Painlevé-type asymptotics for the defocusing Manakov system with nonzero boundary conditions

P ainlev ´ e-t yp e asymptotics for the defo cusing Manak o v system with nonzero b oundary conditions Haibing Zhang a , Xianguo Geng a,b ∗ , Ruomeng Li a , Huan Liu a a Sc ho ol of Mathematics and Statistics, Zhengzhou Univ ersit y , 100 Kexue Road, Zhengzhou, Henan 450001, P eople’s Republic of China b Institute of Mathematics, Henan Academ y of Sciences, Zhengzhou, Henan 450046, P eople’s Republic of China Abstract W e in v estigate the long-time asymptotic b eha vior of a class of solutions to the defo- cusing Manak o v system under nonzero b oundary conditions. These solutions are c harac- terized by a 3 × 3 matrix Riemann Hilb ert problem. W e ﬁnd that they exhibit interesting asymptotic b ehavior within a narro w transition zone in the x - t plane. W e determine the leading-order asymptotic term and the error b ound in this region, and we demonstrate that the leading term can be expressed in terms of the Hastings–McLeo d solution of the P ainlev ´ e I I equation. The pro of is rigorously established by applying the Deift-Zhou non- linear steep est descen t metho d to the asso ciated Riemann Hilb ert problem. Keyw ords the defo cusing Manak ov; nonzero bac kground; long-time asymptotics; P ainlev´ e I I equation. Mathematics Sub ject Classiﬁcation 35G25, 35Q15, 76B15. Con ten ts 1 In tro duction 2 1.1 Statemen t of results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 ∗ Corresp onding author. Email address : xggeng@zzu.edu.cn 1 2 Analysis of the mo del RH problem 7 2.1 Deﬁnition of the mo del problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.2 Pro of of Theorem 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 3 Review of the IST for the defo cusing Manak o v system with NZBCs 14 4 Long time asymptotics 19 4.1 The ﬁrst transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 4.2 The second transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 4.3 The third transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 4.4 The local parametrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 4.5 Final transformation and the small norm RH problem . . . . . . . . . . . . . . . 46 5 Pro of of Theorem 1.4 49 6 Concluding Remarks 51 A Pro of of Lemma 4.6 53 B Painlev ´ e I I mo del problem 58 C Error function mo del problem 60 References 62 1 In tro duction The Manak ov system (see e.g. [ 1 , 2 ]) i q t + q xx − 2 σ qq † q = 0 , q = ( q 1 , q 2 ) ⊤ , σ = ± 1 , (1.1) is a signiﬁcan t researc h topic in the ﬁeld of mathematical physics. When q 2 ( x, t ) ≡ 0, the system reduces to the well-kno wn cubic nonlinear Sc hr¨ odinger (NLS) equation [ 3 , 4 ]; hence, the Manak o v system is also referred to as the tw o-comp onent NLS equation. The cases σ = 1 and σ = − 1 corresp ond to the defo cusing and fo cusing regimes, resp ectively . The Manak o v system ha v e b een deriv ed in man y ph ysical ﬁelds, such as deep w ater w a v es, nonlinear optics, acoustics, and Bose–Einstein condensation (see, for eaxmple, [ 5 – 8 ]). Mathematically sp eaking, 2 the scalar and v ector NLS equations b elong to the class of completely in tegrable systems, whic h p ossess extremely rich mathematical structures. It is well known that the initial v alue problem for these integrable systems can b e solv ed using the inv erse scattering transform (IST). The ma jorit y of works in the literature primarily fo cused on the case of zero b oundary conditions (ZBCs)—where the p oten tial decays to zero as the spatial v ariable tends to inﬁnity . Ho w ev er, recen t exp erimen ts and studies hav e indicated that solutions of the NLS-t yp e equations under the follo wing non-zero boundary conditions (NZBCs) lim x →±∞ q ( x, t ) = q o e i θ ± +2i σ q 2 0 t , ∥ q o ∥ = q 0 > 0 , θ ± ∈ R (1.2) are closely related to mo dulation instability [ 15 – 17 ] and the generation mechanisms of rogue w a v es [ 11 – 14 ]. Consequen tly , researc h on IST for NLS-type equations under NZBCs of t yp e ( 1.2 ) has receiv ed increasing atten tion. W e review the relev an t literature b elow. The IST for the defo cusing scalar NLS equation with NZBCs of type ( 1.2 ) was done as early as in [ 49 , 50 ] and w as recently revisited in [ 51 ]. P artial results for the fo cusing case were presen ted in [ 9 ] and subsequently studied rigorously by Biondini et al. in [ 52 ]. Extending the IST from the scalar to the v ector case requires additional considerations. Although the IST for the vector NLS equation with ZBCs w as settled a long time ago [ 1 , 2 ], the case with NZBCs, ev en for t w o-comp onen t systems (i.e., the Manako v system), has remained an op en problem for nearly three decades. The in v erse scattering analysis for the defo cusing Manak o v system with NZBCs ( 1.2 ) w as ultimately completed in Refs. [ 18 , 19 ], where the authors utilized ideas originally introduced for solving the initial v alue problem of the three-wa v e in teraction equations [ 10 ]. So on after, the fo cusing Manako v system w as studied in [ 20 ] using similar metho ds. Unfortunately , the approac h used in [ 18 – 20 ] cannot b e extended to the v ector NLS equation with more than tw o components. Recen tly , Prinari et al. [ 22 ] ha ve made a signiﬁcant step tow ards the IST for the defo cusing N -comp onen t NLS equation with NZBCs. Ho w ev er, sev eral k ey issues remain open. A further IST c haracterization for the defocusing N -comp onen t NLS equation w as presented in [ 21 ] for N = 3. These results w ere recen tly extended to the defo cusing N -comp onent ( N ≥ 4) [ 24 ] and the fo cusing N -comp onen t cases [ 23 ]. Regarding these adv ances, we suggest that readers refer to [ 25 ] to obtain a comprehensive review on this topic. Despite these signiﬁcan t adv ances, certain asp ects of scalar nonlinear Schr¨ odinger theory remain unresolved for coupled systems. A crucial op en problem—as highlighted in Refs. [ 19 , 21 , 25 ]—is the application of the Deift-Zhou nonlinear steep est descent metho d [ 26 ] to deter- mine the long-time dynamics of the Manako v system under NZBCs ( 1.2 ). Although extensiv e researc h has been conducted on the long-time asymptotic b eha vior of the scalar NLS equation with NZBCs ( 1.2 ) [ 27 – 33 ], the literature remains surprisingly sparse for the tw o-comp onent systems. Indeed, the vector case is muc h more complicated than the scalar case b ecause it requires conducting the steep est descen t analysis for higher-order and more complex Riemann 3 Hilb ert (RH) problems, whic h undoubtedly signiﬁcantly increases the diﬃculty of the theoret- ical deriv ation. In our recent w ork [ 36 ], w e ha v e made signiﬁcant progress on this problem. W e rigorously carry out the steepest descen t analysis for the defo cusing Manak o v system with NZBCs in the so-called soliton region. Ho wev er, the b ehavioral characteristics to solutions in other regions are fundamentally diﬀerent. In particular, w e ﬁnd that they exhibit rather in teresting asymptotic b ehavior in a narrow transition zone of the x - t plane. The main purp ose of this w ork is to study the long-time asymptotics of solutions to the defo cusing Manak ov system i q t + q xx + 2 q 2 0 q − 2 qq † q = 0 , (1.3) with the following NZBCs at inﬁnity: lim x →±∞ q ( x, t ) = q ± = q o e i θ ± , ∀ t ≥ 0 (1.4) in the transitional regions, where q o denote constan t 2-comp onen t v ectors with ∥ q o ∥ = q 0 , and θ ± ∈ R . The nonzero b oundary problem ( 1.3 )–( 1.4 ) is obtained by applying a simple scaling transformation q ( x, t ) → q ( x, t )e − 2i q 2 0 t to the nonzero b oundary problem ( 1.1 )–( 1.2 ) with σ = 1. W e ﬁnd that the leading asymptotic term of the solutions in the transition regions matches the Painlev ´ e I I transcenden t. T o the best of our knowledge, this is the ﬁrst time to establish a connection b et w een the long-time dynamic b eha vior of the defo cusing Manako v system and a sp eciﬁc P ainlev´ e transcenden t. There is a substantial b o dy of literature utilizing the Deift-Zhou nonlinear steep est descen t metho d to study the transition asymptotics of integrable systems; how ever, the analysis is largely conﬁned to cases asso ciated with 2 × 2 RH problems, despite cov ering b oth zero and nonzero b oundary conditions (see e.g. [ 26 , 33 , 37 – 41 ]). F urthermore, although some work has explored the Painlev ´ e asymptotics of in tegrable systems corresp onding to higher-order RH problems [ 42 – 47 ], these studies ha v e exclusiv ely addressed the zero boundary case. This pap er in v estigates the Painlev ´ e asymptotics of in tegrable systems asso ciated with higher-order RH problems under nonzero b oundary conditions, rev ealing nov el analytical features compared to existing results. 1.1 Statemen t of results W e consider the long-time b eha vior of a class of solutions p ossessing high regularit y and de- ca y prop erties (to the non-zero bac kground). More precisely , we analyze the global solutions of ( 1.3 )–( 1.4 ) in the follo wing sense: Deﬁnition 1.1. The function q ( x, t ) is c al le d a glob al solution of the defo cusing Manakov system ( 1.3 ) with the NZBCs ( 1.4 ) if the fol lowing c onditions ar e satisﬁe d: 4 • q : R × [0 , ∞ ) → C 2 is smo oth function of ( x, t ) ∈ R × [0 , ∞ ) . • q ( x, t ) satisfy ( 1.3 ) for ( x, t ) ∈ R × [0 , ∞ ) . • F or every t ≥ 0 , q ( x, t ) satisﬁes the b oundary c ondition ( 1.4 ) , that is, for e ach inte ger N ≥ 1 and e ach T > 0 , sup x ∈ (0 , ∞ ) t ∈ [0 ,T ) N X j =0 (1 + | x | ) N ∥ ∂ j x ( q − q + ) ∥ + sup x ∈ ( −∞ , 0) t ∈ [0 ,T ) N X j =0 (1 + | x | ) N ∥ ∂ j x ( q − q − ) ∥ < ∞ . The p oten tials w e consider constitute merely a subclass of those in [ 19 ] for whic h the inv erse scattering analysis can pro ceed smo othly . This is b ecause our primary ob jectiv e is to rev eal the asymptotic characteristics of solutions to the defo cusing Manak o v system ( 1.3 ) on non- zero bac kground ( 1.4 ), and thus we do not take into account solutions with low regularity . F ollowing [ 19 ], we assume that the initial data generate “generic” scattering data; sp eciﬁcally , the sp ectral function a 11 ( z ) —deﬁned as the (1 , 1)-en try of the scattering matrix A ( z ) given b y ( 3.6 )—satisﬁes the following assumption: Assumption 1.2. The sp e ctr al function a 11 ( z ) p ossesses the fol lowing pr op erties: (a) The zer os of a 11 ( z ) ar e al l simple, ﬁnite in numb er, and none of them lie on the r e al axis. (b) a 11 ( z ) exhibits gener al b ehavior as z appr o aches the br anch p oints ± q 0 , sp e ciﬁc al ly lim z →± q 0 ( z ∓ q 0 ) a 11 ( z )  = 0 . (1.5) W e provide the following brief commen ts on the ab ov e assumption: (i) . The assumption (a) actually ac hiev es t w o goals: Firstly , it excludes the existence of the so-called sp ectral singularities; Secondly , it rules out higher-order p oles solutions of the defo cusing Manak ov system. Although double-p ole solutions for the system are considered in [ 19 , section 4], suc h cases fall outside the scop e of the present w ork. (ii) . As p ointed out in Ref. [ 19 ], all simple zeros of a 11 ( z ) lie either on the upp er semicircle C 0 = { z ∈ C : | z | = q 0 , ℑ z > 0 } or strictly inside it. W e denote these t w o sets of zeros b y { ζ j } N 1 j =1 on C 0 and { z j } N 2 j =1 inside C 0 , resp ectiv ely . (iii) . Assume that equation ( 1.5 ) holds, whic h implies that a 11 ( z ) p ossesses ﬁrst- order singularities at the branch p oints ± q 0 . As indicated in the in v erse scattering analysis of the defo cusing scalar NLS equation with a similar nonzero bac kground, this represents the generic case (see [ 31 , App endix C]). Next, w e imp ose a relatively strict assumption on the initial data. Assumption 1.3. We assume that q ( x, 0) is identic al ly e qual to the b ackgr ounds outside a c omp act set. That is, ther e exists a c onstant L > 0 such that q ( x, 0) = q + for x ≥ L and q ( x, 0) = q − for x ≤ − L . 5 This assumption is made here merely for the sake of conv enience. In fact, ev en if this as- sumption is remo v ed, our main theorem still holds true. How ever, based on this assumption, the reﬂection co eﬃcien ts can b e analytically con tin ued oﬀ the real axis, whic h greatly simpliﬁes our analysis. F or generic initial data, the reﬂection co eﬃcients generally cannot b e analytically con tin ued oﬀ the real axis. How ever, this is not an issue, as one can employ rational appro xi- mation techniques [ 26 , 48 ] to o v ercome this diﬃcult y . T o minimize technical complications, w e adopt the ab ov e assumption. Our main theorem concerns the long-time asymptotic b ehavior of the global solution to the defo cusing Manak ov system ( 1.3 ) with the NZBCs ( 1.4 ) in the transition region P L P L = { ( x, t ) ∈ R × R + : | ξ + q 0 | ≤ C t − 2 / 3 } , where ξ = x/ (2 t ) and C is a p ositive constan t. Theorem 1.4 (P ainlev ´ e asymptotics in P L ) . Supp ose that ther e exists a glob al solution q ( x, t ) to the defo cusing Manakov system ( 1.3 ) with the NZBCs ( 1.4 ) and its initial data satisﬁes Assumptions 1.2 and 1.3 . L et the functions δ 1 ( z ) , δ ♯ ( z ) and P 1 ( z ) b e deﬁne d by ( 4.13 ) , ( 4.22 ) and ( 4.23 ) , r esp e ctively. Then for ( x, t ) ∈ P L and t → ∞ , q ( x, t ) ob eys the fol lowing asymptotic formula q ( x, t ) = δ 1 (0)P 1 (0) q 0 δ ♯ (0) " q 0 + i 2( 3 t 4 q 0 ) 1 / 3  Z ∞ y u 2 H M ( y ′ ) dy ′ + u H M ( y )  # q + + O ( t − 2 / 3 ln t ) , (1.6) wher e the p ar ameter y is given by y = 2  4 3 q 0  1 / 3 ( ξ + q 0 ) t 2 / 3 , and u H M ( y ) is the Hastings–McL e o d solution of Painlev ´ e II e quation u ′′ ( y ) = y u ( y ) + 2 u 3 ( y ) , that is, the classic al solution of the e quation satisfying the fol lowing b oundary c onditions: u H M ( y ) =    Ai( y )(1 + o (1)) , y → + ∞ , p − y / 2(1 + o (1)) , y → −∞ . (1.7) Her e, Ai( · ) denotes the classic al Airy function. Remark 1.5. Our r esults ar e b ase d on the assumption of the glob al existenc e of solutions to ( 1.3 ) – ( 1.4 ) . F or the sc alar c ase, R ef. [ 31 , App endix B] establishe d the glob al existenc e of solutions when the nonzer o b ackgr ound satisﬁes q + = q − = 1 and the initial data satisﬁes q o ( x ) ∈ tanh( x ) + Σ 4 , wher e Σ 4 denotes a c ertain Sob olev sp ac e deﬁne d explicitly in R ef. [ 31 ]. Ther efor e, one might exp e ct similar r esults to hold for the ve ctor c ase as wel l; however, this go es b eyond the sc op e of the pr esent p ap er. 6 Remark 1.6. When ( 1.5 ) br e aks down (i.e., in the non-generic c ase), the r esulting asymptotic formula is analo gous to ( 1.6 ) . However, the le ading asymptotic term of the solution wil l b e r elate d to the Ablowitz-Se gur solution of the Painlev´ e II e quation. Remark 1.7 (On another Painlev ´ e region) . In fact, ther e exists a similar Painlev´ e tr ansition r e gion when ξ ≈ q 0 . Our appr o ach is also applic able to deriving the asymptotic formula within this r e gion. We ne e d to r emind the r e ader that the blow-up of δ 1 ( z ) as z tends to q 0 do es not intr o duc e essential c omplic ations for the Deift–Zhou analysis in this r e gion. This is b e c ause on the p ositive r e al axis, the c orr esp onding jump matrix no longer c ontains δ 1 ( z ) ; inste ad, the function ρ ( z ) app e ars, and ρ ( z ) has a ﬁnite limit as z → q 0 . Conse quently, the c alculations for another Painlev´ e r e gion similar to those in this p ap er. We omit the analysis of that r e gion solely to limit the length of the p ap er. Basic notations. Throughout this pap er, the asterisk indicates complex conjugation, and the sup erscripts ⊤ and † resp ectiv ely represen t the transp ose and the conjugate transp ose of the matrix. C > 0 and c > 0 will denote generic constants that may c hange within a computation. ℜ z and ℑ z denote the real and imaginary parts of the complex num b er z , resp ec- tiv ely . Let R + = (0 , + ∞ ), R − = ( −∞ , 0), and let C + and C − represen t the upp er and lo w er complex half-planes, resp ectively . Let ¯ D denote the closure of a region D in the complex plane. F or an y smo oth curv e Γ, the norm in the space L p (Γ), 1 ≤ p ≤ ∞ , is written as ∥ · ∥ L p (Γ) . F or a 3 × 3 matrix A deﬁned on Γ, we also deﬁne ∥ A ∥ L p (Γ) = max 1 ≤ i,j ≤ 3 ∥ A ij ∥ L p (Γ) . Giv en a function f and an orien tated curve, f + and f − stand for the limiting v alues of f when approac hing the curv e from the left and the right, resp ectively . Moreov er, w e alw a ys set the orientations of in terv als on the real axis to b e directed from the left to the righ t. This paper is organized as follo ws: Section 2 analyzes the mo del problem. W e then review the RH c haracterization of solutions to problem ( 1.3 )–( 1.4 ) in section 3 . Next, in section 4 w e apply the Deift–Zhou steep est descent analysis to RH problem 3.3 , and in section 5 , w e complete the pro of of the main Theorem 1.4 . Finally , some technical details are presen ted in App endices A to C . 2 Analysis of the mo del RH problem 2.1 Deﬁnition of the mo del problem Belo w w e provide the deﬁnition of the mo del RH problem in the case ( x, t ) ∈ P + , where P + = P L ∩ { ξ ≥ − q 0 } . The mo del problem presented in this pap er is a coupling of the Painlev ´ e I I mo del problem and an error function model problem. Let the jump contour X = ∪ 7 j =1 X j b e 7 1 2 4 3 6 5 7 β 2 β 3 β 0 β 1 Figure 1: The jump contour X for the mo del problem N X . deﬁned as follows (see Figure 1 ): X 1 =  β 2 + r e i π 6   0 ≤ r < ∞  , X 2 =  β 2 + r e 5i π 6   0 ≤ r < ∞  , X 3 =  β 3 + r e − 5i π 6   0 ≤ r < ∞  , X 4 =  β 3 + r e − i π 6   0 ≤ r < ∞  , X 5 =  β 0 + r e − 5i π 6   0 ≤ r < ∞  , X 7 =  β 1 + r e i π 6   0 ≤ r < ∞  , X 6 = [ β 0 , β 1 ] . Here, β j is the image of the critical point z j under the conformal mapping z → β (see ( 4.45 )). Then w e deﬁne a 3 × 3 matrix-v alued function V X ( y , s, t ; β ) on X , whic h serv es as the jump matrix for the mo del problem. The parameters y and s will b e sp eciﬁed in ( 4.41 ) and ( 4.48 ), resp ectiv ely . The exact expression of V X is giv en as follo ws: V X 1 =    1 0 0 0 1 0 ie 2i( y β + 4 3 β 3 ) 0 0    , V X 2 =    1 0 0 0 1 0 − ie 2i( y β + 4 3 β 3 ) 0 0    , V X 3 =    1 0 ie − 2i( y β + 4 3 β 3 ) 0 1 0 0 0 0    , V X 4 =    1 0 − ie − 2i( y β + 4 3 β 3 ) 0 1 0 0 0 0    , V X 5 = V X 6 = V X 7 =     1 i s e − i y β +i( 4 3 q 0 ) 2 3 t 1 3 β 2 0 0 1 0 0 − s e i y β +i( 4 3 q 0 ) 2 3 t 1 3 β 2 1     , (2.1) where V X j denotes the restriction of V X to the sub contour X j . RH Problem 2.1 (Mo del RH problem) . N X ( y , s, t ; β ) is a 3 × 3 matrix-value d function on C exc ept for X . • N X ( y , s, t ; β ) satisﬁes the fol lowing jump c onditions: N X + ( y , s, t ; β ) = N X − ( y , s, y ; β ) V X ( y , s, t ; β ) , β ∈ X . 8 • N X ( y , s, t ; β ) has the fol lowing b oundary c ondition: as β → ∞ , N X ( y , s, t ; β ) = I + O ( 1 β ) . • N X ( y , s, t ; β ) has the fol lowing b oundary c ondition: as β → 0 , N X ( y , s, t ; β ) = O (1) . The follo wing theorem further elaborates on the prop erties of this mo del RH problem, whic h is essen tial for subsequen t calculations. Theorem 2.2. Ther e exists a T > 0 such that for t > T RH pr oblem 2.1 admits a unique solution N X ( y , s, t ; β ) whenever ( x, t ) ∈ P + . Mor e over, as β → ∞ , the fol lowing asymptotic exp ansion holds uniformly for arg β ∈ [0 , 2 π ] : N X ( y , s, t ; β ) = I + 1 β  M P 1 ( y ) + E X 1 ( y , s, t )  + O ( 1 β 2 ) , (2.2) wher e E X 1 ( y , s, t ) satisﬁes E X 1 ( y , s, t ) = M P ( y , 0) M E 1 ( s )( M P ( y , 0)) − 1 ( 4 3 q 0 ) 1 / 3 t 1 / 6 + O ( t − 1 / 3 ) , t → ∞ . (2.3) In the ab ove expr ession, the functions M P 1 ( y ) , M P ( y , 0) and M E 1 ( s ) ar e deﬁne d by ( B.3 ) , ( B.4 ) , and ( C.2 ) , r esp e ctively. Since the pro of of this theorem is quite long, w e hav e dev oted a separate subsection to it. 2.2 Pro of of Theorem 2.2 Our proof is based on the analysis of a similar mo del problem in [ 46 , App endix A]. The main idea of the pro of is to subtract the contribution of the outer mo del a w a y from the origin, and to subtract the contribution of the lo cal mo del near the origin, thereby reducing the problem to a small norm RH problem. By analyzing this small norm RH problem, one can obtain the desired result. F or brevit y , we may omit the dep endence on parameters, for instance b y letting N X ( β ) := N X ( y , s, t ; β ). W e ﬁrst note that b y p erforming a simple con tour deformation, w e can (and will) henceforth assume that β 2 = i and β 3 = − i. On the other hand, when ( x, t ) ∈ P + and t → ∞ , β 1 and β 0 will deca y to 0 at a rate of O ( t − 1 / 3 ). Therefore, in the vicinit y of the origin, it is necessary to introduce a suitable scaling transformation so that the contribution near the origin can b e appro ximated by an exactly solv able mo del RH problem. Outside a certain neighborho o d of the origin, we show that the remaining con tribution can b e approximated by the solution of the Painlev ´ e I I mo del RH problem. T o this end, w e deﬁne D ϵ (0) as a small disk centered at the origin with radius ϵ , and its b oundary is orien ted coun terclo c kwise. W e c ho ose ϵ suﬃcien tly small so that ± i are not in the disk. Inside the disk, w e introduce the transformation β → α as α = ( 4 3 q 0 ) 1 / 3 t 1 / 6 β . 9 Our next lemma sho ws that inside D ϵ (0), N X ( β ) is w ell appro ximated b y M E ( α ( β )) as t → ∞ , where M E ( α ) denotes the solution to the error function RH problem C.1 . Let E ϵ := ∪ 7 j =5 E ϵ j , where E ϵ j := D ϵ (0) ∩ X j . Lemma 2.3. The function ˜ M E ( β ) := M E ( α ( β )) satisﬁes the jump c ondition: ˜ M E + ( β ) = ˜ M E − ( β ) ˜ V E ( β ) , β ∈ E ϵ , (2.4) wher e ˜ V E ( β ) = V E ( α ( β )) , β ∈ E ϵ . F or lar ge t , the jump matrix ˜ V E ( β ) satisﬁes the estimates    ∥ V X − ˜ V E ∥ L ∞ ( E ϵ ) ≤ C t − 1 / 6 , ∥ V X − ˜ V E ∥ L 1 ( E ϵ ) ≤ C t − 1 / 3 . (2.5) F urthermor e, as t → ∞ , ∥ ( ˜ M E ) − 1 − I ∥ L ∞ ( ∂ D ϵ (0)) = O ( t − 1 / 6 ) , (2.6) ( ˜ M E ) − 1 ( β ) − I = − M E 1 ( 4 q 0 3 ) 1 / 3 t 1 / 6 β + O ( t − 1 / 2 ) , β ∈ ∂ D ϵ (0) , (2.7) wher e M E 1 is given by ( C.2 ) . Pr o of. F or β ∈ E ϵ , w e ha v e V X − ˜ V E =     0 i s e i( 4 3 q 0 ) 2 3 t 1 3 β 2  e − i y β − 1  0 0 0 0 0 − s e i( 4 3 q 0 ) 2 3 t 1 3 β 2  e i y β − 1  0     . If β ∈ E ϵ 5 ∪ E ϵ 7 , it’s easy to see that    e i( 4 3 q 0 ) 2 3 t 1 3 β 2    ≤ e − ct 1 / 3 β 2 ,   e ± i y β − 1   ≤ C | y || β | ≤ C | β | . Then w e immediately obtain     V X − ˜ V E  12    ≤ C | β | e − ct 1 / 3 β 2 ≤ C t − 1 / 6 , and     V X − ˜ V E  12    L 1 ( E ϵ 5 ∪ E ϵ 7 ) ≤ C Z ∞ 0 τ e − ct 1 / 3 τ 2 d τ ≤ C t − 1 / 3 . The estimate for  V X − ˜ V E  32 on E ϵ 5 ∪ E ϵ 7 is similar. Therefore, ( 2.5 ) holds on the contour E ϵ 5 ∪ E ϵ 7 . If β ∈ E ϵ 6 , then we ha v e    e i( 4 3 q 0 ) 2 3 t 1 3 β 2    ≤ C,   e ± i y β − 1   ≤ C | y || β | ≤ C | β | . 10 This implies that     V X − ˜ V E  12    ≤ C | β | ,     V X − ˜ V E  32    ≤ C | β | , β ∈ E ϵ 6 . Since on E ϵ 6 w e hav e | β | ≤ | β 0 | + | β 1 | ≤ C t − 1 / 3 , estimate ( 2.5 ) therefore also holds on E ϵ 6 . Thus w e complete the pro of of ( 2.5 ). No w it remains only to pro v e ( 2.6 )and ( 2.7 ). The v ariables α go es to inﬁnit y as t → ∞ if β ∈ ∂ D ϵ (0). This is b ecause | α | =  4 q 0 3  1 / 3 t 1 / 6 | β | . Th us ( C.1 ) yields M E ( x, t, α ( β )) = I + M E 1 ( 4 q 0 3 ) 1 / 3 t 1 / 6 β + O ( t − 1 / 2 ) , t → ∞ . (2.8) uniformly for β ∈ D ϵ (0) and ( x, t ) ∈ P + . Since M E 1 is b ounded, we immediately obtain ( 2.6 ) and ( 2.7 ). Based on the previous analysis, w e pro ceed to introduce a transformation to obtain a small- norm RH problem and conduct error analysis. Let us deﬁne E X ( β ) as E X ( β ) =    N X ( β )  M P ( β )  − 1 , β ∈ C \ D ϵ (0) , N X ( β )  ˜ M E ( β )  − 1  M P ( β )  − 1 , β ∈ D ϵ (0) , (2.9) where M P ( β ) is the solution to the Painlev ´ e mo del problem B.1 . Deﬁne the contours ˆ E and E ′ b y ˆ E = ∪ 7 j =5 X j ∪ ∂ D ϵ (0) , E ′ = ˆ E \ D ϵ (0) . Then it can b e directly v eriﬁed that E X ( β ) has no jump on ∪ 4 j =1 X j ; it only has a jump on ˆ E . The jump matrix ˆ v for E X ( β ) is given by the follo wing expression: ˆ v =          M P V X ( M P ) − 1 , β ∈ E ′ , M P ( ˜ M E ) − 1 ( M P ) − 1 , β ∈ ∂ D ϵ (0) , M P ˜ M E − V X ( ˜ M E + ) − 1 ( M P ) − 1 , β ∈ E ϵ . (2.10) Lemma 2.4. L et ˆ w = ˆ v − I . The fol lowing estimates hold uniformly for lar ge t and ( x, t ) ∈ P + : ∥ ˆ w ∥ ( L 1 ∩ L ∞ )( E ′ ) ≤ C e − ct , (2.11) ∥ ˆ w ∥ L 1 ∩ L ∞ ( ∂ D ϵ ) ≤ C t − 1 / 6 , (2.12) ∥ ˆ w ∥ L 1 ( E ϵ ) ≤ C t − 1 / 3 , (2.13) ∥ ˆ w ∥ L ∞ ( E ϵ ) ≤ C t − 1 / 6 . (2.14) 11 Pr o of. When z ∈ E ′ , b oth M P and ( M P ) − 1 are uniformly b ounded, and V X − I decays exp onen tially as t → ∞ ; hence, ( 2.11 ) holds. ( 2.12 ) follo ws directly from ( 2.6 ) and ( 2.10 ). When z ∈ E ϵ , w e ha v e ˆ w = M P ˜ M E −  V X − ˜ V E  ( ˜ M E + ) − 1 ( M P ) − 1 . Com bining the ab ov e equation with ( 2.5 ), w e immediately obtain ( 2.13 ) and ( 2.14 ). The estimates in Lemma 2.4 show that ∥ ˆ w ∥ ( L 1 ∩ L ∞ )( ˆ E ) ≤ C t − 1 / 6 , ( x, t ) ∈ P + . Th us by employing the general inequalit y ∥ f ∥ L p ≤ ∥ f ∥ 1 p L 1 ∥ f ∥ p − 1 p L ∞ , w e immediately get ∥ ˆ w ∥ L p ( ˆ E ) ≤ C t − 1 / 6 , ( x, t ) ∈ P + . (2.15) This indicates that E X ( β ) satisﬁes a small-norm RH problem. Therefore, by the standard theory of RH problems, E X ( β ) exists and is unique for suﬃcien tly large t . T o show this, F or the con tour ˆ E and a function h ( z ) ∈ L 2 ( ˆ E ), w e deﬁne the Cauc hy transform C ( h )( β ) asso ciated with ˆ E b y C ( h )( β ) := 1 2 π i Z ˆ E h ( ζ ) dζ ζ − β . It is ko wn that the left and right non-tangential b oundary v alues C + h and C − h of C ( h ) exist a.e. on ˆ E and belong to L 2 ( ˆ E ). Let B ( L 2 ( ˆ E )) denotes the space of b ounded linear op erators on L 2 ( ˆ E ). Then C ± ∈ B ( L 2 ( ˆ E )) and C + − C − = I , where I denotes the identit y op erator on L 2 ( ˆ E ). F urthermore, we deﬁne the op erator C ˆ w b y C ˆ w ( h ) = C − ( h ˆ w ). F rom the preceding analysis, w e hav e known that ∥ ˆ w ∥ L 2 ( ˆ E ) → 0 as t → ∞ . Consequen tly , there exists a T > 0 such that the op erator I − C ˆ w is inv ertible whenev er t > T and ( x, t ) ∈ P + . Therefore, w e can deﬁne a function ˆ u ( x, t, β ) for β ∈ ˆ E and t > T by u = I + ( I − C ˆ w ) − 1 C ˆ w I ∈ I + L 2 ( ˆ E ) . (2.16) Then b y the standard theory of RH problems (see e.g. [ 57 ]), E X ( β ) deﬁned by E X ( β ) = I + 1 2 π i Z ˆ E ˆ u ( ζ ) ˆ w ( ζ ) dζ ζ − β , β ∈ C \ ˆ E , (2.17) is the unique solution of the small-norm RH problem. F rom ( 2.17 ), we kno w that for eac h N > 0 the follo wing expansion holds: E X ( β ) = I + N X j =1 E X j β j + O ( 1 β N +1 ) , β → ∞ , (2.18) where E X 1 ( s, y , t ) := ∠ lim β →∞ β ( E X ( β ) − I ) = − 1 2 π i Z ˆ E ˆ u ( ζ ) ˆ w ( ζ )d ζ . (2.19) 12 Lemma 2.5. As t → ∞ , E X 1 ( s, y , t ) = − 1 2 π i Z ∂ D ϵ (0) ˆ w ( ζ )d ζ + O ( t − 1 / 3 ) . (2.20) Pr o of. The function E X 1 ( s, y , t ) can b e rewritten as E X 1 ( s, y , t ) = − 1 2 π i Z ∂ D ϵ (0) ˆ w ( ζ ) dζ + Q 1 ( x, t ) + Q 2 ( x, t ) , where Q 1 ( x, t ) = − 1 2 π i Z E ′ ˆ w ( ζ )d ζ , Q 2 ( x, t ) = − 1 2 π i Z ˆ E ( ˆ u ( ζ ) − I ) ˆ w ( ζ )d ζ . Let’s estimate ∥ ˆ u − I ∥ L 2 ( ˆ E ) . A direct calculation yields ∥ ˆ u − I ∥ L 2 ( ˆ E ) ≤ ∥ ( I − C ˆ w ) − 1 C ˆ w I ∥ L 2 ( ˆ E ) ≤ ∞ X j =0 ∥C ˆ w ∥ j B ( L 2 ( ˆ E )) ∥C ˆ w I ∥ L 2 ( ˆ E ) ≤ ∥C − ∥ B ( L 2 ( ˆ E )) ∥ ˆ w ∥ L 2 ( ˆ E ) 1 − ∥C − ∥ B ( L 2 ( ˆ E )) ∥ ˆ w ∥ L ∞ ( ˆ E ) ≤ C t − 1 / 6 , t > T . (2.21) Then the lemma follo ws from Lemma 2.4 and Eq. ( 2.21 ) and straigh tforw ard estimates. No w, it remains to analyze − 1 2 π i R ∂ D ϵ (0) ˆ w ( ζ )d ζ . F or β ∈ ∂ D ϵ (0), w e ha v e ˆ w ( β ) = M P ( β )   ˜ M E ( β )  − 1 − I   M P ( β )  − 1 . (2.22) Th us by ( 2.6 ) and Cauc h y’s form ula, w e obtain − 1 2 π i Z ∂ D ϵ (0) ˆ w ( ζ )d ζ = M P ( y , 0) M E 1 ( s )( M P ( y , 0)) − 1 ( 4 3 q 0 ) 1 / 3 t 1 / 6 + O ( t − 1 / 2 ) . Com bining the ab o v e expression with ( 2.20 ) yields ( 2.3 ). F rom ( 2.9 ), we kno w that for suﬃ- cien tly large β , N X = E X M P . (2.23) Therefore, from ( 2.18 ), ( B.2 ) and ( 2.23 ), w e obtain that ( 2.2 ) holds. 13 3 Review of the IST for the defo cusing Manak o v system with NZBCs In this section, w e review the IST of the defo cusing Manako v system ( 1.3 ) with NZBCs ( 1.4 ). Since these results ha v e b een well established in Ref. [ 19 ], w e will omit the pro of. As usual, the IST for an integrable system is based on its form ulation in terms of a Lax pair. It is w ell-known that the defocusing Manako v system ( 1.3 ) p ossesses a 3 × 3 matrix Lax pair [ 19 ]: Φ x = ˜ XΦ , Φ t = ˜ TΦ , (3.1) where ˜ X ( x, t, k ) = − i k J + Q , ˜ T ( x, t, k ) = 2i k 2 J − i J  Q x − Q 2 + q 2 0  − 2 k Q , J = 1 0 ⊤ 0 − I 2 × 2 ! , Q = 0 q † q 0 2 × 2 ! . Compared to the zero-b oundary case, the in v erse scattering analysis with NZBCs is more complicated due to the app earance of the function λ ( k ) = p k 2 − q 2 0 . T o a v oid working on the Riemann sphere, following the idea in Ref. [ 19 ], we in tro duce the uniformization v ariable by deﬁning z = k + λ. The in verse transformation can b e obtained b y k = 1 2 ( z + q 2 0 z ) , λ = 1 2 ( z − q 2 0 z ) . (3.2) Thereafter, the IST will b e carried out in the z -plane. F ollowing the notation in Ref. [ 19 ], w e denote the orthogonal vector of a t w o-comp onen t complex-v alued vector v = ( v 1 , v 2 ) as v ⊥ = ( v 2 , − v 1 ) † . W e then in tro duce three matrices E ± ( z ) = 1 0 − i q 0 z i q ± z q ⊥ ± q 0 q ± q 0 ! , Λ ( z ) = diag ( − λ, k , λ ) , Ω ( z ) = diag  − 2 k λ, k 2 + λ 2 , 2 k λ  , (3.3) whic h satisfy the relation E − 1 ± ˜ X ± E ± = i Λ , E − 1 ± ˜ T ± E ± = − i Ω , where ˜ X ± = lim x →±∞ ˜ X and ˜ T ± = lim x →±∞ ˜ T . Let ∆ Q ± ( x, t ) = Q ( x, t ) − Q ± , Q ± = 0 q † ± q ± 0 2 × 2 ! . 14 It is easy to see that ∆ Q ± ( x, t ) deca ys to zero suﬃcien tly rapidly as x → ±∞ . W e deﬁne the Jost eigenfunctions as the unique solutions to the follo wing in tegral equations: µ − ( x, t, z ) = E − ( z ) + Z x −∞ E − ( z )e i( x − y ) Λ ( z ) E − 1 − ( z )∆ Q − ( y , t ) µ − ( y , t, z )e − i( x − y ) Λ ( z ) d y , (3.4a) µ + ( x, t, z ) = E + ( z ) − Z + ∞ x E + ( z ) e i( x − y ) Λ ( z ) E − 1 + ( z )∆ Q + ( y , t ) µ + ( y , t, z )e − i( x − y ) Λ ( z ) d y , (3.4b) Since ∆ Q ± ( x, t ) deca ys rapidly as x → ±∞ , then it can b e prov ed that { µ +1 , µ − 3 } and { µ − 1 , µ +3 } are w ell-deﬁned and con tin uous for z ∈ ¯ C − \ { 0 , ± q 0 } and z ∈ ¯ C + \ { 0 , ± q 0 } , resp ectiv ely (see Theorem 2.1 in Ref. [ 19 ]). Moreo v er, these pairs of functions are analytic in C − and C + , resp ectively . µ ± 2 can only b e deﬁned on R \ { 0 , ± q 0 } and are no where analytic. In Ref. [ 19 ], the authors discussed the b ehavior of the Jost eigenfunctions at the branch p oin ts ± q 0 . They ha v e sho wn that the Jost solutions admit a well-deﬁned limit at the branch p oin ts if ∆ Q ± ( x, t ) → 0 suﬃcien tly fast as x → ±∞ . W e deﬁne Φ ± ( x, t, z ) = µ ± ( x, t, z )e i Λ ( z ) x − i Ω ( z ) t . Then, Φ ± ( x, t, z ) are the fundamental solu- tions of Lax pair ( 3.1 ) for z ∈ R \ { 0 , ± q 0 } . This is b ecause det µ ± ( x, t, z ) = det E ± ( z ) = 1 − q 2 0 z 2 := γ ( z ) , (3.5) and γ ( z ) ∈ C \ { 0 } for z ∈ R \ { 0 , ± q 0 } . Therefore, there exists a matrix A ( z ) indep endent of x and t suc h that Φ − ( x, t, z ) = Φ + ( x, t, z ) A ( z ) , z ∈ R \ { 0 , ± q 0 } . (3.6) Deﬁning B ( z ) = A − 1 ( z ) , w e let a ij and b ij denote the ( ij )-th entries of the scattering matrices A ( z ) and B ( z ), resp ectively . The prop erties of the scattering matrices A ( z ) and B ( z ) are crucial for analyzing the reﬂection co eﬃcients. These results are presented in Lemmas 2.13, 2.16 and subsections 2.6, 2.7 in Ref. [ 19 ]; w e omit the details here. No w let us deﬁne the reﬂection co eﬃcien ts r 1 ( z ), r 2 ( z ) and r 3 ( z ) by r 1 ( z ) = a 21 ( z ) a 11 ( z ) , r 2 ( z ) = a 31 ( z ) a 11 ( z ) , r 3 ( z ) = a 23 ( z ) a 33 ( z ) . (3.7) In fact, the reﬂection co eﬃcien ts { r j } 3 j =1 dep end only on the initial data q ( x, 0), since w e may set t = 0 in ( 3.6 ). Moreo v er, one can observe that we hav e only tw o indep endent reﬂection co eﬃcien ts. This is b ecause r 1 ( z ) = i q 0 z r 3 ( ˆ z ), where ˆ z = q 2 0 z . T o simplify the Deift–Zhou steep est-descen t analysis in the subsequent sections, we require that the reﬂection co eﬃcients admit analytic con tin uation oﬀ the real axis. This can b e achiev ed if we assume that the initial v alue q ( x, 0) satisfy Assumption 1.3 . Under Assumption 1.3 , the reﬂection co eﬃcients enjo y “ go o d” properties, whic h w e summarize in the follo wing lemma: Lemma 3.1. Supp ose the initial data q ( x, 0) satisﬁes Assumption 1.2 and 1.3 . Then the asso ciate d r eﬂe ction c o eﬃcients deﬁne d by ( 3.7 ) have the fol lowing pr op erties: 15 • L et S ε denote a neighb orho o d of the r e al axis; mor e pr e cisely, set S ε := { z ∈ C : |ℑ z | ≤ ε } , wher e ε > 0 is an arbitr ary c onstant. Then e ach r j ( z ) is analytic for z ∈ S ε \ { 0 , ± q 0 } , and as z → ∞ within S ε , we have r 1 ( z ) = O ( 1 z 2 ) , r 3 ( z ) = O ( 1 z ) , r 2 ( z ) = O ( 1 z ) . (3.8) • Deﬁne ˜ S ε = S ε \ ( B 1 ∪ B 2 ) , wher e B 1 and B 2 denote the disks c enter e d at i 2 and − i 2 , r esp e ctively, e ach with r adius 1 2 . Then the functions { r j ( z ) } 3 j =1 have wel l-deﬁne d limits as ˜ S ε ∋ z → 0 . F urthermor e, we have r 2 ( z ) = O ( z 2 ) , r 3 ( z ) = O ( z 2 ) , ˜ S ε ∋ z → 0 . (3.9) • The functions { r j ( z ) } 3 j =1 have wel l-deﬁne d limits at the br anch p oints ± q 0 . F urthermor e, we have lim z →± q 0 r 2 ( z ) = ∓ i , lim z →± q 0 r 1 ( z ) = lim z →± q 0 r 3 ( z ) = 0 . (3.10) The proof of this lemma can be found in our recen t paper [ 36 , proof of Lemma 2.2 ]. Remark 3.2. Henc eforth, we shal l denote by r j (0) the limit lim ˜ S ε ∋ z → 0 r j ( z ) . It is worth noting that, although r 2 (0) = r 3 (0) = 0 , in gener al r 1 (0)  = 0 . F ollowing the idea from Ref. [ 19 ], w e deﬁne a piecewise meromorphic function M ( x, t, z ) as follo ws (see [ 19 , Eq.(3.1a) and Eq.(3.1b)]): M ( x, t, z ) =     µ − 1 a 11 , m b 33 , µ +3  , ℑ z > 0 ,  µ +1 , − ¯ m b 11 , µ − 3 a 33  , ℑ z < 0 , (3.11) where ¯ m ( x, t, z ) = − J [ Φ ∗ − 1 × Φ ∗ +3 ]( x, t, z ∗ ) /γ ( z ) , m ( x, t, z ) = − J [ Φ ∗ − 3 × Φ ∗ +1 ]( x, t, z ∗ ) /γ ( z ). Here “ × ” denotes the usual cross pro duct. Then the function M ( x, t, z ) deﬁned by ( 3.11 ) satisﬁes the follo wing residue conditions at the tw o t yp es of discrete eigen v alues { ζ j } N 1 j =1 and { z j } N 2 j =1 : (i) for eac h ζ j , j = 1 , ..., N 1 , Res z = ζ j M ( x, t, z ) = lim z → ζ j M ( x, t, z )    0 0 0 0 0 0 τ j e θ 31 ( x,t,ζ j ) 0 0    , (3.12) where τ j ζ j ∈ R . 16 (ii) for eac h z j , j = 1 , ..., N 2 , Res z = z j M ( x, t, z ) = lim z → z M ( x, t, z )    0 0 0 κ j e θ 21 ( x,t,z j ) 0 0 0 0 0    , Res z = z ∗ j M ( x, t, z ) = lim z → z ∗ j M ( x, t, z )     0 κ ∗ j γ ( z ∗ j ) e θ 12 ( x,t,z ∗ j ) 0 0 0 0 0 0 0     , (3.13) where κ j is arbitrary complex constan t. In ( 3.12 )and ( 3.13 ), θ iℓ ( x, t, z ) = θ i ( x, t, z ) − θ ℓ ( x, t, z ) , 1 ≤ i, ℓ ≤ 3 , where { θ i } 3 i =1 is given b y ( 3.15 ). The scattering data asso ciated with the initial data q ( x, 0) is deﬁned as follows: σ d =  r 1 ( z ) , r 2 ( z ) , { ζ j , τ j } N 1 j =1 , { z j , κ j } N 2 j =1  . The inv erse problem consists primarily in reconstructing the p oten tial from the given scat- tering data. In the mo dern form ulation of inv erse scattering theory , one typically constructs an appropriate RH problem and recov ers the p otential from its solution. The main result of Ref. [ 19 ] provides suc h a RH characterization for the solution of the defo cusing Manako v sys- tem ( 1.3 ) with NZBCs ( 1.4 ). Given the scattering data σ d , one can deﬁne the jump matrix V as follo ws: e V ( x, t, z ) = e Θ ( x,t,z )    1 − 1 γ ( z ) | r 1 ( z ) | 2 − | r 2 ( z ) | 2 1 γ ( z ) ( − r 1 ( z ) + r 2 ( z ) r 3 ( z )) ∗ − r ∗ 2 ( z ) r 1 ( z ) − r 2 ( z ) r 3 ( z ) 1 + 1 γ ( z ) | r 3 ( z ) | 2 − r 3 ( z ) r 2 ( z ) − 1 γ ( z ) r ∗ 3 ( z ) 1    e − Θ ( x,t,z ) , (3.14) where Θ ( x, t, z ) = diag ( θ 1 ( x, t, z ) , θ 2 ( x, t, z ) , θ 3 ( x, t, z )) with          θ 1 ( x, t, z ) = − i λ ( z ) x + 2i k ( z ) λ ( z ) t, θ 2 ( x, t, z ) = i k ( z ) x − i( k 2 ( z ) + λ 2 ( z )) t, θ 3 ( x, t, z ) = i λ ( z ) x − 2i k ( z ) λ ( z ) t. (3.15) RH Problem 3.3. Find a 3 × 3 matrix-value d function M ( x, t, z ) with the fol lowing pr op erties: • M ( x, t, · ) : C \ ( R ∪ Z ) → C 3 × 3 is analytic, wher e Z = Z 1 ∪ Z 2 , with Z 1 = ∪ N 1 j =1  ζ j ∪ ζ ∗ j  , Z 2 = ∪ N 2 j =1  z j ∪ z ∗ j ∪ ˆ z j ∪ ˆ z ∗ j  . (3.16) A cr oss R , M ( x, t, z ) satisﬁes the jump c ondition: M + ( x, t, z ) = M − ( x, t, z ) e V ( x, t, z ) , z ∈ R \ { 0 } . (3.17) 17 • M ( x, t, z ) admits the asymptotic b ehavior: M ( x, t, z ) = I + O ( 1 z ) , z → ∞ ; M ( x, t, z ) = 1 z σ 1 + O (1) , z → 0 , (3.18) wher e σ 1 =    0 0 − i q 0 0 0 0 i q 0 0 0    . (3.19) • M ( x, t, z ) satisﬁes the gr owth c onditions ne ar the br anch p oints ± q 0 :    M 1 ( x, t, z ) = O ( z ∓ q 0 ) , z ∈ C + → ± q 0 , M 3 ( x, t, z ) = O ( z ∓ q 0 ) , z ∈ C − → ± q 0 . (3.20) • M ( x, t, z ) satisﬁes the symmetries M ( x, t, z ) = M ( x, t, ˆ z ) Π ( z ) , ( M − 1 ) ⊤ ( x, t, z ) = − 1 γ ( z ) J M ∗ ( x, t, z ∗ ) Γ ( z ) , (3.21) wher e ˆ z = q 2 0 z and Π ( z ) =    0 0 − i q 0 z 0 1 0 i q 0 z 0 0    , Γ ( z ) =    − 1 0 0 0 γ ( z ) 0 0 0 1    . (3.22) • M ( x, t, z ) satisﬁes the r esidue c onditions ( 3.12 ) and ( 3.13 ) at the discr ete sp e ctr al p oints ζ j , z j , and z ∗ j , r esp e ctively. Mor e over, the r esidue c onditions at the discr ete sp e ctr al p oints ζ ∗ j , ˆ z j and ˆ z ∗ j c an b e derive d fr om the z → ˆ z symmetry in ( 3.21 ) . Note that our RH problem diﬀers slightly from that in Ref. [ 19 ]; ho w ever, this diﬀerence is merely for computational con v enience and not essen tial. Theorem 3.4 (RH characterization of the solution) . If ther e exists a glob al solution q ( x, t ) to the defo cusing Manakov system ( 1.3 ) satisfying the NZBCs ( 1.4 ) and its initial data satisfy Assumption 1.2 and 1.3 , then the RH pr oblem 3.3 admits a unique solution M ( x, t, z ) . Mor over, the solution q ( x, t ) c an b e r e c over e d as fol lows: q ( x, t ) = ˇ M ∞ − i lim z →∞ z M 21 ( x, t, z ) M 31 ( x, t, z ) !! , (3.23) wher e M ij denotes the ( ij ) -entry of the matrix-value d function M , and ˇ M ∞ is deﬁne d by ˇ M ∞ = q ∗ 2 , + q 0 q 1 , + q 0 − q ∗ 1 , + q 0 q 2 , + q 0 ! . 18 0 − q 0 − q 0 0 0 − q 0 Figure 2: F rom left to righ t: The signature tables for ϕ 21 , ϕ 32 and ϕ 31 for ξ = − 1 and q 0 = 1. The purple regions corresp ond to { z : Re ϕ ij < 0 } and the pink regions to { z : Re ϕ ij > 0 } . Pr o of. Let M ( x, t, z ) be the solution of RH problem 3.3 , then the function ˜ M ( x, t, z ) = M ∞ M ( x, t, z ) , M ∞ = 1 0 0 0 q ⊥ + /q 0 q + /q 0 ! , satisﬁes the RH problem in [ 19 , section 3.1 ]. Therefore, the theorem is a direct consequence of [ 19 , Theorem 3.8]. 4 Long time asymptotics In this section, w e study the long-time b eha vior of the RH problem 4.5 in the P ainlev´ e region P L b y using Deift-Zhou nonlinear steepest descent metho d. Recall that the region P L is deﬁned as follo ws: P L = { ( x, t ) ∈ R × R + | | ξ + q 0 | ≤ C t − 2 / 3 } , (4.1) where ξ = x/ (2 t ) and C > 0 is a constant. F or the purp ose of the pro of, it is necessary to split this region into t w o subregions and address them separately . Denote P L = P + ∪ P − , where P + := P L ∩ { ξ ≥ − q 0 } , P − := P L ∩ { ξ ≤ − q 0 } . (4.2) Since the analysis for regions P + and P − is quite similar, w e present only the pro of for region P + . The long-time asymptotics of RH problem 4.5 is aﬀected b y the growth and decay of the oscillatory terms e θ 21 , e θ 31 and e θ 32 in the jump matrix e V ( x, t, z ). Let θ ij ( x, t, z ) = t Φ ij ( ξ , z ). Therefore, w e need to analyze the prop erties of the three phase functions Φ 21 ( ξ , z ), Φ 31 ( ξ , z ) 19 and Φ 32 ( ξ , z ). The expressions for these three phase functions are as follo ws: Φ 32 ( ξ , z ) = − 2i ξ q 2 0 z + i q 4 0 z 2 , Φ 21 ( ξ , z ) = 2i ξ z − i z 2 , Φ 31 ( ξ , z ) = 4i ξ λ ( z ) − 4i k ( z ) λ ( z ) . (4.3) W e need the sign tables of these functions (see Fig 2 ), whic h are crucial in the Deift-Zhou analysis. W e also need information ab out the critical p oints of these phase functions. A straigh tforw ard computation yields the following result: • The function Φ 21 ( ξ , z ) has only one stationary p oin t z 1 = ξ . When ξ ≥ − q 0 and | ξ + q 0 | ≤ C t − 2 / 3 , w e ha v e | z 1 + q 0 | ≤ C t − 2 / 3 . • The function Φ 32 ( ξ , z ) has only one stationary p oint z 0 = q 2 0 ξ . When ξ ≥ − q 0 and | ξ + q 0 | ≤ C t − 2 / 3 , w e ha v e | z 0 + q 0 | ≤ C t − 2 / 3 . • The function Φ 31 ( ξ , z ) has t wo stationary p oin ts that do not lie on the real axis, expressed as z 2 = η + i p 4 q 2 0 − η 2 2 , z 3 = z ∗ 2 = η − i p 4 q 2 0 − η 2 2 , η = ξ − p ξ 2 + 8 q 2 0 2 . (4.4) When ξ ≥ − q 0 and | ξ + q 0 | ≤ C t − 2 / 3 , w e ha v e | z 2 + q 0 | ≤ C t − 1 / 3 , | z 3 + q 0 | ≤ C t − 1 / 3 . The ab ov e analysis shows that these four stationary p oin ts will conv erge to the same p oint − q 0 as t → ∞ . This represen ts a completely diﬀeren t structure compared to the standard P ainlev´ e asymptotic analysis of in tegrable systems, such as those in Refs. [ 37 ] and [ 33 ]. W e give more information ab out the phase function Φ 31 ( ξ , z ). When ξ = − q 0 , w e ha v e ℜ Φ 31 ( − q 0 , z ) = q 0  1 + q 2 0 | z | 2  + ℜ z  1 + q 4 0 | z | 4  . Therefore, ℜ Φ 31 ( − q 0 , z ) = 0 actually deﬁnes a curv e in the complex plane, whic h w e denote as L (see Fig 3 ). W e aim to analyze the prop erties of the tw o tangent lines of curve L at the p oin t − q 0 . Lemma 4.1. The slop es of the two tangent lines of curve L at the p oint − q 0 ar e √ 3 and − √ 3 , r esp e ctively. Pr o of. F or con v enience, let x = ℜ z and y = ℑ z in this pro of. Then the curv e L can b e expressed as L =  ( x, y ) | F ( x, y ) := q 0  1 + q 2 0 x 2 + y 2  + x  1 + q 4 0 ( x 2 + y 2 ) 2  = 0  . (4.5) 20 L 1 L 2 Figure 3: The green curv e denotes L , and the dashed lines L 1 and L 2 represen t the tangent lines at the p oin t − q 0 . When ( x, y )  = ( − q 0 , 0) but lies near this p oint, the implicit function theorem implies that the slop e K ( x, y ) can be given b y K ( x, y ) = − 3 q 4 0 x 2 + q 4 0 y 2 − 2 q 3 0 x 3 − 2 q 3 0 xy 2 + x 6 + 3 x 4 y 2 + 3 x 2 y 4 + y 6 2 q 3 0 y (2 xq 0 + x 2 + y 2 ) := F 1 ( x, y ) F 2 ( x, y ) . (4.6) T o obtain the slop e of the tangent line at − q 0 , we need to study the limit of K ( x, y ) as ( x, y ) approac hes ( − q 0 , 0) along the curv e L . F or conv enience, w e divide the p ortion of curv e L near − q 0 in to four parts, namely L = L a ∪ L b ∪ L c ∪ L d , where L a = L ∩ { x ≥ − q 0 , y ≥ 0 } , L b = L ∩ { x ≤ − q 0 , y ≥ 0 } , L c = L ∩ { x ≥ − q 0 , y ≤ 0 } , L d = L ∩ { x ≤ − q 0 , y ≤ 0 } . Next, we will compute the limit of K ( x, y ) as ( x, y ) ∈ L a approac hes ( − q 0 , 0). ( 4.5 ) implies that q 0  ( x 2 + y 2 ) 2 + q 2 0 ( x 2 + y 2 )  + x  ( x 2 + y 2 ) 2 + q 4 0  = 0 . Simplifying the ab ov e expression leads to x ( x + q 0 ) 2 ( x 2 − q 0 x + q 2 0 ) = y 2  − q 3 0 − 2 q 0 x 2 − 2 x 3  − ( x + q 0 ) y 4 . (4.7) F rom ( 4.7 ), w e know that as ( x, y ) approaches ( − q 0 , 0) along the curv e L a , w e ha v e √ 3( x + q 0 ) = y + o ( y ) . (4.8) No w recall that K ( x, y ) = F 1 ( x,y ) F 2 ( x,y ) , where F 1 ( x, y ) and F 2 ( x, y ) are giv en by ( 4.6 ). As ( x, y ) ap- proac hes ( − q 0 , 0) along the curve L a , b y using ( 4.8 ) and through a straigh tforw ard calculation, 21 w e obtain F 1 ( x, y ) = − 3 q 4 0 x 2 − 2 q 3 0 x 3 + x 6 + O ( y 2 ) = ( x + q 0 ) x 2  x 3 − q 0 x 2 + q 0 x − 3 q 2 0  + O ( y 2 ) = − 2 √ 3 q 5 0 y + o ( y ) . (4.9) Similarly , as ( x, y ) approac hes ( − q 0 , 0) along the curv e L a , w e ha v e F 2 ( x, y ) = − 2 q 5 0 y + o ( y ) . (4.10) Therefore, com bining ( 4.9 ) and ( 4.10 ), we conclude that lim L a ∋ ( x,y ) → ( − q 0 , 0) K ( x, y ) = √ 3 . By a completely similar analysis, one can compute the limit v alues of K ( x, y ) as ( x, y ) ap- proac hes ( − q 0 , 0) along the curves L b , L c , and L d , resp ectiv ely . This completes the pro of of the lemma. Remark 4.2. When ξ  = − q 0 but lies ne ar this p oint, the e quation ℜ Φ 31 ( ξ , z ) = 0 also deﬁnes a curve ˜ L in the c omplex z plane. The stationary p oints z 2 and z 3 lie pr e cisely on this curve. In the tr ansformation detaile d in the subse quent subse ction, we wil l “op en up” the c ontour ne ar these two stationary p oints. It is e asy to se e that the slop es of the tangents to ˜ L at the stationary p oints z 2 and z 3 dep end c ontinuously on ξ . Sinc e ξ → − q 0 as t → ∞ , the slop es of the tangents at z 2 and z 3 appr o ach √ 3 or − √ 3 by L emma 4.1 . Conse quently, the c ontour deformation ne ar the stationary p oints z 2 and z 3 c an always b e r e alize d when t is suﬃciently lar ge (se e Fig. 4 ). 4.1 The ﬁrst transformation This transformation is identical to the one described in [ 36 , section 3.2]. Its main purp ose is to handle the jump matrix on R + so that it can b e analytically con tin ued oﬀ the real axis in a “suitable” manner. Let us now explain this p oin t. W e ﬁrst note that the jump matrix e V admits the following upp er-low er triangular factorization: e V =    1 − 1 γ r ∗ 1 e θ 12 − r ∗ 2 e θ 13 0 1 − r 3 e θ 23 0 0 1       1 0 0 r 1 e θ 21 1 0 r 2 e θ 31 − 1 γ r ∗ 3 e θ 32 1    . (4.11) Ho w ev er, this factorization is not suitable on R + b ecause, as indicated b y the signature tables (Fig. 2 ), the jump matrix extended via the abov e triangular factorization gro ws exp onentially . Therefore, we need to introduce a function ∆ ( z ) suc h that ∆ − 1 − e V ∆ + admits a low er-upp er triangular factorization on R + . This kind of jump matrix is then “suitable” for the next transformation. As we ha v e shown in [ 36 , section 3.2] ho w to construct ∆ , we will omit the details below and presen t the result directly . 22 The ﬁrst transformation is deﬁned by M (1) ( x, t, z ) = ∆ − 1 ∞ M ( x, t, z ) ∆ ( z ) , where ∆ ( z ) =    δ 1 ( z ) 1 δ 1 ( z ) δ 1 ( ˆ z ) δ 1 ( ˆ z )    , (4.12) with δ 1 ( z ) = exp  − 1 2 π i Z R + ln  1 − 1 γ ( s ) | r 1 ( s ) | 2 − | r 2 ( s ) | 2  s − z d s  , , (4.13) ∆ ∞ = diag (1 , 1 /δ 1 (0) , δ 1 (0)) . (4.14) Here δ 1 (0) = exp  − 1 2 π i R R + ln ( 1 − 1 γ ( s ) | r 1 ( s ) | 2 −| r 2 ( s ) | 2 ) s d s  is w ell-deﬁned. Moreo v er, although ∆ has a singularit y at z = q 0 , we hav e already pro v ed in [ 36 , Lemma 3.3] that the limit of M (1) exists as z → q 0 from both the upper and lo wer half-planes. W e now present the jump matrix of M (1) on the real axis. By a direct calculation, we hav e e V (1) =                                 1 − 1 γ r ∗ 1 1 δ 2 1 ( z ) δ 1 ( ˆ z ) e θ 12 − r ∗ 2 δ 1 ( z ) δ 1 ( ˆ z ) e θ 13 0 1 − r 3 δ 2 1 ( ˆ z ) δ 1 ( z )e θ 23 0 0 1           1 0 0 r 1 δ 2 1 ( z ) δ 1 ( ˆ z )e θ 21 1 0 r 2 δ 1 ( ˆ z ) δ 1 ( z ) e θ 31 − 1 γ r ∗ 3 1 δ 2 1 ( ˆ z ) δ 1 ( z ) e θ 32 1      , z ∈ R − ,      1 0 0 − ˜ r 1 e θ 21 1 0 ˜ r 2 e θ 31 1 γ ˜ r ∗ 3 e θ 32 1           1 1 γ ˜ r ∗ 1 e θ 12 − ˜ r ∗ 2 e θ 13 0 1 ˜ r 3 e θ 23 0 0 1      , z ∈ R + , (4.15) where ˜ r 1 ( z ) = ρ 2 ( z ) ρ ( ˆ z ) ˆ r 1 ( z ) , ˜ r 2 ( z ) = ρ ( z ) ρ ( ˆ z ) ˆ r 2 ( z ) , ˜ r 3 ( z ) = ρ ( z ) ρ 2 ( ˆ z ) ˆ r 3 ( z ) , with ˆ r 1 ( z ) = b 21 ( z ) b 11 ( z ) , ˆ r 2 ( z ) = a 31 ( z ) a 33 ( z ) , ˆ r 3 ( z ) = b 23 ( z ) b 33 ( z ) , ρ ( z ) = N 1 Y j =1 z − ζ ∗ j z − ζ j N 2 Y j =1 z − z ∗ j z − z j exp  − 1 2 π i Z R − ln ( a 11 ( ζ ) b 11 ( ζ )) ζ − z d ζ  . F or the deriv ation of jump matrix e V (1) , w e refer the reader to [ 36 , section 3.2]. Additionally , w e remark that the function ρ ( z ) deﬁned here diﬀers sligh tly from the one deﬁned in [ 36 , section 3.2], as we are no w considering the case where b oth types of discrete sp ectra are present. Nev ertheless, the corresp onding computational pro cedure remains essen tially preserv ed. Then it’s easy to verify that M (1) satisﬁes the following RH problem: 23 RH Problem 4.3. Find a 3 × 3 matrix-value d function M (1) ( x, t, z ) with the fol lowing pr op- erties: • M (1) ( x, t, · ) : C \ ( R ∪ Z ) → C 3 × 3 is analytic. M (1) ( x, t, z ) satisﬁes the jump c ondition: M (1) + ( x, t, z ) = M (1) − ( x, t, z ) e V (1) ( x, t, z ) , z ∈ R \ { 0 } . (4.16) • M (1) ( x, t, z ) admits the asymptotic b ehavior: M (1) ( x, t, z ) = I + O ( 1 z ) , z → ∞ ; M (1) ( x, t, z ) = 1 z σ 1 + O (1) , z → 0 . (4.17) • M (1) ( x, t, z ) satisﬁes the gr owth c onditions ne ar the br anch p oints ± q 0 :    M (1) 1 = O ( z + q 0 ) , z ∈ C + → − q 0 , M (1) 3 = O ( z + q 0 ) , z ∈ C − → − q 0 ,    M (1) 3 = O ( z − q 0 ) , z ∈ C + → q 0 , M (1) 1 = O ( z − q 0 ) , z ∈ C − → q 0 . (4.18) • M (1) ( x, t, z ) satisﬁes the symmetries M (1) ( x, t, z ) = M (1) ( x, t, ˆ z ) Π ( z ) , [( M (1) ( x, t, z )) − 1 ] ⊤ = − 1 γ ( z ) J  M (1) ( x, t, z ∗ )  ∗ Γ ( z ) . (4.19) • The r esidue c onditions satisﬁe d by M (1) take the same form as those for M , but with the r esidue c onstant τ j r elate d to ζ j r eplac e d by ˜ τ j := τ j | δ 1 ( ζ j ) | 2 , and the r esidue c onstant κ j r elate d to z j r eplac e d by ˜ κ j := κ j δ 2 1 ( z j ) δ 1 ( ˆ z j ) . Remark 4.4. Note that ∆ ( z ) satisﬁes the symmetries ∆ ( z ) = Π − 1 ( z ) ∆ ( ˆ z ) Π ( z ) and ∆ − 1 ( z ) = ∆ ∗ ( z ∗ ) . Then, thr ough a dir e ct c alculation, one c an verify that the jump matrix e V (1) satisﬁes the symmetries e V (1) ( x, t, z ) = Π − 1 ( z )  e V (1) ( x, t, ˆ z )  − 1 Π ( z ) , D ( z ) e V (1) ( x, t, z ) D − 1 ( z ) =  e V (1) ( x, t, z ∗ )  † , wher e D ( z ) = − γ ( z ) Γ − 1 ( z ) . W e observe that the RH problem 4.3 is singular at the origin. In order to remov e the singularit y , w e deﬁne the follo wing RH problem. RH Problem 4.5. Find a 3 × 3 matrix-value d function N ( x, t, z ) with the fol lowing pr op erties: • N ( x, t, · ) : C \ ( R ∪ Z ) → C 3 × 3 is analytic. A cr oss R , N ( x, t, z ) satisﬁes the jump c ondition: N + ( x, t, z ) = N − ( x, t, z ) e V (1) ( x, t, z ) , z ∈ R . (4.20) 24 • N ( x, t, z ) admits the asymptotic b ehavior: N ( x, t, z ) = I + O ( 1 z ) , z → ∞ ; N ( x, t, z ) = O (1) , z → 0 . • N ( x, t, z ) satisﬁes the same r esidue c onditions as M (1) ( x, t, z ) in the RH pr oblem 4.3 . The follo wing lemma establishes a connection betw een the t wo RH problems. Lemma 4.6 (Gauge T ransformation) . Supp ose that M (1) ( x, t, z ) is the unique solution to RH pr oblem 4.3 and N ( x, t, z ) is the unique solution to RH pr oblem 4.5 . Then ther e exists a function A ( x, t ) such that M (1) ( x, t, z ) =  I + A ( x, t ) z  N ( x, t, z ) , (4.21) wher e A ( x, t ) = σ 1 N − 1 + ( x, t, 0) = σ 1 N − 1 − ( x, t, 0) with N ± ( x, t, 0) := lim ϵ → 0 + N ( x, t, 0 ± i ϵ ) . Pr o of. See App endix A . Next, w e will perform transformations on RH problem 4.5 . 4.2 The second transformation In this subsection, we need to introduce a transformation such that the transformed jump matrix, except at the critical p oin t − q 0 , deca ys uniformly as t → ∞ . In the so-called left soliton region R − sol := { ( x, t ) , − C < x 2 t < − c } , we hav e already demonstrated how to ac hiev e this goal by performing contour deformations (see [ 36 , section 4]). Therefore, when ( x, t ) ∈ P + , w e should deﬁne the transformation matrix in the same spirit. In fact, the transformation matrix deﬁned here diﬀers only slightly from the one deﬁned in [ 36 , section 4]. Therefore, we will omit the calculations and directly presen t the deﬁnition of the transformation matrix G . W e ﬁrst deﬁne the functions ∆ ♯ ( z ) and P ( z ) as follo ws: ∆ ♯ ( z ) =    δ ♯ ( ˆ z ) 1 δ ♯ ( z ) δ ♯ ( ˆ z ) δ ♯ ( z )    , P ( z ) =    P 1 ( z ) 1 P 1 ( z )P 1 ( ˆ z ) P 1 ( ˆ z )    , where δ ♯ ( z ) = exp  1 2 π i Z − q 0 −∞ ln(1 + 1 γ ( s ) | r 3 ( s ) | 2 ) s − z d s  , (4.22) 25 P 1 ( z ) = Y j ∈∇ z − ζ j z − ζ ∗ j Y j ∈ e ∇ z − z j z − z ∗ j , (4.23) with ∇ =  1 ≤ j ≤ N 1 : ℜ ζ j > ξ  , e ∇ =  1 ≤ j ≤ N 2 : ℜ z j > ξ  . Lemma 4.7. δ ♯ ( z ) is wel l-deﬁne d for z ∈ C \ ( −∞ , − q 0 ] . When z → − q 0 along any non- tangential limit of ( −∞ , − q 0 ] , the fol lowing estimate holds:   δ ♯ ( z ) − δ ♯ ( − q 0 )   ≤ C | z + q 0 || ln | z + q 0 || . (4.24) Pr o of. The function ln(1 + 1 γ ( z ) | r 3 ( z ) | 2 ) is smo oth on ( −∞ , − q 0 ) and v anishes at z = − q 0 . Moreo v er, as z → −∞ , ln(1 + 1 γ ( z ) | r 3 ( z ) | 2 ) = O (1 /z 2 ). Therefore, a standard discussion of the endp oin t b ehavior of the Cauc h y integral (see, for example, [ 46 , Lemma 4.7] or [ 56 , Lemma 2.11]) yields ( 4.24 ). Let the contour Σ (1) and the region { D j } 9 j =1 b e as illustrated in Figure 4 , and let T ( z ) := ∆ ♯ ( z ) P ( z ) = diag ( T 1 ( z ) , T 2 ( z ) , T 3 ( z )) , T ♯ ( z ) := ∆ ( z ) ∆ ♯ ( z ) P ( z ) = diag  T ♯ 1 ( z ) , T ♯ 2 ( z ) , T ♯ 3 ( z )  . W e deﬁne the functions G j ( x, t, z ) for z ∈ D j ( j = 1 , 2 , ..., 9) as follows: G 1 = ∆ − 1 e Θ    1 0 0 − r 1 ( z ) 1 0 0 0 1       1 0 0 0 1 0 − r 2 ( z ) − r 1 ( z ) r ∗ 3 ( z ∗ ) γ ( z ) 0 1        1 0 0 0 1 r 3 ( z ) 1+ 1 γ ( z ) r 3 ( z ) r ∗ 3 ( z ∗ ) 0 0 1     e − Θ T ♯ ( z ) , G 5 = ∆ − 1 e Θ    1 − r ∗ 1 ( z ∗ ) γ ( z ) 0 0 1 0 0 0 1       1 0 − r ∗ 2 ( z ∗ ) − r ∗ 1 ( z ∗ ) r 3 ( z ) γ ( z ) 0 1 0 0 0 1        1 0 0 0 1 0 0 1 γ ( z ) r ∗ 3 ( z ∗ ) 1+ 1 γ ( z ) r 3 ( z ) r ∗ 3 ( z ∗ ) 1     e − Θ T ♯ ( z ) , G 3 = ∆ − 1 e Θ    1 0 0 0 1 0 0 r ∗ 3 ( z ∗ ) γ ( z ) 1       1 0 0 0 1 0 − r 2 ( z ) 0 1        1 1 γ ( z ) r ∗ 1 ( z ∗ ) 1 − 1 γ ( z ) r ∗ 1 ( z ∗ ) r 1 ( z ) 0 0 1 0 0 0 1     e − Θ T ♯ ( z ) , G 7 = ∆ − 1 e Θ    1 0 0 0 1 − r 3 ( z ) 0 0 1       1 0 − r ∗ 2 ( z ∗ ) 0 1 0 0 0 1        1 0 0 r 1 ( z ) 1 − 1 γ ( z ) r ∗ 1 ( z ∗ ) r 1 ( z ) 1 0 0 0 1     e − Θ T ♯ ( z ) , G 2 = ∆ − 1 e Θ    1 0 0 0 1 0 − r 2 ( z ) 0 1    e − Θ T ♯ ( z ) , G 6 = ∆ − 1 e Θ    1 0 − r ∗ 2 ( z ∗ ) 0 1 0 0 0 1    e − Θ T ♯ ( z ) , 26 D 1 D 2 D 2 D 6 D 6 D 5 D 3 D 7 D 8 D 9 D 4 Figure 4: The contour Σ (1) (solid), and the region { D j } 9 j =1 . The dashed line corresp onds to the con tour where ℜ Φ 31 ( ξ , z ) = 0. G 4 = e Θ    1 − 1 γ ( z ) ˜ r ∗ 1 ( z ∗ ) ˜ r ∗ 2 ( z ∗ ) + 1 γ ( z ) ˜ r ∗ 1 ( z ∗ ) ˜ r 3 ( z ) 0 1 − ˜ r 3 ( z ) 0 0 1    e − Θ T ( z ) , G 9 = e Θ    1 0 0 − ˜ r 1 ( z ) 1 0 ˜ r 2 ( z ) 1 γ ( z ) ˜ r ∗ 3 ( z ∗ ) 1    e − Θ T ( z ) , G 8 = ∆ − 1 e Θ     1 0 0 r 1 ( z ) 1 − 1 γ ( z ) r ∗ 1 ( z ∗ ) r 1 ( z ) 1 0 0 0 1     e − Θ T ♯ ( z ) . The transformation matrix G ( x, t, z ) is deﬁned by G ( x, t, z ) =    G j ( x, t, z ) , z ∈ D j , j = 1 , 2 , ..., 9 , T ( z ) , elsewhere. (4.25) The following transformation aims to ensure that the jump matrix uniformly approaches the iden tity matrix I everywhere except near the critical point − q 0 as t → ∞ . W e deﬁne N (1) ( x, t, z ) = ( T ∞ ) − 1 N ( x, t, z ) G ( x, t, z ) , where G ( x, t, z ) is deﬁned b y ( 4.25 ) and T ∞ := lim z →∞  ∆ ♯ ( z ) P ( z )  = diag  δ ♯ (0) , 1 /  δ ♯ (0)P 1 (0)  , P 1 (0)  . (4.26) The jump contour Σ (1) for N (1) is illustrated in Fig. 4 . F or con v enience, we deﬁne R 1 ( z ) = r 1 ( z ) 1 − 1 γ ( z ) r 1 ( z ) r ∗ 1 ( z ∗ ) , R 3 ( z ) = r 3 ( z ) 1 + 1 γ ( z ) r 3 ( z ) r ∗ 3 ( z ∗ ) , R 2 ( z ) = r 2 ( z ) + 1 γ ( z ) r 1 ( z ) r ∗ 3 ( z ∗ ) . 27 The jump matrix for N (1) is denoted by V (1) . The expression for V (1) is as follows: V (1) 1 ( z ) = e Θ     1 0 0 0 1 0 r 2 ( z ) T ♯ 1 T ♯ 3 ( z ) 0 1     e − Θ , V (1) 2 ( z ) = e Θ     1 0 0 0 1 0 − r 2 ( z ) T ♯ 1 T ♯ 3 ( z ) 0 1     e − Θ , V (1) 3 ( z ) = e Θ     1 0 − r ∗ 2 ( z ∗ ) T ♯ 3 T ♯ 1 ( z ) 0 1 0 0 0 1     e − Θ , V (1) 4 ( z ) = e Θ     1 0 r ∗ 2 ( z ∗ ) T ♯ 3 T ♯ 1 ( z ) 0 1 0 0 0 1     e − Θ , V (1) 5 ( z ) =      1 − R ∗ 1 ( z ∗ ) γ ( z ) T ♯ 2 T ♯ 1 ( z )e θ 12 0 0 1 0 0 − r ∗ 3 ( z ∗ ) γ ( z ) T ♯ 2 T ♯ 3 ( z )e θ 32 1      , V (1) 8 ( z ) =     1 0 0 − R 1 ( z ) T ♯ 1 T ♯ 2 ( z )e θ 21 1 r 3 ( z ) T ♯ 3 T ♯ 2 ( z )e θ 23 0 0 1     , V (1) 6 ( z ) = e Θ      1 0 0 − r 1 ( z ) T ♯ 1 T ♯ 2 ( z ) 1 R 3 ( z ) T ♯ 3 T ♯ 2 ( z ) ( r 2 ( z ) − R 2 ( z )) T ♯ 1 T ♯ 3 ( z ) 0 1      e − Θ , V (1) 7 ( z ) = e Θ ( T ♯ ) − 1    1 − ( r 1 ( z ∗ )) ∗ γ ( z ) + r ∗ 1 ( z ∗ ) r 3 ( z ) R ∗ 3 ( z ∗ ) γ 2 ( z ) ( r 2 ( z ∗ ) − R 2 ( z ∗ )) ∗ 0 1 0 0 − 1 γ ( z ) R ∗ 3 ( z ∗ ) 1    e − Θ T ♯ , V (1) 9 ( z ) = e Θ ( T ♯ ) − 1    1 − 1 γ ( z ) r 1 ( z ) r ∗ 1 ( z ∗ ) − r ∗ 1 ( z ∗ ) γ ( z ) 0 r 1 ( z ) 1 + 1 γ ( z ) r 3 ( z ) r ∗ 3 ( z ∗ ) − r 3 ( z ) 0 − r ∗ 3 ( z ∗ ) γ ( z ) 1    e − Θ T ♯ , V (1) 13 ( z ) = e Θ      1 0 0 −  R 1 ( z ) + ˜ r 1 ( z ) 1 δ 2 1 ( z ) δ 1 ( ˆ z )  T ♯ 1 T ♯ 2 ( z ) 1 0 ˜ r 2 ( z ) δ 1 ( ˆ z ) δ 1 ( z ) T ♯ 1 T ♯ 3 ( z ) ˜ r ∗ 3 ( z ∗ ) γ ( z ) δ 2 1 ( ˆ z ) δ 1 ( z ) T ♯ 2 T ♯ 3 ( z ) 1      e − Θ , V (1) 12 ( z ) = e Θ      1 0 − r ∗ 2 ( z ∗ ) T ♯ 3 T ♯ 1 ( z ) 0 1 [ − r ∗ 3 ( z ∗ ) + r ∗ 2 ( z ∗ ) R 1 ( z )] T ♯ 3 T ♯ 2 ( z ) 0 0 1      e − Θ , V (1) 10 = G − 1 4 T ♯ , V (1) 11 = ( T ♯ ) − 1 G 3 . Here, V (1) j denotes the jump matrix V (1) restricted to Σ (1) j . W e omit the expressions of the jump matrices on the remaining contours for brevit y . The follo wing lemma pro vides an estimate for the jump matrix near the origin. Lemma 4.8. F or z ∈ Σ (1) { 10 , 11 , 12 , 13 } , we have the estimate     V (1) ( x, t, z ) − I  ij    ≤ C | z | e t ℜ Φ ij ( ξ ,z ) . (4.27) 28 Mor e over, if ( i, j )  = (2 , 1) , we further have     V (1) ( x, t, z ) − I  ij    ≤ C | z | 2 e t ℜ Φ ij ( ξ ,z ) . (4.28) Pr o of. W e ﬁrst prov e that V (1) 12 satisﬁes estimates ( 4.27 ) and ( 4.28 ). If a matrix en try of V (1) 12 − I is zero, then it clearly satisﬁes estimates ( 4.27 ) and ( 4.28 ). It follows from a direct computation that V (1) 12 ( z ) − I =      0 0 − r ∗ 2 ( z ∗ ) T ♯ 3 T ♯ 1 ( z )e θ 13 0 0 [ − r ∗ 3 ( z ∗ ) + r ∗ 2 ( z ∗ ) R 1 ( z )] T ♯ 3 T ♯ 2 ( z )e θ 23 0 0 0      . Therefore, it suﬃces to estimate its (1 , 3) and (2 , 3) en tries. Let G ( z ) := [ − r ∗ 3 ( z ∗ ) + r ∗ 2 ( z ∗ ) R 1 ( z )] T ♯ 3 T ♯ 2 ( z ) . Observ e that b oth r 3 ( z ) and r 2 ( z ) are O ( z 2 ) as Σ (1) 12 ∋ z → 0, so the following estimate holds for z ∈ Σ (1) 12 : | r ∗ 2 ( z ∗ ) | ≤ C | z | 2 , | r 3 ( z ) | ≤ C | z | 2 , | R 1 ( z ) − R 1 (0) | ≤ C | z | . On the other hand, we hav e      T ♯ 3 T ♯ 2 ( z )      < C, z ∈ Σ (1) 12 , whic h implies the estimate |G ( z ) | ≤ C | z | 2 , z ∈ Σ (1) 12 . Th us we get     V (1) 12 ( x, t, z ) − I  23    ≤ C | z | 2 e t ℜ Φ 23 . The estimate for the (1 , 3) en try is entirely analogous, therefore w e conclude that V (1) 12 satisﬁes estimates ( 4.27 ) and ( 4.28 ). No w, let’s turn our attention to V (1) 13 . Direct computation yields V (1) 13 ( z ) − I =      0 0 0 −  R 1 ( z ) + ˜ r 1 ( z ) 1 δ 2 1 ( z ) δ 1 ( ˆ z )  T ♯ 1 T ♯ 2 ( z )e θ 21 0 0 ˜ r 2 ( z ) δ 1 ( ˆ z ) δ 1 ( z ) T ♯ 1 T ♯ 3 ( z )e θ 31 ˜ r ∗ 3 ( z ∗ ) γ ( z ) δ 2 1 ( ˆ z ) δ 1 ( z ) T ♯ 2 T ♯ 3 ( z )e θ 32 0      . Next, w e will sho w ho w to estimate the (2 , 1) entry . Let Q ( z ) := R 1 ( z ) + ˜ r 1 ( z ) 1 δ 2 1 ( z ) δ 1 ( ˆ z ) = R 1 ( z ) + ˆ r 1 ( z ) ρ 2 ( z ) ρ ( ˆ z ) δ 2 1 ( z ) δ 1 ( ˆ z ) . 29 Using the so-called trace formula (see [ 19 , section 3.3]), it is straigh tforw ard to verify that δ 1 ( z ) = s 1 ( z ) ρ ( z ), where s 1 ( z ) =    a 11 ( z ) , z ∈ C + , 1 b 11 ( z ) , z ∈ C − . Based on this relationship, we obtain Q ( z ) = R 1 ( z ) + ˆ r 1 ( z ) b 2 11 ( z ) b 33 ( z ) . A straigh tforward calculation giv es us |Q ( z ) | ≤ | R 1 ( z ) − R 1 (0) | + | ˆ r 1 − ˆ r 1 (0) | +   R 1 (0) + ˆ r 1 (0) b 2 11 ( z ) b 33 ( z )   . (4.29) On the other hand, for z ∈ Σ (1) 13 w e hav e | R 1 (0) − r 1 (0) | ≤ C | z | , | ˆ r 1 (0) b 2 11 ( z ) + r 1 (0) | ≤ C | z | . (4.30) W e need to demonstrate wh y the last inequalit y abov e holds. It’s easy to v erify that ˆ r 1 ( z ) b 2 11 ( z ) = b 12 ( z ) b 11 ( z ) b 2 11 ( z ) = − r 1 ( z ) a 33 ( z ) a 11 ( z ) b 11 ( z ) + a 23 ( z ) a 31 ( z ) b 11 ( z ) . (4.31) Applying the ab ov e equation along with the estimates | a 33 ( z ) a 11 ( z ) b 11 ( z ) − 1 | ≤ C | z | , | a 23 ( z ) a 31 ( z ) b 11 ( z ) | ≤ C | z | 2 , (4.32) w e obtain | ˆ r 1 (0) b 2 11 ( z ) + r 1 (0) | ≤ | ˆ r 1 ( z ) b 2 11 ( z ) + r 1 (0) | + C | ˆ r 1 ( z ) − ˆ r 1 (0) | ≤ C | − r 1 ( z ) + r 1 (0) | + C | a 33 ( z ) a 11 ( z ) b 11 ( z ) − 1 | + | a 23 ( z ) a 31 ( z ) b 11 ( z ) | + C | ˆ r 1 ( z ) − ˆ r 1 (0) | ≤ C | z | . This pro v es the last inequalit y in ( 4.30 ). Let us con tin ue our calculations to estimate Q ( z ). Com bining ( 4.29 ) with ( 4.30 ), w e conclude that Q ( z ) satisﬁes |Q ( z ) | ≤ | R 1 ( z ) − R 1 (0) | + | ˆ r 1 − ˆ r 1 (0) | + | R 1 (0) − r 1 (0) | + | ˆ r 1 (0) b 2 11 ( z ) + r 1 (0) | ≤ C | z | . So w e immediately obtain     V (1) 13 ( x, t, z ) − I  21    ≤ C | z | e t ℜ Φ 21 . The method for estimating the (3 , 1) − and (3 , 2) − entries of  V (1) 13 ( x, t, z ) − I  is en tirely similar to the metho d used for estimating  V (1) 12 ( x, t, z ) − I  23 ; w e omit the details and state directly that they satisfy     V (1) 13 ( x, t, z ) − I  31    ≤ C | z | 2 e t ℜ Φ 31 ,     V (1) 13 ( x, t, z ) − I  32    ≤ C | z | 2 e t ℜ Φ 32 . 30 Th us V (1) 13 satisﬁes estimates ( 4.27 ) and ( 4.28 ). The steps to pro v e that V (1) 10 and V (1) 11 satisfy estimates ( 4.27 ) and ( 4.28 ) are en tirely anal- ogous; for brevity , w e omit them. Let ϵ > 0 b e a small constant, and let D ϵ denote the op en disk of radius ϵ cen tered at the p oin t − q 0 . Lemma 4.9. The jump matrix V (1) c onver ges to the identity matrix I as t → ∞ uniformly for ( x, t ) ∈ P + and z ∈ Σ (1) exc ept ne ar the p oint − q 0 . Mor e over, the fol lowing estimates hold: ∥ V (1) − I ∥ L 1 (Σ (1) \D ϵ ) ≤ C t − 1 , ∥ V (1) − I ∥ L ∞ (Σ (1) \D ϵ ) ≤ C t − 1 / 2 . (4.33) Pr o of. The pro of of this lemma is en tirely analogous to the proof of [ 36 , Lemma 3.10]. Remark 4.10. Sinc e for ( i, j )  = (2 , 1) , we have     1 z  V (1) ( x, t, z ) − I  ij     ≤ C | z | e t ℜ Φ ij . Thus one c an e asily verify that the fol lowing estimates hold:     1 z  V (1) ( x, t, z ) − I  ij     L 1 (Σ (1) \D ϵ ) ≤ C t − 1 , ( i, j )  = (2 , 1) . (4.34) W e note that this transformation aﬀects the residue conditions; below, we present the residue conditions satisﬁed by N (1) ( x, t, z ). Lemma 4.11. At e ach p oint ζ ∈ Z , only one c olumn of N (1) ( x, t, z ) has a simple p ole at ζ , while the other two c olumns ar e analytic. Mor e over, the fol lowing r esidue c onditions hold: (i) – F or j ∈ { 1 , ..., N 1 } ∩ ∇ , we have Res z = ζ j N (1) ( x, t, z ) = lim z → ζ j N (1) ( x, t, z )    0 0 C ♯ j 0 0 0 0 0 0    , (4.35) wher e C ♯ j = τ − 1 j e θ 13 ( x,t,ζ j ) [( T ♯ 3 ) − 1 ] ′ ( ζ j )( T ♯ 1 ) ′ ( ζ j ) . – F or j ∈ { 1 , ..., N 1 } \ ∇ , we have Res z = ζ j N (1) ( x, t, z ) = lim z → ζ j N (1) ( x, t, z )    0 0 0 0 0 0 e C ♯ j 0 0    , (4.36) wher e e C ♯ j = τ j T ♯ 1 ( ζ j ) T ♯ 3 ( ζ j ) e θ 31 ( x,t,ζ j ) . 31 (ii) – F or j ∈ { 1 , ..., N 2 } ∩ e ∇ , we have Res z = z j N (1) ( x, t, z ) = lim z → z j N (1) ( x, t, z )    0 D ♯ j 0 0 0 0 0 0 0    , Res z = z ∗ j N (1) ( x, t, z ) = lim z → z ∗ j N (1) ( x, t, z )    0 0 0 ˇ D ♯ j 0 0 0 0 0    , (4.37) wher e D ♯ j = κ − 1 j e θ 12 ( x,t,z j ) [( T ♯ 2 ) − 1 ] ′ ( z j )( T ♯ 1 ) ′ ( z j ) , ˇ D ♯ j = γ ( z ∗ j )( κ ∗ j ) − 1 e θ 21 ( x,t,z ∗ j ) [( T ♯ 1 ) − 1 ] ′ ( z ∗ j )( T ♯ 2 ) ′ ( z ∗ j ) . – F or j ∈ { 1 , ..., N 2 } \ e ∇ , we have Res z = z j N (1) ( x, t, z ) = lim z → z j N (1) ( x, t, z )    0 0 0 e D ♯ j 0 0 0 0 0    , Res z = z ∗ j N (1) ( x, t, z ) = lim z → z ∗ j N (1) ( x, t, z )    0 D ♯ j 0 0 0 0 0 0 0    , (4.38) wher e e D ♯ j = κ j T ♯ 1 ( z j ) T ♯ 2 ( z j ) e θ 21 ( x,t,z j ) , D ♯ j = κ ∗ j γ ( z ∗ j ) T ♯ 2 ( z ∗ j ) T ♯ 1 ( z ∗ j ) e θ 12 ( x,t,z ∗ j ) . Pr o of. Let X i denote the i -th column of the matrix X . When j ∈ { 1 , ..., N 1 } ∩ ∇ + , it follo ws from ( 3.12 ) and the deﬁnition of P 1 ( z ) that the third column of N (1) has a simple p ole, and w e hav e Res z = ζ j N (1) 3 ( x, t, z ) =  lim z → ζ j ( z − ζ j ) T 3 ( z )  T − 1 ∞ N 3 ( x, t, ζ j ) = ˜ τ − 1 j e θ 13 ( x,t,ζ j )  lim z → ζ j ( z − ζ j ) T 3 ( z )  T − 1 ∞ lim z → ζ j ( z − ζ j ) N 1 ( x, t, z ) = τ − 1 j δ 1 ( ζ ∗ j ) δ 1 ( ζ j ) e θ 13 ( x,t,ζ j )  lim z → ζ j ( z − ζ j ) T 3 ( z )   lim z − ζ j z − ζ j T 1 ( z )  × T − 1 ∞ lim z → ζ j ( N 1 ( x, t, z ) T 1 ( z )) = τ − 1 j e θ 13 ( x,t,ζ j )  lim z → ζ j ( z − ζ j ) T ♯ 3 ( z )   lim z − ζ j z − ζ j T ♯ 1 ( z )  N (1) 1 ( x, t, ζ j ) = τ − 1 j e θ 13 ( x,t,ζ j ) ( T ♯ 1 ) ′ ( ζ j )[( T ♯ 3 ) − 1 ] ′ ( ζ j ) N (1) 1 ( x, t, ζ j ) , where in the last equality we ha v e used the equation lim z → ζ j ( z − ζ j ) T ♯ 3 ( z ) = 1 [( T ♯ 3 ) − 1 ] ′ ( ζ j ) . The pro of for the remaining form ulas is similar. 32 4.3 The third transformation In this subsection, w e remo v e the p oles by in tro ducing small jumps in a neighborho o d of eac h discrete sp ectrum ζ ∈ Z . W e then show that the jumps near the discrete sp ectrum deca y uniformly to the iden tit y matrix at an exp onential rate, allo wing this contribu tion to be absorb ed in to the error term. W e b egin b y introducing the follo wing notation: ζ j =    ζ j , 1 ≤ j ≤ N 1 , z j − N 1 , N 1 + 1 ≤ j ≤ N 1 + N 2 . F or each ζ j , 1 ≤ j ≤ N 1 + N 2 , w e let D j b e a small op en disk centered at ζ j of radius ϵ > 0. W e let ˆ D j , D ∗ j , and ˆ D ∗ j b e the images of D j under the maps z → ˆ z , z → z ∗ , and z → ˆ z ∗ , resp ectiv ely . If 1 ≤ j ≤ N 1 , then ˆ D j = D ∗ j , ˆ D ∗ j = D j . W e choose ϵ suﬃciently small so that these small disks are disjoin t. W e then let D sol = ∪ N 1 + N 2 j =1  D j ∪ ˆ D j ∪ D ∗ j ∪ ˆ D ∗ j  . Based on the residue condition satisﬁed b y N (1) (see Lemma 4.11 ), w e introduce the follo wing 33 matrix: H ( x, t, z ) =                                                                                                         0 0 C ♯ j 0 0 0 0 0 0      , z ∈ D j , j ∈ { 1 , ..., N 1 } ∩ ∇ ,      0 0 0 0 0 0 e C ♯ j 0 0      , z ∈ D j j ∈ { 1 , ..., N 1 } \ ∇ ,      0 D ♯ j 0 0 0 0 0 0 0      , z ∈ D j , j − N 1 ∈ { 1 , ..., N 2 } ∩ e ∇ ,      0 0 0 e D ♯ j 0 0 0 0 0      , z ∈ D j , j − N 1 ∈ { 1 , ..., N 2 } \ e ∇ ,      0 0 0 ˇ D ♯ j 0 0 0 0 0      , z ∈ D ∗ j , j − N 1 ∈ { 1 , ..., N 2 } ∩ e ∇ ,      0 D ♯ j 0 0 0 0 0 0 0      , z ∈ D ∗ j , j − N 1 ∈ { 1 , ..., N 2 } \ e ∇ . (4.39) Note that H ( x, t, z ) has not y et b een deﬁned in ˆ D j for 1 ≤ j ≤ N 1 + N 2 and ˆ D ∗ j for N 1 + 1 ≤ j ≤ N 1 + N 2 . W e extend the domain of deﬁnition of H to these regions via the following iden tit y: H ( x, t, z ) = − ˆ ζ ζ j Π − 1 ( ˆ ζ j ) H ( x, t, ˆ z ) Π ( ˆ ζ j ) . Th us H ( x, t, z ) is deﬁned on the en tire region D sol . The second transformation is deﬁned as follo ws: N (2) ( x, t, z ) = N (1) ( x, t, z ) ˜ H ( x, t, z ) , where ˜ H ( x, t, z ) =                      I − 1 z − ζ j H ( x, t, z ) , z ∈ D j , 1 ≤ j ≤ N 1 + N 2 , I − 1 z − ζ ∗ j H ( x, t, z ) , z ∈ D ∗ j , 1 ≤ j ≤ N 1 + N 2 , I − 1 z − ˆ ζ j H ( x, t, z ) , z ∈ ˆ D j , N 1 + 1 ≤ j ≤ N 1 + N 2 , I − 1 z − ˆ ζ ∗ j H ( x, t, z ) , z ∈ ˆ D ∗ j , N 1 + 1 ≤ j ≤ N 1 + N 2 , I , z ∈ C \ D sol . 34 Let ∂ D j and ∂ ˆ D ∗ j and b e oriented counterclockwise, and let D ∗ j and ˆ D j b e orien ted clo ckwise. W e further deﬁne ∂ D sol b y ∂ D sol = ∪ N 1 + N 2 j =1  ∂ D j ∪ ∂ ˆ D j ∪ ∂ D ∗ j ∪ ∂ ˆ D ∗ j  . Lemma 4.12. The function N (2) ( x, t, z ) is analytic at e ach discr ete sp e ctr al p oint ζ j , 1 ≤ j ≤ N 1 + N 2 , and satisﬁes the jump c ondition N (2) + ( x, t, z ) = N (2) − ( x, t, z ) V (2) ( x, t, z ) , z ∈ Σ (2) := Σ (1) ∪ ∂ D sol , wher e V (2) ( x, t, z ) =          V (1) ( x, t, z ) , z ∈ Σ (1) , ˜ H ( x, t, z ) , z ∈ D j ∪ ˆ D ∗ j , j = 1 , ..., N 1 + N 2 , ˜ H − 1 ( x, t, z ) , z ∈ D ∗ j ∪ ˆ D j , j = 1 , ..., N 1 + N 2 . Mor e over, the fol lowing estimate holds uniformly as t → ∞ and ( x, t ) ∈ P + : ∥ V (2) ( x, t, · ) − I ∥ ( L 1 ∩ L ∞ )( ∂ D sol ) = O (e − ct ) . (4.40) The proof of this lemma is a straightforw ard calculation and is omitted for brevit y . 4.4 The lo cal parametrix In the previous sub ections, w e ha v e shown that the jump matrix V (2) deca ys uniformly to I except in the vicinity of the p oint − q 0 . Now we sho w that N (2) can b e approximated lo cally near − q 0 b y solution of the mo del RH problem 2.1 . The ﬁrst step inv olves constructing appro ximations for the phase function θ 13 ( x, t, z ), whic h is deﬁned by θ 13 ( x, t, z ) = i( − 2 z + 2 q 2 0 z ) x + i(2 z 2 − 2 q 4 0 z 2 ) t, and app ears in the exp onential of the (1 , 3) entry of the jump matrix V (2) . The b ehavior of this function is essential for matching the Painlev ´ e I I mo del problem. F or this purp ose, by p erforming a T a ylor expansion of this function at z = − q 0 , w e obtain θ 13 ( x, t, z ) = − 2i  ( x + 2 tq 0 )( z + q 0 ) + ( x 2 q 0 + t )( z + q 0 ) 2 + x + 4 tq 0 2 q 2 0 ( z + q 0 ) 3  + ˜ S 1 ( x, t, z ) , where the remainder term ˜ S 1 ( x, t, z ) is giv en by ˜ S 1 ( x, t, z ) = t ∂ 4 k Φ 13 ( ξ , z ⋆ ) 4! ( z + q 0 ) 4 . Here z ⋆ denotes a p oint lies on the line segmen t b etw een z and − q 0 . 35 In tro duce the v ariables y and β by y = 2  4 3 q 0  1 / 3 ( ξ + q 0 ) t 2 / 3 , β =  3 t 4 q 0  1 / 3 ( z + q 0 ) , (4.41) and deﬁne function S 1 ( x, t, z ) b y S 1 ( x, t, z ) = − 2i( x 2 q 0 + t )( z + q 0 ) 2 − i  x + 2 tq 0 q 2 0  ( z + q 0 ) 3 + ˜ S 1 ( x, t ) . (4.42) Then the function θ 13 ( x, t, z ) can b e expressed as follo ws : θ 13 ( x, t, z ) = − 2i  y β + 4 3 β 3  + S 1 ( x, t, z ) . (4.43) F or the sake of subsequent calculations, w e also need to examine the functions θ 12 ( x, t, z ) and θ 32 ( x, t, z ). By applying T a ylor expansions to these functions around the p oin t − q 0 , w e arriv e at the follo wing expressions: θ 32 ( x, t, z ) = θ 32 ( x, t, − q 0 ) + i( x + 2 q 0 t )( z + q 0 ) +  i t + i 2 t q 0 ( ξ + q 0 )  ( z + q 0 ) 2 +  i q 2 0 ( x + 2 q 0 t ) + 2i t q 0  ( z + q 0 ) 3 + ˜ S 2 ( x, t, z ) , θ 12 ( x, t, z ) = θ 12 ( x, t, − q 0 ) − i( x + 2 tq 0 )( z + q 0 ) + i t ( z + q 0 ) 2 , where the remainder term is deﬁned as ˜ S 2 ( x, t, z ) = t ∂ 4 k Φ 32 ( ξ , z ⋆ ) 4! ( z + q 0 ) 4 . Here z ⋆ represen ts a p oint situated on the line segmen t connecting z and − q 0 . Let S 2 ( x, t, z ) = i 2 t q 0 ( ξ + q 0 )( z + q 0 ) 2 +  i q 2 0 ( x + 2 q 0 t ) + 2i t q 0  ( z + q 0 ) 3 + ˜ S 2 ( x, t, z ) . (4.44) Then the functions θ 12 ( x, t, z ) and θ 32 ( x, t, z ) can b e expressed as θ 32 ( x, t, z ) = θ 32 ( x, t, − q 0 ) + i y β ( z ) + i  4 3 q 0  2 / 3 t 1 / 3 β 2 ( z ) + S 2 ( x, t, z ) and θ 12 ( x, t, z ) = θ 12 ( x, t, − q 0 ) − i y β ( z ) + i  4 3 q 0  2 / 3 t 1 / 3 β 2 ( z ) , resp ectiv ely . Recall that D ϵ denote the op en disk of radius ϵ centered at the p oin t − q 0 . Then z → β is a biholomorphism from D ϵ on to the op en disk of radius (3 t/ (4 q 0 )) 1 / 3 ϵ cen tered at the 36 Figure 5: The contour X ϵ is the solid line within the dashed circle; the region enclosed b y the dashed circle is D ϵ . origin. In particular, the images of the critical p oints { z j } 3 j =0 of the phase functions Φ 21 ( ξ , z ), Φ 32 ( ξ , z ) and Φ 31 ( ξ , z ) in the β -plane are { β j } 3 j =0 , and their expressions are as follo ws: β j =  3 t 4 q 0  1 / 3 ( z j + q 0 ) . (4.45) Let X ϵ = Σ (2) ∩ D ϵ b e as sho wn in Fig 5 . Note that the angle b et w een X ϵ j ( j = 1 , 2 , ..., 8) and the real axis is either 1 6 π or 5 6 π . T o demonstrate ho w to construct mo del RH problem 2.1 w e need to study the b ehavior of the reﬂection co eﬃcients { r j } 3 j =1 at the branch p oint − q 0 . First, it follo ws directly from ( 3.10 ) that lim z →− q 0 r 2 ( z ) = i. No w, it remains to analyze the other t w o reﬂection co eﬃcien ts. One can observ e that r ∗ 3 ( z ) v anishes at the branch p oint − q 0 , while γ ( z ) has a simple zero at − q 0 ; thus, their ratio is w ell-deﬁned there. W e deﬁne ˜ s to be this ratio, i.e., ˜ s := lim z →− q 0 r ∗ 3 ( z ∗ ) γ ( z ) . (4.46) Then, b y symmetry , we immediately obtain lim z →− q 0 r ∗ 1 ( z ∗ ) γ ( z ) = − i ˜ s. F urthermore, w e need to consider the b ehavior of the functions T ♯ 2 T ♯ 1 ( z ) and T ♯ 2 T ♯ 3 ( z ) as z → − q 0 . F rom Lemma 4.7 , we know that T ♯ 2 T ♯ 1 ( − q 0 ) and T ♯ 2 T ♯ 3 ( − q 0 ) are well-deﬁned and satisfy T ♯ 2 T ♯ 1 ( − q 0 ) = T ♯ 2 T ♯ 3 ( − q 0 ) = 1  δ ♯ ( − q 0 ) δ 1 ( − q 0 )P 1 ( − q 0 )  3 . 37 Lemma 4.13. F or z ne ar − q 0 , the functions T ♯ 2 T ♯ 1 ( z ) and T ♯ 2 T ♯ 3 ( z ) satisfy the fol lowing estimate:      T ♯ 1 T ♯ 3 ( z ) − 1      ≤ C ( | ln | z + q 0 || ) | z + q 0 | , (4.47a)      T ♯ 2 T ♯ 1 ( z ) − T ♯ 2 T ♯ 1 ( − q 0 )      ≤ C ( | ln | z + q 0 || ) | z + q 0 | , (4.47b)      T ♯ 2 T ♯ 3 ( z ) − T ♯ 2 T ♯ 3 ( − q 0 )      ≤ C ( | ln | z + q 0 || ) | z + q 0 | . (4.47c) Pr o of. The lemma is a direct consequence of Lemma 4.7 . No w we deﬁne s as s := ˜ s T ♯ 2 T ♯ 3 ( − q 0 ) = ˜ s  δ ♯ ( − q 0 ) δ 1 ( − q 0 )P 1 ( − q 0 )  3 , (4.48) whic h is the parameter that app ears in mo del RH problem 2.1 . In order to relate N (2) to N X , w e deﬁne Y ( x, t ) =    e θ 1 ( x,t, − q 0 ) e θ 2 ( x,t, − q 0 ) e θ 3 ( x,t, − q 0 )    =    1 e − i q 0 x − i q 2 0 t 1    . (4.49) Let ˜ N ( x, t, z ) = N (2) ( x, t, z ) Y ( x, t ) , z ∈ D ϵ . Clearly , the jump matrix ˜ V = Y − 1 V (2) Y . Let ˜ V | X ϵ denotes the restriction of ˜ V to X ϵ . F or a ﬁxed β , as t → ∞ (this leads to z → − q 0 ), w e ha v e r 2 ( z ) T ♯ 1 T ♯ 3 ( z ) → i , r ∗ 3 ( z ∗ ) γ ( z ) T ♯ 2 T ♯ 3 ( z ) → s, r ∗ 1 ( z ∗ ) γ ( z ) T ♯ 2 T ♯ 1 ( z ) → − i s. This suggests that ˜ V | X ϵ ( x, t, z ( β )) tends to the jump matrix V X ( x, t, β ) deﬁned in ( 2.1 ) for large t . The ab ov e analysis suggests that we can appro ximate N (2) ( x, t, z ) in D ϵ b y the 3 × 3 matrix-v alued function N loc ( x, t, z ) deﬁned b y N loc ( x, t, z ) = Y N X ( y , s, t ; β ( z )) Y − 1 , z ∈ D ϵ , (4.50) where N X ( y , s, t ; β ) is the solution of the mo del RH problem 2.1 . In order to sho w that N loc appro ximates N (2) w ell in D ϵ , w e need to carry out some preparatory w ork. 38 Lemma 4.14. L et ( x, t ) ∈ P + . Then the fol lowing estimate holds for lar ge t :    e − 2i ( y β ( z )+ 4 3 β 3 ( z ) )    ≤ C e − ct | z + q 0 | 3 , z ∈ X ϵ 3 ∪ X ϵ 4 , (4.51a)    e 2i ( y β ( z )+ 4 3 β 3 ( z ) )    ≤ C e − ct | z + q 0 | 3 , z ∈ X ϵ 1 ∪ X ϵ 2 , (4.51b)    exp n ± i y β ( z ) + i (4 / 3 q 0 ) 2 / 3 t 1 / 3 β 2 ( z ) o    ≤ C e − ct | z + q 0 | 2 , z ∈ X ϵ 5 ∪ X ϵ 7 , (4.51c)    exp n ± i y β ( z ) − i ((4 / 3) q 0 ) 2 / 3 t 1 / 3 β 2 ( z ) o    ≤ C e − ct | z + q 0 | 2 , z ∈ X ϵ 6 ∪ X ϵ 8 , (4.51d) wher e c > 0 and C > 0 ar e indep endent of ξ and z . Pr o of. W e ﬁrst prov e ( 4.51a ). F or z ∈ X ϵ 3 ∪ X ϵ 4 and ( x, t ) ∈ P + , w e hav e ℜ ( − 2i y β ( z )) ≤ 0 . It immediately follo ws that    e − 2i ( y β ( z )+ 4 3 β 3 ( z ) )    ≤ e ℜ ( − i 8 3 β 3 ( z ) ) ≤ C e − ct | z + q 0 | 3 for z ∈ X ϵ 3 ∪ X ϵ 4 and ( x, t ) ∈ P + . The pro of of ( 4.51b ) proceeds analogously . Next, w e prov e ( 4.51c ). Let z ∈ X ϵ 5 . First, supp ose | z − z 1 | ≥ 3 | ξ + q 0 | . In this case, w e ha v e | z + q 0 | ≥ 3 | ξ + q 0 | . Observe that ± i y β ( z ) + i (4 / 3 q 0 ) 2 / 3 t 1 / 3 β 2 ( z ) = i t ( z + q 0 ) 2  ± 2( ξ + q 0 ) z + q 0 + 1  . (4.52) Since    ξ + q 0 z + q 0    ≤ 1 / 3 when | z − z 1 | ≥ 3 | ξ + q 0 | , it follows that     ± 2( ξ + q 0 ) z + q 0 + 1     ≥ 1 / 3 , z ∈ X ϵ 5 , | z − z 1 | ≥ 3 | ξ + q 0 | . (4.53) Consequen tly , com bining ( 4.52 ) and ( 4.53 ), w e deduce that there exists a c > 0 such that ℜ  ± i y β ( z ) + i (4 / 3 q 0 ) 2 / 3 t 1 / 3 β 2 ( z )  ≥ − ct | z + q 0 | 2 , z ∈ X ϵ 5 , | z − z 1 | ≥ 3 | ξ + q 0 | . Therefore, ( 4.51c ) holds for | z − z 1 | ≥ 3 | ξ + q 0 | , z ∈ X ϵ 5 . On the other hand, if | z − z 1 | ≤ 3 | ξ + q 0 | , then | z + q 0 | ≤ | z − z 1 | + | z 1 + q 0 | ≤ 4 | ξ + q 0 | ≤ C t − 2 / 3 , ( x, t ) ∈ P + . It follo ws that, for z ∈ X ϵ 5 satisfying | z − z 1 | ≤ 3 | ξ + q 0 | ,    exp n ± i y β ( z ) + i (4 / 3 q 0 ) 2 / 3 t 1 / 3 β 2 ( z ) o    ≤ e 2 t | ξ + q 0 || z + q 0 |    e i t ( z + q 0 ) 2    ≤ e C t − 1 / 3    e i t ( z + q 0 ) 2    ≤ C e − ct | z + q 0 | 2 . Therefore, ( 4.51c ) holds for all z ∈ X ϵ 5 and ( x, t ) ∈ P + . By an en tirely similar argumen t, the same conclusion also holds for all z ∈ X ϵ 7 . The pro of of ( 4.51d ) follo ws in the same w a y as that of ( 4.51c ), and w e omit the details. 39 Lemma 4.15. If ϵ > 0 is suﬃciently smal l, then for suﬃciently lar ge t we have | e t Φ 31 ( ξ ,z ) | ≤ C e − ct | z + q 0 | 3 , ( x, t ) ∈ P + , z ∈ X ϵ 1 ∪ X ϵ 2 ∪ X ϵ 6 , (4.54a) | e t Φ 13 ( ξ ,z ) | ≤ C e − ct | z + q 0 | 3 , ( x, t ) ∈ P + , z ∈ X ϵ 3 ∪ X ϵ 4 ∪ X ϵ 7 , (4.54b) | e t Φ 12 ( ξ ,z ) | + | e t Φ 32 ( ξ ,z ) | ≤ C e − ct | z + q 0 | 2 , ( x, t ) ∈ P + , z ∈ X ϵ 5 ∪ X ϵ 7 , (4.54c) | e t Φ 21 ( ξ ,z ) | + | e t Φ 23 ( ξ ,z ) | ≤ C e − ct | z + q 0 | 2 , ( x, t ) ∈ P + , z ∈ X ϵ 6 ∪ X ϵ 8 . (4.54d) wher e c > 0 and C > 0 ar e indep endent of ξ and z . Pr o of. W e b egin by pro ving ( 4.54a ). W e treat only the case z ∈ X ϵ 2 , since the other cases are analogous. Fix a > 0. Expanding Φ 31 ( ξ , z ) at z = − q 0 and using T a ylor’s form ula, w e obtain Φ 31 ( ξ , z ) = 2i  2( ξ + q 0 )( z + q 0 ) + 1 q 0 ( ξ + q 0 )( z + q 0 ) 2 + ξ + 2 q 0 q 2 0 ( z + q 0 ) 3  + ∂ 4 k Φ 31 ( ξ , z ⋆ ) 4! ( z + q 0 ) 4 , for ξ ∈ [ − q 0 , − q 0 + a ] and z ∈ X ϵ 2 , where z ⋆ lies b etw een z and − q 0 . Since z ⋆ → − q 0 as a, ϵ → 0 and ∂ 4 k Φ 31 dep ends contin uously on ( ξ , z ), the quantit y ∂ 4 k Φ 31 ( ξ , z ⋆ ) is uniformly b ounded for ξ ∈ [ − q 0 , − q 0 + a ] and z ∈ X ϵ 2 . Hence, for suﬃcien tly small ϵ > 0, ℜ Φ 31 ( ξ , z ) = − 2 ℑ  2( ξ + q 0 )( z + q 0 ) + 1 q 0 ( ξ + q 0 )( z + q 0 ) 2 + ξ + 2 q 0 q 2 0 ( z + q 0 ) 3  + O ( | z + q 0 | 4 ) , (4.55) uniformly for ( x, t ) ∈ P + . Next, recall that the critical p oint z 2 is giv en b y z 2 = η + i p 4 q 2 0 − η 2 2 , η = ξ − p ξ 2 + 8 q 2 0 2 , ( x, t ) ∈ P + . (4.56) Since ξ → − q 0 as t → ∞ in P + , it follows from ( 4.56 ) that arg( z 2 + q 0 ) → π 2 . Therefore, by elemen tary geometry , there exists T > 0 suc h that whenev er t ≥ T and z ∈ X ϵ 2 satisﬁes | z − z 2 | ≥ 3 | z 2 + q 0 | , one has arg( z + q 0 ) ∈  3 π 4 , 5 π 6  . (4.57) On the other hand, using ( 4.56 ) and ( 4.57 ), w e also obtain     ξ + q 0 z + q 0     ≤     ξ + q 0 z 2 + q 0     ≤ C p | ξ + q 0 | ≤ C t − 1 / 3 ≤ 1 2 , (4.58) for ( x, t ) ∈ P + , z ∈ X ϵ 2 with | z − z 2 | ≥ 3 | z 2 + q 0 | , and suﬃciently large t . Consequen tly , b y com bining the inequality ℑ ( z + q 0 ) 3 > 0 , z ∈ X ϵ 2 40 with ( 4.58 ) and ( 4.57 ), we infer that for z ∈ X ϵ 2 with | z − z 2 | ≥ 3 | z 2 + q 0 | , ℑ  1 q 0 ( ξ + q 0 )( z + q 0 ) 2 + ξ + 2 q 0 q 2 0 ( z + q 0 ) 3  ≥ ℑ  1 2 q 0 ( z + q 0 ) 3  ≥ √ 2 8 q 0 | z + q 0 | 3 , (4.59) pro vided t is suﬃcien tly large. Moreov er, ℑ  ( ξ + q 0 )( z + q 0 )  > 0 , ( x, t ) ∈ P + , z ∈ X ϵ 2 . Substituting this and ( 4.59 ) in to ( 4.55 ), and shrinking ϵ > 0 if necessary so that the quartic remainder is negligible, w e conclude that there exists c > 0 suc h that ℜ Φ 31 ( ξ , z ) ≤ − c | z + q 0 | 3 , for z ∈ X ϵ 2 satisfying | z − z 2 | ≥ 3 | z 2 + q 0 | , ( x, t ) ∈ P + , and suﬃcien tly large t . This pro ves ( 4.54a ) in the case z ∈ X ϵ 2 , | z − z 2 | ≥ 3 | z 2 + q 0 | . It remains to consider the case | z − z 2 | ≤ 3 | z 2 + q 0 | . In this case, | z + q 0 | ≤ | z − z 2 | + | z 2 + q 0 | ≤ 4 | z 2 + q 0 | ≤ C p | ξ + q 0 | ≤ C t − 1 / 3 , ( x, t ) ∈ P + . Hence t | z + q 0 | 3 ≤ C , and therefore | e t Φ 31 ( ξ ,z ) | = e ℜ Φ 31 ( ξ ,z ) ≤ 1 = e ct | z + q 0 | 3 × e − ct | z + q 0 | 3 ≤ C e − ct | z + q 0 | 3 , for z ∈ X ϵ 2 with | z − z 2 | ≤ 3 | z 2 + q 0 | and ( x, t ) ∈ P + . Com bining the t w o cases, w e obtain ( 4.54a ) for all z ∈ X ϵ 2 . The pro of of ( 4.54b ) is analogous and is therefore omitted. Finally , the pro ofs of ( 4.54c ) and ( 4.54d ) follow in exactly the same w ay as the proof of ( 4.51c ) in Lemma 4.14 . Lemma 4.16. L et ( x, t ) ∈ P + and cho ose ϵ > 0 suﬃciently smal l. Then ther e exists a c onstant C > 0 , indep endent of ξ and z , such that as t → ∞ , we have    e θ 13 ( x,t,z ) − e − 2i ( y β ( z )+ 4 3 β 3 ( z ) )    ≤ C t − 1 / 3 , z ∈ X ϵ 3 ∪ X ϵ 4 , (4.60a)    e θ 31 ( x,t,z ) − e 2i ( y β ( z )+ 4 3 β 3 ( z ) )    ≤ C t − 1 / 3 , z ∈ X ϵ 1 ∪ X ϵ 2 , (4.60b) | F 32 ( x, t, z ) | ≤ C t − 1 / 2 , z ∈ X ϵ 5 ∪ X ϵ 7 , (4.60c) | F 23 ( x, t, z ) | ≤ C t − 1 / 2 , z ∈ X ϵ 6 ∪ X ϵ 8 , (4.60d) wher e F 32 ( x, t, z ) = exp { θ 32 ( x, t, z ) − θ 32 ( x, t, − q 0 ) } − exp n i y β ( z ) + i ((4 / 3) q 0 ) 2 3 t 1 / 3 β 2 ( z ) o , F 23 ( x, t, z ) = exp { θ 23 ( x, t, z ) − θ 23 ( x, t, − q 0 ) } − exp n − i y β ( z ) − i ((4 / 3) q 0 ) 2 3 t 1 / 3 β 2 ( z ) o . 41 Pr o of. W e ﬁrst prov e ( 4.60a ). Using the standard inequality | e w 1 − e w 2 | ≤ ( | e w 1 | + | e w 2 | ) | w 1 − w 2 | , ∀ w 1 , w 2 ∈ C , (4.61) together with the estimates from ( 4.54b ) and ( 4.51a ), w e immediately obtain    e θ 13 ( x,t,z ) − e − 2i ( y β ( z )+ 4 3 β 3 ( z ) )    ≤ C | S 1 ( x, t, z ) | e − ct | z + q 0 | 3 , z ∈ X ϵ 3 ∪ X ϵ 4 . (4.62) Since ∂ 4 k Φ 13 ( ξ ,z ) 4! dep ends contin uously on ( ξ , z ) for ξ near − q 0 and z near − q 0 , by ( 4.42 ) we ha v e | S 1 ( x, t, z ) | ≤ C t 1 / 3 | z + q 0 | 2 + C t 1 / 3 | z + q 0 | 3 + C t | z + q 0 | 4 . (4.63) Com bining the ab ov e estimate with ( 4.62 ), w e obtain    e θ 13 ( x,t,z ) − e − 2i ( y β ( z )+ 4 3 β 3 ( z ) )    ≤ C t − 1 / 3 sup u ≥ 0 u 2 e − cu 3 + C t − 2 / 3 sup u ≥ 0 u 3 e − cu 3 + C t − 1 / 3 sup u ≥ 0 u 4 e − cu 3 ≤ C t − 1 / 3 , z ∈ X ϵ 3 ∪ X ϵ 4 . This completes the pro of of ( 4.60a ), and the pro of for ( 4.60b ) is similar. W e now pro v e ( 4.60c ). Similarly , using ( 4.54c ), ( 4.51c ), and ( 4.61 ), we ha v e | F 32 ( x, t, z ) | ≤ C | S 2 ( x, t, z ) | e − ct | z + q 0 | 2 , z ∈ X ϵ 5 ∪ X ϵ 7 . (4.64) It follo ws from ( 4.44 ) that S 2 ( x, t, z ) satisﬁes | S 2 ( x, t, z ) | ≤ C t 1 / 3 | z + q 0 | 2 + C t 1 / 3 | z + q 0 | 3 + C t | z + q 0 | 3 + C t | z + q 0 | 4 . (4.65) Then, w e obtain | F 32 ( x, t, z ) | ≤ C t − 2 / 3 sup u 2 ≥ 0 u 2 e − cu 2 + C t − 5 / 6 sup u ≥ 0 u e − cu 2 + C t − 1 / 2 sup u ≥ 0 u 2 e − cu 2 + C t − 1 sup u ≥ 0 u 4 e − cu 2 ≤ C t − 1 / 2 . The proof for ( 4.60d ) is similar. Remark 4.17. In fact, thr ough entir ely similar c alculations, one c an estimate the inte gr als of the functions on the left-hand side of ( 4.60a ) – ( 4.60d ) . Mor e pr e cisely, let { f j ( x, t, z ) } 4 j =1 denote the functions on the left-hand side of ine qualities ( 4.60a ) – ( 4.60d ) , r esp e ctively; then the fol lowing estimate holds: Z X ϵ 3 ∪X ϵ 4 f 1 ( x, t, z )d | z | ≤ C t − 2 3 , Z X ϵ 1 ∪X ϵ 2 f 2 ( x, t, z )d | z | ≤ C t − 2 3 , Z X ϵ 5 ∪X ϵ 7 f 3 ( x, t, z )d | z | ≤ C t − 2 3 , Z X ϵ 6 ∪X ϵ 8 f 4 ( x, t, z )d | z | ≤ C t − 2 3 . (4.66) 42 W e are no w ready to pro ve that N (2) can be w ell appro ximated b y N loc in D ϵ . Lemma 4.18. F or e ach ( x, t ) , the function N loc ( x, t, z ) deﬁne d in ( 4.50 ) is an analytic and b ounde d function of z ∈ D ϵ \ X ϵ . A cr oss X ϵ , N loc ob eys the jump c ondition N loc + = N loc − V loc , wher e the jump matrix V loc satisﬁes the fol lowing estimate for suﬃciently lar ge t :    ∥ V (2) − V loc ∥ L ∞ ( X ϵ ) ≤ C t − 1 / 3 ln t, ∥ V (2) − V loc ∥ L 1 ( X ϵ ) ≤ C t − 2 / 3 ln t, ( x, t ) ∈ P + . (4.67) F urthermor e, as t → ∞ , ∥ ( N loc ) − 1 − I ∥ L ∞ ( ∂ D ϵ ) = O ( t − 1 / 3 ) , (4.68) ( N loc ) − 1 ( x, t, z ) − I = − Y  E X 1 + M P 1  Y − 1 ( 3 4 q 0 ) 1 / 3 t 1 / 3 ( z + q 0 ) + O ( t − 2 / 3 ) , z ∈ ∂ D ϵ , (4.69) wher e E X 1 and M P 1 ar e given by ( 2.3 ) and ( B.3 ) , r esp e ctively. Pr o of. W e ha ve V (2) − V loc = Y  ˜ V − V X  Y − 1 , z ∈ X ϵ . Since Y is b ounded, the pro of of ( 4.67 ) reduces to estimating ˜ V − V X . Next we prov e the estimate ( 4.67 ) on the contours X ϵ 1 , X ϵ 7 and X ϵ 9 resp ectiv ely; the proof on the remaining contours can be obtained in the same w ay . On contour X ϵ 1 . Straightforw ard calculation yields ˜ V − V X =    0 0 0 0 0 0 f ( x, t, z ) 0 0    , for z ∈ X ϵ 1 , where f ( x, t, z ) = r 2 ( z ) T ♯ 1 T ♯ 3 ( z )e θ 31 ( x,t,z ) − ie 2i ( y β ( z )+ 4 3 β 3 ( z ) ) . By Lemma 4.13 , w e immediately obtain      r 2 ( z ) T ♯ 1 T ♯ 3 ( z ) − i      ≤ C ( | ln | z + q 0 || ) | z + q 0 | e t ℜ Φ 31 , and therefore | f ( x, t, z ) | ≤ C ( | ln | z + q 0 || ) | z + q 0 | e t ℜ Φ 31 + C    e θ 31 ( x,t,z ) − e 2i ( y β ( z )+ 4 3 β 3 ( z ) )    . Using ( 4.54a ) and ( 4.60b ), w e hav e | f ( x, t, z ) | ≤ C ( | ln | z + q 0 || ) | z + q 0 | e − ct | z + q 0 | 3 + C t − 1 / 3 43 ≤ C t − 1 / 3 ln t  sup u ≥ 0 u | ln u | e − cu 3  + C t − 1 / 3 ≤ C t − 1 / 3 ln t. Similarly , applying ( 4.54a ) together with the estimate ( 4.66 ) in Remark 4.17 , w e obtain Z X ϵ 1 | f ( x, t, z ) | d | z | ≤ C Z ∞ 0 τ | ln τ | e − ctτ 3 d τ + C t − 2 / 3 ≤ C t − 2 / 3 ln t. The abov e analysis sho ws that estimate ( 4.67 ) holds on X ϵ 1 . On contour X ϵ 7 . A direct calculation shows that when z ∈ X ϵ 7 , w e ha v e ˜ V − V X = ( T ♯ ) − 1    0 g 1 ( x, t, z ) g 2 ( x, t, z ) 0 0 0 0 g 3 ( x, t, z ) 0    T ♯ where g 1 ( x, t, z ) = ˜ g 1 ( x, t, z ) T ♯ 2 T ♯ 1 ( z )e θ 12 ( x,t,z ) − θ 12 ( x,t, − q 0 ) − i s e − i y β +i( 4 3 q 0 ) 2 3 t 1 3 β 2 , ˜ g 1 ( x, t, z ) = − ( r 1 ( z ∗ )) ∗ γ ( z ) + r ∗ 1 ( z ∗ ) r 3 ( z ) R ∗ 3 ( z ∗ ) γ 2 ( z ) , g 2 ( x, t, z ) = ( r 2 ( z ∗ ) − R 2 ( z ∗ )) ∗ T ♯ 3 T ♯ 1 ( z )e θ 13 ( x,t,z ) , g 3 ( x, t, z ) = − 1 γ ( z ) R ∗ 3 ( z ∗ ) T ♯ 2 T ♯ 3 ( z )e θ 32 ( x,t,z ) − θ 32 ( x,t, − q 0 ) + s e i y β +i( 4 3 q 0 ) 2 3 t 1 3 β 2 . W e ﬁrst consider g 1 . It’s easy to verify the follo wing estimates hold: | ˜ g 1 ( x, t, z ) − ˜ g 1 ( x, t, − q 0 ) | ≤ C | z 0 + q 0 | , z ∈ X ϵ 7 . Then b y the expression of g 1 ( x, t, z ) and ( 4.51c ), we hav e | g 1 ( x, t, z ) | ≤ C | ˜ g 1 ( z ) − ˜ g 1 ( − q 0 ) | e t ℜ Φ 12 + C | z + q 0 | | ln | z + q 0 || e t ℜ Φ 12 ≤ C ( | ln | z + q 0 || ) | z + q 0 | e − ct | z − z 0 | 2 ≤ t − 1 / 2 ln t  sup u ≥ 0 u | ln u | e − u 2  ≤ C t − 1 / 2 ln t, z ∈ X ϵ 7 . (4.70) Based on ( 4.70 ), w e can further estimate the L 1 norm of g 1 . More precisely , w e ha v e Z X ϵ 7 | g 1 ( x, t, z ) | d | z | ≤ C Z ∞ 0 τ | ln τ | e − ctτ 2 d τ ≤ C t − 1 ln t. (4.71) W e now turn to estimating g 3 ( x, t, z ). It is not diﬃcult to see that g 3 satisﬁes | g 3 ( x, t, z ) | ≤ C ( | ln | z + q 0 || ) | z + q 0 | e t ℜ Φ 32 + C | F 32 ( x, t, z ) | . (4.72) 44 Then b y ( 4.54c ) and ( 4.60c ), we hav e | g 3 ( x, t, z ) | ≤ C ( | ln | z + q 0 || ) | z + q 0 | e − ct | z + q 0 | 2 + C t − 1 / 2 ≤ C t − 1 / 2 ln t  sup u ≥ 0 u | ln u | e − u 2  + C t − 1 / 2 ≤ C t − 1 / 2 ln t, z ∈ X ϵ 7 . Using ( 4.72 ) and ( 4.66 ), a calculation similar to that of ( 4.71 ) yields Z X ϵ 7 | g 3 ( x, t, z ) | d | z | ≤ C Z ∞ 0 τ | ln τ | e − ctτ 2 d τ + C t − 2 / 3 ≤ C t − 2 / 3 . No w it remains to estimate g 2 . W e note that for z ∈ X ϵ 7 , | g 2 ( x, t, z ) | ≤ C ( | ln | z + q 0 || ) | z + q 0 | e t ℜ Φ 13 . Based on ( 4.54b ), a similar calculation giv es | g 2 ( x, t, z ) | ≤ C ( | ln | z + q 0 || ) | z + q 0 | e − ct | z + q 0 | 3 ≤ C t − 1 / 3 ln t, z ∈ X ϵ 7 and Z X ϵ 7 | g 2 ( x, t, z ) | d | z | ≤ Z ∞ 0 τ | ln τ | e − ctτ 3 d τ ≤ C t − 2 / 3 ln t. (4.73) The abov e analysis sho ws that estimate ( 4.67 ) holds on X ϵ 7 . On contour X ϵ 9 . F or z ∈ X ϵ 9 , w e ha v e ˜ V − V X =     − 1 γ ( z ) | r 1 ( z ) | 2 h 1 0 r 1 ( z ) T ♯ 1 T ♯ 2 ( z )e θ 21 ( x,t,z ) − θ 21 ( x,t, − q 0 ) 1 γ ( z ) | r 3 ( z ) | 2 − r 3 ( z ) T ♯ 3 T ♯ 2 ( z )e θ 23 ( x,t,z ) − θ 23 ( x,t, − q 0 ) 0 h 2 0     , where h 1 := − 1 γ ( z ) r ∗ 1 ( z ) T ♯ 2 T ♯ 1 ( z )e θ 12 ( x,t,z ) − θ 12 ( x,t, − q 0 ) − i s e − i y β +i( 4 3 q 0 ) 2 3 t 1 3 β 2 , h 2 := − 1 γ ( z ) r ∗ 3 ( z ) T ♯ 2 T ♯ 3 ( z )e θ 32 ( x,t,z ) − θ 32 ( x,t, − q 0 ) + s e i y β +i( 4 3 q 0 ) 2 3 t 1 3 β 2 . Let’s ﬁrst analyze h 1 . Note that θ 12 ( x, t, z ) − θ 12 ( x, t, − q 0 ) = − i y β ( z ) + i( 4 3 q 0 ) 2 3 t 1 3 β 2 ( z ) , hence | h 1 | =      − 1 γ ( z ) r ∗ 1 ( z ) T ♯ 2 T ♯ 1 ( z ) − i s        e θ 12 ( x,t,z ) − θ 12 ( x,t, − q 0 )   ≤ C ( | ln | z + q 0 || ) | z + q 0 | . 45 On the other hand, we observe that | z + q 0 | ≤ C | ξ + q 0 | ≤ C t − 2 / 3 , z ∈ X ϵ 9 , ( x, t ) ∈ P + . Therefore, w e immediately obtain that h 1 satisﬁes | h 1 | ≤ C t − 2 / 3 ln t, z ∈ X ϵ 9 . Then clearly ∥ h 1 ( x, t, · ) ∥ L 1 ( X ϵ 9 ) is at least O ( t − 4 / 3 ln t ) as t → ∞ . The estimates for the remaining matrix entries are completely analogous; therefore, w e conclude that estimate ( 4.67 ) holds on X ϵ 9 . No w it remains only to prov e ( 4.68 )and ( 4.69 ). The v ariables β go es to inﬁnit y as t → ∞ if z ∈ ∂ D ϵ . This is b ecause | β | =  3 t 4 q 0  1 / 3 | z + q 0 | . Th us ( 2.2 ) yields N X ( x, t, β ( z )) = I + E X 1 + M P 1 ( 3 4 q 0 ) 1 / 3 t 1 / 3 ( z + q 0 ) + O ( t − 2 / 3 ) , t → ∞ (4.74) uniformly for z ∈ D ϵ and ( x, t ) ∈ P + . Recalling ( 4.50 ) and noting the b oundedness of Y , E X 1 and M P 1 , w e ha v e prov ed ( 4.68 ). ( 4.69 ) can b e obtained from the deﬁnition of N loc and the asymptotic expansion ( 4.74 ). 4.5 Final transformation and the small norm RH problem W e deﬁne the ﬁnal transformation to obtain a small-norm RH problem as follows: E ( x, t, z ) =    N (2) ( x, t, z ) , z ∈ C \ D ϵ , N (2) ( x, t, z )( N loc ) − 1 ( x, t, z ) , z ∈ D ϵ . (4.75) W e will sho w that E ( x, t, z ) is close to I for large t and ( x, t ) ∈ P + . Let Σ E = Σ (2) ∪ ∂ D ϵ , and deﬁne the matrix-v alued function V E ( x, t, z ) for z ∈ Σ E as follo ws: V E ( x, t, z ) =          V (2) , z ∈ Σ E \ D ϵ , ( N loc ) − 1 , z ∈ ∂ D ϵ , N loc − V (7) ( N loc + ) − 1 , z ∈ X ϵ . (4.76) The function E ( x, t, z ) satisﬁes the following RH problem. 46 RH Problem 4.19. Find a 3 × 3 matrix-value d function E ( x, t, z ) with the fol lowing pr op erties: • E ( x, t, · ) : C \ Σ E → C 3 × 3 is analytic. • E + ( x, t, z ) = E − ( x, t, z ) V E ( z ) , z ∈ Σ E . • E ( x, t, z ) admits the fol lowing asymptotic b ehavior E ( x, t, z ) = I + O ( 1 z ) , z → ∞ . Lemma 4.20. L et w E = V E − I . The fol lowing estimates hold uniformly for lar ge t and ( x, t ) ∈ P + : ∥ w E ∥ L ∞ (Σ E \D ϵ ) ≤ C t − 1 / 2 , (4.77) ∥ w E ∥ L 1 (Σ E \D ϵ ) ≤ C t − 1 , (4.78) ∥ w E ∥ L 1 ∩ L ∞ ( ∂ D ϵ ) ≤ C t − 1 / 3 , (4.79) ∥ w E ∥ L 1 ( X ϵ ) ≤ C t − 2 / 3 ln t, (4.80) ∥ w E ∥ L ∞ ( X ϵ ) ≤ C t − 1 / 3 ln t. (4.81) Pr o of. It follo ws from ( 4.33 ), ( 4.40 ) and ( 4.76 ) that ( 4.77 ) and ( 4.78 ) hold. F rom ( 4.68 ), w e obtain ( 4.79 ) directly . F or z ∈ X ϵ , w e ha v e w E = N loc −  V (7) − V loc  ( N loc + ) − 1 . Based on the b oundedness of N loc in D ϵ and estimate ( 4.67 ), ( 4.81 ) follows immediately . The estimates in Lemma 4.20 show that    ∥ w E ∥ L 1 (Σ E ) ≤ C t − 1 / 3 ∥ w E ∥ L ∞ (Σ E ) ≤ C t − 1 / 3 ln t ( x, t ) ∈ P + , t > T 0 . Th us by employing the general inequalit y ∥ f ∥ L p ≤ ∥ f ∥ 1 p L 1 ∥ f ∥ p − 1 p L ∞ , w e immediately get ∥ w E ∥ L p (Σ E ) ≤ C t − 1 / 3 (ln t ) ( p − 1) /p , for ( x, t ) ∈ P + and large t. (4.82) F or the con tour Σ E and a function h ( z ) ∈ L 2 (Σ E ), we deﬁne the Cauch y transform C ( h )( z ) asso ciated with Σ E b y C ( h )( z ) := 1 2 π i Z Σ E h ( z ′ ) dz ′ z ′ − z . F urthermore, we deﬁne the op erator C w E b y C w E ( h ) = C − ( h w E ). F rom the preceding analysis, w e ha ve known that ∥ w E ∥ L 2 (Σ E ) → 0 as t → ∞ . Consequen tly , there exists a T ∗ > 0 suc h that 47 the op erator I − C w E is inv ertible whenev er t > T ∗ and ( x, t ) ∈ P + . Therefore, w e can deﬁne a function u E ( x, t, z ) for z ∈ Σ E and t > T ∗ b y u E = I + ( I − C w E ) − 1 C w E I ∈ I + L 2 (Σ E ) . (4.83) W e need the estimate of ∥ u E − I ∥ L 2 (Σ E ) . F rom ( 4.82 ) and ( 4.83 ), it follo ws that ∥ u E − I ∥ L 2 (Σ E ) ≤ ∥ ( I − C w E ) − 1 C w E I ∥ L 2 (Σ E ) ≤ ∞ X j =0 ∥C w E ∥ j B ( L 2 (Σ E )) ∥C w E I ∥ L 2 (Σ E ) ≤ ∥C − ∥ B ( L 2 (Σ E )) ∥ w E ∥ L 2 (Σ E ) 1 − ∥C − ∥ B ( L 2 (Σ E )) ∥ w E ∥ L ∞ (Σ E ) ≤ C t − 1 / 3 (ln t ) 1 / 2 , t > T ∗ . (4.84) According to the standard theory of RH problems (see e.g. [ 57 ]), E ( x, t, z ) can b e expressed as E ( x, t, z ) = I + 1 2 π i Z Σ E u E ( z ′ ) w E ( z ′ ) dz ′ z ′ − z , z ∈ C \ Σ E . (4.85) F rom ( 4.85 ), w e know the follo wing nontangen tial limit is well-deﬁned: L ( x, t ) := ∠ lim z →∞ z ( E ( x, t, z ) − I ) = − 1 2 π i Z Σ E u E ( x, t, ζ ) w E ( x, t, ζ )d ζ . (4.86) Lemma 4.21. As t → ∞ , L ( x, t ) = − 1 2 π i Z ∂ D ϵ w E ( x, t, ζ )d ζ + O ( t − 2 / 3 ln t ) . (4.87) Pr o of. The function L ( x, t ) can be rewritten as L ( x, t ) = − 1 2 π i Z ∂ D ϵ w E ( x, t, z ) dz + L 1 ( x, t ) + L 2 ( x, t ) , where L 1 ( x, t ) = − 1 2 π i Z Σ E \ ∂ D ϵ w E ( x, t, z )d z , L 2 ( x, t ) = − 1 2 π i Z Σ E ( u E ( x, t, z ) − I ) w E ( x, t, z )d z . Then the lemma follo ws from Lemma 4.20 and Eq. ( 4.84 ) and straigh tforw ard estimates. Recall that when z ∈ ∂ D ϵ , w e ha v e w E ( x, t, z ) = ( N loc ) − 1 ( x, t, z ) − I , th us by ( 4.68 ) and Cauc h y’s form ula, w e obtain − 1 2 π i Z ∂ D ϵ w E ( x, t, z ) dz = Y ( E X 1 + M P 1 ) Y − 1 ( 3 4 q 0 ) 1 / 3 t 1 / 3 + O ( t − 2 / 3 ) . 48 This implies L ( x, t ) = Y ( E X 1 + M P 1 ) Y − 1 ( 3 4 q 0 ) 1 / 3 t 1 / 3 + O ( t − 2 / 3 ln t ) . (4.88) Note that near z = 0, we ha v e w E ( x, t, z ) = V (1) − I , and V (1) − I is known to b e 0 at z = 0 from ( 4.27 ); th us w E ( x, t, z ) is zero at the origin. This observ ation sho ws that E ( x, t, 0) is w ell-deﬁned and w e ha v e E ( x, t, 0) = I + 1 2 π i Z Σ E u E ( x, t, z ′ ) w E ( x, t, z ′ ) z ′ d z ′ . (4.89) The follo wing lemma will b e crucial in the subsequen t computations. Lemma 4.22. The fol lowing estimates hold uniformly for lar ge t and ( x, t ) ∈ P + : E j ( x, t, 0) = e j + 1 2 π i Z ∂ D ϵ w E j ( x, t, ζ ) ζ d ζ + O ( t − 2 / 3 ln t ) , j = 2 , 3 , (4.90) wher e E j ( x, t, 0) , e j and w E j ( x, t, ζ ) denote the j-th c olumns of matric es E ( x, t, 0) , I and w E ( x, t, ζ ) , r esp e ctively. Pr o of. By ( 4.89 ), the function E j ( x, t, 0) can b e rewritten as E j ( x, t, 0) = e j + 1 2 π i Z ∂ D ϵ w E j ( x, t, ζ ) ζ d ζ + Q 1 ( x, t ) + Q 2 ( x, t ) , where Q 1 ( x, t ) = 1 2 π i Z Σ E \ ∂ D ϵ w E j ( x, t, ζ ) ζ d ζ , Q 2 ( x, t ) = 1 2 π i Z Σ E ( u E ( x, t, ζ ) − I ) w E j ( x, t, ζ ) ζ d ζ . Then the lemma follows from Remark 4.10 , Lemma 4.20 , Eq. ( 4.84 ) and straigh tforw ard esti- mates. 5 Pro of of Theorem 1.4 F rom Lemma 4.6 , we obtain M ( x, t, z ) = ∆ ∞  I + A ( x, t ) z  N ( x, t, z ) ∆ − 1 ( z ) , (5.1) where A ( x, t ) = σ 1 N − 1 + ( x, t, 0) = σ 1 N − 1 − ( x, t, 0). Recalling reconstruction formula ( 3.23 ), we therefore need to study the large- z b ehavior of M . Supp ose M and N admit the follo wing expansions as z → ∞ resp ectively: M ( x, t, z ) = I + M (1) z + O ( 1 z 2 ) , N ( x, t, z ) = I + N (1) z + O ( 1 z 2 ) . 49 In this section, for a matrix C , w e deﬁne the notation ( C ) rc as the t w o-dimensional column v ector formed by the (2 , 1) and (3 , 1) en tries of matrix C . Then b y ( 5.1 ), w e ha v e ( M (1) ) rc =  ∆ ∞ A ∆ − 1 ∞  rc +  ∆ ∞ N (1) ∆ − 1 ∞  rc =  σ 1 ∆ (0) N − 1 + ( x, t, 0) ∆ − 1 ∞  rc +  ∆ ∞ N (1) ∆ − 1 ∞  rc , (5.2) where ∆ (0) = diag ( δ 1 (0) , 1 /δ 1 (0) , 1). Recall the transformations introduced in Section 4 . In the neigh b orho o ds of z = 0 and z = ∞ , w e ha ve N ( x, t, z ) = T ∞ E ( x, t, z ) G − 1 ( x, t, z ) . Let the region D c = C \ ∪ 9 j =1 D j , then when z ∈ D c , we ha v e G ( x, t, z ) = T ( z ). Thus, as z tends to inﬁnity and to zero resp ectiv ely from inside region D c , w e obtain N + ( x, t, 0) = T ∞ E ( x, t, 0)( T ( x, t, 0)) − 1 , (5.3) N (1) ( x, t ) = T ∞ L ( x, t )( T ∞ ) − 1 . (5.4) F or an in v ertible 3 × 3 matrix C = ( C 1 , C 2 , C 3 ), w e ha v e C − 1 = 1 det C    ( C 2 × C 3 ) ⊤ ( C 3 × C 1 ) ⊤ ( C 1 × C 2 ) ⊤    , where “ × ” denotes the usual cross pro duct. Using this identit y , w e immediately obtain  σ 1 ∆ (0) N − 1 + ( x, t, 0) ∆ − 1 ∞  rc = 0 i q 0 δ 1 (0)  N + , 2 ( x, t, 0) × N + , 3 ( x, t, 0)  11 ! , (5.5) where N + ,j ( x, t, 0) denotes the j -th column of N + ( x, t, 0), and ( ⋆ ) 11 denotes the ﬁrst entry of the corresponding v ector. Com bining ( 5.4 ), ( 4.26 ), ( 5.5 ) with ( 5.2 ), we conclude that ( M (1) ) rc ( x, t ) = 0 i q 0 δ 1 (0)  N + , 2 ( x, t, 0) × N + , 3 ( x, t, 0)  11 ! + 1 δ 1 (0)( δ ♯ (0)) 2 P 1 (0) L 21 ( x, t ) δ 1 (0)P 1 (0) δ ♯ (0) L 31 ( x, t ) ! . (5.6) Firstly , let us consider the ﬁrst term on the righ t-hand side of the ab ov e equation. F rom ( 5.3 ), w e know that w e need to estimate E j ( x, t, 0). Recall that when z ∈ ∂ D ϵ , w e ha v e w E ( x, t, z ) = ( N loc ) − 1 ( x, t, z ) − I . Th us by ( 4.68 ) and Cauc h y’s form ula, w e obtain 1 2 π i Z ∂ D ϵ w E 2 ( x, t, ζ ) ζ dζ = 1 q 0  Y M P 1 Y − 1  2 ( 3 4 q 0 ) 1 / 3 t 1 / 3 + 1 q 0  Y E X 1 Y − 1  2 ( 3 4 q 0 ) 1 / 3 t 1 / 3 + O ( t − 2 / 3 ) , (5.7) 50 1 2 π i Z ∂ D ϵ w E 3 ( x, t, ζ ) ζ dζ = 1 q 0  Y M P 1 Y − 1  3 ( 3 4 q 0 ) 1 / 3 t 1 / 3 + 1 q 0  Y E X 1 Y − 1  3 ( 3 4 q 0 ) 1 / 3 t 1 / 3 + O ( t − 2 / 3 ) . (5.8) In the expressions ab o ve, [ A ] j denotes the j -th column of matrix A . Then, from the expression for M P 1 , Eq. ( 2.3 ), and Lemma 4.22 , a straightforw ard calculation yields E 2 ( x, t, 0) =    0 1 0    − e π 4 i s e U H M ( y ) e − θ 2 ( x,t, − q 0 ) 2 q 0 √ π t 1 / 2    1 0 i    + O ( t − 2 / 3 ln t ) , (5.9) E 3 ( x, t, 0) =    0 0 1    + 1 q 0 ( 3 t 4 q 0 ) 1 / 3    − u H M ( y ) 2 0 − i 2 R y ∞ u 2 H M ( y ′ )d y ′    + O ( t − 2 / 3 ln t ) . (5.10) Th us we hav e i q 0  N + , 2 ( x, t, 0) × N + , 3 ( x, t, 0)  11 = i q 0 δ 1 (0)  T ∞ E ( x, t, 0) T − 1 ( x, t, 0)  2 ×  T ∞ E ( x, t, 0) T − 1 ( x, t, 0)  3  11 = δ 1 (0)P 1 (0) δ ♯ (0) " i q 0 − 1 2( 3 t 4 q 0 ) 1 / 3 Z ∞ y u 2 H M ( y ′ )d y ′ # + O ( t − 2 / 3 ln t ) . (5.11) W e hav e completed the estimation of the ﬁrst term on the right-hand side of ( 5.13 ). Next, we pro ceed to analyze its second term. Note that the ﬁrst column of E X 1 satisﬁes the estimate [ E X 1 ] 1 ( x, t ) = O ( t − 1 / 3 ) , t → ∞ , and then, combining the expression for M P 1 and ( 4.88 ), we obtain    L 21 ( x, t ) = O ( t − 2 / 3 ln t ) , L 31 ( x, t ) = − u H M ( y ) 2( 3 t 4 q 0 ) 1 / 3 + O ( t − 2 / 3 ln t ) , t → ∞ . (5.12) Th us, from ( 5.13 ), ( 5.11 ), and ( 5.12 ), it follows that ( M (1) ) rc ( x, t ) =   0 i q 0 δ 1 (0)P 1 (0) δ ♯ (0) − δ 1 (0)P 1 (0) 2 δ ♯ (0)( 3 t 4 q 0 ) 1 / 3  u H M ( y ) + R ∞ y u 2 H M ( y ′ )d y ′    + O ( t − 2 / 3 ln t ) . (5.13) Substituting the asymptotic estimate ( 5.13 ) of ( M (1) ) rc ( x, t ) in to reconstruction formula ( 3.23 ), a straigh tforward calculation yields asymptotic formula ( 1.6 ). 6 Concluding Remarks In this work, w e inv estigate the P ainlev ´ e asymptotics of solutions to the defo cusing Manako v system under a class of non-zero b oundary conditions. The leading term of the long-time 51 asymptotics for these solutions is shown to b e gov erned by the Painlev ´ e I I equation. By applying the Deift–Zhou steep est descent analysis to the asso ciated 3 × 3 matrix RH problem, w e establish rigorous justiﬁcation for this b eha vior. Since the Manako v system represents a tw o-comp onent generalization of the scalar NLS equation, our results should b e compared with the scalar case. It is found that our asymptotic leading term reduces to that of the scalar case (see [ 33 , Eq. 1.5]) up on setting q 2 ( x, t ) ≡ 0 and q + = (1 , 0) ⊤ . Ho w ever, it should b e noted that the corresp onding error term in [ 33 ] is O ( t − 1 / 2 ), whereas our error term is O ( t − 2 / 3 ln t ). F rom our p ersp ectiv e, the main reason is that the initial data considered in [ 33 ] is muc h weak er than that considered here, so it is natural that the error term is not as go o d as the one obtained in this work. Nev ertheless, we b eliev e our results remain consistent with those in the scalar case. On the other hand, we emphasize that the generalization from the scalar case to the v ector case is non trivial. In fact, although the RH formulation for the vector NLS equation under nonzero boundary conditions has b een established for at least a decade, the corresp onding Deift-Zhou analysis has yet to be carried out. Our work partially addresses this problem. One particularly in triguing outcome of our analysis is that the mo del problem app ears to exhibit nov el characteristics. In the scalar case, the corresp onding mo del problem can b e di- rectly matched to the RH problem for P ainlev ´ e I I. Ho w ev er, in this pap er, it is a coupling of the Painlev ´ e I I mo del problem and an error function mo del problem. Notably , a similar mo del problem has recen tly emerged in the study of transition asymptotics for Sch wartz-class solutions to the “bad” Boussinesq equation [ 46 ]. W e b eliev e this nov el mo del problem is not restricted to these tw o integrable systems. In fact, our results suggest it also arises in the P ainlev ´ e asymptotic analysis of other coupled in tegrable systems with NZBCs—such as the coupled mKdV equations [ 53 ], coupled Hirota equations [ 54 ], and coupled Gerdjik o v–Iv anov equations [ 55 ]—all of whic h share a similar int egrable structure to the defo cusing Manako v sys- tem. F urthermore, we an ticipate that the results of this work can b e extended to N -comp onent coupled systems [ 24 ], where a generalized version of this new model problem should naturally emerge. An in teresting direction for future researc h is to extend our results to N -comp onen t systems within the Sob olev space framew ork. In this setting, a ¯ ∂ -generalization of the nonlinear steep est descen t metho d [ 31 , 34 , 35 ] would likely oﬀer a more eﬃcient approac h, although this is nontrivial and sev eral tec hnical c hallenges remain to b e addressed. Another comp elling op en question is whether other types of transitional asymptotics exist b etw een the Painlev ´ e and soliton regions. This is a highly challenging problem, whic h is left for further inv estigation. 52 A Pro of of Lemma 4.6 It is easy to verify from ( 4.15 ) and ( 4.20 ) that N + ( x, t, 0) = N − ( x, t, 0)    1 0 0 ⋆ 1 0 0 0 1    , (A.1) where ⋆ denotes an unsp eciﬁed entry . Then we know that det N is entire on C and det N ≡ 1. Th us N − 1 ± ( x, t, 0) is well-deﬁned. Using ( A.1 ), a straightforw ard computation yields σ 1 N − 1 + ( x, t, 0) = σ 1 N − 1 − ( x, t, 0) . W e now pro ceed to prov e ( 4.21 ). Our strategy is to show that the matrix-v alued function on the right-hand side of ( 4.21 ) satisﬁes the singular RH problem 4.3 ; then, b y the uniqueness result, ( 4.21 ) follo ws directly . Let N ( x, t, z ) denote the unique solution of the RH problem 4.5 . It is not diﬃcult to verify that ˜ M deﬁned b y ˜ M ( x, t, z ) =  I + A ( x, t ) z  N ( x, t, z ) satisﬁes the jump condition ( 4.16 ), the asymptotic b ehavior ( 4.17 ), and the residue condi- tions. It now remains to verify ˜ M satisﬁes the gro wth condition ( 4.18 ) and the symmetry conditions ( 4.19 ). Step 1. W e ﬁrst prov e that ˜ M satisﬁes the ﬁrst symmetry in ( 4.19 ), namely the z → ˆ z symmetry . Since the jump matrix V (1) satisﬁes (see Remark 4.4 ) e V (1) ( x, t, z ) = Π − 1 ( z )  e V (1) ( x, t, ˆ z )  − 1 Π ( z ) , b oth N ( x, t, z ) and N ( x, t, ˆ z ) Π ( z ) satisfy the same jump condition. Deﬁne the function F ( x, t, z ) = N ( x, t, ˆ z ) Π ( z ) N − 1 ( x, t, z ) . Then the function F ( x, t, z ) has no jump on the real axis. Therefore, we can determine F ( x, t, z ) b y analyzing its asymptotic b ehavior as z → 0 and as z → ∞ . W e rewrite Π ( z ) as Π ( z ) = σ 2 + 1 z σ 1 , where σ 2 =    0 0 0 0 1 0 0 0 0    . Letting z → ∞ , we ﬁnd that F ( x, t, z ) is at most O (1), and F ( x, t, ∞ ) = N + ( x, t, 0) σ 2 = N − ( x, t, 0) σ 2 . Letting z → 0, we ﬁnd that F ( x, t, z ) is O (1 /z ) with leading term lim z → 0 z F ( x, t, z ) = σ 1 N − 1 + ( x, t, 0) = σ 1 N − 1 − ( x, t, 0) . 53 Th us we hav e F ( x, t, z ) = 1 z σ 1 N − 1 + ( x, t, 0) + N + ( x, t, 0) σ 2 = 1 z σ 1 N − 1 − ( x, t, 0) + N − ( x, t, 0) σ 2 . (A.2) Then w e conclude that N ( x, t, ˆ z ) Π ( z ) N − 1 ( x, t, z ) = 1 z σ 1 N − 1 + ( x, t, 0) + N + ( x, t, 0) σ 2 . Multiplying the matrix function N ( x, t, ˆ z ) Π ( z ) N − 1 ( x, t, z ) on the left by I + z q 2 0 σ 1 N − 1 + ( x, t, 0) and on the righ t by N ( x, t, z ) , and applying the ab o v e equalit y , yields: I + z q 2 0 σ 1 N − 1 + ( x, t, 0) N ( x, t, ˆ z ) Π ( z ) =  I + z q 2 0 σ 1 N − 1 + ( x, t, 0)  ×  1 z σ 1 N − 1 + ( x, t, 0) + N + ( x, t, 0) σ 2  N ( x, t, z ) . This equation can b e written as ˜ M ( x, t, ˆ z ) Π ( z ) =  N + ( x, t, 0) σ 2 + 1 q 2 0 σ 1 N − 1 + ( x, t, 0) σ 1 N − 1 + ( x, t, 0) + 1 z σ 1 N − 1 + ( x, t, 0)  × N ( x, t, z ) . Therefore, to pro v e that ˜ M ( x, t, z ) satisﬁes the ﬁrst symmetry in ( 4.19 ), it suﬃces to verify that N + ( x, t, 0) σ 2 + 1 q 2 0 σ 1 N − 1 + ( x, t, 0) σ 1 N − 1 + ( x, t, 0) = I . (A.3) Let X 1 and X 2 denote the co eﬃcients in the following tw o asymptotic expansions: N − 1 ( x, t, z ) = I − X 1 z + O ( 1 z 2 ) , z → ∞ ; N ( x, t, ˆ z ) = N + ( x, t, 0) + X 2 z + O ( z 2 ) , C − ∋ z → ∞ . It is straigh tforw ard to compute that the co eﬃcien t of the 1 /z term in the asymptotic expansion of the function F ( x, t, z ) as C − ∋ z → ∞ is N + ( x, t, 0) σ 1 + X 2 σ 2 − N + ( x, t, 0) σ 2 X 1 . Com bining with ( A.2 ), w e obtain N + ( x, t, 0) σ 1 + X 2 σ 2 − N + ( x, t, 0) σ 2 X 1 = σ 1 N − 1 + ( x, t, 0) . Multiplying both sides of the ab o v e equation by σ 1 N − 1 + ( x, t, 0) from the left yields σ 2 1 + σ 1 N − 1 + ( x, t, 0) X 2 σ 2 = σ 1 N − 1 + ( x, t, 0) σ 1 N − 1 + ( x, t, 0) . 54 The abov e expression can b e rewritten as q 2 0    1 ⋆ 0 0 0 0 0 ⋆ 1    = σ 1 N − 1 + ( x, t, 0) σ 1 N − 1 + ( x, t, 0) , (A.4) where ⋆ denotes an unsp eciﬁed entry . Let matrices N + ( x, t, 0) and N − 1 + ( x, t, 0) b e expressed as N + ( x, t, 0) =    A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33    , N − 1 + ( x, t, 0) =    B 11 B 12 B 13 B 21 B 22 B 23 B 31 B 32 B 33    . F rom ( A.4 ), it can b e seen that only the entries (1 , 2) and (3 , 2) of matrix σ 1 N − 1 + ( x, t, 0) σ 1 N − 1 + ( x, t, 0) are undetermined. W e now pro ceed to determine them. Before that, we need to establish the relationship b etw een B 13 and B 31 . Considering the (1 , 3) en try of the matrices on b oth sides of ( A.4 ), a straigh tforw ard calculation yields B 13 = B 31 . Let the (1 , 2) and (3 , 2) en tries of matrix σ 1 N − 1 + ( x, t, 0) σ 1 N − 1 + ( x, t, 0) b e denoted as Y 12 and Y 32 , resp ectiv ely . Using the equalit y B 13 = B 31 , direct calculation yields Y 12 = − q 2 0 B 31 B 32 + q 2 0 B 33 B 12 = q 2 0 ( B 33 B 12 − B 13 B 32 ) = − q 2 0 A 12 , Y 32 = q 2 0 B 11 B 32 − q 2 0 B 12 B 13 = q 2 0 ( B 11 B 32 − B 12 B 31 ) = − q 2 0 A 32 . Th us we conclude that σ 1 N − 1 + ( x, t, 0) σ 1 N − 1 + ( x, t, 0) = q 2 0    1 − A 12 0 0 0 0 0 − A 32 1    . Finally , Considering the (1 , 1) en try of the matrices on b oth sides of ( A.4 ), we ha v e q 2 0 = − q 2 0 B 2 31 + q 2 0 B 11 B 33 = q 2 0 ( B 11 B 33 − B 31 B 13 ) = q 2 0 A 22 . Hence, w e get A 22 = 1. Therefore, N + ( x, t, 0) σ 2 + 1 q 2 0 σ 1 N − 1 + ( x, t, 0) σ 1 N − 1 + ( x, t, 0) =    0 A 12 0 0 1 0 0 A 32 0    +    1 − A 12 0 0 0 0 0 − A 32 1    = I , and w e ha v e thus prov en ( A.3 ). Step 2. No w we pro ceed to prov e that ˜ M ( x, t, z ) satisﬁes gro wth condition ( 4.18 ) at the branc h p oin ts ± q 0 .W e present the proof only for the branc h p oin t q 0 , and lea v e the analogous pro of for − q 0 . 55 Recalling the b ehavior of the reﬂection co eﬃcien ts near the branc h points, namely lim z → q 0 ˜ r 2 ( z ) = i , lim z → q 0 ˜ r 1 ( z ) = lim z → q 0 ˜ r 3 ( z ) = 0 , w e consequently hav e ˜ M + ( x, t, q 0 ) = ˜ M − ( x, t, q 0 )    1 ⋆ i 0 1 0 i ⋆ 0    . Considering the third column of the ab o v e expression yields ˜ M +3 ( x, t, q 0 ) = i ˜ M − 1 ( x, t, q 0 ) . On the other hand, w e ha v e already prov ed that ˜ M ( x, t, z ) = ˜ M ( x, t, ˆ z ) Π ( z ) in Step 1, hence the third column yields ˜ M +3 ( x, t, q 0 ) = − i ˜ M − 1 ( x, t, q 0 ) . Therefore, w e conclude that ˜ M +3 ( x, t, q 0 ) = ˜ M − 1 ( x, t, q 0 ) = 0 . This sho ws that ˜ M ( x, t, z ) satisﬁes the gro wth condition ( 4.18 ) at q 0 . Step 3. Next, w e v erify the second symmetry in ( 4.19 ). Let D ( z ) = − γ ( z ) Γ − 1 ( z ). W e note that the jump matrix V (1) satisﬁes D ( z ) e V (1) ( x, t, z ) D − 1 ( z ) =  e V (1) ( x, t, z ∗ )  † . Therefore, ˜ M ( x, t, z ) D − 1 ( z ) and [ ˜ M † ( x, t, z ∗ )] − 1 satisfy the same jump conditions on R . If w e deﬁne K ( x, t, z ) = ˜ M ( x, t, z ) D − 1 ( z ) ˜ M † ( x, t, z ∗ ) , then K has no jump across R . It is clear that K ( x, t, z ) is analytic on C \ ( { 0 , ± q 0 } ∪ Z ). Next, w e prov e that K ( x, t, z ) is also analytic for z ∈ { 0 , ± q 0 } ∪ Z . First, note that D − 1 ( z ) =    1 γ ( z ) − 1 − 1 γ ( z )    =    O ( z 2 ) − 1 O ( z 2 )    , z → 0 , and that ˜ M ( x, t, z )    z 1 z    = O (1) , z → 0;    z 1 z    ˜ M † ( x, t, z ∗ ) = O (1) , z → 0 . 56 Then w e immediately obtain K ( x, t, z ) = O (1) as z → 0. Next, we analyze the b ehavior of K ( x, t, z ) near the branc h p oin ts ± q 0 . Let ˜ M j denote the j -th column of the matrix-v alued function ˜ M . Therefore, K ( x, t, z ) can b e rewritten as K ( x, t, z ) =  1 γ ( z ) ˜ M 1 ( x, t, z ) − ˜ M 2 ( x, t, z ) ˜ M 3 ( x, t, z )     ˜ M † 1 ( x, t, z ∗ ) ˜ M † 2 ( x, t, z ∗ ) − 1 γ ( z ) ˜ M † 3 ( x, t, z ∗ )    =  ˜ M 1 ( x, t, z ) − ˜ M 2 ( x, t, z ) − 1 γ ( z ) ˜ M 3 ( x, t, z )     1 γ ( z ) ˜ M † 1 ( x, t, z ∗ ) ˜ M † 2 ( x, t, z ∗ ) ˜ M † 3 ( x, t, z ∗ )    F rom the ab o v e tw o expressions for K , com bined with the gro wth condition satisﬁed b y ˜ M , one can easily see that K is O (1) near the branc h p oints ± q 0 . Now it only remains to analyze the b eha vior of K at the discrete sp ectrum. Due to symmetry , it suﬃces to examine the prop erties of K at z = ζ j and z = z j . T aking z = z j as an example, we prov e that K is O (1) near this p oin t; the analysis for z = ζ j is similar. Based on the residue conditions satisﬁed by ˜ M , we can assume ˜ M ( x, t, z ) and ˜ M † ( x, t, z ∗ ) ha ve the follo wing asymptotic b ehavior as z → z j ˜ M ( x, t, z ) = ˜ κ j e θ 21 ( x, t, z j ) z − z j    a 1 0 0 a 2 0 0 a 3 0 0    +    ⋆ a 1 ⋆ ⋆ a 2 ⋆ ⋆ a 3 ⋆    + O ( z − z j ) , ˜ M † ( x, t, z ∗ ) = ˜ κ j γ ( z j ) e θ 21 ( x,t,z j ) z − z j    0 0 0 A 1 A 2 A 3 0 0 0    +    A 1 A 2 A 3 ⋆ ⋆ ⋆ ⋆ ⋆ ⋆    + O ( z − z j ) , for some functions { a j ( x, t ) } 3 j =1 and { A j ( x, t ) } 3 j =1 . Therefore, ˜ M D − 1 has the following asymp- totic behavior: ˜ M ( x, t, z ) D − 1 ( z ) = ˜ κ j γ ( z j ) e θ 21 ( x, t, z j ) z − z j    a 1 0 0 a 2 0 0 a 3 0 0    +    ⋆ − a 1 ⋆ ⋆ − a 2 ⋆ ⋆ − a 3 ⋆    + O ( z − z j ) . Com bining the ab ov e form ula with the asymptotic expansion of ˜ M † ( x, t, z ∗ ), by a simple cal- culation one can ﬁnd that K has no negative pow ers as z → z j i.e., z j is a remo v able singularit y of K . Based on the ab o ve analysis, one can conclude that K is holomorphic on C . As z → ∞ , w e hav e K ( x, t, z ) = J + O (1 /z ). Liouville’s theorem implies that K ( x, t, z ) = ˜ M ( x, t, z ) D − 1 ( z ) ˜ M † ( x, t, z ∗ ) = J . P erforming a simple transformation on the ab o v e equation yields ( ˜ M − 1 ) ⊤ ( x, t, z ) = − 1 γ ( z ) J ˜ M ∗ ( x, t, z ∗ ) Γ ( z ) , 57 1 2 3 4 i − i Figure 6: The jump contour P = ∪ 4 j =1 X j for the RH problem for M P . whic h is the second symmetry in ( 4.19 ). In summary , ˜ M satisﬁes the RH problem 4.3 . F rom the uniqueness result, we obtain ˜ M = M (1) . This completes the pro of of Lemma 4.6 . B P ainlev ´ e I I mo del problem Consider the sub contour P = ∪ 4 j =1 X j of X , see Figure 6 . Deﬁne the jump matrix V P ( y , β ) b y V P 1 =    1 0 0 0 1 0 ie 2i( y β + 4 β 3 3 ) 0 1    , V P 2 =    1 0 0 0 1 0 − ie 2i( y β + 4 β 3 3 ) 0 1    , V P 3 =    1 0 ie − 2i( y β + 4 β 3 3 ) 0 1 0 0 0 1    , V P 4 =    1 0 − ie − 2i( y β + 4 β 3 3 ) 0 1 0 0 0 1    , (B.1) where V P j denotes the restriction of V P to X j . RH Problem B.1 (RH problem for M P ) . Find a 3 × 3 -matrix value d function M P ( y , β ) with the fol lowing pr op erties: • M P ( y , · ) : C \ ∪ 4 j =1 X j → C 3 × 3 is analytic. • M P ( y , · ) has c ontinuous b oundary values on ∪ 4 j =1 X j \ {± i } satisfying the jump r elation M P + ( y , β ) = M P − ( y , β ) V P ( y , β ) , β ∈ ∪ 4 j =1 X j \ {± i } . • M P ( y , β ) = I + O ( β − 1 ) as β → ∞ and M P ( y , β ) = O (1) as β → ± i . Lemma B.2. F or e ach y ∈ R , RH pr oblem B.1 has a unique solution M P ( y , β ) with the fol lowing pr op erties: 58 Ω Ω Figure 7: The con tour E (solid black line) and the contour ˜ E (solid blue line). • Ther e ar e smo oth functions { M P j ( y ) } ∞ j =1 of y ∈ R such that, for e ach inte ger N ≥ 0 , M P ( y , β ) = I + N X j =1 M P j ( y ) β j + O ( β − N − 1 ) , β → ∞ , (B.2) uniformly for y in c omp act subsets of R and for arg β ∈ [0 , 2 π ] . • The le ading term is given by M P 1 ( y ) =    i 2 R y ∞ u H M ( y ′ ) 2 dy ′ 0 − u H M ( y ) 2 0 0 0 − u H M ( y ) 2 0 − i 2 R y ∞ u H M ( y ′ ) 2 dy ′    , (B.3) wher e u H M is the Hasting–McL e o d solution of Painlev´ e II, that is, the classic al solution of the e quation satisfying the fol lowing b oundary c onditions ( 1.7 ) . • A t β = 0 , we have M P ( y , 0) =    cosh U H M ( y ) 0 − i sinh U H M ( y ) 0 1 0 i sinh U H M ( y ) 0 cosh U H M ( y )    , (B.4) wher e U H M ( y ) := R y ∞ u H M ( y ′ ) dy ′ . Pr o of. Since the jump contour can b e arranged to in tersect at the origin through a simple con- tour deformation, RH problem B.1 is reduced to the RH problem asso ciated with the Hastings- McLeo d solution of the P ainlev ´ e I I equation. Th us all the assertions ab ov e are well-kno wn results. F or details of the pro ofs, we refer the reader to [ 58 , Prop osition 5.2 & Theorem 11.7]. 59 C Error function mo del problem Let the countor E is sho wn in Fig. 7 . Consider the following RH problem. RH Problem C.1 (RH problem for M E ) . Find a 3 × 3 -matrix value d function M E ( s, α ) with the fol lowing pr op erties: • M E ( s, · ) : C \ E → C 3 × 3 is analytic. • M E ( s, · ) has c ontinuous b oundary values on E satisfying the jump r elation M E + ( s, α ) = M E − ( s, α ) V E ( s, α ) , α ∈ E , wher e V E =    1 i s e i α 2 0 0 1 0 0 − s e i α 2 1    . • M E ( s, α ) = I + O ( α − 1 ) as α → ∞ . The follo wing lemma con tains the results we need ab out M E ( s, α ). Lemma C.2. RH pr oblem C.1 has a unique solution M E ( s, α ) . F or e ach inte ger N ≥ 0 , M E ( s, α ) = I + N X j =0 M E 2 j +1 ( s ) α 2 j +1 + O ( α − 2 N − 3 ) , α → ∞ , (C.1) uniformly for arg α ∈ [0 , 2 π ] , wher e the le ading c o eﬃcient is given by M E 1 ( s ) = − se πi 4 2 √ π    0 1 0 0 0 0 0 i 0    . (C.2) Pr o of. First, w e transform the jump contour from E to ˜ E b y introducing the transformation M ˜ E = M E ˜ G , ˜ G =    ( V E ) − 1 , α ∈ Ω , I , el sew her e. The con tour ˜ E and the region Ω are sho wn in Figure 7 . It is easy to verify that M ˜ E has a jump across ˜ E , and its jump matrix V ˜ E is equal to V E . Moreov er, from the expression of V E , it can b e seen that ˜ G exhibits the follo wing asymptotic behavior: ˜ G ( s, α ) =    1 O (e − c | α | ) 0 0 1 0 0 O (e − c | α | ) 1    , α → ∞ . (C.3) 60 Therefore, it suﬃces to study the asymptotic b ehavior of M ˜ E as α → ∞ . First, it is easy to see that the ﬁrst and third columns of M ˜ E are constan t and coincide with the corresp onding columns of the 3 × 3 identit y matrix, resp ectiv ely . F urthermore, we ﬁnd that M ˜ E has the following form: M E ( s, α ) =    1 X 12 0 0 1 0 0 X 32 1    , (C.4) where X 12 ( s, α ) and X 32 ( s, α ) are functions to b e determined. In fact, the jump condition M ˜ E + = M ˜ E − V ˜ E is equiv alent to    ( X 12 ) + − ( X 12 ) − = i s e i α 2 , α ∈ R e π 4 i , X 12 → 0 , α → ∞ ,    ( X 32 ) + − ( X 32 ) − = − s e i α 2 , α ∈ R e π 4 i , X 32 → 0 , α → ∞ . (C.5) Then, from the Plemelj’s formula, it follo ws that X 12 ( s, α ) = 1 2 π i Z R e π 4 i i s e i ζ 2 ζ − α d ζ = i s 2 " 1 π i Z ∞ −∞ e − τ 2 τ − e − π 4 i α d τ # , X 32 ( s, α ) = i X 12 ( s, α ) . (C.6) Using the following in tegral represen tation for the error function erf ( z ) (see [ 59 , Eq.(7.7.2)] ): e − z 2 (1 ∓ erf ( − i z )) = ± 1 π i Z ∞ −∞ e − t 2 dt t − z , ℑ z ≷ 0 , w e ﬁnd X 12 =    i s 2 e i α 2  1 + erf ( α e − 3 π 4 i )  , arg α ∈ ( π 4 , 5 π 4 ) , i s 2 e i α 2  erf ( α e π 4 i ) − 1  , arg α ∈ ( − 3 π 4 , π 4 ) . (C.7) F urthermore, for eac h ε > 0, the error function erf z satisﬁes the asymptotic form ula erf ( z ) ∼ 1 − e − z 2 √ π ∞ X j =0 ( 1 2 − 1)( 1 2 − 2) · · · ( 1 2 − j ) z 2 j +1 = 1 − e − z 2 √ π  1 z − 1 2 z 3 + O ( z − 5 )  (C.8) as z → ∞ uniformly for arg z ∈ [ − 3 π 4 + ε, 3 π 4 − ε ]. Using ( C.7 ), the asymptotics ( C.8 ) for erf ( z ), and the relation erf ( − z ) = − erf ( z ), we ﬁnd the asymptotic expansion of M ˜ E as α → ∞ . 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Painlevé-type asymptotics for the defocusing Manakov system with nonzero boundary conditions

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