**3. Spectral Analysis**

From the previous paragraph, we know the Lax pair of the mNLS equation. By introducing

$$Q = \left( \begin{array}{cc} 0 & q \\ -\overline{q} & 0 \end{array} \right) \iota^{\sigma\_3} \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) . \tag{16}$$

where the *q*¯ denotes the conjugation of *q*, the Lax pair (15) can be rewritten in this form

$$\begin{cases} \psi\_x + i\xi^2 \sigma\_3 \psi - i\sigma\_3 \psi = \xi Q \psi\_r \\ \psi\_t + 2i\xi^4 \sigma\_3 + 2i\sigma\_3 - 4i\xi^2 \sigma\_3 = -i\xi^2 Q^2 + 2i\xi^3 Q - 2\xi Q + \xi Q^3 + i\xi Q^3 + i\xi\sigma\_3 Q\_\mathbf{x}. \end{cases} \tag{17}$$

In our analysis, we assume that *q* decays to zero sufficiently fast as *x* → ±<sup>∞</sup>. So, it is correct to extend the column vector *ψ* to a 2 × 2 matrix. For simplicity, we substitute *λ*<sup>2</sup> for *ξ*2 − 1. Letting *ψ* = <sup>Ψ</sup>*e*<sup>−</sup>*<sup>i</sup>*(*ξ*<sup>2</sup>*x*+2*ξ*<sup>4</sup>*t*)*<sup>σ</sup>*<sup>3</sup> , then the Lax pair (17) becomes

$$\begin{aligned} \Psi\_{\mathbf{x}} + i\mathfrak{J}^{2}[\sigma\_{3\star}\Psi] &= \mathfrak{J}Q\Psi, \\ \Psi\_{t} + 2i\mathfrak{J}^{4}[\sigma\_{3\star}\Psi] &= (-i\mathfrak{J}^{2}Q^{2}\sigma\_{3} + 2\mathfrak{J}^{3}Q - 2\mathfrak{J}Q + \mathfrak{J}Q^{3} + i\mathfrak{J}\sigma\_{3}Q\_{\mathbf{x}})\Psi \end{aligned} \tag{18}$$

We can write (18) in full derivative form

$$d\left(e^{i\left(\xi^2 x + 2\xi^4 t\right)\vartheta\_3} \Psi(x, t; \xi)\right) = e^{i\left(\lambda^2 x + 2\lambda^4 t\right)\vartheta\_3} \mathcal{U}(x, t; \xi)\Psi. \tag{19}$$

where

$$\begin{split} \mathcal{U} &= \mathcal{U}\_1 \mathbf{dx} + \mathcal{U}\_2 \mathbf{d}t \\ &= \mathfrak{J} \mathbf{Q} \mathbf{dx} + (-i\mathfrak{J}^2 \mathbf{Q}^2 \sigma\_3 + 2\mathfrak{J}^3 \mathbf{Q} - 2\mathfrak{J} \mathbf{Q} + \mathfrak{J} \mathbf{Q}^3 + i\mathfrak{J} \sigma\_3 \mathbf{Q}\_x) \mathbf{d}t. \end{split} \tag{20}$$

and *σ*ˆ3 denotes the operator to matrix by *σ*ˆ3*M* = [*<sup>σ</sup>*3, *<sup>M</sup>*], hence it is easy to prove *e<sup>σ</sup>*ˆ3 *M* = *eσ*<sup>3</sup> *Me*−*σ*<sup>3</sup> , where *M* is a 2 × 2 matrix.

### *3.1. The Reconstruction of Lax Pair*

Expanding the solution of (19) in this way

$$\Psi = D + \frac{\Psi\_1}{\tilde{\xi}} + \frac{\Psi\_2}{\tilde{\xi}^2} + \frac{\Psi\_3}{\tilde{\xi}^3} + \mathcal{O}(\frac{1}{\tilde{\xi}^4}), \tilde{\xi} \to \infty,\tag{21}$$

where *D*, Ψ1, Ψ2, Ψ3 are independent of *ξ*. Bringing this equation into the first equation of (18), and comparing the same order of *ξ*'s frequency, we obtain the following equations

$$\begin{aligned} O(1)&: D\_{\mathcal{X}} + i[\sigma\_{3\prime}\Psi\_2] - i[\sigma\_{3\prime}D] = Q\Psi\_1; \\ O(\xi)&: i[\sigma\_{3\prime}\Psi\_1] = QD; \\ O(\xi^2)&: i[\sigma\_{3\prime}D] = 0. \end{aligned} \tag{22}$$

Using the same method, taking (21) into another equation of (18), we have

$$\begin{aligned} O(1)&: D\_l - 4i[\sigma\_3, \Psi\_2] + 2i[\sigma\_3, D] = -iQ^2 \sigma\_3 \Psi\_2 + 2Q\Psi\_3 - 2Q\Psi\_1 + Q^3 \Psi\_1 + i\sigma\_3 Q\_2 \Psi\_1; \\ O(\xi)&: 2i[\sigma\_3, \Psi\_3] - 4i[\sigma\_3, \Psi\_1] = -iQ^2 \sigma\_3 \Psi\_1 + 2Q\Psi\_2 - 2QD + Q^3 D + i\sigma\_3 Q\_2 D; \\ O(\xi^2)&: 2i[\sigma\_3, \Psi\_2] - 4i[\sigma\_3, D] = -iQ^2 \sigma\_3 D + 2Q\Psi\_1; \\ O(\xi^3)&: 2i[\sigma\_3, \Psi\_1] = 2QD; \\ O(\xi^4)&: 2i[\sigma\_3, D] = 0. \end{aligned} \tag{23}$$

For (22), We find that *D* is a diagonal matrix from *<sup>O</sup>*(*ξ*<sup>2</sup>). Without loss of generality, we suppose

$$D = \begin{pmatrix} D\_0^{11} & 0 \\ 0 & D\_0^{22} \end{pmatrix}. \tag{24}$$

From *O*(*ξ*) we have

$$
\Psi\_1^\sigma = \frac{i}{2} Q D \sigma\_3. \tag{25}
$$

where Ψ*o* 1denotes the off-diagonal part of Ψ1. So, we can ge<sup>t</sup> *Dx* from *O*(1) easily

$$D\_x = \frac{i}{2} Q^2 \sigma\_3 D. \tag{26}$$

For (23), after a lengthy calculation, we ge<sup>t</sup>

$$\begin{split} D\_l &= \frac{3i}{4} Q^4 \sigma\_3 D + \frac{1}{2} Q Q\_x D - \frac{1}{2} Q\_x Q D \\ &= (\frac{3i}{4} |q|^4 + \bar{q} q\_x - q \bar{q}\_x) \sigma\_3 D. \end{split} \tag{27}$$

The mNLS equation admits the conservation law

$$2(|q|^2)\_t = (2iq\_x\overline{q} - 3|q|^4 - 2iq\overline{q}\_x)\_x. \tag{28}$$

From the above results, we find (26) and (27) admit the conservation law. Define

$$D(x,t) = e^{i\int\_{(0,0)}^{(x,t)} \Lambda(x,t)\sigma\_3}.\tag{29}$$

where Δ is the differential one-form, and it is given by

$$
\Delta(\mathbf{x},t) = \Delta\_1 d\mathbf{x} + \Delta\_2 dt = -\frac{1}{2} q \vec{q} dx + [\frac{3}{4} q^2 \vec{q}^2 + \frac{i}{2} (q \vec{q}\_x - q\_x \vec{q})] dt. \tag{30}
$$

It is not difficult to find that the integral is path independent. So, we introduce

$$\Psi(\mathbf{x},t;\xi) = e^{i\int\_{(0,0)}^{(\mathbf{x},t)} \Delta \mathbf{r}\_3} \mu(\mathbf{x},t;\xi) D(\mathbf{x},t), 0 < \mathbf{x} < \infty, 0 < t < T. \tag{31}$$

Then the form of the Lax pair (18) can be replaced with

$$d(e^{i(\lambda^2 \mathbf{x} + 2\lambda^4 t)\hat{\sigma}\_3} \mu(\mathbf{x}, t; \xi^\mathbf{})) = \mathcal{W}(\mathbf{x}, t; \xi^\mathbf{}). \tag{32}$$

where

$$\begin{split} \mathcal{W}(\mathbf{x}, t; \xi) &= e^{i(\lambda^2 \mathbf{x} + 2\lambda^4 t)\hat{\sigma}\_3} V(\mathbf{x}, t; \xi)\mu\_\prime \\ V = V\_1 d\mathbf{x} + V\_2 dt &= e^{-i\int\_{(0,0)}^{(\mathbf{x}, t)} \Delta \theta\_3} (\mathcal{U} - i\Delta \sigma\_3) . \end{split} \tag{33}$$

Considering the definitions of *U* and Δ, we have

$$\begin{split} V\_{1}(\mathbf{x},t;\xi^{\mathsf{T}}) &= \begin{pmatrix} \frac{i}{2}q\bar{q} & \frac{\mathsf{Z}\_{1}q^{-2i}\int\_{(0,0)}^{(\mathsf{x},t)}\Lambda \\ -\frac{\mathsf{x}}{\xi^{2}q\bar{q}}e^{2i\int\_{(0,0)}^{(\mathsf{x},t)}\Lambda} & -\frac{i}{2}q\bar{q} \end{pmatrix}, \\ V\_{2}(\mathbf{x},t;\xi) &= \begin{pmatrix} V\_{2}^{11}(\mathbf{x},t;\xi^{\mathsf{T}}) & V\_{2}^{12}(\mathbf{x},t;\xi^{\mathsf{T}}) \\ V\_{2}^{21}(\mathbf{x},t;\xi^{\mathsf{T}}) & V\_{2}^{22}(\mathbf{x},t;\xi^{\mathsf{T}}) \end{pmatrix}. \end{split} \tag{34}$$

where

$$\begin{split} &V\_{2}^{11}(\mathbf{x},t;\xi) = i\xi^{2}q\bar{q} - \frac{3i}{4}q^{2}\bar{q}^{2} + \frac{1}{2}(q\bar{q}\_{x} - q\_{x}\bar{q}), \\ &V\_{2}^{12}(\mathbf{x},t;\xi) = (2\xi^{3} - 2\xi\bar{q} - \xi\bar{q}|q|^{2} + i\xi\bar{q}\_{x})e^{-2i\int\_{(0,0)}^{(\mathbf{x},t)}\Lambda}, \\ &V\_{2}^{21}(\mathbf{x},t;\xi) = (-2\xi^{3}\bar{q} + 2\xi\bar{q} + \xi\bar{q}|q|^{2} + i\xi\bar{q}\_{x})e^{2i\int\_{(0,0)}^{(\mathbf{x},t)}\Lambda}, \\ &V\_{2}^{22}(\mathbf{x},t;\xi) = -i\xi^{2}q\bar{q} + \frac{3i}{4}q^{2}\bar{q}^{2} - \frac{1}{2}(q\bar{q}\_{x} - q\_{x}\bar{q}). \end{split}$$

Thus, (32) changes into

$$\begin{cases} \mu\_{\rm x} + i\lambda^2 [\sigma\_{\rm 3}, \mu] = V\_1 \mu, \\ \mu\_t + 2i\lambda^4 [\sigma\_{\rm 3}, \mu] = V\_2 \mu. \end{cases} \tag{35}$$

### *3.2. The Riemann-Hilbert Problem And Some Relations*

Supposing that *q*(*<sup>x</sup>*, *t*) is smooth function in the domain *D* = {0 < *x* < <sup>∞</sup>, 0 < *t* < *<sup>T</sup>*}. Then we define the eigenfunctions *<sup>μ</sup>j*(*<sup>x</sup>*, *t*, ; *ξ*)(*j* = 1, 2, 3) of (34) as follows

$$\mu\_{\vec{f}}(\mathbf{x}, t; \xi\_{\mathfrak{z}}^{\mathfrak{z}}) = I + \int\_{(\mathbf{x}\_{\mathfrak{f}}, t\_{\mathfrak{f}})}^{(\mathbf{x}, t)} e^{-i(\lambda \mathbf{x} + 2\lambda^2 t)\hat{\sigma}\_{\mathfrak{z}}} \mathcal{W}(\mathbf{x}', t'; \xi\_{\mathfrak{z}}^{\mathfrak{z}}), 0 < \mathbf{x} < \infty, 0 < t < T. \tag{36}$$

The integral curve is from (*xj*, *tj*) to (*<sup>x</sup>*, *t*), where (*<sup>x</sup>*1, *<sup>t</sup>*1)=(0, *<sup>T</sup>*),(*<sup>x</sup>*2.*t*2)=(0, 0) and (*<sup>x</sup>*3, *<sup>t</sup>*3) = (<sup>∞</sup>, *t*). Furthermore, the point (*<sup>x</sup>*, *t*) is an arbitrary point in the domain *D*. We know that the integral of (36) is independent of the path of integration. Without loss of generality, we will consider the particular integral paths as follows, see Figure 1.

**Figure 1.** Integral paths.

By this method, we ge<sup>t</sup>

$$\begin{cases} \mu\_{1}(\mathbf{x},t;\xi) = I + \int\_{0}^{\mathbf{x}} e^{i\lambda^{2}(\mathbf{x}'-\mathbf{x})\cdot\xi\_{3}} (V\_{1}\mu\_{1})(\mathbf{x}',t;\lambda) d\mathbf{x}' \\ \quad - e^{-i\lambda^{2}\mathbf{x}\hat{\mathbf{y}}\_{3}} \int\_{t}^{t} e^{2i\lambda^{4}(t'-t)\hat{\mathbf{c}}\_{3}} (V\_{2}\mu\_{1})(0,t';\xi) dt', \\ \mu\_{2}(\mathbf{x},t;\xi) = I + \int\_{0}^{\mathbf{x}} e^{i\lambda^{2}(\mathbf{x}'-\mathbf{x})\hat{\mathbf{c}}\_{3}} (V\_{1}\mu\_{2})(\mathbf{x}',t;\xi) d\mathbf{x}' \\ \quad - e^{-i\lambda^{2}\mathbf{x}\hat{\mathbf{y}}\_{3}} \int\_{0}^{t} e^{2i\lambda^{4}(t'-t)\hat{\mathbf{c}}\_{3}} (V\_{2}\mu\_{2})(0,t';\xi) dt', \\ \mu\_{3}(\mathbf{x},t,\xi) = I - \int\_{\infty}^{\mathbf{x}} e^{i\lambda^{2}(\mathbf{x}'-\mathbf{x})\hat{\mathbf{c}}\_{3}} (V\_{1}\mu\_{3})(\mathbf{x}',t;\xi) d\mathbf{x}'. \end{cases} \tag{37}$$

Noting that the first column of *μj* includes *<sup>e</sup>*<sup>−</sup>2*<sup>i</sup>*[*λ*<sup>2</sup>(*x*−*x*)+2*λ*<sup>4</sup>(*t*−*t*)]. So, in different integral paths, we have the following inequalities

$$\begin{cases} \ell\_1 : (\mathbf{x}\_1, t\_1) \to (\mathbf{x}, t) : 0 < \mathbf{x}' < \mathbf{x}, t < t' < T\_\prime \\\ell\_2 : (\mathbf{x}\_2, t\_2) \to (\mathbf{x}, t) : 0 < \mathbf{x}' < \mathbf{x}, 0 < t' < t\_\prime \\\ell\_3 : (\mathbf{x}\_3, t\_3) \to (\mathbf{x}, t) : \mathbf{x} < \mathbf{x}' < \mathbf{x}. \end{cases} \tag{38}$$

Due to the exponential function decaying sufficiently, these inequalities imply that the first of the functions *<sup>μ</sup>j*(*<sup>x</sup>*, *t*; *ξ*),(*j* = 1, 2, 3) are analytic if

$$\mu\_1^{(1)}(\mathbf{x}, t; \xi) : \mathfrak{F} \in \{Im \mathfrak{g}^2 \ge 0\} \cap \{Im \mathfrak{g}^4 \le 0\},$$

$$\mu\_2^{(1)}(\mathbf{x}, t; \xi) : \mathfrak{F} \in \{Im \mathfrak{g}^2 \ge 0\} \cap \{Im \mathfrak{g}^4 \ge 0\},\tag{39}$$

$$\mu\_3^{(1)}(\mathbf{x}, t; \xi) : \mathfrak{F} \in \{Im \mathfrak{g}^2 \le 0\}.$$

At the same time, the second column of the functions *<sup>μ</sup>j*(*<sup>x</sup>*, *t*; *ξ*),(*j* = 1, 2, 3) are analytic if

$$\mu\_1^{(2)}(\mathbf{x}, t; \mathfrak{z}^{\mathfrak{z}}) : \mathfrak{z} \in \{Im \mathfrak{z}^2 \le 0\} \cap \{Im \mathfrak{z}^4 \ge 0\},$$

$$\mu\_2^{(2)}(\mathbf{x}, t; \mathfrak{z}^{\mathfrak{z}}) : \mathfrak{z} \in \{Im \mathfrak{z}^2 \le 0\} \cap \{Im \mathfrak{z}^4 \le 0\},\tag{40}$$

$$\mu\_3^{(2)}(\mathbf{x}, t; \mathfrak{z}^{\mathfrak{z}}) : \mathfrak{z} \in \{Im \mathfrak{z}^2 \ge 0\}.$$

Hence, we ge<sup>t</sup>

$$\begin{aligned} \mu\_1(\mathbf{x}, t; \xi) &= (\mu\_1^{D\_2}(\mathbf{x}, t; \xi), \mu\_1^{D\_3}(\mathbf{x}, t; \xi)), \\ \mu\_2(\mathbf{x}, t; \xi) &= (\mu\_2^{D\_1}(\mathbf{x}, t; \xi), \mu\_2^{D\_4}(\mathbf{x}, t; \xi)), \\ \mu\_3(\mathbf{x}, t; \xi) &= (\mu\_3^{D\_3 \cup D\_4}(\mathbf{x}, t; \xi), \mu\_3^{D\_1 \cup D\_2}(\mathbf{x}, t; \xi)). \end{aligned} \tag{41}$$

where *μDi j* stands for *μj* is analytic if *ξ* ∈ *Di*, where *Di* = *ωi* ∪ (−*ωi*), −*ωi* = {−*ξ* ∈ *C*|*ξ* ∈ *<sup>ω</sup>i*}, *ωi* = {*ξ* ∈ *C*| *i*−14 *π* < *ξ* < *i*<sup>4</sup>*<sup>π</sup>*}, see Figure 2.

**Figure 2.** *Dj*, *j* = 1, 2, 3, 4.

The eigenfunctions *<sup>μ</sup>j*(*j* = 1, 2, 3) possess these properties.


where the eigenfunctions *<sup>μ</sup>j*(*<sup>x</sup>*, *t*; *ξ*) = *μ*11*j* (*<sup>x</sup>*, *t*; *ξ*) *μ*12*j* (*<sup>x</sup>*, *t*; *ξ*) *μ*21*j* (*<sup>x</sup>*, *t*; *ξ*) *μ*22*j* (*<sup>x</sup>*, *t*; *ξ*) (*j* = 1, 2, <sup>3</sup>).

For the purpose of formulating a Riemann-Hilbert problem, our main task is to find the jump matrices for every *Di*(*i* = 1, 2, 3, 4) to any other regions. Then we define the spectral functions *s*(*ξ*) and *S*(*ξ*)

$$\begin{cases} \mu\_3(\mathbf{x}, t; \mathfrak{z}^{\mathfrak{z}}) = \mu\_2(\mathbf{x}, t; \mathfrak{z}^{\mathfrak{z}}) e^{-i(\lambda^2 \mathbf{x} + 2\lambda^4 t)\vartheta\_3} S(\mathfrak{z}^{\mathfrak{z}}), \\ \mu\_1(\mathbf{x}, t; \mathfrak{z}^{\mathfrak{z}}) = \mu\_2(\mathbf{x}, t; \mathfrak{z}^{\mathfrak{z}}) e^{-i(\lambda^2 \mathbf{x} + 2\lambda^4 t)\vartheta\_3} S(\mathfrak{z}^{\mathfrak{z}}). \end{cases} \tag{42}$$

According to the above definition, we have

$$
\mu\_1(\mathbf{x}, t; \xi) = \mu\_3(\mathbf{x}, t; \xi) e^{-i(\lambda^2 \mathbf{x} + 2\lambda^4 t)\theta\_3} (\mathbf{s}(\xi))^{-1} \mathbf{S}(\xi). \tag{43}
$$

Combining (37) with (42), we acquire

$$s(\xi) = \mu \circ (0, 0; \xi),\\ S(\xi) = \left(e^{2i\lambda^4 T \mathfrak{d}\_3} \mu \mathbf{2}(0, T; \xi)\right)^{-1}.\tag{44}$$

Owing to (37), it is clear to see that

$$\begin{aligned} \mu\_1(0, t; \xi) &= I - \int\_t^T e^{2i\lambda^4 (t'-t)\beta\_3} (V\_2 \mu\_1)(0, t'; \xi) dt', \\ \mu\_2(0, t; \xi) &= I + \int\_0^t e^{2i\lambda^4 (t'-t)\beta\_3} (V\_2 \mu\_2)(0, t'; \xi) dt', \\ \mu\_3(x, 0; \xi) &= I + \int\_\infty^x e^{i\lambda^2 (x'-x)\hat{\sigma}\_3} (V\_1 \mu\_3)(x', 0; \xi) dx', \\ \mu\_2(x, 0; \xi) &= I + \int\_0^x e^{i\lambda^2 (x'-x)\hat{\sigma}\_3} (V\_1 \mu\_2)(x', 0; \xi) dx', \end{aligned} \tag{45}$$

Considering the initial values *q*(*<sup>x</sup>*, 0) = *<sup>u</sup>*0(*x*), *q*(0, *t*) = *g*0(*t*), boundary values *q*(0, *t*) = *g*0(*t*) and *qx*(0, *t*) = *g*1(*t*). For convenience, the initial-boundary values of *q*¯(*<sup>x</sup>*, *t*) can be written in this form, namely *q*¯(*<sup>x</sup>*, 0) = *<sup>u</sup>*¯0(*x*), *q*¯(0, *t*) = *g*¯0(*t*), and *q*¯*x*(0, *t*) = *g*¯1(*t*). Then *<sup>V</sup>*2(0, *t*; *ξ*) and *<sup>V</sup>*1(*<sup>x</sup>*, 0; *ξ*) can be expressed with

$$\begin{split} V\_{1}(\mathbf{x},0;\boldsymbol{\xi}) &= \begin{pmatrix} \frac{\dot{l}}{2}|u\_{0}|^{2} & \xi\_{0}u\_{0}e^{i\int\_{0}^{x}|u\_{0}|^{2}dx'} \\ -\xi\_{0}^{T}u\_{0}e^{-i\int\_{0}^{x}|u\_{0}|^{2}dx'} & -\frac{\dot{l}}{2}|u\_{0}|^{2} \end{pmatrix}, \\ V\_{2}(0,t;\boldsymbol{\xi}) &= \begin{pmatrix} V\_{2}^{11}(0,t;\boldsymbol{\xi}) & V\_{2}^{12}(0,t;\boldsymbol{\xi}) \\ V\_{2}^{21}(0,t;\boldsymbol{\xi}) & V\_{2}^{22}(0,t;\boldsymbol{\xi}) \end{pmatrix}. \end{split} \tag{46}$$

where

$$\begin{split} V\_{2}^{11}(0,t;\mathfrak{boldsymbol}) &= i\mathfrak{z}^{2}|\mathfrak{g}\_{0}|^{2} - \frac{3i}{4}|\mathfrak{g}\_{0}|^{4} - \frac{1}{2}(\mathfrak{g}\_{0}\mathfrak{g}\_{1} - \mathfrak{g}\_{1}\mathfrak{g}\_{0}) = -V\_{2}^{22}(\mathfrak{x},t;\mathfrak{ζ}),\\ V\_{2}^{12}(0,t;\mathfrak{boldsymbol}) &= (2\mathfrak{z}^{3}\mathfrak{g}\_{0} - 2\mathfrak{z}\mathfrak{g}\_{0} - \mathfrak{z}\mathfrak{g}\_{0}|\mathfrak{g}\_{0}|^{2} + i\mathfrak{z}\mathfrak{g}\_{1})e^{-2i\int\_{0}^{t}\Delta\_{2}(0,t')dt'},\\ V\_{2}^{21}(0,t;\mathfrak{boldsymbol}) &= (-2\mathfrak{z}^{3}\bar{\mathfrak{g}}\_{0} + 2\mathfrak{z}\bar{\mathfrak{g}}\_{0} + \mathfrak{z}\bar{\mathfrak{g}}\_{0}|\mathfrak{g}\_{0}|^{2} + i\mathfrak{z}\bar{\mathfrak{g}}\_{1})e^{2i\int\_{0}^{t}\Delta\_{2}(0,t')dt'}.\end{split}$$

with

$$
\Delta\_2(0, t') = \frac{3}{4} |\lg\_0|^4 + \frac{1}{2} (\lg\_0 \overline{\varrho}\_1 - \lg\_1 \overline{\varrho}\_0).
$$

Due to *μj* have symmetry, the *s*(*ξ*) and *S*(*ξ*) also have symmetry

$$\begin{array}{c} s\_{11}(\xi) = \overline{s\_{22}(\xi)}, s\_{21}(\xi) = \overline{s\_{12}(\xi)},\\ S\_{11}(\xi) = \overline{S\_{22}(\xi)}, S\_{21}(\xi) = \overline{S\_{12}(\xi)}.\end{array}$$

Without loss of generality, we assume

$$s(\xi) = \left(\begin{array}{cc}\overline{a(\frac{\xi}{\xi})} & b(\xi) \\ \overline{b(\frac{\xi}{\xi})} & a(\xi) \end{array}\right), S(\xi) = \left(\begin{array}{cc}\overline{A(\frac{\xi}{\xi})} & B(\xi) \\ \overline{B(\frac{\xi}{\xi})} & A(\xi) \end{array}\right). \tag{47}$$

According to (42) and (44), we have

$$\begin{split} s(\xi) &= I - \int\_{\infty}^{0} e^{i\lambda^2 (\mathbf{x}' - \mathbf{x}) \theta\_3} (V\_1 \mu\_3)(\mathbf{x}', 0; \xi) d\mathbf{x}', \\ S(\xi) &= (I + \int\_{0}^{T} e^{2i\lambda^4 t' \theta\_3} (V\_2 \mu\_2)(0, t'; \xi) dt')^{-1}. \end{split} \tag{48}$$

The spectral functions *<sup>s</sup>*(*ξ*), *S*(*ξ*) have the following properties

$$\begin{array}{c} \bullet \quad \left(\begin{array}{c} b(\xi) \\ a(\xi) \end{array}\right) = \left(\begin{array}{c} \mu\_{3}^{12}(0,0;\xi) \\ \mu\_{3}^{22}(0,0;\xi) \end{array}\right) = \mu\_{3}^{(2)}(0,0;\xi), \\\ \left(\begin{array}{c} e^{-4i\lambda^{4}T}B(\xi) \\ \overline{A(\xi)} \end{array}\right) = \left(\begin{array}{c} \mu\_{2}^{12}(0,T;\xi) \\ \mu\_{2}^{22}(0,T;\xi) \end{array}\right) = \mu\_{2}^{(2)}(0,T;\xi). \end{array}$$


$$\begin{array}{ll} \bullet & a(\xi) = 1 + O(\frac{1}{\xi}), b(\xi) = O(\frac{1}{\xi}), \xi \to \infty, Im\xi^2 \ge 0, \\\ A(\xi) = 1 + O(\frac{1}{\xi}), B(\xi) = O(\frac{1}{\xi}), \xi \to \infty, Im\xi^4 \ge 0. \end{array}$$

These spectral functions do not exist independently, but depend on each other and satisfy certain relationships, we call it global relation

$$B(\xi)a(\xi) - A(\xi)b(\xi) = e^{4i\lambda^4T}c^+(\xi), \\ Im \xi^2 \ge 0. \tag{49}$$

where

$$\varepsilon^{+}\left(\xi\right) = \int\_{0}^{\infty} e^{2i\lambda^{2}\chi'} \left(V\_{1}\mu\_{3}\right) \left(\varkappa', T; \xi\right) d\varkappa'.$$

> For simplicity, we define *<sup>M</sup>*(*<sup>x</sup>*, *t*; *ξ*)

$$\begin{split} M\_{+} &= (\frac{\mu\_{2}^{D\_{1}}}{d(\frac{\pi}{3})}, \mu\_{3}^{D\_{1} \cup \Box D\_{2}}), \mathfrak{x} \in D\_{1}, \\ M\_{-} &= (\frac{\mu\_{1}^{D\_{2}}}{d(\frac{\pi}{3})}, \mu\_{3}^{D\_{1} \cup \Box D\_{2}}), \mathfrak{x} \in D\_{2}, \\ M\_{+} &= (\mu\_{3}^{D\_{3} \cup D\_{4}}, \frac{\mu\_{1}^{D\_{3}}}{d(\frac{\pi}{3})}), \mathfrak{x} \in D\_{3}, \\ M\_{-} &= (\mu\_{3}^{D\_{3} \cup D\_{4}}, \frac{\mu\_{2}^{D\_{4}}}{d(\frac{\pi}{3})}), \mathfrak{x} \in D\_{4}. \end{split} \tag{50}$$

where

$$d(\xi) = a((\xi)\overline{A(\xi)} - b(\xi)\overline{B(\xi)}), \xi \in D\_2. \tag{51}$$

Synthesizing the above definitions, we can ge<sup>t</sup>

$$\det M(\mathbf{x}, \mathbf{t}; \xi) = \mathbf{1},\tag{52}$$

and

$$M(\mathbf{x}, t; \xi) = I + O(\frac{1}{\xi}), \xi \to \infty. \tag{53}$$

**Theorem 2.** *Given a smooth function q*(*<sup>x</sup>*, *t*)*. Define <sup>M</sup>*(*<sup>x</sup>*, *t*; *ξ*) *as (50), and define <sup>μ</sup>j*(*<sup>x</sup>*, *t*; *ξ*)(*j* = 1, 2, 3) *like (37). Then the jump matrices can be derived through*

$$M\_{+}(\mathbf{x},t;\xi) = M\_{-}(\mathbf{x},t;\xi)f(\mathbf{x},t;\xi), \xi^{\mathbf{4}} \in \mathbb{R},\tag{54}$$

*where*

$$J = \begin{cases} J\_1(\mathbf{x}, t; \xi), & \arg \xi^2 = 0, \\ J\_2(\mathbf{x}, t; \xi), & \arg \xi^2 = \frac{\pi}{2}, \\ J\_3(\mathbf{x}, t; \xi) = J\_2 J\_1^{-1} J\_4, & \arg \xi^2 = \pi, \\ J\_4(\mathbf{x}, t; \xi), & \arg \xi^2 = \frac{3}{2}\pi. \end{cases} \tag{55}$$

*and*

$$\begin{aligned} J\_1 &= \begin{pmatrix} \frac{1}{\alpha} & \frac{b}{d} e^{-2i\theta(\xi)} \\ \frac{b}{d} e^{2i\theta(\xi)} & 1 \end{pmatrix}, \\\ J\_2 &= \begin{pmatrix} 1 & 0 \\ -\Gamma(\frac{x}{\xi}) e^{2i\theta(\xi)} & 1 \end{pmatrix}, \\\ J\_4 &= \begin{pmatrix} 1 & \overline{\Gamma(\frac{x}{\xi})} e^{2i\theta(\xi)} \\ 0 & 1 \end{pmatrix}. \end{aligned} \tag{56}$$

*with*

$$\begin{aligned} \theta(\emptyset) &= \lambda^2 \mathbf{x} + 2\lambda^4 t = (\mathfrak{J}^2 - 1)\mathbf{x} + 2(\mathfrak{J}^2 - 1)^2 t, \\ \Gamma(\mathfrak{J}) &= \frac{\overline{B(\mathfrak{J})}}{\overline{a(\mathfrak{J})d(\mathfrak{J})}}. \end{aligned} \tag{57}$$

According to definition, we have to consider the residue conditions of *<sup>M</sup>*(*<sup>x</sup>*, *t*; *ξ*). By analyzing, we can know that both *a*(*ξ*) and *d*(*ξ*) have an even zero. Hence, we suppose that

1. *a*(*ξ*) has 2*n* simple zeros {*<sup>ε</sup>j*}<sup>2</sup>*nj*=1, 2*n* = 2*n*1 + 2*n*2. Furthermore, *<sup>ε</sup>j*(*j* = 1, 2, ..., <sup>2</sup>*n*1) lie in *D*1, *ε* ¯ *j*(*j* = 1, 2, ..., <sup>2</sup>*n*2) lie in *D*2.


**Theorem 3.** *For convenience, the mark* [*M*(*<sup>x</sup>*, *t*; *ξ*)]1 *denotes the first column of <sup>M</sup>*(*<sup>x</sup>*, *t*; *ξ*)*. Similarly,* [*M*(*<sup>x</sup>*, *t*; *ξ*)]2 *denotes the second column. At the same time, we let a*˙(*ξ*) = *dadξ . Then, we get the residue condition as follows:*

$$\mathbf{A}(\mathbf{i}) \qquad \text{Res}\{ [M(\mathbf{x}, t; \xi\_{\mathbf{i}}^{\tau})]\_{1}, \varepsilon\_{\mathbf{j}} \} = \frac{1}{\mathfrak{a}(\varepsilon\_{\mathbf{j}}) b(\varepsilon\_{\mathbf{j}})} \varepsilon^{2i\theta(\varepsilon\_{\mathbf{j}})} [M(\mathbf{x}, t; \varepsilon\_{\mathbf{j}})]\_{2\nu} \mathbf{j} = 1, 2, \dots, 2n\_{1\nu}$$

$$\begin{aligned} \text{(iii)} \quad \text{Res}\{ [M(\mathbf{x},t;\xi)]\_{2}, \overline{\varepsilon}\_{j} \} &= \frac{1}{\overline{a(\varepsilon\_{j})b(\overline{\varepsilon}\_{j})}} e^{-2i\theta(\overline{\varepsilon}\_{j})} [M(\mathbf{x},t;\overline{\varepsilon}\_{j})]\_{1}, j = 1,2,\dots,2u\_{2}, \\ &\qquad \overline{B(\overline{\varepsilon}\_{j})} \quad\_{\text{2-a}} \end{aligned} \tag{58}$$

$$\begin{aligned} \text{(iii)} \quad &\text{Res}\{ [M(\mathbf{x},t;\xi)]\_1, \gamma\_{\bar{j}} \} = \frac{B(\bar{\gamma}\_{\bar{j}})}{a(\gamma\_{\bar{j}})d(\gamma\_{\bar{j}})}e^{2i\theta(\gamma\_{\bar{j}})}[M(\mathbf{x},t;\gamma\_{\bar{j}})]\_{2\nu} \mathbf{j} = 1,2,\dots,2N\_1, \\\ &\text{(iv)} \quad &B(\bar{\gamma}\_{\bar{j}}) \quad &B(\bar{\gamma}\_{\bar{j}})[M(\mathbf{x},t;\mathbf{x})]\_{\bar{j}} \text{ (i.e., } \mathbf{x} \text{ )} = 1,\dots,2N\_1 \end{aligned}$$

$$\mathbf{P}(i\nu) \quad \text{Res}\{ [M(\mathbf{x}, \mathbf{t}; \tilde{\xi})]\_2, \tilde{\gamma}\_{\tilde{\jmath}} \} = \frac{\mathbf{e}^{\mathbf{B}(\tilde{\gamma}\_{\tilde{\jmath}})}}{a(\tilde{\gamma}\_{\tilde{\jmath}}) \dot{d}(\tilde{\gamma}\_{\tilde{\jmath}})} \mathbf{e}^{2i\theta(\tilde{\gamma}\_{\tilde{\jmath}})} [M(\mathbf{x}, \mathbf{t}; \tilde{\gamma}\_{\tilde{\jmath}})]\_1, j = 1, 2, \dots, 2N\_2.$$

**Proof.** Just prove (*i*), and the other proof can be proved in the same way. Firstly, we take account of *<sup>M</sup>*(*<sup>x</sup>*, *t*; *ξ*)=( *μ <sup>D</sup>*1 2 *a*(*ξ*) , *μ D*1∪*D*2 3 ), the simple zeros *<sup>ε</sup>j*(*j* = 1, 2, ..., <sup>2</sup>*n*1) of *a*(*ξ*) are the simple poles of *μ <sup>D</sup>*1 2 *a*(*ξ*). Then we ge<sup>t</sup>

 $\text{Res}\{\frac{\mu\_2^{D\_1}(\mathbf{x},t;\boldsymbol{\xi}^x\_2)}{a(\boldsymbol{\xi})},\varepsilon\_j\} = \lim\_{\boldsymbol{\xi}\to\boldsymbol{\varepsilon}\_j} (\boldsymbol{\xi}-\varepsilon\_j)$  $\frac{\mu\_2^{D\_1}(\mathbf{x},t;\boldsymbol{\xi}^x\_2)}{a(\boldsymbol{\xi})} = \lim\_{\boldsymbol{\xi}\to\boldsymbol{\varepsilon}\_j} \frac{\mu\_2^{D\_1}(\mathbf{x},t;\boldsymbol{\varepsilon}\_j)}{\frac{a(\boldsymbol{\xi})-a(\boldsymbol{\varepsilon}\_j)}{\boldsymbol{\xi}-\varepsilon\_j}} = \frac{\mu\_2^{D\_1}(\mathbf{x},t;\boldsymbol{\varepsilon}\_j)}{a(\boldsymbol{\varepsilon}\_j)}.$ 

Then taking *ξ* = *εj* into the equation

$$
\mu\_3^{D\_1 \cup D\_2} = e^{-2i\theta(\xi)} b(\xi) \mu\_2^{D\_1} + a(\xi) \mu\_2^{D\_4},
$$

we obtain

$$
\mu\_3^{D\_1 \cup D\_2}(\mathfrak{x}, t; \varepsilon\_j) = e^{-2i\theta(\varepsilon\_j)} b(\varepsilon\_j) \mu\_2^{D\_1}(\mathfrak{x}, t; \varepsilon\_j) + a(\varepsilon\_j) \mu\_2^{D\_4}(\mathfrak{x}, t; \varepsilon\_j),
$$

Finally,

$$\text{Res}\left\{\frac{\mu\_2^{D\_1}}{a(\xi)}, \varepsilon\_j\right\} = \frac{\mathfrak{c}^{2i\vartheta(\varepsilon\_j)}}{a(\varepsilon\_j)b(\varepsilon\_j)} \mu\_3^{D\_1 \cup D\_2} \left(\mathfrak{x}, t; \varepsilon\_j\right).$$

Now, we discuss how to derive the potential *q*(*<sup>x</sup>*, *t*)from the spectral functions *<sup>μ</sup>j*(*<sup>x</sup>*, *t*; *ξ*) (*j* = 1, 2, <sup>3</sup>). Reviewing what we did before, when (21) is a solution of (19), we have <sup>Ψ</sup>*<sup>o</sup>*1= *i*2*QD<sup>σ</sup>*3. Suppose

$$\mu = I + \frac{m^{(1)}}{\xi} + \frac{m^{(2)}}{\xi^2} + \frac{m^{(3)}}{\xi^3} + O(\xi^4), \xi \to \infty$$

is a solution of (32).

As *ξ* → <sup>∞</sup>, letting *<sup>m</sup>*(*<sup>x</sup>*, *t*) = *m*(1) 12 (*<sup>x</sup>*, *t*), namely

$$m(\mathfrak{x},t) = \lim\_{\substack{\mathfrak{z} \to \infty}} (\mathfrak{z}\mu\_{\mathfrak{j}}(\mathfrak{x},t;\mathfrak{\zeta}))\_{12}.$$

By direct calculation, we have

$$q(\mathbf{x},t) = 2im(\mathbf{x},t)e^{2i\int\_{(0,0)}^{(\mathbf{x},t)}\Lambda}m(\mathbf{x},t),\tag{59}$$

After that, it is clear to find that

$$q\eta = 4|m|^2,\\ q\overline{q}\_x - q\_x\overline{q} = 4(m\overline{n}\_x - m\_x\eta\overline{n}) - 32i|m|^4. \tag{60}$$

and

$$
\Delta = -2|m|^2 d\mathbf{x} + [2i(m\ddot{m}\_\mathbf{x} - m\_\mathbf{x}\ddot{m}) + 28|m|^4]dt.\tag{61}
$$

Eventually, we can ge<sup>t</sup> the final form of the potential *q*(*<sup>x</sup>*, *t*).

### **4. The Spectral Map and the Regular Riemann-Hilbert Problem**

### *4.1. The Spectral Map*

**Definition 1.** *For initial values q*0(*x*) = *q*(*<sup>x</sup>*, <sup>0</sup>)*, the map* S *can be defined by*

$$\mathbb{S}: \{q\_0(\mathfrak{x})\} \to \{a(\mathfrak{f}), b(\mathfrak{f})\}$$

*with*

$$\left(\begin{array}{c} b(\xi) \\ a(\xi) \end{array}\right) = \mu\_3^{(2)}(x,0;\xi), \operatorname{Im}\xi^2 \ge 0,$$

*where μ*3(*<sup>x</sup>*, 0; *ξ*) *is the unique solution of the Volterra linear integral equation*

$$\mu\_{\mathcal{B}}(\mathbf{x},0;\mathfrak{z}^{\mathbf{x}}) = I + \int\_{\infty}^{\infty} e^{i\lambda^{2}(\mathbf{x}'-\mathbf{x})\theta\_{\mathfrak{z}}} V\_{1}(\mathbf{x}',0;\mathfrak{z}^{\mathbf{x}}) \mu\_{\mathfrak{z}}(\mathbf{x}',0;\mathfrak{z}^{\mathbf{x}}) d\mathbf{x}'$$

*and <sup>V</sup>*1(*<sup>x</sup>*, 0; *ξ*) *is given by Equation (46).*

**Proposition 1.** *a*(*ξ*) *and b*(*ξ*) *possess these properties.*



$$q\_0(\mathbf{x}) = 2im(\mathbf{x})e^{4i\int\_0^{\mathbf{x}} |m(\mathbf{x}')|^2 d\mathbf{x}'},\\m(\mathbf{x}) = \lim\_{\substack{\mathbf{j} \rightarrow \infty \\ \mathbf{j}' \rightarrow \infty}} (\not{p}M^{(\mathbf{x})}(\mathbf{x}, \mathbf{j}'))\_{12}.\tag{62}$$

*where <sup>M</sup>*(*x*)(*<sup>x</sup>*, *ξ*) *is the unique solution of the following Riemann-Hilbert problem.*


$$J^{(\mathbf{x})}(\mathbf{x},\xi) = \begin{pmatrix} \frac{1}{a(\xi)a(\xi)} & \frac{b(\xi)}{a(\xi)}e^{-2i\lambda^2 x} \\ -\frac{b(\xi)}{a(\xi)}e^{2i\lambda^2 x} & 1 \end{pmatrix}, \xi^2 \in \mathbb{R}. \tag{63}$$


$$\text{Res}\{ [M^{(\mathbf{x})}(\mathbf{x}, \boldsymbol{\xi})]\_{1}, \boldsymbol{\varepsilon}\_{\boldsymbol{j}} \} = \frac{\boldsymbol{\varepsilon}^{2i(\boldsymbol{\varepsilon}\_{\boldsymbol{j}}^{2} - 1)} \boldsymbol{\chi}}{\boldsymbol{\dot{a}}(\boldsymbol{\varepsilon}\_{\boldsymbol{j}}) \boldsymbol{b}(\boldsymbol{\varepsilon}\_{\boldsymbol{j}})} [M^{(\mathbf{x})}(\mathbf{x}, \boldsymbol{\varepsilon}\_{\boldsymbol{j}})]\_{2}, \boldsymbol{j} = 1, 2, \dots, 2n\_{1} \tag{64}$$

$$\text{Res}\{\left[M^{(\mathbf{x})}(\mathbf{x},\boldsymbol{\varepsilon}\_{\mathbf{s}}^{\mathbf{r}})\right]\_{2\prime}\boldsymbol{\varepsilon}\_{\mathbf{j}}\} = \frac{e^{-2i(\boldsymbol{\varepsilon}\_{\mathbf{j}}^{2}-1)}\mathbf{x}}{\overline{\boldsymbol{d}(\boldsymbol{\varepsilon}\_{\mathbf{j}})\boldsymbol{b}(\boldsymbol{\varepsilon}\_{\mathbf{j}})}}[M^{(\mathbf{x})}(\mathbf{x},\boldsymbol{\varepsilon}\_{\mathbf{j}})]\_{1\prime};j=1,2,...,2n\_{2}.\tag{65}$$

**Definition 2.** *For boundary values g*0(*t*) = *q*(0, *t*), *g*1(*t*) = *qx*(0, *t*)*, the map* S ˜ *can be defined by* ˜

S : {*g*0(*t*), *g*1(*t*)}→{*A*(*ξ*), *B*(*ξ*)}

*with*

$$\left( \begin{array}{c} B(\xi) \\ A(\xi) \end{array} \right) = \mu\_1^{(2)}(0, t; \xi), \operatorname{Im} \xi^2 \ge 0,$$

*where μ*1(0, *t*; *ξ*) *is the unique solution of the Volterra linear integral equation*

$$\mu\_1(0, t; \mathfrak{f}) = I - \int\_t^T \mathfrak{e}^{2i\lambda^4(t'-t)\hat{\sigma}\_3} V\_2(0, t'; \mathfrak{f}) \,\mu\_1(0, t', \mathfrak{f}) dt'$$

*and <sup>V</sup>*2(0, *t*; *ξ*) *is given by (46).*

**Proposition 2.** *A*(*ξ*) *and B*(*ξ*) *possess these properties.*


$$\begin{split} g\_0(t) &= 2im\_{12}^{(1)}(t)e^{2i\int\_0^t \mathbf{A}\_2(t')dt'}, \\ g\_1(t) &= (4m\_{12}^{(2)} + |g\_0(t)|^2m\_{12}^{(1)}(t))e^{2i\int\_0^t \mathbf{A}\_2(t')dt'} + ig\_0(t)(2m\_{22}^{(2)}(t) + |g\_0(t)|^2), \end{split} \tag{66}$$

*where*

$$\Delta\_2(t) = 4|m\_{12}^{(1)}|^4 + 8(\operatorname{Re}[m\_{12}^{(1)}\bar{m}\_{12}^{(3)}] - |m\_{12}^{(1)}|^2 \operatorname{Re}[m\_{22}^{(2)}]),$$

*with the functions m*(*i*)(*t*)(*<sup>i</sup>* = 1, 2, 3.) *are depend on*

$$M^{(t)}(t,\xi^t) = I + \frac{m^{(1)}(t)}{\xi} + \frac{m^{(2)}(t)}{\xi^2} + \frac{m^{(3)}(t)}{\xi^3} + O(\frac{1}{\xi^4}), \xi \to \infty$$

*where <sup>M</sup><sup>t</sup>*(*<sup>t</sup>*, *ξ*) *is the unique solution of the following Riemann-Hilbert problem*


$$J^{(t)}(t,\xi) = \begin{pmatrix} \frac{1}{A(\xi)\overline{A(\xi)}} & \frac{B(\xi)}{A(\xi)}e^{-4i\lambda^4 t} \\ -\frac{B(\xi)}{A(\xi)}e^{4i\lambda^4 t} & 1 \end{pmatrix}, \xi^4 \in \mathbb{R}.\tag{67}$$


$$\text{Res}\{ [M^{(t)}(t,\zeta)]\_1, \gamma\_{\dot{\gamma}} \} = \frac{e^{4i(\gamma\_{\dot{\gamma}}^2 - 1)^2 t}}{A(\gamma\_{\dot{\gamma}})B(\gamma\_{\dot{\gamma}})} [m^{(t)}(t,\gamma\_{\dot{\gamma}})]\_{2\prime} j = 1,2,...,2N\_1. \tag{68}$$

$$\text{Res}\{[M^{(t)}(t,\xi)]\_{2},\bar{\gamma}\_{j}\} = \frac{e^{-4i(\gamma\_{j}^{2}-1)^{2}t}}{\dot{A}(\bar{\gamma}\_{j})b(\bar{\gamma}\_{j})}[M^{(t)}(t,\bar{\gamma}\_{j})]\_{1},j=1,2,...,2N\_{2}.\tag{69}$$

### *4.2. The Regular Riemann-Hilbert Problem*

**Theorem 4.** *Given the smooth function q*0(*x*)*, which is compatible with g*0(*t*) *and g*1(*t*)*. The spectral functions <sup>a</sup>*(*ξ*)*,b*(*ξ*)*,<sup>A</sup>*(*ξ*)*, and B*(*ξ*) *are defined according to the previous definitions. Furthermore, they satisfy the global relation (49). Clearly, it becomes <sup>B</sup>*(*ξ*)*a*(*ξ*) − *A*(*ξ*)*b*(*ξ*) = 0 *when ξ* → ∞*. Define the <sup>M</sup>*(*<sup>x</sup>*, *t*; *ξ*) *as the solution of this following Riemann-Hilbert problem.*


$$M\_+(\mathfrak{x}, \mathfrak{t}; \mathfrak{f}) = M\_-(\mathfrak{x}, \mathfrak{t}; \mathfrak{f}) f(\mathfrak{x}, \mathfrak{t}; \mathfrak{f}), \mathfrak{f}^4 \in R\_+$$

*where the jump matrices are defined by (55)–(57).*

• *<sup>M</sup>*(*<sup>x</sup>*, *t*; *ξ*) = *I* + *O*( 1*ξ* ), *ξ* → ∞.

*Then, <sup>M</sup>*(*<sup>x</sup>*, *t*; *ξ*) *not only exists but is unique. In this way, the solution of the mNLS equation can be derived, which can be defined by*

$$\begin{aligned} q(\mathbf{x},t) &= 2im(\mathbf{x},t)e^{2i\int\_{(0,0)}^{(\mathbf{x},t)} \Delta}, \\ m(\mathbf{x},t) &= \lim\_{\xi \to \infty} (\not\!\!\!\!\mu\_{j}(\mathbf{x},t;\xi))\_{12\prime} \\ \Delta &= -2|m|^{2}dx + [2i(m\mathfrak{m}\_{x} - m\_{x}m) + 28|m|^{4}]dt. \end{aligned} \tag{70}$$

*Besides, q*(*<sup>x</sup>*, *t*) *also satisfies the initial-boundary values condition*

$$q(\mathbf{x},0) = q\_0(\mathbf{x}), \\ q(0,t) = \mathbb{g}\_0(t), \\ \text{and } q\_\mathbf{x}(0,t) = \mathbb{g}\_1(t).$$

**Proof.** Actually, if there are no zeros of *a*(*ξ*) and *d*(*ξ*), then the 2 × 2 function *<sup>M</sup>*(*<sup>x</sup>*, *t*; *ξ*) satisfies a non-singular Riemann-Hilbert problem. Due to the jump matrices *J*(*<sup>x</sup>*, *t*; *ξ*) possessing symmetry, we can find that this problem has a unique solution. On the other hand, when *a*(*ξ*) and *d*(*ξ*) have a certain number of zeros, by specific mapping, the singular Riemann-Hilbert problem can become no zeros with a system of algebraic equations; the unique solvability can be proved by the following theorem.

**Theorem 5.** *The Riemann-Hilbert problem in Theorem 4 with the vanishing boundary condition*

$$M(\mathfrak{x}, t; \xi) \to 0, \xi \to \infty,$$

*has only the zero solution.*

**Proof.** Firstly, we suppose that the matrix function *<sup>M</sup>*(*<sup>x</sup>*, *t*; *ξ*) is a solution of the Riemann-Hilbert problem in Theorem 4. At the same time, *A*† means the complex conjugate transpose of *A*, where *A* is a 2 × 2 matrix. We define

¯

$$\begin{aligned} H\_{+}(\lhd) &= M\_{+}(\lhd)M\_{-}^{\dagger}(-\lkern-1.1mu\lkern-1.1mu\lkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1mu\mkern-1.1$$

where the *x* and *t* are dependent with each other. *H*+(*ξ*) and *<sup>H</sup>*−(*ξ*) are analytic in {*ξ* ∈ *C*|*Imξ*<sup>4</sup> > 0} and {*ξ* ∈ *C*|*Imξ*<sup>4</sup> < <sup>0</sup>}, respectively. Due to the symmetry, we can obtain from (54) and (55)

$$J\_1^\uparrow(-\underline{\xi}) = J\_1(\underline{\xi}), \\ J\_3^\uparrow(-\underline{\xi}) = J\_3(\underline{\xi}), \\ J\_2^\uparrow(-\underline{\xi}) = J\_4(\underline{\xi}). \tag{72}$$

Then

$$\begin{aligned} H\_{+}\left(\xi\right) &= M\_{-}\left(\xi\right)I\left(\xi\right)M\_{-}^{\sharp}\left(-\frac{\overline{\varsigma}}{\xi}\right), Im\xi^{4} \in \mathbb{R}, \\ H\_{-}\left(\xi\right) &= M\_{-}\left(\overline{\varsigma}\right)I^{\dagger}\left(-\frac{\overline{\varsigma}}{\overline{\varsigma}}\right)M\_{-}^{\sharp}\left(-\frac{\overline{\varsigma}}{\overline{\varsigma}}\right), Im\xi^{4} \in \mathbb{R}. \end{aligned} \tag{73}$$

From the above two equations, it is easy to find that *H*+(*ξ*) = *<sup>H</sup>*−(*ξ*). This means that *H*+(*ξ*) and *<sup>H</sup>*−(*ξ*) define an entire function decaying at infinity, hence the *H*+(*ξ*) ≡ 0 and *<sup>H</sup>*−(*ξ*) ≡ 0. Finding *J*3(*ih*¯)(*h*¯ ∈ *R*) is a 2 × 2 unit Hermitian matrix for any *h*¯ ∈ *R*. It is not difficult to see that *J*3(*ih*¯)(*h*¯ ∈ *R*) is a positive definite matrix. Now that *<sup>H</sup>*−(*h*¯) = 0 for ¯*h* ∈ *iR*, we have

$$(M\_+\left(i\hbar\right))\_3(i\hbar)M\_+^\dagger(i\hbar) = 0.\tag{74}$$

After simple calculation, we have *<sup>M</sup>*+(*ih*¯) = 0 for ¯*h* ∈ *R*. Therefore, *M*+(*ξ*) = 0, *<sup>M</sup>*−(*ξ*) = 0.

### **Remark 1.** *q*(*<sup>x</sup>*, *t*) *satisfies the mNLS equation.*

*In fact, if <sup>M</sup>*(*<sup>x</sup>*, *t*; *ξ*) *is the solution of the Riemann-Hilbert problem defined by Theorem 4 and q*(*<sup>x</sup>*, *t*) *is defined as the previous definition, with the help of the dressing method [45], we can find that q*(*<sup>x</sup>*, *t*) *satisfies the Lax pair (18). Hence, q*(*<sup>x</sup>*, *t*) *satisfies the mNLS equation.*

**Remark 2.** *Using the same proof method in Reference [32] can we prove that q*(*<sup>x</sup>*, *t*) *satisfies the initial values q*(*<sup>x</sup>*, 0) = *q*0(*x*) *and boundary values q*(0, *t*) = *g*0(*t*), *qx*(0, *t*) = *g*1(*t*)*, so in this paper, we leave this proof out.*
