Nonlinear Schrödinger equation under a magnetic field kept ... · Nonlinear Schrödinger equation...

Nonlinear Schrödinger equation under amagnetic field kept in the foreground

Jean VAN SCHAFTINGEN

Institut de Recherche enMathématique et Physique

Workshop in Nonlinear PDEsBrussels, September 7–11, 2015

J. Van Schaftingen (UCLouvain) Magnetic nonlinear Schrödinger Brussels, 2015-09-11 1 / 23

Nonlinear Schrödinger equation when themagnetic field is kept in the foreground

1 The magnetic nonlinear Schrödinger equation

2 Semiclassical results

3 Constant magnetic field problem


Magnetic nonlinear Schödinger functional

∫

Ω

ϵ2|DA/ϵu|2

2+

V|u|2

2−|u|p

p

where

u : Ω→ C ' R2 is a wave-function,−|u|p−2 is an interaction by δ-functions,V is an (adimensionalized) electric potential,ϵ is a characteristic length (adimensionalized Planckconstant),DA/ϵ is a covariant derivative.


Covariant derivative and connectionDifferential geometry meets physics

DAu4= Du− iuA,

where A : Ω→∧1

R3 ' R3.1 differential calculus:

DA(ηu) = ηDAu + uDη, D|u|2 = 2(u|DAu),

2 the curvature is the magnetic field B = dA ' ∇ × B,

KA[v,w]u4= DA(DAu[w])[v]− DA(DAu[v])[w]

= −idA[v,w]u ' −i(∇ × A|v × w)u,

3 gauge invariance: if A′4= A + dΛ ' A + ∇Λ and u′

4= eiΛu,

thenDA′u′ = eiΛDAu.


Magnetic nonlinear Schrödinger equation

The Euler–Lagrange equation associated to

∫

Ω

ϵ2|DA/ϵu|2

2+

V|u|2

2−|u|p

p

is−ϵ2∆A/ϵu + Vu = |u|p−2u,

where the magnetic Laplacian is defined as




∫

Ω

ϵ2|DA/ϵu|2

2+

V|u|2

2−|u|p

p



−ϵ2∆A/ϵu = −ϵ2∆u + 2iϵ⟨A,Du⟩+ (|A|2 − id∗A)u

' −ϵ2∆u + 2iϵA · ∇u + (|A|2 − i∇ · A)u.




∫

Ω

ϵ2|DA/ϵu|2

2+

V|u|2

2−|u|p

p



−ϵ2∆A/ϵu4= −ϵ2∆u + 2iϵ⟨A,Du⟩+ |A|2u

' −ϵ2∆u + 2iϵA · ∇u + |A|2u,

in the Coulomb/Lorenz gauge (d∗A ' ∇ · A = 0).


Existence of groundstatesSeminal work of Esteban and Lions

Groundstate solutions are minimizers of∫

Ω

ϵ2|DA/ϵu|2 + V|u|2

∫

Ω

|u|p

2p

.

Theorem (M. Esteban and P.-L. Lions, 1989)Groundstates exist when either

1 Ω is bounded,2 Ω = R3, and V and dA are constant,3 a strict inequality is satisfied.


Magnetic field going into the backgroundMagnetic field in the background

Theorem (Kurata (2000))

If 12 −

13 <

1p <

12 and V0

4= infV < lim inf∞V, then there exist

(xϵ)ϵ>0 in R3 and (ωϵ)ϵ>0 such that V(xϵ)→ V0 and

uϵ(x) ' αϵUx− xϵ

ϵ

ei

A(xϵ)[x−xϵ ]

ϵ +ωϵ

,

where U is the unique positive radial groundstate of thelimiting problem

−∆U + V0U = |U|p−2U.

See also Cingolani (2003), Cingolani & Secchi (2002), Secchi &Squassina (2005), Cao & Tang (2006), Barile (2008), Cingolani,Jeanjean & Secchi (2009), Cingolani & Clapp (2009, 2010), Barile,Cingolani & Secchi (2011), Ding & Liu (2013). . .




If 12 −

13 <

1p <

12 and V0




ϵ

ei

A(xϵ)[x−xϵ ]

ϵ +ωϵ

,


−∆U + V0U = |U|p−2U.

The behavior of the concentration points and of the densitiesare independent of the magnetic field:

|uϵ(x)|2 '

Ux− xϵ

ϵ

2.




If 12 −

13 <

1p <

12 and V0




ϵ

ei

A(xϵ)[x−xϵ ]

ϵ +ωϵ

,


−∆U + V0U = |U|p−2U.

Asymptoticaly consistent with the Lorentz force when v = 0:

0 = F = −q(dV + vùdA) ' q(−∇V + v × (∇ × A)).




If 12 −

13 <

1p <

12 and V0




ϵ

ei

A(xϵ)[x−xϵ ]

ϵ +ωϵ

,


−∆U + V0U = |U|p−2U.

Question: How to have interaction with the magnetic field inthe semiclassical limit?


Magnetic semiclassical soliton dynamicsThe Lorentz force is validated in the dynamics

Theorem (Selvitella (2008) & Squassina (2009))

If 2 < p < 2 + 43 , then there exists solutions to

iϵ∂tψ = −ϵ2∆A/ϵψ+ Vψ − |ψ|p−2ψ,

of the form

ψϵ(x, t) ' Ux− x(t)

ϵ

ei

A(x(t))[x−x(t)]+x(t)·(x−x(t))

ϵ +ωϵ(t)

.

with

x(t) = −dV(x(t))− x(t)ùdA(x(t))

' −∇V(x(t)) + x(t)× (∇ × A)(x(t)).


Why is the magnetic field fading out?

By the Heisenberg uncertainty principle, a groundstate mighthave a magnetic momentum μ ∈

∧2R3 ' R3. It is thus subject

to a force of

−qdV + d⟨dA, μ⟩ ' −q∇V + ∇(∇ × A · μ).

There is no asymptotic magnetic effect:

q =

∫

R3|∇uϵ|2 ' ϵ3, μ =

∫

R3((x− xϵ)∧ (−iϵ∇A/ϵuϵ(x))|uϵ(x)) dx ' ϵ4.

Idea: Interesting phenomena should happen in the strongmagnetic field régime A ' ϵ−1.


The new problem

We study solutions of the Euler–Lagrange equation associatedto

∫

Ω

ϵ2|DA/ϵ2u|2

2+

V|u|2

2−|u|p

p

which is−ϵ2∆A/ϵ2u + Vu = |u|p−2u,

where

−ϵ2∆A/ϵu = −ϵ2∆u− 2i⟨A,Du⟩+|A|2 − id∗A

ϵ2u

' −ϵ2∆u− 2iA · ∇u +|A|2 − i∇ · A

ϵ2u.


The new problem

We study solutions of the Euler–Lagrange equation associatedto

∫

Ω

ϵ2|DA/ϵ2u|2

2+

V|u|2

2−|u|p

p

which is−ϵ2∆A/ϵ2u + Vu = |u|p−2u,

where

−ϵ2∆A/ϵu4= −ϵ2∆u + 2i⟨A,Du⟩+

|A|2

ϵ2u

' −ϵ2∆u + 2iA · ∇u +|A|2

ϵ2u,

in the Coulomb/Lorenz gauge (d∗A ' ∇ · A = 0).


The concentration function

For V ∈ R and B ∈∧2

R3 ' R3, define

A(x)[v] =B(x,v)

2'

(B× x) · v

2,

so that dA = B

and set

E(V,B) =1

2−

1

p

infu∈C∞c (R3;C)

∫

R3|DAu|2 + V|u|2

p

p−2

∫

R3|u|p

2

p−2

.

1 E is continuous,2 diamagnetic inequality:

E(V,B) ≥ E(V,0).




R3 ' R3, define

A(x)[v] =B(x,v)

2'

(B× x) · v

2,

so that dA = B and set

E(V,B) =1

2−

1

p

infu∈C∞c (R3;C)

∫

R3|DAu|2 + V|u|2

p

p−2

∫

R3|u|p

2

p−2

.

1 E is continuous,

2 diamagnetic inequality:

E(V,B) ≥ E(V,0).




R3 ' R3, define

A(x)[v] =B(x,v)

2'

(B× x) · v

2,

so that dA = B and set

E(V,B) =1

2−

1

p

infu∈C∞c (R3;C)

∫

R3|DAu|2 + V|u|2

p

p−2

∫

R3|u|p

2

p−2

.

1 E is continuous,2 diamagnetic inequality:

E(V,B) ≥ E(V,0).


Asymptotic result

Theorem (Di Cosmo and Van Schaftingen (2015))

If 12 −

13 <

1p <

12 and Ω is bounded, then there exist (xϵ)ϵ>0 in

R3 such thatlim

R→∞ϵ→0

‖uϵ‖L∞(Ω\BϵR(xϵ)) = 0,

limϵ→0

ϵ−NFϵ(uϵ) = limϵ→0

E(V(xϵ),dA(xϵ)) = infΩ

E(V,dA).

If V(xϵn)→ V∗, dA(xn)→ B∗ and A∗(x)[v] =B∗(x,v)

2 , up to asubsequence

uϵn ' UV∗,A∗

x− xn

ϵn

ei

A(xϵn )[x−xϵn ]

ϵ2n+ωn

,

with UV∗,A∗ a groundstate of

−∆A∗UV∗,A∗ + V∗UV∗,A∗ = |UV∗,A∗ |p−2UV∗,A∗ .

See also Fournais and Raymond (2015).


Asymptotic result

Theorem (Di Cosmo and Van Schaftingen (2015))

If 12 −

13 <

1p <

12 and Ω is bounded, then there exist (xϵ)ϵ>0 in

R3 such thatlim

R→∞ϵ→0

‖uϵ‖L∞(Ω\BϵR(xϵ)) = 0,

limϵ→0

ϵ−NFϵ(uϵ) = limϵ→0

E(V(xϵ),dA(xϵ)) = infΩ

E(V,dA).

If V(xϵn)→ V∗, dA(xn)→ B∗ and A∗(x)[v] =B∗(x,v)

2 , up to asubsequence

uϵn ' UV∗,A∗

x− xn

ϵn

ei

A(xϵn )[x−xϵn ]

ϵ2n+ωn

,

See also Fournais and Raymond (2015).


Forces in the semiclassical limit

If x∗ ∈ Ω and

E(V(x∗),dA(x∗)) = infΩ

E(V,dA),

and UV∗,A∗ is the corresponding groundstate, then

qdV(x∗) = d⟨dA, μ⟩(x∗),

with

q =

∫

R3|UV∗,A∗ |

2

and

μ =

∫

R3(y∧ (−i∇AUV∗,A∗(y))|UV∗,A∗(y)) dy.


About the proof

Challenges:1 uniqueness and nondegeneracy of solutions of the

limiting problem are unknown,

2 the coefficients of the rescaled problem are not locallyuniformly bounded coefficients,

3 natural function space of the limiting problem at x∗depends on x∗.


About the proof


limiting problem are unknown,2 the coefficients of the rescaled problem are not locally

uniformly bounded coefficients,

3 natural function space of the limiting problem at x∗depends on x∗.


About the proof



uniformly bounded coefficients,3 natural function space of the limiting problem at x∗

depends on x∗.


About the proof



uniformly bounded coefficients,3 natural function space of the limiting problem at x∗

depends on x∗.

Also covers local minimizers (del Pino & Felmer penalization),with slow decay at infinity

lim|x|→∞

V(x)|x|2 > 0.


The constant electromagnetic potential problem

What happens to the groundstate of

−∆A + Vu = |u|p−2u

when V = 1 and A(x)[v] = 12B[x,v] ' 1

2(B× x) · v for someB ∈

∧2R3?

How does the solution depend on B?

Is the solution symmetric?How does the solution decay?




−∆A + Vu = |u|p−2u

when V = 1 and A(x)[v] = 12B[x,v] ' 1


∧2R3?

How does the solution depend on B?Is the solution symmetric?

How does the solution decay?




−∆A + Vu = |u|p−2u

when V = 1 and A(x)[v] = 12B[x,v] ' 1


∧2R3?

How does the solution depend on B?Is the solution symmetric?How does the solution decay?


Free electron in a magnetic fieldThe first Landau level

Groundstate of the linear Schrödinger equation satisfy

−∆Aψ = Eψ.

If A(x)[v] = 12B(x,v) ' 1

2(B× x) · v, then a groundstate is givenby

ψ(x) = e−|B×x|2

4|B| ,

withE = |B|.

PropertiesE increases with |B|,

ψ is axially symmetric with respect to B,ψ has Gaussian decay perpendiculary to B.




−∆Aψ = Eψ.

If A(x)[v] = 12B(x,v) ' 1


ψ(x) = e−|B×x|2

4|B| ,

withE = |B|.

PropertiesE increases with |B|,ψ is axially symmetric with respect to B,

ψ has Gaussian decay perpendiculary to B.




−∆Aψ = Eψ.

If A(x)[v] = 12B(x,v) ' 1


ψ(x) = e−|B×x|2

4|B| ,

withE = |B|.

PropertiesE increases with |B|,ψ is axially symmetric with respect to B,ψ has Gaussian decay perpendiculary to B.


Groundstates under a small constant magneticfield

Theorem (Bonheure, Nys and Van Schaftingen, 2015)

If V ' 1 and |A|2 ® 1 then1 UA,V is real and radially symmetric,

2 UA,V(x) ® e−|xùB|24|B| ' e−

|B×x|24|B| ,

3 E(V,B) ' E(1,0) +V − 1

2

∫

R3|U|2 +

|B|2

4 · 3

∫

R3|y|2|U(y)|2 dy

.

“Essentially” means up to translations in R3 and rotation in C.





2 UA,V(x) ® e−|xùB|24|B| ' e−

|B×x|24|B| ,

3 E(V,B) ' E(1,0) +V − 1

2

∫

R3|U|2 +

|B|2

4 · 3

∫

R3|y|2|U(y)|2 dy

.

By scaling, we cover in fact |A|2/V ® 1.





2 UA,V(x) ® e−|xùB|24|B| ' e−

|B×x|24|B| ,

3 E(V,B) ' E(1,0) +V − 1

2

∫

R3|U|2 +

|B|2

4 · 3

∫

R3|y|2|U(y)|2 dy

.

In R2, the Gaussian decay follows from the linear theory (L.Erdos, 1996).


Proving the properties by essential uniqueness

Idea: Prove that when |A|2 ® 1, then the groundstate isessentially unique.

Good news: When A = 0, U is essentially nondegenerate(Weinstein, 1985).

Problem: UA lives naturally in the space

H1A(R3) = u ∈ L2(R3) : DAu ∈ L2(R3);

the natural functional space depends on the parameter A.


Working across functional spaces

the gap between U1 ∈ H1A1

(R3) and U2 ∈ H1A2

(R3) can bemeasured by

∫

R3|DA1U1 − DA2U2|2 + |U1 − U2|2,

the uniqueness part of the implicit function theorem justuses nondegeneracy and continuity of the linearizedoperator.


Thank you for your attention.


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