1 Introduction and main result

We consider the following stationary Hall-MHD system on $\mathbb{R}^{3}$ :

$\left\{ \begin{align} & (u.\nabla )u-(\nabla \times B)\times B-\Delta u+\nabla \pi =0, \\ & -\nabla \times (u\times B)+\nabla \times [(\nabla \times B)\times B]-\Delta B=0, \\ & \nabla .u=\nabla .B=0, \\ & (u,B)(x,0)=({{u}_{0}}(x),{{B}_{0}}(x)), \\ \end{align} \right.$

(1.1)

where $x\in \mathbb{R}^{3}$. Here $u=u(x, t)\in \mathbb{R}^{3}$, $% B=B(x, t)\in \mathbb{R}^{3}$ and $\pi =\pi (x, t)$ are non-dimensional quantities corresponding to the flow velocity, the magnetic field and the pressure at the point $\left( x, t\right)$, while $u_{0}(x)$ and $B_{0}(x)$ are the given initial velocity and initial magnetic field with $\nabla \mathbf{.}u_{0}=0$ and $\nabla \mathbf{.}B_{0}=0$, respectively. An explanation of the mathematical and physical background of equations (1.1) is given for example in ^[1] (see also ^{[4,5,6,7,9,10,11,12,13]} and the references therein).

In their famous paper ^[2], Chae-Degond-Liu proved (Theorem 2.5, p. 558) (see also ^[14]) the following Liouville-type theorem for the smooth solutions of (1.1) :

Theorem 1.1. Let $\left( u, B\right) \in C^{2}\left( \mathbb{R}^{3}\right)$ be a smooth solution of the stationary Hall-MHD system (1.1) such that

(i) $(u, B)\in L^{\frac{9}{2}}(\mathbb{R}^{3}),$

(ii) $(u, B)\in L^{\infty }(\mathbb{R}^{3}),$

(iii) the (weak and then by classical) solution $(u, B):\mathbb{R}% ^{3}\rightarrow \mathbb{R}^{3}$ is of finite energy in the sense that

$\int_{\mathbb{R}^{3}}\left\vert \nabla u\right\vert ^{2}dx+\int_{\mathbb{R}% ^{3}}\left\vert \nabla B\right\vert ^{2}dx ＜\infty .$

Then,

$u=B=0.$

The purpose of this note is to get rid of hypothesis (ii) and (iii) in theorem 1.1. More precisely, we shall prove the following result.

Theorem 1.2. Let $\left( u, B\right) \in C^{2}\left( \mathbb{R}^{3}\right)$ be a smooth solution of the Hall-MHD equations (1.1}) such that

$(u, B)\in L^{\frac{9}{2}}(\mathbb{R}^{3})\text{ and }\int_{\mathbb{R}% ^{3}}\left\vert \nabla B\right\vert ^{2}dx ＜\infty .$

Then,

$u=B=0\text{ in }\mathbb{R}^{3}.$

Remark 1.1. As mentioned in ^[3], if we set $B=0$ in the Hall-MHD system, the above theorem reduces to the well-known Galdi result ^[8] for the Navier--Stokes equations (see Theorem X.9.5, pp.729-730).

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2 Proof of Theorem 1.2

In order to prove our main result, we introduce some basic identifies in the fluid dynamic.

Lemma 2.1.

$\Delta u =\nabla \text{div}u-\nabla \times (\nabla \times u), \\ u\times (\nabla \times u) =\frac{1}{2}\nabla \left\vert u\right\vert ^{2}-(u\mathbf{.}\nabla )u, \\ \nabla \times (u\times B) =(B\mathbf{.}\nabla )u-(u\mathbf{.}\nabla )B+u% \text{div}B-B\text{div}u.$

Remark 2.1. Based on $\nabla \mathbf{.}B=0$ and Lemma 2.1, we get

$(\nabla \times B)\times B=\text{div}(B\otimes B-\frac{1}{2}\left\vert B\right\vert ^{2}I)=-\nabla \left\vert B\right\vert ^{2}-(B\mathbf{.}\nabla )B,$

(2.1)

where $I$ is the identical matrix.

We are now in a position to the proof of our main result.

Proof: Let $\left( u, B\right) \in C^{2}\left( \mathbb{R}^{3}\right)$ be a smooth solution of the Hall-MHD equations (1.1) satisfies

$(u, B)\in L^{\frac{9}{2}}(\mathbb{R}^{3})\text{ and }\int_{\mathbb{R}% ^{3}}\left\vert \nabla B\right\vert ^{2}dx ＜\infty .$

We shall first estimate the pressure in (1.1)$_{1}$. Taking the divergence of (1.1)$_{1}$ and using the identity (1.1), we have

$\Delta \left( \pi +\frac{\left\vert B\right\vert ^{2}}{2}\right) =-\sum_{j, k=1}^{3}\partial _{j}\partial _{k}(u_{j}u_{k}-B_{j}B_{k}),$

from which we have the representation formula of the pressure, using the Riesz transforms in $\mathbb{R}^{3}$:

$\pi =\sum_{j, k=1}^{3}\mathcal{R}_{j}\mathcal{R}_{k}(u_{j}u_{k}-B_{j}B_{k})-%\frac{\left\vert B\right\vert ^{2}}{2}.$

(2.2)

Using (2.2) and Calderón--Zygmund estimate, one has that

$\left\Vert \pi \right\Vert _{L^{q}}\leq C(\left\Vert u\right\Vert _{L^{2q}}^{2}+\left\Vert B\right\Vert _{L^{2q}}^{2}), \text{ }1＜q ＜\infty .$

(2.3)

For $\tau >0$, let $\varphi _{\tau }$ be a real nonincreasing smooth function defined in $\mathbb{R}^{3}$ such that

${{\varphi }_{\tau }}(x)=\left\{ \begin{align} & 1\ \text{for}\ \ \left| \text{x} \right|\le \tau , \\ & 0\ \text{for}\ \ \left| \text{x} \right|\ge \text{2}\tau , \\ \end{align} \right.$

and satisfying

$\left\Vert \nabla ^{k}\varphi _{\tau }\right\Vert _{L^{\infty }}\leq C\tau ^{-k}\text{ for }k=0, 1, 2, 3,$

for some positive constant $C$ independent of $x\in \mathbb{R}^{3}$.

Multiplying (1.1)$_{1}$ by $u\varphi _{\tau }$ and (1.1)$% _{2}$ by $B\varphi _{\tau }$, respectively, integrating by parts over $% \mathbb{R}^{3}$ and taking into acount (1.1)$_{3}$, add the result together, we obtain

$\int_{\mathbb{R}^{3}}\left\vert \nabla u\right\vert ^{2}\varphi _{\tau }dx+\int_{\mathbb{R}^{3}}\left\vert \nabla B\right\vert ^{2}\varphi _{\tau }dx \notag \\ =\frac{1}{2}\int_{\mathbb{R}^{3}}\left\vert u\right\vert ^{2}(u\mathbf{.}% \nabla )\varphi _{\tau }dx+\int_{\mathbb{R}^{3}}\pi (u\mathbf{.}\nabla )\varphi _{\tau }dx-\int_{\mathbb{R}^{3}}(u\times B).(\nabla \varphi _{\tau }\times B)dx \notag \\ +\int_{\mathbb{R}^{3}}[(\nabla \times B)\times B].(\nabla \varphi _{\tau }\times B)dx+\frac{1}{2}\int_{\mathbb{R}^{3}}\left\vert u\right\vert ^{2}\Delta \varphi _{\tau }dx+\frac{1}{2}\int_{\mathbb{R}^{3}}\left\vert B\right\vert ^{2}\Delta \varphi _{\tau }dx \notag \\ =\sum_{k=1}^{6}A_{k},$

(2.4)

where we have used the fact

$-\int_{\mathbb{R}^{3}}(\Delta w)w\varphi _{\tau }dx =\int_{\mathbb{R}% ^{3}}\left\vert \nabla w\right\vert ^{2}\varphi _{\tau }dx+\int_{\mathbb{R}% ^{3}}(w\nabla w).\nabla \varphi _{\tau }dx \\ =\int_{\mathbb{R}^{3}}\left\vert \nabla w\right\vert ^{2}\varphi _{\tau }dx+\frac{1}{2}\int_{\mathbb{R}^{3}}\nabla w^{2}.\nabla \varphi _{\tau }dx \\ =\int_{\mathbb{R}^{3}}\left\vert \nabla w\right\vert ^{2}\varphi _{\tau }dx-\frac{1}{2}\int_{\mathbb{R}^{3}}w^{2}\Delta \varphi _{\tau }dx.$

In the following, we will estimate all the terms on the right-hand side of (% 2.4). For the first integral $A_{1}$, Hólder's inequality yields

$\left\vert A_{1}\right\vert \leq C\int_{\tau \leq \left\vert x\right\vert \leq 2\tau }\left\vert u\right\vert ^{3}\left\vert \nabla \varphi _{\tau }\right\vert dx \\ \leq \frac{1}{2\tau }\left\Vert \nabla \varphi \right\Vert _{L^{\infty }}\left( \int_{\tau \leq \left\vert x\right\vert \leq 2\tau }\left\vert u\right\vert ^{\frac{9}{2}}dx\right) ^{\frac{2}{3}}\left( \int_{\tau \leq \left\vert x\right\vert \leq 2\tau }dx\right) ^{\frac{1}{3}} \\ \leq C\left\Vert u\right\Vert _{L^{\frac{9}{2}}(\tau \leq \left\vert x\right\vert \leq 2\tau )}^{3}\rightarrow 0\text{ as }\tau \rightarrow +\infty .$

As for $A_{2}$, using the Hólder inequality, it follows according to (2.3) that

$\left\vert A_{2}\right\vert \leq \int_{\tau \leq \left\vert x\right\vert \leq 2\tau }\left\vert \pi \right\vert \left\vert u\right\vert \left\vert \nabla \varphi _{\tau }\right\vert dx \\ \leq \frac{1}{\tau }\left\Vert \nabla \varphi \right\Vert _{L^{\infty }}\left( \int_{\mathbb{R}^{3}}\left\vert \pi \right\vert ^{\frac{9}{4}% }dx\right) ^{\frac{4}{9}}\left( \int_{\tau \leq \left\vert x\right\vert \leq 2\tau }\left\vert u\right\vert ^{\frac{9}{2}}dx\right) ^{\frac{2}{9}}\left( \int_{\tau \leq \left\vert x\right\vert \leq 2\tau }dx\right) ^{\frac{1}{3}} \\ \leq C(\left\Vert u\right\Vert _{L^{\frac{9}{2}}}^{2}+\left\Vert B\right\Vert _{L^{\frac{9}{2}}}^{2})\left\Vert u\right\Vert _{L^{\frac{9}{2}% }(\tau \leq \left\vert x\right\vert \leq 2\tau )}\rightarrow 0\text{ as }% \tau \rightarrow +\infty .$

Analogously to $A_{1}$, an application of the Hólder inequality shows that

$\left\vert A_{3}\right\vert \leq \int_{\tau \leq \left\vert x\right\vert \leq 2\tau }\left\vert u\right\vert \left\vert B\right\vert ^{2}\left\vert \nabla \varphi _{\tau }\right\vert dx \\ \leq \frac{1}{\tau }\left\Vert \nabla \varphi \right\Vert _{L^{\infty }}\left( \int_{\tau \leq \left\vert x\right\vert \leq 2\tau }\left\vert B\right\vert ^{\frac{9}{2}}dx\right) ^{\frac{4}{9}}\left( \int_{\tau \leq \left\vert x\right\vert \leq 2\tau }\left\vert u\right\vert ^{\frac{9}{2}% }dx\right) ^{\frac{2}{9}}\left( \int_{\tau \leq \left\vert x\right\vert \leq 2\tau }dx\right) ^{\frac{1}{3}} \\ \leq C\left\Vert B\right\Vert _{L^{\frac{9}{2}}(\tau \leq \left\vert x\right\vert \leq 2\tau )}^{2}\left\Vert u\right\Vert _{L^{\frac{9}{2}}(\tau \leq \left\vert x\right\vert \leq 2\tau )}\rightarrow 0\text{ as }\tau \rightarrow +\infty .$

Similar to the treatment of $A_{3}$, $A_{4}$ can be estimated as

$\left\vert A_{4}\right\vert \leq \int_{\tau \leq \left\vert x\right\vert \leq 2\tau }\left\vert \nabla B\right\vert \left\vert B\right\vert ^{2}\left\vert \nabla \varphi _{\tau }\right\vert dx \\ \leq \frac{1}{\tau }\left\Vert \nabla \varphi \right\Vert _{L^{\infty }}\left( \int_{\tau \leq \left\vert x\right\vert \leq 2\tau }\left\vert \nabla B\right\vert ^{2}dx\right) ^{\frac{1}{2}}\left( \int_{\tau \leq \left\vert x\right\vert \leq 2\tau }\left\vert B\right\vert ^{6}dx\right) ^{% \frac{1}{3}}\left( \int_{\tau \leq \left\vert x\right\vert \leq 2\tau }dx\right) ^{\frac{1}{6}} \\ \leq \frac{C}{\sqrt{\tau }}\left\Vert \nabla \varphi \right\Vert _{L^{\infty }}\left\Vert \nabla B\right\Vert _{L^{2}}\left\Vert B\right\Vert _{L^{6}}^{2} \\ \leq \frac{C}{\sqrt{\tau }}\left\Vert \nabla \varphi \right\Vert _{L^{\infty }}\left\Vert \nabla B\right\Vert _{L^{2}}^{3}\rightarrow 0\text{ as }\tau \rightarrow +\infty .$

Finally, calculating $A_{5}+A_{6}$ we obtain

$\left\vert A_{5}\right\vert +\left\vert A_{6}\right\vert \leq C\int_{\tau \leq \left\vert x\right\vert \leq 2\tau }(\left\vert u\right\vert ^{2}+\left\vert B\right\vert ^{2})\left\vert \Delta \varphi _{\tau }\right\vert dx \\ \leq C\frac{1}{\tau ^{2}}\left\Vert \Delta \varphi \right\Vert _{L^{\infty }}\left( \int_{\tau \leq \left\vert x\right\vert \leq 2\tau }(\left\vert u\right\vert ^{2}+\left\vert B\right\vert ^{2})^{\frac{9}{4}}dx\right) ^{% \frac{4}{9}}\left( \int_{\tau \leq \left\vert x\right\vert \leq 2\tau }dx\right) ^{\frac{5}{9}} \\ \leq C\frac{1}{\tau ^{\frac{1}{3}}}\left\Vert \Delta \varphi \right\Vert _{L^{\infty }}\left( \int_{\tau \leq \left\vert x\right\vert \leq 2\tau }(\left\vert u\right\vert ^{\frac{9}{2}}+\left\vert B\right\vert ^{\frac{9}{2% }})dx\right) ^{\frac{4}{9}} \\ \leq C\frac{1}{\tau ^{\frac{1}{3}}}\left\Vert \Delta \varphi \right\Vert _{L^{\infty }}\left( \left\Vert u\right\Vert _{L^{\frac{9}{2}% }}^{2}+\left\Vert B\right\Vert _{L^{\frac{9}{2}}}^{2}\right) \rightarrow 0% \text{ as }\tau \rightarrow +\infty .$

Here we have used the Cauchy inequality. Consequently, letting $\tau \rightarrow +\infty$ in (2.4), we obtain

$\underset{\tau \rightarrow +\infty }{\lim }\left( \int_{\mathbb{R}% ^{3}}\left\vert \nabla u\right\vert ^{2}\varphi _{\tau }dx+\int_{\mathbb{R}% ^{3}}\left\vert \nabla B\right\vert ^{2}\varphi _{\tau }dx\right) =0$

On the other hand, by means of the monotone convergence theorem, we deduce

$\int_{\mathbb{R}^{3}}\left\vert \nabla u\right\vert ^{2}dx+\int_{\mathbb{R}% ^{3}}\left\vert \nabla B\right\vert ^{2}dx=\underset{\tau \rightarrow +\infty }{\lim }\left( \int_{\mathbb{R}^{3}}\left\vert \nabla u\right\vert ^{2}\varphi _{\tau }dx+\int_{\mathbb{R}^{3}}\left\vert \nabla B\right\vert ^{2}\varphi _{\tau }dx\right) =0,$

and thus $u=$const and $B=$const. Since $(u, B)\in L^{\frac{9}{2}}(\mathbb{R}% ^{3})$, this latter condition delivers

$u=B=0.$

This completes the proof of Theorem 1.2.

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Acknowledgments

The author would like to express gratitude to Professor G.P. Galdi for valuable discussions on the results and suggestions to the improvement of this work.

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Conflict of Interest

We declare no conflicts of interest in this paper.

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