Analytic Left Inversion of SISO Lotka-Volterra Models Luis A. - PowerPoint PPT Presentation

Analytic Left Inversion of SISO Lotka-Volterra Models † Luis A. Duffaut Espinosa George Mason University Joint work with W. Steven Gray (ODU) and K. Ebrahimi-Fard (ICMAT) † This research was supported by the BBVA Foundation Grant to Researchers, Innovators and Cultural Creators (Spain).

ICMAT Overview 1. Introduction 2. Preliminaries on Fliess Operators and Their Inverses 3. Left System Inversion of Lotka-Volterra Input-Output Systems 4. Numerical Simulations 5. Conclusions 2

ICMAT 1. Introduction • Given an operator F , describing the dynamics of a system, and a function y in its range, the left inversion problem (LIP) is to determine a unique input u such that y = F [ u ]. • A sufficient condition (but not necessary) for solving this problem in a state space setting is to have well defined relative degree (Isidori, 1995). • The solution of LIP does not require a state space realization. • Fliess operators provide an explicit analytical solution. • Introduced by M. Fliess in 1983, Fliess operators are analytic input-output systems described by coefficients and iterated integrals of the inputs. • Fliess operators can be viewed as a functional generalization of a Taylor series. For example, any Volterra operator with analytic kernels has a Fliess operator representation. 3

ICMAT 2. Preliminaries Population models: • Here we apply the method to the population dynamical system: n � z i = β i z i + ˙ α ij z i z j , i = 1 , 2 , . . . , n, (Lotka-Volterra model) j =1 where z i ∝ to the i -th population, β i is the growth rate for the i -th population, and α ij weights the effect of the j -th species on the i -th species. • Input-output models are obtained by introducing time dependence on the β i ( t )’s or α ij ( t )’s (inputs), and assuming y = h ( z ) (outputs). • For n = 2,      ˙ z 1 β 1 z 1 − α 12 z 1 z 2  = (Predator-Prey model)   − β 2 z 2 + α 21 z 1 z 2 z 2 ˙ The vector fields are complete within the first quadrant giving concentric periodic trajectories about z e = ( β 2 /α 21 , β 1 /α 12 ). 4

ICMAT Fliess Operators: • Let X = { x 0 , x 1 , . . . , x m } be an alphabet and X ∗ the set of all words over X (including the empty word ∅ ). • A formal power series is any mapping c : X ∗ → R ℓ . Typically, c is written as a formal sum c = � η ∈ X ∗ ( c, η ) η , and the set of all such series is R ℓ �� X �� . • For a measurable function u : [ a, b ] → R m with finite L 1 -norm, define E η : L m 1 [ t 0 , t 0 + T ] → C [ t 0 , t 0 + T ] by E ∅ [ u ] = 1, and t � E x i η ′ [ u ]( t, t 0 ) = u i ( τ ) E η ′ [ u ]( τ, t 0 ) dτ, t 0 where x i ∈ X , η ′ ∈ X ∗ and u 0 � 1. • Note that to each letter x i has been assigned a function u i . • For each c ∈ R ℓ �� X �� ⇒ F c [ u ]( t ) = � η ∈ X ∗ ( c, η ) E η [ u ]( t, t 0 ), which is called a Fliess operator (Fliess, 1983). 5

ICMAT Fliess Operator Inverses: F c ⇒ F c F d = F c ⊔ ⊔ d , where ⊔ denotes the ⊔ u y × shuffle product. F d Fig. 2.1: Product connection. ⇒ F c ◦ F d = F c ◦ d , where c ◦ d denotes v the composition product of c ∈ R ℓ �� X �� u F d F c y and d ∈ R m �� X �� (Gray et al., 2014) Fig. 2.2: Cascade connection. Given c, d ∈ R m �� X �� , y satisfies v u F c y y = F c [ v ] = F c [ u + F d [ y ]] . F d If there exists e so that y = F e [ u ], then Fig. 2.3: Feedback connection. F e [ u ] = F c [ u + F d ◦ e [ u ]] . (contraction!) 6

ICMAT ✞ ☎ On the other hand, v = u + F d ◦ c [ v ] ⇒ ( I + F − d ◦ c ) [ v ] = u. ✝ ✆ Apply the compositional inverse to both sides of this equation: v = ( I + F − d ◦ c ) − 1 [ u ] := � � I + F ( − d ◦ c ) ◦− 1 [ u ] . In which case, F c @ d [ u ] = F c [ v ] = F c ◦ ( δ − d ◦ c ) ◦− 1 [ u ] , (explicit formula!) or equivalently, c @ d = c ◦ ( δ − d ◦ c ) ◦− 1 , where F δ := I . The set of operators F δ = { I + F c : c ∈ R �� X ��} , forms a group under composition, in particular, a Fa` a di Bruno Hopf algebra with antipode, α , satisfying ( δ + c ) ◦− 1 := δ + c ◦− 1 = δ + � ( α a η )( c ) η, η ∈ X ∗ where c ◦− 1 denotes the composition inverse of c and a η : R �� X �� → R : c �→ ( c, η ) . Remark: The antipode has an explicit series representation (Gray & Duffaut Espinosa, 2011, 2014). 7

ICMAT Now observe that any c ∈ R �� X �� can be written as c = c N + c F , where k ≥ 0 ( c, x k 0 ) x k c N := � 0 and c F := c − c N . Definition 2.1: Given c ∈ R �� X �� , let r ≥ 1 be the largest integer such that supp( c F ) ⊆ x r − 1 X ∗ . Then c has relative degree r if the linear word 0 x r − 1 x 1 ∈ supp( c ), otherwise it is not well defined. 0 Remark: This definition coincides with the usual definition of relative degree given in a state space setting. But this definition is independent of the state space setting. Definition 2.2: Given ξ ∈ X ∗ , the corresponding left-shift operator is � η ′ η = ξη ′ : ξ − 1 : X ∗ → R � X � : η �→ 0 : otherwise . ⊔ − 1 , Remark: The operation F c /F d = F c/d is given by c/d := c ⊔ ⊔ d ⊔ ∞ ⊔ − 1 := ( c, ∅ ) − 1 ⊔ k , and c ′ = 1 − c/ ( c, ∅ ) is proper. � ( c ′ ) ⊔ where c ⊔ k =0 8

ICMAT  v − F ( x r 0 ) − 1 ( c ) [ u ] ( y ( r ) = v ) y = F c [ u ] u =    F ( x r − 1 x 1 ) − 1 ( c ) [ u ] y (1) = F x − 1   ( c ) [ u ]  0   0 .  − F ( x r 0 ) − 1 ( c − c y ) [ u ] .   .  x 1 ) − 1 ( c ) [ u ] = − F d [ u ] , = y ( r − 1) = F ( x r − 1 F ( x r − 1 ) − 1 ( c ) [ u ] ☛ ✟ 0   0  y ( r ) = F ( x r  d = ( x r 0 ) − 1 ( c − c y ) 0 ) − 1 ( c ) [ u ]    x 1 ) − 1 ( c ) .   ( x r − 1  + u F ( x r − 1 x 1 ) − 1 ( c ) [ u ] .  ✡ ✠ 0  0 Theorem 2.3: Suppose c ∈ R �� X �� has relative degree r . Let y be analytic at t = 0 with generating series c y ∈ R LC [[ X 0 ]] satisfying ( c y , x k 0 ) = ( c, x k 0 ), k = 0 , . . . , r − 1. (Here X 0 := { x 0 } .) Then the input ∞ 0 ) t k � ( c u , x k u ( t ) = k ! , k =0 0 ) − 1 ( c − c y ) / ( x r − 1 x 1 ) − 1 ( c )) ◦− 1 , is the unique solution to where c u = (( x r 0 F c [ u ] = y on [0 , T ] for some T > 0. Remark: The condition on c y ensures that y is in the range of F c . 9

ICMAT 3. Left System Inversion of LV Input-Output Systems Four SISO predator-prey systems with output y = z 1 (prey): I/0 map state space realization rel. degree � � � � − α 12 z 1 z 2 z 1 F c : β 1 �→ y g 0 ( z ) = , g 1 ( z ) = 1 − β 2 z 2 + α 21 z 1 z 2 0 � � � � β 1 z 1 − z 1 z 2 F c : α 12 �→ y g 0 ( z ) = , g 1 ( z ) = 1 − β 2 z 2 + α 21 z 1 z 2 0 � � � � 0 β 1 z 1 − α 12 z 1 z 2 F c : β 2 �→ y g 0 ( z ) = , g 1 ( z ) = 2 α 21 z 1 z 2 − z 2 � � � � 0 β 1 z 1 − α 12 z 1 z 2 F c : α 22 �→ y g 0 ( z ) = , g 1 ( z ) = 2 − β 2 z 2 z 1 z 2 10

ICMAT The population system under study:        ˙ z 1 − α 12 z 1 z 2  z 1  =  u,  + y = z 1  z 2 ˙ − β 2 z 2 + α 21 z 1 z 2 0 with z 1 (0) = z 1 , 0 and z 2 (0) = z 2 , 0 . Make α 12 = α 21 = β 2 = 1. 4 Initial orbit 4.5 Final orbit Vector field 3.5 Vector field 4 3 3.5 Predator population Predator population 3 2.5 2.5 2 2 1.5 • ← Equilibrium (1 , 1 . 5) 1.5 1 1 • ← Equilibrium (1 , 1) 0.5 0.5 0 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 0 0.5 1 1.5 2 2.5 3 Prey population Prey population Fig. 3.1: u = 1. Fig. 3.2: u = 1 . 5. c = z 1 , 0 − α 12 z 2 , 0 z 1 , 0 x 0 + z 1 , 0 ✄ � 2 z 2 , 0 2 z 1 , 0 − α 12 α 21 z 2 , 0 z 1 , 0 2 � x 1 + α 12 ✂ ✁ + α 12 β 2 z 2 , 0 z 1 , 0 ) x 2 0 − α 12 z 2 , 0 z 1 , 0 x 0 x 1 − α 12 z 2 , 0 z 1 , 0 x 1 x 0 + z 1 , 0 x 2 1 + · · · . ✞ ☎ Relative degree r = 1. ✝ ✆ 11

ICMAT One must select an output function ∞ 0 ) t k � ( c y , x k y ( t ) = k ! , k =0 where c y is the generating series of y . It is sufficient to consider a polynomial of degree 3, so let ( c y , ∅ ) = v 0 , ( c y , x i 0 ) = v i for i = 1 , 2 , 3. Thus, 0 ) − 1 ( c − c y ) d := ( x r � x 1 ) − 1 ( c ) = − α 12 z 2 , 0 − v 1 α 12 β 2 z 2 , 0 − v 2 z 1 , 0 + ( x r − 1 z 1 , 0 0 � − α 12 α 21 z 1 , 0 z 2 , 0 − v 1 α 12 z 2 , 0 x 0 + v 1 z 1 , 0 x 1 z 1 , 0 2 z 2 , 0 2 � − v 1 α 12 + v 1 α 12 β 2 z 2 , 0 + − v 1 α 12 α 21 z 2 , 0 z 1 , 0 z 1 , 0 − 2 v 2 α 12 z 2 , 0 2 + 2 α 12 β 2 α 21 z 1 , 0 z 2 , 0 2 α 21 z 1 , 0 z 2 , 0 + α 12 z 1 , 0 � − v 3 2 z 2 , 0 − α 12 α 21 2 z 1 , 0 2 z 2 , 0 x 2 z 1 , 0 − α 12 β 2 0 + · · · 12

ICMAT In which case, � �  − v 1 v 1 z 1 , 0 − α 12 z 2 , 0 N = v 1 + v 1 α 12 z 2 , 0 d ◦− 1 � � c u = z 1 , 0 + α 12 z 2 , 0 +  z 1 , 0 z 1 , 0 � + v 2 z 1 , 0 − α 12 β 2 z 2 , 0 + α 12 α 21 z 1 , 0 z 2 , 0 x 0 + · · · Design example: Given ◮ [ t 1 , t 2 ] = [12 . 5 , 12 . 7] (∆ t = 0 . 2), ◮ u ( t 1 ) = 1, u ( t 2 ) = 1 . 5, ◮ initial orbit exit point [ z 1 (12 . 5) , z 2 (12 . 5)] T = [3 . 82 , 2 . 25] T , ◮ y ( t 2 ) = 2, find a smooth u ( t ) for t ∈ ( t 1 , t 2 ) so that all constraints are satisfied. 13