Symmetry, Functional Analisis and Non-linear...

Symmetry, Functional Analisis and Non-linearPhysics

Renato Alvarez-Nodarse

IMUS & Dpto. Analisis Matematico, Universidad de Sevilla

v ∼ ǫ21ǫ2 cos(φ+ θ)

E.I.T.A. 2014, Alquezar, 17-19 de octubre de 2014

Renato Alvarez-Nodarse Symmetry, Functional Analisis and Non-linear Physics

Centenario del nacimiento de Luis Vigil


El Teorema de Galileo

Galileo: La filosofıa [natural]esta escrita en ese grandioso libroque tenemos abierto ante los ojos,(quiero decir, el universo), perono se puede entender si antes nose aprende a entender la lengua, aconocer los caracteres en los queesta escrito. Esta escrito en lenguamatematica y sus caracteres sontriangulos, cırculos y otras figurasgeometricas, sin las cuales esimposible entender ni una palabra;sin ellos es como girar vanamenteen un oscuro laberinto.


El Teorema de Galileo

Galileo: La filosofıa [natural]esta escrita en ese grandioso libroque tenemos abierto ante los ojos,(quiero decir, el universo), perono se puede entender si antes nose aprende a entender la lengua, aconocer los caracteres en los queesta escrito. Esta escrito en lenguamatematica y sus caracteres sontriangulos, cırculos y otras figurasgeometricas, sin las cuales esimposible entender ni una palabra;sin ellos es como girar vanamenteen un oscuro laberinto.

⇒ Corolario: Si queremos conocer el mundo que nos rodeatenemos que saber Matematicas.


Invariance (Symmetry)



Invariance: a thing is symmetric if there is something we can

do to it so that after we have done it, it looks the same as it did

before.

Let us consider a .

Transformations: We can do thing to our : T : 7→ ′: To donothing, cut it, compress it, deform it, rotate it ...



Invariance: a thing is symmetric if there is something we can

do to it so that after we have done it, it looks the same as it did

before.

Let us consider a .

Transformations: We can do thing to our : T : 7→ ′: To donothing, cut it, compress it, deform it, rotate it ...

Invariance: The way the new ′ looks should be identical to theoriginal


Playing with the symmetry



A

B

C



B

C

A



A

B

C


A very important Symmetry: The time-shift symmetry

What time-shift invariant of the measurement means?

Time-shift invariant: The experiments can be repeated at anytime. Under the same conditions, the same results are obtained.


Motivation: A problem from Physics

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!

!!!!!!!!!!!!!!!!!!

system

π

−3.12.3

System: device, equations, experiments, simulations...

Input: harmonic (sinusoidal) functions

Output: a number, which is an average measurement


Motivation: A problem from Physics

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!

!!!!!!!!!!!!!!!!!!

system

π

−3.12.3

System: device, equations, experiments, simulations...

Input: harmonic (sinusoidal) functions

Output: a number, which is an average measurement

It has been found in experiments, simulations, . . . that theoutput has the same functional dependence on the parameters ofthe input

Why?


Math. formulation of the problem

The output is a functional of an arbitrary number, n, of externalforces fi (t), i = 1, 2, . . . , n

In the following we will denote the output (functional) byv [f(t)], and the input will the the vector f(t).

An special emphasis will be done when fi (t) = ǫi cos(qiωt + φi ),i = 1, 2, . . . , n )






v [f] is time-shift invariant functional, i.e., for any τ

v [f(t)] = v [f(t + τ)]






v [f] is time-shift invariant functional, i.e., for any τ

v [f(t)] = v [f(t + τ)]

Implications in Physics: the experiments can be repeated at anytime. Under the same conditions, the same results are obtained.

Implications in Mathematics: the dependence of v of theamplitudes and phases of f are fixed and can be calculated as longas v is sufficiently smooth.


Paradigmatic example: The ratchet effect

Ratchet effect: net motion of particles or solitons induced byperiodic or random, zero-average external forces f.




The motion of cold atoms in symmetric optical lattices, (Schiavoni

et al. PRL 2003; Gommers, et al. PRL 2005; PRL 2006; PRA 2007; PRL 2008). The motion offluxons in the Josephson junctions (Ustinov et al. PRL 2004; Ooi et al. PRL 2007).




The motion of cold atoms in symmetric optical lattices, (Schiavoni

et al. PRL 2003; Gommers, et al. PRL 2005; PRL 2006; PRA 2007; PRL 2008). The motion offluxons in the Josephson junctions (Ustinov et al. PRL 2004; Ooi et al. PRL 2007).

Measurement of x(t) (position of particles or non-linear waves)

Ratchet velocity v 6= 0 is an average velocity induced by 〈f〉 = 0

v := v [f(t)] = lımt→+∞

1

t

∫ t

0x(τ) dτ = lım

t→+∞

x(t)− x(0)

t.


Universality: “Same cause” ⇒ “same effect”

A fact: Very different non-linear systems driven by the forcef (t) = ǫ1 cos(pΩt) + ǫ2 cos(qΩt + φ)

Experiments: optical lattices, ferrofluids, semiconductors,Josephson junctions . . .

Simulations: dissipative dynamics of particles in symmetricpotentials

Simulations: dissipative dynamics of non-linear Schrodingersolitons in symmetric potentials

Theory: dissipative dynamics of sine-Gordon and φ4 solitons

Theory: overdamped dynamics of particles in symmetricpotentials

have the same behaviour:

v [f(t)] = Aǫp1ǫq2 cos(pφ+ θ0)


A special case: f (t) = ǫ1 cos(Ωt) + ǫ2 cos(2Ωt + φ)

A first attempt for understanding this: The method of moments . . .

which is based on the assumption that v [f(t)] can be spanned asthe Taylor series

x(t) =∞∑

k=1

ck(f (t))2k+1 ⇒

v [f(t)] = 〈x〉 =∞∑

k=1

c2k+1〈(f (t))2k+1〉,

where here, we use the notation 〈X 〉 := 1/T∫ T

0 X (t) dt.





x(t) =∞∑

k=1

ck(f (t))2k+1 ⇒

v [f(t)] = 〈x〉 =∞∑

k=1

c2k+1〈(f (t))2k+1〉,


0 X (t) dt.

First approximation: v [f(t)] = A〈(f (t))3〉 = Aǫ21ǫ2 cos(φ)





x(t) =∞∑

k=1

ck(f (t))2k+1 ⇒

v [f(t)] = 〈x〉 =∞∑

k=1

c2k+1〈(f (t))2k+1〉,


0 X (t) dt.

First approximation: v [f(t)] = A〈(f (t))3〉 = Aǫ21ǫ2 cos(φ)

What happens with the phase! v [f(t)] = Aǫ21ǫ2 cos(φ+ θ0)


Why the methods of moments does not work?

Because the assumption that v [f(t)] can be spanned as Taylorseries is wrong: v [f(t)] is a functional, thus we should use afunctional series expansion: f(t) = (f1(t), f2(t))

v [f] =

∞∑

n1=0

∞∑

n2=0

〈cn1,n2(t11, · · · , t1n1 , t21, . . . , t2n2)

× f1(t11) · · · f1(t1n1)f2(t21) · · · f2(t2n2)〉,

where the kernels cn1,n2 are real, T -periodic and symmetric. Here

〈Ω(t1, . . . , tr )〉 =1

T r

∫ T

0dt1 · · ·

∫ T

0dtr Ω(t1, . . . , tr )


But the situation is worst ...

The method of moments is wrong. Just one example: Let be asystem defined by (x is the position, u is the velocity)

d

dt

Mu√1− u2

= −f (t)− γu√

1− u2,

dx

dt= u(t),

f (t) = ǫ1 cos(ωt+φ1)+ǫ2 cos(2ωt+φ2), u(0) = u0, x(0) = x0,




d

dt

Mu√1− u2

= −f (t)− γu√

1− u2,

dx

dt= u(t),


It has an exact solution: u(t) =

∞∑

k=0

(−1)k(1/2)kk!M2k+1

[P(t)]2k+1

donde P(t) = · · · and v = Bǫ21ǫ2 cos(2φ1 − φ2 + θ0︸︷︷︸

6=0

).




d

dt

Mu√1− u2

= −f (t)− γu√

1− u2,

dx

dt= u(t),



∞∑

k=0

(−1)k(1/2)kk!M2k+1

[P(t)]2k+1


6=0

).

If we use the force f (t) =

ǫ1 if 0< t<Tl

0 if Tl < t<T−Tl

−ǫ1 if T−Tl < t<T

⇒




d

dt

Mu√1− u2

= −f (t)− γu√

1− u2,

dx

dt= u(t),



∞∑

k=0

(−1)k(1/2)kk!M2k+1

[P(t)]2k+1


6=0

).

If we use the force f (t) =

ǫ1 if 0< t<Tl

0 if Tl < t<T−Tl

−ǫ1 if T−Tl < t<T

⇒〈f 2k+1〉 = 0

⇓v = 0


A little math ...

We recall that v [f] not an arbitrary functional. It is

a time-shift invariant functional, ∀τ

v [f(t)] = v [f(t + τ)]


A little math ...



v [f(t)] = v [f(t + τ)]

How the kernels are affected by this symmetry?

v [f] =

∞∑

n1=0

∞∑

n2=0

〈cn1,n2(t11, · · · , t1n1 , t21, . . . , t2n2)

× f1(t11) · · · f1(t1n1)f2(t21) · · · f2(t2n2)〉.

⇓


A little math ...



v [f(t)] = v [f(t + τ)]

How the kernels are affected by this symmetry?

v [f] =

∞∑

n1=0

∞∑

n2=0

〈cn1,n2(t11, · · · , t1n1 , t21, . . . , t2n2)

× f1(t11) · · · f1(t1n1)f2(t21) · · · f2(t2n2)〉.

⇓Fundamental property ∀τ ∈ (0,T ):

cn1,n2(t11−τ, · · · , t1n1−τ, t21−τ, . . . , t2n2−τ) = cn1,n2(t11, · · · , t1n1 , t21, . . . , t2n2)


Functional Taylor expansion of time-shift invariant functional v [f1, f2, · · · ]

Theorem: Let f(t) = ǫ1 cos(q1ωt + φ1), ǫ2 cos(q2ωt + φ2).If ∀τ v [f(t + τ)] = v [f(t)] and v [f] smooth enough then

v [f1, f2] = C0(ǫ1, ǫ2) +∑

x∈D+

ǫ|x1|1 ǫ

|x2|2 Cx(ǫ1, ǫ2) cos(x ·φ+ θx(ǫ1, ǫ2)).

x := (x1, x2) ∈ D+: set of the solutions of the DiophantineEq. q1x1 + q2x2 = 0 with x1 > 0.

φ = (φ1, φ2).

Cx and θx are functions of ǫ21 and ǫ22.


Functional Taylor expansion of time-shift invariant functional v [f1, f2, · · · ]

Theorem: Let f(t) = ǫ1 cos(q1ωt + φ1), ǫ2 cos(q2ωt + φ2).If ∀τ v [f(t + τ)] = v [f(t)] and v [f] smooth enough then

v [f1, f2] = C0(ǫ1, ǫ2) +∑

x∈D+

ǫ|x1|1 ǫ

|x2|2 Cx(ǫ1, ǫ2) cos(x ·φ+ θx(ǫ1, ǫ2)).

x := (x1, x2) ∈ D+: set of the solutions of the DiophantineEq. q1x1 + q2x2 = 0 with x1 > 0.

φ = (φ1, φ2).

Cx and θx are functions of ǫ21 and ǫ22.

Equivalently

v=C0(ǫ1, ǫ2)+∞∑

k=1

(ǫq21 ǫq12 )kCk(ǫ1, ǫ2) cos(k(q1φ2−q2φ1)+θk(ǫ1, ǫ2))


Theory vs. Experiments: going down to the Earth

small amplitude limit: v = C0ǫ21ǫ2 cos(φ+ θ0), agrees with the

simulations and experiments e.g. optical lattices, bi-harmonic force.





intermediate amplitudes: v = vmax(ǫ1, ǫ2) cos(φ+ θ1(ǫ1, ǫ2))(agrees with experiments on optical lattices)





intermediate amplitudes: v = vmax(ǫ1, ǫ2) cos(φ+ θ1(ǫ1, ǫ2))(agrees with experiments on optical lattices)

large amplitudes: Prediction of the Theory: By increasing theamplitude of the forces, the direction of the motion can bechanged! This has been observed in experiments: optical lattices,bi-harmonic force. Cubero, et. al., PRE 2010; also in simulations ofthe NLKG, etc.

+ a lot of interesting Math Corollaries ...


An astonishing phenomena: Not math model at all!

(PRL, Noblin et. al), fv = ǫ1 cos(ωt + φ), fh = ǫ2 cos(ωt).

0 π/2 π 3π/2 2πφ

-1

-0.5

0

0.5

1

v (cm/s)

v [−fv , fh] = −v [fv , fh]




0 π/2 π 3π/2 2πφ

-1

-0.5

0

0.5

1

v (cm/s)

v [−fv , fh] = −v [fv , fh]

v = vmax(ǫ1, ǫ2) cos(φ+ θ1(ǫ1, ǫ2)) + v2(ǫ1, ǫ2) cos(3φ+ θ2(ǫ1, ǫ2)) + · · · large amplitude regime: v is no longer a sinusoidal function




0 π/2 π 3π/2 2πφ

-1

-0.5

0

0.5

1

v (cm/s)

v [−fv , fh] = −v [fv , fh]

v = vmax(ǫ1, ǫ2) cos(φ+ θ1(ǫ1, ǫ2)) + v2(ǫ1, ǫ2) cos(3φ+ θ2(ǫ1, ǫ2)) + · · · large amplitude regime: v is no longer a sinusoidal function

Theory vs experiment: v = a cos(φ+ b) + c cos(3φ+ d)


A curious phenomena: Could dissipation enhance the transport?

v =∞∑

k=1(odd)

(ǫq21 ǫq12 )kCk(ǫ1, ǫ2) cos(kq1φ+ θk(ǫ1, ǫ2)).



v =∞∑

k=1(odd)


Time-reversal symmetry v [f1(−t), f2(−t)] = −v [f1(t), f2(t)]⇐⇒ θk = ±π/2 (Hamiltonian limit)

Time-reversal symmetry v [f1(−t), f2(−t)] = v [f1(t), f2(t)] ⇐⇒θk = 0 or π (Overdamped systems)

Time-reversal symmetries fix the “phase lag”



v =∞∑

k=1(odd)


Time-reversal symmetry v [f1(−t), f2(−t)] = −v [f1(t), f2(t)]⇐⇒ θk = ±π/2 (Hamiltonian limit)

Time-reversal symmetry v [f1(−t), f2(−t)] = v [f1(t), f2(t)] ⇐⇒θk = 0 or π (Overdamped systems)

Time-reversal symmetries fix the “phase lag”

By setting φ = 0, close to the Hamiltonian limit, v = 0. Byincreasing the damping, the transport is induced. We have beensee this in the equation for a relativistic particle discussed before!


Main ingredients of the proof: fi = ǫi cos(qiωt + φi)

v [f] =

∞∑

n1=0

∞∑

n2=0

〈cn1,n2(t11, · · · , t1n1 , t21, . . . , t2n2)

× f1(t11) · · · f1(t1n1)f2(t21) · · · f2(t2n2)〉,

∀τ, v [f(t + τ)] = v [f(t)] ⇒ cn(t1+ τ, · · · , tn + τ) = cn(t1, · · · , tn)



v [f] =

∞∑

n1=0

∞∑

n2=0

〈cn1,n2(t11, · · · , t1n1 , t21, . . . , t2n2)

× f1(t11) · · · f1(t1n1)f2(t21) · · · f2(t2n2)〉,


v [f] =∑

k,l∈N20

ǫk1+l11 ǫk2+l2

2 e i(k−l)·φA(k, l),

time-shift invariace ⇒ A(k, l) = 0 if q · (k− l) 6= 0

q · (k− l) = 0 Diophantine Eq.



v [f] =

∞∑

n1=0

∞∑

n2=0

〈cn1,n2(t11, · · · , t1n1 , t21, . . . , t2n2)

× f1(t11) · · · f1(t1n1)f2(t21) · · · f2(t2n2)〉,


v [f] =∑

k,l∈N20

ǫk1+l11 ǫk2+l2




The solutions of the Eq. q · (k− l) = 0 determine k, l



v [f] =

∞∑

n1=0

∞∑

n2=0

〈cn1,n2(t11, · · · , t1n1 , t21, . . . , t2n2)

× f1(t11) · · · f1(t1n1)f2(t21) · · · f2(t2n2)〉,


v [f] =∑

k,l∈N20

ǫk1+l11 ǫk2+l2




The solutions of the Eq. q · (k− l) = 0 determine k, l

For arbitrary n see (Cuesta, Quintero, Renato, PRX 2013)Renato Alvarez-Nodarse Symmetry, Functional Analisis and Non-linear Physics

A simple proof ... Why another proof?



Input: fj(t) = ǫj cos [ωj(t − t0) + ϕj ]




This set of periodic functions is invariant under these twotransformations

T1 : t0,ω,ϕ, ǫ 7−→ t0,ω,ϕ+ π (j), ǫ(j) ,T2 : t0,ω,ϕ, ǫ 7−→ 0,ω,ϕ− t0ω, ǫ ,

where

(j) canonical basis of Rn and the vector ǫ(j) is obtainedfrom the vector ǫ by replacing its jth component by −ǫj .




This set of periodic functions is invariant under these twotransformations

T1 : t0,ω,ϕ, ǫ 7−→ t0,ω,ϕ+ π (j), ǫ(j) ,T2 : t0,ω,ϕ, ǫ 7−→ 0,ω,ϕ− t0ω, ǫ ,

where

(j) canonical basis of Rn and the vector ǫ(j) is obtainedfrom the vector ǫ by replacing its jth component by −ǫj .Let Υ represent a certain (physical) quantity of the system:

Υ(t0,ϕ, ǫ) = Υ(t0,ϕ+ π (j), ǫ(j)) = Υ(0,ϕ− ωt0, ǫ) ,

By applying the first equality in the above equation twice, we seethat Υ is periodic with respect to all the components of the vectorϕ with period 2π.


A simple proof ... Υ can be expanded in Fourier series as

Υ(ζ0,ϕ, ǫ) =∑

∈ZN

υ

(ǫ)e i(ϕ−Ωζ0)· (in real form)

! "

! "



Υ(ζ0,ϕ, ǫ) =∑

∈ZN

υ


=∑

∈ZN

c

(ǫ) cos [(ϕ− ωt0) · + χ

(ǫ)] , (1)

If Υ is time-shift invariant ⇒ ω · = 0.

! "

! "



Υ(ζ0,ϕ, ǫ) =∑

∈ZN

υ


=∑

∈ZN

c

(ǫ) cos [(ϕ− ωt0) · + χ

(ǫ)] , (1)

If Υ is time-shift invariant ⇒ ω · = 0.If ω = (q1ω, · · · , qsω) weobtain the Diophantine Eq.

! "

! "



Υ(ζ0,ϕ, ǫ) =∑

∈ZN

υ


=∑

∈ZN

c

(ǫ) cos [(ϕ− ωt0) · + χ

(ǫ)] , (1)

If Υ is time-shift invariant ⇒ ω · = 0.If ω = (q1ω, · · · , qsω) weobtain the Diophantine Eq.

Case t0 = 0: fj(t) = ǫj cos [ωj t + ϕj ] = αj cos(ωj t) + βj sin(ωj t)

and analytic output: Υ =∑

! ,"∈NN0a! ,"

∏Nj=1 α

ljj β

rjj , then computing

the Fourier coeff.

υ

(ǫ) =

∫ π

−π. . .

∫ π

−π

dNϕ

(2π)NΥ(0,ϕ, ǫ)e−iϕ· .

givesRenato Alvarez-Nodarse Symmetry, Functional Analisis and Non-linear Physics

Υ can be expanded as the following series

Υ(ϕ, ǫ) =∑

k∈D+

|γ

(ǫ)|

N∏

j=1

ǫ|kj |j

cos [ϕ · + χ

(ǫ)] ,

γ

(ǫ) =∑

!∈NN0

b ,!

N∏

j=1

ǫ2pjj

that is the same formula obtained using the functional analysis.


Thank you for your attention!

Cuesta, Quintero, RAN, Physical Review X 3, 041014 (2013).


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