The ${\bar{\partial }}$-dressing method applied to nonlinear defocusing Hirota equation with nonzero boundary conditions

Abstract

We study the defocusing Hirota equation with nonzero boundary condition by using the ${\bar{\partial }}$-dressing method. The ${\bar{\partial }}$-problem with non-canonical normalization conditions is introduced. The Lax pair of the defocusing Hirota equation with nonzero boundary condition is derived from an asymptotic expansion method. The N-soliton solutions are constructed under the selected special spectral transformation matrix, and the dynamic behavior of soliton solutions isanalyzed.

A Appendix

Derivation of the temporal linear spectrum problem of the defocusing Hirota equation with NZBCs.

As $z\rightarrow \infty$, from (24), we have

$$\begin{aligned}&{{\psi _{t}}}=\frac{i}{2}\left( \beta \sigma _{3}z^{3}-(\alpha \sigma _{3}-\beta a_{1}\sigma _{3})z^{2}+(\beta a_{2}\sigma _{3} \right. \nonumber \\&\qquad -\alpha a_{1} \sigma _{3}+3\beta q_{0}^{2}\sigma _{3})z+(\beta a_{3}\sigma _{3}+3\beta q_{0}^{2}a_{1}\sigma _{3}\nonumber \\&\qquad \left. -\alpha a_{2}\sigma _{3})+O\left( \frac{1}{z}\right) \right) e^{i\theta (z)\sigma _{3}},\end{aligned}$$

(65)

$$\begin{aligned}&{-4i\beta } k^{3}\sigma _{3}\psi = \frac{i}{2}\left( \beta \sigma _{3} z^{3}+\beta \sigma _{3} a_{1}z^{2}\right. \nonumber \\&\quad \left. +(\beta \sigma _{3} a_{2}+3\beta \sigma _{3}q_{0}^{2})z \right. \nonumber \\&\qquad \left. +\,\beta \sigma _{3} a_{3}+3\beta \sigma _{3} q_{0}^{2} a_{1}+O\left( \frac{1}{z}\right) \right) e^{i\theta (z)\sigma _{3}}, \end{aligned}$$

(66)

$$\begin{aligned}&-2i\alpha k^{2}\sigma _{3}\psi +4\beta k^{2}Q\psi \nonumber \\&\quad =\frac{1}{2}\left( (-i\alpha \sigma _{3}+2\beta Q)z^{2}\right. \nonumber \\&\qquad -\,(i\alpha \sigma _{3}a_{1} -2\beta Qa_{1})z+2i\alpha \sigma _{3} q_{0}^{2}-i\alpha \sigma _{3}a_{2}\nonumber \\&\qquad \left. +\,2\beta Qa_{2}+4\beta q_{0}^{2}Q+O\left( \frac{1}{z}\right) \right) e^{i\theta (z)\sigma _{3}},\end{aligned}$$

(67)

$$\begin{aligned}&k(2\alpha Q-2i\beta (Q_{x}+Q^{2})\sigma _{3})\psi \nonumber \\&\quad = \left( (i\beta Q_{x}\sigma _{3}\right. \nonumber \\&\qquad +\,i\beta Q^{2}\sigma _{3}-\alpha Q)z-\alpha Qa_{1}+i\beta (Q_{x}+Q^{2})\nonumber \\&\qquad \left. \sigma _{3}a_{1}+O\left( \frac{1}{z}\right) \right) e^{i\theta (z)\sigma _{3}},\end{aligned}$$

(68)

$$\begin{aligned}&(-i\alpha (Q_{x}+Q^{2}- q_{0}^{2})\sigma _{3}+\beta (Q_{x}Q\nonumber \\&\qquad -QQ_{x}+{2}Q^{3}-Q_{xx}))\psi \nonumber \\&\quad =\left( -i\alpha (Q_{x}+Q^{2}- q_{0}^{2})\sigma _{3}\right. \nonumber \\&\qquad \left. +\,\beta (Q_{x}Q-QQ_{x}+2Q^{3}-Q_{xx})\right. \nonumber \\&\qquad \left. +O\left( \frac{1}{z}\right) \right) e^{i\theta (z)\sigma _{3}}. \end{aligned}$$

(69)

Further we obtain

$$\begin{aligned}&(-4i\beta k^{3}\sigma _{3}-2i\alpha k^{2}\sigma _{3}+4\beta k^{2}Q+(2\alpha Q-2i\beta (Q_{x}\nonumber \\&\qquad +Q^{2})\sigma _{3})k-i\alpha (Q_{x}+Q^{2}- q_{0}^{2})\sigma _{3}\nonumber \\&\qquad +\beta (Q_{x}Q-QQ_{x}+2Q^{3}-Q_{xx}))\psi \nonumber \\&\quad =\left( \frac{i}{2}\beta \sigma _{3} z^{3}+\left( \frac{i}{2}\beta \sigma _{3}a_{1}-\frac{i}{2}\alpha \sigma _{3}+\beta Q\right) z^{2}\right. \nonumber \\&\qquad +\left( \frac{i}{2}\beta \sigma _{3}a_{2}+\frac{3}{2}i\beta q_{0}^{2}\sigma _{3}-\frac{i}{2}\alpha \sigma _{3}a_{1}+\beta Qa_{1}+i\beta Q_{x}\sigma _{3}\right. \nonumber \\&\qquad \left. +i\beta Q^{2}\sigma _{3}-\alpha Q\right) z+\frac{i}{2}\beta \sigma _{3} a_{3}+\frac{3}{2}i\beta q_{0}^{2}\sigma _{3}a_{1}-i\alpha q_{0}^{2}\sigma _{3}\nonumber \\&\qquad -\frac{i}{2}\alpha \sigma _{3}a_{2}+\beta Qa_{2}+2\beta q_{0}^{2}Q-\alpha Qa_{1}+i\beta (Q_{x}+Q^{2})\nonumber \\&\qquad \sigma _{3}a_{1} -i\alpha (Q_{x}+Q^{2}- q_{0}^{2})\sigma _{3}+\beta (Q_{x}Q-QQ_{x}\nonumber \\&\qquad \left. +2Q^{3}-Q_{xx})\right) e^{i\theta (z)\sigma _{3}}. \end{aligned}$$

(70)

From (44), the following relation can be calculated.

$$\begin{aligned} {\frac{i}{2}\beta \sigma _{3}a_{1}-\frac{i}{2}\alpha \sigma _{3}+\beta Q=\frac{i}{2}(\beta a_{1}\sigma _{3}-\alpha \sigma _{3})} \end{aligned}$$

(71)

Using (45) and considering the relationship between Q and $a_{1}$, we get

$$\begin{aligned}&{\frac{i}{2}\beta \sigma _{3}a_{2}+\frac{3}{2}i\beta q_{0}^{2}\sigma _{3}-\frac{i}{2}\alpha \sigma _{3}a_{1}+\beta Qa_{1}+i\beta Q_{x}\sigma _{3}}\nonumber \\&\quad {+i\beta Q^{2}\sigma _{3}-\alpha Q =\frac{i}{2}\beta a_{2}\sigma _{3}+\frac{3}{2}i\beta q_{0}^{2}\sigma _{3}-\frac{i}{2}\alpha a_{1}\sigma _{3}.} \end{aligned}$$

(72)

Using (46)–(48), considering the relationship between Q and $a_{1}$, we get

$$\begin{aligned}&{\frac{i}{2}}{\beta \sigma _{3} a_{3}+\frac{3}{2}i\beta q_{0}^{2}\sigma _{3}a_{1}-i\alpha q_{0}^{2}\sigma _{3}-\frac{i}{2}\alpha \sigma _{3}a_{2}+\beta Qa_{2}}\nonumber \\&\qquad +\,2\beta q_{0}^{2}Q-\alpha Qa_{1}+i\beta (Q_{x}+Q^{2})\sigma _{3}a_{1}-i\alpha (Q_{x}\nonumber \\&\qquad +\,Q^{2}- q_{0}^{2})\sigma _{3}+\beta (Q_{x}Q-QQ_{x}+2Q^{3}-Q_{xx})\nonumber \\&\quad =\frac{i}{2}\beta a_{3}\sigma _{3}+\frac{3}{2}i\beta q_{0}^{2}a_{1}\sigma _{3}-\frac{i}{2}\alpha a_{2}\sigma _{3}. \end{aligned}$$

(73)

The formula (70) leads to

$$\begin{aligned}&(-4i\beta k^{3}\sigma _{3}-2i\alpha k^{2}\sigma _{3}+4\beta k^{2}Q+(2\alpha Q-2i\beta (Q_{x}\nonumber \\&\qquad +\,Q^{2})\sigma _{3})k-i\alpha (Q_{x}+Q^{2}- q_{0}^{2})\sigma _{3}\nonumber \\&\qquad +\beta (Q_{x}Q-QQ_{x}+2Q^{3}-Q{xx}))\psi \nonumber \\&\quad =\frac{i}{2}\left( \beta \sigma _{3}z^{3}-(\alpha \sigma _{3}-\beta a_{1}\sigma _{3})z^{2}\right. \nonumber \\&\qquad +\,(\beta a_{2}\sigma _{3}-\alpha a_{1}\sigma _{3}+3\beta q_{0}^{2}\sigma _{3})z \nonumber \\&\qquad +(\beta a_{3}\sigma _{3}+3\beta q_{0}^{2}a_{1}\sigma _{3}\nonumber \\&\qquad \left. -\,\alpha a_{2}\sigma _{3})+O\left( \frac{1}{z}\right) \right) e^{i\theta (z)\sigma _{3}}. \end{aligned}$$

(74)

Equations (65) and (74) imply the Laurent series of $\psi _{t}$ and $(-4i\beta k^{3}\sigma _{3}-2i\alpha k^{2}\sigma _{3}+4\beta k^{2}Q+(2\alpha Q-2i\beta ($ $Q_{x}+$ $Q^{2})\sigma _{3})k-i\alpha (Q_{x}+Q^{2}- q_{0}^{2})\sigma _{3}+\beta (Q_{x}Q-QQ_{x}+2Q^{3}-Q_{xx}))\psi$ at $z\rightarrow \infty$ share the same principal part. Using (7) and (25) and considering the relationship between $a_{l}$ and $b_{m}$, we find

$$\begin{aligned} \psi _{t}&=\left( -\frac{i}{2}\beta q_{0}^{6}b_{-1}\sigma _{3}z^{-4}+\left( \frac{i}{2}\alpha q_{0}^{4}b_{-1}\sigma _{3}-\frac{i}{2}\beta q_{0}^{6}b_{0}\sigma _{3}\right) z^{-3}\right. \nonumber \\&+\left( \frac{i}{2}\alpha q_{0}^{4}b_{0}\sigma _{3}-\frac{3}{2}i\beta q_{0}^{4}b_{-1}\sigma _{3}-\frac{i}{2}\beta q_{0}^{6}b_{1}\sigma _{3}\right) z^{-2}+\left( \frac{i}{2}\alpha q_{0}^{4}\right. \nonumber \\&\left. \left. b_{1}\sigma _{3}-\frac{i}{2}\beta q_{0}^{6}b_{2}\sigma _{3}-\frac{3}{2}i\beta q_{0}^{4}b_{0}\sigma _{3}\right) z^{-1}+O(1)\right) e^{i\theta (z)\sigma _{3}}, \end{aligned}$$

(75)

and

$$\begin{aligned}&(-4i\beta k^{3}\sigma _{3}-2i\alpha k^{2}\sigma _{3}+4\beta k^{2}Q+(2\alpha Q\nonumber \\&\qquad -\,2i\beta (Q_{x}+Q^{2})\sigma _{3})k-i\alpha (Q_{x}+Q^{2}- q_{0}^{2})\sigma _{3}\nonumber \\&\qquad +\beta (Q_{x}Q-QQ_{x}+2Q^{3}-Q_{xx}))\psi \nonumber \\&\quad =\left( -\frac{i}{2}\beta q_{0}^{6}b_{-1}\sigma _{3}z^{-4} \right. \nonumber \\&\qquad +\,\left( \frac{i}{2}\alpha q_{0}^{4}b_{-1}\sigma _{3}-\frac{i}{2}\beta q_{0}^{6}b_{0}\sigma _{3}\right) z^{-3}+\left( \frac{i}{2}\alpha q_{0}^{4}b_{0}\sigma _{3}\right. \nonumber \\&\qquad \left. -\,\frac{3}{2}i\beta q_{0}^{4}b_{-1}\sigma _{3}-\frac{i}{2}\beta q_{0}^{6}b_{1}\sigma _{3}\right) z^{-2}+\left( \frac{i}{2}\alpha q_{0}^{4}b_{1}\sigma _{3}\right. \nonumber \\&\qquad \left. \left. -\,\frac{i}{2}\beta q_{0}^{6}b_{2}\sigma _{3}-\frac{3}{2}i\beta q_{0}^{4}b_{0}\sigma _{3}\right) z^{-1}+O(1)\right) e^{i\theta (z)\sigma _{3}}, \end{aligned}$$

(76)

which imply that the Laurent series of $\psi _{t}$ and $(-4i\beta k^{3}\sigma _{3}-2i\alpha k^{2}\sigma _{3}+4\beta k^{2}Q+(2\alpha Q-2i\beta (Q_{x}+Q^{2})\sigma _{3})k-i\alpha (Q_{x}+Q^{2}- q_{0}^{2})\sigma _{3}+\beta (Q_{x}Q-QQ_{x}+2Q^{3}-Q_{xx}))\psi$ at $z\rightarrow 0$ share the same principal part. From (74) and (76), we find that

$$\begin{aligned} {\mathcal {N}}(\psi _{t})&={\mathcal {N}}((-4i\beta k^{3}\sigma _{3} -2i\alpha k^{2}\sigma _{3}+4\beta k^{2}Q+(2\alpha Q\nonumber \\&\quad -2i\beta (Q_{x}+Q^{2})\sigma _{3})k-i\alpha (Q_{x}+Q^{2}- q_{0}^{2})\sigma _{3}\nonumber \\&\quad +\beta (Q_{x}Q-QQ_{x}+2Q^{3}-Q_{xx}))\psi ). \end{aligned}$$

(77)

According to Lemma 1, the temporal linear spectral problem of the defocusing Hirota equation with NZBCs can be obtained

$$\begin{aligned} {\psi _{t}}&=(-4i\beta k^{3}\sigma _{3}-2i\alpha k^{2}\sigma _{3}+4\beta k^{2}Q+(2\alpha Q\nonumber \\&\quad -2i\beta (Q_{x}+Q^{2})\sigma _{3})k-i\alpha (Q_{x}+Q^{2}- q_{0}^{2})\sigma _{3}\nonumber \\&\quad +\beta (Q_{x}Q-QQ_{x}+2Q^{3}-Q_{xx}))\psi . \end{aligned}$$

(78)