geometric integration for damped hamiltonian pdes€¦ · geometric integration for damped...

Geometric Integration forDamped Hamiltonian PDEs

Brian E. Moore, Laura Norena, and Constance Schober

bmoore@math.ucf.edu

University of Central Florida

SIAM Conference on Nonlinear Waves and Coherent Structures, University of

Washington, Seattle, June 13, 2011

Thanks to the National Science Foundation for partial funding

Jul 2011 – p.1/16

Multi-Symplectic PDE, Bridges (1997)

These equations can be written

Kzt + Lzx = ∇zS(z)

K and L are skew-symmetric matrices

z = z(x, t) is the vector of state variables

S is smooth

Examples include KdV, Boussinesq, Zakharov-Kuznetsov,nonlinear Schrödinger, nonlinear wave equations, etc.

Conservation laws are derived directly from the equation.An example is the multi-symplectic conservation law∂t 〈KU, V 〉+ ∂x 〈LU, V 〉 = 0, where U and V satisfy thevariational equation.

Jul 2011 – p.2/16

Conformal Multi-Symplectic PDE, M. (2009)

These equations can be written

Kzt + Lzx = ∇zS(z)−a

2Kz −

b

2Lz + F (x, t)

a and b are positive real constants

S is smooth and may depend on a and b

F is a forcing term

Equation satisfies a conformal multi-symplectic conservation law

∂t 〈KU, V 〉+ ∂x 〈LU, V 〉 = −a 〈KU, V 〉 − b 〈LU, V 〉.

Methods that preserve this property behave similarly tostandard multi-symplectic methods for small a and b.

Jul 2011 – p.3/16

General Conformal Conservation Laws

IfKzt + Lzx = ∇zS(z)

has a conservation law

∂tP + ∂xQ = 0,

then

Kzt + Lzx = ∇zS(z)−a

2Kz −

b

2Lz

has a conformal conservation law if it satisfies

∂tP + ∂xQ = −aP − bQ.

Jul 2011 – p.4/16

Examples of Conservation Laws for

Kzt + Lzx = ∇zS(z)

Energy: inner product of the equation with zt

∂t(

S(z) + 12(z

Tx Lz)

)

+ ∂x12(z

TLzt) = 0

Jul 2011 – p.5/16

Examples of Conservation Laws for

Kzt + Lzx = ∇zS(z)

Energy: inner product of the equation with zt

∂t(

S(z) + 12(z

Tx Lz)

)

+ ∂x12(z

TLzt) = 0

Momentum: inner product of the equation with zx

∂x(

S(z) + 12(z

Tt Kz)

)

+ ∂t12(z

TKzx) = 0

Jul 2011 – p.5/16

Examples of Conservation Laws for

Kzt + Lzx = ∇zS(z)

Energy: inner product of the equation with zt

∂t(

S(z) + 12(z

Tx Lz)

)

+ ∂x12(z

TLzt) = 0

Momentum: inner product of the equation with zx

∂x(

S(z) + 12(z

Tt Kz)

)

+ ∂t12(z

TKzx) = 0

Linear Symmetries: inner product with Bz

∂t(

zTKBz)

+ ∂x(

zTLBz)

= 0

Jul 2011 – p.5/16

Examples of Conformal Conservation Laws for

Kzt + Lzx = ∇zS(z)−a2Kz − b

2Lz

Energy: inner product of the equation with zt

∂t(

S(z) + 12(z

Tx Lz)

)

+ ∂x12(z

TLzt) =

−b2 (zTLzt)

Momentum: inner product of the equation with zx

∂x(

S(z) + 12(z

Tt Kz)

)

+ ∂t12(z

TKzx) =

−a2 (zTKzx)

Linear Symmetries: inner product with Bz

∂t(

zTKBz)

+ ∂x(

zTLBz)

= −a(

zTKBz)

− b(

zTLBz)

Jul 2011 – p.6/16

Numerical Preservation

Suppose we have the following PDE and Conformal CL

Kzt + Lzx = ∇S(z)−a

2Kz ∂tP + ∂xQ = −aP

Jul 2011 – p.7/16

Numerical Preservation

Suppose we have the following PDE and Conformal CL

Kzt + Lzx = ∇S(z)−a

2Kz ∂tP + ∂xQ = −aP

Integrating with appropriate boundary conditions, yields

∂tP = −aP ⇐⇒ P (t) = exp(−at)P (0) with P =

∫

Pdx.

Jul 2011 – p.7/16

Numerical Preservation

Suppose we have the following PDE and Conformal CL

Kzt + Lzx = ∇S(z)−a

2Kz ∂tP + ∂xQ = −aP

Integrating with appropriate boundary conditions, yields

∂tP = −aP ⇐⇒ P (t) = exp(−at)P (0) with P =

∫

Pdx.

A numerical method preserves this property if it satisfies

P i+1 = exp(−a∆t)P i with P i =∑

n

Pn,i∆x.

Jul 2011 – p.7/16

Numerical Preservation

Suppose we have the following PDE and Conformal CL

Kzt + Lzx = ∇S(z)−a

2Kz ∂tP + ∂xQ = −aP

Integrating with appropriate boundary conditions, yields

∂tP = −aP ⇐⇒ P (t) = exp(−at)P (0) with P =

∫

Pdx.

A numerical method preserves this property if it satisfies

P i+1 = exp(−a∆t)P i with P i =∑

n

Pn,i∆x.

Note: ∂tP < 0 is typically considered preserved if P i+1 < P i

Jul 2011 – p.7/16

Splitting Methods, McLachlan and Quispel (2002)

zt = FH(z) + FD(z)

Solve zt = fH(z) with conservative method zi+1 = Ψ∆t(zi).

Solve zt = Dz exactly with flow map Φt(z) = exp(Dt)z.

Original system is solved by composing these flow maps.

Jul 2011 – p.8/16

Splitting Methods, McLachlan and Quispel (2002)

zt = FH(z) + FD(z)

Solve zt = fH(z) with conservative method zi+1 = Ψ∆t(zi).

Solve zt = Dz exactly with flow map Φt(z) = exp(Dt)z.

Original system is solved by composing these flow maps.

Example: Define non-standard difference/average operators

Dat z =

zi+1 − e−a∆tzi

∆tand Aa

t z =zi+1 + e−a∆tzi

2

Implicit midpoint for zt = FH(z) is D0t z = FH

(

A0t z)

.

Splitting method for zt = FH(z) + FD(z) is Da/2t z = FH

(

Aa/2t z

)

Jul 2011 – p.8/16

Structure-Preservation

A non-standard 1-step FD method preserves a conformal CL, ifit has a discrete product rule and the corresponding CL ispreserved by the conservative method upon which it is based.

Example: The conformal box scheme

K

(

Da/2t Ab/2

x z)

+ L

(

Db/2x A

a/2t z

)

= ∇S(

Aa/2t Ab/2

x z)

satisfies Dat

⟨

Ab/2x z,KBAb/2

x z⟩

+Dbx

⟨

Aa/2t z,LBA

a/2t z

⟩

= 0.

Sum over spatial index with b = 0 and appropriate BCs gives

P i+1 = exp(−a∆t)P i with P i =∑

n

⟨

A0xz,KBA0

xz⟩

∆x.

Jul 2011 – p.9/16

Nonlinear Schrodinger with Dissipation

iψt + ψxx + V ′(|ψ|2)ψ + ia

2ψ = 0

Can be written Kzt + Lzx = ∇zS(z)−a2Kz with

J =

[

0 −1

1 0

]

K =

[

−J 0

0 0

]

, L =

[

0 −I

I 0

]

,

z = [v,w, p, q]T , S(z) = 12(p

2 + q2 + V (v2 + w2))

The conformal norm conservation law

∂t(w2 + v2) + 2∂x(vwx − wvx) = −a(w2 + v2)

Jul 2011 – p.10/16

Residual in Local Conformal Conservation Law

a = 0.01, h = 0.01,

05

1015

2025

−10

−5

0

5

100

0.2

0.4

0.6

0.8

1

1.2

x 10−13

MS−CC−CL−1: Γ = 0.010, |RaMAX

| = 1.19904e−13

Jul 2011 – p.11/16

Residual in Global Conformal Conservation

10 20 30 40 50 60 70 80 90 10010

−14

10−13

10−12

TIME

ln(D

)

PreissmanConformal

Jul 2011 – p.12/16

Dissipative Semi-Linear Wave Equation

utt = uxx − aut − f ′(u)

Can be written Kzt + Lzx = ∇zS(z)−a2Kz with

J =

[

0 −1

1 0

]

K =

[

J 0

0 J

]

, L =

[

0 −I

I 0

]

,

z = [u, v, w, p]T , S(z) = 12

(

a(uv + wp) + v2 − w2 + 2f(u))

momentum:∂t(vux+pwx)+∂x

(

f(u) + v2

2 − w2

2 − vut − pwt

)

= −a(vux+pwx)

angular momentum: ∂t(u× ut) + ∂x(ux × u) = −a(u× ut)

others: ∂t(vw + cup) + 1

2∂x

(

v2 + w2− c(u2 + p2)

)

= −a(vw + cup)

Jul 2011 – p.13/16

Error in dissipation rates for utt = uxx − aut − cu

c = 1, a = 34 , ∆t = 0.025, 80 grid points on [−π, π].

0 5 10 15 20 25 30 35 40 45 50−0.15

−0.1

−0.05

0

0.05

0.1

0.15

0.2

t

d

standard box schemeconformal box scheme

Jul 2011 – p.14/16

Solution Behavior for utt = uxx − aut − cu

c = 1, a = 34 , ∆t = 0.025, 80 grid points on [−π, π].

−4 −3 −2 −1 0 1 2 3 4−1

−0.5

0

0.5

1x 10

−8

x

uT = 50

−4 −3 −2 −1 0 1 2 3 4−1

−0.5

0

0.5

1x 10

−8

x

u

exactstandard

exactconformal

Jul 2011 – p.15/16

Solution Behavior for utt = uxx − aut − cu

c = 1, a = 34 , ∆t = 0.025, 80 grid points on [−π, π].

−4 −3 −2 −1 0 1 2 3 4−4

−2

0

2

4x 10

−14

x

uT = 100

exactstandard

−4 −3 −2 −1 0 1 2 3 4−1

−0.5

0

0.5

1x 10

−16

x

u

exactconformal

Jul 2011 – p.16/16