Instituto Nacional de Matem´
atica pura e aplicada
Doutorado em Matem´
atica
Well-Posedness for a Generalized Nonlinear Derivative
Schr¨
odinger Equation
Gleison do Nascimento Santos
Thesis presented to the Post-graduate Pro-gram in Mathematics at Instituto Nacional de Matem´atica Pura e Aplicada as partial fulfillment of the requirements for the degree of Doctor in Mathematics.
Advisor : Jose Felipe Linares Ramirez
Co-advisor : Gustavo Alberto Ponce Rugero
Rio de Janeiro August, 2014
Agradecimentos
Gostaria primeiramente de agradecer `a minha m˜ae, Dona Socorro, meus irm˜aos pelos por estarem sempre ao meu lado me apoiando, e me motivando e me confortando durante os momentos mais dif´ıceis dessa longa etapa que passei durante o doutorado.
Gostaria de agradecer a meu orientador, Felipe Linares, pela orienta¸c˜ao, por sua paciˆencia, disponibilidade para tirar d´uvidas, por ter confiado em mim como aluno, e ter me guiado no caminho certo para poder resolver o problema da tese.
Gostaria de agradecer a meu co-orientador Gustavo Ponce, por ter me feito enxergar qual era o caminho correto a seguir para finalizar meu trabalho, por sua simplicidade, e maneira acolhedora como me recebeu durante minha estadia na Universidade de Santa Barbara UCSB. Gostaria de agradecer aos professores Roger Peres Moura e Didier Pilod, que foram muito importantes da minha forma¸c˜ao e que me possibilitaram chegar ao doutorado no IMPA.
Agrade¸co, aos valiosos amigos do IMPA e da UFRJ onde fiz mestrado. A Deus, por ter posto todos os que eu citei acima em minha vida. Agrade¸co `a CAPES, pelo suporte financeiro.
Abstract
In this work we study the well-posedness for the initial value problem associated to a generalized derivative Schr¨odinger equation for small size initial data in weighted Sobolev space. The techniques used include parabolic regularization method combined with sharp linear estimates. An important point in our work is that the contraction principle is likely to fail but gives us inspiration to obtain certain uniform estimates that are crucial to obtain the main result. To prove such uniform estimates we assume smallness on the initial data in weighted Sobolev spaces.
Resumo
Neste trabalho estudamos a boa coloca¸c˜ao para o problema de valor inicial associado a equa¸c˜ao de Schr¨odinger com derivada generalizada com dado inicial pequeno em espa¸co de Sobolev com peso. A t´ecnica que empregamos para conseguir tal resultado ´e o m´etodo da regulariza¸c˜ao parab´olica mais estimativas ´otimas da solu¸c˜ao do problema linear. Um ponto importante no nosso trabalho ´e que o argumento do princ´ıpio de contra¸c˜ao tende a falhar mas ele nos da inspira¸c˜ao para obter certas estimativas uniformes que s˜ao cruciais na obten¸c˜ao do resultado principal. Para estabelecer tais estimativas uniformes, n´os assumimos dados iniciais pequenos em espa¸cos com peso.
Notations
• The notation A . B means there is a constant c > 0 such that A ≤ cB. And we say, A∼ B when A . B and B . A.
• For a real number r we shall denote r+ instead of r + , whenever is a positive number
whose value is not important. Also we write T+, to denote T raised to some positive
power.
• We denote hxi = (1 + x2)1/2, called Japanese bracket.
• If f ∈ L1(R) then we denote ˆf the Fourier transform of f , and ˇf denotes its inverse
Fourier transform.
• For a s ∈ R, Js, and Ds stand the Bessel and the Riesz potentials of order s, given via
Fourier transform by the formulas d
Jsf = hξisf , and db Dsf =|ξ|sf .b
• Hs(R) denotes the Sobolev space of order s defined by
Hs(R) = {f ∈ S0(R); Jsf ∈ L2(R)} endowed with the norm
kfkHs =kJsfkL2;
• We also consider the Sobolev spaces
Lpα(R) = {f ∈ S0(R); Jαf ∈ Lp(R)} for p≥ 1.
• For a function f = f(x, t), with (x, t) ∈ R × [0, T ], we denote kfkLqTLpx = Z T 0 Z R|f(x, t)| pdx q p dt 1 q and kfkLqTHs x = Z T 0 kf(·, t)kq Hsdt 1 q . • Similarly we denote kfkLpxLqT = Z R Z T 0 |f(x, t)|qdt p q dx 1 p .
• When f(x, t) is defined for time t in the whole line R we shall consider the notations kfkLqtLpx,kfkLqtHxs and kfkL
p xLqt.
Contents
1 Preliminaries 13
1.1 Basic results . . . 13
1.2 The Schr¨odinger propagator . . . 15
1.3 Two technical lemmas . . . 16
2 The viscosity argument 23 2.1 Solution of the -nonlinear problem in H3/2 . . . . 23
2.2 Uniform estimates for the -linear problem . . . . 34
3 Uniform estimate for the solutions u 47 3.1 Estimate for the norms Ω1 and Ω3 . . . 47
3.2 Estimates for the norms Ω2 and Ω4 . . . 50
3.2.1 Linear terms . . . 51
3.2.2 Nonlinear terms - Part I . . . 51
3.2.3 Nonlinear terms - Part II . . . 52
3.3 Uniform time of definition and uniform estimate . . . 54
4 Sending to zero 57 4.1 Convergence in L2 . . . . 57 4.2 Existence of solution . . . 61 4.3 Uniqueness . . . 64 5 Further results 70 6 Additional Remarks 79
Introduction
We shall study the following initial value problem (IVP)
i∂tu + ∂x2u + i|u|1+a∂xu = 0
u(·, 0) = u0
(1)
where u is a complex valued function of (x, t)∈ R × R and 0 < a < 1.
The equation in (1) is a generalization of the derivative nonlinear Schr¨odinger equation, (DNLS)
i∂tu + ∂x2u + i(|u|2u)x = 0, (x, t)∈ R × R. (2)
The DNLS equation appears in physics as a model that describes the propagation of Alfv´en waves in plasma (see [22], [23], [25]). In mathematics, this equation has also been extensively studied in regard of well-posedness of its associated IVP ([6], [8], [11], [12], [13], [14], [30], [31]). Tsutsumi and Fukuda [32], using parabolic regularization, proved local well-posedness in Sobolev spaces Hs(R), s > 3/2. Hayashi [12] proved well-posedness for initial data u
0 ∈ H1(R)
satisfying the smallness condition
ku0kL2 <
√
2π. (3)
His idea was to use a gauge transformation to turn the DNLS equation into a system of nonlinear Schr¨odinger equations without derivative in the nonlinearity. This system, in turn, can be treated using Strichartz estimates. It is known that DNLS can be writen as a Hamiltonian system
d
dtu(t) =−iE
0(u(t)) (4)
where E(u) the energy of u defined by
E(u)(t) = 1 2 Z |∂xu|2dx + 1 4Im Z |u|2u∂¯ xudx. (5)
As a consequence of (4) it follows that E is a conserved quantity. In particular, the result of Hayashi is global in time. Later on Hayashi and Ozawa [13] based on the same gauge transformation proved global well-posedness for initial condition in Hm(R), m ∈ N,
also satisfying the smallness condition (3). The best result regarding local well-posedness was obtained by Takaoka in [30]. He proved the IVP associated to DNLS was well-posed in Hs(R),
for s≥ 1/2. He considered the gauge transformation
v(x, t) = u(x, t) exp i 2 Z |u(y, t)|2dy .
to convert the DNLS into
i∂tv + ∂x2v =−iv 2∂ xv¯− 1 2|v| 4v (6)
and used the Fourier restriction norm method introduced by Bourgain [3]. Biagioni and Linares [2] proved that the IVP associated to DNLS is not well-posed in Hs(R) for s < 1/2, which
implies that Takaoka’s result is sharp. Using the I-method Colliander, Keel, Staffilani, Takaoka and Tao ([5], [6]) showed that the IVP associated to DNLS is global well-posed for s > 1/2.
The main difficulty to deal with DNLS is the presence of the derivative in the nonlinearity, which causes the so called loss of derivatives. This means that the standard way of proving existence of solution of
u(t) = U (t)u0−
Z t
0
U (t− t0)∂x(|u|2u)dt0 (7)
can not be accomplished only using the property of unitary group and Strichartz (Lemma 1.2.1) of the Schr¨odinger propagator U (t) = eit∂2
x. In fact the right hand side of (7) has less derivative
than the left hand side and Strichartz estimates do not provide us gain of derivatives. This is one point that makes the study of DNLS more difficulty than the corresponding cubic nonlinear Schr¨odinger equation (NLS), namely
i∂tu + ∂x2u + λ|u|
2u = 0.
The equation in (1) admits a family of solitary waves solutions given explicitily by
ψ(x, t) = ϕω,c(x− ct) exp i ωt + c 2(x− ct) − 1 a + 3 Z x−ct −∞ ϕ1+aω,c (y)dy , where ω > c2/4 and ϕω,c(y)1+a = (3 + a)(4ω− c2) 4√ω cosh(a+12 √4ω− c2y)− c 2√ω .
Liu, Simpson and Sulem in [21] studied the orbital stability of these solitary waves for the equation in (1). More precisely,
Definition 0.0.1. The solitary wave ψω,c is said to be orbitally stable for (1) if for each initial
data u0 ∈ H1(R) sufficiently close to ψω,c(0) corresponds a unique u global solution of (1) and
sup t≥0 inf (θ,y)∈R×Rku(t) − e iθψ ω,c(t,· − y)kH1
is sufficiently small. Otherwise ψω,c is said to be orbitally unstable.
Results of orbital stability or instability were obtained according with the value of a. There it was assumed the existence of solution of the IVP (1) for arbitrary a >−1, and initial data
u0 ∈ H1(R). However, besides the case a = 1 there is only a few well-posedness results for (1).
For integer powers a≥ 4 Hao [10] proved local well-posedness in H1/2(R).
In the present work we carry out the case a > 0. The main result regards the most interesting case 0 < a < 1.
Theorem 0.0.2. Let X = H3/2(R) ∩ {f ∈ S0(R); xf ∈ H1/2(R)}. If u
0 ∈ X and
ku0kX =ku0kH3/2+kxu0kH1/2
is sufficiently small, then there exist T = T (ku0kX) > 0 and an unique
u∈ C([0, T ]; H3/2−)∩ L∞([0, T ]; X)
solution of the integral equation
u(t) = U (t)u0− i
Z t
0
U (t− t0)(|u|1+a∂xu)(t0)dt0. (8)
Our strategy to prove this theorem will be based on the technique known as parabolic regularization (or viscosity argument) introduced by T. Kato [16]. It consists in the following: First we prove that for each positive parameter (viscosity) the initial value problem
i∂tu+ ∂x2u+ i|u|1+a∂xu = i∂x2u
u(·, 0) = u0
(9)
has solution in [0, T]. The second step is to prove that the solutions u converge as goes to
term i∂2
xu. As is usual we can obtain solution u ∈ C([0, T]; Hs(R)), with T = /ku0kHs,
in some Sobolev space Hs(R). The difficulty lies in the second step. But before considering the problem of convergence we have to argue that all the solutions can be extended to a same interval of time. This uniform extention will be possible if
sup
>0 ku
kL∞THs
x <∞. (10)
The estimate (10) permits to extend all the solutions to an interval of time [0, T ] independent of
and the extension still satisfies sup>0kukL∞THs
x <∞. Using compactness, for each t ∈ [0, T ]
we have weak convergence u(t) of some subsequence to an element u(t) in Hs(R). So, u is the
candidate to be a solution. Thus we have a great chance to be successful with the parabolic regularization if we can prove the uniform estimate (10). In [15] it is already presented how (10) can be obtained in Hs(R), s > 3/2, for nonlinearities like um∂
xu, m∈ N. There, the argument
can also be adapted to the nonlinearity |u|1+a∂xu whenever we have some smoothness of the
nonlinearity, for example a≥ 1, that allows us to prove
k|u|1+ak
Hs ≤ ckuk1+a
Hs . (11)
However, (11) is no longer true for low powers a since |z|1+a does not have enough
regular-ity. In this work we obtain a uniform estimate like (10) in H3/2(R) when 0 < a < 1 (See Theorem 3.20). Our argument was inspired trying to use the contraction mapping princi-ple. Using sharp smoothing properties associated to the linear equation we define a subspace
E ⊂ C([0, T ]; Hs(R)) and prove that the associated integral operator
Ψ(u) = U (t)u0−
Z t
0
U (t− t0)(|u|1+a∂xu)dt0
is well defined, i.e, Ψ : E → E for small data. However Ψ is not a contraction. This is why we use parabolic regularization to study this problem.
This work is organized as follows:
In Chapter 1 we present a list of results that will be used along the work.
In Chapter 2 we consider the regularized equation (2.1). We prove the existence of solutions
udefined in an interval of time that depends on the parameter . Aiming to extend the solution
to an interval of time independent of the parameter we establish some uniform estimates for the linear solutions of the regularized equation.
We prove that the solutions u of the regularized problem can be all extended to an interval
[0, T ] where T depends only on the size of the initial data. We also prove that the solutions
u are uniformly bounded in H3/2 with respect to the parameter . To achieve this uniform
estimate we impose that the initial data belongs to a weighted Sobolev space and has small size on it.
In Chapter 4 the convergence of the solutions defined in [0, T ] will be studied. First it is proved that this sequence converges strongly in L∞T L2
x to some function u. Then using the
uniform estimate provided in the third chapter and compactness argument, it is shown that for each t ∈ [0, T ] there is some subsequence {u0(t)} converging weakly in H3/2(R) and the weak
limit is in fact u(t). Finally, we prove that u is solution of the integral equation (8) and it is the unique possible solution.
In Chapter 5 we study the case a > 1. In particular, we establish well-posedness for the IVP (1) in Hs(R), s > 1, for small data via contraction argument.
Chapter 1
Preliminaries
1.1
Basic results
In this section we list without proof some elementary inequalities and some commutator esti-mates useful in our analysis below.
Lemma 1.1.1. (Gronwall inequality) Let f be a nonnegative absolutely continuous function
satisfying the differential inequality
f0(t)≤ ϕ(t)f(t) + ψ(t)
almost everywhere, with ϕ and ψ also nonegative. Then we have
f (t)≤ eR0tϕ(s)ds f (0) + Z t 0 ψ(s)ds .
Proof. See for example [26].
Lemma 1.1.2. Given s0 < s < s1, we have
kfkHs ≤ kfkθHs0kfk1H−θs1
where θ = s1 − s s1− s0
.
Proof. See for example [20].
Lemma 1.1.3. (Sobolev embeddings)
i) For 2≤ p < ∞, there exists a constant c > 0 such that,
for all f ∈ Hs(R), where s = 1/2 − 1/p;
ii) When p =∞, there exists a constant c > 0 such that,
kfkL∞ ≤ ckfkH1/2+
or all f ∈ H1/2+(R).
Proof. See [20] Chapter 3.
Lemma 1.1.4. (Leibniz rule for fractional derivatives)
(i) Let 0 < s < 1, s1, s2 ∈ [0, s] with s = s1 + s2 and 1 < p, p1, p2 < ∞, such that 1/p =
1/p1+ 1/p2. Then, for some constant c > we have
kDs
(f g)− fDsg− gDsfkLp ≤ ckDs1fkLp1kDs2gkLp1.
(ii) For f = f (x, t), g = g(x, t) we have
kDs(f g)− fDsg− gDsfk LpxLqT ≤ ckD s1fk Lp1x Lq1TkD s2gk Lp1x Lq1T (1.1)
for 1 < p, p1, p2, q, q1, q2 <∞, such that 1/p = 1/p1+ 1/p2 and 1/q = 1/q1 + 1/q2. Moreover,
for s1 = 0 the value q1 =∞ is allowed.
(iii) If p = 1 and q = 2 , there exists some constant c > 0 such that
kDs(f g)− fDsg− gDsfk L1 xL2T ≤ ckD s1fk Lp1x Lq1TkD s2gk Lp1x Lq1T
for 1 < p1, p2, q1, q2 <∞, such that 1 = 1/p1+ 1/p2 and 1/2 = 1/q1+ 1/q2.
Lemma 1.1.5. (Chain rule) Let 0 < s < 1 and F :C → C be given such that F is C1 (regarded as a function in R2).
(i) For all 1 < p, p1, p2 <∞ such that 1/p = 1/p1+ 1/p2, there is a constant c > 0 such that
kDs(F (f ))k
Lp ≤ ckF0(f )kLp1kDsfkLp2.
(ii) In the case of functions f = f (x, t) it holds
kDs(F (f ))k LpxLqT ≤ ckF 0(f )k Lp1x Lq1T kD sfk Lp2x Lq2T
for 1 < p, p1, p2, q, q1, q2 <∞, such that 1/p = 1/p1+ 1/p2 and 1/q = 1/q1 + 1/q2.
The proof of Lemma 1.1.4 and Lemma 1.1.5 were established by Kenig, Ponce e Vega [17]. Remark 1. Visan and Killip [33] established a chain rule when F is only a H¨older continuous function of order α∈ (0, 1).
1.2
The Schr¨
odinger propagator
Consider the IVP associated to the free Schr¨odinger equation i∂tu + ∂x2u = 0, x ∈ R, t > 0 u(·, 0) = f (1.2)
whose solutions is given by
u(x, t) ={e−itξ2fˆ}∨(x)
and is denoted by U (t)f . The family {U(t)}t∈R forms a unitary group in Hs(R), for all s ∈ R.
In the following we will list some estimates satisfied for solutions of IVP (1.2). We begin by the so called LqLp estimates or Strichartz estimates.
Lemma 1.2.1. (Strichartz estimates) For all pair (p, q) satisfying 2≤ p ≤ ∞ and 2/q = 1/2 − 1/p we have kU(t)fkLqTLpx ≤ ckfkL2 (1.3) and Z0tU (t− t0)F (x, t0)dt0 LqTLpx ≤ kF kLq0 TL p0 x (1.4)
where 1/p0+ 1/p = 1/q0+ 1/q = 1, for some constant c > 0.
Proof. See Ginibre and Velo [7]
Next we have the smoothing effects estimates
Lemma 1.2.2. (homogeneous smoothing effect) There exists a constant c > 0, such that
kD1/2U (t)fk
L∞xL2T ≤ ckfkL2
for all f ∈ L2(R).
By duality it follows from Lemma 1.2.2:
Lemma 1.2.3. There exists a constant c > 0 such that D1/2 Z R U (t0)F (·, t0)dt0 L∞TL2 x ≤ ckF kL1 xL2T for all F ∈ L1xL2T.
For the inhomogeneous problem i∂tu + ∂x2u = F (x, t) u(·, 0) = 0 (1.5) whose solution is u(x, t) = Z t 0 U (t− t0)F (·, t0)dt0 (1.6) we have:
Lemma 1.2.4. (Inhomogeneus smoothing effect) For u given as (1.6) we have
k∂xukL2
TL∞x . kF kL1xL2T
The estimates above were proved by Kenig, Ponce and Vega [17]. For a detailed proof of Lemmas 1.2.1 to 1.2.4 see [20] Chapter 4.
Finally, we present some maximal function estimates for the Schr¨odinger propagator. More precisely,
Lemma 1.2.5. (Maximal L2 estimate) Given s > 1/2 there exists a constant c > 0 such that
kU(t)fkL2
xL∞T ≤ ckfkHs
for all f ∈ Hs(R).
Proof. See Kenig, Ponce and Vega in [18] .
Lemma 1.2.6. (Maximal L4 estimate) There exists a constant c > 0 such that
kU(t)fkL4
xL∞t ≤ ckD
1/4fk
L2
for all f ∈ H1/4(R).
Proof. See Kenig and Ruiz [19] .
1.3
Two technical lemmas
Lemma 1.3.1. (Interpolation) Let θ ∈ [0, 1] be given. There exists some constant c > 0 such that, kJ1/2+θ(hxi1−θf )k L2 ≤ ckJ1/2(hxif)k1−θ L2 kJ 3/2fkθ L2.
Lemma 1.3.2. Given 0 < s < 1 we have
kJs(hxif)k
L2 ≤ ckJs(xf )kL2 +kJsfkL2
for some constant c > 0.
Before going through the proof of these lemmas we shall present some facts that will help us in their proofs. To prove Lemma 1.3.1 we consider for 0 < α < 2 the following derivative
Dαf (x) = lim →0 Z |y|> f (x + y)− f(x) |y|1+α dy. (1.7)
Theorem 1.3.3 (Characterization I). Let 0 < α < 2 and 1 < p <∞. Then f ∈ Lp
α(R) if, and
only if, f ∈ Lp(R) and the limit defined in (1.7) converges in Lp norm. In this case
kJαfk
Lp ≈ kfkLp+kDαfkLp.
Proof. See Stein [27] or [29].
Lemma 1.3.4. Let 0 < α < 1 and 1 < p <∞. Then there exists a constant c > 0 such that (c(1 +|t|))−1kJα(h·iitf )kLp ≤ kJαfkLp ≤ c(1 + |t|)kJα(h·iitf )kLp
for all t∈ R.
Proof. Denote ϕ(x) = loghxi.
Dα(h·iitf )(x) = Dα(eitϕf )(x) = lim →0 Z |y|> eitϕ(x+y)f (x + y)− eitϕ(x)f (x) |y|1+α dy = eitϕ(x)lim →0 Z |y|> eit(ϕ(x+y)−ϕ(x))− 1 |y|1+α f (x + y)dy +eitϕ(x)lim →0 Z |y|> f (x + y)− f(x) |y|1+α dy = eitϕ(x)φα,t(f )(x) + eitϕ(x)Dαf (x). Thus kDα(h·iitf )(x)kLp ≤ kφα,t(f )kLp+kDαfkLp.
Let us estimate kφα,t(f )kLp. φα,t(f )(x) = lim →0 Z <|y|<1 eit(ϕ(x+y)−ϕ(x))− 1 |y|1+α f (x + y)dy + Z |y|>1 eit(ϕ(x+y)−ϕ(x))− 1 |y|1+α f (x + y)dy = I0+ I∞.
Since |eiθ− 1| ≤ 2 we have
kI∞kLp ≤ 2 Z |y|>1 1 |y|1+αkf(· + y)kLpdy = 2kfkLp Z |y|>1 1 |y|1+αdy = cαkfkLp.
To take care of I0 we first note that
|eiθ − 1| = |eiθ/2− e−iθ/2| = 2| sin(θ/2)| ≤ |θ|.
So
|eit(ϕ(x+y)−ϕ(x))− 1| ≤ |t||ϕ(x + y) − ϕ(x)|.
Since ϕ is Lipschitz we have
|eit(ϕ(x+y)−ϕ(x))− 1| ≤ |t||y|. (1.8) Then |I0| ≤ lim →0 Z <|y|<1 |t||y| |y|1+α|f(x + y)|dy = |t| lim →0 Z <|y|<1 1 |y|α|f(x + y)|dy. (1.9)
Applying Minkowsky’s inequality in the integral (1.9) we get
kI0kLp ≤ |t| lim →0 Z <|y|<1 1 |y|αkf(· + y)kLpdy = |t|kfkLplim →0 Z <|y|<1 1 |y|αdy = cα|t|kfkLp. We conclude that kφα,t(f )kLp . (1 + |t|)kfkLp.
Then applying Theorem 1.3.3 we obtain kDα(h·iitf )kLp . (1 + |t|)(kfkLp+kDαfkLp) . (1 + |t|)kJα fkLp. Therefore kJα (h·iitf )kLp . kDα(h·iitf )kLp+kfkLp . (1 + |t|)kJα fkLp. (1.10)
The opposite inequality follows immediately by applying (1.10) to the function h·i−itf instead
of f .
Now we are ready to prove Lemma 1.3.1 Proof of Lemma 1.3.1 :
Given g ∈ L2(R) such that kgk
L2 = 1 define Fg : S −→ C by
Fg(z) = ez
2−1Z
R
J1/2+z(hxi1−zf )(x)¯g(x)dx
where S is the strip S = {z ∈ C; 0 ≤ Re(z) ≤ 1}. Using the Cauchy-Schwarz inequality and Lemma 1.3.4
|Fg(iy)| = e−(y
2+1)
Z
R
J1/2+iy(hxi1−iyf )(x)¯g(x)dx ≤ e−(y2+1)
kJ1/2+iy
(hxi1−iyf )kL2k¯gkL2
= e−(y2+1)kJ1/2(hxi1−iyf )kL2
. (1 + |y|)e−(y2+1)
kJ1/2
(hxif)kL2.
Since the function (1 +|y|)e−(y2+1) is bounded it follows
|Fg(iy)| . kJ1/2(hxif)kL2. Similarly, we have |Fg(1 + iy)| = e−y 2 Z R
J3/2+iy(hxi−iyf )(x)¯g(x)dx ≤ e−y2
kJ3/2+iy(hxi−iyfk
L2kgkL2
Before applying Lemma 1.3.4 to (1.11) we make the following computation
kJ3/2
(hxi−iyf )kL2 ∼ khxi−iyfkL2 +kD3/2(hxi−iyf )kL2
= kfkL2 +kD1/2∂x(hxi−iyf )kL2 = kfkH3/2+|y| D1/2 hxi−iy x hxi2f L2 . (1.12)
Now we can use Lemma 1.3.4 to obtain D1/2 hxi−iy x hxi2f L2 . (1 + |y|) J1/2 x hxi2f L2 . (1.13)
It turns out we can bound (1.13) as J1/2 x hxi2f L2 ≤ J1 x hxi2f L2 ∼ hxix2f L2 + ∂x x hxi2f L2 ≤ kfkL2 + ∂x x hxi2 L∞ kfkL2 + hxix2 L∞ k∂xfkL2 . kfkL2 +k∂xfkL2. (1.14)
We conclude from the estimates (1.12), (1.13) and (1.14)
kJ3/2(hxi−iyf )k
L2 . (1 + |y|)2kfkH3/2
and then we finally get
|Fg(1 + iy)| . kfkH3/2.
Therefore using the Three Lines Theorem (see [28]) we obtain
|Fg(θ)| . kJ1/2(hxif)k1L−θ2 kJ
3/2fkθ
L2 (1.15)
for all θ ∈ [0, 1]. Taking the supremum over all g ∈ L2(R) such that kgk
L2 = 1 in (1.15) we obtain eθ2−1kJ1/2+θ(hxi1−θf )kL2 . kJ1/2(hxif)k1−θ L2 kJ 3/2fkθ L2. So we finally conclude kJ1/2+θ(hxi1−θf ))k L2 . kJ1/2(hxif)k1−θ L2 kJ 3/2fkθ L2. To prove Lemma 1.3.2 it will be more convenient to consider the characterization of Sobolev
spaces in terms of the following derivative Dsf (x) = Z R |f(x + y) − f(x)|2 |y|1+2s dy 1/2 . (1.16)
We have the following characterization
Theorem 1.3.5. Let s∈ (0, 1) and 2/(1 + 2s) < p < ∞. Then f ∈ Lps(R) if, and only if f and
Dsf belong to Lp(R). Moreover
kJs
fkLp ∼ kfkLp+kDsfkLp
for all f ∈ Lp s(R).
Proof. See Stein [27] or [29].
Remark 2. When p = 2 we have the following product rule
kDs(f g)k
L2 ≤ kfDs(g)kL2 +kgDsfkL2. (1.17)
Proof of Lemma 1.3.2: Consider the cut-off function χ ∈ Cc∞(R), supported in [−2, 2] and identically 1 in [−1, 1]. Before going any further notice the following
CLAIM 1: If ϕ is Lipschitz and bounded then Dsϕ is bounded as well.
In fact, (Dsϕ(x))2 = Z R |ϕ(x + y) − ϕ(x)|2 |y|1+2s dy . Z |y|<1 |y|2 |y|1+2sdy + 2kϕk 2 L∞ Z |y|>1 1 |y|1+2s < ∞.
To prove Lemma 1.3.2 it is enough to prove
kDs
(hxif)kL2 . kJs(xf )kL2 +kJsfkL2.
We have
kDs
(hxif)kL2 ≤ kDs(hxifχ(x)f)kL2 +kDs(hxif(1 − χ(x))f)kL2
We are going to use the characterization in Theorem 1.3.5 to estimate I and II. In fact, using Theorem 1.3.5, together with (1.17) and CLAIM 1 with ϕ(x) =hxiχ(x) we have
kDs(hxifχ(x)f)k L2 . kϕfkL2 +kDs(ϕf )kL2 . kϕfkL2 +kϕDsfkL2 +kfDsϕkL2 . kϕkL∞(kfkL2+kDsfkL2) +kDsϕkL∞kfkL2 . kfkL2 +kDsfkL2. Then we conclude I . kJsfkL2.
Similarly, using Theorem 1.3.5, (1.17) and CLAIM 1 for the function ϕ(x) = hxix (1− χ(x)) we have kDs(hxi(1 − χ(x))f)k L2 = kDs(ϕxf )kL2 . kϕxfkL2 +kDs(ϕxf )kL2 . kϕxfkL2 +kxfDsϕkL2 +kϕDs(xf )kL2 . kϕkL∞(kxfkL2 +kDs(xf )kL2) +kDsϕkL∞kxfkL2 . kxfkL2 +kDs(xf )kL2. Then II . kJs(xf )kL2.
Chapter 2
The viscosity argument
In this chapter we are going to perform the viscosity argument. That is, for every > 0 we are going to solve the (IVP) problem
i∂tu + ∂x2u + i|u|1+a∂xu = i∂x2u
u(·, 0) = u0
(2.1)
where (x, t) ∈ R × [0, ∞) and a ∈ (0, 1). This problem is simpler to solve than (1) since the additional term i∂2
xu provides us with the propagator eit(1−i)∂
2
x which has stronger decay
properties than the original eit∂2x. These decay properties translate into gain of derivatives
which are very convenient in handling the loss of derivative introduced by the nonlinear term. A crucial point here is to establish estimates for eit(1−i)∂2x uniformly in the parameter . These
uniform estimates, in turn, will allow us to pass to the limit as goes to zero. In the next chapter we will prove that the solutions u can be uniformly estimated in H3/2(R). Therefore
in this chapter we will solve the reguralized problem (2.1) for initial data u0 in H3/2(R), even
though we could do so for u0 in a Sobolev space of lower regularity.
2.1
Solution of the -nonlinear problem in H
3/2In our work we are going to denote by U(t) the linear propagator eit(1−i)∂
2 x, which describes the solution of i∂tu + ∂x2u = i∂x2u, u(·, 0) = u0, (2.2)
defined via Fourier transform by
U(t)u0 ={e−(i+)tξ
2
ˆ
u0}∨(x)
for all t≥ 0. We will solve the integral equation
u(t) = U(t)u0−
Z t
0
U(t− t0)(|u|1+a∂xu)(t0)dt0, (2.3)
and later on we will justify that the solution of (2.3) is actually solution of the diffential equation (2.1). Denoting by Ψ the right hand side of (2.3) we are looking for u such that Ψ(u) = u.
Following the contraction principle argument we have to prove in a first step that there exist a time T > 0 and a constant A > 0 such that the integral operator Ψ maps the space
EA,T ={u ∈ C([0, T ] : H3/2(R)); kukL∞ TH
3/2
x ≤ A} (2.4)
to itself for all 0 < T ≤ T, then we have to prove that Ψ : EA,T −→ EA,T is a contraction, for
all those T . Consequently the Banach Theorem for Contraction Mappings assures the existence and uniqueness of u ∈ EA,T satisfying Ψ(u) = u. In this task we will use the following lemma:
Lemma 2.1.1. Let s≥ 0. Then,
i) There exists cs> 0 such that
kU(t)fkHs x ≤ cs 1 + 1 |t|s/2 kfkL2, (2.5)
for all f ∈ L2(R), and t > 0.
ii) for all f ∈ L2(R) the map
t∈ (0, ∞) 7−→ U(t)f ∈ Hs(R) (2.6)
is continuous. Moreover, if f ∈ Hs(R) then the map in (2.6) is continuous to the right at t = 0.
Proof. Indeed, kDsU (t)fkL2 x = k|ξ| se−(+i)ξ2tb fkL2 ξ = k|ξ|se−ξ2tfbkL2 ξ = 1 |t|s/2k|ξ 2t|s/2e−ξ2tfbk L2 ξ.
Since the function ys/2e−y is bounded in {y ∈ R; y > 0} we obtain kDs U(t)fkL2 x ≤ 1 |t|s/2 sup y>0{|y| s/2 e−y}k bfkL2 = 1 |t|s/2 sup y>0{|y| s/2 e−y}kfkL2. (2.7)
Estimate (2.7) with s = 0 gives us
kU(t)fkL2
x ≤ kfkL2. (2.8)
Then, using (2.7) and (2.8) we obtain (2.5). Now we are going to prove the property ii). Let
t0 > 0 be given and t > t0. We have from (2.5) that
kU(t)f − U(t0)fkHs x = kU(t0)(U(t− t0)f− f)kHxs . 1 + 1 |t0|s/2 kU(t − t0)f − fkL2 x. Turns out kU(t − t0)f − fkL2 x =k[e −(+i)(t−t0)ξ2 − 1] bfk L2 ξ (2.9)
Since for each t0 < t the function
ξ7−→ |(e−(+i)(t−t0)ξ2 − 1) bf (ξ)|2
is integrable and bounded by the function 2| bf (ξ)|2, which is integrable, it follows from the Dominated Convergence Theorem that
lim
t&t0
k(e−(+i)(t−t0)ξ2 − 1) bfk
L2
ξ = 0.
Therefore we have continuity to right. To prove the continuity to the left we consider 0 < t < t0.
Using (2.5) we have already proved we get
kU(t)f− U(t0)fkHs x = kU(t)(U(t0− t)f − f)kHsx (2.10) . 1 + 1 |t|s/2 kU(t0− t)f − fkL2 x. (2.11)
Then we proceed using the Dominated Convergence Theorem to conclude lim
t%t0kU(t
0− t)f − fkL2
x = 0.
Moreover, supposing f ∈ Hs(R) we have that for each t > 0 the function
is integrable and bounded by 2|hξisfb|2 which is integrable as well. Using one more time the
Dominated Convergence Theorem it follows lim t&0kU(t)f− fkH s x = lim t&0k(e −(i+)tξ2 − 1)hξisb fkL2 ξ = 0.
We start the contraction principle argument by controlling Ψ(u) in the norm k · kL∞ TH
3/2
x
for a function u belonging to the class EA,T defined in (2.4). Indeed, Lemma 2.1.1 with s = 0
implies kU(t)u0kH3/2 x = kU(t)(J 3/2u 0)kL2 x ≤ kJ3/2u 0kL2. Then we have kΨ(u)kH3/2 x ≤ kU(t)u0kHx3/2+k Z t 0
U(t− t0)(|u|1+a∂xu)(t0)dt0kH3/2
x
≤ ku0kH3/2 + (N L).
Lemma 2.1.1 also implies (N L) ≤
Z t
0
kU(t− t0)(|u|1+a∂xu)(t0)kH3/2
x dt 0 . Z t 0 1 + 1 (|t − t0|)3/4 k|u|1+a∂ xu(t0)kL2 xdt 0 . T + 4T 1/4 3/4 k|u|1+a∂ xukL∞TL2 x.
Using the Sobolev embedding H1/2+(R) ⊂ L∞(R) we have
k|u|1+a∂ xukL∞TL2 x ≤ kuk a+1 L∞TL∞x k∂xukL∞TL2x . kuka+2 L∞THx3/2 . We conclude kΨ(u)kL∞ TH 3/2 x ≤ ku0kH3/2 + c T + 4cT 1/4 3/4 kuka+2 L∞THx3/2 , (2.12)
for some constant c. If we take T= T (,ku0kH3/2) > 0 sufficiently small such that
T+
4T1/4
3/4 ≤
1
and A = 2ku0kH3/2 it follows from the estimate (2.12) that Ψ(u)∈ L∞([0, T ]; H3/2(R)) and
kΨ(u)kL∞ TH
3/2
x ≤ A
for all 0 < T ≤ T. To conclude Ψ(u) ∈ EA,T it remains to prove Ψ(u) is continuous with
respect to t. We already know from Lemma 2.1.1 the map
t∈ [0, T ] 7−→ U(t)u0 ∈ H3/2(R)
is continuous. It remains to prove the continuity of
v(t) =−
Z t
0
U(t− t0)(|u|1+a∂xu)dt0.
Denote F =|u|1+a∂
xu and let t0 ∈ [0, T ]. For each t > t0 we have
v(t)− v(t0) = − Z t0 0 U(t0− t0)(U(t− t0)F (t0)− F (t0))dt0 − Z t t0 U(t− t0)F (t0)dt0 = I + II.
Before estimating I and II notice that F ∈ L∞([0, T ]; L2(R)). In fact, using Sobolev embedding
kF kL∞TL2 x ≤ kuk 1+a L∞TL∞x k∂xukL∞TL2x . kuk1+a L∞THx1/2+kukL ∞ THx1 ≤ kuk2+a L∞THx3/2 .
Now using Lemma 2.1.1 we conclude
kIIkH3/2 x ≤ Z t t0 kU(t− t0)F (t0)kH3/2 x dt 0 . kF kL∞TL2 x Z t t0 1 + 1 ((t− t0))3/4 dt0. (2.13)
The integral in (4.25) goes to zero as t goes to t0, because the function
t0 ∈ [0, T ] 7−→ 1 + 1
((t− t0))3/4
is integrable. Next, we look at the term I. Also using Lemma 2.1.1 we have
kIkH3/2 ≤ Z t0 0 kU(t0− t0)(U(t− t0)F (t0)− F (t0))kH3/2 x dt 0 . Z t0 0 1 + 1 ((t0− t0))3/4 kU(t− t0)F (t0)− F (t0)kL2 xdt 0. (2.14)
Since kU(t− t0)F (t0)− F (t0)kL2 x ≤ 2kF (t 0)k L2 x
we can apply the Dominated Convergence Theorem to the integral in (2.14) and conclude
kIkH3/2 goes to zero as t goes to t0. This concludes the proof that v is continuous to the right.
The continuity on the left is proved similarly. Therefore, Ψ(EA,T)⊂ EA,T.
Now let us prove that we can still choose T> 0 such that the map
Ψ: EA,T −→ EA,T
is a contraction with respect to the norm k · kL∞
TH
3/2
x for all 0 < T < T. Consider u, v ∈ EA,T
and denote G(u, v) =|u|1+a∂
xu− |v|1+a∂xv. We have
Ψ(u)− Ψ(v) = −
Z t
0
U(t− t0)G(u, v)(t0)dt0.
Then using Lemma 2.1.1
kΨ(u)− Ψ(v)kH3/2 x = k Z t 0 U(t− t0)G(u, v)(t0)dt0kL∞ TH 3/2 x . Z t 0 1 + 1 (|t − t0|)3/2 kG(u, v)(t0)k L2 xdt 0 . T + 4cT 1/4 3/4 kG(u, v)kL∞TL2 x. To estimate kG(u, v)kL∞TL2
x first note that G(u, v) can be written as
G(u, v) = |u|1+a− |v|1+a∂xu +|v|1+a∂x(u− v).
Then using
|u|1+a− |v|1+a .(|u|a+|v|a)|u − v| (2.15) we have kG(u, v)kL∞TL2 x ≤ k |u| 1+a− |v|1+a∂ xukL∞TL2 x+k|v| 1+a∂ x(u− v)kL∞TL2 x
≤ k|u|1+a− |v|1+ak
L∞TL∞x k∂xukL∞TL2x+kvk 1+a L∞TL∞x k∂x(u− v)kL∞TL2x . kuka L∞TL∞x +kvk a L∞TL∞x ku − vkL∞TL∞x k∂xukL∞TL2x + kvk1+aL∞ TL∞x k∂x(u− v)kL∞TL2x. (2.16)
Applying Sobolev embedding in (2.16) we conclude kG(u, v)kL∞TL2 x . kukL∞ TH 3/2 x +kvkL∞TH 3/2 x 1+a ku − vkL∞ TH 3/2 x . 4ku0kH3/2 1+a ku − vkL∞ TH 3/2 x . Therefore kΨ(u)− Ψ(v)kL∞ TH 3/2 x ≤ c T + 4cT 1/4 3/4 4ku0kH3/2 1+a ku − vkL∞ TH 3/2 x , (2.17)
for some constant c > 0. Take T such that
T+ 4cT1/4 3/4 ≤ 1 2c(4ku0kH3/2)1+a .
Thus from (2.17) we end up with
kΨ(u)− Ψ(v)kL∞ TH 3/2 x ≤ 1 2ku − vkL∞THx3/2.
Remark 3. We can choose
T =
3
cku0k1+aH3/2
for some constant c sufficiently large.
We have proved the following
Theorem 2.1.2. Let > 0 and u0 ∈ H3/2(R) be given. Then for all 0 < T ≤
3 cku0k1+aH3/2
there exists one, and only one, u ∈ C([0, T ]; H3/2(R)) satisfying the integral equation (2.3).
Next we prove that the solution u in Theorem 2.1.2 is in fact solution of the differential
equation (2.1).
Theorem 2.1.3. Let u be the solution of (2.3) given in Theorem 2.1.2. Then,
u ∈ C((0, T]; H2(R)) (2.18)
and ∂tu exists in the topology of H−1(R) and satisfies
Proof. We already know from Lemma 2.1.1 that U(t)u0 ∈ C((0, T]; H2(R)). So we have just
to investigate the continuity of
v(t) =−
Z t
0
U(t− t0)F(t0)dt0
where F =|u|1+a∂xu. From Lemma 2.1.1 we have,
kv(t)kH2 x ≤ Z t 0 kU(t− t0)F(t0)kH2 xdt 0 . kFkL∞ TH 1/2 x Z t 0 1 + 1 ((t− t0))3/4 dt0 . kFkL∞ TH 1/2 x . So, if we assume F ∈ L∞([0, T ]; H1/2(R)) (2.20) we obtain v ∈ L∞([0, T ]; H2(R)). (2.21)
Lets keep the assumption (2.20) and prove (2.18). We claim that (2.21) together with the continuity in H3/2 gives us (2.18). Indeed, choose some s ∈ (2, 5/2). Using Lemma 1.1.2 we
have kv(t)− v(t0)kH2 x ≤ kv(t)− v(t0)k θ Hx3/2kv (t)− v(t0)k1H−θs0 x (2.22) for θ = 2s− 4
2s− 3. Using (2.21) and (2.22) we obtain
kv(t)− v(t0)kHs
x . kv(t)− v(t0)k
θ Hx3/2
. (2.23)
Since v ∈ C([0, T]; H3/2(R)) it follows from (2.23) that v ∈ C([0, T]; H2(R)). To finish the
proof of (2.18) we need to justify (2.20). Indeed, using the Leibniz rule (Lemma 1.1.4) we have
kD1/2F kL2 x . kD 1/2(|u |1+a)kLpxk∂xukLqx +k|u| 1+ak L∞x kD 1/2∂ xukL2 x, (2.24)
for 2 < p, q <∞ satisfying 1/2 = 1/p + 1/q. Using Sobolev embedding in (2.24) we obtain
kD1/2F kL2 x . k|u| 1+ak H1 xkukHx3/2+kuk 2+a Hx3/2 . (2.25)
Computing ∂x(|u|1+a) and using Sobolev embedding we obtain
k|u|1+akH1 x . kuk 1+a L2+2ax +k|u| a∂ xukL2 x . kuk1+a Hx3/2 +kukaL∞x k∂xukL2x . kuk2+a Hx3/2 . (2.26)
Combining (2.26) and (2.25) we finally obtain (2.20). Now we are going to prove (2.19). Consider t∈ (0, T ] and h sufficiently small. Regarding the linear term U(t)u0 we have
U(t + h)u0− U(t)u0 h − (i + )∂ 2 xU(t)u0 ={g(t; h, ξ)bu0}∨ where g(t; h, ξ) = e −(i+)ξ2(t+h) − e−(i+)ξ2t h + (i + )ξ 2e−(i+)ξ2t . Notice that |g(t; h, ξ)| ≤ 2(1 + )ξ2.
Then we can use the Dominated Convergence Theorem and conclude that lim h→0kg(t; h, ξ)hξi −1bu 0kL2 ξ = 0, which means lim h→0 U(t + h)u0− U(t)u0 h − (i + )∂ 2 xU(t)u0 Hx−1 = 0. (2.27)
Next we analyse the nonlinear part v. For all h > 0 we have,
v(t + h)− v(t) h = − 1 h Z t 0 U(t + h− t0)− U(t− t0) F(t0)dt0 −1 h Z t+h t U(t + h− t0)F(t0)dt0 = I1(t, h) + I2(t, h). We have I1(t, h) = − 1 h Z t 0 U(h)U(t− t0)F(t0)dt0 −1 h Z t 0 U(t− t0)F(t0)dt0 = U(h)v(t)− v(t) h . (2.28)
Since v(t)∈ H2(R), it follows from (2.28) that
lim
h&0kI1(t, h)− (i + )∂
2
Regarding the term I2(t, h) we have kI2(t, h)− F(t)kHx−1 ≤ 1 h Z t+h t kU(t + h− t0)(F(t0)− F(t))kHx−1dt 0 +1 h Z t+h t kU(t + h− t0)F(t)− F(t)kH−1x dt 0 . 1 h Z t+h t kF(t0)− F(t)kHx−1dt 0 +1 h Z t+h t kU(t + h− t0)F(t)− F(t)kH−1x dt 0. (2.30)
Using that F belongs to the class C([0, T ]; H−1(R)) as well as the map
τ ∈ [0, T ] 7−→ U(τ )F(t)
we conclude from (2.30) that
lim h&0kI2(t, h)− F(t)kHx−1 = 0. (2.31) Consequently lim h&0 u(t + h)− u(t) h = (i + )∂ 2 xu(t)− |u|1+a∂xu(t)
in the H−1(R) topology. To conclude our proof, we study the limit lim
h%0
v(t + h)− v(t)
h .
Indeed, for each h < 0 we write
v(t + h)− v(t) h = − 1 h Z t+h 0 U(t + h− t0)− U(t− t0) F(t0)dt0 +1 h Z t t+h U(t− t0)F(t0)dt0 = I˜1(t, h) + ˜I2(t, h).
The argument we used to prove (2.31) can also be applied to the term ˜I2(h, t). Thus we also
obtain
lim
h&0k˜I2(t, h)− F(t)kHx−1 = 0. (2.32)
Finally, we consider ˜I1(t, h). Using the definition of v we can write
˜
where ˜ I11(t, h) =−(i + ) Z t t+h ∂x2U(t− t0)F(t0)dt0, and ˜ I12(t, h) = Z t+h 0 U(t + h− t0)− U(t− t0) h F(t 0)− (i + )∂2 xU(t− t0)F(t0) dt0.
To prove both ˜I11(t, h) and ˜I12(t, h) converge to zero as h goes to zero, notice that
F ∈ L∞([0, T ]; H1(R)) (2.33)
This follows from the fact u ∈ L∞([0, T ]; H2(R)). Using the boundness property of U we have
k˜I11(t, h)kHx−1 . Z t t+h k∂2 xU(t− t0)F(t0)kHx−1dt0 . Z t t+h kF(t0)kH1 xdt 0 . −hkFkL∞TH1 x. Hence lim h%0k˜I11(t, h)kHx−1 = 0.
It is remaining to analyse ˜I12(t, h). Notice we can write
U(t + h− t0)− U(t− t0) h F(t 0)− (i + )∂2 xU(t− t0)F(t0) ={q(h, t0, ξ) [F(t0)}∨, where q(h, t0, ξ) = e −(i+)(t+h−t0)ξ2 − e−(i+)(t−t0)ξ2 h + (i + )ξ 2 e−(i+)(t−t0)ξ2. Thus k˜I12(t, h)kHx−1 ≤ Z t+h 0 kq(t, t0, ξ) [F (t0)hξi−1kL2 ξdt 0. Since |q(h, t0, ξ)χ [0,t+h](t0)| . ξ2, (2.34)
it follows the function
t0 ∈ [0, t] 7−→ khξi−1q(h, t0, ξ)χ[0,t+h](t0) [F(t0)kL2ξ
is bounded by
t0 7−→ khξi [F(t0)kL2
which, in turn, is integrable because F ∈ L∞([0, T ]; H1(R)). Consequently, we can apply the
Dominated Convergence Theorem and get lim h%0k˜I12(t, h)kHx−1 = Z t 0 lim h%0khξi −1q(h, t0, ξ)χ [0,t+h](t0) [F(t0)kL2 ξdt 0 (2.35)
Using (2.34) again we can apply the Dominated Convergence Theorem to conclude that for each t0 ∈ [0, t] lim h%0khξi −1q(h, t0, ξ)χ [0,t+h](t0) [F(t0)kL2 ξ = k limh%0hξi −1q(h, t0, ξ)χ [0,t+h](t0) [F(t0)kL2 ξ = 0. This finishes our proof.
Remark 4. The argument we presented above in the proof of (2.19) implies that
u ∈ C([0, T ]; Hs(R))
whenever
F ∈ L∞([0, T ]; Hx(s−2)+) (2.36)
Since we proved F ∈ L∞([0, T ]; H1(R)), it is possible to prove u ∈ C([0, T ]; H3−(R)). But
we will not need this in our work. Unfortunately, we can not derivate the nonlinearity F =
|u|1+a∂xu twice when 0 < a < 1. This means we do not have (2.36) for s ≥ 4. Consequently
we can not guarantee u is sufficiently regular.
2.2
Uniform estimates for the -linear problem
In this section we establish properties that are true for U (t) and still hold for U(t) uniformly
in the parameter . All these properties will be important to prove uniform estimates for the solution of the problem (2.1). We are looking for estimates independent of because we intend to take the limit when goes to zero later on.
Proposition 2.2.1. There exists a constant c > 0 independent of such that
kU(t)fkLqTLpx ≤ ckfkL2
for all pairs (p, q) satisfying
2≤ p ≤ ∞, 1 2 = 2 q + 1 p.
Proof. Notice that U(t) = E(t)U (t) where
E(t)g ={e−tξ
2
ˆ
g}∨.
Let ϕ(x) = e−πx2 and denote ϕρ the function ϕρ(x) = ρϕ(ρx). Since ˇϕ = ϕ, we have from the
properties of the Fourier transform that
{e−tξ2 }∨(x) = nϕ( r t π·) o∨ (x) = r π tϕˇ rπ tx = r π tϕ rπ tx = ϕρ,t(x) where ρ,t = pπ t. So we can write E(t) as E(t)g(·, t) = ϕρ,t ∗ g. (2.37)
Finally, we combine the Young inequality and Strichartz estimate for the Schr¨odinger group to conclude kU(t)fkLqTLxp = kE(t)U (t)fkLqTL p x . kϕρ,t ∗ U(t)fkLqTLpx . kkϕρ,tkL1xkU(t)fkLpxkLqT. (2.38) Using that Z ϕρ,t(x)dx = Z ϕ(x)dx in (2.38) we conclude that kU(t)fkLqTLpx . kU(t)fkLqTL p x . kfkL2.
Next we are concerned with the question whether or not the smoothing effects for the Schr¨odinger propagator are still true for U(t). If we knew, for example, that
with c independent of we would obtain sup
>0 kD
1/2
U(t)fkL∞x L2T ≤ ckfkL2.
Unfortunately we do not know if (2.39) is true. Looking for a replacement we introduce a norm somehow similar to k · kL∞
x L2T but in which we are able to prove that the operators E(t) are
bounded uniformly in . The norm is
k · kl∞j (L2(Q
j)) = sup
j∈Z k · kL
2(Q
j)
where Qj is the rectangle Qj = [0, T ]× [j, j + 1], j ∈ Z. This norm is weaker than k · kL∞x L2T, i.e.
k · kl∞j (L2(Q
j))≤ k · kL∞xL2T. (2.40)
In this weaker norm we are able to prove analogous of the Lemmas 1.2.2, 1.2.3 and 1.2.4 for U(t) with bounds independents of .
Proposition 2.2.2. Let T > 0 and > 0 satisfying 0 < < π/T . Then there exists c > 0
independent of such that
kD1/2U
(t)fkl∞j (L2(Q
j)) ≤ ckfkL2
for all f ∈ L2(R).
To prove this proposition we need to prove the following lemma.
Lemma 2.2.3. Let T > 0 and > 0 satisfying 0 < < π/T . Then there exists c > 0
independent of such that
kE(t)gkl∞j (L2(Q
j)) ≤ ckgkl∞j (L2(Qj))
for all g∈ lj∞(L2(Q
j)).
Proof. Writing E(t) as the convolution in (2.37) we have
E(t)g(·, t)(x) = Z R ϕρ,t(y)g(x− y, t)dy = Z |y|<1 ϕρ,t(y)g(x− y, t)dy + Z |y|>1 ϕρ,t(y)g(x− y, t)dy = I0(x, t) + I∞(x, t).
By using Minkowsky inequality, a change of variables and [j− y, j + 1 − y] ⊂ [j − 1, j + 2] for all|y| < 1, we have that
kI0(·, t)kL2([j,j+1]) ≤ Z |y|<1 ϕρ,t(y)kg(· − y, t)kL2([j,j+1])dy = Z |y|<1 ϕρ,t(y)kg(·, t)kL2([j−y,j−y+1])dy = Z |y|<1 ϕρ,t(y)kg(·, t)kL2([j−1,j+2])dy = kg(·, t)kL2([j−1,j+2]) Z R ϕ(y)dy = kg(·, t)kL2([j−1,j+2])
for all j ∈ Z. So we obtain
kI0kl∞j (L2(Q j)) = sup j∈Z kkI0kL2([j,j+1])kL2([0,T ]) = sup j∈Z kgkL2([j−1,j+2]×[0,T ]) ≤ 3kgkl∞j (L2(Q j)).
Now let us estimate kI∞klj∞(L2(Q
j)). In order to do so notice that ρ,t ≤ ρ
2 ,t whenever ρ,t > 1, i.e. < πT. Then ϕρ,t(y) = ρ,te −π|ρ,ty|2 ≤ ρ2 ,te−π|ρ,t y|2 .
This allows us to conclude
|I∞(x, t)| ≤ Z |y|>1 ϕρ,t(y)|g(x − y, t)|dy ≤ Z |y|>1 ρ2,te−|ρ,ty|2|g(x − y, t)|dy = Z |y|>1 1 y2|yρ,t| 2e−|ρ,ty|2|g(x − y, t)|dy ≤ sup{δe−δ : δ > 0}Z |y|>1 1 y2|g(x − y, t)|dy.
Using change of variables we have
kI∞kL2(Q j) . Z |y|>1 1 y2kg(· − y, ·)kL2(Qj)dy . Z |y|>1 1 y2kgkL2([j−y,j+1−y]×[0,T ])dy . Z |y|>1 1 y2kgkL2([mj,y,mj,y+2]×[0,T ])dy
where mj,y denotes the integer number such that mj,y ≤ j − y < mj,y+ 1. Then we conclude kI∞kL2(Q j) . sup m∈ZkgkL 2([m,m+1]×[0,T ]) Z |y|>1 1 y2dy . kgkl∞j (L2(Q j)). Therefore kI∞kl∞j (L2(Q j)) . kgkl∞j (L2(Qj)).
Proof of Proposition 2.2.2 : Since
D1/2U(t)f = D1/2E(t)U (t)f
= E(t)D1/2U (t)f
we have from Lemma 2.2.3 that
kD1/2 U(t)fkl∞j (L2(Q j))≤ ckD 1/2 U (t)fkl∞j (L2(Q j)).
To complete the proof we recall (2.40) as well as the smoothing effect Lemma 1.2.2 to conclude
kD1/2 U(t)fkl∞j (L2(Q j)) ≤ ckD 1/2 U (t)fkL∞x L2T ≤ ckfkL2.
Next we establish the smoothing effect for the inhomogeneous problem associated to (2.2). Proposition 2.2.4. Let T > 0 and > 0 , such that 0 < π/T . Then there exists c > 0
independent of such that
kD1/2 Z t 0 U(t− t0)F (·, t0)dt0kL∞TL2 x ≤ cT 1/2kF k l1 j(L2(Qj)).
Proof. By duality it is equivalent to prove
sup G Z R D1/2 Z t 0 U(t− t0)F (·, t0)(x)dt0 ! G(x)dx ≤ckF kl1 j(L2(Qj))
where the supremum is taken over all G∈ L2(R), such that kGk
Parserval’s identity and the definition of U(t) it follows Z R D1/2 Z t 0 U(t− t0)F (·, t0)(x)dt0 ! G(x)dx = Z R Z t 0 D1/2U(t− t0)F (·, t0)(x)dt0 ! G(x)dx = Z t 0 Z R D1/2U(t− t0)F (·, t0)(x)G(x)dx ! dt0 = Z t 0 Z R|ξ| 1/2e−(t−t0)(i−)ξ2 \ F (·, t0)(ξ) bG(ξ)dξdt0 = Z t 0 Z R F (x, t0)D1/2E (t− t0)U (t0− t)G(x)dxdt0. (2.41)
Splitting the integral with respect to the variable x in (2.41) into a sum of integrals over the intervals [j, j + 1] we obtain Z R D1/2 Z t 0 U(t− t0)F (·, t0)(x)dt0 ! G(x)dx ≤ X j∈Z Z j+1 j Z t 0 |F (x, t0)||D1/2E (t− t0)U (t0− t)G(x)|dt0dx. (2.42)
Using H¨older’s inequality we have that (2.42) is bounded by X j∈Z kF kL2([j,j+1]×[0,t])kD1/2E(t− ·)U(· − t)GkL2([j,j+1]×[0,t]) ≤ sup j∈Z kD 1/2E (t− ·)U(· − t)GkL2([j,j+1]×[0,t]) X j∈Z kF kL2([j,j+1]×[0,t]).
Notice that by the change of variables s7−→ t − t0 and Lemma 2.2.3
kD1/2E (t− ·)U(· − t)GkL2([j,j+1]×[0,t]) = kD1/2E(s)U (−s)GkL2([j,j+1]×[0,t]) = kE(s)D1/2U (−s)GkL2([j,j+1]×[0,t]) . kD1/2U (−s)Gk lj∞(L2(Q j)) = sup k∈Z kD1/2U (−s)Gk L2([k,k+1]×[0,T ]) ≤ kD1/2 U (−s)GkL∞x L2 T . kGkL2.
Then we conclude that Z R Z t 0 U(t− t0)F (·, t0)(x)dt0 ! G(x)dx . kFkl1 j(L2(Qj))
Proposition 2.2.5. Let T > 0 and > 0 , such that 0 < π/T . Then there exists c > 0
independent of such that k∂x
Z t
0
U(t− t0)F (·, t0)dt0kl∞j (L2(Q
j)) ≤ ckF kl1j(L(Qj)).
To prove this proposition we basically follow the ideas presented by Kenig, Ponce e Vega [17]. Our additional effort consists basically in proving the following:
Lemma 2.2.6. For each w ∈ C define the function fw(x) =
x
x2+ w2 . Then there exists a
constant c > 0 such that
kcfwkL∞ ≤ c
for all w∈ C.
Proof. We consider three cases
Case 1: Re(w) > 0.
Consider for each ξ ∈ R the complex function
F+(z) =
ei|ξ|z z2+ w2
and Γ+R the boundary of
DR+={z ∈ C; |z| < R, Im(z) > 0}.
Since F+ is holomorphic in D+R\ {iw} , we have from the Residue Formula
Z Γ+R F+(z)dz = 2πiRes(F+, iw) = 2πie −w|ξ| 2iw = π we −w|ξ|.
On the other hand, Z Γ+R F+(z)dz = Z R −R eiξx x2 + w2dx + Z π 0 F+(γR(θ))γR0 (θ)dθ = IR++ IIR+
where γR is the curve θ∈ [0, π] 7−→ Reiθ. Let us estimate the integral IIR+.
IIR+ = Z π 0
ei|ξ|R(cos θ+i sin θ)iReiθ
(Reiθ)2 + w2 dθ ≤ Z π 0 e−|ξ|R sin θR |(Reiθ)2+ w2|dθ ≤ Z π 0 e−|ξ|R sin θR |R2− |w|2|dθ. (2.43)
Taking R >√2|w| we have that the integral (2.43) is less than 1 2R2 Z π 0 e−|ξ|R sin θRdθ.
Since sin θ is positive in the range 0 < θ < π we conclude
|II+
R| ≤
π
2R and then IIR+ −→ 0 as R −→ +∞. Therefore
Z R eix|ξ| x2+ w2dx = R→+∞lim I + R + II + R = lim R→+∞ Z Γ+R F+(z)dz = π we −w|ξ|
To finish this case notice that Z R cos(ξx) x2 + w2dx = Z R cos(|ξ|x) x2+ w2 dx
because the function cos(ξx)
x2+ w2 is even and Z R sin(ξx) x2+ w2dx = 0 = Z R sin(|ξ|x) x2+ w2 dx
because the function sin(ξx)
x2+ w2 is odd. Then Z R e−iξx x2+ w2dx = Z R cos(ξx) x2+ w2dx− i Z R sin(ξx) x2+ w2dx = Z R cos(|ξ|x) x2+ w2 dx + i Z R sin(|ξ|x) x2+ w2dx = Z R ei|ξ|x x2+ w2dx = π we −w|ξ|.
We have just proven that
n 1 x2+ w2 o∧ (ξ) = π we −w|ξ|.
Then using the relation between Fourier transform and differentiation we finally obtain c fw(ξ) = x x2+ w2 ∧ (ξ) = i d dξ 1 x2+ w2 ∧ (ξ) = iπ w d dξ e−w|ξ| = −iπsgn(ξ)e−w|ξ|.
