Universidade Federal da Paraíba

Universidade Federal de Campina Grande

Programa Associado de Pós-Graduação em Matemática

Doutorado em Matemática

Controle hierárquico via estratégia deStackelbergNash para controlabilidadede sistemas parabólicos e hiperbólicos

por

Luciano Cipriano da Silva

João Pessoa - PB

Março/2017

Controle hierárquico via estratégia deStackelbergNash para controlabilidadede sistemas parabólicos e hiperbólicos

por

Luciano Cipriano da Silva †

sob orientação do

Prof. Dr. Fágner Dias Araruna

e co-orientação do

Prof. Dr. Enrique Fernández 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

Março/2017

†Este trabalho contou com apoio nanceiro da ......

ii

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

Prof. Dr. Enrique Fernández Cara

Co-orientador

Prof. Dr. Damião Junio Gonçalves de Araújo

Prof. Dr. Felipe Wallison Chaves Silva

Prof. Dr. Juan Bautista Límaco Ferrel

Prof. Dr. Maurício Cardoso Santos

Prof. Dr. Aldo Trajano Lourêdo (Suplente)

Prof. Dr. Diego Araújo de Souza (Suplente)

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.

Março/2017

iii

Resumo

Nesta tese apresentamos resultados sobre controlabilidade exata de Equações Diferenciais

Parciais (EDPs) dos tipos parabólico e hiperbólico, no contexto de controle hierárquico,

usando a estratégia de Stackelberg-Nash. Em todos os problemas consideramos um con-

trole principal (líder) e dois controles secundários (seguidores). Para cada líder obtemos

um equilíbrio de Nash correspondente, associado a um problema de controle ótimo bi-

objetivo; então buscamos o líder de custo que resolve o problema de controlabilidade.

Para os problemas parabólicos temos controles distribuídos e na fronteira, já nos hiper-

bólico todos os controles são distribuídos. Consideramos casos lineares e semilineares, os

quais resolvemos usando desigualdade de observabilidade obtidas combinando desigual-

dades de Carleman adequadas. Também usamos um método de ponto xo.

Palavras-chave: Controlabilidade; Controle hierárquico; Estratégia de Stackelberg-Nash;

Desigualdade de observabilidade; Desigualdade de Carleman; Equação do calor; Equação

da onda; Equação de Burgers.

iv

Abstract

In this thesis we presents results on the exact controllability of the partial Dierential

Equations (PDEs) of the parabolic and hyperbolic type, in the context of hierarchic

control, using the Stackelberg-Nash strategy. In every problems we consider a main control

(leader) and two secondary controls (followers). To each leader we obtain a correnponding

Nash equilibrium, associated to a bi-objective optimal control problem; then we look for

a leader of minimal cost that solves the exact controllability problem. For the parabolic

problems we have distributed and boundary controls, now in the hyperbolics every controls

are distributed. We consider linear and semilinear cases, which we solve using observability

inequality obtained combining right Carleman inequalities. Also we use a xed point

method.

Keywords: Controllability; Hierarchic control; Stackelberg-Nash strategy; Observability

inequality; Carleman inequality; Heat equation; Wave equation; Burgers' equation.

v

Agradecimentos

A Deus, força ...

A minha mãe, ...

A minha esposa ...

vi

Dedicatória

À minha mãe Lucia e à minha esposa

Joseane.

vii

Sumário

Introdução . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1 Hierarchic control for exact controllability of parabolic equations with

distributed and boundary controls 20

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.1.1 Main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

1.2 The exact controllability with distributed leader and boundary followers . . 27

1.2.1 The linear case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.2.2 The semilinear case . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

1.3 Hierarchic control with boundary leader and distributed followers . . . . . 46

1.3.1 The linear case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

1.3.2 The semilinear case . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

1.4 Final comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

1.4.1 The case with all controls on the boundary . . . . . . . . . . . . . . 55

1.4.2 The assumption on the µi . . . . . . . . . . . . . . . . . . . . . . . 56

1.4.3 The hypotheses on the Oi,d and Γi,d . . . . . . . . . . . . . . . . . . 56

1.4.4 Hierarchic control of the wave equation . . . . . . . . . . . . . . . . 56

2 Hierarchic control for the wave equation 58

2.1 Statement of the problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

2.1.1 Main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

2.2 The linear case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

2.2.1 The existence and uniqueness of a Nash equilibrium . . . . . . . . . 63

2.2.2 The Optimality system . . . . . . . . . . . . . . . . . . . . . . . . . 65

2.2.3 Null controllability . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

2.3 The semilinear case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

2.3.1 Equilibrium and Quasi-equilibrium . . . . . . . . . . . . . . . . . . 75

2.4 Final comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

2.4.1 Boundary controllability . . . . . . . . . . . . . . . . . . . . . . . . 78

2.4.2 On the nonlinearity . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

3 Hierarchic control for Burgers equation via StackelbergNash strategy 80

3.1 Introducton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

3.1.1 Statement of the problem . . . . . . . . . . . . . . . . . . . . . . . 81

3.1.2 Main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

3.2 Existence and uniqueness of Nash equilibrium . . . . . . . . . . . . . . . . 85

3.3 Null controllability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

3.4 Final comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

3.4.1 StackelbergNash strategy with controls in the boundary . . . . . . 95

3.4.2 The NavierStokes system . . . . . . . . . . . . . . . . . . . . . . . 95

Referências 97

ix

Introdução

Nesta tese apresentamos alguns resultados sobre controlabilidade de sistemas as-

sociados a equações diferenciais parciais parabólicas e hiperbólicas. Vamos resolver os

problemas de controlabilidade no contexto do que Lions [50] chamou de controle hierár-

quico e aplicaremos a estratégia de StackelbergNash.

A grosso modo, controlar um sistema de equações diferenciais ordinárias ou parciais,

como o próprio nome sugere, é atuar sobre ele para obter um comportamento desejado.

Um problema de controle para sistemas governados por uma equação parabólica ou hi-

perbólica pode ser formulado como segue: Dados um intervalo de tempo [0, T ], um estado

inicial e um estado nal, buscamos por um controle que atua na trajetória do sistema (o

lado direito da equação diferencial ou a condição de fronteira) , de modo que, a solução

do sistema associado a tal controle seja igual ao estado inicial no tempo t = 0 e ao estado

nal no tempo t = T .

Para denir os tipos de controlabilidade vamos considerar um problema linear abs-

trato: yt + Ay = Bu em (0, T ),

y(·, 0) = y0 em Ω,(1)

onde A e B são operadores lineares, u : [0, T ] → U é o controle e y : [0, T ] → H é o

estado. Por H e U denotamos dois espaços de funções adequados. Então xado T > 0,

podemos denir vários tipos de problemas de controlabilidade no tempo T .

Controlabilidade exata: Dados y0 e y1 estados possíveis do sistema, obter um controle

u tal que a correspondente solução y de (1) satisfaz y(T ) = y1.

Controlabilidade aproximada: Dados y0 e y1 estados possíveis do sistema e um nú-

mero ε > 0, obter um controle u tal que a correspondente solução y de (1) satisfaz

||y(T )− y1||H < ε.

Controlabilidade nula: Dado y0, provar que existe um controle u tal que a correspon-

dente solução y de (1) satisfaz y(T ) = 0.

Controlabilidade exata por trajetórias: Dado y0 e uma trajetória arbitrária y do

sistema (1) (solução associada a um controle u), obter um controle u tal que a

correspondente solução y de (1) satisfaz y(T ) = y(T ).

A controlabilidade exata é um conceito mais forte do que o de controlabilidade

aproximada, nula e exata por trajetórias. No caso de sistemas lineares como (1) contro-

labilidade nula é equivalente a controlabilidade exata por trajetórias. Estes dois últimos

tipos de controlabilidade são frequentemente analisados quando trabalhamos com siste-

mas não reversíveis e com efeito regularizante, por exemplo sistemas parabólicos, já que

neste caso não podemos esperar a controlabilidade exata, veja [13], [22], [25], [54].

O controle de equações diferenciais parciais tem sido intensivamente estudado nas

últimas décadas. Um dos trabalhos que podemos destacar é o de D.L.Russel de 1978, [60],

onde o autor desenvolveu importantes ferramentas para resolver problemas de controlabi-

lidade, fazendo relação com outros métodos de equações diferenciais parciais: multiplica-

dores, problemas de momento, etc. Outros que destacamos são os trabalhos de J-L. Lions

de 1988, [46], [47], [48], onde o autor introduziu o Método da Unicidade de Hilbert (em

inglês Hilbert Uniqueness MethodH.U.M.), que logo se tornou uma ferramenta muito

importante para o desenvolvimento da Teoria de Controle.

A ferramenta usada para resolver problemas de controlabilidade de sistemas lineares

é a desigualdade de observabilidade para o sistema adjunto. Considerando, por exemplo

o sistema linear abstrato, a controlabilidade no tempo T > 0 é equivalente a analisar

sobrejetividade de um certo operador linear FT : L2(0, T ;U) → H que associa cada

u ∈ L2(0, T ;U) a um elemento FT (u) := y(T, ·), onde y ∈ C0([0, T ];H) é a solução de

(1) correspondente a u com y0 := 0. Por sua vez, devido a resultados clássicos de Análise

Funcional, essa sobrejetividade é equivalente a existência de uma constante C > 0 tal que

||F∗T (zT )||L2(0,T ;U) ≥ C||zT ||H zT ∈ H, (2)

2

onde F∗T : H → L2(0, T ;U) é o adjunto de FT . A desigualdade (2) é chamada desigual-

dade de observabilidade. Para os detalhes sobre a construção desses operadores pode-se

consultar [13].

Quando falamos em equações parabólicas e hiperbólicas merecem destaque as equa-

ções do calor e da onda, pois as técnicas usadas para análise dessas equações, em parti-

cular nos problemas de controlabilidade, geralmente podem ser adaptadas para outras do

mesmo tipo. Merece destaque também a equação de Burgers (que é parabólica), muito

importante entre as equações que descrevem uxo de uidos, que tem sido usada como

uma versão unidimensional da equação de NavierStokes, que por sua vez tem sido inten-

sivamente estudada nos últimos anos.

Nos problemas de controlabilidade para sistemas não lineares, primeiro resolvemos

o problema de controlabilidade nula para o sistema linear associado. Então, para obter

a desigualdade de observabilidade usamos as chamadas desigualdades de Carleman, que

são estimativas de normas L2 ponderadas com funções peso convenientes. O uso dessas

desigualdades para resolver problemas de controlabilidade teve notória popularização após

o trabalho de Fursikov e Imanuvilov [34]. Também podemos destacar: [38], [39], [40], [41]

e [44].

Uma parte importante em Teoria de Controle é a Otimização. Neste caso, além

de resolver o problema de controlabilidade, buscamos o controle ótimo no sentido de

minimizar custos (ou maximizar os benefícios). Isto é, considere novamente o sistema (1),

se denotarmos por Uad o espaço de todos os controles admissíveis e por Y o espaço dos

estados associados, assumindo que ambos sejam espaços de Banach com as normas || · ||Uade || · ||Y , respectivamente, então podemos considerar o problema de obter um controle que

minimiza um funcional da forma:

J(u) :=1

2||y(u)− yd||2Y ∀u ∈ Uad,

onde yd ∈ Y é dada. Ou ainda, mais geralmente, podemos considerar o funcional

J(u) :=1

2||y(u)− yd||2Y +

µ

2||u||Uad ∀u ∈ Uad, (3)

com µ ≥ 0. A situação a ser analisada com o funcional (3) é: além de atingir o estado

nal desejado, com o controle de menor custo, esperamos que a solução do sistema esteja

próxima de uma função dada.

3

A Teoria de Controle é uma parte da Matemática muito rica em aplicações, pois

problemas de diversas áreas, como engenharia, biologia, medicina, economia entre outras,

quando formulados matematicamente dão origem a problemas de controle. Por exem-

plo, questões de meio ambiente como despoluir um lago ou rio podem ser formuladas

como problemas de controles de certas equações parabólicas, como equação do calor ou

de NavierStokes. Pensando em controlar estruturas exíveis surgem problemas de con-

trole que envolvem equações hiperbólicas, já para controle de uxo de tráfego aparecem

problemas de controle, por exemplo, para equação de Burgers.

Para resolver um problema de controle ótimo, Lions em [50] introduziu o conceito

de controle hierárquico. A motivação veio da noção de otimização introduzida e usada em

Economia pelo economista H. Von Stackelberg, [61]. Neste processo tem-se pelo menos

dois controles atuando, porém apenas um controle, chamado líder, é independente e os

outros são considerados a partir de cada escolha do líder, estes são chamados seguidores.

Ele considerou um conjunto aberto Ω ⊂ Rn com fronteira suave Γ e um sistema da formayt + Ay = v11O1 + v21O2 em Q = Ω× (0, T ),

y = 0 sobre Σ = Γ× (0, T ),

y(·, 0) = 0 em Ω.

(4)

onde,

Ay = −n∑

i,j=1

∂

∂xi

(aij(x)

∂y

∂xj

)+

n∑i=1

ai(x)∂y

∂xi,

com

−n∑

i,j=1

aij(x)ξiξj ≥ αn∑i=1

ξ2i , α > 0.

onde 1Oidenota a função característica do conjunto aberto Oi ⊂ Ω e O1 ∩ O2 = ∅. Os

objetivos cumpridos foram: aproximar y(·, T ), onde y é solução de (4), de um estado

desejado y1 ∈ L2(Ω), e considerar a solução de (4) que não se afaste de uma função dada

y2 ∈ L2(Ω×(0, T )). Para conseguir esses objetivos foram denidos os seguintes funcionais

custo

J1(v1) :=1

2

∫∫O1×(0,T )

|v1|2dxdt, (5)

e

J2(v1, v2) =1

2

∫∫Ω×(0,T )

|y(v1, v2)− y2|2dxdt+β

2

∫∫O2×(0,T )

|v2|2, β > 0. (6)

4

Então foi usada a estratégia de Stackelberg, com o líder sendo v1 e o seguidor v2. Ou seja,

para cada escolha do controle v1, o controle v2 é escolhido de forma a minimizar o funcional

(6). Após obter o controle seguidor ótimo, v2 = F(v1), o qual ca caracterizado em um

sistema de otimalidade, o problema de controlabilidade aproximada é resolvido usando

o controle independente v1, que minimiza o funcional (5), pois agora tem-se y(v1, v2) =

y(v1,F(v1)). Para obter a controlabilidade foi usado o Teorema de continuação única de

Mizohata (veja [55]) e teoremas de regularidade.

Em [51], Lions estudou o controle hierárquico para um sistema governado pela equa-

ção da onda usando a estratégia de Stackelberg e analisando a controlabilidade aproximada

com controles na fronteira. Ele considerou um sistema como segueytt −∆y = 0 em Q,

y = v11Γ1 + v21Γ2 sobre Σ,

y(·, 0) = 0 em Ω.

(7)

Com Q = Ω× (0, T ), Σ = Γ× (0, T ) e Ω um conjunto aberto do Rn com fronteira suave Γ.

Onde neste problema 1Γidenota a função característica do subconjunto Γi de Γ, i = 1, 2,

vi é o controle em L2(Γi× (0, T )). Além disso, considerou v1 sendo o líder e v2 o seguidor.

Agora em [16], Diaz e Lions resolveram o problema de controlabilidade aproximada

para um sistema parabólico, como (4), onde consideraram um líder e um número n de se-

guidores. Neste trabalho, pela primeira vez, usaram a estratégia de Stackelberg associada

a idéia de equilíbrio de Nash, [56], a qual consiste em obter n seguidores que minimizem

simultaneamente n funcionais custo, similares a (6). Então, eles chamaram esse processo

de estratégia de StackelbergNash. Ou seja, nesse trabalho para cada controle líder f foi

encontrado um equilíbrio de Nash para os seus funcionais custo Ji, isto é, uma n-úpla

(w1, · · · , wn) que resolve o problema de minimização da forma

Ji(f ; w1, . . . , wi, . . . , wn) = minwi

Ji(f ; w1, . . . , wi, . . . , wn), i = 1, . . . , n.

Esse equilíbrio de Nash ca caracterizado em um sistema de otimalidade, então eles con-

cluem a controlabilidade aproximada aplicando o Teorema de HahnBanach e o Teorema

de continuação única de Mizohata.

Em [16], [50] e [51] o foco foi a controlabilidade aproximada, já em [3] os autores

5

consideraram sistema distribuído com a equação do calor não linearyt −∆y + a(x, t)y = F (y) + f1O + v11O1 + v21O2 em Q,

y = 0 sobre Σ,

y(·, 0) = y0 em Ω.

(8)

Com Ω um domínio limitado do Rn com fronteira suave Γ. Onde a ∈ L∞(Q), F é

uma função contínua localmente lipschitziana e y0 é dada, O, O1, O2 ⊂ Ω são conjuntos

abertos e 1A denota a função característica do conjunto A. Os controles são f, v1 e v2,

sendo f o líder e v1, v2 os seguidores. Os funcionais considerados são

J(f) =1

2

∫∫O×(0,T )

|f |2dxdt (9)

e

Ji(f ; v1, v2) =αi2

∫∫Oi,d×(0,T )

|y − yi,d|2dxdt+µi2

∫∫Oi×(0,T )

|vi|2dxdt, (10)

para i = 1, 2. Então, eles resolveram o problema de controlabilidade exata por trajetó-

rias usando controle hierárquico via estratégia de StackelbergNash. A desigualdade de

observabilidade para o adjunto do sistema de otimalidade linear foi obtida combinando

as desigualdades de Carleman associadas a cada equação do sistema. Depois concluíram

o problema de controlabilidade usando o Teorema do ponto xo de Shauder.

Por m, em relação a equação de Burgers e NavierStokes existem vários trabalhos

importantes sobre controlabilidade nula, podemos destacar [26], [27] e [34]. Contudo,

ainda não existem trabalhos sobre controle hierárquico e estratégia de StackelbergNash

para controlabilidade nula dessas equações.

O objetivo desta tese é analisar o controle hierárquico aplicando a estratégia de

StackelbergNash para resolver problemas de controlabilidade exata para as equações do

calor e da onda, e ainda, controlabilidade nula para equação de Burgers.

Descrição dos resultados

Nesta tese resolvemos os problemas de controle exato por trajetórias para sistemas

como (8) com controles tanto na fronteira como distribuídos. Além disso, vamos resolver

o problema de controle exato para um sistema como (7), mas com controles distribuídos.

No que segue detalharemos os trabalhos desenvolvidos.

6

Capítulo 1: Controle hierárquico para controlabilidade exata de

equações parabólicas com controles distribuídos e na fronteira.

Seja Ω ⊂ Rn (n ≥ 1) um domínio limitado com fronteira Γ sucientemente regular.

Considere O, O1, O2 ⊂ Ω conjuntos abertos não vazios e sejam S, S1 e S2 subconjuntos

disjuntos fechados de Γ. Dado T > 0, vamos considerar o domínio cilíndricoQ = Ω×(0, T )

cuja fronteira lateral é Σ = Γ × (0, T ). Denotaremos por ν(x) o vetor normal unitário

exterior a Ω no ponto x ∈ Γ.

Vamos considerar sistemas da formayt −∆y + a(x, t)y = F (y) + f1O em Q,

y = v11S1 + v21S2 sobre Σ,

y(·, 0) = y0 em Ω,

(11)

e pt −∆p+ a(x, t)p = F (p) + u11O1 + u21O2 em Q,

p = g1S sobre Σ,

p(·, 0) = p0 em Ω,

(12)

onde y0, p0, f, g, vi e ui, são dadas em espaços apropriados, F é uma função localmente

lipschitziana e por 1A denotamos a função característica do conjunto A.

Os problemas que vamos resolver tem os seguintes objetivos: obter o controle ótimo,

no contexto de controlabilidade exata por trajetórias para os sistemas (11) e (12), impedir

que a solução do sistema se distancie de uma certa função dada.

Neste sentido, focando inicialmente no sistema (11), denimos os funcionais custo:

Ji(f, v1, v2) :=αi2

∫∫Oi,d×(0,T )

|y − ξi,d|2 dx dt+µi2

∫∫Si×(0,T )

|vi|2 dσ dt, i = 1, 2, (13)

onde ξi,d = ξi,d(x, t) são dadas em L2(Oi,d × (0, T )) e αi, µi são constantes positivas.

Também denimos o funcional principal

J(f) :=1

2

∫∫O×(0,T )

|f |2 dx dt.

Então aplicamos a estratégia de StackelbergNash, com f sendo o controle líder e v1, v2 os

seguidores. Neste sentido, para cada f vamos encontrar um par (v1, v2) = (v1(f), v2(f))

que minimiza simultaneamente os funcionais custo Ji, isto é, resolvem o problema

J1(f ; v1, v2) = minv1

J1(f ; v1, v2), J2(f ; v1, v2) = minv2

J2(f ; v1, v2). (14)

7

Chamamos o par (v1(f), v2(f)) assim obtido de equilíbrio de Nash para os funcionais Ji.

Lembremos que uma trajetória do sistema (11) é a solução do sistemayt −∆y + a(x, t)y = F (y) in Q,

y = 0 on Σ,

y(·, 0) = y0 in Ω.

(15)

Depois de provar a existência de equilíbrio de Nash, xada uma trajetória y, buscamos

um controle ótimo f ∈ L2(O × (0, T )), que resolve o problema

J(f) = minfJ(f), (16)

sujeito a condição de controlabilidade exata

y(·, T ) = y(·, T ) in Ω. (17)

Em relação ao sistema (12), o líder é o controle g e os seguidores são u1 e u2. Os

funcionais custo são da forma:

Ki(g, u1, u2) :=

αi2

∫∫Σi,d

∣∣∣∣∂p∂ν − ζi,d∣∣∣∣2 dσ dt+

µi2

∫∫Oi×(0,T )

|ui|2 dx dt, i = 1, 2, (18)

e

K(g) :=1

2||g||2

H32 , 3

4 (S×(0,T )),

onde Σi,d = Γi,d × (0, T ) e Γi,d são subconjuntos fechados não vazios de Γ, ζi,d = ζi,d(x, t)

são funções dadas e αi, µi são constantes positivas. Aplicando a estratégia de Stackelberg

Nash, para g obtemos um equilíbrio de Nash, isto é, um par (u1(g), u2(g)) que resolve

simultaneamente o problema de minimização

K1(g;u1, u2) = minu1

K1(g; u1, u2), K2(g;u1, u2) = minu2

K2(g;u1, u2). (19)

Vamos considerar a seguinte trajetória do sistema (1.2):pt −∆p+ a(x, t)p = F (p) in Q,

p = 0 on Σ,

p(·, 0) = p0 in Ω.

(20)

Então, uma vez provada a existência de equilíbrio de Nash para cada controle líder,

encontramos um controle g ∈ H 32, 34 (S × (0, T )) tal que

K(g) = mingK(g), (21)

8

sujeito a condição de controlabilidade

p(·, T ) = p(·, T ) in Ω. (22)

Observe que o sistema (15) estudado em [3] tem apenas controles distribuídos. Investiga-

mos questões semelhantes a esse trabalho, mas agora a diculdade é que temos controles

tanto na fronteira quanto distribuídos. Resolver problemas de controle hierárquico para

controlabilidade exata por trajetórias para equação do calor é a principal contribuição

deste artigo.

Vamos enunciar os principais resultados deste trabalho no que segue.

Teorema (Caso linear): Suponha Oi,d∩O 6= ∅, i = 1, 2. Assuma que uma das seguintes

condições é verdadeira:

O1,d = O2,d := Od, (23)

ou

O1,d ∩ O 6= O2,d ∩ O. (24)

Assuma F ≡ 0 e as constantes µi > 0 (i = 1, 2) são sucientemente grandes. Então existe

uma função positiva ρ = ρ(t), que decai exponencialmente para zero quando t tende a T ,

tal que se y é a solução de (15) (com F ≡ 0) associada ao estado inicial y0 ∈ L2(Ω) e as

funções ξi,d ∈ L2(Oi,d × (0, T )) satisfazem a condição:∫∫Oi,d×(0,T )

ρ−2|y − ξi,d|2 dx dt < +∞, i = 1, 2, (25)

então, dado y0 ∈ L2(Ω), existem um controle f ∈ L2(O× (0, T )) e um equilíbrio de Nash

associado (v1(f), v2(f)) tais que a solução de (11) correspondente (com F ≡ 0) satisfaz

(17).

A hipótese (23) é natural, pois desejamos obter (17) sem que a solução de (11) se afaste da

função ξi,d, então é bastante razoável que essa função se aproxime da trajetória y quando

t tende a T . Além disso, como o sistema é linear os funcionais Ji são convexos, então a

condição (14) é equivalente a

J ′i(f ; v1, v2) · vi = 0, ∀vi ∈ L2(Si × (0, T )), i = 1, 2. (26)

Logo, no caso linear um par (v1, v2) é um equilíbrio de Nash para os funcionais Ji se, e

somente se, satisfaz (26).

Para demonstrar o Teorema no caso linear, transformamos o problema de controla-

bilidade exata por trajetórias em um problema de controlabilidade nula. Caracterizamos

9

o equilíbrio de Nash em um sistema de otimalidade e obtemos a desigualdade de obser-

vabilidade para o sistema adjunto combinando desigualdades de Carleman adequadas.

No caso semilinear, como a convexidade dos Ji é perdida, não temos mais a equi-

valência entre (14) e (26). Neste caso, usamos a denição de quase equilíbrio de Nash:

Um par (v1, v2) é um quase equilíbrio de Nash para os funcionais Ji se a condição (26) é

satisfeita.

Com esta denição temos os seguintes resultados para o caso semilinear:

Teorema (Caso semilinear): Suponha Oi,d e µi (i = 1, 2) nas condições do Teorema

do caso linear, seja F ∈ W 1,∞(R) e que as funções ξi,d ∈ L2(Oi,d× (0, T )) (i = 1, 2) tem a

propriedade: existe uma função ρ, também como no caso linear, tal que se y é a solução

de (15) para um estado inicial y0 ∈ L2(Ω) (25) vale. Então dado y0 ∈ L2(Ω), existem um

controle f ∈ L2(O × (0, T )) e um quase equilíbrio de Nash associado (v1, v2) tais que a

solução de (11) correspondente satisfaz (17).

Para demonstrar este resultado usamos os resultados dos caso linear e aplicamos o Teorema

do ponto xo de Shauder.

Podemos, sob certas condições, obter uma equivalência entre as denições de equi-

líbrio e quase equilíbrio de Nash. Este é o conteúdo do seguinte resultado:

Teorema (Relação entre equilíbrio e quase equilíbrio): Suponha F ∈ W 2,∞(R) e

ξi,d ∈ L∞(Oi,d × (0, T )) (i = 1, 2). Se y0 ∈ L2(Ω) e n ≤ 6, então existe C > 0 tal que, se

f ∈ L2(O × (0, T )) e µi satisfazem

µi ≥ C(1 + ‖f‖L2(O×(0,T ))

),

as condições (14) e (26) são equivalentes.

Para demonstrar este resultado basicamente vericamos as condições para que a segunda

derivada dos Ji tenha sinal positivo.

Em relação ao sistema (12), temos o seguinte resultado para o caso linear:

Teorema (Caso linear sistema (12)): Suponha

Γi,d ∩ S = ∅, i = 1, 2, (27)

e que as constantes µi > 0 (i = 1, 2) são sucientemente grandes. Então existe uma

função positiva ρ = ρ(t), que é nula em t = T , tal que se p é a solução de (20) (com

F ≡ 0) associada ao estado inicial p0 ∈ V e as funções ζi,d ∈ L2(Γi,d × (0, T )) satisfazem

a condição: ∫∫Γi,d×(0,T )

ρ−2

∣∣∣∣∂p∂ν − ζi,d∣∣∣∣2 dσ dt < +∞, i = 1, 2, (28)

10

então, para todo p0 ∈ V existem um controle g ∈ H 32, 34 (S×(0, T )) e um equilíbrio de Nash

(u1, u2) associado tais que a correspondente solução de (12) satisfaz (22).

A hipótese (28) tem justicativa análoga a (23). E mais, como os funcionais Ki são

convexos neste caso temos que um par (u1, u2) é um equilíbrio de Nash para os funcionais

Ki se, e somente se, satisfazem

K ′i(g;u1, u2) · ui = 0, ∀ ui ∈ L2(Oi × (0, T )), i = 1, 2. (29)

Para demonstrar o teorema no caso linear para sistema (12), fazemos um prolongamento

do domínio e resolvemos o problema com controles distribuídos.

Usando a denição de quase equilíbrio para os funcionais Ki obtemos o seguinte

resultado para o caso semilinear:

Teorema (Caso semilinear sistema (12)): Suponha Γi,d, µi > 0 (i = 1, 2) como no

caso linear (sistema (12)) e F ∈ W 1,∞(R). Então existe uma função ρ tal que, se p é a

solução de (20) para um estado inicial p0 ∈ V e as funções ζi,d ∈ L2(Γi,d×(0, T )) (i = 1, 2)

são tais que (28) vale, para cada p0 ∈ V dado, existem um controle g ∈ H 32, 34 (S×(0, T )) e

um quase equilíbrio de Nash (u1, u2) associado tais que a correspondente solução de (12)

satisfaz (22).

Para demonstrar este resultado também usamos o Teorema do ponto xo de Shauder.

Por m, obtemos a equivalência entre equilíbrio e quase equilíbrio de Nash para os

funcionais Ki:

Teorema (Relação entre equilíbrio e quase equilíbrio, sistema (12)): Suponha

que F ∈ W 2,∞(R) e ζi,d ∈ H12, 14 (Σi,d) (i = 1, 2). Se p0 ∈ V e N ≤ 6, então existe C > 0

tal que, se g ∈ H 32, 34 (S × (0, T )) e µi satisfazem

µi ≥ C(

1 + ‖g‖H

32 , 3

4 (S×(0,T ))

),

as condições (19) e (29) são equivalentes.

Capítulo 2: Controle hierárquico para equação da onda.

Seja Ω ⊂ Rn um domínio limitado com fronteira Γ de classe C2 e suponha T > 0.

Vamos considerar pequenos conjuntos abertos não vazios e disjuntos O, O1, O2 ⊂ Ω.

Usaremos a notação Q = Ω × (0, T ), cuja fronteira lateral é Σ = Γ × (0, T ); por ν(x) o

vetor normal unitário exterior a Ω no ponto x ∈ Γ.

11

Consideramos o seguinte sistemaytt −∆y + a(x, t)y = F (y) + f1O + v11O1 + v21O2 in Q,

y = 0 on Σ,

y(·, 0) = y0, yt(·, 0) = y1 in Ω,

(30)

onde a ∈ L∞(Q), f ∈ L2(O× (0, T )), vi ∈ L2(Oi× (0, T )), F : R→ R é uma função local-

mente lipschitziana, (y0, y1) ∈ H10 (Ω) × L2(Ω) e com 1A indicamos função característica

de A.

Resolvemos um problema com dois objetivos: obter um controle ótimo que resolve

o problema de controlabilidade exata e impedir que a solução do sistema se afaste de uma

função dada. Para conseguir estes objetivos vamos considerar o controle hierárquico, com

f sendo o controle líder v1, v2 os seguidores, e aplicaremos a estratégia de Stackelberg

Nash.

Consideramos o funcional principal

1

2

∫∫O×(0,T )

|f |2 dx dt,

e os funcionais custo secundários

Ji(f, v1, v2) :=

αi2

∫∫Oi,d×(0,T )

|y − yi,d|2 dx dt+µi2

∫∫Oi×(0,T )

|vi|2 dx dt, (31)

onde Oi,d ⊂ Ω são abertos não vazios, yi,d ∈ L2(Oi,d × (0, T )) são funções dadas e αi e µi

são constantes positivas. A estratégia de StackelbergNash funciona como segue. Para

cada escolha de um líder f , buscamos um equilíbrio de Nash para os funcionais Ji, ou seja,

um par (v1(f), v2(f)) ∈ L2(O1× (0, T ))×L2(O2× (0, T )) que resolve simultaneamente o

seguinte problema de minimização:

J1(f ; v1, v2) = minv1

J1(f ; v1, v2), J2(f ; v1, v2) = minv2

J2(f ; v1, v2). (32)

No caso em que os funcionais Ji são convexos, (32) é equivalente a

J ′i(f ; v1, v2) · vi = 0, ∀ vi ∈ L2(Oi × (0, T )), i = 1, 2. (33)

Este é caso quando o sistema (30) é linear, no entanto, geralmente isto não acontece no

caso não linear.

12

Dado (y0, y1) ∈ H10 (Ω)× L2(Ω), uma trajetória para (30) é a solução do sistema

ytt −∆y + a(x, t)y = F (y) in Q,

y = 0 on Σ,

y(·, 0) = y0, yt(·, 0) = y1 in Ω.

(34)

Depois de provar que para cada líder f existe um par equilíbrio de Nash (v1, v2) =

(v1(f), v2(f)), xamos uma trajetória y e buscamos um líder f ∈ L2(O × (0, T )) tal que

J(f) := minfJ(f), (35)

sujeito a condição de controlabilidade exata

y(·, T ) = y(·, T ), yt(·, T ) = yt(·, T ) in Ω. (36)

Em [51], o autor provou resultados sobre controle hierárquico aplicando a estratégia de

Stackelberg, no contexto da controlabilidade aproximada para equação da onda, com

controles distribuídos. A principal contribuição deste trabalho é entender a estratégia

de Stackelberg associada a noção de equilíbrio de Nash, trabalhando no contexto da

controlabilidade exata para equação da onda com controles distribuídos.

Antes de enunciar os resultados obtidos neste trabalho façamos algumas considera-

ções. Dado x0 ∈ Rn \ Ω denimos o conjunto

Γ+ := (x− x0) · ν(x) > 0

e as funções d : Ω → R, com d(x) = |x − x0|2 para todo x ∈ Ω. Vamos supor que existe

δ > 0 tal que

O ⊃ Oδ(Γ+) ∩ Ω, (37)

onde

Oδ(Γ+) = x ∈ Rn; |x− x′| < δ, x′ ∈ Γ+.

Devido a velocidade de propagação da equação da onda ser nita, o tempo T > 0

tem que ser sucientemente grande. Assim, no que segue vamos supor T > 2R1, onde

R1 := max√d(x) : x ∈ Ω.

Os resultados obtidos são os seguintes.

13

Teorema (Caso linear): Suponha F ≡ 0 e que as constantes µi > 0 (i = 1, 2) suciente-

mente grandes. Então dado (y0, y1) ∈ H10×L2(Ω), existem um controle f ∈ L2(O×(0, T ))

e um equilíbrio de Nash (v1, v2) = (v1(f), v2(f)) tais que a correspondente solução de (30)

satisfaz (36).

Para demonstrar este resultado transformamos o problema de controlabilidade exata

em um de controlabilidade nula equivalente, depois caracterizamos o equilíbrio de Nash

em um sistema de otimalidade. Concluímos a controlabilidade nula usando a desigualdade

de observabilidade que obtemos por meio de desigualdades de Carleman apropriadas e da

energia.

Considerando o caso semilinear com F localmente lipschitziana, como os funcionais

Ji não são convexos, em geral (32) and (33) não são equivalentes. Desse modo, usamos a

denição quase equilíbrio de Nash: dado f , um par (v1, v2) é um quase equilíbrio de Nash

para os funcionais Ji se (33) é satisfeita. Com esta denição temos o seguinte resultado:

Teorema (Caso semilinear): Suponha que (37) vale, F ∈ W 1,∞(R) e os µi > 0 (i =

1, 2) são sucientemente grandes. Então, dado qualquer par (y0, y1) ∈ H10 (Ω) × L2(Ω),

existem um controle f ∈ L2(O×(0, T )) e um quase equilíbrio de Nash associado (v1, v2) =

(v1(f), v2(f)) tais que a correspondente solução de (30) satisfaz (36).

Demonstramos este Teorema usando o caso linear, resultados de compacidade e

ponto xo.

Analisamos ainda em que condições as denições de equilíbrio e quase equilíbrio de

Nash são equivalentes, a resposta esta no seguinte resultado:

Teorema (Relação entre equilíbrio e quase equilíbrio): Suponha F ∈ W 2,∞(R) e

que yi,d ∈ C0([0, T ];H10 (Oi,d))∩C1([0, T ];L2(Oi,d)). Se (y0, y1) ∈ H1

0 (Ω)×L2(Ω) e n ≤ 8,

então existe C > 0 tal que, se f ∈ L2(O × (0, T )) e µi satisfazem

µi ≥ C(1 + ‖f‖L2(O×(0,T ))),

as condições (32) e (33) são equivalentes.

Para demonstrar este resultado analisamos a derivada segunda dos funcionais Ji.

14

Capítulo 3: Controle hierárquico para a equação de Burgers via

estratégia de StackelbergNash.

Seja T > 0 e considere os conjuntos abertos não vazios O, O1, O2 ⊂ (0, 1), com

0 6∈ O. Introduzimos o seguinte sistema para a equação de Burgers:yt − yxx + yyx = f1O + v11O1 + v21O2 em (0, 1)× (0, T ),

y(0, t) = y(1, t) = 0 sobre (0, T ),

y(x, 0) = y0(x) em (0, 1),

(38)

onde f, v1, v2 são os controles e y é o estado e com a notação 1A representamos a função

característica do conjunto A.

Existem vários trabalhos importante sobre a controlabilidade de sistemas como (38),

entre os quais podemos citar [26], [34] e [53]. No Capítulo 3 vamos analisar a controlabi-

lidade nula do sistema (34) no contexto do controle hierárquico e aplicando a estratégia

de StackelbergNash, com f sendo o líder e v1, v2 os seguidores.

Seguindo metodologia dos capítulos anteriores, consideramos os funcionais custo

para os seguidores:

Ji(f ; v1, v2) :=αi2

∫∫Oi,d×(0,T )

|y − yi,d|2 dx dt+µi2

∫∫Oi×(0,T )

|vi|2 dx dt, (39)

e o funcional principal

J(f) :=1

2

∫∫O×(0,T )

|f |2 dx dt,

onde Oi,d ⊂ (0, 1) é um conjunto aberto não vazio, αi, µi > 0 são constantes e yi,d =

yi,d(x, t) são funções dadas em L2(Oi,d × (0, T )).

Para cada líder f escolhido obtemos um par Equilíbrio de Nash associado (v1(f), v2(f),

ou seja, um par (v1(f), v2(f) que minimiza, simultaneamente, os funcionais Ji

J1(f ; v1, v2) = minv1

J1(f ; v1, v2) e J2(f ; v1, v2) = minv2

J2(f ; v1, v2). (40)

Quando os Ji são convexos, i = 1, 2, (40) é equivalente a

J ′i(f ; v1, v2) · vi = 0 ∀ vi ∈ L2(Oi × (0, T )), (41)

Mas como (38) é não linear, em geral, a equivalência entre (40) e (41) não vale. Então,

para provar a existência e unicidade de equilíbrio de Nash, devemos mostrar que existe

um único par v1, v2 que satisfaz (41) e⟨J ′′(f ; v1, v2), vi

⟩> 0 ∀ vi ∈ L2(Oi × (0, T )). (42)

15

Depois de provar que existe um único par de Nash para cada controle líder em L2(O ×

(0, T )), buscamos um controle f ∈ L2(O × (0, T )) tal que

J(f) = minfJ(f) (43)

e a correspondente solução de (38) satisfaz a condição de controlabilidade nula:

y(·, T ) = 0 in (0, 1). (44)

Agora vamos apresentar os resultados do Capítulo 3. O primeiro resultado arma

sobre a existência e unicidade de equilíbrio de Nash.

Teorema (Existência e unicidade de equilíbrio de Nash): Suponha µi suciente-

mente grande. Então, para cada f ∈ L2(O × (0, T )) dado, existe um único equilíbrio de

Nash (v1(f), v2(f)) para os funcionais Ji.

O seguinte resultado garante a controlabilidade nula local com dado inicial em

H10 (0, 1):

Teorema (caso y0 ∈ H10 (0, 1)): Suponha y0 ∈ H1

0 (0, 1) com ||y0||H10 (0,1) ≤ r para algum

r > 0 e Oi,d ∩ O 6= ∅, i = 1, 2. Assuma uma das seguintes condições valem:

O1,d = O2,d := Od, (45)

ou

O1,d ∩ O 6= O2,d ∩ O. (46)

Se as constantes µi > 0 (i = 1, 2) são sucientemente grandes e existe uma função

positiva ρ = ρ(t), que decai exponencialmente para 0 quando t→ T−, tal que, as funções

yi,d ∈ L2(Oi,d × (0, T )) (i = 1, 2) satisfazem∫∫Oi,d×(0,T )

ρ−2 y2i,d dx dt < +∞, i = 1, 2, (47)

então, existe um controle f ∈ L2(O × (0, T )) e um equilíbrio de Nash (v1(f), v2(f))

associado tal que tem-se (43) e a correspondente solução de (38) satisfaz (44).

Aqui temos uma interpretação para a hipótese (46) parecida com a (24) no caso da

equação do calor. Como desejamos que o estado seja nulo no tempo T > 0, sem que nos

afastemos de yi,d, é esperado que estas funções quem perto de se anular quando t tende

a T .

Supor que y0 é limitada em H10 (0, 1) é uma condição razoável, pois devido aos resul-

tados em [26] e [34], sabemos que para o sistema (38) não podemos ter controlabilidade

16

nula global. Além disso, de [26] segue que para y0 ∈ L2(0, 1) com ||y0||L2(0,1) ≤ r, r > 0,

temos um tempo mínimo para controlabilidade nula T (r) > 0, com

C0 log(1/r)−1 ≤ T (r) ≤ C1 log(1/r)−1,

onde C0, C1 > 0 são constantes adequadas.

Em relação à controlabilidade nula local com dado inicial em L2(0, 1) temos o resul-

tado:

Teorema (caso y0 ∈ L2(0, 1)): Suponha y0 ∈ L2(0, 1) com ||y0||L2(0,1) ≤ r para algum

r > 0 e T ≥ T (r). Assumindo, como no teorema do caso y0 ∈ H10 (0, 1), que uma das

condições (45), (46) é satisfeita, µi sucientemente grande e que existe uma função ρ tal

que a hipótese (47) vale, então, existe um controle f ∈ L2(O × (0, T )) e um equilíbrio de

Nash associado (v1(f), v2(f)) tal que correspondente solução de (38) satisfaz (44).

Questões em aberto e trabalhos futuros

Problema parabólico com todos os controles na fronteira.

Considere o problema linearzt −∆z + a(x, t)z = 0 in Q,

z = v1S + v11S1 + v21S2 on Σ,

z(·, 0) = z0 in Ω,

(48)

Podemos pensar em funcionais custo como os Ji ou Ki denidos no Capítulo 1. Então,

seguindo as ideias do Capítulo 1 obtemos um equilíbrio de Nash para cada líder e um

sistema de otimalidade. No entanto, para a controlabilidade nula aparece a diculdade

de obter desigualdades de Carleman para provar a desigualdade de observabilidade. A

controlabilidade nula para o sistema (48) usando a estratégia de StackelbergNash é uma

questão em aberto.

Controle hierárquico aplicando a estratégia de StackelbergNash para o sis-

tema de NavierStokes

O sistema (38) pode ser considerado como uma versão unidimensional simplicada

do sistema de NavierStokes. Seja Ω um domínio limitado do RN (N = 2 ou N = 3),

17

cuja fronteira Γ é sucientemente regular. Considere o sistema de NavierStokes

yt −∆y + (y · ∇)y +∇p = f1O + v11O1 + v21O2 in Q,

∇ · y = 0 in Q,

y = 0 on Σ,

y(·, 0) = y0 in Ω,

(49)

onde T > 0 é dado, Q = Ω× (0, T ), Σ = Γ× (0, T ) e os conjuntos O, O1, O2 ⊂ Ω abertos

não vazios.

Consideramos y0 no espaço

H := z ∈ L2(Ω)N : ∇ · z = 0 in Ω, z · ν on Γ,

onde ν(x) denota o vetor normal unitário exterior a Ω no ponto x ∈ Γ.

A controlabilidade do sistema (49) tem sido bastante estudada nos últimos anos,

veja por exemplo, [27] e [34], onde podemos encontrar resultados sobre controlabilidade

com controles distribuídos e na fronteira, respectivamente.

No contexto do controle hierárquico, em [2] os autores analizaram o sistema (49),

aplicando a estratégia de StackelbergNash, onde eles já conseguiram resultados sobre

existência e unicidade de equilíbrio de Nash para funcionais similares aos Ji denidos em

(39). Mas controlabilidade nula para (49) ainda é uma questão em aberto.

Controle hierárquico para equação da onda com controles na fronteira.

No Capítulo 2 resolvemos o problema de controle hierárquico para a controlabili-

dade exata da equação da onda, com um líder e dois seguidores, com todos os controles

distribuídos. Em [50], o autor analisou o controle hierárquico com um líder e um segui-

dor, ambos na fronteira, onde resolveu o problema de controlabilidade aproximada para

equação da onda. Um problema que surgiu nessa direção foi o da controlabilidade exata

com um líder e pelo menos dois seguidores, com todos os controles na fronteira.

Considere um sistema do tipo:ytt −∆y + a(x, t)y = 0 in Q,

y = f1Γ0 + v11Γ1 + v21Γ2 on Σ,

y(·, 0) = y0, yt(·, 0) = y1 in Ω,

(50)

18

Então, a parte de encontrar um equilíbrio de Nash para cada líder é análoga ao que é

feito no Capítulo 2, mas a diculdade aparece para demonstrar a desigualdade de obser-

vabilidade. Esta ainda é uma questão em aberto.

19

Capítulo 1

Hierarchic control for exact

controllability of parabolic equations

with distributed and boundary controls

Hierarchic control for the exact

controllability of parabolic equations

with distributed and boundary controls

F. D. Araruna, E. Fernández-Cara, L. C. Silva

Abstract. We present some results of exact controllability for parabolic equations in the

context of hierarchic control through Stackelberg-Nash strategy. We analyze two cases:

in the rst one, the main control (the leader) acts in the interior of the domain and the

secondary controls (the followers) act on small parts of boundary; in the second one, we

consider a leader acting on the boundary while the followers are of the distributed kind. In

both cases, for each leader, an associated Nash equilibrium pair is found; then, we obtain

a leader that leads the system exactly to a prescribed (but arbitrary) trajectory. We will

consider linear and semilinear problems.

1.1 Introduction

Let Ω ⊂ Rn (n ≥ 1) be a bounded domain with suciently regular boundary

Γ = ∂Ω. Let O,O1,O2 ⊂ Ω be (small) nonempty open sets and let S, S1, and S2 be

nonempty closed and disjoints subsets of Γ. Given T > 0, let us consider the cylinder

domain Q = Ω× (0, T ) with lateral boundary Σ = Γ× (0, T ). We will denote by ν(x) the

outward unit normal to Ω at the point x ∈ Γ.

We will consider parabolic systems of the formyt −∆y + a(x, t)y = F (y) + f1O in Q,

y = v11S1 + v21S2 on Σ,

y(·, 0) = y0 in Ω,

(1.1)

and pt −∆p+ a(x, t)p = F (p) + u11O1 + u21O2 in Q,

p = g1S on Σ,

p(·, 0) = p0 in Ω,

(1.2)

where y0, p0, f, g, vi, and ui, are given in appropriate spaces and F : R→ R is a locally

Lipschitz-continuous function. In this paper, the notation 1A denotes the characteristic

function of the set A.

21

We will analyze the exact controllability to the trajectories of the systems (1.1)

and (1.2) following the process of control called, by J.-L. Lions [50], hierarchic control.

More precisely, we will apply the Stackelberg-Nash strategy, which combines cooperative

optimization techniques by Stackelberg and non-cooperative optimization techniques by

Nash.

The control of the hyperbolic equation has been studied in [50], with two controls

(one leader and one follower). hierarchic control results based on StackelbergNash stra-

tegy have been obtained in [3], [16] and [37].

The main goal this article is to analyze the exact controllability (to the trajecto-

ries) for the systems (1.1) and (1.2) in the context of hierarchic control, applying the

StackelbergNash strategy. In order to explain the methodology we will be initially con-

cerned with system (1.1).

Let O1,d and O2,d be non-empty open subsets of Ω and let us dene the cost functi-

onals

Ji(f, v1, v2) =αi2

∫∫Oi,d×(0,T )

|y − ξi,d|2 dx dt+µi2

∫ ∫∫Si×(0,T )

|vi|2 dσ dt, i = 1, 2, (1.3)

where ξi,d = ξi,d(x, t) are given functions in L2(Oi,d × (0, T )) and αi, µi are positive

constants. Let us also introduce the main functional

J(f) :=1

2

∫∫O×(0,T )

|f |2 dx dt.

For each choice of the leader f , we look for controls v1 and v2, depending on f , which

minimize simultaneously the cost functionals J1 and J2, that is,

J1(f ; v1, v2) = minv1

J1(f ; v1, v2), J2(f ; v1, v2) = minv2

J2(f ; v1, v2). (1.4)

This pair (v1, v2) is called a Nash equilibrium.

Let us consider a trajectory of the system (1.1):yt −∆y + a(x, t)y = F (y) in Q,

y = 0 on Σ,

y(·, 0) = y0 in Ω.

(1.5)

Assuming that a Nash equilibrium for each leader, we then look for a control f ∈ L2(O×

(0, T )), such that

J(f) = minfJ(f), (1.6)

22

subject to condition of exact controllability

y(·, T ) = y(·, T ) in Ω. (1.7)

In this article, we will use the Hilbert space

W (Q) :=u ∈ L2(0, T ;H1(Ω)); ut ∈ L2(0, T ;H−1(Ω))

,

which is equipped with the norm

||u||W =(||u||2L2(0,T ;H1(Ω)) + ||ut||2L2(0,T ;H−1(Ω))

) 12,

see [52]. We will also use the Hilbert spaces Hr,s(Q) = L2(0, T ;Hr(Ω))∩Hs(0, T ;L2(Ω)),

for r, s ≥ 0, with norms

||u||Hr,s(Q) :=(||u||2L2(0,T ;Hr(Ω)) + ||u||2Hs(0,T ;L2(Ω))

) 12,

and their boundary counter parts Hr,s(Σ) := L2(0, T ;Hr(Γ)) ∩ Hs(0, T ;L2(Γ)). Finally,

let us introduce the space

V := u ∈ H2(Ω); u = 0 on Γ \ S.

In relation to system (1.2), the secondary cost functionals are dened as follows:

Ki(g, u1, u2) :=

αi2

∫∫Σi,d

∣∣∣∣∂p∂ν − ζi,d∣∣∣∣2 dσ dt+

µi2

∫∫Oi×(0,T )

|ui|2 dx dt, i = 1, 2, (1.8)

where Σi,d = Γi,d × (0, T ) and Γi,d are nonempty closed subset of Γ, ζi,d = ζi,d(x, t) are

given functions and αi, µi are positive constants. The main functional is the following:

K(g) :=1

2||g||2

H32 , 3

4 (S×(0,T )).

For each control leader g, we will nd a Nash equilibrium for the cost functionals Ki, that

is, a couple (u1, u2) such that

K1(g;u1, u2) = minu1

K1(g; u1, u2), K2(g;u1, u2) = minu2

K2(g;u1, u2). (1.9)

Let the following trajectory for (1.2) be given:pt −∆p+ a(x, t)p = F (p) in Q,

p = 0 on Σ,

p(·, 0) = p0 in Ω.

(1.10)

23

Then, we look for a control g ∈ H 32, 34 (S × (0, T )) verifying

K(g) = mingK(g), (1.11)

subject to condition of exact controllability

p(·, T ) = p(·, T ) in Ω. (1.12)

The motivation for these problems comes, for example, from environmental pro-

blems. We can view y = y(x, t) as the concentration of a chemical product in a lake Ω.

Then, in relation to (1.1), O can be regarded as a part of the lake where we can apply a

control f which tries to approach y to a desired y at time T . But, at the same time, we

do not want y to be too far from ξi,d in Oi,d, for i = 1, 2 and, to this purpose, we use the

controls vi on parts Si of the shore of lake. A similar interpretation can be given in the

context of (1.2).

1.1.1 Main results

First, we will consider the system (1.1) with F ≡ 0. In this linear case, we have the

following result:

Theorem 1 Suppose Oi,d ∩O 6= ∅, i = 1, 2. Assume that one of the following conditions

holds:

O1,d = O2,d := Od, (1.13)

or

O1,d ∩ O 6= O2,d ∩ O. (1.14)

If the constants µi > 0 (i = 1, 2) are large enough, there exists a positive function ρ = ρ(t),

blowing up at t = T such that, if y is the unique solution to (1.5) (with F ≡ 0) associated

to the initial state y0 ∈ L2(Ω) and the functions ξi,d ∈ L2(Oi,d× (0, T )) (i = 1, 2) are such

that ∫∫Oi,d×(0,T )

ρ2|y − ξi,d|2 dx dt < +∞, i = 1, 2, (1.15)

then for any y0 ∈ L2(Ω), there exist a control f ∈ L2(O × (0, T )) and a associated Nash

equilibrium (v1, v2) such that the corresponding solution to (1.1) (with F ≡ 0) satises

(1.7).

The assumptions in (1.15) are natural, because we want to get (1.7) with a state y

not too far from ξi,d and, consequently, it is reasonable that ξi,d approach to trajectory y

24

as t goes to T . Note that in this linear case, the cost functionals Ji are convex. So, (1.4)

is equivalent to

J ′i(f ; v1, v2) · vi = 0, ∀vi ∈ L2(Si × (0, T )), i = 1, 2. (1.16)

Contranly, in the semilinear one, we cannot guarantee the convexity of the functio-

nals Ji. This motivates the following denition:

Denition 1 The pair (v1, v2) is a Nash quasi-equilibrium for the functionals Ji if the

condition (1.16) is satised.

With this denition, we have the following results:

Theorem 2 Suppose that Oi,d and µi (i = 1, 2) are in the conditions of Theorem 1

and F ∈ W 1,∞(R). There exists a function ρ as in Theorem 1 such that, if y is the

unique solution of (1.5) associated to the initial state y0 ∈ L2(Ω) and the functions

ξi,d ∈ L2(Oi,d × (0, T )) (i = 1, 2) are such that (1.15) holds, then for any y0 ∈ L2(Ω),

there exist controls f ∈ L2(O × (0, T )) and associated Nash quasi-equilibria (v1, v2) such

that the corresponding solutions to (1.1) satisfy (1.7).

In the semilinear case, there are situations where we have in fact equivalence of the

denitions of Nash equilibrium and Nash quasi-equilibrium. This is seen in the following

result:

Proposition 1 Assume that F ∈ W 2,∞(R) and ξi,d ∈ L∞(Oi,d × (0, T )) (i = 1, 2). Sup-

pose that y0 ∈ L2(Ω) and n ≤ 6. There exists C > 0 such that, if f ∈ L2(O× (0, T )) and

the µi satisfy

µi ≥ C(1 + ‖f‖L2(O×(0,T ))

),

the conditions (1.4) and (1.16) are equivalent.

Let us now turn to systems of the kind (1.2). Assuming that F ≡ 0 we get following

result:

Theorem 3 Suppose that

Γi,d ∩ S = ∅, i = 1, 2, (1.17)

and the constants µi > 0 (i = 1, 2) are large enough. Then, there exists a positive

function ρ = ρ(t), blowing up at t = T such that, if p is the solution to (1.10) (with

F ≡ 0) associated to the initial state p0 ∈ V and the ζi,d ∈ L2(Γi,d × (0, T )) are such that∫∫Γi,d×(0,T )

ρ2

∣∣∣∣∂p∂ν − ζi,d∣∣∣∣2 dσ dt < +∞, i = 1, 2, (1.18)

then, for any p0 ∈ V , there exist a control g ∈ H 32, 34 (S × (0, T )) and a associated Nash

equilibrium (u1, u2) such that the corresponding solution to (1.2) satisfy (1.12).

25

As in the case of (1.15), the assumption (1.18) means that ∂p/∂ν approaches

zetai,d on Γi,d× (0, T ) as t goes to T . In this linear case, the functionals Ki given in (1.8)

are convex. So, (1.9) is equivalent to

K ′i(g;u1, u2) · ui = 0, ∀ ui ∈ L2(Oi × (0, T )), i = 1, 2. (1.19)

Thus, a pair (u1, u2) is a Nash equilibrium if and only if it satises (1.19).

As we know, in the semilinear case we lose the convexity of the Ki. This implies

that (1.9) and (1.19) are not equivalent. Accordingly, we give the following

Denition 2 The pair (u1, u2) is a Nash quasi-equilibrium for the functionals Ki if the

condition (1.19) is satised.

The following result holds:

Theorem 4 Suppose that the Γi,d and the µi > 0 (i = 1, 2) as in Theorem 3 and F ∈W 1,∞(R). There exists a function ρ as in Theorem 3 such that, if p is the solution to (1.10)

associated to the initial state p0 ∈ V and the functions ζi,d ∈ L2(Γi,d × (0, T )) (i = 1, 2)

are such that (1.18) holds, then for each p0 ∈ V , there exist controls g ∈ H 32, 34 (S× (0, T ))

and associated Nash quasi-equilibria (u1, u2) such that the corresponding solutions to (1.2)

satisfy (1.12).

Analogously to Proposition 7, we can analyze the equivalence between the Nash

equilibrium and Nash quasi-equilibrium for the functionals Ki. This analysis leads to the

following result:

Proposition 2 Let us assume that F ∈ W 2,∞(R) and ζi,d ∈ H12, 14 (Σi,d) (i = 1, 2). Sup-

pose that p0 ∈ V and n ≤ 6. There exists C > 0 such that, if g ∈ H 32, 34 (S × (0, T )) and

the µi satisfy

µi ≥ C(

1 + ‖g‖H

32 , 3

4 (S×(0,T ))

),

the conditions (1.9) and (1.19) are equivalent.

The rest of this paper is organized as follows. In Section 1.2, we analyze the exact

controllability of the system (1.1), that we divide in two cases, corresponding to linear

and semilinear systems. In the linear case, we prove Theorem 1. In the semilinear case,

we establish Theorem 2 using a xed point argument. In Section 1.3, we investigate the

control of (1.2), following ideas similar to those in Section 2. Finally, in Section 1.4 we

dedicate to some additional comments.

26

1.2 The exact controllability with distributed leader

and boundary followers

1.2.1 The linear case

In the linear case, for (1.1) (with F ≡ 0), we can reduce the problem of exact con-

trollability to the trajectory to a problem of null controllability. Indeed, let us introduce

the change of variable z = y − y, with y the solution (1.5). Then z is the solution to the

system zt −∆z + a(x, t)z = f1O in Q,

z = v11S1 + v21S2 on Σ,

z(·, 0) = z0 in Ω

(1.20)

where z0 = y0 − y0 and (1.7) is equivalent to

z(·, T ) = 0 in Ω. (1.21)

We can rewrite the functionals Ji (1.3) as follows:

Ji(f, v1, v2) =αi2

∫∫Oi,d×(0,T )

|z − zi,d|2 dx dt+µi2

∫∫Si×(0,T )

|vi|2 dσ dt, i = 1, 2, (1.22)

where zi,d = ξi,d − y, i = 1, 2.

Existence and uniqueness of Nash Equilibrium

Let f ∈ L2(O × (0, T )) be xed. Let us consider the spaces Hi = L2(Si × (0, T )),

i = 1, 2, and H = H1 ×H2. Then, since the Ji are convex, (v1, v2) is a Nash equilibrium

if and only if

αi

∫∫Oi,d×(0,T )

(z − zi,d)wi dx dt+ µi

∫∫Si×(0,T )

vivi dσ dt = 0 ∀vi ∈ Hi, (1.23)

where wi is the solution to the systemwit −∆wi + a(x, t)wi = 0 in Q,

wi = vi1Sion Σ,

wi(·, 0) = 0 in Ω.

(1.24)

27

Let us dene the operator Li ∈ L(Hi;L2(Q)), with Livi = wi, where wi is the solution of

(1.24) associated to vi. Let z be the solution to the systemzt −∆z + a(x, t)z = f1O in Q,

z = 0 on Σ,

z(·, 0) = z0 in Ω.

Then we can write the solution to (1.20) in the form z = L1v1 + L2v

2 + z. Therefore, by

(1.23), we have that (v1, v2) is a Nash equilibrium if and only if

αiL∗i ((L1v

1 + L2v2)1Oi,d

) + µivi = αiL

∗i ((zi,d − z)1Oi,d

) in Hi, i = 1, 2. (1.25)

In the above equality, L∗i ∈ L(L(Q);Hi) is the adjoint of Li, i = 1, 2. Let us dene the

operator L : H → H with L(v1, v2) := (α1L∗1((L1v

1 + L2v2)1O1,d

) + µ1v1, α2L

∗2((L1v

1 + L2v2)1O2,d

) + µ2v2)

∀(v1, v2

)∈ H.

We have L ∈ L(H) and we can choose µi large enough, such that

(L(v1, v2), (v1, v2))H > µ0||(v1, v2)||2H, (1.26)

where

µ0 = maxi=1,2

µi −

(α1 + α2

2

)δi

, δi = min

i=1,2

||Li||2L(Hi,L2(Oi,d)), ||Li||2L(Hi,L2(O3−i,d))

.

Now, let us consider the following bilinear form on H:

A((v1, v2), (v1, v2)) := (L(v1, v2), (v1, v2))H ∀(v1, v2

),(v1, v2

)∈ H.

From the denition of L and the inequality (1.26), follows that A(·, ·) is continuous and

coercive. Hence by Lax-Milgram's Theorem, for each Φ ∈ H′ there exists exactly one pair

(v1, v2) satisfying

A((v1, v2), (v1, v2)) = 〈Φ, (v1, v2)〉H′×H ∀(v1, v2) ∈ H. (1.27)

In particular, if Φ ∈ H′ is given by⟨Φ, (v1, v2)

⟩H′×H :=

((α1L

∗1((z1,d − z)1O1,d

), α2L∗2((z2,d − z)1O2,d

)), (v1, v2))H

∀(v1, v2

)∈ H.

we get (1.25). This proves the following result:

28

Proposition 3 If

µi −(α1 + α2

2

)δi > 0, i = 1, 2,

then, for each f ∈ L2(O × (0, T )), there exists a unique Nash equilibrium (v1, v2) for the

functionals Ji.

Remark 1 Since the bilinear form A(·, ·) is coercive, we obtain a constant C > 0 such

that

||z||L2(0,T ;H1(Ω)) + ||zt||L2(0,T ;H−1(Ω)) ≤ C(1 + ||f ||L2(O×(0,T ))

),

where z is the solution to (1.20).

The optimality system

In this subsection we will obtain the optimality system, that characterizes the Nash

equilibrium for the cost functionals Ji. Thus, let f ∈ L2(O × (0, T )) be given and let

(v1, v2) be the associated Nash equilibrium. Let us introduce the adjoint system to (1.24):−φit −∆φi + a(x, t)φi = αi(z − zi,d)1Oi,d

in Q,

φi = 0 on Σ,

φi(·, T ) = 0 in Ω.

(1.28)

Then, for all vi ∈ Hi with wi = Livi, from systems (1.24) and (1.28), we have∫∫

Oi,d×(0,T )

αi(z − zi,d)wi dx dt = −∫∫

Si×(0,T )

vi∂φi

∂νdσ dt.

By (1.23), we deduce that

vi =1

µi

∂φi

∂ν

∣∣∣∣∣Si×(0,T )

, i = 1, 2.

Consequently, we get the optimality system

zt −∆z + a(x, t)z = f1O in Q,

−φit −∆φi + a(x, t)φi = αi(z − zi,d)1Oi,din Q,

z =1

µ1

∂φ1

∂ν1S1 +

1

µ2

∂φ2

∂ν1S2 , φ

i = 0 on Σ,

z(., 0) = z0, φi(., T ) = 0 in Ω.

(1.29)

These ideas indicate that our task is just to nd a control f ∈ L2(O × (0, T )) such that

the solution to (2.20) satises (2.10).

29

Null Controllability

The null controllability of (2.20) will be a consequence of an observability inequality

for the solutions to the adjoint, given by

−ψt −∆ψ + a(x, t)ψ =2∑i=1

αiγi1Oi,d

in Q,

γit −∆γi + a(x, t)γi = 0 in Q,

ψ = 0, γi =1

µi

∂ψ

∂ν1Si

on Σ,

ψ(·, T ) = ψT , γi(·, 0) = 0 in Ω.

(1.30)

From (2.20) and (1.30), we have:∫∫Q

f1Oψ dx dt =

∫Ω

z(x, T )ψT (x) dx−∫

Ω

z0(x)ψ(x, 0) dx

+2∑i=1

αi

∫∫Q

zi,dγi1Oi,d

dx dt.(1.31)

Therefore, to prove the null controllability is equivalent to nd, for each z0 ∈ L2(Ω), a

control f such that, for all ψT ∈ L2(Ω), we have:∫∫O×(0,T )

fψ dx dt = −∫

Ω

z0(x)ψ(x, 0) dx+2∑i=1

αi

∫∫Q

zi,dγi1Oi,d

dx dt

The following result furnishes an observability inequality for the adjoint system (1.30):

Proposition 4 Suppose that Oi,d ∩ O 6= ∅, i = 1, 2. Assume that one of the following

conditions holds:

O1,d = O2,d = Od, (1.32)

or

O1,d ∩ O 6= O2,d ∩ O. (1.33)

Then, there exist C > 0 and a weight ρ = ρ(t) (depending on ||a||L∞(Q)) such that, for

any ψT ∈ L2(Ω), the solution (ψ, γ1, γ2) to (1.30), satises∫Ω

|ψ(x, 0)|2 dx+2∑i=1

∫∫Q

ρ−2|γi|2 dx dt ≤ C

∫∫O×(0,T )

|ψ|2 dx dt. (1.34)

The proof of Proposition 4 is given below. Let us see how this result leads to the

null controllability of (2.20). The argument is standard. Thus, for each ε > 0, let us

introduce the functional

Fε(ψT ) :=

1

2

∫∫O×(0,T )

|ψ|2 dx dt+ ε||ψT ||L2(Ω) +

∫Ω

z0(x)ψ(x, 0) dx

−2∑i=1

αi

∫∫Oi,d×(0,T )

zi,dγi dx dt.

30

Obviously Fε : L2(Ω)→ R is continuous and strictly convex. We also have, by (1.34), that

Fε is coercive. Therefore, Fε possesses a unique minimum. Let us denote by ψTε ∈ L2(Ω)

the minimizer. If ψTε = 0, arguing as in [21], we deduce that the control f = 0 furnishes

a state satisfying

||zε(·, T )|| ≤ ε. (1.35)

If ψTε 6= 0,

F ′ε(ψTε ) · ψT = 0 ∀ψT ∈ L2(Ω), (1.36)

whence,∫∫O×(0,T )

ψεψ dx dt+ ε

(ψTε||ψTε ||

, ψT)

+

∫Ω

z0(x)ψ(x, 0) dx

−2∑i=1

αi

∫∫Oi,d×(0,T )

zi,dγi dx dt = 0 ∀ψT ∈ L2(Ω),

(1.37)

where (ψε, γ1ε , γ

2ε ) is the solution to (1.30) corresponding to ψTε . Let us set fε := ψε1O×(0,T )

and let us denote by zε the associated state. Then, comparing (1.37) and (1.31) we obtain∫Ω

(zε(x, T ) + ε

ψTε||ψTε ||

)ψT (x) dx = 0 ∀ψT ∈ L2(Ω),

and, again, one has (1.35). Furthermore, from (1.34) and (1.37), it follows that

||fε||L2(O×(0,T )) ≤ C

(2∑i=1

∫∫Oi,d×(0,T )

ρ2|zi,d|2 dx dt+ ||z0||2L2(Ω)

) 12

.

The controls fε are uniformly bounded in L2(O×(0, T )) and, therefore, it can be assumed

that fε → f weakly in L2(O × (0, T )). Accordingly, one has we see that

zε → z weakly in W (Q),

φiε → φi weakly in H2,1(Q), i = 1, 2,(1.38)

where (z, φ1, φ2) is the solution to (2.20) associated to f . In particular, we have weak

convergence for (zε(·, T )) in L2(Ω). Thus, from (1.35), we get z(·, T ) = 0.

Let us now turn to the proof of Proposition 4.

Proof of Proposition 4.

Case 1: Let us assume (1.32). Then, there exists a nonempty open set ω with ω ⊂⊂

Od ∩ O. Let η0 ∈ C2(Ω) be a function verifying

η0 > 0 in Ω, η0 = 0 on Γ, |∇η0| > 0 in Ω\ω.

31

The proof of the existence of such function can be found in [34]. Let l ∈ C∞([0, T ]) be a

function satisfying

l(t) > 0 ∀t ∈ (0, T ), l(t) = t ∀t ∈[0,T

4

], l(t) = T − t ∀t ∈

[3T

4, T

]and let us introduce

ϕ(x, t) :=eλ(η0(x)+m1)

lk(t), α(x, t) :=

eλ(η0(x)+m1) − eλ(||η0||L∞(Ω)+m2)

lk(t), (1.39)

where k ≥ 2, λ ≥ 1, m1 and m2 are constants chosen such that m1 ≤ m2 and there exists

C > 0 such that, for all λ ≥ 1, we have

|αt| ≤ Cϕk+1k . (1.40)

Let us set h := α1γ1 + α2γ

2. Note that, in view of the density of H10 (Ω) in L2(Ω), it can

be assumed that ψT ∈ H10 (Ω). Then ψ ∈ H2,1(Q) and h ∈ W (Q).

Now, let us recall the Carleman estimates (see [34], [39])

I1(h) ≤ C

s− 12

2∑i=1

α2i

µ2i

∣∣∣∣∣∣∣∣∣∣ϕ− 1

4∂ψ

∂νesα

∣∣∣∣∣∣∣∣∣∣2

H12 , 1

4 (Si×(0,T ))

+ sλ2

∫∫ω×(0,T )

ϕe2sα|h|2 dx dt) (1.41)

and

I2(ψ) ≤ C

(∫∫Q

e2sα|h|2 dx dt+ s3λ4

∫∫ω×(0,T )

ϕ3e2sα|ψ|2 dx dt), (1.42)

where

I1(h) :=

∫∫Q

(s−1ϕ−1|∇h|2 + sλ2ϕ|h|2)e2sα dx dt

and

I2(ψ) :=

∫∫Q

[s−1ϕ−1(|ψt|2 + |∆ψ|2) + sλ2ϕ|∇ψ|2 + s3λ4ϕ3|ψ|2

]e2sα dx dt.

Note that, in ω × (0, T ) thanks to (1.32), one has h = −ψt −∆ψ + a(x, t)ψ. Let ω′

be an open set such that ω ⊂⊂ ω′ ⊂⊂ Od∩O and let ζ be a function in C20(ω′) satisfying

ζ = 1 in ω and 0 ≤ ζ ≤ 1. (1.43)

32

From (1.30), integrating by parts and using (1.40) and (1.43), we obtain:∫∫ω×(0,T )

sλ2ϕe2sα|h|2 dx dt

≤ sλ2

∫∫ω′×(0,T )

ζϕe2sαh(−ψt −∆ψ + a(x, t)ψ) dx dt

≤ C

s3λ4

∫∫ω′×(0,T )

ϕ3e2sαhψ dx dt+ s2λ3

∫∫ω′×(0,T )

ϕ2e2sα|∇h|ψ dx dt

≤ C

εI1(h) +

1

εs5λ6

∫∫ω′×(0,T )

ϕ5e2sα|ψ|2 dx dt.

(1.44)

Then, choosing 0 < ε < 1/(2C), by (1.41) we have

I1(h) ≤ C

s− 12

2∑i=1

α2i

µ2i

∣∣∣∣∣∣∣∣∣∣ϕ− 1

4∂ψ

∂νesα

∣∣∣∣∣∣∣∣∣∣2

H12 , 1

4 (Si×(0,T ))

+ s5λ6

∫∫ω′×(0,T )

ϕ5e2sαψ2 dx dt

).

(1.45)

On the other hand, arguing as in [40], we have the following for k ≥ 4,

s−12

2∑i=1

αiµi

∣∣∣∣∣∣∣∣∣∣ϕ− 1

4∂ψ

∂νesα

∣∣∣∣∣∣∣∣∣∣2

H12 , 1

4 (Si×(0,T ))

≤ C

(∫∫Q

s32ϕ2e2sα|ψ|2 dx dt

+ s−12

∫∫Od×(0,T )

e2sα|h|2 dx dt),

(1.46)

where C > 0 depends of µi, (i = 1, 2). Therefore, if the µi and s are large enough, by

(1.41), (1.42), (1.45) and (1.46) we obtain

I1(h) + I2(ψ) ≤ C

∫∫ω′×(0,T )

ϕ5e2sα|ψ|2 dx dt. (1.47)

At this point, we will consider new weight functions:

ϕ(x, t) :=eλ(η0(x)+m1)

lk(t), α(x, t) :=

eλ(η0(x)+m1) − eλ(||η0||L∞(Ω)+m2)

lk(t), (1.48)

where l : [0, T ]→ R is dened as follows

l(t) =

l(t0), if t ∈ [0, t0],

l(t), if [t0, T ](1.49)

and t0 ∈ (T/4, 3T/4) is such that l(t0) is a local maximum.

Arguing as in [27, Lemma 1], using the equations in (1.30) and the inequality (1.47),

we deduce that there exists a constant C > 0 such that

||ψ(0)||2L2(Ω) +

∫∫Q

ϕe2sα|∇ψ|2 dx dt+

∫∫Q

ϕ3e2sα|ψ|2 dx dt+ I1(h)

≤ C

∫∫ω′×(0,T )

ϕ5e2sα|ψ|2 dx dt,(1.50)

33

where I1(h) is the analog to I1(h), where we have changed α and ϕ by α and ϕ, respectively.

Let us set

α∗(t) := minx∈Ωα(x, t).

Then, γi := esα∗γi is the solution to

γit −∆γi + a(x, t)γi = sα∗t esα∗γi in Q,

γi =esα

∗

µi

∂ψ

∂ν1Si

on Σ,

γi(·, 0) = 0 in Ω

and, consequently,

||γi||2L2(Q) + ||∇γi||2L2(Q) ≤ C

∣∣∣∣∣∣∣∣∣∣esα

∗

µi

∂ψ

∂ν

∣∣∣∣∣∣∣∣∣∣2

H12 , 1

4 (Si×(0,T ))

. (1.51)

On the other hand, by the same argument used in (1.46), we have that∣∣∣∣∣∣∣∣∣∣esα

∗

µi

∂ψ

∂ν1Si

∣∣∣∣∣∣∣∣∣∣2

H12 , 1

4 (Si×(0,T ))

≤ C

(∫∫Q

s2ϕk+1k e2sα∗|ψ|2 dx dt

+

∫∫Od×(0,T )

e2sα∗|h|2 dx dt).

(1.52)

Hence, by (1.51) and (1.52), taking k ≥ 4, one has:∫∫Q

|γi|2 dx dt ≤ C

(∫∫Od×(0,T )

e2sα∗h2 dx dt+ s2

∫∫Q

ϕ52 e2sα∗ |ψ|2 dx dt

). (1.53)

Using the denition of α∗ and taking µi large enough, we can estimate the right hand side

in (1.53) by the left hand side of (1.50). Hence, from (1.50) and (1.53), we see that there

exists a constant C > 0 such that

||ψ(0)||2L2(Ω) +2∑i=1

∫∫Q

|γi|2 dx dt ≤ C

∫∫ω′×(0,T )

ϕ5e2sα|ψ|2 dx dt,

which yields the inequality (1.34), with ρ(t) = e−sα∗(t) for all t ∈ (0, T ).

Case 2: Now, let us assume that (1.33) holds. Let O ⊂⊂ O be a nonempty connected

open set. By (1.33), there exist non-empty open sets ωi ⊂⊂ Oi,d∩O (i = 1, 2), such that

ω1 ∩ ω2 = ∅. Hence, we have to analyze the following situations:

ω1 ∩ O2,d = ∅ and ω2 ∩ O1,d = ∅ (1.54)

34

or

ωi ⊂ Oj,d and ωj ∩ Oi,d = ∅, with (i, j) = (1, 2) or (i, j) = (2, 1). (1.55)

Now, there exist functions η1, η2 ∈ C2(Ω) such that ηi > 0 in Ω, ηi = 0 on Γ, |∇ηi| > 0 in Ω \ ωi,

η1 = η2 in Ω \ O and ||η1||∞ = ||η2||∞.(1.56)

The proof of the existence of these functions can to be found in [1]. Let us dene the

following weight functions:

ϕi(x, t) =eλ(ηi(x)+m1)

lk(t), σi(x, t) =

eλ(ηi(x)+m1) − eλ(||ηi||L∞(Ω)+m2)

lk(t)(1.57)

and let us introduce the notation

I i1(γj) =

∫∫Q

(s−1ϕ−1i |∇γj|2 + sλ2ϕi|γj|2)e2sσi dx dt. (1.58)

By [40], the following inequality holds:

I i1(γi) ≤ C

s− 12

1

µ2i

∣∣∣∣∣∣∣∣∣∣ϕ− 1

4i

∂ψ

∂νesσi

∣∣∣∣∣∣∣∣∣∣2

H12 , 1

4 (Si×(0,T ))

+

∫∫ω×(0,T )

sλ2ϕi|γi|2e2sσi dx dt

, (1.59)

for i = 1, 2.

Let ζ be a function in C2(Ω) such that

ζ = 0 in O, ζ = 1 in Ω \ O.

Then it is clear that ψ = ζψ is the solution to the system−ψt −∆ψ + a(x, t)ψ =

2∑i=1

αiζγi1Oi,d

− 2(∇ζ · ∇ψ)− ψ∆ζ in Q,

ψ = 0 on Σ,

ψ(·, T ) = ζψT in Ω.

By [41], we have the existence of a positive constant C > 0 such that

L1(ψ) ≤ C

(s2λ3

∫∫ω1×(0,T )

ϕ21e

2sσ1 |ψ|2 dx dt

+ s−1λ−1

∫∫O1,d×(0,T )

ϕ−11 e2sσ1|γ1|2 dx dt (1.60)

+ s−1λ−1

∫∫O2,d×(0,T )

ϕ−11 e2sσ1 |ζγ2|2 dx dt+ sλ

∫∫O×(0,T )

ϕ1e2sσ1|ψ|2 dx dt

),

35

where

Li(ψ) := λ

∫∫Q

ϕ2i e

2sσi|∇ψ|2 dx dt+ s2λ3

∫∫Q

ϕ2i e

2sσi |ψ|2 dx dt (1.61)

On the other hand, since ζ = 1 in Ω \ O, we also have

L1(ψ) ≥ s2λ3

∫∫Q

ϕ21e

2sσ1|ψ|2 dx dt− s2λ3

∫∫O×(0,T )

ϕ21e

2sσ1|ψ|2 dx dt.

Let us rst assume that (1.54) holds. Combining this last inequality, (1.59) and (1.60), it

follows that

I11 (γ1) + I2

1 (γ2) + s2λ3

∫∫Q

ϕ21e

2sσ1|ψ|2 dx dt

≤ C

2∑i=1

s−12

1

µ2i

∣∣∣∣∣∣∣∣ϕ− 14

i

∂ψ

∂νe2sσi

∣∣∣∣∣∣∣∣2H

12 , 1

4 (Si×(0,T ))

+2∑i=1

sλ2

∫∫ωi×(0,T )

ϕie2sσi |γi|2 dx dt+ s−1λ−1

∫∫O1,d×(0,T )

ϕ−11 e2sσ1|γ1|2 dx dt

+s−1λ−1

∫∫O2,d×(0,T )

ϕ−11 e2sσ1|ζγ2|2 dx dt+ s2λ3

∫∫O×(0,T )

ϕ21e

2sσ1|ψ|2 dx dt

.

(1.62)

Since that η1 = η2 = 0 on Σ, and ||η1||∞ = ||η2||∞, we have σ1 = σ2 and φ1 = φ2 on Σ.

Hence arguing as in [40], for k ≥ 4, we get

s−12

2∑i=1

∣∣∣∣∣∣∣∣ϕ− 14

i

1

µi

∂ψ

∂νe2sσi

∣∣∣∣∣∣∣∣2H

12 , 1

4 (Si×(0,T ))

≤ C

(s−

12

2∑i=1

∫∫Oi,d×(0,T )

ϕ− 1

2i e2sσi |γi|2 dx dt

+ s32

∫∫Q

ϕ21e

2sσ1|ψ|2 dx dt),

(1.63)

where C > 0 depends of µi.

For s and λ large enough, the third and fourth terms in the right hand side in (1.62)

and the right hand side in (1.63) can be absorbed by left hand side in the (1.62). Thus

we get:

I11 (γ1) + I2

1 (γ2) + s2λ3

∫∫Q

ϕ21e

2sσ1|ψ|2 dx dt

≤ C

(sλ2

2∑i=1

∫∫ωi×(0,T )

ϕie2sσi |γi|2 dx dt

+ s2λ3

∫∫O×(0,T )

ϕ21e

2sσ1|ψ|2 dx dt).

(1.64)

To eliminate the rst term in right-hand side of (1.64), let us consider open sets ωi ⊂ Ω

such that ωi ⊂⊂ ωi ⊂⊂ Oi,d∩O and ωi∩O3−i,d = ∅ and let us dene functions δi ∈ C20(ωi),

36

with

δi = 1 in ωi, 0 ≤ δi ≤ 1. (1.65)

Then, by (1.30) and (1.58), for s and λ large enough, we have for i = 1, 2,

sλ2

∫∫ωi×(0,T )

ϕie2sσi |γi|2 dx dt

≤ sλ2

∫∫ωi×(0,T )

δiϕie2sσi |γi|2 dx dt

≤ sλ2

αi

∫∫ωi×(0,T )

δiϕie2sσiγi (−ψt −∆ψ + a(x, t)ψ) dx dt

≤ C

ε I i1(γi) +

1

εs5λ6

∫∫ω1×(0,T )

ϕ51e

2sσ1|ψ|2 dx dt.

(1.66)

Thus, by (1.64) and (1.66), we see that the following inequality holds:

I11 (γ1) + I2

1 (γ2) + s2λ3

∫∫Q

ϕ21e

2sσ1 |ψ|2 dx dt

≤ Cs5λ6

∫∫O×(0,T )

(ϕ51e

2sσ1 + ϕ52e

2sσ2)|ψ|2 dx dt.(1.67)

Let us now consider the weight functions

ϕi(x, t) =eλ(ηi(x)+m1)

lk(t), σi(x, t) =

eλ(ηi(x)+m1) − eλ(||ηi||L∞(Ω)+m2)

lk(t), (1.68)

where l is dened as in (1.49). Let us denote by I i1 an expression similar to (1.58), where

we replace σi and ϕi by σi and ϕi, respectively. Then, arguing as we did to get the

inequality (1.50), we obtain:

||ψ(0)||2L2(Ω) + I11 (γ1) + I2

1 (γ2) + s2λ3

∫∫Q

ϕ21e

2sσ1|ψ|2 dx dt

≤ C s5λ6

∫∫O×(0,T )

(ϕ51e

2sσ1 + ϕ52e

2sσ2)|ψ|2 dx dt.(1.69)

Note that, since that ||η1||∞ = ||η2||∞, for each t ∈ [0, T ) we have

σ := minx∈Ω

σ1(x, t)

= min

x∈Ω

σ2(x, t)

. (1.70)

Then by (1.69),

||ψ(0)||2L2(Ω) +2∑i=1

∫∫Q

e2sσ(t)|γi|2 dx dt ≤ C

∫∫O×(0,T )

(ϕ5

1e2sσ1 + ϕ5

2e2sσ2)|ψ|2 dx dt

and we get the inequality (1.34) with ρ(t) = e−sσ(t) for all t ∈ (0, T ).

37

Let us now consider the case in (1.55). To x ideas, let us take (i, j) = (2, 1). For

h := α1γ1 + α2γ2, we have

I21 (h) = s−1

∫∫Q

ϕ−12 e2sσ2|∇h|2 dx dt+ sλ2

∫∫Q

ϕ2e2sσ2|h|2 dx dt (1.71)

whence,

I21 (h) ≤ C

2∑i=1

s−12α2i

µ2i

∣∣∣∣∣∣∣∣∣∣ϕ− 1

4i

∂ψ

∂νesσ2

∣∣∣∣∣∣∣∣∣∣2

H12 , 1

4 (Si×(0,T ))

+ sλ2

∫∫ω2×(0,T )

ϕ2e2sσ2 |h|2 dx dt

).

(1.72)

Combining (1.59), (1.60) and (1.72), it follows that

I11 (γ1) + I2

1 (h) + s2λ3

∫∫Q

ϕ21e

2sσ1|ψ|2 dx dt

≤ C

(sλ2

∫∫ω1×(0,T )

ϕ1e2sσ1 |γ1|2 dx dt+ s−

12

1

µ21

∣∣∣∣∣∣∣∣ϕ− 14

1

∂ψ

∂νe2sσ1

∣∣∣∣∣∣∣∣2H

12 , 1

4 (S1×(0,T ))

+2∑i=1

s−12α2i

µ2i

∣∣∣∣∣∣∣∣ϕ− 14

2

∂ψ

∂νe2sσ2

∣∣∣∣∣∣∣∣2H

12 , 1

4 (Si×(0,T ))

+ sλ2

∫∫ω2×(0,T )

ϕ2e2sσ2|h|2 dx dt+ s−1λ−1

∫∫O1,d×(0,T )

ϕ−11 e2sσ1|γ1|2 dx dt

+ s−1λ−1

∫∫O2,d×(0,T )

ϕ−11 e2sσ1|ζγ2|2 dx dt+ s2λ3

∫∫O×(0,T )

ϕ21e

2sσ1|ψ|2 dx dt

).

(1.73)

Let us analyze each term on the right hand side of (1.73). By (1.66), we obtain

sλ2

∫∫ω1×(0,T )

ϕ1e2sσ1 |γ1|2 dx dt

≤ C

(εI1

1 (γ1) +1

εs5λ6

∫∫ω1×(0,T )

ϕ51e

2sσ1|ψ|2 dx dt).

(1.74)

Analogously to (1.44) or (1.66), considering the nonempty conected open set ω2 ⊂⊂

O2,d ∩ O with ω2 ⊂⊂ ω2 and the function δ2 ∈ C20(ω2) satisfying (1.65) for i = 2,

δ2 = 1 in ω2, 0 ≤ δ2 ≤ 1, (1.75)

we obtain

sλ2

∫∫ω2×(0,T )

ϕ2e2sσ2|h|2 dx dt

≤ C

(εI2

1 (h) +1

εs5λ6

∫∫ω2×(0,T )

ϕ52e

2sσ2|ψ|2 dx dt).

(1.76)

38

Since that η1 = η2 in supp(ζ) ∩ O2,d, then for s and λ large enough, we have

s−1λ−1

∫∫O2,d×(0,T )

ϕ−11 e2sσ1|ζγ2|2 dx dt

≤ C s−1λ−1

(∫∫O2,d×(0,T )

ϕ−12 e2sσ2|ζh|2 dx dt+

∫∫O2,d×(0,T )

ϕ−11 e2sσ1|ζγ1|2 dx dt

)≤ ε

(I2

1 (h) + I11 (γ1)

).

(1.77)

Moreover, note that ϕ1 = ϕ2 and σ1 = σ2 on Σ. Hence, arguing as in [40], for k ≥ 4,

we get

s−12

1

µ21

∣∣∣∣∣∣∣∣ϕ− 14

1

∂ψ

∂νe2sσ1

∣∣∣∣∣∣∣∣2H

12 , 1

4 (S1×(0,T ))

+2∑i=1

s−12α2i

µ2i

∣∣∣∣∣∣∣∣ϕ− 14

2

∂ψ

∂νe2sσ2

∣∣∣∣∣∣∣∣2H

12 , 1

4 (Si×(0,T ))

≤ C

s−

12

∫∫O1,d×(0,T )

ϕ− 1

21 e2sσ1|γ1|2 dx dt

+ s−12

∫∫O2,d×(0,T )

ϕ− 1

21 e2sσ1|ζγ2|2 dx dt

+ s32

∫∫Q

ϕ21e

2sσ1|ψ|2 dx dt.

(1.78)

So, considering s and λ large enough, by (1.78) we have

s−12

1

µ21

∣∣∣∣∣∣∣∣ϕ− 14

1

∂ψ

∂νe2sσ1

∣∣∣∣∣∣∣∣2H

12 , 1

4 (S1×(0,T ))

+2∑i=1

s−12α2i

µ2i

∣∣∣∣∣∣∣∣ϕ− 14

2

∂ψ

∂νe2sσ2

∣∣∣∣∣∣∣∣2H

12 , 1

4 (Si×(0,T ))

≤ ε

(I2

1 (h) + I11 (γ1) + s2λ3

∫∫Q

ϕ21e

2sσ1|ψ|2 dx dt).

(1.79)

Finally, for the fth term of the right hand side in (1.73), for s and λ large enough we

have

s−1λ−1

∫∫O1,d×(0,T )

ϕ−11 e2sσ1|γ1|2 dx dt ≤ ε I1

1 (γ1). (1.80)

Hence, combining (1.66), (1.73) and (1.76)(1.80), we obtain

I11 (γ1) + I2

1 (h) + s2λ3

∫∫Q

ϕ21e

2sσ1|ψ|2 dx dt

≤ C s5λ6

∫∫O×(0,T )

(ϕ5

1e2sσ1 + ϕ5

2e2sσ2)|ψ|2 dx dt.

(1.81)

Now, considering ϕi and σi as in (1.68), arguing as in (1.69), we get:

||ψ(·, 0)||2L2(Ω) + I11 (γ1) + I2

1 (h) + s2λ3

∫∫Q

ϕ21e

2sσ1|ψ|2 dx dt

≤ C s5λ6

∫∫O×(0,T )

(ϕ51e

2sσ + ϕ52e

2sσ2)|ψ|2 dx dt.(1.82)

39

Considering the function σ dened in (1.70), since that h = α1γ1 + α2γ

2, by (1.82), we

see that

||ψ(0)||2L2(Ω) +2∑i=1

∫∫Q

e2sσ(t)|γi|2 dx dt ≤ C

∫∫O×(0,T )

(ϕ51e

2sσ1 + ϕ52e

2sσ2)|ψ|2 dx dt

and, again, we get the inequality (1.34) with ρ(t) = e−sσ(t) for all t ∈ (0, T ). This

concludes the proof.

1.2.2 The semilinear case

In this subsection we will prove Theorem 2 and Proposition 7. Previously, we will

obtain the optimality system characterizing a Nash quasi-equilibrium.

The optimality system in the semilinear case

Let us consider the semilinear system (1.5). According to denition (1), if a pair

(v1, v2) is a Nash quasi-equilibrium, then

αi

∫∫Od×(0,T )

(y − ξi,d)yi dx dt+ µi

∫∫Si×(0,T )

vivi dx dt = 0 ∀vi ∈ Hi, i = 1, 2, (1.83)

where yi is solution to yit −∆yi + a(x, t)yi = F ′(y)yi in Q,

yi = vi1Sion Σ,

yi(·, 0) = 0 in Ω.

(1.84)

The adjoint system to (1.84) is given by−φit −∆φi + a(x, t)φi = αi(y − ξi,d)1Od

+ F ′(y)φi in Q,

φi = 0 on Σ,

φi(·, T ) = 0 in Ω.

(1.85)

Therefore, we get from (1.83)(1.85) that∫∫Si×(0,T )

(−∂φ

i

∂ν+ µiv

i

)vi dσ dt = 0 ∀vi ∈ Hi, (1.86)

then,

vi =1

µi

∂φi

∂ν

∣∣∣∣Si×(0,T )

, i = 1, 2. (1.87)

40

Consequently, if (v1, v2) is a Nash quasi-equilibrium for the functionals Ji, i = 1, 2, it

must be given by (1.87), where the φi satisfy the following optimality system:

yt −∆y + a(x, t)y = F (y) + f1O in Q,

−φit −∆φi + a(x, t)φi = αi(y − ξi,d)1Od+ F ′(y)φi in Q,

y =2∑i=1

1

µi

∂φi

∂ν1Si, φi = 0 on Σ,

y(·, 0) = y0, φi(·, T ) = 0 in Ω.

(1.88)

Proof of Theorem 2

We will proof Theorem 2 using arguments similar to those in [3]. Let us take

z = y − y, where y is solution to (1.5) and let us rewrite (1.88) in the form

zt −∆z + a(x, t)z = G(x, t; z)z + f1O in Q,

−φit −∆φi + a(x, t)φi = αi(z − zi,d)1Od+ F ′(z + y)φi in Q,

z =2∑i=1

1

µi

∂φi

∂ν1Si, φi = 0 on Σ,

z(·, 0) = z0, φi(·, T ) = 0 in Ω,

(1.89)

where zi,d = ξi,d − y, z0 = y0 − y0, and

G(x, t; z) =

∫ 1

0

F ′ (y + τz) dτ. (1.90)

Let us consider, for each z ∈ L2(Q) and each f ∈ L2(O × (0, T )), the linear system

wt −∆w + a(x, t)w = G(x, t; z)w + f1O in Q,

−φit −∆φi + a(x, t)φi = αi(w − zi,d)1Od+ F ′(z + y)φi in Q,

w =2∑i=1

1

µi

∂φi

∂ν1Si, φi = 0 on Σ,

w(·, 0) = z0, φi(·, T ) = 0 in Ω.

(1.91)

Since F ∈ W 1,∞(R), there exists a constant M > 0 such that

|G(x, t; s)|+ |F ′(s)| ≤M ∀ (x, t, s) ∈ Q× R. (1.92)

From (1.91), arguing as in the proof of Proposition 3 and Remark 3, we see that there

exists a constant C > 0 such that

||w||L2(0,T ;H1(Ω)) + ||wt||L2(0,T ;H−1(Ω)) ≤ C(1 + ||f ||L2(O×(0,T ))

). (1.93)

41

Now, let us x z. Let us denote by (wz, φ1z, φ

2z) the solution to (1.91) corresponding to z

and let us consider the adjoint system of (1.91):

−ψz,t −∆ψz + a(x, t)ψz = G(x, t; z)ψz +2∑i=1

αiγiz1Od

in Q,

γiz,t −∆γiz + a(x, t)γiz = F ′(z + y)γiz in Q,

ψz = 0, γiz =1

µi

∂ψz∂ν

1Sion Σ,

ψz(·, T ) = ψT , γiz(·, 0) = 0 in Ω.

(1.94)

Then from (1.91) and (1.94), we have∫∫Q

f1Oψz dx dt =

∫Ω

wz(x, T )ψT (x) dx−∫

Ω

z0(x)ψz(x, 0) dx

+2∑i=1

αi

∫∫Od×(0,T )

γizzi,d dx dt.

Hence, we get the null controllability if and only if∫∫Q

f1Oψz dx dt = −∫

Ω

z0(x)ψz(x, 0) dx+2∑i=1

αi

∫∫Od×(0,T )

γizzi,d dx dt.

As in the linear case, let us dene the functional

Fz,ε(ψT ) :=

∫∫O×(0,T )

|ψz|2 dx dt+ ε||ψT ||L2(Ω) +

∫Ω

z0(x)ψz(x, 0) dx

−2∑i=1

αi

∫∫Od×(0,T )

zi,dγiz dx dt.

Arguing as in Proposition 4, we deduce the following observability inequality:

||ψz(·, 0)||2L2(Ω) +2∑i=1

∫∫Q

ρ2|γiz|2 dx dt ≤ C

∫∫O×(0,T )

|ψz|2 dx dt, (1.95)

where C > 0 is independent of z. This inequality allows minimize Fz,ε in L2(Ω). Then,

taking the limit as ε goes to zero, we get a (unique) leader fz ∈ L2(O× (0, T )) such that

the associate solution (wz, φ1z, φ

2z) to (1.91), satises

wz(·, T ) = 0 in Ω. (1.96)

Furthermore,

||fz||L2(O×(0,T )) ≤ C ∀z ∈ L2(Q). (1.97)

At this point, we can apply a standard xed point argument to the mapping z 7→ wz

and deduce that (1.89) is null-controllable. Hence, exact controllability to the trajectories

holds for (1.88) and Theorem 2 is proved.

42

Equilibrium and quasi-equilibrium

In this section we will prove Proposition 7. Let us suppose that F ∈ W 2,∞(R) and

let us consider, for f ∈ L2(O×(0, T )), the associated Nash quasi-equilibrium pair (v1, v2).

For all w1, w2 ∈ H1 and any s ∈ R, we denote by ys the solution to the systemyst −∆ys + a(x, t)ys = F (ys) + f1O in Q,

ys = (v1 + sw1)1S1 + v21S2 on Σ,

ys(·, 0) = y0 in Ω.

(1.98)

Let us use the notation ys|s=0 = y. Then

D1J1(f ; v1 + sw1, v2) · w2 −D1J1(f ; v1, v2) · w2

= α1

∫∫Od×(0,T )

(ys − ξ1,d)qs dx dt− α1

∫∫Od×(0,T )

(y − ξ1,d)q dx dt

+ sµ1

∫∫S1×(0,t)

w1w2 dσ dt,

(1.99)

where qs is the derivative of ys in respect to v1 + sw1 in the direction of w2. Thus, qs is

the solution to qst −∆qs + a(x, t)qs = F ′(ys)qs in Q,

qs = w21S1 on Σ,

qs(·, 0) = 0 in Ω.

(1.100)

Here, we also use the notation qs|s=0 = q. Let us introduce the adjoint of (1.100):−φst −∆φs + a(x, t)φs = α1(ys − ξ1,d)1Od

+ F ′(ys)φs in Q,

φs = 0 on Σ,

φs(·, 0) = 0 in Ω.

(1.101)

Then, from (1.100) and (1.101), we get the identity

α1

∫∫Od

(ys − ξ1,d)qs dx dt =

∫∫Σ

w21S1

∂φs

∂νdσ dt. (1.102)

Replacing in (1.99), we get:

D1J1(f ; v1 + sw1, v2) · w2 −D1J1(f ; v1, v2)

=

∫∫S1×(0,T )

(∂φs

∂ν− ∂φ

∂ν

)w2 dσ dt+ sµ1

∫∫S1×(0,T )

w1w2 dσ dt.(1.103)

Note that the following limits exist

h = lims→0

1

s(ys − y), η = lim

s→0

1

s(φs − φ) and

∂η

∂ν= lim

s→0

1

s

(∂φs

∂ν− ∂φ

∂ν

).

43

Note also that the pair (h, η) solves the system

ht −∆h+ a(x, t)h = F ′(y)h in Q,

−ηt −∆η + a(x, t)η = α1h1Od+ F ′′(y)hφ+ F ′(y)η in Q,

h = w11S1 , η = 0 on Σ,

h(·, 0) = 0, η(·, T ) = 0 in Ω.

(1.104)

Thus, from (1.99) and (1.104), we obtain

D21J1(f ; v1, v2)·(w1, w2) =

∫∫S1×(0,T )

∂η

∂νw2 dσ dt+µ1

∫∫S1×(0,T )

w1w2 dσ dt ∀w1, w2 ∈ H1.

In particular, for all w1 ∈ H1, we have

D21J1(f ; v1, v2) · (w1, w1) =

∫∫S1×(0,T )

∂η

∂νw1 dσ dt+ µ1

∫∫S1×(0,T )

|w1|2 dσ dt. (1.105)

Let us show that there exists a positive constant C1 depending on ||a||∞, ||F ′′||∞, M and

||y0||L2(Ω), such that∣∣∣∣∫∫S1×(0,T )

∂η

∂νw1 dσ dt

∣∣∣∣ ≤ C1

(1 + ||f ||L2(O×(0,T ))

)||w1||2H1

∀w1 ∈ H1. (1.106)

In fact, given w1 ∈ H1, we have h ∈ H12, 14 (Q). Moreover, there exists a constant C > 0,

independent of w1, such that

||h||H

12 , 1

4 (Q)≤ C||w1||H1 . (1.107)

From (1.104), we obtain∫∫S1×(0,T )

∂η

∂νw1 dσ dt =

∫∫Q

(α1h

21Od+ F ′′(y)h2φ

)dx dt. (1.108)

Let us check that the right hand side of (1.108) is bounded by the right hand side of

(1.106). To this end, let us obtain r and s such that

φ ∈ Lr(0, T ;Ls(Ω)) and h ∈ L2r′(0, T ;L2s′(Ω)),

where r′ and s′ are conjugate of r and s, respectively.

Since h ∈ H12, 14 (Q), by interpolation we get that h ∈ La(0, T ;Lb(Ω)), with

1

a=

1 + θ

4and

1

b=a(n− 2) + 4

2an,

44

where θ ∈ (0, 1). Accordingly, taking a = 2r′ and b = 2s′, it follows that appropriate

values of r and s are

r =a

a− 2and s =

an

2a− 4.

On the other hand, vi ∈ H12, 14 (Si × (0, T )) and y0 ∈ L2(Ω), then y ∈ L2(0, T ;H1(Ω)) ∩

L∞(0, T ;L2(Ω)) and, by interpolation, y ∈ La(0, T ;Lb(Ω)), where

1

a=θ

2, and

1

b=a(n− 2) + 4

2an,

with θ ∈ (0, 1). Thus, from the regularity results for solution of heat equation, we get

φ ∈ La(0, T ;w2,b(Ω)) → La(0, T ;Lbn

n−2b (Ω)) = La(0, T ;L2an

a(n−6)+4 (Ω)).

So, taking a = r, it follows that φ ∈ Lr(0, T ;L2an

a(n−2)−8 (Ω)) and, in order to have φ ∈

Lr(0, T ;Ls(Ω)), we needan

2a− 4≤ 2an

a(n− 2)− 8,

which is true if and only if n ≤ 6. In this case, we can guarantee by (1.88), (1.98), (1.101),

(1.108), that∣∣∣∣∫∫S1×(0,T )

∂η

∂νw1 dσ dt

∣∣∣∣ ≤ C(||h||2L2(Od×(0,T )) + ||F ′′||∞||h||2L2r′ (0,T ;L2s′ (Ω))

||φ||Lr(0,T ;Ls(Ω))

)≤ C

(1 + ||φ||Lr(0,T ;Ls(Ω))

)||w1||2H1

≤ C(1 + ||y||L2(Q)

)||w1||2H1

≤ C(1 + ||f ||L2(O×(0,T ))

)||w1||2H1

,

which proves (1.106).

As a consequence of (1.105) and (1.106), it follows that

D21J1(f ; v1, v2) · (w1, w1) ≥ [µ1 − C1(1 + ||f ||L2(O×(0,T )))]||w1||2H1

∀w1 ∈ H1.

Analogously, we can prove that there exists another constant C2 > 0 such that

D22J2(f ; v1, v2) · (w2, w2) ≥ [µ2 − C2(1 + ||f ||L2(O×(0,T )))]||w2||2H2

∀w2 ∈ H2.

Note that the constants Ci are independent of µ1 and µ2. Then, for µi large enough, we

conclude that µi − Ci(1 + ||f ||L2(O×(0,T ))

)is a positive number for i = 1, 2 and the pair

(v1, v2) is in fact a Nash equilibrium.

45

1.3 Hierarchic control with boundary leader and distri-

buted followers

In this section we will analyze the hierarchic control via StackelbergNash strategy

in the context of the exact controllability to the trajectoriesy of (1.2). As in Section 1.2,

we will rst analyze the linear (Section 1.3.1) and then semilinear case (Section 1.3.2).

1.3.1 The linear case

Let us set z = p− p, where p and p are the solutions of (1.2) and (1.10) with F ≡ 0,

respectively. Then, z is the solution to the systemzt −∆z + a(x, t)z = u11O1 + u21O2 in Q,

z = g1S on Σ,

z(·, 0) = z0 in Ω,

(1.109)

where z0 = p0 − p0. Thus, (1.12) is equivalent to

z(·, T ) = 0 in Ω. (1.110)

We can rewrite the cost functionals Ki (i = 1, 2), given in (1.8), as follows:

Ki(g, u1, u2) :=

αi2

∫∫Σi,d

∣∣∣∣∂z∂ν − zi,d∣∣∣∣2 dσ dt+

µi2

∫∫Oi×(0,T )

|ui|2 dx dt, i = 1, 2,

where zi,d = ζi,d − ∂p/∂ν.

Existence and uniqueness of Nash equilibrium

Let g ∈ H 32, 34 (S × (0, T )) be xed. Let us consider the spaces Hi = L2(Oi × (0, T ))

and H = H1×H2. Then, since the Ki are convex, (1.9) and (1.19) are equivalent. Hence,

(u1, u2) is a Nash equilibrium for the Ki if and only if

αi

∫∫Σi,d

(∂z

∂ν− zi,d

)∂zi

∂νdσ dt+ µi

∫∫Oi×(0,T )

uiui dx dt = 0 ∀ui ∈ Hi, (1.111)

where the zi solve zit −∆zi + a(x, t)zi = ui1Oi

in Q,

zi = 0 on Σ,

zi(·, 0) = 0 in Ω.

(1.112)

46

Let us dene the operators Li : Hi → H2,1(Q), with Li(ui) = zi, where zi is the

solution of (1.112). Let z be the solution of the systemzt −∆z + a(x, t)z = 0 in Q,

z = g1S on Σ,

z(·, 0) = z0 in Ω.

Then z = L1(u1) + L2(u2) + z and, by (1.111), it follows thatαi

∫∫Σi,d

(γ1(L1u

1 + L2u2 + z)− zi,d

)γ1 Li(ui) dσ dt+ µiαi

∫∫Oi×(0,T )

uiui dx dt = 0

∀ui ∈ Hi,

where γ1 denote the usual trace mapping. Then, proceeding as in Section 1.2.1, applying

the Lax-Milgram's Theorem, we get the following result:

Proposition 5 If

µi − ||γ1||2||Li||2(

3αi + α3−i

2

)> 0, i = 1, 2,

then, for each g ∈ H 32, 34 (S × (0, T )), there exists a unique Nash Equilibrium (u1, u2) for

the functionals Ki.

Remark 2 From the proof of Proposition 5, we can deduce that there exists C > 0 such

that

||z||L2(0,T ;H2(Ω)) + ||zt||L2(Q) ≤ C(

1 + ||g||H

32 , 3

4 (S×(0,T ))

), (1.113)

where z is the solution of (1.109).

Optimality system

Let g ∈ H 32, 34 (S × (0, T )) be given. For any (u1, u2) = (u1(g), u2(g)) ∈ H, let z be

the associated solutions to (1.109). Let us consider the adjoint system to (1.112):−φit −∆φi + a(x, t)φi = 0 in Q,

φi = −αi(∂z

∂ν− zi,d

)1Γi,d

on Σ,

φi(·, T ) = 0 in Ω,

(1.114)

where (1.114)2 is motivated by (1.111). From (1.111), (1.112) and (1.114) we have∫∫Oi×(0,T )

(φiui + µiu

i)ui dx dt = 0 ∀ui ∈ L2(Oi × (0, T )).

47

Hence, the Nash equilibrium is characterized by

ui = − 1

µiφi∣∣∣∣Oi×(0,T )

. (1.115)

Thus, the optimality system in this case is:

zt −∆z + a(x, t)z = −2∑i=1

1

µiφi1Oi

in Q,

−φit −∆φi + a(x, t)φi = 0 in Q,

z = g1S, φi = −αi

(∂z

∂ν− zi,d

)1Γi,d

on Σ,

z(·, 0) = z0, φi(·, T ) = 0 in Ω,

(1.116)

Now, let us introduce a bounded regular domain Ω such that Ω ⊂ Ω, ∂Ω∩ Γ = Γ \ S and

Γ := ∂Ω ∈ C2. Let Q = Ω×(0, T ) be a new cylinder with lateral boundary Σ = Γ×(0, T ).

By z0 we denote the prolongment of z0 to Ω. We will consider the following extended

system

zt −∆z + a(x, t)z = g1O −2∑i=1

1

µiφi1Oi

in Q,

−φit −∆φi + a(x, t)φi = 0 in Q,

z = 0, φi = −αi(∂z

∂ν− zi,d

)1Γi,d

on Σ,

z(·, 0) = z0, φi(·, T ) = 0 in Ω,

(1.117)

where g ∈ L2(O × (0, T )) and O ⊂ Ω \Ω. Then, we will proof that there exists a control

g ∈ L2(O × (0, T )) such that the solution of (1.117) satisfy (1.110).

Null controllability

In this section we will prove the null controllability of (1.117) as a consequence of

an observability inequality. Let us consider the following adjoint system to (1.117):

−ψt −∆ψ + a(x, t)ψ = 0 in Q,

γit −∆γi + a(x, t)γi =1

µiψ1Oi

in Q,

ψ =2∑i=1

αi∂γi

∂ν1Γi,d

, γi = 0 on Σ,

ψ(·, T ) = ψT , γi(·, 0) = 0 in Ω.

(1.118)

48

Hence, by (1.117) and (1.118), it follows that

(z(T ), ψT )− (z0, ψ(0))−2∑i=1

αi

∫∫Σi,d

zi,d∂γi

∂νdσ dt =

∫∫O×(0,T )

g ψ dx dt, (1.119)

Thus, we see that the null controllability to (1.117) is equivalent prove that, for each

z0 ∈ L2(Ω), there exists g ∈ L2(O× (0, T )) such that the following equality holds for each

ψT ∈ L2(Ω):

−(z0, ψ(0))−2∑i=1

αi

∫∫Σi,d

zi,d∂γi

∂νdσ dt =

∫∫O×(0,T )

g ψ dx dt. (1.120)

To this end, we have to prove an observability inequality, which is given in the following

result:

Proposition 6 Suppose that Γi,d ∩ S = ∅. There exist a constant C > 0 and a weight

function ρ = ρ(t), such that

||ψ(0)||2L2(Ω)

+2∑i=1

∫∫Σi,d

ρ2

∣∣∣∣∂γi∂ν

∣∣∣∣2 dσ dt ≤ ∫∫O×(0,T )

ψ2 dx dt, (1.121)

where (ψ, γ1, γ2) is the solution of (1.118) associated to ψT .

With help of Proposition 6, arguing as in the Section 1.2, we deduce the desired null

controllability result. Thus, taking g = z∣∣S×(0,T )

, we have that g ∈ H32, 34 (S × (0, T )) and

the corresponding solution (z, φ1, φ2) of the system (1.116) veries z(·, T ) = 0.

Proof of Proposition 6. Let us consider ω ⊂ O a nonempty open set and a function

η0 ∈ C2(Ω) such that

η0 > 0 in Ω, η0 = 0 on Γ, |∇η0| > 0 in Ω\ω.

The proof of the existence of η0 can be found in [34]. Then, we dene weight functions

α = α(x, t) and ϕ = ϕ(x, t) as in (1.39). Let us dene h = α1γ1 + α2γ

2. In sequel we will

use the Carleman estimates, which the proof can be found in [34] and [40],

I1(ψ) ≤ C

(s−

12

2∑i=1

∣∣∣∣∣∣∣∣ϕ− 14 esα

∂γi

∂ν

∣∣∣∣∣∣∣∣2H

12 , 1

4 (Σi,d)

+ s

∫∫ω×(0,T )

ϕ e2sα ψ2 dx dt

)(1.122)

and, since γi = 0 in O,

I2(γi) ≤ Cα2i

µ2i

∫∫Q

e2sα ψ212Oidx dt, (1.123)

49

where

I1(ψ) =

∫∫Q

(sϕ)−1 e2sα |∇ψ|2 dx dt+ s

∫∫Q

ϕ e2sα ψ2 dx dt

and

I2(γi) =

∫∫Q

(sϕ)−1 e2sα(|γit|2 + |∆γi|2

)dx dt+ s

∫∫Q

ϕ e2sα |∇γi|2 dx dt

+ s3

∫∫Q

ϕ3e2sα |γi|2 dx dt.(1.124)

Proceeding as in [40], we have

s−12

∣∣∣∣∣∣∣∣ϕ− 14 esα

∂γi

∂ν

∣∣∣∣∣∣∣∣2H

12 , 1

4 (Σi,d)

≤ C

(C1 s

− 12

∫∫Q

ϕ−12 e2sα ψ2 dx dt

+ s32

∫∫Q

ϕ2e2sα |γi|2 dx dt),

(1.125)

where C1 > 0 depends on µi. Then, taking s > 1 large enough, the inequalities (1.122)

(1.125) imply that

I1(ψ) + I2(γ1) + I2(γ2) ≤ C s

∫∫ω×(0,T )

ϕ e2sα ψ2 dx dt. (1.126)

Thus, arguing as in the proof of Proposition 4, for µi large enough we get a constant

C > 0 such that

||ψ(0)||2L2(Ω)

+2∑i=1

∫∫Σi,d

e2sα

∣∣∣∣∂γi∂ν

∣∣∣∣2 dσ dt ≤ C

∫∫ω×(0,T )

ϕ e2sα ψ2 dx dt (1.127)

where α, ϕ are dened in (1.48) and α(t) = minα(x, t) : x ∈ Ω for all t ∈ (0, T ). Thus,

the inequality (1.121) follows from (1.127) with ρ(t) = e−sα(t), for all t ∈ (0, T ). This

ends the proof.

1.3.2 The semilinear case

In this section we will prove Theorem 4 and Proposition 2. The proofs follow the

same ideas in Section 1.2.2.

Semilinear optimality system

We will obtain the semilinear optimality system which characterize the Nash quasi-

equilibrium for the functionals Ki. A pair (u1, u2) ∈ H is a Nash quasi-equilibrium if and

only if satises (1.19), that is,

αi

∫∫Σi,d

(∂p

∂ν− ζi,d

)∂pi

∂νdσ dt+ µi

∫∫Oi×(0,T )

uiui dx dt = 0 ∀u ∈ Hi, (1.128)

50

where pi is the solution of the systempt −∆p+ a(x, t)p = F ′(p)pi + ui1Oi

in Q,

p = 0 on Σ,

p(·, 0) = 0 in Ω.

(1.129)

Motivated by (1.128), we dene the adjoint system for (1.129) as follows:−φit −∆φi + a(x, t)φi = F ′(p)φi in Q,

φi = αi

(∂p

∂ν− ζi,d

)1Γi,d

on Σ,

φi(·, T ) = 0 in Ω.

(1.130)

By (1.128), (1.129) and (1.130) we deduce that∫∫Oi×(0,T )

(φi + µiu

i)ui dx dt = 0 ∀u ∈ Hi. (1.131)

This give us the following characterization Nash quasi-equilibrium:

ui = − 1

µiφi∣∣∣∣Oi×(0,T )

, i = 1, 2. (1.132)

Therefore, we get the following optimality system:

pt −∆p+ a(x, t)p = F (p)−2∑i=1

1

µiφiOi

in Q,

−φit −∆φi + a(x, t)φi = F ′(p)φi in Q,

p = g1S, φi = −αi(∂p

∂ν− ζi,d

)1Γi,d

on Σ,

p(·, 0) = p0, φi(·, T ) = 0 in Ω.

(1.133)

Sketch of the proof of Theorem 4

To prove Theorem 4, we will use again the ideas in [3]. Considering the change of

variables z = p− p, where p is the solution to (1.2) and p is the solution to (1.10), we can

rewrite (1.133) as follows:

zt −∆z + a(x, t)z = G(x, t; z)z −2∑i=1

1

µiφiOi

in Q,

−φit −∆φi + a(x, t)φi = F ′(z + p)φi in Q,

z = g1S, φi = −αi(∂z

∂ν− zi,d

)1Γi,d

on Σ,

z(·, 0) = z0, φi(·, T ) = 0 in Ω,

(1.134)

51

where z0 = p0 − p0, zi,d = ζi,d − ∂p/∂ν, and

G(x, t; z) =

∫ 1

0

(F ′(p) + τz) dτ.

For each z ∈ L2(Q) and each g ∈ H 34, 32 (S × (0, T )), let us consider the linear system

wt −∆w + a(x, t)w = G(x, t; z)w −2∑i=1

1

µiφiOi

in Q,

−φit −∆φi + a(x, t)φi = F ′(z + p)φi in Q,

w = g1S, φi = −αi(∂w

∂ν− zi,d

)1Γi,d

on Σ,

w(·, 0) = z0, φi(·, T ) = 0 in Ω.

(1.135)

Since F ∈ W 1,∞(R), there exists a constant M > 0 such that

|G(x, t; s)|+ |F (s)| ≤M, ∀(x, t, s) ∈ Q× R. (1.136)

Furthermore, by (1.113), there exists a constant C > 0 such that

||wz||L2(0,T ;H2(Ω)) + ||wz,t||L2(Q) ≤(

1 + ||g||H

32 , 3

4 (S×(0,T ))

). (1.137)

Let us x z and let us denote by (wz, φ1z, φ

2z) the associated solution to (1.135). Now, as in

the linear case, we can extend the system (1.135). This way, we get the following system:

wz,t −∆wz + a(x, t)wz = G(x, t; z)wz −2∑i=1

1

µiφiOi

+ g1O in Q,

−φit −∆φi + a(x, t)φi = F ′(z + p)φi in Q,

wz = 0, φi = −αi(∂wz∂ν− zi,d

)1Γi,d

on Σ,

wz(·, 0) = z0, φi(·, T ) = 0 in Ω.

(1.138)

Let us consider the corresponding adjoint:

−ψz,t −∆ψz + a(x, t)ψz = G(x, t; z)ψz in Q,

γiz,t −∆γiz + a(x, t)γiz = F ′(z + p)γiz +1

µiψz1Oi

in Q,

ψz =2∑i=1

αi∂γiz∂ν

1Γi,d, γiz = 0 on Σ,

ψz(·, T ) = ψTz , γiz(·, 0) = 0 in Ω.

(1.139)

Then, by (1.138)(1.139), we have

(wz(T ), ψTz )− (z0, ψz(0))−2∑i=1

αi

∫∫Σi,d

zi,d∂γiz∂ν

dσ dt =

∫∫O×(0,T )

g ψz dx dt,

52

and the null controllability property holds if and only if, for each z0, there exists a control

g ∈ L2(O × (0, T )) such that

−(z0, ψz(0))−2∑i=1

αi

∫∫Σi,d

zi,d∂γiz∂ν

dσ dt =

∫∫O×(0,T )

ψzf dx dt ∀ψTz ∈ L2(Ω).

The reminder of the proof is very similar to the nal part of the proof of Theorem 2. For

brevity, it will be omitted.

Equilibrium and quasi-equilibrium with distributed followers

In this section we will prove Proposition 2, we will follow the ideas of the Section

1.2.2. Let us assume F ∈ W 2,∞(R), let g be the leader control and (u1, u2) the associated

Nash quasi-equilibrium pair. For all w1 ∈ H1 and s ∈ R, we denote by ps the solution topst −∆ps + a(x, t)ps = F (ps) + (u1 + sw1)1O1 + u21O2 in Q,

ps = g1S on Σ,

ps(·, 0) = p0 in Ω,

(1.140)

and we set ps∣∣s=0

= p. For all w2 ∈ H1 we have

D1K1(g;u1 + sw1, u2) · w2 −D1K1(g;u1, u2) · w2

= α1

∫∫Σ1,d

(∂ps

∂ν− ζ1,d

)∂qs

∂νdσ dt− α1

∫∫Σ1,d

(∂p

∂ν− ζ1,d

)∂q

∂νdσ dt

+α1

∫∫O1×(0,T )

w1w2 dx dt,

(1.141)

where qs, with qs∣∣s=0

= q, is the solution to the systemqst −∆qs + a(x, t)qs = F ′(ps)qs + w21O1 in Q,

qs = 0 on Σ,

qs(·, 0) = 0 in Ω.

(1.142)

Let us introduce the associated adjoint system:−φst −∆φs + a(x, t)φs = F ′(ps)φs in Q,

φs = −α1

(∂ps

∂ν− ζ1,d

)1Γi,d

on Σ,

φs(·, T ) = 0 in Ω,

(1.143)

where also we dene φ∣∣s=0

= φ. Then, by (1.141), (1.142) and (1.143), we deduce that

α1

∫∫Σ1,d

(∂ps

∂ν− ζ1,d

)∂qs

∂νdσ dt =

∫∫O1×(0,T )

φsw2 dx dt. (1.144)

53

Hence substituting (1.144) in (1.141), we have

D1K1(g;u1 + sw1, u2) · w2 −D1K1(g;u1, u2) · w2

=

∫∫O1×(0,T )

(φs − φ)w2 dx dt+ µ1s

∫∫O1×(0,T )

w1w2 dx dt.(1.145)

Note that

−(φs − φ)t −∆(φs − φ) + a(x, t)(φs − φ) = [F ′(ps)− F ′(p)]φs + F ′(p)(φs − φ)

and

(φs − φ)∣∣Σ

= −α1

(∂ps

∂ν− ∂p

∂ν

)1Γ1,d

.

Therefore, the following limits exist:

h = lims→0

1

s(ps − p), η = lim

s→0

1

s(φs − φ)

and the couple (h, η) solves the system

ht −∆h+ a(x, t)h = F ′(p)h+ w21O1 in Q,

−ηt −∆η + a(x, t)η = F ′′(p)hφ+ F ′(p)η in Q,

h = 0, η = −α1∂h

∂ν1Γ1,d

on Σ,

h(·, 0) = 0, η(·, T ) = 0 in Ω.

(1.146)

Thus, by (1.145) and (1.146), we see that

D21K1(g;u1, u2) · (w1, w2) =

∫∫O1×(0,T )

η w2 dx dt+ µ1

∫∫O1×(0,T )

w1w2 dx dt ∀w2 ∈ H1,

In particular, for all w1 ∈ H1, we have

D21K1(g;u1, u2) · (w1, w1) =

∫∫O1×(0,T )

η w1 dx dt+ µ1

∫∫O1×(0,T )

|w1|2 dx dt. (1.147)

Arguing as in the proof of (1.106), we deduce that, if n ≤ 6, there exist C > 0 such that∣∣∣∣∫∫O1×(0,T )

η w1 dx dt

∣∣∣∣ ≤ C(

1 + ||g||2H

32 , 3

4 (S×(0,T ))

)||w1||2H1

. (1.148)

This way, by (1.147) and (1.148), one also has

D21K1(g;u1, u2) · (w1, w1) ≥

[µ1 − C

(1 + ||g||

H32 , 3

4 (S×(0,T ))

)]||w1||2H1

∀w1 ∈ H1.

Analogously, we get

D22K2(g;u1, u2) · (w2, w2) ≥

[µ2 − C

(1 + ||g||

H32 , 3

4 (S×(0,T ))

)]||w2||2H2

∀w2 ∈ H2.

54

In both inequalities the constant C > 0 is independent of µi. Then, taking µi large

enough, we get a constant C > 0 such that

D2iKi(g;u1, u2) · (wi, wi) ≥ C ||wi||2Hi

∀wi ∈ Hi.

We deduce at once that (u1, u2) is a Nash equilibrium for the functionals Ki.

1.4 Final comments

1.4.1 The case with all controls on the boundary

The third situation that we can analyze corresponds to take all the controls (the

leader and the followers) on the boundary. Thus, let us consider the linear systemzt −∆z + a(x, t)z = 0 in Q,

z = v1S + v11S1 + v21S2 on Σ,

z(·, 0) = z0 in Ω,

(1.149)

where S, S1, S2 ∈ Γ are non-empty closed and disjoints subsets. We can introduce cost

functionals as in (1.3). Then, arguing as in Sections 1.2.1 and 1.3.1, we can prove the

existence and uniqueness of a Nash equilibrium for each leader. Furthermore, we get the

optimality system

zt −∆z + a(x, t)z = 0 in Q,

−φit −∆φi + a(x, t)φi = αi(z − zi,d)1Oi,din Q,

z = v1S +2∑i=1

1

µi

∂φi

∂ν1Si, φi = 0 on Σ,

z(·, 0) = z0, φi(·, T ) = 0 in Ω,

(1.150)

whose adjoint is given as follows:

−ψt −∆ψ + a(x, t)ψ =2∑i=1

αiγi1Oi,d

in Q,

γit −∆γi + a(x, t)γi = 0 in Q,

ψ = 0, γi =1

µi

∂ψ

∂ν1Si

on Σ,

ψ(·, T ) = ψT , γi(·, 0) = 0 in Ω.

(1.151)

In the present situation there is a diculty to obtain the observability inequality. Conse-

quently, the Stacelberg Nash null controllability of (1.150) is an open question.

55

1.4.2 The assumption on the µi

In the proof of Propositions 3 and 5, we assumed that the µi, i = 1, 2, are large

enough. This assumption allowed to conclude the results thanks to Lax-Milgram's Theo-

rem. However, we can prove the existence and uniqueness of a Nash equilibrium without

using Lax-Milgram's Theorem.

For instance, it can be proved that a sequence µn exists with the following properties:

µ1 ≥ µ2 ≥ · · · ≥ µn ≥ · · · , µn > 0 ∀n, µn → 0

and, for µ1 = µ2 = µ dierent from all µn, we can assign to any leader control exactly one

Nash equilibrium pair. An interesting question remains: can we prove the null controlla-

bility of the associated system when for these µi? This will be analyzed in a forthcoming

paper.

1.4.3 The hypotheses on the Oi,d and Γi,d

In this article, we have used (1.13) and (1.14) in Theorem 1 only in the proof of

Proposition 4 to allow for the connection between the Carleman inequalities for ψ and

h = α1γ1 + α2γ2, in the rst case, and the Carleman inequalities for ψ and γi, i = 1, 2,

in the second one. A question that remains unknown is the null controllability of (1.1)

when we assume that O1,d 6= O2,d and O1,d ∩ O = O2,d ∩ O.

On the other hand, the hypothesis Γi,d ∩ S = ∅, i = 1, 2, was used only in the

proof of Proposition 6. Therefore, another interesting open question concerns case where

Γi,d ∩ S 6= ∅, i = 1, 2.

1.4.4 Hierarchic control of the wave equation

It makes sense to consider hierarchic controllability problems for the wave equation.

For instance, we can consider the systemztt −∆z + a(x, t)z = f1O + v11O1 + v21O2 in Q,

z = 0 on Σ,

z(·, 0) = z0, zt(·, 0) = z1 in Ω.

(1.152)

56

where the O, O1, O2 ⊂ Ω are nonempty open sets. Let us introduce the secondary cost

functionals

Pi(f ; v1, v2) :=αi2

∫∫Oi,d×(0,T )

|z − zi,d|2 dx dt+µi2

∫∫Oi×(0,T )

|v|2 dx dt, i = 1, 2,

and the main functional

P (f) :=1

2

∫∫O×(0,T )

|f |2 dx dt,

where the Oi,d ⊂ Ω are nonempty open sets and the zi,d ∈ L2(Oi,d), i = 1, 2. In this

context, appropriate problem is the following: for any given (z1, z2) and (w1, w2), in

appropriate spaces, for each leader f , we look for a Nash equilibrium (v1, v2) associated

to the Pi; we look for a control f ∈ L2(O × (0, T )) such that

P (f) = minfP (f),

subject to

z(·, T ) = w1, zt(·, T ) = w2 in Ω.

Arguing as in Sections 1.2.1 and 1.3.1, we get the following optimality system:

ztt −∆z + a(x, t)z = f1O −2∑i=1

1

µiφi1Oi

in Q,

−φitt −∆φi + a(x, t)φi = αi(z − zi,d)1Oi,din Q,

z = 0, φi = 0 on Σ,

z(·, 0) = z0, zt(·, 0) = z1, φi(·, T ) = 0, φit(·, T ) = 0 in Ω.

(1.153)

The associated adjoint system is

−ψtt −∆ψ + a(x, t)ψ =2∑i=1

αiγi1Oi,d

in Q,

γitt −∆γi + a(x, t)γi = − 1

µi1Oi

in Q,

z = 0, φi = 0 on Σ,

ψ(·, T ) = ψT , ψt(·, T ) = ψT1 , γi(·, 0) = 0, γit(·, 0) = 0 in Ω.

(1.154)

Thus, the tasks are in this case to prove an existence and uniqueness for (1.153) and an

observability inequality for (1.154). This problem will be also analized in a forthcoming

paper.

57

Capítulo 2

Hierarchic control for the wave

equation

Hierarchic control for the wave

equation

F. D. Araruna, E. Fernández-Cara, L. C. Silva

Abstract. We present some results on the exact controllability of the wave PDE in the

context of hierarchic control through Stackelberg-Nash strategy. We consider one leader and

two followers. To each leader we associate a Nash equilibrium corresponding to a bi-objective

optimal control problem; then we look for a leader that solves the exact controllability

problem. We consider linear and semilinear equations; for the latter, we use a xed point

method.

2.1 Statement of the problem

Let Ω ⊂ Rn be a bounded domain with boundary Γ of class C2 and assume that

T > 0. Let us consider small disjoints open nonempty sets O, O1, O2 ⊂ Ω. We will use

the notation Q = Ω × (0, T ), which lateral boundary Σ = Γ × (0, T ); by ν(x) we denote

the outward unit normal to Ω at the point x ∈ Γ.

Let us consider the following systemytt −∆y + a(x, t)y = F (y) + f1O + v11O1 + v21O2 in Q,

y = 0 on Σ,

y(·, 0) = y0, yt(·, 0) = y1 in Ω,

(2.1)

where a ∈ L∞(Q), f ∈ L2(O × (0, T )), vi ∈ L2(Oi × (0, T )), F : R 7→ R is a locally

Lipschitz-continuous function, (y0, y1) ∈ H10 (Ω) × L2(Ω) and the notation 1A indicates

the characteristic function of A.

In [50], Lions analyzed the approximate controllability for a hyperbolic PDE and

introduced, in the context of controllability, the concept of hierarchic control, where he

considered a main control independent, called the leader, and a control dependent on

leader, called the follower. In this paper, we will analyze a related problem for (2.1),

where f is the leader and v1 and v2 are the followers. We will apply the Stackelberg-Nash

rule, which combines the strategies of cooperative optimization of Stackelberg and the

non-cooperative strategy of Nash.

59

In [16] and [37], the hierarchic control of a parabolic PDE and the Stokes systems

have been analyzed and used to solve a approximate controllability problem. In [3] and [6],

the hierarchic control of a parabolic PDE has been used to provide exact controllability

(to the trajectories); the problem was analyzed in [3] with distributed leader and follower

controls, while [6] deals with distributed and boundary controls.

The main goal this article is to analyze the hierarchic control of (2.1) and, in parti-

cular, to prove that the Stackelberg-Nash strategy allows to solve the exact controllability

problem.

Let us dene the following main cost functional

1

2

∫∫O×(0,T )

|f |2 dx dt,

and the secondary cost functionals

Ji(f, v1, v2) :=

αi2

∫∫Oi,d×(0,T )

|y − yi,d|2 dx dt+µi2

∫∫Oi×(0,T )

|vi|2 dx dt, (2.2)

where the Oi,d ⊂ Ω are nonempty open sets, yi,d ∈ L2(Oi,d × (0, T )) are given functions

and αi and µi are positive constants.

Now, let us describe the Stackelberg-Nash strategy. Thus, for each choice of the

leader f , we try to nd a Nash equilibrium pair for cost functionals Ji; that is, we look

for controls v1 ∈ L2(O1 × (0, T )) and v2 ∈ L2(O1 × (0, T )) depending on f , satisfying

simultaneously

J1(f ; v1, v2) = minv1

J1(f ; v1, v2), J2(f ; v1, v2) = minv2

J2(f ; v1, v2). (2.3)

If the functionals Ji are convex, (2.3) is equivalent to

J ′i(f ; v1, v2) · vi = 0, ∀ vi ∈ L2(Oi × (0, T )), i = 1, 2. (2.4)

Note that this is the case if the PDE in (2.1) is linear, but this is not true in general.

Given (y0, y1) in H10 (Ω)× L2(Ω), a trajectory to (2.1) is a solution to the system

ytt −∆y + a(x, t)y = F (y) in Q,

y = 0 on Σ,

y(·, 0) = y0, yt(·, 0) = y1 in Ω.

(2.5)

60

After proving that, for each leader f , there exists at least one Nash equilibrium (v1, v2) =

(v1(f), v2(f)) and after xing a trajectory y, we will look for a leader f ∈ L2(O× (0, T )),

such that,

J(f) := minfJ(f), (2.6)

subject to the exact controllability condition

y(·, T ) = y(·, T ), yt(·, T ) = yt(·, T ) in Ω. (2.7)

As an application we can consider in the exact controllability of a exible structure

that we wish to make attain a desired state at time T in such a way that the state does

not separate too much from yi,d in Oi,d along (0, T ).

2.1.1 Main results

Let x0 ∈ Rn \ Ω be given and let us consider the following set

Γ+ := (x− x0) · ν(x) > 0

and the function d : Ω 7→ R, with d(x) = |x − x0|2 for all x ∈ Ω. We will impose of the

following assumption: there exists δ > 0 such that

O ⊃ Oδ(Γ+) ∩ Ω, (2.8)

where

Oδ(Γ+) = x ∈ Rn; |x− x′| < δ, x′ ∈ Γ+.

Due to the nite propagation speed of the solutions to the wave PDE, the time

T > 0 has to be large enough. Therefore, we will also assume in the sequel that the

estimate T > 2R1, where R1 := max√d(x) : x ∈ Ω.

In the linear case, for F ≡ 0, we have the following result on the exact controllability

of (2.1):

Theorem 5 Suppose that F ≡ 0 and the constants µi > 0 (i = 1, 2) are suciently

large. Then, for any (y0, y1) ∈ H10 × L2(Ω), there exist a control f ∈ L2(O × (0, T )) and

an associated Nash equilibrium pair (v1, v2) = (v1(f), v2(f)) such that the corresponding

solution to (2.1) satises (2.7).

61

In the linear case, the functionals Ji are convex whence, as we had already observed,

a pair (v1, v2) is a Nash equilibrium if and only if satises (2.4). However, in the semilinear

case, with F being a locally Lipschitz continuous function, we do not have the convexity

of the Ji. This motivates the following weaker denition:

Denition 3 Let f be given. The pair (v1, v2) is a Nash quasi-equilibrium for the func-

tionals Ji associated to f if (2.4) is satised.

The following results hold in the semilinear case:

Theorem 6 Assume that (2.8) holds, F ∈ W 1,∞(R) and the µi > 0 (i = 1, 2) are

suciently large. Then, for any (y0, y1) ∈ H10 (Ω) × L2(Ω), there exist a control f ∈

L2(O × (0, T )) and an associated Nash quasi-equilibrium pair (v1, v2) = (v1(f), v2(f))

such that the corresponding solutions to (2.1) satises (2.7).

In the semilinear case, we can prove that, under certain conditions, the denitions

of Nash equilibrium and Nash quasi-equilibrium are equivalent. More precisely, we have

the following result:

Proposition 7 Assume that

F ∈ W 2,∞(R) and yi,d ∈ C0([0, T ];H10 (Oi,d)) ∩ C1([0, T ];L2(Oi,d)).

Suppose that (y0, y1) ∈ H10 (Ω)× L2(Ω) and n ≤ 8. Then, there exists C > 0 such that, if

f ∈ L2(O × (0, T )) and the µi satisfy

µi ≥ C(1 + ‖f‖L2(O×(0,T ))

),

the conditions (2.3) and (2.4) are equivalent.

The contents of this article is organized as follows. In Section 2.2, we analyze

the linear system, we prove the existence and uniqueness of a Nash equilibrium pair for

each leader, we obtain an associated optimality system and we prove Theorem 5 using

a standard technique based on a observability estimate, which we get using Carleman

inequalities. In Section 2.3, we analyze the semilinear case, we deduce another optimality

system, we prove the Theorem 6 using xed point techniques and we prove Proposition

7 by adapting some arguments from [3] and [6]. Finally, in Section 2.4, we present some

additional comments.

62

2.2 The linear case

In this section we consider the linear case, (F = 0 in (2.1)). The goal is to prove of

the Theorem 5.

Since the system is linear, we have that the exact controllability is equivalent to null

controllability. In fact, we can introduce the change of variable z = y − y, where y is the

solution to system (2.1); then, we see that z is the solution toztt −∆z + a(x, t)z = f1O + v11O1 + v21O2 in Q,

z = 0 on Σ,

z(·, 0) = z0, zt(·, 0) = z1 in Ω,

(2.9)

where z0 = y0 − y0 ∈ H10 (Ω) and z1 = y1 − y1 ∈ L2(Ω). We will apply the Stackelberg-

Nash strategy to solve the null controllability to system (2.9), that is, we will look for a

leader f and an associated Nash equilibrium (v1(f), v2(f)) such that the solution to (2.9)

satises

(z(·, T ), zt(·, T )) = (0, 0). (2.10)

With this change of variables, the cost functionals become

Ji(f, v1, v2) :=

αi2

∫∫Oi,d×(0,T )

|z − zi,d|2dxdt+µi2

∫∫Oi×(0,T )

|vi|2 dx dt, (2.11)

where zi,d = yi,d − y. We will solve the null controllability problem for (2.9) with f and

(v1(f), v2(f)) in the following three subsections.

2.2.1 The existence and uniqueness of a Nash equilibrium

Let us prove that for each f , a unique Nash pair (v1(f), v2(f)) exists. The proof

is implied by Lax-Milgram's Theorem, following some arguments introduced by [?]. Let

f ∈ L2(O × (0, T )) be xed. Let us dene the spaces

Hi = L2(Oi × (0, T )) and H = H1 ×H2.

The pair (v1, v2) is a Nash equilibrium for the functionals Ji if and only if

αi

∫∫Oi,d×(0,T )

(z − zi,d)wi dx dt+ µi

∫∫Oi×(0,T )

vivi dx dt = 0 ∀vi ∈ Hi. (2.12)

63

where wi is the solution to the systemwitt −∆wi + a(x, t)wi = vi1Oi

in Q,

wi = 0 on Σ,

wi(·, 0) = 0, wit(·, 0) = 0 in Ω.

(2.13)

Now, we introduce the operators Li : Hi 7→ L2(Q), with Li(vi) = wi, where wi is the

solution to (2.13). Let u be the solution of toutt −∆u+ a(x, t)u = f1O in Q,

u = 0 on Σ,

u(·, 0) = z0, ut(·, 0) = z1 in Ω.

(2.14)

Then, we can write the solution to system (2.9) in the form z = L1(v1) + L2(v2) + u.

Consequently, (v1, v2) is a Nash equilibrium for Ji if and only if

αi

∫∫Oi,d×(0,T )

(L1v1 + L2v

2 − (zi,d − u))Livi dx dt+ µi

∫∫Oi×(0,T )

vivi dx dt = 0 ∀vi ∈ Hi,

that is to say∫∫Oi×(0,T )

[αiL∗i

((L1v

1 + L2v2 − (zi,d − u))1Oi,d

)+ µiv

i]vi dx dt = 0 ∀vi ∈ Hi,

where L∗i ∈ L(L2(Q),Hi) is the adjoint of Li. Obviously, this is equivalent to

αiL∗i ((L1v

1 + L2v2)1Oi,d

) + µivi = αiL

∗i ((zi,d − u)1Oi,d

), i = 1, 2. (2.15)

Let us introduce the operator L : H 7→ H, with

L(v1, v2) := (α1L∗1((L1v

1 + L2v2)1O1,d

) + µ1v1, α2L

∗2((L1v

1 + L2v2)1O2,d

) + µ2v2).

We have that L ∈ L(H). Now, let us assume that the µi > 0 satisfy

δi = µi −(

3α3−i + αi2

)||L3−i||2 > 0, i = 1, 2.

Then

(L(v1, v2), (v1, v2))H ≥ δ||((v1, v2))||2H ∀(v1, v2) ∈ H, (2.16)

where δ = minδ1, δ2.

Hence, from Lax-Milgram's Theorem, for each Φ ∈ H′ there exists exactly one pair

(v1, v2) satisfying

(L(v1, v2), (v1, v2))H = 〈Φ, (v1, v2)〉H′×H ∀(v1, v2) ∈ H. (2.17)

64

In particular, if we take Φ ∈ H′ with

〈Φ, (v1, v2)〉H′×H := ((α1L∗1((z1,d − u)1O1,d

), α2L∗2((z2,d − u)1O2,d

)), (v1, v2))H,

we deduce the existence and uniqueness of a solution to (??).

Let us summarize what we have been able to prove:

Proposition 8 Suppose that the µi > 0 satisfy

µi −(

3α3−i + αi2

)||L3−i||2 > 0, i = 1, 2.

Then, for each f ∈ L2(O × (0, T )), there exists a unique Nash equilibrium pair

(v1(f), v2(f)) for the Ji.

Remark 3 As a consequence of the proof of Proposition 8 we obtain a constant C > 0,

such that

||z||C0([0,T ];H1(Ω)) + ||zt||C0([0,T ];L2(Ω)) ≤ C(1 + ||f ||L2(O×(0,T ))),

where z is the solution of (2.9) corresponding to f , v1 = v1(f) and v2 = v2(f).

2.2.2 The Optimality system

In this section we obtain an optimality system, which characterizes the Nash equili-

brium (v1(f), v2(f)). Multiplying in (2.13) by a function φi and integrating by parts, we

have ∫∫Q

(φitt −∆φi + a(x, t)φi)wi dx dt+

∫Ω

wit(x, T )φi(x, T ) dx

−∫

Ω

wi(x, T )φit(x, T ) dx−∫∫

Σ

∂wi

∂νφi dσ dt =

∫∫Oi×(0,T )

viφi dx dt,(2.18)

This leads us to dene the adjoint systemsφitt −∆φi + a(x, t)φi = αi(z − zi,d)1Oi,d

in Q,

φi = 0 on Σ,

φi(·, T ) = 0, φit(·, T ) = 0 in Ω.

(2.19)

From (2.18) and (2.19), we have that∫∫Oi,d×(0,T )

αi(z − zi,d)wi dx dt =

∫∫Oi×(0,T )

viφi dx dt,

and, replacing in (2.12), we see that∫∫Oi×(0,T )

(φi + µivi)vi dx dt = 0 ∀vi ∈ Hi.

65

As a conclusion, we get the following optimality system for (v1(f), v2(f)):

ztt −∆z + a(x, t)z = f1O −2∑i=1

1

µiφi1Oi

in Q,

φitt −∆φi + a(x, t)φi = αi(z − zi,d)1Oi,din Q,

z = 0, φi = 0 on Σ,

z(·, 0) = z0, zt(·, 0) = z1, φi(·, T ) = 0, φit(·, T ) = 0 in Ω,

(2.20)

vi(f) = − 1

µiφi∣∣∣∣Oi×(0,T )

. (2.21)

The system (2.20)-(2.21) characterizes the Nash equilibrium (v1(f), v2(f)) for each

f ∈ L2(O × (0, T )). The next objective is to nd a leader f ∈ L2(O × (0, T )) such that

the solution to system (2.20) satises (2.10).

2.2.3 Null controllability

We will prove in this section the null controllability of the system (2.20). The proof

is based on an observability inequality for the adjoint to (2.20), which is the following:

ψtt −∆ψ + a(x, t)ψ =2∑i=1

αiγi1Oi,d

in Q,

γitt −∆γi + a(x, t)γi = − 1

µiψ1Oi

in Q,

ψ = 0, γi = 0 on Σ,

z(·, T ) = ψT0 , ψt(·, T ) = ψT1 , γi(·, 0) = 0, γit(·, 0) = 0 in Ω.

(2.22)

Note that from the equations in (2.20) and (2.22), one has

2∑i=1

αi

∫∫Oi,d

γiz dx dt+

∫Ω

zt(x, T )ψT0 (x) dx− 〈ψ(0), z1〉 − 〈ψT1 , z(T )〉

+

∫Ω

z0ψt(x, 0) dx =

∫∫O×(0,T )

fψ dx dt+2∑i=1

αi

∫∫Oi,d×(0,T )

(z − zi,d)γi dx dt,

where we have denoted by 〈·, ·〉 the duality pairing H−1(Ω)×H10 (Ω). Then∫∫

O×(0,T )

fψ dx dt =

∫Ω

zt(x, T )ψT0 (x) dx− 〈ψT1 , z(T )〉+

∫Ω

z1(x)ψ(x, 0) dx

−〈ψt(0), z0〉+2∑i=1

αi

∫∫Oi,d×(0,T )

zi,dγi dx dt,

(2.23)

66

and consequently, the null controllability of (2.20) holds if and only if, for each (z0, z1)

given in H10 (Ω)× L2(Ω), there exists f ∈ L2(O × (0, T )) such that

∫∫O×(0,T )

fψ dx dt =

∫Ω

z1(x)ψ(x, 0) dx− 〈ψt(0), z0〉+2∑1

αi

∫∫Oi,d×(0,T )

zi,dγi dx dt

∀(ψT0 , ψT1 ) ∈ L2(Ω)×H−1(Ω).

Given (z0, z1) ∈ H10 (Ω)×L2(Ω) and ε > 0, we introduce Fε : L2(Ω)×H−1(Ω) 7→ R,

as follows

Fε(ψT0 , ψ

T1 ) :=

∫∫O×(0,T )

|ψ|2 dx dt+ ε||(ψT0 , ψT1 )||L2(Ω)×H−1(Ω)

+ 〈〈(ψ(0), ψt(0)), (z0, z1)〉〉+2∑i=1

αi

∫∫Oi,d×(0,T )

zi,dγi dx dt,

(2.24)

where we have set by denition

〈〈(ψ(0), ψt(0)), (z0, z1)〉〉 :=

∫Ω

z1(x)ψ(x, 0) dx− 〈ψt(0), z0〉.

Note that Fε is continuous, strictly convex and coercive in L2(Ω)×H−1(Ω).

In the sequel, we associate to each ψ, γ1, γ2 the "energy"Eψ, with

Eψ(t) :=1

2

[||ψ(·, t)||2L2(Ω) + ||ψt(·, t)||2H−1(Ω)

].

The following result furnishes the desired observability inequality:

Proposition 9 Suppose that (2.8) holds. Then there exists C > 0 such that, for all

(ψT0 , ψT1 ) ∈ L2(Ω)×H−1(Ω), the solution (ψ, γ1, γ2) to (2.22) satises

Eψ(0) +2∑i=1

∫∫Q

eR20λ|γi|2 dx dt ≤ C

∫∫O×(0,T )

ψ2 dx dt, (2.25)

where R0 := min√d(x) : x ∈ Ω.

Let us assume for the moment that Proposition 9 holds. Since that Fε is continuous,

strictly convex and coercive, it has a unique minimum (ψTε,0, ψTε,1) ∈ L2(Ω) × H−1(Ω).

Moreover, for any h > 0 and (ψT , ψT1 ) ∈ L2(Ω)×H−1(Ω), one has the following:

• either (ψTε,0, ψTε,1) = 0, or

• for all (ψT0 , ψT1 ) ∈ L2(Ω)×H−1(Ω),∫∫

O×(0,T )

ψεψ dx dt+ε

||(ψTε,0, ψTε,1)||L2×H−1

((ψTε,0, ψTε,1), (ψT0 , ψ

T1 ))L2×H−1

+⟨⟨

(ψ(0), ψt(0)), (z0, z1)⟩⟩

+2∑i=1

αi

∫∫Oi,d×(0,T )

γizi,d dx dt = 0(2.26)

67

In both cases, taking

fε = ψε

∣∣∣O×(0,T )

, (2.27)

and denoting by (zε, φ1ε, φ

1ε) the corresponding solution to (2.20), we have

∣∣∣⟨⟨(zε(·, T ), zε,t(·, T )), (ψT , ψT1 )⟩⟩∣∣∣ ≤ ε

∣∣∣∣(ψT0 , ψT1 )∣∣∣∣L2(Ω)×H−1(Ω)

∀(ψT0 , ψT1 ) ∈ L2(Ω)×H−1(Ω).(2.28)

Therefore, ∣∣∣∣(zε(·, T ), zε,t(·, T ))∣∣∣∣H1

0×L2 ≤ ε. (2.29)

Furthermore, from (2.25) and (2.27), we get:

||fε||L2(O×(0,T )) ≤ C

||(z0, z1)||2H1

0×L2 +2∑i=1

α2i

∫∫Oi,d×(0,T )

e−2λϕ|zi,d|2 dx dt

12

,

whence (fε) is uniformly bounded in L2(O × (0, T )) and possesses a subsequence weakly

convergent to some function f ∈ L2(O × (0, T )). It is then clear that

(zε(·, T ), zε,t(·, T ))→ ((z(·, T ), zt(·, T ))) weakly in H10 (Ω)× L2(Ω),

where (z, φ1, φ2) is the solution to (2.20) associated to f . In view of (2.29), we conclude

that z satises (2.10) and the null controllability of (2.20) is satised.

Proof of the Proposition 9. The proof relies on the arguments of [19] and [33]. We

introduce the function ϕ : Q 7→ R, with

ϕ(x, t) := d(x)− c(t− T/2)2, (2.30)

where c ∈ (0, 1). Let us suppose that (2.8) holds and let us set ω := Oδ(Γ+) ∩ Ω ⊂ O.

Then, there exists λ0 ≥ 1 such that, for any λ ≥ λ0 and any function u ∈ C0([0, T ];L2(Ω))

satisfying

u(·, 0) = u(·, T ) = 0, (utt −∆u) ∈ H−1(Q); (2.31)

(u, ηtt −∆η)L2(Q) = 〈utt −∆u, η〉H−1(Q),H10 (Q) ∀η ∈ H1

0 (Q), (2.32)

we have the Carleman inequality

λ

∫∫Q

e2λϕ|u|2 dx dt ≤ C

(∣∣eλϕ(utt −∆u)∣∣2H−1(Q)

+ λ2

∫∫ω×(0,T )

e2λϕ|u|2 dx dt), (2.33)

where C > 0 is independent of λ and u. The proof of (2.33) can to be found at [19] or

[33].

68

Recall that we have assumed that T > 2R1. Then we can choose c in (2.30) such

that (2R1

T

)2

< c <2R1

T. (2.34)

Now, we can also choose T0, T1 ∈ (0, T ) such that

ϕ(x, t) ≤ R21

2− cT

2

8< 0 ∀(x, t) ∈ Ω× ((0, T1) ∪ (T ′1, T )) (2.35)

and

ϕ(x, t) ≥ R20

2, ∀(x, t) ∈ Ω× (T0, T

′0). (2.36)

where we have set T ′i := T − Ti for i = 0, 1.

Let us consider a function ξ ∈ C∞0 (0, T ) so that ξ(t) ≡ 1 in (T1, T′1). Then

(ψ, γ1, γ1) = (ξψ, ξγ1, ξγ2) is the unique solution of the system

ψtt −∆ψ + a(x, t)ψ =2∑i=1

αiξγi1Oi,d

+ 2ξtψt + ξttψ in Q,

γitt −∆γi + a(x, t)γi = − 1

µiξψ1Oi

+ 2ξtγit + ξttγ

i in Q,

ψ = 0, γi = 0 on Σ,

ψ(·, T ) = ψ(·, 0) = ψt(·, T ) = γi(·, 0) = γi(·, T ) = γit(·, 0) = 0 in Ω.

(2.37)

Since ψ ∈ C0([0, T ];L2(Ω)) satises (2.31) and (2.32), for any λ ≥ λ0 we have

λ

∫∫Q

e2λϕ|ξψ|2 dx dt ≤ C

∣∣∣∣∣eλϕ(

2∑i=1

αiξγi1Oi,d

+ 2ξtψt + ξttψ

)∣∣∣∣∣2

H−1(Q)

+λ2

∫∫ω×(0,T )

e2λϕ|ψ|2 dx dt].

(2.38)

Then, using the inequality (2.38) and the equations in (2.37), arguing as in [33], we get:

λ

∫∫Q

e2λϕ|ψ|2 dx dt ≤ C

[λ2

∫∫ω×(0,T )

e2λϕ|ψ|2 dx dt+N

+λ2∑i=1

∫∫Oi,d×(0,T )

e2λϕ|γi|2 dx dt

],

(2.39)

where C > 0 is independent of λ e

N := λ2e(R21−cT 2/4)λ

(||ψ||2L2(Ω×(0,T1)) + ||ψ||2L2(Ω×(T ′1,T ))

). (2.40)

69

From (2.39), the following is found:

λ

∫∫Q

e2λϕ|ψ|2 dx dt+ λ

2∑i=1

∫∫Oi,d×(0,T )

eR20λ|γi|2 dx dt

≤ C

[λ2

∫∫ω×(0,T )

e2λϕ|ψ|2 dx dt + N

+2∑i=1

∫∫Oi,d×(0,T )

(λeR20λ + e2λϕ)|γi|2 dx dt

] (2.41)

From (2.35), we deduce that there is a λ1 ≥ λ0 such that λ2e(R21−cT/4)λ < 1 for all λ ≥ λ1.

Consequently, by (2.40) and (2.41),

λ

∫∫Q

e2λϕ|ψ|2 dx dt+2∑i=1

λ

∫∫Oi,d×(0,T )

eR20λ|γi|2 dx dt

≤ C

[λ2

∫∫ω×(0,T )

e2λϕ|ψ|2 dx dt+ ||ψ||2L2(Ω×(0,T1))

+ ||ψ||2L2(Ω×(T ′1,T )) +2∑i=1

∫∫Oi,d×(0,T )

(λeR20λ + e2λϕ)|γi|2 dx dt

],

(2.42)

where C > 0 is independent of λ.

Taking into account (2.36) we have:∫∫Q

e2λϕ|ψ|2 dx dt ≥ eR20λ

∫ T ′0

T0

∫Ω

|ψ|2 dx dt. (2.43)

On the other hand, from the usual energy method (see for example [33], lemmas 3.3 and

3.4), it follows that, for any S0 ∈ (T0, T/2) and S ′0 ∈ (T/2, T ′0), we also have∫ S′0

S0

Eψ(t) dt ≤ C

∫ T ′0

T0

∫Ω

|ψ|2 dx dt+2∑i=1

∫ T ′0

T0

∫Oi,d

|γi|2 dx dt, (2.44)

||ψ||2L2(Ω×(0,T1)) + ||ψ||2L2(Ω×(T ′1,T )) ≤ CEψ(0) + C2∑i=1

∫∫Oi,d×(0,T )

|γi|2 dx dt (2.45)

and also

CEψ(0) ≤∫ S′0

S0

Eψ(t) dt+2∑i=1

∫ S′0

0

∫Oi,d

|γi|2 dx dt. (2.46)

Then, combining (2.42)(2.46) we arrive at the inequality

C1 λEψ(0) eR20λ+C1 +

2∑i=1

∫∫Q

eR20λ|γi|2 dx dt ≤ Cλ2

∫∫ω×(0,T )

e2λϕ|ψ|2 dx dt

+C2 eC2Eψ(0) +

C

minµ1, µ2e2λR1Eψ(0),

70

where C, C1 and C2 are independent of λ. Thus, choosing λ and µi such that

C1λ > C C2, λR20 + C1 > C2,

Ce2λR1

minµ1, µ2< 1,

we get (2.25).

2.3 The semilinear case

In this section we consider the semilinear case, with F ∈ W 1,∞(R) in (2.1). The

main is to prove the Theorem 6. We will follow arguments similar to those in [3] and [6].

First, let us deduce the optimality system, which in this case characterizes the Nash

quasi-equilibrium pair. Thus, given f ∈ L2(O × (0, T )), let (v1(f), v2(f)) ∈ H be a Nash

quasi-equilibrium corresponding to the functionals Ji. Then, we have by denition

J ′i(f, v1, v2) · vi = 0 ∀vi ∈ Hi.

Note that this is equivalent to

αi

∫∫Oi,d×(0,T )

(y − yi,d)yi dx dt+ µi

∫∫Oi×(0,T )

vivi dx dt = 0 ∀vi ∈ Hi, (2.47)

where yi is the solution to the systemyitt −∆yi + a(x, t)yi = F ′(y)yi + vi1Oi

in Q,

yi = 0, on Σ,

yi(·, 0), yit(·, 0) = 0, in Ω.

(2.48)

Multiplying in (2.48) by a function φi, integrating in Q with respect to x and t and using

integration by parts, we see that∫∫Q

(φitt −∆φi + a(x, t)φi)yi dx dt−∫∫

Σ

φi∂yi

∂νdσ dt+

⟨yit(T ), φi(T )

⟩−(yi(T ), φi(T )

)=

∫∫Q

F ′ (y) yiφi dx dt+

∫∫Oi×(0,T )

viφi dx dt.(2.49)

Then, we dene the following adjoint systems to (2.48)φitt −∆φi + a(x, t)φi = F ′(y)φi + αi(y − yi,d)1Oi,d

in Q,

φi = 0, on Σ,

φi(·, T ), φit(·, T ) = 0, in Ω.

(2.50)

71

From (2.48) - (2.50), we have

αi

∫∫Oi,d×(0,T )

(y − yi,d)yi dx dt =

∫∫Oi×(0,T )

viφi dx dt ∀vi ∈ Hi. (2.51)

Hence, replacing (2.51) in (2.47), it follows that∫∫Oi×(0,T )

(φi + µivi)vi dx dt = 0 ∀vi ∈ Hi.

This leads to the optimality system

ytt −∆y + a(x, t)y = F (y) + f1O −2∑i=1

1

µiφi1Oi

in Q,

φitt −∆φi + a(x, t)φi = F ′(y)φi + αi(y − yi,d)1Oi,din Q,

y = 0, φi = 0 on Σ,

y(·, 0) = y0, yt(·, 0) = y1, φi(·, T ) = 0, φit(·, T ) = 0 in Ω,

(2.52)

vi = − 1

µiφi∣∣∣∣Oi×(0,T )

. (2.53)

Now, given (y0, y1) ∈ H10 (Ω)× L2(Ω), let y be a free trajectory of the system (2.1),

that is, the solution to (2.5). Let us consider the change of variables z = y − y. From

(2.52) and (2.5), we see that (z, φ1, φ2) is the solution to

ztt −∆z + a(x, t)z = G(x, t; z)z + f1O −2∑i=1

1

µiφi1Oi

in Q,

φitt −∆φi + a(x, t)φi = F ′(y + z)φi + αi(z − zi,d)1Oi,din Q,

z = 0, φi = 0 on Σ,

z(·, 0) = z0, zt(·, 0) = z1, φi(·, T ) = 0, φit(·, T ) = 0 in Ω,

(2.54)

where, z0 = y0 − y0, z1 = y1 − y1, zi,d = yi,d − y and

G(x, t; z) =

∫ 1

0

F ′(y + τz) dτ.

With this change of variable, it becomes clear that the exact controllability property for

(2.52) is equivalent to the null controllability property for (2.54). We will prove the latter

by a xed point method.

For each z ∈ L2(Q) and each f ∈ L2(O × (0, T )), let us consider the linear system

wtt −∆w + a(x, t)w = G(x, t; z)w + f1O −2∑i=1

1

µiφi1Oi

in Q,

φitt −∆φi + a(x, t)φi = F ′(y + z)φi + αi(w − zi,d)1Oi,din Q,

w = 0, φi = 0 on Σ,

w(·, 0) = z0, wt(·, 0) = z1, φi(·, T ) = 0, φit(·, T ) = 0 in Ω.

(2.55)

72

Since F ∈ W 1,∞(R), there exists M > 0 such that

|G(x, t; s)|+ |F ′(s)| ≤M ∀(x, t; s) ∈ Q× R. (2.56)

Furthermore, there exists C > 0 such that

||w||L∞(0,T ;H10 (Ω)) + ||wt||L∞(0,T ;L2(Ω)) ≤ (1 + ||f ||L2(O×(0,T ))). (2.57)

For any xed z ∈ L2(Q), let us denote by (wz, φ1z, φ

2z) the solution to (2.55) associated to

z. Then, by multiplying in (2.55)1 and (2.55)2 by functions ψz and γiz (respectively) and

integrating by parts in Q, we get:∫∫Q

(ψz,tt −∆ψz + a(x, t)ψz)wz dx dt+

∫Ω

wz,t(x, T )ψz(x, T ) dx

−∫

Ω

z1(x)ψz(x, 0) dx− 〈wz(T ), ψz,t(T )〉+⟨z0, ψz,t(0)

⟩=

∫∫Σ

ψz∂wz∂ν

dσ dt+

∫∫Q

G (x, t; z)wzψz dx dt

−2∑i=1

1

µi

∫∫Oi×(0,T )

φizψz dx dt+

∫∫O×(0,T )

fψz dx dt

(2.58)

and ∫∫Q

(ψz,tt −∆ψz + a(x, t)ψz)wz dx dt+ 〈φiz(0), γiz,t(0)〉

−∫

Ω

φiz(x, 0)γiz(x, 0) dx−∫∫

Σ

γiz∂φiz∂ν

dσ dt

=

∫∫Q

F ′(y + z)φizγiz dx dt+ αi

∫∫Oi,d×(0,T )

(wz − zi,d)γiz dx dt.

(2.59)

In view of (2.58) and (2.59), we introduce the following adjoint system

ψz,tt −∆ψz + a(x, t)ψz = G(x, t; z)ψz +2∑i=1

αiγiz1Oi,d

in Q,

γiz,tt −∆γiz + a(x, t)γiz = F ′(y + z)γiz − 1µiψz1Oi

in Q,

ψz = 0, γiz = 0 on Σ,

ψz(·, T ) = ψT0 , ψz,t(·, T ) = ψT1 , γiz(·, 0) = 0, γiz,t(·, 0) = 0 in Ω.

(2.60)

Then,

− 1

µi

∫∫Oi×(0,T )

φizψz dx dt = αi

∫∫Oi,d×(0,T )

(wz − zi,d)γiz dx dt,

whence

−2∑i=1

1

µi

∫∫Oi×(0,T )

φizψz dx dt =2∑i=1

αi

∫∫Oi,d×(0,T )

(wz − zi,d)γiz dx dt (2.61)

73

and, by (2.58) and (2.60), we see that∫Ω

wz,t(x, T )ψT0 (x) dx−∫

Ω

z1(x)ψz(x, 0) dx−⟨wz(T ), ψT1

⟩+⟨z0, ψz,t(0)

⟩+

2∑i=1

αi

∫∫Oi,d×(0,T )

wzγiz dx dt

= −2∑i=1

1

µi

∫∫Oi×(0,T )

φizψz dx dt+

∫∫O×(0,T )

fψz dx dt.

(2.62)

Finally, replacing (2.61) in (2.62), the following identity is found:∫∫O×(0,T )

fψz dx dt =

∫Ω

wz,t(x, T )ψT0 (x) dx−∫

Ω

z1(x)ψz(x, 0) dx

−⟨wz(T ), ψT1

⟩+⟨z0, ψz,t(0)

⟩+

2∑i=1

αi

∫∫Oi,d×(0,T )

zi,dγiz dx dt.

As a consequence, we have the null controllability of (2.55) if and only if, for each (z0, z1) ∈

H10 (Ω)× L2(Ω), there exists f ∈ L2(O × (0, T )) such that

∫∫O×(0,T )

fψz dx dt =⟨z0, ψz,t(0)

⟩+

2∑i=1

αi

∫∫Oi,d×(0,T )

zi,dγiz dx dt

−∫

Ω

z1(x)ψz(x, 0) dx ∀(ψT0 , ψT1 ) ∈ L2(Ω)×H−1(Ω).

As in the previous section, let us dene the functional Fz,ε, with

Fz,ε(ψT0 , ψ

T1 ) :=

∫∫O×(0,T )

|ψz|2 dx dt+ ε||(ψT , ψT1 )||

+⟨(ψz(0), ψz,t(0)), (z0, z1)

⟩+

2∑i=1

αi

∫∫Oi,d×(0,T )

zi,dγiz dx dt.

As in the linear case, we get the following observability inequality

Eψz(0) +2∑i=1

∫∫Q

eR20λ|γiz|2dxdt ≤ C

∫∫O×(0,T )

|ψz|2dxdt, (2.63)

where C > 0 is a constant independent of z and Eψz denotes the L2 × H−1 energy

associated to ψz. Arguing as in the linear case, the inequality (2.63) allows us to prove

the existence of minimizer (ψT0,ε, ψT1,ε) of Fz,ε in L2(Ω) × H−1(Ω). Thus, taking limits as

ε → 0, we get a unique control fz ∈ L2(O × (0, T )) such that the associated solution

(wz, φ1z, φ

2z) to system (2.55), satisfy

wz(·, T ) = 0 in Ω, (2.64)

74

and

||fz||L2(O×(0,T )) ≤ C, (2.65)

where C is independent of z. The remainder of the proof is completely standard and

well known. It sues to introduce the mapping Λ : L2(Q) → L2(Q), with Λ(z) := wz,

where wz is, together with φiz, the solution to (2.55) satisfying (2.64) that we have just

constructed. From the properties of fz and the usual energy estimates, it is clear that Λ

satises the assumptions of Schauder's Theorem and, consequently, possesses at least one

xed point. This ends the proof of Theorem 6.

2.3.1 Equilibrium and Quasi-equilibrium

In this section we will prove Proposition 7 that says that under some conditions,

the denitions of Nash equilibrium and Nash quasi-equilibbrium are equivalent. Similar

questions are analyzed in [3] and [6] for parabolic systems.

Suppose that F ∈ W 2,∞(R). Let f ∈ L2(O × (0, T )) be given and let (v1, v2) ∈ H

be the Nash quasi-equilibrium associated to f . For all v1 ∈ H1 and s ∈ R, let ys be the

solution toystt −∆ys + a(x, t)ys = F (ys) + f1O + (v1 + sv1)1O1 + v21O2 in Q,

ys = 0 on Σ,

ys(·, 0) = y0, yst (·, 0) = y1 in Ω.

(2.66)

Let us consider the notation ys|s=0 = y. Now, for all v1 ∈ H1, we have:

D1J1(f ; v1 + sv1, v2) · v1 −D1J1(f ; v1, v2) · v1

= α1

∫∫O1,d×(0,T )

(ys − y1,d)qs dx dt− α1

∫∫O1,d×(0,T )

(y − y1,d)q dx dt

+ sµ1

∫∫O1×(0,T )

v1v1 dx dt,

(2.67)

where qs is the solution toqstt −∆qs + a(x, t)qs = F (ys)qs + v11O1 in Q,

qs = 0 on Σ,

qs(·, 0) = 0, qst (·, 0) = 0 in Ω,

(2.68)

and we denote used the notation q := qs|s=0. Multiplying in (2.68) by a function φs and

75

integrating by parts in Q, we have∫∫Q

(φstt −∆φs + a(x, t)φs)qs dx dt−∫∫

Σ

φs∂qs

∂νdσ dt

+ 〈qst (T ), φs(T )〉 −∫

Ω

qs(x, T )φs(x, T ) dx

=

∫∫Q

F ′(ys)qsφs dx dt+

∫∫Oi×(0,T )

v1φs dx dt.

(2.69)

Let us introduce the adjointφstt −∆φs + a(x, t)φs = F (ys)φs + α1(ys − y1,d)1O1,d

in Q,

φs = 0 on Σ,

φs(·, T ) = 0, φst(·, T ) = 0 in Ω.

(2.70)

By (2.69) and (2.70), we have

α1

∫∫Oi,d×(0,T )

(ys − y1,d)qs dx dt =

∫∫Oi×(0,T )

v1φs dx dt.

Replacing in (2.67), we get

D1J1(f ; v1 + sv1, v2) · v1 −D1J1(f ; v1, v2) =

∫∫O1×(0,T )

v1(φs − φ) dx dt

+ sµ1

∫∫O1×(0,T )

v1v1 dx dt ∀v1, v1 ∈ H1.(2.71)

Note that

(φs−φ)tt−∆(φs−φ)+ a(x, t)(φs−φ) = [F ′(ys)−F (y)]φs+F ′(y)(φs−φ)+α1(ys−y)1O1,d,

and the following limits exist

h = lims→0

1

s(ys − y), η = lim

s→0

1

s(φs − φ),

moreover, (h, η) is the solution to the system

htt −∆h+ a(x, t)h = F ′(y)h+ v11O1 in Q,

ηtt −∆η + a(x, t)η = F ′′(y)hφ+ F ′(y)η + α1h1O1,din Q,

h = 0, η = 0 on Σ,

h(·, 0) = 0, ht(·, 0) = 0, η(·, T ) = 0, ηt(·, T ) = 0 in Ω.

(2.72)

Thus, by (2.71) and (2.72), we deduce thatD2

1J1(f ; v1, v2) · (v1, v1) =

∫∫O1×(0,T )

v1η dx dt+ µ1

∫∫O1×(0,T )

v1v1 dx dt

∀v1, v1 ∈ H1.

76

In particular,D2

1J1(f ; v1, v2) · (v1, v1) =

∫∫O1×(0,T )

v1η dx dt+ µ1

∫∫O1×(0,T )

|v1|2 dx dt

∀v1 ∈ H1.

(2.73)

Let us see that for some C1 > 0 independent of f, η, h, v1, one has∣∣∣∣∫∫O1×(0,T )

v1η dx dt

∣∣∣∣ ≤ C1

(1 + ||f ||L2(O×(0,T ))

)||v1||H1 ∀v1 ∈ H1. (2.74)

Indeed, since F ′ ∈ W 1,∞(R), from the energy estimates, we rst see that∫Ω

|ht|2 dx+

∫Ω

|∇h|2 dx ≤ C||v1||2H1, t ∈ [0, T ] a. e. (2.75)

Also, ∫∫Q

v1η dx dt =

∫∫Q

F ′′(y)φh2 dx dt+ α1

∫∫Oi,d×(0,T )

h2 dx dt. (2.76)

Let us nd r and s such that

φ ∈ Lr(0, T ;Ls(Ω)) and h ∈ Lr′(0, T ;Ls′(Ω)),

where r′ and s′ are conjugate of r and s, respectively. In fact, as h ∈ C0([0, T ];H10 (Ω))

and ht ∈ C0([0, T ];L2(Ω)), we have in particular, by Sobolev embedding

h ∈ L∞(0, T ;L2nn−2 ),

that is, r′ = +∞ and s′ = n/(n− 2). Thus r = 1 and s = n/2.

Now, since

y, y1,d ∈ C0([0, T ];H10 (Ω)) ∩ C1([0, T ];L2(Ω)),

we have φ ∈ C1([0, T ];H10 (Ω)) ∩ C0([0, T ], H2(Ω) ∩H1

0 (Ω)),

φt ∈ C1([0, T ];L2(Ω)) ∩ C0([0, T ], H10 (Ω)).

As a consequence, φ ∈ L1(0, T ;La(Ω)), where a = 2n/(n − 4). Thus, we have that

φ ∈ L1(0, T ;Ls(Ω)) if, and only if,

n

2≤ 2n

n− 4,

that is, if and only if n ≤ 8.

77

Then, if n ≤ 8 by (2.66), (2.70), (2.75) and (2.76), we obtain∣∣∣∣∫∫O1×(0,T )

v1η dx dt

∣∣∣∣ ≤ C(||h||2O1,d×(0,T ) + ||F ′′||∞||h||2L∞(0,T ;L2s′ (Ω))

||φ||L1(0,T ;Ls(Ω))

)≤ C(1 + ||φ||L1(0,T ;Ls(Ω)))||v1||H1

≤ C(1 + ||y||L2(Q))||v1||H1

≤ C1(1 + ||f ||L2(O×(0,T )))||v1||H1 ,

which proves (2.74).

Thus, by (2.73) and (2.74), we have

D21J1(f ; v1, v2) · (v1, v1) ≥ [µ1 − C1(1 + ||f ||L2(O×(0,T )))]||v1||2H1

∀v1 ∈ H1.

Analogously, we get a constant C2 > 0, such that

D22J2(f ; v1, v2) · (v2, v2) ≥ [µ2 − C2(1 + ||f ||L2(O×(0,T )))]||v2||2H2

∀v2 ∈ H2.

Finally, noting that C1 and C2 are independent of µi, for µi large enough we have

that µi −Ci(1 + ||f ||L2(O×(0,T ))) is a positive constant. In other words, (v1, v2) minimizes

simultaneously the functionals Ji and (v1, v2) is a Nash equilibrium.

2.4 Final comments

In this section we indicate several problems related to hierarchic control and exact

controllability that can be analyzed.

2.4.1 Boundary controllability

In [50], the author analyzes the hierarchic control of a hyperbolic equation with one

leader and one follower and he solves the problem of approximate controllability. The

boundary exact controllability problem with at least two followers is still not completely

understood.

Let Ω ⊂ Rn be a bounded domain with boundary Γ of class C2 and let Γ0, Γ1, Γ2 ⊂ Γ

be nonempty subsets. Let us consider the linear systemytt −∆y + a(x, t)y = 0 in Q,

y = f1Γ0 + v11Γ1 + v21Γ2 on Σ,

y(·, 0) = y0, yt(·, 0) = y1 in Ω,

(2.77)

78

the cost functionals

Ji(f, v1, v2) :=

αi2

∫∫Γi,d×(0,T )

|y − yi,d|2 dx dt+µi2

∫∫Γi×(0,T )

|vi|2 dx dt, (2.78)

where the Γi,d ⊂ Γ are nonempty yi,d ∈ L2(Γi,d× (0, T )) are given and αi, µi > 0 and the

main functional

J(f) :=1

2

∫∫Γ0×(0,T )

|f |2 dx dt, (2.79)

Following similar arguments to those in Section 2.2, we can prove the existence and

uniqueness of a Nash equilibrium for each f . We can also deduce an optimality system

similar to (2.20). However, at present we do not know whether an approppriate observa-

bility inequality holds for the solutions of the corresponding adjoint. The hierarchic exact

controllability of (2.77)(2.78) is therefore an open question.

2.4.2 On the nonlinearity

In this work we have considered nonlinear terms locally Lischitz-continuous. We

could think of other assumptions on F . For example, in [33] the authors consider nonlinear

F ∈ C1(R) satisfying the following condition

lim sups→+∞

F (s)

|s| · log12 |s|

= 0.

Following the arguments in Section 2.3, we can obtain an optimality system similar to

(1.88). However, in this case, will show the function F ′, on which we have no information.

The analysis of hierarchic control in the context of the exact controllability for hyperbolic

equations with nonlinear terms of this type is an open question.

79

Capítulo 3

Hierarchic control for Burgers equation

via StackelbergNash strategy

Hierarchic control for Burgers equation

via StackelbergNash strategy

F. D. Araruna, E. Fernández-Cara, L. C. Silva

Abstract. We present some results on the null controllability of the Burgers equation. We

analyze the hierarchic control through StackelbergNash strategy, where we consider one

leader and two followers. To each leader we associate a Nash equilibrium corresponding

to a bi-objective optimal control problem; then we look for a leader that solves the null

controllability problem. We prove linear case and we use a xed point method to solve the

semilinear problems.

3.1 Introducton

The Burgers' equation is the simplest of evolution PDE's that describe a uid ux.

This equation was introduced in the paper [11] by Burgers. The study of this equation

is important to investgate proprieties of systems governed by more complex equations in

higher dimensions. Por example, the analysis of the controllability for Burgers' equation

has been used to try understand questions on the controllability of NavierStokes system.

3.1.1 Statement of the problem

Let T > 0 be and consider the nonempty open sets O, O1, O2 ⊂ (0, 1), with 0 6∈ O.

We introduce the Burgers' system:yt − yxx + yyx = f1O + v11O1 + v21O2 in (0, 1)× (0, T ),

y(0, t) = y(1, t) = 0 on (0, T ),

y(x, 0) = y0(x) in (0, 1),

(3.1)

where the controls are f, v1, v2, the state is y and with the notation 1A we denote the

characteristic function of the set A.

The controllability of systems as in (3.1) has been extensively analyzed in the last

years. In [34], it is shown that, in general, a stationary solution of system as (3.1)

with large L2-norm cannot be reached (not even approximately) at any time T. Similar

questions are analyzed in [15].

81

The local null conrollability for Burgers equation is investigate in [26], where was

obtained explicit sharp estimates of the minimal time of controllability T (r) of initial data

with L2-norm less or igual r. Now in [7] is obtained results on the local null controllability

of the Burgers-alpha model.

In a more recent work [53] the author, considering two controls forces, proves that

one can drive the solution of the corresponding Burgers equation from any initial state y0

in L2(0, 1) to null state in any time T > 0.

When we have a problem multi-objective, for example, beyond to obtain that solves

the controllability problem we want that the associate solution not go away too of a

given function. For to study problem of that type generaly we use the hierarchic control

introduced by Lions in [50], which cosists in the process of leader (independent control)

and follower (dependent on leader). In [50] was considered one leader and one follower

and applied the Stackelberg optimization, [61], to solve a bi-objective optimal approximate

controllability problem for a hyperbolic equation. In [51], similar results are obtained for

a parabolic system.

The hierarchic control using the StackelbergNash strategy was introduced in [16],

where was associated the strategies of cooperative optimization of Stackelberg and of non-

cooperative optimization of Nash. The authors has been considered this strategy to solve

a approximate controllability problem for a parabolic system with a leader and a number

N of followers. Another references on this strategy, but in the framework of the exact

controllability for trajectories of the parabolic system are [1], [3] and [6].

The objective this paper is analyze the null controllability of the system (3.1) in the

context of the hierarchic control applying the StackelbergNash strategy with leader f

and followers v1 and v2.

In sequel, to describe the methodology, we consider the cost functionals for the

followers:

Ji(f ; v1, v2) :=αi2

∫∫Oi,d×(0,T )

|y − yi,d|2 dx dt+µi2

∫∫Oi×(0,T )

|vi|2 dx dt, (3.2)

and main functional

J(f) :=1

2

∫∫O×(0,T )

|f |2 dx dt,

where Oi,d ⊂ (0, 1) is a open nonempty set, αi, µi > 0 are constants and yi,d = yi,d(x, t) is

a given function in L2(Oi,d × (0, T )).

82

For each leader control f choosing we look for an pair (v1, v2) that minimize, simul-

taneously, the functionals Ji, that is,

J1(f ; v1, v2) = minv1

J1(f ; v1, v2) and J2(f ; v1, v2) = minv2

J2(f ; v1, v2). (3.3)

An pair (v1, v2) = (v1(f), v2(f)) that satises (3.3) is called Nash equilibrium for the

functionals Ji.

Note that when Ji is convex, i = 1, 2, (3.3) is equivalent to

J ′i(f ; v1, v2) · vi = 0 ∀ vi ∈ L2(Oi × (0, T )), (3.4)

but, in general, as (3.1) is nonlinear the equivalence between (3.3) and (3.4) is not true.

Then, we must to prove that an pair (v1, v2) satisfying (3.4) is unique and we have⟨J ′′(f ; v1, v2), vi

⟩> 0 ∀ vi ∈ L2(Oi × (0, T )). (3.5)

Thus, (v1, v2) is a Nash equilibrium if and only if satises (3.4) and (3.5).

After proved the existence and uniqueness of Nash equilibrium for each leader f ∈

L2(O × (0, T )), we look for to prove the null controllability for the system (3.1). More

precisely, we prove that there exists f ∈ L2(O × (0, T )) such that

J(f) = minfJ(f) (3.6)

and

y(·, T ) = 0 in (0, 1). (3.7)

In (3.1), we can think in the state y(x, t) as a one dimensional velocity of a uid

and y0(x) as a initial datum. Then, roughly speaking, the problem that we will analyze

has the following interpretation: we want to act with minimal control in a set of paticles

of the uid of such a way that the velocity of the all uid, which is y0(x) initialy, vanish

at time T > 0, but we also want to act with minimal control, during the all time, for that

determined set of particles of uid the velocity y(x, t) does not go away too much from

the velocity yi,d.

3.1.2 Main results

In the rst reult we have conditons to guarantee the existence and uniqueness of the

Nash equilibrium.

83

Theorem 7 Suppose µi large enough. Then, for each f ∈ L2(O × (0, T )) given, there

exists a Nash Equilibrium pair (v1(f), v2(f)) for the functionals Ji.

The local null controllability with the initial data in H10 (0, 1) given by the result:

Theorem 8 Suppose y0 ∈ H10 (0, 1) with ||y0||H1

0 (0,1) ≤ r for some r > 0 and Oi,d∩O 6= ∅,i = 1, 2. Assume that one of the following conditions holds:

O1,d = O2,d := Od, (3.8)

or

O1,d ∩ O 6= O2,d ∩ O. (3.9)

If the constants µi > 0 (i = 1, 2) are large enough, and there exists a positive function

ρ = ρ(t), which decay exponentialy to 0 when t→ T−, such that, functions yi,d ∈ L2(Oi,d×(0, T )) (i = 1, 2) satises∫∫

Oi,d×(0,T )

ρ−2 y2i,d dx dt < +∞, i = 1, 2, (3.10)

then, there exist a control f ∈ L2(O×(0, T )) and a associated Nash equilibrium (v1(f), v2(f))

such that has one (3.6) and the corresponding solution to (3.1) satises (3.7).

Since that we want to drive the solution of (3.1) to null state in the time T > 0 without

that the state y go away of the functions yi,d, the hypothesis (3.10) is natural.

In view of the result obtained in [26] for y0 ∈ L2(0, 1) with ||y0||L2(0,1) ≤ r, r > 0,

we must have a minimal time for null controllability T (r) > 0 with

C0 log(1/r)−1 ≤ T (r) ≤ C1 log(1/r)−1,

for suitables constants C0, C1 > 0. Now we present a result on local null controllability

with initial data in L2(0, 1).

Theorem 9 Suppose y0 ∈ L2(0, 1) with ||y0||L2(0,1) ≤ r for some r > 0 and T ≥ T (r).

Assuming as in the Theorem 7 that the conditions (3.8), (3.9) holds, µi is large enough

and there exists a function ρ such that the hypothesis (3.10), then there exists a control f ∈L2(O×(0, T )) and a associated Nash equilibrium (v1(f), v2(f)) such that the corresponding

solution to (3.1) satises (3.7).

The rest this paper is organized as follows. In Section 3.2, we prove the Theorem 7.

In Section 3.3, we concern in the prove of the Theorems 8 and 9. Finally, in Section 3.4,

we comment on some open problems.

84

3.2 Existence and uniqueness of Nash equilibrium

In this section we will prove the Theorem 7, that is, for each f ∈ L2(O) we will

obtain a unique Nash equilibrium pair (v1(f), v2(f)) for the functionals Ji. By dention

this mean that (v1(f), v2(f)) is the solution to minimal problem (3.3), which is equivalent

to satisfy (3.4) and (3.5), as we has mentioned above. Firstly, we will characterize the

pair satisfying (3.4) by a optimality system. Then, after to prove the well posedness that

system, we will verify that the pair satises (3.5).

Suppose that there exists an pair (v1, v2) satisfying (3.4). Then, for i = 1, 2, we

have αi

∫∫Oi,d×(0,T )

(y − yi,d)wi dx dt+ µi

∫∫Oi×(0,T )

vivi dx dt = 0

∀ vi ∈ L2(Oi × (0, T )), i = 1, 2,

(3.11)

where wi is the Gateaux derivative of y in relation to vi in the direction of vi. Thus, wi

is solution to wit − wixx + ywix + wiyx = vi1Oi

in (0, 1)× (0, T ),

y(0, t) = y(1, t) = 0 on (0, T ),

y(x, 0) = 0 in (0, 1).

(3.12)

We introduce the adjoint to (3.12)−φit − φixx − yφix = αi(y − yi,d)1Oi,d

in (0, 1)× (0, T ),

φ(0, t) = φ(1, t) = 0 on (0, T ),

φ(x, T ) = 0 in (0, 1),

(3.13)

where the right-hand side in (3.13)1 is motivated from (3.11). Using the PDE's in (3.12)

and (3.13) we deduce

αi

∫∫Oi,d×(0,T )

(y − yi,d)wi dx dt = µi

∫∫Oi×(0,T )

φivi dx dt ∀vi ∈ L2(Oi × (0, T )),

then, replacing in (3.11), we see that∫∫Oi×(0,T )

(µivi + φi)vi dx dt = 0 ∀vi ∈ L2(Oi × (0, T )). (3.14)

85

Thus, we get the following optimality system

yt − yxx + yyx = f1O −2∑i=1

1

µiφi1Oi

, in (0, 1)× (0, T ),

−φit − φixx − yφix = αi(y − yi,d)1Oi,d, in (0, 1)× (0, T ),

y(0, t) = y(1, t) = φi(0, t) = φi(1, t) = 0, on (0, T ),

y(x, 0) = y0(x), φi(x, T ) = 0 in (0, 1),

(3.15)

vi = − 1

µiφi∣∣∣∣Oi×(0,T )

. (3.16)

We will prove that the system (3.15) is well posed. Fix y ∈ L2(0, T ;H10 (0, 1)) and consider

the system

yt − yxx + yyx = f1O −2∑i=1

1

µiφi1Oi

, in (0, 1)× (0, T ),

−φit − φixx − yφix = αi(y − yi,d)1Oi,d, in (0, 1)× (0, T ),

y(0, t) = y(1, t) = φi(0, t) = φi(1, t) = 0, on (0, T ),

y(x, 0) = y0(x), φi(x, T ) = 0 in (0, 1),

(3.17)

Note that in (3.17) the equation are not conjugate, we can obtain a unique solution(y, φ1, φ2

)∈[L∞(0, T ;L2(0, 1)) ∩ L2(0, T ;H1

0 (0, 1))]3.

Multiplying in (3.17)1 by y and integrating in (0, 1), using integration by parts, we have

1

2

d

dt|y|2L2(0,1) + |yx|2L2(0,1) ≤ |y|2L4(0,1) +

1

4|yx|2L2(0,1) +

1

2

(|f |2L2(O) + |y|2L2(0,1)

)+

2∑i=1

1

µ2i

|φi|2L2(Oi)+

1

2|y|2L2(0,1),

(3.18)

on the other hand

|y|2L4(0,1) ≤ C |yx|L2(0,1) · |y|L2(0,1) ≤1

4|yx|2L2(0,1) + C 2 |y|2L2(0,1),

then, replacing in (3.18)

d

dt|y|2L2(0,1) + |yx|2L2(0,1) ≤ 2

(C2 + 1

)|y|2L2(0,1) + |f |2L2(O) +

2∑i=1

1

µ2i

|φi|2L2(Oi). (3.19)

Integrating in (0, t), t ∈ (0, T ] we obtain

|y(t)|2L2(0,1) +

∫ t

0

|yx(s)|2L2(0,1) ds ≤ |y0|2L2(0,1) + 2(C2 + 1

) ∫ t

0

|y(s)|2L2(0,1) ds

+ ||f ||2L2(O×(0,T )) +2∑i=1

1

µ2i

||φi||2L2(Oi×(0,T )),

86

thus, by Gronwall's lemma we get

||y||2L∞(0,T ;L2(0,1)) + ||y||2L2(0,T ;H10 (0,1)) ≤ C1 + C

2∑i=1

1

µ2i

||φi||2L2(Oi×(0,T )), (3.20)

where C1 ≥(|y0|2L2(0,1) + ||f ||2L2(O×(0,T ))

).

Now multiplying in (3.17)2 by φi and integrating in (0, 1), using integration by parts,

we have

−1

2

d

dt|φi|2L2(0,1) + |φix|2L2(0,1) −

∫ 1

0

yφixφi dx = αi

∫Oi,d

(y − yi,d)φidx, (3.21)

but, ∫ 1

0

yφixφi dx

1

4|φix|2L2(0,1) +

C

2|yx|2L2(0,1) · |φi|2L2(0,1)

and

αi

∫Oi,d

(y − yi,d)φi dx ≤ Cα2i

(∫ 1

0

|y|2dx+

∫Oi,d

|yi,d|2dx

)+

1

4

∫ 1

0

|φix|2dx,

then, replacing in (3.21) and integrating in (t, T ), t ∈ [0, T ), we get

|φi(t)|2L2(0,1) +

∫ T

t

|φix(s)|2L2(0,1) ds ≤ C(||y||2L2(0,1)×(0,T ) + ||yi,d||2L2(Oi,d×(0,T ))

)+

∫ T

t

|yx(s)|2L2(0,1) · |φi(s)|2L2(0,1) ds,

thus, by Gronwall's lemma, we obtain

||φi||2L∞(0,T ;L2(0,1)) + ||φi||2L2(0,T ;H10 (0,1))

≤ CC(||y||2L2((0,1)×(0,T )) + ||yi,d||2L2(Oi,d×(0,T ))

) (3.22)

where C = e||y||2

L2(0,T ;H10(0,1)) . From (3.20) and (3.22) we deduce

||y||2L∞(0,T ;L2(0,1)) + ||y||2L2(0,T ;H10 (0,1))

≤ C1 +C

minµ1, µ2

(C2 + C||y||2L2((0,1)×(0,T ))

),

(3.23)

where C2 ≥ C||yi,d||2L2(Oi,d×(0,T )), i = 1, 2.

Now we consider r >√C1 and the set

K := z ∈ L2(0, T ;H10 (0, 1)); ||z||L2(0,T ;H1

0 (0,1)) ≤ r.

Then, taking

minµ1, µ2 ≥

√(C2 + C r2) er2

r2 − C1

,

87

we can dene the operator Λ : K → K, with Λ(y) := y, where (y, φ1, φ2) is solu-

tion to (3.17). Note that Λ is a compact operator, in fact, K is a bounded subset of

L2(0, T ;H10 (0, 1)) which is compactly immerse in L2(0, T ;L2(0, 1)) = L2((0, 1) × (0, T ))

and y depends continuously on φi, then, Λ is the composition od the operators

KF1−→ L2((0, 1)× (0, T ))

F2−→ K,

y 7→ φi 7→ y,

where F1 is a compact operator and F2 is a continuous operator.

Hence, by Schauder's Theorem, there exists a xed point y ∈ K to Λ, that is,

Λ(y) = y. Thus, (y, φ1, φ2) is solution to (3.15). To prove the uniqueness of solution,

suppose that (y1, φ1,1, φ1,2) and (y2, φ2,1, φ2,2) are solutions to (3.15). Then, considering

y = y1 − y2 and φi = φ1,i − φ2,i, we have that (y, φ1, φ2) is solution to

yt − yxx +1

2

(yy1 + yy2

)x

= −2∑i=1

1

µiφi1Oi

, in (0, 1)× (0, T ),

−φit − φixx − yφix = αi y 1Oi,d, in (0, 1)× (0, T ),

y(0, t) = y(1, t) = φ(0, t) = φ(1, t) = 0, on (0, T ),

y(x, 0) = 0, φ(x, T ) = 0 in (0, 1),

(3.24)

Multiplying in (3.24)1 by y and integrating in (0, 1), applying integration by parts, we

have1

2

d

dt|y(t)|2L2(0,1) + |yx(t)|2L2(0,1) −

1

2

∫ 1

0

y(y1 + y2

)yx dx

≤2∑i=1

1

2µ2i

∫Oi

|φi|2 dx+1

2

∫ 1

0

|y|2 dx

+1

4

∫ 1

0

|yx|2 dx+1

4

∫ 1

0

|y1 + y2|2|y|2 dx,

Since y1, y2 ∈ L2(0, T ;H10 (0, 1)), follows that:

t 7→∣∣y1(·, t) + y2(·, t)

∣∣2L∞(0,1)

,

belongs to L1(0, T ), then, integrating in (0, t), with t ∈ (0, T ), we get:

1

2|y(t)|2L2(0,1) +

3

4

∫ t

0

|yx(s)|2L2(0,1)

≤2∑i=1

1

2µ2i

∣∣∣∣φi∣∣∣∣2L2(Oi×(0,T ))

+1

2

∫ t

0

|y(s)|2L2(0,1) ds

+1

2

∫ t

0

∣∣y1(s) + y2(s)∣∣2L∞(0,1)

|y(s)|2L2(0,1) ds

(3.25)

88

Analogously to we did to get (3.22), we obtain:

||φi||2L2(Oi×(0,T )) ≤ C(||y||2L∞(0,T ;L2(0,1)) + ||y||2L2(0,T ;H1

0 (0,1))

). (3.26)

Hence, replacing (3.26) in (3.25) and using the Gronwall's lemma, we get

||y||2L∞(0,T ;L2(0,1)) + ||y||2L2(0,T ;H10 (0,1))

≤ C

minµ21, µ

22

(||y||2L∞(0,T ;L2(0,1)) + ||y||2L2(0,T ;H1

0 (0,1))

),

(3.27)

thus, considering minµ21, µ

22 > C, we conclude that y ≡ 0 and, as consequence, φi ≡ 0,

i = 1, 2. This prove the uniqueness of solution to (3.15).

As the system (3.15) is well posed, there exists a unique pair (v1, v2) ∈ L2(O1 ×

(0, T ))×L2(O2× (0, T )) to satises (3.11). We will prove that the pair (v1, v2) is a Nash

equilibrium, for this, we will prove that this pair satises (3.5). That is, already we obtain

the critical point (v1, v2) for the functionals Ji, now using the second derivative test we

will prove that it is a minimum point to Ji.

In fact, given v1 ∈ L(O1 × (0, T )) and s ∈ R, we denote by ys the solution of the

following systemyst − ysxx + ysysx = f1O +

(v1 + v1

)1O1 + v21O2 , in (0, 1)× (0, T ),

y(0, t) = y(1, t) = 0, on (0, T ),

y(x, 0) = 0, in (0, 1),

(3.28)

and let us use the notation ys|s=0 = y. For each v1 ∈ L2(O × (0, T )) we have

D1J1(f ; v1 + sv1, v2) · v1 −D1J1(f ; v1, v2)

:= α1

∫∫Oi,d×(0,T )

(ys − yi,d)w1,s dx dt− α1

∫∫O1,d×(0,T )

(y − y1,d)w1 dx dt

+ s µ1

∫∫O1,d×(0,T )

v1v1 dx dt,

(3.29)

where w1,s is the derivative of ys in relation to v1 + sv1 in the direction of v1. Then, w1,s

is the solution tow1,st − w1,s

xx + w1,syx + yw1,sx = v11O1 , in (0, 1)× (0, T ),

w1,s(0, t) = w1,s(1, t) = 0, on (0, T ),

w1,s(x, 0) = 0, in (0, 1),

(3.30)

89

and let us use the notation w1,s|s=0 = w1. We have the following adjoint system to (3.30)−φ1,s

t − φ1,sxx − yφ1,s

x = α1 (ys − y1,d) 1O1,d, in (0, 1)× (0, T ),

φ1,s(0, t) = φ1,s(1, t) = 0, on (0, T ),

φ1,s(x, T ) = 0, in (0, 1),

(3.31)

where the right hand side in (3.31)1 s motivated by (3.11). From (3.29), (3.30) and (3.31),

we get

D1J1(f ; v1 + sv1, v2) · v1 −D1J1(f ; v1, v2) :=

∫∫O1×(0,T )

(φ1,s − φ1

)v1 dx dt

+ s µ1

∫∫O1×(0,T )

v1v1 dx dt,(3.32)

then, taking the limit as s→ 0, we obtainD2

1(f ; v1, v2) · (v1, v1) =

∫∫O1×(0,T )

ηv1 dx dt+ µ1

∫∫O1×(0,T )

v1v1 dx dt

∀v1, v1 ∈ L2(O × (0, T )),

in particular,D2

1(f ; v1, v2) · (v1, v1) =

∫∫O1×(0,T )

η v1 dx dt+ µ1

∫∫O1×(0,T )

|v1|2 dx dt

∀v1 ∈ L2(O × (0, T )),

(3.33)

where

η := lims→0

1

s

(φ1,s − φ1

).

Note that also exists the limit

h := lims→0

1

s(ys − y) ,

and the pair (h, η) is solution to

ht − hxx + yhx + hyx = v11O1 , in (0, 1)× (0, T ),

−ηt − ηxx − yηx = α1h1O1,d, in (0, 1)× (0, T ),

h(0, t) = h(1, t) = η(0, t) = η(1, t) = 0, on (0, T ),

h(x, 0) = 0, η(x, T ) = 0 in (0, 1).

(3.34)

Multiplying in (3.34)1 by η and integrating in (0, 1) × (0, T ), using integration by parts,

we get ∫∫O1×(0,T )

η v1 dx dt = α1

∫∫O1,d×(0,T )

|h|2 dx dt,

90

then, replacing in (3.33), we obtainD2

1(f ; v1, v2) · (v1, v1) = α1

∫∫O1,d×(0,T )

|h|2 dx dt,+µ1

∫∫O1×(0,T )

|v1|2 dx dt,

∀v1 ∈ L2(O1 × (0, T )).

(3.35)

Hence,

D21(f ; v1, v2) · (v1, v1) ≥ C||v1||2L2(O1×(0,T )) ∀v1 ∈ L2(O1 × (0, T )),

and analogously,

D21(f ; v1, v2) · (v1, v1) ≥ C||v1||2L2(O1×(0,T )) ∀v1 ∈ L2(O1 × (0, T )).

Thus, the pair (v1, v2) is a Nash equilibrium to functionals Ji.

3.3 Null controllability

In this section we will prove the Theorems 7 and 8. In other words, we will obtain the

local null controllability to system (3.15) considering initial data in H10 (0, 1) and L2(0, 1),

respectively.

Initially, for each y ∈ L∞((0, 1) × (0, T )), let us consider the linearized system

corresponding (3.15)

yt − yxx + y(x, t)yx = f1O −2∑i=1

1

µiφi1Oi

, in (0, 1)× (0, T ),

−φit − φixx − y(x, t)φix = αi(y − yi,d)1Oi,d, in (0, 1)× (0, T ),

y(0, t) = y(1, t) = φi(0, t) = φi(1, t) = 0, on (0, T ),

z(x, 0) = y0(x), φi(x, T ) = 0 in (0, 1).

(3.36)

We have the following adjoint system for (3.36)

−ψt − ψxx − (y(x, t)ψ)x =2∑i=1

αiγi1Oi,d

, in (0, 1)× (0, T ),

γit − γixx + (y(x, t)γi)x = − 1

µiψ1Oi

, in (0, 1)× (0, T ),

ψ(0, t) = ψ(1, t) = γi(0, t) = γi(1, t) = 0, on (0, T ),

ψ(x, T ) = ψT (x), γi(x, 0) = 0 in (0, 1).

(3.37)

91

From (3.36) and (3.37), we get∫∫O×(0,T )

f ψ dx dt =2∑i=1

αi

∫∫Oi,d×(0,T )

γiyi,d dx dt

+(y(·, T ), ψT

)L2(0,1)

−(y0, ψ(·, 0)

)L2(0,1)

,

(3.38)

then, we have null controllability if and only if, for each y0 ∈ L2(0, 1) given, there exists

a control f ∈ L2(O × (0, T )), such that∫∫O×(0,T )

fψ dx dt =2∑i=1

αi

∫∫Oi,d×(0,T )

γiyi,d dx dt−(y0, ψ(·, 0)

)L2(0,1)

.

Now, for each ε > 0, we dene the functional

Fε(ψT ) :=

1

2

∫∫O×(0,T )

|ψ|2 dx dt+ ε|ψT |L2(0,1)

+(y0, ψ(·, 0)

)L2(0,1)

−2∑i=1

αi

∫∫Oi,d×(0,T )

γiyi,d dx dt.

from [1] and [6], we have that the following result for observability inequality:

Proposition 10 Suppose that Oi,d ∩ O 6= ∅, i = 1, 2. Assume that one of the following

conditions holds:

O1,d = O2,d = Od, (3.39)

or

O1,d ∩ O 6= O2,d ∩ O. (3.40)

Then, in both cases, for all ψT ∈ L2(Ω), there are C = C(Ω,O,Oi,d, αi, µi, T, ||y||L∞(Q)) >

0 and a weight function ρ = ρ(t), which decay exponentially to 0 when t→ T−, such that

for the solution (ψ, γ1, γ2) of (3.37), we have∫ 1

0

|ψ(x, 0)|2 dx+2∑i=1

∫∫(0,1)×(0,T )

ρ2|γi|2 dx dt ≤ C

∫∫O×(0,T )

|ψ|2 dx dt. (3.41)

By Proposition 10, Fε is continuous, strictly convex and coercive in L2(0, 1). Then it has

a unique minimum ψTε ∈ L2(0, 1). Then, either ψTε = 0 or for all ψT ∈ L2(0, 1), we have∫∫O×(0,T )

ψεψ dx dt+ ε

(ψTε||ψTε ||

, ψT)L2(0,1)

+(y0(x), ψ(x, 0)

)L2(0,1)

−2∑i=1

αi

∫∫Oi,d×(0,T )

yi,dγi dx dt = 0,

(3.42)

where (ψε, γ1ε , γ

2ε ) is the solution of (3.37) corresponding to ψTε . Thus, considering f =

ψε|O×(0,T ), by (3.38) and (3.42), we get∫Ω

(yε(x, T ) + ε

ψTε||ψTε ||

)ψT (x) dx = 0 ∀ψT ∈ L2(0, 1).

92

This way, follows that

||yε(·, T )|| ≤ ε. (3.43)

Moreover, from (3.41) and (3.42), we deduce

||fε||L2(O×(0,T )) ≤ C

(2∑i=1

∫∫Oi,d×(0,T )

ρ−2|yi,d|2 dx dt+ ||y0||2L2(0,1)

) 12

,

where we has used that (3.10) holds.

So, fε is uniformly bounded in L2(O × (0, T )), then, it possesses a subsequence

weakly convergent for some f ∈ L2(O× (0, T )). From regularity of solution of the system

(3.36), taking the limit as ε→ 0, by inequality (3.43) and Banach-Steinhaus' Theorem, we

conclude that correponding solution (y, φ1, φ2) to (3.36) satises the null controllability

condition y(·, T ) = 0.

Now we will use this result and xed point method to prove the Theorem 6.

Proof of Theorem 8. From the linear case, for each y ∈ L∞((0, 1)×(0, T )) xed, given

y0 ∈ H10 (0, 1) with ||y0||H1

0 (0,1) ≤ r, for some r > 0, if the constants µi are large enough

we get a minimal control f = fy ∈ L2(O × (0, T )) such that the corresponding solution

(y, φ1, φ2) = (yy, φ1y, φ

2y) to (3.36) satises y(·, T ) = 0. Moreover, we have that

||fy||L2(O×(0,T )) ≤ C r ∀ y ∈ L∞((0, 1)× (0, T ))

and y ∈ L2(0, T ;H2(0, 1)) ∩ L∞(0, T ;H10 (0, 1)) with

||y||L∞(0,T ;H10 (0,1)) + ||y||L2(0,T ;H2(0,1)) ≤ C r eC||y||L∞((0,1)×(0,T )) ,

where C > 0. So, in partcular y = yy ∈ L∞((0, 1)× (0, T )) and we can dene the map

Λ : L∞((0, 1)× (0, T ))→ L∞((0, 1)× (0, T )), Λ(y) := yy.

Let K > 0 be a constant, we introduce the set

Y := y ∈ L∞((0, 1)× (0, T )) : ||y||L∞((0,1)×(0,T )) ≤ K,

then, for all y ∈ Y , if

r ≤ K/C eCK (3.44)

we obtain y ∈ Y . That is, Λ(Y ) ⊂ Y .

93

Note that Λ is continuous and since that L2(0, T ;H2(0, 1))∩L∞(0, T ;H10 (0, 1)) → Y

with compact embedding, Λ maps the space Y into compact subset. Hence, by Shauder's

xed point theorem, there exists y ∈ L2(0, T ;H2(0, 1)) ∩ L∞(0, T ;H10 (0, 1)) such that

Λ(y) = y. Thus, there exists a minimal control f ∈ L2(O × (0, T )) such that the cor-

responding solution (y, φ1, φ2) to (3.15), with y0 ∈ H10 (0, 1), satises y(·, T ) = 0. This

proves the Theorem 8.

In sequel we will prove the Theorem 9 using the result of the Theorem 8.

Proof of Theorem 9. We assume that y0 ∈ L2(0, 1), with ||y0||L2(0,1) ≤ r. Then,

considering the trajectory y of the system (3.1), that is, the solution toyt − yxx + yyx = 0 in (0, 1)× (0, T ),

y(0, t) = y(1, t) = 0 on (0, T ),

y(x, 0) = y0(x) in (0, 1),

(3.45)

from parabolic regularity, we have y(·, t) ∈ H10 (0, 1) for any t > 0 and there exists cons-

tants τ and M such that

||y(·, t)||L∞(0,1) ≤Mt−14 ||y0||L2(0,1) ∀t ∈ (0, τ).

Then, considering t0 = M4 and taking y0 = y(·, t0), we have y0 ∈ H10 (0, 1) and ||y0||L∞(0,1) ≤

r. Now we consider the systemyt − yxx + yyx = f1O + v11O1 + v21O2 in (0, 1)× (t0, t0 + t1),

y(0, t) = y(1, t) = 0 on (t0, t0 + t1),

y(x, 0) = y0(x) in (0, 1),

(3.46)

where t1 = C/ log(1/r), where C > 0 is a suitable constant such that allow as to obtain

a estimate as in (3.44) . Hence, considering the functionals

Ji(f ; v1, v2) :=αi2

∫∫Oi,d×(t0,t0+t1)

|y − yi,d|2 dx dt+µi2

∫∫Oi×(t0,t0+t1)

|vi|2 dx dt, (3.47)

and main functional

J(f) :=1

2

∫∫O×(t0,t0+t1)

|f |2 dx dt,

we can apply the Stackelbrg-Nash strategy is this case, where we obtain a leader f ∈

L2(O×(t0, t0 + t1)) a associate Nash equilibrium (v1, v2) ∈ L2(O1×(t0, t0 + t1))×L2(O2×

(t0, t0 + t1)) such that the corresponding solution to (3.46) satises y(·, t0 + t1) = 0.

94

Thus, we dene the control

f(x, t) :=

0, if (x, t) ∈ O × (0, t0);

f(x, t), if (x, t) ∈ O × (t0, t0 + t1),(3.48)

we have f ∈ L2(O× (0, t0 + t1)) and obtain a Nash equilibrium (v1(f), v2(f)) ∈ L2(O1 ×

(0, t0 + t1))×L2(O2 × (0, t0 + t1)) such that the corresponding solution to (3.15) satises

y(·, t0 + t1) = 0. Hence, the Theorem 9 is proved.

The control f dened in (3.48) is not optimal. But as mentioned it before, by [26]

we not have null contrallability for any small time T > 0, then is rasonable to consider

the control null in a suitable interval.

3.4 Final comments

3.4.1 StackelbergNash strategy with controls in the boundary

A natural and interesting question is the hierarchic control applying Stackelberg

Nash stategy for null controllability of the burgers equation with at last a control, leader

or follower, in the boundary. For example, we can consider systems in the formyt − yxx + yyx = v11O1 + v21O2 in (0, 1)× (0, T ),

y(0, t) = 0, y(1, t) = f(t) on (0, T ),

y(x, 0) = y0(x) in (0, 1),

(3.49)

or yt − yxx + yyx = f1O in (0, 1)× (0, T ),

y(0, t) = v1(t), y(1, t) = v2(t) on (0, T ),

y(x, 0) = y0(x) in (0, 1),

(3.50)

where f is the leader and v1, v2 are the followers.

Then, we apply the StackebergNash strategy using cost functionals similar to (3.2).

This way for each leader f xed, we obtain a Nash equilibrium arguing as in the Section

3.2, but in both cases, the dicult is combine suitables Carleman inequalities to obtain

the observability. This situations will be analyzed in a forthcoming paper.

3.4.2 The NavierStokes system

The system (3.1) is a one dimensional simplication of the control system associated

to the NavierStokes system. Let us consider a bounded domain Ω of RN (N = 2 or

95

N = 3), which boundary Γ is regular anough. Let us consider the NavierStokes system

yt −∆y + (y · ∇)y +∇p = f1O + v11O1 + v21O2 in Q,

∇ · y = 0 in Q,

y = 0 on Σ,

y(·, 0) = y0 in Ω,

(3.51)

where T > 0 is given, Q = Ω × (0, T ), Σ = Γ × (0, T ) and the sets O, O1, O2 ⊂ Ω are

nonempty open.

We consider y0 in the space

H := z ∈ L2(Ω)N : ∇ · z = 0 in Ω, z · ν on Γ,

where ν(x) denotes the outward unit normal to Ω at the point x ∈ Γ.

The controllability to system (3.51) has been extensively studied these last years,

see for example, [27] and [34], where we can nd results on the exact controllability with

distributed and boundary controls, respectively.

In the context of the hierarchic control, in [2] the authors has been analized the

system (3.51), applying the StackelbergNash strategy, where they already obtained po-

sitives results on existence and uniqueness of Nash equilibrium for functionals similar to

Ji dened in (3.2). But the null controllability for the (3.51) is still a open question.

96

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