考虑粘性依赖于密度的一维等熵可压缩Navier-Stokes方程的初值问题, 在Lagrange坐标系下方程的形式如下

(1.1) $ \begin{eqnarray} \left\{\begin{array}{ll} \rho_{t}+\rho^{2}u_{x} = 0, (x, t)\in{{\Bbb R}} \times{\Bbb R_{+}}, \\ u_{t}+p(\rho)_{x} = (\rho\mu(\rho)u_{x})_{x}, \end{array}\right. \end{eqnarray} $

其中$ \rho(x, t)\geq0, u(x, t) $分别表示流体的密度和速度, $ p(\rho) $和$ \mu(\rho) $分别为压强和粘性系数. 对于等熵流体$ p(\rho) = A\rho^{\gamma}, A>0, \gamma>1 $. 假设粘性系数为

(1.2) $ \begin{eqnarray} \mu(\rho) = B\rho^{\alpha}, \alpha\in(0, 1]. \end{eqnarray} $

不失一般性, 假设$ A = B = 1 $. 本文考虑方程$ (1.1) $的强解的整体存在性, 假设初始值在无穷远处有不同的极限

(1.3) $ \begin{eqnarray} (\rho, u)(x, 0) = (\rho_{0}(x), u_{0}(x))\rightarrow (\rho^{\pm}, u^{\pm}), \; \; |x|\rightarrow \infty, \end{eqnarray} $

其中$ \rho^{\pm}>0 $和$ u^{\pm} $是4个常数, $ u^{+}{\geq}u^{-} $. 令$ \overline{\rho}(x)>0 $和$ \overline{u}(x) $是定义在$ {{\Bbb R}} $上的两个光滑单调函数,

$ \overline{\rho}(x) = \left\{\begin{array}{ll} \rho^{+}, x\geq1, \\ \rho^{-}, x\leq-1, \end{array}\right.{\nonumber} $

$ \overline{u}(x) = \left\{\begin{array}{ll} u^{+}, x\geq1, \\ u^{-}, x\leq-1, \end{array}\right.{\nonumber} $

使得$ \overline{\rho}(x)>0, 0\leq\overline{u}(x)_{x}<\infty $. 假设初值满足

(1.4) $ \begin{eqnarray} \left\{\begin{array}{ll} \rho_{0}-\overline{\rho}\in{H^{1}({{\Bbb R}} ), u_{0}-\overline{u}\in H^{1}({{\Bbb R}} )}, \\ 0<c_{0}\leq\rho_{0}(x)\leq{c^{0}}<+\infty, (\rho_{0})_{x}\in{L^{\infty}({{\Bbb R}} )}, \end{array}\right. \end{eqnarray} $

其中$ c_{0}, c^{0} $为常数.

近些年来, 关于等熵可压缩Navier-Stokes方程的解的研究受到学者们广泛关注. 文献[1-2]研究了方程(1.1)中粘性系数$ \mu $为常数时的情形. 文献[3-4]考虑了等熵粘性流体的物理特性, 这时会出现粘性系数与密度函数$ \rho $的次幂成比例的情形, 如: $ \mu(\rho) = \rho^{\alpha}, \alpha>0 $.文献[5-12]研究了流体连续连接真空或不连续连接真空时不同范围的$ \alpha $的自由边界问题. 在关于方程(1.1)的自由边界问题中, 自由边界条件对于密度函数的下界的推导起到关键作用. 但是, 在关于方程(1.1)的初边值问题或Cauchy问题中情况是不同的. 文献[13]研究了初边值问题弱解的整体存在性和真空区域的动力学性质, 其中$ \alpha>\frac12 $. 文献[14]研究了Cauchy问题的整体强解的存在唯一性, 其中粘性系数$ \mu $满足: 存在常数$ \nu>0, C>0 $, 使得当$ \rho\geq1 $时$ \mu(\rho)\geq\nu $; 当$ \rho\leq1 $时$ \mu(\rho)\geq\nu\rho^{\alpha}, \alpha\in[0, \frac12) $; 任意的$ \rho\geq0 $都有$ \mu(\rho)\leq{C}(1+p(\rho)) $. 后来, 文献[15]拓展了$ \alpha $的范围, 得到$ \mu(\rho) = \rho^{\alpha}, \alpha\in(0, 1) $时方程(1.1)的Cauchy问题强解的整体存在性. 那么, 在保证强解的整体存在性的前提下, 对于粘性系数$ \mu(\rho) = \rho^{\alpha} $中指数$ \alpha $的范围的极大值的研究是有意义的.

本文的主要目的是证明$ \mu(\rho) = \rho^{\alpha}, \alpha\in(0, 1] $时方程(1.1)的强解的整体存在性. 为此, 需要证明密度函数$ \rho $是有正上、下界的, 然后, 利用文献[16]中关于解的局部存在性结论可知在任意时间$ t>0 $都不会出现真空区域或聚集区域, 从而也就证明了解的整体存在性. 本文的主要困难在于对密度函数$ \rho $在$ (x, t)\in{{\Bbb R}} \times[0, T] $时正下界的估计(见引理3.6), 为了克服这个困难本文首先考虑了在任意有界区域$ K\times[0, T] $上的情形, 然后利用$ \rho\rightarrow \overline{\rho}, |x|\rightarrow \infty $得到密度函数在$ {{\Bbb R}} \times[0, T] $上的正下界的估计.

接下来, 首先给出本文的主要结论, 然后得到一些预备引理, 最后借助文献[16]中的结论(见引理4.1)给出主要定理的证明.

定理2.1  假设初值$ (\rho_{0}, u_{0}) $满足$ (1.4) $式, 那么初值问题(1.1)–(1.3) 存在整体强解$ (\rho, u) $, 使得任意的$ T>0 $都有

(2.1) $ \begin{equation} \rho-\overline{\rho}\in{L^{\infty}(0, T;H^{1}({{\Bbb R}} ))}, \rho_{t}\in{L^{2}(0, T;L^{2}({{\Bbb R}} ))}, \end{equation} $

(2.2) $ \begin{equation} u-\overline{u}\in{L^{\infty}(0, T;H^{1}({{\Bbb R}} ))}\bigcap{L^{2}(0, T;H^{2}({{\Bbb R}} ))}, u_{t}\in{L^{2}(0, T;L^{2}({{\Bbb R}} ))}. \end{equation} $

另外, 存在常数$ C(T)>0 $使得

(2.3) $ \begin{eqnarray} 0<C^{-1}(T)\leq\rho(x, t)\leq{C(T)}<\infty, \forall(x, t)\in{{\Bbb R}} \times(0, T). \end{eqnarray} $

为使得到的解$ (\rho, u) $在无穷远处收敛到$ (\rho^{\pm}, u^{\pm}) $, 参考文献[14] 引入相对熵为

$ H = {\frac12}(u-\overline{u})^{2}+p(\rho|\overline{\rho}), {\nonumber} $

其中$ p(\rho|\overline{\rho}) $满足

$ \begin{eqnarray} p(\rho|\overline{\rho}) = \frac{1}{\gamma-1}\rho^{\gamma-1}+\frac{(\overline{\rho})^{\gamma}}{\rho}-\frac{\gamma}{\gamma-1}(\overline{\rho})^{\gamma-1}, {} \end{eqnarray} $

注2.1  $ {\rm (H_{1})} $易知$ {\rho}p(\rho|\overline{\rho}) $是严格凸函数$ , {\rho}p(\rho|\overline{\rho})\geq0 $, 当且仅当$ \rho = \overline{\rho} $时$ {\rho}p(\rho|\overline{\rho}) = 0 $;

$ {\rm (H_{2})} $易知存在依赖于$ \rho^{\pm} $的常数$ C>0 $, 使得$ \mathop{\lim \inf }\limits_{\rho\rightarrow \infty} p(\rho|\overline{\rho})\geq{C} $;

$ {\rm (H_{3})} $下文中出现的$ C $表示依赖于初值的常数, $ C(\cdot) $表示依赖于初值和括号中的量"$ \cdot $"的常数.

本节的先验估计包含对密度函数$ \rho $的正则性估计和上、下界估计, 还有对速度$ u $的正则性估计. 引理3.1, 3.2和3.4由文献[15]得到, 虽然文献[15]中考虑的是$ \mu(\rho) = \rho^{\alpha}, \alpha\in(0, 1) $的情形, 但是对于$ \mu(\rho) = \rho^{\alpha}, \alpha\in(0, 1] $的情形引理3.1, 3.2和3.4仍然适用. 本节的重难点是对密度函数$ \rho $的上、下界估计(见引理3.3和引理3.6), 借助引理3.1和引理3.2可以得到引理3.3, 利用引理3.4中关于$ (\rho^{\alpha})_{x}, (u-\overline{u}) $的高阶可积性和引理3.5中在任意有界区域$ (x, t){\in}K\times[0, T] $上$ \int_{0}^{t}||\rho^{2n(\gamma-\alpha)}u^{2n}||_{L^{\infty}([-M, M])}{\rm d}s\leq{C(T, M)}, \int_{0}^{t}(\rho^{\gamma})_{x}^{2n}{\rm d}s\leq{C(T, M)}, \forall{x}\in[-M, M] $的结论得到密度函数$ \rho $在任意有界区域$ (x, t){\in}K\times[0, T] $上是有正下界的, 再结合$ \rho\rightarrow \overline{\rho} $, $ |x|\rightarrow \infty $证得引理3.6. 最后参考文献[14]给出了关于密度函数$ \rho $和速度$ u $的正则性估计(见引理3.7).

引理3.1  若$ (\rho, u) $是初值问题(1.1)–(1.3)的强解, 则任意的$ T>0 $, 都有常数$ C(T)>0 $使得

$ \int_{{{\Bbb R}} }(u-\overline{u})^{2}+p(\rho|\overline{\rho}){\rm d}x+\int_{0}^{T}\int_{{{\Bbb R}} }\rho\mu(\rho)u_{x}^{2}{\rm d}x{\rm d}t\leq{C(T)}.{\nonumber} $

证  证明过程参考文献[15].

引理3.2  若$ (\rho, u) $是初值问题(1.1)–(1.3)的强解, 则任意的$ T>0 $, 都有常数$ C(T)>0 $使得

$ \int_{{{\Bbb R}} }(u-\overline{u}+(\rho^{\alpha})_{x}/\alpha)^{2}+p(\rho|\overline{\rho}){\rm d}x+\int_{0}^{T}\int_{{{\Bbb R}} }\gamma\rho^{\gamma+\alpha-2}\rho_{x}^{2}{\rm d}x{\rm d}t\leq{C(T)}.{\nonumber} $

证   证明过程参考文献[15].

引理3.3  若$ (\rho, u) $是初值问题(1.1)–(1.3)的强解, 则任意的$ T>0 $, 都有常数$ C(T)>0 $使得

$ \rho(x, t)\leq{C(T)}, \forall(x, t)\in{{\Bbb R}} \times[0, T].{\nonumber} $

证   根据注$ 2.1{\rm (H_{2})} $可知存在$ \eta>0 $使得

$ p(\rho|\overline{\rho})\geq\frac{C}{2}, \forall\rho>\eta. $

下面证明任意的$ (x_{0}, t_{0})\in{{\Bbb R}} \times[0, T] $都有$ \rho(x_{0}, t_{0})\leq{C(T)}. $若任意的$ r(T)>0 $都有

$ \inf\limits_{x\in[x_{0}-r(T), x_{0}]} \rho(x, t_{0})>\eta, $

则

(3.1) $ \begin{eqnarray} \int_{{{\Bbb R}} }p(\rho|\overline{\rho}){\rm d}x\geq\int_{x_{0}-r(T)}^{x_{0}}p(\rho|\overline{\rho}){\rm d}x\geq\frac{C}{2}r(T), \end{eqnarray} $

可以取适当的$ r(T) $使得(3.1)式与引理3.1矛盾, 因此存在$ r(T)>0, x_{1}\in[x_{0}-r(T), x_{0}] $使得$ \rho(x_{1}, t_{0})\leq\eta. $又根据引理3.1和引理3.2易知$ \int_{{{\Bbb R}} }(\rho^{\alpha})_{x}^{2}{\rm d}x\leq{C(T)} $, 所以

$ \rho^{\alpha}(x_{0}, t_{0})-\rho^{\alpha}(x_{1}, t_{0}) = \int_{x_{1}}^{x_{0}}(\rho^{\alpha})_{x}{\rm d}x \leq\int_{x_{0}-r(T)}^{x_{0}}(1+(\rho^{\alpha})_{x}^{2})/2{\rm d}x{\nonumber} \leq{C(T)}, $

即

$ \rho(x_{0}, t_{0})\leq{C(T)}, \; \; \forall(x_{0}, t_{0})\in{{\Bbb R}} \times[0, T].{\nonumber} $

引理3.3得证.

引理3.4  若$ (\rho, u) $是初值问题(1.1)–(1.3)的强解, 则$ \forall{T}>0, \forall{n}\in{\Bbb Z}^{+} $ (正整数集), 都有常数$ C(T, n)>0 $使得

$ \int_{{\Bbb R}} (\rho^{\alpha})_{x}^{2n}{\rm d}x+\int_{{\Bbb R}} (u-\overline{u})^{2n}{\rm d}x\leq{C(T, n)}.{\nonumber} $

证   证明过程参考文献[15].

引理3.5  若$ (\rho, u) $是初值问题(1.1)–(1.3)的强解, 则任意的$ {T}>0, 0<M<+\infty $, $ n\geq\frac{1+\alpha}{4(\gamma-\alpha)} $, 都有依赖于初值, 时间$ T $, 常数$ M, n $的常数$ C(T, M, n)>0 $, 使得

$ \int_{0}^{t}||\rho^{2n(\gamma-\alpha)}u^{2n}||_{L^{\infty}([-M, M])}{\rm d}s\leq{C(T, M, n)}, $

$ \int_{0}^{t}(\rho^{\gamma})_{x}^{2n}{\rm d}s\leq{C(T, M, n)}, \; \; \forall{x}\in[-M, M].{\nonumber} $

证   利用Sobolev嵌入定理, 引理3.1–3.4和Cauchy不等式, 有

(3.2) $ \begin{eqnarray} &&||\rho^{2n(\gamma-\alpha)}u^{2n}||_{L^{\infty}([-M, M])}{}\\ &\leq&{C}\int_{-M}^{M}\rho^{2n(\gamma-\alpha)}u^{2n}{\rm d}x+C\int_{-M}^{M}|(\rho^{2n(\gamma-\alpha)}u^{2n})_{x}|{\rm d}x{}\\ &\leq&{C(T, n)}\int_{-M}^{M}u^{2n}{\rm d}x+C(n)\int_{-M}^{M}|\rho^{2n(\gamma-\alpha)-1}\rho_{x}u^{2n}|{\rm d}x{}\\ &&+C(n)\int_{-M}^{M}|\rho^{2n(\gamma-\alpha)}u^{2n-1}u_{x}|{\rm d}x{}\\ &\leq&{C(T, n)}\int_{-M}^{M}(u-\overline{u})^{2n}{\rm d}x+C(T, n)\int_{-M}^{M}(\overline{u})^{2n}{\rm d}x+C(T, n)\int_{-M}^{M}|(\rho^{\alpha})_{x}u^{2n}|{\rm d}x{}\\ &&+C(n)\int_{-M}^{M}\rho^{\alpha+1}u_{x}^{2}{\rm d}x+C(n)\int_{-M}^{M}\rho^{4n(\gamma-\alpha)-(1+\alpha)}u^{4n-2}{\rm d}x{}\\ &\leq&{C(T, n)}\int_{-M}^{M}(\rho^{\alpha})_{x}^{2}{\rm d}x+{C(n)}\int_{-M}^{M}\rho^{\alpha+1}u_{x}^{2}{\rm d}x+C(T, M, n){}\\ &\leq&{C(n)}\int_{-M}^{M}\rho^{\alpha+1}u_{x}^{2}{\rm d}x+C(T, M, n), \end{eqnarray} $

再由(3.2)式在$ (0, t) $上积分, 利用引理3.1, 在取定$ n $ (使得$ n\geq\frac{(1+\alpha)}{4(\gamma-\alpha)} $)后可得

$ \begin{eqnarray} \int_{0}^{t}||\rho^{2n(\gamma-\alpha)}u^{2n}||_{L^{\infty}([-M, M])}{\rm d}s\leq{C(T, M, n)}.{} \end{eqnarray} $

将$ (1.1)_{1} $式代入$ (1.1)_{2} $式得到

(3.3) $ \begin{eqnarray} (\rho^{\alpha})_{xt} = -\alpha{u_{t}}-\alpha(\rho^{\gamma})_{x}, \end{eqnarray} $

再由(3.3)式在$ (0, t) $上积分可得

(3.4) $ \begin{eqnarray} (\rho^{\alpha})_{x} = (\rho_{0}^{\alpha})_{x}+\alpha(u_{0}-u)-\alpha\int_{0}^{t}(\rho^{\gamma})_{x}{\rm d}s, \end{eqnarray} $

利用(3.4)式, Cauchy不等式和H$ \ddot{\rm o} $lder不等式, 有

$ \begin{eqnarray*} \int_{0}^{t}(\rho^{\gamma})_{x}^{2n}{\rm d}s & = &({\gamma}/{\alpha})^{2n}\int_{0}^{t}\rho^{2n(\gamma-\alpha)}(\rho^{\alpha})_{x}^{2n}{\rm d}s{\nonumber} \\ & = &({\gamma}/{\alpha})^{2n}\int_{0}^{t}\rho^{2n(\gamma-\alpha)}((\rho_{0}^{\alpha})_{x}+\alpha(u_{0}-u)-\alpha\int_{0}^{s}(\rho^{\gamma})_{x}{\rm d}l)^{2n}{\rm d}s{\nonumber} \\ &\leq&{C(n)}\int_{0}^{t}\rho^{2n(\gamma-\alpha)}((\rho_{0}^{\alpha})_{x}^{2n}+u_{0}^{2n}+u^{2n}){\rm d}s\\ &&+C(T, n)\int_{0}^{t}\rho^{2n(\gamma-\alpha)}\int_{0}^{s}(\rho^{\gamma})_{x}^{2n}{\rm d}l{\rm d}s{\nonumber} \\ &\leq&{C(T, M, n)}+C(T, n)\int_{0}^{t}\int_{0}^{s}(\rho^{\gamma})_{x}^{2n}{\rm d}l{\rm d}s, {\nonumber} \end{eqnarray*} $

再利用Gronwall不等式, 在取定$ n $充分大(使得$ n\geq\frac{(1+\alpha)}{4(\gamma-\alpha)} $) 后, 有

$ \int_{0}^{t}(\rho^{\gamma})_{x}^{2n}{\rm d}s\leq{C(T, M, n)}, \; \; {x}\in[-M, M].{\nonumber} $

引理3.5得证.

引理3.6  若$ (\rho, u) $是初值问题(1.1)–(1.3)的强解, 则任意的$ T>0 $, 都有常数$ C(T)>0 $使得

$ \rho(x, t)\geq{C(T)}, (x, t)\in{{\Bbb R}} \times[0, T].{\nonumber} $

证   利用引理3.1和引理3.2可知$ \int_{{{\Bbb R}} }(\rho^{\alpha})_{x}^{2}{\rm d}x\leq{C(T)} $, 再结合引理3.3和$ \alpha\in(0, 1] $得到$ \int_{{{\Bbb R}} }\rho_{x}^{2}{\rm d}x\leq{C(T)} $, 因此

$ \begin{eqnarray*} \int_{{{\Bbb R}} }({\rho}p(\rho|\overline{\rho}))_{x}^{2}{\rm d}x&\leq&2(\frac{\gamma}{\gamma-1})^{2}\int_{{{\Bbb R}} }(\rho^{\gamma-1}-(\overline{\rho})^{\gamma-1})^{2}\rho_{x}^{2}{\rm d}x +2\gamma^{2}\int_{{{\Bbb R}} }(\overline{\rho})^{2(\gamma-2)}(\rho-\overline{\rho})^{2}(\overline{\rho})_{x}^{2}{\rm d}x{\nonumber} \\ &\leq&{C}\int_{{{\Bbb R}} }(\rho^{\gamma-1}-(\overline{\rho})^{\gamma-1})^{2}\rho_{x}^{2}{\rm d}x+C\int_{-1}^{1}(\rho-\overline{\rho})^{2}(\overline{\rho})_{x}^{2}{\rm d}x{\nonumber} \\ &\leq&{C(T)}, {\nonumber} \end{eqnarray*} $

又根据引理3.1和引理3.3可知$ \int_{{{\Bbb R}} }{\rho}p(\rho|\overline{\rho}){\rm d}x\leq{C(T)} $, 所以

$ {\rho}p(\rho|\overline{\rho})\rightarrow0, |x|\rightarrow \infty, {\nonumber} $

而由注$ {\rm 2.1(H_{1})} $可知$ {\rho}p(\rho|\overline{\rho})\geq{0} $, 当且仅当$ \rho = \overline{\rho} $时$ {\rho}p(\rho|\overline{\rho}) = 0 $, 所以

$ \rho\rightarrow \overline{\rho}, |x|\rightarrow \infty.{\nonumber} $

由此可知, 可以取$ \epsilon>0 $, 则存在$ N>0 $, 使得当$ |x|\geq{N} $时

(3.5) $ \begin{eqnarray} \rho(x, t)\geq\overline{\rho}(x)-\epsilon>0. \end{eqnarray} $

令$ K = [-N, N] $, 设$ \upsilon(x, t) = \frac{1}{\rho(x, t)}, \forall(x, t)\in{K}\times[0, T]. $根据$ (1.1)_{1} $式易知

(3.6) $ \begin{eqnarray} (\upsilon^{b})_{t} = b\upsilon^{b-1}\upsilon_{t} = b\upsilon^{b-1}u_{x}, \forall{b>1}. \end{eqnarray} $

由(3.6)式在$ K\times(0, t) $上积分, 利用(3.4)式有

$ \begin{eqnarray*} \int_{K}\upsilon^{b}{\rm d}x& = &\int_{K}\upsilon_{0}^{b}{\rm d}x+\int_{0}^{t}\int_{K}b\upsilon^{b-1}u_{x}{\rm d}x{\rm d}s{\nonumber} \\ & = &\int_{K}\upsilon_{0}^{b}{\rm d}x+\frac{b(b-1)}{\alpha}\int_{0}^{t}\int_{K}\upsilon^{b-1+\alpha}(\rho^{\alpha})_{x}u{\rm d}x{\rm d}s+\int_{0}^{t}b(\upsilon^{b-1}u)|_{-N}^{N}{\rm d}s{\nonumber} \\ & = &\int_{K}\upsilon_{0}^{b}{\rm d}x+\frac{b(b-1)}{\alpha}\int_{0}^{t}\int_{K}\upsilon^{b-1+\alpha}u((\rho_{0}^{\alpha})_{x}+\alpha(u_{0}-u)-\alpha\int_{0}^{s}(\rho^{\gamma})_{x}{\rm d}l){\rm d}x{\rm d}s{\nonumber} \\ &&+\int_{0}^{t}b(\upsilon^{b-1}u)|_{-N}^{N}{\rm d}s, {\nonumber} \end{eqnarray*} $

上式可等价变形为

(3.7) $ \begin{eqnarray} &&\int_{K}\upsilon^{b}{\rm d}x+b(b-1)\int_{0}^{t}\int_{K}\upsilon^{b-1+\alpha}u^{2}{\rm d}x{\rm d}s{}\\ & = &b\int_{0}^{t}(\upsilon^{b-1}u)|_{-N}^{N}{\rm d}s+\int_{K}\upsilon_{0}^{b}{\rm d}x+(b(b-1)/\alpha)\int_{0}^{t}\int_{K}\upsilon^{b-1+\alpha}u(\rho_{0}^{\alpha})_{x}{\rm d}x{\rm d}s{}\\ &&+b(b-1)\int_{0}^{t}\int_{K}\upsilon^{b-1+\alpha}uu_{0}{\rm d}x{\rm d}s-b(b-1)\int_{0}^{t}\int_{K}\upsilon^{b-1+\alpha}u\int_{0}^{s}(\rho^{\gamma})_{x}{\rm d}l{\rm d}x{\rm d}s{}\\ :& = &I_{1}+I_{2}+I_{3}+I_{4}+I_{5}. \end{eqnarray} $

在$ (x, t)\in{K}\times[0, T] $上, 设$ w(T) = \max\{\upsilon(x, t)\} $. 由引理3.3知$ \rho(x, t)\leq{C(T)} $, 不妨设$ w(T)\geq1 $, 则由Sobolev不等式, Hölder不等式和引理3.4, 有

$ \begin{eqnarray*} ||u||_{L^{\infty}(K)}^{4}& = &||u^{4}||_{L^{\infty}(K)}\leq{C}\int_{K}u^{4}{\rm d}x+{C}\int_{K}|(u^{4})_{x}|{\rm d}x{\nonumber} \\ &\leq&{C}\int_{K}(u-\overline{u})^{4}+(\overline{u})^{4}{\rm d}x+C\int_{K}|u^{3}u_{x}|{\rm d}x{\nonumber} \\ &\leq&{C(T, N)}+C\bigg(\int_{K}u^{6}{\rm d}x\bigg)^{1/2}\bigg(\int_{K}u_{x}^{2}{\rm d}x\bigg)^{1/2}{\nonumber} \\ &\leq&{C(T, N)}+C(T, N)\bigg(\int_{K}u_{x}^{2}{\rm d}x\bigg)^{1/2}, {\nonumber} \end{eqnarray*} $

因此

(3.8) $ \begin{eqnarray} \int_{0}^{t}||u||_{L^{\infty}(K)}{\rm d}s &\leq&{C(T, N)}+C(T, N)\int_{0}^{t}\bigg(\int_{K}u_{x}^{2}{\rm d}x\bigg)^{1/8}{\rm d}s{} \\ &\leq&{C(T, N)}+C(T, N)(w(T))^{(1+\alpha)/8}\int_{0}^{t}\bigg(\int_{K}\rho^{1+\alpha}u_{x}^{2}{\rm d}x\bigg)^{1/8}{\rm d}s{} \\ &\leq&{C(T, N)}+C(T, N)(w(T))^{(1+\alpha)/8}. \end{eqnarray} $

根据(3.8)式和(3.5)式易知

$ \begin{eqnarray*} I_{1}& = &b\int_{0}^{t}(\upsilon^{b-1}u)|_{-N}^{N}{\rm d}s{\nonumber} \\ &\leq&{b(v(N, t))^{b-1}}\int_{0}^{t}||u||_{L^{\infty}(K)}{\rm d}s+{b(v(-N, t))^{b-1}}\int_{0}^{t}||u||_{L^{\infty}(K)}{\rm d}s{\nonumber} \\ &\leq&{C(T, N)}+{C(T, N)}(w(T))^{(1+\alpha)/8}.{\nonumber} \end{eqnarray*} $

另外

$ I_{2} = \int_{K}\upsilon_{0}^{b}{\rm d}x\leq{C(N)}, $

这里的$ C(N) $表示与初值和$ N $有关的常数. 利用$ (1.4) $式, 引理3.3和$ 0<\alpha\leq1 $可得

$ \begin{eqnarray*} I_{3}&\leq&(b(b-1)/4)\int_{0}^{t}\int_{K}\upsilon^{b-1+\alpha}u^{2}{\rm d}x{\rm d}s+C(T)\int_{0}^{t}\int_{K}\upsilon^{b-1+\alpha}(\rho_{0}^{\alpha})_{x}^{2}{\rm d}x{\rm d}s{\nonumber} \\ & = &(b(b-1)/4)\int_{0}^{t}\int_{K}\upsilon^{b-1+\alpha}u^{2}{\rm d}x{\rm d}s+C(T)\int_{0}^{t}\int_{K}\upsilon^{b}\rho^{1-\alpha}\alpha^{2}(\rho_{0})^{2(\alpha-1)} (\rho_{0})_{x}^{2}{\rm d}x{\rm d}s{\nonumber} \\ &\leq&(b(b-1)/4)\int_{0}^{t}\int_{K}\upsilon^{b-1+\alpha}u^{2}{\rm d}x{\rm d}s+C(T)||\rho||_{L^{\infty}({{\Bbb R}} \times[0, T])}^{1-\alpha}||\rho_{0x}||_{L^{\infty}({{\Bbb R}} )}^{2}\int_{0}^{t}\int_{K}\upsilon^{b}{\rm d}x{\rm d}s{\nonumber} \\ &\leq&(b(b-1)/4)\int_{0}^{t}\int_{K}\upsilon^{b-1+\alpha}u^{2}{\rm d}x{\rm d}s+C(T)\int_{0}^{t}\int_{K}\upsilon^{b}{\rm d}x{\rm d}s.{\nonumber} \end{eqnarray*} $

利用$ (1.4) $式和Sobolev不等式易知

$ ||u_{0}||_{L^{\infty}({{\Bbb R}} )}\leq||u_{0}-\overline{u}||_{L^{\infty}({{\Bbb R}} )}+||\overline{u}||_{L^{\infty}({{\Bbb R}} )}\leq ||u_{0}-\overline{u}||_{H^{1}({{\Bbb R}} )}+C\leq{C}, $

因此

$ \begin{eqnarray} I_{4}&\leq&(b(b-1)/4)\int_{0}^{t}\int_{K}\upsilon^{b-1+\alpha}u^{2}{\rm d}x{\rm d}s+C(T)\int_{0}^{t}\int_{K}\upsilon^{b}u_{0}^{2}{\rm d}x{\rm d}s{} \\ &\leq&(b(b-1)/4)\int_{0}^{t}\int_{K}\upsilon^{b-1+\alpha}u^{2}{\rm d}x{\rm d}s+C(T)||u_{0}||_{L^{\infty}({{\Bbb R}} )}^{2}\int_{0}^{t}\int_{K}\upsilon^{b}{\rm d}x{\rm d}s{} \\ &\leq&(b(b-1)/4)\int_{0}^{t}\int_{K}\upsilon^{b-1+\alpha}u^{2}{\rm d}x{\rm d}s+C(T)\int_{0}^{t}\int_{K}\upsilon^{b}{\rm d}x{\rm d}s.{} \end{eqnarray} $

由引理3.5可知当$ n\geq\frac{(1+\alpha)}{4(\gamma-\alpha)} $时

$ \bigg(\int_{0}^{t}(\rho^{\gamma})_{x}{\rm d}s\bigg)^{2}\leq{C(T)} \bigg(\int_{0}^{t}(\rho^{\gamma})_{x}^{2n}{\rm d}s\bigg)^{1/n}\leq{C(T, N, n)}, \; \; x\in{K}, {\nonumber} $

因此

$ \begin{eqnarray} I_{5}&\leq&(b(b-1)/4)\int_{0}^{t}\int_{K}\upsilon^{b-1+\alpha}u^{2}{\rm d}x{\rm d}s+C\int_{0}^{t} \int_{K}\upsilon^{b-1+\alpha}\bigg(\int_{0}^{s}(\rho^{\gamma})_{x}{\rm d}l\bigg)^{2}{\rm d}x{\rm d}s{} \\ &\leq&(b(b-1)/4)\int_{0}^{t}\int_{K}\upsilon^{b-1+\alpha}u^{2}{\rm d}x{\rm d}s+C(T, N)\int_{0}^{t}\int_{K}\upsilon^{b}{\rm d}x{\rm d}s.{} \end{eqnarray} $

将$ I_{1}-I_{5} $的估计式代入(3.7)式得到

$ \int_{K}\upsilon^{b}{\rm d}x+\int_{0}^{t}\int_{K}\upsilon^{b-1+\alpha}u^{2}{\rm d}x{\rm d}s\leq{C(T, N)}(1+(w(T))^{\frac{(1+\alpha)}{8}})+{C(T, N, n)}\int_{0}^{t}\int_{K}\upsilon^{b}{\rm d}x{\rm d}s, $

再由Gronwall不等式可得

(3.9) $ \begin{eqnarray} \int_{K}\upsilon^{b}{\rm d}x&\leq{C(T, N, n)}+{C(T, N)}(w(T))^{\frac{(1+\alpha)}{8}}, \forall{b>1}. \end{eqnarray} $

根据引理3.1可知$ \int_{{{\Bbb R}} }p(\rho|\overline{\rho}){\rm d}x\leq{C(T)} $, 因此$ \int_{K}{\upsilon}{\rm d}x\leq{C(T, N)} $. 假设任意的$ (x, t)\in{K}\times[0, T] $都有$ \upsilon(x, t)\geq{C(T, N)} $, 则$ \int_{K}{\upsilon}{\rm d}x\geq{C(T, N)} $, 显然矛盾. 因此必存在$ x_{1}\in{K} $使得$ \upsilon(x_{1}, t)\leq{C(T, N)} $. 再利用引理3.1–3.2和(3.9)式, 有

$ \begin{eqnarray} \upsilon^{b}(x, t)-\upsilon^{b}(x_{1}, t)& = &\int_{x_{1}}^{x}(\upsilon^{b})_{y}{\rm d}y\leq(b/\alpha)\int_{K}|\upsilon^{b-1}\rho^{-1-\alpha}(\rho^{\alpha})_{x}|{\rm d}x{} \\ &\leq&{C}\bigg(\int_{K}\upsilon^{2(b+\alpha)}{\rm d}x\bigg)^{1/2} \bigg(\int_{K}(\rho^{\alpha})_{x}^{2}{\rm d}x\bigg)^{1/2}{} \\ &\leq&{C(T, N, n)}+C(T, N)(w(T))^{\frac{1+\alpha}{8}}, {} \end{eqnarray} $

又因为$ b>1, 0<\alpha\leq1, $所以

(3.10) $ \begin{eqnarray} \upsilon(x, t)\leq{C(T, N, n)}, \forall(x, t)\in{K\times[0, T]}. \end{eqnarray} $

结合(3.5)式和(3.10)式可得

$ \rho(x, t)\geq{C(T)}, \forall(x, t)\in{{\Bbb R}} \times[0, T].{\nonumber} $

引理3.6得证.

引理3.7  若$ (\rho, u) $是初值问题(1.1)–(1.3)的强解, 则任意的$ T>0 $, 都有常数$ C(T)>0 $使得

$ ||\rho-\overline{\rho}||_{L^{\infty}(0, T;H^{1}({{\Bbb R}} ))}\leq{C(T)}, $

$ ||u-\overline{u}||_{L^{\infty}(0, T;H^{1}({{\Bbb R}} ))}\leq{C(T)}, $

$ ||\rho_{t}||_{L^{2}(0, T;L^{2}({{\Bbb R}} ))}\leq{C(T)}, $

$ ||u-\overline{u}||_{L^{2}(0, T;H^{2}({{\Bbb R}} ))}\leq{C(T)}, $

$ ||u_{t}||_{L^{2}(0, T;L^{2}({{\Bbb R}} ))}\leq{C(T)}. $

证   证明过程参考文献[14].

在证明定理2.1之前先给出下述关于解的局部存在唯一性的引理.

引理4.1  若初值$ (\rho_{0}, u_{0}) $满足$ (1.4) $式, $ \mu(\rho)\geq\theta>0 $, 则存在$ T_{0}>0 $($ T_{0} $仅依赖于初值), 使得$ t\in(0, T_{0}) $时, 初值问题(1.1)–(1.3)存在唯一的解$ (\rho, u) $满足

$ \rho-\overline{\rho}\in{L^{\infty}(0, T_{1};H^{1}({{\Bbb R}} ))}, \rho_{t}\in{L^{2}((0, T_{1})\times{{\Bbb R}} )}, \forall{T_{1}}\in(0, T_{0}), {\nonumber} $

$ u-\overline{u}\in{L^{2}(0, T_{1};H^{2}({{\Bbb R}} ))}{\bigcap}L^{\infty}(0, T_{1};H^{1}({{\Bbb R}} )), u_{t}\in{L^{2}((0, T_{1})\times{{\Bbb R}} )}, \forall{T_{1}}\in(0, T_{0}), {\nonumber} $

同时, 存在$ c_{0}(t), c^{0}(t)>0 $使得

$ c_{0}(t)\leq\rho(x, t)\leq{c^{0}(t)}, \forall{t}\in(0, T_{0}).{\nonumber} $

证   参考文献[16].

下面根据上述引理给出定理2.1的证明.

证   首先, 变换粘性系数为

$ \mu_{n}(\rho) = \max\{\mu(\rho), 1/n\}\geq1/n>0.{\nonumber} $

借助引理4.1可以假设$ (\rho_{n}, u_{n}) $是$ \mu = \mu_{n}(\rho) $时初值问题(1.1)–(1.3)在$ {t}\in(0, T_{0}) $上的强解. 其次, 利用引理3.1–3.7可知$ (\rho_{n}, u_{n}) $在$ {t}\in(0, T_{0}) $上满足

$ \rho_{n}-\overline{\rho}\in{L^{\infty}(0, T_{0};H^{1}({{\Bbb R}} )}, \; \; u_{n}-\overline{u}\in{L^{\infty}(0, T_{0};H^{1}({{\Bbb R}} )}, \; \; C^{-1}(T)\leq\rho_{n}\leq{C(T)}.{\nonumber} $

最后, 由引理4.1可知当$ n $充分大($ 1/n\leq{C^{-1}(T)} $) 时$ (\rho_{n}, u_{n}) $是初值问题(1.1)–(1.3)的一个整体解, 即$ T_{0}(n)\rightarrow \infty\; (n\rightarrow \infty $), 此时有$ \mu_{n}(\rho)\rightarrow \mu(\rho)\; (n\rightarrow \infty $), 即初值问题(1.1)–(1.3)存在整体强解$ (\rho, u) $, 使得$ C^{-1}(T)\leq\rho\leq{C(T)} $. $ (\rho, u) $的正则性由引理3.1–3.7和引理4.1给出, 定理2.1得证.