Simple Periodic Boundary Data and Riemann-Hilbert Problem for Integrable Model of the Stimulated Raman Scattering
We consider the initial-boundary value (IBV) problem for nonlinear equations related to the integrable model of the stimulated Raman scattering in the quarter xt-plane with vanishing at infinity initial conditions and single-frequency periodic boundary data. We propose a matrix Riemann-Hilbert probl...
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irk-123456789-1065342016-10-01T03:01:47Z Simple Periodic Boundary Data and Riemann-Hilbert Problem for Integrable Model of the Stimulated Raman Scattering Moskovchenko, E.A. We consider the initial-boundary value (IBV) problem for nonlinear equations related to the integrable model of the stimulated Raman scattering in the quarter xt-plane with vanishing at infinity initial conditions and single-frequency periodic boundary data. We propose a matrix Riemann-Hilbert problem, which provides the existence of the solution of the IBV problem for all t and allows us to obtain an explicit formula for the asymptotics of the solution, using the steepest descent method for the oscillatory matrix RH problem introduced by P. Deift and X. Zhou [6]. 2009 Article Simple Periodic Boundary Data and Riemann-Hilbert Problem for Integrable Model of the Stimulated Raman Scattering / E.A. Moskovchenko // Журнал математической физики, анализа, геометрии. — 2009. — Т. 5, № 1. — С. 82-103. — Бібліогр.: 14 назв. — англ. 1812-9471 http://dspace.nbuv.gov.ua/handle/123456789/106534 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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We consider the initial-boundary value (IBV) problem for nonlinear equations related to the integrable model of the stimulated Raman scattering in the quarter xt-plane with vanishing at infinity initial conditions and single-frequency periodic boundary data. We propose a matrix Riemann-Hilbert problem, which provides the existence of the solution of the IBV problem for all t and allows us to obtain an explicit formula for the asymptotics of the solution, using the steepest descent method for the oscillatory matrix RH problem introduced by P. Deift and X. Zhou [6]. |
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Article |
| author |
Moskovchenko, E.A. |
| spellingShingle |
Moskovchenko, E.A. Simple Periodic Boundary Data and Riemann-Hilbert Problem for Integrable Model of the Stimulated Raman Scattering Журнал математической физики, анализа, геометрии |
| author_facet |
Moskovchenko, E.A. |
| author_sort |
Moskovchenko, E.A. |
| title |
Simple Periodic Boundary Data and Riemann-Hilbert Problem for Integrable Model of the Stimulated Raman Scattering |
| title_short |
Simple Periodic Boundary Data and Riemann-Hilbert Problem for Integrable Model of the Stimulated Raman Scattering |
| title_full |
Simple Periodic Boundary Data and Riemann-Hilbert Problem for Integrable Model of the Stimulated Raman Scattering |
| title_fullStr |
Simple Periodic Boundary Data and Riemann-Hilbert Problem for Integrable Model of the Stimulated Raman Scattering |
| title_full_unstemmed |
Simple Periodic Boundary Data and Riemann-Hilbert Problem for Integrable Model of the Stimulated Raman Scattering |
| title_sort |
simple periodic boundary data and riemann-hilbert problem for integrable model of the stimulated raman scattering |
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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2009 |
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http://dspace.nbuv.gov.ua/handle/123456789/106534 |
| citation_txt |
Simple Periodic Boundary Data and Riemann-Hilbert Problem for Integrable Model of the Stimulated Raman Scattering / E.A. Moskovchenko // Журнал математической физики, анализа, геометрии. — 2009. — Т. 5, № 1. — С. 82-103. — Бібліогр.: 14 назв. — англ. |
| series |
Журнал математической физики, анализа, геометрии |
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AT moskovchenkoea simpleperiodicboundarydataandriemannhilbertproblemforintegrablemodelofthestimulatedramanscattering |
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2025-07-07T18:36:52Z |
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2025-07-07T18:36:52Z |
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1837014359114514432 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2009, vol. 5, No. 1, pp. 82�103
Simple Periodic Boundary Data and Riemann�Hilbert
Problem for Integrable Model of the Stimulated Raman
Scattering
E.A. Moskovchenko
Mathematical Division, B. Verkin Institute for Low Temperature Physics and Engineering
National Academy of Sciences of Ukraine
47 Lenin Ave., Kharkiv, 61103, Ukraine
E-mail:kuznetsova@ilt.kharkov.ua
Received March 24, 2008
We consider the initial-boundary value (IBV) problem for nonlinear equa-
tions related to the integrable model of the stimulated Raman scattering in
the quarter xt-plane with vanishing at in�nity initial conditions and single-
frequency periodic boundary data (pei!t). We propose a matrix Riemann�
Hilbert problem, which provides the existence of the solution of the IBV
problem for all t and allows us to obtain an explicit formula for the asymp-
totics of the solution, using the steepest descent method for the oscillatory
matrix RH problem introduced by P. Deift and X. Zhou [6].
Key words: nonlinear equations, Riemann�Hilbert problem, the steepest
descent method, asymptotics.
Mathematics Subject Classi�cation 2000: 37K15, 35Q15, 35B40.
1. Introduction
The phenomenon of stimulated Raman scattering (SRS) is described by three
coupled PDEs. In the transient limit these equations are integrable [1, 2], i.e.
they admit a Lax pair formulation. Paper [1] is devoted to the Raman soliton
generation from laser inputs in the SRS model. In [2], the authors studied the
asymptotic behavior of the solution of the initial-boundary-value (IBV) problem
in the semistrip (x 2 [0;1), t 2 [0; 1]) by using the method [3] based on the
simultaneous spectral analysis of the two parts forming the Lax pair and a ma-
trix Riemann�Hilbert problem on the complex k-plane. This method includes
more boundary values than required for a well-posed IBV problem. This over-
determination of the boundary data implies the so-called global relation [3, 4]
c
E.A. Moskovchenko, 2009
Simple Periodic Boundary Data and Riemann�Hilbert Problem...
between the corresponding spectral functions. Fortunately, the initial boundary
value problem for nonlinear SRS equations considered below is a nice model of
PDEs, which can be solved by using the matrix Riemann�Hilbert problem with-
out restrictions caused by global relation. In this case all spectral functions are
uniquely de�ned by given initial and boundary data only. For the �nite domain
[0; L]x[0;T] the IBV problem for the SRS equations was studied in [5], where the
di�culties on the presence of two essential singularities in the matrix Riemann-
Hilbert problem have been overcome. In the present paper, we consider the IBV
problem for the SRS equations in the quarter xt-plane with vanishing at in�nity
initial function and simple periodic boundary data. In general, one can propose
di�erent matrix RH problems suitable for the given IBV problem. We propose
a matrix Riemann�Hilbert problem, which provides the existence of the solution
for all t and allows us to obtain an explicit formula for the asymptotics of the
solution, using the steepest descent method for the oscillatory matrix RH problem
introduced by P. Deift and X. Zhou [6]. To make the asymptotic analysis more
transparent we restrict our attention to the special case when boundary data take
the single-frequency periodic form, and the initial function is identically equal
to zero. We show that in the region x > !
2
t, where ! is the frequency of the
boundary data, see (3) below, the asymptotics has a quasi-linear dispersive cha-
racter and is described by Zakharov�Manakov type formula. In other regions the
asymptotic analysis turns to be more complicated and will be presented elsewhere.
The IBV problem under consideration is
2iqt = �; �x = 2i�q; �x = i(�q�� q��); x 2 (0;1); t 2 (0;1); (1)
with the initial function
q(x; 0) = u(x); x 2 (0;1); (2)
and the boundary condition
�(0; t) = pei!t; p > 0; (3)
�(0; t) = l = const; l 2 R: (4)
We suppose that the function u(x) is absolutely continuous, xu(x) and u0x(x) 2
L
1(0;1):
1Z
0
[xju(x)j+ ju0x(x)j]dx <1: (5)
Let the absolutely continuous in x and t functions q(x; t); �(x; t) 2 C ; �(x; t) 2 R
satisfy the SRS equations (1) on the semi-in�nite domain x; t 2 ((0;1)� (0;1)),
initial (2) and boundary (3) conditions. Since (1) implies
@
@x
�
�
2(x; t) + j�(x; t)j2
�
= 0;
Journal of Mathematical Physics, Analysis, Geometry, 2009, vol. 5, No. 1 83
E.A. Moskovchenko
in what follows we assume that
�
2(x; t) + j�(x; t)j2 � 1
and, particularly, p2 + l
2 = 1. All considerations of the paper are valid if the
boundary conditions (3), (4) are replaced by
�(0; t) = pei!t + v(t); �(0; t) = l + w(t); (6)
where v(t) and w(t) are given functions decreasing fast as t ! 1. The IBV
problem of this type was considered in [7], where the generation of asymptotic
solitons by boundary data (6) was studied using the Marchenko integral equations.
Notice that, if q(x; t) is real and 2q = vx, � = i sinv, � = cos v, then the SRS
equations are reduced to the sine-Gordon equation
vxt = sin v: (7)
The asymptotic behavior of the rapidly decreasing (as jxj ! 1) solution was
studied in [8].
2. Basic Solutions of Linear Over-Determined Equations
For studying the initial boundary value problem (1)�(3), we will use
the simultaneous spectral analysis [3] of the linear x-equation
�x + ik�3� = Q(x; t)�; (8)
�3 =
�
1 0
0 �1
�
; Q(x; t) =
�
0 q(x; t)
��q(x; t) 0
�
(9)
and the linear t-equation
�t =
i
4k
bQ(x; t)�; (10)
bQ(x; t) = � �(x; t) i�(x; t)
�i��(x; t) ��(x; t)
�
; (11)
where �(x; t; k) is a 2� 2 matrix-valued function and k 2 C is a parameter. Let
us rewrite equations (8), (10) in the equivalent form:
Wx = U(x; t; k)W; U(x; t; k) = Q(x; t)� ik�3; (12)
Wt = V (x; t; k)W; V (x; t; k) =
i
4k
bQ(x; t): (13)
84 Journal of Mathematical Physics, Analysis, Geometry, 2009, vol. 5, No. 1
Simple Periodic Boundary Data and Riemann�Hilbert Problem...
It is easy to verify that the over-determined system of di�erential equations (12),
(13) is compatible (i.e., @2W
@x@t
= @2W
@t@x
) if and only if the matrices U(x; t; k) and
V (x; t; k) satisfy the compatibility condition
Ut(x; t; k)� Vx(x; t; k) + U(x; t; k)V (x; t; k)� V (x; t; k)U(x; t; k) = 0; k 2 C ;
(14)
which is equivalent to the SRS equations (1) on the functions q(x; t), �(x; t),
�(x; t).
Below we will use the following lemma.
Lemma 2.1. Let the compatibility condition (14) be ful�lled for all k 2 C .
Let W (x; t; k) be a matrix satisfying the x-equation (12) for all t (the t-equation
(13) for all x). Assume that W (x0; t; k) satis�es the t-equation (13) for some
x = x0 (W (x; t0; k) satis�es the x-equation (12) for some t = t0), including the
case when x0 =1 ( t0 =1). Then W (x; t; k) satis�es the t-equation (13) for all
x (satis�es the x-equation (12) for all t).
P r o o f. Let W = W (x; t; k) be a solution to (12). Then, due to the com-
patibility condition the matrix Ŵ (x; t; k) =Wt � V (x; t; k)W is also the solution
to (12). Indeed, Ŵx = U(x; t; k)Ŵ +(Ut�Vx+UV �V U)W = U(x; t; k)Ŵ : Since
the matrices W and Ŵ are the solutions of the same equation (12), it follows that
Ŵ (x; t; k) =W (x; t; k)C(t; k) for some C(t; k) independent of x. By assumption,
Ŵ (x0; t; k) = 0. Hence C(t; k) � 0 and thus Ŵ (x; t; k) � 0, what means that
W (x; t; k) satis�es the t-equation (13) for all x. The proof of the statement with
x and t interchanged is similar.
To introduce the basic solutions of the over-determined equations we have to
�nd the exact solution of the t-equation for x = 0. It takes the form
�t =
i
4k
�
l ipei!t
�ipe�i!t �l
�
�: (15)
The following matrix
E(t; k) = 1
2
ei!�3t=2
0B@
1
{(k)
+ {(k)
1
{(k)
� {(k)
1
{(k)
� {(k)
1
{(k)
+ {(k)
1CA e�i
(k)�3t (16)
is a solution of this equation if
{(k) = 4
r
k �E
k � �E
;
Journal of Mathematical Physics, Analysis, Geometry, 2009, vol. 5, No. 1 85
E.A. Moskovchenko
where
E =
l
2!
+
ip
2j!j = E1 + iE2;
�E = E1 � iE2;
and
(k) =
j!j
2k
q
(k �E)(k � �E):
Without loss of generality we assume here and in what follows that
! > 0. Indeed, we obtain the case ! < 0 if we take the complex conjugated func-
tions �q(x; t); ��(x; t); �(x; t) instead of q(x; t); �(x; t); �(x; t). To �x the branches
of the roots we choose a cut in the complex k-plane along the curve
[ �
, where
Im
(k) = 0, and de�ne {(k) and
(k) as
{(k) = 1 +O(k�1);
(k) =
!
2
+O(k�1); k !1:
A simple analysis shows that the set � := fk 2 C j Im
(k) = 0g consists of the
real line Im k = 0 and the circle arc
̂, which is de�ned by�
k1 �
jEj2
2E1
�2
+ k
2
2 =
� jEj2
2E1
�2
;
k
2
1 + k
2
2 � jEj2
(see Fig. 1). If
�(k), {�(k) are boundary values of the functions
(k), {(k) on
the cut
[ �
from the right (+) and left (-) sides of the cut, then
+(k) = �
�(k); {�(k) = i{+(k):
Then the matrix-valued function E(t; k) is analytic when being away from the
point 0, where it has an essential singularity, and the circle arc
̂.
The function
(k) has the following asymptotics:
(k) =
8><>:
!
2
� l
4k
+O(k�2); k !1;
� 1
4k
� l!
2
+O(k); k ! 0; sign(l!) = �1:
In the present paper we consider the case l < 0 and ! > 0. The matrix E(t; k)
behaves as follows:
E(t; k) = I +O(k�1); k !1
and
E(t; k)eit�3=4k = E0(t) +O(k); k ! 0;
where
E0(t) = ei!t�3=2
�
cos(argE
2
) �i sin(argE
2
)
�i sin(argE
2
) cos(argE
2
)
�
ei!lt�3=2:
86 Journal of Mathematical Physics, Analysis, Geometry, 2009, vol. 5, No. 1
Simple Periodic Boundary Data and Riemann�Hilbert Problem...
Im k
Re k
E
0
��l
���
�
�
�
�
E
Fig. 1: Set �; !l < 0.
Now we introduce the basic solutions (eigenfunctions) of equations (8) and
(10). The �rst eigenfunction has the form:
�1(x; t; k) =
�
e�ikx�3 +
xZ
�x
K(x; y; t)e�iky�3dy
�
E(t; k); (17)
where the kernel K(x; y; t) is chosen to be so that the �rst factor satis�es the x-
equation (8) for all t, and the second factor satis�es the t-equation (10) for x = 0.
By Lemma 2.1, �1(x; t; k) satis�es both equations (8) and (10). The existence of
the solution represented by the transformation operators with the kernelK(x; y; t)
is proved in [9].
If the functions q(x; t), �(x; t) and �(x; t) are absolutely continuous in x and t
and satisfy the initial and boundary conditions (2)�(3) and the di�erential equa-
tions (1) almost everywhere, then the matrix valued function (17) has the following
properties:
1) �1(x; t; k) satis�es the x- and t-equations (8)�(10) for k 2 C n(f0g[fEg[f �Eg);
2) �1(x; t; k) = � ��1(x; t; �k)�
�1, k 2 C n (f0g[fEg[f �Eg), where � =
�
0 1
�1 0
�
;
3) det�1(x; t; k) � 1; k 2 C ;
4) for k 6= 0; E; �E the map (x; t) 7�! �1(x; t; k) is absolutely continuous together
with its partial derivatives;
5) the map k 7�! �1(x; t; k) is analytic in k 2 C n (f0g [
[ �
) and it has the
forth root singularities at the points E and �E;
Journal of Mathematical Physics, Analysis, Geometry, 2009, vol. 5, No. 1 87
E.A. Moskovchenko
6) �1(x; t; k)e
ikx�3 = I +O(k�1) +O(k�1e2ikx�3); k !1;
7) �1(x; t; k)e
it�3
4k = �0(x; t) +O(k); k ! 0;
�0(x; t) =
0@I + xZ
�x
K(x; y; t)dy
1A E0(t):
The eigenfunction �2(x; t; k) normalized by the condition
�2(0; 0; k) = I
has the form
�2(x; t; k) =
�
e�ikx�3 +
xZ
�x
K(x; y; t)e�iky�3dy
�
E(t; k)E�1(0; k): (18)
It is related to �1(x; t; k) by
�1(x; t; k) = �2(x; t; k)E(0; k):
The eigenfunction �2(x; t; k) satis�es the properties 1)�6) and 7) with
�0(x; t)E�10 (0) instead of �0(x; t).
Finally, we chose the eigenfunction �3(x; t; k) in the form:
�3(x; t; k) =
�
e�
it�3
4k +
i
4k
tZ
�t
L(x; t; s)e�
is�3
4k ds
�
(19)
�
�
e�ikx�3 +
1Z
x
N(x; y)e�iky�3dy
�
;
where the kernels L(x; y; s) and N(x; y) are such that the �rst factor satis�es
(10) for any x, and the second factor satis�es (8) for t = 0. Due to Lem. 2.1,
�3(x; t; k) satis�es both equations (8) and (10). The matrix �3(x; t; k) possesses
the properties 1)�4) for k 2 R n f0g. Other important properties �3(x; t; k) are
as follows:
5) the matrix columns [�3]1(x; t; k) and [�3]2(x; t; k) are analytic in k 2 C � ;
respectively;
6) at in�nity they have the asymptotics:
[�3]1(x; t; k)e
ikx =
�
1
0
�
+O(k�1); k !1; Imk � 0;
88 Journal of Mathematical Physics, Analysis, Geometry, 2009, vol. 5, No. 1
Simple Periodic Boundary Data and Riemann�Hilbert Problem...
[�3]2(x; t; k)e
�ikx =
�
0
1
�
+O(k�1); k !1; Im k � 0;
7) for k ! 0 and Imk 6= 0 they have the following asymptotics:
[�3]1(x; t; k)e
it
4k = [�̂3]1(x; t) +O(k); k ! 0; Im k < 0;
[�3]2(x; t; k)e
� it
4k = [�̂3]2(x; t) +O(k); k ! 0; Im k > 0;
where [�̂3]1(x; t) and [�̂3]2(x; t) are some absolutely continuous vector-functions
depending on the entries of matrices L(x; t; s) and N(x; y). The existence of the
transformation operators with kernels K(x; y; t), L(x; t; s) and N(x; y) can be
proved in the same way as in [9].
Since all the introduced matrix valued functions �j(x; t; k), j = 1; 2; 3, are the
solutions to the x- and t-equations (8)�(10), they are linear dependent, so there
exist transition matrices S(k), s(k) and R(k) independent of x and t such that
�1(x; t; k) = �2(x; t; k)S(k); �2(x; t; k) = �3(x; t; k)s(k); (20)
�1(x; t; k) = �3(x; t; k)R(k): (21)
They can be written as follows:
S(k) = E(0; k); s(k) = ��13 (0; 0; k); R(k) = s(k)S(k):
The transition matrices have the following representations:
s
�1(k) =
�
�a(�k) b(k)
��b(�k) a(k)
�
; (22)
S(k) =
�
�A(�k) B(k)
� �B(�k) A(k)
�
; (23)
where
a(k) = 1 +
1Z
0
N22(0; y)e
iky
dy; b(k) =
1Z
0
N12(0; y)e
iky
dy;
2A(k) = {(k) +
1
{(k)
= 2 �A(�k); 2B(k) =
� 1
{(k)
� {(k)
�
= �2 �B(�k):
The functions N12(0; y) and N22(0; y) are absolutely continuous and their deriva-
tives belong to the space L1(0;1). For R(k) = s(k)S(k) we have
R(k) =
�
�aR(�k) bR(k)
��bR(�k) aR(k)
�
; (24)
where aR(k) = �a(�k)A(k) + �b(�k)B(k) and bR(k) = a(k)B(k) � b(k)A(k).
Journal of Mathematical Physics, Analysis, Geometry, 2009, vol. 5, No. 1 89
E.A. Moskovchenko
Further we prove a one-to-one correspondence between the initial function
u(x) and spectral data a(k) and b(k). Namely, let u(x) be absolutely continu-
ous, xu(x); u0x(x) 2 L
1(0;1). Then the vector-function (x; k) := [�3]2(x; 0; k)
(the second column of the matrix �3(x; t; k)), which satis�es the equation
x + ik�3 =
�
0 u(x)
��u(x) 0
�
; 0 < x <1; (25)
and the boundary condition
lim
x!1
(x; k)e�ikx =
�
0
1
�
:
de�nes the direct map
S : fu(x)g ! fa(k); b(k)g (26)
by the formula �
b(k)
a(k)
�
= (0; k):
The spectral data a(k) and b(k) possess the following properties:
1) a(k) and b(k) are analytic in k 2 C + and continuous in k 2 C + functions
represented in the form
a(k) = 1 +
1Z
0
�(y)eikydy; b(k) =
1Z
0
�(y)eikydy;
where �(y); �(y) are absolutely continuous and �
0
y(y); �
0
y(y) 2 L
1(0;1);
2) ja(k)j2 + jb(k)j2 � 1; k 2 R;
3) a(k) = 1 +O(k�1); b(k) = O(k�1); k !1.
The inverse map Q is given by
u(x) = 2i lim
k!1
kM
(x)
12 (x; k); (27)
where M
(x)
12 (x; k) is the entry (12) of matrixM (x)(x; k). This matrix is the unique
solution of the following Riemann�Hilbert problem:
� M
(x)(x; k) is a sectionally analytic matrix valued function in k 2 C n�, where
the oriented contour � is a union of the real line R and the circle S1 = fk 2 C :
jkj = jS1jg, where jS1j is a su�ciently large positive number. The orientation
of � is chosen so that k-plane is a union of the two open domains
� and their
common boundary � (Fig. 2).
� M
(x)(x; k) = I +O(k�1); k !1.
� M
(x)
+ (x; k) =M
(x)
� (x; k)J (x)(x; k); k 2 �;
90 Journal of Mathematical Physics, Analysis, Geometry, 2009, vol. 5, No. 1
Simple Periodic Boundary Data and Riemann�Hilbert Problem...
0
S 8
�
�
�
�
�
�
�
�
Fig. 2: The oriented contour � for the x-problem.
where M
(x)
+ (x; k), M
(x)
� (x; k) are the boundary values of matrix M
(x)(x; k) on
contour � from domains
+,
�, and
J
(x)(x; k) =
8>>>><>>>>:
1 0
0 1
!
; k 2 R; jkj < jS1j;
1 b(k)
a(k)
e�2ikx
�b(�k)
�a(�k)
e2ikx 1
ja(k)j2
!
; k 2 R; jkj > jS1j;
(28)
J
(x)(x; k) =
8>>>><>>>>:
1 b(k)
a(k)
e�2ikx
0 1
!
; jkj = jS1j; arg k 2 (0; �);
1 0
�b(�k)
�a(�k)
e2ikx 1
!
; jkj = jS1j; arg k 2 (�; 2�):
(29)
This RH problem is uniquely solvable [12].
3. Main Matrix Riemann�Hilbert Problem: Reconstruction
of the SRS Model
Under the assumption that x- and t-equations (8) and (10), respectively, are
compatible, the relations (20) between their solutions can be written in the form
of the matrix Riemann�Hilbert problem.
Journal of Mathematical Physics, Analysis, Geometry, 2009, vol. 5, No. 1 91
E.A. Moskovchenko
Let q(x; t); �(x; t); �(x; t) be absolutely continuous functions with respect to
x 2 [0;1) and t 2 [0;1) satisfying the SRS equations (1), the initial (2) and
boundary (3) conditions. Then the relations (20) de�ne a map
SR : fq(x; t); �(x; t); �(x; t)g ! fa(k); b(k); A(k); B(k)g: (30)
In fact the spectral functions fa(k); b(k)g are de�ned by initial function u(x) =
q(x; 0); and the spectral functions fA(k); B(k)g are de�ned by boundary data.
In our case they take the explicit form
2A(k) = {(k) +
1
{(k)
= 2 �A(�k); 2B(k) =
� 1
{(k)
� {(k)
�
= �2 �B(�k):
To describe the map inverse to (30) we additionally use the auxiliary spectral
functions aR(k) = �a(�k)A(k) + �b(�k)B(k) and bR(k) = a(k)B(k) � b(k)A(k) which
are the entries of the transition matrix R(k) = s(k)S(k).
The inverse (to (30)) map QR is de�ned by
q(x; t) = 2im1
12(x; t); (31)
�(x; t) = m11(x; t); (32)
�(x; t) = �im12(x; t); (33)
where
m
1(x; t) = lim
k!1
kM(x; t; k);
m(x; t) = �m0(x; t)�3m
�1
0 (x; t);
m0(x; t) = lim
k!0
M(x; t; k);
and the matrix M(x; t; k) is the solution of the following Riemann�Hilbert prob-
lem RHxt:
� M(x; t; k) is sectionally analytic for k 2 C n �; the oriented contour � is
de�ned as follows: � = R [ S1 [
[ �
(Fig. 3);
� M(x; t; k) has the fourth-root singularities at the points E and �E;
� M�(x; t; k) =M+(x; t; k)J(x; t; k); k 2 �, where
92 Journal of Mathematical Physics, Analysis, Geometry, 2009, vol. 5, No. 1
Simple Periodic Boundary Data and Riemann�Hilbert Problem...
0
S 8
�
�
�
�
�
�
�
�
�
��
Fig. 3: The contour � for the xt-problem.
J(x; t; k) =
8>>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>>:
0@ aR(k) 0
�bR(�k)e
2it�(k) 1
aR(k)
1A ; jkj = jS1j; Im k < 0;
0BB@ 1
bR(k)
aR(k)
e�2it�(k)
�bR(�k)
�aR(�k)
e2it�(k)
1
jaR(k)j2
1CCA ; jkj > jS1j; Im k = 0;
0@�aR(�k) bR(k)e
�2it�(k)
0
1
�aR(�k)
1A ; jkj = jS1j; Im k > 0;
J(x; t; k) =
8>>>>>><>>>>>>:
1 0
0 1
!
; jkj < jS1j; Im k = 0;
0 �ie�2it�(k)
�ie2it�(k) 0
!
; k 2
[ �
;
with �(k) =
1
4k
+ k
x
t
;
� M(x; t; k) = I +O(k�1); k !1;
� M(x; t; k) = m0(x; t) +O(k); k ! 0:
Journal of Mathematical Physics, Analysis, Geometry, 2009, vol. 5, No. 1 93
E.A. Moskovchenko
P r o o f. To construct the Riemann�Hilbert problem RHxt, we de�ne the
following matrices:
M(x; t; k) =
8>>><>>>:
�
[�1]1(x; t; k)
�aR(�k)
; [�3]2(x; t; k)
�
eit�(k)�3 ; jkj > jS1j; Imk > 0;
�1(x; t; k)e
it�(k)�3 ; jkj < jS1j; Imk > 0;
M(x; t; k) =
8>>><>>>:
�1(x; t; k)e
it�(k)�3 ; jkj < jS1j; Imk < 0;
�
[�3]1(x; t; k);
[�1]2(x; t; k)
aR(k)
�
eit�(k)�3 ; jkj > jS1j; Imk < 0;
where [�1]1;2(x; t; k), [�3]1;2(x; t; k), are the vector columns of the matrices
�1(x; t; k) = ([�1]1(x; t; k); [�1]2(x; t; k));
�3(x; t; k) = ([�3]1(x; t; k); [�3]2(x; t; k)):
The radius jS1j of the circle S1 is su�ciently large so that aR(k) 6= 0 (�aR(�k) 6= 0)
for jkj > jS1j, Im k < 0, Imk > 0. Then the matrices M�(x; t; k) are analytic
functions in the domains
�. They have the forth root singularities at the points
E and �E, because the matrix �1(x; t; k) as well as the matrix E(t; k) has the same
singularities at these points. The determinants of these matrices are equal to one,
which follows from the vector relations
[�1]1(x; t; k) = �aR(�k)[�3]1(x; t; k) � �bR(�k)[�3]2(x; t; k);
[�1]2(x; t; k) = bR(k)[�3]1(x; t; k) + aR(k)[�3]2(x; t; k)
arising from (20). Direct calculation gives the form of the jump matrix J(x; t; k)
on di�erent parts of �. The asymptotic formulas for M(x; t; k) as k ! 1 and
k ! 0 follow from the corresponding equations for the eigenfunctions, see Sect. 2,
and from the asymptotic behavior of the spectral function aR(k). In particular,
we have M�(x; t; k) = �0(x; t) +O(k); k ! 0: Therefore m0(x; t) = �0(x; t):
Using general ideas of [11] and the results of [12] for contours with self-
intersections, we prove the following theorem.
Theorem 3.1. Let u(x) be an absolutely continuous function satisfying (5).
Let �(0; t) = l, l < 0, �(0; t) = pe
2i!t (!; p > 0, l2+p2 = 1). Let fa(k); b(k); A(k);
B(k)g be the corresponding spectral functions. Then the Riemann�Hilbert problem
RHxt has the unique solution M(x; t; k). The functions q(x; t), �(x; t) and �(x; t),
de�ned by the equations
q(x; t) = 2i lim
k!1
kM12(x; t; k); (34)
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Simple Periodic Boundary Data and Riemann�Hilbert Problem...
�(x; t) = m11(x; t); �(x; t) = �im12(x; t)
with the matrix
m(x; t) = �M(x; t; 0)�3M
�1(x; t; 0);
satisfy the SRS equations (1), the initial condition
q(x; 0) = u(x); x 2 (0;1);
and the boundary conditions
�(0; t) = l; �(0; t) = pei!t; t 2 (0;1):
The proof of this theorem is performed in the same way as in [5] and [10].
4. Asymptotic Behaviour of the Solution
in the Zakharov�Manakov Region
In this section we study the asymptotic behavior of the solution to the IBV
problem (1)�(3) as t ! 1. To �x the ideas of asymptotic analysis and to make
it more transparent we restrict our attention to a special case when the initial
function is equal to zero identically. We will use the steepest descent method [6]
by P. Deift and X. Zhou; many technical details of this method become much
more simple in this special case. We describe the asymptotics of the solution in
the sector x > !
2
t, where it has a quasilinear dispersive character. In the adjacent
sector x < !
2
t of the quarter xt-plane the asymptotics is more complicated and
will be studied elsewhere.
For the case u(x) � 0, �(0; t) = pei!t and �(0; t) = l the corresponding spectral
functions are as follows:
a(k) � 1; b(k) � 0; (35)
aR(k) = A(k) =
1
2
�
{(k) +
1
{(k)
�
; bR(k) = B(k) =
1
2
�
1
{(k)
� {(k)
�
;
(36)
where {(k) = 4
r
k �E
k � �E
, E =
l + ip
2!
(!; p > 0; l < 0; l2 + p
2 = 1). These formulas
show that the spectral data A(k) and B(k) are analytic functions everywhere with
the exception of arc
[ �
, and that A(k) 6= 0. We recall that the complex k-plane
is cut along the contour
[ �
. Therefore the main Riemann�Hilbert problem
RHxt can be reduced to the equivalent one:
Journal of Mathematical Physics, Analysis, Geometry, 2009, vol. 5, No. 1 95
E.A. Moskovchenko
� matrix valued function M
(1)(x; t; k) is analytic in the domain C + n
and
C � n �
;
� M
(1)(x; t; k) has the fourth-root singularities at the points E and �E;
� M
(1)
� (x; t; k) =M
(1)
+ (x; t; k)J (1)(x; t; k); k 2 R [
[ �
, where
J
(1)(x; t; k) =
8>>>>>>>>>>>>>><>>>>>>>>>>>>>>:
1 %(k)e�2it�(k)
�%(k)e2it�(k) 1� %
2(k)
!
; k 2 R;
1 0
f(k)e2it�(k) 1
!
; k 2
;
1 f(k)e�2it�(k)
0 1
!
; k 2 �
;
� M
(1)(x; t; k) = I +O(k�1); k !1;
� M
(1)(x; t; k) = ~m0(x; t) +O(k); k ! 0;
where %(k) :=
B(k)
A(k)
and f(k) := %�(k)� %+(k) = � 1
B+(k)A+(k)
.
� The functions q(x; t), �(x; t) and �(x; t) are determined by M
(1)(x; t; k) in
the same way as in (34).
P r o o f. Since aR(k) � A(k) 6= 0 for all k, the RHxt problem can be
simpli�ed. Indeed, let us transform the initial matrix M(x; t; k) to the following
one
M
(1)(x; t; k) =M(x; t; k)G(1)(x; t; k);
where G(1)(x; t; k) =
�
1 0
0 1
�
for jkj > jS1j and
G
(1)(x; t; k) =
8>>>>>><>>>>>>:
1
A(k)
�B(k)e�2it�(k)
0 A(k)
!
; jkj < jS1j; Im k > 0;
A(k) 0
�B(k)e2it�(k) 1
A(k)
!
; jkj < jS1j; Im k < 0:
The transformation eliminates the circle S1, where the jump matrix J(x; t; k) =
G
(1)(x; t; k) is unbounded as t ! 1. It is easy to see that the matrix valued
function M
(1)(x; t; k) is analytic in the domains C + n
and C � n �
and has the
forth-root singularities at branch points. The new jump matrix coincides with
96 Journal of Mathematical Physics, Analysis, Geometry, 2009, vol. 5, No. 1
Simple Periodic Boundary Data and Riemann�Hilbert Problem...
the jump matrix J
(1)(x; t; k). Furthermore, since
G
(1)(x; t; k) =
8>>>>>><>>>>>>:
1
A(k)
O(e�t Imk=2jkj2)
0 A(k)
!
; k ! 0; Im k > 0;
A(k) 0
O(et Imk=2jkj2) 1
A(k)
!
; k ! 0; Im k < 0
becomes diagonal in the limit as t!1, then
lim
k!0
M1(x; t; k)�3M
�1
1 (x; t; k) = lim
k!0
M(x; t; k)�3M
�1(x; t; k)
and, thus, �(x; t) and �(x; t) given by M1(x; t; k) according to (34) are similar to
the ones given byM(x; t; k). Finally, since M1(x; t; k) =M(x; t; k) for jkj > jS1j;
we have that the same is true for q(x; t).
�
�
�
�
�
Fig. 4: The signature table of the function Im �(k).
To study the asymptotic behavior of the Riemann�Hilbert problem RHxt in
the region x > !
2
t we use the well-known technics from [6, 13, 14]. In what
follows, a signi�cant role is played by the decomposition of the complex k-plane
according to the signature table of the imaginary part of the phase function �(k) =
1
4
�
1
k
+ k
�2
�
, where �2 = t=4x. The stationary points of the phase function �(k)
are real and equal to ��. We have
Im �(k) =
jkj2 � �
2
4jkj2�2 Im k:
Journal of Mathematical Physics, Analysis, Geometry, 2009, vol. 5, No. 1 97
E.A. Moskovchenko
Thus Im �(k) > 0 (Im �(k) < 0) for k lying in the lower (upper) half-disk and out
of the upper (lower) half-disk de�ned by the circle jkj = � (Fig. 4). For � < jEj =
1=2! (that is, for x > !
2
t), the jump matrix J
(1)(x; t; k) approaches the identity
matrix as t ! 1 for k 2
[ �
. Hence the contour
[ �
does not contribute to
the main term of asymptotics, which is de�ned by the stationary points �� and
has the order O(t�1=2). The contour
[ �
plays a crucial role in the description
of asymptotics in the region x < !
2
t. We conjecture that in this region the
asymptotics is of order O(1) and takes the form of an elliptic modulated wave
for jEj < � < �0 and a plane wave for �0 < � < 1. In this paper we study the
asymptotics in the region x > !
2
t only.
To study the asymptotic behavior of the RH problem for the matrix
M
(1)(x; t; k) let us use the transform
M
(2)(x; t; k) =M
(1)(x; t; k)��3 (k);
where the function Æ(k) is equal to (cf.[6])
Æ(k) = exp
8><>: 1
2�i
�Z
��
log(1� %
2(s))ds
s� k
9>=>; ; k 2 C n [��; �];
and � =
p
t=4x > 0. Then the jump matrix J
(2)(x; t; k) has a lower/upper
factorization for jkj < � and an upper/lower factorization for jkj > �
J
(2)(x; t; k) =
8>>>>>>>>><>>>>>>>>>:
1 A(k)B(k)Æ2+(k)e
�2it�(k)
0 1
!
1 0
�A(k)B(k)�2� (k)e2it�(k) 1
!
;
jkj < �;
1 0
�%(k)�2(k)e2it�(k) 1
!
1 %(k)Æ2(k)e�2it�(k)
0 1
!
; jkj > �;
where we use the identity
%(k)
1� %2(k)
= A(k)B(k):
The jump matrices on the contours
[ �
are
J
(2)(x; t; k) =
8>>>>>><>>>>>>:
1 0
f(k)�2(k)e2it�(k) 1
!
; k 2
;
1 �f(k)Æ2(k)e�2it�(k)
0 1
!
; k 2 �
:
98 Journal of Mathematical Physics, Analysis, Geometry, 2009, vol. 5, No. 1
Simple Periodic Boundary Data and Riemann�Hilbert Problem...
Let us de�ne a decomposition of the complex k-plane into six domains
D1; : : : ;D6 as shown in Fig. 5. The contours L2 and L5 lie in the disk jkj < �;
the contours L1, L6 (L3, L4) range from the point � (��) to in�nity along the
rays arg k = ��=4 (arg k = � � �=4). Then the next transformation is
M
(3)(x; t; k) =M
(2)(x; t; k)G(2)(k);
where
G
(2)(k) =
8>>>>>>>>>>>>>><>>>>>>>>>>>>>>:
1 0
�%(k)�2(k)e2it�(k) 1
!
; k 2 D1 [D3;
1 0
0 1
!
; k 2 D2 [D5;
1 �%(k)Æ2(k)e�2it�(k)
0 1
!
; k 2 D4 [D6;
(37)
G
(2)(k) =
8>>>>>><>>>>>>:
1 A(k)B(k)Æ2(k)e2it�(k)
0 1
!
; k 2 D8;
1 0
A(k)B(k)�2(k)e�2it�(k) 1
!
; k 2 D7:
(38)
D
D
D
D
D
D
D
D
1
2
3
4
5
6
7
8
�
�
_
-
L1
L6
L5
L2
L
3
L4
Fig. 5: The contour � for the M (3)(x; t; k)-problem.
Journal of Mathematical Physics, Analysis, Geometry, 2009, vol. 5, No. 1 99
E.A. Moskovchenko
Remark 4.1. The transformation M
(2)(x; t; k) ! M
(3)(x; t; k) has the form
as above due to the fact that %(k), corresponding to the initial and boundary data
considered here, is in fact analytic outside
[ �
(see (35), (36)). An analogous
transformation in the case of more general initial and boundary conditions (i.e.
when, for example, the initial function is fast decreasing as x ! 1) requires
an analytic approximation of the corresponding spectral functions (cf. [6]).
The G-transformation leads to the following RH problem
M
(3)
� (x; t; k) =M
(3)
+ (x; t; k)J (3)(x; t; k)
on the contour depicted in Fig. 5 with the jump matrices J (3)(x; t; k) which are
equal to the identity matrix on real axis, they coincide with the matrices G(2)(k)
from (37)�(38) chosen for the contours k 2 Lj , j = 1; 2; : : : ; 6. Moreover, the
jump matrix J (3)(x; t; k) is equal to the identity matrix on the arc
[ �
because
f(k) = %�(k)� %+(k). Hence, in the region x > !
2
t (� < 1=2!) the jump across
the arc
[ �
does not contribute to the asymptotics of the solution. Furthermore,
it is easy to see that J (3)(x; t; k) = I + O(e��t) as t ! 1 and k 2 Lj with the
exception of some neighborhoods of the stationary points ��. Since
M
(3)(x; t; k) = I +
m
(3)
1 (x; t)
k
+O(k�2); k !1;
we have
q(x; t) = 2i[m
(3)
1 (x; t)]12 +O(e�"t); " > 0:
Now we have to evaluate the main contributions from neighborhoods of the
stationary points k0 = �� of the phase function �(k) = 1=4k + k=4�2. To do this
we use the scaling operators
F (k)! [N�F ](z) = F (z
p
�3t�1 + k0)jk0=��;
which for the matrices M (4)(x; t; z)� = [N�M
(3)](z
p
�3t�1 + k0) imply
[m
(3)
1 (x; t)]12 =
r
�3
t
[m
(4)
1 (x; t)]�12:
The scaling operators act on the product Æ(k)e�it�(k) as follows:
[N�Æe
�it�] = Æ
(0)
� (�; t) Æ
(1)
� (z; �; t);
where
Æ
(0)
� (�; t) =
�
4t
�
��i�(k0)=2
e
�it=2�
e
�(k0);
100 Journal of Mathematical Physics, Analysis, Geometry, 2009, vol. 5, No. 1
Simple Periodic Boundary Data and Riemann�Hilbert Problem...
Æ
(1)
� (z; �; t) = (�z)�i�(k0) e�iz2=4
�
1 +O
�
zp
t
��
with the functions
�(k) =
1
2�
log[1� %
2(k)] =
1
2�
log[1 + j%(k)j2] > 0
and
�(k) = � 1
2�i
�Z
��
log js� k0jd log[1� %
2(s)]:
The function Æ
(0)
� (�; t) does not depend on z, but the functions (�z)�i�(k0) e�iz2=4
do and lead to the model RH problems
H+(z) = H�(z)e
�iz2�3=4J0(k0)e
iz2�3=4;
where
J0(k0) =
�
1� %
2(k0) �%(k0)
%(k0) 1
�
; k0 = ��;
and
H(z) =
�
I +
m
�
1 (�; t)
z
+O(z�2)
�
z
i�(k0)�3 ; z !1:
These problems can be solved explicitly in the terms of parabolic cylinder func-
tions [13]. Thus we have
q(x; t)
=
r
�3
t
��
Æ
(0)
+ (�; t)
�2
2i[m+
1 (�; t)]12 +
�
Æ
(0)
� (�; t)
�2
2i[m�
1 (�; t)]12
�
+ o
�
1p
t
�
;
where
[m�
1 (�; t)]12 = �i
p
2�ei�=4e���(k0)=2
%(k0)�(�i�(k0)))
; k0 = ��;
and �(z) denotes Euler's gamma-function. Finally, using the basic identity
j�(�i�)j2 = j�(i�)j2 = �
� sinh�
;
we come to the following statement.
Journal of Mathematical Physics, Analysis, Geometry, 2009, vol. 5, No. 1 101
E.A. Moskovchenko
Theorem 4.1. Let q(x; t), �(x; t) and �(x; t) be the solution of the
SRS equations (1) with the initial function satisfying (5) and the boundary con-
ditions (3)�(4). Then in the region x > !
2
t, the function q(x; t) has a quasilinear
dispersive character, i.e., it is described by the Zakharov�Manakov type formulas
q(x; t) = 2
r
�3�(�)
t
exp
n
2i
p
xt� i�(�) log
p
xt+ i�(�)
o
+2
r
�3�(��)
t
exp
n
�2i
p
xt+ i�(��) log
p
xt+ i�(��)
o
+ o(t�1=2); t!1;
where the functions �(k) and �(k) are given by the equations
�(k) =
1
2�
log
�
1� %
2(k)
�
; �
2 =
t
4x
;
�(k) =
�
4
� 3�(k) log 2� arg %(k)� arg �(�i�(k))
+
1
�
�Z
��
log js� kjd log[1� %
2(s)]:
Here �(�i�(k)) is the Euler gamma-function, and %(k) = i tan[arg{(k)].
The asymptotics of functions �(x; t) and �(x; t) can be found by formulas
�(x; t) = 2iqt(x; t); �(x; t) =
p
1� j�(x; t)j2:
It is easy to �nd that the residual of this asymptotic solution in the SRS equations
has the order O(log t=t3=2) as t!1 and x > !
2
t.
Remark 4.2. Qualitatively this result is valid for the case u(x) 6= 0 and the
general boundary conditions (6). As we have noticed above, in this case the corre-
sponding transformations of the RH problem include the suitable analytic appro-
ximations of spectral functions.
Acknowledgment. The author thanks V.P. Kotlyarov and D.G. Shepelsky
for helpful discussions, comments and remarks.
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