Normal forms and geometric numerical integration of ...Normal forms and geometric numerical integration of Hamiltonian PDEs Part I: Linear equations Erwan Faou INRIA & ENS Cachan Bretagne

Normal forms and geometric numericalintegration of Hamiltonian PDEs

Part I: Linear equations

Erwan Faou

INRIA & ENS Cachan Bretagne

Beijing, 21 May 2009

Joint works with Guillaume Dujardin (Cambridge)Arnaud Debussche (ENS Cachan Bretagne)

E. Faou (INRIA) Normal forms & numerics for PDEs Beijing, 21 May 2009 1 / 35

Time-dependent Schrodinger equation

i∂

∂tϕ(x , t) = −∆ϕ(x , t) + V (x)ϕ(x , t), ϕ(x , 0) = ϕ0(x).

x ∈ Td (d = 1). Laplace operator ∆ = ∂xx .

V (x) ∈ R analytic function.

Conservation properties : if H = −∆ + V

∀ s > 0, 〈ϕ(t)|Hs |ϕ(t)〉 = 〈ϕ(0)|Hs |ϕ(0)〉.

s = 1 : energy. s = 0 : L2 norm.

This implies the conservation of the regularity over long time.


Splitting methods

Th = exp(ih∆) exp(−ihV )

where

e it∆ϕ0 :

{i∂tϕ(t, x) = −∆ϕ(t, x) (t, x) ∈ R× Tϕ(0, x) = ϕ0(x) x ∈ T

and

e−itVϕ0 :

{i∂tϕ(t, x) = V (x)ϕ(t, x) (t, x) ∈ R× Tϕ(0, x) = ϕ0(x) x ∈ T

Stepsize : h > 0


Splitting methods

Properties :

Easy to compute using FFT.

Order 1, effective order 2 (Jahnke & Lubich, 2000)

L2 norm conservation.

Practical computations : Resonances.

Long time behaviour in the infinite dimensional case ?

Same long time behavior as the Strang splitting

ϕ1 = exp(−ihV /2) exp(ih∆) exp(−ihV /2)ϕ0

.


Numerical test

V (x) =0.03

5− 4 cos(x)and ψ0(x) = sin(x)

Stepsizes :

resonant : h =2π

62 − 22= 0.196 . . .

non resonant : h = 0.2

We plot the energies errors∣∣|ϕn|2k − |ϕ0|2k

∣∣ where

|ϕ|2k = |ϕk |2 + |ϕ−k |2.


Conservation of energies

0 2 4 6 8 10x 104

!10

!8

!6

!4

!2

0

Iterations

log

of e

nerg

ies

0 2 4 6 8 10x 104

!10

!8

!6

!4

!2

0

Iterations

log

of e

nerg

ies

Fig.: Energies error. Non resonant stepsize (left) and resonant stepsize (right)


Modified energy ?

BCH formula :

Th = exp(ih∆) exp(−ihV )

' exp(ih(−∆ + V − 12 (ih)[−∆,V ] + · · ·+ (ih)kHk · · · ))

For all k, Hk is an operator of order k.

Does not converge for h > 0.No standard backward error analysis available.


Analytic functions and operatorsWe identify a function ϕ(x) with its Fourier coefficientsϕn = 1

2π

∫T e−inxϕ(x)dx

Analytical norm for functions : ‖ϕ‖ρ

= supn∈Z

(eρ|n||ϕk |

)Operators : S = (Sij)i ,j∈Z. Action : (Sϕ)i =

∑j∈Z Sijϕj .

Product of two operators : (AB)ij =∑

k∈Z AikBkj

Analytical norm for operators : ‖S‖ρ

= supk,`∈Z

(eρ|k−`||Sk`|

)‖AB‖

ρ≤ C

δ‖A‖

ρ‖B‖

ρ+δ

Hypothesis on V : ‖V ‖ρV<∞. As operator : Vij = Vi−j .

V real implies V symmetric (Sij = S∗ji )

Aρ = { S | ‖S‖ρ<∞}


A family of schemesThe idea is consider V as a small perturbation of −∆.We embed Th into the family of unitary propagators :

L(λ) = exp(ih∆) exp(−ihλV ), λ > 0.

λ = 0 : Free Schrodinger operator.e ih∆ diagonal operator with entries e−ihk2

.Conservation of |ψn| for all n ∈ Z.

Is it possible to find a normal form for L(λ) (λ small) :

Q(λ)L(λ)Q(λ)∗ = Σ(λ)

Q(0) = Id, Q(λ) unitary : Q(λ)Q(λ)∗ = Id

Σ(0) = e ih∆, Σ(λ) unitary and “nice” ( ! ?) (with conservationproperties)

Q(λ) and Σ(λ) are in Aρ for some ρ


Formal series equations

To control the unitarity of Q(λ) and Σ(λ) we introduce

S(λ) = Q(λ)∗(i∂λQ(λ)) and X (λ) = Σ(λ)∗(i∂λΣ(λ))

S and X symmetric implies Q and Σ unitary.

Equation : S(λ)− L∗(λ)S(λ)L(λ) = hV − Q(λ)∗X (λ)Q(λ)

Formal series : S(λ) =∑

n≥0 λnSn, X (λ) =

∑n≥0 λ

nXn.

Recursive equations

Sn − e−ih∆Sneih∆ + Xn = Gn(V ,Si ,Xi | i = 1, . . . , n − 1)

Gn symmetric if Si and Xi symmetric.


Homological equation

Given an operator G symmetric, is it possible to find S and Xsymmetric (and nice !) such that

S − e−ih∆Se ih∆ + X = G

In coordinates

∀ (k, `) ∈ Z2, (1− e ih(k2−`2))Sk` + Xk` = Gk`

Problems when h(k2 − `2) ' 2πm, m ∈ Z (resonances)

Non-resonance condition : for all k ∈ Z, k 6= 0,∣∣∣∣1− e ihk

h

∣∣∣∣ ≥ γ|k |−ν , γ > 0, ν > 1.

(see Shang 2000 ; Hairer, Lubich, Wanner (GNI) Chap. X.)Generic condition on h


X-shaped operators

(1− e ih(k2−`2))Sk` + Xk` = Gk`

Under the non-resonance condition, we can solve this equation by :

|k | = |`| :

{Sk` = 0,Xk` = Gk`,

|k | 6= |`| :

{Sk` = 1

1−e ih(k2−`2)Gk`,

Xk` = 0,

(S and X symmetric)X-shaped operators :

Xk` 6= 0 =⇒ |k| = |`|


Solution of the homological equation

Problem : using the diophantine condition, we do not stay in Aρ.

With the preceding definition, we have

|Sk`| ≤ γ−1h−1|k2 − `2|ν |Gk`|

G ∈ Aρ implies Gk` ≤ Ce−ρ|k−`| but this does not implies S ∈ Aρ.Possible unbounded k + `.

For a given K > 0 we define the set of indices

IK = {(k , `) ∈ Z | |k | ≤ K or |`| ≤ K}

(k , `) ∈ IK =⇒ |k + `| ≤ 2K + |k − `|


IK -solution of the homological equation

(1− e ih(k2−`2))Sk` + Xk` = Gk`

Under the non resonance condition, we can solve this equation by :

|k | = |`| or (k , `) /∈ IK :

{Sk` = 0,Xk` = Gk`,

(k , `) ∈ IK such that |k | 6= |`| :

{Sk` = 1

1−e ih(k2−`2)Gk`,

Xk` = 0,

Bounds

‖S‖ρ−δ ≤

K ν

γh

(4ν

eδ

)2ν

‖G‖ρ

and ‖X‖ρ≤ ‖G‖

ρ.


Nekoroshev machinery

We can prove (α = 2ν and β = 4ν + 3).

‖SJ‖ρV /3+ ‖QJ‖ρV /3

≤(C0K

αJβ)J

and

‖XJ‖ρV /3≤ h

(C0K

αJβ)J,

Optimal truncations : S [N](λ) =∑N

j=0 λjSj , etc.

K ' λ−σ and N ' λ−µ

We have (C0λKαNβ

)N' exp(−cλ−σ)

Almost X-shaped operators : X ∈ Xρ if

Xk` 6= 0 =⇒(|k | = |`| or (k , `) /∈ IK

)E. Faou (INRIA) Normal forms & numerics for PDEs Beijing, 21 May 2009 15 / 35

A normal form theorem

Theorem [Dujardin & Faou, 2007]∃Q(λ) ∈ AρV /4 and Σ(λ) ∈ XK

ρV /4 with K = λ−σ where

σ = 1/16(ν + 1) < 1 satisfying for λ ∈ (0, λ0)

‖Q(λ)− Id‖ρV /4

≤ C1λ1/2 and ‖Σ(λ)− e ih∆‖

ρV /4≤ C2hλ

1/2

and such that the following relations hold :

Q(λ)∗Q(λ) = Id, and Σ(λ)∗Σ(λ) = Id,

andQ(λ)L(λ)Q(λ)∗ = Σ(λ) + R(λ)

with‖R(λ)‖

ρV /5≤ C3 exp(−cλ−σ).

The constants depend only on V , h0, γ and ν.


Long time behavior ?

New variables : ψ = Q(λ)ϕ.

Action of Σ(λ) : if ψ1 = Σ(λ)ψ0, then for |k | ≤ λ−σ,(ψ1

k

ψ1−k

)=

(ak(λ) bk(λ)ck(λ) dk(λ)

)(ψ0

k

ψ0−k

)The 2× 2 matrix in this equation is unitary . This implies

∀ |k | ≤ λ−σ |ψ1k |2 + |ψ1

−k |2 = |ψ0k |2 + |ψ0

−k |2,

Notation : For k ≥ 0,

|ψ|2k := |ψk |2 + |ψ−k |2


Corollaries

For n ∈ N, let ϕn = L(λ)nϕ0 (in the new variables : ψn = (Σ + R)nψ0).(i) Assume that ϕ0 is in L2 . With the notations of the previous theorem,we have for all n ≤ exp(cλ−σ/2) and all λ ∈ (0, λ0),

∀ |k| ≤ λ−σ∣∣ |ϕn|k − |ϕ0|k

∣∣ ≤ Cλ1/2‖ϕ0‖ ,

for a constant C that depend only on V , h0, γ and ν.Key estimate : (global L2 preservation)∣∣ |ψn+1|k − |ψn|k

∣∣ ≤ C‖R(λ)‖ρV /5‖ψn‖

≤ C exp(−cλ−σ)‖ϕ0‖ .


Conservation of regularity

For s > 0, we introduce the norm :

‖ϕ‖s,∞ = sup

k≥0((1 + k)s |ϕ|k)

For all λ ∈ (0, λ0), all n ≤ exp(cλ−σ/2) we have(ii) Let s > 1/2 be given, and let s ′ be such that s − s ′ ≥ 1/2.

sup0≤k≤λ−σ

((1 + k)s′

∣∣ |ϕn|k − |ϕ0|k∣∣ ) ≤ Csλ

1/2‖ϕ0‖s,∞ ,

(iii) Let ρ ∈ (0, ρV ), there exists µ0 ∈ (0, ρ) such that for all µ < µ0,

sup0≤k≤λ−σ

(eµk

∣∣ |ϕn|k − |ϕ0|k∣∣ ) ≤ Cρ,µλ

1/2‖ϕ0‖ρ

The constant depend only on V , h0, γ and ν.


Numerical test

V (x) =3

5− 4 cos(x)and ψ0(x) = sin(x)

Stepsizes :

bad : h =2π

62 − 22= 0.196 . . .

good : h = 0.2

We plot the energies errors∣∣|ϕn|2k − |ϕ0|2k

∣∣.



0 2 4 6 8 10x 105

!12

!10

!8

!6

!4

!2

0

Iterations

Log(

erro

r)

0 2 4 6 8 10x 105

!20

!15

!10

!5

0

Iterations

Log(

erro

r)

Fig.: Energies error for the 5 first modes, λ = 0.1. Non resonant stepsize (left)and resonant stepsize (right)



0 2 4 6 8 10x 105

!10

!8

!6

!4

!2

0

Iterations

Log(

erro

r)

0 2 4 6 8 10x 105

!20

!15

!10

!5

0

Iterations

Log(

erro

r)

Fig.: Energies error for the 5 first modes, λ = 0.01. Non resonant stepsize (left)and resonant stepsize (right)


Implicit-explicit integrators

Can we find numerical schemes without resonances ?

Can we do backward error analysis for PDEs in some cases ?

Positive answer in the linear case.



Linear Schrodinger equation on the torus

i∂tu(x , t) = −∆u(x , t) + V (x)u(x , t)

Mid-split scheme :

exp(−ih(−∆ + V )) ' R(ih∆) exp(−ihV )

where

R(z) =1 + z/2

1− z/2' exp(z)

Next figures : plot of the maximal oscillations of the H1 normsbetween t = 0 and t = 50.



0.02 0.04 0.06 0.08 0.10.30.40.50.60.70.80.9

h0.02 0.04 0.06 0.08 0.11

1.2

1.4

1.6

1.8

2

h

Left : mid-split. Right : classical splitting scheme with resonances.



Operator exp(−ih∆). Resonances reflect the control of

exp(ih(|k |2 − |`|2)) 6= 1

for all k and ` in Zd , |k | 6= |`|.Mid-split integrators :

R(ih∆) =1 + ih∆/2

1− ih∆/2= exp(2i arctan(h∆/2))

Control of

exp(2i arctan(h|k |2/2)− 2i arctan(h|`|2/2)) 6= 1

Always satisfied !


Search for a modified energy

Framework : for operators S = (Sk`)k,`∈Zd ,

‖S‖α

= supk,`|Sk`(1 + |k − `|α)|

We have (α > d)‖AB‖

α≤ Cα‖A‖α ‖B‖α

Search for a function t 7→ Z (t), t ∈ [0, h] such that

exp(−itV )R(ih∆) = exp(iZ (t))

Z (0) = Z0 = 2 arctan(h∆/2), k , ` ∈ Zd ,

(Z0)k` = −δk`2 arctan(h|k |2/2).



We differentiate exp(−itV )R(ih∆) = exp(iZ (t)) with respect to t :

iV exp(−itV )R(ih∆) = i(d expiZ(t) Z ′(t)) exp(iZ (t))

equivalent to

Z ′(t) = (d expiZ(t))−1V =∑k≥0

Bk

k!adk

iZ(t)(V )

adA(B) = [A,B] = AB − BA

Bk Bernouilli numbers.

∀ |z | < 2π,∑k≥0

Bk

k!zk =

z

ez − 1.



Formal series : Z (t) =∑

`≥0 t`Z`Plugging into the previous one :∑

`≥1

`t`−1Z` =∑k≥0

Bk

k!

(i∑`≥0

t`adZ`

)k(V )

=∑`≥0

t`∑k≥0

Bk

k!ik

∑`1+···+`k=`

adZ`1· · · adZ`k

(V ).

Identifying the coefficients in the formal series, we obtain

∀ ` ≥ 1, (`+ 1)Z`+1 =∑k≥0

Bk

k!ik

∑`1+···+`k=`

adZ`1· · · adZ`k

(V ).



For ` = 0, this equation yields

Z1 =∑k≥0

Bk

k!ikadk

Z0(V ).

Crucial estimate :‖adZ0W ‖α ≤ π‖W ‖α

Proof :

(adZ0W )k` = −(2 arctan(h|k |2/2)− 2 arctan(h|`|2/2)

)Wk`

and ∣∣2 arctan(h|k |2/2)− 2 arctan(h|`|2/2)∣∣ ≤ π!!



Hence we have

‖Z1‖α ≤∑k≥0

Bk

k!πk‖V ‖

α< +∞

Convergent series !

By induction we can show

‖Z`‖α ≤ (C‖V ‖α

)`

Z (h) well defined as a convergent series for

|h| < 1

C‖V ‖α

.


Modified energy

Theorem [Debussche & Faou, 2009]There exists a symmetric operator S(h) such that for all h ≤ h0

R(ih∆) exp(−ihV ) = exp(ihS(h))

Moreover

S(h) =2

harctan(h∆/2) + V (h)

V (h) modified potential

〈u|S(h)|u〉 invariant of the numerical scheme

No residual term.

Backward error analysis result.


Long time behavior

S(h) =2

harctan(h∆/2) + V (h)

We have for the numerical solution un :

〈un|S(h)|un〉 = 〈u0|S(h)|u0〉

Control of the H1 norm for low modes and L2 norm for high modes .


Control of the solution

For low modes : |k |2 . 1/h, arctan(x) ' x

2

harctan(h|k |2/2)|uk |2 ' c |k |2|uk |2

For high modes : |k|2 & 1/h, arctan(x) ' π/2,

2

harctan(h|k |2/2)|uk |2 '

c

h|uk |2

Corollary : for all n we have∑|k|≤1/

√h

|k |2|unk |2 +

1

h

∑|k|>1/

√h

|unk |2 ≤ C0‖u0‖2

H1 .


Conclusions

In the linear case :

Resonances for pure splitting methodsGenerically no problem

Use of implicit integrator for the unbounded part :No resonancesBackward error analysis resultVery specific (midpoint rule) because of the constant π !


Documents

Normal forms and geometric numerical integration of ...Normal forms and geometric numerical integration of Hamiltonian PDEs Part I: Linear equations Erwan Faou INRIA & ENS Cachan Bretagne