Carleman estimate for Zaremba boundary condition Pierre Cornilleau and Luc Robbiano Lycée du parc des Loges, Évry and Université de Versailles

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1 Carleman estimate for Zaremba boundary condition Pierre Cornilleau and Luc Robbiano Lycée du parc des Loges, Évry and Université de Versailles Saint-Quentin LMV, UMR CNRS 8100

2 N 8 >< (@ 2 t )u = 0 in n u + a(x)@ t u = 0 N u = 0 D where a(x) 0, a 6 0, and a supported N.

3 KNOWN RESULTS 2 t + a(x)@ t )u = 0 in ; (Lebeau 93) u = 0 (@ 2 t )u = 0 in ; (Lebeau-Robbiano 95) E (u(t)) = 1 n u + a(x)@ t u = 0 Z j@ t u(t)j 2 + j@ x u(t)j 2 dx: C We have E (u(t)) log 2, (for smooth u). (2 + t) Remind: a(x) 0 and a 6 0.

4 RESULTS Theorem For Zaremba problem we have E (u(t)) C log 2 (2 + t). Related result: Let v solution of 8 < ( + 2 )v = f 0 in (@ n + i a(x))v = f 1 : N v = 0 D Theorem (Burq) If there exists C > 0 such that kvk Ce C jj (kf 0 k L 2 + jf 1 j L 2) then C we have E (u(t)) log 2 (2. + t) This means estimate on resolvent ) stabilisation estimate.

5 ESTIMATE ON THE DISSIPATIVE REGION 8 < ( + 2 )v = f 0 in v (@ n + i a(x))v = f 1 : N v = 0 D We obtain, Z Z f 0 vdx = ( v + 2 v )vdx Z Z = ( jrvj jvj 2 )dx + (@ n v Z Z = ( jrvj jvj 2 )dx + Taking the imaginary part, we obtain (jj > 1), Z a>0 ( i av + f 1 jvj 2 d C (kf 0 k L 2 + jf 1 j L 2)kvk H 1

6 INTERPOLATION INEQUALITY 8 < ( + 2 )v = f 0 in (@ n + i a(x))v = f 1 : N v = 0 D Let w (x) = e x 0 v (x). We x k 0 w = k v thus 8 < ( x 2 0 )w = e x 0 f 0 in (@ n + ia(x)@ x0 )w = e x 0 f 1 : N v = 0 D Denote by X = [0; 3], Y = [1; 2], = [0; 3], where is a part Assume that 9C > 0; 0 2 (0; 1), 8 2 (0; 0 ], 8w satisfying an adapted boundary condition, we have, kwk H 1 (Y ) C k(@ 2 x 0 + )wk L 2 (X ) +j@ n wj L 2 ( ) +jwj L 2 ( ) kwk 1 H 1 (X )

7 ESTIMATE IN INTERIOR n + ia(x)@ x0 )w = e x 0 f 1, ( 2 x 0 )w = e x 0 f 0, w (x) = e x 0 v (x). For C > 0, > 0 we have the estimates, kwk H 1 (X ) Ce C kvk H 1 (), kvk H 1 () Ce C kwk H 1 (Y ), j@ n wj L 2 ( ) Ce C (jf 1 j L 2 + jvj L 2 () ). Remind the interpolation inequality, kwk H 1 (Y ) C k(@ 2 x 0 + )wk L 2 (X ) +j@ n wj L 2 ( ) +jwj L 2 ( ) kwk 1 We obtain, kvk H 1 () Ce C kf 0 k L 2 () + jf 1 j L 2 + jvj L 2 () H 1 (X ) :

8 ESTIMATE ON THE RESOLVENT Remind kvk 2 H 1 () CeC kf 0 k 2 L 2 () + jf 1j 2 L 2 + jvj 2 L 2 () : Z a>0 jvj 2 d C (kf 0 k L 2 + jf 1 j L 2)kvk H 1 () e C 1 2C kvk2 H 1 () + 10C 0 e C (kf 0 k 2 L 2 + jf 1 j 2 L 2 ): Taking = fx; a(x) g we have, kvk 2 H 1 () CeC kf 0 k 2 L 2 + jf 1 j 2 L 2 :

9 ' a real function. WHAT IS CARLEMAN ESTIMATE? ' = cst D Carleman estimate has the following form, 9h 0 > 0, C > 0, 8u supported in D, 8h 2 (0; h 0 ] h 1=2 ke '=h uk L 2 (D) + h 3=2 ke '=h ruk L 2 (D) Cke '=h h 2 uk L 2 (D) :

10 CARLEMAN ESTIMATES AT THE BOUNDARY (I) First without boundary conditions, ' = cst D + 9h 0 > 0, C > 0, 8u supported in D, 8h 2 (0; h 0 ] h 1=2 ke '=h uk L 2 (D + ) + h 3=2 ke '=h ruk L 2 (D + ) Cke '=h h 2 uk L 2 (D + ) +Ch 1=2 je '=h uj L 2 (@D + ) +Ch 3=2 je '=h ruj L 2 (@D + ) :

11 CARLEMAN ESTIMATES AT THE BOUNDARY (II) With boundary conditions, ' = cst D h 1=2 ke '=h uk L 2 (D ) + h3=2 ke '=h ruk L 2 (D ) + Ch1=2 je '=h uj L 2 (@D ) +Ch 3=2 je '=h ruj L 2 (@D )Cke '=h h 2 uk L 2 (D ) +Cje'=h Buj L 2 (@D ) ; where B is a boundary condition, for instance, Bu = u j@d or Bu n u j@d.

12 CARLEMAN ESTIMATES ) INTERPOLATION INEQUALITY (I) = 0 ' = C 5 ' = C 4 ' = C 3 = 1 = 0 For simplicity, I give an idea for interior and local interpolation estimate and I shall forget the x 0 variable. We have v and a cut-off function, we set u = v. we apply local Carleman estimate to u i.e. ' = C 2 ' = C 1 ke '=h uk H 1. ke '=h uk L 2: We obtain, ke '=h vk H 1 (C3 'C 4 ). ke '=h vk L 2 ('C5 ) + ke'=h vk H 1 (C4 'C 5 ) + ke'=h vk H 1 ('C2 ) :

13 CARLEMAN ESTIMATES ) INTERPOLATION INEQUALITY (II) ke '=h vk H 1 (C3 'C 4 ). ke '=h vk L 2 ('C5 ) + ke'=h vk H 1 (C4 'C 5 ) + ke'=h vk H 1 ('C2 ) e C 3=h kvk H 1 (C3 'C 4 ). e C 5=h k vk L 2 + kvk H 1 (C4 'C 5 ) + e C 2=h kvk H 1 kvk H 1 (C3 'C 4 ). e A=h k vk L 2 + kvk H 1 (C4 'C 5 ) +e c=h kvk H 1:

14 CARLEMAN ESTIMATES ) INTERPOLATION INEQUALITY (III) kvk H 1 (C3 'C 4 ). e A=h k vk L 2 + kvk H 1 (C4 'C 5 ) +e c=h kvk H 1: We choose h such that we obtain e A=h k vk L 2 + kvk H 1 (C4 'C 5 ) = e c=h kvk H 1; kvk H 1 (C3 'C 4 ). k vk L 2 + kvk H 1 (C4 'C 5 ) We need h small but h large means k vk L 2 + kvk H 1 (C4 'C 5 ) & kvk H 1. In this case interpolation inequality is trivial. kvk 1 H 1 :

15 CARLEMAN ESTIMATES ) INTERPOLATION INEQUALITY (IV) = 0 = 1 = 0 ' = C 5 ' = C 4 ' = C 3 ' = C 2 ' = C 1 kvk H 1 (C3 'C 4 ). k vk L 2 + kvk H 1 (C4 'C 5 ) kvk 1 H 1 :

16 CARLEMAN FOR ZAREMBA BOUNDARY CONDITION ' = cst D h 1=2 ke '=h uk L 2 (D ) + h3=2 ke '=h ruk L 2 (D ) + Ch1=2 je '=h uj H 1=2 (@D ) +Ch 3=2 je xn uj H 1=2 (@D ) Cke'=h h 2 uk L 2 (D ) +Cje'=h Buj H ; where B is the Zaremba boundary condition and H adapted Hilbert space.

17 CONJUGAISON BY THE WEIGHT x n u = 0 x 1 > 0 u = 0 x 1 < 0 ' = cst = fx n > 0g We set v = e '=h u. x n We have h 2 e '=h u = h 2 e '=h (e '=h v ) = P ' v, where the semi-classical symbol of P ' is given by p ' (x; ) = 2 n + 2i '0 x n n + q 2 (x; 0 ) + 2iq 1 (x; 0 ) where q 2 (x; 0 ) = j 0 j 2 j@ x '(x)j 2 q 1 (x; 0 ) = x 0'(x)

18 THE PROPERTIES OF ROOTS Informations on the roots: important for boundary value problems. p ' (x; ) = 2 n + 2i '0 x n n + q 2 (x; 0 ) + 2iq 1 (x; 0 ). There exists a real root if and only if 2 n + q 2(x; 0 ) = 0 and ' 0 x n n + q 1 (x; 0 ) = 0, thus (x; 0 ) = q 2 (x; 0 ) + q2 1 (x; 0 ) ' 0 2 xn (x) = 0. The roots are simple if (x; 0 ) > (' 0 x n (x)) 2 and in this case, Im 1 (x; 0 ) > ' 0 x n (x) > Im 2 (x; 0 ), where j are the roots. We distinguish two cases, < 0 and > (' 0 x n ) 2.

19 CASE < 0 We have Im j < 0 if ' 0 x n (x) > 0 (it is our choice of '). P ' v = f in x n > 0. Denote v = v if xn > 0 0 if x n < 0 P ' v = f x n=0 + 1 xn=0, where j depend on v jxn=0 and D xn v jxn=0. Choose Q a parametrix of P! QP = Id + R where R smoothing. v = Qf + C C Rf where C j are operators on traces. On x n = 0, (i.e. x n = 0 + ) v jxn=0 = Qf jxn=0 + C C Rf jxn=0, and similar relation xn v jxn=0.

20 CALDERON OPERATORS v jxn=0 = Qf jxn=0 + C C Rf jxn=0. C 0, C 1 have the following form (principal symbol) Z e ix 0 0 ^( 0 ) Z e ixnn n p ' (x; ) d nd 0 = 0 WHY?! By residues formula and Im j < 0. n comes from Fourier transform of () x n=0. We obtain, v jxn=0 = Qf jxn=0 + Rf jxn=0 + ~ R( 0 ; 1 ), (where ~ R is smoothing). Thus we can estimates the two traces v jxn=0 xn v jxn=0 microlocally in < 0. Important: in this region we do not need boundary conditions.

21 CASE > (' 0 x n ) 2 We have p ' (x; ) = ( n 2 (x; 0 )) ( n 1 (x; 0 )). Modulo some remainder terms, we have P ' v = (D n op( 2 )) (D n op( 1 )) v. We quantify symbols in semi-classical sense, i.e. op() = D = h Let z = (D n op( 1 )) v, we have, (D n op( 2 )) z = f + h i z jx n=0 xn=0. The root of n 2 (x; 0 ) is in fim < 0g we can perform as in previous case and we can estimate z jxn=0 = D n v jxn=0 op( 1 )v jxn=0 = g, where g depends on f.

22 D n v jxn=0 op( 1 )v jxn=0 = g EQUATION ON TRACES (I) We recall v = e '=h u u jxn=0 = 0 on x 1 < 0 and D xn u jxn=0 = 0 on x 1 > 0, then D xn u = D xn e '=h v = e '=h D xn v + i ' 0 x n v. Let v 0 = v jxn=0 and v 1 = D xn v + i ' 0 x n v jx n=0. We obtain on v 0 and v 1 the equation v 1 op( 1 + i ' 0 x n )v 0 = g on x n = 0; with v 0 = 0 on x 1 < 0; v 1 = 0 on x 1 > 0:

23 EQUATION ON TRACES (II) v 1 op( 1 + i ' 0 x n )v 0 = g on x n = 0; with v 0 = 0 on x 1 < 0; v 1 = 0 on x 1 > 0: Let r x1 >0 the restriction to x 1 > 0. We can write r x1 >0 op( 1 + i ' 0 x n )v 0 = r x1 >0g on x n = 0: We have a pseudo-differential boundary problem, studied by Eskin, Boutet de Monvel, Rempel-Schulze, Grubb, Harutjunjan-Schulze... Problem: 1 + i ' 0 x n does not satisfy the transmission condition. Before taking the restriction we must transform the equation on the traces.

24 EQUATION ON TRACES (III) v 1 op( 1 + i ' 0 x n )v 0 = g on x n = 0; with v 0 = 0 on x 1 < 0; v 1 = 0 on x 1 > 0: Let s ( 0 ) = 1 + i p s, j 00 j 2 + " 2 s + (0 ) = 1 i p s. j 00 j 2 + " 2 The kernel of op( s ) is supported in x 1 0. Let z 0 = op( 1=2 )v + 0 is supported in x 1 0, z 1 = op( 1=2 )v 1 is supported in x 1 0. z 1 op( 1=2 ) op( 1 + i ' 0 x n ) op( 1=2 )z + 0 = g 0 where g 0 is known. We obtain r x1 >0 op( 1=2 ) op( 1 + i ' 0 x n ) op( 1=2 + )z 0 = r x1 >0g 0 :

25 EQUATION ON TRACES (IV) ir x1 >0 op( 1=2 ) op( 1 + i ' 0 x n ) op( 1=2 + )z 0 = r x1 >0g 0 : Computation of the symbol when x = 0. The remainder are treated as errors terms. First what is 1? p ' (x; ) = 2 n + 2i '0 x n n + q 2 (x; 0 ) + 2iq 1 (x; 0 ). For simplicity x 0'(0) = 0. Thus q 2 (0; 0 ) = j 0 j 2 j' 0 x n (0)j 2 and q 1 (0; 0 ) = 0. We obtain, p ' (x; ) = 2 n + 2i '0 x n n + j 0 j 2 j' 0 x n (0)j 2 = n + i ' 0 x n 2 + j0 j 2, thus 1 (0; 0 ) = i ' 0 x n (0) + ij 0 j. Symbol of op( 1=2 ) op( 1 + i ' 0 x n ) op( 1=2 + ) is (on x = 0) 1=2 ij 0 j 1=2 1=2 1=2 + = 1 + i qj 00 j 2 + " 2 ij 0 j 1 i qj 00 j 2 + " 2

26 EQUATION ON TRACES (V) r x1 >0 op( 1=2 ) op( 1 + i ' 0 x n ) op( 1=2 + )z 0 = r x1 >0g 0. 1=2 ij 0 j 1=2 1=2 1=2 + = 1 + i qj 00 j 2 + " 2 ij 0 j 1 i qj 00 j 2 + " 2 = ij 0 j p j0 j 2 + " 2 = i + O("): Then we obtain, ir x1 >0z 0 = r x1 >0g 0 + O(")(z 0 ) + O(x 0 )(z 0 ) and we can estimate z 0 (supported in fx 1 0g) by g 0 in suitable norm. From traces we can estimate v in interior by more classical way.

On the Resolvent Estimates of some Evolution Equations and Applications

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