
Citation: Kedong Wang, Xianguo Geng, Mingming Chen, Ruomeng Li. Long-time asymptotics for the generalized Sasa-Satsuma equation[J]. AIMS Mathematics, 2020, 5(6): 7413-7437. doi: 10.3934/math.2020475
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The Sasa-Satsuma equation
ut+uxxx+6|u|2ux+3u(|u|2)x=0, | (1.1) |
so-called high-order nonlinear Schrödinger equation [1], is relevant to several physical phenomena, for example, in optical fibers [2,3], in deep water waves [4] and generally in dispersive nonlinear media [5]. Because this equation describes these important nonlinear phenomena, it has received considerable attention and extensive research. The Sasa-Satsuma equation has been discussed by means of various approaches such as the inverse scattering transform [1], the Riemann-Hilbert method [6], the Hirota bilinear method [7], the Darboux transformation [8], and others [9,10,11]. The initial-boundary value problem for the Sasa-Satsuma equation on a finite interval was studied by the Fokas method [12], which is also effective for the initial-boundary value problems on the half-line [35,36,37]. In Ref. [13], finite genus solutions of the coupled Sasa-Satsuma hierarchy are obtained in the basis of the theory of trigonal curves, the Baker-Akhiezer function and the meromorphic functions [14,15,16]. In Ref. [17], the super Sasa-Satsuma hierarchy associated with the 3×3 matrix spectral problem was proposed, and its bi-Hamiltonian structures were derived with the aid of the super trace identity.
The nonlinear steepest descent method [18], also called Deift-Zhou method, for oscillatory Riemann-Hilbert problems is a powerful tool to study the long-time asymptotic behavior of the solution for the soliton equation, by which the long-time asymptotic behaviors for a number of integrable nonlinear evolution equations associated with 2×2 matrix spectral problems have been obtained, for example, the mKdV equation, the KdV equation, the sine-Gordon equation, the modified nonlinear Schrödinger equation, the Camassa-Holm equation, the derivative nonlinear Schrödinger equation and so on [19,20,21,22,23,24,25,26,27,28,29,30]. However, there is little literature on the long-time asymptotic behavior of solutions for integrable nonlinear evolution equations associated with 3×3 matrix spectral problems [31,32,33]. Usually, it is difficult and complicated for the 3×3 case. Recently, the nonlinear steepest descent method was successfully generalized to derive the long-time asymptotics of the initial value problems for the coupled nonlinear Schrödinger equation and the Sasa-Satsuma equation with the complex potentials [33,34]. The main differences between the 2×2 and 3×3 cases is that the former corresponds to a scalar Riemann-Hilbert problem, while the latter corresponds to a matrix Riemann-Hilbert problem. In general, the solution of the matrix Riemann-Hilbert problem can not be given in explicit form, but the scalar Riemann-Hilbert problem can be solved by the Plemelj formula.
The main aim of this paper is to study the long-time asymptotics of the Cauchy problem for the generalized Sasa-Satsuma equation [38] via nonlinear steepest decent method,
{ut+uxxx−6a|u|2ux−6bu2ux−3au(|u|2)x−3b∗u∗(|u|2)x=0,u(x,0)=u0(x), | (1.2) |
where a is a real constant, b is a complex constant that satisfies a2≠|b|2, the asterisk "∗" denotes the complex conjugate. It is easy to see that the generalized Sasa-Satsuma equation (1.2) can be reduced to the Sasa-Satsuma equation (1.1) when a=−1 and b=0. Suppose that the initial value u0(x) lies in Schwartz space S(R)={f(x)∈C∞(R):supx∈R|xα∂βf(x)|<∞,∀α,β∈N}. The vector function γ(k) is determined by the initial data in (2.15) and (2.19), and γ(k) satisfies the conditions (P1) and (P2), where
(P1):{γ†(k∗)B1γ(k)+|b|24(γ†(k∗)σ3γ(k))2<1,γ†(k∗)B1γ(k)+aγ†(k∗)σ3γ(k)<2,2γ†(k∗)B1γ(k)+|γ(k)|2+|B1γ(k)|2<4, |
with
σ3=(100−1),B1=(ab∗ba); | (1.3) |
(P2): When detB1>0 and a>0, (2a−|B1γ(k)|2) and (2a−detB1|γ(k)|2)/(1−γ†(k)B1γ(k)) are positive and bounded; otherwise, (|B1γ(k)|2−2a) and (detB1|γ(k)|2−2a)/(1−γ†(k)B1γ(k)) are positive and bounded.
The main result of this paper is as following:
Theorem 1.1. Let u(x,t) be the solution of the Cauchy problem for the generalized Sasa-Satsuma equation (1.2) with the initial value u0∈S(R). Suppose that the vector function γ(k) is defined in (2.19), the hypotheses (P1) and (P2) hold. Then, for x<0 and √−xt<C,
u(x,t)=ua(x,t)+O(c(k0)t−1logt). | (1.4) |
where C is a fixed constant, and
ua(x,t)=√ν12tk0γ†(k0)B1γ(k0)(−|γ1(−k0)|ei(ϕ+argγ1(−k0))+|γ2(−k0)|e−i(ϕ+argγ2(−k0))),k0=√−x12t,ν=−12πlog(1−γ†(k0)B1γ(k0)),ϕ=νlog(196tk30)−16tk30+argΓ(−iν)+1π∫k0−k0log|ξ+k0|d(1−γ†(ξ)B1γ(ξ))−π4, |
c(⋅) is rapidly decreasing, Γ(⋅) is the Gamma function, γ1 and γ2 are the first and the second row of γ(k), respectively.
Remark 1.1. The two conditions (P1) and (P2) satisfied by γ(k) are necessary. The condition (P1) guarantees the existence and the uniqueness of the solutions of the basic Riemann-Hilbert problem (2.16) and the Riemann-Hilbert problem (3.1). The boundedness of the function δ(k) defined in subsection 3.1 relies on the condition (P2).
Remark 1.2. In the case of a=−1 and b=0, the generalized Sasa-Satsuma equation (1.2) can be reduced to the Sasa-Satsuma equation. Then it is obvious that the condition (P1) is true, and the condition (P2) is reduced to the case that |γ(k)| is bounded. Therefore, the conditions (P1) and (P2) in this case are equivalent to the condition related to the reflection coefficient in [34], that is, |γ(k)| is bounded for the Sasa-Satsuma equation.
The outline of this paper is as follows. In section 2, we derive a Riemann-Hilbert problem from the scattering relation. The solution of the generalized Sasa-Satsuma equation is changed into the solution of the Riemann-Hilbert problem. In section 3, we deal with the Riemann-Hilbert problem via nonlinear steepest decent method, from which the long-time asymptotics in Theorem 1.1 is obtained at the end.
We begin with the 3×3 Lax pair of the generalized Sasa-Satsuma equation
ψx=(ikσ+U)ψ, | (2.1a) |
ψt=(4ik3σ+V)ψ, | (2.1b) |
where ψ is a matrix function and k is the spectral parameter, σ=diag(1,1,−1),
U=(00u00u∗au∗+buau+b∗u∗0) | (2.2) |
V=4k2U+2ik(U2+Ux)σ+[Ux,U]−Uxx+2U3. | (2.3) |
We introduce a new eigenfunction μ through μ=ψe−ikσx−4ik3σt, where eσ=diag(e,e,e−1). Then (2.1a) and (2.1b) become
μx=ik[σ,μ]+Uμ, | (2.4a) |
μt=4ik3[σ,μ]+Vμ, | (2.4b) |
where [⋅,⋅] is the commutator, [σ,μ]=σμ−μσ. From (2.4a), the matrix Jost solution μ± satisfy the Volterra integral equations
μ±(k;x,t)=I+∫x±∞eikσ(x−ξ)U(ξ,t)μ±(k;ξ,t)e−ikσ(x−ξ)dξ. | (2.5) |
Set μ±L represent the first two columns of μ±, and μ±R denote the third column, i.e., μ±=(μ±L,μ±R). Furthermore, we can infer that μ+R and μ−L are analytic in the lower complex k-plane C−, μ+L and μ−R are analytic in the the upper complex k-plane C+. Then we can introduce sectionally analytic function P1(k) and P2(k) by
P1(k)=(μ−L(k),μ+R(k)),k∈C−,P2(k)=(μ+L(k),μ−R(k)),k∈C+. |
One can find that U is traceless from (2.2), so detμ± are independent of x. Besides, detμ±=1 according to the evolution of detμ± at x=±∞. Because all the μ±eikσx+4ik3σt satisfy the differential equations (2.1a) and (2.1b), they are linear related. So there exists a scattering matrix s(k) that satisfies
μ−=μ+eikσx+4ik3σts(k)e−ikσx−4ik3σt,dets(k)=1. | (2.6) |
In this paper, we denote a 3×3 matrix A by the block form
A=(A11A12A21A22), |
where A11 is a 2×2 matrix and A22 is scalar. Let q=(u,u∗)T and we can rewrite U of (2.2) as
U=(02×2qq†B10), |
where "†" is the Hermitian conjugate. In addition, there are two symmetry properties for U,
B−1U†(k∗)B=−U(k),τU∗(−k∗)τ=U(k), | (2.7) |
B=(B100−1),τ=(σ1001),σ1=(0110), | (2.8) |
where B and τ are represented as block forms. Hence, the Jost solutions μ± and the scattering matrix s(k) also have the corresponding symmetry properties
B−1솱(k∗)B=μ−1±(k),τμ∗±(−k∗)τ=μ±(k); | (2.9) |
B−1s†(k∗)B=s−1(k),τs∗(−k∗)τ=s(k). | (2.10) |
We write s(k) as block form (sij)2×2 and from the symmetry properties (2.10) we have
s22(k)=det[s†11(k∗)],B−11s†21(k∗)=adj[s11(k)]s12(k), | (2.11) |
where adjX denote the adjoint of matrix X. Then we can write s(k) as
s(k)=(s11(k)s12(k)s†12(k∗)adj[s†11(k∗)]B1det[s†11(k∗)]), | (2.12) |
where
σ1s∗11(−k∗)σ1=s11(k),σ1s∗12(−k∗)=s12(k). | (2.13) |
From the evaluation of (2.6) at t=0, one infers
s(k)=limx→+∞e−ikxσμ−(k;x,0)eikxσ, | (2.14) |
which implies that
{s11(k)=I+∫+∞−∞q(ξ,0)μ−21(k;ξ,0)dξ,s12(k)=∫+∞−∞e−2ikξq(ξ,0)μ−22(k;ξ,0)dξ. | (2.15) |
Theorem 2.1. Let M(k;x,t) be analytic for k∈C∖R and satisfy the Riemann-Hilbert problem
{M+(k;x,t)=M−(k;x,t)J(k;x,t),k∈R,M(k;x,t)→I,k→∞, | (2.16) |
where
M±(k;x,t)=limϵ→0+M(k∓iϵ;x,t), | (2.17) |
J(k;x,t)=(I−γ(k)γ†(k∗)B1−e−2itθγ(k)e2itθγ†(k∗)B11) | (2.18) |
θ(k;x,t)=−xtk−4k3,γ(k)=s−111(k)s12(k), | (2.19) |
γ(k) lies in Schwartz space and satisfies
σ1γ∗(−k∗)=γ(k). | (2.20) |
Then the solution of this Riemann-Hilbert problem exists and is unique, the function
q(x,t)=(u(x,t),u∗(x,t))T=−2ilimk→∞(k(M(k;x,t))12) | (2.21) |
and u(x,t) is the solution of the generalized Sasa-Satsuma equation.
Proof. The matrix (J(k;x,t)+J†(k;x,t))/2 is positive definite because of the condition (P1) that γ(k) satisfies, then the solution of the Riemann-Hilbert problem (2.16) is existent and unique according to the Vanishing Lemma [39]. We define M(k;x,t) by
M(k;x,t)={(μ−L(k),μ+R(k)det[a†(k∗)]),k∈C−,(μ+L(k)a(k),μ−R(k)),k∈C+. | (2.22) |
Considering the scattering relation (2.6) and the construction of M(k;x,t), we can obtain the jump condition and the corresponding Riemann-Hilbert problem (2.16) after tedious but straightforward algebraic manipulations. Substituting the large k asymptotic expansion of M(k;x,t) into (2.4a) and compare the coefficients of O(1k), we can get (2.21).
In this section, we compute the Riemann-Hilbert problem (2.16) by the nonlinear steepest decent method and study the long-time asymptotic behavior of the solution. We make the following basic notations. (i) For any matrix M define |M|=(trM†M)12 and for any matrix function A(⋅) define ‖A(⋅)‖p=‖|A(⋅)|‖p. (ii) For two quantities A and B define A≲ if there exists a constant C > 0 such that \vert A\vert\leqslant CB . If C depends on the parameter \alpha we shall say that A\lesssim_{\alpha}B . (iii) For any oriented contour \Sigma , we say that the left side is + and the right side is - .
First of all, it is noteworthy that there are two stationary points \pm k_0 , where \pm k_0 = \pm \sqrt{-\frac{x}{12t}} satisfied \frac{\mathrm{d}\theta}{\mathrm{d}k}\big\vert_{k = \pm k_0} = 0 . The jump matrix J(k; x, t) have a lower-upper triangular factorization and a upper-lower triangular factorization. We can introduce an appropriate Rieman-Hilbert problem to unify these two forms of factorizations. In this process, we have to reorient the contour of the Riemann-Hilbert problem.
The two factorizations of the jump matrix J are
\begin{equation*} J = \begin{cases} \left(\begin{array}{cc} I & -e^{-2it\theta}\gamma(k)\\ 0 & 1\\ \end{array}\right) \left(\begin{array}{cc} I & 0\\ e^{2it\theta}\gamma^{\dagger}(k^{\ast})B_1 & 1\\ \end{array}\right), \\ \left(\begin{array}{cc} I & 0\\ \frac{e^{2it\theta}\gamma^{\dagger}(k^{\ast})B_1}{1-\gamma^{\dagger}(k^{\ast})B_1\gamma(k)} & 1\\ \end{array}\right) \left(\begin{array}{cc} I-\gamma(k)\gamma^{\dagger}(k^{\ast})B_1 & 0\\ 0 & (1-\gamma^{\dagger}(k^{\ast})B_1\gamma(k))^{-1}\\ \end{array}\right) \left(\begin{array}{cc} I & \frac{-e^{-2it\theta}\gamma(k)}{1-\gamma^{\dagger}(k^{\ast})B_1\gamma(k)}\\ 0 & 1\\ \end{array}\right). \end{cases} \end{equation*} |
We introduce a 2 \times 2 matrix function \delta(k) to make the two factorization unified, and \delta(k) satisfies the following Riemann-Hilbert problem
\begin{align} \left\{\begin{array}{rll} \delta_{+}(k)& = \delta_{-}(k)(I-\gamma(k)\gamma^{\dagger}(k^{\ast})B_1), & k\in(-k_0, k_0), \\ & = \delta_{-}(k), & k\in(-\infty, -k_0)\cup(k_0, +\infty), \\ \delta(k)&\to I, & k\to\infty, \end{array}\right. \end{align} | (3.1) |
which implies a scalar Riemann-Hilbert problem
\begin{equation} \left\{\begin{array}{rll} \det\delta_{+}(k)& = \det\delta_{-}(k)(1-\gamma^\dagger(k^\ast)B_1\gamma(k)), & k\in(-k_0, k_0), \\ & = \det\delta_{-}(k), & k\in(-\infty, -k_0)\cup(k_0, +\infty), \\ \det\delta(k)&\to 1, & k\to\infty. \end{array}\right. \end{equation} | (3.2) |
The jump matrix I-\gamma(k)\gamma^{\dagger}(k^{\ast})B_1 of Riemann-Hilbert problem (3.1) is positive definite, so the solution \delta(k) exists and is unique. The scalar Riemann-Hilbert problem (3.2) can be solved by the Plemelj formula,
\begin{equation} \det\delta(k) = \left(\frac{k-k_0}{k+k_0}\right)^{-i\nu}e^{\chi(k)}, \end{equation} | (3.3) |
where
\begin{gather*} \nu = -\frac{1}{2\pi}\log(1-\gamma^\dagger(k_0)B_1\gamma(k_0)), \\ \chi(k) = -\frac{1}{2\pi i}\int_{-k_0}^{k_0}\log\left(\frac{1-\gamma^\dagger(\xi^\ast)B_1\gamma(\xi)}{1-\gamma^\dagger(k_0^\ast)B_1\gamma(k_0)}\right)\, \frac{\mathrm{d}\xi}{\xi-k}. \end{gather*} |
Then we have by uniqueness that
\begin{equation} \delta(k) = B_1^{-1}(\delta^\dagger(k^\ast))^{-1}B_1, \quad \delta(k) = \sigma_1\delta^\ast(-k^\ast)\sigma_1. \end{equation} | (3.4) |
Substituting (3.4) to (3.1), we have
\begin{equation} \delta^\dagger_+(k^\ast)B_1\delta_+(k) = B_1-B_1\gamma(k)\gamma^\dagger(k^\ast)B_1, \end{equation} | (3.5) |
which means that
\begin{equation} \mathrm{tr}[\delta^\dagger_+(k^\ast)B_1\delta_+(k)] = 2a-|B_1\gamma(k)|^2. \end{equation} | (3.6) |
Actually, the condition (P_2) satisfied by \gamma(k) guarantee the boundedness of \delta_\pm(k) and we give a brief proof below. When \det B_1 > 0 , we find that the Hermitian matrix B_1 can be decomposition. In other words, there exists a triangular matrix S that satisfies B_1 = a S^\dagger S . So \mathrm{tr}[\delta^\dagger_+B_1\delta_+] = a|S\delta_+|^2 . When \det B_1 < 0 and |a| > 0 , the matrix B_1 has a decomposition B_1 = S^\dagger DS , where S is a triangular matrix and D is a diagonal matrix and the diagonal elements have opposite signs. In the case of a > 0 , B_1 can be decomposed as below,
\begin{equation} B_1 = \left(\begin{array}{cc} -a&b^\ast\\ 0&1\\ \end{array}\right)^{-1} \left(\begin{array}{cc} a\det B_1&0\\ 0&a\\ \end{array}\right) \left(\begin{array}{cc} -a&0\\ b&1\\ \end{array}\right)^{-1}. \end{equation} | (3.7) |
We denote S\delta_+(k) by (G_{ij})_{3 \times 3} and c_1 = 2a-|B_1\gamma(k)|^2 is negative, then
\begin{equation} a\det B_1(|G_{11}|^2+|G_{21}|^2)+a(|G_{12}|^2+|G_{22}|^2) = c_1. \end{equation} | (3.8) |
Noticing that \det B_1 < 0 , we find a negative constant c_2 that satisfies c_2\leqslant a\det B_1(c_3-1)/(1-\det B_1c_3) , where c_3 is a constant and 0 < c_3 < 1 , which impies
\begin{equation} |S\delta_+(k)|^2\leqslant\frac{c_1}{c_2}\lesssim1. \end{equation} | (3.9) |
The case that a < 0 is similar. In particular, when a = 0 , then |b| > 0 , it is easy to see that B_1 is not definite. For |\mathrm{Re}b| > 0 , we have the decomposition
\begin{equation} B_1 = \left(\begin{array}{cc} \frac{b}{|b|^2+(b^\ast)^2}&\frac{b^\ast}{|b|^2+(b^\ast)^2}\\ \frac{b^\ast}{|b|^2+(b^\ast)^2}&-\frac{b^\ast}{|b|^2+(b^\ast)^2}\\ \end{array}\right)^{-1} \left(\begin{array}{cc} \frac{1}{b+b^\ast}&0\\ 0&-\frac{1}{b+b^\ast}\\ \end{array}\right) \left(\begin{array}{cc} \frac{b^\ast}{|b|^2+b^2}&\frac{b}{|b|^2+b^2}\\ \frac{b}{|b|^2+b^2}&-\frac{b}{|b|^2+b^2}\\ \end{array}\right)^{-1}. \end{equation} | (3.10) |
For |\mathrm{Re}b| = 0 , we have
\begin{equation} B_1 = \left(\begin{array}{cc} \frac{i}{2}&\frac{1-i}{2}\\ -\frac{i}{2}&\frac{1+i}{2}\\ \end{array}\right)^{-1} \left(\begin{array}{cc} -ib/2&0\\ 0&ib/2\\ \end{array}\right) \left(\begin{array}{cc} -\frac{i}{2}&\frac{i}{2}\\ \frac{1+i}{2}&\frac{1-i}{2}\\ \end{array}\right)^{-1}. \end{equation} | (3.11) |
So we get the boundedness of |\delta_+(k)| . The others have the same analysis,
\begin{align} &\delta^\dagger_-(k^\ast)B_1\delta_-(k) = (B_1-\gamma(k)\gamma^\dagger(k^\ast))^{-1}, \quad \mathrm{as} \quad k\in(-k_0, k_0), \end{align} | (3.12) |
\begin{align} &|\delta_+(k)|^2 = |\delta_-(k)|^2 = 2, \quad \mathrm{as} \quad k\in(-\infty, -k_0)\cup(k_0, +\infty), \end{align} | (3.13) |
\begin{align} &|\det\delta_+(k)| = \begin{cases} 1-\gamma^\dagger(k^\ast)B_1\gamma(k), & k\in(-k_0, k_0), \\ 1 , & k\in(-\infty, -k_0)\cup(k_0, +\infty), \\ \end{cases} \end{align} | (3.14) |
\begin{align} &|\det\delta_-(k)| = \begin{cases} \frac{1}{1-\gamma^\dagger(k^\ast)B_1\gamma(k)}, & k\in(-k_0, k_0), \\ 1, & k\in(-\infty, -k_0)\cup(k_0, +\infty).\\ \end{cases} \end{align} | (3.15) |
Hence, by the maximum principle, we have
\begin{equation} \vert\delta(k)\vert\leqslant \mathrm{const} \lt \infty, \quad \vert\det\delta(k)\vert\leqslant \mathrm{const} \lt \infty, \end{equation} | (3.16) |
for all k\in\mathbb{C} . We define the functions
\begin{align} &\rho(k) = \begin{cases} -\gamma(k), & k\in(-\infty, -k_0)\cup(k_0, +\infty), \\ \dfrac{\gamma(k)}{1-\gamma^\dagger(k^\ast)B_1\gamma(k)}, & k\in(-k_0, k_0), \\ \end{cases} \end{align} | (3.17) |
\begin{align} &\Delta(k) = \left(\begin{array}{cc} \delta(k) & 0\\ 0 & (\det\delta(k))^{-1}\\ \end{array}\right). \end{align} | (3.18) |
We reverse the orientation for k\in(-\infty, k_0)\cup(k_0, +\infty) as in Figure 1, and M^\Delta(k; x, t) = M(k; x, t)\Delta^{-1}(k) satisfies the Riemann-Hilbert problem on the reoriented contour
\begin{equation} \begin{cases} M^\Delta_+(k;x, t) = M^\Delta_-(k;x, t)J^\Delta(k;x, t), & k\in\mathbb{R}, \\ M^\Delta(k;x, t)\to I, & k\to\infty, \\ \end{cases} \end{equation} | (3.19) |
where the jump matrix J^\Delta(k; x, t) has a decomposition
\begin{equation} J^\Delta(k;x, t) = (b_-)^{-1}b_+ = \left(\begin{array}{cc} I & 0\\ \dfrac{e^{2it\theta}(k)\rho^\dagger(k^\ast)B_1\delta_-^{-1}(k)}{\det\delta_-(k)} & 1\\ \end{array}\right) \left(\begin{array}{cc} I & -e^{-2it\theta}\delta_+(k)\rho(k)[\det\delta_+(k)]\\ 0 & 1\\ \end{array}\right). \end{equation} | (3.20) |
For the convenience of discussion, we define
\begin{align*} L = \{k_0+\alpha k_0e^{-\frac{3\pi i}{4}}:-\infty \lt \alpha\leqslant\sqrt{2}\}\cup\{-k_0+\alpha k_0e^{-\frac{\pi i}{4}}:-\infty \lt \alpha\leqslant\sqrt{2}\}, \\ L_\epsilon = \{k_0+\alpha k_0e^{-\frac{3\pi i}{4}}:-\epsilon \lt \alpha\leqslant\sqrt{2}\}\cup\{-k_0+\alpha k_0e^{-\frac{\pi i}{4}}:-\epsilon \lt \alpha\leqslant\sqrt{2}\}. \end{align*} |
Theorem 3.1. The vector function \rho(k) has a decomposition
\begin{equation*} \rho(k) = h_1(k)+h_2(k)+R(k), \quad k\in\mathbb{R}, \end{equation*} |
where R(k) is a piecewise-rational function and h_2(k) has a analytic continuation to L . Besides, they admit the following estimates
\begin{gather} |e^{-2it\theta(k)}h_{1}(k)|\lesssim\frac{1}{(1+\vert k\vert^{2})t^{l}}, \quad k\in\mathbb{R}, \end{gather} | (3.21) |
\begin{gather} \vert e^{-2it\theta(k)}h_{2}(k)\vert\lesssim\frac{1}{(1+\vert k\vert^{2})t^{l}}, \quad k\in L, \end{gather} | (3.22) |
\begin{gather} \vert e^{-2it\theta(k)}R(k)\vert\lesssim e^{-16\epsilon^{2}k^{3}_{0}t}, \quad k\in L_{\epsilon}, \end{gather} | (3.23) |
for an arbitrary positive integer l . Considering the Schwartz conjugate
\begin{equation*} \rho^{\dagger}(k^{\ast}) = R^{\dagger}(k^{\ast})+h^{\dagger}_{1}(k^{\ast})+h^{\dagger}_{2}(k^{\ast}), \end{equation*} |
we can obtain the same estimate for e^{2it\theta(k)}h^{\dagger}_{1}(k^{\ast}) , e^{2it\theta(k)}h^{\dagger}_{2}(k^{\ast}) and e^{2it\theta(k)}R^{\dagger}(k^{\ast}) on \mathbb{R}\cup L^{\ast} .
Proof. It follows from Proposition 4.2 in [18].
A direct calculation shows that b_\pm of (3.20) can be decomposed further
\begin{align*} b_{+}& = b^{o}_{+}b^{a}_{+} = (I_{3 \times 3}+\omega^{o}_{+})(I_{3 \times 3}+\omega^{a}_{+})\notag\\ & = \left(\begin{array}{cc} I_{2 \times 2}&-e^{-2it\theta}[\mathrm{det}\delta_+(k)]\delta_+(k)h_{1}(k)\\ 0&1\\ \end{array}\right) \left(\begin{array}{cc} I_{2 \times 2}&-e^{-2it\theta}[\mathrm{det}\delta_+(k)]\delta_+(k)[h_{2}(k)+R(k)]\\ 0&1\\ \end{array}\right), \\ b_{-}& = b^{o}_{-}b^{a}_{-} = (I_{3 \times 3}-\omega^{o}_{-})(I_{3 \times 3}-\omega^{a}_{-})\notag\\ & = \left(\begin{array}{cc} I_{2 \times 2}&0\\ -\dfrac{e^{2it\theta}h^{\dagger}_{1}(k^{\ast})B_1\delta_-^{-1}(k)}{\mathrm{det}\delta_-(k)}&1\\ \end{array}\right) \left(\begin{array}{cc} I_{2 \times 2}&0\\ -\dfrac{e^{2it\theta}[h^{\dagger}_{2}(k^{\ast})+R^{\dagger}(k^{\ast})]B_1\delta_-^{-1}(k)}{\mathrm{det}\delta_-(k)}&1\\ \end{array}\right). \end{align*} |
Define the oriented contour \Sigma by \Sigma = L\cup L^\ast as in Figure 2. Let
\begin{equation} M^\sharp(k;x, t) = \begin{cases} M^\Delta(k;x, t), & k\in\Omega_{1}\cup\Omega_{2}, \\ M^\Delta(k;x, t)(b_+^a)^{-1}, & k\in\Omega_{3}\cup\Omega_{4}\cup\Omega_{5}, \\ M^\Delta(k;x, t)(b_-^a)^{-1}, & k\in\Omega_{6}\cup\Omega_{7}\cup\Omega_{8}.\\ \end{cases} \end{equation} | (3.24) |
Lemma 3.1. M^\sharp(k; x, t) is the solution of the Riemann-Hilbert problem
\begin{equation} \begin{cases} M^\sharp_+(k;x, t) = M^\sharp_-(k;x, t)J^\sharp(k;x, t), & k\in\Sigma, \\ M^\sharp(k;x, t)\to I, & k\to\infty, \end{cases} \end{equation} | (3.25) |
where the jump matrix J^\sharp(k; x, t) satisfies
\begin{equation} J^\sharp(k;x, t) = (b_-^\sharp)^{-1}b_+^\sharp = \begin{cases} I^{-1}b_+^a, & k\in L, \\ (b_-^a)^{-1}I, & k\in L^\ast, \\ (b_-^o)^{-1} b_+^o, & k\in\mathbb{R}.\\ \end{cases} \end{equation} | (3.26) |
Proof. We can construct the Riemann-Hilbert problem (3.25) based on the Riemann-Hilbert problem (3.19) and the decomposition of b_\pm . In the meantime, the asymptotics of M^\sharp(k; x, t) is derived from the convergence of b_\pm as k\to\infty . For fixed x and t , we pay attention to the domain \Omega_{3} . Noticing the boundedness of \delta(k) and \det\delta(k) in (3.16), we arrive at
\begin{equation*} \vert e^{-2it\theta}[\mathrm{det}\delta(k)][h_{2}(k)+R(k)]\delta(k)\vert \lesssim \vert e^{-2it\theta}h_{2}(k)\vert+\vert e^{-2it\theta}R(k)\vert. \end{equation*} |
Consider the definition of R(k) in this domain,
\begin{align*} |e^{-2it\theta}h_2(k)| \lesssim \dfrac{1}{\vert k+i\vert^{2}}, \quad \vert e^{-2it\theta}R(k)\vert \lesssim \dfrac{\vert\sum\limits_{i = 0}^m\mu_i(k-k_0)^i\vert}{\vert(k+i)^{m+5}\vert} \lesssim \dfrac{1}{\vert k+i\vert^{5}}, \end{align*} |
where m is a positive integer and \mu_i is the coefficient of the Taylor series around k_0 . Combining with the boundedness of h_2(k) in Theorem 3.1, we obtain that M^\sharp(k; x, t)\to I when k\in\Omega_3 and k\to\infty . The others are similar to this domain.
The above Riemann-Hilbert problem (3.25) can be solved as follows. Set
\begin{equation*} \omega^\sharp_\pm = \pm(b^\sharp_\pm-I), \quad \omega^\sharp = \omega^\sharp_++\omega^\sharp_-. \end{equation*} |
Let
\begin{equation} (C_\pm f)(k) = \int_\Sigma\frac{f(\xi)}{\xi-k_\pm} \, \frac{\mathrm{d}\xi}{2\pi i}, \quad f\in \mathscr{L}^2(\Sigma) \end{equation} | (3.27) |
denote the Cauchy operator, where C_+f\ (C_-f) denotes the left (right) boundary value for the oriented contour \Sigma in Figure 2. Define the operator C_{\omega^\sharp}:\mathscr{L}^2(\Sigma)+\mathscr{L}^\infty(\Sigma)\to\mathscr{L}^2(\Sigma) by
\begin{equation} C_{\omega^{\sharp}}f = C_+\left(f\omega^\sharp_-\right)+C_-\left(f\omega^\sharp_+\right) \end{equation} | (3.28) |
for the 3 \times 3 matrix function f .
Lemma 3.2 (Beals-Coifman). If \mu^\sharp(k; x, t)\in \mathscr{L}^2(\Sigma)+\mathscr{L}^\infty(\Sigma) is the solution of the singular integral equation
\begin{equation*} \mu^\sharp = I+C^\sharp_{\omega^\sharp}\mu^\sharp. \end{equation*} |
Then
\begin{equation*} M^\sharp(k;x, t) = I+\int_\Sigma\dfrac{\mu^\sharp(\xi;x, t)\omega^\sharp(\xi;x, t)}{\xi-k} \, \frac{\mathrm{d}\xi}{2\pi i} \end{equation*} |
is the solution of the Riemann-Hilbert problem (3.25).
Proof. See [18], P. 322 and [40].
Theorem 3.2. The expression of the solution q(x, t) can be written as
\begin{equation} q(x, t) = (u(x, t), u^\ast(x, t))^T = \frac{1}{\pi}\left(\int_\Sigma\left((1-C_{\omega^\sharp})^{-1}I\right)(\xi)\omega^\sharp(\xi)\, \mathrm{d}\xi\right)_{12}. \end{equation} | (3.29) |
Proof. From (2.21), (3.24) and Lemma 2, the solution q(x, t) of the generalized Sasa-Satsuma equation is expressed by
\begin{align} \begin{split} q(x, t)& = \lim\limits_{k\to\infty}-2i\left[k(M^\sharp(k;x, t))_{12}\right]\\ & = \frac{1}{\pi}\left(\int_\Sigma\mu^\sharp(\xi;x, t)\omega^\sharp(\xi)\, \mathrm{d}\xi\right)_{12}\\ & = \frac{1}{\pi}\left(\int_\Sigma((1-C_{\omega^\sharp})^{-1}I)(\xi)\omega^\sharp(\xi)\, \mathrm{d}\xi\right)_{12}. \end{split} \end{align} |
Set \Sigma^\prime = \Sigma\backslash(\mathbb{R}\cup L_\epsilon\cup L_\epsilon^\ast) oriented as in Figure 3. We will convert the Riemann-Hilbert problem on the contour \Sigma to a Riemann-Hilbert problem on the contour \Sigma^\prime and estimate the errors between the two Riemann-Hilbert problems. Let \omega^\sharp = \omega^e+\omega^\prime = \omega^a+\omega^b+\omega^c+\omega^\prime , where \omega^a = \omega^\sharp |_\mathbb{R} is supported on \mathbb{R} and is composed of terms of type h_1(k) and h_1^\dagger(k^\ast) ; \omega^b is supported on L\cup L^\ast and is composed of contribution to \omega^\sharp from terms of type h_2(k) and h_2^\dagger(k^\ast) ; \omega^c is supported on L_\epsilon\cup L_\epsilon^\ast and is composed of contribution to \omega^\sharp from terms of type R(k) and R^\dagger(k^\ast) .
Lemma 3.3. For arbitrary positive integer l , as t\to\infty ,
\begin{gather} \Vert\omega^a\Vert_{\mathscr{L}^1(\mathbb{R})\cap\mathscr{L}^2(\mathbb{R})\cap\mathscr{L}^\infty(\mathbb{R})}\lesssim t^{-l}, \end{gather} | (3.30) |
\begin{gather} \Vert\omega^b\Vert_{\mathscr{L}^1(L\cup L^\ast)\cap\mathscr{L}^2(L\cup L^\ast)\cap\mathscr{L}^\infty(L\cup L^\ast)}\lesssim t^{-l}, \end{gather} | (3.31) |
\begin{gather} \Vert\omega^c\Vert_{\mathscr{L}^1(L_\epsilon\cup L_\epsilon^\ast)\cap\mathscr{L}^2(L_\epsilon\cup L_\epsilon^\ast)\cap\mathscr{L}^\infty(L_\epsilon\cup L_\epsilon^\ast)}\lesssim e^{-16\epsilon^{2}k^{3}_{0}t}, \end{gather} | (3.32) |
\begin{gather} \Vert\omega^\prime\Vert_{\mathscr{L}^2(\Sigma)}\lesssim (tk_0^3)^{-\frac{1}{4}}, \quad \Vert\omega^\prime\Vert_{\mathscr{L}^1(\Sigma)}\lesssim (tk_0^3)^{-\frac{1}{2}} \end{gather} | (3.33) |
Proof. The proof of estimates (3.30), (3.31), (3.32) follows from Theorem 3.1. Afterwards, we consider the definition of R(k) on the contour \{k = k_0+\alpha k_0e^{\frac{-3\pi i}{4}}\vert -\infty < \alpha < \epsilon\} ,
\begin{equation*} |R(k)|\lesssim (1+|k|^5)^{-1}. \end{equation*} |
Resorting to \mathrm{Re}(i\theta)\geqslant8\alpha^2 k_0^3 and the boundedness of \delta(k) and \det\delta(k) in (3.16), we can obtain
\begin{equation*} \vert e^{-2it\theta}[\det\delta(k)]R(k)\delta(k)\vert \lesssim e^{-16tk_0^3\alpha^2}(1+|k|^5)^{-1}. \end{equation*} |
Then we obtain (3.33) by simple computations.
Lemma 3.4. As t\to\infty , (1-C_{\omega^\prime})^{-1}:\mathscr{L}^2(\Sigma)\to\mathscr{L}^2(\Sigma) exists and is uniformly bounded:
\begin{equation*} \Vert(1-C_{\omega^\prime})^{-1}\Vert_{\mathscr{L}^2(\Sigma)}\lesssim1. \end{equation*} |
Furthermore, \Vert(1-C_{\omega^\sharp})^{-1}\Vert_{\mathscr{L}^2(\Sigma)}\lesssim1 .
Proof. It follows from Proposition 2.23 and Corollary 2.25 in [18].
Lemma 3.5. As t\to\infty ,
\begin{equation} \int_\Sigma((1-C_{\omega^\sharp})^{-1}I)(\xi)\omega^\sharp(\xi) \, \mathrm{d}\xi = \int_\Sigma((1-C_{\omega^\prime})^{-1}I)(\xi)\omega^\prime(\xi) \, \mathrm{d}\xi+O((tk_0^3)^{-l}). \end{equation} | (3.34) |
Proof. A simple computation shows that
\begin{align} \begin{split} \left((1-C_{\omega^\sharp}\right)^{-1}I)\omega^\sharp = &\left((1-C_{\omega^\prime})^{-1}I\right)\omega^\prime+\omega^e+\left((1-C_{\omega^\prime})^{-1}(C_{\omega^e}I)\right)\omega^\sharp\\ &+((1-C_{\omega^\prime})^{-1}(C_{\omega^\prime}I))\omega^e +\left((1-C_{\omega^\prime})^{-1}C_{\omega^e}(1-C_{\omega^\sharp})\right)(C_{\omega^\sharp}I)\omega^\sharp. \end{split} \end{align} | (3.35) |
After a series of tedious computations and utilizing the consequence of Lemma 4, we arrive at
\begin{align*} &\Vert\omega^e\Vert_{\mathscr{L}^1(\Sigma)} \leqslant \Vert\omega^a\Vert_{\mathscr{L}^1(\mathbb{R})}+\Vert\omega^b\Vert_{\mathscr{L}^1(L\cup L^\ast)}+\Vert\omega^c\Vert_{\mathscr{L}^1(L_\epsilon\cup L_\epsilon^\ast)} \lesssim (tk_0^3)^{-l}, \\ &\begin{aligned} \Vert\left((1-C_{\omega^\prime})^{-1}(C_{\omega^e}I)\right)\omega^\sharp\Vert_{\mathscr{L}^1(\Sigma)} &\leqslant \Vert(1-C_{\omega^\prime})^{-1}\Vert_{\mathscr{L}^2(\Sigma)}\Vert C_{\omega^e}I\Vert_{\mathscr{L}^2(\Sigma)}\Vert\omega^\sharp\Vert_{\mathscr{L}^2(\Sigma)}\\ &\lesssim \Vert\omega^e\Vert_{\mathscr{L}^2(\Sigma)}\Vert\omega^\sharp\Vert_{\mathscr{L}^2(\Sigma)} \lesssim (tk_0^3)^{-l-\frac{1}{4}}, \end{aligned}\\ &\begin{aligned} \Vert\left((1-C_{\omega^\prime})^{-1}(C_{\omega^\prime}I)\right)\omega^e\Vert_{\mathscr{L}^1(\Sigma)} &\leqslant \Vert(1-C_{\omega^\prime}^{-1})\Vert_{\mathscr{L}^2(\Sigma)}\Vert C_{\omega^\prime}I\Vert_{\mathscr{L}^2(\Sigma)}\Vert\omega^e\Vert_{\mathscr{L}^2(\Sigma)}\\ &\lesssim \Vert\omega^\prime\Vert_{\mathscr{L}^2(\Sigma)}\Vert\omega^e\Vert_{\mathscr{L}^2(\Sigma)} \lesssim (tk_0^3)^{-l-\frac{1}{4}}, \end{aligned}\\ &\Vert\left((1-C_{\omega^\prime})^{-1}C_{\omega^e}(1-C_{\omega^\sharp})\right)(C_{\omega^\sharp}I)\omega^\sharp\Vert_{\mathscr{L}^1(\Sigma)}\\ &\leqslant \Vert(1-C_{\omega^\prime})^{-1}\Vert_{\mathscr{L}^2(\Sigma)}\Vert(1-C_{\omega^\sharp})^{-1}\Vert_{\mathscr{L}^2(\Sigma)}\Vert C_{\omega^e}\Vert_{\mathscr{L}^2(\Sigma)}\Vert C_{\omega^\sharp}I\Vert_{\mathscr{L}^2(\Sigma)}\Vert\omega^\sharp\Vert_{\mathscr{L}^2(\Sigma)}\\ &\lesssim \Vert\omega^e\Vert_{\mathscr{L}^\infty(\Sigma)}\Vert\omega^\sharp\Vert^2_{\mathscr{L}^2(\Sigma)} \lesssim (tk_0^3)^{-l-\frac{1}{2}}. \end{align*} |
Then the proof is accomplished as long as we substitute the estimates above into (3.35).
Notice that \omega^\prime(k) = 0 when k\in\Sigma\backslash\Sigma^\prime , let C_{\omega^\prime}|_{\mathscr{L}^2(\Sigma^\prime)} denote the restriction of C_{\omega^\prime} to \mathscr{L}^2(\Sigma^\prime) . For simplicity, we write C_{\omega^\prime}|_{\mathscr{L}^2(\Sigma^\prime)} as C_{\omega^\prime} . Then
\begin{equation*} \int_\Sigma((1-C_{\omega^\prime})^{-1}I)(\xi)\omega^\prime(\xi) \, \mathrm{d}\xi = \int_{\Sigma^\prime}((1-C_{\omega^\prime})^{-1}I)(\xi)\omega^\prime(\xi) \, \mathrm{d}\xi. \end{equation*} |
Lemma 3.6. As t\to\infty ,
\begin{equation} q(x, t) = (u(x, t), u^\ast(x, t))^T = \frac{1}{\pi}\left(\int_{\Sigma^\prime}((1-C_{\omega^\prime})^{-1}I)(\xi)\omega^\prime(\xi)\, \mathrm{d}\xi\right)_{12}+O((tk_0^3)^{-l}). \end{equation} | (3.36) |
Proof. From (3.29) and (3.34), we can obtain the result directly.
Let L^\prime = L\backslash L_\epsilon and \mu^\prime = (1-C_{\omega^\prime})^{-1}I . Then
\begin{equation*} M^\prime(k;x, t) = I+\int_{\Sigma^\prime}\frac{\mu^\prime(k;x, t)\omega^\prime(k;x, t)}{\xi-k} \, \frac{\mathrm{d}\xi}{2\pi i} \end{equation*} |
solves the Riemann-Hilbert problem
\begin{equation*} \begin{cases} M_+^\prime(k;x, t) = M_-^\prime(k;x, t)J^\prime(k;x, t), & k\in\Sigma^\prime, \\ M^\prime(k;x, t)\to I, & k\to\infty, \end{cases} \end{equation*} |
where
\begin{gather*} J^\prime = (b_-^\prime)^{-1}b_+^\prime = (I-\omega_-^\prime)^{-1}(I+\omega_+^\prime), \\ \omega^\prime = \omega_+^\prime+\omega_-^\prime, \\ b_+^\prime = \left( \begin{array}{cc} I & -e^{-2it\theta}[\det\delta(k)]\delta(k)R(k)\\ 0 & 1 \\ \end{array} \right), \quad b_-^\prime = I, \quad \mathrm{on}\ L^\prime, \\ b_+^\prime = I, \quad b_-^\prime = \left( \begin{array}{cc} I & 0\\ -\dfrac{e^{2it\theta}R^\dagger(k^\ast)B_1\delta^{-1}(k)}{\det\delta(k)} & 1\\ \end{array} \right), \quad \mathrm{on}\ ({L^\prime})^\ast. \end{gather*} |
Let the contour \Sigma^{\prime} = \Sigma^{\prime}_{A}\cup\Sigma^{\prime}_{B} and \omega^{\prime}_{\pm} = \omega^{\prime}_{A\pm}+\omega^{\prime}_{B\pm} , where
\begin{equation} \omega^{\prime}_{A\pm}(k) = \begin{cases} \omega^{\prime}_{\pm}(k), & k\in\Sigma^{\prime}_{A}, \\ 0, & k\in\Sigma^{\prime}_{B}, \\ \end{cases}\quad \omega^{\prime}_{B\pm}(k) = \begin{cases} 0, & k\in\Sigma^{\prime}_{A}, \\ \omega^{\prime}_{\pm}(k), & k\in\Sigma^{\prime}_{B}.\\ \end{cases} \end{equation} | (3.37) |
Define the operators C_{\omega^{\prime}_{A}} and C_{\omega^{\prime}_{B}} : \mathscr{L}^{2}(\Sigma^{\prime})+\mathscr{L}^{\infty}(\Sigma^{\prime})\rightarrow\mathscr{L}^{2}(\Sigma^{\prime}) as in definition (3.28).
Lemma 3.7.
\begin{equation*} ||C_{\omega^{\prime}_{B}}C_{\omega^{\prime}_{A}}||_{\mathscr{L}^{2}(\Sigma^{\prime})} = || C_{\omega^{\prime}_{A}}C_{\omega^{\prime}_{B}}||_{\mathscr{L}^{2}(\Sigma^{\prime})}\lesssim_{k_{0}}(tk_0^3)^{-\frac{1}{2}}, \end{equation*} |
\begin{equation*} ||C_{\omega^{\prime}_{B}}C_{\omega^{\prime}_{A}}||_{\mathscr{L}^{\infty}(\Sigma^{\prime})\rightarrow\mathscr{L}^{2}(\Sigma^{\prime})}, || C_{\omega^{\prime}_{A}}C_{\omega^{\prime}_{B}}||_{\mathscr{L}^{\infty}(\Sigma^{\prime})\rightarrow\mathscr{L}^{2}(\Sigma^{\prime})}\lesssim_{k_{0}}(tk_0^3)^{-\frac{3}{4}}. \end{equation*} |
Proof. See Lemma 3.5 in [18].
Lemma 3.8. As t\rightarrow\infty ,
\begin{align} \begin{split} \int_{\Sigma^{\prime}}((1-C_{\omega^{\prime}})^{-1}I)(\xi)\omega^{\prime}(\xi)\, \mathrm{d}\xi = &\int_{\Sigma^{\prime}_{A}}((1-C_{\omega^{\prime}_{A}})^{-1}I)(\xi)\omega^{\prime}_{A}(\xi)\, \mathrm{d}\xi\\ &+\int_{\Sigma^{\prime}_{B}}((1-C_{\omega^{\prime}_{B}})^{-1}I)(\xi)\omega^{\prime}_{B}(\xi)\, \mathrm{d}\xi+O(\frac{c(k_{0})}{t}). \end{split} \end{align} | (3.38) |
Proof. From identity
\begin{align*} (1-C_{\omega^{\prime}_{A}}-C_{\omega^{\prime}_{B}})(1+C_{\omega^{\prime}_{A}}(1-C_{\omega^{\prime}_{A}})^{-1}+C_{\omega^{\prime}_{B}}(1-C_{\omega^{\prime}_{B}})^{-1})\\ = 1-C_{\omega^{\prime}_{B}}C_{\omega^{\prime}_{A}}(1-C_{\omega^{\prime}_{A}})^{-1}-C_{\omega^{\prime}_{A}}C_{\omega^{\prime}_{B}}(1-C_{\omega^{\prime}_{B}})^{-1}, \end{align*} |
we have
\begin{align*} (1-C_{\omega^{\prime}})^{-1} = &1+C_{\omega^{\prime}_{A}}(1-C_{\omega^{\prime}_{A}})^{-1}+C_{\omega^{\prime}_{B}}(1-C_{\omega^{\prime}_{B}})^{-1}\\ &+[1+C_{\omega^{\prime}_{A}}(1-C_{\omega^{\prime}_{A}})^{-1}+C_{\omega^{\prime}_{B}}(1-C_{\omega^{\prime}_{B}})^{-1}][1-C_{\omega^{\prime}_{B}}C_{\omega^{\prime}_{A}}(1-C_{\omega^{\prime}_{A}})^{-1}\\ &-C_{\omega^{\prime}_{A}}C_{\omega^{\prime}_{B}}(1-C_{\omega^{\prime}_{B}})^{-1}]^{-1}[C_{\omega^{\prime}_{B}}C_{\omega^{\prime}_{A}}(1-C_{\omega^{\prime}_{A}})^{-1}+C_{\omega^{\prime}_{A}}C_{\omega^{\prime}_{B}}(1-C_{\omega^{\prime}_{B}})^{-1}]. \end{align*} |
Based on Lemma (3.7) and Lemma (3.4), we arrive at (3.38).
For the sake of convenience, we write the restriction C_{\omega^{\prime}_{A}}\mid_{\mathscr{L}^{2}(\Sigma^{\prime}_{A})} as C_{\omega^{\prime}_{A}} , similar for C_{\omega^{\prime}_{B}} . From the consequences of Lemma 3.6 and Lemma 3.8, as t\rightarrow\infty , we have
\begin{equation} \begin{split} q(x, t) = &-\left(\int_{\Sigma^{\prime}_{A}}((1-C_{\omega^{\prime}_{A}})^{-1}I)(\xi)\omega^{\prime}_{A}(\xi)\, \frac{\mathrm{d}\xi}{\pi}\right)_{12}\\ &-\left(\int_{\Sigma^{\prime}_{B}}((1-C_{\omega^{\prime}_{B}})^{-1}I)(\xi)\omega^{\prime}_{B}(\xi)\, \frac{\mathrm{d}\xi}{\pi}\right)_{12}+O(\frac{c(k_{0})}{t}). \end{split} \end{equation} | (3.39) |
Extend the contours \Sigma^{\prime}_{A} and \Sigma^{\prime}_{B} to the contours
\begin{gather} \hat{\Sigma}^{\prime}_{A} = \left\{k = -k_{0}+k_{0}\alpha e^{\pm\frac{\pi i}{4}}:\alpha\in\mathbb{R}\right\}, \end{gather} | (3.40) |
\begin{gather} \hat{\Sigma}^{\prime}_{B} = \left\{k = k_{0}+k_{0}\alpha e^{\pm\frac{3\pi i}{4}}:\alpha\in\mathbb{R}\right\}, \end{gather} | (3.41) |
respectively. We introduce \hat{\omega}^{\prime}_{A} and \hat{\omega}^{\prime}_{B} on \hat{\Sigma}^{\prime}_{A} and \hat{\Sigma}^{\prime}_{B} , respectively, by
\begin{gather} \hat{\omega}^{\prime}_{A\pm} = \begin{cases} \omega^{\prime}_{A\pm}(k), & k\in\Sigma^{\prime}_{A}, \\ 0, & k\in\hat{\Sigma}^{\prime}_{A}\backslash\Sigma^{\prime}_{A}, \end{cases}\quad \hat{\omega}^{\prime}_{B\pm} = \begin{cases} \omega^{\prime}_{B\pm}(k), & k\in\Sigma^{\prime}_{B}, \\ 0, & k\in\hat{\Sigma}^{\prime}_{B}\backslash\Sigma^{\prime}_{B}. \end{cases} \end{gather} | (3.42) |
Let \Sigma_{A} and \Sigma_{B} denote the contours \{k = k_{0}\alpha e^{\pm\frac{\pi i}{4}}:\alpha\in\mathbb{R}\} oriented inward as in \Sigma^{\prime}_{A} , \hat{\Sigma}^{\prime}_{A} , and outward as in \Sigma^{\prime}_{B} , \hat{\Sigma}^{\prime}_{B} , respectively. Define the scaling operators
\begin{gather} \begin{split} N_{A}:\ &\mathscr{L}^{2}(\hat{\Sigma}^{\prime}_{A})\rightarrow\mathscr{L}^{2}(\Sigma_{A}), \\ &f(k)\rightarrow(N_{A}f)(k) = f(\frac{k}{\sqrt{48tk_{0}}}-k_{0}), \end{split} \end{gather} | (3.43) |
\begin{gather} \begin{split} N_{B}:\ &\mathscr{L}^{2}(\hat{\Sigma}^{\prime}_{B})\rightarrow\mathscr{L}^{2}(\Sigma_{B}), \\ &f(k)\rightarrow(N_{B}f)(k) = f(\frac{k}{\sqrt{48tk_{0}}}+k_{0}), \end{split} \end{gather} | (3.44) |
and set
\begin{equation*} \omega_{A} = N_{A}\hat{\omega}^{\prime}_{A}, \quad \omega_{B} = N_{B}\hat{\omega}^{\prime}_{B}. \end{equation*} |
A simple change-of-variable arguments shows that
\begin{equation*} \label{Cwprime definition} C_{\hat{\omega}^{\prime}_{A}} = N^{-1}_{A}C_{\omega_{A}}N_{A}, \quad C_{\hat{\omega}^{\prime}_{B}} = N^{-1}_{B}C_{\omega_{B}}N_{B}, \end{equation*} |
where the operator C_{\omega_{A}}\ (C_{\omega_{B}}) is a bounded map from \mathscr{L}^{2}(\Sigma_{A})\ (\mathscr{L}^{2}(\Sigma_{B})) into \mathscr{L}^{2}(\Sigma_{A})\ (\mathscr{L}^{2}(\Sigma_{B})) . On the part
\begin{equation*} L_{A} = \left\{k = \alpha k_{0}\sqrt{48tk_{0}}e^{\frac{3\pi i}{4}}:-\epsilon \lt \alpha \lt +\infty\right\} \end{equation*} |
of \Sigma_{A} , we have
\begin{equation*} \omega_{A} = \omega_{A+} = \left(\begin{matrix}0&(N_{A}s_{1})(k)\cr 0&0\end{matrix}\right), \end{equation*} |
on L^{\ast}_{A} we have
\begin{equation*} \omega_{A} = \omega_{A-} = \left(\begin{matrix}0&0\cr (N_{A}s_{2})(k)&0\end{matrix}\right), \end{equation*} |
where
\begin{equation*} s_{1}(k) = -e^{-2it\theta(k)}[\det\delta(k)]\delta(k)R(k), \quad s_{2}(k) = \frac{e^{2it\theta}R^{\dagger}(k^{\ast})B_1\delta^{-1}(k)}{\det\delta(k)}. \end{equation*} |
Lemma 3.9. As t\rightarrow\infty , and k\in L_{A} , then
\begin{equation} \left|(N_{A}\tilde{\delta})(k)\right|\lesssim t^{-l}, \end{equation} | (3.45) |
where \tilde{\delta}(k) = e^{-2it\theta(k)}[\delta(k)R(k)-(\mathrm{det}\delta(k))R(k)] .
Proof. It follows from (3.1) and (3.2) that \tilde{\delta} satisfies the following Riemann-Hilbert problem:
\begin{equation} \begin{cases} \tilde{\delta}_{+}(k) = \tilde{\delta}_{-}(k)(1-\gamma^\dagger(k^\ast)B_1\gamma(k))+e^{-2it\theta}f(k), & k\in(-k_{0}, k_{0}), \\ \tilde{\delta}(k)\rightarrow0, & k\rightarrow\infty. \end{cases} \end{equation} | (3.46) |
where f(k) = \delta_{-}(k)[\gamma^\dagger(k^\ast)B_1\gamma(k)I-\gamma(k)\gamma^\dagger(k^\ast)B_1]R(k) . The solution for the above Riemann-Hilbert problem can be expressed by
\begin{gather*} \tilde{\delta}(k) = X(k)\int_{k_{0}}^{-k_{0}}\frac{e^{-2it\theta(\xi)}f(\xi)}{X_{+}(\xi)(\xi-k)}\, \frac{\mathrm{d}\xi}{2\pi i}, \\ X(k) = \mathrm{exp}\left\{{\frac{1}{2\pi i}\int_{k_{0}}^{-k_{0}}\frac{\log(1-\left|\gamma(\xi)\right|^{2})}{\xi-k}}\, \mathrm{d}\xi\right\}. \end{gather*} |
Observing that
\begin{equation*} \begin{split} (\gamma^\dagger(k^\ast)B_1\gamma(k)I-\gamma(k)\gamma^\dagger(k^\ast)B_1)R(k)& = (\gamma^\dagger(k^\ast)B_1\gamma(k)I-\gamma(k)\gamma^\dagger(k^\ast)B_1)(R(k)-\rho(k))\\ & = \mathrm{adj}[B_1]\mathrm{adj}[\gamma(k)\gamma^\dagger(k^\ast)](h_{1}(k)+h_{2}(k)), \end{split} \end{equation*} |
we obtain f(k) = O((k^{2}-k^{2}_{0})^{l}). Similar to the Lemma 3.1, f(k) can be decomposed into two parts: f(k) = f_{1}(k)+f_{2}(k) , and
\begin{equation} \left|e^{-2it\theta(k)}f_{1}(k)\right|\lesssim\frac{1}{(1+\left|k\right|^{2})t^{l}}, \quad k\in\mathbb{R}, \end{equation} | (3.47) |
\begin{equation} \left|e^{-2it\theta(k)}f_{2}(k)\right|\lesssim\frac{1}{(1+\left|k\right|^{2})t^{l}}, \quad k\in L_{t}, \end{equation} | (3.48) |
where f_{2}(k) has an analytic continuation to L_{t} , l is a positive integer and l\geqslant2 ,
\begin{equation*} \begin{split} L_{t} = &\left\{k = k_{0}+k_{0}\alpha e^{-\frac{3\pi i}{4}}:0\leqslant\alpha\leqslant\sqrt{2}(1-\frac{1}{2t})\right\}\\ &\cup\left\{k = \frac{k_{0}}{t}-k_{0}+k_{0}\alpha e^{\frac{-\pi i}{4}}:0\leqslant\alpha\leqslant\sqrt{2}(1-\frac{1}{2t})\right\}, \end{split} \end{equation*} |
(see Figure 5).
As k\in L_{A} , we obtain
\begin{equation*} \begin{split} (N_{A}\tilde{\delta})(k) = &X(\frac{k}{\sqrt{48tk_{0}}}-k_{0})\int^{-k_{0}}_{\frac{k_{0}}{t}-k_{0}}\frac{e^{-2it\theta(\xi)}f(\xi)}{X_{+}(\xi)(\xi+k_{0}-\frac{k}{\sqrt{48tk_{0}}})}\, \frac{\mathrm{d}\xi}{2\pi i}\\ &+X(\frac{k}{\sqrt{48tk_{0}}}-k_{0})\int^{\frac{k_{0}}{t}-k_{0}}_{k_{0}}\frac{e^{-2it\theta(\xi)}f_{1}(\xi)}{X_{+}(\xi)(\xi+k_{0}-\frac{k}{\sqrt{48tk_{0}}})}\, \frac{\mathrm{d}\xi}{2\pi i}\\ &+X(\frac{k}{\sqrt{48tk_{0}}}-k_{0})\int^{\frac{k_{0}}{t}-k_{0}}_{k_{0}}\frac{e^{-2it\theta(\xi)}f_{2}(\xi)}{X_{+}(\xi)(\xi+k_{0}-\frac{k}{\sqrt{48tk_{0}}})}\, \frac{\mathrm{d}\xi}{2\pi i}\\ = &I_{1}+I_{2}+I_{3}. \end{split} \end{equation*} |
\begin{equation*} \left|I_{1}\right|\lesssim\int_{-k_{0}}^{\frac{k_{0}}{t}-k_{0}}\frac{\left|f(\xi)\right|}{|\xi+k_{0}-\frac{k}{\sqrt{48tk_{0}}}|}\, \mathrm{d}\xi\lesssim t^{-l-1}, \\ \end{equation*} |
\begin{equation*} \left|I_{2}\right|\lesssim\int_{\frac{k_{0}}{t}-k_{0}}^{k_{0}}\frac{\left|e^{-2it\theta(\xi)}f_{1}(\xi)\right|}{|\xi+k_{0}-\frac{k}{\sqrt{48tk_{0}}}|}\, \mathrm{d}\xi\leqslant t^{-l}\frac{\sqrt{2}t}{k_{0}}(2k_{0}-\frac{k_{0}}{t})\lesssim t^{-l+1}. \end{equation*} |
As a consequence of Cauchy's Theorem, we can evaluate I_{3} along the contour L_{t} instead of the interval (\frac{k_{0}}{t}-k_{0}, k_{0}) and obtain \left|I_{3}\right|\lesssim t^{-l+1}. Therefore, (3.45) holds.
Corollary 3.1. As t\rightarrow\infty , and k\in L_{A}^\ast , then
\begin{equation} \left|(N_{A}\hat{\delta})(k)\right|\lesssim t^{-l}, \quad t\rightarrow\infty, \quad k\in L^{\ast}_{A}, \end{equation} | (3.49) |
where \hat{\delta}(k) = e^{2it\theta(k)}R^{\dagger}(k^{\ast})B_1[\delta^{-1}(k)-(\mathrm{det}\delta(k))^{-1}I] .
Let J^{A^0} = (I-\omega_{A^0-})^{-1}(I+\omega_{A^0+}) , where
\begin{gather} \omega_{A^0} = \omega_{A^0+} = \begin{cases} \left(\begin{array}{cc} 0 & -(\delta_A^0)^{2}(-k)^{2i\nu}e^{-\frac{ik^2}{2}}\frac{\gamma(-k_0)}{1-\gamma^\dagger(-k_0)B_1\gamma(-k_0)}\\ 0 & 0\\ \end{array}\right), & k\in\Sigma_A^1, \\ \left(\begin{array}{cc} 0 & (\delta_A^0)^{2}(-k)^{2i\nu}e^{-\frac{ik^2}{2}}\gamma(-k_0)\\ 0 & 0\\ \end{array}\right), & k\in\Sigma_A^3, \\ \end{cases} \end{gather} | (3.50) |
\begin{gather} \delta_A^0 = (196tk_0^3)^{-\frac{i\nu}{2}}e^{8itk_0^3}e^{\chi(-k_0)} \end{gather} | (3.51) |
\begin{gather} \omega_{A^0} = \omega_{A^0-} = \begin{cases} \left(\begin{array}{cc} 0 & 0\\ (\delta_A^0)^{-2}(-k)^{-2i\nu}e^{\frac{ik^2}{2}}\frac{\gamma^\dagger(-k_0)B_1}{1-\gamma^\dagger(-k_0)B_1\gamma(-k_0)} & 0\\ \end{array}\right), & k\in\Sigma_A^2, \\ \left(\begin{array}{cc} 0 & 0\\ -(\delta_A^0)^{-2}(-k)^{-2i\nu}e^{\frac{ik^2}{2}}\gamma^\dagger(-k_0)B_1 & 0\\ \end{array}\right), & k\in\Sigma_A^4.\\ \end{cases} \end{gather} | (3.52) |
It follows from (3.78) in [18] that
\begin{equation} \Vert\omega_A-\omega_{A^0}\Vert_{\mathscr{L}^1(\Sigma_A)\cap\mathscr{L}^2(\Sigma_A)\cap\mathscr{L}^\infty(\Sigma_A)} \lesssim_{k_0} \frac{\log{t}}{\sqrt{tk_0^3}}. \end{equation} | (3.53) |
There are similar consequences for k\in\Sigma_B . Let J^{B^0} = (I-\omega_{B^0-})^{-1}(I+\omega_{B^0+}) , where
\begin{gather} \omega_{B^0} = \omega_{B^0+} = \begin{cases} \left(\begin{array}{cc} 0 & (\delta_B^0)^{2}k^{-2i\nu}e^{\frac{ik^2}{2}}\gamma(k_0)\\ 0 & 0\\ \end{array}\right), & k\in\Sigma_B^2, \\ \left(\begin{array}{cc} 0 & -(\delta_B^0)^{2}k^{-2i\nu}e^{\frac{ik^2}{2}}\frac{\gamma(k_0)}{1-\gamma^\dagger(k_0)B_1\gamma(k_0)}\\ 0 & 0\\ \end{array}\right), & k\in\Sigma_B^4, \\ \end{cases} \end{gather} | (3.54) |
\begin{gather} \delta_B^0 = (196tk_0^3)^{\frac{i\nu}{2}}e^{-8itk_0^3}e^{\chi(k_0)} \end{gather} | (3.55) |
\begin{gather} \omega_{B^0} = \omega_{B^0-} = \begin{cases} \left(\begin{array}{cc} 0 & 0\\ -(\delta_B^0)^{-2}k^{2i\nu}e^{-\frac{ik^2}{2}}\gamma^\dagger(k_0)B_1 & 0\\ \end{array}\right), & k\in\Sigma_B^1, \\ \left(\begin{array}{cc} 0 & 0\\ (\delta_B^0)^{-2}k^{2i\nu}e^{-\frac{ik^2}{2}}\frac{\gamma^\dagger(k_0)B_1}{1-\gamma^\dagger(k_0)B_1\gamma(k_0)} & 0\\ \end{array}\right), & k\in\Sigma_B^3.\\ \end{cases} \end{gather} | (3.56) |
Theorem 3.3. As t\to\infty ,
\begin{align} \begin{split} q(x, t) = &(u(x, t), u^\ast(x, t))^T\\ = &\frac{1}{\pi\sqrt{48tk_0}}\left(\int_{\Sigma_A}\left((1-C_{\omega_{A^0}})^{-1}I\right)(\xi)\omega_{A^0}(\xi)\, \mathrm{d}\xi\right)_{12}\\ &+\frac{1}{\pi\sqrt{48tk_0}}\left(\int_{\Sigma_B}\left((1-C_{\omega_{B^0}})^{-1}I\right)(\xi)\omega_{B^0}(\xi)\, \mathrm{d}\xi\right)_{12}+O\left(\frac{c(k_0)\log{t}}{t}\right). \end{split} \end{align} | (3.57) |
Proof. Notice that
\begin{align*} \left((1-C_{\omega_A})^{-1}I\right)\omega_A = &\left((1-C_{\omega_{A^0}})^{-1}I\right)\omega_{A^0}+\left((1-C_{\omega_A})^{-1}I\right)(\omega_A-\omega_{A^0})\notag\\ &+(1-C_{\omega_A})^{-1}(C_{\omega_A}-C_{\omega_{A^0}})(1-C_{\omega_{A^0}})I\omega_{A^0}. \end{align*} |
Utilizing the triangle inequality and the boundedness in (3.53), we have
\begin{align*} \int_{\Sigma_A}\left((1-C_{\omega_A})^{-1}I\right)(\xi)\omega_A(\xi)\, \mathrm{d}\xi = \int_{\Sigma_A}\left((1-C_{\omega_{A^0}})^{-1}I\right)(\xi)\omega_{A^0}(\xi)\, \mathrm{d}\xi+O\left(\frac{\log{t}}{\sqrt{t}}\right). \end{align*} |
According to (3.5) and a simple change-of-variable argument, we have
\begin{align*} \begin{split} &\frac{1}{\pi}\left(\int_{\Sigma^\prime}\left((1-C_{\omega_A^\prime})^{-1}I\right)(\xi)\omega_A^\prime(\xi)\, \mathrm{d}\xi\right)_{12}\\ = &\frac{1}{\pi}\left(\int_{\hat\Sigma_A^\prime}\left(N_A^{-1}(1-C_{\omega_A})^{-1}I\right)(\xi)\omega_A^\prime(\xi)\, \mathrm{d}\xi\right)_{12}\\ = &\frac{1}{\pi}\left(\int_{\hat\Sigma_A^\prime}\left((1-C_{\omega_A})^{-1}I\right)\left((\xi+k_0)\sqrt{48tk_0}\right)(N_A\omega_A^\prime)\left((\xi+k_0)\sqrt{48tk_0}\right)\, \mathrm{d}\xi\right)_{12}\\ = &\frac{1}{\pi\sqrt{48tk_0}}\left(\int_{\Sigma_A}\left((1-C_{\omega_A})^{-1}I\right)(\xi)\omega_A(\xi)\, \mathrm{d}\xi\right)_{12}\\ = &\frac{1}{\pi\sqrt{48tk_0}}\left(\int_{\Sigma_A}\left((1-C_{\omega_{A^0}})^{-1}I\right)(\xi)\omega_{A^0}(\xi)\, \mathrm{d}\xi\right)_{12}+O\left(\frac{c(k_0)\log{t}}{t}\right). \end{split} \end{align*} |
There are similar computations for the other case. Together with (3.39), one can obtain (3.57).
For k\in\mathbb{C}\backslash\Sigma_A , set
\begin{equation} M^{A^0}(k;x, t) = I+\int_{\Sigma_A}\frac{\left((1-C_{\omega_{A^0}})^{-1}I\right)(\xi)\omega_{A^0}(\xi)}{\xi-k} \, \frac{\mathrm{d}\xi}{2\pi i}. \end{equation} | (3.58) |
Then M^{A^0}(k; x, t) is the solution of the Riemann-Hilbert problem
\begin{equation} \begin{cases} M^{A^0}_+(k;x, t) = M^{A^0}_-(k;x, t)J^{A^0}(k;x, t), & k\in\Sigma_A, \\ M^{A^0}(k;x, t)\to I, & k\to\infty. \end{cases} \end{equation} | (3.59) |
In particular
\begin{equation} M^{A^0}(k) = I+\frac{M^{A^0}_1}{k}+O(k^{-2}), \quad k\rightarrow\infty, \end{equation} | (3.60) |
then
\begin{equation} M^{A^0}_1 = -\int_{\Sigma_A}\left((1-C_{\omega_{A^0}})^{-1}I\right)(\xi)\omega_{A^0}(\xi)\, \frac{\mathrm{d}\xi}{2\pi i}. \end{equation} | (3.61) |
There is a analogous Riemann-Hilbert problem on \Sigma_{B} ,
\begin{equation} \begin{cases} M^{B^0}_+(k;x, t) = M^{B^0}_-(k;x, t)J^{B^0}(k;x, t), & k\in\Sigma_B, \\ M^{B^0}(k;x, t)\to I, & k\to\infty, \end{cases} \end{equation} | (3.62) |
where J^{B^0}(k; x, t) is defined in (3.54) and (3.56). In the meantime, we have
\begin{equation} M^{B^0}(k) = I+\frac{M^{B^0}_1}{k}+O(k^{-2}), \quad k\rightarrow\infty. \end{equation} | (3.63) |
Next, we consider the relation between M^{A^0}_1 and M^{B^0}_1 . From the expression (3.50), (3.52), (3.54) and (3.56), we have the symmetry relation
\begin{equation*} J^{A^0}(k) = \tau(J^{B^0}(-k^\ast))^\ast\tau. \end{equation*} |
By the uniqueness of the Riemann-Hilbert problem,
\begin{equation*} M^{A^0}(k) = \tau(M^{B^0}(-k^\ast))^\ast\tau. \end{equation*} |
Combining with the expansion (3.60) and (3.63), one can verify that
\begin{equation*} M^{A^0}_1 = -\tau(M^{B^0}_1)^\ast\tau, \quad (M^{A^0}_1)_{12} = -\sigma_1(M^{B^0}_1)^\ast_{12}. \end{equation*} |
Therefore, from (3.57) and (3.61), we have
\begin{align} \begin{split} q(x, t) = &(u(x, t), u^\ast(x, t))^T\\ = &\frac{-2i}{\sqrt{48tk_0}}\left(M_1^{A^0}+M_1^{B^0}\right)_{12}+O\left(\frac{c(k_0)\log{t}}{t}\right)\\ = &-\frac{i}{\sqrt{12tk_0}}\left((M_1^{A^0})_{12}-\sigma_1(M_1^{A^0})^\ast_{12}\right)+O\left(\frac{c(k_0)\log{t}}{t}\right). \end{split} \end{align} | (3.64) |
In this subsection, we compute (M_1^{A^0})_{12} explicitly. It is important to set
\begin{equation} \Psi(k) = H(k)(-k)^{i\nu\sigma}e^{-\frac{1}{4}ik^2\sigma}, \quad H(k) = (\delta_A^0)^{-\sigma}M^{A^0}(k)(\delta_A^0)^{\sigma}. \end{equation} | (3.65) |
Then it follows from (3.59) that
\begin{equation} \Psi_+(k) = \Psi_-(k)v(-k_0), \quad v = e^{\frac{1}{4}ik^2\sigma}(-k)^{-i\nu\sigma}(\delta_A^0)^{-\sigma}J^{A^0}(k)(\delta_A^0)^{\sigma}(-k)^{i\nu\sigma}e^{-\frac{1}{4}ik^2\sigma}. \end{equation} | (3.66) |
The jump matrix is the constant one on the four rays \Sigma_A^1 , \Sigma_A^2 , \Sigma_A^3 , \Sigma_A^4 , so
\begin{equation} \frac{\mathrm{d}\Psi_+(k)}{\mathrm{d}k} = \frac{\mathrm{d}\Psi_-(k)}{\mathrm{d}k}v(-k_0). \end{equation} | (3.67) |
Then it follows that (\mathrm{d}\Psi/\mathrm{d}k+ik\sigma\Psi)\Psi^{-1} has no jump discontinuity along any of the four rays. Besides, from the relation between \Psi(k) and H(k) , we have
\begin{align*} \frac{\mathrm{d}\Psi(k)}{\mathrm{d}k}\Psi^{-1}(k) = &\frac{\mathrm{d}H(k)}{\mathrm{d}k}H^{-1}(k)-\frac{ik}{2}H(k)\sigma H^{-1}(k)+\frac{i\nu}{k}H(k)\sigma H^{-1}(k)\notag\\ = &O(k^{-1})-\frac{ik\sigma}{2}+\frac{i}{2}(\delta_A^0)^{\sigma}[\sigma, M^{A^0}_1](\delta_A^0)^{-\sigma}. \end{align*} |
It follows by the Liouville's Theorem that
\begin{equation} \frac{\mathrm{d}\Psi(k)}{\mathrm{d}k}+\frac{ik}{2}\sigma\Psi(k) = \beta\Psi(k), \end{equation} | (3.68) |
where
\begin{equation*} \beta = \frac{i}{2}(\delta_A^0)^{\sigma}[\sigma, M^{A^0}_1](\delta_A^0)^{-\sigma} = \left(\begin{array}{cc} 0 & \beta_{12}\\ \beta_{21} & 0 \end{array}\right). \end{equation*} |
Moreover,
\begin{equation} (M_1^{A^0})_{12} = -i(\delta_A^0)^{-2}\beta_{12}. \end{equation} | (3.69) |
Set
\begin{equation*} \Psi(k) = \left(\begin{array}{cc} \Psi_{11}(k) & \Psi_{12}(k)\\ \Psi_{21}(k) & \Psi_{22}(k)\\ \end{array}\right). \end{equation*} |
From (3.68) and its differential, we obtain
\begin{gather*} \frac{\mathrm{d}^{2}\beta_{21}\Psi_{11}(k)}{\mathrm{d}k^{2}}+\left(\frac{i}{2}+\frac{k^2}{4}-\beta_{21}\beta_{12}\right)\beta_{21}\Psi_{11}(k) = 0, \\ \Psi_{21}(k) = \frac{1}{\beta_{21}\beta_{12}}\left(\frac{\mathrm{d}\beta_{21}\Psi_{11}(k)}{\mathrm{d}k}+\frac{ik}{2}\beta_{21}\Psi_{11}(k)\right), \\ \frac{\mathrm{d}^{2}\Psi_{22}(k)}{\mathrm{d}k^{2}}+\left(-\frac{i}{2}+\frac{k^2}{4}-\beta_{21}\beta_{12}\right)\Psi_{22}(k) = 0, \\ \beta_{21}\Psi_{12}(k) = \left(\frac{\mathrm{d}\Psi_{22}(k)}{\mathrm{d}k}-\frac{ik}{2}\Psi_{22}(k)\right). \end{gather*} |
As is well known, the Weber's equation
\begin{equation*} \frac{\mathrm{d}^{2}g(\zeta)}{\mathrm{d}\zeta^{2}}+\left(\varrho+\frac{1}{2}-\frac{\zeta^{2}}{4}\right)g(\zeta) = 0 \end{equation*} |
has the solution
\begin{equation*} g(\zeta) = c_{1}D_{\varrho}(\zeta)+c_{2}D_{\varrho}(-\zeta), \end{equation*} |
where D_{\varrho}(\cdot) denotes the standard parabolic-cylinder function, and c_1 , c_2 are constants. The parabolic-cylinder function satisfies [41]
\begin{gather} \frac{\mathrm{d}D_{\varrho}(\zeta)}{\mathrm{d}\zeta}+\frac{\zeta}{2}D_{\varrho}(\zeta)-\varrho D_{\varrho-1}(\zeta) = 0, \end{gather} | (3.70) |
\begin{gather} D_{\varrho}(\pm\zeta) = \frac{\Gamma(\varrho+1)e^{\frac{i\pi \varrho}{2}}}{\sqrt{2\pi}}D_{-\varrho-1}(\pm i\zeta)+\frac{\Gamma(\varrho+1)e^{-\frac{i\pi \varrho}{2}}}{\sqrt{2\pi}}D_{-\varrho-1}(\mp i\zeta). \end{gather} | (3.71) |
As \zeta\rightarrow\infty , from [42], we have
\begin{equation} D_{\varrho}(\zeta) = \begin{cases} \zeta^{\varrho}e^{-\frac{\zeta^{2}}{4}}(1+O(\zeta^{-2})), & \left|\arg{\zeta}\right| \lt \frac{3\pi}{4}, \\ \zeta^{\varrho}e^{-\frac{\zeta^{2}}{4}}(1+O(\zeta^{-2}))-\frac{\sqrt{2\pi}}{\Gamma(-\varrho)}e^{\varrho\pi i+\frac{\zeta^{2}}{4}}\zeta^{-\varrho-1}(1+O(\zeta^{-2})), & \frac{\pi}{4} \lt \arg{\zeta} \lt \frac{5\pi}{4}, \\ \zeta^{\varrho}e^{-\frac{\zeta^{2}}{4}}(1+O(\zeta^{-2}))-\frac{\sqrt{2\pi}}{\Gamma(-\varrho)}e^{-\varrho\pi i+\frac{\zeta^{2}}{4}}\zeta^{-\varrho-1}(1+O(\zeta^{-2})), & -\frac{5\pi}{4} \lt \arg{\zeta} \lt -\frac{\pi}{4}, \end{cases} \end{equation} | (3.72) |
where \Gamma(\cdot) is the Gamma function. Set \varrho = i\beta_{21}\beta_{12} ,
\begin{gather} \beta_{21}\Psi_{11}(k) = c_1D_\varrho\left(e^{\frac{\pi i}{4}}k\right)+c_2D_\varrho\left(e^{\frac{-3\pi i}{4}}k\right), \end{gather} | (3.73) |
\begin{gather} \Psi_{22}(k) = c_3D_{-\varrho}\left(e^{\frac{-\pi i}{4}}k\right)+c_4D_{-\varrho}\left(e^{\frac{3\pi i}{4}}k\right), \end{gather} | (3.74) |
where a_1, a_2, a_3, a_4 are constants. As \arg{k}\in(-\pi, -\frac{3\pi}{4})\cup(\frac{3\pi}{4}, \pi) and k\rightarrow\infty , we arrive at
\begin{equation*} \Psi_{11}(k)(-k)^{-i\nu}e^{\frac{ik^{2}}{4}}\rightarrow I, \quad \Psi_{22}(k)(-k)^{i\nu}e^{-\frac{ik^{2}}{4}}\rightarrow 1, \end{equation*} |
then
\begin{gather*} \beta_{21}\Psi_{11}(k) = \beta_{21}e^{\frac{\pi\nu}{4}}D_{\varrho}\left(e^{-\frac{3\pi i}{4}}k\right), \quad \nu = \beta_{21}\beta_{12}, \\ \Psi_{22}(k) = e^{\frac{\pi\nu}{4}}D_{-\varrho}\left(e^{\frac{3\pi i}{4}}k\right). \end{gather*} |
Consequently,
\begin{gather*} \Psi_{21}(k) = \beta_{21}e^{\frac{\pi\nu}{4}}e^{-\frac{\pi i}{4}}D_{\varrho-1}\left(e^{-\frac{3\pi i}{4}}k\right), \\ \beta_{21}\Psi_{12}(k) = \varrho e^{\frac{\pi\nu}{4}}e^{-\frac{\pi i}{4}}D_{-\varrho-1}\left(e^{\frac{3\pi i}{4}}k\right). \end{gather*} |
For \arg{k}\in(-\frac{3\pi}{4}, -\frac{\pi}{4}) and k\rightarrow\infty , we have
\begin{equation*} \Psi_{11}(k)(-k)^{-i\nu}e^{\frac{ik^{2}}{4}}\rightarrow I, \quad \Psi_{22}(k)(-k)^{i\nu}e^{-\frac{ik^{2}}{4}}\rightarrow 1, \end{equation*} |
then
\begin{gather*} \beta_{21}\Psi_{11}(k) = \beta_{21}e^{-\frac{3\pi\nu}{4}}D_{\varrho}\left(e^{\frac{\pi i}{4}}k\right), \\ \Psi_{22}(k) = e^{\frac{\pi\nu}{4}}D_{-\varrho}\left(e^{\frac{3\pi i}{4}}k\right). \end{gather*} |
Consequently,
\begin{gather*} \Psi_{21}(k) = \beta_{21}e^{-\frac{3\pi\nu}{4}}e^{\frac{3\pi i}{4}}D_{\varrho-1}\left(e^{\frac{\pi i}{4}}k\right), \\ \beta_{21}\Psi_{12}(k) = \varrho e^{\frac{\pi\nu}{4}}e^{-\frac{\pi i}{4}}D_{-\varrho-1}\left(e^{\frac{3\pi i}{4}}k\right). \end{gather*} |
Along the ray \arg k = -\frac{3\pi}{4},
\begin{equation} \Psi_{+}(k) = \Psi_{-}(k) \left(\begin{array}{cc} I & 0\\ -\gamma^\dagger(-k_0)B_1 & 1\\ \end{array}\right). \end{equation} | (3.75) |
Notice the (2, 1) entry of the Riemann-Hilbert problem,
\begin{align*} &\beta_{21}e^{\frac{\pi(\nu-i)}{4}}D_{\varrho-1}(e^{-\frac{3\pi i}{4}}k)\\ = &\beta_{21}e^{\frac{\pi(3i-3\nu)}{4}}D_{\varrho-1}(e^{\frac{\pi i}{4}}k)-e^{\frac{\pi\nu}{4}}D_{-\varrho}(e^{\frac{3\pi i}{4}}k)\gamma^\dagger(-k_0)B_1. \end{align*} |
It follows from (3.71) that
\begin{equation*} D_{-\varrho}(e^{\frac{3\pi i}{4}}k) = \frac{\Gamma(-\varrho+1)e^{\frac{\pi\nu}{2}}}{\sqrt{2\pi}}D_{\varrho-1}(e^{-\frac{3\pi i}{4}}k)+\frac{\Gamma(-\varrho+1)e^{-\frac{\pi\nu}{2}}}{\sqrt{2\pi}}D_{\varrho-1}(e^{\frac{\pi i}{4}}k). \end{equation*} |
Then we separate the coefficients of the two independent functions and obtain
\begin{gather} \beta_{21} = e^{-\frac{3\pi i}{4}}e^{\frac{\pi\nu}{2}}\frac{\Gamma(-\varrho+1)}{\sqrt{2\pi}}\gamma^\dagger(-k_0)B_1. \end{gather} | (3.76) |
Noting that B^{-1}(J^{A^0}(k^\ast))^\dagger B = (J^{A^0}(k))^{-1} , we have \beta_{12} = -B_1^{-1}\beta_{21}^\dagger , which means that
\begin{equation} \beta_{12} = -B_1^{-1}B_1^\dagger\gamma(-k_0)e^{\frac{3\pi i}{4}}e^{\frac{\pi\nu}{2}}\frac{\Gamma(-\varrho+1)}{\sqrt{2\pi}} = e^{-\frac{\pi i}{4}}e^{\frac{\pi\nu}{2}}\nu\frac{\Gamma(-i\nu)}{\sqrt{2\pi}}\gamma(-k_0). \end{equation} | (3.77) |
Finally, we can obtain (1.4) from (3.64), (3.69) and (3.77).
This work is supported by the National Natural Science Foundation of China (Grant Nos. 11871440 and 11931017).
The authors declare no conflict of interest.
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