引言

时滞现象广泛存在于物理学、化学、力学、经济学等研究领域。一般来说，这些现象不仅依赖于当前的状态，而且依赖于过去的状态^[1-3]。近几年来，具有时滞现象的二阶发展方程相关性质的研究引起了国内外许多数学同行们的广泛关注，并且成为研究热点。

文献[2]的研究结果表明，时滞作用是使得系统不稳定的一个重要因素，即使是一个比较小的时滞作用也可以使原来稳定的系统变得不稳定，必须添加一些额外的耗散条件才可使系统再稳定。Datko^[3]考虑了一维时滞波动方程

$ {u_u} = {u_{xx}} - 2{u_t}\left( {x,t - \tau } \right),\left( {x,t} \right) \in \left( {0,1} \right) \times \left( {0, + \infty } \right) $

的初边值问题，并得到时滞作用可破坏原本稳定波动方程的结论^[2]。

对于如下的n-维波动方程初边值问题

$ \left\{ \begin{array}{l} {u_u}\left( {x,t} \right) - \Delta u\left( {x,t} \right) + {\mu _1}\left( {x,t} \right) + {\mu _2}\left( {x,t - \tau } \right) = 0,\\ \;\;\;\;\;\;\;\;\;\left( {x,t} \right) \in \mathit{\Omega } \times \left( {0, + \infty } \right)\\ u\left( {x,0} \right) = {u_0}\left( x \right),{u_t}\left( {x,0} \right) = {u_1}\left( x \right),x \in \mathit{\Omega }\\ u\left( {x,t} \right) = 0,\left( {x,t} \right) \in {\mathit{\Gamma }_0} \times \left( {0, + \infty } \right)\\ \frac{{\partial u}}{{\partial \upsilon }}\left( {x,t} \right) = 0,\left( {x,t} \right) \in {\mathit{\Gamma }_1} \times \left( {0, + \infty } \right) \end{array} \right. $

没有时滞项(μ₂=0) 时，系统呈指数型稳定状态^[4]；有时滞项(μ₂ > 0) 时，Nicaise等^[5]的研究结果表明, 当μ₂ < μ₁时系统稳定, 当μ₂≥μ₁时存在一列时滞使系统不稳定。

Cavalcanti等^[6]首次研究了如式(1) 所示的拟线性粘弹性波动方程

$ \begin{array}{l} \;\;\;\;\;\;{\left| {{u_t}} \right|^\rho }{u_{tt}} - \Delta u - \Delta {u_{tt}} + \int_0^t {g\left( {t - s} \right)\Delta u\left( s \right){\rm{d}}s} - \\ \gamma \Delta {u_t} = F\left( {x,u,{u_t}} \right) \end{array} $

(1)

进一步，当F(x, u, u_t)=0，并假设ρ > 0满足H₀¹(Ω)→ L^2(ρ+1)(Ω)时，得到了γ≥0时解的整体存在性及γ > 0时能量指数型稳定性的结果。

由于式(1) 来源于式(2)

$ f\left( {{u_t}} \right){u_{tt}} - \Delta u - \Delta {u_{tt}} = 0 $

(2)

当f(u_t)为常数时，Δu_tt项的存在使式(2) 不同于D'Alembert型波动方程^[7]；当f(u_t)不为常数时，式(2) 表示材料的密度、浓度依赖于速率u_t^[8]。而当ρ=0、μ₂=0及去掉Δu_tt项时，式(1) 变为经典的粘弹性波动方程^[9-10]。

Kirane等^[11]研究了方程

$ \begin{array}{l} \;\;\;\;\;\;\;\;{u_{tt}} - \Delta u + \int_0^t {g\left( {t - s} \right)\Delta u\left( s \right){\rm{d}}s} + {\mu _1}{u_t}\left( {x,t} \right) + \\ {\mu _2}{u_t}\left( {x,t - \tau } \right) = 0 \end{array} $

的初边值问题，当μ₂≤μ₁时，利用Lyapunov泛函法得到了能量的衰减估计，该方程在文献[12-14]中得到了进一步研究。

此外，Messaoudi等^[15]首次研究了如下具强时滞的波动方程问题

$ \left\{ \begin{array}{l} {u_{tt}}\left( {x,t} \right) - \Delta u\left( {x,t} \right) - {\mu _1}\Delta {u_t}\left( {x,t} \right) + {\mu _2}\Delta {u_t}\left( {x,t} \right) = 0,\\ \;\;\;\;\;\;\;\mathit{\Omega } \times \left( {0, + \infty } \right)\\ u\left( {x,t} \right) = 0,\\ \;\;\;\;\;\;\;\;\partial \mathit{\Omega } \times \left( {0, + \infty } \right)\\ {u_t}\left( {x,t - \tau } \right) = {f_0}\left( {x,t} \right),\\ \;\;\;\;\;\;\;\;\;t \in \left( {0,\tau } \right)\\ u\left( {x,0} \right) = {u_0}\left( x \right),{u_t}\left( {x,0} \right) = {u_1}\left( x \right),\\ \;\;\;\;\;\;\;\;t \in \mathit{\Omega } \end{array} \right. $

利用半群理论得到了该系统的整体适定性以及在一定条件下能量的衰减估计。

然而，同时具有强阻尼和强时滞性的粘弹性波动方程解的研究尚处于起步阶段，故本文在上述研究结果的基础上，考虑如下具有强阻尼与强时滞作用的拟线性粘弹性方程的初边值问题：

$ \left\{ \begin{array}{l} {\left| {{u_t}} \right|^\rho }{u_{tt}} - \Delta u - \Delta {u_{tt}} + \int_0^t {g\left( {t - s} \right)\Delta u\left( s \right){\rm{d}}s - } \\ \;\;\;\;{\mu _1}\Delta {u_t} - {\mu _2}\Delta {u_t}\left( {t - \tau } \right) = 0,\\ \;\;\;\;\left( {x,t} \right) \in \mathit{\Omega } \times \left( {0, + \infty } \right)\\ u\left( {x,0} \right) = {u_0}\left( x \right),{u_t}\left( {x,0} \right) = {u_1}\left( x \right),x \in \mathit{\Omega }\\ u\left( {x,t} \right) = 0,\left( {x,t} \right) \in \partial \mathit{\Omega } \times \left( {0, + \infty } \right)\\ u\left( {x,t - \tau } \right) = {f_0}\left( {x,t - \tau } \right),x \in \mathit{\Omega },t \in \left( {0,\tau } \right) \end{array} \right. $

(3)

其中Ω⊂Rⁿ为一个具有光滑边界∂Ω的有界集，τ > 0代表时滞，μ₁、μ₂为实数，(u₀, u₁, f₀)为给定的合适初始数据。本文的目的是通过构造合适的Lyapunov泛函来研究式(1) 解的能量衰减估计。

1 预备知识

对于式(3)，假设ρ满足

$ 0 < \rho \le \frac{2}{{N - 2}}\left( {N \ge 3} \right),\;aaa\;\rho > 0\left( {N = 1,2} \right) $

(4)

对于记忆核g(t)，假设满足条件(H)

(H):g:R⁺→R⁺是一个连续可微函数，满足$ 1 - \int_0^\infty {g\left( s \right){\rm{d}}s = l > 0} $，并且存在正的非增函数ξ(t)，使得式(5) 成立。

$ g'\left( t \right) \le - \xi \left( t \right)g\left( t \right),\int_0^{ + \infty } {\xi \left( s \right){\rm{d}}s} = + \infty $

(5)

引理1  (Sobolev-Poincaré不等式)

若$ 2 \le p \le \frac{{2N}}{{N - 2}} $，u∈H₀¹(Ω)，则存在c_s > 0，使得‖u‖_p≤c_s‖Δu‖₂。

引理2  对任意的g∈C¹(R)和ϕ∈H¹(0, T)有

$ \begin{array}{l} - 2\int_0^t {\int_\mathit{\Omega } {g\left( {t - s} \right)\phi {\phi _t}{\rm{d}}x{\rm{d}}s} } = \frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\left( {g \triangleleft \phi } \right)\left( t \right) - } \right.\\ \left. {\int_0^t {g\left( s \right){\rm{d}}s\left\| \phi \right\|_2^2} } \right) + g\left( t \right)\left\| \varphi \right\|_2^2 - \left( {g' \triangleleft \phi } \right)\left( t \right) \end{array} $

其中(g◁ϕ)(t)=$ \int_0^t {g\left( {t - s} \right){\rm{d}}s} \int_\mathit{\Omega} {{{\left| {\phi \left( s \right) - \phi \left( t \right)} \right|}^2}{\rm{d}}x{\rm{d}}s} $。

进一步，定义新的变量z, 满足

$ z\left( {x,k,t} \right) = {u_t}\left( {t - \tau k} \right),x \in \mathit{\Omega ,k} \in \left( {0,1} \right) $

则有

$ \tau {z_t}\left( {x,k,t} \right) + {z_k}\left( {x,k,t} \right) = 0,\left( {x,k,t} \right) \in \mathit{\Omega } \times \left( {0,1} \right) \times \left( {0, + \infty } \right)。$

因此式(3) 可转化为式(6)

$ \left\{ \begin{array}{l} {\left| {{u_t}} \right|^\rho }{u_{tt}} - \Delta u - \Delta {u_{tt}} + \int_0^t {g\left( {t - s} \right)\Delta u\left( s \right){\rm{d}}s - } \\ \;\;\;\;{\mu _1}\Delta {u_t}\left( {x,t} \right) - {\mu _2}\Delta z\left( {x,1,t} \right) = 0,\\ \;\;\;\;\;\;\;\;\;\;\;\left( {x,t} \right) \in \mathit{\Omega } \times \left( {0, + \infty } \right)\\ {z_t}\left( {x,k,t} \right) + {z_k}\left( {x,k,t} \right) = 0,\\ \;\;\;\;\;\;\;\;\;\;\;\;x \in \mathit{\Omega ,k} \in \left( {0,1} \right),t > 0\\ z\left( {x,0,t} \right) = {u_t}\left( {x,t} \right),x \in \mathit{\Omega },t > 0\\ z\left( {x,k,0} \right) = {f_0}\left( {x, - \tau k} \right),x \in \mathit{\Omega }\\ u\left( {x,0} \right) = {u_0}\left( x \right),{u_t}\left( {x,0} \right) = {u_1}\left( x \right),x \in \mathit{\Omega }\\ u\left( {x,t} \right) = 0,x \in \mathit{\Omega },t \ge 0 \end{array} \right. $

(6)

基于引理1和引理2，结合文献[6]和[11]，可以不加证明地给出解的整体存在性定理。

定理1  假设|μ₂| < μ₁，且式(4) 和条件(H)成立，则当u₀, u₁∈H₀¹(Ω)，f₀∈L²((0，1)，H₀¹(Ω))时，式(6) 至少存在一弱解(u, z)，且满足

$ \begin{array}{l} \;\;\;\;\;\;u,{u_t} \in C\left( {\left[ {0,T} \right],H_0^1\left( \mathit{\Omega } \right)} \right),z \in C\left( {\left[ {0,T} \right],{L^2}} \right.\\ \left. {\left( {\left( {0,1} \right),H_0^1\left( \mathit{\Omega } \right)} \right)} \right),T > 0。\end{array} $

2 能量的衰减估计

能量泛函的定义如式(7)

$ \begin{array}{l} E\left( t \right) = E\left( {t,u,z} \right) = \frac{1}{{\rho + 2}}\left\| {{u_t}} \right\|_{\rho + 2}^{\rho + 2} + \frac{1}{2}\left( {1 - } \right.\\ \left. {\int_0^t {g\left( {t - s} \right){\rm{d}}s} } \right)\left\| {\nabla u\left( t \right)} \right\| + \frac{1}{2}\left( {g \triangleleft \nabla u} \right)\left( t \right) + \\ \frac{1}{2}\left\| {\nabla \nabla {u_t}\left( t \right)} \right\|_{\rho + 2}^{\rho + 2} + \frac{\xi }{2}\int_\mathit{\Omega } {\int_0^1 {{{\left| {\nabla z\left( {x,k,t} \right)} \right|}^2}{\rm{d}}k{\rm{d}}x} } \end{array} $

(7)

其中常数ξ满足

$ \tau \left| {{\mu _2}} \right| \le \xi \le \tau \left( {2{\mu _1} - \left| {{\mu _2}} \right|} \right) $

(8)

下面基于能量泛函的定义, 进行相关的估计和证明。

引理3  设E(t)是[0, T]上的一个非增函数，且

$ \begin{array}{l} \;\;\;\;\;E'\left( t \right) \le - {c_1}{\left\| {\nabla {u_t}} \right\|^2} - {c_2}\int_\mathit{\Omega } {{{\left| {\nabla z\left( {x,1,t} \right)} \right|}^2}{\rm{d}}x} + \\ \frac{1}{2}\left( {g' \triangleleft \nabla u} \right)\left( t \right) - \frac{1}{2}g\left( t \right){\left\| {\nabla u\left( t \right)} \right\|^2} \le \frac{1}{2}\left( {g' \triangleleft } \right.\\ \left. {\nabla u} \right)\left( t \right) - \frac{1}{2}g\left( t \right){\left\| {\nabla u\left( t \right)} \right\|^2} \le 0,t \ge 0 \end{array} $

(9)

其中

$ {c_1} = \left\{ \begin{array}{l} {\mu _1} - \frac{\xi }{{2\tau }} - \frac{{\left| {{\mu _2}} \right|}}{2} > 0\;\;\;\;\;\left| {{\mu _2}} \right| < {\mu _1}\\ 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left| {{\mu _2}} \right| = {\mu _1} \end{array} \right. $

$ {c_2} = \left\{ \begin{array}{l} \frac{\xi }{{2\tau }} - \frac{{\left| {{\mu _2}} \right|}}{2} > 0\;\;\;\;\;\left| {{\mu _2}} \right| < {\mu _1}\\ 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left| {{\mu _2}} \right| = {\mu _1} \end{array} \right. $

证明  方程组(6) 的第一个方程两边同乘以u_t，利用引理2得

$ \begin{array}{l} \frac{{\rm{d}}}{{{\rm{d}}t}}\left[ {\frac{1}{{\rho + 2}}\left\| {{u_t}\left( t \right)} \right\|_{\rho + 2}^{\rho + 2} + \frac{1}{2}\left( {1 - \int_0^t {g\left( s \right){\rm{d}}s} } \right)} \right.\\ \left. {\left\| {\nabla u\left( t \right)} \right\|_2^2 + \frac{1}{2}\left\| {\nabla {u_t}\left( t \right)} \right\|_2^2 + \frac{1}{2}\left( {g \triangleleft \nabla u} \right)\left( t \right)} \right] + \\ \frac{1}{2}g\left( t \right)\left\| {\nabla u\left( t \right)} \right\|_2^2 - \frac{1}{2}\left( {g' \triangleleft \nabla u} \right)\left( t \right) + {\mu _1}\left\| {\nabla {u_t}} \right.\\ \left. {\left( t \right)} \right\|_2^2 + {\mu _2}\int_\mathit{\Omega } {\nabla z\left( {x,1,t} \right) \cdot \nabla {u_t}\left( t \right){\rm{d}}x} = 0 \end{array} $

(10)

方程组(6) 的第二个方程两边同乘以$ \frac{\xi }{\tau }z $并积分得

$ \begin{array}{l} \;\;\;\;\;\;\frac{\xi }{2}\frac{{\rm{d}}}{{{\rm{d}}t}}\int_\mathit{\Omega } {\int_0^1 {{{\left| {\nabla z\left( {x,k,t} \right)} \right|}^2}{\rm{d}}k{\rm{d}}x} } + \frac{\xi }{{2\tau }}\int_\mathit{\Omega } {\left| {\nabla z\left( {x,1,} \right.} \right.} \\ {\left. {\left. t \right)} \right|^2}{\rm{d}}x - \frac{\xi }{{2\tau }}\int_\mathit{\Omega } {{{\left| {\nabla z\left( {x,0,t} \right)} \right|}^2}{\rm{d}}x} = 0 \end{array} $

(11)

联立式(10)、(11)，并根据式(7) 可得

$ \begin{array}{l} E'\left( t \right) = - \left( {{\mu _1} - \frac{\xi }{{2\tau }}} \right)\left\| {\nabla {u_t}\left( s \right)} \right\|_2^2 - \frac{\xi }{{2\tau }}\int_\mathit{\Omega } {\left| {\nabla {z_n}} \right.} \\ {\left. {\left( {x,1,s} \right)} \right|^2}{\rm{d}}x - {\mu _2}\int_0^t {\int_\mathit{\Omega } {\nabla z\left( {x,1,s} \right) \cdot \nabla {u_t}\left( s \right){\rm{d}}x{\rm{d}}s} } - \frac{1}{2}\\ g\left( t \right)\left\| {\nabla u\left( t \right)} \right\|_2^2 + \frac{1}{2}\left( {g' \triangleleft \nabla u} \right)\left( t \right) \end{array} $

(12)

进一步，对式(12) 右端第三项利用杨不等式

$ \left( {ab \le \frac{1}{2}{a^2} + {\rm{ }}\frac{1}{2}{b^2}} \right) $进行估计，即可得式(9)。证毕。

注：由E(t)的定义式(7) 和引理3可知，能量是一致有界的，故式(6) 的解是整体存在的。

为了本文主要结论证明的需要和便利，现给出如下几个公式的定义。

$ L\left( t \right) = ME\left( t \right) + {\varepsilon _1}\varphi \left( t \right) + {\varepsilon _2}I\left( t \right) + \chi \left( t \right) $

其中M、ε₂、ε₁是待定的正常数。

$ \varphi \left( t \right) = \frac{1}{{\rho + 1}}\int_\mathit{\Omega } {{{\left| {{u_t}} \right|}^\rho }{u_t}u{\rm{d}}x} + \int_\mathit{\Omega } {\nabla {u_t}\left( t \right) \cdot \nabla u\left( t \right){\rm{d}}x} $

(13)

$ I\left( t \right) = \tau \int_0^1 {\int_\mathit{\Omega } {{{\rm{e}}^{ - k\tau }}{{\left| {\nabla z\left( {x,1,t} \right)} \right|}^2}{\rm{d}}x{\rm{d}}k} } $

(14)

$ \chi \left( t \right) = \int_\mathit{\Omega } {\left( {\nabla {u_t} - \frac{1}{{\rho + 1}}{{\left| {{u_t}} \right|}^\rho }{u_t}} \right)} \int_0^t {g\left( {t - s} \right)\left( {u\left( t \right) - u\left( s \right)} \right){\rm{d}}s{\rm{d}}x} $

(15)

进一步，可以得到引理4的结论。

引理4  若(u, z)为式(4) 的解，则L(t)~E(t)，即存在α₁、α₂ > 0，使得M足够大时，有

$ {\alpha _1}E\left( t \right) \le L\left( t \right) \le {\alpha _2}E\left( t \right),\forall t \ge 0。$

证明  首先，由E(t)的定义式(7)、引理3和条件(H)，有

$ l\left\| {\nabla u\left( t \right)} \right\|_2^2 + \left\| {\nabla {u_t}} \right\|_2^2 \le 2E\left( t \right) \le 2E\left( 0 \right),\forall t \ge 0 $

(16)

$ \begin{array}{l} \;\;\;\;\;\;\left| {\frac{1}{{\rho + 1}}\int_\mathit{\Omega } {{{\left| {{u_t}} \right|}^\rho }{u_t}u{\rm{d}}x} } \right| \le \frac{1}{{\rho + 2}}\left\| {\nabla {u_t}} \right\|_2^2 + \\ \frac{{C_s^{\rho + 2}}}{{\left( {\rho + 2} \right)\left( {\rho + 1} \right)}}{\left( {\frac{{2E\left( 0 \right)}}{l}} \right)^{\frac{\rho }{2}}}\left\| {\nabla u} \right\|_2^2 \end{array} $

(17)

$ \left| {\int_\mathit{\Omega } {\nabla {u_t}\left( t \right) \cdot \nabla u\left( t \right){\rm{d}}x} } \right| \le \frac{1}{2}\left\| {\nabla {u_t}} \right\|_2^2 + \frac{1}{2}\left\| {\nabla u} \right\|_2^2 $

(18)

$ \left| {I\left( t \right)} \right| \le {c_3}\int_\mathit{\Omega } {\int_0^1 {{{\left| {\nabla z\left( {x,k,t} \right)} \right|}^2}{\rm{d}}k{\rm{d}}x} } ,{c_3} > 0 $

(19)

其次，利用杨不等式估计χ(t)，有

$ \begin{array}{l} \;\;\;\;\;\;\left| { - \int_\mathit{\Omega } {\nabla {u_t}} \int_0^t {g\left( {t - s} \right)\left( {\nabla u\left( t \right) - \nabla u\left( s \right)} \right){\rm{d}}s{\rm{d}}x} } \right| \le \\ \frac{1}{2}\left\| {\nabla {u_t}\left( t \right)} \right\|_2^2 + \frac{{1 - l}}{2}\left( {g \triangleleft \nabla u} \right)\left( t \right) \end{array} $

(20)

$ \begin{array}{l} \;\;\left| { - \frac{1}{{\rho + 1}}\int_\mathit{\Omega } {{{\left| {{u_t}} \right|}^\rho }{u_t}} \int_0^t {g\left( {t - s} \right)\left( {\nabla u\left( t \right) - \nabla u\left( s \right)} \right)} } \right.\\ \left. {{\rm{d}}s{\rm{d}}x} \right| \le \frac{1}{{\rho + 2}}\left( {\left\| {\nabla {u_t}} \right\|_2^2 + \frac{{{{\left( {1 - l} \right)}^{\rho + 1}}C_s^{\rho + 2}}}{{\rho + 1}}} \right.\\ {\left( {\frac{{8E\left( 0 \right)}}{l}} \right)^{\frac{\rho }{2}}}\left( {g \triangleleft \nabla u} \right)\left( t \right) \end{array} $

(21)

进而，联立式(17)~(21)，有

$ \begin{array}{l} \left| {L\left( t \right) - ME\left( t \right)} \right| = {\varepsilon _1}\varphi \left( t \right) + {\varepsilon _2}I\left( t \right) + \chi \left( t \right) \le \\ C\left( {\left\| {{u_t}} \right\|_{\rho + 2}^{\rho + 2} + \left\| {\nabla u} \right\|_2^2 + \left\| {\nabla {u_t}} \right\|_2^2 + \left( {g \triangleleft \nabla u} \right)\left( t \right) + } \right.\\ \left. {\int_0^1 {\int_\mathit{\Omega } {{{\left| {\nabla z\left( {x,k,t} \right)} \right|}^2}{\rm{d}}x{\rm{d}}k} } } \right) \le {c_4}E\left( t \right)。\end{array} $

最后，由式(7) 知，取M足够大时，存在正数α₁和α₂，使得

$ {\alpha _1}E\left( t \right) \le L\left( t \right) \le {\alpha _2}E\left( t \right) $

成立。证毕。

基于上述所有引理和定理的结果，下面给出本文的核心结论。

定理2  设u₀, u₁∈H₀¹(Ω)，f₀∈L²((0, 1)，H₀¹(Ω))，并且条件(H)、式(4)、式(8) 及|μ₂|≤μ₁成立，则对于∀t₀ > 0，存在正常数K、ω，使得式(22) 成立

$ E\left( t \right) \le K{{\rm{e}}^{ - \omega }}\int_{{t_0}}^t {\xi \left( s \right){\rm{d}}s} ,\forall t > {t_0} $

(22)

证明  第一步，证明L′(t)的估计。分步证明如下。

1) 估计φ′(t)由式(13) 及式(6) 得

$ \begin{array}{l} \;\;\;\;\;\;\varphi '\left( t \right) = - \left\| {\nabla u} \right\|_2^2 + \int_\mathit{\Omega } {\nabla u\left( t \right)} \cdot \int_0^t {g\left( {t - s} \right)\nabla u} \\ \left( s \right){\rm{d}}s{\rm{d}}x - {\mu _1}\int_\mathit{\Omega } {\nabla u \cdot \nabla {u_t}{\rm{d}}x} - {\mu _2}\int_\mathit{\Omega } {\nabla u \cdot \nabla z\left( {x,1,t} \right)} \\ {\rm{d}}x + \frac{1}{{\rho + 1}}\left\| {{u_t}} \right\|_{\rho + 2}^{\rho + 2} + \left\| {\nabla {u_t}} \right\|_2^2 \end{array} $

(23)

现估计式(23) 中第2、3、4项，不妨分别记为I₁、I₂、I₃，则有

$ \begin{array}{l} \;\;\;\;\;\;\left| {{I_1}} \right| = \left| {\int_\mathit{\Omega } {\nabla u\left( t \right)} \cdot \int_0^t {g\left( {t - s} \right)\left( {\nabla u\left( s \right) - \nabla u\left( t \right) + } \right.} } \right.\\ \left. {\left. {\nabla u\left( t \right)} \right){\rm{d}}s{\rm{d}}x} \right| \le \eta \left\| {\nabla u} \right\|_2^2 + \frac{1}{{4\eta }}\int_\mathit{\Omega } {\left( {\int_0^t {g\left( {t - s} \right)} } \right.} \\ {\left. {\left| {\nabla u\left( s \right) - \nabla u\left( t \right)} \right|{\rm{d}}s} \right)^2}{\rm{d}}x + \int_0^t {g\left( {t - s} \right){\rm{d}}s\left\| {\nabla u} \right\|_2^2} \le \eta \\ \left\| {\nabla u\left( t \right)} \right\|_2^2 + \frac{1}{{4\eta }}\left( {1 - l} \right)\left( {g \triangleleft \nabla u} \right)\left( t \right) + \left( {1 - l} \right)\\ \left\| {\nabla u} \right\|_2^2; \end{array} $

$ \left| {{I_2}} \right| \le \eta \left\| {\nabla u} \right\|_2^2 + \frac{{\mu _1^2}}{{4\eta }}\left\| {\nabla {u_t}} \right\|_2^2; $

$ \left| {{I_3}} \right| \le \eta \left\| {\nabla u} \right\|_2^2 + \frac{{\mu _2^2}}{{4\eta }}\left\| {\nabla z\left( {x,1,t} \right)} \right\|_2^2。$

取$ \eta = \frac{l}{4} $，则可得

$ \begin{array}{l} \;\;\;\;\;\;\;\varphi '\left( t \right) \le - \frac{l}{4}\left\| {\nabla u} \right\|_2^2 + \frac{{1 - l}}{l}\left( {g \triangleleft \nabla u} \right)\left( t \right) + \\ \left( {\frac{{\mu _1^2}}{l} + 1} \right)\left\| {\nabla {u_t}} \right\|_2^2 + \frac{{\mu _2^2}}{4}\left\| {\nabla z\left( {x,1,t} \right)} \right\|_2^2 + \frac{1}{{\rho + 2}}\left\| {{u_t}} \right\|_{\rho + 2}^{\rho + 2} \end{array} $

(24)

2) 估计I′(t)由式(14) 和文献[16], 容易得式(25) 成立

$ \begin{array}{l} \;\;\;\;\;I'\left( t \right) \le \left\| {\nabla {u_t}} \right\|_2^2 - {{\rm{e}}^{ - \tau }}\left\| {\nabla z\left( {x,1,t} \right)} \right\|_2^2 - \\ \tau {{\rm{e}}^{ - \tau }}\int_0^1 {\int_\mathit{\Omega } {{{\left| {\nabla z\left( {x,1,t} \right)} \right|}^2}{\rm{d}}x{\rm{d}}k} } \end{array} $

(25)

3) 估计χ′(t)由方程(6) 得

$ \begin{array}{l} \;\;\;\;\;\chi '\left( t \right) = \int_\mathit{\Omega } {\nabla u\left( t \right)} \cdot \int_0^t {g\left( {t - s} \right)\left( {\nabla u\left( t \right) - \nabla u\left( s \right)} \right)} \\ {\rm{d}}s{\rm{d}}x - \int_\mathit{\Omega } {\left( {\int_0^t {g\left( {t - s} \right)\nabla u\left( s \right){\rm{d}}s} } \right)} \cdot \left( {\int_0^t {g\left( {t - s} \right)\left( {\nabla u} \right.} } \right.\\ \left. {\left. {\left( t \right) - \nabla u\left( s \right)} \right){\rm{d}}s} \right){\rm{d}}x + {\mu _1}\int_\mathit{\Omega } {\nabla u\left( t \right)} \cdot \int_0^t {g\left( {t - s} \right)\left( {\nabla u} \right.} \\ \left. {\left( t \right) - \nabla u\left( s \right)} \right){\rm{d}}s{\rm{d}}x + {\mu _2}\int_\mathit{\Omega } {\nabla z\left( {x,1,t} \right)} \cdot \int_0^t {g\left( {t - s} \right)\left( {\nabla u} \right.} \\ \left. {\left( t \right) - \nabla u\left( s \right)} \right){\rm{d}}s{\rm{d}}x - \int_\mathit{\Omega } {\nabla {u_t}} \cdot \int_0^t {g'\left( {t - s} \right)\left( {\nabla u\left( t \right) - \nabla u} \right.} \\ \left. {\left( s \right)} \right){\rm{d}}s{\rm{d}}x - \frac{1}{{\rho + 1}}\int_\mathit{\Omega } {{{\left| {{u_t}} \right|}^\rho }{u_t}} \int_0^t {g'\left( {t - s} \right)\left( {\nabla u\left( t \right) - \nabla u} \right.} \\ \left. {\left( s \right)} \right){\rm{d}}s{\rm{d}}x - \left( {\int_0^t {g\left( s \right){\rm{d}}s} } \right)\left\| {\nabla {u_t}} \right\|_2^2 - \frac{1}{{\rho + 1}}\left( {\int_0^t {g\left( s \right)} } \right.\\ \left. {{\rm{d}}s} \right)\left\| {{u_t}} \right\|_{\rho + 2}^{\rho + 2} \end{array} $

(26)

式(26) 等号右边共有7项，下面对前6项分别进行估计如下：

第1项

$ \begin{array}{l} \;\;\;\;\;\;\;\left| {\int_\mathit{\Omega } {\nabla {u_t}} \cdot \int_0^t {g\left( {t - s} \right)\left( {\nabla u\left( t \right) - \nabla u\left( s \right)} \right){\rm{d}}s{\rm{d}}x} } \right| \le \\ \delta \left\| {\nabla u} \right\|_2^2 + \frac{{1 - l}}{{4\delta }}\left( {g \triangleleft \nabla u} \right)\left( t \right); \end{array} $

第2项

$ \begin{array}{l} \;\;\;\;\;\;\;\left| {\int_\mathit{\Omega } {\left( {\int_0^t {g\left( {t - s} \right)\nabla u\left( s \right){\rm{d}}s} } \right)} \cdot \left( {\int_0^t {g\left( {t - s} \right)\left( {\nabla u} \right.} } \right.} \right.\\ \left. {\left. {\left. {\left( t \right) - \nabla u\left( s \right)} \right){\rm{d}}s} \right){\rm{d}}x} \right| \le 2\delta \left( {1 - l} \right)\left\| {\nabla u} \right\|_2^2 + \left( {2\delta + } \right.\\ \left. {\frac{1}{{4\delta }}} \right)\left( {1 - l} \right)\left( {g \triangleleft \nabla u} \right)\left( t \right),\delta > 0; \end{array} $

第3项

$ \begin{array}{l} \;\;\;\;\;\;\;\left| {{\mu _1}} \right.\int_\mathit{\Omega } {\nabla {u_t}\left( t \right)} \cdot \int_0^t {g\left( {t - s} \right)\left( {\nabla u\left( t \right) - \nabla u\left( s \right)} \right)} \\ \left. {{\rm{d}}s{\rm{d}}x} \right| \le {\delta _1}\left\| {\nabla {u_t}} \right\|_2^2 + \frac{{\mu _1^2\left( {1 - l} \right)}}{{4{\delta _1}}}\left( {g \triangleleft \nabla u} \right)\left( t \right),{\delta _1} > 0; \end{array} $

第4项

$ \begin{array}{l} \;\;\;\;\;\;\;\left| {{\mu _2}} \right.\int_\mathit{\Omega } {\nabla z\left( {x,1,t} \right)} \cdot \int_0^t {g\left( {t - s} \right)\left( {\nabla u\left( t \right) - \nabla u\left( s \right)} \right)} \\ \left. {{\rm{d}}s{\rm{d}}x} \right| \le {\delta _2}\left\| {\nabla z\left( {x,1,t} \right)} \right\|_2^2 + \frac{{\mu _1^2\left( {1 - l} \right)}}{{4{\delta _2}}}\left( {g \triangleleft \nabla u} \right)\left( t \right),\\ {\delta _2} > 0; \end{array} $

第5项

$ \begin{array}{l} \;\;\;\;\;\;\;\;\left| {\int_\mathit{\Omega } {\nabla {u_t}} \cdot \int_0^t {g'\left( {t - s} \right)\left( {\nabla u\left( t \right) - \nabla u\left( s \right)} \right){\rm{d}}s{\rm{d}}x} } \right| \le \\ {\delta _3}\left\| {\nabla {u_t}} \right\|_2^2 + \frac{1}{{4{\delta _3}}}\int_\mathit{\Omega } {\left( {\int_0^t {g'\left( {t - s} \right)\left( {\nabla u\left( t \right) - \nabla u\left( s \right)} \right)} } \right.} \\ {\left. {{\rm{d}}s} \right)^2}{\rm{d}}x \le {\delta _3}\left\| {\nabla {u_t}} \right\|_2^2 + \frac{{g\left( 0 \right)}}{{4{\delta _3}}}\left( {g' \triangleleft \nabla u} \right)\left( t \right),{\delta _3} > 0; \end{array} $

第6项

$ \begin{array}{l} \;\;\;\;\;\;\;\;\left| {\frac{1}{{\rho + 1}}\int_\mathit{\Omega } {{{\left| {{u_t}} \right|}^\rho }{u_t}} \int_0^t {g'\left( {t - s} \right)\left( {\nabla u\left( t \right) - \nabla u\left( s \right)} \right)} } \right.\\ \left. {{\rm{d}}s{\rm{d}}x} \right| \le \frac{1}{{\rho + 1}}\left( {{\delta _4}\left\| {{u_t}} \right\|_{2\left( {\rho + 1} \right)}^{2\left( {\rho + 1} \right)} + \frac{1}{{4{\delta _4}}}\int_\mathit{\Omega } {\left( {\int_0^t {g'\left( {t - s} \right)} } \right.} } \right.\\ \left. {\left. {{{\left( {\nabla u\left( t \right) - \nabla u\left( s \right)} \right)}^2}{\rm{d}}x} \right){\rm{d}}s} \right) \le \frac{{{\delta _4}c_s^{2\left( {\rho + 1} \right)}}}{{\rho + 1}}{\left( {2E\left( 0 \right)} \right)^\rho }\\ \left\| {\nabla {u_t}} \right\|_2^2 - \frac{{g\left( 0 \right)c_s^2}}{{4{\delta _4}\left( {\rho + 1} \right)}}\left( {g' \triangleleft \nabla u} \right)\left( t \right),{\delta _4} > 0。\end{array} $

第二步，分别估计式(26) 等号右边的每一项。将第1~6项的估计式代入χ′(t)得

$ \begin{array}{l} \;\;\;\;\;\;\chi '\left( t \right) \le \delta {c_4}\left\| {\nabla u} \right\|_2^2 + {c_5}\left( {g \triangleleft \nabla u} \right)\left( t \right) - {c_6}\left( {g' \triangleleft } \right.\\ \left. {\nabla u} \right)\left( t \right) + \left( {{c_7} - \int_0^t {g\left( s \right){\rm{d}}s} } \right)\left\| {\nabla {u_t}} \right\|_2^2 + {\delta _2}\left\| {\nabla z\left( {x,1,} \right.} \right.\\ \left. {\left. t \right)} \right\|_2^2 - \frac{1}{{\rho + 1}}\left( {\int_0^t {g\left( s \right){\rm{d}}s} } \right)\left\| {{u_t}} \right\|_2^2 \end{array} $

(27)

其中

$ {c_4} = 1 + 2{\left( {1 - l} \right)^2}, $

$ {c_5} = \left( {2\delta + \frac{1}{{2\delta }} + \frac{{\mu _1^2}}{{4{\delta _1}}} + \frac{{\mu _2^2}}{{4{\delta _2}}}} \right)\left( {1 - l} \right), $

$ {c_6} = \frac{{g\left( 0 \right)c_s^2}}{{4{\delta _4}\left( {\rho + 1} \right)}} + \frac{{g\left( 0 \right)}}{{4{\delta _3}}}, $

$ {c_7} = {\delta _1} + {\delta _3} + \frac{{{\delta _4}c_s^{2\left( {\rho + 1} \right)}}}{{\rho + 1}}{\left( {2E\left( 0 \right)} \right)^\rho }。$

由g(0) > 0以及条件(H)得，∀t₀ > 0，

$ \int_0^t {g\left( s \right){\rm{d}}s} \ge \int_0^{{t_0}} {g\left( s \right){\rm{d}}s} = {g_0},\forall t \ge {t_0}。$

根据引理3，将式(24)~(27) 代入L′(t)得

$ \begin{array}{l} \;\;\;\;\;\;\;\;L'\left( t \right) = ME'\left( t \right) + {\varepsilon _1}\varphi '\left( t \right) + {\varepsilon _2}I'\left( t \right) + \chi '\left( t \right) \le \\ \left( {\frac{M}{2} - {c_6}} \right)\left( {g' \triangleleft \nabla u} \right)\left( t \right) - \frac{{{g_0} - {\varepsilon _1}}}{{\rho + 1}}\left\| {{u_t}} \right\|_{\rho + 2}^{\rho + 2} - \left( {\frac{l}{4}{\varepsilon _1} - } \right.\\ \left. {\delta {c_4}} \right)\left\| {\nabla u} \right\|_2^2 - \left( {{g_0} - {c_7} - {\varepsilon _2} - \left( {\frac{{\mu _1^2}}{l} + 1} \right){\varepsilon _1}} \right)\left\| {\nabla u} \right\| - \\ \left( {{\varepsilon _2}{{\rm{e}}^{ - \tau }} - {\varepsilon _1}\frac{{\mu _2^2}}{4} - {\delta _2}} \right)\left\| {\nabla z\left( {x,1,t} \right)} \right\|_2^2 - \tau {{\rm{e}}^{ - \tau }}\\ \int_0^1 {\int_\mathit{\Omega } {\left| {\nabla z\left( {x,k,t} \right)} \right|{\rm{d}}x{\rm{d}}s} } + \left( {\frac{{\left( {1 - l} \right){\varepsilon _1}}}{l} + {c_5}} \right)\left( {g \triangleleft } \right.\\ \left. {\nabla u} \right)\left( t \right)。\end{array} $

现选取合适的ε_i(i=1, 2) 和δ_j(j=1, 2, 3, 4)，使得存在正常数β₁和β₂满足式(28)

$ L'\left( t \right) \le - {\beta _1}E\left( t \right) + {\beta _2}\left( {g \triangleleft \nabla u} \right)\left( t \right),\forall t \ge {t_0} $

(28)

取δ₁=δ₃=δ₄足够小，使得

$ {c_7} = {\delta _1}\left( {2 + \frac{{c_s^{2\left( {\rho + 1} \right)}}}{{\rho + 1}}{{\left( {2E\left( 0 \right)} \right)}^\rho }} \right) \le \frac{{{g_0}}}{2}; $

然后取ε₂ > 0足够小，使得$ {\varepsilon _2} \le \frac{{{g_0}}}{8} $，则

g₀-c₇-ε₂≥$ \frac{3}{8}{g_0} $成立；ε取定后，选取δ₂≤$ \frac{{{\varepsilon _2}{{\rm{e}}^{{\rm{ - }}\tau }}}}{2} $。进一步选取ε足够小，使得

$ {\varepsilon _1} < \min \left\{ {{g_0},\frac{{\frac{{{g_0}}}{8}}}{{\frac{{\mu _1^2}}{l} + 1}},\frac{{{\varepsilon _2}{{\rm{e}}^{ - \tau }}}}{{\mu _2^2}}} \right\}; $

当ε固定后，选取δ > 0足够小，使得$ \delta < \frac{{l{\varepsilon _1}}}{{8{c_4}}} $；最后选取M足够大，使得M > 4c₆。最后由式(7) 可得式(28) 成立。

第三步，用ξ(t)同乘以式(27) 的两边，进而可得

$ \xi \left( t \right)L'\left( t \right) \le - {\beta _1}\xi \left( t \right)E\left( t \right) + {\beta _2}\xi \left( t \right)\left( {g \triangleleft \nabla u} \right)\left( t \right) $

由引理3及g′(t)≤-ξ(t)g(t)，易得-(g◁∇u)(t)≤－2E′(t)，进而可有

$ \begin{array}{l} \;\;\;\;\;\;\xi \left( t \right)L'\left( t \right) \le - {\beta _1}\xi \left( t \right)E\left( t \right) - {\beta _2}\left( {g \triangleleft \nabla u} \right)\left( t \right) \le \\ - {\beta _1}\xi \left( t \right)E\left( t \right) - 2{\beta _2}E'\left( t \right),\forall t \ge {t_0}。\end{array} $

现定义H(t)=ξ(t)L(t)+2β₂E(t)，易证H(t)等价于E(t)。由式(26) 可知

$ \begin{array}{l} \;\;\;\;\;H'\left( t \right) = \xi '\left( t \right)L\left( t \right) - {\beta _1}\xi \left( t \right)E\left( t \right) \le - {\beta _1}\xi \left( t \right)E\\ \left( t \right) \le - {\beta _3}\xi \left( t \right)H\left( t \right),\forall t \ge {t_0} \end{array} $

(29)

将式(29) 在(t₀, t)上积分，得

$ H\left( t \right) \le H\left( 0 \right){{\rm{e}}^{ - {\beta _3}\int_{{t_0}}^t {\xi \left( s \right){\rm{d}}s} }},\forall t \ge {t_0}。$

故由关系式H(t)等价于E(t)得

$ E\left( t \right) \le K{{\rm{e}}^{ - \omega \int_{{t_0}}^t {\xi \left( s \right){\rm{d}}s} }},\forall t \ge {t_0}。$

其中K和ω为正常数。证毕。

3 举例说明

1) 当g(t)=a(1+t)^-ν(a > 0，ν > 1) 时，ξ(t)=$ \frac{\nu }{{1 + t}} $满足式(5)，则由式(22) 可知

$ E\left( t \right) \le K{\left( {1 + t} \right)^{ - \omega }}; $

2) 当g(t)=ae^-b(1+t)ν, 且a、b > 0, 0 < ν≤1时，ξ(t)=bγ(1+t)^ν-1满足式(5)，由式(22) 可知

$ E\left( t \right) \le K{{\rm{e}}^{ - \omega \left( {1 + t} \right)v}}; $

3) 当g(t)=ae^-blnν(1+t), 且a、b > 0，ν > 1时，ξ(t)=$ \frac{{b\nu {\rm{l}}{{\rm{n}}^{\nu - 1}}\left( {1 + t} \right)}}{{1 + t}} $满足式(5)，由式(22) 可知

$ E\left( t \right) \le K{{\rm{e}}^{ - \omega {{\ln }^v}\left( {1 + t} \right)}}; $

4) 当g(t)=$ \frac{a}{{\left( {1 + t} \right){\rm{l}}{{\rm{n}}^\nu }\left( {1 + t} \right)}} $, 且a > 0，ν > 1时，ξ(t)=$ \frac{{\nu + {\rm{ln}}\left( {1 + t} \right)}}{{\left( {1 + t} \right){\rm{l}}{{\rm{n}}^\nu }\left( {1 + t} \right)}} $满足式(5)，由式(22) 可知

$ E\left( t \right) \le K{\left( {\left( {1 + t} \right){{\ln }^v}\left( {1 + t} \right)} \right)^{ - \omega }}。$

4 结束语

本文通过构造合适的Lyapunov泛函，研究了一类具有强阻尼和强时滞作用的粘弹性方程的初边值问题，得到了该系统能量的衰减估计。