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DEPARTAMENTO DEECUACIONES DIFERENCIALES Y ANALISIS NUMERICO

UNIVERSIDAD DE SEVILLA

CONTROLABILIDAD DE ALGUNAS

ECUACIONES EN DERIVADAS PARCIALES

NO LINEALES DE TIPO PARABOLICO

E HIPERBOLICO

Memoria presentada porSergio Guerrero Rodrıguezpara optar al grado de Doctoren Matematicas.

Sevilla, Mayo 2005.

Fdo.: Sergio Guerrero Rodrıguez

V.B de los directores

Fdo: Dr. D. Enrique Fernandez Cara Fdo.: Dr. D. Manuel Gonzalez BurgosProfesores de la Universidad de Sevilla.

Agradecimientos

En primer lugar, me gustarıa agradecer a todos los miembros del Departamento EDAN porla gentileza y disponibilidad que han demostrado, lo cual ha hecho la convivencia diaria muchomas llevadera.

En particular pienso en Ara. Siempre ha estado y sigue estando ahı, pendiente del masmınimo papeleo necesario.

Gratitud especial muestro hacia Enrique y Manolo, directores de esta memoria, por habersabido darme acertados consejos y por su dedicacion y perseverancia.

Destaco tambien el buen ambiente creado en torno a nosotros, los becarios: Monica y Pedro,el tiempo que hemos coincidido, con la ayuda indispensable de Antonio Suarez.

Agradezco particularmente la amistad y generosidad demostrada por Anna, Blanca, Chari,Kisko y Rosa.

Agradezo a Manuel Delgado Delgado y a Francisco Guillen Gonzalez que hayan aceptadoavalar este trabajo y alabo la efectividad con la que lo han hecho.

Por ultimo, reconocimiento especial dedico a mis amigos de Lucena y a mi familia: mi padre,mi madre y mi hermana por creer en mı.

Indice general

Introduccion 7

1. Null controllability of the heat equation with Fourier boundary conditions:The linear case 371. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382. Proof of theorem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413. Controllability of the linear system . . . . . . . . . . . . . . . . . . . . . . . . . . 49

2. Exact Controllability to the trajectories of the heat equation with Fourierboundary conditions: The semilinear case 691. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 702. A null controllability result for the linear system . . . . . . . . . . . . . . . . . . 733. Controllability of the nonlinear system . . . . . . . . . . . . . . . . . . . . . . . . 76

3.1. The case in which F and f are C1 . . . . . . . . . . . . . . . . . . . . . . 793.2. The general case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

3. Some controllability results for the N-dimensional Navier-Stokes and Boussi-nesq systems with N − 1 scalar controls 931. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 942. Some previous results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

2.1. New Carleman estimates for system (3.18) . . . . . . . . . . . . . . . . . . 992.2. New Carleman estimates for system (3.19) . . . . . . . . . . . . . . . . . . 1032.3. An observability estimate for system (3.20) . . . . . . . . . . . . . . . . . 108

3. Null controllability of the linearized systems (3.13), (3.14) and (3.15) . . . . . . . 1123.1. Null controllability of (3.13) . . . . . . . . . . . . . . . . . . . . . . . . . . 1123.2. Null controllability of system (3.14) . . . . . . . . . . . . . . . . . . . . . 1153.3. Null controllability of system (3.15) . . . . . . . . . . . . . . . . . . . . . 116

4. Proofs of the controllability results for the nonlinear systems . . . . . . . . . . . 1194.1. Proof of Theorem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1194.2. Proof of theorem 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1204.3. Proof of theorem 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

6

4. Remarks on the controllability of the anisotropic Lame system 1251. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1262. A Carleman estimate for the anisotropic Lame system . . . . . . . . . . . . . . . 129

2.1. First Case: rµ(γ∗) = 0, rλ+2µ(γ∗) 6= 0. . . . . . . . . . . . . . . . . . . . 1382.2. Second Case: rµ(γ∗) 6= 0, rλ+2µ(γ∗) = 0 . . . . . . . . . . . . . . . . . . 1452.3. Third Case: rµ(γ∗) 6= 0 and rλ+2µ(γ∗) 6= 0 or rµ(γ∗) = rλ+2µ(γ∗) = 0. 150

3. Controllability of the Lame system. . . . . . . . . . . . . . . . . . . . . . . . . . . 152

Apendice 159

Introduccion

En esta memoria se presentan varios resultados de controlabilidad para sistemas asociados aalgunas ecuaciones en derivadas parciales de tipo parabolico e hiperbolico. En lo que a problemasparabolicos se refiere, vamos a considerar la ecuacion del calor con condiciones de contorno detipo Fourier (lineales y no lineales), sistemas de Navier-Stokes con condiciones de contorno detipo Dirichlet y de tipo Navier y el sistema de Boussinesq con condiciones de Dirichlet. Porotro lado, como ejemplo no trivial de problema hiperbolico, trataremos tambien el sistema deelasticidad de Lame anisotropico.

A lo largo de este trabajo, Ω ⊂ RN (N ≥ 1) denotara un abierto acotado, ω, O ⊂ Ω seranabiertos no vacıos (los dominios de control), 1U denotara la funcion caracterıstica de U y T > 0sera el tiempo final de evolucion del sistema.

En general, para un sistema como los anteriores, la tarea consistira en encontrar un controlv que, actuando de algun modo sobre el sistema, haga que alguna solucion de este tenga uncomportamiento deseado en el instante final de tiempo T . Diremos que tenemos la controlabilidadaproximada del sistema si la solucion puede conducirse arbitrariamente cerca (en cierta norma) deun estado deseado arbitrario. Por otra parte, la controlabilidad exacta indicara que la solucionpuede llevarse exactamente a todo estado deseado. Como caso particular de controlabilidadexacta, se dira que el sistema posee la propiedad de controlabilidad nula si, partiendo de unestado inicial arbitrario, puede siempre lograrse conducir la solucion a cero. Finalmente, otroejemplo interesante de controlabilidad exacta es la controlabilidad exacta a trayectorias, queindica que podemos hacer que una solucion de nuestro sistema controlado coincida con unatrayectoria del mismo sistema, es decir, con una solucion no controlada.

En los ultimos tiempos se ha progresado considerablemente en la controlabilidad de pro-blemas como los que preceden. Con respecto a resultados de controlabilidad aproximada, desta-camos [13], [8], [15] y [46]. Algunos trabajos relevantes sobre controlabilidad exacta o controla-bilidad nula son [52], [27], [25] y [35].

Para probar resultados de controlabilidad de problemas no lineales, demostraremos primerola controlabilidad nula de problemas linealizados adecuados. La herramienta principal para es-tablecer el control nulo son las llamadas desigualdades globales de Carleman para los correspon-dientes problemas adjuntos asociados. Estas desigualdades han sido aplicadas en el contexto dela controlabilidad en muchos trabajos, entre otros: [34], [14], [49], [37] y [36]. Para otros aspectosy consecuencias de las desigualdades de Carleman, vease [33]. Las desigualdades de Carleman

8

son estimaciones de normas L2 ponderadas y, en su forma mas simple, obedecen a la estructura∫∫

Ω×(0,T )ρ21(x, t)|ϕ|2 dx dt ≤ C

∫∫

ω×(0,T )ρ22(x, t)|ϕ|2 dx dt,

donde ϕ es la solucion de un problema de evolucion retrogrado con condicion final ϕ(T ) = ϕ0

(el problema adjunto) y la constante C > 0 y las funciones positivas ρ1 y ρ2 son independientesde ϕ0.

En lo que a la ecuacion del calor se refiere, el control exacto a cero del problema lineal

yt −∆y = v1ω en Q = Ω× (0, T ),

y = 0 sobre Σ = ∂Ω× (0, T ),

y(0) = y0 en Ω

(1)

fue probado en [34] y [43]. En el primero de estos trabajos, la herramienta fundamental es unadesigualdad de tipo Carleman para el problema adjunto asociado a (1). En [43] se usa otrometodo que aprovecha el caracter disipativo de (1) y solo tiene validez en el contexto de la EDPclasica del calor. Ademas, desde el punto de vista de la metodologıa, los argumentos en [34] sonpuramente Hilbertianos mientras que los argumentos utilizados en [43] se basan en el Analisisde Fourier.

Cuando aparecen terminos no lineales en la ecuacion del calor, la tarea es mucho mas compli-cada. En [25] se prueba que, incluso para algunas no linealidades “explosivas” que solo dependendel estado, el sistema

yt −∆y + f(y) = v1ω en Q,

y = 0 sobre Σ,

y(0) = y0 en Ω,

es controlable. Mas precisamente, se prueba el control exacto a trayectorias de este sistema parafunciones f localmente lipschitzianas que cumplen f(0) = 0 y

lım|s|→∞

f(s)s log3/2(1 + |s|) = 0.

Mas recientemente, como aplicacion de los resultados probados en [37], se prueba en [11] quede nuevo tenemos el control exacto a trayectorias del sistema

yt −∆y + F (y,∇y) = v1ω en Q,

y = 0 sobre Σ,

y(0) = y0 en Ω,

siempre que F sea localmente lipschitziana y verifique F (0, 0) = 0 y, esencialmente, que laderivada parcial de F respecto de su primera variable (resp. sus N ultimas variables) crezca maslentamente en el infinito que log3/2(1 + |s|+ |p|) (resp. log1/2(1 + |s|+ |p|)).

9

Todo lo anterior es valido cuando las condiciones de contorno son de tipo Dirichlet. Cuandohablamos de condiciones de contorno de tipo Fourier, la tarea es mas complicada. A. V. Fursikovy O. Yu. Imanuvilov probaron en [27] que, si β ∈ L∞(Σ) y βt ∈ L∞(Σ), entonces el sistema

yt −∆y = v1ω en Q,

∂y

∂n+ β y = 0 sobre Σ,

y(0) = y0 en Ω

(2)

es controlable a cero. Como consecuencia de este resultado, en [10] se prueba el control local acero del sistema no lineal

yt −∆y = v1ω en Q,

∂y

∂n+ g(y) = 0 sobre Σ,

y(0) = y0 en Ω,

(3)

donde g : R 7−→ R es una funcion de clase C3 y g(0) = 0.En los dos primeros capıtulos de esta memoria, consideraremos los sistemas (2) y (3) respec-

tivamente con β solo en L∞(Σ) y g solo globalmente lipschitziana. El objetivo es mejorar losresultados mencionados y probar el control global del sistema no lineal (3). Como explicaremosmas adelante, esto pasara por eliminar la condicion βt ∈ L∞(Σ).

En el primer capıtulo, consideraremos el sistema lineal

yt −∆y + B · ∇y + a y = v1ω en Q,

∂y


y(0) = y0 en Ω

(4)

y probaremos que, si a, B y β son de clase L∞, (4) es exactamente controlable a cero en elinstante T , es decir, para cada T > 0 y cada y0 ∈ L2(Ω), existen controles v ∈ L2(ω × (0, T ))tales que la correspondiente solucion de (4) verifica

y(T ) = 0 en Ω.

Ademas, seremos capaces de construir el control dependiendo continuamente de y0 y verificandouna estimacion

‖v‖L2(ω×(0,T )) ≤ C‖y0‖L2(Ω),

con una constante C conocida explıcitamente en funcion de T , ‖a‖∞, ‖B‖∞ y ‖β‖∞. Estecapıtulo corresponde al artıculo [17].

El segundo capıtulo de la memoria esta dedicado al estudio de las propiedades de contro-labilidad del problema no lineal (5). En primer lugar probamos la existencia de controles enL∞(Q) que conducen el sistema (4) a cero. Ademas obtenemos una estimacion explıcita de la

10

norma L∞ del control respecto de T y de la norma L∞ de los coeficientes. Como consecuencia,deducimos el control global a trayectorias del sistema

yt −∆y + F (y,∇y) = v1ω en Q,

∂y

∂n+ f(y) = 0 sobre Σ,

y(0) = y0 en Ω,

(5)

bajo condiciones adecuadas sobre F y f (ver (31)–(33) mas abajo). Concretamente, dada y unasolucion ‘regular’ de (5) con v ≡ 0 y con condicion inicial y0, probamos la existencia de controlesv tal que y(·, T ) = y(·, T ) en Ω, para todo y0.

Todo esto se encuentra desarrollado en el trabajo [18].

Con respecto a sistemas de tipo Navier-Stokes, mencionemos que los primeros resultados decontrolabilidad se encuentran en [24] y [16]. En estos trabajos, se prueba que el espacio vectorialgenerado por los estados finales alcanzables es denso (en el sentido de la norma L2) en el espaciode Hilbert H constituido por los v ∈ L2(Ω)N que verifican ∇ · v = 0 en Ω y v · n = 0 sobre∂Ω. Por otro lado, la cuestion fue considerada por A. V. Fursikov y O. Yu. Imanuvilov (veansepor ejemplo [26] y [28]), donde las condiciones de contorno son de tipo Navier o periodicas, loque hace el problema mas sencillo. La continuacion unica para problemas de tipo Stokes fueestablecida por C. Fabre y G. Lebeau en [14] a partir de desigualdades de Carleman localesadecuadas. De esta propiedad, C. Fabre dedujo la controlabilidad aproximada. En el caso decondiciones de Navier, el resultado mas interesante fue probado en [8] usando el metodo deretorno.

Por ultimo, el control local exacto a trayectorias para el sistema de Navier-Stokes

yt −∆y + (y,∇)y +∇p = v1ω en Q,

∇ · y = 0 en Q,

y = 0 sobre Σ,

y(0) = y0 en Ω

(6)

fue establecido por O. Yu. Imanuvilov en [35] a partir de desigualdades de Carleman globales.Una mejora de este resultado se ha establecido en el reciente trabajo [21].

Basandonos en esta ultima referencia, probamos en el capıtulo 3 en primer lugar el controllocal exacto a trayectorias del sistema N -dimensional (6) con N − 1 controles escalares. Enconcreto, bajo ciertas hipotesis sobre el dominio de control ω y suponiendo que partimos de unacondicion inicial y0 cercana (en cierta norma) a una trayectoria de (6), se prueba la existenciade controles v con al menos una componente nula de modo que y coincide con dicha trayectoriaen el instante final T .

Tambien recientemente (y siguiendo las ideas de [21]), se ha probado en [31] el control local

11

a trayectorias del sistema de Boussinesq

yt −∆y + (y,∇)y +∇p = v11ω + θ eN en Q,

∇ · y = 0 en Q,

θt −∆θ + y · ∇θ = v21ω en Q,

y = 0 sobre Σ,

y(0) = y0 en Ω.

(7)

Aquı, hemos denotado eN el N -esimo vector de la base canonica de RN . Usando este resultado,establecemos en el capıtulo 3 un resultado de controlabilidad similar pero con ayuda de soloN − 1 controles escalares.

Por ultimo, en este capıtulo se recoge un resultado de controlabilidad global de un sistematruncado de Navier-Stokes bidimensional:

yt −∆y + (y,∇)TM (y) +∇p = v1O en Q,

∇ · y = 0 en Q,

∇× y = 0, y · n = 0 sobre Σ,

y(0) = y0 en Ω,

(8)

con M > 0. Aquı, TM (y) = (TM (y1), TM (y2)) y

TM (s) =

−M si s ≤ −M,

s si −M < s < M,

M si s ≥ M.

Observese que las condiciones de contorno aquı utilizadas son de tipo Navier (veanse [29] y[26] por ejemplo). Al igual que en [8], [26] and [28], el estudio de la controlabilidad de estesistema es menos complicado que el de (6), debido a las condiciones de contorno. En concreto,las condiciones de contorno de (8) nos dicen que la vorticidad (ω = ∇× y) verifica una ecuaciondel calor con condiciones de contorno de tipo Dirichlet.

Mediante la aplicacion de un punto fijo sobre el sistema (8), se deduce el control a cero(global) del mismo. Ası pues, este resultado supone una primera aproximacion al intento decontrolar el sistema de Navier-Stokes de forma global (es decir, sin suponer que partimos de unestado cercano a la trayectoria que queremos alcanzar). Todo esto es motivo del artıculo [23].

Para mayor claridad, hemos incluido un Apendice al final de la memoria donde presentamosun resumen de la demostracion de la desigualdad de Carleman para problemas de tipo Stokescon condiciones de Dirichlet.

En cuanto al sistema de elasticidad se refiere, la gran mayorıa de los trabajos de controlabi-lidad toman como modelo el sistema de Lame isotropico. De manera general, el sistema de

12

Lame tridimensional se escribe como sigue

ρ∂2x0

ui −3∑

j=1

∂xj (σij) = f, 1 ≤ i ≤ 3. (9)

En (9), x0 denota la variable temporal, x′ = (x1, x2, x3) es la variable espacial y σ es el tensorde esfuerzos, que tiene la forma

σ = R + (∇u)R + λ(trε)I + 2µε + β1(trε)(trR)I + β2(trR)ε

+ β3((trε)R + tr(εR)I) + β4(εR + Rε),(10)

donde R = R(x) es un tensor simetrico que verifica ∇ ·R = 0 y ε = 12(∇u +∇tu) es el tensor de

deformaciones.En (10), ρ = ρ(x′) es la densidad y λ = λ(x′) y µ = µ(x′) son los coeficientes de Lame. El

sistema de Lame isotropico corresponde a tomar R ≡ 0 en (10). Esto proporciona propiedadessimplificadoras, pues las variables z0 = ∇·u y z′ = ∇×u verifican entonces ecuaciones independi-entes. Las herramientas fundamentales mas utilizadas para obtener resultados de controlabilidadpara este sistema han sido (de nuevo) desigualdades de Carleman. Para desplazamientos u consoporte compacto, estas desigualdades fueron obtenidas en [9] y [51] para el sistema estacionarioy en [47], [44] y [42] para el sistema de evolucion. Como consecuencia, en estos trabajos seprueban resultados de continuacion unica y de control aproximado. Para desplazamientos u ver-ificando (9) (con R ≡ 0) y completados con condiciones de contorno de tipo Dirichlet, vease [41]en el caso estacionario y [38]–[40] en el caso de evolucion.

En el capıtulo 4 presentamos el sistema de Lame anisotropico. Por simplicidad, hemossupuesto que βj ≡ 0 (1 ≤ j ≤ 4) en (10). En el caso general el analisis serıa similar, de-biendose imponer hipotesis suplementarias a los coeficientes de Lame y al tensor residual R(vease (68)).

En primera instancia, la ecuacion (9) se completa con condiciones de contorno de tipo Dirich-let y condiciones iniciales. Por tanto, el sistema de control queda como sigue:

ρ(x′)∂2x0

u− µ∆u− (λ + µ)∇(∇ · u) + [(R,∇)∇t]u = f + vχω en Q,

u = 0 sobre Σ,

u(0) = u0, ∂x0u(0) = u1 en Ω,

(11)

donde las funciones ρ, µ, λ y R son dadas. Con respecto a este tipo de sistemas, se prueba en[1] un resultado de controlabilidad exacta frontera basado en el metodo de los multiplicadores,utilizado previamente en [45] tambien para el sistema de Lame. Este resultado se establece paraun tiempo suficientemente grande y es valido para operadores no necesariamente isotropicospero con tensores R ‘pequenos’.

En este ultimo capıtulo probamos el control exacto de (11), como consecuencia de unadesigualdad de Carleman global. Concretamente, fijados dos estados finales u2, u3, se prueba laexistencia de un control v tal que la solucion u de (11) verifica

u(·, T ) = u2 y ux0(·, T ) = u3 en Ω

13

para todo tiempo T > 0 que permita la existencia de una ciertas funciones peso (veanse Condi-ciones A y B en el ultimo capıtulo).

El contenido de este capıtulo corresponde al artıculo [32].

A continuacion, vamos a explicar con mas detalle el desarrollo de cada uno de los capıtulosde esta memoria.

Controlabilidad a cero de la ecuacion del calor lineal con condi-ciones de contorno de Fourier

En la primera parte de esta memoria trabajaremos con el sistema

yt −∆y + B · ∇y + a y = v1ω en Q,

∂y


y(0) = y0 en Ω

(12)

y probaremos la controlabilidad a cero del mismo cuando y0 ∈ L2(Ω) y los coeficientes a, B y βson de clase L∞.

El primer resultado importante que demostramos es una desigualdad global de Carlemanpara las soluciones debiles de problemas retrogrados de la forma

−ϕt −∆ϕ = f1 +∇ · f2 en Q,

(∇ϕ + f2) · n = f3 sobre Σ,

ϕ(T ) = ϕ0 en Ω,

(13)

con f1 ∈ L2(Q), f2 ∈ L2(Q)N y f3 ∈ L2(Σ). Observese que se le puede dar ‘a priori’ un sentidoa la condicion frontera, puesto que la solucion debil de (13) verifica ∇ϕ + f2 ∈ L2(Q)N y∇ · (∇ϕ + f2) ∈ H−1(0, T ; L2(Ω)).

Esta desigualdad es la siguiente:

Teorema 1 Bajo las condiciones anteriores, existen constantes λ, σ1, σ2 y C dependientes deΩ y ω tales que, para todo λ ≥ λ, para todo s ≥ s = σ1(eσ2λT + T 2) y para todo ϕ0 ∈ L2(Ω), lasolucion debil de (13) satisface

∫∫

Qe−2sα(s λ2ξ |∇ϕ|2 + s3 λ4 ξ3 |ϕ|2)dx dt

+s2λ3

∫∫

Σe−2sα ξ2 |ϕ|2 dσ dt

≤ C

(∫∫

Qe−2sα(|f1|2 + s2 λ2 ξ2 |f2|2)dx dt

+sλ

∫∫

Σe−2sα ξ |f3|2 dσ dt + s3λ4

∫∫

ω×(0,T )e−2sα ξ3 |ϕ|2 dx dt

).

(14)

14

Aquı, los pesos ξ = ξ(x, t) y α = α(x, t) vienen dados por

ξ(x, t) =eλη0(x)

t(T − t), α(x, t) =

e2λ‖η0‖∞ − eλη0(x)

t(T − t), (15)

donde η0 = η0(x) verifica

η0 ∈ C2(Ω), η0(x) > 0 en Ω, η0(x) = 0 sobre ∂Ω,

|∇η0(x)| > 0 en Ω \ ω′(16)

y ω′ ⊂⊂ ω es un abierto no vacıo.La prueba de este resultado esta inspirada en las ideas de [37]. Mas concretamente, seguimos

una estrategia de dualidad que nos permite relajar las hipotesis de regularidad impuestas sobrelos datos. Esto nos permite imponer solo la hipotesis L∞ sobre el coeficiente β = β(x, t).

Para utilizar esta tecnica, hay que basarse en una desigualdad de Carleman para un sistemahomogeneo asociado a (13). En concreto, se puede probar que las soluciones del problema

−qt −∆q = g en Q,

∂q

∂n= 0 sobre Σ,

q(T ) = q0 en Ω,

con g ∈ L2(Q) y q0 ∈ L2(Ω), satisfacen

Is,λ(q) ≤ C

(∫∫

Qe−2sα |g|2 dx dt + s3λ4

∫∫

ω×(0,T )e−2sα ξ3 |q|2 dx dt

), (17)

para elecciones de s y de λ apropiadas. El termino Is,λ(q) esta dado por

Is,λ(q) =∫∫

Qe−2sα

((sξ)−1(|qt|2 + |∆q|2) + sλ2ξ |∇q|2 + s3λ4ξ3 |q|2) dx dt.

Una vez establecida esta desigualdad, miramos ϕ como la solucion por transposicion de (13),es decir, la unica funcion de L2(Q) que satisface

∫∫

Qϕhdx dt =

∫∫

Qf1(x, t) z dx dt−

∫∫

Qf2(x, t) · ∇z dx dt

+∫∫

Σf3(x, t) z dσ dt +

∫

Ωϕ0(x) z(x, T ) dx ∀h ∈ L2(Q),

(18)

donde z es, para cada h ∈ L2(Q), la solucion del problema

zt −∆z = h en Q,

∂z

∂n= 0 sobre Σ,

z(0) = 0 en Ω.

15

Consideramos ahora el siguiente problema de mınimos:

Minimizar12

(∫∫

Qe2sα |z|2 + s−3λ−4

∫∫

ω×(0,T )e2sα ξ−3 |v|2 dx dt

)

sujeto a v ∈ L2(Q) y

zt −∆z = s3λ4e−2sαξ3ϕ + v1ω en Q,

∂z

∂n= 0 sobre Σ,

z(0) = 0, z(T ) = 0 en Ω.

(19)

Siguiendo las ideas de [37], llegamos al sistema de optimalidad

L(e−2sα L∗p) + s3λ4e−2sαξ3p1ω = s3λ4e−2sαξ3ϕ en Q,

∂p

∂n= 0,

∂

∂n(e−2sα L∗p) = 0 sobre Σ,

(e−2sαL∗p)|t=0 = (e−2sαL∗p)|t=T = 0 en Ω,

(20)

donde L y L∗ son los operadores L = ∂t−∆ y L∗ = −∂t−∆. Entonces, debido a la desigualdadde Carleman (17), podemos probar que (20) (y por tanto (19)) admite una unica solucion p yque

v = −s3 λ4 e−2sα ξ3 p1ω y z = e−2sαL∗pes la solucion de (19). Se prueba que esta solucion verifica la desigualdad

∫∫

Qe2sα |z|2 dx dt + s−2λ−2

∫∫

Qe2sα ξ−2 |∇z|2 dx dt

+s−1λ−1

∫∫

Σe2sαξ−1 |z|2 dσ dt + s−3λ−4

∫∫


≤ C(Ω, ω) s3λ4

∫∫

Qe−2sα ξ3 |ϕ|2 dx dt,

(21)

para ciertos s y λ.Elegimos ahora

h = s3λ4e−2sαξ3ϕ + v1ω

en (18) y, teniendo en cuenta (21), deducimos que

s3λ4

∫∫

Qe−2sα ξ3 |ϕ|2 dx dt ≤ C(Ω, ω)

(∫∫

Qe−2sα |f1|2 dx dt

+s2λ2

∫∫

Qe−2sα ξ2 |f2|2 dx dt + sλ

∫∫

Σe−2sα ξ |f3|2 dσ dt

+s3λ4

∫∫


)(22)

16

para s y λ apropiados. Por ultimo, deducimos (14) directamente a partir de (22).

Como consecuencia del teorema 1, se deduce una desigualdad de observabilidad para el sistemaadjunto asociado a (12), esto es,

−ϕt −∆ϕ−∇ · (ϕB) + aϕ = 0 en Q,

(∇ϕ + ϕB) · n + β ϕ = 0 sobre Σ,

ϕ(T ) = ϕ0 en Ω.

(23)

De hecho, las soluciones de (23) verifican∫∫

Ω×(T/4,3T/4)|ϕ|2 dx dt ≤ K

∫∫

ω×(0,T )|ϕ|2 dx dt, (24)

con una constante K de la forma

K = expC(1 +1T

+ ‖a‖2/3∞ + ‖B‖2

∞ + ‖β‖2∞). (25)

El segundo resultado importante de este capıtulo es la controlabilidad nula del sistema (12) :

Teorema 2 Supongamos que a ∈ L∞(Q), B ∈ L∞(Q)N y β ∈ L∞(Σ). Entonces (12) es con-trolable a cero para todo T > 0 con controles v tales que

‖v‖L2(ω×(0,T )) ≤ K ‖y0‖L2(Ω) , (26)

con una constante K de la forma

K = exp

C

(1 +

1T

+ ‖a‖2/3∞ + ‖B‖2

∞ + ‖β‖2∞ + T (‖a‖∞ + ‖B‖2

∞ + ‖β‖2∞)

), (27)

donde C = C(Ω, ω) es una constante positiva.

La deduccion de este teorema a partir de la desigualdad de observabilidad (24) es “standard”.

En el segundo capıtulo de la memoria presentamos resultados de controlabilidad exacta atrayectorias para una ecuacion del calor no lineal con condiciones de contorno de tipo Fourierno lineales. Este constituye una continuacion natural del primero. De hecho, los resultadospresentados en los teoremas 1 y 2 seran cruciales.

Controlabilidad exacta a las trayectorias de la ecuacion del calorno lineal con condiciones de contorno de tipo Fourier no lineales

Con ayuda de los resultados enunciados en el capıtulo anterior, vamos a deducir la controla-bilidad exacta (global) a las trayectorias del sistema

yt −∆y + F (y,∇y) = v1ω en Q,

∂y


y(0) = y0 en Ω,

(28)

17

bajo ciertas condiciones de crecimiento en el infinito para las funciones F y f . En esta situacion,consideraremos datos iniciales y0 ∈ L∞(Ω). Algunos resultados de existencia, unicidad, regular-idad y otras propiedades de las soluciones de problemas con la estructura de (28) pueden serencontrados, por ejemplo, en [2], [3] y [12].

Fijamos una trayectoria no controlada del sistema (28), es decir, una solucion y de

yt −∆y + F (y,∇y) = 0 en Q,

∂y


y(0) = y0 en Ω.

(29)

Imponemos las siguientes hipotesis:

y ∈ L2(0, T ; H1(Ω)) ∩ C0([0, T ]; L2(Ω)) ∩ L∞(Q), y0 ∈ L∞(Ω). (30)

El resultado principal de este trabajo es el siguiente:

Teorema 3 Supongamos que F y f son funciones localmente lipschitzianas y verifican

lım|s|→∞

|F (s, p)− F (r, p)||s− r| log3/2(1 + |s− r|) = 0, (31)

uniformemente en (r, p) ∈ [−K, K]×RN para cada K > 0,

∀L > 0, ∃M > 0 tal que

|F (s, p)− F (r, p)| ≤ M |s− r|, |F (s, p)− F (s, q)| ≤ M |p− q|∀(s, r, p, q) ∈ [−L,L]2 ×RN ×RN

(32)

y

lım|s|→∞

|f(s)− f(r)||s− r| log1/2(1 + |s− r|) = 0 (33)

uniformemente en r ∈ [−K, K] para todo K > 0. Entonces, fijado T > 0, tenemos la controla-bilidad exacta a las trayectorias y que verifican (30), con controles v ∈ L∞(ω × (0, T )).

Antes de resumir la prueba del teorema 3, necesitaremos un resultado de controlabilidadnula para el sistema lineal (12) con controles en L∞(ω × (0, T )):

Proposicion 1 Sea T > 0. Entonces, el sistema (12) es exactamente controlable a cero concontroles v ∈ L∞(ω × (0, T )). Ademas, podemos encontrar controles v satisfaciendo

‖v‖L∞(ω×(0,T )) ≤ eC(Ω,ω)K(T,‖a‖∞,‖B‖∞,‖β‖∞) ‖y0‖L2(Ω) , (34)

dondeK = 1 + 1/T + ‖a‖2/3

∞ + ‖B‖2∞ + ‖β‖2

∞ + T (1 + ‖a‖∞ + ‖B‖2∞ + ‖β‖2

∞). (35)

18

La demostracion que presentamos de este resultado esta inspirada en la tecnica utilizada en[7] para regularizar controles. Indicaremos a continuacion las ideas principales de este argumento.

Sea y0 ∈ L2(Ω) y sean ω′ y ω′′ dos abiertos con ω′′ ⊂⊂ ω′ ⊂⊂ ω. Entonces, utilizandoel teorema 2 con dominio de control ω′′, deducimos que existe v ∈ L2(ω′′ × (0, T )) tal que lasolucion y de (12) verifica y(T ) = 0 y tenemos la estimacion

‖v‖L2(ω′′×(0,T )) ≤ eC(Ω,ω) K(T,‖a‖∞,‖B‖∞,‖β‖∞) ‖y0‖L2(Ω), (36)

con K de la forma (35). Definimos ahora una funcion truncante η ∈ C∞([0, T ]) cumpliendo

η(t) = 1 en (0, T/4), η(t) = 0 en (3T/4, T ), 0 ≤ η(t) ≤ 1 en (0, T ).

Consideremos la solucion χ del sistema

χt −∆χ + a χ + B · ∇χ = 0 en Q,

∂χ

∂n+ β χ = 0 sobre Σ,

χ(0) = y0 en Ω.

Entonces w = y − ηχ cumple

wt −∆w + a w + B · ∇w = −η′(t) χ + v1ω′′ en Q,

∂w

∂n+ β w = 0 sobre Σ,

w(0) = 0, w(T ) = 0 en Ω.

(37)

Introduzcamos ahora un abierto ω0 con ω′ ⊂⊂ ω0 ⊂⊂ ω y una nueva funcion cortante ξ queverifica

ξ ∈ C20 (ω0), ξ ≡ 1 en ω′.

Entonces la funcion w = (1− ξ) w verifica

wt −∆w + aw + B · ∇w = −η′(t)χ + v1ω en Q,

∂w

∂n+ β w = 0 sobre Σ,

w(0) = 0, w(T ) = 0 en Ω,

conv = η′ ξ χ + 2∇ξ · ∇w + ∆ξ w −B · ∇ξ w. (38)

Usando resultados de regularidad local para las soluciones del sistema (37), no es difıcil deducirque v ∈ L∞(ω × (0, T )) y que la funcion y = w + ηχ resuelve (con este control v) el problemade controlabilidad a cero de (12) del modo indicado en la proposicion 1.

Otro resultado que se utilizara para demostrar el teorema 2 es una estimacion L∞ de lassoluciones del sistema (12):

19

Proposicion 2 Supongamos que v ∈ L∞(Q), y0 ∈ L∞(Ω) y los coeficientes a, B, β estan enL∞. Entonces y ∈ L∞(Q) y verifica

‖y‖∞ ≤ eC T(1+‖a‖∞+‖B‖2∞+‖β‖2∞) (‖y0‖∞ + ‖f‖∞). (39)

con una constante C > 0 solo dependiente de Ω.

Volvamos a la demostracion del teorema 2. Consideremos el sistema (28) que, tras el cambiode variable y = y + w, puede escribirse en la forma

wt −∆w + F1(w,∇w;x, t)w + F2(∇w; x, t) · ∇w = v1ω en Q,

∂w

∂n+ F3(w; x, t)w = 0 sobre Σ,

w(0) = y0 − y(0) en Ω,

(40)

donde

F1(s, p;x, t) =F (y(x, t) + s,∇y(x, t) + p)− F (y(x, t),∇y(x, t) + p)

s, (41)

F2 = (F21, . . . , F2N ), F2j(p; x, t) =∫ 1

0

∂F

∂pj(y(x, t),∇y(x, t) + λp) dλ (42)

y

F3(s; x, t) =f(y(x, t) + s)− f(y(x, t))

s(43)

para s ∈ R y p ∈ RN .De este modo, nuestra tarea consiste en demostrar la controlabilidad nula del sistema (40)

y, para hacer esto, nos restringimos (por densidad) al caso en el que las funciones Fj tienen laprimera derivada continua.

La idea de la prueba es clasica y se basa en un argumento de punto fijo en el espacioZ = L2(0, T ; H1(Ω)). Estas tecnicas fueron introducidas en [52], en el contexto de la controla-bilidad exacta de la ecuacion de ondas. Desde entonces, han sido aplicadas a un gran numerode problemas diferentes. En nuestro caso, para cada R > 0, consideramos la funcion MR, con

MR(s) =

−R si s < −R,

s si −R ≤ s ≤ R,

R si s > R

y los coeficientes aR,z, Bz y βR,z, con

aR,z(x, t) = F1(MR(z(x, t)),∇z(x, t);x, t),

Bz(x, t) = F2(∇z(x, t);x, t)

yβR,z(x, t) = F3(MR(z(x, t));x, t).

20

Sabemos que el problema de controlabilidad nula

wt −∆w + aR,z w + Bz · ∇w = v1ω en Q,

∂w

∂n+ βR,z w = 0 sobre Σ,

w(0) = y0 − y(0), w(T ) = 0 en Ω,

(44)

tiene solucion, en virtud de la proposicion 1.A continuacion, elegimos una solucion particular del problema (44), procediendo como en

[25], es decir, trabajando en un intervalo temporal [0, Tz] que hace que el control y el estadoasociado verifiquen

‖vR,z‖L∞(ω×(0,T )) ≤ CR ‖w0‖L2(Ω), (45)

‖wR,z‖Z ≤ CR ‖w0‖L2(Ω), (46)

y‖wR,z‖∞ ≤ CR ‖w0‖L∞(Ω), (47)

conCR = exp

C(Ω, ω, T )

(1 + a

2/3R + B

2 + β2R

),

dondeaR = sup

|s|≤R, p∈RN

ess sup(x,t)∈Q

|F1(s, p; x, t)|,

B = supp∈RN

ess sup(x,t)∈Q

|F2(p;x, t)|

yβR = sup

|s|≤Ress sup(x,t)∈Σ

|F3(s; x, t)|.

Finalmente, definimos la aplicacion multivaluada ΛR que a cada z ∈ Z asocia ΛR(z) quees el conjunto de estados wR,z que satisfacen las estimaciones (46) y (47) correspondientes alos controles vR,z ∈ L∞(ω × (0, T )) que verifican (45) y hacen que wR,z(T ) = 0. Para terminarla demostracion, basta probar la existencia de puntos fijos de ΛR. Dividimos la prueba en dosetapas:

Demostramos primero que, para cada R > 0, ΛR tiene al menos un punto fijo wR. Paraello, utilizamos el teorema de Kakutani (vease [4]). La parte mas delicada de la aplicacionde este resultado consiste en comprobar que existe un compacto K (que depende de R)tal que ΛR(z) ⊂ K, para cada z ∈ Z.

Por ultimo, demostramos la existencia de un R suficientemente grande tal que MR(wR) =wR. Esto probarıa el resultado.

21

Controlabilidad de sistemas N-dimensionales de Navier-Stokes yde Boussinesq con N − 1 controles escalares

En la tercera parte de esta memoria, probaremos diversos resultados de controlabilidad paraproblemas no lineales de tipo Navier-Stokes y Boussinesq.

• En primer lugar, trabajamos con el sistema controlado de Navier-Stokes

yt −∆y + (y,∇)y +∇p = v1O, ∇ · y = 0 en Q,

y = 0 sobre Σ,

y(0) = y0 en Ω(48)

y probamos el control local exacto a trayectorias del mismo con la ayuda de N − 1 controlesescalares.

Para poder imponer las hipotesis de regularidad necesarias sobre las condiciones iniciales,definimos los espacios funcionales H, E y V , dados por:

H = w ∈ L2(Ω)N : ∇ · w = 0 en Ω, w · n = 0 sobre ∂Ω , (49)

E =

H si N = 2,

L4(Ω)3 ∩H si N = 3

yV = w ∈ H1

0 (Ω)N : ∇ · w = 0 en Ω.Tambien deberemos suponer que el dominio de control O sea adyacente a la frontera ∂Ω, esdecir:

∃x0 ∈ ∂Ω, ∃ε > 0 tal que O ∩ ∂Ω ⊃ B(x0; ε) ∩ ∂Ω (50)

(B(x0; ε) es la bola de centro x0 y radio ε).Recientemente se ha probado que, con N controles escalares, se puede conseguir la contro-

labilidad local exacta a trayectorias (y, p) del sistema (48) que verifican

y ∈ L∞(Q)N , yt ∈ L2(0, T ; Lσ(Ω)N )

(σ > 1 si N = 2σ > 6/5 si N = 3

)(51)

(vease [20] y [21]). Logicamente, estas hipotesis tambien deberan ser impuestas aquı.Mas precisamente, probaremos que para cualquier par (y, p) que verifique

yt −∆y + (y,∇)y +∇p = 0, ∇ · y = 0 en Q,

y = 0 sobre Σ,(52)

y (51), existe δ > 0 tal que, para cualquier y0 ∈ E satisfaciendo

‖y0 − y(0)‖E ≤ δ,

22

encontramos controles v de clase L2 con vk ≡ 0 para al menos un k y estados asociados (y, p)verificando

y(T ) = y(T ) en Ω. (53)

Observese que, en esta situacion, tras el tiempo t = T podemos dejar evolucionar el sistemalibremente; en efecto, este seguira la trayectoria ‘ideal’ (y, p).

Para mayor claridad, enunciamos el resultado a continuacion:

Teorema 4 Supongamos que O satisface (50). Entonces, para todo T > 0, (48) es localmenteexactamente controlable en el tiempo T a las trayectorias (y, p) que verifican (51), con controlesv ∈ L2(O × (0, T ))N tales que vk ≡ 0 para al menos un k.

La demostracion sigue los mismos pasos del resultado principal de [21] y se basa princi-palmente en una desigualdad de Carleman adecuada para el sistema adjunto asociado a unalinealizacion de (48):

−ϕt −∆ϕ− (Dϕ) y +∇π = g, ∇ · ϕ = 0 en Q,

ϕ = 0 sobre Σ,

ϕ(T ) = ϕ0 en Ω,

(54)

con Dϕ = ∇ϕ + ∇ϕt. Recordemos que fue probado en [21] que, si y cumple (51), existe unaconstante C > 0 tal que la siguiente desigualdad es cierta para todo ϕ0 ∈ L2(Ω)N :

∫∫

Qe

−2α

t4(T−t)4(t−12(T − t)−12 |ϕ|2 + t−4(T − t)−4 |∇ϕ|2) dx dt

+∫∫

Qe

−2αt4(T−t)4 t4(T − t)4

(|∆ϕ|2 + |ϕt|2)

dx dt

≤ C

(∫∫

Qe−4α+2αt4(T−t)4 t−30(T − t)−30 |g|2 dx dt

+∫∫

O0×(0,T )e−8α+6αt4(T−t)4 t−64(T − t)−64 |ϕ|2 dx dt

)

(55)

(α y α son dos constantes positivas). Aquı, hemos debido elegir O0 ⊂ O, de modo que para todopunto x ∈ O0, la recta x + Rek corte a ∂Ω en un punto de ∂O0 para cierto k ∈ 1, · · · , N.Observese que esto es posible gracias a (50).

La desigualdad (55) usada en combinacion con la condicion de incompresibilidad de ϕ, per-mite probar una nueva desigualdad de Carleman donde no aparece el termino |ϕk|2 a la derecha(vease el lema 4 del capıtulo 3).

Con la ayuda de esta ultima desigualdad de Carleman, podemos probar la controlabilidadnula de un sistema linealizado con segundo miembro no necesariamente nulo:

yt −∆y + (y,∇)y + (y,∇)y +∇p = f + v1O, ∇ · y = 0 en Q,

y = 0 sobre Σ,

y(0) = y0 en Ω.

(56)

23

En concreto, en la proposicion 5 del capıtulo 3 se prueba que, fijados f e y0 adecuados, tenemosel control nulo de (56) con controles v cuya k-esima componente es nula. Ademas, se consigue eneste resultado que el campo de velocidades asociado y decrezca exponencialmente a cero cuandot → T−. Para esto ultimo se usan la desigualdad de Carleman probada en el lema 4 del capıtulo3 y ciertos argumentos debidos a O. Yu. Imanuvilov (veanse por ejemplo [35] y [21]).

Por ultimo, con un argumento local basado en un teorema de la funcion inversa (vease elteorema 8 del capıtulo 3), se demuestra la controlabilidad local a trayectorias enunciada en elteorema 4.

• En la segunda parte de este capıtulo estudiamos las propiedades de controlabilidad delsistema de Boussinesq:

yt −∆y + (y,∇)y +∇p = v1O + θ eN , ∇ · y = 0 en Q,

θt −∆θ + y · ∇θ = h1O en Q,

y = 0, θ = 0 sobre Σ,

y(0) = y0, θ(0) = θ0 en Ω

(57)

en dimension N = 2 y en dimension N = 3.Probaremos un resultado analogo al teorema 4. Ası, fijemos una trayectoria no controlada

(y, p, θ), esto es, una solucion de

yt −∆y + (y,∇)y +∇p = θ eN , ∇ · y = 0 en Q,

θt −∆θ + y · ∇θ = 0 en Q,

y = 0, θ = 0 sobre Σ,

y(0) = y0, θ(0) = θ0 en Ω.

(58)

Al igual que antes, supondremos que y verifica (51). Supondremos tambien que

θ ∈ L∞(Q), θt ∈ L2(0, T ;Lσ(Ω))

(σ > 1 if N = 2σ > 6/5 if N = 3

). (59)

El resultado que probamos es el siguiente:

Teorema 5 Supongamos que O satisface (50) con nk(x0) 6= 0 para cierto k < N . Entonces, paracualquier T > 0, (57) es localmente exactamente controlable en el instante T a las trayectorias(y, p, θ) que cumplen (51) y (59), con controles v y h de clase L2 tales que vk ≡ vN ≡ 0.En particular, si N = 2, tenemos el control exacto local a trayectorias con controles v ≡ 0 yh ∈ L2(O × (0, T )).

Analogamente a como ocurre en el paragrafo anterior, todo reposa sobre una desigualdad deCarleman ‘apropiada’ para el problema adjunto

−ϕt −∆ϕ− (Dϕ) y +∇π = g + θ∇ψ, ∇ · ϕ = 0 en Q,

−ψt −∆ψ − y · ∇ψ = q + ϕN en Q,

ϕ = 0, ψ = 0 sobre Σ,

ϕ(T ) = ϕ0, ψ(T ) = ψ0 en Ω.

(60)

24

De hecho, se probo en [31] que si y y θ verifican respectivamente (51) y (59), entonces existeuna constante C > 0 de modo que, para todo (ϕ0, ψ0) ∈ L2(Ω)N × L2(Ω), se tiene:

∫∫

Qe

−2αt4(T−t)4 t−12(T − t)−12 (|ϕ|2 + |ψ|2) dx dt

+∫∫

Qe

−2αt4(T−t)4 t−4(T − t)−4 (|∇ϕ|2 + |∇ψ|2) dx dt

+∫∫

Qe

−2αt4(T−t)4 t4(T − t)4

(|∆ϕ|2 + |∆ψ|2 + |ϕt|2 + |ψt|2)

dx dt

≤ C

(∫∫

Qe−4α+2αt4(T−t)4 t−30(T − t)−30 (|g|2 + |q|2) dx dt

+∫∫

O0×(0,T )e−8α+6αt4(T−t)4 t−64(T − t)−64 (|ϕ|2 + |ψ|2) dx dt

).

(61)

En esta desigualdad, se puede elegir de nuevo el mismo abierto O0 de antes. Luego, usando (50),podemos obtener una desigualdad analoga, sin el termino |ϕk|2 a la derecha. Ademas, en virtudde la ecuacion del calor que aparece en (60), podemos poner la integral de |ϕN |2 en funcion deintegrales de |ψ|2 y de terminos dependientes de ϕN que pueden ser compensados con lo que haya la izquierda. La conclusion es una desigualdad en la cual solo aparecen como terminos locales|ψ|2 y (en dimension 3) |ϕ2|2 (si k = 1) o |ϕ1|2 (si k = 2) (vease el lema 7 del capıtulo 3).

Ahora podemos seguir el mismo razonamiento que para el sistema de Navier-Stokes y deducir,en primer lugar, que tenemos la controlabilidad nula de un sistema linealizado asociado a (57)(proposicion 6) y, por ultimo, la controlabilidad local exacta a trayectorias del sistema (57).

• El ultimo resultado de este capıtulo concierne al siguiente sistema truncado de Navier-Stokes con N = 2:

yt −∆y + (y,∇)TM (y) +∇p = v1O, ∇ · y = 0 en Q,

y · n = 0, ∇× y = 0 sobre Σ,

y(0) = y0 en Ω,

(62)

donde M > 0 es una constante.El objetivo es probar la controlabilidad nula global para este sistema con controles en un

cierto espacio de Hilbert.En concreto, para cada y0 ∈ H, probaremos que existen controles v1O, con v perteneciendo

al espacioW = ∇ × z = (∂2z,−∂1z) : z ∈ L2(0, T ;H1(O)) ,

tales que las soluciones asociadas (y, p) verifican

y(T ) = 0 en Ω. (63)

Por el momento, no se sabe si es cierto este mismo resultado con condiciones de contorno deDirichlet.

El resultado que probamos es el siguiente:

25

Teorema 6 Sea N = 2. Entonces, para todo T > 0 y todo M > 0, (62) es exactamentecontrolable a cero en el tiempo T con controles de la forma v1O, donde v ∈ W .

La estrategia que seguimos para probar este resultado difiere de las utilizadas en las de-mostraciones de los teoremas 4 y 5, pues en este caso resolveremos el problema no lineal a travesde un argumento de punto fijo. Mas concretamente, consideremos para cada y ∈ L∞(Q)2 elsistema

yt −∆y + (y,∇)y +∇p = v1O, ∇ · y = 0 en Q,

y · n = 0, ∇× y = 0 sobre Σ,

y(0) = y0 en Ω(64)

y su adjunto, escrito en la forma

−ρt −∆ρ−∇× ((y · ∇×)∇γ) = 0, ∆γ = ρ en Q,

γ = 0, ρ = 0 sobre Σ,

ρ(T ) = ρ0 en Ω.

(65)

Para las soluciones de este problema se puede de nuevo probar una desigualdad de tipo Carleman(vease el lema 10 del capıtulo 3), que nos conduce a una desigualdad de observabilidad:

‖(∇γ)(0)‖2L2 ≤ C

∫∫

O×(0,T )|∇γ|2 dx dt. (66)

Basandonos en esta desigualdad, se prueba la controlabilidad nula de (64). Ademas, la solu-cion asociada tiene la regularidad siguiente:

y ∈ L2(0, T ; H1(Ω)2) ∩H1(0, T ; H−1(Ω)2)

(proposicion 7 del capıtulo 3). Esto permite probar que una adecuada ecuacion de punto fijo enL2(Q)2 posee soluciones. En consecuencia, nuestro problema de controlabilidad nula para (62)posee solucion.

Controlabilidad del sistema de Lame anisotropico

En la cuarta y ultima parte de esta memoria, probaremos la controlabilidad exacta delsistema de Lame tridimensional anisotropico siguiente:

ρ(x′)∂2x0

u− µ∆u− (λ + µ)∇(∇ · u) + [(R,∇)∇t]u = f + v1ω en Q,

u = 0 sobre Σ,

u(0) = u0, ∂x0u(0) = u1 en Ω,

(67)

donde las funciones f , u0 y u1 son dadas y ω ⊂ Ω es un abierto no vacıo.Como se explico en la introduccion, este sistema proviene del mas general (9)–(10) donde se

supone que los βj = 0 y por tanto

σ = R + (∇u)R + λ(trε)I + 2µε.

26

En este trabajo, necesitamos imponer varias hipotesis de regularidad y positividad de R y loscoeficientes de Lame. En concreto, supondremos que

ρ, λ, µ, Rij ∈ C2(Ω), ∇ ·R = 0, ρ > 0, µ−R33 > 0, y λ + 2µ−R33 > 0 en Ω, (68)

para i, j = 1, 2, 3.La herramienta clave para establecer este resultado de controlabilidad es una desigualdad de

Carleman para las soluciones de sistemas analogos de la forma

ρ(x′)∂2x0

u− µ∆u− (λ + µ)∇(∇ · u) + [(R,∇)∇t]u = f en Q,

u = 0 sobre Σ,

u(0) = ∂x0u(0) = u(T ) = ∂x0u(T ) = 0 en Ω.

(69)

A continuacion destacamos los puntos fundamentales de la prueba de esta estimacion.Definimos la funcion peso

φ(x) = eτψ(x),

donde τ > 0 y suponemos que ψ verifica las hipotesis detalladas en la Condicion A que apareceal principio de la segunda seccion del capıtulo 4. En la practica, esta condicion juega un papelanalogo al desempenado por la condicion geometrica de Bardos, Lebeau y Rauch para la ecuacionde ondas (vease [6]). Tenemos el siguiente resultado:

Teorema 7 Sea f ∈ H1(Q)3 y supongamos que existe una funcion ψ = ψ(x) con ∂ψ∂n sufi-

cientemente grande verificando la Condicion A y que los coeficientes de Lame verifican (68).Entonces existe τ∗ > 0 tal que para todo τ > τ∗, existe s∗ > 0 de modo que toda solucionu ∈ L2(0, T ; H2(Ω)3) ∩H1(Q)3 de (69) verifica

‖u‖Yφ(Q) ≤ C(‖esφf‖H1,s(Q)3 + ‖u‖Bφ(Qω)) ∀s > s∗ (70)

para cierta constante C > 0 independiente de s.

En el teorema 7, hemos utilizado la notacion Qω para designar el dominio de control (Qω =ω × (0, T ) con ω ⊂ Ω abierto no vacıo). Tambien hemos denotado

‖u‖2H1,s(Q) = s2

∫

Q|u|2 dx +

∫

Q|∇u|2 dx,

‖u‖2Bφ(Qω) =

∫

Qω

e2sφ(∑2

|α|=0 s4−2|α||Dαu|2 + s|∇(∇× u)|2

+ s3|∇ × u|2 + s|∇(∇ · u)|2 + s3|∇ · u|2)

dx

(71)

y

‖u‖2Yφ(Q) = ‖u‖2

Bφ(Q) + s

∥∥∥∥esφ ∂u

∂n

∥∥∥∥2

H1,s(Σ)

+ s

∥∥∥∥esφ ∂2u

∂n2

∥∥∥∥2

L2(Σ)

. (72)

27

Explicaremos a continuacion cual es la idea de la demostracion del teorema 7. Empezaremosescribiendo las ecuaciones que verifican las funciones ∇× u y ∇ · u:

P (x,D)(∇× u) ≡ ∂2x0

(∇× u)− µ∆(∇× u) + (R,∇)∇T (∇× u)

= ∇× f + P1u(73)

yP (x,D)(∇ · u) ≡ ∂2

x0(∇ · u)− (λ + 2µ)∆(∇ · u) + (R,∇)∇T (∇ · u)

= ∇ · f + P2u,(74)

con P1 y P2 operadores diferenciales de segundo orden.Observese que no tenemos condiciones de contorno para ∇ × u ni ∇ · u. Utilizaremos una

desigualdad de Carleman probada en [49] que puede ser aplicada a las ecuaciones (73) y (74).Esta desigualdad, escrita para ∇× u y para ∇ · u, nos dice que

s‖esφ(∇× u)‖2H1,s(Q)3 + s‖esφ(∇ · u)‖2

H1,s(Q) ≤ C(‖esφf‖2

H1,s(Q)3

+s


∂n

∥∥∥∥2

H1,s(Σ)3+ s

∥∥∥∥esφ ∂2u

∂n2

∥∥∥∥2

L2(Σ)3+ ‖u‖2

Bφ(Qω)

)∀s ≥ s0.

(75)

Usando ahora la identidad∆u = −∇× (∇× u) +∇(∇ · u)

y la condicion de contorno u = 0 sobre Σ, deducimos directamente que

‖u‖2Yφ(Q) ≤ C

(‖esφf‖2

H1,s(Q)3 + s


∂n

∥∥∥∥2

H1,s(Σ)3+ s

∥∥∥∥esφ ∂2u

∂n2

∥∥∥∥2

L2(Σ)3

+‖u‖2Bφ(Qω)

), ∀s ≥ s0.

(76)

Para obtener (70), sera suficiente estimar los terminos frontera de la expresion anterior. Coneste objetivo, definimos una nueva funcion peso ϕ que verifica ϕ = φ sobre Σ:

ϕ = eτψ, con ψ = ψ − 1Z2

`1 + Z`21,

donde Z es un numero suficientemente grande y `1 es una funcion regular que verifica

`1 = 0 sobre ∂Ω, `1 > 0 en Ω y ∇`1 6= 0 sobre ∂Ω.

En concreto, si denotamos Ω1/Z2 al conjunto de los puntos de Ω que se encuentran a una distanciade ∂Ω inferior a 1/Z2, podemos suponer que

ϕ(x) < φ(x) ∀x ∈ Ω1/Z2 × (0, T ). (77)

Para una funcion peso de este tipo podemos probar el siguiente resultado:

28

Lema 1 Con la notacion anterior, tenemos que

‖u‖Yϕ(Q) ≤ C(‖esϕf‖H1,s(Q)3 + ‖u‖Bϕ(Qω)

) ∀s ≥ s0, (78)

para las soluciones u de (69) con soporte en Ω1/Z2 × [0, T ].

Supongamos por un momento que hemos probado el lema 1 y comprobemos que esto bastapara concluir la prueba del teorema 7.

Tomemos ε ∈ (0, 1/Z2), con Z tal que se cumple (77). Entonces

ϕ(x) < φ(x) ∀x ∈ (Ωε \ Ωε/2)× [0, T ]. (79)

Definimos ahora una funcion truncante θ ∈ C2c (Ωε) verificando θ = 1 en Ωε/2. Obtenemos que

la funcion θu verifica

P (θu) ≡ θf + [P, θ]u en Q,

θu = 0 sobre Σ,

(θu)(T ) = ∂x0(θu)(T ) = (θu)(0) = ∂x0(θu)(0) = 0 en Ω.

Aplicamos pues la desigualdad (78) a θu y deducimos que

s


∂n

∥∥∥∥2

H1,s(∂Ω×(0,T ))3+ s

∥∥∥∥esφ ∂2u

∂n2

∥∥∥∥2

L2(∂Ω×(0,T ))3

≤ C(‖esϕf‖2

H1,s(Q)3 + ‖esϕ[P, θ]‖2H1,s(Q)3 + ‖u‖2

Bϕ(Qω)

)∀s ≥ s0,

(80)

pues ϕ = φ sobre la frontera. Como el soporte de [P, θ]u esta contenido en Ωε \ Ωε/2 × [0, T ] yϕ < φ en este conjunto (vease (77)), tenemos que

‖esϕ[P, θ]u‖H1,s(Q)3 ≤ C2∑

|α|=0

‖esφDαu‖L2(Q)3 ∀s > 0.

Esto termina la prueba del teorema 7.

Por lo tanto, en la prueba del lema 1 podemos restringirnos a la situacion supp u ⊂ Bδ ∩(Ω1/Z2 × [0, T ]). En primer lugar, realizamos el cambio de variables

y1 = x1,

y2 = x2,

y3 = x3 − `(x1, x2),

que nos permite situarnos en un pequeno entorno del punto y∗ = (y0, 0, 0, 0). Escribimos ahorael sımbolo principal de las ecuaciones satisfechas por ∇× u y ∇ · u en las nuevas coordenadas:

pβ(y, ξ) = −ξ20 + (β −R11)ξ2

1 + (β −R22)ξ22 + [(βE3 −R)Gt]Gξ2

3

−2R12ξ1ξ2 − 2∑2

j=1(Rj3 + `yj (β −Rjj)− `y3−jR12)ξjξ3,(81)

con

29

β =

µ para ∇× u,

λ + 2µ para ∇ · uy

G = (−`y1 ,−`y2 , 1)t. (82)

Consideremos ahora un recubrimiento finito

ζ ∈ S3 : |ζ − ζ∗ν | < δ11≤ν≤M(δ1),

de la esfera unidad de R4

S3 = ζ = (s, ξ′) : s2 + ξ20 + ξ2

1 + ξ22 = 1.

Aquı, ζ∗ν representa un punto de S3. A este recubrimiento le asociamos una particion de la unidadχν1≤ν≤M , extendiendo χν fuera de S3 por una funcion homogenea de orden 0 con soporte en

O(δ1) =

ζ :∣∣∣∣

ζ

|ζ| − ζ∗ν

∣∣∣∣ < δ1

.

El lema 1 es entonces consecuencia del lema siguiente:

Lema 2 Sea γ∗ = (y∗, ζ∗) ∈ ∂G×S3. Entonces, si δ y δ1 son suficientemente pequenos, tenemos

s‖zν‖2H1,s(G)4 + s(‖zν‖2

H1,s(∂G)4 + ‖∂y3zν‖2L2(∂G)4)

≤ C(‖esϕf‖2H1,s(G)3 + ‖z‖2

H1,s(G)4).(83)

En esta desigualdad, hemos introducido el conjunto G = R3 × [0, 1/Z2] y la funcion zν =χν(s,D′)z, donde z = esϕw. y w = (w′, w0) es el par (∇× u,∇ · u) escrito en la nueva variabley.

Para demostrar el lema 2, distinguimos varios casos, segun que los sımbolos principales de∇×u y ∇·u (definidos en (81)) se anulen o no en γ∗. En cada uno de los casos, las desigualdadesbuscadas son, esencialmente, consecuencia de la desigualdad de Garding (vease, por ejemplo,[50]). Todos los detalles de la prueba se encuentran en la segunda seccion del capıtulo 4.

Explicaremos ahora brevemente como se puede deducir el lema 1 a partir del lema 2. De ladefinicion de zν , tenemos directamente que

s‖z‖2H1,s(G)4 + s(‖z‖2

H1,s(∂G)4 + ‖∂y3z‖2L2(∂G)4)

≤ CsM∑

ν=1

(‖zν‖2H1,s(G)4 + ‖zν‖2

H1,s(∂G)4 + ‖∂y3zν‖2L2(∂G)4)

≤ C(‖esϕf‖2H1,s(G)3 + ‖esϕu‖2

H2,s(G)3).

(84)

Ahora usamos el resultado enunciado en la proposicion 4.2 de [39] y encontramos que

Z2∑

|α|=0

s4−2|α|‖esϕDαy′u‖2

L2(G)3 ≤ C(‖esϕf‖2H1,s(G)3 + ‖esϕu‖2

H2,s(G)3),

30

con C una constante positiva independiente de s y de Z.Usando la ecuacion diferencial en (69) junto con desigualdades de interpolacion, deducimos

que las normas ‖esϕ∂2y0

u‖2L2(G)3 y ‖esϕ∂2

y0y1u‖2

L2(G)3 pueden anadirse a la izquierda de la ultimadesigualdad. Por tanto, combinando esto con (84), obtenemos que

2∑

|α|=0

s4−2|α|‖esϕDαu‖2L2(G)3 + s(‖z‖2

H1,s(∂G)4 + ‖∂y3z‖2L2(∂G)4) ≤ C‖esϕf‖2

H1,s(G)3 . (85)

Finalmente, veamos que la desigualdad (78) se puede deducir a partir de (85), es decir, quepodemos acotar apropiadamente los terminos frontera:

s ‖esϕ∂y3u‖2H1,s(∂G)3 y s

∥∥esϕ∂2y3

u∥∥2

L2(∂G)3. (86)

Para ello, solo usaremos la definicion de las variables yj y las condiciones de contorno de tipoDirichlet para u, En primer lugar, tenemos que

|∂y3uj | ≤ |(∇× u)3−j |+ ε(δ)|∂y3u| sobre ∂G j = 1, 2,

|∂y3u3| ≤ |∇ · u|+ ε(δ)|∂y3u| sobre ∂G,

lo cual nos dice que|esϕ∂y3u| ≤ |z1|+ |z2|+ |z4| sobre ∂G. (87)

Ademas, del mismo modo deducimos que

|∂2yjy3

uk| ≤ |∂yj (∇× u)3−k|+ ε(δ)|∂yjy3u| sobre ∂G j = 0, 1, 2, k = 1, 2

y|∂2

yjy3u3| ≤ |∂yj (∇ · u)|+ ε(δ)|∂yjy3u| sobre ∂G j = 0, 1, 2.

Luego|esϕ∇tg

y ∂y3u| ≤ |∇tgy z1|+ |∇tg

y z2|+ |∇tgy z4| sobre ∂G. (88)

Por ultimo,

|∂2y3

uj | ≤ |∂y3(∇× u)3−j |+ |∂yj (∇ · u)|+ ε(δ)|∂y3∇yu| sobre ∂G j = 1, 2

y|∂2

y3u3| ≤ |∂y1(∇× u)2|+ |∂y2(∇× u)1|+ |∂y3(∇ · u)|+ ε(δ)|∂y3∇yu| sobre ∂G,

lo cual conduce a la desigualdad

|esϕ∂2y3

u| ≤ |∇yz1|+ |∇yz2|+ |∇yz4|+ ε(δ)|esϕ∂y3∇tgy u| sobre ∂G. (89)

Podemos ver ya directamente que las estimaciones (87)–(89) implican la desigualdad deseadapara los terminos (86).

Con esto, queda justificada la prueba del lema 1 a partir de la del lema 2.

Podemos ahora formular el resultado principal de este capıtulo:

31

Teorema 8 En las condiciones del teorema 7, si la funcion ψ verifica ademas la Condicion B(vease la seccion 3 del capıtulo 4), se tienen las propiedades de controlabilidad siguientes:• Dados f ∈ L2(Q), (u0, u1) ∈ H1

0 (Ω) × L2(Ω) y (u2, u3) ∈ H10 (Ω) × L2(Ω), existe un control

v ∈ L2(ω × (0, T )) tal que la correspondiente solucion u de (67) verifica

u(T ) = u2, ∂x0u(T ) = u3 en Ω. (90)

• Dados f ∈ H−1(Q), (u0, u1) ∈ L2(Ω)×H−1(Ω) y (u2, u3) ∈ L2(Ω)×H−1(Ω), existe un controlv ∈ H−1(ω × (0, T )) tal que la correspondiente solucion de (67) verifica (90).

La demostracion del teorema 8 reposa sobre una desigualdad de observabilidad adecuada que esconsecuencia directa del teorema 7. Para mas detalles, vease el capıtulo 4.

Otros resultados

En esta seccion mencionaremos otros resultados relacionados con los que preceden (aunquedistintos), en cuya obtencion tambien ha intervenido el autor de la memoria.

En [19], se presenta una panoramica actualizada del papel que juegan las desigualdadesglobales de Carleman en el contexto de la controlabilidad de ecuaciones en derivadas parciales. Sepresta especial atencion a las demostraciones de dichas desigualdades para funciones que verificanecuaciones parabolicas de tipo calor o Stokes. Se tratan distintos casos segun la regularidaddel segundo miembro y las condiciones de contorno impuestas sobre la solucion. Tambien serecuerdan algunos de los ultimos avances con respecto a la controlabilidad de dichas ecuaciones.

En el trabajo [21] se prueba el control local a las trayectorias del sistema de Navier-Stokes.Los resultados principales se habıan expuesto previamente en [20]. Estos trabajos constituyenun paso previo a los resultados de [23] (que fueron anunciados en la Nota [22]), recogidos comose ha dicho en el capıtulo 3.

Tambien, como etapa previa a uno de los resultados establecidos en [23], se prueba en [31] elcontrol local a las trayectorias del sistema de Boussinesq con ayuda de N +1 controles escalares.

Por ultimo, otro trabajo a resenar es [30], donde el autor prueba el control local a lastrayectorias de un sistema de Navier-Stokes con condiciones de contorno de tipo Navier nolineales.

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[39] O. Yu. Imanuvilov y M. Yamamoto, Carleman estimates for the three-dimensional non-stationary Lame system and the application to an inverse problem, to appear in Contemp.Math.

[40] O.Yu. Imanuvilov y M. Yamamoto, Carleman estimates for the Lame system withstress boundary condition and the application to an inverse problem, (Preprint: Universityof Tokyo Graduate School of Mathematical Sciences UTMS 2003-2).

36

[41] O. Yu. Imanuvilov y M. Yamamoto, Carleman estimate for stationary isotropicLame system, 83, (2004), 243-270.

[42] V. Isakov, G. Nakamura y J.-N. Wang Uniqueness and stability in the Cauchy problemfor the elasticity system with residual stress, Contemp. Math. 333 ( 2003), 99–113.

[43] G. Lebeau y L. Robbiano, Controle exact de l’equation de la chaleur. (French) [Exactcontrol of the heat equation], Comm. Partial Differential Equations 20 (1995), no. 1-2, 335–356.

[44] C.-L. Lin y J.-N. Wang, Uniqueness in inverse problems for an elasticity system withresidual stress by a single measurement, Inverse Problems 19 (2003), no. 4, 807–820.

[45] J.-L. Lions, Exact controllability, stabilization and perturbations for distributed systems,SIAM Review 30 (1988) no. 1, 1–68.

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[47] G. Nakamura y J.-N. Wang, Unique continuation for an elasticity system with the resid-ual stress and its applications, SIAM J. Math. Anal 35 (2003), no 2, 304-317.

[48] L. Tartar, An introduction to Sobolev spaces and interpolation spaces, Course (2000),http://www.math.cmu.edu/cna/publications/SOB+Int.pdf.

[49] D. Tataru, Carleman estimates and unique continuation for solutions to boundary valueproblems, J. Math. Pures Appl., 75, 1996, 367-408.

[50] M. E. Taylor, Pseudodifferential operators, Princeton Mathematical Series, 34. PrincetonUniversity Press, Princeton, N.J. (1981).

[51] N. Weck, Unique continuation for systems with Lame principal part, Math. Meth. Appl.Sci., 24 (2001), no. 9, 595-605.

[52] E. Zuazua, Exact boundary controllability for the semilinear wave equation, in “NonlinearPartial Differential Equations and their Applications”, Vol. X, H. Brezis and J.L. Lionseds., Pitman, 1991, 357–391.

Capıtulo 1

Null controllability of the heatequation with Fourier boundaryconditions: The linear case

38

Null controllability of the heat equationwith Fourier boundary conditions: The

linear case

E. Fernandez-Cara, M. Gonzalez-Burgos, S. Guerrero y J.-P. Puel

Abstract

In this paper, we prove the global null controllability of the linear heat equation completedwith linear Fourier boundary conditions of the form ∂y

∂n + β y = 0. We consider distributedcontrols with support in a small set and nonregular coefficients β = β(x, t). For the proof ofnull controllability, a crucial tool will be a new Carleman estimate for the weak solutions of theclassical heat equation with nonhomogeneous Neumann boundary conditions.

1. Introduction

Let Ω ⊂ RN be a bounded connected open set whose boundary ∂Ω is regular enough(N ≥ 1). Let ω ⊂ Ω be a (small) nonempty open subset and let T > 0. We will use the notationQ = Ω × (0, T ) and Σ = ∂Ω × (0, T ) and we will denote by n(x) the outward unit normal toΩ at x ∈ ∂Ω. On the other hand, we will denote by C, C1 , C2 , . . . generic positive constants(usually depending on Ω and ω).

We will consider the linear heat equation with linear Fourier (or Robin) conditions

yt −∆y + B(x, t) · ∇y + a(x, t) y = v(x, t)1ω in Q,

∂y

∂n+ β(x, t) y = 0 on Σ,

y(x, 0) = y0(x) in Ω.

(1.1)

Here, it will be assumed that the coefficients a, B and β satisfy

a ∈ L∞(Q), B ∈ L∞(Q)N , β ∈ L∞(Σ). (1.2)

On the other hand, we suppose that v ∈ L2(ω × (0, T )), 1ω is the characteristic function ofω and y0 ∈ L2(Ω). In (1.1), y = y(x, t) is the state and v = v(x, t) is the control. It is assumedthat we can act on the system only through ω × (0, T ).

An illustrative interpretation of the data and variables in (1.1) is the following. The functiony can be viewed as the relative temperature of a body (with respect to the exterior surroundingair). The parabolic equation in (1.1) means, among other things, that a heat source v1ω acts

39

on a part of the body. On the boundary, − ∂y∂n must be viewed as the normal heat flux, directed

inwards, up to a positive coefficient. Thus, the equality

−∂y

∂n= β y

means that this flux is a linear function of the temperature. Thus, it is reasonable to supposethat β ≥ 0 (although this assumption will not be imposed in this paper).

The main goal of this paper is to analyze the controllability properties of (1.1). It willbe said that this system is null controllable at time T if, for each y0 ∈ L2(Ω), there existsv ∈ L2(ω × (0, T )) such that the associated solution satisfies

y(x, T ) = 0 in Ω. (1.3)

The null controllability of linear parabolic equations has been intensively studied these lastyears; see for instance [8], [7], [5], [4] and [1].

In this paper, we will be concerned with (1.1), where the main difficulties arise from theparticular form of the boundary condition. Indeed, it has been shown in [5] and [2] that this ismore difficult to analyze than the case of Dirichlet boundary conditions, considered in [7], [5]and [4].

More precisely, what has been proved until now is that (1.1) is null controllable with B ≡ 0under the assumptions (1.2) whenever βt ∈ L∞(Σ). This was shown in [5]. However, it wouldbe important to prove the null controllability of (1.1) without this regularity hypothesis on βt

in view of applications to control systems with nonlinear boundary conditions.

The first main result in this paper concerns a Carleman inequality for a general (adjoint)system of the form

−ϕt −∆ϕ = f1(x, t) +∇ · f2(x, t) in Q,

(∇ϕ + f2(x, t)) · n = f3(x, t) on Σ,

ϕ(x, T ) = ϕ0(x) in Ω,

(1.4)

where f1 ∈ L2(Q), f2 ∈ L2(Q)N and f3 ∈ L2(Σ). Observe that, as long as ϕ ∈ L2(Q), ∇ϕ+f2 ∈L2(Q)N and ∇ · (∇ϕ + f2) ∈ H−1(0, T ;L2(Ω)), we can give a sense to the boundary conditionin the space H−1(0, T ; H−1/2(∂Ω)).

We present now this result:

Theorem 1 Under the previous assumptions on f1 , f2 and f3 , there exist λ, σ1, σ2 and C,only depending on Ω and ω, such that, for any λ ≥ λ, any s ≥ s = σ1(eσ2λ T + T 2) and any

40

ϕ0 ∈ L2(Ω), the weak solution to (1.4) satisfies∫∫

Qe−2sα(s λ2ξ |∇ϕ|2 + s3 λ4 ξ3 |ϕ|2)dx dt

+s2λ3

∫∫


≤ C

(∫∫

Qe−2sα(|f1|2 + s2 λ2 ξ2 |f2|2)dx dt

+sλ

∫∫

Σe−2sα ξ |f3|2 dσ dt + s3λ4

∫∫


).

(1.5)

Here, α = α(x, t) and ξ = ξ(x, t) are appropriate positive functions, again only dependingon Ω and ω. They are given below; see (1.13)–(1.14).

As a consequence of theorem 1, we can deduce an observability inequality for the adjointsystem associated to (1.1). More precisely, let us consider the backward in time system

−ϕt −∆ϕ−∇ · (ϕB(x, t)) + a(x, t) ϕ = 0 in Q,

(∇ϕ + ϕB(x, t)) · n + β(x, t)ϕ = 0 on Σ,

ϕ(x, T ) = ϕ0(x) in Ω,

(1.6)

where ϕ0 ∈ L2(Ω). It will be seen that, for some K of the form

K = eC(1+ 1T

+‖a‖2/3∞ +‖B‖2∞+‖β‖2∞), (1.7)

the solutions of (1.6) satisfy∫∫

Ω×(T/4,3T/4)|ϕ|2 dx dt ≤ K

∫∫

ω×(0,T )|ϕ|2 dx dt. (1.8)

Remark 1 In fact, (1.8) is not the unique way of saying that (1.6) is observable. It is indeedmore frequent to use other inequalities of the form

‖ϕ(·, 0)‖2L2(Ω) ≤ C

∫∫

ω×(0,T )|ϕ|2 dx dt (1.9)

for some C. The estimates (1.9) can be easily deduced from (1.8) and the energy inequalitiessatisfied by ϕ.

The second main result in this paper concerns the null controllability of (1.1). It is thefollowing:

Theorem 2 Let us assume that (1.2) is satisfied. Then, for each T > 0, (1.1) is null controllableat time T with controls v ∈ L2(ω × (0, T )). Moreover, one can find v such that

‖v‖L2(ω×(0,T )) ≤ H ‖y0‖L2 , (1.10)

41

with a constant H of the form

H = eC(1+ 1T

+‖a‖2/3∞ +‖B‖2∞+‖β‖2∞+T (‖a‖∞+‖B‖2∞+‖β‖2∞)) (1.11)

for some C = C(Ω, ω).

In the proof of theorem 2, the main tool is the estimate (1.8). This arises from a generalprinciple that asserts that the null controllability of (1.1) with controls in L2(ω × (0, T )) (de-pending continuously on the data) is equivalent to the observability of (1.6). More details willbe given below.

In a second part of this work, which will appear in a forthcoming paper, we will consider con-trollability questions for semilinear heat equations completed with nonlinear Fourier boundaryconditions of the form

∂y

∂n+ f(y) = 0 on Σ,

where f : R 7→ R is locally Lipschitz-continuous. For the analysis of these systems, theorems 1and 2 of the present paper will be crucial.

The rest of this paper is organized as follows. Section 2 is devoted to the proof of theorem 1. Insection 3, we deduce the observability inequality (1.8) and we prove theorem 2. For completeness,we have included an Appendix, where we give a detailed proof of the standard Carleman estimatefor the solutions of the heat equation with homogeneous Neumann boundary conditions. (thisestimate was already proved in [5]; however, in this paper, a careful study of the dependence ofthe constants on s, λ and T is needed).

2. Proof of theorem 1

The main arguments used below are similar to those in [6]. This is related to a generalstrategy which is used to relax the regularity assumptions on the various coefficients involved inthe problem. Here, it will allow us to proceed without any kind of regularity on the coefficientβ = β(x, t).

Let us recall the definition of a weak solution: we say that ϕ is a weak solution to (1.4) if itsatisfies

ϕ ∈ L2(0, T ;H1(Ω)) ∩ C0([0, T ]; L2(Ω)),

−〈ϕt, v〉(H1(Ω))′,H1(Ω) +∫

Ω∇ϕ · ∇v dx =

∫

Ωf1(x, t) v dx

−∫

Ωf2(x, t) · ∇v dx +

∫

∂Ωf3(x, t) v dσ

a.e. in (0, T ), ∀v ∈ H1(Ω),

ϕ(x, T ) = ϕ0(x) in Ω.

(1.12)

It is well known that, for f1 ∈ L2(Q), f2 ∈ L2(Q)N , f3 ∈ L2(Σ) and ϕ0 ∈ L2(Ω), (1.4) possessesexactly one weak solution ϕ.

42

To prove the Carleman inequality (1.5), we will need two weight functions:

ξ(x, t) =eλη0(x)

t(T − t), α(x, t) =

e2λ‖η0‖∞ − eλη0(x)

t(T − t). (1.13)

Here, λ ≥ 1 is a parameter to be chosen below and η0 = η0(x) is a function satisfying

η0 ∈ C2(Ω), η0(x) > 0 in Ω, η0(x) = 0 on ∂Ω,

|∇η0(x)| > 0 in Ω \ ω′,(1.14)

where ω′ ⊂⊂ ω is a nonempty open set. The existence of η0 satisfying (1.14) is proved in [5].For the proof of theorem 1, we will need an auxiliary result: a Carleman inequality for the

solutions to the heat equation with homogeneous Neumann boundary conditions. This is givenin the following result:

Lemma 1 Let f ∈ L2(Q) be given. There exist λ∗, σ∗ and C only depending on Ω and ω suchthat, for any λ ≥ λ∗, any s ≥ s∗(λ) = σ∗(e4λ‖η0‖∞ T +T 2) and any q0 ∈ L2(Ω), the weak solutionto

−qt −∆q = f(x, t) in Q,

∂q

∂n= 0 on Σ,

q(x, T ) = q0(x) in Ω

satisfies

Is,λ(q) ≤ C

(∫∫

Qe−2sα |f |2 dx dt + s3λ4

∫∫


), (1.15)

where we have used the notation

Is,λ(q) =∫∫

Qe−2sα

((sξ)−1(|qt|2 + |∆q|2) + sλ2ξ |∇q|2 + s3λ4ξ3 |q|2) dx dt.

This result is a particular case of lemma 1.2 of Chapter I in [5]. For completeness and also inorder to explain and justify the particular form of the constants λ∗ and s∗(λ), we give a completeproof in the Appendix, at the end of this paper.

Let us continue with the proof of theorem 1. We can view ϕ as a solution by transpositionof (1.4). This means that ϕ is the unique function in L2(Q) satisfying

∫∫

Qϕhdx dt =

∫∫

Qf1(x, t) z dx dt−

∫∫

Qf2(x, t) · ∇z dx dt

+∫∫

Σf3(x, t) z dσ dt +

∫

Ωϕ0(x) z(x, T ) dx ∀h ∈ L2(Q),

(1.16)

43

where we have denoted by z the (strong) solution of the following problem:

zt −∆z = h(x, t) in Q,

∂z

∂n= 0 on Σ,

z(x, 0) = 0 in Ω.

We will argue as follows. Let us first estimate the second term in the left hand side of (1.5),i.e.

s3λ4

∫∫

Qe−2sα ξ3 |ϕ|2 dx dt. (1.17)

To this end, we will deal with techniques inspired by the arguments in [6].Thus, let us see that the term in (1.17) can be bounded by the right hand side of (1.5), i.e.

s3λ4

∫∫

Qe−2sα ξ3 |ϕ|2 dx dt ≤ C(Ω, ω)

(∫∫


+s2λ2

∫∫


∫∫


+s3λ4

∫∫


)(1.18)

for a good choice of the parameters λ and s.

Let us consider the following constrained extremal problem:

Minimize12

(∫∫

Qe2sα |z|2 + s−3λ−4

∫∫


)

subject to v ∈ L2(Q) and

zt −∆z = s3λ4e−2sαξ3ϕ + v1ω in Q,

∂z

∂n= 0 on Σ,

z(x, 0) = 0, z(x, T ) = 0 in Ω.

(1.19)

Here, s and λ are chosen like in lemma 1.By virtue of Lagrange’s principle and arguing as in [6], we are led from (1.19) to the next

optimality system, which is of fourth order in space and second order in time:

L(e−2sα L∗p) + s3λ4e−2sαξ3p1ω = s3λ4e−2sαξ3ϕ in Q,

∂p

∂n= 0,

∂

∂n(e−2sα L∗p) = 0 on Σ,

(e−2sαL∗p)|t=0 = (e−2sαL∗p)|t=T = 0 in Ω.

(1.20)

Here, L = ∂t −∆ is the heat operator and L∗ = −∂t −∆ is its formal adjoint. If p is a solutionto (1.20) (in an appropriate sense), then

v = −s3 λ4 e−2sα ξ3 p1ω and z = e−2sαL∗p (1.21)

44

solve (1.19).Let us show that (1.20) has a unique weak solution. To this end, we are going to rewrite this

problem as a Lax-Milgram variational equation. Let us introduce the space

X0 = z ∈ C2(Q) :∂z

∂n= 0 on Σ

and the norm ‖ · ‖X , with

‖q‖2X =

∫∫

Qe−2sα |L∗q|2 dx dt + s3λ4

∫∫


for all q ∈ X0 .Due to lemma 1, ‖ · ‖X is indeed a norm in X0 . Let X be the completion of X0 for the norm

‖ · ‖X . Then X is a Hilbert space for the scalar product (· , ·)X , with

(p, q)X =∫∫

Qe−2sα (L∗p)(L∗q) dx dt + s3λ4

∫∫

ω×(0,T )e−2sα ξ3 p q dx dt.

With this notation, system (1.20) is equivalent to find a function p ∈ X such that

(p, q)X = `(q) ∀q ∈ X, (1.22)

where`(q) = s3 λ4

∫∫

Qe−2sα ξ3 ϕ q dx dt ∀q ∈ X.

Of course, (1.22) is equivalent to another extremal problem

Minimize12(q, q)X − `(q)

subject to q ∈ X.

By virtue of lemma 1, one can easily check that ` ∈ X ′. Consequently, one can apply Lax-Milgram lemma and deduce that there exists a unique solution to (1.20).

Let us now takeh = s3λ4e−2sαξ3ϕ + v1ω

in (1.16). This gives

s3λ4

∫∫

Qe−2sα ξ3 |ϕ|2 dx dt =

∫∫

Qf1 z dx dt−

∫∫

Qf2 · ∇z dx dt

+∫∫

Σf3 z dσ dt−

∫∫

ω×(0,T )ϕ v dx dt

(1.23)

(recall that v and z are given by (1.21)). The idea of the proof of (1.18) is to bound z, ∇z andv in Q and the trace of z on Σ in terms of the left hand side of (1.23). For this purpose, we firstmultiply the equation in (1.20) by p and integrate in Q, which gives

‖p‖2X ≤ ‖`‖X′ ‖p‖X

45

and, consequently,

‖p‖2X =

∫∫


∫∫


≤ Cs3λ4

∫∫

Qe−2sα ξ3 |ϕ|2 dx dt,

(1.24)

for λ ≥ λ(Ω, ω), s ≥ σ(Ω, ω)(e4λ‖η0‖∞ T + T 2). This provides the desired bounds of z and v1ω .Let us now multiply the equation satisfied by z by s−2λ−2e2sαξ−2z and let us integrate in

Q. After integration by parts, we obtain:

12s−2λ−2

∫∫

Qe2sαξ−2 ∂

∂t|z|2 dx dt + s−2λ−2

∫∫

Qe2sαξ−2 |∇z|2 dx dt

−s−1λ−1

∫∫

Qe2sα ξ−1∇η0 · ∇|z|2 dx dt

−2s−2λ−1

∫∫

Qe2sα ξ−2 (∇η0 · ∇z) z dx dt

= sλ2

∫∫

Qξ ϕ z dx dt + s−2λ−2

∫∫

ω×(0,T )e2sα ξ−2 v z dx dt,

whence

s−2λ−2

∫∫

Qe2sα ξ−2 |∇z|2 dx dt− s−1λ−1

∫∫

Σe2sαξ−1 ∂η0

∂n|z|2 dσ dt

=12s−2λ−2

∫∫

Q

∂

∂t(e2sα ξ−2) |z|2 dx dt

−s−1λ−1

∫∫

Q∇ · (e2sα ξ−1∇η0) |z|2 dx dt

+2s−2λ−1

∫∫

Qe2sα ξ−2∇η0 · ∇z z dx dt + sλ2

∫∫

Qξ ϕ z dx dt

+s−2λ−2

∫∫

ω×(0,T )e2sα ξ−2 v z dx dt.

(1.25)

We need now some estimates concerning the weight functions in order to preserve explicitbounds in s, λ and T . Notice that

∂

∂t(e2sα ξ−2) = −2(T − 2t) e−λη0

e2sα(s e−λη0(e2λ‖η0‖∞ − eλη0

)− ξ−1)

≤ C T e2sα(e2λ‖η0‖∞ s + ξ−1) ≤ C T s e2sα e2λ‖η0‖∞ ,

where we have taken s ≥ C T 2. More generally, observe that, for any fixed m, one also has

|∇(e2sα ξm)| ≤ Cm(Ω, ω) s λ e2sα ξm+1 (1.26)

46

whenever s ≥ C T 2. Indeed, we have

∇(e2sα ξm) = e2sα λ∇η0 ξm(2s ξ + m) ≤ C(Ω, ω) e2sα λ ξm(s ξ + 1)

and, taking into account that

C s ξ ≥ 1 for s ≥ T 2

4C, (1.27)

we directly get (1.26).Turning back to (1.25), we obtain

s−2λ−2

∫∫


∫∫


∂n|z|2 dσ dt

≤ C(Ω, ω)(

T s−1λ−2 e2λ‖η0‖∞∫∫

Qe2sα |z|2 dx dt

+∫∫


∫∫

Qe2sα ξ−1 |z|2 dx dt

+s−2

∫∫

Qe2sα ξ−2 |z|2 dx dt + s2λ4

∫∫

Qe−2sα ξ2 |ϕ|2 dx dt

+s−4λ−4

∫∫

Qe2sα ξ−4 |v|2 dx dt

)+

12s−2λ−2

∫∫

Qe2sα ξ−2 |∇z|2 dx dt,

where we have taken s ≥ C T 2. Now, we take into account (1.27) and we deduce that

s−2λ−2

∫∫


∫∫


∂n|z|2 dσ dt

≤ C(Ω, ω)(∫∫

Qe2sα |z|2 dx dt + s3λ4

∫∫


s−3λ−4

∫∫

Qe2sα ξ−3 |v|2 dx dt

)

for any λ ≥ C(Ω, ω) and any s ≥ C(Ω, ω)(e2λ‖η0‖∞ T + T 2).From (1.14), this gives an estimate of the gradient and the trace of z in terms of z, v1ω and

ϕ. In view of (1.24), we now have∫∫


∫∫

Qe2sα ξ−2 |∇z|2 dx dt

+s−1λ−1

∫∫

Σe2sαξ−1 |z|2 dσ dt + s−3λ−4

∫∫


≤ C(Ω, ω) s3λ4

∫∫


for λ ≥ C(Ω, ω), s ≥ C(Ω, ω)(e2λ‖η0‖∞ T + T 2).

47

It suffices to combine this inequality and the identity (1.23) to deduce (1.18).

Let us now show that

sλ2

∫∫

Qe−2sα ξ |∇ϕ|2 dx dt ≤ C(Ω, ω)

(∫∫


+s2λ2

∫∫


∫∫


+s3λ4

∫∫


).

(1.28)

To this end, we now have to use not only that ϕ is a solution by transposition but a weaksolution as well. More precisely, let us take

v = sλ2e−2sα(·,t)ξ(·, t)ϕ(·, t)

in (1.12). Then, let us integrate in (0, T ) and let us perform integrations by parts similarly aswe did before. We get:

−12sλ2

∫∫

Qe−2sαξ

∂

∂t|ϕ|2 dx dt + sλ2

∫∫

Qe−2sαξ |∇ϕ|2 dx dt

+sλ2

∫∫

Q∇ϕ · ∇(e−2sα ξ)ϕ dx dt

= sλ2

∫∫

Qe−2sα ξ f1 ϕdx dt− sλ2

∫∫

Qf2 · ∇(e−2sα ξ ϕ) dx dt

+sλ2

∫∫

Σe−2sα ξ f3 ϕdσ dt.

We integrate by parts again and we obtain

sλ2

∫∫


= −12sλ2

∫∫

Q(e−2sαξ)t |ϕ|2 dx dt− sλ2

∫∫

Q∇ϕ · ∇(e−2sα ξ) ϕdx dt

+sλ2

∫∫

Qe−2sα ξ f1 ϕdx dt− sλ2

∫∫

Qf2 · ∇(e−2sα ξ) ϕ dx dt

−sλ2

∫∫

Qf2 · ∇ϕe−2sα ξ dx dt + sλ2

∫∫

Σe−2sα ξ f3 ϕdσ dt.

48

In view of (1.26), we find:

sλ2

∫∫


≤ C(Ω, ω)(

T s2 λ2 e2λ‖η0‖∞∫∫


+s3λ4

∫∫

Qe−2sα ξ3 |ϕ|2 dx dt +

∫∫


+s2λ4

∫∫

Qe−2sα ξ2 |ϕ|2 dx dt + sλ2

∫∫

Qe−2sα ξ |f2|2 dx dt

sλ

∫∫

Σe−2sα ξ |f3|2 dσ dt + sλ3

∫∫

Σe−2sα ξ |ϕ|2 dσ dt

)

+12sλ2

∫∫

Qe−2sαξ |∇ϕ|2 dx dt,

where we have taken s ≥ CT 2 and λ ≥ C. Making several simplifications, we easily see that

sλ2

∫∫

Qe−2sαξ |∇ϕ|2 dx dt ≤ C

(s3λ4

∫∫


+∫∫

Qe−2sα |f1|2 dx dt + s2λ2

∫∫

Qe−2sα ξ2 |f2|2 dx dt

+sλ

∫∫

Σe−2sα ξ |f3|2 dσ dt + sλ3

∫∫

Σe−2sα ξ |ϕ|2 dσ dt

),

(1.29)

for s ≥ C(e2λ‖η0‖∞ T + T 2) and λ ≥ C, whence (1.28) follows easily.

Let us finally estimate the trace of ϕ in terms of ϕ and ∇ϕ. Notice that

−s2λ3

∫∫

Qe−2sα ξ2 (∇η0 · ∇ϕ) ϕ dx dt

= −12s2λ3

∫∫

Σe−2sα ξ2 ∂η0

∂n|ϕ|2 dσ dt

+12s2λ3

∫∫

Q∇ · (e−2sα ξ2∇η0) |ϕ|2 dx dt.

Taking into account (1.14), the following is found:

s2λ3

∫∫

Σe−2sα ξ2 |ϕ|2 dσ dt ≤ C s2 λ3

∫∫

Q|∇ · (e−2sα ξ2∇η0)| |ϕ|2 dx dt

+C

(s3λ4

∫∫


∫∫


)

≤ C

(s3λ4

∫∫


∫∫


),

49

with s ≥ C(T + T 2) and λ ≥ C.This last inequality, together with (1.18) and (1.29), provides (1.5) and permits to achieve

the proof of theorem 1.

3. Controllability of the linear system

This section is devoted to prove theorem 2. This will be a consequence of the Carlemaninequality (1.5).

We will start with an explicit bound of the weak solution to the linear problem

yt −∆y + B(x, t) · ∇y + a(x, t) y = f(x, t) in Q,

∂y

∂n+ β(x, t)y = 0 on Σ,

y(x, 0) = y0(x) in Ω,

(1.30)

where f ∈ L2(Q), y0 ∈ L2(Ω) and (1.2) is fulfilled. Then, we will use this result in combinationwith (1.5) to deduce the observability inequality (1.8) for the solutions to (1.6). Finally, we willend the proof of theorem 2 in a classical way, using this observability inequality.

Proposition 1 Under the previous assumptions, the weak solution to (1.30) satisfies the esti-mate

‖y‖Y ≤ eC T (1+‖a‖∞+‖B‖2∞+‖β‖2∞) (‖f‖L2(Q) + ‖y0‖L2(Ω)) (1.31)

for some constant C > 0. Here, Y is the usual energy space:

Y = L2(0, T ;H1(Ω)) ∩ C0([0, T ];L2(Ω)).

Proof: The existence and uniqueness of a solution to (1.30) is well known. Furthermore, thefollowing identity can be deduced for each t ∈ (0, T ) in a standard way:

12

d

dt

∫

Ω|y(x, t)|2 dx +

∫

Ω|∇y(x, t)|2 dx +

∫

∂Ωβ(x, t) |y(x, t)|2 dσ

+∫

ΩB(x, t) · ∇y(x, t) y(x, t) dx +

∫

Ωa(x, t) |y(x, t)|2 dx

=∫

Ωf(x, t) y(x, t) dx.

(1.32)

We will now use the following trace estimate for the functions in H1(Ω):

∫

∂Ω|u|2 dσ ≤ C

(∫

Ω(|u|2 + |∇u|2) dx

)1/2 (∫

Ω|u|2 dx

)1/2

∀u ∈ H1(Ω),(1.33)

for some positive C = C(Ω). This inequality can be proved arguing first for regular functions ina dense subspace of H1(Ω) and then passing to the limit. For a regular function u, (1.33) is very

50

easy to establish when Ω = RN+ . Then, a standard localization argument leads to the proof in

the case of a general domain Ω.In view of (1.32) and (1.33), we have:

12

d

dt

∫

Ω|y(x, t)|2 dx +

∫

Ω|∇y(x, t)|2 dx

≤ −∫

ΩB(x, t) · ∇y(x, t) y(x, t) dx−

∫

Ωa(x, t) |y(x, t)|2 dx

+∫

Ωf(x, t) y(x, t) dx + C ‖β‖∞ ‖y(·, t)‖H1(Ω) ‖y(·, t)‖L2(Ω).

Combining this and Young’s inequality, we obtain:

d

dt‖y(·, t)‖2

L2(Ω) + ‖y(·, t)‖2H1(Ω)

≤ C((1 + ‖a‖∞ + ‖B‖2∞ + ‖β‖2

∞)‖y(·, t)‖2L2(Ω) + ‖f(·, t)‖2

L2(Ω))

for all t ∈ (0, T ). From these estimates, it is not difficult to obtain (1.31).This ends the proof.

The announced observability estimate is proved in the following result:

Proposition 2 For every ϕ0 ∈ L2(Ω), the associated solution to (1.6) satisfies the observabilityinequality ∫∫

Ω×(T/4,3T/4)|ϕ|2 dx dt ≤ K

∫∫

ω×(0,T )|ϕ|2 dx dt. (1.34)

for a constant K of the form

K = expC(1 +1T

+ ‖a‖2/3∞ + ‖B‖2

∞ + ‖β‖2∞). (1.35)

Proof: Let ϕ0 ∈ L2(Ω) be given. Notice that the corresponding ϕ solves (1.4) with

f1 = −aϕ ∈ L2(Q), f2 = ϕB ∈ L2(0, T ;L2(Ω)N ), f3 = −β ϕ ∈ L2(Σ).

Thus, we can apply theorem 1 to ϕ and deduce that∫∫

Qe−2sα(s λ2 ξ |∇ϕ|2 + s3 λ4 ξ3 |ϕ|2)dx dt + s2λ3

∫∫


≤ C(Ω, ω)(‖a‖2

∞

∫∫

Qe−2sα |ϕ|2 dx dt + s2λ2‖B‖2

∞

∫∫


+s λ ‖β‖2∞

∫∫

Σe−2sα ξ |ϕ|2 dσ dt + s3λ4

∫∫


)

for any λ ≥ λ and any s ≥ σ(e4λ‖η0‖∞T + T 2).

51

We will now try to eliminate the global terms in the right hand side of this inequality bymaking a convenient choice of the parameter s.

Taking s ≥ C T 2 (‖a‖2/3∞ + ‖B‖2∞), we see that

C

(s2λ2‖B‖2

∞

∫∫

Qe−2sα ξ2 |ϕ|2 dx dt + ‖a‖2

∞

∫∫

Qe−2sα |ϕ|2 dx dt

)

≤ 12

s3 λ4

∫∫

Qe−2sα ξ3 |ϕ|2 dx dt.

On the other hand, taking s ≥ C T 2 ‖β‖2∞ , we find that

C s λ ‖β‖2∞

∫∫

Σe−2sα ξ |ϕ|2 dσ dt ≤ 1

2s2 λ3

∫∫

Σe−2sα ξ2 |ϕ|2 dσ dt.

All this leads to the estimate∫∫

Qe−2sα ξ3 |ϕ|2 dx dt ≤ C

∫∫

ω×(0,T )e−2sα ξ3 |ϕ|2 dx dt,

which holds for λ ≥ λ and s ≥ σ(e4λ‖η0‖∞ T + T 2(1 + ‖a‖2/3∞ + ‖B‖2∞ + ‖β‖2∞)).

Taking into account the properties of the weight functions as well as the choice of s and λwe have made, it is not difficult to realize that the function

t 7→ exp(−2smax

x∈Ωα(t)

)mınx∈Ω

ξ(t)3

reaches its minimum in (T/4, 3T/4) at t = T/4 and that the function

t 7→ exp(−2smın

x∈Ωα(t)

)maxx∈Ω

ξ(t)3

reaches its maximum in (0, T ) at t = T/2. With this, the previous Carleman inequality directlygives ∫∫

Ω×(T/4,3T/4)|ϕ|2 dx dt ≤ exp

−2s

(mınx∈Ω

α(x,T

2)−max

x∈Ωα(x,

T

4))

×mınx∈Ω

ξ(x,T

4)−3 max

x∈Ωξ(x,

T

2)3

∫∫

ω×(0,T )|ϕ|2 dx dt,

for the same choice of the parameters s and λ.Now, taking λ = λ and s = s = σ(e4λ‖η0‖∞ T + T 2(1 + ‖a‖2/3

∞ + ‖B‖2∞ + ‖β‖2∞)), we have∫∫

Ω×(T/4,3T/4)|ϕ|2 dx dt ≤ C(Ω, ω) eC(Ω,ω) s/T 2

∫∫

ω×(0,T )|ϕ|2 dx dt,

which gives (1.35) and (1.34).This ends the proof of proposition 2.

52

Let us now finish the proof of theorem 2. We will apply a well known argument that hasalready been used in several similar situations (see [3] and [5]).

Let us introduce a function η ∈ C∞(0, T ), with

η(t) = 1 for t ∈ (0, T/4), η(t) = 0 for t ∈ (3T/4, T )

and|η′(t)| ≤ C/T for t ∈ (0, T ).

Let χ be the weak solution of

χt −∆χ + B(x, t) · ∇χ + a(x, t) χ = 0 in Q,

∂χ

∂n+ β(x, t) χ = 0 on Σ,

χ(x, 0) = y0(x) in Ω.

and let us put y = w + ηχ. If y is the state associated to v, i.e. the solution to (1.1), then wsatisfies

wt −∆w + B(x, t) · ∇w + a(x, t) w = −η′(t) χ + v1ω in Q,

∂w

∂n+ β(x, t)w = 0 on Σ,

w(x, 0) = 0 in Ω.

(1.36)

Our task is to find a control v ∈ L2(ω × (0, T )) such that the associated solution to (1.36)satisfies

w(x, T ) = 0 in Ω. (1.37)

After this, just taking y = w+ηχ we will have proved our result with a control in L2(ω×(0, T )).For each ε > 0, let us consider the functional Jε , with

Jε(ϕ0) =12

∫∫

ω×(0,T )|ϕ|2 dx dt + ε‖ϕ0‖L2(Ω) −

∫∫

Qη′ χϕdx dt

∀ϕ0 ∈ L2(Ω),

where, for each ϕ0 ∈ L2(Ω), ϕ is the solution to (1.6) associated to ϕ0.It is clear that

ϕ0 7→ Jε(ϕ0)

is a continuous, strictly convex and (in view of (1.8)) coercive function on L2(Ω). Consequently,it possesses exactly one minimizer ϕ0

ε and it is not difficult to check that ϕ0ε = 0 if and only if

the solution w to (1.36) associated to v = 0 satisfies ‖w(·, T )‖L2(Ω) ≤ ε.Let us denote by ϕε the solution to (1.6) associated to ϕ0

ε , let us put

vε = ϕε1ω

and let us denote by wε the solution to (1.36) associated to the control vε. Then

‖wε(·, T )‖L2(Ω) ≤ ε. (1.38)

53

Indeed, it is not restrictive to assume that ϕ0ε 6= 0. Accordingly, Jε is differentiable at ϕ0

ε and

(J ′ε(ϕ0ε), ϕ

0)L2(Ω) = 0 ∀ϕ0 ∈ L2(Ω).

That is to say,

∫∫

ω×(0,T )ϕε ϕdx dt + (ε

ϕ0ε

‖ϕ0ε‖L2

, ϕ0)L2(Ω) −∫∫

Qη′ χϕdx dt = 0

∀ϕ0 ∈ L2(Ω).

Since ∫∫

ω×(0,T )ϕε ϕdx dt−

∫∫

Qη′ χϕ dx dt = (wε(·, T ), ϕ0)L2(Ω) ,

we have

(wε(·, T ), ϕ0)L2(Ω) = −(εϕ0

ε

‖ϕ0ε‖L2(Ω)

, ϕ0)L2(Ω) ∀ϕ0 ∈ L2(Ω),

which implies (1.38).

Since Jε(ϕ0ε) ≤ Jε(0) = 0, we also have

‖vε‖2L2(ω×(0,T ))

≤(∫∫

Ω×(T/4,3T/4)|ϕε|2 dx dt

)1/2(∫∫

Ω×(T/4,3T/4)|η′χ|2 dx dt

)1/2

.

From proposition 2 and the definition of vε, we deduce now that

‖vε‖2L2(ω×(0,T )) ≤

C

TK1/2 ‖vε‖L2(Q)

(∫∫

Ω×(T/4,3T/4)|χ|2 dx dt

)1/2

and, using proposition 1, we have

‖vε‖L2(ω×(0,T )) ≤ C K1/2 ‖χ‖Y ≤ H ‖y0‖L2(Ω) , (1.39)

where the constant H is as in (1.11).Consequently, vε1ω and wε are uniformly bounded in the spaces L2(ω × (0, T )) and

Z = w ∈ L2(0, T ; H1(Ω)) : wt ∈ L2(0, T ;H−1(Ω)),

respectively. Obviously, we can extract sequences converging weakly to a control v1ω and theassociated solution w of (1.36), with

w(x, T ) = 0 in Ω.

We have thus proved the existence of a control v ∈ L2(Q) such that (1.10) and (1.37) are fulfilled.This ends the proof of theorem 2.

54

Appendix: Proof of lemma 1

We divide the proof in three steps:1 - First, we set ψ = e−sα q and we prove the following inequality:

∫∫

Q(s−1 ξ−1(|ψt|2 + |∆ψ|2) + s λ2 ξ |∇ψ|2 + s3 λ4 ξ3 |ψ|2) dx dt

−2s3 λ3

∫∫

Σ|∇η0|2 ξ3 ∂η0

∂n|ψ|2 dσ dt− 4s λ2

∫∫

Σ|∇η0|2 ξ

∂ψ

∂nψ dσ dt

−4s λ

∫∫

Σ

∂η0

∂nξ

∣∣∣∣∂ψ

∂n

∣∣∣∣2

dσ dt + 2s λ

∫∫

Σ

∂η0

∂nξ |∇ψ|2 dσ dt

+2∫∫

Σ

∂ψ

∂nψt dσ dt− 2s2 λ

∫∫

Σαt

∂η0

∂nξ |ψ|2 dσ dt

≤ C

(∫∫

Qe−2sα |f |2 dx dt + s3 λ4

∫∫

ω×(0,T )ξ3 |ψ|2 dx dt

)

(1.40)

for λ ≥ C(Ω, ω) and s ≥ C(Ω, ω)(T e2λ‖η0‖∞ + T 2).

2 - Then, we set ψ = e−sα q and we prove that∫∫

Q(s−1 ξ−1(|ψt|2 + |∆ψ|2) + s λ2 ξ |∇ψ|2 + s3 λ4 ξ 3 |ψ|2)dx dt

+2s3 λ3

∫∫

Σ|∇η0|2 ξ 3 ∂η0


∫∫

Σ|∇η0|2 ξ

∂ψ

∂nψ dσ dt

+4s λ

∫∫

Σ

∂η0

∂nξ

∣∣∣∣∣∂ψ

∂n

∣∣∣∣∣2

dσ dt− 2s λ

∫∫

Σ

∂η0

∂nξ |∇ψ|2 dσ dt

+2∫∫

Σ

∂ψ

∂nψt dσ dt + 2s2 λ

∫∫

Σαt

∂η0

∂nξ |ψ|2 dσ dt

≤ C

(∫∫


∫∫

ω×(0,T )ξ 3 |ψ|2 dx dt

)

(1.41)

for any λ ≥ C(Ω, ω) and s ≥ C(Ω, ω)(e4λ‖η0‖∞ T + T 2). Here, ξ and α stand for the functions

ξ(x, t) =e−λη0(x)

t(T − t), α(x, t) =

e2λ‖η0‖∞ − e−λη0(x)

t(T − t).

3 - Finally, we add the previous two inequalities and we come back to the original variable ϕ.This will give the desired inequality (1.15).

55

Step 1: Let us put ψ = e−sα q. Since −qt −∆q = f , we also have

M1ψ + M2ψ = F, (1.42)

whereM1ψ = 2s λ2 ξ |∇η0|2 ψ + 2s λ ξ∇η0 · ∇ψ − ψt ,

M2ψ = −s2 λ2 ξ2 |∇η0|2 ψ −∆ψ − sαt ψ,

F = e−sα f − s λ ξ ∆η0 ψ + s λ2 ξ |∇η0|2 ψ.

From (1.42), we have that

‖M1ψ‖2L2(Q) + ‖M2ψ‖2

L2(Q) + 2(M1ψ, M2ψ)L2(Q) = ‖F‖2L2(Q) . (1.43)

The main idea is to expand the term 2(M1ψ,M2ψ)L2(Q) and use the particular structureof α and the fact that s is large enough in order to obtain large positive terms in this scalarproduct. Denoting by (Miψ)j (1 ≤ i ≤ 2, 1 ≤ j ≤ 3) the j-th term in the above expression ofMiψ, we find that

(M1ψ, M2ψ)L2(Q) =∑

1≤i,j≤3

((M1ψ)i, (M2ψ)j)L2(Q) .

Let us compute each of these terms.

First, we have

((M1ψ)1, (M2ψ)1)L2(Q) = −2s3 λ4

∫∫

Q|∇η0|4 ξ3 |ψ|2 dx dt = A.

Then,

((M1ψ)2, (M2ψ)1)L2(Q) = −2s3 λ3

∫∫

Q|∇η0|2 ξ3 (∇η0 · ∇ψ) ψ dx dt

= 3s3 λ4

∫∫

Q|∇η0|4 ξ3 |ψ|2 dx dt + s3 λ3

∫∫

Q∆η0 |∇η0|2 ξ3 |ψ|2 dx dt

+2s3 λ3

∫∫

Q∂iη

0 ∂ijη0 ∂jη

0 ξ3 |ψ|2 dx dt

−s3 λ3

∫∫

Σ|∇η0|2 ξ3 ∂η0

∂n|ψ|2 dσ dt = B1 + B2 + B3 + B4 .

We clearly have that A+B1 is a positive term. As a consequence of the properties of η0 (see(1.14)), we have

s3 λ4

∫∫

Q|∇η0|4 ξ3 |ψ|2 dx dt ≥ C s3 λ4

∫∫

Qξ3 |ψ|2 dx dt

−C s3 λ4

∫∫

ω′×(0,T )ξ3 |ψ|2 dx dt

56

for some C = C(Ω, ω). The first of these last two integrals will stay in the left hand side andthe second one will go to the right.

The boundary term B4 will also stay in the left hand side, while B2 and B3 will be absorbedby simply taking λ ≥ C(Ω, ω).

We also have

((M1ψ)3, (M2ψ)1)L2(Q) = s2 λ2

∫∫

Q|∇η0|2 ξ2 ψt ψ dx dt

= −s2 λ2

∫∫

Q|∇η0|2 ξ ξt |ψ|2 dx dt ≤ C s2 λ2 T

∫∫

Qξ3 |ψ|2 dx dt,

which is also absorbed by taking λ ≥ 1 and s ≥ C(Ω, ω)T .Consequently, we obtain

(M1ψ, (M2ψ)1)L2(Q) = ((M1ψ)1 + (M1ψ)2 + (M1ψ)3, (M2ψ)1)L2(Q)

≥ C s3 λ4

∫∫

Qξ3 |ψ|2 dx dt− s3 λ3

∫∫

Σ|∇η0|2 ξ3 ∂η0

∂n|ψ|2 dσ dt

−C s3 λ4

∫∫

ω′×(0,T )ξ3 |ψ|2 dx dt,

(1.44)

for any λ ≥ C(Ω, ω) and s ≥ C(Ω, ω) T .On the other hand, we have

((M1ψ)1, (M2ψ)2)L2(Q) = −2s λ2

∫∫

Q|∇η0|2 ξ ∆ψ ψ dx dt

= −2s λ2

∫∫

Σ|∇η0|2 ξ

∂ψ

∂nψ dσ dt + 2s λ2

∫∫

Q|∇η0|2 ξ |∇ψ|2 dx dt

+4s λ2

∫∫

Q∂iη

0 ∂ijη0 ξ ∂jψ ψ dx dt + s λ3

∫∫

Q|∇η0|2 ξ∇η0 · ∇|ψ|2 dx dt

= C1 + C2 + C3 + C4 .

We will keep C1 and C2 in the left hand side. For C3 and C4 , we have

C3 ≤ C sλ4

∫∫

Qξ |ψ|2 dx dt + C s

∫∫

Qξ |∇ψ|2 dx dt

andC4 ≤ C s2 λ4

∫∫

Qξ2 |ψ|2 dx dt + C λ2

∫∫

Q|∇ψ|2 dx dt.

Therefore, by taking s ≥ C T 2, we find that

C1 + C2 + C3 + C4 ≥ −2s λ2

∫∫

Σ|∇η0|2 ξ

∂ψ

∂nψ dσ dt

+2s λ2

∫∫

Q|∇η0|2 ξ |∇ψ|2 dx dt− C s2 λ4

∫∫

Qξ2 |ψ|2 dx dt

−C

∫∫

Q(s ξ + λ2) |∇ψ|2 dx dt.

(1.45)

57

We also have

((M1ψ)2, (M2ψ)2)L2(Q) = −2s λ

∫∫

Qξ (∇η0 · ∇ψ)∆ψ dx dt

= −2s λ

∫∫

Σ

∂η0

∂nξ

∣∣∣∣∂ψ

∂n

∣∣∣∣2

dσ dt + 2s λ

∫∫

Q∂ijη

0 ξ ∂iψ ∂jψ dx dt

+2s λ2

∫∫

Qξ |∇η0 · ∇ψ|2 dx dt + s λ

∫∫

Qξ∇η0 · ∇|∇ψ|2 dx dt

= D1 + D2 + D3 + D4 .

As before, we will keep the boundary integral D1 in the left hand side. Also,

D2 ≤ C s λ

∫∫

Qξ |∇ψ|2 dx dt.

Moreover, D3 ≥ 0. After some additional computations, we also see that

D4 = s λ

∫∫

Qξ∇η0 · ∇|∇ψ|2 dx dt = s λ

∫∫

Σξ

∂η0

∂n|∇ψ|2 dσ dt

−s λ2

∫∫

Q|∇η0|2 ξ |∇ψ|2 dx dt− s λ

∫∫

Q∆η0 ξ |∇ψ|2 dx dt

= D41 + D42 + D43 .

Now, we keep once more D41 in the left and we notice that D43 can be bounded in the sameform as D2 .

Consequently,

D1 + D2 + D3 + D4 ≥ −2s λ

∫∫

Σ

∂η0

∂nξ

∣∣∣∣∂ψ

∂n

∣∣∣∣2

dσ dt

+s λ

∫∫

Σ

∂η0

∂nξ |∇ψ|2 dσ dt− s λ2

∫∫

Q|∇η0|2 ξ |∇ψ|2 dx dt

−C s λ

∫∫

Qξ |∇ψ|2 dx dt.

(1.46)

Additionally, we find that

((M1ψ)3, (M2ψ)2)L2(Q) =∫∫

Qψt ∆ψ dx dt

=∫∫

Σ

∂ψ

∂nψt dσ dt = E,

(1.47)

which will stay in the left hand side.

58

From (1.45)-(1.47), we deduce that


≥ s λ2

∫∫

Q|∇η0|2 ξ |∇ψ|2 dx dt− 2s λ2

∫∫

Σ|∇η0|2 ξ

∂ψ

∂nψ dσ dt

−2s λ

∫∫

Σ

∂η0

∂nξ

∣∣∣∣∂ψ

∂n

∣∣∣∣2

dσ dt + s λ

∫∫

Σ

∂η0

∂nξ |∇ψ|2 dx dt

+∫∫

Σ

∂ψ

∂nψt dσ dt− C s2 λ4

∫∫

Qξ2 |ψ|2 dx dt

−C

∫∫

Q(s λ ξ + λ2) |∇ψ|2 dx dt

for any λ ≥ 1. Hence, we have the following for any λ ≥ C(Ω, ω) and any s ≥ C(Ω, ω) T 2:

(M1ψ, (M2ψ)2)L2(Q) ≥ C s λ2

∫∫

Qξ |∇ψ|2 dx dt

−2s λ2

∫∫

Σ|∇η0|2 ξ

∂ψ

∂nψ dσ dt− 2s λ

∫∫

Σ

∂η0

∂nξ

∣∣∣∣∂ψ

∂n

∣∣∣∣2

dσ dt

+s λ

∫∫

Σ

∂η0

∂nξ |∇ψ|2 dx dt +

∫∫

Σ

∂ψ

∂nψt dσ dt

−C s2 λ4

∫∫

Qξ2 |ψ|2 dx dt− C s λ2

∫∫

ω′×(0,T )ξ |∇ψ|2 dx dt.

(1.48)

Let us now consider the scalar product

((M1ψ)1, (M2ψ)3)L2(Q) = −2s2 λ2

∫∫

Q|∇η0|2 αt ξ |ψ|2 dx dt

≤ C(Ω, ω) e2λ‖η0‖∞ s2 λ2 T

∫∫

Qξ3 |ψ|2 dx dt,

(1.49)

Obviously, this will be absorbed by the term in s3 λ4 in (1.44) if we take λ ≥ 1 and s ≥C(Ω, ω) e2λ‖η0‖∞ T .

Furthermore,

((M1ψ)2, (M2ψ)3)L2(Q) = −2s2 λ

∫∫

Qαt ξ (∇η0 · ∇ψ) ψ dx dt

= −s2 λ

∫∫

Σαt

∂η0

∂nξ |ψ|2 dσ dt + s2 λ2

∫∫

Qαt |∇η0|2 ξ |ψ|2 dx dt

+ s2 λ

∫∫

Q∇αt · ∇η0 ξ |ψ|2 dx dt + s2 λ

∫∫

Qαt ∆η0 ξ |ψ|2 dx dt.

With λ ≥ 1, the last three terms in the left hand side can be bounded by

C(Ω, ω) e2λ‖η0‖∞ s2 λ2 T

∫∫

Qξ3 |ψ|2 dx dt.

59

Thus, we have

((M1ψ)2, (M2ψ)3)L2(Q) ≥ −s2 λ

∫∫

Σαt

∂η0

∂nξ |ψ|2 dx dt

−C e2λ‖η0‖∞ s2 λ2 T

∫∫

Qξ3 |ψ|2 dx dt

(1.50)

Finally, we have

((M1ψ)3, (M2ψ)3)L2(Q) = s

∫∫

Qαt ψt ψ dx dt

≤ C e2λ‖η0‖∞ s T 2

∫∫

Qξ3 |ψ|2 dx dt,

(1.51)

since

αtt ≤ C e2λ‖η0‖∞ ξ2(1 + T 2 ξ) ≤ C e2λ‖η0‖∞ T 2 ξ3.

From (1.49)-(1.51), we deduce that, for λ ≥ C(Ω, ω) and s ≥ C(Ω, ω) e2λ‖η0‖∞ T , one has


≥ G− C s3 λ2

∫∫

Qξ3 |ψ|2 dx dt,

(1.52)

where

G = −s2 λ

∫∫

Σαt

∂η0

∂nξ |ψ|2 dσ dt.

Taking into account (1.44), (1.48) and (1.52), we obtain

(M1ψ, M2ψ)L2(Q) ≥ C

∫∫

Q(s λ2 ξ |∇ψ|2 + s3 λ4 ξ3 |ψ|2) dx dt

−s3 λ3

∫∫

Σ|∇η0|2 ξ3 ∂η0


∫∫

Σ|∇η0|2 ξ

∂ψ

∂nψ dσ dt

−2s λ

∫∫

Σ

∂η0

∂nξ

∣∣∣∣∂ψ

∂n

∣∣∣∣2

dσ dt + s λ

∫∫

Σ

∂η0

∂nξ |∇ψ|2 dx dt

+∫∫

Σ

∂ψ

∂nψt dσ dt− s2 λ

∫∫

Σαt

∂η0

∂nξ |ψ|2 dσ dt

−C

∫∫

ω′×(0,T )(s λ2 ξ |∇ψ|2 + s3 λ4 ξ3 |ψ|2) dx dt

60

for any λ ≥ C(Ω, ω) and s ≥ C(Ω, ω)(e2λ‖η0‖∞ T + T 2). Using (1.43), this gives

‖M1ψ‖2L2(Q) + ‖M2ψ‖2

L2(Q) +∫∫

Q(s λ2 ξ |∇ψ|2 + s3 λ4 ξ3 |ψ|2) dx dt

+2(B4 + C1 + D1 + D41 + E + G) ≤ C(‖F‖2

L2(Q)

+s λ2

∫∫

ω′×(0,T )ξ |∇ψ|2 dx dt + s3 λ4

∫∫

ω′×(0,T )ξ3 |ψ|2) dx dt

)

≤ C

(∫∫


∫∫

Qξ2 |ψ|2 dx dt

+s λ2

∫∫

ω′×(0,T )ξ |∇ψ|2 dx dt + s3 λ4

∫∫

ω′×(0,T )ξ3 |ψ|2) dx dt

).

Thus, we also have

‖M1ψ‖2L2(Q) + ‖M2ψ‖2

L2(Q) +∫∫

Q(s λ2 ξ |∇ψ|2 + s3 λ4 ξ3 |ψ|2) dx dt

+2(B4 + C1 + D1 + D41 + E + G) ≤ C

(∫∫

Qe−2sα |f |2 dx dt

+s λ2

∫∫

ω′×(0,T )ξ |∇ψ|2 dx dt + s3 λ4

∫∫

ω′×(0,T )ξ3 |ψ|2) dx dt

)(1.53)

for λ ≥ C(Ω, ω) and s ≥ C(Ω, ω)(e2λ‖η0‖∞ T + T 2).

The next step is to try to add integrals of |∆ψ|2 and |ψt|2 to the left hand side of (1.53).This can be made using the expressions of Miψ (i = 1, 2). Indeed, we have

s−1

∫∫

Qξ−1 |ψt|2 dx dt ≤ C

(s λ2

∫∫

Qξ |∇ψ|2 dx dt

+s λ4

∫∫

Qξ |ψ|2 dx dt + ‖M1ψ‖2

L2(Q)

)

and

s−1

∫∫

Qξ−1 |∆ψ|2 dx dt ≤ C

(s3 λ4

∫∫

Qξ3 |ψ|2 dx dt

+s T 2 e4λ‖η0‖∞∫∫

Qξ3 |ψ|2 dx dt + ‖M2ψ‖2

L2(Q)

)

61

for s ≥ C T 2. Accordingly, we deduce from (1.53) that∫∫

Q

((sξ)−1(|ψt|2 + |∆ψ|2) + s λ2 ξ |∇ψ|2 + s3 λ4 ξ3 |ψ|2) dx dt

+2(B4 + C1 + D1 + D41 + E + G) ≤ C

(∫∫


+s λ2

∫∫

ω′×(0,T )ξ |∇ψ|2 dx dt + s3 λ4

∫∫

ω′×(0,T )ξ3 |ψ|2) dx dt

)(1.54)

for λ ≥ C(Ω, ω) and s ≥ C(Ω, ω)(e2λ‖η0‖∞ T + T 2).We are now ready to eliminate the second integral in the right hand side. To this end, let us

introduce a function θ = θ(x), with

θ ∈ C2c (ω), θ ≡ 1 in ω′, 0 ≤ θ ≤ 1

and let us make some computations:

s λ2

∫∫

ω′×(0,T )ξ |∇ψ|2 dx dt ≤ s λ2

∫∫

ω×(0,T )θ ξ |∇ψ|2 dx dt

= −s λ2

∫∫

ω×(0,T )θ ξ ∆ψ ψ dx dt− s λ2

∫∫

ω×(0,T )ξ (∇θ · ∇ψ) ψ dx dt

−s λ3

∫∫

ω×(0,T )θ ξ (∇η0 · ∇ψ) ψ dx dt ≤ ε s−1

∫∫

ω×(0,T )ξ−1 |∆ψ|2 dx dt

+C

(s3 λ4

∫∫

ω×(0,T )ξ3 |ψ|2 dx dt + s λ4

∫∫

ω×(0,T )ξ |ψ|2 dx dt

),

where we have used that λ ≥ 1. In view of this estimate, we deduce that the integral on |∇ψ|2of the right hand side of (1.54) can be suppressed if the last integral is performed in the slightlygreater set ω × (0, T ). From (1.54) and this remark, we deduce (1.40).

Step 2: The proof of (1.41) is very similar to the proof of (1.40). We will only sketch the mainpoints.

We start from the identityM1ψ + M2ψ = F ,

whereM1ψ = 2s λ2 |∇η0|2 ξ ψ − 2s λ ξ∇η0 · ∇ψ − ψt ,

M2ψ = −s2 λ2 |∇η0|2 ξ 2 ψ −∆ψ − s αt ψ,

F = e−sα f + s λ ξ ∆η0 ψ + s λ2 |∇η0|2 ξ ψ.

We then have

‖M1ψ‖2L2(Q) + ‖M2ψ‖2

L2(Q) + 2(M1ψ, M2ψ)L2(Q) = ‖F‖2L2(Q) . (1.55)

62

After a lengthy computation, we find that

(M1ψ, M2ψ)L2(Q) ≥ C

∫∫

Q(s λ2 ξ |∇ψ|2 + s3 λ4 ξ3 |ψ|2) dx dt

+2(B4 + C1 + D1 + D41 + E + G)

−C

∫∫

ω′×(0,T )(s λ2 ξ |∇ψ|2 + s3 λ4 ξ 3 |ψ|2) dx dt

for any λ ≥ C(Ω, ω) and s ≥ C(Ω, ω)(e2λ‖η0‖∞ T + T 2), where

B4 = s3 λ3

∫∫

Σ|∇η0|2 ξ 3 ∂η0

∂n|ψ|2 dσ dt,

C1 = −2s λ2

∫∫

Σ|∇η0|2 ξ

∂ψ

∂nψ dσ dt,

D1 = 2s λ

∫∫

Σ

∂η0

∂nξ

∣∣∣∣∣∂ψ

∂n

∣∣∣∣∣2

dσ dt, D41 = −s λ

∫∫

Σ

∂η0

∂nξ |∇ψ|2 dx dt,

E =∫∫

Σ

∂ψ

∂nψt dσ dt, G = s2 λ

∫∫

Σαt

∂η0

∂nξ |ψ|2 dσ dt.

This, together with (1.55), gives

‖M1ψ‖2L2(Q) + ‖M2ψ‖2

L2(Q) +∫∫

Q(s λ2 ξ |∇ψ|2 + s3 λ4 ξ 3 |ψ|2) dx dt

+2(B4 + C1 + D1 + D41 + E + G) ≤ C(‖F‖2

L2(Q)

+s λ2

∫∫

ω′×(0,T )ξ |∇ψ|2 dx dt + s3 λ4

∫∫

ω′×(0,T )ξ3 |ψ|2) dx dt

).

(1.56)

With similar arguments to those in the first step, we can now assume that, in (1.56), ‖F‖2L2(Q)

is replaced by ∫∫


and ‖M1ψ‖2L2(Q) + ‖M2ψ‖2

L2(Q) is replaced by

∫∫

Qs−1 ξ−1(|ψt|2 + |∆ψ|2) dx dt.

Finally, for λ ≥ C(Ω, ω) and s ≥ C(Ω, ω) T 2 large enough, we can replace the integrals of |∇ψ|2and |ψ|2 in the right hand side by

s3 λ4

∫∫

ω×(0,T )ξ 3 |ψ|2 dx dt.

63

This yields the estimate (1.41).

Step 3: Now, let us add the inequalities (1.40) and (1.41) and let us check that all the integralson Σ can be simplified, so that there will only remain integrals in Q.

Since η0 = 0 on ∂Ω, we have

ξ = ξ, α = α and ψ = ψ on Σ. (1.57)

Consequently, B4 + B4 = 0 and G + G = 0.Let us see that

∂ψ

∂n≡ −∂ψ

∂non Σ. (1.58)

From the definitions of ψ and ψ, we have

∂iψ = e−sα(∂iq + s λ ∂iη

0 ξ q), ∂iψ = e−sα

(∂iq − s λ ∂iη

0 ξ q)

, (1.59)

whence∂ψ

∂n= s λ

∂η0

∂nξ e−sα q,

∂ψ

∂n= −s λ

∂η0

∂nξ e−sα q on Σ

and we certainly have (1.58).We deduce from (1.57) and (1.58) that C1 + C1 = 0, D1 + D1 = 0 and E + E = 0.On the other hand, since ϕ satisfies a zero Neumann condition and η0 = 0 on ∂Ω, we also

have|∇ψ|2 = |∇ψ|2 on Σ,

whence D41 + D41 = 0.With all this, we obtain

s−1

∫∫

Q(ξ−1(|ψt|2 + |∆ψ|2) + ξ−1(|ψt|2 + |∆ψ|2)) dx dt

+s λ2

∫∫

Q(ξ |∇ψ|2 + ξ |∇ψ|2) dx dt

+s3 λ4

∫∫

Q(ξ3 |ψ|2 + ξ 3 |ψ|2) dx dt

≤ C

(s3 λ4

∫∫

ω×(0,T )

(ξ3 |ψ|2 + ξ 3 |ψ|2

)dx dt

+∫∫

Q

(e−2sα + e−2sα

) |f |2 dx dt

),

(1.60)

for λ ≥ C(Ω, ω) and s ≥ C(Ω, ω)(e4λ‖η0‖∞ T + T 2).From the definitions of ξ, ξ, α and α, we have

ξ ≤ ξ, e−2sα ≤ e−2sα in Q,

64

so (1.60) yields∫∫

Q

((sξ)−1(|ψt|2 + |∆ψ|2) + s λ2 ξ |∇ψ|2 + s3 λ4 ξ3 |ψ|2) dx dt

≤ C

(∫∫


∫∫

ω×(0,T )ξ3 |ψ|2 dx dt

),

(1.61)

for any λ ≥ C(Ω, ω) and s ≥ C(Ω, ω)(e4λ‖η0‖∞ T + T 2).We finally turn back to ϕ. For the moment, we have

s−1

∫∫

Qξ−1

(|ψt|2 + |∆ψ|2) dx dt

+s λ2

∫∫

Qξ |∇ψ|2 dx dt + s3 λ4

∫∫

Qe−2sα ξ3 |q|2 dx dt

≤ C

(∫∫


∫∫


).

(1.62)

Using (1.59), we find that

s λ2

∫∫

Qe−2sα ξ |∇q|2 dx dt ≤ C s λ2

∫∫

Qξ |∇ψ|2 dx dt

+C s3 λ4

∫∫

Qe−2sα ξ3 |q|2 dx dt.

Accordingly, the previous integral of |∇q|2 can be added to the left hand side of (1.62):

s−1

∫∫

Qξ−1 |ψt|2 dx dt + s−1

∫∫

Qξ−1 |∆ψ|2 dx dt

+s λ2

∫∫

Qξ |∇q|2 dx dt + s3 λ4

∫∫


≤ C

(s3 λ4

∫∫

ω×(0,T )e−2sα ξ3 |q|2 dx dt +

∫∫


).

For ∆q, we use the identity

∆ψ = e−sα(∆q + s λ ∆η0 ξ q + s λ2 |∇η0|2 ξ q

+2s λ ξ∇η0 · ∇q + s2 λ2 |∇η0|2 ξ2 q)

and we obtain

s−1

∫∫

Qe−2sα ξ−1 |∆q|2 dx dt ≤ C

(s−1

∫∫

Qξ−1 |∆ψ|2 dx dt

+s λ2

∫∫

Qe−2sα ξ |q|2 dx dt + s λ4

∫∫

Qe−2sα ξ |q|2 dx dt

+s λ2

∫∫

Qe−2sα ξ |∇q|2 dx dt + s3 λ4

∫∫


).

65

Finally, for qt , we get

s−1

∫∫

Qe−2sα ξ−1 |qt|2 dx dt ≤ C(Ω, ω)

(s−1

∫∫

Qξ−1 |ψt|2 dx dt

+s e4λ‖η0‖C(Ω) T 2

∫∫


),

where we have used the identityqt = esα(ψt + sαt ψ).

Thus, taking λ ≥ 1 and s ≥ C(Ω, ω)(e2λ‖η0‖C(Ω) T + T 2), we are able to introduce all terms ofIs,λ(q) in the left hand side of (1.62). This yields (1.15) and concludes the proof of lemma 1.

Bibliografıa

[1] V. Barbu, Controllability of parabolic and Navier-Stokes equations, Sci. Math. Jpn., No. 1(2002), p. 143–211.

[2] A. Doubova, E. Fernandez-Cara, M. Gonzalez-Burgos, On the controllability ofthe heat equation with nonlinear boundary Fourier conditions, J. Diff. Equ. 196 (2) (2004),p. 385–417.

[3] C. Fabre, J.P. Puel, E. Zuazua, Approximate controllability of the semilinear heatequation, Proc. Royal Soc. Edinburgh, 125A (1995), p. 31–61.

[4] E. Fernandez-Cara, E. Zuazua, The cost of approximate controllability for heat equa-tions: the linear case, Adv. Differential Equations, no. 5 (2000), p. 465–514.

[5] A. Fursikov, O.Yu. Imanuvilov, Controllability of Evolution Equations, Lecture Notes#34, Seoul National University, Korea, 1996.

[6] O.Yu. Imanuvilov, M. Yamamoto, Carleman estimate for a parabolic equation in aSobolev space of negative order and its applications, Lecture Notes in Pure and Appl. Math.,218, Dekker, New York (2001)

[7] G. Lebeau, L. Robbiano, Controle exacte de l’equation de la chaleur (French), Comm.Partial Differential Equations, No. 20 (1995), p. 335–356.

[8] D. L. Russell, A unified boundary controllability theory for hyperbolic and parabolic partialdifferential equations, No. 52 (1973), p. 189–211.

Capıtulo 2

Exact Controllability to thetrajectories of the heat equationwith Fourier boundary conditions:The semilinear case

70

Exact Controllability to the trajectories ofthe heat equation with Fourier boundary

conditions: The semilinear case

E. Fernandez-Cara, M. Gonzalez-Burgos, S. Guerrero y J.-P. Puel

Abstract

This paper is concerned with the global exact controllability of the semilinear heat equation (withnonlinear terms involving the state and the gradient) completed with boundary conditions ofthe form ∂y

∂n + f(y) = 0. We consider distributed controls, with support in a small set. The nullcontrollability of similar linear systems has been analyzed in a previous first part of this work.In this second part we show that, when the nonlinear terms are locally Lipschitz-continuous andslightly superlinear, one has exact controllability to the trajectories.

1. Introduction

Let Ω ⊂ RN (N ≥ 1) be a bounded connected open set whose boundary ∂Ω is regularenough (for instance ∂Ω ∈ C2). Let ω ⊂ Ω be a (small) nonempty open subset and let T > 0.We will use the notation Q = Ω × (0, T ) and Σ = ∂Ω × (0, T ) and we will denote by n(x) theoutward unit normal to Ω at the point x ∈ ∂Ω.

We will consider the semilinear heat equation with nonlinear Fourier (or Robin) boundaryconditions

yt −∆y + F (y,∇y) = v1ω in Q,

∂y

∂n+ f(y) = 0 on Σ,

y(x, 0) = y0(x) in Ω.

(2.1)

Here, we assume that v ∈ L2(ω × (0, T )) (at least), 1ω is the characteristic function of ω,y0 ∈ L∞(Ω) and F : R ×RN 7→ R and f : R 7→ R are given functions. In (2.1), y = y(x, t) isthe state and v = v(x, t) is the control; it is assumed that we can act on the system only throughω × (0, T ).

For the existence, uniqueness, regularity and general properties of the solutions to problemslike (2.1), see for instance [1], [2] and [7]. An illustrative interpretation of the data and variablesin (2.1) is the following. The function y = y(x, t) can be viewed as the relative temperatureof a medium (with respect to the exterior surrounding air) subject to transport and chemicalreactions. The parabolic equation in (2.1) means, among other things, that a heat source v1ω

is applied on a part of the body. On the boundary, − ∂y∂n can be viewed as the normal heat flux,

inwards directed, up to a positive coefficient. Thus, the equality

−∂y

∂n= f(y)

71

means that this flux is a (nonlinear) function of the temperature. Accordingly, it is reasonableto assume that f is nondecreasing and f(0) = 0.

A simplified linear model which was considered in a previous paper [10] is the following:

yt −∆y + a(x, t) y + B(x, t) · ∇y = v1ω in Q,

∂y

∂n+ β(x, t) y = 0 on Σ,

y(x, 0) = y0(x) in Ω,

(2.2)

Here, it is assumed that the coefficients a, B and β satisfy

a ∈ L∞(Q), B ∈ L∞(Q)N , β ∈ L∞(Σ) (2.3)

and, for the reasons above, it is also natural to assume that β ≥ 0 (although this assumptionwas not used in [10]).

The main goal of this paper is to analyze the controllability properties of the nonlinear system(2.1). More precisely, we will try to reach exactly uncontrolled solutions of (2.1), i.e. functionsy = y(x, t) satisfying

yt −∆y + F (y,∇y) = 0 in Q,

∂y

∂n+ f(y) = 0 on Σ,

y(x, 0) = y0(x) in Ω.

(2.4)

It will be said that (2.1) is (globally) exactly controllable to the trajectories at time T if,for any solution of (2.4) with ‘suitable’ regularity and any y0 ∈ L∞(Ω), there exist controlsv ∈ L2(ω × (0, T )) and associated solutions y ∈ C0([0, T ]; L2(Ω)) such that

y(x, T ) = y(x, T ) in Ω. (2.5)

Here, by suitable regularity we mean the following:

y ∈ L2(0, T ; H1(Ω)) ∩ C0([0, T ]; L2(Ω)) ∩ L∞(Q), y0 ∈ L∞(Ω). (2.6)

The controllability properties of semilinear time-dependent systems have been studied in-tensively these last years. See for instance [16], [13], [17], [5], [11] and [11], where nonlinearitiesof the form f(y) are considered. See also the general treatise [14]. In particular, for parabolicsystems completed with Dirichlet boundary conditions, nonlinear terms f(y,∇y) depending onboth the state and the gradient have been taken into account in [9] and [6]. For the similar linearsystem (2.2), the null controllability was analyzed more in detail in [10]. In the case of (2.1),some partial results have been given in [5].

Our main result concerns the global exact controllability to the trajectories of (2.1). It is thefollowing:

Theorem 3 Let us assume that F and f are locally Lipschitz-continuous and satisfy

lım|s|→∞

|F (s, p)− F (r, p)||s− r| log3/2(1 + |s− r|) = 0, (2.7)

72

uniformly in (r, p) ∈ [−K, K]×RN ∀K > 0,

∀L > 0, ∃M > 0 such that

|F (s, p)− F (r, p)| ≤ M |s− r|, |F (s, p)− F (s, q)| ≤ M |p− q|∀(s, r, p, q) ∈ [−L, L]2 ×RN ×RN

(2.8)

andlım|s|→∞

|f(s)− f(r)||s− r| log1/2(1 + |s− r|) = 0 (2.9)

uniformly in r ∈ [−K,K] ∀K > 0. Then, for each T > 0, the nonlinear system (2.1) is exactlycontrollable to the trajectories at time T with L∞ controls.

Remark 2 Conditions (2.7)–(2.9) are satisfied if F and f are globally Lipschitz continuous.Notice that (2.7) means that the function F can only be slightly superlinear in s, uniformly inp. In the similar case of Dirichlet boundary conditions, it is known that conditions like these aresharp. Indeed, for instance, when F does not depend on p and

|F (s)− F (r)| ∼ |s− r| logβ(1 + |s− r|), β > 2,

due to blow-up phenomena, the system fails to be controllable whenever ω 6= Ω (see [11]). On theother hand, (2.9) is also a slightly superlinear growth assumption for f . It would be interestingto know whether a more superlinear f leading to blow up in the absence of control can also bean obstruction for the null controllability of (2.1). But this question does not seem obvious andremains open.

Remark 3 A result proved in [5] says that when F ≡ 0, f is smooth near zero and

f(s) s ≥ 0 ∀s ∈ R, (2.10)

the nonlinear system (2.1) is null controllable for large T . That is to say, under these assumptions,for each y0 ∈ L2(Ω) there exist T (y0) > 0 and controls v in L∞(ω×(0, T ) such that the associatedstates y satisfy

y(x, T (y0)) = 0 in Ω.

By inspection of the proof of theorem 3, we see that the same result holds for (2.1) with F ≡ 0whenever f is locally Lipschitz-continuous and satisfies the good sign condition (2.10).

For the proof of theorem 3, we will first establish a null controllability result for (2.2) (seeproposition 3 below). This will be used, together with an appropriate fixed point argument, todeduce the desired result.

This strategy was introduced in [16] in the framework of the exact controllability of thesemilinear wave equation. See also [5] and [11] for similar results concerning the approximateand null controllability of the semilinear heat equation with Dirichlet or Neumann boundaryconditions.

Our null controllability result for (2.2) is the following:

73

Proposition 3 For every T > 0, system (2.2) is null controllable at time T , with controls inL∞(ω × (0, T )). More precisely, for each y0 ∈ L2(Ω), there exists v ∈ L∞(ω × (0, T )) such thatthe associated solution to (2.2) satisfies y(x, T ) = 0 in Ω. Furthermore, the control v can befound satisfying

‖v‖L∞(ω×(0,T )) ≤ eC(Ω,ω)K(T,‖a‖∞,‖B‖∞,‖β‖∞) ‖y0‖L2(Ω) , (2.11)

where

K = 1 + 1/T + ‖a‖2/3∞ + ‖B‖2

∞ + ‖β‖2∞ + T (1 + ‖a‖∞ + ‖B‖2

∞ + ‖β‖2∞). (2.12)

For the proof of proposition 3, we first introduce a control L2(ω × (0, T )) which leads thesolution of (2.2) to zero at time T . In a second step, arguing as in Section 2 in [4], a regularizingargument will lead to the desired L∞ control.

The rest of this paper is organized as follows. In Section 2, we prove proposition 3. Section 3is devoted to the proof of theorem 3. For completeness, we have also included an Appendixwhere the proof of a rather technical local regularity result is given in detail.

In the sequel, C denotes a generic positive constant only depending on Ω and ω.

2. A null controllability result for the linear system

In this Section we present the proof of proposition 3.Let y0 ∈ L2(Ω) be given and let us introduce two open sets ω′ and ω′′, with ω′′ ⊂⊂ ω′ ⊂⊂ ω.

Then, we can use the main result in [10] (theorem 2) with control region ω′′ × (0, T ) to deducethe existence of a control v ∈ L2(ω′′ × (0, T )) such that the associated solution to (2.2) verifiesy(x, T ) = 0 in Ω and also the estimate

‖v‖L2(ω′′×(0,T )) ≤ eC(Ω,ω) K(T,‖a‖∞,‖B‖∞,‖β‖∞) ‖y0‖L2(Ω), (2.13)

where K is of the form (2.12).Let us denote by y the state associated to v. We now introduce a cut-off function η = η(t)

satisfyingη ∈ C∞([0, T ]), η(t) = 1 in (0, T/4), η(t) = 0 in (3T/4, T )

and

0 ≤ η(t) ≤ 1, |η′(t)| ≤ C

tin (0, T )

and we denote by χ the solution to the system

χt −∆χ + a(x, t) χ + B(x, t) · ∇χ = 0 in Q,

∂χ

∂n+ β(x, t)χ = 0 on Σ,

χ(x, 0) = y0(x) in Ω.

74

Then, the function w = y − ηχ satisfies

wt −∆w + a(x, t) w + B(x, t) · ∇w = −η′(t) χ + v1ω′′ in Q,

∂w

∂n+ β(x, t) w = 0 on Σ,

w(x, 0) = 0, w(x, T ) = 0 in Ω.

Our aim is to construct a control v ∈ L∞(ω × (0, T )) which drives the solution of (2.2) tozero at time t = T . To this end, we will need a local regularity result for the solutions to linearheat equations with L∞ coefficients a and B. This will be used below for the functions χ and wand reads as follows:

Lemma 2 Let us denote by Y the space L2(0, T ;H1(Ω)) ∩ C0([0, T ];L2(Ω)). Let y ∈ Y be asolution to the equation

yt −∆y + a(x, t) y + B(x, t) · ∇y = f, (2.14)

where a ∈ L∞(Q), B ∈ L∞(Q)N and f ∈ L2(Q). Let O ⊂ Ω be a nonempty open set and assumethat f is L∞ in the cylinder O × (0, T ). Then

y ∈ L∞(δ, T ; W 1,∞(O′))

for any δ ∈ (0, T ) and any nonempty open set O′ ⊂⊂ O. Furthermore, there exists a positiveconstant C(O′) such that the following estimate holds:

‖y‖L∞(δ,T ;W 1,∞(O′)) ≤ C(O′) (T 1/2 + TN/2)×(1 + δ−1 + ‖a‖∞ + ‖B‖∞

)N+1 (‖y‖Y + ‖f‖L∞(O×(0,T ))

).

(2.15)

The previous regularity also holds with δ = 0 if, besides (2.14), we have y(x, 0) = 0 in Ω. In thatcase, one has an estimate similar to (2.15) without the term in δ.

This lemma is implied by well known parabolic regularity theory. For completeness, its proofis given in an Appendix, at the end of this paper.

Let us now consider an open set ω0 with ω′ ⊂⊂ ω0 ⊂⊂ ω and a cut-off function ξ, with

ξ ∈ C20 (ω0), ξ ≡ 1 in ω′

and let us set w = (1− ξ) w. Then we have:

wt −∆w + a(x, t) w + B(x, t) · ∇w = −η′(t) χ + v1ω in Q,

∂w

∂n+ β(x, t)w = 0 on Σ,

w(x, 0) = 0, w(x, T ) = 0 in Ω,

withv = η′ ξ χ + 2∇ξ · ∇w + ∆ξ w −B · ∇ξ w. (2.16)

75

Let us remark that supp v ⊂ ω × [0, T ]. Therefore, if we prove that v ∈ L∞(ω × (0, T )), we willhave that the function y = w + ηχ solves (together with v) the null controllability problem for(2.2).

Thus, let us check that v ∈ L∞(ω × (0, T )) and let us estimate its norm in this space:

• The regularity of the first term in the right hand side of (2.16) is implied by the interiorregularity of χ not only in space but in time as well. From lemma 2 with O = ω, we deduce thatχ ∈ L∞(ω0 × (δ, T )) with supp ξ ⊂ ω0 ⊂⊂ ω (we even have χ ∈ L∞(δ, T ; W 1,∞

loc (ω))) and

‖χ‖L∞(ω0×(δ,T )) ≤ C (T 1/2 + TN/2)(1 + δ−1 + ‖a‖∞ + ‖B‖∞

)N+1 ‖χ‖Y ;

recall that Y = L2(0, T ;H1(Ω)) ∩ C0([0, T ];L2(Ω)).Consequently taking for instance δ = T/8, since η′ ≡ 0 in (0, T/4), we get

‖η′ ξ χ‖L∞(ω×(0,T )) ≤ C T−1 (T 1/2 + TN/2)×(1 + T−1 + ‖a‖∞ + ‖B‖∞

)N+1 ‖χ‖Y .

• The regularity of the other three terms in the right hand side of (2.16) is related to theinterior space regularity of w. Thus, let us introduce ω1 with ω0 ⊂⊂ ω1 ⊂⊂ ω and let us applylemma 2 with O = ω1 \ ω′′. This gives w ∈ L∞(0, T ;W 1,∞(ω0 \ ω′)) and the estimate

‖w‖L∞(0,T ;W 1,∞(ω0\ω′)) ≤ C (T 1/2 + TN/2)×(1 + ‖a‖∞ + ‖B‖∞)N+1 (‖w‖Y + ‖η′ χ‖L∞(ω1×(0,T ))),

whence ‖2∇ξ · ∇w + ∆ξ w −B · ∇ξ w‖L∞(ω×(0,T )) ≤ C(T 1/2 + TN/2)×

(1 + ‖a‖∞ + ‖B‖∞)N+2 (‖w‖Y + ‖η′ χ‖L∞(ω1×(0,T ))).

Putting the previous estimates together, we find that v ∈ L∞(ω × (0, T )) and

‖v‖L∞(ω×(0,T )) ≤ C(1 + TN−1)×(1 + T−1 + ‖a‖∞ + ‖B‖∞

)2N+3 (‖w‖Y + ‖χ‖Y ).(2.17)

At this point, notice that for any f ∈ L2(Q) and any y0 ∈ L2(Ω) the solution y to the linearsystem

yt −∆y + a(x, t) y + B(x, t) · ∇y = f in Q,

∂y

∂n+ β(x, t) y = 0 on Σ,

y(x, 0) = y0(x) in Ω

(2.18)

satisfies‖y‖Y ≤ eC T (1+‖a‖∞+‖B‖2∞+‖β‖2∞)(‖f‖L2(Q) + ‖y0‖L2(Ω)).

76

For a detailed proof, see for example proposition 1 in [10].This can be used to estimate ‖w‖Y and ‖χ‖Y in terms of ‖v‖L2(ω×(0,T )) and ‖y0‖L2(Ω). In

view of (2.17), we see that

‖v‖L∞(ω×(0,T )) ≤ L (‖v‖L2(ω′′×(0,T )) + ‖y0‖L2(Ω)), (2.19)

whereL = C T−1(1 + TN−1)

(1 + T−1 + ‖a‖∞ + ‖B‖∞

)2N+3×expC T (1 + ‖a‖∞ + ‖B‖2

∞ + ‖β‖2∞).

Combining this estimate and (2.13), we finally obtain that

‖v‖L∞(ω×(0,T )) ≤ eC K(T,‖a‖∞,‖B‖∞,‖β‖∞) ‖y0‖L2(Ω), (2.20)

where K is given by (2.12).This ends the proof of proposition 3.

3. Controllability of the nonlinear system

In this Section we will prove theorem 3. The following auxiliary result will be needed:

Proposition 4 Let us assume that, in (2.18), we have f ∈ L∞(Q) and y0 ∈ L∞(Ω). Let us alsoassume that the coefficients a, B and β satisfy (2.3). Then y ∈ L∞(Q) and

‖y‖∞ ≤ eC T(1+‖a‖∞+‖B‖2∞+‖β‖2∞) (‖y0‖∞ + ‖f‖∞). (2.21)

for some C = C(Ω).

Proof: We will consider two different situations:

Case 1 - We will first assume that a ≥ 1 and β ≥ 0 and we will establish (2.21) in this case. Infact, we will show that, under these assumptions,

‖y‖∞ ≤ ‖y0‖∞ + ‖f‖∞. (2.22)

To this end, let us introduce the system

zt −∆z + a(x, t) z + B(x, t) · ∇z = h in Q,

∂z

∂n+ β(x, t) z = k on Σ,

z(x, 0) = z0(x) in Ω,

where h ∈ L∞(Q), k ∈ L∞(Σ) and z0 ∈ L∞(Ω) and let us show that, if h, z0 and k arenonnegative, then this is also the case for z.

77

Indeed, by multiplying the equation satisfied by z by z−(·, t) (the negative part of z(·, t)) foreach t ∈ (0, T ) and integrating in Ω, after several simplifications, we find:

12

d

dt

∫

Ω|z−(x, t)|2 dx +

∫

Ω|∇z−(x, t)|2 dx

+∫

∂Ωβ(x, t) (z−(x, t) + k(x, t)) z−(x, t) dσ +

∫

Ωa(x, t) |z−(x, t)|2 dx

= −∫

Ωh(x, t) z−(x, t) dx−

∫

ΩB(x, t) · ∇z−(x, t) z−(x, t) dx.

From this identity, in view of the positiveness of a, h, β and k, we easily deduce that

d

dt

∫

Ω|z−(x, t)|2 dx ≤ ‖B‖2

∞

∫

Ω|z−(x, t)|2 dx,

whence z ≥ 0 in Q.Now, let M > 0 be a large constant (to be chosen below). The function z = M − y satisfies

zt −∆z + a(x, t) z + B(x, t) · ∇z = a(x, t) M − f in Q,

∂z

∂n+ β(x, t) z = β(x, t) M on Σ,

z(x, 0) = M − y0(x) in Ω.

Therefore, if we takeM ≥ max ‖f‖L∞(Q), ‖y0‖L∞(Ω),

we can apply the previous argument and deduce that y ≤ M . In a similar way, one can deducethat y ≥ −M and, consequently, |y| ≤ M . This proves that whenever a ≥ 1 and β ≥ 0, theestimate (2.22) holds.

Case 2 - We will now prove (2.21) for general L∞ coefficients a and β.Let γ ∈ C2(Ω) be a function satisfying

γ ≥ 0 in Ω,∂γ

∂n≤ −‖β‖∞ on ∂Ω, ‖γ‖∞ ≤ 1,

‖∇γ‖∞ ≤ C ‖β‖∞, ‖D2γ‖∞ ≤ C ‖β‖2∞.

(2.23)

We give here a sketch of the proof of the existence of such a function γ. To this end, let δ > 0be a parameter (depending on Ω) such that

x ∈ Ωδ 7→ dist(x, ∂Ω)

is C2, with Ωδ = x ∈ Ω : dist(x, ∂Ω) < δ. We distinguish two cases.Let us first assume that ‖β‖∞ ≥ 1/δ. Then we take γ(x) ≡ 1 in Ω\Ωδ, γ(x) = ‖β‖∞ dist(x, ∂Ω)

in Ωε with ε = 1/(2‖β‖∞) and a regularization of γ in Ωδ \Ωε. This gives the desired propertiesfor γ.

78

On the other hand, if ‖β‖∞ < 1/δ, we take γ(x) = δ ‖β‖∞ in Ω\Ωδ, γ(x) = ‖β‖∞ dist(x, ∂Ω)in Ωδ/2 and a regularization in Ωδ \ Ωδ/2. This also provides a desired function in this case.

Let us now set y = eγ(x) y. Then y satisfies

yt −∆y + a(x, t) y + B(x, t) · ∇y = eγ(x)f in Q,

∂y

∂n+ β(x, t) y = 0 on Σ,

y(x, 0) = eγ(x) y0(x) in Ω,

(2.24)

wherea = a + ∆γ − |∇γ|2 −B · ∇γ,

B = B + 2∇γ, β = β − ∂γ

∂n≥ 0 on Σ.

Notice that, from the inequalities (2.23) satisfied by γ, we know that

|a + ∆γ − |∇γ|2 −B · ∇γ| ≤ C1 (‖a‖∞ + ‖B‖2∞ + ‖β‖2

∞) in Q

for some C1 > 0.Now, let us set

y = e−(C1 (‖a‖∞+‖B‖2∞+‖β‖2∞)+1)t y.

Then y satisfies

yt −∆y + a(x, t) y + (B(x, t) + 2∇γ(x)) · ∇y = f in Q,

∂y

∂n+ β(x, t) y = 0 on Σ,

y(x, 0) = eγ(x) y0(x) in Ω,

(2.25)

wherea = a + ∆γ − |∇γ|2 −B · ∇γ + C1 (‖a‖∞ + ‖B‖2

∞ + ‖β‖2∞) + 1,

f = e−(C1 (‖a‖∞+‖B‖2∞+‖β‖2∞)+1)t+γ(x) f

andβ = β.

Since a ≥ 1 and β ≥ 0, we can apply Case 1 to y. This provides the estimates

‖y‖∞ ≤ ‖y‖∞ ≤ eC T(1+‖a‖∞+‖B‖2∞+‖β‖2∞) (‖y0‖∞ + ‖f‖∞),

whence we deduce (2.21).

79

Let us now start with the proof of theorem 3. Let y0 ∈ L∞(Ω) and y be given and assumethat y satisfies (2.6) and (2.4) in the weak sense. Let us consider the nonlinear system

wt −∆w + F1(w,∇w; x, t)w + F2(∇w;x, t) · ∇w = v1ω in Q,

∂w

∂n+ F3(w; x, t)w = 0 on Σ,

w(x, 0) = y0(x)− y(x, 0) in Ω,

(2.26)

where we have used the notation

F1(s, p;x, t) =F (y(x, t) + s,∇y(x, t) + p)− F (y(x, t),∇y(x, t) + p)

s, (2.27)

F2 = (F21, . . . , F2N ), F2j(p; x, t) =∫ 1

0

∂F

∂pj(y(x, t),∇y(x, t) + λp) dλ (2.28)

andF3(s; x, t) =

f(y(x, t) + s)− f(y(x, t))s

(2.29)

for s ∈ R and p ∈ RN .We will prove that there exist a control v ∈ L∞(ω × (0, T )) and an associated solution to

(2.26) such thatw(x, T ) = 0 in Ω. (2.30)

With this control and the state y = w + y, we will have solved the exact controllability problemfor (2.1) and we will have thus proved theorem 3.

We will first assume that the functions F and f are continuously differentiable. Then, by adensity argument, we will be able to prove the result in the general case.

3.1. The case in which F and f are C1

The idea of the proof is well known: we introduce an appropriate (set-valued) fixed pointmapping and we check that it possesses at least one fixed point; this will be a solution to thenull controllability problem associated to (2.26).

Let R > 0 be given and let us introduce the following function:

MR(s) =

−R if s < −R,

s if −R ≤ s ≤ R,

R if s > R.

Let us denote by Z the Hilbert space Z = L2(0, T ; H1(Ω)) and let us set for each R > 0 andeach z ∈ Z

aR,z(x, t) = F1(MR(z(x, t)),∇z(x, t);x, t),

Bz(x, t) = F2(∇z(x, t);x, t)

andβR,z(x, t) = F3(MR(z(x, t));x, t).

80

Consider the linear null controllability problem

wt −∆w + aR,z(x, t) w + Bz(x, t) · ∇w = v1ω in Q,

∂w

∂n+ βR,z(x, t) w = 0 on Σ,

w(x, 0) = y0(x)− y(x, 0) in Ω,

(2.31)

together with (2.30).From (2.6), (2.8) and the fact that f ∈ C1(R), we have

aR,z ∈ L∞(Q), Bz ∈ L∞(Q)N , βR,z ∈ L∞(Σ).

Consequently, in view of proposition 3, (2.30)–(2.31) can be solved with controls in L∞(ω ×(0, T )).

We are now going to select a particular solution to (2.30)–(2.31) constructed as in [11]. Todo this, we first set TR = mınT, a

−1/3R > 0, where

aR = sup|s|≤R, p∈RN

ess sup(x,t)∈Q

|F1(s, p; x, t)|.

We can follow the steps of Section 2 and construct a control vR,z ∈ L∞(ω × (0, TR)) suchthat the solution wR,z to (2.31) in Ω× (0, TR) verifies

wR,z(x, TR) = 0 in Ω.

The estimates we have been able to establish in propositions 3 and 4 written for vR,z and wR,z

with final time TR will now give

‖vR,z‖L∞(ω×(0,TR)) ≤ CR ‖w0‖L2(Ω), (2.32)

‖wR,z‖L2(0,TR;H1(Ω)) ≤ CR ‖w0‖L2(Ω) (2.33)

and‖wR,z‖L∞(Ω×(0,TR)) ≤ CR ‖w0‖L∞(Ω), (2.34)

whereCR = exp

C(Ω, ω, T )

(1 + a

2/3R + B

2 + β2R

),

B = supp∈RN

ess sup(x,t)∈Q

|F2(p;x, t)|

andβR = sup

|s|≤Ress sup(x,t)∈Σ

|F3(s; x, t)|.

In fact, the estimates obtained in the previous section imply (2.32)–(2.34) with CR replacedby CR(z), where

CR(z) = exp

C(Ω, ω)(1 + T−1

R + TR + ‖aR,z‖2/3∞ + ‖Bz‖2

∞

+‖βR,z‖2∞ + TR

(‖aR,z‖∞ + ‖Bz‖2∞ + ‖βR,z‖2

∞))

;

81

but taking into account the definitions of TR, aR, B and βR it is clear that CR(z) ≤ CR for allz ∈ Z.

At this moment, we can extend by zero the functions vR,z and wR,z for t ∈ (TR, T ). In thisway, we will have built a control vR,z and an associated state wR,z satisfying (2.30)–(2.31) and

‖vR,z‖L∞(ω×(0,T )) ≤ CR ‖w0‖L2(Ω), (2.35)

‖wR,z‖Z ≤ CR ‖w0‖L2(Ω) (2.36)

and‖wR,z‖∞ ≤ CR ‖w0‖L∞(Ω). (2.37)

We will now introduce a set-valued mapping leading to the solution to our controllabilityproblem.

We first consider the set of admissible controls AR(z). By definition, this is the set of controlsvR,z ∈ L∞(ω × (0, T )) which lead the solution to (2.31) to zero at time T and satisfy (2.35).Then, for each z ∈ Z, we denote by ΛR(z) the set of states wR,z associated to the controlsvR,z ∈ AR(z) furthermore satisfying (2.36) and (2.37). In view of the arguments above, ΛR(z)is a nonempty subset of Z.

The plan of the rest of the proof is the following:

We will first see that, for each R > 0, ΛR possesses a fixed point wR. This will be impliedby Kakutani’s theorem.

Then, we will find R > 0 (large enough) such that MR(wR) = wR. At this level, the useof proposition 4 will be crucial.

As a consequence, for large R, the fixed point wR of ΛR will be, together with some vR ∈L∞(ω × (0, T )), a solution to (2.30)–(2.31).

Thus, let us recall Kakutani’s fixed point theorem (see, for instance, [2]):

Theorem 4 Let Z be a Banach space and let Λ : Z 7→ Z be a set-valued mapping satisfying thefollowing assumptions:

1. Λ(z) is a nonempty closed convex set of Z for every z ∈ Z.

2. There exists a nonempty convex compact set K ⊂ Z such that Λ(K) ⊂ K.

3. Λ is upper-hemicontinuous in Z, i.e. for each σ ∈ Z ′ the single-valued mapping

z 7→ supy∈Λ(z)

〈σ, y〉Z′,Z (2.38)

is upper-semicontinuous.

Then Λ possesses a fixed point in the set K, i.e. there exists z ∈ K such that z ∈ Λ(z).

82

Let us check that Kakutani’s theorem can be applied to ΛR.

That ΛR(z) is a nonempty closed convex set of Z for every z ∈ Z is very easy to verify.

Let us prove that ΛR maps a compact set into itself. In fact, let us see that ΛR maps thewhole space Z into a fixed convex compact set KR.

Our argument will be the following: we choose an arbitrary sequence zn in Z and a sequencewn with wn ∈ ΛR(zn) for all n and we prove that wn possesses a strongly convergentsubsequence.

Thus, let the sequences zn and wn be given. From (2.35)–(2.37), the equations satisfiedby the functions wn and the fact that ‖aR,zn‖∞ ≤ aR, ‖Bzn‖∞ ≤ B and ‖βR,zn‖∞ ≤ βR for alln ≥ 1, we deduce the existence of subsequences wn′ and vn′ such that

wn′ → w weakly in Z,

wn′,t → wt weakly in L2(0, T ;H−1(Ω))

and

vn′ → v weakly-∗ in L∞(Q)

as n′ → ∞. We can also assume that the coefficients associated to zn′ converge weakly-∗ inL∞(Q) and L∞(Σ). Thus, we can pass to the limit in the weak formulations satisfied by wn′

and deduce that w and v satisfy

wt −∆w + a(x, t) w + θ(x, t) = v1ω in Q,

∂w

∂n+ β(x, t)w = 0 on Σ,

w(x, 0) = w0(x) in Ω

for some a ∈ L∞(Q) and β ∈ L∞(Σ), where θ ∈ L2(Q) is the weak limit of Bzn′ ·∇wn′ in L2(Q).

After substraction of the equations satisfied by the functions wn′ and w, we find that

(wn′ − w)t −∆(wn′ − w) = a(x, t) w − aR,zn′ (x, t) wn′

+θ(x, t)−Bzn′ (x, t) · ∇wn′ + (vn′ − v)1ω in Q,

∂(wn′ − w)∂n

+ βR,zn′ (x, t) wn′ − β(x, t) w = 0 on Σ,

(wn′ − w)(x, 0) = 0 in Ω.

83

Consequently,

12

∫

Ω|(wn′ − w)(x, T )|2 dx +

∫ T

0

∫

Ω|∇(wn′ − w)(x, s)|2 dx ds

=∫ T

0

∫

∂Ω(β w − βR,zn′ wn′) (wn′ − w)(x, s) dσ ds

+∫ T

0

∫

Ω(aw − aR,zn′ wn′) (wn′ − w)(x, s) dx ds

+∫ T

0

∫

Ω(θ −Bzn′ · ∇wn′) (wn′ − w)(x, s) dx ds

+∫ T

0

∫

ω(vn′ − v) (wn′ − w)(x, s) dx ds.

(2.39)

We are now going to check that all the terms in the right hand side of this last equality tendsto zero. Among other things, this will imply that

wn′ → w strongly in Z. (2.40)

• The first term in the right hand side converges to zero, since

wn′ → w strongly in L2(Σ)

and consequentlyβR,zn′ wn′ → β w weakly in L2(Σ).

Indeed, the strong convergence of wn′ is an immediate consequence of the compact embeddingof the space

z ∈ L2(0, T ;H1(Ω)) : zt ∈ L2(0, T ;H−1(Ω)) in L2(0, T ; Hs(Ω)) for all s ∈ (1/2, 1) and the fact that the lateral trace operator is well defined,linear and continuous from L2(0, T ; Hs(Ω)) into L2(Σ).

• The convergence of the other three terms in the right hand side is a consequence of theweak convergence in L2(Q) of aR,zn′ wn′ and Bzn′ ·∇wn′ , the weak convergence in L2(ω× (0, T ))of vn′ and the strong convergence of wn′ in L2(Q).

We have thus seen that wn possesses a strongly convergent subsequence and, consequently,ΛR maps the space Z into a fixed compact set.

It remains to check that ΛR is upper-hemicontinuous. Thus, assume that σ ∈ Z ′ and let asequence zn be given, with zn → z strongly in Z. We must prove that

lım supn→+∞

supw∈ΛR(zn)

〈σ,w〉Z′,Z ≤ supw∈ΛR(z)

〈σ,w〉Z′,Z .

Let zn′ be a subsequence of zn such that

lım supn→+∞

supw∈ΛR(zn)

〈σ,w〉Z′,Z = lımn′→+∞

supw∈ΛR(zn′ )

〈σ,w〉Z′,Z .

84

Since each ΛR(zn′) is a compact set of Z, for every n′ we have

supw∈ΛR(zn′ )

〈σ,w〉Z′,Z = 〈σ,wn′〉Z′,Z

for some wn′ ∈ ΛR(zn′). On the other hand, since all the states wn′ belong to the same compactset KR, at least for a new subsequence (again indexed by n′), we must have (2.40). We will nowprove that w ∈ ΛR(z). This will achieve the proof of the upper hemicontinuity of ΛR.

Indeed, we can assume that the weak limits of the coefficients associated to zn′ are aR,z, Bz

and βR,z, since zn′ converges strongly in Z towards z and therefore the coefficients aR,zn′ , Bzn′and βR,zn′ converge almost everywhere (observe that we are using here the C1 regularity of Fand f).

On the other hand, it can be assumed that the controls vn′ converge to a function v weakly-∗in L∞(ω × (0, T )). Then, w solves (2.31) and w(T ) = 0. Moreover, since inequality (2.35) isindependent of n, v also satisfies (2.35). Therefore, v ∈ AR(z). Consequently, it is immediatethat w is the solution to (2.31) associated to the control v.

This shows that w ∈ ΛR(z) and, therefore, ΛR is upper hemicontinuous.

In view of these arguments, Kakutani’s theorem can be applied and we deduce that, for eachR > 0, ΛR possesses at least one fixed point wR that belongs to Z and L∞(Q).

Our aim is now to find R > 0 such that

‖wR‖∞ ≤ R.

This will be a consequence of the estimates we know for wR and the properties satisfied by thefunctions Fi.

From (2.37), we obtain

‖wR‖∞ ≤ eC(Ω,ω,T )(1+a2/3R +B

2+β2

R) ‖w0‖∞. (2.41)

On the other hand, from (2.8)–(2.9) it is also clear that, for every ε > 0, there exists Cε > 0such that

(ess sup(x,t)∈Q

|F1(s, p;x, t)|)2/3

≤ ε log(1 + |s|) + Cε ∀s ∈ R, ∀p ∈ RN ,

(ess sup(x,t)∈Q

|F2(p; x, t)|)2

≤ Cε ∀p ∈ RN ,

(ess sup(x,t)∈Σ

|F1(s;x, t)|)2

≤ ε log(1 + |s|) + Cε ∀s ∈ R.

(2.42)

Consequently, it is also true that, for every ε > 0, there exists Cε > 0 (independent of R) suchthat

a2/3R + B

2 + β2R ≤ ε log(1 + R) + Cε.

85

These estimates, together with (2.41) and the definitions of aR, B and βR, lead to thefollowing inequality:

‖wR‖∞ ≤ C(Ω, ω, T, ε) (1 + R)C(Ω,ω,T ) ε ‖w0‖∞.

Accordingly, taking ε > 0 small enough to satisfy C(Ω, ω, T ) ε < 1, we can ensure that, forR > 0 sufficiently large (depending on Ω, ω, T and ‖y0 − y0‖L∞(Ω)), one has

‖wR‖∞ ≤ R.

This ends the proof of theorem 3 when F and f are C1 functions.

3.2. The general case

We will now assume that f and F are locally Lipschitz-continuous functions satisfying (2.7)–(2.9).

Let us introduce the functions ρ1 ∈ C∞c (R × RN ), ρ2 ∈ C∞

c (RN ) and ρ3 ∈ C∞c (R), with

ρj ≥ 0, supp ρ1 ⊂ B((0, 0); 1), supp ρ2 ⊂ B(0; 1), supp ρ3 ⊂ [−1, 1] and∫∫

R×RN

ρ1(s, p) ds dp =∫

RN

ρ2(p) dp =∫

Rρ3(s) ds = 1.

Let us consider, for each n ≥ 1, the associated mollifiers

ρ1n(s, p) = nN+1 ρ1(ns, np), ρ2

n(p) = nNρ2(np) ∀(s, p) ∈ R×RN

andρ3

n(s) = nρ3(ns) ∀s ∈ R

and the regularized functionsFi,n = ρi

n ∗ Fi (i = 1, 2, 3)

(the functions F1, F2 and F3 were defined in (2.27)–(2.29)).These functions satisfy the following:

• For each n ≥ 1, F1n : R×RN ×Q 7→ R, F2n : RN ×Q 7→ RN and F3n : R× Σ 7→ R areCaratheodory functions (respectively continuous in (s, p), p and s and measurable in (x, t)).

• If we setFn(s, p;x, t) = F1,n(s, p; x, t)s + F2,n(p; x, t) · p

andfn(s; x, t) = F3,n(s; x, t)s,

then the asymptotic properties (2.8) and (2.42) remain true uniformly in n. In other words, forany L > 0, there exists M > 0 (independent of n) such that

|Fn(s, p)− Fn(r, p)| ≤ M |s− r|, |Fn(s, p)− Fn(s, q)| ≤ M |p− q|∀(s, r, p, q) ∈ [−L,L]2 ×RN ×RN .

86

Moreover, for each ε > 0, there exists Cε > 0 such that

(ess sup(x,t)∈Q

|F1n(s, p; x, t)|)2/3

≤ ε log(1 + |s|) + Cε ∀s ∈ R, ∀p ∈ RN ,

(ess sup(x,t)∈Q

|F2n(p; x, t)|)2

≤ Cε ∀p ∈ RN ,

(ess sup(x,t)∈Σ

|F3n(s; x, t)|)2

≤ ε log(1 + |s|) + Cε ∀s ∈ R,

for each n ≥ 1.

• From the definitions of Fn and fn, we also have that

Fn(zn,∇zn; ·) → F1(z,∇z; ·)z + F2(∇z; ·) · ∇z weakly in L2(Q)

andfn(zn; ·) → F3(z; ·)z weakly-∗ in L∞(Σ)

wheneverzn → z weakly-∗ in L∞(Q) and strongly in L2(0, T ; H1(Ω)).

As a consequence, we can argue as in Subsection 3.1 and deduce that, for each n, there existsa control vn ∈ L∞(ω × (0, T )) such that

wn,t −∆wn + Fn(wn,∇wn; x, t) = vn1ω in Q,

∂wn

∂n+ fn(wn; x, t) = 0 on Σ,

wn(x, 0) = w0(x) in Ω

(2.43)

andwn(x, T ) = 0 in Ω. (2.44)

In view of the properties satisfied by the functions Fin, the estimates we have established inSubsection 3.1 are independent of n. Accordingly, at least for a subsequence, we also have

vn → v weakly-∗ in L∞(ω × (0, T )),

wn,t → wt weakly in L2(0, T ;H−1(Ω))

andwn → w weakly-∗ in L∞(Q) and strongly in L2(0, T ; H1(Ω)).

Thus, we can pass to the limit in (2.43) and find a control v ∈ L∞(ω× (0, T )) such that theassociated solution to (2.26) satisfies (2.30).

This ends the proof of theorem 3.

87

Remark 4 The proof of theorem 3 can also be achieved by applying another fixed point argu-ment. More precisely, we can first introduce a small parameter ε > 0 and find a control vε suchthat the solution of (2.1) satisfies

‖y(·, T )− y(·, T )‖L2 ≤ ε.

This can be made by previously solving an approximate controllability problem for the linearsystem (2.31) with the control of minimal norm in Lp(ω × (0, T )) for p large enough and, then,using Schauder’s theorem. Since all the estimates we can establish are uniform in ε, we can passto the limit as ε → 0 and deduce the desired result.

Appendix: Proof of lemma 2

Let us introduce N + 1 open subsets of O satisfying

ON = O′ ⊂⊂ ON−1 ⊂⊂ ... ⊂⊂ O1 ⊂⊂ O0 ⊂⊂ O.

We will also consider subintervals of (0, T ) of the form (δ/(i + 1), T ) for 0 ≤ i ≤ N .We will strict our considerations to the proof of lemma 2 in the case where no initial condition

is imposed. The result concerning a vanishing initial condition will follow readily from theargument below.

We are first going to see that

y ∈ L∞(δ/(N + 1), T ;H1(O0)) and ∆y ∈ L2(δ/(N + 1), T ; L2(O0)),

with an estimate of the associated norms independent of T . To this end, let ξ0 ∈ C2c (O) and

η0 ∈ C1([0, T ]) be two functions satisfying

ξ0(x) = 1 in O0, η0(t) = 1 in [δ

N + 1, T ], η0(0) = 0, |η0,t(t)| ≤ C

δin (0, T )

(of course, C depends on N) and let us introduce the function y0 = η0 ξ0 y. Then

y0,t −∆y0 = f0 in Q,

y0 = 0 on Σ,

y0(x, 0) = 0 in Ω,

wheref0 = η0 ξ0 f + η0,t ξ0 y − 2η0∇ξ0 · ∇y − η0 ∆ξ0 y

− a η0 ξ0 y − η0 ξ0 B · ∇y − η0(B · ∇ξ0)y.

We have f0 ∈ L2(Q). Consequently, ∆y0 ∈ L2(Q), y0 ∈ C0([0, T ]; H10 (Ω)) and appropri-

ate estimates are satisfied. Indeed, by multiplying the equation satisfied by y0 by −∆y0 andintegrating with respect to x in Ω, we find

12

d

dt‖∇y0(·, t)‖2

L2 +∫

Ω|∆y0(x, t)|2 dx = −

∫

Ωf0(x, t)∆y0(x, t) dx. (2.45)

88

Since‖f0‖L2 ≤ C

(‖f‖L2 + (1 + δ−1 + ‖a‖∞ + ‖B‖∞) ‖y‖Y

),

we easily obtain from (2.45) that

‖y0‖C0([0,T ];H10 (Ω)) + ‖∆y0‖L2(Q) ≤ C

(‖f‖L2 +(1 + δ−1 + ‖a‖∞ + ‖B‖∞

) ‖y‖Y

). (2.46)

Clearly, the same estimate holds for

‖y‖L∞(δ/(N+1),T ;H1(O0)) + ‖∆y‖L2(O0×(δ/(N+1),T )).

Let us now try to improve the local space regularity properties of y. To this end, we will usethe following lemma:

Lemma 3 Let us set p0 = 2, let pi be defined by

1pi

=1

pi−1− 1

2N

for 1 ≤ i ≤ N − 1 and let us set pN = +∞. Let us denote by Xi the space

Xi = L∞((i + 1)δ/(N + 1), T ; W 1,pi(Oi))

for 0 ≤ i ≤ N and suppose that y ∈ Xj−1 for some j. Then we also have y ∈ Xj and

‖y‖Xj ≤ C(O′)(T 1/2 ‖f‖∞ + D(T, δ, ‖a‖∞, ‖B‖∞) ‖y‖Xj−1),

whereD(T, δ, ‖a‖∞, ‖B‖∞) = (T 1/4 + T 1/2)(1 + δ−1 + ‖a‖∞ + ‖B‖∞). (2.47)

Proof of lemma 3: Let us introduce ξj ∈ C2c (Oj−1) and ηj ∈ C1([0, T ]), with

ξj(x) = 1 in Oj , ηj(t) = 1 in [(j + 1)δ/(N + 1), T ],

ηj(t) = 0 in [0, jδ/(N + 1)], |ηj,t(t)| ≤ C

δin (0, T )

and let us put yj = ηj ξj y. Then yj satisfies the following:

yj,t −∆yj = fj in Q,

yj = 0 on Σ,

yj(x, 0) = 0 in Ω(2.48)

withfj = fj,1 + fj,2 + fj,3 ,

wherefj,1 = ηj ξj f, fj,2 = ηj,t ξj y − ηj ∆ξj y − a ηj ξj y − ηj(B · ∇ξj)y,

fj,3 = −2ηj ∇ξj · ∇y − ηj ξj B · ∇y.

89

From the fact that the system (2.48) is linear, we see that yj can be written as the sum of threesolutions to similar systems with right hand sides fj,1 , fj,2 and fj,3 . Let us respectively denotethem by yj,1 , yj,2 and yj,3 . We are now going to deduce estimates of yj,k in Xj for 1 ≤ k ≤ 3.

To this end, we will use the usual representation of yj,k provided by the semigroup S(t)associated to the heat equation with homogeneous Dirichlet conditions, say

yj,k(·, t) =∫ t

0S(t− s) fj,k(·, s) ds

for all t ∈ (0, T ).Since f ∈ L∞(Q), we can write

‖yj,1(·, t)‖W 1,pj (Ω)≤ C

∫ t

0(t− s)−1/2 ‖fj,1(·, s)‖Lpj (Ω) ds.

Therefore, from Young’s inequality we find that yj,1 ∈ L∞(0, T ; W 1,pj (Ω)) and

‖yj,1‖L∞(0,T ;W 1,pj (Ω))≤ C T 1/2 ‖fj,1‖L∞(0,T ;Lpj (Ω))

≤ C(O′) T 1/2 ‖f‖L∞(O×(0,T )) .

Taking into account that fj,2 ∈ L∞(0, T ;Lp∗j−1(Ω)) with

p∗j−1 =

∞ if j > N − 1,

p (arbitrary in (1, +∞)) if j = N − 1,

2N

N − j − 1if j < N − 1,

we see that fj,2 is not worse than fj,1 and, again,

‖yj,2(·, t)‖W 1,pj (Ω)≤ C

∫ t

0(t− s)−1/2 ‖fj,2(·, s)‖Lpj (Ω) ds

for all t. From Young’s inequality and the assumption y ∈ Xj−1 , we also get yj,2 ∈L∞(0, T ; W 1,pj (Ω)) and

‖yj,2‖L∞(0,T ;W 1,pj (Ω))≤ C T 1/2 ‖fj,2‖

L∞(0,T ;Lp∗j−1 (Ω))

≤ C(O′) T 1/2 (1 + δ−1 + ‖a‖∞) ‖y‖Xj−1 .

In the definition of fj,3, we find ∇y.Consequently, we can only ensure that fj,3 ∈ L∞(0, T ;Lpj−1(Ω)). Since

−N

2

(1

pj−1− 1

pj

)− 1

2= −3

4,

we have

‖yj,3(·, t)‖W 1,pj (Ω)≤ C

∫ t

0(t− s)−3/4 ‖fj,3(·, s)‖Lpj−1 (Ω) ds

90

and now Young’s inequality gives yj,3 ∈ L∞(0, T ; W 1,pj (Ω)) and

‖yj,3‖L∞(0,T ;W 1,pj (Ω))≤ C T 1/4 ‖fj,3‖L∞(0,T ;Lpj−1 (Ω))

≤ C(O′) T 1/4 (1 + ‖B‖∞) ‖y‖Xj−1 .

Putting the estimates of ‖yj,k‖L∞(0,T ;W 1,pj (Ω))together and taking into account the defini-

tions of ηj and ξj , we obtain the desired inequality for ‖y‖Xj .This concludes the proof of lemma 3. ¤

Since we already had y ∈ X0, we deduce from lemma 3 that y ∈ XN and

‖y‖XN≤ C(O′)(T 1/2 ‖f‖∞ + D(T, δ, ‖a‖∞, ‖B‖∞) ‖y‖XN−1

),

where, D is given by (2.47).We can apply lemma 3 subsequently for j = N, N − 1, . . . , 1. The estimates we find yield

‖y‖XN≤ C(T 1/2(1 + DN−1)‖f‖L∞(O×(0,T )) + DN ‖y‖X0).

This, together with (2.46), yields

‖y‖XN≤ C(T 1/2 + TN/2)D(T, δ, ‖a‖∞, ‖B‖∞)N+1(‖f‖L∞(O×(0,T )) + ‖y‖Y ),

which is exactly (2.15).

Bibliografıa

[1] H. Amann, Parabolic evolution equations and nonlinear boundary conditions, J. Diff. Equ.72 (1988), p. 201–269.

[2] J. Arrieta, A. Carvalho and A. Rodrıguez-Bernal, Parabolic problems with non-linear boundary conditions and critical nonlinearities, J. Diff. Equ. 156 (1999), p. 376–406.

[3] J.P. Aubin, L’Analyse non Lineaire et ses Motivations Economiques, Masson, Paris 1984.

[4] O. Bodart, M. Gonzalez-Burgos, R. Perez-Garcıa, Insensitizing controls for a semi-linear heat equation with a superlinear nonlinearity, C. R. Math. Acad. Sci. Paris 335 (2002),no. 8, p. 677–682.

[5] A. Doubova, E. Fernandez-Cara and M. Gonzalez-Burgos, On the controllabilityof the heat equation with nonlinear boundary Fourier conditions, J. Diff. Equ. 196 (2)(2004), p. 385–417.

[6] A. Doubova, E. Fernandez-Cara, M. Gonzalez-Burgos and E. Zuazua, On thecontrollability of parabolic systems with a nonlinear term involving the state and the gradi-ent, SIAM J. Control Optim. Vol. 41, No. 3 (2002), p. 798–819.

[7] L. Evans, Regularity properties of the heat equation subject to nonlinear boundary con-straints, Nonlinear Anal. 1 (1997), p. 593–602.

[8] C. Fabre, J.P. Puel and E. Zuazua, Approximate controllability of the semilinear heatequation, Proc. Royal Soc. Edinburgh, 125A (1995), 31–61.

[9] L.A. Fernandez, E. Zuazua, E., Approximate controllability for the semilinear heatequation involving gradient terms, J. Optim. Theory Appl., 101 (1999), no. 2, 307–328.

[10] E. Fernandez-Cara, M. Gonzalez-Burgos, S. Guerrero and J.P. Puel, Null con-trollability of the heat equation with boundary Fourier conditions: The linear case, to appear.

[11] E. Fernandez-Cara and E. Zuazua, Null and approximate controllability for weaklyblowing up semilinear heat equations, Ann. Inst. H. Poincare, Anal. non lineaire, 17 (2000),no. 5, 583–616.

[12] A. Fursikov, O.Yu. Imanuvilov, Controllability of Evolution Equations, Lecture Notes#34, Seoul National University, Korea, 1996.

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[13] I. Lasiecka and R. Triggiani, Exact controllability of semilinear abstract systems withapplications to waves and plates boundary control, Appl. Math. & Optim., 23 (1991), 109–154.

[14] I. Lasiecka and R. Triggiani, Control Theory for Partial Differential Equations: Con-tinuous and Approximation Theories, Cambridge University Press, Cambridge 2000.

[15] J.L. Lions, Controlabilite Exacte, Perturbations et Stabilisation de Systemes Distribues,Vol. 1 et 2, Masson, Collection RMA, Paris 1988.

[16] E. Zuazua, Exact boundary controllability for the semilinear wave equation, in “NonlinearPartial Differential Equations and their Applications”, Vol. X, H. Brezis and J.L. Lionseds., Pitman, 1991, 357–391.

[17] E. Zuazua, Exact controllability for the semilinear wave equation in one space dimension,Ann. I.H.P., Analyse non Lineaire, 10 (1993), 109–129.

Capıtulo 3

Some controllability results for theN-dimensional Navier-Stokes andBoussinesq systems with N − 1 scalarcontrols

94

Some controllability results for theN-dimensional Navier-Stokes and

Boussinesq systems with N − 1 scalarcontrols

E. Fernandez-Cara, S. Guerrero, O. Yu. Imanuvilov y J.-P. Puel

Abstract

In this paper we deal with some controllability problems for systems of the Navier-Stokes andBoussinesq kind with distributed controls supported in small sets. Our main aim is to controlN -dimensional systems (N +1 scalar unknowns in the case of the Navier-Stokes equations) withN − 1 scalar control functions. In a first step, we present some global Carleman estimates forsuitable adjoint problems of linearized Navier-Stokes and Boussinesq systems. In this way, weobtain null controllability properties for these systems. Then, we deduce results concerning thelocal exact controllability to the trajectories. We also present (global) null controllability resultsfor some (truncated) approximations of the Navier-Stokes equations.

1. Introduction

Let Ω ⊂ RN (N = 2 or 3) be a bounded connected open set whose boundary ∂Ω is regularenough (for instance of class C2). Let O ⊂ Ω be a (small) nonempty open subset and let T > 0.We will use the notation Q = Ω × (0, T ) and Σ = ∂Ω × (0, T ) and we will denote by n(x) theoutward unit normal to Ω at the point x ∈ ∂Ω.

On the other hand, we will denote by C, C1 , C2 , . . . various positive constants (usuallydepending on Ω and O).

We will be concerned with the following controlled Navier-Stokes and Boussinesq systems:

yt −∆y + (y,∇)y +∇p = v1O, ∇ · y = 0 in Q,

y = 0 on Σ,

y(0) = y0 in Ω(3.1)

and

yt −∆y + (y,∇)y +∇p = v1O + θ eN , ∇ · y = 0 in Q,

θt −∆θ + y · ∇θ = h1O in Q,

y = 0, θ = 0 on Σ,

y(0) = y0, θ(0) = θ0 in Ω

(3.2)

95

(in both dimensions N = 2 and N = 3).For N = 2, we will also consider the following approximation of the Navier-Stokes system

with boundary conditions of the Navier kind:

yt −∆y + (y,∇)TM (y) +∇p = v1O, ∇ · y = 0 in Q,

y · n = 0, ∇× y = 0 on Σ,

y(0) = y0 in Ω,

(3.3)

where M > 0, TM (y) = (TM (y1), TM (y2)) and TM is given by

TM (s) =

−M if s ≤ −M,s if −M ≤ s ≤ M,M if s ≥ M.

In systems (3.1), (3.2) and (3.3), v = v(x, t) and h = h(x, t) stand for the control functions.They act during the whole time interval (0, T ) over the set O. The symbol 1O stands for thecharacteristic function of O and eN is the N -th vector of the canonical basis of RN .

The controllability of Navier-Stokes systems has been the objective of considerable workalong the last years. Until now, the best results have been given in [6], where a strategy basedon the methods in [12] and [13] has been followed. Recently, the techniques in [6] have beenadapted in [11] to cover Boussinesq systems (see also [3], [4], [7] and [9] for other results).

This paper can be viewed as a continuation of [6]. We will present some new results thatindicate that the N -dimensional systems (3.1) and (3.2) can be controlled, at least under somegeometrical assumptions, with only N − 1 scalar controls in L2(O × (0, T )). We will also provethat the two-dimensional system (3.3) can be controlled with controls of the form v1O where vis the curl of a function in L2(0, T ;H1(O)).

Along this paper, we will have to impose some regularity assumptions on the initial data. Tothis purpose, we introduce the spaces H, E and V , with

H = w ∈ L2(Ω)N : ∇ · w = 0 in Ω, w · n = 0 on ∂Ω , (3.4)

E =

H if N = 2,

L4(Ω)3 ∩H if N = 3

andV = w ∈ H1

0 (Ω)N : ∇ · w = 0 in Ω.

• For system (3.1), we will assume that the control region O is adjacent to the boundary∂Ω (see assumption (3.10) below) and we will deal with the local exact controllability to thetrajectories. More precisely, our task will be to prove that, for any bounded and sufficientlyregular solution (y, p) of the uncontrolled Navier-Stokes equations

yt −∆y + (y,∇)y +∇p = 0, ∇ · y = 0 in Q,

y = 0 on Σ,

y(0) = y0 in Ω,

(3.5)

96

there exists δ > 0 such that, whenever y0 ∈ E and

‖y0 − y0‖E ≤ δ,

we can find L2 controls v with vk ≡ 0 for at least one k and associated states (y, p) satisfying

y(T ) = y(T ) in Ω. (3.6)

Notice that, under these circumstances, after time t = T we can switch off the control andlet the system follow the ‘ideal’ trajectory (y, p).

• For the Boussinesq system (3.2), we will assume that O is adjacent to ∂Ω near a pointx0 such that nk(x0) 6= 0 for some k < N . We will also be concerned with the local exactcontrollability to the trajectories. Now, a trajectory is a bounded and sufficiently regular solution(y, p, θ) of

yt −∆y + (y,∇)y +∇p = θ eN , ∇ · y = 0 in Q,

θt −∆θ + (y · ∇θ = 0 in Q,

y = 0, θ = 0 on Σ,

y(0) = y0, θ(0) = θ0 in Ω.

(3.7)

The goal will be to prove that there exists δ > 0 such that, whenever (y0, θ0) ∈ E × L2(Ω)and

‖(y0, θ0)− (y0, θ0)‖E×L2 ≤ δ,

we can find L2 controls v and h with vk ≡ vN ≡ 0 and associated states (y, p, θ) satisfying

y(T ) = y(T ) and θ(T ) = θ(T ) in Ω. (3.8)

In this context, the results established in [11] will be fundamental.Notice that, in particular, when N = 2, we try to control the whole system (3.2) with just

one scalar control h.

• As long as (3.3) is concerned, our goal will be to prove the (global) null controllability.That is to say, for each y0 ∈ H, we will try to find controls of the form v1O, where v belongs tothe Hilbert space

W = ∇ × z = (∂2z,−∂1z) : z ∈ L2(0, T ;H1(O)) ,such that the associated solutions (y, p) satisfy

y(T ) = 0 in Ω. (3.9)

Observe that in this system the boundary conditions are of the Navier kind (for their physicalmeaning, see for instance [10]). This and the fact that N = 2 will be essential in the argumentspresented below.

Similarly to the previous situation, an extension by zero of the control after time t = T willkeep (y, p) at rest.

97

As mentioned above, some hypotheses will be imposed on the control domain and the tra-jectories. More precisely, we will frequently assume that

∃x0 ∈ ∂Ω, ∃ε > 0 such that O ∩ ∂Ω ⊃ B(x0; ε) ∩ ∂Ω (3.10)

(B(x0; ε) is the ball centered at x0 of radius ε),

y ∈ L∞(Q)N , yt ∈ L2(0, T ;Lσ(Ω)N )

(σ > 1 if N = 2σ > 6/5 if N = 3

)(3.11)

and

θ ∈ L∞(Q), θt ∈ L2(0, T ;Lσ(Ω))

(σ > 1 if N = 2σ > 6/5 if N = 3

). (3.12)

Let us now present our main results in a precise form. The first one concerns the local exactcontrollability to the trajectories of system (3.1):

Theorem 5 Assume that O satisfies (3.10). Then, for any T > 0, (3.1) is locally exactly con-trollable at time T to the trajectories (y, p) satisfying (3.11) with controls v ∈ L2(O × (0, T ))N

such that vk ≡ 0 for at least one k.

The second main result concerns the controllability of (3.2). It is the following:

Theorem 6 Assume that O satisfies (3.10) with nk(x0) 6= 0 for some k < N . Then, for anyT > 0, (3.2) is locally exactly controllable at time T to the trajectories (y, p, θ) satisfying (3.11)–(3.12) with L2 controls v and h such that vk ≡ vN ≡ 0. In particular, if N = 2, we have localexact controllability to the trajectories with controls v ≡ 0 and h ∈ L2(O × (0, T )).

The last main result we present in this paper is the following:

Theorem 7 Let N = 2. Then, for any T > 0 and any M > 0, (3.3) is null controllable at timeT with controls of the form v1O, where v ∈ W .

For the proofs of these results, following a standard approach, we will first deduce nullcontrollability results for suitable linearized versions of (3.1), (3.2) and (3.3), namely:

yt −∆y + (y,∇)y + (y,∇)y +∇p = f + v1O, ∇ · y = 0 in Q,

y = 0 on Σ,

y(0) = y0 in Ω,

(3.13)

yt −∆y + (y,∇)y + (y,∇)y +∇p = f + v1O + θ eN in Q,

∇ · y = 0 in Q,

θt −∆θ + y · ∇θ + y · ∇θ = k + h1O in Q,

y = 0, θ = 0 on Σ,

y(0) = y0, θ(0) = θ0 in Ω

(3.14)

98

and

yt −∆y + (y,∇)y +∇p = v1O, ∇ · y = 0 in Q,

y · n = 0, ∇× y = 0 on Σ,

y(0) = y0 in Ω.

(3.15)

Then, appropriate arguments will be used to deduce the controllability of the nonlinearsystems (3.1)–(3.3).

Remark 5 When N = 3, it is very natural to ask whether a result similar to theorem 5 holdswith controls having two zero components. In general, the answer is no. In fact, it seems difficultto identify the open sets Ω and O such that one has null controllability for all T > 0 withcontrols of this kind. This is unknown even for the classical Stokes equations for which, up tonow, the unique known results concern approximate controllability; see [15].

Remark 6 Assume that N = 2. The arguments in [6] implicitly show that, under hypotheses(3.11), we can find controls v1O with v ∈ W such that the associated solutions to (3.1) satisfyy(T ) = y(T ). Observe that the assumption (3.10) on the control domain is not necessary here.

This paper is organized as follows. We will first establish all the technical results needed inthis work in Section 2. Section 3 will deal with null controllability results for the linear controlsystems (3.13)–(3.15). Finally, the proofs of theorems 5, 6 and 7 will be given in Section 4.

2. Some previous results

In this Section we will establish all the technical results needed in this paper. More precisely,we will present and prove the required Carleman estimates for the backward systems (3.18),(3.19) and (3.20), given below.

To do this, let us first introduce some weight functions:

α(x, t) =e5/4λm‖η0‖∞ − eλ(m‖η0‖∞+η0(x))

t4(T − t)4,

ξ(x, t) =eλ(m‖η0‖∞+η0(x))

t4(T − t)4,

α(t) = mınx∈Ω

α(x, t) =e5/4λm‖η0‖∞ − eλ(m+1) ‖η0‖∞

t4(T − t)4,

α∗(t) = maxx∈Ω

α(x, t) =e5/4λm‖η0‖∞ − eλm‖η0‖∞

t4(T − t)4,

ξ(t) = maxx∈Ω

ξ(x, t) =eλ(m+1)‖η0‖∞

t4(T − t)4, ξ∗(t) = mın

x∈Ωξ(x, t) =

eλm‖η0‖∞

t4(T − t)4,

(3.16)

where m > 4 is a fixed real number. Here, η0 is a function verifying

η0 ∈ C2(Ω), |∇η0| > 0 in Ω \ O0, η0 > 0 in Ω and η0 ≡ 0 on ∂Ω (3.17)

99

with O0 a nonempty open subset of O that will be determined below. For any O0, the existenceof such a function η0 is proved in [8]. Remark that these weights have already been used in [6]and [11].

We will be dealing in this Section with the adjoint systems to (3.13), (3.14) and (3.15), thatis to say

−ϕt −∆ϕ− (Dϕ) y +∇π = g, ∇ · ϕ = 0 in Q,

ϕ = 0 on Σ,

ϕ(T ) = ϕ0 in Ω,

(3.18)

−ϕt −∆ϕ− (Dϕ) y +∇π = g + θ∇ψ, ∇ · ϕ = 0 in Q,

−ψt −∆ψ − y · ∇ψ = q + ϕN in Q,

ϕ = 0, ψ = 0 on Σ,

ϕ(T ) = ϕ0, ψ(T ) = ψ0 in Ω

(3.19)

(where Dϕ = ∇ϕ +∇ϕt) and

−ρt −∆ρ−∇× ((y · ∇×)∇γ) = 0, ∆γ = ρ in Q,

γ = 0, ρ = 0 on Σ,

ρ(T ) = ρ0 in Ω,

(3.20)

respectively. Here, g ∈ L2(Q)N , q ∈ L2(Q), ϕ0 ∈ H, ψ0 ∈ L2(Ω) and ρ0 ∈ H−1(Ω) (of course,ϕN stands for the last component of the vector field ϕ).

2.1. New Carleman estimates for system (3.18)

In this paragraph, we will establish some new Carleman estimates for the solutions of (3.18).We will assume that O and y satisfy (3.10)–(3.11). To fix ideas, we will also assume for themoment that N = 3 and n1(x0) 6= 0 (x0 appears in assumption (3.10)).

The desired Carleman inequalities will have the form

I(ϕ) ≤ C

(∫∫

Qρ21 |g|2 dx dt +

∫∫

O×(0,T )ρ22

(|ϕ2|2 + |ϕ3|2)

dx dt

),

where I(ϕ) contains global weighted integrals of |ϕ|2, |∇ϕ|2, etc. and ρ1 and ρ2 are appropriateweights that vanish exponentially as t → T . This will suffice to prove in Section 3 the nullcontrollability of (3.13) with controls v1O satisfying v1 ≡ 0.

Lemma 4 Assume that N = 3, n1(x0) 6= 0 and O and y verify (3.10)–(3.11). Then there existsa positive constant C such that, for any g ∈ L2(Q)3 and any ϕ0 ∈ H, the associated solution to

100

(3.18) satisfies:

I(ϕ) :=∫∫

Qe

−2αt4(T−t)4 t−12(T − t)−12 |ϕ|2 dx dt

+∫∫

Qe

−2αt4(T−t)4 t−4(T − t)−4 |∇ϕ|2) dx dt

+∫∫

Qe

−2αt4(T−t)4 t4(T − t)4

(|∆ϕ|2 + |ϕt|2)

dx dt

≤ C

(∫∫


+∫∫

O×(0,T )e−16α+14αt4(T−t)4 t−132(T − t)−132 (|ϕ2|2 + |ϕ3|2) dx dt

).

(3.21)

Here, α and α are constants only depending on Ω, O, T and y satisfying 0 < α < α and8α− 7α > 0.

Proof: Let us first recall a Carleman inequality for the solutions of (3.18) which has been provedin [6] whenever (3.11) is fulfilled:

s3λ4

∫∫

Qe−2sαξ3|ϕ|2 dx dt + sλ2

∫∫

Qe−2sαξ|∇ϕ|2 dx dt

+s−1

∫∫

Qe−2sαξ−1 (|ϕt|2 + |∆ϕ|2) dx dt

≤ C0(1 + T 2)(

s15/2λ20

∫∫

Qe−4sα+2sα∗ ξ15/2|g|2 dx dt

+s16λ40

∫∫

O0×(0,T )e−8sα+6sα∗ ξ16|ϕ|2 dx dt

).

(3.22)

Here, s ≥ s0 and λ ≥ λ0 are arbitrarily large and C0, s0 and λ0 are suitable constants dependingon Ω, O0, T and y ; see theorem 1 in [6].

Recall that an inequality like (3.22) had already been proved in [12] using stronger propertieson y than (3.11).

It is immediate from (3.22) that, for some C1, α and α depending on Ω, O0, T and y, wehave:

∫∫

Qe

−2α

t4(T−t)4(t−12(T − t)−12 |ϕ|2 + t−4(T − t)−4 |∇ϕ|2) dx dt

+∫∫

Qe

−2α

t4(T−t)4 t4(T − t)4(|∆ϕ|2 + |ϕt|2

)dx dt

≤ C1

(∫∫


+∫∫

O0×(0,T )e−8α+6αt4(T−t)4 t−64(T − t)−64 |ϕ|2 dx dt

).

(3.23)

101

Indeed, it suffices to choose

α = s0

(e5/4λm‖η0‖∞ − eλm‖η0‖∞

),

α = s0

(e5/4λm‖η0‖∞ − eλ(m+1)‖η0‖∞

) (3.24)

and C1 = C0(1+T 2)s160 λ40e16λ(m+1)‖η0‖∞ with λ ≥ λ0. Notice that 0 < α < α. Moreover, it can

be assumed that 8α− 7α > 0 (it suffices to take λ large enough in (3.24)).We will apply (3.23) for the open set O0 ⊂ O defined as follows. We choose κ > 0 such that

n1(x) 6= 0 ∀x ∈ B(x0; κ) ∩ ∂O ∩ ∂Ω

and we denote this set by Γκ. Then, we define

O0 = x ∈ Ω : x = w + τ e1, w ∈ Γκ, |τ | < τ0 , (3.25)

with κ, τ0 > 0 small enough so that we still have

O0 ⊂ O and d0 := dist(O0, ∂O ∩ Ω) > 0. (3.26)

Observe that, with this choice, each P ∈ O0 verifies that one of the two points where thestraight line P + R e1 intersects ∂Ω belongs to ∂O0.

Once O0 is defined, we apply inequality (3.23) in this open set and we try to bound the term∫∫

O0×(0,T )e−8α+6α

t4(T−t)4 t−64(T − t)−64 |ϕ1|2 dx dt

in terms of local integrals of ϕ2 and ϕ3.To this end, for each (x, t) ∈ O0 × (0, T ) we denote by l(x, t) (resp. l(x, t)) the segment that

starts from (x, t) with direction e1 in the positive (resp. negative) sense and ends at ∂O0. Then,since ϕ is divergence-free, it is not difficult to see that

ϕ1(x, t) =∫

l(x,t)(∂2ϕ2 + ∂3ϕ3)(y1, x2, x3, t) dy1

for each (x, t) ∈ O0 × (0, T ). For simplicity, let us introduce the notation

β(t) = e−8α+6αt4(T−t)4 t−64(T − t)−64 ∀t ∈ (0, T ).

Applying at this point Holder’s inequality and Fubini’s formula, we obtain∫∫

O0×(0,T )β(t) |ϕ1|2 dx dt

≤ C2

∫∫

O0×(0,T )β(t)

(∫

l(x,t)(|∂2ϕ2|2 + |∂3ϕ3|2) dy1

)dx dt

= C2

∫∫

O0×(0,T )(|∂2ϕ2|2 + |∂3ϕ3|2)

(∫

l(y1)β(t) dx1

)dy1 dx2 dx3 dt

≤ C3

∫∫

O0×(0,T )β(t)(|∂2ϕ2|2 + |∂3ϕ3|2) dx dt,

(3.27)

102

where l(y1) stands for the segment l(y1, x2, x3, t). Then, we introduce a function ζ ∈ C2(O) suchthat

ζ ≡ 1 in O0, 0 ≤ ζ ≤ 1

and ζ(x) = 0 at any point x ∈ O satisfying dist(x, ∂O ∩ Ω) ≤ d0/2 (d0 was defined in (3.26)).This and the fact that ϕ|Σ ≡ 0 imply

∫∫

O0×(0,T )β(t) |∂iϕi|2 dx dt ≤

∫∫

O×(0,T )ζ β(t) |∂iϕi|2 dx dt

=12

∫∫

O×(0,T )∂2

iiζ β(t) |ϕi|2 dx dt−∫∫

O×(0,T )ζ β(t) ∂2

iiϕi ϕi dx dt

for i = 2, 3. Finally, in view of Young’s inequality and regularity estimates for ϕi in Ω (ϕi ∈H2(Ω) and ‖ϕi‖H2 ≤ C‖∆ϕi‖L2), we also have:

∫∫

O0×(0,T )β(t) |∂iϕi|2 dx dt

≤ C4

∫∫

O×(0,T )e−16α+14α

t4(T−t)4 t−132(T − t)−132 |ϕi|2 dx dt

+1

2C1C3

∫∫

Qe

−2αt4(T−t)4 t4(T − t)4 |∆ϕi|2 dx dt,

which, combined with (3.23) and (3.27), yields (3.21).

Let us now present another Carleman inequality for (3.18) with weight functions not van-ishing at time t = 0:

Lemma 5 Assume that N = 3, n1(x0) 6= 0 and O and y verify (3.10)–(3.11). Then there existpositive constants C, α and α with 0 < α < α and 8α − 7α > 0 depending on Ω, O, T and ysuch that, for any g ∈ L2(Q)3 and any ϕ0 ∈ H, the associated solution to (3.18) satisfies:

∫∫

Qe−2α`(t)4

(`(t)−12 |ϕ|2 + `(t)−4 |∇ϕ|2) dx dt

≤ C

(∫∫

Qe−4α+2α

`(t)4 `(t)−30 |g|2 dx dt

+∫∫

O×(0,T )e−16α+14α

`(t)4 `(t)−132 (|ϕ2|2 + |ϕ3|2) dx dt

),

(3.28)

where ` is the C1 function given by

`(t) =

T 2

4for 0 ≤ t ≤ T/2,

t(T − t) for T/2 ≤ t ≤ T.

(3.29)

103

To prove (3.28), it suffices to use (3.21) and the classical parabolic estimates for the Stokessystem satisfied by ϕ. The argument has already been used in [8], [12] and [6] in several similarsituations, so we omit it for simplicity.

For completeness, let us state the similar result that can be established when N = 2. Here,we assume again that n1(x0) 6= 0.

Lemma 6 Assume that N = 2, n1(x0) 6= 0 and O and y verify (3.10)–(3.11). Then there existpositive constants C, α and α with 0 < α < α and 8α − 7α > 0 depending on Ω, O, T and ysuch that, for any g ∈ L2(Q)2 and any ϕ0 ∈ H, the associated solution to (3.18) satisfies:

∫∫

Qe−2α

`(t)4(`(t)−12 |ϕ|2 + `(t)−4 |∇ϕ|2) dx dt

≤ C

(∫∫

Qe−4α+2α

`(t)4 `(t)−30 |g|2 dx dt

+∫∫

O×(0,T )e−16α+14α

`(t)4 `(t)−132 |ϕ2|2 dx dt

),

(3.30)

where ` is the function given by (3.29).

2.2. New Carleman estimates for system (3.19)

In this paragraph, we will establish suitable Carleman inequalities for the solutions of (3.19).To this end, our approach will be similar to the one in subsection 2.1.

Thus, we will assume again that N = 3 and n1(x0) 6= 0 and we will prove an estimate of theform

K(ϕ,ψ) ≤ C

(∫∫

Qρ23 (|g|2 + |q|2) dx dt +

∫∫

O×(0,T )ρ24

(|ϕ2|2 + |ψ|2) dx dt

),

where K(ϕ,ψ) = I(ϕ) + I(ψ) (I(ϕ) has been given in (3.21)) and ρ3 and ρ4 are appropriateweights. This will be used in Section 3 to find controls v1O and h1O with v1 ≡ v3 ≡ 0 leadingto the null controllability of (3.14).

Lemma 7 Assume that N = 3, n1(x0) 6= 0 and O and (y, θ) satisfy (3.10)–(3.12). Then, thereexist positive constants C, α and α depending on Ω, O, T , y and θ with 0 < α < α and16α − 15α > 0 such that, for any g ∈ L2(Q)3, q ∈ L2(Q), ϕ0 ∈ H and ψ0 ∈ L2(Ω), the

104

associated solution to (3.19) satisfies:

I(ϕ) + I(ψ) ≤ C

(∫∫


+∫∫

Qe−32α+30αt4(T−t)4 t−252(T − t)−252 |q|2 dx dt

+∫∫

O×(0,T )e−16α+14αt4(T−t)4 t−132(T − t)−132 |ϕ2|2 dx dt

+∫∫

O×(0,T )e−32α+30αt4(T−t)4 t−268(T − t)−268 |ψ|2 dx dt

).

(3.31)

Proof: Let us first recall a Carleman inequality for the solutions of (3.19) which has recentlybeen proved in [11] whenever (3.11)–(3.12) are fulfilled:

s3λ4

∫∫

Qe−2sαξ3(|ϕ|2 + |ψ|2) dx dt

+sλ2

∫∫

Qe−2sαξ(|∇ϕ|2 + |∇ψ|2) dx dt

+s−1

∫∫

Qe−2sαξ−1 (|ϕt|2 + |ψt|2 + |∆ϕ|2 + |∆ψ|2) dx dt

≤ C5(1 + T 2)(

s15/2λ28

∫∫

Qe−4sα+2sα∗ ξ15/2(|g|2 + |q|2) dx dt

+s16λ56

∫∫

O0×(0,T )e−8sα+6sα∗ ξ16(|ϕ|2 + |ψ|2) dx dt

).

(3.32)

Here, s ≥ s1 and λ ≥ λ1 are arbitrarily large and C5, s1 and λ1 are suitable constants dependingon Ω, O0, T , y and θ; see proposition 1 in [11]. The proof of this inequality follows the samearguments employed in [6] to prove (3.22) and can be achieved without any further regularityon y or θ.

It is clear from (3.32) that, for some C6, α and α depending on Ω, O0, T , y and θ, we have:∫∫

Qe

−2αt4(T−t)4 t−12(T − t)−12 (|ϕ|2 + |ψ|2) dx dt

+∫∫

Qe

−2αt4(T−t)4 t−4(T − t)−4 (|∇ϕ|2 + |∇ψ|2) dx dt

+∫∫

Qe

−2αt4(T−t)4 t4(T − t)4

(|∆ϕ|2 + |∆ψ|2 + |ϕt|2 + |ψt|2)

dx dt

≤ C6

(∫∫

Qe−4α+2αt4(T−t)4 t−30(T − t)−30 (|g|2 + |q|2) dx dt

+∫∫

O0×(0,T )e−8α+6αt4(T−t)4 t−64(T − t)−64 (|ϕ|2 + |ψ|2) dx dt

).

(3.33)

105

Indeed, it suffices to take α and α as in (3.24) and

C6 = C5(1 + T 2)s161 λ56e16λ(m+1)‖η0‖∞

with λ ≥ λ1. We thus obtain 0 < α < α and, choosing λ large enough, 16α− 15α > 0.We apply (3.33) for the open set O0 defined in (3.25). Then we can argue as in paragraph 2.1

and deduce that∫∫

O0×(0,T )e−8α+6αt4(T−t)4 t−64(T − t)−64 |ϕ1|2 dx dt

≤ C7

∫∫

O1×(0,T )e−16α+14αt4(T−t)4 t−132(T − t)−132 (|ϕ2|2 + |ϕ3|2) dx dt

+ε

∫∫

Qe

−2α

t4(T−t)4 t4(T − t)4 (|∆ϕ2|2 + |∆ϕ3|2) dx dt,

where O1 is an appropriate nonempty open set verifying

O0 ⊂ O1 ⊂ O, d1 := dist(O1, ∂O ∩ Ω) > 0.

This inequality combined with (3.33) yields:∫∫

Qe

−2αt4(T−t)4 t−12(T − t)−12 (|ϕ|2 + |ψ|2) dx dt

+∫∫

Qe

−2αt4(T−t)4 t−4(T − t)−4 (|∇ϕ|2 + |∇ψ|2) dx dt

+∫∫

Qe

−2αt4(T−t)4 t4(T − t)4

(|∆ϕ|2 + |∆ψ|2 + |ϕt|2 + |ψt|2)

dx dt

≤ C8

(∫∫

Qe−4α+2α

t4(T−t)4 t−30(T − t)−30 (|g|2 + |q|2) dx dt

+∫∫

O1×(0,T )e−16α+14α

t4(T−t)4 t−132(T − t)−132 (|ϕ2|2 + |ϕ3|2) dx dt

+∫∫

O×(0,T )e−8α+6αt4(T−t)4 t−64(T − t)−64 |ψ|2 dx dt

).

(3.34)

Our last task will be to estimate the integral∫∫

O1×(0,T )e−16α+14αt4(T−t)4 t−132(T − t)−132 |ϕ3|2 dx dt

in terms of εI(ϕ3) and local integrals of ψ and q. To do this, we set

β1(t) = e−16α+14αt4(T−t)4 t−132(T − t)−132

and we introduce a function ζ ∈ C2(O) such that

ζ ≡ 1 in O1, 0 ≤ ζ ≤ 1

106

and ζ(x) = 0 at any point x ∈ O satisfying dist(x, ∂O∩Ω) ≤ d1/2. From the differential equationsatisfied by ψ (see (3.19)), we have

∫∫

O1×(0,T )β1(t) |ϕ3|2 dx dt ≤

∫∫

O×(0,T )β1(t) ζ |ϕ3|2 dx dt

=∫∫

O×(0,T )β1(t) ζ ϕ3(−ψt −∆ψ − y · ∇ψ − q) dx dt.

(3.35)

To end the proof, we perform integrations by parts in the last integral and pass all the derivativesfrom ψ to ϕ3:

• First, we integrate by parts in time taking into account that β1(0) = β1(T ) = 0:

−∫∫

O×(0,T )β1(t) ζ ϕ3 ψt dx dt

=∫∫

O×(0,T )β1,t(t) ζ ϕ3 ψ dx dt +

∫∫

O×(0,T )β1(t) ζ ϕ3,t ψ dx dt

≤ εI(ϕ3) + C9(ε)∫∫

O×(0,T )e−32α+30αt4(T−t)4 t−268(T − t)−268 |ψ|2 dx dt.

(3.36)

• Next, we integrate by parts twice in space. Here, we use the properties of the cut-offfunction ζ and the Dirichlet boundary conditions for ϕ3 and ψ:

−∫∫

O×(0,T )β1(t) ζ ϕ3 ∆ψ dx dt

=∫∫

O×(0,T )β1(t) (−∆ζ ϕ3 − 2∇ζ · ∇ϕ3 − ζ ∆ϕ3) ψ dx dt

≤ εI(ϕ3) + C10(ε)∫∫


(3.37)

• We also integrate by parts in the third term with respect to x and we use the incompress-ibility condition on y:

−∫∫

O×(0,T )β1(t) ζ ϕ3 y · ∇ψ dx dt

=∫∫

O×(0,T )β1(t) y · (ϕ3∇ζ + ζ∇ϕ3)ψ dx dt

≤ εI(ϕ3) + C11(ε)∫∫


(3.38)

107

• We finally apply Young’s inequality in the last term and we have:

−∫∫

O×(0,T )β1(t) ζ ϕ3 q dx dt

≤ εI(ϕ3) + C12(ε)∫∫

O×(0,T )e−32α+30αt4(T−t)4 t−252(T − t)−252 |q|2 dx dt.

(3.39)

From (3.34), (3.35) and (3.36)–(3.39), it is easy to deduce the desired inequality (3.31).

Arguing as in subsection 2.1, that is to say, combining the previous result and the classicalenergy estimates satisfied by ϕ and ψ, we can deduce the following Carleman inequality:

Lemma 8 Assume that N = 3, n1(x0) 6= 0 and O and (y, θ) satisfy (3.10)–(3.12). Then, thereexist positive constants C, α and α depending on Ω, O, T , y and θ with 0 < α < α and16α − 15α > 0 such that, for any g ∈ L2(Q)3, q ∈ L2(Q), ϕ0 ∈ H and ψ0 ∈ L2(Ω), theassociated solution to (3.19) satisfies:

∫∫

Qe−2α

`(t)4(`(t)−12 (|ϕ|2 + |ψ|2) + `(t)−4 (|∇ϕ|2 + |∇ψ|2)) dx dt

≤ C

(∫∫

Qe−4α+2α

`(t)4 `(t)−30 |g|2 dx dt

+∫∫

Qe−32α+30α

`(t)4 `(t)−252 |q|2 dx dt

+∫∫

O×(0,T )e−16α+14α

`(t)4 `(t)−132 |ϕ2|2 dx dt

+∫∫

O×(0,T )e−32α+30α

`(t)4 `(t)−268 |ψ|2 dx dt

),

(3.40)

where the function ` was defined in (3.29).

The similar result that can be established when N = 2 is the following:

Lemma 9 Assume that N = 2, n1(x0) 6= 0 and O and (y, θ) satisfy (3.10)–(3.12). Then, thereexist positive constants C, α and α depending on Ω, O, T , y and θ with 0 < α < α and16α − 15α > 0 such that, for any g ∈ L2(Q)2, q ∈ L2(Q), ϕ0 ∈ H and ψ0 ∈ L2(Ω), the

108

associated solution to (3.19) satisfies:

∫∫

Qe−2α`(t)4

(`(t)−12 (|ϕ|2 + |ψ|2) + `(t)−4 (|∇ϕ|2 + |∇ψ|2)) dx dt

≤ C

(∫∫

Qe−4α+2α

`(t)4 `(t)−30 |g|2 dx dt

+∫∫

Qe−32α+30α

`(t)4 `(t)−252 |q|2 dx dt

+∫∫

O×(0,T )e−32α+30α

`(t)4 `(t)−268 |ψ|2 dx dt

).

(3.41)

2.3. An observability estimate for system (3.20)

In this paragraph, we will prove an observability estimate for the system

−ρt −∆ρ−∇× ((y · ∇×)∇γ) = 0, ∆γ = ρ in Q,

γ = 0, ρ = 0 on Σ,

ρ(T ) = ρ0 in Ω.

(3.42)

This estimate will be implied by a Carleman inequality of the form

S(∇γ) ≤ C

∫∫

O×(0,T )|∇γ|2 dx dt,

where S(∇γ) contains several global weighted integrals involving ∇γ (see (3.43)).

Lemma 10 Assume that N = 2 and y ∈ L∞(Q)2. There exist three positive constants C, s andλ depending on Ω, O, T and y such that, for any ρ0 ∈ H−1(Ω), the associated solution to (3.42)satisfies:

S(∇γ) := s4λ4

∫∫

Qe−2sα ξ4|∇γ|2 dx dt

sλ2

∫∫

Qe−2sα ξ|∇ρ|2 dx dt + s3λ4

∫∫

Qe−2sαξ3|ρ|2 dx dt

≤ C s5λ6

∫∫

O×(0,T )e−2sαξ5|∇γ|2 dx dt,

(3.43)

for any s ≥ s and any λ ≥ λ. Recall that α and ξ were defined in (3.16).

Proof: Along the proof, sj and λj (j ≥ 2) will denote various positive constants that caneventually depend on Ω, O, T and y.

109

Let O0 be a nonempty open set satisfying O0 ⊂⊂ O and let us apply to ρ the Carlemaninequality for parabolic systems with right hand sides in L2(0, T ; H−1(Ω)) proved in [14]:

sλ2

∫∫


∫∫


≤ C13

(s2λ2‖y‖2

∞

∫∫

Qe−2sαξ2|∇(∇× γ)|2 dx dt

+s3λ4

∫∫

O0×(0,T )e−2sαξ3|ρ|2 dx dt

),

(3.44)

for any s ≥ s2 and λ ≥ λ2.Observe that, here, the assumption ρ0 ∈ H−1(Ω) may seem too weak to apply this result.

Indeed, (3.44) can be proved as in [14] whenever ρ ∈ C1(Q) and, by a continuity argument, alsofor the solutions of problem (3.42) for which the left hand side of (3.44) is finite. This is ourcase, since one can ensure that ρ ∈ L2(Q) as soon as ρ0 ∈ H−1(Ω) (for instance, taking intoaccount the definition of ρ as the solution by transposition of (3.42)).

Once (3.44) has been justified, let us first estimate the last integral in its right hand side.Thus, let ζ ∈ C2(O) be a cut-off function satisfying

ζ ≡ 1 in O0, 0 ≤ ζ ≤ 1 and ζ = 0 on ∂O.

We have:

s3λ4

∫∫

O0×(0,T )e−2sαξ3|∆γ|2 dx dt ≤ s3λ4

∫∫

O×(0,T )ζ e−2sαξ3|∆γ|2 dx dt

= −s3λ4

∫∫

O×(0,T )e−2sαξ3(∇ζ · ∇γ)∆γ dx dt

−3s3λ5

∫∫

O×(0,T )ζ e−2sαξ3(∇η0 · ∇γ)∆γ dx dt

+2s4λ5

∫∫

O×(0,T )ζ e−2sαξ4(∇η0 · ∇γ)∆γ dx dt

−s3λ4

∫∫

O×(0,T )ζ e−2sαξ3(∇∆γ · ∇γ) dx dt.

Now, we apply Young’s inequality several times and we obtain

s3λ4

∫∫

O0×(0,T )e−2sαξ3|∆γ|2 dx dt

≤ C14(ε) s5λ6

∫∫

O×(0,T )e−2sαξ5|∇γ|2 dx dt

+ε

(s3λ4

∫∫

Qe−2sαξ3|∆γ|2 dx dt + sλ2

∫∫

Qe−2sαξ|∇∆γ|2 dx dt

),

110

for s ≥ s3 and λ ≥ λ3 and for any small positive constant ε. Combining this, the fact thatρ = ∆γ and (3.44), we get

sλ2

∫∫


∫∫


≤ C15

(s2λ2‖y‖2

∞

∫∫

Qe−2sαξ2|∇(∇× γ)|2 dx dt

+s5λ6

∫∫


)(3.45)

for any s ≥ s4 and λ ≥ λ4.Finally, we are going to estimate the first integral in the right hand side of (3.44). To this

end, let us notice that, for j = 1 and 2 and almost every t ∈ (0, T ), the function ∂jγ(t) satisfies:

∆(∂jγ)(t) = ∂jρ(t) in Ω.

Let us apply the main result in [13] to ∂jγ. This yields the existence of two numbers τ > 1and λ > 1 such that

τ4λ4

∫

Ωe2τηη4|∂jγ|2(t) dx + τ2λ2

∫

Ωe2τηη2|∇(∂jγ)|2(t) dx

≤ C16

(τ

∫

Ωe2τηη|∂jρ|2(t) dx + τ4λ4

∫

Oe2τηη4|∂jγ|2(t) dx

+τ5/2λ2 e2τ‖∂jγ(t)‖2H1/2(∂Ω)

)(3.46)

for τ ≥ τ and λ ≥ λ. Here, we have introduced the function η, with

η(x) = eλη0(x).

In fact, the inequality one can find in [13] contains local integrals of |∂jγ|2 and |∇(∂jγ)|2 in theright hand side. But it can be written for a smaller set O′ ⊂⊂ O. Using localizing argumentstogether with the fact that we actually have a global weighted integral of |∆(∂jγ)|2 in the lefthand side, (3.46) is easily found.

Following the same steps of [6], we set

τ =s

t4(T − t)4eλ m‖η0‖∞ ,

we multiply (3.46) by

exp

−2s

e5/4λm‖η0‖∞

t4(T − t)4

111

and we integrate in time over (0, T ). This gives

s4λ4

∫∫

Qe−2sαξ4|∂jγ|2 dx dt + s2λ2

∫∫

Qe−2sαξ2|∇(∂jγ)|2 dx dt

≤ C17

(s

∫∫

Qe−2sαξ|∂jρ|2 dx dt + s4λ4

∫∫

O×(0,T )e−2sαξ4|∂jγ|2 dx dt

+s5/2λ2

∫ T

0e−2sα∗(ξ∗)5/2‖∂jγ‖2

H1/2(∂Ω)

)

for s ≥ s5 and λ ≥ λ. Combining this estimate and (3.45), we have

s4λ4

∫∫

Qe−2sαξ4|∂jγ|2 dx dt + sλ2

∫∫

Qe−2sα ξ|∇ρ|2 dx dt

+s3λ4

∫∫

Qe−2sαξ3|ρ|2 dx dt ≤ C18

(s5/2λ2

∫ T

0e−2sα∗(ξ∗)5/2‖∂jγ‖2

H1/2(∂Ω)

+s5λ6

∫∫


)

for any s ≥ s6 and λ ≥ λ5. On the other hand, the boundary term can readily be bounded usingthe continuity of the trace operator:

‖∂jγ(t)‖2H1/2(∂Ω)

≤ C19(‖∂jγ(t)‖2L2 + ‖∇(∂jγ)(t)‖2

L2).

Furthermore, since γ|Σ ≡ 0, we know that there exists a positive constant C20 such that

‖∇(∂jγ)(t)‖L2 ≤ C20‖∆γ(t)‖L2 a.e. in (0, T ) for j = 1, 2.

Consequently,

s4λ4

∫∫

Qe−2sαξ4|∂jγ|2 dx dt + sλ2

∫∫

Qe−2sα ξ|∇ρ|2 dx dt

+s3λ4

∫∫

Qe−2sαξ3|ρ|2 dx dt ≤ C21 s5λ6

∫∫


for s ≥ s6.This implies (3.43) and ends the proof of lemma 12.

Remark 7 An almost immediate consequence of the Carleman estimate (3.43) is the followingobservability inequality:

‖(∇γ)(0)‖2L2 ≤ C

∫∫

O×(0,T )|∇γ|2 dx dt. (3.47)

In fact, this is what will be used in Section 3 to prove the null controllability of system (3.15).

112

3. Null controllability of the linearized systems (3.13), (3.14)and (3.15)

3.1. Null controllability of (3.13)

We are dealing here with the following system:

yt −∆y + (y,∇)y + (y,∇)y +∇p = f + v1O, ∇ · y = 0 in Q,

y = 0 on Σ,

y(0) = y0 in Ω,

(3.48)

where O satisfies (3.10) and y satisfies (3.11). Our goal will be to find a control v such thaty(T ) = 0 in Ω.

Let us introduce some weight functions:

β2(t) = exp

α

`(t)4

`(t)6, β3(t) = exp

2α− α

`(t)4

`(t)15

and

β4(t) = exp

8α− 7α

`(t)4

`(t)66

(recall that ` was defined in (3.29)) where α and α are the constants provided by lemma 5 whenN = 3 and lemma 6 when N = 2. Recall that, in particular, 0 < α < α and 8α− 7α > 0.

Of course, we will need some specific conditions on f and y0 to get the null controllabilityof (3.48). We will use the arguments in [6].

Thus, let us setLy = yt −∆y + (y,∇)y + (y,∇)y (3.49)

and let us introduce the spaces

E2 = (y, p, v) : (y, v) ∈ E0, `−4β2(Ly +∇p− v1O) ∈ L2(0, T ; H−1(Ω)2) when N = 2 and

E3 = (y, p, v) : (y, v) ∈ E0, `−2β1/22 y ∈ L4(0, T ; L12(Ω)3),

`−4β2(Ly +∇p− v1O) ∈ L2(0, T ;W−1,6(Ω)3) when N = 3, where

E0 = (y, v) : β3 y, β4 v1O ∈ L2(Q)N , v1 ≡ 0,

`−2β1/22 y ∈ L2(0, T ; V ) ∩ L∞(0, T ; H) .

It is clear that EN is a Banach space for the norm ‖ · ‖EN, where

‖(y, p, v)‖E2 =(‖β3 y‖2

L2 + ‖β4 v1O‖2L2

+‖`−2β1/22 y‖2

L2(0,T ;V ) + ‖`−2β1/22 y‖2

L∞(0,T ;H)

+‖`−4β2(Ly +∇p− v1O)‖2L2(0,T ;H−1)

)1/2

113

and‖(y, p, v)‖E3 =

(‖β3 y‖2L2 + ‖β4 v1O‖2

L2

+‖`−2β1/22 y‖2

L2(0,T ;V ) + ‖`−2β1/22 y‖2

L∞(0,T ;H)

+‖`−2β1/22 y‖2

L4(0,T ;L12)

+‖`−4β2(Ly +∇p− v1O)‖2L2(0,T ;W−1,6)

)1/2.

Proposition 5 Assume that n1(x0) 6= 0 and O and y verify (3.10)–(3.11). Let y0 ∈ E and letus assume that

`−4β1f ∈

L2(0, T ; H−1(Ω)2) if N = 2,

L2(0, T ; W−1,6(Ω)3) if N = 3.

Then, we can find a control v such that the associated solution (y, p) to (3.48) satisfies (y, p, v) ∈EN . In particular, v1 ≡ 0 and y(T ) = 0.

Sketch of the proof: The proof of this proposition is very similar to the one of proposition 2in [6], so we will just give the main ideas. For simplicity, we will only consider the case N = 3.When N = 2, the proof is even easier.

Following the arguments in [8] and [12], let us introduce the auxiliary optimal control problem

ınf12

(∫∫

Q|β3 y|2dxdt +

∫∫

O×(0,T )|β4 v|2dxdt

)

subject to v ∈ L2(Q)3, supp v ⊂ O × (0, T ), v1 ≡ 0 and

Ly +∇p = f + v1O in Q,

∇ · y = 0 in Q,

y = 0 on Σ,

y(0) = y0, y(T ) = 0 in Ω.

(3.50)

Notice that a solution (y, p, v) to (3.50) is a good candidate to satisfy (y, p, v) ∈ E3.For the moment, let us assume that (3.50) possesses a solution (y, p, v). Then, by virtue of

Lagrange’s principle, there must exist dual variables z and q such that

y = β−23 (L∗z +∇q), ∇ · z = 0 in Q,

v1 ≡ 0, vi = −β−24 zi (i = 2, 3) in O × (0, T ),

z = 0 on Σ,

(3.51)

where L∗ is the adjoint operator of L, i.e.

L∗z = −zt −∆z − (Dz) y.

At least formally, the couple (z, q) satisfies

a((z, q), (w, h)) = 〈G, (w, h)〉 ∀(w, h) ∈ P0, (3.52)

114

where P0 is the space

P0 = (w, h) ∈ C2(Q)4 : ∇ · w = 0, w = 0 on Σ,

∫

Oh(x, t) dx = 0

and we have used the notation

a((z, q), (w, h)) =∫∫

Qβ−2

3 (L∗z +∇q) · (L∗w +∇h) dx dt

+∫∫

O×(0,T )β−2

4 (z2 w2 + z3 w3) dx dt

and

〈G, (w, h)〉 =∫ T

0〈f(t), w(t)〉H−1,H1

0dt +

∫

Ωy0 · w(0) dx.

Conversely, if we are able to “solve” (3.52) and then use (3.51) to define (y, p, v), we willprobably have found a solution to (3.50).

Thus, let us consider the linear space P0. It is clear that a(· , ·) : P0×P0 7→ R is a symmetric,definite positive bilinear form on P0. We will denote by P the completion of P0 for the norminduced by a(· , ·). Then a(· , ·) is well-defined, continuous and again definite positive on P .Furthermore, in view of the Carleman estimate (3.28), the linear form (w, h) 7→ 〈G, (w, h)〉is well-defined and continuous on P . Hence, from Lax-Milgram’s lemma, we deduce that thevariational problem

a((z, q), (w, h)) = 〈G, (w, h)〉∀(w, h) ∈ P, (z, q) ∈ P,

(3.53)

possesses exactly one solution (z, q).Let y and v be given by (3.51). Then, it is readily seen that they verify

∫∫

Qβ2

3 |y|2dx dt +∫∫

O×(0,T )β2

4 |v|2dx dt < +∞

and, also, that y is, together with some pressure p, the weak solution (belonging to L2(0, T ; V )∩L∞(0, T ; H)) of the Stokes system in (3.50) for v = v.

In order to prove that (y, p, v) ∈ E3, it only remains to check that `−2 β1/22 y is, together

with `−2 β1/22 p, a weak solution of a Stokes problem of the kind (3.48) with a right hand side

in L2(0, T ;W−1,6(Ω)3) that belongs to L4(0, T ; L12(Ω)3). To this end, we define the functionsy∗ = `−2 β

1/22 y, p∗ = `−2 β

1/22 p and f∗ = `−2 β

1/22 (f + v1O). Then (y∗, p∗) satisfies

Ly∗ +∇p∗ = f∗ + (`−2β−1/22 )t y, ∇ · y∗ = 0 in Q,

y∗ = 0 on Σ,

y∗(0) = `−2(0)β1/22 (0)y0 in Ω.

(3.54)

From the fact that f∗ ∈ L2(0, T ; H−1(Ω)3) and y0 ∈ H, we have indeed

y∗ ∈ L2(0, T ;V ) ∩ L∞(0, T ; H).

Finally, we deduce that y∗ ∈ L4(0, T ; L12(Ω)3) from lemma 2 in [6]. This ends the sketch of theproof of proposition 8.

115

3.2. Null controllability of system (3.14)

In this paragraph, we will establish the null controllability of the linear system

yt −∆y + (y,∇)y + (y,∇)y +∇p = f + v1O + θ eN in Q,

∇ · y = 0 in Q,

θt −∆θ + y · ∇θ + y · ∇θ = k + h1O in Q,

y = 0, θ = 0 on Σ,

y(0) = y0, θ(0) = θ0 in Ω,

(3.55)

where O satisfies (3.10) and y and θ satisfy (3.11) and (3.12), for suitable right hand sides fand k.

The arguments we present here are completely analogous to those in [11] and subsection3.1 of this paper, so that we will only give a sketch. Thus, we restrict ourselves again to thethree-dimensional case with n1(x0) 6= 0.

Let us introduce the weight functions

β5(t) = exp

α

`(t)4

`(t)6, β6(t) = exp

2α− α

`(t)4

`(t)15,

β7(t) = exp

16α− 15α`(t)4

`(t)126, β8(t) = exp

8α− 7α

`(t)4

`(t)66

and

β9(t) = exp

16α− 15α

`(t)4

`(t)134,

where the constants α and α are furnished by lemma 8 when N = 3 and lemma 9 when N = 2(and, in particular, 0 < α < α and 16α− 15α > 0).

Let us setPθ = θt −∆θ + y · ∇θ (3.56)

and let us introduce the spaces

E2 = (y, p, θ, v, h) : (y, θ, v, h) ∈ E0,

`−4β5(Ly +∇p− v1O) ∈ L2(0, T ;H−1(Ω)2),

`−4β5(Pθ + y · ∇θ − h1O) ∈ L2(0, T ; H−1(Ω))

when N = 2 and

E3 = (y, p, θ, v, h) : (y, θ, v, h) ∈ E0,

`−2β1/25 y ∈ L4(0, T ; L12(Ω)3),

`−4β5(Ly +∇p− v1O) ∈ L2(0, T ;W−1,6(Ω)3),

`−4β5(Pθ + y · ∇θ − h1O) ∈ L2(0, T ; H−1(Ω))

116

when N = 3, where

E0 = (y, θ, v, h) : (β6 y)i, β7 θ, (β8 v1O)i, β9 h1O ∈ L2(Q) (1 ≤ i ≤ N),

v1 ≡ vN ≡ 0, `−2β1/25 y ∈ L2(0, T ; V ) ∩ L∞(0, T ;H),

`−2β1/25 θ ∈ L2(0, T ; H1

0 (Ω)) ∩ L∞(0, T ;L2(Ω)) .

It can be readily seen now that E0, E2 and E3 are Banach spaces for the norms

‖(y, θ, v, h)‖E0

=(‖β6 y‖2

L2 + ‖β7 θ‖2L2 + ‖β8 v‖2

L2

+‖β9 h‖2L2 + ‖`−2β

1/25 y‖2

L2(0,T ;V ) + ‖`−2β1/25 y‖2

L∞(0,T ;H)

+‖`−2β1/25 θ‖2

L2(0,T ;H10 )

+ ‖`−2β1/25 θ‖2

L∞(0,T ;L2)

)1/2,

‖(y, p, θ, v, h)‖E2

=(‖(y, θ, v, h)‖2

E0

+‖`−4β5(Ly +∇p− v1O)‖2L2(0,T ;H−1)

+‖`−4β5(Pθ + y · ∇θ − h1O)‖2L2(0,T ;H−1)

)1/2

and‖(y, p, θ, v, h)‖

E3=

(‖(y, θ, v, h)‖2

E0+ ‖`−2β

1/25 y‖2

L4(0,T ;L12)

+‖`−4β5(Ly +∇p− v1O)‖2L2(0,T ;W−1,6)

+‖`−4β5(Pθ + y · ∇θ − h1O)‖2L2(0,T ;H−1)

)1/2

Proposition 6 Assume that n1(x0) 6= 0 and O and (y, θ) satisfy (3.10)–(3.12). Let y0 ∈ E,θ0 ∈ L2(Ω) and let us assume that

`−4β1(f, k) ∈

L2(0, T ; H−1(Ω)2)× L2(0, T ; H−1(Ω)) if N = 2,

L2(0, T ; W−1,6(Ω)3)× L2(0, T ;H−1(Ω)) if N = 3.

Then, we can find controls v and h such that the associated solution to (3.55) satisfies (y, p, θ, v, h) ∈EN . In particular, v1 ≡ vN ≡ 0 and y(T ) = θ(T ) = 0.

We omit the proof of this proposition, since it is essentially the same as the one of propo-sition 2 in [11] and follows the steps of proposition 5 above. As we have already indicated, themain ideas come from the paper [12].

3.3. Null controllability of system (3.15)

In this paragraph, we will prove the null controllability of the linear system

yt −∆y + (y,∇)y +∇p = v1O, ∇ · y = 0 in Q,

y · n = 0, ∇× y = 0 on Σ,

y(0) = y0 in Ω,

(3.57)

117

where N = 2 and y ∈ L∞(Q)2.For this purpose, we first rewrite this system using the streamline-vorticity formulation.

Thus, setting ω = ∇× y, we have

ωt −∆ω +∇× ((∇× ψ,∇)y) = ∇× (v1O), ∆ψ = ω in Q,

ψ = 0, ω = 0 on Σ,

ω(0) = ∇× y0 in Ω.

(3.58)

Proposition 7 Assume that y0 ∈ H and y ∈ L∞(Q)2. Then, there exist controls v1O withv ∈ W such that

‖v‖L2 ≤ C‖y0‖H (3.59)

and the associated solutions of (3.57) satisfy

y ∈ L2(0, T ; H1(Ω)2) ∩ C0([0, T ];L2(Ω)2), yt ∈ L2(0, T ;H−1(Ω)2), (3.60)

and y(T ) = 0, with

‖y‖L2(0,T ;H1) + ‖y‖C0([0,T ];L2) + ‖yt‖L2(0,T ;H−1) ≤ C‖y0‖H . (3.61)

Proof: We first establish the null controllability property for y. This can be done in severalways. One of them is the following. We first define for each ε > 0 the functional

Jε(γ0) =12

∫∫

O×(0,T )|∇ × γ|2 dx dt + ε‖∇γ0‖L2 + ((∇× γ)(0), y0)L2

∀γ0 ∈ H10 (Ω),

where γ is given by (3.42) with ρ0 = ∆γ0 ∈ H−1(Ω).It is not difficult to see from the observability inequality (3.47) that this functional possesses

a unique minimizer γ0ε ∈ H1

0 (Ω) (see, for instance, [5]). Now, from the necessary conditions forJε to reach a minimum, we have

∫∫

Q((∇× γε)1O) · (∇× γ) dx dt + ε(

∇× γ0ε

‖∇ × γ0ε‖L2

,∇× γ0)L2

+((∇× γ)(0), y0)L2 = 0 ∀γ0 ∈ H10 (Ω).

(3.62)

Thus, setting vε = (∇× γε)1O and putting γ0 = γ0ε , we find from (3.47) and (3.62) that (3.59)

holds for vε for some C independent of ε:

‖∇ × γε‖L2(O×(0,T ))2 = ‖vε‖L2(O×(0,T ))2 ≤ C. (3.63)

Let us denote by (ωε, ψε) the solution to (3.58) for v = vε. Then, taking into account thesystems satisfied by (ρ, γ) and (ωε, ψε), we deduce that

∫∫

Q∇× (vε1O) γ dx dt + (∇× γ0, (∇× ψε)(T ))L2

−((∇× γ)(0), y0)L2 = 0 ∀γ0 ∈ H10 (Ω).

118

Combining this and (3.62), we obtain

‖(∇× ψε)(T )‖L2 ≤ ε. (3.64)

From (3.63) and (3.64) written for each ε > 0, we deduce that, at least for a subsequence,vε → v weakly in L2(O× (0, T ))2, where the control v1O is such that the corresponding solution(ω, ψ) to (3.58) satisfies

(∇× ψ)(T ) = y(T ) = 0 in Ω.

Since v ∈ L2(O × (0, T ))2 and ∇ · v = 0 in O × (0, T ), we necessarily have v ∈ W (from DeRham’s lemma applied to (v2,−v1)).

In order to obtain the desired regularity for y, we will consider again the equations satisfiedby ψ and ω and we will check that

ψ ∈ L2(0, T ;H2(Ω)) ∩ C0([0, T ]; H10 (Ω)) and ψt ∈ L2(Q), (3.65)

with appropriate estimates.For simplicity, we will only present the estimates. The rigorous argument relies on introducing

a standard Galerkin approximation of (3.58) with a ‘special’ basis of H10 (Ω) (more precisely, the

basis formed by the eigenfunctions of the Laplacian-Dirichlet operator in Ω) and deducing forthe associated approximate solutions the estimates below.

Thus, let us multiply the first equation in (3.58) by ψ and let us integrate by parts. We findthat

12

∫

Ω|∇ψ(t)|2 dx +

∫ t

0

∫

Ω|∆ψ|2 dx dτ =

∫ t

0

∫

Ov · (∇× ψ) dx dτ

−∫ t

0

∫

Ω(((∇× ψ),∇) y) · (∇× ψ) dx dτ +

12‖(∇ψ)(0)‖2

L2

for all t ∈ (0, T ). If we integrate by parts in the last integral, we also have

−∫ t

0

∫

Ω(((∇× ψ) · ∇) y) · (∇× ψ) dx dτ

=∫ t

0

∫

Ω((∇× ψ) · ∇)(∇× ψ) · y dx dτ.

Since ψ|Σ ≡ 0, we deduce that

ψ ∈ L2(0, T ; H2(Ω)) ∩ L∞(0, T ; H10 (Ω)) (3.66)

and‖ψ‖L2(0,T ;H2) + ‖ψ‖L∞(0,T ;H1

0 ) ≤ C‖y0‖L2 . (3.67)

Now, let us introduce for each t the function ψ∗(t) = ∆−1ψt(t), i.e. the solution to −∆ψ∗(t) = ψt(t) in Ω

ψ∗(t) = 0 on ∂Ω.

119

Observe that, whenever ψt(t) ∈ L2(Ω), this function satisfies ψ∗(t) ∈ H2(Ω) ∩H10 (Ω) and

‖ψ∗(t)‖H2 ≤ C‖ψt(t)‖L2 . (3.68)

Then, we multiply the first equation of (3.58) by ψ∗ and we integrate by parts. This gives∫∫

Q|ψt|2 dx dt =

∫∫

Q(∆ψ)ψt dx dt−

∫∫

Q((∇× ψ) · ∇)(∇× ψ∗) · y dx dt

+∫∫

O×(0,T )v · (∇× ψ∗) dx dt.

Using that v ∈ L2(Q)2 and we already have (3.67) and (3.68), we conclude that ψt ∈ L2(Q) and

‖ψt‖L2 ≤ C‖y0‖L2 . (3.69)

From (3.67) and (3.69), we immediately obtain (3.65), (3.60) and (3.61).This ends the proof of proposition 9.

4. Proofs of the controllability results for the nonlinear systems

In this last Section, we will give the proofs of theorems 5, 6 and 7. For the proofs of theorems 5and 6 we employ an inverse mapping theorem, while a fixed point argument is used for theorem 7.

4.1. Proof of Theorem 5

We also follow here the steps in [6].Thus, we set y = y+z and p = p+χ and we use these identities in (3.1). Taking into account

that (y, p) solves (3.5), we find:

Lz + (z,∇)z +∇χ = v1O, ∇ · z = 0 in Q,

z = 0 on Σ,

z(0) = y0 − y0 in Ω(3.70)

(recall that L was defined in (3.49)).This way, we have reduced our problem to a local null controllability result for the solution

(z, χ) to the nonlinear problem (3.70).We will use the following inverse mapping theorem (see [1]):

Theorem 8 Let B1 and B2 be two Banach spaces and let A : B1 7→ B2 satisfy A ∈ C1(B1; B2).Assume that b0 ∈ B1, A(b0) = d0 and also that A′(b0) : B1 7→ B2 is surjective. Then thereexists δ > 0 such that, for every d ∈ B2 satisfying ‖d− d0‖B2 < δ, there exists a solution of theequation

A(b) = d, b ∈ B1.

We will apply this result with B1 = EN ,

120

B2 =

L2(`−4β2; 0, T ; H−1(Ω)2)×H if N = 2,

L2(`−4β2; 0, T ; W−1,6(Ω)3)× (H ∩ L4(Ω)3) if N = 3

andA(z, χ, v) = (Lz + (z,∇)z +∇χ− v1O, z(0)) ∀(z, χ, v) ∈ EN .

From the facts that `−2β1/22 y ∈ L4(0, T ; L12(Ω)3) and A is bilinear, it is not difficult to check

that A ∈ C1(B1; B2); more details can be found in [12] or [6].Let b0 be the origin in B1. Notice that A′(0, 0, 0) : B1 7→ B2 is given by

A′(0, 0, 0)(z, χ, v) = (Lz +∇χ− v1O, z(0)) ∀(z, χ, v) ∈ EN

and is surjective, in view of the null controllability result for (3.13) given in proposition 8.Consequently, we can indeed apply theorem 8 with these data and there exists δ > 0 such

that, if ‖z(0)‖E ≤ δ, then we find a control v satisfying v1 ≡ 0 such that the associated solutionto (3.70) verifies z(T ) = 0 in Ω.

This concludes the proof of theorem 5.

4.2. Proof of theorem 6

Again, we follow here the ideas of [11].Therefore, we set y = y + z, p = p + χ and θ = θ + ρ, so from (3.2) and (3.7), we find:

Lz + (z,∇)z +∇χ = v1O + ρ eN , ∇ · z = 0 in Q,

Pρ + (z,∇)ρ + z · ∇θ = h1O in Q,

z = 0, ρ = 0 on Σ,

z(0) = y0 − y0, ρ(0) = θ0 − θ(0) in Ω

(3.71)

(L and P were respectively defined in (3.49) and (3.56)).We are thus led to prove the local null controllability of (3.71). To this end, we will use

again theorem 8, which was presented in subsection 4.1. Using the same notation as there, weset B1 = EN ,

B2 = L2(`−4β5; 0, T ;H−1(Ω)3)×H × L2(Ω)

if N = 2 and

B2 = L2(`−4β5; 0, T ;W−1,6(Ω)3 ×H−1(Ω))× (L4(Ω)3 ∩H)× L2(Ω).

if N = 3.Let us introduce A, with

A(z, χ, ρ, v, h) = (A1(z, χ, ρ, v),A2(z, ρ, h), z(0), ρ(0)),

A1(z, χ, ρ, v) = Lz + (z,∇)z +∇χ− v1O − ρeN

andA2(z, ρ, h) = Pρ + (z,∇)ρ + z · ∇θ − h1O

121

for every (z, χ, ρ, v, h) ∈ EN .Using the fact that `−2β

1/25 z ∈ L4(0, T ;L12(Ω)3), it can be checked that A1 is C1. Then,

since `−2β1/25 ρ ∈ L2(0, T ; H1(Ω))∩L∞(0, T ; L2(Ω)) and this space is continuously embedded in

L4(0, T ;L3(Ω)), we deduce that

`−4β5(z,∇)ρ = ∇ · (z ρ) ∈ L2(0, T ; W−1,12/5(Ω)) ⊂ L2(0, T ;H−1(Ω))

and, consequently, A is well-defined and satisfies A ∈ C1(B1; B2).The fact that A′(0, 0, 0, 0, 0) : B1 7−→ B2 is onto is an immediate consequence of the result

given in proposition 6.As a conclusion, we can apply theorem 8 and the null controllability for system (3.71) holds.

4.3. Proof of theorem 7

Let us recall the nonlinear system we are dealing with:

yt −∆y + (y,∇)TM (y) +∇p = v1O in Q,

∇ · y = 0 in Q,

y · n = 0, ∇× y = 0 on Σ,

y(0) = y0 in Ω.

In this case, we are going to apply Kakutani’s fixed point theorem (see for instance [2]):

Theorem 9 Let Z be a Banach space and let Λ : Z 7→ Z be a set-valued mapping satisfying thefollowing assumptions:

• Λ(z) is a nonempty closed convex set of Z for every z ∈ Z.

• There exists a convex compact set K ⊂ Z such that Λ(K) ⊂ K.

• Λ is upper-hemicontinuous in Z, i.e. for each σ ∈ Z ′ the single-valued mapping

z 7→ supy∈Λ(z)

〈σ, y〉Z′,Z (3.72)

is upper-semicontinuous.Then Λ possesses a fixed point in the set K, i.e. there exists z ∈ K such that z ∈ Λ(z).

In order to apply this result, we set Z = L2(Q)2 and, for each z ∈ Z, we consider thefollowing system:

yt −∆y + (y,∇)TM (z) +∇p = v1O in Q,

∇ · y = 0 in Q,

y · n = 0, ∇× y = 0 on Σ,

y(0) = y0 in Ω.

(3.73)

Then, for each z ∈ Z, we denote by A(z) the set of controls v1O with v ∈ W that drivesystem (3.73) to zero and satisfy (3.59). Finally, our set-valued mapping is given as follows: for

122

each z ∈ Z, Λ(z) is the set of functions y that solve, together with some p, the linear system(3.73) corresponding to a control v ∈ A(z).

Let us check that the assumptions of theorem 9 are satisfied in this setting. The first oneholds easily, so we omit the proof. Next, the estimates (3.60) and (3.61) tell us that the wholespace Z is actually mapped into a compact set.

Let us finally see that Λ is upper-hemicontinuous in Z. Assume that σ ∈ Z ′ and let zn bea sequence in Z such that zn → z in Z. We have to prove that

lım supn→∞

supy∈Λ(zn)

〈σ, y〉Z′,Z ≤ supy∈Λ(z)

〈σ, y〉Z′,Z . (3.74)

Let us choose a subsequence zn′ such that

lım supn→∞

supy∈Λ(zn)

〈σ, y〉Z′,Z = lımn′→∞

supy∈Λ(zn′ )

〈σ, y〉Z′,Z . (3.75)

From the fact that Λ(zn′) is a compact set of Z, for each n′ we have

supy∈Λ(zn′ )

〈σ, y〉Z′,Z = 〈σ, yn′〉Z′,Z

for some yn′ ∈ Λ(zn′). Obviously, it can be assumed that

zn′(x, t) → z(x, t) a.e. (x, t) ∈ Q (3.76)

andvn′ v weakly in L2(Q)2 (3.77)

with v ∈ A(z). Furthermore, since all the yn′ belong to a fixed compact set, we can also assumethat

yn′ → y in Z.

This, together with (3.75)–(3.77) implies that y ∈ Λ(z). As a conclusion, (3.74) holds and theproof of theorem 7 is achieved.

Bibliografıa

[1] V. M. Alekseev, V. M. Tikhomirov, S. V. Fomin, Optimal Control, Translated fromthe Russian by V. M. Volosov, Contemporary Soviet Mathematics. Consultants Bureau,New York 1987.

[2] J.P. Aubin, L’Analyse non Lineaire et ses Motivations Economiques, Masson, Paris 1984.

[3] J.-M. Coron, On the controllability of the 2-D incompressible Navier-Stokes equations withthe Navier slip boundary conditions, ESAIM Control Optim. Calc. Var., 1 (1996), 35–75.

[4] C. Fabre, Uniqueness results for Stokes equations and their consequences in linear andnonlinear control problems, ESAIM Control Optim. Calc. Var., 1 (1995/96), 267–302.

[5] C. Fabre, J.P. Puel, E. Zuazua, Approximate controllability of the semilinear heatequation, Proc. Royal Soc. Edinburgh, 125A (1995), p. 31–61.

[6] E. Fernandez-Cara, S. Guerrero, O. Yu. Imanuvilov and J.-P. Puel, Local exactcontrollability of the Navier-Stokes system, to appear in J. Math. Pures Appl.

[7] A. Fursikov, O. Yu. Imanuvilov, Local exact controllability for 2-D Navier-Stokes equa-tions, Sbornik: Mathematics 187, #9, 1996, p. 1355–1390.

[8] A. Fursikov, O. Yu. Imanuvilov, Controllability of Evolution Equations, Lecture Notes#34, Seoul National University, Korea, 1996.

[9] A. Fursikov, O. Yu. Imanuvilov, Exact controllability of the Navier-Stokes and Boussi-nesq equations, Russian Math. Surveys, 54 (1999), no. 3, 565–618.

[10] G. Geymonat, E. Sanchez-Palencia, On the vanishing viscosity limit for acoustic phe-nomena in a bounded region, Arch. Rational Mech. Anal. 75 (1981), 257–268.

[11] S. Guerrero, Local exact controllability to the trajectories of the Boussinesq system, inpreparation.


[13] O. Yu. Imanuvilov, J.-P. Puel, Global Carleman estimates for weak elliptic non homo-geneous Dirichlet problem, Int. Math. Research Notices, 16 (2003), 883-913.

124

[14] O.Yu. Imanuvilov, M. Yamamoto, Carleman estimate for a parabolic equation in aSobolev space of negative order and its applications, Lecture Notes in Pure and Appl. Math.,218, Dekker, New York 2001.

[15] J.-L. Lions and E. Zuazua, A generic uniqueness result for the Stokes system and its con-trol theoretical consequences, Partial Differential Equations and Applications, 177 (1996),221–235.

Capıtulo 4

Remarks on the controllability of theanisotropic Lame system

126

Remarks on the controllability of theanisotropic Lame system

S. Guerrero y O. Yu. Imanuvilov

Abstract

In this paper we establish a Carleman estimate for the elasticity system with residual stresses.As an application of this estimate we obtain exact controllability results for the same systemwith locally distributed control.

1. Introduction

Let us denote x = (x0, x′), where x0 (resp. x′) stands for the time (resp. spatial) variable.

This paper is concerned with global Carleman estimates for the Lame system

ρ∂2x0

ui −3∑

j=1

∂xi(σij) = fi in Q, 1 ≤ i ≤ 3, (4.1)

where Ω is a bounded domain with boundary ∂Ω ∈ C3, Q = (0, T ) × Ω, u(x) = (u1, u2, u3) isthe displacement, f = (f1, f2, f3) is the density of external forces and σij is the stress tensor:

σij = aijhk(x)∂xhuk.

On the boundary, we equip the Lame system with zero Dirichlet boundary conditions:

u = 0 on Σ,

where we have denoted Σ = (0, T )× ∂Ω.We introduce the following standard assumptions on the coefficients aijhk

aijhk = ajikh = ahkij ,

aijhkXijXkh ≥ αXijXij ∀X ∈ R9 with Xij = Xji,(4.2)

where α is some positive number.In this paper we will strict to the case of the anisotropic Lame system with residual free

stresses R:σ = R + (∇u)R + λ(trε)I + 2µε + β1(trε)(trR)I + β2(trR)ε

+ β3((trε)R + tr(εR)I) + β4(εR + Rε),(4.3)

127

whereε =

12(∇u + (∇u)t)

and∇ ·R = 0 in Q. (4.4)

We will assume for simplicity that β1 = β2 = β3 = β4 = 0 and therefore

σ = R + (∇u)R + λ(trε)I + 2µε, (4.5)

for some λ, µ and R. We will impose the following regularity and positivity assumptions on theLame coefficients

ρ, λ, µ, Rij ∈ C2(Ω), ρ > 0, µ−R33 > 0, and λ + 2µ−R33 > 0 in Ω, (4.6)

for i, j = 1, 2, 3.

The first goal of this paper is to establish appropriate global Carleman estimates for theLame system. For displacements u with compact support, such estimates were obtained in theprevious works [27], [25], [18]. More results are available for the isotropic Lame system. Thus,in the stationary case we refer to [8] and [31] for displacements with compact support and[17] for displacements satisfying Dirichlet boundary conditions. For the nonstationary isotropicLame system, see [10] for displacements with compact support and [14]–[16] in the other case.In this paper we have extended the techniques in [15] to consider the anisotropic Lame system(4.1) with σ given by (4.5). Our Carleman estimates will hold for displacements u satisfying zeroDirichlet coditions on Σ.

The last section of this paper is devoted to the exact controllability of the Lame system. Toour best knowledge, the first observability result for the Lame system was proved in [24] usingmultipliers of the form (xi − x0

i )∂u∂xi

, which led to the observability inequality

E(t) =∫

Ω(µ|∇u|2 + (λ + µ)|div u|2)dx ≤ C

∫

(0,T )×Γ0

∣∣∣∣∂u∂n

∣∣∣∣2

dσ

when the Lame coefficients µ and λ are constants. Here, E(t) is the energy. The control is exertedat (0, T )× Γ0 where Γ0 ⊂ ∂Ω and homogeneous Dirichlet boundary conditions are assumed foru. Further results have been deduced in [1] for the anisotropic Lame system by essentially thesame multipliers method. Several questions concerning the approximate controllability and theuniqueness of the Lame system were studied in [10] by means of Carleman estimates. Theyobtained approximate controllability with a control distributed over any open subset of theboundary for a sufficiently large time. A series of important results have been obtained quiterecently in [5]–[7]. In particular, a “logarithmic type” energy decay estimate has been provedin the case where the geometrical control condition of Bardos, Lebeau, Rauch is not fulfilled.An interesting result was also proved in [32], for the isotropic Lame system with the locallydistributed control v of the form χωv , where v = (v1, . . . , vn) and vn ≡ 0. Under some geometricassumptions on the domain D he established the approximate controllability for the isotropicLame system.

128

Several works are devoted to the construction of dissipative “feedback” boundary conditionsfor the Lame system. In [2], dissipative boundary conditions of the form σ(u)n+Au+B∂tu = 0were introduced on the controlled part of the boundary Γ0. Under some geometric conditionson Γ0, they established the exponential decay of the energy

E(t) =12

∫

Ω(|∂tu|2 + σijεij(u))dx +

12

∫

Γ0

A|u|2dS ≤ e−ωt

for the anisotropic Lame system. In the isotropic case, stabilizing boundary feedbacks were alsoconstructed in [13] and [26] but under less restrictive geometrical assumptions.

A closely related question is the control and stabilization of layered plate models. Concerningthe control of thermoelastic systems, the uniform stabilization of thermoelastic Reissner plateswere proved in [21] using thermal and mechanical boundary feedbacks. In [20], the same au-thor proved the exact controllability of the mechanical component of a thermoelastic Kirchhoffplate using mechanical boundary controls. In one-dimensional cases, this was improved to exactnull controllability (of thermal and mechanical components) by the mechanical variable on theboundary in [12]. In [22], this result was extended to the case of a three-dimensional thermoelas-tic Lame system with a mechanical distributed control on a neighborhood of the boundary. In[4], a related result for the boundary control of a thermoelastic Kirchhoff plate was established,but with no restriction on the size of the coupling constant.

In the last part of this paper, we will present a controllability result for the anisotropicLame system with distributed controls supported by Qω = ω×(0, T ), where ω ⊂ Ω is a nonemptyopen set.

Notation: Let x = (x0, x′), where x0 = t stands for the time variable. We denote by H2,s(Q)

and H1,s(Q) the Hilbert spaces H2(Q) and H1(Q) with the norms

‖u‖2H2,s(Q) =

2∑

|α|=0

s4−2|α|‖Dαu‖2L2(Q)

and‖u‖2

H1,s(Q) = s2

∫

Q|u|2 dx +

∫

Q|∇u|2 dx,

respectively.We also introduce the following norms:

‖u‖2Bφ(Q) =

∫

Qe2sφ

(∑2|α|=0 s4−2|α||Dαu|2 + s|∇(∇× u)|2

+s3|∇ × u|2 + s|∇(∇ · u)|2 + s3|∇ · u|2) dx

(4.7)

and

‖u‖2Yφ(Q) = ‖u‖2

Bφ(Q) + s

∥∥∥∥esφ ∂u∂n

∥∥∥∥2

H1,s(Σ)

+ s

∥∥∥∥esφ ∂2u∂n2

∥∥∥∥2

L2(Σ)

, (4.8)

where φ is a given function.In the sequel, we will denote by ε(δ) a positive function such that

limδ→0+ε(δ) = 0.

129

2. A Carleman estimate for the anisotropic Lame system

Let us consider the following anisotropic Lame system completed with Dirichlet boundaryconditions:

P (x,D)u ≡ ρ(x′)∂2x0

u− L(x,D)u = f in Q,

u = 0 on Σ,

u(·, T ) = ∂x0u(·, T ) = u(·, 0) = ∂x0u(·, 0) = 0 in Ω,

(4.9)

whereL(x,D)u = µ∆u + (λ + µ)∇(∇ · u)− [(R,∇)∇t]u.

Here, f and u are three-dimensional vector fields functions and R is a symmetric matrix-valuedfunction, whose components will be denoted by

u = (uj)3j=1, f = (fj)3j=1 and R = (Rjk)3j,k=1.

Let ξ = (ξ0, ξ1, ξ2, ξ3) = (ξ0, ξ′). We introduce the symbols

p1(x, ξ) = ρ(x′)ξ2

0 − µ(x′)(ξ21 + ξ2

2 + ξ23) + (R(x′)ξ)ξ,

p2(x, ξ) = ρ(x′)ξ20 − (λ(x′) + 2µ(x′))(ξ2

1 + ξ22 + ξ2

3) + (R(x′)ξ)ξ.

For any two smooth functions w(x, ξ) and z(x, ξ), let us introduce the formula for the Poissonbracket

w, z =3∑

j=0

(∂ξjw ∂xjz − ∂ξjz ∂xjw).

Through this paper we will assume the existence of a function ψ satisfying the followingcondition:

Condition A There exists a function ψ ∈ C3(Q) such that• |∇x′ψ|Q\Qω

6= 0

• pj , pj , ψ(x, ξ) > 0 for all (x, ξ) ∈ (Q \Qω)×R4 with ξ 6= 0 such that

pj(x, ξ) = 〈∇ξpj ,∇ψ〉 = 0 for j = 1, 2

and• pj(x, ξ − is∇ψ(x)), pj(x, ξ + is∇ψ(x))/(2is) > 0 for all ξ ∈ R4 \ 0 and s ∈ R \ 0

satisfying

pj(x, ξ + is∇ψ(x)) = 〈∇ξpj(x, ξ + is∇ψ(x)),∇ψ(x)〉 = 0, x ∈ Q \Qω, j = 1, 2.

On the boundary, the following is required:

p1(x,∇ψ(x)) < 0 ∀x ∈ ∂Ω× (0, T )

130

and

(λ + 2µ)∂ψ

∂n−

3∑

j=1

Ri3∂ψ

∂xi< 0 and µ

∂ψ

∂n−

3∑

j=1

Ri3∂ψ

∂xi< 0

∀x ∈ (∂Ω \ ∂ω)× (0, T ).

Let thus fix a function ψ satisfying the previous conditions and let us set

φ(x) = eτψ(x) ∀x ∈ Q

for some positive parameter τ which will be chosen later on.

Theorem 10 Let f ∈ H1(Q)3 and let (4.2), (4.4) and (4.6) hold. Suppose there exists a functionψ satisfying condition A and such that its normal derivative is large enough. Then, there existsτ∗ > 0 such that, for any τ > τ∗, there exists s∗ > 0 such that

‖u‖Yφ(Q) ≤ C(‖esφf‖H1,s(Q)3 + ‖u‖Bφ(Qω)) ∀s > s∗ (4.10)

for some C > 0 independent of s and for any solution u ∈ L2(0, T ; H2(Ω)3)∩H1(Q)3 of system(4.9).

Proof: Let us first write down the equations verified by ∇× u and ∇ · u:

Pµ(x,D)(∇× u) ≡ ∂2x0

(∇× u)− µ∆(∇× u) + (R,∇)∇T (∇× u)

= ∇× f + P1(x,D)u(4.11)

and

Pλ+2µ(x,D)(∇ · u) ≡ ∂2x0

(∇ · u)− (λ + 2µ)∆(∇ · u) + (R,∇)∇T (∇ · u)

= ∇ · f + P2(x,D)u,(4.12)

where the Pj are second order differential operators.In this situation, we are able to apply the Carleman estimates obtained in [28] and [9] to the

equations (4.11) and (4.12). Combining them, we obtain

s3τ3(‖φ3/2esφ(∇× u)‖2L2(Q)3 + ‖φ3/2esφ(∇ · u)‖2

L2(Q)3)

+sτ(‖φ1/2esφ(∇ · u)‖2H1(Q)3 + ‖φ1/2esφ(∇× u)‖2

H1(Q)3)

≤ C

‖esφf‖2

H1(Q)3 +∑

|α|≤2

‖esφDαu‖2L2(Q)3 + sτ

∥∥∥∥φ1/2esφ ∂u∂n

∥∥∥∥2

L2(Σ)3

+s3τ3

∥∥∥∥φ3/2esφ ∂(∂tgu)∂n

∥∥∥∥2

L2(Σ)3+ sτ‖φ1/2esφ∇(∇× u)‖2

L2(Qω)9

+sτ‖φ1/2esφ∇(∇ · u)‖2L2(Qω)3) + s3τ3‖φ3/2esφ(∇× u,∇ · u)‖2

L2(Qω)4

+2∑

|α|=0

(sτ)4−2|α|‖φ2−|α|esαDαu‖2L2(Qω)3

∀τ ≥ τ0, ∀s ≥ s0(τ),

(4.13)

131

where C is independent of s and τ . We recall that the definition of ‖ · ‖Bφ(Q) was given in (4.7).Now, from the well known identity

∆u = −∇× (∇× u) +∇(∇ · u),

we can use the Carleman inequality proved in [11] for the case of a (simpler) elliptic equationwith homogeneous Dirichlet conditions and combine this with (4.13), so we deduce in a standardway that

s4τ5‖φ2esφu‖L2(Q)3 + s2τ3‖φesφ∇u‖2L2(Q)9 + τ‖esφD2u‖2

L2(Q)3

≤ C

‖esφf‖2

H1(Q)3 +∑

|α|≤2

‖esφDαu‖2L2(Q)3 + sτ

∥∥∥∥φ1/2esφ ∂u∂n

∥∥∥∥2

L2(Σ)3

+s3τ3

∥∥∥∥φ3/2esφ ∂(∂tgu)∂n

∥∥∥∥2

L2(Σ)3+ sτ‖φ1/2esφ∇(∇× u)‖2

L2(Qω)9

+sτ‖φ1/2esφ∇(∇ · u)‖2L2(Qω)3) + s3τ3‖φ3/2esφ(∇× u,∇ · u)‖2

L2(Qω)4

+2∑

|α|=0

(sτ)4−2|α|‖φ2−|α|esαDαu‖2L2(Qω)3

∀τ ≥ τ1, s ≥ s1(τ).

(4.14)

Taking τ large enough the global term in the right hand side concerning u can be absorbed.On the other hand, from this point of the proof we forget about the dependence on τ and

possible powers of φ (which is a regular function) in our inequalities, since it will not be crucial.Consequently, for the moment, we have

‖u‖Yφ(Q) ≤ C(‖esφf‖2

H1,s(Q)3 + s


∥∥∥∥2

H1,s(Σ)3+ s

∥∥∥∥esφ ∂2u∂n2

∥∥∥∥2

L2(Σ)3

+‖u‖2Bφ(Qω)

)∀τ ≥ τ1, s ≥ s1(τ).

(4.15)

The norm ‖ · ‖Yφ(Q) was introduced in (4.8).

The goal will be now to estimate the previous boundary terms. For this purpose, we consideranother weight function ϕ such that ϕ = φ on Σ. For instance, let us take

ϕ = eτψ with ψ = ψ − 1Z2

`1 + Z`21,

where Z is a large positive number and `1 is a regular function verifying

`1 = 0 on ∂Ω, `1 > 0 in Ω and ∇`1 6= 0 on ∂Ω.

Moreover, if we set

Ω1/Z2 = x′ = (x1, x2, x3) ∈ Ω : dist(x′, ∂Ω) < 1/Z2,

132

we can suppose that the function `1 is chosen such that

ϕ(x) < φ(x) ∀x ∈ Ω1/Z2 × (0, T )

provided Z is large enough.For this new weight function, we will be able to prove the following lemma:

Lemma 11 Under the previous conditions, the following inequality holds

‖u‖Yϕ(Q) ≤ C(‖esϕf‖H1,s(Q)3 + ‖u‖Bϕ(Qω)

) ∀s ≥ s0, (4.16)

for functions u verifyingsuppu ⊂ Ω1/Z2 × [0, T ].

Let us suppose that lemma 11 holds and let us deduce theorem 10 from it. Suppose that Zis already fixed such that (4.16) holds and take ε ∈ (0, 1/Z2). Then we have

ϕ(x) < φ(x) ∀x ∈ (Ωε \ Ωε/2)× [0, T ]. (4.17)

Let us now introduce a cut-off function θ ∈ C2c (Ωε) such that θ ≡ 1 in Ωε/2. Then, it readily

follows that the function θu fulfills the following system

P (x,D)(θu) ≡ θf + [P, θ](x,D)u in Q,

θu = 0 on Σ,

(θu)(·, T ) = ∂x0(θu)(·, T ) = (θu)(·, 0) = ∂x0(θu)(·, 0) = 0 in Ω.

Consequently, we can apply estimate (4.16) to θu and deduce that

s


∥∥∥∥2

H1,s(∂Ω×(0,T ))3+ s

∥∥∥∥esφ ∂2u∂n2

∥∥∥∥2

L2(∂Ω×(0,T ))3

≤ C(‖esϕf‖2

H1,s(Q)3 + ‖esϕ[P, θ]u‖2H1,s(Q)3 + ‖u‖2

Bϕ(Qω)

)∀s ≥ s0,

(4.18)

since ϕ ≡ φ on the boundary. Next, we observe that the support of the function [P, θ]u iscontained in Ωε \ Ωε/2 × [0, T ], while ϕ < φ in that set (see (4.17)). Thus,

‖esϕ[P, θ]u‖H1,s(Q)3 ≤ C2∑

|α|=0

‖esφDαu‖L2(Q)3 ∀s > 0.

Finally, we put this together with (4.18) and (4.14) and we obtain the desired inequality (4.10).This ends the proof of theorem 10.

Proof of Lemma 11: Assume that suppu ⊂ Bδ ∩ ([0, T ] × Ω1/Z2) and that the small part ofthe boundary where we are working on is given by the equation

x3 = `(x1, x2).

133

Without lack of generality, the function ` ∈ C3 can be taken to satisfy

∇′`(0, 0) = (`x1 , `x2)(0, 0) = 0.

In order to work in an appropriate frame, we perform the change of variables

y1 = x1,

y2 = x2,

y3 = x3 − `(x1, x2)

and we set y∗ = (y0, 0, 0, 0). Let us denote by w = (w′, w4) the functions ∇×u and ∇ ·u in thenew coordinates. Then, some simple computations show that, in the new variables, the mainsymbols of equations (4.11) and (4.12) are

pβ(y, ξ) = −ξ20 + (β −R11)ξ2

1 + (β −R22)ξ22 + [(βE3 −R)Gt]Gξ2

3

−2R12ξ1ξ2 − 2∑2

j=1(Rj3 + `yj (β −Rjj)− `y3−jR12)ξjξ3,

where

β =

µ for ∇× u,

λ + 2µ for ∇ · uand

G = (−`y1 ,−`y2 , 1)t. (4.19)

Let us set some other standard ingredients in the microlocal analysis frame. Thus, we considerthe unit sphere in R4, say

S3 = ζ = (s, ξ′) : s2 + ξ20 + ξ2

1 + ξ22 = 1

and the following associated finite covering:

ζ ∈ S3 : |ζ − ζ∗ν | < δ11≤ν≤M(δ1),

with ζ∗ν ∈ S3. To this covering we associate the partition of unity χν1≤ν≤M , extending χν outof S3 like a homogenous function of order 0 with support contained in the conic neighborhood

O(δ1) =

ζ :∣∣∣∣

ζ

|ζ| − ζ∗ν

∣∣∣∣ < δ1

.

In order to finish the proof, we need another lemma:

Lemma 12 Let γ∗ = (y∗, ζ∗) ∈ ∂G × S3 be fixed and suppχν ⊂ O(δ1). Then, for sufficientlysmall δ and δ1, we have

s‖zν‖2H1,s(G)4 + s(‖zν‖2

H1,s(∂G)4 + ‖∂y3zν‖2L2(∂G)4) ≤ C(‖esϕf‖2

H1,s(G)3 + ‖z‖2H1,s(G)4). (4.20)

Here, we have denoted G = R3 × [0, 1/Z2] and zν = χν(s,D′)z, with z = esϕw.

134

Let us suppose that lemma 12 holds. Then we have

s‖z‖2H1,s(G)4 + s(‖z‖2

H1,s(∂G)4 + ‖∂y3z‖2L2(∂G)4)

≤ CsM∑

ν=1

(‖zν‖2H1,s(G)4 + ‖zν‖2

H1,s(∂G)4 + ‖∂y3zν‖2L2(∂G)4)

≤ C(‖esϕf‖2H1,s(G)3 + ‖esϕu‖2

H2,s(G)3).

(4.21)

Let us now use the result stated in proposition 4.2 of [14]:

Z

2∑

|α|=0

s4−2|α|‖esϕDαy′u‖2

L2(G)3 ≤ C(‖esϕf‖2H1,s(G)3 + ‖esϕu‖2

H2,s(G)3),

where C is independent of s and Z.From the differential equation in (4.9), we deduce that ‖esϕ∂2

y0u‖2

L2(G)3 can be added to theleft hand side of the previous inequality. The same can be said of ‖esϕ∂2

y0y1u‖2

L2(G)3 in view ofwell known interpolation arguments. Consequently, we have

Z2∑

|α|=0

s4−2|α|‖esϕDαu‖2L2(G)3 ≤ C‖esϕf‖2

H1,s(G)3 .

Combining this with (4.21), we obtain:

2∑

|α|=0

s4−2|α|‖esϕDαu‖2L2(G)3 + s(‖z‖2

H1,s(∂G)4 + ‖∂y3z‖2L2(∂G)4) ≤ C‖esϕf‖2

H1,s(G)3 . (4.22)

Let us finally see that we can deduce inequality (4.16) from (4.22). To do this, we just haveto provide ‘good’ estimates for the terms

s ‖esϕ∂y3u‖2H1,s(∂G)3 and s

∥∥esϕ∂2y3

u∥∥2

L2(∂G)3.

From the definition of the y variables in terms of x and the Dirichlet boundary condition onu, we see that the following estimates hold:

|∂y3uj | ≤ |(∇× u)3−j |+ ε(δ)|∂y3u| on ∂G for j = 1, 2,

|∂y3u3| ≤ |∇ · u|+ ε(δ)|∂y3u| on ∂G.

These two estimates tell that

|esϕ∂y3u| ≤ |z1|+ |z2|+ |z4| on ∂G. (4.23)

Additionally,

|∂2yjy3

uk| ≤ |∂yj (∇× u)3−k|+ ε(δ)|∂yjy3u| on ∂G for j = 0, 1, 2, k = 1, 2

135

and|∂2

yjy3u3| ≤ |∂yj (∇ · u)|+ ε(δ)|∂yjy3u| on ∂G for j = 0, 1, 2,

whence we deduce that

|esϕ∇tgy ∂y3u| ≤ |∇tg

y z1|+ |∇tgy z2|+ |∇tg

y z4| on ∂G. (4.24)

Finally, we have

|∂2y3

uj | ≤ |∂y3(∇× u)3−j |+ |∂yj (∇ · u)|+ ε(δ)|∂y3∇yu| on ∂G for j = 1, 2

and|∂2

y3u3| ≤ |∂y1(∇× u)2|+ |∂y2(∇× u)1|+ |∂y3(∇ · u)|+ ε(δ)|∂y3∇yu| on ∂G,

which lead to the estimate

|esϕ∂2y3

u| ≤ |∇yz1|+ |∇yz2|+ |∇yz4|+ ε(δ)|esϕ∂y3∇tgy u| on ∂G. (4.25)

One can readily see that (4.23)–(4.25) imply that

‖esϕ∂y3u‖2H1,s(∂G)3 +

∥∥esϕ∂2y3

u∥∥2

L2(∂G)≤ C(‖z‖2

H1,s(∂G)4 + ‖∂y3z‖2L2(∂G)4),

as we wanted to prove.

As a conclusion it suffices to prove lemma 12, so the rest of this section will be dedicated toit.

Proof of lemma 12: Let us introduce the notation

ID = D + is∇ϕ.

Then the main symbol of the differential operator Pβ(y, ID) is

pβ,s(y, s, ξ) = −(ξ0 + isϕy0)2 + (β −R11)(ξ1 + isϕy1)

2

+(β −R22)(ξ2 + isϕy2)2 + [(βE3 −R)Gt]G(ξ3 + isϕy3)

2

−2R12(ξ1 + isϕy1)(ξ2 + isϕy2)

−2∑2

j=1(Rj3 + `yj (β −Rjj)− `y3−jR12)(ξj + isϕyj )(ξ3 + isϕy3).

(4.26)

We recall that G = (−`y1 ,−`y2 , 1)t. The roots of this polynomial with respect to the ξ3 variableare

Γ±β (y, s, ξ′) = −isϕy3 + α±β (y, s, ξ′), (4.27)

where

α±β =

∑2j=1(Rj3 + `yj (β −Rjj)− `y3−jR12)(ξj + isϕyj )±

√rβ(y, s, ξ′)

[(βE3 −R)Gt]G

136

andrβ(y, s, ξ′) =

(∑2j=1(Rj3 + `yj (β −Rjj)− `y3−jR12)(ξj + isϕyj )

)2

−[(βE3 −R)Gt]G(−(ξ0 + isϕy0)2 + (β −R11)(ξ1 + isϕy1)

2

+(β −R22)(ξ2 + isϕy2)2 − 2R12(ξ1 + isϕy1)(ξ2 + isϕy2)).

(4.28)

It will be useful for the sequel to factorize Pβ,s as the product of two first order operators.This is made in the following proposition:

Proposition 8 Let β ∈ µ, λ + 2µ and |rβ(γ)| ≥ C > 0 for all γ ∈ (Bδ ∩ G)×O(2δ1). Then,for any function v such that supp v ⊂ Bδ ∩ G, we have

Pβ,s(y, D)vν = [(βE3 −R)Gt]G(Dy3 − Γ−β (y, s,D′))(Dy3 − Γ+β (y, s, D′))vν + Tβ,svν ,

where Tβ,s is a continuous operator

Tβ,s : H1,s(G)3 7−→ L2(G)3.

Once this decomposition can be done, one would desire to obtain appropriate estimates of somenorms of vν . More precisely, let v satisfy

(Dy3 − Γ−β (y, s,D′))vν = q, v|y3=1/Z2 = 0, supp v ⊂ Bδ ∩ G.

We can then prove the following:

Proposition 9 Let β ∈ µ, λ + 2µ and |rβ(γ)| ≥ C > 0 for all γ ∈ Bδ ×O(2δ1). Then,

s‖vν‖2L2(∂G) ≤ C‖q‖2

L2(∂G),

for some positive constant C independent of s and Z.

Proposition 8 and proposition 9 can both be proved as in [14].

The next step will be to obtain a Carleman inequality for a function satisfying our (secondorder hyperbolic) differential equation but with no imposed boundary conditions. This will befundamental to obtain the desired estimate (4.20). Indeed, let w satisfy

Pβ(y, ID)w = g in G, suppw ⊂ Bδ × [0, 1/Z2).

Let us denote by P ∗β (·, ID) the adjoint operator of Pβ(·, ID) (β ∈ µ, λ + 2µ) and let us set

L+,β(·, ID) =Pβ(·, ID) + P ∗

β (·, ID)2

, L−,β(·, ID) =Pβ(·, ID)− P ∗

β (·, ID)2

.

We haveL+,β(·, ID)w + L−,β(·, ID)w = g.

After several computations involving integration by parts, we get

‖L+,β(·, ID)w‖2L2(G) + ‖L−,β(·, ID)w‖2

L2(G) + Re∫

G[L+,β, L−,β]w dy + Σβ(w) = ‖g‖2

L2(G), (4.29)

137

with Σβ(w) = (Σ1β + Σ2

β)(w), where Σ1β can be written as Σ1,1

β + Σ1,2β + Σ1,3

β with

Σ1,1β (w) = s

∫

∂G(β −R33)2(y∗)ϕy3(y

∗)(|∂y3w|2 + s2ϕ2y3|w|2) dy′, (4.30)

Σ1,2β (w) = s

∫

∂G(β −R33)(y∗)ϕy3(y

∗)[|∂y0w|2 − s2ϕ2

y0(y∗)|w|2

−(

β −R11 − R213

β −R33

)(y∗)(|∂y1w|2 − s2ϕ2

y1(y∗)|w|2)

−(

β −R22 − R223

β −R33

)(y∗)(|∂y2w|2 − s2ϕ2

y2(y∗)|w|2)

+2(

R12 +R13R23

β −R33

)(y∗)(∂y1w ∂y2w − s2ϕy1(y

∗)ϕy2(y∗)|w|2)

]dy′

(4.31)

and

Σ1,3β (w) = −2sRe

∫

∂G((β −R33)(y∗)∂y3w −R13∂y1w −R23∂y2w)

(ϕy0(y∗)∂y0w −

(β −R11 − R2

13

β −R33

)(y∗)ϕy1(y∗)∂y1w

−(

β −R22 − R223

β −R33

)(y∗)ϕy2(y∗)∂y2w

+(

R12 +R13R23

β −R33

)(y∗)(ϕy2(y∗)∂y1w + ϕy1(y∗)∂y2w)

)dy′

(4.32)

and

Σ2β(w) ≤ ε(δ)s

(‖w‖2

H1,s(∂G) + ‖∂y3w‖2L2(∂G)

), (4.33)

with ε(δ) → 0 when δ → 0+. In (4.30) and (4.31), we have used the notation

∂y3 = − R13

β −R33(y∗)∂y1 −

R23

β −R33(y∗)∂y2 + ∂y3 . (4.34)

In fact, the expressions of L+,β(·, ID)w and L−,β(·, ID)w are

L+,β(y, ID)w = ∂2y0

w + s2ϕ2y0

w − (β −R11)(∂2y1

w + s2ϕ2y1

w)

− (β −R22)(∂2y2

w + s2ϕ2y2

w)− (β −R33)(∂2y3

w + s2ϕ2y3

w)

+ 2R12(∂2y1y2

w + s2ϕy1ϕy2w) + 2R13(∂2y1y3

w + s2ϕy1ϕy3w)

+ 2R23(∂2y2y3

w + s2ϕy2ϕy3w)

138

and

L−,β(y, ID)w = s (−∂y0(ϕy0w)− ϕy0∂y0w + (β −R11)(∂y1(ϕy1w) + ϕy1∂y1w)

+ (β −R22)(∂y2(ϕy2w) + ϕy2∂y2w) + (β −R33)(∂y3(ϕy3w) + ϕy3∂y3w)

−R12(∂y2(ϕy1w) + ϕy1∂y2w + ∂y1(ϕy2w) + ϕy2∂y1w)

−R13(∂y3(ϕy1w) + ϕy1∂y3w + ∂y1(ϕy3w) + ϕy3∂y1w)

−R23(∂y2(ϕy3w) + ϕy3∂y2w + ∂y3(ϕy2w) + ϕy2∂y3w)) .

One just has to integrate by parts, keeping the boundary terms in order to conclude that

Σβ(w) = Σ1β(w) + Σ2

β(w),

where Σ1β(w) is given by (4.30)–(4.32) and Σ2

β verifies the estimate (4.33).Using (4.29), one can prove in the same way as in Appendix II of [14] that the following

inequality holds:

s‖w‖H1,s(G) ≤ ‖L+,β(·, ID)w‖2L2(G) + ‖L−,β(·, ID)w‖2

L2(G) + Re ([L+,β, L−,β]w, w)L2(G)

+sC‖w‖L2(∂G)‖∂y3w‖L2(∂G) ∀s ≥ s0.

Combining this with (4.29), we get

C1s‖w‖2H1,s(G) + Σβ(w) ≤ C2(‖g‖2

L2(G) + s‖w‖L2(∂G)‖∂y3w‖L2(∂G)) ∀s ≥ s0. (4.35)

Having this inequality in mind, we will prove lemma 12 distinguishing several cases accordingto the values of rβ(γ∗) (recall that γ∗ = (y∗, ζ∗)):

• First Case: rµ(γ∗) = 0, rλ+2µ(γ∗) 6= 0.

• Second case: rλ+2µ(γ∗) = 0, rµ(γ∗) 6= 0.

• Third Case: rµ(γ∗) 6= 0, rλ+2µ(γ∗) 6= 0 or rµ(γ∗) = rλ+2µ(γ∗) = 0.

2.1. First Case: rµ(γ∗) = 0, rλ+2µ(γ∗) 6= 0.

In this situation, taking δ and δ1 small enough, one can suppose that

|rλ+2µ(γ)| ≥ C > 0 ∀γ = (y, ζ) ∈ Bδ × (O(δ1) ∩ |ζ| ≥ 1). (4.36)

Let us start applying estimate (4.35) to z′ν = χν(s,D′)esϕ(∇×u). Recall that w = (w′, w4) =(∇× u,∇ · u), where the differential operators are taken in the y variables. This yields

s‖z′ν‖2H1,s(G)3 + Σ1

µ(z′ν) ≤ ε(δ)s(‖z′ν‖2H1,s(∂G)3 + ‖∂y3z

′ν‖2

L2(∂G)3)

+ C(‖esϕf‖2H1,s(G)3 + ‖z‖2

H1,s(G)4),(4.37)

with Σ1µ(z′ν) given by (4.30)–(4.32). Let us rewrite the boundary terms in the form

Σ1µ(z′ν) = Σ1,1

µ (z′ν) + Σ1,2µ (z′ν) + Σ1,3

µ (z′ν), (4.38)

139

withΣ1,1

µ (z′ν) = s

∫

∂G(µ−R33)2(y∗)ϕy3(y

∗)(|∂y3z′ν |2 + s2ϕ2

y3|z′ν |2) dy′,

Σ1,2µ (z′ν) = s

∫

∂G(µ−R33)(y∗)ϕy3(y

∗)[|∂y0z

′ν |2 − s2ϕ2

y0(y∗)|z′ν |2

−(

µ−R11 − R213

µ−R33

)(y∗)(|∂y1z

′ν |2 − s2ϕ2

y1(y∗)|z′ν |2)

−(

µ−R22 − R223

µ−R33

)(y∗)(|∂y2z

′ν |2 − s2ϕ2

y2(y∗)|z′ν |2)

+2(

R12 +R13R23

µ−R33

)(y∗)(∂y1z

′ν ∂y2z

′ν − s2ϕy1(y

∗)ϕy2(y∗))|z′ν |2

]dy′

and

Σ1,3µ (z′ν) = −2sRe

∫

∂G

((µ−R33)(y∗)∂y3z

′ν −R13(y∗)∂y1z

′ν −R23(y∗)∂y2z

′ν

)

×(

ϕy0(y∗)∂y0z′ν −(

µ−R11 − R213

µ−R33

)(y∗)ϕy1(y∗)∂y1z′ν

−(

µ−R22 − R223

µ−R33

)(y∗)ϕy2(y∗)∂y2z′ν

+(

R12 +R13R23

µ−R33

)(y∗)(ϕy2(y∗)∂y1z′ν + ϕy1(y∗)∂y2z′ν)

)dy′.

(4.39)

Taking into account (4.28), we observe that

0 = Re rµ(γ∗) = R213[(ξ

∗1)

2 − (s∗ϕy1(y∗))2] + R2

23[(ξ∗2)

2 − (s∗ϕy2(y∗))2]

+2R13(y∗)R23(y∗)[ξ∗1ξ∗2 − (s∗)2ϕy1(y

∗)ϕy2(y∗)]

−(µ−R33)(y∗)[−(ξ∗0)2 + (s∗ϕy0(y

∗))2] + (µ−R11)(y∗)[(ξ∗1)2 − (s∗ϕy1(y

∗))2]

+(µ−R22)(y∗)[(ξ∗2)2 − (s∗ϕy2(y

∗))2]− 2R12(y∗)[ξ∗1 ξ∗2 − (s∗)2ϕy1(y∗)ϕy2(y

∗)].Consequently,

|Re rµ(γ)| ≤ ε(δ1)(s2 + |ξ0|2 + |ξ1|2 + |ξ2|2) ∀γ = (y∗, s, ξ′) ∈ O(δ1).

SinceΣ1,2

µ (z′ν) = s

∫

R3

(µ−R33)(y∗)ϕy3(y∗)Re rµ(y∗, s, ξ′)|z′ν |2 dξ′,

we readily deduce thatΣ1,2

µ (z′µ) ≤ ε(δ1)s‖z′ν‖2H1,s(∂G)3 . (4.40)

Also, from rµ(γ∗) = 0 we find that

|ξ0|2 ≤ C(s2 + |ξ1|2 + |ξ2|2) ∀(s, ξ0, ξ1, ξ2) ∈ O(δ1) (4.41)

140

for a positive constant C, provided δ1 is small enough.

In order to estimate Σ1,3µ (z′ν), we will have to distinguish again whether s∗ is equal to zero

or not.

Taking into account (4.36), an application of proposition 8 provides the identity

Pλ+2µ,s(y, D)z4,ν = [(λ + 2µ)E3 −R)Gt]G(Dy3 − Γ−λ+2µ(y, s, D′))z+4,ν + Tz4,ν ,

for some T ∈ L(H1,s(G);L2(G)), where we have put

z+4,ν = (Dy3 − Γ+

λ+2µ(y, s, D′))z4,ν .

Then, proposition 9 applied to z+4,ν yields

s‖(IDy3 − α+λ+2µ)z4,ν‖2

L2(∂G) ≤ (‖esϕf‖2H1,s(G)3 + ‖z‖2

H1,s(G)4). (4.42)

Case 2.1.1: s∗ 6= 0We recall here the expression of Im rµ(γ∗):

0 = Im rµ(γ∗) = 2s∗R213(y

∗)ϕy1(y∗)ξ∗1 + R2

23(y∗)ϕy2(y

∗)ξ∗2+R13(y∗)R23(y∗)(ϕy2(y

∗)ξ∗1 + ϕy1(y∗)ξ∗2)

−(µ−R33)(y∗)[−ϕy0(y∗)ξ∗0 + (µ−R11)(y∗)(ϕy1(y

∗)ξ∗1)

+(µ−R22)(y∗)(ϕy2(y∗)ξ∗2)−R12(y∗)(ϕy2(y

∗)ξ∗1 + ϕy1(y∗)ξ∗2)],

so we have

|Im rµ(γ)| ≤ ε(δ1)(s2 + |ξ0|2 + |ξ1|2 + |ξ2|2) ∀γ = (y∗, s, ξ′) ∈ O(δ1).

Then, a similar argument to the one above leads us to estimate the term Σ1,3µ (z′ν):

Σ1,3µ (z′ν) ≤ ε(δ1)s(‖z′ν‖2

H1,s(∂G)3 + ‖∂y3z′ν‖2

L2(∂G)3). (4.43)

Therefore, from the expression of Σ1µ(z′ν) (see (4.30)–(4.32)), (4.40), (4.43) and the positiveness

of Σ1,1µ (z′ν), we deduce that

Σ1ν(z

′µ) ≥ Cs(‖z′ν‖2

H1,s(∂G)3 + ‖∂y3z′ν‖2

L2(∂G)3)

for some positive constant C. Here, we have used Condition A (at the beginning of section 2)on ψ, µ − R33 > 0 (see (4.6)) and the fact that s2 ≥ C(ξ2

0 + ξ21 + ξ2

2) in O(δ1). Combining thisand (4.37), we get

s(‖z′ν‖2H1,s(G)3 + ‖z′ν‖2

H1,s(∂G)3 + ‖∂y3z′ν‖2

L2(∂G)3) ≤ C(‖esϕf‖2H1,s(G)3 + ‖z‖2

H1,s(G)4), (4.44)

In order to estimate the boundary norms of z4,ν , we will use the boundary Dirichlet conditionsand the equations of the Lame system written on ∂G. Indeed, from the two first ones, it is notdifficult to deduce that

|IDyjz4,ν | ≤ C(|esϕf |+ s|z′ν |+ |∇yz′ν |) + ε(δ)(s|zν |+ |∇tgy zν |+ |∂y3zν |)

on ∂G, for j = 1, 2

141

and from the third one

|IDy3z4,ν | ≤ C(|esϕf |+ s|z′ν |+ |∇tgy z′ν |+ |IDy1z4,ν |+ |IDy2z4,ν |)

+ε(δ)(s|zν |+ |∇tgy zν |+ |∂y3zν |) on ∂G.

Now, using that λ + 2µ−R33 > 0 along with the Dirichlet boundary conditions, we have

‖IDy3z4,ν‖2L2(∂G) ≤ C(‖esϕf‖2

L2(∂G)3 + ‖z′ν‖2H1,s(∂G)3 + ‖∂y3z

′ν‖2

L2(∂G)4)

+ε(δ)(‖zν‖2H1,s(∂G)4 + ‖∂y3zν‖2

L2(∂G)4).(4.45)

In addition, combining (4.45) and (4.42), we find an estimate of the L2 norm of z4 and itstangential derivatives on ∂G:

s‖z4,ν‖2H1,s(∂G) ≤ C(‖esϕf‖2

H1,s(G)3 + s‖z‖2H1,s(G)4 + s‖z′ν‖2

H1,s(∂G)3 + s‖∂y3z′ν‖2

L2(∂G)4)

+ε(δ)s(‖zν‖2H1,s(∂G)4 + ‖∂y3zν‖2

L2(∂G)4).(4.46)

Indeed, in view of (4.36), we can apply Garding’s inequality after (4.42) and obtain (4.46).Finally, (4.46) and (4.45) give an estimate of the normal derivative of z4,ν in the L2 norm:

s‖z4,ν‖2H1,s(∂G) + s‖∂y3z4,ν‖2

L2(∂G) ≤ C(‖esϕf‖2H1,s(G)3 + s‖z‖2

H1,s(G)4 + s‖z′ν‖2H1,s(∂G)3

+s‖∂y3z′ν‖2

L2(∂G)4) + ε(δ)s(‖zν‖2H1,s(∂G)4 + ‖∂y3zν‖2

L2(∂G)4).(4.47)

This and (4.44) provide the desired inequality (4.20). This ends the proof of lemma 12 in thiscase.

Case 2.1.2: s∗ = 0We first remark that, thanks to the Dirichlet boundary conditions, we have

s‖z3,ν‖2H1,s(∂G) ≤ ε(δ)s(‖zν‖2

H1,s(∂G)4 + ‖∂y3zν‖2L2(∂G)4). (4.48)

Once the tangential derivatives of z3,ν are bounded, an application of (4.35) for w = z3,ν alsogives an estimate for its normal derivative (see the expression of Σ1

µ(z3,ν) in (4.30)–(4.32), (4.34)and (4.33)):

s(‖z3,ν‖2H1,s(∂G) + ‖∂y3z3,ν‖2

L2(∂G)) ≤ ε(δ)s(‖zν‖2H1,s(∂G)4 + ‖∂y3zν‖2

L2(∂G)4)

+ C(‖esϕf‖2H1,s(G) + ‖z‖2

H1,s(G)4).(4.49)

We next deal with the terms Σ1µ(z1,ν) and Σ1

µ(z2,ν). In fact, (4.40) together with (4.41) andthe fact that ϕy3 is large enough yields

Σ1µ(zj,ν) ≥ s(‖(µ−R33)∂y3zj,ν − (R13∂y1zj,ν + R23∂y2zj,ν)‖2

L2(∂G) + s2‖zj,ν‖2L2(∂G))

− C(‖esϕf‖2H1,s(G)3 + ‖z‖2

H1,s(G)4)− ε(δ)s‖∇′yzj,ν‖2L2(∂G)

for j = 1, 2.

142

An application of (4.37) taking into account (4.49), leads to the inequality

s2∑

j=1

(‖(µ−R33)∂y3zj,ν − (R13∂y1zj,ν + R23∂y2zj,ν)‖2L2(∂G) + s2‖zj,ν‖2

L2(∂G))

+s‖z′ν‖2L2(G)3 + s(‖z3,ν‖2

H1,s(∂G) + ‖∂y3z3,ν‖2L2(∂G))

≤ C(‖esϕf‖2H1,s(G)3 + ‖z‖2

H1,s(G)4) + ε(δ)s2∑

j=1

‖∇′yzj,ν‖2L2(∂G).

(4.50)

At this point of the proof, we divide it into two subcases:

• Case 2.1.2: A) R13(y∗)ξ∗1 + R23(y∗)ξ∗2 = 0

In this situation, inequality (4.50) readily provides an estimate for the normal derivatives ofz1,ν and z2,ν in L2(G), so we deduce that

s

2∑

j=1

(‖∂y3zj,ν‖2L2(∂G) + s2‖zj,ν‖2

L2(∂G)) + s‖z′ν‖2L2(G)3

+s(‖z3,ν‖2H1,s(∂G) + ‖∂y3z3,ν‖2

L2(∂G))

≤ C(‖esϕf‖2H1,s(G)3 + ‖z‖2

H1,s(G)4) + ε(δ)s2∑

j=1

‖∇′yzj,ν‖2L2(∂G).

(4.51)

On the other hand, the two first equations of our Lame system provide

−(µ−R33)∂y3z2,ν + 2R13∂y1z2,ν + 2R23∂y2z2,ν

= F1 + (λ + 2µ−R33)∂y1z4,ν ,

(µ−R33)∂y3z1,ν − 2R13∂y1z1,ν − 2R23∂y2z1,ν

= F2 + (λ + 2µ−R33)∂y2z4,ν ,

(4.52)

with F1 and F2 satisfying estimate

s‖Fj‖2L2(∂G) ≤ C(‖esϕf‖2

H1,s(G)3 + ‖z‖2H1,s(G)4)

+ε(δ)s(‖zν‖2H1,s(∂G)4 + ‖∂y3zν‖2

L2(∂G)4) j = 1, 2.(4.53)

Consequently, this combined with (4.41) show that the term

s‖z4,ν‖2H1,s(∂G)

can be added to the left hand side of (4.51). The same happens for the normal derivative of z4,ν ,after an application of (4.42).

143

In this way, we have

s

2∑

j=1

(‖∂y3zj,ν‖2L2(∂G) + s2‖zj,ν‖2

L2(∂G)) + s‖z′ν‖2L2(G)3

+s4∑

j=3

(‖zj,ν‖2H1,s(∂G) + ‖∂y3zj,ν‖2

L2(∂G))

≤ C(‖esϕf‖2H1,s(G)3 + ‖z‖2

H1,s(G)4) + ε(δ)s2∑

j=1

‖∇′yzj,ν‖2L2(∂G).

(4.54)

Next, in order to estimate the tangential derivatives of z1,ν and z2,ν , we use (4.41) togetherwith the fact that estimate (4.49) also holds for

∂y1z1,ν + ∂y2z2,ν and ∂y2z1,ν − ∂y1z2,ν . (4.55)

Let us briefly say why (4.49) holds for ∂y1z1,ν +∂y2z2,ν and ∂y2z1,ν−∂y1z2,ν . For the first one,it suffices to realize that the definition of z = ∇× u implies that ∇ · z = 0 while for the secondone we use the fact that we already have estimates for all the norms of z4,ν and the followingexpression for the third equation of the Lame system:

(λ + 2µ−R33)∂y3z4,ν − 2R13∂y1z4,ν − 2R23∂y2z4,ν

= F3 + (µ−R33)(∂y1z2,ν − ∂y2z1,ν),(4.56)

where F3 satisfies estimate (4.53).Finally, since the determinant of (

ξ∗1 ξ∗2ξ∗1 −ξ∗2

)

is nonzero (observe that the situation ξ∗1 = ξ∗2 = 0 is not possible in case 2.1.2), we can boundthe tangential derivatives of z1,ν and z2,ν as in (4.54). As a conclusion, we deduce inequality(4.20) and we conclude the proof of lemma 12 also in this case.

144

• Case 2.1.2: B) R13(y∗)ξ∗1 + R23(y∗)ξ∗2 6= 0In this second subcase we will first obtain estimates for the norms of z4,ν (getting profit of

the estimates of ∂y3z′ν given by (4.50)) and then for those of z′ν .

Applying (4.42), we have that

s‖(∂y3 −√rλ+2µ+(y, s, D′))z4,ν‖2

L2(∂G) ≤ (‖esϕf‖2H1,s(G)3 + ‖z‖2

H1,s(G)4)

+ε(δ)s(‖zν‖2H1,s(∂G)4 + ‖∂y3zν‖2

L2(∂G)4).(4.57)

Then, the third equation of the Lame system (see (4.56)) implies that

(√

rλ+2µ+(y, s, D′)−R13∂y1 −R23∂y2)z4,ν = (µ−R33)(∂y1z2,ν − ∂y2z1,ν) + F4, (4.58)

with F4 satisfying estimate (4.53).We are now going to express z1,ν and z2,ν in terms of z4,ν , using the two first equations of

the Lame systems, written in the form (4.52):

R13∂y1z2,ν + R23∂y2z2,ν = F5 + (λ + 2µ−R33)∂y1z4,ν ,

R13∂y1z1,ν + R23∂y2z1,ν = F6 − (λ + 2µ−R33)∂y2z4,ν .(4.59)

Here, F5 and F6 satisfy estimate (4.53) and so we have taken into account (4.50) in order toestimate ∂y3z1,ν and ∂y3z2,ν . Since R13(y∗)ξ∗1 + R23(y∗)ξ∗2 6= 0, we have

zk,ν = (−1)k (λ + 2µ−R33)(y∗)ξ3−k

R13(y∗)ξ1 + R23(y∗)ξ2z4,ν + F7.

Plugging this into (4.58), we obtain(√

rλ+2µ+(s∗, ξ′)−R13ξ1 −R23ξ2 − (λ + 2µ−R33)(µ−R33)(ξ2

1 + ξ22)

R13ξ1 + R23ξ2

)z4,ν = F8.

This directly gives an estimate of the the L2 norm of z4,ν and its tangential derivatives in thefollowing way:

s‖z4,ν‖2H1,s(∂G) ≤ (‖esϕf‖2

H1,s(G)3 + ‖z‖2H1,s(G)4) + ε(δ)s(‖zν‖2

H1,s(∂G)4 + ‖∂y3zν‖2L2(∂G)4).

Now, (4.57) also provides an estimate for the normal derivative of z4,ν , so for the moment wehave:

s2∑

j=1

(‖∂y3zj,ν‖2L2(∂G) + s2‖zj,ν‖2

L2(∂G)) + s‖z′ν‖2L2(G)3

+s4∑

j=3

(‖zj,ν‖2H1,s(∂G) + ‖∂y3zj,ν‖2

L2(∂G))

≤ C(‖esϕf‖2H1,s(G)3 + ‖z‖2

H1,s(G)4) + ε(δ)s(‖zν‖2H1,s(∂G)4 + ‖∂y3zν‖2

L2(∂G)4).

(4.60)

Finally, we use the same argument as in (4.55) in order to deduce that the tangential deriva-tives of z1,ν and z2,ν are also bounded. Since ∂y3z1,ν and ∂y3z2,ν are already bounded, the normalderivatives of z1,ν and z2,ν are too.

As a conclusion, we also obtain (4.20) in this situation.

145

2.2. Second Case: rµ(γ∗) 6= 0, rλ+2µ(γ∗) = 0

In a way similar to above, one can take δ and δ1 to be small enough so that

|rµ(γ)| ≥ C > 0 ∀γ = (y, ζ) ∈ Bδ ×O(δ1). (4.61)

In this situation, condition rλ+2µ(γ∗) = 0 also provides the estimate (4.41).We distinguish now two different situations depending on s∗.

Case 2.2.1: s∗ = 0First, we can suppose that

ϕy3(y∗) > Im(rµ(γ∗)/s)/(2

√Re(rµ(γ∗))),

since ϕy3 is large enough.Under these hypotheses, we can take δ and δ1 small enough and suppose that

−ImΓ±µ (y, ζ) ≥ Cs ∀γ = (y, ζ) ∈ Bδ ×O(δ1),

since ϕy3 > 0 (see the expression of Γ±µ in (4.27)).Consequently, we can apply proposition 8 in two different ways and deduce that

Pµ,s(y,D)z′ν = [(µE3 −R)Gt]G(Dy3 − Γ−µ (y, s,D′))w′+ν + T+

µ,sw′+ν

= [(µE3 −R)Gt]G(Dy3 − Γ+µ (y, s, D′))w′−

ν + T−µ,sw′−ν ,

with T±µ,s ∈ L(H1,s(G)3, L2(G)3). Here, we have set

w′±ν = (Dy3 − Γ±µ (y, s,D′))z′ν .

Next, we apply proposition 9 and we get the following estimates:

s‖w′±ν ‖2

L2(∂G) ≤ C(‖esϕf‖2H1,s(G)3 + ‖z‖2

H1,s(G)4). (4.62)

On the other hand, since

w′−ν −w′+

ν = 2√

rµ(y, s, D′)z′ν on ∂Gand rµ(γ∗) 6= 0, we can apply Garding’s inequality and obtain from (4.62) that

s‖z′ν‖2H1,s(∂G)3 ≤ C(‖esϕf‖2

H1,s(G)3 + ‖z‖2H1,s(G)4).

Again, (4.62) indicates that the normal derivative of z′ν is also bounded:

s(‖z′ν‖2H1,s(∂G)3 + ‖∂y3z

′ν‖2

L2(∂G)3) ≤ C(‖esϕf‖2H1,s(G)3 + ‖z‖2

H1,s(G)4). (4.63)

Next, we can estimate ∂y1z4,ν and ∂y2z4,ν by means of the two first equations of the Lame system(see (4.52)). This, together with (4.41), provides the estimate

s‖z4,ν‖2H1,s(∂G) ≤ C(‖esϕf‖2

H1,s(G)3 + ‖z‖2H1,s(G)4)

+ ε(δ)s(‖zν‖2H1,s(∂G)4 + ‖∂y3zν‖2

L2(∂G)4).(4.64)

146

Finally, the third equation of the Lame system (see (4.56)) gives an estimate of the normalderivative of z4,ν . Combining this, (4.64) and (4.63), we deduce the desired inequality (4.20).

Case 2.2.2: s∗ 6= 0We first apply estimate (4.37) to z4,ν and get

s‖z4,ν‖2H1,s(G) + Σ1

λ+2µ(z4,ν) ≤ C(‖esϕf‖2H1,s(G)3 + ‖z‖2

H1,s(G)4)

+ ε(δ)s(‖z4,ν‖2H1,s(∂G) + ‖∂y3z4,ν‖2

L2(∂G)).(4.65)

We recall that we may write

Σ1λ+2µ(z4,ν) = Σ1,1

λ+2µ(z4,ν) + Σ1,2λ+2µ(z4,ν) + Σ1,3

λ+2µ(z4,ν),

where Σ1,iλ+2µ(z4,ν) (1 ≤ i ≤ 3) corresponds to the expressions given just after (4.38) with λ+2µ

instead of µ and z4,ν instead of z′ν .In order to estimate Σ1,2

λ+2µ(z4,ν), we use that rλ+2µ(γ∗) = 0 and we obtain

|Σ1,2λ+2µ(z4,ν)| ≤ ε(δ1)s‖z4,ν‖2

H1,s(∂G).

Also, from rλ+2µ(γ∗) = 0 and (4.41), we obtain for small δ1 that

s

∣∣∣∣ϕy0(y∗)ξ0 −

(λ + 2µ−R11 − R2

13

λ + 2µ−R33

)(y∗)ϕy1(y

∗)ξ1

−(

λ + 2µ−R22 − R223

λ + 2µ−R33

)ϕy2(y

∗)ξ2

+(

R12 +R13R23

λ + 2µ−R33

)(y∗) (ϕy2(y

∗)ξ1 + ϕy1(y∗)ξ2)

∣∣∣∣≤ ε(δ1)(s2 + ξ2

1 + ξ22) ∀ζ ∈ O(δ1),

which yields|Σ1,3

λ+2µ(z4,ν)| ≤ ε(δ1)s(‖z4,ν‖2H1,s(∂G) + ‖∂y3z4,ν‖2

L2(∂G)).

Consequently, we have

Σ1λ+2µ(z4,ν) ≥ −ε(δ1)s(‖z4,ν‖2

H1,s(∂G) + ‖∂y3z4,ν‖2L2(∂G))

+s(s2‖z4,ν‖2L2(∂G) + ‖∂y3z4,ν‖2

L2(∂G)).

From the fact that s∗ 6= 0, we also deduce that

Σ1λ+2µ(z4,ν) ≥ s(‖z4,ν‖2

H1,s(∂G) + ‖∂y3z4,ν‖2L2(∂G)),

which, together with (4.65), gives

s‖z4,ν‖2H1,s(G) + s(‖z4,ν‖2

H1,s(∂G) + ‖∂y3z4,ν‖2L2(∂G))

≤ C(‖esϕf‖2H1,s(G)3 + ‖z‖2

H1,s(G)4).(4.66)

147

In this situation, estimate (4.48) also holds. Furthermore, we can apply proposition 9 to z3,ν , sowe deduce

s(‖z3,ν‖2H1,s(∂G) + ‖∂y3z3,ν‖2

L2(∂G)) ≤ C(‖esϕf‖2H1,s(G)3 + ‖z‖2

H1,s(G)4)

+ε(δ)s(‖zν‖2H1,s(∂G)4 + ‖∂y3zν‖2

L2(∂G)4).(4.67)

The next step will be to estimate the functions z1,ν and z2,ν . To this end, we first observethat the case where

√rµ(γ∗)−R13(ξ∗1 + is∗ϕy1)−R23(ξ∗2 + is∗ϕy2) 6= 0 (4.68)

directly provides estimate (4.20); indeed, the two first equations in (4.9) can be rewritten as

(√

rµ(y, s, D′)−R13(Dy1 + isϕy1)−R23(Dy2 + isϕy2))z2,ν = F9

and(√

rµ(y, s, D′)−R13(Dy1 + isϕy1)−R23(Dy2 + isϕy2))z1,ν = F10

respectively, with F9 and F10 satisfying estimate (4.53).To prove this, it suffices to take into account (4.66) and proposition 9 applied to z1,ν and

z2,ν . Now, from (4.68), we can apply Garding’s inequality to the two previous equations andwe find out estimates of the H1,s norm of z1,ν and z2,ν on ∂G. Finally, again an application ofproposition 9 also provides the estimate of the normal derivative of z1,ν and z2,ν .

Outside the particular situation (4.68), we distinguish another two cases:

• If (ξ∗1 + isϕy1(y∗))2 + (ξ∗2 + isϕy2(y

∗))2 6= 0In this case, we are going to represent (z1,ν , z2,ν) in terms of z4,ν and we will then use estimate

(4.66).In order to do this, let us introduce the following differential operator:

M(y, s,D′)(z1,ν , z2,ν) = (IDy1z1,ν + IDy2z2,ν , (µ−R33)(IDy1z2,ν − IDy2z1,ν)) (4.69)

From the third equation of our Lame system, we have

(µ−R33)(IDy1z2,ν − IDy2z1,ν)

= (λ + 2µ−R33)IDy3z4,ν − 2R13IDy1z4,ν − 2R23IDy2z4,ν + F11

(4.70)

where F11 satisfies estimate (4.53).Then, since the divergence of a curl is identically zero and we have estimate (4.49) for ∂y3z3,ν ,

we find that

s‖IDy1z1,ν + IDy2z2,ν‖2L2(∂G) ≤ ε(δ)s(‖zν‖2

H1,s(∂G)4 + ‖∂y3zν‖2L2(∂G)4)

+ C(‖esϕf‖2H1,s(G) + ‖z‖2

H1,s(G)4).(4.71)

The principal symbol of the operator M is(

ξ1 + isϕy1 ξ2 + isϕy2

−(µ−R33)(ξ2 + isϕy2) (µ−R33)(ξ1 + isϕy1)

),

148

which has a nonzero determinant at the point γ∗ because of the hypothesis made in this case.Therefore, we deduce that there exists a parametrix of this operator such that

(z1,ν , z2,ν) = M−1(y, s, D′)(IDy1z1,ν + IDy2z2,ν , (µ−R33)(IDy1z2,ν − IDy2z1,ν)) + T (z1,ν , z2,ν),

with T ∈ S−1.Consequently, if we use the identity (4.70) and we combine estimates (4.71) with (4.66), we

find

s2∑

j=1

‖zj,ν‖2H1,s(∂G) ≤ C(‖esϕf‖2

H1,s(G)3 + ‖z‖2H1,s(G)4) + ε(δ)s(‖zν‖2

H1,s(∂G)4 + ‖∂y3zν‖2L2(∂G)4).

Finally, the estimate of ∂y3zj,ν (j = 1, 2) comes from proposition 9 applied to z1,ν and z2,ν . As aconclusion, we have

s2∑

j=1

(‖zj,ν‖2H1,s(∂G) + ‖∂y3zj,ν‖2

L2(∂G)) ≤ C(‖esϕf‖2H1,s(G)3 + ‖z‖2

H1,s(G)4)

+ ε(δ)s(‖zν‖2H1,s(∂G)4 + ‖∂y3zν‖2

L2(∂G)4)

which, combined with (4.66) and (4.67), gives (4.20).

• If (ξ∗1 + is∗ϕy1(y∗))2 + (ξ∗2 + is∗ϕy2(y

∗))2 = 0After an application of estimate (4.37) to z1,ν and z2,ν , we obtain that

s‖zj,ν‖2H1,s(G) + Σ1

µ(zj,ν) ≤ ε(δ)s(‖zj,ν‖2H1,s(∂G) + ‖∂y3zj,ν‖2

L2(∂G))

+C(‖esϕf‖2H1,s(G)3 + ‖z‖2

H1,s(G)4) for j = 1, 2,(4.72)

where Σ1µ(zj,ν) are given just after (4.38).

Let us estimate Σ1,2µ (zj,ν). First, we rewrite it in the form

Σ1,2µ (zj,ν) = s

∫

∂G(µ−R33)(y∗)ϕy3(y

∗)[|∂y0zj,ν |2 − s2ϕ2

y0(y∗)|zj,ν |2

−(

λ + 2µ−R11 − R213

λ + 2µ−R33

)(y∗)(|∂y1zj,ν |2 − s2ϕ2

y1(y∗)|zj,ν |2)

−(

λ + 2µ−R22 − R223

λ + 2µ−R33

)(y∗)(|∂y2zj,ν |2 − s2ϕ2

y2(y∗)|zj,ν |2)

+2(

R12 +R13R23

λ + 2µ−R33

)(y∗)(∂y1zj,ν ∂y2zj,ν − s2ϕy1(y

∗)ϕy2(y∗))|zj,ν |2

]dy′

+s

∫

∂G(µ−R33)(y∗)ϕy3

[(λ + µ)

(|∂y1zj,ν |2 + |∂y2zj,ν |2 − s2ϕ2y1

(y∗)|zj,ν |2

−s2ϕ2y2

(y∗)|zj,ν |2)− (λ + 2µ−R33)

−1 (y∗)(R2

13(y∗)

(|∂y1zj,ν |2

−s2ϕ2y1

(y∗)|zj,ν |2)

+ R223(y

∗)(|∂y2zj,ν |2 − s2ϕ2

y2(y∗)|zj,ν |2

)

+2R13(y∗)R23(y∗)(∂y1zj,ν∂y2zj,ν − s2ϕy1ϕy2 |zj,ν |2

))]dy′

= J4(zj,ν) + J5(zj,ν),

149

where

J4(zj,ν) = s

∫

∂G(µ−R33)(y∗)ϕy3(y

∗)[|∂y0zj,ν |2 − s2ϕ2

y0(y∗)|zj,ν |2

−(

λ + 2µ−R11 − R213

λ + 2µ−R33

)(y∗)(|∂y1zj,ν |2 − s2ϕ2

y1(y∗)|zj,ν |2)

−(

λ + 2µ−R22 − R223

λ + 2µ−R33

)(y∗)(|∂y2zj,ν |2 − s2ϕ2

y2(y∗)|zj,ν |2)

+2(

R12 +R13R23

λ + 2µ−R33

)(y∗)(∂y1zj,ν ∂y2zj,ν − s2ϕy1(y

∗)ϕy2(y∗))|zj,ν |2

]dy′

andJ5(zj,ν) = Σ1,2

µ (zj,ν)− J4(zj,ν).

From Re(rλ+2µ(γ∗)) = 0, we deduce that

|J4(zj,ν)| ≤ ε(δ1)s‖zj,ν‖2H1,s(∂G). (4.73)

Let us now estimate J5(zj,ν); we observe that

J5(zj,ν) = s

∫

R2

(µ−R33)(y∗)ϕy3(y∗)

[(λ + µ)

(ξ21 + ξ2

2 − s2ϕ2y1

(y∗)

−s2ϕ2y2

(y∗))− (λ + 2µ−R33)

−1 (y∗) Re (R13(y∗)(ξ1 + isϕy1(y∗))

+R23(y∗)(ξ2 + isϕy2(y∗)))2

]|zj,ν |2 dξ1 dξ2.

We first realize that, since rλ+2µ(γ∗) = 0 and√

rµ(γ∗)−R13(ξ∗1 + is∗ϕy1)−R23(ξ∗2 + is∗ϕy2) = 0,

we have

Re (R13(y∗)(ξ∗1 + isϕy1(y∗)) + R23(y∗)(ξ∗2 + isϕy2(y

∗)))2

= (λ + 2µ−R33)(λ + µ)(y∗) Re((ξ∗1 + isϕy1(y

∗))2 + (ξ∗2 + isϕy2(y∗))2

).

Then, it is readily seen that J5(zj,ν) can be bounded as in (4.73). Consequently,

|Σ1,2µ (zj,ν)| ≤ ε(δ1)s‖zj,ν‖2

H1,s(∂G). (4.74)

On the other hand, the term Σ1,1µ (zj,ν) provides estimates for

C∗s(‖∂y3zj,ν‖2L2(∂G) + s2‖zj,ν‖2

L2(∂G)),

with C∗ large (since ϕy3 is too). Then, since s∗ 6= 0, we also have estimates for the tangentialderivatives of zj,ν (that is to say, for ‖zj,ν‖2

H1,s(∂G), j = 1, 2) and hence for the normal derivativetoo.

Finally, the fact that C∗ is large is used for estimating Σ1,3µ (zj,ν).

As a conclusion, from (4.66)–(4.67) and (4.72)–(4.74), we deduce the desired inequality (4.20).

150

2.3. Third Case: rµ(γ∗) 6= 0 and rλ+2µ(γ∗) 6= 0 or rµ(γ∗) = rλ+2µ(γ∗) = 0.

It is in this paragraph where we will use Calderon’s method. The ideas we develop here aresimilar to those in [15].

Let us introduce the variables U = (Uj)6j=1, given by

(U1, U2, U3) = Λ(s,D′)(esϕu), (U4, U5, U6) = IDy3(esϕu),

where Λ is the pseudodifferential operator with symbol (1 + s2 + |ξ′|2). With these notations,we can rewrite the Lame system (4.9) as

Dy3U = K(y, s,D′)U + F in R3 × [0, 1/Z2],

(U1, U2, U3) = 0 on y3 ≡ 0,U = 0 on y3 ≡ 1/Z2,

(4.75)

where F = (0, esϕf) and K(y, s, D′) is a pseudodifferential operator whose principal symbol is

K1(γ) =(

0 Λ1E3

A−1K11Λ−1 A−1K12

)− isϕy3E6.

Here, we have set

A = (λ + µ)GGt + [((µE3 −R)Gt)G]E3,

K11 = −(λ + µ)θθt + [(ξ0 + isϕy0)2 − ((µE3 −R)θt)θ]E3,

K12 = −(λ + µ)(θGt + Gθt)− 2((µE3 −R)θt)GE3,

(4.76)

with θ = (ξ1 + isϕy1 , ξ2 + isϕy2 , 0)t. Recall that G was defined (also as a column vector field) in(4.19).

In this context, one can check that the eigenvalues of the matrix K1 coincide with Γ±β (givenin (4.27)).

Let us consider again several situations:

Case 2.3.1: rµ(γ∗) = 0 = rλ+2µ(γ∗)Then, the expression of rβ indicates directly that Im(Γ±β ((γ∗)) < 0 for β ∈ µ, λ + 2µ; this

comes from the fact that ϕy3 > 0. Consequently, the eigenvalues of the matrix K1 have negativeimaginary parts and we can suppose that

Im(Γ±β (γ)) < −C|ζ| ∀γ ∈ Bδ ×O(δ1).

Now, using standard arguments (see, for instance, Chapter 7 in [19]), we obtain that

‖χνU‖2H2,s(G)6 ≤ C(‖esϕf‖2

H1,s(G)3 + ‖esϕu‖2H1,s(G)3).

From this inequality, estimate (4.20) is readily deduced.

Case 2.3.2: rµ(γ∗) 6= 0, rλ+2µ(γ∗) 6= 0 and rµ(γ∗) 6= rλ+2µ(γ∗)

151

In this situation, the matrix K1 has four eigenvalues which are given by (4.27) and thecorresponding eigenvectors:

v±1 = ((θt + α±λ+2µGt)Λ−11 , α±λ+2µ(θt + α±λ+2µGt)Λ−2

1 )t,

v±2 = (v±21, α±µ v±21Λ

−11 )t, v±3 = (v±31, α

±µ v±31Λ

−11 )t,

(4.77)

withv±21 = (`y2α

±µ − θ2,−`y1α

±µ + θ1, 0)Λ−1

1

and

v±31 = (α±µ (θ1 − `y1α±µ ), α±µ (θ2 − `y2α

±µ ),−

2∑

j=1

(θj − `yjα±µ )2)Λ−2

1 .

Observe that v±2 , v±3 is a basis of the orthogonal space to the vector θ + α±µ G.Let us take the symbol S(γ) in the form S = v+

1 , v+2 , v+

3 , v−1 , v−2 , v−3 and let us extend it asan homogeneous function of order 0 and of class C3 in the ζ variables. Now, U is determinedby (4.75). Let us introduce W = S−1(y, s, D′)U . Then, system (4.75) takes the form

Dy3W = K(y, s, D′)W + T (y, s, D′)W + F,

where K is a diagonal matrix and T ∈ L∞(0, 1;L(H1,s(R3)3);H1,s(R3)3). A standard argumentfor pseudodifferential systems allows to estimate the last three components of W as follows (seefor instance [19]):

s‖(W4, W5,W6)‖2H1,s(∂G)3 ≤ C(‖esϕf‖2

H1,s(G)3 + ‖esϕu‖2H2,s(G)3). (4.78)

Finally, the first three components can be bounded in terms of the last three by means of thefirst-squared 3× 3 matrix inside S, which is

Λ−11

θ1 − α+λ+2µ`y1 −θ2 + `y2α

+µ α+

µ (θ1 − `y1α+µ )Λ−1

1

θ2 − α+λ+2µ`y2 θ1 − `y1α

+µ α+

µ (θ2 − `y2α+µ )Λ−1

1

α+λ+2µ 0 −∑2

j=1(θj − `yjα+µ )2Λ−1

1

This yieldss‖(W1, W2,W3)‖2

H1,s(∂G)3 ≤ C(‖esϕf‖2H1,s(G)3 + ‖esϕu‖2

H2,s(G)3),

which, in combination with (4.78) and (4.35), provides (4.20).

Case 2.3.3: rµ(γ∗) = rλ+2µ(γ∗) 6= 0The argument needed to prove estimate (4.20) in this case coincides exactly with the one

developed in [15].

This ends the proof of Lemma 12.

152

3. Controllability of the Lame system.

In this section we obtain some exact controllability results for the Lame system with controlslocally distributed over the cylinder Qω = ω × (0, T ).

First, we need a Carleman estimate similar to (4.10) but with the right hand side in thespaces L2(Q) and H−1(Q). In order to prove such an estimate, we need a pseudoconvex functionψ which satisfies the additional condition:

∂x0ψ(·, 0) > 0, ∂x0ψ(·, T ) < 0 ∀x ∈ Ω. (4.79)

We have the following result:

Theorem 11 Let f ∈ L2(Q)3 and assume that (4.2), (4.4), (4.6) and (4.79) hold true. Supposethere exists a function ψ which satisfies condition A. Then, there exists τ∗ > 0 such that, forany τ > τ∗, there exists s∗ > 0 such that

‖u‖H1,s(Q)3 ≤ C(‖esφf‖L2(Q)3 + ‖u‖H1,s(Qω)3) ∀s > s∗ (4.80)

for some positive constant C independent of s and for any solution u ∈ H1(Q)3 of (4.9).

Next, we consider a situation with the function f in the space H−1(Q).

Theorem 12 Let f = f−1 +∑3

k=0 ∂xkfk, where f0, f1, f2, f3 ∈ L2(Q)3 and assume that (4.2),

(4.4), (4.6) and (4.79) hold true. Suppose there exists a function ψ which satisfies condition A.Then, there exists τ∗ > 0 such that, for any τ > τ∗, there exists s∗ > 0 such that

‖u‖L2(Q)3 ≤ C(‖esφf−1‖H−1(Q)3 +3∑

k=0

‖esφfk‖L2(Q)3 + ‖u‖L2(Qω)3) ∀s > s∗ (4.81)

for some positive constant C independent of s and for any solution u ∈ L2(Q)3 of (4.9).

The proofs of theorems 11 and 12 rely on theorem 10 and are similar to the proofs of the similarresults presented in [14]).

Next, we consider the controllability problem

Pu = f + χωv in Q, u = 0 on Σ,

u(·, 0) = u0, ∂x0u(·, 0) = u1 in Ω.(4.82)

Here f ,u0,u1 are given functions and v is a control. Recall that the operator P was defined in(4.9). Suppose that two target functions u2 and u3 are given. We have to find a control v suchthat

u(·, T ) = u2, ∂x0u(·, T ) = u3 in Ω. (4.83)

Condition B. Suppose that there exists t∗ ∈ (0, T ) such that

mınx′∈Ω\ω

ψ(x′, t∗) > max maxx′∈Ω\ω

ψ(x′, 0), maxx′∈Ω\ω

ψ(x′, T ).

153

Before presenting the main result of this section, let us remind a well known result on thesolvability of the Lame system

Pp = q in Q, p = 0 on Σ,

p(·, 0) = p0, ∂x0p(·, 0) = p1 in Ω.(4.84)

We have the following:

Theorem 13 Assume that (4.2) and (4.6) hold. Then, if q ∈ L2(Q)3, p0 ∈ H10 (Ω)3 and p1 ∈

L2(Ω)3, problem (4.84) possesses exactly one solution p ∈ H1(Q)3.Furthermore, if q ∈ H−1(Q)3, p0 ∈ L2(Ω)3 and p1 ∈ H−1(Ω)3, problem (4.84) possesses

exactly one solution p ∈ L2(Q)3.

Now, we state the main result of this paper:

Theorem 14 Assume that (4.2), (4.4) and (4.6) hold. Suppose that there exists a function ψwhich satisfies condition A, condition B and (4.79). Then if f ∈ L2(Q)3, u0 ∈ H1

0 (Ω)3 andu1 ∈ L2(Ω)3, there exists a solution (u,v) ∈ H1(Q)3 × L2(Qω)3 to the controllability problem(4.82)–(4.83).

Furthermore, if f ∈ H−1(Q)3, u0 ∈ L2(Ω)3 and u1 ∈ H−1(Ω)3, there exists a solution(u,v) ∈ L2(Q)3 ×H−1(Qω)3 to (4.82)–(4.83).

Proof. Let ε be a sufficiently small positive number such that t∗ ∈ (ε, T − ε) and set

M = mınx′∈Ω\ω

ψ(x′, t∗) > max maxx′∈Ω\ω

ψ(x′, ε), maxx′∈Ω\ω

ψ(x′, T − ε)

We introduce a cut-off function χ(x0) ∈ C∞0 ([0, T ]) such that χ|[ε,T−ε] = 1. Let p be a solution

to the system Pp(x) ≡ ρ(x′)∂2

x0p− Lp = q in Q,

p = 0 on Σ.(4.85)

Then the function p = χ(x0)p satisfies the equation

P p(x) = χq− [χ, ∂2x0

]p in Q,

p = 0 on Σ.(4.86)

Note that [χ, ∂2x0

] is a first order operator with coefficients having support in [0, ε]∪ [T−ε, T ]×Ω.Therefore, we have the following a priori estimates:

‖[χ, ∂2x0

]pesφ‖L2(Q)3 ≤ eδM‖p‖H1(Q)3 , ‖[χ, ∂2x0

]pesφ‖H−1(Q)3 ≤ eδM‖p‖L2(Q)3 .

Additionally, we apply Carleman estimate (4.10) to the equation (4.86) and we obtain that

‖p‖Hk,s(Q)3 ≤ C(‖esφq‖Hk−1(Q)3 + ‖p‖Hk,s(Qω)3 + esδM‖p‖H1−k(Q)3).

Note that, for any sufficiently small ε > 0 there exists a δ1 > δ such that

esδ1M‖(p(·, t∗), ∂x0p(·, t∗))‖Hk(Ω)3×Hk−1(Ω)3

≤ C(‖p‖Hk,s(Q)3 + ‖esφq‖Hk−1(Q)3 + ‖p‖Hk,s(Qω)3 + eδM‖p‖H1−k(Q)3).

154

Since δ1 > δ, taking the parameter s sufficiently large in the previous inequality thanks tothe theorem 13, we arrive at the observability estimate

‖p‖Hk(Q)3 ≤ C(‖q‖Hk−1(Q)3 + ‖p‖Hk(Qω)3).

This observability estimate can be readily converted into the controllability result stated inour theorem by the well known HUM method (see [24]).

This ends the proof of Theorem 14.

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Apendice: desigualdad de Carlemanpara el sistema de Stokes

160

Desigualdad de Carleman para el sistema deStokes

En esta seccion presentamos un resumen de la demostracion de la desigualdad de Carlemanpara el problema de Stokes con condiciones de Dirichlet que ha sido probada con detalle en [2].Hemos visto que es esencial para el desarrollo de la tercera parte de esta memoria, correspondi-ente al trabajo [3].

Para a ∈ L∞(Q)N , introducimos el siguiente sistema retrogrado de ecuaciones de tipo Stokes:

−ϕt −∆ϕ−Dϕ a +∇π = g en Q

∇ · ϕ = 0 en Q,

ϕ = 0 sobre Σ,

ϕ(T ) = ϕ0 en Ω.

(A.1)

Aquı hemos denotado

Dϕ = ∇ϕ +∇ϕt.

Las siguientes hipotesis de regularidad tendran que ser impuestas sobre a:

a ∈ L∞(Q)N , at ∈ L2(0, T ;Lσ(Ω)N )

(σ > 1 si N = 2

σ > 6/5 si N = 3

). (A.2)

Probaremos una desigualdad del tipo

∫∫

Qρ21(x, t) |ϕ|2 dx dt ≤ C

∫∫

ω×(0,T )ρ22(t) |ϕ|2 dx dt

para toda ϕ0 ∈ H, donde C > 0 es una constante que solo depende de Ω, ω, T y a.Para probar esto, en primer lugar definimos las funciones peso que utilizaremos. La funcion

peso basica es η0 ∈ C2(Ω), que verifica

η0 > 0 en Ω, η0 = 0 sobre ∂Ω, |∇η0| > 0 en Ω \ ω1,

con ω1 ⊂⊂ ω un abierto no vacıo. La existencia de una tal η0 esta probada en [4]. Ahora, para

161

ciertos numeros positivos s y λ, definimos

α(x, t) =e5/4λm‖η0‖∞ − eλ(m‖η0‖∞+η0(x))

t4(T − t)4,

ξ(x, t) =eλ(m‖η0‖∞+η0(x))

t4(T − t)4,

α(t) = mınx∈Ω

α(x, t) =e5/4λm‖η0‖∞ − eλ(m+1) ‖η0‖∞

t4(T − t)4,

α∗(t) = maxx∈Ω

α(x, t) =e5/4λm‖η0‖∞ − eλm‖η0‖∞

t4(T − t)4,

ξ(t) = maxx∈Ω

ξ(x, t) =eλ(m+1)‖η0‖∞

t4(T − t)4, ξ∗(t) = mın

x∈Ωξ(x, t) =

eλm‖η0‖∞

t4(T − t)4,

θ(t) = sλe−sαξ, θ(t) = s15/4e−2sα+sα∗ ξ15/4,

(A.3)

donde m > 4 es un numero fijo.Por comodidad, durante la prueba usaremos la notacion

I(s, λ; ϕ) = s−1

∫∫

Qe−2sαξ−1(|ϕt|2 + |∆ϕ|2) dx dt

+sλ2

∫∫

Qe−2sαξ|∇ϕ|2 dx dt + s3λ4

∫∫

Qe−2sαξ3|ϕ|2 dx dt.

El resultado principal de esta seccion es el siguiente:

Teorema 1 Supongamos que a verifica (A.2). Entonces existen tres constantes positivas C, s0

y λ0 tales que para todo ϕ0 ∈ H y toda g ∈ L2(Q)N , la solucion de (A.1) verifica

I(s, λ; ϕ) ≤ C

(s15/2λ20

∫∫

Qe−4sα+2sα∗ ξ15/2|g|2 dx dt

+ s16λ40

∫∫

ω×(0,T )e−8sα+6sα∗ ξ16|ϕ|2 dx dt

),

(A.4)

para todo λ ≥ λ0(Ω, ω, T, a) y todo s ≥ s0(Ω, ω, T ).

Observacion 1 En el teorema anterior es posible conocer de que forma dependen λ0 y s0 de Ty de a. En concreto, se prueba en [2] que

λ0 = λ(1 + ‖a‖∞ + ‖at‖2L2(0,T ;Lσ(Ω)N ) + eλT‖a‖2∞)

ys0 = s(T 4 + T 8),

donde λ y s son constantes positivas que dependen de Ω y de ω.

162

La estimacion (A.4) es una nueva desigualdad global de Carleman que ha sido demostrada en[2] con un metodo novedoso, si bien la prueba obedece al esquema marcado por O. Yu. Imanuviloven [6]:

Primera parte: Consideramos (A.1) como un sistema de N ecuaciones parabolicas y uti-lizamos la desigualdad de Carleman para la ecuacion del calor con condiciones de Dirichlet. Unaprueba de esta se puede encontrar en [4]. Esto proporciona una estimacion de la velocidad enfuncion de la presion:

I(s, λ;ϕ) ≤ C

(∫∫

ω×(0,T )ρ23(x, t)|ϕ|2 dx dt +

∫∫

Qρ24(x, t) (|π|2 + |g|2) dx dt

)

(vease (A.7) mas abajo).Segunda parte: Tomando el operador divergencia en la ecuacion satisfecha por ϕ, obte-

nemos la ecuacion elıptica verificada por la presion π. Usando la desigualdad de Carleman paralas soluciones debiles de problemas elıpticos probada en [7], deducimos una estimacion de lapresion en funcion de su traza sobre la frontera y de la velocidad:

∫∫

Qρ24(x, t) |π|2 dx dt ≤ C

(∫∫

ω×(0,T )ρ25(x, t)|π|2 dx dt

∫∫

Qρ26(x, t)(|∇ϕ|2 + |g|2) dx dt +

∫ T

0ρ27(t)‖π(t)‖2

H1/2(∂Ω)dt

)

(vease (A.11)). Seguimos ahora las ideas de [6] y usamos resultados de regularidad para el sistemade Stokes para acotar la traza de la presion. Combinado con lo anterior, esto nos dara

I(s, λ;ϕ) ≤ C

(∫∫

ω×(0,T )ρ28(x, t)(|ϕ|2 + |π|2) dx dt +

∫∫

Qρ29(x, t) |g|2 dx dt

)

(desigualdad (A.15)).

En lo que queda de prueba, nos centraremos en estimar el termino local de la presion. Envirtud de la ecuacion que verifica ϕ, la tarea consiste en conseguir estimaciones adecuadas deintegrales locales de ∆ϕ y ϕt.

tercera parte: En un primer apartado, indicaremos como se acota localmente ∆ϕ. Estaes una desigualdad clasica y no conlleva argumentos de dificultad importantes, de ahı que nohagamos la prueba detallada.

cuarta parte: A continuacion, demostraremos la estimacion local de ϕt. En ella, des-compondremos nuestra solucion ϕ (salvo un peso dependiente de t) en otras dos soluciones deproblemas de tipo Stokes. En ambas podremos razonar para realizar acotaciones locales de susderivadas en tiempo, valiendonos exclusivamente de resultados de regularidad de las solucionesdel problema de Stokes.

Antes de comenzar la prueba del teorema 1, haremos un comentario que nos sera de utilidad.

163

Observacion 2 Sean ϕ0 ∈ H y g ∈ L2(Q)N dados y sea (ϕ, π) la correspondiente solucionde (A.1). Sabemos que ϕ ∈ L2(0, T ; V ) ∩ C0([0, T ];H). Sea γ ∈ C1([0, T ]) una funcion tal queγ(T ) = 0. Entonces (ϕ, π) := (γϕ, γπ) resuelve el sistema

−ϕt −∆ϕ−Dϕ a +∇π = γg − γ′ϕ en Q,

∇ · ϕ = 0 en Q,

ϕ = 0 sobre Σ,

ϕ(T ) = 0 en Ω.

(A.5)

Por tanto, podemos decir que (ϕ, π) es una solucion fuerte de (A.5). En particular,

π(t) ∈ H1(Ω), ϕ(t) ∈ H2(Ω)N , y ϕt(t) ∈ L2(Ω)N

para casi todo t ∈ (0, T ).

Estimacion de Carleman para la ecuacion del calorAplicamos la desigualdad de Carleman habitual para la ecuacion del calor con segundo

miembro en L2(Q) a la ecuacion satisfecha por ϕi, que tiene como segundo miembro

Gi = gi + (Dϕa)i − ∂iπ.

La consecuencia es que existen tres constantes positivas C1(Ω, ω), λ0(Ω, ω) ≥ 1 y s0(Ω, ω) > 0tales que

I(s, λ; ϕ) ≤ C1

(∫∫

Qe−2sα(|g|2 + |Dϕ a|2 + |∇π|2) dx dt

+s3λ4

∫∫

ω1×(0,T )e−2sαξ3|ϕ|2 dx dt

) (A.6)

para todo λ ≥ λ0 y todo s ≥ s0(T 7 +T 8). De hecho, para probar (A.6) basta con seguir las ideasde [4] teniendo en cuenta que

|ξ−1| ≤ C T 8 and |αt| ≤ C T ξ5/4

para cierta C > 0 independiente de λ y despues razonar como en [1].Ahora, eliminamos el termino de Dϕ a en la derecha de (A.6), usando que

C1 |Dϕ a|2 ≤ Cs‖a‖2∞ξ|∇ϕ|2 ≤ 1

2sλ2ξ|∇ϕ|2,

para λ ≥ λ1(Ω, ω)‖a‖∞ y s ≥ s1(Ω, ω)T 8. Deducimos que

I(s, λ;ϕ) ≤ C2

(∫∫

Qe−2sα(|g|2 + |∇π|2) dx dt

+s3λ4

∫∫


) (A.7)

164

para todo λ ≥ λ2(Ω, ω)(1 + ‖a‖∞) y todo s ≥ s2(Ω, ω)(T 7 + T 8).

Localizacion de la presionEn este apartado acotaremos la integral de |∇π|2 a la derecha de (A.7) en funcion de un

termino local de |π|2, la norma en H1/2 de la traza de π y otros dos terminos globales de |g|2 y|∇ϕ|2; este ultimo sera absorbido mas adelante con I(s, λ; ϕ).

Esto sera conseguido con la ayuda de una desigualdad de Carleman aplicada a la ecuacionelıptica que verifica la presion; esta desigualdad se probo en [7].

Concretamente, aplicando el operador divergencia a la primera ecuacion de (A.1), tenemos

∆π(t) = ∇ · (Dϕa + g)(t) en Ω, (A.8)

para casi todo t ∈ (0, T ). Observese que el segundo miembro de (A.8) pertenece a H−1(Ω).Usando la Observacion 2, tambien sabemos que π(t) ∈ H1(Ω). Aplicamos entonces el resultadoprincipal contenido en [7] (vease la desigualdad (2.10) en dicha referencia), el cual nos dice queexiste una constante C1(Ω, ω) > 0 y dos numeros λ > 1, τ > 1 tal que

∫

Ωe2τη|∇π(t)|2 dx ≤ C1

(τ

∫

Ωe2τηη(|Dϕ a|2 + |g|2)(t) dx

+τ1/2e2τ‖π(t)‖2H1/2(∂Ω)

+∫

ω1

e2τη(|∇π|2 + τ2λ2η2|π|2)(t) dx

) (A.9)

para todo λ ≥ λ y todo τ ≥ τ , donde η esta dada por

η(x) = eλη0(x) ∀x ∈ Ω.

La integral local de |∇π|2 en (A.9) puede ser acotada a costa de tener |π|2 en un abierto ω2 masgrande: ω1 ⊂⊂ ω2 ⊂⊂ ω. Para ello, consideremos una funcion ζ ∈ C2

c (ω2) tal que

ζ(x) = 1 en ω1, 0 ≤ ζ ≤ 1

e integremos por partes varias veces:∫

ω1

e2τη|∇π(t)|2 dx ≤∫

ω2

e2τηζ∇π(t) · ∇π(t) dx

= −12

∫

ω2

∇(e2τηζ) · ∇|π(t)|2 dx− 〈e2τη∆π(t), ζπ(t)〉H−1(ω2),H10 (ω2)

=12

∫

ω2

∆(e2τηζ)|π(t)|2 dx

−〈e2τη∇ · (Dϕa + g)(t), ζπ(t)〉H−1(ω2),H10 (ω2).

(A.10)

Como|∆(e2τηζ)| ≤ 2C2τ

2λ2η2e2τη en ω2

para λ ≥ λ0(Ω, ω) y cierta constante C2(Ω, ω) > 0, podemos acotar el primer termino de laderecha de (A.10) por

C2τ2λ2

∫

ω2

e2τηη2|π(t)|2 dx.

165

Integramos de nuevo por partes en el otro termino. Obtenemos que

−〈e2τη∇ · (Dϕa + g)(t), ζπ(t)〉H−1(ω2),H10 (ω2)

=∫

ω2

∇(e2τηζ) · (Dϕ a + g)(t)π(t) dx

+∫

ω2

e2τηζ(Dϕ a + g)(t) · ∇π(t) dx

≤ C4

(τ2λ2

∫

ω2

e2τηη2|π(t)|2 dx +∫

ω2

e2τη(|Dϕ a|2 + |g|2)(t) dx

)

+12

∫

ω2

e2τηζ|∇π(t)|2 dx

para cierta constante C4(Ω, ω) > 0, pues

|∇(e2τηζ)| ≤ C3(Ω, ω)τληe2τη en ω2

para λ ≥ λ1(Ω, ω). A partir de (A.10), tenemos ahora que

∫

ω1

e2τη|∇π(t)|2 dx ≤ C5

(τ2λ2

∫

ω2

e2τηη2|π(t)|2 dx

+∫

ω2

e2τη(|Dϕ a|2 + |g|2)(t) dx

)

para λ ≥ λ2(Ω, ω), lo cual, junto con (A.9), nos da

∫

Ωe2τη|∇π(t)|2 dx ≤ C6

(τ

∫

Ωe2τηη(|Dϕ y|2 + |g|2)(t) dx

+τ1/2e2τ‖π(t)‖2H1/2(∂Ω)

+ τ2λ2

∫

ω2

e2τηη2|π(t)|2 dx

),

para λ ≥ λ2 y τ ≥ τ .Para conectar esta estimacion elıptica con (A.7), tomamos

τ =s

t4(T − t)4eλm‖η0‖∞ ,

multiplicamos la expresion por

exp

−2s

e5/4λm‖η0‖∞

t4(T − t)4

e integramos con respecto a t en (0, T ). Observese que, para que esta eleccion de τ sea mayor

166

que τ , bastara tomar s ≥ (τ/28) T 8 y entonces obtenemos∫∫

Qe−2sα|∇π|2 dx dt ≤ C7

(s

∫∫

Qe−2sαξ|g|2 dx dt

+s

∫∫

Qe−2sαξ|Dϕa|2 dx dt + s1/2

∫ T

0e−2sα∗(ξ∗)1/2‖π(t)‖2

H1/2(∂Ω)dt

+s2λ2

∫∫

ω2×(0,T )e−2sαξ2|π|2 dx dt

)(A.11)

para todo λ ≥ λ2 y todo s ≥ s0T8.

Tal y como comentamos al comienzo, seguimos ahora las ideas de [6] para acotar la traza deπ. En concreto, usaremos estimaciones fuertes para las soluciones de sistemas de Stokes.

Definamos las funciones

ϕ∗ = s1/4e−sα∗(ξ∗)1/4ϕ, π∗ = s1/4e−sα∗(ξ∗)1/4π,

que satisfacen el sistema

−ϕ∗t −∆ϕ∗ +∇π∗ = g∗ en Q,

∇ · ϕ∗ = 0 en Q,

ϕ∗ = 0 sobre Σ,

ϕ∗(T ) = 0 en Ω,

cong∗ = s1/4e−sα∗(ξ∗)1/4g + s1/4e−sα∗(ξ∗)1/4Dϕ a− s1/4(e−sα∗(ξ∗)1/4)tϕ.

Usando propiedades de regularidad bien conocidas para el sistema de Stokes (vease porejemplo [8]), deducimos que ϕ∗ ∈ L2(0, T ;H2(Ω)N ∩ V ) ∩ L∞(0, T ; V ), ϕ∗t ∈ L2(0, T ; H), π∗ ∈L2(0, T ;H1(Ω)) y, tambien, que estas funciones estan acotadas en esos espacios por la norma L2

del segundo miembro. En particular,∫∫

Q(|π∗|2 + |∇π∗|2) dx dt ≤ C8

∫∫

Q|g∗|2 dx dt

y, en consecuencia, tenemos que∫ T

0‖π∗(t)‖2

H1/2(∂Ω)dt

≤ C9

(s1/2

∫∫

Qe−2sα∗(ξ∗)1/2|g|2 dx dt

+s1/2‖a‖2∞

∫∫

Qe−2sα∗(ξ∗)1/2|∇ϕ|2 dx dt

+s1/2

∫∫

Q|(e−sα∗(ξ∗)1/4)t|2|ϕ|2 dx dt

).

(A.12)

167

Teniendo en cuenta la definiciıon de α∗ y de ξ∗ (vease (A.3)), vemos que las dos primerasintegrales de la derecha de (A.12) pueden acotarse por

s

∫∫

Qe−2sαξ|g|2 dx dt + s‖a‖2

∞

∫∫

Qe−2sαξ|∇ϕ|2 dx dt,

si s ≥ s1T8.

Finalmente, obtenemos una estimacion de la derivada en tiempo del peso e−sα∗(ξ∗)1/4:

(e−sα∗(ξ∗)1/4)t = e−sα∗(−sα∗t (ξ∗)1/4 + 1/4(ξ∗)−3/4ξ∗t )

≤ C10e−sα∗(sT (ξ∗)3/2 + T (ξ∗)1/2) ≤ C11e

−sα∗sT (ξ∗)3/2,

para cierta constante C11 > 0 independiente de λ; aquı se ha elegido s ≥ s2T8. Con esto,

podemos acotar el ultimo termino de (A.12). Teniendo en cuenta (A.11), obtenemos:∫∫

Qe−2sα|∇π|2 dx dt ≤ C12

(s

∫∫


+s‖a‖2∞

∫∫

Qe−2sαξ|∇ϕ|2 dx dt + s5/2T 2

∫∫

Qe−2sαξ3|ϕ|2 dx dt

+s2λ2

∫∫

ω2×(0,T )e−2sαξ2|π|2 dx dt

)(A.13)

para todo λ ≥ λ0 y todo s ≥ s3T8.

Ahora, combinamos esta desigualdad con (A.7) y llegamos a que

I(s, λ; ϕ) ≤ C3

(s3λ4

∫∫


+s2λ2

∫∫

ω2×(0,T )e−2sαξ2|π|2 dx dt + s

∫∫

Qe−2sαξ|g|2 dx dt ,

+s5/2T 2

∫∫

Qe−2sαξ3|ϕ|2 dx dt + s‖a‖2

∞

∫∫

Qe−2sαξ|∇ϕ|2 dx dt

)(A.14)

para λ ≥ λ3(1 + ‖a‖∞) y s ≥ s3(T 7 + T 8).A continuacion, absorbemos los dos ultimos terminos de (A.14) tomando s ≥ s4T

4 y λ ≥λ4‖a‖∞ de modo que

C3s5/2T 2 ≤ 1

2s3, C3‖a‖2

∞ ≤ 12λ2.

Por tanto, obtenemos la desigualdad

I(s, λ; ϕ) ≤ C4

(s3λ4

∫∫


+s2λ2

∫∫

ω2×(0,T )e−2sαξ2|π|2 dx dt + s

∫∫


) (A.15)

168

para λ ≥ λ5(1 + ‖a‖∞) y s ≥ s5(T 4 + T 8).

Como anunciamos al comienzo, el resto de la demostracion tiene como objetivo la eliminaciondel termino local de la presion que aparece a la derecha de (A.15). Podemos mencionar dosgrandes dificultades para llevar a cabo esta tarea: el hecho de estimar localmente la presion deun problema de Stokes como (A.1) en funcion de terminos locales que solo dependen del campode velocidades es bastante delicado; tambien lo es que tengamos que tratar con funciones pesoque dependen de la variable x.

Por consiguiente, en primer lugar reemplazaremos la funcion peso por otra que solo dependade t. Esto nos permitira reducirnos a estimar una integral de |∇π|2 en vez de |π|2. Entonces,teniendo en cuenta la ecuacion que verifican ϕ y π, el objetivo sera acotar integrales locales de|∆ϕ|2 y |ϕt|2.

De hecho, las definiciones de α, ξ y θ (vease (A.3)) dan directamente

s2λ2

∫∫

ω2×(0,T )e−2sαξ2|π|2 dx dt ≤

∫∫

ω2×(0,T )|θ|2|π|2 dx dt.

Podemos tomar π(t) cumpliendo ∫

ω2

π(t) dx = 0

para cada t ∈ (0, T ). Ası, usando la desigualdad de Poincare-Wirtinger, sabemos que existeC5 > 0 tal que ∫∫

ω2×(0,T )|θ|2|π|2 dx dt ≤ C5

∫∫

ω2×(0,T )|θ|2|∇π|2 dx dt.

De la primera ecuacion de (A.1), deducimos por tanto que

s2λ2

∫∫

ω2×(0,T )e−2sαξ2|π|2 dx dt ≤ C6

(∫∫

ω2×(0,T )|θ|2|g|2 dx dt

+∫∫

ω2×(0,T )|θ|2|∆ϕ|2 dx dt +

∫∫

ω2×(0,T )|θ|2|ϕt|2 dx dt

+‖a‖2∞

∫∫

ω2×(0,T )|θ|2|∇ϕ|2 dx dt

).

(A.16)

Estimacion local de |∆ϕ|2En este apartado vamos a indicar como se puede acotar el segundo termino a la derecha de

(A.16). Sea ω3 un abierto no vacıo tal que

ω2 ⊂⊂ ω3 ⊂⊂ ω

y sea ρ ∈ D(ω3) una funcion verificando ρ ≡ 1 en ω2.Pongamos

u(x, t) = θ(t)ρ(x)∆ϕ(x, t) en RN × (0, T ).

169

Aquı, hemos supuesto que u se extiende por cero fuera de ω3. El objetivo sera acotar la integral∫∫

ω2×(0,T )|u|2 dx dt.

Veamos que ecuacion del calor verifica u. Para ello, aplicamos el operador de Laplace a laecuacion satisfecha por ϕ y tenemos en cuenta (A.8); obtenemos

(∆ϕ)t −∆(∆ϕ) = f en Q, (A.17)

dondef = ∆(Dϕ a) + ∆g −∇(∇ · (Dϕ a))−∇(∇ · g).

A partir de (A.17), deducimos

ut −∆u = F en RN × (0, T ),

u(0) = 0 en RN ,(A.18)

conF = θρf + θ ′ρ∆ϕ− 2θ∇ρ · ∇∆ϕ− θ∆ρ∆ϕ.

Observese que F ∈ L2(0, T ; H−2(RN )N ) y que a priori sabemos que u ∈ L2(RN × (0, T ))N

(Observacion 2). De la propia ecuacion de (A.18), deducimos que ut ∈ L2(0, T ; H−2(RN )N ),luego u(0) tiene sentido. Ademas, es posible probar que el sistema (A.18) posee una unicasolucion en esta clase.

Ahora, reescribimos F como la suma de derivadas segundas de g, Dϕy y ϕ mas el resto determinos. En concreto, escribimos F = F1 + F2, con

F1 = θ∆(ρ(Dϕ a)) + θ∆(ρg)− θ∇(∇ · (ρ(Dϕ a)))

−θ∇(∇ · (ρ g)) + θ ′∆(ρϕ)

yF2 = −2θ∇ρ · ∇(Dϕa)− θ∆ρ(Dϕ a)− 2θ∇ρ · ∇g

−θ∆ρg + θ∇(∇ρ · (Dϕa)) + θ∇ρ(∇ · (Dϕ a))

θ∇(∇ρ · g) + θ∇ρ(∇ · g)− 2θ ′∇ρ · ∇ϕ

−θ ′∆ρϕ− 2θ∇ρ · ∇∆ϕ− θ∆ρ∆ϕ.

Notese que F1 ∈ L2(0, T ; H−2(RN )N ), mientras que F2 ∈ L2(0, T ;H−1(RN )N ). A conti-nuacion, definiremos y estimaremos dos funciones u1 y u2 en L2(RN × (0, T ))N cumpliendo

ui

t −∆ui = Fi en RN × (0, T ),

ui(0) = 0 en RN(A.19)

para i = 1, 2. Por tanto, bastara con estimar las integrales∫∫

ω2×(0,T )|ui|2 dx dt.

170

Para la primera de ellas, definimos u1 ∈ L2(RN × (0, T )) como la solucion por transposicionde (A.19) para i = 1. En otras palabras, u1 es la unica funcion de L2(RN × (0, T )) que verificapara cada h ∈ L2(RN × (0, T )) la igualdad

∫∫

RN×(0,T )u1 · h dx dt

=∫∫

RN×(0,T )(θρ (g + Dϕ a)) ·∆z) dx dt

−∫∫

RN×(0,T )θρ(Dϕ a) · ∇(∇ · z) dx dt

−∫∫

RN×(0,T )θρg · ∇(∇ · z) dx dt

+∫∫

RN×(0,T )θ ′ρϕ ·∆z dx dt,

(A.20)

donde z es la solucion de zt −∆z = h en RN × (0, T ),

z(0) = 0 en RN .(A.21)

A partir de la definicion, no es difıcil obtener la estimacion L2 de u1 (para mas detalles,vease [2]): ∫∫

ω2×(0,T )|u1|2 dx dt ≤

∫∫

RN×(0,T )|u1|2 dx dt

≤ C1

(∫∫

ω3×(0,T )|θg|2 dx dt +

∫∫

ω3×(0,T )|θ ′ϕ|2 dx dt

+∫∫

ω3×(0,T )|θDϕ a|2 dx dt

),

(A.22)

con C1(ω) > 0.

Para realizar la estimacion de u2, basta definirla como la solucion debil de (A.19) para i = 2(puesto que F2 ∈ L2(0, T ;H−1(RN ))). Directamente, deducimos

∫∫

ω2×(0,T )(|u2|2 + |∇u2|2) dx dt ≤ C2

(∫∫

ω3×(0,T )|θ ′ϕ|2 dx dt

+∫∫

ω3×(0,T )|θ|2(|g|2 + |Dϕ a|2 + |ϕ|2) dx dt

).

(A.23)

Naturalmente, esto tambien es consecuencia de que u2 es la solucion de (A.19) en el sentido delsemigrupo asociado a la ecuacion del calor en RN ; para mas detalles, vease [2].

171

En definitiva, combinando esto con (A.22), llegamos a que

∫∫

ω2×(0,T )|θ|2|∆ϕ|2 dx dt ≤ C3

(∫∫

ω3×(0,T )|θ ′|2|ϕ|2 dx dt

+∫∫

ω3×(0,T )|θ|2(|g|2 + |Dϕa|2 + |ϕ|2) dx dt

).

(A.24)

Estimacion local de |ϕt|2Este paragrafo contiene la parte mas complicada de la prueba. Resumiremos el argumento

que esta desarrollado en su completitud en [2].Sean (ψ1, q1) y (ψ2, q2) las soluciones respectivas de

−ψ1,t −∆ψ1 −Dψ1 a +∇q1 = θg en Q,

∇ · ψ1 = 0 en Q,

ψ1 = 0 sobre Σ,

ψ1(T ) = 0 en Ω

(A.25)

y

−ψ2,t −∆ψ2 −Dψ2 a +∇q2 = −θ′ϕ en Q,

∇ · ψ2 = 0 en Q,

ψ2 = 0 sobre Σ,

ψ2(T ) = 0 en Ω

(A.26)

(θ se definio en (A.3)). Por la unicidad del sistema de Stokes, tenemos que

θϕ = ψ1 + ψ2 y θπ = q1 + q2.

Luego debemos acotar el termino∫∫

ω2×(0,T )|θ|2|ϕt|2 dx dt =

∫∫

ω2×(0,T )θ−2|θ|2|θ ϕt|2 dx dt

= s−11/2λ2

∫∫

ω2×(0,T )e−2sα∗+2sαξ−11/2|ψ1,t + ψ2,t − θ′ϕ|2 dx dt.

(A.27)

A continuacion, acotamos las integrales locales asociadas a ψ1,t y ψ2,t.

La estimacion del termino e−2sα∗+2sαξ−11/2|ψ1,t|2 se hara en realidad en Ω× (0, T ) y sin laayuda de la funcion peso. En concreto, es facil ver que la funcion peso

e−2sα∗+2sαξ−11/2

esta uniformemente acotada en (0, T ). Ahora, aplicamos estimaciones de regularidad para elsistema de Stokes (vease, por ejemplo, [8]) a (A.25) y deducimos que

ψ1 ∈ L2(0, T ; H2(Ω)N ) ∩H1(0, T ;L2(Ω)N )

172

y‖ψ1,t‖2

L2(Q)N + ‖ψ1‖2L2(0,T ;H2(Ω)N ≤ C∗

1 (1 + ‖a‖2∞eC∗2T‖a‖2∞)‖θg‖2

L2(Q)N .

Luego, tenemos

s−11/2λ2

∫∫

ω2×(0,T )e−2sα∗+2sαξ−11/2|ψ1,t|2 dx dt

≤ C∗3 (1 + ‖a‖2

∞eC∗2T‖a‖2∞)∫∫

Q|θ|2|g|2 dx dt.

(A.28)

En lo que se refiere a la estimacion de ψ2,t, indiquemos que en el transcurso de la pruebanecesitaremos acotar la segunda derivada en tiempo de ψ2. Esto sera posible bajo las hipotesisde regularidad sobre a impuestas en (A.2).

En concreto, integrando por partes dos veces con respecto a t, obtenemos

s−11/2λ2

∫∫

ω2×(0,T )e−2sα∗+2sαξ−11/2|ψ2,t|2 dx dt

=12s−11/2λ2

∫∫

ω2×(0,T )(e−2sα∗+2sαξ−11/2)tt|ψ2|2 dx dt

−s−11/2λ2

∫∫

ω2×(0,T )e−2sα∗+2sαξ−11/2ψ2,tt · ψ2 dx dt.

(A.29)

De nuevo, no es difıcil ver que la funcion peso

(s−11/2λ2e−2sα∗+2sαξ−11/2)tt

esta acotada uniformemente.Introduzcamos la funcion

θ∗ = s−11/2λ−4e−2sα∗+2sαξ−11/2

y acotemos el segundo termino de la derecha de (A.29). Usando la desigualdad de Holder,deducimos que

−λ6

∫∫

ω2×(0,T )θ∗ψ2,ttψ2 dx dt

≤ λ6‖θ∗ψ2,tt‖L2(0,T ;Lr(ω2)N )‖ψ2‖L2(0,T ;Lr′ (ω2)N )

≤ 12‖θ∗ψ2,tt‖2

L2(0,T ;Lr(ω2)N ) +12λ12‖ψ2‖2

L2(0,T ;Lr′ (ω2)N ),

(A.30)

donde 6/5 < r < σ si N = 3 y 1 < r < σ si N = 2 (σ se definio en (A.2)).Acotemos en primer lugar el ultimo termino de (A.30). Sea ζ ∈ C2(ω3) tal que

supp ζ ⊂ ω3 y ζ = 1 en ω2.

173

Entonces,‖ψ2‖2

L2(0,T ;Lr′ (ω2)N )≤ C∗

4‖∆(ζψ2)‖2L2(0,T ;L2(ω3)N )

= C∗4‖ψ2∆ζ + 2∇ζ · ∇ψ2 + ζ∆ψ2‖2

L2(0,T ;L2(ω3)N ),

pues H2(ω3)N ∩H10 (ω3)N se inyecta continuamente en Lr′(ω3)N , para todo r′ < ∞.

En este punto, podemos aplicar la estimacion local obtenida en el paragrafo anterior a‖ζ∆ψ2‖2

L2(0,T ;L2(ω3)N ). De hecho, como ψ2(T ) = 0, basta aplicar la desigualdad (A.24) con

ϕ = ψ2, θ = 1 y g = −θ′ϕ. Esto da

‖ψ2‖2L2(0,T ;Lr′ (ω2)N )

≤ C∗5 (1 + ‖a‖2

∞)∫∫

ω3×(0,T )(|ψ2|2 + |∇ψ2|2 + |θ′ϕ|2) dx dt.

De las definiciones de ψ1 y ψ2, tambien deducimos que

|ψ2|2 + |∇ψ2|2 ≤ 2(|ψ1|2 + |∇ψ1|2 + |θ|2(|ϕ|2 + |∇ϕ|2)).

Como consecuencia, si vemos ψ1 como la solucion debil de (A.25) y usamos estimaciones globales,llegamos a

‖ψ2‖2L2(0,T ;Lr′ (ω2)N )

≤ C∗6 (1 + ‖a‖2

∞)

(eC∗7T‖a‖2∞‖θg‖2

L2(Q)N

+∫∫

ω3×(0,T )(|θ|2 + |θ′|2)|ϕ|2 dx dt +

∫∫

ω3×(0,T )|θ|2|∇ϕ|2 dx dt

).

(A.31)

Por ultimo, digamos como se puede obtener una cota del termino de ψ2,tt en (A.30).Consideramos el par (ψ, q) := (θ∗ψ2,t, θ

∗q2,t). Se puede probar (por ejemplo, usando unargumento de densidad sobre a) que este par verifica

−ψt −∆ψ −Dψ a +∇q = G en Q,

∇ · ψ = 0 en Q,

ψ = 0 sobre Σ,

ψ(T ) = 0 en Ω,

(A.32)

dondeG = −θ∗θ′′ϕ− θ∗θ′ϕt + θ∗Dψ2 at − (θ∗)′ψ2,t.

Con el objetivo de estimar ψt en L2(0, T ; Lr(Ω)N ), primero acotamos el termino de transporte(Dψ a) en el mismo espacio. De hecho, si miramos ψ como la solucion debil de (A.32), deducimosque ψ ∈ L2(0, T ; V ) y

‖ψ‖L2(0,T ;V ) ≤ C∗8eC∗9T‖a‖2∞‖G‖L2(0,T ;H−1(Ω)N ), (A.33)

174

luego tenemos la misma cota para ‖Dψ‖L2(Q)N .Tras diversas estimaciones de cierta complejidad, se puede demostrar que

θ∗Dψ2 at ∈ L2(0, T ; Lr(Ω)N )

y que

‖θ∗Dψ2 at‖L2(0,T ;Lr(Ω)N )

≤ C∗10 ‖at‖L2(0,T ;Lσ(Ω)N ) T−1/2+1/k′3(1 + ‖a‖2

∞)(‖θ∗∆ψ2‖L2(Q)N

+‖(θ∗)′∆ψ2‖L2(Q)N + ‖(θ∗ψ2)t‖L2(Q)N + ‖((θ∗)′ψ2)t‖L2(Q)N

+‖θ∗θ′∆ϕ‖L2(Q)N + ‖(θ∗θ′ϕ)t‖L2(Q)N ).

(A.34)

La prueba detallada de esto se encuentra en la etapa 6 de la prueba del teorema 1 de [2].Bajo estas condiciones, podemos aplicar el teorema 2.8 de [5] para deducir que ψt ∈

L2(0, T ;Lr(Ω)N ), junto con la estimacion

‖ψt‖L2(0,T ;Lr(Ω)N ) ≤ C∗11‖G‖L2(0,T ;Lr(Ω)N ) (A.35)

para cierta constante C∗11 > 0 que depende de Ω pero no de T . Como Lr(Ω) se inyecta continua-

mente en H−1(Ω), de (A.33) y (A.35) deducimos que

‖θ∗ψ2,tt‖L2(0,T ;Lr(Ω)N ) ≤ C∗12(1 + ‖a‖∞)eC∗13T‖a‖2∞(‖θ∗θ′′ϕ‖L2(Q)N

+‖θ∗θ′ϕt‖L2(Q)N + ‖(θ∗)′ψ2,t‖L2(Q)N + ‖θ∗Dψ2 at‖L2(0,T ;Lr(Ω)N )).

Usando (A.29)–(A.31) y esta ultima desigualdad, tenemos

s−11/2λ2

∫∫

ω2×(0,T )e−2sα∗+2sαξ−11/2|ψ2,t|2 dx dt

≤ C∗14(1 + ‖a‖2

∞)eC∗15T‖a‖2∞(λ12(‖θg‖2

L2(Q)N

+‖θϕ‖2L2(0,T ;L2(ω5)N ) + ‖θ′ϕ‖2

L2(0,T ;L2(ω5)N ) + ‖θ∇ϕ‖2L2(0,T ;L2(ω5)N ))

+‖θ∗θ′′ϕ‖2L2(Q)N + ‖θ∗θ′ϕt‖2

L2(Q)N + ‖(θ∗)′ψ2,t‖2L2(Q)N

+‖θ∗Dψ2 at‖2L2(0,T ;Lr(Ω)N )

).

(A.36)

Para obtener la estimacion de la derivada en tiempo de ϕ, combinamos (A.27), (A.28) y(A.36) y obtenemos que

∫∫

ω2×(0,T )|θ|2|ϕt|2 dx dt

≤ C∗16(1 + ‖a‖2

∞)eC∗17T‖a‖2∞(λ12(‖θg‖2

L2(Q)N

+‖θϕ‖2L2(0,T ;L2(ω5)N ) + ‖θ′ϕ‖2

L2(0,T ;L2(ω5)N ) + ‖θ∇ϕ‖2L2(0,T ;L2(ω5)N ))

+‖θ∗θ′′ϕ‖2L2(Q)N + ‖θ∗θ′ϕt‖2

L2(Q)N + ‖(θ∗)′ψ2,t‖2L2(Q)N

+‖θ∗Dψ2 at‖2L2(0,T ;Lr(Ω)N )

).

(A.37)

175

Para acabar, en primer lugar usamos la expresion de ψ2 = −ψ1 + θϕ junto con la estimacion(A.28) para ψ1. Esto nos lleva a la desigualdad

∫∫

ω2×(0,T )|θ|2|ϕt|2 dx dt

≤ C∗18(1 + ‖a‖6

∞)‖at‖2L2(0,T ;Lσ(Ω)N )e

C∗19T‖a‖2∞(λ12(‖θ g‖2

L2(Q)N + ‖θϕ‖2L2(0,T ;L2(ω5)N )

+‖θ′ϕ‖2L2(0,T ;L2(ω5)N ) + ‖θ∇ϕ‖2

L2(0,T ;L2(ω5)N ))

+(1 + T 1/2)(‖θ∗θ′ϕ‖2L2(Q)N + ‖(θ∗)′θ′ϕ‖2

L2(Q)N

+‖(θ∗)′′θϕ‖2L2(Q)N + ‖θ∗θ′′ϕ‖2

L2(Q)N + ‖θ∗θ∆ϕ‖2L2(Q)N

+‖θ∗θ′∆ϕ‖L2(Q)N + ‖(θ∗)′θ∆ϕ‖2L2(Q)N + ‖θ∗θϕt‖2

L2(Q)N

+‖θ∗θ′ϕt‖2L2(Q)N + ‖(θ∗)′θϕt‖2

L2(Q)N ))

.

(A.38)

Teniendo en cuenta la definicion de las funciones peso (vease A.3) y haciendo una buenaeleccion de los parametros s y λ, obtenemos

C4C6

∫∫

ω2×(0,T )|θ|2|ϕt|2 dx dt

≤ C∗20λ

20(‖θg‖2L2(Q)N + ‖θϕ‖2

L2(0,T ;L2(ω5)N ) + ‖θϕ‖2L2(0,T ;L2(ω5)N )

+‖θ∇ϕ‖2L2(0,T ;L2(ω5)N )) +

12I(s, λ; ϕ).

Finalmente, combinamos esta desigualdad con (A.15), (A.16) y (A.24) y llegamos a que

I(s, λ;ϕ) ≤ C7λ20

(‖θg‖2

L2(Q)N + ‖θϕ‖2L2(0,T ;L2(ω5)N )

+‖θϕ‖2L2(0,T ;L2(ω5)N ) + ‖θ∇ϕ‖2

L2(0,T ;L2(ω5)N )

),

(A.39)

de donde deducimos (A.4) casi directamente.

Bibliografıa

[1] E. Fernandez-Cara, S. Guerrero, Global Carleman inequalities for parabolic systemsand applications to null controllability, aparecera en SIAM J. Control Optimization.

[2] E. Fernandez-Cara, S. Guerrero, O. Yu. Imanuvilov y J.-P. Puel, Local nullcontrollability of the Navier-Stokes system, Journal de Mathematiques Pures et Appliquees,Vol 83/12, pp 1501-1542.

[3] E. Fernandez-Cara, S. Guerrero, O. Yu. Imanuvilov and J.-P. Puel, Some con-trollability results for the N -dimensional Navier-Stokes and Boussinesq systems with N − 1scalar controls, enviado a SIAM J. Control Optimization.


[5] Y. Giga, H. Sohr, Abstract Lp estimates for the Cauchy problem with applications to theNavier-Stokes equations in exterior domains, Journal of Functional Analysis, 102 (1991),72–94.


[7] O. Yu. Imanuvilov, J.-P. Puel, Global Carleman estimates for weak elliptic non homo-geneous Dirichlet problem, Int. Math. Research Notices, 16 (2003), 883-913.

[8] R. Temam, Navier-Stokes equations. Theory and numerical analysis. Studies in Mathemat-ics and its applications, 2; North Holland Publishing Co., Amsterdam-New York-Oxford(1977).

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