logf (t))00 = 1f2

�4Rj�u (x; t)j2 dx

Rju (x; t)j2 dx�

��2Rjru (x; t)j2 dx

�2�� 0 by Cauchy-Schwarz inequality,

which impliesZ

ju (x; t)j2 dx ��Z

ju (x; T )j2 dx�t=T �Z

ju (x; 0)j2 dx�1�t=T

.

This estimate has similar form than a quantitative Hölder continuous depen-dence. 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 butwith

f (t) =

Z

ju (x; t)j2G (x; t) dx ,

where G is a suitable weighted function.

Step 1 .- Make appear Br � .- Let Br be the ball of radius r and centerx0, and contained in . Denote m0 = max

x2jx� x0j2. Let � > 0. Take G (x; t) =

e�jx�x0j2T�t+� . If logf is a convex function, then

Rju (x; t)j2 e�

jx�x0j2T�t+� dx

��Rju (x; T )j2 e�

jx�x0j2� dx

�t=T �Rju (x; 0)j2 e�

jx�x0j2T+� dx

�1�t=T��

1"(1��)=�

Rju (x; T )j2 e�

jx�x0j2� dx

�� "Rju (x; 0)j2 e�

jx�x0j2T+� dx

�1�� 1

"(1��)=�

Rju (x; T )j2 e�

jx�x0j2� dx+ "

Rju (x; 0)j2 e�

jx�x0j2T+� dx 8" > 0 .

On the other hand,

Rju (x; T )j2 e�

jx�x0j2� dx

=RBrju (x; T )j2 e�

jx�x0j2� dx+

RnBr

ju (x; T )j2 e�jx�x0j2

� dx

�RBrju (x; T )j2 dx+ e� r2

�

RnBr

ju (x; T )j2 dx

�RBrju (x; T )j2 dx+ e� r2

� em0T�t

Rju (x; t)j2 e�

jx�x0j2T�t+� dx .

10

Take t = T=2 i.e., � = 1=2 and � > 0 such that 1"(1��)=�

e�r2

� em0T=2 = 1

2 . ThenRju (x; T=2)j2 e�

jx�x0j2T=2+� dx � 1

"

RBrju (x; T )j2 dx

+ 12

Rju (x; T=2)j2 e�

jx�x0j2T=2+� dx

+"Rju (x; 0)j2 e�

jx�x0j2T+� dx 8" > 0 .

It impliesRju (x; T=2)j2 dx � e

m0T=2Rju (x; T=2)j2 e�

jx�x0j2T=2+� dx

� 2em0T=2 1

"

RBrju (x; T )j2 dx+ 2e

m0T=2 "

Rju (x; 0)j2 dx ,

for any " > 0. Take " = 12

hRju (x; T=2)j2 dx

i h2e

m0T=2Rju (x; 0)j2 dx

i�1. ThenR

ju (x; T )j2 dx �

Rju (x; T=2)j2 dx

� 4em0T=2

hRBrju (x; T )j2 dx

i1=2 hRju (x; 0)j2 dx

i1=2.

Step 2 .- Get good sign for a derivative. The choice of the weighted functionG can be seen as follows. The quantityZ

(@t ��)�ju (x; t)j2

�G (x; t) dx+

Z

ju (x; t)j2 (@t +�)G (x; t) dx

where G 2 C1, has two expressionsZ

d

dt

�ju (x; t)j2G (x; t)

�dx�

Z@

@�

�ju (x; t)j2

�G (x; t)�ju (x; t)j2 @�G (x; t) d�

and

�2Z

jru (x; t)j2G (x; t) dx+Z

ju (x; t)j2 (@t +�)G (x; t) dx .

When (@t +�)G (x; t) = 0 and u = 0 on @,

1

2

d

dt

Z

ju (x; t)j2G� (x; t) dx+Z

jru (x; t)j2G� (x; t) dx = 0 .

This can be compared with the �rst energy identity. The natural choice for Gis given by

G� (x; t) =1

(T � t+ �)n=2e�

jx�x0j24(T�t+�) ,

which satis�es (@t +�)G� (x; t) = 0 and G� (x; T ) = 1�n=2

e�jx�x0j2

4� .

Of course, we will need to compute ddt

Rjru (x; t)j2G� (x; t) dx.

11

Conclusion .- There are two parts: "Make appear Br � " and "Get goodsign for a derivative". With the weighted logarithmic convexity method, "Makeappear Br � " was easy but "Get good sign for a derivative" is di¢ cult andactually unsolved.

Extension .- The above computation can be extended to the heat equationwith second member. We consider the heat equation in the solution U = U(x; t)with a suitable second member F = F (x; t)8<: @tU ��U = F in � (0; T ) ,

U = 0 on @� (0; T ) ,U (�; 0) 2 L2() .

The two energy identities become as follows. For t 2 (0; T ],

1

2

d

dt

Z

jU (x; t)j2 dx+Z

jrU (x; t)j2 dx =Z

F (x; t)U (x; t) dx ,

1

2

d

dt

Z

jrU (x; t)j2 dx+Z

j�U (x; t)j2 dx = �Z

F (x; t)�U (x; t) dx .

Further,

12ddt

RjU (x; t)j2G� (x; t) dx+

RjrU (x; t)j2G� (x; t) dx

=RU (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 ],

N (t) =

Rjru (x; t)j2 dxRju (x; t)j2 dx

, wheneverZ

ju (x; t)j2 dx 6= 0 .

Then the �rst energy identity becomes as follows.

1

2

d

dt

Z

ju (x; t)j2 dx+N (t)Z

ju (x; t)j2 dx = 0 .

On the other hand, we study the sign N 0 (t).

d

dtN (t) = 2

�Rj�u (x; t)j2 dx

Rju (x; t)j2 dx+

�Rjru (x; t)j2 dx

�2�R

ju (x; t)j2 dx

�2 � 0

12

by Cauchy-Schwarz inequality. Therefore, for t 2 [�; T ],

N (T ) � N (t) � N (�) .

Backward uniqueness uses this technique when u (�; 0) 2 H10 () with N (t) �

N (0). We have to solve

0 � 1

2

d

dt

Z

ju (x; t)j2 dx+N (0)Z

ju (x; t)j2 dx .

That is

0 � d

dt

�e2tN(0)

Z

ju (x; t)j2 dx�.

Integrating over (0; t),Z

ju (x; 0)j2 dx � e2tN(0)Z

ju (x; t)j2 dx .

In our problem, we use N (T ) � N (t). We have to solve

1

2

d

dt

Z

ju (x; t)j2 dx+N (T )Z

ju (x; t)j2 dx � 0 .

That isd

dt

�e2tN(T )

Z

ju (x; t)j2 dx�� 0 .

Integrating over (0; T ),

e2TN(T )Z

ju (x; T )j2 dx �Z

ju (x; 0)j2 dx .

That is

2TN (T ) � logRju (x; 0)j2 dxR

ju (x; T )j2 dx

.

Extension .- The above computation can be extended to the heat equationwith second member. We consider the heat equation in the solution U = U(x; t)with a suitable second member F = F (x; t)8<: @tU ��U = F in � (0; T ) ,

U = 0 on @� (0; T ) ,U (�; 0) 2 L2() .

The following inequality holds (see [BT]).

d

dtN (t) �

RjF (x; t)j2 dxR

jU (x; t)j2 dx

.

13

Indeed,

ddtN (t) = 2

[�Rj�U j2�

RF�Udx]

RjU j2dx+(

RjrU j2dx)

2�RFUdx

RjrU j2dx

(RjU j2dx)

2 .

But �RjrU j2 dx

�2�RFUdx

RjrU j2 dx

=�R

jrU j2 dx� 1

2

RFUdx

�2� 1

4

�RFUdx

�2=��R�UUdx� 1

2

RFUdx

�2 � 14

�RFUdx

�2=��R

��U + 1

2F�Udx

�2 � 14

�RFUdx

�2�R

��U + 12F��2 dx R

jU j2 dx� 1

4

�RFUdx

�2by Cauchy-Schwarz inequality. Finally,�R

jrU j2 dx

�2�RFU (x; t) dx

RjrU j2 dx

�R

��U + 12F��2 dx R

jU j2 dx

=Rj�U j2 dx

RjU j2 dx+

R�UFdx

RjU j2 dx+

R

�� 12F��2 dx R

jU j2 dx .

5.2 The frequency function with weight

Let � > 0. Recall that for t 2 [0; T ]

G� (x; t) =1

(T � t+ �)n=2e�

jx�x0j24(T�t+�) .

De�ne for t 2 (0; T ],

N� (t) =

Rjru (x; t)j2G� (x; t) dxRju (x; t)j2G� (x; t) dx

, wheneverZ

ju (x; t)j2 dx 6= 0 .

Step 1 .- Get good sign for a derivative. We claim that if is convex orstar-shaped w.r.t. x0, then

d

dtN� (t) �

1

T � t+ �N� (t) .

That isd

dt[(T � t+ �)N� (t)] � 0 .

Integrating over (t; T ),

�N� (T ) � (T + �)N� (t) .

14

Recall that

1

2

d

dt

Z

ju (x; t)j2G� (x; t) dx+Z

jru (x; t)j2G� (x; t) dx = 0 .

That is

1

2

d

dt

Z

ju (x; t)j2G� (x; t) dx+N� (t)Z

ju (x; t)j2G� (x; t) dx = 0 .

Therefore,

d

dt

Z

ju (x; t)j2G� (x; t) dx+2�N� (T )

T + �

Z

ju (x; t)j2G� (x; t) dx � 0 .

That isd

dt

�e2t�N�(T )

T+�

Z

ju (x; t)j2G� (x; t) dx�� 0 .

Integrating over (0; T=2),

eT�N�(T )

T+�

Z

ju (x; T=2)j2G� (x; T=2) dx �Z

ju (x; 0)j2G� (x; 0) dx .

But Rju (x; T )j2 dx �

Rju (x; T=2)j2 dx

� em02T

Rju (x; T=2)j2 e�

jx�x0j24(T=2+�) dx .

where m0 = maxx2

jx� x0j2. Therefore,

eT�N�(T )

T+� �Rju(x;0)j2G�(x;0)dxR

ju(x;T=2)j2G�(x;T=2)dx

�Rju(x;0)j2dxR

ju(x;T=2)j2e�

jx�x0j24(T=2+�) dx

� em02T

Rju(x;0)j2dxR

ju(x;T )j2dx .

That is

�N� (T ) ��1 +

�

T

�logem02T

Rju (x; 0)j2 dxR

ju (x; T )j2 dx

.

On the other hand,

n

4� n

4

�1 +

�

T

�loge1+

m02T

Rju (x; 0)j2 dxR

ju (x; T )j2 dx

.

Finally,

n

4+ �N� (T ) �

�n4+ 1��1 +

�

T

�loge1+

m02T

Rju (x; 0)j2 dxR

ju (x; T )j2 dx

.

15

Now, we prove the claim saying that if is convex or star-shaped w.r.t. x0,then

d

dtN� (t) �

1

T � t+ �N� (t) .

First, by an integration by parts and by usingrG� (x; t) = � x�x02(T�t+�)G� (x; t),

we haveRjruj2G�dx = �

R(�uuG� +ruurG�) dx+

R@@�uuG�d�

= �R

�@tu� x�x0

2(T�t+�) � ru�uG�dx .

Next, we compute ddt

Rjruj2G�dx.

ddt

Rjruj2G�dx =

R

�2ru@truG� + jruj2 @tG�

�dx

= �2R(�u@tuG� +ru@turG�) dx�

Rjruj2�G�dx

= �2Rj@tuj2G�dx+

R@tu

x�x0T�t+� � ruG�dx

+Rr�jruj2

�rG�dx+

R@jruj2 x�x0

2(T�t+�) � �G�d� .

But Rr�jruj2

�rG�dx =

R@i

�j@juj2

�@iG�dx

= 2R@ju@

2iju@iG�dx

= �2R

�@2j u@iu@iG� + @ju@iu@

2ijG�

�dx

+2R@@ju@iu�j@iG�d�

= �2R

��ururG� + @ju@iu

h�@j(xi�x0i)

2(T�t+�) +(xi�x0i)(xj�x0j)

4(T�t+�)2

iG�

�dx

�R@j@�uj2 x�x0

(T�t+�) � �G�d�

=R@tu

x�x0T�t+� � ruG�dx+

1(T�t+�)

Rjruj2G�dx� 2

R

�x�x0

2(T�t+�) � ru�2G�dx


(T�t+�) � �G�d� .

Therefore,

ddt

Rjruj2G�dx

= �2Rj@tuj2G�dx+ 2

R@tu

x�x0T�t+� � ruG�dx� 2

R

�x�x0

2(T�t+�) � ru�2G�dx

+ 1(T�t+�)

Rjruj2G�dx


(T�t+�) � �G�d� +R@jruj2 x�x0

2(T�t+�) � �G�d� .

That is

ddt

Rjruj2G�dx = �2

R

��@tu� x�x02(T�t+�) � ru

��2G�dx+ 1(T�t+�)

Rjruj2G�dx


2(T�t+�) � �G�d� .

16

Finally, we compute ddtN� (t)

ddtN� (t) =

ddt

Rjruj2G�dx

Rjuj2G�dx�

Rjruj2G�dx

ddt

Rjuj2G�dx

(Rjuj2G�dx)

2

= 1

(Rjuj2G�dx)

2

��2R

��@tu� x�x02(T�t+�) � ru

��2G�dx R juj2G�dx�+ 1

(Rjuj2G�dx)

2

�1

(T�t+�)Rjruj2G�dx

Rjuj2G�dx

�+ 1

(Rjuj2G�dx)

2

��R@j@�uj2 x�x0

2(T�t+�) � �G�d�Rjuj2G�dx

�+ 1

(Rjuj2G�dx)

2 2�R

jruj2G�dx

�2= 1

(Rjuj2G�dx)

2

��2R

��@tu� x�x02(T�t+�) � ru

��2G�dx R juj2G�dx�+ 1(T�t+�)N� (t)�

R@j@�uj2 x�x0

2(T�t+�) ��G�d�Rjuj2G�dx

+ 1

(Rjuj2G�dx)

2 2��R

�@tu� x�x0

2(T�t+�) � ru�uG�dx

�2� 1

(T�t+�)N� (t)

by Cauchy-Schwarz inequality and the fact that (x� x0) � � � 0 for convexdomain .

Step 2 .- Make appear Br � .Rju (x; T )j2 e�

jx�x0j24� dx

�RBrju (x; T )j2 e�

jx�x0j24� dx+

RnBr

ju (x; T )j2 e�jx�x0j2

4� dx

�RBrju (x; T )j2 dx+ 1

r2

Rjx� x0j2 ju (x; T )j2 e�

jx�x0j24� dx .

On the other hand,Rjx� x0j2 ju (x; T )j2 e�

jx�x0j24� dx

=R(x� x0) ju (x; T )j2 � (�2�)re�

jx�x0j24� dx

= �2�R@((x� x0) � �) ju (x; T )j2 e�

jx�x0j24� d� + 2�n

Rju (x; T )j2 e�

jx�x0j24� dx

+4�R(x� x0)u (x; T ) � ru (x; T ) e�

jx�x0j24� dx by integration by parts

� 2�nRju (x; T )j2 e�

jx�x0j24� dx

+ 12

R16�2 jru (x; T )j2 e�

jx�x0j24� dx+ 1

2

Rjx� x0j2 ju (x; T )j2 e�

jx�x0j24� dx ,

by Cauchy-Schwarz inequality. Therefore,Rju (x; T )j2 e�

jx�x0j24� dx

�RBrju (x; T )j2 dx

+ 1r2

�4�n

Rju (x; T )j2 e�

jx�x0j24� dx+

R16�2 jru (x; T )j2 e�

jx�x0j24� dx

��RBrju (x; T )j2 dx+ 16�

r2

�n4 + �N� (T )

� Rju (x; T )j2 e�

jx�x0j24� dx .

17

By step 1, if is convex,Rju (x; T )j2 e�

jx�x0j24� dx

�RBrju (x; T )j2 dx

+ 16�r2

��n4 + 1

� �1 + �

T

�log

e1+m02T

Rju(x;0)j2dxR

ju(x;T )j2dx

� Rju (x; T )j2 e�

jx�x0j24� dx .

Take

� = �T +

264T 2 + 4 r216

T

2�n4 + 1

�log

e1+m02T

Rju(x;0)j2dxR

ju(x;T )j2dx

3751=2

in order that

16�

r2

"�n4+ 1��1 +

�

T

�loge1+

m02T

Rju (x; 0)j2 dxR

ju (x; T )j2 dx

#=1

2.

Then Rju (x; T )j2 dx � e

m04�

Rju (x; T )j2 e�

jx�x0j24� dx

� 2em04�

RBrju (x; T )j2 dx .

But

m0

4� = m0

4

0B@�T +24T 2 + 4 r216 T

2(n4+1)loge1+

m02T

Rju(x;0)j2dxR

ju(x;T )j2dx

351=21CA�1

� m0

4

��1 + r2

(n+4)m0

�1=2+ 1

�2r2 (n+ 4) log

e1+m02T

Rju(x;0)j2dxR

ju(x;T )j2dx .

Finally,Rju (x; T )j2 dx

� 2�e1+

m02T

Rju(x;0)j2dxR

ju(x;T )j2dx

�(n4+1)��1+ r2

(n+4)m0

�1=2+1

�2m0r2 R

Brju (x; T )j2 dx .

That isZ

ju (x; T )j2 dx ��e1+

m02T

Z

ju (x; 0)j2 dx��

2

ZBr

ju (x; T )j2 dx�1��

,

where

� =

�n4 + 1

��1 + r2

(n+4)m0

�1=2+ 1

�2m0

r2

1 +�n4 + 1

��1 + r2

(n+4)m0

�1=2+ 1

�2m0

r2

.

18

It impliesRju (x; T )j2 dx

��"Rju (x; 0)j2 dx

�� 1

"�=(1��)2e(1+

m02T )

�1��

RBrju (x; T )j2 dx

�1�� 1

"�2e(1+

m02T )�

RBrju (x; T )j2 dx+ "

Rju (x; 0)j2 dx 8" > 0 ,

where

� =�

1� � =�n4+ 1� �

1 +r2

(n+ 4)m0

�1=2+ 1

!2m0

r2.

Conclusion .- There are two parts: "Make appear Br � " and "Get goodsign for a derivative". With the frequency function with weight, "Make appearBr � " is unusual "Get good sign for a derivative" is more usual.

Now, recall that

ku (�; T )kL2() �C1" eC2T ku (�; T )kL2(!) + " ku (�; 0)kL2() 8" > 0 ,

implies

ku (�; T )kL2() �3

T

r + 1

+ 2

C1C2e4C2(1+

1T )(2+ )

2

Z T

0

ku (�; t)kL2(!) dt .

Take ! = Br, = �,

C1 = 2e� and C2 =

m0

2� =

�n4+ 1� �

1 +r2

(n+ 4)m0

�1=2+ 1

!m20

r2

then using the fact that r2 � m0 = maxx2

jx� x0j2,

ku (�; T )kL2() � C(n)1

T

1

m0eC(n)m0(1+ 1

T )(1+m0r2)3Z T

0

ku (�; t)kL2(Br)dt ,

whenever is convex.

5.3 The heat equation with space-time potential

The above computation can be extended to the heat equation with second mem-ber. We consider the heat equation in the solution U = U(x; t) with a suitablesecond member F = F (x; t)8<: @tU ��U = F in � (0; T ) ,

U = 0 on @� (0; T ) ,U (�; 0) 2 L2() .

19

The following inequality holds (see [PW]). If is convex,

d

dtN� (t) �

1

T � t+ �N� (t) +RjF (x; t)j2G� (x; t) dxR

jU (x; t)j2G� (x; t) dx

.

Consider the heat equation with potentials a 2 L1 (� (0; T )) and b 2L1 (� (0; T ))n .8<: @tv ��v + av + b � rv = 0 in � (0; T ) ,

v = 0 on @� (0; T ) ,v (�; 0) 2 L2() .

Take F = av + b � rv. Then

d

dtN� (t) �

1

T � t+ �N� (t) + kakL1(�(0;T )) + kbkL1(�(0;T ))N� (t) ,

whenever is convex. We are able to reproduce the above treatment to get are�ned observability estimate (see [PW2]).

In order to drop the convexity hypothesis, a local study is necessary (seenext section). Take w = �u where � 2 C10 (BR) with 0 � � � 1 and � = 1 inBR=2. Then8<: @tw ��w = �2r�ru��u in BR � (0; T ) ,

w = 0 on @BR � (0; T ) ,w (�; 0) 2 L2(BR) .

Take F = �2r�ru��u. Then

d

dtN� (t) �

1

T � t+ �N� (t) + 2Rj2r�ruj2G�dxRj�uj2G�dx

+ 2

Rj��uj2G�dxRj�uj2G�dx

.

We need to bound Rj2r�ruj2G�dxRj�uj2G�dx

+

Rj��uj2G�dxRj�uj2G�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 alreadymay 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 followingquantity

rRB0;r

jrv (y)j2 dyR@B0;r

jv (y)j2 d� (y).

However, it seems more natural in our context to consider the monotonicityproperties of the frequency function (see [Ze]) de�ned byR

B0;rjrv (y)j2

�r2 � jyj2

�dyR

B0;rjv (y)j2 dy

.

6.1 Monotonicity formula

Following the ideas of I. Kukavica ([Ku2], [Ku], [KN], see also [E], [AE]), oneobtains the following three lemmas. Detailed proofs are given in [Ph3].

Lemma A .- Let D � RN+1, N � 1, be a connected bounded open setsuch that Byo;Ro � D with yo 2 D and Ro > 0. If v = v (y) 2 H2 (D) is asolution of �yv = 0 in D, then

� (r) =

RByo;r

jrv (y)j2�r2 � jy � yoj2

�dyR

Byo;rjv (y)j2 dy

is non-decreasing on 0 < r < Ro , and

d

drlnZByo;r

jv (y)j2 dy = 1

r(N + 1 + � (r)) .

Lemma B .- Let D � RN+1, N � 1, be a connected bounded open setsuch that Byo;Ro

� D with yo 2 D and Ro > 0. Let r1, r2, r3 be three realnumbers such that 0 < r1 < r2 < r3 < Ro. If v = v (y) 2 H2 (D) is a solutionof �yv = 0 in D, thenZ

Byo;r2

jv (y)j2 dy � Z

Byo;r1

jv (y)j2 dy!� Z

Byo;r3

jv (y)j2 dy!1��

,

21

where � = 1ln r2r1

�1

ln r2r1+ 1

ln r3r2

��12 (0; 1).

The above two results are still available when we are closed to a part � ofthe boundary @ under the homogeneous Dirichlet boundary condition on �,as follows.

Lemma C .- Let D � RN+1, N � 1, be a connected bounded open setwith boundary @D. Let � be a non-empty Lipschitz open subset of @D. Letro, 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 H2 (D) is a solution of �yv = 0 in D and v = 0 on �, then

ZByo;r2

\Djv (y)j2 dy �

ZByo;r1

jv (y)j2 dy!� Z

Byo;r3\Djv (y)j2 dy

!1��


�1

ln r2r1+ 1

ln r3r2

��12 (0; 1).

6.1.1 Proof of Lemma B

Let

H (r) =

ZByo;r

jv (y)j2 dy .

By applying Lemma A, we know that

d

drlnH (r) =

1

r(N + 1 + � (r)) .

Next, from the monotonicity property of �, one deduces the following two in-equalities

ln�H(r2)H(r1)

�=R r2r1

N+1+�(r)r dr

� (N + 1 + � (r2)) ln r2r1 ,

ln�H(r3)H(r2)

�=R r3r2

N+1+�(r)r dr

� (N + 1 + � (r2)) ln r3r2 .

Consequently,

ln�H(r2)H(r1)

�ln r2r1

� (N + 1) + � (r2) �ln�H(r3)H(r2)

�ln r3r2

,

22

and therefore the desired estimate holds

H (r2) � (H (r1))� (H (r3))1�� ,


�1

ln r2r1+ 1

ln r3r2

��1.

6.1.2 Proof of Lemma A

We introduce the following two functions H and D for 0 < r < Ro :

H (r) =RByo;r

jv (y)j2 dy ,D (r) =

RByo;r

jrv (y)j2�r2 � jy � yoj2

�dy .

First, the derivative of H (r) =R r0

RSNjv (�s+ yo)j2 �Nd�d� (s) is given by

H 0 (r) =R@Byo;r

jv (y)j2 d� (y). Next, recall the Green formula

R@Byo;r

jvj2 @�Gd� (y)�R@Byo;r

@�

�jvj2�Gd� (y)

=RByo;r

jvj2�Gdy �RByo;r

��jvj2�Gdy .

We apply it with G (y) = r2 � jy � yoj2 where Gj@Byo;r= 0, @�Gj@Byo;r

= �2r,and �G = �2 (N + 1). It gives

H 0 (r) = 1r

RByo;r

(N + 1) jvj2 dy + 12r

RByo;r

��jvj2��r2 � jy � yoj2

�dy

= N+1r H (r) + 1

r

RByo;r

div (vrv)�r2 � jy � yoj2

�dy

= N+1r H (r) + 1

r

RByo;r

�jrvj2 + v�v

��r2 � jy � yoj2

�dy .

Consequently, when �yv = 0,

H 0 (r) =N + 1

rH (r) +

1

rD (r) , (A.1)

that is H0(r)H(r) =

N+1r + 1

rD(r)H(r) the second equality in Lemma A.

Now, we compute the derivative of D (r).

D0 (r) = ddr

�r2R r0

RSN

��(rv)j�s+yo ��2 �Nd�d� (s)��RSNr2��(rv)jrs+yo ��2 rNd� (s)

= 2rR r0

RSN

��(rv)j�s+yo ��2 �Nd�d� (s)= 2r

RByo;r

jrvj2 dy .

(A.2)

23

On the other hand, we have by integrations by parts that

2rRByo;r

jrvj2 dy = N+1r D (r) + 4

r

RByo;r

j(y � yo) � rvj2 dy� 1r

RByo;r

rv � (y � yo)�v�r2 � jy � yoj2

�dy .

(A.3)

Therefore,

(N + 1)RByo;r

jrvj2�r2 � jy � yoj2

�dy

= 2r2RByo;r

jrvj2 dy � 4RByo;r

j(y � yo) � rvj2 dy+2RByo;r

(y � yo) � rv�v�r2 � jy � yoj2

�dy ,

and this is the desired estimate (A.3). Consequently, from (A.2) and (A.3), weobtain, when �yv = 0, the following formula

D0 (r) =N + 1

rD (r) +

4

r

ZByo;r

j(y � yo) � rvj2 dy . (A.4)

The computation of the derivative of � (r) = D(r)H(r) gives

�0 (r) =1

H2 (r)[D0 (r)H (r)�D (r)H 0 (r)] ,

which implies using (A.1) and (A.4) that

H2 (r) �0 (r) =1

r

4

ZByo;r

j(y � yo) � rvj2 dyH (r)�D2 (r)

!� 0 ,

indeed, thanks to an integration by parts and using Cauchy-Schwarz inequality,we have

D2 (r) = 4�R

Byo;rvrv � (y � yo) dy

�2� 4

�RByo;r

j(y � yo) � rvj2 dy��R

Byo;rjvj2 dy

�� 4

�RByo;r

j(y � yo) � rvj2 dy�H (r) .

Therefore, we have proved the desired monotonicity for � and this completesthe proof of Lemma A.

6.1.3 Proof of Lemma C

Under the assumption Byo;r \ @D � � for any r 2 [ro; Ro), we extend v by zeroin Byo;Ro nD and denote by v its extension. Since v = 0 on �, we have8<: v = v1D in Byo;Ro ,

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, r = Byo;r,when 0 < r < ro. We introduce the following three functions:

H (r) =Rrjv (y)j2 dy ,

D (r) =Rrjrv (y)j2

�r2 � jy � yoj2

�dy ,

and

� (r) =D (r)

H (r)� 0 .

Our goal is to show that � is a non-decreasing function. Indeed, we will provethat the following equality holds

d

drlnH (r) = (N + 1)

d

drlnr +

1

r� (r) . (C.1)

Therefore, from the monotonicity of �, we will deduce (in a similar way than inthe proof of Lemma A) that

ln�H(r2)H(r1)

�ln r2r1

� (N + 1) + � (r2) �ln�H(r3)H(r2)

�ln r3r2

,

and this will imply the desired estimateZr2

jv (y)j2 dy � Z

Byo;r1

jv (y)j2 dy!� Z

r3

jv (y)j2 dy!1��

,


�1

ln r2r1+ 1

ln r3r2

��1.

First, we compute the derivative of H (r) =RByo;r

jv (y)j2 dy.

H 0 (r) =RSNjv (rs+ yo)j2 rNd� (s)

= 1r

RSNjv (rs+ yo)j2 rs � srNd� (s)

= 1r

RByo;r

div�jv (y)j2 (y � yo)

�dy

= 1r

RByo;r

�(N + 1) jv (y)j2 +r jv (y)j2 � (y � yo)

�dy

= N+1r H (r) + 2

r

Rrv (y)rv (y) � (y � yo) dy .

(C.2)

Next, when �yv = 0 in D and vj� = 0, we remark that

D (r) = 2

Zr

v (y)rv (y) � (y � yo) dy , (C.3)

indeed,Rrjrvj2


�dy

=Rrdivhvrv


�idy �

Rrvdiv

hrv�r2 � jy � yoj2

�idy

= �Rrv�v


�dy �

Rrvrv � r


�dy

because on @Byo;r, r = jy � yoj and vj� = 0= 2

Rrvrv � (y � yo) dy because �yv = 0 in D .

25

Consequently, from (C.2) and (C.3), we obtain

H 0 (r) =N + 1

rH (r) +

1

rD (r) , (C.4)

and this is (C.1).

On another hand, the derivative of D (r) is

D0 (r) = 2rR r0

RSN

��(rv)j�s+yo ��2 �Nd�d� (s)= 2r

Rrjrv (y)j2 dy .

(C.5)

Here, when �yv = 0 in D and vj� = 0, we will remark that

2rRrjrv (y)j2 dy = N+1

r D (r) + 4r

RByo;r

j(y � yo) � rv (y)j2 dy+ 1r

R�\Byo;r

j@�vj2�r2 � jy � yoj2

�(y � yo) � �d� (y)

(C.6)indeed,

(N + 1)Rrjrvj2


�dy

=Rrdiv�jrvj2


�(y � yo)

�dy

�Rrr�jrvj2


�� (y � yo) dy

=R�\Byo;r


�(y � yo) � �d� (y)

�Rr@yi

�jrvj2


��(yi � yoi) dy

=R�\Byo;r


�(y � yo) � �d� (y)

�Rr2rv@yirv


�(yi � yoi) dy

+2Rrjrvj2 jy � yoj2 dy ,

and �Rr@yjv@

2yiyjv


�(yi � yoi) dy

= �Rr@yj

�(yi � yoi) @yjv@yiv


��dy

+Rr@yj (yi � yoi) @yjv@yiv


�dy

+Rr(yi � yoi) @2yjv@yiv


�dy

+Rr(yi � yoi) @yjv@yiv@yj


�dy

= �R�\Byo;r

�j

�(yi � yoi) @yjv@yiv


��d� (y)

+Rrjrvj2


�dy

+0 because �yv = 0 in D�Rr2 j(y � yo) � rvj2 dy .

26

Therefore, when �yv = 0 in D, we have

(N + 1)Rrjrvj2


�dy

=R�\Byo;r


�(y � yo) � �d� (y)

�2R�\Byo;r

@yjv�j ((yi � yoi) @yiv)�r2 � jy � yoj2

�d� (y)

+2r2Rrjruj2 dy � 4

Rrj(y � yo) � rvj2 dy .

By using the fact that vj� = 0, we get rv = (rv � �) � on � and we deduce that

(N + 1)Rrjrvj2


�dy

= �R�\Byo;r


�(y � yo) � �d� (y)

+2r2Rrjrvj2 dy � 4

Rrj(y � yo) � rvj2 dy ,

and this is (C.6). Consequently, from (C.5) and (C.6), when �yv = 0 in D andvj� = 0, we have

D0 (r) = N+1r D (r) + 4

r

Rrj(y � yo) � rv (y)j2 dy

+ 1r

R�\Byo;r


�(y � yo) � �d� (y) .

(C.7)

The computation of the derivative of � (r) = D(r)H(r) gives

�0 (r) =1

H2 (r)[D0 (r)H (r)�D (r)H 0 (r)] ,

which implies from (C.4) and (C.7), that

H2 (r) �0 (r) = 1r

�4Rrj(y � yo) � rv (y)j2 dy H (r)�D2 (r)

�+H(r)

r

R�\Byo;r


�(y � yo) � �d� (y)

Thanks to (C.3) and Cauchy-Schwarz inequality, we obtain that

0 � 4Zr

j(y � yo) � rv (y)j2 dy H (r)�D2 (r) .

The inequality 0 � (y � yo) � � on � holds when Byo;r \D is star-shaped withrespect to yo for any r 2 (0; Ro). Therefore, we get the desired monotonicityfor � which completes the proof of Lemma C.

27

6.2 Quantitative unique continuation property for the Lapla-cian

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 inD, with a homogeneous Dirichlet boundary condition on � � @:8<: �yv = 0 in D ,

v = 0 on � ,v = v (y) 2 H2 (D) .

(D.1)

The goal of this section is to describe interpolation inequalities associated tosolutions v of (D.1).

Theorem D .- Let ! be a non-empty open subset of D. Then, for anyD1 � D such that @D1 \ @D b � and D1 n(� \ @D1) � D, there exist C > 0and � 2 (0; 1) such that for any v solution of (D.1), we haveZ

D1

jv (y)j2 dy � C�Z

!

jv (y)j2 dy��Z

D

jv (y)j2 dy�1��

.

Or in a equivalent way,

Theorem D�.- Let ! be a non-empty open subset of D. Then, for anyD1 � D such that @D1 \ @D b � and D1 n(� \ @D1) � D, there exist C > 0and � 2 (0; 1) such that for any v solution of (D.1), we haveZ

D1

jv (y)j2 dy � C�1

"

� 1��Z!

jv (y)j2 dy + "ZD

jv (y)j2 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. Moreprecisely, we claim that for any non-empty compact sets in D, K1 and K2, suchthat meas(K1) > 0, there exists � 2 (0; 1) such that for any v = v (y) 2 H2 (D),solution of �yv = 0 in D, we haveZ

K2

jv (y)j2 dy ��Z

K1

jv (y)j2 dy��Z

D


. (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 ofD, 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 anyv = v (y) 2 H2 (D) such that �yv = 0 on D and v = 0 on �, we haveZ

D1

jv (y)j2 dy � C�Z

!

jv (y)j2 dy��Z

D


.

This completes the proof.

6.3 Quantitative unique continuation property for the el-liptic operator @2t +�

In this section, we present the following result.

Theorem E .- Let be a bounded open set in Rn, n � 1, either convexor C2 and connected. We choose T2 > T1 and � 2 (0; (T2 � T1) =2). Letf 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<: @2tw +�w = f in � (T1; T2) ,

w = 0 on @� (T1; T2) ,w = w (x; t) 2 H2 (� (T1; T2)) .

(E.1)

Then, for any ' 2 C10 (� (T1; T2)), ' 6= 0, there exist C > 0 and � 2 (0; 1)such that for any w solution of (E.1), we haveR T2��

T1+�

Rjw (x; t)j2 dxdt

� C�R T2

T1

Rjw (x; t)j2 dxdt

�1��R T2T1

Rj'w (x; t)j2 dxdt+

R T2T1

Rjf (x; t)j2 dxdt

��.

Proof .- First, by a di¤erence quotient technique and a standard extension at� fT1; T2g, we check the existence of a solution u 2 H2 (� (T1; T2)) solving�

@2t u+�u = f in � (T1; T2) ,u = 0 on @� (T1; T2) [ � fT1; T2g ,

such thatkukH2(�(T1;T2)) � c kfkL2(�(T1;T2)) ,

for some c > 0 only depending on (; T1; T2). Next, we apply Theorem D withD = � (T1; T2), � (T1 + �; T2 � �) � D1, y = (x; t), �y = @2t + �, andv = w � u.

29

6.4 The heat equation and the Hölder continuous depen-dence from one point in time

In this section, we presents the following result.

Theorem F .- Let be a bounded open set in Rn, n � 1, either convexor C2 and connected. Let ! be a nonempty open subset of , and T > 0. Thenthere are C > 0 and � 2 (0; 1) such that any solution to8<: @tu��u = 0 in � (0; T ) ,

u = 0 on @� (0; T ) ,u (�; 0) 2 L2 () ,

(F.1)

satis�es, for any to 2 (0; T ),Z

ju (x; to)j2 dx � C�eCto

Z

ju (x; 0)j2 dx�1��Z

!

ju (x; to)j2 dx��

.

Proof .- We divide the proof into three steps.

Step 1 .- Let �1,�2,� � � and e1,e2,� � � be the eigenvalues and eigenfunctionsof the Laplacian �� in H1

0 (), constituting an orthonormal basis in L2 ().

For any uo = u (�; 0) =Pj�1

�jej in L2 () with �j =Ruoejdx, the solution

u to (F.1), can be written as u (x; t) =Pj�1

�jej (x) e��jt. Let to 2 (0; T ). We

introduce (see [L] or [CRV]) the function

w (x; t) =Xj�1

�jej (x) e��jtoch

�p�jt�� (x)

Xj�1

�jej (x) e��jto ,

where � 2 C10 (!), � = 1 in e! b !. Recall that cht = (et + e�t) =2. Then, wsolves 8>><>>:

@2tw +�w = �f in � (0; T ) ,w = 0 on @� (0; T ) ,w = @tw = 0 in e! � f0g ,w = w (x; t) 2 H2 (� (0; T )) ,

where f = ��u (�; to) 2 H20 ().

We denote by w the extension of w to zero in e! � (�T; 0). Then, w solves8<: @2tw +�w = �f1�(0;T ) in � (0; T ) [ e! � (�T; 0) ,w = 0 in e! � (�T; 0) ,w = 0 on @� (0; T ) .

Now, we construct D, an open connected set in RN+1, satisfying the followingsix 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 C2 when is of class C2 and connected ;vi). D is convex with an adequate choice of the pair (�; To) when is convex

.

In particular, w 2 H2 (D).

Step 2 .- We claim that there is g 2 H2 (� (�T; T ))\H10 (� (�T; T )) �

H2 (D) such that�@2t g +�g = �f1�(0;T ) in � (�T; T ) ,g = 0 on @ (� (�T; T )) = @� (�T; T ) [ � f�Tg ,

andkgkL2(D) � kfkL2(�(0;T )) . (F.2)

Indeed, we will proceed in six substep when is of class C2 and connected (thecase where is convex is well-known because � (�T; T ) is then convex). Wede�ne h = �f1�(0;T ) 2 L2 (� (�T; T )).Substep 1: recall that h 2 L2 (� (�T; T )) implies the existence of such

g 2 H10 (� (�T; T )).

Substep 2: thanks to the interior regularity theorem of elliptic systems, forany D0 b � (�T; T ), g 2 H2 (D0).Substep 3: thanks to the boundary regularity theorem of elliptic systems,

but far for � f�T; Tg, g is locally in H2 because is of class C2.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�

@2t g +�g = h in � (�T; 3T ) ,g = 0 on @ (� (�T; 3T )) = @� (�T; 3T ) [ � f�T; 3Tg .

By applying the boundary regularity theorem of elliptic systems as in substep3, we get g 2 H2 (� (0; 2T )). In particular, g 2 H2 (� (0; T )).Substep 5: we extend in a similar way the solution at t = �T in order to

conclude that g 2 H2 (� (�T; 0)).Substep 6: Finally, we multiply @2t g+�g = h by (��)

�1g and integrate by

parts over � (�T; T ), to getR T�T k@tgk

2H�1() dt+ kgk

2L2(�(�T;T ))

=R T0

R��f (x) (��)�1 g (x; t) dxdt

=R T0

Rf (x) (��) (��)�1 g (x; t) dxdt because f 2 H2

0 ()� kfkL2(�(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 = @2t +�, v = w � g in Dwhere � = @ � (�; T � �) and � (�=2 + T=4; 3T=4� �=2) � D1 � D suchthat @D1 \ @D b � and D1 n(� \ @D1) � D in order that

ZD1

jw � gj2 dy � C�1

"

� 1��Z!o

jw � gj2 dy + "ZD

jw � gj2 dy 8" > 0 ,

which impliesZD1

jwj2 dy � C�1

"

� 1��ZD

jgj2 dy + "ZD

jwj2 dy 8" 2 (0; 1) ,

where we used the fact that w = 0 in !o. By (F.2), we conclude that there existC > 0 and � 2 (0; 1) such thatR 3T=4��=2

�=2+T=4

Rjw (x; t)j2 dxdt � C

�1"

� 1��R T0

Rjf (x)j2 dxdt

+"R T0

Rjw (x; t)j2 dxdt 8" > 0 .

Consequently, there exist � 2 (0; T=2), C > 0 and � 2 (0; 1) such thatR T��

Rjw (x; t)j2 dxdt � C

�1"

� 1��R T0

Rjf (x)j2 dxdt

+"R T0

Rjw (x; t)j2 dxdt 8" > 0 .

On the other hand, the following four inequalities hold.Z T

0

Z

jf (x)j2 dxdt � TZ!

j� (x)u (x; to)j2 dx ,

Z T

0

Z

jw (x; t)j2 dxdt � 2TXj�1

�2je�2(�jto�

p�jT) + 2T

Z!

j� (x)u (x; to)j2 dx ,

Pj�1


p�jT) =

Pfj�1;

p�j� T

tog�2je

�2(�jto�p�jT)

+P

fj�1;p�j>

Ttog�2je

�2(�jto�p�jT)

� e2T2

to

Pj�1

�2j ,

R T��

R

��Pj�1�jej (x) e��jtoch �p�jt��2

dxdt � 2R T��

Rjw (x; t)j2 dxdt

+2TR!j� (x)u (x; to)j2 dx ,

32

We deduce from the above �ve inequalities that for any " > 0,

(T � 2�)Pj�1

�2je�2�jto � (T � 2�)

Pj�1


p�j�)

�R T��

R

��Pj�1�jej (x) e��jtoch �p�jt��2

dxdt

� 2TC�1"

� 1��R!j� (x)u (x; to)j2 dx

+4T"

e2

T2

to

Pj�1

�2j +R!j� (x)u (x; to)j2 dx

!+2T

R!j� (x)u (x; to)j2 dx .

Finally, there is C > 0 such that for any to > 0,Rju (x; to)j2 dx =

Pj�1

�2je�2�jto

� C�1"

� 1��R!ju (x; to)j2 dx+ "e

Cto

Rju (x; 0)j2 dx 8" 2 (0; 1) ,

which implies the desired estimate of Theorem F,Z

ju (x; to)j2 dx � C�eC(1��)

to

�Z

ju (x; 0)j2 dx�1��Z

!

ju (x; to)j2 dxdt��

.

References

[AE] V. Adolfsson and L. Escauriaza, C1;� domains and unique continuationat the boundary, Comm. Pure Appl. Math. 50 (1997) 935-969.

[BT] C. Bardos, L. Tartar, Sur l�unicité retrograde des équations paraboliqueset quelques questions voisines, Arch. Rational Mech. Anal. 50 (1973)10-25.

[CRV] B. Canuto, E. Rosset, S. Vessella, Quantitative estimates of unique con-tinuation for parabolic equations and inverse initial-boundary value prob-lems with unknown boundaries, Trans. Amer. Math. Soc. 354 (2001) 491-535.

[E] L. Escauriaza, Doubling property and linear combinations of eigenfunc-tions, manuscript.

[EFV] L. Escauriaza, F.J. Fernandez, S. Vessella, Doubling properties of caloricfunctions, Applicable analysis 85 (1-2) (2006) 205-223.

33

[FI] A.V. Fursikov, O.Yu. Imanuvilov, Controllability of evolution equations.Lecture Notes Series, 34. Seoul National University, Research Instituteof Mathematics, Global Analysis Research Center, Seoul, 1996.

[FEF] X-L. Feng, L. Eldén, C-L Fu, Stability and regularization of a backwardparabolic PDE with variable coe¢ cients, Journal of Inverse and Ill-PosedProblems 18 (2010) 217-243.

[GaL] N. Garofalo and F H. Lin, Monotonicity properties of variational inte-grals, Ap-weights and unique continuation, Indiana Univ. Math. J. 35(1986) 245-268.

[K] C. Kenig, Quantitative unique continuation, logarithmic convexity ofgaussian means and Hardy�s uncertainty principle, Proc. of Symp. ofPure Math., volume in celebration of V. Mazya�s 70 birthday, 79, 2008,p.207-227.

[KT] H. Koch, D. Tataru, Carleman estimates and unique continuation forsecond order parabolic equations with non smooth coe¢ cients, Comm.Partial Di¤erential Equations 34 (4) (2009) 305-366.

[Ku] I. Kukavica, Level sets for the stationary Ginzburg-Landau equation,Calc. Var. 5 (1997) 511-521.

[Ku2] I. Kukavica, Quantitative uniqueness for second order elliptic operators,Duke Math. J. 91 (1998) 225-240.

[KN] I. Kukavica, K. Nyström, Unique continuation on the boundary for Dinidomains, Proc. of Amer. Math. Soc. 126 (1998) 441-446.

[Le] M. Léautaud, Spectral inequalities for non-selfadjoint elliptic operatorsand application to the null-controllability of parabolic systems, Journalof Functional Analysis 258 (2010) 2739�2778.

[LR] G. Lebeau, L. Robbiano, Contrôle exact de l�équation de la chaleur,Comm. Part. Di¤. Eq. 20 (1995) 335-356.

[Li] F.H. Lin, Remarks on a backward parabolic problem, Methods and Ap-plications of Analysis 10 (2) (2003) 245-252.

[L] F.H. Lin, A uniqueness theorem for the parabolic equations, Comm. PureAppl. Math. 43 (1990) 127-136.

[M] L.Miller, A direct Lebeau-Robbiano strategy for the observability of heat-like semigroups, Discrete and Continuous Dynamical Systems Series B14 (4) (2010) 1465-1485.

[PW] K. D. Phung , G. Wang , Quantitative unique continuation for the semi-linear heat equation in a convex domain, Journal of Functional Analysis259 (5) (2010) 1230-1247.

34

[PW2] K. D. Phung , G. Wang, An observability for parabolic equations frommeasurable set in time and its applications, in preparation.

[P] C.C. Poon, Unique continuation for parabolic equations, Comm. PartialDi¤erential Equations 21 (3-4) (1996) 521-539.

[Ph3] K.-D. Phung, Observation et stabilisation d�ondes: géométrie et coût ducontrôle, Habilitation à diriger des recherches, (2007).

[Ro] L. Robbiano, Théorème d�unicité adapté au contrôle des solutions desproblèmes hyperboliques, Comm. Part. Di¤. Eq. 16 (1991) 789-800.

[W] G. Wang, L1-null controllability for the heat equation and its conse-quences for the time optimal control problem, SIAM J. Control Optim.47 (2008) 1701-1720.

[Ze] S. Zelditch, Local and global analysis of eigenfunctions on Riemannianmanifolds, arXiv:09033420.

35

observability for heat equations · 2018-12-10 · this talk describes di⁄erent approaches to get...

Documents