Completeness theorem for the system of eigenfunctions of the complex Schrödinger operator \mathscr{L}_c=-d^2/dx^2+cx^α
aa r X i v : . [ m a t h . SP ] F e b Completeness theorem for the system ofeigenfunctions of the complex Schr¨odinger operator L c,α = − d /dx + cx α Sergey Tumanov ∗ Moscow Center of Fundamental and Applied Mathematics at LomonosovMoscow State University, Russian Federation
February 25, 2021
Abstract
The completeness of the system of eigenfunctions of the complex Schr¨odingeroperator L c,α = − d /dx + cx α on the semi-axis with Dirichlet boundary conditionsis proved for an arbitrary α ∈ (0 ,
2) and | arg c | < πα/ ( α + 2) + ∆ t ( α ) with some∆ t ( α ) > We consider the operator L c,α = − d dx + cx α in L ( R + ) with Dirichlet boundary conditions with c ∈ C , | arg c | < π , α > L c,α has a compact inverse, the spectrum is discrete, root subspaces areone-dimensional [1].For 0 < | arg c | < π it is not self-adjoint, moreover, it has bad spectral properties: thenorm of the resolvent [2, 3] and the norms of spectral projectors [4] grow exponentially.Under these conditions, the operator cannot be similar to self-adjoint, its eigenfunctionsdo not form a Riesz basis in L ( R + ). Nevertheless, the completeness of the system ofeigenfunctions (S.E.) in L ( R + ) is an open problem, which is what our work is devotedto. ∗ [email protected] α ≥ L c,α is fully explored [2, 5, 6]: thesystem is complete for all c ∈ C : | arg c | < π .For α ∈ (0 ,
2) completeness is proved in case | arg c | < t ( α ) = 2 πα/ ( α + 2) [6]. Atthe same time, the case t ( α ) ≤ | arg c | < π is almost not studied, since it is a much moredifficult one. The corresponding arguments are given in [1, 6].Perhaps the first result for the case t ( α ) ≤ | arg c | < π was the study of Savchuk andShkalikov [6] of the complex Airy operator ( α = 1). The authors proved the completenessof S.E. of L c, in case | arg c | < t (1) + π/ π/ L c, / in case | arg c | < t (2 /
3) + ∆ t = π/ t , where ∆ t > π/
10 is the only zero ofsome transcendental equation.It turns out that these results generalize for all α ∈ (0 , t = ∆ t ( α ) > α , such that the S.E. of L c,α is complete in case | arg c | 3? The answer is affirmative if θ ( α ) = t ( α ) + ∆ t ( α ) > π/ 2, inparticular, for α = 2 / 3. Taking into account the continuity of θ ( α ), there exists α < / L i,α is complete for all α > α . This hypothesis was expressed bySavchuk and Shkalikov [6]. In our paper it is a simple Corollary to the main Theorem.For complex ζ = | ζ | e i arg ζ , − π < arg ζ ≤ π and real β we use the term ζ β to denotethe main branch: ζ β = | ζ | β e iβ arg ζ .Let θ ∈ [ t ( α ) , π ) ∩ [ t ( α ) , πα ), and ζ ( θ ) = e i ( t ( α ) − θ ) /α , Z ( θ ) = (sin t ( α ) / sin θ ) /α , we determine ρ ( θ ) = Re n Z ( θ ) Z p e iθ ζ α − e it ( α ) dζ − ζ ( θ ) Z p e iθ ζ α − e it ( α ) dζ o == Re n Z ( θ ) Z ζ ( θ ) p e iθ ζ α − e it ( α ) dζ − ζ ( θ ) Z p e iθ ζ α − e it ( α ) dζ o , where the integration is performed over segments and the branch of the square root ischosen so thatRe Z ( θ ) Z p e iθ ζ α − e it ( α ) dζ > , Re ζ ( θ ) Z p e iθ ζ α − e it ( α ) dζ > . .20.40.60.81.01.21.41.61.82.02.22.42.62.83.0 (cid:1) (cid:2) ( (cid:3) ) t ( (cid:0) ) (cid:4) Figure 1: Schematic chart t ( α ) and θ ( α ). Theorem 1. Given α ∈ (0 , , the function ρ ( θ ) has the only zero θ ( α ) within theinterval: θ ( α ) ∈ ( t ( α ) , π ) ∩ ( t ( α ) , πα ) ; θ ( α ) = t ( α ) + ∆ t ( α ) , ∆ t ( α ) > .The function θ ( α ) is continuous for α ∈ (0 , .For | arg c | < θ ( α ) the S.E. of L c,α is complete in L ( R + ) . Corollary. There exists α < / such that the S.E. of L i,α is complete in L ( R + ) forall α > α . The previously known completeness boundary t ( α ) and the one obtained in thepresent work θ ( α ) are shown in the figure 1. The operator L c,α in L ( R + ) with Dirichlet boundary conditions is defined by differ-ential expression l c,α ( y ) = − d ydx + cx α y, x ∈ [0 , + ∞ ) , and is considered on the domainn D ( L c,α ) = (cid:8) y ∈ L ( R + ) (cid:12)(cid:12) y ∈ W , loc ( R + ) , l c,α ( y ) ∈ L ( R + ) , y (0) = 0 (cid:9) . The following statements underlie the proof of the Theorem 1. They correspond toProposition A.1 and Lemmas 1 and 2 of [1]. Lemma 1. Given | arg c | < π , the operator L c,α is closed, has compact inverse. Theeigenvalues { λ n } , n ∈ N are simple (the root subspaces are one-dimensional), and havethe form of λ n = c / ( α +2) τ n , n ∈ N , where τ n > don’t depend on c . y ′′ = ( cx α − λ ) y. (1)The existence of the so-called Weyl solution — the solution in L ( R + ) is known [10, ChII] for any homogeneous Sturm–Liouville equation with real locally integrable potential.This theory was generalized by Lidskii to the case of some complex potentials [11]. Thefollowing Lemma states for the equation (1): Lemma 2 (On the Weyl solution) . Given < | arg c | < π , there exists Y ( x, λ ) ( x ∈ R + , λ ∈ C ) — non-trivial solution to (1) with properties: • the function Y ( x, λ ) is continuous of two variables ( x, λ ) ∈ R + × C ; • for any λ ∈ C , Y ( x, λ ) ∈ L ( R + ) as a function of x ; • for any x ≥ , Y ( x, λ ) is an entire function of λ of the order of growth of ≤ ( α + 2) / (2 α ) ; • the zeros of Y (0 , λ ) coincide with the eigenvalues of L c,α ; • for an arbitrary f ∈ L ( R + ) the integral G ( λ ) = + ∞ Z Y ( x, λ ) f ( x ) dx (2) is an entire function of the order of growth of ≤ (2 + α ) / (2 α ) ; • the function F ( λ ) = G ( λ ) / Y (0 , λ ) is bounded in any closed sector for which λ Λ = { λ ∈ C (cid:12)(cid:12)(cid:12) arg λ ∈ [0 , arg c ] } . Referring to Levin [12], the system of eigenfunctions { y n } , n ∈ N of the operator L c,α is generated by Y ( x, λ ) and the sequence of eigenvalues { λ n } : y n ( x ) = Y ( x, λ n ). Itturns out that Y ( x, λ ) is a closed kernel in terms of Levin: if for some f ∈ L ( R + ) holds G ( λ ) ≡ 0, then f ≡ 0. Moreover, Lemma 3 (On the closed kernel) . Given < | arg c | < π and some f ∈ L ( R + ) . If thefunction F ( λ ) = 1 Y (0 , λ ) + ∞ Z Y ( x, λ ) f ( x ) dx (3) is a constant, then f ≡ . The proof of the Theorem 1 In our proof we use some notation.In estimates, C denotes arbitrary positive constants, possibly different on both sidesof the inequalities.If for some a > a ≪ a ≫ < a < a ( a > a ) for some a > φ ≡ arg λ , θ ≡ arg c .Let fix some α ∈ (0 , c > | c | = 1 — with complexconjugation and scaling in x one can cover any c ∈ C : 0 < | arg c | < π .Restrict ourselves to the case arg c ∈ [ t ( α ) , π ) ∩ [ t ( α ) , πα ).Let fix some δ > 0: 0 < δ < ( π min { , α } − θ ) / 2. Additional restrictions on δ > l δ = { λ ∈ C (cid:12)(cid:12) t ( α ) − δ < arg λ < t ( α ) } . With our restrictions, arg c > arg λ > λ ∈ l δ .Let’s walk through the main idea of the proof and its stages.Assuming that for some c the S.E. { y n } , n ∈ N is incomplete, one can find f f ∈ L ( R + ), such that F ( λ ) (3) turns out to be an entire function. If we show that F ( λ )is constant, then in view of the Lemma 3 we come to a contradiction with f F ( λ ) grows not faster than a polynomial just in the smallsector l δ . Indeed, on the one hand, the order of growth of F ( λ ) is not higher than π/t ( α ),and on the other, — F ( λ ) is bounded in closed sectors outside Λ (Lemma 2). The centralangles of the sectors complementing l δ to Λ are strictly less than t ( α ), so referring toPhragm´en-Lindel¨of Principle, we obtain that F ( λ ) is a polynomial. Taking into accountthe boundedness of F ( λ ) outside Λ, we obtain that F ( λ ) is a constant.The condition 0 < θ < θ ( α ) turns out to be sufficient for such estimate of F ( λ ) in l δ .We split the proof into several steps: • Step 1 . The independent variable and parameter substitution in (1). • Step 2 . Construction of the Weyl solution in terms of new variable and parameter,investigation of its uniform in x ∈ R + asymptotics for l δ ∋ λ → ∞ . • Step 3 . Proving that the condition ρ (arg c ) < L c,α . • Step 4 . Verification of the sufficient condition. The proof of the existence of acritical ∆ t ( α ) > α such that the condition ρ ( θ ) < θ : t ( α ) ≤ θ < t ( α ) + ∆ t ( α ). This completes the proof of theTheorem 1. 5 .1 Step 1. The independent variable and parameter substitu-tion Let λ ∈ l δ . We fix some arg λ , at the same time | λ | > y ( x ) = y ( x, λ ) to the equation (1), let w ( t ) = y ( | λ | /α t ) , k = | λ | / /α . The equation is converted to the form: w ′′ = k ( ct α − µ ) w, (4)with µ = e iφ , c = e iθ , 0 < φ < θ . Proposition 1. The following inequalities hold: < arg c/µ < π , < arg( c/µ ) /α < π . (5) Proof. Only the upper bounds are of interest.For 0 < α < δ + arg c < πα . Thusarg c/µ < πα − t ( α ) = πα α + 2 < π/ , arg( c/µ ) /α < παα + 2 < π/ , For 1 ≤ α < δ + arg c < π . Thusarg c/µ < π − t ( α ) = π (2 − α ) α + 2 ≤ π/ , arg( c/µ ) /α < π (2 − α ) α ( α + 2) ≤ π/ . In each case, the estimates reach their maximum at α = 1. (cid:3) Further, let µ ∈ l δ , µ = e iφ .Determine a multivalued analytic function in C r = { z ∈ C | Re z > } : S ( z ) = z Z p cζ α − µ dζ , Z l l l l ζ R Δ Figure 2: The paths of monotonicity of Re S ( z ). The wavy line corresponds to the cut.integrating over rectifiable paths in the right half-plane that do not pass through the onlysingularity ζ = ( µ/c ) /α in the IVth quarter (5).Determine the horizontal cut ∆ = { z = ζ + t, t ≥ } (see fig.2).For z ∈ C r we define q ( z ) = cz α − µ .We also define the main branches of q / ( z ) and S ( z ) in the simply connected domain C r \ ∆ by the conditions: S (0) = 0 (by continuity); Re q / ( z ) > S ( z ) > z > z ≪ 1. In what follows, √ cz α − µ and S ( z ) will denote the main branches of q / ( z ) and S ( z ). Cases of the continuation of these branches through the cut ∆ will bespecially noted.We also fix in C r \ ∆ any branch of q / ( z ), which we call the main one.Denote Z = (Im µ/ Im c ) /α > Proposition 2. There exist ε > , R > , the value Re S ( z ) is monotone along each ofthe following paths for R > R : l = { ζ e − iε t, ≤ t < + ∞} ,l = { Z + it, t ≥ } ∪ [ Z , + ∞ ) ,l ,R = { Z + it, t ≥ } ∪ [ Z , R ] ∪ Γ R ∪ { z ∈ l , | z | ≥ R } ,l = [0 , Z ] ∪ { Z + it, t ≥ } , where Γ R = { Re it , t ∈ [arg ζ − ε, } (see fig.2). The analytic continuation of the mainbranch of S ( z ) is considered along l ,R , since l ,R is crossing ∆ .When z moves from the origin to an infinite point along each of the paths l and l ,the value of Re S ( z ) increases monotonically from to + ∞ .When z moves from + ∞ to Z + i ∞ along l , the value of Re S ( z ) increases mono-tonically from −∞ to + ∞ . hen z moves from ζ e − iε ∞ to Z + i ∞ along l ,R , the value of Re S ( z ) decreasesmonotonically from + ∞ to −∞ . The continuation of S ( z ) through ∆ is considered here.Along the real axis the value Re S ( z ) : is increasing on [0 , Z ] , has a single maximumat z = Z , is decreasing to −∞ on ( Z , + ∞ ) with asymptotics: S ( z ) ∼ − c / α + 2 z α/ , as z → + ∞ . (6) For all z > the value Im q / ( z ) < for the main branch of q / .The inequality holds: Re S ( ζ ) > . Proof. We use the asymptotic formula following from the integral representation of S ( z )in neighborhoods of the infinite parts of the curves l j : S ( z ) ∼ z Z p cζ α dζ ∼ ± c / z α/ α/ , z → ∞ , (7)the choice of the sign will be further specified for each curve l j .It also follows from the integral representation of S ( z ): S ( z ) ∼ z Z √− µ dζ ∼ − iµ / z , as z → , Re z > , (8)the choice of the sign is determined by the condition Re S ( z ) > z > z ≪ S ( z ) < z ≪ S ( z ) along the real axis for z = t ≥ ddt Re S ( t ) = Re √ ct α − µ = 0 ⇔ √ ct α − µ = iβ for some t ≥ β ∈ R \ { } , that is equivalent to ct α − µ = − β < 0. Considering theimaginary part, t = Z = (Im µ/ Im c ) /α . For the real part,Re cZ α − Re µ = Im { µc } Im c = | c | Im c Im µc < , taking (5) into account. Thus there is a unique extremum of Re S ( z ) at Z > z ≥ S ( z ) for z = t ≥ ddt Im S ( t ) = 0 ⇔ √ ct α − µ = β ∈ R \ { } , ct α − µ = β > 0. Considering the imaginary part, again we get t = Z , but cZ α − µ = − β < 0, thus, Im S ( z ) has no extrema for z ≥ z ≥ S ( z ) increases, reaches its maximum at z = Z , then decreases. The function Im S ( z ) decreases for all z ≥ 0. In particular forthe main branch of q / ( z ) we obtain Im q / ( z ) = Im S ′ ( z ) < z > S ( z ) along the vertical rays { a + it, t ≥ } for a > ddt Re S ( a + it ) = 0 ⇔ c ( a + it ) α − µ = β > ⇔ ( a + it ) α = µc + β c . Both terms on the right-hand side lie in the lower half-plane (5). At the same time,since α ∈ (0 , a + it ) α lies in the upper half-plane, or on the real axis. Inother words, there are no extrema, Re S ( z ) is strictly monotone. In particular, along thevertical ray l ∩ l .Let’s move along l from the origin and observe the curve in the image of S ( z ). Thevalue Re S ( z ) is strictly increasing, and the value Im S ( z ) is strictly decreasing on thesegment [0 , Z ]. The tangent to the image S ([0 , Z ]) at S ( Z ) is strictly vertical anddirected downward. Since S ( z ) is univalent in Z , the angles are preserved: when movingalong l from Z to Z + i ∞ , the value Re S ( z ) will increase.The same reasoning shows that Re S ( z ) is monotone along l .We take ε > l lies in the IVth quarter. We reduce ε > S ( z ) liesin the IVth quarter for arbitrarily small points of l . This is possible, as δ < ( πα − θ ) / t → +0 arg S ( ζ e − iε t ) = arg (cid:8) − iζ e − iε µ / (cid:9) == − π t ( α )2 − δ t ( α ) α − δ α − θα − ε = π − θα − δ α + 22 α − ε >> π − θα − δ α + 22 α − ε (9)where δ = t ( α ) − φ ∈ (0 , δ ). Since α ∈ (0 , π − θα − δ α + 22 α > π − θα − 12 ( πα − θ ) α + 22 α > π − θα − α + 2 ( πα − θ ) α + 22 α = − π , therefore, we reduce ε > − π/ 2. On theother hand, as π/ < t ( α ) /α ≤ θ/α , we havelim t → +0 arg S ( ζ e − iε t ) = π − θα − δ α + 22 α − ε < . Further we explore the extrema of Re S ( z ) along l : ddt Re S ( ζ e − iε t ) = 0 ⇔ Re (cid:8) ζ e − iε µ / p e − iεα t α − (cid:9) = 0 . (10)9he value ζ e − iε µ / lies in the Ist quarter (see the first equality in (9)); for t > S ( z ) has noextrema along l .With ε = 0, the same reasoning proves the monotonic growth of Re S ( ζ t ), along0 ≤ t < 1. Whence Re S ( ζ ) > S ( z ) is monotonic along Γ R = { Re it , t ∈ [arg ζ − ε, } : ddt Re S ( Re it ) = 0 ⇔ e it ( cR α e itα − µ ) = β > . (11)For R ≫ 1, the argument of the left part of the last equality in (11) is determined bythe argument of ce it ( α +2) , butarg { ce it ( α +2) } = θ + t ( α + 2) ≥ θ + (cid:0) φ − θα − ε (cid:1) ( α + 2) == 2 πα − θα − δ α + 2 α − ε ( α + 2) , (12)where δ = t ( α ) − φ ∈ (0 , δ ). As δ < ( πα − θ ) / 2, we have:2 πα − θα − δ α + 2 α > πα − θα − δ α + 2 α = α + 2 α (cid:16) α + 2 ( πα − θ ) − δ (cid:17) >> α + 2 α (cid:16) 12 ( πα − θ ) − δ (cid:17) > , reducing ε > { ce it ( α +2) } > { ce it ( α +2) } ≤ arg c < π for t ≤ 0, i.e. ce it ( α +2) lies in the upperhalf-plane, therefore, for sufficient large R > R > z = Re it ∈ Γ R the value ce it ( α +2) lies in the upper half-plane, c / e it ( α/ — inthe Ist quarter, in particular for t = arg ζ − ε . Thus, for z ∈ l , z = | z | e it , | z | → ∞ : S ( z ) ∼ c / e it ( α/ | z | α/ α/ , the value S ( z ) has to lie in the right half-plane due to the monotonic growth of Re S ( z )along l , this determines the choice of the sign in the last formula, and Re S ( z ) → + ∞ as l ∋ z → ∞ .Let z ∈ Γ R , consider the analytic continuation of S ( z ) through the cut ∆ from below,(7) takes the form: S ( z ) ∼ c / z α/ α/ , uniformly in z ∈ Γ R , R → + ∞ . (13)10or R ≫ z ∈ Γ R the value Re S ( z ) lies in the Ist quarter. Taking into account themonotonicity of Re S ( z ) along Γ R , it follows from (13) that Re S ( z ) decreases with z moving along Γ R counterclockwise for R ≫ 1. In view of the previous considerationsalong the rest parts of l ,R , this justifies the monotonicity of Re S ( z ) along the entire l ,R for R ≫ 1, without loss of generality, for R > R > S ( z ) is unboundedfor z → + ∞ . Since Re S ( z ) decreases on z ∈ [ Z , + ∞ ), we obtain Re S ( z ) → −∞ , andthe sign “minus” is selected in (6).Let z ∈ l , z → ∞ . Then z = | z | e iπ/ o (1) ∼ | z | e iπ/ . Using the asymptotics of theintegral S ( z ), we obtain: S ( z ) ∼ − c / e iπ ( α/ / | z | α/ α/ , the term e iπ ( α/ / lies in the IInd quarter, as α ∈ (0 , c / lies in the Istquarter. The product c / e iπ ( α/ / lies in the left half-plane, thus Re S ( z ) is unbounded,Re S ( z ) → + ∞ , and the sign “minus” is used in the last asymptotic formula. (cid:3) The next Proposition is the main one in the current section. Proposition 3. For k ≫ and t ≥ there is the solution W ( t, k ) to (4) , for which theuniform asymptotic formulas are valid: W ( t, k ) = ( ct α − µ ) − / (cid:8) − ie k ( S ( t ) − S ( ζ )) [1] + e − kS ( t ) [1] (cid:9) , t ∈ [0 , Z ] , W ( t, k ) = ( ct α − µ ) − / (cid:8) − ie k ( S ( t ) − S ( ζ )) [1] (cid:9) , t > Z , (14) where [1] = 1 + O ( k − ) is uniform in t on the corresponding interval.For k ≫ , W ( t, k ) ∈ L ( R + ) as a function of t . Proof. Further, let z ∈ C r ∪ { } .The function Re S ( z ) increases along l . For k ≫ 1, we construct the subordinate solution to (4) at the infinite point of l , according to [13, Ch.6 § W ( z, k ) = ( cz α − µ ) − / e − kS ( z ) (1 + ǫ ( z, k )) , (15)where | ǫ ( z, k ) | ≤ exp n V z, ∞ ( F )2 k o − , F = Z q / d dz (cid:16) q / (cid:17) dz. The symbol V z, ∞ ( F ) denotes to the variation of the error-control function F ( z ) along theunbounded part of l , starting from z ∈ l .We estimate: V z, ∞ ( F ) = Z [ z, ∞ ) ⊂ l (cid:12)(cid:12)(cid:12) q / d dz (cid:16) q / (cid:17)(cid:12)(cid:12)(cid:12) | dζ | ≤ C Z [ z, ∞ ) ⊂ l | dζ | | ζ | α/ ≤ C | z | α/ , ǫ ( z, k ) = O ( k − ) uniformly in z ∈ l .According to [13, Ch.5 § W ( z, k ) can be continued analytically to z ∈ C r .Consider l ,R for R > R > l ,R subordinate at the infinite point of the ray l ∩ l ,R . By construction, it coincides with W ( z, k ) along l ∩ l ,R , thus it coincides with W ( z, k ) in C r due to the uniqueness ofthe analytic continuation.The formula (15) remains along the entire l ,R with the remark, that when passingthrough the cut ∆ from below, one should consider the analytic continuation of the mainbranches q / ( z ) and S ( z ).The new branches of q / ( z ) and S ( z ) obtained as a result of this continuation will bealso considered in the domain C r \ ∆ (on the second sheet). Eventually, for z ∈ C r \ ∆,the new branch of q / ( z ) takes the form of iq / ( z ). The new branch of S ( z ) (denoted as˜ S ( z )) takes the form of: ˜ S ( z ) = 2 S ( ζ ) − S ( z ) . (16)In terms of new branches, for z ∈ ( l ∩ l ,R ) ∪ [ Z , R ], k ≫ W ( z, k ) = − i ( cz α − µ ) − / e − k ˜ S ( z ) (1 + ǫ ( z, k )) , (17) | ǫ ( z, k ) | ≤ exp n V z, ∞ ( F )2 k o − , where the variation is considered along the unbounded part of l ,R with the end at theinfinite point of the ray l ∩ l ,R . Obviously: V z, ∞ ( F ) < C, for z ∈ l ,R , V z, ∞ ( F ) < C (cid:16) R Z z + Z Γ R + Z [ z, ∞ ) ⊂ l (cid:17) | dζ || ζ | α/ ≤ Cz α/ + CR α/ , for z ∈ [ Z , R ] , where the constant C > R > R .Since the choice of R > R is arbitrary, the formula (17) defines a representation for W ( z, k ) along l for k ≫ z ∈ l estimate ǫ ( z, k ) = O ( k − ). Thisproves the second formula in (14).We turn to the path l , along which we construct two solutions to (4): u ( z, k ) —subordinate at z = 0 and v ( z, k ) — subordinate at the infinite point of l . Like W ( z, k ),both u ( z, k ) and v ( z, k ) can be analytically continued to C r . The following asymptoticsare valid for k ≫ z ∈ l : u ( z, k ) = ( cz α − µ ) − / e kS ( z ) (1 + O ( k − )) ,v ( z, k ) = ( cz α − µ ) − / e − kS ( z ) (1 + O ( k − )) . (18)12oreover, (18) guarantees the linear independence of u ( z, k ) and v ( z, k ) for k ≫ k ≫ A ( k ) and B ( k ): W ( z, k ) = A ( k ) u ( z, k ) + B ( k ) v ( z, k ) . (19)To find A ( k ) and B ( k ), we use (15), (17), (18) at z = 0 and at some arbitrary large inabsolute value point z ∗ ∈ { Z + it, t ≥ } ⊂ l : ( A ( k )[1] + B ( k )[1] = 1 A ( k ) e kS ( z ∗ ) [1] + B ( k ) e − kS ( z ∗ ) [1] = − ie − k ˜ S ( z ∗ ) , where [1] = 1 + O ( k − ) for k ≫ z ∈ { Z + it, t ≥ } .Taking into account (16), the limit Re S ( z ) → + ∞ for l ∋ z → ∞ , the inequalityRe S ( ζ ) > A ( k ) = − ie − kS ( ζ ) [1] , B ( k ) = [1] . Combining this result with the formula (19) and the asymptotics (18), we get the firstformula in (14).For k ≫ 1, taking into account (6) and (14), we obtain that W ( t, k ) is bounded andexponentially decaying as t → + ∞ , thus W ( t, k ) ∈ L ( R + ) as a function of t . (cid:3) Note that the function S ( z ), the critical points Z = (Im µ/ Im c ) /α > ζ = ( µ/c ) /α depend on θ = arg c ∈ [ t ( α ) , π ) ∩ [ t ( α ) , πα ), and on φ = arg µ ∈ ( t ( α ) − δ, t ( α )).We determine: ρ ( θ, φ ) = Re (cid:8) S ( Z ) − S ( ζ ) (cid:9) . For each θ ∈ [ t ( α ) , π ) ∩ [ t ( α ) , πα ) the function ρ ( θ, φ ) can be continuously extenderto the point φ = t ( α ). Determine ρ ( θ ) = ρ ( θ, t ( α )) = Re (cid:8) S ( Z ) − S ( ζ ) (cid:9)(cid:12)(cid:12)(cid:12) φ = t ( α ) . Proposition 4. The condition ρ ( θ ) < , θ = arg c ∈ [ t ( α ) , π ) ∩ [ t ( α ) , πα ) is sufficientfor the completeness of S.E. of the operator L c,α . Proof. Let ρ ( θ ) < 0. We take δ > δ < ( π min { , α } − θ ) / ρ ( θ, φ ) < µ ∈ l δ .Recently with the substitution of the variable and the parameter, we constructed thesolution W ( t, k ) ∈ L ( R + ) to (4) for k ≫ 1. The function Y ( | λ | /α t, λ ) ∈ L ( R + ) is alsothe solution to (4). Hereinafter k = | λ | / /α .13he eigenvalues of L c,α lie on the ray arg λ = 2 θ/ ( α +2) (Lemma 1). As δ < ( πα − θ ) / θ/ ( α + 2) < t ( α ) − δ , thus, there are no eigenvalues inside l δ . Therefore for λ ∈ l δ , the solution y ( x ) to (1) with the initial conditions y (0) = 0, y ′ (0) = 1 is not in L ( R + ); the subspace of solutions in L ( R + ) is one-dimensional. The same is true for theequation (4). Thus for k ≫ 1, there exists the function A ( k ) such that Y ( | λ | /α t, λ ) = A ( k ) W ( t, k ) . (20)Let us assume the contrary, that the S.E. of L c,α is incomplete. Then there exists f ∈ L ( R + ), f 0, and for all eigenvalues { λ n } of the operator L c,α , G ( λ n ) = 0 (2). Inthis case F ( λ ) (3) is an entire function. We estimate it for | λ | ≫ l ⊂ l δ with fixed φ = arg λ . |F ( λ ) | ≤ C |Y (0 , λ ) | ∞ Z |Y ( x, λ ) | dx = C |Y (0 , λ ) | | λ | /α + ∞ Z |Y ( | λ | /α t, λ ) | dt == C |Y (0 , λ ) | | λ | /α + ∞ Z | A ( k ) W ( t, k ) | dt = C |W (0 , k ) | | λ | /α + ∞ Z |W ( t, k ) | dt ≤≤ C | λ | /α n Z Z ( e k Re( S ( t ) − S ( ζ )) + e − k Re S ( t ) ) dt + + ∞ Z Z e k Re( S ( t ) − S ( ζ )) dt o ≤≤ C | λ | /α (cid:8) e k Re( S ( Z ) − S ( ζ )) + 1 (cid:9) = C | λ | /α (cid:8) e kρ ( θ,φ ) + 1 (cid:9) < C | λ | /α . The first estimation follows from the Cauchy–Bunyakovsky inequality and the condi-tion f ∈ L ( R + ); then we make the substitution x = | λ | α t ; then we use (20) and bound-edness of | ct α − µ | − / along l . We take into account that Re S ( t ) ≥ t ∈ [0 , Z ]. Thefinal estimation follows from Laplace method. Everywhere C does not depend on k .By construction 0 < t ( α ) − δ < φ < t ( α ) ≤ θ . As δ < πα − θ , it follows: δ < πα/ ( α + 2) − θ , thus θ − t ( α ) + δ < t ( α ).In view of these estimates, we can form two adjacent sectors Λ , Λ with a commonboundary passing along the ray l , so that 1) the central angle of each sector is less than t ( α ), 2) the second boundary of each sector lies outside Λ = { arg λ ∈ [0 , θ ] } .The order of growth of F ( λ ) is at most π/t ( α ) (Lemma 2). The function F ( λ ) isbounded on the rays arg λ Λ and grows no faster than a polynomial on the ray l . Itfollows from the Phragm´en-Lindel¨of Principle that in each of the sectors Λ j ( j = 1 , F ( λ ) grows no faster than a polynomial, and therefore in the whole plane C = Λ ∪ Λ ∪ Λ . In this way, the entire function F ( λ ) is a polynomial itself. Because ofits boundedness on the rays arg λ Λ, it follows that F ( λ ) ≡ const.Applying Lemma 3, we conclude that f ≡ 0. We arrived at a contradiction by assumingthat there is no completeness with ρ ( θ ) < (cid:3) .4 Step 4. Verification of the sufficient condition Proposition 5. The function ρ ( θ ) increases for θ ∈ [ t ( α ) , π ) ∩ [ t ( α ) , πα ) , takes valuesof different signs in the neighborhoods of the boundaries of this interval.The only zero of ρ ( θ ) , θ ( α ) = t ( α ) + ∆ t ( α ) lies strictly inside the interval, in par-ticular, ∆ t ( α ) > , and is a continuous function for α ∈ (0 , .Thus, the sufficient conditions for the completeness of S.E. of the operator L c,α aresatisfied for all arg c ∈ [ t ( α ) , θ ( α )) . Proof. Denote, µ = e it ( α ) .The function ρ ( θ ) is continuous at θ = t ( α ) and ρ ( t ( α )) = − Re S ( ζ ) < 0, since inthis case Z = ζ .Exploring the second boundary value, consider two cases: 0 < α < ≤ α < < α < θ → πα − 0. We substitute the variable in the integral: S ( ζ ) = − i µ /α +1 / c /α Z p − ξ α dξ = ic /α Z p − ξ α dξ, Obviously, Re S ( ζ ) → 0, therefore ρ ( θ ) ∼ Re S ( Z ) (cid:12)(cid:12)(cid:12) θ = πα > ≤ α < θ → π − 0. Then Im c → 0, Re c → − 1, Re c / ∼ Im c/ Z → + ∞ . In this case Re S ( ζ ) is bounded.Let us estimate Re S ( Z ):Re S ( Z ) > Z Z Z / Re p cζ α − µ dζ = Z Z / Re r cξ α Im µ Im c − µ dξ >> CZ (Im µ ) / (Im c ) / Re c / Z / ξ α/ (1 + O (Im c )) dξ > C (Im c ) /α − / ;the first estimate is legal, since Re S ( z ) is positive and increasing in the interval [0 , Z ].Then we substitute the variable ζ = Z ξ ; the explicit expression is used for Z =(Im µ / Im c ) /α . The two final inequalities are valid for Im c ≪ 1. The estimation O (Im c )is uniform in ξ ∈ [1 / , α < 2, we conclude Re S ( Z ) → + ∞ , thus ρ ( θ ) → + ∞ .Now we prove the strict increase of ρ ( θ ) on the interval θ ∈ ( t ( α ) , π ) ∩ ( t ( α ) , πα ).15ote, ρ ( θ ) = Re n Z Z ζ p cζ α − µ dζ − ζ Z p cζ α − µ dζ o == Re Z Z ζ p cζ α − µ dζ − sin θα Z p − ξ α dξ. (21)We turn to the first term in (21), denoting I ( θ ) = Re Z Z ζ p cζ α − µ dζ . For I ( θ ) we will carry out the integration along the path γ = { ζ = ζ ( t ) , t ∈ [0 , τ ] } , ζ ( t ) = (( µ − t ) /c ) /α , τ = Im( c/µ ) / Im c > µ − t ) /c – image of the interval t ∈ [0 , τ ] is a segment in the IVth quarter (dueto the location of endpoints — Proposition 1). The IVth quarter also contains the path γ — this is true for endpoints by virtue of the same Proposition 1, and taking into accountthe continuity of the arg function — for the whole path γ .For the main branch q / ( ζ ) = √ cζ α − µ = − it / ( ζ ) along γ . Indeed, according toProposition 2, Im q / ( Z ) < 0, i.e q / ( Z ) = − iτ / ; for other points of γ the sign ispreserved due to the continuity of the square root.To shorten the notation, we denote the partial derivatives by indices with respectto the corresponding variables. For example, ζ t is the derivative of the parameterizationfunction for γ .We have: I ( θ ) = τ Z t / Im ζ t dt, I θ ( θ ) = τ / Im ζ t ( τ ) τ θ + τ Z t / Im ζ θt dt. (22)Obviously, c θ = ic , τ θ = − (Re c/ Im c ) θ Im µ = − (Re c θ Im c − Im c θ Re c ) Im µ / (Im c ) ,thus τ θ = Im µ / (Im c ) , as well as | c | = 1.Further, ζ t ( τ ) = − (1 /α ) ζ ( τ ) / ( µ − τ ). As ζ ( τ ) = Z and µ − τ = cZ α = c Im µ / Im c ,we obtain Im ζ t ( τ ) τ θ = Z /α . Also note, that ζ θ = ζ c c θ = icζ c = − iζ /α .Combining these results with (22), we get: I θ ( θ ) = 1 α τ / Z − α τ Z t / Re ζ t dt. ϑ ∈ [0 , τ ] for which I θ ( θ ) = 1 α τ / Z − α τ / (Re ζ ( τ ) − Re ζ ( ϑ )) = 1 α τ / Re ζ ( ϑ ) > , we have used the facts that all points of γ lie strictly in the right half-plane, and τ > J ( θ ) = sin θα Z p − ξ α dξ. Since t ( α ) < θ < πα , then π/ < π/ ( α + 2) < θ/α < π , therefore: J θ ( θ ) = 1 α cos θα Z p − ξ α dξ < , after all we obtain that ρ θ ( θ ) = I θ ( θ ) − J θ ( θ ) > 0. The function ρ ( θ ) is strictly increasing,has the only zero θ = θ ( α ) = t ( α ) + ∆ t ( α ) ∈ ( t ( α ) , π ) ∩ ( t ( α ) , πα ). For all θ ∈ [ t ( α ) , θ ( α )), the value ρ ( θ ) < α ∈ (0 , θ = θ ( α ), consider ρ ( θ ) = R ( α, θ ) as a function of two variables: α and θ in the small neighborhood Υ of ( α , θ ).From the representation (21) and the first formula of (22) we obtain the existence ofa small neighborhood Υ = Υ( α , θ ), in which R ( α, θ ) is continuous as a function of twovariables.Since R ( α , θ ) = 0 and R ( α , θ ) is strictly monotone as a function of θ for ( α , θ ) ∈ Υ,the Implicit Function Theorem implies the continuity of θ ( α ) in some small neighborhoodof the point α . Since α is arbitrary, it’s valid for all α ∈ (0 , (cid:3) We have proved the completeness of the of S.E. of the operator L c,α in case t ( α ) ≤| arg c | < θ ( α ). As already noted, the completeness in case | arg c | < t ( α ) is a knownfact [6]. This completes the proof of the Theorem 1. We turn to the complex Airy operator to evaluate θ (1) as an example. Let α = 1;taking into account (21), (22) and the evaluation ζ t ( t ) = − /c , we obtain: ρ ( θ ) = τ ( θ ) Z t / sin θ dt − 23 sin θ = 23 ( τ / ( θ ) − 1) sin θ. 17e solve the equation ρ ( θ ) = 0 for θ = θ (1) ∈ ( t (1) , π ) = (2 π/ , π ). The onlypossible case — τ ( θ ) = 1. As well as τ ( θ ) = Im( c/µ ) / Im c = sin( θ − π/ / sin θ , theequation on θ takes the form of:sin( θ − π/ 3) = sin θ , and it has the only solution θ = 5 π/ 6, which exactly corresponds to the result of Savchukand Shkalikov [6].The author expresses his deep appreciation to Andrei Andreyevich Shkalikov for hisattention to the work and valuable advice, as well as to the team of the scientific seminar”Operator Models in Mathematical Physics“ for their support.This research was supported by RFBR grant No 19-01-00240. References [1] Tumanov S., Completeness theorem for the system of eigenfunctions of the complexSchr¨odinger operator L c = − d /dx + cx / , J. Funct. Anal., :7 (2021), 108820.[2] Davies E. 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