Regularization of optimal control problems Daniel Wachsmuth (RICAM - PowerPoint PPT Presentation

Regularization of optimal control problems Daniel Wachsmuth (RICAM Linz) joint work with Gerd Wachsmuth (TU Chemnitz)

1. Control-constrained problems 2. Tychonov regularization and regularization error estimates 3. Necessary conditions for convergence rates 4. Parameter choice: a discrepancy principle 5. Parameter choice & discretization RICAM special semester Daniel Wachsmuth

Optimization problem 1 Optimization problem: min � Su − z � 2 Y + β � u � L 1 ( Ω ) subject to u ∈ U ad . Here: U = L 2 ( Ω ) , Y Hilbert spaces, S : U → Y linear and compact operator. U ad ⊂ U convex, closed, and non-empty, β ≥ 0 fixed. Motivation: Actuator placement problem RICAM special semester Daniel Wachsmuth

Optimization problem 1 Optimization problem: min � Su − z � 2 Y + β � u � L 1 ( Ω ) subject to u ∈ U ad . Here: U = L 2 ( Ω ) , Y Hilbert spaces, S : U → Y linear and compact operator. U ad ⊂ U convex, closed, and non-empty, β ≥ 0 fixed. Motivation: Actuator placement problem Solvability: If z ∈ R ( S ) or U ad is bounded then problem is solvable. Uniqueness of solutions: If in addition S is injective then solution is unique. RICAM special semester Daniel Wachsmuth

Ill-posedness, difference to inverse problems 2 Ill-posedness: Solutions are not stable with respect to perturbations, i.e. in case where only z δ ≈ z is available. Problem is difficult to solve numerically. ( β > 0 does not help) Difference to inverse problems: Non-attainability: z �∈ S ( U ad ) The constraint u ∈ U ad is not extra information about solution, but a serious constraint: it may hinder us to reach the goal z . RICAM special semester Daniel Wachsmuth

Model problem 3 Elliptic PDE: S is the mapping from u �→ y , where y is the weak solution of − ∆y = u in Ω y = 0 on ∂Ω Optimal control problem: Minimize 1 2 � Su − z � 2 L 2 ( Ω ) + β � u � L 1 ( Ω ) subject to u a ≤ u ≤ u b a.e. on Ω. Here: u a , u b ∈ L 2 ( Ω ), u a ≤ 0 ≤ u b , Y = L 2 ( Ω ). Standing assumption on S : S : U = L 2 ( Ω ) → Y and S ∗ : Y ∗ �→ L ∞ ( Ω ) linear and continuous. RICAM special semester Daniel Wachsmuth

Ill-posedness 4 Recall standing assumption on S : S : L 2 ( Ω ) → Y and S ∗ : Y ∗ �→ L ∞ ( Ω ) linear and continuous. Theorem: Under this assumptions the range of S is closed if and only if it is finite-dimensional. Proof by closed-range theorem and a result of [Grothendieck ’54]. RICAM special semester Daniel Wachsmuth

1. Control-constrained problems 2. Tychonov regularization and regularization error estimates 3. Necessary conditions for convergence rates 4. Parameter choice: a discrepancy principle 5. Parameter choice & discretization RICAM special semester Daniel Wachsmuth

Regularized problem 5 Noisy data: z δ with � z − z δ � Y ≤ δ (prototype for discretization error) Take α > 0. Regularized problem: Minimize min 1 Y + β � u � L 1 ( Ω ) + α 2 � Su − z δ � 2 2 � u � 2 L 2 subject to u a ≤ u ≤ u b . u δ α , y δ α := Su δ Unique solution in case α > 0: α . Solution of original problem denoted by u 0 , y 0 := Su 0 . Questions: Convergence for ( α, δ ) → 0 ? Choice α = α ( δ )? RICAM special semester Daniel Wachsmuth

Optimality condition 6 Necessary optimality condition: There exists λ δ α ∈ ∂ � · � L 1 ( Ω ) ( u δ α ) α := S ∗ ( z δ − y δ such that with p δ α ) it holds ( αu δ α − p δ α + βλ δ α , u − u δ α ) ≥ 0 ∀ u ∈ U ad . Consequence 1: � y 0 − y δ α � 2 Y + α � u 0 − u δ α � 2 L 2 ≤ α ( u 0 , u 0 − u δ α ) L 2 + δ � y 0 − y δ α � Y Consequence 2: Noise error estimate � y α − y δ α � Y ≤ δ α � Y ≤ 1 δ � u α − u δ √ α 2 RICAM special semester Daniel Wachsmuth

Structure of solutions 7 Bang-bang solutions in case β = 0 : Control u 0 a.e. on the bounds. In general: u 0 discontinuous with jumps across { x ∈ Ω : | p 0 | = β } . RICAM special semester Daniel Wachsmuth

Source conditions 8 Observed convergence rates: � u 0 − u α � L 2 ∼ α 1 / 2 Explanation wanted! Source conditions: e.g. [Neubauer][Engl, Hanke, Neubauer] u 0 �∈ R ( S ∗ ), u 0 can be in R (( S ∗ S ) ν/ 2 ) only for small ν . Source conditions alone cannot explain these convergence rates. Idea: Exploit the structure (discontinuity). RICAM special semester Daniel Wachsmuth

Regularity condition 9 Assumption: There exist a set I ⊂ Ω , a function w ∈ Y , and positive constants κ, c such that it holds: 1. (source condition) I ⊃ { x ∈ Ω : | p 0 ( x ) | = β } and  Proj [ u a , 0] ( S ∗ w ) on { x ∈ Ω : p 0 ( x ) = − β }  u 0 = Proj [0 ,u b ] ( S ∗ w ) on { x ∈ Ω : p 0 ( x ) = + β }  2. (structure of active set) A = Ω \ I and for all ǫ > 0 � < ǫ } � � ≤ c ǫ κ . � � { x ∈ A : 0 < � | p 0 ( x ) | − β meas [Wachsmuth 2 2010] RICAM special semester Daniel Wachsmuth

Regularity condition - comments 10 Source condition: • Gives smoothness of u 0 on I . • Weaker than u 0 = Proj U ad ( S ∗ w ) [Wachsmuth 2 2010] Structure of active set: • Allows to control the size of almost-active sets. • If p 0 ∈ C 1 ( ¯ Ω ) and ∇ p 0 � = 0 on { x : | p 0 ( x ) | = β } then Assumption is fulfilled with κ = 1. [Deckelnick, Hinze 2010] References: • Case I = ∅ : [Felgenhauer 2003][Deckelnick, Hinze 2010][Wachsmuth 2 2009] • Case A = ∅ : ( p 0 = 0) with stronger source condition u 0 = Proj U ad ( S ∗ w ) [Neubauer 1988][Lorenz, R¨ osch 2010] RICAM special semester Daniel Wachsmuth

Regularization error 11 Regularization error estimate: Let the assumption be satisfied. Then we have � y 0 − y α � Y ≤ c α d , � u 0 − u α � L 2 ≤ c α d − 1 2 with  1 if κ ≤ 1 ,  2 − κ   d = 1 if κ > 1 and A � = Ω and w � = 0 ,  κ +1  if κ > 1 and A = Ω or w = 0 .  2 [For κ = 1 we get � u 0 − u α � L 2 = O ( α 1 / 2 ).] Question: What are necessary conditions to obtain these convergence rates? RICAM special semester Daniel Wachsmuth

Regularization error 12 Optimal a-priori choice: With α ∼ δ 1 /d we get � y 0 − y δ α � Y ≤ c δ, α � L 2 ≤ c δ 1 − 1 � u 0 − u δ 2 d . Problem: How to choose α if κ is unknown? RICAM special semester Daniel Wachsmuth

Regularization error - L p -case 13 Weaker assumption: Let S ∗ : Y → L p , 2 < p < ∞ . Regularization error estimate: Let the assumption be satisfied. Then we have � y 0 − y α � Y ≤ c α d , � u 0 − u α � L 2 ≤ c α d − 1 2 , with � 1 + κ � − 1 d = min p , d 1 , d 2 , 2 where d 1 , d 2 are given by   p + κ 2 p if A � = Ω if p ≤ 4 or κ ≤ 1 p − 4 ,   2 p +4 κ − κ p d 1 = , d 2 = + ∞ + ∞ if A = Ω otherwise .   Consistency: no rate for p ց 2, rates for p → ∞ coincide with p = ∞ . RICAM special semester Daniel Wachsmuth

1. Control-constrained problems 2. Tychonov regularization and regularization error estimates 3. Necessary conditions for convergence rates 4. Parameter choice: a discrepancy principle 5. Parameter choice & discretization RICAM special semester Daniel Wachsmuth

Convergence rate d = 1 14 Suppose � y 0 − y α � Y + � p 0 − p α � L ∞ ≤ c α . Difference quotients y 0 − y α are bounded in Y . α With a special test function  { u 0 } on { x ∈ Ω : | p 0 | � = β }    u ∈ ˜ [ u a , 0] on { x ∈ Ω : p 0 = − β }   [0 , u b ] on { x ∈ Ω : p 0 = β }  it holds ( αu α − ( p α − p 0 ) , ˜ u − u α ) ≥ 0 . Hence ( u 0 + S ∗ ˙ y 0 , ˜ u − u 0 ) ≥ 0 y 0 of y 0 − y α for each weak accumulation point ˙ . α RICAM special semester Daniel Wachsmuth

Convergence rate d = 1 15 Due to construction of ˜ u , we have  Proj [ u a , 0] ( S ∗ ˙ y 0 ) on { x ∈ Ω : p 0 ( x ) = − β }  u 0 = Proj [0 ,u b ] ( S ∗ ˙ on { x ∈ Ω : p 0 ( x ) = + β } y 0 )  Necessary condition to obtain d ≥ 1 : If � y 0 − y α � Y ≤ c α d , d ≥ 1, y 0 ∈ Y such that then there is ˙  Proj [ u a , 0] ( S ∗ ˙ y 0 ) on { x ∈ Ω : p 0 ( x ) = − β }  u 0 = Proj [0 ,u b ] ( S ∗ ˙ y 0 ) on { x ∈ Ω : p 0 ( x ) = + β }  In addition, � y 0 − y α � Y = o ( α ) implies ˙ y 0 = 0 and u 0 = 0 on { x ∈ Ω : | p 0 ( x ) | = β } . • resembles source-condition part of our assumption see also [Neubauer 1988] • source condition realized by derivative of α �→ y α at α = 0. RICAM special semester Daniel Wachsmuth

Convergence rate d > 1 16 Abbreviation: � < ǫ } � � � � µ ( ǫ ) := meas { x ∈ A : 0 < � | p 0 ( x ) | − β Define auxiliary function ˜ u α as solution of u ∈ U ad − ( p 0 , u ) + β � u � L 1 ( Ω ) + α 2 � u � 2 min L 2 . Suppose there exists M, σ > 0 such that − M ≤ u a ≤ − σ ≤ 0 ≤ σ ≤ u b ≤ M. Then � σ � p � σ � L p ( Ω ) ≤ M p µ ( Mα ) . u α � p ≤ � u 0 − ˜ µ 2 α 2 Hence we have the implication u α � 2 L 2 ( Ω ) ≤ Cα d − 1 / 2 µ ( ǫ ) ≤ ˜ C ǫ 2 d − 1 � u 0 − ˜ ⇒ RICAM special semester Daniel Wachsmuth