Machine-translated from Chinese.
椭圆方程
回忆上节说明了$Ax=b$的存在性等价于$A^\ast y=0$的唯一性.
若$A$是$n\times n$矩阵, 那么$A^\ast y=0$的唯一性等价于$Ay=0$的唯一性.
因此$Ax=b$的存在性等价于唯一性.
对于椭圆型方程, 考虑齐次Dirichlet边界问题$Lu=f\in H^{-1}(\Omega).$
我们有Garding不等式:
$$
\left<{}-Lu,u\right>_{L^2}\ge C_1\Vert u\Vert_1^2-C_2\Vert u\Vert^2.
$$
当$C_2\le 0$时, 我们可以说明问题的唯一性. 至于存在性,
按原来的方法需要考虑对偶算子$L^\ast $对应的Garding不等式中是否有$C_2^\ast \le 0,$
从而通过对偶问题的唯一性说明原问题存在性.
但事实上通过$C_2\le 0$直接可以说明原问题解的存在性.
考虑伽辽金(Galerkin)方法.
对偶问题唯一性导出存在性的方法只说明了理论上的存在性,
无法具体给出解的构造. 而该方法能够给出解的构造方法,
是微分方程数值解中有限元方法的理论基础.
回忆$H^{m,p}$为可分空间, 因为它能等距嵌入到$(L^p)^N$中,
而后者是可分空间. 因此, $H_0^1(\Omega)$为可分Hilbert空间,
有一组完备正交基$\{\phi_i\}.$ 理论上,
可以取特征函数系$-\Delta \phi_j=\lambda_j\phi_j,$
$\phi_j|_{\partial\Omega}=0.$ 算子自共轭时特征函数系当然正交.
由于$C_c^\infty(\Omega)$在$H_0^1(\Omega)$中稠密,
我们当然可以取$\{\phi_i\}\subset C_c^\infty(\Omega).$
考虑$u_m=\sum_{i=1}^ma_i\phi_i,$ $f=\sum_{i=1}^\infty f_i\phi_i,$
$f_m=\sum_{i=1}^m f_i\phi_i\rightharpoonup f.$ 为了解$Lu=f,$
一种期待是通过解$Lu_m=f_m$得到$u_m$给出$u.$ 但这种期望往往是落空的,
因为对于方程
$$
Lu_m=\sum_{i=1}^m a_iL\phi_i=\sum_{i=1}^mf_i\phi_i=f_m,
$$
当$\{\phi_i\}$不取特征函数系时, 这将是一个无穷维方程,
无法针对有限个未知量求解.
Galerkin的方法是, 我们考虑如下问题:
$$
\left<{}Lu_m,\phi_j\right>=\left<{}f_m,\phi_j\right>,\quad j=1,\cdots,m.
$$
也就是$\sum_{i=1}^m a_i\left<{}L\phi_i,\phi_j\right>=\sum_{i=1}^mf_i\left<{}\phi_i,\phi_j\right>,$
$j=1,\cdots,m.$ 这是矩阵方程$Ax=b$的形式. 由一开始所述,
为了证明解的存在性, 只需说明解的唯一性. 为此采用椭圆估计即可.
考虑$\left<{}Lu_m-f_m,u_m\right>=0,$ 由Garding不等式, 当$C_2\le 0$时,
$$
\begin{aligned}
C_1\Vert u_m\Vert^2_1&\le \left<{}-f_m,u_m\right>\le \Vert f_m\Vert_{-1}\Vert u_m\Vert_1\le \frac{1}{2C_1}\Vert f_m\Vert_{-1}^2+\frac{C_1}{2}\Vert u_m\Vert_1^2\\
\frac{C_1}{2}\Vert u_m\Vert^2_1&\le \frac{1}{2C_1}\Vert f_m\Vert_{-1}^2\\
\Vert u_m\Vert_1&\le \frac{1}{C_1}\Vert f_m\Vert_{-1}\le \frac{1}{C_1}\Vert f\Vert_{H^{-1} }+M
\end{aligned}
$$
最后一个不等号是因为$f_m\rightharpoonup f$弱$\ast$收敛,
推出$\Vert f_m-f\Vert_{H^{-1} }$有界(共鸣定理).
代入$f_m=0$即得到方程解的唯一性, 从而解也是存在的, 矩阵$A$可逆,
我们可以找到$u_m$为问题解.
特别地, 根据最后一行, 我们可以看到$\{u_m\}\subset H_0^1(\Omega)$有界.
它是Hilbert空间, 当然是某个空间(自身)的对偶空间(自反),
从而$\{u_m\}$是弱紧的, 即其有子列弱收敛, 不妨设自身弱收敛到$u.$
断言$u$即为弱解.
$\,\forall\,v\in H_0^1(\Omega),$ 取$v_m=\sum_{i=1}^m\xi_i\phi_i,$
那么$v_m\xrightarrow{H^1}v.$
由于$\left<{}Lu_m,\phi_j\right>=\left<{}f_m,\phi_j\right>,$
$j=1,\cdots,m.$ 我们有$\left<{}Lu_m,v_m\right>=\left<{}f_m,v_m\right>.$
注意到$\left<{}f_m,v_m\right>=\left<{}f_m,v_m-v\right>+\left<{}f_m,v\right>\rightarrow \left<{}f,v\right>.$
其中
$$
|\left<{}f_m,v_m-v\right>|\le \Vert f_m\Vert_{-1}\Vert v_m-v\Vert_1\le (\Vert f\Vert_{-1}+M)\Vert v_m-v\Vert_1\rightarrow 0.
$$
同时,
$$
\left<{}Lu_m,v_m\right>=-\sum_{i,j=1}^n\int_{\Omega}\partial_ju_m\partial_i(a_{ij}(x)v_m)dx-\sum_{i=1}^n \int_\Omega u_m\partial_i(b_i(x)v_m)dx+\int_\Omega c(x)u_mv_mdx.
$$
类似地,
$\left<{}Lu_m,v_m\right>=\left<{}Lu_m,v_m-v\right>+\left<{}Lu_m,v\right>,$
其中根据强收敛与分部积分,
$$
|\left<{}Lu_m,v_m-v\right>|\le \alpha\Vert u_m\Vert_1\Vert v_m-v\Vert_1\rightarrow 0;
$$
$$
\left<{}Lu_m,v\right>=\left<{}u_m,L^\ast v\right>\rightarrow \left<{}u,L^\ast v\right>=\left<{}Lu,v\right>.
$$
因此$\left<{}Lu_m,v_m\right>\rightarrow \left<{}Lu,v\right>.$
于是$\left<{}Lu,v\right>=\left<{}f,v\right>,$
$\,\forall\,v\in H_0^1(\Omega).$ 这就说明了$u$为弱解,
由此证明了原问题解的存在性, 同时给出了具体得到解的方法.
双曲型方程
考虑双曲型方程
$$
Mu=\frac{\partial^2 u}{\partial t^2}-\left(\sum_{i,j=1}^n\frac{\partial}{\partial x_i}\left(a_{ij}(t,x)\frac{\partial u}{\partial x_j}\right)+\sum_{i=1}^nb_i(t,x)\frac{\partial u}{\partial x_i}+c(t,x)u\right)=f(t,x).
$$
其中设$a_{ij}$满足一致椭圆性条件: $\,\exists\,\alpha>0,$ 使得
$$
\sum_{i,j=1}^n a_{ij}(t,x)\xi_i\xi_j\ge \alpha\sum_{i=1}^n \xi_i^2, \quad\,\forall\,\xi\in \mathbb{R}^n\setminus\{0\}.
$$
我们考虑Galerkin方法: 对Dirichlet齐次边界问题,
取$H_0^1(\Omega)$的一组$C_c^\infty$完备正交基$\{\phi_i(x)\},$
那么记$u_m(t,x)=\sum_{i=1}^m a_i(t)\phi_i(x),$ 给定初值解方程
$$
\left<{}Mu_m,\phi_j\right>=\left<{}f,\phi_j\right>, \quad j=1,\cdots,m,
$$
求系数$a_i(t).$ 这是$m$个二阶常微分方程构成的ODE方程组, 总是可解的.
为了说明$\{u_m\}$有子列弱收敛于弱解, 我们只需讨论其有界性.
可以考虑差分方法, 此处我们给出能量不等式, 采用能量方法来解决.
利用分部积分, 由$u$具齐次边界推出$u_t$也是, 考虑
$$
\int_\Omega u_tMudx=\frac{1}{2}\partial_t \int_\Omega u_t^2dx+\int_\Omega \sum_{i,j=1}^nu_{it}a_{ij}u_jdx-\int_\Omega\sum_i u_tb_iu_idx-\int_\Omega u_tcudx=\int_\Omega u_tfdx.
$$
利用对称性, 可以得到
$$
\int_\Omega \sum_{i,j=1}^nu_{it}a_{ij}u_jdx= \frac{1}{2}\partial_t\int_\Omega \sum_{i,j=1}^n a_{ij}u_iu_jdx-\frac{1}{2}\int_\Omega \sum_{i,j=1}^n\partial_t a_{ij}u_iu_jdx.
$$
因此,
$$
\frac{1}{2}\partial_t \int_\Omega u_t^2dx+\frac{1}{2}\partial_t\int_\Omega \sum_{i,j=1}^n a_{ij}u_iu_jdx=\int_\Omega u_tfdx+\frac{1}{2}\int_\Omega \sum_{i,j=1}^n\partial_t a_{ij}u_iu_jdx+\int_\Omega\sum_i u_tb_iu_idx+\int_\Omega u_tcudx.
$$
在$\overline{\Omega_T}=[0,T]\times \overline{\Omega}$上考虑,
可以取到$C>0,$ 使得
$$
\partial_t \int_\Omega u_t^2+\sum_{i,j=1}^n a_{ij}u_iu_j dx\le C(\Vert u\Vert_1^2+\Vert u_t\Vert^2+\Vert f\Vert^2)
$$
两边积分, 那么在$t$时刻, 记初值为$u=g,$ $u_t=h,$ 则
$$
\Vert u_t\Vert^2+\Vert u\Vert^2_1\le \Vert g\Vert_1^2+\Vert h\Vert^2+C\int_0^t\Vert u_t\Vert^2+\Vert u\Vert_1^2+\Vert f\Vert^2d\tau
$$
记$E(t)=\Vert u_t\Vert^2+\Vert u\Vert^2_1,$
$E_0=\Vert g\Vert_1^2+\Vert h\Vert^2+C\int_0^T\Vert f\Vert^2d\tau,$ 则
$$
E(t)\le E_0+C\int_0^tE(\tau)d\tau.
$$
总结上面的方法,
我们通过取模的方式, 将PDE化为ODE, 得到了一个常微分不等式.
那么利用Gronwall不等式即可得到能量不等式.
引理 1 (Gronwall不等式). 设非负$E(t)\in C[0,T],$ 满足$E(t)\le E_0+C\int_0^tE(\tau)d\tau,$ $C,E_0>0,$ 则
$$
E(t)\le E_0e^{Ct}.
$$
证: 记$I(t)=\int_0^tE(\tau)d\tau,$
则原式等于$\partial_t I(t)\le E_0+CI(t).$ 因此,
$$
\partial_t(e^{-Ct}I(t))\le E_0e^{-Ct}\,\Rightarrow\, e^{-Ct}I(t)\le \frac{E_0}{C}(1-e^{-Ct})\,\Rightarrow\, I(t)\le \frac{E_0}{C}(e^{Ct}-1).
$$
代入上式, 即有
$$
E(t)=\partial_t I(t)\le E_0e^{Ct}.
$$
特别地, 我们有$E(t)\le E_0e^{CT},$ $\,\forall\,t\in [0,T].$ 因此局部上,
$u$是有界的, 进而是弱紧的. 进一步, 和椭圆方程类似地,
$$
\left<{}f,v\right>\leftarrow \left<{}Mu_m,v\right>=\left<{}u_m,M^\ast v\right>\rightarrow \left<{}u,M\ast v\right>=\left<{}Mu,v\right>
$$
从而$\{u_m\}$弱收敛到的$u$为弱解. 当然,
也可以通过对偶问题的能量估计说明唯一性, 从而给出原问题解的存在性.
但同样的, 这只能给出理论上的说明. 令$E_0=0$即可说明唯一性,
对一般的$E_0,$ 能量方法给出了稳定性的说明.
抛物型方程
抛物型方程与双曲型方程类似, 在此只做能量估计的推导.
$$
\left<{}Mu,u\right>=\frac{1}{2}\partial_t\int_\Omega u^2dx+\int_\Omega \sum_{i,j=1}^n a_{ij}u_iu_jdx-\int_\Omega \sum_{i=1}^n b_iu_iudx-\int_\Omega cu^2dx=\int_\Omega fudx
$$
通过$ab\le \varepsilon{a^2}+\frac{b^2}{\varepsilon},$ 可以说明:
$$
\frac{1}{2}\partial_t\Vert u\Vert^2\le \frac{1}{2}\partial_t\Vert u\Vert^2+\frac{\lambda}{2} \Vert\nabla u\Vert^2\le C(\Vert u\Vert^2+\Vert f\Vert^2)
$$
因此, 设初值$u=g,$ 则
$$
\Vert u\Vert^2\le \Vert g\Vert^2+C\int_0^T\Vert f\Vert d\tau + C\int_0^t \Vert u\Vert^2d\tau.
$$
记$E(t)=\Vert u\Vert^2,$
$E_0=\Vert g\Vert^2+C\int_0^T\Vert f\Vert d\tau,$
则同样由Gronwall不等式, 我们有
$$
E(t)\le E_0e^{Ct}.
$$
从而通过Galerkin方法能够得到弱解, 同时也讨论了解的唯一性与稳定性.
算子半群
另一种非常实用的方法为算子半群方法. 如对抛物型问题,
考虑解齐次Dirichlet边界初值问题:
$$
\left\{\begin{aligned}
&u_t=Au=\sum_{i,j=1}^n\partial_j(a_{ij}u_i)\\
&u(0,x)=g(x)\\
&u|_{\partial\Omega\times [0,T]}=0
\end{aligned}\right.
$$
那么形式上,
$u(t,x)=e^{At}g(x)$即给出了问题解, 因为$u_t=Ae^{At}g(x)=Au,$
且满足初值与边界条件. 这里$e^{At}:=\sum_n \frac{A^n}{n!}t^n.$
为了给出严格说明, 对算子取模, 那么
$$
\Vert e^{At}\Vert\le \sum_n\frac{\Vert A^n\Vert}{n!}t^n\le \sum_n\frac{\Vert A\Vert^n}{n!}t^n
$$
这里利用了相容条件$\Vert AB\Vert\le \Vert A\Vert\Vert B\Vert.$
从而倘若$\Vert A\Vert\le 1,$ 那么$\Vert e^{At}\Vert\le e^t,$
算子是有意义的.
然而一般地, 显然$\Vert A\Vert\le 1$不见得成立.
回忆$(\lambda-A)^{-1}$对充分大的$\lambda$是有界线性算子,
因此对其讨论是有希望的.
模仿$a=\lim\limits_{\lambda\rightarrow\infty}\lambda a(\lambda-a)^{-1},$
定义
$$
A_\lambda=\lambda A(\lambda-A)^{-1}=\lambda(A-\lambda+\lambda)(\lambda-A)^{-1}=-\lambda+\lambda^2(\lambda-A)^{-1}.
$$
我们可以期待如下极限成立:
$$
e^{At}=\lim\limits_{\lambda\rightarrow \infty}e^{A_\lambda t}=\lim\limits_{\lambda\rightarrow \infty}e^{-\lambda t}e^{\lambda^2(\lambda-A)^{-1}t}
$$
此时$\Vert e^{A_\lambda t}\Vert=e^{-\lambda t}\Vert e^{\lambda^2(\lambda-A)^{-1}t}\Vert.$
而$\Vert(\lambda-A)^{-1}\Vert\le \frac{1}{\lambda},$
因此$\Vert e^{\lambda^2(\lambda-A)^{-1}t}\Vert\le e^{\lambda t},$
$\Vert e^{A_\lambda t}\Vert\le 1,$ 算子是有意义的.
文章最后更新于 2021-12-13 11:11:02
Elliptic equation
Recall from the previous section that the existence of $Ax=b$ is equivalent to the uniqueness of $A^\ast y=0$.
If $A$ is the $n\times n$ matrix, then the uniqueness of $A^\ast y=0$ is equivalent to the uniqueness of $Ay=0$.
Therefore, the existence of $Ax=b$ is equivalent to uniqueness.
For elliptic equations, consider the homogeneous Dirichlet boundary problem $Lu=f\in H^{-1}(\Omega).$
We have Garding's inequality:
$$
\left<{}-Lu,u\right>_{L^2}\ge C_1\Vert u\Vert_1^2-C_2\Vert u\Vert^2.
$$
When $C_2\le 0$, we can explain the uniqueness of the problem. As for the existence,
According to the original method, it is necessary to consider whether there is $C_2^\ast \le 0,$ in the Garding inequality corresponding to the dual operator $L^\ast $.
Thus, the existence of the original problem is explained through the uniqueness of the dual problem.
But in fact, $C_2\le 0$ can directly illustrate the existence of the solution to the original problem.
consider Galerkin method.
The method of deriving existence from the uniqueness of the dual problem only explains the theoretical existence.
The construction of the solution cannot be given specifically. However, this method can give the construction method of the solution,
is the numerical solution of the differential equation finite element method theoretical basis.
Recall that $H^{m,p}$ is a separable space because it can be equidistantly embedded in $(L^p)^N$,
The latter is a separable space. Therefore, $H_0^1(\Omega)$ is a separable Hilbert space,
There is a set of complete orthonormal bases $\{\phi_i\}.$. Theoretically,
Can take the characteristic function system $-\Delta \phi_j=\lambda_j\phi_j,$
$\phi_j|_{\partial\Omega}=0.$ When the operator is self-conjugated, the characteristic function system is of course orthogonal.
Since $C_c^\infty(\Omega)$ is dense in $H_0^1(\Omega)$,
Of course we can take $\{\phi_i\}\subset C_c^\infty(\Omega).$
Consider $u_m=\sum_{i=1}^ma_i\phi_i,$ $f=\sum_{i=1}^\infty f_i\phi_i,$
$f_m=\sum_{i=1}^m f_i\phi_i\rightharpoonup f.$ To understand $Lu=f,$
One expectation is that $u_m$ will give $u.$ by solving $Lu_m=f_m$, but this expectation is often disappointed.
Because for the equation
$$
Lu_m=\sum_{i=1}^m a_iL\phi_i=\sum_{i=1}^mf_i\phi_i=f_m,
$$
When $\{\phi_i\}$ does not take the characteristic function system, this will be an infinite-dimensional equation,
It cannot be solved for a finite number of unknown quantities.
Galerkin's approach is that we consider the following questions:
$$
\left<{}Lu_m,\phi_j\right>=\left<{}f_m,\phi_j\right>,\quad j=1,\cdots,m.
$$
That is $\sum_{i=1}^m a_i\left<{}L\phi_i,\phi_j\right>=\sum_{i=1}^mf_i\left<{}\phi_i,\phi_j\right>,$
$j=1,\cdots,m.$ This is the form of the matrix equation $Ax=b$. As stated at the beginning,
In order to prove the existence of the solution, it is only necessary to explain the uniqueness of the solution. For this purpose, elliptic estimation can be used.
Consider $\left<{}Lu_m-f_m,u_m\right>=0,$ by Garding's inequality, when $C_2\le 0$,
$$
\begin{aligned}
C_1\Vert u_m\Vert^2_1&\le \left<{}-f_m,u_m\right>\le \Vert f_m\Vert_{-1}\Vert u_m\Vert_1\le \frac{1}{2C_1}\Vert f_m\Vert_{-1}^2+\frac{C_1}{2}\Vert u_m\Vert_1^2\\
\frac{C_1}{2}\Vert u_m\Vert^2_1&\le \frac{1}{2C_1}\Vert f_m\Vert_{-1}^2\\
\Vert u_m\Vert_1&\le \frac{1}{C_1}\Vert f_m\Vert_{-1}\le \frac{1}{C_1}\Vert f\Vert_{H^{-1} }+M
\end{aligned}
$$
The last inequality sign is because $f_m\rightharpoonup f$ is weakly convergent $\ast$,
Deduced that $\Vert f_m-f\Vert_{H^{-1} }$ is bounded (resonance theorem).
Substituting into $f_m=0$, we get the uniqueness of the solution of the equation, so the solution also exists. The matrix $A$ is reversible,
We can find $u_m$ as the solution to the problem.
In particular, according to the last line, we can see that $\{u_m\}\subset H_0^1(\Omega)$ is bounded.
It is a Hilbert space, which is of course the dual space (reflexive) of a certain space (self),
Therefore $\{u_m\}$ is weakly compact, that is, it has subsequences that are weakly convergent. We might as well assume that it is weakly convergent to $u.$
Assert $u$ is the weak solution.
$\,\forall\,v\in H_0^1(\Omega),$ takes $v_m=\sum_{i=1}^m\xi_i\phi_i,$
Then $v_m\xrightarrow{H^1}v.$
Due to $\left<{}Lu_m,\phi_j\right>=\left<{}f_m,\phi_j\right>,$
$j=1,\cdots,m.$ We have $\left<{}Lu_m,v_m\right>=\left<{}f_m,v_m\right>.$
Notice$\left<{}f_m,v_m\right>=\left<{}f_m,v_m-v\right>+\left<{}f_m,v\right>\rightarrow \left<{}f,v\right>.$
Among them
$$
|\left<{}f_m,v_m-v\right>|\le \Vert f_m\Vert_{-1}\Vert v_m-v\Vert_1\le (\Vert f\Vert_{-1}+M)\Vert v_m-v\Vert_1\rightarrow 0.
$$
At the same time,
$$
\left<{}Lu_m,v_m\right>=-\sum_{i,j=1}^n\int_{\Omega}\partial_ju_m\partial_i(a_{ij}(x)v_m)dx-\sum_{i=1}^n \int_\Omega u_m\partial_i(b_i(x)v_m)dx+\int_\Omega c(x)u_mv_mdx.
$$
Similarly,
$\left<{}Lu_m,v_m\right>=\left<{}Lu_m,v_m-v\right>+\left<{}Lu_m,v\right>,$
Among them, according to strong convergence and integral by parts,
$$
|\left<{}Lu_m,v_m-v\right>|\le \alpha\Vert u_m\Vert_1\Vert v_m-v\Vert_1\rightarrow 0;
$$
$$
\left<{}Lu_m,v\right>=\left<{}u_m,L^\ast v\right>\rightarrow \left<{}u,L^\ast v\right>=\left<{}Lu,v\right>.
$$
Therefore $\left<{}Lu_m,v_m\right>\rightarrow \left<{}Lu,v\right>.$
So$\left<{}Lu,v\right>=\left<{}f,v\right>,$
$\,\forall\,v\in H_0^1(\Omega).$ This shows that $u$ is a weak solution,
This proves the existence of the solution to the original problem and provides a specific method to obtain the solution.
hyperbolic equation
Consider the hyperbolic equation
$$
Mu=\frac{\partial^2 u}{\partial t^2}-\left(\sum_{i,j=1}^n\frac{\partial}{\partial x_i}\left(a_{ij}(t,x)\frac{\partial u}{\partial x_j}\right)+\sum_{i=1}^nb_i(t,x)\frac{\partial u}{\partial x_i}+c(t,x)u\right)=f(t,x).
$$
Where $a_{ij}$ satisfies consistent ellipticity condition: $\,\exists\,\alpha>0,$ makes
$$
\sum_{i,j=1}^n a_{ij}(t,x)\xi_i\xi_j\ge \alpha\sum_{i=1}^n \xi_i^2, \quad\,\forall\,\xi\in \mathbb{R}^n\setminus\{0\}.
$$
We consider the Galerkin method: For the Dirichlet homogeneous boundary problem,
Take a set of $C_c^\infty$ complete orthonormal bases $\{\phi_i(x)\},$ of $H_0^1(\Omega)$
Then record $u_m(t,x)=\sum_{i=1}^m a_i(t)\phi_i(x),$ to solve the equation given the initial value
$$
\left<{}Mu_m,\phi_j\right>=\left<{}f,\phi_j\right>, \quad j=1,\cdots,m,
$$
Find the coefficient $a_i(t).$ This is a system of ODE equations composed of $m$ second-order ordinary differential equations, which is always solvable.
In order to illustrate that $\{u_m\}$ has subsequences that weakly converge to weak solutions, we only need to discuss its boundedness.
The difference method can be considered. Here we give the energy inequality and use the energy method to solve it.
Using integrals by parts, $u$ has a homogeneous boundary to deduce $u_t$ as well. Consider
$$
\int_\Omega u_tMudx=\frac{1}{2}\partial_t \int_\Omega u_t^2dx+\int_\Omega \sum_{i,j=1}^nu_{it}a_{ij}u_jdx-\int_\Omega\sum_i u_tb_iu_idx-\int_\Omega u_tcudx=\int_\Omega u_tfdx.
$$
Using symmetry, we can get
$$
\int_\Omega \sum_{i,j=1}^nu_{it}a_{ij}u_jdx= \frac{1}{2}\partial_t\int_\Omega \sum_{i,j=1}^n a_{ij}u_iu_jdx-\frac{1}{2}\int_\Omega \sum_{i,j=1}^n\partial_t a_{ij}u_iu_jdx.
$$
Therefore,
$$
\frac{1}{2}\partial_t \int_\Omega u_t^2dx+\frac{1}{2}\partial_t\int_\Omega \sum_{i,j=1}^n a_{ij}u_iu_jdx=\int_\Omega u_tfdx+\frac{1}{2}\int_\Omega \sum_{i,j=1}^n\partial_t a_{ij}u_iu_jdx+\int_\Omega\sum_i u_tb_iu_idx+\int_\Omega u_tcudx.
$$
Consider on $\overline{\Omega_T}=[0,T]\times \overline{\Omega}$,
We can get $C>0,$ such that
$$
\partial_t \int_\Omega u_t^2+\sum_{i,j=1}^n a_{ij}u_iu_j dx\le C(\Vert u\Vert_1^2+\Vert u_t\Vert^2+\Vert f\Vert^2)
$$
Integrate both sides, then at the moment $t$, record the initial value as $u=g,$ $u_t=h,$, then
$$
\Vert u_t\Vert^2+\Vert u\Vert^2_1\le \Vert g\Vert_1^2+\Vert h\Vert^2+C\int_0^t\Vert u_t\Vert^2+\Vert u\Vert_1^2+\Vert f\Vert^2d\tau
$$
Remember$E(t)=\Vert u_t\Vert^2+\Vert u\Vert^2_1,$
$E_0=\Vert g\Vert_1^2+\Vert h\Vert^2+C\int_0^T\Vert f\Vert^2d\tau,$ then
$$
E(t)\le E_0+C\int_0^tE(\tau)d\tau.
$$
To summarize the above methods,
We transform PDE into ODE by taking modulo, and obtain an ordinary differential inequality.
Then use Gronwall's inequality to get the energy inequality.
Lemma 1 (Gronwall's inequality). Assume that non-negative $E(t)\in C[0,T],$ satisfies $E(t)\le E_0+C\int_0^tE(\tau)d\tau,$ $C,E_0>0,$, then
$$
E(t)\le E_0e^{Ct}.
$$
Certificate: Remember$I(t)=\int_0^tE(\tau)d\tau,$
Then the principle formula is equal to $\partial_t I(t)\le E_0+CI(t).$ Therefore,
$$
\partial_t(e^{-Ct}I(t))\le E_0e^{-Ct}\,\Rightarrow\, e^{-Ct}I(t)\le \frac{E_0}{C}(1-e^{-Ct})\,\Rightarrow\, I(t)\le \frac{E_0}{C}(e^{Ct}-1).
$$
Substituting into the above formula, we have
$$
E(t)=\partial_t I(t)\le E_0e^{Ct}.
$$
In particular, we have $E(t)\le E_0e^{CT},$ $\,\forall\,t\in [0,T].$ so locally,
$u$ is bounded and therefore weakly compact. Furthermore, similar to the elliptic equation,
$$
\left<{}f,v\right>\leftarrow \left<{}Mu_m,v\right>=\left<{}u_m,M^\ast v\right>\rightarrow \left<{}u,M\ast v\right>=\left<{}Mu,v\right>
$$
Therefore, $\{u_m\}$ weakly converges to $u$, which is a weak solution. Of course,
The uniqueness can also be explained through the energy estimation of the dual problem, thereby giving the existence of the solution to the original problem.
But again, this can only give a theoretical explanation. Let $E_0=0$ illustrate the uniqueness,
A stability statement is given for the general $E_0,$ energy method.
parabolic equation
Parabolic equations are similar to hyperbolic equations, and only the derivation of energy estimates is done here.
$$
\left<{}Mu,u\right>=\frac{1}{2}\partial_t\int_\Omega u^2dx+\int_\Omega \sum_{i,j=1}^n a_{ij}u_iu_jdx-\int_\Omega \sum_{i=1}^n b_iu_iudx-\int_\Omega cu^2dx=\int_\Omega fudx
$$
Through $ab\le \varepsilon{a^2}+\frac{b^2}{\varepsilon},$ it can be explained:
$$
\frac{1}{2}\partial_t\Vert u\Vert^2\le \frac{1}{2}\partial_t\Vert u\Vert^2+\frac{\lambda}{2} \Vert\nabla u\Vert^2\le C(\Vert u\Vert^2+\Vert f\Vert^2)
$$
Therefore, assuming the initial value $u=g,$ is
$$
\Vert u\Vert^2\le \Vert g\Vert^2+C\int_0^T\Vert f\Vert d\tau + C\int_0^t \Vert u\Vert^2d\tau.
$$
Remember$E(t)=\Vert u\Vert^2,$
$E_0=\Vert g\Vert^2+C\int_0^T\Vert f\Vert d\tau,$
Then similarly from Gronwall's inequality, we have
$$
E(t)\le E_0e^{Ct}.
$$
Therefore, the weak solution can be obtained through the Galerkin method, and the uniqueness and stability of the solution are also discussed.
operator semigroup
Another very practical method is the operator semigroup method. For example, for parabolic problems,
Consider solving the homogeneous Dirichlet boundary initial value problem:
$$
\left\{\begin{aligned}
&u_t=Au=\sum_{i,j=1}^n\partial_j(a_{ij}u_i)\\
&u(0,x)=g(x)\\
&u|_{\partial\Omega\times [0,T]}=0
\end{aligned}\right.
$$
So formally,
$u(t,x)=e^{At}g(x)$ gives the solution to the problem, because $u_t=Ae^{At}g(x)=Au,$
And satisfy the initial value and boundary conditions. Here $e^{At}:=\sum_n \frac{A^n}{n!}t^n.$
In order to give a strict explanation, take modulo the operator, then
$$
\Vert e^{At}\Vert\le \sum_n\frac{\Vert A^n\Vert}{n!}t^n\le \sum_n\frac{\Vert A\Vert^n}{n!}t^n
$$
The compatibility condition $\Vert AB\Vert\le \Vert A\Vert\Vert B\Vert.$ is used here
Thus if $\Vert A\Vert\le 1,$ then $\Vert e^{At}\Vert\le e^t,$
The operator is meaningful.
However, in general, it is obvious that $\Vert A\Vert\le 1$ does not necessarily hold.
Recall that $(\lambda-A)^{-1}$ is a bounded linear operator for sufficiently large $\lambda$,
Therefore, discussion on it is promising.
Imitation$a=\lim\limits_{\lambda\rightarrow\infty}\lambda a(\lambda-a)^{-1},$
definition
$$
A_\lambda=\lambda A(\lambda-A)^{-1}=\lambda(A-\lambda+\lambda)(\lambda-A)^{-1}=-\lambda+\lambda^2(\lambda-A)^{-1}.
$$
We can expect the following limits to hold:
$$
e^{At}=\lim\limits_{\lambda\rightarrow \infty}e^{A_\lambda t}=\lim\limits_{\lambda\rightarrow \infty}e^{-\lambda t}e^{\lambda^2(\lambda-A)^{-1}t}
$$
At this time $\Vert e^{A_\lambda t}\Vert=e^{-\lambda t}\Vert e^{\lambda^2(\lambda-A)^{-1}t}\Vert.$
And $\Vert(\lambda-A)^{-1}\Vert\le \frac{1}{\lambda},$
Therefore $\Vert e^{\lambda^2(\lambda-A)^{-1}t}\Vert\le e^{\lambda t},$
$\Vert e^{A_\lambda t}\Vert\le 1,$ operators are meaningful.
The article was last updated on 2021-12-13 11:11:02
