Bridging steady states with renormalization group analysisfaculty.missouri.edu/~liyan/PDE.pdf · 2014-08-29 · Outline Main contents 1 Introduction Physics and heteroclinic connections

Outline

Bridging steady states with renormalizationgroup analysis

Yueheng LanDepartment of Physics

Tsinghua University

April, 2013

Yueheng Lan Bridging steady states with RG analysis

Outline

Main contents

1 IntroductionPhysics and heteroclinic connectionsRenormalization groupThe RG and differential equations

2 An extension of the RG analysis

3 Several examplesThe Lotka-Volterra model of competitionThe Gray-Scott modelThe Kuramoto-Sivashinsky equation

4 Summary


Outline

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4 Summary


Outline

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4 Summary


Outline

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4 Summary


IntroductionAn extension of the RG analysis

Several examplesSummary

Physics and heteroclinic connectionsRenormalization groupThe RG and differential equations

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4 Summary





Nonlinear dynamics and phase space

State and dynamics

x1 = f1(x1 , x2 , · · · , xn)x2 = f2(x1 , x2 , · · · , xn)· · · = · · ·xn = fn(x1 , x2 , · · · , xn)

The phase space - a geometricrepresentationVector field and trajectoriesInvariant set and organizationof trajectories





Connections





Pendulum orbits





State transition in chemical reactions





Transition orbit of Chang E I





Computation of heteroclinic connections

Exact analytic solutions under certain conditions: nonlinearintegrable systems, partially integrable systems.

Asymptotic methods for analytic approximations: localstability analysis plus interpolation.

Numerical methods: two-point boundary problem; shootingmethod; relaxation method.

Challenge:(1) Need to know orbit existence and both end points;(2) Need to know the local behavior near two ends;(3) Hard to represent the dynamics on the connection;(4) Hard to derive analytic expressions.
































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4 Summary





Renormalization group in physics

RG investigates changes ofphysical laws at differentscales.RG and scale invariance: arenormalizable system at onescale consists of self-similarcopies of itself at a smallerscale, with convergent couplingparameters when scaled up.In statistical physics: blockspin; In quantum physics:renormalization equation∂g/∂ lnµ = β(g); In nonlineardynamics: the universal routeto chaos.

Block spin renormalizationgroup for a spin systemdescribed byH(T, J):(T, J) → (T ′, J ′) →(T ′′, J ′′). Resummation?


























Main contents




4 Summary





One example

The Van der Pol equation is

d2y

dt2+ y = ε[

dy

dt− 1

3(dy/dt)3] .

A naive expansion

y = y0 + εy1 + ε2y2 + · · ·gives

y(t) = R0 sin(t + Θ0) + ε[−R30

96cos(t + Θ0) +

R02 (1− R2

0

4)(t− t0) sin(t + Θ0) +

R30

96cos 3(t + Θ0)] + O(ε2) ,

where R0 ,Θ0 are determined by the initial conditions.The expansion breaks down when ε(t− t0) > 1. Thearbitrary initial time t0 may be treated as the ultravioletcutoff in the usual field theory.





One example


d2y

dt2+ y = ε[

dy

dt− 1

3(dy/dt)3] .

A naive expansion

y = y0 + εy1 + ε2y2 + · · ·gives

y(t) = R0 sin(t + Θ0) + ε[−R30

96cos(t + Θ0) +

R02 (1− R2

0

4)(t− t0) sin(t + Θ0) +

R30

96cos 3(t + Θ0)] + O(ε2) ,






One example


d2y

dt2+ y = ε[

dy

dt− 1

3(dy/dt)3] .

A naive expansion

y = y0 + εy1 + ε2y2 + · · ·gives

y(t) = R0 sin(t + Θ0) + ε[−R30

96cos(t + Θ0) +

R02 (1− R2

0

4)(t− t0) sin(t + Θ0) +

R30

96cos 3(t + Θ0)] + O(ε2) ,






Renormalization?

Split t− t0 as t− τ + τ − t0 and absorb the terms containingτ − t0 into the renormalized counterparts R ,Θ of R0 andΘ0.Assume R0(t0) = Z1(t0, τ)R(τ) ,Θ0(t0) = Θ(τ) + Z2(t0, τ)where Z1 = 1 +

∑∞1 anεn , Z2 =

∑∞1 bnεn. The choice

a1 = −(1/2)(1−R2/4)(τ − t0) , b1 = 0 removes the secularterm to order ε:

y(t) = [R + εR

2(−R2

4)(t− τ)] sin(t + Θ)−

εR3

96cos(t + Θ) + ε

R3

96cos 3(t + Θ) + O(ε2) ,

where R ,Θ are functions of τ .





Renormalization?

Split t− t0 as t− τ + τ − t0 and absorb the terms containingτ − t0 into the renormalized counterparts R ,Θ of R0 andΘ0.Assume R0(t0) = Z1(t0, τ)R(τ) ,Θ0(t0) = Θ(τ) + Z2(t0, τ)where Z1 = 1 +

∑∞1 anεn , Z2 =

∑∞1 bnεn. The choice

a1 = −(1/2)(1−R2/4)(τ − t0) , b1 = 0 removes the secularterm to order ε:

y(t) = [R + εR

2(−R2

4)(t− τ)] sin(t + Θ)−

εR3

96cos(t + Θ) + ε

R3

96cos 3(t + Θ) + O(ε2) ,

where R ,Θ are functions of τ .





Renormalization?

The solution should not depend on τ . Therefore(∂y/∂τ)t = 0:

dR

dτ= ε

R

2(1− R2

4) + O(ε2) ,

dΘdτ

= O(ε2) .

The initial condition R(0) = 2a , Θ(0) = 0 gives

y(t) = R(t) sin(t) +ε

96R(t)3[cos(3t)− cos(t)] + O(ε2) .





Renormalization?

The solution should not depend on τ . Therefore(∂y/∂τ)t = 0:

dR

dτ= ε

R

2(1− R2

4) + O(ε2) ,

dΘdτ

= O(ε2) .

The initial condition R(0) = 2a , Θ(0) = 0 gives

y(t) = R(t) sin(t) +ε

96R(t)3[cos(3t)− cos(t)] + O(ε2) .





Series expansion of a differential equation

Suppose that we have a set of n-dimensional ODEs

x = Lx + εN(x)

We may make the expansion

x = u0 + εu1 + ε2u2 + · · ·

which results in

u0 = Lu0

u1 = Lu1 + N(u0)u2 = Lu2 + N2(u0,u1)

...





Series expansion of a differential equation

Suppose that we have a set of n-dimensional ODEs

x = Lx + εN(x)

We may make the expansion

x = u0 + εu1 + ε2u2 + · · ·

which results in

u0 = Lu0

u1 = Lu1 + N(u0)u2 = Lu2 + N2(u0,u1)

...





From the naive solution to the RG equation

This series of equations can be solved as

u0(t, t0) = eL(t−t0)A(t0)

u1(t, t0) = eL(t−t0)

∫ t

t0

e−L(τ−t0)N(eL(t−t0)A)dτ

u2(t, t0) = eL(t−t0)

∫ t

t0

e−L(τ−t0)N2(eL(t−t0)A,u1(t, t0))dτ .

The series expansion gives x = x(t; t0,A(t0)).

The RG equation is a set of equations for dA(t0)/dt0derived from

dx(t; t0,A(t0))dt0

|t=t0 = 0





One simple examle

Consider the simple example

x = y , y = −x ,

which can be solved exactly with

x = R sin(t− t0 + θ) , y = R cos(t− t0 + θ) ,

where R = R(t0) , θ = θ(t0) specify the initial condition.In phase space, orbits of the equation are circles with radiusR and azimuth angle θ.

The RG equation derived from

∂x(t;R(t0), θ(t0), t0)∂t0

|t=t0 = 0,∂y(t;R(t0), θ(t0), t0)

∂t0|t=t0 = 0

is dR(t0)/dt0 = 0, dθ(t0)/dt0 = 1 as expected.





One simple examle


x = y , y = −x ,





∂x(t;R(t0), θ(t0), t0)∂t0

|t=t0 = 0,∂y(t;R(t0), θ(t0), t0)

∂t0|t=t0 = 0






One simple examle


x = y , y = −x ,





∂x(t;R(t0), θ(t0), t0)∂t0

|t=t0 = 0,∂y(t;R(t0), θ(t0), t0)

∂t0|t=t0 = 0






The RG analysis as a coordinate transformation

Hamiltonian dynamics: action-angle variables. For aharmonic oscillator H = 1/2p2 + 1/2q2 = I.In a general nonlinear dynamical system,

x = f(x)

which has the general solution x(t) = φ(t;A0(t0), t0). Theequation

∂φ(t;A0(t0), t0)∂t0

|t=t0 = 0

gives an equation for dA0(t0)/dt0, which governs theevolution of the new coordinates A0.The RG analysis is equivalent to a coordinatetransformation in this sense, but often associated withapproximations in the nonlinear case.







x = f(x)


∂φ(t;A0(t0), t0)∂t0

|t=t0 = 0








x = f(x)


∂φ(t;A0(t0), t0)∂t0

|t=t0 = 0





Further development of the RG analysis

It can be used for nonlinear partial equations and is able toderive the phase or amplitude equations.

The invariance condition has been extended to the analysisof maps.

It is also used to determine the center manifold near abifurcation point.

Problem: for the dynamics on a submanifold, the ninvariance equations contain less than n unknowns!?




























Dynamics on a submanifold

Without loss of generality, we will concentrate on the 1-dsubmanifold. Higher dimensional ones can be treated in asimilar way.

The initial vector A should be taken asA = (A1 , 0 , 0 , · · · , 0)t.

The i-th (i 6= 1) component of u1 can be computed as

u1,i(t, t0) = eλi(t−t0)

∫ t

e−λi(τ−t0)N(eL(t−t0)A)dτ ,

where∫ t denotes integration without constant term.

The first componentdx1(t; t0, A1(t0))

dt0|t=t0 = 0 (1)

is enough to derive the RG equation for dA1(t0)/dt0, whichalso satisfies other component equations.








u1,i(t, t0) = eλi(t−t0)

∫ t




dt0|t=t0 = 0 (1)









u1,i(t, t0) = eλi(t−t0)

∫ t




dt0|t=t0 = 0 (1)









u1,i(t, t0) = eλi(t−t0)

∫ t




dt0|t=t0 = 0 (1)





A proof by mathematical induction

it is easy to write down an integral equation from its i-th(i 6= 1) component

xi(t; t0, A1(t0)) = εeλi(t−t0)

∫ t

e−λi(τ−t0)N(x(t; t0, A1(t0)))dτ .

Take t0-derivatives on both sides and impose t → t0

∂xi(t; t0, A1(t0))∂t0

|t=t0 = ε

∫ t0

e−λi(τ−t0)∇N(x(t0; t0, A1(t0)))

· ∂x(t; t0, A1(t0))∂t0

|t=t0dτ ∼ O(εm+1) .

Our assertion is surely true for m = 0. By induction, it istrue for all values of m.







∫ t

e−λi(τ−t0)N(x(t; t0, A1(t0)))dτ .


∂xi(t; t0, A1(t0))∂t0

|t=t0 = ε

∫ t0

e−λi(τ−t0)∇N(x(t0; t0, A1(t0)))

· ∂x(t; t0, A1(t0))∂t0

|t=t0dτ ∼ O(εm+1) .








∫ t

e−λi(τ−t0)N(x(t; t0, A1(t0)))dτ .


∂xi(t; t0, A1(t0))∂t0

|t=t0 = ε

∫ t0

e−λi(τ−t0)∇N(x(t0; t0, A1(t0)))

· ∂x(t; t0, A1(t0))∂t0

|t=t0dτ ∼ O(εm+1) .





The Lotka-Volterra model of competitionThe Gray-Scott modelThe Kuramoto-Sivashinsky equation

Main contents




4 Summary





The Lotka-Volterra model

The Lotka-Volterra model ofcompetition is

x = x(3− x− 2y)y = y(2− x− y)

The model describes thecompetition between therabbits and the sheep fed onthe grass of the same lawn.Vector field and trajectories

four equilibriaP1 = (0, 0) , P2 = (0, 2) , P3 = (1, 1) , P4 = (3, 0).Their approximation is(1, 1) , (2.908,−0.003) , (−0.113, 2.105).







x = x(3− x− 2y)y = y(2− x− y)









x = x(3− x− 2y)y = y(2− x− y)









x = x(3− x− 2y)y = y(2− x− y)









x = x(3− x− 2y)y = y(2− x− y)







Series solution

Around the saddle P3, we take a coordinate transformation

x = 1−√

23z +

√23w , y = 1 +

√13z +

√13w .

Assume

z = εz1 + ε2z2 + ε3z3 + O(ε4) (2)w = εw1 + ε2w2 + ε3w3 + O(ε4) , (3)

we have

L ◦ z1 = 0 , M ◦ w1 = 0L ◦ z2 = F2(z1, w1) , M ◦ w2 = G2(z1, w1)L ◦ z3 = F3(z1, w1, z2, w2) , M ◦ w3 = G3(z1, w1, z2, w2) ,

where L ≡ 1−√

2 + ddt ,M ≡ 1 +

√2 + d

dt andF2 , G2 , F3 , G3 are polynomial functions of their arguments.





Series solution


x = 1−√

23z +

√23w , y = 1 +

√13z +

√13w .

Assume


we have


where L ≡ 1−√

2 + ddt ,M ≡ 1 +

√2 + d






Series solution


x = 1−√

23z +

√23w , y = 1 +

√13z +

√13w .

Assume


we have


where L ≡ 1−√

2 + ddt ,M ≡ 1 +

√2 + d






Series solution (continued)

The general solution is

z1(t) = a(t0)e(√

2−1)(t−t0) , w1(t) = b(t0)e−(1+√

2)(t−t0) .

Set b(t0) = 0 and the solution is

z = εa(t0)e(√

2−1)(t−t0) +√

3ε2a2(t0)6

(√

2− 1)

(e2(√

2−1)(t−t0) − e(√

2−1)(t−t0)) + O(ε3)

w =√

3ε2a2(t0)102

(1 + 3√

2)e2(√

2−1)(t−t0) + O(ε3) .

From ∂z(t, t0)/∂t0 = 0, we get

da(t0)dt0

= a

(√

2− 1− 17√

3(3− 2√

2)102

εa

).







z1(t) = a(t0)e(√

2−1)(t−t0) , w1(t) = b(t0)e−(1+√

2)(t−t0) .


z = εa(t0)e(√

2−1)(t−t0) +√

3ε2a2(t0)6

(√

2− 1)

(e2(√

2−1)(t−t0) − e(√

2−1)(t−t0)) + O(ε3)

w =√

3ε2a2(t0)102

(1 + 3√

2)e2(√

2−1)(t−t0) + O(ε3) .


da(t0)dt0

= a

(√

2− 1− 17√

3(3− 2√

2)102

εa

).







z1(t) = a(t0)e(√

2−1)(t−t0) , w1(t) = b(t0)e−(1+√

2)(t−t0) .


z = εa(t0)e(√

2−1)(t−t0) +√

3ε2a2(t0)6

(√

2− 1)

(e2(√

2−1)(t−t0) − e(√

2−1)(t−t0)) + O(ε3)

w =√

3ε2a2(t0)102

(1 + 3√

2)e2(√

2−1)(t−t0) + O(ε3) .


da(t0)dt0

= a

(√

2− 1− 17√

3(3− 2√

2)102

εa

).





Main contents




4 Summary





The Dependence on β

The Gray-Scott model represents the cubic autocatalyticchemical reactions for two chemical species. The stationarypatterns are described by

u′′ = uv2 − λ(1− v)γv′′ = v − uv2 .

The equation is invariant under x → −x. Two heteroclinicorbits exist at γ = 2/9 and λ = 9/2, together with threeequilibria P1 = (1, 0) , P2 = (1/3, 3) , P3 = (2/3, 3/2)It can be converted to a 4-d dynamical system

u′ = p

p′ = uv2 − λ(1− v)v′ = w

w′ = v − uv2 .







u′′ = uv2 − λ(1− v)γv′′ = v − uv2 .


u′ = p

p′ = uv2 − λ(1− v)v′ = w

w′ = v − uv2 .







u′′ = uv2 − λ(1− v)γv′′ = v − uv2 .


u′ = p

p′ = uv2 − λ(1− v)v′ = w

w′ = v − uv2 .





Series solution

The stability exponents of P1 are ±3√

2/2, both beingdoubly degenerate. We have to use two parametersr0(x0) , r1 to parametrize the initial position.we obtain

u = 1 + ε(−√

2r0f(x, x0)3

) + ε24

243(f2(x , x0)− f(x , x0))r2

0r21 + · · ·

p = εf(x , x0)r0 − ε22√

281

(2f2(x , x0)− f(x , x0))r20r

21 + · · ·

v = ε(−√

2r0r1f(x, x0)3

)− ε2227

(f2(x , x0)− f(x , x0))r20r

21 + · · ·

w = εf(x , x0)r0r1 − ε2√

29

(2f2(x , x0)− f(x , x0))r20r

21 + · · · ,

where

f(x , x0) = exp(−3√

22

(x− x0)) .





Series solution

The stability exponents of P1 are ±3√

2/2, both beingdoubly degenerate. We have to use two parametersr0(x0) , r1 to parametrize the initial position.we obtain

u = 1 + ε(−√

2r0f(x, x0)3

) + ε24

243(f2(x , x0)− f(x , x0))r2

0r21 + · · ·

p = εf(x , x0)r0 − ε22√

281

(2f2(x , x0)− f(x , x0))r20r

21 + · · ·

v = ε(−√

2r0r1f(x, x0)3

)− ε2227

(f2(x , x0)− f(x , x0))r20r

21 + · · ·

w = εf(x , x0)r0r1 − ε2√

29

(2f2(x , x0)− f(x , x0))r20r

21 + · · · ,

where

f(x , x0) = exp(−3√

22

(x− x0)) .





The RG equation

The RG equation for r0(x0) is given by setting∂u(x, x0)/∂x0 = 0 followed by taking x → x0

dr0(x0)dx0

= −3r0√2

+227

εr20r

21 +

r0

21870(−45

√2r2

1(9 + 2r1)ε2r20

+ 8r31(9 + 2r1)ε3r3

0) + · · · .

By setting r1 = −9/2, we have

dr0(x0)dx0

= − 3√2r0 +

32r20 .

which has the solution

r0(x0) =√

22

(1− tanh3x0

2√

2) .





The RG equation

The RG equation for r0(x0) is given by setting∂u(x, x0)/∂x0 = 0 followed by taking x → x0

dr0(x0)dx0

= −3r0√2

+227

εr20r

21 +

r0

21870(−45

√2r2

1(9 + 2r1)ε2r20

+ 8r31(9 + 2r1)ε3r3

0) + · · · .

By setting r1 = −9/2, we have

dr0(x0)dx0

= − 3√2r0 +

32r20 .

which has the solution

r0(x0) =√

22

(1− tanh3x0

2√

2) .





The profile of the exact solution

The exact analytic solution of the original equation is thus

u(x) = 1−√

23

r0(x) =13(2 + tanh

3x

2√

2)

v(x) =3√2r0(x) =

32(1− tanh

3x

2√

2) .





Main contents




4 Summary





The Kuramoto-Sivashinsky equation

The Kuramoto-Sivashinsky equation is an importantphysics model

ut = (u2)x − uxx − νuxxxx ,

where ν > 0 is the hyper-viscosity parameter.With periodic boundary condition on [0 , 2π], we mayexpand

u(t, x) = i

∞∑k=−∞

akeikx .

For the antisymmetric solution u(t,−x) = −u(t, x), ak isreal and a−k = −ak. The equation becomes

ak = (k2 − νk4)ak − k

∞∑m=−∞

amak−m .









u(t, x) = i

∞∑k=−∞

akeikx .


ak = (k2 − νk4)ak − k

∞∑m=−∞

amak−m .









u(t, x) = i

∞∑k=−∞

akeikx .


ak = (k2 − νk4)ak − k

∞∑m=−∞

amak−m .





Perturbation analysis

Assumeak = εak,1 + ε2ak,2 + ε3ak,3 + · · · .

For the 1− d unstable manifold of the origin at ν < 1, wemay get

a1,1(t, t0) = r(t0)e(1−ν)(t−t0) , ak,1 = 0 for k > 1 .

where r(t0) is the renormalization parameter.The RG equation for r(t0) is obtained fromda1(t, t0)/dt0|t=t0 = 0:

dr0

dt0= (1− v)r0 +

2r30

1− 7ν− 6r5

0

(1− 7ν)2(−1 + 13ν)+ · · · .










dr0

dt0= (1− v)r0 +

2r30

1− 7ν− 6r5

0

(1− 7ν)2(−1 + 13ν)+ · · · .










dr0

dt0= (1− v)r0 +

2r30

1− 7ν− 6r5

0

(1− 7ν)2(−1 + 13ν)+ · · · .





The time evolution on the connection

ν = 0.5





The manifold and physical observable

ν = 0.5





The time evolution on the connection

ν = 0.3





The manifold and physical observable

ν = 0.3

[Y. Lan, Phys. Rev. E 87, 012914(2013)]




Summary

An extension of the RG method has been proposed and wassuccessfully used for the determination of heteroclinic orbitsin the phase space.The method was applied to three typical physical systems:the Lotka-Volterra model of competition, the Gray-Scottmodel and the Kuramoto-Sivasshinsky equation.There seems no obstacles to generalize the currenttechnique to the treatment of dynamics on invariantsubmanifolds of dimension higher than one.Problems and challenges:(1) How to adapt the current scheme to the oscillatory caseis an interesting problem.(2) What if eigen-directions are not known.(3) How to treat more complex connections.




Summary





Summary





Summary



Documents

Bridging steady states with renormalization group analysisfaculty.missouri.edu/~liyan/PDE.pdf · 2014-08-29 · Outline Main contents 1 Introduction Physics and heteroclinic connections