Well-posedness for the $\bar\partial$-problem relevant to the AKNS spectral problem

W ell-p osedness for the ¯ ∂ -problem relev an t to the AKNS sp ectral problem Jun yi Zh u ∗ , Huan Liu † School of Mathematics and Statistics, Zhengzhou Universit y , Zhengzhou, Henan 450001, China Abstract The well-posedness for the Dbar problem asso ciated with the AKNS sp ec- tral problem is considered. In general, the relev an t Dbar equation with nor- malization condition is equiv alent to an in tegral equation, where the kernel in volv es exp onen ts e ± 2 ikx with physical v ariable x as a parameter. W e de- v elop a decomp osition technique to con trol the conv ergence of the in tegral by defining a new in tegral operator RT C ( k ; x ). The small norm condition of the op erator is obtained to sho w that there exists a unique solution for the Dbar problem. Moreo ver, the Dbar dressing metho d is extended to construct the AKNS sp ectral problem and the p oten tial construction is presented via the Dbar data. Prior estimates are given to show that the map from the Dbar data to the AKNS p oten tial is Lipschitz con tinuous. Keyw ords: w ell-p osedness; Dbar problem; AKNS spectral problem; de- comp osition tec hnique Con ten ts 1 In tro duction 2 2 Preliminary for the in tegral op erator to the Dbar problem on b ounded domain 6 ∗ E-mail: jyzhu@zzu.edu.cn † E-mail: liuhuan@zzu.edu.cn 1 3 Dbar problem relev an t to the AKNS sp ectral problem 11 4 P otential reconstruction of AKNS sp ectral problem 17 5 Conclusions and Discussions 22 1 In tro duction The ¯ ∂ (Dbar) problem can b e regarded as a generalization of the Cauch y-Riemann equation for non-analytical functions. Since Beals and Coifman introduced the Dbar problem to study the in tegrable system [ 1 – 3 ], Dbar problem has play ed imp ortan t roles to construct explicit solutions [ 4 – 33 ], and to discuss the asymptotic b ehaviors of the nonlinear in tegrable system [ 34 – 46 ]. In particular, it is effectiv e in case where the eigenfunction is no where analytic, failure for the classical Riemann-Hilb ert problem, but the Dbar problem remains v alid [ 11 ]. F or in tegrable system, one needs to consider the follo wing nonhomogeneous Dbar equation ¯ ∂ ψ ( k , ¯ k ) := ∂ ψ ( k , ¯ k ) ∂ ¯ k = ψ ( k , ¯ k ) R ( k , ¯ k ) , (1.1) where R ( k , ¯ k ) is called the sp ectral transformation matrix function, or a distribu- tion. Under the normalization conditon ψ ( k , ¯ k ) → I , k → ∞ , the Dbar problem is equiv alen t to a in tegral equation ψ ( k , ¯ k ) = I + ψ RT C ( k ) , (1.2) where T C ( k ) acting to the left is called Cauch y-Green op erator ψ RT C ( k ) = 1 2 iπ Z Z C ψ ( z , ¯ z ) R ( z , ¯ z ) z − k d z ∧ d ¯ z . (1.3) It is worth noting that a matrix integrable op erator K [ ν ]( z , ¯ z ) = Z Z D K ( z , ¯ z ; w , ¯ w ) ν ( w ) d ¯ w ∧ d w 2 i , z , ¯ z ∈ D K ( z , ¯ z ; w , ¯ w ) = f T ( z , ¯ z ) g ( w , ¯ w ) z − w , f T ( z , ¯ z ) g ( z , ¯ z ) = 0 , 2 on a b ounded domain D is considered in [ 47 ], and the resolven t of the in tegral op erator I d − K is obtained through the solution of a Dbar-Problem ¯ ∂ Γ( z , z ) = Γ( z , z ) M ( z , ¯ z ); Γ → I , z → ∞ , M ( z , ¯ z ) = π f ( z , ¯ z ) g T ( z , ¯ z ) χ D ( z ) . In the follo wing, w e will omit the v ariable ¯ k and ¯ z in functions, just write them as ψ ( k ) and R ( k ). F or a certain Lax integrable system with linear sp ectral problem ∂ x ϕ ( x ; k ) = U ( x ; k ) ϕ ( x ; k ), the potential matrix U is an imp ortant c haracteristic. F or conv e- nience, the v ariable x ∈ R is called a ph ysical v ariable, and k ∈ C is a sp ectral parameter. W e note that the sp ectral transformation matrix R ( k ) to the relev an t Dbar problem will tak e the same role as the p otential matrix U to c haracter the non- linear integrable PDE. The so called relev ant Dbar problem means that the sp ectral transformation matrix R ( k ) should satisfy an certain evolution equation ab out the ph ysical v ariable. In this pap er, we will discuss the AKNS sp ectral problem [ 48 ] or Zakharo v-Shabat sp ectral problem [ 49 ] with U = − ik σ 3 + Q ( x ) , σ 3 = 1 0 0 − 1 ! , Q = 0 u v 0 ! . The relev ant Dbar problem with R ( k ; x ) admitting ∂ x R = − ik [ σ 3 , R ], which implies that there are exp onents e ± 2 ikx , x ∈ R , k ∈ C in R , and they will en ter the kernel of the integral ( 1.2 ). There is one key question: In what space the in tegral in ( 1.2 ) with exp onen ts in its kernel and with the whole k plane as its region of integration is conv ergent? W e remark ed that the prop erties of the in tegral ( 1.3 ) without exp onen ts and ph ysical parameters hav e b een well studied in [ 50 ]. Although the Dbar problem has b een in tro duced to inv estigate nonlinear in tegrable PDEs, the conv ergence of the in tegral with exp onents and physical parameters was rarely considered. In the ex- isting applications of Dbar problem to in tegrable system, an imp ortan t prerequisite is the existence of the in v erse operator ( I − R T C ) − 1 [ 3 – 5 , 12 , 13 , 51 ]. T o ensure this assumption makes sense, a small op erator norm for the op erator R T C should b e v alid. How ev er, the exp onen tial factors are not alw ays b ounded, for example, see | e ± 2 ikx | = e ∓ 2 x Im k , x ∈ R , k ∈ C . One needs to giv e a decomp osition for k ∈ C and x ∈ R and split the sp ectral transformation matrix R ( k ; x ). 3 W e giv e the following assumptions: i. The sp ectral transformation matrix R ( k ; x ) is off-diagonal matrix R ( k ; x ) = 0 r + ( k )e − 2 ikx r − ( k )e 2 ikx 0 ! . Since ∂ x diag R = 0, we assume that the diagonal part of the sp ectral transfor- mation matrix R ( k ; x ) is zero. ii. There are no Dirac δ functions in the sp ectral transformation matrix R ( k ; x ). The soliton solutions for in tegrable equation are giv en by the discrete sp ectra whic h are related to the Dirac δ functions in the sp ectral transformation matrix R ( k ; x ). In general, R is related to the discrete sp ectrum and contin uous sp ectrum [ 4 ]. If R only contains the Dirac- δ function, then in tegral equation ( 1.2 ) can b e reduced to a closed linear system. The solv abilit y of the Dbar problem or the equation ( 1.2 ) will b e simple. Ho w ever, there will b e a logical problem. In fact, the condition of R only con taining the Dirac- δ function implies that the sp ectral analysis is conducted only with discrete sp ectrum, whic h is inconsisten t with the in v erse scattering transform. Otherwise, the equation ( 1.2 ) will con tain the in tegral part giv en b y the associated con tinuous sp ectrum. Moreo ver, the Dirac δ function is a generalized function, neither H¨ older con tinuous and nor L p function. F or simplicity , w e exclude the Dirac δ functions in R . In this pap er, w e shall develop a decomp osition technique to discuss the well- p osedness of the Dbar problem relev ant to the AKNS sp ectral problem. T o this end, w e introduce tw o nilp otent matrices w ± ( k ; x ) admitting R = w − + w + and define a new op erator R T C in the following ψ RT C ( k ; x ) =[ ψ w − T C + ( k ; x ) + ψ w + T C − ( k ; x )] χ x> 0 + [ ψ w − T C − ( k ; x ) + ψ w + T C + ( k ; x )] χ x< 0 , (1.4) where χ is the indicator function of the half lines x ≷ 0 and the region of in tegrations C ± b e upper and lo w er half-planes. W e note that the exp onen tial factors in integrals ψ w ± T C ∓ ( k ; x ) χ x> 0 and ψ w ± T C ± ( k ; x ) χ x< 0 are b ounded. 4 No w we introduce a function set H α ( G ) = C ( G ) + h α ( G ) , 0 < α ≤ 1 , where G is a b ounded domain and C ( G ) = { f ( k ) : sup k ∈ G | f ( k ) | < ∞} , h α ( G ) =  f ( k ) : sup k 1 ,k 2 ∈ G | f ( k 1 ) − f ( k 2 ) | | k 1 − k 2 | α < ∞ , ( k 1  = k 2 )  . F or f ( k ) ∈ H α ( G ), the norm is defined by ∥ f ∥ H α ( G ) = ∥ f ∥ L ∞ ( G ) + ∥ f ∥ h α ( G ) , (1.5) where ∥ f ∥ L ∞ ( G ) = sup k ∈ G | f ( k ) | , ∥ f ∥ h α ( G ) = sup k 1 ,k 2 ∈ G | f ( k 1 ) − f ( k 2 ) | | k 1 − k 2 | α . Th us H α ( G ) is a Banach space. Let f ( k ) b e giv en on the k ∈ C plane, and satisfies the conditions f ( k ) ∈ L p ( E 1 ) , f ( ν ) ( k ) = | k | − ν f ( k − 1 ) ∈ L p ( E 1 ) , (1.6) where E 1 = { k : | k | ≤ 1 } and ν is a real num b er. The set of such functions will b e denoted by L p,ν ( C ). F or f ( k ) ∈ L p,ν ( C ), we define the norm b y ∥ f ∥ L p,ν ( C ) = ∥ f ∥ L p ( E 1 ) + ∥ f ( ν ) ∥ L p ( E 1 ) . (1.7) It is verified that the space L p,ν ( C ) is a Banach space. T o discuss the Dbar problem relev ant to the AKNS prob lem, one need to consider matrix functions. F or the matrix ϕ ( k ), w e let ∥ ϕ ( k ) ∥ b e a certain matrix norm, and ∥ ϕ ( k ) ∥ L p b e the L p norm related to the matrix norm. T o give the prior estimates for the integrals ψ w ± T C ∓ ( k ; x ) χ ( x > 0) and ψ w ± T C ± ( k ; x ) χ ( x < 0), we need to further decomp ose the unit domain and its external region. The unit domain is split into t wo b ounded domains E ± 1 = { k : | k | ≤ 1 , Im k ≷ 0 } , (1.8) and the counterparts of the split external region are mapp ed into E ∓ 1 . So the norm of ϕ ∈ L p,ν ( C ± ) can b e defined by t w o norms on bounded domains in the following form ∥ ϕ ∥ L p,ν ( C ± ) = ∥ ϕ ∥ L p ( E ± 1 ) + ∥ ϕ ( ν ) ∥ L p ( E ∓ 1 ) . (1.9) 5 In section 2, we present the prop erties of the integral on b ounded domain. In section 3, w e dev elop a decomp osition technique, and define a new integral op erator on k ∈ C with parameter x ∈ R relev ant to the AKNS sp ectral problem. The new in tegral op erator can b e decomp osed in to several op erators on C ± and x ≷ 0, and further b e decomp osed in to E ± 1 and x ≷ 0. Using the prop erties on the b ounded domain and the b oundedness of the exp onential factors, w e discussed the prop erties for the new and decomp osed op erators, and find the exist conditions of the inv erse op erator ( I − RT C ) − 1 ( k ; x ). In section 4, we extend the Dbar dressing metho d based on the existence of the inv erse op erator ( I − R T C ) − 1 ( k ; x ), and give the AKNS p oten tial reconstruction. Then we show that the maps L q , 2 ( C ) ∋ r ± ( k ) → u ( x ) , v ( x ) ∈ L 2 ( R ) are Lipsc hitz contin uous. In section 5, w e give some conclusions and discussions. 2 Preliminary for the in tegral op erator to the Dbar problem on b ounded domain Let G b e a b ounded domain with b oundary Γ, and ¯ G = G ∪ Γ. Suppose the functions ϕ ( k ) ∈ C ( ¯ G ) ∩ C 1 ( G ) , f ( k ) ∈ C ( ¯ G ) admitting the Dbar problem ¯ ∂ ϕ ( k ) = ∂ ϕ ( k ) ∂ ¯ k = f ( k ) , (2.1) then the Pompeiu formula implies that ϕ ( k ) = Φ( k ) + f ˆ T G ( k ) , (2.2) where Φ( k ) = 1 2 π i Z Γ ϕ ( z ) z − k d z , f ˆ T G ( k ) = 1 2 π i Z Z G f ( z ) z − k d z ∧ d ¯ z . (2.3) Theorem 1 [ 50 ] L et G b e a b ounde d domain, and f ( k ) ∈ L 1 ( ¯ G ) , then the inte gr al f ˆ T G ( k ) = 1 2 π i Z Z G f ( z ) z − k d z ∧ d ¯ z (2.4) exist for al l p oints k outside ¯ G , f ˆ T G ( k ) is holomorphic outside ¯ G , and vanish at k = ∞ . 6 Theorem 2 [ 50 ] L et G b e a b ounde d domain, and f ( k ) ∈ L 1 ( ¯ G ) , then the function f ˆ T G ( k ) = 1 2 π i Z Z G f ( z ) z − k d z ∧ d ¯ z , ( 2.4 ) r e gar de d as a function of p oint k ∈ G , exist almost everywher e, and f ˆ T G ( k ) ∈ L p ( ¯ G ∗ ) , 1 ≤ p < 2 , wher e G ∗ is an arbitr ary b ounde d domain of the plane. The t wo theorems tell us that f ˆ T G ( k ) is a function in the k plane ev en though f ( k ) is defined on the b ounded domain G . Lemma 1 L et G b e a b ounde d domain, and k 1 , k 2 ∈ C ar e differ ent p oints. L et I ( µ, ν ) = Z Z G 1 | z − k 1 | µ | z − k 2 | ν d σ z , 0 < µ, ν < 2 , with d σ z = | 1 2 i d z ∧ d z | = d z R d z I , ( z = z R + iz I ) , then I ( µ, ν ) ≤      M 1 ( µ, ν, G ) , µ + ν < 2; M 2 ( µ, ν, G ) + 8 π | ln | k 1 − k 2 || , µ + ν = 2; M 3 ( µ, ν ) | k 1 − k 2 | 2 − µ − ν , µ + ν > 2 , (2.5) wher e M 1 ( µ, ν, G ) , M 2 ( µ, ν, G ) ar e p ositive c onstants dep ending on µ, ν and G , and M 3 ( µ, ν ) is a p ositive c onstant on µ, ν . Pro of . T aking k 1 as the cen ter, dra w a circle G 1 with radius β = 2 | k 1 − k 2 | , and then draw a concentric circle G 0 with radius 2 β 0 , suc h that G ⊂ G 0 . In the annulus G 0 − G 1 , we hav e | z − k 1 | ≤ | z − k 2 | + | k 2 − k 1 | ≤ 2 | z − k 2 | . W e note that there are three cases on the p osition ab out the p oints k 1 , k 2 and the domain G : (i). k 1 , k 2 lo cate outside of the domain G ; (ii). k 1 , k 2 lo cate in the domain G ; (iii). k 1 outside of G , and k 2 in G . 7 Figure 1: Case i. (left); case ii.(meddle);case iii. (righ t). W e consider the integral I 0 := Z Z ( G 0 − G 1 ) ∩ G 1 | z − k 1 | µ | z − k 2 | ν d σ z ≤ Z Z ( G 0 − G 1 ) ∩ G 1 | z − k 1 | µ (2 − 1 | z − k 1 | ) ν d σ z ≤ 2 ν ( θ 1 − θ 0 ) Z 2 β 0 β r 1 − µ − ν d r , (2.6) where θ 1 − θ 0 is the cen tral angle (measured counterclockwise) subtended by the region ( G 0 − G 1 ) ∩ G with resp ect to k 1 . It is noted that 0 < θ 1 − θ 0 < 2 π for the cases (i) and (iii), and 0 ≤ θ 1 − θ 0 ≤ 2 π for the case (ii). If µ + ν < 2, then ( 2.6 ) implies that I 0 ≤ π 2 1+ ν 2 − µ − ν ((2 β 0 ) 2 − µ − ν − β 2 − µ − ν ) ≤ π 2 3 − µ 2 − µ − ν β 2 − µ − ν 0 . (2.7) If µ + ν = 2, then we hav e from ( 2.6 ) that I 0 ≤ π 2 1+ ν ln r | 2 β 0 β ≤ π 2 1+ ν ( | ln β 0 | + | ln | k 1 − k 2 || ) . (2.8) And for µ + ν > 2, we ha ve I 0 ≤ π 2 1+ ν µ + ν − 2 (2 | k 1 − k 2 | ) 2 − µ − ν = π 2 3 − µ µ + ν − 2 | k 1 − k 2 | 2 − µ − ν . (2.9) Next, we consider the in tegral I 1 := Z Z G 1 ∩ G 1 | z − k 1 | µ | z − k 2 | ν d σ z , 8 and let d 0 = inf  | k 1 − k | | k 1 − k 2 | : ∀ k ∈ G 1 ∩ G  . Th us 0 < d 0 < 2 for the cases (i) and (iii), 0 ≤ d 0 ≤ 2 for the case (ii). F or z ∈ G 1 ∩ G , w e let z − k 1 = ω ( k 1 − k 2 ) , | ω | = w , r = | z − k 1 | = w | k 1 − k 2 | , then d 0 ≤ w ≤ 2, and z − k 2 = ( ω + 1)( k 1 − k 2 ) , | z − k 2 | = | ω + 1 || k 1 − k 2 | , d σ z = r d r d θ = | k 1 − k 2 | 2 w d w d θ = | k 1 − k 2 | 2 d σ ω . Th us we find that I 1 ≤ | k 1 − k 2 | 2 − µ − ν Z Z d 0 ≤ w ≤ 2 w − µ | ω + 1 | − ν d σ ω ≤ M ( µ, ν ) | k 1 − k 2 | 2 − µ − ν . (2.10) Here the integral   Z Z 0 ≤ w ≤ 2 w − µ | ω + 1 | − ν d σ ω   ≤   Z Z 0 ≤ w ≤ 1 w − µ | ω + 1 | − ν d σ ω   +   Z Z 1 ≤ w ≤ 2 w − µ | ω + 1 | − ν d σ ω   ≤ 2 π Z 1 0 w 1 − µ (1 − w ) − ν d w +   Z Z 1 ≤ w ≤ 2 w − µ | ω + 1 | − ν d σ ω   , and the righ t integral is b ounded for 0 < µ < 2 , 0 < ν < 2 b y virtue of the theories of the Euler integral and the h yp ergeometric function. Since I ( µ, ν ) = I 0 + I 1 , with M 1 ( µ, ν, G ) = M ( µ, ν ) | k 1 − k 2 | 2 − µ − ν + π 2 3 − µ 2 − µ − ν β 2 − µ − ν 0 ; M 2 ( µ, ν, G ) = M ( µ, ν ) | k 1 − k 2 | 2 − µ − ν + π 2 1+ ν | ln β 0 | ; M 3 ( µ, ν ) = M ( µ, ν ) + π 2 3 − µ µ + ν − 2 . then the Lemma is prov ed. □ 9 Theorem 3 L et G b e a b ounde d domain and f ( k ) ∈ L p ( ¯ G ) , p > 2 , then the function g ( k ) = f T G ( k ) satisfying the c onditions | g ( k ) | ≤ M 4 ( p, G ) ∥ f ∥ L p ( G ) , k ∈ C , (2.11) | g ( k 1 ) − g ( k 2 ) | ≤ M 5 ( p ) ∥ f ∥ L p ( G ) | k 1 − k 2 | γ , (2.12) wher e γ = p − 2 p , k 1 , k 2 ∈ C . Her e the p ositive c onstant M 4 ( p, G ) dep ends on p and G , the p ositive c onstant M 5 ( p ) dep ends on p . Pro of . T o pro v e the inequality ( 2.11 ), making use of the H¨ older inequality , we find that | g ( k ) | ≤ 1 π Z Z G | f ( z ) | | z − k | dσ z ≤ 1 π  Z Z G | f ( z ) | p dσ z  1 p  Z Z G | z − k | − q dσ z  1 q , (2.13) where q = p p − 1 . F or k ∈ ¯ G , w e let z − k = r e iθ , and d b e the redius of the domain G , then  Z Z G | z − k | − q d σ z  1 q ≤  Z 2 π 0 Z d 0 r 1 − q d r d θ  1 q =  2 π 2 − q  1 q d 2 − q q = ( 2 π 2 − q ) 1 q d γ , (2.14) where the identit y γ = p − 2 p = 2 − q q has been used. F or k / ∈ ¯ G , letting d 0 b e the distance from k to the domain G , then w e hav e  Z Z G | z − k | − q d σ z  1 q ≤  Z 2 π 0 Z d 0 + d d 0 r 1 − q d r d θ  1 q =  2 π 2 − q  ( d 0 + d ) 2 − q − d 2 − q 0   1 q ≤  2 π 2 − q  (1 + d 0 d ) 2 − q − ( d 0 d ) 2 − q  1 q d γ . (2.15) Th us the equality ( 2.11 ) is obtained from ( 2.13 ),( 2.14 ) and ( 2.15 ). Next we prov e the condition ( 2.12 ). F or p > 2 , q = p p − 1 , then 1 < q < 2. By 10 virtue of the H¨ older inequality and the Lemma 1 , we obtain that | g ( k 1 ) − g ( k 2 ) | ≤ | k 1 − k 2 | π Z Z G | f ( z ) | | z − k 1 || z − k 2 | d σ z ≤ | k 1 − k 2 | π ∥ f ∥ L p ( ¯ G )  Z Z G ( | z − k 1 || z − k 2 | ) − q d σ z  1 q ≤ | k 1 − k 2 | π ∥ f ∥ L p ( ¯ G )  M 3 ( q , q ) | k 1 − k 2 | 2 − 2 q  1 q ≤ M 5 ( p ) ∥ f ∥ L p ( ¯ G ) | k 1 − k 2 | γ . The theorem is prov ed. □ It is remarked that, from the conditions ( 2.11 ) and ( 2.12 ), ˆ T G is a linear com- pletely contin uous op erator in the space L p ( ¯ G ), and ˆ T G : L p ( ¯ G ) → H γ ( ¯ G ) , 0 < γ < 1 , γ = p − 2 p , p > 2 . 3 Dbar problem relev an t to the AKNS sp ectral problem In this section, we will dev elop a decomp osition tec hnique for the Dbar problem and extend the ab o ve prop erties to consider the AKNS sp ectral problem. F or con ve- nience, the k plane will b e called sp ectral space, and x, t b e ph ysical v ariables. F or Lax in tegrable system with linear sp ectral problem ϕ x = U ϕ , the p oten tial matrix U = U ( x ; k ) can b e regarded as an imp ortan t c haracteristic to certain nonlinear in tegrable PDE. Now applying the Dbar problem ¯ ∂ ψ ( k ) = ψ ( k ) R ( k ) (3.1) to in tegrable system, the sp ectral transformation matrix or the distribution R ( k ) will take the same role as the p otential matrix U to c haracter the certain nonlinear in tegrable PDE. It is known that the Dbar problem with normalization condition ψ ( k ) → I as k → ∞ is equiv alen t to the Pompeiu formula ψ ( k ) = I + ψ RT C ( k ) , (3.2) where the op erator T C acting to the left is defined in the following form ψ RT C ( k ) = 1 2 iπ Z Z C ψ ( z ) R ( z ) z − k d z ∧ d ¯ z . (3.3) 11 T o discuss the AKNS sp ectral problem by the Dbar problem, one need to intro- duce the physical v ariable x into R ( k ) as following ∂ x R = − ik [ σ 3 , R ] , σ 3 = diag(1 , − 1) , (3.4) whic h implies that R ( k ; x ) = e − ikx ˆ σ 3 r ( k ) = 0 r + ( k )e − 2 ikx r − ( k )e 2 ikx 0 ! , (3.5) where the off-matrix function r ( k ) or r ± ( k ) are indep endent of x , [ σ 3 , A ] = σ 3 A − Aσ 3 and e α ˆ σ 3 A = e ασ 3 A e − ασ 3 for some 2 × 2 matrix A . Th us the eigenv alue function ψ in ( 3.1 ) will dependent on the ph ysical v ariable x , that is ψ = ψ ( k ; x ). It is emphasized that, substituting R ( k ; x ) into the integral equation ( 3.2 ) directly , it ma y lead to uncertain ty . Since the exp onentials e ± 2 ikx with x ∈ R and k ∈ C are not b ounded for x Im k ≶ 0, the function ψ ( · , x ) R ( · , x ) ˆ T C ( k ) ma y not con vergen t in view of the in tegral ( 3.2 ). T o control the sp ectral transformation matrix R ( k ; x ) in ( 3.5 ), we split it in to t wo nilp otent matrices w ± ( k ; x ), that is R = w − + w + , where w − = 0 0 r − ( k )e 2 ikx 0 ! , w + = 0 r + ( k )e − 2 ikx 0 0 ! . (3.6) Then we define a new in tegral op erator R T C ( k ; x ) by giving the follo wing decomp o- sition ψ RT C ( k ; x ) :=[ ψ w − T C + ( k ; x ) + ψ w + T C − ( k ; x )] χ x> 0 + [ ψ w − T C − ( k ; x ) + ψ w + T C + ( k ; x )] χ x< 0 , (3.7) where χ is the indicator function of the half lines x ≷ 0. Lemma 2 L et ψ ∈ L ∞ x ( R + , L p, 0 k ( C ± )) , r ∓ ∈ L q , 2 k ( C ± ) , and ψ w ∓ ∈ L ∞ x ( R + , L µ k ( E ± 1 )) as wel l as ψ (0) w (2) , ∓ ∈ L ∞ x ( R + , L µ k ( E ∓ 1 )) , µ > 2 , wher e E ± 1 = { k : | k | ≤ 1 , Im k ≷ 0 } and ψ (0) , w (2) , ∓ ar e define d in ( 1.6 ) . L et matrix functions g ± ( k ; x ) = ψ w ∓ T C ± ( k ; x ) χ x> 0 , (3.8) then their norms satisfy the fol lowing c onditions ∥ g ± ( k ; x ) ∥ ≤ M p,q ± ∥ r ∓ ∥ L q, 2 ( C ± ) sup x> 0 ∥ ψ ∥ L p, 0 k ( C ± ) , (3.9) 12 ∥ g ± ( k 1 ) − g ± ( k 2 ) ∥ ≤ ˜ M p,q ± ∥ r ∓ ∥ L q, 2 ( C ± ) sup x> 0 ∥ ψ ∥ L p, 0 k ( C ± ) ∥ k 1 − k 2 | α , (3.10) wher e k 1 , k 2 ∈ C , 1 p + 1 q = 1 µ , and α = 1 − 2  1 p + 1 q  . Her e the p ositive c onstants M p,q ± , ˜ M p,q ± dep end on p and q . Pro of . F or conv enience, we introduce the follo wing domains E ± 2 = { k : | k | ≥ 1 , Im k ≷ 0 } . W e split the infinite domain C ± in to E ± 1 and E ± 2 , and let g ± ( k ; x ) = g ± 1 ( k ; x ) + g ± 2 ( k ; x ) , (3.11) where g ± 1 ( k ; x ) = ψ w ∓ T E ± 1 ( k ; x ) χ x> 0 , g ± 2 ( k ; x ) = ψ w ∓ T E ± 2 ( k ; x ) χ x> 0 . (3.12) Since E ± 1 is a b ounded domain, using the H¨ older inequalit y and the Theorem 3 , w e hav e ∥ g ± 1 ( k ; x ) ∥ ≤ M µ 4 , ± sup x> 0 ∥ ψ w ∓ ∥ L µ k ( E ± 1 ) ≤ M p,q 4 , ± ∥ r ∓ ( k ) ∥ L q ( E ± 1 ) sup x> 0 ∥ ψ ∥ L p k ( E ± 1 ) , (3.13) where µ > 2 and 1 p + 1 q = 1 µ . F or the infinite domains E ± 2 , they can b e mapp ed in to E ∓ 1 via the transformation k → k − 1 . Th us we find that g ± 2 ( k ; x ) = 1 2 iπ Z Z E ± 2 ψ ( z ; x ) w ∓ ( z ; x ) z − k d z ∧ d ¯ z χ x> 0 = 1 2 iπ Z Z E ∓ 1 ψ ( z − 1 ; x ) w ∓ ( z − 1 ; x ) ¯ z 2 ( 1 z − 1 z − k − 1 )d z ∧ d ¯ z χ x> 0 := g ∓ 0 (0; x ) − g ∓ 0 ( k − 1 ; x ) , (3.14) where g ∓ 0 ( λ ; x ) = 1 2 iπ Z Z E ∓ 1 ψ ( z − 1 ; x ) ¯ z − 2 w ∓ ( z − 1 ; x ) z − λ d z ∧ d ¯ z χ x> 0 = 1 2 iπ Z Z E ∓ 1 ψ (0) ( z ; x ) w (2) , ∓ ( z ; x ) z − λ d z ∧ d ¯ z χ x> 0 . 13 Here w e ha ve use the notation in ( 1.6 ). It is noted that the exp onents e ± 2 iz − 1 x in w (2) , ∓ ( z ; x ) are b ounded in E ∓ 1 for x > 0. Applying same discussion to g ∓ 0 ( λ ; x ) as g ± 1 ( k ; x ) in ( 3.13 ), and using the triangle inequalit y , w e hav e ∥ g ± 2 ( k ; x ) ∥ ≤ ˜ M p,q 4 , ± ∥ r (2) , ∓ ∥ L q k ( E ∓ 1 ) sup x> 0 ∥ ψ (0) ∥ L p k ( E ∓ 1 ) . (3.15) Then using the triangle inequality again, the condition ( 3.9 ) is obtained b y the substitution of estimates ( 3.13 ) and ( 3.15 ) in to ( 3.11 ). Next, we pro ve the condition ( 3.10 ). F or g ± 1 ( k ; x ) and k 1 , k 2 ∈ C , using the H¨ older inequality and the Theorem 3 , w e hav e ∥ g ± 1 ( k 1 ; x ) − g ± 1 ( k 2 ; x ) ∥ ≤ M µ 5 , ± sup x> 0 ∥ ψ w ∓ ∥ L µ k ( E ± 1 ) ∥ k 1 − k 2 | µ − 2 µ ≤ M p,q 5 , ± ∥ r ± ∥ L q k ( E ± 1 ) sup x> 0 ∥ ψ ∥ L p k ( E ± 1 ) ∥ k 1 − k 2 | α , (3.16) where µ > 2, 1 p + 1 q = 1 µ and α = µ − 2 µ . No w, w e give the same estimation for g ± 2 ( k ; x ) giv en in ( 3.14 ). T o this end, we find that ∥ g ± 2 ( k 1 ; x ) − ∥ g ± 2 ( k 2 ; x ) ∥ = ∥ g ∓ 0 ( k − 1 1 ; x ) − ∥ g ∓ 0 ( k − 1 2 ; x ) ∥ ≤ | k 1 − k 2 | π Z Z E ∓ 1 ∥ ψ (0) ( z ; x ) ∥| r (2) , ∓ ( z ) | 1 | 1 − z k 1 || 1 − z k 2 | d σ z . (3.17) W e will consider the following cases ab out the k ernel in ( 3.17 ): (1). If | k j | ≤ 1 2 , j = 1 , 2, then for | z | ≤ 1, | z k j | ≤ 1 2 and 1 | 1 − z k j | ≤ 2. In this case, w e can get ∥ g ± 2 ( k 1 ; x ) − g ± 2 ( k 2 ; x ) ∥ ≤ | k 1 − k 2 | π ˜ M µ 5 , ∓ sup x> 0 ∥∥ ψ (0) ( z ; x ) ∥| r (2) , ∓ ( z ) |∥ L µ k ( E ∓ 1 ) ≤ | k 1 − k 2 | ˜ M p,q 5 , ∓ sup x> 0 ∥ ψ (0) ∥ L p k ( E ∓ 1 ) ∥ r (2) , ∓ ∥ L q k ( E ∓ 1 ) ≤ ˜ M p,q 5 , ∓ sup x> 0 ∥ ψ (0) ∥ L p k ( E ∓ 1 ) ∥ r (2) , ∓ ∥ L q k ( E ∓ 1 ) | k 1 − k 2 | α , (3.18) in view of | k 1 − k 2 | < 1. (2). If | k 1 | < 1 2 , | k 2 | ≥ 1 2 , then 1 | 1 − z k 1 | ≤ 2. Thus we ha ve ∥ g ± 2 ( k 1 ; x ) − g ± 2 ( k 2 ; x ) ∥ ≤ | k 1 − k 2 | k 2 π ˜ M µ 6 , ∓ sup x> 0 ∥∥ ψ (0) ( z ; x ) ∥| r (2) , ∓ ( z ) |∥ L µ z ( E ∓ 1 )   1 z − k − 1 2   L ν z ( E ∓ 1 ) , 14 where ν = µ − 2 µ . Since | k − 1 2 | ≤ 2, then | z − k − 1 2 | < | z | + | k − 1 2 | < 3 and   1 z − k − 1 2   L ν k ( E ∓ 1 ) ≤  π Z 3 0 r 1 − ν d r  1 ν =  π 3 2 − ν 2 − ν  1 ν . (3.19) In addition, using | k 1 k 2 − 1 | < 2, we can get that | k 1 − k 2 | k 2 = | k 1 − k 2 | µ − 2 µ | k 1 − k 2 | 2 µ | k 2 | µ − 2 µ | k 2 | 2 µ = | k 1 k 2 − 1 | 2 µ | k 2 | µ − 2 µ | k 1 − k 2 | µ − 2 µ ≤ 2 | k 1 − k 2 | α . Th us we hav e ∥ g ± 2 ( k 1 ; x ) − g ± 2 ( k 2 ; x ) ∥ ≤ ˜ M p,q 7 , ∓ sup x> 0 ∥ ψ (0) ∥ L p k ( E ∓ 1 ) ∥ r (2) , ∓ ∥ L q k ( E ∓ 1 ) | k 1 − k 2 | α . (3.20) (3). If | k j | ≥ 1 2 , j = 1 , 2, then using ( 3.19 ), we find that ∥ g ± 2 ( k 1 ; x ) − g ± 2 ( k 2 ; x ) ∥ = ∥ g ∓ 0 ( k − 1 1 ; x ) − ∥ g ∓ 0 ( k − 1 2 ; x ) ∥ ≤ ˜ M p,q 8 , ∓ sup x> 0 ∥ ψ (0) ∥ L p k ( E ∓ 1 ) ∥ r (2) , ∓ ∥ L q k ( E ∓ 1 ) | 1 k 1 − 1 k 2 | α ≤ ˜ M p,q 9 , ∓ sup x> 0 ∥ ψ (0) ∥ L p k ( E ∓ 1 ) ∥ r (2) , ∓ ∥ L q k ( E ∓ 1 ) | k 1 − k 2 | α . (3.21) F rom ( 3.16 )-( 3.21 ), w e can get that ∥ g ± ( k 1 ) − g ± ( k 2 ) ∥ ≤ ˜ M p,q ± | k 1 − k 2 | α  sup x> 0 ∥ ψ ∥ L p k ( E ± 1 ) ∥ r ± ∥ L q k ( E ± 1 ) + sup x> 0 ∥ ψ (0) ∥ L p k ( E ∓ 1 ) ∥ r (2) , ∓ ∥ L q k ( E ∓ 1 )  ≤ ˜ M p,q ± | k 1 − k 2 | α  sup x> 0 ∥ ψ ∥ L p k ( E ± 1 ) + sup x> 0 ∥ ψ (0) ∥ L p k ( E ∓ 1 )   ∥ r ± ∥ L q k ( E ± 1 ) + ∥ r (2) , ∓ ∥ L q k ( E ∓ 1 )  . Th us the condition ( 3.10 ) is pro ved in view of the definition ( 1.9 ). □ Lemma 3 L et ψ ∈ L ∞ x ( R − , L p, 0 k ( C ± )) , r ± ∈ L q , 2 k ( C ± ) , and ψ w ± ∈ L ∞ x ( R − , L µ k ( E ± 1 )) as wel l as ψ (0) w (2) , ± ∈ L ∞ x ( R − , L µ k ( E ∓ 1 )) , µ > 2 , wher e E ± 1 = { k : | k | ≤ 1 , Im k ≷ 0 } . L et matrix functions h ± ( k ; x ) = ψ w ± T C ± ( k ; x ) χ x< 0 , (3.22) then their matrix norms admit the fol lowing c onditions ∥ h ± ( k ; x ) ∥ ≤ N p,q ± ∥ r ± ∥ L q, 2 ( C ± ) sup x< 0 ∥ ψ ∥ L p, 0 k ( C ± ) , (3.23) 15 ∥ h ± ( k 1 ; x ) − h ± ( k 2 ; x ) ∥ ≤ ˜ N p,q ± ∥ r ± ∥ L q, 2 ( C ± ) sup x< 0 ∥ ψ ∥ L p, 0 k ( C ± ) ∥ k 1 − k 2 | α , (3.24) wher e 1 p + 1 q = 1 µ , and α = 1 − 2  1 p + 1 q  . Her e the p ositive c onstants N p,q ± , ˜ N p,q ± dep end on p and q . Pro of . The lemma can b e prov ed similarly . □ Theorem 4 L et ψ ∈ L ∞ x ( R , L p, 0 k ( C )) , r ± ∈ L q , 2 k ( C ) , and ψ w ± ∈ L ∞ x ( R , L µ k ( E 1 )) as wel l as ψ (0) w (2) , ± ∈ L ∞ x ( R , L µ k ( E 1 )) , µ > 2 , wher e E 1 = { k : | k | ≤ 1 } . Then the inte gr al ψ RT C ( k ; x ) define d in ( 3.7 ) satisfies the fol lowing c onditions ∥ ψ RT C ( k ; x ) ∥ ≤ M p,q ( ∥ r + ∥ L q, 2 ( C ) + ∥ r − ∥ L q, 2 ( C ) ) sup x ∈ R ∥ ψ ∥ L p, 0 k ( C ) , (3.25) and for k 1 , k 2 ∈ C , ∥ ψ RT C ( k 1 ; x ) − ψ R T C ( k 2 ; x ) ∥ ≤ ˜ M p,q ( ∥ r + ∥ L q, 2 ( C ) + ∥ r − ∥ L q, 2 ( C ) ) sup x ∈ R ∥ ψ ∥ L p, 0 k ( C ) ∥ k 1 − k 2 | α , (3.26) wher e 1 p + 1 q = 1 µ , and α = 1 − 2  1 p + 1 q  . Her e the p ositive c onstants M p,q , ˜ M p,q dep end on p and q . Pro of . W e note that ∥ r ± ∥ L q, 2 ( C ± ) ≤ ∥ r ± ∥ L q, 2 ( C ) and sup x ≷ 0 ∥ ψ ∥ L p, 0 k ( C ± ) ≤ sup x ∈ R ∥ ψ ∥ L p, 0 k ( C ) . In addition, ψ RT C ( k ; x ) = g + ( k ; x ) + g − ( k ; x ) + h + ( k ; x ) + h − ( k ; x ) , and collect the conditions in Lemma 2 and Lemma 3 , the theorem can b e pro v ed. □ Corollary 1 L et ψ ( k ; x ) ∈ L ∞ x ( R , L p, 0 k ( C )) , r ± ( k ) ∈ L q , 2 ( C ) , and ψ w ± ( k ; x ) ∈ L ∞ x ( R , L µ k ( E 1 )) as wel l as ψ (0) w (2) , ± ∈ L ∞ x ( R , L µ k ( E 1 )) , µ > 2 , wher e E 1 = { k : | k | ≤ 1 } . Then ψ RT C ( k ; x ) ∈ H α k ( C ) is H¨ older c ontinuous in the k plane, wher e 1 p + 1 q = 1 µ and α = 1 − 2  1 p + 1 q  . In addition, ∥ RT C ( k ; x ) ∥ L p, 0 k → H α k ≤ M p,q ( ∥ r + ∥ L q, 2 k ( C ) + ∥ r − ∥ L q, 2 k ( C ) ) . (3.27) 16 In summary , if r ± ∈ L q , 2 k ( C ) then the op erator R T C ( k ; x ) maps the space L ∞ x ( R , L p, 0 k ( C )) in to L ∞ x ( R , H α k ( C )). F urthermore, for the p ositiv e constan t M p,q in ( 3.27 ) related to the p, q and the k plane, if let M p,q < 1 ∥ r + ∥ L q, 2 k ( C ) + ∥ r − ∥ L q, 2 k ( C ) , then the inv erse op erator ( I − RT C ) − 1 on L ∞ ( R , H α k ( C )) is exist. 4 P oten tial reconstruction of AKNS sp ectral prob- lem No w w e consider the Dbar problem relev an t to the AKNS sp ectral problem, that is the sp ectral transformation function R ( k ; x ) admits ( 3.5 ). Under the normalization condition, if r ± ∈ L q , 2 k ( C ), then we hav e ψ ( k ; x ) = I + ψ R T C ( k ; x ) , (4.1) and ψ ( k ; x ) = I ( I − RT C ) − 1 ( k ; x ) , (4.2) whic h imply that ψ ∈ L ∞ x ( R , L p, 0 k ( C )) and ψ − I ∈ L ∞ x ( R , H α k ( C )). It is remarked that H α k ( C ) ⊂ L p, 0 k ( C ) [ 50 ], the inclusion map I = I L p, 0 k → H α k can b e regarded as a iden tity op erator. Theorem 5 If ψ ∈ C 1 x ( R ) ∩ L ∞ x ( R , L p, 0 k ( C )) ∩ L ∞ x ( R , L p, 1 k ( C )) , and r ± ∈ L q , 2 k ( C ) , 1 p + 1 q < 1 2 , then AKNS sp e ctr al pr oblem c an b e obtaine d ∂ x ψ ( k ; x ) = − ik [ σ 3 , ψ ( k ; x )] + Q ( x ) ψ ( k ; x ) , (4.3) wher e the p otential matrix Q is given by the sp e ctr al tr ansformation matrix R and the eigenfunction ψ as fol lowing Q = − i [ σ 3 , ⟨ ψ R ⟩ ] := 0 u v 0 ! , (4.4) with ⟨ ψ R ⟩ = ⟨ ψ w − ⟩ + + ⟨ ψ w + ⟩ − + ⟨ ψ w + ⟩ + + ⟨ ψ w − ⟩ − . 17 Her e ⟨ ψ w ∓ ⟩ ± = 1 2 π i Z Z C ± ψ ( k ; x ) w ∓ ( k ; x )d k ∧ d ¯ k χ x> 0 , ⟨ ψ w ± ⟩ ± = 1 2 π i Z Z C ± ψ ( k ; x ) w ± ( k ; x )d k ∧ d ¯ k χ x< 0 . (4.5) Pro of . Since ψ ∈ C 1 x ( R ), then differen tiating of equation ( 4.1 ) with resp ect to x , and using ( 4.2 ), ( 3.4 ), we find that ∂ x ψ = − ik ψ σ 3 RT C ( I − RT C ) − 1 + ik ψ RT C σ 3 ( I − RT C ) − 1 . (4.6) F or the first term on the right, since RT C = I − I ( I − RT C ), then [ 51 ] RT C ( I − R T C ) − 1 = ( I − RT C ) − 1 − I . (4.7) F or the second term on the right of ( 4.6 ), we know, from ( 3.7 ), that k ψ RT C =[ k ψ w − T C + ( k ; x ) + k ψ w + T C − ( k ; x )] χ x> 0 + [ k ψ w − T C − ( k ; x ) + k ψ w + T C + ( k ; x )] χ x< 0 . If k ψ ∈ L ∞ x ( R , L p, 0 k ( C )) or ψ ∈ L ∞ x ( R , L p, 1 k ( C )) and r ± ∈ L q , 2 k ( C ), then, from theorem 4 , the integral k ψ RT C is well defined. Since k ψ w ∓ T C ± ( k ; x ) χ x> 0 = 1 2 iπ Z Z C ± z ψ ( z ; x ) w ∓ ( z ; x ) z − k d z ∧ d ¯ z χ x> 0 = k [ ψ w ∓ T C ± ( k ; x ) χ x> 0 ] + ⟨ ψ w ∓ ⟩ ± , and k ψ w ± T C ± ( k ; x ) χ x< 0 = k [ ψ w ± T C ± ( k ; x ) χ x> 0 ] + ⟨ ψ w ± ⟩ ± , then we obtain that k ψ RT C = k [ ψ RT C ] + ⟨ ψ R ⟩ , whic h can b e reduced to k ψ RT C = k ψ − k I + ⟨ ψ R ⟩ , (4.8) in terms of ( 4.1 ). Substituting ( 4.7 ) and ( 4.8 ) in to ( 4.6 ), and using ( 4.2 ), we can get that ∂ x ψ = ik ψ σ 3 + i ⟨ ψ R ⟩ σ 3 ψ − iσ 3 k ( I − RC k ) − 1 . (4.9) 18 In addition, from ( 4.8 ) and ( 4.2 ), w e obtain that k ( I − RC k ) − 1 = k ψ + ⟨ ψ R ⟩ ( I − RC k ) − 1 = k ψ + ⟨ ψ R ⟩ ψ . It is remark ed that ⟨ ψ R ⟩ is indep endent of k , so the last term can b e obtained via ( 4.2 ). The pro of of the theorem is finished. □ Lemma 4 L et r ± ∈ L q , 2 k ( C ) and ψ 11 , ψ 22 ∈ L ∞ x ( R , L p, 0 k ( C )) , wher e ψ = ( ψ mn ) 2 × 2 . L et ⟨ ψ 11 r + ⟩ ± = 1 2 π i Z Z C ± ψ 11 ( k ; x ) r + ( k )e − 2 ikx d k ∧ d ¯ k χ x ≶ 0 , ⟨ ψ 22 r − ⟩ ± = 1 2 π i Z Z C ± ψ 22 ( k ; x ) r − ( k )e 2 ikx d k ∧ d ¯ k χ x ≷ 0 , then |⟨ ψ 11 r + ⟩ ± | ≤ M p,q ± , 12 sup x ≶ 0 ∥ ψ 11 ∥ L p, 0 k ( C ± ) ∥ r + ∥ L q, 2 k ( C ± ) , (4.10) |⟨ ψ 22 r − ⟩ ± | ≤ M p,q ± , 13 sup x ≷ 0 ∥ ψ 22 ∥ L p, 0 k ( C ± ) ∥ r − ∥ L q, 2 k ( C ± ) , (4.11) wher e 1 p + 1 q < 1 2 . Pro of . W e only give the pro of of the condition ( 4.10 ). T o this end, w e give the follo wing decomp osition ⟨ ψ 11 r + ⟩ ± = 1 2 π i Z Z E ± 1 ψ 11 ( k ; x ) r + ( k )e − 2 ikx d k ∧ d ¯ k χ x ≶ 0 + 1 2 π i Z Z E ± 2 ψ 11 ( k ; x ) r + ( k )e − 2 ikx d k ∧ d ¯ k χ x ≶ 0 = A ± 1 + A ± 2 . Then using the H¨ older inequality , we find that | A ± 1 | ≤ 1 π Z Z E ± 1 sup x ≶ 0 | ψ 11 ( k ; x ) || r + ( k ) | d σ k ≤ 1 π Z Z E ± 1 (sup x ≶ 0 | ψ 11 ( k ; x ) || r + ( k ) | ) µ d σ k ! 1 µ Z Z E ± 1 d σ k ! 1 ν ≤ 1 π ( π 2 ) 1 ν sup x ≶ 0 ∥ ψ 11 ( k ; x ) ∥ L p k ( E ± 1 ) ∥ r + ∥ L q k ( E ± 1 ) , 19 where 1 µ + 1 ν = 1 , µ > 2 and 1 µ = 1 p + 1 q . In addition, for A ± 2 , we hav e A ± 2 = 1 2 π i Z Z E ± 2 ψ 11 ( k ; x ) r + ( k )e − 2 ikx d k ∧ d ¯ k χ x ≶ 0 = 1 2 π i Z Z E ∓ 1 ψ 11 ( k − 1 ; x ) r + ( k − 1 )e − 2 ik − 1 x k − 2 ¯ k − 2 d k ∧ d ¯ k χ x ≶ 0 . Similarly , w e obtain that | A ± 2 | ≤ 1 π Z Z E ∓ 1 sup x ≶ 0 | ψ (0) , 11 ( k ; x ) || r (2) , + ( k ) || k | − 2 d σ k ≤ 1 π Z Z E ∓ 1  sup x ≶ 0 | ψ (0) , 11 ( k ; x ) || r (2) , + ( k ) |  µ d σ k ! 1 ν Z Z E ∓ 1 | k | − 2 ν d σ k ! 1 ν , where 1 µ + 1 ν = 1 , µ > 2, then 0 < ν < 1. Since Z Z E ∓ 1 | k | − 2 ν d σ k = π Z 1 0 r 1 − 2 ν d r = π 2 − 2 ν , then | A ± 2 | ≤ 1 2 − 2 ν sup x ≶ 0 ∥ ψ (0) , 11 ( k ; x ) ∥ L p k ( E ∓ 1 ) ∥ r (2) , + ∥ L q k ( E ∓ 1 ) . Th us, we hav e |⟨ ψ 11 r + ⟩ ± | ≤ M p,q 12 (sup x ≶ 0 ∥ ψ 11 ( k ; x ) ∥ L p k ( E ± 1 ) ∥ r + ∥ L q k ( E ± 1 ) + sup x ≶ 0 ∥ ψ (0) , 11 ∥ L p k ( E ∓ 1 ) ∥ r (2) , + ∥ L q k ( E ∓ 1 ) ) ≤ M p,q 12 (sup x ≶ 0 ∥ ψ 11 ∥ L p k ( E ± 1 ) + sup x ≶ 0 ∥ ψ (0) , 11 ∥ L p k ( E ∓ 1 ) ) × ( ∥ r + ∥ L q k ( E ± 1 ) + ∥ r (2) , + ∥ L q k ( E ∓ 1 ) )) , whic h pro ves the condition ( 4.10 ). The condition ( 4.11 ) can b e pro v ed similarly . □ Theorem 6 L et r ± ∈ L q , 2 k ( C ) , then the p otentials u and v define d in ( 4.4 ) satisfy the c onditions | u ( x ) | ≤ M p,q 14 ∥ r + ∥ L q, 2 k ( C ) 1 − M p,q 16 ( ∥ r + ∥ L q, 2 k ( C ) + ∥ r − ∥ L q, 2 k ( C ) ) , | v ( x ) | ≤ M p,q 15 ∥ r − ∥ L q, 2 k ( C ) 1 − M p,q 16 ( ∥ r + ∥ L q, 2 k ( C ) + ∥ r − ∥ L q, 2 k ( C ) ) . (4.12) 20 Pro of . F rom ( 4.4 ) and ( 4.5 ), we know that u = ⟨ ψ 11 r + ⟩ + + ⟨ ψ 11 r + ⟩ − , v = ⟨ ψ 22 r − ⟩ + + ⟨ ψ 22 r − ⟩ − . Then using the conditions ( 4.10 ) and ( 4.11 ), w e can get that | u ( x ) | ≤ M p,q 14 sup x ∈ R ∥ ψ 11 ∥ L p, 0 k ( C ) ∥ r + ∥ L q, 2 k ( C ) , | v ( x ) | ≤ M p,q 15 sup x ∈ R ∥ ψ 22 ∥ L p, 0 k ( C ) ∥ r − ∥ L q, 2 k ( C ) . (4.13) W e hav e prov ed that the inv erse op erator ( I − RT C ) − 1 is exist based on the small norm condition. According to the identit y ( 4.7 ), the inv erse op erator admits the relation ( I − R T C ) − 1 = I + RT C ( I − R T C ) − 1 , whic h maps the space H α k ( C ) into L p, 0 k ( C ). Th us, we find that ∥ ( I − R T C ) − 1 ∥ H α k ( C ) → L p, 0 k ( C ) ≤ 1 1 − ∥ R T C ∥ L p, 0 k ( C ) → H α k ( C ) ≤ 1 1 − M p,q 16 ( ∥ r + ∥ L q, 2 k ( C ) + ∥ r − ∥ L q, 2 k ( C ) ) , in terms of the condition ( 3.27 ). In addition, the representation of ψ ( k ; x ) is given via the inv erse op erator in ( 4.2 ). Since I ∈ H α ( C ), then we hav e sup x ∈ R ∥ ψ ( k ; x ) ∥ L p, 0 k ( C ) ≤ 1 1 − M p,q 16 ( ∥ r + ∥ L q, 2 k ( C ) + ∥ r − ∥ L q, 2 k ( C ) ) . (4.14) Hence, the conditions ( 4.12 ) can b e prov ed via ( 4.13 ) and ( 4.14 ). □ Corollary 2 The map L q , 2 ( C ) ∋ r ± ( k ) → u ( x ) , v ( x ) ∈ L ∞ ( R ) is Lipschitz c ontinuous. Pro of . Let ˜ r ± ( k ) and ˜ ψ ( k ; x ) = ( ˜ ψ mn ) are relev ant to the p otential ˜ u ( x ) , ˜ v ( x ) admitting the relation ( 4.4 ), and ˜ r ± ( k ), ˜ ψ ( k ; x ) = ( ˜ ψ mn ) satisfy the relation ( 4.2 ), then u − ˜ u = ⟨ ψ 11 ( r + − ˜ r + ) ⟩ + + ⟨ ( ψ 11 − ˜ ψ 11 ) ˜ r + ⟩ + + ⟨ ψ 11 ( r + − ˜ r + ) ⟩ − + ⟨ ( ψ 11 − ˜ ψ 11 ) ˜ r + ⟩ − . (4.15) 21 Using ( 4.10 ) and ( 4.11 ), w e can get | u − ˜ u | ≤ M p,q 14  sup x ∈ R ∥ ψ 11 ∥ L p, 0 k ( C ) ∥ r + − ˜ r + ∥ L q, 2 k ( C ) + sup x ∈ R ∥ ψ 11 − ˜ ψ 11 ∥ L p, 0 k ( C ) ∥ ˜ r + ∥ L q, 2 k ( C )  . F rom the relation ( 4.2 ), we find that ψ − ˜ ψ = I ( I − R T C ) − 1 − I ( I − ˜ RT C ) − 1 = I ( I − R T C ) − 1 ( R − ˜ R ) T C ( I − ˜ RT C ) − 1 . In a similar wa y , we ha ve sup x ∈ R ∥ ψ − ˜ ψ ∥ L p, 0 k ( C ) ≤ M p,q ( ∥ r + − ˜ r + ∥ L q, 2 k + ∥ r − − ˜ r − ∥ L q, 2 k ) [1 − M p,q 16 ( ∥ r + ∥ L q, 2 k ( C ) + ∥ r − ∥ L q, 2 k ( C ) )][1 − ˜ M p,q 16 ( ∥ ˜ r + ∥ L q, 2 k ( C ) + ∥ ˜ r − ∥ L q, 2 k ( C ) )] . If let ∥ r ± ∥ L q, 2 k , ∥ ˜ r ± ∥ L q, 2 k < B , then, from ( 4.15 ), we can obtain | u − ˜ u | ≤ M ( p, q , B )  ∥ r + − ˜ r + ∥ L q, 2 k + ∥ r − − ˜ r − ∥ L q, 2 k  . (4.16) Similar discussion can deriv e the same condition as ( 4.16 ) for the p oten tial v . □ 5 Conclusions and Discussions W e presen ted prop erties for the in tegral op erator ˆ T G asso ciated with the Dbar equa- tion ¯ ∂ ϕ ( k ) = f ( k ) on the b ounded domain G , and sho wed that ˆ T G w as a com- pletely contin uous op erator on the space L p ( ¯ G ) , p > 2 and mapp ed it into the space H γ ( ¯ G ) , γ = p − 2 p . Since the relev an t constan ts were dep endent on the radius of the domain, so it was difficult to let the op erator was small norm, which implied that the condition for the exist of the inv erse op erator ( I − ˆ T G ) − 1 w as difficult to achiev e. T o apply the Dbar problem to integrable system, we considered the matrix Dbar problem ¯ ∂ ψ ( k ) = ψ ( k ) R ( k ) on the complex k plane C , and let the sp ectral trans- formation off-diagonal matrix R admitting the linear ev olution equation about the ph ysical v ariable x as ∂ x R ( k ; x ) = − ik [ σ 3 , R ( k ; x )]. The Dbar problem with normal- ization condition was equiv alen t to the integral equation ψ ( k ; x ) = I + ψ R T C ( k ; x ), where the kernel contained the exp onen tial factors e ± 2 ikx . 22 T o con trol the conv ergence of the integral, we split the sp ectral transformation function R in to t wo nilp oten t matrices w ± ( k ; x ) containing the factors r ± ( k )e ∓ 2 ikx , and the complex plane C into low er and upp er half planes C ± . So the in tegral ψ RT C w as decomp osed in to the sum of the factors ψ w ∓ T C ± ( k ; x ) χ x> 0 and ψ w ± T C ± ( k ; x ) χ x< 0 . W e further split the lo w er and upp er half planes C ± in to lo w er/upp er inner half unit domain E ± 1 and lo w er/upp er outer half unit domain E ± 2 . By in tro duc- ing the space L p,ν ( C ± ), w e gav e the prior estimates for the decomposed integrals ψ w ∓ T C ± ( k ; x ) χ x> 0 and ψ w ± T C ± ( k ; x ) χ x< 0 , and sho wed that if ψ ∈ L ∞ x ( R , L p, 0 k ( C )), r ± ∈ L q , 2 k ( C ), and ψ w ± ∈ L ∞ x ( R , L µ k ( E 1 )) as well as ψ (0) w (2) , ± ∈ L ∞ x ( R , L µ k ( E 1 )), µ > 2, where E 1 = { k : | k | ≤ 1 } . Then ψ RT C ( k ; x ) ∈ L ∞ x ( R , H α k ( C )) is H¨ older con tinuous in the k plane. W e found that small norm condition for the op erator RT C , and show ed that the inv erse op erator ( I − RT C ) − 1 w as exist. In this pap er, w e consider the no soliton case, that is no Dirac delta factors in the sp ectral transformation R . W e sho w ed that, if r ± ∈ L q , 2 k ( C ), there exists a unique solution ψ = I ( I − R T C ) − 1 ∈ L ∞ x ( R , L p, 0 k ( C )) for the Dbar problem. By using the Dbar dressing metho d, we constructed the AKNS sp ectral problem ∂ x ψ = − ik [ σ 3 , ψ ] + Qψ , and established the p otential reconstruction Q = − i [ σ 3 , ⟨ ψ R ⟩ ]. By decomp osing ⟨ ψ R ⟩ into ⟨ ψ w ∓ ⟩ ± and ⟨ ψ w ± ⟩ ± , we sho wed that the map from r ± ∈ L q , 2 k ( C ) to u, v ∈ L ∞ x ( R ) is Lipschitz contin uous. It is remark ed that, in the direct scattering analysis for the AKNS sp ectral problem, the p otentials u, v are usually b e taken as L 2 ( R ) functions. So, for the this condition, we find from ( 4.13 ) that ∥ u ( x ) ∥ L 2 ( R ) ≤ M p,q 14 ∥ ψ 11 ∥ L 2 x ( R ,L p, 0 k ( C )) ∥ r + ∥ L q, 2 k ( C ) , ∥ v ( x ) ∥ L 2 ( R ) ≤ M p,q 15 ∥ ψ 22 ∥ L 2 x ( R ,L p, 0 k ( C )) ∥ r − ∥ L q, 2 k ( C ) . (5.1) Moreo ver, if ψ ∈ L 2 x ( R , L p, 0 k ( C )), then we can get, from ( 3.25 ) and ( 3.26 ), that ∥ ψ RT C ( k ; x ) ∥ L 2 x ( R ) ≤ M p,q 10 ( ∥ r + ∥ L q, 2 k ( C ) + ∥ r − ∥ L q, 2 k ( C ) ) ∥ ψ ∥ L 2 x ( R ,L p, 0 k ( C )) , (5.2) and for k 1 , k 2 ∈ C , ∥ ψ RT C ( k 1 ; x ) − ψ R T C ( k 2 ; x ) ∥ L 2 x ( R ) ≤ M p,q 11 ( ∥ r + ∥ L q, 2 k ( C ) + ∥ r − ∥ L q, 2 k ( C ) ) ∥ ψ ∥ L 2 x ( R ,L p, 0 k ( C )) ∥ k 1 − k 2 | α , (5.3) 23 where 1 p + 1 q < 1 2 , and α = 1 − 2  1 p + 1 q  . So the conditions ( 5.2 ) and ( 5.3 ) imply that the op erator R T C is also a map from space L 2 x ( R , L p, 0 k ( C )) to L 2 x ( R , H α k ( C )), and ∥ RT C ( k ; x ) ∥ L 2 x ( R , L p, 0 k ( C )) → L 2 x ( R , H α k ( C )) ≤ M p,q ( ∥ r + ∥ L q, 2 k ( C ) + ∥ r − ∥ L q, 2 k ( C ) ) . (5.4) Under the condition of small norm, the in verse op erator ( I − RT C ) − 1 is also exist, and it can b e used to define the eigenfunction ψ in ( 4.2 ). W e similar hav e ∥ u ( x ) ∥ L 2 ( R ) ≤ M p,q 14 ∥ r + ∥ L q, 2 k ( C ) 1 − M p,q 16 ( ∥ r + ∥ L q, 2 k ( C ) + ∥ r − ∥ L q, 2 k ( C ) ) , ∥ v ( x ) ∥ L 2 ( R ) ≤ M p,q 15 ∥ r − ∥ L q, 2 k ( C ) 1 − M p,q 16 ( ∥ r + ∥ L q, 2 k ( C ) + ∥ r − ∥ L q, 2 k ( C ) ) , whic h imply that the map L q , 2 ( C ) ∋ r ± ( k ) → u ( x ) , v ( x ) ∈ L 2 ( R ) , is Lipschitz contin uous. It is also remarked that, the pap er only considered the well-posedness for the AKNS sp ectral problem. If consider the Dbar problem asso ciated with a nonlin- ear in tegrable equation, one needs to introduce a time evolution equation for the sp ectral transformation function R , for example ∂ t R = A ( k )[ σ 3 , R ], where A ( k ) is a p olynomial. So the exp onents will take the form e ± 2 iθ ( k ; x,t ) . T o ensure the conv er- gence of the relev an t integral, one needs to give a more complicated decomposition ab out the domain. These cases will b e considered in our subsequent articles. Statemen ts and Declarations Authorship con tribution statemen ts Junyi Zhu: Conceptualization, Inv estiga- tion, Metho dology , V alidation, W riting-review and editing; Huan Liu: Inv estigation, Metho dology , V alidation, Visualization, Pro ject administration. Conflict of interest The authors declare that they ha ve no conflicts of in terest. F undings This work was supp orted b y National Natural Science F oundation of China (Grant No. 12571268), Natural Science F oundation of Henan Province (Grant No. 262300421239), and Program for Science & T echnology Inno v ation T alents in Univ ersities of Henan Province (Gran t No. 26HASTIT049). 24 References [1] R. Beals, R. R. Coifman, Scattering, transformations sp ectrales et equations d’ev olution nonlineare, I, seminaire Goulaouic-Mey er-Sch wartz, exp.22, T ec h. rep., Ecole Polytec hnique, P alaiseau (1981). [2] R. Beals, R. R. 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