Maximum Principle of State-Constraint Optimal Control Governed by - PowerPoint PPT Presentation

Maximum Principle of State-Constraint Optimal Control Governed by Navier-Stokes Equations in 2-D PhD student: Hanbing LIU Supervisor: Viorel BARBU Marie Curie Initial Training Network (ITN) Alexandru Ioan Cuza University Iasi, Romania

INTRODUCTION In this work we consider the optimal problem: � T Min 1 | y ( t ) − y 0 ( t ) | 2 + | u ( t ) | 2 � � dt ; (1) U 2 0 y ( t ) ∈ K , K is a closed convex subset in H . Here y 0 ( t ) ∈ L 2 (0 , T ; H ) , and ( y ( t ) , u ( t )) is the solution to the following equation: � y ′ ( t ) + ν Ay ( t ) + By ( t ) = Du ( t ) + f , (2) y (0) = y 0 f ( t ) ∈ L 2 (0 , T ; H ) , u ( t ) ∈ L 2 (0 , T ; U ) , y 0 ∈ V

INTRODUCTION H = { y ( t ); y ( t ) ∈ ( L 2 (Ω)) 2 , ∇ · y ( t ) = 0 , y ( t ) · n = 0 on ∂ Ω } V = { y ( t ); y ( t ) ∈ ( H 1 0 (Ω)) 2 , ∇ · y ( t ) = 0 } and V ′ is the dual space of V , D ( A ) = ( H 2 (Ω)) 2 ∩ V , Ω is a bounded open subset with smooth boundary in R 2 , n is the outward vector to ∂ Ω and A = − P △ , By = P [( ∇ · y ) y ] where P is the projection to H . We shall denote by the symbol | · | the norm in R 2 , H and ( L 2 (Ω)) 2 , and � · � the norm of the space V . Define the trilinear function b ( y , z , w ) by 2 � � y i D i z j w j dx , ∀ y , z , w ∈ V b ( y , z , w ) = Ω i , j =1 U is a Hilbert space and D ∈ L ( U , H ). We denote by | · | U the norm in U , and ( · , · ) U the scalar product in U .

INTRODUCTION Lemma (1) b ( y , z , w ) = − b ( y , w , z ) and there exists a positive constant C, s.t. | b ( y , z , w ) | ≤ C � y � m 1 � z � m 2 +1 � w � m 3 where m 1 , m 2 , m 3 are positive number, satisfy the inequality: m 1 + m 2 + m 3 ≥ 1 , m i � = 1 m 1 + m 2 + m 3 > 1 , ∃ m i = 1 (2)there exists a positive constant C, s.t. � y � m ≤ C � y � 1 − α � y � α l +1 l where α = m − l ∈ (0 , 1) . Here � · � m i denotes the norm of the space H m i (Ω) .

INTRODUCTION Definition Let E be a Banach space, and E ∗ is it’s dual space. ∀ ω ( t ) ∈ BV (0 , T ; E ∗ ), we define the continuous functional µ ω on C ([0 , T ]; E ) by � T µ ω ( z ( t )) = ( z ( t ) , d ω ( t )) ( E , E ∗ ) 0 ( · , · ) ( E , E ∗ ) denotes the dual product between E and E ∗ , the integral is the Riemann-Steiljes integral. Denote M (0 , T ; E ∗ ) the dual space of C ([0 , T ]; E ). For the closed convex subset K in E , denote K by K = { y ( t ) ∈ C ([0 , T ]; E ); y ( t ) ∈ E , ∀ t ∈ [0 , T ] } , and the normal cone of K on y ( t ) is N K ( y ( t )) = { µ ∈ M (0 , T ; E ∗ ); µ ( y ( t ) − z ( t )) ≥ 0 , ∀ z ( t ) ∈ K}

INTRODUCTION The main results of this work is about the maximum principle of the optimal control problem governed by Navier-Stokes equations with state constraint in 2-D. To get the results, we make some assumptions as following: (A) ∃ ˜ z ( t ) ∈ int K , for t in a dense subset of z ( t ) , ˜ u ( t ) such that ˜ [0 , T ], where ˜ z ( t ) , ˜ u ( t ) satisfies the following equation: � ˜ z ′ ( t ) + ν A ˜ z ( t ) + ( B ′ ( y ∗ ( t )))˜ z ( t ) = B ( y ∗ ( t )) + D ˜ u ( t ) + f ( t ) , ˜ z (0) = y 0 (3) Here y ∗ ( t ) is the optimal state function for the optimal control problem (1),(2). (A’) ∃ ˜ z ( t ) , ˜ u ( t ) such that ˜ z ( t ) ∈ int V K , for t in a dense subset of [0 , T ], where ˜ z ( t ) , ˜ u ( t ) satisfies the equation (3)

MAIN RESULTS Theorem Suppose that the pair ( y ∗ ( t ) , u ∗ ( t )) is solution for optimal control problem (1),(2). Then under the assumption (A), there are p ( t ) ∈ L ∞ (0 , T ; H ) and ω ( t ) ∈ BV (0 , T ; H ) , such that: D ∗ p ( t ) = u ∗ ( t ) a . e . [0 , T ] (4) where p ( t ) satisfies the following equation � p ′ ( t ) = ν Ap ( t ) + ( B ′ ( y ∗ ( t )) ∗ ) p ( t ) + y ∗ ( t ) − y 0 ( t ) + d ω ( t ) , p ( T ) = 0 (5)

MAIN RESULTS Theorem The latter equation holds in the sense of � T � p ′ ( s ) − ν Ap ( s ) − ( B ′ ( y ∗ ( s )) ∗ ) p ( s ) , ψ ( s ) � ds t � T � T � y ∗ ( s ) − y 0 ( s ) , ψ ( s ) � ds + = � d ω ( s ) , ψ ( s ) � t t ∀ ψ ( t ) ∈ C 1 ([0 , T ]; D ( A )) . Moreover, µ ω ∈ N K ( y ∗ ( t )) (6) where µ ω and N K ( y ∗ ( t )) are defined as in definition 1 in the case that E = H. Here B ′ ( y ) is the operator defined by � B ′ ( y ) z , w � = b ( y , z , w ) + b ( z , y , w ) , ∀ z , w ∈ V

MAIN RESULTS Theorem Suppose the pair ( y ∗ ( t ) , u ∗ ( t )) is the solution for optimal control problem (1),(2), then under (A’) there are p ( t ) ∈ L ∞ (0 , T ; V ′ ) , ω ( t ) ∈ BV (0 , T ; V ′ ) , such that (4) holds, and (5) holds in the sense of � T ( p ′ ( s ) − ν Ap ( s ) − ( B ′ ( y ∗ ( s )) ∗ ) p ( s ) , ψ ( s )) ( V ′ , V ) ds t � T � T ( y ∗ ( s ) − y 0 ( s ) , ψ ( s )) ( V ′ , V ) ds + = ( d ω ( s ) , ψ ( s )) ( V ′ , V ) t t ∀ ψ ( t ) ∈ C 1 ([0 , T ]; D ( A )) , here ( · , · ) ( V ′ , V ) is the dual product between V ′ and V . Moreover,(6) also holds, where µ ω and N K ( y ∗ ( t )) are defined as in definition 1 in the case that E = V

PROOF Before the proof of the two theorems we define the approximating cost function to the original one F ( y , u ) which is defined by (1) as � T 1 2[ | y ( t ) − y 0 ( t ) | 2 + | u ( t ) | 2 + | u ( t ) − u ∗ ( t ) | 2 F ε ( y , u ) = U ]+ ϕ ε ( y ε ( t )) dt 0 (7) where ϕ ε ( y ) is the regularization of ϕ , which is the characteristic function of K , and the function ϕ ε ( y ) is defined by ϕ ε ( y ) = inf {| y − x | 2 + ϕ ( x ); x ∈ H } (8) 2 ε Define C = { ( y , u ) ∈ C ([0 , T ]; H ) × L 2 (0 , T ; U ); ( y ( t ) , u ( t ))is the solution to (2) } .

PROOF Lemma There exists at least one optimal pair for the optimal control problem: Min { F ε ( y , u ); ( y , u ) ∈ C } (9) Lemma Suppose z ε ( t ) is the solution to the equation: � z ′ ε ( t ) + ν Az ε ( t ) + ( B ′ ( y ε ( t ))) z ε ( t ) = B ( y ε ( t )) + D ˜ u ( t ) + f ( t ) , z ε (0) = y 0 (10) z ( t ) strongly in C ([0 , T ]; H ) ∩ L 2 (0 , T ; V ) , where then z ε ( t ) → ˜ z (( t ) , ˜ ˜ u (( t ) is defined in equation (3), and y ε ( t ) is the the optimal solution in lemma 2.

PROOF Proof of theorem 1: step 1: (first order necessary condition for approximate problem) Since ( y ε , u ε ) minimize the functional F ε ( y , u ) , we know that F ε ( u ε + hu ) − F ε ( u ε ) lim = 0 , ∀ u ∈ U h h → 0 and this yields � y ε − y 0 , w ε � + ( u ε , u ) U + ( u ε − u ∗ , u ) U + � ∂ϕ ε ( y ε ) , w ε � = 0 (11) y h ε − y ε , ( y h ε , u ε + hu ) ∈ C and w ε ( t ) is the where w ε = lim h → 0 h solution to the equation w ′ ε ( t ) + ν Aw ε ( t ) + B ′ ( y ε ( t )) w ε ( t ) = Du , w ε (0) = 0 (12)

PROOF suppose p ε ( t ) is the solution to the backward equation � p ′ ε ( t ) = ν Ap ε ( t ) + ( B ′ ( y ε ( t )) ∗ ) p ε ( t ) + y ε ( t ) − y 0 ( t ) + ∂ϕ ε ( y ε ( t )) p ε ( T ) = 0 (13) By (11) together with (12),(13), we get by calculation that � p ′ ε ( t ) , w ε ( t ) � + �− Ap ε ( t ) − ( B ′ ( y ε ( t )) ∗ ) p ε ( t ) , w ε ( t ) � +( u ε − u ∗ , u ) U = 0 . Hence we have ( − D ∗ p ε ( t ) + 2 u ε − u ∗ , u ) U = 0 , ∀ u ∈ U so, we get D ∗ p ε ( t ) = 2 u ε ( t ) − u ∗ ( t ) , a . e . t ∈ [0 , T ] (14)

PROOF step 2: (pass ( y ε , u ε ) to limit) By lemma 2, ∃ ( y ε , u ε ) ∈ C , s . t . F ε ( u ε , y ε ) = inf F ε ( u , y ) = d ε . since F ε ( y ε , u ε ) ≤ F ε ( y ∗ , u ∗ ) = F ( y ∗ , u ∗ ) = d so | d ε | ≤ C , ∀ ε > 0,hence � u ε � L 2 (0 , T ; H ) ≤ C (15) Multiply the equation y ′ ε ( t ) + ν Ay ε ( t ) + By ε ( t ) = Du ε ( t ) + f ( t ) (16) by y ε ( t ), Ay ε ( t ), integrate from 0 to t , we get

PROOF � T � T � T � y ε ( t ) � 2 + | Ay ε ( t ) | 2 dt + | By ε ( t ) | 2 dt + | ( y ε ( t )) ′ | 2 dt ≤ C 0 0 0 (17) hence, on a subsequence convergent to 0, again denoted by λ , we have y ε ( t ) → y 1 ( t ) strongly in C ([0 , T ; H ]) ∩ L 2 (0 , T ; V ) Ay ε ( t ) → Ay 1 ( t ) , ( y ε ( t )) ′ → y ′ 1 ( t ) weakly in L 2 (0 , T ; H ) u ε ( t ) → u 1 ( t ) weakly in L 2 (0 , T ; U ) By ε ( t ) → By 1 ( t ) strongly in L 2 (0 , T ; H ) so ( y 1 ( t ) , u 1 ( t )) is a solution to equation (2)

PROOF ϕ ( y ε ) = ε ε ( y ε )) ≥ ε 2 | ∂ϕ ε ( y ε ) | 2 + ϕ ( J ϕ 2 | ∂ϕ ε ( y ε ) | 2 so { ε | ∂ϕ ε ( y ε ) | 2 } is bounded in L 1 (0 , T ) and since ε ( y ε − J ϕ ε ( y ε )), where J ϕ ∂ϕ ε ( y ε )= 1 ε ( y ε ) is the function satisfies J ϕ ε ( y ε ) − y ε + ∂ϕ ε ( J ϕ ε ( y ε )) ∋ 0, we have � T � T | y ε − J ϕ ε | ∂ϕ ε ( y ε ) | 2 dt → 0 as ε → 0 ε ( y ε ) | dt ≤ ε T 0 0 so y ε − J ϕ ε ( y ε ) → 0 a . e . (0 , T ) . since J ϕ ε ( y ε ) ∈ K , ∀ t ∈ [0 , T ] , so y 1 ( t ) ∈ K . ∀ t ∈ [0 , T ], Inasmuch as ε → 0 F ε ( y ∗ , u ∗ ) = F ( y ∗ , u ∗ ) lim inf ε → 0 F ε ( y ε , u ε ) ≤ lim we have u 1 = u ∗ , y 1 = y ∗ and u ε → u ∗ strongly in L 2 (0 , T ; H ).