Estimativas de Carleman para uma classe de problemas parabólicos degenerados e aplicações à controlabilidade multi-objetivo

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1 Universidade Federal da Paraíba Universidade Federal de Campina Grande Programa Associado de Pós-Graduação em Matemática Doutorado em Matemática Estimativas de Carleman para uma classe de problemas parabólicos degenerados e aplicações à controlabilidade multi-objetivo por Bruno Sérgio Vasconcelos de Araújo João Pessoa - PB Junho/217

2 Estimativas de Carleman para uma classe de problemas parabólicos degenerados e aplicações à controlabilidade multi-objetivo por Bruno Sérgio Vasconcelos de Araújo sob orientação do Prof. Dr. Fágner Dias Araruna sob co-orientação do Prof. Dr. Enrique Fernandez-Cara Tese apresentada ao Corpo Docente do Programa Associado de Pós-Graduação em Matemática - UFPB/UFCG, como requisito parcial para obtenção do título de Doutor em Matemática. João Pessoa - PB Junho/217 Este trabalho contou com apoio nanceiro da CAPES ii

3 Universidade Federal da Paraíba Universidade Federal de Campina Grande Programa Associado de Pós-Graduação em Matemática Doutorado em Matemática Área de Concentração: Análise Aprovada em: Prof. Dr. Fágner Dias Araruna (Orientador) UFPB Prof. Dr. Enrique Fernandez-Cara (Co-Orientador) US Prof. Dr. Diego Araújo de Souza UFPE Prof. Dr. Manuel Antolino Milla Miranda UEPB Prof. Claudianor Oliveira Alves UFCG Prof. Dr. Pablo Gustavo Albuquerque Braz e Silva UFPE Prof. Marco Aurélio Soares Souto (Suplente) UFCG Prof. Maurício Cardoso Santos (Suplente) UFPE Tese apresentada ao Corpo Docente do Programa Associado de Pós-Graduação em Matemática - UFPB/UFCG, como requisito parcial para obtenção do título de Doutor em Matemática. Junho/217 iii

4 Resumo Neste trabalho apresentamos estimativas de Carleman para uma classe de problemas parabólicos degenerados sobre um quadrado (no caso bidimensional) ou sobre um intervalo limitado (no caso unidimensional). Consideramos um operador diferencial que degenera apenas em uma parte da fronteira. Provamos resultados de existência, unicidade e estimativas de energia via teoria do semigrupo. Em seguida usamos funções peso adequadas para obter estimativas de Carleman e, como aplicações, resultados de controlabilidade multi-objetivo. Palavras-chave: Controle nulo de Stackelberg-Nash; Desigualdade de Carleman; Observabilidade; Equações com coeciente degenerados. iv

5 Abstract This work presents Carleman estimates to a class of degenerate parabolic problems over a square (in the two dimensional case) or a bounded interval (in the one dimensional case). We consider a dierential operator that degenerate only in a part of the boundary. Using semigroup theory, we prove well posedness results. Then, using suitables weight functions, we prove Carleman estimates and, as application, results on multi-objective controllability. Keywords: Stackelberg-Nash null controllability; Carleman estimates; Observability; Degenerate equations. v

6 Agradecimentos O primeiro trabalho cientíco é um grande desao. Dado o desenvolvimento da ciência nos dias atuais, não é nada fácil extrapolar seus limites. Mas quando se tem uma gama de agentes apoiando, guiando o estudante, a meta deixa de parecer tão distante. Por ter me conduzido até aqui, não ter me deixado cair uma queda que eu não pudesse me levantar, me ajudado a sempre se erguer, agradeço ao Criador da vida: Deus. Por toda a paciência e por terem acreditado no meu trabalho, agradeço aos meus orientadores, Fagner e Enrique. Me sinto muito honrado por ter tido a oportunidade de trabalhar com estes pesquisadores tão talentosos. Agradeço também a toda minha família pelo apoio que me foi dado. Finalizo esta nota agradecendo a Capes pelo apoio nanceiro. vi

7 Se o conhecimento pode criar problemas, não é através da ignorância que podemos solucioná-los. Isaac Asimov vii

8 Dedicatória Para Elon Lages Lima, o grande escritor de livros matemáticos brasileiros. viii

9 Sumário Introdução Controlabilidade nula de equações parabólicas degeneradas Controlabilidade nula de equações parabólicas degeneradas Controlabilidade multi-objetivo Contribuições e organização do trabalho Controlabilidade nula de Stackelberg-Nash para algumas equações parabólicas degeneradas lineares e semilineares Introduction Notations and preliminary results Notations and spaces Existence and estimates Carleman estimates for (1.6) and (1.7) Carleman estimate for (1.8) Carleman estimate for (1.12) Stackelberg-Nash null controllability Proof of Theorem Similar results for semilinear problems Additional comments and open questions On the assuption O 1,d = O 2,d Higher dimensions Wave equations Appendix A: Proof of Theorem Appendix B: Proof of Theorem

10 1.7 Appendix C: Proof of Theorem Estimativas de Carleman para algumas EDPs parabólicas degeneradas em dimensão 2 e aplicações Introduction Preliminar results and well-posedness Notations, spaces and operators Trace operators Well posedness Carleman estimates and null controllability results Application to null controllability Stackelberg-Nash null controllability Proof of Theorem Further extensions and open questions On systems with gradients On other degenerate operators On spatial dimension On the assumptions on ω Appendix A: Proof of Theorem Estimativas de Carleman para alguns operadores parabólicos degenerados em dimensões superiores Introduction Notations, spaces and preliminaries results Carleman estimates and null controllability results Application to null controllability Extension to higher dimensions Extension to more general degenerate operators Open questions on others extensions On systems with gradient On semilinear problems On others degenerate operators On assumptions on the observations domains x

11 3.8 Appendix A: Proof of Theorem Appendix B: Scketch of the proof of the Theorem Referências 91 xi

12 Introdução.1 Controlabilidade nula de equações parabólicas degeneradas Com o desenvolvimento do Calculo Diferencial e Integral em meados do século XVII, a ciência provocou um grande impacto na sociedade. Desde então, um sem número de fenômenos foram analisados, com resultados que permitiram avanços tecnológicos sem precedentes na história da humanidade. Dentre as principais ferramentas responsáveis por tamanho progresso, destacam-se as equações diferenciais. Capazes de modelar diversos problemas, a resolução dessas equações permitiam prever o comportamento futuro de várias variáveis. Infelizmente (ou felizmente) muitas destas equações são tão difíceis de se resolver, que mesmo 4 séculos depois do inicio do Cálculo, ainda existem inúmeras equações passíveis de análises. Não por menos, o campo da matemática que se dedica a encontrar soluções aproximadas destas equações se desenvolveu tanto no último século. Certamente, depois de se prever o comportamento de um determinado fenômeno, um passo seguinte é inuenciá-lo. É nesse sentido que atua a teoria de controle: atuar e inuenciar o comportamento de certas variáveis em tais fenômenos. Um sistema de controle é uma equação de evolução (EDO ou EDP) que depende de um parâmetro u, que escreveremos da seguinte forma: y = f(t, y, u), onde t [, T é o tempo, y : [, T Y é a função estado, u : [, T U é o controle e Y e U são espaços de funções adequados. Na equação acima, y representa a derivada

13 de y em relação ao tempo t. O problema de controle consiste em encontrar um controle u tal que a função estado se comporta de uma forma desejada. Exemplicaremos alguns, dentre os vários, problemas de controlabilidade presentes na literatura. Controle Ótimo: Encontrar um controle que minimiza algum funcional custo, por exemplo, J(u) = y(t ; u) ȳ 2 U + u 2 U, em que ȳ é um alvo desejado e y(t ; u) é o estado alcançado pelo sistema no tempo nal T. Controlabilidade Exata: Dado dois tempos T < T 1 e y, y 1 dois possíveis estados do sistema, encontrar u : [T, T 1 U tal que y = f(y, u) em [T, T 1 y(t ) = y, y(t 1 ) = y 1. Em outras palavras, partindo de qualquer conguração inicial y, podemos conduzir a solução y para o estado y 1 sob a ação do controle u. Controlabilidade Aproximada: Dados T < T 1, dois possíveis estados y, y 1 e ɛ >, encontrar u : [T, T 1 U tal que y = f(y, u) em [T, T 1 y(t ) = y, y(t 1 ) y 1 U < ɛ. A controlabilidade aproximada é uma versão mais fraca se comparada a controlabilidade exata. De fato, em vez de pedirmos que a função estado seja exatamente y 1 em T 1, pedimos apenas que o estado esteja arbitrariamente perto de y 1. Controlabilidade Nula: Dados dois tempos T < T 1 e y um estado do sistema, encontrar u : [T, T 1 U tal que y = f(y, u) em [T, T 1 y(t ) = y, y(t 1 ) =. Controlabilidade Exata para as Trajetórias: Dados T < T 1, y Y e ȳ uma trajetória (uma solução com controle ū : [T, T 1 U), encontrar um controle u : [T, T 1 U tal que y = f(y, u) em [T, T 1 y(t ) = y, y(t 1 ) = ȳ(t 1 ). 2

14 Os conceitos de controlabilidade nula e controlabilidade exata para as trajetórias são de especial importância em sistemas não reversíveis e sistemas com efeito regularizante. Nestes casos, a controlabilidade exata não é esperada..2 Controlabilidade nula de equações parabólicas degeneradas O estudo da controlabilidade de equações diferenciais parciais atraiu o interesse de vários cientistas nas últimas décadas. Diversos resultados foram desenvolvidos sobre problemas semi-lineares, problemas em domínios ilimitados, sistemas de dinâmica dos uidos entre outros. Nessas direções, alguns trabalhos notáveis são [15, 18, 2, 21, 24. Por outro lado, no caso particular de equações parabólicas degeneradas, ainda pouco se sabe, veja [7, 19, 27. Primeiramente, consideremos o modelo mais básico de equação parabólica degenerada estudado na última década: u t (x α u x ) x + b (x, t)u = g1 O em (, 1) (, T ), u(, ) = se α (, 1) u(1, ) = e em (, T ), (x α u x )(, ) = se α [1, 2) u(, ) = u em (, 1), onde α (, 2), ω (, 1) é um aberto e 1 ω sua função característica associada, T >, b L (, 1), g L 2 (ω (, T )) e u L 2 (Ω). Dizemos que (1) é nulamente controlável quando dado u L 2 (, 1), existe g L 2 ((, 1) (, T )) tal que a solução u de (1) satisfaz (1) u(, T ) =. (2) Na década de 9, no trabalho [2, o método HUM começa a ser popularizado e se consagra como o principal método para provar a controlabilidade nula de equações diferenciais parciais. Tal método consiste em reduzir o problema da controlabilidade nula ao problema de obter uma certa desigualdade para o estado adjunto do sistema original. Tal desigualdade, chamada de desigualdade de observabilidade, viria a se tornar na década seguinte, o principal método para provar a controlabilidade nula de problemas parabólicos degenerados. Mais especicamente, em [7, os autores consideraram o 3

15 sistema adjunto de (1), isto é, v t + (x α v x ) x + b (x, t)v = h em (, 1) (, T ), v(, ) = se α (, 1) v(1, ) = e em (, T ), (x α v x )(, ) = se α [1, 2) v(, T ) = v T em (, 1). e provaram a seguinte desigualdade de observabilidade para (3): (3) Proposition.2.1 Sejam α (, 2) e T > dados e seja ω um subintervalo aberto e não vazio de (, 1). Então existe C > tal que, para todo v T L 2 (, 1), a solução v de (3) satisfaz 1 x α v x (, x) 2 dx C ω (,T ) v(x, t) 2 dx dt. (4) Desde sua popularização na década de 9, a principal ferramenta usada para obter desigualdades de observabilidade tem sido as famigeradas Desigualdades de Carleman. Na década seguinte, com as EDPs degeneradas, não foi diferente, apesar exigirem outras ferramentas adicionais. Em [7 os autores provaram a seguinte desigualdade de Carleman: Proposition.2.2 Assuma α (, 2). Existem constantes positivas s e C tais que, para qualquer s s e qualquer solução v de (3), vale: e [ 2sσ sθx α v x 2 + s 3 θ 3 x 2 α v 2 [ dx dt C e sσ h 2 + s T e 2sσ θ v x 2 dt. (5) x=1 A inclusão de um termo de primeira ordem espacial na equação de (1) é uma questão delicada que ainda não foi totalmente solucionada. Em [19 os autores estenderam os resultados de [7 para o seguinte problema u t (x α u x ) x + (x q b 1 (x, t)u) x + b (x, t)u = g1 O em (, 1) (, T ), u(, ) = se α (, 1) u(1, ) = and em (, T ), (x α u x )(, ) = se α [1, 2) u(, ) = u em (, 1), onde q α/2 e b 1 L (, 1). Neste caso a Proposição.2.2 ainda vale para o sistema adjunto: v t + (x α v x ) x + x q b 1 v x + b v = h em (, 1) (, T ), v(, ) = se α (, 1) v(1, ) = and em (, T ), (x α v x )(, ) = se α [1, 2) v(, ) = v em (, 1). 4 (6) (7)

16 A controlabilidade nula de (6) sem o peso x q, isto é, do problema u t (x α u x ) x + (b 1 (x, t)u) x + b (x, t)u = g1 O em (, 1) (, T ), u(, ) = se α (, 1) u(1, ) = and em (, T ), (x α u x )(, ) = se α [1, 2) u(, ) = u em (, 1), (8) até onde sabemos, permanece aberta. Em [27, considerando α (, 1/2), os autores zeram modicações nas já clássicas funções peso introduzidas em [7 e conseguiram provar estimativas de Carleman semelhantes para o problema adjunto: v t + (x α v x ) x + b 1 v x + b v = h em (, 1) (, T ), v(, ) = se α (, 1) v(1, ) = and em (, T ), (x α v x )(, ) = se α [1, 2) v(, ) = v em (, 1). Com isso, a controlabilidade nula de (8) foi estabelecida para α (, 1/2). O caso α = 2 é interessante do ponto de vista das aplicações, pois a equação de (6), com α = 2, tem como caso particular a célebre equação de Black Scholes [6, que modela o preço de opções de compra de ativos nanceiros. Porém, já se sabe desde [7, que o problema (1) com α = 2, isto é, o problema u t (x 2 u x ) x + b (x, t)u = g1 O em (, 1) (, T ), u(1, ) = e (x 2 u x )(, ) = em (, T ), u(, ) = u em (, 1), não é, em geral, nulamente controlável. Passando a dimensões espaciais superiores, até onde sabemos, o único trabalho publicado é [1, onde os autores obtiveram controlabilidade nula (usando novamente estimativas de Carleman) do seguinte sistema: u t div(a u) + b u = g 1 ω em, u = if α (, 1) u ν = if sobre α [1, 2) Σ, u(, ) = u em Ω, onde := Ω (, T ), T >, Ω R 2 é um dominio limitado com fronteira Γ de classe C 4, Σ := Γ (, T ), ω Ω é aberto, u L 2 (Ω ), g L 2 ( ), b L ( ), α (, 2), e A : Ω M 2 2 (R) satisfaz as seguintes condições: 5 (9) (1) (11)

17 (i) a ij C 3 (Ω ; R) C (Ω ; R), onde A (x) = (a ij (x)); (ii) A (x) é simétrica x Ω ; (iii) A (x) é positiva denida x Ω ; (iv) Sejam r i (x), os autovalores e ε i (x) os autovetores unitários correspondentes à A (x), i = 1, 2. Denotemos por P Γ (x) a projeção de x até a fronteira Γ e O(Γ ; δ) := {x Ω : d(x, Γ ) < δ}. Existe δ > tal que 1. r 1 (x) = d(x, Γ ) α, x O(Γ ; δ), 2. r 2 (x) > x Ω \O(Γ ; δ); 3. ε 1 (x) = ν(p Γ (x)) x O(Γ ; δ). Vale salientar a discrepância entre o sistema (11), cujo operador diferencial degenera em toda a fronteira, e o sistema (1), cujo operador diferencial degenera em apenas uma parte da fronteira. Tal discrepância é fruto da diculdade de construir pesos adequados para a desigualdade de Carleman..3 Controlabilidade multi-objetivo Diferentemente dos conceitos de controlabilidade usuais ja descritos acima, a controlabilidade multi-objetivo, como o próprio nome sugere, consiste em buscar um ou mais controles que façam o estado atender mais de um requisito. Neste trabalho, temos como objetivo principal provar resultados de controlabilidade hierárquica. Neste tipo de controlabilidade o objetivo é, além de controlar o estado no instante nal T, controlar o estado também ao longo do processo evolutivo, pelo menos em uma parte do domínio. Para sermos mais claros considere agora o sistema u t u = f1 ω + v 1 1 ω1 + v 2 1 ω2 em u = sobre Σ (12) u(x, ) = u (x) em Ω, onde Ω R n é um domínio limitado com fronteira Γ suave, = Ω (, T ), Σ = Γ (, T ) e ω, ω i Ω são abertos. Fixemos abertos ω i,d Ω e funções u i,d L 2 (Ω) e 6

18 introduzimos os funcionais J i (f; v 1, v 2 ) = β i 2 u u i,d 2 dx dt + µ i ω i,d (,T ) v i 2 dx dt, (13) ω i (,T ) onde β i, µ i > são constantes xadas e u é o estado de (12) associado à terna (f; v 1, v 2 ). O problema da controlabilidade nula hierárquica consiste em buscar controles f (o líder) e (v 1, v 2 ) (os seguidores) que levem o estado u de (12) ao estado nulo em T e, ademais, mantenham o estado u o mais próximo possível dos estados u i,d ao longo de todo o processo evolutivo, pelo menos nas regiões de controle ω i,d. Em termos mais técnicos isto signica que os controles f e (v 1, v 2 ) devem ser tais que a solução de (12) satisfaz u(x, T ) = em Ω e minimizam (em um certo sentido) os funcionais J i. Diferentes formas de minimizar os funcionais J i conduzem a diferentes tipos de controlabilidade hierárquica. Denição: Dado f L 2 (), dizemos que um par (v 1, v 2 ) L 2 (ω 1 (, T ), ω 2 (, T )) é um equilíbrio de Nash associado à f se J 1 (g; f 1, f 2 ) J 2 (g; f 1, f 2 ), J 2 (g; f 1, f 2 ) J 2 (g; f 1, f 2 ), f 1 L 2 (), f 2 L 2 (). (14) Denição: Dado f L 2 (O (, T )), dizemos que um par (v 1, v 2 ) L 2 (ω 1 (, T ), ω 2 (, T )) é um equilíbrio de Pareto associado à f se, para qualquer (h 1, h 2 ) L 2 (ω 1 (, T ), ω 2 (, T )) tivermos 1. J 1 (f; h 1, h 2 ) J 1 (f; v 1, v 2 ) J 2 (f; v 1, v 2 ) J 2 (f; h 1, h 2 ), 2. J 2 (f; h 1, h 2 ) J 2 (f; v 1, v 2 ) J 1 (f; v 1, v 2 ) J 1 (f; h 1, h 2 ). (15) A principal diferença entre os conceitos de equilíbrio de Nash e Pareto é que o equilíbrio de Pareto é cooperativo, enquanto o de Nash não. Em [4, os autores provaram a existência e unicidade do equilíbrio de Nash, assim como resultados de controlabilidade nula hierárquica..4 Contribuições e organização do trabalho Este trabalho está organizado da seguinte maneira: Capítulo 1: Neste capítulo nos dedicamos inteiramente à controlabilidade nula hierárquica de Stackelberg-Nash para o sistema parabólico degenerado unidimensional. 7

19 A principal ferramenta para obter resultados dessa classe, utilizando as técnicas desenvolvidas em [4, são boas estimativas de Carleman. Boas no sentido de que, entre outras coisas, a desigualdade precisa ter derivadas temporal de primeira ordem e derivadas espaciais de segunda ordem no lado esquerdo da desigualdade. Apesar desse não ser o caso da desigualdade na Proposição.2.2, não é difícil incluir estes termos na mesma. O real problema da desigualdade na Proposição 1, reside no fato do termo de observação atuar na fronteira. Para obter a desigualdade de observabilidade com observação no interior do domínio, em [7, os autores contornaram esse inconveniente usando a desigualdade de Cacciopolli. Infelizmente essa desigualdade não ajuda para obter a observabilidade necessária que conduz a controlabilidade nula hierárquica. Dessa forma, construímos novos pesos para obter uma estimativa de Carleman com observação atuando no interior do domínio. Esta nova estimativa de Carleman se estende naturalmente ao sistema (7). Os novos pesos também podem ser alterados como em [27 para obter estimativas de Carleman para o sistema (9) no caso α (, 1/2). Com estas novas estimativas de Carleman, somos capazes de aplicar as técnicas de [4 e provar a controlabilidade nula hierarquica de Stackelberg-Nash para os sistemas (1), (6) e (8). uanto ao sistema (1), a questão é um pouco mais delicada, pois até já se sabe que o sistema sequer é, em geral, nulamente controlável. Não obstante, apresentamos hipóteses geométricas sobre o domínio de controle, sobre as quais é possível obter boas estimativas de Carleman e resultados de controlabilidade nula hierárquica de Stackelberg Nash. Capítulo 2: Neste capítulo começamos a estender os resultados do Capítulo 1 para duas dimensões espaciais. Motivados por problemas nanceiros, Consideramos o seguinte sistema u t div(a u) + bu = g1 ω em, B.C. sobre Σ, u(, ) = u em Ω, (16) onde Ω = (, 1) (, 1), Γ := Ω, T >, = Ω (, T ), Σ := Γ (, T ), ω Ω é aberto e 1 ω é a função característica, b L (), g L 2 (), u L 2 (Ω), A : Ω M 2 2 (R) é dada por A(x) = diag(x α 1 1, x α 2 2 ), 8

20 B.C. := u = sobre Σ se α 1, α 2 [, 1), u = sobre Σ 3,4 e (A u)ν = sobre Σ 1,2 se α 1, α 2 [1, 2, u = sobre Σ 1,3,4 e (A u)ν = sobre Σ 2 se α 1 [, 1) e α 2 [1, 2, u = sobre Σ 2,3,4 e (A u)ν = sobre Σ 1 se α 1 [1, 2 e α 2 [, 1), α = (α 1, α 2 ) [, 2 [, 2, Σ i,j,l := (Γ i Γ j Γ l ) (, T ), e Γ 1 := {} [, 1, Γ 2 := [, 1 {}, Γ 3 := {1} [, 1, Γ 4 := [, 1 {1}. Vale notar que, comparando com o sistema (11), o sistema (16) é uma extensão mais el para duas dimensões do sistema unidimensional (1). Neste capítulo, provamos a boa colocação do sistema (16) e, sob certas condições geométricas sobre o domínio de controle, estimativas de Carleman para o caso α 1, α 2 (, 2). Como consequência obtemos resultados de controlabilidade nula hierárquica de Stackelberg-Nash. Capítulo 3: No capítulo anterior, provamos estimativas de Carleman para o sistema (16) apenas para o caso α 1, α 2 (, 2). A técnica usada para construir os pesos necessários para a desigualdade de Carleman enfrenta diculdades técnicas nos demais casos de combinações de valores de α i. Neste capítulo propomos outra técnica, com outros pesos, para contemplar as demais combinações de valores de α i. Dessa forma, no caso em que α 1 = ou α 2 =, obtemos resultados com hipóteses geométricas menos restritivas, enquanto que nos demais casos precisamos de hipóteses mais restritivas às consideradas no Capítulo 2. Ademais, estendemos os resultados para sistemas em dimensões espaciais superiores. 9

21 Capítulo 1 Controlabilidade nula de Stackelberg-Nash para algumas equações parabólicas degeneradas lineares e semilineares Stackelberg-Nash null controllability for some linear and semilinear degenerate parabolic equations F. D. ARARUNA, B. S. V. de ARAÚJO, and E. FERNÁNDEZ-CARA Abstract This paper deals with the application of Stackelberg-Nash strategies to the null controllability of degenerate parabolic equations. We assume that we can act on the system through a hierarchy of controls. A rst control (the leader) is assumed to determine the policy; then, a Nash equilibrium pair (corresponding to a noncooperative multi-objective optimization strategy) is found; this governs the action of other controls (the followers). This way, the state of the system is driven to zero and, consequently, we solve a hierarchical null controllability problem. The main novelty in this paper is that the physical systems are governed by linear or semilinear 1D heat equations with degenerate coecients.

22 Keywords: Null controllability, Stackelberg-Nash strategies, degenerate parabolic equations, Carleman inequalities. Mathematics Subject Classication: 34K35, 49J2, 35K Introduction The study of the controllability of partial dierential equations and systems has attracted the interest of many authors. The theory has been extended to semilinear problems, equations in unbounded domains, and systems in uid dynamics, among others; see for instance [15, 18, 2, 21, 24. In the particular case of degenerate parabolic equation, still not many things are known, see [7, 19, 27. In this paper, we assume that α [, 2 is an exponent. Let us rst consider the following degenerate systems: u t (x α u x ) x + b (x, t)u = g1 O in, u(, ) = if α [, 1) u(1, ) = and on (, T ), (x α u x )(, ) = if α [1, 2 u(, ) = u in (, 1), u t (x α u x ) x + (x q b 1 (x, t)u) x + b (x, t)u = g1 O in, u(, ) = if α [, 1) u(1, ) = and on (, T ), (x α u x )(, ) = if α [1, 2 u(, ) = u in (, 1), (1.1) (1.2) (with q α/2), and u t (x α u x ) x + (b 1 (x, t)u) x + b (x, t)u = g1 O in, u(, ) = if α [, 1) u(1, ) = and on (, T ), (x α u x )(, ) = if α [1, 2 u(, ) = u in (, 1), (1.3) where = (, 1) (, T ), O (, 1) is a non-empty open subset and 1 O is its associated characteristic function, b L (), b 1 L (, T ; H 1 (, 1)), g L 2 (), and u L 2 (, 1). When α [, 1), we say that the problems (1.1), (1.2) and (1.3) are weakly degenerate; contrarily, we say that they are strongly degenerate if α [1, 2). When 11

23 α = 2, an appropriate change of variables show that (1.1), (1.2) and (1.3) are equivalent to some nondegenerate problems in an unbounded domain. In particular, (1.2) and (1.3) can be viewed as generalizations of the celebrated Black-Scholes equation u t x2 2 σu xx rxu x + ru =, (1.4) which models the behavior of an option u = u(x, t) as a function of the portfolio cotization and time. Here, σ is the volatility and r is the risk free rate [6. From the control viewpoint, a relevant question is whether or not these systems are null-controllable. In other words, for each u L 2 (, 1), we try to elucidate if there exist controls g L 2 () such as the associated states u (the corresponding solutions to (1.1), (1.2) or (1.3)) satisfy u(x, T ) = in (, 1). (1.5) In the nondegenerate case, that is, when the spatial domain is replaced by an interval (a, b), with a >, the null controllability of these systems is a trivial consequence of the (classical) results in [2 and [25. The null controllability of systems (1.1), (1.2) and (1.3) have been proved, repectively, in [7 (for α (, 2)), [12 (for α (, 2)), and [27 (for α (, 1/2)). These works are based on classical duality arguments, which reduces a null controllability property to an observability inequality for the solutions of the adjoint system. In these cases, the adjoint system of (1.1), (1.2) and (1.3) are given respectively by v t + (x α v x ) x + b (x, t)v = h in, v(, ) = if α [, 1) v(1, ) = and on (, T ), (x α v x )(, ) = if α [1, 2) v(, ) = v in (, 1), (1.6) v t + (x α v x ) x + x q b 1 v x + b v = h in, v(, ) = if α [, 1) v(1, ) = and on (, T ), (x α v x )(, ) = if α [1, 2) v(, ) = v in (, 1), (1.7) and 12

24 v t + (x α v x ) x + b 1 v x + b v = h in, v(, ) = if α [, 1) v(1, ) = and on (, T ), (x α v x )(, ) = if α [1, 2) v(, ) = v in (, 1). (1.8) The observability inequalities for (1.6), (1.7), and (1.8) were obtained making use of suitable Carleman estimates for themselves. The inclusion of a rst-order term in the equation in (1.1) is a delicate question that, in general, remains open. If it is a multiple of an appropriate power of x, like in (1.2), it is easy to show that the Carleman estimates proved in [7 again hold for its adjoint system and, consequently, null controllability holds too, see [19. However, the more general situation presented in (1.3) has been solved only for α [, 1/2), see [27. Now, taking α = 2 in (1.1), we have the following system: u t (x 2 u x ) x + b (x, t)u = g1 O in, u x (1, ) = u x (, ) = on (, T ), (1.9) u(, ) = u in (, 1). In this case, an additional diculty is found: there is no known Carleman estimate for the solution to the associated adjoint. To overcome this diculty, we introduce the change of variables y = log(1/x), U(y, t) = x 1/2 u(x, t), (1.1) which transform (1.9) into U t U yy + B(y, t)u = G(y, t)1 ϑ in, (U y + 12 ) U (, ) = and lim (U y + 12 ) U (y, ) = on (, T ), y + U(, ) = U in (, + ), (1.11) where and = (, + ) (, T ), B(y, t) = b (x, t) + 1/4, G = e y/2 g, ϑ = {y (, + ) : e y O}. Since we are dealing here with a problem in unbounded domain, the null controllability properties depend on the choice of ϑ. Indeed, in [21 the authors present a Carleman 13

25 estimate for solutions of the adjoint system of a similar one to (1.11), but with Dirichlet boundary conditions. It is well know that the boundary conditions have an important role in the Carleman inequality, hence, to deal with (1.11), a new inequality must be proved for its adjoint system: w t + w yy + B(y, t)w = F in, (w y + 12 ) w (, ) = and lim (w y + 12 ) w (y, ) = on (, T ), y w(, T ) = w T in (, + ). (1.12) As mentioned in [21, there are few situations where we can get a Carleman estimate for systems in unbounded domain. The most simple of this situations is when the non-controllable region is a bounded set. To deal with this case, (, + )\ϑ must be a bounded set; and this happens if and only if O. This paper deals with Stackelberg-Nash strategies for the null controllability of degenerate parabolic systems similar to those above. To be more specic, let us x the non-empty open subsets O i (, 1) (i = 1, 2), and for each triplet (g, f 1, f 2 ) [L 2 () 3, let us consider systems (1.1), (1.3) and (1.9) with g1 O replaced by g1 O + f 1 1 O1 + f 2 1 O2, i.e. u t (x α u x ) x + b (x, t)u = g1 O + f 1 1 O1 + f 2 1 O2 in, u(, ) = if α [, 1) u(1, ) = and on (, T ), (x α u x )(, ) = if α [1, 2) u(, ) = u in (, 1), (1.13) u t (x α u x ) x + (x q b 1 (x, t)u) x + b (x, t)u = g1 O + f 1 1 O1 + f 2 1 O2 in, u(1, ) = and u(, ) = if α [, 1) (x α u x )(, ) = if α [1, 2) on (, T ), (1.14) u(, ) = u in (, 1), u t (x α u x ) x + (b 1 (x, t)u) x + b (x, t)u = g1 O + f 1 1 O1 + f 2 1 O2 in, u(, ) = if α [, 1) u(1, ) = and on (, T ), (x α u x )(, ) = if α [1, 2) u(, ) = u in (, 1), (1.15) 14

26 and u t (x 2 u x ) x + b (x, t)u = g1 O + f 1 1 O1 + f 2 1 O2 in, u x (1, ) = u x (, ) = on (, T ), u(, ) = u in (, 1). (1.16) For simplicity, we will assume that only three controls are applied (one leader and two followers), but very similar considerations hold for systems with a higher number of controls. Now, for i = 1, 2, let us introduce the non-empty open sets O i,d (, 1), the functions u i,d L 2 (O i,d (, T )), and the functionals J i : L 2 (O (, T )) U R given by J i (g; f 1, f 2 ) := β i 2 u u i,d 2 dx dt + µ i O i,d (,T ) f i 2 dx dt, (1.17) O i (,T ) where U := U 1 U 2, with U i := L 2 (O (, T )), β i and µ i are positive constants, and u is the associated state to (1.13), (1.14), (1.15) or (1.16). For a xed g L 2 (O (, T )), we say that a pair (f 1, f 2 ) U is a Nash equilibrium for (J 1, J 2 ) associated to g when J 1 (g; f 1, f 2 ) J 2 (g; f 1, f 2 ), J 2 (g; f 1, f 2 ) J 2 (g; f 1, f 2 ), f 1 L 2 (), f 2 L 2 (). (1.18) Our goal is to prove that, for any u L 2 (, 1), there exist a control g L 2 (O (, T )) (called leader) and a associated Nash equilibrium (f 1, f 2 ) U (called followers) such that the associated state u of (1.13), (1.14), (1.15) or (1.16) satises (1.5). Results of this type, i.e. Stackelberg-Nash null controllability, were proved for the rst time in [4, in the context of nondegenerate parabolic equations. After that, we can mention [23. Most of the works dealing with Stackelberg-Nash strategy are in the context of the approximate controllability. In this issue, we cite [22. The rest of this paper is organized as follows. In Section 2 we present the notations, the denitions of some spaces and some preliminary results. We also provide in this section the main tool of this work: Carleman estimates for degenerate parabolic equations with control regions in the interior of the domain. In Section 3, we establish and prove our main results on Stackelberg-Nash null controllability. In Section 4, we make some additional comments, discuss open questions and advance some future 15

27 work. The paper ends with three appendices containing the proofs of the Carleman estimates. 1.2 Notations and preliminary results Notations and spaces The usual norm and inner product in L 2 (, 1) and in L 2 () will be denoted respectvely by and (, ), and and ((, )). The norms in L (, 1) and in L () will be denoted respectively by and. Let us consider the sets H α := { u L 2 (, 1) : u is absolutely continuous in [, 1, x α/2 u x L 2 (, 1), u() = u(1) = }, for α [, 1) and H α := { u L 2 (, 1); u is locally absolutely continuous in (, 1, x α/2 u x L 2 (, 1), u(1) = } for α [1, 2. In the sequel, for any two Banach spaces X and Y, the notation X Y indicates that X Y and, moreover, the embedding X Y is continuous. It is easy see that L 2 (, 1) H α for all α [, 2, so that L 2 () L 2 (, T ; H α). Moreover, it is shown in [1 that the embbeding H α L 2 (, 1) is compact. Hence, from the well known Aubin-Lions Compactness Theorem, we see that the Hilbert space W α := { y L 2 (, T ; H α ) : y t L 2 (, T ; H α) } is compactly embedded in L 2 () Existence and estimates The following result has been proved in [1, 19, 27: Proposition For any g L 2 () and any u L 2 (, 1), there exists exactly one solution u to each of the systems (1.1), (1.2), (1.3) and (1.9), with u L 2 (, T ; H α ) C ([, T ; L 2 (, 1)). 16

28 Furthermore, there exists a constant C > only depending on T, α, b and b 1, such that sup u(, t) 2 + x α u x 2 dx dt C( g 2 + u 2 ). t [,T Hardy inequalities are a standard tool in the analysis of degenerate equations. The following result is proved in [1: Proposition (Hardy's inequality) Assume that α < 2 and α 1. Let z : [, 1 R be locally absolutely continuous in (, 1, with Then, for all δ (, 1, one has δ x α 2 z 2 4 dx (1 α) 2 1 x α 2 z 2 dx 4 (1 α) 2 1 δ 1 x α z x 2 dx < +. x α z x 2 dx if α < 1 and lim x + z(x) = x α z x 2 dx if α (1, 2) and lim z(x) =. x Carleman estimates for (1.6) and (1.7) Let us introduce the functions θ, p and σ with θ(t) := 1 (t(t t)) 4, p (x) := 1 x2 α (2 α) 2 and σ (x, t) := θ(t)p (x). The following Carleman estimate is proved in [7: Proposition There exists positive constants s and C such that, for any s s and any solution v to (1.6), one has: e [ 2sσ sθx α v x 2 + s 3 θ 3 x 2 α v 2 [ T dx dt C e sσ h 2 + s e 2sσ θ v x 2 dt (1.19). x=1 We will need below a Carleman estimate with the previous boundary integral replaced by an interior observation term and time derivatives and second-order spatial derivatives in the left hand side. The goal of this section is to present such an estimate. The main idea is to modify the weight functions and then proceed as in [7; a similar result was recently proved in [1, but there the equation is slightly dierent (the coecient degenerate at both x = and x = 1). Let us x α [, 2) and the non-empty open sets ω (, 1) and ω = (a, c ) ω. We will use the following notations: ω T := ω (, T ), ω T := ω (, T ) and := (, a ) (, T ). 17

29 Let η C (, 1) be a function such that η(x) = x2 α 2 α in [, a and η(x) = x2 α 2 α in [c, 1. Now, for λ λ > and s R, we introduce p(x) := e λ(2 η +η(x)), ξ(x, t) := p(x)θ(t), σ(x, t) := θ(t)e 4λ η ξ, γ 1 (λ) := 1 α + λ 1/4, and γ 2 (s) := 1 α + s 1/2. The main result in this section is the following: Theorem There exist positive constants C, s and λ, depending only on ω, ω, b, T and α such that, for any s s, any λ λ and any solution v to (1.6), one has: e [ 2sσ s 1 γ 1 (λ)ξ 1 ( v t 2 + (x α v x ) x 2 ) + sλx α ξ v x 2 + sλ 2 x 2α η 2 ξ v x 2 dx dt + e 2sσ v [ 2 s 2 λ 2 γ 1 (λ)ξ 2 γ 2 (sξ) + s 3 λ 3 x 2 α ξ 3 + s 3 λ 4 x 2α η 4 ξ 3 dx dt C [ e sσ h 2 + s 3 λ 3 e 2sσ ξ 3 v 2 dx dt. (1.2) ω T The proof of Theorem is presented in Appendix A. It is possible to rene the estimate in Theorem by replacing the powers of η by powers of x at the price of increasing the power of λ in the local term: Corollary There exist positive constants C, s and λ, depending only on ω, ω, b, T and α such that, for any s s, any λ λ, and any solution v to (1.6), one has: e [ 2sσ s 1 γ 1 (λ)ξ 1 ( v t 2 + (x α v x ) x 2 ) + sλx α ξ v x 2 + sλ 2 x 2 ξ v x 2 dx dt + e 2sσ v [ 2 s 2 λ 2 γ 1 (λ)ξ 2 γ 2 (sξ) + s 3 λ 3 x 2 α ξ 3 + s 3 λ 4 x 4 2α ξ 3 dx dt C [ e sσ h 2 + s 3 λ 4 e 2sσ ξ 3 v 2 dx dt. (1.21) ω T If v is a solution to (1.7) and we introduce h = h x q b 1 v x, we see that v solves a system of the kind (1.6). Thus, the following result holds. Corollary Let us assume that q α/2 and b, b 1 L (). Then, there exists positive constants C, s, and λ, depending only on ω, b, b 1, T, and α such that, for any s s, any λ λ, and any solution v for (1.7), the estimates (1.2) and (1.21) still hold. 18

30 1.2.4 Carleman estimate for (1.8) Let us assume that α [, 1/2). To our knowledge, the arguments in the proofs of the Carleman estimates in Section 2.3 cannot be adapted to the solutions to (1.8). Therefore, we have to establish other estimates with other weight functions. Let η C (, 1) be a function such that η(x) = 3x(4 2α)/3 4 2α in [, a and η(x) = 3x(4 2α)/3 4 2α in [c, 1. For simplicity, we will keep the same notation for the functions σ and ξ with the function η replaced by η. With this function, we get the following result: Theorem There exist positive constants C, s and λ, depending only on ω, ω, b, b 1, T and α such that, for any s s, any λ λ and any solution v to (1.8) one has e [ 2sσ s 1 γ 1 (λ)ξ 1 ( v t 2 + (x α v x ) x 2 ) + sλx (4α 2)/3 ξ v x 2 + sλ 2 x 2α η 2 ξ v x 2 dx dt + e 2sσ v [ 2 sλ 2 ξx (2α 4)/3 + s 3 λ 3 ξ 3 + s 3 λ 4 x 2α η 4 ξ 3 dx dt C [ e sσ h 2 + s 3 λ 3 e 2sσ ξ 3 v 2 dx dt. (1.22) ω T The proof of Theorem is given in Appendix B. An estimate similar to those in Corollary can also be obtained Carleman estimate for (1.12) In general, there is no estimate of the Carleman kind for the adjoint system of (1.9). This is expected since, in general, this system is not null-controllable. However, if the control domain is of the form ω = (, c) for some c (, 1), then in the system (1.11) obtained after the change of variables (1.1), the new control domain ϑ possesses a bounded complementary set in (, + ). In view of the results in [21, null controllability can be expected in this case and, consequently, it is reasonable to try to prove a Carleman estimate for the solutions to (1.12). Thus, let us assume that ω = (, c) and let us introduce ϑ := (a, + ) ϑ, ϑ T := ϑ (, T ) and ϑ T := ϑ (, T ). Let ˆη C 2 ([, )) be a function such 19

31 that ˆη(y) = y in [, a, ˆη, ˆη and ˆη are bounded in [, + ) and ˆη C > in [, + )\ϑ. Finally, for λ λ >, let us set ˆp(y) := e λ(2 ˆη +ˆη(y)), p(y) := e λ(2 ˆη ˆη(y)), ˆξ(y, t) := θ(t)ˆp(y), ξ(y, t) := θ(t) p(y), ˆσ(y, t) := θ(t)e 4λ ˆη ˆξ, σ(y, t) := θ(t)e 4λ ˆη ξ and ϱ := e 2sˆσ + e 2s σ. The Carleman estimates that we can get for (1.12) are given in the following result: Theorem There exist positive constants C, s and λ, depending only on ω, B and T such that, for any s s, any λ λ and any solution w to (1.12), one has: ϱ [s 1 ˆξ 1 ( w t 2 + w yy 2 ) + sλ 2 ˆξ wy 2 + s 3 λ 4 ˆξ3 w 2 dy dt C [ ϱ 1/2 F 2 + s 3 λ 4 ϱˆξ 3 w 2 dy dt. (1.23) ϑ T The proof of Theorem is given in Appendix C. 1.3 Stackelberg-Nash null controllability In this section we will prove the Stackelberg-Nash null controllabilty of (1.13), (1.14), (1.15) and (1.16). From the linearity of the systems and the convexity of the functionals J i, it is clear that (f 1, f 2 ) is a Nash equilibrium for (J 1, J 2 ) if and only if J 1(g; f 1, f 2 )(f, ) =, f U 1 and J 2(g; f 1, f 2 )(, f) =, f U 2. Arguing as in [4, 22 the following result holds. Proposition There exist a constant µ > such that, if µ i µ (i = 1, 2), for each g L 2 ()) there exists a unique associated Nash equilibrium (f 1, f 2 ) for (J 1, J 2 ). Furthermore, there exists a constant C > such that (f 1, f 2 ) C(1 + g ). In particular, the corresponding state u satises u L (,T ;L 2 (,1)) + u L 2 (,T ;H α) + u t L 2 (,T ;H 1 (,1)) C(1 + g ). 2

32 To establish the Stackelberg-Nash null controllability, in the sequel we will impose the following assumptions: O 1,d = O 2,d ; the common observability set will be denoted O d. O d O. θ 2 u i,d 2 dx dt < + for i = 1, 2. O d (,T ) In the case of system (1.3) α [, 1/2). In the case of system (1.9) O = (, c) for some c [, 1). The main result in this section is the following. (1.24) Theorem Assume that (1.24) holds. There exists µ µ such that, if µ 1, µ 2 µ, for every u L 2 (, 1) there exist a leader control g L 2 () and a unique associated Nash equilibrium (f 1, f 2 ) such that the corresponding state, i.e. the solution to (1.13), (1.14), (1.15) or (1.16), satises (1.5) Proof of Theorem The strategy will be the following: First, we will characterize the Nash equilibrium associated to g as the solution, together with u, to an appropriate coupled system. Then, we will prove that, if µ 1 and µ 2 are large enough, any solution to the corresponding adjoint system satises an observability estimate. As a consequence of a well known duality argument, this will imply the desired result. This strategy is very similar to systems (1.13), (1.14), (1.15) and (1.16). Hence we will only consider the system (1.13). Arguing as in [4, 22, the following result can be easily proved. Proposition Let g L 2 (O (, T )) be given. Then (f 1, f 2 ) is a Nash equilibrium of (1.13) associated to g if and only if f i = 1 φ i Oi (,T ), µ i 21

33 where the φ i (i = 1, 2) solve, together with u, the following coupled system: u t (x α u x ) x + b (x, t)u = 1 O f 1 φ 1 1 O1 1 φ 2 1 O2 in, µ 1 µ 2 (φ i ) t (x α (φ i ) x ) x + b (x, t)φ i = α i (u u i,d )1 Oi,d in, t) = φ i (1, t) = { u(1, on (, T ), u(, t) = φ i (, t) = if α [, 1) (x α u x )(, t) = (x α (φ i ) x )(, t) = if α [1, 2) on (, T ), u(x, ) = u (x), φ i (x, T ) = in (, 1). (1.25) As we said previously, to prove Theorem it is sucient to nd an observability estimate for the adjoint system of (??), which is given by z t (x α z x ) x + b (x, t)z = β 1 ϕ 1 1 O1,d + β 2 ϕ 2 1 O2,d in, (ϕ i ) t (x α (ϕ i ) x ) x + b (x, t)ϕ i = 1 µ i z1 Oi (i = 1, 2) in, z(1, t) = ϕ i (1, t) = on (, T ), z(, t) = ϕ i (, t) = if α [, 1) on (, T ), (x α z x )(, t) = (x α (ϕ i ) x )(, t) = if α [1, 2) z(x, T ) = z T (x), ϕ i (x, ) = in (, 1). For this, we consider the following result. (1.26) Theorem Let assumptions in (1.24) hold. Then there exist positive constants µ and C, and a weight function ρ = ρ(t) blowing up at t = T such that, if µ i µ (i = 1, 2), for any z T L 2 (, 1) and any solution (z, ϕ i ) to (1.26), the following inequality holds: z(, ) i=1 Proof. Let us x ω and ω 1 with ρ 2 ϕ i 2 dx dt C O (,T ) z 2 dx dt. (1.27) ω = (a, c ) ω 1 = (a 1, c 1 ) O O d and a function ψ C (, 1) such that ψ 1, ψ = 1 in ω and supp(ψ) ω 1. We will use the notations ω T := ω (, T ), ω 1T := ω 1 (, T ) and we will denote by E(v) the left hand side of (1.2). 22

34 Let us assume that (z, ϕ 1, ϕ 2 ) solves (1.26) and let us introduce h := β 1 ϕ 1 +β 2 ϕ 2. It is clear that Theorem can be applied to z and also to h. Thus, there exist positive constants λ, s, and C such that, for s s and λ λ, one has [ E(z) C e 2sσ h 2 dx dt + s 3 λ 4 e 2sσ ξ 3 z 2 dx dt ω T and [ E(h) C e 2sσ z 2 dx dt + s 3 λ 4 e 2sσ ξ 3 h 2 dx dt. ω T Then, for s large enough, we get E(z) + E(h) C [s 3 λ 4 e 2sσ ξ 3 z 2 dx dt + s 3 λ 4 e 2sσ ξ 3 h 2 dx dt. (1.28) ω T ω T Since h = z t (x α z x ) x + bz in ω 1, it follows that s 3 λ 4 e 2sσ ξ 3 h 2 dx dt s 3 λ ω 4 e 2sσ ξ 3 ψh [ z t (x α z x ) x + bz dx dt (1.29) 1T ω T Using integration by parts and Young's inequality, we nd s 3 λ 4 e 2sσ ξ 3 ψhz t dx dt Cs 1 E(h) + s 8 λ 8 e 2sσ ξ 7 z 2 dx dt, (1.3) ω 1T ω 1T s 3 λ 4 e 2sσ ξ 3 ψh(x α z x ) x dx dt Cs 1 E(h) + Cs 8 λ 8 e 2sσ ξ 7 z 2 dx dt,(1.31) ω 1T ω 1T and s 3 λ 4 e 2sσ ξ 3 ψhb z dx dt Cs 1 E(h) + Cs 4 λ 4 e 2sσ ξ 3 z 2 dx dt. (1.32) ω 1T ω 1T Combining the estimates in (1.29), (1.3), (1.31) and (1.32) we see that s 3 λ 4 e 2sσ ξ 3 h 2 dx dt Cs 1 E(h) + Cs 8 λ ω 8 e 2sσ ξ 7 z 2 dx dt 1T ω T and, from (1.28), E(z) + E(h) Cs 8 λ 8 ω 1T e 2sσ ξ 7 z 2 dx dt, (1.33) for large enough s and λ. From now on, the constant C may depend on s and λ. 23

35 Let ψ 1 C 1 ([, T ) be such that ψ 1 1, ψ 1 = 1 in [, T/2 and ψ 1 = in [3T/4, T and let us modify the functions θ, ξ and σ as follows: (4/T θ(t) 2 ) 4 if t [, T/2, := ξ := p θ, θ(t) if t [T/2, T, and σ := θe λ η ξ. Then Z := ψ 1 z is a solution to the following system Z t (x α Z x ) x + b (x, t)z = ψ 1 (t)h + ψ 1(t)z in Z(, ) = if α [, 1) Z(1, ) = and on (, T ) (x α Z x )(, ) = if α [1, 2) and, from the energy estimate in Proposition 1.2.1, the following estimate holds: T [ z(, ) [ + z 2 + x α z x 2 3T 4 1 3T dx dt C h dx dt + z 2 dx dt. Using this inequality and the fact that e 2s σ and ξ are bounded from above in [, T/2, we obtain T/2 1 z(, ) 2 + e [ 2s σ ξ 3 x 2 α z 2 + ξx α z x 2 dx dt [ 3T/4 1 3T/4 1 C h 2 dx dt + z 2 dx dt. Now, the fact that e 2sσ and ξ are bounded from below in [T/2, 3T/4, we get z(, ) 2 + T/2 1 T/2 On the other hand, since ξ = ξ in [T/2, T, we have T T/2 1 From (1.34) and (1.35), a new estimate is found: e 2s σ [ ξ3 x 2 α z 2 + ξx α z x 2 dx dt [ 3T/4 1 C h 2 dx dt + E(z). (1.34) e 2s σ [ ξ3 x 2 α z 2 + ξx α z x 2 dx dt E(z). (1.35) z(, ) 2 + e 2s σ [ ξ3 x 2 α z 2 + ξx α z x 2 dx dt C 24 [ 3T 4 1 T 2 h 2 dx dt + E(z).(1.36)

36 Using again Proposition and the inequalities e 2s σ, ξ C > in [, 3T/4, we see that 3T/4 1 h 2 dx dt C and, from (1.36) and (1.37), we conclude that z(, ) 2 + [ β1 + β [ 2 e 2s σ [ ξ3 x 2 α z 2 + µ 1 µ ξx α z x 2 dx dt (1.37) 2 e 2s σ [ ξ3 x 2 α z 2 + ξx α z x 2 dx dt CE(z), (1.38) whenever µ 1 and µ 2 are large enough. By (1.33) and (1.38), it follows that z(, ) 2 + E(z) + E(h) C Let us introduce the functions ω T e 2sσ ξ 7 z 2 dx dt. (1.39) σ (t) := max σ(x, t) = C θ(t) and ρ(t) := e sσ(t). x [,1 Note that ρ is a positive nondecreasing function in C 1 ([, T ) that blows up at t = T. Multiplying the equation satised by ϕ i by ρ 2 ϕ i and integrating in space, we nd that 1 d 2 dt ρ 1 ϕ i 2 + ρ 2 x α/2 (ϕ i ) x 2 Cρ 2 z 2 + C ρ 1 ϕ i 2 and, from Gronwall's Lemma, ρ 1 ϕ i can be bounded as follows: ρ(t) 1 ϕ i (, t) 2 C Using (1.39) and the fact that ρ 2 e 2s σ, we get the estimates ρ 2 z 2 dx dt CE(z) C Finally, in view of (1.39) and (1.4), (1.27) holds. ρ 2 z 2 dx dt, t [, T. (1.4) ω T e 2sσ ξ 7 z 2 dx dt. As mentioned above, the observability estimate (1.27) implies the null controllability of (1.13). We still have an estimate of the control: [ 2 g 2 dx dt C u 2 + ρ 2 u i,d 2 dx dt, i=1 O d (,T ) where C is the constant in (1.27). This ends the proof of Theorem 4. 25

37 1.3.2 Similar results for semilinear problems In this section, we extend Theorem to semilinear systems of the form u t (x α u x ) x + b (x, t)u = F (u) + g1 O + f 1 1 O1 + f 2 1 O2 in, u(, t) = if α [, 1) u(1, t) = and on (, T ), (x α u x )(, t) = if α [1, 2) u(x, ) = u (x) in (, 1), (1.41) where F : R R is a globally Lipschitz-continuous function satisfying F () =. (1.41). Note that an existence-uniqueness result like Proposition still holds for In this semilinear framework, the functionals J i in (1.17) are not convex in general. Accordingly, we must consider a weaker concept of Nash equilibrium: Denition Let g L 2 (O (, T )) be given. The pair (f 1, f 2 ) U is called a Nash quasi-equilibrium of (1.41) associated to g if J 1(g; f 1, f 2 )(f, ) =, f L 2 (ω 1 (, T )), J 2(g; f 1, f 2 )(, f) =, f L 2 (ω 2 (, T )). The main result in this section is the following: Theorem Assume (1.24) holds and µ 1 and µ 2 are large enough. Then, for each u L 2 (, 1), there exist a leader control g L 2 (ω (, 1)) and an associated Nash quasi-equilibrium (f 1, f 2 ) such that the corresponding solution to (1.41) satises (1.5). The proof relies on a standard and well known xed-point argument. For details, the reader is referred to [4, where the same result is proved for a nondegenerate semilinear parabolic PDE. Completely similar results can be obtained for semilinear systems of the kind (1.2), (1.3) and (1.9). 1.4 Additional comments and open questions On the assuption O 1,d = O 2,d The assumption (1.24) 1 can be suppressed if we assume that O i,d O = for i = 1, 2 and O 1,d O O 2,d O, see [5 for details. However, when O 1,d O 2,d, but O 1,d O = O 2,d O, the Stackelberg-Nash null controllability is open. 26

38 1.4.2 Higher dimensions Little is known on the control of degenerate parabolic PDEs in higher dimensions. Recently, some results have been established on the global null controllability of a model similar to (1.1) that degenerate at the whole border, see [1. They rely on suitable Carleman estimates similar to (1.2), whence it is reasonable to think that the results in this work can be extended to this model. In a forthcoming paper we will present Carleman estimates for some models that degenerates in only a part of the border Wave equations An interesting question is whether Stackelberg-Nash null controllability can be proved for wave equations. This question is open; one of the main diculties is that, to our best knowledge, the few Carleman estimates with interior control region already in the literature do not have all the necessary terms in the left hand side. 1.5 Appendix A: Proof of Theorem 1 It is sucient to prove Theorem 3 assuming that b =. Let v be the solution to (1.6) (where v T L 2 (, 1) and g L 2 ()). For any s s >, we set z = e sσ v. By a density argument we can assume without loss of generality that v is regular enough. We have: v t = e sσ [sσ t z + z t, (x α v x ) x = e sσ [s 2 σxx 2 α z + 2sσ x x α z x + s(σ x x α ) x z + (x α z x ) x and, consequently, P + z + P z = e sσ g, where P + z = sσ t z + s 2 x α σ 2 xz + (x α z x ) x and P z = z t + s(x α σ x ) x z + 2sx α σ x z x. This gives: e sσ g 2 = P + z 2 + P z 2 + 2((P + z, P z)). (1.42) 27

39 We have that ((P + z, P z)) = I I 4, where I 1 = ((sσ t z + s 2 σxx 2 α z + (x α z x ) x, z t )) I 2 = s 2 ((σ t z, (x α σ x ) x z + 2σ x x α z x )) I 3 = s 3 ((σ xx 2 α z, (x α σ x ) x z + 2σ x x α z x )) I 4 = s(( (x α z x ) x, (x α σ x ) x z + 2σ x x α z x )). The next step is to compute I 1, I 2, I 3 and I 4. To this purpose, we will use that z = z x = at t = and t = T and, also, (x α z x ) x z t dx dt =. After integrating by parts, we deduce easily that I 1 = s 2 I 2 = s 2 z 2 (σ tt + 2sσ x σ xt x α ) dx dt x α σ x σ xt z 2 dx dt; I 3 = s 3 x α σ x (x α σx) 2 x z 2 dx dt; I 4 = s (x α σ x ) xx x α zz x dx dt 2s Consequently, +sα T σ x x 2α 1 z x 2 dx dt + s (x α σ x ) x x α z x 2 dx dt σ x x 2α z x 2 dt T x=1 ((P + z, P z)) = s σ x x 2α z x 2 dt s 3 x α σ x (x α σx) 2 x z 2 dx dt x= 2s (x α σ x ) x x α z x 2 dx dt 2s 2 σ x σ xt x α z 2 dx dt +αs σ x x 2α 1 z x 2 dx dt s (x α σ x ) xx x α zz x dx dt s σ tt z 2 dx dt. (1.43) 2 x=1 x=. 28

40 Using the denitions of the functions σ and ξ, we see that σ x σ xt x α = λ 2 x α η 2 ξξ t x α σ x (x α σ 2 x) x = λ 3 x α η (x α η 2 ) x ξ 3 2λ 4 x 2α η 4 ξ 3 (x α σ x ) x x α = λx α (x α η ) x ξ λ 2 x 2α η 2 ξ σ x x 2α 1 = λx 2α 1 η ξ (x α σ x ) xx x α = λx α (x α η ) xx ξ 2λ 2 x α η (x α η ) x ξ λ 2 x 2α η η ξ λ 3 x 2α (η ) 3 ξ σ x x 2α = λx 2α η ξ. With this information, we will now estimate each term in the right hand side of (1.43). For s and λ large enough, we have: T s σ x x 2α z x 2 dt s 3 x=1 x=, (1.44) x α σ x (x α σx) 2 x z 2 dx dt C [s 3 λ 4 x 2α η 4 ξ 3 z 2 dx dt + s 3 λ 3 x 2 α ξ 3 z 2 dx dt Cs 3 λ 3 ω T ξ 3 z 2 dx dt, (1.45) 2s and 2s 2 x α (x α σ x ) x z x 2 dx dt Csλ 2 x 2α η 2 ξ z x 2 dx dt + 2sλ x α ξ z x 2 dx dt Csλ ξ z x 2 dx dt, (1.46) ω T [ x α σ x σ xt z 2 dx dt Cs 2 λ 2 x 2α η 4 ξ 3 z 2 dx dt + x 2 α ξ 3 z 2 dx dt + ξ 3 z 2 dx dt, (1.47) ω T αs x 2α 1 σ x z x 2 dx dt αsλ x α ξ z x 2 dx dt Csλ ξ z x 2 dx dt,(1.48) ω T where C only depends on a, T and α. On the other hand, s x α (x α σ x ) xx zz x dx dt = sλ x α (x α η ) xx ξzz x dx dt + sλ 2 x 2α η η ξzz x dx dt +2sλ 2 x α η (x α η ) x ξzz x dx dt + sλ 3 x 2α (η ) 3 ξzz x dx dt.(1.49) 29

41 In order to estimate this last term, we note that ( sλ x α (x α η ) xx ξzz x dx dt Csλ ξ 3 z 2 + ξ z x 2) dx dt, (1.5) ω T sλ 2 2sλ 2 and sλ 3 ( x 2α η η ξzz x dx dt C s 2 λ 4 x 2α η 4 ξ 3 z 2 + x 2α η 2 ξ z x 2) dx dt ( C s 2 λ 3 x 2 α ξ 3 z 2 + λx α ξ z x 2) dx dt ( C s 2 λ 3 ξ 3 z 2 + λξ z x 2) dx dt, (1.51) ω T ( x α η (x α η ) x ξzz x dx dt C s 2 λ 4 x 2α η 4 ξ 3 z 2 + x 2α η 2 ξ z x 2) dx dt ( C s 2 λ 3 x 2 α ξ 3 z 2 + λx α ξ z x 2) dx dt ( C s 2 λ 3 ξ 3 z 2 + λξ z x 2) dx dt, (1.52) ω T x 2α (η ) 3 ξzz x dx dt C ( λ 2 x 2α η 2 ξ z x 2 + s 2 λ 4 x 2α η 4 ξ 3 z 2) dx dt. (1.53) From (1.49)-(1.53), we conclude that ( s x α (x α σ x ) xx zz x dx dt C s 2 λ 4 x 2α η 4 ξ 3 z 2 + λ 2 x 2α η 2 ξ z x 2) dx dt ( C s 2 λ 3 x 2 α ξ 3 z 2 + λx α ξ z x 2) dx dt ( C s 2 λ 3 ξ 3 z 2 + sλξ z x 2) dx dt. (1.54) ω T From (1.43)-(1.48) and (1.54), we conclude that [ ( ((P + z, P z)) C s 3 λ 4 x 2α η 4 ξ 3 z 2 + sλ 2 x 2α η 2 ξ z x 2) dx dt ( + s 3 λ 3 x 2 α ξ 3 z 2 + sλx α ξ z x 2) dx dt [ ( C s 3 λ 3 ξ 3 z 2 + sλξ z x 2) dx dt + s ξ 3/2 z 2 dx dt ω T and, using (1.42), we nd that ( P + z 2 + P z 2 + s 3 λ 4 x 2α η 4 ξ 3 z 2 + sλ 2 x 2α η 2 ξ z x 2) dx dt ( + s 3 λ 3 x 2 α ξ 3 z 2 + sλx α ξ z x 2) dx dt [ ( C e sσ g 2 + s 3 λ 3 ξ 3 z 2 + sλξ z x 2) dx dt + s ξ 3/2 z 2 dx dt.(1.55) ω T 3

42 Furthermore, we see that s 3 λ 3 ξ 3 x 2 α z 2 dx dt Cs 3 λ 3 ξ [ 3 x 2 α z 2 dx dt + ξ 3 x 2α η 4 z 2 dx dt + ξ 3 z 2 dx dt ω T and [ sλ ξx α z x 2 dx dt Csλ ξx α z x 2 dx dt + ξx 2α η 2 z x 2 dx dt + ξ z x 2 dx dt. ω T Hense, from (1.55) we obtain P + z 2 + P z 2 + ( s 3 λ 4 x 2α η 4 ξ 3 z 2 + sλ 2 x 2α η 2 ξ z x 2) dx dt ( + s 3 λ 3 x 2 α ξ 3 z 2 + sλx α ξ z x 2) dx dt [ ( C e sσ g 2 + s 3 λ 3 ξ 3 z 2 + sλξ z x 2) dx dt + s ξ 3/2 z 2 dx dt.(1.56) ω T Let us denote by L(z) all the terms in the left hand side of (1.56). For instance, assume that α 1. Using Young and Hardy's Inequality we deduce that s 2 λ 2 ξ 2 z 2 dx dt s 3 λ 3 ξ 3 x 2 α z 2 dx dt + sλ x α 2 ξ 1/2 z 2 dx dt s 3 λ 3 ξ 3 x 2 α z 2 dx dt + sλ x α (ξ 1/2 z) x 2 dx dt CL(z). (1.57) Now assume that α = 1. Using Hölder, Young and Hardy's inequality we see that s 3/2 λ 7/4 ξ 3/2 z 2 dx dt = (s 3 λ 4 x 2 ξ 3 z 2 ) 1/4.(sλx 2/3 ξ z 2 ) 3/4 dx dt Cs 3 λ 4 1 T +s 3/2 λ 7/4 ξ 3/2 z 2 dx dt a ξ 3 x 2α η 4 z 2 dx dt + sλ x 2+4/3 (ξ 1/2 z) 2 dx dt CL(z). (1.58) From (1.56), (1.57) and (1.58) we deduce that ( P + z 2 + P z 2 + s 3 λ 4 x 2α η 4 ξ 3 z 2 + sλ 2 x 2α η 2 ξ z x 2) dx dt ( + s 3 λ 3 x 2 α ξ 3 z 2 + s 2 λ 2 γ 1 (λ)ξ 2 γ 2 (sξ) z 2 + sλx α ξ z x 2) dx dt [ ( C e sσ g 2 + s 3 λ 3 ξ 3 z 2 + sλξ z x 2) dx dt. (1.59) ω T 31

43 Now, we will work to include the terms with a rst-order time derivative and second order spatial derivatives in the left hand side. Using the estimate (1.59) and the denitions of P z and P + z, we see that and s 1 γ 1 (λ) ξ 1 z t 2 dx dt C s 1 γ 1 (λ) ξ 1 (x α z x ) x 2 dx dt C [ ( e sσ g 2 + s 3 λ 3 ξ 3 z 2 + sλξ z x 2) dx dt ω T (1.6) [ ( e sσ g 2 + s 3 λ 3 ξ 3 z 2 + sλξ z x 2) dx dt (1.61). ω T From (1.59), (1.6) and (1.61), it is now clear that ( s 1 γ(λ)ξ [ 1 z t 2 + (x α z x ) x 2 + sλx α ξ z x 2 + sλ 2 x 2α η 2 ξ z x 2) dx dt + ( s 2 λ 2 γ 1 (λ)ξ 2 γ 2 (sξ) z 2 + s 3 λ 3 x 2 α ξ 3 z 2 + s 3 λ 4 x 2α η 4 ξ 3 z 2) dx dt C [ ( e sσ g 2 + s 3 λ 3 ξ 3 z 2 + sλξ z x 2) dx dt. ω T Coming back to the original variable v and using the estimate e 2sσ ξ v x 2 dx dt Cs 2 λ 2 e 2sσ ξ 3 v 2 dx dt ω T ω T +Cs 2 λ 2 e 2sσ ξ 1 (x α v x ) x 2 dx dt we nd the desired inequality (1.2). 1.6 Appendix B: Proof of Theorem 2 Again, we will assume that b =. Let g = g b 1 v x. Arguing as before, we can deduce that T x=1 ((P + z, P z)) = s σ x x 2α z x 2 dt s 3 x α σ x (x α σx) 2 x z 2 dx dt x= 2s (x α σ x ) x x α z x 2 dx dt 2s 2 σ x σ xt x α z 2 dx dt +αs x x σ 2α 1 z x 2 dx dt s σ tt z 2 dx dt 2 s (x α σ x ) xx x α zz x dx dt. (1.62) Furthermore, 32

44 T s x 2α σ x z x 2 dt x=1 x=, (1.63) s 3 x α σ x (x α σx) 2 x z 2 dx dt C [s 3 λ 4 x 2α η 4 ξ 3 z 2 dx dt + s 3 λ 3 ξ 3 z 2 dx dt Cs 3 λ 3 ω T ξ 3 z 2 dx dt, (1.64) 2s x α (x α σ x ) x z x 2 dx dt Csλ 2 x 2α η 2 ξ z x 2 dx dt Csλ ξ z x 2 dx dt ω T (1 + α)sλ x (4α 2)/3 ξ z x 2 dx dt, (1.65) 2s 2 x α σ x σ xt z 2 dx dt Cs 2 λ 2 ξ 3 z 2 dx dt Cs 2 λ 2 ξ 3 z 2 dx dt ω T Cs 2 λ 2 x 2α η 4 ξ 3 z 2 dx dt, (1.66) αs x 2α 1 σ x z x 2 dx dt αsλ x (4α 2)/3 ξ z x 2 dx dt Csλ ξ z x 2 dx dt,(1.67) ω T and s 2 σ tt z 2 dx dt Cs ξ 3 z 2 dx dt Cs ξ 3 z 2 dx dt ω T Cs x 2α η 4 ξ 3 z 2 dx dt. (1.68) Note that in this case we can control the term on the left hand side of (1.68) much more easily than in the previous case. However, to estimate the last term in the right hand side of (1.62) is more dicult. It is just here where the restriction α < 1/2 is required. We have: s x α (x α σ x ) xx zz x dx dt = sλ x α (x α η ) xx ξzz x dx dt +sλ 3 x 2α (η ) 3 ξzz x dx dt + 2sλ 2 x α η (x α η ) x ξzz x dx dt +sλ 2 x 2α η η ξzz x dx dt. (1.69) In the previous case we had (x α η ) xx = in (, 1)\ω. Here, this is lost and, therefore, we have to work dierently. 33

45 Using integration by parts, we have that sλ x α (x α η ) xx ξzz x dx dt = In view of Proposition 1.2.2, we have sλ [x α (x α η ) xx x ξ z 2 dx dt 2 Furthermore, sλ 2 ξ z 2 [[x α (x α η ) xx x + λx α η (x α η ) xx dx dt. (1.7) sλ (1 + α)(2 α)(5 4α) 54 Csλ ξ z 2 dx dt Csλ ω T 2 + α)(2 α) sλ( α Csλ ξ 3 z 2 dx dt Csλ ω T 2 + α)(2 α) sλ( α = m(α)sλ x (4α 2)/3 x 4α (ξ 1/2 z) 2 dx dt T 1 b ξ z 2 dx dt x (4α 2)/3 (ξ 1/2 z) 2 x dx dt x 2α η 4 ξ 3 z 2 dx dt. (1.71) x (4α 2)/3 (ξ 1/2 z) 2 x dx dt [ξ z x 2 + λx (1 2α)/3 ξzz x + λ2 4 x(2 4α)/3 ξ z 2 dx dt, (1.72) where m(α) := 2(1 + α)(2 α)(5 3 4α) 1 and ( Csλ 2 x (2α 1)/3 ξzz x dx dt Cλ λx (4α 2)/3 ξ z x 2 + s 2 λ 3 ξ 3 z 2) dx dt. (1.73) Then from (1.71), (1.72) and (1.73), we get sλ [x α (x α η ) xx x ξ z 2 dx dt m(α)sλ x (4α 2)/3 ξ z x 2 dx dt 2 Cs 2 λ 3 ξ 3 z 2 dx dt Csλ x 2α η 4 ξ 3 z 2 dx dt Cλ x (4α 2)/3 ξ z x 2 dx dt Csλ ξ 3 z 2 dx dt. (1.74) ω T On the other hand, sλ2 x α η (x α η ) xx ξ z 2 dx dt 2 Csλ 2 (1 + α)(2 α) sλ 2 x 18 (2α 4)/3 ξ z 2 dx dt ω T ξ z 2 dx dt Csλ 2 34 T 1 b x (2α 4)/3 ξ z 2 dx dt (1.75)

46 and, from (1.7), (1.74) and (1.75), we see that sλ x α (x α η ) xx ξzz x dx dt Csλ 2 x (2α 4)/3 ξ z 2 dx dt Cs 2 λ 3 ξ 3 z 2 dx dt m(α)sλ x (4α 2)/3 ξ z x 2 dx dt Cλ x (4α 2)/3 ξ z x 2 dx dt Csλ 2 x 2α η 4 ξ 3 z 2 dx dt Csλ ω 2 ξ 3 z 2 dx dt. (1.76) T Note that the new strategy makes appear a new term in the right hand side of (1.76). For the other terms in (1.69), we have that ( 2sλ 2 x α η (x α η ) x ξzz x dx dt C s 2 λ 4 x 2α η 4 ξ 3 z 2 + x 2α η 2 ξ z x 2) dx dt ( C s 2 λ 3 ξ 3 z 2 + λx (4α 2)/3 ξ z x 2) dx dt ( C s 2 λ 3 ξ 3 z 2 + λξ z x 2) dx dt, (1.77) ω T sλ 2 and sλ 3 ( x 2α η η ξzz x dx dt C s 2 λ 4 x 2α η 4 ξ 3 z 2 + x 2α η 2 ξ z x 2) dx dt ( C s 2 λ 3 ξ 3 z 2 + λx (4α 2)/3 ξ z x 2) dx dt ( C s 2 λ 3 ξ 3 z 2 + λξ z x 2) dx dt, (1.78) ω T x 2α (η ) 3 ξzz x dx dt C ( λ 2 x 2α η 2 ξ z x 2 + s 2 λ 4 x 2α η 4 ξ 3 z 2) dx dt.(1.79) From (1.69), (1.76), (1.77), (1.78) and (1.79), we conclude that ( s x α (x α σ x ) xx zz x dx dt C λ 2 x 2α η 2 ξ z x 2 + s 2 λ 4 x 2α η 4 ξ 3 z 2) dx dt ( +Csλ 2 x (2α 4)/3 ξ z 2 dx dt C s 2 λ 3 ξ 3 z 2 + sλξ z x 2) dx dt ω T ( C s 2 λ 3 ξ 3 z 2 + λx (4α 2)/3 ξ z x 2) dx dt m(α)sλ x (4α 2)/3 ξ z x 2 dx dt. (1.8) It is important to take care of the constants accompanying the term sλ x (4α 2)/3 ξ z x 2 dx dt 35

47 in the estimates (1.65), (1.67) and (1.8). The sum of these constants is that is only positive for α [, 1/2). that (2 α)(1 2α), 5 4α From (1.62), (1.63), (1.64), (1.65), (1.66), (1.67), (1.8) and (1.68), we deduce ( P + z 2 + P z 2 + s 3 λ 4 x 2α η 4 ξ 3 z 2 + sλ 2 x 2α η 2 ξ z x 2) dx dt ( + s 3 λ 3 ξ 3 z 2 + sλ 2 x (2α 4)/3 ξ z 2 + sλx (4α 2)/3 ξ z x 2) dx dt [ ( C e sσ g 2 + s 3 λ 3 ξ 3 z 2 + sλξ z x 2) dx dt. ω T Arguing as in Appendix A, we can replaced the integral in in the left hand side by integrals in. Moreover, thanks to the additional term in the left hand side of this inequality, we can incorporate terms with time derivatives and second-order spatial derivatives more easily and also return to the variable v: e [ 2sσ s 1 ξ [ 1 v t 2 + (x α v x ) x 2 + sλx (4α 2)/3 ξ v x 2 + sλ 2 x 2α η 2 ξ v x 2 dx dt + e 2sσ v [ 2 sλ 2 x (2α 4)/3 ξ + s 3 λ 3 ξ 3 + s 3 λ 4 x 2α η 4 ξ 3 v 2 dx dt [ C e sσ (g b 1 v x ) 2 + e [ 2sσ s 3 λ 3 ξ 3 z 2 + sλξ z x 2 dx dt. (1.81) ω T Since the power of x in the local term with rst-order spatial derivatives is negative, and we can deduce that (1.81) remains true with (g b 1 v x ) replaced by g. Finally, to eliminate the term with derivatives in the right hand side, it suces to work as in the case considered in Appendix A. 1.7 Appendix C: Proof of Theorem 3 Again we will prove Theorem when B =. In view of the presence of Robin conditions in (1.12), we will perform another change of variables: v(y, t) = e η(y) w(y, t). Now, (1.12) becomes 36

48 v t + v yy = g in, ( vx 1v) (, t) = and lim (v 2 y 12 ) v (y, t) = on (, T ), y v(x, T ) = v T (x) in (, 1), (1.82) where g := e η F (η + η 2 )v 2η v y. Let v be a solution to (1.82). For any s s >, we set z = e sσ v and z = e s σ v. We have that z and z satisfy the following initial, nal and boundary conditions: ( ( z y + sλξ 1 ) 2 ( ( z y + sλ ξ 1 ) z 2 ) z z = z = z y = z y = at t = and t = T, (, t) = and lim z y(y, t) = y + ) (, t) = and lim z y(y, t) = y + Again, we assume that v is sucintly regular. We have: lim z(y, t) = on [, T, y + lim z(y, t) = on [, T. y + v t = e sσ [sσ t z + z t, v yy = e sσ [z yy + s 2 σ 2 yz + 2sσ y z y + sσ yy z and, consequently, M 1 z + M 2 z = g 1, with M 1 z := I 11 + I 12 + I 13 := 2sλ 2 η 2 ξz 2sλ η ξz y + z t, M 2 z := I 21 + I 22 + I 23 := s 2 λ 2 η 2 ξ 2 z + z yy + sσ t z and g 1 := e sσ g + sλη ξz sλ 2 η 2 ξz. This gives M 1 z 2 + M 2 z 2 + 2((M 1 z, M 2 z)) = g 1 2. (1.83) Let us estimate ((M 1 z, M 2 z)): ((I 11, I 21 )) = 2s 3 λ 4 η 4 ξ 3 z 2 dy dt, ((I 12, I 21 )) = 3s 3 λ 3 η 2 η ξ 3 z 2 dy dt + 3s 3 λ 4 η 4 ξ 3 z 2 dy dt T + s 3 λ 3 (η ) 3 ξ 3 z 2 dt, y= ((I 13, I 21 )) = s 2 λ 2 η 2 ξ 2 θ(2t T ) z 2 dy dt, 37

49 ((I 11, I 22 )) = 4sλ 2 η η ξzz y dy dt sλ 3 (3 η 2 η + λ η 4 )ξ z 2 dy dt T T +2sλ 2 η 2 ξ z y 2 dy dt sλ 3 (η ) 3 ξ z 2 dt + 2sλ 2 η 2 ξzz y dt, y= y= T ((I 12, I 22 )) = sλ η ξ z y 2 dy dt + sλ 2 η 2 ξ z y 2 dy dt + s λ η ξ z y 2 dt, y= ((I 13, I 22 )) = T zz ty dt, y= ((I 11, I 23 )) = 2s 2 λ 2 η 2 ξσ t z 2 dy dt, and ((I 12, I 23 )) = s 2 λ η ξσ t z 2 dy dt + s 2 λ 2 η 2 ξσ t z 2 dy dt T +s 2 λ η ξσ yt z 2 dy dt + s 2 λ η ξσ t z 2 dt, y= ((I 13, I 23 )) = s 2 T σ tt z 2 dy dt. Adding all theses inequalities, we obtain after some work that ( ((M 1 z, M 2 z)) C s 3 λ 4 ξ 3 z 2 + sλ 2 ξ z x 2) dy dt ( C s 3 λ 4 ξ 3 z 2 + sλ 2 ξ z y 2) dy dt ω T T [ + s 3 λ 3 (η ) 3 ξ 3 z 2 sλ 3 (η ) 3 ξ z 2 + s 2 λη ξσ t z 2 dt + Working similarly with the function z, we obtain y= [ sλη ξ z y 2 + 2sλ 2 η 2 ξzz y + zz yt dt. (1.84) y= M 1 z + M 2 z = g 1 with M 1 z := Ĩ11 + Ĩ12 + Ĩ13 := 2sλ 2 η 2 ξ z + 2sλη ξ zy + z t, M2 z := Ĩ21 + Ĩ22 + Ĩ23 := s 2 λ 2 η 2 ξ2 z + z yy + s σ t z and g 1 := e s σ g sλη ξ z sλ 2 η 2 ξ z. This gives: g 1 2 = M 1 z 2 + M 2 z 2 + 2(( M 1 z, M 2 z)) (1.85) 38

50 and (( M 1 z, M 2 z)) C s 3 λ 4 ξ3 z 2 + sλ 2 ξ zy 2 dy dt C s 3 λ 4 ξ3 z 2 + sλ 2 ξ zy 2 dy dt ω T T + [ s 3 λ 3 (η ) 3 ξ3 z 2 + sλ 3 (η ) 3 ξ z 2 s 2 λη ξ σt z 2 dt y= T + sλη 2 ξ z y + 2sλ 2 η 2 ξ z zy + z z yt dt. (1.86) y= Note that z = z, ξ = ξ, σ = σ and σ t = σ t for y =. Hence, from (1.84) and (1.86), we nd ((M 1 z, M 2 z)) + (( M 1 z, M 2 z)) C C see that + T ω T [ s 3 λ 4 (ξ 3 z 2 + ξ 3 z 2 ) + sλ 2 (ξ z y 2 + ξ z y 2 ) dy dt [ s 3 λ 4 (ξ 3 z 2 + ξ 3 z 2 ) + sλ 2 (ξ z y 2 + ξ z y 2 ) dy dt [2sλ 2 η 2 ( ξ z z y + ξzz y ) sλη ( ξ z 2y ξz 2y) + z z ty + zz ty dt.(1.87) y= On the other hand, in view of the boundary conditions satised by z and z we = T T [zz yt + z z yt dt y= [sλξ t ( z 2 z 2 ) + 14 ( z 2 + z 2 ) t + 12 sλξ( z 2 z 2 ) t dt =, y= T T 2sλ 2 η 2 ξ[zz y + z z y dt = 2sλ 2 η 2 ξ z 2 dt, y= y= and T sλ T η ξ[ z y 2 zy 2 dt = 2s 2 λ 2 ξ 2 z 2 dt. y= y= As a consequence, from (1.87), (1.85) and (1.83), we deduce that 2 ( M i z 2 + M i z 2 ) + i=1 C [ sλ 2 (ξ z y 2 + ξ z y 2 ) + s 2 λ 4 (ξ 3 z 2 + ξ 3 z 2 ) dy dt [ [ g g sλ 2 (ξ z y 2 + ξ z y 2 ) + s 3 λ 4 (ξ 3 z 2 + ξ 3 z 2 ) ω T dy dt (1.88). 39

51 Using (1.88) and the denitions of g 1, g 1, M i z and M i z, we see that, for s and λ large enough, C s 1 + [ ϱ 1/2 g 2 + [ ξ 1 ( z t 2 + z yy 2 ) + ξ 1 ( z t 2 + z yy 2 ) dx dt [ sλ 2 (ξ z y 2 + ξ z y 2 ) + s 2 λ 4 (ξ 3 z 2 + ξ 3 z 2 ) ω T dy dt [ sλ 2 (ξ z y 2 + ξ z y 2 ) + s 3 λ 4 (ξ 3 z 2 + ξ 3 z 2 ) dy dt. From classical arguments, we can eliminate the terms with derivatives in the right hand side of the inequality and nd that [ ξ 1 ( z t 2 + z yy 2 ) + ξ 1 ( z t 2 + z yy 2 ) dx dt s 1 + C [ ϱ 1/2 g 2 + s 3 λ 4 [ sλ 2 (ξ z y 2 + ξ z y 2 ) + s 2 λ 4 (ξ 3 z 2 + ξ 3 z 2 ) ω T [ξ 3 z 2 + ξ 3 z 2 dy dt dy dt. We then conclude the proof coming back to the original variable w and using the denition of g and the fact that ξ C ξ Cξ. 4

52 Capítulo 2 Estimativas de Carleman para algumas EDPs parabólicas degeneradas em dimensão 2 e aplicações Carleman estimates for some 2-D degenerate parabolic PDEs and applications F. D. ARARUNA, B. S. V. de ARAÚJO, and E. FERNÁNDEZ-CARA Abstract This paper deals with a class of two-dimensional degenerate parabolic equations in a squere. The goal is to obtain Carleman estimates to obtain controllability results. In order to be more faithfull with the 1D degenerate problem, we consider that the degeneracy ocour only in a part of the boundary. Then, we present well posedness results and, under some geometrical assumptions, we build suitable weight functions that allows to deduce a global Carleman estimates. As an application, we prove some null controllability and Stackelberg-Nash null controllability results. Keywords: Null controllability, degenerate parabolic equations, Carleman inequalities, Stackelber-Nash strategies. Mathematics Subject Classication: 34K35, 49J2, 35K1.

53 2.1 Introduction The study of the controllability of partial dierential equations and systems has attracted the interest of many authors. The theory has been extended to semilinear problems, equations in unbounded domains of some kinds, and systems in uid dynamics, among others; see for instance [15, 18, 2, 21, 24. The study of controllability of degenerate parabolic equations started in the last decade with the works [1, 7, 8, 9, 12, 13, 14. In this paper, we will analyze the following problem in two spatial dimensions: u t div(a u) + bu = g1 ω in, B.C. on Σ, u(, ) = u in Ω, (2.1) where Ω = (, 1) (, 1), Γ := Ω, T >, = Ω (, T ), Σ := Γ (, T ), ω Ω is a non-empty open set, b L (), g L 2 (), u L 2 (Ω), A : Ω M 2 2 (R) is given by A(x) = diag(x α 1 1, x α 2 2 ), and the boundary conditions are given by u = on Σ if α 1, α 2 [, 1), u = on Σ 3,4 and (A u)ν = on Σ 1,2 if α 1, α 2 [1, 2, B.C. := u = on Σ 1,3,4 and (A u)ν = on Σ 2 if α 1 [, 1), α 2 [1, 2, u = on Σ 2,3,4 and (A u)ν = on Σ 1 if α 1 [1, 2, α 2 [, 1), with α = (α 1, α 2 ) [, 2 [, 2, Σ i,j,l := (Γ i Γ j Γ l ) (, T ), and Γ 1 := {} [, 1, Γ 2 := [, 1 {}, Γ 3 := {1} [, 1, Γ 4 := [, 1 {1}. In previous papers, the main model is the following: u t (x α u x ) x = g(x, t)1 O in (, 1) (, T ), u(, ) = if α [, 1) u(1, ) = and on (, T ), (x α u x )(, ) = if α [1, 2) u(, ) = u in (, 1), (2.2) where α (, 2), T >, u L 2 (, 1), g L 2 ((, 1) (, T )), O (, 1) is a non-empty open set, and 1 O is the associated characteristic function. The global null controllability 42

54 of this system is proved using Carleman estimates with appropriate weight functions [1. Recently, some results on 2D degenerate parabolic equations have appeared in [1, 11. There, the authors study the well-posedness and the global null controllability of the following system: u t div(a u) + b u = g 1 ω in, u = if α (, 1) u ν = if on α [1, 2) Σ, u(, ) = u in Ω, (2.3) where := Ω (, T ), T >, Ω R 2 is a bounded domain with boundary Γ of class C 4, Σ := Γ (, T ), ω Ω is open, u L 2 (Ω ), g L 2 ( ), b L ( ), α (, 2), and A : Ω M 2 2 (R) satises the following conditions: (i) A (x) = {a ij (x)}, with the a ij C 3 (Ω ; R) C (Ω ; R); (ii) A (x) is simetric for all x Ω and positive denite for all x Ω ; (iii) Let r i (x) be the eigenvalues and let ε i (x) be the associated unit-norm eigenvectors of A (x) for i = 1, 2. Let us denote by P Γ (x) the projection of x onto the boundary Γ and O(Γ ; δ) := {x Ω : d(x, Γ ) < δ}. There exists δ > such that the following holds: 1. r 1 (x) = d(x, Γ ) α x O(Γ ; δ), 2. r 2 (x) > x Ω \O(Γ ; δ); 3. ε 1 (x) = ν(p Γ (x)) x O(Γ ; δ). It is convenient to enhance the main dierences between systems (2.3) and (2.1). First, contrarily to (2.1), (2.3) degenerates on the whole border. This is crucial, because Carleman estimates need weight functions with a specic behavior near the part of the boundary where the degeneracy occurs. The techniques used in [1 can be extended to problems where the degeneracy appears on a part of the boundary, but this part must be separated from the rest. A typical example is when Ω is an annulus and the coecients degenerate just one of the components of the boundary. Hence, a new thecnique is required to deal with (2.1). 43 Here, the main idea is to use the control

55 domain to "separate" the part of the boundary where the degeneracy does not occur from a neighborhood of the origin. Note that not only one but the two eigenvalues of the matrix A can degenerate. In this work, we will prove the well-posedness of (2.1) using semigroup theory in all cases α i [, 2. However, the proof of Carleman estimates requires dierent techniques and weights in each case; therefore, for this purpose, we will strict to the case α 1, α 2 (, 2). The other cases will be considered in a forthcoming paper. It will be seen that, as an application of Carleman estimates, we can prove results on the null controllability and Stackelberg-Nash controllability for linear and semilinear degenerate parabolic systems. This paper is organized as follows. In Section 2.2, we will present some results on the degenerate operator associated to (2.1) and we will deduce the well posedness. In Section 2.3, we will present Carleman estimates for the solutions to the adjoint of (2.1) when α i (, 2). This will allow to prove results concerning the null controllability of linear and semilinear problems of the kind (2.1). In Section 2.4, we will use these Carleman estimates to prove the Stackelberg-Nash controllability. In Section 2.5, we will present some extensions of the results, open questions and future work. Finally, the paper contains an Appendix where the proof of the Carleman estimate presented in the Section 2.3 is given. 2.2 Preliminar results and well-posedness Notations, spaces and operators The usual norm and inner product in L 2 (Ω) and in L 2 () will be denoted respectively by and (, ), and and ((, )). The norms in L (Ω) and L () will be denoted respectively by and. Now, let us introduce some matrices, spaces, and operators. A r (x) := diag(x α 1r 1, x α 2r 2 ), with r R, H 1 α(ω) := {u L 2 (Ω) : ua u L 1 (Ω)}, H 2 α(ω) := {u H 1 α(ω) : div(a u) L 2 (Ω)}, 44

56 H div (Ω) := {w L 2 (Ω) 2 : div(w) L 2 (Ω)}, L 2 α 1 (Ω) := {w L 2 (Ω) 2 : wa 1 w L 1 (Ω)}, Hα div (Ω) := {w L 2 α (Ω) : div(w) L 2 (Ω)}, 1 u := A 1/2 u, u H 1 α(ω), u := div(a u), u H 2 α(ω). Now, let us consider the following norms: u α := ( u 2 + u 2 ) 1/2, u H 1 α(ω), u 2,α := ( u 2 α + u 2 ) 1/2, u Hα(Ω), 2 ( ) 1/2 w div,α := wa 1 w dx + div(w) 2, Ω w Hα div (Ω). Note that, for these (natural) norms, Hα(Ω), 1 Hα(Ω) 2 and Hα div (Ω) are Hilbert spaces and one has the following continuous embbedings: H 1 (Ω) Hα(Ω) 1 L 2 (Ω), Hα(Ω) 2 Hα(Ω), 1 Hα div (Ω) H div (Ω). Furthermore, H 1 α(ω) H 1 loc (Ω) and H2 α(ω) H 2 loc (Ω). Lemma C (Ω) is dense in H 1 α(ω). Proof. We know that H 1 (Ω) H 1 α(ω) and that C (Ω) is dense in H 1 (Ω). Hence, it is sucient to prove that H 1 (Ω) is dense in H 1 α(ω). Let us x u H 1 α(ω). For λ > 1 let us introduce the dieomorphism f λ : Ω Ω λ where f λ (x) = x λ + x,λ and x,λ = 1 (λ 1, λ 1). 2λ It is clear that Ω λ Ω and Ω λ Ω, as λ 1. Now let us introduce u(x) if x Ω λ u λ (x) := u(f λ (x)) if x Ω\Ω λ. Using that Hα(Ω) 1 Hloc 1 (Ω) and Ω λ Ω we deduce that u λ H 1 (Ω). Furthermore, u λ u 2 α = u(f λ (x)) u(x) 2 + (u f λ u)(x) 2 dx Ω\Ω λ (λ 2 u 2 + u 2 ) dx + ( u 2 + u 2 ) dx as λ 1. f λ (Ω\Ω λ ) Ω\Ω λ 45

57 Therefore, H 1 (Ω) is dense in Hα(Ω). 1 Lemma leads to the following denition Hα,(Ω) 1 Hα := D 1 (Ω), (2.4) where the denition of the space D depends on α: {v C (Ω) : supp(v) Ω} if α 1, α 2 [, 1), {v C (Ω) : δ > ; supp(v) (, 1 δ) (, 1 δ)} if α 1, α 2 [1, 2, D := {v C (Ω) : δ > ; supp(v) (δ, 1 δ) (, 1 δ)} if α 1 [, 1), α 2 [1, 2, {v C (Ω) : δ > ; supp(v) (, 1 δ) (δ, 1 δ)} if α 1 [1, 2, α 2 [, 1) Trace operators We know that the trace operator T : H 1 (Ω) L 2 (Γ) is continuous and H 1/2 (Γ) := T (H 1 (Ω)) is a Hilbert space for the norm v H 1/2 (Γ) := inf{ u H 1 (Ω) : u H 1 (Ω), T (u) = v}. Moreover, there exists a unique normal trace operator T ν : H div (Ω) H 1/2 (Γ), which is continuous and satises T ν (w) = (w ν) Γ w C (Ω) 2, (div(w)u + w u) dx = T ν (w), T (u) w H div (Ω) u H 1 (Ω). Ω Given w H div (Ω), we can introduce a continuous linear form T w : H 1 (Ω) R by putting T w (u) := (div(w)u + w u) dx u H 1 (Ω). Moreover, we have T w (u) = T ν (w), T (u). Now, for any δ >, we introduce Ω δ := {x Ω : d(x, Γ) > δ} and Γ δ := Ω δ, r δ : Hα(Ω) 1 H 1 (Ω), with r δ (u) := u Ωδ, 46 Ω

58 R δ : H div α (Ω) H div (Ω), with R δ (w) = w Ωδ, T δ := T δ r δ, where T δ : H 1 (Ω δ ) H 1/2 (Γ δ ) is the trace operator, T δ ν := T ν,δ R δ, where T ν,δ : H div (Ω δ ) H 1/2 (Γ δ ) is the normal trace operator. Lemma If w Hα div (Ω), then the functional T w can be continuously extended to the space Hα(Ω). 1 Moreover, Proof. For u Hα(Ω) 1 we have (div(w)u + w u)dx = Ω T w (u) = lim δ T δ ν (w), T δ (u) u H 1 α(ω). Ω (div(w)u + [(A 1/2 ) 1 w[a 1/2 u)dx div(w) u + (A 1/2 ) 1 w A 1/2 u ( (A 1/2 ) 1 w + div(w) )( u + A 1/2 u ) w div,α u α. This conclude the proof. In view to obtain appropriate results on the normal traces in Hα div (Ω), we will consider the cases α i [, 1) and α i [1, 2 separately. The case α 1, α 2 [, 1) The classical results on the normal trace theory remain true for the spaces Hα div (Ω) and H 1 α(ω). Lemma There exists a unique trace operator T α : Hα(Ω) 1 L 2 (Γ) which extends T : H 1 (Ω) L 2 (Γ). Moreover T α is continuous and Hα 1/2 (Ω) := T α (Hα(Ω)) 1 is a Hilbert space equipped with the norm v := inf{ u 1/2 H α (Γ) α : u Hα(Ω) 1 and T α (u) = v}. Proof. We know that C (Ω) is dense in H 1 α(ω). Consequently, it is sucient to prove that T : (C (Ω), α ) L 2 (Γ) is continuous. First let us consider the following open cover U,..., U 4 of Ω, where U := B 1/4 ((1/2, 1/2)), U 1 := B 1/2 ((, 1/2)), U 2 := B 1/2 ((1/2, )), U 3 := B 1/2 ((1, 1/2)) and U 4 := B 1/2 ((1/2, 1)). Let ϕ,..., ϕ 4 C (Ω) be a partition of unity subordinate to the cover {U,..., U 4 }, i.e. ϕ ϕ 4 = 1, 47

59 ϕ 1, supp ϕ is a compact set and supp ϕ U i. Given u C (Ω), let u i = ϕ 1/2 u. Using that α i [, 1) we deduce that Hence u 1 (, x 2 ) = 3/4 x α 1/2 u 1 1 x α 1/2 1 dx 1 C x 1 u 1 2 ds C Γ 1 In a similar way we conclude that Ω x α 1 1 u 1 x 1 2 ( 1 x α 1 u 1 x 1 dx C u 2 α. Γ i u i 2 ds C u 2 α, i = 1,..., 4. 2 dx 1 ) 1/2. Therefore T (u) 2 = u 2 ds = Γ 4 ϕ i u 2 ds = i= Γ 4 i= Γ i u i 2 ds C u 2 α. This conclude the proof. The following result is a fundamental tool, not only to deduce the existence of a normal trace operator, but also to prove the Carleman estimates present in the next section. Lemma (Hardy inequality I) positive constant C = C(α 1, α 2 ) such that Ω x α i 2 i u 2 dx C x α i i Ω Assume that α 1, α 2 [, 1). There exists a u x i 2 dx u ker(t α ). Before proving this Hardy Inequality, we consider the following result: Lemma Assume α 1, α 2 [, 1). There exist C = C(α) > such that, for any u C (Ω), one has Ω x α u(x) u(x 1, ) 2 dx C x α 1 1 Ω u x 1 2 dx and Ω x α u(x) u(, x 2 ) 2 dx C x α 2 2 Ω u x 2 2 dx. 48

60 Proof. Let us x max{α 1, α 2 } < a < b < 1. We have that [ 1 1 ( x2 ) 2 x α u(x) u(, x 2 ) 2 b/2 u dx = t (x 1, t)t b/2 dt x α dx 2 x 2 Ω and Thus t 1 = b 1 1 b [ ( 1 x2 ) 2 ( t b u x2 (x 1, t) x 2 dt [ 1 ( 1 x2 ) 2 t b u (x 1, t) x 2 dt [ ( t b u 1 (x 1, t) x 2 x α 2 1 b 2 dx 2 = 1 t x α 2 a 2.x a 1 b 2 dx 2 t α 2 a = tα 2 a a b (1 ta b ) tα2 b b a. u(x) u(, x 2 ) 2 x α dx C x α 2 2 Ω The other inequality can be proved in a similar way. t dx 1 ) t b dt x α dx 2 dx 1 x α 2 1 b 2 dx 2 dx 1 ) x α 2 1 b 2, dx 2 dt dx 1 1 t u x 2 Now, we will return to the Lemma to establish its proof. Proof. [Proof of Lemma x a 1 b 2 dx 2 Now, let us x u ker(t α ) and δ >. There exists a sequence u n C (Ω) such that u n u in H 1 α(ω). Let us introduce two auxiliary sequences given by u 1,n (x) = u n (x 1, ) and u 2,n (x) = u n (, x 2 ). From T α (u n ) in L 2 (Γ) we deduce that u i,n in L 2 (Ω). Using Lemma we have that u n u i,n 2 x α i 2 i Ω δ dx C Passing to the limit n + we obtain that u 2 x α 2 i dx C u Ω δ Ω x i This conclude the proof. From Lemma the following result holds: 2 Ω u n x i 2 x α i i dx. x α i dx, δ >. Lemma ker (T α ) = H 1 α,(ω) (recall the denition of this space in (2.4)). dx. 49

61 Proof. It is clear that H 1 α,(ω) ker(t α ). Let us x u ker(t α ) and introduce f(x) := x 1 x 2, f n (x) := min{nf(x), 1}, u n := f n u and Ω n := {x Ω : f n (x) = 1}. It is clear that f n = on Γ, Ω n Ω n+1 and Ω n Ω. Furthermore, ( ) ( u n 2 + u n 2 ) dx C(n) u 2 dx + u 2 dx < + Ω Ω Ω and u n H 1 (Ω). On the other hand T (u n ) = T α (u n ) = f n Γ T α (u) =, then u n ker(t ) = H 1 (Ω). Moreover, we have that Ω ( u u n 2 + (u u n ) 2 ) dx = ( u f n u 2 + (u f n u) 2 ) dx Ω\Ω n C ( u 2 + u 2 ) dx + n Ω\Ω 2 f 2 u 2 dx. (2.5) n Ω\Ω n The rst integral in the right hand side of (2.5) tends to as n +. With respect to the second integral, using Lemma we have that n 2 Ω\Ω n f 2 u 2 dx (x 1 x 2 ) 2 (x α 1 1 x x α 2 2 x 2 1) u 2 dx Ω\Ω n (x α x α ) u 2 dx. Ω We conclude that u n u in H 1 α(ω) and, consequently, ker(t α ) H 1 (Ω) H1 α(ω) H 1 α,(ω) H1 α(ω) = H 1 α (Ω). As a consequence of Lemmas and 2.2.6, the space H 1 α,(ω) can be endowed with the norm u α, := that is equivalent to the norm α on H 1 α,(ω). Ω u 2 dx, Lemma There exists a unique normal trace operator Tν α that is continuous and satises : Hα div (Ω) H 1/2 (Γ) T α ν (w) = (w ν) Γ w C (Ω) 2, T w (u) = Tν α (w), T α (u) w Hα div (Ω) u Hα(Ω). 1 To prove Lemma 2.2.7, before we need to stablish some preliminaries results. Lemma Assume that α i [, 1). Then H 1 α(ω) W 1,1 (Ω). 5

62 Proof. From Hölder inequality [ u Ω x i dx x α i i dx. Ω Ω x α i i u x i 2 dx < u H 1 α(ω). This conclude the proof. Lemma Assume α i [, 1). Given u Hα(Ω), 1 let us introduce { u(x) if x Ω ũ(x) := if x R 2 \Ω. If ũ W 1,1 (R 2 ), then u H 1 α,(ω). Proof. From Lemma 2.2.8, we have that H 1 α,(ω) = H 1 α(ω) W 1,1 (Ω). Hence, it is sucient to show that u W 1,1 (Ω). Given λ > 1 let us introduce the dieomorphism f λ : Ω Ω λ where f λ (x) = λx x 1,λ and x,λ := 1 (λ 1, λ 1). 2λ It is clear that Ω Ω λ and Ω λ Ω as λ 1. Now let us introduce u λ (x) := u(x) if x f 1 λ (Ω) ũ(f λ (x)) if x f 1 λ (Ω). It is clear that supp(u λ ) Ω, whence it remains to conclude that u λ W 1,1 (Ω) and u λ u in W 1,1 (Ω). In fact, we have that u λ W 1,1 (Ω) = ( u + u ) dx + f 1 λ (Ω) u α + f λ (Ω\f 1 λ (Ω)) ( u(f λ (x)) + u(f λ (x)) ) dx Ω\f 1 λ (Ω) ( u λ 2 + u λ 3 ) dx < +. Hence, u λ W 1,1 (Ω). Moreover, arguing as in the proof of Lemma we conclude that u λ u in W 1,1 (Ω). Lemma Assume that the α i [, 1). Then C (Ω) 2 is dense in Hα div (Ω). Proof. Let us denote by (, ) α,div the inner product of H div α (Ω). Let us x w (C (Ω) 2 ) := {w Hα div (Ω) : (w, v) α,div = v C (Ω) 2 } and u = div(w) L 2 (Ω). We have that Ω udiv(v) dx = (A 1 w)v dx v C (Ω) 2. Ω 51

63 Thus u = A 1 w in the sense of distributions. Consequently, A u = w L 2 α 1 (Ω), that is to say, ua u L 1 (Ω). Then u H 1 α(ω). Moreover, since α i [, 1), we deduce that u = A 1 w L 1 (Ω) and à 1 w L 1 (R 2 ) 2. Using that ũ = à 1 w in the sense of distributions we conclude that ũ W 1,1 (R 2 ). From Lemma A.3, we see that u H 1 α,(ω). Consequently there exists a sequence u n C (Ω) such that u n u in H 1 α(ω). Using that Hα div (Ω) H div (Ω), from standard normal trace theory we get (v u n + div(v)u n ) dx = v Hα div (Ω). Ω Consequently, for any xed v Hα div (Ω), we have that (w, v) α,div = (wa 1 v + div(w)div(v)) dx = (v u + udiv(v)) dx Ω Ω = (v (u u n ) + (u u n )div(v)) dx Ω ( A 1/2 (u u n ) (A 1/2 ) 1 v + u u n div(v) ) dx Ω u u n α v div,α. Therefore w = and thus (C (Ω) 2 ) = {}. Finally, we are ready to give the proof of Lemma Proof. [Proof of Lemma Let T ν : H div (Ω) H 1/2 (Γ) be the standard normal trace operator. We know that H div α (Ω) H div (Ω). Then, from Lemma A.4 it is sucient to prove that T ν : (C (Ω) 2, div,α ) H 1/2 α (Γ) is continuous. For this purpose, let us x w C (Ω) 2. Given u H 1 α(ω) there exists a sequence u n C (Ω) such that u n u in H 1 α(ω). We have that T w (u n ) = T ν (w), T α (u n ) and from Lemma we get T w (u) = T ν (w), T α (u). This way, we can deduce that T ν (w), T α (u) w div,α. u α u H 1 α(ω). (2.6) Given v Hα 1/2 (Γ), there exists a sequence u n Hα(Ω) 1 such that T α (u n ) = v and u n α v 1/2 H. From (2.6) we get: α (Γ) T ν (w), v w div,α. u n α. Consequently T ν (w), v w div,α. v 1/2 H v H1/2 α (Γ) α (Γ) and therefore T ν (w) 1/2 H α w div,α. Hence T ν : (C (Ω) 2, div,α ) L 2 (Γ) is continuous. 52 (Γ)

64 The case where α i [1, 2 for some i Now, let us assume that α i [1, 2 for some i {1, 2}. It is no possible to prove the existence of a trace operator, but we have the following: Lemma For any w Hα div (Ω), one has (w u + div(w)u) dx = u Hα,(Ω). 1 Ω Before to prove Lemma , we will stablish the following result: Lemma Assume w = (w, 1, w 2 ) H div (Ω) and u D. The functions f, g : [, 1/2 R, given by f(δ) := T δ u 2 ds and g(δ) := Tν δ (w), T δ (u) Γ δ are continuous in δ =. Proof. There exists δ > such that supp(u) (, 1 δ ) (, 1 δ ). Thus, f(δ) = f 1 (δ) + f 2 (δ) δ [, δ /2 where f 1 (δ) := 1 δ δ u(δ, x 2 ) 2 dx 2 and f 2 (δ) := 1 δ δ u(x 1, δ) 2 dx 1. Given δ 1, δ 2 [, δ /2, with δ 1 < δ 2, we have that 1 δ2 δ2 f 1 (δ 1 ) f 1 (δ 2 ) ( u(δ 1, x 2 ) 2 u(δ 2, x 2 ) 2 ) dx 2 + u(δ 1, x 2 ) 2 dx 2 δ 2 δ 1 1 δ1 + u(δ 1, x 2 ) 2 dx 2 1 δ 2 1 δ2 δ2 u L (Ω) δ 1 δ 2 + ( u 2 ) dx 1 dx 2 δ 2 δ 1 x 1 ( u L (Ω) + u L (Ω)) δ 1 δ 2. Hence, f 1 is continuous. In a similar way we deduce that f 2 is continuous. Therefore f is continuous in [, δ /2. Now let us x δ 1, δ 2 [, δ /2 with δ 1 < δ 2. From standard normal trace theory, g(δ 1 ) g(δ 2 ) (div(w)u + w. u) dx w H div (Ω δ1 \Ω δ2 ). u H 1 (Ω δ1 \Ω δ2 ). Ω δ1 \Ω δ2 Thus, since Ω δ1 \Ω δ2 as δ 1 δ 2 the conclusion follows. 53

65 Now we will to come back to proof of Lemma proof of Lemma To x ideas, we will present the proof only in the case α 1, α 2 [1, 2. The other cases are similar. Let us x w = (w 1, w 2 ) C (Ω) 2, u D 1, δ > such that supp(u) (, 1 δ ) (, 1 δ ) and δ 1, δ 2 (, δ /2) with δ 1 < δ 2. We have that δ2 δ 1 1 δ T δ ν (w), T δ (u) 2 dδ δ2 ( 1 1 δ δ 1 δ δ 1 δ + δ u 2 L (Ω) 1 δ w 1 (δ, x 2 ) 2 dx 2. u(δ, x 2 ) 2 dx 2 δ 1 δ ) w 2 (x 1, δ) 2 dx 1. u(x 1, δ) 2 dx 1 dδ ( δ ) x 1 1 w 1 2 dx + x 1 2 w 2 2 dx. (δ 1,δ 2 ) (,1) (,1) (δ 1,δ 2 ) We know that C (Ω) 2 is dense in H div (Ω). Hence, for all w H div (Ω) one has δ2 ( ) 1 δ 1 δ T ν δ (w), T δ (u) 2 dδ u 2 L (Ω) x 1 1 w 1 2 dx + x 1 2 w 2 2 dx. (δ 1,δ 2 ) (,1) (,1) (δ 1,δ 2 ) Moreover, if w Hα div (Ω) we deduce that δ2 1 δ 1 δ T ν δ (w), T δ (u) 2 dδ u 2 L (Ω) w div.α < δ 1 < δ 2 < δ /2. This gives δ2 1 δ T δ ν (w), T δ (u) 2 dδ u 2 L (Ω) w div.α < +. From the continuity of g in δ = we deduce that From Lemma we conclude that g() = lim δ g(δ) =. T w (u) = w H div α (Ω) and u D 1. Now the result follows from the density of D 1 in H 1 α,1(ω). We end this section with a Hardy-like inequality corresponding to this case. Lemma (Hardy inequality II) Assume that α i 1 for i = 1, 2. There exist C = C(α 1, α 2 ) such that 2 x α i 2 i u 2 dx C x α i u i Ω Ω x i dx u Hα,(Ω). 1 54

66 Proof. Again, we will present the proof only for the case α 1, α 2 (1, 2. The other cases could be deduced using these ideas combined with the present in the case α 1, α 2 [, 1). Let us x v D, x 1 (, 1) and put β = 2 α 1. We have that 1 v(x 1, x 2 ) 2 = 1 (3 β)/4 v s x 1 1 s (3 β)/2 v x 1 1 s (3 β)/2 v x 1 1 s (3 β)/2 v x 1 2 s v(s, x 2)s (3 β)/4 ds 2 s (s, x 1 2) ds. s (3 β)/2 ds x 1 2 s (s, x 2) 2 ds. 1 β (x(β 1)/2 1 1) 2 s (s, x 2) 2 ds. α 1 x(β 1)/2 1. (2.7) Multiplying (2.7) by x β 1 and integrating x 1 in (, 1), we get: ( x β 1 v(x 1, x 2 ) ) 2 dx 1 s (3 β)/2 v α 1 1 x 1 s (s, x 2) ds x ( 1 β)/2 1 dx ( = s (3 β)/2 v α 1 1 s (s, x s ) 2) x ( 1 β)/2 1 dx 1 ds = s (3 β)/2 v (α 1 1) 2 s (s, x 2) s (1 β)/2 ds = x α 1 v (α 1 1) 2 1 (x 1, x 2 ) x 1 dx 1. (2.8) Now, integrating (2.8) with respect to x 2 in (, 1) we obtain: Ω x α v 2 dx C x α 1 1 Ω v x 1 2 dx v D. In a similar way we deduce that Ω x α v 2 dx C x α 2 2 Ω v x 2 2 dx v D. The result follows by density argument Well posedness Let us x α = (α 1, α 2 ) in [, 2 [, 2. Lemma The operator : D( ) L 2 (Ω), where D( ) := Hα(Ω) 2 Hα,(Ω), 1 is m-dissipative and self-adjoint. Moreover, if α 1 1 α 2, then is strictly dissipative. 55

67 Proof. [Proof of Lemma Let us consider the bilinear form q : Hα(Ω) 1 2 R given by q(u, v) := ua v dx. It is clear that q is simmetric and q(u, u). Furthermore q(u, v) u α. v α. Ω Hence q is continuous. Moreover, from Lemmas and we deduce that q is coercive on H 1 α,(ω) if α 1. have that Now supose that (u, v) H 2 α(ω) H 1 α,(ω). From Lemmas and we q(u, v) = u.v dx. Ω From the properties of q, the result follows. As consequence of Lemma , is the innitesimal generator of a strongly continuous semigroups. Thus, using standard techniques, we can prove the following well-posedness result: Theorem For any g L 2 () and any u L 2 (Ω), there exists a unique solution u C ([, T ; L 2 (Ω)) L 2 (, T ; Hα,(Ω)) 1 to (2.1). Furthermore, there exists a positive constant C such that T sup u(t) 2 + u(t) 2 α dt C ( u 2 + g 2). t [,T (2.1): 2.3 Carleman estimates and null controllability results In this section we will assume that α 1, α 2 (, 2). Let us consider the adjoint of w t w + bw = f in, B.C. on Σ, w(, ) = w in Ω. (2.9) In order to establish a Carleman estimate for the solutions to (2.9), we will assume the following on the observabillity domain: 56

68 δ >, a, b (2δ, 1 2δ ) such that ω := B δ (P 1 ) B δ (P 2 ) ω, where B δ (P i ) := {x Ω : x P i < δ }, P 1 = (, b ), P 2 (a, ). (2.1) Lemma Assume that (2.1) holds. There exist C >, c 1 (b δ, b ), c 2 (a δ, a ), d 1 (b, b + δ ), d 2 (a, a + δ ), and η C (Ω) C (Ω) such that, for V := {x Ω : c 1 x 1 + c 2 (x 2 c 1 ) < } and W := {x Ω : d 1 x 1 + d 2 (x 2 d 1 ) > }, one has: (i) η(x) = 2 i=1 x 2 α i i 2 α i in V, η(x) = 2 i=1 (ii) η, ηa η, A η C > in (Ω\ω )\V, (iii) A η ( ηa η), A (div(a η)), 2 i=1 x 2 α i i 2 α i in W, ( A x α i η x i ), 2 i=1 x 2α 1 i η C (Ω). x i Proof. Figura 2.1: illustration of the sets V and W. For instance, let us x V 1 = V with c 1 = b δ /2 and c 2 = a δ /2 and W 1 = W with d 1 = b + δ /2 and d 2 = a + δ /2. Let us consider h C (Ω) such that h(x) = x 1 2 α 1 + x 2 2 α 2 in V 1 and h(x) = x 1 2 α 1 x 2 2 α 2 in W 1. 57