We suppose that \((\rho ,u)\) is a classical solution to (1.1)-(1.3) over the interval \([0,T]\) with \(T\in(0,\infty)\). For simplicity reasons, we may choose \(\|\rho _{0}\|_{L^{1}}=1\), which implies
$$ \int_{-N_{0}}^{N_{0}}\rho _{0}(x) \,dx\ge \frac{1}{2}\int_{\mathbb{R}} \rho _{0}(x)\,dx= \frac{1}{2} $$
(2.1)
for some large number \(N_{0}\).
First of all, multiplying (1.1)2 by u and integrating by parts, we get
Lemma 2.1
We have
$$ \sup_{0\le t\le T} \bigl(\bigl\| \rho ^{1/2} u\bigr\| _{L^{2}}+\|P\| _{L^{1}} \bigr)+\mu\int_{0}^{T} \|u_{x}\|_{L^{2}}^{2}\,dt\le C, $$
(2.2)
where as below
C
denotes a generic constant depending on
μ, γ, K, a, and the initial data; particularly, the expression
\(C(\alpha )\)
emphasizes that
C
depends on
α.
The next lemma derives the bound on the density \(\rho (x,t)\) from above.
Lemma 2.2
We have
$$ \rho (x,t)\le C,\quad (x,t)\in \mathbb{R}\times[0,+\infty). $$
(2.3)
Proof
The proof is almost exactly the same as that in [19], Lemma 2.3. Here we state the details for completeness. By (1.1)1, one has for \(t\in[0,\infty)\)
$$ \bigl\| \rho (t) \bigr\| _{L^{1}}=\|\rho _{0}\|_{L^{1}}=1. $$
(2.4)
Put
$$ \xi(x,t)=\int_{-\infty}^{x}\rho u(y,t) \,dy. $$
(2.5)
We express (1.1)2 in the form
$$\begin{aligned} \xi_{xt}+ \bigl(\rho u^{2} \bigr)_{x}= (\mu u_{x}-P )_{x}, \end{aligned}$$
which, along with (1.1)1, provides us after integration in variable x with
$$\begin{aligned} \xi_{t}+ \rho u^{2}=\mu u_{x}-P=-\mu \frac{1}{\rho }(\rho _{t}+u\rho _{x})-P. \end{aligned}$$
Hence
$$\begin{aligned} \frac{d}{dt}\xi \bigl(X(t,x),t \bigr)&=- \mu\frac {d}{dt} \ln \rho \bigl(X(t,x),t \bigr)-P \bigl(X(t,x),t \bigr) \\ &\le- \mu\frac{d}{dt}\ln \rho \bigl(X(t,x),t \bigr), \end{aligned}$$
(2.6)
where \(X(t,x)\) is the particle trajectory satisfying
$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{d}{dt}X(t,x)=u(X(t,x),t), &t>0, \\ X(0,x)=x. \end{array}\displaystyle \right . \end{aligned}$$
Integration of (2.6) in time gives
$$\begin{aligned} \mu\ln \rho \bigl(X(t,x),t \bigr) +\xi \bigl(X(t,x),t \bigr)\le\mu\ln \rho (x,0)+\xi (x,0). \end{aligned}$$
Making use of (1.5), (2.2), (2.4), and (2.5), we find
$$\begin{aligned} \mu\ln \rho \bigl(X(t,x),t \bigr)&\le\mu\ln \rho (x,0)+ \xi(x,0)+ \xi \bigl(X(t,x),t \bigr) \\ &\le\mu\ln \rho _{0}(x)+ \int_{\mathbb{R}}\rho _{0} |u_{0}|\,dx+\int_{\mathbb{R}}\rho |u|\,dx \\ &\le\mu \rho _{0}+ \bigl\| \rho _{0}^{1/2} u_{0} \bigr\| _{L^{2}}\|\rho _{0}\|_{L^{1}}^{1/2} +\bigl\| \rho ^{1/2} u\bigr\| _{L^{2}}\|\rho \|_{L^{1}}^{1/2} \\ &\le C. \end{aligned}$$
This finishes the proof. □
Lemma 2.3
We have
$$ \| u_{x}\|_{L^{2}} + \int_{0}^{T} \bigl\| \rho ^{1/2} \dot {u}\bigr\| _{L^{2}}^{2}\,dt\le C(T). $$
(2.7)
Proof
Multiplying (1.1)2 by \(\dot{u}\) and integrating by parts give rise to
$$\begin{aligned} & \frac{d}{dt}\int_{\mathbb{R}} \frac{\mu}{2}|u_{x}|^{2}\,dx+ \int_{\mathbb{R}} \rho |\dot {u} |^{2}\,dx \\ &\quad = \int_{\mathbb{R}} P\dot{u}_{x}\,dx-\mu\int _{\mathbb{R}} u_{x}(uu_{x})_{x}\,dx \\ &\quad = \int_{\mathbb{R}} P\dot{u}_{x}\,dx- \frac{\mu}{2} \int_{\mathbb{R}}u_{x}^{3}\,dx \\ &\quad=\frac{d}{dt}\int_{\mathbb{R}} Pu_{x}\,dx-\int _{\mathbb{R}} \bigl(P_{t}+ (Pu)_{x} \bigr) u_{x}\,dx+\int_{\mathbb{R}} P u_{x}^{2} \,dx-\frac{\mu}{2}\int_{\mathbb{R}}u_{x}^{3} \,dx \\ &\quad = \frac{d}{dt}\int_{\mathbb{R}} Pu_{x}\,dx + \gamma \int_{\mathbb{R}} Pu_{x}^{2}\,dx- \frac {\mu}{2}\int_{\mathbb{R}}u_{x}^{3}\,dx, \end{aligned}$$
(2.8)
where the last equality comes from
$$ P_{t}+P_{x}u+\gamma Pu_{x}=0, $$
(2.9)
owing to (1.1)1. By virtue of (2.2), (2.3), and the Sobolev inequality, it satisfies
$$\begin{aligned} &\|u_{x}\|_{L^{\infty}}+ \|P\|_{L^{\infty }} \\ &\quad\le C \|\mu u_{x}-P\|_{L^{\infty}} +C\|P\|_{L^{\infty}} \\ &\quad\le C \|u_{x}\|_{L^{2}}+ C\bigl\| (\mu u_{x}-P)_{x} \bigr\| _{L^{2}} +C\|P\| _{L^{2}} +C\|P\|_{L^{\infty}} \\ &\quad\le C \bigl( \|u_{x}\|_{L^{2}}+ \bigl\| \rho ^{1/2} \dot{u} \bigr\| _{L^{2}} \bigr)+C. \end{aligned}$$
(2.10)
So, it follows from (2.8) that
$$\begin{aligned} \frac{d}{dt}B(t)+\bigl\| \rho ^{1/2}\dot{u} \bigr\| _{L^{2}}^{2} &\le C \bigl( 1+ \|u_{x}\|_{L^{\infty}}+ \|P\|_{L^{\infty}} \bigr)\| u_{x}\| _{L^{2}}^{2} \\ &\le C\|u_{x}\|_{L^{2}}^{2}+C\|u_{x} \|_{L^{2}}^{4}+\frac{1}{2}\bigl\| \rho ^{1/2}\dot{u} \bigr\| _{L^{2}}^{2}, \end{aligned}$$
(2.11)
where
$$ B(t)= \int_{\mathbb{R}} \biggl(\frac{\mu}{2}|u_{x}|^{2}- Pu_{x} \biggr)\,dx $$
(2.12)
satisfies
$$ \frac{\mu}{4}\|u_{x}\|_{L^{2}}^{2}-C \|P\| _{L^{2}}^{2}\le B(t)\le\mu\|u_{x} \|_{L^{2}}^{2}+C\|P\| _{L^{2}}^{2}. $$
(2.13)
Noting (2.2), (2.3), and (2.13), we integrate (2.11) to get
$$\begin{aligned} \|u_{x}\|_{L^{2}(\mathbb{R})}^{2}+\int_{0}^{T} \bigl\| \rho ^{1/2}\dot{u}\bigr\| _{L^{2}(\mathbb{R})}^{2}\,dt \le C(T) +C\int _{0}^{T}\|u_{x}\|_{L^{2}(\mathbb{R})}^{4} \,dt, \end{aligned}$$
which completes the proof in view of the Gronwall inequality and (2.2). □
Lemma 2.4
For all
\(t\in[0,+\infty)\),
$$ \int_{-A}^{A} \rho (x,t)\,dx\ge \frac{1}{4}, $$
(2.14)
where
\(A=N_{0}(1+t)\ln (e+t)\), and
\(N_{0}\)
is taken from (2.1).
Proof
The idea is borrowed from [17]. Using (2.2) and (2.4), we multiply (1.1)1 by \(|x|\) to obtain
$$\begin{aligned} \frac{d}{dt} \int_{\mathbb{R}} \rho |x|\,dx\le C\int _{\mathbb{R}} \rho |u|\,dx\le C\|\rho \| _{L^{1}}^{1/2}\bigl\| \rho ^{1/2}u\bigr\| _{L^{2}}\le C. \end{aligned}$$
This together with (1.5) brings about
$$ \int_{\mathbb{R}} \rho (x,t)|x|\,dx\le C(1+t). $$
(2.15)
Define a cut-off function \(\varphi_{1}(x)\in C_{0}^{1}(\mathbb{R})\) satisfying
$$ 0\le\varphi_{1}(x)\le1,\qquad \varphi _{1}(x)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} 1, & |x|< 1,\\ 0, & |x|>2, \end{array}\displaystyle \displaystyle \right . \qquad \bigl| \varphi_{1}' \bigr|\le C. $$
(2.16)
Multiplied by \(\varphi_{1}(y)\) with \(y= \eta x [(1+t)\ln (e+t) ]^{-1}\), from (1.1)1
$$\begin{aligned} \frac{d}{dt} \int_{\mathbb{R}} \rho \varphi_{1}(y) \,dx&=\int_{\mathbb{R}} \rho \varphi_{1}' y_{t}\,dx+\int_{\mathbb{R}} \rho u\varphi_{1}' y_{x}\,dx \\ &\ge- \frac{2\eta}{(1+t)^{2}\ln(e+t)}\int_{\mathbb{R}} \rho |x|\,dx- \frac {\eta }{(1+t)\ln(e+t)}\int_{\mathbb{R}} \rho |u|\,dx \\ &\ge- \frac{C\eta}{(1+t) \ln(e+t)}, \end{aligned}$$
where the last inequality is due to (2.15) and (2.2). Integrating the above inequality in time we conclude that
$$\begin{aligned} \int_{\mathbb{R}} \rho \varphi_{1}(y)\,dx \ge\int _{\mathbb{R}} \rho _{0} \varphi_{1}(\eta x)\,dx -C \eta\ge\frac{1}{4}. \end{aligned}$$
This proves (2.14) by choosing \(\eta =(N_{0}+4C)^{-1}\). □
Our next object is to derive some weighted \(L^{p}\) estimate on u, which plays a key role in our analysis.
Lemma 2.5
We have for all
\(p\in[2,\infty)\)
$$ \sup_{0\le t\le T}\bigl\| u\bar{x}^{-2/p}\bigr\| _{L^{p}}^{p}\le C(p,T) \bigl(\bigl\| \rho ^{1/p}u \bigr\| _{L^{p}}^{p}+\bigl\| \bigl(u^{p/2} \bigr)_{x}\bigr\| _{L^{2}}^{2} \bigr). $$
(2.17)
Proof
Multiplying (1.1)2 by \(p |u|^{p-2}u\) with \(p\ge2\), we obtain
$$\begin{aligned} &\frac{d}{dt}\int_{\mathbb{R}} \rho |u|^{p}\,dx+ \frac{4\mu (p-1)}{p}\int_{\mathbb{R}}\bigl| \bigl(u^{p/2} \bigr)_{x}\bigr|^{2}\,dx \\ &\quad=2(p-1)\int_{\mathbb{R}}Pu^{(p-2)/2} \bigl(u^{p/2} \bigr)_{x}\,dx \\ &\quad\le \frac{2\mu(p-1)}{p}\int_{\mathbb{R}}\bigl| \bigl(u^{p/2} \bigr)_{x}\bigr|^{2}\,dx+C(p)\int_{\mathbb{R}}P^{2}|u|^{p-2} \,dx \\ &\quad\le \frac{2\mu(p-1)}{p}\int_{\mathbb{R}}\bigl| \bigl(u^{p/2} \bigr)_{x}\bigr|^{2}\,dx+C(p)\int_{\mathbb{R}} \rho |u|^{p}\,dx+C, \end{aligned}$$
where in the last inequality we have used (2.2) and (2.3). Integrating in time leads to
$$ \bigl\| \rho ^{1/p}u \bigr\| _{L^{p}}+ \int_{0}^{T} \bigl\| \bigl(u^{p/2} \bigr)_{x}\bigr\| _{L^{2}}^{2} \,dt \le C(T,p). $$
(2.18)
Let
$$\overline{u^{p}}=\frac{1}{2N_{0}}\int_{-N_{0}}^{N_{0}}u^{p} \,dx. $$
On account of the Poincaré inequality and (2.14), we infer
$$\begin{aligned} \overline{u^{p}}&\le4\int_{-N_{0}}^{N_{0}} \rho \overline{u^{p}}\,dx \le4 \int_{-N_{0}}^{N_{0}} \rho \bigl|u^{p}- \overline{u^{p}}\bigr|\,dx+4\int_{-N_{0}}^{N_{0}} \rho |u|^{p}\,dx \\ &\le C\int_{-N_{0}}^{N_{0}} \bigl|u^{p}- \overline{u^{p}}\bigr|\,dx+C\int_{\mathbb{R}} \rho |u|^{p}\,dx \\ &\le\frac{1}{2}\overline{u^{p}}+C(N_{0})\int _{\mathbb{R}} \bigl| \bigl(u^{p/2} \bigr)_{x}\bigr|^{2} \,dx+C\int_{\mathbb{R}} \rho |u|^{p}\,dx, \end{aligned}$$
that is,
$$ \int_{-N_{0}}^{N_{0}}u^{p}\,dx \le C(N_{0}) \int_{\mathbb{R}} \bigl| \bigl(u^{p/2} \bigr)_{x}\bigr|^{2}\,dx+C(N_{0})\int _{\mathbb{R}} \rho |u|^{p}\,dx. $$
(2.19)
Next, a straight computation shows for an even number \(p=2,4,\ldots \)
$$\begin{aligned} 2 \int_{\mathbb{R}} \frac{(1-\varphi_{1}) u^{p}}{x^{2}\ln^{2} |x|}\,dx & \le\int _{\mathbb{R}} \biggl(\frac{-1}{x\ln|x|} \biggr)' (1- \varphi_{1})u^{p} \,dx \\ &=2 \int_{\mathbb{R}} \frac{(1-\varphi_{1})}{x\ln|x|} u^{p/2} \bigl(u^{p/2} \bigr)_{x}\,dx-\int_{\mathbb{R}} \frac{\varphi_{1}'}{x\ln|x|} u^{p} \,dx \\ &\le\int_{\mathbb{R}} \frac{(1-\varphi_{1})u^{p}}{x^{2}\ln^{2} |x|}\,dx +C \int _{\mathbb{R}} \bigl| \bigl(u^{p/2} \bigr)_{x}\bigr|^{2} \,dx+C\int_{\{1\le|x|\le2\}}|u|^{p} \,dx, \end{aligned}$$
where \(\varphi_{1}\) is defined in (2.16). This together with (2.19) gives rise to
$$ \int_{\mathbb{R}} \frac{(1-\varphi_{1}) u^{p}}{x^{2}\ln^{2} |x|}\,dx \le C \int _{\mathbb{R}} \bigl| \bigl(u^{p/2} \bigr)_{x}\bigr|^{2} \,dx+C\int_{\mathbb{R}} \rho |u|^{p}\,dx,\quad p=2,4,\ldots. $$
(2.20)
In terms of the Cauchy inequality, (2.20) holds true for all \(p\ge2\). □
Lemma 2.6
We have
$$ \sup_{0\le t\le T} {\bigl\| \bar{x}^{a}\rho \bigr\| }_{L^{1}\cap H^{1}} \le C(T). $$
(2.21)
Proof
It follows from (1.1)1 that
$$ \bigl(\rho \bar{x}^{a} \bigr)_{t}+ \bigl(u\rho \bar{x}^{a} \bigr)_{x} = a u\rho \bar{x}^{a} (\ln\bar{x})_{x}. $$
(2.22)
Observe (2.18) and the following simple facts:
$$\begin{aligned} &|\partial _{x}\bar{x}|+|\partial_{xx} \bar{x}| \le C\ln \bigl(e+x^{2} \bigr), \\ &\bigl\| \bar{x}^{(a-p)/p}\ln \bigl(e+x^{2} \bigr)\bigr\| _{L^{\infty}} \le C, \quad\mbox{for } a< p, \end{aligned}$$
(2.23)
and integrate (2.22) to obtain for \(p\ge2\)
$$\begin{aligned} \frac{d}{dt}\int_{\mathbb{R}}\rho \bar{x}^{a}\,dx&\le C \int_{\mathbb{R}} \rho |u|\bar{x}^{(a-1)}\ln \bigl(e+x^{2} \bigr)\,dx \\ &\le C\bigl\| \rho ^{1/p}u\bigr\| _{L^{p}}\bigl\| \rho ^{(p-1)/p} \bar{x}^{(a-1)}\ln \bigl(e+x^{2} \bigr)\bigr\| _{L^{p/(p-1)}} \\ &\le C\bigl\| \rho ^{1/p}u\bigr\| _{L^{p}}\bigl\| \bar{x}^{(a-p)/p}\ln \bigl(e+x^{2} \bigr)\bigr\| _{L^{\infty }} \biggl(\int_{\mathbb{R}} \rho \bar{x}^{a}\,dx \biggr)^{(p-1)/p} \\ &\le C \biggl(\int_{\mathbb{R}}\rho \bar{x}^{a}\,dx \biggr)^{(p-1)/p}, \end{aligned}$$
and hence
$$ \sup_{0\le t\le T} {\bigl\| \rho \bar {x}^{a}\bigr\| }_{L^{1}}\le C(T,p). $$
(2.24)
Differentiating (2.22) in variable x once more we get
$$\begin{aligned} & \bigl(\rho \bar{x}^{a} \bigr)_{xt}+ u \bigl(\rho \bar{x}^{a} \bigr)_{xx}+ 2u_{x} \bigl(\rho \bar{x}^{a} \bigr)_{x}+\rho \bar{x}^{a} \bigl(u_{xx}-\mu^{-1}P_{x} \bigr) \\ &\quad = -\mu^{-1}P_{x}\rho \bar{x}^{a}+a \bigl[u_{x}\rho \bar{x}^{a} (\ln\bar{x})_{x}+u \bigl(\rho \bar{x}^{a} \bigr)_{x}(\ln\bar{x})_{x}+u\rho \bar{x}^{a}(\ln\bar{x})_{xx} \bigr]. \end{aligned}$$
(2.25)
Notice that
$$\begin{aligned} P_{x}\rho \bar{x}^{a}= K\gamma \rho ^{\gamma } \bigl[ \bigl(\rho \bar{x}^{a} \bigr)_{x} -a\rho \bar{x}^{a}(\ln \bar{x})_{x} \bigr] \end{aligned}$$
and
$$\begin{aligned} \bigl\| u_{xx}-\mu^{-1}P_{x}\bigr\| _{L^{2}} \le C\|\rho \dot{u}\|_{L^{2}}\le C \bigl\| \rho ^{1/2}\dot{u}\bigr\| _{L^{2}}, \end{aligned}$$
it satisfies from (2.25) after multiplication by \(2 (\rho \bar{x}^{a})_{x}\)
$$\begin{aligned} \frac{d}{dt}{\bigl\| \bigl(\rho \bar{x}^{a} \bigr)_{x}\bigr\| }_{L^{2}} \le{}& C \bigl(1+{ \|u_{x} \|}_{L^{\infty}}+ \bigl\| u(\ln\bar{x})_{x}\bigr\| _{L^{\infty}} \bigr){\bigl\| \bigl(\rho \bar{x}^{a} \bigr)_{x} \bigr\| }_{L^{2}} \\ &{} +C\bigl\| \rho \bar{x}^{a}\bigr\| _{L^{\infty}} \bigl\| \rho ^{1/2}\dot{u}\bigr\| _{L^{2}} \\ &{} +C\bigl\| \rho \bar{x}^{a}\bigr\| _{L^{2}} \bigl(1+\bigl\| u_{x}(\ln \bar{x})_{x}\bigr\| _{L^{\infty }}+\bigl\| u(\ln\bar{x})_{xx} \bigr\| _{L^{\infty}} \bigr). \end{aligned}$$
(2.26)
Inequalities (2.7), (2.17)-(2.18), and (2.23) ensure that for \(p\in(2,\infty)\)
$$\begin{aligned} &\bigl\| u(\ln\bar{x})_{x}+u(\ln \bar{x})_{xx} \bigr\| _{L^{\infty}} \\ &\quad \le C(p) {\bigl\| u\bar{x}^{-2/p}\bigr\| }_{L^{\infty}} \\ &\quad\le C(p) \bigl({\bigl\| u\bar{x}^{-2/p}\bigr\| }_{L^{p}}+ {\bigl\| \bigl(u \bar{x}^{-2/p} \bigr)_{x}\bigr\| }_{L^{2}} \bigr) \\ &\quad \le C(p,T) \bigl(\bigl\| \rho ^{1/p}u\bigr\| _{L^{p}}+ \bigl\| \bigl(u^{p/2} \bigr)_{x}\bigr\| _{L^{2}}^{2/p}+ {\|u_{x}\|}_{L^{2}}+{\bigl\| u\bar{x}^{-1}\bigr\| }_{L^{2}} \bigr) \\ &\quad \le C(p,T) \bigl(1+{\bigl\| \bigl(u^{p/2} \bigr)_{x} \bigr\| }_{L^{2}}^{2/p} \bigr) \end{aligned}$$
(2.27)
and that for \(p=2\)
$$ {\bigl\| u\bar{x}^{-1} \bigr\| }_{L^{\infty}} \le C(T) \bigl({\bigl\| \rho ^{1/2}u\bigr\| }_{L^{2}} +{ \|u_{x}\|}_{L^{2}} \bigr) \le C(T). $$
(2.28)
Thanks to (2.27), (2.7), (2.10), and (2.24), we conclude from (2.26) that
$$\begin{aligned} \frac{d}{dt} {\bigl\| \bigl(\rho \bar{x}^{a} \bigr)_{x} \bigr\| }_{L^{2}}\le C \bigl(1+ \bigl\| \rho ^{1/2}\dot{u}\bigr\| _{L^{2}} +\bigl\| \bigl(u^{p/2}\bigr)_{x}\bigr\| _{L^{2}}\bigr)\bigl(1+{\bigl\| \bigl(\rho \bar{x}^{a} \bigr)_{x}\bigr\| }_{L^{2}} \bigr), \end{aligned}$$
which implies by using (2.7), (2.18), and the Gronwall inequality
$$\begin{aligned} {\bigl\| \bigl(\rho \bar{x}^{a} \bigr)_{x}\bigr\| }_{L^{2}}\le C (T). \end{aligned}$$
This together with (2.24) finishes the proof. □
Lemma 2.7
We have
$$ \bigl\| \rho^{1/2} \dot{u} \bigr\| _{L^{2}} + \int _{0}^{T} \| \dot{u}_{x}\| _{L^{2}}^{2}\,dt \le C(T) $$
(2.29)
and
$$ \|u_{x}\|_{L^{\infty}} + \|u_{xx} \|_{L^{2}} \le C(T). $$
(2.30)
Proof
Operating \(\partial _{t}+\partial _{x} (u\cdot)\) to (1.1)2, using (2.9) and (1.1)1, we obtain
$$\begin{aligned} \rho \dot{u}_{t}+\rho u \dot{u}_{x}-\mu \dot{u}_{xx} &= -\mu \bigl(|u_{x}|^{2} \bigr)_{x}- \bigl(P_{t}+ (Pu)_{x} \bigr)_{x}+ (u_{x}P)_{x} \\ &= -\mu \bigl(|u_{x}|^{2} \bigr)_{x}+ \gamma (u_{x}P)_{x}. \end{aligned}$$
(2.31)
Integration of (2.31) after multiplication by \(\dot{u}\) leads to
$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\bigl\| \rho^{1/2} \dot{u} \bigr\| _{L^{2}}^{2}+\mu\| \dot {u}_{x}\|_{L^{2}}^{2} =\mu\int_{\mathbb{R}} \dot{u}_{x}|u_{x}|^{2} \,dx -\gamma \int_{\mathbb{R}} \dot{u}_{x} u_{x}P \,dx, \end{aligned}$$
which combining with (2.2), (2.3), (2.7), and (2.10) lets us conclude that
$$\begin{aligned} \frac{d}{dt}\bigl\| \rho ^{1/2} \dot{u}\bigr\| _{L^{2}}^{2}+ \| \dot{u}_{x} \|_{L^{2}}^{2} & \le C \| u_{x}\|_{L^{4}}^{4}+C \|P\|_{L^{4}}^{4} \\ &\le C \| u_{x}\|_{L^{\infty}}^{2} \| u_{x} \|_{L^{2}}^{2}+C \|P\| _{L^{1}} \\ &\le C \bigl\| \rho ^{1/2}\dot{u}\bigr\| _{L^{2}}^{2} +C . \end{aligned}$$
(2.32)
Remembering (1.6), we integrate (2.32) to obtain (2.29).
Next, by virtue of (2.21) and (2.29), it follows from (1.1)2 that
$$\begin{aligned} \|u_{xx} \|_{L^{2}}&\le C(\mu) \bigl\| u_{xx}- \mu^{-1}P_{x}\bigr\| _{L^{2}}+\| P_{x} \|_{L^{2}} \\ &\le C\bigl\| \rho ^{1/2}\dot{u}\bigr\| _{L^{2}}+C\bigl\| \rho \bar{x}^{a} \bigr\| _{H^{1}}\le C, \end{aligned}$$
which together with (2.10) yields (2.30). □