Nonlinear Schrödinger equation with nonzero boundary conditions revisited: Dbar approach

Abstract

The Dbar dressing method is extended to study the focusing/defocusing nonlinear Schrödinger (NLS) equation with nonzero boundary condition. A special type of complex function is considered. The function is meromorphic outside an annulus with center 0 and satisfies a local Dbar problem inside the annulus. The theory of such function is extended to construct the Lax pair of the NLS equation with nonzero boundary condition. In this procedure, the relation between the NLS potential and the solution of the Dbar problem is established. A certain distribution for the Dbar problem is introduced to obtain the focusing/defocusing NLS equation and the conservation laws.

Appendix

Proof of Lemma 1

Suppose C is a simple closed contour in D and encloses the circle $\Gamma _\varepsilon$. For all z in the region with boundary C, and admitting $|z-z_0|>\varepsilon$, we have

$$\begin{aligned} \frac{1}{2\pi i}\int _C\frac{f(k)}{k-z}\textrm{d} k=f(z)+\frac{1}{(m-1)!}\lim \limits _{k\rightarrow z_0}\frac{\textrm{d}^{m-1}}{\textrm{d} k^{m-1}}\frac{f(k)(k-z_0)^m}{k-z}. \end{aligned}$$

(A.1)

Since $z_0$ is a mth-order pole of f, and $0<|k-z_0|<\varepsilon$, then

$$\begin{aligned} (k-z_0)^mf(k)=\sum \limits _{j=0}^{m-1} a_{j}(k-z_0)^{j}+\sum \limits _{j=m}^\infty \tilde{a}_{j-m}(k-z_0)^{j}. \end{aligned}$$

In addition, for $|z-z_0|>\varepsilon$, we have

$$\begin{aligned} \frac{1}{k-z}=\frac{-1}{z-z_0}\frac{1}{1-\frac{k-z_0}{z-z_0}} =-\sum \limits _{l=0}^\infty \frac{(k-z_0)^l}{(z-z_0)^{l+1}},\quad 0<|k-z_0|<\varepsilon . \end{aligned}$$

Thus,

$$\begin{aligned} \frac{1}{(m-1)!}\lim \limits _{k\rightarrow z_0}\frac{\textrm{d}^{m-1}}{\textrm{d} k^{m-1}}\frac{f(k)(k-z_0)^m}{k-z} =-\sum \limits _{j=1}^{m}\frac{a_{m-j}}{(z-z_0)^{j}}. \end{aligned}$$

(A.2)

Substituting (A.2) into (A.1) implies that

$$\begin{aligned} \frac{1}{2\pi i}\int _C\frac{f(k)}{k-z}\textrm{d} k=f(z)-\sum \limits _{j=1}^{m}\frac{a_{m-j}}{(z-z_0)^{j}}. \end{aligned}$$

(A.3)

On the other hand, for the region with boundary C and $\Gamma _\varepsilon$, we apply the Cauchy’s formula and find

$$\begin{aligned} f(z)=\frac{1}{2\pi i}\int _{C+\Gamma _\varepsilon ^-}\frac{f(k)}{k-z}\textrm{d} k=\frac{1}{2\pi i}\int _C\frac{f(k)}{k-z}\textrm{d} k+\frac{1}{2\pi i}\int _{\Gamma _\varepsilon ^-}\frac{f(k)}{k-z}\textrm{d} k. \end{aligned}$$

(A.4)

The Lemma 1 is proved by comparing equations (A.3) and (A.4). It is noted that

$$\begin{aligned} \int _{\Gamma _\varepsilon ^-}\frac{f(k)}{k-z}\textrm{d} k=\lim \limits _{\varepsilon \rightarrow 0}\int _{\Gamma _\varepsilon ^-}\frac{f(k)}{k-z}\textrm{d} k. \end{aligned}$$

Proof of Lemma 2

To consider the contour integral on the circle $\Gamma _r=\{k:|k|=r, \epsilon<r<R\}$, we let $z=\zeta ^{-1}$ and make the transformation $k=\mu ^{-1}, f(\mu ^{-1})=g(\mu )$. Then the circle $\Gamma _r$ is transformed to the circle $C=\{\mu :|\mu |=r^{-1}\}$. Noting the direction of the contours will change after the transformation, we have

$$\begin{aligned} \frac{1}{2\pi i}\oint _{\Gamma _r^-}\frac{f(k)}{k-z }\textrm{d}k=\frac{\zeta }{2\pi i}\oint _C\frac{g(\mu )}{\mu (\mu -\zeta )}\textrm{d}\mu . \end{aligned}$$

(A.5)

Since $0<|\zeta |<r^{-1}$ and

$$\begin{aligned} g(\mu )=\sum \limits _{j=0}^n\frac{b_j}{\mu ^j}+O(\mu ), \quad \mu \rightarrow 0, \end{aligned}$$

then

$$\begin{aligned} \frac{1}{2\pi i}\oint _C\frac{g(\mu )}{\mu (\mu -\zeta )}\textrm{d}\mu =\frac{g(\zeta )}{\zeta }+\frac{1}{n!}\lim \limits _{\mu \rightarrow 0}\frac{\textrm{d}^n}{\textrm{d}\mu ^n}\bigg [\frac{g(\mu )\mu ^n}{\mu -\zeta }\bigg ]. \end{aligned}$$

(A.6)

For fixed $\zeta$, we know that

$$\begin{aligned} \frac{g(\mu )\mu ^n}{\mu -\zeta }=-\sum \limits _{l=0}^n\bigg (\sum \limits _{j=0}^l\frac{b_{n-l+j}}{\zeta ^{j+1}}\bigg )\mu ^l+O(\mu ^{n+1}), \quad \mu \rightarrow 0. \end{aligned}$$

(A.7)

Substituting (A.7) into (A.6), we find

$$\begin{aligned} \frac{1}{2\pi i}\oint _C\frac{g(\mu )}{\mu (\mu -\zeta )}\textrm{d}\mu =\frac{g(\zeta )}{\zeta }-\sum \limits _{j=0}^n\frac{b_{j}}{\zeta ^{j+1}}. \end{aligned}$$

(A.8)

Then, from (A.8) and (A.5), we have

$$\begin{aligned} \frac{1}{2\pi i}\oint _{\Gamma _r^-}\frac{f(k)}{k-z }\textrm{d}k=f(z)-\sum \limits _{j=0}^nb_jz^j. \end{aligned}$$

(A.9)

On the other hand, in the annulus $r<|z|<R$, the Cauchy’s formula implies that

$$\begin{aligned} f(z)=\frac{1}{2\pi i}\oint _{\Gamma _R}\frac{f(k)}{k-z }\textrm{d}k+\frac{1}{2\pi i}\oint _{\Gamma _r^-}\frac{f(k)}{k-z }\textrm{d}k. \end{aligned}$$

(A.10)

Lemma 2 is proved by comparing (A.9) and (A.10).

Lemmas 1 and  2 give the proof the Lemma 3 in the Introduction.