近年来, 对于波动方程解的爆破问题有很多研究, 其中关于非线性耦合波动方程解的存在和爆破问题已有许多结果[1-11].然而, 对于波动方程爆破时间的研究还不是很多.爆破的准确时间一般很难确定, 因此在实践中计算出爆破时间的上、下确界具有重要意义.在关于波动方程爆破问题的文章中, 一般从结果中都可以得到爆破时间的上确界, 而下确界的确定却还没有被普遍关注.文献[12]研究了耦合波动方程, 在适当假设情况下证明了解的爆破, 并且得到了爆破时间的上确界.文献[13]研究了两类非线性波动方程爆破时间的问题, 在解爆破的前提下精确地计算出了爆破时间的下确界.文献[14]利用辅助函数在爆破的前提下得到了爆破时间的下确界.
本文考虑如下的问题
$ \left\{ \begin{array}{l} {u_{tt}} + {u_t} + {\left| {{u_t}} \right|^{m - 1}}{u_t} = {\rm{div}}\left( {{\rho _1}\left( {{{\left| {\nabla u} \right|}^2}} \right)\nabla u} \right) + {f_1}\left( {u,\upsilon } \right),\;\;\;\;\left( {x,t} \right) \in \mathit{\Omega } \times \left( {0,T} \right),\\ {\upsilon _{tt}} + {\upsilon _t} + {\left| {{\upsilon _t}} \right|^{\gamma - 1}}{\upsilon _t} = {\rm{div}}\left( {{\rho _2}\left( {{{\left| {\nabla \upsilon } \right|}^2}} \right)\nabla \upsilon } \right) + {f_2}\left( {u,\upsilon } \right),\;\;\;\left( {x,t} \right) \in \mathit{\Omega } \times \left( {0,T} \right),\\ u = \upsilon = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {x,t} \right) \in \partial \mathit{\Omega } \times \left( {0,T} \right),\\ u\left( {x,0} \right) = {u_0}\left( x \right),u\left( {x,0} \right) = {u_1}\left( x \right),\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \mathit{\Omega },\\ \upsilon \left( {x,0} \right) = {\upsilon _0}\left( x \right),\upsilon \left( {x,0} \right) = {\upsilon _1}\left( x \right),\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x \in \mathit{\Omega }\mathit{.} \end{array} \right. $ | (1) |
其中Ω是R2中的有界区域, 且具有光滑边界∂Ω, m, γ≥1, T > 0;文章将在解爆破的前提下得到问题(1) 爆破时间的下确界.
为了获得问题(1) 解的爆破结果, 作如下假设.
(A1) 假设fi(·, ·):R2→R2为给定的函数, F(u, v)=a|u+v|p+1+2b
(A2) 存在正常数c0, c1使得∀(u, v)∈R2,F(u, v) 满足
$ {c_0}\left( {{{\left| u \right|}^{p + 1}} + {{\left| \upsilon \right|}^{p + 1}}} \right) \le F\left( {u,\upsilon } \right) \le {c_1}\left( {{{\left| u \right|}^{p + 1}} + {{\left| \upsilon \right|}^{p + 1}}} \right). $ | (2) |
(A3) 给定函数ρ1, ρ2对所有s > 0满足
$ {\rho _i}\left( s \right) \in {C^1},{\rho _i}\left( s \right) > 0,{\rho _i}\left( s \right) + 2s{{\rho '}_i}\left( s \right) > 0,{\rho _i}\left( s \right) = {b_1} + {b_2}{s^{{q_i}}},{q_i} > 0,{b_1} + {b_2} > 0. $ | (3) |
文章中记‖·‖p为Lp(Ω) 的范数, 用E(t) 表示系统(1) 的能量.
1 相关定理及引理首先给出解的局部存在性定理和解的爆破结果.
定理1[15-16] 假设(A1)~(A3) 成立, 对于任意初值u0∈W02q1+2(Ω)∩Lp+1(Ω), v0∈ W02q2+2 (Ω)∩Lp+1(Ω) 并且u1, v1∈L2(Ω), 则问题(1) 存在唯一的局部弱解(u, v): u∈L∞([0, T); W02q1+2 (Ω)∩Lp+1(Ω), v∈L∞([0, T); W02q2+2 (Ω)∩Lp+1(Ω) ut∈L∞([0, T); L2(Ω)∩Lm+1(Ω×(0, T)), vt∈L∞[0, T); L2(Ω)∩Lr+1(Ω×(0, T)) 其中, T > 0.
定理2[12] 假设(u, v) 时系统(1) 的解, 且满足假设(A1)~(A3).p > max{2q1+1, 2q2+1, m, r}, 初始能量E(0) < 0, 并且存在常数τ和B使得b1q >
$ T \le \frac{{1 - \alpha }}{{\gamma \alpha }}{\psi ^{\frac{\alpha }{{\alpha - 1}}}}\left( 0 \right). $ |
其中, 0 < α <
为了获得系统(1) 爆破时间的下确界, 有以下引理.
引理1[14] 假设Ω是Rn(n=2, 3) 中的有界区域, 且具有光滑边界∂Ω.如果u(x) 在Ω上满足u(x)∈C1而且在∂Ω上u(x)=0, 则有不等式
$ \int_\mathit{\Omega } {{u^{2p}}{\rm{d}}x < \delta {{\left( {\int_\mathit{\Omega } {{{\left| {\nabla \upsilon } \right|}^2}{\rm{d}}x} } \right)}^p}} . $ |
其中:当n=2时, p > 1;当n=3时, 1 < p <
定理3 假设u(t) 是系统(1) 的解, 且u(t) 在有限时间T爆破, 则
$ T \ge \int_{G\left( 0 \right)}^\infty {\frac{{{\rm{d}}y}}{{\frac{{p + 1}}{2}\delta {{\left( {2{c_2}E\left( 0 \right) + 2{c_2}{c_1}y} \right)}^p} + E\left( 0 \right) + {c_1}y}}} . $ |
其中
$ G\left( 0 \right) = \int_\mathit{\Omega } {\left( {{{\left| {{u_0}} \right|}^{p + 1}} + {{\left| {{\upsilon _0}} \right|}^{p + 1}}} \right){\rm{d}}x,\delta = {{\left( {\frac{1}{{{2^{3/2}}}}} \right)}^{2p}}\left| \mathit{\Omega } \right|} . $ |
证明 首先, 对系统(1) 的前两个等式两边分别同乘ut与vt,并且在Ω上积分,得
$ \int_\mathit{\Omega } {\left( {{u_{tt}}{u_t} + {u_t}{u_t} + {{\left| {{u_t}} \right|}^{m - 1}}{u_t}{u_t}} \right){\rm{d}}x} = \int_\mathit{\Omega } {\left( {{\rm{div}}\left( {{\rho _1}\left( {{{\left| {\nabla u} \right|}^2}} \right)\nabla u} \right){u_t} + {f_1}\left( {u,\upsilon } \right){u_t}} \right){\rm{d}}x} , $ | (4) |
$ \int_\mathit{\Omega } {\left( {{\upsilon _{tt}}{\upsilon _t} + {\upsilon _t}{\upsilon _t} + {{\left| {{\upsilon _t}} \right|}^{m - 1}}{\upsilon _t}{\upsilon _t}} \right){\rm{d}}x} = \int_\mathit{\Omega } {\left( {{\rm{div}}\left( {{\rho _1}\left( {{{\left| {\nabla \upsilon } \right|}^2}} \right)\nabla \upsilon } \right){\upsilon _t} + {f_1}\left( {u,\upsilon } \right){\upsilon _t}} \right){\rm{d}}x} . $ | (5) |
则式(4) 等号右边第一项可化简为
$ \begin{array}{*{20}{c}} {\int_\mathit{\Omega } {\left( {{\rm{div}}\left( {{\rho _1}\left( {{{\left| {\nabla u} \right|}^2}} \right)\nabla u} \right){u_t}{\rm{d}}x = } \right.} \int_{\partial \mathit{\Omega }} {{\rho _1}\left( {{{\left| {\nabla u} \right|}^2}} \right)\nabla u \cdot \mathit{\boldsymbol{v}} \cdot {u_t}{\rm{d}}} \mathit{\Gamma } - \int_\mathit{\Omega } {{\rho _1}\left( {{{\left| {\nabla u} \right|}^2}} \right)\nabla u\nabla {u_t}{\rm{d}}x} = }\\ { - \int_\mathit{\Omega } {{\rho _1}\left( {{{\left| {\nabla u} \right|}^2}} \right)\nabla u\nabla {u_t}{\rm{d}}x} .} \end{array} $ |
其中, ν为边界的外法向量.令Pi=
$ \int_\mathit{\Omega } {{\rho _1}\left( {{{\left| {\nabla u} \right|}^2}} \right)\nabla u\nabla {u_t}{\rm{d}}x} = \frac{1}{2}\frac{{\rm{d}}}{{{\rm{d}}t}}\int_\mathit{\Omega } {{P_1}\left( {{{\left| {\nabla u} \right|}^2}} \right){\rm{d}}x} . $ |
同理,
$ \int_\mathit{\Omega } {{\rho _2}\left( {{{\left| {\nabla u} \right|}^2}} \right)\nabla u\nabla {u_t}{\rm{d}}x} = \frac{1}{2}\frac{{\rm{d}}}{{{\rm{d}}t}}\int_\mathit{\Omega } {{P_2}\left( {{{\left| {\nabla u} \right|}^2}} \right){\rm{d}}x} . $ |
所以, 式(4)~(5) 可化简为
$ \frac{1}{2}\frac{{\rm{d}}}{{{\rm{d}}t}}\int_\mathit{\Omega } {{{\left| {{u_t}} \right|}^2}{\rm{d}}x} + \int_\mathit{\Omega } {{u_t}{u_t}{\rm{d}}x} + \int_\mathit{\Omega } {{{\left| {{u_t}} \right|}^{m - 1}}{u_t}{u_t}{\rm{d}}x} = - \frac{1}{2}\frac{{\rm{d}}}{{{\rm{d}}t}}\int_\mathit{\Omega } {{P_1}\left( {{{\left| {\nabla u} \right|}^2}} \right){\rm{d}}x} + \int_\mathit{\Omega } {\frac{{\partial F}}{{\partial u}}{u_t}{\rm{d}}x} , $ | (6) |
$ \frac{1}{2}\frac{{\rm{d}}}{{{\rm{d}}t}}\int_\mathit{\Omega } {{{\left| {{\upsilon _t}} \right|}^2}{\rm{d}}x} + \int_\mathit{\Omega } {{\upsilon _t}{\upsilon _t}{\rm{d}}x} + \int_\mathit{\Omega } {{{\left| {{\upsilon _t}} \right|}^{\gamma - 1}}{\upsilon _t}{\upsilon _t}{\rm{d}}x} = - \frac{1}{2}\frac{{\rm{d}}}{{{\rm{d}}t}}\int_\mathit{\Omega } {{P_1}\left( {{{\left| {\nabla \upsilon } \right|}^2}} \right){\rm{d}}x} + \int_\mathit{\Omega } {\frac{{\partial F}}{{\partial \upsilon }}{\upsilon _t}{\rm{d}}x} . $ | (7) |
将式(6) 与式(7) 相加, 整理可得
$ \begin{array}{l} \frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\frac{1}{2}\left\| {{u_t}} \right\|_2^2 + \frac{1}{2}\left\| {{\upsilon _t}} \right\|_2^2 + \frac{1}{2}\int_\mathit{\Omega } {\left( {{P_1}\left( {{{\left| {\nabla u} \right|}^2}} \right) + {P_2}\left( {{{\left| {\nabla \upsilon } \right|}^2}} \right)} \right){\rm{d}}x} - \int_\mathit{\Omega } {F\left( {u,\upsilon } \right){\rm{d}}x} } \right) = \\ - \int_\mathit{\Omega } {{u_t}{u_t}{\rm{d}}x} - \int_\mathit{\Omega } {{{\left| {{u_t}} \right|}^{m - 1}}{u_t}{u_t}{\rm{d}}x} - \int_\mathit{\Omega } {{\upsilon _t}{\upsilon _t}{\rm{d}}x} + \int_\mathit{\Omega } {{{\left| {{\upsilon _t}} \right|}^{\gamma - 1}}{\upsilon _t}{\upsilon _t}{\rm{d}}x} . \end{array} $ |
所以可以得到能量及其导数的表达式, 即
$ E\left( t \right) = \frac{1}{2}\left( {\left\| {{u_t}} \right\|_2^2 + \left\| {{\upsilon _t}} \right\|_2^2} \right) + \frac{1}{2}\int_\mathit{\Omega } {\left( {{P_1}\left( {{{\left| {\nabla u} \right|}^2}} \right) + {P_2}\left( {{{\left| {\nabla \upsilon } \right|}^2}} \right)} \right){\rm{d}}x} - \int_\mathit{\Omega } {F\left( {u,\upsilon } \right){\rm{d}}x} , $ | (8) |
$ \frac{{{\rm{d}}E\left( t \right)}}{{{\rm{d}}t}} = E'\left( t \right) = - \int_\mathit{\Omega } {{u_t}{u_t}{\rm{d}}x} - \int_\mathit{\Omega } {{{\left| {{u_t}} \right|}^{m - 1}}{u_t}{u_t}{\rm{d}}x} - \int_\mathit{\Omega } {{\upsilon _t}{\upsilon _t}{\rm{d}}x} + \int_\mathit{\Omega } {{{\left| {{\upsilon _t}} \right|}^{\gamma - 1}}{\upsilon _t}{\upsilon _t}{\rm{d}}x} . $ | (9) |
由式(9) 得
$ \frac{{{\rm{d}}E\left( t \right)}}{{{\rm{d}}t}} < 0, $ |
则
$ E\left( t \right) \le E\left( 0 \right). $ | (10) |
所以由式(8) 和式(10) 可得
$ E\left( 0 \right) + \int_\mathit{\Omega } {F\left( {u,\upsilon } \right){\rm{d}}x} \ge \frac{1}{2}\left( {\left\| {{u_t}} \right\|_2^2 + \left\| {{\upsilon _t}} \right\|_2^2} \right) + \frac{1}{2}\int_\mathit{\Omega } {\left( {{P_1}\left( {{{\left| {\nabla u} \right|}^2}} \right) + {P_2}\left( {{{\left| {\nabla \upsilon } \right|}^2}} \right)} \right){\rm{d}}x} . $ | (11) |
接下来设辅助函数为
$ G\left( t \right) = \int_\mathit{\Omega } {\left( {{{\left| u \right|}^{p + 1}} + {{\left| \upsilon \right|}^{p + 1}}} \right){\rm{d}}x} , $ |
则关于t求导可得
$ G'\left( t \right) = \left( {p + 1} \right)\int_\mathit{\Omega } {\left( {{{\left| u \right|}^p}{u_t} + {{\left| \upsilon \right|}^p}{\upsilon _t}} \right){\rm{d}}x} . $ |
利用Cauchy不等式得
$ G'\left( t \right) \le \frac{{\left( {p + 1} \right)}}{2}\left( {\int_\mathit{\Omega } {{{\left| u \right|}^{2p}}{\rm{d}}x} + \int_\mathit{\Omega } {{{\left| {{u_t}} \right|}^2}{\rm{d}}x} + \int_\mathit{\Omega } {{{\left| \upsilon \right|}^{2p}}{\rm{d}}x} + \int_\mathit{\Omega } {{{\left| {{\upsilon _t}} \right|}^2}{\rm{d}}x} } \right). $ |
由式(11) 可得
$ G'\left( t \right) \le \frac{{\left( {p + 1} \right)}}{2}\left( {\int_\mathit{\Omega } {{{\left| u \right|}^{2p}}{\rm{d}}x} + \int_\mathit{\Omega } {{{\left| \upsilon \right|}^{2p}}{\rm{d}}x} } \right) + E\left( 0 \right) + \int_\mathit{\Omega } {F\left( {u,\upsilon } \right){\rm{d}}x} . $ |
再利用引理1可得
$ \begin{array}{*{20}{c}} {G'\left( t \right) \le \frac{{\left( {p + 1} \right)}}{2}\delta \left( {{{\left( {\int_\mathit{\Omega } {{{\left| {\nabla u} \right|}^2}} {\rm{d}}x} \right)}^p} + {{\left( {\int_\mathit{\Omega } {{{\left| {\nabla \upsilon } \right|}^2}} {\rm{d}}x} \right)}^p} + E\left( 0 \right) + \int_\mathit{\Omega } {F\left( {u,\upsilon } \right){\rm{d}}x} \le } \right.}\\ {\frac{{\left( {p + 1} \right)}}{2}\delta {{\left( {\int_\mathit{\Omega } {{{\left| {\nabla u} \right|}^2}} {\rm{d}}x + \int_\mathit{\Omega } {{{\left| {\nabla \upsilon } \right|}^2}} {\rm{d}}x} \right)}^p} + E\left( 0 \right) + \int_\mathit{\Omega } {F\left( {u,\upsilon } \right){\rm{d}}x} .} \end{array} $ | (12) |
由式(3) 可知, 存在一个常数c2使得
$ {c_2}\int_0^s {{\rho _i}\left( \xi \right){\rm{d}}\xi } \ge s $ |
成立,即
$ {c_2}{P_i}\left( s \right) \ge s, $ |
所以有
$ {c_2}{P_1}\left( {{{\left| {\nabla u} \right|}^2}} \right) \ge {\left| {\nabla u} \right|^2},{c_2}{P_2}\left( {{{\left| {\nabla \upsilon } \right|}^2}} \right) \ge {\left| {\nabla \upsilon } \right|^2}. $ | (13) |
将式(13) 代入式(12) 可得
$ G'\left( t \right) \le \frac{{\left( {p + 1} \right)}}{2}\delta {\left( {\int_\mathit{\Omega } {{c_2}{P_1}\left( {{{\left| {\nabla u} \right|}^2}} \right){\rm{d}}x} + \int_\mathit{\Omega } {{c_2}{P_2}\left( {{{\left| {\nabla \upsilon } \right|}^2}} \right){\rm{d}}\upsilon } } \right)^p} + E\left( 0 \right) + \int_\mathit{\Omega } {F\left( {u,\upsilon } \right){\rm{d}}x} . $ |
再由式(11) 可得
$ G'\left( t \right) \le \frac{{\left( {p + 1} \right)}}{2}\delta {\left( {2{c_2}E\left( 0 \right) + 2{c_2}\int_\mathit{\Omega } {F\left( {u,\upsilon } \right){\rm{d}}x} } \right)^p} + E\left( 0 \right) + \int_\mathit{\Omega } {F\left( {u,\upsilon } \right){\rm{d}}x} . $ |
根据式(2) 可以得到
$ \begin{array}{l} G'\left( t \right) \le \frac{{\left( {p + 1} \right)}}{2}\delta {\left( {2{c_2}E\left( 0 \right) + 2{c_2}{c_1}\left( {{{\left| u \right|}^{p + 1}} + {{\left| \upsilon \right|}^{p + 1}}} \right)} \right)^p} + E\left( 0 \right) + {c_1}\left( {{{\left| u \right|}^{p + 1}} + {{\left| \upsilon \right|}^{p + 1}}} \right) \le \\ \;\;\;\;\;\;\;\;\;\;\;\frac{{\left( {p + 1} \right)}}{2}\delta {\left( {2{c_2}E\left( 0 \right) + 2{c_2}{c_1}G\left( t \right)} \right)^p} + E\left( 0 \right) + {c_1}G\left( t \right). \end{array} $ | (14) |
因为
$ T \ge \int_{G\left( 0 \right)}^\infty {\frac{{{\rm{d}}y}}{{\frac{{p + 1}}{2}\delta {{\left( {2{c_2}E\left( 0 \right) + 2{c_2}{c_1}y} \right)}^p} + E\left( 0 \right) + {c_1}y}}} . $ |
证毕.
3 结束语计算波动方程解爆破时间的下确界的关键是G(t) 的选取以及得到最终的微分不等式.而且G(t) 一定要是非线性的, 否则会与前面的爆破结果相矛盾, 而利用引理中的方法就可保证G(t) 非线性.本文尽可能使得爆破时间的下确界更精确, 但由于技术和方法的局限, 求解爆破时间的更为精确的方法还有待进一步研究.
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