Dispersive Properties of Schrödinger Equations Citation Cai, Kaihua (2005) Dispersive Properties of Schrödinger Equations. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/EGNB-FZ41. https://resolver.caltech.edu/CaltechETD:etd-06022005-153453 Abstract This thesis mainly concerns the dispersive properties of Schrodinger equations with certain potentials, and some of their consequences. First, we consider the charge transfer models in R n with n > 2. In this case, the potential is a sum of several individual real-valued potentials, each moving with constant velocities. We get an L 1 to L ∞ estimate for the evolution and the asymptotic completeness of the evolution in any Sobolev space. Second, we derive the L 1 to L ∞ estimate for the Schrodinger operators with a Lame potential. The Lame potential is spatially periodic and its spectrum has the structure of finite bands. We obtain a dispersive estimate with a decay rate t -1/3 . Item Type: Thesis (Dissertation (Ph.D.)) Subject Keywords: Dispersive; Schrodinger Degree Grantor: California Institute of Technology Division: Physics, Mathematics and Astronomy Major Option: Mathematics Thesis Availability: Public (worldwide access) Research Advisor(s): Schlag, Wilhelm Thesis Committee: Schlag, Wilhelm (chair) Pramanik, Malabika Goldberg, Michael Killip, Rowan Defense Date: 19 May 2005 Record Number: CaltechETD:etd-06022005-153453 Persistent URL: https://resolver.caltech.edu/CaltechETD:etd-06022005-153453 DOI: 10.7907/EGNB-FZ41 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 2392 Collection: CaltechTHESIS Deposited By: Imported from ETD-db Deposited On: 03 Jun 2005 Last Modified: 10 Dec 2020 20:16 Thesis Files Preview PDF - Final Version See Usage Policy. 512kB Repository Staff Only: item control page

Dispersive Properties of Schrödinger Equations Thesis by Kaihua Cai In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy IFORNIA I L A CHNOLOG TE UTE O TIT F S N 1891 Y C California Institute of Technology Pasadena, California 2005 (Defended May 19, 2005) ii c 2005 ° Kaihua Cai All Rights Reserved iii Acknowledgements I wish to express my deep gratitude to my advisor and mentor, Wilhelm Schlag, for leading me through my graduate study, and for his inspiring discussions and many valuable suggestions. Having him as an advisor has been the best part of my life at the California Institute of Technology. I would also like to thank many other people in the math department, especially the department staff and the graduate students in my year. There are too many to list without forgetting someone. They have been so much helpful in all aspects of my mathematical education. I have had a great time at Caltech. Caltech has provided me with a wonderful environment for my life and study. The people on campus are all extremely nice. Finally, I want to thank my family for their support all along my pursuit of mathematics. iv Abstract This thesis mainly concerns the dispersive property of Schrödinger equations with certain potentials, and some of their consequences. We denote the evolution generated by the Hamiltonian H(t) = −∆ + V (x, t) as U (t). First, we consider the charge transfer models in Rn with n ≥ 3. In this case, the potential V (x, t) is a sum of several individual real-valued potentials, each moving with constant velocities. Our results are motivated by considering the asymptotic stability of noninteracting multisoliton states. For suitable initial data ψ0 , we obtain 1 1 kU (t)ψ0 kLp < C(n, p)t−n( 2 − p ) kψ0 kLq , p−1 + q −1 = 1, 2 ≤ p ≤ +∞. (0.0.1) Second, we derive the same estimate as (0.0.1) for the derivatives of U (t)ψ0 and prove the asymptotic completeness for charge transfer models in the Sobolev space H κ (Rn ). Another kind of potential we consider is the spatially periodic case. In this case, the Hamiltonian H(x) is called a Hill’s operator. Generally, its spectrum is union of infinitely many intervals (bands). To get started, we focus on one class of finite band potentials, i.e., the Lamé operators. We derive a dispersive estimate with a decay 1 rate t− 3 . v Contents Acknowledgements iii Abstract iv 1 Background 1 1.1 Schrödinger operators in dimension n ≥ 3 . . . . . . . . . . . . . . . . 2 1.2 Dispersive estimates for Schrödinger operators in dimension n = 1, 2 . 4 2 Charge Transfer Models 7 2.1 Definitions and main results . . . . . . . . . . . . . . . . . . . . . . . 7 2.2 Preparation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.3 Proof of the decay estimates . . . . . . . . . . . . . . . . . . . . . . . 21 2.3.1 Bound states . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 2.3.2 Low and high velocity estimates . . . . . . . . . . . . . . . . . 33 2.4 Decay estimates of the derivatives of U (t) . . . . . . . . . . . . . . . 41 2.5 Boundedness of the Sobolev norm of U (t, s)ψ0 . . . . . . . . . . . . . 48 2.6 Asymptotic completeness in Sobolev spaces . . . . . . . . . . . . . . . 56 2.6.1 Existence of wave operators . . . . . . . . . . . . . . . . . . . 57 2.6.2 Asymptotic completeness . . . . . . . . . . . . . . . . . . . . . 63 3 Schrödinger Operators with Lamé Potentials on R 72 3.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 3.2 Analysis of K1 (t, x, x0 ) . . . . . . . . . . . . . . . . . . . . . . . . . . 80 3.3 Analysis of K2 (t, x, x0 ) . . . . . . . . . . . . . . . . . . . . . . . . . . 85 vi 3.4 Optimality of the decay factor . . . . . . . . . . . . . . . . . . . . . 94 3.5 Appendix: Elements of Weierstrass functions . . . . . . . . . . . . . . 98 Bibliography 101 1 Chapter 1 Background The well-known free Schrödinger equation is of the form 1 ∂t ψ(t, x) = −∆ψ(t, x), i (1.0.1) where the space variable x ∈ Rn and the time variable t ∈ R. It is well-known that kψ(t, ·)kL2 (Rn ) = kψ(0, ·)kL2 (Rn ) for any t ∈ R and the solution that satisfies the initial condition ψ(0, x) = ψ0 (x) is given by the following: Z ψ(t, x) = Cn t −n 2 ei |x−y|2 4t ψ0 (y)dy. (1.0.2) kψ(t, ·)kL∞ (Rn ) ≤ Cn t− 2 kψ0 kL1 (Rn ) . (1.0.3) Rn This implies the L1 → L∞ estimate n By interpolation, we derive the dispersive estimate 1 1 kψ(t, ·)kLp (Rn ) ≤ Cn t−n( 2 − p ) kψ0 kLq (Rn ) 1 1 + = 1 2 ≤ p ≤ +∞. p q (1.0.4) The constant Cn depends on the dimension n and may vary from line to line. Many efforts have been made to obtain an analog of (1.0.4) for Schrödinger equations with a potential 2 1 ∂t ψ(t, x) = −∆ψ(t, x) + V (x)ψ(t, x), i (1.0.5) where V (x) is a time-independent and real-valued function defined on Rn . Write H = −∆ + V (x). For n ≥ 3 and small V , we employ a perturbative argument. When V is large, H may have bound states (eigenfunctions). It is well-known that bound states can arise for arbitrarily small potentials in dimensions n = 1, 2 (see Theorem XIII.11 in Reed and Simon [30]). Denote Pc (H) as the projection onto the continuous part of the spectrum of self-adjoint operator H. The estimate should take the form 1 1 1 + = 1 2 ≤ p ≤ +∞. (1.0.6) p q 1 keitH Pc (H)ψ0 kLp (Rn ) ≤ Cn,p t−n( 2 − p ) kψ0 kLq (Rn ) 1.1 Schrödinger operators in dimension n ≥ 3 In dimension n = 3, Kato [24] showed that −∆ + V is unitarily equivalent to −∆, 1 provided that kV kRoll < 4π , where Z kV k2Roll = V (x)V (y) dxdy. 2 R6 |x − y| Similar results are known for unitary equivalence for d ≥ 4. For large V in dimension n = 3, Rauch [28] and Jensen and Kato [22] obtained weighted L2 estimates, 3 kweitH wf kL2 ≤ C|t|− 2 kf kL2 , (1.1.1) where w(x) = e−ρhxi with some ρ > 0 and V exponentially decaying (Rauch) or w(x) = hxi−σ with some σ > 0 and V decaying at a power rate (Jensen, Kato). We 1 denote hxi = (1 + |x|2 ) 2 . In addition, they assume that zero energy is neither an eigenvalue nor a resonance, i.e., lim sup kw(H − (λ ± i0))−1 wkL2 →L2 < ∞. λ→0 (1.1.2) 3 Journeé, Soffer and Sogge [23] proved (1.0.6) for dimension n ≥ 3 under suitable power decay and regularity condition on V and under the assumption (1.1.2). Specifically, in dimension n = 3, they assume that |V | < Chxi−7−² and V̂ ∈ L1 (R3 ). These requirements were later relaxed by Yajima [40],[41] and [42]. Yajima also proved the Lp boundedness of the wave operators in 1 ≤ p ≤ ∞. A different approach introduced by Rodnianski, Schlag [29] and by Goldberg, Schlag [16], leads to even weaker conditions on V . Finally, Goldberg [15] obtains (1.0.6) in dimension n = 3 under the assumption that (1.1.2) and |V (x)| ≤ Chxi−2−² . For more details and references, see Schlag’s survey [33]. For a general time-dependent potential V (t, x), it is not clear how to introduce an analog of bound states and the spectral projection. In the case of small potentials, Rodnianski, Schlag treat the potential as a perturbation [29]. The case of large time-dependent potentials is still under investigation. In this thesis, we will consider one class of time-dependent potential, namely, the charge transfer model. The charge transfer model has been devised to describe the motion of a light particle in a collision between heavy ones. In this model, only the light particle is subject to quantum dynamics, while the heavy ones follow assigned classical trajectories, which are asymptotically inertial (see Graf [17]). This leads to the Hamiltonian m X 1 H(t) = − 4 + Vj (x − ~vj t). 2 j=1 (1.1.3) The charge transfer model has been extensively studied in the literature in connection with the question of asymptotic completeness, which is first established by Yajima [43]. Later, Graf [17] obtains a new proof containing a crucial argument showing that the energy associated with a solution of (1.1.4) remains bounded in time. Similar results are also obtained by Wüller [38] and Zielinski [44]. The charge transfer model also arises in the study of asymptotic stability of noninteracting multi-soliton states. This refers to solving an NLS 1 i∂t ψ + 4ψ + β(|ψ|2 )ψ = 0 2 4 in Rn , n ≥ 3 with initial data ψ0 = Pm j=1 wj (0, ·) + R0 , where wj are special standing wave solutions called solitons and R0 is a small perturbation. In [32], Rodnianski, Schlag and Soffer show that if the solitons are sufficiently separated at time t = 0, and if R0 is sufficiently small in a suitable norm, then the solution ψ evolves like a sum of solitons with time-dependent parameters approaching a limit, plus a radiation term that goes to zero in L∞ (R3 ). This leads to the problem of establishing dispersive estimates for the linear problem, which is closely related to the charge transfer model (see [31] and [32]): m X 1 1 ∂t ψ = − 4ψ + Vj (· − ~vj t)ψ. i 2 j=1 (1.1.4) In [31] a weak version of the dispersive estimate for (1.1.4) is derived. Motivated by their approach, we will obtain a dispersive estimate for (1.1.4) in Chapter 2. 0 Furthermore, we derive a W κ,p → W κ,p estimate for the solution of (1.1.4). As an application, we will obtain the asymptotic completeness in H κ (Rn ). 1.2 Dispersive estimates for Schrödinger operators in dimension n = 1, 2 In dimension n = 1, 2, even a small potential may produce bound states, thus we cannot treat the potential as a perturbation. The dispersive estimates (1.0.6) in dimension n = 1 with a spatially decaying potential were first obtained by Weder [37] and Artbazar, Yajima [3]. These authors express the resolvent via the Jost solutions, namely, for =z > 0 (−∂x2 + V − z 2 )−1 (x, y) = f+ (x, z)f− (y, z) , x>y W (z) and symmetrically if x < y. Here f± are the Jost solutions defined as solutions of −f±00 (·, z) + V f± (·, z) = z 2 f± (·, z) 5 with the asymptotics f+ (x, z) ∼ eixz f− (x, z) ∼ e−ixz as x → ∞ as x → −∞, and W (z) is the Wronskian of f+ (·, z), f− (·, z). In this case, we say zero energy is a resonance if and only if W (0) = 0. Note that the free case V = 0 has a resonance at zero energy. Later, Goldberg and Schlag [16] proved (1.0.6) with a slightly less restrictive condition on V . They assumed hxiV (x) ∈ L1 (R) and the zero energy is not a resonance. Note that in term of pointwise decay, this is in agreement with the hxi−2 threshold. If zero is a resonance, they proved the same estimate assuming that hxi2 V (x) ∈ L1 (R). The dispersive estimates in dimension n = 2 were established by Schlag [34]. One crucial ingredient of the arguments is an asymptotic expansion of the resolvent around zero energy from Jensen and Nenciu [21]. In odd dimensions, the free resolvent (−∆ − z 2 )−1 is analytic for all z 6= 0. When =z > 0, (−∆ + V − z 2 )−1 = z −2 A−2 + z −1 A−1 + A0 + zA1 + O(z 2 ) as z → 0 (1.2.1) where the O-term is understood in the operator norm on a suitable weighted L2 space. The free resolvent (−∆ − z 2 )−1 is analytic for all z 6= 0. In even dimensions the Riemann surface of the free resolvent is that of the logarithm. Thus, (3.3.11) needs to include powers of log z in even dimensions. In Chapter 3, we will study the dispersive property of Schrödinger operators with a periodic potentials on the real line. When V (x) is periodic real function defined on R1 , the spectrum of the Schrödinger operator −(d2 /dx2 ) + V (x) acting on L2 (R1 ) is a union of intervals carrying a purely absolutely continuous spectrum. The absence of a point spectrum and a singular spectrum suggests the dispersion of the solution. 6 We will illustrate this dispersive phenomenon for a special analytic potential, whose spectrum is a union of two intervals (bands); namely, all gaps but one are degenerate. It is known that a one-gap potential V (x) must be an elliptic function ([19]). With such a potential, we have y 00 (x) − 2℘(x + ω3 )y(x) = −Ey(x), (1.2.2) where ℘(z) is the Weierstrass elliptic function with periods 2ω1 , 2ω3 , satisfying the following differential equations: ℘0 (a)2 = 4℘3 (a) − g2 ℘(a) − g3 , g2 ℘00 (a) = 6℘2 (a) − , 2 (1.2.3) (1.2.4) where g2 , g3 are the invariants of ℘(z) defined by (3.5.1). Known as Lamé’s equation, (1.2.2) arises from the theory of the potential of an ellipsoid ([39],[11]). We assume ω1 = ω > 0, ω3 = iω 0 and ω 0 > 0 to guarantee that ℘(x + ω3 ) is real-valued for x ∈ R. If we choose the potential in (1.2.2) to be n(n + 1)℘, instead of 2℘ (n is any positive integer), then the spectrum of Lamé’s equation consists of n + 1 bands ([27]). In Chapter 3, we give a dispersive estimate similar to (1.0.4) for the following equation: 1 d2 ∂t ψ(x, t) = − 2 ψ(x, t) + 2℘(x + ω3 )ψ(x, t), i dx ψ(x, 0) = ψ0 (x). 7 Chapter 2 Charge Transfer Models In Section 2.1, we give our main results about the charge transfer models, and some necessary definitions. In Section 2.2, we present some lemma that are useful for proving our results. In Section 2.3, we prove the dispersive estimate for the charge transfer models. In Section 2.4, the dispersive estimates for the derivatives of the evolutions of the charge transfer models are proved by further investigating the argument in Section 2.3. In Section 2.5, we prove that the Sobolev norms of the evolution of charge transfer models remain bounded. In Section 2.6, the existence of the wave operator and the asymptotic completeness for the charge transfer models are established. 2.1 Definitions and main results Definition 2.1.1. By a charge transfer model we mean a Schrödinger equation 1 ∂ ψ − 21 4ψ + i t Pm vκ t)ψ = 0 κ=1 Vκ (x − ~ (2.1.1) ψ|t=0 = ψ0 , x ∈ Rn , where ~vκ are distinct vectors in Rn , n ≥ 3, and the real potentials Vκ are such that for every 1 ≤ κ ≤ m, 1. Vκ is time independent and has compact support (or fast decay), Vκ , ∇Vκ ∈ L∞ ; 8 2. 0 is neither an eigenvalue nor a resonance of the operators 1 Hκ = − 4 + Vκ (x). 2 Recall that ψ is a resonance if it is a distributional solution of the equation Hκ ψ = 0, which belongs to the space L2 (hxi−σ dx) for any σ > 21 , but not for σ = 0. The conditions in the above definition are always assumed when we prove and apply the dispersive estimates, i.e., Theorem 2.1.3 and Theorem 2.1.4. The conditions are not optimal, but for convenience. This definition is standard (see [17], [42]). The Schrödinger group e−itHκ is known to satisfy the decay estimates (see Journé, Soffer, Sogge [23], and Yajima [41]) n ke−itHκ Pc (Hκ )ψ0 kL∞ . |t|− 2 kψ0 kL1 (2.1.2) for n ≥ 3 under various conditions on the potential. Here Pc (Hκ ) is the spectral projection onto the continuous spectrum of Hκ and . denotes bounds involving multiplicative constants independent of ψ0 and t. For n = 3, [16] proved (2.1.2) under the assumption that |Vκ (x)| ≤ C(1 + |x|)−β , for some β > 3. For n > 3, (2.1.2) holds ([23]) under the additional assumption: F(Vκ ) ∈ L1 . Yajima [41] proved (2.1.2) with slightly weaker conditions than [23]. In this chapter we shall assume that F(Vκ ) ∈ L1 to guarantee the estimate (2.1.2), except in Section 2.5. To establish similar dispersive estimates for time-dependent Schrödinger equations is more involved. Heuristically, we can’t project away the bounded states as they are moving in different directions. Rodnianski and Schlag [29] proved dispersive estimates for small time-dependent potentials. In this paper, we will focus on the charge transfer model. An indispensable tool in the study of the charge transfer model is the Galilean transforms |~ v |2 g~v,y (t) = e−i 2 t e−ix·~v ei(y+t~v)·~p , (2.1.3) ~ Under g~v,y (t), the Schrödinger equation transforms as cf. [17], where p~ = −i∇. 9 follows: 4 4 g~v,y (t)eit 2 = eit 2 g~v,y (0) (2.1.4) and moreover, with H = − 12 4 + V , ψ(t) := g~v,y (t)−1 e−itH g~v,y (0)φ0 , , g~v,y (t)−1 = e−iy·~v g−~v,−y (t) (2.1.5) solves 1 ∂ ψ − 12 4ψ + V (· − t~v − y)ψ = 0 i t (2.1.6) ψ|t=0 = φ0 . Since in our case y = 0 always, we set g~v (t) := g~v,0 (t). Note that the transformations g~v,y (t) are isometries on all Lp spaces and ge~1 (t)−1 = g−e~1 (t) because of (2.1.5). In the following, we shall assume that the number of potentials is m = 2 and that the velocities are ~v1 = 0, ~v2 = (1, 0, . . . , 0) = ~e1 . The arguments generalize easily to m ≥ 3. We now introduce the appropriate analog to project away bounded states for the problem 1 1 ∂t ψ − 4ψ + V1 ψ + V2 (· − te~1 )ψ = 0 i 2 ψ|t=0 = ψ0 (2.1.7) with compactly supported potentials V1 , V2 . Let u1 , . . . , um and w1 , . . . , w` be the normalized bound states of H1 and H2 corresponding to the negative eigenvalues λ1 , . . . , λm and µ1 , . . . , µ` , respectively (recall that we are assuming that 0 is not an eigenvalue). We denote by Pb (H1 ) and Pb (H2 ) the corresponding projections onto the bound states of H1 and H2 , respectively, and let Pc (Hκ ) = Id − Pb (Hκ ), κ = 1, 2. 10 The projections Pb (H1,2 ) have the form m X Pb (H1 ) = h·, ui iui , Pb (H2 ) = i=1 X̀ h·, wj iwj . j=1 We introduce the following orthogonality condition in the context of the charge transfer Hamiltonian: Definition 2.1.2. Let U (t)ψ0 = ψ(t, x) be the solutions of (2.1.7). We say that ψ0 (or also ψ(t, ·)) is asymptotically orthogonal to the bound states of H1 and H2 if kPb (H1 )U (t)ψ0 kL2 + kPb (H2 , t)U (t)ψ0 kL2 → 0 as t → +∞. (2.1.8) Here Pb (H2 , t) := g−e~ 1 (t)Pb (H2 ) ge~1 (t) (2.1.9) for all times t. Remark 2.1.1. From Corollary 2.2.3, kU (t)ψ0 kLp ≤ Ct kψ0 kLp0 , we know that U (t)ψ0 ∈ 0 Lp is well-defined for ψ0 ∈ Lp . As the bound states ui , wj are exponentially decaying 0 at infinity, Definition 2.1.2 makes sense for any initial data ψ0 ∈ Lp for p0 ∈ [1, 2]. Remark 2.1.2. Clearly, Pb (H2 , t) is again an orthogonal projection for every t. It gives the projection onto the bound states of H2 that have been translated to the position of the potential V2 (· − te~1 ). Equivalently, one can think of it as translating the solution of (2.1.7) from that position to the origin, projecting onto the bound states of H2 , and then translating back. Remark 2.1.3. From Proposition 3.1 of [31], the decay rate of (2.1.8) is actually exponential. More precisely, the following holds: kPb (H1 )U (t)ψ0 kL2 + kPb (H2 , t)U (t)ψ0 kL2 . e−αt kψ0 kL2 , for some α > 0. (2.1.10) 11 Remark 2.1.4. It is clear that all ψ0 that satisfy (2.1.8) form a closed subspace. This subspace coincides with the space of scattering states for the charge transfer problem. The latter is well-defined by Graf ’s asymptotic completeness result [17]. We can only expect the dispersive estimate for (2.1.7) for the initial data satisfying Definition 2.1.2, just as we have to project away the bound states for (2.1.2). Rodnianski, Schlag, Soffer [31] established the following estimate: n kU (t)ψ0 kL2 +L∞ . hti− 2 kψ0 kL1 ∩L2 (2.1.11) with initial data ψ0 ∈ L1 ∩ L2 satisfying (2.1.8), where U (t) is the evolution of the 1 charge transfer model and hti = (1+t2 ) 2 . By definition, kf kL2 +L∞ := inf f =h+g (khkL2 + kgkL∞ ) and kf kL1 ∩L2 = kf kL1 + kf kL2 . (2.1.11) has an important application to the asymptotic stability and asymptotic completeness for the small perturbation of noncolliding solitons for NLS ([32]). [31] decomposes the evolution into different channels according to each potential. Every channel splits into a large velocity part and a low velocity part. For the large velocity part, Kato’s smoothing estimate was employed; for the low velocity part, a propagation estimate was used. In this paper, we will combine the methods from [23] and [31] and obtain the following: Theorem 2.1.3. Consider the charge transfer model as in Definition 2.1.1 with two potentials, cf. (2.1.7). Assume Vb1 , Vb2 ∈ L1 (Rn ). Let U (t) denote the propagator of (2.1.7). Then for any initial data ψ0 ∈ L1 , which is asymptotically orthogonal to the bound states of H1 and H2 in the sense of Definition 2.1.2, one has the decay estimates n kU (t)ψ0 kL∞ . |t|− 2 kψ0 kL1 . (2.1.12) An analogous statement holds for any number of potentials, i.e., with arbitrary m in (2.1.1). Inspection of the argument in the following sections shows that it applies, say, to exponentially decaying potentials. But sufficiently fast power decay at infinity is also 12 allowed. We shall prove (3.3.1) by means of a bootstrap argument. More precisely, we prove that the bootstrap assumption n kU (t)ψ0 kL∞ ≤ C0 |t|− 2 kψ0 kL1 for all 0 ≤ t ≤ T (2.1.13) C0 − n |t| 2 kψ0 kL1 for all 0 ≤ t ≤ T. 2 (2.1.14) implies that kU (t)ψ0 kL∞ ≤ Here T is any given fixed large constant and (2.1.13) holds for C0 some sufficiently large constant because of Corollary 2.2.3. C0 may depend on T in the beginning. The above implication (2.1.13) =⇒ (2.1.14) holds as long as C0 is larger than some universal constant independent of the time T . Thus iterating (2.1.13) =⇒ (2.1.14) then yields a constant that does not depend on T . The theorem follows by letting T → +∞. As the L2 norm of U (t)ψ0 remains constant, by interpolation, the following holds: 1 1 kU (t)ψ0 kLp . Cp |t|−n( 2 − p ) kψ0 kLp0 1 1 + 0 = 1. p p p ≥ 2, (2.1.15) Our next theorem is about the decay estimates of ∂ α U (t)ψ0 , where α = (α1 , · · · , αn ) α is an n-tuple of nonnegative integers and ∂ α = ∂xα1∂···∂xαn . We write |α| = α1 +· · ·+αn . 1 n Theorem 2.1.4. Let U (t) denote the propagator of the equation (2.1.7). Assume (2.1.2) holds for H1 and H2 . Let Vj ∈ C0κ+1 where κ is a positive integer and j = 1, 2. β V ∈ L1 (Rn ). Then for any Moreover, assume that for ∀|β| ≤ κ and j = 1, 2, ∂d j 0 initial data ψ0 ∈ W κ,p , which is asymptotically orthogonal to the bound states of Hj (j = 1, 2) in the sense of Definition 2.1.2, one has the decay estimates 1 1 kU (t)ψ0 kW κ,p . |t|−n( 2 − p ) kψ0 kW κ,p0 , (2.1.16) where 2 ≤ p < ∞ and p1 + p10 = 1. 2n Remark 2.1.5. It suffices to prove Theorem 2.1.4 for p > n−2 , because interpolating with Theorem 2.1.5, which holds under the assumption of Theorem 2.1.4, we derive 13 2n Theorem 2.1.4 for any p ∈ [2, +∞]. p > n−2 guarantees that R∞ 1 1 1 |t|−n( 2 − p ) < ∞. We need to exclude the case p = ∞, since part of our proof relies on singular integrals and we do not know whether or not (2.1.16) holds for p = ∞. The second part of this paper is motivated by Graf [17]. Graf proved energy boundedness for U (t, s) by a geometric method, where U (t, s) is the solution operator corresponding to the time-dependent Schrödinger equation m X 1 4 ∂t ψ − ψ + Vj (x − ~vj t)ψ = 0, i 2 j=1 (2.1.17) ψ|t=s = ψ0 , i.e., ψ(t, ·) = U (t, s)ψ0 . Graf proved that kU (t, s)ψ0 kH 1 is bounded as t → ∞ provided that the initial data ψ0 ∈ H 1 (Rn ), n ≥ 1. For the higher degree Sobolev norms, J. Bourgain [6] proved the following for the general time-dependent Hamiltonian H(t): Suppose the time-dependent potential V (x, t) is bounded, real, and supt∈R |V (x, t)| is compactly supported. Moreover, for any n-tuple α sup kDxα V (t)k∞ < Cα . t∈R Then for ∀² > 0 and κ > 0, kU (t, 0)ψ0 kH κ ≤ C²,κ |t|² kψ0 kH κ for all t. (2.1.18) An example ([6]) is given to show that we cannot remove the |t|² growth for general time-dependent potentials. In this paper, it is shown that (2.1.18) does hold with ² = 0 for the case of the charge transfer Hamiltonian. More precisely, in Section 4, the time-boundedness of kU (t, s)ψ0 kH κ for charge transfer models is established by the same geometric method as in [17] for any real number κ. We write dxe as the least integer no less than x. The precise statement is as follows: Theorem 2.1.5. Let U (t, s) be the evolution operator for (2.1.17), and let κ ∈ R 14 d|κ|e and the dimension n ≥ 1. Furthermore, suppose Vj ∈ C0 (Rn ), (j = 1, 2, · · · , m), i.e., Vj has derivatives up to degree d|κ|e, which are all continuous and compactly supported. Then for ∀t, s ∈ R kU (t, s)ψ0 kH κ ≤ Cκ kψ0 kH κ , where Cκ depends on κ and the potentials Vj . Remark 2.1.6. By interpolation, it clearly suffices to consider the case where κ is an integer. By duality, it suffices to prove the case where κ is a positive integer. Indeed, assuming κ < 0, due to the fact that U (t, s) is unitary on L2 (Rn ), we have kU (t, s)ψ0 kH κ = sup hU (t, s)ψ0 , φiL2 kφkH −κ =1 = sup hψ0 , U (s, t)φiL2 ≤ C−κ kψ0 kH κ . kφkH −κ =1 No assumption is made on the spectra of the subsystems Hj . The assumption of compact support of Vj is for convenience only and the proof works for sufficiently fast polynomial decay at infinity without essential change ([17]). Suppose all assumptions of both Theorem 2.1.4 and Theorem 2.1.5 hold; then by interpolation, the estimate (2.1.16) holds for 2 ≤ p < ∞. Remark 2.1.7. It follows from Duhamel’s formula and Gronwall’s inequality, that kU (t, s)ψ0 kH κ ≤ C(I)kψ0 kH κ t, s ∈ I, (2.1.19) for any compact interval I. Therefore, it suffices to prove Theorem 2.1.5 when |t| or |s| is large. As an important consequence, we apply Theorem 2.1.4 and Theorem 2.1.5 to obtain the following asymptotic completeness for the charge transfer model in the H κ sense: Theorem 2.1.6. Let u1 , . . . , um and w1 , . . . , w` be the eigenfunctions of H1 = − 42 + V1 (x) and H2 = − 42 + V2 (x), respectively, corresponding to the negative eigenvalues 15 λ1 , · · · , λm and µ1 , · · · , µ` . Assume that Vj ∈ C0n+2κ+2 (Rn ), (n ≥ 3, j = 1, 2), and that 0 is neither an eigenvalue nor a resonance of H1 , H2 , where κ is a nonnegative integer. Then for any initial data ψ0 ∈ H κ , the solution U (t)ψ0 of the charge transfer problem (2.1.7) can be written in the form U (t)ψ0 = m X Ar e−iλr t ur + X̀ r=1 4 Bk e−iµk t g−~e1 (t)wk + e−it 2 φ0 + R(t), k=1 for some choice of the constants Ar , Bk and the function φ0 ∈ H κ . The remainder term R(t) satisfies the estimate kR(t)kH κ −→ 0, as t → ∞. Remark 2.1.8. The above theorem holds for m potentials. We are not aiming to give the optimal regularity condition on the potentials. The theorem is equivalent to claiming that H κ (Rn ) is the sum of the ranges of the wave operators Ω− l , (l = 0, 1, 2) defined in Section 6.1. [17] proved that the ranges of the wave operators are orthogonal κ to each other in the L2 sense. Therefore, H κ (Rn ) again is a direct sum of Ω− l (H ). 2.2 Preparation In this section, we present some lemma that are indispensable for the proof of the above theorems. The first ingredient of our proof is the notion of cancellation. In this section, U (t) will denote the evolution operator of (2.1.7) or (2.1.1). It is clear from their proofs that the following lemmas also hold for the general time-dependent Hamiltonian H0 + V (t). Lemma 2.2.1. sup −∞<t<∞ keit∆ V e−it∆ kp→p ≤ kVb k1 , (2.2.1) where p ∈ [1, ∞] and k · kp→p means the operator norm from Lp to Lp . For the proof of the lemma, notice that equation (2.1.4) implies [eit∆ eiζx e−it∆ f ](x) = 2 g−ζ (2t)f (x) = e−it|ζ| eixζ f (x − 2ζt). 16 Lemma 2.2.2. Suppose t, s ∈ R, then we have the following: n sup ke−i(t−s)H0 V (r)U (s)k1→∞ < |t|− 2 CM eM |s| , (2.2.2) r∈R [ where M = maxr∈R kV (r)k1 < ∞. Proof. Let’s write Ψ(t, s) := supr∈R ke−i(t−s)H0 V (r)U (s)k1→∞ . Without loss of generality, we suppose that s > 0. By Duhamel’s formula, −i(t−s)H0 e V (r)U (s) = e −i(t−s)H0 −isH0 V (r){e Z s −i e−i(s−τ )H0 V (τ )U (τ )dτ }, 0 it follows that ke−i(t−s)H0 V (r)U (s)k1→∞ V (r)e −isH0 Z s k1→∞ + ke−i(t−s)H0 V (r)e−i(s−τ )H0 V (τ )U (τ )k1→∞ dτ 0 Z s −n b b 2 ke−i(t−τ )H0 V (τ )U (τ )k1→∞ dτ ≤ CkV (r)k1 |t| + kV (r)k1 0 Z s n ≤ CM |t|− 2 + M Ψ(t, τ )dτ. ≤ ke −i(t−s)H0 0 n Taking the supremum over r, we get Ψ(t, s) ≤ CM |t|− 2 + M Rs 0 Ψ(t, τ )dτ . By Gron- wall’s inequality, n Ψ(t, s) ≤ CM |t|− 2 eM s . Note that the lemma still holds with other constants C and M on the right-hand side if we replace V (r) with Vj (r) or replace U (s) with another evolution, say e−isHj . Another observation is that the lemma can be generalized to the following by the same proof: sup ke−i(t−s)H0 V (r)U (s)ψ0 kp . |t|−γ M eM s kψ0 kp0 r∈R (2.2.3) 17 where γ = n( 12 − p1 ) and 2 ≤ p ≤ ∞, p1 + p10 = 1. This will be useful in Section 4. Corollary 2.2.3. Suppose U (t) is the evolution operator of (2.1.7) or (2.1.1). Assume t > 0, then 1 1 kU (t)kp0 →p . t−n( 2 − p ) eM t 1 1 + 0 = 1, 2 ≤ p ≤ ∞ p p (2.2.4) Rt Proof. By Duhamel’s formula, U (t) = e−itH0 − i 0 e−i(t−τ )H0 V (τ )U (τ )dτ . Write γ = n( 21 − p1 ), then by Lemma 2.2.2, we have kU (t)kp0 →p ≤ Ct −γ Z t + Ψ(t, τ )dτ ≤ Ct −γ Z t + 0 Ct−γ M eM τ dτ ≤ Ct−γ eM t . 0 From the corollary, the bootstrap assumption (2.1.13) holds for any time T if we take C0 = CeM T . Lemma 2.2.4. Suppose m ≥ 1 and ² > 0. If u1 , u2 , . . . , um are either all positive or P all negative, satisfying | m j=1 uj | > ², then there exists a constant C = C(m, ²) such that k m−1 Y (e iuj H0 ium H0 V (sj ))e k1→∞ ≤ CM m−1 j=1 k m Y n huj i− 2 (2.2.5) j=1 m−1 Y m Y j=1 j=1 (eiuj H0 V (sj ))U (um )k1→∞ ≤ CM m−1 n huj i− 2 eM um (2.2.6) where sj is any real number and M = sups∈R (kV (s)k1 + kVb (s)k1 ). Proof. The first inequality is from [23]. Assume that u1 , u2 , . . . , um are all positive without loss of generality. We apply the dispersive estimate for eiuj H0 repeatedly Q −n 2 u and the left-hand side is dominated by CM m−1 m j=1 j , which is dominated by the right-hand side up to a constant, provided each uj > ². If some uj ≤ ², it is inefficient 18 to use a dispersive estimate for eiuj H0 . Instead, we apply the cancellation lemma 2.2.1 and obtain Z e iuj H0 V (sj )e iuj+1 H0 = \ (sj )(ζ) dζei(uj+1 +uj )H0 , eiuj H0 eixζ e−iuj H0 V where eiuj H0 eixζ e−iuj H0 is the Galilean transform g−ζ (−uj ) according to (2.1.4). If again uj + uj+1 < ², we can repeat this procedure until uj−l + · · · + uj + · · · + uj+k > ² P which always happens because | m j=1 uj | > ². Then we apply the dispersive estimate to obtain the inequality. We sketch the proof of the second equation. When m = 1, it is just (2.2.4) provided that um > ². When m = 2, if u1 > 2² and u2 > 2² , n k(eiu1 H0 V (s1 ))U (u2 )k1→∞ . |u1 |− 2 kV (s1 )U (u2 )k1→1 n . |u1 |− 2 kU (u2 )k1→∞ n n n n . |u1 |− 2 |u2 |− 2 eM u2 . hu1 i− 2 hu2 i− 2 eM u2 If u1 ≤ 2² or u2 ≤ 2² , we apply Lemma 2.2.2 n n n keiu1 H0 V (s1 )U (u2 )k1→∞ . (|u1 | + |u2 |)− 2 eM u2 . hu1 i− 2 hu2 i− 2 eM u2 The case where m > 2 follows exactly as the first inequality using Lemma 2.2.1. Next we list a variant of Kato’s 12 -smoothing estimate. It appears in [31]. We give the proof for completeness. Lemma 2.2.5. Let H = − 21 4 + V , kV k∞ < ∞. Then for all T, R, M ≥ 1, Z T sup BR 0 kF (|~p| ≥ M )e−itH f kL2 (BR ) dt ≤ C(n, V ) TR 1 M2 kf kL2 . (2.2.7) Here the supremum ranges over all balls BR of radius R ≥ 1 and C(n, V ) is a constant that depends only on kV k∞ and the dimension n. 19 Proof. We first prove the following estimate, which will then imply the lemma. Let ψ(t) = e−itH ψ0 with H = − 12 4 + V , kV k∞ < ∞. Then for all T > 0 and 0 < α, Z TZ 1 |∇h∇i− 2 ψ(x, t)|2 sup x0 ∈Rn 0 Rn 1 (1 + |x − x0 |α ) α +1 dxdt ≤ Cα,n T (1 + kV k∞ )kψ0 k22 . 1 1 (2.2.8) 1 The multiplier ∇h∇i− 2 corresponds to the symbol ξhξi− 2 = ξ(1 + |ξ|2 )− 4 . It suffices to prove this with x0 = 0 fixed. The proof is based on taking the commutator of H ∇ with m := w(x)x · h∇i , where 1 w(x) = (1 + |x|α )− α , α > 0. One has, with ψ = ψ(t) for simplicity, d hmψ, ψi = −ih[m, H]ψ, ψi dt Z T h−[m, H]ψ(t), ψ(t)i dt = ihmψ(0), ψ(0)i − ihmψ(T ), ψ(T )i 0 h−[m, H]ψ, ψi D E 1D E ∂j ∂j = −h[m, V ]ψ, ψi + ∂` (w(x) xj ) ψ, ∂` ψ + 4(w(x) xj ) ψ, ψ (2.2.9) h∇i 2 h∇i Z Z 1 1 1 = (2.2.10) w|∇h∇i− 2 ψ|2 dx + (∂` w)(x)xj ∂j h∇i− 2 ψ ∂` h∇i− 2 ψ̄ dx Rn Rn E D E D 1 1 1 1 ∇ ∂j ψ, ∇h∇i− 2 ψ − [h∇i 2 , (∂` w)(x)xj ] ψ, ∂` h∇i− 2 ψ (2.2.11) − [h∇i 2 , w] h∇i h∇i D E 1 ∂j + 4(w(x) xj ) ψ, ψ − h[m, V ]ψ, ψi. (2.2.12) 2 h∇i One now checks easily that the two terms in (2.2.10) satisfy Z Z w(x)|h∇i Z ≥ Rn − 12 1 2 1 (∂` w)(x)xj h∇i− 2 ∂j ψ(t, x) h∇i− 2 ∂` ψ̄(t, x) dx ∇ψ(t, x)| dx + Rn 1 w(x) |h∇i− 2 ∇ψ(t, x)|2 dx = α Rn 1 + |x| Z 1 |h∇i− 2 ∇ψ(t, x)|2 Rn 1 (1 + |x|α ) α +1 dx. (2.2.13) Notice that (2.2.13) is precisely the space integral in the desired lower bound from (2.2.8). 20 1 There are several ways to bound the commutators A1 := [h∇i 2 , w] and A2 := 1 [h∇i 2 , (∂` w)(x)xj ]. For example, one can use the Kato square root formula as in [23]. Alternatively, one can invoke the standard composition formula from ΨDO calculus. 1 This gives A1 = T{h∇i 12 ,w} +Ra1 where {h∇i 2 , w} is the Poisson bracket of the symbols 1 h∇i 2 and w. Moreover, T{h∇i 12 ,w} is an associated ΨDO operator and Ra1 another 3 ΨDO operator with symbol a1 ∈ S − 2 . A similar expression holds for A2 . One checks 1 that |{h∇i 2 , w}(ξ, x)| . |∇w(x)| for all x, ξ. Therefore, the two terms in (2.2.11) both satisfy ¯­ ° h∇i− 12 ∇ψ(t) ° E¯ 1 ∇ ¯ ¯ ° ° − 12 ψ(t), ∇h∇i ψ(t) ¯ . kψ(t)k2 ° ¯ Ai ° + kψ(t)k2 kRai ∇h∇i− 2 ψ(t)k2 1 +1 h∇i 2 (1 + |x|α ) α 1 ° 1° ° h∇i− 2 ∇ψ(t) °2 (2.2.14) ≤ Ckψ(t)k22 + ° °. 4 (1 + |x|α ) α1 +1 2 Finally, the two terms in (2.2.12) are bounded by (1 + kV k∞ )kψ(t)k22 ≤ (1 + kV k∞ )kψ0 k22 . (2.2.15) ∂ j In the above estimate we have used the boundedness of the multipliers m and 4(w(x) xj ) h∇i on L2 . Integrating (2.2.13), (2.2.14), and (2.2.15) in time, inserting the resulting bounds into (2.2.9), and finally using |hmψ(0), ψ(0)i| + |hmψ(T ), ψ(T )i| ≤ 2kψ0 k22 , one obtains (2.2.8). To pass from (2.2.8) to (2.2.7), let χR be a smooth cutoff to the ball BR , so that χ cR has compact support in a ball of size ∼ R−1 . Then, by (2.2.8) 21 with α = 1 Z T −itH kF (|~p| ≥ M )e 0 ψ0 k2L2 (BR ) dt ≤ Z T Z T 0 kχR F (|~p| ≥ M )e−itH ψ0 k2L2 dt Z T kF (|~p| ≥ M )χR e−itH ψ0 k2L2 dt + k[χR , F (|~p| ≥ M )]k22→2 ke−itH ψ0 k2L2 dt 0 0 Z T 1 k∇h∇i− 2 F (|~p| ≥ M )χR e−itH ψ0 k2L2 dt + T (M R)−2 kψ0 k2L2 . M −1 0 Z T 1 . M −1 kF (|~p| ≥ M )χR ∇h∇i− 2 e−itH ψ0 k2L2 dt 0 Z T 1 + M −1 k[∇h∇i− 2 , χR ]k22→2 ke−itH ψ0 k2L2 + T (M R)−2 kψ0 k22 . 0 Z TZ 1 |∇h∇i− 2 e−itH ψ0 |2 . M R dx dt + T M −1 R−1 kψ0 k2L2 2 (1 + |x|) 0 Rn −1 2 ≤ C(n, V ) T M R kψ0 k2L2 . −1 2 The lemma follows. 2.3 Proof of the decay estimates Theorem 2.1.3 will be proven in this section by a bootstrap argument. By Corollary 2.2.3, we can assume that t is large enough in Theorem 2.1.3. More precisely, t will be bigger than any constant appearing in our estimate, except the bootstrap constant C0 in (2.3.1). By assumption, H1 , H2 can only admit finitely many negative eigenvalues. Let α > 0 satisfy: −α is bigger than any eigenvalue of H1 , H2 . For technical reasons, we will assume that the initial data ψ belong to L1 ∩ L2 and employ the following bootstrap argument: Specifically, we will show that n αT kU (t)ψ0 kL∞ ≤ C0 |t|− 2 (kψ0 kL1 + e− 2 kψ0 kL2 ) for all 0 ≤ t ≤ T, (2.3.1) implies that kU (t)ψ0 kL∞ ≤ αT C0 − n |t| 2 (kψ0 kL1 + e− 2 kψ0 kL2 ) for all 0 ≤ t ≤ T, 2 (2.3.2) 22 provided that C20 remains larger than some constant that does not depend on T . The logic here is that for arbitrary but fixed T , the assumption (2.3.1) can be made to hold for some C0 depending on T , because of Corollary 2.2.3. Iterating the implication (2.3.1) =⇒ (2.3.2) then yields a constant that does not depend on T . So we can let T → +∞ to eliminate kψ0 kL2 on the right-hand side. Since L1 ∩ L2 is dense in L1 and U (t) is a linear operator, we get the dispersive estimate (3.3.1) for any initial αT data ψ0 ∈ L1 . To simplify the notation, we write kψ0 kL1 + e− 2 kψ0 kL2 as |||ψ0 |||(T ) or |||ψ0 |||. We proceed by expanding U (t) via Duhamel’s formula with respect to the free evolution H0 : Z t e−i(t−s)H0 V (s)U (s)ψ0 ds φ0 − i 0 Z t = e−itH0 ψ0 − i e−i(t−s)H0 V (s)e−isH0 ψ0 ds Z tZ s 0 − e−i(t−s)H0 V (s)e−i(s−τ )H0 V (τ )U (τ )ψ0 dτ ds. −itH0 U (t)φ0 = e 0 (2.3.3) (2.3.4) 0 n Note that ke−itH0 ψ0 k∞ . |t|− 2 kψ0 k1 . For the second term in (2.3.4), we divide the integration interval (0, t) into three pieces and handle them by means of the cancellation lemma. Firstly, Z 1 k n n e−i(t−s)H0 V (s)e−isH0 ψ0 dsk∞ . |t|− 2 sup keisH0 V (s)e−isH0 k1→1 kψ0 k1 . |t|− 2 kψ0 k1 . s 0 Similarly, we have Z t k n n e−i(t−s)H0 V (s)e−isH0 ψ0 dsk∞ . |t|− 2 sup keisH0 V (s)e−isH0 k1→1 kψ0 k1 . |t|− 2 kψ0 k1 . t−1 The third piece is s 23 Z t−1 k e −i(t−s)H0 −isH0 V (s)e Z t−1 ψ0 dsk∞ . 1 n n n |t−s|− 2 sup kV (s)k1 |s|− 2 dskψ0 k1 . |t|− 2 kψ0 k1 , s 1 where we observed that Z t−1 n n n |t − s|− 2 |s|− 2 ds . t− 2 given n ≥ 3. (2.3.5) 1 The third term in (2.3.4) is Z t Z s ds 0 dτ e−i(t−s)H0 V (s)e−i(s−τ )H0 V (τ )U (τ )ψ0 . (2.3.6) 0 We will decompose the domain of integration Rt Rs ds dτ into several pieces and 0 0 treat each piece separately. We fix A > 0 as a large constant and ² > 0 as a small constant. Write min{s, A} = s ∧ A. Then Lemma 2.2.4 and (2.3.5) implies that Z s∧A Z t ds k 0 0 Z t Z s∧A . ds 0 .t dτ e−i(t−s)H0 V (s)e−i(s−τ )H0 V (τ )U (τ ) dτ dsk1→∞ n n n dτ ht − si− 2 hs − τ i− 2 hτ i− 2 eAM 0 −n 2 . By k · k1→∞ , we mean the operator norm from L1 to L∞ . However, when we apply αT the bootstrap assumption, kψ0 kL1 has to be modified to kψ0 kL1 +e− 2 kψ0 kL2 := |||ψ0 |||. An application of Lemma 2.2.1 and the bootstrap assumption show that Z t k Z s ds t−² t−² Z t . dτ e−i(t−s)H0 V (s)e−i(s−τ )H0 V (τ )U (τ ) dτ dsψ0 k∞ Z s dτ ke−i(t−s)H0 V (s)e−i(s−τ )H0 k1→∞ kV (τ )U (τ )ψ0 k1 dτ ds ds t−² t−² Z t . Z t dτ t−² τ n ds|t − τ |− 2 max kU (τ )ψ0 k∞ . τ ∈(t−²,t) If n = 3, then the above is dominated by 24 Z t 1 n dτ |t − τ |− 2 C0 t− 2 |||ψ0 ||| . . √ n ²C0 t− 2 |||ψ0 |||. t−² n 1 Taking ² small enough, the above term can be dominated by 100 C0 t− 2 |||ψ0 |||. When n n > 3, we need to expand U (t) further to remove the singularity of |t − τ |− 2 at τ = t, (see [23] Section 2 for details). The following is another piece of (2.3.6): Z t−A Z s e−i(t−s)H0 V (s)e−i(s−τ )H0 V (τ )U (τ )ψ0 dτ dsk∞ A A Z t−A Z s n n . ds dτ ht − si− 2 hs − τ i− 2 kV (τ )U (τ )ψ0 k1 A A Z t−A Z s n n n . ds dτ ht − si− 2 hs − τ i− 2 C0 τ − 2 |||ψ0 ||| A A Z t−A n n dsht − si− 2 hsi− 2 . C0 |||ψ0 ||| k A n . C0 |||ψ0 |||t− 2 κA ≤ where κA < R +∞ A n 1 C0 |||ψ0 |||t− 2 , 100 n dshsi− 2 → 0 as A → ∞. Lemma 2.2.4 and the bootstrap assump- tion are applied in turn in the above. The last line of above inequality holds provided that A is large enough. By Corollary 2.2.3, we can assume t >> A. Similarly, the following piece in (2.3.6) also requires that A is large: Z t Z s−A k t−A Z t A . e−i(t−s)H0 V (s)e−i(s−τ )H0 V (τ )U (τ )ψ0 dτ dsk∞ Z s−A ds t−A n n dτ ht − si− 2 hs − τ i− 2 kU (τ )ψ0 k∞ ≤ A n 1 C0 |||ψ0 |||t− 2 . 100 So what remains in (2.3.6) is m Z t X j=1 t−A Z s∧(t−²) ds s−A dτ e−i(t−s)H0 V (s)e−i(s−τ )H0 Vj (· − τ~vj )U (τ )ψ0 . (2.3.7) 25 For the term containing Vj in (2.3.7), U (τ ) will be expanded with respect to Hj by Duhamel’s formula. Abusing the notation, we will write V1 (· − τ~v1 ) as V1 (τ ). In the following, we only deal with the term containing V1 , which will be decomposed into two parts by U (τ ) = Pb (H1 , τ )U (τ ) + Pc (H1 , τ )U (τ ). 2.3.1 Bound states Proposition 2.3.1. Let ψ(t, x) = (U (t)ψ0 )(x) be a solution of (2.1.7), which is asymptotically orthogonal to the bound states of Hj , j = 1, 2 in the sense of Definition 2.1.2. Provided the bootstrap assumption (2.3.1), we have for any t ∈ (0, T ) αt n αT kPb (H1 , t)U (t)ψ0 k∞ . C0 e− 4 t− 2 (kψ0 kL1 + e− 2 kψ0 kL2 ), (2.3.8) where C0 is the constant in the bootstrap assumption. Proof. Let Ũ (t) := ge~1 (t)U (t) and φ(t) = Ũ (t)ψ0 . Then φ(t) solves 1 ∂ φ − 12 4φ + V (· + tv~1 )φ = 0, i t (2.3.9) φ|t=0 (x) = ( ge~1 (0)ψ0 )(x). Then kPb (H1 , t)U (t)ψ0 k∞ = kPb (H1 )Ũ (t)φ0 k∞ so without loss of generality, we can assume that v~1 is the zero vector. Suppose that the bound states of H1 are u1 , u2 , . . . , ul and we decompose U (t)ψ0 = l X ai (t)ui + ψ1 (t, x) (2.3.10) i=1 with respect to H1 so that Pc (H1 )ψ1 = ψ1 and Pb (H1 )ψ1 = 0. By the asymptotic orthogonality assumption, l X i=1 |ai (t)|2 → 0 as t → ∞. 26 Substituting (2.3.10) into (2.1.7) yields 1 1 ∂t ψ1 − 4ψ1 + V1 ψ1 + V2 (· − te~1 )ψ1 + i 2 ¸ l · X 1 1 + ȧj (t)uj − 4uj aj (t) + V1 uj aj (t) + V2 (· − te~1 )uj aj (t) = 0. i 2 j=1 (2.3.11) Since Pc (H1 )ψ1 = ψ1 , we have 1 (− 4 + V1 )ψ1 = H1 ψ1 = Pc (H1 )H1 ψ1 , 2 In particular, µ Pb (H1 ) 1 1 ∂t ψ1 − 4ψ1 + V1 ψ1 i 2 ∂t ψ1 = Pc (H1 )∂t ψ1 . ¶ = 0. Thus, taking an inner product of the equation (2.3.11) with uκ and using the fact that huκ , uj i = δjκ as well as the identity 1 − 4uj + V1 uj = λj uj , 2 we obtain the ODE m X 1 ȧκ (t) + λκ aκ (t) + hV2 (· − te~1 )ψ1 , uκ i + aj (t)hV2 (· − te~1 )uj , uκ i = 0 i j=1 for each aκ with the condition that aκ (t) → 0 as t → +∞. Recall that uκ is an eigenfunction of H1 = − 12 4 + V1 with eigenvalue λκ < 0. It is well-known (e.g., Agmon [1]) that such eigenfunctions are exponentially localized, i.e., Z e2α|x| |uκ (x)|2 dx ≤ C = C(V1 , n) < ∞ for some positive α. Rn (2.3.12) 27 Therefore, the assumption that V2 has compact support implies kV2 (· − te~1 )uκ k2 . e−αt for all t ≥ 0. (2.3.13) The implicit constant in (2.3.13) depends on the size of the support of V2 and kV2 kL∞ . By the bootstrap assumption, fκ (t) := hV2 (· − te~1 )ψ1 , uκ i satisfies |fκ (t)| . kψ1 k∞ kV2 (· − te~1 )uκ k1 . e−αt kψ1 k∞ . e−αt k(Id − Pb (H1 ))U (t)ψ0 k∞ −αt − n 2 .e t C0 (kψ0 k L1 − αT 2 +e kψ0 k ) + e L2 −αt l X |ai (t)| kui k∞ , (2.3.14) i=1 where t ∈ (0, T ). Notice that (2.3.14) fails for t > T because the bootstrap assumption only applies to 0 < t < T . Instead, we have the following for t > T : |fκ (t)| < kV2 (· − te~1 )uκ k2 kψ1 k2 . e−αt kψ0 k2 . (2.3.15) In view of (2.3.1), aκ solves the equation 1 ȧ (t) + λκ aκ (t) + i κ Pm j=1 aj (t)Cjκ (t) + fκ (t) = 0, (2.3.16) aκ (∞) = 0, where Cjκ (t) = Cκj (t) = hV2 (· − te~1 )uj , uκ i. By (2.3.13), maxj,κ |Cjκ (t)| . e−αt . Solving (2.3.16) explicitly, we obtain ~a(t) = ie R −i 0t B(s) ds Z ∞ t where Bjκ (t) = λj δjκ + Cjκ (t). Rs ei 0 B(τ ) dτ f~(s) ds, 28 Rs By (2.3.14), (2.3.15) and the unitarity of ei 0 B(τ ) dτ , we conclude that Z T |~a(t)| ≤ Z T . + t −αs − n 2 e s Z ∞ Z T C0 ds|||ψ0 ||| + |f~(s)| ds T −αs e t t l X Z ∞ |aj (s)|kui k∞ ds + e−αs dskψ0 kL2 . T j=1 Choose a large constant t0 > 0 such that for all t1 > t0 , the following holds: Z T e −αs t1 l X |aj (s)|kui k∞ ds ≤ j=1 1 sup |~a(t)|, 2 t1 <t<T (2.3.17) then −n αt1 −n sup |~a(t)| . e−αt1 t1 2 C0 |||ψ0 ||| + e−αT kψ0 kL2 . e− 4 t1 2 C0 |||ψ0 |||. t1 <t<T Remark 2.3.1. In the above proof, if we change (2.3.14) into the following: |fκ (t)| . kψ1 (t)kp kV2 (· − te~1 )uκ kp0 . e−αt kψ1 (t)kp . e−αt k (Id − Pb (H1 ))U (t)ψ0 kp αT . e−αt t−γ C0 (kψ0 kp0 + e− 2 kψ0 kL2 ) + e−αt l X |ai (t)| kui kp , i=1 where γ = n( 12 − p1 ) > 1, and follow the same arguments, we see that for large t, αT αT kPb (H1 , t)U (t)ψ0 kp . t−γ C0 e− 4 (kψ0 kp0 + e− 2 kψ0 kL2 ). If the potential V1 is smooth enough, it is known (e.g., [1]) that the bound state uj of H1 is differentiable. Moreover, its derivatives decay exponentially at infinity. Thus, k∂Pb (H1 , t)U (t)ψ0 kp ≤ l X i=1 αT αT |ai (t)|k∂ui kp . t−γ C0 e− 4 (kψ0 kp0 + e− 2 kψ0 kL2 ). (2.3.18) 29 In addition, the above claims hold with H1 replaced by Hj , j = 2, · · · , m. These results will be used to prove Theorem 2.1.4 in Section 4. ¤ With Proposition 2.3.1, the Pb (H1 , τ )U (τ ) part of (2.3.7) can be estimated by the following: Z t Z s∧(t−²) ds t−A s−A Z t . Z s∧(t−²) ds t−A .A2 dτ ke−i(t−s)H0 V (s)e−i(s−τ )H0 V1 (τ )Pb (H1 , τ )U (τ )ψ0 k∞ n n dτ ht − si− 2 hs − τ i− 2 kV1 (τ )Pb (H1 , τ )U (τ )ψ0 k1 s−A sup kPb (H1 , τ )U (τ )ψ0 k∞ τ ∈(t−2A,t) < αT C0 − n t 2 (kψ0 kL1 + e− 2 kψ0 kL2 ). 100 For the Pc (H1 , τ )U (τ ) part of (2.3.7), we need to apply Duhamel’s formula again and expand (2.3.7) further with respect to H1 . We assume that v~1 = 0 and m = 2 to simplify our notation. Specifically, we plug the following Pc (H1 , τ )U (τ ) = Pc (H1 )U (τ ) = Pc (H1 )e −iτ H1 Z τ − iPc (H1 ) e−i(τ −r)H1 V2 (r)U (r) dr 0 into (2.3.7). For the term containing Pc (H1 )e−iτ H1 , we apply the dispersive decay for Pc (H1 )e−iτ H1 : Z t Z s∧(t−²) ds t−A Z t . s−A Z s∧(t−²) ds t−A dτ ke−i(t−s)H0 V (s)e−i(s−τ )H0 V1 (τ )Pc (H1 )e−iτ H1 ψ0 k∞ dτ ht − si−n/2 hs − τ i−n/2 hτ i−n/2 kψ0 k1 . t−n/2 kψ0 k1 . s−A The second term is Z t Z s∧(t−²) ds t−A Z τ dτ s−A 0 dr e−i(t−s)H0 V (s)e−i(s−τ )H0 V1 (τ )Pc (H1 )e−i(τ −r)H1 V2 (r)U (r)ψ0 . (2.3.19) Now take a small constant δ > 0 and a large constant A1 > 0 to be specified later. 30 We decompose the integral Z δ Z τ dr = 0 Rτ 0 dr in (2.3.19) as follows: Z A1 dr + 0 Z τ −A1 dr + δ Z τ −δ dr + A1 Z τ dr + τ −A1 dr. (2.3.20) τ −δ To simplify the notation, we will write A1 as A. Our goal is to estimate each term in (2.3.20). The second term of (2.3.20) is estimated as follows: Z t Z s∧(t−²) ds t−A Z t . Z A dτ s−A Z s∧(t−²) ds Z A dτ t−A s−A −n/2 kψ0 k1 . .t δ drke−i(t−s)H0 V (s)e−i(s−τ )H0 V1 (τ )Pc (H1 )e−i(τ −r)H1 V2 (r)U (r)ψ0 k∞ drht − si−n/2 hs − τ i−n/2 hτ − ri−n/2 hri−n/2 erM kψ0 k1 δ The implicit constant above depends on A, δ. The third term of (2.3.20) is estimated as follows: Z t Z s∧(t−²) ds t−A Z t . Z τ −A dτ s−A Z s∧(t−²) ds t−A Z τ −A dτ s−A drht − si−n/2 hs − τ i−n/2 hτ − ri−n/2 kU (r)ψ0 k∞ A Z s∧(t−²) ds t−A Z τ −A dτ s−A Z t . A drke−i(t−s)H0 V (s)e−i(s−τ )H0 V1 (τ )Pc (H1 )e−i(τ −r)H1 V2 (r)U (r)ψ0 k∞ drht − si−n/2 hs − τ i−n/2 hτ − ri−n/2 hri−n/2 C0 |||ψ0 ||| A 1 C0 t−n/2 |||ψ0 |||, . t−n/2 C0 κA |||ψ0 ||| ≤ 100 where κA → 0 as A → ∞. So the above inequality holds for large enough A. 31 For the fourth term in (2.3.20), we have Z t Z s∧(t−²) ds Z τ t−A s−A Z t τ −δ Z s∧(t−²) . ds t−A Z t . Z τ dτ s−A Z s∧(t−²) ds t−A τ −A Z τ dτ s−A Z t . drke−i(t−s)H0 V (s)e−i(s−τ )H0 V1 (τ )e−i(τ −r)H1 Pc (H1 )V2 (r)U (r)ψ0 k∞ dτ t−A drht − si−n/2 hs − τ i−n/2 kU (r)ψ0 k∞ τ −A Z s∧(t−²) ds drht − si−n/2 hs − τ i−n/2 kV1 (τ )Pc (H1 )e−i(τ −r)H1 V2 (r)U (r)ψ0 k1 Z τ dτ s−A drht − si−n/2 hs − τ i−n/2 hri−n/2 C0 |||ψ0 ||| τ −δ 1 C0 t−n/2 |||ψ0 |||, . t−n/2 C0 κδ |||ψ0 ||| ≤ 100 where κδ → 0 as δ → 0. So the above inequality holds for δ small enough. Rδ For the 0 dr part of (2.3.20), we expand −i(τ −r)H1 e −i(τ −r)H0 =e Z τ −r −i e−i(τ −r−β)H1 V1 e−iβH0 dβ. 0 Here we put H0 after H1 in the integral because we want H0 to appear immediately Rδ before U (r) and apply Lemma 2.2.4. Substitute this expansion into the 0 dr part of (2.3.20) and we get two terms. The first one is Z t Z s∧(t−²) ds t−A Z δ dτ s−A dre−i(t−s)H0 V (s)e−i(s−τ )H0 V1 (τ )Pc (H1 )e−i(τ −r)H0 V2 (r)U (r)ψ0 . 0 (2.3.21) Notice that Pc (H1 ) = Id−Pb (H1 ), and because kPb (H1 )kp→p is bounded, kPc (H1 )kp→p is bounded as well. Therefore the L∞ norm of (2.3.21) is estimated as follows: Z t . Z s∧(t−²) ds t−A Z δ dτ s−A drht − si−n/2 hs − τ i−n/2 hτ i−n/2 eM r kψ0 k1 . t−n/2 kψ0 k1 . 0 The second term of the Rδ 0 dr part of (2.3.20) after substitution is 32 Z s∧(t−²) Z t ds t−A Z δ dτ s−A Z τ −r Pc (H1 ) dr e−i(t−s)H0 V (s)e−i(s−τ )H0 V1 (τ ) 0 e−i(τ −r−β)H1 V1 e−iβH0 dβ V2 (r)U (r)ψ0 . (2.3.22) 0 R τ −r Decompose 0 dβ so we can rewrite (2.3.22)= J1 + J2 + J3 , where J1 , J2 , and J3 Rδ R τ −r−1 R τ −r correspond to 0 dβ, δ dβ, and τ −r−1 dβ, respectively. We proceed to estimate J1 as follows: Z δ Z δ dr dβ kPc (H1 )e−i(τ −r−β)H1 V1 e−iβH0 V2 (r)U (r)ψ0 k∞ 0 0 Z δ Z δ dβ hτ − r − βi−n/2 ke−iβH0 V2 (r)U (r)ψ0 k∞ dr . 0 0 Z δ Z δ . hτ i−n/2 dr dβ(β + r)−n/2 eM r kψ0 k1 . 0 0 In the above expression, when n = 3, Rδ Rδ dr dβ(β + r)−n/2 eM r is integrable. 0 0 When n > 3, we need to further expand e−i(τ −r−β)H1 to remove the singularity of (β + r)−n/2 at β + r = 0. In either case, we can conclude that kJ1 k∞ . t−n/2 kψ0 k1 . For J2 , our estimate is the following: Z τ −r−1 Z δ dr dβkPc (H1 )e−i(τ −r−β)H1 V1 e−iβH0 V2 (r)U (r)ψ0 k∞ 0 δ Z δ Z τ −r−1 . dr dβhτ − r − βi−n/2 kV1 e−iβH0 V2 (r)U (r)ψ0 k1 0 δ Z δ Z τ −r−1 . dr dβhτ − r − βi−n/2 ke−iβH0 V2 (r)U (r)ψ0 k∞ 0 δ Z δ Z τ −r−1 . dr dβhτ − r − βi−n/2 (β + r)−n/2 eM r kψ0 k1 0 δ Z δ Z τ −r−1 . dr dβhτ − r − βi−n/2 hβ + ri−n/2 kψ0 k1 0 δ . τ −n/2 kψ0 k1 . 33 The implicit constant above depends on δ and is independent of t and ψ0 . Plugging the above estimate into J1 , we derive that kJ2 k∞ . t−n/2 kψ0 k1 . To estimate J3 , we notice that Z τ −r k dβV1 (τ )Pc (H1 )e−i(τ −r−β)H1 V1 e−iβH0 V2 (r)U (r)ψ0 k1 τ −r−1 Z τ −r ≤ dβkV1 (τ )k2 kPc (H1 )e−i(τ −r−β)H1 k2→2 kV1 e−iβH0 V2 (r)U (r)ψ0 k2 τ −r−1 Z τ −r . τ −r−1 Z τ −r . dβkV1 k2 ke−iβH0 V2 (r)U (r)ψ0 k∞ n dβ|β|− 2 eM r kψ0 k1 . τ −r−1 Observe that r is small and β ≃ τ . Plugging the above estimate into J3 , we derive Rδ n that kJ3 k∞ . t− 2 kψ0 k1 . Thus, we finish the estimate of the 0 dr part of (2.3.20). 2.3.2 Low and high velocity estimates So far we have estimated four parts of (2.3.20). This subsection is devoted to deriving R τ −δ the estimate of the τ −A dr part of (2.3.20), which will be decomposed as follows: Z t Z s∧(t−²) ds t−A Z t = dτ s−A Z s∧(t−²) ds t−A Z τ −δ dre−i(t−s)H0 V (s)e−i(s−τ )H0 V1 Pc (H1 )e−i(τ −r)H1 V2 (r)U (r)ψ0 τ −A Z τ −δ dτ s−A dre−i(t−s)H0 V (s)e−i(s−τ )H0 · τ −A V1 Pc (H1 )e−i(τ −r)H1 (F (|~p| ≥ N ) + F (|~p| ≤ N ))V2 (r)U (r)ψ0 = Jhigh + Jlow . F (|~p| ≤ N ) and F (|~p| ≥ N ) denote smooth projections onto the frequencies |~p| ≤ N and |~p| ≥ N , respectively. For the low velocity part Jlow , firstly, (t − s) + (s − τ ) ≥ ² and Lemma 2.2.4 imply 34 n n kei(t−s)H0 V (s)ei(s−τ )H0 k1→∞ . ht − si− 2 hs − τ i− 2 . (2.3.23) Secondly, we need the following proposition from [31]: Proposition 2.3.2. Let χ1 (t, x) be a smooth cut of the ball B(0, δt) with respect to x, where δ is a small constant only depending on e~1 and B(0, δt) is a ball centered at 0 with radius δt. Let A, M be large positive constants and A, M << t. Then sup kχ1 (t, ·)e−i(t−s)H1 Pc (H1 )F (|~p| ≤ M )V2 (· − se~1 )kL2 →L2 ≤ |t−s|≤A AM . δt The idea behind Proposition 2.3.2 can be explained as the following: The support of V2 (· − se~1 ) is contained in B(se~1 , R). Here R is the size of the support of V2 . The operator e−i(t−s)H1 Pc (H1 )F (|~p| ≤ M ) can “propagate” B(se~1 , R) into B(0, tδ) only if (t − s)M ≥ dist(B(se~1 , R), B(0, tδ)) according to the classical picture. However if |t−s| < A, tA, M ¿ t, then (τ −r)M ¿ dist(B(re~1 , R), B(0, τ δ)). This proposition appears in [31]. We give a proof here for completeness. Proof. The proof is a commutator argument. Let χ1 = χ(t, x). Firstly, we claim that k[χ1 , Pc (H1 )]kL2 →L2 . e−δαt . (2.3.24) Clearly, [χ1 , Pc (H1 )] = [χ1 , I − Pb (H1 )] = −[χ1 , Pb (H1 )]. Recall that u1 , . . . , um are the exponentially decaying eigenvalues of H1 . Therefore, m X [χ1 , Pb (H1 )]f = (χ1 ui hf, ui i − ui hf χ1 , ui i) = i=1 m X i=1 ((−1 + χ1 )ui hf, ui i − ui hf, (χ1 − 1)ui i) . 35 In of the support of χ1 − 1 we have that k(1 − χ1 )ui k2 . e−αδt and thus k[χ1 , Pb (H1 )]f kL2 ≤ e−αδt kf kL2 , as desired. Secondly, we claim that k[χ1 , e−i(t−s)H1 ]F (|~p| ≤ M )kL2 →L2 . AM . δt (2.3.25) We write [χ1 , e−i(t−s)H1 ] = e−i(t−s)H1 (ei(t−s)H1 χ1 e−i(t−s)H1 − χ1 ) and e i(t−s)H1 −i(t−s)H1 χ1 e Z t−s − χ1 = i Z0 t−s = i 0 eiτ H1 [H1 , χ1 ]e−iτ H1 dτ 1 eiτ H1 (−∇χ1 ∇ − 4χ1 )e−iτ H1 dτ. 2 Observe now that |∇χ1 (t, x)| . 1 , δt |4χ1 (t, x)| . 1 . (δt)2 Therefore, k[χ1 , e−(t−s)H1 ]F (|~p| ≤ M )kL2 →L2 ³ ´ . |t − s| k∇χ1 ∇e−iτ H1 F (|~p| ≤ M )kL2 →L2 + k∆χ1 e−iτ H1 F (|~p| ≤ M )kL2 →L2 . A A k∇e−iτ H1 F (|~p| ≤ M )kL2 →L2 + ke−iτ H1 F (|~p| ≤ M )kL2 →L2 . δt (δt)2 Since the potential V1 is bounded it is standard that sup k∇e−iτ H1 f kL2 . k∇f kL2 + kf kL2 . τ (2.3.26) 36 Indeed, sup k4e−iτ H1 f kL2 ≤ sup ke−iτ H1 H1 f kL2 + sup kV1 e−iτ H1 f kL2 . k∇2 f kL2 + kf kL2 , τ τ τ and (2.3.26) follows by interpolation with L2 . Therefore, k∇e−iτ H1 F (|~p| ≤ M )kL2 →L2 ≤ M. Combining terms we obtain the bound k[χ1 , e−i(t−s)H1 ]F (|~p| ≤ M )kL2 →L2 . AM AM A ≤ 2 + δt (δt)2 δt since M << t, which is (2.3.25). Finally, we invoke one more standard fact, namely k[χ1 , F (|~p| ≤ M )]kL2 →L2 . M −1 k∇χ1 k∞ . 1 . δM t (2.3.27) To see this, write F (|~p| ≤ M )f = [η̂(ξ/M )fˆ(ξ)]∨ with some smooth bump function η. Hence the kernel K of [χ1 , F (|~p| ≤ M )] is K(x, y) = M n η(M (x − y))(χ1 (x) − χ1 (y)), and (2.3.29) follows from Schur’s test. One concludes from estimates (2.3.24), (2.3.25), (2.3.29) that ° ° °χ1 (t, ·)e−i(t−s)H1 Pc (H1 )F (|~p| ≤ M )V2 (· − se~1 ) −e −i(t−s)H1 ° AM ° . . Pc (H1 )F (|~p| ≤ M )χ1 (t, ·)V2 (· − se~1 )° → 2 δt L L It remains to be seen that χ1 (t, ·)V2 (· − se~1 ) = 0 since the supports of χ1 (t, ·) and V2 (· − se~1 ) are disjoint. To apply Proposition 2.3.2 to Jlow , note that χτ V1 = V1 . Let χ2 be a smooth cut of the support of V2 and f be any function in L∞ (Rn ). Then it follows from 37 Proposition 2.3.2 that kV1 Pc (H1 )e−i(τ −r)H1 F (|~p| ≤ N )V2 (r)f k1 = kV1 χτ Pc (H1 )e−i(τ −r)H1 F (|~p| ≤ N )V2 (r)χ2 (· − r~v2 )f k1 ≤ kV1 k2 kχτ Pc (H1 )e−i(τ −r)H1 F (|~p| ≤ N )V2 (r)k2→2 kχ2 k2 kf k∞ . AN M 2 kf k∞ . δt Combining the above estimate with (2.3.23) and noting A, M, N ¿ t, we conclude Z t Z s∧(t−²) kJlow k∞ . ds t−A Z τ −δ dτ s−A drht − si−n/2 hs − τ i−n/2 τ −A AN M 2 kU (r)ψ0 k∞ δt C0 −n/2 ≤ t |||ψ0 |||. 100 From the above estimate for Jlow , it is worth remarking that the purpose of the multiple expansions by Duhamel’s formula is to prepare a cushion (the potentials V1 and V2 ) to apply the L2 → L2 estimate (Prop 2.3.2) between the L1 → L∞ estimates. For the high velocity part Jhigh , we shall further expand U (r) with respect to H0 , followed by a commutator argument. By Duhamel’s formula −irH0 U (r) = e Z r −i e−i(r−α)H0 V (α)U (α) dα, 0 we write Jhigh = Jhigh,1 − iJhigh,2 , where Z t Jhigh,1 = t−A Z s∧(t−²) Z τ −δ ds dτ dr e−i(t−s)H0 V (s)e−i(s−τ )H0 V1 Pc (H1 )e−i(τ −r)H1 · s−A τ −A · F (|~p| ≥ N )V2 (r)e−irH0 ψ0 , 38 and Z t Jhigh,2 = Z s∧(t−²) ds Z τ −δ dr e−i(t−s)H0 V (s)e−i(s−τ )H0 V1 Pc (H1 )e−i(τ −r)H1 · τ −A Z r · F (|~p| ≥ N )V2 (r) e−i(r−α)H0 V (α)U (α)ψ0 dα. dτ t−A s−A 0 The decay of Jhigh,1 will come easily from e−irH0 . Indeed, we apply Lemma 2.2.4 to e−i(t−s)H0 V (s)e−i(s−τ )H0 as in (2.3.23) and notice that kPc (H1 )e−i(τ −r)H1 F (|~p| ≥ N )kL2 →L2 ≤ 1. (2.3.28) Then it is clear that kJhigh,1 k∞ is dominated by Z t Z s∧(t−²) ds t−A Z τ −η dτ s−A drht − si−n/2 hs − τ i−n/2 hri−n/2 kψ0 k1 . t−n/2 kψ0 k1 . τ −A 1 2 3 Jhigh,2 will be decomposed into three parts Jhigh,2 , Jhigh,2 , and Jhigh,2 , corresponding RB R r−B Rr to 0 dα, B dα, and r−B dα, respectively, where B > 0 is a large constant to be specified. 1 For Jhigh,2 , the decay comes from e−i(r−α)H0 . Indeed, it follows from Lemma 2.2.2 and 0 < α < B that ke−i(r−α)H0 V (α)U (α)k1→∞ . r−n/2 eM α . hri−n/2 . 1 Hence, it follows from (2.3.23), (2.3.28), and the above inequality that kJhigh,2 k∞ is dominated by Z t Z s∧(t−²) ds t−A Z τ −δ dτ s−A drht − si−n/2 hs − τ i−n/2 hri−n/2 kψ0 k1 . hti−n/2 kψ0 k1 . τ −A 2 Jhigh,2 will be estimated by an application of the bootstrap assumption, and the smallness comes from choosing B to be sufficiently large. Indeed, it follows from (2.3.23), (2.3.28), Lemma 2.2.4 and the bootstrap assumption that 39 2 kJhigh,2 k∞ . Z t Z s∧(t−²) ds t−A Z r−B Z τ −δ dτ s−A drht − si−n/2 hs − τ i−n/2 · τ −A hr − αi−n/2 hαi−n/2 dα C0 |||ψ0 ||| B Z s∧(t−²) Z τ −δ ds dτ drht − si−n/2 hs − τ i−n/2 hri−n/2 κB C0 |||ψ0 ||| · Z t . t−A ≤ s−A τ −A 1 C0 t−n/2 |||ψ0 |||. 100 In the above inequality, B is chosen to be sufficiently large, because κB = R∞ B hαi−n/2 dα → 0 when B → ∞. 3 The decay of Jhigh,2 can only come from U (α). As usual we need to generate 1 1 the smallness 100 for the bootstrap assumption. Here the smallness 100 comes from the high velocity and a commutator argument. Write F (|~p| ≥ N )V2 (r) = [F (|~p| ≥ 3,1 3 N ), V2 (r)] + V2 (r)F (|~p| ≥ N ) and correspondingly, we decompose Jhigh,2 = Jhigh,2 + 3,2 3,1 3,2 3 Jhigh,2 . That is to say Jhigh,2 and Jhigh,2 are just Jhigh,2 with F (|~p| ≥ N )V2 (r) replaced by [F (|~p| ≥ N ), V2 (r)] and V2 (r)F (|~p| ≥ N ). 3,1 1 Specifically, the smallness 100 for Jhigh,2 comes from the following standard fact, namely k[F (|~p| ≤ N ), V2 ]kL2 →L2 . N −1 k∇V2 k∞ . (2.3.29) To see this, write F (|~p| ≤ N )f = [η̂(ξ/N )fˆ(ξ)]∨ with some smooth bump function η. Hence the kernel K of [F (|~p| ≤ N ), V2 ] is K(x, y) = N n η(N (x − y))(V2 (y) − V2 (x)), and (2.3.29) follows from Schur’s test and supx kK(x, ·)kL1 = supy kK(·, y)kL1 . N −1 k∇V2 k∞ . It follows from (2.3.23), kPc (H1 )e−i(τ −r)H1 k2→2 ≤ 1, (2.3.29) and the bootstrap assumption that 40 3,1 kJhigh,2 k∞ Z t Z s∧(t−²) Z τ −δ kV1 k2 k∇V2 k∞ . ds dτ dr(ht − sihs − τ i)−n/2 · N t−A s−A τ −A Z r · ke−i(r−α)H0 V (α)U (α)ψ0 k2 dα r−B 1 C0 −n/2 C0 −n/2 . sup kU (t)ψ0 k∞ . t |||ψ0 ||| ≤ t |||ψ0 |||, N t−3A−B<α<t N 100 where N1 is chosen sufficiently small to dominate the implicit constant in “ . ” which only depends on n, V, ~v2 and ², δ, A, B. 3,2 The smallness for Jhigh,2 comes from the following version of Kato’s 21 -smoothing estimate: Z α+B k α BR χ2 (· − r~v2 )F (|~p| ≥ N )e−i(r−α)H0 drk2→2 . √ , N (2.3.30) where χ2 (·) is a smooth cut around the support of V2 and R is radius of the support of χ2 . The implicit constant only depends on n, V2 . We refer to Section 3.5 in [31] for its proof and further references. R τ −δ R r−B Now observe that the region of integration τ −A dr r dα is contained in that of R τ −δ R α+B dα α dr and kPc (H1 )e−i(τ −r)H1 k2→2 ≤ 1. It follows from (2.3.23), (2.3.30), τ −A−B and the above observation that 3,2 kJhigh,2 k∞ . Z t Z s∧(t−²) ds t−A Z τ −δ dτ s−A BR dαht − si−n/2 hs − τ i−n/2 √ kU (α)ψ0 k∞ N τ −A−B C0 −n/2 . t |||ψ0 |||, 100 where √1N is chosen to be sufficiently small to dominate the implicit constant, which only depends on n, V, ~v2 and ², δ, A, B. Therefore, we conclude that (2.3.1) implies (2.3.2), from which Theorem 2.1.3 follows. 41 2.4 Decay estimates of the derivatives of U (t) In this section we prove Theorem 2.1.4 by induction on κ by following the same scheme as the proof of Theorem 2.1.3. The first step is to set up the cancellation lemma for ∂U (t)ψ0 . Lemma 2.4.1. Let κ be a nonnegative integer. Assume that β V (r)k 1 < M. sup sup k∂\ L 0≤β≤κ r∈R Let α be a nonnegative integer n-tuple with |α| = κ. Suppose U (t) is the evolution operator of (2.1.7) as before. Then sup ke−i(t−s)H0 V (r)∂ α U (s)ψ0 kp < |t|−γ M e(κ+1)M s kψ0 kW κ,p0 , (2.4.1) r∈R where γ = n( 12 − p1 ) and 2 ≤ p < ∞, p1 + p10 = 1. Proof. Write the left-hand side of (2.4.1):= Ψ(t, s). When κ = 0, (2.4.1) is just the inequality (2.2.3). Note that the inequality (2.2.3) holds with V replaced by its β V lies in L1 (Rn ). Assume κ = 1 and apply Duhamel’s derivative ∂ β V , as long as ∂d formula: ke−i(t−s)H0 V (r)∂U (s)ψ0 kp ≤ ke −i(t−s)H0 V (r)∂e −isH0 Z s ψ0 kp + ke−i(t−s)H0 V (r)e−i(s−τ )H0 ∂V (τ )U (τ )ψ0 kp dτ 0 Z s ≤ CkVb (r)k1 t k∂ψ0 kp0 + kVb (r)k1 ke−i(t−τ )H0 (∂V )(τ )U (τ )ψ0 kp dτ 0 Z s + kVb (r)k1 ke−i(t−τ )H0 V (τ )∂U (τ )ψ0 kp dτ 0 Z s Z s −γ τ M −γ t e dτ kψ0 kp0 + M Ψ(t, τ )dτ ≤ CM t kψ0 kW 1,p0 + M 0 0 Z s −γ sM ≤ CM t e kψ0 kW 1,p0 + M Ψ(t, τ )dτ. −γ 0 Taking supremum over r, we get Ψ(t, s) ≤ CM t−γ esM kψ0 kW 1,p0 + M Rs 0 Ψ(t, τ )dτ . By 42 Gronwall’s inequality, Ψ(t, s) ≤ CM t−γ e2M s . For κ > 1, the above argument goes through by induction, provided that the Fourier transforms of the derivatives up to degree κ of V (r) are uniformly bounded in L1 (Rn ). The following is an analog of Corollary 2.2.3: Corollary 2.4.2. With the same notations and assumptions as in Lemma 2.4.1, we have kU (t)ψ0 kW κ,p . t−γ e(1+κ)M t kψ0 kW κ,p0 . (2.4.2) Proof. By Duhamel’s formula, Lemma 2.4.1, and the fact that ∂ commutes with e−itH0 , we have the following estimate: Z t ke−i(t−τ )H0 (∂ β V )(τ )∂ α−β U (τ )ψ0 kp dτ ∂ ψ0 kp + Σβ≤α 0 Z t . t−γ kψ0 kW κ,p0 + Σβ≤α t−γ e(|β|+1)M τ dτ kψ0 kW κ,p0 α k∂ U (t)ψ0 kp . ke −itH0 α 0 −γ (κ+1)M t ≤ Ct e kψ0 kW κ,p0 . Similarly, the following lemma generalizes Lemma 2.2.4: Lemma 2.4.3. Let α be an n-tuple with |α| = κ and U (t) be the evolution operator of (2.1.7). For each m ≥ 1 and ² > 0, u1 , u2 , . . . , um are all in either R+ or R− , P satisfying | m j=1 uj | > ², then there exists constant C = C(m, ², κ, p) such that k m−1 Y iuj H0 (e α V (sj ))∂ U (um )ψ0 kp ≤ CM j=1 m−1 m Y huj i−γ e(κ+1)M um kψ0 kW κ,p0 , j=1 2n where sj is any real number, n−2 < p < ∞, p1 + p10 = 1 and β V (s)k ). M = Σ0≤β≤α sup(k∂ β V (s)k1 + k∂d 1 s∈R (2.4.3) 43 Using Lemma 2.4.1 and Corollary 2.4.2, the proof of Lemma 2.4.3 is exactly the same as that of Lemma 2.2.4. We only prove Theorem 2.1.4 for the case κ = 1, 2. The case κ > 2 can be proven by induction. Specifically, we prove the following implication: For any fixed sufficiently large time T , αT kU (t)ψ0 kW κ,p ≤ C0 |t|−γ (kψ0 kW κ,p0 + e− 2 kψ0 kL2 ) for 0 ≤ t ≤ T, κ = 1, 2 (2.4.4) implies that kU (t)ψ0 kW κ,p ≤ αT C0 −γ |t| (kψ0 kW κ,p0 + e− 2 kψ0 kL2 ) for 0 ≤ t ≤ T, κ = 1, 2 (2.4.5) 2 provided that C20 remains larger than some constant that does not depend on T . The assumption (2.4.4) can be made to hold for some C0 depending on T , because of Corollary 2.4.2. Letting T → +∞ to eliminate kψ0 kL2 , Theorem 2.1.4 follows from the iteration of the above implication. We will first prove (2.4.5) for κ = 1. For technical reasons (see (2.4.15)), we need the above bootstrap assumption (2.4.4) for κ + 2. To simplify the notation, we write ∂ α = ∂ and αT kψ0 kW 1,p0 + e− 2 kψ0 kL2 := |||ψ0 |||(1,p0 ) . With these cancellation lemmas for ∂U (t)ψ0 , the proof of Theorem 2.1.4 follows the scheme of that of Theorem 2.1.3. The difference is that now we need to commute ∂x with operators such as eitH0 , V and eitH1 to apply the cancellation lemma and the bootstrap assumption. We proceed by expanding U (t) with Duhamel’s formula: 44 ∂U (t)ψ0 = ∂e −itH0 Z t ψ0 − i Z t ∂e−i(t−s)H0 V (s)U (s)ψ0 ds 0 = ∂e−itH0 ψ0 − i e−i(t−s)H0 (∂V )(s)U (s)ψ0 ds 0 Z t −i e−i(t−s)H0 V (s)∂U (s)ψ0 ds. (2.4.6) 0 Notice that [∂, V ] = (∂V )· is a multiplication operator, which can be viewed as another potential and Theorem 2.1.3 can be applied to the second term of (2.4.6). This idea has appeared in the proof of Lemma 2.4.1. Specifically, it follows from the proof of Theorem 2.1.3 and an interpolation with the L2 conservation of U (t) that Z t k e−i(t−s)H0 V (s)U (s)ψ0 dskp . t−γ kψ0 kp0 . 0 By assumption, ∂Vj satisfies the regularity and smoothness conditions for Vj in Theorem 2.1.3, and we conclude that Z t k e−i(t−s)H0 (∂V )(s)U (s)ψ0 dskp . t−γ kψ0 kp0 . 0 We expand the last term of (2.4.6) by Duhamel’s formula just as in Section 2.3 and perform the same decomposition. With the cancellation lemma for ∂U (t) and Remark 2.3.1, the last term (2.4.6) is reduced to the following: 2 Z t X j=1 t−A Z s∧(t−²) ds dτ e−i(t−s)H0 V (s)e−i(s−τ )H0 Vj (· − τ~vj )∂Pc (H1 , τ )U (τ )ψ0 . (2.4.7) s−A Before we proceed, we observe that our assumptions guarantee kPc (H1 )e−itH1 ψ0 kLq ≤ Cq |t|−γ kψ0 kLq0 . (2.4.8) 45 This implies that kH1 Pc (H1 )e−itH1 ψ0 kLq = kPc (H1 )e−itH1 H1 ψ0 kLq ≤ Cq |t|−γ kH1 ψ0 kLq0 ≤ Cq |t|−γ kψ0 kW 2,q0 . As V1 ∈ L∞ (Rn ) and double Riesz transforms are bounded on Lq (Rn ) 1 < q < +∞, the above inequality in the case of 1 < q < +∞, implies that kPc (H1 )e−itH1 ψ0 kW 2,q . |t|−γ kψ0 kW 2,q0 . (2.4.9) Interpolating between (2.4.8) and (2.4.9) (Theorem 6.4.5 [5]), we conclude that kPc (H1 )e−itH1 ψ0 kW 1,q ≤ Cq |t|−γ kψ0 kW 1,q0 , (2.4.10) where 2 ≤ q < ∞, 1q + q10 = 1 and γ = n( 12 − 1q ). Because double Riesz transforms are unbounded on L∞ (Rn ), we exclude p = ∞ in Theorem 2.1.4. Rτ We write Pc (H1 )U (τ ) = Pc (H1 )e−iτ H1 − iPc (H1 ) 0 e−i(τ −r)H1 V2 (r)U (r) dr and (2.4.7) is broken into two terms. It follows from (2.4.10), among other things, that the first term of (2.4.7), which contains Pc (H1 )e−iτ H1 , is dominated by |t|−γ kψ0 kW 1,p . The second term of (2.4.7) is decomposed as follows: Z τ Z δ dr = 0 Z A dr + 0 Z τ −A dr + δ Z τ −δ dr + A Z τ dr + τ −A dr. (2.4.11) τ −δ We estimate each term in (2.4.11) with similar methods as those for (2.3.20). BeRA R τ −A cause of (2.4.10), the terms containing δ dr and A dr in (2.4.11) can be estimated exactly as there is no derivative before P (H1 ), and we omit the details here. Again Rτ by (2.4.10) with q = 2, the term containing τ −δ dr in (2.4.11) is estimated as follows: 46 Z s∧(t−²) Z t ds t−A Z τ drke−i(t−s)H0 V (s)e−i(s−τ )H0 V1 (τ )∂e−i(τ −r)H1 Pc (H1 )V2 (r)U (r)ψ0 kp τ −δ Z τ . sup drk∂Pc (H1 )e−i(τ −r)H1 V2 (r)U (r)ψ0 k2 t−2A<τ <t τ −δ Z τ . sup drkV2 (r)U (r)ψ0 kW1,2 t−2A<τ <t τ −δ Z τ . sup drkU (r)ψ0 kW1,p dτ s−A t−2A<τ <t τ −δ . t−γ C0 δ|||ψ0 |||(1,p0 ) ≤ C0 −γ t |||ψ0 |||(1,p0 ) . 100 Here δ > 0 is chosen sufficiently small. Rδ The 0 dr term in (2.4.11) is expanded by Duhamel’s formula: −i(τ −r)H1 e −i(τ −r)H0 =e Z τ −r −i e−i(τ −r−β)H1 V1 e−iβH0 dβ. 0 Plugging the above expression into the Rδ 0 dr term, we get two terms. The first one containing e−i(τ −r)H0 is Z t Z s∧(t−²) ds t−A Z δ dτ s−A dre−i(t−s)H0 V (s)e−i(s−τ )H0 V1 (τ )∂Pc (H1 )e−i(τ −r)H0 V2 (r)U (r)ψ0 . 0 (2.4.12) Since Pc (H1 ) = Id − Pb (H1 ) and Pb (H1 ) is a bounded operator from Lp to Lp , Pc (H1 ) is bounded from Lp to Lp . It follows from Lemma 2.4.1, 0 < r < δ, and the Leibnitz rule that kH1 Pc (H1 )e−i(τ −r)H0 V2 (r)U (r)ψ0 kp = kPc (H1 )H1 e−i(τ −r)H0 V2 (r)U (r)ψ0 kp ≤ CkH1 e−i(τ −r)H0 V2 (r)U (r)ψ0 kp ≤ CkV1 k∞ ke−i(τ −r)H0 V2 (r)U (r)ψ0 kp + ke−i(τ −r)H0 ∆V2 (r)U (r)ψ0 kp ≤ Cτ −γ kψ0 kW 2,p0 . 47 Since H1 = H0 + V1 and V1 is bounded, we see that k∆Pc (H1 )e−i(τ −r)H0 V2 (r)U (r)ψ0 kp ≤ Cτ −γ kψ0 kW 2,p0 . Because the double Riesz transforms are bounded on Lp (Rn ) 1 < p < ∞, it follows that kPc (H1 )e−i(τ −r)H0 V2 (r)U (r)ψ0 kW 2,p ≤ Cτ −γ kψ0 kW 2,p0 . Therefore, by complex interpolation, we see that kPc (H1 )e−i(τ −r)H0 V2 (r)U (r)ψ0 kW 1,p ≤ Cτ −γ kψ0 kW 1,p0 , (2.4.13) which implies that k(2.4.12)kW 1,p . t−γ kψ0 kW 1,p0 . R δ R τ −r For the term containing 0 dr 0 dβ , we perform the exact same decomposition as in (2.3.22) and each step there goes through provided (2.4.10) and (2.4.13). R τ −δ The term containing τ −A dr in (2.4.11) is Z t Z s∧(t−²) ds t−A Z τ −δ dτ s−A dr ke−i(t−s)H0 V (s)e−i(s−τ )H0 V1 ∂Pc (H1 )e−i(τ −r)H1 V2 (r)U (r)ψ0 kp . τ −A (2.4.14) The proof of Theorem 2.1.3 showed that ∀² > 0, the following holds: n kV1 Pc (H1 )e−i(τ −r)H1 V2 (r)U (r)ψ0 k∞ < ²C0 t− 2 kψ0 k1 , given t sufficiently large. Going through the proof, we see that the same argument also shows kV1 Pc (H1 )e−i(τ −r)H1 V2 (r)U (r)ψ0 kp < ²C0 t−γ kψ0 kp0 . Furthermore, the above inequality holds if V1 or V2 is replaced by its derivative. Another observation is that, given our new cancellation lemma for ∂U (r)ψ0 , kV1 Pc (H1 )e−i(τ −r)H1 V2 (r)∂ β U (r)ψ0 kp < ²C0 t−γ |||ψ0 |||(|β|,p0 ) . 48 Indeed, to prove the above inequality, we decompose the left-hand side into a high velocity part and a low velocity part. Each part generates the small constant ² for the same reason as in Section 3.3. The same argument with the bootstrap assumption (2.4.4) implies kV1 H1 Pc (H1 )e−i(τ −r)H1 V2 (r)U (r)ψ0 kp = kV1 Pc (H1 )e−i(τ −r)H1 H1 V2 (r)U (r)ψ0 kp . ²C0 t−γ |||ψ0 |||(2,p0 ) . (2.4.15) It follows from the above inequality and an elementary calculation that kV1 Pc (H1 )e−i(τ −r)H1 V2 (r)U (r)ψ0 kW 2,p . ²C0 t−γ |kψ0 |k(2,p0 ) . (2.4.16) Hence, by complex interpolation, for ∀ ² > 0, kV1 Pc (H1 )e−i(τ −r)H1 V2 (r)U (r)ψ0 kW 1,p . ²C0 t−γ |kψ0 |k(1,p0 ) , given t sufficiently large. (2.4.17) This implies that k(2.4.14)kW 1,p can be estimated by 1 C t−γ |||ψ0 |||1,p0 . 100 0 Therefore, we have proven (2.4.5) for κ = 1. The same procedure also proves (2.4.5) for κ = 2. Thus, we finish the bootstrap argument and conclude that kU (t)ψ0 kW κ,p . kψ0 kW (κ,p0 ) , by letting T → ∞. The proof for κ > 2 is similar by induction. Thus, we have proven Theorem 2.1.4. 2.5 Boundedness of the Sobolev norm of U (t, s)ψ0 The goal of this section is to prove Theorem 2.1.5 when κ is a positive integer. The intuition comes from the case κ = 1 ([17]). To bound the kinetic energy (the H 1 P norm), we look at the observable K(t) = 12 (p − xt )2 + m l=1 Vl (t). hK(t)i will decrease if the particle is far away from any potential, since the observable (p − xt )2 decreases 49 like t−2 for the free motion (the pseudo-conformal identity). If the particle is close to the center of potential Vl , then xt ≈ v~l and hK(t)i ≈ h 12 (p − v~l )2 + Vl (x − v~l t)i, which clearly is the total energy of this one potential stationary subsystem up to a Galilean transform. To carry this boundedness from hK(t)i to hp2 i, we need to replace the vector field xt by ν(x, t), such that ν(x, t) is uniformly bounded and is equal to v~l in an increasingly large neighborhood of x = v~l t. Rigorously, consider a smooth, uniformly bounded vector field ν(x, t) : Rn × (−∞, −T ] ∪ [T, +∞) → Rn and let m X 1 2 K0 (t) = (p − ν(x, t)) + Vl (t), 2 l=1 where T is a large positive constant, p = (p1 , · · · , pn ), and pj = −i ∂x∂ j . Note p2 = H0 and 12 (p − ν(x, t))2 is a well-defined self-adjoint positive operator. In [17], Graf constructed ν(x, t) and proved kU (t, s)ψ0 kH 1 is bounded as t → ∞ by bounding dtd hK0 (t)i from above by a time-integrable function, where hK0 (t)i = (U (t, s)ψ0 , K0 (t)U (t, s)ψ0 )L2 . We write (f, g) as the inner product of f, g in the L2 (R) sense. To prove Theorem 2.1.5, we need to define the proper analog of K0 (t) suitable to the H κ norm of U (t, s)ψ0 to match the intuition given by the classical system. Fortunately the following observable works: m X 1 1 ( (p − ν(x, t))2 + Vl (t))κ − (m − 1)( (p − ν(x, t))2 )κ . K(t) = 2 2 l=1 Notice that K(t) = K0 (t) if κ = 1. Because ν(x, t) and its derivatives are bounded uniformly in space-time and Vj ∈ C0κ (Rn ), we have the following, writing hK(t, s)i = (U (t, s)ψ0 , K(t)U (t, s)ψ0 ): kU (t, s)ψ0 k2H κ . hK(t, s)i + kU (t, s)ψ0 k2H κ−1 ; hK(t, s)i . kU (t, s)ψ0 k2H κ . 50 By induction on κ, it suffices to show hK(t, s)i is bounded uniformly in t and s. Expand K(t) as a polynomial of ( 12 (p − ν(x, t))2 . Though ( 21 (p − ν(x, t))2 and Vl (t) do not commute with each other, viewing K(t) as a differential operator, the term of highest degree is (( 21 (p − ν(x, t))2 )κ , which is positive, self-adjoint. The other terms in K(t) are of degree no bigger than 2κ − 2 with bounded and smooth enough coefficients. Correspondingly, hK(t, s)i breaks into two parts. The part (U (t, s)ψ0 , ( 21 (p − ν(x, t))2 )κ U (t, s)ψ0 ) is always nonnegative. The other part containing the low degree terms can be dominated by kU (t, s)ψ0 k2H κ−1 . By the induction hypothesis, it follows that hK(t, s)i is bounded from below. To bound hK(t, s)i from above, it suffices to show that for t > T , d hK(t, s)i ≤ t−(1+δ) C(hK(t, s)i + kU (t, s)ψ0 k2H κ−1 ). dt (2.5.1) For t < −T , the opposite of the above inequality should hold: d hK(t, s)i ≥ t−(1+δ) C(hK(t, s)i + kU (t, s)ψ0 k2H κ−1 ). dt (2.5.2) First let’s consider t > T , integrating (2.5.1), Z t2 hK(t2 , s)i − hK(t1 , s)i ≤ C t1 t−(1+δ) hK(t, s)idt + C sup kU (t, s)ψ0 k2H κ−1 . t,s∈R Choosing T > 0 large enough such that C R ∞ −1−δ t dt < 12 , then T hK(t2 , s)i ≤ hK(t1 , s)i + C sup kU (t, s)ψ0 k2H κ−1 + t,s∈R 1 max hK(t, s)i. 2 t1 <t<t2 By (2.1.19), maxt1 <t<t2 hK(t, s)i < ∞. This implies that max hK(t, s)i ≤ 2hK(t1 , s)i + Ckψ0 k2H κ−1 . t1 <t<t2 Letting t2 → +∞ and t1 = T , it follows that maxt>T hK(t, s)i < ChK(T, s)i + Ckψ0 k2H κ−1 . Hence hK(t, s)i ≤ CT kψ0 kH κ for t > T and s ∈ [−T, T ]. For t < −T , 51 we integrate (2.5.2) to bound hK(t, s)i from above and Theorem 2.1.5 follows in this case by the same argument given that (2.5.2) holds. Before we proceed to proving (2.5.1) and (2.5.2), let’s specify some properties of the vector field ν(x, t). It is convenient to describe ν(x, t) in the rescaled coordinates y vl |. When |y| > u0 , ν(x, t) = u0 |y| . When y ∈ Bl , y = xt . Let u0 = 2 max1≤l≤m |~ we specify ν(x, t) = v~l , where Bl is a fixed ball centered at v~l . We suppose that Bl (l = 1, · · · , m) lie in the big ball B0 centered at the origin with radius u0 and that they are disjoint from each other. When y ∈ B0 − ∪m l=1 Bl , we specify ν(x, t) = y. To make the vector field smooth, we modify and smooth the vector field in the scale of |t|1−γ , where γ is a small positive number. In the rescaled coordinates y, the scale is |t|−γ . Specifically, consider ω(s, α) = s ϕ( u0 − s u0 − s ) + u0 (1 − ϕ( )), α α where ϕ ∈ C ∞ (R) with ϕ0 ≥ 0 and ϕ(x) = 0 for x ≤ 0 ϕ(x) = 1 for x > 1. Then writing y = xt , we define ω (0) (x, t) = ω(|y|, |t|−γ ) y |y| and ω (`) (x, t) = −(y − v~` )ϕ(2 − |t|δ |y − v~` |), where ` = 1, 2, · · · , m. Finally, ν(x, t) := Pm `=0 ω (`) The properties of the vector field ν(x, t) that concern us are as follows: 1. ν is bounded in space time. The k-th space derivatives of ν uniformly decay as |t|−k(1−γ) as t → ∞. 2. (νi,j )n×n as a matrix is symmetric and positive semidefinite when t > 0, negative semidefinite when t < 0, where νi is the i-th component of vector ν and the indices following a comma stand for partial derivatives in space. As νk,j = νj,k , pk − νk and pj − νj commute with each other, i.e., [pk − νk , pj − νj ] = 0. 52 i 3. kνi,j νj + ∂ν k ≤ C|t|−(1+δ) . We make the choice 1 + δ = min{1+ γ, 2 − 2γ} > 1. ∂t ∞ Summation over double indices is understood. These properties can be shown by a direct calculation ([17]). Now we are going to prove (2.5.1) and (2.5.2) and proceed by observing that ∂ U (t, s)ψ0 = H(t)U (t, s)ψ0 and ∂t ∂ − i U (t, s)ψ0 = U (t, s)H(s)ψ0 . ∂s i (2.5.3) (2.5.4) It follows from the above that dtd hK(t, s)i = (U (t, s)ψ0 , (i[H(t), K(t)]+ ∂K )U (t, s)ψ0 ). ∂t A straightforward calculation shows that m κ−1 ∂K X X 1 d 1 1 = ( (p − ν(x, t))2 + Vl (t))k ( (p − ν)2 + Vl (t))( (p − ν)2 + Vl (t))κ−1−k ∂t 2 dt 2 2 l=1 k=0 − (m − 1) κ−1 X 1 d1 1 ( (p − ν)2 )k (p − ν(x, t))2 ( (p − ν)2 )κ−1−k := J1 + J2 , 2 dt 2 2 k=0 and the commutator m m X X 1 1 1 ( (p − ν)2 + Vl (t))κ − (m − 1)( (p − ν)2 )κ ] Vl (t), [H(t), K(t)] = [ p2 + 2 2 2 l=1 l=1 κ−1 m X X 1 1 1 1 ( (p − ν)2 + Vl (t))k [ p2 , (p − ν)2 + Vl (t)]( (p − ν)2 + Vl (t))κ−1−k = 2 2 2 2 l=1 k=0 − (m − 1) + κ−1 X 1 1 1 1 ( (p − ν)2 )k [ p2 , (p − ν)2 ]( (p − ν)2 )κ−1−k 2 2 2 2 k=0 m X κ−1 X 1 1 1 ( (p − ν)2 + Vl (t))k [Vj (t), (p − ν)2 + Vl (t)]( (p − ν)2 + Vl (t))κ−1−k 2 2 2 l,j=1 k=0 m X κ−1 X 1 1 1 − (m − 1) ( (p − ν)2 )k [Vj (t), (p − ν)2 ]( (p − ν)2 )κ−1−k := J3 + J4 + J5 + J6 . 2 2 2 j=1 k=0 53 First, let’s consider J1 + iJ3 + iJ5 = m X κ−1 X 1 1 ( (p − ν)2 + Vl (t))k M1 ( (p − ν)2 + Vl (t))κ−1−k , 2 2 l=1 k=0 where M1 = i[ 12 p2 + (2.5.5) Pm 1 d 1 2 2 j=1 Vj (t), 2 (p − ν) + Vl (t)] + dt ( 2 (p − ν) + Vl (t)). Another elementary calculation gives the following: X 1 1 ∂νi 1 ∂νi M1 = i[ Vj , (p − ν)2 + Vl ] + A − pi (νi,j νj + ) − (νi,j νj + )pi 2 2 ∂t 2 ∂t j6=l + νi (νi,j νj + ν ∂νi 1 ) + νi,ijj + (ν − v~l ) · ∇Vl , ∂t 4 (2.5.6) +ν where A = −(pi − νi ) i,j 2 j,i (pj − νj ) is a symmetric, semidefinite negative operator when t > 0 and a semidefinite positive operator when t < 0. It follows from the properties of νx,t that kνi,j νj + ∂νi k∞ ≤ C|t|−(1+δ) , ∂t kνi,ijj k∞ ≤ C|t|−1−δ (2.5.7) and that the L∞ norm of derivatives of these terms decay even faster because each P space derivative gains a factor |t|δ−1 . Moreover, m ~l ) · ∇Vl vanishes as |t| > T l=1 (ν − v is sufficiently large, since ν − v~l vanishes on an increasing neighborhood of x = t~ vl , which will eventually contain the support of ∇Vl . Plugging the expression of M1 into expression (2.5.5), we claim that the decaying terms listed in equation (2.5.7) only produce time integrable term. We calculate the i term containing 12 pi (νi,j νj + ∂ν ) as an example to illustrate this point: ∂t 1 ∂νi 1 1 |(U (t, s)ψ0 , ( (p − ν)2 + Vl (t))k pi (νi,j νj + )( (p − ν)2 + Vl (t))κ−1−k U (t, s)ψ0 )| 2 2 ∂t 2 ∂νi 1 1 1 )( (p − ν)2 + Vl (t))κ−1−k U (t, s)ψ0 )|. = |( pi ( (p − ν)2 + Vl (t))k U (t, s)ψ0 , (νi,j νj + 2 2 ∂t 2 (2.5.8) 54 If 2k + 1 = κ or 2k + 2 = κ, (2.5.8) can be dominated by 1 1 C|t|−1−δ kpi ( (p − ν)2 + Vl (t))k U (t, s)ψ0 kL2 k( (p − ν)2 + Vl (t))κ−1−k U (t, s)ψ0 kL2 2 2 ≤ C|t|−1−δ kU (t, s)ψ0 k2H κ ≤ C|t|−1−δ (hK(t, s)i + kU (t, s)ψ0 k2H κ−1 ). If κ 6= 2k + 1 or 2k + 2, first consider 2k + 2 < κ and κ = 2d + 1, an odd integer. i We need to commute νi,j νj + ∂ν with ( 12 (p − ν)2 + Vl )d−k . Specifically, we claim that ∂t 1 ∂νi pi 1 ( (p − ν)2 + Vl (t))d−k (νi,j νj + ) ( (p − ν)2 + Vl (t))k 2 ∂t 2 2 is an differential operator of degree 2d + 1, whose coefficients are of magnitude t−1−δ . i and its derivatives decay at least as |t|−1−δ . Hence, This is clear because νi,j νj + ∂ν ∂t (2.5.8) is dominated by C|t|−1−δ (hK(t, s)i + kU (t, s)ψ0 k2H κ−1 ). In the case that 2k + 2 < κ and κ = 2d or 2k + 1 > κ, (2.5.8) is dominated by C|t|−1−δ (hK(t, s)i + kU (t, s)ψ0 k2H κ−1 ) for the same reason. Therefore, it remains to estimate the following in expression (2.5.5) : κ−1 m X X X 1 1 1 ( (p−ν)2 +Vl (t))k (i[ Vj , (p−ν)2 +Vl ]+A)( (p−ν)2 +Vl (t))κ−1−k . (2.5.9) 2 2 2 l=1 k=0 j6=l Observe that for given time t, ν(x, t) is a constant vector on a ball centered at t~vl with radius growing linearly in |t| approximately. So, as long as |t| is large, ν(x, t) will be constant on the support of Vl (t). This implies that νj,i , νi,j both vanish on the ν +ν support of Vl (t). Hence it follows from A = −(pi − νi ) i,j 2 j,i (pj − νj ) that A Vl = 0 and Vl A = 0. Moreover, for j 6= l, Vj (t), Vl (t) have disjoint supports given that t is large. So the expression (2.5.9) is reduced to the following: 55 κ−1 m X X X 1 1 1 ( (p − ν)2 )k (i[ Vj , (p − ν)2 ] + A)( (p − ν)2 )κ−1−k 2 2 2 l=1 k=0 j6=l = κ−1 X X 1 1 1 ( (p − ν)2 )k (mA + (m − 1)i[ Vj , (p − ν)2 ])( (p − ν)2 )κ−1−k 2 2 2 j k=0 (2.5.10) (2.5.11) Secondly, we consider J2 + iJ4 + iJ6 , which equals κ−1 m X 1 2 X 1 d1 1 1 2 k −(m − 1) ( (p − ν) ) (i[ p + Vj (t), (p − ν)2 ] + (p − ν)2 )( (p − ν)2 )κ−1−k . 2 2 2 dt 2 2 j=1 k=0 (2.5.12) Setting all potentials Vl = 0 in (2.5.6), we see that 1 1 d1 i[ p2 , (p − ν)2 ] + (p − ν)2 = A + time integrable terms, 2 2 dt 2 where the time-integrable terms are equal to 1 ∂νi 1 ∂νi ∂νi 1 A − pi (νi,j νj + ) − (νi,j νj + )pi + νi (νi,j νj + ) + νi,ijj 2 ∂t 2 ∂t ∂t 4 and can be estimated exactly as those in J1 + iJ3 + iJ5 . We are left to estimate in J2 + iJ4 + iJ6 : κ−1 m X X 1 1 1 2 k ( (p − ν) ) (A + i[ −(m − 1) Vj (t), (p − ν)2 ])( (p − ν)2 )κ−1−k 2 2 2 j=1 k=0 (2.5.13) Now adding (2.5.11) and (2.5.13) together, we see that hi[H(t), K(t)] + ∂K i is ∂t simplified as some time-integrable terms plus the following: κ−1 X 1 1 ( (p − ν)2 )k A( (p − ν)2 )κ−1−k , 2 2 k=0 which is a differential operator of degree 2κ. (2.5.14) 56 ν +ν First we observe that [pk − νk , pj − νj ] = 0 and (pk − νk ) i,j 2 j,i = ν νi,j +νj,i (pk − 2 +ν νk ) + i,jk 2 j,ik . Second, νi,jk + νj,ik and its derivatives decay at least as fast as |t|−1−δ when t → ∞ and thus is integrable in time. Hence if we commute A with (p − ν)2 or pj − νj , the commutator is time-integrable. If κ = 2d + 1, an odd integer, then 1 1 1 1 ( (p−ν)2 )k A( (p−ν)2 )κ−1−k = ( (p−ν)2 )d A( (p−ν)2 )d +time-integrable terms. 2 2 2 2 The first summand is negative (positive) definite when t > 0 (t < 0). If κ = 2d, an even integer, then ( 12 (p−ν)2 )k A( 21 (p−ν)2 )κ−1−k = ( 12 (p−ν)2 )h−1 21 (pj − νj )A(pj − νj )( 21 (p − ν)2 )h−1 + time-integrable terms. Again the first summand is negative (positive) definite if t > 0 (t < 0). Hence, we have written dtd hK(t, s)i as a sum of a negative (positive if t < 0) term and other time-integrable terms. More precisely, the time-integrable terms decay at least as fast as |t|−1−δ . Therefore, we have proven (2.5.1) for t > T and (2.5.2) for t < −T . Finally, we deal with the case where |t| < T, s > T by time reversal. Write r = s−t e (r, s) = U (s − r, s), H(r) e e (r, s) = −H(r) e U e (r, s). and U = H(s − r). Then we have i∂r U Define the corresponding observable: e K(r) = m X 1 1 ( (p + ν(x, s − r))2 + Vl (x − s~vl + ~vl r))κ − (m − 1)( (p + ν(x, s − r))2 )κ . 2 2 l=1 e (r, s) is a bounded operator from H κ to itself by the same It can be shown that U e (r, s). The case of |t| < T, s < −T is similar. argument with U (t, s) replaced by U 2.6 Asymptotic completeness in Sobolev spaces Recall that we are considering (2.1.7). V1 is stationary (we denote its velocity as ~e0 = 0) and V2 is moving with velocity ~e1 . There are two approaches to prove The- 57 orem 2.1.6. Graf ([17]) proved the asymptotic completeness for the charge transfer model in the L2 sense by proving a RAGE theorem. Our first option to prove Theorem 2.1.6 is to generalize Graf’s idea. We find that this approach works, provided that each individual subsystem (i.e., p2 + Vl ) is asymptotically complete in the H κ sense. However, the only direct way to prove this fact, as we know, is by the dispersive estimate. The advantage of this approach is that it requires less restrictive conditions on the potentials and the spectrum of the individual subsystem, given that nontrivial fact. Our second option to prove Theorem 2.1.6 is to apply the dispersive estimate (Theorem 2.1.4) directly. To illustrate both of these ideas, the following proof is somehow a combination of these two options. Specifically, we follow [17] to prove the existence of the wave operators and then apply Theorem 2.1.4 to prove Theorem 2.1.6. 2.6.1 Existence of wave operators The well-known wave operators are defined as follows: −i(t−s)H0 Ω− , 0 (s) = s − lim U (s, t)e t→+∞ −i(t−s)H1 Ω− Pb (H1 ), 1 (s) = s − lim U (s, t)e t→+∞ −i(t−s)H2 Ω− Pb (H2 ) g~e1 (s). 2 (s) = s − lim U (s, t) g−~e1 (t)e t→+∞ Theorem 2.6.1. Under the assumption of Theorem 2.1.5, the above wave operators exist in the space H κ . More precisely, for l = 0, 1, 2 and ∀ψ0 ∈ H κ , the limits converge κ n in the H κ sense and Ω− l (s)ψ0 lies in H (R ). Remark 2.6.1. The above theorem can be proven by Cook’s method together with Theorem 2.1.4 and Theorem 2.1.5 if we are willing to impose more regularity on the potentials and the spectrum condition. The following proof originated in [17], which, we believe, requires the least conditions on the system. 58 We present some preliminary facts before proceeding: Lemma 2.6.2. Let g ∈ C0∞ (Rn ) and ν > 0. Suppose 1. g(p) = 0 for |p| ≥ ν and fix α > 1. Then for R > 0, t > 0 and any N > 0, p2 kF (|x| > α(R + νt))e−i 2 t g(p)F (|x| < R)ψkH κ ≤ CN,κ (R + νt)−N kψkL2 . 2. g(p) = 0 for |p| ≤ ν and ν0 > 0, 0 < α < 1. Then for t > 0 and any N > 0, p2 kF (|x| < α(ν − ν0 )t)e−i 2 t g(p)F (|x| < ν0 t)ψkH κ ≤ CN,κ t−N kψkL2 . These estimates are fairly common for κ = 0 and may be proven by the stationary phase methods (e.g., [10], Lemma (6.3)). For the case κ ≥ 1, the above lemma follows from a commutator argument and the fact that the derivative on the left-hand side can be absorbed into g(p) because g ∈ C0∞ (Rn ). The next lemma represents to some extent the counterpart of Lemma 2.6.2 for Hl = H0 + Vl . Lemma 2.6.3. Let g ∈ C0∞ (R) and v > 0. Suppose g(e) = 0 for e ≥ v 2 /2 and fix α > 1. Then for l = 1, 2, R > 0 and t ≥ 0, we have kF (|x| > α(R + vt))e−iHl t g(Hl )F (|x| < R)ψ(x)kH κ ≤ CN,κ (R + vt)−² kψkL2 . (2.6.1) When κ = 0, the lemma is just Lemma 4.2 of [17]. For κ ≥ 1, the left-hand side of (2.6.1) is dominated up to a constant by κ k(Hl + M ) 2 e−iHl t g(Hl )F (|x| < R)ψ(x)kL2 (|x|>α(R+vt)) , where M is chosen so large that Hl + M is a positive operator. If we define ge(Hl ) = κ (Hl + M ) 2 g(Hl ), then ge ∈ C0∞ (R). The above is of the form κ = 0 and the lemma follows from the case κ = 0. 59 Lemma 2.6.4. 1. Let 0 < v0 < v and g ∈ C0∞ (Rn ) with g(p) = 0 for {|p| < S v} {|p − ~e1 | < v}. Then for any s ∈ R, lim sup k(U (t2 , t1 )−e −iH0 (t2 −t1 ) t1 →+∞t2 >t1 )e −iH0 (t1 −s) g(p) 1 Y F (|x−~el s| < v0 (t1 −s))kL2 →H κ = 0. l=0 2. Let v0 , v > 0 with v0 + v < |~e1 | and g ∈ C0∞ (R) with g(p) = 0 for p > v 2 /2. Then lim sup k(U (t2 , t1 ) − e−iH1 (t2 −t1 ) )g(H1 )F (|x| < v0 t1 )kL2 →H κ = 0. t1 →+∞t2 >t1 For κ = 0, the lemma was proven in [17]. We will follow the approach there to prove the case κ > 0. Proof. Part (1): Take α < α1 < 1 and let f ∈ C0∞ (Rn ) with f (y) = 0 if |y−~el | > α(v− P v0 ) for both l = 0, 1. Since αt < α1 (t − s), we have |f (x/t)| ≤ |f (x/t)| 1l=0 F (|x − ~el t| < α1 (v − v0 )(t − s)) for large enough t. 1 Y x −iH0 (t−s) F (|x − ~el s| < v0 (t − s))kL2 →H κ kf ( )e g(p) t l=0 . 1 XX k∂ β f k∞ kF (|x − ~el t| < α1 (v − v0 )(t − s)) |β|<κ l=0 e−iH0 (t−s) g β (p)F (|x − ~el s| < v0 (t − s))kL2 →L2 ≤C 1 XX (2.6.2) kF (|x| < α1 (v − v0 )(t − s))e−iH0 (t−s) g β (p + ~el )F (|x| < v0 (t − s))k |β|<κ l=0 ≤ C(t − s)−N , where g β (p) = P γ |β+γ|=κ p g(p). The above inequality follows by commuting the derivative through f (x/t), by applying a Galilean transform to the second expression, and by Lemma 2.6.2. By (2.6.2) and Theorem 2.1.5, it suffices to show that 60 ° sup °(U (t2 , t1 )(1 − f (x/t1 )) − (1 − f (x/t2 ))e−iH0 (t2 −t1 ) )e−iH0 (t1 −s) g(p)· t2 >t1 · 1 Y ° F (|x − ~el s| < v0 (t1 − s))°L2 →H κ → 0. (2.6.3) l=0 Substituting (U (t2 , t1 )(1−f (x/t1 ))−(1−f (x/t2 ))e −iH0 (t2 −t1 ) Z t2 )= t1 d (U (t2 , t)(1−f (x/t))e−iH0 (t−t1 ) )dt dt into (2.6.3), it follows from Theorem 2.1.5 that the left-hand side of (2.6.3) is dominated by Z +∞ t1 ° °[iH(t)(1 − f (x/t)) − i(1 − f (x/t))H0 − ∂ f (x/t)]e−iH0 (t−s) g(p)· ∂t 1 Y ° F (|x − ~el s| < v0 (t1 − s))°L2 →H κ dt. · l=0 The expression within the square brackets consists of (1) − (3), which are estimated as follows: 1. Suppose t is sufficiently large, then Vl (t)(1−f (x/t)) = 0, because Vl is compactly supported, where we take f (y) = 1 for |y − ~el | < α(v − v0 )/2; 2. H0 f (x/t) − f (x/t)H0 = − 12 t−2 (4f )(x/t) − it−1 (∇f )(x/t)p; and ∂ 3. ∂t f (x/t) = −t−1 (x/t)(∇f )(x/t) are treated using (2.6.2). Part (2): Choose α > 1 and v1 with α(v + v0 ) < v1 < |e1 | and let f ∈ C0∞ (Rn ) with f (y) = 1 for |y| < α(v + v0 ) and f (y) = 0 for |y| > v1 . We first claim that lim supk(1 − f (x/t))e−iH1 (t−t1 ) )g(H1 )F (|x| < v0 t1 )kL2 →H κ = 0. t1 →+∞ t>t1 (2.6.4) 61 Since 1 − f (x/t) is supported in |x| > α(v + v0 )t > α(v0 t1 + v(t − t1 )), it follows from Lemma 2.6.3 that kF (|x| > α(v0 t1 +v(t−t1 ))e−iHl (t−t1 ) g(Hl )F (|x| < v0 t1 )kL2 →H κ ≤ CN,κ (v0 t1 +v(t−t1 ))−² . (2.6.5) Now by Theorem 2.1.5 and (U (t2 , t1 )f (x/t1 ) − f (x/t2 )e −iH1 (t2 −t1 ) Z t2 )= t1 d (U (t2 , t)f (x/t)e−iH1 (t−t1 ) )dt, dt it suffices to estimate that sup k(U (t2 , t1 )f (x/t1 ) − f (x/t2 )e−iH1 (t2 −t1 ) )g(H1 )F (|x| < v0 t1 )kL2 →H κ t2 >t1 Z +∞ ∂ ≤ dt k[iH(t)f (x/t) − if (x/t)H1 + f (x/t)]e−iH1 (t−t1 ) g(H1 )F (|x| < v0 t1 )kL2 →H κ . ∂t t1 As in Part (1), a discussion of terms (a)-(d) in the square brackets now follows: (a) V1 (x)f (x/t) − f (x/t)V1 (x) = 0. (b) V2 (x − e1 t)f (x/t) = 0 if t is large enough because V2 is compactly supported and f (y) = 0 for |y| > v1 and |e1 | > v1 . (c) [H0 , f (x/t)] = 12 t−2 4f (x/t) − ipt ∇f (x/t). Since V1 ∈ C0κ , we can take M large enough so that the corresponding term can be dominated by κ 1 ip k(M + H1 ) 2 ( t−2 4f (x/t) − ∇f (x/t))e−iH1 (t−t1 ) g(H1 )F (|x| < v0 t1 )kL2 →L2 . 2 t κ Commute (M + H1 ) 2 through ( 21 t−2 4f (x/t) + 1t ∇f (x/t)∇) and the commutators generated will decay at least as fast as t−2 , hence they are time-integrable. Note that k(p2 + 1)σ g(H1 )kL2 →L2 < Cσ . The only term that does not decay as fast as t−2 is 62 k(M + H1 )−1 ( κ ip ∇f (x/t))e−iH1 (t−t1 ) (M + H1 ) 2 +1 g(H1 )F (|x| < v0 t1 )kL2 →L2 , t which is integrable, due to the fact that (M + H1 )−1 p is a bounded operator from L2 κ to L2 and due to (2.6.5) (with g(H1 ) replaced by (M + H1 ) 2 +1 g(H1 )), and due to the support property of ∇f (x/t). ∂ (d) ∂t f (x/t) = − xt f (x/t)t−1 , which can be treated as part (c), using (2.6.5) with f (x) replaced by xf (x). Proof of Theorem 2.6.1. Since U (s, t)e−i(t−s)H0 and U (s, t)e−i(t−s)H1 are uniformly bounded operators from H κ to H κ , it suffices to prove the existence of the strong limits Ω− 0 (s) and Ω− 1 (s) on a dense set D: D = {g(p)f (x)ψ : g ∈ C0∞ (Rn \{0, e1 }), f ∈ C0∞ (Rn ), ψ ∈ L2 (Rn )}. g(p) satisfies the hypothesis of Lemma 2.6.4 Part (1), with a suitable v > 0. Take 0 < v0 < v and note that 2 Y F (|x − ~el | < v0 (t1 − s))f (x) = f (x) l=1 for t1 big enough. For t2 > t1 , it follows from Theorem 2.1.5 that k(U (s, t1 )e−iH0 (t1 −s) − U (s, t2 )e−iH0 (t2 −s) )g(p)f (x)ψkH κ −iH0 (t2 −t1 ) .k(U (t2 , t1 ) − e )e −iH0 (t1 −s) g(p) 1 Y F (|x − ~el s| < v0 (t1 − s))kL2 →H κ kf (x)ψkL2 . l=0 Hence Lemma 2.6.4 implies that U (s, t)e−iH0 (t−s) g(p)f (x)ψ is Cauchy sequence in H κ (Rn ) as t → +∞, which is equivalent to the existence of Ω− 0 (s) . − We will only show the existence of Ω− 1 (s). The existence of Ω2 (s) follows from the same argument up to a Galilean transform ([17]). Since the eigenfunctions of 63 H1 span the range of Pb (H1 ), it suffices to prove convergence on the eigenfunctions ψ : H1 ψ = Eψ. Due to our assumptions on the potentials, the positive eigenvalues are excluded. Thus for any v > 0, we can find a suitable g as in Lemma 2.6.4 part (2) with g(H1 )P (H1 ) = P (H1 ). More precisely, we take v, v0 > 0 with v + v0 < |e1 |. For t2 ≥ t1 , k(U (s, t1 )e−iH1 (t1 −s) P (H1 ) − U (s, t2 )e−iH1 (t2 −s) )ψkH κ =kU (s, t2 )(U (t2 , t1 ) − e−iH1 (t2 −t1 ) )e−iH1 (t1 −s) g(H1 )(F (|x| < v0 t1 ) + F (|x| > v0 t1 ))ψkH κ .k(U (t2 , t1 ) − e−iH1 (t2 −t1 ) )e−iH1 (t1 −s) g(H1 )F (|x| < v0 t1 )kL2 →H κ kψkL2 + kF (|x| > v0 t1 )ψkH κ , since U (s, t), H1 (s) and g(H1 ) are bounded operators on H κ (Rn ) with a uniform bound. Lemma 2.6.4 part (2) and the fact that kF (|x| > v0 t1 )ψkH κ → 0 when t1 → +∞ imply that U (s, t)e−iH1 (t−s) P (H1 )ψ is a Cauchy sequence in H κ . 2.6.2 Asymptotic completeness In this section we will apply theorems 2.1.4 and 2.1.5 to prove Theorem 2.1.6. For the case κ = 0, we refer the reader to [31]. 0 Proof of Theorem 2.1.6. First let us assume that ψ0 ∈ W κ,2 ∩ W κ,p for some 2n . Decompose 1 < p0 < 2+n ψ(t) := U (t)ψ0 = Pb (H1 )U (t)ψ0 + Pb (H2 , t)U (t)ψ0 + R(t). By construction, we clearly have Pb (H2 , t)U (t)ψ0 + R(t) ∈ Ran(Pc (H1 )), Pb (H1 )U (t)ψ0 + R(t) ∈ Ran(Pc (H2 , t)). We further write Pb (H1 )U (t)ψ0 = m X r=1 e−iλr t ar (t)ur (x) (2.6.6) 64 for some choice of unknown functions ar (t). Due to the smoothness of the potentials, ur belongs to H κ (Rn ). It follows from (3.0.3) that, similar to (2.3.1), ȧr + i hV2 (· − t~e1 )ψ(t), ur i = 0 for all 1 ≤ r ≤ m. The exponential localization of ur implies that |hV2 (· − t~e1 )ψ(t), ur i| . e−αt . Therefore, ar (t) has a limit, writing limt→+∞ ar (t) = Ar , and m ° ° X ° −iλr t ° Ar e ur ° → 0, °Pb (H1 )U (t)ψ0 − κ r=1 H t → +∞. (2.6.7) We next define the functions vr = lim U (t)−1 e−iλr t ur . The existence of vr and vr ∈ t→+∞ H κ is guaranteed by Theorem 2.6.1. By Theorem 2.1.5, we have m m ° ° ¢ X ¡X ° −iλr t ° A e u A v − U (t) → 0, ° r r° r r κ r=1 r=1 H t → +∞. (2.6.8) t → +∞. (2.6.9) We then infer from (2.6.7) that m ° ° ¡X ¢ ° ° U (t) A v − P (H )U (t)ψ → 0, ° r r b 1 0° κ r=1 H The above arguments apply to Pb (H2 , t)U (t)ψ0 in a similar fashion. More precisely, we write U (t)ψ0 = Pb (H2 , t)U (t)ψ0 + Γ(t) = g−~e1 (t)Pb (H2 ) g~e1 (t)U (t)ψ0 + Γ(t). Therefore, g~e1 (t)U (t)ψ0 = Pb (H2 ) g~e1 (t)U (t)ψ0 + g~e1 (t)Γ(t). (2.6.10) Recall that the function ψ̃(t) = g~e1 (t)U (t)ψ0 is a solution of the problem 4 1 ∂t ψ̃ − ψ̃ + V2 (x)ψ̃ + V1 (x + t~e1 )ψ̃ = 0, i 2 ψ̃|t=0 = g~e1 (0)ψ0 . (2.6.11) According to (2.6.10), ψ̃(t) = Pb (H2 )ψ̃(t) + Γ1 (t), where Γ1 (t) = g~e1 (0)Γ(t). In 65 particular, Γ1 (t) ∈ Ran(Pc (H2 )). Decompose Pb (H2 )ψ̃(t) = X̀ bs (t)e−iµs t ws s=1 for some choice of unknown functions bs (t). Again, due to the smoothness of the potentials, ws ∈ H κ (Rn ). After substituting the decomposition in (2.6.11) we obtain the equation ḃs (t) + i hV1 (· + t~e1 )ψ̃, ws i = 0 for all 1 ≤ s ≤ `. Using exponential localization of ws we conclude the existence of the limit bs (t) → Bs P as t → +∞. Thus kPb (H2 )ψ̃(t) − `s=1 Bs e−iµs t ws kH κ → 0, t → ∞. Equivalently, after applying g−~e1 (t), we have ° ° X̀ ° ° −iµj t Bs e g−~e1 (t)ws ° → 0. °Pb (H2 , t)U (t)ψ0 − κ H s=1 (2.6.12) Now Theorem 2.6.1 allows us to define −1 ωs := Ω− g−~e1 (t)e−itH2 Pb (H2 )ws ∈ H κ . 2 ws = s − lim U (t) t→+∞ Moreover, ° ° ¡ X̀ ¢ X̀ ° ° −iµs t Bs ω s − Bs e g−~e1 (t)ws ° → 0, °U (t) κ s=1 H s=1 t → +∞. (2.6.13) It then follows from (2.6.12) that ¡ X̀ ¢ kPb (H2 , t)U (t)ψ0 − U (t) Bs ωs kH κ → 0, s=1 t → +∞. (2.6.14) 66 We now define the function φ := ψ0 − m X A r vr − X̀ r=1 Bs ω s , (2.6.15) s=1 which will lead to the initial data φ0 for the free channel. We have that m ¢ ¢ ¡ X̀ ¡X Bs ωs . Ar vr − Pb (H1 )U (t) Pb (H1 )U (t)φ = Pb (H1 )U (t)ψ0 − Pb (H1 )U (t) s=1 r=1 It follows from (2.6.9) and the identity Pb2 (H1 ) = Pb (H1 ) that m ° ¡X ¢° ° ° →0 P (H )U (t)ψ − P (H )U (t) A v ° b 1 0 b 1 r r ° κ H r=1 as t → +∞. (2.6.16) Furthermore, Pb (H1 ) X̀ Bs e −iµs t g−~e1 (t)wj = s=1 m X̀ X Bs e−iµs t h g−~e1 (t)wj , ur i ur → 0 (2.6.17) r=1 s=1 in the H κ sense as t → +∞, due to the exponential localization of the eigenfunctions ur . We infer from (2.6.16), (2.6.13), and (2.6.17) that kPb (H1 )U (t)φkH κ → 0. Similarly, kPb (H2 , t)U (t)φkH κ → 0. Thus, U (t)φ is asymptotically orthogonal to the bound states of H1 and H2 . Vj ∈ C0n+2κ+2 implies that (1 + |ξ|)κ+1+ 2 Vbj (ξ) ∈ L2 (Rn ). n So (1 + |ξ|)κ Vbj (ξ) ∈ L1 (Rn ). Therefore, according to Theorem 2.1.4, U (t)φ satisfies the estimate 1 1 kU (t)φkW κ,p . |t|−n( 2 − p ) kφkW κ,p0 , (2.6.18) 2n where n−2 < p < +∞. In order to be able to apply the estimate (2.6.18), one 0 0 needs to verify that φ ∈ W κ,p . By assumption, ψ0 ∈ W κ,p . Thus it remains to 0 0 check vr ∈ W κ,p , r = 1, · · · , m and ωs ∈ W κ,p , s = 1, · · · , `, which is guaranteed by Lemma 2.6.5 below. Assuming this lemma for the moment, we now consider the 67 expression −it 4 2 e Z t U (t)φ = φ − i 4 e−is 2 (V1 (x) + V2 (x − s~e1 )) U (s)φ ds. 0 2p Writing p−2 = r, we have the following estimate: Z +∞ 4 ke−is 2 (V1 (x) + V2 (x − s~e1 )) U (s)φkH κ ds t Z +∞ . (kV1 kW κ,r + kV2 kW κ,r ) kU (s)φkW κ,p ds t Z +∞ 1 1 . |s|−n( 2 − p ) kφkW κ,p0 (kV1 kW κ,r + kV2 kW κ,r ) → 0, as t → +∞. t Here we note that −n( 21 − p1 ) < −1. This allows us to show the existence of the limit 4 φ0 := lim eit 2 U (t)φ ∈ H κ . t→∞ It follows that 4 kU (t)φ − e−it 2 φ0 kH κ → 0, t → +∞. (2.6.19) Combining (2.6.8), (2.6.13), (2.6.15), and (2.6.19) we infer that m ° ° X X̀ ° ° −iλr t −iµs t −it 4 2 U (t)ψ − A e u − B e g φ (t)w − e → 0, ° 0 r r s −~e1 0° s κ r=1 H s=1 as t → +∞, 0 as claimed. Because W κ,2 ∩ W κ,p is dense in W κ,2 , for any ψ0 ∈ W κ,2 , there is a 0 sequence ψl ∈ W κ,2 ∩ W κ,p converging to ψ0 in the W κ,2 norm. Then for each ψl , we have the following decomposition: U (t)ψl = m X r=1 Alr e−iλr t ur + X̀ 4 Bkl e−iµk t g−~e1 (t)wk + e−it 2 φl + Rl (t). k=1 It follows from Theorem 2.1.5 that ψl = Pm l − r=1 Ar Ω1 ur + P` − l − k=1 Bk Ω2 wk + Ω0 φl . 2 n Since the ranges of Ω− 0,1,2 are orthogonal to each other in L (R ) ([17]), the fact that ψl converges as l → +∞, implies that each component in the above equation 68 converges. Hence, liml→+∞ Alr = A0r , liml→+∞ Bkl = Bk0 . These imply that Ω− 0 φl converges in H κ , since all other terms in the above identity converge in H κ . Write κ liml→+∞ Ω− 0 φl = f0 ∈ H . By the asymptotic completeness theorem for L2 ([17]), there are φ0 ∈ L2 such that the following holds: ψ0 = m X A0r Ω− 1 ur + r=1 X̀ − Bk0 Ω− 2 wk + Ω0 φ0 k=1 in the L2 sense. This implies that f0 = Ω− 0 φ0 . Then by the definition of the wave operator, U (0, t)e−itH0 φ0 − f0 → 0 as t → +∞ in L2 (Rn ). Since U (t, 0) and eitH0 are uniformly bounded operators on H κ (Rn ) and L2 (Rn ), we see that φ0 = lim eitH0 U (t, 0)f0 in L2 (Rn ). We claim that this implies t→+∞ φ0 ∈ H κ (Rn ). It suffices to prove the following: Assume gn is a sequence in H κ (Rn ) and kgn kH κ < 1. Moreover, gn converges to g in the L2 norm. Then g lies in H κ (Rn ). To see this, note that on Fourier side, H κ (Rn ) is just a weighted L2 (Rn ) space. More precisely, kgˆn −ĝkL2 (Rn ) → 0 implies that for the ball BR with radius R, centered at the origin, κ k(1 + |ξ|2 ) 2 (gˆn (ξ) − ĝ(ξ))kL2 (BR ) → 0 as n → +∞. κ This implies that k(1 + |ξ|2 ) 2 ĝ(ξ)kL2 (BR ) is uniformly bounded by sup kgn kH κ ≤ 1. Let R → +∞, we see that kgkH κ ≤ 1. Now it is clear that the following decomposition holds in the space H κ for any ψ0 ∈ H κ : ψ0 = m X r=1 A0r Ω− 1 ur + X̀ − Bk0 Ω− 2 wk + Ω0 φ0 . k=1 To complete the proof of Theorem 2.1.6, it remains to prove the following lemma: Lemma 2.6.5. Assume that the potentials V1 (x), V2 ∈ C0n+2κ+2 (Rn ). Let U (t) be the evolution operator of (2.1.7) and Ω− 1,2 the wave operators corresponding to U (t), as 69 0 κ,p defined at the beginning of this section. Then for ∀f ∈ L2 (Rn ), Ω− , 1,2 f lies in W 2n where 1 < p0 < n+2 . Proof. The proof is essentially contained in [31] Section 4. For the reader’s convenience, we present the details here. Without loss of generality we only consider the 2 wave operator Ω− 1 . For an arbitrary L function f Ω− 1f = m X fr lim U (t)−1 e−itH1 ur , r=1 where Pb (H1 )f = t→+∞ Pm r=1 fr ur for some constants fr . It follows from Duhamel’s formula that −1 −itH1 U (t) e Z t ur =ur + i Z0 t =ur + i U (s)−1 V2 (· − s~e1 )e−isH1 ur ds U (s)−1 V2 (· − s~e1 )e−iλr s ur ds, (2.6.20) 0 since ur is an eigenfunction of H1 corresponding to an eigenvalue λr . The function ur is exponentially localized in L2 together with its n + 2 derivatives 1 X 0≤|γ|≤n+2 Z Rn e2α|x| |∂xγ ur (x)|2 dx ≤ C for some positive constant α appearing in (2.3.12). This implies that the function Gr (s, x) := e−iλr s V2 (x − s~e1 )ur (x) has the property that for any k ≥ 0 and multi-index γ, 0 ≤ |γ| ≤ n + 2 khxik ∂xγ Gr (s, ·)kL2x ≤ c(r, |γ|, k)hsi−3j0 −2−κ . 1 The localization of higher derivatives of ur follows from the localization of ur stated in (2.3.12) and the equation − 4 2 ur + V1 (x)ur = λr ur with potential V1 (x), which is bounded together with all its derivatives of order ≤ (n + 2). 70 0 2p By Hölder’s inequality, we have, writing q = 2−p 0, k∂xγ U (−s)G(s, x)kLp0 ≤ khxi−j0 kLq khxij0 ∂xγ U (−s)G(s, x)kL2 . (2.6.21) 0 Take j0 = [ 2−p n] + 1 > nq , then khxi−j0 kLq < +∞. 2p0 To prove the desired conclusion, it would then suffice to show that for any |γ| = κ, there exists a positive constant k such that for any function g(x) X khxij ∂ γ U (t)gkL2 . hti3j0 +κ khxik ∂xβ gkL2 , ∀t ≥ 0. (2.6.22) |β|≤j0 +κ We note that the estimates of the type (2.6.22) for problems with time-independent potentials are well-known. They have been proved in the paper by Hunziker [18]. In the time-dependent case the argument is essentially the same. More precisely, define the functions Φj,|γ| (t) := |γ| j X X 0 0 khxij ∂xγ U (t)gkL2 j 0 =0 |γ 0 |=0 for any index j ≥ 0 and any multi-index γ. Using equation (2.1.7) we obtain £4 ¤ d khxij ∂xγ U (t)gk2L2 = i(−1)|γ| h − V (t, x), ∂xγ hxi2j ∂xγ U (t)g, U (t)gi. dt 2 Computing the commutator we obtain the recurrence relation Φj,|γ| (t) .Φj,|γ| (0) + hti2 ° X ° ° hxi γ 0 ° ∂x V ° ∞ sup Φj−1,|γ|+1 (τ ) ≤ ° hti Lt,x 0≤τ ≤t 0 |γ |≤2|γ| ¶ µX j−1 2k 2j hti Φj−k,|γ|+k (0) + hti Φ0,|γ|+j (τ ) , C(V ) k=0 where C(V ) is a constant depending on X |γ 0 |≤2(|γ|+j−1) ° hxi ° ° γ0 ° ∂ V° ∞. ° hti x Lt,x (2.6.23) In addition, differentiating the equation (2.1.7) |γ| + j times with respect to x and 71 using the standard L2 estimate, we have Φ0,|γ|+j (τ ) ≤ C(V )(1 + |τ ||γ|+j )Φ0,|γ|+j (0). Therefore, Φj,|γ| (t) ≤ C(V )(1 + |t|)3j+|γ| Φj,|γ|+j (0). Now setting j = j0 and |γ| = κ we obtain the desired estimate (2.6.22) with k = j0 . Observe that the assumption V1 , V2 ∈ C0n+2κ+2 (Rn ) controls the constant C(V ) in (2.6.23) for the potential V (t, x) = V1 (x) + V2 (x − t~e1 ). 72 Chapter 3 Schrödinger Operators with Lamé Potentials on R Even though the dispersive properties of Schrödinger operators have been extensively studied for potentials vanishing at infinity, they are little known in the case that the 2 d potential is periodic in space. Assuming V (x) = V (x + 2ω), − dx 2 + V (x) is called Hill’s operator. The spectrum of Hill’s operator is purely continuous and a union of infinitely many intervals (bands), generically. A potential V (x) is called a finite band 2 d potential if the spectrum of − dx 2 + V (x) is a union of finitely many intervals, one of which is a half axis. One example of finite band potentials is the so-called Lamé 2 d The Laḿe potential. The corresponding − dx 2 + V (x) is called a Lamé operator. operator has a rich history (see [39]) and the properties of its eigenfunctions are still of interest for current research. In this chapter, we explore the dispersive property of Schrödinger operator with a Lamé potential. We assume that ψ0 ∈ L1 (R) and denote U (t)ψ0 as the solution of the following problem d2 1 ∂t ψ(x, t) = − 2 ψ(x, t) + 2℘(x + ω3 )ψ(x, t), i dx ψ(x, 0) = ψ0 (x). (3.0.1) Theorem 3.0.6. Generically, for almost all ω, ω 0 ∈ R, there exists a constant C > 0 such that for t > 1 73 1 kU (t)ψ0 kL∞ (R) < C t− 3 kψ0 kL1 (R) . (3.0.2) Moreover, for all nonzero ω, ω 0 ∈ R, there exists a constant C > 0 such that for t > 1 1 kU (t)ψ0 kL∞ (R) < C t− 4 kψ0 kL1 (R) . (3.0.3) (3.0.2) is optimal in the sense that for any nonzero ω, ω 0 ∈ R, there exist constants c > 0 and T > 0, depending only on ω, ω 0 such that for t > T sup 1 kU (t)ψ0 kL∞ (R) > c t− 3 . (3.0.4) ψ0 :kψ0 kL1 (R) =1 1 We require t > 1 to be large only to exclude t → 0. The decay rates t− 3 and 1 1 t− 4 are different from t− 2 in (1.0.4) because phase function is nonquadratic, which 1 is a natural outcome of the periodic potential. The decay factor t− 3 as t → ∞ has appeared in the analysis of the Modified KdV equation ([8]), where the nonlinear phase of the main term is cubic. In our case, the analytic phase function, roughly speaking, satisfies a cubic relation up to a change of variables. This cubic relation comes from the differential equations satisfied by the Weierstrass ℘ function. We denote P (x) to be the real-coefficient cubic polynomial 2x3 + 6ζ(ω) 2 g2 g2 ζ(ω) x + x + g3 − . ω 2 2ω (3.0.5) We shall prove (3.0.2) under the assumption that P (x) has no double root in (−∞, ℘(ω3 )]. (3.0.6) If (3.0.6) does not hold, then we shall prove (3.0.3). In this case, by Lemma 3.1.2, P (x) has no root of degree 3. Our proof implies that (3.0.3) is optimal in the sense stated in Theorem 3.0.6. However, we are unable to give an explicit example such that P (x) does have a double root in (−∞, ℘(ω3 )]. Finally, we prove that assumption (3.0.6) holds for almost all ω, ω 0 ∈ R. 74 3.1 Preliminaries Eigenfunctions of (1.2.2) are expressed in terms of the Weierstrass σ-function and ζ-function ([25], [14]) as follows: fa (x) = σ(x + iω 0 + a) −ζ(a)x−ζ(iω0 )a e , σ(x + iω 0 ) (3.1.1) where the energy E = −℘(a). (3.1.1) can be verified by noticing that ([39]) fa0 (x) = (ζ(x + ω3 + a) − ζ(x + ω3 ) − ζ(a))fa (x), (3.1.2) and (ζ(x + y) − ζ(x) − ζ(y))2 = ℘(x + y) + ℘(x) + ℘(y). Some basic properties of Weierstrass functions are listed in the appendix. fa is periodic when a is one of the half periods ω1 , ω2 = ω1 + ω3 or ω3 . f−a and fa are the two Floquet-type solutions of (1.2.2). We write fa (x) = ma (x)eik(a)x , where ma (x) = σ(x + iω 0 + a) −aζ(iω0 )−a x ζ(ω) ω e σ(x + iω 0 ) is periodic with period 2ω. Denote Σ = [−℘(ω1 ), −℘(ω2 )] ∪ [−℘(ω3 ), +∞), (3.1.3) 75 and the quasimomentum k(a) = iω −1 (ωζ(a) − aζ(ω)) is real-valued for E ∈ Σ. fa is bounded when E ∈ Σ and is unbounded otherwise, which implies that Σ is the spectrum of (1.2.2) ([27]). It is known that U (t) is an integral operator with kernel Z 0 eitE Pa.c. (E, x, x0 )dE. K(t, x, x ) = Σ Namely, Z Z eitE Pa.c. (E, x, x0 )ψ0 (x0 )dx0 dE. ψ(x, t) = Σ R The absolutely continuous spectral projection is Pa.c. (E, x, x0 ) = 1 [(H − (E + i0))−1 (x, x0 ) − (H − (E − i0))−1 (x, x0 )], 2πi and by definition (H − (E ± i0))−1 = lim+ (H − (E ± i²))−1 , ²→0 which can be expressed by fa and f−a . Hence, we obtain for x > x0 Z 0 eitE (f−a (x0 )fa (x) + f−a (x)fa (x0 )) K(t, x, x ) = ZΣ 0 dE W (E) 0 eitE (eik(a)(x−x ) m−a (x0 )ma (x) + e−ik(a)(x−x ) m−a (x)ma (x0 )) = Σ dE , W (E) (3.1.4) 0 − fa0 f−a , called the Wronskian of fa , f−a , where W (E) = W (a) = W (fa , f−a ) = fa f−a is independent of x. Because the spectral projection Pa.c. is self-adjoint, Pa.c. (E, x, x0 ) = Pa.c. (E, x0 , x). 76 Therefore, when x < x0 Z 0 eitE (f−a (x0 )fa (x) + f−a (x)fa (x0 )) K(t, x, x ) = Σ dE . W (E) (3.1.5) The proof of (3.0.2) and (3.0.3) shall be reduced to proving that 1 1 sup |K(t, x, x0 )| < Ct− 3 and Ct− 4 . x,x0 It follows from (3.1.2) that W (fa , f−a ) = fa (x)f−a (x)(ζ(x + ω3 − a) + 2ζ(a) − ζ(x + ω3 + a)). Since W (fa , f−a ) is independent of x, we set x = 0 and obtain W (fa , f−a ) = fa (0)f−a (0)(ζ(ω3 − a) + 2ζ(a) − ζ(ω3 + a)). By the addition formula for Weierstrass functions ([2], §15) ζ(u + v) − ζ(u − v) − 2ζ(v) = − ℘0 (v) , ℘(u) − ℘(v) we have W (E) = σ(iω 0 + a)σ(iω 0 − a) ℘0 (a) . σ 2 (iω 0 ) ℘(iω 0 ) − ℘(a) (3.1.6) Hence, dE σ 2 (iω 0 )(℘(iω 0 ) − ℘(a)) =− da. W (E) σ(iω 0 + a)σ(iω 0 − a) (3.1.7) Remark 3.1.1. Because ℘0 (iω 0 ) = 0 and zeroes of σ are the lattice points {n1 2ω1 + 0 )−℘(a) n2 2ω3 : n1 , n2 ∈ Z}, all of which are of degree 1, it follows that σ(iω℘(iω 0 +a)σ(iω 0 −a) is bounded and smooth when a → iω 0 . 0 0 )−℘(a) )−℘(a) −2 ) when a → 0, and that σ(iω℘(iω Also, it is clear that σ(iω℘(iω 0 +a)σ(iω 0 −a) = O(a 0 +a)σ(iω 0 −a) 77 and its ∂a -derivatives are bounded on [ω1 , ω2 ], where [ω1 , ω2 ] denotes the set {λω1 + (1 − λ)ω2 : λ ∈ [0, 1]}. By (3.1.3), ma , m−a , and their ∂a -derivatives are bounded uniformly for a ∈ [ω1 , ω2 ] ∪ [0, ω3 ], x ∈ R. Since Σ is a union of two intervals, we shall decompose the integral of K(t, x, x0 ) into two parts. Namely, K(t, x, x0 ) = K1 (t, x, x0 ) + K2 (t, x, x0 ), where Z −℘(ω2 ) 0 K1 (t, x, x ) = eitE Pa.c. (E, x, x0 )dE, −℘(ω1 ) 0 Z +∞ K2 (t, x, x ) = eitE Pa.c. (E, x, x0 )dE. −℘(ω3 ) Before we proceed to analyze K1 (t, x, x0 ) and K2 (t, x, x0 ), we prove two technical lemmas. Lemma 3.1.1. Let F (x) be a real-valued and smooth function on (a, b), 1. Suppose |F 0 (x)| ≥ ², |F 00 (x)| ≤ M for all x ∈ (a, b), then Z b ¯Z b ¯ h i ¯ ¯ −itF (x) −1 −1 e ψ(x)dx¯ ≤ c ² |t| |ψ(b)| + (|ψ 0 (x)| + |ψ(x)|)dx , ¯ a a where c depends on M . 2. Suppose k ≥ 2, k ∈ Z, and |F (k) (x)| ≥ ² for all x ∈ (a, b), then Z b ¯ ¯Z b h i ¯ ¯ − k1 −itF (x) − k1 0 e ψ(x)dx¯ ≤ c ² |t| |ψ(b)| + |ψ (x)|dx , ¯ a where c depends on k. a 78 The first part of Lemma 3.1.1 follows from integration by parts. The second part is proved in [36] (p.334). Lemma 3.1.2. Let ej = ℘(ωj ), j = 1, 2, 3. Then P (x) has a unique simple root in [e2 , e1 ], and P (ej ), j = 1, 2, 3, are nonzero. Also P (x) has no root of degree 3 in R. Moreover, − ζ(ω) ∈ (e3 , e2 ). ω Proof. P (x) = 0 if and only if 4x3 − g2 x − g3 = (6x2 − g22 )(x + ζ(ω) ). Denote p1 (x) = ω 4x3 − g2 x − g3 and p2 (x) = (6x2 − g22 )(x + ζ(ω) ). We shall examine the roots of p1 (x) ω and p2 (x) on the real line. It follows from (3.5.2) that p1 (x) = 4(x − e1 )(x − e2 )(x − e3 ), where ej = ℘(ωj ), j = 1, 2, 3. Because there is no quadratic term in p1 (x), e1 + e2 + e3 = 0. Since e3 < e2 < e1 , we have e3 < 0 < e1 . Observe that ℘00 (ω1 ) > 0, ℘00 (ω2 ) < 0 and ℘00 (ω3 ) > 0, and by Eq (1.2.4), we obtain ℘(ω2 )2 < g2 < min{℘(ω1 )2 , ℘(ω3 )2 }. 12 Now we shall prove ζ(ω) ∈ (−e2 , −e3 ). Indeed, let y1 (x, E) and y2 (x, E) be the ω solutions of (1.2.2), which satisfy y1 (0, E) = y20 (0, E) = 1, y10 (0, E) = y2 (0, E) = 0. And we introduce the discriminant ∆(E) = y1 (2ω, E) + y20 (2ω, E). Recall a and E are related by E = −℘(a), and as E → +∞ on the real line, a → 0 on the positive imaginary axis. Therefore iζ(a) and k(a) go to +∞ on the real line when E → +∞. By Lemma 2.1 of [12], ∆(E) = 2 cos k(a) and k(E) = k(a(E)) is the conformal map from the upper half plane to a slit quarter plane Ω = {<z > 0, =z > 0}\T , with π the slit T = { 2ω + iy : 0 < y ≤ h}, where h is some positive real number. Moreover, π k(−e1 ) = 0 and k(−e2 ) = k(−e3 ) = 2ω . π Denote Q0 to be the preimage of the tip 2ω + ih of the slit T under the map k(E). 79 π π Then −℘(ω2 ) < Q0 < −℘(ω3 ), and k(E) sends [−℘(ω2 ), Q0 ] to [ 2ω , 2ω + ih], and π π [Q0 , −℘(ω3 )] to [ 2ω + ih, 2ω ], respectively. Thus when E ∈ (−℘(ω2 ), Q0 ), 1i ∂E k(E) ≥ 0. We observe that ∂E k(E) = ´ 1 i ³ ζ(ω) ∂ k(a) = + ℘(a) , a −℘0 (a) ℘0 (a) ω which implies that ℘(a) + ζ(ω)/ω ≥ 0. ℘0 (a) Since ℘0 (a) > 0 when a ∈ (ω3 , ω2 ), we conclude that E = −℘(a) ≤ ζ(ω)/ω for any E ∈ (−℘(ω2 ), Q0 ). Hence, Q0 ≤ ζ(ω)/ω. On the other hand, 1i ∂E k(E) ≤ 0 when E ∈ (Q0 , −℘(ω3 )). Following a similar argument, Q0 ≥ ζ(ω)/ω. Therefore ζ(ω)/ω = Q0 ∈ (−℘(ω2 ), −℘(ω3 )). π In fact, k(E) maps E = ζ(ω) to the tip 2ω + ih of the slit T and ∆(E) reaches its ω . minimum at E = ζ(ω) ω p g2 and − ζ(ω) . From the above analysis, we The three roots of p2 (x) are ± 12 ω p g2 p g2 ζ(ω) have that 12 ∈ (e2 , e1 ) and − 12 , − ω ∈ (e3 , e2 ), which implies p2 (e1 ) > 0 and p2 (e2 ) < 0. Hence P (x) has either one or three zeroes in (e2 , e1 ) and clearly P (ej ), j = 1, 2, 3, are nonzero. To verify that P (x) has no root of degree 3, we consider P 0 (x) = 6x2 + 12 ζ(ω) g2 x+ . ω 2 ∈ (e3 , e2 ) and is equal to g22 − 6( ζ(ω) )2 . The minimum of P 0 (x) is reached at x = − ζ(ω) ω ω p g2 Notice that − ζ(ω) < e2 < 12 always holds. ω p g2 If − ζ(ω) > − 12 , then P 0 (x) > 0 holds for all x ∈ R. P (x) has no root of degree ω greater or equal to 2. p g2 If − ζ(ω) = − , then P 0 (x) has a double root − ζ(ω) ∈ (e3 , e2 ). Since P (x) has ω 12 ω a root in (e2 , e1 ), we conclude that P (x) has no root of degree 3. p g2 If − ζ(ω) < − 12 , then P 0 (x) has no double root. Hence P (x) has no root of ω 80 degree 3 on the whole real line. If P (x) has three zeroes in (e2 , e1 ), then P 0 (x) has two roots in (e2 , e1 ), which is impossible because − ζ(ω) < e2 . Therefore, P (x) has unique simple root in (e2 , e1 ). ω 3.2 Analysis of K1(t, x, x0) We first consider K1 (t, x, x0 ). We proceed by making the following observation: Lemma 3.2.1. Let b = 2ω2 − a for a ∈ [ω1 , ω2 ]. Write W (a) = W (fa , f−a ). Then for x, x0 ∈ R fa (x0 )f−a (x) fb (x)f−b (x0 ) =− . W (a) W (b) (3.2.1) Proof. It is clear that ℘(a) = ℘(b) and ℘0 (a) = −℘0 (b). We prove (3.2.1) by direct calculation. By definition, fa (x0 )f−a (x) = σ(x0 + ω3 + a)σ(x + ω3 − a) ζ(a)(x−x0 ) e . σ(x + ω3 )σ(x0 + ω3 ) By (3.5.4) and (3.5.5), this equals σ(x0 + ω3 − b)σ(x + ω3 + b) ζ(b)(x0 −x) 4η3 (ω3 −b) e e = fb (x)f−b (x0 )e4η3 (ω3 −b) . σ(x + ω3 )σ(x0 + ω3 ) Also by (3.1.6) and (3.5.5), W (a) = σ(iω 0 − b)σ(iω 0 + b) exp (4η3 (ω3 − b)) −℘0 (b) = −W (b)e4η3 (ω3 −b) . 2 0 σ (iω ) ℘(ω3 ) − ℘(b) Combining them, (3.2.1) follows. 81 It follows from Lemma 3.2.1 that Z ω2 0 −it℘(a) fa (x )f−a (x) e W (a) ω1 Z ω2 +iω0 d℘(a) = e−it℘(b) ω2 fb (x)f−b (x0 ) d℘(b). W (b) Hence we have that for x > x0 Z −℘(ω2 ) 0 K1 (t, x, x ) = eitE (f−a (x0 )fa (x) + f−a (x)fa (x0 )) −℘(ω1 ) Z ω1 +i2ω0 = e−it℘(a) ω1 Z ω1 +i2ω0 = dE W (E) fa (x)f−a (x0 ) d(−℘(a)) W (a) 0 e−it℘(a)+i(x−x )k(a) ma (x)m−a (x0 ) ω1 −℘0 (a)da . W (a) (3.2.2) Note k(a) is real-valued and by (3.1.5), we have that for x < x0 0 Z ω1 +i2ω0 K1 (t, x, x ) = 0 e−it℘(a)+i(x−x )k(a) ma (x0 )m−a (x) ω1 −℘0 (a)da . W (a) 0 ∈ R and To simplify notation, we set τ = x−x t Fτ (a) = ℘(a) − iτ (ζ(a) − a ζ(ω)). ω Moreover, we write 0 Z ω1 +i2ω0 K1 (t, x, x ) = e−itFτ (a) ϕ(a, x, x0 )da, (3.2.3) ω1 0 (a) where ϕ(a, x, x0 ) = ma (x)m−a (x0 ) −℘ when x > x0 , and ϕ(a, x, x0 ) = ϕ(a, x0 , x) W (a) when x < x0 . Without losing clarity, ϕ(a, x, x0 ) will be written simply as ϕ(a). By Remark 3.1.1, ϕ(a, x, x0 ) and its ∂a -derivatives are bounded uniformly for a ∈ [ω1 , ω2 ] and x, x0 ∈ R. To apply Lemma 3.1.1 to (3.2.3), we analyze the ∂a -derivatives of Fτ (a). Our plan is to decompose the integral in (3.2.3) into several regions and on each region, Lemma 3.1.1 for some exponent k will be applied. We observe that 82 ∂a Fτ (a) = ℘0 (a) + τ i(ζ(ω)/ω + ℘(a)), (3.2.4) ∂a2 Fτ (a) = ℘00 (a) + τ i℘0 (a), (3.2.5) ∂a3 Fτ (a) = ∂a3 ℘(a) + τ i℘00 (a). (3.2.6) Let c1 = min{ζ(ω)/ω + ℘(a) : a ∈ [ω1 , ω1 + 2iω 0 ]}. Then c1 = ζ(ω)/ω + ℘(ω2 ) and by Lemma 3.1.2, c1 > 0. Also we denote M1 = 1 + max{|℘0 (a)|, |℘00 (a)|, ζ(ω)/ω + ℘(a) : a ∈ [ω1 , ω1 + 2iω 0 ]}. (3.2.7) 1 When |τ | > 2M , we have for a ∈ [ω1 , ω1 + 2iω 0 ] c1 1 |∂a Fτ (a)| > |τ |c1 − M1 > |τ |c1 , 2 and |∂a2 Fτ (a)| < M1 (|τ | + 1). Integrating by parts and recalling ϕ(a) and its derivatives are uniformly bounded, we obtain Z ω1 +i2ω0 ¯ ϕ(a) ¯ d e−itFτ (a) ¯ ∂a Fτ (a) ω1 Z 0 ¯ ³ 0 ω 1 +i2ω 4ω 0 kϕ(a)kL∞ [ω1 ,ω2 ] 1 ¯ ϕ(a)Fτ00 (a) ´ ¯¯ −itFτ (a) ϕ (a) ≤ − da¯ + ¯ e tτ c1 t ω1 Fτ0 (a) (Fτ0 (a))2 1 ¯¯ |K1 (t, x, x )| = ¯ t 0 ≤Ct−1 . (3.2.8) 83 1 We now estimate K1 (t, x, x0 ) when |τ | ≤ 2M . Suppose both (3.2.4) and (3.2.5) c1 1 2M1 , c1 ]. Then vanish for a = a0 ∈ [ω1 , ω2 ] and τ = τ0 ∈ [− 2M c1 ℘0 (a0 )2 = ℘00 (a0 )( ζ(ω) + ℘(a0 )). ω By (1.2.3) and (1.2.4), this is equivalent to 2℘(a0 )3 + 6ζ(ω) g2 g2 ζ(ω) ℘(a0 )2 + ℘(a0 ) + g3 − = 0. ω 2 2ω Thus ℘(a0 ) is the simple root of P (x) in [℘(ω2 ), ℘(ω1 )] and ℘0 (a0 ) 6= 0 by Lemma 3.1.2. Observe that (3.2.4) and (3.2.5) also vanish when (a, τ ) = (2ω2 − a0 , −τ0 ). The analysis of (2ω2 − a0 , −τ0 ) is the same as that of (a0 , τ0 ) and we will focus on (a0 , τ0 ). Also, we observe that ∂a3 Fτ (a) vanishes at (a0 , τ0 ) if and only if  det   ℘0 (a0 ) ∂a3 ℘(a0 ) ζ(ω) + ℘(a0 ) ω  ℘00 (a0 ) = 0; namely,  ∂a det   0 ℘ (a) ℘00 (a) ζ(ω) + ℘(a) ω  ℘0 (a) = 0. a=a0 Since ℘0 (a0 ) 6= 0, that ∂a3 Fτ0 (a0 ) = 0 is equivalent to the fact that ℘(a0 ) is a double root of P (x). By Lemma 3.1.2, P (x) has no double root in [℘(ω2 ), ℘(ω1 )]. Hence, ∂a3 Fτ0 (a0 ) 6= 0 and there exists ² > 0 such that min{Σ3j=1 |∂aj Fτ (a)| : a ∈ [ω1 , ω1 + 2iω 0 ], τ ∈ [−2M1 /c1 , 2M1 /c1 ]} > ² > 0. 1 2M1 Let χ3 (a, τ ) be a smooth function defined on [ω1 , ω1 + 2iω 0 ] × [− 2M , c1 ] such c1 that 0 ≤ χ3 ≤ 1, χ3 (a, τ ) = 1 when |∂a Fτ (a)| + |∂a2 Fτ (a)| ≤ 13 ², and χ3 (a, τ ) = 0 when |∂a Fτ (a)| + |∂a2 Fτ (a)| ≥ 32 ². Similarly, let χ2 (a, τ ) to be a smooth function defined on 1 2M1 [ω1 , ω1 + 2iω 0 ] × [− 2M , c1 ], such that 0 ≤ χ2 ≤ 1, χ2 (a, τ ) = 1 when |∂a2 Fτ (a)| ≥ 61 ², c1 84 and χ2 (a, τ ) = 0 when |∂a2 Fτ (a)| ≤ 19 ². On the support of χ3 , |∂a3 Fτ (a)| ≥ 13 ². It follows from Lemma 3.1.1 that ¯ Z ω1 +i2ω0 ¯ 1 ¯ ¯ e−itFτ (a) χ3 (a, τ )ϕ(a)da¯ ≤ C3 (τ )t− 3 . ¯ (3.2.9) ω1 On the support of χ2 (1 − χ3 ), |∂a2 Fτ (a)| ≥ 19 ². And similarly ¯ ¯ Z ω1 +i2ω0 1 ¯ ¯ e−itFτ (a) χ2 (a, τ )(1 − χ3 (a, τ ))ϕ(a)da¯ ≤ C2 (τ )t− 2 . ¯ (3.2.10) ω1 On the support of (1−χ2 )(1−χ3 ), |∂a2 Fτ (a)| ≤ 16 ² and |∂a Fτ (a)| ≥ 16 ². Lemma 3.1.1 yields ¯ Z ω1 +i2ω0 ¯ ¯ ¯ −itFτ (a) e (1 − χ2 (a, τ ))(1 − χ3 (a, τ ))ϕ(a)da¯ ≤ C1 (τ )t−1 . ¯ (3.2.11) ω1 Note that Cj (τ ), j = 1, 2, 3, are continuous functions of τ ∈ [−2M1 /c1 , 2M1 /c1 ]. Let C = Σ3j=1 max{Cj (τ ) : τ ∈ [−2M1 /c1 , 2M1 /c1 ]}. 1 Then |K1 (t, x, x0 )| ≤ Ct− 3 for large t, because χ3 + χ2 (1 − χ3 ) + (1 − χ2 )(1 − χ3 ) = 1. Consequently, we have proven that for large t ¯ ¯ 1 sup ¯K1 (t, x, x0 )¯ < Ct− 3 , x,x0 where C only depends on ω, ω 0 . (3.2.12) 3.3 85 0 Analysis of K2(t, x, x ) We now consider K2 (t, x, x0 ). Let b = 2ω3 − a for a ∈ (0, ω3 ), and the proof of Lemma 3.2.1 gives fa (x0 )f−a (x) fb (x)f−b (x0 ) =− . W (a) W (b) Then for x > x0 Z +∞ 0 eitE (f−a (x0 )fa (x) + f−a (x)fa (x0 )) K2 (t, x, x ) = −℘(ω3 ) Z +∞ 0 −it℘(a) fa (x)f−a (x ) = e W (a) −℘(ω3 ) Z i2ω0 −itFτ (a) e = 0 dE W (E) d(−℘(a)) −℘0 (a) ma (x)m−a (x ) da, W (a) 0 0 where τ = x−x . For x < x0 , K2 (t, x, x0 ) can be written in a similar form. Therefore t 0 Z i2ω0 K2 (t, x, x ) = e−itFτ (a) ϕ(a, x, x0 )da, 0 where ϕ(a, x, x0 ) was defined in the previous section. Step 1. The analysis of the nonlinear phase in K2 (t, x, x0 ) is similar to that of K1 (t, x, x0 ). However, by Remark 3.1.1, ϕ(a, x, x0 ) in K2 (t, x, x0 ) is unbounded when a → 0 and a → 2iω 0 , contrary to the case of K1 (t, x, x0 ). Our strategy then is to change variables to remove this singularity. Define λ2 = ℘(ω3 ) − ℘(a) such that λ > 0 when a ∈ (0, ω3 ) and λ < 0 when a ∈ (ω3 , 2ω3 ). Then the map a → λ is one-to-one, onto and analytic from (0, 2ω3 ) to R. Note that λ(2iω 0 − a) = −λ(a) and the behavior of λ(a) as a → 2iω 0 is the same as that when a → 0. ∂a We claim that ∂λ = −℘2λ0 (a) is never zero when a ∈ (0, 2ω3 ). In fact, the claim is obvious for a 6= ω3 . When a = ω3 , by L’Hôpital’s rule, 86 ∂a 2λ 2 (0) = lim = lim , 0 λ→0 −℘ (a) λ→0 −℘00 (a) ∂a ∂λ ∂λ which implies that ¯ ∂a ¯ ¯ ¯ ¯ (0)¯ = ∂λ s 2 ℘00 (ω3 ) > 0. Observe that when λ → ±∞, λϕ(λ) = λ3 · O(1) and | − ℘0 (a)| = |λ|3 + O(λ2 ). λϕ(λ) 0 Hence, −℘ 0 (a) and its λ-derivatives are bounded uniformly for x, x , λ ∈ R. After changing the variables, we obtain Z i2ω0 Z e −itFτ (a) 0 e−itFτ (λ) ϕ(λ) ϕ(a)da = R ∂a dλ, ∂λ (3.3.1) where Fτ (λ) = Fτ (a(λ)) and ϕ(λ) = ϕ(a(λ)). We will decompose (3.3.1) into different integral regions and estimate them separately. Define χ(·) to be a smooth function supported in (−2, 2) such that χ(x) = 1 when x ∈ [−1, 1], and let M be a large number to be specified. Step 2. We claim that ¯Z ¯ ¯ e−itFτ (λ) ϕ(λ) R ¯ 2λ ¯ − 41 − 13 χ(λ/M )dλ < C t or C t , ¯ M M −℘0 (a) (3.3.2) depending on whether P (x) has a double root in (−∞, ℘(ω3 )] or not. The proof of (3.3.2) will follow the lines of the proof of (3.2.12). Recall that the map λ → a is one-to-one from R onto (0, 2ω3 ), and satisfies λ2 = ℘(ω3 ) − ℘(a). Also, we observe that ∂a 0 (℘ (a) + τ i(ζ(ω)/ω + ℘(a))), ∂λ ³ ∂a ´2 ∂2a 0 (℘ (a) + τ i(ζ(ω)/ω + ℘(a))) + (℘00 (a) + τ i℘0 (a)). ∂λ2 Fτ (λ) = ∂λ2 ∂λ ∂λ Fτ (λ) = (3.3.3) (3.3.4) 87 By Lemma 3.1.2, inf{|ζ(ω)/ω + ℘(a)| : a ∈ (0, ω3 )} = |ζ(ω)/ω + ℘(ω3 )| = c2 > 0. Denote M2 = max{|℘0 (a)| + |℘00 (a)| + |ζ(ω)/ω + ℘(a)| : λ(a) ∈ [−2M, 2M ]}. ∂a ∂a Since ∂λ is smooth and never zero, there exist c3 and M3 such that 0 < c3 < | ∂λ |< √ ∂2a M3 for all λ ∈ [−2M, 2M ]. Moreover, suppose | ∂λ 2 | < M3 for λ ∈ [−2M, 2M ]. Then for λ ∈ [−2M, 2M ] and |τ | ≥ 2M2 /c2 1 |∂λ Fτ (λ)| > c2 c3 |τ |, 2 |∂λ2 Fτ (λ)| < 2M3 M2 (1 + |τ |). Integrating by parts, an argument similar to (3.2.8) shows that for |τ | ≥ 2M2 /c2 , ¯Z ¯ ¯ e−itFτ (λ) ϕ(λ) R ¯ 2λ ¯ χ(λ/M )dλ¯ < CM t−1 . 0 −℘ (a) To prove (3.3.2) for |τ | ≤ 2M2 /c2 , we first suppose that P (x) has no double root ∂a ∂Fτ (a) ∂a τ (λ) = ∂λ and ∂λ 6= 0 for a ∈ (0, 2ω3 ), it follows from in (−∞, ℘(ω3 )]. Since ∂F∂λ ∂a 2 Fτ (λ) τ (λ) (3.3.3) and (3.3.4) that ∂F∂λ and ∂ ∂λ vanish at (λ0 , τ0 ) if and only if (3.2.4) and 2 (3.2.5) vanish at (a0 , τ0 ), where λ20 = ℘(ω3 ) − ℘(a0 ). Therefore, the fact that P (x) has no double root implies that there exists ² such that min 3 nX o |∂λj Fτ (λ)| : |λ| ≤ 2λ|τ | < 2M2 /c2 > ² > 0. j=1 Just as in the case of K1 (t, x, x0 ), we define χ2 (λ, τ ), χ3 (λ, τ ) on [−2M, 2M ] × [−2M2 /c2 , 2M2 /c2 ]. Namely, χ2 (λ, τ ) = 1 when |∂λ2 Fτ (λ)| > 61 ², and χ2 (λ, τ ) = 0 when |∂λ2 Fτ (λ)| < 19 ². χ3 (λ, τ ) = 1 when |∂λ Fτ (λ)| + |∂λ2 Fτ (λ)| < 31 ², and χ3 (λ, τ ) = 0 when |∂λ Fτ (λ)| + |∂λ2 Fτ (λ)| > 32 ². 88 Decompose the integral in (3.3.2) according to 1 = χ3 + χ2 (1 − χ3 ) + (1 − χ2 )(1 − χ3 ). The same arguments as those in (3.2.9), (3.2.10), and (3.2.11) yield for |τ | ≤ 2M2 /c2 , ¯Z ¯ ¯ e−itFτ (λ) ϕ(λ) R ¯ 2λ ¯ − 13 χ(λ/M )dλ < C t . ¯ M −℘0 (a) In the case that P (x) has a double root ℘(a0 ) ∈ (−∞, ℘(ω3 )], there is τ0 ∈ R, such that ∂λj Fτ (λ), j = 1, 2, 3, vanish at (λ0 , τ0 ), where λ20 = ℘(ω3 ) − ℘(a0 ). The fact that P (x) has no root of degree 3 implies that ∂λ4 Fτ0 (λ0 ) 6= 0. Therefore, there exists ² such that min 4 nX o |∂λj Fτ (λ)| : λ ∈ [−2M, 2M ], τ ∈ [−2M2 /c2 , 2M2 /c2 ] > 2² > 0. j=1 Define smooth function χ4 (λ, τ ) : [−2M, 2M ] × [−2M2 /c2 , 2M2 /c2 ] → [0, 1], such that χ4 = 1 when Σ3j=1 |∂λj Fτ (λ)| ≤ ², and χ4 = 0 when Σ3j=1 |∂λj Fτ (λ)| ≥ 23 ². Hence on the support of χ4 , |∂λ4 Fτ (λ)| ≥ 21 ². It follows from Lemma 3.1.1 that ¯Z ¯ ¯ e−itFτ (λ) ϕ(λ) R ¯ 1 2λ ¯ χ(λ/M )χ4 (λ, τ )dλ¯ < C4 (τ )t− 4 . 0 −℘ (a) We decompose the integral in (3.3.2) by using χ4 + (1 − χ4 )χ3 + χ2 (1 − χ3 )(1 − χ4 ) + (1 − χ2 )(1 − χ3 )(1 − χ4 ) = 1. The analysis of the terms containing (1 − χ4 )χ3 , χ2 (1 − χ3 )(1 − χ4 ), and (1 − χ2 )(1 − χ3 )(1 − χ4 ) is similar to (3.2.9), (3.2.10), and (3.2.11) respectively. Therefore, under the assumption that P (x) has a double root in (−∞, ℘(ω3 )], we have proven that 89 ¯Z ¯ ¯ e−itFτ (λ) ϕ(λ) R ¯ 1 2λ ¯ χ(λ/M )dλ ¯ < C M t− 4 . 0 −℘ (a) Step 3. It now remains to estimate Z e−itFτ (λ) ϕ(λ) R 2λ (1 − χ(λ/M ))dλ, −℘0 (a) which by definition equals Z e−itFτ (λ) ϕ(λ) lim N →+∞ R 2λ (χ(λ/N ) − χ(λ/M ))dλ. −℘0 (a) (3.3.5) Since (3.3.5) are not integrable on the support of 1 − χ(λ/M ), Lemma 3.1.1 cannot be applied to (3.3.5) directly. We shall explore the oscillation of the phase e−itFτ (λ) and perform integration by parts to bound (3.3.5), which requires us to exclude the zeroes of ∂λ Fτ (λ). 1 g2 a2 + O(a4 ), and ℘(a) = ℘(iω 0 ) − λ2 , hence By definition, ℘(a) = a−2 + 20 λ(a) = i + α1 a + O(a3 ) as a → 0, a ∈ (0, ω3 ), a which is an meromorphic function of a. It follows that ζ(a) = −iλ + O(λ−1 ) as a → 0, a ∈ (0, ω3 ). Consequently τ Fτ (λ) = −λ2 − τ λ + ℘(iω 0 ) + O( ), λ ∂λ Fτ (λ) = −2λ − τ + O(τ λ−2 ), ∂λ2 Fτ (λ) = −2 + O(τ λ−3 ), λ → ±∞; λ → ±∞; λ → ±∞. We require M large enough such that o n ∂Fτ (λ) ∂ 2 Fτ (λ) , both vanish at (τ , λ ) . M > 1 + max |λ0 | : 0 0 ∂λ ∂λ2 2 Fτ (λ) τ (λ) Therefore, if |λ| > M , ∂F∂λ and ∂ ∂λ cannot vanish at the same (τ, λ). 2 (3.3.6) (3.3.7) (3.3.8) 90 When |λ| > M and |λ + τ2 | > 1, we claim that τ 1 |∂λ Fτ (λ)| > |λ + | − . 2 2 (3.3.9) In fact, by (3.3.7) τ |∂λ Fτ (λ)| > 2|λ + | − O(τ λ−2 ). 2 |τ | |τ | We choose M large enough such that O(τ λ−2 ) < 100|λ| . If |λ + τ2 | > 100 , (3.3.9) clearly |τ | |τ | holds. If |λ + τ2 | ≤ 100 , then 100|λ| < 12 and (3.3.9) also follows. When (− τ2 − 1, − τ2 + 1) is not contained in (−M, M ), by (3.3.8), |∂λ2 Fτ (λ)| > 1 for λ ∈ (− τ2 − 2, − τ2 + 2) as long as M is large enough. By Lemma 3.1.1, we have Z e−itFτ (λ) ϕ(λ) R 1 2λ χ(λ + τ /2)dλ < Ct− 2 , 0 −℘ (a) where χ(x) = 1 when |x| < 1, and χ(x) = 0 when |x| > 2. To estimate (3.3.5), we first consider the case when |τ10| > M and estimate Z e−itFτ (λ) R 2λϕ(λ) (χ(10λ/|τ |) − χ(λ/M ))dλ, −℘0 (a) which equals 1 it Z −itFτ (λ) d e R ³ 2λϕ(λ) χ(10λ/|τ |) − χ(λ/M ) ´ dλ −℘0 (a) ∂λ F (λ) dλ. (3.3.10) It follows from (3.3.7) that on the support of χ(10λ/|τ |) − χ(λ/M ) |∂λ Fτ (λ)| > |τ | − 2|λ| − O(τ λ−2 ) > |τ |/2, and |∂λ2 Fτ (λ)| < |τ |/4, as long as M is large enough. Hence |∂λ (∂λ Fτ (λ))−1 | < 1 . |τ | (3.3.11) 91 Since (χ(10λ/|τ |) − χ(λ/M )), 2λϕ(λ) and their λ-derivatives are uniformly bounded, −℘0 (a) we have ¯ d ³ 2λϕ(λ) χ(10λ/|τ |) − χ(λ/M ) ´¯ C ¯ ¯ , ¯ ¯< 0 dλ −℘ (a) ∂λ F (λ) |τ | from which it follows that |(3.3.10)| < Ct−1 . To complete the estimate on (3.3.5) when |τ |/10 > M , it remains to bound Z e−itFτ (λ) ϕ(λ) R 2λ (χ(λ/N ) − χ(10λ/|τ |) − χ(λ + τ /2))dλ. −℘0 (a) Integrating by parts, this equals 2 it Z e−itFτ (λ) R d ³ λϕ(λ) (χ(λ/N ) − χ(10λ/τ ) − χ(λ + τ /2)) ´ dλ := J1 + J2 , dλ −℘0 (a) ∂λ Fτ (λ) where 2 J1 = it Z e−itFτ (λ) (χ(λ/N ) − χ(10λ/|τ |) − χ(λ + τ /2)) d λϕ(λ) dλ, ∂λ Fτ (λ) dλ −℘0 (a) e−itFτ (λ) λϕ(λ) d (χ(λ/N ) − χ(10λ/|τ |) − χ(λ + τ /2)) dλ. −℘0 (a) dλ ∂λ Fτ (λ) R and 2 J2 = it Z R In J2 , ∂λ (χ(λ/N ) − χ(10λ/|τ |) − χ(λ + τ /2)) = 10 0 1 0 χ (λ/N ) − χ (10λ/|τ |) − χ0 (λ + τ /2). N |τ | (3.3.12) On the support of (3.3.12), we have |λ + τ /2| > 1 and |λ| > M . Thus |(∂λ Fτ (λ))−1 | < C by (3.3.9). Consequently, 92 ¯2 Z ³1 ´ ¯ λϕ(λ) 10 0 ¯ ¯ 0 0 −itFτ (λ) χ (λ/N ) − χ (10λ/|τ |) − χ (λ + τ /2) dλ¯ < Ct−1 . e ¯ 0 it R −℘ (a)∂λ Fτ (λ) N |τ | On the support of χ(λ/N ) − χ(10λ/|τ |) − χ(λ + τ /2) we have |λ| > |τ /10|. It follows from (3.3.8) that |∂λ2 Fτ (λ)| < 3. Combining it with (3.3.9), we obtain τ ∂λ (∂λ Fτ (λ))−1 < C(λ + )−2 , 2 which is integrable. Therefore ¯2 Z ¯ ´ λϕ(λ) ³ 1 ¯ ¯ e−itFτ (λ) ∂ (χ(λ/N ) − χ(10λ/|τ |) − χ(λ + τ /2))dλ ¯ ¯ < Ct−1 . λ 0 it R −℘ (a) ∂λ Fτ (λ) This completes the estimate on J2 . As for J1 , integrating by parts again, we obtain 4 J1 = − 2 t Z e −itFτ (λ) d R ³ (χ(λ/N ) − χ(10λ/|τ |) − χ(λ + τ /2)) d λϕ(λ) ´ dλ (∂λ Fτ (λ))2 dλ −℘0 (a) dλ. d Applying Leibnitz’s rule, we are left with three terms. Two terms come from dλ hitting χ(λ/N ) − χ(10λ/|τ |) − χ(λ + τ /2) and (∂λ Fτ (λ))−2 , and the analysis is analogous to d d λϕ(λ) hits dλ , we obtain the third term: that of J2 . When dλ −℘0 (a) 4 − 2 t Z e−itFτ (λ) R χ(λ/N ) − χ(10λ/|τ |) − χ(λ + τ /2) d2 λϕ(λ) dλ. (∂λ Fτ (λ))2 dλ2 −℘0 (a) Because |(∂λ Fτ (λ))2 | > |λ+τ /2|2 /4 on the support of χ(λ/N )−χ(10λ/|τ |)−χ(λ+τ /2) 2 d λϕ(λ) −2 and dλ , where the 2 −℘0 (a) is uniformly bounded, the above term is dominated by Ct constant C is independent of N . This completes the estimate of (3.3.5) when |τ |/10 > M . The analysis is similar and even simpler when |τ |/10 ≤ M . Therefore, when P (x) has no double root in (−∞, ℘(ω3 )], we have 93 1 sup |K2 (t, x, x0 )| < Ct− 3 . x,x0 1 1 The decay factor t− 3 is replaced by t− 4 when P (x) has a double root in (−∞, ℘(ω3 )]. 2 Combining the estimates on K1 (t, x, x0 ) and K2 (t, x, x0 ), we have proven (3.0.2) under the assumption (3.0.6). We have also proven (3.0.3) for all nonzero ω, ω 0 ∈ R. It remains to prove that (3.0.6) holds for almost all ω, ω 0 ∈ R. Suppose P (x) has a double root x0 ∈ (−∞, ℘(ω3 )]. Then x0 is a root of P 0 (x). Recall that P 0 (x) = 6x2 + with its roots ζ(ω) ± r+ , r− = − ω 12ζ(ω) g2 x+ , ω 2 r ³ ζ(ω) ´2 ω − g2 . 12 That P (x) has a double root in (−∞, ℘(ω3 )] implies that (ζ(ω)/ω)2 − g2 /12 > 0 and P (r− ) = 0. By (3.5.1), g2 and g3 are real analytic for ω, ω 0 ∈ R+ . By (3.5.3), ζ(ω) is also real analytic for ω, ω 0 ∈ R+ . Therefore, r+ , r− are analytic when ω, ω 0 ∈ R+ , ´2 ³ g2 = 0. To prove (3.0.6) for almost all ω, ω 0 ∈ R+ , it − 12 with branches at ζ(ω) ω suffices to show that P (r− ) is nonzero at one point. This can be done by direct numerical calculation. For example, take ω = 5.5 and ω 0 = 2. Then we have g2 = 0.507343, g3 = −0.0695438, r+ , r− = 0.0628169 ± 0.195787i, P (r+ ), P (r− ) = −0.0386656 ± 0.0300201i. This indicates that P (r− ) is nonzero for almost all ω, ω 0 ∈ R. Therefore, (3.0.6) 94 holds for almost all ω, ω 0 ∈ R. 3.4 Optimality of the decay factor So far we have proven the first part of Theorem 3.0.6. To verify (3.0.4), we first reduce it to showing that there exist constants c > 0 and T > 0 such that for t > T 1 kK(t, x, x0 )kL∞ > c t− 3 . (3.4.1) Accepting (3.4.1) temporarily, we obtain that for any given large t, there exist 1 (x0 , x00 ) such that x0 6= x00 and |K(t, x0 , x00 )| > ct− 3 . Without loss of generality, suppose that c 1 <(K(t, x0 , x00 )) > t− 3 . 2 As K(t, x, x0 ) is smooth away from x = x0 , there exists δ > 0 such that for any (x, x0 ) ∈ (x0 − δ, x0 + δ) × (x00 − δ, x00 + δ) c 1 <(K(t, x, x0 )) > t− 3 . 4 Take the initial data ψ0 (x0 ) = 2δ1 χ(x00 −δ,x00 +δ) (x0 ). Then kψ0 kL1 = 1 and for any x ∈ (x0 − δ, x0 + δ) ¯Z ¯ c 1 ¯ 0 0 0¯ |ψ(t, x)| = ¯ K(t, x, x )ψ0 (x )dx ¯ > t− 3 . 4 To prove (3.4.1), we need the following lemma (Prop. 3, Chap. 8 [36]): Lemma 3.4.1. Suppose k ≥ 2, and φ(x0 ) = φ0 (x0 ) = · · · = φ(k−1) (x0 ) = 0, while φ(k) (x0 ) 6= 0. If ψ is supported on a sufficiently small neighborhood of x0 and ψ(x0 ) 6= 0, then 95 Z 1 1 1 eiλφ(x) ψ(x)dx = ak ψ(x0 )(φ(k) (x0 ))− k λ− k + O(λ− k −1 ), R 1 where ak 6= 0 only depends on k. The implicit constant in O(λ− k −1 ) depends on only finitely many derivatives of φ and ψ at x0 . By Lemma 3.1.2, P (a) has a unique simple root in (℘(ω2 ), ℘(ω1 )), thus we can choose a0 ∈ (ω1 , ω2 ) and a corresponding τ0 such that both ∂a Fτ (a) and ∂a2 Fτ (a) vanish at (a0 , τ0 ). First, we denote I = [0, i2ω 0 ] ∪ [ω1 , ω1 + i2ω 0 ] and assume that for any a ∈ I, a 6= a0 , at least one of ∂a Fτ0 (a) and ∂a2 Fτ0 (a) does not vanish. Then we take δ > 0 small enough such that for a ∈ / (a0 − δ, a0 + δ) ⊂ I, |∂a Fτ0 (a)| + |∂a2 Fτ0 (a)| is greater than some positive constant. 0 Given any large t, take (x, x0 ) such that x−x = τ0 and t ³ Z i2ω0 0 K(t, x, x ) = Z ω1 +i2ω0 ´ + 0 The R i2ω0 0 −itFτ0 (a) e ω1 −℘0 (a)da . ma (x)m−a (x ) W (a) 0 1 -term is bounded by C t− 2 using an argument analogous to that in Section 4, because |∂a Fτ0 (a)| + |∂a2 Fτ0 (a)| is uniformly greater than some positive constant for a ∈ (0, i2ω 0 ). We decompose the R ω1 +i2ω0 ω1 Z ω1 +i2ω0 -term as follows e−itFτ0 (a) ma (x)m−a (x0 ) ω1 −℘0 (a)da := J3 + J4 , W (a) where Z ω1 +i2ω0 J3 = e−itFτ0 (a) ρ(a)ma (x)m−a (x0 ) −℘0 (a)da , W (a) e−itFτ0 (a) ρ̃(a)ma (x)m−a (x0 ) −℘0 (a)da . W (a) ω1 and Z ω1 +i2ω0 J4 = ω1 Here ρ(a) is a smooth cutoff function supported on (a0 −δ, a0 +δ) and ρ̃(a) = 1−ρ(a). 1 Under our assumption, |J4 | < C t− 2 , following the same reasoning as that in 96 Section 3. Considering J3 , the phase function Fτ0 (a) satisfies ∂a Fτ0 (a0 ) = ∂a2 Fτ0 (a0 ) = 0 and 0 (a) ∂a3 Fτ0 (a0 ) 6= 0. ma (x) and m−a (x0 ) do not vanish when a ∈ (ω1 , ω2 ) by (3.1.3). −℘ W (a) 0 (a0 ) is nonzero when a ∈ (ω1 , ω2 ). Therefore ma0 (x)m−a0 (x0 ) −℘ is nonzero. W (a0 ) Since ρ(a) is supported in a sufficiently small neighborhood of a0 , by Lemma 3.4.1, there exist c1 > 0 and T > 0 such that for t > T 1 |J3 | > c1 t− 3 , where c1 is independent of t. 1 1 1 Combining these estimates, we have |K(t, x, x0 )| > c1 t− 3 − 2Ct− 2 > c1 /2 t− 3 for any (x, x0 ) satisfying (x − x0 )/t = τ0 , which implies (3.4.1). Second, suppose there are other a1 , a2 ∈ I such that a1 , a2 , a0 are distinct and ∂a Fτ0 (a), ∂a2 Fτ0 (a) both vanish at a = a1 , a2 . Then P (x) vanishes at ℘(aj ), j = 0, 1, 2. Since a0 ∈ (ω1 , ω2 ), we have that −i℘0 (a0 ) > 0 and ζ(ω)/ω + ℘(a0 ) > 0 by Lemma 3.1.2. Thus τ0 < 0 by (3.2.4). Similar analysis shows that when a ∈ (ω2 , ω2 + iω 0 ) ∪ (iω 0 , 2iω 0 ), ∂a Fτ0 (a) 6= 0. Therefore, a1 , a2 ∈ (0, iω 0 ). Thus ℘(aj ), j = 0, 1, 2, are distinct and are the three roots of P (x). This implies that there is no other a ∈ I such that ∂a Fτ0 (a) = ∂a2 Fτ0 (a) = 0. S We again set δ > 0 small enough such that for a ∈ / 3j=0 (aj − δ, aj + δ) ⊂ I, |∂a Fτ0 (a)| + |∂a2 Fτ0 (a)| is uniformly greater than some positive constant. Given any 0 = τ0 . The earlier argument implies that large t, take (x, x0 ) such that x−x t 0 K(t, x, x ) = 2 Z X j=0 e−itFτ0 (a) ρj (a)ma (x)m−a (x0 ) I 1 −℘0 (a)da + O(t− 2 ), W (a) where ρj (a) = 1 when |a − aj | < δ and ρj (a) = 0 when |a − aj | > 2δ. By Lemma 3.4.1, 0 K(t, x, x ) = a3 t − 13 2 0 X 1 0 −℘ (aj ) − 13 m (x)m (x ) + O(t− 2 ). (Fτ(3) (a )) a −a i j j 0 W (aj ) j=0 Recall that x and x0 are related by (x − x0 )/t = τ0 . maj (x)m−aj (x0 ), j = 0, 1, 2, are 97 linearly independent as functions of x ∈ R and their nontrivial linear combination is a nonzero function. Therefore, there exist x0 and x00 , satisfying (x0 − x00 )/t = τ0 and 1 1 K(t, x0 , x00 ) = c t− 3 + O(t− 2 ), where c is nonzero. Thus there exists some T such that for t > T |K(t, x0 , x00 )| > c −1 t 3. 2 Finally, suppose that a1 = a2 in the second case, which is equivalent to ℘(a1 ) being a double root of P (x) in (−∞, ℘(ω3 )). Similarly, we have for t > T 0 K(t, x, x ) = 1 Z X j=0 e−itFτ0 (a) ρj (a)ma (x)m−a (x0 ) I 1 1 = a3 t− 3 (Fτ(3) (a0 ))− 3 ma0 (x)m−a0 (x0 ) 0 1 1 −℘0 (a)da + O(t− 2 ) W (a) −℘0 (a0 ) + W (a0 ) 1 a4 t− 4 (Fτ(4) (a1 ))− 4 ma1 (x)m−a1 (x0 ) 1 1 −℘0 (a1 ) + O(t− 2 ). W (a1 ) 1 1 Therefore, there exists (x0 , x00 ) such that |K(t, x, x0 )| > c t− 4 > c t− 3 . This completes the proof of (3.4.1). Our proof also gives the optimality of (3.0.3) in the case that P (x) has a double root in (−∞, ℘(ω3 )]. By the proof of Lemma 3.1.2, we have the following corollary: Corollary 3.4.2. Suppose that (ζ(ω)/ω)2 ≤ g2 /12. Then for t > 1, 1 kU (t)ψ0 kL∞ < Ct− 3 kψ0 kL1 (R) . g2 − (ζ(ω)/ω)2 , as a function of Set ω = 1; it follows from (3.5.1) and (3.5.3) that 12 ω 0 > 0, is analytic and ω 0 = 0 is its essential singular point. Numerical experiment indicates that g2 /12 − (ζ(ω)/ω)2 ≈ 0.966104 when ω = 1 and ω 0 > 5. When ω = 1 and ω 0 → 0+ , g2 /12 − (ζ(ω)/ω)2 assumes each real number infinitely many times. 98 3.5 Appendix: Elements of Weierstrass functions Here we list some elementary properties of Weierstrass functions ([39], [7], [2], [13]). A doubly-periodic function that is meromorphic is called an elliptic function. Suppose that 2ω1 and 2ω3 are two periods of an elliptic function f (z) and =(ω3 /ω1 ) 6= 0. Join in succession the points 0, 2ω1 , 2ω1 + 2ω3 , 2ω3 , 0 and we obtain a parallelogram. If there is no point ω inside or on the boundary of this parallelogram (the vertices excepted) such that f (z + ω) = f (z) for all values of z, this parallelogram is called a fundamental period-parallelogram for an elliptic function with periods 2ω1 and 2ω3 . As a set, we assume this parallelogram only includes one of four vertices and two edges adjacent to it. In this way, the z-plane can be covered with the translations of this parallelogram without any overlap. It can be shown that for any c ∈ C, the number of roots (counting multiplicity) of the equation f (z) = c that lie in the fundamental period-parallelogram do not depend on c. This number is called the order of the elliptic function f (z) and it equals the number of poles of f inside a fundamental period-parallelogram. Given ω1 , ω3 ∈ C with =(ω3 /ω1 ) 6= 0, the Weierstrass elliptic function is defined as ℘(z) = 1 + z2 X © ª (z − 2mω1 − 2nω3 )−2 − (2mω1 + 2nω3 )−2 . (m,n)6=(0,0) The summation extends over all integer values of m and n, simultaneous zero values of m and n excepted. ℘(z) is doubly-periodic, namely ℘(z) = ℘(z + 2ω1 ) = ℘(z + 2ω3 ). ℘(z) is an elliptic function of order 2, with poles Ωm,n = 2mω1 + 2nω3 . Each pole Ωm,n is of degree 2. ℘(z) is an even function, ℘(z) = ℘(−z). The Laurent’s expansion of ℘(z) at z = 0 is written as 99 ℘(z) = z −2 + 1 1 g2 z 2 + g3 z 4 + O(z 6 ), 20 28 where g2 , g3 are the constants in (1.2.3) and (1.2.4). Explicitly, we have g2 = 60 X Ω−4 m,n , g3 = 140 (m, n)6=(0,0) X Ω−6 m,n . (3.5.1) (m, n)6=(0,0) Here g2 and g3 are called the invariants of ℘ and they uniquely characterize ℘. Since ℘0 is odd and elliptic of order 3, it has three zeroes in its fundamental periodparallelogram. It is clear that these zeroes are the half periods ω1 , ω2 = ω1 + ω3 , and ω3 . Denote ej = ℘(ωj ), j = 1, 2, 3. The fact that ℘(z) is of order 2 implies that e1 , e2 , e3 are distinct and that ℘00 does not vanish at ωj , j = 1, 2, 3. Furthermore, (1.2.3) implies that e1 , e2 , e3 are the roots of the cubic polynomial 4x3 − g2 x − g3 = 0. (3.5.2) The function ζ(z) is defined by the equation d ζ(z) = −℘(z), dz coupled with the condition limz→0 (ζ(z) − z −1 ) = 0. ζ(z) may also be represented as ζ(z) = 1 + z X (m, n)6=(0,0) n o 1 1 z + + . (3.5.3) z − 2mω1 − 2nω3 2mω1 + 2nω3 (2mω1 + 2nω3 )2 ζ(z) is an odd meromorphic function of z over the whole complex plane except at the simple poles Ωm,n . The residue at each pole is 1. Write ζ(ω1 ) = η1 and ζ(ω3 ) = η3 ; then 1 η1 ω3 − η3 ω1 = πi. 2 ζ(z) is not doubly-periodic; however, it satisfies the following equations: 100 ζ(z + 2ω1 ) = ζ(z) + 2η1 , ζ(z + 2ω3 ) = ζ(z) + 2η3 . (3.5.4) Next we define σ(z) by the equation d log σ(z) = ζ(z), dz coupled with the condition limz→0 σ(z)/z = 1. σ(z) is an odd entire function with simple zeroes at Ωm,n . 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