The vector derivative nonlinear Schrödinger equation on the half-line

The vector derivative nonlinear Schrödinger equation on the half-line Abstract In this paper, we analyse initial-boundary problems for the vector derivative nonlinear Schrödinger equation on the semi-infinite strip $$(x, t) \in (0,\infty )\times (0, T)$$ via the unified transform method of Fokas. Even though additional technical complications arise in the vector case compared with scalar ones, we show that it also can be expressed in terms of the solution of a matrix Riemann–Hilbert problem. The Riemann–Hilbert problem involves a jump matrix, uniquely defined in terms of four matrix functions called spectral functions and denoted by {a(λ), b(λ), A(λ), B(λ)} that depend on the initial data and all boundary values, respectively. A key role is played by the so-called global relation which involves the known and unknown boundary values. By analysing the global relation, we present an effective characterization of the latter two spectral functions in terms of the given initial and boundary data. 1. Introduction One of the most remarkable developments in the theory of integrable systems is inverse scattering transform, which was introduced to solve the initial value problems of nonlinear integrable evolution equations (Gardner et al., 1967, 1974). Following that, a natural issue is to study the initial-boundary value (IBV) problems. A general approach to IBV problems for integrable nonlinear equations originated in the seminal work of Fokas (1997), which is usually referred to as the unified method, or the Fokas method, and it has been actively developed since then (Fokas, 2002, 2008, 2005; Fokas et al., 2005, Fokas & Lennels, 2010, 2012; Lenells, 2008, 2011; Lenells & Fokas, 2009; Boutet De Monvel et al., 2004, 2006, 2009). Interestingly, there exist several integrable vector extensions of scalar classical integrable equations such as the vector NLS equation (Fordy & Kulish, 1983), the vector DNLS equation (Fordy, 1984) and the vector modified KdV equation Kaup & Newell (1978), which describe the wave interactions in the case of that two or more wave packets of different carrier frequencies appear simultaneously. Considering the applications in some physical situations, we believe that it is significant to investigate multi-component cases. Therefore, in this paper we study IBV problems for the vector DNLS equation,   $$ i q_{t}+q_{xx}+i(|q|^{2}q)_{x}=0, \quad q=(q_{1},q_{2},\ldots,q_{n})^{T}, $$ (1.1) where |q|2 = q*q, the superscript “*” denotes Hermitian conjugation. The scalar case, known as the celebrated Kaup–Newell equation (Kaup & Newell, 1978), is relevant to Alvén waves in plasma physics and the propagation of nonlinear pulses in optical fibres (Mio et al., 1976; Mjolhus, 1976; Kodama, 1985). The global existence and long-time behaviour of scalar DNLS have been investigated by inverse scattering transform method Liu et al. (2016, 2017). The unified transform method has formerly been applied to the IBV problems for the scalar DNLS equation (Lenells, 2008) and integrable evolutions with 3 × 3 Lax pairs (Lenells, 2012, 2013; Caudrelier & Zhang 2012; Xu & Fan, 2013, 2016; Geng et al., 2015; Liu & Geng, 2016). Although the analysis of (1.1) is inspired by the preceding references, it also involves some novelties: (a) In order to ensure that the normalization of RH problem is canonical, i.e. the solution of RH problem is the identity matrix at infinity, we transform the Lax pair to a new one by introducing the matrix functions Y (x, t) and $$\tilde {Y}(x)$$. Indeed, provided that q(x, t) solves the equation (1.1), both of them cannot be explicitly expressed in terms of a closed 1-form as in the scalar case (Lenells, 2008) but are well defined. (b) All of the higher-order square matrices in this paper can be rewritten as 2 × 2 block ones, thus we can directly formulate the higher-order matrix RH problem by the combinations of the entries of three eigenfunctions. Compared with the idea of that constructing the 3 × 3 matrix RH problem by the Fredholm integral equation in Lenells (2012) (also see Beals & Coifman, 1984), this procedure is more convenient for analysis of the multi-component evolution equations. (c) By introducing Sherman–Morrison formula, we derive the jump matrices and residue conditions of the RH problem. The outline of the paper is as follows. In Section 2 we define three eigenfunctions of the Lax pair for spectral analysis. In addition, we investigate the matrix-valued spectral functions a(λ), b(λ), A(λ) and B(λ) further in Section 3. In Section 4 we show that q(x, t) can be expressed in terms of the unique solution of a matrix RH problem which involves the aforementioned spectral functions. In Section 5 we present the so-called Dirichlet to Neumann map and discuss the vanishing Dirichlet boundary data further. 2. Spectral analysis Basic assumptions and notations Denote $$\Omega =\{(x,t)|0<x<\infty ,0<t<T\}$$, where T is a given positive constant, unless otherwise specified, we suppose that $$T<\infty $$. Let q(x, t) be a complex vector-valued function of (x, t) ∈Ω, assume that $$q(x,t)\in C^{2,1}(\mathbb {R}^{+}\times \mathbb {R}^{+})$$, for all t as $$x\rightarrow \infty $$, $$q(x,t)\rightarrow 0$$, $$q_{x}(x,t)\rightarrow 0$$. The initial data q0(x), the Dirichlet boundary data g0(t) and the Neumann boundary data g1(t) of equation (1.1) are denoted by  \begin{align*} q_{0}(x)&=q(x,0),\quad 0<x<\infty,\\ g_{0}(t)&=q(0,t),\quad g_{1}(t)= q_{x}(0,t),\quad 0<t<T. \end{align*} Let $$D_{k}=\left \{\lambda \in \mathbb {C}:\arg \lambda ^{2}\in \left (\frac {(k-1)\pi }{2},\frac {k\pi }{2}\right )\right \}$$, k = 1, …, 4, see Fig. 2; $$\mathbb {C}_{\pm }=\{\lambda \in \mathbb {C}:\mathrm {Im}\lambda ^{2}\gtrless 0\}$$. Denote the matrix A = (AL, AR), where AL is the first column of A, and AR is the rest of columns. The notation A(λ), λ ∈ (D1, D2), means that AL and AR hold for λ ∈ D1 and D2, respectively. In this paper, without otherwise specified, let each matrix A be rewritten as a block form $$A=\left({{A_{11}}\atop{A_{21}}} \quad {{A_{12}} \atop{A_{22}}} \right)$$, where A11 is scalar. Let In denote n × n identity matrix. In this paper, $$\lambda \rightarrow \lambda _{0}$$ or $$\lambda \rightarrow \infty $$ non-tangentially on $$\mathbb {C}$$. 2.1 Lax pair The vector DNLS equation (1.1) admits the Lax pair Fordy (1984)  $$ \begin{aligned} \psi_{x}&=i\lambda^{2}\sigma\psi+\lambda Q\psi,\\ \psi_{t}&=2i\lambda^{4}\sigma \psi+(2\lambda^{3} Q+i\lambda^{2}Q^{2}\sigma+i\lambda Q_{x}\sigma+\lambda Q^{3})\psi, \end{aligned} $$ (2.1) where ψ(x, t, λ) is a (n + 1) × (n + 1) matrix-valued function and $$\lambda \in \mathbb {C}$$ is the spectral parameter,  $$ \sigma=\left(\begin{array}{ccc} 1&0\\0&-I_{n} \end{array}\right),\quad Q(x,t)=-i\left(\begin{array}{cc} 0&q^{\ast}(x,t)\\ q(x,t)&0 \end{array}\right).$$ In order to seek solutions of the spectral problem which approach the identity matrix In+1 as $$\lambda \rightarrow \infty $$ and ensure that the spectral function S(λ) defined in the following is independent of the initial data, following the similar steps as in Lenells (2008), we define two (n + 1) × (n + 1) matrix-valued functions Y (x, t) and $$\tilde {Y}(x)$$ by  $$ Y(x,t)=\left(\begin{array}{cc} \mathrm{e}^{i\int_{(0,0)}^{(x,t)}\Delta(x^{\prime},t^{\prime})}&0 \\0 &W(x,t) \end{array}\right),\quad \tilde{Y}(x)=\left(\begin{array}{cc} \mathrm{e}^{\frac{i}{2}\int^{x}_{\infty}|q_{0}|^{2}(x^{\prime})\,\mathrm{d}x^{\prime}}&0 \\0 &\tilde{W}(x) \end{array}\right), $$ (2.2) where Δ(x, t) is the closed real-valued 1-form  $$ \Delta=\Delta_{1}+\Delta_{2},\quad \Delta_{1}=\frac{1}{2}|q |^{2} \mathrm{d}x,\quad \Delta_{2}=-\frac14\left(3|q |^{4}-2iq^{\ast} q_{x} +2iq^{\ast}_{x} q \right)\mathrm{d}t, $$ (2.3) which follows by a conservation law of equation (1.1)  $$ \frac{i}{2} \left(|q|^{2}\right)_{t}+\frac14\left(3i|q|^{4}+2q^{\ast}q_{x}-2q^{\ast}_{x}q\right)_{x}=0, $$ and the n × n matrix-valued functions W(x, t) and $$\tilde {W}(x)$$ are the unique solutions of  $$ \begin{cases} W_{x}=-\frac{i}{2}qq^{\ast}W,\\ W_{t}=\frac14\left(3i|q|^{2}qq^{\ast}-2qq^{\ast}_{x}+2q_{x}q^{\ast}\right)W,\\ W(0,0)=I_{n}, \end{cases}\quad \begin{cases} \tilde{W}_{x}=-\frac{i}{2}q_{0}q_{0}^{\ast}\tilde{W},\\ \tilde{W}(\infty)=I_{n}. \end{cases} $$ (2.4) Provided that q(x, t) solves equation (1.1), it can be shown that Wxt = Wtx, thus the solution W exists. In other words, equation (2.4) is equivalent to the Volterra integral equations  $$ \begin{aligned} W(x,t)&=I_{n}+\int_{(0,0)}^{(x,t)}\omega(x^{\prime},t^{\prime}),\\ \omega(x,t)&= -\frac{i}{2}qq^{\ast}W\mathrm{d}x+\frac14\left(3i|q|^{2}qq^{\ast}-2qq^{\ast}_{x}+2q_{x}q^{\ast}\right)W\mathrm{d}t, \end{aligned} $$ (2.5)  $$ \tilde{W}(x)=I_{n}+\frac i2\int_{x}^{\infty} q_{0}(x^{\prime})q_{0}^{\ast}(x^{\prime})\tilde{W}(x^{\prime})\mathrm{d}x^{\prime}. $$ (2.6) Also, W*(x, t) and W−1(x, t) satisfy the same linear differential equations and initial condition, thus they are identical. Introducing a new eigenfunction μ(x, t, λ) by  $$ \psi(x,t,\lambda)=Y(x,t)\mu(x,t,\lambda)\tilde{Y}(0)\mathrm{e}^{i(\lambda^{2}x+2\lambda^{4}t)\sigma}, $$ (2.7) we obtain the Lax pair equations  $$ \mu_{x}-i\lambda^{2}[\sigma,\mu]=U\mu, $$ (2.8a)  $$ \mu_{t}-2i\lambda^{4}[\sigma,\mu]=V\mu, $$ (2.8b) where  \begin{align*}U(x,t,\lambda)&=\lambda Y^{-1}QY+\frac{i}{2}Y^{-1}Q^{2}\sigma Y,\\ V(x,t,\lambda)&=Y^{-1}\left(2\lambda^{3} Q+i\lambda^{2}Q^{2}\sigma+i\lambda Q_{x}\sigma+\lambda Q^{3}+\frac{3i}{4}Q^{4}\sigma-\frac{1}{2}[Q,Q_{x}]\right)Y. \end{align*} Equation (2.8) can be rewritten in differential form as  $$ \mathrm{d}\left(\mathrm{e}^{-i(\lambda^{2}x+2\lambda^{4}t)\hat{\sigma}}\mu\right)=\tilde{\omega}, $$ (2.9) where the exact 1-form $$\tilde {\omega }(x,t,\lambda )$$ is defined by  $$ \tilde{\omega}=\mathrm{e}^{-i(\lambda^{2}x+2\lambda^{4}t)\hat{\sigma}}(U\mu\,\mathrm{d}x+V\mu\,\mathrm{d}t), $$ (2.10) and $$\hat {\sigma }$$ acts on a matrix X by $$\hat {\sigma }X=[\sigma , X]$$, then $$\mathrm {e}^{\hat {\sigma }}X=\mathrm {e}^{\sigma }X\mathrm {e}^{-\sigma }$$. 2.2 Bounded and analytic eigenfunctions Three eigenfunctions $$\{\mu _{j}\}_{1}^{3}$$ of equation (2.8) are defined by the Volterra integral equations  $$ \mu_{j}(x,t ,\lambda)=I_{n+1}+\int_{\gamma_{j}}\mathrm{e}^{i\left(\lambda^{2}x+2\lambda^{4}t\right)\hat{\sigma}}\tilde{\omega}_{j}(x^{\prime},t^{\prime},\lambda),\quad j=1,2,3, $$ (2.11) where $$\tilde {\omega }_{j}$$ is defined by equation (2.10) with μ replaced by μj, the contour γj denotes a smooth curve from (xj, tj) to (x, t), and (x1, t1) = (0, T), (x2, t2) = (0, 0), $$(x_{3},t_{3})=(\infty ,t)$$. Since the 1-form is exact, $$\{\mu _{j}\}_{1}^{3}$$ are independent of the path of integration, specially we can choose the contours shown in Fig. 1, and this choice implies the following inequalities on the contours: Fig. 1. View largeDownload slide The contours γ1, γ2 and γ3 in the (x′, t′)-plane. Fig. 1. View largeDownload slide The contours γ1, γ2 and γ3 in the (x′, t′)-plane.   $$ \begin{aligned} \gamma_{1}:x-x^{\prime}&\geqslant 0,\quad t-t^{\prime}\leqslant 0,\\ \gamma_{2}:x-x^{\prime}&\geqslant 0,\quad t-t^{\prime}\geqslant 0,\\ \gamma_{3}:x-x^{\prime}&\leqslant 0.\\ \end{aligned} $$ (2.12) The first column of the matrix equation (2.11) involves the exponential $$\mathrm {e}^{2i\lambda ^{2}(x^{\prime }-x)+4i \lambda ^{4}(t^{\prime }-t)}$$. Using the inequalities (2.12) it follows that the exponential is bounded in the following regions of the complex λ-plane:  $$ \begin{aligned} \gamma_{1}&: \{\mathrm{Im}\,\lambda^{2}<0 \}\cap\{ \mathrm{Im}\,\lambda^{4}>0\},\\ \gamma_{2}&: \{\mathrm{Im}\,\lambda^{2}<0 \}\cap\{ \mathrm{Im}\,\lambda^{4}<0\},\\ \gamma_{3}&:\{\mathrm{Im}\,\lambda^{2}>0\}. \end{aligned} $$ (2.13) Also, similar to that in Kaup & Newell (1978), we assume that  $$ \int_{0}^{\infty} \left(|q|+|q|^{2}+|q_{x}|\right)\mathrm{d}x<\infty,$$ where $$|q|=\sum _{j=1}^{n} |q_{j}|.$$ Since the equations in (2.11) are Volterra integral equations, these boundedness properties imply that μ1L, μ2L and μ3L are bounded and analytic for λ ∈ D3, D4 and $$\mathbb {C}_{+}$$, respectively. Similar conditions are valid for $$\{\mu _{jR}\}_{1}^{3}$$. Thus,  $$ \begin{aligned} &\mu_{1}\ \textrm{is bounded and analytic for}\ \lambda\in(D_{3},D_{2}),\\ &\mu_{2}\ \textrm{is bounded and analytic for} \ \lambda\in(D_{4},D_{1}),\\ &\mu_{3}\ \textrm{is bounded and analytic for}\ \lambda\in(\mathbb{C}_{+},\mathbb{C}_{-}), \end{aligned} $$ (2.14) and all of them admit a bounded and continuous extension to the boundary. For T finite, μ1 and μ2 are entire functions of λ. Moreover, the column vectors of μj approach the corresponding column vectors of the identity matrix as $$\lambda \rightarrow \infty $$ in their corresponding regions of boundedness, e.g.  $$ \left(\mu_{3L},\mu_{2R}\right)=I_{n+1}+O\left(\frac{1}{\lambda}\right),\quad \lambda\in D_{1}\rightarrow \infty. $$ 2.3 The spectral functions The solutions of the system of differential equations (2.1) must be related by a matrix independent of x and t, therefore  $$ \begin{aligned} \mu_{3}(x,t,\lambda)&=\mu_{2}(x,t,\lambda)\mathrm{e}^{i(\lambda^{2}x+2\lambda^{4}t)\hat{\sigma}}s(\lambda),\\ \mu_{1}(x,t,\lambda)&=\mu_{2}(x,t,\lambda)\mathrm{e}^{i(\lambda^{2}x+2\lambda^{4}t)\hat{\sigma}}S(\lambda). \end{aligned} $$ (2.15) Evaluating the above equations at (x, t) = (0, 0) and (x, t) = (0, T), respectively, we note that the following (n + 1) × (n + 1) matrix-valued spectral functions s(λ) and S(λ):  $$ s(\lambda)=\mu_{3}(0,0,\lambda),\quad S(\lambda)=\mu_{1}(0,0,\lambda)=\mathrm{e}^{2i\lambda^{4}T\hat{\sigma}}\mu^{-1}_{2}(0,T,\lambda). $$ (2.16) Hence, functions s(λ) and S(λ) can be obtained from the evaluations at x = 0 and at t = 0 of functions μ3(x, 0, λ) and μ1(0, t, λ), respectively, which satisfy the linear integral equations  $$ \mu_{3}(x,0,\lambda)=I_{n+1}-\int_{x}^{\infty} \mathrm{e}^{i\lambda^{2}(x-x^{\prime})\hat{\sigma}}(U\mu_{3})(x^{\prime},0,\lambda)\mathrm{d}x^{\prime}, $$ (2.17)  $$ \mu_{1}(0,t,\lambda)=I_{n+1}-{{{\int_{t}^{T}}}} \mathrm{e}^{2i\lambda^{4}(t-t^{\prime})\hat{\sigma}}(V\mu_{1})(0,t^{\prime},\lambda)\mathrm{d}t^{\prime}. $$ (2.18) For x = 0, μ1 and μ2 have the following enlarged domains of boundedness: μ1(0, t, λ) is bounded for λ ∈ (D1 ∪ D3, D2 ∪ D4), μ2(0, t, λ) is bounded for λ ∈ (D2 ∪ D4, D1 ∪ D3). Proposition 2.1 For j = 1, 2, 3, function μj(x, t, λ) satisfies the symmetry relations  $$ \mu^{-1}_{j}(x,t,\lambda)=\mu_{j}^{\ast}(x,t,\lambda^{\ast}),\quad \mu_{j}(x,t,\lambda)=\sigma\mu_{j}(x,t,-\lambda)\sigma. $$ (2.19) Proof. It is easy to see that this proof follows by the symmetry relations  $$ Y^{-1}=Y^{\ast},\quad Q^{\ast}=-Q,\quad [\sigma, Y]=0, \quad\sigma Q\sigma=-Q,$$ and the initial conditions μj(xj, tj, λ) = In+1, j = 1, 2, 3. □ The fact that U and V are traceless implies $$\det [\mu (x, t,\lambda )] $$ is independent of x and t. Evaluation of $$\det [\mu _{j}(x, t,\lambda )]$$ at (xj, tj) shows that $$\det [\mu _{j}(x, t,\lambda )]=1,j=1,2,3$$. Consequently,  $$ \det [s(\lambda)]=\det [S(\lambda)]=1. $$ (2.20) It follows from(2.16), (2.19) and (2.20) that  $$ s^{\ast}(\lambda^{\ast})=\mathrm{adj}[s(\lambda)],\quad S^{\ast}(\lambda^{\ast})=\mathrm{adj}[S(\lambda)], $$ (2.21)  $$ s^{\ast}(\lambda^{\ast})s(\lambda)=I,\quad S^{\ast}(\lambda^{\ast})S(\lambda)=I, $$ (2.22) where adj(A) denotes the classical adjoint matrix of A. From (2.22), we obtain  $$ (s^{\ast}(\lambda^{\ast})s(\lambda))_{12}=0,\quad (S^{\ast}(\lambda^{\ast})S(\lambda))_{12}=0. $$ (2.23) Therefore, from (2.21) and (2.23), we find  \begin{align*} s^{\ast}_{11}(\lambda^{\ast})&=\det [s_{22}(\lambda)], \quad s^{\ast}_{21}(\lambda^{\ast})=-s_{12}(\lambda)\mathrm{adj}[s_{22}(\lambda)],\\ S^{\ast}_{11}(\lambda^{\ast})&=\det [S_{22}(\lambda)], \quad S^{\ast}_{21}(\lambda^{\ast})=-S_{12}(\lambda)\mathrm{adj}[S_{22}(\lambda)]. \end{align*} Thus s(λ) and S(λ) can be written as the following form:  $$ s(\lambda)=\left(\begin{array}{cc} \det [a^{\ast}(\lambda^{\ast})]&b(\lambda)\\ -\mathrm{adj}\left[a^{\ast}(\lambda^{\ast})\right]b^{\ast}(\lambda^{\ast})&a(\lambda) \end{array}\right),\quad S(\lambda)=\left(\begin{array}{cc} \det[ A^{\ast}(\lambda^{\ast})]&B(\lambda)\\ -\mathrm{adj}\left[A^{\ast}(\lambda^{\ast})\right]B^{\ast}(\lambda^{\ast})&A(\lambda) \end{array}\right). $$ (2.24) Also, the symmetry relations (2.19) imply that  $$ a(-\lambda)=a(\lambda),\quad b(-\lambda)=-b(\lambda),\quad A(-\lambda)=A(\lambda),\quad B(-\lambda)=-B(\lambda). $$ (2.25) The definitions of s(λ) and S(λ) imply  $$ \left(\begin{array}{c} b(\lambda)\\a(\lambda) \end{array}\right)=\mu_{3R}(0,0,\lambda),\quad \left(\begin{array}{c} B(\lambda)\\A(\lambda) \end{array}\right)=\mu_{1R}(0,0,\lambda)=\left[\mathrm{e}^{2i\lambda^{4}T\hat{\sigma}}\mu^{-1}_{2}(0,T,\lambda)\right]_{R}.$$ The above definitions imply the following properties of the spectral functions: a(λ) and b(λ) are analytic for $$\lambda \in \mathbb {C}_{-}$$ and continuous and bounded for $$\lambda \in \bar {\mathbb {C}}_{-}$$, A(λ) and B(λ) are entire functions and bounded for $$\lambda \in \bar {D}_{2}\cup \bar {D}_{4}$$. a*(λ*)a(λ) + b*(λ*)b(λ) = In, A*(λ*)A(λ) + B*(λ*)B(λ) = In, $$\lambda \in \mathbb {C}$$. $$a(\lambda )=I_{n}+O\left (\frac {1}{\lambda }\right )$$, $$b(\lambda )=O\left (\frac {1}{\lambda }\right ),\ \lambda \in \mathbb {C}_{-}\rightarrow \infty $$, $$A(\lambda )=I_{n}+O\left (\frac {1}{\lambda }\right )$$, $$B(\lambda )=O\left (\frac {1}{\lambda }\right ),\ \lambda \in D_{2}\cup D_{4}\rightarrow \infty $$. All of these properties follow from the analyticity and boundedness of μ3(x, 0, λ) and μ1(0, t, λ), from the relation s*(λ*)s(λ) = S*(λ*)S(λ) = In+1 and from the large λ asymptotics of these eigenfunctions. (ii) shows that whereas a(λ) and b(λ) are bounded for $$\lambda \in \bar {\mathbb {C}}_{-}$$, the expression appearing on the left-hand side of equation is identity matrix for all $$\lambda \in \mathbb {C}$$. Also, (ii) imply that  $$ \det [a^{\ast}(\lambda^{\ast})a(\lambda)]+b(\lambda)b^{\ast}(\lambda^{\ast})=1,\quad\det [A^{\ast}(\lambda^{\ast})A(\lambda)]+B(\lambda)B^{\ast}(\lambda^{\ast})=1.$$ 2.4 The Riemann–Hilbert problem We introduce the following proposition to prepare for formulation of the RH problem. Proposition 2.2 (Sherman–Morrison formula; Sherman & Morrison, 1950; Press, 2007; Bartlett, 1951) Suppose that $$X\in \mathbb {R}^{n}\times \mathbb {R}^{n}$$ is an invertible square matrix, x and y are n-dimensional column vectors, if X + xyT is invertible, thus  $$ (X+xy^{T})^{-1}=X^{-1}-\frac{X^{-1}xy^{T}X^{-1}}{1+y^{T}X^{-1}x}.$$ Corollary 2.3 Suppose that $$X\in \mathbb {R}^{n}\times \mathbb {R}^{n}$$, x, y and z are n-dimensional column vectors, thus  $$ \mathrm{adj}(X+xy^{T})x=\mathrm{adj}(X)x, $$ (2.26)  $$ \mathrm{adj}[X+\det( X)xy^{T}]=[1+y^{T}\mathrm{adj} (X) x ]\mathrm{adj}(X)-\mathrm{adj} (X) xy^{T} \mathrm{adj} (X), $$ (2.27) In particular, as $$\det X=0$$,  $$ \mathrm{adj} (X) xy^{T} \mathrm{adj} (X)=\left[y^{T}\mathrm{adj} (X) x\right ]\mathrm{adj}(X), $$ (2.28)  $$ \frac{\mathrm{adj} (X) x }{y^{T}\mathrm{adj} (X) x } =\frac{\mathrm{adj} (X) z }{y^{T}\mathrm{adj} (X) z } . $$ (2.29) Proof. As X, X + xyT and $$X+\det ( X)xy^{T}$$ are invertible, it is easy to verify that this corollary follows by Proposition 2.2. There exist n distinct real number α1, …, αn, such that X + αjI, X + αjI + xyT and $$X+\alpha _{j}I+\det ( X)xy^{T}$$ are invertible. Therefore, this corollary holds for all $$\alpha \in \mathbb {R}$$ when X is replaced by X + αI. In particular, when α = 0, this corollary is proved. □ Equation (2.15) can be rewritten in a form expressing the jump condition of a (n + 1) × (n + 1) matrix RH problem. This involves only tedious but straightforward algebraic manipulations. The final form is  $$ M_{-}(x,t,\lambda)=M_{+}(x,t,\lambda)J(x,t,\lambda),\quad \lambda^{4}\in\mathbb{R}, $$ (2.30) where the matrices M+, M− and J are defined as follows:  $$ \begin{aligned} M_{+}&=\left(\mu_{3L},\mu_{2R}\,\left[a^{\ast}(\lambda^{\ast})\right]^{-1}\right),\quad \lambda\in \bar{D}_{1},\quad M_{-}=\left(\mu_{3L},\mu_{1R}\,\left[d^{\ast}(\lambda^{\ast})\right]^{-1}\right), \quad \lambda\in \bar{D}_{2},\\ M_{+}&=\left(\frac{\mu_{1L}}{\det [d(\lambda)]},\mu_{3R} \right), \quad \lambda\in \bar{D}_{3},\quad M_{-}=\left(\frac{\mu_{2L}}{\det [a(\lambda)]},\mu_{3R}\right), \quad \lambda\in \bar{D}_{4}, \end{aligned} $$ (2.31)  $$ d(\lambda)=A^{\ast}(\lambda^{\ast})a(\lambda)+B^{\ast}(\lambda^{\ast})b(\lambda),$$  $$ J=\begin{cases} J_{1},&\arg \lambda^{2}=\frac{\pi}{2},\\[1ex] J_{2}=J_{3}J_{4}^{-1}J_{1},&\arg \lambda^{2}=\pi,\\[1ex] J_{3},&\arg \lambda^{2}=\frac{3\pi}{2},\\[1ex] J_{4},&\arg \lambda^{2}=0, \end{cases} $$ (2.32) with  $$ J_{1}=\left(\begin{array}{cc} 1&\mathrm{e}^{2i\theta}\Gamma^{\ast}(\lambda^{\ast})\\0&I_{n} \end{array}\right),\ J_{3}=\left(\begin{array}{cc} 1&0\\[1ex] \mathrm{e}^{-2i\theta}\Gamma(\lambda)&I_{n} \end{array}\right),\ J_{4}=\left(\begin{array}{cc} 1+\gamma^{\ast}(\lambda^{\ast})\gamma(\lambda)&\mathrm{e}^{2i\theta}\gamma^{\ast}(\lambda^{\ast})\\ \mathrm{e}^{-2i\theta}\gamma(\lambda)&I_{n} \end{array}\right), $$ (2.33)  $$ \theta(x,t,\lambda)= \lambda^{2}x+2 \lambda^{4}t, \quad \gamma(\lambda)=\frac{b^{\ast}(\lambda^{\ast})}{\det [a(\lambda)]},\quad \Gamma(\lambda)=\frac{\mathrm{adj}[d(\lambda)]B^{\ast}(\lambda^{\ast})}{\det [d(\lambda)]\det [a(\lambda)]}. $$ (2.34) It follows from (2.26) that Γ(λ) also can be written as  $$ \Gamma(\lambda)=\frac{\mathrm{adj}\left[A^{\ast}(\lambda^{\ast})a(\lambda)\right]B^{\ast}(\lambda^{\ast})}{\det [d(\lambda)]\det [a(\lambda)]}.$$ Also,  $$ \det [d(\lambda)]=\det\left[A^{\ast}(\lambda^{\ast})\right]\det[a(\lambda)]+b(\lambda)\mathrm{adj}\left[A^{\ast}(\lambda^{\ast})a(\lambda)\right]B^{\ast}(\lambda^{\ast}). $$ (2.35) The contour for this RH problem is depicted in Fig. 2. The matrix M(x, t, λ) defined by (2.31) is a meromorphic function of λ in $$ \mathbb {C}\backslash \{\lambda ^{4}\in \mathbb {R}\}$$. It follows from (2.31) that M(x, t, λ) can only have singularities at the zeros of $$\det [a(\lambda )]$$, $$\det [d(\lambda )]$$ and the complex conjugate of these zeros. Since a(λ) is an even function, each zero λj of $$\det [ a(\lambda )]$$ is accompanied by another zero at − λj. Similarly, each zero κj of $$\det [d(\lambda )]$$ is accompanied by a zero at − κj. In particular, both $$\det [a(\lambda )]$$ and $$\det [d(\lambda )]$$ have an even number of zeros. Assumption 2.4 We assume that $$\det [a(\lambda )]$$ has 2N simple zeros $$\{\lambda _{j}\}_{1}^{2N}$$ in $$ \mathbb {C}_{-}$$, N = n1 + n2, where λj ∈ D4, j = 1, …, 2n1; λj ∈ D3, j = 2n1 + 1, …, 2N. $$\det [d(\lambda )]$$ has 2K simple zeros $$\{\kappa _{j}\}_{1}^{2K}$$ in D3. None of the zeros of $$\det [a(\lambda )]$$ for λ ∈ D3 coincides with a zero of $$\det [d(\lambda )]$$. None of these functions has zeros on $$\{\lambda ^{4}\in \mathbb {R}\}$$. For a function f(λ), we denote $$\dot {f}(\lambda )=\frac {\mathrm {d} f(\lambda )}{\mathrm {d} \lambda }$$. Proposition 2.5 Let M be the eigenfunction defined by (2.31) and assume that the set of the singularities are as in Assumption 2.4. Then the following residue conditions hold:  \begin{align} \underset{\lambda_{j}}{\mathrm{Res}}\,M_{L}(x,t,\lambda)=&\mathrm{e}^{ -2i\theta(\lambda_{j})}M_{R}(x,t,\lambda_{j}) \Lambda_{j},\ j=1,\ldots, 2n_{1}, \end{align} (2.36a)  \begin{align} \underset{\lambda^{\ast}_{j}}{\mathrm{Res}}\,M_{R}(x,t,\lambda)=&-\mathrm{e}^{2i\theta\left(\lambda_{j}^{\ast}\right)}M_{L}\left(x,t,\lambda^{\ast}_{j}\right)\Lambda_{j}^{\ast},\ j=1,\ldots, 2n_{1}, \end{align} (2.36b)  \begin{align} \underset{\kappa_{j}}{\mathrm{Res}}\, M_{L}(x,t,\lambda)=&-\mathrm{e}^{-2i\theta(\kappa_{j})}M_{R}(x,t,\kappa_{j})\tilde{\Lambda}_{j},\ j=1,\ldots,2K, \end{align} (2.36c)  \begin{align} \underset{\kappa^{\ast}_{j}}{\mathrm{Res}}\, M_{R}(x,t,\lambda)=& \mathrm{e}^{2i\theta\left(\kappa_{j}^{\ast}\right)}M_{L}\left(x,t,\kappa_{j}^{\ast}\right)\tilde{\Lambda}_{j}^{\ast},\ j=1,\ldots,2K. \end{align} (2.36d)  $$ \Lambda_{j}=\frac{\mathrm{adj}[a(\lambda_{j})]\tilde{b}(\lambda_{j})}{\dot{\det}[a(\lambda_{j})] b(\lambda_{j})\mathrm{adj}[a(\lambda_{j})]\tilde{b}(\lambda_{j})},\quad \tilde{\Lambda}_{j}=\underset{\kappa_{j}}{\mathrm{Res}}\, \Gamma(\lambda)=\frac{\mathrm{adj}[d(\kappa_{j})]B^{\ast}\left(\kappa_{j}^{\ast}\right)}{\dot{\det} [d(\kappa_{j})]\det [a(\kappa_{j})]}, $$ (2.37) where the column vector $$\tilde {b}(\lambda )$$ is analytic for $$\lambda \in \mathbb {C}_{-}$$, and satisfies  $$ b(\lambda_{j})\mathrm{adj}[a(\lambda_{j})]\tilde{b}(\lambda_{j})\neq 0,\quad j=1,\ldots,2n_{1}.$$ Proof. In the view of the expressions for s(λ) and S(λ) given in (2.24), the equation (2.15) reads  \begin{align} \mu_{3L}(\lambda)&=\mu_{2L}(\lambda)\det[a^{\ast}(\lambda^{\ast})]-\mathrm{e}^{-2i\theta(\lambda)}\mu_{2R}(\lambda)\mathrm{adj}[a^{\ast}(\lambda^{\ast})]b^{\ast}(\lambda^{\ast}), \end{align} (2.38a)  \begin{align} \mu_{3R}(\lambda)&=\mathrm{e}^{2i\theta(\lambda)}\mu_{2L}(\lambda)b (\lambda)+\mu_{2R}(\lambda)a(\lambda), \end{align} (2.38b)  \begin{align} \mu_{1L}(\lambda)&=\mu_{2L}(\lambda)\det[A^{\ast}(\lambda^{\ast})]-\mathrm{e}^{-2i\theta(\lambda)}\mu_{2R}(\lambda)\mathrm{adj}[A^{\ast}(\lambda^{\ast})]B^{\ast}(\lambda^{\ast}), \end{align} (2.38c)  \begin{align} \mu_{1R}(\lambda)&=\mathrm{e}^{2i\theta(\lambda)}\mu_{2L}(\lambda)B(\lambda)+\mu_{2R}(\lambda)A(\lambda). \end{align} (2.38d) where, for simplicity of notation, we have suppressed the x and t dependence. Multipling (2.38b) by $$a^{-1}(\lambda )\tilde {b}(\lambda )$$ from right, we find  \begin{align*} \frac{\mu_{3R}(\lambda) \mathrm{adj}[a(\lambda)]\tilde{b}(\lambda)}{\textrm{det} [a(\lambda)]}=\frac{\mathrm{e}^{2i\theta(\lambda)}\mu_{2L}(\lambda) b(\lambda)\mathrm{adj}[a(\lambda)]\tilde{b}(\lambda)}{\textrm{det} [a(\lambda)]}+ \mu_{2R}(\lambda) \tilde{b}(\lambda).\end{align*} Taking the residues of this equation at $$\{\lambda _{j}\}_{1}^{2n_{1}}$$, we get (2.36a). Multipling (2.38a) by $$\tilde {b}^{\ast }(\lambda ^{\ast })\mathrm {adj}[a^{\ast }(\lambda ^{\ast })]$$ from right, we find  \begin{align*} \mu_{3L}(\lambda)\tilde{b}^{\ast}(\lambda^{\ast})\mathrm{adj}[a^{\ast}(\lambda^{\ast})]&=\mu_{2L}(\lambda)\det [a^{\ast}(\lambda^{\ast})]\tilde{b}^{\ast}(\lambda^{\ast})\mathrm{adj}[a^{\ast}(\lambda^{\ast})]\\ &\quad-\mathrm{e}^{-2i\theta(\lambda)}\mu_{2R}(\lambda)\mathrm{adj}[a^{\ast}(\lambda^{\ast})]b^{\ast}(\lambda^{\ast})\tilde{b}^{\ast}(\lambda^{\ast})\mathrm{adj}[a^{\ast}(\lambda^{\ast})]. \end{align*} Evaluating this equation at $$ \{\lambda ^{\ast }_{j}\}_{1}^{2n_{1}}$$, and using (2.28), we find  $$ \mu_{3L}\left(\lambda_{j}^{\ast}\right)\tilde{b}^{\ast}(\lambda_{j})\mathrm{adj}[a^{\ast}(\lambda_{j})]= -\mathrm{e}^{-2i\theta(\lambda_{j})}\left\{\tilde{b}^{\ast}(\lambda_{j})\mathrm{adj}[a^{\ast}(\lambda_{j})]b^{\ast}(\lambda_{j})\right\}\mu_{2R}\left(\lambda_{j}^{\ast}\right)\mathrm{adj}[a^{\ast}(\lambda_{j})]. $$ Thus, (2.36b) is proved. In order to derive (2.36d), we note that M−R = M+J1R yields  $$ \mu_{1R}(\lambda)[d^{\ast}(\lambda^{\ast})]^{-1}=\mathrm{e}^{2i\theta(\lambda)}\mu_{3L}(\lambda)\Gamma^{\ast}(\lambda^{\ast})+\mu_{2R}(\lambda)[a^{\ast}(\lambda^{\ast})]^{-1}.$$ Taking the residues of this equation at $$\{\kappa ^{\ast }_{j}\}_{1}^{2K}$$, we get (2.36d). The proof of (2.36c) is similar. □ Remark 2.6 We should consider the existence of $$\tilde {b}(\lambda )$$. Indeed, for the fixed j ∈{1, …, 2N}, thus $$\det [a(\lambda _{j})]=0$$, it follows from the nonsingularity of s(λ) that b(λj)adj[a(λj)]≠0, without loss of generality, suppose that the first entry is not zero, then we define  $$ \tilde{b}(\lambda)=\mathrm{diag}(1,0,\ldots 0)\mathrm{adj}[a^{T}(\lambda)]b^{T}(\lambda).$$ Specially, for each j ∈{1, …, 2N}, |b(λj)adj[a(λj)]|2≠0, although adj[a*(λ)]b*(λ) is not analytic for $$\lambda \in \mathbb {C}_{-}$$, it follows from (2.29) that $$\tilde {b}(\lambda _{j})$$ can be replaced by adj[a*(λj)]b*(λj). Consequently, Λj can be rewritten as  $$ \Lambda_{j}=\frac{\mathrm{adj}[a^{\ast}(\lambda_{j})a(\lambda_{j})]b^{\ast}(\lambda_{j})}{\dot{\det}[a(\lambda_{j})]| b(\lambda_{j})\mathrm{adj}[a(\lambda_{j})]|^{2}}. $$ (2.39) 2.5 Reconstructing the potential q(x, t) The potential q(x, t) can be reconstructed from the eigenfunctions μ(x, t, λ). Substituting  $$ \mu=I_{n+1}+\frac{m^{(1)}}{\lambda}+O\left(\frac{1}{\lambda^{2}}\right),\quad \lambda\rightarrow \infty,$$ into the x-part of (2.8) and considering terms of O(λ), we find that  $$ q(x,t)=-2\mathrm{e}^{-i\int_{(0,0)}^{(x,t)}\Delta(x^{\prime},t^{\prime})}W(x,t)m(x,t), $$ (2.40) where Δ and W are defined by (2.3) and (2.5), respectively, and we write m(x, t) for $$m^{(1)}_{21}(x,t)$$. Following from (2.40) and its Hermitian conjugate, and recalling that W* = W−1, we obtain  \begin{align*}|q|^{2}&=4|m|^{2}, \quad q^{\ast}q_{x}-q^{\ast}_{x}q=4\left(m^{\ast}m_{x}-m_{x}^{\ast}m\right)-32i|m|^{4},\\ qq^{\ast}&=4Wmm^{\ast}W^{\ast},\quad qq^{\ast}_{x}-q_{x}q^{\ast}=32i|m|^{2}Wmm^{\ast}W^{\ast}-4W\left(m_{x}m^{\ast}-mm_{x}^{\ast}\right)W^{\ast}. \end{align*} Thus, the 1-form Δ(x, t) can be expressed in terms of m(x, t) as  $$ \Delta=2|m|^{2}\,\mathrm{d}x+\left(4|m|^{4}+2im^{\ast}m_{x}-2im_{x}^{\ast}m\right)\,\mathrm{d}t, $$ (2.41) and W(x, t) is defined by  $$ \begin{aligned} W(x,t)&=I_{n}+\int_{(0,0)}^{(x,t)}\omega(x^{\prime},t^{\prime}),\\ \omega(x,t)&=-2iWmm^{\ast}\,\mathrm{d}x+2W\left(-2i|m|^{2}mm^{\ast}+m_{x}m^{\ast}-mm^{\ast}_{x}\right)\,\mathrm{d}t. \end{aligned} $$ (2.42) The potential q(x, t) can now be reconstructed as follows: Compute m according to  $$ m(x,t)=\lim_{\lambda\rightarrow \infty}(\lambda \mu(x,t,\lambda))_{21}.$$ Determine Δ(x, t) and W(x, t) from (2.41) and (2.42), respectively. q(x, t) is given by (2.40). 2.6 The global relation The spectral functions s(λ) and S(λ) are not independent but satisfy an important relation. Indeed, it follows from (2.15) that  $$ \mu_{1}\mathrm{e}^{i(\lambda^{2}x+2\lambda^{4}t)\hat{\sigma}}(S^{-1}s)=\mu_{3}. $$ (2.43) Since μ1(0, T, λ) = In+1, evaluating at (0, T) and recalling the definition (2.11) of μ3(x, t, λ), we find  $$ S^{-1}(\lambda)s(\lambda)=I_{n+1}-\mathrm{e}^{-2i \lambda^{4}T\hat{\sigma}}\int_{0}^{\infty}\mathrm{e}^{-i\lambda^{2}x^{\prime}\hat{\sigma}} (U\mu_{3})(x^{\prime},T,\lambda)\,\mathrm{d} x^{\prime}, $$ (2.44) The (12) entry of this equation yields the following global relation:  $$ \det[A(\lambda)]b(\lambda)-B(\lambda)\mathrm{adj}[A(\lambda)]a(\lambda) =\mathrm{e}^{-4i\lambda^{4}T}f(T,\lambda),\quad \lambda\in D_{4}, $$ (2.45)  $$ f(T,\lambda)=-\int_{0}^{\infty}\mathrm{e}^{-2i\lambda^{2}x^{\prime}} (U\mu_{3})_{12}(x^{\prime},T,\lambda)\,\mathrm{d} x^{\prime}.$$ It follows that f(T, λ) is analytic and bounded for λ ∈ D4. Also,  $$ f(T,\lambda)=O\left(\frac{1}{\lambda}\right),\quad \lambda\in D_{4}\rightarrow \infty. $$ In particular, if $$T=\infty $$, the global relation becomes  $$ \det[A(\lambda)]b(\lambda)-B(\lambda)\mathrm{adj}[A(\lambda)]a(\lambda) =0,\quad \lambda\in D_{4}. $$ (2.46) 3. The matrix-valued spectral function The analysis of Section 2 motivates the following definitions for the matrix-valued spectral functions. Definition 3.1 (The matrix-valued spectral function a(λ) and b(λ)). Given $$q_{0}(x)\in S(\mathbb {R}^{+})$$, we define the map  $$ \mathscr{F}:q_{0}(x)\mapsto (a(\lambda),b(\lambda)) $$ as follows:  $$ \left(\begin{array}{c} b(\lambda)\\ a(\lambda) \end{array}\right)=\phi_{R}(0,\lambda), $$ where $$S(\mathbb {R}^{+})$$ denotes the Schwartz space on $$\mathbb {R}^{+}$$, and ϕ(x, λ) is the unique solution of equation (2.8a) evaluated at t = 0 defined by the Volterra integral equation  $$ \phi(x,\lambda)=I_{n+1}+\int^{x}_{\infty} \mathrm{e}^{i \lambda^{2}(x-x^{\prime})\hat{\sigma}}[U(x^{\prime},0,\lambda)\phi(x^{\prime},\lambda)]\, \mathrm{d}x^{\prime}, $$ (3.1) where U(x, 0, λ) is given in terms of W(x, 0) and Q(x, 0), which are both determined by initial data q0(x). Proposition 3.2 The spectral functions a(λ) and b(λ) have the following properties: a(λ) and b(λ) are analytic for $$\lambda \in \mathbb {C}_{-}$$ and continuous and bounded for $$\lambda \in \bar {\mathbb {C}}_{-}$$. a*(λ*)a(λ) + b*(λ*)b(λ) = In, $$\lambda \in \mathbb {C}$$. a(λ) = a(−λ), b(λ) = −b(−λ). $$a(\lambda )=I_{n}+O\left (\frac {1}{\lambda }\right ),\ b(\lambda )=O\left (\frac {1}{\lambda }\right ),\ \lambda \in \mathbb {C}_{-}\rightarrow \infty $$. The map $$\mathscr {H}: \{a(\lambda ),b(\lambda )\}\mapsto q_{0}(x)$$, inverse to $$\mathscr {F}$$, is defined as follows  \begin{align} \begin{aligned} q_{0}(x)&=-2\mathrm{e}^{-2i{{{\int_{0}^{x}}}}|\check m(x^{\prime})|^{2}\,\mathrm{d}x^{\prime}}\check{W}(x)\check m(x),\\ \check m(x)&=\lim_{\lambda\rightarrow \infty}(\lambda M^{(x)}(x,\lambda))_{21}, \end{aligned} \end{align} (3.2) and $$\check {W}(x)$$ satisfies the following Volterra integral equation  $$ \check{W}(x)=I_{n}-2i{{{\int_{0}^{x}}}} \check{W}(x^{\prime})\check m(x^{\prime})\check m^{\ast}(x^{\prime})\,\mathrm{d}x^{\prime},$$ where M(x)(x, λ) is the unique solution of the following matrix RH problem: $$M^{(x)}(x,\lambda )=\left\{\begin {array}{ll} M_{+}^{(x)}(x,\lambda ), &\mathrm {Im} \lambda ^{2}\geqslant 0,\\ M_{-}^{(x)}(x,\lambda ), &\mathrm {Im} \lambda ^{2} \leqslant 0, \end {array}\right.$$ is a meromorphic function of λ in $$ \mathbb {C}\backslash \{\lambda ^{2}\in \mathbb {R}\}$$. M(x)(x, λ) satisfies the jump condition  $$ M_{-}^{(x)}(x,\lambda)=M_{+}^{(x)}(x,\lambda)J^{(x)}(x,\lambda), \quad \lambda^{2}\in\mathbb{R}, $$ (3.3) where  $$ J^{(x)}(x,\lambda)=\begin{pmatrix} \dfrac{1}{\det [a(\lambda)a^{\ast}(\lambda^{\ast})]}&\dfrac{\mathrm{e}^{2i\lambda^{2}x}b(\lambda)}{\det[a^{\ast}(\lambda^{\ast})]}\\[2ex] \dfrac{\mathrm{e}^{-2i\lambda^{2}x}b^{\ast}(\lambda^{\ast})}{\det[a(\lambda)]}&I_{n} \end{pmatrix}. $$ (3.4) If $$\det [a(\lambda )]$$ has 2N simple zeros $$\{\lambda _{j}\}_{1}^{2N}$$ in $$ \mathbb {C}_{-}$$, N = n1 + n2, where λj ∈ D4, j = 1, …, 2n1; λj ∈ D3, j = 2n1 + 1, …, 2N. $$M_{L}^{(x)}$$ can have simple poles at $$\{\lambda _{j}\}_{1}^{2N}$$, and $$M_{R}^{(x)}$$ can have simple poles at $$\{\lambda ^{\ast }_{j}\}_{1}^{2N}$$. The associated residues are given by  $$ \begin{aligned} \underset{\lambda_{j}}{\mathrm{Res}}\,M^{(x)}_{L}(x,\lambda)&=\mathrm{e}^{ -2i{{{\lambda_{j}^{2}}}}x}M^{(x)}_{R}(x,\lambda_{j}) \Lambda_{j},\ j=1,\ldots,2N,\\ \underset{\lambda^{\ast}_{j}}{\mathrm{Res}}\,M^{(x)}_{R}(x,\lambda)&=-\mathrm{e}^{2i \lambda^{*2}_{j}x}M^{(x)}_{L}\left(x,\lambda_{j}^{\ast}\right)\Lambda^{\ast}_{j},\ j=1,\ldots,2N, \end{aligned} $$ (3.5) where Λj is defined by (2.39). $$M^{(x)}(x,\lambda )=I_{n+1}+O\left (\frac {1}{\lambda }\right ),\quad \lambda \rightarrow \infty $$. Proof. (i)–(iv) follow from the discussion in Section 2.3. Let $$\tilde {\phi }(x,\lambda )$$ be the unique solution of equation (2.8a) evaluated at t = 0 defined by the Volterra integral equation  \begin{align} \tilde{\phi}(x,\lambda)=I_{n+1}+{{{\int^{x}_{0}}}} \mathrm{e}^{i \lambda^{2}(x-x^{\prime})\hat{\sigma}}\left[U(x^{\prime},0,\lambda)\tilde{\phi}(x^{\prime},\lambda)\right]\, \mathrm{d}x^{\prime}. \end{align} (3.6) Define  \begin{align*} M^{(x)}_{+}(x,\lambda)&=\left(\phi_{L}(x,\lambda),\tilde{\phi}_{R}(x,\lambda)[a^{\ast}(\lambda^{\ast})]^{-1}\right),\quad \mathrm{Im} \lambda^{2}\geqslant 0,\\ M^{(x)}_{-}(x,\lambda)&=\left(\frac{\tilde{\phi}_{L}(x,\lambda)}{\det[a(\lambda)]},\phi_{R}(x,\lambda)\right),\quad \mathrm{Im} \lambda^{2}\leqslant 0. \end{align*} Equation $$\phi (x,\lambda )=\tilde {\phi }(x,\lambda )\mathrm {e}^{i\lambda ^{2}x\hat {\sigma }}s(\lambda )$$ can be rewritten as (3.3). By the similar arguments for M(x, t, λ) and Theorem 4.1, it is straightforward to prove (v) (also see Fokas et al., 2005; Lenells, 2008). □ Definition 3.3 (The matrix-valued spectral function A(λ) and B(λ)). Let g0(t) and g1(t) be smooth functions, we define the map  $$ \tilde{\mathscr{F}}:(g_{0}(t),g_{1}(t))\mapsto (A(\lambda),B(\lambda)) $$ as follows:  $$ \left(\begin{array}{c} B(\lambda)\\A(\lambda) \end{array}\right)=\varphi_{R}(0,\lambda), $$ where φ(t, λ) is the unique solution of equation (2.8b) evaluated at x = 0 defined by the Volterra integral equation  $$ \varphi(t,\lambda)=I_{n+1}+{{{\int_{T}^{t}}}} \mathrm{e}^{2i \lambda^{4}(t-t^{\prime})\hat{\sigma}}\left[V(0,t^{\prime},\lambda)\varphi(t^{\prime},\lambda)\right]\, \mathrm{d}t^{\prime}. $$ (3.7) Here V (0, t, λ) is given in terms of W(0, t), Q(0, t) and Qx(0, t), which all are determined by boundary data g0(t) and g1(t). Proposition 3.4 The spectral functions A(λ) and B(λ) have the following properties: A(λ) and B(λ) are entire functions and bounded for $$\lambda \in \bar {D}_{2}\cup \bar {D}_{4}$$. $$A^{\ast }(\lambda ^{\ast })A(\lambda )+B^{\ast }(\lambda ^{\ast })B(\lambda )=I_{n},\ \lambda \in \mathbb {C}$$. A(λ) = A(−λ), B(λ) = −B(−λ). $$A(\lambda )=I_{n}+O\left (\frac {1}{\lambda }\right ),\ B(\lambda )=O\left (\frac {1}{\lambda }\right ),\ \lambda \in D_{2}\cup D_{4}\rightarrow \infty $$. The map $$\tilde {\mathscr {H}}: \{A(\lambda ),B(\lambda )\}\mapsto \{g_{0}(t),g_{1}(t)\}$$, inverse to $$\tilde {\mathscr {F}}$$, is defined as follows  $$ \begin{aligned} g_{0}(t)&=-2\mathrm{e}^{-i{{{\int_{0}^{t}}}} \hat{\Delta}(t^{\prime})}\hat{W}(t)m^{(1)}_{21}(t),\\ g_{1}(t)&=-\frac{i}{2}|g_{0}|^{2}g_{0}(t)+2im^{(2)}_{11}(t)g_{0}(t)+4i\mathrm{e}^{-i{{{\int_{0}^{t}}}} \hat{\Delta}(t^{\prime})}\hat{W}(t)m^{(3)}_{21}(t),\\ \hat{\Delta}(t)&=\left[-4\left|m^{(1)}_{21}\right|^{4}+8\mathrm{Re} \left(m^{(1)*}_{21}m^{(3)}_{21}-\left|m^{(1)}_{21}\right|^{2}m^{(2)}_{11}\right)\right]\,\mathrm{d}t, \end{aligned} $$ (3.8) and $$\hat {W}(t)$$ satisfies the following Volterra integral equation  $$ \hat{W}(t)=I_{n}+4i{{{\int_{0}^{t}}}} \hat{W}(t^{\prime})\left[\left|m^{(1)}_{21}\right|^{2}m^{(1)}_{21}m^{(1)*}_{21}+2\mathrm{Re}\left(m^{(2)}_{11}m^{(1)}_{21}m^{(1)*}_{21} -m^{(3)}_{21}m^{(1)*}_{21}\right)\right]\mathrm{d}t^{\prime}. $$ Here functions $$m^{(1)}_{21},m^{(2)}_{11}$$ and $$m^{(3)}_{21}$$ are determined by the asymptotic expansion  $$ M^{(t)}(t,\lambda)=I_{n+1}+\frac{m^{(1)}(t)}{\lambda}+\frac{m^{(2)}(t)}{\lambda^{2}}+\frac{m^{(3)}(t)}{\lambda^{3}}+O\left(\frac{1}{\lambda^{4}}\right),\quad \lambda\rightarrow\infty.$$ where M(t)(t, λ) is the unique solution of the following matrix RH problem: $$M^{(t)}(t,\lambda )=\begin {cases} M_{+}^{(t)}(t,\lambda ), &\mathrm {Im} \lambda ^{4} > 0,\\ M_{-}^{(t)}(t,\lambda ), &\mathrm {Im} \lambda ^{4} < 0, \end {cases}$$ is a meromorphic function of λ in $$ \mathbb {C}\backslash \{\lambda ^{4}\in \mathbb {R}\}$$. M(t)(x, λ) satisfies the jump condition  $$ M_{-}^{(t)}(t,\lambda)=M_{+}^{(t)}(t,\lambda)J^{(t)}(t,\lambda), \quad \lambda^{4}\in\mathbb{R}, $$ (3.9) where  $$ J^{(t)}(t,\lambda)=\begin{pmatrix} \dfrac{1}{\det [A(\lambda)A^{\ast}(\lambda^{\ast})]}&\dfrac{\mathrm{e}^{4i\lambda^{4}t}B(\lambda)}{\det[A^{\ast}(\lambda^{\ast})]}\\[2ex] \dfrac{\mathrm{e}^{-4i\lambda^{4}t}B^{\ast}(\lambda^{\ast})}{\det [A(\lambda)]}&I_{n} \end{pmatrix}. $$ (3.10) If $$\det [A(\lambda )]$$ has 2n0 simple zeros $$\{\zeta _{j}\}_{1}^{2n_{0}}$$ in D2 ∪ D4. $$M_{L}^{(t)}$$ can have simple poles at $$\{\zeta _{j}\}_{1}^{2n_{0}}$$, and $$M_{R}^{(t)}$$ can have simple poles at $$\{\zeta ^{\ast }_{j}\}_{1}^{2n_{0}}$$. The associated residues are given by  $$ \begin{aligned} \underset{\zeta_{j}}{\mathrm{Res}}\,M^{(t)}_{L}(t,\lambda)&=\mathrm{e}^{ -4i{{{\zeta_{j}^{4}}}}t}M^{(t)}_{R}(t,\zeta_{j}) \hat{\Lambda}_{j},\ j=1,\ldots,2n_{0},\\ \underset{\zeta^{\ast}_{j}}{\mathrm{Res}}\,M^{(t)}_{R}(t,\lambda)&=-\mathrm{e}^{4i \zeta^{*4}_{j}t}M^{(t)}_{L}(t,\zeta^{\ast}_{j})\hat{\Lambda}^{\ast}_{j},\ j=1,\ldots,2n_{0}, \end{aligned} $$ (3.11) where $$\hat {\Lambda }_{j}=\frac {\mathrm {adj}[A^{\ast }(\zeta _{j})A(\zeta _{j})]B^{\ast }(\zeta _{j})}{\dot {\det }[A(\zeta _{j})]| B(\zeta _{j})\mathrm {adj}[A(\zeta _{j})]|^{2}}$$. $$M^{(t)}(t,\lambda )=I_{n+1}+O\left (\frac {1}{\lambda }\right ),\quad \lambda \rightarrow \infty $$. Proof. (i)–(iv) follow from the discussion in Section 2.3. Let $$\tilde {\varphi }(t,\lambda )$$ be the unique solution of equation (2.8b) evaluated at x = 0 defined by the Volterra integral equation  \begin{align} \tilde{\varphi}(t,\lambda)=I_{n+1}+{{{\int_{0}^{t}}}} \mathrm{e}^{2i \lambda^{4}(t-t^{\prime})\hat{\sigma}}\left[V(0,t^{\prime},\lambda)\tilde{\varphi}(t^{\prime},\lambda)\right]\, \mathrm{d}t^{\prime}. \end{align} (3.12) Define  \begin{align*} M_{+}^{(t)}(t,\lambda)&=\left(\varphi_{L}(t,\lambda),\tilde{\varphi}_{R}(t,\lambda)[A^{\ast}(\lambda^{\ast})]^{-1}\right),\quad \mathrm{Im} \lambda^{4}> 0,\\ M_{-}^{(t)}(t,\lambda)&=\left(\frac{\tilde{\varphi}_{L}(t,\lambda)}{\det [A(\lambda)]},\varphi_{R}(t,\lambda)\right),\quad \mathrm{Im} \lambda^{4}<0. \end{align*} Equation $$\varphi (t,\lambda )=\tilde {\varphi }(t,\lambda )\mathrm {e}^{2i\lambda ^{4} t\hat {\sigma }}S(\lambda )$$ can be rewritten as (3.10). By the similar arguments for M(x, t, λ) and Theorem 4.1, it is straightforward to prove (v) (also see Fokas et al., 2005; Lenells, 2008). □ Definition 3.5 (An admissible set of functions). Given $$q_{0}(x)\in S(\mathbb {R}^{+})$$, define a(λ) and b(λ) by definition 3.1. Suppose that there exist smooth function g0(t) and g1(t) such that The associated A(λ) and B(λ) defined by definition 3.3 satisfy the global relation (2.45). The initial and boundary data are compatible with equation (1.1) at x = t = 0, i.e.  $$ g_{0}(0)=q_{0}(0),\quad g_{1}(0)=q^{\prime}_{0}(0),$$  $$ ig_{1}(0)+q^{\prime\prime}_{0}(0)+i[q^{\prime\ast}_{0}(0)q_{0}(0)+q^{\ast}_{0}(0)q^{\prime}_{0}(0)]q_{0}(0)+i|q_{0}(0)|^{2}q^{\prime}_{0}(0)=0,$$ where $$q_{0}^{\prime }(x)=\frac {\mathrm {d} q_{0}(x)}{\mathrm {d} x}$$. Then we call {g0(t), g1(t)} an admissible set of functions with respect to q0(x). 4. The main result Theorem 4.1 Let $$q_{0}(x)\in S(\mathbb {R}^{+})$$, suppose that the functions g0(t) and g1(t) are admissible with respect to q0(x) (see definition 3.5). Define the spectral functions {a(λ), b(λ)} and {A(λ), B(λ)} in terms of q0(x) and {g0(t), g1(t)} according to definitions 3.1 and 3.3. Assume that the possible zeros $$\{\lambda _{j}\}_{1}^{2N}$$ of $$\det [a(\lambda )]$$ and $$\{\kappa _{j}\}_{1}^{2K}$$ of $$\det [d(\lambda )]$$ are as in assumption 2.4. Define M(x, t, λ) as the solution of the following matrix RH problem: M is sectionally meromorphic in $$\mathbb {C}\backslash \{\lambda ^{4}\in \mathbb {R}\}$$. ML can have simple poles at $$\{\lambda _{j}\}_{1}^{2n_{1}}$$ and $$\{\kappa _{j}\}_{1}^{2K}$$, MR can have simple poles at $$\{\lambda ^{\ast }_{j}\}_{1}^{2n_{1}}$$ and $$\{\kappa ^{\ast }_{j}\}_{1}^{2K}$$. The associated residues of M satisfy the relations in (2.36). M satisfies the jump condition  $$ M_{-}(x,t,\lambda)=M_{+}(x,t,\lambda)J(x,t,\lambda),\quad \lambda^{4}\in\mathbb{R}, $$ (4.1) where M is M+ for λ ∈ D1 ∪ D3, M is M− for λ ∈ D2 ∪ D4 and J is defined in terms of {a, b, A, B} by (2.32) and (2.33), see Fig. 2. M(x, t, λ) has the following asymptotics as $$\lambda \rightarrow \infty $$,  $$ M(x,t,\lambda)=I_{n+1}+O\left(\frac{1}{\lambda}\right),\ \lambda\rightarrow \infty. $$ (4.2) Then M(x, t, λ) exists and is unique. Define q(x, t) in terms of M(x, t, λ) by  $$ q(x,t)=-2\mathrm{e}^{-i\int_{(0,0)}^{(x,t)}\Delta(x^{\prime},t^{\prime})}W(x,t)m(x,t), $$ (4.3) with  $$ m(x,t)=\lim_{\lambda\rightarrow \infty} (\lambda M(x,t,\lambda))_{21}, $$ (4.4) where Δ(x, t) and W(x, t) are defined by (2.41) and (2.42), respectively. Then q(x, t) solves equation (1.1). Furthermore,  \begin{align*} q(x,0)=q_{0}(x),\quad q(0,t)=g_{0}(t),\quad q_{x}(0,t)=g_{1}(t).\end{align*} Proof. The existence and uniqueness of the solution for the above RH problem are established by the vanishing lemma, see Zhou (1989). Moreover, it follows from standard arguments via the dressing method (Zakharov & Shabat, 1974, 1979) that if M solves the above RH problem and q(x, t) is defined by (4.3), then q(x, t) solves equation (1.1). In order to prove that q(x, 0) = q0(x), we define M(x)(x, λ) by  \begin{align} \begin{aligned} M^{(x)}(x,\lambda)&=M(x,0,\lambda),\quad \lambda\in \bar{D}_{1}\cup \bar{D}_{4},\\ M^{(x)}(x,\lambda)&=M(x,0,\lambda)J_{1}^{-1}(x,0,\lambda),\quad \lambda\in \bar{D}_{2},\\ M^{(x)}(x,\lambda)&=M(x,0,\lambda)J_{3}(x,0,\lambda),\quad \lambda\in \bar{D}_{3}. \end{aligned} \end{align} (4.5) If the sets {λj} and {κj} are empty, then the function M(x) is analytic in $$\mathbb {C}\backslash \{\lambda ^{2}\in \mathbb {R}\}$$. Furthermore,  \begin{align*} M_{-}^{(x)}(x,\lambda)&=M_{+}^{(x)}(x,\lambda)J^{(x)}(x,\lambda), \quad \lambda^{2}\in\mathbb{R},\\ M^{(x)}(x,\lambda)&=I_{n+1}+O\left(\frac{1}{\lambda}\right),\quad \lambda\rightarrow \infty, \end{align*} where J(x)(x, t) is defined by (3.4). Comparing (3.2) with (4.3) evaluated at t = 0, we conclude that q0(x) = q(x, 0). If the sets {λj} and {κj} are not empty. ML(x, t, λ) has poles at $$\{\lambda _{j}\}_{1}^{2n_{1}}$$ in D4 and has poles at $$\{\kappa _{j}\}_{1}^{2K}$$ in D3. On the other hand, $$M^{(x)}_{L}(x,\lambda )$$ has poles at $$\{\lambda _{j}\}_{1}^{2N}$$ in $$\mathbb {C}_{-}$$. We will now show that the transformation defined by (4.5) maps the former poles to the latter ones. Since M(x)(x, λ) = M(x, 0, λ) for λ ∈ D4, M(x) has poles at $$\{\lambda _{j}\}_{1}^{2n_{1}}$$ with the correct residue condition. Equation (4.5) can be rewritten as  $$ M^{(x)}(x,\lambda)=\left(M_{L}(x,0,\lambda)+\mathrm{e}^{-2i\lambda^{2}x}M_{R}(x,0,\lambda)\Gamma(\lambda),M_{R}(x,0,\lambda)\right),\quad \lambda\in D_{3}. $$ (4.6) The residue condition (2.36c) at κj implies that M(x) has no poles at κj. On the other hand, equation (4.6) shows that M(x) has poles $$\{\lambda _{j}\}_{2n_{1}+1}^{2N}$$ at with residues given by  $$ \underset{\lambda_{j}}{\mathrm{Res}}\,M_{L}^{(x)}(x,\lambda)=\mathrm{e}^{-2i{{{\lambda_{j}^{2}}}}x}M_{R}^{(x)}(x,\lambda)\underset{\lambda_{j}}{\mathrm{Res}}\,\Gamma(\lambda),\quad j=2n_{1}+1,\ldots, 2N.$$ Recalling the definition of Γ(λ), (2.29) and (2.35), this becomes the residue condition (3.5). Similar considerations apply to $$\{\lambda _{j}^{\ast }\}_{1}^{2n_{1}}$$ and $$\{\kappa _{j}^{\ast }\}_{1}^{2K}$$. In what follows, we prove that q(0, t) = g0(t) and qx(0, t) = g1(t). Define M(t)(t, λ) by  $$ M^{(t)}(t,\lambda)=M(0,t,\lambda)G_{j}(t,\lambda),\quad \lambda\in \bar{D}_{j},\ j=1,\ldots,4, $$ (4.7) with  $$ G_{1}=\begin{pmatrix} \frac{\det[A^{\ast}(\lambda^{\ast})]}{\det[a^{\ast}(\lambda^{\ast})]}&0\\\mathrm{e}^{4i\lambda^{4}(T-t)}f^{\ast}(T,\lambda^{\ast})& a^{\ast}(\lambda^{\ast})[A^{\ast}(\lambda^{\ast})]^{-1} \end{pmatrix},$$  $$ G_{2}=\begin{pmatrix} \frac{1}{\det[d^{\ast}(\lambda^{\ast})]}&0\\ \mathrm{e}^{-4i\lambda^{4} t}\frac{b^{\ast}(\lambda^{\ast})}{\det[A(\lambda)]}&d^{\ast}(\lambda^{\ast}) \end{pmatrix}, $$ (4.8)  $$ G_{3}=\begin{pmatrix} \det[d(\lambda)]&-\mathrm{e}^{4i\lambda^{4}t}b(\lambda)\mathrm{adj}[a(\lambda)][A^{\ast}(\lambda^{\ast})]^{-1}\\ 0&[d(\lambda)]^{-1} \end{pmatrix},$$  $$ G_{4}=\begin{pmatrix} \frac{\det [a(\lambda)]}{\det[A(\lambda)]}&\mathrm{e}^{4i\lambda^{4}(t-T)}f(T,\lambda)\frac{\operatorname{adj}[a(\lambda)]A(\lambda)}{\det [A(\lambda)]}\\ 0&a^{-1}(\lambda)A(\lambda) \end{pmatrix}.$$ where f(T, λ) is defined by (2.45). After a tedious but straightforward calculation by virtue of the global relation (2.45), it follows that  $$ \begin{aligned} J_{1}(0,t,\lambda)G_{2}(t,\lambda)&=G_{1}(t,\lambda)J^{(t)}(t,\lambda),\quad \lambda\in \bar{D}_{1}\cap\bar{D}_{2},\\ J_{2}(0,t,\lambda)G_{2}(t,\lambda)&=G_{3}(t,\lambda)J^{(t)}(t,\lambda),\quad \lambda\in \bar{D}_{2}\cap\bar{D}_{3},\\ J_{3}(0,t,\lambda)G_{4}(t,\lambda)&=G_{3}(t,\lambda)J^{(t)}(t,\lambda),\quad \lambda\in \bar{D}_{3}\cap\bar{D}_{4},\\ J_{4}(0,t,\lambda)G_{4}(t,\lambda)&=G_{1}(t,\lambda)J^{(t)}(t,\lambda),\quad \lambda\in \bar{D}_{1}\cap\bar{D}_{4}, \end{aligned} $$ (4.9) where J(t)(t, λ) is defined by (3.9). Equations (4.1), (4.7) and (4.9) imply that M(t) satisfies of the RH problem defined in proposition 3.4 except the residue conditions. If the sets {λj} and {κj} are empty, this immediately yields the desired result. Otherwise, we should verify that M(t) satisfies the residue conditions (3.11). This follows similar arguments in the proof of Theorem 4.1 in Fokas et al. (2005). □ 5. The Dirichlet to Neumann map We consider the Dirichlet boundary value problem for the vector DNLS equation (1.1) posed on the half-line. 5.1 The global relation Evaluating (2.15) at x = 0, we find  $$ \mu_{2}(0,t,\lambda)\mathrm{e}^{2i \lambda^{4}t\hat{\sigma}}s(\lambda)=\mu_{3}(0,t,\lambda),\quad \lambda\in(\mathbb{C}_{+}, \mathbb{C}_{-}). $$ (5.1) Denoting  $$ \Phi(t,\lambda)=(Y\mu_{2}Y^{-1})(0,t,\lambda),\quad c(t,\lambda)=(Y\mu_{3}Y^{-1})_{12}(0,t,\lambda)a^{-1}(\lambda),$$ consequently, Φ−1(t, λ) =Φ*(t, λ*), thus we rewrite Φ(t, λ) as  $$ \Phi(t,\lambda)=\begin{pmatrix} \det \Phi_{2}^{\ast}(t,\lambda^{\ast})&\Phi_{1}(t,\lambda)\\ -\mathrm{adj}[\Phi_{2}^{\ast}(t,\lambda^{\ast})]\Phi_{1}^{\ast}(t,\lambda^{\ast})&\Phi_{2}(t,\lambda) \end{pmatrix}, $$ (5.2) and rewrite the (12) entry of equation (5.1) as  $$ \Phi_{1}(t,\lambda)+ \mathrm{e}^{4i \lambda^{4}t+i{{{\int_{0}^{t}}}}\Delta_{2}(0,t^{\prime})}\det(\Phi_{2}^{\ast}(t,\lambda^{\ast}))b(\lambda)W^{-1}(0,t)a^{-1}(\lambda)=c(t,\lambda),\quad \lambda\in \mathbb{C}_{-}. $$ (5.3) The function c(t, λ) is analytic and bounded in $$\mathbb {C}_{-}$$ away from the possible zeros of $$\det [a(\lambda )]$$ and of order O(1/λ) as $$\lambda \rightarrow \infty $$. 5.2 Asymptotics Substituting the asymptotic expansion  $$ \mu_{j}(x,t,\lambda)=I_{n+1}+\sum_{k\geqslant 1}\frac{\mu_{j}^{(k)}(x,t)}{\lambda^{k}},\quad j=1,2,3,\ \lambda\rightarrow \infty,$$ into (2.8), here $$\lambda \rightarrow \infty $$ means that λ in the bounded domain of μj and approaches to infinity, we find that  $$ \begin{aligned} Y\mu_{j}^{(1)}Y^{-1}&=\frac{i}{2}\sigma Q, \quad Y\mu_{j}^{(3)}Y^{-1}=\frac{i}{2}\sigma QY\mu_{j}^{(2)}Y^{-1}+\frac{i}{8}\sigma Q^{3}+\frac{1}{4}Q_{x},\\ Y\mu_{jx}^{(2)}Y^{-1}&=-\frac{i}{8}\sigma Q^{4}+\frac{1}{4}QQ_{x},\\ Y\mu_{jt}^{(2)}Y^{-1}&=-\frac{i}{4}\sigma Q^{6}+\frac{i}{4}\sigma(QQ_{xx}-{{{Q_{x}^{2}}}})+\frac{1}{2}Q^{3}Q_{x}+\frac{1}{4}QQ_{x}Q^{2}+\frac{1}{8}[Q^{2},Q_{x}Q]. \end{aligned} $$ (5.4) Thus, as $$\lambda \rightarrow \infty $$,  $$ \begin{aligned} (Y\mu_{j}Y^{-1})_{22}(x,t,\lambda)&=I_{n}+\frac{1}{\lambda^{2}}\int_{(x_{j},t_{j})}^{(x,t)} \tilde{\Delta}(x^{\prime},t^{\prime})+O\left(\frac{1}{\lambda^{4}}\right),\\ (Y\mu_{j}Y^{-1})_{12}(x,t,\lambda)&=\frac{q^{\ast}}{2\lambda}-\frac{1}{\lambda^{3}}\left(\frac{1}{8}|q|^{2}q^{\ast}+\frac{i}{4}q^{\ast}_{x}-\frac{1}{2}q^{\ast}\int_{(x_{j},t_{j})}^{(x,t)}\tilde{\Delta}(x^{\prime},t^{\prime})\right)+O\left(\frac{1}{\lambda^{5}}\right), \end{aligned} $$ (5.5) where the closed 1-form $$\tilde {\Delta }(x,t)$$ is defined by  $$ \tilde{\Delta}=\tilde{\Delta}_{1}+\tilde{\Delta}_{2},\, \tilde{\Delta}_{1}=\frac{1}{8}(i|q|^{2}qq^{\ast}-2qq^{\ast}_{x})\,\mathrm{d}x,$$  $$ \tilde{\Delta}_{2}=\left[\frac{i}{4}(-|q|^{4}qq^{\ast}-q_{x}q^{\ast}_{x}+qq^{\ast}_{xx})+\frac12|q|^{2}qq^{\ast}_{x}+\frac14(|q|^{2})_{x}qq^{\ast}-\frac18(q^{\ast}q_{x}qq^{\ast}+|q|^{2}q_{x}q^{\ast})\right]\mathrm{d}t.$$ Hence,  $$ \begin{aligned} \Phi_{2}(t,\lambda)&=I_{n}+\frac{\Phi_{2}^{(1)}(t)}{\lambda^{2}}+O\left(\frac{1}{\lambda^{4}}\right),\quad \lambda\in D_{1}\cup D_{3}\rightarrow\infty,\\ \Phi_{1}(t,\lambda)&=\frac{\Phi_{1}^{(1)}(t)}{\lambda}+\frac{\Phi_{1}^{(2)}(t)}{\lambda^{3}}+O\left(\frac{1}{\lambda^{5}}\right),\quad \lambda\in D_{1}\cup D_{3}\rightarrow\infty, \end{aligned} $$ (5.6) where  $$ \Phi_{2}^{(1)}(t)={{{\int_{0}^{t}}}} \tilde\Delta_{2}(0,t^{\prime}),\quad \Phi_{1}^{(1)}(t)=\frac12g^{\ast}_{0}(t),$$  $$ \Phi_{1}^{(2)}(t)=-\frac18|g_{0}|^{2}g_{0}^{\ast}(t)-\frac i4 g^{\ast}_{1}(t)+\frac 12g_{0}^{\ast}(t){{{\int_{0}^{t}}}} \tilde\Delta_{2}(0,t^{\prime}).$$ In particular, we find the following expressions for the boundary values:  $$ g^{\ast}_{0}(t)=2\Phi_{1}^{(1)}(t),\quad g_{1}^{\ast}(t)=\frac i2|g_{0}|^{2}g^{\ast}_{0}(t)-2ig^{\ast}_{0}(t)\Phi_{2}^{(1)}(t)+4i\Phi_{1}^{(2)}(t). $$ (5.7) The global relation (5.3) also implies that  $$ c(t,\lambda)=\frac{\Phi_{1}^{(1)}(t)}{\lambda}+\frac{\Phi_{1}^{(2)}(t)}{\lambda^{3}}+O\left(\frac{1}{\lambda^{5}}\right),\quad\lambda\in \mathbb{C}_{-}\cap(D_{2}\cup D_{4})\rightarrow\infty. $$ (5.8) 5.3 The Dirichlet problem The following theorem expresses the spectral functions A(λ) and B(λ) in terms of the Dirichlet boundary data and initial data via the solution of a system of nonlinear integral equations. Let ∂Dj, j = 1, …, 4, denote the boundary of the Dj, oriented so that Dj lies to the left of ∂Dj, moreover, let $$D_{4}^{-}=D_{4}\cap \{\mathrm {Im}\lambda\! \leqslant\! 0\}$$. Theorem 5.1 Let $$T<\infty $$, $$q_{0}(x)\in S(\mathbb {R}^{+})$$. Assume that the function g0(t), $$0\leqslant t < T $$, has sufficient smoothness and is compatible with q0(x) at x = t = 0. Suppose that $$\det [a(\lambda )]$$ has a finite (possibly empty) set of simple zeros in D4, which are denoted by $$\{\lambda _{j}\}_{1}^{2n_{1}}$$, and assume that no zeros of $$\det [a(\lambda )]$$ occur on ∂D4. Then the matrix-valued spectral functions A(λ) and B(λ) are given by  $$ \begin{aligned} A(\lambda)&=W^{-1}(0,T)\Phi^{\ast}_{2}(T,\lambda^{\ast})W(0,T),\\ B(\lambda)&=-\mathrm{e}^{4i\lambda^{4}T-i{{{\int_{0}^{T}}}} \Delta_{2}(0,t)}\Phi_{1}(T,\lambda)\mathrm{adj}[\Phi_{2}(t,\lambda)]W(0,T), \end{aligned} $$ (5.9) where  $$ \Delta_{2}(0,t)=-\frac14\left(3|g_{0} |^{4}-2ig_{0}^{\ast} g_{1} +2ig_{1}^{\ast} g_{0} \right)\mathrm{d}t, $$ (5.10) and W(0, t) satisfies the Volterra integral equation  $$ W(0,t)=I_{n}+{{{\int_{0}^{t}}}}\tilde\omega(t^{\prime})W(0,t^{\prime})\,\mathrm{d}t^{\prime}, $$ (5.11)  $$ \tilde\omega=\frac14\left(3i|g_{0}|^{2}g_{0}g_{0}^{\ast}-2g_{0}g_{1}^{\ast}+2g_{1}g_{0}^{\ast}\right)(t),$$ and Φ1(t, λ) and Φ2(t, λ) satisfies the Volterra integral equation  $$\! \begin{pmatrix} \Phi_{1}(t,\lambda)\\\Phi_{2}(t,\lambda) \end{pmatrix}\!\!=\!\!\begin{pmatrix} 0\\I_{n} \end{pmatrix}\!+\!\!{{{\int_{0}^{t}}}}\!\!\begin{pmatrix}\! \mathrm{e}^{4i\lambda^{4}(t-t^{\prime})}\left[-\left(i\lambda^{2}|g_{0}|^{2}\Phi_{1}+\Phi_{1} \tilde\omega\right)+i\lambda\left(|g_{0}|^{2}g_{0}^{\ast}-2\lambda^{2}g_{0}^{\ast}+ig_{1}^{\ast}\right)\Phi_{2}\right](t^{\prime},\lambda)\\ \left[i\lambda(|g_{0}|^{2}g_{0}-2\lambda^{2}g_{0}-ig_{1})\Phi_{1}+(i\lambda g_{0}g_{0}^{\ast}\Phi_{2}-\Phi_{2}\tilde\omega)\right](t^{\prime},\lambda) \!\end{pmatrix}\!\mathrm{d}t^{\prime}, $$ (5.12) with 0 < t < T. For a function f(λ), let f+(λ) and f−(λ) denote  $$ f_{+}(\lambda)=f(\lambda)+if(i\lambda),\quad f_{-}(\lambda)=f(\lambda)-f(i\lambda). $$ The unknown Neumann boundary value g1(t) is given by  $$ \begin{aligned} g_{1}^{\ast}(t)&=\frac i2|g_{0}|^{2}g^{\ast}_{0}(t)-\frac{4g_{0}^{\ast}(t)}{\pi}\int_{\partial D_{4}^{-}}\lambda\Phi_{2-}(t,\lambda)\,\mathrm{d}\lambda+\frac 4\pi\int_{\partial D_{4}^{-}}\left(\lambda g^{\ast}_{0}(t)-\lambda^{2}\Phi_{1+}(t,\lambda)\right)\mathrm{d}\lambda\\ &\quad-\frac 8\pi\mathrm{e}^{i{{{\int_{0}^{t}}}} \Delta_{2}(0,t^{\prime})}\int_{\partial D_{4}^{-}}\lambda^{2}\mathrm{e}^{4i \lambda^{4}t}\det\left(\Phi_{2}^{\ast}(t,\lambda^{\ast})\right)b(\lambda)W^{-1}(0,t)a^{-1}(\lambda)\,\mathrm{d}\lambda. \end{aligned} $$ (5.13) Proof. Equations (5.9)–(5.12) follow from (2.2)–(2.5), (2.16), (2.24) and (5.2). In order to derive (5.13) we note that (5.7) expresses g1(t) in terms of $$\Phi _{2}^{(1)}(t)$$ and $$\Phi _{1}^{(2)}(t)$$. Furthermore, it follows from (2.19), (5.2), (5.6) and Cauchy’s theorem that  $$ \begin{aligned} i\pi\Phi_{2}^{(1)}(t)&=-\int_{\partial D_{1}\cup\partial D_{3} }\lambda\left[\Phi_{2}(t,\lambda)-I_{n}\right]\mathrm{d} \lambda=\int_{\partial D_{2}\cup \partial D_{4}}\lambda\left[\Phi_{2}(t,\lambda)-I_{n}\right]\mathrm{d} \lambda\\ &=\int_{\partial D_{4}}\lambda\left[\Phi_{2}(t,\lambda)-I_{n}\right]\mathrm{d} \lambda-\int_{\partial D_{4}}\lambda\left[\Phi_{2}(t,i\lambda)-I_{n}\right]\mathrm{d} \lambda\\ &=\int_{\partial D_{4}}\lambda\Phi_{2-}(t,\lambda)\,\mathrm{d} \lambda=2\int_{\partial D_{4}^{-}}\lambda\Phi_{2-}(t,\lambda)\,\mathrm{d} \lambda, \end{aligned} $$ (5.14)  $$ \begin{aligned} i\pi \Phi_{1}^{(2)}(t)&=-\int_{\partial D_{1}\cup\partial D_{3}}\left[\lambda^{2}\Phi_{1}(t,\lambda)-\lambda\Phi_{1}^{(1)}(t)\right]\mathrm{d}\lambda\\ &= \int_{\partial D_{2}\cup\partial D_{4}}\left[\lambda^{2}\Phi_{1}(t,\lambda)-\lambda\Phi_{1}^{(1)}(t)\right]\mathrm{d}\lambda\\ &=\left(\int_{\partial D_{2}}-\int_{\partial D_{4}}\right)\left[\lambda^{2}\Phi_{1}(t,\lambda)-\lambda\Phi_{1}^{(1)}(t)\right]\mathrm{d}\lambda+F(t)\\ &=\int_{\partial D_{4}}\left[2\lambda\Phi_{1}^{(1)}(t)-\lambda^{2}\Phi_{1+}(t,\lambda)\right]\mathrm{d}\lambda+F(t),\\ &=2\int_{\partial D_{4}^{-}}\left[2\lambda\Phi_{1}^{(1)}(t)-\lambda^{2}\Phi_{1+}(t,\lambda)\right]\mathrm{d}\lambda+F(t), \end{aligned} $$ (5.15) where F(t) is defined by  $$ F(t)=2\int_{\partial D_{4}}\left[\lambda^{2}\Phi_{1}(t,\lambda)-\lambda\Phi_{1}^{(1)}(t)\right]\mathrm{d}\lambda=4\int_{\partial D_{4}^{-}} \left[\lambda^{2}\Phi_{1}(t,\lambda)-\lambda\Phi_{1}^{(1)}(t)\right]\mathrm{d}\lambda. $$ Using the global relation (5.3) and the asymptotics (5.8) of c(t, λ), we find  $$ \begin{aligned} F(t)&=4\int_{\partial D_{4}^{-}}\left[\lambda^{2} c(t,\lambda)-\lambda\Phi_{1}^{(1)}(t)\right]\mathrm{d}\lambda-4\int_{\partial D_{4}^{-}}\lambda^{2}\left[c(t,\lambda)-\Phi_{1}(t,\lambda)\right]\mathrm{d}\lambda\\ &=-i\pi\Phi_{1}^{(2)}(t)-4\mathrm{e}^{i{{{\int_{0}^{t}}}} \Delta_{2}(0,t^{\prime})}\int_{\partial D_{4}^{-}}\lambda^{2}\mathrm{e}^{4i \lambda^{4}t}\det(\Phi_{2}^{\ast}(t,\lambda^{\ast}))b(\lambda)W^{-1}(0,t)a^{-1}(\lambda) \,\mathrm{d}\lambda. \end{aligned} $$ (5.16)From the above arguments we prove (5.13). □ Note: In the case that $$\det [a(\lambda )]$$ has a finite number of simple zeros in $$D_{4}^{-}$$, the representation (5.13) is modified by adding a term  $$ -\frac8\pi\mathrm{e}^{i{{{\int_{0}^{t}}}} \Delta_{2}(0,t^{\prime})}\sum_{\lambda_{j}\in D_{4}^{-}}{{{\lambda_{j}^{2}}}}\mathrm{e}^{4i{{{\lambda_{j}^{4}}}}t}\det(\Phi_{2}^{\ast} (t,\lambda_{j}^{\ast}))b(\lambda_{j})W^{-1}(0,t)\frac{\mathrm{adj}[a(\lambda_{j})]}{\dot{\mathrm{det}}[a(\lambda_{j})]} $$ provided that the integration contour $$\partial D_{4}^{-}$$ is replaced with τ, where τ denotes the contour obtained by deforming $$\partial D_{4}^{-}$$ such that it does not surround the possible zeros of $$\det [a(\lambda )]$$, see Fig. 3. Fig. 2. View largeDownload slide The contour for the RH problem in the complex λ-plane. Fig. 2. View largeDownload slide The contour for the RH problem in the complex λ-plane. Fig. 3. View largeDownload slide The deformed contour τ in the complex λ-plane. The zeros of det[a(λ)] are indicated by “.” in this figure. Fig. 3. View largeDownload slide The deformed contour τ in the complex λ-plane. The zeros of det[a(λ)] are indicated by “.” in this figure. Substitution of the expression (5.13) into (5.10)–(5.12) yield a system of nonlinear integral equations involving the functions Δ2(0, t), W(0, t), Φ1(t, λ) and Φ2(t, λ). Assuming that this system has a unique solution, A(λ) and B(λ) can be determined from (5.9). In fact, for the particular boundary problem, we can present an effective characterizations of A(λ) and B(λ). 6. Conclusions As was mentioned in the introduction, we implement the Fokas method to analyse IBV problems for the vector DNLS. In particular, as n = 1,  $$ Y(x,t)=\mathrm{e}^{i\int_{(0,0)}^{(x,t)}\Delta(x^{\prime},t^{\prime})\sigma_{3}}, \quad \tilde{Y}(x)=\mathrm{e}^{i\int_{\infty}^{x}\Delta_{1}(x^{\prime},0)\sigma_{3}},\quad \sigma_{3}=\mathrm{diag}\{1,-1\}.$$ we derive the same conclusions as in Lenells (2008, 2011). While the main ingredients of the Fokas method can be used for the case of $$n\geqslant 2$$, the actual implementation of this method presents some additional technical difficulties. In particular, the formulation of the RH problem is more complicated because the Lax pairs of multi-component equations involve (n + 1) × (n + 1) matrices. As a consequence, we transform these matrices to the 2 × 2 block ones, and construct the RH problem as the scalar case by using Sherman–Morrison formula. 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The vector derivative nonlinear Schrödinger equation on the half-line

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Abstract

Abstract In this paper, we analyse initial-boundary problems for the vector derivative nonlinear Schrödinger equation on the semi-infinite strip $$(x, t) \in (0,\infty )\times (0, T)$$ via the unified transform method of Fokas. Even though additional technical complications arise in the vector case compared with scalar ones, we show that it also can be expressed in terms of the solution of a matrix Riemann–Hilbert problem. The Riemann–Hilbert problem involves a jump matrix, uniquely defined in terms of four matrix functions called spectral functions and denoted by {a(λ), b(λ), A(λ), B(λ)} that depend on the initial data and all boundary values, respectively. A key role is played by the so-called global relation which involves the known and unknown boundary values. By analysing the global relation, we present an effective characterization of the latter two spectral functions in terms of the given initial and boundary data. 1. Introduction One of the most remarkable developments in the theory of integrable systems is inverse scattering transform, which was introduced to solve the initial value problems of nonlinear integrable evolution equations (Gardner et al., 1967, 1974). Following that, a natural issue is to study the initial-boundary value (IBV) problems. A general approach to IBV problems for integrable nonlinear equations originated in the seminal work of Fokas (1997), which is usually referred to as the unified method, or the Fokas method, and it has been actively developed since then (Fokas, 2002, 2008, 2005; Fokas et al., 2005, Fokas & Lennels, 2010, 2012; Lenells, 2008, 2011; Lenells & Fokas, 2009; Boutet De Monvel et al., 2004, 2006, 2009). Interestingly, there exist several integrable vector extensions of scalar classical integrable equations such as the vector NLS equation (Fordy & Kulish, 1983), the vector DNLS equation (Fordy, 1984) and the vector modified KdV equation Kaup & Newell (1978), which describe the wave interactions in the case of that two or more wave packets of different carrier frequencies appear simultaneously. Considering the applications in some physical situations, we believe that it is significant to investigate multi-component cases. Therefore, in this paper we study IBV problems for the vector DNLS equation,   $$ i q_{t}+q_{xx}+i(|q|^{2}q)_{x}=0, \quad q=(q_{1},q_{2},\ldots,q_{n})^{T}, $$ (1.1) where |q|2 = q*q, the superscript “*” denotes Hermitian conjugation. The scalar case, known as the celebrated Kaup–Newell equation (Kaup & Newell, 1978), is relevant to Alvén waves in plasma physics and the propagation of nonlinear pulses in optical fibres (Mio et al., 1976; Mjolhus, 1976; Kodama, 1985). The global existence and long-time behaviour of scalar DNLS have been investigated by inverse scattering transform method Liu et al. (2016, 2017). The unified transform method has formerly been applied to the IBV problems for the scalar DNLS equation (Lenells, 2008) and integrable evolutions with 3 × 3 Lax pairs (Lenells, 2012, 2013; Caudrelier & Zhang 2012; Xu & Fan, 2013, 2016; Geng et al., 2015; Liu & Geng, 2016). Although the analysis of (1.1) is inspired by the preceding references, it also involves some novelties: (a) In order to ensure that the normalization of RH problem is canonical, i.e. the solution of RH problem is the identity matrix at infinity, we transform the Lax pair to a new one by introducing the matrix functions Y (x, t) and $$\tilde {Y}(x)$$. Indeed, provided that q(x, t) solves the equation (1.1), both of them cannot be explicitly expressed in terms of a closed 1-form as in the scalar case (Lenells, 2008) but are well defined. (b) All of the higher-order square matrices in this paper can be rewritten as 2 × 2 block ones, thus we can directly formulate the higher-order matrix RH problem by the combinations of the entries of three eigenfunctions. Compared with the idea of that constructing the 3 × 3 matrix RH problem by the Fredholm integral equation in Lenells (2012) (also see Beals & Coifman, 1984), this procedure is more convenient for analysis of the multi-component evolution equations. (c) By introducing Sherman–Morrison formula, we derive the jump matrices and residue conditions of the RH problem. The outline of the paper is as follows. In Section 2 we define three eigenfunctions of the Lax pair for spectral analysis. In addition, we investigate the matrix-valued spectral functions a(λ), b(λ), A(λ) and B(λ) further in Section 3. In Section 4 we show that q(x, t) can be expressed in terms of the unique solution of a matrix RH problem which involves the aforementioned spectral functions. In Section 5 we present the so-called Dirichlet to Neumann map and discuss the vanishing Dirichlet boundary data further. 2. Spectral analysis Basic assumptions and notations Denote $$\Omega =\{(x,t)|0<x<\infty ,0<t<T\}$$, where T is a given positive constant, unless otherwise specified, we suppose that $$T<\infty $$. Let q(x, t) be a complex vector-valued function of (x, t) ∈Ω, assume that $$q(x,t)\in C^{2,1}(\mathbb {R}^{+}\times \mathbb {R}^{+})$$, for all t as $$x\rightarrow \infty $$, $$q(x,t)\rightarrow 0$$, $$q_{x}(x,t)\rightarrow 0$$. The initial data q0(x), the Dirichlet boundary data g0(t) and the Neumann boundary data g1(t) of equation (1.1) are denoted by  \begin{align*} q_{0}(x)&=q(x,0),\quad 0<x<\infty,\\ g_{0}(t)&=q(0,t),\quad g_{1}(t)= q_{x}(0,t),\quad 0<t<T. \end{align*} Let $$D_{k}=\left \{\lambda \in \mathbb {C}:\arg \lambda ^{2}\in \left (\frac {(k-1)\pi }{2},\frac {k\pi }{2}\right )\right \}$$, k = 1, …, 4, see Fig. 2; $$\mathbb {C}_{\pm }=\{\lambda \in \mathbb {C}:\mathrm {Im}\lambda ^{2}\gtrless 0\}$$. Denote the matrix A = (AL, AR), where AL is the first column of A, and AR is the rest of columns. The notation A(λ), λ ∈ (D1, D2), means that AL and AR hold for λ ∈ D1 and D2, respectively. In this paper, without otherwise specified, let each matrix A be rewritten as a block form $$A=\left({{A_{11}}\atop{A_{21}}} \quad {{A_{12}} \atop{A_{22}}} \right)$$, where A11 is scalar. Let In denote n × n identity matrix. In this paper, $$\lambda \rightarrow \lambda _{0}$$ or $$\lambda \rightarrow \infty $$ non-tangentially on $$\mathbb {C}$$. 2.1 Lax pair The vector DNLS equation (1.1) admits the Lax pair Fordy (1984)  $$ \begin{aligned} \psi_{x}&=i\lambda^{2}\sigma\psi+\lambda Q\psi,\\ \psi_{t}&=2i\lambda^{4}\sigma \psi+(2\lambda^{3} Q+i\lambda^{2}Q^{2}\sigma+i\lambda Q_{x}\sigma+\lambda Q^{3})\psi, \end{aligned} $$ (2.1) where ψ(x, t, λ) is a (n + 1) × (n + 1) matrix-valued function and $$\lambda \in \mathbb {C}$$ is the spectral parameter,  $$ \sigma=\left(\begin{array}{ccc} 1&0\\0&-I_{n} \end{array}\right),\quad Q(x,t)=-i\left(\begin{array}{cc} 0&q^{\ast}(x,t)\\ q(x,t)&0 \end{array}\right).$$ In order to seek solutions of the spectral problem which approach the identity matrix In+1 as $$\lambda \rightarrow \infty $$ and ensure that the spectral function S(λ) defined in the following is independent of the initial data, following the similar steps as in Lenells (2008), we define two (n + 1) × (n + 1) matrix-valued functions Y (x, t) and $$\tilde {Y}(x)$$ by  $$ Y(x,t)=\left(\begin{array}{cc} \mathrm{e}^{i\int_{(0,0)}^{(x,t)}\Delta(x^{\prime},t^{\prime})}&0 \\0 &W(x,t) \end{array}\right),\quad \tilde{Y}(x)=\left(\begin{array}{cc} \mathrm{e}^{\frac{i}{2}\int^{x}_{\infty}|q_{0}|^{2}(x^{\prime})\,\mathrm{d}x^{\prime}}&0 \\0 &\tilde{W}(x) \end{array}\right), $$ (2.2) where Δ(x, t) is the closed real-valued 1-form  $$ \Delta=\Delta_{1}+\Delta_{2},\quad \Delta_{1}=\frac{1}{2}|q |^{2} \mathrm{d}x,\quad \Delta_{2}=-\frac14\left(3|q |^{4}-2iq^{\ast} q_{x} +2iq^{\ast}_{x} q \right)\mathrm{d}t, $$ (2.3) which follows by a conservation law of equation (1.1)  $$ \frac{i}{2} \left(|q|^{2}\right)_{t}+\frac14\left(3i|q|^{4}+2q^{\ast}q_{x}-2q^{\ast}_{x}q\right)_{x}=0, $$ and the n × n matrix-valued functions W(x, t) and $$\tilde {W}(x)$$ are the unique solutions of  $$ \begin{cases} W_{x}=-\frac{i}{2}qq^{\ast}W,\\ W_{t}=\frac14\left(3i|q|^{2}qq^{\ast}-2qq^{\ast}_{x}+2q_{x}q^{\ast}\right)W,\\ W(0,0)=I_{n}, \end{cases}\quad \begin{cases} \tilde{W}_{x}=-\frac{i}{2}q_{0}q_{0}^{\ast}\tilde{W},\\ \tilde{W}(\infty)=I_{n}. \end{cases} $$ (2.4) Provided that q(x, t) solves equation (1.1), it can be shown that Wxt = Wtx, thus the solution W exists. In other words, equation (2.4) is equivalent to the Volterra integral equations  $$ \begin{aligned} W(x,t)&=I_{n}+\int_{(0,0)}^{(x,t)}\omega(x^{\prime},t^{\prime}),\\ \omega(x,t)&= -\frac{i}{2}qq^{\ast}W\mathrm{d}x+\frac14\left(3i|q|^{2}qq^{\ast}-2qq^{\ast}_{x}+2q_{x}q^{\ast}\right)W\mathrm{d}t, \end{aligned} $$ (2.5)  $$ \tilde{W}(x)=I_{n}+\frac i2\int_{x}^{\infty} q_{0}(x^{\prime})q_{0}^{\ast}(x^{\prime})\tilde{W}(x^{\prime})\mathrm{d}x^{\prime}. $$ (2.6) Also, W*(x, t) and W−1(x, t) satisfy the same linear differential equations and initial condition, thus they are identical. Introducing a new eigenfunction μ(x, t, λ) by  $$ \psi(x,t,\lambda)=Y(x,t)\mu(x,t,\lambda)\tilde{Y}(0)\mathrm{e}^{i(\lambda^{2}x+2\lambda^{4}t)\sigma}, $$ (2.7) we obtain the Lax pair equations  $$ \mu_{x}-i\lambda^{2}[\sigma,\mu]=U\mu, $$ (2.8a)  $$ \mu_{t}-2i\lambda^{4}[\sigma,\mu]=V\mu, $$ (2.8b) where  \begin{align*}U(x,t,\lambda)&=\lambda Y^{-1}QY+\frac{i}{2}Y^{-1}Q^{2}\sigma Y,\\ V(x,t,\lambda)&=Y^{-1}\left(2\lambda^{3} Q+i\lambda^{2}Q^{2}\sigma+i\lambda Q_{x}\sigma+\lambda Q^{3}+\frac{3i}{4}Q^{4}\sigma-\frac{1}{2}[Q,Q_{x}]\right)Y. \end{align*} Equation (2.8) can be rewritten in differential form as  $$ \mathrm{d}\left(\mathrm{e}^{-i(\lambda^{2}x+2\lambda^{4}t)\hat{\sigma}}\mu\right)=\tilde{\omega}, $$ (2.9) where the exact 1-form $$\tilde {\omega }(x,t,\lambda )$$ is defined by  $$ \tilde{\omega}=\mathrm{e}^{-i(\lambda^{2}x+2\lambda^{4}t)\hat{\sigma}}(U\mu\,\mathrm{d}x+V\mu\,\mathrm{d}t), $$ (2.10) and $$\hat {\sigma }$$ acts on a matrix X by $$\hat {\sigma }X=[\sigma , X]$$, then $$\mathrm {e}^{\hat {\sigma }}X=\mathrm {e}^{\sigma }X\mathrm {e}^{-\sigma }$$. 2.2 Bounded and analytic eigenfunctions Three eigenfunctions $$\{\mu _{j}\}_{1}^{3}$$ of equation (2.8) are defined by the Volterra integral equations  $$ \mu_{j}(x,t ,\lambda)=I_{n+1}+\int_{\gamma_{j}}\mathrm{e}^{i\left(\lambda^{2}x+2\lambda^{4}t\right)\hat{\sigma}}\tilde{\omega}_{j}(x^{\prime},t^{\prime},\lambda),\quad j=1,2,3, $$ (2.11) where $$\tilde {\omega }_{j}$$ is defined by equation (2.10) with μ replaced by μj, the contour γj denotes a smooth curve from (xj, tj) to (x, t), and (x1, t1) = (0, T), (x2, t2) = (0, 0), $$(x_{3},t_{3})=(\infty ,t)$$. Since the 1-form is exact, $$\{\mu _{j}\}_{1}^{3}$$ are independent of the path of integration, specially we can choose the contours shown in Fig. 1, and this choice implies the following inequalities on the contours: Fig. 1. View largeDownload slide The contours γ1, γ2 and γ3 in the (x′, t′)-plane. Fig. 1. View largeDownload slide The contours γ1, γ2 and γ3 in the (x′, t′)-plane.   $$ \begin{aligned} \gamma_{1}:x-x^{\prime}&\geqslant 0,\quad t-t^{\prime}\leqslant 0,\\ \gamma_{2}:x-x^{\prime}&\geqslant 0,\quad t-t^{\prime}\geqslant 0,\\ \gamma_{3}:x-x^{\prime}&\leqslant 0.\\ \end{aligned} $$ (2.12) The first column of the matrix equation (2.11) involves the exponential $$\mathrm {e}^{2i\lambda ^{2}(x^{\prime }-x)+4i \lambda ^{4}(t^{\prime }-t)}$$. Using the inequalities (2.12) it follows that the exponential is bounded in the following regions of the complex λ-plane:  $$ \begin{aligned} \gamma_{1}&: \{\mathrm{Im}\,\lambda^{2}<0 \}\cap\{ \mathrm{Im}\,\lambda^{4}>0\},\\ \gamma_{2}&: \{\mathrm{Im}\,\lambda^{2}<0 \}\cap\{ \mathrm{Im}\,\lambda^{4}<0\},\\ \gamma_{3}&:\{\mathrm{Im}\,\lambda^{2}>0\}. \end{aligned} $$ (2.13) Also, similar to that in Kaup & Newell (1978), we assume that  $$ \int_{0}^{\infty} \left(|q|+|q|^{2}+|q_{x}|\right)\mathrm{d}x<\infty,$$ where $$|q|=\sum _{j=1}^{n} |q_{j}|.$$ Since the equations in (2.11) are Volterra integral equations, these boundedness properties imply that μ1L, μ2L and μ3L are bounded and analytic for λ ∈ D3, D4 and $$\mathbb {C}_{+}$$, respectively. Similar conditions are valid for $$\{\mu _{jR}\}_{1}^{3}$$. Thus,  $$ \begin{aligned} &\mu_{1}\ \textrm{is bounded and analytic for}\ \lambda\in(D_{3},D_{2}),\\ &\mu_{2}\ \textrm{is bounded and analytic for} \ \lambda\in(D_{4},D_{1}),\\ &\mu_{3}\ \textrm{is bounded and analytic for}\ \lambda\in(\mathbb{C}_{+},\mathbb{C}_{-}), \end{aligned} $$ (2.14) and all of them admit a bounded and continuous extension to the boundary. For T finite, μ1 and μ2 are entire functions of λ. Moreover, the column vectors of μj approach the corresponding column vectors of the identity matrix as $$\lambda \rightarrow \infty $$ in their corresponding regions of boundedness, e.g.  $$ \left(\mu_{3L},\mu_{2R}\right)=I_{n+1}+O\left(\frac{1}{\lambda}\right),\quad \lambda\in D_{1}\rightarrow \infty. $$ 2.3 The spectral functions The solutions of the system of differential equations (2.1) must be related by a matrix independent of x and t, therefore  $$ \begin{aligned} \mu_{3}(x,t,\lambda)&=\mu_{2}(x,t,\lambda)\mathrm{e}^{i(\lambda^{2}x+2\lambda^{4}t)\hat{\sigma}}s(\lambda),\\ \mu_{1}(x,t,\lambda)&=\mu_{2}(x,t,\lambda)\mathrm{e}^{i(\lambda^{2}x+2\lambda^{4}t)\hat{\sigma}}S(\lambda). \end{aligned} $$ (2.15) Evaluating the above equations at (x, t) = (0, 0) and (x, t) = (0, T), respectively, we note that the following (n + 1) × (n + 1) matrix-valued spectral functions s(λ) and S(λ):  $$ s(\lambda)=\mu_{3}(0,0,\lambda),\quad S(\lambda)=\mu_{1}(0,0,\lambda)=\mathrm{e}^{2i\lambda^{4}T\hat{\sigma}}\mu^{-1}_{2}(0,T,\lambda). $$ (2.16) Hence, functions s(λ) and S(λ) can be obtained from the evaluations at x = 0 and at t = 0 of functions μ3(x, 0, λ) and μ1(0, t, λ), respectively, which satisfy the linear integral equations  $$ \mu_{3}(x,0,\lambda)=I_{n+1}-\int_{x}^{\infty} \mathrm{e}^{i\lambda^{2}(x-x^{\prime})\hat{\sigma}}(U\mu_{3})(x^{\prime},0,\lambda)\mathrm{d}x^{\prime}, $$ (2.17)  $$ \mu_{1}(0,t,\lambda)=I_{n+1}-{{{\int_{t}^{T}}}} \mathrm{e}^{2i\lambda^{4}(t-t^{\prime})\hat{\sigma}}(V\mu_{1})(0,t^{\prime},\lambda)\mathrm{d}t^{\prime}. $$ (2.18) For x = 0, μ1 and μ2 have the following enlarged domains of boundedness: μ1(0, t, λ) is bounded for λ ∈ (D1 ∪ D3, D2 ∪ D4), μ2(0, t, λ) is bounded for λ ∈ (D2 ∪ D4, D1 ∪ D3). Proposition 2.1 For j = 1, 2, 3, function μj(x, t, λ) satisfies the symmetry relations  $$ \mu^{-1}_{j}(x,t,\lambda)=\mu_{j}^{\ast}(x,t,\lambda^{\ast}),\quad \mu_{j}(x,t,\lambda)=\sigma\mu_{j}(x,t,-\lambda)\sigma. $$ (2.19) Proof. It is easy to see that this proof follows by the symmetry relations  $$ Y^{-1}=Y^{\ast},\quad Q^{\ast}=-Q,\quad [\sigma, Y]=0, \quad\sigma Q\sigma=-Q,$$ and the initial conditions μj(xj, tj, λ) = In+1, j = 1, 2, 3. □ The fact that U and V are traceless implies $$\det [\mu (x, t,\lambda )] $$ is independent of x and t. Evaluation of $$\det [\mu _{j}(x, t,\lambda )]$$ at (xj, tj) shows that $$\det [\mu _{j}(x, t,\lambda )]=1,j=1,2,3$$. Consequently,  $$ \det [s(\lambda)]=\det [S(\lambda)]=1. $$ (2.20) It follows from(2.16), (2.19) and (2.20) that  $$ s^{\ast}(\lambda^{\ast})=\mathrm{adj}[s(\lambda)],\quad S^{\ast}(\lambda^{\ast})=\mathrm{adj}[S(\lambda)], $$ (2.21)  $$ s^{\ast}(\lambda^{\ast})s(\lambda)=I,\quad S^{\ast}(\lambda^{\ast})S(\lambda)=I, $$ (2.22) where adj(A) denotes the classical adjoint matrix of A. From (2.22), we obtain  $$ (s^{\ast}(\lambda^{\ast})s(\lambda))_{12}=0,\quad (S^{\ast}(\lambda^{\ast})S(\lambda))_{12}=0. $$ (2.23) Therefore, from (2.21) and (2.23), we find  \begin{align*} s^{\ast}_{11}(\lambda^{\ast})&=\det [s_{22}(\lambda)], \quad s^{\ast}_{21}(\lambda^{\ast})=-s_{12}(\lambda)\mathrm{adj}[s_{22}(\lambda)],\\ S^{\ast}_{11}(\lambda^{\ast})&=\det [S_{22}(\lambda)], \quad S^{\ast}_{21}(\lambda^{\ast})=-S_{12}(\lambda)\mathrm{adj}[S_{22}(\lambda)]. \end{align*} Thus s(λ) and S(λ) can be written as the following form:  $$ s(\lambda)=\left(\begin{array}{cc} \det [a^{\ast}(\lambda^{\ast})]&b(\lambda)\\ -\mathrm{adj}\left[a^{\ast}(\lambda^{\ast})\right]b^{\ast}(\lambda^{\ast})&a(\lambda) \end{array}\right),\quad S(\lambda)=\left(\begin{array}{cc} \det[ A^{\ast}(\lambda^{\ast})]&B(\lambda)\\ -\mathrm{adj}\left[A^{\ast}(\lambda^{\ast})\right]B^{\ast}(\lambda^{\ast})&A(\lambda) \end{array}\right). $$ (2.24) Also, the symmetry relations (2.19) imply that  $$ a(-\lambda)=a(\lambda),\quad b(-\lambda)=-b(\lambda),\quad A(-\lambda)=A(\lambda),\quad B(-\lambda)=-B(\lambda). $$ (2.25) The definitions of s(λ) and S(λ) imply  $$ \left(\begin{array}{c} b(\lambda)\\a(\lambda) \end{array}\right)=\mu_{3R}(0,0,\lambda),\quad \left(\begin{array}{c} B(\lambda)\\A(\lambda) \end{array}\right)=\mu_{1R}(0,0,\lambda)=\left[\mathrm{e}^{2i\lambda^{4}T\hat{\sigma}}\mu^{-1}_{2}(0,T,\lambda)\right]_{R}.$$ The above definitions imply the following properties of the spectral functions: a(λ) and b(λ) are analytic for $$\lambda \in \mathbb {C}_{-}$$ and continuous and bounded for $$\lambda \in \bar {\mathbb {C}}_{-}$$, A(λ) and B(λ) are entire functions and bounded for $$\lambda \in \bar {D}_{2}\cup \bar {D}_{4}$$. a*(λ*)a(λ) + b*(λ*)b(λ) = In, A*(λ*)A(λ) + B*(λ*)B(λ) = In, $$\lambda \in \mathbb {C}$$. $$a(\lambda )=I_{n}+O\left (\frac {1}{\lambda }\right )$$, $$b(\lambda )=O\left (\frac {1}{\lambda }\right ),\ \lambda \in \mathbb {C}_{-}\rightarrow \infty $$, $$A(\lambda )=I_{n}+O\left (\frac {1}{\lambda }\right )$$, $$B(\lambda )=O\left (\frac {1}{\lambda }\right ),\ \lambda \in D_{2}\cup D_{4}\rightarrow \infty $$. All of these properties follow from the analyticity and boundedness of μ3(x, 0, λ) and μ1(0, t, λ), from the relation s*(λ*)s(λ) = S*(λ*)S(λ) = In+1 and from the large λ asymptotics of these eigenfunctions. (ii) shows that whereas a(λ) and b(λ) are bounded for $$\lambda \in \bar {\mathbb {C}}_{-}$$, the expression appearing on the left-hand side of equation is identity matrix for all $$\lambda \in \mathbb {C}$$. Also, (ii) imply that  $$ \det [a^{\ast}(\lambda^{\ast})a(\lambda)]+b(\lambda)b^{\ast}(\lambda^{\ast})=1,\quad\det [A^{\ast}(\lambda^{\ast})A(\lambda)]+B(\lambda)B^{\ast}(\lambda^{\ast})=1.$$ 2.4 The Riemann–Hilbert problem We introduce the following proposition to prepare for formulation of the RH problem. Proposition 2.2 (Sherman–Morrison formula; Sherman & Morrison, 1950; Press, 2007; Bartlett, 1951) Suppose that $$X\in \mathbb {R}^{n}\times \mathbb {R}^{n}$$ is an invertible square matrix, x and y are n-dimensional column vectors, if X + xyT is invertible, thus  $$ (X+xy^{T})^{-1}=X^{-1}-\frac{X^{-1}xy^{T}X^{-1}}{1+y^{T}X^{-1}x}.$$ Corollary 2.3 Suppose that $$X\in \mathbb {R}^{n}\times \mathbb {R}^{n}$$, x, y and z are n-dimensional column vectors, thus  $$ \mathrm{adj}(X+xy^{T})x=\mathrm{adj}(X)x, $$ (2.26)  $$ \mathrm{adj}[X+\det( X)xy^{T}]=[1+y^{T}\mathrm{adj} (X) x ]\mathrm{adj}(X)-\mathrm{adj} (X) xy^{T} \mathrm{adj} (X), $$ (2.27) In particular, as $$\det X=0$$,  $$ \mathrm{adj} (X) xy^{T} \mathrm{adj} (X)=\left[y^{T}\mathrm{adj} (X) x\right ]\mathrm{adj}(X), $$ (2.28)  $$ \frac{\mathrm{adj} (X) x }{y^{T}\mathrm{adj} (X) x } =\frac{\mathrm{adj} (X) z }{y^{T}\mathrm{adj} (X) z } . $$ (2.29) Proof. As X, X + xyT and $$X+\det ( X)xy^{T}$$ are invertible, it is easy to verify that this corollary follows by Proposition 2.2. There exist n distinct real number α1, …, αn, such that X + αjI, X + αjI + xyT and $$X+\alpha _{j}I+\det ( X)xy^{T}$$ are invertible. Therefore, this corollary holds for all $$\alpha \in \mathbb {R}$$ when X is replaced by X + αI. In particular, when α = 0, this corollary is proved. □ Equation (2.15) can be rewritten in a form expressing the jump condition of a (n + 1) × (n + 1) matrix RH problem. This involves only tedious but straightforward algebraic manipulations. The final form is  $$ M_{-}(x,t,\lambda)=M_{+}(x,t,\lambda)J(x,t,\lambda),\quad \lambda^{4}\in\mathbb{R}, $$ (2.30) where the matrices M+, M− and J are defined as follows:  $$ \begin{aligned} M_{+}&=\left(\mu_{3L},\mu_{2R}\,\left[a^{\ast}(\lambda^{\ast})\right]^{-1}\right),\quad \lambda\in \bar{D}_{1},\quad M_{-}=\left(\mu_{3L},\mu_{1R}\,\left[d^{\ast}(\lambda^{\ast})\right]^{-1}\right), \quad \lambda\in \bar{D}_{2},\\ M_{+}&=\left(\frac{\mu_{1L}}{\det [d(\lambda)]},\mu_{3R} \right), \quad \lambda\in \bar{D}_{3},\quad M_{-}=\left(\frac{\mu_{2L}}{\det [a(\lambda)]},\mu_{3R}\right), \quad \lambda\in \bar{D}_{4}, \end{aligned} $$ (2.31)  $$ d(\lambda)=A^{\ast}(\lambda^{\ast})a(\lambda)+B^{\ast}(\lambda^{\ast})b(\lambda),$$  $$ J=\begin{cases} J_{1},&\arg \lambda^{2}=\frac{\pi}{2},\\[1ex] J_{2}=J_{3}J_{4}^{-1}J_{1},&\arg \lambda^{2}=\pi,\\[1ex] J_{3},&\arg \lambda^{2}=\frac{3\pi}{2},\\[1ex] J_{4},&\arg \lambda^{2}=0, \end{cases} $$ (2.32) with  $$ J_{1}=\left(\begin{array}{cc} 1&\mathrm{e}^{2i\theta}\Gamma^{\ast}(\lambda^{\ast})\\0&I_{n} \end{array}\right),\ J_{3}=\left(\begin{array}{cc} 1&0\\[1ex] \mathrm{e}^{-2i\theta}\Gamma(\lambda)&I_{n} \end{array}\right),\ J_{4}=\left(\begin{array}{cc} 1+\gamma^{\ast}(\lambda^{\ast})\gamma(\lambda)&\mathrm{e}^{2i\theta}\gamma^{\ast}(\lambda^{\ast})\\ \mathrm{e}^{-2i\theta}\gamma(\lambda)&I_{n} \end{array}\right), $$ (2.33)  $$ \theta(x,t,\lambda)= \lambda^{2}x+2 \lambda^{4}t, \quad \gamma(\lambda)=\frac{b^{\ast}(\lambda^{\ast})}{\det [a(\lambda)]},\quad \Gamma(\lambda)=\frac{\mathrm{adj}[d(\lambda)]B^{\ast}(\lambda^{\ast})}{\det [d(\lambda)]\det [a(\lambda)]}. $$ (2.34) It follows from (2.26) that Γ(λ) also can be written as  $$ \Gamma(\lambda)=\frac{\mathrm{adj}\left[A^{\ast}(\lambda^{\ast})a(\lambda)\right]B^{\ast}(\lambda^{\ast})}{\det [d(\lambda)]\det [a(\lambda)]}.$$ Also,  $$ \det [d(\lambda)]=\det\left[A^{\ast}(\lambda^{\ast})\right]\det[a(\lambda)]+b(\lambda)\mathrm{adj}\left[A^{\ast}(\lambda^{\ast})a(\lambda)\right]B^{\ast}(\lambda^{\ast}). $$ (2.35) The contour for this RH problem is depicted in Fig. 2. The matrix M(x, t, λ) defined by (2.31) is a meromorphic function of λ in $$ \mathbb {C}\backslash \{\lambda ^{4}\in \mathbb {R}\}$$. It follows from (2.31) that M(x, t, λ) can only have singularities at the zeros of $$\det [a(\lambda )]$$, $$\det [d(\lambda )]$$ and the complex conjugate of these zeros. Since a(λ) is an even function, each zero λj of $$\det [ a(\lambda )]$$ is accompanied by another zero at − λj. Similarly, each zero κj of $$\det [d(\lambda )]$$ is accompanied by a zero at − κj. In particular, both $$\det [a(\lambda )]$$ and $$\det [d(\lambda )]$$ have an even number of zeros. Assumption 2.4 We assume that $$\det [a(\lambda )]$$ has 2N simple zeros $$\{\lambda _{j}\}_{1}^{2N}$$ in $$ \mathbb {C}_{-}$$, N = n1 + n2, where λj ∈ D4, j = 1, …, 2n1; λj ∈ D3, j = 2n1 + 1, …, 2N. $$\det [d(\lambda )]$$ has 2K simple zeros $$\{\kappa _{j}\}_{1}^{2K}$$ in D3. None of the zeros of $$\det [a(\lambda )]$$ for λ ∈ D3 coincides with a zero of $$\det [d(\lambda )]$$. None of these functions has zeros on $$\{\lambda ^{4}\in \mathbb {R}\}$$. For a function f(λ), we denote $$\dot {f}(\lambda )=\frac {\mathrm {d} f(\lambda )}{\mathrm {d} \lambda }$$. Proposition 2.5 Let M be the eigenfunction defined by (2.31) and assume that the set of the singularities are as in Assumption 2.4. Then the following residue conditions hold:  \begin{align} \underset{\lambda_{j}}{\mathrm{Res}}\,M_{L}(x,t,\lambda)=&\mathrm{e}^{ -2i\theta(\lambda_{j})}M_{R}(x,t,\lambda_{j}) \Lambda_{j},\ j=1,\ldots, 2n_{1}, \end{align} (2.36a)  \begin{align} \underset{\lambda^{\ast}_{j}}{\mathrm{Res}}\,M_{R}(x,t,\lambda)=&-\mathrm{e}^{2i\theta\left(\lambda_{j}^{\ast}\right)}M_{L}\left(x,t,\lambda^{\ast}_{j}\right)\Lambda_{j}^{\ast},\ j=1,\ldots, 2n_{1}, \end{align} (2.36b)  \begin{align} \underset{\kappa_{j}}{\mathrm{Res}}\, M_{L}(x,t,\lambda)=&-\mathrm{e}^{-2i\theta(\kappa_{j})}M_{R}(x,t,\kappa_{j})\tilde{\Lambda}_{j},\ j=1,\ldots,2K, \end{align} (2.36c)  \begin{align} \underset{\kappa^{\ast}_{j}}{\mathrm{Res}}\, M_{R}(x,t,\lambda)=& \mathrm{e}^{2i\theta\left(\kappa_{j}^{\ast}\right)}M_{L}\left(x,t,\kappa_{j}^{\ast}\right)\tilde{\Lambda}_{j}^{\ast},\ j=1,\ldots,2K. \end{align} (2.36d)  $$ \Lambda_{j}=\frac{\mathrm{adj}[a(\lambda_{j})]\tilde{b}(\lambda_{j})}{\dot{\det}[a(\lambda_{j})] b(\lambda_{j})\mathrm{adj}[a(\lambda_{j})]\tilde{b}(\lambda_{j})},\quad \tilde{\Lambda}_{j}=\underset{\kappa_{j}}{\mathrm{Res}}\, \Gamma(\lambda)=\frac{\mathrm{adj}[d(\kappa_{j})]B^{\ast}\left(\kappa_{j}^{\ast}\right)}{\dot{\det} [d(\kappa_{j})]\det [a(\kappa_{j})]}, $$ (2.37) where the column vector $$\tilde {b}(\lambda )$$ is analytic for $$\lambda \in \mathbb {C}_{-}$$, and satisfies  $$ b(\lambda_{j})\mathrm{adj}[a(\lambda_{j})]\tilde{b}(\lambda_{j})\neq 0,\quad j=1,\ldots,2n_{1}.$$ Proof. In the view of the expressions for s(λ) and S(λ) given in (2.24), the equation (2.15) reads  \begin{align} \mu_{3L}(\lambda)&=\mu_{2L}(\lambda)\det[a^{\ast}(\lambda^{\ast})]-\mathrm{e}^{-2i\theta(\lambda)}\mu_{2R}(\lambda)\mathrm{adj}[a^{\ast}(\lambda^{\ast})]b^{\ast}(\lambda^{\ast}), \end{align} (2.38a)  \begin{align} \mu_{3R}(\lambda)&=\mathrm{e}^{2i\theta(\lambda)}\mu_{2L}(\lambda)b (\lambda)+\mu_{2R}(\lambda)a(\lambda), \end{align} (2.38b)  \begin{align} \mu_{1L}(\lambda)&=\mu_{2L}(\lambda)\det[A^{\ast}(\lambda^{\ast})]-\mathrm{e}^{-2i\theta(\lambda)}\mu_{2R}(\lambda)\mathrm{adj}[A^{\ast}(\lambda^{\ast})]B^{\ast}(\lambda^{\ast}), \end{align} (2.38c)  \begin{align} \mu_{1R}(\lambda)&=\mathrm{e}^{2i\theta(\lambda)}\mu_{2L}(\lambda)B(\lambda)+\mu_{2R}(\lambda)A(\lambda). \end{align} (2.38d) where, for simplicity of notation, we have suppressed the x and t dependence. Multipling (2.38b) by $$a^{-1}(\lambda )\tilde {b}(\lambda )$$ from right, we find  \begin{align*} \frac{\mu_{3R}(\lambda) \mathrm{adj}[a(\lambda)]\tilde{b}(\lambda)}{\textrm{det} [a(\lambda)]}=\frac{\mathrm{e}^{2i\theta(\lambda)}\mu_{2L}(\lambda) b(\lambda)\mathrm{adj}[a(\lambda)]\tilde{b}(\lambda)}{\textrm{det} [a(\lambda)]}+ \mu_{2R}(\lambda) \tilde{b}(\lambda).\end{align*} Taking the residues of this equation at $$\{\lambda _{j}\}_{1}^{2n_{1}}$$, we get (2.36a). Multipling (2.38a) by $$\tilde {b}^{\ast }(\lambda ^{\ast })\mathrm {adj}[a^{\ast }(\lambda ^{\ast })]$$ from right, we find  \begin{align*} \mu_{3L}(\lambda)\tilde{b}^{\ast}(\lambda^{\ast})\mathrm{adj}[a^{\ast}(\lambda^{\ast})]&=\mu_{2L}(\lambda)\det [a^{\ast}(\lambda^{\ast})]\tilde{b}^{\ast}(\lambda^{\ast})\mathrm{adj}[a^{\ast}(\lambda^{\ast})]\\ &\quad-\mathrm{e}^{-2i\theta(\lambda)}\mu_{2R}(\lambda)\mathrm{adj}[a^{\ast}(\lambda^{\ast})]b^{\ast}(\lambda^{\ast})\tilde{b}^{\ast}(\lambda^{\ast})\mathrm{adj}[a^{\ast}(\lambda^{\ast})]. \end{align*} Evaluating this equation at $$ \{\lambda ^{\ast }_{j}\}_{1}^{2n_{1}}$$, and using (2.28), we find  $$ \mu_{3L}\left(\lambda_{j}^{\ast}\right)\tilde{b}^{\ast}(\lambda_{j})\mathrm{adj}[a^{\ast}(\lambda_{j})]= -\mathrm{e}^{-2i\theta(\lambda_{j})}\left\{\tilde{b}^{\ast}(\lambda_{j})\mathrm{adj}[a^{\ast}(\lambda_{j})]b^{\ast}(\lambda_{j})\right\}\mu_{2R}\left(\lambda_{j}^{\ast}\right)\mathrm{adj}[a^{\ast}(\lambda_{j})]. $$ Thus, (2.36b) is proved. In order to derive (2.36d), we note that M−R = M+J1R yields  $$ \mu_{1R}(\lambda)[d^{\ast}(\lambda^{\ast})]^{-1}=\mathrm{e}^{2i\theta(\lambda)}\mu_{3L}(\lambda)\Gamma^{\ast}(\lambda^{\ast})+\mu_{2R}(\lambda)[a^{\ast}(\lambda^{\ast})]^{-1}.$$ Taking the residues of this equation at $$\{\kappa ^{\ast }_{j}\}_{1}^{2K}$$, we get (2.36d). The proof of (2.36c) is similar. □ Remark 2.6 We should consider the existence of $$\tilde {b}(\lambda )$$. Indeed, for the fixed j ∈{1, …, 2N}, thus $$\det [a(\lambda _{j})]=0$$, it follows from the nonsingularity of s(λ) that b(λj)adj[a(λj)]≠0, without loss of generality, suppose that the first entry is not zero, then we define  $$ \tilde{b}(\lambda)=\mathrm{diag}(1,0,\ldots 0)\mathrm{adj}[a^{T}(\lambda)]b^{T}(\lambda).$$ Specially, for each j ∈{1, …, 2N}, |b(λj)adj[a(λj)]|2≠0, although adj[a*(λ)]b*(λ) is not analytic for $$\lambda \in \mathbb {C}_{-}$$, it follows from (2.29) that $$\tilde {b}(\lambda _{j})$$ can be replaced by adj[a*(λj)]b*(λj). Consequently, Λj can be rewritten as  $$ \Lambda_{j}=\frac{\mathrm{adj}[a^{\ast}(\lambda_{j})a(\lambda_{j})]b^{\ast}(\lambda_{j})}{\dot{\det}[a(\lambda_{j})]| b(\lambda_{j})\mathrm{adj}[a(\lambda_{j})]|^{2}}. $$ (2.39) 2.5 Reconstructing the potential q(x, t) The potential q(x, t) can be reconstructed from the eigenfunctions μ(x, t, λ). Substituting  $$ \mu=I_{n+1}+\frac{m^{(1)}}{\lambda}+O\left(\frac{1}{\lambda^{2}}\right),\quad \lambda\rightarrow \infty,$$ into the x-part of (2.8) and considering terms of O(λ), we find that  $$ q(x,t)=-2\mathrm{e}^{-i\int_{(0,0)}^{(x,t)}\Delta(x^{\prime},t^{\prime})}W(x,t)m(x,t), $$ (2.40) where Δ and W are defined by (2.3) and (2.5), respectively, and we write m(x, t) for $$m^{(1)}_{21}(x,t)$$. Following from (2.40) and its Hermitian conjugate, and recalling that W* = W−1, we obtain  \begin{align*}|q|^{2}&=4|m|^{2}, \quad q^{\ast}q_{x}-q^{\ast}_{x}q=4\left(m^{\ast}m_{x}-m_{x}^{\ast}m\right)-32i|m|^{4},\\ qq^{\ast}&=4Wmm^{\ast}W^{\ast},\quad qq^{\ast}_{x}-q_{x}q^{\ast}=32i|m|^{2}Wmm^{\ast}W^{\ast}-4W\left(m_{x}m^{\ast}-mm_{x}^{\ast}\right)W^{\ast}. \end{align*} Thus, the 1-form Δ(x, t) can be expressed in terms of m(x, t) as  $$ \Delta=2|m|^{2}\,\mathrm{d}x+\left(4|m|^{4}+2im^{\ast}m_{x}-2im_{x}^{\ast}m\right)\,\mathrm{d}t, $$ (2.41) and W(x, t) is defined by  $$ \begin{aligned} W(x,t)&=I_{n}+\int_{(0,0)}^{(x,t)}\omega(x^{\prime},t^{\prime}),\\ \omega(x,t)&=-2iWmm^{\ast}\,\mathrm{d}x+2W\left(-2i|m|^{2}mm^{\ast}+m_{x}m^{\ast}-mm^{\ast}_{x}\right)\,\mathrm{d}t. \end{aligned} $$ (2.42) The potential q(x, t) can now be reconstructed as follows: Compute m according to  $$ m(x,t)=\lim_{\lambda\rightarrow \infty}(\lambda \mu(x,t,\lambda))_{21}.$$ Determine Δ(x, t) and W(x, t) from (2.41) and (2.42), respectively. q(x, t) is given by (2.40). 2.6 The global relation The spectral functions s(λ) and S(λ) are not independent but satisfy an important relation. Indeed, it follows from (2.15) that  $$ \mu_{1}\mathrm{e}^{i(\lambda^{2}x+2\lambda^{4}t)\hat{\sigma}}(S^{-1}s)=\mu_{3}. $$ (2.43) Since μ1(0, T, λ) = In+1, evaluating at (0, T) and recalling the definition (2.11) of μ3(x, t, λ), we find  $$ S^{-1}(\lambda)s(\lambda)=I_{n+1}-\mathrm{e}^{-2i \lambda^{4}T\hat{\sigma}}\int_{0}^{\infty}\mathrm{e}^{-i\lambda^{2}x^{\prime}\hat{\sigma}} (U\mu_{3})(x^{\prime},T,\lambda)\,\mathrm{d} x^{\prime}, $$ (2.44) The (12) entry of this equation yields the following global relation:  $$ \det[A(\lambda)]b(\lambda)-B(\lambda)\mathrm{adj}[A(\lambda)]a(\lambda) =\mathrm{e}^{-4i\lambda^{4}T}f(T,\lambda),\quad \lambda\in D_{4}, $$ (2.45)  $$ f(T,\lambda)=-\int_{0}^{\infty}\mathrm{e}^{-2i\lambda^{2}x^{\prime}} (U\mu_{3})_{12}(x^{\prime},T,\lambda)\,\mathrm{d} x^{\prime}.$$ It follows that f(T, λ) is analytic and bounded for λ ∈ D4. Also,  $$ f(T,\lambda)=O\left(\frac{1}{\lambda}\right),\quad \lambda\in D_{4}\rightarrow \infty. $$ In particular, if $$T=\infty $$, the global relation becomes  $$ \det[A(\lambda)]b(\lambda)-B(\lambda)\mathrm{adj}[A(\lambda)]a(\lambda) =0,\quad \lambda\in D_{4}. $$ (2.46) 3. The matrix-valued spectral function The analysis of Section 2 motivates the following definitions for the matrix-valued spectral functions. Definition 3.1 (The matrix-valued spectral function a(λ) and b(λ)). Given $$q_{0}(x)\in S(\mathbb {R}^{+})$$, we define the map  $$ \mathscr{F}:q_{0}(x)\mapsto (a(\lambda),b(\lambda)) $$ as follows:  $$ \left(\begin{array}{c} b(\lambda)\\ a(\lambda) \end{array}\right)=\phi_{R}(0,\lambda), $$ where $$S(\mathbb {R}^{+})$$ denotes the Schwartz space on $$\mathbb {R}^{+}$$, and ϕ(x, λ) is the unique solution of equation (2.8a) evaluated at t = 0 defined by the Volterra integral equation  $$ \phi(x,\lambda)=I_{n+1}+\int^{x}_{\infty} \mathrm{e}^{i \lambda^{2}(x-x^{\prime})\hat{\sigma}}[U(x^{\prime},0,\lambda)\phi(x^{\prime},\lambda)]\, \mathrm{d}x^{\prime}, $$ (3.1) where U(x, 0, λ) is given in terms of W(x, 0) and Q(x, 0), which are both determined by initial data q0(x). Proposition 3.2 The spectral functions a(λ) and b(λ) have the following properties: a(λ) and b(λ) are analytic for $$\lambda \in \mathbb {C}_{-}$$ and continuous and bounded for $$\lambda \in \bar {\mathbb {C}}_{-}$$. a*(λ*)a(λ) + b*(λ*)b(λ) = In, $$\lambda \in \mathbb {C}$$. a(λ) = a(−λ), b(λ) = −b(−λ). $$a(\lambda )=I_{n}+O\left (\frac {1}{\lambda }\right ),\ b(\lambda )=O\left (\frac {1}{\lambda }\right ),\ \lambda \in \mathbb {C}_{-}\rightarrow \infty $$. The map $$\mathscr {H}: \{a(\lambda ),b(\lambda )\}\mapsto q_{0}(x)$$, inverse to $$\mathscr {F}$$, is defined as follows  \begin{align} \begin{aligned} q_{0}(x)&=-2\mathrm{e}^{-2i{{{\int_{0}^{x}}}}|\check m(x^{\prime})|^{2}\,\mathrm{d}x^{\prime}}\check{W}(x)\check m(x),\\ \check m(x)&=\lim_{\lambda\rightarrow \infty}(\lambda M^{(x)}(x,\lambda))_{21}, \end{aligned} \end{align} (3.2) and $$\check {W}(x)$$ satisfies the following Volterra integral equation  $$ \check{W}(x)=I_{n}-2i{{{\int_{0}^{x}}}} \check{W}(x^{\prime})\check m(x^{\prime})\check m^{\ast}(x^{\prime})\,\mathrm{d}x^{\prime},$$ where M(x)(x, λ) is the unique solution of the following matrix RH problem: $$M^{(x)}(x,\lambda )=\left\{\begin {array}{ll} M_{+}^{(x)}(x,\lambda ), &\mathrm {Im} \lambda ^{2}\geqslant 0,\\ M_{-}^{(x)}(x,\lambda ), &\mathrm {Im} \lambda ^{2} \leqslant 0, \end {array}\right.$$ is a meromorphic function of λ in $$ \mathbb {C}\backslash \{\lambda ^{2}\in \mathbb {R}\}$$. M(x)(x, λ) satisfies the jump condition  $$ M_{-}^{(x)}(x,\lambda)=M_{+}^{(x)}(x,\lambda)J^{(x)}(x,\lambda), \quad \lambda^{2}\in\mathbb{R}, $$ (3.3) where  $$ J^{(x)}(x,\lambda)=\begin{pmatrix} \dfrac{1}{\det [a(\lambda)a^{\ast}(\lambda^{\ast})]}&\dfrac{\mathrm{e}^{2i\lambda^{2}x}b(\lambda)}{\det[a^{\ast}(\lambda^{\ast})]}\\[2ex] \dfrac{\mathrm{e}^{-2i\lambda^{2}x}b^{\ast}(\lambda^{\ast})}{\det[a(\lambda)]}&I_{n} \end{pmatrix}. $$ (3.4) If $$\det [a(\lambda )]$$ has 2N simple zeros $$\{\lambda _{j}\}_{1}^{2N}$$ in $$ \mathbb {C}_{-}$$, N = n1 + n2, where λj ∈ D4, j = 1, …, 2n1; λj ∈ D3, j = 2n1 + 1, …, 2N. $$M_{L}^{(x)}$$ can have simple poles at $$\{\lambda _{j}\}_{1}^{2N}$$, and $$M_{R}^{(x)}$$ can have simple poles at $$\{\lambda ^{\ast }_{j}\}_{1}^{2N}$$. The associated residues are given by  $$ \begin{aligned} \underset{\lambda_{j}}{\mathrm{Res}}\,M^{(x)}_{L}(x,\lambda)&=\mathrm{e}^{ -2i{{{\lambda_{j}^{2}}}}x}M^{(x)}_{R}(x,\lambda_{j}) \Lambda_{j},\ j=1,\ldots,2N,\\ \underset{\lambda^{\ast}_{j}}{\mathrm{Res}}\,M^{(x)}_{R}(x,\lambda)&=-\mathrm{e}^{2i \lambda^{*2}_{j}x}M^{(x)}_{L}\left(x,\lambda_{j}^{\ast}\right)\Lambda^{\ast}_{j},\ j=1,\ldots,2N, \end{aligned} $$ (3.5) where Λj is defined by (2.39). $$M^{(x)}(x,\lambda )=I_{n+1}+O\left (\frac {1}{\lambda }\right ),\quad \lambda \rightarrow \infty $$. Proof. (i)–(iv) follow from the discussion in Section 2.3. Let $$\tilde {\phi }(x,\lambda )$$ be the unique solution of equation (2.8a) evaluated at t = 0 defined by the Volterra integral equation  \begin{align} \tilde{\phi}(x,\lambda)=I_{n+1}+{{{\int^{x}_{0}}}} \mathrm{e}^{i \lambda^{2}(x-x^{\prime})\hat{\sigma}}\left[U(x^{\prime},0,\lambda)\tilde{\phi}(x^{\prime},\lambda)\right]\, \mathrm{d}x^{\prime}. \end{align} (3.6) Define  \begin{align*} M^{(x)}_{+}(x,\lambda)&=\left(\phi_{L}(x,\lambda),\tilde{\phi}_{R}(x,\lambda)[a^{\ast}(\lambda^{\ast})]^{-1}\right),\quad \mathrm{Im} \lambda^{2}\geqslant 0,\\ M^{(x)}_{-}(x,\lambda)&=\left(\frac{\tilde{\phi}_{L}(x,\lambda)}{\det[a(\lambda)]},\phi_{R}(x,\lambda)\right),\quad \mathrm{Im} \lambda^{2}\leqslant 0. \end{align*} Equation $$\phi (x,\lambda )=\tilde {\phi }(x,\lambda )\mathrm {e}^{i\lambda ^{2}x\hat {\sigma }}s(\lambda )$$ can be rewritten as (3.3). By the similar arguments for M(x, t, λ) and Theorem 4.1, it is straightforward to prove (v) (also see Fokas et al., 2005; Lenells, 2008). □ Definition 3.3 (The matrix-valued spectral function A(λ) and B(λ)). Let g0(t) and g1(t) be smooth functions, we define the map  $$ \tilde{\mathscr{F}}:(g_{0}(t),g_{1}(t))\mapsto (A(\lambda),B(\lambda)) $$ as follows:  $$ \left(\begin{array}{c} B(\lambda)\\A(\lambda) \end{array}\right)=\varphi_{R}(0,\lambda), $$ where φ(t, λ) is the unique solution of equation (2.8b) evaluated at x = 0 defined by the Volterra integral equation  $$ \varphi(t,\lambda)=I_{n+1}+{{{\int_{T}^{t}}}} \mathrm{e}^{2i \lambda^{4}(t-t^{\prime})\hat{\sigma}}\left[V(0,t^{\prime},\lambda)\varphi(t^{\prime},\lambda)\right]\, \mathrm{d}t^{\prime}. $$ (3.7) Here V (0, t, λ) is given in terms of W(0, t), Q(0, t) and Qx(0, t), which all are determined by boundary data g0(t) and g1(t). Proposition 3.4 The spectral functions A(λ) and B(λ) have the following properties: A(λ) and B(λ) are entire functions and bounded for $$\lambda \in \bar {D}_{2}\cup \bar {D}_{4}$$. $$A^{\ast }(\lambda ^{\ast })A(\lambda )+B^{\ast }(\lambda ^{\ast })B(\lambda )=I_{n},\ \lambda \in \mathbb {C}$$. A(λ) = A(−λ), B(λ) = −B(−λ). $$A(\lambda )=I_{n}+O\left (\frac {1}{\lambda }\right ),\ B(\lambda )=O\left (\frac {1}{\lambda }\right ),\ \lambda \in D_{2}\cup D_{4}\rightarrow \infty $$. The map $$\tilde {\mathscr {H}}: \{A(\lambda ),B(\lambda )\}\mapsto \{g_{0}(t),g_{1}(t)\}$$, inverse to $$\tilde {\mathscr {F}}$$, is defined as follows  $$ \begin{aligned} g_{0}(t)&=-2\mathrm{e}^{-i{{{\int_{0}^{t}}}} \hat{\Delta}(t^{\prime})}\hat{W}(t)m^{(1)}_{21}(t),\\ g_{1}(t)&=-\frac{i}{2}|g_{0}|^{2}g_{0}(t)+2im^{(2)}_{11}(t)g_{0}(t)+4i\mathrm{e}^{-i{{{\int_{0}^{t}}}} \hat{\Delta}(t^{\prime})}\hat{W}(t)m^{(3)}_{21}(t),\\ \hat{\Delta}(t)&=\left[-4\left|m^{(1)}_{21}\right|^{4}+8\mathrm{Re} \left(m^{(1)*}_{21}m^{(3)}_{21}-\left|m^{(1)}_{21}\right|^{2}m^{(2)}_{11}\right)\right]\,\mathrm{d}t, \end{aligned} $$ (3.8) and $$\hat {W}(t)$$ satisfies the following Volterra integral equation  $$ \hat{W}(t)=I_{n}+4i{{{\int_{0}^{t}}}} \hat{W}(t^{\prime})\left[\left|m^{(1)}_{21}\right|^{2}m^{(1)}_{21}m^{(1)*}_{21}+2\mathrm{Re}\left(m^{(2)}_{11}m^{(1)}_{21}m^{(1)*}_{21} -m^{(3)}_{21}m^{(1)*}_{21}\right)\right]\mathrm{d}t^{\prime}. $$ Here functions $$m^{(1)}_{21},m^{(2)}_{11}$$ and $$m^{(3)}_{21}$$ are determined by the asymptotic expansion  $$ M^{(t)}(t,\lambda)=I_{n+1}+\frac{m^{(1)}(t)}{\lambda}+\frac{m^{(2)}(t)}{\lambda^{2}}+\frac{m^{(3)}(t)}{\lambda^{3}}+O\left(\frac{1}{\lambda^{4}}\right),\quad \lambda\rightarrow\infty.$$ where M(t)(t, λ) is the unique solution of the following matrix RH problem: $$M^{(t)}(t,\lambda )=\begin {cases} M_{+}^{(t)}(t,\lambda ), &\mathrm {Im} \lambda ^{4} > 0,\\ M_{-}^{(t)}(t,\lambda ), &\mathrm {Im} \lambda ^{4} < 0, \end {cases}$$ is a meromorphic function of λ in $$ \mathbb {C}\backslash \{\lambda ^{4}\in \mathbb {R}\}$$. M(t)(x, λ) satisfies the jump condition  $$ M_{-}^{(t)}(t,\lambda)=M_{+}^{(t)}(t,\lambda)J^{(t)}(t,\lambda), \quad \lambda^{4}\in\mathbb{R}, $$ (3.9) where  $$ J^{(t)}(t,\lambda)=\begin{pmatrix} \dfrac{1}{\det [A(\lambda)A^{\ast}(\lambda^{\ast})]}&\dfrac{\mathrm{e}^{4i\lambda^{4}t}B(\lambda)}{\det[A^{\ast}(\lambda^{\ast})]}\\[2ex] \dfrac{\mathrm{e}^{-4i\lambda^{4}t}B^{\ast}(\lambda^{\ast})}{\det [A(\lambda)]}&I_{n} \end{pmatrix}. $$ (3.10) If $$\det [A(\lambda )]$$ has 2n0 simple zeros $$\{\zeta _{j}\}_{1}^{2n_{0}}$$ in D2 ∪ D4. $$M_{L}^{(t)}$$ can have simple poles at $$\{\zeta _{j}\}_{1}^{2n_{0}}$$, and $$M_{R}^{(t)}$$ can have simple poles at $$\{\zeta ^{\ast }_{j}\}_{1}^{2n_{0}}$$. The associated residues are given by  $$ \begin{aligned} \underset{\zeta_{j}}{\mathrm{Res}}\,M^{(t)}_{L}(t,\lambda)&=\mathrm{e}^{ -4i{{{\zeta_{j}^{4}}}}t}M^{(t)}_{R}(t,\zeta_{j}) \hat{\Lambda}_{j},\ j=1,\ldots,2n_{0},\\ \underset{\zeta^{\ast}_{j}}{\mathrm{Res}}\,M^{(t)}_{R}(t,\lambda)&=-\mathrm{e}^{4i \zeta^{*4}_{j}t}M^{(t)}_{L}(t,\zeta^{\ast}_{j})\hat{\Lambda}^{\ast}_{j},\ j=1,\ldots,2n_{0}, \end{aligned} $$ (3.11) where $$\hat {\Lambda }_{j}=\frac {\mathrm {adj}[A^{\ast }(\zeta _{j})A(\zeta _{j})]B^{\ast }(\zeta _{j})}{\dot {\det }[A(\zeta _{j})]| B(\zeta _{j})\mathrm {adj}[A(\zeta _{j})]|^{2}}$$. $$M^{(t)}(t,\lambda )=I_{n+1}+O\left (\frac {1}{\lambda }\right ),\quad \lambda \rightarrow \infty $$. Proof. (i)–(iv) follow from the discussion in Section 2.3. Let $$\tilde {\varphi }(t,\lambda )$$ be the unique solution of equation (2.8b) evaluated at x = 0 defined by the Volterra integral equation  \begin{align} \tilde{\varphi}(t,\lambda)=I_{n+1}+{{{\int_{0}^{t}}}} \mathrm{e}^{2i \lambda^{4}(t-t^{\prime})\hat{\sigma}}\left[V(0,t^{\prime},\lambda)\tilde{\varphi}(t^{\prime},\lambda)\right]\, \mathrm{d}t^{\prime}. \end{align} (3.12) Define  \begin{align*} M_{+}^{(t)}(t,\lambda)&=\left(\varphi_{L}(t,\lambda),\tilde{\varphi}_{R}(t,\lambda)[A^{\ast}(\lambda^{\ast})]^{-1}\right),\quad \mathrm{Im} \lambda^{4}> 0,\\ M_{-}^{(t)}(t,\lambda)&=\left(\frac{\tilde{\varphi}_{L}(t,\lambda)}{\det [A(\lambda)]},\varphi_{R}(t,\lambda)\right),\quad \mathrm{Im} \lambda^{4}<0. \end{align*} Equation $$\varphi (t,\lambda )=\tilde {\varphi }(t,\lambda )\mathrm {e}^{2i\lambda ^{4} t\hat {\sigma }}S(\lambda )$$ can be rewritten as (3.10). By the similar arguments for M(x, t, λ) and Theorem 4.1, it is straightforward to prove (v) (also see Fokas et al., 2005; Lenells, 2008). □ Definition 3.5 (An admissible set of functions). Given $$q_{0}(x)\in S(\mathbb {R}^{+})$$, define a(λ) and b(λ) by definition 3.1. Suppose that there exist smooth function g0(t) and g1(t) such that The associated A(λ) and B(λ) defined by definition 3.3 satisfy the global relation (2.45). The initial and boundary data are compatible with equation (1.1) at x = t = 0, i.e.  $$ g_{0}(0)=q_{0}(0),\quad g_{1}(0)=q^{\prime}_{0}(0),$$  $$ ig_{1}(0)+q^{\prime\prime}_{0}(0)+i[q^{\prime\ast}_{0}(0)q_{0}(0)+q^{\ast}_{0}(0)q^{\prime}_{0}(0)]q_{0}(0)+i|q_{0}(0)|^{2}q^{\prime}_{0}(0)=0,$$ where $$q_{0}^{\prime }(x)=\frac {\mathrm {d} q_{0}(x)}{\mathrm {d} x}$$. Then we call {g0(t), g1(t)} an admissible set of functions with respect to q0(x). 4. The main result Theorem 4.1 Let $$q_{0}(x)\in S(\mathbb {R}^{+})$$, suppose that the functions g0(t) and g1(t) are admissible with respect to q0(x) (see definition 3.5). Define the spectral functions {a(λ), b(λ)} and {A(λ), B(λ)} in terms of q0(x) and {g0(t), g1(t)} according to definitions 3.1 and 3.3. Assume that the possible zeros $$\{\lambda _{j}\}_{1}^{2N}$$ of $$\det [a(\lambda )]$$ and $$\{\kappa _{j}\}_{1}^{2K}$$ of $$\det [d(\lambda )]$$ are as in assumption 2.4. Define M(x, t, λ) as the solution of the following matrix RH problem: M is sectionally meromorphic in $$\mathbb {C}\backslash \{\lambda ^{4}\in \mathbb {R}\}$$. ML can have simple poles at $$\{\lambda _{j}\}_{1}^{2n_{1}}$$ and $$\{\kappa _{j}\}_{1}^{2K}$$, MR can have simple poles at $$\{\lambda ^{\ast }_{j}\}_{1}^{2n_{1}}$$ and $$\{\kappa ^{\ast }_{j}\}_{1}^{2K}$$. The associated residues of M satisfy the relations in (2.36). M satisfies the jump condition  $$ M_{-}(x,t,\lambda)=M_{+}(x,t,\lambda)J(x,t,\lambda),\quad \lambda^{4}\in\mathbb{R}, $$ (4.1) where M is M+ for λ ∈ D1 ∪ D3, M is M− for λ ∈ D2 ∪ D4 and J is defined in terms of {a, b, A, B} by (2.32) and (2.33), see Fig. 2. M(x, t, λ) has the following asymptotics as $$\lambda \rightarrow \infty $$,  $$ M(x,t,\lambda)=I_{n+1}+O\left(\frac{1}{\lambda}\right),\ \lambda\rightarrow \infty. $$ (4.2) Then M(x, t, λ) exists and is unique. Define q(x, t) in terms of M(x, t, λ) by  $$ q(x,t)=-2\mathrm{e}^{-i\int_{(0,0)}^{(x,t)}\Delta(x^{\prime},t^{\prime})}W(x,t)m(x,t), $$ (4.3) with  $$ m(x,t)=\lim_{\lambda\rightarrow \infty} (\lambda M(x,t,\lambda))_{21}, $$ (4.4) where Δ(x, t) and W(x, t) are defined by (2.41) and (2.42), respectively. Then q(x, t) solves equation (1.1). Furthermore,  \begin{align*} q(x,0)=q_{0}(x),\quad q(0,t)=g_{0}(t),\quad q_{x}(0,t)=g_{1}(t).\end{align*} Proof. The existence and uniqueness of the solution for the above RH problem are established by the vanishing lemma, see Zhou (1989). Moreover, it follows from standard arguments via the dressing method (Zakharov & Shabat, 1974, 1979) that if M solves the above RH problem and q(x, t) is defined by (4.3), then q(x, t) solves equation (1.1). In order to prove that q(x, 0) = q0(x), we define M(x)(x, λ) by  \begin{align} \begin{aligned} M^{(x)}(x,\lambda)&=M(x,0,\lambda),\quad \lambda\in \bar{D}_{1}\cup \bar{D}_{4},\\ M^{(x)}(x,\lambda)&=M(x,0,\lambda)J_{1}^{-1}(x,0,\lambda),\quad \lambda\in \bar{D}_{2},\\ M^{(x)}(x,\lambda)&=M(x,0,\lambda)J_{3}(x,0,\lambda),\quad \lambda\in \bar{D}_{3}. \end{aligned} \end{align} (4.5) If the sets {λj} and {κj} are empty, then the function M(x) is analytic in $$\mathbb {C}\backslash \{\lambda ^{2}\in \mathbb {R}\}$$. Furthermore,  \begin{align*} M_{-}^{(x)}(x,\lambda)&=M_{+}^{(x)}(x,\lambda)J^{(x)}(x,\lambda), \quad \lambda^{2}\in\mathbb{R},\\ M^{(x)}(x,\lambda)&=I_{n+1}+O\left(\frac{1}{\lambda}\right),\quad \lambda\rightarrow \infty, \end{align*} where J(x)(x, t) is defined by (3.4). Comparing (3.2) with (4.3) evaluated at t = 0, we conclude that q0(x) = q(x, 0). If the sets {λj} and {κj} are not empty. ML(x, t, λ) has poles at $$\{\lambda _{j}\}_{1}^{2n_{1}}$$ in D4 and has poles at $$\{\kappa _{j}\}_{1}^{2K}$$ in D3. On the other hand, $$M^{(x)}_{L}(x,\lambda )$$ has poles at $$\{\lambda _{j}\}_{1}^{2N}$$ in $$\mathbb {C}_{-}$$. We will now show that the transformation defined by (4.5) maps the former poles to the latter ones. Since M(x)(x, λ) = M(x, 0, λ) for λ ∈ D4, M(x) has poles at $$\{\lambda _{j}\}_{1}^{2n_{1}}$$ with the correct residue condition. Equation (4.5) can be rewritten as  $$ M^{(x)}(x,\lambda)=\left(M_{L}(x,0,\lambda)+\mathrm{e}^{-2i\lambda^{2}x}M_{R}(x,0,\lambda)\Gamma(\lambda),M_{R}(x,0,\lambda)\right),\quad \lambda\in D_{3}. $$ (4.6) The residue condition (2.36c) at κj implies that M(x) has no poles at κj. On the other hand, equation (4.6) shows that M(x) has poles $$\{\lambda _{j}\}_{2n_{1}+1}^{2N}$$ at with residues given by  $$ \underset{\lambda_{j}}{\mathrm{Res}}\,M_{L}^{(x)}(x,\lambda)=\mathrm{e}^{-2i{{{\lambda_{j}^{2}}}}x}M_{R}^{(x)}(x,\lambda)\underset{\lambda_{j}}{\mathrm{Res}}\,\Gamma(\lambda),\quad j=2n_{1}+1,\ldots, 2N.$$ Recalling the definition of Γ(λ), (2.29) and (2.35), this becomes the residue condition (3.5). Similar considerations apply to $$\{\lambda _{j}^{\ast }\}_{1}^{2n_{1}}$$ and $$\{\kappa _{j}^{\ast }\}_{1}^{2K}$$. In what follows, we prove that q(0, t) = g0(t) and qx(0, t) = g1(t). Define M(t)(t, λ) by  $$ M^{(t)}(t,\lambda)=M(0,t,\lambda)G_{j}(t,\lambda),\quad \lambda\in \bar{D}_{j},\ j=1,\ldots,4, $$ (4.7) with  $$ G_{1}=\begin{pmatrix} \frac{\det[A^{\ast}(\lambda^{\ast})]}{\det[a^{\ast}(\lambda^{\ast})]}&0\\\mathrm{e}^{4i\lambda^{4}(T-t)}f^{\ast}(T,\lambda^{\ast})& a^{\ast}(\lambda^{\ast})[A^{\ast}(\lambda^{\ast})]^{-1} \end{pmatrix},$$  $$ G_{2}=\begin{pmatrix} \frac{1}{\det[d^{\ast}(\lambda^{\ast})]}&0\\ \mathrm{e}^{-4i\lambda^{4} t}\frac{b^{\ast}(\lambda^{\ast})}{\det[A(\lambda)]}&d^{\ast}(\lambda^{\ast}) \end{pmatrix}, $$ (4.8)  $$ G_{3}=\begin{pmatrix} \det[d(\lambda)]&-\mathrm{e}^{4i\lambda^{4}t}b(\lambda)\mathrm{adj}[a(\lambda)][A^{\ast}(\lambda^{\ast})]^{-1}\\ 0&[d(\lambda)]^{-1} \end{pmatrix},$$  $$ G_{4}=\begin{pmatrix} \frac{\det [a(\lambda)]}{\det[A(\lambda)]}&\mathrm{e}^{4i\lambda^{4}(t-T)}f(T,\lambda)\frac{\operatorname{adj}[a(\lambda)]A(\lambda)}{\det [A(\lambda)]}\\ 0&a^{-1}(\lambda)A(\lambda) \end{pmatrix}.$$ where f(T, λ) is defined by (2.45). After a tedious but straightforward calculation by virtue of the global relation (2.45), it follows that  $$ \begin{aligned} J_{1}(0,t,\lambda)G_{2}(t,\lambda)&=G_{1}(t,\lambda)J^{(t)}(t,\lambda),\quad \lambda\in \bar{D}_{1}\cap\bar{D}_{2},\\ J_{2}(0,t,\lambda)G_{2}(t,\lambda)&=G_{3}(t,\lambda)J^{(t)}(t,\lambda),\quad \lambda\in \bar{D}_{2}\cap\bar{D}_{3},\\ J_{3}(0,t,\lambda)G_{4}(t,\lambda)&=G_{3}(t,\lambda)J^{(t)}(t,\lambda),\quad \lambda\in \bar{D}_{3}\cap\bar{D}_{4},\\ J_{4}(0,t,\lambda)G_{4}(t,\lambda)&=G_{1}(t,\lambda)J^{(t)}(t,\lambda),\quad \lambda\in \bar{D}_{1}\cap\bar{D}_{4}, \end{aligned} $$ (4.9) where J(t)(t, λ) is defined by (3.9). Equations (4.1), (4.7) and (4.9) imply that M(t) satisfies of the RH problem defined in proposition 3.4 except the residue conditions. If the sets {λj} and {κj} are empty, this immediately yields the desired result. Otherwise, we should verify that M(t) satisfies the residue conditions (3.11). This follows similar arguments in the proof of Theorem 4.1 in Fokas et al. (2005). □ 5. The Dirichlet to Neumann map We consider the Dirichlet boundary value problem for the vector DNLS equation (1.1) posed on the half-line. 5.1 The global relation Evaluating (2.15) at x = 0, we find  $$ \mu_{2}(0,t,\lambda)\mathrm{e}^{2i \lambda^{4}t\hat{\sigma}}s(\lambda)=\mu_{3}(0,t,\lambda),\quad \lambda\in(\mathbb{C}_{+}, \mathbb{C}_{-}). $$ (5.1) Denoting  $$ \Phi(t,\lambda)=(Y\mu_{2}Y^{-1})(0,t,\lambda),\quad c(t,\lambda)=(Y\mu_{3}Y^{-1})_{12}(0,t,\lambda)a^{-1}(\lambda),$$ consequently, Φ−1(t, λ) =Φ*(t, λ*), thus we rewrite Φ(t, λ) as  $$ \Phi(t,\lambda)=\begin{pmatrix} \det \Phi_{2}^{\ast}(t,\lambda^{\ast})&\Phi_{1}(t,\lambda)\\ -\mathrm{adj}[\Phi_{2}^{\ast}(t,\lambda^{\ast})]\Phi_{1}^{\ast}(t,\lambda^{\ast})&\Phi_{2}(t,\lambda) \end{pmatrix}, $$ (5.2) and rewrite the (12) entry of equation (5.1) as  $$ \Phi_{1}(t,\lambda)+ \mathrm{e}^{4i \lambda^{4}t+i{{{\int_{0}^{t}}}}\Delta_{2}(0,t^{\prime})}\det(\Phi_{2}^{\ast}(t,\lambda^{\ast}))b(\lambda)W^{-1}(0,t)a^{-1}(\lambda)=c(t,\lambda),\quad \lambda\in \mathbb{C}_{-}. $$ (5.3) The function c(t, λ) is analytic and bounded in $$\mathbb {C}_{-}$$ away from the possible zeros of $$\det [a(\lambda )]$$ and of order O(1/λ) as $$\lambda \rightarrow \infty $$. 5.2 Asymptotics Substituting the asymptotic expansion  $$ \mu_{j}(x,t,\lambda)=I_{n+1}+\sum_{k\geqslant 1}\frac{\mu_{j}^{(k)}(x,t)}{\lambda^{k}},\quad j=1,2,3,\ \lambda\rightarrow \infty,$$ into (2.8), here $$\lambda \rightarrow \infty $$ means that λ in the bounded domain of μj and approaches to infinity, we find that  $$ \begin{aligned} Y\mu_{j}^{(1)}Y^{-1}&=\frac{i}{2}\sigma Q, \quad Y\mu_{j}^{(3)}Y^{-1}=\frac{i}{2}\sigma QY\mu_{j}^{(2)}Y^{-1}+\frac{i}{8}\sigma Q^{3}+\frac{1}{4}Q_{x},\\ Y\mu_{jx}^{(2)}Y^{-1}&=-\frac{i}{8}\sigma Q^{4}+\frac{1}{4}QQ_{x},\\ Y\mu_{jt}^{(2)}Y^{-1}&=-\frac{i}{4}\sigma Q^{6}+\frac{i}{4}\sigma(QQ_{xx}-{{{Q_{x}^{2}}}})+\frac{1}{2}Q^{3}Q_{x}+\frac{1}{4}QQ_{x}Q^{2}+\frac{1}{8}[Q^{2},Q_{x}Q]. \end{aligned} $$ (5.4) Thus, as $$\lambda \rightarrow \infty $$,  $$ \begin{aligned} (Y\mu_{j}Y^{-1})_{22}(x,t,\lambda)&=I_{n}+\frac{1}{\lambda^{2}}\int_{(x_{j},t_{j})}^{(x,t)} \tilde{\Delta}(x^{\prime},t^{\prime})+O\left(\frac{1}{\lambda^{4}}\right),\\ (Y\mu_{j}Y^{-1})_{12}(x,t,\lambda)&=\frac{q^{\ast}}{2\lambda}-\frac{1}{\lambda^{3}}\left(\frac{1}{8}|q|^{2}q^{\ast}+\frac{i}{4}q^{\ast}_{x}-\frac{1}{2}q^{\ast}\int_{(x_{j},t_{j})}^{(x,t)}\tilde{\Delta}(x^{\prime},t^{\prime})\right)+O\left(\frac{1}{\lambda^{5}}\right), \end{aligned} $$ (5.5) where the closed 1-form $$\tilde {\Delta }(x,t)$$ is defined by  $$ \tilde{\Delta}=\tilde{\Delta}_{1}+\tilde{\Delta}_{2},\, \tilde{\Delta}_{1}=\frac{1}{8}(i|q|^{2}qq^{\ast}-2qq^{\ast}_{x})\,\mathrm{d}x,$$  $$ \tilde{\Delta}_{2}=\left[\frac{i}{4}(-|q|^{4}qq^{\ast}-q_{x}q^{\ast}_{x}+qq^{\ast}_{xx})+\frac12|q|^{2}qq^{\ast}_{x}+\frac14(|q|^{2})_{x}qq^{\ast}-\frac18(q^{\ast}q_{x}qq^{\ast}+|q|^{2}q_{x}q^{\ast})\right]\mathrm{d}t.$$ Hence,  $$ \begin{aligned} \Phi_{2}(t,\lambda)&=I_{n}+\frac{\Phi_{2}^{(1)}(t)}{\lambda^{2}}+O\left(\frac{1}{\lambda^{4}}\right),\quad \lambda\in D_{1}\cup D_{3}\rightarrow\infty,\\ \Phi_{1}(t,\lambda)&=\frac{\Phi_{1}^{(1)}(t)}{\lambda}+\frac{\Phi_{1}^{(2)}(t)}{\lambda^{3}}+O\left(\frac{1}{\lambda^{5}}\right),\quad \lambda\in D_{1}\cup D_{3}\rightarrow\infty, \end{aligned} $$ (5.6) where  $$ \Phi_{2}^{(1)}(t)={{{\int_{0}^{t}}}} \tilde\Delta_{2}(0,t^{\prime}),\quad \Phi_{1}^{(1)}(t)=\frac12g^{\ast}_{0}(t),$$  $$ \Phi_{1}^{(2)}(t)=-\frac18|g_{0}|^{2}g_{0}^{\ast}(t)-\frac i4 g^{\ast}_{1}(t)+\frac 12g_{0}^{\ast}(t){{{\int_{0}^{t}}}} \tilde\Delta_{2}(0,t^{\prime}).$$ In particular, we find the following expressions for the boundary values:  $$ g^{\ast}_{0}(t)=2\Phi_{1}^{(1)}(t),\quad g_{1}^{\ast}(t)=\frac i2|g_{0}|^{2}g^{\ast}_{0}(t)-2ig^{\ast}_{0}(t)\Phi_{2}^{(1)}(t)+4i\Phi_{1}^{(2)}(t). $$ (5.7) The global relation (5.3) also implies that  $$ c(t,\lambda)=\frac{\Phi_{1}^{(1)}(t)}{\lambda}+\frac{\Phi_{1}^{(2)}(t)}{\lambda^{3}}+O\left(\frac{1}{\lambda^{5}}\right),\quad\lambda\in \mathbb{C}_{-}\cap(D_{2}\cup D_{4})\rightarrow\infty. $$ (5.8) 5.3 The Dirichlet problem The following theorem expresses the spectral functions A(λ) and B(λ) in terms of the Dirichlet boundary data and initial data via the solution of a system of nonlinear integral equations. Let ∂Dj, j = 1, …, 4, denote the boundary of the Dj, oriented so that Dj lies to the left of ∂Dj, moreover, let $$D_{4}^{-}=D_{4}\cap \{\mathrm {Im}\lambda\! \leqslant\! 0\}$$. Theorem 5.1 Let $$T<\infty $$, $$q_{0}(x)\in S(\mathbb {R}^{+})$$. Assume that the function g0(t), $$0\leqslant t < T $$, has sufficient smoothness and is compatible with q0(x) at x = t = 0. Suppose that $$\det [a(\lambda )]$$ has a finite (possibly empty) set of simple zeros in D4, which are denoted by $$\{\lambda _{j}\}_{1}^{2n_{1}}$$, and assume that no zeros of $$\det [a(\lambda )]$$ occur on ∂D4. Then the matrix-valued spectral functions A(λ) and B(λ) are given by  $$ \begin{aligned} A(\lambda)&=W^{-1}(0,T)\Phi^{\ast}_{2}(T,\lambda^{\ast})W(0,T),\\ B(\lambda)&=-\mathrm{e}^{4i\lambda^{4}T-i{{{\int_{0}^{T}}}} \Delta_{2}(0,t)}\Phi_{1}(T,\lambda)\mathrm{adj}[\Phi_{2}(t,\lambda)]W(0,T), \end{aligned} $$ (5.9) where  $$ \Delta_{2}(0,t)=-\frac14\left(3|g_{0} |^{4}-2ig_{0}^{\ast} g_{1} +2ig_{1}^{\ast} g_{0} \right)\mathrm{d}t, $$ (5.10) and W(0, t) satisfies the Volterra integral equation  $$ W(0,t)=I_{n}+{{{\int_{0}^{t}}}}\tilde\omega(t^{\prime})W(0,t^{\prime})\,\mathrm{d}t^{\prime}, $$ (5.11)  $$ \tilde\omega=\frac14\left(3i|g_{0}|^{2}g_{0}g_{0}^{\ast}-2g_{0}g_{1}^{\ast}+2g_{1}g_{0}^{\ast}\right)(t),$$ and Φ1(t, λ) and Φ2(t, λ) satisfies the Volterra integral equation  $$\! \begin{pmatrix} \Phi_{1}(t,\lambda)\\\Phi_{2}(t,\lambda) \end{pmatrix}\!\!=\!\!\begin{pmatrix} 0\\I_{n} \end{pmatrix}\!+\!\!{{{\int_{0}^{t}}}}\!\!\begin{pmatrix}\! \mathrm{e}^{4i\lambda^{4}(t-t^{\prime})}\left[-\left(i\lambda^{2}|g_{0}|^{2}\Phi_{1}+\Phi_{1} \tilde\omega\right)+i\lambda\left(|g_{0}|^{2}g_{0}^{\ast}-2\lambda^{2}g_{0}^{\ast}+ig_{1}^{\ast}\right)\Phi_{2}\right](t^{\prime},\lambda)\\ \left[i\lambda(|g_{0}|^{2}g_{0}-2\lambda^{2}g_{0}-ig_{1})\Phi_{1}+(i\lambda g_{0}g_{0}^{\ast}\Phi_{2}-\Phi_{2}\tilde\omega)\right](t^{\prime},\lambda) \!\end{pmatrix}\!\mathrm{d}t^{\prime}, $$ (5.12) with 0 < t < T. For a function f(λ), let f+(λ) and f−(λ) denote  $$ f_{+}(\lambda)=f(\lambda)+if(i\lambda),\quad f_{-}(\lambda)=f(\lambda)-f(i\lambda). $$ The unknown Neumann boundary value g1(t) is given by  $$ \begin{aligned} g_{1}^{\ast}(t)&=\frac i2|g_{0}|^{2}g^{\ast}_{0}(t)-\frac{4g_{0}^{\ast}(t)}{\pi}\int_{\partial D_{4}^{-}}\lambda\Phi_{2-}(t,\lambda)\,\mathrm{d}\lambda+\frac 4\pi\int_{\partial D_{4}^{-}}\left(\lambda g^{\ast}_{0}(t)-\lambda^{2}\Phi_{1+}(t,\lambda)\right)\mathrm{d}\lambda\\ &\quad-\frac 8\pi\mathrm{e}^{i{{{\int_{0}^{t}}}} \Delta_{2}(0,t^{\prime})}\int_{\partial D_{4}^{-}}\lambda^{2}\mathrm{e}^{4i \lambda^{4}t}\det\left(\Phi_{2}^{\ast}(t,\lambda^{\ast})\right)b(\lambda)W^{-1}(0,t)a^{-1}(\lambda)\,\mathrm{d}\lambda. \end{aligned} $$ (5.13) Proof. Equations (5.9)–(5.12) follow from (2.2)–(2.5), (2.16), (2.24) and (5.2). In order to derive (5.13) we note that (5.7) expresses g1(t) in terms of $$\Phi _{2}^{(1)}(t)$$ and $$\Phi _{1}^{(2)}(t)$$. Furthermore, it follows from (2.19), (5.2), (5.6) and Cauchy’s theorem that  $$ \begin{aligned} i\pi\Phi_{2}^{(1)}(t)&=-\int_{\partial D_{1}\cup\partial D_{3} }\lambda\left[\Phi_{2}(t,\lambda)-I_{n}\right]\mathrm{d} \lambda=\int_{\partial D_{2}\cup \partial D_{4}}\lambda\left[\Phi_{2}(t,\lambda)-I_{n}\right]\mathrm{d} \lambda\\ &=\int_{\partial D_{4}}\lambda\left[\Phi_{2}(t,\lambda)-I_{n}\right]\mathrm{d} \lambda-\int_{\partial D_{4}}\lambda\left[\Phi_{2}(t,i\lambda)-I_{n}\right]\mathrm{d} \lambda\\ &=\int_{\partial D_{4}}\lambda\Phi_{2-}(t,\lambda)\,\mathrm{d} \lambda=2\int_{\partial D_{4}^{-}}\lambda\Phi_{2-}(t,\lambda)\,\mathrm{d} \lambda, \end{aligned} $$ (5.14)  $$ \begin{aligned} i\pi \Phi_{1}^{(2)}(t)&=-\int_{\partial D_{1}\cup\partial D_{3}}\left[\lambda^{2}\Phi_{1}(t,\lambda)-\lambda\Phi_{1}^{(1)}(t)\right]\mathrm{d}\lambda\\ &= \int_{\partial D_{2}\cup\partial D_{4}}\left[\lambda^{2}\Phi_{1}(t,\lambda)-\lambda\Phi_{1}^{(1)}(t)\right]\mathrm{d}\lambda\\ &=\left(\int_{\partial D_{2}}-\int_{\partial D_{4}}\right)\left[\lambda^{2}\Phi_{1}(t,\lambda)-\lambda\Phi_{1}^{(1)}(t)\right]\mathrm{d}\lambda+F(t)\\ &=\int_{\partial D_{4}}\left[2\lambda\Phi_{1}^{(1)}(t)-\lambda^{2}\Phi_{1+}(t,\lambda)\right]\mathrm{d}\lambda+F(t),\\ &=2\int_{\partial D_{4}^{-}}\left[2\lambda\Phi_{1}^{(1)}(t)-\lambda^{2}\Phi_{1+}(t,\lambda)\right]\mathrm{d}\lambda+F(t), \end{aligned} $$ (5.15) where F(t) is defined by  $$ F(t)=2\int_{\partial D_{4}}\left[\lambda^{2}\Phi_{1}(t,\lambda)-\lambda\Phi_{1}^{(1)}(t)\right]\mathrm{d}\lambda=4\int_{\partial D_{4}^{-}} \left[\lambda^{2}\Phi_{1}(t,\lambda)-\lambda\Phi_{1}^{(1)}(t)\right]\mathrm{d}\lambda. $$ Using the global relation (5.3) and the asymptotics (5.8) of c(t, λ), we find  $$ \begin{aligned} F(t)&=4\int_{\partial D_{4}^{-}}\left[\lambda^{2} c(t,\lambda)-\lambda\Phi_{1}^{(1)}(t)\right]\mathrm{d}\lambda-4\int_{\partial D_{4}^{-}}\lambda^{2}\left[c(t,\lambda)-\Phi_{1}(t,\lambda)\right]\mathrm{d}\lambda\\ &=-i\pi\Phi_{1}^{(2)}(t)-4\mathrm{e}^{i{{{\int_{0}^{t}}}} \Delta_{2}(0,t^{\prime})}\int_{\partial D_{4}^{-}}\lambda^{2}\mathrm{e}^{4i \lambda^{4}t}\det(\Phi_{2}^{\ast}(t,\lambda^{\ast}))b(\lambda)W^{-1}(0,t)a^{-1}(\lambda) \,\mathrm{d}\lambda. \end{aligned} $$ (5.16)From the above arguments we prove (5.13). □ Note: In the case that $$\det [a(\lambda )]$$ has a finite number of simple zeros in $$D_{4}^{-}$$, the representation (5.13) is modified by adding a term  $$ -\frac8\pi\mathrm{e}^{i{{{\int_{0}^{t}}}} \Delta_{2}(0,t^{\prime})}\sum_{\lambda_{j}\in D_{4}^{-}}{{{\lambda_{j}^{2}}}}\mathrm{e}^{4i{{{\lambda_{j}^{4}}}}t}\det(\Phi_{2}^{\ast} (t,\lambda_{j}^{\ast}))b(\lambda_{j})W^{-1}(0,t)\frac{\mathrm{adj}[a(\lambda_{j})]}{\dot{\mathrm{det}}[a(\lambda_{j})]} $$ provided that the integration contour $$\partial D_{4}^{-}$$ is replaced with τ, where τ denotes the contour obtained by deforming $$\partial D_{4}^{-}$$ such that it does not surround the possible zeros of $$\det [a(\lambda )]$$, see Fig. 3. Fig. 2. View largeDownload slide The contour for the RH problem in the complex λ-plane. Fig. 2. View largeDownload slide The contour for the RH problem in the complex λ-plane. Fig. 3. View largeDownload slide The deformed contour τ in the complex λ-plane. The zeros of det[a(λ)] are indicated by “.” in this figure. Fig. 3. View largeDownload slide The deformed contour τ in the complex λ-plane. The zeros of det[a(λ)] are indicated by “.” in this figure. Substitution of the expression (5.13) into (5.10)–(5.12) yield a system of nonlinear integral equations involving the functions Δ2(0, t), W(0, t), Φ1(t, λ) and Φ2(t, λ). Assuming that this system has a unique solution, A(λ) and B(λ) can be determined from (5.9). In fact, for the particular boundary problem, we can present an effective characterizations of A(λ) and B(λ). 6. Conclusions As was mentioned in the introduction, we implement the Fokas method to analyse IBV problems for the vector DNLS. In particular, as n = 1,  $$ Y(x,t)=\mathrm{e}^{i\int_{(0,0)}^{(x,t)}\Delta(x^{\prime},t^{\prime})\sigma_{3}}, \quad \tilde{Y}(x)=\mathrm{e}^{i\int_{\infty}^{x}\Delta_{1}(x^{\prime},0)\sigma_{3}},\quad \sigma_{3}=\mathrm{diag}\{1,-1\}.$$ we derive the same conclusions as in Lenells (2008, 2011). While the main ingredients of the Fokas method can be used for the case of $$n\geqslant 2$$, the actual implementation of this method presents some additional technical difficulties. In particular, the formulation of the RH problem is more complicated because the Lax pairs of multi-component equations involve (n + 1) × (n + 1) matrices. As a consequence, we transform these matrices to the 2 × 2 block ones, and construct the RH problem as the scalar case by using Sherman–Morrison formula. 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Published: Feb 1, 2018

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