göteborg january 28, 2019 long-time asymptotics jonatan ... · jonatan lenells kth royal institute...

Göteborg January 28, 2019

Long-time asymptotics for nonlinear integrable PDEs

Jonatan LenellsKTH Royal Institute of Technology

Outline

‣ sine-Gordon equation

‣ Integrable systems

‣ Solitons

‣Asymptotics

Outline

‣ Solitons

‣Asymptotics

utt � uxx + sinu = 0

u(x, t) real-valued

sine-Gordon equation

u(x, t) real-valued

‣ Name is a pun on “Klein-Gordon”:utt � uxx + u = 0 (Klein-Gordon equation, 1926)

u(x, t) real-valued

Oskar KleinBorn: 1894 MörbyDied: 1977 Stockholm

u(x, t) real-valued

‣ Discovered as the “Gauss-Codazzi” equationfor surfaces of constant negative curvature (1862)

u(x, t) real-valued

‣ Model in condensed matter (1932)

u(x, t) real-valued

‣ Model in condensed matter (1932)‣ Magnetic flux propagation in Josephson junctions

u(x, t) real-valued

‣ Model in condensed matter (1932)‣ Magnetic flux propagation in Josephson junctions‣ Nonlinear optics

‣ First equation for which Bäcklund transformationswere discovered (~1880).

Albert Victor BäcklundBorn: 1845 HöganäsDied: 1922 Lund

Rector Lund University 1907-1909

sine-Gordon in the quarter plane

u(x, t) real-valued

u0(x) = u(x, 0)u1(x) = ut(x, 0)

g0(t) = u(0, t) utt � uxx + sinu = 0

sine-Gordon in the quarter plane

u(x, t) real-valued

Asymptotics

Sector I(rapid decay)Sect

or II (

transi

Sector III(solitons & radiation)

(with L. Huang)

Outline

‣ Solitons

‣Asymptotics

Integrable systemsIn finite dimensions:

is completely integrable if there exist independentPoisson commuting conserved quantities

pi = �@H

i = 1, . . . , n,

A Hamiltonian system

Integrable systemsIn finite dimensions:

is completely integrable if there exist independentPoisson commuting conserved quantities

pi = �@H

i = 1, . . . , n,

Origin of name: Solution can be found by integration

A Hamiltonian system

Integrable systemsExamples include:

‣ The harmonic oscillator

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‣ Two-body problem in Newtonian mechanics

‣ n-dimensional body rotating about its center of mass

‣ Lagrange and Kovalevskaya tops

Integrable systems

Action-Angle variablesOriginal variables {Ji, ✓i}ni=1{pi, qi}ni=1

Nonlinear time evolution Linear time evolution

Integrable systems

{pi(0), qi(0)}ni=1

Possible solution strategy:

initial data

Integrable systems

{pi(t), qi(t)}ni=1

{pi(0), qi(0)}ni=1

initial data

Integrable systems

{pi(t), qi(t)}ni=1

{pi(0), qi(0)}ni=1 {Ji(0), ✓i(0)}ni=1

Change variablesinitial data

Integrable systems

{pi(t), qi(t)}ni=1

{pi(0), qi(0)}ni=1 {Ji(0), ✓i(0)}ni=1

Change variables

Ji(t) = Ji(0)

✓i(t) = ✓i(0) + vit

initial data

? Lineartime evolution

Integrable systems

{pi(t), qi(t)}ni=1

{pi(0), qi(0)}ni=1 {Ji(0), ✓i(0)}ni=1

Change variables

Ji(t) = Ji(0)

✓i(t) = ✓i(0) + vit

Change back

initial data

Integrable systems

{pi(t), qi(t)}ni=1

{pi(0), qi(0)}ni=1 {Ji(0), ✓i(0)}ni=1

Change variables

Ji(t) = Ji(0)

✓i(t) = ✓i(0) + vit

Change back

Easy!Hard!

initial data

Integrable systems

What about infinite-dimensional

systems?

xu0(x) = u(x, 0)

satisfies anu(x, t)integrable PDE for t > 0

Initial value problem

Inverse Scattering TransformPossible solution strategy:

initial datau0(x)

initial data

?u0(x)

u(x, t)

nonlinearPDE

Inverse Scattering Transform

Change variables

{kj , cj}N1

u(x, t)

initial datau0(x)

nonlinearPDE

Change variables

Lineartime

{kj , cj}N1

u(x, t)

r(k)ei✓(k)t

{kj , cjei✓(kj)t}N1

initial datau0(x)

nonlinearPDE

evolution

Change variables

Change back

{kj , cj}N1

u(x, t)

r(k)ei✓(k)t

initial datau0(x)

nonlinearPDE

Lineartime evolution

Change variables

Change back

{kj , cj}N1

u(x, t)

r(k)ei✓(k)t

initial datau0(x)

nonlinearPDE

Reflectioncoefficient

“Nonlinear” Fourier

{kj , cj}N1

u(x, t)

r(k)ei✓(k)t

initial datau0(x)

transform

Inverse “nonlinear”Fourier transform

nonlinearPDE

Possible solution strategy:⇢

Inverse Scattering

{kj , cj}N1

u(x, t)

r(k)ei✓(k)t

initial datau0(x)

nonlinearPDE

Transform

“Nonlinear” Fouriertransform

Initial-value problems for integrable PDEs

‣ 1967 - KdV equation

can be solved via the Inverse Scattering Transform

‣ 1972 - Nonlinear Schrödinger equation

‣ 1972 - Nonlinear Schrödinger equation‣ 1973 - Sine-Gordon equation

can be solved via the Inverse Scattering TransformInitial-value problems for integrable PDEs

‣ 1972 - Nonlinear Schrödinger equation‣ 1973 - Sine-Gordon equation

......

What is an integrable system?

In infinite dimensions:

‣ 1991 book “What is integrability?” Editor V. E. Zakharov

What is an integrable system?

In infinite dimensions:

‣ 1991 book “What is integrability?” Editor V. E. Zakharov

‣ Percy Deift 2017: A problem is integrable if youcan solve it.

Outline

‣ Solitons

‣Asymptotics

Solitons‣ First discovered by John Scott Russell 1834

‣ Very stable

‣ Particle-like (origin of the name)

‣ Interact (almost) linearly although equation

‣ Very stable

‣ Particle-like (origin of the name)

is nonlinear

Solitons

sine-Gordon 1-soliton

sine-Gordon 1-solitonu(x, t) = 4 arctan

⇣e� x�vtp

1�v2

� = 1

v 2 (�1, 1)Velocity

⇣e� x�vtp

1�v2

� = 1

� = �1

antikink

⇣e� x�vtp

1�v2

� = 1

� = �1

uTopological charge

antikink

sine-Gordon solitons

u(x, t)

antikink

u(x, t)

breather

antikink

u(x, t)

breather

antikink

Topological charge(or winding number)

u(x, t)

Outline

‣ Solitons

‣Asymptotics

xu0(x) = u(x, 0)

satisfies anu(x, t)integrable PDE

xu0(x) = u(x, 0)

Initial value problemInverse Scattering Transform [GGKM 1967, …]

xu0(x) = u(x, 0)

r(k)Reflection coefficient

Inverse Scattering Transform [GGKM 1967, …]

Initial-boundary value problem

u0(x) = u(x, 0)

u(0,t)=

g 0(t)

The unified transform method [Fokas 1997, …]

u0(x) = u(x, 0)

u(0,t)=

g 0(t)

The unified transform method [Fokas 1997, …]

Initial-boundary value problemThe unified transform method [Fokas 1997, …]

u0(x) = u(x, 0)

u(0,t)=

g 0(t)

Solution can be found by solving au(x, t)

Riemann-Hilbert problem

Riemann-Hilbert problemCONSTRUCTION OF SOLUTIONS AND ASYMPTOTICS 7

Figure 2. The contour � and the domains {Dj}41 in the complex k-plane.

2. Spectral functions

This section introduces several spectral functions and reviews how these functionscan be combined to set up a RH problem suitable for solving equation (1.1) in thequarter plane.

2.1. Lax pair. Equation (1.1) is the compatibility condition of the Lax pair(µx +

4(k � 1k)[�3, µ] = Q(x, t, k)µ,

µt +i

4(k + 1k)[�3, µ] = Q(x, t,�k)µ,

where k 2 C is the spectral parameter, µ(x, t, k) is a 2⇥2-matrix valued eigenfunction,and Q is defined by

Q(x, t, k) = Q0(x, t) +Q1(x, t)

k, (2.2)

Q0(x, t) = � i(ux + ut)

4�2, Q1(x, t) =

i sin u

4�1 +

i(cosu� 1)

4�3.

2.2. Spectral functions. Let m � 1, n � 1, Nx 2 Z, and Nt 2 Z be integers.Let u0(x), u1(x), g0(t), and g1(t) be functions satisfying the following regularity anddecay assumptions (see (1.4)):

(1 + x)n(u0(x)� 2⇡Nx) 2 L1([0,1)),

(1 + x)n@iu0(x) 2 L1([0,1)), i = 1, . . . ,m+ 2,

(1 + x)n@iu1(x) 2 L1([0,1)), i = 0, 1, . . . ,m+ 1,

(2.3a)

and 8><

(1 + t)n(g0(t)� 2⇡Nt) 2 L1([0,1)),

(1 + t)n@ig0(t) 2 L1([0,1)), i = 1, . . . ,m+ 2,

(1 + t)n@ig1(t) 2 L1([0,1)), i = 0, 1, . . . ,m+ 1.

(2.3b)

Riemann-Hilbert problem:M(x, t, k) k 2 C \ �

M = I +O(1/k) k ! 1

is analytic for

asM+ = M�J

(�a.e. on

complex k-plane

Jump matrixCONSTRUCTION OF SOLUTIONS AND ASYMPTOTICS 11

satisfies the normalization condition limk!1 M(x, t, k) = I as well as the jump con-dition M+ = M�J across the contour � = R [ {|k| = 1}, where the jump matrix Jis given by

J(x, t, k) =

8>>>>>>>>>>>>>><

>>>>>>>>>>>>>>:

�h(k)e2i✓ 1

!, k 2 D1 \ D2,

1 �r(k)e�2i✓

�r(k)e2i✓ 1 + |r(k)|2

!, k 2 D2 \ D3,

1 �h(k)e�2i✓

!, k 2 D3 \ D4,

1 + |r1(k)|2 r1(k)e�2i✓

r1(k)e2i✓ 1

!, k 2 D4 \ D1.

(2.18)

In particular, if a(k) and d(k) have no zeros, then M satisfies the RH problem(M(x, t, ·) 2 I + E2(C \ �),M+(x, t, k) = M�(x, t, k)J(x, t, k) for a.e. k 2 �.

(2.19)

Roughly speaking, the functions r1(k) and r(k) play the roles of ‘reflection coe�-cients’ for the initial half-line {x � 0, t = 0} and for the union {x � 0, t = 0}[ {x =0, t � 0} of the initial half-line and the boundary, respectively.

3. Main results

This section presents the four main theorems of the paper in the pure radiationcase. The theorems are extended to the case when solitons are present in Section 10.

3.1. Construction of solutions.

Assumption 3.1. Suppose r1 : R ! C and h : @D2 ! C are continuous functionswith the following properties:

(a) There exist complex constants {r1,j}2j=1 such that

r1(k) =r1,1k

+r1,2k2

+O⇣ 1

⌘, |k| ! 1, k 2 R. (3.1)

(b) r1(k) and h(k) obey the symmetries

r1(k) = r1(�k), h(k) = h(�k). (3.2)

(c) The function r : [�1, 1] ! C defined by r(k) = r1(k) + h(k) satisfies r(±1) = 0and r(k) = O(k3) as k ! 0.

Theorem 3.2 (Construction of quarter-plane solutions). Let r1 : R ! C and h :@D2 ! C be functions satisfying Assumption 3.1. Define the jump matrix J(x, t, k)by (2.18). Then the RH problem (2.19) has a unique solution for each (x, t) 2[0,1)⇥ [0,1). Moreover, the nontangential limit

m(x, t) :=\limk!0

M(x, t, k) (3.3)

✓ =1

✓k � 1

✓k +

Jump matrixCONSTRUCTION OF SOLUTIONS AND ASYMPTOTICS 11

satisfies the normalization condition limk!1 M(x, t, k) = I as well as the jump con-dition M+ = M�J across the contour � = R [ {|k| = 1}, where the jump matrix Jis given by

J(x, t, k) =

8>>>>>>>>>>>>>><

>>>>>>>>>>>>>>:

�h(k)e2i✓ 1

!, k 2 D1 \ D2,

1 �r(k)e�2i✓

�r(k)e2i✓ 1 + |r(k)|2

!, k 2 D2 \ D3,

1 �h(k)e�2i✓

!, k 2 D3 \ D4,

1 + |r1(k)|2 r1(k)e�2i✓

r1(k)e2i✓ 1

!, k 2 D4 \ D1.

(2.18)

In particular, if a(k) and d(k) have no zeros, then M satisfies the RH problem(M(x, t, ·) 2 I + E2(C \ �),M+(x, t, k) = M�(x, t, k)J(x, t, k) for a.e. k 2 �.

(2.19)

Roughly speaking, the functions r1(k) and r(k) play the roles of ‘reflection coe�-cients’ for the initial half-line {x � 0, t = 0} and for the union {x � 0, t = 0}[ {x =0, t � 0} of the initial half-line and the boundary, respectively.

3. Main results

This section presents the four main theorems of the paper in the pure radiationcase. The theorems are extended to the case when solitons are present in Section 10.

3.1. Construction of solutions.

Assumption 3.1. Suppose r1 : R ! C and h : @D2 ! C are continuous functionswith the following properties:

(a) There exist complex constants {r1,j}2j=1 such that

r1(k) =r1,1k

+r1,2k2

+O⇣ 1

⌘, |k| ! 1, k 2 R. (3.1)

(b) r1(k) and h(k) obey the symmetries

r1(k) = r1(�k), h(k) = h(�k). (3.2)

(c) The function r : [�1, 1] ! C defined by r(k) = r1(k) + h(k) satisfies r(±1) = 0and r(k) = O(k3) as k ! 0.

Theorem 3.2 (Construction of quarter-plane solutions). Let r1 : R ! C and h :@D2 ! C be functions satisfying Assumption 3.1. Define the jump matrix J(x, t, k)by (2.18). Then the RH problem (2.19) has a unique solution for each (x, t) 2[0,1)⇥ [0,1). Moreover, the nontangential limit

m(x, t) :=\limk!0

M(x, t, k) (3.3)

All (x,t)-dependencein exponential

✓ =1

✓k � 1

✓k +

Method of nonlinear steepest descent[Deift & Zhou 1993, …]

is small when is negative

Asymptotics for sine-Gordon

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is small when is negative

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steepest descentcontouris small when is negative

Main contributionfrom here

steepest descentcontouris small when is negative

Expect main

from herecontribution

u0(x) = u(x, 0)u1(x) = ut(x, 0)

g0(t) = u(0, t) utt � uxx + sinu = 0

u0(x) = u(x, 0)u1(x) = ut(x, 0)

g0(t) = u(0, t) utt � uxx + sinu = 0

t ! 1as

x ! 1as

g0 ! 2⇡Nt

u0 ! 2⇡Nx

Nx, Nt integers

Sector I(rapid decay)Sect

or II (

transi

Sector III(solitons & radiation)

Theorem [Asymptotics]Under the assumptions … , the sG solution u(x, t)

u(x, t) =

(usol(x, t; j) + uconst(j) + (�1)Nurad(x, t), ⇣ 2 (vj � ✏, vj + ✏),

uconst(j) + (�1)Nurad(x, t), ⇣ 2 (vj + ✏, vj�1 � ✏),

In Sector I: u(x, t) = O�x�N

In Sector II: u(x, t) = O�(1� x/t)N + t

�N�

In Sector III:

has the following asymptotics:

Results for sG (with L. Huang)

uconst(j) = �2⇡j�1X

sgn(Im ci)

usol(x, t; j) is a kink/antikink of speed if �j 2 iR :

usol(x, t; j) = �4 arctan

⇢e��j(x�vjt)Im cj

2|�j |

j�1Y

��j � �l

�j � �l

��2�

urad(x, t) = �2

s2(1 + k20)⌫

k0tsin ↵

sgn(Im ci)

2|�j |

j�1Y

��j � �l

�j � �l

��2�

soliton-radiationinteraction

urad(x, t) = �2

s2(1 + k20)⌫

k0tsin ↵

sgn(Im ci)

2|�j |

j�1Y

��j � �l

�j � �l

��2�

soliton-solitoninteractionurad(x, t) = �2

s2(1 + k20)⌫

k0tsin ↵

soliton-radiationinteraction

66 CONSTRUCTION OF SOLUTIONS AND ASYMPTOTICS

�2�3

�6�7�8

Re k1�1

v 4⇣=v 2

⇣=v 1

Figure 18. The left figure displays a possible distribution of the �j. The associatedsine-Gordon solution is dominated asymptotically by solitons traveling in the threedirections shown on the right. The line ⇣ = 1 is also shown (dashed). The pureimaginary poles �1 and �4 generate kinks/antikinks traveling with speeds v1 and v4,respectively. The pair (�2,�3 = ��2) generates a breather traveling with speed v2 =v3. Only those �j with |�j| < 1 generate solitons.

imaginary pole �j with |�j| < 1 gives rise to a kink/antikink, whereas each pair(�j,��j) with |�j| < 1 gives rise to a breather. The speeds of these solitons satisfy

0 < v⇤ · · · v2 v1 < 1

with strict inequality vj+1 < vj except when �j+1 = ��j.

Theorem 10.9 (Asymptotics of quarter-plane solutions). Let r1 : R ! C and h :D2 ! C satisfy Assumption 3.5, let {�j}N1 ⇢ C+\� satisfy Assumption 10.8, and let{cj}N1 be nonzero complex numbers satisfying (10.13). Let u(x, t) be the associatedsine-Gordon quarter-plane solution of Theorem 10.5.Then u 2 C1([0,1)⇥ [0,1),R) and there exists a choice of the branch of arg in

(3.4) such that u(x, 0) ! 0 as x ! 1. For this choice of branch, u(x, t) satisfiesthe asymptotic formulas (3.8a) and (3.8b) in Sectors I and II. In Sector III, theasymptotics of u(x, t) is given by adding the multi-soliton determined by {�j, cj}N1 tothe radiation solution of Theorem 3.6. More precisely, for each 1 j ⇤ and each✏ > 0, the asymptotics in the narrow sector ⇣ 2 (vj � ✏, vj + ✏) centered on the line⇣ = vj is given uniformly by

u(x, t) = usol(x, t; j) + uconst(j) + (�1)Nurad,as(x, t)

✓ln t

◆, ⇣ 2 (vj � ✏, vj + ✏), t > 2, (10.22a)

while the asymptotics outside these sectors is given uniformly by

u(x, t) = uconst(j) + (�1)Nurad,as(x, t) +O

✓ln t

k3/20 t

⇣ 2 (vj + ✏, vj�1 � ✏), t > 2, j = 1, . . . ,⇤+ 1, (v⇤+1 + ✏ ⌘ 0, v0 � ✏ ⌘ 1), (10.22b)

vj =1� |�j |2

1 + |�j |2Point spectrum

Winding number =X

|�j |<1,�j2iRsgn(Im cj)

Winding number =

Only solitons can generate a=)nonzero winding number

|�j |<1,�j2iRsgn(Im cj)

Thank you!

göteborg january 28, 2019 long-time asymptotics jonatan ... · jonatan lenells kth royal institute...

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