Solvability of Matrix-Exponential Equations Joël Ouaknine, Amaury Pouly, João Sousa-Pinto, James Worrell University of Oxford July 8, 2016
Related work in the discrete case Input: A , C ∈ Q d × d matrices Output: ∃ n ∈ N such that A n = C ? Example: ∃ n ∈ N such that � n � � � 1 1 1 100 = ? 0 1 0 1
Related work in the discrete case Input: A , C ∈ Q d × d matrices Output: ∃ n ∈ N such that A n = C ? � Decidable (PTIME) Example: ∃ n ∈ N such that � n � � � 1 1 1 100 = ? 0 1 0 1
Related work in the discrete case Input: A , C ∈ Q d × d matrices Output: ∃ n ∈ N such that A n = C ? � Decidable (PTIME) Input: A , B , C ∈ Q d × d matrices Output: ∃ n , m ∈ N such that A n B m = C ? Example: ∃ n , m ∈ N such that � n � 1 � m � � � 1 2 3 1 60 2 2 = ? 0 1 0 1 0 1
Related work in the discrete case Input: A , C ∈ Q d × d matrices Output: ∃ n ∈ N such that A n = C ? � Decidable (PTIME) Input: A , B , C ∈ Q d × d matrices Output: ∃ n , m ∈ N such that A n B m = C ? � Decidable Example: ∃ n , m ∈ N such that � n � 1 � m � � � 1 2 3 1 60 2 2 = ? 0 1 0 1 0 1
Related work in the discrete case Input: A , C ∈ Q d × d matrices Output: ∃ n ∈ N such that A n = C ? � Decidable (PTIME) Input: A , B , C ∈ Q d × d matrices Output: ∃ n , m ∈ N such that A n B m = C ? � Decidable Input: A 1 , . . . , A k , C ∈ Q d × d matrices i =1 A n i Output: ∃ n 1 , . . . , n k ∈ N such that � k i = C ? Example: ∃ n , m , p ∈ N such that � n � 1 � m � � p � � � 1 2 3 2 5 81 260 2 2 = ? 0 1 0 1 0 1 0 1
Related work in the discrete case Input: A , C ∈ Q d × d matrices Output: ∃ n ∈ N such that A n = C ? � Decidable (PTIME) Input: A , B , C ∈ Q d × d matrices Output: ∃ n , m ∈ N such that A n B m = C ? � Decidable Input: A 1 , . . . , A k , C ∈ Q d × d matrices i =1 A n i Output: ∃ n 1 , . . . , n k ∈ N such that � k i = C ? � Decidable if A i commute × Undecidable in general Example: ∃ n , m , p ∈ N such that � n � 1 � m � � p � � � 1 2 3 2 5 81 260 2 2 = ? 0 1 0 1 0 1 0 1
Related work in the discrete case Input: A , C ∈ Q d × d matrices Output: ∃ n ∈ N such that A n = C ? � Decidable (PTIME) Input: A , B , C ∈ Q d × d matrices Output: ∃ n , m ∈ N such that A n B m = C ? � Decidable Input: A 1 , . . . , A k , C ∈ Q d × d matrices i =1 A n i Output: ∃ n 1 , . . . , n k ∈ N such that � k i = C ? � Decidable if A i commute × Undecidable in general Input: A 1 , . . . , A k , C ∈ Q d × d matrices Output: C ∈ � semi-group generated by A 1 , . . . , A k � ? Semi-group: � A 1 , . . . , A k � = all finite products of A 1 , . . . , A k Examples: A 8 3 A 2 A 3 1 A 42 A 1 A 3 A 2 A 1 A 2 A 1 A 2 3
Related work in the discrete case Input: A , C ∈ Q d × d matrices Output: ∃ n ∈ N such that A n = C ? � Decidable (PTIME) Input: A , B , C ∈ Q d × d matrices Output: ∃ n , m ∈ N such that A n B m = C ? � Decidable Input: A 1 , . . . , A k , C ∈ Q d × d matrices i =1 A n i Output: ∃ n 1 , . . . , n k ∈ N such that � k i = C ? � Decidable if A i commute × Undecidable in general Input: A 1 , . . . , A k , C ∈ Q d × d matrices Output: C ∈ � semi-group generated by A 1 , . . . , A k � ? � Decidable if A i commute × Undecidable in general Semi-group: � A 1 , . . . , A k � = all finite products of A 1 , . . . , A k Examples: A 8 3 A 2 A 3 1 A 42 A 1 A 3 A 2 A 1 A 2 A 1 A 2 3
Hybrid/Cyber-physical systems ◮ physics: continuous dynamics ◮ electronics: discrete states state continuous dynamics guard φ ( x ) x ′ = F 1 ( x ) x ′ = F 2 ( x ) x ← R ( x ) discrete update
Hybrid/Cyber-physical systems ◮ physics: continuous dynamics ◮ electronics: discrete states state continuous dynamics guard φ ( x ) x ′ = F 1 ( x ) x ′ = F 2 ( x ) x ← R ( x ) discrete update Some classes: Typical questions ◮ F i ( x ) = 1: timed automata ◮ reachability ◮ F i ( x ) = c i : rectangular hybrid automata ◮ safety ◮ F i ( x ) = A i x : linear hybrid automata ◮ controllability
Recap on linear differential equations Let x : R + → R n function, A ∈ Q n × n matrix x 1 ( t ) a 11 · · · a 1 n . . . ... . . . x ( t ) = A = . . . x n ( t ) · · · a n 1 a nn Linear differential equation: x ′ ( t ) = Ax ( t ) x (0) = x 0
Recap on linear differential equations Let x : R + → R n function, A ∈ Q n × n matrix x 1 ( t ) a 11 · · · a 1 n . . . ... . . . x ( t ) = A = . . . x n ( t ) · · · a n 1 a nn Linear differential equation: x ′ ( t ) = Ax ( t ) x (0) = x 0 Examples: � x ′ 1 ( t )= x 2 ( t ) x ′ ( t ) = 7 x ( t ) x ′ 2 ( t )= − x 1 ( t ) ❀ x ( t ) = e 7 t � x 1 ( t )= sin( t ) ❀ x 2 ( t )= cos( t )
Recap on linear differential equations Let x : R + → R n function, A ∈ Q n × n matrix x 1 ( t ) a 11 · · · a 1 n . . . ... . . . x ( t ) = A = . . . x n ( t ) · · · a n 1 a nn Linear differential equation: x ′ ( t ) = Ax ( t ) x (0) = x 0 Examples: � ′ � � � � � � x ′ 1 ( t )= x 2 ( t ) 0 1 x 1 x 1 x ′ ( t ) = 7 x ( t ) ⇔ = x ′ 2 ( t )= − x 1 ( t ) x 2 − 1 0 x 2 ❀ x ( t ) = e 7 t � x 1 ( t )= sin( t ) ❀ x 2 ( t )= cos( t )
Recap on linear differential equations Let x : R + → R n function, A ∈ Q n × n matrix x 1 ( t ) a 11 · · · a 1 n . . . ... . . . x ( t ) = A = . . . x n ( t ) · · · a n 1 a nn Linear differential equation: x ′ ( t ) = Ax ( t ) x (0) = x 0 General solution form: x ( t ) = exp( At ) x 0 ∞ M n � where exp( M ) = n ! n =0
Switching system x ′ = A 1 x x ′ = A 2 x Restricted hybrid system: ◮ linear dynamics ◮ no guards (nondeterministic) x ′ = A 4 x x ′ = A 3 x ◮ no discrete updates switch x 1 ( t ) t x ′ = A 1 x x ′ = A 2 x x ′ = A 3 x x ′ = A 4 x t 1 t 2 t 3 t 4
Switching system x ′ = A 1 x x ′ = A 2 x Restricted hybrid system: ◮ linear dynamics ◮ no guards (nondeterministic) x ′ = A 4 x x ′ = A 3 x ◮ no discrete updates switch x 1 ( t ) t x ′ = A 1 x x ′ = A 2 x x ′ = A 3 x x ′ = A 4 x t 1 t 2 t 3 t 4 Dynamics: e A 4 t 4 e A 3 t 3 e A 2 t 2 e A 1 t 1
Switching system x ′ = A 1 x x ′ = A 2 x Restricted hybrid system: ◮ linear dynamics ◮ no guards (nondeterministic) x ′ = A 4 x x ′ = A 3 x ◮ no discrete updates switch x 1 ( t ) t x ′ = A 1 x x ′ = A 2 x x ′ = A 3 x x ′ = A 4 x t 1 t 2 t 3 t 4 Problem: e A 4 t 4 e A 3 t 3 e A 2 t 2 e A 1 t 1 = C ? What we control: t 1 , t 2 , t 3 , t 4 ∈ R +
Switching system x ′ = A 1 x x ′ = A 2 x What about a loop ? x ′ = A 4 x x ′ = A 3 x
Switching system x ′ = A 1 x x ′ = A 2 x What about a loop ? x ′ = A 4 x x ′ = A 3 x x 1 ( t ) A 1 A 2 A 3 A 4 A 1 A 2 A 3 A 4 t t 1 t 2 t 3 t 4 t ′ t ′ t ′ t ′ 1 2 3 4 Dynamics: e A 4 t ′ 4 e A 3 t ′ 3 e A 2 t ′ 2 e A 1 t ′ 1 e A 4 t 4 e A 3 t 3 e A 2 t 2 e A 1 t 1
Switching system x ′ = A 1 x x ′ = A 2 x Loop ⇔ clique x ′ = A 4 x x ′ = A 3 x x 1 ( t ) A 1 A 4 A 3 A 2 t t 1 t 4 t 3 t 2 t 2 = t 3 =0 t 1 = t 2 =0 t 4 = t 1 =0 Remark: zero time dynamics ( t i = 0) are allowed
Switching system x ′ = A 1 x x ′ = A 2 x x ′ = A 4 x x ′ = A 3 x x 1 ( t ) A 1 A 4 A 3 A 2 t t 1 t 4 t 3 t 2 Dynamics: any finite product of e A i t semigroup! ❀
Switching system x ′ = A 1 x x ′ = A 2 x x ′ = A 4 x x ′ = A 3 x x 1 ( t ) A 1 A 4 A 3 A 2 t t 1 t 4 t 3 t 2 Problem: C ∈ G ? where G = � semi-group generated by e A i t for all t � 0 �
Main results Input: A 1 , . . . , A k , C ∈ Q d × d matrices Output: ∃ t 1 , . . . , t k � 0 such that n e A i t i = C � ? i =1 Input: A 1 , . . . , A k , C ∈ Q d × d matrices Output: C ∈ � semigroup generated by e A 1 t , . . . , e A k t : t � 0 � ? Theorem Both problems are: ◮ Undecidable in general ◮ Decidable when all the A i commute
Some words about the proof (commuting case) Product Problem Semigroup Problem equivalent ∃ t 1 , . . . , t k � 0 s.t. C ∈ � e A 1 t , . . . , e A k t : t � 0 � ? i =1 e A i t i = C � n ? reduce Integer Linear Programming ∃ n ∈ Z d s.t. π Bn � s
Some words about the proof (commuting case) Product Problem Semigroup Problem equivalent ∃ t 1 , . . . , t k � 0 s.t. C ∈ � e A 1 t , . . . , e A k t : t � 0 � ? i =1 e A i t i = C � n ? reduce ! s of the form: a 0 + log( a 1 ) + · · · + log( a k ) Integer Linear Programming ∃ n ∈ Z d s.t. π Bn � s � B , a 0 , . . . , a k are algebraic
Some words about the proof (commuting case) Product Problem Semigroup Problem equivalent ∃ t 1 , . . . , t k � 0 s.t. C ∈ � e A 1 t , . . . , e A k t : t � 0 � ? i =1 e A i t i = C � n ? reduce ! s of the form: a 0 + log( a 1 ) + · · · + log( a k ) Integer Linear Programming ∃ n ∈ Z d s.t. π Bn � s � B , a 0 , . . . , a k are algebraic How did we get from reals to integers with π ? e it = α ⇔ t ∈ log( α ) + 2 π Z
Recommend
More recommend