· controle hierárquico via estratégia de stackelberg nash para controlabilidade de sistemas...
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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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