带粘性的含不活泼项Cahn-Hilliard方程柯西问题形式如下:

$ \begin{equation} \left\{ \begin{aligned} &\ {u}_{tt}+{u}_t+\Delta^{2}u-\Delta\partial_{t}u-\Delta f(u) = 0, \ \ x\in {R}^{n}, \; \; t>0, \\ &\ {u}|_{t = 0} = u_{0}(x), {u}_t|_{t = 0} = u_1(x), \\ \end{aligned} \right. \end{equation} $

(1.1)

此处$ n $是空间维数, $ n\geq1 $; 未知函数$ u $表示一个相的相对浓度; 非线性项$ \Delta f(u) $中$ f(u) $取成$ u^2 $. ($ 1.1 $)式中若去掉$ u_{tt}-\Delta{\partial_{t}}u $, 方程为

$ \begin{equation} u_t+\Delta^{2}u-\Delta f(u) = 0. \end{equation} $

(1.2)

(1.2)式是著名的Cahn-Hilliard方程^[1].近年来, Galenko等^[2-5]为模拟在某些玻璃中有深过冷产生的非平衡分解提出在(1.2)式中加上不活泼项$ u_{tt} $, 得

$ \begin{equation} u_{tt}+u_t+\Delta^{2}u-\Delta f(u) = 0. \end{equation} $

(1.3)

(1.3)式详细的物理背景可参看文献[4-6]. (1.3)式是一个带松弛项的双曲方程, 它在有限的时间内的不正则化导致很难得到其解的整体存在性.对(1.3)式, 前面的工作主要集中在弱解和拟强解.如Grasselli等^{[7, 8]}得到2维和3维情况下拟强解的存在性; Wang和Wu ^[9]得到(1.3)式在$ n\geq3 $时小初值情况下经典解的存在性.为克服不正则化困难, 在数学上经常是对方程加一粘性项, 增加耗散性, 如是得到(1.1)式.本文考虑(1.1)式解的整体存在性.

本文中用$ C $表示常数, $ L_{p}({R}^{n}) $, $ H^l(R^n) $表示通用的龙贝格可测函数空间和Sobolev空间, 其模为$ \|\cdot\|_{L_p}, \|\cdot\|_{H^l} $.

本文先分析(1.1)式的格林函数, 再用压缩映像原理证明(1.1)式解的整体存在性.

定义函数$ {f} $关于变量$ {x} $的傅里叶变换为

$ \begin{equation*} \hat{f}(\xi, t) = \int f(x, t)e^{-\sqrt{-1}x\cdot {\xi}}\, dx, \end{equation*} $

则对$ \hat{f} $关于变量$ {\xi} $的逆傅里叶变换为

$ \begin{equation*} f(x, t) = F^{-1}(\hat{f})(x, t) = (2\pi)^{-\frac{n}{2}}\int \hat{f}(\xi, t)e^{\sqrt{-1}x\cdot {\xi}}\, d\xi. \end{equation*} $

方程(1.1)格林函数定义如下:

$ \begin{equation} \left\{ \begin{aligned} &\ (\partial_{t}^{2}+\partial_{t}+\Delta^2-\Delta\partial_{t})G = 0, \\ &\ {G}|_{t = 0} = 0, \\ &\ {G_t}|_{t = 0} = \delta(x), \\ \end{aligned} \right. \end{equation} $

(2.1)

其中$ \delta(x) $为常用Dirac函数.对(2.1)式关于变量$ {x} $作傅里叶变换, 得常微分方程

$ \begin{equation} \left\{ \begin{aligned} &\ (\partial_{t}^{2}+\partial_{t}+{\left| \xi \right|}^4+{\left|\xi\right|}^{2}\partial_t)\hat{G} = 0, \\ &\ \hat{G}|_{t = 0} = 0, \\ &\ \hat{G_t}|_{t = 0} = 1.\\ \end{aligned} \right. \end{equation} $

(2.2)

($ 2.2 $)式的解为

$ \begin{equation} \hat{G}(\xi, t) = {\frac{1}{\sqrt{(1+{\left|\xi \right|}^{2})^{2}-4{\left| \xi \right|}^4 }}}(e^{\lambda_{+}t}-e^{\lambda_{-}t}), \end{equation} $

(2.3)

其中

$ \begin{equation} \lambda_{\pm} = \frac{1}{2}(-(1+{\left|\xi\right|}^2)\pm\sqrt{(1+{\left|\xi\right|}^2)^2-4{\left|\xi\right|}^4}). \end{equation} $

(2.4)

由($ 2.4 $)式可知

$ \begin{equation} {\rm Re}{\lambda_{-}}\leq-\frac{1}{2}(1+{\left|\xi\right|}^2). \end{equation} $

(2.5)

当$ {\left|\xi\right|}^2<1 $时, 有

$ \begin{equation} \begin{aligned} \sqrt{(1+{\left|\xi\right|}^2)^2-4{\left|\xi\right|}^4}& = (1+{\left|\xi\right|}^2)\sqrt{1-\frac{4{\left|\xi\right|}^4}{(1+{\left|\xi\right|}^2)^2}} \\ & = (1+{\left|\xi\right|}^2)(1-\frac{2{\left|\xi\right|}^{4}}{(1+{\left|\xi\right|}^{2})^2}+O({\left|\xi\right|}^6))\\ & = (1+{\left|\xi\right|}^2)-2{\left|\xi\right|}^{4}(1-{\left|\xi\right|}^2)+O({\left|\xi\right|}^6)\\ & = 1+{\left|\xi\right|}^2-2{\left|\xi\right|^4}+O({\left|\xi\right|}^6). \end{aligned} \end{equation} $

(2.6)

当$ \left|{\left|\xi\right|^2-1}\right|<\varepsilon $, 其中$ \varepsilon $是一很小的正数, 有$ \lambda_{+}-\lambda_{-}\rightarrow0 $, 此时有

$ \begin{equation} \left|{\frac{e^{\lambda_{+}t}-e^{\lambda_{-}t}}{\lambda_{+}-\lambda_{-}}}\right| = \left|{\frac{e^{\lambda_{+}t}(e^{(\lambda_{+}+\lambda_{-})t}-1)}{\lambda_{+}-\lambda_{-}}}\right| \leq Ce^{\lambda_{+}t}, \end{equation} $

(2.7)

$ \begin{equation} \begin{aligned} \lambda_{+}& = \frac{1}{2}\left(-(1+{\left|\xi\right|}^2)+\sqrt{(1+{\left|\xi\right|}^2)^2-4{\left|\xi\right|}^4}\right) \\ & = \frac{1}{2}\frac{(1+{\left|\xi\right|}^2)^2-4{\left|\xi\right|}^4-(1+{\left|\xi\right|}^2)^2}{\sqrt{(1+{\left|\xi\right|}^2)^2-4{\left|\xi\right|}^4}+(1+{\left|\xi\right|}^2)} \leq\frac{1}{2}\frac{-4{\left|\xi\right|}^4}{2(1+{\left|\xi\right|}^2)} \leq-a{\left|\xi\right|}^2, \end{aligned} \end{equation} $

(2.8)

其中$ a>0. $当$ {\left|\xi\right|}^2>1+\varepsilon $时,

$ \begin{equation} {\rm Re}\lambda_{+} = -\frac{1}{2}(1+{\left|\xi\right|}^2). \end{equation} $

(2.9)

由以上分析得到下述定理.

定理2.1  对任意多重指标$ \alpha $, 存在常数$ C_\alpha $, 有

$ \|\partial_{x}^{\alpha}G\|_{L_2}\leq C_{\alpha}(1+t)^{-\frac{n}{8}-\frac{\left|\alpha\right|}{4}} \; , \; \|{\partial_{x}^\alpha} \partial_{t} G\|_{L_2}\leq C(1+t)^{-\frac{n}{8}-\frac{\left|\alpha\right|}{4}}. $

证  由Plancherel等式,

$ \begin{equation} \begin{aligned} {\|{\partial_{x}^\alpha}G\|}_{L_2}^2 = &{\|{\xi}^{\alpha}\hat{G}\|}_{L_2}^2\\ \leq&\left(\int_{\left|\xi\right|^2\leq{1-\varepsilon}} {{\left|\frac{\xi^{\alpha}e^{{\lambda_+}t}}{\lambda_{+}-\lambda_{-}}\right|}^2} \, d\xi\right) +\left(\int_{\left|{{\left|\xi\right|}^2-1}\right|\leq\varepsilon}{\left|\frac{{\xi^\alpha}e^{{\lambda_+}t}(e^{(\lambda_{-}-\lambda_{+})t}-1)}{\lambda_{+}-\lambda_{-}}\right|}^2 \, d\xi\right)\\ &+\int_{\left|\xi\right|^2\geq{1+\varepsilon}}{{\left|\frac{\xi^{\alpha}e^{{\lambda_+}t}}{\lambda_{+}-\lambda_{-}}\right|}^2} \, d\xi +\int_{\left|{{\left|\xi\right|}^2-1}\right|>\varepsilon}{{\left|\frac{\xi^{\alpha}e^{{\lambda_-}t}}{\lambda_{+}-\lambda_{-}}\right|}^2} \, d\xi. \end{aligned} \end{equation} $

(2.10)

由($ 2.5 $)式,

$ \begin{equation} \begin{aligned} \int_{\left|{{\left|\xi\right|}^2-1}\right|>\varepsilon}{{\left|\frac{\xi^{\alpha}e^{{\lambda_-}t}}{\lambda_{+}-\lambda_{-}}\right|}^2} \, d\xi \leq\frac{1}{\varepsilon^2} \int{{\left|\xi\right|}^{2\left|\alpha\right|}}e^{-{({1+{\left|\xi\right|}^2})t}} \, d\xi \leq Ce^{-t}. \end{aligned} \end{equation} $

(2.11)

由($ 2.6 $)式,

$ \begin{equation} \begin{aligned} \int_{|\xi|^{2} \leq 1-\varepsilon}\left|\frac{\xi^{\alpha} e^{\lambda+t}}{\lambda_{+}-\lambda_{-}}\right|^{2} d \xi \leq \frac{1}{\varepsilon^{2}} \int|\xi|^{2|\alpha|} e^{2 \lambda+t} d \xi\\ \leq \frac{1}{\varepsilon^{2}} \int|\xi|^{2|\alpha|} e^{-4|\xi|^{4} t} d \xi \leq C(1+t)^{-\frac{n}{4}-\frac{|\alpha|}{2}}. \end{aligned} \end{equation} $

(2.12)

由($ 2.9 $)式,

$ \begin{equation} \begin{aligned} \int_{\left|\xi\right|^2\geq{1+\varepsilon}}{{\left|\frac{\xi^{\alpha}e^{{\lambda_+}t}}{\lambda_{+}-\lambda_{-}}\right|}^2} \, d\xi \leq\frac{1}{\varepsilon^2} \int{{{\left|\xi\right|}^{2\left|\alpha\right|}}e^{-({1+{\left|\xi\right|}^2})t}} \, d\xi \leq Ce^{-t}. \end{aligned} \end{equation} $

(2.13)

由($ 2.7 $), ($ 2.8 $)式得

$ \begin{equation} \begin{aligned} & \int_{{\left|{{\left|\xi\right|}^2-1}\right|}\leq\varepsilon}{\left|{\frac{\xi^{\alpha}e^{{\lambda_+}t} (e^{-(\lambda_{+}-\lambda_{-})t}-1)}{\lambda_{+}-\lambda_{-}}}\right|^2} \, d\xi \leq \int_{{\left|{{\left|\xi\right|}^2-1}\right|}\leq \varepsilon}{{\left|\xi\right|}^{2\left|\alpha\right|}}e^{2\lambda_{+}t} \, d\xi\\ \leq& \int_{{\left|{{\left|\xi\right|}^2-1}\right|}\leq \varepsilon}{{\left|\xi\right|}^{2\left|\alpha\right|}}e^{-2a{\left|\xi\right|^2}t} \, d\xi\leq C(1+t)^{-\frac{n}{2}-\left|\alpha\right|}. \end{aligned} \end{equation} $

(2.14)

由式($ 2.10 $)–($ 2.14 $), 得到$ \|{\partial_{x}^{\alpha}}G\|_{L_2}\leq C(1+t)^{-\frac{n}{8}-\frac{\left|\alpha\right|}{4}}. $

因为

$ \begin{equation*} \partial_{t}\hat{G} = \frac{\lambda_{+}e^{\lambda_{+}t}-\lambda_{-}e^{\lambda_{-}t} }{\lambda_{+}-\lambda_{-}} = \lambda_{+}e^{\lambda_+t}\frac{1-e^{(\lambda_{+}-\lambda_{-})t}}{\lambda_{+}-\lambda_{-}}+e^{\lambda_-t} . \end{equation*} $

用同样的方法, 得到$ \|{\partial_{x}^{\alpha}}{\partial_t}G\|_{L_2}\leq C(1+t)^{-\frac{n}{8}-\frac{\left|\alpha\right|}{4}}. $

下面将由不动点定理证明(1.1)解的存在性.

令

$ \begin{equation} T(u) = G(\cdot, t)\ast(u_0+u_1-\Delta u_0)+\partial_t G(\cdot, t)\ast u_0+\int_{0}^{t} G(\cdot, t-\tau)\ast \Delta f(u)(\tau)\, d\tau, \end{equation} $

(3.1)

其中$ \ast $是对变量$ {x} $的卷积.

令$ E = \max[\|u_0\|_{L_1}, \|u_1\|_{L_1}] $, 记$ X = \{u(x, t)\in H^1({R}^n)|D_X(u)\leq CE\} $, 其中$ D_X(u) = \sup\limits_{t\geq0}(1+t)^{\frac{n}{8}}\|u\|_{H^1} $.显然$ X $为一非空完备度量空间, $ X $内任两函数$ u_1 $, $ u_2 $距离为$ \rho(u_1, u_2) = D_X{(u_1, u_2)} $, 则当$ u\in X $时, 有$ \|u\|_{H^1}\leq C(1+t)^{-\frac{n}{8}} $.

定理3.1  $ T $是从$ X $到$ X $的压缩映射.

证  由定理2.1, 当$ \left|\alpha\right|\leq1 $, 有

$ \begin{eqnarray*} \|\partial_x^\alpha G(\cdot, t)\ast(u_0+u_1-\Delta u_0)\|_{L_2} &\leq&\|\partial_x^\alpha G\|_{L_2}(\|u_0\|_{L_1}+\|u_1\|_{L_1})+\|\Delta\partial_x^\alpha G\|_{L_2}\|u_0\|_{L_1}\\ &\leq &C(1+t)^{-\frac{n}{8}}E, \\ \|\partial_x^\alpha\partial_ t G\ast u_0\|_{L_2}&\leq& \|\partial_x^\alpha\partial_ t G\|_{L_2}\|u_0\|_{L_1}\leq C(1+t)^{-\frac{n}{8}}E. \end{eqnarray*} $

由定理2.1, Minkowski不等式及$ f(u) = u^2 $, 当$ \left|\alpha\right|\leq1 $, 有

$ \begin{equation*} \begin{aligned} &\left\|\int_{0}^{t}\partial_x^\alpha G(\cdot, t-\tau)\ast \Delta f(u)(\tau)\, d\tau \right\|_{L_2} \leq \left( \int_{0}^{t}{\|\partial_x^\alpha\Delta G(\cdot, t-\tau)\|^2_{L_2}}{\|f(u)\|^2_{L_1}}\, d\tau \right)^{\frac{1}{2}}\\ \leq& \left( \int_{0}^{t}(1+t-\tau)^{-\frac{n}{4}-1}\|u\|_{L_2}^4\, d\tau \right)^{\frac{1}{2}}\leq \left( \int_{0}^{t}(1+t-\tau)^{-\frac{n}{4}-1}(1+\tau)^{-\frac{n}{2}}\, d\tau \right)^{\frac{1}{2}}E^2\\ \leq& CE^2(1+t)^{-\frac{n}{8}-\frac{1}{8}}. \end{aligned} \end{equation*} $

所以$ \|\partial_x^\alpha T(u)\|_{L_2}\leq CE(1+t)^{-\frac{n}{8}}+CE^2(1+t)^{-\frac{n}{8}-\frac{1}{8}} $.得$ \sup(1+t)^{\frac{n}{8}}\|\partial_x^\alpha T(u)\|_{L_2}\leq CE+CE^2(1+t)^{-\frac{1}{8}}\leq CE $.所以$ T $是$ X $到$ X $的映射.当$ u_1, u_2\in X $时, 若$ \left|\alpha\right|\leq1 $,

$ \begin{equation*} \begin{aligned} \left\|\partial_x^\alpha (T(u_1)-T(u_2))\right\|_{L_2}&\leq \left(\int_{0}^{t}\|\partial_x^\alpha\Delta G(t-\tau)\|^2_{L_2}\; \|{u_1}^2-{u_2}^2\|^2_{L_1}\, d\tau \right)^{\frac{1}{2}}\\ &\leq \left(\int_{0}^{t}(1+t-\tau)^{-\frac{n}{4}-1} \; \left\|u_1-u_2\right\|^2_{L_2} \left\|u_1+u_2\right\|^2_{L_2} \, d\tau\right)^\frac{1}{2}\\ &\leq C\left(\int_{0}^{t}( 1+t-\tau)^{-\frac{n}{4}-1}E^2\; \left\|u_1-u_2\right\|^2_{H^1}(1+\tau)^{-\frac{n}{4}}\, d\tau\right)^\frac{1}{2}\\ &\leq C(1+t)^{-\frac{n}{8}}E\|u_1-u_2\|_{H^1}. \end{aligned} \end{equation*} $

所以$ \rho(T(u_1), T(u_2))\leq CE(1+t)^{-\frac{n}{4}}\rho(u_1, u_2) $, 当$ t $充分大, 有$ CE(1+t)^{-\frac{n}{4}}<1 $.所以$ T $是$ X $到$ X $的压缩映射.定理得证.

因为方程(1.1)的解为$ u = T(u) $, $ X $又为完备度量空间, 于是得到本文结论.

定理3.2  若$ u_0, u_1\in L_1({R}^{n}) $, 则方程(1.1)有整体经典解存在, 且$ u\in L_\infty(0, +\infty;H^1({R}^n)), $ $ \|u\|_{H^1(R)^n}\leq C(1+t)^{-\frac{n}{8}}. $