Dbar dressing method to nonlinear Schrödinger equation with nonzero boundary conditions

Nonlinear integrable equations with nonzero boundary condition (NZBC) have been well studied. Among the methods, the inverse scattering transform or the Riemann-Hilbert problem play an important role [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 29, 30, 28]. It is worth noting that Jaulent, Manna, et al. introduced the spatial transform method based on certain Dbar equation to study the integrable systems, such as KdV, Toda and AKNS hierarchy [31, 32, 33, 34].

The $\bar{\partial}$ (Dbar) problem is a powerful tools to study the nonlinear integrable equations, such as multidimensional equations [35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47], differential-difference equations [48, 49, 50, 47], (1+1) dimensional equations [38, 51, 52, 53, 54, 55, 56]. The Dbar-steepest descent method is developed to study the the asymptotic behavior [57, 58, 59]. The Dbar problem can also be used to consider the well-posedness of integrable equations [60, 61]. To our knowledge, very few of the nonlinear integrable equations with NZBCs are considered by the Dbar problem. we note that multi-lump solutions of KP equation with integrable boundary $u_{y}|_{y=0}=0$ via $\bar{\partial}$-dressing method were given in [62].

In this paper, we give a different view to know about the nonlinear integrable equations with NZBCs. As an example, we extend the Dbar approach to discuss the focusing and defocusing nonlinear Schrödinger (NLS) equation with NZBC. The associated theory is developed, and can also be used to discuss other nonlinear integrable equations with NZBCs. For convenience, we consider the NLS equation with nonzero boundary condition in the following form [63]

$$iq_{t}+q_{xx}-2\nu(|q|^{2}-q_{0}^{2})q=0,\quad\nu=\pm 1,$$

(1.1)

and

$$q(x,t)\rightarrow\rho,\quad|x|\to\infty,$$

(1.2)

where $\rho$ is a constant and $|\rho|=q_{0}\neq 0$. Equation (1.1) is the compatibility condition of the linear system

$$\varphi_{x}=U\varphi,\quad\varphi_{t}=V\varphi,$$

(1.3)

where

$$U=\left(\begin{matrix}ik&q\\ \nu\bar{q}&-ik\end{matrix}\right),\quad V=\left(\begin{matrix}-2ik^{2}-i\nu(|q|^{2}-q_{0}^{2})&-2kq+iq_{x}\\ -2k\nu\bar{q}-i\nu\bar{q}_{x}&2ik^{2}+i\nu(|q|^{2}-q_{0}^{2})\end{matrix}\right).$$

(1.4)

It is noted that the eigenvalues of the matrix $U_{0}=U(q=\rho)$ have double branches, and the associated spectral space for the nonlinear Schrödinger (NLS) equation with NZBC is multi-sheeted Riemann surface [9, 8]. To use the Dbar approach solving the NLS equation with nonzero boundary condition, one needs to transform the multi-sheeted Riemann surface into a Riemann sphere. This can be done by introducing the the uniformization variable $z$ defined by $z=k+\lambda$ and

$$\lambda(z)=\frac{1}{2}(z-\nu\frac{q_{0}^{2}}{z}),\quad k(z)=\frac{1}{2}(z+\nu\frac{q_{0}^{2}}{z}).$$

(1.5)

Hence, the eigenfunction of the spectral problem (1.3) as $q=\rho$ can be given as

$$\left(I+\frac{i}{z}\sigma_{3}Q_{0}\right)e^{i\theta(z;x,t)\sigma_{3}},$$

(1.6)

where

$\displaystyle Q_{0}=\left(\begin{matrix}0&\rho\\ \nu\bar{\rho}&0\end{matrix}\right),\quad\theta(z;x,t)=\lambda(z)(x-2k(z)t).$

(1.7)

In the following, we consider the Dbar problem in the extended complex $z$ plane. To do this, we construct an annulus with center at 0, that is, 0 and $\infty$ are outside the annulus. In Section 2, we introduce a special complex function which satisfies a Dbar problem in the annulus, and is meromorphic outside the annulus. Thus the Laurent series near the points 0 and $\infty$ play the role of non-canonical normalization conditions to the Dbar problem. The Dbar problem with normalization conditions is equivalent to an inhomogeneous integral equation, and the inhomogeneous terms are given by the normalization conditions. We present the following theorem proved in Appendix to fullfill the Dbar dressing.

Theorem    Suppose that $f(z)$ admits $\bar{\partial}f(z)\neq 0$ in $\mathbb{C}^{0}=\mathbb{C}\setminus\{0\}$. If $f(z)$ satisfies the following asymptotic behaviors in $\mathbb{C}^{*}=\mathbb{C}\cup\{\infty\}$

	$\displaystyle f(z)=\sum\limits_{j=1}^{m}\frac{a_{m-j}}{z^{j}}+O(1),\quad z\to 0,$		(1.8)
	$\displaystyle f(z)=\sum\limits_{j=0}^{n}b_{j}z^{j}+O(1/z),\quad z\to\infty,$

then for the circles $\Gamma_{R}=\{z:|z|=R>0\}$ and $\Gamma_{\varepsilon}=\{z:|z|=\varepsilon>0\}$

$$\frac{1}{2\pi i}\oint_{\Gamma_{R}+\Gamma_{\varepsilon}^{-}}\frac{f(k)}{k-z}{\rm d}k=\sum\limits_{j=0}^{n}b_{j}z^{j}+\sum\limits_{j=1}^{m}\frac{a_{m-j}}{z^{j}},\quad\varepsilon<|z|<R,$$

(1.9)

where $R\to\infty$ and $\varepsilon\to 0$.

The Dbar dressing method is based on the hypothesis that the homogeneous integral equation has only zero solution. To establish the relation between the NLS potential and the solution of the Dbar problem, we construct, in Section 3, the Lax pair of the NLS equation with NZBC. To this end, it is important to find two sets of operator which have same normalization conditions.

A special distribution (or spectral transform matrix) for the Dbar problem is introduced in Section 4 to construct the NLS equation under the nonzero boundary conditions and the conservation laws. The determinant of the associated eigenfunction (or the solution of the Dbar problem) is shown to be analytic in the annulus. In this procedure, we introduce a symmetry matrix function about the eigenfunction, and give its evolution equation in terms of the Lax pair. By substituting the expansion of the symmetry matrix function into the evolution equation and taking the $O(z^{l})$ terms, we find the NLS equation with nonzero boundary condition from the off-diagonal parts, and the conservation laws from the diagonal parts. We note that the AKNS hierarchy and infinite conservation laws are shown in [31].

In Section 5, the explicit solutions of the focusing and defocusing NLS equation with NZBC are obtained from two special distributions, which make sure the small norm of the operator in the integral equation associated with the Dbar problem. $N$-soliton solutions of the NLS equation with NZBC are given, and for the defocusing NLS equation, dark one-soliton and dark two-soliton are presented. We show that that the collision angle of dark two-soliton in $x$-$t$ plane is determined not only by the eigenvalues but also by the boundary condition.

The conclusions are given in Section 6. At last, the theory of the normalization part of the associated eigenfunctions are presented in the Appendixes.

We consider the following $\bar{\partial}$(Dbar)-problem

$$\bar{\partial}\chi(z;x,t):=\frac{\partial\chi(z;x,t)}{\partial\bar{z}}=\chi(z;x,t)r(z),\quad z\in\mathbb{C}^{0},$$

(2.1)

where $\chi(z;x,t),r(z)$ are $2\times 2$ matrices, the distribution $r(z)$ is independent of $x$ and $t$. To study the NLS equation with NZBC, we introduce the following normalization condition

	$\displaystyle\chi(z;x,t)$	$\displaystyle\sim e^{i\theta(z;x,t)\sigma_{3}},\quad z\rightarrow\infty,$		(2.2)
	$\displaystyle\chi(z;x,t)$	$\displaystyle\sim\frac{i}{z}\sigma_{3}Q_{0}e^{i\theta(z;x,t)\sigma_{3}},\quad z\rightarrow 0.$

We note that in the procedure of the inverse scattering transform, one can introduce the Jost functions which tend to (1.6) as $|x|\to\infty$. It is remarked that the condition (1.6) implies the normalization condition (2.2) for the Dbar problem.

For simplicity, we introduce a new function

$$\hat{\chi}(z;x,t)=\chi(z;x,t)e^{-i\theta(z;x,t)\sigma_{3}},$$

(2.3)

then it satisfies the asymptotic behavior

$$\hat{\chi}(z;x,t)\sim I,\quad z\rightarrow\infty,$$

(2.4)

$$\hat{\chi}(z;x,t)\sim\frac{i}{z}\sigma_{3}Q_{0},\quad z\rightarrow 0.$$

(2.5)

and the generalised Cauchy integral formula

$$\hat{\chi}(z)=\frac{1}{2\pi i}\int_{\Gamma_{R}+\Gamma_{\varepsilon}^{-}}\frac{\hat{\chi}(k)}{k-z}\mathrm{d}k+\frac{1}{2\pi i}\iint_{\varepsilon<|k|<R}\frac{\bar{\partial}\hat{\chi}(k)}{k-z}\mathrm{d}k\wedge\mathrm{d}{\bar{k}},$$

(2.6)

where $\Gamma_{R}$ and $\Gamma_{\varepsilon}$ are oriented circle with center at origin of $z$ plane and radius $R$ and $\varepsilon$, respectively. Here, $R\to\infty$ and $\varepsilon\to 0$. For simplicity, we define the first Cauchy integral on the right hand side of (2.6) as the normalization part of $\chi$, and denote it as $\mathcal{N}(\chi)$, that is

$$\mathcal{N}(\chi)=\frac{1}{2\pi i}\int_{\Gamma_{R}+\Gamma_{\varepsilon}^{-}}\frac{\hat{\chi}(k)}{k-z}\mathrm{d}k.$$

(2.7)

Then for $\chi$ in (2.2), we find from the Theorem that

$$\mathcal{N}(\chi)=I+\frac{i}{z}\sigma_{3}Q_{0}.$$

(2.8)

As $R\rightarrow\infty$ and $\varepsilon\rightarrow 0$, (2.6) reduces to

$$\hat{\chi}(z)=I+\frac{i}{z}\sigma_{3}Q_{0}+J\hat{\chi}(z),$$

(2.9)

where

$$J\hat{\chi}(z)=\frac{1}{2\pi i}\iint_{\mathbb{C}^{0}}\frac{\hat{\chi}(k)e^{i\theta(k)\sigma_{3}}r(k)e^{-i\theta(k)\sigma_{3}}}{k-z}\mathrm{d}{k}\wedge\mathrm{d}{\bar{k}}.$$

(2.10)

It is important to assume that the homogeneous equation of (2.9) only has zero solution, that is,

$$(I-J)f=0\Rightarrow f=0.$$

(2.11)

It is valid for small norm of the operator $J$.

Now we introduce the following solution space of the Dbar-problem (2.1) as

$$\mathcal{F}=\{\chi(z;x,t)|\bar{\partial}\chi(z;x,t)=\chi(z;x,t)r(z),\quad z\in\mathbb{C}^{0}\}.$$

(2.12)

In particular, let $\psi(x,;z)\in\mathcal{F}$ and $\mathcal{N}(\psi)=I+\frac{i}{z}\sigma_{3}Q_{0}$. Note that the distribution $r(z)$ is independent of the variables $x$ and $t$. To study the NLS equation with NZBC via the Dbar-problem (2.1), one needs to introduce certain constraint in the physic space. Here, we suppose

$$\psi(z;x,t)\sim\left(I+\frac{i}{z}\sigma_{3}Q_{0}\right)e^{i\theta(z;x,t)\sigma_{3}},\quad|x|\to\infty,$$

(2.13)

and

$$\psi(z;x,t)=\frac{i}{z}\psi(\nu\frac{q_{0}^{2}}{z};x,t)\sigma_{3}Q_{0}.$$

(2.14)

Then from (2.9), we know that $\psi(z;x,t)$ has the following asymptotic behaviors

$$\psi(z;x,t)=\left(I+\sum_{l=1}^{\infty}a_{l}(x,t)z^{-l}\right)e^{i\theta(z;x,t)\sigma_{3}},\quad z\to\infty,$$

(2.15)

$$\psi(z;x,t)=\left(\sum_{m=-1}^{\infty}b_{m}(x,t)z^{m}\right)e^{i\theta(z;x,t)\sigma_{3}},\quad z\to 0,$$

(2.16)

where

$$a_{l}(x,t)=\delta_{l,1}\cdot i\sigma_{3}Q_{0}-\frac{1}{2\pi i}\iint_{D}\psi(z;x,t)r(z)e^{-i\theta(z)\sigma_{3}}z^{l-1}\mathrm{d}z\wedge\mathrm{d}{\bar{z}},$$

(2.17)

and

$$b_{m}(x,t)=\left\{\begin{array}[]{ll}\delta_{m,0}+\frac{1}{2\pi i}\iint_{D}\psi(z;x,t)r(z)e^{-i\theta(z)\sigma_{3}}z^{-m-1}\mathrm{d}z\wedge\mathrm{d}{\bar{z}},&m\geq 0,\\ i\sigma_{3}Q_{0},&m=-1.\end{array}\right.$$

(2.18)

It is remarked that the coefficients $a_{l}(x,t)$ and $b_{m}(x,t)$ are not independent in terms of the symmetry condition (2.14). In fact,

$$b_{-1}=i\sigma_{3}Q_{0},\quad b_{m-1}(x,t)=\frac{i}{\nu^{m}q_{0}^{2m}}a_{m}(x,t)\sigma_{3}Q_{0}.$$

(2.19)

We note that equations (2.11) and (2.8) imply that, for $\chi_{1},\chi_{2}\in\mathcal{F}$,

$$\mathcal{N}(\chi_{1})=\mathcal{N}(\chi_{2})\Leftrightarrow\chi_{1}=\chi_{2}.$$

(3.1)

This result can be used to construct the Lax pair of NLS equation (1.1). In fact, Since the distribution $r(z)$ is independent of the variables $x$ and $t$, we know that $\alpha(z;x,t)\partial_{x}\psi+\beta(z;x,t)\partial_{t}\psi+A(z;x,t)\psi\in\mathcal{F}$, if $\psi\in\mathcal{F}$. Here $\alpha(z;x,t),\beta(z;x,t)$ and $A(z;x,t)$ are some $2\times 2$ matrices.

To obtain the spatial linear spectral problem, a little manipulation is needed. Let $\psi\in\mathcal{F}$. Here and after, $\psi=\psi(z;x,t)$. From (2.15), we find, at $z\to\infty$, that

	$\displaystyle\psi_{x}=$	$\displaystyle\left(\frac{i}{2}\sigma_{3}z+\frac{i}{2}a_{1}\sigma_{3}+\frac{i}{2}\sum_{l=1}^{\infty}a_{l+1}\sigma_{3}z^{-l}\right.$		(3.2)
		$\displaystyle\quad\left.+\frac{iq_{0}^{2}}{2}(\frac{\sigma_{3}}{z}+\sum_{l=2}^{\infty}a_{l-1}\sigma_{3}z^{-l})+\sum_{l=1}^{\infty}a_{l,x}z^{-l}\right)e^{i\theta(z)\sigma_{3}},$

$$ik\sigma_{3}\psi=\frac{i}{2}\left(\sigma_{3}z+\sigma_{3}a_{1}+\sum_{l=1}^{\infty}\sigma_{3}a_{l+1}z^{-l}-q_{0}^{2}(\frac{\sigma_{3}}{z}+\sum_{l=2}^{\infty}\sigma_{3}a_{l-1}z^{-l})\right)e^{i\theta(z)\sigma_{3}},$$

(3.3)

$$-\frac{i}{2}[\sigma_{3},a_{1}]\psi=\frac{i}{2}\left(-[\sigma_{3},a_{1}]-\sum_{l=1}^{\infty}[\sigma_{3},a_{1}]a_{l}z^{-l}\right)e^{i\theta(z)\sigma_{3}},$$

(3.4)

where $k=k(z)$ is defined in (1.5), then

$$\psi_{x}=\left(\frac{i}{2}\sigma_{3}z+\frac{i}{2}a_{1}\sigma_{3}+O(\frac{1}{z})\right)e^{i\theta(z)\sigma_{3}},\quad z\rightarrow\infty,$$

(3.5)

$\displaystyle ik\sigma_{3}\psi-\frac{i}{2}[\sigma_{3},a_{1}]\psi$

$\displaystyle=\left(\frac{i}{2}\sigma_{3}z+\frac{i}{2}a_{1}\sigma_{3}+O(\frac{1}{z})\right)e^{i\theta(z)\sigma_{3}},\quad z\rightarrow\infty,$

(3.6)

with $[\sigma_{3},a_{1}]=\sigma_{3}a_{1}-a_{1}\sigma_{3}$. Equations (3.5) and (3.6) imply that the Laurent series of $\partial_{x}\psi$ and $ik\sigma_{3}\psi-\frac{i}{2}[\sigma_{3},a_{1}]\psi$ at $z\to\infty$ share the same principal part.

Similarly, from (2.16), we know that, as $z\to 0$

	$\displaystyle\psi_{x}=\left({b_{-1,x}z^{-1}-\frac{i}{2}\nu q_{0}^{2}(b_{0}\sigma_{3}z^{-1}+b_{-1}\sigma_{3}z^{-2})+\sum_{m=0}^{\infty}b_{m,x}z^{m}}\right.$		(3.7)
	$\displaystyle\left.{+\frac{i}{2}\sum_{m=0}^{\infty}b_{m-1}\sigma_{3}z^{m}-\frac{i}{2}\nu q_{0}^{2}\sum_{m=0}^{\infty}b_{m+1}\sigma_{3}z^{m}}\right)e^{i\theta(z)\sigma_{3}},$

	$\displaystyle ik\sigma_{3}\psi=$	$\displaystyle\left(\frac{i}{2}\nu q_{0}^{2}(\sigma_{3}b_{-1}z^{-2}+\sigma_{3}b_{0}z^{-1})+\frac{i}{2}\sum_{m=0}^{\infty}\sigma_{3}b_{m-1}z^{m}\right.$		(3.8)
		$\displaystyle\quad\left.+\frac{i}{2}\nu q_{0}^{2}\sum_{m=0}^{\infty}\sigma_{3}b_{m+1}z^{m}\right)e^{i\theta(z)\sigma_{3}},$

		$\displaystyle-\frac{i}{2}\nu q_{0}^{2}(\sigma_{3}b_{0}+b_{0}\sigma_{3})b_{-1}^{-1}\psi=\left(-\frac{i}{2}\nu q_{0}^{2}(\sigma_{3}b_{0}+b_{0}\sigma_{3})z^{-1}\right.$		(3.9)
		$\displaystyle\qquad\left.-\frac{i}{2}\nu q_{0}^{2}(\sigma_{3}b_{0}+b_{0}\sigma_{3})b_{-1}^{-1}\sum_{m=0}^{\infty}b_{m}z^{m}\right)e^{i\theta(z)\sigma_{3}},$

and further

$$\psi_{x}=\left(-\frac{i}{2}\nu q_{0}^{2}(b_{-1}\sigma_{3}z^{-2}+b_{0}\sigma_{3}z^{-1})+O(1)\right)e^{i\theta(z)\sigma_{3}},\quad z\to 0,$$

(3.10)

		$\displaystyle ik\sigma_{3}\psi-\frac{i}{2}\nu q_{0}^{2}(\sigma_{3}b_{0}+b_{0}\sigma_{3})b_{-1}^{-1}\psi$		(3.11)
		$\displaystyle\quad=\left(-\frac{i}{2}\nu q_{0}^{2}(b_{-1}\sigma_{3}z^{-2}+b_{0}\sigma_{3}z^{-1})+O(1)\right)e^{i\theta(z)\sigma_{3}},\quad z\to 0,$

which implies that the Laurent series of $\partial_{x}\psi$ and $ik\sigma_{3}\psi+\frac{i}{2}q_{0}^{2}(\sigma_{3}b_{0}+b_{0}\sigma_{3})b_{-1}^{-1}\psi$ at $z=0$ share the same principal part.

Now, using the relation (2.19), we find that the coefficient of second item on the left hand side of (3.11) is equivalent to that of (3.6),

$$\frac{i}{2}\nu q_{0}^{2}(\sigma_{3}b_{0}+b_{0}\sigma_{3})b_{-1}^{-1}=\frac{i}{2}[\sigma_{3},a_{1}].$$

(3.12)

Since $\partial_{x}\psi$ defined by (3.5) and (3.10) belongs to the space $\mathcal{F}$, then, from (2.7) and the Theorem, we find

$$\mathcal{N}(\psi_{x})=\frac{i}{2}\sigma_{3}z+\frac{i}{2}a_{1}\sigma_{3}+\frac{i}{2}q_{0}^{2}(b_{0}\sigma_{3}z^{-1}+b_{-1}\sigma_{3}z^{-2}).$$

(3.13)

Similarly, for $ik\sigma_{3}\psi-\frac{i}{2}[\sigma_{3},a_{1}]\psi\in\mathcal{F}$ in terms of (3.6) and (3.11), we get

$$\mathcal{N}(\psi_{x})=\mathcal{N}(ik\sigma_{3}\psi-\frac{i}{2}[\sigma_{3},a_{1}]\psi),$$

(3.14)

which gives the spatial linear spectral problem

$$\psi_{x}=ik\sigma_{3}\psi+Q\psi,\quad Q=-\frac{i}{2}[\sigma_{3},a_{1}],$$

(3.15)

in view of the identity (3.1).

Substituting the expansion (2.15) into (3.15) and taking the $O(z^{-1})$ terms, we get the following equation

$$-\frac{i}{2}a_{2}\sigma_{3}=-i\nu q_{0}^{2}\sigma_{3}-\frac{i}{2}\sigma_{3}a_{2}+a_{1,x}-Qa_{1},$$

(3.16)

which can also be derived from equations (3.2)-(3.4). Equations (3.15) and (3.16) imply that

$$a_{1,x}=i\sigma_{3}(Q_{x}-Q^{2}+\nu q_{0}^{2}).$$

(3.17)

Similarly, substituting the expansion (2.16) into (3.15), and taking the $O(z^{0})$ terms, we get

$$\frac{i}{2}\nu q_{0}^{2}b_{1}\sigma_{3}=b_{0,x}+Q_{0}-\frac{i}{2}\nu q_{0}^{2}\sigma_{3}b_{1}-Qb_{0}.$$

(3.18)

Here $b_{-1}=i\sigma_{3}Q_{0}$ has been used.

Next, we will derive the temporal linear spectral problem. As $z\to\infty$, from (3.14), we have

	$\displaystyle\psi_{t}=$	$\displaystyle\left({-\frac{i}{2}\sigma_{3}z^{2}-\frac{i}{2}a_{1}\sigma_{3}z-\frac{i}{2}a_{2}\sigma_{3}+\sum_{l=1}^{\infty}a_{l,t}z^{-l}}\right.$		(3.19)
		$\displaystyle\left.{-\frac{i}{2}\sum_{l=1}^{\infty}a_{l+2}\sigma_{3}z^{-l}+\frac{i}{2}q_{0}^{4}(\frac{\sigma_{3}}{z^{2}}+\sum_{l=3}^{\infty}a_{l-2}\sigma_{3}z^{-l})}\right)e^{i\theta(z)\sigma_{3}},$

	$\displaystyle-2ik^{2}\sigma_{3}\psi=\left({-\frac{i}{2}\sigma_{3}z^{2}-\frac{i}{2}\sigma_{3}a_{1}z-i\nu q_{0}^{2}\sigma_{3}-\frac{i}{2}\sigma_{3}a_{2}-\frac{i}{2}\sigma_{3}\sum_{l=1}^{\infty}a_{l+2}z^{-l}}\right.$		(3.20)
	$\displaystyle\left.{-i\nu q_{0}^{2}\sigma_{3}\sum_{l=1}^{\infty}a_{l}z^{-l}-\frac{i}{2}q_{0}^{4}\sigma_{3}(\frac{1}{z^{2}}+\sum_{l=3}^{\infty}a_{l-2}z^{-l})}\right)e^{i\theta(z)\sigma_{3}},$

$$-2kQ\psi=\left(-Qz-Qa_{1}-\sum_{l=1}^{\infty}Qa_{l+1}z^{-l}-\nu q_{0}^{2}(\frac{Q}{z}+\sum_{l=2}^{\infty}Qa_{l-1}z^{-l})\right)e^{i\theta(z)\sigma_{3}},$$

(3.21)

$$i\sigma_{3}(Q_{x}-Q^{2}+\nu q_{0}^{2})\psi=\left(i\sigma_{3}(Q_{x}-Q^{2}-q_{0}^{2})+\sum_{l=1}^{\infty}i\sigma_{3}(Q_{x}-Q^{2}-q_{0}^{2})a_{l}z^{-l}\right)e^{i\theta(z)\sigma_{3}},$$

(3.22)

which imply that

$$\psi_{t}=\left(-\frac{i}{2}\sigma_{3}z^{2}-\frac{i}{2}a_{1}\sigma_{3}z-\frac{i}{2}a_{2}\sigma_{3}+O(\frac{1}{z})\right)e^{i\theta(z)\sigma_{3}},\quad z\rightarrow\infty,$$

(3.23)

and

		$\displaystyle-2ik^{2}\sigma_{3}\psi-2kQ\psi+i\sigma_{3}(Q_{x}-Q^{2}+\nu q_{0}^{2})\psi$		(3.24)
		$\displaystyle\quad=\left(-\frac{i}{2}\sigma_{3}z^{2}-\frac{i}{2}\sigma_{3}a_{1}z-Qz-Qa_{1}-i\nu q_{0}^{2}\sigma_{3}\right.$
		$\displaystyle\quad\qquad\left.-\frac{i}{2}\sigma_{3}a_{2}+i\sigma_{3}(Q_{x}-Q^{2}-q_{0}^{2})+O(\frac{1}{z})\right)e^{i\theta(z)\sigma_{3}},\quad z\rightarrow\infty.$

Using (3.15)-(3.17), equation (3.24) can be further reduced to

		$\displaystyle-2ik^{2}\sigma_{3}\psi-2kQ\psi+i\sigma_{3}(Q_{x}-Q^{2}+\nu q_{0}^{2})\psi$		(3.25)
		$\displaystyle\quad=\left(-\frac{i}{2}\sigma_{3}z^{2}-\frac{i}{2}a_{1}\sigma_{3}z-\frac{i}{2}a_{2}\sigma_{3}+O(\frac{1}{z})\right)e^{i\theta(z)\sigma_{3}},\quad z\rightarrow\infty.$

Equations (3.23) and (3.25) imply that the Laurent series of $\partial_{t}\psi$ and $-2ik^{2}\sigma_{3}\psi-2kQ\psi+i\sigma_{3}(Q_{x}-Q^{2}+\nu q_{0}^{2})\psi$ at $z\to\infty$ share the same principal part.

Using (2.19), (3.14) and (3.15), we find from (2.16) that

$$\psi_{t}=\left(\frac{i}{2}q_{0}^{4}(b_{1}\sigma_{3}z^{-1}+b_{0}\sigma_{3}z^{-2}+b_{-1}\sigma_{3}z^{-3})+O(1)\right)e^{i\theta(z)\sigma_{3}},\quad z\rightarrow 0,$$

(3.26)

and

		$\displaystyle-2ik^{2}\sigma_{3}\psi-2kQ\psi+i\sigma_{3}(Q_{x}-Q^{2}+\nu q_{0}^{2})\psi$		(3.27)
		$\displaystyle=\left(\frac{i}{2}q_{0}^{4}(b_{1}\sigma_{3}z^{-1}+b_{0}\sigma_{3}z^{-2}+b_{-1}\sigma_{3}z^{-3})+O(1)\right)e^{i\theta(z)\sigma_{3}},\quad z\rightarrow 0.$

Since $\partial_{t}\psi\in\mathcal{F}$ and has the asymptotic behaviors given by (3.23) and (3.26), then

	$\displaystyle\mathcal{N}(\psi_{t})=$	$\displaystyle-\frac{i}{2}\sigma_{3}z^{2}-\frac{i}{2}a_{1}\sigma_{3}z-\frac{i}{2}a_{2}\sigma_{3}$		(3.28)
		$\displaystyle+\frac{i}{2}q_{0}^{4}(b_{1}\sigma_{3}z^{-1}+b_{0}\sigma_{3}z^{-2}+b_{-1}\sigma_{3}z^{-3}).$

From (3.25) and (3.27), we find that

$$\mathcal{N}(\psi_{t})=\mathcal{N}\left(-2ik^{2}\sigma_{3}\psi-2kQ\psi+i\sigma_{3}(Q_{x}-Q^{2}+\nu q_{0}^{2})\psi\right).$$

(3.29)

From this equation and the fact that $-2ik^{2}\sigma_{3}\psi-2kQ\psi+i\sigma_{3}(Q_{x}-Q^{2}-q_{0}^{2})\psi\in\mathcal{F}$, we obtain the temporal linear spectral problem

$$\psi_{t}=-2ik^{2}\sigma_{3}\psi-2kQ\psi+i\sigma_{3}(Q_{x}-Q^{2}+\nu q_{0}^{2})\psi,$$

(3.30)

in view of identity (3.1).

To derive the focusing/defocusing NLS equation (1.1) with NZBC (1.2), we need to introduce the symmetry condition on the off-diagonal matrix $Q$ in (3.15) as

$$\sigma_{\nu}\bar{Q}\sigma_{\nu}=Q,\quad\sigma_{\nu}=\left\{\begin{matrix}\sigma_{2},&\nu=-1,\\ \sigma_{1},&\nu=1,\end{matrix}\right.,$$

(3.31)

which implies that the potential $Q$ takes the following form

$$Q=\left(\begin{matrix}0&q\\ \nu\bar{q}&0\end{matrix}\right),$$

(3.32)

and $\sigma_{\nu}\overline{U(x,t;\bar{z})}\sigma_{\nu}=U(z;x,t)$, where $U(z;x,t)=ik(z)\sigma_{3}+Q$. For the linear system (3.15) and (3.30) with the boundary condition (2.13), in addition to the first symmetry condition (2.14), the matrix eigenfunction $\psi(z;x,t)$ admits another symmetry condition

$$\psi(z;x,t)=\sigma_{\nu}\overline{\psi(x,t;\bar{z})}\sigma_{\nu}.$$

(3.33)

It is remarked that the first symmetry condition (2.14) plays the role of construction of the linear spectral problems (3.15) and (3.30), and the second symmetry condition (3.33) is about to curb the potential matrix $Q$. As a result, the focusing NLS equation (1.1), is equivalent to the compatibility condition of the linear system (3.15) and (3.30) with $Q$ given by (3.32).

In this section, we consider that the $2\times 2$ distribution $R(z;x,t)$ admits the following properties

(i) The matrix $R(z;x,t)$ has zero diagonal part;

(ii) The time evolution of $R(z;x,t)$ is $\partial_{t}R(z;x,t)=p(z;x,t)\sigma_{3}R(z;x,t)$, where $p$ is a scaler function

$$p(z;x,t)=2i\partial_{t}\theta(z)=-4i\lambda(z)k(z).$$

Here $\lambda(z),k(z)$ and $\theta(z)=\theta(z;x,t)$ are defined in (1.7).

Given the distribution, we can define a new Dbar problem

$$\bar{\partial}\hat{\chi}(z;x,t)=\hat{\chi}(z;x,t)R(z;x,t),\quad z\in\mathbb{C}^{0}.$$

(4.1)

and denote the associated solution space by $\mathcal{\hat{F}}$. It is verified that if $\hat{\chi}\in\mathcal{\hat{F}}$, then

$$V(z;x,t)\hat{\chi}\in\mathcal{\hat{F}},\quad\hat{\chi}_{t}+\frac{1}{2}p(z;x,t)\hat{\chi}\sigma_{3}\in\mathcal{\hat{F}},$$

(4.2)

for some matrix $V(z;x,t)=\sum_{n=-M}^{N}V_{n}(x,t)z^{n}$, where $N,M\in\mathbb{N}$. In fact, the first property is obviously. To prove the second property, we let $\hat{X}(z;x,t)=\partial_{t}\hat{\chi}+\frac{1}{2}p(z;x,t)\hat{\chi}\sigma_{3}$, then, using the properties (i) and (ii), we find for $z\in\mathbb{C}^{0}$

	$\displaystyle\bar{\partial}\hat{X}=$	$\displaystyle(\hat{\chi}_{t})R+p\hat{\chi}\sigma_{3}R+\frac{1}{2}p\hat{\chi}R\sigma_{3}$
	$\displaystyle=$	$\displaystyle(\hat{\chi}_{t})R+p\hat{\chi}\sigma_{3}R-\frac{1}{2}p\hat{\chi}\sigma_{3}R$
	$\displaystyle=$	$\displaystyle(\hat{\chi}_{t}+\frac{1}{2}p\hat{\chi}\sigma_{3})R=\hat{X}R.$