Symmetry Results in Two-Dimensional Inequalities for Aharonov–Bohm Magnetic Fields

Abstract

This paper is devoted to the symmetry and symmetry breaking properties of a two-dimensional magnetic Schrödinger operator involving an Aharonov–Bohm magnetic vector potential. We investigate the symmetry properties of the optimal potential for the corresponding magnetic Keller–Lieb–Thirring inequality. We prove that this potential is radially symmetric if the intensity of the magnetic field is below an explicit threshold, while symmetry is broken above a second threshold corresponding to a higher magnetic field. The method relies on the study of the magnetic kinetic energy of the wave function and amounts to study the symmetry properties of the optimal functions in a magnetic Hardy–Sobolev interpolation inequality. We give a quantified range of symmetry by a non-perturbative method. To establish the symmetry breaking range, we exploit the coupling of the phase and of the modulus and also obtain a quantitative result.

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Acknowledgements

This research has been partially supported by the project EFI, contract ANR-17-CE40-0030 (D.B., J.D.) of the French National Research Agency (ANR), by the NSF grant DMS-1600560 (M.L.), and by the PDR (FNRS) grant T.1110.14F and the ERC AdG 2013 339958 “Complex Patterns for Strongly Interacting Dynamical Systems - COMPAT” Grant (D.B.). The authors thank the referees for a careful reading which helped to remove some typos and improve the notations. © 2019 by the authors. This paper may be reproduced, in its entirety, for non-commercial purposes.

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Appendix

Appendix

Optimal constants in the symmetric case

It is known from [25] that

$$\begin{aligned} {\mathsf {K}}_\mathrm {A}^\star =\frac{p}{2}\,|\mathrm {A}|^{1+\frac{2}{p}}\left( \frac{2\,\sqrt{\pi }\,\varGamma \big (\frac{p}{p-2}\big )}{(p-2)\,\varGamma \big (\frac{p}{p-2}+\frac{1}{2}\big )}\right) ^{1-\frac{2}{p}} \end{aligned}$$

is the optimal constant in the inequality

$$\begin{aligned} \int _{{{\mathbb {R}}}}{|w'|^2}\,ds+\mathrm {A}^2\int _{{{\mathbb {R}}}}{|w|^2}\,ds\ge {\mathsf {K}}_\mathrm {A}^\star \left( \int _{{{\mathbb {R}}}}{|w|^p}\,ds\right) ^{2/p}\quad \forall \,w\in {\mathrm {H}}^1({{\mathbb {R}}}). \end{aligned}$$

For any given \(p>2\), the optimal constant in (13) is also \({\mathsf {K}}_\mathrm {A}^\star \) in the particular case of symmetric functions, because \(d\sigma \) is the uniform probability measure on \({{\mathbb {S}}}^1\). See for instance [7] for details.

The optimal constants in (10) and (13) are related by (14). In the symmetry range, we have \({\mathsf {K}}_\mathrm {A}={\mathsf {K}}_\mathrm {A}^\star =(2\,\pi )^{\frac{2}{p}-1}\,{\mathsf {C}}_\mathrm {A}^\star =(2\,\pi )^{\frac{2}{p}-1}\,{\mathsf {C}}_\mathrm {A}\), where

$$\begin{aligned} {\mathsf {C}}_\mathrm {A}^\star =\frac{p}{2}\,(2\,\pi )^{1-\frac{2}{p}}\,|\mathrm {A}|^{1+\frac{2}{p}}\left( \frac{2\,\sqrt{\pi }\,\varGamma \big (\frac{p}{p-2}\big )}{(p-2)\,\varGamma \big (\frac{p}{p-2}+\frac{1}{2}\big )}\right) ^{1-\frac{2}{p}}. \end{aligned}$$

This expression can be recovered by writing that the equality case in (10) is achieved by the function \(v_\star \). Indeed, with the change of variables \((r,\theta )\mapsto (s,\theta )\) with \(s=r^{-\mathrm {A}\,(p-2)/2}\) and \(n=2\,p/(p-2)\) as in [13, Section 3.1], and \(f_\star (s)=v_\star (r)\), this means that

$$\begin{aligned} {\mathsf {C}}_\mathrm {A}^\star =(2\,\pi )^{1-\frac{2}{p}}\left( \frac{\mathrm {A}}{2}\,(p-2)\right) ^{1+\frac{2}{p}}\,\frac{\int _0^{+\infty }|f_\star '|^2\,s^{n-1}\,ds}{\left( \int _0^{+\infty }|f_\star |^p\,s^{n-1}\,ds\right) ^{2/p}} \end{aligned}$$

where \(f_\star \) is the Aubin-Talenti function \(f_\star (s)=\left( 1+s^2\right) ^{-(n-2)/2}\).

Ground state eigenvalues of the quadratic form

\(\bullet \) Linearization and eigenvalues: the one-dimensional case

Let us summarize some classical results on the linearization of the Gagliardo-Nirenberg inequalities in the one-dimensional case, based on [9, Appendix A.2]. According, e.g., to [10], the function \({\overline{w}}(s)=(\cosh s)^{-\frac{2}{p-2}}\) is the unique positive solution of

$$\begin{aligned} -\,(p-2)^2\,{\overline{w}}''+4\,{\overline{w}}-2\,p\,{\overline{w}}^{p-1}=0 \end{aligned}$$

on \({{\mathbb {R}}}\), up to translations. The function \(w(s):=\alpha \,{\overline{w}}(\beta \,s)\) solves

$$\begin{aligned} -\,w''+\frac{4\,\beta ^2}{(p-2)^2}\,w-\frac{2\,p\,\beta ^2}{(p-2)^2}\,\alpha ^{2-p}\,w^{p-1}=0. \end{aligned}$$

With \(\beta =\frac{p-2}{2}\,\sqrt{\kappa }\) and \(\alpha =(\frac{p}{2}\,\kappa )^\frac{1}{p-2}\), \(w=\alpha \,w_\star \) is given by

$$\begin{aligned} w(s)=\left( \frac{p}{2}\,\kappa \right) ^\frac{1}{p-2}\left[ \cosh \left( \frac{p-2}{2}\,\sqrt{\kappa }\,s\right) \right] ^{-\frac{2}{p-2}}\quad \forall \,s\in {{\mathbb {R}}}\end{aligned}$$

and solves

$$\begin{aligned} -\,w''+\kappa \,w=|w|^{p-2}\,w. \end{aligned}$$

The ground state energy \(\lambda _1({\mathcal {H}}_\kappa )\) of the Pöschl-Teller operator

$$\begin{aligned} {\mathcal {H}}_\kappa :=-\,\frac{d^2}{ds^2}+\,\kappa -\,(p-1)\,w^{p-2} \end{aligned}$$

is characterized as follows. The function

$$\begin{aligned} \varphi _1(s):=\alpha ^\frac{p}{2}\,\big (\cosh (\beta \,s)\big )^{-\frac{p}{p-2}}=w^\frac{p}{2} \end{aligned}$$

solves

$$\begin{aligned} -\,\varphi _1''+\frac{1}{4}\,\kappa \,p^2\,\varphi _1-\,(p-1)\,w^{p-2}\,\varphi _1=0 \end{aligned}$$

and therefore provides the principal eigenvalue of \({\mathcal {H}}_\kappa \),

$$\begin{aligned} \lambda _1({\mathcal {H}}_\kappa )=-\,\frac{\kappa }{4}\left( p^2-\,4\right) . \end{aligned}$$

The Sturm-Liouville theory guarantees that \(\varphi _1\) generates the ground state.

\(\bullet \) The lowest non-radial mode on the cylinder

Let us consider the operator

$$\begin{aligned} {\mathcal {H}}_{\kappa ,\nu }:=-\,\frac{\partial ^2}{\partial s^2}-\,\nu \,\frac{\partial ^2}{\partial \theta ^2}+\,\kappa -\,(p-1)\,w^{p-2} \end{aligned}$$

on the cylinder \({{\mathbb {R}}}\times {{\mathbb {S}}}^1\ni (s,\theta )\). By separation of variables, the lowest non-symmetric eigenvalue is associated with the function \(\varphi (s,\theta )=\varphi _1(s)\,\cos \theta \), so that the ground state of \({\mathcal {H}}_{\kappa ,\nu }\) is

$$\begin{aligned} \lambda _1({\mathcal {H}}_{\kappa ,\nu })=\nu -\,\frac{\kappa }{4}\left( p^2-\,4\right) . \end{aligned}$$

\(\bullet \) Lowest eigenvalues and threshold for the linear instability

Whenever the optimal function in (3) is radially symmetric, we get that \(\mu (\lambda )={\mathsf {C}}_\mathrm {A}^\star \) with \(\mathrm {A}=\sqrt{a^2+\lambda }\), i.e., \(\mu (\lambda )=(2\,\pi )^{1-\frac{2}{p}}\,{\mathsf {k}}^\star (\lambda )\), or

$$\begin{aligned} \mu (\lambda )=\frac{p}{2}\,(2\,\pi )^{1-\frac{2}{p}}\left( \lambda +a^2\right) ^{\frac{1}{2}+\frac{1}{p}}\left( \frac{2\,\sqrt{\pi }\,\varGamma \big (\frac{p}{p-2}\big )}{(p-2)\,\varGamma \big (\frac{p}{p-2}+\frac{1}{2}\big )}\right) ^{1-\frac{2}{p}}. \end{aligned}$$
(18)

With \({\mathcal {F}}_{\kappa ,\nu }\) defined by (15), let us consider a Taylor expansion of \({\mathcal {F}}_{\kappa ,\nu }[w_\varepsilon ]\) with \(w_\varepsilon (s,\theta ):=w_\star (s)+\varepsilon \,\varphi (s,\theta )\) at order two with respect to \(\varepsilon \). For \(\varepsilon >0\) small enough, the sign of \({\mathcal {F}}_{\kappa ,\nu }[w_\varepsilon ]\) is determined by the sign of the quadratic form

$$\begin{aligned}&\varphi \mapsto \iint _{{{\mathbb {R}}}\times {{\mathbb {S}}}^1}{\left( |\partial _s\varphi |^2+\nu \,|\partial _\theta \varphi |^2+\kappa \,\varphi ^2\right) }\,ds\,d\sigma \\&\quad -\,(2\,\pi )^{\frac{2}{p}-1}\,(p-1)\,\mu \left( \iint _{{{\mathbb {R}}}\times {{\mathbb {S}}}^1}{w_\star ^p}\,ds\,d\sigma \right) ^{\frac{2}{p}-1}\iint _{{{\mathbb {R}}}\times {{\mathbb {S}}}^1}{w_\star ^{p-2}\,|\varphi |^2}\,ds\,d\sigma . \end{aligned}$$

Hence \({\mathcal {F}}_{\kappa ,\nu }\) can be made negative by choosing \(\varphi =\varphi _1\), which shows that \(w_\star \) is an instable critical point of \({\mathcal {F}}_{\kappa ,\nu }\) if and only if \(\lambda _1({\mathcal {H}}_{\kappa ,\nu })<0\). Notice that Lemma 3 states the reverse result, which is the difficult part of the result: whenever \(\lambda _1({\mathcal {H}}_{\kappa ,\nu })\ge 0\), the minimum of \({\mathcal {F}}_{\kappa ,\nu }\) is achieved by \(w_\star \) so that \({\mathcal {F}}_{\kappa ,\nu }\ge {\mathcal {F}}_{\kappa ,\nu }[w_\star ]\ge 0\).

Applied with \(\kappa =\lambda +a^2\) and \(\nu =1\), we recover the computation of [19], which determines \(\lambda _{\mathrm {FS}}\) as in (17). Applied with \(\kappa =\lambda +a^2\) and \(\nu =1-\,4\,a^2\), we obtain that

$$\begin{aligned} \lambda _1({\mathcal {H}}_{\lambda +a^2,1})-\,4\,a^2=1-\,\frac{1}{4}\left( \lambda +a^2\right) \left( p^2-\,4\right) -\,4\,a^2 \end{aligned}$$

and observe that it is negative if and only if \(\lambda >\lambda _\star (a)\), with

$$\begin{aligned} \lambda _\star (a)=\frac{4\left( 1-3\,a^2\right) -\,a^2\,p^2}{p^2-4}. \end{aligned}$$

Using \(\mu (\lambda )=(2\,\pi )^{1-\frac{2}{p}}\,{\mathsf {k}}^\star (\lambda )\) in the symmetry range, we obtain that \(\mu _\star (a)=\mu \big (\lambda _\star (a)\big )\) with \(\mu \) given by (18), i.e.,

$$\begin{aligned} \mu _\star (a)=2\,p\left( \frac{1-4\,a^2}{p^2-4}\right) ^{\frac{1}{2}+\frac{1}{p}}\pi ^{\frac{3}{2}-\frac{3}{p}} \left( \frac{2\,\varGamma \big (\frac{p}{p-2}\big )}{(p-2)\,\varGamma \big (\frac{p}{p-2}+\frac{1}{2}\big )}\right) ^{1-\frac{2}{p}}. \end{aligned}$$

Computation of \(\lambda _\bullet \)

Let us give some details on the computation of \(\lambda _\bullet \). An expansion of \({\mathcal {Q}}[\varphi ,\chi ]\) as defined in Section 3.1 computed with the ansatz (16) shows that it has the sign of

$$\begin{aligned} \textstyle {\mathsf {q}}(\lambda ):=-\,\lambda ^2-\,2\left( 4\,\frac{p^2+4\,p-4}{(p-2)^3\,(p+2)}+\,a^2\right) \lambda +\,8\,\frac{2\,(3\,p-2)-a^2\left( p^3+2\,p^2+12\,p-8\right) }{(p-2)^3\,(p+2)}-a^4. \end{aligned}$$

Since \({\mathsf {q}}(\lambda _\star )=\big (\frac{8\,a}{p^2-4}\big )^2\,(1-4\,a^2)\) is positive for any \(a\in (0,1/2)\) and since \(\lim _{\lambda \rightarrow \infty }{\mathsf {q}}(\lambda )=-\infty \), we know that \(\lambda _\bullet \) defined by \({\mathsf {q}}(\lambda _\bullet )=0\) is such that \(\lambda _\bullet >\lambda _\star \). Notice that the other root of \({\mathsf {q}}(\lambda )=0\) is in the range \((-\infty ,-a^2)\), and that the discriminant \(p^4-a^2\,(p-2)^2\,(p+2)\,(3\,p-2)\) is positive for any \((a,p)\in (0,1/2)\times (2,+\infty )\). Additionally, we obtain by direct computation that

$$\begin{aligned} \textstyle \lambda _\bullet -\lambda _\star =\frac{8}{(p-2)^3\,(p+2)}\left( \sqrt{p^4-a^2\,(p-2)^2\,(3\,p^2+4\,p-4)}+2\,a^2\,(p-2)^2-p^2\right) \end{aligned}$$

is positive for any \(a\in (0,1/2)\). Numerically this difference turns out to be very small: see Figs. 1 and 2.

Fig. 1
figure1

Here we assume that \(a\in (0,1/2)\) and consider the case \(p=4\). Left: The region of symmetry is the dark grey area which lies between the curves \(a\mapsto -\,a^2\) and \(a\mapsto \lambda _\star (a)\). The light grey area above \(a\mapsto \lambda _\bullet (a)\) is the region of symmetry breaking. The curve \(a\mapsto \lambda _{\mathrm {FS}}(a)\) is the dashed curve, above which the symmetry breaking is shown by considering only a perturbation of the modulus. It is a poor estimate away from a neighborhood of \(a=0\). Right: An enlargement of the boxed area shows that \(\lambda _\bullet \) and \(\lambda _\star \) do not coincide. Also see Fig. 2

Fig. 2
figure2

The curve \(a\mapsto \lambda _\bullet (a)-\lambda _\star (a)\) with \(p=4\) shows that there is a little gap between the symmetry and the symmetry breaking region, which is to be expected because \(\lambda _\star \) is determined by a non-optimal test function

The range of linear instability of the magnetic interpolation inequality

The high level of accuracy shown in Fig. 2 deserves some comments. For any given \((a,p)\in (0,1/2)\times (2,+\infty )\), the threshold between the symmetry region and the symmetry breaking region is in the interval \((\lambda _\star ,\lambda _\bullet )\). Since \(\lambda _\bullet \) is determined by the choice of (16), one has to understand why this test function gives such a precise estimate.

Let us consider an ansatz in which only the angular dependence is fixed. With

$$\begin{aligned} \varphi (s,\theta )=H(z)\,\cos \theta \quad \text{ and }\quad \chi (s,\theta )=G(z)\,\frac{\sin \theta }{w_\star (s)} \end{aligned}$$

and the change of variables

$$\begin{aligned} z=\tanh \left( \omega \,s\right) \quad \text{ and }\quad \omega =\frac{p-2}{2}\,\sqrt{\lambda +a^2}, \end{aligned}$$

the computation of \({\mathcal {Q}}[\varphi ,\chi ]\) is reduced to the computation of

$$\begin{aligned} \textstyle&{\mathsf {Q}}(G,H)\\&\quad :=\textstyle \int _{-1}^{+1}{\left( \omega ^2\left( 1-z^2\right) \,|G'|^2+\left( 1+\frac{4\,\omega ^2}{(p-2)^2}\right) \frac{G^2}{1-z^2}-\frac{2\,p\,\omega ^2}{(p-2)^2}\,G^2-\, \frac{4\,a}{1-z^2}\,G\,H\right) }\,dz\\&\quad \textstyle +\int _{-1}^{+1}{\left( \omega ^2\left( 1-z^2\right) \,|H'|^2+\left( 1+\frac{4\,\omega ^2}{(p-2)^2}\right) \frac{H^2}{1-z^2}-(p-1)\,\frac{2\,p\,\omega ^2}{(p-2)^2}\,H^2\right) }\,dz \end{aligned}$$

using

$$\begin{aligned} \frac{dz}{ds}=\omega \left( 1-z^2\right) ,\quad \lambda +a^2=\frac{4\,\omega ^2}{(p-2)^2}\quad \text{ and }\quad w_\star ^{p-2}=\frac{2\,p\,\omega ^2}{(p-2)^2}\left( 1-z^2\right) . \end{aligned}$$

We recover the expression of \({\mathsf {q}}(\lambda )\) with the choice

$$\begin{aligned} G[z]=\zeta \,H[z]\quad \text{ and }\quad H(z)=\left( 1-z^2\right) ^{\frac{p}{2\,(p-2)}}, \end{aligned}$$

after optimizing on \(\zeta \). All computations done, the optimal value of \(\zeta \) is

$$\begin{aligned} \zeta =\frac{a\,(p+2)\,(3\,p-2)}{p^2+\sqrt{p^4-a^2\,(p-2)^2 \,(p+2)\,(3\,p-2)}}, \end{aligned}$$

and we find that \({\mathsf {q}}(\lambda )<0\) for \(\lambda \) in the admissible range if and only if \(\lambda >\lambda _\bullet \).

A minimization of \({\mathsf {Q}}(G,H)\) under the constraint

$$\begin{aligned} \int _{-1}^{+1}{\frac{G^2+H^2}{1-z^2}}\,dz=1 \end{aligned}$$

reduces the problem to the identification of the ground state energy \(\varLambda \) in the eigenvalue problem

$$\begin{aligned} \left\{ \begin{array}{l} -\,\omega ^2\left( \left( 1-z^2\right) G'\right) '+\left( 1+\frac{4\omega ^2}{(p-2)^2}\right) \frac{G}{1-z^2}- \frac{2p\omega }{(p-2)^2} G-\frac{2a}{1-z^2}H=\frac{\varLambda }{1-z^2}G,\\ -\,\omega ^2\left( \left( 1-z^2\right) H'\right) '+\left( 1+\frac{4\omega ^2}{(p-2)^2}\right) \frac{H}{1-z^2}-(p-1)\frac{2p\omega }{(p-2)^2}H-\frac{2a}{1-z^2}G=\frac{\varLambda }{1-z^2}H. \end{array}\right. \end{aligned}$$

For given \((a,p)\in (0,1/2)\times (2,+\infty )\), the linear instability range \({\mathcal {I}}\) is the set of the parameters \(\lambda \) for which \(\varLambda \) is negative. We know that

$$\begin{aligned} (\lambda _\bullet ,+\infty )\subset {\mathcal {I}}\subset (\lambda _\star ,+\infty ) \end{aligned}$$

but we do not even know whether \({\mathcal {I}}\) is an interval or not. Notice that for \(a=1/2\), we find that \(\zeta =1\) and \(G=H\) is a good test function for any \(\omega >0\): this means that there is symmetry breaking for any \(\lambda >1/4\).

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Bonheure, D., Dolbeault, J., Esteban, M.J. et al. Symmetry Results in Two-Dimensional Inequalities for Aharonov–Bohm Magnetic Fields. Commun. Math. Phys. 375, 2071–2087 (2020). https://doi.org/10.1007/s00220-019-03560-y

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