Koopman Operators and Dynamic Mode Decomposition Shubhendu Trivedi - - PowerPoint PPT Presentation

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Koopman Operators and Dynamic Mode Decomposition Shubhendu Trivedi - - PowerPoint PPT Presentation

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Koopman Operators and Dynamic Mode Decomposition Shubhendu Trivedi The University of Chicago Toyota Technological Institute Chicago, IL - 60637 Shubhendu Trivedi (TTI-C) Koopman Operators 1 / 50 On White Board (fill later) Intro to

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Koopman Operators and Dynamic Mode Decomposition

Shubhendu Trivedi

The University of Chicago Toyota Technological Institute Chicago, IL - 60637

Shubhendu Trivedi (TTI-C) Koopman Operators 1 / 50

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On White Board (fill later)

Intro to dynamical systems and Poincair´ e’s geometric picture

Shubhendu Trivedi (TTI-C) Koopman Operators 2 / 50

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On White Board (fill later)

Intro to dynamical systems and Poincair´ e’s geometric picture Definitions: Fixed points, Limit cycles, Invariant sets, Attractors, Bifurcations

Shubhendu Trivedi (TTI-C) Koopman Operators 2 / 50

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On White Board (fill later)

Intro to dynamical systems and Poincair´ e’s geometric picture Definitions: Fixed points, Limit cycles, Invariant sets, Attractors, Bifurcations Two results: Poincair´ e’s recurrence theorem and Bendixson’s criterion

Shubhendu Trivedi (TTI-C) Koopman Operators 2 / 50

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On White Board (fill later)

Intro to dynamical systems and Poincair´ e’s geometric picture Definitions: Fixed points, Limit cycles, Invariant sets, Attractors, Bifurcations Two results: Poincair´ e’s recurrence theorem and Bendixson’s criterion Problems with the geometric picture

Shubhendu Trivedi (TTI-C) Koopman Operators 2 / 50

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On White Board (fill later)

Intro to dynamical systems and Poincair´ e’s geometric picture Definitions: Fixed points, Limit cycles, Invariant sets, Attractors, Bifurcations Two results: Poincair´ e’s recurrence theorem and Bendixson’s criterion Problems with the geometric picture Alternative picture: Dynamics of observables

Shubhendu Trivedi (TTI-C) Koopman Operators 2 / 50

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On White Board (fill later)

Intro to dynamical systems and Poincair´ e’s geometric picture Definitions: Fixed points, Limit cycles, Invariant sets, Attractors, Bifurcations Two results: Poincair´ e’s recurrence theorem and Bendixson’s criterion Problems with the geometric picture Alternative picture: Dynamics of observables Two dual operators in the ”dynamics of observables” picture: Perron-Frobenius operator and Koopman Operator

Shubhendu Trivedi (TTI-C) Koopman Operators 2 / 50

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On White Board (fill later)

Intro to dynamical systems and Poincair´ e’s geometric picture Definitions: Fixed points, Limit cycles, Invariant sets, Attractors, Bifurcations Two results: Poincair´ e’s recurrence theorem and Bendixson’s criterion Problems with the geometric picture Alternative picture: Dynamics of observables Two dual operators in the ”dynamics of observables” picture: Perron-Frobenius operator and Koopman Operator Next: Koopman Operator

Shubhendu Trivedi (TTI-C) Koopman Operators 2 / 50

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Dynamical Systems

Denote the state space by M

Shubhendu Trivedi (TTI-C) Koopman Operators 3 / 50

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Dynamical Systems

Denote the state space by M M can be an arbitrary set with no structure

Shubhendu Trivedi (TTI-C) Koopman Operators 3 / 50

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Dynamical Systems

Denote the state space by M M can be an arbitrary set with no structure The dynamics on M are specified by an iterated map T : M → M

Shubhendu Trivedi (TTI-C) Koopman Operators 3 / 50

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Dynamical Systems

Denote the state space by M M can be an arbitrary set with no structure The dynamics on M are specified by an iterated map T : M → M The abstract dynamical system is specified by the pair (M, T)

Shubhendu Trivedi (TTI-C) Koopman Operators 3 / 50

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Measure Preserving Dynamical Systems

M is a measurable space with a σ-algebra B

Shubhendu Trivedi (TTI-C) Koopman Operators 4 / 50

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Measure Preserving Dynamical Systems

M is a measurable space with a σ-algebra B T is B measurable

Shubhendu Trivedi (TTI-C) Koopman Operators 4 / 50

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Measure Preserving Dynamical Systems

M is a measurable space with a σ-algebra B T is B measurable T is measure preserving:

Shubhendu Trivedi (TTI-C) Koopman Operators 4 / 50

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Measure Preserving Dynamical Systems

M is a measurable space with a σ-algebra B T is B measurable T is measure preserving: ∃ an invariant measure µ, such that for any S ∈ B

Shubhendu Trivedi (TTI-C) Koopman Operators 4 / 50

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Measure Preserving Dynamical Systems

M is a measurable space with a σ-algebra B T is B measurable T is measure preserving: ∃ an invariant measure µ, such that for any S ∈ B µ(S) = µ(T −1S)

Shubhendu Trivedi (TTI-C) Koopman Operators 4 / 50

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Observables on State Space

Want to study the behaviour of observables on the state space

Shubhendu Trivedi (TTI-C) Koopman Operators 5 / 50

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Observables on State Space

Want to study the behaviour of observables on the state space Observable: Some f : M → C

Shubhendu Trivedi (TTI-C) Koopman Operators 5 / 50

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Observables on State Space

Want to study the behaviour of observables on the state space Observable: Some f : M → C f ∈ F (F is a function space, of unspecified structure)

Shubhendu Trivedi (TTI-C) Koopman Operators 5 / 50

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Observables on State Space

Want to study the behaviour of observables on the state space Observable: Some f : M → C f ∈ F (F is a function space, of unspecified structure) Concrete interpretation: Sensor probe for the dynamical system

Shubhendu Trivedi (TTI-C) Koopman Operators 5 / 50

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Observables on State Space

Want to study the behaviour of observables on the state space Observable: Some f : M → C f ∈ F (F is a function space, of unspecified structure) Concrete interpretation: Sensor probe for the dynamical system Instead of tracking p → T(p) → T 2(p) → T(p3) . . .

Shubhendu Trivedi (TTI-C) Koopman Operators 5 / 50

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Observables on State Space

Want to study the behaviour of observables on the state space Observable: Some f : M → C f ∈ F (F is a function space, of unspecified structure) Concrete interpretation: Sensor probe for the dynamical system Instead of tracking p → T(p) → T 2(p) → T(p3) . . . Track: f(p) → f(T(p)) → f(T 2(p)) → f(T(p3)) . . .

Shubhendu Trivedi (TTI-C) Koopman Operators 5 / 50

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Observables on State Space

Want to study the behaviour of observables on the state space Observable: Some f : M → C f ∈ F (F is a function space, of unspecified structure) Concrete interpretation: Sensor probe for the dynamical system Instead of tracking p → T(p) → T 2(p) → T(p3) . . . Track: f(p) → f(T(p)) → f(T 2(p)) → f(T(p3)) . . . Can describe the dynamics as: pn+1 = T(pn) and vn = f(pn)

Shubhendu Trivedi (TTI-C) Koopman Operators 5 / 50

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Koopman Operator

Discrete time Koopman Operator UT : F → F [UT f](p) = f(T(p))

Shubhendu Trivedi (TTI-C) Koopman Operators 6 / 50

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Koopman Operator

Discrete time Koopman Operator UT : F → F [UT f](p) = f(T(p)) Is a composition: UT f = f ◦ T

Shubhendu Trivedi (TTI-C) Koopman Operators 6 / 50

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Koopman Operator

Discrete time Koopman Operator UT : F → F [UT f](p) = f(T(p)) Is a composition: UT f = f ◦ T When F is a vector space, UT is a linear operator

Shubhendu Trivedi (TTI-C) Koopman Operators 6 / 50

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Koopman Operator

Discrete time Koopman Operator UT : F → F [UT f](p) = f(T(p)) Is a composition: UT f = f ◦ T When F is a vector space, UT is a linear operator M is a finite set = ⇒ U is finite dimensional, represented by a matrix

Shubhendu Trivedi (TTI-C) Koopman Operators 6 / 50

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Koopman Operator

Discrete time Koopman Operator UT : F → F [UT f](p) = f(T(p)) Is a composition: UT f = f ◦ T When F is a vector space, UT is a linear operator M is a finite set = ⇒ U is finite dimensional, represented by a matrix Generally U is infinite-dimensional

Shubhendu Trivedi (TTI-C) Koopman Operators 6 / 50

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Koopman Operator

Usually only have access to a collection of observables {f1, . . . , fK} ⊂ F

Shubhendu Trivedi (TTI-C) Koopman Operators 7 / 50

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Koopman Operator

Usually only have access to a collection of observables {f1, . . . , fK} ⊂ F f1, . . . , fK could be physically relevant observables or part of the function basis for F

Shubhendu Trivedi (TTI-C) Koopman Operators 7 / 50

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Extended Koopman Operator

Can extend the Koopman operator to this larger space

Shubhendu Trivedi (TTI-C) Koopman Operators 8 / 50

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Extended Koopman Operator

Can extend the Koopman operator to this larger space Denote F = (f1, . . . , fK)T ∈ FK

Shubhendu Trivedi (TTI-C) Koopman Operators 8 / 50

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Extended Koopman Operator

Can extend the Koopman operator to this larger space Denote F = (f1, . . . , fK)T ∈ FK Then UK : FK → FK [UKF](p) :=    [Uf1](p) . . . [UfK](p)   

Shubhendu Trivedi (TTI-C) Koopman Operators 8 / 50

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Extended Koopman Operator

Can extend the Koopman operator to this larger space Denote F = (f1, . . . , fK)T ∈ FK Then UK : FK → FK [UKF](p) :=    [Uf1](p) . . . [UfK](p)    Then UK =

K

  • 1

U

Shubhendu Trivedi (TTI-C) Koopman Operators 8 / 50

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Extended Koopman Operator

Can extend the Koopman operator to this larger space Denote F = (f1, . . . , fK)T ∈ FK Then UK : FK → FK [UKF](p) :=    [Uf1](p) . . . [UfK](p)    Then UK =

K

  • 1

U FK is the space of CK-valued observables on the state space M

Shubhendu Trivedi (TTI-C) Koopman Operators 8 / 50

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Extended Koopman Operator

Can extend the Koopman operator to this larger space Denote F = (f1, . . . , fK)T ∈ FK Then UK : FK → FK [UKF](p) :=    [Uf1](p) . . . [UfK](p)    Then UK =

K

  • 1

U FK is the space of CK-valued observables on the state space M More generally: F : M → V where V is a vector space

Shubhendu Trivedi (TTI-C) Koopman Operators 8 / 50

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Koopman Operators in Continuous Time D.S.

Consider the continuous time dynamical system ˙ p = T(p)

Shubhendu Trivedi (TTI-C) Koopman Operators 9 / 50

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Example: Cyclic Group

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Setup

Reminder: Group that can be obtained by a single generator

Shubhendu Trivedi (TTI-C) Koopman Operators 11 / 50

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Setup

Reminder: Group that can be obtained by a single generator Let M = {e, a, a2} be a cyclic group of order 3

Shubhendu Trivedi (TTI-C) Koopman Operators 11 / 50

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Setup

Reminder: Group that can be obtained by a single generator Let M = {e, a, a2} be a cyclic group of order 3 Define T : M → M as T(p) = a · p

Shubhendu Trivedi (TTI-C) Koopman Operators 11 / 50

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Setup

Reminder: Group that can be obtained by a single generator Let M = {e, a, a2} be a cyclic group of order 3 Define T : M → M as T(p) = a · p Entire state space is a periodic orbit with period 3

Shubhendu Trivedi (TTI-C) Koopman Operators 11 / 50

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Setup

Reminder: Group that can be obtained by a single generator Let M = {e, a, a2} be a cyclic group of order 3 Define T : M → M as T(p) = a · p Entire state space is a periodic orbit with period 3 Let F be C-valued functions on M

Shubhendu Trivedi (TTI-C) Koopman Operators 11 / 50

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Setup

Reminder: Group that can be obtained by a single generator Let M = {e, a, a2} be a cyclic group of order 3 Define T : M → M as T(p) = a · p Entire state space is a periodic orbit with period 3 Let F be C-valued functions on M Space of observables is C3

Shubhendu Trivedi (TTI-C) Koopman Operators 11 / 50

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Setup

Let f1, f2, f3 be indicator functions on e, a, a2: f1(p) =

  • 1

if p = e if p = e f2(p) =

  • 1

if p = a if p = a f3(p) =

  • 1

if p = a2 if p = a2

Shubhendu Trivedi (TTI-C) Koopman Operators 12 / 50

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Setup

Let f1, f2, f3 be indicator functions on e, a, a2: f1(p) =

  • 1

if p = e if p = e f2(p) =

  • 1

if p = a if p = a f3(p) =

  • 1

if p = a2 if p = a2 Form a basis for F

Shubhendu Trivedi (TTI-C) Koopman Operators 12 / 50

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Example: Cyclic Group

Action of the Koopman operator on this basis: [Uf1](p) = f1(a · p) = f3(p) [Uf2](p) = f2(a · p) = f1(p) [Uf3](p) = f3(a · p) = f2(p)

Shubhendu Trivedi (TTI-C) Koopman Operators 13 / 50

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Example: Cyclic Group

Action of the Koopman operator on this basis: [Uf1](p) = f1(a · p) = f3(p) [Uf2](p) = f2(a · p) = f1(p) [Uf3](p) = f3(a · p) = f2(p) Consider arbitrary observable f ∈ F i.e. f = c1f1 + c2f2 + c3f3

Shubhendu Trivedi (TTI-C) Koopman Operators 13 / 50

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Example: Cyclic Group

Action of the Koopman operator on this basis: [Uf1](p) = f1(a · p) = f3(p) [Uf2](p) = f2(a · p) = f1(p) [Uf3](p) = f3(a · p) = f2(p) Consider arbitrary observable f ∈ F i.e. f = c1f1 + c2f2 + c3f3 Consider the action of the Koopman operator on f: Uf = U(c1f1 + c2f2 + c3f3) = c1f3 + c2f1 + c3f2

Shubhendu Trivedi (TTI-C) Koopman Operators 13 / 50

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Example: Cyclic Group

Matrix representation of the Koopman operator U in the {f1, f2, f3} basis: U   c1 c2 c3   =   1 1 1     c1 c2 c3  

Shubhendu Trivedi (TTI-C) Koopman Operators 14 / 50

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Example: Linear Diagonalizable Systems

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Setup

Let M = Rd, and define T : M → M as : (T(x))i = µixi

Shubhendu Trivedi (TTI-C) Koopman Operators 16 / 50

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Setup

Let M = Rd, and define T : M → M as : (T(x))i = µixi x = (x1, . . . , xd)T ∈ M and µi ∈ R

Shubhendu Trivedi (TTI-C) Koopman Operators 16 / 50

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Setup

Let M = Rd, and define T : M → M as : (T(x))i = µixi x = (x1, . . . , xd)T ∈ M and µi ∈ R Let F denote space of functions Rd → C

Shubhendu Trivedi (TTI-C) Koopman Operators 16 / 50

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Setup

Let M = Rd, and define T : M → M as : (T(x))i = µixi x = (x1, . . . , xd)T ∈ M and µi ∈ R Let F denote space of functions Rd → C Let {b1 . . . , bd} ⊂ M be a basis for M; define fi(x) = bi, x

Shubhendu Trivedi (TTI-C) Koopman Operators 16 / 50

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Example: Linear Diagonalizable Systems

The action of the Koopman operator U : F → F on fi is [Ufi](x) = bi, T(x) =

  • bi,1

. . . bi,d

  µ1x1 . . . µdxd   

Shubhendu Trivedi (TTI-C) Koopman Operators 17 / 50

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Example: Linear Diagonalizable Systems

The action of the Koopman operator U : F → F on fi is [Ufi](x) = bi, T(x) =

  • bi,1

. . . bi,d

  µ1x1 . . . µdxd    [Ufi](x) =

  • bi,1

. . . bi,d

    µ1 . . . µ2 . . . . . . . . . ... . . . . . . µd         x1 . . . xd   

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Example: Linear Diagonalizable Systems

Recall Fd =

d

  • 1

F, define Ud as earlier, then for F = (f1, . . . , fd)T

Shubhendu Trivedi (TTI-C) Koopman Operators 18 / 50

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Example: Linear Diagonalizable Systems

Recall Fd =

d

  • 1

F, define Ud as earlier, then for F = (f1, . . . , fd)T Then the action of the extended Koopman operator

Shubhendu Trivedi (TTI-C) Koopman Operators 18 / 50

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Example: Linear Diagonalizable Systems

Recall Fd =

d

  • 1

F, define Ud as earlier, then for F = (f1, . . . , fd)T Then the action of the extended Koopman operator [UdF](x) =    b1,1 . . . b1,d . . . ... . . . bd,1 . . . bd,d         µ1 . . . µ2 . . . 0 . . . . . . ... . . . . . . µd         x1 . . . xd   

Shubhendu Trivedi (TTI-C) Koopman Operators 18 / 50

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Example: Linear Diagonalizable Systems

Recall Fd =

d

  • 1

F, define Ud as earlier, then for F = (f1, . . . , fd)T Then the action of the extended Koopman operator [UdF](x) =    b1,1 . . . b1,d . . . ... . . . bd,1 . . . bd,d         µ1 . . . µ2 . . . 0 . . . . . . ... . . . . . . µd         x1 . . . xd    This is the action of the Koopman operator on the particular

  • bservable F, not the entire observable space F

Shubhendu Trivedi (TTI-C) Koopman Operators 18 / 50

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Example: Heat equation with periodic boundary conditions

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Mode Analysis

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Eigenfunctions and Koopman Modes

We have put no structure on F so far

Shubhendu Trivedi (TTI-C) Koopman Operators 21 / 50

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Eigenfunctions and Koopman Modes

We have put no structure on F so far When F is a vector space, the Koopman operator is linear

Shubhendu Trivedi (TTI-C) Koopman Operators 21 / 50

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Eigenfunctions and Koopman Modes

We have put no structure on F so far When F is a vector space, the Koopman operator is linear Interest: Study spectral properties of the Koopman Operator to probe into the dynamics of the system

Shubhendu Trivedi (TTI-C) Koopman Operators 21 / 50

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Eigenfunctions and Koopman Modes

We have put no structure on F so far When F is a vector space, the Koopman operator is linear Interest: Study spectral properties of the Koopman Operator to probe into the dynamics of the system Assume: F is a Banach space

Shubhendu Trivedi (TTI-C) Koopman Operators 21 / 50

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Eigenfunctions and Koopman Modes

We have put no structure on F so far When F is a vector space, the Koopman operator is linear Interest: Study spectral properties of the Koopman Operator to probe into the dynamics of the system Assume: F is a Banach space Assume: U is a bounded, continuous operator on F

Shubhendu Trivedi (TTI-C) Koopman Operators 21 / 50

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Eigenfunctions and Koopman Modes

Let {φ1, . . . , φn} be a set of eigenfunctions of U, where n = 1, 2, .., ∞

Shubhendu Trivedi (TTI-C) Koopman Operators 22 / 50

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Eigenfunctions and Koopman Modes

Let {φ1, . . . , φn} be a set of eigenfunctions of U, where n = 1, 2, .., ∞ For the discrete case:

Shubhendu Trivedi (TTI-C) Koopman Operators 22 / 50

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Eigenfunctions and Koopman Modes

Let {φ1, . . . , φn} be a set of eigenfunctions of U, where n = 1, 2, .., ∞ For the discrete case: [Uφi](p) = λiφi(p)

Shubhendu Trivedi (TTI-C) Koopman Operators 22 / 50

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Eigenfunctions and Koopman Modes

Let {φ1, . . . , φn} be a set of eigenfunctions of U, where n = 1, 2, .., ∞ For the discrete case: [Uφi](p) = λiφi(p) For the continuous case:

Shubhendu Trivedi (TTI-C) Koopman Operators 22 / 50

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Eigenfunctions and Koopman Modes

Let {φ1, . . . , φn} be a set of eigenfunctions of U, where n = 1, 2, .., ∞ For the discrete case: [Uφi](p) = λiφi(p) For the continuous case: [U tφi](p) = eλitφi(p)

Shubhendu Trivedi (TTI-C) Koopman Operators 22 / 50

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Eigenfunctions and Koopman Modes

Let {φ1, . . . , φn} be a set of eigenfunctions of U, where n = 1, 2, .., ∞ For the discrete case: [Uφi](p) = λiφi(p) For the continuous case: [U tφi](p) = eλitφi(p) λ’s are the eigenvalues of the generator U, and {eλi} of the Koopman semi-group

Shubhendu Trivedi (TTI-C) Koopman Operators 22 / 50

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Algebraic Structure of Eigenfunctions

Assume that F is a subset of all C valued functions on M

Shubhendu Trivedi (TTI-C) Koopman Operators 23 / 50

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Algebraic Structure of Eigenfunctions

Assume that F is a subset of all C valued functions on M Also assume that it forms a vector space that is closed under pointwise products of functions

Shubhendu Trivedi (TTI-C) Koopman Operators 23 / 50

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Algebraic Structure of Eigenfunctions

Assume that F is a subset of all C valued functions on M Also assume that it forms a vector space that is closed under pointwise products of functions = ⇒ set of eigenfunctions forms an abelian semigroup under pointwise products of functions

Shubhendu Trivedi (TTI-C) Koopman Operators 23 / 50

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Algebraic Structure of Eigenfunctions

Assume that F is a subset of all C valued functions on M Also assume that it forms a vector space that is closed under pointwise products of functions = ⇒ set of eigenfunctions forms an abelian semigroup under pointwise products of functions Concretely: If φ1, φ2 ∈ F are eigenfunctions of U with eigenvalues λ1 and λ2, then φ1φ2 is an eigenfunction of U with eignevalue λ1λ2

Shubhendu Trivedi (TTI-C) Koopman Operators 23 / 50

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Algebraic Structure of Eigenfunctions

If p > 0 and φ is an eigenfunction with eigenvalue λ, then φp is an eigenfunction with eigenvalue λp

Shubhendu Trivedi (TTI-C) Koopman Operators 24 / 50

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Algebraic Structure of Eigenfunctions

If p > 0 and φ is an eigenfunction with eigenvalue λ, then φp is an eigenfunction with eigenvalue λp If φ is an eigenfunction that vanishes nowhere and r ∈ R, then φr is an eigenfunction with eigenvalue λr

Shubhendu Trivedi (TTI-C) Koopman Operators 24 / 50

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Algebraic Structure of Eigenfunctions

If p > 0 and φ is an eigenfunction with eigenvalue λ, then φp is an eigenfunction with eigenvalue λp If φ is an eigenfunction that vanishes nowhere and r ∈ R, then φr is an eigenfunction with eigenvalue λr Eigenfunctions that vanish nowhere form an Abelian group

Shubhendu Trivedi (TTI-C) Koopman Operators 24 / 50

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SLIDE 83

Spectral Equivalence of Topologically Conjugate Systems

Proposition Let S : M → M and T : N → N be topologically conjugate; i.e. ∃ a homomorphism h : N → M such that S ◦ h = h ◦ T. If φ is an eigenfunction of US with eigenvalue λ, then φ ◦ h is an eigenfunction of UT at eigenvalue λ

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SLIDE 84

Example: Linear Diagonalizable Systems

Let y(k) = (y(k)

1 , y(k) 2 )T ((k) indexes time)

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SLIDE 85

Example: Linear Diagonalizable Systems

Let y(k) = (y(k)

1 , y(k) 2 )T ((k) indexes time)

Let y(k+1) = Ty(k)

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SLIDE 86

Example: Linear Diagonalizable Systems

Let y(k) = (y(k)

1 , y(k) 2 )T ((k) indexes time)

Let y(k+1) = Ty(k) T is a matrix with eigenvectors v1, v2 at eigenvalues λ1, λ2 with vi = ej

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SLIDE 87

Example: Linear Diagonalizable Systems

Let y(k) = (y(k)

1 , y(k) 2 )T ((k) indexes time)

Let y(k+1) = Ty(k) T is a matrix with eigenvectors v1, v2 at eigenvalues λ1, λ2 with vi = ej If V = [v1v2], then with new coordinates x(k) = (xk

1, x(k) 2 )T = V −1y(k)

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SLIDE 88

Example: Linear Diagonalizable Systems

Let y(k) = (y(k)

1 , y(k) 2 )T ((k) indexes time)

Let y(k+1) = Ty(k) T is a matrix with eigenvectors v1, v2 at eigenvalues λ1, λ2 with vi = ej If V = [v1v2], then with new coordinates x(k) = (xk

1, x(k) 2 )T = V −1y(k)

  • x(k+1)

1

x(k+1)

2

  • =

λ1 λ2 x(k)

1

x(k)

2

  • := Λ
  • x(k)

1

x(k)

2

  • Shubhendu Trivedi (TTI-C)

Koopman Operators 26 / 50

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SLIDE 89

Example: Linear Diagonalizable Systems

Let y(k) = (y(k)

1 , y(k) 2 )T ((k) indexes time)

Let y(k+1) = Ty(k) T is a matrix with eigenvectors v1, v2 at eigenvalues λ1, λ2 with vi = ej If V = [v1v2], then with new coordinates x(k) = (xk

1, x(k) 2 )T = V −1y(k)

  • x(k+1)

1

x(k+1)

2

  • =

λ1 λ2 x(k)

1

x(k)

2

  • := Λ
  • x(k)

1

x(k)

2

  • Maps Λ and T are topologically conjugate by ΛV −1 = V −1T

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SLIDE 90

Example: Linear Diagonalizable Systems

Let y(k) = (y(k)

1 , y(k) 2 )T ((k) indexes time)

Let y(k+1) = Ty(k) T is a matrix with eigenvectors v1, v2 at eigenvalues λ1, λ2 with vi = ej If V = [v1v2], then with new coordinates x(k) = (xk

1, x(k) 2 )T = V −1y(k)

  • x(k+1)

1

x(k+1)

2

  • =

λ1 λ2 x(k)

1

x(k)

2

  • := Λ
  • x(k)

1

x(k)

2

  • Maps Λ and T are topologically conjugate by ΛV −1 = V −1T

V −1 is now the h from the proposition

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SLIDE 91

Koopman Modes

Assume f ∈ F is an observable in the linear span of a set of eigenfunctions {φi}n

1, then for ci(f) ∈ C:

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SLIDE 92

Koopman Modes

Assume f ∈ F is an observable in the linear span of a set of eigenfunctions {φi}n

1, then for ci(f) ∈ C:

f(p) =

n

  • i=1

ci(f)φi(p)

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SLIDE 93

Koopman Modes

Assume f ∈ F is an observable in the linear span of a set of eigenfunctions {φi}n

1, then for ci(f) ∈ C:

f(p) =

n

  • i=1

ci(f)φi(p) Dynamics of f have a simple form: [Uf](p) = f(T(p)) =

n

  • i=1

ci(f)φi(T(p)) =

n

  • i=1

ci(f)[Uφi](p)

Shubhendu Trivedi (TTI-C) Koopman Operators 27 / 50

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SLIDE 94

Koopman Modes

Assume f ∈ F is an observable in the linear span of a set of eigenfunctions {φi}n

1, then for ci(f) ∈ C:

f(p) =

n

  • i=1

ci(f)φi(p) Dynamics of f have a simple form: [Uf](p) = f(T(p)) =

n

  • i=1

ci(f)φi(T(p)) =

n

  • i=1

ci(f)[Uφi](p) [Uf](p) = f(T(p)) =

n

  • i=1

λici(f)φi(p)

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SLIDE 95

Koopman Modes

Dynamics of f have a simple form: [Uf](p) = f(T(p)) =

n

  • i=1

ci(f)φi(T(p)) =

n

  • i=1

ci(f)[Uφi](p)

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SLIDE 96

Koopman Modes

Dynamics of f have a simple form: [Uf](p) = f(T(p)) =

n

  • i=1

ci(f)φi(T(p)) =

n

  • i=1

ci(f)[Uφi](p) [Uf](p) = f(T(p)) =

n

  • i=1

λici(f)φi(p)

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SLIDE 97

Koopman Modes

Dynamics of f have a simple form: [Uf](p) = f(T(p)) =

n

  • i=1

ci(f)φi(T(p)) =

n

  • i=1

ci(f)[Uφi](p) [Uf](p) = f(T(p)) =

n

  • i=1

λici(f)φi(p) Likewise [U mf](p) =

n

  • i=1

λm

i ci(f)φi(p)

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SLIDE 98

Koopman Modes

Extension to vector valued observables F = (f1, . . . , fK)T , with each fi in the closed linear span of eigenfunctions:

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SLIDE 99

Koopman Modes

Extension to vector valued observables F = (f1, . . . , fK)T , with each fi in the closed linear span of eigenfunctions: [U kF](p) =

n

  • i=1

λm

i φi(p)

   ci(f1) . . . ci(fK)   

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SLIDE 100

Koopman Modes

Extension to vector valued observables F = (f1, . . . , fK)T , with each fi in the closed linear span of eigenfunctions: [U kF](p) =

n

  • i=1

λm

i φi(p)

   ci(f1) . . . ci(fK)    Written compactly:

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SLIDE 101

Koopman Modes

Extension to vector valued observables F = (f1, . . . , fK)T , with each fi in the closed linear span of eigenfunctions: [U kF](p) =

n

  • i=1

λm

i φi(p)

   ci(f1) . . . ci(fK)    Written compactly: [U kF](p) =

n

  • i=1

λm

i φiCi(F)

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SLIDE 102

Koopman Mode

Definition Let φi be an eigenfunction for the Koopman operator corresponding to eigenvalue λi. For a vector valued observable F : M → V , the Koopman mode Ci(F), corresponding to φi is the vector of coefficients

  • f the projection of F onto span{φi}

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SLIDE 103

Computation of Koopman Modes: Theory

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SLIDE 104

Theorem (Yosida) Let F be a Banach space and U : F → F. Assume U ≤ 1. Let λ be an eigenvalue of U such that |λ| = 1. Let ˜ U = λ−1U, and define: AK( ˜ U) = 1 K

K−1

  • k=0

˜ U k Then AK converges in the strong operator topology to the projection

  • perator on the subspace of U-invariant function; i.e. onto the

eigenspace Eλ corresponding to λ. That is, for all f ∈ F, lim

K→∞ AKf = lim K→∞

1 K

K−1

  • k=0

˜ U kf = Pλf where Pλ : F → Eλ is a projection operator.

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SLIDE 105

Special Case

Consider the case when the eigenvalues are simple and |λ1| = · · · = |λℓ| = 1 and |λn| < 1 for n > ℓ

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SLIDE 106

Special Case

Consider the case when the eigenvalues are simple and |λ1| = · · · = |λℓ| = 1 and |λn| < 1 for n > ℓ Then, λj = ei2πωj for some real ωj, when j ≤ ℓ

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SLIDE 107

Special Case

Consider the case when the eigenvalues are simple and |λ1| = · · · = |λℓ| = 1 and |λn| < 1 for n > ℓ Then, λj = ei2πωj for some real ωj, when j ≤ ℓ For vector valued observables, the projections take the form:

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SLIDE 108

Special Case

Consider the case when the eigenvalues are simple and |λ1| = · · · = |λℓ| = 1 and |λn| < 1 for n > ℓ Then, λj = ei2πωj for some real ωj, when j ≤ ℓ For vector valued observables, the projections take the form: φjCj(F) = lim

K→∞

1 K

K−1

  • k=0

ei2πωjk[U kF] for j = 1, . . . , ℓ

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SLIDE 109

Special Case

Consider the case when the eigenvalues are simple and |λ1| = · · · = |λℓ| = 1 and |λn| < 1 for n > ℓ Then, λj = ei2πωj for some real ωj, when j ≤ ℓ For vector valued observables, the projections take the form: φjCj(F) = lim

K→∞

1 K

K−1

  • k=0

ei2πωjk[U kF] for j = 1, . . . , ℓ Previous theorem reduces to Fourier analysis for those eigenvalues on the unit circle.

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SLIDE 110

Special Case

Consider the case when the eigenvalues are simple and |λ1| = · · · = |λℓ| = 1 and |λn| < 1 for n > ℓ Then, λj = ei2πωj for some real ωj, when j ≤ ℓ For vector valued observables, the projections take the form: φjCj(F) = lim

K→∞

1 K

K−1

  • k=0

ei2πωjk[U kF] for j = 1, . . . , ℓ Previous theorem reduces to Fourier analysis for those eigenvalues on the unit circle. When an observable is a linear combination of a finite collection of eigenfunctions corresponding to simple eigenvalues, we have an extension of the previous theorem

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SLIDE 111

Theorem (Generalized Laplace Analysis) Let {λ1, . . . , λm} be a finite set of simple eigenvalues for U, ordered so that |λ1| ≥ · · · ≥ |λm| and let φi be an eigenfunction corresponding to λi. For each n ∈ {1, . . . , N}, assume fn : M → C and fn ∈ span{φ1, . . . , φm}. Define the vector-valued observable F = (f1, . . . , fN)T . Then the Koopman modes can be computed via: φjCj(F) = lim

K→∞

1 K

K−1

  • k=0

λ−k

j

  • U kF −

j−1

  • i=1

λk

i φiCi(F)

  • A simple consequence of the theorem of Yosida

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SLIDE 112

A Numerical Algorithm: Dynamic Mode Decomposition

Shubhendu Trivedi (TTI-C) Koopman Operators 35 / 50

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SLIDE 113

Introduction

Problem: Don’t usually have access to an explicit representation

  • f the Koopman operator

Shubhendu Trivedi (TTI-C) Koopman Operators 36 / 50

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SLIDE 114

Introduction

Problem: Don’t usually have access to an explicit representation

  • f the Koopman operator

Can only understand its behaviour by looking at its action on an

  • bservable at only a finite number of initial conditions

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SLIDE 115

Introduction

Problem: Don’t usually have access to an explicit representation

  • f the Koopman operator

Can only understand its behaviour by looking at its action on an

  • bservable at only a finite number of initial conditions

Data driven approach: Have a sequence of observations of a vector-valued observable along a trajectory {T kp}

Shubhendu Trivedi (TTI-C) Koopman Operators 36 / 50

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SLIDE 116

Introduction

Problem: Don’t usually have access to an explicit representation

  • f the Koopman operator

Can only understand its behaviour by looking at its action on an

  • bservable at only a finite number of initial conditions

Data driven approach: Have a sequence of observations of a vector-valued observable along a trajectory {T kp} Dynamic mode decomposition: Data driven approach to approximate the modes and eigenvalues of the Koopman operator without numerically implementing a laplace transform

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SLIDE 117

Introduction

Main idea: Find the best approximation of U on some finite-dimensional subspace and compute the eigenfunctions of this finite-dimensional operator

Shubhendu Trivedi (TTI-C) Koopman Operators 37 / 50

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SLIDE 118

Introduction

Main idea: Find the best approximation of U on some finite-dimensional subspace and compute the eigenfunctions of this finite-dimensional operator How do we define best?

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SLIDE 119

Introduction

Fix observable F : M → Cm and consider the cyclic subspace K∞ = span{U kF}∞

k=0

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SLIDE 120

Introduction

Fix observable F : M → Cm and consider the cyclic subspace K∞ = span{U kF}∞

k=0

Fix r < ∞ and consider the Krylov subspace Kr = span{U kF}r−1

k=0

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SLIDE 121

Introduction

Fix observable F : M → Cm and consider the cyclic subspace K∞ = span{U kF}∞

k=0

Fix r < ∞ and consider the Krylov subspace Kr = span{U kF}r−1

k=0

Assume {U kF}r−1

k=0 is a linearly independent set, and forms a

basis for Kr

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SLIDE 122

Introduction

Fix observable F : M → Cm and consider the cyclic subspace K∞ = span{U kF}∞

k=0

Fix r < ∞ and consider the Krylov subspace Kr = span{U kF}r−1

k=0

Assume {U kF}r−1

k=0 is a linearly independent set, and forms a

basis for Kr Let Pr : Fm → Kr be a projection of observations onto Kr

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SLIDE 123

Introduction

Fix observable F : M → Cm and consider the cyclic subspace K∞ = span{U kF}∞

k=0

Fix r < ∞ and consider the Krylov subspace Kr = span{U kF}r−1

k=0

Assume {U kF}r−1

k=0 is a linearly independent set, and forms a

basis for Kr Let Pr : Fm → Kr be a projection of observations onto Kr Then PrU|Kr : Kr → Kr is a finite dimensional linear operator

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SLIDE 124

Introduction

PrU|Kr has a matrix representation Ar : Cr → Cr in the {U kF}r−1

k=1

basis

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SLIDE 125

Introduction

PrU|Kr has a matrix representation Ar : Cr → Cr in the {U kF}r−1

k=1

basis The matrix Ar depends on:

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SLIDE 126

Introduction

PrU|Kr has a matrix representation Ar : Cr → Cr in the {U kF}r−1

k=1

basis The matrix Ar depends on:

The observable (vector valued)

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SLIDE 127

Introduction

PrU|Kr has a matrix representation Ar : Cr → Cr in the {U kF}r−1

k=1

basis The matrix Ar depends on:

The observable (vector valued) Dimension of the Krylov subspace r

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SLIDE 128

Introduction

PrU|Kr has a matrix representation Ar : Cr → Cr in the {U kF}r−1

k=1

basis The matrix Ar depends on:

The observable (vector valued) Dimension of the Krylov subspace r The projection operator Pr

Shubhendu Trivedi (TTI-C) Koopman Operators 39 / 50

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SLIDE 129

Introduction

PrU|Kr has a matrix representation Ar : Cr → Cr in the {U kF}r−1

k=1

basis The matrix Ar depends on:

The observable (vector valued) Dimension of the Krylov subspace r The projection operator Pr

If (λ, v) is an eigenpair for Ar with v, then φ =

r−1

  • j=0

vj[U jF] is an eigenfunction of PrU|Kr

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SLIDE 130

Introduction

PrU|Kr has a matrix representation Ar : Cr → Cr in the {U kF}r−1

k=1

basis The matrix Ar depends on:

The observable (vector valued) Dimension of the Krylov subspace r The projection operator Pr

If (λ, v) is an eigenpair for Ar with v, then φ =

r−1

  • j=0

vj[U jF] is an eigenfunction of PrU|Kr Restricting our attention on a fixed observable F and a Krylov subspace, the problem of finding eigenvalues and Koopman modes is reduced to finding eigenvalues and eigenvectors for matrix Ar

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SLIDE 131

Arnoldi Recap

If we had Ar, we could just use the Arnoldi algorithm

Shubhendu Trivedi (TTI-C) Koopman Operators 40 / 50

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SLIDE 132

Arnoldi Recap

If we had Ar, we could just use the Arnoldi algorithm Given A ∈ Cm×m, want to compute eigenvectors and eigenvalues

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SLIDE 133

Arnoldi Recap

If we had Ar, we could just use the Arnoldi algorithm Given A ∈ Cm×m, want to compute eigenvectors and eigenvalues Procedure:

Consider random b ∈ Cm with b = 1

Shubhendu Trivedi (TTI-C) Koopman Operators 40 / 50

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SLIDE 134

Arnoldi Recap

If we had Ar, we could just use the Arnoldi algorithm Given A ∈ Cm×m, want to compute eigenvectors and eigenvalues Procedure:

Consider random b ∈ Cm with b = 1 Form the Krylov subspace Kr = {b, Ab, A2b, . . . , Ar−1b}

Shubhendu Trivedi (TTI-C) Koopman Operators 40 / 50

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SLIDE 135

Arnoldi Recap

If we had Ar, we could just use the Arnoldi algorithm Given A ∈ Cm×m, want to compute eigenvectors and eigenvalues Procedure:

Consider random b ∈ Cm with b = 1 Form the Krylov subspace Kr = {b, Ab, A2b, . . . , Ar−1b} Apply Gram-Schmidt to {Ajb}j=r−1

j=0

to obtain orthonormal basis {qj}r

j=1, arranged into an orthonormal matrix Qr

Shubhendu Trivedi (TTI-C) Koopman Operators 40 / 50

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SLIDE 136

Arnoldi Recap

If we had Ar, we could just use the Arnoldi algorithm Given A ∈ Cm×m, want to compute eigenvectors and eigenvalues Procedure:

Consider random b ∈ Cm with b = 1 Form the Krylov subspace Kr = {b, Ab, A2b, . . . , Ar−1b} Apply Gram-Schmidt to {Ajb}j=r−1

j=0

to obtain orthonormal basis {qj}r

j=1, arranged into an orthonormal matrix Qr

Normalize and orthonormalize at every step j

Shubhendu Trivedi (TTI-C) Koopman Operators 40 / 50

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SLIDE 137

Arnoldi Recap

If we had Ar, we could just use the Arnoldi algorithm Given A ∈ Cm×m, want to compute eigenvectors and eigenvalues Procedure:

Consider random b ∈ Cm with b = 1 Form the Krylov subspace Kr = {b, Ab, A2b, . . . , Ar−1b} Apply Gram-Schmidt to {Ajb}j=r−1

j=0

to obtain orthonormal basis {qj}r

j=1, arranged into an orthonormal matrix Qr

Normalize and orthonormalize at every step j

Hr = Q∗

rAQr is the orthonormal projection of A onto Kr

Shubhendu Trivedi (TTI-C) Koopman Operators 40 / 50

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SLIDE 138

Arnoldi Recap

If we had Ar, we could just use the Arnoldi algorithm Given A ∈ Cm×m, want to compute eigenvectors and eigenvalues Procedure:

Consider random b ∈ Cm with b = 1 Form the Krylov subspace Kr = {b, Ab, A2b, . . . , Ar−1b} Apply Gram-Schmidt to {Ajb}j=r−1

j=0

to obtain orthonormal basis {qj}r

j=1, arranged into an orthonormal matrix Qr

Normalize and orthonormalize at every step j

Hr = Q∗

rAQr is the orthonormal projection of A onto Kr

The top r eigenvalues of Hr approximate that of Ar

Shubhendu Trivedi (TTI-C) Koopman Operators 40 / 50

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SLIDE 139

Relevance to Koopman Modes?

By using Arnoldi we have an implicit assumption

Shubhendu Trivedi (TTI-C) Koopman Operators 41 / 50

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SLIDE 140

Relevance to Koopman Modes?

By using Arnoldi we have an implicit assumption ∃ matrix A, whose evolution Akb ∈ Cm, matches that of [U kF](p) ∈ Cm

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SLIDE 141

Relevance to Koopman Modes?

By using Arnoldi we have an implicit assumption ∃ matrix A, whose evolution Akb ∈ Cm, matches that of [U kF](p) ∈ Cm Don’t have an explicit representation of the Koopman operator, so can’t use standard Arnoldi

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SLIDE 142

Relevance to Koopman Modes?

By using Arnoldi we have an implicit assumption ∃ matrix A, whose evolution Akb ∈ Cm, matches that of [U kF](p) ∈ Cm Don’t have an explicit representation of the Koopman operator, so can’t use standard Arnoldi Why?

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SLIDE 143

Relevance to Koopman Modes?

By using Arnoldi we have an implicit assumption ∃ matrix A, whose evolution Akb ∈ Cm, matches that of [U kF](p) ∈ Cm Don’t have an explicit representation of the Koopman operator, so can’t use standard Arnoldi Why? Need to normalize and orthonormalize at each step = ⇒ need to change observables F at each time step p

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SLIDE 144

Relevance to Koopman Modes?

By using Arnoldi we have an implicit assumption ∃ matrix A, whose evolution Akb ∈ Cm, matches that of [U kF](p) ∈ Cm Don’t have an explicit representation of the Koopman operator, so can’t use standard Arnoldi Why? Need to normalize and orthonormalize at each step = ⇒ need to change observables F at each time step p Another interpretation: ?

Shubhendu Trivedi (TTI-C) Koopman Operators 41 / 50

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SLIDE 145

Dynamic Mode Decomposition

Only require a sequence of vectors {bk}r

k=0

Shubhendu Trivedi (TTI-C) Koopman Operators 42 / 50

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SLIDE 146

Dynamic Mode Decomposition

Only require a sequence of vectors {bk}r

k=0

Where bk := U kF(p) ∈ Cm

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SLIDE 147

Dynamic Mode Decomposition

Only require a sequence of vectors {bk}r

k=0

Where bk := U kF(p) ∈ Cm This is for some fixed F : M → Cm and fixed p ∈ M

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SLIDE 148

Dynamic Mode Decomposition

Only require a sequence of vectors {bk}r

k=0

Where bk := U kF(p) ∈ Cm This is for some fixed F : M → Cm and fixed p ∈ M Let Kr = [b0, b1, . . . , br−1]

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SLIDE 149

Dynamic Mode Decomposition

Only require a sequence of vectors {bk}r

k=0

Where bk := U kF(p) ∈ Cm This is for some fixed F : M → Cm and fixed p ∈ M Let Kr = [b0, b1, . . . , br−1] Think of them as point evaluations of the {U kF} basis for the Krylov subspace Kr at point p ∈ M

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slide-150
SLIDE 150

Dynamic Mode Decomposition

br will not be in the span of columns of Kr

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slide-151
SLIDE 151

Dynamic Mode Decomposition

br will not be in the span of columns of Kr Let br =

r−1

  • j=0

cjbj + ηr

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slide-152
SLIDE 152

Dynamic Mode Decomposition

br will not be in the span of columns of Kr Let br =

r−1

  • j=0

cjbj + ηr The cj’s are chosen to minimize the residual ηr

Shubhendu Trivedi (TTI-C) Koopman Operators 43 / 50

slide-153
SLIDE 153

Dynamic Mode Decomposition

br will not be in the span of columns of Kr Let br =

r−1

  • j=0

cjbj + ηr The cj’s are chosen to minimize the residual ηr = ⇒ choosing projection PrU rF of U rF at point p ∈ M

Shubhendu Trivedi (TTI-C) Koopman Operators 43 / 50

slide-154
SLIDE 154

Dynamic Mode Decomposition

br will not be in the span of columns of Kr Let br =

r−1

  • j=0

cjbj + ηr The cj’s are chosen to minimize the residual ηr = ⇒ choosing projection PrU rF of U rF at point p ∈ M [U rF](p) − Pr[U rF](p)Cm = br −

r−1

  • j=0

cjbjCm

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slide-155
SLIDE 155

Dynamic Mode Decomposition

br will not be in the span of columns of Kr Let br =

r−1

  • j=0

cjbj + ηr The cj’s are chosen to minimize the residual ηr = ⇒ choosing projection PrU rF of U rF at point p ∈ M [U rF](p) − Pr[U rF](p)Cm = br −

r−1

  • j=0

cjbjCm Minimize over cj

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slide-156
SLIDE 156

Dynamic Mode Decomposition

Since br = Krc + ηr, we have: UKr = [b1, . . . , br] = [b1, . . . , br−1, Krc + ηr]

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slide-157
SLIDE 157

Dynamic Mode Decomposition

Since br = Krc + ηr, we have: UKr = [b1, . . . , br] = [b1, . . . , br−1, Krc + ηr] UKr = KrAr + ηreT

Shubhendu Trivedi (TTI-C) Koopman Operators 44 / 50

slide-158
SLIDE 158

Dynamic Mode Decomposition

Since br = Krc + ηr, we have: UKr = [b1, . . . , br] = [b1, . . . , br−1, Krc + ηr] UKr = KrAr + ηreT With e = (0, . . . , 0, 1)T ∈ Cm and Ar =        . . . c0 1 . . . c1 1 . . . c2 . . . . . . ... . . . . . . . . . 1 cr−1       

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slide-159
SLIDE 159

Dynamic Mode Decomposition

Since br = Krc + ηr, we have: UKr = [b1, . . . , br] = [b1, . . . , br−1, Krc + ηr] UKr = KrAr + ηreT With e = (0, . . . , 0, 1)T ∈ Cm and Ar =        . . . c0 1 . . . c1 1 . . . c2 . . . . . . ... . . . . . . . . . 1 cr−1        Ar is the companion matrix ; a representation of PrU in the {U kF}r−1

k=0 basis

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slide-160
SLIDE 160

Dynamic Mode Decomposition

Diagonalize Ar = V −1ΛV

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slide-161
SLIDE 161

Dynamic Mode Decomposition

Diagonalize Ar = V −1ΛV Recall: UKr = KrAr + ηreT

Shubhendu Trivedi (TTI-C) Koopman Operators 45 / 50

slide-162
SLIDE 162

Dynamic Mode Decomposition

Diagonalize Ar = V −1ΛV Recall: UKr = KrAr + ηreT Substitute for Ar and multiply with V −1

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slide-163
SLIDE 163

Dynamic Mode Decomposition

Diagonalize Ar = V −1ΛV Recall: UKr = KrAr + ηreT Substitute for Ar and multiply with V −1 UKrV −1 = KrV −1Λ + ηreT V −1

Shubhendu Trivedi (TTI-C) Koopman Operators 45 / 50

slide-164
SLIDE 164

Dynamic Mode Decomposition

Diagonalize Ar = V −1ΛV Recall: UKr = KrAr + ηreT Substitute for Ar and multiply with V −1 UKrV −1 = KrV −1Λ + ηreT V −1 Define E := KrV −1, to get UE = EΛ + ηreT V −1

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slide-165
SLIDE 165

Dynamic Mode Decomposition

For large m, we hope that ηreT V −1 is small

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slide-166
SLIDE 166

Dynamic Mode Decomposition

For large m, we hope that ηreT V −1 is small Then UE ≈ EΛ, and columns of E approximate some eigenvectors of U

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slide-167
SLIDE 167

Dynamic Mode Decomposition

For large m, we hope that ηreT V −1 is small Then UE ≈ EΛ, and columns of E approximate some eigenvectors of U Procedure described is tied to initialization

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slide-168
SLIDE 168

Dynamic Mode Decomposition

For large m, we hope that ηreT V −1 is small Then UE ≈ EΛ, and columns of E approximate some eigenvectors of U Procedure described is tied to initialization Different initial conditions will reveal different parts of the spectrum

Shubhendu Trivedi (TTI-C) Koopman Operators 46 / 50

slide-169
SLIDE 169

Dynamic Mode Decomposition

For large m, we hope that ηreT V −1 is small Then UE ≈ EΛ, and columns of E approximate some eigenvectors of U Procedure described is tied to initialization Different initial conditions will reveal different parts of the spectrum If F / ∈ span{φi} for some eigenfunction φi, then DMD will not reveal that mode

Shubhendu Trivedi (TTI-C) Koopman Operators 46 / 50

slide-170
SLIDE 170

Dynamic Mode Decomposition

For large m, we hope that ηreT V −1 is small Then UE ≈ EΛ, and columns of E approximate some eigenvectors of U Procedure described is tied to initialization Different initial conditions will reveal different parts of the spectrum If F / ∈ span{φi} for some eigenfunction φi, then DMD will not reveal that mode The version described is numerically ill-conditioned (columns of Kr can become linearly dependent)

Shubhendu Trivedi (TTI-C) Koopman Operators 46 / 50

slide-171
SLIDE 171

Standard DMD

Arrange data {b0, . . . , br} into matrices

Shubhendu Trivedi (TTI-C) Koopman Operators 47 / 50

slide-172
SLIDE 172

Standard DMD

Arrange data {b0, . . . , br} into matrices X = [b0, . . . , br−1], Y = [b1, . . . , br] Compute SVD X = UΣV ∗

Shubhendu Trivedi (TTI-C) Koopman Operators 47 / 50

slide-173
SLIDE 173

Standard DMD

Arrange data {b0, . . . , br} into matrices X = [b0, . . . , br−1], Y = [b1, . . . , br] Compute SVD X = UΣV ∗ Define matrix ˜ A = U ∗Y V Σ−1

Shubhendu Trivedi (TTI-C) Koopman Operators 47 / 50

slide-174
SLIDE 174

Standard DMD

Arrange data {b0, . . . , br} into matrices X = [b0, . . . , br−1], Y = [b1, . . . , br] Compute SVD X = UΣV ∗ Define matrix ˜ A = U ∗Y V Σ−1 Compute eigenvalues and eigenvectors of ˜ A; ˜ Aw = λw

Shubhendu Trivedi (TTI-C) Koopman Operators 47 / 50

slide-175
SLIDE 175

Standard DMD

Arrange data {b0, . . . , br} into matrices X = [b0, . . . , br−1], Y = [b1, . . . , br] Compute SVD X = UΣV ∗ Define matrix ˜ A = U ∗Y V Σ−1 Compute eigenvalues and eigenvectors of ˜ A; ˜ Aw = λw DMD mode corresponding to eigenvalue λ is Uw

Shubhendu Trivedi (TTI-C) Koopman Operators 47 / 50

slide-176
SLIDE 176

Exact DMD

Limitation of previous approach: Order of vectors is critical

Shubhendu Trivedi (TTI-C) Koopman Operators 48 / 50

slide-177
SLIDE 177

Exact DMD

Limitation of previous approach: Order of vectors is critical Such that the vectors approximately satisfy zk+1 = Azk for unknown A

Shubhendu Trivedi (TTI-C) Koopman Operators 48 / 50

slide-178
SLIDE 178

Exact DMD

Limitation of previous approach: Order of vectors is critical Such that the vectors approximately satisfy zk+1 = Azk for unknown A Now we relax this constraint and restrict ourselves to data pairs (x1, y1), . . . , (xm, ym)

Shubhendu Trivedi (TTI-C) Koopman Operators 48 / 50

slide-179
SLIDE 179

Exact DMD

Limitation of previous approach: Order of vectors is critical Such that the vectors approximately satisfy zk+1 = Azk for unknown A Now we relax this constraint and restrict ourselves to data pairs (x1, y1), . . . , (xm, ym) Define X and Y as before

Shubhendu Trivedi (TTI-C) Koopman Operators 48 / 50

slide-180
SLIDE 180

Exact DMD

Limitation of previous approach: Order of vectors is critical Such that the vectors approximately satisfy zk+1 = Azk for unknown A Now we relax this constraint and restrict ourselves to data pairs (x1, y1), . . . , (xm, ym) Define X and Y as before Define operator A = Y X†

Shubhendu Trivedi (TTI-C) Koopman Operators 48 / 50

slide-181
SLIDE 181

Exact DMD

Limitation of previous approach: Order of vectors is critical Such that the vectors approximately satisfy zk+1 = Azk for unknown A Now we relax this constraint and restrict ourselves to data pairs (x1, y1), . . . , (xm, ym) Define X and Y as before Define operator A = Y X† The DMD modes and eigenvalues are eigenvalues and eigenvectors of A

Shubhendu Trivedi (TTI-C) Koopman Operators 48 / 50

slide-182
SLIDE 182

Exact DMD

Arrange the data pairs in matrices X, Y as before

Shubhendu Trivedi (TTI-C) Koopman Operators 49 / 50

slide-183
SLIDE 183

Exact DMD

Arrange the data pairs in matrices X, Y as before Compute the SVD of X, write X = UΣV ∗

Shubhendu Trivedi (TTI-C) Koopman Operators 49 / 50

slide-184
SLIDE 184

Exact DMD

Arrange the data pairs in matrices X, Y as before Compute the SVD of X, write X = UΣV ∗ Define matrix ˜ A = U ∗Y V Σ−1

Shubhendu Trivedi (TTI-C) Koopman Operators 49 / 50

slide-185
SLIDE 185

Exact DMD

Arrange the data pairs in matrices X, Y as before Compute the SVD of X, write X = UΣV ∗ Define matrix ˜ A = U ∗Y V Σ−1 Computer eigenvalues and eigenvectors of ˜ A, writing ˜ Aw = λw. Every nonzero eigenvalue is a DMD eigenvalue

Shubhendu Trivedi (TTI-C) Koopman Operators 49 / 50

slide-186
SLIDE 186

Exact DMD

Arrange the data pairs in matrices X, Y as before Compute the SVD of X, write X = UΣV ∗ Define matrix ˜ A = U ∗Y V Σ−1 Computer eigenvalues and eigenvectors of ˜ A, writing ˜ Aw = λw. Every nonzero eigenvalue is a DMD eigenvalue The DMD mode corresponding to λ is given as: Φ = 1 λY V Σ−1w

Shubhendu Trivedi (TTI-C) Koopman Operators 49 / 50

slide-187
SLIDE 187

Kernel Trick and Learning the Subspace

Kernels Neural Networks

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