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Hardy inequalities with Aharonov-Bohm type magnetic field on the Heisenberg group

Journal of Inequalities and Applications20152015:95

https://doi.org/10.1186/s13660-015-0617-4

• Accepted: 3 March 2015
• Published:

Abstract

We introduce an Aharonov-Bohm type magnetic field on three-dimensional Heisenberg group and show this quadratic form satisfy an improved Hardy inequality with weights.

Keywords

• Hardy inequality
• Heisenberg group
• Aharonov-Bohm magnetic field

• 22E25
• 35H20

1 Introduction

The classical Hardy inequality states that, for $$N\geq3$$ and for all $$u\in C^{\infty}_{0}(\mathbb{R}^{N})$$,
$$\int_{\mathbb{R}^{N}}|\nabla u|^{2}\, dx\geq \biggl(\frac{N-2}{2} \biggr)^{2}\int_{\mathbb{R}^{N}} \frac{u^{2}}{|x|^{2}}\, dx$$
(1.1)
and $$(\frac{N-2}{2} )^{2}$$ is the best constant in (1.1). If $$N=2$$, the classical Hardy inequality fails. However, for some magnetic forms in dimension two, the Hardy inequality becomes possible. In fact, if β a is the Aharonov-Bohm magnetic field
$$\beta\mathbf{a}=\beta \biggl(-\frac{x_{2}}{x^{2}_{1}+x^{2}_{2}},\frac {x_{1}}{x^{2}_{1}+x^{2}_{2}} \biggr),$$
then for all $$u\in C^{\infty}_{0}(\mathbb{R}^{2}\setminus\{0\})$$ (cf. ),
$$\int_{\mathbb{R}^{2}}\bigl\vert (\nabla+i\beta\mathbf{a}) u\bigr\vert ^{2}\, dx\geq \min_{k\in\mathbb{Z}}|k- \beta|^{2}\int_{\mathbb{R}^{2}}\frac{u^{2}}{|x|^{2}}\, dx.$$
(1.2)
Recently, a version of the Aharonov-Bohm magnetic field for a Grushin subelliptic operator has been introduced by Aermark and Laptev . Furthermore, such quadratic form also satisfies an improved Hardy inequality. In the same paper, they asked the following question: does there exist a similar result for the Heisenberg quadratic form?
Recall that the three-dimension Heisenberg group $$\mathbb {H}_{1}=(\mathbb{R}^{2}\times\mathbb{R},\circ)$$ is a step-two nilpotent group whose group structure is given by
$$(x,y,t)\circ\bigl(x',y',t'\bigr)= \biggl(x+x',y+y',t+t'-\frac{1}{2} \bigl(xy'-yx'\bigr) \biggr).$$
The vector fields
$$X_{=}\frac{\partial}{\partial x}+\frac{y}{2}\frac{\partial}{\partial t},\qquad Y= \frac{\partial}{\partial y}-\frac{x}{2}\frac{\partial}{\partial t}$$
are left invariant and generate the Lie algebra of $$\mathbb{H}_{1}$$. The Kohn sub-Laplacian on $$\mathbb{H}_{1}$$ is
$$\Delta_{\mathbb{H}}=X^{2}+Y^{2}=\frac{\partial^{2}}{\partial x^{2}}+ \frac{\partial^{2}}{\partial y^{2}}+ \frac{x^{2}+y^{2}}{4}\frac{\partial^{2}}{\partial t^{2}}+\frac{1}{4} \biggl(y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y} \biggr)\frac{\partial}{\partial t}$$
and the subgradient is the vector given by $$\nabla_{\mathbb{H}}=(X,Y)$$. For simplicity, we let $$z=x+yi$$. Then $$|z|=\sqrt{x^{2}+y^{2}}$$. Denote
$$\rho:=\rho(z,t)=\bigl(|z|^{4}+16t^{2}\bigr)^{\frac{1}{4}}.$$
Similar as in [1, 2], we define an Aharonov-Bohm type magnetic field $$\mathcal{A}$$ on $$\mathbb{H}_{1}$$:
$$\mathcal{A}=(\mathcal{A}_{1},\mathcal{A}_{2})= \biggl(-\frac{Y\rho}{\rho },\frac{X\rho}{\rho} \biggr).$$
(1.3)
To our surprise, for such magnetic field (1.3), we cannot deal with the Hardy inequality
$$\int_{\mathbb{H}_{1}}|\nabla_{\mathbb{H}} u|^{2}\, dz\, dt\geq\int_{\mathbb{H}_{1}}\frac{|u|^{2}}{\rho^{2}}| \nabla _{\mathbb{H}} \rho|^{2}\, dz\, dt,\quad u\in C^{\infty}_{0}(\mathbb{H}_{1}),$$
(1.4)
but the Hardy inequality with weight
$$\int_{\mathbb{H}_{1}}\frac{|\nabla_{\mathbb{H}} u|^{2}}{|\nabla_{\mathbb{H}} \rho|^{2}}\, dz\, dt\geq\int _{\mathbb{H}_{1}}\frac{|u|^{2}}{\rho^{2}}\, dz\, dt,\quad u\in C^{\infty}_{0}(\mathbb{H}_{1}).$$
For the reasons, see Remark 2.2. For more Hardy inequalities on Heisenberg groups, we refer to .

The main result is the following theorem.

Theorem 1.1

We have, for $$u\in C^{\infty}_{0}(\mathbb{H}_{1})$$,
$$\int_{\mathbb{H}_{1}}\frac{|(\nabla_{\mathbb{H}}+i\beta\mathcal{A}) u|^{2}}{|\nabla_{\mathbb{H}}\rho|^{2}}\, dz\, dt\geq \Bigl(1+\min_{k\in \mathbb{Z}}|k-\beta|^{2} \Bigr)\int _{\mathbb{H}_{1}}\frac{|u|^{2}}{\rho ^{2}}\, dz\, dt.$$
(1.5)

2 Proof of Theorem 1.1

Before the proof of Theorem 1.1, we need a polar coordinate associated with ρ on $$\mathbb{H}_{1}$$. We describe it as follows. For each real number $$\lambda>0$$, there is a dilation naturally associated with the group structure which is usually denoted $$\delta_{\lambda}(x,y,t)=(\lambda x, \lambda y,\lambda^{2}t)$$. The Jacobian determinant of $$\delta_{\lambda}$$ is $$\lambda^{Q}$$, where $$Q=4$$ is the homogeneous dimension of $$\mathbb{H}_{1}$$. For simplicity, we use the notation $$\lambda (x,y,t)=\delta_{\lambda}(x,y,t)$$. Given any $$\xi=(x,y,t)\in \mathbb{H}_{1}$$, set $$x^{\ast}=\frac{x}{\rho}$$, $$y^{\ast}=\frac{y}{\rho}$$, $$t^{\ast}=\frac{t}{\rho^{2}}$$, and $$\xi^{\ast}=(x^{\ast},y^{\ast},t^{\ast})$$ if $$\rho(\xi)\neq0$$. The polar coordinate on $$\mathbb{H}_{1}$$ associated with ρ is the following (cf. , Proposition (1.15)):
$$\int_{\mathbb{H}_{1}}f(\xi)\, dz\, dt=\int^{\infty}_{0} \int_{\Sigma}f\bigl(\lambda \xi^{\ast}\bigr) \lambda^{3}\, d\sigma\, d\lambda,\quad f\in L^{1}( \mathbb{H}_{1}),$$
where $$\Sigma=\{(x,y,t)\in\mathbb{H}_{1}: \rho(x,y,t)=1\}$$ is the unit sphere associated with ρ. Moreover, there is a parametrization of this polar coordinate (cf. , Theorem 5.12):
$$\left \{ \begin{array}{l} x=\rho\sqrt{\cos\alpha}\cos\theta; \\ y=\rho\sqrt{\cos\alpha}\sin\theta; \\ t=\frac{1}{4}\rho^{2}\sin\alpha, \end{array} \right .$$
(2.1)
where $$\alpha\in[-\pi/2,\pi/2)$$, $$\theta\in[0,2\pi)$$, and $$0\leq\rho <\infty$$. Using this parametrization, we can rewrite the polar coordinate as follows:
$$\int_{\mathbb{H}_{1}}f(\xi)\, dz\, dt=\frac{1}{4} \int^{\infty}_{0}\int^{\pi /2}_{-\pi/2} \int^{2\pi}_{0}f(\xi)\rho^{3}\, d\rho\, d \alpha\, d\theta,\quad f\in L^{1}(\mathbb{H}_{1}).$$
(2.2)

Proof of Theorem 1.1

Using the identity
$$\bigl(a^{2}+b^{2}\bigr) \bigl(|z_{1}|^{2}+|z_{2}|^{2} \bigr)=|az_{1}+bz_{2}|^{2}+|az_{2}-bz_{1}|^{2}, \quad a,b\in\mathbb{R}, z_{1},z_{2}\in\mathbb{C},$$
we have
\begin{aligned}& |\nabla_{\mathbb{H}}\rho|^{2}\bigl\vert ( \nabla_{\mathbb{H}}+i\beta\mathcal{A}) u\bigr\vert ^{2} \\& \quad = \bigl(|X \rho|^{2}+|Y\rho|^{2}\bigr) \biggl(\biggl\vert Xu-i\beta \frac{Y\rho}{\rho}\biggr\vert ^{2}+\biggl\vert Yu+i\beta \frac {X\rho}{\rho}\biggr\vert ^{2} \biggr) \\& \quad = |X\rho\cdot Xu+Y\rho\cdot Yu|^{2}+\biggl\vert X\rho\cdot Yu-Y\rho \cdot Xu+i\beta\frac{|\nabla_{\mathbb{H}}\rho|^{2}u}{\rho}\biggr\vert ^{2}. \end{aligned}
(2.3)
By (2.1),
$$\frac{\partial}{\partial\theta}= \frac{\partial x}{\partial\theta }\frac{\partial}{\partial x}+ \frac{\partial y}{\partial\theta} \frac{\partial}{\partial y} =x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}$$
(2.4)
and
\begin{aligned} \frac{\partial}{\partial\alpha} =&\frac{\partial x}{\partial \alpha}\frac{\partial}{\partial x}+ \frac{\partial y}{\partial \alpha}\frac{\partial}{\partial y} +\frac{\partial t}{\partial\alpha}\frac{\partial}{\partial t} \\ =&-\frac{\rho\sin\alpha\cos\theta}{2\sqrt{\cos\alpha}}\frac{\partial }{\partial x}-\frac{\rho\sin\alpha\sin\theta}{2\sqrt{\cos\alpha}}\frac{\partial }{\partial y} + \frac{1}{4}\rho^{2}\cos\alpha\frac{\partial}{\partial t} \\ =&-\frac{2xt}{|z|^{2}}\frac{\partial}{\partial x}-\frac {2yt}{|z|^{2}}\frac{\partial}{\partial y}+ \frac{|z|^{2}}{4}\frac {\partial}{\partial t}. \end{aligned}
(2.5)
Therefore, by (2.4) and (2.5)
\begin{aligned} \begin{aligned}[b] X\rho\cdot Yu-Y\rho\cdot Xu&=\frac{|z|^{2}x+4yt}{\rho^{3}} \biggl( \frac {\partial u}{\partial y}-\frac{x}{2}\frac{\partial u}{\partial t} \biggr)- \frac{|z|^{2}y-4xt}{\rho^{3}} \biggl(\frac{\partial u}{\partial x}+\frac {y}{2}\frac{\partial u}{\partial t} \biggr) \\ &=\frac{|z|^{2}}{\rho^{3}} \biggl(x\frac{\partial}{\partial y}-y\frac {\partial}{\partial x} \biggr)+ \frac{4xt}{\rho^{3}} \frac{\partial u}{\partial x}+\frac{4yt}{\rho^{3}} \frac{\partial u}{\partial y}- \frac{|z|^{4}}{2\rho^{3}}\frac{\partial u}{\partial t} \\ &=\frac{|z|^{2}}{\rho^{3}}\frac{\partial u}{\partial\theta}+\frac {|z|^{2}}{\rho^{3}} \biggl( \frac{4xt}{|z|^{2}}\frac{\partial}{\partial x}+\frac{4yt}{|z|^{2}}\frac {\partial}{\partial y}- \frac{|z|^{2}}{2}\frac{\partial}{\partial t} \biggr) \\ &=\frac{|z|^{2}}{\rho^{3}} \biggl(\frac{\partial u}{\partial\theta }-2\frac{\partial u}{\partial\alpha} \biggr)= \frac{|\nabla_{\mathbb {H}}\rho|^{2}}{\rho} \biggl(\frac{\partial u}{\partial\theta}-2\frac {\partial u}{\partial\alpha} \biggr). \end{aligned} \end{aligned}
(2.6)
To get the last inequality above, we use the fact $$|\nabla_{\mathbb {H}}\rho|=\frac{|z|}{\rho}$$. Combining (2.3) and (2.6) yields
\begin{aligned} \int_{\mathbb{H}_{1}}\frac{|(\nabla_{\mathbb{H}}+i\beta\mathcal{A}) u|^{2}}{|\nabla_{\mathbb{H}}\rho|^{2}}\, dz\, dt =& \int _{\mathbb{H}_{1}}\frac{|X\rho\cdot Xu+Y\rho\cdot Yu|^{2}}{|\nabla _{\mathbb{H}}\rho|^{4}}\, dz\, dt \\ &{}+\int_{\mathbb{H}_{1}} \frac{|\frac{\partial u}{\partial\theta}-2\frac{\partial u}{\partial\alpha}+i\beta u|^{2}}{\rho^{2}}\, dz\, dt \\ =&(I)+(\mathit{II}), \end{aligned}
(2.7)
where
$$(I):= \int_{\mathbb{H}_{1}}\frac{|X\rho\cdot Xu+Y\rho\cdot Yu|^{2}}{|\nabla_{\mathbb{H}}\rho|^{4}}\, dz\, dt$$
and
\begin{aligned} (\mathit{II})&:= \int_{\mathbb{H}_{1}}\frac{|\frac{\partial u}{\partial \theta}-2\frac{\partial u}{\partial\alpha}+i\beta u|^{2}}{\rho^{2}}\, dz\, dt \\ &= \frac{1}{4}\int^{\infty}_{0}\int ^{\pi/2}_{-\pi/2}\int^{2\pi}_{0} \biggl\vert \frac{\partial u}{\partial\theta}-2\frac{\partial u}{\partial\alpha}+i\beta u\biggr\vert ^{2}\rho \, d\rho\, d\alpha \, d\theta. \end{aligned}
If we represent u by the Fourier series
$$u(\rho,\alpha,\theta)=\sum^{+\infty}_{n=-\infty}u_{n}( \rho,\alpha )e^{in\theta}/\sqrt{2\pi},$$
then
$$(\mathit{II})=\sum^{+\infty}_{n=-\infty} \frac{1}{4}\int^{\infty}_{0}\int ^{\pi /2}_{-\pi/2}\biggl\vert 2\frac{\partial u_{n} (\rho,\alpha)}{\partial \alpha}-i( \beta+n)u_{n}(\rho,\alpha)\biggr\vert ^{2}\rho\, d\rho \, d \alpha.$$
(2.8)
Similarly, representing $$u_{n}(\rho,\alpha)$$ by the Fourier series
$$u_{n}(\rho,\alpha)=\sum^{+\infty}_{k=-\infty}u_{n,k}( \rho)e^{i2k\alpha }/\sqrt{\pi},$$
we have
\begin{aligned} (\mathit{II}) =&\sum^{+\infty}_{n=-\infty} \sum^{+\infty}_{k=-\infty}\frac{|4k-n-\beta |^{2}}{4}\int ^{\infty}_{0}\bigl\vert u_{n,k}(\rho)\bigr\vert ^{2}\rho\, d\rho \\ \geq&\min_{k\in \mathbb{Z}}|k-\beta|^{2}\cdot\frac{1}{4} \sum^{+\infty}_{n=-\infty}\sum ^{+\infty}_{k=-\infty}\int^{\infty}_{0} \bigl\vert u_{n,k}(\rho)\bigr\vert ^{2}\rho \, d\rho \\ =&\min_{k\in \mathbb{Z}}|k-\beta|^{2}\cdot\frac{1}{4} \int^{\infty}_{0}\int^{\pi /2}_{-\pi/2} \int^{2\pi}_{0}\bigl\vert u(\rho,\alpha,\theta) \bigr\vert ^{2}\rho \, d\rho\, d\alpha\, d\theta \\ =&\min_{k\in \mathbb{Z}}|k-\beta|^{2}\int_{\mathbb{H}_{1}} \frac{|u|^{2}}{\rho^{2}}\, dz\, dt. \end{aligned}
(2.9)
To finish the proof, it is enough to show
$$(I)=\int_{\mathbb{H}_{1}}\frac{|X\rho\cdot Xu+Y\rho\cdot Yu|^{2}}{|\nabla_{\mathbb{H}}\rho|^{4}}\, dx\, dy\, dt \geq \int _{\mathbb{H}_{1}}\frac{|u|^{2}}{\rho^{2}}\, dx\, dy\, dt.$$
This will be done in Lemma 2.1. The proof of Theorem 1.1 is therefore completed. □
Before we turn to the proof of Lemma 2.1, we need the horizontal polar coordinates on $$\mathbb{H}_{1}$$ which have been introduced by Korányi and Reimann  (see also , pp.110-112). Set
$$\gamma_{\xi}(\rho)= \biggl(sze^{4i\frac{t}{|z|^{2}}\log \rho},\frac{1}{4} \rho^{2}t \biggr),\quad \xi=(z,t)\in\Sigma.$$
The horizontal polar coordinate on $$\mathbb{H}_{1}$$ is
$$\int_{\mathbb{H}_{1}}f(z,t)\, dz\, dt=\int^{\infty}_{0} \int_{\Sigma}f\bigl(\gamma _{\xi}(\rho)\bigr) \rho^{3}\, ds\, d\sigma,\quad f\in L^{1}( \mathbb{H}_{1}).$$
Furthermore, we can also give a parametrization of this polar coordinate through setting (, pp.111-112)
$$\Phi: \left \{ \begin{array}{l} x=\rho\sqrt{\cos\alpha}\cos(\theta+4\tan\alpha\log\rho); \\ y=\rho\sqrt{\cos\alpha}\sin(\theta+4\tan\alpha\log\rho); \\ t=\rho^{2}\sin\alpha. \end{array} \right .$$
The Jacobian determinant of Φ is $$\rho^{3}$$ so that
\begin{aligned} \int_{\mathbb{H}_{1}}f(z,t)\, dz\, dt =&\int ^{\infty}_{0}\int_{\Sigma}f\bigl(\gamma _{\xi}(\rho)\bigr)\rho^{3}\, d\rho \, d\xi \\ =&\frac{1}{4}\int^{\infty}_{0}\int ^{\pi/2}_{-\pi/2}\int^{2\pi }_{0}f \bigl(\gamma_{\xi}(\rho)\bigr)\rho^{3}\, d\rho \, d\alpha \, d \theta. \end{aligned}
(2.10)
Using this parametrization, we have (see , p.112)
\begin{aligned} \frac{d}{d\rho}f\bigl(\gamma_{\xi}(\rho)\bigr) =& \frac{1}{\rho|z|^{2}} \bigl(\bigl(x|z|^{2}-4yt\bigr)Xf+ \bigl(y|z|^{2}+4xt\bigr)Yf \bigr) \\ =&\frac{1}{4}\frac{\langle\nabla_{\mathbb{H}}\rho^{4},\nabla_{\mathbb{H}} f\rangle}{\rho|z|^{2}}=\frac{\rho^{2}}{|z|^{2}}\langle \nabla_{\mathbb {H}}\rho,\nabla_{\mathbb{H}} f\rangle \\ =&\frac{\langle\nabla_{\mathbb{H}}\rho,\nabla_{\mathbb{H}} f\rangle}{|\nabla_{\mathbb{H}}\rho|^{2}}. \end{aligned}
(2.11)

Lemma 2.1

We have, for $$u\in C^{\infty}_{0}(\mathbb{H}_{1})$$,
$$\int_{\mathbb{H}_{1}}\frac{|X\rho\cdot Xu+Y\rho\cdot Yu|^{2}}{|\nabla_{\mathbb{H}}\rho|^{4}}\, dz\, dt \geq \int _{\mathbb{H}_{1}}\frac{|u|^{2}}{\rho^{2}}\, dz\, dt.$$

Proof

Notice that
\begin{aligned} \begin{aligned}[b] &\int^{\infty}_{0}\biggl\vert \frac{d}{d\rho}f\bigl(\gamma_{\xi}(\rho)\bigr) \biggr\vert ^{2}\rho^{3}\, d\rho- \int^{\infty}_{0} \bigl\vert f\bigl(\gamma_{\xi}(\rho)\bigr)\bigr\vert ^{2} \rho\, d\rho \\ &\quad =\int^{\infty}_{0}\biggl\vert \frac{d(\rho f(\gamma_{\xi}(\rho)))}{d\rho}\biggr\vert ^{2}\rho \, d\rho\geq0. \end{aligned} \end{aligned}
(2.12)
Integrating over $$-\pi/2\leq\alpha\leq\pi/2$$ and $$0\leq\theta\leq 2\pi$$ and using (2.10) yields
$$\int_{\mathbb{H}_{1}}\biggl\vert \frac{d}{d\rho}f\bigl( \gamma_{\xi}(\rho)\bigr) \biggr\vert ^{2}\, dz\, dt\geq \int _{\mathbb{H}_{1}}\frac{|f(\gamma_{\xi}(\rho))|^{2}}{\rho^{2}}\, dz\, dt.$$
Combining the inequality above and (2.11) gives
$$\int_{\mathbb{H}_{1}}\frac{|X\rho\cdot Xu+Y\rho\cdot Yu|^{2}}{|\nabla_{\mathbb{H}}\rho|^{4}}\, dz\, dt \geq \int _{\mathbb{H}_{1}}\frac{|u|^{2}}{\rho^{2}}\, dz\, dt.$$
This completes the proof of Lemma 2.1. □

Remark 2.2

If one considers the Hardy inequality (1.4) with the Aharonov-Bohm type magnetic field $$\mathcal{A}$$ on $$\mathbb{H}_{1}$$, then, following the proof above, one needs to show
$$\int^{\pi/2}_{-\pi/2}\biggl\vert \frac{\partial u}{\partial\alpha}-i\beta u\biggr\vert ^{2}\cos\alpha \, d\alpha\geq \min_{k\in\mathbb{Z}}|2k-\beta|^{2} \int^{\pi/2}_{-\pi/2} \vert u\vert ^{2}\cos\alpha\, d\alpha.$$
(2.13)
However, to the best of our knowledge, it is not known whether inequality (2.13) is valid. The reason is that, in this case, $$\{e^{2ki\pi}/\sqrt{\pi } \}$$ is not an orthonormal basis because there exists a weight cosα in (2.13).

Declarations

Acknowledgements

This research is supported by the key project of Hubei provincial education department in P.R. China (No. D20142701) and the National Natural Science Foundation of China (No. 11201346).

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Authors’ Affiliations

(1)
School of Mathematics and Statistics, Hubei Engineering University, Xiaogan, Hubei, 432000, People’s Republic of China

References 