L 2 MOHAMMEDHICHEMMORTAD ON L -ESTIMATESFORTHETIMEDEPENDENTSCHRÖDINGEROPERATORON p

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L^p-Estimates for the Schrödinger Operator Mohammed Hichem Mortad vol. 8, iss. 3, art. 80, 2007

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ON L

^p

-ESTIMATES FOR THE TIME DEPENDENT SCHRÖDINGER OPERATOR ON L

²

MOHAMMED HICHEM MORTAD

Département de Mathématiques Université d’Oran (Es-senia)

B.P. 1524, El Menouar, Oran. Algeria.

EMail:mortad@univ-oran.dz

Received: 07 July, 2007

Accepted: 29 August, 2007

Communicated by: S.S. Dragomir

2000 AMS Sub. Class.: Primary 35B45; Secondary 35L10.

Key words: Schrödinger Equation, Strichartz Estimates and Self-adjointness.

Abstract: LetLdenote the time-dependent Schrödinger operator innspace variables. We consider a variety of Lebesgue norms for functionsuonRⁿ⁺¹, and prove or disprove estimates for such norms ofuin terms of theL² norms ofuandLu.

The results have implications for self-adjointness of operators of the formL+V whereV is a multiplication operator. The proofs are based mainly on Strichartz- type inequalities.

Acknowledgements: I would like to express all my gratitude to my ex-Ph.D.-supervisor Professor Alexander M. Davie for his helpful comments, especially in the counterexamples section.

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1. Introduction

Let (x, t) ∈ Rⁿ⁺¹ wheren ≥ 1. The Schrödinger equation ^∂u_∂t = i4_xu has been much studied using spectral properties of the self-adjoint operator 4_x. When a multiplication operator (potential) V is added, it becomes important to determine whether4x+V is a self-adjoint operator, and there is a vast literature on this question (see e.g. [9]).

One can also, however, regard the operator L = −i_∂t^∂ − 4_x as a self-adjoint operator onL²(Rⁿ⁺¹), and that is the point of view taken in this paper. We ask what can be said about the domain ofL, more specifically, we ask whichL^q spaces, and more generally mixedL^q_t(L^r_x)space, a functionumust belong to, given thatuis in the domain ofL(i.e. uandLuboth belong toL²(Rⁿ⁺¹)). We answer this question and, using the Kato-Rellich theorem, deduce sufficient conditions onV forL+V to be self-adjoint.

Our approach is based on the fact that any sufficiently well-behaved function u onRⁿ⁺¹can be regarded as a solution of the initial value problem (IVP)

(1.1)

( −iu_t− 4_xu=g(x, t), u(x, α) = f(x)

whereα∈R,f(x) =u(x, α)andg =Lu.

To apply this, we will use estimates for u based on given bounds for f and g.

A number of such estimates are known and generally called Strichartz inequalities, after [12] which obtained such anL^qbound foru. This has since been generalized to give inequalities for mixed norms [13, 4]. The specific inequalities we use concern the caseg = 0of (1.1) and give bounds foruin terms ofkfk_L²_(Rⁿ₎- see (3.2) below.

The precise range of mixedL^q_t(L^r_x)norms for which the bound (3.2) holds is known as a result of [13,4] and the counterexample in [6].

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In Section2we prove a special case of our main theorem, namely a bound foru inL^∞_t (L²_x), which does not require Strichartz estimates, only elementary arguments using the Fourier transform. The main theorem, givingL^q_t(L^r_x)bounds for the largest possible set of (q, r) pairs, is proved in Section 3. In fact, we prove a somewhat stronger bound, in a smaller spaceL_2,q,r defined below. The fact that the set of pairs (q, r)covered by Theorem3.1is the largest possible is shown in Section4.

Some results on a similar question for the wave operator can be found in [7]. For Strichartz-type inequalities for the wave operator, see e.g. [11,12,2,3,4].

We assume notions and definitions about the Fourier Transform and unbounded operators and for a reference one may consult [8], [5] or [10]. We also use on several occasions the well-known Duhamel principle for the Schrödinger equation (see e.g.

[1]).

Notation. The symbol uˆ stands for the Fourier transform of u in the space (x) variable while the inverse Fourier transform will be denoted either byF⁻¹uoru.ˇ

We denote byC₀^∞(Rⁿ⁺¹)the space of infinitely differentiable functions with compact support.

We denote byR⁺the set of all positive real numbers together with+∞.

For 1 ≤ p ≤ ∞, k · kp is the usual L^p-norm whereas k · k_L^p

t(L^qx) stands for the mixed spacetime Lebesgue norm defined as follows

kuk_L^q

t(L^r_x) = Z

R

ku(t)k^q_Lr xdt

¹_q .

We also define some modified mixed norms. First we define, for any integerk, kuk_L^q

t,k(L^r_x) =

Z k+1

k

ku(t)k^q_Lr xdt

¹_q ,

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and then

kuk_L_p,q,r = X

k∈Z

kuk^p_Lq t,k(L^r_x)

!¹_p .

We note that kukLp,q1,r ≥ kukLp,q2,r if q₁ ≥ q₂, and that kuk_L^q

t(L^r_x) ≤ kukLp,q,r if q≥p.

Finally we define

M_Lⁿ={f ∈L²(Rⁿ⁺¹) :Lf ∈L²(Rⁿ⁺¹)},

whereLis defined as in the abstract and where the derivative is taken in the distri- butional sense. We note thatM_Lⁿ=D(L), the domain ofL, and also thatC₀^∞(Rⁿ⁺¹) is dense inM_Lⁿin the graph normkuk_L²_(R_n+1₎+kLuk_L²_(R_n+1₎.

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2. L

^∞_t

(L

²_x

) Estimates.

Before stating the first result, we are going to prepare the ground for it. Take the Fourier transform of the IVP (1.1) in the space variable to get

( −iˆu_t+η²uˆ= ˆg(η, t), ˆ

u(η, α) = ˆf(η) which has the following solution (valid for allt∈R):

(2.1) u(η, t) = ˆˆ f(η)e^−iη²^t+i Z t

α

e^−iη²^(t−s)g(η, s)ds,ˆ

whereη∈Rⁿ.

Duhamel’s principle gives an alternative way of writing the part of the solution depending ong. Taking the casef = 0, the solution of (1.1) can be written as

(2.2) u(x, t) =i

Z t

α

u_s(x, t)ds,

whereu_sis the solution of

( Lu_s= 0, t > s, u_s(x, s) = g(x, s).

Now we state a result which we can prove using (2.1). In the next section we prove a more general result using Strichartz inequalities and Duhamel’s principle (2.2).

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Proposition 2.1. For alla >0, there existsb >0such that kukL2,∞,2 ≤akLuk²_L2(Rⁿ⁺¹)+bkuk²_L2(Rⁿ⁺¹)

for allu∈M_Lⁿ.

Proof. We prove the result foru ∈ C₀^∞(Rⁿ⁺¹)and a density argument allows us to deduce it foru∈M_Lⁿ.

We use the fact that any such uis, for any α ∈ R, the unique solution of (1.1), wheref(x) = u(x, α)andg =Lu, and therefore satisfies (2.1).

Let k ∈ Z and let t and α be such that k ≤ t ≤ k + 1and k ≤ α ≤ k + 1.

Squaring (2.1), integrating with respect toηinRⁿ, and using Cauchy-Schwarz (and the fact that|t−α| ≤1), we obtain

(2.3) kˆu(·, t)k²_L2(Rⁿ) ≤2 Z

Rⁿ

|ˆu(η, α)|²dη+ 2 Z

Rⁿ

Z t

α

|ˆg(η, s)|²dsdη.

Now integrating againstαin[k, k+ 1]allows us to say that ku(·, t)k²_L2(Rⁿ)≤2

Z k+1

k

Z

Rⁿ

|ˆu(η, α)|²dηdα+ 2 Z k+1

k

Z

Rⁿ

|ˆg(η, s)|²dηds.

Now take the essential supremum of both sides in t over [k, k + 1],then sum in k overZto get (recalling thatg =Lu)

∞

X

k=−∞

ess sup

k≤t≤k+1

ku(·, t)k²_L2(Rⁿ)≤2kLuk²_L2(Rⁿ⁺¹)+ 2kuk²_L2(Rⁿ⁺¹).

Finally to get an arbitrarily small constant in theLuterm we use a scaling argument: letmbe a positive integer and letv(x, t) =u(mx, m²t). Then we find

kvk_L²_(Rⁿ⁺¹₎ =m^−1−n/2kuk_L²_(Rⁿ⁺¹₎

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and

kLvk_L²₍_Rⁿ⁺¹₎=m^1−n/2kLuk_L²₍_Rⁿ⁺¹₎. Also,

kv(·, t)k_L²₍_Rⁿ₎ =m^−n/2ku(·, m²t)k_L²₍_Rⁿ₎ and so

sup

k≤t≤k+1

kv(·, t)k²_L2(Rⁿ)=m⁻ⁿ sup

m²k≤t≤m²(k+1)

ku(·, t)k²_L2(Rⁿ)

≤m⁻ⁿ

m²(k+1)−1

X

j=m²k

sup

j≤t≤j+1

ku(·, t)k²_L2(Rⁿ).

Summing overkgives

kvk²_L_2,∞,2 ≤m⁻ⁿkuk²_L_2,∞,2

≤m⁻ⁿ

2kLuk²_L2(Rⁿ⁺¹)+ 2kuk²_L2(Rⁿ⁺¹)

≤2m⁻²kLvk²_L2(Rⁿ⁺¹)+ 2m²kvk²_L2(Rⁿ⁺¹)

and choosingmso that2m⁻² < acompletes the proof.

Now we recall the Kato-Rellich theorem which states that if L is a self-adjoint operator on a Hilbert space andV is a symmetric operator defined onD(L), and if there are positive constantsa < 1and b such that kV uk ≤ akLuk+bkuk for all u∈ D(L),thenL+V is self-adjoint onD(L)(see [9]).

Corollary 2.2. LetV be a real-valued function inL∞,2,∞. ThenL+V is self-adjoint onD(L) = M_Lⁿ.

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Proof. One can easily check that

kV uk_L²₍_Rⁿ⁺¹₎ ≤ kVkL∞,2,∞kukL2,∞,2.

Choose a < kVk⁻¹_L_∞,2,∞ and then Proposition 2.1 shows that L+V satisfies the hypothesis of the Kato-Rellich theorem.

In particular, it follows thatL+V is self-adjoint wheneverV ∈L²_t(L^∞_x ).

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3. L

^q_t

(L

^r_x

) Estimates.

Now we come to the main theorem in this paper, which depends on the following Strichartz-type inequality. Suppose n ≥ 1 and q and r are positive real numbers (possibly infinite) such thatq≥2and

(3.1) 2

q + n r = n

2.

Whenn = 2we exclude the case q = 2, r = ∞. Then there is a constant C such that iff ∈L²(Rⁿ)andg = 0, the solutionuof (1.1) satisfies

(3.2) kuk_L^q

t(L^r_x)≤Ckfk_L²₍_Rⁿ₎.

This result can be found in [13] for q > 2; the more difficult ‘end-point’ case whereq = 2, n ≥ 3is treated in [4]. That (3.2) fails in the exceptional casen = 2, q= 2, r =∞is shown in [6].

Forn ≥1we define a regionΩn∈R⁺×R⁺as follows: forn 6= 2,

(3.3) Ωn =

(q, r)∈R⁺×R⁺: 2 q +n

r ≥ n

2, q≥2, r ≥2

and forn = 2,Ω₂ is defined by the same expression, with the omission of the point (2,∞).

The sets Ω_n are probably most easily visualized in the (¹_q,¹_r)-plane. Then Ω₁ is a quadrilateral with vertices (¹₄,0),(¹₂,0),(0,¹₂),(¹₂,¹₂) and for n ≥ 2, Ω_n is a triangle with vertices(¹₂,ⁿ⁻²_2n ),(0,¹₂),(¹₂,¹₂), the point (¹₂,0)being excluded in the casen = 2.

Theorem 3.1. Letn≥ 1, and let(q, r)∈Ω_n. Then for alla > 0, there existsb > 0 such that

(3.4) kukL_2,q,r ≤akLuk_L²₍_Rⁿ⁺¹₎+bkuk_L²₍_Rⁿ⁺¹₎ for allu∈M_Lⁿ.

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Proof. By the inclusion L_2,q₁_,r ⊆ L_2,q₂_,r,when q₁ ≥ q₂ it suffices to treat the case where ²_q +ⁿ_r = ⁿ₂, for which (3.2) holds.

Letk ∈ Z and letα ∈ [k, k+ 1]. As in the proof of Proposition2.1 we use the fact thatuis the solution of (1.1) withf =u(·, α)andg =Lu. Now we splituinto two partsu=u1+u2,whereu1,u2 are the solutions of

( Lu₁ =g, u₁(x, α) = 0,

( Lu₂ = 0, u₂(x, α) = f.

The estimate foru₂is deduced from (3.2):

(3.5) ku₂k_L^q

t(L^r_x)≤Ckfk_L²_(Rⁿ₎≤Cku(·, α)k_L²_(Rⁿ₎. Foru₁we apply (2.2) to obtain

(3.6) u₁(x, t) = i

Z t

α

u_s(x, t)ds, from which we deduce

ku₁(·, t)k_L^r_(Rⁿ₎≤ Z k+1

k

ku_s(·, t)k_L^r_(Rⁿ₎ds

fort ∈[k, k+ 1], and hence ku₁k_L^q

t,k(L^r_x)≤ Z k+1

k

ku_sk_L^q

t(L^r_x)ds

≤C Z k+1

k

kg(·, s)k_L²_(Rⁿ₎ds

≤Ckgk_L²_(Rⁿ_×[k,k+1]).

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Combining this with (3.5) we have kuk²_L^q

t,k(L^r_x) ≤2C²ku(·, α)k²_L2(Rⁿ)+ 2C²kLuk²_L2(Rⁿ×[k,k+1]). Integrating w.r.t.αfromktok+ 1gives

kuk²_L^q

t,k(L^r_x) ≤2C²kuk²_L2(Rⁿ×[k,k+1])+ 2C²kLuk²_L2(Rⁿ×[k,k+1]). Summing overk, we obtain

kuk²_L_2,q,r ≤2C²kuk_L²_(Rⁿ⁺¹₎+ 2C²kLuk_L²_(Rⁿ⁺¹₎,

and the proof is completed by a similar scaling argument to that used in Proposition 2.1.

Using the inclusionL_2,q,r ⊆L^q_t(L^r_x)forq≥2we deduce

Corollary 3.2. Letn ≥1, and let(q, r)∈Ω_n. Then for alla >0, there existsb >0 such that

(3.7) kuk_L^q

t(L^r_x) ≤akLuk_L²₍_Rⁿ⁺¹₎+bkuk_L²₍_Rⁿ⁺¹₎ for allu∈M_Lⁿ.

In particular, we get such a bound forkuk_L^q_(Rⁿ⁺¹₎whenever2≤q≤(2n+ 4)/n.

By applying the Kato-Rellich theorem we can deduce a generalization of Corol- lary2.2from Theorem3.1. We first define

(3.8) Ω^∗_n=

(p, s)∈R⁺×R⁺ : 2 p+ n

s ≤1, p≥2, s≥2

forn6= 2, and forn= 2,Ω₂is defined by the same expression, with the omission of the point(2,∞).

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Corollary 3.3. Let n ≥ 1 and let (p, s) ∈ Ω^∗_n. Let V be a real-valued function belonging toL∞,p,s. ThenL+V is self-adjoint onM_Lⁿ.

Proof. Let q = _p−2^2p and r = _s−2^2s . Then (q, r) ∈ Ω_n and the conclusion (3.4) of Theorem3.1applies. Now we have

Z k+1

k

kV u(·, t)k²_L2(Rⁿ) ≤ Z k+1

k

ku(·, t)k²_Lr(Rⁿ)kV(·, t)k²_Ls(Rⁿ)

≤ kuk²_L^q

t,k(L^r_x)kVk²_L^p

t,k(L^s_x)

and summation overkgives

kV uk_L²_(Rⁿ⁺¹₎≤ kukL2,q,rkVkL∞,p,s.

Then, using (3.4), the result follows in the same way as Corollary2.2.

It follows from Corollary 3.3 thatL+V is self-adjoint whenever V ∈ L^p_t(L^s_x) for(p, s) ∈ Ω^∗_n. Taking the cases = p,we find thatL+V is self-adjoint if V ∈ L^p(Rⁿ⁺¹)for somep≥n+ 2.

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4. Counterexamples

Now we show that Theorem3.1is sharp, as far as the allowed set ofq, ris concerned.

Proposition 4.1. Letn≥1and letqandrbe positive real numbers, possibly infinite, such that(q, r) ∈/ Ω_n. Then there are no constantsaandb such that (3.7) holds for allu∈M_Lⁿ.

Proof. For (q, r) to fail to be in Ω_n one of the following three possibilities must occur: (i)q <2orr <2; (ii) ²_q+ⁿ_r < ⁿ₂; (iii)n = 2,q= 2andr=∞. We consider these cases in turn.

(i) Ifq < 2, choose a sequence(β_k)k∈Z which is inl² but not inl^q. Letφ(x, t)be a smooth function of compact support onRⁿ⁺¹which vanishes fortoutside[0,1], and letu(x, t) = P

k∈Zβkφ(x, t−k). Thenu∈M_Lⁿ,butu /∈L^q_t(L^r_x)for anyr.

The caser < 2can be treated similarly. We chose a sequenceβk which is inl² but not l^r, and a smooth φ which vanishes forx₁ outside [0,1], then set u(x, t) = P

k∈Zβ_kφ(x−ke₁, t),wheree₁is the unit vector(1,0, . . . ,0)inRⁿ. Thenu∈M_Lⁿ, butu /∈L^q_t(L^r_x)for anyq.

(ii) In this case we use the scaling argument which shows that the Strichartz estimates fail, together with a cutoff to ensureuandLuare inL².

We start with a non-zero f ∈ L²(Rⁿ), and let u be the solution of (1.1) with α = 0andg = 0. (An explicit example would bef(x) = e^−|x|² and thenu(x, t) = (1 + 4it)^−n/2e^−|x|²^/(1+4it)). Choose a smooth functionφ onR such that φ(0) 6= 0 and such thatφandφ⁰are inL². Then forλ >0define

v_λ(x, t) =λ^n/2u(λx, λ²t)φ(t).

Then (using Lu = 0) we find Lv(x, t) = −iλ^n/2u(λx, λ²t)φ⁰(t). We calculate

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kv_λk_L²_(Rⁿ⁺¹₎ =kfk_L²_(Rⁿ₎kφk_L² andkLv_λk_L²_(Rⁿ⁺¹₎ =kfk_L²_(Rⁿ₎kφ⁰k_L². Also kv_λk_L^q

t(L^r_x) =λ^β Z

R

ku(·, t)k^q_Lr(Rⁿ)|φ(λ⁻²t)|^qdt ¹_q

,

whereβ = ⁿ₂−ⁿ_r−²_q >0. Soλ^−βkv_λk_L^q

t(L^r_x)→ |φ(0)|kuk_L^q

t(L^r_x)(note that the norm on the right may be infinite) and hencekv_λk_L^q

t(L^r_x)tends to∞asλ → ∞, completing the proof.

(iii) This exceptional case we treat in a similar fashion to (ii), but we need the result from [6], that the Strichartz inequality fails in this case. We start by fixing a smooth functionφonRsuch thatφ= 1on[−1,1]andφandφ⁰ are inL².

Now letM > 0be given and we use [6] to findf ∈ L²(R²)withkfk_L²_(R²₎ = 1 such that the solutionu of (1.1) withα = 0 andg = 0satisfieskuk_L²

t(L^∞_x ) > M. Then we can find R > 0so that RR

−Rku(·, t)k²_L∞(R²)dt > M². Let λ = R^1/2 and definev(x, t) =λ^n/2u(λx, λ²t)φ(t). Thenkvk_L²_(R³₎=kφk_L²,kLvk_L²_(R³₎ =kφ⁰k_L² and

kvk²_L2

t(L^∞_x) ≥ Z 1

−1

kv(·, t)k²_L∞(R²)dt > M², which completes the proof, sinceM is arbitrary.

We remark that [6] also gives an example off ∈L²(R²)such thatu /∈L²_t(BM O_x) and the argument of part (iii) can then be applied to show that no inequality

kuk_L²

t(BM Ox) ≤akLuk_L²_(R³₎+bkuk_L²_(R³₎ can hold.

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5. Question

We saw as a result of Corollary3.3 that if(p, s) ∈ Ω^∗, thenL+V is self-adjoint on M_Lⁿ whenever V ∈ L^p_t(L^s_x). One can ask whether this can be extended to a larger range of(p, s)withp, s ≥ 2. If one asks whether L+V is defined onM_Lⁿ, then we would require a boundkV uk_L²_(Rⁿ⁺¹₎≤akLuk_L²_(Rⁿ⁺¹₎+bkukto hold for all u∈M_Lⁿ. If such a bound is to hold for allV ∈L^p_t(L^s_x),then, in fact, we require (3.7) to hold forq= _p−2^2p andr= _s−2^2s , which we know cannot hold unless(p, s)∈Ω^∗.

One can instead ask forL+V, defined on sayC₀^∞(Rⁿ⁺¹), to be essentially self- adjoint. This is equivalent to saying that the only (distribution) solution inL²(Rⁿ⁺¹) of the PDE

−iu_t− 4_xu+V u=±iu isu= 0(see e.g. [8]).

We do not know if there are any values of(p, s)not inΩ^∗_nsuch that this holds for allV ∈ L^p_t(L^s_x). The analogous question for the Laplacian is extensively discussed in [9].

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