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Orbital stability of standing waves for the nonlinear Schr¨ odinger equation with attractive delta potential and double power repulsive nonlinearity

Ram´on G. Plaza

Institute of Applied Mathematics (IIMAS)

Universidad Nacional Aut´onoma de M´exico (UNAM) July 3, 2019

ZumbrunFest - Stability of Nonlinear Waves: Analysis and Computation. Institut Henri Poincar´e.

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Joint work with: Jaime Angulo Pava(Univ. Sao Paulo) and C´esar Hern´andez Melo(Univ. Maring´a)

Sponsors:

• DGAPA-UNAM, program PAPIIT, grant no. IN-100318.

• FAPESP, S˜ao Paulo, processo 12543-4.

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Table of contents

1. Introduction

2. Local and global well posedness for the NLS 3. Existence of standing waves

4. Orbital stability

2

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Introduction

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Nonlinear Schr¨ odinger equation (NLS)

NLS equation withattractive delta potentialandrepulsive double power nonlinearity:

iut+uxx+Zδ(x)u+λ1u|u|p−12u|u|2p−2= 0

• Unknown: u=u(x,t)∈C, forx,t∈R.

• Parameters: λ1≤0,λ2<0,Z>0,p>1.

• δ :H1(R)→C,hδ,gi=g(0) (Dirac deltacentered atx= 0.)

• Linear interaction: ∂x2+Zδ(x).

• Nonlinear term: λ1u|u|p−12u|u|2p−2.

• i2=−1.

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Physical applications

• We recall that the general NLS model

iut+uxx+V(x)u+f(|u|2)u= 0,

represents a trapping (wave-guiding) structure for light beams, induced by an inhomogeneity of the local refractive index.

• The delta-function termV(x) =Zδ(x) represents anarrow trap which is able to capture broad solitonic beams.

• It models a spatially localized point defectof the medium in which the soliton travels (localized attractive “impurity”).

• The non linear termf(x) =λ1x(p−1)/22xp−1is well known in optical media.

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Standing waves

Standing wavesare solutions to the NLS model of the form u(x,t) =e−iωtφ(x),

whereω∈Rand theprofile of the waveφ:R→Rsatisfies (

φ00+Zδ(x)φ+ω φ+f(|φ|2)φ= 0,

φ∈H1(R), (ODE)

wheref =f(·) is an arbitrary function satisfying f ∈C1((0,+∞);R) with f(0) = 0,

f0(x)<0 for allx>0. (Hf)

Example: if 1<p<∞,λ1≤0 andλ2<0 then f(x) =λ1x(p−1)/22xp−1 satisfies (Hf).

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Standing waves

Standing wavesare solutions to the NLS model of the form u(x,t) =e−iωtφ(x),

whereω∈Rand theprofile of the waveφ:R→Rsatisfies (

φ00+Zδ(x)φ+ω φ+f(|φ|2)φ= 0,

φ∈H1(R), (ODE)

wheref =f(·) is an arbitrary function satisfying f ∈C1((0,+∞);R) with f(0) = 0,

f0(x)<0 for allx>0. (Hf) Example: if 1<p<∞,λ1≤0 andλ2<0 then

f(x) =λ1x(p−1)/22xp−1 satisfies (Hf).

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δ -interaction quantum operator

Theδ-interaction quantum operatorAZ is defined as AZ:=−∂x2−Zδ(x) (AZf(x) =−f00(x), x6= 0,

D(AZ) ={f ∈H1(R)∩H2(R\{0}) :f0(0+)−f0(0−) =−Zf(0)},

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Orbital stability

Definition (orbital stability)

The standing wavee−iωtφω is orbitally stableby the flow of the NLS equation nH1(R), if for anyε>0 there existsδ>0 such that if ku0−φωkH1<δ then

inf

θ∈R

ku(t)−eφωkH1<ε, for all t∈R,

whereu(t) denotes the solution to the NLS equation with initial data u(0) =u0∈H1(R). Otherwise,e−iωtφω is said to beorbitally unstablein H1(R).

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Main theorem

Theorem (Angulo Pava, Hernandez Melo, P (2019))

Let1<p<∞,λ1≤0,λ2<0and Z>0 in the NLS equation. Then for all values ofω<0 satisfying

− pλ12 (p+ 1)2λ2

<−ω<Z2 4

the family of standing wave solutions, u(x,t) =e−iωtφω, withφω given by

φω= α

−ω +

√v

−ωsinh

(p−1)√

−ω

|x|+R11 Z

2√

−ω

p−11

where v=ω β−α2, areorbitally stablesolutions in H1(R)under the flow of the NLS equation.

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Previous results

• CaseZ = 0: Ohta (1995), double power nonlinearity;Maeda (2008), multiple power nonlinearity.

• Repulsive δ potential: Fukuizumi, Jeanjean (2008)

• Attractiveδ potential: Fukuizumi et al. (2008)

• Kaminaga, Ohta (2009): attractiveδ with repulsive single power nonlinearity.

Multibody interactions ofsame sign(repulsive, double power

nonlinearity) appear in the study ofBose-Einstein condensates: Brazhnyi, Konotop (2004); Belobo Belobo et al. (2014); Kamchatnov, Salerno (2009); Kamchatnov, Korneev (2010) (dark solitons).

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Concentration-compactness method: Cazenave, Lions (1982)

• The Cauchy problem: The initial value problem associated to the NLS equation isglobally well-posedinH1(R) for 1<p<+∞, λ1≤0,λ2<0 andZ>0.

• Existence of profile solution φω: for parameter valuesp,λ12,Z andω satisfying the assumptions there exists a profile solutionφω of the elliptic equation (ODE) (explicit construction).

• The stationary problem: The setAω ofnon-trivial solutionsof the equation for the profiles inH1(R) will be characterized, via

uniqueness, by

Aω={v:Gω0(v) = 0,v6= 0}={eφω:θ∈R}, where

Gω(v) =E(v)−ω 2kvk2L2, E(v) :=1

2kvxk2L2−Z

2|v(0)|2− λ1

p+ 1kvkp+1

Lp+1−λ2

2pkvk2p

L2p, forv∈H1(R).

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Concentration-compactness method (ii)

• The minimization problem: Forp,λ12,Z andω satisfying the assumptions of our main theorem, the quantity

m(ω) = inf{Gω(v) :v∈H1(R)}, satisfies the following properties:

(a) (boundedness below) −∞<m(ω)<0; and, (b) (compactness) any sequencehn∈H1(R) such that

limn→∞Gω(hn) =m(ω) admits a subsequence converging to some h∈H1(R) withGω(h) =m(ω).

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Local and global well posedness

for the NLS

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Preliminaries (i)

The formal expressionAZ :=−∂x2−Zδ(x) represents all the self-adjoint extensions (von Neumann theory) associated to the following closed, symmetric, densely defined linear operator:

( A0=−∂x2

D(A0) ={g∈H2(R) :g(0) = 0}.

More precisely, the quantum operatorAZ =−∂x2−Zδ(x) is given by (AZf(x) =−f00(x) x6= 0,

D(AZ) ={f ∈H1(R)∩H2(R\{0}) :f0(0+)−f0(0−) =−Zf(0)}.

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Preliminaries (ii)

Upon application of theFirst Representation Form Theorem(cf. Kato), it is possible to show that the associated form toAZ is given by

FZ[u,v] = Re Z +

−∞

u0(x)v0(x)dx−ZRe(u(0)v(0)),

where (u,v)∈D(FZ) =H1(R)×H1(R). The bilinear form defined above is closed and bounded below. In addition, operatorAZ =−∂x2−Zδ(x) can be extended as a linear bounded operatoru→AZufromH1(R) to H1(R). This action is defined by

hAZu,vi=FZ[u,v], for u,v∈H1(R).

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Spectral properties of A

Z

• Essential spectrum: Σess(AZ) = [0,+∞), for allZ∈R.

• Discrete spectrum:

Σdis(AZ) = (

∅, Z≤0, −Z2/4 , Z>0, ForZ>0, ΨZ(x) =

qZ

2eZ2|x| is thenormalized eigenfunction

associated to theunique negative simple eigenvalue−Z2/4. In addition, the operatorsAZ are bounded from below:

(AZ≥ −Z2/4, Z>0, AZ≥0, Z<0 Ref.: Albeverio, Gesztesy, Høegh-Kron (2005).

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The Cauchy problem

Consider theCauchy problem,

( iut−AZu+ (λ1|u|p−12|u|2p−2)u= 0, u(0) =u0∈H1(R).

Ref. Cazenave, Courant LN, vol. 10 (2003).

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Local well-posedness

Theorem (local well-posedness)

For any u0∈H1(R)and Z∈R, there exists T>0and a unique solution u∈C([−T,T];H1(R))∩C1([−T,T];H1(R))to the NLS equation with u(0) =u0such that

lim

t→Tku(t)kH1= +∞, if T <∞.

Moreover, the solution u(t)satisfies conservation of charge and energy:

ku(t)kL2=ku0kL2, E(u(t)) =E(u0), for all t∈[−T,T], where the energy functional E is defined as

E(v) :=1

2kvxk2L2−Z

2|v(0)|2− λ1

p+ 1kvkpL+1p+1−λ2

2pkvk2Lp2p, for v∈H1(R).

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Proof sketch

• The nonnegative self-adjoint operatorA ≡AZ+β on the space X =L2(R), withβ =Z2/4 for Z>0 andβ= 0 forZ ≤0, and domainD(A) =D(AZ), induces a norm

kuk2X

A =kuxk2L2+ (β+ 1)kuk2L2−Z|u(0)|2, which is equivalent to the usual norm inH1(R).

• The self-adjoint operator AZ generates a strongly continuous group of unitary operatorsT(t)g=e−itAZg.

• Duhamel integral u(t) =T(t)u0+

Z t 0

T(t−s)(λ1|u(s)|p−1u(s) +λ2|u(s)|2p−2u(s))ds Direct application of Thm. 3.7.1 in Cazenave.

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Remark: Gagliardo-Nirenberg interpolation inequality

For anyp>1,H1(R)⊂L2(R)∩Lp+1(R)∩L2p(R), inasmuch as the Gagliardo-Nirenberg interpolation inequality(cf. Leoni, 2017) yields

kukLp+1≤C1kukθL12kuxk1L−θ2 1, kukL2p≤C2kukθ2

L2kuxk1−θ2

L2 ,

with uniform constantsCj>0 andθ1= (p+ 2)/(2p+ 2)∈(0,1), θ2= (p+ 1)/2p∈(0,1).

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Conservation of charge/energy

• Conservation of charge:

d

dt||u(t)||2L2= 2Re Z

R

utudx

= 2Re Z

R

−i(AZu)u+if(|u|2)|u|2dx= 0.

• Conservation of Energy: NLS equation can be written in the Hamiltonian form ut=−iE0(u(t)), then

d

dtE(u(t)) = Re Z

R

E0(u(t))utdx= Re Z

R

i|E0(u(t))|2dx= 0.

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Auxiliary bound

Let us define the followingC1functional inH1(R), R(v) :=1

2kvxk2L2−Z

2|v(0)|2−λ2

2pkvk2p

L2p =E(v) + λ1

p+ 1kvkp+1

Lp+1.

Lemma (Auxiliary bound)

Let1<p<∞,λ1≤0,λ2<0and Z>0. Then there exists a uniform constant C=C(p,Z)>0such that

Z

2|v(0)|2≤R(v) +C, for all v∈H1(R).

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Proof sketch

By Sobolev and Young’s inequalities, for anyZ>0 there exists C1=C1(Z)>0 such that forv∈H1(R)

Z|v(0)|2≤1

2kvxk2L2+C1kvk2L2(1,1). Apply H¨older’s and Young’s inequalities to estimate

kvk2L2(1,1)≤2(p−1)/p Z 1

1

|v|2pdx 1/p

≤δkvk2p

L2p(1,1)+ 2Cδ, for anyδ >0. Since λ2<0, chooseδ =−λ2/(2pC1)>0 to obtain

Z|v(0)|2≤1

2kvxk2L2−λ2

2pkvk2p

L2p+ 2C1Cδ.

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Global well-posedness

Theorem (global well-posedness)

For every p>1, Z>0,λ1≤0andλ2<0the Cauchy problem is globally well-posed in H1(R).

Proof. Letu∈C([−T,T];H1(R))∩C1([−T,T];H1(R)) be the local solution to the Cauchy problem fort∈(−T,T).

1

2kuxk2L2=E(u) +Z

2|u(t)|2+ λ1

p+ 1kukp+1

Lp+12

2pkuk2p

L2p

≤E(u(t)) +Z 2|u(t)|2

≤E(u(t)) +R(u(t)) +C

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Thus, we arrive at 1

2kux(t)k2L2≤E(u(t)) +R(u(t)) +C≤2E(u(t)) +C.

In view thatu conserves charge and energy we finally conclude that ku(t)k2H1≤4E(u(0)) +ku(0)k2L2+ 2C,

which implies, together with lim

t→Tku(t)kH1= +∞, if T<∞, that the time of existence of the solutionuisT = +∞.

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Existence of standing waves

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ODE problem

Recall theprofile equation (

φ00+Zδ(x)φ+ω φ+f(|φ|2)φ= 0,

φ∈H1(R), (ODE)

Hypothesis onf:

f ∈C1((0,+∞);R) with f(0) = 0,

f0(x)<0 for allx>0. (Hf) φ∈H1(R) is a solution in thedistributional senseif for everyχ∈H1(R)

0 = Re Z +

−∞ φ0(x)χ0(x)dx−Zφ(0)χ(0)−ω Z +

−∞

φ(x)χ(x)dx

− Z +

−∞

f(|φ|2(x))φ(x)χ(x)dx

.

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Analysis of the (ODE) (i)

Lemma

Letφ∈H1(R), withφ00+Zδ(x)φ+ω φ+f(|φ(x)|2)φ(x) = 0in the distributional sense, then

φ∈Cj(R\{0})∩C(R), j= 1,2. (1a) φ00(x) +ω φ(x) +f(|φ(x)|2)φ(x) = 0, for x6= 0. (1b)

φ0(0+)−φ0(0−) =−Zφ(0). (1c)

φ0(x),φ(x)→0, if |x| →∞. (1d)

0(x)|2+ω|φ(x)|2+g(|φ(x)|2) = 0, for x6= 0. (1e) where g(s) =

Z s 0

f(s)ds.

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Analysis of the (ODE) (ii)

Lemma

Let p>1,ω,λ12∈Rand Z∈R\ {0}. Letφ be a non-trivial solution to(1a) -(1e). Thenφ(x)6= 0for all x∈Rand|φ|>0. −φ is also a solution.

Lemma (Useful)

Let p>1,ω,λ12∈Rand Z∈R\ {0}. Letφ be a non-trivial solution to(1a) -(1e). Then we have either one of the following:

(i) Im(φ(x)) = 0 for all x∈R; or,

(ii) there exists c∈Rsuch that Re(φ(x)) =c Im(φ(x))for all x∈R.

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Explicit profile construction for ω 6= 0, Z = 0, λ

1

≤ 0, λ

2

< 0

By usingφ,φ0→0 asx→∞we obtain

0]2+ω φ2+ 2α φp+1+β φ2p= 0, withα=λ1/(p+ 1),β=λ2/p. Then,

φ(x) =

"

−α ω +

pω β−α2

ω sinh (p−1)√

−ωx

#p−11

,

is the profile of the standing wave solution provided that

− pλ12 (p+ 1)2λ2

<−ω.

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Explicit construction

The function φ1(x) :=φ(−|x| −d), −l<d,

satisfies all the properties of our first lemma except possibly the jump condition: φ0(0+)−φ0(0+) =−Zφ(0). If we consider

R1: (−l,∞)→(1,∞) the diffeomorphism defined by R1(d) =

pω β−α2cosh((p−1)√

−ωd) pω β−α2sinh((p−1)√

−ωd) +α .

then, we get d=R11

Z 2√

−ω

, with Z>0 and −ω<Z2 4 .

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Profile existence theorem; case ω 6= 0

Theorem

Let p>1,λ1≤0,λ2<0and Z>0 in the NLS equation. Then for all values ofω<0satisfying

− pλ12 (p+ 1)2λ2

<−ω<Z2 4

the familiy of standing wave solutions, u(x,t) =e−iωtφω, withφω given by

φω= α

−ω +

√v

−ωsinh

(p−1)√

−ω

|x|+R11 Z

2√

−ω

p−11

are solutions to the NLS equation. Here v=ω β−α2.

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���

���

���

���

���

ϕω

Figure 1: Profile functionφωω(x) for parameter valuesω=−0.25,Z= 2, p= 3,λ12=−1.

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Figure 2: Time evolution of the standing wave solutionu(x,t) =e−iωtφω(x) withω=−0.25,Z= 2 and in the case of a quintic/cubic (p= 3), doubly repulsive (λ12=−1) nonlinearity.

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���

���

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Figure 3: Dynamics in the (φ,φ0)-plane forf(x) =−x(1 +x2), that is, for λ12=−1 andω=−0.25, in the case of a quintic/cubic nonlinearity with p= 3.

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Orbital stability

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Critical points

Let us consider the functionalGω:H1(R)→Rfor valuesω≤0, defined as

Gω(v) =1

2kvxk2L2−Z

2|v(0)|2−ω

2kvk2L2−1 2

Z

−∞

g(|v(x)|2)dx, and theset of critical pointsassociated toGω as

Aω={v∈H1(R) :Gω0(v) = 0,v6= 0}.

Hereg =g(·) is the antiderivative off =f(·). Forφ∈Aω we have the relation

Gω0(φ) =AZφ−ω φ−f(|φ|2

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Properties of the set of critical points (i)

Lemma

Let1<p<∞, Z>0and let ω∈Rbe such that ω+Z42 ≤0. Then the setAω is empty.

Proof. If there existsh∈H1(R)\ {0} satisfyingGω0(h) = 0, then 0 = d

dsG(sh)

s=1, and since hAZh,hi ≥ −Z2 4 khk2L2

for allh∈H1(R), we then obtain

0 =khxk2L2−Z|h(0)|2−ωkhk2L2− Z

−∞

f(|h(x)|2)|h(x)|2dx

≥ −(Z2/4 +ω)khk2L2− Z

−∞

f(|h(x)|2)|h(x)|2dx

≥ − Z

−∞

f(|h(x)|2)|h(x)|2dx>0,

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Properties of the set of critical points (ii)

Lemma

Let1<p<∞and Z∈R. Ifω>0then Aω=∅. Lemma

Letω∈Rand Z<0. Then, we have thatAω=∅.

Proofs by contradiction.

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Properties of the set of critical points (ii)

Lemma

Let1<p<∞and Z∈R. Ifω>0then Aω=∅. Lemma

Letω∈Rand Z<0. Then, we have thatAω=∅. Proofs by contradiction.

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Properties of the set of critical points (iii)

Lemma

Let p>1,λ1<0,λ2<0, Z >0andω such that −(p+1)122

λ2 <−ω<Z42. Considering f(x) =λ1x(p−1)/22xp−1, then

Aω={eφω:θ∈R}.

Proof. It is clear that for allθ∈R, eφω∈Aω. Conversely, ifg∈Aω, theng satisfies all the necessary conditions to be a solution of the Euler-Lagrange equation and|g|>0. Goal: to show that there exist θ∈Rsuch that g(x) =eφω(x) for allx∈R.

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• φω∈D(AZ) is the unique positive solution of the Euler-Lagrange equation. Indeed, ifv∈H1(R) is a positive solution thenv satisfies the IVP

(−ψ00(x) =ω ψ(x) +f(ψ2(x))ψ(x) :=H(ψ(x)), x>0, ψ(0) =c0, ψ0(0) =−Zc0/2,

where c0 is the unique positive root of Φω(c,Zc/2) =Z2

4 c2+ωc2+g(c2).

SinceHis locally Lipschitz around zero the IVP has a unique positive solution given byφω. Thus,v≡φω on (0,∞). Similar arguments show that v≡φω on (−∞,0). Hence,v(x) =φω(x) for allx∈R.

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• Ifg(x) =e(x)ρ(x) then θ,ρ>0 satisfy

( θ00ρ+ 2θ0ρ0= 0, x>0,

−(θ0)2ρ+ρ00+ω ρ+f(|ρ|2)ρ= 0, x>0.

The first equation together with the boundedness of|g0|imply that g(x) =e0ρ(x) for allx∈(0,+∞). Then, from second equation and by the analysis above we necessarily have thatg(x) =e0φω(x) for allx∈(0,∞). A similar analysis shows thatg(x) =e1φω(x) for all x∈(−∞,0). Hence,

g(x) =e0φω(x), for allx∈R.

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The minimization problem

Let us suppose that 1<p<∞,Z>0,λ1≤0,λ2<0,ω is such that

− pλ12 (p+ 1)2λ2

<−ω<Z2

4 , and f(x) =λ1x(p−1)/22xp−1. Minimization problemassociated toGω:

m(ω) = inf{Gω(v) :v∈H1(R)}, and theminimal set

M(ω) ={u∈H1(R) :Gω(u) =m(ω)}.

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The set of minima

Lemma

−∞<m(ω)<0and M(ω)⊂Aω.

Proof. First verify that−∞<m(ω). Write Gω(v) =R(v)−ω

2kvk2L2− λ1

p+ 1kvkp+1

Lp+1, v∈H1(R).

Then, by the auxiliary bound lemma we get Gω(v)≥R(v)≥Z

2|v(0)|2−C≥ −C,

for allv∈H1(R) and some uniformC>0, yielding−∞<m(ω).

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To show thatm(ω)<0, letv(x) :=sh(x)∈H1(R) withs>0 and where h(x) =eZ|x|2 is the eigenfunction of the operatorAZ associated to the eigenvalue −Z42. Therefore

Gω(v) =−s2 2

Z2 4 +ω

khk2L2−1 2

Z

−∞

g(s2h2(x))dx.

Since −g(s2h2(x))<−f(s2)s2h2(x), Gω(v)≤ −s2

2khk2L2

Z2

4 +ω+f(s2) .

SinceZ2/4 +ω>0 and lims→0+f(s2) = 0 we conclude that there exists s0>0 such thatZ2/4 +ω>−f(s2)>0 for 0<s≤s0and so

Gω(s0h)<0. Lastly, supposeM(ω)6=∅. Then since forh∈M(ω) we haveh6= 0 andGω0(h) = 0, then by previous Lemmata we obtain M(ω)⊂Aω.

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Auxiliary result: Br´ ezis-Lieb lemma

A refinement of Fatou’s lemma:

Lemma (Br´ezis-Lieb, 1983)

Let2≤q<∞and{uj} be a bounded sequence in Lq(R)such that uj(x)→u(x)a.e. in x∈Ras j→∞. Then,

kujkqLq− kuj−ukqLq− kukqLq→0, as j→∞.

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Compactness

Lemma

Let hn∈H1(R)be such thatlimn→∞Gω(hn) =m(ω). Then there exists a subsequence hnj and h∈H1(R)such that limnj→∞hnj=h in H1(R)and Gω(h) =m(ω).

Proof. First, notice that for allv∈H1(R) Iω(v) :=1

2kvxk2L2−ω 2kvk2L2

=Gω(v) +Z

2|v(0)|2+ λ1

p+ 1kvkp+1

Lp+12

2pkvk2p

L2p. Sinceω<0, it follows that Iω(v) is equivalent tokvk2H1. From the fact thatλ12<0, we obtain

1

2kvxk2L2−ω

2kvk2L2≤Gω(v) +R(v) +C≤2Gω(v) +C, for some uniformC>0.

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Hence, it is clear that if the sequenceGω(hn) converges then the sequencehn is bounded inH1(R). Thus, there exists a subsequencehnj andh∈H1(R) such that {hnj}converges wealky tohinH1(R). Since H1(−1,1) is compactly embedded in C[−1,1], we deduce that

hnj(0)→h(0). Thus,

m(ω)≤Gω(h)≤lim inf

nj→∞Gω(hnj) =m(ω), which implies thath∈M(ω).

Now, sincehnj*hweakly inH1(R) we have that hnj(x)→h(x) a.e. in x∈Rand also that

khnj−hk2L2+khk2L2=khnjk2L2+o(1), k∂xhnj−hxk2L2+khxk2L2=k∂xhnjk2L2+o(1), asnj→∞.

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khnjkH1 uniformly bounded⇒ khnjkLp+1 andkhnjkL2p are uniformly bounded (by Gagliardo-Nirenberg interpolation inequalities). As hnj(x)→h(x) a.e. inx∈R, by Br´ezis-Lieb lemma we get

khnj−hkp+1

Lp+1+khkp+1

Lp+1=khnjkp+1

Lp+1+o(1), khnj−hk2p

L2p+khk2p

L2p=khnjk2p

L2p+o(1), asnj→∞.

Combining yields

Gω(hnj−h) +Gω(h) =Gω(hnj) +o(1), as nj→∞.

From the def. ofIω,

0≤Iω(hnj−h)≤Iω(hnj−h)− λ1

p+ 1khnj−hkp+1

Lp+1−λ2

2pkhnj−hk2p

L2p

=Gω(hnj−h) +Z

2|hnj(0)−h(0)|2

=Gω(hnj)−Gω(h) +o(1),

inasmuch ashnj(0)→h(0). This yieldshnj→hinH1(R).

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Characterization of the minimal set

Lemma

M(ω) =Aω={eφω:θ∈R}, whereφω denotes the standing wave profile.

Proof. From the previous lemmas, we infer thatM(ω)6=∅. Then there existsh∈H1(R) such thatGω(h) =m(ω), that is,h∈M(ω). Since M(ω)⊂Aω, h∈Aω. Thus, there existsθ0∈Rsuch thath=e0φω. Now, sinceφω∈H1(R) and

Gωω) =Gω(h) =m(ω),

thenφω∈M(ω). This implies thatAω⊂M(ω). The other inclusion was already proved above.

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Proof of the main theorem

Suppose that the standing wavee−iωtφω is orbitally unstable. Then there existsε0>0, a sequence{hn(t)}of solutions of the NLS equation and a sequencetn>0, such that

n→∞limkhn(0)−φωkH1= 0, (2a)

θ∈infR

khn(tn)−eφωkH1≥ε0. (2b) SinceGω is conserved by the flow of the NLS equation, we get that Gω(hn(tn)) =Gω(hn(0)) for alln∈N. Then (2a) and continuity ofGω yield

n→∞limGω(hn(tn)) =Gωω) =m(ω).

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Henceforth, from the former results there exists a subsequencehnj such thathnj(tnj)→hwithGω(h) =m(ω). Thenh∈Aω andh=e0φω for someθ0∈R. Therefore,

nlimj→∞hnj(tnj) =e0φω, inH1(R), which contradicts

θ∈infR

khn(tn)−eφωkH1≥ε0. Hence, we conclude thate−iωtφω is orbitally stable.

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Reference

• J. Angulo Pava, C. A. Hern´andez Melo, R. G. P.,J. Math. Phys.

(2019), in press.

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Happy birthday Kevin!

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Figure

Figure 1: Profile function φ ω = φ ω (x ) for parameter values ω = −0.25, Z = 2, p = 3, λ 1 = λ 2 = −1.
Figure 2: Time evolution of the standing wave solution u(x,t) = e −iωt φ ω (x ) with ω = −0.25, Z = 2 and in the case of a quintic/cubic (p = 3), doubly repulsive (λ 1 = λ 2 = −1) nonlinearity.
Figure 3: Dynamics in the (φ, φ 0 )-plane for f (x ) = −x (1 + x 2 ), that is, for λ 1 = λ 2 = −1 and ω = −0.25, in the case of a quintic/cubic nonlinearity with p = 3.

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