Formal solution Chen-Fliess series u a ( t ) a , S = S ( t ) S - - PowerPoint PPT Presentation

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Formal solution Chen-Fliess series u a ( t ) a , S = S ( t ) S - - PowerPoint PPT Presentation

= Title Intro: Control Series expansions Exp-prod, H Sinusoids Feedback Outlook 1 Formal solution Chen-Fliess series u a ( t ) a , S = S ( t ) S ( 0 ) = 1 | Z | = m a Z on algebra A ( Z ) of formal


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SLIDE 1

Title Intro: Control Series expansions Exp-prod, ξH Sinusoids Feedback Outlook ζ = π′

1 ◦ ξ

Formal solution – Chen-Fliess series

˙ S = S(t) ·

  • a∈Z

ua(t) a, S(0) = 1 |Z| = m

  • n algebra ˆ

A(Z) of formal power series in aset Z of noncommuting indeterminates (letters) has the unique solution CF(T, u) =

  • w∈Z ∗

T t1 · · · tp−1 uap(tp) . . . ua1(t1) dt1 . . . dtp

  • Υw(T,u)

a1 . . . ap

  • =w

Use as asymptotic expansion for evolution of output y = ϕ(x) along solution of ˙ x =

a∈Z uafa(x).

ϕ(x(T, u)) ∼

  • w∈Z ∗

T t1 · · · tp−1 uap(tp). . .ua1(t1) dt1 . . . dtp (fa1 . . . aipϕ)(

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Title Intro: Control Series expansions Exp-prod, ξH Sinusoids Feedback Outlook ζ = π′

1 ◦ ξ

Series solution by iteration

φ(x(t, u)) = 1 · φ(x0) + t

0ua(s)ds (faφ)(x0)

+ t

0ub(s)ds (fbφ)(x0)

+ 1

2

t s1 s1

0 ua(s1)ua(s2)ds2ds1 (fafaφ)(x0)

+ 1

2

t s1

0 ua(s1)ub(s2)ds2ds1 (fafbφ)(x0)

+ 1

2

t s1

0 ub(s1)ua(s2)ds2ds1 (fbfaφ)(x0)

+ 1

2

t s1

0 ub(s1)ub(s2)ds2ds1 (fbfbφ)(x0)

+ 1

6

t s1 s2

0 ua(s1)ua(s2)ua(s3)ds3ds2ds1 (fafafaφ)(x0)

+ . . . Objective: Collect first order differential operators, and minimize number of higher order differential operators involved

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Title Intro: Control Series expansions Exp-prod, ξH Sinusoids Feedback Outlook ζ = π′

1 ◦ ξ

Integrate by parts: The wrong way to do it

t s1

0 ua(s1)ub(s2)ds2ds1 fafb+

t s1

0 ub(s1)ua(s2)ds2ds1 fbfa =

= t s1

0 ua(s1)ub(s2)ds2ds1 fafb −

t s1

0 ua(s1)ub(s2)ds2ds1 fbfa

+ t s1

0 ua(s1)ub(s2)ds2ds1 fbfa +

t s1

0 ub(s1)ua(s2)ds2ds1 fbfa

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SLIDE 4

Title Intro: Control Series expansions Exp-prod, ξH Sinusoids Feedback Outlook ζ = π′

1 ◦ ξ

Integrate by parts: The wrong way to do it

t s1

0 ua(s1)ub(s2)ds2ds1 fafb+

t s1

0 ub(s1)ua(s2)ds2ds1 fbfa =

= t s1

0 ua(s1)ub(s2)ds2ds1 fafb −

t s1

0 ua(s1)ub(s2)ds2ds1 fbfa

+ t s1

0 ua(s1)ub(s2)ds2ds1 fbfa +

t s1

0 ub(s1)ua(s2)ds2ds1 fbfa

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SLIDE 5

Title Intro: Control Series expansions Exp-prod, ξH Sinusoids Feedback Outlook ζ = π′

1 ◦ ξ

Integrate by parts: The wrong way to do it

t s1

0 ua(s1)ub(s2)ds2ds1 fafb+

t s1

0 ub(s1)ua(s2)ds2ds1 fbfa =

= t s1

0 ua(s1)ub(s2)ds2ds1 fafb −

t s1

0 ua(s1)ub(s2)ds2ds1 fbfa

+ t s1

0 ua(s1)ub(s2)ds2ds1 fbfa +

t s1

0 ub(s1)ua(s2)ds2ds1 fbfa

= t s1

0 ua(s1)ub(s2)ds2ds1 (fafb − fbfa)

+ t

  • ua(s1)

s1

0 ub(s2) + ub(s1)

s1

0 ua(s2)ds2

  • ds1 fbfa
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SLIDE 6

Title Intro: Control Series expansions Exp-prod, ξH Sinusoids Feedback Outlook ζ = π′

1 ◦ ξ

Integrate by parts: The wrong way to do it

t s1

0 ua(s1)ub(s2)ds2ds1 fafb+

t s1

0 ub(s1)ua(s2)ds2ds1 fbfa =

= t s1

0 ua(s1)ub(s2)ds2ds1 fafb −

t s1

0 ua(s1)ub(s2)ds2ds1 fbfa

+ t s1

0 ua(s1)ub(s2)ds2ds1 fbfa +

t s1

0 ub(s1)ua(s2)ds2ds1 fbfa

= t s1

0 ua(s1)ub(s2)ds2ds1 (fafb − fbfa)

+ t

  • ua(s1)

s1

0 ub(s2) + ub(s1)

s1

0 ua(s2)ds2

  • ds1 fbfa

= t s1

0 ua(s1)ub(s2)ds2ds1 (fafb − fbfa)

+ t

0ua(s) ds

  • ·

t

0ub(s) ds

  • fbfa
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SLIDE 7

Title Intro: Control Series expansions Exp-prod, ξH Sinusoids Feedback Outlook ζ = π′

1 ◦ ξ

Integrate by parts: The wrong way to do it

t s1

0 ua(s1)ub(s2)ds2ds1 fafb+

t s1

0 ub(s1)ua(s2)ds2ds1 fbfa =

= t s1

0 ua(s1)ub(s2)ds2ds1 fafb −

t s1

0 ua(s1)ub(s2)ds2ds1 fbfa

+ t s1

0 ua(s1)ub(s2)ds2ds1 fbfa +

t s1

0 ub(s1)ua(s2)ds2ds1 fbfa

= t s1

0 ua(s1)ub(s2)ds2ds1 (fafb − fbfa)

+ t

  • ua(s1)

s1

0 ub(s2) + ub(s1)

s1

0 ua(s2)ds2

  • ds1 fbfa

= t s1

0 ua(s1)ub(s2)ds2ds1 (fafb − fbfa)

+ t

0ua(s) ds

  • ·

t

0ub(s) ds

  • fbfa

Lie brackets together w/ iterated integrals in right order higher order deriv’s (wrong order) w/ pointwise prod’s of int’s

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Title Intro: Control Series expansions Exp-prod, ξH Sinusoids Feedback Outlook ζ = π′

1 ◦ ξ

Integrate by parts, smart way

Do not manipulate iterated integrals and iterated Lie brackets of vector fields by hand – work on level of “words” (their indices)

  • I∈{a,b}∗ I ⊗ I =

1 ⊗ 1 + a ⊗ a + b ⊗ b + 1

2

aa ⊗ aa + 1

2

ab ⊗ ab + 1

2

ba ⊗ ba + 1

2

bb ⊗ bb + 1

6

aaa ⊗ aaa + . . . = 1 ⊗ 1 + a ⊗ a + b ⊗ b + 1

2

aa ⊗ aa + 1

2

ab ⊗ (ab − ba) + 1

2

(ab + ba) ⊗ ba + 1

2

bb ⊗ bb + 1

6

aaa ⊗ aaa + . . .

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SLIDE 9

Title Intro: Control Series expansions Exp-prod, ξH Sinusoids Feedback Outlook ζ = π′

1 ◦ ξ

Drop everything except the indices - maps

  • The iterated integral

Υa1a2...an = t t1 · · · tn−1 ua1(t1)ua2(t2) · · · uan(tn) dtn dtn−1 · · · dt1 is uniquely identified by the multi-index (“word”) a1a2 . . . an

  • The n-th order partial differential operator fan ◦ fan−1 ◦ . . . fa1

is uniquely identified by the multi-index (“word”) a1a2 . . . an

  • The Chen series is identified with the identity map on free

associative algebra A(Z) over set Z of with |Z| = m CF ∼ IdA(Z) =

  • n≥0
  • w∈Z n

w ⊗ w ∈ ˆ A(Z) ⊗ A(Z) with shuffle product on left and concatenation on right

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Title Intro: Control Series expansions Exp-prod, ξH Sinusoids Feedback Outlook ζ = π′

1 ◦ ξ

Homomorphisms I

  • For fixed smooth vector fields fi

F : A(Z) → partial diff operators on C∞(M) F : (a1a2 . . . an) → fa1 ◦ fa2 ◦ . . . fan associative algebras: concatenation → composition

  • For fixed control u ∈ UZ

Υ(u): A(Z) → AC([0, T], R) Υ(u): (a1a2 . . . an) → T t1 · · · tp−1 uap(tp) . . . ua1(t1) dt1 . . . dtp associative algebras (Ree’s theorem): shuffle of words → pointwise multiplication of functions

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Title Intro: Control Series expansions Exp-prod, ξH Sinusoids Feedback Outlook ζ = π′

1 ◦ ξ

Recall: definition of the shuffle SKIP

Combinatorially: for words w, z ∈ Z ∗ and letters a, b ∈ Z ( w a ) X ( z b ) = (( w a ) X z ) b + ( w X ( z b )) a Example: (ab) X (cd) = a b c d + a c b d + c a b d + a c d b + c a d b + c d a b Algebraically: transpose of the coproduct ∆ < v X w , z > = < v ⊗ w , ∆(z) > where ∆: A(Z) → A(Z) ⊗ A(Z) by ∆(a) = 1⊗a+a⊗1 for a ∈ Z

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SLIDE 12

Title Intro: Control Series expansions Exp-prod, ξH Sinusoids Feedback Outlook ζ = π′

1 ◦ ξ

Shuffles and simplices SKIP

On permutations algebras Duchamp and Agrachev consider partially commutative and noncommutative shuffles. Illustration:

✲ ✲ ✲ ✻ ✻ ✻

  • σ1x 2

σ1 σ2 = σ21 ∪ σ12

= ∪ ∪

σ(12)x 3 = σ312 ∪ σ132 ∪

E.g. σ(12)x 3 = {t : 0 ≤ t1 ≤ t2 ≤ 1, 0 ≤ t3 ≤ 1}

For multiplicative integrands f(x, y, z) = f1(x) · f2(y) · f3(z)

1 y ( · )dx dy

  • ·

1 (·)dz = 1 y x ( · ) dz dx dy+ 1 y z ( · ) dx dz dy+ 1 z y ( · ) dx dy dz

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SLIDE 13

Title Intro: Control Series expansions Exp-prod, ξH Sinusoids Feedback Outlook ζ = π′

1 ◦ ξ

Homomorphisms II

  • Restriction is Lie algebra homomorphism

F : L(Z) ⊆ A(Z) → Γ∞(M) (vector fields)

  • Do not fix controls: iterated integral functionals

Υ: ∈ A(Z) → IIF(UZ) Υ: (a1a2 . . . an) →

  • u →

T t1 · · · tp−1 uap(tp) . . . ua1(t1) dt1 . . . dt associative algebras: shuffle of words → pointwise multiplication of iterated integral functionals

  • Much better: Theorem: If U = L1([0, T], [−1, 1]) then

Υ: (A(Z), ∗) → IIF(UZ) is a Zinbiel algebra isomorphism.

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SLIDE 14

Title Intro: Control Series expansions Exp-prod, ξH Sinusoids Feedback Outlook ζ = π′

1 ◦ ξ

Zinbiel: The product & algebra of iterated integrals

Abstractly, right Zinbiel identity U ∗ (V ∗ W) = (U ∗ V) ∗ W + (V ∗ U) ∗ W Concrete examples in control:

  • polynomials

X n ∗ X m =

m n+mX n+m

and X n ⋆ X m = 1

nX n+m

  • AC([0, ∞)):

(U ∗ V)(t) = t

0 U(s) V ′(s) ds

and (U ⋆ V)(t) = t

0 U(s)ds V(t)

  • iterated integrals functionals
  • subsets, e.g. exponentials

eimt ∗ eint =

m n+mei(m+n)t

and eimt ⋆ eint = 1

mei(m+n)t

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SLIDE 15

Title Intro: Control Series expansions Exp-prod, ξH Sinusoids Feedback Outlook ζ = π′

1 ◦ ξ

Abstract solution of DE by iteration SKIP

Rewrite universal control system ˙ S = S · Φ, S(0) = 1 with Φ =

m

  • i=1

uiXi or Φ =

  • a∈Z

a ⊗ a as integral equation using right Zinbiel product S = 1 + S ∗ Φ and iterate to obtain Chen Fliess series S = 1 + (1 + S ∗ Φ) ∗ Φ = 1 + Φ + ((1 + S ∗ Φ) ∗ Φ) ∗ Φ = 1 + Φ + (Φ ∗ Φ) + (((1 + S ∗ Φ) ∗ Φ) ∗ Φ) ∗ Φ = 1 + Φ + (Φ ∗ Φ) + ((Φ ∗ Φ) ∗ Φ) + ((((1 + S ∗ Φ) ∗ Φ) ∗ Φ) ∗ Φ) ∗ Φ . . . = 1 + Φ + (Φ ∗ Φ) + ((Φ ∗ Φ) ∗ Φ) + (((Φ ∗ Φ) ∗ Φ) ∗ Φ) . . .

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SLIDE 16

Title Intro: Control Series expansions Exp-prod, ξH Sinusoids Feedback Outlook ζ = π′

1 ◦ ξ

Rewriting the Chen series

  • Recall: Chen series is an exponential Lie series
  • w∈Z ∗

w ⊗ w = exp

  • b∈B

ζb ⊗ b

  • =

  • b∈B

exp(ξb ⊗ b) where B is ordered basis of free Lie algebra L(Z) ⊂ A(Z)

  • Coord’s of 2nd kind (classic)

ξHK = ξH ∗ ξK

  • Coord’s of the 1st kind (new)

ζ = π′

1 ◦ ξ

  • Use in control / geometric integration: explicit formula for

iterated integral functionals Υζb and Υξb.

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SLIDE 17

Title Intro: Control Series expansions Exp-prod, ξH Sinusoids Feedback Outlook ζ = π′

1 ◦ ξ

Hall words

Hall words (in narrow sense as in Bourbaki)

  • a ∈ Z ⇒ a ∈ ˜

H

  • w, z ∈ ˜

H, u < v ⇒ |u| < |v|

  • If a ∈ Z, then

(ua) ∈ ˜ H ⇔ u < a and u < (ua)

  • (u, (vw)) ∈ ˜

H ⇔ u, (vw) ∈ ˜ H and v ≤ u < (vw), u < (u(vw)) ((a(ab))(b(ab))) ((ab)(b(b(ab)))) ((ab)(b(a(ab)))) ((ab)(a(a(ab)))) (b(b(b(b(ab))))) (b(b(b(a(ab))))) (b(b(a(a(ab))))) (b(a(a(a(ab))))) ((ab)(b(ab))) ((ab)(a(ab))) (b(b(b(ab)))) (b(b(a(ab)))) (b(a(a(ab)))) (a(a(a(ab)))) (b(b(ab))) (b(a(ab))) (a(a(ab))) (b(ab)) (a(ab)) (ab) b a

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SLIDE 18

Title Intro: Control Series expansions Exp-prod, ξH Sinusoids Feedback Outlook ζ = π′

1 ◦ ξ

Iterated integral functionals ξH (coord.’s of 2nd kind

Presented as normal form for free nilpotent system Normal from for a free nilpotent system (of rank r = 5) using a typical Hall set on the alphabet Z = {a, b}

˙ ξa = ua ˙ ξb = ub ˙ ξab = ξa · ˙ ξb = ξa ub ˙ ξaab = ξa · ˙ ξab = ξ2

a ub

using ψ−b(aab) = (a(ab)) ˙ ξbab = ξb · ˙ ξab = ξbξa ub using ψ−b(bab) = (b(ab)) ˙ ξaaab = ξa · ˙ ξaab = ξ3

a ub

using ψ−b(aaab) = (a(a(ab))) ˙ ξbaab = ξb · ˙ ξaab = ξbξ2

a ub

using ψ−b(baab) = (b(a(ab))) ˙ ξbbab = ξb · ˙ ξbab = ξ2

bξa ub

using ψ−b(bbab) = (b(b(ab))) ˙ ξaaaab = ξa · ˙ ξaaab = ξ4

a ub

using ψ−b(aaaab) = (a(a(a(ab)))) ˙ ξbaaab = ξb · ˙ ξaaab = ξbξ3

a ub

using ψ−b(baaab) = (b(a(a(ab)))) ˙ ξbbaab = ξb · ˙ ξbaab = ξ2

bξ2 a ub

using ψ−b(bbaab) = (b(b(a(ab)))) ˙ ξabaab = ξab · ˙ ξaab = ξabξ3

a ub

using ψ−b(abaab) = ((ab)(a(ab))) ˙ ξabbab = ξab · ˙ ξbab = ξabξ2

bξa ub

using ψ−b(abbab) = ((ab)(b(ab)))

Analyze orbits of all nilpotent systems under feedback group