Asymptotics for the focusing integrable
discrete nonlinear Schrödinger equation
Hideshi YAMANE
Kwansei Gakuin University

March 5, 2019
Himeji1. Nonlinear Schrödinger equation and a soliton
focusing NLS
iut + uxx +2|u|2 u = 0soliton
22u(x, t) =2ηe2iξx−4i(ξ −η )t+i(ψ0 +π/2)
× sech(2ηx − 8ξηt − 2δ0 )carrier wave (exp, oscillatory)×traveling solitary wave (sech)2. Long-time asymptotics of soliton equations
As t → ∞, the solution is asymptoticallya sum of solitons plus a small perturbation.
NLS: Fokas-Its ‘96 (IBVP), Kamvissis ‘95 (IVP)
Toda lattice: Krüger-Teschl ‘09 (IVP)
KdV: Tanaka ‘75 (IVP), Grunert-Teschl ‘09 (IVP)
SOLITON RESOLUTION in recent terminology(e.g. Terence Tao’s “Why are solitons stable?”, 2009)
Valid for non-integrable equations as well,but INTEGRABLE ones are particularly important because• they are model cases• phase shift can be written down in the inverse scatteringparlance.

3. Integrable Discrete NLS (IDNLS) 1
NLS (focusing)iut + uxx +2|u|2 u = 0
Ablowitz-Ladik (’75)
integrable discrete nonlinear Schrödinger equation (focusing)id
Rn +(Rn+1 −2Rn +Rn−1 )+|Rn |2 (Rn+1 +Rn−1 ) = 0dt

Both have solitons:
carrier wave (exp, oscillatory)×traveling solitary
wave (sech)4. Integrable discrete NLS (IDNLS) 2

z1 eigenvalue, |z1 | > 1,C1 (0) norming constanta parameter (to be explained laer)bright soliton
BS(n, t; z1 , C1 (0))= (exp carrier wave) × (sech traveling wave)
If we multiply C1 (0) by another constant⇒ PHASE SHIFT in exp and sech.
It happens when solitons collide with one another.

5. Soliton
Rn (t) = BS (n, t; z1 , C1 (0)), soliton
z1 = exp(α1 + iβ1 ), α1 > 0: eigenvalue
C1 (0):norming constant (at t = 0)
BS(n, t; z1 , C1 (0)) = carrier wave × traveling wave
= exp −i[2β1 (n + 1) − 2w1 t − arg C1 (0)]× sinh(2α1 )sech[2α1 (n + 1) − 2v1 t − θ1 ].
v1 , w1 : written in terms of α1 , β1 .
θ1 . : written in terms of |C1 (0)|, α1 .
If we multiply C1 (0) by another number⇒ PHASE SHIFT in exp and sech.
It happens when solitons with different velocities collide.

5. Soliton
Rn (t) = BS (n, t; z1 , C1 (0)), soliton
z1 = exp(α1 + iβ1 ), α1 > 0: eigenvalue
C1 (0):norming constant (at t = 0)
BS(n, t; z1 , C1 (0)) = carrier wave × traveling wave
= exp −i[2β1 (n + 1) − 2w1 t − arg C1 (0)]× sinh(2α1 )sech[2α1 (n + 1) − 2v1 t − θ1 ].
v1 , w1 : written in terms of α1 , β1 .
θ1 . : written in terms of |C1 (0)|, α1 .
If we multiply C1 (0) by another number⇒ PHASE SHIFT in exp and sech.
It happens when solitons with different velocities collide.

5. Soliton
Rn (t) = BS (n, t; z1 , C1 (0)), soliton
z1 = exp(α1 + iβ1 ), α1 > 0: eigenvalue
C1 (0):norming constant (at t = 0)
BS(n, t; z1 , C1 (0)) = carrier wave × traveling wave
= exp −i[2β1 (n + 1) − 2w1 t − arg C1 (0)]× sinh(2α1 )sech[2α1 (n + 1) − 2v1 t − θ1 ].
v1 , w1 : written in terms of α1 , β1 .
θ1 . : written in terms of |C1 (0)|, α1 .
If we multiply C1 (0) by another number⇒ PHASE SHIFT in exp and sech.
It happens when solitons with different velocities collide.

6. Soliton collision and phase shift
Popular topic in Integrable systems.
A faster soliton overtakes a slower one.
Velocity and shape are preserved after overtaking.
There may be phase shift, like f (x − ct) 7→ f (x − ct + x0 ).
Multiplication of the norming constant by another constant⇒phase shift
Different constants as t → ∞ and t → −∞⇒phase shift due to collision
Studying phase shift is reduced to studying these constants.
TWO GOALS• Soliton resolution.
Is the solution is asymptotically a sum of 1-solitons?
• Calculation of phase shift (as t → ∞)

6. Soliton collision and phase shift
Popular topic in Integrable systems.
A faster soliton overtakes a slower one.
Velocity and shape are preserved after overtaking.
There may be phase shift, like f (x − ct) 7→ f (x − ct + x0 ).
Multiplication of the norming constant by another constant⇒phase shift
Different constants as t → ∞ and t → −∞⇒phase shift due to collision
Studying phase shift is reduced to studying these constants.
TWO GOALS• Soliton resolution.
Is the solution is asymptotically a sum of 1-solitons?
• Calculation of phase shift (as t → ∞)

6. Soliton collision and phase shift
Popular topic in Integrable systems.
A faster soliton overtakes a slower one.
Velocity and shape are preserved after overtaking.
There may be phase shift, like f (x − ct) 7→ f (x − ct + x0 ).
Multiplication of the norming constant by another constant⇒phase shift
Different constants as t → ∞ and t → −∞⇒phase shift due to collision
Studying phase shift is reduced to studying these constants.
TWO GOALS• Soliton resolution.
Is the solution is asymptotically a sum of 1-solitons?
• Calculation of phase shift (as t → ∞)

7. IDNLS and its Lax pair
idRn +(Rn+1 −2Rn +Rn−1 )+|Rn |2 (Rn+1 +Rn−1 ) = 0 (IDNLS)
dt...........................................................
z Rnn-part : Xn+1 =Xn
Rn z −1"#dt-part :Xn = a complicated matrix Xndt(IDNLS) is the compatibility condition.
z ∈ C \ {0} is called the spectral parameter,Eigenvalues are its special values.

8. Eigenfunctions of the n-part
If Rn → 0 (rapidly) as n → ±∞, then approximately n 
z0z
0Xn+1 ≈Xn .
‘solutions’, −n0z0 z −1
There exist eigenfunctions
φn (z), ψn (z) in |z| ≥ 1 and ψn∗ (z) in |z| ≤ 1
which behave like z ±n as n → ±∞.
3 solutions in the 2-dimensional solution space.
(There’s another, but we omit it.)

8. Eigenfunctions of the n-part
If Rn → 0 (rapidly) as n → ±∞, then approximately n z0z
0
Xn+1 ≈Xn .
‘solutions’, −n0z0 z −1
There exist eigenfunctions
φn (z), ψn (z) in |z| ≥ 1 and ψn∗ (z) in |z| ≤ 1
which behave like z ±n as n → ±∞.
3 solutions in the 2-dimensional solution space.
(There’s another, but we omit it.)

8. Eigenfunctions of the n-part
If Rn → 0 (rapidly) as n → ±∞, then approximately n z0z
0Xn+1 ≈Xn .
‘solutions’, −n0z0 z −1
There exist eigenfunctions
φn (z), ψn (z) in |z| ≥ 1 and ψn∗ (z) in |z| ≤ 1
which behave like z ±n as n → ±∞.
3 solutions in the 2-dimensional solution space.
(There’s another, but we omit it.)9. Eigenvalues and the reflection coefficient
On |z| = 1, ∃a(z), b(z) = b(z, t) such thatφn(z) = b(z)ψn(z) + a(z)ψn∗ (z),If a(zj ) = 0 , then a(−zj ) = 0.
{±zj , ±z̄j−1 } is called a quartet of eigenvalues .
It corresponds to a soliton .
On |z| = 1, the reflection coefficient r(z) isr(z) :=b(z)a(z)

10. Reflection coefficient
On |z| = 1, the reflection coefficient r(z) isr(z) :=b(z)a(z)

  0∗n 1
Recall: ψn ∼ z, ψn ∼ z
10r is characterized by nz∗rψn + ψn ∼ const.
(n → −∞).
0−nas n → ∞.

time evolutionr(z, t) = r(z) exp (it(z − z −1 )2 ), where r(z) = r(z, 0).

11. Scattering data
Assume a(zj ) = 0 (order 1). ±zj is an eigenvalue.
φn (zj ) = ∃bj ψn (zj ).
bj
The norming constant is defined by Cj := da(zj )dzScattering Data{(±zj , ±z¯j −1 , Cj )}Jj=1 ,r(z)The potential Rn is said to be reflectionless if r(z) = 0.
Inverse Scattering Transform
The potential Rn is reconstructed from the scattering data.
Done by using a Riemann-Hilbert problem with poles.

12. Riemann-Hilbert Problem (RHP)
Boundary value problem on the complex plane
Γ: curve (the left-hand side is the + side).
m(z): unknown matrix, components are holomorphic in C \ Γ
Example: 1. Γ = R, m(z) is holomorphic in ±Im z > 0 .
2. Γ = {|z| = 1}, m(z) is holomorphic in |z| =6 1.
m+ , m− : boundary values on Γ from ± sides
RHP: m+ = m− v on Γ (v : given, JUMP MATRIX)
No jump if v = I. m is analytically continued.
Normalization: m(z) → I as z → ∞.
(ensures uniqueness)12. Riemann-Hilbert Problem (RHP)
Boundary value problem on the complex plane
Γ: curve (the left-hand side is the + side).
m(z): unknown matrix, components are holomorphic in C \ Γ
Example: 1. Γ = R, m(z) is holomorphic in ±Im z > 0 .
2. Γ = {|z| = 1}, m(z) is holomorphic in |z| =6 1.
m+ , m− : boundary values on Γ from ± sides
RHP: m+ = m− v on Γ (v : given, JUMP MATRIX)
No jump if v = I. m is analytically continued.
Normalization: m(z) → I as z → ∞.
(ensures uniqueness)

13. RHPs are like integrals
RHP: m+ = m− v on Γ
Deift-Zhou’s nonlinear steepest descent is based on• Contour deformation, introduction of a new unknown
and a new jump matrices
original RHP ⇔ new RHP.
• Continuity v 7→ m is continuous.
(justification of perturbation analysis)• Removing a part of the contour
1. If v = I (no jump) on Γ̂ ⊂ Γ ,m[original] = m[with Γ̂ deleted]
2. If v ≈ I on Γ̂, m[original] ≈ m[with Γ̂ deleted]13. RHPs are like integrals
RHP: m+ = m− v on Γ
Deift-Zhou’s nonlinear steepest descent is based on• Contour deformation, introduction of a new unknown
and a new jump matrices
original RHP ⇔ new RHP.
• Continuity v 7→ m is continuous.
(justification of perturbation analysis)• Removing a part of the contour
1. If v = I (no jump) on Γ̂ ⊂ Γ ,m[original] = m[with Γ̂ deleted]
2. If v ≈ I on Γ̂, m[original] ≈ m[with Γ̂ deleted]14. RHP with polesm(z): unknown matrix, components are meromorphic in C \ Γ.
RHP: m+ = m− v on Γm(z) has poles. Residue conditions imposed.
Inverse Scattering• The jump matrix written in terms of the reflection
coefficient.
• The poles of m(z) are the eigenvalues.
• Residue conditions written in terms of the norming
constants.
The potential can be reconstructed from the solution m(z) ofthe RHP.

14. RHP with polesm(z): unknown matrix, components are meromorphic in C \ Γ.
RHP: m+ = m− v on Γm(z) has poles. Residue conditions imposed.
Inverse Scattering• The jump matrix written in terms of the reflection
coefficient.
• The poles of m(z) are the eigenvalues.
• Residue conditions written in terms of the norming
constants.
The potential can be reconstructed from the solution m(z) ofthe RHP.

15. ‘blowup’of polesRes(m(z); zj ) = lim m(z)z→zj00.
zj−2n Cj (t) 0

Replace a pole with a circle.
Then one can use the RHPs-are-like-integrals technique.
In the disk D(zj , ε),subtract the singular part of m.
New unknown m̂.
holomorphic at zj .
jump along C(zj , ε) instead.
If the jump matrix is close to the identity, it can be ignored.
Decrease ] of poles ⇒ reduction to the 1-soliton case

16. IVP of IDNLSTime evolution of the scattering dataEigenvalues are independent of time.
Cj (t) = Cj (0) exp it(zj − zj−1 )2 ,r(z, t) = r(z) exp it(z − z −1 )2 on |z| = 1,where r(z) := r(z, 0)Initial value problem

Rn (0) determines the scattering data at t = 0.
The scattering data in t > 0 are determined.
Formulate RHP with poles (involving the scattering data).
The potential Rn (t) (t > 0) is reconstructed from the solution of the RHP.






17. Reflectionless CaseIf r(z) = r(z, 0) = 0 , Rn (t) = multi-soliton.
It approaches a sum of 1-solitons as t → ∞.
phase shift (formal proof in Ablowitz-Prinari-Trubatch ’04)
Each term is of the form BS(n, t, zj , pj T (zj )−2 Cj (0))

phase shift
Phase shift is determined by the eigenvalues:
Y zk2 (zj2 − z̄k−2 )
Y2 −2
pj :=
zk z̄k ,T (zj ) :=zj2 − zk−2
k>jk>jThe j-th soliton is faster than the (j − 1)-th.
Q1 Is soliton resolution valid even if r(z) 6≡ 0?
Q2 How does the reflection affect the phase shift?

17. Reflectionless CaseIf r(z) = r(z, 0) = 0 , Rn (t) = multi-soliton.
It approaches a sum of 1-solitons as t → ∞.
phase shift (formal proof in Ablowitz-Prinari-Trubatch ’04)
Each term is of the form BS(n, t, zj , pj T (zj )−2 Cj (0))

phase shift
Phase shift is determined by the eigenvalues:
Y zk2 (zj2 − z̄k−2 )Y2 −2pj :=zk z̄k ,T (zj ) :=zj2 − zk−2k>jk>j
The j-th soliton is faster than the (j − 1)-th.
Q1 Is soliton resolution valid even if r(z) 6≡ 0?
Q2 How does the reflection affect the phase shift?

18. Main results (sketched)
What happens as t → ∞ if there is reflection (r(z) 6≡ 0)?
Some generic assumptions (finite number of eigenvalues, ...)SOLITON RESOLUTION
A sum of 1-solitons plus a small perturbation
A new PHASE SHIFT formula involving the REFLECTION
COEFFICIENT r(z).
|n|/t < 2 (the ’timelike’ region)
There is a new factor written in terms of r(z):BS(n, t, zj , New·pj T (zj )−2 Cj (0))|n|/t ≥ 2
Leading term is the same as in the reflectionless case.

18. Main results (sketched)
What happens as t → ∞ if there is reflection (r(z) 6≡ 0)?
Some generic assumptions (finite number of eigenvalues, ...)SOLITON RESOLUTION
A sum of 1-solitons plus a small perturbation
A new PHASE SHIFT formula involving the REFLECTION
COEFFICIENT r(z).
|n|/t < 2 (the ’timelike’ region)
There is a new factor written in terms of r(z):BS(n, t, zj , New·pj T (zj )−2 Cj (0))|n|/t ≥ 2
Leading term is the same as in the reflectionless case.

19. Asymptotic Behavior: r(z) 6≡ 0
♣ tw is the velocity of the soliton (traveling wave).
|tw(zj )| < 2 Timelike Region: New Phase Shift FormulaRn (t) = BS n, t; zj , δ(0)δ(zj )−2 pj T (zj )−2 Cj (0) +O(t−1/2 ).
δ(z) determined by r(z).
pj , T (zj ) determined by zk ’s (k ≥ j).
zk ’s correspond to the j-th and faster solitons.
|tw(zj )| = 2 Leading term remains the sameRn (t) = BS n, t; zj , pj T (zj )−2 Cj (0) + O(t−1/3 ).
|tw(zj )| > 2 Leading term remains the same
As |n| → ∞,Rn (t) = BS n, t; zj , pj T (zj )−2 Cj (0) + O(n−k ),∀k.

19. Asymptotic Behavior: r(z) 6≡ 0
♣ tw is the velocity of the soliton (traveling wave).
|tw(zj )| < 2 Timelike Region: New Phase Shift FormulaRn (t) = BS n, t; zj , δ(0)δ(zj )−2 pj T (zj )−2 Cj (0) +O(t−1/2 ).
δ(z) determined by r(z).
pj , T (zj ) determined by zk ’s (k ≥ j).
zk ’s correspond to the j-th and faster solitons.
|tw(zj )| = 2 Leading term remains the sameRn (t) = BS n, t; zj , pj T (zj )−2 Cj (0) + O(t−1/3 ).
|tw(zj )| > 2 Leading term remains the same
As |n| → ∞,
Rn (t) = BS n, t; zj , pj T (zj )−2 Cj (0) + O(n−k ),∀k.

20. function δ(z) in the timelike region
In |n| < 2t (the timelike region), pp
A := 122 + n/t − i 2 − n/t .
S1 := e−πi/4 A, S2 := e−πi/4 Ā, S3 := −S1 , S4 := −S2 .
all on |z| = 1.

Saddle points of the phase function to be explained later.
δ(z) := exp−12πi

Z S2Z S4 (τ − z)+S1−12log(1 + |r(τ )| ) dτS3δ(z) is determined by the reflection coefficient.
δ(z) ≡ 1 in the reflectionless case.

21. RHP with poles and phase function
m+ (z) = m− (z)v(z) on |z| = 1

1 + |r(z)|2 e−2ϕ r̄(z)v(z) =JUMP MATRIX
e2ϕ r(z)11ϕ = it(z − z −1 )2 − n log z PHASE FUNCTION!
2“Nonlinear Fourier-Laplace analysis”
Residue conditions at the poles of m(z) written in terms ofthe norming constants.
dPotential reconstruction Rn (t) = −m(z)21dz
z=0IVP solved. Asymptotic expansion calculated.

22. Different behaviors in different regions. Why?
ϕ = 12 it(z − z −1 )2 − n log z phase function{Re ϕ(z) = 0} ⊃ {|z| = 1}. Red dots are stationary points.
|n| < 2t|n| = 2t|n| > 2t

22. Different behaviors in different regions. Why?
ϕ = 12 it(z − z −1 )2 − n log z phase function{Re ϕ(z) = 0} ⊃ {|z| = 1}. Red dots are stationary points.
|n| < 2t|n| = 2t

Steepest descent paths.

|n| > 2t

Thank you very much.
• Long-time asymptotics for the integrable discrete
nonlinear Schrödinger equation: the focusing case,
to appear in Funkcialaj Ekvacioj,
arXiv:1512.01760 [math-ph]• (related work) Riemann-Hilbert factorization of matrices
invariant under inversion in a circle,
to appear in Proc. AMS, Volume 147, Number 5, May
2019,
arXiv:1805.12366 [math-ph]