1. Introduction
For nonlinear partial differential equations, solutions of spatial infinite decay and spatially periodic solutions are the two most widely studied classes of solutions. In the first of these classes, solutions of the initial value problem of integrable evolution equations can be solved by linear operations [1]. To be precise, the inverse scatter transform, which was introduced about half a century ago, solved this problem groundbreakingly [2–6].
For the above-mentioned second classes-the spatially periodic solution, the finite gap integration method was introduced in the 1970s [7]. Considering rational combinations of theta functions, which are defined on Riemann surfaces that the number of branch cuts is finite, the approach generates exact solutions named finite-gap solutions. If the given initial data belongs to the finite gap class, according to the Jacobian inversion problem on the finite Riemann surface, we can retrieve the solution of the initial value problem. But nevertheless, if the initial data does not belong to the finite gap potential energy category, then an additional theory is needed. Although the method can be extended to infinite Riemann surfaces [8–10], the solutions of the Jacobian inversion problems on infinite surfaces are quite complicated. In fact, solutions with finite gaps are not easy to obtain computationally [2, 11].
In 1997, Fokas proposed a novel method called the unified transform method (UTM) for solving the initial boundary value problem of integrable systems [12, 13]. The initial boundary value problems of many integrable equations have been studied, such as the nonlinear Schrödinger equation, sine-Gordon equation, and the mKdV equation [14–19]. In 2013, Lenells generalized the UTM to the initial boundary value problem of an integrable system with pairs of third-order matrices Lax on the half-line [20]. In 2014, Fan Engui and Xu Jian applied the UTM to the initial boundary value problem on the half-line of the Sasa-Satsuma equation and three-wave equation [21, 22]. Recently, Yan Zhenya extended the UTM to the fourth-order matrix spectral problem [23].
In 2004, Fokas and Its realized the UTM to the nonlinear Schrödinger equation on the finite interval [24]. They show the solution of the nonlinear Schrödinger equation can be obtained by the related Riemann–Hilbert problem involving spectral functions defined in terms of initial datum and boundary values. Facts have proved that the boundary conditions, which correspond to the x-periodic problem, are linearizable. Recently, using the NLS equation as an example, Fokas and Lenells proposed a new method based on the unified transform to obtain the solution of the NLS equation with periodic initial conditions [25, 26]. Their method provides a new perspective on the initial value problem on the circle, and creatively they get the expression of the solution of a Riemann–Hilbert problem only involving the initial data.
For the inverse scattering method, the unified transformation method is a generalization of the initial value problem to the initial boundary value problem. Through the simultaneous spectral analysis of the x-part and the t-part of the Lax pair, the solution of the initial boundary value problem of the integrable equation can be represented by the solution of the matrix RH problem on the complex plane. The jump matrix of the RH problem of the initial boundary value problem can be explicitly expressed by the spectral matrix corresponding to the initial conditions and boundary conditions.
It is well known that the nonlinear Schrödinger equation can well describe the wave evolution in deep water and optical fibers [27, 28]. In addition, the nonlinear Schrödinger equation is also a general model for some multiplicity of nonlinear phenomena in plasma waves, Bose–Einstein condensates, and other physical phenomena [29–31]. One of the most important applications is the soliton in optical fibers [32]. Nonetheless, several phenomena can not be described by low order dispersion NLS equation. When considering the propagation of ultrashort pulses through the fiber, that is, pulses less than 100-fs, it has been shown that the model should include high-order dispersion, self-steepening, self-frequency, and high-order effects. The high-order nonlinear Schrödinger equation (1.1 ) is particularly important for describing the more complex phenomena. In this paper, we study the high-order nonlinear Schrödinger equation1.1 ) for $(x,t)\in {\mathbb{R}}\times [0,\infty )$ which is x-periodic of period L > 0, i.e. q(x + L, t) = q(x, t). We consider the following initial datum involving a single exponential:
$\begin{eqnarray}\begin{array}{l}{\rm{i}}{q}_{t}+{q}_{{xx}}+2q| q{| }^{2}+{q}_{{xxxx}}+6{{q}_{x}}^{2}\bar{q}+4q| {q}_{x}{| }^{2}\\ \quad +8| q{| }^{2}{q}_{{xx}}+2| q{| }^{2}{\bar{q}}_{{xx}}+6| q{| }^{4}q=0,\end{array}\end{eqnarray}$
suppose q(x, t) is a smooth solution of equation ( $\begin{eqnarray}q(x,0)={q}_{0}{{\rm{e}}}^{\tfrac{2{\rm{i}}\pi N}{L}x},\quad x\in [0,L],\end{eqnarray}$
where N is an integer and the constant q0 > 0. The high-order nonlinear Schrödinger equation has been studied through many methods, such as Darboux Transformation, the Inverse scattering transform and so on [33–41]. However, to the best of our knowledge, the high-order nonlinear Schrödinger equation with periodic initial conditions was not reported before. Although we used the same method as in [25, 26], it is worth noting that the RH problem we studied was more complex. It lays the foundation for us to further explore a certain class of nonlinear integrable equations, such as the KdV equation, the mKdV equation, and the Hirota equation.The outline of this paper is as follows: in section 2 , we apply the UTM to a high-order nonlinear Schrödinger equation posed on a finite interval $x\in [0,L]\subset {\mathbb{R}}$. In section 3 , the high-order nonlinear Schrödinger equation on the periodic problem with a single exponential initial data $(i.e.q(x,0)={q}_{0}{{\rm{e}}}^{\tfrac{2{\rm{i}}\pi N}{L}x},x\in [0,L])$ was considered. Moreover, by solving the associated RH problem involving the initial datum alone, the exact solution is obtained, which corresponds to the one-gap solution.
2. The UTM to the high-order nonlinear Schrödinger equation on a finite interval
In this section, we apply the UTM to the high-order nonlinear Schrödinger equation with the following data,
$\begin{eqnarray*}\begin{array}{l}q(x,0)={q}_{0}(x),\quad x\in [0,L],\\ q(0,t),{q}_{x}(0,t),{q}_{{xx}}(0,t),{q}_{{xxx}}(0,t),q(L,t),\\ {q}_{x}(L,t),{q}_{{xx}}(L,t),{q}_{{xxx}}(L,t),\quad t\in [0,T].\end{array}\end{eqnarray*}$
According to the AKNS formalism, the high-order nonlinear Schrödinger equation has a lax pair given by2.1 ), we get
$\begin{eqnarray}\left\{\begin{array}{l}{{\rm{\Psi }}}_{x}+{\rm{i}}\lambda [{\sigma }_{3},{\rm{\Psi }}]=Q{\rm{\Psi }},\\ {{\rm{\Psi }}}_{t}+{\rm{i}}(2{\lambda }^{2}-8{\lambda }^{4})[{\sigma }_{3},{\rm{\Psi }}]=V{\rm{\Psi }},\end{array}\right.\end{eqnarray}$
with $\begin{eqnarray*}\begin{array}{rcl}V & = & {\rm{i}}{V}_{3}{\lambda }^{3}+{\rm{i}}{V}_{2}{\lambda }^{2}+{\rm{i}}{V}_{1}\lambda +{\rm{i}}{V}_{0},\\ {V}_{0} & = & (6{Q}^{2}-{Q}_{x}-{Q}_{{xxx}}+{Q}_{x}^{2}+3{Q}^{4}-{{QQ}}_{{xx}}-{Q}_{{xx}}Q){\sigma }_{3},\\ {V}_{1} & = & 2{\rm{i}}(-{{QQ}}_{x}+{Q}_{x}Q-Q+2{Q}^{3}-{Q}_{{xx}}),\\ {V}_{2} & = & (4{Q}^{2}+4{Q}_{x}){\sigma }_{3},{V}_{3}=8{\rm{i}}Q,\end{array}\end{eqnarray*}$
where $\Psi$(x, t, λ) is a 2 × 2 matrix-valued function, $\lambda \in {\mathbb{C}}$ is an auxiliary (spectra) parameter, the potential matrix Q and σ3 are defined by $\begin{eqnarray}Q=\left(\begin{array}{cc}0 & q(x,t)\\ -\bar{q}(x,t) & 0\end{array}\right),{\sigma }_{3}=\left(\begin{array}{cc}1 & 0\\ 0 & -1\end{array}\right),\end{eqnarray}$
the W(x, t, λ) and θ are defined by $W(x,t,\lambda )\,={{\rm{e}}}^{{\rm{i}}\theta \hat{{\sigma }_{3}}}(Q{\rm{\Psi }}{\rm{d}}x+V{\rm{\Psi }}{\rm{d}}t),\theta =\lambda x+(2{\lambda }^{2}-8{\lambda }^{4})t.$ According to equation ( $\begin{eqnarray}{\rm{\Psi }}(x,t,\lambda )=I+{\int }_{({x}_{* },{t}_{* })}^{(x,t)}{{\rm{e}}}^{-{\rm{i}}\theta \hat{{\sigma }_{3}}}W(y,\tau ,\lambda ),\end{eqnarray}$
where (x*, t*) ∈ [0, L] × [0, T] is an arbitrary point and integral means the line integral that smoothly connects the indicated points. We choose the point (x*, t*) as each of the corners of the polygonal domain. Therefore we define four solutions $\Psi$1, $\Psi$2, $\Psi$3, $\Psi$4, corresponding to the four points (0, T), (0, 0), (L, 0), (L, T). $\begin{eqnarray}\begin{array}{l}{{\rm{\Psi }}}_{1}(x,t,\lambda )=I+{\displaystyle \int }_{0}^{x}{{\rm{e}}}^{-{\rm{i}}\lambda (x-y)\hat{{\sigma }_{3}}}(Q{{\rm{\Psi }}}_{1})(y,t,\lambda ){\rm{d}}y-{{\rm{e}}}^{-{\rm{i}}\lambda x\hat{{\sigma }_{3}}}\\ \quad \times {\displaystyle \int }_{t}^{T}{{\rm{e}}}^{-{\rm{i}}(2{\lambda }^{2}-8{\lambda }^{4})(t-\tau )\hat{{\sigma }_{3}}}(V{{\rm{\Psi }}}_{1})(0,\tau ,\lambda ){\rm{d}}\tau ,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{{\rm{\Psi }}}_{2}(x,t,\lambda )=I+{\displaystyle \int }_{0}^{x}{{\rm{e}}}^{-{\rm{i}}\lambda (x-y)\hat{{\sigma }_{3}}}(Q{{\rm{\Psi }}}_{2})(y,t,\lambda ){\rm{d}}y+{{\rm{e}}}^{-{\rm{i}}\lambda x\hat{{\sigma }_{3}}}\\ \quad \times {\displaystyle \int }_{0}^{t}{{\rm{e}}}^{-{\rm{i}}(2{\lambda }^{2}-8{\lambda }^{4})(t-\tau )\hat{{\sigma }_{3}}}(V{{\rm{\Psi }}}_{2})(0,\tau ,\lambda ){\rm{d}}\tau ,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{{\rm{\Psi }}}_{3}(x,t,\lambda )=I-{\displaystyle \int }_{x}^{L}{{\rm{e}}}^{-{\rm{i}}\lambda (x-y)\hat{{\sigma }_{3}}}(Q{{\rm{\Psi }}}_{3})(y,t,\lambda ){\rm{d}}y\\ \quad +{{\rm{e}}}^{-{\rm{i}}\lambda (x-L)\hat{{\sigma }_{3}}}\\ \quad \times {\displaystyle \int }_{0}^{t}{{\rm{e}}}^{-{\rm{i}}(2{\lambda }^{2}-8{\lambda }^{4})(t-\tau )\hat{{\sigma }_{3}}}(V{{\rm{\Psi }}}_{3})(L,\tau ,\lambda ){\rm{d}}\tau ,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{{\rm{\Psi }}}_{4}(x,t,\lambda )=I-{\int }_{x}^{L}{{\rm{e}}}^{-{\rm{i}}\lambda (x-y)\hat{{\sigma }_{3}}}(Q{{\rm{\Psi }}}_{4})(y,t,\lambda ){\rm{d}}y\\ \quad -{{\rm{e}}}^{-{\rm{i}}\lambda (x-L)\hat{{\sigma }_{3}}}{\int }_{t}^{T}{{\rm{e}}}^{-{\rm{i}}(2{\lambda }^{2}-8{\lambda }^{4})(t-\tau )\hat{{\sigma }_{3}}}(V{{\rm{\Psi }}}_{4})(L,\tau ,\lambda ){\rm{d}}\tau .\end{array}\end{eqnarray}$
Further, we can show that $\Psi$j(x, t, λ), j = 1, 2, 3, 4 satisfies the following bounded and analytic properties in the complex λ plane $\begin{eqnarray}\begin{array}{l}{[{{\rm{\Psi }}}_{2}]}_{1},{[{{\rm{\Psi }}}_{4}]}_{2}:\quad {D}_{1},\quad {[{{\rm{\Psi }}}_{1}]}_{1},{[{{\rm{\Psi }}}_{3}]}_{2}:\quad {D}_{2},\\ \quad {[{{\rm{\Psi }}}_{3}]}_{1},{[{{\rm{\Psi }}}_{1}]}_{2}:\quad {D}_{3},\quad {[{{\rm{\Psi }}}_{4}]}_{1},{[{{\rm{\Psi }}}_{2}]}_{2}:\quad {D}_{4},\end{array}\end{eqnarray}$
where [A]1 and [A]2 denote the first and second columns of a matrix A, respectively. The Dj (j = 1, ⋯ ,4) is defined as follows $\begin{eqnarray*}\begin{array}{rcl}{D}_{1} & = & \{\lambda \in {\mathbb{C}}:\mathrm{Im}\lambda \geqslant 0,\mathrm{Re}\lambda [1-8{\left(\mathrm{Re}\lambda \right)}^{2}\\ & & +8{\left(\mathrm{Im}\lambda \right)}^{2}]\geqslant 0\},\\ {D}_{2} & = & \{\lambda \in {\mathbb{C}}:\mathrm{Im}\lambda \geqslant 0,\mathrm{Re}\lambda [1-8{\left(\mathrm{Re}\lambda \right)}^{2}\\ & & +8{\left(\mathrm{Im}\lambda \right)}^{2}]\leqslant 0\},\\ {D}_{3} & = & \{\lambda \in {\mathbb{C}}:\mathrm{Im}\lambda \leqslant 0,\mathrm{Re}\lambda [1-8{\left(\mathrm{Re}\lambda \right)}^{2}\\ & & +8{\left(\mathrm{Im}\lambda \right)}^{2}]\leqslant 0\},\\ {D}_{4} & = & \{\lambda \in {\mathbb{C}}:\mathrm{Im}\lambda \leqslant 0,\mathrm{Re}\lambda [1-8{\left(\mathrm{Re}\lambda \right)}^{2}\\ & & +8{\left(\mathrm{Im}\lambda \right)}^{2}]\geqslant 0\}.\end{array}\end{eqnarray*}$
Note that all the $\Psi$j are entire functions of λ, since the matrices Q and V are traceless, thus we have
$\begin{eqnarray*}\begin{array}{rcl}\det {{\rm{\Psi }}}_{j}(x,t,\lambda ) & = & 1,\quad j=1,\cdots ,4,\\ {{\rm{\Psi }}}_{j}(x,t,\lambda ) & = & I+O\left(\displaystyle \frac{1}{\lambda }\right),\quad \lambda \to \infty .\end{array}\end{eqnarray*}$
The asymptotics here and later are in the corresponding domains of boundedness. See figure 1 for details.
The eigenfunctions $\Psi$j are related by the equations2.9 ) and (2.11 ) imply2.10 ), we find $S(\lambda )\,={\left({{\rm{e}}}^{{\rm{i}}(2{\lambda }^{2}-8{\lambda }^{4})T\hat{{\sigma }_{3}}}{{\rm{\Psi }}}_{2}(0,T,\lambda )\right)}^{-1}$. Evaluating equation (2.12 ) at the point (0, T) and writing $\Psi$2(0, T, λ) in terms of S(λ), we find2.13 ) by ${{\rm{e}}}^{{\rm{i}}(2{\lambda }^{2}-8{\lambda }^{4})T\hat{{\sigma }_{3}}}$ and using equation (2.7 ), we get
$\begin{eqnarray}{{\rm{\Psi }}}_{3}(x,t,\lambda )={{\rm{\Psi }}}_{2}(x,t,\lambda ){{\rm{e}}}^{-{\rm{i}}\theta \hat{{\sigma }_{3}}}s(\lambda ),\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Psi }}}_{1}(x,t,\lambda )={{\rm{\Psi }}}_{2}(x,t,\lambda ){{\rm{e}}}^{-{\rm{i}}\theta \hat{{\sigma }_{3}}}S(\lambda ),\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Psi }}}_{4}(x,t,\lambda )={{\rm{\Psi }}}_{3}(x,t,\lambda ){{\rm{e}}}^{-{\rm{i}}(\theta -\lambda L)\hat{{\sigma }_{3}}}{S}_{L}(\lambda ).\end{eqnarray}$
Through the above equation, we get $\begin{eqnarray*}\begin{array}{rcl}s(\lambda ) & = & {{\rm{\Psi }}}_{3}(0,0,\lambda ),\\ S(\lambda ) & = & {{\rm{\Psi }}}_{1}(0,0,\lambda ),\\ {S}_{L}(\lambda ) & = & {{\rm{\Psi }}}_{4}(L,0,\lambda ).\end{array}\end{eqnarray*}$
Meanwhile, equations ( $\begin{eqnarray}{{\rm{\Psi }}}_{4}(x,t,\lambda )={{\rm{\Psi }}}_{2}(x,t,\lambda ){{\rm{e}}}^{-{\rm{i}}\theta \hat{{\sigma }_{3}}}(s(\lambda ){{\rm{e}}}^{{\rm{i}}\lambda L\hat{{\sigma }_{3}}}{S}_{L}(\lambda )).\end{eqnarray}$
When x = 0, t = T, from equation ( $\begin{eqnarray}{{\rm{\Psi }}}_{4}(0,T,\lambda )={{\rm{e}}}^{-{\rm{i}}(2{\lambda }^{2}-8{\lambda }^{4})T\hat{{\sigma }_{3}}}({S}^{-1}(\lambda )s(\lambda ){{\rm{e}}}^{{\rm{i}}\lambda L\hat{{\sigma }_{3}}}{S}_{L}(\lambda )).\end{eqnarray}$
Multiplying equation ( $\begin{eqnarray}\begin{array}{l}-I+{S}^{-1}s({{\rm{e}}}^{{\rm{i}}\lambda L\hat{{\sigma }_{3}}}{S}_{L})+{{\rm{e}}}^{{\rm{i}}(2{\lambda }^{2}-8{\lambda }^{4})T\hat{{\sigma }_{3}}}\\ \quad \times {\displaystyle \int }_{0}^{L}{{\rm{e}}}^{{\rm{i}}\lambda y\hat{{\sigma }_{3}}}(Q{{\rm{\Psi }}}_{4})(y,T,\lambda ){\rm{d}}y=0.\end{array}\end{eqnarray}$
The (1, 2)th element of above equation is $\begin{eqnarray}\begin{array}{l}(a{ \mathcal A }-\bar{b}{{\rm{e}}}^{2{\rm{i}}\lambda L}{ \mathcal B })B-(b{ \mathcal A }+\bar{a}{{\rm{e}}}^{2{\rm{i}}\lambda L}{ \mathcal B })A\\ \quad ={{\rm{e}}}^{2{\rm{i}}(2{\lambda }^{2}-8{\lambda }^{4})T}{c}^{+}(\lambda ),\end{array}\end{eqnarray}$
where $\begin{eqnarray}{c}^{+}(\lambda )=O\left(\displaystyle \frac{1}{\lambda }\right)+O\left(\displaystyle \frac{{{\rm{e}}}^{2{\rm{i}}\lambda L}}{\lambda }\right),\quad \lambda \to \infty .\end{eqnarray}$
According to symmetry properties, we have $\begin{eqnarray}\begin{array}{rcl}s(\lambda ) & = & \left(\begin{array}{cc}\bar{a}(\lambda ) & b(\lambda )\\ -\bar{b}(\lambda ) & b(\lambda )\end{array}\right),S(\lambda )=\left(\begin{array}{cc}\bar{A}(\lambda ) & B(\lambda )\\ -\bar{B}(\lambda ) & A(\lambda )\end{array}\right),\\ {S}_{L}(\lambda ) & = & \left(\begin{array}{cc}\bar{{ \mathcal A }}(\lambda ) & { \mathcal B }(\lambda )\\ -\bar{{ \mathcal B }}(\lambda ) & { \mathcal A }(\lambda )\end{array}\right).\end{array}\end{eqnarray}$
Here, and in what follows, a bar over a function denotes that the complex conjugate is taken, not only of the function but also of its argument; this is called the Schwarz conjugate. We can easily get the following properties: $\begin{eqnarray}\begin{array}{rcl}a\bar{a}+b\bar{b} & = & 1,\quad A\bar{A}+B\bar{B}=1,\\ { \mathcal A }\bar{{ \mathcal A }}+{ \mathcal B }\bar{{ \mathcal B }} & = & 1,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}a(\lambda ) & = & 1+O\left(\displaystyle \frac{1}{\lambda }\right)+O\left(\displaystyle \frac{{{\rm{e}}}^{2{\rm{i}}\lambda L}}{\lambda }\right),\\ b(\lambda ) & = & O\left(\displaystyle \frac{1}{\lambda }\right)+O\left(\displaystyle \frac{{{\rm{e}}}^{2{\rm{i}}\lambda L}}{\lambda }\right),\quad \lambda \to \infty ,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}A(\lambda ) & = & 1+O\left(\displaystyle \frac{1}{\lambda }\right)+O\left(\displaystyle \frac{{{\rm{e}}}^{2{\rm{i}}(2{\lambda }^{2}-8{\lambda }^{4})T}}{\lambda }\right),\\ B(\lambda ) & = & O\left(\displaystyle \frac{1}{\lambda }\right)+O\left(\displaystyle \frac{{{\rm{e}}}^{2{\rm{i}}(2{\lambda }^{2}-8{\lambda }^{4})T}}{\lambda }\right),\quad \lambda \to \infty .\end{array}\end{eqnarray}$
Figure 2. The contour $\tilde{{\rm{\Sigma }}}$ in the complex λ-plane and the associated jump matrices for RH problem for $\tilde{m}$. |
Now we construct the RH problem for M(x, t, λ)
$\begin{eqnarray}M=\left\{\begin{array}{ll}\left(\displaystyle \frac{{[{{\rm{\Psi }}}_{2}]}_{1}}{\alpha },{[{{\rm{\Psi }}}_{4}]}_{2}\right),\quad & \lambda \in {D}_{1},\\ \left(\displaystyle \frac{{[{{\rm{\Psi }}}_{1}]}_{1}}{d},{[{{\rm{\Psi }}}_{3}]}_{2}\right),\quad & \lambda \in {D}_{2},\\ \left({[{{\rm{\Psi }}}_{3}]}_{1},\displaystyle \frac{{[{{\rm{\Psi }}}_{1}]}_{2}}{\bar{d}}\right),\quad & \lambda \in {D}_{3},\\ \left({[{{\rm{\Psi }}}_{4}]}_{1},\displaystyle \frac{{[{{\rm{\Psi }}}_{2}]}_{2}}{\bar{\alpha }}\right),\quad & \lambda \in {D}_{4}.\end{array}\right.\end{eqnarray}$
The M satisfies1.1 ) can be recovered from
$\begin{eqnarray}{M}_{-}(x,t,\lambda )={M}_{+}(x,t,\lambda )J(x,t,\lambda ),\lambda \in {\rm{\Sigma }},\end{eqnarray}$
$\begin{eqnarray}M(x,t,\lambda )=I+O(\displaystyle \frac{1}{\lambda }),\lambda \to \infty ,\end{eqnarray}$
where $\begin{eqnarray}{\rm{\Sigma }}={{\rm{\Sigma }}}_{1}\cup {{\rm{\Sigma }}}_{2}\cup {{\rm{\Sigma }}}_{3}\cup {{\rm{\Sigma }}}_{4},\end{eqnarray}$
$\begin{eqnarray*}\begin{array}{rcl}{{\rm{\Sigma }}}_{1} & = & \{\lambda \in {\mathbb{C}}:\arg \lambda =\displaystyle \frac{\pi }{2}\}\cup \{\lambda \in {\mathbb{C}}:\mathrm{Im}\lambda \geqslant 0,1\\ & & -8{\left(\mathrm{Re}\lambda \right)}^{2}+8{\left(\mathrm{Im}\lambda \right)}^{2}=0\},\\ {{\rm{\Sigma }}}_{2} & = & \{\lambda \in {\mathbb{C}}:\mathrm{Im}\lambda =0,-\infty \lt \mathrm{Re}\lambda \leqslant -\displaystyle \frac{\sqrt{2}}{4}\}\cup \\ & & \{\lambda \in {\mathbb{C}}:\mathrm{Im}\lambda =0,0\leqslant \mathrm{Re}\lambda \leqslant \displaystyle \frac{\sqrt{2}}{4}\},\\ {{\rm{\Sigma }}}_{3} & = & \{\lambda \in {\mathbb{C}}:\arg \lambda =\displaystyle \frac{3\pi }{2}\}\cup \{\lambda \in {\mathbb{C}}:\mathrm{Im}\lambda \leqslant 0,1\\ & & -8{\left(\mathrm{Re}\lambda \right)}^{2}+8{\left(\mathrm{Im}\lambda \right)}^{2}=0\},\\ {{\rm{\Sigma }}}_{4} & = & \{\lambda \in {\mathbb{C}}:\mathrm{Im}\lambda =0,\displaystyle \frac{\sqrt{2}}{4}\leqslant \mathrm{Re}\lambda \lt +\infty \}\cup \\ & & \{\lambda \in {\mathbb{C}}:\mathrm{Im}\lambda =0,-\displaystyle \frac{\sqrt{2}}{4}\leqslant \mathrm{Re}\lambda \leqslant 0\},\end{array}\end{eqnarray*}$
and the jump matrix J is defined by $\begin{eqnarray}J=\left\{\begin{array}{l}{J}_{1}=\left(\begin{array}{cc}\displaystyle \frac{\delta }{d} & -{ \mathcal B }{{\rm{e}}}^{2{\rm{i}}\lambda L}{{\rm{e}}}^{-2{\rm{i}}\theta }\\ \displaystyle \frac{-\bar{B}{{\rm{e}}}^{2{\rm{i}}\theta }}{d\alpha } & \displaystyle \frac{a}{\alpha }\end{array}\right),\quad \lambda \in {{\rm{\Sigma }}}_{1},\\ {J}_{2}=\left(\begin{array}{cc}1 & -\displaystyle \frac{\beta {{\rm{e}}}^{-2{\rm{i}}\theta }}{\bar{\alpha }}\\ \displaystyle \frac{-\bar{\beta }{{\rm{e}}}^{2{\rm{i}}\theta }}{\alpha } & \displaystyle \frac{1}{\alpha \bar{\alpha }}\end{array}\right),\quad \lambda \in {{\rm{\Sigma }}}_{2},\\ {J}_{3}=\left(\begin{array}{cc}\displaystyle \frac{\bar{\delta }}{\bar{d}} & -\displaystyle \frac{B{{\rm{e}}}^{-2{\rm{i}}\theta }}{\bar{d}\bar{\alpha }}\\ -\bar{{ \mathcal B }}{{\rm{e}}}^{-2{\rm{i}}\lambda L}{{\rm{e}}}^{2{\rm{i}}\theta } & \displaystyle \frac{\bar{a}}{\bar{\alpha }}\end{array}\right),\quad \lambda \in {{\rm{\Sigma }}}_{3},\\ {J}_{4}={J}_{3}{J}_{2}^{-1}{J}_{1},\quad \lambda \in {{\rm{\Sigma }}}_{4},\end{array}\right.\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{rcl}\alpha (\lambda ) & = & a{ \mathcal A }-\bar{b}{ \mathcal B }{{\rm{e}}}^{2{\rm{i}}\lambda L},\\ \beta (\lambda ) & = & b{ \mathcal A }+\bar{a}{ \mathcal B }{{\rm{e}}}^{2{\rm{i}}\lambda L},\\ d(\lambda ) & = & a\bar{A}+b\bar{B},\quad \delta (\lambda )=\alpha \bar{A}+\beta \bar{B}.\end{array}\end{eqnarray}$
The solution q(x, t) of equation ( $\begin{eqnarray}q(x,t)=2{\rm{i}}\mathop{\mathrm{lim}}\limits_{\lambda \to \infty }\lambda {M}_{12}(x,t,\lambda ).\end{eqnarray}$
Note that jump matrix J depends on the initial datum q(x, 0) and the boundary values q(0, t), qx(0, t), qxx(0, t), qxxx(0, t), q(L, t), qx(L, t), qxx(L, t), qxxx(L, t). If all values are known, we can obtain the q(x, t) via the solution of the RH problem.3. The high-order nonlinear Schrödinger equation on the periodic problem with a single exponential initial data
Suppose q(x, t) is a smooth solution of equation (1.1 ) for $(x,t)\in {\mathbb{R}}\times [0,\infty )$ which is x-periodic of period L > 0, i.e. q(x + L, t) = q(x, t). We consider the following initial datum involving a single exponential:2.25 ) become
$\begin{eqnarray}q(x,0)={q}_{0}{{\rm{e}}}^{\tfrac{2{\rm{i}}\pi N}{L}x},\quad x\in [0,L],\end{eqnarray}$
where N is an integer and the constant q0 > 0. In this case, we clearly have ${ \mathcal A }=A$ and ${ \mathcal B }=B$. Therefore the jump matrices Ji defined in equation ( $\begin{eqnarray}\begin{array}{rcl}{J}_{1} & = & \left(\begin{array}{cc}\displaystyle \frac{\delta }{d} & -B{{\rm{e}}}^{2{\rm{i}}\lambda L}{{\rm{e}}}^{-2{\rm{i}}\theta }\\ \displaystyle \frac{-\bar{B}{{\rm{e}}}^{2{\rm{i}}\theta }}{d\alpha } & \displaystyle \frac{a}{\alpha }\end{array}\right),\quad {J}_{2}=\left(\begin{array}{cc}1 & -\displaystyle \frac{\beta {{\rm{e}}}^{-2{\rm{i}}\theta }}{\bar{\alpha }}\\ \displaystyle \frac{-\bar{\beta }{{\rm{e}}}^{2{\rm{i}}\theta }}{\alpha } & \displaystyle \frac{1}{\alpha \bar{\alpha }}\end{array}\right),\\ {J}_{3} & = & \left(\begin{array}{cc}\displaystyle \frac{\bar{\delta }}{\bar{d}} & -\displaystyle \frac{B{{\rm{e}}}^{-2{\rm{i}}\theta }}{\bar{d}\bar{\alpha }}\\ -\bar{B}{{\rm{e}}}^{-2{\rm{i}}\lambda L}{{\rm{e}}}^{2{\rm{i}}\theta } & \displaystyle \frac{\bar{a}}{\bar{\alpha }}\end{array}\right),\quad {J}_{4}={J}_{3}{J}_{2}^{-1}{J}_{1},\end{array}\end{eqnarray}$
where $\begin{eqnarray}\begin{array}{rcl}\alpha (\lambda ) & = & {aA}-\bar{b}B{{\rm{e}}}^{2{\rm{i}}\lambda L},\quad \beta (\lambda )={bA}+\bar{a}B{{\rm{e}}}^{2{\rm{i}}\lambda L},\\ d(\lambda ) & = & a\bar{A}+b\bar{B},\quad \delta (\lambda )=\alpha \bar{A}+\beta \bar{B}.\end{array}\end{eqnarray}$
Similarly, we have2.18 ), equation (2.19 ) and equation (2.20 ). According to equation (2.27 ), the solution q(x, t) of equation (1.1 ) can be recovered by
$\begin{eqnarray}s(\lambda )=\left(\begin{array}{cc}\bar{a}(\lambda ) & b(\lambda )\\ -\bar{b}(\lambda ) & b(\lambda )\end{array}\right),\quad S(\lambda )=\left(\begin{array}{cc}\bar{A}(\lambda ) & B(\lambda )\\ -\bar{B}(\lambda ) & A(\lambda )\end{array}\right),\end{eqnarray}$
where $\begin{eqnarray}s(\lambda )={{\rm{\Psi }}}_{3}(0,0,\lambda ),\quad S(\lambda )={{\rm{\Psi }}}_{1}(0,0,\lambda ).\end{eqnarray}$
Clearly, S(λ) depends on T, while s(λ) does not. We have the global relation: $\begin{eqnarray}\begin{array}{l}({aA}-\bar{b}{{\rm{e}}}^{2{\rm{i}}\lambda L}B)B-({bA}+\bar{a}{{\rm{e}}}^{2{\rm{i}}\lambda L}B)A\\ \quad ={{\rm{e}}}^{2{\rm{i}}(2{\lambda }^{2}-8{\lambda }^{4})T}{c}^{+}(\lambda ),\end{array}\end{eqnarray}$
where $\begin{eqnarray}{c}^{+}(\lambda )=O\left(\displaystyle \frac{1}{\lambda }\right)+O\left(\displaystyle \frac{{{\rm{e}}}^{2{\rm{i}}\lambda L}}{\lambda }\right),\quad \lambda \to \infty .\end{eqnarray}$
Meanwhile, a, b, A, B satisfy the equation ( $\begin{eqnarray}q(x,t)=2{\rm{i}}\mathop{\mathrm{lim}}\limits_{\lambda \to \infty }\lambda {M}_{12}(x,t,\lambda ).\end{eqnarray}$
3.1. The RH problem for m
It can be formulated, which involves an RH problem where the ratio Γ = B/A is used to define the matrix jump, relative to A and B. Now we consider the piecewise meromorphic function m(x, t, k) defined by2.21 ) is as follows:
$\begin{eqnarray}m=\left\{\begin{array}{l}\left(\displaystyle \frac{A{[{{\rm{\Psi }}}_{2}]}_{1}}{\alpha },\displaystyle \frac{{[{{\rm{\Psi }}}_{4}]}_{2}}{A}\right),\quad \lambda \in {D}_{1},\\ \left(\displaystyle \frac{{[{{\rm{\Psi }}}_{1}]}_{1}}{d},{[{{\rm{\Psi }}}_{3}]}_{2}\right),\quad \lambda \in {D}_{2},\\ \left({[{{\rm{\Psi }}}_{3}]}_{1},\displaystyle \frac{{[{{\rm{\Psi }}}_{1}]}_{2}}{\bar{d}}\right),\quad \lambda \in {D}_{3},\\ \left(\displaystyle \frac{{[{{\rm{\Psi }}}_{4}]}_{1}}{\bar{A}},\displaystyle \frac{\bar{A}{[{{\rm{\Psi }}}_{2}]}_{2}}{\bar{\alpha }}\right),\quad \lambda \in {D}_{4}.\end{array}\right.\end{eqnarray}$
The relationship between functions m and M defined in equation ( $\begin{eqnarray}m={MH},\end{eqnarray}$
where the piecewise meromorphic function H is defined by $\begin{eqnarray}\begin{array}{rcl}{H}_{1} & = & \left(\begin{array}{cc}A & 0\\ 0 & \displaystyle \frac{1}{A}\end{array}\right),\quad {H}_{2}={H}_{3}=I,\\ {H}_{4} & = & \left(\begin{array}{cc}\displaystyle \frac{1}{\bar{A}} & 0\\ 0 & \bar{A}\end{array}\right),\end{array}\end{eqnarray}$
where Hj is defined in the region Dj, j = 1, ⋯ ,4.We can get a RH problem for m as follows:
The jump matrix &ugr;(x, t, λ) is defined by
| • | m(x, t, λ) is analytic in ${\mathbb{C}}\setminus {\rm{\Sigma }}$, and continuous on ${\rm{\Sigma }}\setminus \{0,\pm \tfrac{\sqrt{2}}{4}\}$. |
| • | m−(x, t, λ)=m+(x, t, λ)&ugr;(x, t, λ), λ ∈ Σ$\setminus \{0,\pm \tfrac{\sqrt{2}}{4}\}$. |
| • | $m(x,t,\lambda )=I\,+\,O(\tfrac{1}{\lambda }),\quad \lambda \to \infty $. |
| • | m(x, t, λ) = O(1) λ → 0. |
$\begin{eqnarray}\begin{array}{l}{\upsilon }_{1}=\left(\begin{array}{cc}\displaystyle \frac{a+b\bar{{\rm{\Gamma }}}+{\rm{\Gamma }}(\bar{a}\bar{{\rm{\Gamma }}}-\bar{b}){{\rm{e}}}^{2{\rm{i}}\lambda L}}{a+b\bar{{\rm{\Gamma }}}} & -{\rm{\Gamma }}{{\rm{e}}}^{2{\rm{i}}\lambda L}{{\rm{e}}}^{-2{\rm{i}}\theta }\\ \displaystyle \frac{-\bar{{\rm{\Gamma }}}{{\rm{e}}}^{2{\rm{i}}\theta }}{(a+b\bar{{\rm{\Gamma }}})(a-\bar{b}{\rm{\Gamma }}{{\rm{e}}}^{2{\rm{i}}\lambda L})} & \displaystyle \frac{a}{a-\bar{b}{\rm{\Gamma }}{{\rm{e}}}^{2{\rm{i}}\lambda L}}\end{array}\right),\\ \quad \lambda \in {{\rm{\Sigma }}}_{1},\\ {\upsilon }_{2}=\left(\begin{array}{cc}1+{\rm{\Gamma }}\bar{{\rm{\Gamma }}} & \displaystyle \frac{-(\bar{a}{\rm{\Gamma }}{{\rm{e}}}^{2{\rm{i}}\lambda L}+b){{\rm{e}}}^{-2{\rm{i}}\theta }}{\bar{a}-b\bar{{\rm{\Gamma }}}{{\rm{e}}}^{-2{\rm{i}}\lambda L}}\\ \displaystyle \frac{-(\bar{b}+{{\rm{e}}}^{-2{\rm{i}}\lambda L}a\bar{{\rm{\Gamma }}}){{\rm{e}}}^{2{\rm{i}}\theta }}{a-{{\rm{e}}}^{2{\rm{i}}\lambda L}\bar{b}{\rm{\Gamma }}} & \displaystyle \frac{1}{(a-\bar{b}{\rm{\Gamma }}{{\rm{e}}}^{2{\rm{i}}\lambda L})(\bar{a}-b\bar{{\rm{\Gamma }}}{{\rm{e}}}^{-2{\rm{i}}\lambda L})}\end{array}\right),\\ \quad \lambda \in {{\rm{\Sigma }}}_{2},\\ {\upsilon }_{3}=\left(\begin{array}{cc}\displaystyle \frac{\bar{a}+\bar{b}{\rm{\Gamma }}+\bar{{\rm{\Gamma }}}(a{\rm{\Gamma }}-b){{\rm{e}}}^{-2{\rm{i}}\lambda L}}{\bar{a}+\bar{b}{\rm{\Gamma }}} & -\displaystyle \frac{{\rm{\Gamma }}{{\rm{e}}}^{-2{\rm{i}}\theta }}{(\bar{a}+\bar{b}{\rm{\Gamma }})(\bar{a}-b\bar{{\rm{\Gamma }}}{{\rm{e}}}^{-2{\rm{i}}\lambda L})}\\ -\bar{{\rm{\Gamma }}}{{\rm{e}}}^{-2{\rm{i}}\lambda L}{{\rm{e}}}^{2{\rm{i}}\theta } & \displaystyle \frac{\bar{a}}{\bar{a}-b\bar{{\rm{\Gamma }}}{{\rm{e}}}^{-2{\rm{i}}\lambda L}}\end{array}\right),\\ \quad \lambda \in {{\rm{\Sigma }}}_{3},\\ {\upsilon }_{4}=\left(\begin{array}{cc}\displaystyle \frac{1+{\rm{\Gamma }}\bar{{\rm{\Gamma }}}}{(a+b\bar{{\rm{\Gamma }}})(\bar{a}+\bar{b}{\rm{\Gamma }})} & \displaystyle \frac{(b-a{\rm{\Gamma }}){{\rm{e}}}^{-2{\rm{i}}\theta }}{\bar{a}+\bar{b}{\rm{\Gamma }}}\\ \displaystyle \frac{(\bar{b}-\bar{a}\bar{{\rm{\Gamma }}}){{\rm{e}}}^{2{\rm{i}}\theta }}{a+b\bar{{\rm{\Gamma }}}} & 1\end{array}\right),\\ \quad \lambda \in {{\rm{\Sigma }}}_{4}.\end{array}\end{eqnarray}$
Furthermore, the function m admit the symmetry relations $\begin{eqnarray}\begin{array}{rcl}{m}_{11}(x,t,\lambda ) & = & {\bar{m}}_{22}(x,t,\bar{\lambda }),\\ {m}_{21}(x,t,\lambda ) & = & -{\bar{m}}_{12}(x,t,\bar{\lambda }).\end{array}\end{eqnarray}$
3.2. The RH problem for $\tilde{m}$
We defined the function $\tilde{{\rm{\Gamma }}}$ by
$\begin{eqnarray}\tilde{{\rm{\Gamma }}}=-\displaystyle \frac{1}{2{{\rm{e}}}^{{\rm{i}}\lambda L}\bar{b}}(\bar{a}{{\rm{e}}}^{{\rm{i}}\lambda L}-a{{\rm{e}}}^{-{\rm{i}}\lambda L}-{\rm{i}}\sqrt{4-{\bigtriangleup }^{2}}),\end{eqnarray}$
where △(λ) is given by $\begin{eqnarray}\bigtriangleup =a{{\rm{e}}}^{-{\rm{i}}\lambda L}+\bar{a}{{\rm{e}}}^{{\rm{i}}\lambda L},\quad \lambda \in {\mathbb{C}}.\end{eqnarray}$
We define a(λ) and b(λ) in terms of the initial datum (3.1 ),
$\begin{eqnarray}a(\lambda )={{\rm{\Psi }}}_{22}(0,\lambda ),\quad b(\lambda )={{\rm{\Psi }}}_{12}(0,\lambda ),\end{eqnarray}$
where $\Psi$12(x, λ) and $\Psi$22(x, λ) satisfy the following equation: $\begin{eqnarray}\left\{\begin{array}{l}{{\rm{\Psi }}}_{12x}+2{\rm{i}}\lambda {\rm{\Psi }}12={q}_{0}{{\rm{e}}}^{\tfrac{2{\rm{i}}\pi N}{L}x}{{\rm{\Psi }}}_{22},\quad x\in [0,L],\quad \lambda \in {\mathbb{C}},\\ {{\rm{\Psi }}}_{22x}=-{q}_{0}{{\rm{e}}}^{\tfrac{-2{\rm{i}}\pi N}{L}x}{{\rm{\Psi }}}_{12},\end{array}\right.\end{eqnarray}$
with $\Psi$22(L, λ) = 1, $\Psi$22x(L, λ) = 0. By direct integration of the above equation, we get the expressions of the spectral functions a(λ) and b(λ): $\begin{eqnarray}\begin{array}{rcl}a(\lambda ) & = & \displaystyle \frac{{{\rm{e}}}^{{\rm{i}}(\lambda L+\pi N)}({Lr}(\lambda )\cos ({Lr}(\lambda ))-{\rm{i}}(\lambda L+\pi N)\sin ({Lr}(\lambda )))}{{Lr}(\lambda )},\\ b(\lambda ) & = & \displaystyle \frac{-{q}_{0}{{\rm{e}}}^{{\rm{i}}(\lambda L+\pi N)}\sin ({Lr}(\lambda ))}{r(\lambda )},\end{array}\end{eqnarray}$
where $\begin{eqnarray}r(\lambda )=\sqrt{{\left(\lambda +\displaystyle \frac{\pi N}{L}\right)}^{2}+{{q}_{0}}^{2}}.\end{eqnarray}$
It follows that $\begin{eqnarray}\bigtriangleup (\lambda )=2{\left(-1\right)}^{N}\cos ({Lr}(\lambda )).\end{eqnarray}$
Even though r(λ) has a branch cut, we note that a(λ), b(λ) and △ are entire functions. The zeros of $4-{\bigtriangleup }^{2}\,=4(\sin ({Lr}(\lambda ))$ consist of the two simple zeros λ± defined by $\begin{eqnarray}{\lambda }^{\pm }=-\displaystyle \frac{\pi N}{L}\pm {\rm{i}}{q}_{0},\end{eqnarray}$
as well as the infinite double zeros $\begin{eqnarray}-\displaystyle \frac{\pi N}{L}\pm \sqrt{\displaystyle \frac{{n}^{2}{\pi }^{2}}{{L}^{2}}-{{q}_{0}}^{2}},\quad n\in {\mathbb{Z}}\setminus \{0\}.\end{eqnarray}$
The zeros are real for $| n| \geqslant \tfrac{{{Lq}}_{0}}{\pi }$ and non-real for $| n| \leqslant \tfrac{{{Lq}}_{0}}{\pi }$. The function $\sqrt{4-{\bigtriangleup }^{2}}$ is single-valued on ${\mathbb{C}}\setminus { \mathcal C }$, where ${ \mathcal C }$ consists of the single branch cut ${ \mathcal C }=({\lambda }^{-},{\lambda }^{+})$. We fix the branch in the definition of r(λ) so that r(λ): ${\mathbb{C}}\setminus { \mathcal C }\to {\mathbb{C}}$ is ananlytic and $r(\lambda )=\lambda +\tfrac{\pi N}{L}+O(\tfrac{1}{\lambda })$ as λ → ∞ . Then we have $\begin{eqnarray}\sqrt{4-{\bigtriangleup }^{2}}=2{\left(-1\right)}^{N}\sin ({Lr}(\lambda )),\end{eqnarray}$
and by definition of the function $\tilde{{\rm{\Gamma }}}$: ${\mathbb{C}}\setminus { \mathcal C }\to {\mathbb{C}}$, we get $\begin{eqnarray}\tilde{{\rm{\Gamma }}}(\lambda )=\displaystyle \frac{{\rm{i}}(\lambda -r(\lambda )+\tfrac{\pi N}{L})}{{q}_{0}},\end{eqnarray}$
note that $\tilde{{\rm{\Gamma }}}(\lambda )$ has no poles.In order to define a RH problem which depends on $\tilde{{\rm{\Gamma }}}$ instead of Γ, where $\tilde{{\rm{\Gamma }}}$ is independent of T, we defined the function g(λ) by
$\begin{eqnarray}\begin{array}{rcl}{g}_{1} & = & \left(\begin{array}{cc}\displaystyle \frac{a-\bar{b}{\rm{\Gamma }}{{\rm{e}}}^{2{\rm{i}}\lambda L}}{a-\bar{b}\tilde{{\rm{\Gamma }}}{{\rm{e}}}^{2{\rm{i}}\lambda L}} & (\tilde{{\rm{\Gamma }}}-{\rm{\Gamma }}){{\rm{e}}}^{2{\rm{i}}\lambda L}{{\rm{e}}}^{-2{\rm{i}}\theta }\\ 0 & \displaystyle \frac{a-\bar{b}\tilde{{\rm{\Gamma }}}{{\rm{e}}}^{2{\rm{i}}\lambda L}}{a-\bar{b}{\rm{\Gamma }}{{\rm{e}}}^{2{\rm{i}}\lambda L}}\end{array}\right),\\ {g}_{2} & = & \left(\begin{array}{cc}1 & 0\\ \displaystyle \frac{(\bar{{\rm{\Gamma }}}-\bar{\tilde{{\rm{\Gamma }}}}){{\rm{e}}}^{2{\rm{i}}\theta }}{(a+b\bar{{\rm{\Gamma }}})(a+b\bar{\tilde{{\rm{\Gamma }}}})} & 1\end{array}\right),\\ {g}_{3} & = & \left(\begin{array}{cc}1 & \displaystyle \frac{(\tilde{{\rm{\Gamma }}}-{\rm{\Gamma }}){{\rm{e}}}^{-2{\rm{i}}\theta }}{(\bar{a}+\bar{b}{\rm{\Gamma }})(\bar{a}+\bar{b}\tilde{{\rm{\Gamma }}})}\\ 0 & 1\end{array}\right),\\ {g}_{4} & = & \left(\begin{array}{cc}\displaystyle \frac{\bar{a}-b\bar{\tilde{{\rm{\Gamma }}}}{{\rm{e}}}^{-2{\rm{i}}\lambda L}}{\bar{a}-b\bar{{\rm{\Gamma }}}{{\rm{e}}}^{-2{\rm{i}}\lambda L}} & 0\\ (\bar{{\rm{\Gamma }}}-\bar{\tilde{{\rm{\Gamma }}}}){{\rm{e}}}^{-2{\rm{i}}\lambda L}{{\rm{e}}}^{2{\rm{i}}\theta } & \displaystyle \frac{\bar{a}-b\bar{{\rm{\Gamma }}}{{\rm{e}}}^{-2{\rm{i}}\lambda L}}{\bar{a}-b\bar{\tilde{{\rm{\Gamma }}}}{{\rm{e}}}^{-2{\rm{i}}\lambda L}}\end{array}\right),\end{array}\end{eqnarray}$
where gj is defined in the region Dj, for j = 1, ⋯ ,4. We defined $\tilde{m}(x,t,\lambda )$ by $\begin{eqnarray}\tilde{m}={mg}.\end{eqnarray}$
The $\tilde{m}$ satisfies the jump relation ${\tilde{m}}_{-}={\tilde{m}}_{+}\tilde{\upsilon }$ on ${\rm{\Sigma }}\setminus { \mathcal C }$. We can get the jump matrix $\tilde{\upsilon }(x,t,\lambda )$, it follows that $\begin{eqnarray}\begin{array}{rcl}{\tilde{\upsilon }}_{1} & = & \left(\begin{array}{cc}\displaystyle \frac{-2r(\lambda -r+\pi N/L)}{{q}_{0}^{2}} & \displaystyle \frac{-{\rm{i}}(\lambda -r+\pi N/L){{\rm{e}}}^{-2{\rm{i}}(\theta -\lambda L)}}{{q}_{0}}\\ \displaystyle \frac{{\rm{i}}(\lambda -r+\pi N/L){{\rm{e}}}^{2{\rm{i}}(\theta -\lambda L+{Lr})}}{{q}_{0}} & \displaystyle \frac{(\lambda +\pi N/L)(1-{{\rm{e}}}^{2{\rm{i}}{Lr}})+r(1+{{\rm{e}}}^{2{\rm{i}}{Lr}})}{2r}\end{array}\right),\,\lambda \in {{\rm{\Sigma }}}_{1},\\ {\tilde{\upsilon }}_{2} & = & \left(\begin{array}{cc}\displaystyle \frac{-2r(\lambda -r+\pi N/L)}{{q}_{0}^{2}} & \displaystyle \frac{-{\rm{i}}(\lambda -r+\pi N/L){{\rm{e}}}^{-2{\rm{i}}(\theta -\lambda L+{Lr})}}{{q}_{0}}\\ \displaystyle \frac{{\rm{i}}(\lambda -r+\pi N/L){{\rm{e}}}^{2{\rm{i}}(\theta -\lambda L+{Lr})}}{{q}_{0}} & 1\end{array}\right),\,\lambda \in {{\rm{\Sigma }}}_{2},\\ {\tilde{\upsilon }}_{3} & = & \left(\begin{array}{cc}\displaystyle \frac{-2r(\lambda -r+\pi N/L)}{{q}_{0}^{2}} & \displaystyle \frac{-{\rm{i}}(\lambda -r+\pi N/L){{\rm{e}}}^{-2{\rm{i}}(\theta -\lambda L+{Lr})}}{{q}_{0}}\\ \displaystyle \frac{{\rm{i}}(\lambda -r+\pi N/L){{\rm{e}}}^{2{\rm{i}}(\theta -\lambda L)}}{{q}_{0}} & \displaystyle \frac{(\lambda +\pi N/L)(1-{{\rm{e}}}^{-2{\rm{i}}{Lr}})+r(1+{{\rm{e}}}^{-2{\rm{i}}{Lr}})}{2r}\end{array}\right),\,\lambda \in {{\rm{\Sigma }}}_{3},\\ {\tilde{\upsilon }}_{4} & = & \left(\begin{array}{cc}\displaystyle \frac{-2r(\lambda -r+\pi N/L)}{{q}_{0}^{2}} & \displaystyle \frac{-{\rm{i}}(\lambda -r+\pi N/L){{\rm{e}}}^{-2{\rm{i}}(\theta -\lambda L)}}{{q}_{0}}\\ \displaystyle \frac{{\rm{i}}(\lambda -r+\pi N/L){{\rm{e}}}^{2{\rm{i}}(\theta -\lambda L)}}{{q}_{0}} & 1\end{array}\right),\,\lambda \in {{\rm{\Sigma }}}_{4}.\end{array}\end{eqnarray}$
However, because of the square root $\sqrt{4-{\bigtriangleup }^{2}}$, the $\tilde{m}$ also have jumps across the contours ${ \mathcal C }\cap {D}_{j},j=1,\cdots ,4$, we assume that N ≠ 0 in the following section. The case of the N = 0 see the remark 1. Then, we have $\begin{eqnarray}\begin{array}{rcl}{\tilde{\upsilon }}_{{{\rm{D}}}_{1}}^{\mathrm{cut}} & = & \left(\begin{array}{cc}{{\rm{e}}}^{-2{\rm{i}}{Lr}} & \displaystyle \frac{2{\rm{i}}\rho (\lambda ){{\rm{e}}}^{-2{\rm{i}}(\theta -\lambda L)}}{{q}_{0}}\\ 0 & {{\rm{e}}}^{2{\rm{i}}{Lr}}\end{array}\right),\,\lambda \in { \mathcal C }\cap {D}_{1},\\ {\tilde{\upsilon }}_{{{\rm{D}}}_{2}}^{\mathrm{cut}} & = & \left(\begin{array}{cc}1 & 0\\ \displaystyle \frac{2{\rm{i}}\rho (\lambda ){{\rm{e}}}^{2{\rm{i}}(\theta -\lambda L)}}{{q}_{0}} & 1\end{array}\right),\,\lambda \in { \mathcal C }\cap {D}_{2},\\ {\tilde{\upsilon }}_{{{\rm{D}}}_{3}}^{\mathrm{cut}} & = & \left(\begin{array}{cc}1 & \displaystyle \frac{2{\rm{i}}\rho (\lambda ){{\rm{e}}}^{-2{\rm{i}}(\theta -\lambda L)}}{{q}_{0}}\\ 0 & 1\end{array}\right),\,\lambda \in { \mathcal C }\cap {D}_{3},\\ {\tilde{\upsilon }}_{{{\rm{D}}}_{4}}^{\mathrm{cut}} & = & \left(\begin{array}{cc}{{\rm{e}}}^{-2{\rm{i}}{Lr}} & 0\\ \displaystyle \frac{2{\rm{i}}\rho (\lambda ){{\rm{e}}}^{2{\rm{i}}(\theta -\lambda L)}}{{q}_{0}} & {{\rm{e}}}^{2{\rm{i}}{Lr}}\end{array}\right),\,\lambda \in { \mathcal C }\cap {D}_{4},\end{array}\end{eqnarray}$
where $\rho (\lambda )=\sqrt{{{q}_{0}}^{2}+{\left(\lambda +\pi N/L\right)}^{2}}\geqslant 0$.We draw the following figure according to all possible distributions of branch cut ${ \mathcal C }$, combined with the direction of the jump curve of the RH problem. See figure 2 for details.
If $\tfrac{-\pi N}{L}\in [\tfrac{-\sqrt{2}}{4},0)\cup (0,\tfrac{\sqrt{2}}{4}]$, we show that in the case of the single exponential initial profile (3.1 ), the RH problem for $\tilde{m}$ can be constructed as follows:
If $\tfrac{-\pi N}{L}\in [\tfrac{-\sqrt{2}}{4},0)$,
| • | $\tilde{m}(x,t,\lambda )$ is analytic in ${\mathbb{C}}\setminus \tilde{{\rm{\Sigma }}}$, and continuous on $\tilde{{\rm{\Sigma }}}\setminus \{0,\pm \tfrac{\sqrt{2}}{4},\tfrac{-\pi N}{L},{\lambda }^{\pm }\}$. |
| • | ${\tilde{m}}_{-}(x,t,\lambda )={\tilde{m}}_{+}(x,t,\lambda )\tilde{\upsilon }(x,t,\lambda ),\lambda \in \tilde{{\rm{\Sigma }}}\setminus \{0,\pm \tfrac{\sqrt{2}}{4},\tfrac{-\pi N}{L},{\lambda }^{\pm }\}$. |
| • | $\tilde{m}(x,t,\lambda )=I\,+\,O(\tfrac{1}{\lambda }),\lambda \to \infty ,\lambda \in {\mathbb{C}}\setminus {\bigcup }_{n\in {\mathbb{Z}}}{{ \mathcal E }}_{n}$. |
| • | $\tilde{m}(x,t,\lambda )=O(1),\lambda \to \{0,\pm \tfrac{\sqrt{2}}{4},\tfrac{-\pi N}{L},{\lambda }^{\pm }\}.$ |
$\begin{eqnarray}\tilde{\upsilon }=\left\{\begin{array}{l}{\tilde{\upsilon }}_{1},\quad \lambda \in {{\rm{\Sigma }}}_{1},\\ {\tilde{\upsilon }}_{2},\quad \lambda \in {{\rm{\Sigma }}}_{2},\\ {\tilde{\upsilon }}_{3},\quad \lambda \in {{\rm{\Sigma }}}_{3},\\ {\tilde{\upsilon }}_{4},\quad \lambda \in {{\rm{\Sigma }}}_{4},\\ {\tilde{\upsilon }}_{{{\rm{D}}}_{2}}^{\mathrm{cut}},\quad \lambda \in (-\displaystyle \frac{\pi N}{L},-\displaystyle \frac{\pi N}{L}+{\rm{i}}{q}_{0}),\\ {\tilde{\upsilon }}_{{{\rm{D}}}_{3}}^{\mathrm{cut}},\quad \lambda \in (-\displaystyle \frac{\pi N}{L}-{\rm{i}}{q}_{0},-\displaystyle \frac{\pi N}{L}).\end{array}\right.\end{eqnarray}$
If $\tfrac{-\pi N}{L}\in (0,\tfrac{\sqrt{2}}{4}]$,3.27 ) and equation (3.28 ). It is worth noting that all these jumps can be completely represented by a(λ) and b(λ).
$\begin{eqnarray}\tilde{\upsilon }=\left\{\begin{array}{l}{\tilde{\upsilon }}_{1},\quad \lambda \in {{\rm{\Sigma }}}_{1},\\ {\tilde{\upsilon }}_{2},\quad \lambda \in {{\rm{\Sigma }}}_{2},\\ {\tilde{\upsilon }}_{3},\quad \lambda \in {{\rm{\Sigma }}}_{3},\\ {\tilde{\upsilon }}_{4},\quad \lambda \in {{\rm{\Sigma }}}_{4},\\ {\tilde{\upsilon }}_{{{\rm{D}}}_{1}}^{\mathrm{cut}},\quad \lambda \in (-\displaystyle \frac{\pi N}{L},-\displaystyle \frac{\pi N}{L}+{\rm{i}}{q}_{0}),\\ {\tilde{\upsilon }}_{{{\rm{D}}}_{4}}^{\mathrm{cut}},\quad \lambda \in (-\displaystyle \frac{\pi N}{L}-{\rm{i}}{q}_{0},-\displaystyle \frac{\pi N}{L}),\end{array}\right.\end{eqnarray}$
where $\tilde{{\rm{\Sigma }}}={\rm{\Sigma }}\bigcup { \mathcal C }$, ${{ \mathcal E }}_{n}=\{\lambda \in {\mathbb{C}}:| \lambda -\tfrac{n\pi }{L}| \lt \tfrac{\pi }{4L}\}$. ${\{{\tilde{\upsilon }}_{j}\}}_{1}^{4}$, ${\tilde{\upsilon }}_{{{\rm{D}}}_{1}}^{\mathrm{cut}}$ ${\tilde{\upsilon }}_{{{\rm{D}}}_{2}}^{\mathrm{cut}}$ ${\tilde{\upsilon }}_{{{\rm{D}}}_{3}}^{\mathrm{cut}}$ and ${\tilde{\upsilon }}_{{{\rm{D}}}_{4}}^{\mathrm{cut}}$ are defined by equation (If $-\tfrac{\pi N}{L}\in {\mathbb{R}}\setminus [\tfrac{-\sqrt{2}}{4},\tfrac{\sqrt{2}}{4}]$, therefore the formulation involves the jump matrices ${\tilde{\upsilon }}_{{{\rm{D}}}_{1}}^{\mathrm{cut}}$ ${\tilde{\upsilon }}_{{{\rm{D}}}_{2}}^{\mathrm{cut}}$ ${\tilde{\upsilon }}_{{{\rm{D}}}_{3}}^{\mathrm{cut}}$ and ${\tilde{\upsilon }}_{{{\rm{D}}}_{4}}^{\mathrm{cut}}$. We show that in the case of the single exponential initial profile (3.1 ), the RH problem for $\tilde{m}$ can be constructed as follows:
Here ${\lambda }_{1}^{\pm }=\pm \tfrac{16{N}^{2}{\pi }^{2}-2{L}^{2}}{4L}$,
| • | $\tilde{m}(x,t,\lambda )$ is analytic in ${\mathbb{C}}\setminus \tilde{{\rm{\Sigma }}}$, and continuous on $\tilde{{\rm{\Sigma }}}\setminus \{0,\pm \tfrac{\sqrt{2}}{4},\tfrac{-\pi N}{L},{\lambda }^{\pm },{\lambda }_{1}^{\pm }\}$. |
| • | ${\tilde{m}}_{-}(x,t,\lambda )={\tilde{m}}_{+}(x,t,\lambda )\tilde{\upsilon }(x,t,\lambda ),\,\lambda \in \tilde{{\rm{\Sigma }}}\setminus \{0,\pm \tfrac{\sqrt{2}}{4},\tfrac{-\pi N}{L},{\lambda }^{\pm },{\lambda }_{1}^{\pm }\}$. |
| • | $\tilde{m}(x,t,\lambda )=I\,+\,O(\tfrac{1}{\lambda }),\lambda \to \infty ,\lambda \in {\mathbb{C}}\setminus {\bigcup }_{n\in {\mathbb{Z}}}{{ \mathcal E }}_{n}$. |
| • | $\tilde{m}(x,t,\lambda )=O(1),\lambda \to \{0,\pm \tfrac{\sqrt{2}}{4},\tfrac{-\pi N}{L},{\lambda }^{\pm },{\lambda }_{1}^{\pm }\}.$ |
$\begin{eqnarray}\tilde{\upsilon }=\left\{\begin{array}{l}{\tilde{\upsilon }}_{1},\quad \lambda \in {{\rm{\Sigma }}}_{1},\\ {\tilde{\upsilon }}_{2},\quad \lambda \in {{\rm{\Sigma }}}_{2},\\ {\tilde{\upsilon }}_{3},\quad \lambda \in {{\rm{\Sigma }}}_{3},\\ {\tilde{\upsilon }}_{4},\quad \lambda \in {{\rm{\Sigma }}}_{4},\\ {\tilde{\upsilon }}_{{{\rm{D}}}_{1}}^{\mathrm{cut}},\quad \lambda \in { \mathcal C }\cap {D}_{1},\\ {\tilde{\upsilon }}_{{{\rm{D}}}_{2}}^{\mathrm{cut}},\quad \lambda \in { \mathcal C }\cap {D}_{2},\\ {\tilde{\upsilon }}_{{{\rm{D}}}_{3}}^{\mathrm{cut}},\quad \lambda \in { \mathcal C }\cap {D}_{3},\\ {\tilde{\upsilon }}_{{{\rm{D}}}_{4}}^{\mathrm{cut}},\quad \lambda \in { \mathcal C }\cap {D}_{4}.\end{array}\right.\end{eqnarray}$
The above RH problem has a unique solution $\tilde{m}(x,t,\lambda )$ for each (x, t) ∈ [0, L] × [0, ∞ ). The solution q(x, t) of the equation (1.1 ) can be obtained from $\tilde{m}$ via the equation
$\begin{eqnarray}q(x,t)=2{\rm{i}}\mathop{\mathrm{lim}}\limits_{\lambda \to \infty }\lambda {\tilde{m}}_{12}(x,t,\lambda ).\end{eqnarray}$
3.3. Solution of the RH problem for $\tilde{m}$
In this section, We will show that the RH problem of $\tilde{m}$ can be clearly solved by transforming it to a RH problem, which has a constant off-diagonal across on the branch cut ${ \mathcal C }=(-\tfrac{\pi N}{L}-{\rm{i}}{q}_{0},-\tfrac{\pi N}{L}+{\rm{i}}{q}_{0})$.
We get that the jump matrices ${\tilde{\upsilon }}_{1}$ and ${\tilde{\upsilon }}_{3}$ satisfy the following factorizations:3.27 ) and (3.28 ), straightforward computations show that $\hat{m}$ only has a jump across $({\lambda }^{-},{\lambda }^{+})=(-\tfrac{\pi N}{L}-{\rm{i}}{q}_{0},-\tfrac{\pi N}{L}+{\rm{i}}{q}_{0})$.
$\begin{eqnarray}\begin{array}{rcl}{\tilde{\upsilon }}_{1} & = & \left(\begin{array}{cc}1 & 0\\ \displaystyle \frac{-{\rm{i}}{q}_{0}{{\rm{e}}}^{2{\rm{i}}(\theta -\lambda L+{Lr})}}{2r} & 1\end{array}\right)\\ & & \times \left(\begin{array}{cc}\displaystyle \frac{-2r(\lambda -r+\pi N/L)}{{q}_{0}^{2}} & \displaystyle \frac{-{\mathrm{ie}}^{-2{\rm{i}}(\theta -\lambda L)(\lambda -r+\pi N/L)}}{{q}_{0}}\\ 0 & \displaystyle \frac{\lambda +r+\pi N/L}{2r}\end{array}\right),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{\tilde{\upsilon }}_{3} & = & \left(\begin{array}{cc}1 & 0\\ \displaystyle \frac{-{\rm{i}}{q}_{0}{{\rm{e}}}^{2{\rm{i}}(\theta -\lambda L)}}{2r} & 1\end{array}\right)\\ & & \times \left(\begin{array}{cc}\displaystyle \frac{-2r(\lambda -r+\pi N/L)}{{q}_{0}^{2}} & \displaystyle \frac{-{\mathrm{ie}}^{-2{\rm{i}}(\theta -\lambda L+{Lr})(\lambda -r+\pi N/L)}}{{q}_{0}}\\ 0 & \displaystyle \frac{\lambda +r+\pi N/L}{2r}\end{array}\right).\end{array}\,\end{eqnarray}$
Then we define a new solution $\hat{m}$ by $\begin{eqnarray}\begin{array}{l}\hat{m}=\tilde{m}\times \\ \left\{\begin{array}{l}\left(\begin{array}{cc}1 & 0\\ \displaystyle \frac{-{\rm{i}}{q}_{0}{{\rm{e}}}^{2{\rm{i}}(\theta -\lambda L+{Lr})}}{2r} & 1\end{array}\right),\quad \quad \quad \quad \quad \quad \quad \quad \quad \lambda \in {D}_{1},\\ {\left(\begin{array}{cc}\displaystyle \frac{-2r(\lambda -r+\pi N/L)}{{q}_{0}^{2}} & \displaystyle \frac{-{\mathrm{ie}}^{-2{\rm{i}}(\theta -\lambda L)(\lambda -r+\pi N/L)}}{{q}_{0}}\\ 0 & \displaystyle \frac{\lambda +r+\pi N/L}{2r}\end{array}\right)}^{-1},\lambda \in {D}_{2},\\ \left(\begin{array}{cc}1 & 0\\ \displaystyle \frac{-{\rm{i}}{q}_{0}{{\rm{e}}}^{2{\rm{i}}(\theta -\lambda L)}}{2r} & 1\end{array}\right),\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \lambda \in {D}_{3},\\ {\left(\begin{array}{cc}\displaystyle \frac{-2r(\lambda -r+\pi N/L)}{{q}_{0}^{2}} & \displaystyle \frac{-{\mathrm{ie}}^{-2{\rm{i}}(\theta -\lambda L+{Lr})(\lambda -r+\pi N/L)}}{{q}_{0}}\\ 0 & \displaystyle \frac{\lambda +r+\pi N/L}{2r}\end{array}\right)}^{-1},\lambda \in {D}_{4}.\end{array}\right.\end{array}\,\end{eqnarray}$
Using the equations ($\hat{m}$ satisfies the following RH problem:
By performing another transformation, we find that the jump matrix equation (3.36 ) can be made constant (i.e. independent of λ). Define δ(λ) ≡ δ(x, t, λ) by
The unique solution of this problem is given explicitly by
| • | $\hat{m}(x,t,\lambda ):{\mathbb{C}}\setminus ({\lambda }^{-},{\lambda }^{+})\to {{\mathbb{C}}}^{2\times 2}$ is analytic. |
| • | $\hat{m}$ satisfies the jump condition $\begin{eqnarray}\begin{array}{l}{\hat{m}}_{-}(x,t,\lambda )={\hat{m}}_{+}(x,t,\lambda )\left(\begin{array}{cc}0 & f\\ -\displaystyle \frac{1}{f} & 0\end{array}\right),\\ \quad \lambda \in ({\lambda }^{-},{\lambda }^{+}),\end{array}\end{eqnarray}$ where f(λ) ≡ f(x, t, λ) is defined by $\begin{eqnarray}\begin{array}{l}f(\lambda )=\displaystyle \frac{2{\rm{i}}{r}_{+}(\lambda )}{{q}_{0}}{{\rm{e}}}^{-2{\rm{i}}(\theta -\lambda L)}\\ \quad =-\displaystyle \frac{2{\rm{i}}\sqrt{| {\left(\lambda +\pi N/L\right)}^{2}+{q}_{0}^{2}| }{{\rm{e}}}^{-2{\rm{i}}(\theta -\lambda L)}}{{q}_{0}}.\end{array}\end{eqnarray}$ |
| • | $\hat{m}=I+O(\tfrac{1}{\lambda }),\quad \lambda \to \infty $. |
$\begin{eqnarray}\delta (\lambda )={{\rm{e}}}^{\tfrac{r(\lambda )}{2\pi {\rm{i}}}{\int }_{{\lambda }^{-}}^{{\lambda }^{+}}\displaystyle \frac{\mathrm{ln}f(s)}{{r}_{+}(s)(s-\lambda )}{\rm{d}}s},\quad \lambda \in {\mathbb{C}}\setminus ({\lambda }^{-},{\lambda }^{+}).\end{eqnarray}$
The function δ satisfies the jump relation δ+δ− = f on (λ−, λ+) and ${\mathrm{lim}}_{\lambda \to \infty }\delta ={\delta }_{\infty }={{\rm{e}}}^{-\tfrac{1}{2\pi {\rm{i}}}{\int }_{{\lambda }^{-}}^{{\lambda }^{+}}\tfrac{\mathrm{ln}f(s)}{{r}_{+}(s)}{\rm{d}}s}$, where the boundary values of a function h on a contour from the left and right will be denoted by h+ and h−, respectively. Moreover, we get $\begin{eqnarray}\begin{array}{rcl}\delta (\lambda ) & = & O({\left(\lambda -{\lambda }^{\pm }\right)}^{1/4}),\\ \delta {\left(\lambda \right)}^{-1} & = & O({\left(\lambda -{\lambda }^{\pm }\right)}^{-1/4}),\quad \lambda \to {\lambda }^{\pm }.\end{array}\end{eqnarray}$
Consequently, we defined the $\check{m}$ by $\check{m}={{\delta }_{\infty }}^{-{\sigma }_{3}}\hat{m}{\delta }^{{\sigma }_{3}}$, the $\check{m}$ satisfies the following RH problem:| • | $\check{m}(x,t,\lambda ):{\mathbb{C}}\setminus ({\lambda }^{-},{\lambda }^{+})\to {{\mathbb{C}}}^{2\times 2}$ is analytic. |
| • | $\check{m}$ satisfies the jump condition ${\check{m}}_{-}(x,t,\lambda )\,={\check{m}}_{+}(x,t,\lambda )\left(\begin{array}{cc}0 & 1\\ -1 & 0\end{array}\right),\quad \lambda \in ({\lambda }^{-},{\lambda }^{+}).$ |
| • | $\check{m}=I+O(\tfrac{1}{\lambda }),\quad \lambda \to \infty $, $\quad \check{m}=O({\left(\lambda -{\lambda }^{\pm }\right)}^{1/4}),\quad \lambda \to {\lambda }^{\pm }.$ |
$\begin{eqnarray}\begin{array}{rcl}\check{m} & = & \displaystyle \frac{1}{2}\left(\begin{array}{cc}\check{Q}+{\check{Q}}^{-1} & {\rm{i}}(\check{Q}-{\check{Q}}^{-1})\\ -{\rm{i}}(\check{Q}-{\check{Q}}^{-1}) & \check{Q}+{\check{Q}}^{-1}\end{array}\right),\\ \check{Q}(\lambda ) & = & {\left(\displaystyle \frac{\lambda -{\lambda }^{+}}{\lambda -{\lambda }^{-}}\right)}^{1/4},\end{array}\end{eqnarray}$
where the branch of $\check{Q}:{\mathbb{C}}\setminus ({\lambda }^{-},{\lambda }^{+})\to {\mathbb{C}}$ is such that $\check{Q}\sim 1$ as λ → ∞ . We can easily get that $\tilde{m}$ by inverting the transformations $\tilde{m}\mapsto \hat{m}\mapsto \check{m}$, and this provides an explicit solution of the RH problem for $\tilde{m}$.By the equation (3.32 ) and the explicit formula equation (3.40 ), we can obtain the3.42 ) involving x − L and t can be obtained by opening the contour and deforming the contour to infinity:3.41 ), we find that the solution q(x, t) of equation (1.1 ) corresponding to the initial datum $q(x,0)={q}_{0}{{\rm{e}}}^{\tfrac{2{\rm{i}}\pi N}{L}x}$ is given by 3.48 ) is consistent with equation (3.47 ) when N = 0.
$\begin{eqnarray}\begin{array}{l}q(x,t)=2{\rm{i}}\mathop{\mathrm{lim}}\limits_{\lambda \to \infty }\lambda \,{\hat{m}}_{12}(x,t,\lambda )\\ \quad =2{\rm{i}}{\delta }_{\infty }^{2}\mathop{\mathrm{lim}}\limits_{\lambda \to \infty }\lambda \,{\check{m}}_{12}(x,t,\lambda )=\displaystyle \frac{{\lambda }^{+}-{\lambda }^{-}}{2}{\delta }_{\infty }^{2}.\end{array}\end{eqnarray}$
In order to calculate δ∞, we note that $\begin{eqnarray}{\delta }_{\infty }={{\rm{e}}}^{{c}_{0}}{{\rm{e}}}^{{\int }_{{\lambda }^{-}}^{{\lambda }^{+}}\displaystyle \frac{s(x-L)}{\pi {r}_{+}(s)}{\rm{d}}s}{{\rm{e}}}^{{\int }_{{\lambda }^{-}}^{{\lambda }^{+}}\displaystyle \frac{(2{s}^{2}-8{s}^{4})t}{\pi {r}_{+}(s)}{\rm{d}}s},\end{eqnarray}$
where c0 is a constant $\begin{eqnarray}{c}_{0}=-\displaystyle \frac{1}{2\pi {\rm{i}}}{\int }_{{\lambda }^{-}}^{{\lambda }^{+}}\displaystyle \frac{\mathrm{ln}(2{\rm{i}}{r}_{+}(s))-\mathrm{ln}{q}_{0}}{{r}_{+}(s)}{\rm{d}}s.\end{eqnarray}$
The integrals in equation ( $\begin{eqnarray}\begin{array}{l}{\displaystyle \int }_{{\lambda }^{-}}^{{\lambda }^{+}}\displaystyle \frac{s(x-L)}{\pi {r}_{+}(s)}{\rm{d}}s=\displaystyle \frac{{\rm{i}}\pi N(x-L)}{L},\\ \quad {\displaystyle \int }_{{\lambda }^{-}}^{{\lambda }^{+}}\displaystyle \frac{(2{s}^{2}-8{s}^{4})t}{\pi {r}_{+}(s)}{\rm{d}}s={\rm{i}}({q}_{0}^{2}(1+3{q}_{0}^{2})+\displaystyle \frac{8{\pi }^{4}{N}^{4}}{{L}^{4}}\\ \quad -\displaystyle \frac{2{\pi }^{2}{N}^{2}}{{L}^{2}}(1+12{q}_{0}^{2}))t.\end{array}\end{eqnarray}$
Let $s=\tfrac{-\pi N}{L}+{\rm{i}}{q}_{0}\sin \theta $, we get $\begin{eqnarray}{c}_{0}=\displaystyle \frac{1}{\pi }{\int }_{0}^{\tfrac{\pi }{2}}\mathrm{ln}(-2\mathrm{icos}\theta ){\rm{d}}\theta =\displaystyle \frac{-\pi {\rm{i}}}{4}.\end{eqnarray}$
We finally get the expression of δ∞ as follows $\begin{eqnarray}{\delta }_{\infty }={{\rm{e}}}^{\tfrac{-\pi {\rm{i}}}{4}+\displaystyle \frac{{\rm{i}}\pi N(x-L)}{L}+{\rm{i}}({q}_{0}^{2}(1+3{q}_{0}^{2})+\displaystyle \frac{8{\pi }^{4}{N}^{4}}{{L}^{4}}-\displaystyle \frac{2{\pi }^{2}{N}^{2}}{{L}^{2}}(1+12{q}_{0}^{2}))t}.\end{eqnarray}$
Substituting the above expression of δ∞ into the equation ( $\begin{eqnarray}q(x,t)={q}_{0}{{\rm{e}}}^{\tfrac{2{\rm{i}}\pi {Nx}}{L}+2{\rm{i}}({q}_{0}^{2}(1+3{q}_{0}^{2})+\displaystyle \frac{8{\pi }^{4}{N}^{4}}{{L}^{4}}-\displaystyle \frac{2{\pi }^{2}{N}^{2}}{{L}^{2}}(1+12{q}_{0}^{2}))t}.\end{eqnarray}$
Meanwhile, it is easy to verify that this q(x, t) indeed satisfies the correct initial value problem. In the terminology of the finite-gap approach, this corresponds to one-gap solutions.We consider the case of constant initial data (N = 0, in equation (
$\begin{eqnarray}q(x,t)={q}_{0}{{\rm{e}}}^{2{\rm{i}}({q}_{0}^{2}(1+3{q}_{0}^{2}))t}.\end{eqnarray}$
It is worth pointing out that equation (