A THEOREM OF PALEY-WIENER TYPE FOR SCHRODINGER EVOLUTIONS

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A THEOREM OF PALEY-WIENER TYPE FOR SCHRÖDINGEREVOLUTIONSC. E. KENIG, G. PONCE, AND L. VEGAAbstract. We prove unique continuation principles for solutions of evolutionSchrödinger equations with time dependent potentials. These correspond touncertainly principles of Paley-Wiener type for the Fourier transform. Ourresults extends to a large class of semi-linear Schrödinger equation.Titre : Un théoreme de type Paley-Wiener pour les évolutions de Schrodinger.Résumé : On prouve des principes de prolongement unique pour les solutions d’équations d’évolution de Schrödinger avec potentiels dépendant dutemps. Ceux-ci correspondent á des principes d’incertitude de type PaleyWiener pour la transformée de Fourier. Nos résultats se généralisent á unelarge classe d’équations de Schrödinger semi-linéaires.1. IntroductionIn this paper we study unique continuation properties of solutions of Schrödingerequations of the form(1.1) t u i(4u V (x, t)u),(x, t) Rn [0, T ],T 0.The goal is to obtain sufficient conditions on the behavior of the solution u at twodifferent times and on the potential V which guarantee that u 0 in Rn [0, T ].Under appropriate assumptions this result will extend to the difference v u1 u2of two solutions u1 , u2 of semi-linear Schrödinger equation(1.2) t u i(4u F (u, u)),from which one can conclude that u1 u2 .Defining the Fourier transform of a function f asfb(ξ) (2π) n/2Ze iξ·x f (x)dx.Rn1991 Mathematics Subject Classification. Primary: 35Q55.Key words and phrases. Schrödinger evolutions, unique continuation.The first and second authors are supported by NSF grants DMS-0968472 and DMS-1101499respectively. The third author is supported by grants MTM2011-24054 and IT-305-07.1

2C. E. KENIG, G. PONCE, AND L. VEGAone hasu(x, t) eit 2Zu0 (x) Rn2(1.3)ei x /4t(4πit)n/2 Zei x y /4tu0 (y) dy(4πit)n/22e 2ix·y/4t ei y /4tu0 (y) dyRn2 x ei x /4t2 /4t\i · (eu),02t(2it)n/2 where eit u0 (x) denotes the free solution of the Schrödinger equation with data u0 t u i4u,u(x, 0) u0 (x),(x, t) Rn R.The identity (1.3) tells us that this kind of results for the free solution of theSchrödinger equation are closely related to uncertainty principles for the Fouriertransform. In this regard, one has the well known result of G. H. Hardy [9]:2If f (x) O(e x/β 222), fb(ξ) O(e 4 ξ /α ) and αβ 4, x2 /β 2and if αβ 4, then f (x) c ethen f 0,.Its extension to higher dimensions n 2 was obtained in [15]. The followinggeneralization in terms of the L2 -norm was established in [3]:Ife x 2β2f (x), e4 ξ 2α2fb(ξ) L2 (Rn ), and α β 4, then f 0.In terms of the free solution of the Schrödinger equation the L2 -version of HardyUncertainty Principle says :(1.4)If e x 2β2u0 (x), e x 2α2eit u0 (x) L2 (Rn ), and α β 4t, then u0 0.In [6] the following result was proven:Theorem. ([6]) Given any solution u C([0, T ] : L2 (Rn )) of(1.5) t u i (4u V (x, t)u) ,(x, t) Rn [0, T ],with V L (Rn [0, T ]),(1.6)lim kV kL1 ([0,T ]:L (Rn \Bρ )) 0.ρ ande x 2β2u0 ,e x 2α2eiT u0 L2 (Rn ),with α β 4T , then u0 0.Notice that the above Theorem recovers the L2 -version of the Hardy UncertaintyPrinciple (1.4) for solutions of the IVP (1.5), except for the limiting case α β 4Tfor which the corresponding result was proven to fail, see [6]. For further results inthis direction concerning other uncertainty principles we refer to [8] and referencestherein.Some previous results concerning uniqueness properties of solutions of the Schrödingerequation were not directly motivated by the formula (1.3).

UNIQUE CONTINUATION3For solutions u u(x, t) of the 1-D cubic Schrödinger equation t u i( x2 u u 2 u),(1.7)B. Y. Zhang [17] showed :If u(x, t) 0 for (x, t) ( , a) {0, 1} (or (x, t) (a, ) {0, 1}) for somea R, then u 0.The proof is based on the inverse scattering method, which uses the fact thatthe equation in (1.7) is a completely integrable model.In [13], under general assumptions on F in (1.2), it was proven that :If u1 , u2 C([0, 1] : H s (Rn )), with s max{n/2; 2} are solutions of theequation (1.2) with F as in (1.2) such thatu1 (x, t) u2 (x, t), (x, t) Γcx0 {0, 1},where Γcx0 denotes the complement of a cone Γx0 with vertex x0 Rn and opening 1800 , then u1 u2 .For further results in this direction see [12], [13], [10], [11], and references therein.Note that in [8] a unified approach was given to both kinds of results, using Lemma3 and Corollary 1 below.Returning to the uncertainty principle for the Fourier transform one has :If f L1 (Rn ) is non-zero and has compact support, then fb cannot satisfy acondition of the type fb(y) O(e y ) for any 0.This is due to the fact that fb(y) O(e y ) implies that f has an analyticextension to the strip {z Cn : Im(z) }.In this regard the Paley-Wiener Theorem [14] gives a characterization of a function or distribution with compact support in term of the analyticity properties ofits Fourier transform.Our main result in this work is the following:Theorem 1. Let u C([0, 1] : L2 (Rn )) be a strong solution of the equation(1.8) t u i( u V (x, t)u),(x, t) Rn [0, 1].Assume thatZ(1.9)sup0 t 1Z(1.10) u(x, t) 2 dx A1 ,Rne2a1 x1 u(x, 0) 2 dx A2 ,for some a1 0,Rn(1.11)supp u(·, 1) {x Rn : x1 a2 },for some a2 ,with(1.12)V L (Rn [0, 1]),kV kL (Rn [0,1]) M0 ,and(1.13)Then u 0.lim kV kL1 ([0,1]:L (Rn \Bρ )) 0.ρ

4C. E. KENIG, G. PONCE, AND L. VEGARemarks: (a) Note that in order to prove Theorem 1, by translation in x1 , we canchoose who a2 is. We will show that there exists m 0 (small) with the propertythat if (1.9), (1.10), (1.12), (1.13) hold and (1.11) holds with a2 m, thenu(x, 1) 0for x Rn such that m/2 x1 m.This clearly yields the desired result. Without loss of generality we will assumem 1.(b) By rescaling it is clear that the result in Theorem 1 applies to any timeinterval [0, T ].(c) We recall that in Theorem 1 there are no hypotheses on the size of thepotential V in the given class or on its regularity.(d) A weaker version of Theorem 1 was announced in [8].As a direct consequence of Theorem 1 we get the following result regarding theuniqueness of solutions for non-linear equations of the form (1.2).Theorem 2. Givenu1 , u2 C([0, T ] : H k (Rn )),0 T , strong solutions of (1.2) with k Z , k n/2, F : C2 C, F C k andF (0) u F (0) ū F (0) 0 such thatsupp (u1 (·, 0) u2 (·, 0)) {x Rn : x1 a2 },(1.14)a2 .If for some t (0, T ) and for some 0u1 (·, t) u2 (·, t) L2 (e x1 dx),(1.15)then u1 u2 .Remarks: (a) In particular, by taking u2 0, Theorem 2 shows that if u1 (·, 0)has compact support, then for any t (0, T ) u1 (·, t) cannot decay exponentially.(b) In the case F (u, u) u α 1 u, with α n/2 if α is not an odd integer, wehave that if ϕ is the unique non-negative, radially symmetric solution of ϕ ω ϕ ϕ α 1 ϕ,ω 0,then(1.16)u1 (x, t) eiωt ϕ(x)is a solution (“standing wave”) of(1.17) t u i( u u α 1 u).It was established in [16], [1] that there exist constants c0 , c1 0 such that(1.18)ϕ(x) c0 e c1 x .Therefore, if we denote by u2 (x, t) the solution of the equation (1.17) with α n/2and data u2 (x, 0) ϕ(x) φ(x), φ H s (Rn ), s n/2 having compact support itfollows from Theorem 2, (1.16) and (1.18) that for any t 6 0(1.19)u2 (·, t) / L2 (e x dx),for any 0.In general, the same result (1.19) applies (in the time interval [0, T ]) if one assumesthat u1 is a solution of (1.16) having exponential decay u1 (x, t) c0 e c1 x ,c0 , c1 0(x, t) Rn [0, T ],

UNIQUE CONTINUATION5and u2 is the solution of (1.16) corresponding to an initial datau2 (x, 0) u1 (x, 0) φ(x),φ H s (Rn ), s n/2 with compact support.The rest of this paper is organized as follows: section 2 contains all the preliminary results to be used in the proof of Theorem 1. A version of them has beenproved in [7], [6], [8]. However, in some cases modifications are needed to applythem in the setting considered here. Hence, some of their proofs will be sketched.Section 3 contains the proof of Theorem 1.2. Preliminary EstimatesIn this section we describe the estimates to be used in the proof of Theorem 1.First we recall a key step in the uniform exponential decay estimate establishedin [13] :Lemma 1. There exists n 0 such that ifV : Rn [0, 1] C,(2.1)withkVkL1t L n ,xand u C([0, 1] : L2 (Rn )) is a strong solution of the IVP t u i( V(x, t))u G(x, t),(2.2)u(x, 0) u0 (x),with(2.3)u0 , u1 u( · , 1) L2 (e2λ·x dx),G L1 ([0, 1] : L2 (e2λ·x dx)),for some λ Rn , then there exists cn independent of λ such thatsup keλ·x u( · , t)kL2 (Rn )0 t 1(2.4)Z cn keλ·x u0 kL2 (Rn ) keλ·x u1 kL2 (Rn ) 1 keλ·x G(·, t)kL2 (Rn ) dt .0Notice that in Lemma 1 one assumes the existence of a reference L2 - solution uof the equation (2.2) and gets a control on the decay of the solution in the wholetime interval in terms of that at the end points and that of the “external force”. Ingeneral, under appropriate assumptions on the potential V (x, t) in (1.1) one writesV (x, t)u χρ V (x, t)u (1 χρ )V (x, t)u V(x, t)u G(x, t),with χρ C0 , χρ (x) 1, x ρ, supported in x 2ρ, and obtains the estimate(2.4) by fixing ρ sufficiently large. Also under appropriate hypotheses on F and ua similar argument can be used for the semi-linear equation in (1.2).Next, we recall the conformal or Appell transformation:Lemma 2. If u(y, s) verifies(2.5) s u i (4u V (y, s)u F (y, s)) ,(y, s) Rn [0, 1],and α and β are positive, then n2 (α β) x 2 αβαβ xβtu α(1 t) βt, α(1 t) βte 4i(α(1 t) βt) ,(2.6)ue(x, t) α(1 t) βtverifies(2.7) t ue i 4eu Ve (x, t)eu Fe(x, t) ,(x, t) Rn [0, 1],

6C. E. KENIG, G. PONCE, AND L. VEGAwith(2.8)Ve (x, t) αβ(α(1 t) βt)2V αβ xβtα(1 t) βt , α(1 t) βt ,and(2.9)Fe(x, t) αβα(1 t) βt n2 2F αβ xβtα(1 t) βt , α(1 t) βt (α β) x 2e 4i(α(1 t) βt) .The following result is a modified version of the one in [4] (Lemma 3.1, page1818). It will provide a needed lower bound of the L2 -norm of the solution ofthe equation (1.1) and its first order derivatives in the x1 -variable in the domain{x : R 1 x1 R} [0, 1].Lemma 3. Assume that R 0 large enough and that ϕ : [0, 1] R is a smoothfunction. Then, there exists c c(n; kϕ0 k kϕ00 k ) 0 such that the inequality(2.10)σ 3/2R2eσ x1 x0,1R ϕ(t) 2gL2 (dxdt) c eσ x1 x0,1R ϕ(t) 2(i t )gL2 (dxdt)holds when σ cR2 and g C0 (Rn 1 ) is supported on the setx1 x0,1{(x, t) (x1 , ., xn , t) Rn 1 : ϕ(t) 1}.RProof. As it was remarked above this result is a variation of the one given in detailin [4], hence a sketch will suffice.By translation, without loss of generality, we can assume x0,1 0. Letx12f (x, t) eσ R ϕ(t) g(x, t).Then,x12eσ R ϕ(t) (i t )g Sσ f 4σAσ f,(2.11)whereSσ i t Aσ 1Rx1R4σ 2 x1R2 R ϕ 2 , ϕ x1 12R2 i ϕ02x1R ϕ .Thus,Sσ Sσ ,(2.12)A σ Aσ ,and integrating by parts (possible since g C0 (Rn 1 ) ) one sees thatx12keσ R ϕ (i t )gk22 hSσ f 4σAσ f, Sσ f 4σAσ f i 4σh(Sσ Aσ Aσ Sσ )f, f i 4σh[Sσ , Aσ ]f, f i .A calculation shows that[Sσ , Aσ ] 2 2R2 x1 4σ 2 x1R4 R ϕ 2 12 [( xR1 ϕ)ϕ00 (ϕ0 )2 ] 2iϕ0R x1 .From this it follows thatx1(2.13)2keσ R ϕ (i t )gk22ZZ16σ 38σ22x1 ϕ f dxdt x1 f 2 dxdtRR4R2ZZ8σi000 22x1 2σ [( R ϕ)ϕ (ϕ ) ] f dxdt (ϕ0 x1 f f dxdt) .R

UNIQUE CONTINUATION7Now, when σ cR2 one hasσ3 c2 σ,R4so by taking c large enough, depending on kϕ0 k and kϕ00 k , and using that xR1 ϕ(t) 1 on the supp(f ) supp(g), we can hide the third term on the righthand side (r.h.s.) in the inequality (2.13) in the first term in the r.h.s. Also, sinceZ8σi ϕ0 x1 f f dxdt RZZZ4σ8σ 0 x1 f 2 dxdt,kϕ k f x1 f 4σkϕ0 k2 f 2 dxdt 2 RRthe contribution of this term in (2.13) can be hidden by the first and second termin the r.h.s. of (2.13) if c is large. This concludes the proof. Note that the same proof works by taking c a bit larger, if we only assume xR1 ϕ(t) 1/2 on supp(g).In the proof of Theorem 1 we shall need the following extension of Lemma 3.Corollary 1. Assume g L2 (Rn 1 ) with x1 , t on supp(g) bounded,supp(g) {(x, t) (x1 , ., xn , t) Rn 1 : x1 x0,1 ϕ(t) 1}Rand (i t )g L2 (Rn 1 ), then the inequality (2.10) holds.Proof. We can again assume that x0,1 0. We introduce the notation x (x1 , x0 ) R Rn 1 . Let η1 C0 (R), η1 0, supp(η1 ) { x1 1} and η2 C0 (Rn 1 ),η2 0, supp(η2 ) { x0 1} withZZη1 (x1 )dx1 1andη2 (x0 )dx0 1.Rn 1RFor δ 0 small define1hδ (x, t) n 2 η1 (t/δ 2 )η1 (x1 /δ)η2 (x0 /δ n 1 )δandgδ hδ g.Let θ C0 (Rn 1 ), θ(x0 ) 1, x0 1, and supp(θ) { x0 2}. For l large,definegδ,l (x, t) θ(x0 /l) gδ (x, t).Note that for δ 0 small,supp(gδ ) {(x, t) : xR1 ϕ(t) 2 1/2},and the same holds for gδ,l . Moreover, gδ,l C0 (Rn 1 ).We apply Lemma 3 to gδ,l to obtain:(2.14)σ 3/2R2x1x12eσ R ϕ(t) gδ,lL2 (dxdt)2 c eσ R ϕ(t) (i t )gδ,lL2 (dxdt)Next, we fix δ 0 small and see that(2.15)(i t )gδ,l θ(x0 /l)(i t )gδ21 0 θ(x0 /l) · 0 gδ (x, t) 2 θ(x0 /l)gδ (x, t).ll.

8C. E. KENIG, G. PONCE, AND L. VEGATherefore, by taking l the L2 (dxdt)-norm of the the last two terms on ther.h.s. of (2.15) tend to zero. Hence, inserting this in (2.14) we obtain the sameestimate for gδ . Next, we have that(i t )gδ (i t )(hδ g) hδ (i t )g.Using the supremum in δ (non-isotropic maximal function) and its boundedness,together with the boundedness of the support in (x1 , t) of g, so that2eσ x1 /R ϕ(t) cσ,R ,by the dominated convergence theorem we can pass to the limit as δ 0 to obtainthe desired result. 3. Proof of Theorem 1We divide our argument into six steps:Step 1: We claim thatZ(3.1)sup0 t 1e2a1 x1 u(x, t) 2 dx A3 .RnProof of Step 1 : Using (1.13) in Theorem 1 we choose ρ so large such thatkV χ{ x ρ} kL1 ([0,1]:L (Rn )) n ,with n as in Lemma 1. From (1.9)-(1.11) we haveZe2a1 x1 u(x, 0) 2 dx A2 ,RnandZe2a1 x1 u(x, 1) 2 dx A1 e2a1 m A1 e2a1 .RnWe apply Lemma 1, with G(x, t) χ{ x ρ} V (x, t)u(x, t), using thatZ 1k ea1 x1 χ{ x ρ} V uk2 dt ea1 ρ M0 A1 ,0which gives step 1 with A3 A3 (A1 ; A2 ; a1 ; M0 ; ρ).Step 2: Define δ 0 as n,M0 1with M0 as in (1.12) and n as in Lemma 1. Note that δ 1, andZ 1(3.3)kV (·, t)k dt n .(3.2)δ 1 δLet(3.4)v(x, t) u(δ 1/2 x, δt 1 δ).We shall show that under the hypothesis of Theorem 1ZZ2(3.5) v(x, 1) dx u(x, 1) 2 dx 0mm x1 1/22δ 1/2δas desired.m2 x1 m

UNIQUE CONTINUATION9DefiningVδ (x, t) δ V (δ 1/2 x, δt 1 δ)(3.6)we see that v(x, t) satisfies the equation(x, t) Rn [0, 1]. t v i( v Vδ v),We notice, using (3.3), thatZkVδ kL (Rn [0,1]) M0 δ n ,(3.7)1kVδ (·, t)k dt n ,0andZ v(x, t) 2 dx RnZ1δ n/2 u(y, δt 1 δ) 2 dy RnA1,δ n/2withsupp(v(·, 1)) {x1 m/δ 1/2 }.Thus, from (3.1)ZZ1/2e2a1 x1 δ v(x, 0) 2 dx Rne2a1 x1 δ1/2 u(δ 1/2 x, 1 δ) 2 dx RnA3.δ n/2We remark that δ was fixed in (3.2) (independent of m), and that we can still choosem small.Step 3: Using the Appell (conformal) transformation Lemma 2 we have that if s v i( v Vδ v),(y, s) Rn [0, 1],then for any α, β 0 n2 (α β) x 2αβαβ xβt(3.8)ve(x, t) α(1 t) βtv α(1 t) βt, α(1 t) βte 4i(α(1 t) βt) ,verifies t ve i( ev Ve ve),(x, t) Rn [0, 1],with αβαβxβtV,.δ(α(1 t) βt)2α(1 t) βt α(1 t) βtFor λ 0 given we will choose α α(λ, δ), β β(λ, δ). We recall that1/2A3kea1 x1 δ v(·, 0)k22 n/2 ,δand from the support hypothesisVe (x, t) 1/2keλx1 v(·, 1)k22 e2mλ/δ A1.δ n/2We want γ γ(λ, δ) such that1/2keγx1 ve(x, 0)k2 keγ(α/β)x1v(x, 0)k2 kea1 x1 δ1/21/2v(·, 0)k2 A3,δ n/4and1/21/2keγx1 ve(x, 1)k2 keγ(β/α)x1v(x, 1)k2 keλx1 v(·, 1)k2 Thus, we choose(3.9)γ(α/β)1/2 δ 1/2 a1 ,γ(β/α)1/2 λ,1/2eλm/δ A1.δ n/4

10C. E. KENIG, G. PONCE, AND L. VEGAi.e.γ (λδ 1/2 a1 )1/2 ,(3.10)Next, using the change of variableβbt ,α(1 t) βtit follows thatZ 1k Ve (·, t)k dtα δ 1/2 a1 .β λ,αβdt,(α(1 t) βt)2dbt 0Z1 0Z αβαβxβtkVδ (,)k dt2(α(1 t) βt)α(1 t) βt α(1 t) βt1kVδ (·, bt)k dbt n , 0using (3.7). So we can apply Lemma 1 again, this time with G 0, to obtain that1/2 A1/2 1/2A13sup keγx1 ve(·, t)k2 cn n/4 n/4eλm/δδδ0 t 1(3.11)1/2 cδ,a1 ,A1 ,A3 eλm/δ c eλm/δ1/2,if λ 0 is large and A1 6 0, (how large λ is for this depends on m, A1 , A3 and δ,but this will not matter). Note thatA1βt)k2 n/2 ,(3.12)kev (·, t)k22 kv(·,α(1 t) βt 2δhence1/2sup kev (·, t)k2 (3.13)0 t 1A1.δ n/4Now, we denote by φ0 (x1 ) 0 a C convex function such that 0,x1 0,(3.14)φ0 (x1 ) x1 1/4, x1 1/2,and defineφ(x1 ) (1 (φ0 (x1 ))2 )1/2 .Since γ (λδ 1/2 a1 )1/2 , from (3.11) and for large λ we havesup keγφ(x1 ) ve(·, t)k2 cδ,a1 eλm/δ(3.15)1/2.0 t 1A computation shows thatφ0 (x1 ) andφ0 (x1 ) φ00 (x1 ),(1 (φ0 (x1 ))2 )1/200(φ00 (x1 ))2φ0 (x1 ) φ0 (x1 )φ (x1 ) .(1 (φ0 (x1 ))2 )3/2(1 (φ0 (x1 ))2 )3/200

UNIQUE CONTINUATION11Thus, for x1 1/2 one has that00φ (x1 ) (3.16)1 111 .23/24 (1 x1 )4 hx1 i3We now follow an argument similar to that in [7] section 2. Letf (x, t) eγφ(x1 ) ve(x, t).Then f verifies(3.17) t f Sf Af i eγφ F,in Rn [0, 1],with symmetric and skew-symmetric operators S and A S iγ 2 x1 φ x1 x21 φ , A i 4 γ 2 x1 φ 2 .(3.18)andF Ve ve.A calculation shows that,(3.19)hi0000St [S, A] γ 4 x1 φ x1 4γ 2 φ (φ0 )2 φ(4) .By Lemma 2 in [7](3.20) t2 H t2 (f, f ) 2 t Re ( t f Sf Af, f ) 2 (St f [S, A] f, f ) k t f Af Sf k2 k t f Af Sf k2 ,so(3.21) t2 H 2 t Re ( t f Sf Af, f ) 2 (St f [S, A] f, f ) k t f Af Sf k2 .Multiplying (3.21) by t(1 t) and integrating in t we obtainZ2(3.22)1t(1 t) (St f [S, A] f, f ) dt0 cn sup keγ φ ve(t)k22 cn sup keγ φ F (t) 2 .[0,1][0,1]This computation can be justified by parabolic regularization using the fact thatwe already know the decay estimate for ve, see [5]. Note that for λ sufficiently large(3.23)kVe k β αkVδ k λδ 1/2 a1δM0 λδ 1/2 M0.a1

12C. E. KENIG, G. PONCE, AND L. VEGAHence, combining (3.11), (3.19), and (3.23) it follows thatZ 1Z00t(1 t)φ (x1 ) x1 f 2 dxdt8γ0 8 γ3Z1Z00t(1 t) φ (x1 ) (φ0 (x1 ))2 f 2 dxdt0(3.24) cn γ sup kf (·, t)k22 cδ,M0 ,a1 ,n λ sup kf (·, t)k22 cn sup kf (·, t)k22[0,1][0,1] cδ,M0 ,a1 ,n λ e2λm/δ1/2[0,1].We recall that x1 f eγφ(x1 ) x1 ve γ eγφ(x1 ) φ0 (x1 ) ve,thusγ x1 f 2 γe2γφ(x1 ) x1 ve γφ0 (x1 )ev 2 e2γφ(x1 ) (γ x1 ve 2 2γ 2 φ0 (x1 )ev x1 ve γ 3 (φ0 (x1 ))2 ev 2 ),with1γ x1 ve 2 2γ 3 (φ0 (x1 ))2 ev 2 .2Inserting these estimates in (3.24) for λ large one getsZ 1Z001/24γt(1 t) φ (x1 ) e2γφ(x1 ) x1 ve 2 dxdt cδ,M0 ,a1 ,n λ e2λm/δ . 2γ 2 φ0 (x1 )ev x1 ve 0Hence, for x1 1/2 from (3.16) one has thatZ 1Z1/21e2γφ(x1 ) x1 ve 2 dxdt cδ,M0 ,a1 ,n λ e2λm/δ ,γt(1 t)3hx1 i0for λ large. Collecting the above information, (3.11), and (3.24) we conclude thatZ 1Z1γφ(x1 )2e2γφ(x1 ) x1 ve 2 dxdtt(1 t)sup keve(·, t)k2 γ31hx1i0 t 10x1 2(3.25) cδ,M0 ,a1 ,n λ e2λm/δ1/2.Step 4 : We will give lower bounds forZZ 5/8Φ ev (x, t) 2 dtdx,2 x1 R/23/8for R large to be chosen.First, we recall thatZZ 5/8 n2 2αβαβ xβtΦ v,dxdt.α(1 t) βtα(1 t) βt α(1 t) βt2 x1 R/23/8Next, for t [3/8, 5/8] we see thats(t) βt,α(1 t) βt

UNIQUE CONTINUATION13satisfies that(α(1 t) βt)2 β 2βds ds ds,αβαβα(where A B means : C (0, 1) s.t. CA B C 1 A) withdt s(3/8) 3β (1/2, 1),5α 3βs(5/8) 5β (1/2, 1).3α 5βandTherefore2αβ α, (5α 3β)(3α 5β)βs(5/8) s(3/8) for large λ, ands(5/8) s(3/8) 1In the x-variable we haveas λ . y αβx,α(1 t) βso for t [3/8, 5/8] and 2 x1 R/2 one basically has thatqqRαy1 [2 αβ, 2β ] A.Thus,βΦ cnα(3.26)Z ZAwith v(y, s)

Titre : Un th eoreme de type Paley-Wiener pour les evolutions de Schrodinger. R esum e : On prouve des principes de prolongement unique pour les so-lutions d’ equations d’ evolution de Schrodinger avec potentiels d ependant du temps. Ceux-ci correspondent a des principes d’incertitude de type Paley-Wiener pour la transform ee de Fourier.

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Stokes’ and Gauss’ Theorems Math 240 Stokes’ theorem Gauss’ theorem Calculating volume Stokes’ theorem Theorem (Green’s theorem) Let Dbe a closed, bounded region in R2 with boundary C @D. If F Mi Nj is a C1 vector eld on Dthen I C Mdx Ndy ZZ D @N @x @M @y dxdy: Notice that @N @x @M @y k r F: Theorem (Stokes’ theorem)

Nov 19, 2018 · Theorem 5-4 Angle Bisector Theorem Theorem If a point is on the bisector of an angle, then the point is equidistant from the sides of the angle. If. . . QS bisects PQR, 1 QP, and SR 1 QR P, S Then. SP SR You will prove Theorem 5-4 in Exercise 34. Theorem 5-5 Converse of the Angle Bisector Theorem Theorem

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This section describe the comparative study of various research work presented up till now. 1) A Modified Wiener Filter FOR THE RESTORATION OF BLURRED IMAGES a) Wiener filters give the linear least mean square estimate of the object image from the observations and have been used extensively for the restoration of noisy and Observation and suggestion:blurred images. b) The essential idea behind .

application of linear filter techniques. These filtering techniques are most easily understood in the frequency domain. Wiener, Constrained . Series with Engineering Applications" Norbert Wiener DIP Lecture 16 10. Wiener Filter Original image (left), . where the system transfer function is H(u,v) T sinπ(αu βv) π(αu βv) e iπ .

The objective of speech enhancement process is to improve the quality and intelligibility of speech in noisy environments. The problem has been widely discussed over the years. Many approaches have been proposed like subtractive type [1-4], Perceptual Wiener filtering algorithms. Among them spectral subtraction and the Wiener filtering

Layout of the Book The Substance Abuse & Recovery Workbook is designed to be used either independently or as part of an integrated curriculum. You may administer one of the assessments and the journaling exercises