Parametrizing Linear Systems Daniel Robertz (joint work with F. - PowerPoint PPT Presentation

Parametrizing Linear Systems Daniel Robertz (joint work with F. Chyzak, A. Quadrat) Lehrstuhl B f¨ ur Mathematik 29.10.2010

Outline ◮ Module-theoretic approach to linear systems ◮ Parametrizing linear systems ◮ Injective parametrizations

1. Module-theoretic approach to linear systems

Linear System D ring (field, integral domain, Ore algebra, . . . ) R ∈ D q × p , F left D -module y ∈ F p . R y = 0 , P ∈ D p × r “optimal” answer for us: s.t. ker( R . ) = im( P . ) not possible in general Example. D = k a (skew) field. ◮ Gaussian elimination singles out parameters ◮ injective parametrization

Example de Rham complex: Ω ⊆ R 3 convex F = C ∞ (Ω), D = R [ ∂ x , ∂ y , ∂ z ], e.g. 0 1 0 1 ∂ x 0 ∂ z − ∂ y A . A . ∂ y − ∂ z 0 ∂ x ` ∂ x @ @ ´ ∂ z ∂ y − ∂ x 0 ∂ y ∂ z . → F 3 × 1 → F 3 × 1 0 → R → F − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − → F → 0 0 1 0 1 ∂ x 0 ∂ z − ∂ y . ∂ y . − ∂ z 0 ∂ x ` ∂ x @ A @ A ´ ∂ z ∂ y − ∂ x 0 . ∂ y ∂ z − D 1 × 3 − D 1 × 3 0 ← M ← D ← − − − − − ← − − − − − − − − − − − − − − ← − − − − − − − − − − D ← 0

Module-theoretic approach to linear systems R ∈ D q × p Σ : R y = 0, D = ring of functional operators M = D 1 × p / D 1 × q R independent of eq.’s chosen for Σ F = signal space If F is an injective cogenerator for D M then hom D ( · , F ) � Sol F ( M ) M is a categorical duality. Malgrange, Oberst, Pommaret, Quadrat, Willems, Zerz, . . .

� � � � Module-theoretic approach to linear systems M = D 1 × p / D 1 × q R R ∈ D q × p , Σ : R y = 0, F injective cogenerator for D M : . R . S D 1 × p D 1 × q D 1 × r 0 M exact if and only if (apply hom D ( · , F )) � Sol F ( M ) � F p R . � F q S . � F r 0 exact. Fundamental principle (Ehrenpreis, Malgrange, Palamodov) for D = C [ ∂ 1 , . . . , ∂ n ] acting by differentiation on F : F ∈ { complex-valued C ∞ -functions on R n , e.g. complex-valued distributions on R n , formal / convergent power series }

2. Parametrizing linear systems

� � � � � � � � � � � Parametrization and torsion-freeness Let M be given by the finite presentation ρ . R 0 D 1 × p D 1 × q . M Assume P is a parametrization, i.e. . P . R D 1 × r D 1 × p D 1 × q is an exact sequence of left D -modules. Then M is torsion-free: . P . R F r D 1 × r D 1 × p D 1 × q � ��������� ι ∗ ι ρ Sol F ( M ) M � �������� 0 0 0

Example F. Dubois, N. Petit, and P. Rouchon, Motion Planning and Nonlinear Simulations for a Tank Containing a Fluid , Proc. ECC, Karlsruhe (Germany), 1999.  φ 1 ( t − 2) + φ 2 ( t ) − 2 ˙  φ 3 ( t − 1) = 0 ,  φ 1 ( t ) + φ 2 ( t − 2) − 2 ˙ φ 3 ( t − 1) = 0 . D := Q [ ∂, δ ] (differential time-delay operators) � � δ 2 1 − 2 δ ∂ ∈ D 2 × 3 , R := δ 2 1 − 2 δ ∂ M := D 1 × 3 / D 1 × 2 R

� � � � Parametrizability test M = D 1 × p / D 1 × q R M ⊤ = D q × 1 / R D p × 1 � 0 1 2 δ ∂ „ δ 2 A . « 2 δ ∂ 1 − 2 δ ∂ @ . δ 2 + 1 δ 2 1 − 2 δ ∂ � D 3 × 1 � D 2 × 1 � M ⊤ � 0 D 1 × 1 0 1 2 δ ∂ „ δ 2 . 2 δ ∂ 1 − 2 δ ∂ « @ A . δ 2 + 1 δ 2 1 − 2 δ ∂ D 1 × 1 D 1 × 3 D 1 × 2 not exact 0 1 2 δ ∂ „ δ 2 « . 2 δ ∂ 1 − 2 δ ∂ @ A . δ 2 + 1 δ 2 1 − 2 δ ∂ D 1 × 1 D 1 × 3 D 1 × 2 � ������������������ „ 1 « − 1 0 R ′ := − δ 2 − 1 0 2 δ ∂ D 1 × 2

� � Parametrizability test 0 1 2 δ ∂ „ δ 2 2 δ ∂ « . 1 − 2 δ ∂ @ A . δ 2 + 1 δ 2 1 − 2 δ ∂ D 1 × 1 D 1 × 3 D 1 × 2 � ������������������ „ 1 « − 1 0 R ′ := − δ 2 − 1 0 2 δ ∂ D 1 × 2 t ( M ) = D 1 × 2 R ′ / D 1 × 2 R � = 0. We have ( δ 2 − 1) ( φ 1 ( t ) − φ 2 ( t )) = 0, In particular, φ 1 − φ 2 is 2-periodic.   2 δ ∂   2 δ ∂ is a parametrization of the subsystem δ 2 + 1   φ 1 ( t ) − φ 2 ( t ) = 0 ,  − φ 2 ( t − 2) − φ 2 ( t ) + 2 ˙ φ 3 ( t − 1) = 0 .

Parametrizability test D ( M ⊤ , D ) ∼ ext 1 In fact, we compute = t ( M ). 0 ↓ t ( M ) 0 0 0 ↓ ↓ ↓ ↓ hom D ( M ⊤ , D ) 0 ← − M ← − F 0 ← − F 1 ← − ← − 0 ↓ ↓ ↓ ↓ hom D ( M ⊤ , K ) 0 ← − K ⊗ D M ← − K ⊗ D F 0 ← − K ⊗ D F 1 ← − ← − 0 ↓ ↓ ↓ ↓ − hom D ( M ⊤ , K / D ) ← 0 ← − ( K / D ) ⊗ D M ← − ( K / D ) ⊗ D F 0 ← − ( K / D ) ⊗ D F 1 ← − 0 ↓ ↓ ↓ ↓ ext 1 D ( M ⊤ , D ) 0 0 0 ↓ 0

Parametrizability test We can compute � � � � δ 2 1 − 1 0 − 1 R ′ := R ′′ := , − δ 2 − 1 0 2 ∂ δ 1 − 1 R = R ′′ R ′ . ker( R ′ ) = 0. which satisfy Here we have t ( M ) ∼ = D 1 × 2 / ( D 1 × 2 R ′′ ), M / t ( M ) ∼ = D 1 × 3 / ( D 1 × 2 R ′ ). ⇒ M / t ( M ) corresponds to the parametrizable subsystem   φ 1 ( t ) − φ 2 ( t ) = 0 ,  − φ 2 ( t − 2) − φ 2 ( t ) + 2 ˙ φ 3 ( t − 1) = 0 .

System Module Homological Algebra autonomous ext 1 D ( M T , D ) � = 0 t ( M ) � = 0 elements controllable, ext 1 D ( M T , D ) = 0 t ( M ) = 0 parametrizable ext i D ( M T , D ) = 0 , parametrization reflexive is parametrizable i = 1 , 2 . . . . . . . . . ext i D ( M T , D ) = 0 , . . . projective 1 ≤ i ≤ gld( D ) flatness free . . . Contributions to this classification: Pommaret-Quadrat, Oberst, Fliess.

Presentations M = D 1 × p / D 1 × q R = D 1 × q ′ R ′ / D 1 × q R t ( M ) ∼ R = R ′′ R ′ , R ′′ ∈ D q × q ′ ker( . R ′ ) = D 1 × r ′ R 2 = D 1 × p / D 1 × q ′ R ′ M / t ( M ) ∼ � R ′′ � t ( M ) ∼ = D 1 × q ′ / D 1 × ( q + r ′ ) R 2

Block-triangular presentation � � R ′′ = D 1 × p / D 1 × q ′ R ′ t ( M ) ∼ M / t ( M ) ∼ = D 1 × q ′ / D 1 × ( q + r ′ ) , R ′ 2 Then, 0 → t ( M ) → M → M / t ( M ) → 0 and   R ′ − I q ′ 0 → t ( M ) → D 1 × ( p + q ′ ) / D 1 × ( q ′ + q + r ′ )   → M / t ( M ) → 0 R ′′ 0 R ′ 0 2 are equivalent extensions. Can integrate R y = 0 in cascade:  R ′ ζ = η,    R ′′ η R y = 0 ⇐ ⇒ = 0 ,    R ′ 2 η = 0

3. Injective parametrizations

Injective parametrizations In general, we do not get an injective parametrization for Sol F ( M ). Moreover:   x 2 ( t ) − u 2 ( t ) ˙ = 0 ,  x 1 ( t ) − sin( t ) u 1 ( t ) ˙ = 0 has an injective parametrization   u 1 ( t ) = x 1 ( t ) / sin( t ) , ˙  u 2 ( t ) = x 2 ( t ) , ˙ but it is singular at t = 0.

Injective parametrizations Sol F ( M ) has an injective parametrization ⇐ ⇒ M is free Flatness (Fliess, L´ evine, Martin, Rouchon et al., Pomet, . . . ) Hence, we study how to compute bases of free left D -modules. Remark ∃ R ∈ D q × p , M is free of rank p − q ⇐ ⇒ U ∈ GL( p , D ) M ∼ = D 1 × p / D 1 × q R . s.t. R U = ( I q 0) and Theorem (Quillen-Suslin; Stafford) ◮ D = K [ ∂ 1 , . . . , ∂ n ]: projective ⇒ free ◮ D = Weyl algebra (char. 0): stably free of rank ≥ 2 ⇒ free

Computation of bases of free left A n -modules − D 1 × q ← . R − D 1 × p 0 ← − M ← ← − 0 split exact, p − q ≥ 2 . ˜ → D 1 × q − R → ker( . ˜ → D 1 × p 0 − R ) − − → 0 (apply involution of D ) find E ∈ GL( p , D ) such that   R )) = D 1 × ( p − q ) (0 ker( . ( E ˜ 1 ⋆ . . . ⋆ ⇒ I p − q )  .  .   0 1 ⋆ . R ) = D 1 × ( p − q ) (0 ker( . ˜   ⇒ I p − q ) E   0 0 ... ⋆     Q := � E ˜   I p − q ) T , 0 0 ... 1 define E (0 R =       0 0 . . . 0 − D 1 × q ← . Q . R   − D 1 × ( p − q ) − D 1 × p 0 ← ← ← − 0   . . . . . . . . .   . . . I p − q ) � basis of M : the rows of (0 E − 1 mod R 0 0 . . . 0

Stafford’s Theorem D = A n ( k ) = k [ x 1 , . . . , x n ][ ∂ 1 , . . . , ∂ n ], where Q ⊆ k Theorem (Stafford, 1978) For any a , b , c ∈ D there exist λ, µ ∈ D such that D a + D b + D c = D ( a + λ c ) + D ( b + µ c ) . Example. n = 3, D = k [ x 1 , x 2 , x 3 ][ ∂ 1 , ∂ 2 , ∂ 3 ] For a = ∂ 1 , b = ∂ 2 , c = ∂ 3 may choose λ = 0, µ = x 1 : ( − x 1 ∂ 3 − ∂ 2 ) ( a + 0 · c ) + ∂ 1 ( b + x 1 · c ) = c

Stafford’s Theorem – algorithmic! D = A n ( k ) = k [ x 1 , . . . , x n ][ ∂ 1 , . . . , ∂ n ], where Q ⊆ k Theorem (Stafford) For any a , b , c ∈ D there exist λ, µ ∈ D such that D a + D b + D c = D ( a + λ c ) + D ( b + µ c ) . ◮ Hillebrand & Schmale (2001) ◮ Leykin (2004) � Maple package Stafford developed for OreModules