RETRACTED ARTICLE: Boundary value behaviors for solutions of the equilibrium equations with angular velocity

Abstract

This work is concerned with a mixed boundary value problems for the slow equilibrium equations with prescribed angular velocity. As an application, we find sufficient conditions for the existence and uniqueness of blow-up solutions under weaker conditions.

Introduction

In 4-D space, the equilibrium equations for a self-gravitating fluid rotating about the $x_{4}$ axis with prescribed velocity $\Omega(r)$ can be written as

$$ \left \{ \textstyle\begin{array}{@{}l} \nabla P = \rho\nabla ( -\Phi+\int_{0}^{r} s \Omega^{2} (s)\,ds ),\\ \Delta\Phi=4\pi g \rho. \end{array}\displaystyle \right . $$

(1.1)

Here ρ, g, and Φ denote the density, gravitational constant, and gravitational potential, respectively, P is the pressure of the fluid at a point $x\in {\mathbb{R}}^{4}$, and $r=\sqrt{x_{1}^{2} +x_{2}^{2}}$. We want to find axisymmetric equilibria and therefore always assume that $\rho(x)=\rho(r,x_{4})$.

For the density ρ, from (1.1)₂ we can obtain the induced potential

$$ \Phi_{\rho}(x)= -g \int\frac{\rho(y)}{|x-y|}\,dy, $$

(1.2)

Obviously, $\Phi_{\rho}$ is decreasing when ρ is increasing.

In the study of this model, Auchmuty [1] proved the existence of an equilibrium solution if the angular velocity satisfied certain decay conditions. For a constant angular velocity, Miyamoto [2] has proved that there exists an equilibrium solution if the angular velocity is less than certain constant and that there is no equilibrium for large velocity. Pang et al. [3] talked about the exact numbers of the stationary solutions. For many other interesting results, see references [4–6].

Under more general conditions than in [2], we prove that there exists an equilibrium solution under the following constraint set

$$\begin{aligned} {\mathcal{A}}_{M}:=\biggl\{ \rho \Bigm| \rho\geq0, \rho \mbox{ is axisymmetric}, \int\rho \,dx =M \biggr\} . \end{aligned}$$

(1.3)

A standard method to obtain steady states is prescribing the minimizer of the stellar energy functional. The main problem is to show that the steady state has finite mass and compact support. To approach this problem, we define the energy functional

$$ F(\rho):= \int Q(\rho)\,dx- \int\rho J(r)\,dx -\frac{g}{2} \int \int \frac{\rho(x) \rho(y)}{|x-y|}\,dy\,dx. $$

(1.4)

Here

$$ Q(\rho)=\frac{1}{\gamma-1}P,\qquad J(r)= \int_{0}^{r} s\Omega^{2} (s)\,ds, $$

(1.5)

In this paper, we assume that $J(r)$ is nonnegative, continuous, and bounded on $[0,+\infty)$ and P is nonnegative, continuous, and strictly increasing for $s>0$ and satisfies:

$$P: \lim_{\rho\rightarrow0}P(\rho)\rho^{-1}=0,\qquad \lim_{\rho\rightarrow+\infty}P(\rho)\rho^{-\frac{4}{3}}=+\infty. $$

In Section 2, first we prove the existence of a minimizer of the energy functional F in ${\mathcal{A}}_{M}$. Then we give the properties of minimizers; they are stationary solutions of equation (1.1) with finite mass and compact support. The main difficulty in the proof is the loss of compactness due to the unboundedness of ${\mathbb{R}}^{4}$. To prevent the mass from running off to spatial infinity along a minimizing sequence, our variational approach is related to the concentration-compactness principle due to Fang and Li [4]. For many other interesting results, see references [6–8].

Throughout this paper, for simplicity of presentation, we use ∫ to denote $\int_{{\mathbb{R}}^{4}}$ and use $\| \cdot\|_{p}$ to denote $\| \cdot\|_{L^{p} ({\mathbb{R}}^{4})}$. Define

$$\begin{aligned} &B_{R} (x):=\bigl\{ y \in{{\mathbb{R}}^{4}} \mid |y-x|\leq R \bigr\} , \qquad B_{R,K}(x) :=\bigl\{ y \in{{\mathbb{R}}^{4}} \mid R\leq |y-x|\leq K \bigr\} , \\ &F_{\mathrm{pot}}(\rho):=-\frac{g}{2} \int \int\frac{\rho(x)\rho(y)}{|x-y|}\,dy\,dx =-\frac{1}{8\pi g} \int|\nabla\Phi_{\rho}|^{2}\,dx < 0 . \end{aligned}$$

(1.6)

We denote by C a generic positive constant and by χ the indicator function.

Minimizer of the energy

In this section, we present some properties of the functional F and prove the existence of a minimizer. It is easy to verify that the function F is invariant under any vertical shift, that is, if $\rho\in{{\mathcal{A}}_{M}}$, then $T\rho(x):=\rho(x+ae_{3})\in {\mathcal{A}}_{M}$ and $F(T\rho)=F(\rho)$ for any $a\in\mathbb{R}$. Here $e_{3} =(0,0,1)$. Therefore, if $(\rho_{n})$ is a minimizing sequence of F in ${\mathcal{A}}_{M}$, then $(T\rho_{n})$ is a minimizing sequence of F in ${\mathcal{A}}_{M}$ too. First, we give some estimates.

Lemma 2.1

Let $\rho\in L^{1} \cap L^{\gamma}({\mathbb{R}}^{4})$. If $1\leq \gamma\leq\frac{3}{2}$, then $\Phi\in L^{r} ({\mathbb{R}}^{4})$for $3< r<\frac{3\gamma}{3-2\gamma}$, and

$$ \|\Phi\|_{r}^{\rho} \leq C \bigl( \|\rho \|_{1}^{\alpha}\|\rho\|_{\gamma}^{1-\alpha} +\|\rho \|_{1}^{\beta}\|\rho\|_{\gamma}^{1-\beta} \bigr) $$

(2.1)

for $0<\alpha,\beta<1$. If $\gamma>\frac{3}{2}$, then Φ is bounded and continuous and satisfies (2.1) with $r=+\infty$.

Proof

The proof can be found in [1]. □

Lemma 2.2

For $\rho\in L^{1} \cap L^{\frac{4}{3}} ({\mathbb{R}}^{4})$, we have $\nabla\Phi\in L^{2} ({\mathbb{R}}^{4})$.

Proof

Interpolation inequality [9] implies

$$\|\rho\|_{\frac{6}{5}} \leq\|\rho\|_{1}^{1/3} \|\rho \|_{4/3}^{2/3}. $$

By Sobolev’s theorem, $\|\Phi\|_{6} \leq C \|\rho\|_{\frac{6}{5}}$. So

$$\|\nabla\Phi\|_{2}^{2} =4\pi g \|\rho\Phi\|_{1} \leq C \|\rho\|_{\frac{6}{5}}\|\Phi\|_{6} \leq C \|\rho \|_{\frac{6}{5}}^{2} . $$

From the above estimates we can complete our proof. □

Lemma 2.3

Assume that $P_{1}$holds. Then there exists a nonnegative constantC, depending only on $\frac{1}{|x|}$, M, and $J(r)$, such that $F\geq-C$.

Proof

For $\rho\in{\mathcal{A}}_{M}$, since $P_{1}$ holds, similarly to [2], we know that there exists a constant $S_{1}>0$ such that

$$\begin{aligned} F(\rho) \geq& \int_{\rho< S_{1}} Q(\rho)+ \int_{\rho\geq S_{1}} Q(\rho) -M\|J\|_{\infty} -CM^{2/3} \int\rho^{4/3} \\ \geq& \int_{\rho< S_{1}} Q(\rho)+\frac{1}{2} \int_{\rho\geq S_{1}} Q(\rho) -M\|J\|_{\infty} -CM^{2/3} \int_{\rho< S_{1}} \rho^{4/3} \\ \geq& \frac{1}{2} \int Q(\rho) - M\|J\|_{\infty}-CM^{5/3}S_{1}^{1/3}. \end{aligned}$$

So $F\geq-C_{1}$ with $C_{1} =M\|J\|_{\infty}-CM^{5/3}S_{1}^{1/3}$. □

Let $h_{M}=\inf_{{\mathcal{A}}_{M}} F$. A simple scaling argument shows that $h_{M} <0$: let $\overline{\rho}(x)=\varepsilon^{3}\rho(\varepsilon x)$, then $\int\overline{\rho}=\int\rho$. Since $\lim_{\rho \rightarrow0}Q(\rho)\rho^{-1}=0$, it is easy to see that for ε small enough, $\int Q(\overline{\rho})=\int\varepsilon^{-3}Q(\varepsilon^{3} \rho )\rightarrow0$. Therefore, $h_{M}<0$.

Lemma 2.4

Assume that $P_{1}$holds. Then for every $0<\widetilde{M}\leq M$, we have $h_{\widetilde{M}}\geq ( \frac{\widetilde{M}}{M})^{\frac{5}{3}} h_{M}$.

Proof

Let $\widetilde{\rho}(x)=\rho(ax)$ and $\widetilde{J}(r)=J(ax)$, where $a=(M/\overline{M})^{1/3}\geq1$. So, for any $\rho\in\mathcal{A}_{M}$ and $\widetilde{\rho}\in\mathcal{A}_{\overline{M}}$, we have

$$ F(\widetilde{\rho}) = \int Q(\widetilde{\rho}) - \int\widetilde{\rho }\widetilde{J} +F_{\mathrm{pot}}(\widetilde{\rho}) \geq b^{-3}F(\rho). $$

(2.2)

The mappings ${\mathcal{A}}_{M} \rightarrow{\mathcal {A}}_{\widetilde{M}}$, $\rho\rightarrow\widetilde{\rho}$, $J\rightarrow\widetilde{J}$ are all one-to-one and onto, which completes our proof. □

From Lemma 2.3 we immediately obtain that any minimizing sequence $(\rho_{n})\in{\mathcal{A}}_{M}$ of F satisfies

$$\int\rho_{n}^{4/3} = \int_{\rho_{n}< S_{1}} \rho_{n}^{4/3}+ \int_{\rho_{n}\geq S_{1}} \rho_{n}^{4/3}< MS_{1}^{1/3}+ \int c Q(\rho_{n})< 2cF(\rho_{n})+C+MS_{1}^{1/3}. $$

Lemma 2.5

Let $(\rho_{n})$be bounded in $L^{4/3}({\mathbb{R}}^{4})$and $\rho_{n}\rightharpoonup\rho_{0}$weakly in $L^{4/3}({\mathbb{R}}^{4})$. Then, for any $R>0$,

$$\int|\nabla\Phi_{\chi_{B_{R}} \rho_{n}}|^{2}\,dx \rightarrow \int|\nabla \Phi_{\chi_{B_{R}}\rho_{0}}|^{2}\,dx. $$

Proof

By Sobolev theorem and Lemma 2.1 we can complete the proof. □

Lemma 2.6

Assume that $P_{1}$holds. Let $(\rho_{n})_{n=1}^{\infty}\subset{\mathcal {A}}_{M}$be a minimizing sequence of $F(\rho)$. Then there exist a sequence $(a_{n})_{n=1}^{\infty}\subset{\mathbb{R}}^{4}$and $\delta_{0} >0$, $R_{0}> 0$such that

$$\int_{a_{n} + B_{R}} \rho_{n} (x)\,dx\geq\delta_{0},\quad R \geq R_{0} , $$

for all sufficiently large $n\in\mathbb{N}$.

Proof

Split the potential energy:

$$\begin{aligned} -\frac{2}{g}F_{\mathrm{pot}} :=& \int \int_{|x-y|\leq1/R} \frac{\rho_{n} (x) \rho_{n} (y)}{|x-y|}\,dy\,dx + \int \int_{1/R< |x-y|< R } +\cdots+ \int \int_{|x-y|\geq R} \cdots\\ :=& I_{1}+ I_{2} +I_{3} . \end{aligned}$$

From Lemma 2.2 we easily see that $I_{1} \leq\frac{C}{R}$. The estimates for $I_{2}$ and $I_{3}$ are straightforward:

$$\begin{aligned}& I_{2} \leq R \int \int_{|x-y|< R}\rho_{n} (x) \rho_{n} (y)\,dx\,dy \leq MR \sup_{a \in{\mathbb{R}}^{4}} \int_{a+ B_{R}}\rho_{n} (x)\,dx ; \\& I_{3} = \int \int_{|x-y|\geq R} \frac{\rho(x) \rho(y)}{|x-y|}\,dy\,dx \leq \frac{M^{2}}{R}. \end{aligned}$$

Therefore,

$$ \sup_{a \in{\mathbb{R}}^{4}} \int_{a+ B_{R}}\rho_{n} (x)\,dx \geq\frac{1}{M R} \biggl(-\frac{2}{g}F_{\mathrm{pot}}-\frac{M^{2}}{R}-\frac {C}{R} \biggr). $$

(2.3)

We know that $F_{\mathrm{pot}}(\rho_{n}) <0$ from (1.6). Thus, when R large enough, $-F_{\mathrm{pot}}>0$ dominates the sign of (2.3), so that there exist $\delta_{0} >0$, $R_{0}> 0$ as required. □

We are now ready to show the existence of a minimizer of $h_{M}$, provided that $P_{1}$ holds.

Theorem 2.1

Assume that $P_{1}$holds. Let $(\rho_{n})\in{\mathcal{A}}_{M}$be a minimizing sequence ofF. Then there exist a subsequence, still denoted by $(\rho_{n} )$, and a sequence of translations $T\rho_{n} :=\rho_{n} (\cdot+a_{n} e_{3} )$with constant $a_{n}$and $e_{3}=(0,0,1)$such that

$$F(\rho_{0})=\inf_{{\mathcal{A}}_{M}}F ( \rho_{n}) =h_{M} $$

and $T\rho_{n} \rightharpoonup\rho_{0}$weakly in $L^{\frac{4}{3}}({\mathbb{R}}^{4})$. For the induced potentials, we have $\nabla\Phi_{T\rho_{n} }\rightarrow\nabla\Phi_{\rho_{0} }$strongly in $L^{2}({\mathbb{R}}^{4})$.

Remark 2.1

Without admitting the spatial shifts, the assertion of the theorem is false: Given a minimizer $\rho_{0}$ and a sequence of shift vectors $(a_{n} e_{3} )\in{\mathbb{R}}^{4}$, the functional F is translation invariant, that is, $F(T\rho)=F(\rho)$. But if $|a_{n} e_{3} |\rightarrow \infty$, then this minimizing sequence converges weakly to zero, which is not in ${\mathcal{A}}_{M}$.

Proof

Split $\rho\in{\mathcal{A}}_{M}$ into three different parts:

$$\rho=\chi_{B_{R_{1}}}\rho+ \chi_{B_{R_{1},R_{2}}}\rho+ \chi_{B_{R_{2},\infty}}\rho:= \rho_{1} + \rho_{2} + \rho_{3} $$

with

$$I_{lm}:= \int \int\frac{\rho_{l}(x) \rho_{m} (y)}{|x-y|}\,dy\,dx, \quad l,m=1,2,3. $$

Thus,

$$F (\rho):= F (\rho_{1})+ F (\rho_{2})+F ( \rho_{3})-I_{12}- I_{13}-I_{23}. $$

If we choose $R_{2} >2R_{1}$, then

$$I_{13}\leq2 \int_{B_{R_{1}}}\rho(x)\,dx \int_{B_{R_{2},\infty}}|y|^{-1}\rho(y)\,dy \leq\frac{C_{1}}{R_{2}}. $$

Next we estimate $I_{12}$ and $I_{23}$:

$$\begin{aligned} I_{12}+ I_{23} =& - \int\rho_{1} \Phi_{2}\,dx- \int\rho_{2} \Phi_{3}\,dx = \frac{1}{4\pi g} \int\nabla(\Phi_{1} +\Phi_{3})\cdot\nabla \Phi_{2} \,dx \\ \leq& C_{2} \|\rho_{1} +\rho_{3}\|_{\frac{6}{5}} \|\nabla \Phi_{2}\|_{2} \leq C_{3} \|\nabla \Phi_{2}\|_{2}, \end{aligned}$$

where $\Phi_{l}=\Phi_{\rho_{l}}$.

Denote $M_{l} =\int\rho_{l}$, $l=1,2,3$. Then $M=M_{1}+ M_{2} +M_{3}$. Using the above estimates and Lemma 2.4, we have

$$\begin{aligned} h_{M}-F(\rho) \leq& \biggl( 1- \biggl( \frac{M_{1}}{M} \biggr)^{5/3} - \biggl( \frac{M_{2}}{M} \biggr)^{5/3} - \biggl( \frac{M_{3}}{M} \biggr)^{5/3} \biggr) h_{M} + \frac{C_{1}}{R_{2}}+ C_{3} {\|}\nabla\Phi_{2}\|_{2} \\ \leq& C_{4} h_{M} M_{1} M_{3} + C_{5}\biggl( \frac{1}{R_{2}}+\|\nabla\Phi_{2} \|_{2}\biggr), \end{aligned}$$

(2.4)

where $C_{4}$, $C_{5}$ are positive and depend on M but not on $R_{1}$ or $R_{2}$. Let $(\rho_{n})\in{\mathcal{A}}_{M}$ be a minimizing sequence and $(a_{n} e_{3})\in{\mathbb{R}}^{4}$ such that Lemma 2.6 holds. Since F is translation invariant, the sequence $(T\rho_{n} )$ is a minimizing sequence too. So, $||T\rho_{n}||_{1}\leq M$. Thus, there exists a subsequence, denoted by $(T\rho_{n} )$ again, such that $T\rho_{n} \rightharpoonup\rho_{0}$ weakly in $L^{\frac{4}{3}} ( {\mathbb{R}}^{4})$. By Mazur’s lemma and Fatou’s lemma,

$$ \int Q(\rho_{0})\,dx \leq\liminf_{n\rightarrow\infty} \int Q(T\rho_{n})\,dx. $$

(2.5)

Now we want to show that

$$\begin{aligned} \nabla\Phi_{T\rho_{n} }\rightarrow\nabla \Phi_{\rho_{0} } \mbox{ strongly in } L^{2}\bigl({\mathbb{R}}^{4}\bigr). \end{aligned}$$

(2.6)

Due to Lemma 2.5, $\nabla\Phi_{T\rho_{n,1} +T\rho_{n,2} }$ converge strongly in $L^{2}(B_{R_{2}})$. Therefore, we only need to show that for any $\varepsilon>0$,

$$\int| \nabla\Phi_{T\rho_{n,3} }|^{2}\,dx< \varepsilon. $$

By Lemmas 2.1 and 2.2 it suffices to prove that

$$ \int T\rho_{n,3}\,dx< \varepsilon. $$

(2.7)

Choosing $R_{0} < R_{1}$, we obtain that $M_{n,1} \geq\delta_{0}$ for n large enough from Lemma 2.6. By (2.4) we have

$$\begin{aligned} -C_{4} h_{M} \delta_{0} M_{n,3} \leq& -C_{4} h_{M} M_{n,1} M_{n,3} \\ \leq& \frac{C_{5} }{R_{2}} + C_{5}\|\nabla\Phi_{0,2} \|_{2} +C_{5} \|\nabla\Phi _{n,2}-\nabla \Phi_{0,2}\|_{2} + \bigl| F(T \rho_{n})- h_{M}\bigr|, \end{aligned}$$

(2.8)

where $\Phi_{n,l}$ is the potential induced by $T\rho_{n,l}$, which in turn has mass $M_{n,l}$, $n \in{\mathbb{N}} \cup\{ 0 \}$, and the index $l = 1, 2, 3$ refers to the splitting.

Given any $\varepsilon> 0$, by Lemma 2.6 we can increase $R_{1}> R_{0}$ so that $C_{5} \|\nabla\Phi_{0,2}\|_{2}<\varepsilon/4$. Next, choose $R_{2} > 2R_{1}$ such that the first term in (2.8) is less than $\varepsilon/4$. Now, since $R_{1}$ and $R_{2}$ are fixed, the third term converges to zero by Lemma 2.5. Since $( T\rho_{n})$ is a minimizing sequence, we have $|F( T\rho_{n})- h_{M}|< \varepsilon/4$ for suitable n. So, for n large enough,

$$-C_{4} h_{M} \delta_{0} M_{n,3} \leq \varepsilon, \quad\textit{i.e.}, M_{n,3} \leq \varepsilon; $$

thus, (2.7) holds, (2.6) follows, and

$$M \geq \int_{a_{n} + B_{R_{2}}} T\rho_{n} =M-M_{n,3}\geq M- \varepsilon. $$

Since $T\rho_{n} \rightharpoonup\rho_{0}$ weakly in $L^{1} ( {\mathbb{R}}^{N})$, it follows that for any $\varepsilon>0$, there exists $R>0$ such that

$$M \geq \int_{B_{R}}\rho_{0} \geq M-\varepsilon; $$

thus,

$$\rho_{0} \in L^{1} ({\mathbb{R}}^{N}) \mbox{ with } \int\rho_{0} \,dx =M, $$

so that$\rho_{0} \in{\mathcal{A}}_{M}$. Together with (2.5), we obtain

$$F(\rho_{0})=\inf_{{\mathcal{A}}_{M}}F =h_{M}. $$

The proof is completed. □

Next, we show that the minimizers obtained are steady states of equation (1.1).

Theorem 2.2

Let $\rho_{0}\in{\mathcal{A}}_{M}$be a minimizer of $F(\rho)$with induced potential $\Phi_{0}$. Then

$$\Phi_{0} + Q'(\rho_{0})-J(r)=K_{0}\textit{ on the support of }\rho_{0}, $$

where $K_{0}$is a constant. Furthermore, $\rho_{0}$satisfies (1.1).

Proof

We will derive the Euler-Lagrange equation for the variational problem. Let $\rho_{0} \in{\mathcal{A}}_{M}$ be a minimizer with induced potential $\Phi_{0}$. For any $\epsilon>0$, we define

$$V_{\epsilon}:= \biggl\{ x\in{\mathbb{R}}^{4} \Bigm| \epsilon \leq \rho_{0} \leq\frac{1}{\epsilon} \biggr\} . $$

For a test function $\omega\in L^{\infty}({\mathbb{R}}^{4})$ that has compact support and is nonnegative on $V_{\epsilon}^{c}$, define

$$\rho_{\tau}:= \rho_{0}+ \tau\omega- \tau\frac{\int\omega \,dy}{\operatorname{meas}(V_{\epsilon})} \chi_{V_{\epsilon}}, $$

where $\tau\geq0$ is small such that

$$\rho_{\tau}\geq0,\qquad \int\rho_{\tau}= \int\rho_{0} =M. $$

Therefore, $\rho_{\tau} \in{\mathcal{A}}_{M}$. Since $\rho_{0}$ is a minimizer of $F(\rho)$, we have

$$\begin{aligned} 0 \leq& F(\rho_{\tau})- F(\rho_{0}) \\ =& \int Q (\rho_{\tau})- Q (\rho_{0})\,dx- \int J(r) (\rho_{\tau}-\rho_{0}) +\frac{1}{2} \int(\rho_{\tau}\Phi_{\tau}-\rho_{0} \Phi_{0})\,dx \\ \leq& \int\bigl( Q' (\rho_{0})-J(r)\bigr) ( \rho_{\tau} - \rho_{0})\,dx + \int(\rho_{\tau}\Phi_{0}-\rho_{0} \Phi_{0})\,dx+o(\tau) \\ = &\tau \int\bigl(Q' (\rho_{0})-J(r)+\Phi_{0} \bigr) \biggl(\omega- \frac{\int\omega \,dy}{\operatorname{meas}(V_{\epsilon})}\chi_{V_{\epsilon}} \biggr)\,dx+ o(\tau). \end{aligned}$$

Hence,

$$\int\biggl[Q' (\rho_{0})-J(r)+ \Phi_{0} -\frac{1}{\operatorname{meas}(V_{\epsilon})}\biggl( \int_{V_{\epsilon}}Q' (\rho_{0})-J(r)+ \Phi_{0} \,dy\biggr) \biggr]\omega \,dx \geq0. $$

This holds for all test functions ω positive and negative on $V_{\epsilon}$ as specified above; hence, for all $\epsilon>0$ small enough,

$$ Q' (\rho_{0})-J(r)+\Phi_{0}= K_{\epsilon}\quad\mbox{on } V_{\epsilon}, \quad\mbox{and}\quad Q' ( \rho_{0})-J(r)+\Phi_{0}\geq K_{\epsilon}\quad\mbox{on } V_{\epsilon}^{c}, $$

(2.9)

where $K_{\epsilon}$ is a constant. Taking the limit as $\epsilon \rightarrow0$, we get

$$ Q' (\rho_{0})-J(r)+\Phi_{0}= K_{0} \mbox{ on the support of } \rho_{0}. $$

(2.10)

By taking the gradient of both sides of (2.10) we can prove that $\rho_{0}$ satisfies the equilibrium equation (1.1). □