3.1 The expression of magnetic potential
Combining (2.14) and \(\frac{\partial}{\partial\varphi}=0\), we have
$$\begin{aligned} \nabla\varphi_{0}= \biggl(\frac{\partial\varphi_{0}}{\partial r}, \frac{1}{r}\frac{\partial\varphi_{0}}{\partial\theta},\frac {1}{r\sin\theta}\frac{\partial\varphi_{0}}{\partial\varphi} \biggr) = \biggl(\frac{\partial\varphi_{0}}{\partial r}, \frac{1}{r}\frac {\partial\varphi_{0}}{\partial\theta},0 \biggr). \end{aligned}$$
(3.1)
Using the divergence formula (2.15), from (2.21), (2.24), and (3.1) we can deduce that
$$ \operatorname {div}\biggl(\frac{1}{r^{2}\sin^{2}\theta}\nabla\varphi _{0} \biggr)=\frac{m}{r\sin\theta}=\frac{W}{\mu_{0}}\rho. $$
(3.2)
Let
$$ \varphi_{0}(x_{1},x_{2},x_{3})= \varphi_{0}(r,\theta). $$
(3.3)
Then (3.2) is equivalent to
$$ \frac{1}{x_{1}^{2}+x_{2}^{2}} \biggl(\frac{\partial^{2} \varphi _{0}}{\partial x_{1}^{2}}+ \frac{\partial^{2} \varphi_{0}}{\partial x_{2}^{2}}+\frac{\partial^{2} \varphi_{0}}{\partial x_{3}^{2}}-\frac {2x_{1}}{x_{1}^{2}+x_{2}^{2}}\frac{\partial\varphi_{0}}{\partial x_{1}}- \frac{2x_{2}}{x_{1}^{2}+x_{2}^{2}}\frac{\partial\varphi _{0}}{\partial x_{2}} \biggr)=\frac{W}{\mu_{0}}\rho, $$
(3.4)
where
$$ \rho(x_{1},x_{2},x_{3})=\rho(r,\theta). $$
Define the operator \(\mathcal{L}\) as follows:
$$ \mathcal{L}=\frac{1}{x_{1}^{2}+x_{2}^{2}} \biggl(\frac{\partial ^{2}}{\partial x_{1}^{2}}+ \frac{\partial^{2}}{\partial x_{2}^{2}}+\frac{\partial^{2}}{\partial x_{3}^{2}}-\frac {2x_{1}}{x_{1}^{2}+x_{2}^{2}}\frac{\partial}{\partial x_{1}}- \frac {2x_{2}}{x_{1}^{2}+x_{2}^{2}}\frac{\partial}{\partial x_{2}} \biggr). $$
(3.5)
It is easy to observe that \(\mathcal{L}\) is a symmetric operator. Hence, it has a Green’s function \(G(x,y)\) for \(x,y\in\mathbb{R}^{3}\). The following lemma gives an exact expression of \(G(x,y)\).
Lemma 3.1
The Green’s function
\(G(x,y)\)
for
\(\mathcal{L}\)
has the expression
$$ G(x,y)=-\frac{\Lambda ((x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2} )}{ ((x_{1}-y_{1})^{2} +(x_{2}-y_{2})^{2}+(x_{3}-y_{3})^{2} )^{\frac{3}{2}}}, $$
(3.6)
where is a positive constant.
Proof
We know that (3.2) is equivalent to (3.4). Let \(\varphi_{0}=r^{2}\overline{\varphi}\sin^{2}\theta\). Inserting this into (3.2), we get
$$ \frac{\partial^{2}\overline{\varphi}}{\partial r^{2}}+\frac{4}{r}\frac {\partial\overline{\varphi}}{\partial r}+ \frac{3\cos\theta }{r^{2}\sin\theta}\frac{\partial\overline{\varphi}}{\partial\theta }+\frac{1}{r^{2}}\frac{\partial^{2}\overline{\varphi}}{\partial\theta ^{2}}= \frac{W\rho}{\mu_{0}}, $$
(3.7)
and it is not difficult to observe that the Green’s function of the operator
$$\frac{\partial^{2}}{\partial r^{2}}+\frac{4}{r}\frac{\partial }{\partial r}+\frac{3\cos\theta}{r^{2}\sin\theta} \frac{\partial }{\partial\theta}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}} $$
is given by
$$ G(r,\theta)=-\frac{\Lambda}{r^{3}}, $$
(3.8)
where Λ is a positive constant.
Hence, the Green’s function \(G(x,y)\) of \(\mathcal{L}\) can be expressed as
$$ G(x,y)=-\frac{\Lambda r^{2} \sin^{2}\theta}{r^{3}}=\frac{\Lambda ((x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2} )}{ ((x_{1}-y_{1})^{2} +(x_{2}-y_{2})^{2}+(x_{3}-y_{3})^{2} )^{\frac{3}{2}}}. $$
(3.9)
□
3.2 Variational formulation
In this paper, we assume that the pressure function \(p(\rho)\) satisfies the α-law (1.2) for some constant \(\alpha> 1\). Let
$$ I(\rho)=\frac{\rho^{\alpha}}{\alpha-1}. $$
(3.10)
Then
$$ n(\rho)=I'(\rho). $$
(3.11)
Moreover, the Newtonian potential operator is given by (1.3), and we can denote
$$ \Psi(x)=- \int_{\mathbb{R}^{3}}\frac{\rho(y)}{\vert x-y\vert }\,dy:=-E(\rho). $$
(3.12)
Notice that the magnetic potential \(\varphi_{0}\) satisfies (3.2). By Lemma 3.1 we see that \(G(x,y)\) for \(x,y\in\mathbb{R}^{3}\) is the Green’s function for the operator \(\mathcal{L}\) defined in (3.5), that is,
$$ \mathcal{L}G=\delta(x-y), $$
(3.13)
where \(\delta(x-y)\) is the Dirac measure giving the unit mass to the point x. Since \(\mathcal{L}\) is symmetric, we have
$$ \langle\mathcal{L}\varphi_{0}, G \rangle=\langle \varphi_{0}, \mathcal {L}G \rangle= \bigl\langle \varphi_{0}, \delta(x-y) \bigr\rangle =\varphi_{0}(y), $$
(3.14)
where the inner product \(\langle\cdot, \cdot\rangle\) is taken in \(L^{2}\). Thus, we have the following integral representation for \(\varphi_{0}\):
$$ \varphi_{0}(x)=D(\rho), $$
(3.15)
where the integral operator D is given by
$$ D(\rho)=\frac{W}{\mu_{0}} \int_{\mathbb{R}^{3}}G(x,y)\rho(y)\,dy. $$
(3.16)
By (3.11), (3.12), and (3.16) it is obvious that equation (2.27) can be written as
$$ I'(\rho)+E(\rho)+WD(\rho)=\lambda\quad\text{for } \rho>0. $$
(3.17)
According to (3.17), we define the energy functional F as follows:
$$ F(\rho)= \int_{\mathbb{R}^{3}} \biggl[I(\rho) +\frac{1}{2}\rho E(\rho)+ \frac{1}{2}\rho WD(\rho) \biggr]\,dx, $$
(3.18)
where \(I(\rho)\) is the function given in (3.10). The energy functional \(F(\rho)\) means that solving (3.17) with the total mass constraint (2.28) is equivalent to proving that (3.18) has a minimizer in some function space X.
Now, we review the results for stationary solution (1.5). For \(0< M<\infty\), we define \(X_{M}\) by
$$\begin{aligned} \begin{aligned}[b] X_{M}&= \biggl\{ \rho: \mathbb{R}^{3}\rightarrow\mathbb{R}, \rho >0 \mbox{ a.e.}, \int_{\mathbb{R}^{3}}\rho \,dx=M, \mbox{and} \\ &\quad \int_{\mathbb{R}^{3}} \biggl[I(\rho)+\frac{1}{2}\rho E(\rho ) \biggr] \,dx< \infty \biggr\} . \end{aligned} \end{aligned}$$
(3.19)
For \(\rho\in X_{M}\), we define the energy function \(F_{1}\) for the nonrotating nonmagnetic by
$$ F_{1}(\rho)= \int_{\mathbb{R}^{3}} \biggl[I(\rho)+\frac{1}{2}\rho E(\rho) \biggr] \,dx. $$
(3.20)
Thanks to the Lemma 2 in [1], Federbush et al. [4] obtained the following useful lemma for the minimizer of the functional \(F_{1}\).
Lemma 3.2
Suppose that the pressure
\(p(\rho)=\rho^{\alpha}\) (\(\alpha>\frac {4}{3}\)). Let
\(\rho^{\ast}\)
be a minimizer of (3.20) in
\(X_{M}\), and let
$$ \Gamma_{M}= \bigl\{ x\in\mathbb{R}^{3}: \rho^{\ast}>0 \bigr\} . $$
(3.21)
Then there exists a constant
\(\lambda_{1}\)
such that
$$\begin{aligned} \textstyle\begin{cases} I(\rho)+ \frac{1}{2}\rho E(\rho)=\lambda_{1}, \quad x\in\Gamma_{M}, \\ E(\rho)\geq\lambda_{1}, \quad x\in\mathbb{R}- \Gamma_{M}. \end{cases}\displaystyle \end{aligned}$$
(3.22)
Remark 3.3
The variational problem is unusual, in that a solution turns out to have compact support. The reason is that the functional one seeks to minimize is not lower semicontinuous on the class of all admissible functions. Auchmuty and Richard [1] first restrict their considerations to functions with support in the ball of radius \(R_{M}\). Hence, we should find the radius \(R_{M}\). Note that, for \(\alpha>\frac{4}{3}\), Luo and Smoller [8] have proved that a local minimizer \(\rho^{\ast}\) of the function \(F_{1}\) in \(X_{M}\) exists. Also, they showed that the minimizer \(\rho^{\ast}\) is actually radial and unique and has compact support, that is, for given total mass M, there exists a unique constant \(R_{M}>0\) such that
$$\begin{aligned} \textstyle\begin{cases} \rho^{\ast}(x)>0 &\text{if } \vert x\vert < R_{M}, \\ \rho^{\ast}(x)=0&\text{if } \vert x\vert \geq R_{M}. \end{cases}\displaystyle \end{aligned}$$
(3.23)
Notice that \(\rho^{\ast}\) satisfying (3.22) in \(X_{M}\) is called a nonrotating nonmagnetic star solution, and \(R_{M}\) is called the radius of the nonrotating nonmagnetic star solution with total mass M.
Based on (3.23), we define the function spaces \(Y_{M}\) and \(Y_{\mathit {MR}_{0}}^{\alpha}\) as follows:
$$\begin{aligned}& \begin{aligned}[b] Y_{M}^{\alpha}&= \biggl\{ \rho:\mathbb{R}^{3} \rightarrow\mathbb{R}, \rho(x)=\rho(r,\theta), \rho\geq0, \mbox{a.e.}, \rho\in L^{1} \bigl(\mathbb{R}^{3} \bigr)\cap L^{\alpha} \bigl( \mathbb{R}^{3} \bigr), \\ &\quad \alpha>\frac{4}{3}, \int_{\mathbb{R}^{3}}\rho (x)\,dx=M \biggr\} , \end{aligned} \end{aligned}$$
(3.24)
$$\begin{aligned}& Y_{\mathit {MR}_{0}}^{\alpha}= \bigl\{ \rho\in Y_{M}^{\alpha}, \rho=0 \textrm{ for } r\geq R_{0} \bigr\} , \end{aligned}$$
(3.25)
where \(r=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}\), \(\theta=\arccos\frac {x_{3}}{r}\), \(R_{0}\geq R_{M}\) is a constant, and \(R_{M}\) is the radius of the nonrotating nonmagnetic star solution with prescribed total mass M.
We want to apply Theorem 3.2 in [4]. It is easy to see that a minimizer of the functional F as defined in(3.18) in \(Y_{\mathit {MR}_{0}}^{\alpha}\) solves equation (3.17).
Theorem 3.4
Let
\(\rho_{1}^{\ast}\)
be a minimizer of the energy functional
F
in
\(Y_{\mathit {MR}_{0}}^{\alpha}\), and
$$ \Gamma_{M}= \bigl\{ x\in\mathbb{R}^{3}: \rho_{1}^{\ast}(x)>0 \bigr\} . $$
(3.26)
If
\(\alpha>\frac{6}{5}\), then
\(\rho_{1}^{\ast}\in C(\mathbb {R}^{3})\cap C^{1}(\Gamma_{M})\). Moreover, there exists a constant
\(\lambda_{M}^{\ast}\)
such that
$$ I' \bigl(\rho_{1}^{\ast} \bigr)+E \bigl(\rho_{1}^{\ast} \bigr)+WD \bigl(\rho_{1}^{\ast } \bigr)=\lambda_{M}^{\ast}, \quad x\in\Gamma_{M}. $$
(3.27)
Proof
Let
$$ F_{2}(\rho)=\frac{1}{2}W \int_{\mathbb{R}^{3}}\rho D(\rho)\,dx. $$
(3.28)
Then \(F(\rho)\) can be written in two parts:
$$ F(\rho)=F_{1}(\rho)+F_{2}(\rho), $$
(3.29)
where \(F_{1}(\rho)\) is defined by (3.20).
Let \(\rho+t\sigma\in Y_{\mathit {MR}_{0}}^{\alpha}\) for any \(t\in\mathbb{\mathbb{R}}\) under the condition \(\int _{\mathbb{R}^{3}}\sigma \,dx=0\). Let us note carefully that
$$\begin{aligned} &\lim_{t\rightarrow0} \frac{F_{1}(\rho+t\sigma)-F_{1}(\rho)}{t} \\ &\quad = \lim_{t\rightarrow0}\frac{1}{t} \int_{\mathbb {R}^{3}} \biggl(I(\rho+t\sigma) +\frac{1}{2}(\rho+t \sigma)E(\rho+t\sigma) -I(\rho)-\frac{1}{2}\rho E(\rho) \biggr)\,dx \\ &\quad = \lim_{t\rightarrow0} \int_{\mathbb{R}^{3}}\frac {I(\rho+t\sigma)-I(\rho)}{t}\,dx+\lim_{t\rightarrow0} \int_{\mathbb {R}^{3}}\frac{\rho E(\rho+t\sigma)-\rho E(\rho)}{2t}\,dx \\ &\quad\quad{} + \int_{\mathbb{R}^{3}}\frac{1}{2}\sigma E(\rho)\,dx \\ &\quad = \int_{\mathbb{R}^{3}}I'(\rho)\sigma \,dx+\lim _{t\rightarrow0}\frac{1}{t} \int_{\mathbb{R}^{3}}\frac{\rho}{2} \biggl( \int_{\mathbb{R}^{3}}\frac{-\rho-t\sigma}{ \vert x-y\vert }\,dy+ \int _{\mathbb{R}^{3}} \frac{\rho}{ \vert x-y\vert }\,dy \biggr)\,dx \\ &\quad\quad{} + \int_{\mathbb{R}^{3}}\frac{1}{2}\sigma E(\rho)\,dx \\ &\quad = \int_{\mathbb{R}^{3}} \biggl(I'(\rho)\sigma+ \frac {1}{2}\sigma E(\rho) \biggr)\,dx + \int_{\mathbb{R}^{3}}\frac{\sigma(y)}{2} \biggl( \int_{\mathbb{R}^{3}}\frac{-\rho(x)}{\vert x-y\vert }\,dx \biggr)\,dy \\ &\quad = \int_{\mathbb{R}^{3}} \bigl(I'(\rho)+E(\rho) \bigr)\sigma \,dx \end{aligned}$$
(3.30)
if \(\rho\in Y_{\mathit {MR}_{0}}^{\alpha}\).
For \(F_{2}(\rho)\), we get
$$\begin{aligned} &\lim_{t\rightarrow0} \frac{F_{1}(\rho+t\sigma)-F_{1}(\rho)}{t} \\ &\quad = \lim_{t\rightarrow0}\frac{1}{t} \int_{\mathbb {R}^{3}} \biggl(\frac{W}{2}(\rho+t\sigma)D(\rho+t \sigma) -\frac{W}{2}\rho D(\rho) \biggr)\,dx \\ &\quad =\lim_{t\rightarrow0}\frac{1}{t} \int_{\mathbb {R}^{3}}\frac{W\rho}{2} \bigl(D(\rho+t\sigma)-D(\rho) \bigr)\,dx+ \int _{\mathbb{R}^{3}} \frac{W\sigma}{2}D(\rho)\,dx \\ &\quad =\lim_{t\rightarrow0} \int_{\mathbb{R}^{3}}\frac {W\rho}{2t} \biggl( \int_{\mathbb{R}^{3}} \bigl(G(x,y) (\rho+t\sigma )-G(x,y)\rho \bigr)\,dy \biggr)\,dx + \int_{\mathbb{R}^{3}}\frac{W\sigma}{2}D(\rho)\,dx \\ &\quad = \int_{\mathbb{R}^{3}}\frac{W\rho}{2} \biggl( \int _{\mathbb{R}^{3}}G(x,y)\sigma \,dy \biggr)\,dx+ \int_{\mathbb{R}^{3}}\frac {W\sigma}{2}D(\rho)\,dx \\ &\quad = \int_{\mathbb{R}^{3}}\frac{W\sigma}{2} \biggl( \int _{\mathbb{R}^{3}}G(x,y) \rho(x) \,dx \biggr)\,dy+ \int_{\mathbb {R}^{3}}\frac{W\sigma}{2}D(\rho)\,dx \\ &\quad = \int_{\mathbb{R}^{3}}W\sigma D(\rho)\,dx, \end{aligned}$$
(3.31)
where \(G(x,y)\) is defined by (3.6).
Hence, from (3.30) and (3.31) we have
$$\begin{aligned} \begin{aligned}[b] &\lim_{t\rightarrow0} \frac{F(\rho+t\sigma)-F(\rho)}{t} \\ &\quad = \int_{\mathbb{R}^{3}} \bigl[I'(\rho)+E(\rho)+D(\rho) \bigr] \sigma \,dx=0 \end{aligned} \end{aligned}$$
(3.32)
for all σ satisfying \(\int_{\mathbb{R}^{3}}\sigma \,dx=0\). Then we can prove the theorem using a similar argument as in [1]. □
Now, we give the main theorem of this paper.
Theorem 3.5
Suppose that
\(\alpha>\frac{4}{3}\). Then the following statements hold:
-
1.
\(\inf_{Y_{\mathit {MR}_{0}}^{\alpha}}F(\rho)<0\),
-
2.
\(F(\rho)\geq C_{1}\int_{\mathbb{R}^{3}}\rho^{\alpha }\,dx-C_{2}\), \(\rho\in Y_{\mathit {MR}_{0}}^{\alpha}\), for some positive constants
\(C_{1}\)
and
\(C_{2}\)
independent of
ρ,
-
3.
F
has a minimizer
\(\rho^{\ast}\)
in
\(Y_{\mathit {MR}_{0}}^{\alpha}\).
Remark 3.6
In comparison with Theorem 3.3 with adiabatic exponent \(\alpha>2\) in [4], we only need the adiabatic exponent \(\alpha>\frac{4}{3}\).
3.3 The proof of Theorem 3.5
Before giving the proof of Theorem 3.5, we introduce the following lemma, which ensures that the functional F is bounded on the set \(Y_{\mathit {MR}_{0}}^{\alpha}\) if \(\alpha>\frac{4}{3}\).
Let
$$ F_{M}=\inf_{\rho\in Y_{\mathit {MR}_{0}}^{\alpha}}F(\rho). $$
Lemma 3.7
Let
\(\alpha>\frac{4}{3}\). If
\(\rho\in Y_{\mathit {MR}_{0}}^{\alpha} \), then there exist two positive constants
\(C_{1}\)
and
\(C_{2}\)
depending only on
α
and
M
such that
$$ C_{1} \int_{\mathbb{R}^{3}}\rho^{\alpha}\,dx-C_{2}< F(\rho) \quad \textit{and}\quad F_{M}< 0. $$
Proof
Let \(F(\rho)\) be defined by (3.29). For \(F_{1}(\rho)\), Lemma 2.4 in [8] implies that \(C_{1}\int_{\mathbb{R}^{3}}\rho^{\alpha}\,dx-C_{2}< F_{1}(\rho)\) for two positive constants \(C_{1}\) and \(C_{2}\) depending only on α and M. Here, we only prove that \(C_{1}\int_{\mathbb{R}^{3}}\rho ^{\alpha}\,dx-C_{2}< F_{2}(\rho)\). By Hölder’s inequality we have
$$\begin{aligned} F_{2}(\rho)&=\frac{1}{2}W \int_{\mathbb{R}^{3}}\rho D(\rho)\,dx \\ &\leq C\Vert \rho \Vert _{2-\varepsilon} \bigl\Vert D(\rho) \bigr\Vert _{\frac{2-\varepsilon }{1-\varepsilon}}. \end{aligned}$$
Note that
$$ D(\rho)=-\frac{\Lambda W}{\mu_{0}} \int_{\mathbb{R}^{3}}\frac {(x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2}}{ ((x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2} +(x_{3}-y_{3})^{2} )^{\frac{3}{2}}}\rho(y)\,dy. $$
By the Riesz potential estimate [16], Lemma 7.12, p.159, we get
$$ \bigl\Vert D(\rho) \bigr\Vert _{\frac{2-\varepsilon}{1-\varepsilon}}\leq C\vert \Omega \vert ^{\mu-\delta} \Vert \rho \Vert _{p} \quad \text{if } p< \frac{3(2-\varepsilon )}{7-5\varepsilon} $$
(3.33)
for \(\mu=\frac{2}{3}\) and \(\delta=\frac{1}{p}-\frac{1-\varepsilon}{2-\varepsilon}\), where Ω is the compact support of ρ.
By the interpolation inequality (Theorem 2.11 in [17]), if \(f\in L^{q}\cap L^{r}\) (\(1\leq q < p< r<\infty\)), then
$$ \Vert f\Vert _{p}\leq \Vert f\Vert _{q}^{a}\Vert f\Vert _{r}^{1-a} $$
(3.34)
for \(a=\frac{p^{-1}-r^{-1}}{q^{-1}-r^{-1}}\).
Inserting \(q=1\), \(r=2-\varepsilon\), and \(a=\frac{(2-\varepsilon )/p-1}{1-\varepsilon}<1 \) into (3.34), we get that (3.33) implies
$$ \biggl\vert \frac{1}{2} \int_{\mathbb{R}^{3}}\rho D(\rho)\,dx \biggr\vert \leq C\Vert \rho \Vert _{2-\varepsilon}^{2-a}\Vert \rho \Vert _{1}^{a}. $$
(3.35)
By the interpolation inequality [16], (7.6), p.145, we obtain
$$ \int_{\mathbb{R}^{3}}\rho^{2-\varepsilon}\,dx\leq\omega \int _{\mathbb{R}^{3}}\rho^{\alpha}\,dx+\omega^{-\frac{2-\varepsilon }{\alpha-2+\varepsilon}} \vert \Omega \vert ^{\frac{\alpha-2+\varepsilon }{\alpha}} \leq\omega \int_{\mathbb{R}^{3}}\rho^{\alpha}\,dx+\omega ^{-\frac{2-\varepsilon}{\alpha-2+\varepsilon}} \vert \Omega \vert ^{\frac {\alpha-2+\varepsilon}{\alpha}}, $$
(3.36)
where \((\alpha>2-\varepsilon)\), and Ω is the compact support of ρ.
Together (3.35) with (3.36), we have
$$ \biggl\vert \frac{1}{2} \int_{\mathbb{R}^{3}}\rho D(\rho)\,dx \biggr\vert \leq \Vert \rho \Vert _{1}^{a} \biggl( \int_{\mathbb{R}^{3}}\omega\rho^{\alpha }\,dx+C(\omega) \biggr)^{\frac{2-a}{2-\varepsilon}}, $$
where \(C(\lambda)=\lambda^{-\frac{2-\varepsilon}{\alpha -2+\varepsilon}} \vert \Omega \vert ^{\frac{\alpha-2+\varepsilon}{\alpha}}\).
Choosing \(a=\varepsilon\), it is obvious that
$$ p=\frac{2-\varepsilon}{1+\varepsilon-\varepsilon^{2}}>\frac {3(2-\varepsilon)}{7-5\varepsilon}, $$
which implies that \(7-5\varepsilon>3+3\varepsilon-3\varepsilon^{2}\), \(\varepsilon<\frac{2}{3}\).
It follows from \(\alpha>2-\varepsilon\) that
$$ \alpha>\frac{4}{3}\quad \Longleftrightarrow\quad \varepsilon< \frac{2}{3}. $$
Hence, for \(\alpha>\frac{4}{3} \), we have
$$ \biggl\vert \frac{1}{2} \int_{\mathbb{R}^{3}}\rho D(\rho)\,dx \biggr\vert \leq \omega M^{a} \int_{\mathbb{R}^{3}}\rho^{\alpha}\,dx+C(\omega)M^{a}. $$
Letting ω be sufficiently small, then we have
$$ F(\rho)>C \int_{\mathbb{R}^{3}}\rho^{\alpha}\,dx-C_{1}. $$
□
Proof of Theorem 3.5
Note that Lemma 3.7 proves conclusion (2) of Theorem 3.5. Conclusion (3) in Theorem 3.5 can be proved by using the same method as in [8].
Notice that
$$\begin{aligned} F_{2}(\rho)&=-\frac{1}{2}W \int_{\mathbb{R}^{3}}\rho D(\rho)\,dx \\ &=-\frac{\Lambda W^{2}}{\mu_{0}}\rho \int_{\mathbb{R}^{3}}\frac {(x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2}}{ ((x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2} +(x_{3}-y_{3})^{2} )^{\frac{3}{2}}}\rho(y)\,dy< 0. \end{aligned}$$
Also, by the argument in [8] we get
$$ F_{1}(\rho)= \int_{\mathbb{R}^{3}} \biggl[I(\rho)+\frac{1}{2}\rho E(\rho) \biggr] \,dx< 0. $$
Hence, conclusion (1) of Theorem 3.5 is established. □