1. Introduction
In 1967, Gardner, Greene, Kruskal and Miüra [1] first proposed the classical inverse scattering transform method when they analyzed the initial value problem of the Korteweg–de Vries (KdV) equation. In 1997, Fokas [2] proposed the unified transformation approach when studying the initial-boundary value problems (IBVPs) of nonlinear partial differential equations (PDEs), which can be used to analyze the IBVPs of linear and nonlinear PDEs on the half-line or the finite interval, such as the sine-Gordon equation [3], the nonlinear Schrödinger (NLS) equation [4], the KdV equation [5], the derivative NLS equation [6], etc (see [7–12] and references therein).
In 2012, Lenells popularized the Fokas method [13] and analyzed the IBVPs of the integrable evolution equations with a 3 × 3 matrix Lax pair on the half-line and the finite interval, such as the Degasperis-Procesi equation [14], the Sasa-Satsuma equation [15], the Ostrovsky-Vakhnenko equation [16], the coupled NLS equations [17, 18], etc. After that, the idea was extended to study IBVPs of some integrable nonlinear evolution equations with a 4 × 4 matrix Lax pair on the half-line or the finite interval, such as the matrix Lakshmanan-Porsezian-Daniel system [19], the three-coupled Hirota system [20], the new two-component generalized Sasa-Satsuma equation [21], and the integrable spin-1 Gross–Pitaevskii equations [22] on the half-line, the integrable spin-1 Gross–Pitaevskii equations [23] and the general three-component NLS equation [24] on the finite interval.
It is well known that the modified KdV (mKdV) equation is one of the most important completely integrable nonlinear PDE [25], which has the form1.1 ) can describe internal ocean waves, waves in quantized films, transmission lines in the Schottky barrier, magnetohydrodynamic waves in plasmas, an Alfvén wave in a cold collision-free plasma, acoustic waves in certain anharmonic lattices, ultra-short pulses in nonlinear optics and other aspects [26–29]. Because of these important applications in physics, a series of results for the mKdV equation (1.1 ) have been reported, such as the Hirota bilinear technique [30], the inverse scattering transform [31], the Darboux transformation (DT) [32], etc.
$\begin{eqnarray}{u}_{t}+\alpha (6{u}^{2}{u}_{x}+{u}_{{xxx}})=0,\end{eqnarray}$
where u = u(x, t) is a real function with transverse variable x and evolution variable t. Equation (In 1990, Marchant and Smyth [33] proposed the following extended mKdV equation1.2 ) can describe the evolution of steeper waves with shorter wavelengths than the KdV equation [34]. The Painlevé test, the multi-soliton solutions [35], the infinitely many conservation laws, the periodic, the rational solutions [36], the long-time asymptotic behavior [37], the breather-soliton molecules and the breather-positons [38], the Painlevé-type asymptotics [39], the rational positons and rogue waves [40] for equation (1.2 ) have been studied, when ϵ = 0, the emKdV equation (1.2 ) reduces to the following fifth-order mKdV equation
$\begin{eqnarray}\begin{array}{l}{u}_{t}+\varepsilon (6{u}^{2}{u}_{x}+{u}_{{xxx}})+\kappa (30{u}^{4}{u}_{x}+10{u}_{x}^{3}\\ +\,40{{uu}}_{x}{u}_{{xx}}+10{u}^{2}{u}_{{xxx}}+{u}_{{xxxxx}})=0,\end{array}\end{eqnarray}$
where ϵ ≪ 1 and κ ≪ 1 stand for the third-order and fifth-order dispersion coefficients matching with the relevant nonlinear terms, respectively. Equation ( $\begin{eqnarray}\begin{array}{l}{u}_{t}+\kappa (30{u}^{4}{u}_{x}+10{u}_{x}^{3}+40{{uu}}_{x}{u}_{{xx}}\\ +10{u}^{2}{u}_{{xxx}}+{u}_{{xxxxx}})=0.\end{array}\end{eqnarray}$
In 2008, Kwon discussed the initial value problem of the fifth-order mKdV equation on the Sobolev spaces [41]. In 2017, Cheng et al studied the consistent Riccati expansion and nonlocal symmetry of the fifth-order mKdV equation [42]. In 2018, Kwak considered the low regularity of the Cauchy problem for the fifth-order mKdV equation on [0, 2π] [43]. Recently, the soliton and breather solutions with a nonzero background [44], the long-time asymptotic behavior in the quarter plane [45] and in low regularity spaces [46] for equation (1.3 ) have been discussed. In this note, our aim is to construct the spectral analysis for equation (1.3 ) based on its Lax pair, and to derive the limit form solution of the IBVPs through the following Fokas method.
The design structure of this paper is as follows. In section 2 , we will perform a spectral analysis of the Lax pair for the fifth-order mKdV equation (1.3 ). In section 3 , we will discuss two pairs of spectral functions y(θ), z(θ) and Y(θ), Z(θ). In section 4 , we will present the RH problem of the fifth-order mKdV equation (1.3 ). The last sections feature some conclusions and discussions.
2. The spectral analysis
The fifth-order mKdV equation (1.3 ) admits the Lax pair formulation [45]
$\begin{eqnarray}{{\rm{\Psi }}}_{x}=M(x,t,\theta ){\rm{\Psi }},\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Psi }}}_{t}=N(x,t,\theta ){\rm{\Psi }},\end{eqnarray}$
where ${\rm{\Psi }}={({{\rm{\Psi }}}_{1},{{\rm{\Psi }}}_{2})}^{{\rm{T}}}$ is the vector eigenfunction and complex number θ is a spectral parameter and $\begin{eqnarray}M(x,t,\theta )=\left(\begin{array}{cc}{\rm{i}}\theta & u\\ -u & -{\rm{i}}\theta \end{array}\right)={\rm{i}}\theta {\sigma }_{3}+U,\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}N(x,t,\theta )=\kappa \left(\begin{array}{cc}{N}_{11} & {N}_{12}\\ {N}_{21} & -{N}_{11}\end{array}\right)\\ =-16{\rm{i}}\kappa {\theta }^{5}{\sigma }_{3}-16\kappa {\theta }^{4}U\\ \,-8{\rm{i}}\kappa {\theta }^{3}({U}^{2}{\sigma }_{3}-{\sigma }_{3}{U}_{x})+\kappa {\theta }^{2}(4{U}_{{xx}}-8{U}^{3})\\ \,+{\rm{i}}\kappa \theta {\sigma }_{3}(12{U}^{2}{U}_{x}-2{U}_{{xxx}})\\ \,+2{\rm{i}}\kappa \theta (2{{UU}}_{{xx}}-{U}_{x}^{2}-3{U}^{4}){\sigma }_{3}\\ \,+\kappa (-6{U}^{5}+10{U}^{2}{U}_{{xx}}+10{{UU}}_{x}^{2}-{U}_{{xxxx}}),\end{array}\end{eqnarray}$
with $\begin{eqnarray}\begin{array}{l}{\sigma }_{3}=\left(\begin{array}{cc}1 & 0\\ 0 & -1\end{array}\right),\,U=\left(\begin{array}{cc}0 & u\\ -u & 0\end{array}\right),\\ {N}_{11}=-16{\rm{i}}{\theta }^{5}+8{\rm{i}}{u}^{2}{\theta }^{3}-(6{\rm{i}}{u}^{4}+4{\rm{i}}{{uu}}_{{xx}}-2{\rm{i}}{u}_{x}^{2})\theta ,\\ {N}_{12}=-16u{\theta }^{4}+8{\rm{i}}{u}_{x}{\theta }^{3}+(8{u}^{3}+4{u}_{{xx}}){\theta }^{2}\\ \,\,-(12{\rm{i}}{u}^{2}{u}_{x}+2{\rm{i}}{u}_{{xxx}})\theta \\ -6{u}^{5}-10{u}^{2}{u}_{{xx}}-10{{uu}}_{x}^{2}-{u}_{{xxxx}},\\ {N}_{21}=16u{\theta }^{4}+8{\rm{i}}{u}_{x}{\theta }^{3}-(8{u}^{3}+4{u}_{{xx}}){\theta }^{2}\\ \,-(12{\rm{i}}{u}^{2}{u}_{x}+2{\rm{i}}{u}_{{xxx}})\theta \\ +6{u}^{5}+10{u}^{2}{u}_{{xx}}+10{{uu}}_{x}^{2}+{u}_{{xxxx}}.\end{array}\end{eqnarray}$
2.1. The exact 1-form
One can rewrite equations (2.1a )–(2.1b ) as2.6a )-(2.6b ) can be written as the following full differential
$\begin{eqnarray}{{\rm{\Psi }}}_{x}-{\rm{i}}\theta {\sigma }_{3}{\rm{\Psi }}=U{\rm{\Psi }},\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Psi }}}_{t}+16{\rm{i}}\kappa {\theta }^{5}{\sigma }_{3}{\rm{\Psi }}=V{\rm{\Psi }},\end{eqnarray}$
with $\begin{eqnarray*}\begin{array}{l}V=-16\kappa {\theta }^{4}U+8{\rm{i}}\kappa {\theta }^{3}({\sigma }_{3}{U}_{x}-{U}^{2}{\sigma }_{3})\\ +4\kappa {\theta }^{2}({U}_{{xx}}-2{U}^{3})+10\kappa ({U}^{2}{U}_{{xx}}+{{UU}}_{x}^{2})\\ +2{\rm{i}}\kappa \theta [{\sigma }_{3}(6{U}^{2}{U}_{x}-{U}_{{xxx}})\\ +(2{{UU}}_{{xx}}-{U}_{x}^{2}-3{U}^{4}){\sigma }_{3}]-6\kappa {U}^{5}-\kappa {U}_{{xxxx}},\end{array}\end{eqnarray*}$
and introduce Φ(x, t, θ) by $\begin{eqnarray}{\rm{\Phi }}(x,t,\theta )={\rm{\Psi }}(x,t,\theta ){{\rm{e}}}^{(-{\rm{i}}\theta x+16{\rm{i}}\kappa {\theta }^{5}t){\sigma }_{3}},\end{eqnarray}$
then, one can get the following equivalent Lax pair $\begin{eqnarray}{{\rm{\Phi }}}_{x}-{\rm{i}}\theta [{\sigma }_{3},{\rm{\Phi }}]=U{\rm{\Phi }},\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Phi }}}_{t}+16{\rm{i}}\kappa {\theta }^{5}[{\sigma }_{3},{\rm{\Phi }}]=V{\rm{\Phi }},\end{eqnarray}$
where [σ3, Φ] = σ3Φ − Φσ3, obviously, equation ( $\begin{eqnarray}\begin{array}{l}{\rm{d}}({{\rm{e}}}^{(-{\rm{i}}\theta x+16{\rm{i}}\kappa {\theta }^{5}t){\hat{\sigma }}_{3}}{\rm{\Phi }}(x,t,\theta ))\\ =\,{{\rm{e}}}^{(-{\rm{i}}\theta x+16{\rm{i}}\kappa {\theta }^{5}t){\hat{\sigma }}_{3}}(A(x,t,\theta ){\rm{\Phi }}(x,t,\theta )),\end{array}\end{eqnarray}$
where A(x, t, θ) = U(x, t)dx + V(x, t, θ)dt, and ${\hat{\sigma }}_{3}$ is a matrix operator (see [12]).2.2. The three important eigenfunctions ${\{{{\rm{\Phi }}}_{j}(x,t,\theta )\}}_{1}^{3}$
For (x, t) ∈ D = {(x, t)∣0 < x < + ∞ , 0 < t < T}, let $u(x,t)\in {\mathbb{S}}$. We define three solutions ${\{{{\rm{\Phi }}}_{j}(x,t,\theta )\}}_{1}^{3}$ of equation (2.6a )–(2.6b ) by2.8 ) is path independent, we choose the special curves depicted in figure 1, then we have
$\begin{eqnarray}\begin{array}{l}{{\rm{\Phi }}}_{j}(x,t,\theta )={\rm{I}}+{\displaystyle \int }_{({x}_{j},{t}_{j})}^{(x,t)}{{\rm{e}}}^{({\rm{i}}\theta x-16{\rm{i}}\kappa {\theta }^{5}t){\hat{\sigma }}_{3}}\,F(\zeta ,\tau ,\theta ),\\ j=1,2,3,\end{array}\end{eqnarray}$
where $F(x,t,\theta )={{\rm{e}}}^{(-{\rm{i}}\theta x+16{\rm{i}}\kappa {\theta }^{5}t){\hat{\sigma }}_{3}}A(x,t,\theta ){\rm{\Phi }}(x,t,\theta )$, (x1, t1) = (0, 0), (x2, t2) = (0, T), (x3, t3) = ( ∞ , t). Because the 1-form F(x, t, θ) is exact, the integral of equation ( $\begin{eqnarray}\begin{array}{l}{{\rm{\Phi }}}_{1}(x,t,\theta )={\rm{I}}+{\displaystyle \int }_{0}^{x}{{\rm{e}}}^{{\rm{i}}\theta (x-\zeta ){\hat{\sigma }}_{3}}(U{{\rm{\Phi }}}_{1})(\zeta ,t,\theta ){\rm{d}}\zeta \\ +{{\rm{e}}}^{{\rm{i}}\theta x{\hat{\sigma }}_{3}}{\displaystyle \int }_{0}^{t}{{\rm{e}}}^{-16{\rm{i}}\kappa {\theta }^{5}(t-\tau ){\hat{\sigma }}_{3}}(V{{\rm{\Phi }}}_{1})(0,\tau ,\theta ){\rm{d}}\tau ,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{{\rm{\Phi }}}_{2}(x,t,\theta )={\rm{I}}+{\displaystyle \int }_{0}^{x}{{\rm{e}}}^{{\rm{i}}\theta (x-\zeta ){\hat{\sigma }}_{3}}(U{{\rm{\Phi }}}_{2})(\zeta ,t,\theta ){\rm{d}}\zeta \\ -{{\rm{e}}}^{{\rm{i}}\theta x{\hat{\sigma }}_{3}}{\displaystyle \int }_{t}^{T}{{\rm{e}}}^{-16{\rm{i}}\kappa {\theta }^{5}(t-\tau ){\hat{\sigma }}_{3}}(V{{\rm{\Phi }}}_{2})(0,\tau ,\theta ){\rm{d}}\tau ,\end{array}\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Phi }}}_{3}(x,t,\theta )={\rm{I}}-{\int }_{x}^{\infty }{{\rm{e}}}^{{\rm{i}}\theta (x-\zeta ){\hat{\sigma }}_{3}}(U{{\rm{\Phi }}}_{3})(\zeta ,t,\theta ){\rm{d}}\zeta .\end{eqnarray}$
Figure 1. The three contours η1, η2, η3 in the (x, t)-domain. |
The choice of these integration paths shows that the following inequalities hold on the curves,2.8 ) contains exp [−2iθ(x − ζ) + 32iκθ5(t − τ)], then, the bounded and analytical areas of eigenfunctions ${\{{{\rm{\Phi }}}_{j}(x,t,\theta )\}}_{1}^{3}$ are as follows2.8 ) involves exp $[2\mathrm{i}\theta(x-\zeta)-32\mathrm{i}\kappa\theta^5(t-\tau)]$. Let ${[{{\rm{\Phi }}}_{j}]}_{k}(x,t,\theta ),k\,=\,1,2$ be k-column of Φj(x, t, θ), we have
$\begin{eqnarray}{\eta }_{1}:x-\zeta \geqslant 0,\quad t-\tau \geqslant 0,\end{eqnarray}$
$\begin{eqnarray}{\eta }_{2}:x-\zeta \geqslant 0,\quad t-\tau \leqslant 0,\end{eqnarray}$
$\begin{eqnarray}{\eta }_{3}:x-\zeta \leqslant 0.\end{eqnarray}$
It is not difficult to find that the first column of equation ( $\begin{eqnarray}{[{{\rm{\Phi }}}_{1}]}_{1}(x,t,\theta ):\{\mathrm{Im}\theta \leqslant 0\}\cap \{\mathrm{Im}{\theta }^{5}\geqslant 0\},\end{eqnarray}$
$\begin{eqnarray}{[{{\rm{\Phi }}}_{2}]}_{1}(x,t,\theta ):\{\mathrm{Im}\theta \leqslant 0\}\cap \{\mathrm{Im}{\theta }^{5}\leqslant 0\},\end{eqnarray}$
$\begin{eqnarray}{[{{\rm{\Phi }}}_{3}]}_{1}(x,t,\theta ):\{\mathrm{Im}\theta \geqslant 0\}.\end{eqnarray}$
Similarly, we have $\begin{eqnarray}{[{{\rm{\Phi }}}_{1}]}_{2}(x,t,\theta ):\{\mathrm{Im}\theta \geqslant 0\}\cap \{\mathrm{Im}{\theta }^{5}\leqslant 0\},\end{eqnarray}$
$\begin{eqnarray}{[{{\rm{\Phi }}}_{2}]}_{2}(x,t,\theta ):\{\mathrm{Im}\theta \geqslant 0\}\cap \{\mathrm{Im}{\theta }^{5}\geqslant 0\},\end{eqnarray}$
$\begin{eqnarray}{[{{\rm{\Phi }}}_{3}]}_{2}(x,t,\theta ):\{\mathrm{Im}\theta \leqslant 0\},\end{eqnarray}$
because the second column of equation ( $\begin{eqnarray}\begin{array}{l}{{\rm{\Phi }}}_{1}(x,t,\theta )=({\left[{{\rm{\Phi }}}_{1}\right]}_{1}^{{{\rm{\Omega }}}_{4}}(x,t,\theta ),{\left[{{\rm{\Phi }}}_{1}\right]}_{2}^{{{\rm{\Omega }}}_{2}}(x,t,\theta )),\\ {{\rm{\Phi }}}_{2}(x,t,\theta )=({\left[{{\rm{\Phi }}}_{2}\right]}_{1}^{{{\rm{\Omega }}}_{3}}(x,t,\theta ),{\left[{{\rm{\Phi }}}_{2}\right]}_{2}^{{{\rm{\Omega }}}_{1}}(x,t,\theta )),\\ {{\rm{\Phi }}}_{3}(x,t,\theta )=({\left[{{\rm{\Phi }}}_{3}\right]}_{1}^{{{\mathbb{C}}}_{+}}(x,t,\theta ),{\left[{{\rm{\Phi }}}_{3}\right]}_{2}^{{{\mathbb{C}}}_{-}}(x,t,\theta )),\end{array}\end{eqnarray}$
where ${{\rm{\Phi }}}_{j}^{{{\rm{\Omega }}}_{i}}$ denotes that ωi is the bounded and analytical region of ${\{{{\rm{\Phi }}}_{j}\}}_{1}^{3}$, ${{\mathbb{C}}}_{+}={{\rm{\Omega }}}_{1}\cup {{\rm{\Omega }}}_{2}$ is the upper complex half-plane, ${{\mathbb{C}}}_{-}={{\rm{\Omega }}}_{3}\cup {{\rm{\Omega }}}_{4}$ is the lower complex half-plane, and ωi, i = 1, 2, 3, 4 are depicted in figure 2.
Figure 2. The areas ωi, i = 1,…,4 division on the complex θ-plane. |
In order to establish the RH problem of the fifth-order mKdV equation (1.3 ), we must define f(θ) and g(θ) by the following relations2.14a )–(2.14b ) at (x, t) = (0, 0) and (x, t) = (0, T), respectively, we obtain2.14a )–(2.14b ) and equation (2.15 ), we have
$\begin{eqnarray}{{\rm{\Phi }}}_{3}(x,t,\theta )={{\rm{\Phi }}}_{1}(x,t,\theta ){{\rm{e}}}^{({\rm{i}}\theta x-16{\rm{i}}\kappa {\theta }^{5}t){\hat{\sigma }}_{3}}f(\theta ),\end{eqnarray}$
$\begin{eqnarray}{{\rm{\Phi }}}_{2}(x,t,\theta )={{\rm{\Phi }}}_{1}(x,t,\theta ){{\rm{e}}}^{({\rm{i}}\theta x-16{\rm{i}}\kappa {\theta }^{5}t){\hat{\sigma }}_{3}}g(\theta ).\end{eqnarray}$
By evaluation of equation ( $\begin{eqnarray}\begin{array}{l}g(\theta )={{\rm{\Phi }}}_{2}(0,0,\theta )=({{\rm{e}}}^{16{\rm{i}}\kappa {\theta }^{5}T{\hat{\sigma }}_{3}}{{\rm{\Phi }}}_{1}{\left(0,T,\theta )\right)}^{-1},\\ f(\theta )={{\rm{\Phi }}}_{3}(0,0,\theta ),\end{array}\end{eqnarray}$
thus, from equation ( $\begin{eqnarray}{{\rm{\Phi }}}_{2}(x,t,\theta )={{\rm{\Phi }}}_{3}(x,t,\theta ){{\rm{e}}}^{({\rm{i}}\theta x-16{\rm{i}}\kappa {\theta }^{5}t){\hat{\sigma }}_{3}}{\left(f(\theta )\right)}^{-1}g(\theta ),\end{eqnarray}$
Furthermore, we also get Φ1(x, t, θ), Φ2(x, t, θ) at x = 0
$\begin{eqnarray}\begin{array}{l}{{\rm{\Phi }}}_{1}(0,t,\theta )=({\left[{{\rm{\Phi }}}_{1}\right]}_{1}^{{{\rm{\Omega }}}_{1}\cup {{\rm{\Omega }}}_{4}}(0,t,\theta ),{\left[{{\rm{\Phi }}}_{1}\right]}_{2}^{{{\rm{\Omega }}}_{2}\cup {{\rm{\Omega }}}_{3}}(0,t,\theta ))\\ ={\rm{I}}+{\displaystyle \int }_{0}^{t}{{\rm{e}}}^{-16{\rm{i}}\kappa {\theta }^{5}(t-\tau ){\hat{\sigma }}_{3}}(V{{\rm{\Phi }}}_{1})(0,\tau ,\theta ){\rm{d}}\tau ,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{{\rm{\Phi }}}_{2}(0,t,\theta )=({\left[{{\rm{\Phi }}}_{2}\right]}_{1}^{{{\rm{\Omega }}}_{2}\cup {{\rm{\Omega }}}_{3}}(0,t,\theta ),{\left[{{\rm{\Phi }}}_{2}\right]}_{2}^{{{\rm{\Omega }}}_{1}\cup {{\rm{\Omega }}}_{4}}(0,t,\theta ))\\ ={\rm{I}}-{\displaystyle \int }_{t}^{T}{{\rm{e}}}^{-16{\rm{i}}\kappa {\theta }^{5}(t-\tau ){\hat{\sigma }}_{3}}(V{{\rm{\Phi }}}_{2})(0,\tau ,\theta ){\rm{d}}\tau ,\end{array}\end{eqnarray}$
and get Φ1(x, t, θ), Φ3(x, t, θ) at t = 0 $\begin{eqnarray}\begin{array}{l}{{\rm{\Phi }}}_{1}(x,0,\theta )=({\left[{{\rm{\Phi }}}_{1}\right]}_{1}^{{{\mathbb{C}}}_{-}}(x,0,\theta ),{\left[{{\rm{\Phi }}}_{1}\right]}_{2}^{{{\mathbb{C}}}_{+}}(x,0,\theta )),\\ =\,{\rm{I}}+{\displaystyle \int }_{0}^{x}{{\rm{e}}}^{{\rm{i}}\theta (x-\zeta ){\hat{\sigma }}_{3}}(U{{\rm{\Phi }}}_{1})(\zeta ,0,\theta ){\rm{d}}\zeta ,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{{\rm{\Phi }}}_{3}(x,0,\theta )=({\left[{{\rm{\Phi }}}_{3}\right]}_{1}^{{{\mathbb{C}}}_{+}}(x,0,\theta ),{\left[{{\rm{\Phi }}}_{3}\right]}_{2}^{{{\mathbb{C}}}_{-}}(x,0,\theta ))\\ =\,{\rm{I}}-{\displaystyle \int }_{x}^{\infty }{{\rm{e}}}^{{\rm{i}}\theta (x-\zeta ){\hat{\sigma }}_{3}}(U{{\rm{\Phi }}}_{3})(\zeta ,0,\theta ){\rm{d}}\zeta ,\end{array}\end{eqnarray}$
Assume that u0(x) = u(x, t = 0), v0(t) = u(x = 0, t), v1(t) = ux(x = 0, t), v2(t) = uxx(x = 0, t), v3(t) = uxxx(x = 0, t), v4(t) = uxxxx(x = 0, t) are the initial and boundary values, then, we have
$\begin{eqnarray}\begin{array}{l}U(x,0,\theta )=\left(\begin{array}{cc}0 & {u}_{0}\\ -{u}_{0} & 0\end{array}\right),\\ V(0,t,\theta )=\left(\begin{array}{cc}{V}_{11}(0,t,\theta ) & {V}_{12}(0,t,\theta )\\ {V}_{21}(0,t,\theta ) & -{V}_{11}(0,t,\theta )\end{array}\right).\end{array}\end{eqnarray}$
with $\begin{eqnarray*}\begin{array}{l}{V}_{11}(0,t,\theta )={\rm{i}}\kappa \left[8{v}_{0}^{2}{\theta }^{3}-(6{v}_{0}^{4}+4{v}_{0}{v}_{2}-2{v}_{1}^{2})\theta \right],\\ {V}_{12}(0,t,\theta )=\kappa \left[-16{v}_{0}{\theta }^{4}\right.\\ +8{\rm{i}}{v}_{1}{\theta }^{3}+(8{v}_{0}^{3}+4{v}_{2}){\theta }^{2}-(12{\rm{i}}{v}_{0}^{2}{v}_{1}+2{\rm{i}}{v}_{3})\theta \\ \left.-6{v}_{0}^{5}-10{v}_{0}^{2}{v}_{2}-10{v}_{0}{v}_{1}^{2}-{v}_{4}\right],\\ {V}_{21}(0,t,\theta )=\kappa \left[16{v}_{0}{\theta }^{4}+8{\rm{i}}{v}_{1}{\theta }^{3}-(8{v}_{0}^{3}+4{v}_{2}){\theta }^{2}\right.\\ -(12{\rm{i}}{v}_{0}^{2}{v}_{1}+2{\rm{i}}{v}_{3})\theta \\ \left.+6{v}_{0}^{5}+10{v}_{0}^{2}{v}_{2}+10{v}_{0}{v}_{1}^{2}+{v}_{4}\right],\end{array}\end{eqnarray*}$
2.3. The other properties of the eigenfunctions
(Symmetries) Let ${\rm{\Phi }}(x,t,\theta )={{\rm{\Phi }}}_{j}$$(x,t,\theta ),j=1,2,3,$ then ${\rm{\Phi }}(x,t,\theta )$ admits the following symmetry relations
$\begin{eqnarray}\begin{array}{l}\overline{{\rm{\Phi }}(x,t,\bar{\theta })}={\rm{\Phi }}(x,t,-\theta )={\sigma }_{2}{\rm{\Phi }}(x,t,\theta ){\sigma }_{2},\\ {\sigma }_{2}=\left(\begin{array}{cc}0 & -{\rm{i}}\\ {\rm{i}} & 0\end{array}\right),\end{array}\end{eqnarray}$
as well as $\begin{eqnarray}\begin{array}{l}{{\rm{\Phi }}}_{11}(x,t,\theta )=\overline{{{\rm{\Phi }}}_{22}(x,t,\bar{\theta })},\\ {{\rm{\Phi }}}_{21}(x,t,\theta )=-\overline{{{\rm{\Phi }}}_{12}(x,t,\bar{\theta })}\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{{\rm{\Phi }}}_{11}(x,t,-\theta )=-{{\rm{\Phi }}}_{11}(x,t,\theta ),\\ {{\rm{\Phi }}}_{22}(x,t,-\theta )=-{{\rm{\Phi }}}_{22}(x,t,\theta ),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{{\rm{\Phi }}}_{12}(x,t,-\theta )={{\rm{\Phi }}}_{12}(x,t,\theta ),\\ {{\rm{\Phi }}}_{21}(x,t,-\theta )={{\rm{\Phi }}}_{21}(x,t,\theta ),\end{array}\end{eqnarray}$
The eigenfunctions ${{\rm{\Phi }}}_{j}(x,t,\theta )\,=({[{{\rm{\Phi }}}_{j}]}_{1}(x,t,\theta ),$${[{{\rm{\Phi }}}_{j}]}_{2}(x,t,\theta )),j\,=\,1,2,3$ admit the following properties
| • | $\det {{\rm{\Phi }}}_{j}(x,t,\theta )=1,j\,=\,1,2,3$, |
| • | ${[{{\rm{\Phi }}}_{1}]}_{1}$ is analytical for $\theta \in {{\rm{\Omega }}}_{4}$, as well as continuous to ${\overline{{\rm{\Omega }}}}_{4},$ ${[{{\rm{\Phi }}}_{1}]}_{2}$ is analytical for $\theta \in {{\rm{\Omega }}}_{2},$ as well as continuous to ${\overline{{\rm{\Omega }}}}_{2},$ |
| • | ${[{{\rm{\Phi }}}_{2}]}_{1}$ is analytical for $\theta \in {{\rm{\Omega }}}_{3}$, as well as continuous to ${\overline{{\rm{\Omega }}}}_{3},$ ${[{{\rm{\Phi }}}_{2}]}_{2}$ is analytical for $\theta \in {{\rm{\Omega }}}_{1},$ as well as continuous to ${\overline{{\rm{\Omega }}}}_{1},$ |
| • | ${[{{\rm{\Phi }}}_{3}]}_{1}$ is analytical for $\theta \in {{\mathbb{C}}}_{+}$, as well as continuous to ${{\mathbb{C}}}_{+}\cup \mathrm{Re}\theta ,$ ${[{{\rm{\Phi }}}_{3}]}_{2}$ is analytical for $\theta \in {{\mathbb{C}}}_{-}$, as well as continuous to ${{\mathbb{C}}}_{-}\cup \mathrm{Re}\theta ,$ |
| • | ${[{{\rm{\Phi }}}_{j}]}_{1}{(x,t,\theta )\to (1,0)}^{T}$, ${[{{\rm{\Phi }}}_{j}]}_{2}{(x,t,\theta )\to (0,1)}^{T},$ as $\theta \to \infty $. |
Indeed, according to equation (2.15 ), we can get the expressions for f(θ) and g(θ) as follows:2.1 , we can define
$\begin{eqnarray}f(\theta )={\rm{I}}-{\int }_{0}^{\infty }{{\rm{e}}}^{-{\rm{i}}\theta (\zeta -x){\hat{\sigma }}_{3}}(U{{\rm{\Phi }}}_{3})(\zeta ,0,\theta ){\rm{d}}\zeta ,\end{eqnarray}$
$\begin{eqnarray}{g}^{-1}(\theta )={\rm{I}}+{\int }_{0}^{T}{{\rm{e}}}^{16{\rm{i}}\kappa {\theta }^{5}(\tau -t){\hat{\sigma }}_{3}}(V{{\rm{\Phi }}}_{1})(0,\tau ,\theta ){\rm{d}}\tau ,\end{eqnarray}$
then, by the symmetry of Proposition $\begin{eqnarray}\begin{array}{l}f(\theta )=\left(\begin{array}{cc}\overline{y(\bar{\theta })} & z(\theta )\\ \overline{-z(\bar{\theta })} & y(\theta )\end{array}\right),\\ g(\theta )=\left(\begin{array}{cc}\overline{Y(\bar{\theta })} & Z(\theta )\\ \overline{-Z(\bar{\theta })} & Y(\theta )\end{array}\right).\end{array}\end{eqnarray}$
It follows from equations (
| • | $\left(\begin{array}{c}z(\theta )\\ y(\theta )\end{array}\right)={\left[{{\rm{\Phi }}}_{3}\right]}_{2}^{{{\mathbb{C}}}_{-}}(0,0,\theta )=\left(\begin{array}{c}{\left({{\rm{\Phi }}}_{3}\right)}_{12}^{{{\mathbb{C}}}_{-}}(0,0,\theta )\\ {\left({{\rm{\Phi }}}_{3}\right)}_{22}^{{{\mathbb{C}}}_{-}}(0,0,\theta )\\ \end{array}\right).$ |
| • | $\left(\begin{array}{c}-{{\rm{e}}}^{32{\rm{i}}\kappa {\theta }^{5}T}Z(\theta )\\ \overline{Y(\bar{\theta })})\end{array}\right)={\left[{{\rm{\Phi }}}_{1}\right]}_{2}^{{{\rm{\Omega }}}_{2}\cup {{\rm{\Omega }}}_{3}}(0,T,\theta )=\left(\begin{array}{c}{\left({{\rm{\Phi }}}_{1}\right)}_{12}^{{{\rm{\Omega }}}_{2}\cup {{\rm{\Omega }}}_{3}}(0,T,\theta )\\ {\left({{\rm{\Phi }}}_{1}\right)}_{22}^{{{\rm{\Omega }}}_{2}\cup {{\rm{\Omega }}}_{3}}(0,T,\theta )\\ \end{array}\right).$ |
| • | $y(-\theta )$ $=\overline{y(\bar{\theta })}$, $z(-\theta )=\overline{z(\bar{\theta })}$. |
| • | $Y(-\theta )$ $=\overline{Y(\bar{\theta })}$, $Z(-\theta )=\overline{Z(\bar{\theta })}$. |
| • | $\det f(\theta )$ $=y(\theta )\overline{y(\bar{\theta })}+z(\theta )\overline{z(\bar{\theta })}=1,$ for $\theta \in {\mathbb{R}}.$ |
| • | $\det g(\theta )=Y(\theta )\overline{Y(\bar{\theta })}+Z(\theta )\overline{Z(\bar{\theta })}=1,$ for $\theta \in {\mathbb{C}}(\mathrm{Im}{\theta }^{5}=0,{if}\,T=\infty ).$ |
| • | $y(\theta )$ $=1+{\rm{O}}({\theta }^{-1}),\,z(\theta )={\rm{O}}({\theta }^{-1}),$ as $\theta \to \infty ,\mathrm{Im}\theta \gt 0.$ |
| • | $Y(\theta )$ $=1+{\rm{O}}({\theta }^{-1}),\,Z(\theta )={\rm{O}}({\theta }^{-1}),$ as $\theta \to \infty ,\mathrm{Im}{\theta }^{5}\gt 0.$ |
2.4. The basic RH problem
For the convenience of calculation, we introduce the following expressions
Next, we define the function H(x, t, θ) by
$\begin{eqnarray}\begin{array}{l}\mu (\theta )=-\theta x+16\kappa {\theta }^{5}t,\\ \alpha (\theta )=y(\theta )\overline{Y(\bar{\theta })}+z(\theta )\overline{Z(\bar{\theta })},\quad \theta \in {\bar{{\rm{\Omega }}}}_{2},\\ \beta (\theta )=y(\theta )Z(\theta )-z(\theta )Y(\theta ),\quad \theta \in {\bar{{\rm{\Omega }}}}_{4},\\ S(\theta )=\displaystyle \frac{\overline{Z(\bar{\theta })}}{y(\theta )\alpha (\theta )},\quad \theta \in {\bar{{\rm{\Omega }}}}_{3},\\ s(\theta )=\displaystyle \frac{\overline{\beta (\bar{\theta })}}{\alpha (\theta )}=\displaystyle \frac{\overline{z(\bar{\theta })}}{y(\theta )}-S(\theta ),\quad \theta \in {\mathbb{R}},\end{array}\end{eqnarray}$
then, we get| • | $S(-\theta )=\overline{S(\bar{\theta })}$, $s(-\theta )=\overline{s(\bar{\theta })}$. |
| • | ${g}^{-1}(\theta )f(\theta )=\left(\begin{array}{cc}\overline{\alpha (\bar{\theta })} & \beta (\theta )\\ -\overline{\beta (\bar{\theta })} & \alpha (\theta )\end{array}\right)$, |
| • | $\alpha (-\theta )=\overline{\alpha (\bar{\theta })}$, $\beta (-\theta )=\overline{\beta (\bar{\theta })}$. |
| • | $\det [{g}^{-1}(\theta )f(\theta )]=\alpha (\theta )\overline{\alpha (\bar{\theta })}+\beta (\theta )\overline{\beta (\bar{\theta })}=1$, |
| • | $\alpha (\theta )=1\,+\,O\left(\tfrac{1}{\theta }\right),\beta (\theta )=O\left(\tfrac{1}{\theta }\right){as}\,\theta \to \infty $. |
$\begin{eqnarray}\begin{array}{l}{H}_{+}^{(1)}(x,t,\theta )=\left({\left[{{\rm{\Phi }}}_{3}\right]}_{1}^{{{\mathbb{C}}}_{+}}(x,t,\theta ),\displaystyle \frac{{\left[{{\rm{\Phi }}}_{2}\right]}_{2}^{{{\rm{\Omega }}}_{1}}(x,t,\theta )}{\overline{\alpha (\bar{\theta })}}\right),\\ \theta \in {{\rm{\Omega }}}_{1},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{H}_{-}^{(1)}(x,t,\theta )=\left({\left[{{\rm{\Phi }}}_{3}\right]}_{1}^{{{\mathbb{C}}}_{+}}(x,t,\theta ),\displaystyle \frac{{\left[{{\rm{\Phi }}}_{1}\right]}_{2}^{{{\rm{\Omega }}}_{2}}(x,t,\theta )}{\overline{y(\bar{\theta })}}\right),\\ \theta \in {{\rm{\Omega }}}_{2},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{H}_{+}^{(2)}(x,t,\theta )=\left(\displaystyle \frac{{\left[{{\rm{\Phi }}}_{1}\right]}_{1}^{{{\rm{\Omega }}}_{4}}(x,t,\theta )}{y(\theta )},{\left[{{\rm{\Phi }}}_{3}\right]}_{2}^{{{\mathbb{C}}}_{-}}(x,t,\theta )\right),\\ \theta \in {{\rm{\Omega }}}_{4},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{H}_{-}^{(2)}(x,t,\theta )=\left(\displaystyle \frac{{\left[{{\rm{\Phi }}}_{2}\right]}_{1}^{{{\rm{\Omega }}}_{3}}(x,t,\theta )}{\alpha (\theta )},{\left[{{\rm{\Phi }}}_{3}\right]}_{2}^{{{\mathbb{C}}}_{-}}(x,t,\theta )\right),\\ \theta \in {{\rm{\Omega }}}_{3}.\end{array}\end{eqnarray}$
Therefore, we have $\begin{eqnarray}\begin{array}{ll}\det H(x,t,\theta )=1,H(x,t,\theta )\to {\rm{I}}, & \theta \to \infty .\end{array}\end{eqnarray}$
Let $u(x,t)\in {\mathbb{S}}$, on curve ${\bar{{\rm{\Omega }}}}_{i},i=1,\ldots ,4$, the function $H(x,t,\theta )$ defined by equation (
$\begin{eqnarray}\begin{array}{l}{H}_{+}(x,t,\theta )={H}_{-}(x,t,\theta )L(x,t,\theta ),\\ \theta \in {\rm{\Pi }}=\{\theta \in {\mathbb{C}}| \mathrm{Im}{\theta }^{5}=0\},\end{array}\end{eqnarray}$
where $\begin{eqnarray}L(x,t,\theta )=\left\{\begin{array}{l}{L}_{1}(x,t,\theta ),\quad \theta \in {\mathbb{R}},\\ {L}_{2}(x,t,\theta ),\quad \theta \in \partial {{\rm{\Omega }}}_{2},\\ {L}_{3}(x,t,\theta ),\quad \theta \in \partial {{\rm{\Omega }}}_{4},\end{array}\right.\end{eqnarray}$
with $\begin{eqnarray*}\begin{array}{l}{L}_{1}(x,t,\theta )=\left(\begin{array}{cc}1 & -\overline{s(\bar{\theta })}{{\rm{e}}}^{-2{\rm{i}}\mu (\theta )}\\ s(\theta ){{\rm{e}}}^{2{\rm{i}}\mu (\theta )} & 1-s(\theta )\overline{s(\bar{\theta })}\end{array}\right),\\ {L}_{2}(x,t,\theta )=\left(\begin{array}{cc}1 & \overline{S(\bar{\theta })}{{\rm{e}}}^{-2{\rm{i}}\mu (\theta )}\\ 0 & 1\end{array}\right),\\ {L}_{3}(x,t,\theta )=\left(\begin{array}{cc}1 & 0\\ S(\theta ){{\rm{e}}}^{2{\rm{i}}\mu (\theta )} & 1\end{array}\right).\end{array}\end{eqnarray*}$
The contour for this RH problem is shown in figure 3.
Figure 3. The contour for the RH problem on the complex θ-plane. |
Proof. It follows from the equations (2.14a )–(2.14b ) and equation (2.23 ) that2.29a )–(2.30b ) and equation (2.24 ) show that2.25a )–(2.25d ) and equation (2.27 ), we have2.32a )–(2.32c ) give rise to the jump matrices ${\{{L}_{j}(x,t,\theta )\}}_{1}^{3}$ defined by equation (2.28 ).
$\begin{eqnarray}\begin{array}{l}\overline{y(\bar{\theta })}{[{{\rm{\Phi }}}_{1}]}_{1}^{{{\rm{\Omega }}}_{4}}(x,t,\theta )-\overline{z(\bar{\theta })}{{\rm{e}}}^{2{\rm{i}}\mu (\theta )}\\ \times \,{[{{\rm{\Phi }}}_{1}]}_{2}^{{{\rm{\Omega }}}_{2}}(x,t,\theta )={[{{\rm{\Phi }}}_{3}]}_{1}^{{{\mathbb{C}}}_{+}}(x,t,\theta ),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}z(\theta ){{\rm{e}}}^{-2{\rm{i}}\mu (\theta )}{[{{\rm{\Phi }}}_{1}]}_{1}^{{{\rm{\Omega }}}_{4}}(x,t,\theta )+y(\theta )\\ \times \,{[{{\rm{\Phi }}}_{1}]}_{2}^{{{\rm{\Omega }}}_{2}}(x,t,\theta )={[{{\rm{\Phi }}}_{3}]}_{2}^{{{\mathbb{C}}}_{-}}(x,t,\theta ),\end{array}\end{eqnarray}$
and $\begin{eqnarray}\begin{array}{l}\overline{Y(\bar{\theta })}{\left[{{\rm{\Phi }}}_{1}\right]}_{1}^{{{\rm{\Omega }}}_{4}}(x,t,\theta )-\overline{Z(\bar{\theta })}{{\rm{e}}}^{2{\rm{i}}\mu (\theta )}\\ \times \,[{{\rm{\Phi }}}_{1}{]}_{2}^{{{\rm{\Omega }}}_{2}}(x,t,\theta )={\left[{{\rm{\Phi }}}_{2}\right]}_{1}^{{{\rm{\Omega }}}_{3}}(x,t,\theta ),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}Z(\theta ){{\rm{e}}}^{-2{\rm{i}}\mu (\theta )}{\left[{{\rm{\Phi }}}_{1}\right]}_{1}^{{{\rm{\Omega }}}_{4}}(x,t,\theta )+Y(\theta )\\ \times \,[{{\rm{\Phi }}}_{1}{]}_{2}^{{{\rm{\Omega }}}_{2}}(x,t,\theta )={\left[{{\rm{\Phi }}}_{2}\right]}_{2}^{{{\rm{\Omega }}}_{1}}(x,t,\theta ),\end{array}\end{eqnarray}$
then, the equations ( $\begin{eqnarray}\begin{array}{l}\alpha (\theta ){\left[{{\rm{\Phi }}}_{3}\right]}_{1}^{{{\mathbb{C}}}_{+}}(x,t,\theta )-\overline{\beta (\bar{\theta })}{{\rm{e}}}^{2{\rm{i}}\mu (\theta )}{\left[{{\rm{\Phi }}}_{3}\right]}_{2}^{{{\mathbb{C}}}_{-}}(x,t,\theta )\\ \,=\,{\left[{{\rm{\Phi }}}_{2}\right]}_{1}^{{{\rm{\Omega }}}_{3}}(x,t,\theta ),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}-\beta (\theta ){{\rm{e}}}^{-2{\rm{i}}\mu (\theta )}{\left[{{\rm{\Phi }}}_{3}\right]}_{1}^{{{\mathbb{C}}}_{+}}(x,t,\theta )+\overline{\alpha (\bar{\theta })}{\left[{{\rm{\Phi }}}_{3}\right]}_{2}^{{{\mathbb{C}}}_{-}}(x,t,\theta )\\ \,=\,{\left[{{\rm{\Phi }}}_{2}\right]}_{2}^{{{\rm{\Omega }}}_{1}}(x,t,\theta ).\end{array}\end{eqnarray}$
By equations ( $\begin{eqnarray}\begin{array}{l}\left({\left[{{\rm{\Phi }}}_{3}\right]}_{1}^{{{\mathbb{C}}}_{+}}(x,t,\theta ),\displaystyle \frac{{\left[{{\rm{\Phi }}}_{2}\right]}_{2}^{{{\rm{\Omega }}}_{1}}(x,t,\theta )}{\overline{\alpha (\bar{\theta })}}\right)\\ =\left(\displaystyle \frac{{\left[{{\rm{\Phi }}}_{2}\right]}_{1}^{{{\rm{\Omega }}}_{3}}(x,t,\theta )}{\alpha (\theta )},{\left[{{\rm{\Phi }}}_{3}\right]}_{2}^{{{\mathbb{C}}}_{-}}(x,t,\theta )\right){L}_{1}(x,t,\theta ),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}\left({\left[{{\rm{\Phi }}}_{3}\right]}_{1}^{{{\mathbb{C}}}_{+}}(x,t,\theta ),\displaystyle \frac{{\left[{{\rm{\Phi }}}_{2}\right]}_{2}^{{{\rm{\Omega }}}_{1}}(x,t,\theta )}{\overline{\alpha (\bar{\theta })}}\right)\\ =\left({\left[{{\rm{\Phi }}}_{3}\right]}_{1}^{{{\mathbb{C}}}_{+}}(x,t,\theta ),\displaystyle \frac{{\left[{{\rm{\Phi }}}_{1}\right]}_{2}^{{{\rm{\Omega }}}_{2}}(x,t,\theta )}{\overline{y(\bar{\theta })}}\right){L}_{2}(x,t,\theta ),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}\left(\displaystyle \frac{{\left[{{\rm{\Phi }}}_{1}\right]}_{1}^{{{\rm{\Omega }}}_{4}}(x,t,\theta )}{y(\theta )},{\left[{{\rm{\Phi }}}_{3}\right]}_{2}^{{{\mathbb{C}}}_{-}}(x,t,\theta )\right)\\ =\left(\displaystyle \frac{{\left[{{\rm{\Phi }}}_{2}\right]}_{1}^{{{\rm{\Omega }}}_{3}}(x,t,\theta )}{\alpha (\theta )},{\left[{{\rm{\Phi }}}_{3}\right]}_{2}^{{{\mathbb{C}}}_{-}}(x,t,\theta )\right){L}_{3}(x,t,\theta ).\end{array}\end{eqnarray}$
Therefore, the equations (Furthermore, we can get the jump relation between ${H}_{+}^{(2)}(x,t,\theta )$ and ${H}_{-}^{(1)}(x,t,\theta )$ as follows
$\begin{eqnarray*}\begin{array}{l}{H}_{+}^{(2)}(x,t,\theta )={H}_{-}^{(1)}(x,t,\theta ){L}_{4}(x,t,\theta )\\ \,=\,{H}_{-}^{(1)}({L}_{3}^{-1}{L}_{1}{L}_{2}^{-1})(x,t,\theta ),\end{array}\end{eqnarray*}$
where $\begin{eqnarray*}{L}_{4}(x,t,\theta )=\left(\begin{array}{cc}1 & -\displaystyle \frac{z(\theta )}{\overline{y(\bar{\theta })}}{{\rm{e}}}^{-2{\rm{i}}\mu (\theta )}\\ -\displaystyle \frac{\overline{z(\bar{\theta })}}{y(\theta )}{{\rm{e}}}^{2{\rm{i}}\mu (\theta )} & \displaystyle \frac{1}{y(\theta )\overline{y(\bar{\theta })}}\end{array}\right).\end{eqnarray*}$
Suppose that
| • | $y(\theta )$ enjoys λ simple zeros ${\{{\gamma }_{k}\}}_{k=1}^{\lambda }$, $\lambda =2{\lambda }_{1}$, as well as ${\{{\gamma }_{k}\}}_{1}^{\lambda }\in {{\rm{\Omega }}}_{4}$, and ${\{{\bar{\gamma }}_{k}\}}_{1}^{\lambda }\in {{\rm{\Omega }}}_{2}$. |
| • | $\alpha (\theta )$ enjoys ξ simple zeros ${\{{\delta }_{k}\}}_{k=1}^{\xi }$, $\xi =2{\xi }_{1}+{\xi }_{2}$, as well as ${\{{\delta }_{k}\}}_{1}^{\xi }\in {{\rm{\Omega }}}_{3}$, and ${\{{\bar{\delta }}_{k}\}}_{1}^{\xi }\in {{\rm{\Omega }}}_{1}$. |
| • | None of the zeros of $y(\theta )$ coincides with a zero of $\alpha (\theta )$. |
(The residue formula) Let $\dot{y}(\theta )=\tfrac{{dy}}{d\theta }$, it holds that the following residue formulae:
$\begin{eqnarray}\begin{array}{l}\mathrm{Res}\{{\left[H(x,t,\theta )\right]}_{1},{\gamma }_{k}\}=\displaystyle \frac{1}{z({\gamma }_{k})\dot{y}({\gamma }_{k})}{{\rm{e}}}^{2{\rm{i}}\mu ({\gamma }_{k})}{\left[H(x,t,{\gamma }_{k})\right]}_{2},\\ k=1,\cdots ,2{\lambda }_{1}.\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}\mathrm{Res}\{{\left[H(x,t,\theta )\right]}_{2},{\bar{\gamma }}_{k}\}=-\displaystyle \frac{1}{\overline{z({\gamma }_{k})}\overline{\dot{y}({\gamma }_{k})}}{{\rm{e}}}^{-2{\rm{i}}\mu ({\bar{\gamma }}_{k})}\\ \,\times \,{\left[H(x,t,{\bar{\gamma }}_{k})\right]}_{1},k=1,\cdots ,2{\lambda }_{1}.\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}\mathrm{Res}\{{\left[H(x,t,\theta )\right]}_{1},{\delta }_{k}\}=-\displaystyle \frac{\overline{Z({\bar{\delta }}_{k})}}{y({\delta }_{k})\dot{\alpha }({\delta }_{k})}{{\rm{e}}}^{2{\rm{i}}\mu ({\delta }_{k})}{\left[H(x,t,{\delta }_{k})\right]}_{1},\\ k=1,\cdots ,\xi .\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}\mathrm{Res}\{{\left[H(x,t,\theta )\right]}_{2},{\bar{\delta }}_{k}\}=\displaystyle \frac{Z({\bar{\delta }}_{k})}{\overline{y({\delta }_{k})}\overline{\dot{\alpha }({\delta }_{k})}}{{\rm{e}}}^{-2{\rm{i}}\mu ({\bar{\delta }}_{k})}{\left[H(x,t,{\bar{\delta }}_{k})\right]}_{2},\\ k=1,\cdots ,\xi .\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}\mathrm{Res}\left\{\displaystyle \frac{{{\rm{\Phi }}}_{1}^{{{\rm{\Omega }}}_{4}}(x,t,\theta )}{y(\theta )},{\gamma }_{k}\right\}=\mathop{\mathrm{lim}}\limits_{\theta \to {\gamma }_{k}}(\theta -{\gamma }_{k})\displaystyle \frac{{\left[{{\rm{\Phi }}}_{1}\right]}_{1}^{{{\rm{\Omega }}}_{4}}(x,t,\theta )}{y(\theta )}\\ \,=\,\displaystyle \frac{{\left[{{\rm{\Phi }}}_{1}\right]}_{1}^{{{\rm{\Omega }}}_{4}}(x,t,{\gamma }_{k})}{\dot{y}({\gamma }_{k})},\end{array}\end{eqnarray}$
taking θ = γk into the equation ( $\begin{eqnarray}{\left[{{\rm{\Phi }}}_{1}\right]}_{1}^{{{\rm{\Omega }}}_{4}}(x,t,{\gamma }_{k})=\displaystyle \frac{1}{z({\gamma }_{k})}{{\rm{e}}}^{2{\rm{i}}\mu ({\gamma }_{k})}{\left[{{\rm{\Phi }}}_{3}\right]}_{2}^{{{\mathbb{C}}}_{-}}(x,t,{\gamma }_{k}),\end{eqnarray}$
thus, equation ( $\begin{eqnarray}\begin{array}{l}\mathrm{Res}\left\{\displaystyle \frac{{\left[{{\rm{\Phi }}}_{1}\right]}_{1}^{{{\rm{\Omega }}}_{4}}(x,t,\theta )}{y(\theta )},{\gamma }_{k}\right\}\\ =\displaystyle \frac{1}{z({\gamma }_{k})\dot{y}({\gamma }_{k})}{{\rm{e}}}^{2{\rm{i}}\mu ({\gamma }_{k})}{\left[{{\rm{\Phi }}}_{3}\right]}_{2}^{{{\mathbb{C}}}_{-}}(x,t,{\gamma }_{k}),\end{array}\end{eqnarray}$
which can lead to equation (2.5. The global relation
In this subsection, we show that y(θ), z(θ), Y(θ), Z(θ) are not independent but admit a significant relationship. In fact, the integral of the 1-form A(ζ, τ, θ) defined by the equation (2.7 ) is vanished for the boundary of the region (ζ, τ): 0 < ζ < ∞ , 0 < τ < t. Let Φ(ζ, τ, θ) = Φ3(ζ, τ, θ) in the 1-form A(ζ, τ, θ), we have2.18b ) mean that2.14a ) at x = 0 gives the expressions2.39 ) and equation (2.17a ) imply that2.37 ) becomes2.40 ) is valid for $\theta \in {{\mathbb{C}}}_{-}$, the second column of equation (2.40 ) is valid for $\theta \in {{\mathbb{C}}}_{-}$ and2.40 ) becomes2.41 ) is2.42 ) and equation (2.43 ) are the so-called global relation.
$\begin{eqnarray}\begin{array}{l}{\displaystyle \int }_{\infty }^{0}{{\rm{e}}}^{-{\rm{i}}\theta \zeta {\hat{\sigma }}_{3}}(U{{\rm{\Phi }}}_{3})(\zeta ,0,\theta ){\rm{d}}\zeta \\ +{\displaystyle \int }_{0}^{t}{{\rm{e}}}^{16{\rm{i}}\kappa {\theta }^{5}\tau {\hat{\sigma }}_{3}}(V{{\rm{\Phi }}}_{3})(0,\tau ,\theta ){\rm{d}}\tau \\ +{{\rm{e}}}^{16{\rm{i}}\kappa {\theta }^{5}t{\hat{\sigma }}_{3}}{\displaystyle \int }_{0}^{\infty }{{\rm{e}}}^{-{\rm{i}}\theta \zeta {\hat{\sigma }}_{3}}(U{{\rm{\Phi }}}_{3})(\zeta ,t,\theta ){\rm{d}}\zeta \\ =\mathop{\mathrm{lim}}\limits_{x\to \infty }{{\rm{e}}}^{-{\rm{i}}\theta x{\hat{\sigma }}_{3}}{\displaystyle \int }_{0}^{t}{{\rm{e}}}^{16{\rm{i}}\kappa {\theta }^{5}\tau {\hat{\sigma }}_{3}}(V{{\rm{\Phi }}}_{3})(x,\tau ,\theta ){\rm{d}}\tau .\end{array}\end{eqnarray}$
Since f(θ) = Φ3(0, 0, θ), equation ( $\begin{eqnarray*}{\int }_{\infty }^{0}{{\rm{e}}}^{-{\rm{i}}\theta \zeta {\hat{\sigma }}_{3}}(U{{\rm{\Phi }}}_{3})(\zeta ,0,\theta ){\rm{d}}\zeta =f(\theta )-{\rm{I}}.\end{eqnarray*}$
Evaluation of equation ( $\begin{eqnarray}{{\rm{\Phi }}}_{3}(0,\tau ,\theta )={{\rm{\Phi }}}_{1}(0,\tau ,\theta ){{\rm{e}}}^{-16{\rm{i}}\kappa {\theta }^{5}\tau {\hat{\sigma }}_{3}}f(\theta ),\end{eqnarray}$
as well as $\begin{eqnarray}{{\rm{e}}}^{16{\rm{i}}\kappa {\theta }^{5}\tau {\hat{\sigma }}_{3}}(V{{\rm{\Phi }}}_{3})(0,\tau ,\theta )=[{{\rm{e}}}^{16{\rm{i}}\kappa {\theta }^{5}\tau {\hat{\sigma }}_{3}}(V{{\rm{\Phi }}}_{1})(0,\tau ,\theta )]f(\theta ).\end{eqnarray}$
Furthermore, equation ( $\begin{eqnarray*}\begin{array}{l}{\displaystyle \int }_{0}^{t}{{\rm{e}}}^{16{\rm{i}}\kappa {\theta }^{5}\tau {\hat{\sigma }}_{3}}(V{{\rm{\Phi }}}_{3})(0,\tau ,\theta ){\rm{d}}\tau \\ =[{{\rm{e}}}^{16{\rm{i}}\kappa {\theta }^{5}t{\hat{\sigma }}_{3}}{{\rm{\Phi }}}_{1}(0,t,\theta )-{\rm{I}}]f(\theta ).\end{array}\end{eqnarray*}$
Let $u(x,t)\in {\mathbb{S}}$ for x → ∞ , we find that equation ( $\begin{eqnarray}\begin{array}{l}{g}^{-1}(t,\theta )f(\theta )+{{\rm{e}}}^{16{\rm{i}}\kappa {\theta }^{5}t{\hat{\sigma }}_{3}}\\ \times {\displaystyle \int }_{0}^{\infty }{{\rm{e}}}^{-{\rm{i}}\theta \zeta {\hat{\sigma }}_{3}}(U{{\rm{\Phi }}}_{3})(\zeta ,t,\theta ){\rm{d}}\zeta ={\rm{I}},\end{array}\end{eqnarray}$
where the first column of equation ( $\begin{eqnarray*}{g}^{-1}(t,\theta )={{\rm{e}}}^{16{\rm{i}}\kappa {\theta }^{5}t{\hat{\sigma }}_{3}}{{\rm{\Phi }}}_{1}(0,t,\theta ).\end{eqnarray*}$
Due to g(θ) = g(T, θ) and denoting t = T, equation ( $\begin{eqnarray}\begin{array}{l}{g}^{-1}(\theta )f(\theta )+{{\rm{e}}}^{16{\rm{i}}\kappa {\theta }^{5}T{\hat{\sigma }}_{3}}\\ \times {\displaystyle \int }_{0}^{\infty }{{\rm{e}}}^{-{\rm{i}}\theta \zeta {\hat{\sigma }}_{3}}(U{{\rm{\Phi }}}_{3})(\zeta ,T,\theta ){\rm{d}}\zeta ={\rm{I}}.\end{array}\end{eqnarray}$
Hence, the (12)-component of equation ( $\begin{eqnarray}y(\theta )Z(\theta )-Y(\theta )z(\theta )={{\rm{e}}}^{32{\rm{i}}\kappa {\theta }^{5}T}J(\theta ),\end{eqnarray}$
where $\begin{eqnarray}J(\theta )={\int }_{0}^{\infty }{{\rm{e}}}^{-2{\rm{i}}\theta \zeta }{\left(U{{\rm{\Phi }}}_{3}\right)}_{12}(\zeta ,T,\theta ){\rm{d}}\zeta ,\end{eqnarray}$
Indeed, equation (3. The functions y(θ), z(θ) and Y(θ), Z(θ)
($y(\theta )$ and $z(\theta )$) Given ${u}_{0}(x)=u(x,0)\in {\mathbb{S}}$, one defines the mapping
$\begin{eqnarray*}{\varphi }_{1}:\{{u}_{0}(x)\}\to \{y(\theta ),z(\theta )\},\end{eqnarray*}$
by $\begin{eqnarray*}{\left(z(\theta ),y(\theta )\right)}^{T}={[{{\rm{\Phi }}}_{3}]}_{2}^{{{\mathbb{C}}}_{-}}(x,0,\theta ),\end{eqnarray*}$
where ${{\rm{\Phi }}}_{3}(x,0,\theta )$ is the unique solution of the following Volterra integral equation $\begin{eqnarray*}{{\rm{\Phi }}}_{3}(x,0,\theta )={\rm{I}}-{\int }_{x}^{\infty }{{\rm{e}}}^{-{\rm{i}}\theta (\zeta -x){\hat{\sigma }}_{3}}(U{{\rm{\Phi }}}_{3})(\zeta ,0,\theta ){\rm{d}}\zeta ,\end{eqnarray*}$
where $U(x,0,\theta )$ is given in terms of ${u}_{0}(t)$ expressed by equation (The $y(\theta )$ and $z(\theta )$ possess the following properties
| i | (i) $y(\theta ),z(\theta )$ are analytical for $\mathrm{Im}\theta \gt 0$ as well as bounded and continuous for $\mathrm{Im}\theta \geqslant 0$. |
| ii | (ii) $y(\theta )=1+O\left(\tfrac{1}{\theta }\right),z(\theta )=O\left(\tfrac{1}{\theta }\right)$, $\theta \to \infty $, $\mathrm{Im}\theta \geqslant 0$. |
| iii | (iii) $y(\theta )\overline{y(\bar{\theta })}+z(\theta )\overline{z(\bar{\theta })}=1$, $\theta \in {\mathbb{R}}$. |
| iv | (iv) $y(-\theta )=\overline{y(\bar{\theta })},z(-\theta )=-\overline{z(\bar{\theta })}$, $\mathrm{Im}\theta \geqslant 0$. |
| v | (v)The mapping ${\varphi }_{1}^{-1}={\phi }_{1}:\{y(\theta ),z(\theta )\}\to \{{u}_{0}(x)\}$, inverse to ${\varphi }_{1}$, is defined by $\begin{eqnarray*}{u}_{0}(x)=-2{\rm{i}}\mathop{\mathrm{lim}}\limits_{\theta \to \infty }{\left(\theta {H}^{(x)}(x,\theta )\right)}_{12},\end{eqnarray*}$ where ${H}^{(x)}(x,\theta )$ is the unique solution to the RH problem as follows |
| • | ${H}^{(x)}(x,\theta )=\left\{\begin{array}{l}{H}_{-}^{(x)}(x,\theta ),\mathrm{Im}\theta \leqslant 0,\\ {H}_{+}^{(x)}(x,\theta ),\mathrm{Im}\theta \geqslant 0,\end{array}\right.$ is a sectionally analytical function. |
| • | ${H}_{+}^{(x)}(x,\theta )={H}_{-}^{(x)}(x,\theta ){L}^{(x)}(x,\theta )$, $\theta \in {\mathbb{R}}$, where $\begin{eqnarray}{L}^{(x)}(x,\theta )=\left(\begin{array}{cc}1 & -\displaystyle \frac{z(\theta )}{\overline{y(\bar{\theta })}}{{\rm{e}}}^{-2{\rm{i}}\theta x}\\ -\displaystyle \frac{\overline{z(\bar{\theta })}}{y(\theta )}{{\rm{e}}}^{2{\rm{i}}\theta x} & \displaystyle \frac{1}{y(\theta )\overline{y(\bar{\theta })}}\end{array}\right).\end{eqnarray}$ |
| • | ${H}^{(x)}(x,\theta )={\rm{I}}+O(\tfrac{1}{\theta }),\theta \to \infty .$ |
| • | y(θ) has λ simple zeros ${\{{\gamma }_{k}\}}_{1}^{\lambda }$, λ = 2λ1, such that $\mathrm{Im}{\gamma }_{k}\lt 0,k=1,2,\cdots ,\lambda $, where ${\{{\gamma }_{k}\}}_{1}^{2{\lambda }_{1}}\in {{\rm{\Omega }}}_{4}$. |
| • | The first column of ${H}_{+}^{(x)}(x,\theta )$ possesses simple poles at $\theta ={\{{\bar{\gamma }}_{k}\}}_{1}^{2{\lambda }_{1}}$. The second column of ${H}_{-}^{(x)}(x,\theta )$ possesses simple poles at $\theta ={\{{\gamma }_{k}\}}_{1}^{2{\lambda }_{1}}$. The associated residues expression are $\begin{eqnarray}\begin{array}{l}\mathrm{Res}\{{\left[{H}^{(x)}(x,\theta )\right]}_{1},{\gamma }_{k}\}=\displaystyle \frac{{{\rm{e}}}^{2{\rm{i}}{\gamma }_{k}x}}{\dot{y}({\gamma }_{k})z({\gamma }_{k})}{\left[{H}^{(x)}(x,{\gamma }_{k})\right]}_{2},\\ k=1,2,\cdots ,2{\lambda }_{1},\end{array}\end{eqnarray}$ $\begin{eqnarray}\begin{array}{l}\mathrm{Res}\{{\left[{H}^{(x)}(x,\theta )\right]}_{2},{\bar{\gamma }}_{k}\}=-\displaystyle \frac{{{\rm{e}}}^{-2{\rm{i}}{\overline{\gamma }}_{k}x}}{\overline{\dot{y}({\gamma }_{k})}\overline{z({\gamma }_{k})}}{\left[{H}^{(x)}(x,{\bar{\gamma }}_{k})\right]}_{1},\\ k=1,2,\cdots ,2{\lambda }_{1}.\end{array}\end{eqnarray}$ |
($Y(\theta )$ and $Z(\theta )$). Let ${v}_{0}(t),{\{{v}_{k}(t)\}}_{1}^{4}\in {\mathbb{S}}$, the mapping
$\begin{eqnarray*}{\varphi }_{2}:\{{v}_{0}(t),{v}_{1}(t),{v}_{2}(t),{v}_{3}(t),{v}_{4}(t)\}\to \{Y(\theta ),Z(\theta )\},\end{eqnarray*}$
in terms of $\begin{eqnarray*}{\left(Z(\theta ),Y(\theta )\right)}^{T}={[{{\rm{\Phi }}}_{1}]}_{2}^{{{\rm{\Omega }}}_{2}}(0,t,\theta ),\end{eqnarray*}$
where ${{\rm{\Phi }}}_{1}(0,t,\theta )$ is the unique solution of the following Volterra integral equation $\begin{eqnarray*}{{\rm{\Phi }}}_{1}(0,t,\theta )={\rm{I}}-{\int }_{t}^{T}{{\rm{e}}}^{16{\rm{i}}\kappa {\theta }^{5}(\tau -t){\hat{\sigma }}_{3}}(V{{\rm{\Phi }}}_{1})(0,\tau ,\theta ){\rm{d}}\tau ,\end{eqnarray*}$
and $V(0,t,\theta )$ is given in terms of $\{{v}_{0}(t),{v}_{1}(t),{v}_{2}(t),{v}_{3}(t),{v}_{4}(t)\}$ expressed by equation (The $Y(\theta )$ and $Z(\theta )$ possess the following properties
| i | (i) $Y(\theta ),Z(\theta )$ are bounded for $\mathrm{Im}\kappa {\theta }^{5}\geqslant 0$, if $T=\infty $, as well as the $Y(\theta ),Z(\theta )$ are defined only for $\mathrm{Im}\kappa {\theta }^{5}\geqslant 0$. |
| ii | (ii) $Y(\theta )=1\,+\,$ $O\left(\tfrac{1}{\theta }\right),Z(\theta )$ $=O\left(\tfrac{1}{\theta }\right)$, $\theta \to \infty $, $\mathrm{Im}\kappa {\theta }^{5}\geqslant 0$. |
| iii | (iii) $Y(\theta )\overline{Y(\bar{\theta })}+Z(\theta )\overline{Z(\bar{\theta })}=1$, $\theta \in {\mathbb{C}}(\kappa {\theta }^{5}\in {\mathbb{R}},$ if $T=\infty )$. |
| iv | (iv) $Y(-\theta )=\overline{Y(\bar{\theta })},Z(-\theta )=-\overline{Z(\bar{\theta })}$, $\mathrm{Im}\kappa {\theta }^{5}\geqslant 0$. |
| v | (v)The mapping ${\varphi }_{2}^{-1}={\phi }_{2}:\{Y(\theta ),Z(\theta )\}\to \{{v}_{0}(t),{v}_{1}(t),{v}_{2}(t),{v}_{3}(t),{v}_{4}(t)\}$, inverse to ${\varphi }_{2}$, is defined by $\begin{eqnarray}\begin{array}{rcl}{v}_{0}(t) & = & {v}_{0}(t)=-2{\rm{i}}{\phi }_{12}^{(1)}(t),\\ {v}_{1}(t) & = & 4{\phi }_{12}^{(2)}(t)-2{\rm{i}}{v}_{0}(t){\phi }_{22}^{(1)}(t)+{v}_{0}^{2}(t),\\ {v}_{2}(t) & = & 8{\rm{i}}{\phi }_{12}^{(3)}(t)+4{v}_{0}(t){\phi }_{22}^{(2)}(t)-2{\rm{i}}{v}_{1}(t){\phi }_{22}^{(1)}(t)-{v}_{0}^{3}(t),\\ {v}_{3}(t) & = & -16{\phi }_{12}^{(4)}(t)+8{\rm{i}}{v}_{0}(t){\phi }_{22}^{(3)}(t)+4{v}_{1}(t){\phi }_{22}^{(2)}(t)\\ & & -2{\rm{i}}({v}_{2}(t)+{v}_{0}^{3}(t)){\phi }_{22}^{(1)}(t)-5{v}_{0}^{2}(t){v}_{1}(t),\\ {v}_{4}(t) & = & -32{\rm{i}}{\phi }_{12}^{(5)}(t)-16{v}_{0}(t){\phi }_{22}^{(4)}(t)+8{\rm{i}}{v}_{0}(t){\phi }_{22}^{(3)}(t)\\ & & +\,4({v}_{2}(t)+{v}_{0}^{3}(t)){\phi }_{22}^{(2)}(t)-2{\rm{i}}(5{v}_{0}^{2}(t){v}_{1}(t)\\ & & +\,{v}_{3}(t)){\phi }_{22}^{(1)}(t)\\ & & amp;-(10{v}_{0}^{2}(t){v}_{1}(t)+10{v}_{0}(t){v}_{1}^{2}(t)+6{v}_{0}^{5}(t)),\end{array}\end{eqnarray}$ where the functions ${\phi }^{(j)}(t),j\,=\,1,2,3,4,5$ are determined by $\begin{eqnarray*}{H}^{(t)}(t,\theta )={\rm{I}}+\displaystyle \sum _{j=1}^{5}\displaystyle \frac{{\phi }^{(j)}(t)}{{\theta }^{j}}+O\left(\displaystyle \frac{1}{{\theta }^{6}}\right),\quad \theta \to \infty ,\end{eqnarray*}$ where ${H}^{(t)}(t,\theta )$ is the unique solution to the RH problem as follows |
| • | ${H}^{(t)}(t,\theta )=\left\{\begin{array}{cc}{H}_{-}^{(t)}(t,\theta ), & \mathrm{Im}\kappa {\theta }^{5}\leqslant 0,\\ {H}_{+}^{(t)}(t,\theta ), & \mathrm{Im}\kappa {\theta }^{5}\geqslant 0,\end{array}\right.$ is a sectionally analytical function. |
| • | ${H}_{-}^{(t)}(t,\theta )={H}_{+}^{(t)}(t,\theta ){L}^{(t)}(t,\theta )$, ${\theta }^{5}\in {\mathbb{R}}$, where $\begin{eqnarray}{H}^{(t)}(t,\theta )=\left(\begin{array}{cc}1 & -\displaystyle \frac{Z(\theta )}{\overline{Y(\overline{\theta })}}{{\rm{e}}}^{-32{\rm{i}}\kappa {\theta }^{5}t}\\ -\displaystyle \frac{\overline{Z(\overline{\theta })}}{Y(\theta )}{{\rm{e}}}^{32{\rm{i}}\kappa {\theta }^{5}t} & \displaystyle \frac{1}{Y(\theta )\overline{Y(\overline{\theta })}}\end{array}\right).\end{eqnarray}$ |
| • | ${H}^{(t)}(t,\theta )={\rm{I}}+O\left(\tfrac{1}{\theta }\right),\theta \to \infty .$ |
| • | Y(θ) has 2m simple zeros ${\{{\omega }_{k}\}}_{1}^{2m}$ such that $\mathrm{Im}{\omega }_{k}^{5}\gt 0,k=1,2,\cdots ,2m$. |
| • | The first column of ${H}_{+}^{(t)}(t,\theta )$ possesses simple poles at $\theta ={\{{\bar{\omega }}_{k}\}}_{1}^{2m}$, the second column of ${H}_{-}^{(t)}(t,\theta )$ possesses simple poles at $\theta ={\{{\omega }_{k}\}}_{1}^{2m}$. The associated residues expression are $\begin{eqnarray}\begin{array}{l}\mathrm{Res}\{{\left[{H}^{(t)}(t,\theta )\right]}_{1},{\omega }_{k}\}=\displaystyle \frac{{{\rm{e}}}^{32{\rm{i}}\kappa {\omega }_{k}^{5}t}}{\dot{Y}({\omega }_{k})Z({\omega }_{k})}{\left[{H}^{(t)}(t,{\omega }_{k})\right]}_{2},\\ k=1,2,\cdots ,2m,\end{array}\end{eqnarray}$ $\begin{eqnarray}\begin{array}{l}\mathrm{Res}\{{\left[{H}^{(t)}(t,\theta )\right]}_{2},{\bar{\omega }}_{k}\}=-\displaystyle \frac{{{\rm{e}}}^{-32{\rm{i}}\kappa {\overline{\omega }}_{k}^{5}t}}{\overline{\dot{Y}({\bar{\omega }}_{j})}\overline{Z({\bar{\omega }}_{k})}}{\left[{H}^{(t)}(t,{\bar{\omega }}_{k})\right]}_{1},\\ k=1,2,\cdots ,2m.\end{array}\end{eqnarray}$ |
4. The RH problem
Given ${u}_{0}(x)\in {\mathbb{S}}({{\rm{R}}}^{+})$, we define two pairs of spectral functions $y(\theta ),z(\theta )$ and $Y(\theta ),Z(\theta )$ by ${u}_{0}(x),{v}_{0}(t)$ and ${\{{v}_{k}(t)\}}_{1}^{4}$ according to definitions 3.1 and definitions 3.3, as well as the matrix-value functions $f(\theta )$ and $g(\theta )$ are defined by equation (
| • | $H(x,t,\theta )$ is a sectionally analytical function for $\theta \in {\mathbb{C}}\setminus \{{\theta }^{5}\in {\mathbb{R}}\}$. |
| • | $H(x,t,\theta )$ has the jump relation as in theorem 2.4, i.e $\begin{eqnarray*}{H}_{+}(x,t,\theta )={H}_{-}(x,t,\theta )L(x,t,\theta ),\quad {\theta }^{5}\in {\mathbb{R}},\end{eqnarray*}$ where ${H}_{\pm }(x,t,\theta )$ are defined by equations ( |
| • | $H(x,t,\theta )={\rm{I}}+{\rm{O}}\left(\tfrac{1}{\theta }\right),\theta \to \infty $. |
| • | $H(x,t,\theta )$ associated residues satisfy the relations in proposition |
$\begin{eqnarray}u(x,t)=-2{\rm{i}}\mathop{\mathrm{lim}}\limits_{\theta \to \infty }{\left(\theta H(x,t,\theta )\right)}_{12}.\end{eqnarray}$
Furthermore, $u(x,0)={u}_{0}(x)$, $u(0,t)={v}_{0}(t)$, ${u}_{x}(0,t)\,={v}_{1}(t),$ ${u}_{{xx}}(0,t)={v}_{2}(t)$, ${u}_{{xxx}}(0,t)={v}_{3}(t),$ ${u}_{{xxxx}}(0,t)={v}_{4}(t).$Proof. Firstly, it is not difficult to find that equation (4.1 ) can be obtained by the large θ asymptotic property of the characteristic functions in the appendix . Secondly, when y(θ) and α(θ) have no zeros, H(x, t, θ) can be reduced to a non-regular RH problem, so we can get that H(x, t, θ) is unique from the following Vanishing Lemma. Finally, when y(θ) and α(θ) have zeros, H(x, t, θ) is a regular RH problem, which can be mapped to an algebraic equation without zeros, and the uniqueness can be obtained by solving this algebraic equation.
Therefore, we have the Vanishing lemma as follows
(Vanishing Lemma) Assume that $H(x,t,\theta )\to 0$ as $\theta \to \infty $, then the 2 × 2 matrix RH problem in theorem 4.1 has only the zero solution.
Proof Assume that H(x, t, θ) is the solution to the RH problem given by Theorem 4.1 , and H±(x, t, θ) → 0 when θ → ∞ . Define2.3 and equation (2.28 ), we have4.3 ) and equation (4.4 ) mean that W+(θ) = W−(θ). Thus, W+(θ) and W−(θ) are defined by an entire function vanishing at infinity, and W±(θ) = 0. Since ${L}_{1}({\rm{i}}\varepsilon )(\varepsilon \in {\mathbb{R}})$ is a Hermitian matrix with unit determinant and (2, 2) entry 1 for any $\varepsilon \in {\mathbb{R}}$. Therefore, ${L}_{1}({\rm{i}}\varepsilon )(\varepsilon \in {\mathbb{R}})$ is a positive definite matrix, and because W+(ϵ) vanishes identically for $\varepsilon \in {\rm{i}}{\mathbb{R}}$, this shows that4.5 ) means H−(iϵ) = 0 for $\varepsilon \in {\mathbb{R}}$. Thus, H+(θ) and H−(θ) are vanish identically.
$\begin{eqnarray}\begin{array}{rcl}{W}_{+}(\theta ) & = & {H}_{+}(\theta ){H}_{-}^{\dagger }(-\bar{\theta }),\mathrm{Im}\kappa {\theta }^{5}\geqslant 0,\\ {W}_{-}(\theta ) & = & {H}_{-}(\theta ){H}_{+}^{\dagger }(-\bar{\theta }),\mathrm{Im}\kappa {\theta }^{5}\leqslant 0,\end{array}\end{eqnarray}$
where ${H}_{\pm }^{\dagger }$ denote the complex conjugate transposition of the 2 × 2 matrix H±, the x and t are dependent. Then, W+(θ) is analytic in $\{\theta \in {\mathbb{C}}\setminus \mathrm{Im}\kappa {\theta }^{5}\gt 0\}$, and W−(θ) is analytic in $\{\theta \in {\mathbb{C}}\setminus \mathrm{Im}\kappa {\theta }^{5}\lt 0\}$. Combining the symmetry property $y(-\theta )=\overline{y(\bar{\theta })},z(-\theta )=\overline{z(\bar{\theta })},$ $Y(-\theta )=\overline{Y(\bar{\theta })},Z(-\theta )=\overline{Z(\bar{\theta })}$ in because the RH problem is equivalent to Proposition $\begin{eqnarray}{L}_{4}^{\dagger }(-\bar{\theta })={L}_{1}(\theta ),{L}_{3}^{\dagger }(-\bar{\theta })={L}_{3}(\theta ),{L}_{2}^{\dagger }(-\bar{\theta })={L}_{2}(\theta ).\end{eqnarray}$
Therefore $\begin{eqnarray}\begin{array}{rcl}{W}_{+}(\theta ) & = & {H}_{-}(\theta )L(\theta ){H}_{-}^{\dagger }(-\bar{\theta }),\mathrm{Im}\kappa {\theta }^{5}\in {\mathbb{R}},\\ {W}_{-}(\theta ) & = & {H}_{-}(\theta ){L}^{\dagger }(-\bar{\theta }){H}_{-}^{\dagger }(-\bar{\theta }),\mathrm{Im}\kappa {\theta }^{5}\in {\mathbb{R}}.\end{array}\end{eqnarray}$
For $\mathrm{Im}\kappa {\theta }^{5}\in {\mathbb{R}}$, equation ( $\begin{eqnarray}{H}_{-}({\rm{i}}\varepsilon ){L}_{1}({\rm{i}}\varepsilon ){H}_{-}^{\dagger }({\rm{i}}\varepsilon )=0,\varepsilon \in {\mathbb{R}}.\end{eqnarray}$
Furthermore, equation (5. Conclusions and discussions
In [45], the RH problem of the fifth-order mKdV equation (1.3 ) in the quarter plane was presented and the asymptotic behavior of the solution using the Deift-Zhou method (see [47, 48]) was analyzed. In this paper, we have discussed the IBVPs of the fifth-order mKdV equation (1.3 ) on the half-line. Compared with the results in [45], the difference is that we get the two pairs of spectral functions y(θ), z(θ) and Y(θ), Z(θ) admitting the global relation equation (2.42 ) is not the same, and we can also investigate the IBVPs of the fifth-order mKdV equation (1.3 ) on a finite interval. Furthermore, because the RH problem is equivalent to Gel’fand–Levitan–Marchenko (GLM) theory, we can get the soliton solution of the fifth-order mKdV equation (1.3 ) by solving the GLM equation following [49], these questions are the subject of our future research.
Acknowledgments
This work is supported by the National Natural Science Foundation of China under Grant Nos. 12147115 and 11835011, the Natural Science Foundation of Anhui Province under Grant No. 2108085QA09, the University Natural Science Research Project of Anhui Province under Grant No. KJ2021A1094, China Postdoctoral Science Foundation under Grant No. 2022M712833, the Program for Science and Technology Innovation Talents in Universities of Henan Province under Grant No. 22HASTIT019, and the Natural Science Foundation of Henan Province under Grant No. 202300410524.
Conflict of interest
The authors declare that they have no conflict of interest.
Appendix. Recovering v0(t) and ${\{{v}_{j}(t)\}}_{1}^{4}$
In this appendix, we will give a proof of equation (3.3 ), i.e., derive v0(t) and vj(t), j = 1, 2, 3, 4 from H(t). Assume that Φ(x, t, θ) is a solution of equation (2.6a )–(2.6b ) and Φ(x, t, θ) has the following asymptotic expansionA.1 ) into equation (2.6a ) and comparing the coefficients for θk, we get
$\begin{eqnarray}\begin{array}{rcl}{\rm{\Phi }}(x,t,\theta ) & = & {{\rm{\Phi }}}_{0}+\displaystyle \frac{{{\rm{\Phi }}}_{1}}{\theta }+\displaystyle \frac{{{\rm{\Phi }}}_{2}}{{\theta }^{2}}+\displaystyle \frac{{{\rm{\Phi }}}_{3}}{{\theta }^{3}}+\displaystyle \frac{{{\rm{\Phi }}}_{4}}{{\theta }^{4}}+\displaystyle \frac{{{\rm{\Phi }}}_{5}}{{\theta }^{5}}\\ & & +O\left(\displaystyle \frac{1}{{\theta }^{6}}\right),\theta \to \infty ,\end{array}\end{eqnarray}$
substituting equation ( $\begin{eqnarray}\begin{array}{l}O({\theta }^{1}):{\rm{i}}[{\sigma }_{3},{{\rm{\Phi }}}_{0}]=0,\\ O({\theta }^{0}):{{\rm{\Phi }}}_{0x}-{\rm{i}}[{\sigma }_{3},{{\rm{\Phi }}}_{1}]=U{{\rm{\Phi }}}_{0}.\end{array}\end{eqnarray}$
From O(θ1) and O(θ0), one find that Φ0 enjoys a diagonal matrix form denoting as $\begin{eqnarray}\begin{array}{l}{{\rm{\Phi }}}_{0}=\left(\begin{array}{cc}{{\rm{\Phi }}}_{0}^{11} & 0\\ 0 & {{\rm{\Phi }}}_{0}^{22}\end{array}\right),\,{{\rm{\Phi }}}_{1}^{({od})}=\displaystyle \frac{{\rm{i}}}{2}{\sigma }_{3}U{{\rm{\Phi }}}_{0}=\left(\begin{array}{cc} & \displaystyle \frac{{\rm{i}}}{2}u{{\rm{\Phi }}}_{0}^{22}\\ \displaystyle \frac{{\rm{i}}}{2}u{{\rm{\Phi }}}_{0}^{11} & \end{array}\right),\\ {{\rm{\Phi }}}_{0x}=0,\end{array}\end{eqnarray}$
where ${{\rm{\Phi }}}_{1}^{({od})}$ denotes the off-diagonal part of Φ1.At the same time, substituting equation (A.1 ) into equation (2.6b ) and comparing the coefficient for θk yieldsA.5 ), we obtain2.7 ). LetA.6a )–(A.6d ) gives rise toA.3 ) implies thatA.8a )–(A.8d ), (A.10 ) at x = 0 gives the expressions
$\begin{eqnarray}\begin{array}{rcl}O({\theta }^{5}):16{\rm{i}}[{\sigma }_{3},{{\rm{\Phi }}}_{0}] & = & 0,\\ O({\theta }^{4}):16{\rm{i}}[{\sigma }_{3},{{\rm{\Phi }}}_{1}] & = & -16U{{\rm{\Phi }}}_{0},\\ O({\theta }^{3}):16{\rm{i}}[{\sigma }_{3},{{\rm{\Phi }}}_{2}] & = & -16U{{\rm{\Phi }}}_{1}+8{\rm{i}}({\sigma }_{3}{U}_{x}-{U}^{2}{\sigma }_{3}){{\rm{\Phi }}}_{0},\\ O({\theta }^{2}):16{\rm{i}}[{\sigma }_{3},{{\rm{\Phi }}}_{3}] & = & -16U{{\rm{\Phi }}}_{2}+8{\rm{i}}({\sigma }_{3}{U}_{x}-{U}^{2}{\sigma }_{3}){{\rm{\Phi }}}_{1}\\ & & +\,4({U}_{{xx}}-2{U}^{3}){{\rm{\Phi }}}_{0},\\ O({\theta }^{1}):16{\rm{i}}[{\sigma }_{3},{{\rm{\Phi }}}_{4}] & = & -16U{{\rm{\Phi }}}_{3}+8{\rm{i}}({\sigma }_{3}{U}_{x}-{U}^{2}{\sigma }_{3}){{\rm{\Phi }}}_{2}\\ & & +\,4({U}_{{xx}}-2{U}^{3}){{\rm{\Phi }}}_{1}\\ & & +\,2{\rm{i}}[{\sigma }_{3}(6{U}^{2}{U}_{x}-{U}_{{xxx}})\\ & & +(2{{UU}}_{{xx}}-{U}_{x}^{2}-3{U}^{4}){\sigma }_{3}]{{\rm{\Phi }}}_{0},\\ O({\theta }^{0}):{{\rm{\Phi }}}_{0t}+16{\rm{i}}\kappa [{\sigma }_{3},{{\rm{\Phi }}}_{5}] & = & \kappa \{-16U{{\rm{\Phi }}}_{4}\\ & & +8{\rm{i}}({\sigma }_{3}{U}_{x}-{U}^{2}{\sigma }_{3}){{\rm{\Phi }}}_{3}\\ & & +\,4({U}_{{xx}}-2{U}^{3}){{\rm{\Phi }}}_{2}\\ & & +\,2{\rm{i}}[{\sigma }_{3}(6{U}^{2}{U}_{x}-{U}_{{xxx}})\\ & & +(2{{UU}}_{{xx}}-{U}_{x}^{2}-3{U}^{4}){\sigma }_{3}]{{\rm{\Phi }}}_{1}\\ & & +[10({U}^{2}{U}_{{xx}}\\ & & +{{UU}}_{x}^{2})-6{U}^{5}-{U}_{{xxxx}}]{{\rm{\Phi }}}_{0}\}.\end{array}\end{eqnarray}$
Through tedious calculation, we have $\begin{eqnarray*}\begin{array}{rcl}{{\rm{\Phi }}}_{2}^{12} & = & \displaystyle \frac{{\rm{i}}}{2}u{{\rm{\Phi }}}_{1}^{22}+\displaystyle \frac{1}{4}{u}_{x}{{\rm{\Phi }}}_{0}^{22},\qquad {{\rm{\Phi }}}_{2}^{21}=\displaystyle \frac{{\rm{i}}}{2}u{{\rm{\Phi }}}_{1}^{11}-\displaystyle \frac{1}{4}{u}_{x}{{\rm{\Phi }}}_{0}^{11},\\ {{\rm{\Phi }}}_{3}^{12} & = & \displaystyle \frac{{\rm{i}}}{2}u{{\rm{\Phi }}}_{2}^{22}+\displaystyle \frac{1}{4}{u}_{x}{{\rm{\Phi }}}_{1}^{22}-\displaystyle \frac{{\rm{i}}}{8}({u}_{{xx}}+{u}^{3}){{\rm{\Phi }}}_{0}^{22},\\ {{\rm{\Phi }}}_{3}^{21} & = & \displaystyle \frac{{\rm{i}}}{2}u{{\rm{\Phi }}}_{2}^{11}-\displaystyle \frac{1}{4}{u}_{x}{{\rm{\Phi }}}_{1}^{11}-\displaystyle \frac{{\rm{i}}}{8}({u}_{{xx}}+{u}^{3}){{\rm{\Phi }}}_{0}^{11},\\ {{\rm{\Phi }}}_{4}^{12} & = & \displaystyle \frac{{\rm{i}}}{2}u{{\rm{\Phi }}}_{3}^{22}+\displaystyle \frac{1}{4}{u}_{x}{{\rm{\Phi }}}_{2}^{22}\\ & & -\displaystyle \frac{{\rm{i}}}{8}({u}_{{xx}}+{u}^{3}){{\rm{\Phi }}}_{1}^{22}-\displaystyle \frac{1}{16}(5{u}^{2}{u}_{x}+{u}_{{xxx}}){{\rm{\Phi }}}_{0}^{22},\\ {{\rm{\Phi }}}_{4}^{21} & = & \displaystyle \frac{{\rm{i}}}{2}u{{\rm{\Phi }}}_{3}^{11}-\displaystyle \frac{1}{4}{u}_{x}{{\rm{\Phi }}}_{2}^{11}-\displaystyle \frac{{\rm{i}}}{8}({u}_{{xx}}+{u}^{3}){{\rm{\Phi }}}_{1}^{11}\\ & & +\,\displaystyle \frac{1}{16}(5{u}^{2}{u}_{x}+{u}_{{xxx}}){{\rm{\Phi }}}_{0}^{11},\end{array}\end{eqnarray*}$
$\begin{eqnarray}\begin{array}{rcl}{{\rm{\Phi }}}_{5}^{12} & = & \displaystyle \frac{{\rm{i}}}{2}u{{\rm{\Phi }}}_{4}^{22}+\displaystyle \frac{1}{4}{u}_{x}{{\rm{\Phi }}}_{3}^{22}-\displaystyle \frac{{\rm{i}}}{8}({u}_{{xx}}+{u}^{3}){{\rm{\Phi }}}_{2}^{22}\\ & & -\,\displaystyle \frac{1}{16}(5{u}^{2}{u}_{x}+{u}_{{xxx}}){{\rm{\Phi }}}_{1}^{22}\\ & & +\displaystyle \frac{{\rm{i}}}{32}(10{u}^{2}{u}_{{xx}}+10{{uu}}_{x}^{2}+6{u}^{5}+{u}_{{xxxx}}){{\rm{\Phi }}}_{0}^{22},\\ {{\rm{\Phi }}}_{5}^{21} & = & \displaystyle \frac{{\rm{i}}}{2}u{{\rm{\Phi }}}_{4}^{11}-\displaystyle \frac{1}{4}{u}_{x}{{\rm{\Phi }}}_{3}^{11}-\displaystyle \frac{{\rm{i}}}{8}({u}_{{xx}}+{u}^{3}){{\rm{\Phi }}}_{2}^{11}\\ & & +\,\displaystyle \frac{1}{16}(5{u}^{2}{u}_{x}+{u}_{{xxx}}){{\rm{\Phi }}}_{1}^{11}\\ & & +\displaystyle \frac{{\rm{i}}}{32}(10{u}^{2}{u}_{{xx}}+10{{uu}}_{x}^{2}+6{u}^{5}+{u}_{{xxxx}}){{\rm{\Phi }}}_{0}^{11},\\ {{\rm{\Phi }}}_{0t} & = & 0.\end{array}\end{eqnarray}$
Let Φ0 = I, according to equation ( $\begin{eqnarray}{\sigma }_{3}{U}_{x}=4{\sigma }_{3}{{\rm{\Phi }}}_{2}^{({od})}-2{\rm{i}}U{{\rm{\Phi }}}_{1}^{(d)}+{U}^{2}{\sigma }_{3},\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{U}_{{xx}}=8{\rm{i}}{\sigma }_{3}{{\rm{\Phi }}}_{3}^{({od})}+4U{{\rm{\Phi }}}_{2}^{(d)}-2{\rm{i}}{\sigma }_{3}{U}_{x}{{\rm{\Phi }}}_{1}^{(d)}\\ +2{\rm{i}}{U}^{2}{\sigma }_{3}{{\rm{\Phi }}}_{1}^{({od})}+2{U}^{3},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{\sigma }_{3}{U}_{{xxx}} & = & 16{\sigma }_{3}{{\rm{\Phi }}}_{4}^{({od})}+8{\rm{i}}U{{\rm{\Phi }}}_{3}^{(d)}+4{\sigma }_{3}{U}_{x}{{\rm{\Phi }}}_{2}^{(d)}-4{U}^{2}{\sigma }_{3}{{\rm{\Phi }}}_{2}^{({od})}\\ & & -2{\rm{i}}({U}_{{xx}}-2{U}^{3}){{\rm{\Phi }}}_{1}^{(d)}\\ & & +6{\sigma }_{3}{U}^{2}{U}_{x}+(2{{UU}}_{{xx}}-{U}_{x}^{2}-3{U}^{4}){\sigma }_{3},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{U}_{{xxxx}} & = & -32{\rm{i}}{\sigma }_{3}{{\rm{\Phi }}}_{5}^{({od})}-16U{{\rm{\Phi }}}_{4}^{(d)}+8{\rm{i}}{\sigma }_{3}{U}_{x}{{\rm{\Phi }}}_{3}^{(d)}\\ & & -8{\rm{i}}{U}^{2}{\sigma }_{3}{{\rm{\Phi }}}_{3}^{({od})}\\ & & +\,4({U}_{{xx}}-2{U}^{3}){{\rm{\Phi }}}_{2}^{(d)}+2{\rm{i}}{\sigma }_{3}(6{U}^{2}{U}_{x}-{U}_{{xxx}}){{\rm{\Phi }}}_{3}^{(d)}\\ & & +2{\rm{i}}(2{{UU}}_{{xx}}-{U}_{x}^{2}-3{U}^{4}){\sigma }_{3}{{\rm{\Phi }}}_{1}^{({od})}\\ & & +10({U}^{2}{U}_{{xx}}+{{UU}}_{x}^{2})-6{U}^{5},\end{array}\end{eqnarray}$
where Φ(x, t, θ) is the solution of equation ( $\begin{eqnarray}\begin{array}{l}{\rm{\Phi }}(x,t,\theta )={\rm{I}}+\displaystyle \frac{{\phi }^{(1)}}{\theta }+\displaystyle \frac{{\phi }^{(2)}}{{\theta }^{2}}+\displaystyle \frac{{\phi }^{(3)}}{{\theta }^{3}}\\ +\displaystyle \frac{{\phi }^{(4)}}{{\theta }^{4}}+\displaystyle \frac{{\phi }^{(5)}}{{\theta }^{5}}+O(\displaystyle \frac{1}{{\theta }^{6}}),\theta \to \infty ,\end{array}\end{eqnarray}$
the (12)-entry of equation ( $\begin{eqnarray}{u}_{x}=4{\phi }_{12}^{(2)}-2{\rm{i}}u{\phi }_{22}^{(1)}+{u}^{2},\end{eqnarray}$
$\begin{eqnarray}{u}_{{xx}}=8{\rm{i}}{\phi }_{12}^{(3)}+4u{\phi }_{22}^{(2)}-2{\rm{i}}{u}_{x}{\phi }_{22}^{(1)}-{u}^{3},\end{eqnarray}$
$\begin{eqnarray}\begin{array}{l}{u}_{{xxx}}=-16{\phi }_{12}^{(4)}+8{\rm{i}}u{\phi }_{22}^{(3)}+4{u}_{x}{\phi }_{22}^{(2)}-2{\rm{i}}({u}_{{xx}}+{u}^{3}){\phi }_{22}^{(1)}\\ -5{u}^{2}{u}_{x},\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{u}_{{xxxx}} & = & -32{\rm{i}}{\phi }_{12}^{(5)}-16u{\phi }_{22}^{(4)}+8{\rm{i}}u{\phi }_{22}^{(3)}+4({u}_{{xx}}+{u}^{3}){\phi }_{22}^{(2)}\\ & & -2{\rm{i}}(5{u}^{2}{u}_{x}+{u}_{{xxx}}){\phi }_{22}^{(1)}-(10{u}^{2}{u}_{{xx}}+10{{uu}}_{x}^{2}+6{u}^{5}).\end{array}\end{eqnarray}$
Furthermore, equation ( $\begin{eqnarray}u(x,t)=-2{\rm{i}}\phi (x,t)=-2{\rm{i}}\mathop{\mathrm{lim}}\limits_{\theta \to \infty }{\left(\theta {\rm{\Phi }}(x,t,\theta )\right)}_{12},\end{eqnarray}$
then, we have $\begin{eqnarray}u(x,t)=-2{\rm{i}}{\phi }_{12}^{(1)}(x,t).\end{eqnarray}$
Evaluating equations ( $\begin{eqnarray}{v}_{0}(t)=-2{\rm{i}}{\phi }_{12}^{(1)}(t),\end{eqnarray}$
$\begin{eqnarray}{v}_{1}(t)=4{\phi }_{12}^{(2)}(t)-2{\rm{i}}{v}_{0}(t){\phi }_{22}^{(1)}(t)+{v}_{0}^{2}(t),\end{eqnarray}$
$\begin{eqnarray}{v}_{2}(t)=8{\rm{i}}{\phi }_{12}^{(3)}(t)+4{v}_{0}(t){\phi }_{22}^{(2)}(t)-2{\rm{i}}{v}_{1}(t){\phi }_{22}^{(1)}(t)-{v}_{0}^{3}(t),\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{v}_{3}(t) & = & -16{\phi }_{12}^{(4)}(t)+8{\rm{i}}{v}_{0}(t){\phi }_{22}^{(3)}(t)+4{v}_{1}(t){\phi }_{22}^{(2)}(t)\\ & & -2{\rm{i}}({v}_{2}(t)+{v}_{0}^{3}(t)){\phi }_{22}^{(1)}(t)-5{v}_{0}^{2}(t){v}_{1}(t),\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{v}_{4}(t) & = & -32{\rm{i}}{\phi }_{12}^{(5)}(t)-16{v}_{0}(t){\phi }_{22}^{(4)}(t)+8{\rm{i}}{v}_{0}(t){\phi }_{22}^{(3)}(t)\\ & & +4({v}_{2}(t)+{v}_{0}^{3}(t)){\phi }_{22}^{(2)}(t)-2{\rm{i}}(5{v}_{0}^{2}(t){v}_{1}(t)\\ & & +{v}_{3}(t)){\phi }_{22}^{(1)}(t)\\ & & -(10{v}_{0}^{2}(t){v}_{1}(t)+10{v}_{0}(t){v}_{1}^{2}(t)+6{v}_{0}^{5}(t)),\end{array}\end{eqnarray}$
where φ(j)(t), j = 1, 2, 3, 4, 5 are determined by $\begin{eqnarray}{H}^{(t)}(t,\theta )={\rm{I}}+\displaystyle \sum _{j=1}^{5}\displaystyle \frac{{\phi }^{(j)}(t)}{{\theta }^{j}}+O\left(\displaystyle \frac{1}{{\theta }^{6}}\right),\quad \theta \to \infty .\end{eqnarray}$