守恒律方程是偏微分方程和流体动力学的重要研究领域之一,其一维问题解的稳定性和大时间行为对方程理论的完善和工程实践的应用有着重大的意义,受到众多学者的关注。自从Ilyin等[1]提出守恒律方程以来,不断有学者进行相关研究[2-7]。其中Shi等[6]分析了当黏性为速度梯度的非线性函数时小强度、小扰动可压缩Navier-Stokes方程组稀疏波解的稳定性;Matsumura等[7]证明了守恒律方程为非凸情形下,相应牛顿黏性守恒律方程Cauchy问题稀疏波和接触间断解的渐近稳定性,其理论结果可以适用于更广泛的范围。但文献[6-7]中只考虑了单一非牛顿流体或单一非凸方程的情形。
分析这一类问题的主要困难在于:首先,黏性是与速度梯度相关的非线性函数,这给所构造的近似解的收敛性带来了困难;其次,方程的非凸性使得运用通常的能量估计方法得不到解的基本能量不等式。但以上文献都没有解决这两个问题。因此本文在文献[6-7]的基础上,进一步讨论具有Carreau型非牛顿黏性的非凸守恒律方程的冲击波解的渐近稳定性,运用非线性单调算子理论并结合加权能量估计解决了上述困难。
1 冲击波解的构造及主要定理考虑在欧拉坐标系下的一维Carreau型黏性非牛顿流体非线性守恒律方程的柯西问题:
$\left\{ {\begin{array}{*{20}{l}} {{\partial _t}u - {\partial _x}\left( {f\left( u \right)} \right) = {\partial _x}(\mu ({u_x}){u_x})} & {\left( {t,x} \right) \in \mathit{\Omega }}\\ {u\left( {0,x} \right) = {u_0}\left( x \right)} & {x \in R} \end{array}} \right.$ | (1) |
其中,Ω={(t, x)|t>0, x∈R};u=u(t, x)是Ω上的未知函数;Carreau型黏性μ=μ(ux)关于速度梯度ux是单调的,形如μ(ux)=μ∞+(μ0-μ∞)(1+ux2)
f=f(u)是关于u的完全非线性函数(f为非凸函数),如图 1所示。
对于(α, β)⊂[u+, u-]⊂(a, b)有
$\left\{ {\begin{array}{*{20}{l}} {f\left( u \right) > 0\quad {\rm{ }}u \in (\left. {a,\alpha } \right] \cup \left[ {\beta ,b} \right)}\\ {f\left( u \right) < f\left( {{u_ + }} \right) + \frac{{f\left( {{u_ - }} \right) - f\left( {{u_ + }} \right)}}{{{u_ - } - {u_ + }}}\left( {u - {u_ + }} \right),}\\ {\quad \quad u \in \left( {\alpha ,\beta } \right)} \end{array}} \right.$ | (2) |
下面考虑式(1) 的黏性冲击波解。假设速度u在正负无穷远的状态分别为u+, u-,且满足u->u+,由此可以得到冲击波解的波速为s;由Rankine-Hugoniot(R-H)条件决定,即满足s(u+-u-)=f(u+)-f(u-),引入新变量ξ=x-st,可以推出式(1) 的冲击波解U(ξ)满足式(3)
$\left\{ {\begin{array}{*{20}{l}} { - s{U_\xi } + {\partial _\xi }f\left( U \right) = {\partial _\xi }\left( {\mu \left( {{U_\xi }} \right){U_\xi }} \right),\;{\rm{ }}\xi \in R}\\ {\mathop {{\rm{lim}}}\limits_{\xi \to \pm \infty } U\left( \xi \right) = {u_ \pm }} \end{array}} \right.$ | (3) |
式(3) 中的第一式两边同时对ξ积分可得
$\mu ({U_\xi }){U_\xi } = - s(U - {u_ + }) + f\left( U \right) - f({u_ + })$ | (4) |
由黏性μ(ux)的单调有界性μ∞≤|μ|≤μ0易得到U(ξ)的存在性。
假设式(1) 的初值u0满足相容性条件
$\left\{ {\begin{array}{*{20}{l}} {\mathop {{\rm{lim}}}\limits_{x \to \pm \infty } {u_0}\left( x \right) \to {u_ \pm },} & {{\rm{ }}x \to \pm \infty }\\ {\int_{_{ - \infty }}^0 {({u_0}\left( x \right) - {u_ - }){\rm{d}}x,{\rm{ }}} } & {\int_0^{ + \infty } {({u_0}\left( x \right) - {u_ + }){\rm{d}}x < \infty } } \end{array}} \right.$ | (5) |
并且初值u0满足
$\left\{ {\begin{array}{*{20}{l}} {{u_0} - U \in {H^2}}\\ {\int_{ - \infty }^{ + \infty } {({u_0}\left( y \right) - U(y + {x_0})){\rm{d}}y = 0} } \end{array}} \right.$ | (6) |
其中
定理1 假设初值u0满足式(5)、(6),则存在一个正常数ε0,使得当‖u0-U‖H2≤ε0时,柯西问题(1) 存在唯一的全局解u(t, x), 满足u-U∈
C0(0, +∞; H2)∩L2(0, +∞; H3)并且有
通过变量替换(t, x)→(t, ξ), ξ=x-st,可以将式(1) 转化为如式(7) 的形式
$\left\{ {\begin{array}{*{20}{l}} {{\partial _t}u - s{\partial _\xi }u + {\partial _\xi }\left( {f\left( u \right)} \right) = {\partial _\xi }(\mu ({u_\xi }){u_\xi })}\\ {u\left( {0,\xi } \right) = {u_0}\left( \xi \right) \in {H^2}} \end{array}} \right.$ | (7) |
为了处理非凸函数f(u),引入函数
$h\left( u \right) = f\left( u \right) - f({u_ + }) - \frac{{f({u_ - }) - f({u_ + })}}{{{u_ - } - {u_ + }}}(u - {u_ + })$ |
函数h(u)图像如图 2所示。
同时构造新的未知函数
$\psi \left( {t,\xi } \right) = \int_{ - \infty }^x {(u\left( {t,\xi } \right) - U(\xi + {x_0})){\rm{d}}\xi } $ | (8) |
将式(8) 与式(3) 相减并积分可得
$\begin{array}{l} \quad \quad {\psi _t} - s{\psi _\xi } + f(U + {\psi _\xi }) - f\left( U \right) = \mu \prime ({U_\xi }){U_\xi } + \\ \mu ({U_\xi })]{\psi _{\xi \xi }} + O(\psi _{\xi \xi }^2) \end{array}$ |
结合μ(ux)的表达式、式(4) 以及函数h(u)的表达式,可将式(1) 重构为
$\left\{ {\begin{array}{*{20}{l}} {{\psi _t} + h\prime \left( U \right){\psi _\xi } - G({U_\xi }){\psi _{\xi \xi }} = F(\psi _\xi ^2,\psi _{\xi \xi }^2)}\\ {\psi \left( {0,\xi } \right) = {\psi _0}\left( \xi \right)} \end{array}} \right.$ | (9) |
其中
$\begin{array}{l} \quad F(\psi _\xi ^2,\psi _{\xi \xi }^2) = - (f(U + {\psi _\xi }) - f\left( U \right) - f\prime \left( U \right){\psi _\xi }) + \\ O(\psi _{\xi \xi }^2) = O(\psi _\xi ^2) + {\rm{ }}O(\psi _{\xi \xi }^2)\\ \quad G({U_\xi }) = \mu {\prime _{{U_\xi }}}({U_\xi }){U_\xi } + \mu ({U_\xi }) \end{array}$ |
为了克服方程的完全非线性(非凸性)给能量估计带来的困难,构造加权函数
$w\left( U \right) = \frac{{(U - {u_ + })(U - {u_ - })}}{{h\left( U \right)}}$ | (10) |
经过计算可知加权函数满足条件
${{\left( wh \right)}^{\prime \prime }}\left( U \right) = 2 > 0,{C^{ - 1}} < w\left( U \right) < C$ |
定义加权函数空间
$\begin{array}{l} \quad L_w^2 = \left\{ {\psi \in {L^2}{{\left| \psi \right|}_{{L^{2,w}}}} = {{\left( {\int_{ - \infty }^{ + \infty } {w{\psi ^2}{\rm{d}}\xi } } \right)}^{\frac{1}{2}}} < } \right.\\ \left. { + \infty } \right\} \end{array}$ |
定义式(9) 的解函数空间
$X\left( I \right)=\left\{ \begin{array}{*{35}{l}} \psi \in {{C}^{0}}(I;{{H}^{3}}\cap L_{w}^{2})\cap {{L}^{2}}(I;{{H}^{4}}\cap L_{w}^{2}) \\ \underset{\text{ }t\in I}{\mathop{\text{sup}}}\,\|\psi {{\|}_{{{H}^{3}}}}\le {{\varepsilon }_{0}} \\ \end{array} \right\}$ |
其中I=(t0, t1),0≤t0<t1≤+∞。
定理2 (全局存在性定理) 假设初值ψ0(ξ)∈H3,则存在正常数ε0, C0,使得‖ψ0‖H3≤ε0时,式(9) 存在唯一的全局解ψ∈X(0, +∞), 且满足
$\begin{align} &\quad \|\psi \left( t \right)\|_{{{H}^{3}}}^{2}\left| \psi \left( t \right) \right|_{{{L}^{2,w}}}^{2}+\int_{0}^{+\infty }{(\|{{\psi }_{\xi }}\left( \tau \right)\|_{{{H}^{3}}}^{2}+\text{ }} \\ &\left| \psi \left( \tau \right) \right|_{{{L}^{2,w}}}^{2})\text{d}\tau \le {{C}_{0}}(\|{{\psi }_{0}}\|_{{{H}^{3}}}^{2}+|{{\psi }_{0}}|_{{{L}^{2,w}}}^{2}) \\ \end{align}$ | (11) |
由式(11) 和解的惟一性,定理1可由定理2直接推出。而根据单调算子理论和不动点定理[8], 容易推出式(11) 局部解的存在性。因此为了证明定理2,只需证明先验估计即可。
命题1(先验估计) 假设ψ∈X(0, T)为式(9) 的解,则存在与T无关的正常数ε1和C1,使得当
$\begin{align} &\quad \quad \|\psi \left( t \right)\|_{{{H}^{3}}}^{2}+\left| \psi \left( t \right) \right|_{{{L}^{2,w}}}^{2}+\int_{0}^{t}{(\|{{\psi }_{\xi }}\left( \tau \right)\|_{{{H}^{3}}}^{2}+} \\ &\text{ }\left| \psi \right|_{{{L}^{2,w}}}^{2})\text{d}\tau \le C(\|{{\psi }_{0}}\|_{{{H}^{3}}}^{2}+|{{\psi }_{0}}|_{{{L}^{2,w}}}^{2})~ \\ \end{align}$ | (12) |
首先考虑冲击波解U(ξ)的性质。由式(7) 与函数h(u)的构造可得
$\begin{array}{l} \quad \quad \mu ({U_\xi }){U_\xi } = h\left( U \right) \simeq - c(U - {u_ + }) + o(|U - \\ {u_ + }|) \end{array}$ |
其中,c=s-f′(u+)是大于零的常数。
因此可知
由μ(Uξ)的有界性直接可以推出引理1。
引理1 假设u->u+,当U→u+时,h(U)=O(|U-u+|),则式(1) 的冲击波解U(ξ)满足:|U(ξ)-u+|=O(e-c|ξ|),Uξ=O(e-c|ξ|),Uξξ=O(e-c|ξ|)。
由引理1并经过简单计算,可以得到以下估计:
|U|≤C; μ∞≤|G|≤C;
|μ′(Uξ)|≤m; |μ″(Uξ)|≤C; |G′|≤m
其中,0<m
引理2 若ψ∈X(0, T)为式(9) 的解,则存在一个正常数C,使得对任意的t∈[0, T],有
$\begin{align} &\quad \left| \psi \left( t \right) \right|_{{{L}^{2,w}}}^{2}+\int_{0}^{t}{(\|{{\psi }_{\xi }}\left( \tau \right)\|_{{{L}^{2}}}^{2}+\left| \psi \left( \tau \right) \right|_{{{L}^{2,w}}}^{2})\text{d}\tau \le \text{ }} \\ &C\int_{0}^{t}{\|{{\psi }_{\xi \xi }}\|_{{{L}^{2}}}^{2}\text{d}\tau +|{{\psi }_{0}}|_{{{L}^{2,w}}}^{2}} \\ \end{align}$ | (13) |
证明 将式(9) 中的第一式乘以2wψ,则
$\begin{array}{l} \quad \quad \frac{{\rm{d}}}{{{\rm{d}}t}}(w{\psi ^2}) + ( - \left( {wh} \right){_U}{U_\xi }{\psi ^2}) + 2wG\psi _\xi ^2 - w{\prime _U}h\frac{{\rm{d}}}{{{\rm{d}}\xi }}{\psi ^2} + \\ {\rm{ }}2w{\prime _U}G{U_\xi }\psi {\psi _\xi } + 2w\psi G{\prime _{{U_\xi }}}{U_{\xi \xi }}{\psi _\xi } = 2w\psi F(\psi _\xi ^2,\psi _{\xi \xi }^2) \end{array}$ |
由此可得
$\begin{array}{l} \quad \frac{{\rm{d}}}{{{\rm{d}}t}}\smallint (w{\psi ^2}){\rm{d}}\xi + \smallint ( - 2{U_\xi }){\psi ^2}{\rm{d}}\xi + \smallint 2wG{\psi ^2}_\xi {\rm{d}}\xi = {\rm{ }}\\ - \smallint 2(wG{\prime _{{U_\xi }}}{U_{\xi \xi }}\psi {\psi _\xi } + w\prime \mu \prime U_\xi ^2)\psi {\psi _\xi }{\rm{d}}\xi + {\rm{ }}\smallint 2w\psi F(\psi _\xi ^2,\\ \psi _{\xi \xi }^2){\rm{d}}\xi \end{array}$ |
结合式(9) 及Sobolev不等式
$\left\{ \begin{array}{*{35}{l}} \|{{\psi }_{\xi }}{{\|}_{{{L}^{\infty }}}}\le C\|{{\psi }_{\xi }}\|_{{{L}^{2}}}^{1/2}\|{{\psi }_{\xi \xi }}\|_{{{L}^{2}}}^{1/2} \\ \|{{\psi }_{\xi \xi }}{{\|}_{{{L}^{\infty }}}}\le C\|{{\psi }_{\xi \xi }}\|_{{{L}^{2}}}^{1/2}\|{{\psi }_{\xi \xi \xi }}\|_{{{L}^{2}}}^{1/2} \\ \end{array} \right.$ |
可得
$\begin{align} &\quad \frac{\text{d}}{\text{d}t}\int (w{{\psi }^{2}})\text{d}\xi +\left( \frac{m}{2}-2{{\varepsilon }_{0}} \right)\int{{{\psi }^{2}}\text{d}\xi +\text{ }(2{{\mu }_{\infty }}-{{\varepsilon }_{0}}-} \\ &2m{{(m+{{m}^{2}})}^{2}})\int{w\psi _{\xi }^{2}\text{d}\xi \le {{\varepsilon }_{0}}\int{w\psi _{\xi \xi }^{2}\text{d}\xi }} \\ \end{align}$ | (14) |
将式(14) 对t积分得
$\begin{align} &\quad \left| \psi \left( t \right) \right|_{{{L}^{2,w}}}^{2}+\left( \frac{m}{2}-2{{\varepsilon }_{0}} \right)\int_{0}^{t}{}\|\psi \left( \tau \right)\|_{{{L}^{2}}}^{2}\text{d}\tau + \\ &\int_{0}^{t}{}(2{{\mu }_{\infty }}-{{\varepsilon }_{0}}-2m{{(m+{{m}^{2}})}^{2}})|{{\psi }_{\xi }}\left( \tau \right)|_{{{L}^{2,w}}}^{2}\text{d}\tau \le \text{ }{{\varepsilon }_{0}}~ \\ &\int_{0}^{t}{}\|{{\psi }_{\xi \xi }}\|_{{{L}^{2}}}^{2}\text{d}\tau +|{{\psi }_{0}}|_{{{L}^{2,w}}}^{2} \\ \end{align}$ | (15) |
只要当ε0≤min{
引理3 若ψ∈X(0, T)为式(9) 的解,则存在一个正常数C,使得对任意的t∈[0, T],有
$\begin{align} &\quad \|{{\psi }_{\xi }}\left( t \right)\|_{{{L}^{2}}}^{2}+\int_{0}^{t}{}\|{{\psi }_{\xi \xi }}\|_{{{L}^{2}}}^{2}\text{d}\tau \le \text{ }C(\|{{({{\psi }_{0}})}_{\xi }}\|_{{{L}^{2}}}^{2}+ \\ &\|{{\psi }_{0}}\|_{{{L}^{2}}}^{2})~ \\ \end{align}$ | (16) |
证明 式(9) 中第一式乘以ψξξ,并对ξ从-∞到+∞积分,得
$\begin{align} &\quad \|{{\psi }_{\xi }}\left( t \right)\|_{{{L}^{2}}}^{2}+{{C}_{0}}\int_{0}^{t}{}\|{{\psi }_{\xi \xi }}\|_{{{L}^{2}}}^{2}\text{d}\tau \le \text{ }{{C}_{1}}(\|{{({{\psi }_{0}})}_{\xi }}\|_{{{L}^{2}}}^{2}+ \\ &\text{ }\|{{\psi }_{0}}\|_{{{L}^{2}}}^{2})~ \\ \end{align}$ | (17) |
其中
${{C}_{0}}={{\mu }_{\infty }}-2{{\varepsilon }_{0}}-\frac{{{\varepsilon }_{0}}(2+2\varepsilon _{0}^{2})}{{{\mu }_{\infty }}(2{{\mu }_{\infty }}-{{\varepsilon }_{0}}-2m{{(m+{{m}^{2}})}^{2}})}$ |
由m
引理4 若ψ∈X(0, T)为式(9) 的解,则存在一个正常数C,使得对任意t∈[0, T], k=2, 3有
$\begin{align} &\quad \|\partial _{\xi }^{k}\psi \left( t \right)\|_{{{L}^{2}}}^{2}+\int_{0}^{t}{\|_{\xi }^{k+1}\psi \|_{{{L}^{2}}}^{2}\text{d}\tau \le C(\|\partial _{\xi }^{k}{{\psi }_{0}}\|_{{{L}^{2}}}^{2}+} \\ &\text{ }\|{{\psi }_{0}}\|_{{{H}^{k-1}}}^{2})~ \\ \end{align}$ | (18) |
证明 考虑高阶导估计,将式(9) 中的第一式对ξ求导,得
${{\psi }_{t\xi }}+{{(h\prime \left( U \right){{\psi }_{\xi }})}_{\xi }}-{{(G({{U}_{\xi }}){{\psi }_{\xi \xi }})}_{\xi }}=F{{({{\psi }_{\xi }},{{\psi }_{\xi \xi }})}_{\xi }}$ | (19) |
式(19) 乘以-ψξξξ并对ξ积分可得
$\begin{align} &\quad \frac{\text{d}}{\text{d}t}\int \psi _{\xi \xi }^{2}\text{d}\xi +({{\mu }_{\infty }}-6{{\varepsilon }_{0}})\int \psi _{\xi \xi \xi }^{2}\text{d}\xi \le \left( \frac{2{{({{m}^{2}}+1)}^{2}}}{{{\mu }_{\infty }}} \right.\text{ }+\text{ } \\ &\frac{2{{\varepsilon }_{0}}}{{{\mu }_{\infty }}(2{{\mu }_{\infty }}-{{\varepsilon }_{0}}-2m{{(m+{{m}^{2}})}^{2}})}+2{{\varepsilon }_{0}}~\int \psi _{\xi \xi }^{2}\text{d}\xi ~ \\ \end{align}$ | (20) |
式(20) 对t积分并应用式(17) 可得
$\begin{align} &\quad \|{{\psi }_{\xi \xi }}\left( t \right)\|_{{{L}^{2}}}^{2}+({{\mu }_{\infty }}-2{{\varepsilon }_{0}})\int_{0}^{t}{}\|{{\psi }_{\xi \xi \xi }}\|_{{{L}^{2}}}^{2}\text{d}\tau \le \text{ } \\ &{{C}_{2}}(\|{{({{\psi }_{0}})}_{\xi \xi }}\|_{{{L}^{2}}}^{2}+\|{{({{\psi }_{0}})}_{\xi \xi }}\|_{{{L}^{2}}}^{2}+\|{{\psi }_{0}}\|_{{{L}^{2}}}^{2}) \\ \end{align}$ | (21) |
将方程(9) 中的第一式对ξ求两阶导,得
$\begin{align} &\quad {{\psi }_{t\xi \xi }}+{{(h\prime \left( U \right){{\psi }_{\xi }})}_{\xi \xi }}-{{(G({{U}_{\xi }}){{\psi }_{\xi \xi }})}_{\xi \xi }}=F({{\psi }_{\xi }}, \\ &{{\psi }_{\xi \xi }}{{)}_{\xi \xi }}~ \\ \end{align}$ | (22) |
式(22) 乘以-ψξξξξ并对ξ积分可得
$\frac{\text{d}}{\text{d}t}\int \psi _{\xi \xi \xi }^{2}\text{d}\xi +({{\mu }_{\infty }}-2{{\varepsilon }_{0}})\int \psi _{\xi \xi \xi \xi }^{2}\text{d}\xi \le {{C}_{3}}\int \psi _{\xi \xi \xi }^{2}\text{d}\xi ~$ | (23) |
式(23) 中对t积分并应用式(21) 可得
$\begin{align} &\quad \|{{\psi }_{\xi \xi \xi }}\left( t \right)\|_{{{L}^{2}}}^{2}+({{\mu }_{\infty }}-2{{\varepsilon }_{0}})\int_{0}^{t}{}\|{{\psi }_{\xi \xi \xi \xi }}\|_{{{L}^{2}}}^{2}d\tau \le \text{ }{{C}_{4}} \\ &(\|{{({{\psi }_{0}})}_{\xi \xi }}\|_{{{L}^{2}}}^{2}+\|{{({{\psi }_{0}})}_{\xi \xi }}\|_{{{L}^{2}}}^{2}+\|{{({{\psi }_{0}})}_{\xi \xi }}\|_{{{L}^{2}}}^{2}+\|{{\psi }_{0}}\|_{{{L}^{2}}}^{2})\le \text{ } \\ &C\|{{\psi }_{0}}\|_{{{H}^{3}}}^{2} \\ \end{align}$ | (24) |
只要当ε0≤
因此,由引理2~4,当ε0满足
本文通过构造权函数,对方程进行加权能量估计,解决了方程本身的非凸性问题,得到了解的全局估计。需要强调的是,本文对冲击波的强度并没有限制条件。
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