Control of the motion of a boat Lionel Rosier Universit e Henri - PowerPoint PPT Presentation

Control of the motion of a boat Lionel Rosier Universit´ e Henri Poincar´ e Nancy 1 Control and Optimization of PDEs Graz, October 10-14, 2011

Joint work with Olivier Glass, Universit´ e Paris-Dauphine

Control of the motion of a boat ◮ We consider a rigid body S ⊂ R 2 with one axis of symmetry, surrounded by a fluid, and which is controlled by two fluid flows, a longitudinal one and a transversal one.

Aims ◮ We aim to control the position and velocity of the rigid body by the control inputs. System of dimension 3+3 with a PDE in the dynamics. Control living in R 2 . No control objective for the fluid flow (exterior domain!!). ◮ Model for the motion of a boat with a longitudinal propeller, and a transversal one (thruster) in the framework of the theory of fluid-structure interaction problems. Rockets and planes could also be concerned.

Bowthruster

What is a fluid-structure interaction problem? ◮ Consider a rigid (or flexible) structure in touch with a fluid. ◮ The velocity of the fluid obeys Navier-Stokes (or Euler) equations in a variable domain ◮ The dynamics of the rigid structure is governed by Newton laws. Great role played by the pressure. ◮ Questions of interest: existence of (weak, strong, global) solutions of the system fluid+solid, uniqueness , long-time behavior , control , inverse problems , optimal design , ...

Some references for perfect fluids (Euler eq.) ◮ Models for potential flows Kirchhoff, Lamb, Marsden (et al.) ,... ◮ Control problems for some models with potential flows N. Leonard [1997], N. Leonard et al.,... Chambrion-Sigalotti [2008] ◮ Cauchy problem J. Ortega, LR, T. Takahashi [2005,2007] C. Rosier, LR [2009] O. Glass, F . Sueur, T. Takahashi [2012],... ◮ Inverse Problems C. Conca, P . Cumsille, J. Ortega, LR [2008] C. Conca, M. Malik, A. Munnier [2010]

Main difficulties 1. The systems describing the motions of the fluid and the solid are nonlinear and strongly coupled ; e.g., the pressure of the fluid gives rise to a force and a torque applied to the solid, and the fluid domain changes when the solid is moving. 2. The fluid domain R N \ S ( t ) is an unknown function of time

Why to consider perfect fluids? 1. Euler equations provide a good model for the motion of boats or submarines in a reasonable time-scale. 2. Explicit computations may be performed with the aid of Complex Analysis when the flow is potential and 2D. 3. There is a natural choice for the boundary conditions u rel · n = 0 for Euler equations. For Navier-Stokes flows, one often takes u rel = 0 4. The controllability of Euler equation is well understood Coron 1996 (2D), Glass 2000 (3D).

System under investigation Ω( t ) = R 2 \ S ( t ) Euler u t + ( u · ∇ ) u + ∇ p = 0 , x ∈ Ω( t ) div u = 0 , x ∈ Ω( t ) n = ( h ′ + r ( x − h ) ⊥ ) · � u · � n + w ( x , t ) , x ∈ ∂ Ω( t ) lim | x |→∞ u ( x , t ) = 0 � m h ′′ ( t ) = p � Newton n d σ ∂ Ω( t ) � ( x − h ) ⊥ · p � J r ′ = n d σ ∂ Ω( t ) System supplemented with Initial Conditions , and with the value of the vorticity at the incoming flow (in Ω( t ) ) for the uniqueness

System in a frame linked to the solid After a change of variables and unknown functions, we obtain in Ω := R 2 \ S ( 0 ) v t + ( v − l − ry ⊥ ) · ∇ v + rv ⊥ + ∇ q = 0 , y ∈ Ω y ⊥ = ( − y 2 , y 1 ) div v = 0 , y ∈ Ω n = ( l ′ + ry ⊥ ) · � � v · � n + w j ( t ) χ j ( y ) , y ∈ ∂ Ω 1 ≤ j ≤ 2 | y |→∞ v ( y , t ) = 0 lim � m l ′ ( t ) = n d σ − mrl ⊥ q � ∂ Ω � J r ′ = qn · y ⊥ d σ ∂ Ω where l ( t ) := Q ( θ ( t )) − 1 h ′ ( t ) , r ( t ) = θ ′ ( t ) » cos θ − sin θ – Q ( θ ) = sin θ cos θ

Potential flows Assuming that the initial vorticity and circulation are null � u 0 · n ⊥ d σ = 0 ω 0 := curl u 0 ≡ 0 , Γ 0 := ∂ Ω and that the vorticity at the inflow part of ∂ Ω is null � ω ( y , t ) = 0 w i ( t ) χ i ( y ) ≤ 0 if i = 1 , 2 then the flow remains potential, i.e. v = ∇ φ where φ solves  ∆ φ = 0 in Ω × [ 0 , T ]   ∂φ  � ∂ n = ( l + ry ⊥ ) · n +  w i ( t ) χ i ( y ) on ∂ Ω × [ 0 , T ] i = 1 , 2    lim | y |→∞ ∇ φ ( y , t ) = 0 on [ 0 , T ] 

Potential flows (continued) v = ∇ φ decomposed as � � ∇ φ = l i ( t ) ∇ ψ i ( y ) + r ( t ) ∇ ϕ ( y ) + w i ( t ) ∇ θ i ( y ) i = 1 , 2 i = 1 , 2 where the functions ϕ, ψ i and θ i are harmonic on Ω and fulfill the following boundary conditions on ∂ Ω ∂ϕ ∂ψ i ∂θ i ∂ n = y ⊥ · n , ∂ n = n i ( y ) , ∂ n = χ i ( y ) This gives the following expression for the pressure i θ i + | v | 2 − l · v − ry ⊥ · v } � l ′ i ψ i + r ′ ϕ + � w ′ q = −{ 2 i = 1 , 2 i = 1 , 2 Plugging this expression in Newton’s law yields a

Control system in finite dimension h ′ = Q l C w ′ + B ( l , w ) J l ′ = ◮ Q = diag ( Q , 1 ) , h = [ h 1 , h 2 , θ ] T , l = [ l 1 , l 2 , r ] T ◮ w = [ w 1 , w 2 ] T is the control input ◮ B ( l , w ) is bilinear in ( l , w ) .  �  m + ψ 1 n 1 0 0 ψ 2 y ⊥ · n � � J = 0 m + ψ 2 n 2   ψ 2 y ⊥ · n ϕ y ⊥ · n � � 0 J +  �  − θ 1 n 1 0 � � � C = 0 − θ 2 n 2 ( = )   θ 2 y ⊥ · n ∂ Ω � 0 −

Toy problem w 2 = 0 , h 2 = l 2 = 0 � h ′ 1 = l 1 ( ∗ ) l ′ α w ′ 1 + β w 1 l 1 + γ w 2 1 = 1 where � � � � ψ 1 n 1 ) − 1 ( ( α, β, γ ) := ( m + θ 1 n 1 , χ 1 ∂ 1 ψ 1 , χ 1 ∂ 1 θ 1 ) ∂ Ω ∂ Ω ∂ Ω ∂ Ω Facts ◮ If we add the equation w 1 ′ = v 1 to ( ∗ ) , the system with state ( h 1 , l 1 , w 1 ) and input v 1 is not controllable ! ◮ In general we cannot impose the condition w 1 ( 0 ) = w 1 ( T ) = 0 when l 1 ( 0 ) = l 1 ( T ) = 0 (i.e. fluid at rest at t = 0 , T ). Actually we can do that if and only if γ + αβ = 0.

Generic assumption We shall assume that c 1 � = 0 and that � c 2 � b 3 � = 0 det ˜ c 2 b 5 where � c 1 = − θ 1 n 1 ∂ Ω � c 2 = − θ 2 n 2 ∂ Ω � θ 2 y ⊥ · n ˜ c 2 = − ∂ Ω � � b 3 = − χ 1 ∂ 2 θ 2 − χ 2 ∂ 2 θ 1 ∂ Ω ∂ Ω � � χ 1 ∇ θ 2 · y ⊥ − χ 2 ∇ θ 1 · y ⊥ b 5 = − ∂ Ω ∂ Ω

Control result for potential flows Thm (O Glass, LR) ◮ If the “generic” assumption holds with m >> 1, J >> 1, then the system h ′ = Q l C w ′ + B ( l , w ) J l ′ = with state ( h , l ) ∈ R 6 and control w ∈ R 2 is locally controllable around 0. ◮ If, in addition, γ + αβ = 0, then we have a global controllability for steady states

Example 1: Elliptic boat with 3 controls Actually, the linearized system around the null trajectory is controllable!

Example 2: Elliptic boat with 2 longitudinal controls Generic condition fulfilled iff b 3 = − 1 � |∇ Ψ | 2 n 2 � = 0 2 ∂ Ω where − ∆Ψ = 0, ∂ Ψ /∂ n = χ 1 y 2 > 0

Step 1. Loop-shaped trajectory We consider a special trajectory of the toy problem ( w 2 ≡ 0) constructed as in the flatness approach due to M. Fliess, J. Levine, P. Martin, P. Rouchon ◮ We first define the trajectory h 1 ( t ) = λ ( 1 − cos ( 2 π t / T )) l 1 ( t ) = λ ( 2 π/ T ) sin ( 2 π t / T ) ◮ We next solve the Cauchy problem ′ − γ w 1 2 − β w 1 l 1 } ′ � = α − 1 { l 1 w 1 w 1 ( 0 ) = 0 to design the control input. ◮ Then w 1 exists on [ 0 , T ] for 0 < λ << 1. ( h 1 , l 1 ) = 0 at t = 0 , T . Nothing can be said about w 1 ( T ) .

Step 2. Return Method We linearize along the above (non trivial) reference trajectory to use the nonlinear terms. We obtain a system of the form x ′ = A ( t ) x + B ( t ) u + Cu ′ h,l t T w t T

Linearization along the reference trajectory ◮ For a system x ′ = A ( t ) x + B ( t ) u we denote by R T ( A , B ) the reachable set in time T from 0, i.e. R T ( A , B ) = { x ( T ); x ′ = A ( t ) x + B ( t ) u , x ( 0 ) = 0 , u ∈ L 2 ( 0 , T ) } ◮ Fact. The reachable set from the origin for the system x ′ = A ( t ) x + B ( t ) u + Cu ′ is R = R T ( A , B + AC ) + C R m + Φ( T , 0 ) C R m where Φ( t , t 0 ) is the resolvent matrix associated with the system x ′ = A ( t ) x . ( ∂ Φ /∂ t = A ( t )Φ , Φ( t 0 , t 0 ) = I )

Silverman-Meadows test of controllability Consider a C ω time-varying control system x ∈ R n , t ∈ [ 0 , T ] , u ∈ R m . x = A ( t ) x + B ( t ) u , ˙ Define a sequence ( M i ( · )) i ≥ 0 by M i ( t ) = dM i − 1 M 0 ( t ) = B ( t ) , − A ( t ) M i − 1 ( t ) i ≥ 1 , t ∈ [ 0 , T ] dt Then for any t 0 ∈ [ 0 , T ] Φ( T , t 0 ) M i ( t 0 ) R m = R T ( A , B ) � i ≥ 0

Proof of the main result (continued) To complete the proof of the theorem we use ◮ the generic assumption to prove that the linearized system is controllable . Compute M 0 , M 1 , M 2 in Silverman-Meadows test (and also M 3 for the global controllability) ◮ the Inverse Mapping Theorem to conclude.