Whitham Approach to Certain Large Fluctuation Problems in Statistical
Mechanics

Eldad Bettelheim

Racah Institute of Physics, Hebrew University of Jerusalem, Edmund J Safta
Campus

91904 Jerusalem, Israel

###### Abstract

We show the relationship between the strongly non-linear limit (also termed
the dispersionless or the Whitham limit) of the macroscopic fluctuation theory
of certain statistical models and the inverse scattering method. We show that
in the strongly non-linear limit the inverse scattering problem can be solved
using the steepest descent method of the associated Riemann–Hilbert problem.
The importance of establishing this connection, is that the equations in the
strongly non-linear limit can often be solved exactly by simple means, the
connection then provides a limit in which one can solve the inverse scattering
problem, thus aiding potentially the exact solution of a particular large
deviation problem.

##  1 Introduction

In this paper we study some mathematical aspects of large deviation problems
that have garnered some interest in recent years. Such problems include the
Kardar-Parisi-Zhang problem[1], the Kipnis-Marchioro-Presutti model[2], the
symmetric exclusion process[3], and more. Particularly interesting in this
regard is the availability of exact solutions [4]. Even in the absence of such
exact solution one may use macroscopic fluctuation theory [5] to study the
question of large deviation. One such case is the Kardar-Parisi-Zhang problem
in which the macroscopic fluctuation theory leads to the following
equations[6]:

$\displaystyle\partial_{t}q=\partial_{x}^{2}q+\frac{1}{4}q^{2}p$ (1.1)
$\displaystyle\partial_{t}p=-\partial_{x}^{2}p-\frac{1}{4}qp^{2}$ (1.2)

where the boundary value to be solved is given by

$\displaystyle p(x,1)=\Lambda\delta(x),\quad q(x,0)=f(x).$ (1.3)

for some given $f(x)$. Typically $f(x)$ is chosen to be either constant (flat
initial conditions) or itself proportional to a delta function, but other
boundary conditions may be considered as well.

Due to the similarity of the formulation of the large deviation problem across
different models, we shall concentrate here on the example where the equations
to be solved are given by Eqs. (1.1, 1.2) and the boundary conditions are
given in Eq. (1.3). Other problems have a very close formulation, other
directly [7, 8] or through non-trivial manipulation [9]. For example, for the
Kipnis-Marchioro-Presutti model one has to consider rather the derivative
nonlinear Schödinger equation, a variant of Eqs. (1.1, 1.2). Nevertheless, in
the formal sense the Eqs. (1.1, 1.2) and the derivative non-linear Schrödinger
equations have so much in common, that the methods provided here are easily
modified to deal with that problem as well. As a result, we prefer, for the
sake of brevity, to describe here the method for the particular case of the
nonlinear Schrödinger equation, Eqs. (1.1, 1.2) , without spelling out how the
method must be generalized to other cases.

To solve the boundary value problem described above, one may apply the inverse
scattering method[10, 11, 9, 7, 8], which we will describe below. Despite
being integrable, or ”completely soluble”, there is no recipe to solve a
particular boundary problem for the nonlinear Schödinger equation, such as the
problem presented in Eq. (1.3) above, and it is often necessary to rely on
luck to find such a solution.

Nevertheless, certain limits are tractable. For example, if the fields $p$ and
$q$ are small, such that the non-linear terms may be neglected, then a
solution may be easily found. Another tractable limit, is the limit where the
fields are large, such that the non-linear term dominates over the dispersive
term, leading to non-viscid equations. In such limits it may be instructive to
obtain a full solution to the large deviation problem, provided one knows how
to perform the scattering transform to the solution obtained in the special
limits. Then one may examine the scattering data, in order to have an educated
guess for the full solution.

The current paper concerns itself with this program, for the case of the
strongly non-linear limit (the linear limit is much easier to deal with, and
is usually also much less instructive). In certain cases in this limit, it is
possible simply to drop the dispersive terms in the equations, and thus deal
with the inviscid equations that arise. This case is most instructive,
although the method we present is not restricted to this case. In fact, often
solitons or oscillatory features may appear in the strongly non-linear limit,
and the method described below are suitable to deal with this situation as
well. Despite this, even in the case of sharp or oscillatory features, the
solution may still feature regions in space-time where the inviscid equations
hold with no dispersive terms, such that the inviscid equations contain a
large amount of information even in the presence of sharp or oscillatory
regions .This situation is well known in the strongly non-linear limit, which
is often also called the ”dispersionless limit”, or the ”Whitham limit”[12].

We proceed to solve the inverse scattering problem in the strongly non-linear
limit. This is done by applying the map from inverse scattering problem to the
Riemann-Hilbert problem, and then solving the Riemann-Hilbert problem thus
obtained using the steepest descent method following closely Ref. [13].

The result of the work is then a relation between the strongly non-linear
equations and the inverse scattering problem. In addition to the potential
utility of this approach in discovering new solutions, the connection thus
obtained also has additional advantages. Among them we list two here. First,
the method allows to find a systematic expansion in a small parameter around
the strongly non-linear solution. This is not done in the current paper, since
to apply this approach requires to introduce even more tools associated with
Riemann-Hilbert problem, and we leave that for future work. Secondly, the
method allows to find solutions which feature oscillatory regimes for the
fields $p$ and $q.$ Admittedly, though, the physical meaning of such solutions
is not known at present.

To demonstrate our approach we work out the invscid flows in the case of the
Kardar-Parisi-Zhang problem with flat initial conditions as worked out in Ref.
[14], and show how these are connected to the exact solution given in Ref.
[10].

##  2 Naïve Inviscid Limit

We first study the inviscid limit of Eqs. (1.1, 1.2). Since these are a
variant of the nonlinear Schrödinger equations (where $p$ and $q$ are now not
complex conjugates of each other but rather real) it is natural to consider
the Madelung transformation to the fields such as to obtain hydrodynamic type
equations for appropriately defined density and velocity fields, familiar from
quantum mechanics. In this case it is also appropriate to call such a
transformation a Cole-Hopf transformation. We apply the following[6]:

$\displaystyle q=e^{-\frac{1}{2}\int^{x}v},\quad p=-2\rho
e^{\frac{1}{2}\int^{x}v}$ (2.1)

to obtain:

$\displaystyle\partial_{t}\rho+\partial_{x}(\rho v)$
$\displaystyle=-\partial_{x}^{2}\rho$ (2.2)
$\displaystyle\partial_{t}v+v\partial_{x}v-\partial_{x}\rho$
$\displaystyle=\partial^{2}_{x}v.$ (2.3)

One may now discard the terms on the right hand side of the equation, which
are of higher derivative, and thus may be termed, dispersive, or viscous terms
(thus obtaining the naïve inviscid limit). The resulting equations are those
of a fluid with density $\rho$ and velocity $v$ and pressure given by
$\frac{-\rho^{2}}{2}$ [15]:

$\displaystyle\partial_{t}\rho+\partial_{x}(\rho v)$ $\displaystyle=0,$ (2.4)
$\displaystyle\partial_{t}v+v\partial_{x}v-\partial_{x}\rho$
$\displaystyle=0.$ (2.5)

These equations may be subjected to another transformation to bring them into
the form of Riemann invariants. The Riemann invariants, $\lambda_{1}$,
$\lambda_{2}$ are defined as follows:

$\displaystyle\lambda_{i}=\frac{\imath v}{2}\mp\sqrt{\rho},$ (2.6)

whereby the equations that result by this transformation are given by:

$\displaystyle\partial_{t}\lambda_{1}=\left(\frac{3}{2}\lambda_{1}+\frac{1}{2}\lambda_{2}\right)\imath\partial_{x}\lambda_{1}$
(2.7)
$\displaystyle\partial_{t}\lambda_{2}=\left(\frac{3}{2}\lambda_{2}+\frac{1}{2}\lambda_{1}\right)\imath\partial_{x}\lambda_{2}.$
(2.8)

These equations are typical of Dispersionless integrable equations and are in
fact the Whitham[12] universal equations written through Riemann
invariants[16, 17].

##  3 The Inverse Scattering Method for Large Fluctuation Problems

Now we turn our attention to the inverse scattering method, in order to
understand how the inviscid solutions translate in terms of the scattering
data.

The inverse scattering method relies on the fact that by introducing an
auxiliary parameter $k$ and defining two $k,$ $x$ and $t$ dependent matrices
[18]

$\displaystyle U(k)=\begin{pmatrix}-\frac{\imath k}{2}&-\frac{p}{8}\\\
q&\frac{\imath k}{2}\\\ \end{pmatrix},\quad
V(k)=\begin{pmatrix}\frac{k^{2}}{2}-\frac{pq}{8}&\frac{p^{\prime}-\imath
kp}{8}\\\ p^{\prime}+\imath kp&-\frac{k^{2}}{2}+\frac{pq}{8}\\\
\end{pmatrix},$ (3.1)

one may recast the nonlinear Schrödinger equations, Eqs. (1.1,1.2), in the
following form:

$\displaystyle\partial_{t}U-\partial_{x}V+[U,V]=0.$ (3.2)

Indeed if these equations are to hold for any $k,$ $x$ and $t$ then Eqs.
(1.1,1.2) must be satisfied. Eq. (3.2) has the form of a consistency condition
for the following equations for a $2\times 2$ matrix, $P$:

$\displaystyle\partial_{x}P=UP,\quad\partial_{t}P=VP.$ (3.3)

Indeed, if Eq. (3.2) is satisfied then a solution for $P$ exists.

If one assumes the $p$ and $q$ tend to $0$ at $x\to\pm\infty$ for any $t$,
then $P$ has solutions in that region in the form of plane waves $P\sim
e^{-\sigma_{z}\frac{\imath kx}{2}}T$, for any $x-$independent matrix $T$. This
leads naturally to consider the scattering problem. The problem reads as
follows: given $k$ and a plane wave solution at $x\to-\infty$ of the form
$P\sim e^{-\sigma_{z}\frac{\imath kx}{2}}$, find $T(k,t)$ as featured in the
asymptotic behavior of $P$ at $x\to+\infty$ as follows $P\sim
e^{-\sigma_{z}\frac{\imath kx}{2}}T(k,t)$.

Writing

$\displaystyle T(k,t)=\begin{pmatrix}a_{+}(k,t)&\tilde{b}(k,t)\\\
b(k,t)&a_{-}(k,t)\\\ \end{pmatrix},$ (3.4)

one can easily find the time dependence of the elements as follows:
$a_{\pm}(k,t)=a_{\pm}(k,0),$ $b(k,t)=e^{-k^{2}t}b(k,0)$ and
$\tilde{b}(k,t)=e^{k^{2}t}\tilde{b}(k,0).$ This can be shown by advancing the
plane wave solutions in time at $x=\pm\infty$. One obtains that at large $|x|$
the solution behaves as $P\sim
e^{\sigma_{z}\left(\frac{k^{2}t}{2}-\frac{\imath kx}{2}\right)},$ namely
advancing the plane waves in time amounts to applying the matrix
$e^{\sigma_{z}\frac{k^{2}t}{2}}.$ To obtain then the scattering matrix
$T(k,t)$ at any finite time $t$ it is enough to know the scattering matrix at
time $t=0$. Indeed one may deduce the scattering at time $t$ by first
considering the plane waves at $x=-\infty$ at time $t$ and then rewinding
these plane waves from time $t$ to time $0$ by applying the matrix
$e^{-\sigma_{z}\frac{k^{2}t}{2}},$ then one lets the waves scatter from
$-\infty$ to $\infty$ by applying the matrix $T(k,0)$ and finally one advances
the plane waves from time $t=0$ back to time $t$ by applying the matrix
$e^{\sigma_{z}\frac{k^{2}}{t}}$ . The application of the matrices is always
from the left such that we get
$T(k,t)=e^{\sigma_{z}\frac{k^{2}t}{2}}T(k,0)e^{-\sigma_{z}\frac{k^{2}t}{2}}$.
Furthermore, it can be shown that $\det T(k)=a_{+}a_{-}+b\tilde{b}=1$ and
$a_{\pm}$ can be shown to be analytic in the upper and lower half planes,
respectively.

##  4 The Riemann-Hilbert Approach to Inverse Scattering

Let us find a matrix solution $P(x,y)$ to Eq. (3.3) with boundary conditions:

$\displaystyle\partial_{x}P(x,y;k)=U(k)P(x,y;k),\quad
P(x,x;k)=\begin{pmatrix}e^{\frac{\imath kx}{2}}&\\\ &e^{-\frac{\imath
kx}{2}}\\\ \end{pmatrix}$ (4.1)

This function has the property

$\displaystyle P(x,y;k)P(y,z;k)=P(x,z;k).$ (4.2)

This allows one to define:

$\displaystyle T(k)\equiv\lim_{y,-x\to\infty}\begin{pmatrix}e^{-\frac{\imath
kx}{2}}&\\\ &e^{\frac{\imath kx}{2}}\\\
\end{pmatrix}P(x,y;k)\begin{pmatrix}e^{-\frac{\imath ky}{2}}&\\\
&e^{\frac{\imath kx}{2}}\\\ \end{pmatrix},$ (4.3) $\displaystyle
P^{+}(x;k)=\lim_{y\to\infty}P(x,y;k)\begin{pmatrix}e^{-\frac{\imath
ky}{2}}&\\\ &e^{\frac{\imath ky}{2}}\\\ \end{pmatrix},$ (4.4) $\displaystyle
P^{-}(x;k)=\lim_{y\to-\infty}P(x,y;k)\begin{pmatrix}e^{-\frac{\imath
ky}{2}}&\\\ &e^{\frac{\imath ky}{2}}\\\ \end{pmatrix}.$ (4.5)

Using the property in Eq. (4.2) one obtains:

$\displaystyle P^{+}(x;k)=P^{-}(x;k)T(k).$ (4.6)

It is then possible to take this equation and rearrange the elements[18] such
as to write the following:

$\displaystyle G_{-}=G_{+}G,$ (4.7)

where the different objects are defined as follows:

$\displaystyle G=\begin{pmatrix}1&\frac{\tilde{b}}{a_{-}}\\\
\frac{b}{a_{+}}&\frac{1}{a_{+}a_{-}}\\\ \end{pmatrix},\quad
G_{+}=\begin{pmatrix}\frac{P^{-}_{11}}{a_{+}}&P^{+}_{12}\\\
\frac{P^{-}_{21}}{a_{+}}&P^{+}_{22}\\\ \end{pmatrix},\quad
G_{-}=\begin{pmatrix}P^{+}_{11}&\frac{P^{-}_{12}}{a_{-}}\\\
P^{+}_{21}&\frac{P^{-}_{22}}{a_{-}}\\\ \end{pmatrix},$ (4.8)

and we have dropped denoting the dependence on $k$ of $a_{\pm}$ and
$b,\tilde{b}$ for brevity and it is implicitly assumed that these objects are
to be evaluated at $t=0$. The matrices $G_{+}$ and $G_{-}$ can then be shown
to be analytic in the upper half and lower half planes, respectively. Such
that we obtain the following Riemann-Hilbert problem:

  * •

$G_{\pm}$ are analytic in the upper and lower half planes, respectively.

  * •

On the real axis $G_{-}=G_{+}G$.

  * •

$G_{\pm}(z)\to\mathds{1}+O(1/z)$ as $z\to\infty$

Given $G$, if one is able to solve the Riemann-Hilbert problem, then one
obtains immediately the fields $p$ and $q.$ Indeed, $G_{\pm}$ satisfy:

$\displaystyle\partial_{x}G_{\pm}G_{\pm}^{-1}=\begin{pmatrix}-\frac{\imath
k}{2}&-p\\\ \frac{q}{8}&\frac{\imath k}{2}\\\ \end{pmatrix},$ (4.9)

such that $p$ and $q$ are easily extracted from this equation. However, the
Riemann-Hilbert problem is often intractable. Nevertheless, it has asymptotes
of the solution may be found by using the method of steepest descent presented
in the following.

###  4.1 The Riemann-Hilbert Problem in Macroscopic Fluctuation Theory

Let us first study some particular properties of the Riemann-Hilbert problem
in the case of macroscopic fluctuation theory. In this case $p$ is
proportional to a delta function, so we write $p(x,1)=-8\Lambda\delta(x).$

$\displaystyle\partial_{x}P=\begin{pmatrix}-\frac{\imath
k}{2}&\Lambda\delta(x)\\\ q&\frac{\imath k}{2}\\\ \end{pmatrix}P.$ (4.10)

This is easily solved, for example, by:

$\displaystyle P^{-}=\begin{cases}\begin{pmatrix}e^{-\imath\frac{kx}{2}}&0\\\
e^{\imath\frac{kx}{2}}I(-\infty,x)&e^{\imath\frac{kx}{2}}\\\
\end{pmatrix}&x<0\\\ \begin{pmatrix}e^{-\imath\frac{kx}{2}}\left(1+\Lambda
I(-\infty,0))\right)&\Lambda e^{-\imath\frac{kx}{2}}\\\
e^{\imath\frac{kx}{2}}\left(I(-\infty,x)+\Lambda
I(-\infty,0)I(0,x)\right)&e^{\imath\frac{kx}{2}}\left(1+\Lambda
I(0,x))\right)\\\ \end{pmatrix}&x>0\\\ \end{cases}$ (4.11)

where $I(a,b)=\int_{a}^{b}e^{-\imath kx^{\prime}}q(x^{\prime})dx^{\prime}$. So
that we get:

$\displaystyle T(k,1)=\begin{pmatrix}a_{+}(k)&\tilde{b}(k,1)\\\
b(k,1)&a_{-}(k)\\\ \end{pmatrix}=\begin{pmatrix}1+\Lambda
I(-\infty,0)&\Lambda\\\ I(-\infty,\infty)+\Lambda
I(-\infty,0)I(0,\infty)&1+\Lambda I(0,\infty)\\\ \end{pmatrix}$ (4.12)

where we have thus obtained:

$\displaystyle a_{+}(k,t)=1+\Lambda I(-\infty,0),\quad a_{-}(k,t)=1+\Lambda
I(0,\infty)\quad\tilde{b}(k,1)=\Lambda$ (4.13)

Thus given the time dependence of the elements of $T$ discussed above, we
have:

$\displaystyle T(k,t)=\begin{pmatrix}a_{+}(k)&\Lambda e^{-\imath
kx+k^{2}(t-1)}\\\ \frac{e^{\imath
kx-(t-1)k^{2}}(a_{+}(k)a_{-}(k)-1)}{\Lambda}&a_{-}(k)\\\ \end{pmatrix}$ (4.14)

We find $G,$ the jump matrix associated with the Riemann-Hilbert problem,
according to Eq. (4.8)

$\displaystyle G=\begin{pmatrix}1&\Lambda e^{-\imath
kx+(t-1)k^{2}-\log(a_{-})}\\\ -\frac{e^{\imath
kx-(t-1)k^{2}+\log(a_{-})}}{\Lambda}\left(1-\frac{1}{a_{+}a_{-}}\right)&\frac{1}{a_{+}a_{-}}\\\
\end{pmatrix}$ (4.15)

At $t=1$ one can find $G_{\pm}$ explicitly through the Fourier transform of
$q$. This is done by solving Eq. (4.10) and inserting that solution into Eqs.
(4.4,4.5,4.8). This gives the following expressions:

$\displaystyle G_{+}=\begin{cases}\begin{pmatrix}1&0\\\ \frac{e^{\imath
kx}}{\Lambda}\left(1+\Lambda I(0,x)-a^{-1}_{+})\right)&1\\\
\end{pmatrix}&x>0\\\ \begin{pmatrix}\frac{1}{a_{+}}&-\Lambda e^{-\imath kx}\\\
\frac{e^{\imath kx}I(-\infty,x)}{a_{+}}&1-\Lambda I(0,x)\\\
\end{pmatrix}&x<0\end{cases}$ (4.16) $\displaystyle
G_{-}=\begin{cases}\begin{pmatrix}1&\frac{\Lambda e^{-\imath kx}}{a_{-}}\\\
-e^{\imath kx}I(x,\infty)&\frac{1+\Lambda I(0,x)}{a_{-}}\\\
\end{pmatrix}&x>0\\\ \begin{pmatrix}a_{-}&0\\\ e^{\imath kx}\left(\Lambda
I(0,\infty)I(0,x)-I(x,\infty)\right)&\frac{1}{a_{-}}\\\
\end{pmatrix}&x<0\end{cases}$ (4.17)

It is then easy to ascertain that the matrices $G_{\pm}$ are indeed analytic
in the upper and lower half planes respectively and that $G_{-}=G_{+}G$.

##  5 The Steepest Descent Method

We now want to derive the strongly non-linear limit of the inverse scattering
problem, this limit is in fact the one considered in Ref. [13] where the
steepest descent method to the Riemann-Hilbert problem may be used, but before
going on to describe the steepest descent method for the Riemann-Hilbert
problem relevant for all time $t$, we concentrate on its form in the case
$t=1$. At this time-point, the Riemann-Hilbert problem is solved making use of
Fourier transforms of $q(x)$ and its restriction to subintervals of
$\mathds{R},$ see Eqs. (4.16, 4.17). Such that the steepest descent method
here is really nothing but the usual steepest descent method for the Fourier
integral, which is intimately related to the Legendre transform.

A key to finding the solution in the inverse scattering method is to find
$b(k,t)$ from which all other elements of the scattering matrix $T$ can be
found. Indeed, since we know $\tilde{b}(k,t)=\Lambda e^{k^{2}(t-1)}$ and due
to the unimodularity of the scattering matrix, which results in the equation
$a_{+}a_{-}-b\tilde{b}=1$, one may easily find $a_{+}a_{-}$ given $b(k,t)$.
Then using the fact that $a_{\pm}$ are analytic in the upper and lower half
$k-$planes respectively, one can write:

$\displaystyle a_{\pm}(k)=\mp\int\frac{\log\left(1+\Lambda
b(\mu,1)\right)}{k\pm\imath 0^{+}-\mu}\frac{d\mu}{2\pi\imath}$ (5.1)

Now we assume

$\displaystyle\log(\Lambda b(k,1))=m(k)$ (5.2)

and that $|m(k)|$ is large. Let $S=\\{k\in\mathds{R}|{\rm Re}(m(k))>0\\}$ . We
assume that $S$ is a union of a finite number of intervals. We also have the
condition $m(\mu)=m^{*}(-\mu)$, which is required to obtain a real solution.

Then we may write approximately

$\displaystyle\log a_{\pm}=\mp\int_{S}\frac{m(\mu)}{k\pm\imath
0^{+}-\mu}\frac{d\mu}{2\pi\imath}.$ (5.3)

At $t=1,$ the solution of the Riemann-Hilbert problem is solved by the Fourier
transform of $q(x),$ which is denoted by $I(-\infty,\infty)$. And the steepest
descent method yields nothing but the saddle point equations for this
integral. Let us assume $k$ is in the lower half plane, since $a_{-}$ tends to
one in this plane then $a_{+}$ must be large we may conclude:

$\displaystyle a_{+}(k)=1+\Lambda I(-\infty,0)\sim\Lambda
I(-\infty,0)\sim\Lambda I(-\infty,\infty).$ (5.4)

So that $q(x)$ is the inverse Fourier transform of $a_{+}$ and we find the
saddle point equations:

$\displaystyle
x=-\imath\log^{\prime}a_{\pm}(k^{*}(x))=\pm\int_{S}\frac{m^{\prime}(\mu)}{k^{*}\pm\imath
0^{+}-\mu}\frac{d\mu}{2\pi}.$ (5.5)

Solving this equation for $k$ one may then find $q(x)$ by substituting

$\displaystyle
q=\exp{\imath\left[k^{*}(x)x-\imath\log(a_{\pm}(k^{*}(x))\right]}.$ (5.6)

The term in the square brackets is of course the Legendre transform of the
$\imath\log a_{\pm}$, which we may denote by $\Phi(x)$. Of course
$\Phi^{\prime}(x)=k^{*}(x)$ so that we can also write
$q=\exp\imath\int^{x}k^{*}(x)$. Making the identification
$k^{*}(x)=\imath\frac{v(x)}{2}$ one obtains one half of the Cole-Hopf
transformation, Eq. (2.1), $q(x)=e^{-\frac{1}{2}\int^{x}v}.$ Since $\rho=0$ at
time $t=1$ for all $x\neq 0$ then $p(x)=0$ and one cannot identify the other
half of the Cole-Hopf transformation. At this point the identification
$k^{*}=\imath\frac{v}{2}$ is merely suggestive, but later we shall be able to
make this identification on more general grounds.

This rather trivial steepest descent approach for $t=1$ does not rely on the
Riemann-Hilbert problem at all, since for this specific time
$\tilde{b}(k,1)=\Lambda$ is a constant and the problem becomes essentially
linear. Nevertheless, during the full time evolution The Riemann-Hilbert
problem becomes more complex and requires a more involved solution which we
discuss in the next sub-section.

###  5.1 Reduction of the Problem

The steepest descent method of the Riemann-Hilbert problem includes first
deforming the contour where the matrix $G_{\pm}$ have a jump discontinuity
(the ”jump contour”, or the ”Riemann-Hilbert contour”), such as to simplify
the Riemann-Hilbert problem greatly. This in analogy to the steepest descent
method for integrals, but here this step includes some matrix manipulation and
a doubling or tripling of the jump contour, unlike the simpler steepest
descent methods for integrals. After this is achieved the simpler Riemann-
Hilbert problem is solved. This sub-section concerns itself with the first
step, namely the reduction of the Riemann-Hilbert problem to a simpler one. It
is already in this step that we will be able to identify the Riemann
invariants $\lambda_{1},$ $\lambda_{2},$ or equivalently, the fields $\rho$
and $v$, but only in the inviscid case. In the more general case, the
identification of the fields is more complicated and may only be deduced after
performing the second step of actually solving the reduced Riemann-Hilbert
problem.

The steepest descent method can be applied when have $\log(a_{+}a_{-})$ is
large. In this case the matrix $G$ takes the approximate form:

$\displaystyle G\simeq\begin{pmatrix}1&\Lambda e^{-\imath
kx+k^{2}(t-1)-\log(a_{-})}\\\ -\frac{e^{\imath
kx-k^{2}(t-1)+\log(a_{-})}}{\Lambda}&\frac{1}{a_{+}a_{-}}\\\ \end{pmatrix}$
(5.7)

Now we follow the classic method of Ref. [13] to solve this problem
approximately. The first step is to note that if we find two functions,
$g_{\pm}$ analytic in the upper and lower half planes, respectively, and
tending to $1$ at infinity at both half-planes, then the following
transformation of $G_{\pm}$ and $G$ leads to a new Riemann-Hilbert problem
which has the same formulation as the original one. The transformation reads:

$\displaystyle\tilde{G}=\begin{pmatrix}e^{g_{+}}&\\\ &e^{-g_{+}}\\\
\end{pmatrix}G\begin{pmatrix}e^{-g_{-}}&\\\ &e^{g_{-}}\\\ \end{pmatrix}=$
(5.8) $\displaystyle=\begin{pmatrix}e^{g_{+}-g_{-}}&\Lambda e^{-\imath
kx-\log(a_{-})+g_{+}+g_{-}}\\\ -\frac{1}{\Lambda}e^{\imath
kx+\log(a_{-})-g_{+}-g_{-}}&e^{g_{-}-g_{+}-2\log\Lambda+k^{2}}\\\
\end{pmatrix}$ (5.9)
$\displaystyle\tilde{G}_{+}=G_{+}\begin{pmatrix}e^{-g_{+}}&\\\ &e^{g_{+}}\\\
\end{pmatrix},\quad\tilde{G}_{-}=G_{-}\begin{pmatrix}e^{-g_{-}}&\\\
&e^{g_{-}}\\\ \end{pmatrix}.$ (5.10)

Let us denote:

$\displaystyle\imath h(k)=\imath
kx-\log(a_{-}(k))-g_{+}(k)-g_{-}(k)-\log(\Lambda).$ (5.11)

One can search for such functions $g_{\pm}$ such that on some part of the jump
contour, $g_{+}-g_{-}=0$ and that there $h^{\prime}(k)>0$ on the contour where
the jump occurs. In this case, one may make use of the following decomposition
of $\tilde{G}$:

$\displaystyle\tilde{G}=\begin{pmatrix}1&e^{-\imath h(k)}\\\ -e^{\imath
h(k)}&\frac{1}{a_{+}a_{-}}\\\ \end{pmatrix}\simeq G_{1}G_{2}$ (5.12)
$\displaystyle\tilde{G}_{1}=\begin{pmatrix}1&0\\\ -e^{\imath h(k)}&1\\\
\end{pmatrix},\quad\tilde{G}_{2}=\begin{pmatrix}1&e^{-\imath h(k)}\\\ 0&1\\\
\end{pmatrix},$ (5.13)

where one have used that $a_{+}a_{-}$ is large. One may then separate the jump
contour into two contour, one in which the matrix $\tilde{G}_{+}$ jumps by
$\tilde{G}_{1}$ and the other where the matrix $\tilde{G}_{+}$ jumps by
$\tilde{G}_{2}$ . As a second step we deform each of the contours, that of
$\tilde{G}_{1}$ towards the upper-half plane and that of $\tilde{G}_{2}$
towards the lower half plane. The condition $h^{\prime}(k)>0$ then ensures
that the off-diagonal terms in both $\tilde{G}_{1}$ and $\tilde{G}_{2}$ tends
quickly to zero, and as such the jump matrix becomes the identity matrix,
namely there is no jump. The procedure is illustrated in Fig.1. Thus the part
of the jump contour in which we were successful to find $g_{+}-g_{-}=0$ and
$h^{\prime}(k)>0$ simply disappears.

Figure 1: We have dropped the tildes in this figure. On the left the real axis
is denoted by a heavy line and the matrices $G_{+}$ and $G_{-}$ are shown
above and below it respectively. The jump matrix is $G$ and it can be
decomposed into $G_{1}G_{2}.$ The jump can then be separated into contours
each one with its own jump matrix, $G_{1}$ and $G_{2}$. The jump then proceeds
in two steps $G_{+}$ jump over to $G_{m}$ on the upper contour and $G_{m}$
jumps over to $G_{-}$ on the lower one. However, since the matrices $G_{1}$
and $G_{2}$ tend to the identity, the jump in fact disappears and $G_{+}$
becomes smoothly $G_{-}$ as the real axis is crossed .

Secondly, with the same $g_{\pm}$ there may be a region where
$g_{-}-g_{+}=\log(a_{+}a_{-})$ and $h^{\prime}(k)<0,$ allowing us to write
$\tilde{G}$:

$\displaystyle\tilde{G}\simeq\begin{pmatrix}0&e^{-\imath h(k)}\\\ -e^{\imath
h(k)}&1\\\ \end{pmatrix}=\begin{pmatrix}1&e^{-\imath h(k)}\\\ 0&1\\\
\end{pmatrix}\begin{pmatrix}1&0\\\ -e^{\imath h(k)}&1\\\ \end{pmatrix}.$
(5.14)

and by the same manipulation of the jump contour as before, simply remove that
region where the conditions above are met.

Finally, there may be a region where neither of the conditions above can be
made to be met. In this case one may apply the following condition
$h^{\prime}(k)=0$ and $-\log(a_{+}a_{-})<{\rm Re}(g_{+}-g_{-})<0.$ In this
region one has

$\displaystyle\tilde{G}\simeq\begin{pmatrix}0&e^{-\imath h_{0}}\\\ -e^{\imath
h_{0}}&0\\\ \end{pmatrix}.$ (5.15)

The simplification afforded by this Riemann-Hilbert problem of this form is
that we have segments, each of which the jump matrix is constant and of a
specific off-diagonal form, while on other segments there is no jump. Such a
problem can be solved by making use of Riemann theta functions, as shall be
made explicit below.

Another point to be made is that for $x>0$ and large $k$ the conditions
$h^{\prime}>0,$ $g_{+}-g_{-}=0$ can be met trivially since $\log a_{-}\to 0$.
For $x<0$ it is much more convenient to apply the transformation $G\to
e^{-\sigma_{z}\log a_{+}}Ge^{-\sigma_{z}\log a_{-}},$ $G_{+}\to
G_{+}e^{\sigma_{z}\log a_{+}}$ $G_{-}\to G_{-}e^{-\sigma_{z}\log a_{-}}$ to
first bring $G$ to the form:

$\displaystyle G=\begin{pmatrix}\frac{1}{a_{+}a_{-}}&\Lambda e^{-\imath
kx-\log(a_{+})}\\\ -\frac{1}{\Lambda}e^{\imath
kx+\log(a_{+})}&1\end{pmatrix},$ (5.16)

whereupon $h^{\prime}<0,$ $g_{+}-g_{-}=0$ can be applied rather than letting
$g_{-}-g_{+}=\log(a_{+}a_{-}).$ Of course both points of view are equivalent,
but the latter one is more symmetric with respect to $x\leftrightarrow-x$.

The fact that $g_{\pm}$ can be found is related to the fact that it satisfies
a certain scalar Riemann-Hilbert problem. We shall not go over this here, as
the method is described in Ref. [13] in full. We do mention here, that
practically $g_{\pm}$ can constructed in a self consistent manner which will
be described presently. The method thus described then actually provides the
solution to the inviscid equations, Eqs. (2.4, 2.5) in the case where the
final Riemann-Hilbert problem contains just one non-trivial segment with jump
matrix of the form of Eq. (5.15). Then the endpoints of the segments are
$\lambda_{1},$ $\lambda_{2}$ that obey Eqs. (2.7, 2.8), which are equivalent
to Eqs. (2.4, 2.5).

Since the case where only one segment is already very instructive, and since
additional segments only encumber the notation, we concentrate on this case.
However, it is quite easy, once the case of one segment is understood to
generalize the procedure to several segments. We thus use notations in this
sub-section which suggest a single segment, with the understanding that the
generalization is straightforward.

It further turns out that, in order to achieve all the above conditions, the
segment $[\lambda_{1},\lambda_{2}]$ does not lie on the real axis. This is
already suggested in the transformation to $\rho$ and $v$ in Eq. (2.6), where
$\rho$ and $v$ are real but then $\lambda_{i}$ is not. In fact the segment is
of the form $[\frac{\imath v}{2}-\sqrt{\rho},\frac{\imath v}{2}+\sqrt{\rho}],$
where $v$ and $\rho$ are real, and $\rho\geq 0$. This means that the jump
contour is also deformed to the complex plane. We give a description of the
contour in Fig. 2.

Figure 2: The jump contour after deformation. The actual jump is only on the
segment $[\lambda_{1},\lambda_{2}]$, the rest of the contour, which is denoted
in a dashed line has no jump on it, due to the procedure which removes it by
decomposing $G$ into $G_{1}G_{2}$.

Let us denote the endpoints of the segment by $\lambda_{1}$ and $\lambda_{2}$.
We definte:

$\displaystyle R_{2}(k)=\sqrt{k-\lambda_{1}}\sqrt{k-\lambda_{2}}.$ (5.17)

In fact $g_{\pm}^{\prime}$ must obeys:

$\displaystyle g_{+}+g_{-}=\imath kx-\log(a_{-}(k))-\log(\Lambda)$
$\displaystyle k\in[\lambda_{1},\lambda_{2}],$ (5.18) $\displaystyle
g_{+}-g_{-}=0$ $\displaystyle\mbox{otherwise}.$ (5.19)

One may solve these conditions by taking $g_{\pm}$ to be given by a single
function $g$ defined on the complex with a cut on the segment
$[\lambda_{1},\lambda_{2}],$ while $g_{\pm}$ are just the values of this
single function above and below the cut. Such a function may be given by
naturally associating with it a differential and writing

$\displaystyle g^{\prime}(k)dk=\frac{dw(k)}{R_{2}(k)}$ (5.20)

where $dw(k)$ itself has a jump discontinuity of value $R_{2}(\mu)\left(\imath
x-(t-1)2\mu+\frac{a_{\pm}^{\prime}(\mu)}{a_{\pm}(\mu)}\right)$ over the branch
cut at $[\lambda_{1},\lambda_{2}],$ and is smooth everywhere else. Such a
differential can be written as:

$\displaystyle
dw(k)=dk\int_{\lambda_{1}}^{\lambda_{2}}\frac{\sqrt{(\mu-\lambda_{1})(\mu-\lambda_{2})}}{\mu\mp\imath
0^{+}-k}\left(\imath
x-(t-1)2\mu+\frac{a_{\sigma}^{\prime}(\mu)}{a_{\sigma}(\mu)}\right)\frac{d\mu}{2\pi\imath},$
(5.21)

where we have allowed for using either $a_{+}$ or $a_{-}$ by making use of
$a_{\sigma}$ where $\sigma=\pm$, respectively. or alternatively as:

$\displaystyle dw(k)=R_{2}(k)\left(\imath
x-(t-1)2k+\frac{a_{\sigma}^{\prime}(k)}{a_{\sigma}(k)}\right)dk-dw_{\rm
sing}(k)$ (5.22)

where $dw_{\rm sing}(k)$, which is adorned with the subscript $\rm sing$,
denotes the singular part of terms which precede it, namely all the
singularities of those terms away from the branch cut when this term is
treated as a differential . These singularities are the poles of the
expression at infinity and the branch cut on $S$. Explicitly we have:

$\displaystyle\frac{dw_{\rm sing}(k)}{dk}=x(k-\imath
v/2)+2\imath(t-1)\left(k(k-\imath
v/2)-\frac{\rho}{2}\right)+\sigma\int_{S}\frac{m^{\prime}(\mu)R_{2}(\mu)}{k-\mu}\frac{d\mu}{2\pi\imath}$
(5.23)

In order for $g^{\prime}$ defined in Eq. (5.20) not to diverge at the branch
points $\lambda_{i}$, we must demand:

$\displaystyle dw_{\rm sing}\left(\lambda_{i}\right)=0.$ (5.24)

Setting $\lambda_{i}=\frac{\imath v}{2}\pm\sqrt{\rho}$ in this equation, one
gets more explicitly:

$\displaystyle\pm\sqrt{\rho}\left(x-v(t-1)\right)+\imath\rho(t-1)+\sigma\int_{S}\frac{m^{\prime}(\mu)R_{2}(\mu)}{\frac{\imath
v}{2}\pm\sqrt{\rho}-\mu}\frac{d\mu}{2\pi\imath}=0.$ (5.25)

Summing the two equations and subtracting them (followed by a division by
$\sqrt{\rho}$) leads to the the following:

$\displaystyle
x-v(t-1)=\sigma\int_{S}\frac{m^{\prime}(\mu)R_{2}(\mu)}{(\mu-\frac{\imath
v}{2})^{2}-\rho{}}\frac{d\mu}{2\pi\imath},$ (5.26)
$\displaystyle\rho(t-1)=\imath\sigma\int_{S}\frac{(\mu-\frac{\imath
v}{2})m^{\prime}(\mu)R_{2}(\mu)}{(\mu-\frac{\imath
v}{2})^{2}-\rho}\frac{d\mu}{2\pi\imath}.$ (5.27)

Note that the second equation always has the solution $\rho=0$, as the
integral on the right hand side vanishes at $\rho=0$ due the symmetry
$\mu\to-\mu^{*}$, which $m(\mu)$ and the set $S$ respect, but with respect to
which the integral is antisymmetric. The solution of these equations are
solutions the inviscid equations, Eqs. (2.4, 2.5), above as shall be shown
below.

If set $t=1$ and assume $\rho$ to be small, we obtain from these equations the
two conditions:

$\displaystyle x=\sigma\int_{S}\frac{m(\mu)}{(\mu-\imath
v/2)^{2}}\frac{d\mu}{2\pi\imath},$ (5.28) $\displaystyle
0=\int_{S}\frac{m(\mu)}{(\mu-\imath v/2)^{3}}\frac{d\mu}{2\pi\imath}\mbox{ or
}\rho=0.$ (5.29)

In fact, we know that at $t=1$ we have $\rho=0,$ for any $x\neq 0$, Namely,
the second condition, given in Eq. (5.29), may be only solved by $\rho=0$ for
$x\neq 0.$ To show that this indeed the case, we take a derivative with
respect to $x$ of the first condition, and obtain:

$\displaystyle 1=\imath\partial_{x}v\int_{S}\frac{m(\mu)}{(\mu-\imath
v/2)^{3}}\frac{d\mu}{2\pi\imath}$ (5.30)

Namely, if the integral in the second condition, Eq. (5.29), is to vanish then
then $\partial_{x}v$ must diverge. Since this is not the case for any finite
$x$ by assumption, we must conclude that $\rho(x,1)=0$ for all $x\neq 0$.

More generally and for any $t$ there exists a region in which $\rho=0,$ for
this region the Hodograph equations (Eqs.(5.26, 5.27) above) reduce to

$\displaystyle x+\frac{\imath
v}{2}(t-1)=\sigma\int_{S}\frac{m(\mu)}{(\mu-\imath
v/2)^{2}}\frac{d\mu}{2\pi\imath}.$ (5.31)

It is often the case that one may solve the large $\kappa$ (or, equivalently,
large $\Lambda$) limit at $t=1$, where $\rho=0$ for $x\neq 0$ and obtain that
the left hand side of Eq. (5.31) is equal to a given function of $v$. Then it
is a matter of solving a singular integral equation in order to find $m(\mu)$.
This $m(\mu)$ plays also a role in an exact solution of the problem, where it
is defined by Eq. (5.2). Of course the exact $m(\mu)$ may have small
corrections which are more difficult to obtain by examining the approximate
solution (although it is in principle possible to develop a systematic
expansion), nonetheless, the approximate $m(\mu)$ may either prove exact, or
may supply a valuable first guess to find an exact solution. We demonstrate
this in section 6 below.

Eqs. (5.26,5.27) are general equations for $v$ and $\rho$, or equivalently,
for $\lambda_{1}$ and $\lambda_{2}$. We wish to show now that these equations
are solutions to Eqs. (2.4, 2.5) or, equivalently, to Eqs. (2.7, 2.8).

The requirement that $g^{\prime}$ does not diverge suggests that around
$\lambda_{i}$ it has the form $g^{\prime}(k)\sim\sqrt{k-\lambda_{i}}.$ Let us
define $d\Omega_{x}$ and $d\Omega_{t}$ as follows:

$\displaystyle\frac{\partial g^{\prime}}{\partial x}dk=\imath
d\Omega_{x},\quad\frac{\partial g^{\prime}}{\partial t}dk=-2d\Omega_{t}.$
(5.32)

The compatibility condition for these two equations read:

$\displaystyle\imath\partial_{t}d\Omega_{x}=2\imath\partial_{x}d\Omega_{t}.$
(5.33)

Now $d\Omega_{x}$ and $d\Omega_{t}$ can be easily seen to be meromorphic
differential on the genus-0 Riemann surface associated with $R_{2}(k)$ with
pole of order $2$ and $3$ respectively, and residue $\pm 1$ at infinity at the
upper and lower sheets of the Riemann surface respectively, which, by
uniqueness of such differentials, means that we may immediately write:

$\displaystyle
d\Omega_{x}=\frac{k-\frac{\lambda_{1}+\lambda_{2}}{2}}{\sqrt{k-\lambda_{1}}\sqrt{k-\lambda_{2}}}dk,\quad
d\Omega_{t}=\frac{k\left(k-\frac{\lambda_{1}+\lambda_{2}}{2}\right)-\frac{(\lambda_{2}-\lambda_{1})^{2}}{2}}{\sqrt{k-\lambda_{1}}\sqrt{k-\lambda_{2}}}dk.$
(5.34)

Indeed expanding at infinity we get $d\Omega_{x}\sim\left(\pm
1+O(1/k^{2})\right)dk$ and $d\Omega_{t}\sim\left(\pm k+O(1/k^{2})\right)dk$ on
the upper and lower sheet, respectively.

If the explicit expressions for $d\Omega_{x}$ and $d\Omega_{t}$ are
substituted into the compatibility equation, Eq. (5.34), one obtains equations
for $\lambda_{i}$ by examining the behavior around $\lambda_{i}$ of both sides
of the compatibility condition, Eq. (5.33). Indeed, one obtains a term that
diverges as $(k-\lambda_{i})^{-3/2}$ around $\lambda_{i}$ and the residue of
that divergence must coincide on both sides of the equation. This condition
then reads:

$\displaystyle\left(\lambda_{i}-\frac{\lambda_{1}+\lambda_{2}}{2}\right)\partial_{t}\lambda_{i}=2\imath\left(\lambda_{i}\left(\lambda_{i}-\frac{\lambda_{1}+\lambda_{2}}{2}\right)-\frac{(\lambda_{2}-\lambda_{1})^{2}}{8}\right)\partial_{x}\lambda_{i}$
(5.35)

Dividing both sides by $\lambda_{i}-\frac{\lambda_{1}+\lambda_{2}}{2}$ yields
immediately the inviscid equation encountered above for the Riemann invariants
$\lambda_{1},$ $\lambda_{2}$ of Eqs. (2.4, 2.5), namely equations (2.7, 2.8).
These equations read as follows:

$\displaystyle\partial_{t}\lambda_{i}=\left(\frac{3}{2}\lambda_{i}+\frac{1}{2}\lambda_{3-i}\right)\imath\partial_{x}\lambda_{i}.$
(5.36)

Note that, although we have concentrated on the case where only two Riemann
invariants are present, the same procedure will yield equations with any
(even) number of Riemann invariants, which generalize equations (2.7, 2.8), to
the case where the reduced Riemann-Hilbert contour is the union
$\cup_{j=1}^{g+1}[\lambda_{2j-1},\lambda_{2j}]$, where $g$ is the genus of the
Riemann surface associated with
$R_{2g+2}(k)\equiv\prod_{j=1}^{g+1}\sqrt{(k-\lambda_{2j-1})(k-\lambda_{2j})}$.
The $\lambda_{i}$’s thus obtained are then moduli of oscillatory solutions of
the original nonlinear Schrödinger equation, Eqs. (1.1, 1.2). These
oscillatory solutions are given in Eq. (5.50,5.51) below. When two of the
$\lambda_{i}$’s coincide, namely when $\lambda_{2j-1}\to\lambda_{2j},$such an
oscillatory solution takes on a solitonic nature. The only added component
here is that the differentials $d\Omega_{x}$ and $d\Omega_{t}$ may be shown to
be normalized as to have null $a-$cycles.

###  5.2 Solution of the Reduced Problem

The solution of the reduced Riemann-Hilbert problem is achieved by making use
of the Riemann theta functions. One can find more details about this solution
in Refs [13, 19], while here we give merely a very rapid exposition. It should
be noted that this section is given here only for completeness, since if one
is only interested in the case of two $\lambda_{i}$’s, then the identification
of the Riemann invariants as the end-points of the reduced Riemann-Hilbert
contour has already been made, albeit without justification (it was merely
suggestive that the two objects, the Riemann invariants of Eqs.(2.4, 2.5) and
the endpoints of the reduced Riemann-Hilbert contour, obey the same
differential equations, Eqs. (2.7, 2.8)), and, furthermore, the solution of
the Riemann-Hilbert problem is from this point on standard[13, 19], namely,
there are no special features associated with the peculiar formulation of the
large deviation problem, except the fact that the Riemann-Hilbert contour lies
rather unconventionally away from the real axis, the reflection across the
imaginary axis being the symmetry that is obeyed by the contour in this case.
This situation due to the non-conventional real section of the non-linear
Schrödinger equation afforder by two real fields, $p$ and $q,$ rather than by
two fields which are complex conjugates of each other.

In order to introduce the solution assume that the Riemann surface at hand is
given by

$\displaystyle R_{2g+2}(k)=\prod_{i=1}^{2(g+1)}\sqrt{k-\lambda_{i}}.$ (5.37)

In this case we have $g$ holomorphic differentials $\omega_{i}$ which may be
normalized as follows:

$\displaystyle\oint_{a_{j}}\omega_{i}=2\pi\imath\delta_{ij},$ (5.38)

Figure 3: The cycles over the Riemann surface associated with the function
$R_{2g+2}(k)$. Dashed lines describe the part of the cycle that lies on the
lower sheet.

where $a_{i}$ denotes an $a-$ cycle conventionally defined as shown in Fig. 3.

Then one defines the Abel map:

$\displaystyle A_{i}(P)=\int_{\infty_{+}}^{P}\omega_{i}.$ (5.39)

Namely the integral is taken from $\infty$ on the upper sheet to the point
$k$. Thus $\bm{A}(k^{\pm})$ denotes that the point $P$ is to be taken as the
point $k$ on the upper or lower sheets respectively. One further defines the
Riemann matrix

$\displaystyle B_{ij}=\oint_{b_{j}}\omega_{i}$ (5.40)

The Riemann theta function is defined as[19]:

$\displaystyle\theta(\bm{A})=\sum_{\bm{m}\in\mathds{Z}^{g}}\exp\left[\frac{1}{2}\bm{m}\bm{B}\bm{m}+\bm{m}\bm{A}\right].$
(5.41)

We also define $\bm{A}_{g}$ to be a vector whose $i-$th element is given by
$\oint_{b_{i}}g^{\prime}dk.$ Lastly, we define a differential $d\Omega_{N}$ to
be a meromorphic differential with pole at infinity on the upper and lower
sheets and of residue $\pm 1,$ respectively. Associated with this definition
is the vector $\bm{V}$ a vector whose $i-$th element is given by
$\oint_{b_{i}}d\Omega_{N}.$ In the lore of algebraic Riemann surfaces, the
following is a standard identity:

$\displaystyle\bm{V}=\oint_{\bm{b}}d\Omega_{N}=\int_{\infty_{+}}^{\infty_{-}}d\bm{\omega}=\bm{A}(\infty_{-}).$
(5.42)

Is convenient to find a solution to the matrix $M_{\pm}$ instead of $G_{\pm}$
, where $M_{\pm}$ is defined below:

$\displaystyle M_{\pm}=G_{\pm}\begin{pmatrix}k&0\\\ 0&1\\\ \end{pmatrix},$
(5.43)

where this change of variable results in the following asymptotics as
$k\to\infty$ : $M_{\pm}\to\begin{pmatrix}k&k^{-1}\\\ 1&1\\\ \end{pmatrix}.$
But otherwise the matrix $M_{\pm}$ solves the same Riemann-Hilbert problem as
$G_{\pm}$, which becomes the reduced Riemann-Hilbert problem in the limit of
large $\Lambda$.

The solution reads:

$\displaystyle
M_{\pm}=\begin{pmatrix}\frac{\theta(\bm{D}-\bm{V})\theta(\bm{A}_{g}+\bm{A}(k^{+})-\bm{V}+\bm{D})}{\theta(\bm{A}(k^{+})-\bm{V}+\bm{D})\theta(\bm{A}_{g}-\bm{V}+\bm{D})}e^{g(k)-\int^{k}\Omega_{N}}&\frac{\theta(\bm{D}-\bm{V})\theta(\bm{A}_{g}+\bm{A}(k^{-})-\bm{V}+\bm{D})}{\theta(\bm{A}(k^{-})-\bm{V}+\bm{D})\theta(\bm{A}_{g}-\bm{V}+\bm{D})}e^{-g(k)+\int^{k}\Omega_{N}}\\\
\frac{\theta(\bm{D})\theta(\bm{A}_{g}+\bm{A}(k^{+})+\bm{D})}{\theta(\bm{A}(k^{+})+\bm{D})\theta(\bm{A}_{g}+\bm{D})}e^{g(k)}&\frac{\theta(\bm{D})\theta(\bm{A}_{g}+\bm{A}(k^{-})+\bm{D})}{\theta(\bm{A}(k^{-})+\bm{D})\theta(\bm{A}_{g}+\bm{D})}e^{-g(k)}\\\
\end{pmatrix},$ (5.44)

here $\bm{D}$ is a constant vector depending neither on time, space or
auxiliary spectral parameter, $k$.

The asymptote is therefore:

$\displaystyle M_{\pm}=$
$\displaystyle\left(\mathds{1}+k^{-1}\Psi_{1}+\dots\right)\begin{pmatrix}0&e^{\imath
xk-tk^{2}}\\\ ke^{\imath xk-tk^{2}}&0\\\ \end{pmatrix}$ (5.45)

where

$\displaystyle\Psi_{1}=\begin{pmatrix}\dots&\frac{\rho_{0}\theta(\bm{A}_{g}-\bm{V}+\bm{D})\theta(\bm{D})e^{-\int\imath
k_{0}dx-\omega_{0}dt}}{\theta(\bm{D}-\bm{V})\theta(\bm{A}_{g}+\bm{D})}\\\
\frac{\theta(\bm{D})\theta(\bm{A}_{g}+\bm{V}+\bm{D})e^{\int\imath
k_{0}dx-\omega_{0}dt}}{\theta(\bm{V}+\bm{D})\theta(\bm{A}_{g}+\bm{D})}&\dots\\\
\end{pmatrix},$ (5.46)

and the ellipsis denotes terms unimportant for the sequel. This expression
makes use of the following definitions of $v_{0}$, $\omega_{0}$ and
$\rho_{0}$, connected to the asymptotes of
$\Omega_{i}(k)\equiv\int_{\lambda_{1}}^{k}d\Omega_{i},$ with $i\in\\{x,t,N\\}$

$\displaystyle\Omega_{x}(z)=\pm\left(z-\imath\frac{v_{0}}{2}\right)+O(1/z),\quad\Omega_{t}(z)=\pm(2z^{2}-\omega_{0})+O(1/z),$
(5.47) $\displaystyle\Omega_{N}(z)=\pm\log\frac{2z}{\sqrt{\rho_{0}}}+O(1/z)$
(5.48)

Substituting Eqs. (5.45, 5.46) into Eq. (4.9), one may deduce:

$\displaystyle\Psi_{1}=\begin{pmatrix}\dots&-\frac{p}{8}\\\ q&\dots\\\
\end{pmatrix}$ (5.49)

Thus we have:

$\displaystyle
q=\frac{\theta(\bm{D})\theta(\bm{A}_{g}+\bm{V}+\bm{D})e^{\int\imath
k_{0}dx-\omega_{0}dt}}{\theta(\bm{V}+\bm{D})\theta(\bm{A}_{g}+\bm{D})}$ (5.50)
$\displaystyle
p=-2\rho_{0}\frac{\theta(\bm{A}_{g}-\bm{V}+\bm{D})\theta(\bm{D})e^{-\int\imath
k_{0}dx-\omega_{0}dt}}{\theta(\bm{D}-\bm{V})\theta(\bm{A}_{g}+\bm{D})}$ (5.51)

In the genus $0$ case ($g=0$), the theta function degenerates to a constant.
And computing the asymptotes in Eqs. (5.47,5.47) explicitly gives $v_{0}=v,$
$\rho_{0}=\rho$ and $\omega_{0}=2\rho-\frac{v^{2}}{4}$ which recovers the
Cole-Hopf transformation in Eq. (2.1) $q=e^{-\frac{1}{2}\int^{x}v},$ $p=-2\rho
e^{\frac{1}{2}\int^{x}v}$ .

##  6 Example for Flat Initial Conditions

In Ref. [14] the boundary value problem:

$\displaystyle\rho(x,1)=\frac{16\kappa^{3}}{3\pi}\delta(x),\quad v(x,0)=0,$
(6.1)

was considered in the inviscid limit. This is term ”flat” initial conditions
since $v$ is constant at initial time. In Ref. [14] Eqs. (2.4,2.5) were solved
with these boundary conditions. The solution may be written by first solving
$\beta(t)$ from the equation:

$\displaystyle
t=\frac{2}{\pi\beta(t)}\sqrt{\beta(t)-1}+\frac{2}{\pi}\arctan\sqrt{\beta(t)-1}.$
(6.2)

Then $\rho$ and $v$ can be written explicitly as:

$\displaystyle\rho(x,t)=\begin{cases}\beta\left(\kappa^{2}-\frac{\pi^{2}\beta^{2}}{16}x^{2}\right)&|x|<\frac{4\kappa}{\pi\beta}\\\
0&|x|>\frac{4\kappa}{\pi\beta}\\\ \end{cases}$ (6.3) $\displaystyle
v(x,t)=\begin{cases}-\frac{\pi}{2}\beta\sqrt{\beta-1}x&|x|<\frac{4\kappa}{\pi\beta}\\\
\mbox{solves
}x-vt=-\frac{2}{\pi}\left(2\kappa+v\arctan\frac{v}{2\kappa}\right)\sigma_{v}&\frac{4\kappa}{\pi}>|x|>\frac{4\kappa}{\pi\beta}\\\
0&\frac{4\kappa}{\pi}<|x|\end{cases}.$ (6.4)

To identify the asymptotic Riemann-Hilbert problem associated with this large
$\Lambda$ limit, we may take the solution for $v(x,1)$ identify it as Eq.
(5.31), with $\sigma_{v}=-\sigma$ and $k=\frac{\imath v}{2}$. Then one has:

$\displaystyle\int^{\kappa}_{-\kappa}\frac{m^{\prime}(\mu)}{\mu-\frac{\imath
v}{2}}\frac{d\mu}{2\pi}=v-\frac{2}{\pi}\left(2\kappa+v\arctan\frac{v}{2\kappa}\right).$
(6.5)

This is a singular integral equation which can be solved by standard means,
we, however, may guess the solution, namely $m^{\prime}(\mu)=-4\mu,$ which
suggests $m(\mu)=2(\kappa^{2}-\mu^{2})$. This coincides with the result of
Ref. [10] where it was shown that within an exact solution one obtains
$m(\mu)=2(\kappa^{2}-\mu^{2})-2\log(\mu)$, the last term being a logarithmic
correction to out approximate solution where the large parameter is
$\mu\sim\kappa.$

Having identified $m(\mu)$, we may now check to see if Eqs. (5.26,5.27), which
may be considered as the result of integration of the inviscid differential
equations, are satisfied in this case. This is a matter of substituting
$m^{\prime}(\mu)=-4\mu$ in those equations and performing the integrals. One
find the following:

$\displaystyle
x-v\left(t+\frac{2}{\pi}\arg\frac{{}\sqrt{\kappa-\lambda_{2}}-\sqrt{\kappa-\lambda_{1}}}{\sqrt{\kappa-\lambda_{1}}+{}\sqrt{\kappa-\lambda_{2}}}\right)-\frac{4}{\pi}{\rm
Re}\left[R_{2}(\kappa)\right]=0$ (6.6)
$\displaystyle\pi\rho\left(t+\frac{2}{\pi}\arg\frac{{}\sqrt{\kappa-\lambda_{2}}-\sqrt{\kappa-\lambda_{1}}}{\sqrt{\kappa-\lambda_{1}}+{}\sqrt{\kappa-\lambda_{2}}}\right)-{\rm
Im}\left[\left(2\kappa+\imath v\right)R_{2}(\kappa)\right]=0,$ (6.7)

and the fact that these equations are indeed satisfied can be checked by
direct substitution.

##  7 Conclusion

In this paper we have established a connection between the inverse scattering
method and the Whitham limit in the case of the boundary value problems that
appear in certain large deviation problems. We have tried to give all the
important features of the approach that connects the two problems, while
leaving out many of the specific details that apply to certain cases. First,
we have only dealt with the case where the relevant nonlinear equations to be
solved are the non-linear Schrödinger equations, with real fields. We believe
that the generalization to such systems as the derivative non-linear
Schrödinger equations is not substantially different than the current case.

Furthermore we have mainly dealt with the case where the strongly non-linear
limit leads to inviscid equations, where dispersive terms may simply be
dropped. Although at first site it may seem that this case is rather special,
as it is known that instabilities, such as shocks, can cause oscillations or
solitons to appear in the solution, it is actually quite straightforward to
generalize the method to such cases, since the form of the solution in the
case where the Riemann-Hilbert contour is multi-segmented is written down in
Eqs. (5.50,5.51). The case of solitons appear when two enpoints of the
Riemann-Hilbert contour meet, as is well known. Indeed, in that limit the
theta functions appearing in Eqs. (5.50,5.51) degenerate into hyperbolic
trigonometric functions, from which solitons are easily obtained. In the case
of multi-segmented Riemann-Hilbert contours one can recover a generalization
of the equations for the Riemann invariants, Eqs. (2.7, 2.8), by making use
Eqs. (5.33), by following the procedure outlined in this paper. Such a
procedure is well known from Refrs. [20, 17, 21, 16] .

##  8 Acknowledgement

I wish to thank Baruch Meerson for many useful discussions. I wish to
acknowledge the Binational Science Foundation which has supported this
research through grant number 2020193.

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