Machine-translated from Chinese.
定义
$a^{ij},b^i,c$为在$\Omega_T$上的函数, 记
$$
Lu:=-(a^{ij}u_j)_i+b^iu_i+cu.
$$
若$a^{ij}$对称, 满足一致条件,
则称$\partial_t^2+L$为$\Omega_T$上二阶线性散度型一致双曲算子.
考虑初边值问题
$$
\begin{cases}
u_{tt}+Lu=f&in\:\Omega_T\\
u=0&on\:\partial\Omega\times (0,T]\\
u|_{t=0}=g,u_t|_{t=0}=h&in\:\Omega
\end{cases}
$$
在一致基础上, 考虑条件
$$
a^{ij}\in C^1(\overline\Omega_T),\quad b^i,c\in C(\overline\Omega_T).
$$
$$
f\in L^2(\Omega_T),\quad g\in H_0^1(\Omega),\quad h\in L^2(\Omega).
$$
双曲方程弱解需满足$u\in L^2(0,T;H_0^1(\Omega)),$
$u_{tt}\in L^2(0,T;H^{-1}(\Omega)),$ $\,\forall\,v\in H_0^1(\Omega),$
a.e. $t\in (0,T],$
$$
\left<{}u_{tt},v\right>+B[u,v;t]=\left<{}f(\cdot,t),v\right>
$$
为了满足初始条件, 需要$u,u_t\in C([0,T];\ast ),$
因此需要求$u_t\in L^2(0,T;L^2(\Omega))=L^2(\Omega_T),$ 这样
$$
u\in W^{1,2}(0,T;L^2(\Omega))\hookrightarrow C([0,T];L^2(\Omega)),
$$
$$
u_t\in W^{1,2}(0,T;H^{-1}(\Omega))\hookrightarrow C([0,T];H^{-1}(\Omega)).
$$
初值条件即指按相应模逼近, 未必是$g,h$原本所在的空间.
Galerkin逼近
同理考虑$u_m=\sum_{k=1}^md_m^kw_k,$ 令$v=w_k,$
由$f_k=\left<{}w_k,f\right>\in L^2(0,T),$
$e_l^k=B[w_l,w_k;t]\in C([0,T]),$ 我们有如下定理:
定理 1. $u_m\in H^2(0,T;H_0^1(\Omega))$存在唯一解, $d_m^k\in H^2(0,T),$ $k=1,\cdots,m.$
接下来做能量估计. 首先考虑$u,u_m'.$ 代入$v=u_m',$ 有
$$
(u_m'',u_m')+B[u_m,u_m';t]=(f,u_m')\le \frac{1}{2}\left(\Vert f\Vert_{L^2(\Omega)}^2+\Vert u_m'\Vert_{L^2}^2\right)
$$
$$
B[u_m,u_m';t]:=A(t)+B(t),\quad |B(t)|\le C(L,n)\left(\Vert u_m\Vert^2_{H^1(\Omega)}+\Vert u_m'\Vert_{L^2(\Omega)}\right)
$$
$$
A(t)=\frac{1}{2}\frac{d {} }{d {}t}\int_\Omega a^{ij}{u_m}_i{u_m}_j-\frac{1}{2}\int_\Omega \frac{d {}a^{ij} }{d {}t}{u_m}_i{u_m}_j:=\frac{1}{2}\frac{d {} }{d {}t}A[u_m;t]-\frac{1}{2}A_1[u_m;t],
$$
$$
|A_1|\le C_1(L,n)\Vert u_m\Vert_{H_0^1(\Omega)}^2,\quad \Vert u_m\Vert_{H_0^1(\Omega)}^2\le C(n)\Vert Du_m\Vert_{L^2(\Omega)}^2\le C(\lambda,n)A[u_m;t].
$$
因此, 整理有
$$
\frac{d {} }{d {}t}\left(\Vert u_m'\Vert_{L^2(\Omega)}^2+A[u_m;t]\right)\le C(L,n)\left(\Vert f(t)\Vert_{L^2(\Omega)}^2+A[u_m;t]+\Vert u_m'\Vert_{L^2(\Omega)}^2\right).
$$
记$\eta(t)=\Vert u_m'(t)\Vert_{L^2(\Omega)}^2+A[u_m,t],$
由Gronwall不等式, 我们有
$$
\eta(t)\le C(T,L,n)(\eta(0)+\Vert f\Vert_{L^2(\Omega_T)}^2),\quad a.e. t\in [0,T].
$$
$$
\eta(t)\ge \Vert u_m'\Vert_{L^2(\Omega)}^2+C(\lambda,n)\Vert u_m\Vert_{H_0^1(\Omega)}^2,\quad \eta(0)\le \Vert h\Vert_{L^2(\Omega)}^2+C(L)\Vert g\Vert_{H_0^1(\Omega)}^2
$$
这样我们就给出了
$$
\sup_{0\le t\le T}\left(\Vert u_m(t)\Vert_{H_0^1(\Omega)}^2+\Vert u_m'(t)\Vert_{L^2(\Omega)}^2\right)\le C(L,n,T)\left(\Vert f\Vert_{L^2(\Omega_T)}^2+\Vert g\Vert_{H_0^1(\Omega)}^2+\Vert h\Vert_{L^2(\Omega)}^2\right)
$$
再估计$u_m''.$ 类似地, 记$v_m=\sum_{k=1}^m(v,w_k)w_k,$ 那么
$$
\left<{}u_m'',v\right>=(u_m'',v)=(u_m'',v_m)=(f,v_m)-B[u_m,v_m;t],
$$
$$
|\left<{}u_m'',v\right>|\le \Vert f\Vert_{L^2}\Vert v_m\Vert_{L^2}+C(n,L)\Vert u_m\Vert_{H_0^1}\Vert v_m\Vert_{H_0^1}\le C(n,L)\left(\Vert f\Vert_{L^2}+\Vert u_m\Vert_{H_0^1}\right)\Vert v\Vert_{H_0^1}.
$$
这就给出了$\Vert u_m''(t)\Vert_{H^{-1}(\Omega)}$的估计.
定理 2. 在系数条件满足时, 逼近解$u_m$满足
$$
\sup_{0\le t\le T}\left(\Vert u_m\Vert_{H_0^1}+\Vert u_m'\Vert_{L^2}\right)+\Vert u_m''\Vert_{L^2(0,T;H^{-1}(\Omega))}\le C(\Omega,L,T)\left(\Vert f\Vert_{L^2(\Omega_T)}+\Vert g\Vert_{H_0^1}+\Vert h\Vert_{L^2}\right)
$$
这样之后就有弱收敛子列$\{u_m\},$ 且时间导数也收敛, 给出$u,$
且满足相同的能量估计. (操作上, 为了得到$L^\infty$控制,
需要给出$L^p$控制, $p\rightarrow\infty.$)
下面验证$u$是弱解. 方法同理.
考虑$\eta(t)v_N\xrightarrow{L^2(0,T;H_0^1(\Omega))} \eta(t)v,$
代入到逼近解$u_m$对应的方程中. 先令$m\rightarrow\infty,$
再令$N\rightarrow\infty$即可. 初值条件亦然, 取合适的$\eta$即可.
定理 3. 在系数条件下, 双曲方程存在弱解$u,$ 为Galerkin逼近解的极限, 且满足能量估计.
弱解唯一性
定理 4. 在前面的系数条件基础上, 若$b^i\in C^1([0,T],L^\infty(\Omega)),$ 则双曲方程解唯一.
只需考虑齐次问题. 取弱解$u\in L^2(0,T;H_0^1(\Omega)),$
$u_t\in L^2(\Omega_T),$ $u_{tt}\in L^2(0,T;H^{-1}(\Omega)).$
若$u_t\in L^2(0,T;H_0^1(\Omega)),$ 则取$v=u_t$代入方程即可.
但此时并没有该条件. 注意到,
$$
(u'',w)=\frac{d {} }{d {}t}\left<{}u',w\right>-\left<{}u',w'\right>,
$$
取$w$使得$w'=\pm u,$ $w(T)=0$即可进一步推导. 为此,
$\,\forall\,s\in [0,T],$ 令
$$
w(t)=\begin{cases}
0&s\le t\le T\\
\int_t^s u(\tau)d\tau & 0\le t<s
\end{cases},
$$
则$w$满足条件. 代入, 我们有:
$$
\int_0^s \left<{}u'',w\right>+B[u,w;t]dt=0,\quad \int_0^s \left<{}u'',w\right>ds=\int_0^s \left<{}u',u\right>dt=\frac{1}{2}\Vert u(s)\Vert_{L^2(\Omega)}^2,
$$
$$
\begin{aligned}
\int_0^s B[u,w;t]dt=&-\int_0^sB[w',w;t]dt\\
=&-\int_0^s \frac{d {} }{d {}t}\int_\Omega \frac{1}{2} a^{ij}w_iw_j+b^iw_iw dxdt\\
&+\int_0^s\int_\Omega \frac{1}{2}{a^{ij} }'w_iw_j+{b^i}'w_iw+b^iw_iw'-cw'wdxdt\\
=&\int_\Omega \frac{1}{2}a^{ij}w_iw_j+b^iw_iwdx|_{t=0}+I(w)\\
\ge &\frac{\lambda}{4}\Vert Dw(0)\Vert^2-C(n,L)\left(\Vert w(0)\Vert_{L^2(\Omega)}+\int_0^s \Vert w(t)\Vert_{H^1(\Omega)}^2+\Vert u(t)\Vert_{L^2(\Omega)}^2dt\right)
\end{aligned}
$$
整理得到
$$
\Vert u(s)\Vert^2_{L^2(\Omega)}+\Vert w(0)\Vert_{H_0^1(\Omega)}^2\le C(n,\Omega,L,T)\left(\Vert w(0)\Vert_{L^2(\Omega)}^2+\int_0^s \Vert w(t)\Vert_{H^1(\Omega)}^2+\Vert u(t)\Vert^2_{L^2(\Omega)}dt\right)
$$
接下来希望将其转化为常见的Gronwall不等式.
令$v(t)=\int_0^t u(\tau)d\tau,$ 那么$w(0)=v(s),$ $w(t)=v(s)-v(t),$
$\,\forall\,t\le s.$
$$
\Vert v(s)\Vert_{L^2(\omega)}^2=\int_\Omega \left(\int_0^s u(\tau)d\tau\right)^2dx\le T\int_0^s \Vert u(t)\Vert^2_{L^2(\Omega)}dt,
$$
$$
\int_0^s \Vert w(t)\Vert^2_{H^1}=\int_0^s\Vert v(s)-v(t)\Vert^2_{H^1}\le 2\int_0^s \Vert v(s)\Vert_{H^1}^2+\Vert v(t)\Vert_{H^1}^2\le 2\int_0^s\Vert v(t)\Vert^2_{H^1}dt+2s\Vert v(s)\Vert^2_{H^1}
$$
代入不等式, 我们有$\,\forall\,s\in [0,T],$
$$
\Vert u(s)\Vert^2_{L^2(\Omega)}+(1-2sc)\Vert v(s)\Vert_{H_0^1(\Omega)}^2\le C(n,\Omega,L,T)\left(\int_0^s \Vert u(t)\Vert_{L^2(\Omega)}^2+\Vert v(t)\Vert^2_{H^1(\Omega)}dt\right).
$$
令$T_0=\min\{\frac{1}{4C},T\},$ 那么当$s\in [0,T_0]$时,
记$F(t)=\int_0^t\Vert u(\tau)\Vert^2_{L^2(\Omega)}+\frac{1}{4}\Vert v(\tau)\Vert^2_{H^1(\Omega)}d\tau,$
则
$$
\frac{d {}F(s)}{d {}ds}\le C(\Omega,L,n,T)F(s),\quad \,\forall\,s\in [0,T_0].
$$
由于$F(0)=0,$ 由Gronwanll不等式, $F(s)=0,$ $\,\forall\,s\in [0,T_0],$
因此$u\equiv 0.$ 由分层法我们就知道在$\Omega_T$上均成立.
对于线性方程,
唯一性和解的能量/最大模/Schauder估计就给出了解关于已知数据的稳定性(但关于方程的系数要进一步判断).
解的正则性
简单起见, 设方程系数与时间无关.
二阶正则性
若$f\in H^1(0,T;L^2(\Omega)),$
则$f_k(t)=\left<{}f,w_k\right>\in H^1(0,T).$
此时$e_l^k=B[w_l,w_k;t]$为常数, $u_m=d_m^kw_k,$ ${d_m^k}''\in H^1(0,T),$
故$u_m'''\in L^2(0,T;H_0^1(\Omega)).$ 记$v_m=u_m',$ 对方程两边求导, 有
$$
\left<{}v_m'',v_m'\right>+B[v_m,v_m';t]=(f',v_m')\le \frac{1}{2}\left(\Vert f'\Vert_{L^2(\Omega)}^2+\Vert v_m'\Vert_{L^2(\Omega)}^2\right)
$$
$$
\begin{aligned}
B[v_m,v_m';t]&\ge\frac{d {} }{d {}t}A[v_m,v_m;t]-C(L)\left(\Vert v_m\Vert^2_{H_0^1}+\Vert v_m'\Vert_{L^2(\Omega)}\right)\\
&\ge\frac{d {} }{d {}t}A[v_m,v_m;t]-C(L,n)\left(A(v_m,v_m;t)+\Vert v_m'\Vert_{L^2(\Omega)}\right)
\end{aligned}
$$
记$F(t)=\frac{1}{2}\Vert v_m'\Vert_{L^2(\Omega)}^2+A(v_m,v_m;t),$
我们就有
$$
F'(t)\le C \left(F(t)+\Vert f'\Vert_{L^2(\Omega)}^2\right),\quad F(t)\le C(L,T,n)\left(\Vert f\Vert^2_{H^1(0,T;L^2(\Omega))}+F(0)\right),\quad a.e.\:t\in [0,T]
$$
接下来计算$F(0).$
$$
\begin{aligned}
\Vert v_m'(0)\Vert^2_{L^2(\Omega)}&=(f(0),u_m''(0))-B[u_m(0),u_m''(0);0]\\
&\le \varepsilon\Vert u_m''(0)\Vert_{L^2}^2+\frac{1}{4\varepsilon}\Vert f(0)\Vert^2_{L^2(\Omega)}-2A(u_m(0),u_m''(0);0)-\cdots\\
&\le 3\varepsilon\Vert u_m''(0)\Vert_{L^2(\Omega)}^2-2A(u_m(0),u_m''(0);0)+C(L,\varepsilon)\left(\Vert u_m(0)\Vert_{H_0^1(\Omega)}^2+\Vert f\Vert_{H^1(0,T;L^2(\Omega))}\right)
\end{aligned}
$$
$$
|A(u_m(0),u_m''(0);0)|\le \varepsilon\Vert u_m''(0)\Vert^2_{L^2(\Omega)}+C(L,\varepsilon)\Vert u_m(0)\Vert_{H^2},\quad \Vert u_m(0)\Vert^2_{H^2(\Omega)}\le C\Vert g\Vert^2_{H^2(\Omega)}
$$
令$\varepsilon\le \frac{1}{16},$ 则有
$$
\Vert v_m'(0)\Vert^2_{L^2(\Omega)}\le C(L,n)\left(\Vert f\Vert^2_{H^1(0,T;L^2(\Omega))}+\Vert g\Vert^2_{H^2(\Omega)}\right)
$$
$F(0)$中还有另一项需要处理:
$$
A(v_m(0),v_m(0);0)=A(h_m,h_m;0)\le C(L)\Vert h\Vert^2_{H_0^1(\Omega)}
$$
$$
F(0)\le C(L,n)\left(\Vert f\Vert^2_{H^1(0,T;L^2(\Omega))}+\Vert g\Vert^2_{H^2(\Omega)}+\Vert h\Vert^2_{H_0^1(\Omega)}\right)
$$
令$m\rightarrow\infty,$ 就给出了
$$
\sup_{0\le t\le T}\left(\Vert u''(t)\Vert_{L^2(\Omega)}+\Vert u'(t)\Vert_{H_0^1(\Omega)}\right)\le C(L,n,T)\left(\Vert f\Vert_{H^1(0,T;L^2(\Omega))}+\Vert g\Vert_{H^2(\Omega)}+\Vert h\Vert_{H_0^1(\Omega)}\right)
$$
由$Lu=f-u_{tt},$ 由椭圆方程正则性, 若$\partial\Omega\in C^2,$
$a^{ij}\in C^1(\overline\Omega),$ 有
$$
\Vert u(t)\Vert_{H^2(\Omega)}\le C(\Omega,L,n)\left(\Vert u(t)\Vert_{L^2(\Omega)}+\Vert f(t)\Vert_{L^2(\Omega)}+\Vert u''(t)\Vert_{L^2(\Omega)}\right).
$$
$u'''$的估计和抛物型方程一样. 综上, 我们有如下定理:
定理 5. 设$\Omega\subset \mathbb{R}^n$为有界开集, $L$系数适当, $a^{ij}\in C^1(\overline\Omega),$ $b^i,c\in L^\infty(\Omega),$ $\partial \Omega\in C^2.$ 若$f\in H^1(0,T;L^2(\Omega)),$ $g\in H_0^2(\Omega),$ $h\in H_0^1(\Omega),$ 则$u$满足如下正则性估计:
$$
\begin{aligned}
\sup_{0\le t\le T}\left(\Vert u''(t)\Vert_{L^2(\Omega)}+\Vert u'(t)\Vert_{H_0^1(\Omega)}+\Vert u(t)\Vert_{H^2(\Omega)}\right)&+\Vert u'''\Vert_{L^2(0,T;H^{-1}(\Omega))}\le\\
C(L,n,T,\Omega)&\left(\Vert f\Vert_{H^1(0,T;L^2(\Omega))}+\Vert g\Vert_{H^2(\Omega)}+\Vert h\Vert_{H_0^1(\Omega)}\right)
\end{aligned}
$$
高阶正则性
同理用归纳法. 对于$v=u_t,$ $v|_{t=0}=h,$
$v_t|_{t=0}=u_{tt}|_{t=0}=f(x,0)-Lg.$
此时结合椭圆方程正则性能给出低阶的正则性估计. 为了进一步往下走, 需要
$$
f(0)-Lg\in H_0^1(\Omega),\quad \frac{d {}f}{d {}t}|_{t=0}-Lh\in L^2(\Omega),
$$
由此需引入相容性条件. 记$g_0=g,g_1=h,$
$$
g_{2k}=\frac{d {}^{2k-2}f}{d {}t^{2k-2} }f|_{t=0}-Lg_{2k-2},\quad g_{2k+1}=\frac{d {}^{2k-1}f}{d {}t^{2k-1} }f|_{t=0}-Lg_{2k-1}
$$
若$g_i\in H_0^1(\Omega),$ $0\le i\le m-1,$ $g_m\in L^2(\Omega),$
则称相容性条件($A_m$)满足. 定理类似, 在此不作陈述.
类似地可以推出条件全部光滑时, 弱解也是光滑解.
文章最后更新于 2022-12-24 09:06:53
definition
$a^{ij},b^i,c$ is the function on $\Omega_T$, denoted
$$
Lu:=-(a^{ij}u_j)_i+b^iu_i+cu.
$$
If $a^{ij}$ is symmetrical and satisfies the consistent condition,
Then $\partial_t^2+L$ is called a second-order linear divergence consistent hyperbolic operator on $\Omega_T$.
Consider the initial boundary value problem
$$
\begin{cases}
u_{tt}+Lu=f&in\:\Omega_T\\
u=0&on\:\partial\Omega\times (0,T]\\
u|_{t=0}=g,u_t|_{t=0}=h&in\:\Omega
\end{cases}
$$
On a consistent basis, consider conditions
$$
a^{ij}\in C^1(\overline\Omega_T),\quad b^i,c\in C(\overline\Omega_T).
$$
$$
f\in L^2(\Omega_T),\quad g\in H_0^1(\Omega),\quad h\in L^2(\Omega).
$$
The weak solution of the hyperbolic equation needs to satisfy $u\in L^2(0,T;H_0^1(\Omega)),$
$u_{tt}\in L^2(0,T;H^{-1}(\Omega)),$ $\,\forall\,v\in H_0^1(\Omega),$
a.e. $t\in (0,T],$
$$
\left<{}u_{tt},v\right>+B[u,v;t]=\left<{}f(\cdot,t),v\right>
$$
In order to satisfy the initial conditions, $u,u_t\in C([0,T];\ast ),$ is needed
Therefore, $u_t\in L^2(0,T;L^2(\Omega))=L^2(\Omega_T),$ needs to be like this
$$
u\in W^{1,2}(0,T;L^2(\Omega))\hookrightarrow C([0,T];L^2(\Omega)),
$$
$$
u_t\in W^{1,2}(0,T;H^{-1}(\Omega))\hookrightarrow C([0,T];H^{-1}(\Omega)).
$$
The initial value condition refers to the approximation according to the corresponding module, which is not necessarily the space where $g,h$ originally resides.
Galerkin approaches
Similarly, consider $u_m=\sum_{k=1}^md_m^kw_k,$ and let $v=w_k,$
by$f_k=\left<{}w_k,f\right>\in L^2(0,T),$
$e_l^k=B[w_l,w_k;t]\in C([0,T]),$ We have the following theorem:
Theorem 1. $u_m\in H^2(0,T;H_0^1(\Omega))$ There is a unique solution, $d_m^k\in H^2(0,T),$ $k=1,\cdots,m.$
Next, do the energy estimation. First consider $u,u_m'.$ and substitute $v=u_m',$ to have
$$
(u_m'',u_m')+B[u_m,u_m';t]=(f,u_m')\le \frac{1}{2}\left(\Vert f\Vert_{L^2(\Omega)}^2+\Vert u_m'\Vert_{L^2}^2\right)
$$
$$
B[u_m,u_m';t]:=A(t)+B(t),\quad |B(t)|\le C(L,n)\left(\Vert u_m\Vert^2_{H^1(\Omega)}+\Vert u_m'\Vert_{L^2(\Omega)}\right)
$$
$$
A(t)=\frac{1}{2}\frac{d {} }{d {}t}\int_\Omega a^{ij}{u_m}_i{u_m}_j-\frac{1}{2}\int_\Omega \frac{d {}a^{ij} }{d {}t}{u_m}_i{u_m}_j:=\frac{1}{2}\frac{d {} }{d {}t}A[u_m;t]-\frac{1}{2}A_1[u_m;t],
$$
$$
|A_1|\le C_1(L,n)\Vert u_m\Vert_{H_0^1(\Omega)}^2,\quad \Vert u_m\Vert_{H_0^1(\Omega)}^2\le C(n)\Vert Du_m\Vert_{L^2(\Omega)}^2\le C(\lambda,n)A[u_m;t].
$$
Therefore, sorting out
$$
\frac{d {} }{d {}t}\left(\Vert u_m'\Vert_{L^2(\Omega)}^2+A[u_m;t]\right)\le C(L,n)\left(\Vert f(t)\Vert_{L^2(\Omega)}^2+A[u_m;t]+\Vert u_m'\Vert_{L^2(\Omega)}^2\right).
$$
Remember$\eta(t)=\Vert u_m'(t)\Vert_{L^2(\Omega)}^2+A[u_m,t],$
By Gronwall's inequality, we have
$$
\eta(t)\le C(T,L,n)(\eta(0)+\Vert f\Vert_{L^2(\Omega_T)}^2),\quad a.e. t\in [0,T].
$$
$$
\eta(t)\ge \Vert u_m'\Vert_{L^2(\Omega)}^2+C(\lambda,n)\Vert u_m\Vert_{H_0^1(\Omega)}^2,\quad \eta(0)\le \Vert h\Vert_{L^2(\Omega)}^2+C(L)\Vert g\Vert_{H_0^1(\Omega)}^2
$$
Thus we are given
$$
\sup_{0\le t\le T}\left(\Vert u_m(t)\Vert_{H_0^1(\Omega)}^2+\Vert u_m'(t)\Vert_{L^2(\Omega)}^2\right)\le C(L,n,T)\left(\Vert f\Vert_{L^2(\Omega_T)}^2+\Vert g\Vert_{H_0^1(\Omega)}^2+\Vert h\Vert_{L^2(\Omega)}^2\right)
$$
Then estimate $u_m''.$ similarly, write $v_m=\sum_{k=1}^m(v,w_k)w_k,$ then
$$
\left<{}u_m'',v\right>=(u_m'',v)=(u_m'',v_m)=(f,v_m)-B[u_m,v_m;t],
$$
$$
|\left<{}u_m'',v\right>|\le \Vert f\Vert_{L^2}\Vert v_m\Vert_{L^2}+C(n,L)\Vert u_m\Vert_{H_0^1}\Vert v_m\Vert_{H_0^1}\le C(n,L)\left(\Vert f\Vert_{L^2}+\Vert u_m\Vert_{H_0^1}\right)\Vert v\Vert_{H_0^1}.
$$
This gives an estimate of $\Vert u_m''(t)\Vert_{H^{-1}(\Omega)}$.
Theorem 2. When the coefficient conditions are met, the approximate solution $u_m$ satisfies
$$
\sup_{0\le t\le T}\left(\Vert u_m\Vert_{H_0^1}+\Vert u_m'\Vert_{L^2}\right)+\Vert u_m''\Vert_{L^2(0,T;H^{-1}(\Omega))}\le C(\Omega,L,T)\left(\Vert f\Vert_{L^2(\Omega_T)}+\Vert g\Vert_{H_0^1}+\Vert h\Vert_{L^2}\right)
$$
After this, there is a weakly convergent subsequence $\{u_m\},$ and the time derivative also converges, giving $u,$
and satisfy the same energy estimate. (Operationally, in order to obtain $L^\infty$ control,
Need to give $L^p$ control, $p\rightarrow\infty.$)
Next, verify that $u$ is a weak solution. The method is the same.
Consider$\eta(t)v_N\xrightarrow{L^2(0,T;H_0^1(\Omega))} \eta(t)v,$
Substitute into the equation corresponding to the approximate solution $u_m$. Shilling $m\rightarrow\infty,$
Then set $N\rightarrow\infty$. The same is true for the initial value condition, just pick the appropriate $\eta$.
Theorem 3. Under the coefficient conditions, the hyperbolic equation has a weak solution $u,$ which is the limit of the Galerkin approximation solution and satisfies the energy estimation.
Weak solution uniqueness
Theorem 4. Based on the previous coefficient conditions, if $b^i\in C^1([0,T],L^\infty(\Omega)),$, the solution to the hyperbolic equation is unique.
Just consider the homogeneous problem. Take the weak solution $u\in L^2(0,T;H_0^1(\Omega)),$
$u_t\in L^2(\Omega_T),$ $u_{tt}\in L^2(0,T;H^{-1}(\Omega)).$
If $u_t\in L^2(0,T;H_0^1(\Omega)),$, then take $v=u_t$ and substitute it into the equation.
But this condition does not exist at this time. Note that,
$$
(u'',w)=\frac{d {} }{d {}t}\left<{}u',w\right>-\left<{}u',w'\right>,
$$
Taking $w$ so that $w'=\pm u,$ $w(T)=0$ can be further deduced. For this reason,
$\,\forall\,s\in [0,T],$ Order
$$
w(t)=\begin{cases}
0&s\le t\le T\\
\int_t^s u(\tau)d\tau & 0\le t<s
\end{cases},
$$
Then $w$ satisfies the condition. Substituting, we have:
$$
\int_0^s \left<{}u'',w\right>+B[u,w;t]dt=0,\quad \int_0^s \left<{}u'',w\right>ds=\int_0^s \left<{}u',u\right>dt=\frac{1}{2}\Vert u(s)\Vert_{L^2(\Omega)}^2,
$$
$$
\begin{aligned}
\int_0^s B[u,w;t]dt=&-\int_0^sB[w',w;t]dt\\
=&-\int_0^s \frac{d {} }{d {}t}\int_\Omega \frac{1}{2} a^{ij}w_iw_j+b^iw_iw dxdt\\
&+\int_0^s\int_\Omega \frac{1}{2}{a^{ij} }'w_iw_j+{b^i}'w_iw+b^iw_iw'-cw'wdxdt\\
=&\int_\Omega \frac{1}{2}a^{ij}w_iw_j+b^iw_iwdx|_{t=0}+I(w)\\
\ge &\frac{\lambda}{4}\Vert Dw(0)\Vert^2-C(n,L)\left(\Vert w(0)\Vert_{L^2(\Omega)}+\int_0^s \Vert w(t)\Vert_{H^1(\Omega)}^2+\Vert u(t)\Vert_{L^2(\Omega)}^2dt\right)
\end{aligned}
$$
sorted out
$$
\Vert u(s)\Vert^2_{L^2(\Omega)}+\Vert w(0)\Vert_{H_0^1(\Omega)}^2\le C(n,\Omega,L,T)\left(\Vert w(0)\Vert_{L^2(\Omega)}^2+\int_0^s \Vert w(t)\Vert_{H^1(\Omega)}^2+\Vert u(t)\Vert^2_{L^2(\Omega)}dt\right)
$$
Next we want to convert this into the common Gronwall inequality.
Let $v(t)=\int_0^t u(\tau)d\tau,$ then $w(0)=v(s),$ $w(t)=v(s)-v(t),$
$\,\forall\,t\le s.$
$$
\Vert v(s)\Vert_{L^2(\omega)}^2=\int_\Omega \left(\int_0^s u(\tau)d\tau\right)^2dx\le T\int_0^s \Vert u(t)\Vert^2_{L^2(\Omega)}dt,
$$
$$
\int_0^s \Vert w(t)\Vert^2_{H^1}=\int_0^s\Vert v(s)-v(t)\Vert^2_{H^1}\le 2\int_0^s \Vert v(s)\Vert_{H^1}^2+\Vert v(t)\Vert_{H^1}^2\le 2\int_0^s\Vert v(t)\Vert^2_{H^1}dt+2s\Vert v(s)\Vert^2_{H^1}
$$
Substituting the inequality, we have $\,\forall\,s\in [0,T],$
$$
\Vert u(s)\Vert^2_{L^2(\Omega)}+(1-2sc)\Vert v(s)\Vert_{H_0^1(\Omega)}^2\le C(n,\Omega,L,T)\left(\int_0^s \Vert u(t)\Vert_{L^2(\Omega)}^2+\Vert v(t)\Vert^2_{H^1(\Omega)}dt\right).
$$
Let $T_0=\min\{\frac{1}{4C},T\},$ then when $s\in [0,T_0]$,
Note$F(t)=\int_0^t\Vert u(\tau)\Vert^2_{L^2(\Omega)}+\frac{1}{4}\Vert v(\tau)\Vert^2_{H^1(\Omega)}d\tau,$
rule
$$
\frac{d {}F(s)}{d {}ds}\le C(\Omega,L,n,T)F(s),\quad \,\forall\,s\in [0,T_0].
$$
Since $F(0)=0,$ is determined by Gronwanll's inequality, $F(s)=0,$ $\,\forall\,s\in [0,T_0],$
Therefore, $u\equiv 0.$, we know from the layered method that it is also true on $\Omega_T$.
For linear equations,
The energy/maximum mode/Schauder estimate of the uniqueness solution gives an understanding of the stability of the known data (but further judgment is required regarding the coefficients of the equation).
regularity of solution
For simplicity, let the coefficients of the equation be independent of time.
second-order regularity
If $f\in H^1(0,T;L^2(\Omega)),$
Then $f_k(t)=\left<{}f,w_k\right>\in H^1(0,T).$
At this time $e_l^k=B[w_l,w_k;t]$ is a constant, $u_m=d_m^kw_k,$ ${d_m^k}''\in H^1(0,T),$
Therefore $u_m'''\in L^2(0,T;H_0^1(\Omega)).$ is written as $v_m=u_m',$. Derivative of both sides of the equation, we have
$$
\left<{}v_m'',v_m'\right>+B[v_m,v_m';t]=(f',v_m')\le \frac{1}{2}\left(\Vert f'\Vert_{L^2(\Omega)}^2+\Vert v_m'\Vert_{L^2(\Omega)}^2\right)
$$
$$
\begin{aligned}
B[v_m,v_m';t]&\ge\frac{d {} }{d {}t}A[v_m,v_m;t]-C(L)\left(\Vert v_m\Vert^2_{H_0^1}+\Vert v_m'\Vert_{L^2(\Omega)}\right)\\
&\ge\frac{d {} }{d {}t}A[v_m,v_m;t]-C(L,n)\left(A(v_m,v_m;t)+\Vert v_m'\Vert_{L^2(\Omega)}\right)
\end{aligned}
$$
Note$F(t)=\frac{1}{2}\Vert v_m'\Vert_{L^2(\Omega)}^2+A(v_m,v_m;t),$
we have
$$
F'(t)\le C \left(F(t)+\Vert f'\Vert_{L^2(\Omega)}^2\right),\quad F(t)\le C(L,T,n)\left(\Vert f\Vert^2_{H^1(0,T;L^2(\Omega))}+F(0)\right),\quad a.e.\:t\in [0,T]
$$
Next calculate $F(0).$
$$
\begin{aligned}
\Vert v_m'(0)\Vert^2_{L^2(\Omega)}&=(f(0),u_m''(0))-B[u_m(0),u_m''(0);0]\\
&\le \varepsilon\Vert u_m''(0)\Vert_{L^2}^2+\frac{1}{4\varepsilon}\Vert f(0)\Vert^2_{L^2(\Omega)}-2A(u_m(0),u_m''(0);0)-\cdots\\
&\le 3\varepsilon\Vert u_m''(0)\Vert_{L^2(\Omega)}^2-2A(u_m(0),u_m''(0);0)+C(L,\varepsilon)\left(\Vert u_m(0)\Vert_{H_0^1(\Omega)}^2+\Vert f\Vert_{H^1(0,T;L^2(\Omega))}\right)
\end{aligned}
$$
$$
|A(u_m(0),u_m''(0);0)|\le \varepsilon\Vert u_m''(0)\Vert^2_{L^2(\Omega)}+C(L,\varepsilon)\Vert u_m(0)\Vert_{H^2},\quad \Vert u_m(0)\Vert^2_{H^2(\Omega)}\le C\Vert g\Vert^2_{H^2(\Omega)}
$$
Let $\varepsilon\le \frac{1}{16},$ then have
$$
\Vert v_m'(0)\Vert^2_{L^2(\Omega)}\le C(L,n)\left(\Vert f\Vert^2_{H^1(0,T;L^2(\Omega))}+\Vert g\Vert^2_{H^2(\Omega)}\right)
$$
There is one more item in $F(0)$ that needs to be dealt with:
$$
A(v_m(0),v_m(0);0)=A(h_m,h_m;0)\le C(L)\Vert h\Vert^2_{H_0^1(\Omega)}
$$
$$
F(0)\le C(L,n)\left(\Vert f\Vert^2_{H^1(0,T;L^2(\Omega))}+\Vert g\Vert^2_{H^2(\Omega)}+\Vert h\Vert^2_{H_0^1(\Omega)}\right)
$$
Let $m\rightarrow\infty,$ give
$$
\sup_{0\le t\le T}\left(\Vert u''(t)\Vert_{L^2(\Omega)}+\Vert u'(t)\Vert_{H_0^1(\Omega)}\right)\le C(L,n,T)\left(\Vert f\Vert_{H^1(0,T;L^2(\Omega))}+\Vert g\Vert_{H^2(\Omega)}+\Vert h\Vert_{H_0^1(\Omega)}\right)
$$
From $Lu=f-u_{tt},$ to the regularity of elliptic equations, if $\partial\Omega\in C^2,$
$a^{ij}\in C^1(\overline\Omega),$ Yes
$$
\Vert u(t)\Vert_{H^2(\Omega)}\le C(\Omega,L,n)\left(\Vert u(t)\Vert_{L^2(\Omega)}+\Vert f(t)\Vert_{L^2(\Omega)}+\Vert u''(t)\Vert_{L^2(\Omega)}\right).
$$
The estimation of $u'''$ is the same as the parabolic equation. To sum up, we have the following theorem:
Theorem 5. Assume $\Omega\subset \mathbb{R}^n$ is a bounded open set, $L$ has appropriate coefficients, $a^{ij}\in C^1(\overline\Omega),$ $b^i,c\in L^\infty(\Omega),$ $\partial \Omega\in C^2.$ if $f\in H^1(0,T;L^2(\Omega)),$ $g\in H_0^2(\Omega),$ $h\in H_0^1(\Omega),$ then $u$ satisfies the following regularity estimate:
$$
\begin{aligned}
\sup_{0\le t\le T}\left(\Vert u''(t)\Vert_{L^2(\Omega)}+\Vert u'(t)\Vert_{H_0^1(\Omega)}+\Vert u(t)\Vert_{H^2(\Omega)}\right)&+\Vert u'''\Vert_{L^2(0,T;H^{-1}(\Omega))}\le\\
C(L,n,T,\Omega)&\left(\Vert f\Vert_{H^1(0,T;L^2(\Omega))}+\Vert g\Vert_{H^2(\Omega)}+\Vert h\Vert_{H_0^1(\Omega)}\right)
\end{aligned}
$$
higher order regularity
Use induction in the same way. For $v=u_t,$ $v|_{t=0}=h,$
$v_t|_{t=0}=u_{tt}|_{t=0}=f(x,0)-Lg.$
At this time, a low-order regularity estimate is given based on the regularity performance of elliptic equations. In order to go further, it is necessary to
$$
f(0)-Lg\in H_0^1(\Omega),\quad \frac{d {}f}{d {}t}|_{t=0}-Lh\in L^2(\Omega),
$$
Therefore, compatibility conditions need to be introduced. Note $g_0=g,g_1=h,$
$$
g_{2k}=\frac{d {}^{2k-2}f}{d {}t^{2k-2} }f|_{t=0}-Lg_{2k-2},\quad g_{2k+1}=\frac{d {}^{2k-1}f}{d {}t^{2k-1} }f|_{t=0}-Lg_{2k-1}
$$
If $g_i\in H_0^1(\Omega),$ $0\le i\le m-1,$ $g_m\in L^2(\Omega),$
Then the compatibility condition ($A_m$) is said to be satisfied. The theorem is similar and will not be stated here.
Similarly, it can be deduced that when all conditions are smooth, the weak solution is also a smooth solution.
The article was last updated on 2022-12-24 09:06:53
