Symmetry of the ground state of the nonlinear Schr¨odinger equation in presence of a magnetic field

Schr¨ odinger equation in presence of a magnetic field

Jean Dolbeault

http://www.ceremade.dauphine.fr/ ∼ dolbeaul Ceremade, Universit´e Paris-Dauphine

September 18, 2019 Nonlinear days in Alghero

September 16-20, 2019

Outline

Without magnetic fields: symmetry and symmetry breaking in interpolation inequalities

B Gagliardo-Nirenberg-Sobolev inequalities on the sphere B Keller-Lieb-Thirring inequalities on the sphere

B Caffarelli-Kohn-Nirenberg inequalities With magnetic fields in dimensions 2 and 3 B Interpolation inequalities and spectral estimates

B Estimates, numerics; an open question on constant magnetic fields Magnetic rings: the case of S ¹

B A one-dimensional magnetic interpolation inequality

B Consequences: Keller-Lieb-Thirring estimates, Aharonov-Bohm magnetic fields and a new Hardy inequality in R ²

Aharonov-Bohm magnetic fields in R ² B Aharonov-Bohm effect

B Interpolation and Keller-Lieb-Thirring inequalities in R ² B Aharonov-Symmetry and symmetry breaking

J. Dolbeault Magnetic ground state & symmetry

A joint research program (mostly) with...

M.J. Esteban, Ceremade, Universit´e Paris-Dauphine B symmetry, interpolation, Keller-Lieb-Thirring, magnetic fields

M. Loss, Georgia Institute of Technology (Atlanta) B symmetry, interpolation, Keller-Lieb-Thirring, magnetic fields

A. Laptev, Imperial College London B Keller-Lieb-Thirring, magnetic fields

D. Bonheure, Universit´e Libre de Bruxelles B Aharonov-Bohm magnetic fields

J. Dolbeault Magnetic ground state & symmetry

Definitions in presence of a magnetic field

The magnetic covariant derivative / magnetic gradient

∇ ^A := ∇ + i A The magnetic Dirichlet energy: if ψ = u e ^iS , then

ˆ

R

|∇ ^A ψ | ² dx = ˆ

R

|∇ ψ | ² dx + ˆ

R

|∇ S + A | ² | ψ | ² dx The magnetic Sobolev space

H ¹ _A ( R ^d ) :=

ψ ∈ L ² ( R ^d ) : ∇ ^A ψ ∈ L ² ( R ^d ) The magnetic Laplacian

− ∆ A ψ = − ∆ ψ − 2 i A · ∇ ψ + | A | ² ψ − i (div A) ψ

J. Dolbeault Magnetic ground state & symmetry

Magnetic interpolation, ground state of the nonlinear Schr¨ odinger equation and symmetry

For any p ∈ (2, 2 ^∗ ), the magnetic interpolation inequality k∇ ^A ψ k ² L

( R

) + α k ψ k ² L

( R

) ≥ µ(α) k ψ k ² L

( R

) ∀ ψ ∈ H ¹ _A (R ^d ) is a consequence of the Gagliardo-Nirenberg inequality

k∇ ψ k ² L

( R

) + α k ψ k ² L

( R

) ≥ µ GN (α) k ψ k ² L

( R

) ∀ ψ ∈ H ¹ (R ^d ) and of the diamagnetic inequality

k∇ ^A ψ k ² L

( R

) ≥ k∇ u k ² L

( R

) , u = | ψ | An optimal function solves

− ∆ A ψ + α ψ = | ψ | ^p−2 ψ

If A is invariant under rotation, is there a (complex valued) ground state (minimum of the energy) which is depending only on | x | ?

J. Dolbeault Magnetic ground state & symmetry

Symmetry and symmetry breaking in interpolation inequalities

without magnetic field

Gagliardo-Nirenberg-Sobolev inequalities on the sphere Keller-Lieb-Thirring inequalities on the sphere

Caffarelli-Kohn-Nirenberg inequalities on R ²

J. Dolbeault Magnetic ground state & symmetry

A result of uniqueness on a classical example

On the sphere S ^d , let us consider the positive solutions of

− ∆u + λ u = u ^p−1 p ∈ [1, 2) ∪ (2, 2 ^∗ ] if d ≥ 3, 2 ^∗ = _d−2 ^{2 d}

p ∈ [1, 2) ∪ (2, + ∞ ) if d = 1, 2 Theorem

If λ ≤ d , u ≡ λ ^1/(p−2) is the unique solution

[Gidas & Spruck, 1981], [Bidaut-V´eron & V´eron, 1991]

J. Dolbeault Magnetic ground state & symmetry

Bifurcation point of view and symmetry breaking

2 4 6 8 10

2 4 6 8

Figure: (p − 2) λ 7→ (p − 2) µ(λ) with d = 3

k∇ u k ² L

( S

) + λ k u k ² L

( S

) ≥ µ(λ) k u k ² L

( S

)

Taylor expansion of u = 1 + ε ϕ 1 as ε → 0 with − ∆ϕ 1 = d ϕ 1

µ(λ) < λ if and only if λ > _p−2 ^d

B The inequality holds with µ(λ) = λ = _p−2 ^d [Bakry & Emery, 1985]

[Beckner, 1993], [Bidaut-V´eron & V´eron, 1991, Corollary 6.1]

J. Dolbeault Magnetic ground state & symmetry

The Bakry-Emery method on the sphere

Entropy functional E ^p [ρ] := _p−2 ¹ h ´

S

ρ

dµ − ´

S

ρ dµ

i

if p 6 = 2 E ² [ρ] := ´

S

ρ log

ρ kρk

dµ Fisher information functional

I ^p [ρ] := ´

S

|∇ ρ

| ² d µ

[Bakry & Emery, 1985] carr´ e du champ method: use the heat flow

∂ρ

∂t = ∆ρ and observe that _dt ^d E ^p [ρ] = − I ^p [ρ]

d dt

I ^p [ρ] − d E ^p [ρ]

≤ 0 = ⇒ I ^p [ρ] ≥ d E ^p [ρ]

with ρ = | u | ^p , if p ≤ 2 ^# := _(d−1) ^2d

⁺¹

The evolution under the fast diffusion flow

To overcome the limitation p ≤ 2 ^# , one can consider a nonlinear diffusion of fast diffusion / porous medium type

∂ρ

∂t = ∆ρ ^m

[Demange], [JD, Esteban, Kowalczyk, Loss]: for any p ∈ [1, 2 ^∗ ] K ^p [ρ] := d

dt

I ^p [ρ] − d E ^p [ρ]

≤ 0

1.0 1.5 2.5 3.0

0.0 0.5 1.5 2.0

(p, m) admissible region, d = 5

J. Dolbeault Magnetic ground state & symmetry

References

JD, M. J. Esteban, M. Kowalczyk, and M. Loss. Improved interpolation inequalities on the sphere. Discrete and Continuous Dynamical Systems Series S, 7 (4): 695-724, 2014.

JD, M.J. Esteban, and M. Loss. Interpolation inequalities on the sphere: linear vs. nonlinear flows. Annales de la facult´e des sciences de Toulouse S´er. 6, 26 (2): 351-379, 2017

JD, M.J. Esteban. Improved interpolation inequalities and stability. Preprint, 2019 arXiv:1908.08235

J. Dolbeault Magnetic ground state & symmetry

Optimal inequalities

With µ(λ) = λ = _p−2 ^d : [Bakry & Emery, 1985]

[Beckner, 1993], [Bidaut-V´eron & V´eron, 1991, Corollary 6.1]

k∇ u k ² L

( S

) ≥ d p − 2

k u k ² L

( S

) − k u k ² L

( S

)

∀ u ∈ H ¹ (S ^d )

d ≥ 3, p ∈ [1, 2) or p ∈ (2, _d−2 ^{2 d} ) d = 1 or d = 2, p ∈ [1, 2) or p ∈ (2, ∞ )

p = − 2 = 2 d/(d − 2) = − 2 with d = 1 [Exner, Harrell, Loss, 1998]

k∇ u k ² L

( S

) + 1 4

ˆ

S

1 u ² dµ

−1

≥ 1

4 k u k ² L

( S

) ∀ u ∈ H ¹ + (S ^d )

J. Dolbeault Magnetic ground state & symmetry

Keller-Lieb-Thirring inequalities on the sphere

The Keller-Lieb-Tirring inequality is equivalent to an interpolation inequality of Gagliardo-Nirenberg-Sobolev type

We measure a quantitative deviation with respect to the semi-classical regime due to finite size effects

Joint work with M.J. Esteban and A. Laptev

J. Dolbeault Magnetic ground state & symmetry

An introduction to (Keller)-Lieb-Thirring inequalities in R ^d

(λ k ) k≥1 : eigenvalues of the Schr¨odinger operator H = − ∆ − V on R ^d Euclidean case [Keller, 1961]

| λ 1 | ^γ ≤ L ¹ _γ,d ˆ

R

V ₊ ^γ+

[Lieb-Thirring, 1976]

P

k≥1 | λ k | ^γ ≤ L γ,d

´

R

V ^γ+

d

+

γ ≥ 1/2 if d = 1, γ > 0 if d = 2 and γ ≥ 0 if d ≥ 3 [Weidl], [Cwikel], [Rosenbljum], [Aizenman], [Laptev-Weidl], [Helffer], [Robert], [JD-Felmer-Loss-Paturel], [JD-Laptev-Loss]...[Frank, Hundertmark, Jex, Nam]

Compact manifolds: log Sobolev case: [Federbusch], [Rothaus];

case γ = 0 (Rozenbljum-Lieb-Cwikel inequality): [Levin-Solomyak];

[Lieb], [Levin], [Ouabaz-Poupaud]... [Ilyin]

B How does one take into account the finite size effects on S ^d ?

J. Dolbeault Magnetic ground state & symmetry

H¨ older duality and link with interpolation inequalities

Let p = _q−2 ^q . Consider the Schr¨ odinger energy ˆ

S

|∇ u | ² − ˆ

S

V | u | ² ≥ ˆ

S

|∇ u | ² − µ k u k ² L

( S

)

≥ − λ(µ) k u k ² L

( S

) if µ = k V + k ^L

⁽ S

)

We deduce from k∇ u k ² L

( S

) + λ k u k ² L

( S

) ≥ µ(λ) k u k ² L

( S

) that k∇ u k ² L

( S

) − µ(λ) k u k ² L

( S

) ≥ − λ k u k ² L

( S

)

2 4 6 8 10 12

2 4 6 8 10 12

2 4 6 8 10 12

2 4 6 8 10

µ(λ)

λ(µ)

q = d = 3

J. Dolbeault Magnetic ground state & symmetry

A Keller-Lieb-Thirring inequality on the sphere

Let d ≥ 1, p ∈

max { 1, d/2 } , + ∞

and µ ∗ := ^d ₂ (p − 1) Theorem (JD-Esteban-Laptev)

There exists a convex increasing function λ s.t. λ(µ) = µ if µ ∈ 0, µ ∗ and λ(µ) > µ if µ ∈ µ ∗ , + ∞

and, for any p < d/2,

| λ 1 ( − ∆ − V ) | ≤ λ k V k ^L

⁽ S

)

∀ V ∈ L ^p (S ^d ) This estimate is optimal

For large values of µ, we have λ(µ) ^p−

= L ¹ _p−

2

,d (κ q,d µ) ^p (1 + o(1))

If p = d /2 and d ≥ 3, the inequality holds with λ(µ) = µ iff µ ∈ [0, µ ∗ ]

J. Dolbeault Magnetic ground state & symmetry

A Keller-Lieb-Thirring inequality: second formulation

Let d ≥ 1, γ = p − d/2 Corollary (JD-Esteban-Laptev)

| λ 1 ( − ∆ − V ) | ^γ . L ¹ _γ,d ˆ

S

V ^γ+

as µ = k V k _L

2

( S

) → ∞ if either γ > max { 0, 1 − d /2 } or γ = 1/2 and d = 1 However, if µ = k V k _L

_{( S}

) ≤ µ ∗ , then we have

| λ 1 ( − ∆ − V ) | ^γ+

≤ ˆ

S

V ^γ+

for any γ ≥ max { 0, 1 − d/2 } and this estimate is optimal

L ¹ _γ,d is the optimal constant in the Euclidean one bound state ineq.

| λ 1 ( − ∆ − φ) | ^γ ≤ L ¹ _γ,d ˆ

R

φ ^γ+ +

dx

J. Dolbeault Magnetic ground state & symmetry

References

JD, M. J. Esteban, and A. Laptev. Spectral estimates on the sphere. Analysis & PDE, 7 (2): 435-460, 2014

JD, M.J. Esteban, A. Laptev, M. Loss. Spectral properties of Schr¨ odinger operators on compact manifolds: Rigidity, flows,

interpolation and spectral estimates. Comptes Rendus Math´ematique, 351 (11-12): 437-440, 2013

J. Dolbeault Magnetic ground state & symmetry

Caffarelli-Kohn-Nirenberg, symmetry and symmetry breaking results, and weighted nonlinear flows

Joint work with M.J. Esteban and M. Loss

J. Dolbeault Magnetic ground state & symmetry

Critical Caffarelli-Kohn-Nirenberg inequality

Let D ^a,b := n

v ∈ L ^p R ^d , | x | ^−b dx

: | x | ^−a |∇ v | ∈ L ² R ^d , dx o ˆ

R

| v | ^p

| x | ^{b p} dx ^2/p

≤ C a,b

ˆ

R

|∇ v | ²

| x | ^{2 a} dx ∀ v ∈ D ^a,b holds under conditions on a and b

p = 2 d

d − 2 + 2 (b − a) (critical case) B An optimal function among radial functions:

v ? (x ) =

1 + | x | ^{(p−2) (a}

^−a) −

and C ^? _a,b = k | x | ^−b v _? k ² p

k | x | ^−a ∇ v _? k ² 2

Question: C a,b = C ^? _a,b (symmetry) or C a,b > C ^? _a,b (symmetry breaking) ?

J. Dolbeault Magnetic ground state & symmetry

Critical CKN: range of the parameters

Figure: d = 3 ˆ

R

| v | ^p

| x | ^{b p} dx 2/p

≤ C a,b

ˆ

R

|∇ v | ²

| x | ^{2 a} dx

a b

0 1

− 1

b = a

b = a+ 1

a =

p

p = 2 d

d − 2 + 2 (b − a) a < a c := (d − 2)/2 a ≤ b ≤ a + 1 if d ≥ 3, a + 1/2 < b ≤ a + 1 if d = 1 and a < b ≤ a + 1 < 1,

p = 2/(b − a) if d = 2

[Il’in (1961)]

[Glaser, Martin, Grosse, Thirring (1976)]

[Caffarelli, Kohn, Nirenberg (1984)]

[F. Catrina, Z.-Q. Wang (2001)]

J. Dolbeault Magnetic ground state & symmetry

Linear instability of radial minimizers:

the Felli-Schneider curve

The Felli & Schneider curve b FS (a) := d ( a _c − a )

2 p

(a c − a) ² + d − 1 + a − a c

a b

0

[Smets], [Smets, Willem], [Catrina, Wang], [Felli, Schneider]

v 7→ C ^? _a,b ˆ

R

|∇ v | ²

| x | ^{2 a} dx − ˆ

R

| v | ^p

| x | ^{b p} dx 2/p

is linearly instable at v = v _?

J. Dolbeault Magnetic ground state & symmetry

Symmetry versus symmetry breaking:

the sharp result in the critical case

[JD, Esteban, Loss (2016)]

a b

0

Theorem

Let d ≥ 2 and p < 2 ^∗ . If either a ∈ [0, a c ) and b > 0, or a < 0 and b ≥ b FS (a), then the optimal functions for the critical

Caffarelli-Kohn-Nirenberg inequalities are radially symmetric

B

J. Dolbeault Magnetic ground state & symmetry

The symmetry proof in one slide

A change of variables: v( | x | ^α−1 x ) = w (x), D α v = α ^∂v _∂s , ¹ _s ∇ ^ω v k v k ^L

⁽ R

) ≤ K α,n,p k D α v k ^ϑ L

( R

) k v k ^1−ϑ L

( R

) ∀ v ∈ H ^p _d−n,d−n (R ^d )

Concavity of the R´enyi entropy power: with L ^α = − D ^∗ _α D α = α ² u ⁰⁰ + ⁿ⁻¹ _s u ⁰

+ _s ¹

∆ ω u and ^∂u _∂t = L ^α u ^m

− dt ^d G [u(t , · )] ´

R

u ^m dµ 1−σ

≥ (1 − m) (σ − 1) ´

R

u ^m L ^α P −

´

Rd

´ u |D

P|

dµ

Rd

u

dµ

2

dµ + 2 ´

R

α ⁴ 1 − ¹ n

P ⁰⁰ − ^P s

− α

(n−1) ^∆

^P s

2

+ ^2α _s

∇ ^ω P ⁰ − ^∇

s ^P

2 u ^m d µ + 2 ´

R

(n − 2) α ² _FS − α ²

|∇ ^ω P | ² + c(n , m , d) ^|∇ _P

^P|

u ^m d µ

Elliptic regularity and the Emden-Fowler transformation: justifying the integrations by parts

J. Dolbeault Magnetic ground state & symmetry

The variational problem on the cylinder

B With the Emden-Fowler transformation

v (r, ω) = r ^a−a

ϕ(s, ω) with r = | x | , s = − log r and ω = x r the variational problem becomes

Λ 7→ µ(Λ) := min

ϕ∈H

(C)

k ∂ s ϕ k ² L

(C) + k∇ ^ω ϕ k ² L

(C) + Λ k ϕ k ² L

(C)

k ϕ k ² L

(C)

is a concave increasing function Restricted to symmetric functions, the variational problem becomes

µ ? (Λ) := min

ϕ∈H

( R )

k ∂ s ϕ k ² L

( R

) + Λ k ϕ k ² L

( R

)

k ϕ k ² L

( R

)

= µ ? (1) Λ ^α Symmetry means µ(Λ) = µ ? (Λ)

Symmetry breaking means µ(Λ) < µ ? (Λ)

J. Dolbeault Magnetic ground state & symmetry

Numerical results

20 40 60 80 100

10 20 30 40 50

symmetric non-symmetric asymptotic

bifurcation µ

Λ

µ(Λ)

(Λ) = µ

(1) Λ

Parametric plot of the branch of optimal functions for p = 2.8, d = 5.

Non-symmetric solutions bifurcate from symmetric ones at a bifurcation point Λ

computed by V. Felli and M. Schneider. The branch behaves for large values of Λ as shown by F. Catrina and Z.-Q. Wang

J. Dolbeault Magnetic ground state & symmetry

Three references

Lecture notes on Symmetry and nonlinear diffusion flows...

a course on entropy methods (see webpage)

[JD, Maria J. Esteban, and Michael Loss] Symmetry and symmetry breaking: rigidity and flows in elliptic PDEs

... the elliptic point of view: Proc. Int. Cong. of Math., Rio de Janeiro, 3: 2279-2304, 2018.

[JD, Maria J. Esteban, and Michael Loss] Interpolation inequalities, nonlinear flows, boundary terms, optimality and linearization... the parabolic point of view

Journal of elliptic and parabolic equations, 2: 267-295, 2016.

J. Dolbeault Magnetic ground state & symmetry

Magnetic interpolation inequalities in the Euclidean space

B Three interpolation inequalities and their dual forms B Estimates in dimension d = 2 for constant magnetic fields

Lower estimates

Upper estimates and numerical results

A linear stability result (numerical) and an open question Warning: assumptions are not repeated

Estimates are given only in the case p > 2 but similar estimates hold in the other cases

Joint work with M.J. Esteban, A. Laptev and M. Loss

J. Dolbeault Magnetic ground state & symmetry

Magnetic Laplacian and spectral gap

In dimensions d = 2 and d = 3: the magnetic Laplacian is

− ∆ A ψ = − ∆ ψ − 2 i A · ∇ ψ + | A | ² ψ − i (div A) ψ

where the magnetic potential (resp. field) is A (resp. B = curl A) and H ¹ A (R ^d ) :=

ψ ∈ L ² (R ^d ) : ∇ ^A ψ ∈ L ² (R ^d ) , ∇ ^A := ∇ + i A Spectral gap inequality

k∇ ^A ψ k ² L

( R

) ≥ Λ[B] k ψ k ² L

( R

) ∀ ψ ∈ H ¹ _A ( R ^d ) Λ depends only on B = curl A

Assumption: equality holds for some ψ ∈ H ¹ _A ( R ^d ) If B is a constant magnetic field, Λ[B] = | B |

If d = 2, spec( − ∆ A ) = { (2j + 1) | B | : j ∈ N } is generated by the Landau levels. The Lowest Landau Level corresponds to j = 0

J. Dolbeault Magnetic ground state & symmetry

Magnetic interpolation inequalities

k∇ ^A ψ k ² L

( R

) + α k ψ k ² L

( R

) ≥ µ B (α) k ψ k ² L

( R

) ∀ ψ ∈ H ¹ _A (R ^d ) for any α ∈ ( − Λ[B], + ∞ ) and any p ∈ (2, 2 ^∗ ),

k∇ ^A ψ k ² L

( R

) + β k ψ k ² L

( R

) ≥ ν B (β) k ψ k ² L

( R

) ∀ ψ ∈ H ¹ A ( R ^d ) for any β ∈ (0, + ∞ ) and any p ∈ (1, 2)

k∇ ^A ψ k ² L

( R

) ≥ γ ˆ

R

| ψ | ² log | ψ | ² k ψ k ² L

( R

)

!

dx + ξ B (γ) k ψ k ² L

( R

)

(limit case corresponding to p = 2) for any γ ∈ (0, + ∞ )

C p :=



 

 

min u∈H

( R

)\{0}

k∇uk

)

+kuk

)

kuk

Lp(Rd)

if p ∈ (2, 2 ^∗ ) min _u∈H

( R

)\{0}

k∇uk

)

+kuk

Lp(Rd)

kuk

L2 (Rd)

if p ∈ (1, 2) µ 0 (1) = C p if p ∈ (2, 2 ^∗ ), ν 0 (1) = C p if p ∈ (1, 2)

ξ 0 (γ) = γ log π e ² /γ

if p = 2

J. Dolbeault Magnetic ground state & symmetry

Technical assumptions

A ∈ L ^α _loc ( R ^d ), α > 2 if d = 2 or α = 3 if d = 3 and

σ→+∞ lim σ ^d−2 ˆ

R

| A(x) | ² e ^{−σ |x|} dx = 0 if p ∈ (2, 2 ^∗ )

σ→+∞ lim σ

⁻¹

log σ ˆ

R

| A(x) | ² e ^{−σ |x|}

dx = 0 if p = 2

σ→+∞ lim σ ^d−2 ˆ

|x|<1/σ | A(x ) | ² dx if p ∈ (1, 2) These estimates can be found in [Esteban, Lions, 1989]

J. Dolbeault Magnetic ground state & symmetry

A statement

Theorem

p ∈ (2, 2 ^∗ ): µ B is monotone increasing on ( − Λ[B], + ∞ ), concave and

α→(−Λ[B]) lim

µ B (α) = 0 and lim

α→+∞ µ B (α) α

⁻

= C p

p ∈ (1, 2): ν B is monotone increasing on (0, + ∞ ), concave and

β→0 lim

ν B (β ) = Λ[B] and lim

β→+∞ ν B (β) β ⁻

= C p

ξ B is continuous on (0, + ∞ ), concave, ξ B (0) = Λ[B] and ξ B (γ) = ^d ₂ γ log ^π _γ ^e

(1 + o (1)) as γ → + ∞ Constant magnetic fields: equality is achieved

Nonconstant magnetic fields: only partial answers are known

J. Dolbeault Magnetic ground state & symmetry

2 4 6 8 10 2

4 6 8

Figure: Case d = 2, p = 3, B = 1: plot of α 7→ (2 π)

µ

(α)

J. Dolbeault Magnetic ground state & symmetry

0 1 2 3 4 5 6 0

2 4 6 8

Figure: Case d = 2, p = 1.4, B = 1: plot of β 7→ ν

(β) The horizontal axis is measured in units of (2 π)

β

J. Dolbeault Magnetic ground state & symmetry

Magnetic Keller-Lieb-Thirring inequalities

λ A,V is the principal eigenvalue of − ∆ A + V

α B : (0, + ∞ ) → ( − Λ, + ∞ ) is the inverse function of α 7→ µ B (α) Corollary

(i) For any q = p/(p − 2) ∈ (d /2, + ∞ ) and any potential V ∈ L ^q ₊ ( R ^d ) λ A,V ≥ − α B ( k V k ^L

⁽ R

) )

lim µ→0

α B (µ) = Λ and lim µ→+∞ α B (µ) µ

= − C

2 (q+1) d−2−2q

p

(ii) For any q = p/(2 − p) ∈ (1, + ∞ ) and any 0 < W ⁻¹ ∈ L ^q ( R ^d ) λ A,W ≥ ν B

k W ⁻¹ k ⁻¹ L

( R

)

(iii) For any γ > 0 and any W ≥ 0 s.t. e ^{−W /γ} ∈ L ¹ (R ^d ) λ A,W ≥ ξ B (γ) − γ log ´

R

e ^{−W /γ} dx B

J. Dolbeault Magnetic ground state & symmetry

Proofs

J. Dolbeault Magnetic ground state & symmetry

Interpolation without magnetic field...

Assume that p > 2 and let C p denote the best constant in k∇ u k ² L

( R

) + k u k ² L

( R

) ≥ C p k u k ² L

( R

) ∀ u ∈ H ¹ ( R ^d ) By scaling, if we test the inequality by u · /λ

, we find that

k∇ u k ² L

( R

) +λ ² k u k ² L

( R

) ≥ C p λ ^{2− d (1−}

⁾ k u k ² L

( R

) ∀ u ∈ H ¹ (R ^d ) ∀ λ > 0 An optimization on λ > 0 shows that the best constant in the

scale-invariant inequality k∇ u k ^{d (1−}

2 p

)

L

( R

) k u k ^{2−d (1−}

2 p

)

L

( R

) ≥ S p k u k ² L

( R

) ∀ u ∈ H ¹ ( R ^d ) is given by

S p = ₂ ¹ _p (2 p − d (p − 2)) ^1−d

(d (p − 2))

C p

J. Dolbeault Magnetic ground state & symmetry

... and with magnetic field

Proposition

Let d = 2 or 3. For any p ∈ (2, + ∞ ), any α > − Λ = − Λ[B] < 0

µ B (α) ≥ µ interp (α) :=







S p (α + Λ) Λ ^−d

if α ∈ h

− Λ, ^{Λ (2p−d} _{d (p−2)} ^(p−2)) i C p α ^1−d

if α ≥ ^{Λ (2p−d} d (p−2) ^(p−2))

Diamagnetic inequality: k∇| ψ |k ^L

⁽ R

) ≤ k∇ ^A ψ k ^L

⁽ R

)

Non-magnetic inequality with λ = ^α+Λ _1−t ^t , t ∈ [0, 1]

k∇ ^A ψ k ² L

( R

) + α k ψ k ² L

( R

) ≥ t

k∇ ^A ψ k ² L

( R

) − Λ k ψ k ² L

( R

)

+ (1 − t)

k∇| ψ |k L

( R

) + α + Λ t

1 − t k ψ k ² L

( R

)

≥ C p (1 − t)

(α + t Λ) ^1−d

k ψ k ² L

( R

)

and optimize on t ∈ [max { 0, − α/Λ } , 1]

J. Dolbeault Magnetic ground state & symmetry

The special case of constant magnetic field in dimension d = 2

J. Dolbeault Magnetic ground state & symmetry

Constant magnetic field, d = 2...

Assume that B = (0, B) is constant, d = 2 and choose A 1 = ^B ₂ x 2 , A 2 = − ^B 2 x 1 ∀ x = (x 1 , x 2 ) ∈ R ² Proposition

[Loss, Thaller, 1997] Consider a constant magnetic field with field strength B in two dimensions. For every c ∈ [0, 1], we have

ˆ

R

|∇ ^A ψ | ² dx ≥ 1 − c ² ˆ

R

|∇ ψ | ² dx + c B ˆ

R

ψ ² dx and equality holds with ψ = u e ^iS and u > 0 if and only if

− ∂ 2 u ² , ∂ 1 u ²

= 2 u ²

c (A + ∇ S)

J. Dolbeault Magnetic ground state & symmetry

... a computation (d = 2, constant magnetic field)

ˆ

R

|∇ ^A ψ | ² dx = ˆ

R

|∇ u | ² dx + ˆ

R

| A + ∇ S | ² u ² dx

= 1 − c ² ˆ

R

|∇ u | ² dx + ˆ

R

c ² |∇ u | ² + | A + ∇ S | ² u ² dx

| {z }

≥ ´

R2

2 c |∇u| |A+∇S|u dx

with equality only if c |∇ u | = | A + ∇ S | u

2 |∇ u | | A + ∇ S | u = |∇ u ² | | A + ∇ S | ≥ ∇ u ² ⊥

· (A + ∇ S) where ∇ u ² ⊥

:= − ∂ 2 u ² , ∂ 1 u ² Equality case: − ∂ 2 u ² , ∂ 1 u ²

= γ (A + ∇ S) for γ = 2 u ² /c Integration by parts yields

ˆ

R

c ² |∇ u | ² + | A + ∇ S | ² u ²

dx ≥ B c ˆ

R

u ² dx

J. Dolbeault Magnetic ground state & symmetry

... a lower estimate (d = 2, constant magnetic field)

Proposition

Consider a constant magnetic field with field strength B in two dimensions. Given any p ∈ (2, + ∞ ), and any α > − B, we have

µ B (α) ≥ C p 1 − c ^{2 1−}

(α + c B )

=: µ LT (α) with

c = c(p, η) =

p η ² + p − 1 − η

p − 1 = 1

η + p

η ² + p − 1 ∈ (0, 1) and η = α (p − 2)/(2 B)

J. Dolbeault Magnetic ground state & symmetry

Upper estimate (1): d = 2, constant magnetic field

For every integer k ∈ N we introduce the special symmetry class ψ(x) =

x

+i x

|x|

^k

v( | x | ) ∀ x = (x 1 , x 2 ) ∈ R ² ( C ^k ) [Esteban, Lions, 1989]: if ψ ∈ C ^k , then

1 2 π

ˆ

R

|∇ ^A ψ | ² dx = ˆ +∞

0 | v ⁰ | ² r dr + ˆ +∞

0 k r − ^{B r} 2

²

| v | ² r dr and optimality is achieved in C ^k

Test function v _σ (r) = e ^{− r}

^{/(2 σ)} : an optimization on σ > 0 provides an explicit expression of µ Gauss (α) such that

Proposition If p > 2, then

µ B (α) ≤ µ Gauss (α) ∀ α > − Λ[B]

This estimate is not optimal because v _σ does not solve the Euler-Lagrange equations

J. Dolbeault Magnetic ground state & symmetry

Upper estimate (2): d = 2, constant magnetic field

A more numerical point of view . The Euler-Lagrange equation in C ⁰ is

− v ⁰⁰ − v ⁰ r +

B

4 r ² + α

v = µ EL (α) ˆ +∞

0 | v | ^p r dr

−1

| v | ^p−2 v We can restrict the problem to positive solutions such that

µ EL (α) =

ˆ +∞

0 | v | ^p r dr ¹⁻

and then we have to solve the reduced problem

− v ⁰⁰ − v ⁰ r +

B

4 r ² + α

v = | v | ^p−2 v

J. Dolbeault Magnetic ground state & symmetry

Numerical results and the symmetry issue

J. Dolbeault Magnetic ground state & symmetry

2 4 6 8 10

-1 1 2 3

0.004 0.003 0.002 0.001 0.001 0.002 0.003

-0.02

-0.04

-0.06

-0.08

2 4

-1

6

Figure: Case d = 2, p = 3, B = 1 Upper estimates: α 7→ µ

(α), µ

(α) Lower estimates: α 7→ µ

(α), µ

(α)

The exact value associated with µ

lies in the grey area.

Plots represent the curves log

(µ/µ

) B

J. Dolbeault Magnetic ground state & symmetry

Asymptotics (1): Lowest Landau Level

Proposition

Let d = 2 and consider a constant magnetic field with field strength B. If ψ α is a minimizer for µ B (α) such that k ψ α k ^L

⁽ R

) = 1, then there exists a non trivial ϕ α ∈ LLL such that

α→(−B) lim

k ψ α − ϕ α k H

( R

) = 0

Let ψ α ∈ H ¹ _A (R ² ) be an optimal function such that k ψ α k L

( R

) = 1 and let us decompose it as ψ α = ϕ α + χ α , where ϕ α ∈ LLL and χ α is in the orthogonal of LLL

µ B (α) ≥ (α+B) k ϕ α k ² L

( R

) +(α+3B) k χ α k ² L

( R

) ≥ (α+3B) k χ α k ² L

( R

) ∼ 2B k χ α k ² L

( R

)

as α → ( − B) + because k∇ χ α k ² L

( R

) ≥ 3B k χ α k ² L

( R

)

Since lim α→(−B)

µ B (α) = 0, lim α→(−B)

k χ α k L

( R

) = 0 and

µ B (α) = (α+B) k ϕ α k ² L

( R

) + k∇ ^A χ α k ² L

( R

) +α k χ α k ² L

( R

) ≥ ² 3 k∇ ^A χ α k ² L

( R

)

concludes the proof

J. Dolbeault Magnetic ground state & symmetry

Asymptotics (2): semi-classical regime

Let us consider the small magnetic field regime. We assume that the magnetic potential is given by

A ₁ = ^B ₂ x ₂ , A ₂ = − ^B 2 x ₁ ∀ x = (x 1 , x ₂ ) ∈ R ²

if d = 2. In dimension d = 3, we choose A = ^B ₂ ( − x 2 , x 1 , 0) and observe that the constant magnetic field is B = (0, 0, B), while the spectral gap is Λ[B] = B .

Proposition

Let d = 2 or 3 and consider a constant magnetic field B of intensity B with magnetic potential A

For any p ∈ (2, 2 ^∗ ) and any fixed α and µ > 0, we have

ε→0 lim

µ εB (α) = C p α

⁻

Consider any function ψ ∈ H ¹ _A (R ^d ) and let ψ(x ) = χ( √ ε x ),

√ ε A x / √ ε

= A(x) with our conventions on A

J. Dolbeault Magnetic ground state & symmetry

Numerical stability of radial optimal functions

Let us denote by ψ 0 an optimal function in ( C ⁰ ) such that

− ψ ₀ ⁰⁰ − ψ ⁰ ₀ r +

B

4 r ² + α

ψ 0 = | ψ 0 | ^p−2 ψ 0

and consider the test function

ψ ε = ψ 0 + ε e ^{i θ} v where v = v (r ) and e ^{i θ} = (x 1 + i x 2 )/r As ε → 0 + , the leading order term is

2 π ˆ

R

| v ⁰ | ² dx + ˆ

R

1 r − ^{B r} 2

² + α

| v | ² dx − ^p 2

ˆ +∞

0 | ψ 0 | ^p−2 v ² r dr

ε ²

and we have to solve the eigenvalue problem

− v ⁰⁰ − v ⁰ r +

1 r − ^{B r} 2

² + α

v − ^p 2 | ψ 0 | ^p−2 v = µ v

J. Dolbeault Magnetic ground state & symmetry

2 4 6 8 1

2 3 4

Figure: Case p = 3 and B = 1: plot of the eigenvalue µ as a function of α A careful investigation shows that µ is always positive, including in the limiting case as α → (−B )

, thus proving the numerical stability of the optimal function in C

with respect to perturbations in C

J. Dolbeault Magnetic ground state & symmetry

An open question of symmetry

[Bonheure, Nys, Van Schaftingen, 2016] for a fixed α > 0 and for B small enough, the optimal functions are radially symmetric functions, i.e., belong to C ⁰

This regime is equivalent to the regime as α → + ∞ for a given B, at least if the magnetic field is constant

Numerically our upper and lower bounds are (in dimension d = 2, for a constant magnetic field) numerically extremely close

The optimal function in C ⁰ is linearly stable with respect to perturbations in C ¹

B Prove that the optimality case is achieved among radial function if d = 2 and B is a constant magnetic field

J. Dolbeault Magnetic ground state & symmetry

Reference

JD, M.J. Esteban, A. Laptev, M. Loss. Interpolation inequalities and spectral estimates for magnetic operators. Annales Henri Poincar´e, 19 (5): 1439-1463, May 2018

J. Dolbeault Magnetic ground state & symmetry

Magnetic rings

B A magnetic interpolation inequality on S ¹ : with p > 2 k ψ ⁰ + i a ψ k ² L

( S

) + α k ψ k ² L

( S

) ≥ µ a,p (α) k ψ k ² L

( S

)

B Consequences

A Keller-Lieb-Thirring inequality

A new Hardy inequality for Aharonov-Bohm magnetic fields in R ² Joint work with M.J. Esteban, A. Laptev and M. Loss

J. Dolbeault Magnetic ground state & symmetry

Magnetic flux, a reduction

Assume that a : R → R is a 2π-periodic function such that its restriction to ( − π, π] ≈ S ¹ is in L ¹ (S ¹ ) and define the space

X a :=

ψ ∈ C per ( R ) : ψ ⁰ + i a ψ ∈ L ² ( S ¹ )

A standard change of gauge (see e.g. [Ilyin, Laptev, Loss, Zelik, 2016])

ψ(s) 7→ e ⁱ

´

−π

(a(s)−¯ a) dσ ψ(s) where a ¯ := ´ π

−π a(s) dσ is the magnetic flux, reduces the problem to a is a constant function

For any k ∈ Z , ψ by s 7→ e ^iks ψ(s) shows that µ a,p (α) = µ k+a,p (α) a ∈ [0, 1]

µ a,p (α) = µ 1−a,p (α) because

| ψ ⁰ + i a ψ | ² = | χ ⁰ + i (1 − a) χ | ² =

ψ ⁰ − i a ψ

2 if χ(s) = e ^−is ψ(s) a ∈ [0, 1/2]

J. Dolbeault Magnetic ground state & symmetry

Optimal interpolation

We want to characterize the optimal constant in the inequality k ψ ⁰ + i a ψ k ² L

( S

) + α k ψ k ² L

( S

) ≥ µ a,p (α) k ψ k ² L

( S

)

written for any p > 2, a ∈ (0, 1/2], α ∈ ( − a ² , + ∞ ), ψ ∈ X _a µ a,p (α) := inf

ψ∈X

\{0}

´ π

−π | ψ ⁰ + i a ψ | ² + α | ψ | ² dσ k ψ k ² L

( S

)

p = − 2 = 2 d /(d − 2) with d = 1 [Exner, Harrell, Loss, 1998]

p = + ∞ [Galunov, Olienik, 1995] [Ilyin, Laptev, Loss, Zelik, 2016]

lim _α→− a

µ a,p (α) = 0 [JD, Esteban, Laptev, Loss, 2016]

Using a Fourier series ψ(s) = P

k ∈ Z ψ k e ^iks , we obtain that k ψ ⁰ + i a ψ k ² L

( S

) = X

k ∈ Z

(a + k) ² | ψ k | ² ≥ a ² k ψ k ² L

( S

)

ψ 7→ k ψ ⁰ + i a ψ k ² L

( S

) + α k ψ k ² L

( S

) is coercive for any α > − a ²

J. Dolbeault Magnetic ground state & symmetry

An interpolation result for the magnetic ring

Theorem

For any p > 2, a ∈ R , and α > − a ² , µ a,p (α) is achieved and

(i) if a ∈ [0, 1/2] and a ² (p + 2) + α (p − 2) ≤ 1, then µ a,p (α) = a ² + α and equality is achieved only by the constant functions

(ii) if a ∈ [0, 1/2] and a ² (p + 2) + α (p − 2) > 1, then µ a,p (α) < a ² + α and equality is not achieved by the constant functions

If α > − a ² , a 7→ µ a,p (α) is monotone increasing on (0, 1/2)

-0.2 -0.1 0.1 0.2 0.3 0.4

0.1 0.2 0.3 0.4 0.5

0.2 0.4 0.6 0.8 1.0 1.2 1.4

0.2 0.4 0.6 0.8 1.0 1.2

Figure: α 7→ µ

(α) with p = 4 and (left) a = 0.45 or (right) a = 0.2

J. Dolbeault Magnetic ground state & symmetry

The proof: how to eliminate the phase

J. Dolbeault Magnetic ground state & symmetry

Reformulations of the interpolation problem (1/3)

Any minimizer ψ ∈ X a of µ a,p (α) satisfies the Euler-Lagrange equation (H a + α) ψ = | ψ | ^p−2 ψ , H a ψ = −

d ds + i a

2

ψ (*)

up to a multiplication by a constant and v (s) = ψ(s) e ^ias satisfies the condition

v (s + 2π) = e ^2iπa v (s) ∀ s ∈ R Hence

µ a,p (α) = min

v∈Y

\{0} Q p,α [v]

where Y _a :=

v ∈ C ( R ) : v ⁰ ∈ L ² ( S ¹ ) , (*) holds and Q p,α [v] := k v ⁰ k ² L

( S

) + α k v k ² L

( S

)

k v k ² L

( S

)

J. Dolbeault Magnetic ground state & symmetry

Reformulations of the interpolation problem (2/3)

With v = u e ^iφ the boundary condition becomes

u(π) = u( − π) , φ(π) = 2π (a + k ) + φ( − π) (**) for some k ∈ Z , and k v ⁰ k ² L

( S

) = k u ⁰ k ² L

( S

) + k u φ ⁰ k ² L

( S

)

Hence

µ a,p (α) = min

(u,φ)∈Z

\{0}

k u ⁰ k ² L

( S

) + k u φ ⁰ k ² L

( S

) + α k u k ² L

( S

)

k u k ² L

( S

)

where Z _a :=

(u , φ) ∈ C( R ) ² : u ⁰ , u φ ⁰ ∈ L ² ( S ¹ ) , (**) holds

J. Dolbeault Magnetic ground state & symmetry

Reformulations of the interpolation problem (3/3)

We use the Euler-Lagrange equations

− u ⁰⁰ + | φ ⁰ | ² u + α u = | u | ^p−2 u and (φ ⁰ u ² ) ⁰ = 0

Integrating the second equation, and assuming that u never vanishes , we find a constant L such that φ ⁰ = L / u ² . Taking (*) into account, we deduce from

L ˆ π

−π

ds u ² =

ˆ π

−π

φ ⁰ ds = 2π (a + k ) that

k u φ ⁰ k ² L

( S

) = L ² ˆ π

−π

dσ

u ² = (a + k) ² k u ⁻¹ k ² L

( S

)

Hence

φ(s) − φ(0) = a + k k u ⁻¹ k ² L

( S

)

ˆ s

−π

ds u ²

J. Dolbeault Magnetic ground state & symmetry

Let us define

Q ^a,p,α [u] := k u ⁰ k ² L

( S

) + a ² k u ⁻¹ k ⁻² L

( S

) + α k u k ² L

( S

)

k u k ² L

( S

)

Lemma

For any a ∈ (0, 1/2), p > 2, α > − a ² , µ a,p (α) = min

u∈H

( S

)\{0} Q ^a,p,α [u]

is achieved by a function u > 0

J. Dolbeault Magnetic ground state & symmetry

Proofs

The existence proof is done on the original formulation of the problem using the diamagnetic inequality

To prove that | ψ | 6 = 0, we use ψ(s) e ^ias = v 1 (s) + i v 2 (s), solves

− v _j ⁰⁰ + α v j = (v ₁ ² + v ₂ ² )

⁻¹ v j , j = 1 , 2

and the Wronskian w = (v 1 v ₂ ⁰ − v ₁ ⁰ v 2 ) is constant so that ψ(s) = 0 is incompatible with the twisted boundary condition

if a ² (p + 2) + α (p − 2) ≤ 1, then µ a,p (α) = a ² + α because

k u ⁰ k ² L

( S

) +a ² k u ⁻¹ k ⁻² L

( S

) +α k u k ² L

( S

) = (1 − 4 a ² ) k u ⁰ k ² L

( S

) +α k u k ² L

( S

)

+ 4 a ²

k u ⁰ k ² L

( S

) + ¹ ₄ k u ⁻¹ k ² L

( S

)

if a ² (p + 2) + α (p − 2) > 1, the test function u _ε := 1 + ε w 1

Q ^a,p,α [u ε ] = a ² + α + 1 − a ² (p + 2) − α (p − 2)

ε ² + o (ε ² ) proves the linear instability of the constants and µ a,p (α) < a ² + α

J. Dolbeault Magnetic ground state & symmetry

Q ^a,p,α [u] := ^ku

0

k

+a

ku

k

+α kuk

kuk

Lp(S1 )

,

µ a,p (α) = min

u∈H

( S

)\{0} Q ^a,p,α [u]

Q p,α [u] = Q ^a=0,p,α [u] , ν p (α) := inf

v ∈H

( S

)\{0} Q p,α [v ] Proposition

∀ p > 2, α > − a ² , we have µ a,p (α) < µ 1/2,p (α) ≤ ν p (α) = µ 1/2,p (α)

0.5 1.0 1.5 2.0 2.5 3.0

0.1 0.2 0.3 0.4 0.5 0.6

Figure: p = 4, α = 0, a = 0.40, 0.41,. . . 0.49; u

+ u

= 0

J. Dolbeault Magnetic ground state & symmetry

Consequences: Keller-Lieb-Thirring inequalities and Hardy inequalities for Aharonov-Bohm magnetic fields

J. Dolbeault Magnetic ground state & symmetry

A Keller-Lieb-Thirring inequality

Magnetic Schr¨odinger operator H a − ϕ = − ds ^d + i a ² ψ − ϕ The function α 7→ µ a,p (α) is monotone increasing, concave, and therefore has an inverse, denoted by α a,p : R ⁺ → ( − a ² , + ∞ ), which is monotone increasing, and convex

Corollary

Let p > 2, a ∈ [0, 1/2], q = p/(p − 2) and assume that ϕ is a non-negative function in L ^q (S ¹ ). Then

λ 1 (H a − ϕ) ≥ − α a,p k ϕ k ^L

⁽ S

)

and α a,p (µ) = µ − a ² iff 4 a ² + µ (p − 2) ≤ 1 (optimal ϕ is constant) Equality is achieved

J. Dolbeault Magnetic ground state & symmetry

Aharonov-Bohm magnetic fields

On the two-dimensional Euclidean space R ² , let us introduce the polar coordinates (r, ϑ) ∈ [0, + ∞ ) × S ¹ of x ∈ R ² and consider a magnetic potential a in a transversal (Poincar´e) gauge, or Poincar´e gauge

(a, e r ) = 0 and (a, e _ϑ ) = a _ϑ (r , ϑ) Magnetic Schr¨ odinger energy

ˆ

R

| (i ∇ +a) Ψ | ² dx = ˆ +∞

0

ˆ π

−π

| ∂ r Ψ | ² + 1

r ² | ∂ ϑ Ψ + i r a _ϑ Ψ | ²

r dϑ d r Aharonov-Bohm magnetic fields: a _ϑ (r, ϑ) = a/r for some constant a ∈ R (a is the magnetic flux), with magnetic field b = curl a ˆ

R

| (i ∇ +a) Ψ | ² dx ≥ τ ˆ

R

ϕ(x/ | x | )

| x | ² | Ψ | ² dx ∀ ϕ ∈ L ^q (S ¹ ) , q ∈ (1, + ∞ )

= ⇒ τ = τ a, k ϕ k ^L

⁽ S

)

?

J. Dolbeault Magnetic ground state & symmetry

Hardy inequalities

[Hoffmann-Ostenhof, Laptev, 2015] proved Hardy’s inequality ˆ

R

|∇ Ψ | ² dx ≥ τ ˆ

R

ϕ(x/ | x | )

| x | ² | Ψ | ² dx

where the constant τ depends on the value of k ϕ k L

( S

) and d ≥ 3 Aharonov-Bohm vector potential in dimension d = 2

a(x) = a x ₂

| x | ² , − x ₁

| x | ²

, x = (x 1 , x 2 ) ∈ R ² , a ∈ R and recall the inequality [Laptev, Weidl, 1999]

ˆ

R

| (i ∇ + a) Ψ | ² dx ≥ min

k ∈ Z

(a − k ) ² ˆ

R

| Ψ |

| x | ² dx

J. Dolbeault Magnetic ground state & symmetry

A new Hardy inequality

ˆ

R

| (i ∇ +a) Ψ | ² dx ≥ τ ˆ

R

ϕ(x/ | x | )

| x | ² | Ψ | ² d x ∀ ϕ ∈ L ^q ( S ¹ ) , q ∈ (1, + ∞ ) Corollary

Let p > 2, a ∈ [0, 1/2], q = p/(p − 2) and assume that ϕ is a

non-negative function in L ^q ( S ¹ ). Then the inequality holds with τ > 0 given by

α a,p τ k ϕ k ^L

⁽ S

)

= 0

Moreover, τ = a ² / k ϕ k ^L

⁽ S

) if 4 a ² + k ϕ k ^L

⁽ S

) (p − 2) ≤ 1

For any a ∈ (0, 1/2), by taking ϕ constant, small enough in order that 4 a ² + k ϕ k ^L

⁽ S

) (p − 2) ≤ 1, we recover the inequality

ˆ

R

| (i ∇ + a) Ψ | ² dx ≥ a ² ˆ

R

| Ψ | ²

| x | ² dx

J. Dolbeault Magnetic ground state & symmetry

Proofs (Keller-Lieb-Thirring inequality)

B H¨ older’s inequality

k ψ ⁰ + i a ψ k ² L

( S

) − ˆ _π

−π

ϕ | ψ | ² dσ ≥ k ψ ⁰ + i a ψ k ² L

( S

) − µ k ψ k ² L

( S

)

where µ = k ϕ k L

( S

) and ¹ _q + ² _p = 1: choose µ a,p (α) = µ k ψ ⁰ + i a ψ k ² L

( S

) − µ k ψ k ² L

( S

) ≥ − α k ψ k ² L

( S

)

J. Dolbeault Magnetic ground state & symmetry

Proofs (Hardy inequality)

Let τ ≥ 0, x = (r , ϑ) ∈ R ² be polar coordinates in R ² ˆ

R

| (i ∇ + a) Ψ | ² − τ ϕ

| x | ² | Ψ | ²

dx

= ˆ ∞

0

ˆ

S

r | ∂ r Ψ | ²

| {z }

≥0

+ 1

r | ∂ ϑ Ψ + i a Ψ | ² − τ ϕ r | Ψ | ²

dϑ dr

≥ λ 1 (H a − τ ϕ) ˆ ∞

0

ˆ

S

1

r | Ψ | ² dϑ dr

≥ − α a,p (τ k ϕ k ^L

⁽ S

) ) ˆ ∞

0

ˆ

S

1

r | Ψ | ² dϑ If τ = 0, then α a,p (τ k ϕ k ^L

⁽ S

) ) = α a,p (0) = − a ²

α a,p (τ k ϕ k ^L

⁽ S

) ) > 0 for τ large

= ⇒ ∃ ! τ > 0 such that α a,p (τ k ϕ k ^L

⁽ S

) ) = 0

J. Dolbeault Magnetic ground state & symmetry

Comments

B The region a ² (p + 2) + α (p − 2) < 1 is exactly the set where the constant functions are linearly stable critical points

B The proof of the rigidity result is based

- neither on the carr´ e du champ method, at least directly

- nor on a Fourier representation of the operator as it was the case in earlier proofs (p = + ∞ , or p > 2 and α = 0)

B Magnetic rings: see [Bonnaillie-No¨el, H´erau, Raymond, 2017]

B Deducing Hardy’s inequality applied with Aharonov-Bohm magnetic fields from a Keller-Lieb-Thirring inequality is an extension of

[Hoffmann-Ostenhof, Laptev, 2015] to the magnetic case B Our results are not limited to the semi-classical regime

J. Dolbeault Magnetic ground state & symmetry

Symmetry in Aharonov-Bohm magnetic fields

Aharonov-Bohm effect

Interpolation and Keller-Lieb-Thirring inequalities in R ² B Statements

B Constants and numerics

Symmetry and symmetry breaking

Joint work with D. Bonheure, M.J. Esteban, A. Laptev, & M. Loss

J. Dolbeault Magnetic ground state & symmetry

Aharonov-Bohm effect

A major difference between classical mechanics and quantum mechanics is that particles are described by a non-local object, the wave function. In 1959 Y. Aharonov and D. Bohm proposed a series of experiments intended to put in evidence such phenomena which are nowadays called Aharonov-Bohm effects

One of the proposed experiments relies on a long, thin solenoid which produces a magnetic field such that the region in which the magnetic field is non-zero can be approximated by a line in dimension d = 3 and by a point in dimension d = 2

B [Physics today, 2009] “The notion, introduced 50 years ago, that electrons could be affected by electromagnetic potentials without coming in contact with actual force fields was received with a skepticism that has spawned a flourishing of experimental tests and expansions of the original idea.” Problem solved by considering appropriate weak solutions !

B Is the wave function a physical object or is its modulus ? Decisive experiments have been done only 20 years ago

J. Dolbeault Magnetic ground state & symmetry

The interpolation inequality

Let us consider an Aharonov-Bohm vector potential A(x) = a

| x | ² (x 2 , − x 1 ) , x = (x 1 , x 2 ) ∈ R ² \ { 0 } , a ∈ R Magnetic Hardy inequality [Laptev, Weidl, 1999]

ˆ

R

|∇ ^A ψ | ² dx ≥ min

k∈Z (a − k ) ² ˆ

R

| ψ | ²

| x | ² dx where ∇ ^A ψ := ∇ ψ + i A ψ, so that, with ψ = | ψ | e ^iS

ˆ

R

|∇ ^A ψ | ² dx = ˆ

R

(∂ r | ψ | ) ² + (∂ r S ) ² | ψ | ² + 1

r ² (∂ θ S + A) ² | ψ | ²

dx Magnetic interpolation inequality

ˆ

R

|∇ ^A ψ | ² dx + λ ˆ

R

| ψ | ²

| x | ² dx ≥ µ(λ) ˆ

R

| ψ | ^p

| x | ² dx 2/p

B Symmetrization: [Erd¨ os, 1996], [Boulenger, Lenzmann], [Lenzmann, Sok]

J. Dolbeault Magnetic ground state & symmetry