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Unique continuation for parabolic and elliptic equations and analyticity results for Euler and Navier Stokes equations

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Unique continuation for parabolic and elliptic equations and analyticity results for Euler and Navier Stokes equations

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Content Unique Continuation for Parabolic and Elliptic Equations and Analyticity Results for Euler and Navier Stokes Equations by Guher Camliyurt A Dissertation Presented to the FACULTY OF THE GRADUATE SCHOOL UNIVERSITY OF SOUTHERN CALIFORNIA In Partial Fulfillment of the Requirements for the Degree DOCTOR OF PHILOSOPHY (Applied Mathematics) August 15, 2018 Copyright 2018 Guher Camliyurt Acknowledgements First and foremost, I would like to thank to my supervisor Prof. Igor Kukavica. Being his student equipped me with all the skills I needed to accomplish this dissertation. He has been a great source of intellect, wisdom and enthusiasm. He provided me with great problems to set my academic grounds and constantly supported my work. Profound thanks to Prof. Nabil Ziane for sharing his true opinions wholeheartedly and nurturing our enthusiasm. I have learned so much from him throughout my doctoral education. I am indebted to my old-best friend Sebnem for being there for me at every step of the way. I have always turned to her for comfort and support. I am especially grateful to Melike and Emre who have been by my side throughout the graduate school and contributed their time, support, and advice along with coffee. Special thanks to my close friends Fanhui, Can, Chinmoy, Ezgi, Enes, Duygu, Ilknur and Onur. Their friendship has been vital both within and beyond the academy. I owe a great deal of gratitude to Dan. He has been my main source of inspiration. His encouragement and intellectual insight shaped my thinking process as his humor and musical selection kept my soul hydrated. Lastly, my special thanks to my mother Sevgi Çamlıyurt and my father Bahaettin Çamlıyurt. They have shared my passion all along and supported me in all my pursuits. I am particularly indebted to my bigger family İnci Taşkapan and Orhan Taşkapan for their love and support. I cannot overemphasize their impact on who I am today. ii I would like to extend my gratitude to my sister Bihter and my brother in law Ayhan. Their influence on my life will never diminish. The last but not the least, I would like to acknowledge my favorite person in the whole world–my niece Defne. She found numerous ways to cheer me up. Her continuous energy boost through our regular online meetings is very appreciated. iii Table of Contents Acknowledgements ii Abstract vi Chapter 1: Quantitative unique continuation for a parabolic equation 1 1.1 Notation and the main result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2 Optimizing the frequency function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.3 Similarity variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.4 The proof of the main theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 1.5 The caseR n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 Chapter 2: A local expansion for Stokes system 30 2.1 Notation and the main result on the asymptotic expansion . . . . . . . . . . . . . . . . . 32 2.2 The basic results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 2.3 Proof of the Main Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 Chapter3: Onlocalizationandquantitativeuniquenessfortheellipticpartialdifferential equations 50 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 3.2 A lower bound for the decay of u =Wru +Vu . . . . . . . . . . . . . . . . . . . . . . . 53 3.3 A construction of a localized solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 3.4 A construction of a solution vanishing of high order . . . . . . . . . . . . . . . . . . . . . . 70 3.5 The equation u =Vu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 Chapter 4: On quantitative uniqueness for elliptic equations 84 4.1 Notation and the main result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 4.2 Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 4.3 Proofs of the main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 Chapter 5: The Lagrangian and Eulerian analyticity for the Euler equations 111 5.1 The Euler equations in Lagrangian coordinates . . . . . . . . . . . . . . . . . . . . . . . . 113 5.2 The main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 5.3 A Counterexample . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 Chapter 6: Gevrey regularity for the Navier-Stokes in a half-space 143 6.1 Main result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 6.2 Proof of Theorem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 6.2.1 Local existence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 6.2.2 The Stokes estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 6.2.3 Bounds for the nonlinear term . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 iv 6.2.4 Conclusion of the proof of Theorem 1 . . . . . . . . . . . . . . . . . . . . . . . . . 154 6.3 Space-time analytic estimates for the nonlinear term . . . . . . . . . . . . . . . . . . . . . 155 6.3.1 Gagliardo-Nirenberg inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 6.3.2 Terms with only time derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 6.3.3 Terms with no normal derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 6.3.4 Terms with all mixed derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 Reference List 175 v Abstract This thesis comprises of two parts; the first part is concerned with quantitative uniqueness estimates for several problems including the Stokes system, the parabolic equation @ t u u = W j (x;t)@ j u + V (x;t)u, and the elliptic equations u =W j (x;t)@ j u; and u =V (x;t)u. The last two chapters, which form the second part, discuss analyticity and Gevrey regularity properties for the Euler and Navier-Stokes equations. Chapter 1 utilizes the frequency function method to study the order of vanishing of solutions u which is proven to be bounded by C(kvk 2=3 L 1 +kwk 2 L 1) matching the upper bound previously established in the elliptic case. In Chapter 2, a similar question is addressed for the Stokes system in a neighborhood of the point in whichthevelocityvanishesoforderd. Weprovethatthereexistsadivergence-freepolynomialP inxwith t-dependent coefficients of degree d which approximates the solution u of order d + for certain > 0. The polynomial P satisfies a Stokes equation with a forcing term which is a sum of two polynomials in x of degrees d 1 and d. In Chapter 3 and Chapter 4, we address the spatial decay and the quantitative uniqueness properties for the elliptic equations u =Wru and u =Vu. Firstly, we focus on the question of quantitative uniqueness for the equation u = Vu with either periodic or Dirichlet boundary conditions in a disc. We provide an example of a sequence of solutions u n corresponding to a potential V n such that u n vanishes of orderkV n k 2=3 L 1 up to a constant andkV n k L 1 !1. The example also shows sharpness of recently obtained bounds in the case of a parabolic equation u t u = Vu. Later in Chapter 4, we vi address Landis-Meshkov’s question for the possible rate of decay at infinity of solutions to equation u =Wru; when W is bounded. Based on the former problem, the exponent 2=3 for V in the upper boundC(kvk 2=3 L 1 +kwk 2 L 1)isoptimal, butthequestionremainsiftheexponent 2forW isthebestpossible as well. We prove that there exists a complex valued solution u on the plane, such thatkWk L 1C and ju(x)jC exp(C 1 jxj 2 ). Also, we show that this type of localization cannot be improved upon. In Chapters 5 and 6, we consider analyticity and Gevrey regularity properties of the Euler and the Navier-Stokes systems. In Chapter 5, we address analyticity for the Euler equation in the Lagrangian variables. We prove that the Gevrey class radius is conserved for a positive time when where the norm is based on the supremum of the Taylor coefficients. The previous result by Constantin, Kukavica, and Vicol considered the norm which is based on a summation. Also, we prove preservation of the local space analyticity. For the Euler equations, we provide an elementary counterexample to the preservation of the Eulerian analyticity. In Chapter 6, we consider the Dirichlet problem for the Navier-Stokes system in the half space = x = (x 1 ;:::;x d )2 R d : x d > 0 and prove an instantaneous space-time analytic regularization result, uniformlyuptotheboundaryofthehalfspace. Here, theforceistakenrealanalyticinspace-time. vii Chapter 1 Quantitative unique continuation for a parabolic equation The main purpose of this chapter is to provide quantitative uniqueness estimates for a parabolic equation @ t u u =w j (x;t)@ j u +v(x;t)u (1.0.1) by finding an upper bound for the order of vanishing of a solution in terms of the size of coefficients v and w. Starting with the works of Carleman [C], there have been many results showing that solutions of the equations of parabolic or elliptic type have a finite order of vanishing, in which case we say that the equation has the unique continuation property. For instance, a result of Escauriaza and Vega [EV] shows that forv in a sharp spaceL 1 t L n=2 x (with the natural smallness assumption for the corresponding norm) and w = 0, the equation (1.0.1) has the unique continuation property. Also, Koch and Tataru proved in [KT] that the equation (1.1.1) has this property provided v and w belong to appropriate Lebesgue spaces. A sharp result of this type for the elliptic counterpart was established by Jerison and Kenig [JK] (cf. [K1, K2, K3, V] for reviews on unique continuation for elliptic, parabolic, and dispersive equations). Morerecently, considerableeffortswerededicatedtothequantitativeestimatesofuniquecontinuation, i.e., to estimating the maximal degree to which the solution may vanish at a point. There are many 1 applications of such results; in particular, quantitative uniqueness results yield lower bounds on solutions of the corresponding PDE. Also, estimates on the order of vanishing are an essential tool for obtaining upper bounds on the size of level sets of PDE [DF1, DF2, DF3, Ku2, Ku3, Ku4, L], in inverse problems and control [AE], upper bounds on vortex degrees [Ku2], spectral information on Schrodinger operators [BK], backward uniqueness [EF], Hadamard type theorems [AMRV, B1, B2, BC, CRV, EFV, EVe, LNW], and other topics. The research on quantitative uniqueness was initiated by Donnelly and Fefferman [DF1, DF2, DF3], who proved that the maximal order of vanishingO u of an eigenfunction u of an elliptic operator is bounded by C p , where is the corresponding eigenvalue. They used this estimate to prove that the nodal volume of a zero set of an analytic eigenfunction is bounded by C p . It is natural to ask the following question: Is the order of vanishing of a solution of a boundary value problemu =wru +vu subject to periodic, Dirichlet, or Neumann conditions (or with prescribed constant doubling property) bounded by Ckvk 1=2 L 1 +Ckwk L 1 (1.0.2) [Ku1]. Thisquestionremainsopenandapreciseupperboundiscurrentlyunknown. Basedonanexample by Meshkov [M], who provided a complex valued solution decaying with a certain exponential rate at infinity, Bourgain and Kenig predicted that (when w = 0), the optimal vanishing rate is more likely Ckvk 2=3 L 1. In addition, they proved that the order of vanishing can be estimated by this quantity for any solution of the elliptic equation @ i (a ij @ j u) +vu = 0 (cf. also [KSW] for stronger results in the plane). In the following sections, we address this question for the parabolic equation (1.0.1). As it is the case for the elliptic equation, it is an open problem whether, under natural growth conditions at infinity (for example as those in the present paper), the quantity (1.0.2) provides an upper bound for the order of vanishing for solutions u of (1.0.1). The main result asserts that the order of vanishing bounded by Ckwk 2 L 1 +Ckvk 2=3 L 1: (1.0.3) 2 This result is in agreement with an elliptic upper bound proved by Bourgain and Kenig in [BK] when w = 0. However, our proof is based on a completely completely different approach, which we describe next. The method presented here is based on the parabolic frequency function introduced by Kurata [Kur] and Poon [P], which is in turn inspired by earlier works on elliptic equations by Almgren [Al] and Garofalo and Lin [GL]. It is based on the observation that the frequency (t) = Z u 2 (x;t)G 0 (x;t)dx (1.0.4) is logarithmically convex for solutions of the heat equation. Above, G 0 (x;t) = 1 t n=2 e jxj 2 =4t ; x2R n ; t> 0 (1.0.5) denotes the (2) n=2 -multiple of a backward Gaussian kernel. It was also shown in [Kur, P] that the method can be applied to the equation (1.0.1) yielding a strong unique continuation property for u when v and w are bounded. Following the dependence onkvk L 1 andkwk L 1, we obtain that the degree of vanishing (for example under an assumption of periodicity) is bounded by Ckvk 2 L 1 +Ce Ckwk L 1 : (1.0.6) In order to improve this result and obtain the bound (1.0.3), we use a similarity variable approach from [Ku2] (cf. also [An, Ch]) with the addition of a change of variable introduced in [Ku4] which optimizes the Dirichlet quotient (i.e., replaces Q in (1.3.6) with Q in (1.3.11)). The similarity change of variable was used in many contexts; here the idea is that the parabolic structure and the Dirichlet quotient method ([A, AN, CFNT, FS]) leads to the necessary logarithmic convexity. In addition, we use in an essential 3 way theorems due to Alessandrini and Vessella [AV] on the polynomial approximation of a solution of a parabolic equation (cf. also [H1, H2]). This chapter is structured as follows. In Section 1.1, we state the main result on the maximal order of vanishing for the equation (1.0.1). In Section 1.2 we find a point x and the time, where in a certain sense the frequency is optimized. In Section 1.3, we recall the similarity change of variables and spectral properties of the resulting linear part. Finally, Section 1.4 contains the proof of the main theorem. 1.1 Notation and the main result Our goal is to address the quantitative uniqueness for solutions u of the equation @ t u u =w j (x;t)@ j u +v(x;t)u (1.1.1) defined for (x;t)2R n I, where I is an open interval containing [T 0 ;T 0 +T ], with T 0 2R and T > 0. The solution u as well as the coefficients v;w2L 1 (R n ) are assumed to be periodic in x with respect to = [;] n (for the conditions omitting the periodicity assumption, cf. Remark 1.1.2 and Section 1.4 below). Also, assume that jw j (x;t)jM 1 ; (x;t)2R n I; j = 1;:::;n (1.1.2) and jv(x;t)jM 0 ; (x;t)2R n I: (1.1.3) 4 Since we are interested in the dependence of the order of vanishing on M 0 and M 1 when they are large, we assume M 0 ;M 1 1. For any (x 0 ;t 0 )2R n R and r> 0, denote by Q r (x 0 ;t 0 ) = (x;t)2R n R :jxx 0 j<r;r 2 <tt 0 < 0 ; the parabolic cylinder centered at (x 0 ;t 0 ) with radius r> 0, while the parabolic norm of (x;t)2R n R is given by j(x;t)j = (jxj 2 +jtj) 1=2 : We write W 2;1 1 (Q 1 ) for the Sobolev space of functions whose x-derivatives up to second order and t- derivative up to the first order belong to L 1 (Q 1 ). Denote by O (x0;t0) (u) the order of vanishing of u at (x 0 ;t 0 ), which we define (in the L 2 sense) as the smallest integer d such that kuk L 2 (Qr (x0;t0)) =O(r d+(n+2)=2 ); as r! 0 + : (1.1.4) Let q 0 be an upper bound for the Dirichlet quotients of u on [T 0 ;T 0 +T ], i.e., kru(;t)k 2 L 2 ku(;t)k 2 L 2 q 0 ; t2 [T 0 ;T 0 +T ]: (1.1.5) The following is the main theorem of this chapter. 5 Theorem 1.1.1. Let u2 W 2;1 1 ( I) be a nontrivial solution of (1.1.1) for t in a neighborhood of [T 0 ;T 0 +T ] with w j and v satisfying (1.1.2) and (1.1.3), respectively. Then the order of vanishing of u at (x 0 ;t 0 ) satisfies O (x0;t0) (u)C(M 2 1 +M 2=3 0 ) (1.1.6) for every (x 0 ;t 0 )2 [T 0 +T=2;T 0 +T ] where the constant C depends on T, L, and q 0 . Remark 1.1.2. It is not difficult to check that it is possible to replace thex-periodicity assumption with Z R n u(x;t) 2 dxM Z B1 u(x;t) 2 dx; t2 [T 0 ;T 0 +T ] (1.1.7) whereM is a constant, making an assertion about the doubling at the point (0;T 0 +T ). For the necessary modifications, cf. Section 1.4 below. The proof of Theorem 1.1.1 is divided into several lemmas. Let (x 0 ;t 0 )2 [T 0 +T=2;T 0 +T ]. By translation and rescaling, we may assume that u is defined fort in an open intervalI of [T; 0] and that (x 0 ;t 0 ) = (0; 0): 1.2 Optimizing the frequency function Recall from (1.0.5) that G 0 represents the (4) n=2 -multiple of the backward Gaussian kernel. Lemma 1.2.1. Let u be as above with v and w j satisfying (1.1.2) and (1.1.3). Assume that 2 (0;T ). Then R R n jru(x +y;)j 2 G 0 (y;)dy R R n u(x +y;) 2 G 0 (y;)dy q() (1.2.1) 6 for some x 2 . This lemma shall be used below with = 1=C(M 2 1 +M 2=3 0 ) where C is sufficiently large. Proof of Lemma 1.2.1. Assume that, for some > 0, we have Z R n u(x +y;) 2 G 0 (y;)dy Z R n jru(x +y;)j 2 G 0 (y;)dy; x2R n : (1.2.2) We intend to show that this is not possible if > q() (this is sufficient since due to periodicity and continuity, the minimum of the quotient in (1.2.1) is achieved). The statement (1.2.2) is equivalent to Z R n u(y;) 2 G 0 (yx;)dy 1 Z R n jru(y;)j 2 G 0 (yx;)dy; x2R n : (1.2.3) Integrating the left side of (1.2.3) over and using R R n G 0 (yx;)dx = (4) n=2 , we obtain Z Z R n u(y;) 2 G 0 (yx;)dydx = Z X j2Z n Z j+ u(y;) 2 G 0 (yx;)dydx = Z X j2Z n Z u(y 0 ;) 2 G 0 (y 0 +jx;)dy 0 dx = Z u(y 0 ;) 2 0 @ X j2Z n Z G 0 (y 0 +jx;)dx 1 A dy 0 = Z u(y 0 ;) 2 Z R n G 0 (y 0 x;)dxdy 0 = (4) n=2 Z u(y 0 ;) 2 dy 0 : (1.2.4) Similarly, we have Z Z R n jru(y;)j 2 G 0 (yx;)dydx = (4) n=2 Z jru(y 0 ;)j 2 dy 0 : (1.2.5) 7 Combining (1.2.4) and (1.2.5) with (1.2.3), we get Z u(y;) 2 dy Z jru(y;)j 2 dy =q() Z u(y;) 2 dy: (1.2.6) Sinceu(;) is not identically zero, we obtain a contradiction if >q(). Setting = 2q(), we con- cludethatthereexistsx 2R n suchthat(1.2.1)holds. Byperiodicityofu, wemayassumethatx 2 . We proceed with a change of variables so that at time the solutions starts at the point x from Lemma 1.2.1. Let u(x;t) =e u x x t;t : (1.2.7) We then have e u(x;) =u(x +x ;) (1.2.8) and e u(x; 0) =u(x; 0): (1.2.9) Note also that O (0;0) (e u) =O (0;0) (u): (1.2.10) Furthermore, R jre u(x;t)j 2 dx R e u(x;t) 2 dx = R jru(x;t)j 2 dx R u(x;t) 2 dx =q(t); (1.2.11) 8 for all t2 [; 0), and (1.2.1) may be written as R R n jre u(y;)j 2 G 0 (y;)dy R R n e u(y;) 2 G 0 (y;)dy q(): (1.2.12) The functione u solves a modified equation @ t e u e u = 1 x re u +wre u +ve u: (1.2.13) Observing (1.2.10), we remove tilde from here on, and write u instead ofe u. Setting a = x ; (1.2.13) may then be rewritten as @ t u u =a j @ j u +w j @ j u +vu: (1.2.14) 1.3 Similarity variables Next, we apply a similarity change of variable u(x;t) =e jxj 2 =8(t) U x p t ; log(t) (1.3.1) i.e., U(y;) =e jyj 2 =8 u(ye =2 ;e ) (1.3.2) 9 with = log(t), i.e., t =e for t2 [; 0]. Also, write V (y;) =v(ye =2 ;e ); (y;)2R n [ 0 ;1) (1.3.3) and W j (y;) =w j (ye =2 ;e ); (y;)2R n [ 0 ;1); (1.3.4) for j = 1;:::;n where 0 = log(1=). Denoting HU =U + jyj 2 16 n 4 U (1.3.5) and Q() = (HU;U) L 2 (R n ) kUk 2 L 2 (R n ) ; (1.3.6) the equation (1.2.14) may be written as @ U +HU =e =2 (a j y j U +a j @ j U) +e =2 (y j W j U +W j @ U) +e VU (1.3.7) for 0 , while at = 0 , U(y; 0 ) =e jyj 2 =8 u y p ; : (1.3.8) 10 Then @ U + A() Q()I U + Q()U =e =2 a j @ j U +e =2 (y j W j U +W j @ j U) +e VU (1.3.9) where we denote A()U =HUe =2 a j y j U (1.3.10) and Q() = (A()U;U) L 2 (R n ) kUk 2 L 2 (R n ) =Q() e =2 a j kUk 2 L 2 (R n ) Z R n y j U 2 dy: (1.3.11) A straight-forward change of variables yields kU(;)k 2 L 2 (R n ) = Z R n u(x;t) 2 G 0 (x;t)dx (1.3.12) and (HU(;);U(;)) L 2 (R n ) =jtj Z R n jru(x;t)j 2 G 0 (x;t)dx: (1.3.13) We now observe a simple fact k(H Q()I)vk L 2 (R n ) dist( Q(); sp(H)); kvk L 2 (R n ) = 1: (1.3.14) 11 In order to analyze the asymptotic behavior of eigenvalues of A(), we first recall the spectral properties of H. The Hermite polynomials h k are defined on the real line as h k (x) = (1) k e x 2 d k dx k e x 2 ; while the Hermite functions read e h k (x) =h k (x)e x 2 =2 : By taking the product of one-dimensional Hermite functions, we generalize the definition to R n , (x) = n Y k=1 e h k (x k ); (1.3.15) where = ( 1 ;:::; n )2N n 0 , and x = (x 1 ;:::;x n )2R n . The set f :2N n 0 g (1.3.16) forms a complete orthonormal system for L 2 (R n ) and the functions solve (jxj 2 ) (x) =(2jj +n) : (1.3.17) Then we have H x 2 = jj 2 x 2 ; 2N n 0 (1.3.18) 12 and thus sp(H) = n m 2 :m2N 0 o : (1.3.19) Lemma 1.3.1. For a sufficiently large constant K 0 > 0, set = 1 K 0 (M 2 1 +M 2=3 0 ) : (1.3.20) Then Q() =O 1 ; as !1: (1.3.21) Also, Q()! m 2 ; as !1 (1.3.22) for some m2N 0 for which mC(M 2=3 0 +M 2 1 ) (1.3.23) whereC dependsonq 0 . Moreover, for every> 0, thereexist2 (log(1=); 0) andconstantsA 1 ();A 2 ()> 0 such that A 1 ()jtj m+ Z u 2 (x;t)G 0 (x;t)dxA 2 ()jtj m (1.3.24) for all t2 [; 0). 13 Proof of Lemma 1.3.1. We first divide (1.3.9) bykUk L 2 (R n ) and then take the inner product of the re- sulting equation with (A() Q()I)U=kUk L 2 (R n ) . LetB(U) denote the right side of (1.3.9). We obtain @ U kUk ; (A() Q()I) U kUk + (A() Q()I) U kUk 2 + Q() (U;A()U) kUk 2 = Q 2 () + B(U) kUk ; (A() Q()I) U kUk (1.3.25) where we abbreviate (;) = (;) L 2 (R n ) and kk =kk L 2 (R n ) : Note that Q() (U;A()U) kUk 2 = Q 2 (); (1.3.26) so (1.3.25) becomes @ U kUk ; (A() Q()I) U kUk + (A() Q()I) U kUk 2 = B(U) kUk ; (A() Q()I) U kUk : (1.3.27) On another note, by differentiating (1.3.11), we get Q 0 () = 2 (@ U;A()U) kUk 2 + (U;A 0 ()U) kUk 2 2(@ U;U)(A()U;U) kUk 4 : (1.3.28) 14 Since A() is a symmetric operator whose derivative is given by A 0 ()U = 1 2 e =2 a j y j U (1.3.29) we obtain 1 2 Q 0 () = (@ U;A()U) kUk 2 + 1 4 e =2 (U;a j y j U) kUk 2 (@ U;U)(A()U;U) kUk 4 = (@ U; (A() Q()I)U) kUk 2 + 1 4 e =2 (U;a j y j U) kUk 2 : (1.3.30) Then, substituting (1.3.30) in (1.3.27), we get 1 2 Q 0 () + (A() Q()I) U kUk 2 = 1 4 e =2 (a j y j U;U) kUk 2 + e =2 a j kUk 2 (@ j U; (A() Q()I)U) + 1 kUk 2 e =2 (y j W j U +W j @ j U) +e VU; (A() Q()I)U : (1.3.31) We integrate by parts on the second term on the right hand side and note that R U@ j U = 0 and R U@ j U = 0. Therefore, (1.3.31) becomes 1 2 Q 0 () + (A() Q()I) U kUk 2 = e =2 a j 4kUk 2 Z y j U 2 dy e =2 a j 16kUk 2 Z y j U 2 dy + e jaj 2 2 + e =2 kUk 2 Z y j W j U +W j @ j U +e =2 VU (A() Q()I)Udy: =I 1 +I 2 +I 3 +I 4 : (1.3.32) 15 Now, I 1 = e =2 4kUk 2 a j Z y j U 2 dy e =2 4kUk 2 jaj Z jyjU 2 dy (1.3.33) e =2 4kUk 2 jaj Z jyj 2 U 2 dy 1=2 ; (1.3.34) and similarly I 2 +I 3 e =2 16kUk jaj Z jyj 2 U 2 dy 1=2 + e jaj 2 2 : (1.3.35) As for the last term in (1.3.32), I 4 (A() Q()I) U kUk e =2 kUk ky j W j U +W j @ j Uk +e kVUk : (1.3.36) Hence, for the right side of (1.3.32) we have I 1 +I 2 +I 3 +I 4 C e =2 jaj kUk Z jyj 2 U 2 dy 1=2 +Ce jaj 2 +e =2 (A() Q()I) U kUk M 1 kUk (kyUk +krUk) +e =2 M 0 : (1.3.37) It is easy to check that krUk 2 +kyUk 2 C Q() + +e jaj 2 + 1 kUk 2 (1.3.38) where x + = maxfx; 0g; x2R: (1.3.39) 16 Consequently, 1 2 Q 0 () + (A() Q()I) U kUk 2 Ce =2 jaj Q() + +e jaj 2 + 1 1=2 +Ce jaj 2 +Ce =2 (A() Q()I) U kUk M 1 ( Q() + +e jaj 2 + 1) 1=2 +e =2 M 0 : (1.3.40) Applying Young’s inequality to the last term, we get Q 0 ()Ce =2 jaj Q() + +e jaj 2 + 1 1=2 +Ce jaj 2 +Ce M 2 1 ( Q() + +e jaj 2 + 1) +e M 2 0 Ce M 2 1 Q() + +Ce =2 jaj( Q() + ) 1=2 +Ce =2 jaj +Ce (jaj 2 +M 2 1 ) +Ce 2 M 2 1 jaj 2 +M 2 0 : (1.3.41) Observe that Ce 0 M 2 1 =Ce log(1=) M 2 1 CM 2 1 K 0 (M 2 1 +M 2=3 0 ) 1 2 ; (1.3.42) where the last inequality is obtained by assuming that K 0 is a sufficiently large constant. Thus we may now apply [Ku1, Lemma 2.2] and estimate Q()C max( Q( 0 ); 0) +M 2 1 jaj 2 2 +M 2 1 +M 2 0 2 +jaj 2 +jaj 1=2 ; 0 : (1.3.43) 17 By (1.2.12), (1.3.11), (1.3.12), and (1.3.13) we have Q( 0 )Q() + e 0=2 a j kUk 2 Z y j U 2 dy 2q() + C p (HU( 0 );U( 0 )) 1=2 (U( 0 );U( 0 )) 1=2 2q() + C p (q()) 1=2 2q() +Cq() 1=2 2q 0 +Cq 1=2 0 (1.3.44) whereweused(1.1.5)inthelastinequality. Recallthata =x =, wherex 2 , whichimpliesjajC=. Therefore, Q()C q 0 +q 1=2 0 +M 2 1 + 1 +M 2=3 0 + q M 2 1 +M 2=3 0 C q 0 +M 2 1 + 1 +M 2=3 0 C M 2 1 + 1 +M 2=3 0 =K M0;M1 (1.3.45) by allowing C to depend on q 0 . Going back to (1.3.40) and applying Young’s inequality one more time, we obtain 1 2 Q 0 () + (A() Q()I) U kUk 2 Ce =2 jaj Q() + +e jaj 2 + 1 1=2 +Ce jaj 2 +Ce M 2 1 ( Q() + +e jaj 2 + 1) +M 2 0 e 2 : (1.3.46) Integrating (1.3.46) and using the bound on Q, we get 1 2 Q() Q( 0 ) + Z 0 (A(s) Q(s)I) U(;s) kU(;s)k 2 ds Cjaj K 1=2 M0;M1 ( 1=2 e =2 ) +Cjaj 2 (e ) +C M 2 1 K M0;M1 (e ) +M 2 1 jaj 2 ( 2 e 2 ) +CM 2 0 ( 2 e 2 ): (1.3.47) 18 In order to estimate the second term on the left from below, letkvk L 2 = 1. Then (A(s) QI)v 2 1 2 (H QI)v 2 Ce jaj 2 QCe jaj 2 : (1.3.48) Using this inequality in (1.3.46) and letting !1, we get Z 1 0 (H Q(s)I) U(;s) kU(;s)k 2 ds<1: (1.3.49) Combining this fact with (1.3.14) and lim sup !1 Q 0 () 0; (1.3.50) which follows from (1.3.46), we get dist( Q(); sp(A()))! 0 (1.3.51) as !1. Using (1.3.19), we finally obtain Q()! m 2 (1.3.52) as !1 for some m2N 0 . Going back to (1.3.7), we have (@ U;U) + Q() =e =2 a j (@ j U;U) +e =2 (y j W j U;U) +e =2 (W j @ j U;U) +e (VU;U) which may be rewritten as 1 2 1 kUk 2 d d kUk 2 + Q() = f() kUk 2 (1.3.53) 19 where f() kUk 2 =e =2 a j (@ j U;U) kUk 2 +e =2 (y j W j U;U) kUk 2 +e =2 (W j @ j U;U) kUk 2 +e (VU;U) kUk 2 : Note that (@ j U;U) = 0, and thus f() kUk 2 e =2 M 1 R jyjU 2 dy + R jrUjjUjds kUk 2 ! +e M 0 2e =2 M 1 R jyjU 2 dy + R jrUj 2 ds + R jUj 2 ds kUk 2 ! +e M 0 Ce =2 M 1 j Q()j +e jaj 2 + 1 +e M 0 : As Q() is uniformly bounded, Z 1 f(s)ds kU(;s)k 2 CK M0;M1 M 1 (e 1=2 e =2 ) +CM 1 jaj 2 (e 31=2 e 3=2 ) +M 0 (e 1 e ): Integrating (1.3.53), we get 1 2 logkU(;)k 2 1 2 logkU(; 1 )k 2 = Z 1 Q(s)ds + Z 1 f(s) kU(;s)k 2 ds = m 2 ( 1 ) Z 1 Q(s) m 2 ds +O(e 1=2 e =2 ); as !1: By (1.3.47), we have that for all > 0 there exists t 1 ()> 0 such that ( 1 ) logkU()k 2 logkU( 1 )k 2 +m( 1 )( 1 ); 1 (): 20 Therefore, e (1) e m kU()k 2 e 1m kU( 1 )k 2 e (1) ; 1 : Consequently, for all > 0, there exists 1 0 such that A 1 ()e e m kU()k 2 A 2 ()e ; 1 (1.3.54) for some positive constantsA 1 () andA 2 (). Switching back to the original variables, we obtain (1.3.24), and the proof is concluded. A function f(x;t) is homogeneous of degree d if for any > 0 and (x;t)2R n Rnf(0; 0)g if it satisfies f(x; 2 t) = d f(x;t): A polynomialP (x;t) of degree at mostd2N can be decomposed into a sum of homogeneous polynomials whose degree of homogeneity is at most d. The following elementary results regarding polynomials play an important role in the subsequent argument. Lemma 1.3.2. Let Q = P jj+2l=d C ;l x t l be a homogeneous polynomial of degree d2N 0 . Then Z R n Q(x;t)G 0 (x;t)dx =C 0 jtj d=2 where C 0 is a constant depending on the polynomial. 21 Proof of Lemma 1.3.2. By substituting Q in the integral, we obtain Z R n Q(x;t)G 0 (x;t)dx = X jj+2l=d C ;l Z R n x t l e jxj 2 =4jtj jtj n=2 dx = X jj+2l=d C ;l jtj l+jj=2 Z R n x jtj jj=2 e jxj 2 =4jtj jtj n=2 dx = X jj+2l=d C ;l jtj l+jj=2 Z R Z R x 1 jtj 1=2 1 x n jtj 1=2 n e jxj 2 =4jtj jtj n=2 dx 1 dx n =jtj d=2 X jj+2l=d C ;l ( 1 + 1) ( n + 1) (1.3.55) and the lemma is established. Lemma 1.3.3. Consider the polynomial x t l withjj + 2l = 2d for some integer d 0. If all the coordinates of = ( 1 ; 2 ;:::; n ) are even, then for all r> 0 Z B(0;r) x t l G 0 (x;t)dx =C 0 jtj d +O jtj d+1 ; as t! 0 (1.3.56) where C 0 is a constant. If i is an odd integer for some i2f1; ;ng, then Z B(0;r) x t l G 0 (x;t)dx = 0: (1.3.57) Proof of Lemma 1.3.3. Forjj> 0, we have Z B(0;r) x t l G 0 (x;t)dx Cjtj l+jj=2 n Y i=1 Z r 0 x i jtj 1=2 i e jxij 2 =4jtj jtj 1=2 dx i =Kjtj d n Y i=1 Z r 2 =4jtj 0 z i=21=2 e z dz (1.3.58) 22 where K is a fixed constant. It is easy to check that C 0 K n Y i=1 Z r 2 =4jtj 0 z i=21=2 e z dz =O(jtj); as t! 0 (1.3.59) where C 0 =K n Y i=1 Z 1 0 z i=21=2 e z dz (1.3.60) and (1.3.56) follows. If i is odd for some i, then we have (1.3.57) by symmetry. 1.4 The proof of the main theorem Before concluding with the proof of the main theorem, we need a statement connecting the order of vanishing with the degree of the eigenfunction. Lemma 1.4.1. Consider a solution u2W 2;1 1 ( I) to the equation @ t u + u =a j @ j u +w j (x;t)@ j u +v(x;t)u (1.4.1) for t2 [; 0] with 2 (0; 1) where a j are constants and where w j and v satisfy (1.1.2) and (1.1.3), and suppose that u has finite order of vanishing d2N 0 at (0; 0). With m2N, assume that for all > 0 there exist 2 (0;] and A 1 ();A 2 ()> 0 such that A 1 ()jtj m+ Z R n u 2 (x;t)G 0 (x;t)dxA 2 ()jtj m ; t2 [; 0]: (1.4.2) Then d =m. 23 Proof of Lemma 1.4.1. Since the statement clearly holds when d = 0, we assume d2N. First, assume that (1.4.2) holds for some fixed > 0. Fix > 0. Since the degree of vanishing of u at (0; 0) is d> 0, there exists > 0 such that ju(x;t)j j(x;t)j d = jxj 2 +jtj d=2 (1.4.3) for all (x;t)2Q 1 . Note that for any R> 0 we have Z R n nB(0;R) u 2 (x;t)G 0 (x;t)dxkuk 2 L 1 (R n ) Z 1 R Z @B(0;r) e r 2 =4jtj jtj n=2 dS y drCkuk 2 L 1 jtj 1=2 R e R 2 =8jtj : (1.4.4) By (1.4.2) and (1.4.4), for all > 0 there exist 2 (0;) and A 1 ();A 2 ()> 0 such that 1 2 A 1 ()jtj m+ Z B(0;R) u 2 (x;t)G 0 (x;t)dxA 2 ()jtj m ; t2 [; 0]: (1.4.5) Furthermore, by Lemmas 1.3.2 and 1.3.3, Z B(0;R) u 2 (x;t)G 0 (x;t)dxC 2 Z B(0;R) jxj 2 +jtj d G 0 (x;t)dxCjtj d (1.4.6) where C depends on R. Combining this with (1.4.5), we get 1 2 A 1 ()jtj m+ Cjtj d ; t2 [; 0] which yields dm +: (1.4.7) 24 For the other direction, we use [AV, Theorem 1.1] on a structure of a solution in a neighborhood of a point where u vanishes. Recall that u solves @ t u u =a j @ j u +w j (x;t)@ j u +v(x;t)u =f(x;t); and for all (x;t)2Q 1 , jf(x;t)j jaj +kwk L 1 ( I) max j j@ j u(x;t)j +kv(x;t)k L 1 ( I) ju(x;t)j Cj(x;t)j d1 (1.4.8) where C depends on u, v, and w. This implies that for any r 1, and 2 (0; 1) kfk L 1 (Qr ) Cr d1 : Therefore, by [AV, Theorem 1.1], for the solution u 2 W 2;1 1 (Q 1 ) of (@ t + )u = f, there exists a homogeneous caloric polynomial P d of degree less than or equal to d such that for any 2 (0; 1), ju(x;t)P d (x;t)jCj(x;t)j d+ (1.4.9) for (x;t)2Q 1 . Since the degree of vanishing of u is d, the degree of P d has to be d. Thus we may write P d = X jj+2l=d C ;l x t l 25 where not all C ;d equal zero. Then, by (1.4.2) and (1.4.9), we have for all t2 [; 0], Z B(0;1) P 2 d G 0 (x;t)dx C Z B(0;1) u 2 G 0 (x;t)dx +C Z B(0;1) jxj 2 +jtj d+ G 0 (x;t)dx Cjtj m +C Z R n jxj 2 +jtj d+ G 0 (x;t)dx =C 1 jtj m +O(jtj d+ ): (1.4.10) On the other hand, Z B(0;1) P 2 d G 0 (x;t)dx = X jj+2l=2d C ;l Z B(0;1) x jtj l G 0 (x;t)dx =C 0 jtj d +O jtj d+1 ; as t! 0 (1.4.11) by Lemma 1.3.3 where C 0 > 0. Consequently, we have C 0 jtj d +O jtj d+1 C 1 jtj m +Cjtj d+ ; as t! 0 which implies dm: (1.4.12) By (1.4.7) and (1.4.12), we get mdm +; and since this holds for all > 0, we conclude that d =m. 26 Proof of Theorem 1.1.1. In Lemmas 1.3.1 and 1.4.1, we have shown that Q()C(M 2 1 +M 2=3 0 ) for all 2 [ 0 ;1), and Q()! m 2 ; as !1, where m =O (0;0) (u). Combining these two facts, we arrive at the desired conclusion. 1.5 The case R n Here we present the modifications when the periodicity assumption is removed. Let u be a solution of (1.1.1) defined for (x;t)2 R n I, where I is an open interval containing [T 0 ;T 0 +T ]. Instead of periodicity, we assume (1.1.7), whereC is a constant. The coefficientsv andw satisfy (1.1.2) and (1.1.3). For simplicity, we assume M 0 ;M 1 1. Let q 0 be an upper bound for the Dirichlet quotients of u on [T 0 ;T 0 +T ], i.e., kru(;t)k 2 L 2 ku(;t)k 2 L 2 q 0 ; t2 [T 0 ;T 0 +T ]: (1.5.1) The following is the analog of Theorem 1.1.1. Theorem 1.5.1. Letu2W 2;1 1 ( I) be a nontrivial solution of (1.1.1) withw j andv satisfying (1.1.2) and (1.1.3), respectively. Then the order of vanishing of u at (0;T 0 +T ) satisfies O (0;T0+T) (u)C(M 2 1 +M 2=3 0 ) (1.5.2) where the constant depends on T, M, and q 0 . 27 The proof is the same as in the periodic case except that we need to modify Lemma 1.2.1, stated next. Assume, as above, that T 0 =T so that the time interval is [T; 0]. Lemma 1.5.2. Letu be as above withv andw j satisfying (1.1.2) and (1.1.3). Let2 (0;T ) be such that 1 C (1.5.3) for a sufficiently large constant C. Then R R n jru(x +y;)j 2 G 0 (y;)dy R R n u(x +y;) 2 G 0 (y;)dy CMq() (1.5.4) for some x 2B 2 . Proof of Lemma 1.5.2. Assume that, for some > 0, we have Z R n u(x +y;) 2 G 0 (y;)dy Z R n jru(x +y;)j 2 G 0 (y;)dy; x2B 2 : (1.5.5) We shall show that this cannot hold ifCMq() for a sufficiently large constantC > 0. We integrate (1.5.5) in x over B 2 and change the order of integration to obtain Z R n G 0 (y;)dy Z B2 u(x +y;) 2 dx 1 Z R n G 0 (y;)dy Z B2 jru(x +y;)j 2 dx (1.5.6) Note that the right side of (1.5.6) is bounded from above by 1 kru(;)k 2 L 2 Z G 0 (y;)dy = (2) n=2 kru(;)k 2 L 2 (1.5.7) 28 while the left side of (1.5.6) equals Z R n G 0 (y;)dy Z B2(y) u(x;) 2 dx = Z B 1=2 G 0 (y;)dy Z B2(y) u(x;) 2 dx + Z R n nB 1=2 G 0 (y;)dy Z B2(y) u(x;) 2 dx Z B 1=2 G 0 (y;)dy Z B2(y) u(x;) 2 dx: (1.5.8) The double integral on the far right side is bounded from below by Z B 1=2 G 0 (y;)dy Z B1(0) u(x;) 2 dx 1 2 (2) n=2 Z B1(0) u(x;) 2 dx 1 CM Z R n u(x;) 2 dx (1.5.9) where the first inequality holds by (1.5.3) if C is sufficiently large and the second one holds by (1.1.7). Therefore, we get ku(x;)k 2 L 2 CM kru(;)k L 2: (1.5.10) Using (1.5.1), we get a contradiction if CMq(). 29 Chapter 2 A local expansion for Stokes system In this part, we address local asymptotic development of local solutions for the Stokes equation in the unit cylinder. Namely, given f = (f 1 (x;t);f 2 (x;t);:::;f n (x;t)) we seek a polynomial in x which approximates a local solution u = (u 1 (x;t);u 2 (x;t);:::;u n (x;t)) of the Stokes system u t u +rp =f; (2.0.1) ru = 0; (2.0.2) around a point where the solution vanishes of orderd. The solution is not assumed to have a high degree of regularity and thus the Taylor expansion is not available. The results are then applied to the Oseen and Navier-Stokes systems. Fabre and Lebeau in [FL1, FL2] showed that the system (2.0.1)–(2.0.2) has a unique continuation property,i.e.,localsolutionsof (2.0.1)–(2.0.2)cannotvanishtoinfiniteorderunlesstheyvanishidentically. Having a priori estimates on solutions with respect to their vanishing order is considered a crucial step in many applications. For instance, using a priori estimates on asymptotic polynomials Han improved in [H2] the classical Schauder estimates in a way that the estimates of solutions and their derivatives at one point depend on the coefficient and the nonhomogeneous terms at that particular point. Also, Hardt and Simon [HS] applied an estimate of Donnelly and Fefferman for the order of vanishing of eigenfunctions 30 to find an asymptotic bound of the (n 1)-dimensional measure of v 1 j f0g, where v j is an eigenfunction corresponding to the j-th eigenvalue of the Laplacian on a compact Riemannian manifold. The method we use in proving the main theorem was introduced by Alessandrini and Vessella who found in [AV] an asymptotic development of a solution of a parabolic equation of an arbitrary degree (cf. also [B, H1, H2]). The main idea in [AV] is based on a local expansion of the corresponding funda- mental solution of the global linear equation. There are several difficulties when trying to extend the results to the Stokes system (2.0.1)–(2.0.2). First, due to presence of the pressure, it is not reasonable to expect that the velocity and the pressure would vanish at the same point; for instance, the unique continuation result of Fabre and Lebeau gives a unique continuation property foru and not for the pair (u;p). Thus in our main result we do not require p to vanish. The second difficulty is the lack of smoothing in the time variable in the system, which is a well-known problem for local solutions of the Stokes and Navier-Stokes systems. Indeed, taking the divergence of the evolution equation for the velocity gives p =rf (2.0.3) which is an equation with no smoothing in the time variable. The third difficulty is the nonlocal nature of the Stokes kernel. We note here that there have been many works on unique continuation of elliptic and parabolic equations showing that, under various assumptions on coefficients, no solution can vanish to infinite order (cf. [AE, AMRV, CRV, DF2, EFV, EV, GL, JK, KT, SS1, SS2] for instance); for more complete reviews, see [K1, K2, V]. Unique continuation questions for the Stokes and Navier-Stokes systems were addressed in [CK1, FL1, FL2, Ku1]. This chapter is organized as follows. In Section 2.1, we state the main results, Theorems 2.1.1 and 2.1.3, addressing the forces in standard and divergence forms respectively. We also state a corollary concerning the Oseen system. In Section 2.2, we recall the properties of the Stokes kernel, while the 31 last part contains a construction of a particular solution vanishing of order d as well as the proofs of Theorems 2.1.1 and 2.1.3. 2.1 Notation and the main result on the asymptotic expansion We consider a solution (u;p) of the Stokes system (2.0.1)–(2.0.2) in an open set containing (0; 0) (which can always be assumed using translation). For any (x;t)2 R n R and r > 0 we denote the parabolic cylinder label by (x;t) with radius r> 0 by Q r (x;t) = (y;s)2R n R :jyxj<r;r 2 <st< 0 : The corresponding parabolic norm for (x;t)2R n R is given by j(x;t)j = (jxj 2 +jtj) 1=2 (the homogeneity property being understood in the sense (x;t) = (x; 2 t) for 2 R). Denote by W m;1 q (Q 1 ) the Sobolev space of L q (Q 1 ) functions with x-derivatives up to m-th order and with the t-derivative belonging to L q (Q 1 ). Theorem 2.1.1. Let q> 1 +n=2. Suppose that f j 2L q (Q 1 ), for j = 1; 2;:::;n satisfy kf j k L q (Qr ) r d2++(n+2)=q ; r 1 (2.1.1) 32 for some constants > 0 and 2 (0; 1), where d 2 is an integer. Then for any (local) solution u = (u 1 ;:::;u n )2 W 2;1 q (Q 1 ) of (2.0.1)–(2.0.2) there exists P = (P 1 d;t ;:::;P n d;t ), whose each component P j d;t is a polynomial in x of degree less than or equal to d, such that ju j (x;t)P j d;t (x;t)jC + n X k=1 ku k k W 2;1 q (Q1) ! j(x;t)j d+ ; (x;t)2Q 1=2 ; (2.1.2) where C is a positive constant depending on n, q, d, and . Moreover, P satisfies the Stokes system @ t P j d;t (x) P j d;t (x) +@ j S(x) = d X i=d1 X jj=i E j ;t x ; j = 1;:::;d (2.1.3) rP = 0; (2.1.4) where S is the corresponding pressure. It is clear from the proof that the time regularity of the coefficients ofP depends on the time regularity of u; better regularity cannot be asserted due to the presence of the pressure. Since u2 W 2;1 q (Q 1 ), we obtain that the coefficients of P belong to W 1;q (1=2; 1=2). Above and in the sequel, we denote @ j =@=@x j for j = 1;:::;n. Since we are interested in obtaining estimates in Q 1 , we may assume that f(x;t) = 0; j(x;t)j 1: (2.1.5) Remark 2.1.2. The pressure term S in (2.1.3) is found explicitly in the proof of the main theorem; cf. (2.3.27) below. 33 In the case when the function on the right side of (2.0.1) is in the divergence form, the Stokes system reads as @ t u k u k +@ k p =@ j g jk ; k = 1;:::;n (2.1.6) ru = 0; (2.1.7) for some function g = (g jk ) n j;k=1 2L q (Q 1 ). Here we may also assume that g(x;t) = 0; j(x;t)j> 1: (2.1.8) Then we have the following variant of Theorem 2.1.1. Theorem 2.1.3. Assume that q > n + 2. Let g = [g jk ] 2 L q (Q 1 ) be an n n matrix of functions that satisfies kg jk k L q (Qr ) r d1++(n+2)=q ; r 1; j;k = 1;:::;n (2.1.9) for some constants > 0 and 2 (0; 1). Then for any solution u = (u 1 ;u 2 ;:::;u n ) in W 2;1 q of (2.1.6)– (2.1.7) there exists P = (P 1 d;t ;:::;P n d;t ) whose each component P j d;t is a polynomial in x of degree less than or equal to d such that ju j (x;t)P j d;t (x;t)jC + n X k=1 ku k k W 2;1 q (Q1) ! j(x;t)j d+ ; (2.1.10) for any (x;t)2Q 1=2 , where C is a positive constant depending on n, q, d, . Also, P satisfies the Stokes system 34 @ t P j d;t P j d;t +@ j S(x) = d X i=d1 X jj=i E j ;t x ! ; j = 1;:::;n (2.1.11) rP = 0 (2.1.12) where S is the corresponding pressure. Having a force in the divergence form on the right side of (2.1.7) allows us to apply the above results to the solutions of the Oseen system considered in [FL1]. We note that u is not assumed to be smooth in the space or time variable and thus the assertion cannot be obtained by expanding the solution in the Taylor series. Theorem 2.1.3 may be applied to the Oseen system (2.1.13)–(2.1.14) below considered by Fabre and Lebeau in [FL1], who studied unique continuation properties of solutions. Corollary 2.1.4. Let q> 1 +n=2: Suppose that u = (u 1 ;:::;u n )2W 2;1 q (Q 1 ) solves the Oseen system @ t u u + (ar)u +rp = 0; j = 1;:::;n (2.1.13) ru = 0 (2.1.14) where a = (a 1 ;:::;a n )2L 1 (Q 1 ). Also, assume that u vanishes of the order at least d 2. Then there exists P = (P 1 d;t ;:::;P n d;t ) whose each component P j d;t is a polynomial in x of degree less than or equal to d such that ju j (x;t)P j d;t (x)jCj(x;t)j d+ ; (x;t)2Q 1=2 ; (2.1.15) 35 for j = 1;:::;n, where 2 (0; 1) and C is a positive constant depending on n, d, q, , and u. Moreover, P satisfies the Stokes system @ t P j d P j d +@ j S(x) = d X i=d1 X jj=i E j ;t x ! ; j = 1;:::;n (2.1.16) rP = 0; (2.1.17) where S a suitable pressure term. Remark 2.1.5. It is straight-forward to apply the statements to the forced Navier-Stokes equations @ t u u +r (u u) +rp =f; (2.1.18) ru = 0: (2.1.19) Namely, suppose that u = (u 1 ;:::;u n )2 W 2;1 q (Q 1 ) is a solution of (2.1.18)–(2.1.19) where f2 L q (Q 1 ) andq>n. Also, assume thatu vanishes of the order at leastd 2. Then there existsP = (P 1 d;t ;:::;P n d;t ) whose each component P j d;t is a polynomial in x of degree less than or equal to d such that ju j (x;t)P j d;t (x)jCj(x;t)j d+1 ; (x;t)2Q 1=2 ; (2.1.20) for j = 1;:::;n, where C is a positive constant depending on n, d, q, and u. Moreover, P satisfies the Stokes system @ t P j d P j d +@ j S(x) = d X i=d1 X jj=i E j ;t x ! ; j = 1;:::;n (2.1.21) rP = 0; (2.1.22) 36 whereS a suitable pressure term. The statement follows directly from Theorem 2.1.1 since the condition q>n implies uru2L q (Q 1 ) using Gagliardo-Nirenberg embeddings. 2.2 The basic results We start by briefly recalling pointwise estimates on the derivatives of solutions (local or global) to the homogeneous heat equation @ t u u = 0: (2.2.1) The fundamental solution is given by (x;t) = (4t) n=2 exp jxj 2 =4t for t > 0 and (x;t) = 0 for t 0. Recall that the derivatives are bounded as j@ l t @ x (x;t)j C(;l) (jxj + p t) n+jj+2l e jxj 2 =8t ; l2N 0 ; 2N n 0 : (2.2.2) Also, recall that for any solution u of (2.2.1) in Q R , where R> 0, we have j@ l t @ x u(x;t)j C (Rj(x;t)j) jj+2l sup Q R juj; (x;t)2Q R=2 (2.2.3) where C depends onjj + 2l [Li]. We also briefly recall the basic facts on the fundamental solution to the Stokes system (2.0.1)–(2.0.2). Let u(0;) = u 0 be the initial condition. By uniqueness of solutions and under mild conditions on u at spatial infinity, we have u k (x;t) = Z R n (xy;t)u 0 (y)dy + Z t 0 Z R n (xy;ts)f k (y;s)dyds Z t 0 Z R n (xy;ts)@ k p(y;s)dyds; (2.2.4) 37 for k = 1;:::;n. Using the Fourier transform of both sides in (2.0.3), we get @ k p =R j R k f j ; k = 1;:::;n; where R j g = j ijj ^ g denotes the j-th Riesz transform and where ^ f() = R f(x)e ix dx. Thus (2.2.4) can be written as u k (x;t) = Z R n (xy;t)u 0 (y)dy + Z t 0 Z R n K jk (xy;ts)f j (y;s)dyds; (2.2.5) where K jk (x;t) = jk (x;t) +R j R k (x;t); j;k = 1;:::;n: (2.2.6) For each j;k = 1;:::;n, the function K jk solves the heat equation, i.e., @ t K jk (x;t) K jk (x;t) = 0; t> 0: (2.2.7) Also, @ j K jk (x;t) = 0; j = 1;:::;n: (2.2.8) Furthermore, we have the estimate j@ l t @ x K jk (x;t)j C(;l) j(x;t)j n+jj+2l ; l2N 0 ; 2N n 0 (2.2.9) for all (x;t)2R n R. [FJR, L02, S]. 38 2.3 Proof of the Main Theorem In the next lemma, we construct a solution of the system (2.0.1)–(2.0.2) which vanishes with a certain prescribed degree. Lemma 2.3.1. Assume that f = (f 1 ;:::;f n )2L q (Q 1 ), where q> 1 +n=2, satisfies (2.1.1) and (2.1.5). Then there exists u = (u 1 ;:::;u n )2W 2;1 q (Q 1 ) which solves (2.0.1)–(2.0.2) and satisfies kuk L q (Q1) ;kuk W 2;1 q (Q 3=4 ) C : (2.3.1) Furthermore, ju k (x;t)jC j(x;t)j d+ ; j(x;t)j 1=2; k = 1;:::;n (2.3.2) where is as in (2.1.1) and where the constant C depends on n, d, and . Proof of Lemma 2.3.1. We start by setting w k (x;t) = Z K jk (xy;ts)f j (y;s)dyds = Z j(y;s)j1 K jk (xy;ts)f j (y;s)dyds; k = 1;:::;n (2.3.3) and p = 1 rf = ( j ^ f j =ijj 2 ): (2.3.4) Then we have @ t w k w k +@ k p =f k ; k = 1;:::;n (2.3.5) 39 and, by the standard W 2;1 q estimate for the Stokes equation, kwk W 2;1 q (Q1) Ckfk L q (Q1) C : (2.3.6) Now, we consider the Taylor expansion of K jk (xy;ts) around (0; 0). Letj(y;s)j 1 be such that s6= 0. Denote by K m jk the m-th order terms, i.e., K m jk (x;y;t;s) = X jj+2l=m @ l t @ x K jk (y;s) x t l !l! : (2.3.7) It is easy to check that K m jk solves the heat equation for each j;k, i.e., @ t K m jk (x;y;t;s) x K m jk (x;y;t;s) = 0 for any (y;s). Let v k (x;t) = Z j(y;s)j<1 d X m=0 K m jk (x;y;t;s)f j (y;s)dyds; k = 1;:::;n: (2.3.8) Each v k is a polynomial of (joint in x;t) degree less than or equal to d and it satisfies @ t v k (x;t) v k (x;t) = 0: (2.3.9) Moreover, we have rv = 0: (2.3.10) Indeed, we may write @ k v k = Z j(y;s)j1 d X m=1 X jj+2l=m k >0 @ l t @ x K jk (y;s) x e k t l (e k )!l! f j (y;s)dyds; 40 where e k is the standard k-th unit vector inR n . Note that X jj+2l=m k >0 @ l t @ x K jk (y;s) x e k t l (e k )!l! = X j j+2l=m1 @ k @ l t @ x K jk (y;s) x t l !l! : Using @ k K jk = 0 for j = 1;:::;n, we getrv = 0. Now, set u k (x;t) =w k (x;t)v k (x;t) = Z j(y;s)j1 K jk (xy;ts) d X m=0 K m jk (x;y;t;s) ! f j (y;s)dyds (2.3.11) for k = 1;:::;n and note that we have @ t u k u k +@ k p =f k ; k = 1;:::;n: (2.3.12) We now check the condition (2.0.2). Since @ k w k = Z j(y;s)j1 @ k K jk (xy;ts)f j (y;s)dyds = 0 where we used @ k K jk = 0 for j = 1;:::;n, we getru = 0. Next, we claim that ju(x;t)jC j(x;t)j d+ ; j(x;t)j 1 2 : (2.3.13) 41 Fixingj(x;t)j 1=2, we split the integral on the far right side of (2.3.11) into three parts I 1 = Z j(y;s)j2j(x;t)j K jk (xy;ts)f j (y;s)dyds I 2 = Z j(y;s)j2j(x;t)j d X m=0 K m jk (x;y;t;s)f j (y;s)dyds I 3 = Z 2j(x;t)j<j(y;s)j<1 K jk (xy;ts) d X m=0 K m jk (x;y;t;s) ! f j (y;s)dyds: By a hypothesis, q> 1 +n=2. Therefore, by Hölder’s inequality and (2.2.9), jI 1 jC n X j=1 Z j(y;s)j2j(x;t)j dyds j(xy;ts)j nq 0 ! 1=q 0 Z j(y;s)j2j(x;t)j jf j (y;s)j q dyds ! 1=q C n X j=1 Z j(y;s)j<3j(x;t)j dyds j(y;s)j nq 0 ! 1=q 0 Z j(y;s)j<2j(x;t)j jf j (y;s)j q dyds ! 1=q C j(x;t)j (n+2)=q 0 n j(x;t)j d2++(n+2)=q =C j(x;t)j d+ ; where q 0 = 1=(q 1). Similarly, using (2.2.9) we estimate jI 2 jC n X j=1 d X k=0 j(x;t)j k Z j(y;s)j<2j(x;t)j jf j (y;s)j j(y;s)j n+k dyds C n X j=1 d X k=0 j(x;t)j k 1 X i=0 Z j(x;t)j=2 i <j(y;s)j<j(x;t)j=2 i1 jf j (y;s)j j(y;s)j n+k dyds C d X k=0 j(x;t)j k 1 X i=0 j(x;t)j 2 i dk+ C j(x;t)j d+ : (2.3.14) 42 In order to bound I 3 , we expand K jk (xy;ts) into Taylor series around (x;t) = (0; 0). For each j;k = 1;:::;n, we have K jk (xy;ts) = d X i=0 X jj+l=i @ l t @ x K jk (y;s) x t l !l! + X jj+l=d+1 @ l t @ x K jk (xy;ts) x t l !l! ; (2.3.15) where 0< =(x;t;y;s), =(x;t;y;s)< 1. Therefore, K jk (xy;ts) d X m=0 K m jk (x;y;t;s) = d X i=0 X jj+l=i jj+2ld+1 @ l t @ x K jk (y;s) x t l !l! + X jj+l=d+1 @ l t @ x K jk (xy;ts) x t l !l! : Using the bound onj@ l t @ x K jk (x;t)j, the difference above can be estimated with C 2(d+1) X i=d+1 1 j(y;s)j n+i + 1 j(xy;ts)j n+i j(x;t)j i : (2.3.16) As we assumed 2j(x;t)j<j(y;s)j, we get 2j(xy;ts)j>j(y;s)j: 43 Now, we may estimate jI 3 j Z 2j(x;t)j<j(y;s)j<1 K jk (xy;ts) d X m=0 K m jk (x;y;t;s) jf j (y;s)jdyds; C 2(d+1) X i=d+1 j(x;t)j i Z 2j(x;t)j<j(y;s)j<1 n X j=1 jf j (y;s)j j(y;s)j n+i dyds; C 2(d+1) X i=d+1 j(x;t)j i M X u=1 Z 2 u j(x;t)j<j(y;s)j<2 u+1 j(x;t)j n X j=1 jf j (y;s)j j(y;s)j n+i dyds; C 2(d+1) X i=d+1 j(x;t)j i M X u=1 (2 u j(x;t)j) (n+i) n X j=1 kf j k L q (Q 2 u+1 j(x;t)j ) Z 2 u j(x;t)j<j(y;s)j<2 u+1 dyds from where jI 3 j 2(d+1) X i=d+1 j(x;t)j i M X u=1 C (2 u j(x;t)j) d+i ; C j(x;t)j d+ 2(d+1) X i=d+1 M X u=1 1 (2 id ) u ; C j(x;t)j d+ ; which proves (2.3.2). For (2.3.1), recall that we have the same estimate on kw k k L q (Q1) by (2.3.6). Furthermore, we may show that for k = 1;:::;n jv(x;t)jC ; (x;t)2Q 1 ; (2.3.17) 44 by following the same approach that we took in estimating I 2 . Combining (2.3.17) with (2.2.3) and (2.3.9), we get kvk W 2;1 q (Qr ) C(r) ; r< 1; (2.3.18) which completes the proof of (2.3.1). Proof of Theorem 2.1.1. Suppose (u;p) solves (2.0.1)–(2.0.2). In Lemma 2.3.1, we have already con- structed a solution (e u;e p) of (2.0.1)–(2.0.2) withe u k 2W 2;1 q (Q 1 ) for each k = 1;:::;n, such that je u k (x;t)jC j(x;t)j d+ ; (x;t)2Q 1=2 for each k = 1;:::;n, where C is a constant depending on n, d, and . Also, we have ke u k k L q (Q1) ;ke u k k W 2;1 q (Q 1=2 ) C ; k = 1;:::;n: (2.3.19) Then we set U =ue u. Note that U solves the system U t U +r(pe p) = 0; (2.3.20) rU = 0: (2.3.21) We also have U 2 W 2;1 q (Q 1 ) andr(pe p)2 L q (Q 1 ). Now, we consider the vorticity equation. Let W =rU denote the curl of U, i.e., W i;j = @ i U j @ j U i for i;j = 1;:::;n. Then W = [W i;j ] nn satisfies the heat equation W t W = 0: 45 As W i;j =@ i (u j e u j )@ j (u i e u i ), we obtain n X i;j=1 kW i;j k L q (Qr ) C n X k=1 ku k k W 2;1 q (Qr ) +C n X k=1 ke u k k W 2;1 q (Qr ) (2.3.22) C +C n X k=1 ku k k W 2;1 q (Q1) (2.3.23) for any r2 (0; 1]. (We shall use this estimate for r2 [1=2; 1] below.) By expanding U k into Taylor series in x, we obtain U k (x;t) =P k d;t (x) +R k d;t (x); k = 1;:::;n; (2.3.24) where P k d;t (x) = d X i=0 X jj=i @ x U k (0;t) x ! (2.3.25) and R k d;t (x) = X jj=d+1 @ x U k ((x;t)x;t) x ! (2.3.26) where 0<(x;t) 1. Let (x;t)2Q 1=2 . Note that jR k d;t (x)jC X jj=d+1 j@ x U k ((x;t)x;t)jjx j Cjxj d+1 X jj=d+1 j@ x U k ((x;t)x;t)j Cj(x;t)j d+1 X jj=d+1 j@ x U k ((x;t)x;t)j: 46 Also, selecting 1=2<r 1 <r 2 < 3=4, we get j@ x U k ((x;t)x;t)jk@ x U k (;t)k L 1 x (B 1=2 (0;0)) Ck@ x U k (;t)k 1=2 L q x (B 1=2 (0;0)) k@ +n x U k (;t)k 1=2 L q x (B 1=2 (0;0)) +CkU k (;t)k L q x (B 1=2 (0;0)) CkU k (;t)k L q x (B 1=2 (0;0)) +Ck@ x U k (;t)k L q x (B 1=2 (0;0)) +Ck@ +n x U k (;t)k L q x (B 1=2 (0;0)) Cku k (;t)e u k (;t)k L q x (B 1=2 (0;0)) +C n X i;j=1 k@ 1 x W i;j (;t)k L q x (Br 1 (0;0)) +C n X i;j=1 k@ +n1 x W i;j (;t)k L q x (Br 1 (0;0)) : We may further bound k@ x W i;j k L 1 ([0;1=2];L q x (Br 1 (0;0))) CkW i;j k L q ([0;3=4];L q x (B 3=4 (0;0))) for any multiindex. By (2.3.19) and (2.3.23) we have the desired bound onR k d;t . Going back to (2.3.24), we obtain the extension ju k (x;t)P k d;t (x)jje u k (x;t)j +jR k d;t (x;t)j C j(x;t)j d+ +C + n X k=1 ku k k W 2;1 q (Q1) ! j(x;t)j d+1 ; for anyj(x;t)j 1=2. Furthermore, P satisfies (2.1.3) and (2.1.4). Taking the divergence of (2.3.25) we get @ k P k d;t = d1 X i=0 X jj=i1 @ k @ x U k (0;t) x ! = 0; 47 as U =ue u is divergence free. Similarly, @ t P k d;t (x) P k d;t (x) = d X i=0 X jj=i @ 1 t @ x U k (0;t) x ! d2 X i=0 X jj=i @ 2 k @ x U k (0;t) x ! = 0 @ d2 X i=0 X jj=i @ x (@ k p@ k e p) 1 A + d X i=d1 X jj=i @ t @ x U k (0;t) x ! =@ k 0 @ d2 X i=0 X jj=i @ x (pe p) 1 A + d X i=d1 X jj=i @ t @ x U k (0;t) x ! ; (2.3.27) which proves (2.1.3). Proof of Theorem 2.1.3. The arguments in the proof are similar to those in the proof of Theorem 2.1.1 except for some minor modifications. For the solution (e u;e p) in Lemma 2.3.1, we define w k (x;t) = Z K jk (xy;ts)@ i g ij dyds = (1) Z @ i K jk (xy;ts)g ij dyds (2.3.28) for each k = 1;:::;n, and (2.3.5) becomes @ t w k w k +@ k e p =@ j g jk ; k = 1;:::;n (2.3.29) where e p = 1 r k (@ j g jk ): 48 By (2.3.28), we have kwk W 2;1 q (Q1) Ckgk L q (Q1) C : Similarly, we rewrite v k (x;t) = Z j(y;s)j<1 d X m=0 @ i K m jk (x;y;t;s)g ij (y;s)dyds (2.3.30) where K m jk is as in (2.3.7). Note that we still have @ t v k (x;t) v k (x;t) = 0; k = 1;:::;n: (2.3.31) By (2.3.29) and (2.3.31), we see thate u =wv obeys @ t u k u k +@ k p =@ j g jk ; k = 1;:::;n: (2.3.32) With these changes,e u(x;t) satisfies the same estimate given in (2.3.13); for this, we split the integral into three parts and bound each one by C j(x;t)j d+ : The only difference is that we apply (2.2.9) on @ i K jk instead ofK jk and for that we needq>n+2. The rest of the details are the same as in Theorem 2.1.1. 49 Chapter 3 On localization and quantitative uniqueness for the elliptic partial differential equations 3.1 Introduction In this chapter, we address the spatial decay and the quantitative uniqueness properties for the elliptic equations u =Wru (3.1.1) and u =Vu: (3.1.2) In [M], Meshkov constructed a complex valued solution of (6.0.1) for whichjVj C in the exterior domain B c 1 =fx :jxj 1g, such that u satisfies ju(x)jC exp(jxj 4=3 =C) where C is a sufficiently large constant. This was a counterexample to Landis’s conjecture [KL] which predicted the largest allowable exponent to be 1 rather than 4=3. There are other important aspects of this example: It first demonstrates an optimality of Hörmander’s Carleman inequality (cf. (3.2.4) below). 50 Also, Kenig showed in [K2, p. 34] a connection with the qualitative behavior at a point by constructing a sequence of radii r j ! 0, potentials V j , withkV j k L 1 =M j , and solutions u j to u j +V j u j = 0, with ku j kC 0 andku j k L 1 (B1) 1 such that max jxj<rj ju j (x)jCr CM 2=3 j j . This is obtained by a translation and a dilation of the Meshkov solution. In [BC], Bakri and Casteras constructed a sequence of solutions u j of u j =V j u j : (3.1.3) withkV j k L 1!1 such thatu j vanishes at the point1 of orderkV j k 2=3 L 1, up to a constant. Furthermore, in [CKW], we have constructed solutions of (4.2.6), satisfying the Dirichlet or periodic boundary condi- tions, which vanish at a finite point of orderkV j k 2=3 L 1 (this is higher thankVk 1=2 L 1 conjectured in [Ku1]). We point out that all the examples, starting with Meshkov’s, are complex valued; thus the question of optimality of such upper bound remains open for real valued solutions. In [CK2], two of the authors used the frequency method ([Al, GL, Ku1, Ku3, Kur, P]) to estimate from above the order of vanishing for solutions of the parabolic equation @ t u u =Wru +Vu (3.1.4) with an upper bound CkVk 2=3 L 1 +CkWk 2 L 1 (3.1.5) where C depends on the Dirichlet quotient of u. The primary motivation is to be able to address the 2D Navier-Stokes equations, which in the vorticity formulation read @ t ! +ur! = 0 or the 3D Navier-Stokes equation @ t ! +ur!!ru = 0 51 where u is connected to ! through the Biot-Savart law. Basedon[CKW], theexponent 2=3forV in(3.1.5)isoptimal, butthequestionremainsiftheexponent 2 forW in (3.1.5) is the best possible as well. It is easy to check that for the elliptic type equation (6.0.2), Hörmander’s Carleman inequality leads to the same upper bound (3.1.5) for the order of vanishing of solutions of u =Wru +Vu; (3.1.6) with the Dirichlet boundary condition (for example) where C depends only on the domain. A similar question may be asked for the equivalent of the Landis-Meshkov question for the equation (6.0.2). In this chapter, we solve both questions by proving the following assertions. Regarding the decay at infinity for solutions of (6.0.2), we prove that there exists a solution u in B c 1 , such thatkWk L 1C and ju(x)jC exp(C 1 jxj 2 ): The construction is provided in Section 4.3 below. Also, in Section 6.2, we prove that this type of localization can not be improved upon. Regarding the degree of vanishing, the construction is similar to the one with V provided in [CKW]. For the equation (6.0.2), subject to either periodic or Dirichlet boundary conditions, we construct a solution which achieves the upper bound CkWk 2 L 1 as in [CK2]. Finally in the last section, we recall, without proofs, some recent results on the decay and the order of vanishing for the equation (6.0.1). The field of unique continuation has a very rich history, and we shall not survey it here (cf. instead [K1, K2, K3, V]); however, we point out some key results for the equation (6.0.2), (6.0.1), and (3.1.4). The theory was initiated by Carleman [C], who used a weighted type inequality to obtain a unique continuation result for the equation (6.0.1) in 2D. An extension to a general space dimension to regular elliptic equation was provided by Aronszajn and Cordes [Ar, Co]. In [H], Hörmander provided a Carleman estimate (cf. (3.2.4) below), which gives a strong unique continuation property for the equations (3.1.6) 52 with V and W in certain L p spaces. (Recall that an equation satisfies the strong unique continuation property if every solution which vanishes of infinite order at a point has to vanish identically.) A sharp L p condition on V which guarantees a strong unique continuation property was obtained by Jerison and Kenig in [JK], Koch and Tataru [KT], and Wolff [W1]. For the equations of the type (3.1.4), the sharp L p conditions were obtained in [E, EV, KT2]. For some other results on unique continuation, see [A, AN, AE, AMRV, Al, AV, B1, CK1, EF, EFV, GL, Kur, L, P] while for quantitative uniqueness, cf. [DF1, DF2, EVe, KSW, Ku2, Ku3, Ku4]. The chapter is organized as follows. In Section 6.1, we provide a lower bound for the decay of the equation (3.1.4). In Section 6.2, we construct a complex valued example showing that the rate provided in Section 6.1 for (6.0.2) is optimal. In Section 4.3, we construct a solution of (6.0.2) vanishing of the optimal order (3.1.5). Finally, in the last section, we summarize some known results for (6.0.2). 3.2 A lower bound for the decay of u = Wru +Vu In this section, we provide a lower bound on the decay of a solution of the equation u =Wru +Vu inR N ; (3.2.1) where N 1. Proposition 3.2.1. Let u be a solution to the equation (3.2.1) with u(0) = 1. Assume that jWj;jVj;jujC: (3.2.2) Then for any x 0 2R N withjx 0 j =R 1, we have the lower bound max jxx0j1 ju(x)j 1 C 0 exp (C 0 R 2 logR) 53 for some positive constant C 0 depending on C in (3.2.2). Note that all constants are allowed to depend on the dimension N. The proof follows the approach in [BK]. First, recall Hörmander’s Carleman type inequality from [BK, H]. There exists a (radial) increasing function (r) on [0; 1] for which r C (r)Cr; r2 [0; 1] (3.2.3) such that 3 Z u 2 12 + Z jruj 2 12 C Z (u) 2 22 ; C (3.2.4) for all u2C 1 0 (B 1 nf0g). Proof of Proposition 4.1.1. Since (3.2.1) is translation invariant, we may assume x 0 = 0. Define u(x) =u(KRx) where K > 0 is a large constant to be specified below. Then u satisfies u =KRWru + (KR) 2 Vu where we denote W (x) = W (KRx) and V (x) = V (KRx). From (3.2.2), we obtainjWj;jVj;juj C. Next, by the Sobolev embedding we may estimate kruk L 1 (Br ) Ckuk 1N=2p L p (Br ) kD 2 uk N=2p L p (Br ) + C r 1+N=p kuk L p (Br ) (3.2.5) where r > 0 and p > N=2 is sufficiently large. The interior regularity for elliptic equations (cf. [GT]) gives kD 2 uk L p (Br ) C r 2 kuk L p (B2r ) C r 2N=p kuk L 1 (B2r ) : 54 Using the above estimate in (3.2.5), we arrive at kruk L 1 (Br ) C r kuk L 1 (B2r ) which implies, by rescaling, kruk L 1 (Br ) C r kuk L 1 (B2r ) ; r> 0: (3.2.6) Choose a cut-off function with the properties 0 1; supp 1 4KR jxj 4 ; = 1 in 1 3KR jxj 3 ; j@ jC R jj ifjxj 1 3R ; j@ jC ifjxj 3; (3.2.7) with the last two inequalities holding for every multi-index 2 N N 0 . Applying the Carleman esti- mate (3.2.4) with u, we obtain 3 Z 12 (u) 2 + Z 12 jr(u)j 2 C Z 22 ((u)) 2 C Z 22 (u)) 2 +C Z 22 (rru)) 2 +C Z 22 (u) 2 ; C (3.2.8) for a certain increasing function satisfying r C (r)Cr for0r 4: 55 For the first term on the far right side of (6.3.31), we have Z 22 (u) 2 C(KR) 2 Z 22 2 jWj 2 jruj 2 +C(KR) 4 Z 22 2 jVj 2 u 2 C(KR) 2 Z 22 r(u)ur 2 +C(KR) 4 Z 22 2 jVj 2 u 2 C(KR) 2 Z 12 jr(u)j 2 +C(KR) 4 Z 12 2 jVj 2 u 2 +C(KR) 2 Z 22 ujrj 2 ; (3.2.9) where we used C forjxj 4. Note that suppr; supp 1 4KR jxj 1 3KR [f3jxj 4g: Therefore, the second and the third terms on the right side of (6.3.31) as well as the last term on the right side of (3.2.9) may be decomposed into integrals over the annulif1=4KRjxj 1=3KRg and f3jxj 4g. We first consider the integrals over the inner annulusf1=4KRjxj 1=3KRg. We have Z f1=4KRjxj1=3KRg 22 jruj 2 + (rru) 2 +juj 2 CR 22N R 2 sup jxj1=3KR juj 2 + sup jxj1=3KR jruj 2 +R 2 sup jxj1=3KR juj 2 ! ; (3.2.10) where we used (6.3.30). By (6.3.22), we get sup jxj1=3KR jruj 2 CR 2 sup jxj2=3KR juj 2 : Therefore, we have Z f1=4KRjxj1=3KRg 22 jruj 2 + (rru) 2 +juj 2 CR 2N+2 sup jxj2=3KR juj 2 : (3.2.11) 56 For the outer annulusf3jxj 4g, we use (x) (3); jxj 3 to deduce Z jxj3 22 jurj 2 + (rru) 2 +juj 2 C 22 (3)(C +CR 2+2N=p +C)C 22 (3)R 3 (3.2.12) if we choose p 2N, where we also used 2 2< 0. Combining (6.3.31), (3.2.9), (6.3.23), and (3.2.12) gives 3 Z 12 (u) 2 + Z 12 jr(u)j 2 C(KR) 2 Z 12 jr(u)j 2 +C(KR) 4 Z 12 2 u 2 +CR 2N+2 sup jxj2=3KR juj 2 +C 22 (3)R 3 : (3.2.13) Setting =CR 2 (3.2.14) with C sufficiently large, the first and the second terms on the right side are absorbed by the left side. By the assumptions of the lemma, there exists a2R N withjaj = 1=K such that u(a) = 1: In view of the gradient estimate (6.3.22), we get ju(x)u(a)jCjxaj 1 2 57 ifjxaj 1=2C and C is sufficiently large. Then we obtain a lower bound on the first term on the left side of (6.3.27) which reads 3 Z 12 (u) 2 1 C 3 12 2 K : Let C 0 > 0 be a constant. Using (3.2), we may take K sufficiently large and assure that (2=K) (3)=C 0 . Using this inequality and (3.2.14), we get 1 C 3 2 K 12 1 C R 3 C 0 (3) 1+2 where we used R 6 R 3 . If C 0 is sufficiently large, the last term of the right side of (6.3.27) is absorbed by the first term on the left side, and thus only the third term on the right side of (6.3.27) remains. We obtain 1 C 3 12 2 K CR 2N+2 sup jxj2=3KR juj 2 ; from where, using (3.2) and (3.2.14), sup jxj2=3KR juj 1 C exp (CR 2 logR); concluding the proof. 3.3 A construction of a localized solution In this section, we construct a solution of the equation (6.0.2) with an optimal decay provided in the previous section. 58 Theorem 3.3.1. There exists a complex valued solution u of the equation (6.0.2) inR 2 nB 1 , where W is complex valued with jWjC such that ju(x)jC exp 1 C jxj 2 where C > 0. From here on, denote r 1 r 2 = x2R 2 :r 1 jxjr 2 ; x2R 2 the annulus with inner radius r 1 and outer r 2 . Proof of Theorem 4.1.2. Let m2N be sufficiently large, i.e., mC. Denote n =m 2 and k = 12m + 16: (3.3.1) (so that we have (4.2.3) below). Throughout the proof, we work with complex valued functions and use polar coordinates (r;). First, we define C 1 radially-symmetric cut-off functions 1 and 2 such that 1 = 8 > > > < > > > : 1; rm + 1:5 0; rm + 1:9 59 and 2 = 8 > > > < > > > : 0; rm + 0:1 1; rm + 0:5: In addition, we require j (j) i jC; i = 1; 2; j = 0; 1; 2: Note that the function u 1 = 1 r n e in is harmonic. Also, set u 2 =br n+2k exp i(n + 2k) +i() (3.3.2) whereb> 0 and the real valued function is determined below. In order to define it, first note that the solutions of e in e i(n+2k) = 0; 2 [0; 2): are given by l = l n +k ; l = 0; 1;:::; 2n + 2k 1: Let T ==(n +k). Then is defined as a T-periodic C 1 function such that () =4k( l ); j l j T 5 : It is clear that ( l ) = (lT ) = 0; l = 0; 1;:::; 2n + 2k 1: Furthermore, we may assume that (i)j()j 5kT, 60 (ii)j 0 ()j 5k, (iii)j 00 ()jCkn for 2 [0; 2). By the choice of , the function u 2 is harmonic in the sectorsj l jT=5. Outside of these sectors, we have ju 2 j = (n 2k) 2 r 2 (n + 2k + 0 ) 2 r 2 +i 00 r 2 u 2 Ckn r 2 ju 2 j: (3.3.3) The factor b in (4.2.10) is chosen so that j@ 1 u 1 j =j@ 2 u 2 j; r =m + 1; this gives b = n n 2k 1 (m + 1) 2k : Now, we define u = 1 u 1 + 2 u 2 ; and prove that the ratiojuj=jruj stays bounded on the annulus mm + 2 wheneverjruj > 0. This immediately implies that W is bounded in these regions. For this purpose, we subdividemm + 2 intomm + 0:5[m + 0:5m + 1:5[m + 1:5m + 2 and consider each one of them. (i) On the first annulus mm + 0:5, we have 1 = 1 and 0 2 1. Therefore, we have ru = 0 B B @ u r u r 1 C C A = 0 B B @ i n r u 1 +i 2 n+2k+ 0 r u 2 n r u 1 + n+2k r 2 u 2 + 0 2 u 2 1 C C A : 61 By the bounds on 0 2 and 0 , we obtain j@ r uj n r ju 1 j n 2k r ju 2 j 4ju 2 j: (3.3.4) Note that u 1 u 2 = r n jbjr n+2k = 1 jbjr 2k n 2k n m + 1 m + 1=2 2k : By (6.2.10), we arrive at u 1 u 2 1 2 (1 + 1=m) 24m2C0 (1 + 1=2m) 24m+2C0 1 2 1 1 + e 24 e 12 e 10 2 (3.3.5) for > 0 sufficiently small. Therefore, j@ r uj n r e 10 2 ju 2 j 2(n 2k) r ju 2 j 1 C n r ju 2 j: Moreover, u = ( 2 u 2 ) = 2 u 2 + 2 0 2 @u 2 @r + 1 r 0 2 + 00 2 u 2 : (3.3.6) By the bounds onj 0 2 j andj 00 2 j, we obtain 2 0 2 @u 2 @r = 2j 0 2 j n 2k r ju 2 j Cn r ju 2 j (3.3.7) and 1 r 0 2 +j 00 2 j ju 2 jCju 2 j Cn r ju 2 j (3.3.8) where we also used k=r =O(1) and n=r =O(m). Combining (3.3.3), (3.3.6), (3.3.7), and (3.3.8), we get juj Cnk r 2 ju 2 j: 62 Therefore, juj jruj Cnk r 2 ju 2 j r nju 2 j Ck r =O(1) which completes the proof in the annulus mm + 0:5. (ii) On m + 1:5m + 2, we have 2 = 1 and 0 1 1. Then u = u 2 + 2 0 1 @u 1 @r + 0 1 r + 00 1 u 1 : A similar calculation as (3.3.5) gives u 1 u 2 n 2k n (m + 1) 2k (m + 1:5) 2k e 10 2 : Therefore, juj Ckn r 2 ju 2 j + Cn r ju 1 j +Cju 1 j Ckn r 2 ju 2 j: On the other hand j@ r uj n 2k r ju 2 jj 0 1 jju 1 j n Cr ju 2 j: Hence, juj j@ r uj Ck r =O(1) on m + 1:5m + 2. (iii) On m + 0:5m + 1:5, we have 1 = 2 = 1. Note that on this annulusjru 1 j andjru 2 j are comparable, and thusjruj may vanish. Define, A l = (r;) :m + 0:5rm + 1:5;j l j T 5 63 for l = 0; 1;:::; 2n + 2k 1. We claim that the zeros ofjruj can only be in the sectors A l . In fact, j@ r uj = n r u 1 + n 2k r u 2 = n r r n+2k (m + 1) 2k exp(iS()) (m + 1) 2k r 2k where S() = (2n + 2k) + (). Since S 0 () = 2n + 2k + 0 n> 0, the function S() is monotone on the sector [ l ; l+1 ] taking the values S( l ) = 2l and S( l+1 ) = 2(l + 1) at the end points. If 62A l for all l2f0; 1;:::; 2n + 2k 1g, there exists l such that l + T 5 l + 4T 5 : (3.3.9) Using the monotonicity of S and evaluating it at the end points in (4.2.17), we deduce that 2m + nT 5 S() 2(m + 1) nT 5 : Note that nT=5 =n=5(n +k) and k =O(n 1=2 ), so we may assume nT 5 7 ; which implies 2m + 7 S() 2(m + 1) 7 : Hence, exp(iS()) (m + 1) 2k r 2k sin 7 > 0; from where we arrive at j@ r uj n r r n+2k (m + 1) 2k sin 7 > 0: (3.3.10) 64 Thus we have proven that the zeros ofjruj can only be located in some A l . Since in A l the function u 2 is harmonic, we get u = u 1 + u 2 = 0: In these sectors, we simply set W = 0. Outside of these sectors, we have juj =ju 2 jC kn r ju 2 j: Combining this with (4.2.18), we get juj j@ r uj Ckn r 2 ju 2 j r n sin(=7) ju 2 j = Ck r =O(1) on the annulus m + 0:5m + 1:5. In the next step, we provide u on the annulus m + 2m + 3. We choose a C 1 cut-off function such that = 8 > > > < > > > : 1; rm + 2:5 0; rm + 2:75 with j (j) jC; j = 0; 1; 2: Let u =br n+2k exp i( (r)() + (n + 2k)) : 65 Note that u(r;) =u 2 (r;) for r =m + 2. On m + 2m + 3 we have u = n + 2k r +i 0 2 +i 0 r + 00 (n + 2k + 4 0 ) 2 r 2 + 2 00 r 2 ! u 8nk r 2 juj +j 0 j 2 jj 2 juj + n r j 0 jjjjuj +jjj 00 jjuj + n r 2 j jj 0 jjuj + j j 2 j 0 j 2 juj r 2 + j jj 00 jjuj r 2 : By the assumptions on , we have juj Cnk r 2 juj: On the other hand @ r u = n + 2k r u +i 0 (r)(r)u; which implies j@ r uj n 2k r juj Ck n juj n Cr juj provided n is larger than a constant. Hence, on the annulus m + 2m + 3, we obtain juj j@ r uj =O(1): In the next step, we consider the annulus m + 3m + 4, where we set u =br n+2k e i(n+2k) h(r); where h(r) = (r) + (1 (r))g(r) with = 8 > > > < > > > : 1; rm + 3:3 0; rm + 3:7: (3.3.11) 66 Then define g(r) =dr 4k where d = (m + 3) 4k so that g(m + 3) = 1. A straightforward computation shows that g(r) e 48 on m + 3m + 4, which implies e 48 h(r) 1: Note that @ r u = 8 > > > < > > > : n+2k r u; r2m + 3m + 3:3 n2k r u 3 g; r2m + 3:7m + 4 (3.3.12) where u 3 =br n+2k e i(n+2k) : Then we have j@ r uj n Cr ju 3 j (3.3.13) on the above annuli. On m + 3:3m + 3:7, we havejh 0 jC, which implies j@ r uj n 2k r ju 3 je 48 Cju 3 j n Cr ju 3 j> 0: One may also check that juj Cnk r 2 ju 3 j onm + 3m + 4: Therefore, juj jruj C onm + 3m + 4: 67 Byasimilarconstructionasintheintervalmm + 2,wemayrearrangethefunctionu 4 =bdr (n+2k) e i(n+2k) to u 5 =a 1 r (n+2k) e i(n+k) on m + 4m + 6, where a 1 =bd n + 2k (n +k)(m + 5) k : This completes the first step of the construction, with u = r m 2 e im 2 in a neighborhood offr = mg and u =a 1 r (m+6) 2 e i(m+6) 2 in a neighborhood offr =m + 6g. Before we start with the main construction on the plane, we estimatejuj. Definev(r) = maxfju(r;)j : 0 2g. Then we consider the moduli of the functions u i for i = 1; ; 5 which gives an upper bound for v(r). Note that U(r) =r n ; r2mm + 1 U(r) =br n+2k ; r2m + 1m + 3 U(r) =br n+2k h(r); r2m + 3m + 4 U(r) =bdr n2k ; r2m + 4m + 5 U(r) =a 1 r nk ; r2m + 5m + 6 (3.3.14) yields a piecewise smooth function which satisfies v(r) 2U(r) for r2 [m;m + 6]. Also, v(m) =U(m). Furthermore, d dr lnU(r) = n +O(k) r m 2 4(m + 6) r C : Therefore, lnv(r) lnv(m) ln 2 + Z r m d dt U(t) dt (3.3.15) ln 2 1 C Z r m sds: (3.3.16) 68 In order to extend the construction to the plane we repeat the same procedure on further annuli of the same length, and set m 1 = m, where m was our initial radius choice. Then, we recursively define m j+1 =m j + 6 and set n j =m 2 j . Thus k j =n j+1 n j = 12m j + 36: (3.3.17) Let u j denote the solution of (6.0.2) constructed on m j m j+1 . Then for r m, there exists j2 N such that m j rm j+1 , and a solution of (6.0.2) onfrmg is given by u = j1 i=1 a i u j (r;): For the annulus 0m, we define h(r) = 8 > > > < > > > : r m 2 +1 ; r 1 2 1; r 1 such that h 0 is smooth. Then we consider u = h(r)r m 2 e im 2 . Sincej@ r uj 1 for r 1=2 and j@ uj 1=C for r 1=2, we have juj jruj C sincejujC on 0m. This finishes the construction of the example on the plane. Moreover, given a radius r> 0, there exists l2N such that m + 6lrm + 6(l + 1). Then, lnv(r) = lnv(r) lnv(m + 6l) + l1 X j=0 (v(m + 6(j + 1))v(m + 6j)) + lnv(m): 69 Using the estimate in (3.3.16) we deduce that lnv(r)l ln 2 1 C Z r m sds +v(m)C 1 C r 2 : As a result, v(r)C exp(r 2 =C): The proof of the theorem is thus complete. 3.4 A construction of a solution vanishing of high order The following is the main result on construction of a solution to the problem u =Wru: (3.4.1) Theorem 3.4.1. Let n2N. There exists a smooth complex valued functions W and u such that (4.1.1) holds with jW (x)jCn; (3.4.2) the function u vanishes at 0 of order n 2 , andjuj is constant in B 1 nB 1=2 . The initial step of the construction relies on the following lemma. Lemma 3.4.2. Letn2N be an integer sufficiently large and letk = 2n1. Then there is aC 1 function u defined on the annulus 11 + 8=k such that u =r n 2 e in 2 ; on11 + 0:25=k 70 and u =br (n1) 2 e i(n1) 2 ; 1 + 7:75=k1 + 8=k where b = 1 (1 + 7=k) k (1 + 2=k) 2k : (3.4.3) Moreover, the function u satisfies (3.4.1) on 11 + 8=k with jWjCn where C is a sufficiently large constant. Ontheotherhand, thenextlemmaprovidesthem-thstepoftheiteration, addressingtheconstruction in the annulus m1 m where the radii are defined iteratively by m = m1 + 8 K m ; m2N 0 = 1 where n m =nm + 1 with k m = 2n m 1 and K m = 1 C 1 n 1=2 m n 1=2 + 1 n 1=3 n m + 1 n 5=4 m 1 ; m2N: Note that the expression for K m is comparable with 1 C min n n 1=2 m n 1=2 ;n 1=3 n m ;n 5=4 m o 71 Lemma 3.4.3. Assume that n>m be positive integers such that nmC with C a sufficiently large constant. Then there exists a C 1 function u defined on the annulus m1 m such that u =r (nm+1) 2 e i(nm+1) 2 ; on m1 m1 + 0:25=K m and u =br (nm) 2 e i(nm) 2 on m1 + 7:75=K m m1 + 8=K m where b = n 2 m n 2 m + 2k m 1 ( m1 + 7=K m ) km ( m1 + 2=K m ) 2km : (3.4.4) Moreover, the function u satisfies (3.4.1) on and (3.4.2) on m1 m1 + 8=K m . The next lemma then covers the case when n is smaller than a constant and connects the solution to u =e in . Lemma 3.4.4. Assume that nC. Then there exists a C 1 function u defined on the annulus 23 such that u =r n e in ; on22:3 and u =e in on2:63: Moreover, the function u obeys (3.4.1) and (3.4.2) on 23. Proof of Lemma 3.4.2. Since the proof uses similar ideas from [CKW], we only give a sketch the proof. Assume that n is sufficiently large. 72 1. Construction ofu on 01 + 4=k: In the first step, we connect the harmonic functionsu 1 =r n 2 e in 2 and u 2 =ar n 2 +2k e i((n 2 2k)+()) using u = 1 u 1 + 2 u 2 ; where 1 and 2 are C 1 smooth cut-off functions between 0 and 1 such that 1 = 8 > > > < > > > : 1; r 1 + 3 k 0; r 1 + 3:75 k and 2 = 8 > > > < > > > : 0; r 1 + 0:25 k 1; r 1 + 1 k : (3.4.5) Above the value a = 1 + 2 k 2k 2 (0; 1): is chosen so that the leading terms ofjru 1 j andjru 2 j agree at r = 1 + 2=k. Also, () is a function to be determined below. We shall tacitly assume for all cut-off functions are chosen in an optimal way in terms of bounds on derivatives. Here, we assume (j) i Ck j for j = 0; 1; 2 and i = 1; 2. In order to specify , we point out that the zeros of the equation e in 2 e i(n 2 2k) = 0 in [0; 2) are l = 2l 2n 2 2k ; l = 0; 1;:::; 2n 2 2k 1 73 Now, we define () = 4k( l ) in the neighborhoodj l jd=5 of the point l , where d = n 2 k : Additionally, we may also assume the optimality on the bounds for the derivatives, i.e., (i)j()j 5kd (ii)j 0 ()j 5k (iii)j 00 ()jCkn: (3.4.6) In the sectorsj l jd=5, wherel = 0; 1;:::; 2n 2 2k 1, we have u 2 = 0, while on the complement ju 2 j Cn 2 k r 2 ju 2 j + Cn 2 j 0 jju 2 j r 2 + j 0 j 2 ju 2 j r 2 + j 00 jju 2 j r 2 Cn 2 k r 2 ju 2 j using u = @ 2 u @r 2 + 1 r @u @r + 1 r 2 @ 2 u @ 2 : We need to check thatjuj=jruj is bounded by a constant on 11 + 4=k on the complement of the union of all the sectors. To establish this, we divide 11 + 4=k into three regions as in (3.4) and (3.4.5), but we only provide details for the inner annulus 11 + 1=k. There, we have have 1 = 1 and 0 2 1. Since u 1 is harmonic, we have juj =j(u 1 + 2 u 2 )j =j( 2 u 2 )j = 2 u 2 + 2 0 2 @u 2 @r + 1 r 0 2 + 00 2 u 2 ju 2 j +Ck(n 2 + 2k)ju 2 j +C(k +k 2 )ju 2 jCn 2 kju 2 j: (3.4.7) 74 On the other hand, we have ru = 2 6 6 4 @ r 2 u 2 + 2 @ r u 2 +@ r u 1 2 r 1 @ u 2 +r 1 @ u 1 3 7 7 5 Next we consider the ratio j@ u 1 j j@ u 2 j = 1 ar 2k n 2 n 2 2k + 0 (3.4.8) from where using (3.4.6) we further get j@ u 1 j j@ u 2 j 1 a(1 + 1=k) 2k 1 1 + 1=C = (1 + 2=k) 2k (1 + 1=k) 2k 1 1 + 1 + 1 C for n sufficiently large. We conclude that we may boundjruj from below as jrujj@ u 1 jj@ u 2 j 1 + 1 C j@ u 2 jj@ u 2 j 1 C j@ u 2 j n 2 C ju 2 j: (3.4.9) Using (3.4.7), we getjuj=jrujCk from where juj jruj Cn: (3.4.10) For the proof in the intermediate and outer annuli, we proceed similarly (cf. [CKW] with necessary modifications). 2. Construction of u on 1 + 4=k1 + 5=k: Here we connect the function u 2 from the first step with u 3 =ar n 2 +2k e i(n 2 2k) . Consider a smooth cut-off function such that (r) = 8 > > > < > > > : 1; r 1 + 4:5 k 0; r 1 + 4:75 k : 75 The function u =ar n 2 +2k exp i (r)() +i(n 2 2k) defined in 1 + 4=k1 + 5=k, satisfies juj = (n 2 + 2k) 2 r 2 u + 2(n 2 + 2k) + 1 r i 0 u + i 0 ( 0 ) 2 + i 00 r 2 u (n 2 2k) 2 r 2 u ( 0 ) 2 r 2 u 2 0 (n 2 2k) r 2 u Cn 2 k r 2 juj and j@ r uj n 2 + 2k r jujj 0 jjuj n 2 C juj: Therefore, we may divide and obtain (4.2.14). 3. Construction of u on r2 1 + 5=k1 + 6=k: Here, we let u =ah(r)r n 2 +2k e i(n 2 2k) =h(r)u 3 where h(r) = (r) + (1 (r))g(r) and g(r) = 1=r 4k with the cut-off function which satisfies = 8 > > > < > > > : 1; r 1 + 5 k + 1 3k 0; r 1 + 5 k + 2 3k : Note that jg(r)j 1 + 6 k 4k 1 C : 76 andthusjhj 1=C. Ontheannulus 1 + 5=k1 + 5=k + 1=3k,wehaveh = 1andu =u 3 =r n 2 +2k e i(n 2 2k) from where juj = (n 2 + 2k)(n 2 + 2k 1) r 2 u 3 + n 2 + 2k r 2 u 3 (n 2 2k) 2 r 2 u 3 = (n 2 + 2k) 2 (n 2 2k) 2 r 2 ju 3 jCn 2 kju 3 j: Since jru 3 j n 2 C ju 3 j; we get (3.4.10). In the annulus 1 + 5=k + 1=3k1 + 5=k + 2=3k, the proof is carried on in a similar manner. Finally, on 1 + 5=k + 2=3k1 + 6=k, we have u = 0, and thus we may set W = 0. 4.Constructionofuon 1 + 6=k1 + 8=k: Hereweconnecttheharmonicfunctionsu 4 =ar n 2 2k e i(n 2 2k) and u 5 = br n 2 k e i(n 2 k) , where b = (1 + 7=k) k a. The proof is similar to the arguments as in the construction on 01 + 4=k above and it is thus omitted. Proof of Lemma 3.4.3. Construction ofu on m1 m : Letu = 1 u 1 + 2 u 2 , where 1 and 2 are cut-offs which connect u 1 =r n 2 m e in 2 m and u 2 =a m r n 2 m +2km e i((n 2 m 2km)+()) with () specified below and a m = n 2 m n 2 m + 2k m m1 + 2 K m 2km : (3.4.11) The cut-offs 1 and 2 are monotone and satisfy 1 = 8 > > > < > > > : 1; r 1 + 3 Km 0; r 1 + 3:75 Km 77 and 2 = 8 > > > < > > > : 0; r 1 + 0:25 Km 1; r 1 + 1 Km : Recall that the zeros of the equation e in 2 m e i(n 2 m 2km) = 0 on [0; 2) are l = 2l 2n 2 m 2k m ; l = 0; 1;:::; 2n 2 m 2k m 1: (3.4.12) Now define () = 4k m ( l ) in the neighborhoodj l j d=5 where d = =(n 2 m k m ). As above, we assume thatj()j 5k m d withj 0 ()j 5k m andj 00 ()jCk m n m . In the sectorsj l jd=5, for l = 0; 1;:::; 2n 2 2k 1, u 2 is harmonic, while outside of the sectors ju 2 j = (n 2 m + 2k m ) 2 r 2 (n 2 m 2k m ) 2 r 2 2(n 2 m 2k m ) 0 r 2 j 0 j 2 r 2 +i 00 r 2 u 2 Cn 2 m k m r 2 ju 2 j: (3.4.13) We need to check thatjuj=jruj stays bounded on the annulus 11 + 4=K m where u6= 0. For this purpose, we separate the annulus into three regions, inner, outer, and intermediate as in the proof above. We only show details for the inner annulus m1 m1 + 1=K m . In this inner annulus, we have u =u 1 + 2 u 2 , as 1 = 1 and 0 2 1, whence u = (u 1 + 2 u 2 ) = 2 u 2 + 2 0 2 @u 2 @r + 1 r 0 2 + 00 2 u 2 : 78 Using (4.2.20), we get jujju 2 j +CK m (n 2 m + 2k m )ju 2 j +C(K m +K 2 m )ju 2 j C n 2 m k m +K m (n 2 m + 2k m ) +K m +K 2 m ju 2 j C(K m n 2 m +K 2 m )ju 2 j: (3.4.14) Next, we have j@ r u 1 j j@ r u 2 j = 1 a m r 2km n 2 m n 2 m + 2k m 1 a m (1 + 1=K m ) 2km n 2 m n 2 m + 2k m = (1 + 2=K m ) 2km (1 + 1=K m ) 2km = 1 + 1 1 +K m km 1 + 1 2K m km 1 + k m 4K m 1 + n m 4K m ; (3.4.15) where we used K m C. Therefore, jrujj@ r u 1 jj@ r u 2 jj@ r 2 jju 2 j 1 + 1 C n m K m n 2 m + 2k m r ju 2 j n 2 m + 2k m r ju 2 jCK m ju 2 j = n m K m 1 C n 2 m + 2k m r C K 2 m n m ju 2 j n 3 m CK m ju 2 j assuming r is bounded and nm is sufficiently large, since (n 2 m + 2k m =r) K 2 m =n m C by the definition of K m . By (3.4.9) and (4.2.21), we get juj jruj C K 2 m n m + K 3 m n 3 m C nn m n m +C nn 3 m n 3 m Cn: (3.4.16) 79 The treatment in the other annuli is the same as in the previous proof with the changes shown above. Thus we omit further details. Proof of Lemma 3.4.4. Assume that nC, and let u(x) =h(r)r n e in where h is a C 1 function such that h = 8 > > > < > > > : 1; r 2:3 r n ; r 2:6 : Now, we claim that u = Wru on 23, for some W withjWj C. Since u is harmonic on 22:3, we only need to consider the annulus 2:32:6, where u = r n e in h + 2@ r (r n e in )@ r h which givesjujC. On the other hand, we get @ u r = in r hr n e in ; from wherejruj 1=C and thusjuj=jrujC: Theorem 3.4.1 is based on the steps provided in Lemmas 3.4.2–3.4.4. Proof of Theorem 3.4.1. Due to Lemma 3.4.4, we may assume that nC. First, let u =r n 2 e in 2 ; jxj 1 + 8 k : 80 Denote b 1 = 1 (1 + 7=k) k (1 + 2=k) 2k (the right side of (3.4.3)) and b m = n 2 m n 2 m + 2k m 1 ( m1 + 7=K m ) km ( m1 + 2=K m ) 2km : (the right side of (3.4.4)). Applying Lemma 3.4.2 and then Lemma 3.4.3 successively, we may extend u to the annulus r m , where m is the largest integers such that nmC with C as in Lemma 3.4.3. Note that we have u = (1) m+1 b 1 b m1 r (nm+1) 2 exp (1) m i(nm + 1) 2 on m1 m1 + 0:25=K m for m = 1; 2;:::;m. After the final application of Lemma 3.4.3, we get u = (1) m b 1 b m r (nm) 2 exp (1) m+1 i(nm) 2 on m1 + 7:75=K m m1 + 8=K m : (3.4.17) But then Lemma 3.4.4 allows us to connect the function in (3.4.17) to a e in , wherenC The last step can be performed only if n X m=1 1 K m C; which holds by n X m=1 1 K m =C n X m=1 1 n 1=2 m n 1=2 + 1 n 1=3 n m + 1 n 5=4 m C (3.4.18) and the theorem is established. 81 3.5 The equation u = Vu In this section, we state, without proofs, the vanishing and localization results for the equation u =Vu: Regarding the localization, we have the important Meshkov’s result [M]: Theorem 3.5.1 ([M]). There exist smooth u and V on R 2 nB c 1 such that (3.5) holds, and we have kVk L 1 (B c 1 ) and ju(x)jC exp(jxj 4=3 =C): (3.5.1) for all x2B c 1 . In [CKW], we constructed a solution of (3.5) in B 1 vanishing of high order. Theorem 3.5.2 ([CKW]). For every n2 N, there exists a smooth complex valued function u on B 1 which satisfies (3.5) with the Dirichlet boundary conditions such that jV (x)jCn 3 ; x2B 1 ; such that the function u vanishes at 0 of order n 2 . Fortheproof, cf.[CKW]. Sincetheconstructedin[CKW]iscarriedoutsothatthesolutionisconstant onB 3=4 nB 1=2 , it is easy to adjust the example to be periodic, satisfy Neumann boundary conditions etc. It is also possible to use the solution constructed for Theorem 4.1.3 to obtain a solution of a parabolic equation @ t u u =Vu onR 2 (1; 1), with 1-periodic boundary condition on the sides with the following property. 82 Theorem 3.5.3 ([CKW]). For every n2 N, there exists a smooth complex valued solution u of the problem above, which satisfies sup t2(1;0) q(t)n 3 where q(t) = R (0;1) 2 jru(;t)j 2 R (0;1) 2 ju(;t)j 2 and jV (x;t)jCn 3 : forR 2 (1; 0). Moreover, the function u vanishes of order n 2 at (0; 0). In particular, this theorem shows the optimality of the estimate (3.1.5) (when W = 0) in [CK2]. 83 Chapter 4 On quantitative uniqueness for elliptic equations The main purpose of this chapter is to construct smooth complex valued solutions to the problem u =Vu; (4.0.1) which vanish of the order comparable withkVk 2=3 L 1 in the unit disk of R 2 with Dirichlet boundary con- dition. We start with construction of such solutions in the unit disk so that they satisfy a doubling type property. Then we modify the solutions to be periodic or that they satisfy the Dirichlet boundary condition. Finally, we construct solutions of a parabolic equation @ t u u =Vu (4.0.2) onR 2 (1; 1) under some growth assumptions, showing the sharpness of the estimate in [CK2]. Unique continuation for elliptic and parabolic PDE has a rich history. Here, we can only briefly sum- marize available results and refer to [K1, K2, V, W2] for comprehensive treatments. The classical paper by Carleman [C] established a unique continuation theorem (UCP) for second order elliptic operators 84 without the analyticity assumption, and the method that he introduced has played a central role in the unique continuation theory since then. Using a weighted type inequality, he showed that P (x;D) = +V (x); x2R 2 has the strong UCP whenever V is in L 1 loc (R 2 ). Recall that a strong UCP holds for a PDE operator if every solution which vanishes to infinite order at any x 0 2 , i.e., lim r!0 1 r n Z B(x0;r) juj 2 dx = 0; n 0; vanishes in . On the other hand, we have the UCP if vanishing of a solution in an open subset of implies vanishing in . In 1950’s, Aronszajn [Ar] and Cordes [Co] extended Carleman’s result to higher dimensions. Later, a very powerful Carleman estimate was provided by Hörmander [H]. There have been many results since then under different integrability assumption on the lower order coefficients, and sharp statements have been established by Jerison and Kenig [JK] and Koch and Tataru [KT] for elliptic and by Escauriaza and Vega [E, EV], Koch and Tataru [KT2], and Sogge [S2] for parabolic type equations. For some other results on unique continuation, see [AE, AMRV, AV, B1, B2, CRV, EF, EFV, EVe, IK, K3, KSW, S1], while for an alternative approach to unique continuation using the frequency (also called the logarithmic convexity) method, cf. [A, AN, Al, GL, Ku1, Ku2, Ku3, Kur, L, P]. Recently, there have been more efforts dedicated to the question of quantitative uniqueness. Here, we assume that the coefficients are regular and estimate the order of vanishing in terms of the size of the coefficients. The initial work was by Donnelly and Fefferman, who proved in [DF1] that the order of vanishing of an eigenfunction, denoted byO(u), is estimated by O(u)C p 85 where is the corresponding eigenvalue. It is natural to ask whether the order of vanishing of a solution to the problem u =Vu +Wru; with, say, Dirichlet or periodic boundary conditions, is bounded by C(kWk L 1 +kVk 1=2 L 1) (4.0.3) (cf. [Ku1]). Although the precise best bound is unknown, Kenig predicted in [K2] that the order of vanishing of any solution of the elliptic boundary problemu =V (x;t)u is bounded by CkVk 2=3 L 1 (4.0.4) rather than CkVk 1=2 L 1. In [CK1], two of the authors addressed the question of quantitative uniqueness for the parabolic equation (4.0.2). As it is the case for the elliptic equation, it is an open problem whether, under natural growth conditions at infinity, the quantityC(kWk L 1 +kVk 1=2 L 1) provides an upper bound for the order of vanishing for solutionsu of (4.1.4). Our main result asserts that the order of vanishing may be estimated by C(kWk 2 L 1 +kVk 2=3 L 1) matching the bound resulting from the Carleman estimate in [BK, H, K2, V]. In [M], Meshkov considered the following related question due to Landis on the behavior around the spatial infinity. The basic question is the following: If u is a solution of uVu = 0, with V bounded by a constant, what is the quickest rate by which the function decays? Meshkov proved that there exists a complex valued solution of such an equation for whichju(x)j C exp(jxj 4=3 =C) for some constant C > 0. He also proved that for every > 0, there exists C > 0 such that ifju(x)jC 1 exp(jxj 4=3 ), then u 0. In [K2, p. 234], Kenig used Meshkov’s solution to provide an argument suggesting that (4.0.3) may not be the sharp estimate for the order of vanishing. He namely proved that there exists a sequence of radii r j ! 0, potentials V j ,kV j k L 1 =M j and solutions to u j +V j u j = 0, withku j kC 0 , 86 ku j k L 1 (B1) 1 and max jxj<rj ju j (x)j Cr CM 2=3 j j . (Note that r j depends on u j and thus solutions u j actually do not need to vanish.) In [BC], Bakri and Casteras constructed a solution to (4.0.1), which vanishes at the point infinity of order (4.0.4). As far as we know, solutionsu j of the equation u j =V j u j subject to a boundary condition (or to a suitable doubling condition) which vanish at a finite point with the order of vanishing at least c 0 kV j k 2=3 L 1 (4.0.5) where c 0 > 0 is a constant andkV j k L 1!1 have not been found. In this chapter we provide such an example by constructing a sequence of solutionsfu j g of u j = V j u j withkV j k L 1!1 which vanish of order (4.0.5) and which satisfy a boundary value problem, which is either periodic or Dirichlet (other boundary conditions can be easily obtained as well). As it is the case in [M], the solutions are complex- valued. It remains an open problem whether there exist real-valued solutions with real-valued potentials V, which exhibit a rate of vanishing c 0 kVk L 1 for any 2 (1=2; 2=3]. For other counterexamples for unique continuation withV belonging to a Lebesgue spaces L p or L p w , we refer the readers to [KN, KT3, Ma, W1, W3] and references therein. The chapter is organized as follows. In Section 6.1, we state the main results. In Theorem 4.1.1, we provide an example of a solution vanishing of the above mentioned order which is constant in B 1 nB 1=2 . This allows us to extend the solution periodically, or to adjust it (as we do in Theorem 4.1.2) to a solution of the Dirichlet problem. Theorem 4.1.3 provides an example of a solution of (4.0.2), vanishing of high order. The proofs are based on three lemmas, stated in Section 6.2. The first provides the construction in the initial annulus, the second one gives the construction in higher annuli, while the third one treats the final annulus. The lemmas are finally used in Section 4.3 to provide proofs of all three main theorems in Section 6.1. 87 4.1 Notation and the main result The following is our first main result on construction of a periodic solution to the problem u =Vu: (4.1.1) All the examples are in R 2 and are complex valued. Note that the periodic examples can be easily extended in any number of dimensions. Theorem 4.1.1. For everyn2N, there exists a smooth complex valued function u onB 1 which satisfies (4.1.1) such that jV (x)jCn 3 ; (4.1.2) the function u vanishes at 0 of order n 2 , and u is constant in B 1 nB 1=2 . Consequently, u can be extended to a periodic solution with the same properties. Once the construction as in Theorem 4.1.1 is established, we modify the solution and the equation so thatu satisfies the Dirichlet boundary conditions in a given smooth domain. For simplicity, we assert this in the case of a disk. Theorem 4.1.2. For everyn2N, there exists a smooth complex valued function u onB 1 which satisfies (4.1.1)–(4.1.2) with the boundary condition u @B1 = 0; (4.1.3) and which vanishes at 0 of order n 2 . It is possible to use the solution constructed for Theorem 4.1.1 to obtain a solution of a parabolic equation @ t u u =Vu (4.1.4) 88 onR 2 (1; 1), with 1-periodic boundary condition on the sides with the following property. Theorem 4.1.3. For every n2N, there exists a smooth complex valued solution u of the problem above which satisfies sup t2(1;0) q(t)n 3 where q(t) = R (0;1) 2 jru(;t)j 2 dx R (0;1) 2 ju(;t)j 2 dx (4.1.5) and jV (x;t)jCn 3 forR 2 (1; 0). In addition, the function u vanishes of order n 2 at (0; 0). In [CK2], two of the authors established an upper bound C kWk 2 L 1 +kVk 2=3 L 1 + sup t2(1;0) q(t) 1=2 (4.1.6) for the order of vanishing at (0; 0). Theorem 4.1.3 shows optimality of the bound (6.0.4). 4.2 Construction In this section, we provide lemmas used in the proofs of Theorems 4.1.1–4.1.3. The initial part of the construction in the proof of Theorem 4.1.1 relies on the following statement. Lemma 4.2.1. Let n2N be an integer such that n 7; (4.2.1) and let k = 2n 1: 89 Then there exists a C 1 function u defined on the annulus 1jxj 1 + 8 k which satisfies u =r n 2 e in 2 ; 1jxj 1 + 0:25 k (4.2.2) and u =br (n1) 2 e i(n1) 2 ; 1 + 7:75 k jxj 1 + 8 k where b = 1 (1 + 7=k) k (1 + 2=k) 2k : (4.2.3) Moreover, the function u obeys u =Vu; 1jxj 1 + 8 k where jVjCn 3 : A slight modification of the proof gives the next lemma, which provides the m-th step in the iteration below, i.e., a construction of the solution on the annulus r2 [ m1 ; m ] where 0 = 1 1 = 1 + 8 2n 1 90 and m = m1 + 8 K m ; m2N: Also, n m =nm + 1 with k m = 2n m 1 and K m = 1 C n 1=2 m n 3=2 + 1 nn 1=3 m ! 1 ; m2N; (4.2.4) where C is chosen so that K m k m for m2N. Note that the expression for K m is comparable with 1 C min n 3=2 n 1=2 m ;nn 1=3 m : We point out that the choice (6.0.2) is set due to requirements in (4.2.23) below. Furthermore, it is important below that n2 X m=1 1 K m C: (4.2.5) Lemma 4.2.2. Let n;m2 N be such that n 7 and nm 3. Then there exists a C 1 function u defined on the annulus m1 jxj m which satisfies u =r (nm+1) 2 e i(nm+1) 2 ; m1 jxj m1 + 0:25 K m and u =br (nm) 2 e i(nm) 2 ; m1 + 7:75 K m jxj m1 + 8 K m where b = 1 (1 + 7=K m ) km (1 + 2=K m ) 2km : (4.2.6) 91 Moreover, the function u obeys u =Vu; m1 jxj m1 + 8 K m where jVjCn 3 : Note that the two lemmas only apply when n is sufficiently large. Thus we need the next statement, which allows us to connect the solution of the type (4.2.2), when n is less than a constant, to a constant function. Lemma 4.2.3. Let n be an integer such that nC. Then there exists a C 1 function u defined on the annulus 2jxj 3 which satisfies u =r n e in ; 2jxj 2 + 1 9 and u =M; 3 1 9 jxj 3 where MC is a constant. Moreover, the function u obeys u =Vu; 2jxj 3 where jVjC: Proof of Lemma 4.2.1. First, fix n2f7; 8; 9;:::g. 92 Construction of u for r2 [1; 1 + 4=k]: Define u = 1 u 1 + 2 u 2 ; where 1 and 2 are C 1 cut-off functions which connect the harmonic function u 1 =r n 2 e in 2 with u 2 =ar n 2 +2k e i((n 2 2k)+()) (4.2.7) where () is a function to be determined below and a = 1 + 2 k 2k 2 (0; 1): (4.2.8) The value of a is chosen as in (6.0.1) in order to ensure thatju 1 j andju 2 j agree at r = 1 + 2=k. Smooth radial cut-off functions 1 and 2 take values between 0 and 1 and satisfy 1 = 8 > > > < > > > : 1; r 1 + 3 k 0; r 1 + 3:75 k and 2 = 8 > > > < > > > : 0; r 1 + 0:25 k 1; r 1 + 1 k with j (j) i jCk j ; j = 0; 1; 2; i = 1; 2: (4.2.9) 93 Before specifying , we first find the zeros of the equation e in 2 e i(n 2 2k) = 0 on [0; 2), which are l = 2l 2n 2 2k ; l = 0; 1;:::; 2n 2 2k 1: (4.2.10) We define by setting () = 4k( l ) in the neighborhoodj l jd=5 of the point l , where d = n 2 k (4.2.11) and observe ( l ) = (ld) = 0; l = 0; 1;:::; 2n 2 2k 1: Additionally, we assume: (i)j()j 5kd (ii)j 0 ()j 5k (iii)j 00 ()jCkn. In the sectorsj l jd=5, wherel = 0; 1;:::; 2n 2 2k 1, we have u 2 = 0. Everywhere else, we have ju 2 j = (n 2 + 2k) 2 r 2 (n 2 2k) 2 r 2 2(n 2 2k) 0 r 2 j 0 j 2 r 2 +i 00 r 2 u 2 Cn 2 k r 2 ju 2 j + Cn 2 j 0 jju 2 j r 2 + j 0 j 2 ju 2 j r 2 + j 00 jju 2 j r 2 Cn 2 k r 2 ju 2 j; (4.2.12) 94 where we used the polar representation of the Laplacian u = @ 2 u @r 2 + 1 r @u @r + 1 r 2 @ 2 u @ 2 : We now claim that the ratiojuj=juj stays bounded on [1; 1 + 4=k] as long as u6= 0. (In the regions where u = 0, we simply set V = 0.) In order to prove the claim, we separate the annulus r2 [1; 1 + 4=k] into three regions, treated in (i), (ii), and (iii) below. (i) The inner annulus [1; 1 + 1=k]: Here we have 1 = 1 and 0 2 1, which gives u =u 1 + 2 u 2 . Since u 1 is harmonic, we have u = ( 2 u 2 ) = 2 u 2 + 2 0 2 @u 2 @r + 1 r 0 2 + 00 2 u 2 : Thus, using (4.2.9) and (4.2.12), we have jujju 2 j +Ck(n 2 + 2k)ju 2 j +C(k +k 2 )ju 2 jCn 2 kju 2 j: We check the ratio ju 1 j ju 2 j = r n 2 ar n 2 +2k = 1 ar 2k 1 a(1 + 1=k) 2k = (1 + 2=k) 2k (1 + 1=k) 2k 1 + 1 C : We conclude that we may boundjuj from below as jujju 1 jju 2 j 1 + 1 C ju 2 jju 2 j 1 C ju 2 j: (4.2.13) 95 Note that the last expression is strictly positive. Using (4.2.12), we get juj juj Cn 2 k: (4.2.14) (ii) The outer annulus [1 + 3=k; 1 + 4=k]. Here we have 0 1 1 and 2 = 1 and thus u = 1 u 1 +u 2 . In this annulus, we have ju 2 j ju 1 j =ar 2k 1 + 2 k 2k 1 + 3 k 2k 1 + 1 C ; whence jujju 2 jju 1 j 1 C ju 1 j> 0: Also, jujju 2 j + 2 0 1 @u 1 @r + 1 r 0 1 + 00 1 u 1 Cn 2 kju 2 j +Ckn 2 ju 1 j +C(k +k 2 )ju 1 jCn 2 kju 1 j where we used ju 2 jar 2k ju 1 j 1 + 2 k 2k 1 + 4 k 2k Cju 1 j: Thus we again obtain (4.2.14). (iii) The intermediate annulus [1 + 1=k; 1 + 3=k]. Here we have 1 = 2 = 1 and u =u 1 +u 2 . As above, u = 0 on the sectors A l = (r;) : 1 + 1 k <r< 1 + 3 k ;j l j d 5 ; l = 0; 1;:::; 2n 2 2k + 1 96 where we thus set V = 0. It remains to consider the non-harmonic sectors P l = (r;) : 1 + 1 k r 1 + 3 k ; l + d 5 l + 4d 5 ; l = 0; 1;:::; 2n 2 2k 1: On P l , we have u =u 1 +u 2 =r n 2 e in 2 ar n 2 +2k e i((n 2 2k)+) from where juj =ar n 2 +2k e i((2n 2 2k)+()) 1 ar 2k =ju 2 j e i((2n 2 2k)+()) 1 ar 2k (4.2.15) where l2f0; 1;:::; 2n 2 2k 1g is fixed. We now consider S() = (2n 2 2k) + (): On the segment [ l ; l+1 ], we have ( l ) = 0 and thus, by (4.2.10), S( l ) = (2n 2 2k) l = 2l and S( l+1 ) = 2(l + 1). Moreover, S 0 () = 2n 2 2k + 0 () 2n 2 7k = 1 2 k 2 6k + 1 2 0; which holds since (4.2.1) gives k 12. Therefore, S is nondecreasing on the segment [ l ; l+1 ]. Now, we consider the behavior of S on the interval 2 [ l +d=5; l + 4d=5]. Since S is nondecreasing, we only need to check the values on the boundary of the interval, where we have S l + d 5 = (2n 2 2k) l + (2n 2 2k) d 5 + l + d 5 = 2l + 2n 2 d 5 + 2k d 5 97 and S l + 4d 5 = (2n 2 2k) l + (2n 2 2k) 4d 5 + l + 4d 5 = 2l + 8 5 + l + 4d 5 = 2l + 8 5 4k d 5 = 2(l + 1) 2 5 4k d 5 : Using (4.2.11), we get 2l + 2 5 n 2 +k n 2 k S() 2(l + 1) 2 5 n 2 +k n 2 k and thus 2l + 2 5 S() 2(l + 1) 2 5 (4.2.16) for 2 [ l +d=5; l + 4d=5]. Then for any 0, we have je iS() j = cos(S()) +i sin(S()) = j sin(S())j 2 +j cos(S())j 2 1=2 1 C where we used (4.2.16) and 0 in the last step. Using this in (4.2.15), we obtain juj 1 C ju 2 j while (4.2.12) gives juj =ju 2 jCn 2 kju 2 jCn 3 ju 2 j: Construction of u for r2 [1 + 4=k; 1 + 5=k]: Here we connect the function u 2 in (6.3.2) with u 3 =ar n 2 +2k e i(n 2 2k) : 98 Consider a radial C 1 function such that (r) = 8 > > > < > > > : 1; r 1 + 4:5 k 0; r 1 + 4:75 k with j (j) (r)jCk j ; j = 0; 1; 2: We then set u =ar n 2 +2k exp i (r)() +i(n 2 2k) in the annulus 1 + 4=kr 1 + 5=k. Note that juj = (n 2 + 2k) 2 r 2 u + 2(n 2 + 2k) + 1 r i 0 u + i 00 ( 0 ) 2 + i 00 r 2 u (n 2 2k) 2 r 2 u ( 0 ) 2 r 2 u 2 0 (n 2 2k) r 2 u Cn 2 k r 2 juj and sincejuj> 0, we may divide and obtain juj juj Cn 2 k: (4.2.17) Construction of u for r2 [1 + 5=k; 1 + 6=k]: In this part, let u =ah(r)r n 2 +2k e i(n 2 2k) =h(r)u 3 where h(r) = (r) + (1 (r))g(r) 99 and g(r) = 1 r 4k with a radial cut-off function which satisfies = 8 > > > < > > > : 1; r 1 + 5 k + 1 3k 0; r 1 + 5 k + 2 3k : Also, we may assume that jh (j) jCk j ; j = 1; 2: Note that jg(r)j 1 + 6 k 4k 1 C and thus jhj 1 C : On [1 + 5=k; 1 + 5=k + 1=3k], we have h = 1 and u =u 3 =ar n 2 +2k e i(n 2 2k) and thus juj = (n 2 + 2k)(n 2 + 2k 1) r 2 u 3 + n 2 + 2k r 2 u 3 (n 2 2k) 2 r 2 u 3 = (n 2 + 2k) 2 (n 2 2k) 2 r 2 ju 3 jCn 2 kju 3 j: Asju 3 j> 0, we get (4.2.17). For r2 [1 + 5=k + 1=3k; 1 + 5=k + 2=3k], we have juj = hu 3 + 2h 0 n 2 + 2k r u 3 + h 0 r +h 00 u 3 Cn 2 kju 3 j +Ck(n 2 +k)ju 3 j + (k +k 2 )ju 3 jCn 2 kju 3 j; 100 and thus we obtain (4.2.17). Finally, for r2 [1 + 5=k + 2=3k; 1 + 6=k], we have u =ar n 2 2k e i(n 2 2k) (4.2.18) which satisfies u = 0, and thus we may set V = 0. Construction of u for r2 [1 + 6=k; 1 + 8=k]: Note that the solution at the end of the previous step is given by (4.2.18) for r 1 + 5=k + 2=3k. Here we interpolate between the harmonic functions u 4 =ar n 2 2k e i(n 2 2k) and u 5 =br n 2 k e i(n 2 k) : Choose b> 0 such that u 4 1 + 7 k = u 5 1 + 7 k ; i.e., b = (1 + 7=k) k a. Here we use radial C 1 functions 1 and 2 such that 1 = 8 > > > < > > > : 1; r 1 + 7:5 k 0; r> 1 + 7:75 k and 2 = 8 > > > < > > > : 0; r 1 + 6:25 k 1; r> 1 + 6:5 k : 101 We set u = 1 u 1 + 2 u 2 . It is clear that u is harmonic on [1 + 6:5=k; 1 + 7:5=k]. Similarly, following the arguments on [1; 1 + 1=k] and [1 + 3=k; 1 + 4=k] above, it is easy to check thatjuj=juj remains bounded by Cn 2 k on 1 + 6 k ; 1 + 6:5 k [ 1 + 7:5 k ; 1 + 7 k : This concludes the construction on [1 + 6=k; 1 + 8=k]. Proof of Lemma 4.2.2. Fix n2f7; 8; 9;:::g. Construction of u for r2 [ m1 ; m ]: Similarly to the construction in Lemma 4.2.1, we start with the annulus [ m1 ; m1 + 4=K m ]. Let u = 1 u 1 + 2 u 2 ; where 1 and 2 are cut-offs which connect u 1 =r n 2 m e in 2 m and u 2 =a m r n 2 m +2km e i((n 2 m 2km)+()) with () specified below and a m = m1 + 2 K m 2km : The radial cut-offs 1 and 2 are monotone and satisfy 1 = 8 > > > < > > > : 1; r 1 + 3 Km 0; r 1 + 3:75 Km 102 and 2 = 8 > > > < > > > : 0; r 1 + 0:25 Km 1; r 1 + 1 Km with j (j) i jCK j m ; j = 0; 1; 2; i = 1; 2: (4.2.19) Recall that the zeros of the equation e in 2 m e i(n 2 m 2km) = 0 on [0; 2) are l = 2l 2n 2 m 2k m ; l = 0; 1;:::; 2n 2 m 2k m 1: Now define () = 4k m ( l ) in the neighborhoodj l jd=5 of the point l , where d = n 2 m k m : As above, we assume that (i)j()j 5k m d (ii)j 0 ()j 5k m 103 (iii)j 00 ()jCk m n m . In the sectorsj l jd=5, where l = 0; 1;:::; 2n 2 2k 1, we have u 2 = 0, while elsewhere ju 2 j = (n 2 m + 2k m ) 2 r 2 (n 2 m 2k m ) 2 r 2 2(n 2 m 2k m ) 0 r 2 j 0 j 2 r 2 +i 00 r 2 u 2 Cn 2 m k m r 2 ju 2 j + Cn 2 m j 0 jju 2 j r 2 + j 0 j 2 ju 2 j r 2 + j 00 jju 2 j r 2 Cn 2 m k m r 2 ju 2 j: (4.2.20) We now claim that the ratiojuj=juj stays bounded on [ m1 ; m1 + 4=K m ] as long as u6= 0. (In the regions where u = 0, we simply set V = 0.) In order to prove the claim, we need to separate the annulus r2 [ m1 ; m1 + 4=K m ] into three regions, inner, outer, and intermediate as in the proof above. For the sake of brevity, we show full details for the inner and outer annuli, i.e., r2 [ m1 ; m1 + 1=K m ] and r2 [ m1 + 3=K m ; m1 + 4=K m ]. It is easy to make adjustments for the rest of the proof, and we omit further details. Intheinnerannulusr2 [ m1 ; m1 +1=K m ], wehave 1 = 1and 0 2 1, andthusu =u 1 + 2 u 2 . Therefore u = ( 2 u 2 ) = 2 u 2 + 2 0 2 @u 2 @r + 1 r 0 2 + 00 2 u 2 : Thus, using (4.2.19) and (4.2.20) we have jujju 2 j +CK m (n 2 m + 2k m )ju 2 j +C(K m +K 2 m )ju 2 j C n 2 m k m +K m (n 2 m + 2k m ) +K m +K 2 m ju 2 j C(K m n 2 m +K 2 m )ju 2 j: (4.2.21) 104 Note that (4.2.21) is the first main adjustment to the proof of Lemma 4.2.1. The next adjustment results from the fact that the ratio ofju 1 j andju 2 j cannot be estimated from below by a constant. Instead, we have ju 1 j ju 2 j = r n 2 m a m r n 2 m +2km = 1 a m r 2km 1 a m ( m1 + 1=K m ) 2km = ( m1 + 2=K m ) 2km ( m1 + 1=K m ) 2km = 1 + 1 1 + m1 K m km 1 + 1 CK m km 1 + k m CK m 1 + n m CK m ; (4.2.22) for some constantC 1, where we usedK m 1, m1 C, andk m =CK m 1 as well as the inequality 1 + 1 x ax 1 + 1 C a; x 1; a2 (0; 1): Therefore, jujju 1 jju 2 j 1 + 1 C n m K m ju 2 jju 2 j n m CK m ju 2 j: By the definition of K m and using (4.2.21) with (4.2.13), we get juj juj C K 2 m n m + K 3 m n m C n 3 n m n m +C n 3 n m n m Cn 3 : (4.2.23) Next, we provide modifications for the outer annulus [ m1 +3=K m ; m1 +4=K m ], where 0 1 1 and 2 = 1 whence u = 1 u 1 +u 2 . Then ju 2 j ju 1 j =a m r 2k m1 + 2 K m 2km m1 + 3 K m 2km 1 + n m CK m ; as in (4.2.22). Therefore, jujju 2 jju 1 j n m CK m ju 1 j> 0: 105 Also, jujju 2 j + 2 0 1 @u 1 @r + 1 r 0 1 + 00 1 u 1 Cn 2 m k m ju 2 j +CK m n 2 m ju 1 j +C(K m +K 2 m )ju 1 jC(K m n 2 m +K 2 m )ju 1 j and we again obtain (4.2.23). The treatment in the other annuli is the same as in the previous proof with the changes shown above. Thus we omit further details. Proof of Lemma 4.2.3. Let nC, as in the statement. Denote u 1 (x) =r n e in and u 2 (x) =M where M is a sufficiently large constant, to be determined. On 2jxj 3, we define u = 1 u 1 + 2 u 2 where 1 and 2 are radial C 1 cut-off functions, which satisfy 1 = 8 > > > < > > > : 1; r 2:6 0; r 3 106 and 2 = 8 > > > < > > > : 0; r 2 + 1 9 1; r 2:3: We emphasize that the construction of 2 requires more care. In particular, we require 2 (r) 0 on [2; 2:1], where 0 > 0 is sufficiently small, to be determined, with 2 (r) =A logr +B; 2:1rR 0 whereR 0 2 [2:1; 2:2] and 1 0 2 1 on [R 0 ; 2:3]. This is possible to arrange ifA is chosen sufficiently large. Note that 2 = 0; 2:1jxjR 0 : Now, we claim that u =Vu for r2 [2; 3], for some function V such thatjVjC. Since u 1 and u 2 are harmonic on [2:3; 2:6] and 1 = 2 = 1 there, we may set V = 0. Next, we consider the interval [2; 2:3], where we have 1 = 1 and thus u = 00 2 (r) + 1 r 0 2 (r) u 2 : By the construction of 2 , the expression in the parentheses vanishes for r2 [2:1;R 0 ]. On [R 0 ; 2:3], we have juj =j 2 u 2 +u 1 jju 2 j (1 2 )ju 2 jju 1 j M 0 M 3 n M 4 with the last inequality holding if M 3 n+1 and 0 1=2. On [2; 2:1], on the other hand juj =j 2 u 2 +u 1 jju 1 jj 2 u 2 j 1 0 M 1 2 107 provided 0 is chosen so that 0 1=2M. Finally, on [2:6; 3], we have juj =j 1 u 1 +u 2 jju 2 j 1 ju 1 jM 3 n M 2 if M is chosen so that M 3 n+1 . 4.3 Proofs of the main results In this section, we provide proofs of the main results. Theorem 4.1.1 is based on the steps from Lemmas 4.2.1–4.2.3. Proof of Theorem 4.1.1. By Lemma 4.2.3, we may assume that n 7. Lemma 4.2.1 provides a function u defined on the disk with radius r 1 + 8=k such that u =r n 2 e in 2 ; jxj 1 + 0:25 k and u =br (n1) 2 e i(n1) 2 ; 1 + 7:75 k jxj 1 + 8 k : Denote b 1 = 1 (1 + 7=k) k (1 + 2=k) 2k (the right side of (4.2.3)) and b m = 1 ( m1 + 7=K m ) km ( m1 + 2=K m ) 2km ; m2f2; 3;:::g (the right side of (4.2.6)). 108 Applying Lemma 4.2.2 successively, we may extend u to the annulus r m , where m is the largest integer such that nmC, where C is as in Lemma 4.2.2. Then we have u = (1) m+1 b 1 b m1 r (nm+1) 2 exp i(nm + 1) 2 ; m1 jxj m1 + 0:25 K m for m = 1; 2;:::;m. Note that b j 1 for all j. After the final application of Lemma 4.2.2, we get u = (1) m b 1 b m r (nm) 2 exp i(nm) 2 ; m1 + 7:75 K m jxj m1 + 8 K m : (4.3.1) Lemma 4.2.3 allows us to connect the function in (4.3.1) to a constant function. The last step can be performed only if n X m=1 1 K m C; which holds by (4.2.5). Proof of Theorem 4.1.2. Based on Theorem 4.1.1, we only need to construct a function u, which equals 1 on B 1=2 , is smooth in a neighborhood of B 1 , obeys (4.1.1) withjVjC in a neighborhood of B 1 , and satisfies (4.1.3). In order to accomplish this, consider the function v(x) = logjxj: Then v @B1 = 0. Let 2 (0; 1=4], and choose nonnegative smooth radial cut-off functions 1 ; 2 such that 1 = 8 > > > < > > > : 1; r 1 2 0; r 1 4 109 and 2 = 8 > > > < > > > : 0; r 1 1; r 1 3 4 : It is easy to check that the function u = 1 + 2 v satisfies the desired properties. Proof of Theorem 4.1.3. Let u(x;t) =e u(x), where e u is a function constructed in Theorem 4.1.1. Since u is constant in B 1 nB 1=2 , we may modify it so it is 1-periodic with V = W = 0 in [0; 1] 2 nB 1=2 . Using integration by parts, we have sup t2(1;0) q(t)n 3 and the theorem follows. 110 Chapter 5 The Lagrangian and Eulerian analyticity for the Euler equations The motion of a fluid can be described either in reference to a fixed space-time point (x;t) (Eulerian coordinates) or by following the flow lines of individual fluid particles (Lagrangian coordinates). It is known that these two formulations are significantly different. There are examples of 3D steady flows with complicated particle paths where streamlines have space filling property (cf. [BM]). Conversely, there are cases where Lagrangian formulation is better adapted. In this chapter we study the Lagrangian formulation of the incompressible Euler equations on R d , whered2f2; 3g. WeshowthattheGevrey-classradiusisconservedforasmalltimeintheLagrangianfor- mulation onR d . More precisely, by defining suitable Lagrangian Gevrey-class norms first, we show that if the initial velocity gradient is of Gevrey-classs,s 1 then the Sobolev solutionv(;t)2C([0;T ];H r (R d )) is of Gevrey-classs for allt<T. Moreover, a modification of the above assertion yields a similar result for local solutions of Lagrangian formulation. The latter statement comes along with an additional condition that the Sobolev norm of the velocity stays finite for a small time, i.e., sup t2[0;T] kv(;t)k H r (B(0;R)) <1. As a final note, we demonstrate a contrast to the first assertion in Eulerian setting by constructing a real analytic periodic initial datum which does not yield a solution in the same class (Theorem 5.3.1). The Lagrangian approaches has recently gained a strong interest due to the possibility that the Lagrangian paths could be analytic in time, a fact which was first observed by Serfati [Ser95]. Earlier, Chemin [Che92] proved that the Lagrangian trajectories are C 1 smooth. More recently, in [ZF14, FZ14, 111 FV] the authors developed an elementary theory of analytic fluid particle trajectories. Their work is based on Cauchy’s long-forgotten manuscript (1815) on Lagrangian formulation of 3D incompressible Euler equations in terms of Lagrangian invariants. For further results on time-analyticity of Lagrangian trajectories, we refer the reader to [Shn12, Nad13, CVW14]. The remarkable differences in time analyticity of the two formulations suggests a further discussion on the analytic regularity of the solutions. There has been a quite rich history on the persistence of real-analyticity of the solutions in both two and three dimensions with many works pointing out the contrast between the Lagrangian and Eulerian analytic regularity (cf. [BB, BBZ76, Ba, Be, BG1, BG2]). In [BBZ76], the authors establish a lower bound of the form exp(C exp(CT ))=C for the radius of analyticity in 2D, with the constant C depending on the initial data. Finding explicit rates for the decay of radius of analyticity has been studied further using different methods in [AM1, AM2] for the interior and in [Lb] for the boundary value problem. Also, in [KV11b] the authors extend the results in [AM1, Lb] to the non-analytic Gevrey classes. In space-periodic domains, Levermore and Oliver give a proof of persistence of analyticity for all times. Their proof is based on a characterization of Gevrey classes in terms of decay of Fourier coefficients. Furthermore, the shear flow example (cf. [BT10, DM87]) has generated numerous constructions of explicit solutions to periodic 3D Euler equations whose radius of analyticity decays for all time. In the analytic class (Gevrey-1 class) case, see Remark 1:3 in [KV11b] and [CKV, Theorem 1.3]. Also, one can construct an example in the non-analytic Gevrey classes with s > 1 (cf. [Le]). Moreover, in [CKV] the authors point out the difference of behaviors in two formulations; the radius of analyticity is conserved locally in time for the Lagrangian formulation, whereas it deteriorates instantaneously in the Eulerian one. A similar contrast is also observed in terms of solvability in anisotropic classes. The Lagrangian formulation is locally well-posed in anisotropic classes yet the equations are ill-posed in Eulerian coordinates. 112 The main motivation of this work is to extend the results in [CKV] for a larger class of function spaces. Defining the Gevrey-class norm by the supremum over Taylor coefficients as opposed to Taylor series itself, we pass to a more general functional setting where the same technique in [CKV] continues to be useful. Furthermore, well-posedness of Lagrangian formulation in anisotropic classes seems to hold true for local solutions as well. In order to achieve latter statement we use the same method with a cut-off function. Although it is possible to use the same technique for time analyticity of Lagrangian paths for a short time; unfortunately, it fails when one is interested to prove the time analyticity of local solutions to Lagrangian formulation. This chapter is structured as follows. In Section 5.1, we write the two formulations of the Euler equations and recall the basic concepts in the Lagrangian setting. Section 5.2 consists of the main results and the proofs. Finally, in Section 5.3 we give a counterexample to the Theorem 5.2.1 in the Eulerian setting. 5.1 The Euler equations in Lagrangian coordinates The incompressible homogeneous Euler equations in R d , for d = 2; 3, are given by the system of equations u t +uru +rp = 0 (5.1.1) ru = 0 (5.1.2) u(x; 0) =u 0 (x) (5.1.3) where (x;t)2R d [0;1). The above system models the flow of an incompressible, homogeneous, and inviscid fluid, where u(x;t) = (u 1 ;:::;u d ) denotes the fluid velocity and p(x;t) the pressure. We rewrite the Euler equations using the particle trajectory mapping X(;t): 7! X(;t)2 R d , where t 0. Here the parameter is called the Lagrangian label. The vector X(;t) = (X 1 ; ;X d ) 113 denotes the location of a fluid particle at timet that is initially placed at, and is given by the nonlinear ODE @ t X(;t) =u(X(;t);t) (5.1.4) X(; 0) =: (5.1.5) Composing the velocity and the pressure with X, we obtain the Lagrangian velocity v and the pressure q by v(;t) =u(X(;t);t) (5.1.6) q(;t) =p(X(;t);t): (5.1.7) Also, denote by Y k i the (k;i)-th entry of the inverse of the Jacobian of X, i.e., Y (;t) = (r X(;t)) 1 : (5.1.8) We then write the Lagrangian formulation of the Euler equations as @ t v i +Y k i @ k q = 0; i = 1;:::;d (5.1.9) Y k i @ k v i = 0 (5.1.10) wherethesummationconventionisusedonrepeatedindices. Thesystem(5.1.9)–(5.1.10)issupplemented with the initial conditions v(; 0) =v 0 () =u 0 () (5.1.11) Y (; 0) =I: (5.1.12) 114 Differentiating (5.1.4) with respect to the Lagrangian labels along with using det(rX) = 1, and inverting the matrix in the resulting equation, we get Y t =Y : (rv) :Y (5.1.13) where the symbol : denotes the matrix multiplication. Taking the curl of the equation (5.1.1) and using thatr (uru) =ur!!ru, we obtain @ t ! +ur! =!ru: (5.1.14) Hence, D! Dt =!ru where D=Dt is the convective derivative, i.e., the derivative along the particle trajectories. In 2D flows, we have u = (u 1 (x;y);u 2 (x;y); 0) and so, ! =r ? u = (0; 0;!(x;y)) obeys !ru =!(x;y)@ z u(x;y) = 0; that is, vorticity is conserved as it moves with the flow. In the Lagrangian coordinates, this means that for d = 2, the Lagrangian vorticity (a;t) =!(X(a;t);t) satisfies (;t) =! 0 (a); t 0: (5.1.15) 115 Denote the sign of the permutation (1; 2)7! (i;j) by ij . Using both the conservation of vorticity and the divergence free condition in Lagrangian coordinates, we write the equations for Lagrangian velocity v as ij Y k i @ k v j =Y k 1 @ k v 2 Y k 2 @ k v 1 = =w 0 (5.1.16) Y k i @ k v i =Y k 1 @ k v 1 +Y k 2 @ k v 2 = 0: (5.1.17) In dimension d = 3, (5.1.15) no longer holds. However, the vorticity-transport formula i (;t) =@ k X i (;t)w k 0 () (5.1.18) gives us the 3D version of (5.1.15). As we need time-independent right side, multiplying the above equation by Y i , the i’th coloumn of Y, we get the three-dimensional div-curl system ijk Y m i Y l j @ l v k =Y m i i =w m 0 ; m = 1; 2; 3 (5.1.19) Y k i @ k v i = 0: (5.1.20) 5.2 The main results We start by recalling the definition of Gevrey spaces. Note that for r>d=2, the space H r =H r (R d ) is an algebra. Then, for any s 1, we define the s-Gevrey norm with radius > 0 by kfk G s; = sup jj0 jj jj! s k@ fk H r: 116 When the above norm is finite for s = 1, we say that the function is real-analytic. The anisotropic Gevrey norm is obtained by replacing the multi-index derivative with a directional derivative, i.e., given a direction j2f1;:::;dg, we define the anisotropic s-Gevrey norm with radius > 0 by kfk G (j) s; = sup m0 m m! s k@ m j fk H r and G (j) s; denotes the set of functions f for whichkfk G (j) s; is finite. Theorem 5.2.1. Assume that v 0 2H r+1 (R d ) for a fixed direction j2f1;:::;dg and that rv 0 2G (j) 1; (5.2.1) for some radius > 0. Then there exists T > 0 and a unique solution (v;Y )2 C([0;T ];H r+1 (R d )) C([0;T ];H r (R d )) of the system (5.1.9)–(5.1.10), which further satisfies rv;Y 2L 1 ([0;T ];G (j) 1; ): (5.2.2) Before starting with the proof of the Theorem 5.2.1 we observe that for m2N we have m1 X j=1 1 j 3 (mj) 3 C m 3 (5.2.3) and X 2j(j;k)jm1 1 j 3 k 3 (mjk) 3 C m 3 (5.2.4) where C denotes a positive generic constant. 117 Proof of Theorem 5.2.1. We provide the proof ford = 2 and then outline the necessary changes ford = 3. Without loss of generality, we may assume j = 1. Define m =k@ m 1 rv 0 k H r (5.2.5) for each m2N 0 . Assume that > 0 satisfies 0 ; 1 ; 2 2 M 0 (5.2.6) as well as m m (m 3)! M 0 ; m2f3; 4;:::g (5.2.7) for some M 0 > 0. We follow the setting in [CKV, Section 3]. We fix a time T > 0 and define the quantities V m (T ) = sup t2[0;T] k@ m 1 rv(t)k H r; (5.2.8) Z m (T ) = sup t2[0;T] t 1=2 k@ m 1 (Y (t)I)k H r: (5.2.9) As in [CKV], we bound V m and Z m using the div-curl system (5.1.16)–(5.1.17) and the Lagrangian evolution (5.1.13) in integrated form, respectively. Denote by I =f(j;k)2N 2 0 : 0<j(j;k)jmgnf(m; 0); (0;m)g (5.2.10) 118 the index set. Using the same argument in [CKV], one gets for all m2N V m C m +CT 1=2 Z 0 V m +CT 1=2 Z m V 0 +CT 1=2 m1 X j=1 m j Z j V mj (5.2.11) Z m CT 1=2 (TZ 2 0 V m +TZ m Z 0 V 0 +T 1=2 Z 0 V m +T 1=2 Z m V 0 +V m ) +CT 3=2 X (j;k)2I m j;k Z j Z k V mjk +CT m1 X j=1 m j Z j V mj (5.2.12) for some constant C > 0, while for m = 0 for any t2 (0;T ] we have V 0 (t)C 0 0 +C 0 t 1=2 Z 0 (t)V 0 (t) (5.2.13) Z 0 (t)C 0 t 1=2 sup 2[0;t) (V 0 ()(1 +t 1=2 Z 0 ()) 2 ) (5.2.14) for some C 0 > 0. We then define S m =V m +Z m = sup t2[0;T] (V m (t) +Z m (t)) (5.2.15) for all m 0. By adding (5.2.11) and (5.2.12), we obtain an estimate on S m which reads S m C 1 m +C 1 T 1=2 (1 +S 0 +T 1=2 S 0 +TS 2 0 )S m +C 1 T 1=2 (1 +T 1=2 ) X 0<j<m m j S j S mj +C 1 T 3=2 X (j;k)2I m j;k S j S k S mjk (5.2.16) for all m 1. In view of the initial conditions V 0 (0) = 0 M 0 Z 0 (0) = 0; 119 and the continuity in time of V 0 (t) and Z 0 (t), we assume that T > 0 is small enough that S 0 =V 0 +Z 0 3C 0 M 0 + 1=2: (5.2.17) We may further pick a smaller time 0<T 1 <T such that C 1 T 1=2 1 (1 +S 0 +T 1=2 1 S 0 +T 1 S 2 0 ) 1 2 : (5.2.18) For simplicity, we continue denoting this smaller time choice by T. Using (5.2.17)–(5.2.18) in (5.2.16) we obtain the estimate S m 2C 1 m + 2C 1 T 1=2 (1 +T 1=2 ) X 0<j<m m j S j S mj + 2C 1 T 3=2 X (j;k)2I m j;k S j S k S mjk (5.2.19) (5.2.20) for m2 N. One last arrangement may be done regarding the index set I in (5.2.10). Namely, for all (j;k)2 I with either j = 0 or k = 0 can be absorbed into the summation in the second term, that is using the bound on S 0 in (5.2.17) for any k2f1;:::;m 1g m 0;k S 0 S k S mk 4C 0 M 0 m k S k S mk : (5.2.21) Thus, we may rewrite the estimate (5.2.19) as S m (m 3)! 2C 1 m (m 3)! + 2C 1 (m 3)! T 1=2 (1 +T 1=2 +C 0 M 0 T ) m1 X j=1 m j S j S mj + 2C 1 (m 3)! T 3=2 X (j;k)2 e I m j;k S j S k S mjk (5.2.22) 120 for all m 3, where e I =f(j;k)2N 2 : 0<j(j;k)jmg: (5.2.23) For m = 1; 2, the estimate (5.2.19) reads as S 1 2C 1 1 ; (5.2.24) and S 2 2C 1 2 + 2C 1 T 1=2 (1 +T 1=2 )2S 2 1 + 4C 1 T 3=2 2S 0 S 2 1 : (5.2.25) Using the hypothesis on 1 ; 2 we have S 1 2C 1 M 0 ; (5.2.26) and S 2 2 2C 1 M 0 + 2 5 C 3 1 T 1=2 M 2 (1 +T 1=2 +S 0 T ): (5.2.27) Now, assuming that S j (j 3)! M j ; j = 3;:::;m 1 (5.2.28) 121 for some M > 0, to be determined below, we estimate (5.2.22) from above. By (5.2.7), S m (m 3)! 2C 1 m M 0 + 2C 1 T 1=2 (2 +C 0 M 0 T ) m1 X j=1 M 2 m m! (m 3)! (j 3)! j! (mj 3)! (mj)! + 2C 1 m! (m 3)! T 3=2 X (j;k)2I M 3 m m! (m 3)! (j 3)! j! (k 3)! k! (mjk 3)! (mjk)! : (5.2.29) We use the agreement s! = 1 if s 0. Keeping this in mind, we simply estimate (s 3)! s! 9 s 3 ; s2N: (5.2.30) Observe that m1 X j=1 (j 3)! j! (mj 3)! (mj)! 9 2 m1 X j=1 1 j 3 1 (mj) 3 (5.2.31) and X (j;k)2I (j 3)! j! (k 3)! k! (mjk 3)! (mjk)! 9 3 m1 X j(j;k)j=2 1 j 3 1 k 3 1 (mjk) 3 : (5.2.32) By (5.2.3)–(5.2.4), we finally have an estimate on (5.2.29), that is, S m (m 3)! 2C 1 m M 0 + 2C 1 T 1=2 (2 +C 0 M 0 T )C m 3 m 3 M 2 m + 2C 1 T 3=2 C m 3 m 3 M 3 m = 2C 1 m (M 0 +T 1=2 (2 +M 0 T )CM 2 +T 3=2 CM 3 ) (5.2.33) for all m 3. Setting M > 4C 1 M 0 and choosing a sufficiently small time 0 < T 1=(CC 1 ) 2 M 4 , we obtain the desired result. 122 Remark 5.2.2. (Justification of Theorem 5.2.1 for 3D Case). In 3D the only change arises when estimating V m . As the vorticity is not conserved along particle trajectories, the 2D elliptical div-curl system (5.1.16)–(5.1.17) that we used above is replaced by (5.1.19)–(5.1.20). We then write, (curlv) m = mlk @ l v k =! m 0 + ilk ( im Y m i )@ l v k + mjk ( jl Y l j )@ l v k ijk ( im Y m i )( jl Y l j )@ l v k divv = ( ik Y k i )@ k v i : Further, using the Helmholtz equation we obtain k@ m rvk H rCk@ m ! n 0 k H r +Ck@ m ( ijk )( in Y n i )( jl Y l j )@ l v k k H r +Ck@ m ( njk ( jl Y l j )@ l v k )k H r +Ck@ m ( ijk ( in Y n i )@ j v k )k H r +Ck@ m (( ik Y k i )@ k v i )k H r: Once again applying Leibniz rule and taking the supremum over t2 [0;T ] we get the estimate V m C m +CT 1=2 Z 0 V m +CT 1=2 Z m V 0 +CTZ 2 0 V m +CTZ m Z 0 V 0 CT X (j;k)2I m j;k Z j Z k V mjk +CT 1=2 m1 X j=1 Z j V mj Note that the estimate (5.2.12) for Z m remains the same. Therefore, the bounds for S m changes only slightly; S m 2C 1 m + 2C 1 T 1=2 (1 +T 1=2 ) X 0<j<m m j S j S mj + 2C 1 T (1 +T 1=2 ) X (j;k)2I m j;k S j S k S mjk for m2N. As a result, the conclusion of the Theorem 5.2.1 remains valid for 3d. 123 By using a cut-off function, we have the following result for local solutions of (5.1.1)–(5.1.3). Theorem 5.2.3. Assume that v 0 2H r+1 (B R ) for a fixed direction j2f1;:::;dg and that rv 0 2G (j) s; e (B R ) (5.2.34) for some index sr and radius e > 0. Also, assume that sup t2[0;T1] kv(t)k H r (B R ) <1 (5.2.35) for some timeT 1 > 0. Then there exists 0<T <T 1 and a unique solution (v;Y )2C([0;T ];H r+1 (B R )) C([0;T ];H r (B R )) of the system (5.1.9)–(5.1.10), which satisfies rv;Y 2L 1 ([0;T ];G j s; (B p )); 0<<R 1 (5.2.36) where 0<< e . Proof. Fix > 0 such that + 1 < R. We’ll work in the ball B +1 = B(0; + 1)2R n . Assume that s = s (x) is a cut-off function such that s (x) 1; x2B (5.2.37) s (x) 0; x2R n nB +1=s (5.2.38) and kr s k H rCs r (5.2.39) 124 for some C > 1 independent of s. Similarly, we define m =k@ m 1 rv 0 k H r (B R ) (5.2.40) and assume that (5.2.6)–(5.2.7) hold for some radius> 0 andM 0 > 0. Additionally, by (5.2.35) suppose that sup t2[0;T1] kv(t)k H r (B R ) M 0 : (5.2.41) We use the same argument as in the proof of Theorem 5.2.1 to boundrv locally. Using the Helmholtz decomposition kr((@ m 1 v) m )k H rC 0 k curl((@ m 1 v) m )k H r +C 0 k div((@ m 1 v) m )k H r C 0 k(curl@ m 1 v) m k H r +C 0 k@ m 1 vr m k H r +C 0 k@ m 1 vr m k H r +C 0 k m div@ m 1 vk H r (5.2.42) for some C 0 > 1.From here on we drop the subscript m on , i.e., = m . After swapping @ m 1 with the operators curl and div, we may use (5.1.16)–(5.1.17) on the right side of (5.2.42), that is, kr((@ m 1 v))k H rC 0 k@ m 1 (! 0 + ij ( ik Y k i )@ k v j )k H r +C 0 k@ m 1 (( ik Y k i )@ k v i )k H r + 2C 0 k@ m 1 vk H rkrk H r (5.2.43) Distributing the gradient on the left side of (5.2.43) we get k@ m 1 rvk H rkr((@ m 1 v))k H r +k@ m 1 vrk H r: 125 Keeping in mind that H r is an algebra we apply the Leibniz rule on the right side of (5.2.43), and we obtain k@ m 1 rvk H r 2C 0 kk H r (k@ m 1 ! 0 k H r +kYIk H rk@ m 1 rvk H r +k@ m 1 (YI)k H rkrvk H r + m1 X j=1 m j k@ j 1 (YI)k H rk@ mj 1 rvk H r +k@ m1 1 rvk H rkrk H r : (5.2.44) As all of the terms on the right side of (5.2.44) are multiplied by whose support lies in B +1=m , all of thosekk H r norms are effectively taken in the ball B +1=m . We then fix T > 0 and define fore r> 0 V m (B e r ) = sup t2[0;T] k@ m 1 rv(t)k H r (B e r ) (5.2.45) Z m (B e r ) = sup t2[0;T] t 1=2 k@ m 1 (Y (t)I)k H r (B e r ) (5.2.46) Writing (5.2.44) in terms of (5.2.45)–(5.2.46) we get k@ m 1 rvk H r 2C 2 0 kk H r m +T 1=2 Z m (B m )V 0 (B m ) +T 1=2 Z 0 (B m )V m (B m ) +T 1=2 m1 X j=1 m j Z j (B m )V mj (B m ) +V m1 (B m )m r 1 A (5.2.47) where m = + 1=m. Here, we replace the constant C 0 with a larger one if necessary so that (5.2.39) still holds with C 0 . 126 In order to estimate Z m , we apply @ m 1 to the Lagrangian evolution (5.1.13) in integrated form and then multiply both sides by , that is, @ m 1 (IY (t)) =@ m 1 Z t 0 Y :rv :Y d = Z t 0 X j(j;k)jm m j;k @ j 1 (YI) :@ k 1 rv :@ mjk 1 (YI)d + Z t 0 k X j=0 m j @ j 1 (YI) :@ mj 1 rvd + Z t 0 m X j=0 m j @ j 1 rv :@ mj 1 (YI)d + Z t 0 @ m 1 rvd (5.2.48) for all t2 [0;T ]. Using the notation (5.2.45)–(5.2.46), we arrive at k@ m 1 (Y (t)I)k H rC 0 kk H rT 1=2 TZ 2 0 (B m )V m (B m ) +TZ 0 (B m )Z m (B m )V 0 (B m ) + T 1=2 Z 0 (B m )V m (B m ) +T 1=2 Z m (B m )V 0 (B m ) +V m (B m ) +C 0 kk H r 0 @ T 3=2 X (j;k)2I m j;k Z j (B m )Z k (B m )V mjk (B m ) +T m1 X j=1 m j Z j (B m )V mj (B m ) 1 A (5.2.49) where the index set I is as defined in (5.2.10). We then define S m (B e r ) =V m (B e r ) +Z m (B e r ) = sup t2[0;T] (V m (t;B e r ) +Z m (t;B e r )) (5.2.50) 127 fore r> 0. Adding (5.2.47)–(5.2.49) we write an estimate for S m kk H rS m (B m )C 1 kk H r m +C 1 kk H rT 1=2 2TS 2 0 (B m ) + 2T 1=2 S 0 (B m ) + 1 S m (B m ) +C 1 kk H rT 1=2 (1 +T 1=2 ) m1 X j=1 m j S j (B m )S mj (B m ) C 1 kk H rT 3=2 X (j;k)2I m j;k S j (B m )S k (B m )S mjk (B m ) +C 1 kk H rm r S m1 (B m ) (5.2.51) for some C 1 > 2C 2 0 > 1. Before going into further detail we recall the initial conditions. We have V 0 (0;B R ) =krv 0 k H r (Br ) M 0 Z 0 (0;B R ) = 0: Writing (5.2.42) for m = 0 with a cut-off function = 1 we observe that for all t2 [0;T ] kr(v)k H r (B R ) C 0 kk H r 0 +T 1=2 Z 0 (B R )V 0 (B R ) +krk H r sup t2[0;T] kv(t)k H r (B R ) ! : Note that (5.2.48) is still valid for m = 0. Then, asrv and (IY ) are bounded in time over [0;T ] as well as v by hypothesis (5.2.35), we deduce that both V 0 and Z 0 are continuous in t, and so there exists e T = e T (M 0 )> 0 such that sup t2[0; e T] Z 0 (t;B R ) 1=2 sup t2[0; e T] V 0 (t;B R ) 3C 0 M 0 : (5.2.52) We assume that the time we fixed at the beginning is small enough, i.e., T satisfies T e T: (5.2.53) 128 By (5.2.52) we get S 0 (B R ) 3C 0 M 0 + 1=2: (5.2.54) Having S 0 (B R ) bounded above, we once more assume that 0<T (C;M 0 )<T 1 is sufficiently small that it obeys C 1 T 1=2 2TS 2 0 (B R ) + 2T 1=2 S 0 (B R ) + 1 1=2: (5.2.55) This implies that all of the m-th order terms in (5.2.51) can be absorbed on the left side. Furthermore, as we discussed in the proof of the Theorem 5.2.1, we may switch to the index set e I =f(j;k)2N 2 : 0< j(j;k)j<mg. Together with (5.2.55), this turns (5.2.51) into kk H rS m (B m )C 1 kk H r m +C 1 kk H rT 1=2 (1 +T 1=2 + 4C 0 M 0 T ) m1 X j=1 m j S j (B j )S mj (B mj ) +C 1 kk H rT 3=2 m1 X (j;k)2 e I m j;k S j (B j )S k (B k )S mjk (B mjk ) +C 1 kk H rm r S m1 (B m1 ) (5.2.56) where j = + 1=j. Note that (5.2.56) is very similar to (5.2.19); the only difference is the last term C 1 kk H rm r S m1 (B m1 ). Writing (5.2.56) for m = 1; 2 we get S 1 (B +1 )C 1 1 +C 1 S 0 (B +1 ) and S 2 (B +1=2 )C 1 2 + 2C 1 T 1=2 (1 +T 1=2 +S 0 (B R )T )S 2 1 (B +1 ) +C 1 2 r S 1 (B +1 ) 129 which yield the estimates S 1 (B +1 ) (3C 2 1 M 0 + 2C 1 M 0 ) (5.2.57) and S 2 (B +1=2 ) 2 C 1 M 0 + 2C 1 T 1=2 (1 +T 1=2 + 4C 1 M 0 T )(8C 2 1 + 18C 2 1 )M 2 0 +C 1 2 r (4C 2 1 + 2C 1 )M 0 (5.2.58) upon using (5.2.54). Now, our aim is to find a Gevrey bound on the right side of (5.2.54), (5.2.57) and (5.2.58); that is to show that there exists M(M 0 )> 0 such that S 0 (B R );S 1 (B +1 );S 2 (B +1=2 )M when the Gevrey radius > 0 and the time T > 0 is sufficiently small. Let K > C 1 be a large enough constant which bounds all the multiples of C 1 that we encountered when estimating (5.2.54), (5.2.57) and (5.2.58). Then, selecting > 0 and M > 0 such that they obey K 2 1 (5.2.59) and KM 0 4 M; (5.2.60) we immediately get S 0 (B R );S 1 (B +1 )M: 130 By (5.2.59) and (5.2.60) we estimate S 2 ( + 1=2) S 2 (B +1=2 ) 2 KM 0 +T 1=2 K 3 1 M 2 0 (1 +T 1=2 +KM 0 T ) +M 0 ( 2 K 3 +K 2 ): (5.2.61) Furthermore, selecting T > 0 that satisfies T 1 M 2 K 2 (5.2.62) and using (5.2.59)– (5.2.60) we obtain that (5.2.61) can be bounded from above by M. We now would like to show inductively that S m (B m ) ((m 3)!) s m M holds for all m 3. Assume that S j (B j ) ((j 3)!) s j M; j = 3;:::;m 1: (5.2.63) In view of the notation and the binomial bounds discussed between (5.2.30)–(5.2.32), we derive from (5.2.56) that S m (B m ) ((m 3)!) s m C 1 M 0 + 2C 1 T 1=2 ((m 3)!) s M 2 m1 X j=1 m j ((j 3)!) s ((mj 3)!) s +C 1 T 3=2 ((m 3)!) s M 3 m X j(j;k)j=2; j>0;k>0 m j;k ((j 3)!) s ((k 3)!) s (mjk 3) s +C 1 Mm r ((m 4)!) s ((m 3)!) s : (5.2.64) 131 We shall denote the above sums by I 1 and I 2 , respectively. Note that I 1 = m1 X j=1 m! j!(mj)! (j 3)!(mj 3)! (m 3)! s = m1 X j=1 m 6 j 3 1s (j 3)! j! (mj 3)! (mj)! m! ((m 6)!) 1s ((m 3)!) s =K 1 m 63s m1 X j=1 1 j 3 (mj) 3 K 1 m 33s (5.2.65) where K 1 > 1 is a numerical constant. Similarly, I 2 = m X (j;k)=2; j>0;k>0 m j;k (j 3)!(k 3)!(mjk 3)! (m 3)! s K 2 m X (j;k)=2; j>0;k>0 m 9 j 3;k 3 1s m 96s j 3 k 3 (mjk) 3 K 2 m 66s (5.2.66) with a constant K 2 > 1. Finally, we estimate (5.2.64) by S m (B m ) ((m 3)!) s m 4C 1 M 0 +T 1=2 M 2 0 K 1 m 3s3 +T 3=2 M 3 K 2 m 6s6 +Mm rs : (5.2.67) Noting that sr and in view of (5.2.59)–(5.2.60) and (5.2.62), we arrive at S m (B m ) ((m 3)!) s m M (5.2.68) for all m 3. 132 5.3 A Counterexample In this section, we provide a counterexample to the Theorem 5.2.1 by constructing a solution u(x;t) of the system (5.1.1)–(5.1.3) whose radius of analyticity in Eulerian coordinates is not conserved in time. Theorem 5.3.1. There exists a periodic function u 0 such that ku 0 k G1;1 <1 (5.3.1) divu 0 = 0; for which ku(t)k G1;1 =1 (5.3.2) for any t> 0. In order to build such a solution we appeal to the example introduced in [DM87, BT10]. Denote [;] byT. Recall that given two periodic functions f and g onT, the function u(x 1 ;x 2 ;x 3 ;t) = (f(x 2 ); 0;g(x 1 tf(x 2 ))) (5.3.3) is an explicit solution of the Euler equations onT 3 , with the initial data u 0 (x 1 ;x 2 ;x 3 ;t) = (f(x 2 ); 0;g(x 1 )): The proof of the Theorem 5.3.1 consists of two steps. We start with a real analytic function in a strip whose holomorphic extension has simple poles ati. Under (5.3.3) these poles move closer toward the real axis. Since we would like to control the supremum of the Taylor coefficients, it suffices to integrate such a real valued function twice for the holomorphic extension to haveC 2 regularity up to the boundary. 133 Then we need to multiply the resulting function with a Gaussian kernel in order to obtain an integrable version, which is a necessary condition for periodization. In the second step we show that the initial data constructed yields the solution described in Theorem 5.3.1. Proof of Theorem 5.3.1. First, set f(x) = sinx (5.3.4) in (5.3.3). Asf is entire, we need to construct a 2-periodic functiong as described above withkgk G1;1 < 1 so that we get (5.3.1). Consider H(x) = Z x 0 Z w 0 1 1 +y 2 dydw =x arctanx 1 2 log(1 +x 2 ): (5.3.5) We then define '(x) =H(x)e s 2 x 2 (5.3.6) for some fixed 0 < s < 1. Note that ' is a smooth function that belongs to L 1 (R). Using the Poisson summation formula we periodize the function '(x), and we set (x) = X m2Z '(xm): (5.3.7) 134 Then writing the 1-Gevrey norm for , we have kk G1;1 = sup n0 k d n dx n k H 2 ([0;1]) 1 n n! = sup n0 k d n+2 dx n+2 k L 2 ([0;1]) 1 n n! = sup n0 k X m2Z d n+2 dx n+2 '(xm)k L 2 ([0;1]) 1 n! : (5.3.8) We first find the estimates on the higher order derivatives of the function H(x) and the Gaussiane s 2 x 2 . More precisely, for fixed m2Z and k 2 we get d k dx k H(xm) = 1 2 d k2 dx k2 1 1i(xm) + 1 1 +i(xm) = 1 2 (i) k2 (k 2)! (1i(xm)) k1 + (i) k2 (k 2)! (1 +i(xm)) k1 ; (5.3.9) and d k dx k e s 2 (xm) 2 =e s 2 (xm) 2 s k (1) k H k (sxsm); whereH k (x) denotes thek-th Hermite polynomial. Using the recursion relation on Hermite polynomials one may derive a useful pointwise bound on the real line jH k (x)j 4 k jxj k (k!) 1=2 ; which yields d k dx k e s 2 (xm) 2 e s 2 (xm) 2 s 2k 4 k jxmj k (k!) 1=2 : (5.3.10) 135 Then by the Leibniz rule d n+2 dx n+2 (H(xm)e s 2 (xm) 2 ) = n+2 X j=0 n + 2 j d n+2j dx n+2j H(xm) d j dx j e s 2 (xm) 2 =S 2 + n X j=0 n + 2 j 1 2 d nj dx nj 1 1i(xm) + 1 1 +i(x +m) d j dx j e s 2 (xm) 2 ; where S 2 = (n + 2) d dx H(xm) d n+1 dx n+1 e s 2 (xm) 2 +H(xm) d n+2 dx n+2 e s 2 (xm) 2 : (5.3.11) Using (5.3.9) and (5.3.10) and doing the cancellations in the binomial terms we get d n+2 dx n+2 H(xm)e s 2 (xm) 2 jS 2 j +C e s 2 (xm) 2 (n + 2)! 1 +jxmj n+1 n X j=0 1 (n + 2j)(n + 1j) 1 (j!) 1=2 (4s 2 ) j jxmj 2j (5.3.12) Then we have kk G1;1 sup n X m2Z 0 @ 1 n! kS 2 k L 2 ([0;1]) +k e s 2 (xm) 2 n 2 1 + (xm) n+1 n X j=0 1 (n + 1j) 2 1 (j!) 1=2 (4s 2 ) j (xm) 2j k L 2 ([0;1]) 1 A (5.3.13) Before estimating (5.3.13), we make the following observation: X m2Z e s 2 (xm) 2 jxmj k C(1=s) k k + 1 2 k+1 2 ; x2 [0; 1] (5.3.14) 136 which is obtained from splitting the sum into two partsjmj p k=2s 2 andjmj> p k=2s 2 , where it is increasing and decreasing, respectively. More precisely, X jmj p k=2s 2 e s 2 (xm) 2 jxmj k X jmj p k=2s 2 jxmj k 2 Z p k=2s 2 0 (y + 1) k dy C(1=s 2 ) k=2 k + 1 2 (k+1)=2 ; and X jmj> p k=2s 2 e s 2 (xm) 2 jxmj k 2 Z 1 0 e s 2 y 2 y k dy C(1= p 2s) k k!!: The latter inequality is obtained from the formula for the moments of normal random variable. Recall that the double factorials are defined as follows: (2n)!! = 2 n n! (5.3.15) (2n + 1)!! = (2n + 1)! 2 n n! : (5.3.16) UsingtheStirling’sformulaonthefactorialsweseethat(5.3.15)–(5.3.16)areboundedabovebyaconstant multiple of n n+1=2 (2=e) n . The inequality (5.3.14) then follows from adding the estimates on both sums. 137 Confirming (5.3.14), we go back to the estimation of (5.3.13). We again split the sum into three parts, and recalling the equation (5.3.13) we start with the term S 2 . We first put bounds on H(xm) and H 0 (xm) and then estimate X m2Z S 2 C X m2Z (n + 2) d n+1 dx n+1 e s 2 (xm) 2 +jxmj d n+2 dx n+2 e s 2 (xm) 2 C(n + 2)((n + 1)!) 1=2 (4s 2 ) n X m2Z e s 2 (xm) 2 jxmj n+2 +jxmj n+1 C(4s 2 ) n n 3=2 (n!) 1=2 X m2Z e s 2 (xm) 2 jxmj n+2 : (5.3.17) Then, using (5.3.14) and applying Stirling’s formula on (n!) 1=2 , we see that for every x2 [0; 1] X m2Z S 2 1 n! C(4s 2 ) n n 3=2 1 s n n + 3 2 (n+3)=2 1 n (n=2+1=4) e n=2 C4 n s n (e=2) n=2 n 3+1=4 n + 3 n n=2 ! 0; (5.3.18) provided that we have s 1 4 q 2 e . Secondly, we estimate the terms in the finite sum for smaller j, i.e., X m2Z e s 2 (xm) 2 1 +jxmj n+1 b n 2 c X j=0 n 2 (n + 1j) 2 1 (j!) 1=2 (4s 2 ) j jxmj 2j C X m2Z e s 2 (xm) 2 1 +jxmj n+1 jxmj n 0 @ 1 X j=0 (4s 2 ) j (j!) 1=2 1 A ; (5.3.19) and we note that the right side is uniformly convergent in x and n provided that s 2 1=4. 138 Lastly, we bound the terms withb n 2 c<jn: X m2Z e s 2 (xm) 2 1 +jxmj n+1 n X j=b n 2 c n 2 (n + 1j) 2 1 (j!) 1=2 (4s 2 ) j jxmj 2j Cn 2 n X j=b n 2 c+1 (4s 2 ) j (j!) 1=2 X m2Z e s 2 jxmj 2 jxmj (2jn1) ! : (5.3.20) Then using Stirling’s formula on (j!) 1=2 , and applying (5.3.14) to the sum in m, we find that the right side in (5.3.20) is bounded above by Cn 2 n X j=b n 2 c (4s 2 ) j j j 2 + 1 4 e j=2 1 s 2jn1 j n 2 j n 2 Cn 2 s n n=2 X j=1 (4 p e) j+b n 2 c j j (j +b n 2 c) j 2 + n+1 4 Cn 2 s n (4 p e) n 1 2 j 1 j +b n 2 c n+1 4 j 2 Cn 2 s n (4 p e) n n 2 X j=1 1 2 j : (5.3.21) We again note that if s < 1 4 p e , (5.3.21) converges to 0 uniformly in n. Thus, combining the estimates (5.3.13)–(5.3.21) together, which are uniform in x2 [0; 1], we arrive at kk G1;1 <1; (5.3.22) which concludes the construction needed in the first step. Now, going back to equation (5.3.3), we define g(x) = (2x): (5.3.23) Together with (5.3.22) and (5.3.4), we get ku 0 k G1;1(H 2 (T 3 )) <1: 139 For (5.3.2), we proceed by the way of contradiction , and assume thatku(t)k G1;1 <1 for some t2 (0; 1=10]. Then, define (x 1 ;x 2 ) =@ x1 g(x;t) = 0 (x 1 tf(x 2 )); (5.3.24) for fixed t2 (0; 1=10]. Due to the hypothesisku(t)k G1;1 <1 and the fact that f(x) = sinx is an entire function, we obtain sup 0 k@ k H 2 (T 2 ) (jj + 1)! <1: Note that supremum above is taken over the multi-indices 2N 2 . It follows that is analytic in the closed squarefjx 1 jR;jx 2 jRg for any 0<R< 1 . Then the complex extension defines an analytic function (z 1 ;z 2 ) = (x 1 +iy 1 ;x 2 +iy 2 ) (5.3.25) whose power series converges absolutely infjz 1 j<R;jz 2 j<Rg for 0<R < 1. However, we shall show that lim y1!Rt j (iy 1 ;i log 2)j =1 (5.3.26) for R t = 1 3t 4 2 (0; 1). Note thatjz 2 j = log 2<R t and sin(i log 2) = 3i=4, so (5.3.26) is equivalent to lim y!1 + j 0 (iy)j =1: (5.3.27) 140 which establishes the contradiction to the assertionku(t)k G1;1 <1, and completes the proof. In relation to (5.3.27), we observe that 0 (z) = d dz X m2Z '(zm) = d dz e z 2 =16 H(z) + X m2Znf0g ' 0 (zm); (5.3.28) whereH and' were defined in (5.3.5) and (5.3.6), respectively. Then, noting that H 0 (z) = arctanz and ' 0 (z) =e z 2 =16 (H 0 (z)z=8H(z)), we set H (z) =H 0 (z)z=8H(z) (5.3.29) = arctanz z 2 8 arctanz + z 16 log(1 +z 2 ) (5.3.30) Evaluating (5.3.29) at z =iy and using that arctan(iy) =i arctanhy we get H (iy) =i arctanhy +i y 2 8 arctanhy + iy 16 log(1y 2 ) and taking the limit as y!1 + above, we obtain lim y!1 + jH (iy)j =1; so the first term in (5.3.28) is not finite. The only thing remaining to show is that lim y!1 + X m2Znf0g j' 0 (iym)j<1: (5.3.31) 141 The limit above holds, in fact we have jH (iym)jP (m) (5.3.32) uniformly for 1=2<jyj< 1, where P is a polynomial. In order to see this claim, we use the identity arctanz = 1 2i ln iz i +z +i arg iz i +z together with i (iym) i + (iym) = 1jyj 2 m 2 i2m m 2 + (y + 1) 2 ; and obtain that jarctan(iym)j + j(iym) 2 j 8 jarctan(iym)jC(m 2 + 1); y2 [1=2; 1]: (5.3.33) Similarly, log(1 + (iym) 2 ) = ln(j1 +m 2 y 2 i2ymj) +i arg(1 + (iym) 2 ) C(m 2 + 1) uniformly in y2 [1=2; 1]. By (5.3.29), the estimates above confirm the inequality (5.3.32), and in turn establishes (5.3.31). 142 Chapter 6 Gevrey regularity for the Navier-Stokes in a half-space In this chapter, we consider the Navier-Stokes system @ t u u +uru +rp =f ru = 0; (6.0.1) in the half-space = x = (x 1 ;:::;x d )2R d :x d > 0 (6.0.2) with the no-slip boundary condition u @ = 0 (6.0.3) and the initial condition u(x; 0) =u 0 (x); x2 : (6.0.4) For simplicity, we let d2f2; 3g, but note that higher dimensions may be treated similarly. See e.g. [CF, RSS, T] for the well-posedness and further properties of the solutions to (6.0.1)–(6.0.4). In Theorem 1 below we prove that the solution to (6.0.1)–(6.0.4) immediately becomes space-time real analytic, with analyticity radius which is uniform up to the boundary @ , under the hypothesis that 143 the force is real analytic in space-time. The result only requires finite Sobolev regularity on the initial datum u 0 . Assuming that f is space-time analytic in I, where R 3 and I is a complex neighborhood of (0;T ), Masuda [Ma67] proved that the interior analyticity of a solution u to the Navier-Stokes system follows from that of the external force f (see also [KM]), answering a question posed by Serrin [Se]. Furthermore, in the case that is a bounded domain with analytic boundary@ , assuming thatf is an- alytic uniformly up to the boundary and that the solution (u;p) isC 1 , Komatsu [Ko1, Ko2] showed that (u;rp) is globally analytic inx up to the boundary@ and locally analytic int. His technique is inspired by the previous work by Kinderlehrer and Nirenberg [KN78] for second order parabolic equations, and is based on an induction scheme on the number of derivatives (see also [LG]). A semigroup approach for an- alyticity up to the boundary in (6.0.1)–(6.0.4) was later given by Giga [G] (see also [PSS]), and a complex variables-based proof was given by the second author and Grujić [GK1, GK2] (see also [BGK1, BGK2]). Establishing the analyticity of solutions to (6.0.1)–(6.0.4) on domains with boundaries is particularly im- portant in the context of the vanishing viscosity limit [SC], or equivalently, the infinite Grashof number limit in our context. The proof of the instantaneous space-time analyticity uniformly up to the boundary of the half-space given in this paper is based solely on L 2 x;t energy estimates of the solution and its derivatives (see also [KV2] for the non-homogeneous Stokes system). The main obstacle to energy-based proofs on domains with boundaries is that the normal derivatives of the solution do not obey good boundary conditions. We believe that our approach will be useful in establishing real-analytic and Gevrey-class regularization results for semilinear parabolic PDEs with different types of boundary conditions, by only appealing to energy estimates. We recall that in the case of no boundaries, Foias and Temam have developed in [FoT] a very efficient method to prove analyticity, or more generally Gevrey-type regularity, which has in turn inspired many works (cf. [B05, BF14, BS, BT, FJLRYZ, FT, GK, KV, L00, LO97, OT, OT1] and references therein). 144 The technique in [FoT] is based on Fourier analysis, which is unavailable in the case of domains with boundaries. One of the main aims of this paper is to find a similarly direct approach for establishing analyticity, which is based on the summing Taylor coefficients, rather than on the Fourier techniques. Such methods were introduced in [KV1] for the propagation of analyticity in the Euler equation, and this in turn led to be efficient in estimating the size of the uniform radius in terms of the size of initial data. However, finding an analog in the case of the Navier-Stokes equations proved to be more difficult due to the Laplacian term. In [KV2], the last two authors of the present paper, inspired by Komatsu’s work [Ko2], have used classical energy inequalities for the heat and Laplace equations, to achieve normal, tangential and time derivative reductions on terms of the form t i+j+k3 @ i t @ j d @ k u. Here @ and @ d denote the tangential derivative component and the normal derivative component, respectively. This derivative reduction method works for the heat equation and extends naturally to the inhomogeneous Stokes system, yielding the desired regularization result in [KV2]. In order to address the Navier-Stokes system, we use a Gevrey type norm T (u) = X i+j+k3 i e j k N i;j;k kt i+j+k3 @ i t @ j d @ k uk L 2 ([0;T] ) +kuk H 2 ([0;T] ) where N i;j;k represent certain binomial expressions which account for the possible growth of the Taylor coefficients. Note that the finiteness of the norm T (u) for some T > 0 implies that the function u is real-analytic in space-time on (0;T ) (see e.g. [OT1] and references therein). The main goal is to establish an inequality of the type T (u).K u0 +kfk +K u0 N X j=1 T j ( T (u)) j ; (6.0.5) where j > 0, N 2 N is fixed, and 0 < j 2. Herekfk represents a suitable analytic norm of f, and K u0 is a constant that depends on the Sobolev norm of the initial datum. From (6.0.5) and a standard Grönwall-type barrier argument, we deduce that for short enough time T (u) stays bounded 145 from above by a constant which depends on a Sobolev norm of u 0 and a space-time analytic norm of f, establishing the desired joint space-time analytic regularization. Although establishing (6.0.5) comes with some computational difficulty due to the nonlinear term T (uru), the logic behind the analyticity estimate remains as direct as observed in the case of the nonhomogeneous heat equation. We believe that this method directly generalizes to nonlinear Stokes systems with nonlinearity given by N(x;t;u;ru), a space-time analytic function in each of its variables. The chapter is organized as follows. In Section 2, we introduce some notation, define the Gevrey-class norm T , and state the main result (cf. Theorem 1). Also in Section 2 we recall the derivative reduction estimates from [KV2]. In Section 3, we give the proof Theorem 1, assuming a suitable bound for the nonlinear term, given by terms on the right side of (6.0.5) (cf. Lemma 2). The proof of this nonlinear estimate is finally given in Section 4, and is split into three separate lemmas which each deal with one case of the derivative reduction estimates (cf. Lemmas 5, 6, and 7). 6.1 Main result Before stating the main result of this paper, Theorem 1 below, we first introduce some notation. For r 1 we define the index sets B = (i;j;k) :i;j;k2N 0 ;i +j +kr and B c =N 3 0 nB: For m2N 0 we define the real-analytic binomial coefficient N m = m r m! : All the proofs and statements in this paper carry through for the Gevrey-class s binomial coefficients N m =m r =(m!) s , for s> 1. For simplicity of notation we only discuss the stronger (analytic) case s = 1. 146 For a fixed time horizonT2 (0; 1] and given small parameterse < <2 (0; 1], we consider the sum T (u) = X B N i+j+k i e j k kt i+j+kr @ i t @ j d @ k uk L 2 x;t ([0;T] ) + X B c k@ i t @ j d @ k uk L 2 x;t ([0;T] ) = T (u) + 0;T (u); (6.1.1) where @ denotes the vector of tangential derivatives @ = (@ 1 ;:::;@ d1 ). Above and in the sequel we use the notational agreement that for k2N 0 we use @ k to denote: k@ i t @ j d @ k uk L 2 x;t = X 2N 2 0 ;jj=k k@ i t @ j d @ uk L 2 x;t : Moreover, if the domain in the Lebesgue/Sobolev space is not indicated, it is either or (0;T ), and this will be clear from the context. Throughout the paper we use the symbol a.b to mean that there exists a sufficiently large constant C =C( ;r;d) 1 such that aCb. Remark 1. In (6.1.1) we note that 0;T (u) is the H r1 ([0;T ] ) norm of the solution u of (6.0.1)– (6.0.4). Under suitable smoothness and compatibility conditions on u 0 and f, and for sufficiently small T, it is known (cf. e.g. [Sol, Chapter III]) that 0;T (u) is a priori bounded in terms of Sobolev norms of u 0 and f. Remark 2. The finiteness of the norm T (u) in (6.1.1), for some T > 0, implies that the function u is real-analytic in space-time on (0;T ) (see, e.g. [OT1]). Moreover, for any t 0 2 (0;T ), the finiteness of the sub-sums with i = 0 and i = 1 shows that u(;t 0 ) is real-analytic in space, uniformly up to the boundary of the half space . The radius of analyticity is bounded from below by a constant multiple of t 0 e and the analytic norm is bounded from above by (1 +t 1=2 0 ) T (u). Note also that by changing the binomial weightN m tom r =(m!) s , withs> 1, the finiteness of T (u) implies the Gevrey-classs regularity of u. 147 In [KV2], the last two authors of this paper have showed that the solution u of the Cauchy problem for the inhomogeneous Stokes system @ t u u +rp =f in ru = 0 in u = 0 on @ (6.1.2) satisfies T (u). 0;T (u) +M T (f) (6.1.3) where M T (f) = X i+j+k(r2)+ N i+j+k+2 i e j+2 k kt i+j+k+2r @ i t @ j d @ k fk L 2 x;t ((0;T) ) + X i+k(r2)+ N i+k+2 i k+2 kt i+k+2r @ i t @ k fk L 2 x;t ((0;T) ) + X ir1 N i+1 i+1 kt i+1r @ i t fk L 2 x;t ((0;T) ) ; (6.1.4) provided that 0<e 1 (6.1.5) are suitably chosen small constants depending on T, r, and d. For the sake of completeness, we recall from [KV2] that the constantse ; ; can be chosen as follows: There exists a constant C = C(r;d) 1 such that in addition to (6.1.5) we require 1 C ; T 1 C ; (Te ) 1=2 + e 1 C : (6.1.6) 148 Since in this paper we consider T2 (0; 1], the above condition is satisfied as soon as we impose 1 C ; 1 C ; e 1=2 + e 1 C ; (6.1.7) where as beforeC =C(r;d) 1 is sufficiently large. Throughout this paper fix these values of; ;e which obey (6.1.5) and (6.1.7), and emphasize that their values depend only on r and d. With this notation, the following is our main result. Theorem 1. For d2f2; 3g and r = 3 there exist ;e ; 2 (0; 1], such that the following statement holds: For any divergence-freeu 0 2H 1 0 ( )\H 4 ( ) which satisfies suitable compatibility conditions, and a space- time real-analytic f2L 1 (0; 1;H 3 ( ))\ _ W 1;1 (0; 1;H 1 ( ))\ _ W 2;1 (0; 1;L 2 ( )), for which M 1 (f)<1, there exists T 2 (0; 1] such that the solution u of the Cauchy problem for (6.0.1)–(6.0.4) satisfies the estimate T (u). 1 +M T (f) (6.1.8) for any T2 (0;T ]. The implicit constant only depends on , r, and d. Remark 3. In Theorem 1 the timeT depends on the datum throughku 0 k H 4, and on the force through kfk L 1 t H 3 +k@ t fk L 1 t H 1 x +k@ 2 t fk L 1 t L 2 x +M 1 (f), where M 1 (f) is as defined in (6.1.4). Remark 4. On the initial datum we have imposed, for simplicity, the requirement u 0 2H 1 0 ( )\H 4 ( ), in addition to the usual compatibility conditions at the boundary of . However, if we are only interested in the space-time analyticity of the solution on (t 0 ;T ] , for an arbitrarily smallt 0 > 0, we may simply take u 0 2 H 1 0 ( ). The local existence of the Cauchy problem to (6.0.1)–(6.0.4) with such initial datum is classical, and the H 4 regularity of u(;t 0 =2) follows from the Sobolev smoothing properties of the nonlinear Stokes equation [Sol], which allows us to apply Theorem 1 with initial datum e u 0 =u(;t 0 =2). 149 The main idea in the proof of Theorem 1 is to apply the estimate (6.1.3) withf replaced byfuru, and to perform a nonlinear estimate on T (uru) in terms of 0;T (u) and T (u). The goal is to arrive at an estimate like (6.0.5), which then concludes the proof of the theorem upon choosing a suitable T. The main idea behind the estimate (6.1.3) in [KV2] is to split the sum T in (6.1.1) into several sub-sums, and on each one perform a derivative reduction estimate. For convenience of the reader we recall from [KV2] these derivative reduction estimates, for a solution of the non homogeneous Stokes system (6.1.2) on the half-space. In all the below inequalities, we require i +j +k r. As shown in [KV2, Section 5.1], we may achieve a normal derivative reduction for the Stokes operator kt i+j+kr @ i t @ j d @ k uk L 2 x;t +kt i+j+kr @ i t @ j1 d @ k pk L 2 .kt i+j+kr @ i+1 t @ j2 d @ k uk L 2 x;t +kt i+j+kr @ i t @ j1 d @ k+1 uk L 2 x;t +kt i+j+kr @ i t @ j2 d @ k+2 uk L 2 x;t +kt i+j+kr @ i t @ j2 d @ k uk L 2 x;t +kt i+j+kr @ i t @ j2 d @ k fk L 2 x;t ; j 2 (6.1.9) which allows us to reduce the number of vertical derivatives (@ d ) in the Gevrey (analytic) norm. On the other hand, for j = 1, we have kt i+1+kr @ i t @ d @ k uk L 2 x;t +kt i+1+kr @ i t @ k pk L 2 x;t .kt i+1+kr @ i+1 t @ k1 uk L 2 x;t +kt i+1+kr @ i t @ k1 fk L 2 x;t ; k 1: (6.1.10) For j = 1 and k = 0, we claim kt i+1r @ i t ruk L 2 x;t .kt i+1r @ i t uk 1=2 L 2 x;t kt i+1r @ i+1 t uk 1=2 L 2 x;t +kt i+1r @ i t uk L 2 +kt i+1r @ i t fk L 2 x;t ; ir: (6.1.11) 150 In order to reduce the number of tangential derivatives, we apply the estimate kt i+kr @ i t @ k uk L 2 x;t +kt i+kr @ i t @ k1 pk L 2 x;t .kt i+kr @ i+1 t @ k2 uk L 2 x;t +kt i+kr @ i t @ k2 fk L 2 x;t ; k 2 (6.1.12) given in [KV2, Section 5.2]. Fork = 1, we use a special case (replacer with @) of the inequality (6.1.11): kt i+1r @ i t @uk L 2 x;t .kt i+1r @ i t uk 1=2 L 2 x;t kt i+1r @ i+1 t uk 1=2 L 2 x;t +kt i+1r @ i t uk L 2 +kt i+1r @ i t fk L 2 x;t ; ir: (6.1.13) Lastly, for the pure time derivatives, we have kt ir @ i t uk L 2 x;t . (ir)kt i1r @ i1 t uk L 2 x;t +kt ir @ i1 t fk L 2 x;t ; i 1r (6.1.14) as obtained in [KV2, Section 5.3]. The proofs of these reductions are based on simple H 2 regularity considerations for the linear parabolic type equations. The estimate (6.1.3) is obtained by summing over (i;j;k)2B the estimates (6.1.9)–(6.1.14), and to absorb all the u-dependent terms into the left side of the inequality by choosing ; ;e such that (6.1.5) and (6.1.6) hold. 6.2 Proof of Theorem 1 We appeal to the results in [KV2] by rewriting the Navier-Stokes equation (6.0.1) as a forced Stokes equation @ t u u +rp =uru +f ru = 0 (6.2.1) 151 on the three-dimensional half-space =fx 3 > 0g, with the Dirichlet boundary condition for u on @ . Choosing to work with d = 3 is nonessential, and is convenient only in fixing the Sobolev-embedding exponents in L 1 ( ) H 2 ( ) and L 4 ( ) H 1 ( ). With this choice of dimension, it is possible to set r = 3 in the definition of 0;T (u) and T (u). We note that replacing H 2 and H 1 with higher order Sobolev spaces, and increasing the value of r accordingly, we may treat (6.0.1) in any dimension d 2. 6.2.1 Local existence In order to establish the boundedness of 0 (u) for a finite time horizon, we appeal to a local existence result for the Navier-Stokes equations [Sol, Chapter III] (see also [LM2] for more general local existence resultsforsemi-linearparabolicproblemswithDirichletboundaryconditions): Assumethatu 0 2H 1 0 ( )\ H 4 ( ) is divergence free and obeys suitable compatibility conditions, and that the forcing f lies in L 1 loc ([0;1);H 3 ( )). Then there exists a time T =T (ku 0 k H 4 ( ) ;kfk L 1 loc ([0;1);H 3 ( )) )> 0 (6.2.2) and a unique solution u to the Cauchy problem associated to (6.0.1)–(6.0.4) which obeys sup t2[0;T] ku(t)k H 4 ( ) 2ku 0 k H 4 ( ) : (6.2.3) Without loss of generality, in (6.2.2) we may take the T 1. Furthermore, if we further assume that @ t f 2 L 1 ([0; 1];H 1 ( )) and @ 2 t f 2 L 1 ([0; 1];L 2 ( )) we conclude from (6.0.1) and (6.2.3) that there exists a constant C =C(d; ) 1 such that sup t2[0;T] ku(t)k H 2 ( ) +k@ t u(t)k H 1 ( ) +k@ 2 t u(t)k L 2 ( ) C(1 +ku 0 k H 4 ( ) ) 3 +k@ t fk L 1 ([0;T];H 1 ( )) +k@ 2 t fk L 1 ([0;T];L 2 ( )) =C (u 0 ;f): (6.2.4) 152 The upshot of (6.2.4) is that for any T2 (0;T ] (0; 1] we have 0;T (u)T 1=2 C (u 0 ;f): (6.2.5) When T is sufficiently small, the above estimate implies the smallness of 0;T (u), which is essential in closing the nonlinear argument. 6.2.2 The Stokes estimate From the result in [KV2], namely the estimates (6.1.3)–(6.1.4) for the nonlinear Stokes equation (6.2.1), we obtain T (u). 0;T (u) +M T (f) +M T (uru) = 0;T (u) +M T (f) + X i+j+k(r2)+ N i+j+k+2 i e j+2 k kt i+j+k+2r @ i t @ j d @ k (uru)k L 2 x;t + X i+k(r2)+ N i+k+2 i k+2 kt i+k+2r @ i t @ k (uru)k L 2 x;t + X i(r1)+ N i+1 i+1 kt i+1r @ i t (uru)k L 2 x;t = 0;T (u) +M T (f) +M 1 +M 2 +M 3 (6.2.6) whereM T (f) is as defined in (6.1.4), and the parameters;e ; are fixed as in (6.1.5)–(6.1.7), so that they depend only on d = 3 andr = 3. The bulk of the proof of Theorem 1 is to bound the sums M 1 ;M 2 , and M 3 appearing on the right side of (6.2.6), in terms of 0;T (u) and T (u). 6.2.3 Bounds for the nonlinear term These estimates for M 1 ;M 2 , and M 3 are performed in detail in Section 6.3 below (cf. Lemmas 5, 6, and 7), and may be summarized as follows: 153 Lemma 2. Fix T2 (0; 1], d2f2; 3g and r = 3. Then we have M 1 +M 2 +M 3 . 0;T (u) 1=2 T (u) 3=2 +T 1=2 T (u) 2 ; (6.2.7) where the implicit constant depends only on r, d, and , and is in particular independent of T. From (6.2.6) and (6.2.7) we conclude that there exists C = C(r;d; ) 1 such that T (u) C 0;T (u) 1=2 ( 0;T (u) 1=2 + T (u) 3=2 ) + CM T (f) + CT 1=2 T (u) 2 (6.2.8) for any T2 (0;T ]. 6.2.4 Conclusion of the proof of Theorem 1 In order to complete the proof of Theorem 1, it only remains to combine (6.2.5) with (6.2.8). This is a standard barrier argument, which we sketch briefly. The goal is to prove that for T sufficiently small, we have T (u) 4 C + 4 CM 1 (f) =M (6.2.9) where the constant C is the one given in (6.2.8). Note that M T (f) M 1 (f). In order to prove (6.2.9) for T sufficiently small, first use (6.2.5) and take T 1=C (u 0 ;f) 2 , which ensures that 0;T (u) 1 for 0<TT . Therefore, letting T 1, from (6.2.5) and (6.2.8) we obtain T (u) C + C(C (u 0 ;f)) 1=2 T 1=4 T (u) 3=2 + CM 1 (f) + CT 1=2 T (u) 2 : (6.2.10) 154 If our assertion (6.2.9) does not hold, there exists TT such that(T )<M forT < T and( T ) =M. Then, using (6.2.10) with T = T, we get M C + C(C (u 0 ;f)) 1=2 T 1=4 T (u) 3=2 + CM 1 (f) + C T 1=2 T (u) 2 M 4 + C(C (u 0 ;f)) 1=2 T 1=4 M 3=2 + CT 1=2 M 2 : (6.2.11) RestrictingT so that the last two terms are both less than or equal to M=4 givesM = 0, which leads to a contradiction and thus proves Theorem 1. 6.3 Space-time analytic estimates for the nonlinear term For the remainder of the proof, we omit the T-subindex in the quantities T (u), T (u), and 0;T (u), and simply denote them as (u), (u), and 0 (u). It is convenient to use the notationj(i;j;k)j =i +j +k, which indicates the length of the multi-index, and to denote U i;j;k := 8 > > > < > > > : N i+j+k i e j k kt i+j+kr @ i t @ j d @ k uk L 2 x;t ; i +j +kr; k@ i t @ j d @ k uk L 2 x;t ; 0i +j +kr 1; (6.3.1) where we recall that N i+j+k =j(i;j;k)j r =j(i;j;k)j!. Here the parameters ;e ; are fixed as in (6.1.5)– (6.1.7). With this notation we have (u) = X i+j+kr U i;j;k and 0 (u) = X 0i+j+kr1 U i;j;k : (6.3.2) It shall be convenient to denoter = ( @;@ d ) and u = ( u;u d ), so that uru = u @u +u d @ d u. Remark 5. We emphasize that throughout this last section the implicit constants in the. symbols are allowed to depend on m ,e m , and m , where m2Z is such thatjmj 100. Indeed, since thee ; ; have 155 been fixed solely in terms of , d, and r, cf. (6.1.5)–(6.1.7), they are independent of time and thus any a-priori finite power of these parameters may be hidden in the. symbol. 6.3.1 Gagliardo-Nirenberg inequalities We use a number of well-known space-time Gagliardo-Nirenberg inequalities that we summarize next. For u2H 2 ( ), we shall frequently use the following estimates: kuk L 1 ( ) .kuk d=4 _ H 2 ( ) kuk 1d=4 L 2 ( ) +kuk L 2 ( ) ; u2H 2 ( ); (6.3.3) kuk L 1 ( ) .kuk d=4 _ H 2 ( ) kuk 1d=4 L 2 ( ) ; u2H 2 ( ) with u @ = 0; (6.3.4) kuk L 4 ( ) .kuk d=4 _ H 1 ( ) kuk 1d=4 L 2 ( ) +kuk L 2 ( ) ; u2H 1 ( ); (6.3.5) kuk L 4 ( ) .kuk d=4 _ H 1 ( ) kuk 1d=4 L 2 ( ) ; u2H 1 ( ) with u @ = 0: (6.3.6) For v2H 1 (0;T ) such that vj t=0 = 0, we use Agmon’s inequality kvk L 1 (0;T) .kvk 1=2 L 2 (0;T) k@ t vk 1=2 L 2 (0;T) ; (6.3.7) while in the case vj t=0 6= 0, a lower order term is needed in the above estimate, namely, kvk L 1 (0;T) .kvk 1=2 L 2 (0;T) k@ t vk 1=2 L 2 (0;T) +kvk L 2 (0;T) : (6.3.8) Together, the estimates (6.3.3)–(6.3.8) imply that for u2H 1 (0;T ;H 2 ( )), we have kuk L 1 x;t .k@ t uk 1=2 L 2 t _ H 2 x kuk 1=2 L 2 t _ H 2 x +kuk L 2 t _ H 2 x +k@ t uk L 2 x;t +kuk L 2 x;t : (6.3.9) 156 Similarly, for u2H 1 (0;T ;H 1 ( )), we may bound kuk L 1 t L 4 x .k@ t uk 1=2 L 2 t _ H 1 x kuk 1=2 L 2 t _ H 1 x +kuk L 2 t _ H 1 x +k@ t uk L 2 x;t +kuk L 2 x;t : (6.3.10) The estimates (6.3.9)–(6.3.10) are used repeatedly throughout the next sections. We note that in view of the three derivative loss in the first term on right side of (6.3.9), one in time and two in space, the smallest value we may take for r in the definition of (u) is 3, which justifies our choice r = 3. In order to simplify the computations, we rewrite (6.3.9) and (6.3.10) for a function of the form t `+n+m @ ` t @ n d @ m u. The former inequality becomes kt `+n+m @ ` t @ n d @ m uk L 1 x;t .k@ t (t `+n+m @ ` t @ n d @ m u)k 1=2 L 2 t _ H 2 x kt `+n+m @ ` t @ n d @ m uk 1=2 L 2 t _ H 2 x +kt `+n+m @ ` t @ n d @ m uk L 2 t _ H 2 x +k@ t (t `+n+m @ ` t @ n d @ m u)k L 2 x;t +kt `+n+m @ ` t @ n d @ m uk L 2 x;t . kt `+n+m @ `+1 t @ n d @ m uk 1=2 L 2 t _ H 2 x +j(`;n;m)j 1=2 kt `+n+m1 @ ` t @ n d @ m uk 1=2 L 2 t _ H 2 x kt `+n+m @ ` t @ n d @ m uk 1=2 L 2 t _ H 2 x +kt `+n+m @ ` t @ n d @ m uk L 2 t _ H 2 x +kt `+n+m @ ` t @ n d @ m uk L 2 x;t + kt `+n+m @ `+1 t @ n d @ m uk L 2 x;t +j(`;n;m)jkt `+n+m1 @ ` t @ n d @ m uk L 2 x;t (6.3.11) when ` +n +m 1. Note that since T 1, the second and third term in (6.3.11) is dominated by the first and the last term, respectively. In the following lemma, we express (6.3.11) using the notation U `;n;m . Also, we denote V `;n;m =U `;n;m +U `;n1;m+1 +U `;n2;m+2 : (6.3.12) 157 Lemma 3. For u2H 1 (0;T ;H 2 ( )) and all multi-indicesj(`;n;m)j 1, we have N `+n+m ` e n m kt `+n+m @ ` t @ n d @ m uk L 1 x;t .V 1=2 `+1;n+2;m V 1=2 `;n+2;m T 1=2 j(`;n;m)j 5=2 +V `;n+2;m T 1=2 j(`;n;m)j 5=2 +U `+1;n;m T 1 `+n+m=1 +T 2 1 `+n+m2 j(`;n;m)j +U `;n;m T `+n+m1 1 `+n+m2 +T 2 1 `+n+m3 j(`;n;m)j: Similarly, we write the inequality (6.3.10) for the function t `+n+m1 @ ` t @ n d @ m u with ` +n +m 2, kt `+n+m1 @ ` t @ n d @ m uk L 1 t L 4 x .k@ t (t `+n+m1 @ ` t @ n d @ m u)k 1=2 L 2 t _ H 1 x kt `+n+m1 @ ` t @ n d @ m uk 1=2 L 2 t _ H 1 x +kt `+n+m1 @ ` t @ n d @ m uk L 2 t _ H 1 x +k@ t (t `+n+m1 @ ` t @ n d @ m u)k L 2 x;t +kt `+n+m1 @ ` t @ n d @ m uk L 2 x;t . kt `+n+m1 @ `+1 t @ n d @ m uk 1=2 L 2 t _ H 1 x +j(`;n;m)j 1=2 kt `+n+m2 @ ` t @ n d @ m uk 1=2 L 2 t _ H 1 x kt `+n+m1 @ ` t @ n d @ m uk 1=2 L 2 t _ H 1 x +kt `+n+m1 @ ` t @ n d @ m uk L 2 t _ H 1 x +kt `+n+m1 @ ` t @ n d @ m uk L 2 x;t + kt `+n+m1 @ `+1 t @ n d @ m uk L 2 x;t +j(`;n;m)jkt `+n+m2 @ ` t @ n d @ m uk L 2 x;t : (6.3.13) Using the notation U `;n;m , we rewrite (6.3.13) as follows. 158 Lemma 4. For u2H 1 (0;T ;H 1 ( )) and all multi-indicesj(`;n;m)j 2, we have N `+n+m ` e n m kt `+n+m1 @ ` t @ n d @ m uk L 1 t L 4 x . U `+1;n+1;m +U `+1;n;m+1 1=2 (U `;n+1;m +U `;n;m+1 ) 1=2 T 1=2 j(`;n;m)j 3=2 + (U `;n+1;m +U `;n;m+1 )T 1=2 j(`;n;m)j 3=2 +U `+1;n;m Tj(`;n;m)j +U `;n;m (1 `+n+m=2 +T 1 `+n+m3 )j(`;n;m)j: (6.3.14) We have used once again T 1 to have the second and third term in (6.3.13) dominated by the first and the last term, respectively. 6.3.2 Terms with only time derivatives In this section we estimate M 3 in (6.2.6). From the Leibniz rule we obtain that M 3 X i2 i X `=0 i ` N i+1 i kt i2 @ ` t u@ i` t @uk L 2 x;t +kt i2 @ ` t u d @ i` t @ d uk L 2 x;t : Recalling the notation (6.3.2) we assert that: Lemma 5. For solution u of the Cauchy problem (6.0.1)–(6.0.3), we have M 3 . 0 (u) 1=2 (u) 3=2 +T 1=2 (u) 2 (6.3.15) for 0<T 1. 159 Proof of Lemma 5. We split M 3 into sums M 31 and M 32 corresponding to i = 2 or i 3, respectively. Then M 31 .kuk L 1 x;t k@ 2 t @uk L 2 x;t +k@ 2 t @ d uk L 2 x;t +k@ t uk L 1 t L 4 x k@ t @uk L 2 t L 4 x +k@ t @ d uk L 2 t L 4 x +k@ 2 t uk L 2 t L 4 x k @uk L 1 t L 4 x +k@ d uk L 1 t L 4 x where the three terms correspond to ` = 0; 1; 2. Using the notationbxc = [x] anddxe = [x] + 1, we have M 32 X i3 bi=2c X `=1 i ` N i+1 i kt ` @ ` t uk L 1 x;t kt i`2 @ i` t @uk L 2 x;t +kt i`2 @ i` t @ d uk L 2 x;t + X i3 N i+1 i kuk L 1 x;t kt i2 @ i t @uk L 2 x;t +kt i2 @ i t @ d uk L 2 x;t + X i3 i1 X `=di=2e i ` N i+1 i kt `2 @ ` t uk L 2 t L 4 x kt i` @ i` t @uk L 1 t L 4 x +kt i` @ i` t @ d uk L 1 t L 4 x + X i3 N i+1 i kt i2 @ i t uk L 2 t L 4 x k @uk L 1 t L 4 x +k@ d uk L 1 t L 4 x =M 321 +M 322 +M 323 +M 324 ; (6.3.16) 160 where we separated away ` = 0 and ` = i from the main sums. We start by bounding M 31 . Since @ ` t u @ = 0 for ` 1, we may apply (6.3.9) and (6.3.10) to conclude M 31 . k@ t uk 1=2 L 2 t _ H 2 x kuk 1=2 L 2 t _ H 2 x +kuk L 2 t _ H 2 x +k@ t uk L 2 x;t +kuk L 2 x;t k@ 2 t @uk L 2 x;t +k@ 2 t @ d uk L 2 x;t + k@ 2 t uk 1=2 L 2 t _ H 1 x k@ t uk 1=2 L 2 t _ H 1 x +k@ t uk L 2 t _ H 1 x +k@ 2 t uk L 2 x;t +k@ t uk L 2 x;t k@ t @uk d=4 L 2 t _ H 1 x k@ t @uk 1d=4 L 2 x;t +k@ t @ d uk d=4 L 2 t _ H 1 x k@ t @ d uk 1d=4 L 2 x;t +k@ t @ d uk L 2 x;t + k@ 2 t uk d=4 L 2 t _ H 1 x k@ 2 t uk 1d=4 L 2 x;t k@ t @uk 1=2 L 2 t _ H 1 x k @uk 1=2 L 2 t _ H 1 x +k @uk L 2 t _ H 1 x +k@ t @uk L 2 x;t +k @uk L 2 x;t +k@ t @ d uk 1=2 L 2 t _ H 1 x k@ d uk 1=2 L 2 t _ H 1 x +k@ d uk L 2 t _ H 1 x +k@ t @ d uk L 2 x;t +k@ d uk L 2 x;t : Therefore, M 31 . 0 (u) 1=2 (u) 1=2 + 0 (u) (u) + 0 (u) 1=2 (u) 1=2 + 0 (u) (u) d=4 0 (u) 1d=4 + 0 (u) + (u) d=4 0 (u) 1d=4 (u) 1=2 0 (u) 1=2 + 0 (u) . 0 (u) 2 + 0 (u) 1=2 (u) 3=2 : (6.3.17) 161 ForM 32 , we note that i 3 for each of the sums. We start with the boundary sums M 322 andM 324 , and treat M 321 and M 323 further below. Using (6.3.9), we have M 322 . X i3 N i+1 i k@ t uk 1=2 L 2 t _ H 2 x kuk 1=2 L 2 t _ H 2 x +kuk L 2 t _ H 2 x +k@ t uk L 2 x;t +kuk L 2 x;t kt i2 @ i t @uk L 2 x;t +kt i2 @ i t @ d uk L 2 x;t . ( 0 (u) + 0 (u) 1=2 (u) 1=2 ) X i3 N i+1 i kt i2 @ i t @uk L 2 x;t +kt i2 @ i t @ d uk L 2 x;t . ( 0 (u) + 0 (u) 1=2 (u) 1=2 ) (u). 0 (u)(u) + 0 (u) 1=2 (u) 3=2 : (6.3.18) For M 324 we proceed with the Gagliardo-Nirenberg inequalities (6.3.5)–(6.3.10) and write M 324 . X i3 N i+1 i kt i2 @ i t uk d=4 L 2 t _ H 1 x kt i2 @ i t uk 1d=4 L 2 x;t k@ t @uk 1=2 L 2 t _ H 1 x k @uk 1=2 L 2 t _ H 1 x +k @uk L 2 t _ H 1 x +k@ t @uk L 2 x;t +k @uk L 2 x;t + X i3 N i+1 i kt i2 @ i t uk d=4 L 2 t _ H 1 x kt i2 @ i t uk 1d=4 L 2 x;t k@ t @ d uk 1=2 L 2 t _ H 1 x k@ d uk 1=2 L 2 t _ H 1 x +k@ d uk L 2 t _ H 1 x +k@ t @ d uk L 2 x;t +k@ d uk L 2 x;t : Expressing the estimates in terms of U i;j;k we get M 324 . X i3 N i+1 i (U i;1;0 +U i;0;1 ) d=4 U 1d=4 i;0;0 ( (u) 1=2 0 (u) 1=2 + 0 (u)) (i + 1)! (i + 1) 3 i d=4 i!T i 3 i 1d=4 .T 1d=4 ( (u) 1=2 0 (u) 1=2 + 0 (u)) X i3 (U i;1;0 +U i;0;1 ) d=4 U 1d=4 i;0;0 .T 1d=4 ( (u) 1=2 0 (u) 1=2 + 0 (u))(u) .T 1d=4 0 (u)(u) +T 1d=4 0 (u) 1=2 (u) 3=2 (6.3.19) where we have used 1=Ce 1. 162 For M 321 , we expresskt i`2 @ i` t @uk L 2 x;t andkt i`2 @ i` t @ d uk L 2 x;t in terms of U i`;1;0 and U i`;0;1 and write, using (6.3.11), M 321 . X i3 bi=2c X `=1 kt ` @ ` t uk L 1 x;t (U i`;1;0 +U i`;0;1 ) (i + 1) 2 ` `!(i` + 1) 2 : (6.3.20) Then, we utilize Lemma 3 (which we have derived from (6.3.9)) and `bi=2c to obtain M 321 .T 1=2 X i3 bi=2c X `=1 V 1=2 `+1;2;0 V 1=2 `;2;0 (U i`;0;1 +U i`;1;0 ) +T 1=2 X i3 bi=2c X `=1 V `;2;0 (U i`;0;1 +U i`;1;0 ) + X i3 bi=2c X `=1 U `+1;0;0 (U i`;0;1 +U i`;1;0 ) T 1 `=1 +T 2 1 `2 + X i3 bi=2c X `=1 U `;0;0 (U i`;0;1 +U i`;1;0 ) T `1 1 `2 +T 2 1 `3 : By appealing to the discrete Young’s inequality and the definition of 0 (u), we get M 321 .T 1=2 (u) 2 +T (u) 2 +T 0 (u) (u) +T 2 0 (u) (u) +T 3 (u) 2 + 0 (u) (u) +T 2 (u) 2 : Once again using T 1 and keeping the dominant terms , we obtain: M 321 . 0 (u)(u) +T 1=2 (u) 2 : (6.3.21) 163 Lastly, we treat M 323 in a similar manner. We split the sum into two parts, as done above for M 321 , by appealing to (6.3.6): M 323 . X i3 i1 X `=di=2e i ` N i+1 i kt `2 @ ` t uk d=4 L 2 t _ H 1 x kt `2 @ ` t uk 1d=4 L 2 x;t kt i` @ i` t @uk L 1 t L 4 x + X i3 i1 X `=di=2e i ` N i+1 i kt `2 @ ` t uk d=4 L 2 t _ H 1 x kt `2 @ ` t uk 1d=4 L 2 x;t kt i` @ i` t @ d uk L 1 t L 4 x =M 3231 +M 3232 : For M 3231 we use Lemma 4 for the triple (i`; 0; 1) and obtain M 3231 .T 1=2 X i3 i1 X `=di=2e (U `;1;0 +U `;0;1 ) d=4 U 1d=4 `;0;0 (U i`+1;1;1 +U i`+1;0;2 ) 1=2 (U i`;1;1 +U i`;0;2 ) 1=2 +T 1=2 X i3 i1 X `=di=2e (U `;1;0 +U `;0;1 ) d=4 U 1d=4 `;0;0 (U i`;1;1 +U i`;0;2 ) +T X i3 i1 X `=di=2e (U `;1;0 +U `;0;1 ) d=4 U 1d=4 `;0;0 U i`+1;0;1 + X i3 i1 X `=di=2e (U `;1;0 +U `;0;1 ) d=4 U 1d=4 `;0;0 U i`;0;1 (1 i`=1 +T 1 i`2 ): (6.3.22) Applying the discrete Young’s inequality and using the definition of 0 (u), we get M 3231 .T 1=2 0 (u) 1d=4 (u) 1+d=4 +T 1=2 (u) 2 +T 0 (u) 2d=4 (u) d=4 +T 0 (u) (u) +T 2 0 (u) 1d=4 (u) 1+d=4 +T 2 (u) 2 +T 0 (u) 1d=4 (u) 1+d=4 +T (u) 2 + 0 (u) 2d=4 (u) d=4 + 0 (u) (u) +T 0 (u) 1d=4 (u) 1+d=4 +T (u) 2 : (6.3.23) 164 In estimating M 3232 , we follow the same steps as in (6.3.22), as the only difference is due to having the differential @ d in M 3232 instead of @. We obtain that M 3232 obeys the same exact estimate as M 3231 in (6.3.23), from which we obtain the desired bound for M 323 , namely M 323 . 0 (u)(u) +T 1=2 (u) 2 : (6.3.24) Combining all the terms in (6.3.17), (6.3.18), (6.3.19), (6.3.21), (6.3.24), and selecting the maximal pref- actors in T we obtain the estimate (6.3.15). 6.3.3 Terms with no normal derivatives In this section we estimate M 2 . Lemma 6. For solutions u of the Cauchy problem (6.0.1)–(6.0.3), we get M 2 . 0 (u) 3=2 (u) 1=2 +T 0 (u) 1=2 (u) 3=2 +T 3=4 (u) 2 (6.3.25) for 0<T 1. Proof of Lemma 6. Writinguru = u @u +u d @ d u and separating the terms withi +k = 1, we obtain M 2 X i+k2 i+k X j(`;m)j=0 i ` k m N i+k+2 i k kt i+k1 @ ` t @ m u@ i` t @ k+1m uk L 2 x;t +kt i+k1 @ ` t @ m u d @ i` t @ km @ d uk L 2 x;t +k@ t (u d @ d u) +@ t ( u @u)k L 2 x;t +k @(u d @ d u) + @( u @u)k L 2 x;t =M 21 +M 22 +M 23 : 165 We start with the lower order terms. Using Hölder’s inequality, we get M 22 +M 23 .k@ t uk L 1 t L 4 x k @uk L 2 t L 4 x +k@ d uk L 2 t L 4 x +kuk L 1 x;t k@ t @uk L 2 x;t +k@ t @ d uk L 2 x;t +k @uk L 1 t L 4 x k @uk L 2 t L 4 x +k@ d uk L 2 t L 4 x +kuk L 1 x;t k @ 2 uk L 2 x;t +k @@ d uk L 2 x;t and recalling the definition of (u) and 0 (u), we obtain M 22 +M 23 . (u) 1=2 0 (u) 3=2 + 0 (u) 2 : (6.3.26) Now, we split M 21 into two parts as M 21 . X i+k2 b(i+k)=2c X j(`;m)j=0 i ` k m N i+k+2 i k kt `+m @ ` t @ m uk L 1 x;t kt (i+k)(`+m)1 @ i` t @ k+1m uk L 2 x;t +kt (i+k)(`+m)1 @ i` t @ km @ d uk L 2 x;t + X i+k2 i+k X j(`;m)jd(i+k)=2e i ` k m N i+k+2 i k kt `+m1 @ ` t @ m uk L 1 t L 4 x kt (i+k)(`+m) @ i` t @ k+1m uk L 2 t L 4 x +kt (i+k)(`+m) @ i` t @ km @ d uk L 2 t L 4 x =M 211 +M 212 : (6.3.27) 166 We start with the first sum in (6.3.27), namely M 211 . We apply Lemma 3 onkt `+m @ ` t @ m uk L 1 x;t except whenj(`; 0;m)j = 0. Singling outj(`; 0;m)j = 0, we have M 211 . X i+k2 i k N i+k+2 (U i;0;k+1 +U i;1;k ) T N i+k+1 i k k@ t uk 1=2 L 2 t _ H 2 x kuk 1=2 L 2 t _ H 2 x +kuk L 2 t _ H 2 x +k@ t uk L 2 x;t +kuk L 2 x;t + X i+k2 b(i+k)=2c X `+m=1 i k i ` k m N i+k+2 (U i`;0;k+1m +U i`;1;km ) T N i+k`m+1 i` km (` +m)! ` m T 1=2 (` +m) 1=2 V 1=2 `+1;2;m V 1=2 `;2;m + T 1=2 (` +m) 1=2 V `;2;m + 1 (` +m) 2 U `+1;0;m (T 1 `+m=1 +T 2 1 `+m2 ) +U `;0;m (T `+m1 1 `+m2 +T 2 1 `+m3 ) : Note that whenj(`;m)jb(i +k)=2c, we may bound N i+k+2 (` +m)! N i+k`m+1) (` +m) 1=2 i ` k m . i ` k m i+k `+m . 1: (6.3.28) Then we get M 211 .T 0 (u)(u) +T 0 (u) 1=2 (u) 3=2 +T 3=2 (u) 2 : (6.3.29) 167 Next, we split M 212 into two parts as M 212 = X i+k2 i+k X j(`;m)j=d(i+k)=2e i ` k m N i+k+2 i k kt `+m1 @ ` t @ m uk L 1 t L 4 x kt (i+k)(`+m) @ i` t @ k+1m uk L 2 t L 4 x + X i+k2 i+k X j(`;m)j=d(i+k)=2e i ` k m N i+k+2 i k kt `+m1 @ ` t @ m uk L 1 t L 4 x kt (i+k)(`+m) @ i` t @ km @ d uk L 2 t L 4 x =M 2121 +M 2122 : Both terms are treated analogously, and thus we only bound the first one. M 2121 . X i+k2 i+k X j(`;m)j=d(i+k)=2e i ` k m N i+k+2 i k kt `+m1 @ ` t @ m uk L 1 t L 4 x (U i`;0;k+2m +U i`;1;k+1m ) d=4 (U i`;0;k+1m ) 1d=4 T 1 `+mi+k1 + 1 `+m=i+k N i+k+2`m i` km d=4 T 2 1 `+mi+k2 +T i+k`m 1 `+mi+k1 N i+k+1`m i` km 1d=4 ; (6.3.30) where we used 1=Ce 1. Note that the last two factors in (6.3.30) are bounded from above by T 2d=4 1 `+mi+k2 +T i+k`m 1 `+mi+k1 i` km (i +k`m)! (i +k`m) 2d=4 : (6.3.31) 168 Denote A i;k `;m (u) = (U i`;0;k+2m +U i`;1;k+1m ) d=4 (U i`;0;k+1m ) 1d=4 T 2d=4 1 `+mi+k2 +T i+k`m 1 `+mi+k1 i` km (i +k`m)! (i +k`m) 2d=4 : Applying Lemma 4 on the termkt `+m1 @ ` t @ m uk L 1 t L 4 x , we obtain M 2121 . X i+k2 i+k X j(`;m)j=d(i+k)=2e i ` k m N i+k+2 i k A i;k `;m (u) T 1=2 ` m (` +m)! (` +m) 3=2 (U `+1;1;m +U `+1;0;m+1 ) 1=2 (U `;1;m +U `;0;m+1 ) 1=2 + T 1=2 ` m (` +m)! (` +m) 3=2 (U `;1;m +U `;0;m+1 ) + (` +m)! (` +m) 2 U `+1;0;m T `+1 m + (` +m)! (` +m) 2 U `;0;m 1 `+m=2 +T 1 `+m3 ` m : (6.3.32) For i +k`m 2, we have i ` k m N i+k+2 (i +k`m)! (i +k`m) 2d=4 (` +m)! (` +m) 3=2 . i ` k m i+k `+m . 1: Then, using Young’s inequality in (6.3.32), we deduce M 2121 .T 1=2 0 (u) (u) + 0 (u) 2 +T 0 (u) 1d=4 (u) d=4 T 1=2 (u) + 0 (u) +T 1=2 (u) T 2d=4 (u) +T 0 (u) 1d=4 (u) d=4 + 0 (u) . 0 (u) 2 +T 1=2 0 (u)(u) +T 3=2 0 (u) 1d=4 (u) 1+d=4 +T 0 (u) 2d=4 (u) d=4 +T 3=2d=4 (u) 2 : (6.3.33) 169 SinceM 2122 is nearly identical toM 2121 , the right side of (6.3.33) gives us an estimate for M 212 . Finally, using thatd2f2; 3g, we add the estimates (6.3.26), (6.3.29), and (6.3.33) to get (6.3.25) in Lemma 6. 6.3.4 Terms with all mixed derivatives In this section we estimate M 1 . Lemma 7. For solutions u of the Cauchy problem (6.0.1)–(6.0.3), we have M 1 . 0 (u) 3=2 (u) 1=2 +T 1=2 0 (u)(u) +T 3=2 (u) 2 (6.3.34) for all 0<T 1. Proof of Lemma 7. Using the Leibniz rule we obtain M 1 X i+j+k1 i X `=0 j X n=0 k X m=0 i ` j n k m N i+j+k+2 i e j k kt i+j+k1 @ ` t @ n d @ m u@ i` t @ jn d @ km+1 uk L 2 x;t +kt i+j+k1 @ ` t @ n d @ m u d @ i` t @ jn+1 d @ km uk L 2 x;t : 170 We separate the casej(i;j;k)j = 1 from the sum and then split the rest into two parts, leading to M 1 . X i+j+k=1 kuk L 1 x;t kr@ i t @ j d @ k uk L 2 x;t +k@ i t @ j d @ k uk L 1 t L 4 x kruk L 2 t L 4 x + X i+j+k2 b(i+j+k)=2c X j(`;n;m)j=0 i ` j n k m N i+j+k+2 i e j k kt `+n+m @ ` t @ n d @ m uk L 1 x;t kt (i+j+k)(`+n+m)1 @ i` t @ jn d @ km+1 uk L 2 x;t +kt (i+j+k)(`+n+m)1 @ i` t @ jn+1 d @ km uk L 2 x;t + X i+j+k2 i+j+k X j(`;n;m)j=d(i+j+k)=2e i ` j n k m N i+j+k+2 i e j k kt `+n+m1 @ ` t @ n d @ m uk L 1 t L 4 x kt (i+j+k)(`+n+m) @ i` t @ jn d @ km+1 uk L 2 t L 4 x +kt (i+j+k)(`+n+m) @ i` t @ jn+1 d @ km uk L 2 t L 4 x =M 11 +M 12 +M 13 : The contribution fromj(i;j;k)j = 1 is bounded as M 11 . 0 (u) 2 + 0 (u) 3=2 (u) 1=2 : (6.3.35) For M 12 and M 13 , we follow the same strategy as in the last section. Starting with M 12 , we apply Lemma 3 to estimatekt `+n+m @ ` t @ n d @ m uk L 1 x;t in terms of the analyticity norm (6.3.2). Denote B i;j;k `;n;m (u) =kt (i+j+k)(`+n+m)1 @ i` t @ jn d @ km+1 uk L 2 x;t +kt (i+j+k)(`+n+m)1 @ i` t @ jn+1 d @ km uk L 2 x;t : 171 Next, using the notation (6.3.1) B i;j;k `;n;m (u). (T 1 `+n+mi+j+k2 + 1 `+n+m=i+j+k1 ) (i +j +k`nm + 1)! (i +j +k`nm + 1) 3 U i`;jn;km+1 i` e jn km + U i`;jn+1;km i` e jn km : Estimatingkt `+n+m @ ` t @ n d @ m uk L 1 x;t , the factorial terms obey i ` j n k m (` +n +m)! (i +j +k)! (i +j +k`nm)! (i +j +k`nm) 2 1 (` +n +m) 1=2 1: Therefore, M 12 .T 0 (u) (u) +T 0 (u) 1=2 (u) 3=2 +T 1=2 0 (u) (u) + 0 (u) 2 +T 3=2 (u) 2 +T 1=2 0 (u) 3=2 +T 0 (u) (u) . 0 (u) 2 +T 1=2 0 (u)(u) +T 3=2 (u) 2 : (6.3.36) 172 Lastly, we deal with M 13 . Similarly to M 212 , we split M 13 into two parts M 13 . X i+j+k2 i+j+k X j(`;n;m)j=d(i+j+k)=2e i ` j n k m N i+j+k+2 i e j k kt `+n+m1 @ ` t @ n d @ m uk L 1 t L 4 x kt (i+j+k)(`+n+m) @ i` t @ jn d @ km+1 uk L 2 t L 4 x + X i+j+k2 i+j+k X j(`;n;m)j=d(i+j+k)=2e i ` j n k m N i+j+k+2 i e j k kt `+n+m1 @ ` t @ n d @ m uk L 1 t L 4 x kt (i+j+k)(`+n+m) @ i` t @ jn+1 d @ km uk L 2 t L 4 x =M 131 +M 132 ; (6.3.37) and consider M 131 . Denote A i;j;k `;n;m (u) = kt (i+j+k)(`+n+m) @ i` t @ jn d @ km+1 uk L 2 t L 4 x : Using (6.3.5)–(6.3.6), and (6.3.2) we bound A i;j;k `;n;m (u) from above by A i;j;k `;n;m (u). (U i`;jn+1;km+1 +U i`;jn;km+2 ) d=4 (U i`;jn;km+2 ) 1d=4 (i +j +k`nm)! (i +j +k`nm + 1) 2d=4 1 i` e jn k T 2d=4 1 `+n+mi+j+k2 +T i+j+k`nm 1 `+n+mi+j+k1 +U i`;jn;km+1 (i +j +k`nm)! (i +j +k`nm + 1) 2 1 i` e jn k T 2 1 `+n+mi+j+k2 +T i+j+k`nm 1 `+n+mi+j+k1 : 173 Applying Lemma 4 tokt `+n+m1 @ ` t @ n d @ m uk L 1 t L 4 x and using Young’s inequality, we obtain M 131 . 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Creator Camliyurt, Guher (author)

Core Title Unique continuation for parabolic and elliptic equations and analyticity results for Euler and Navier Stokes equations

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Publication Date 08/08/2018

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