göteborg january 28, 2019 long-time asymptotics jonatan ... · jonatan lenells kth royal institute...
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Göteborg January 28, 2019
Long-time asymptotics for nonlinear integrable PDEs
Jonatan LenellsKTH Royal Institute of Technology
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Outline
‣ sine-Gordon equation
‣ Integrable systems
‣ Solitons
‣Asymptotics
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Outline
‣ sine-Gordon equation
‣ Integrable systems
‣ Solitons
‣Asymptotics
![Page 5: Göteborg January 28, 2019 Long-time asymptotics Jonatan ... · Jonatan Lenells KTH Royal Institute of Technology. Outline ‣sine-Gordon equation ‣Integrable systems ‣Solitons](https://reader035.vdocuments.mx/reader035/viewer/2022063003/5f6c8392e9435218f26ff57c/html5/thumbnails/5.jpg)
utt � uxx + sinu = 0
u(x, t) real-valued
sine-Gordon equation
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utt � uxx + sinu = 0
u(x, t) real-valued
sine-Gordon equation
‣ Name is a pun on “Klein-Gordon”:utt � uxx + u = 0 (Klein-Gordon equation, 1926)
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utt � uxx + sinu = 0
u(x, t) real-valued
‣ Name is a pun on “Klein-Gordon”:utt � uxx + u = 0 (Klein-Gordon equation, 1926)
sine-Gordon equation
Oskar KleinBorn: 1894 MörbyDied: 1977 Stockholm
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utt � uxx + sinu = 0
u(x, t) real-valued
sine-Gordon equation
‣ Name is a pun on “Klein-Gordon”:utt � uxx + u = 0 (Klein-Gordon equation, 1926)
‣ Discovered as the “Gauss-Codazzi” equationfor surfaces of constant negative curvature (1862)
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utt � uxx + sinu = 0
u(x, t) real-valued
‣ Model in condensed matter (1932)
sine-Gordon equation
‣ Name is a pun on “Klein-Gordon”:utt � uxx + u = 0 (Klein-Gordon equation, 1926)
‣ Discovered as the “Gauss-Codazzi” equationfor surfaces of constant negative curvature (1862)
![Page 10: Göteborg January 28, 2019 Long-time asymptotics Jonatan ... · Jonatan Lenells KTH Royal Institute of Technology. Outline ‣sine-Gordon equation ‣Integrable systems ‣Solitons](https://reader035.vdocuments.mx/reader035/viewer/2022063003/5f6c8392e9435218f26ff57c/html5/thumbnails/10.jpg)
utt � uxx + sinu = 0
u(x, t) real-valued
‣ Model in condensed matter (1932)‣ Magnetic flux propagation in Josephson junctions
sine-Gordon equation
‣ Name is a pun on “Klein-Gordon”:utt � uxx + u = 0 (Klein-Gordon equation, 1926)
‣ Discovered as the “Gauss-Codazzi” equationfor surfaces of constant negative curvature (1862)
![Page 11: Göteborg January 28, 2019 Long-time asymptotics Jonatan ... · Jonatan Lenells KTH Royal Institute of Technology. Outline ‣sine-Gordon equation ‣Integrable systems ‣Solitons](https://reader035.vdocuments.mx/reader035/viewer/2022063003/5f6c8392e9435218f26ff57c/html5/thumbnails/11.jpg)
utt � uxx + sinu = 0
u(x, t) real-valued
‣ Model in condensed matter (1932)‣ Magnetic flux propagation in Josephson junctions‣ Nonlinear optics
sine-Gordon equation
‣ Name is a pun on “Klein-Gordon”:utt � uxx + u = 0 (Klein-Gordon equation, 1926)
‣ Discovered as the “Gauss-Codazzi” equationfor surfaces of constant negative curvature (1862)
![Page 12: Göteborg January 28, 2019 Long-time asymptotics Jonatan ... · Jonatan Lenells KTH Royal Institute of Technology. Outline ‣sine-Gordon equation ‣Integrable systems ‣Solitons](https://reader035.vdocuments.mx/reader035/viewer/2022063003/5f6c8392e9435218f26ff57c/html5/thumbnails/12.jpg)
‣ First equation for which Bäcklund transformationswere discovered (~1880).
sine-Gordon equation
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‣ First equation for which Bäcklund transformationswere discovered (~1880).
Albert Victor BäcklundBorn: 1845 HöganäsDied: 1922 Lund
Rector Lund University 1907-1909
sine-Gordon equation
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sine-Gordon in the quarter plane
utt � uxx + sinu = 0
x
t
u(x, t) real-valued
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u0(x) = u(x, 0)u1(x) = ut(x, 0)
g0(t) = u(0, t) utt � uxx + sinu = 0
x
t
sine-Gordon in the quarter plane
u(x, t) real-valued
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Asymptotics
Sector I(rapid decay)Sect
or II (
transi
tion)
Sector III(solitons & radiation)
x=t
x
t
(with L. Huang)
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Outline
‣ sine-Gordon equation
‣ Integrable systems
‣ Solitons
‣Asymptotics
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Integrable systemsIn finite dimensions:
is completely integrable if there exist independentPoisson commuting conserved quantities
(qi =
@H
@pi
pi = �@H
@qi
n
i = 1, . . . , n,
A Hamiltonian system
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Integrable systemsIn finite dimensions:
is completely integrable if there exist independentPoisson commuting conserved quantities
(qi =
@H
@pi
pi = �@H
@qi
n
i = 1, . . . , n,
Origin of name: Solution can be found by integration
A Hamiltonian system
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Integrable systemsExamples include:
‣ The harmonic oscillator
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Integrable systemsExamples include:
‣ The harmonic oscillator
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Integrable systemsExamples include:
‣ Two-body problem in Newtonian mechanics
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Integrable systemsExamples include:
‣ n-dimensional body rotating about its center of mass
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Integrable systemsExamples include:
‣ n-dimensional body rotating about its center of mass
‣ Lagrange and Kovalevskaya tops
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Integrable systems
Action-Angle variablesOriginal variables {Ji, ✓i}ni=1{pi, qi}ni=1
Nonlinear time evolution Linear time evolution
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Integrable systems
{pi(0), qi(0)}ni=1
Possible solution strategy:
initial data
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Integrable systems
{pi(t), qi(t)}ni=1
{pi(0), qi(0)}ni=1
Possible solution strategy:
initial data
?
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Integrable systems
{pi(t), qi(t)}ni=1
{pi(0), qi(0)}ni=1 {Ji(0), ✓i(0)}ni=1
Possible solution strategy:
Change variablesinitial data
?
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Integrable systems
{pi(t), qi(t)}ni=1
{pi(0), qi(0)}ni=1 {Ji(0), ✓i(0)}ni=1
Possible solution strategy:
Change variables
Ji(t) = Ji(0)
✓i(t) = ✓i(0) + vit
⇢
initial data
? Lineartime evolution
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Integrable systems
{pi(t), qi(t)}ni=1
{pi(0), qi(0)}ni=1 {Ji(0), ✓i(0)}ni=1
Possible solution strategy:
Change variables
Ji(t) = Ji(0)
✓i(t) = ✓i(0) + vit
⇢
Change back
initial data
? Lineartime evolution
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Integrable systems
{pi(t), qi(t)}ni=1
{pi(0), qi(0)}ni=1 {Ji(0), ✓i(0)}ni=1
Possible solution strategy:
Change variables
Ji(t) = Ji(0)
✓i(t) = ✓i(0) + vit
⇢
Change back
Easy!Hard!
initial data
? Lineartime evolution
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Integrable systems
What about infinite-dimensional
systems?
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xu0(x) = u(x, 0)
satisfies anu(x, t)integrable PDE for t > 0
Initial value problem
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Inverse Scattering TransformPossible solution strategy:
initial datau0(x)
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Possible solution strategy:
initial data
?u0(x)
u(x, t)
nonlinearPDE
Inverse Scattering Transform
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Possible solution strategy:
Change variables
?r(k)
{kj , cj}N1
u(x, t)
⇢
initial datau0(x)
nonlinearPDE
Inverse Scattering Transform
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Possible solution strategy:
Change variables
Lineartime
⇢
?r(k)
{kj , cj}N1
u(x, t)
⇢
r(k)ei✓(k)t
{kj , cjei✓(kj)t}N1
initial datau0(x)
nonlinearPDE
evolution
Inverse Scattering Transform
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Possible solution strategy:
Change variables
⇢
Change back
?r(k)
{kj , cj}N1
u(x, t)
⇢
r(k)ei✓(k)t
{kj , cjei✓(kj)t}N1
initial datau0(x)
nonlinearPDE
Lineartime evolution
Inverse Scattering Transform
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Possible solution strategy:
Change variables
⇢
Change back
?r(k)
{kj , cj}N1
u(x, t)
⇢
r(k)ei✓(k)t
{kj , cjei✓(kj)t}N1
initial datau0(x)
nonlinearPDE
Lineartime evolution
Inverse Scattering Transform
Reflectioncoefficient
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Possible solution strategy:
“Nonlinear” Fourier
⇢
?r(k)
{kj , cj}N1
u(x, t)
⇢
r(k)ei✓(k)t
{kj , cjei✓(kj)t}N1
initial datau0(x)
transform
Inverse “nonlinear”Fourier transform
nonlinearPDE
Lineartime evolution
Inverse Scattering Transform
Reflectioncoefficient
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Possible solution strategy:⇢
Inverse Scattering
?r(k)
{kj , cj}N1
u(x, t)
⇢
r(k)ei✓(k)t
{kj , cjei✓(kj)t}N1
initial datau0(x)
nonlinearPDE
Lineartime evolution
Transform
Inverse Scattering Transform
“Nonlinear” Fouriertransform
Reflectioncoefficient
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Initial-value problems for integrable PDEs
‣ 1967 - KdV equation
Inverse Scattering Transform
can be solved via the Inverse Scattering Transform
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Initial-value problems for integrable PDEs
‣ 1967 - KdV equation
‣ 1972 - Nonlinear Schrödinger equation
Inverse Scattering Transform
can be solved via the Inverse Scattering Transform
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Initial-value problems for integrable PDEs
‣ 1967 - KdV equation
‣ 1972 - Nonlinear Schrödinger equation‣ 1973 - Sine-Gordon equation
Inverse Scattering Transform
can be solved via the Inverse Scattering Transform
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can be solved via the Inverse Scattering TransformInitial-value problems for integrable PDEs
‣ 1967 - KdV equation
‣ 1972 - Nonlinear Schrödinger equation‣ 1973 - Sine-Gordon equation
......
Inverse Scattering Transform
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What is an integrable system?
In infinite dimensions:
‣ 1991 book “What is integrability?” Editor V. E. Zakharov
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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.
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Outline
‣ sine-Gordon equation
‣ Integrable systems
‣ Solitons
‣Asymptotics
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Solitons‣ First discovered by John Scott Russell 1834
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Solitons‣ First discovered by John Scott Russell 1834
‣ Very stable
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Solitons‣ First discovered by John Scott Russell 1834
‣ Very stable
‣ Particle-like (origin of the name)
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Solitons‣ First discovered by John Scott Russell 1834
‣ Interact (almost) linearly although equation
‣ Very stable
‣ Particle-like (origin of the name)
is nonlinear
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Solitons
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Solitons
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Solitons
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sine-Gordon 1-soliton
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sine-Gordon 1-soliton
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sine-Gordon 1-soliton
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sine-Gordon 1-soliton
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sine-Gordon 1-soliton
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sine-Gordon 1-solitonu(x, t) = 4 arctan
⇣e� x�vtp
1�v2
⌘
� = 1
v 2 (�1, 1)Velocity
2⇡
0 x
u
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sine-Gordon 1-solitonu(x, t) = 4 arctan
⇣e� x�vtp
1�v2
⌘
� = 1
� = �1
v 2 (�1, 1)Velocity
2⇡
0 x
u
2⇡
0 x
u
kink
antikink
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sine-Gordon 1-solitonu(x, t) = 4 arctan
⇣e� x�vtp
1�v2
⌘
� = 1
� = �1
v 2 (�1, 1)Velocity
2⇡
0 x
u
2⇡
0 x
uTopological charge
kink
antikink
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sine-Gordon solitons
x
2⇡
u(x, t)
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sine-Gordon solitons
x
kink
2⇡
u(x, t)
kink
kink
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sine-Gordon solitons
x
kink
antikink
2⇡
u(x, t)
kink
kink
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sine-Gordon solitons
x
breather
kink
antikink
2⇡
u(x, t)
kink
kink
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sine-Gordon solitons
x
breather
kink
antikink
2⇡
Topological charge(or winding number)
u(x, t)
= 2
kink
kink
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Outline
‣ sine-Gordon equation
‣ Integrable systems
‣ Solitons
‣Asymptotics
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xu0(x) = u(x, 0)
satisfies anu(x, t)integrable PDE
Initial value problem
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xu0(x) = u(x, 0)
Initial value problemInverse Scattering Transform [GGKM 1967, …]
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xu0(x) = u(x, 0)
Initial value problem
r(k)Reflection coefficient
Inverse Scattering Transform [GGKM 1967, …]
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Initial-boundary value problem
t
u0(x) = u(x, 0)
u(0,t)=
g 0(t)
x
The unified transform method [Fokas 1997, …]
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Initial-boundary value problem
t
r(k)
u0(x) = u(x, 0)
u(0,t)=
g 0(t)
x
The unified transform method [Fokas 1997, …]
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Initial-boundary value problemThe unified transform method [Fokas 1997, …]
t
r1(k)
r(k)
u0(x) = u(x, 0)
u(0,t)=
g 0(t)
x
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Initial-boundary value problem
Solution can be found by solving au(x, t)
Riemann-Hilbert problem
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Riemann-Hilbert problemCONSTRUCTION OF SOLUTIONS AND ASYMPTOTICS 7
1�1
D1
D2
D3
D4
�
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 +
i
4(k � 1k)[�3, µ] = Q(x, t, k)µ,
µt +i
4(k + 1k)[�3, µ] = Q(x, t,�k)µ,
(2.1)
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)
with
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)):
8><
>:
(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
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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>>>>>>>>>>>>>><
>>>>>>>>>>>>>>:
1 0
�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✓
0 1
!, 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
k3
⌘, |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
4
✓k � 1
k
◆x+
1
4
✓k +
1
k
◆t
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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>>>>>>>>>>>>>><
>>>>>>>>>>>>>>:
1 0
�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✓
0 1
!, 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
k3
⌘, |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
4
✓k � 1
k
◆x+
1
4
✓k +
1
k
◆t
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Method of nonlinear steepest descent[Deift & Zhou 1993, …]
is small when is negative
Asymptotics for sine-Gordon
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Method of nonlinear steepest descent[Deift & Zhou 1993, …]
is small when is negative
Asymptotics for sine-Gordon
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Method of nonlinear steepest descent[Deift & Zhou 1993, …]
steepest descentcontouris small when is negative
Asymptotics for sine-Gordon
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Re� < 0<latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit>
Re� > 0<latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit>
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Method of nonlinear steepest descent[Deift & Zhou 1993, …]
Main contributionfrom here
steepest descentcontouris small when is negative
Asymptotics for sine-Gordon
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Re� < 0<latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit>
Re� > 0<latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit>
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Asymptotics for sine-Gordon
�
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Asymptotics for sine-Gordon
�
Expect main
from herecontribution
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Asymptotics for sine-Gordon
�
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u0(x) = u(x, 0)u1(x) = ut(x, 0)
g0(t) = u(0, t) utt � uxx + sinu = 0
x
t
Asymptotics for sine-Gordon
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u0(x) = u(x, 0)u1(x) = ut(x, 0)
g0(t) = u(0, t) utt � uxx + sinu = 0
x
t
t ! 1as
x ! 1as
g0 ! 2⇡Nt
u0 ! 2⇡Nx
Asymptotics for sine-Gordon
Nx, Nt integers
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Asymptotics for sine-Gordon
Sector I(rapid decay)Sect
or II (
transi
tion)
Sector III(solitons & radiation)
x=t
x
t
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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)
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uconst(j) = �2⇡j�1X
i=1
sgn(Im ci)
where
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
l=1
�����j � �l
�j � �l
����2�
,
vj
‣
‣
‣
urad(x, t) = �2
s2(1 + k20)⌫
k0tsin ↵
Results for sG (with L. Huang)
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uconst(j) = �2⇡j�1X
i=1
sgn(Im ci)
where
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
l=1
�����j � �l
�j � �l
����2�
,
vj
‣
‣
‣
soliton-radiationinteraction
urad(x, t) = �2
s2(1 + k20)⌫
k0tsin ↵
Results for sG (with L. Huang)
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uconst(j) = �2⇡j�1X
i=1
sgn(Im ci)
where
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
l=1
�����j � �l
�j � �l
����2�
,
vj
‣
‣
‣
soliton-solitoninteractionurad(x, t) = �2
s2(1 + k20)⌫
k0tsin ↵
Results for sG (with L. Huang)
soliton-radiationinteraction
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66 CONSTRUCTION OF SOLUTIONS AND ASYMPTOTICS
�1
�2�3
�4
�5
�6�7�8
Re k1�1
x
t
⇣=
v 4⇣=v 2
=v 3
⇣=v 1
⇣=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)
+O
✓ln t
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)
where
vj =1� |�j |2
1 + |�j |2Point spectrum
Results for sG (with L. Huang)
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Winding number =X
|�j |<1,�j2iRsgn(Im cj)
Results for sG (with L. Huang)
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Winding number =
Only solitons can generate a=)nonzero winding number
X
|�j |<1,�j2iRsgn(Im cj)
Results for sG (with L. Huang)
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Thank you!