Coupled Systems: Theory & ExamplesLecture 1

Symmetry andPatterns of Oscillation

Reference: The Symmetry Perspective (G. and Stewart).Japanese translation: Tanaka, Yamada, Takamatsu, and NakagakiMaruzen Publishing, 2003

Martin GolubitskyDepartment of Mathematics

University of Houstonhttp://www.math.uh.edu/emg/

– p. 1/30

Thanks

Ian Stewart Warwick

Luciano Buono Oshawa

Jim Collins Boston University

– p. 2/30

Two Identical Cells1 2 x1 = f(x1, x2)

x2 = f(x2, x1)

σ(x1, x2) = (x2, x1) is a symmetry

Fix(σ) = {x1 = x2} is flow invariantSynchrony is robustExpect synchronous periodic solutions

Time-periodic solutions exist robustly wheretwo cells oscillate a half-period out of phase

x2(t) = x1(t +1

2)

– p. 3/30

Symmetry OverviewA symmetry of a DiffEq x = f(x) is a linear map γ where

γ(sol’n) = sol’n ⇐⇒ f(γx) = γf(x)

Fix(Σ) = {x ∈ Rn : σx = x ∀σ ∈ Σ} is flow invariant

Proof: f(x) = f(σx) = σf(x)

Network symmetries are permutation symmetriesSynchrony is robust in symmetric coupled systems

Symmetry group Γ is a modeling assumption

Network architecture is also a modeling assumption

– p. 4/30

Symmetry OverviewA symmetry of a DiffEq x = f(x) is a linear map γ where

γ(sol’n) = sol’n ⇐⇒ f(γx) = γf(x)

Fix(Σ) = {x ∈ Rn : σx = x ∀σ ∈ Σ} is flow invariant

Proof: f(x) = f(σx) = σf(x)

Network symmetries are permutation symmetriesSynchrony is robust in symmetric coupled systems

Symmetry group Γ is a modeling assumption

Network architecture is also a modeling assumption

– p. 4/30

Symmetry OverviewA symmetry of a DiffEq x = f(x) is a linear map γ where

γ(sol’n) = sol’n ⇐⇒ f(γx) = γf(x)

Fix(Σ) = {x ∈ Rn : σx = x ∀σ ∈ Σ} is flow invariant

Proof: f(x) = f(σx) = σf(x)

Network symmetries are permutation symmetriesSynchrony is robust in symmetric coupled systems

Symmetry group Γ is a modeling assumption

Network architecture is also a modeling assumption

– p. 4/30

Spatio-Temporal SymmetriesLet x(t) be a time-periodic solution• K = {γ ∈ Γ : γx(t) = x(t)} space symmetries

• H = {γ ∈ Γ : γ{x(t)} = {x(t)}} spatiotemporal symm’s

Facts:

• γ ∈ H =⇒ θ ∈ S1 such that γx(t) = x(t + θ)

• H/K is cyclic sinceγ 7→ θ is a homomorphism with kernel K

• Hyperbolic H/K periodic solutions are robust

– p. 5/30

Spatio-Temporal SymmetriesLet x(t) be a time-periodic solution• K = {γ ∈ Γ : γx(t) = x(t)} space symmetries

• H = {γ ∈ Γ : γ{x(t)} = {x(t)}} spatiotemporal symm’s

Facts:

• γ ∈ H =⇒ θ ∈ S1 such that γx(t) = x(t + θ)

• H/K is cyclic sinceγ 7→ θ is a homomorphism with kernel K

• Hyperbolic H/K periodic solutions are robust

– p. 5/30

Three-Cell Unidirectional Ring: Γ = Z3

1

2 3

x1 = f(x1, x3)

x2 = f(x2, x1)

x3 = f(x3, x2)

Discrete rotating waves: H = Z3, K = 1

0 5 10 15−0.2

−0.1

0

0.1

0.2

x 1

0 5 10 15−0.2

−0.1

0

0.1

0.2

x 2

0 5 10 15−0.2

−0.1

0

0.1

0.2

x 3

0 5 10 15−0.2

−0.15

−0.1

−0.05

0

0.05

0.1

0.15

0.2

t

– p. 6/30

Three-Cell Bidirectional Ring: Γ = S3

1

2 3

x1 = f(x1, x2, x3)

x2 = f(x2, x3, x1) f(x2, x1, x3) = f(x2, x3, x1)

x3 = f(x3, x1, x2)

{x1 = x2 = x3} and {xi = xj} are synchrony subspaces

Discrete rotating waves: H = Z3, K = 1

x2(t) = x1

(

t + 1

3

)

and x3(t) = x2

(

t + 1

3

)

Out-of-phase periodic solutions: H = Z2(1 3), K = 1

x3(t) = x1

(

t + 1

2

)

and x2(t) = x2

(

t + 1

2

)

G. and Stewart (1986)

– p. 7/30

Three-Cell Bidirectional Ring: Γ = S3

1

2 3

x1 = f(x1, x2, x3)

x2 = f(x2, x3, x1) f(x2, x1, x3) = f(x2, x3, x1)

x3 = f(x3, x1, x2)

{x1 = x2 = x3} and {xi = xj} are synchrony subspaces

Discrete rotating waves: H = Z3, K = 1

x2(t) = x1

(

t + 1

3

)

and x3(t) = x2

(

t + 1

3

)

Out-of-phase periodic solutions: H = Z2(1 3), K = 1

x3(t) = x1

(

t + 1

2

)

and x2(t) = x2

(

t + 1

2

)

G. and Stewart (1986)

– p. 7/30

A Three-Cell System (2)

0 5 10 15−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

t

– p. 8/30

The H/K TheoremLet Γ be a finite group acting on R

n

There exists hyperbolic periodic soln to someΓ-symmetric system on R

n with space symmetries Kand spatiotemporal symmetries H if and only if

(a) H/K is cyclic

(b) K is an isotropy subgroup

(c) dim Fix(K) ≥ 2If dim Fix(K) = 2, then either H = K or H = N(K)

(d) H fixes a connected component of Fix(K) \ LK

whereLK =

⋃

γ 6∈K

Fix(γ) ∩ Fix(K)

– p. 9/30

The H/K TheoremLet Γ be a finite group acting on R

n

There exists hyperbolic periodic soln to someΓ-symmetric system on R

n with space symmetries Kand spatiotemporal symmetries H if and only if

(a) H/K is cyclic

(b) K is an isotropy subgroup

(c) dim Fix(K) ≥ 2If dim Fix(K) = 2, then either H = K or H = N(K)

(d) H fixes a connected component of Fix(K) \ LK

whereLK =

⋃

γ 6∈K

Fix(γ) ∩ Fix(K)

– p. 9/30

LK (1)γFix(K) = Fix(γKγ−1)

. Suppose h ∈ H ⊂ N(K) then

Suppose kx = x. Then

γkγ−1(γx) = γkx = γx

h Fix(K) = Fix(K)

h (Fix(γ) ∩ Fix(K)) = Fix(hγh−1) ∩ Fix(K)

If γ 6∈ K, then hγh−1 6∈ hKh−1 = K. Since

LK =⋃

γ 6∈K

Fix(γ) ∩ Fix(K)

it follows that h ∈ H implies h : LK → LK

Therefore h permutes connected components ofcomplement of LK in Fix(K)

– p. 10/30

LK (1)γFix(K) = Fix(γKγ−1) . Suppose h ∈ H ⊂ N(K) then

h Fix(K) = Fix(K)

h (Fix(γ) ∩ Fix(K)) = Fix(hγh−1) ∩ Fix(K)

If γ 6∈ K, then hγh−1 6∈ hKh−1 = K. Since

LK =⋃

γ 6∈K

Fix(γ) ∩ Fix(K)

it follows that h ∈ H implies h : LK → LK

Therefore h permutes connected components ofcomplement of LK in Fix(K)

– p. 10/30

LK (1)γFix(K) = Fix(γKγ−1) . Suppose h ∈ H ⊂ N(K) then

h Fix(K) = Fix(K)

h (Fix(γ) ∩ Fix(K)) = Fix(hγh−1) ∩ Fix(K)

since h(Fix(γ) ∩ Fix(K)) = h(Fix(γ)) ∩ h(Fix(K))

If γ 6∈ K, then hγh−1 6∈ hKh−1 = K. Since

LK =⋃

γ 6∈K

Fix(γ) ∩ Fix(K)

it follows that h ∈ H implies h : LK → LK

Therefore h permutes connected components ofcomplement of LK in Fix(K)

– p. 10/30

LK (1)γFix(K) = Fix(γKγ−1) . Suppose h ∈ H ⊂ N(K) then

h Fix(K) = Fix(K)

h (Fix(γ) ∩ Fix(K)) = Fix(hγh−1) ∩ Fix(K)

If γ 6∈ K, then hγh−1 6∈ hKh−1 = K. Since

LK =⋃

γ 6∈K

Fix(γ) ∩ Fix(K)

it follows that h ∈ H implies h : LK → LK

Therefore h permutes connected components ofcomplement of LK in Fix(K)

– p. 10/30

LK (1)γFix(K) = Fix(γKγ−1) . Suppose h ∈ H ⊂ N(K) then

h Fix(K) = Fix(K)

h (Fix(γ) ∩ Fix(K)) = Fix(hγh−1) ∩ Fix(K)

If γ 6∈ K, then hγh−1 6∈ hKh−1 = K. Since

LK =⋃

γ 6∈K

Fix(γ) ∩ Fix(K)

it follows that h ∈ H implies h : LK → LK

Therefore h permutes connected components ofcomplement of LK in Fix(K)

– p. 10/30

LK (2)

��

��

x(t)Fix(K)

�����������������

�����������������

Fix(γ3)Fix(γ2)

Fix(γ1)

��

��

��

��

��

���

BBBBBBBBBBBBB

H fixes conn. comp. of Fix(K) \ LK containing x(t)

Buono and G. (2001)

– p. 11/30

dim Fix(K) = 2; H 6= K

N(K)/K ⊂ O(2) acts on Fix(K)

N(K)/K ∼=

{

Zk k ≥ 2

Dk k ≥ 1

In case Dk, every nonidentity element in N(K)/K

moves x(t) to new connected component of R2 \ LK .

There is a nonidentity element in H/K. Contradiction.So N(K)/K ∼= Zk.In case Zk, let h ∈ H be nontrivial rotation on R

2

h{x(t)} = {x(t)} implies 0 ∈ Int{x(t)}

Then σ ∈ N(K) satisfies σ{x(t)} ∩ {x(t)} 6= ∅.Hence σ{x(t)} = {x(t)} and H = N(K)

– p. 12/30

dim Fix(K) = 2; H 6= K

N(K)/K ⊂ O(2) acts on Fix(K)

N(K)/K ∼=

{

Zk k ≥ 2

Dk k ≥ 1

In case Dk, every nonidentity element in N(K)/K

moves x(t) to new connected component of R2 \ LK .

There is a nonidentity element in H/K. Contradiction.So N(K)/K ∼= Zk.

In case Zk, let h ∈ H be nontrivial rotation on R2

h{x(t)} = {x(t)} implies 0 ∈ Int{x(t)}

Then σ ∈ N(K) satisfies σ{x(t)} ∩ {x(t)} 6= ∅.Hence σ{x(t)} = {x(t)} and H = N(K)

– p. 12/30

dim Fix(K) = 2; H 6= K

N(K)/K ⊂ O(2) acts on Fix(K)

N(K)/K ∼=

{

Zk k ≥ 2

Dk k ≥ 1

In case Dk, every nonidentity element in N(K)/K

moves x(t) to new connected component of R2 \ LK .

There is a nonidentity element in H/K. Contradiction.So N(K)/K ∼= Zk.In case Zk, let h ∈ H be nontrivial rotation on R

2

h{x(t)} = {x(t)} implies 0 ∈ Int{x(t)}

Then σ ∈ N(K) satisfies σ{x(t)} ∩ {x(t)} 6= ∅.Hence σ{x(t)} = {x(t)} and H = N(K)

– p. 12/30

dim Fix(K) = 2; H 6= K

N(K)/K ⊂ O(2) acts on Fix(K)

N(K)/K ∼=

{

Zk k ≥ 2

Dk k ≥ 1

In case Dk, every nonidentity element in N(K)/K

moves x(t) to new connected component of R2 \ LK .

There is a nonidentity element in H/K. Contradiction.So N(K)/K ∼= Zk.In case Zk, let h ∈ H be nontrivial rotation on R

2

h{x(t)} = {x(t)} implies 0 ∈ Int{x(t)}

Then σ ∈ N(K) satisfies σ{x(t)} ∩ {x(t)} 6= ∅.Hence σ{x(t)} = {x(t)} and H = N(K)

– p. 12/30

Polyrhythms1 2

4 5

3

Symmetry group of five-cell system is Z3 × Z2∼= Z6

Coordinates = (x, y) ∈ (Rk)3 × (R`)2

Let σ = (ρ, τ) be generator of Z3 × Z2.

Periodic solutions with (H,K) = (Z6,1) can exist if ` > 1

Fix(1) = {(x, y)} Fix(σ2) = {(x, y) : x1 = x2 = x3}

Fix(σ3) = {(x, y) : y1 = y2} Fix(σ) = Fix(σ2) ∩ Fix(σ3)

– p. 13/30

Polyrhythms (2)(σ2, 1/3) =⇒ 3-cell ring exhibits rotating wave(σ3, 1/2) =⇒ 2-cell ring is out-of-phase(σ , 1/6) =⇒ triple 2-cell freq = double 3-cell freq

0 2 4 6 8 10 12 14 16 18 20−1.5

−1

−0.5

0

0.5

1

1.5

2

t

cells

1−2

−3

0 2 4 6 8 10 12 14 16 18 20−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

t

cells

4−5

0 2 4 6 8 10 12 14 16 18 20−1.5

−1

−0.5

0

0.5

1

1.5

2

t

cells

1−4

−1.5 −1 −0.5 0 0.5 1 1.5 2−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

cell 1

cell 4

– p. 14/30

Standard GaitsBound of the Siberian Souslik

Amble of the Elephant

Trot of the HorseThanks to: Sue Morris at http://www.classicaldressage.co.uk

– p. 15/30

Standard GaitsBound of the Siberian Souslik

Amble of the Elephant

Trot of the HorseThanks to: Sue Morris at http://www.classicaldressage.co.uk

– p. 15/30

Standard GaitsBound of the Siberian Souslik

Amble of the Elephant

Trot of the HorseThanks to: Sue Morris at http://www.classicaldressage.co.uk

– p. 15/30

Standard Gait Phases

0

0 1/2

1/20 1/2

1/4 3/4

0

0

1/2

1/2

0 0

1/2 1/2

0

1/2

0.1

0.6

0

1/2

0 0

0 0

WALK PACE TROT BOUND

TRANSVERSEGALLOP

ROTARYGALLOP

PRONK0.1

0.6

– p. 16/30

Gait SymmetriesGait Spatio-temporal symmetriesTrot (Left/Right, 1

2) and (Front/Back, 1

2)

Pace (Left/Right, 1

2) and (Front/Back, 0)

Walk (Figure Eight, 1

4)

0 1 1/4

JUMP:

0 1/4 1/2 3/4 1

WALK:

0 1/2 1

BOUND:

0 1/2 1

PACE:

0 1

TROT:

0 1/2 1PRONK:

1/2

Collins and Stewart (1993)

– p. 17/30

Central Pattern Generators (CPG)Assumption: There is a network in the nervous system thatproduces the characteristic rhythms of each gait

CPG is network of neurons; neurons modeled by ODEsLocomotor CPG’s modeled by coupled cell systemsKopell and Ermentrout (1986, 1988, 1990);Rand, Cohen, and Holmes (1988); etc.

Design simplest network to produce walk, trot, and pace

Guess at simplest networkOne cell ‘signals’ each leg

1 2

43

– p. 18/30

Central Pattern Generators (CPG)Assumption: There is a network in the nervous system thatproduces the characteristic rhythms of each gait

CPG is network of neurons; neurons modeled by ODEsLocomotor CPG’s modeled by coupled cell systemsKopell and Ermentrout (1986, 1988, 1990);Rand, Cohen, and Holmes (1988); etc.

Design simplest network to produce walk, trot, and pace

Guess at simplest networkOne cell ‘signals’ each leg

1 2

43

– p. 18/30

Four Cells Do Not SufficeΓ = symmetry group of locomotor CPG network

Network produces walk. There is a four-cycle

(1 3 2 4) ∈ Γ

Four-cycle permutes pace to trot

PACE TROT

1 2

3 4

1 2

3 4

CPG cannot be modeled by four-cell networkwhere each cell gives rhythmic pulsing to one leg

– p. 19/30

Simplest Coupled Cell Gait ModelUse gait symmetries to construct coupled network1) walk =⇒ four-cycle ω in symmetry group2) pace or trot =⇒ transposition κ in symmetry group3) Simplest network

LF

LH RH

RF

LH

LF RF

RH

1 2

3 4

5 6

7 8

Γ = Z4(ω) × Z2(κ) is abelian

– p. 20/30

Trot from SymmetriesH = Z4(ω) × Z2(κ) and K = Z4(κω)

x7(t) x8(t)

x5(t) x6(t)

x3(t) x4(t)

x1(t) x2(t)

κω⇒

x2(t) x1(t)

x1(t) x2(t)

x2(t) x1(t)

x1(t) x2(t)

(κ, 12)

⇒

x1(t + 12) x1(t)

x1(t) x1(t + 12)

x1(t + 12) x1(t)

x1(t) x1(t + 12)

– p. 21/30

Primary Gaits: H = Γ = Z4(ω) × Z2(κ)

K Γ/K Phase Diagram Gait

Γ 1

0

@

0 0

0 0

1

A pronk

< ω > Z2

0

@

0 1

2

0 1

2

1

A pace

< κω > Z2

0

@

1

20

0 1

2

1

A trot

< κ, ω2 > Z2

0

@

0 0

1

2

1

2

1

A bound

< κω2 > Z4

0

@

±1

4±

3

4

0 1

2

1

A walk±

< κ > Z4

0

@

0 0

±1

4±

1

4

1

A jump±

• Primary gaits occur by Hopf bifurcation from stand

– p. 22/30

The Jump

Average Right Rear to Right Front = 31.2 frames

Average Right Front to Right Rear = 11.4 frames

31.211.4

= 2.74

– p. 23/30

Secondary Quadrupedal Gaits

H K name LH RH LF RF

κω κω loping trot x1(t) x2(t) x2(t) x1(t)

ω2 rotary gallop x1(t) x2(t) x2(t + 1

2) x1(t + 1

2)

1 rotary x1(t) x2(t) x2(t + 1

4) x1(t + 1

4)

ricocheting jump x1(t) x2(t) x2(t −1

4) x1(t −

1

4)

ω ω loping rack x1(t) x2(t) x1(t) x2(t)

ω2 transverse gallop x1(t) x2(t) x1(t + 1

2) x2(t + 1

2)

1 transverse x1(t) x2(t) x1(t + 1

4) x2(t + 1

4)

ricocheting jump x1(t) x2(t) x1(t −1

4) x2(t −

1

4)

κ, ω2 κ, ω2 loping bound x1(t) x1(t) x3(t) x3(t)

ω2 running walk x1(t) x1(t + 1

2) x3(t) x3(t + 1

2)

– p. 24/30

Primary Versus Secondary Gaits

Distinction should be observable using duty factor

Duty factors of fore legs of walking horse are equal

Duty factors of fore legs of galloping cat are different

– p. 25/30

Primary Gait: Horse Walk

Horizontal bars indicate contact with ground

– p. 26/30

Secondary Gait: Cat Rotary Gallop

Horizontal bars indicate contact with ground

– p. 27/30

Biped Network

LF

LH RH

RF

LH

LF RF

RH

1 2

3 4

5 6

7 8

3 4

1 2

left right

– p. 28/30

Biped Prediction

0 0.5 0.5 0

left right

0 0.5

left right

0 0.5

(b)(a)

– p. 29/30

Biped Gaits: Walk and Run

0 0.5 0.5 0

left right

0 0.5

left right

0 0.5

(b)(a)

Cells control timing of muscle groups

Electromyographic signals from ankle muscles

During walking: gastrocnemius (GA) andtibialis anterior (TA) are activated out-of-phase

During running: (GA) and (TA) are co-activated

– p. 30/30

Biped Gaits: Walk and Run

0 0.5 0.5 0

left right

0 0.5

left right

0 0.5

(b)(a)

Cells control timing of muscle groups

Electromyographic signals from ankle muscles

During walking: gastrocnemius (GA) andtibialis anterior (TA) are activated out-of-phase

During running: (GA) and (TA) are co-activated

– p. 30/30

Biped Gaits: Walk and Run

0 0.5 0.5 0

left right

0 0.5

left right

0 0.5

(b)(a)

Cells control timing of muscle groups

Electromyographic signals from ankle muscles

During walking: gastrocnemius (GA) andtibialis anterior (TA) are activated out-of-phase

During running: (GA) and (TA) are co-activated

– p. 30/30

Biped Gaits: Walk and Run

0 0.5 0.5 0

left right

0 0.5

left right

0 0.5

(b)(a)

Cells control timing of muscle groups

Electromyographic signals from ankle muscles

During walking: gastrocnemius (GA) andtibialis anterior (TA) are activated out-of-phase

During running: (GA) and (TA) are co-activated

– p. 30/30