Machine-translated from Chinese.
经典方程
回忆数理方程, 我们最常接触的是如下方程:
$-\Delta u=f,$ Poisson方程. 讨论边值问题,
边界条件为Dirichlet或Neumann型.
$u_{tt}-\Delta u=f,$ 波动方程. 无界区域上讨论Cauchy问题,
有界区域上讨论初边值问题. 需对$u$与$u_t$给定初始条件,
边界条件为Dirichlet或Neumann型.
$u_t-\Delta u=f,$ 热传导方程. 讨论内容基本同波方程,
初始条件只需对$u$给定.
考虑上述方程的一般化形式, 有如下线性方程,
讨论的问题与上面经典问题都是对应的, 不再重复. 系数都设为光滑的,
且满足椭圆性条件:
$\sum_{ij=1}^n a_{ij(x)}\xi_i\xi_j\ge \alpha \Vert\xi\Vert^2,$
$\,\forall\,\xi\neq 0.$ 不妨设系数是对称的.
$Lu=\sum_{i,j=1}^{n}a_{ij}(x)u_{ij}+\sum_{i=1}^n b_i(x)u_i+c(x)u=f,$
二阶椭圆方程.
$u_{tt}-\sum_{i,j=1}^n\frac{\partial {} }{\partial {}x_i}(a_{ij}(t,x)\frac{\partial {}u}{\partial {}x_j})+\sum_{i=1}^n b_i(t,x)\frac{\partial {}u}{\partial {}x_i}+c(t,x)u=f,$
二阶双曲型方程.
$u_{t}-\sum_{i,j=1}^n\frac{\partial {} }{\partial {}x_i}(a_{ij}(t,x)\frac{\partial {}u}{\partial {}x_j})+\sum_{i=1}^n b_i(t,x)\frac{\partial {}u}{\partial {}x_i}+c(t,x)u=f,$
二阶抛物型方程.
我们从考虑经典解($C^2$解)转移到考虑弱解(广义函数解) (Weak
Formulation). 由Stokes公式:
$$\int_\Omega d\omega=\int_{\partial\Omega} \omega,$$
我们有Gauss-Green公式:
$$\int_\Omega \partial_i fdx=\int_\Omega \nabla \cdot e_i fdx=\int_{\partial\Omega} n\cdot e_i \gamma fdS=\int_{\partial\Omega} n_i\gamma fdS.$$
其中$\gamma$为迹, 使得公式对更多函数成立. 进而我们有分部积分公式:
$$\int_{\Omega} \partial_i fgdx+\int_\Omega f\partial_i gdx=\int_\Omega{\partial_i (fg)}dx=\int_{\partial\Omega} n_i\gamma(fg)dS.$$
对偏微分方程的求解, 一种常用的方法是算子半群方法,
从而将问题转为求解无穷维ODE.
对偶算子
考虑Dirichlet齐次边界问题, $u\in H_0^1(\Omega).$
作用测试函数$v\in H_0^1(\Omega)$到方程上:
$$\begin{aligned}
\int_\Omega vLudx=&{\color{blue}\sum_{i,j=1}^n\int_{\Omega}\frac{\partial {} }{\partial {}x_i}(a_{ij}(x)u_jv)dx}-\sum_{i,j=1}^n\int_{\Omega}u_j\frac{\partial {} }{\partial {}x_i}(a_{ij}(x)v)dx\\
&+{\color{blue}\sum_{i=1}^n \int_\Omega \frac{\partial {} }{\partial {}x_i}(b_i(x)uv)dx}-\sum_{i=1}^n \int_\Omega u\frac{\partial {} }{\partial {}x_i}(b_i(x)v)dx+\int_\Omega c(x)uvdx\\
=&-\sum_{i,j=1}^n\int_{\Omega}u_j\frac{\partial {} }{\partial {}x_i}(a_{ij}(x)v)dx-\sum_{i=1}^n \int_\Omega u\frac{\partial {} }{\partial {}x_i}(b_i(x)v)dx+\int_\Omega c(x)uvdx\\
=&{\color{blue}\sum_{i,j=1}^n\int_{\Omega}\frac{\partial {} }{\partial {}x_j}\left(u\frac{\partial {} }{\partial {}x_i}(a_{ij}(x)v)\right)dx}-\sum_{i,j=1}^n\int_{\Omega}u_j\frac{\partial {} }{\partial {}x_i}(a_{ij}(x)v)dx\\\
&-\sum_{i=1}^n \int_\Omega u\frac{\partial {} }{\partial {}x_i}(b_i(x)v)dx+\int_\Omega c(x)uvdx\\
=&\sum_{i,j=1}^n\int_{\Omega}u\frac{\partial^2 {} }{\partial {}x_i\partial {}x_j}(a_{ij}(x)v)dx-\sum_{i=1}^n \int_\Omega u\frac{\partial {} }{\partial {}x_i}(b_i(x)v)dx+\int_\Omega c(x)uvdx\\
=&\int_\Omega uL^*vdx.
\end{aligned}$$
推导中多次用到Gauss-Green公式.
由于函数具齐次边界条件, 边界项(蓝色项)全部消失.
由此我们得到了对偶算子$L^\ast .$
类似地, 对双曲型, 抛物型方程, 我们也有对偶算子:
$$M^\ast v=\frac{\partial {}^2v}{\partial {}t^2}-\sum_{i,j=1}^n\frac{\partial {} }{\partial {}x_i}(a_{ij}(x)v_j)-\sum_{i=1}^n\frac{\partial {} }{\partial {}x_i}(b_i(x)v)+c(x)v,$$
$$M^\ast v=-\frac{\partial {}v}{\partial {}t}-\sum_{i,j=1}^n\frac{\partial {} }{\partial {}x_i}(a_{ij}(x)v_j)-\sum_{i=1}^n\frac{\partial {} }{\partial {}x_i}(b_i(x)v)+c(x)v.$$
对偶算子的重要性在于, 由泛函分析, 有如下重要命题:
命题 1. 对Hilbert空间$H,$ 有界线性算子$A:H\rightarrow H,$ 问题$Ax=b,$ $\,\forall\,b\in H$的存在性, 等价于问题$A^\ast y=0$的唯一性(即只有零解).
证: 存在性$\Rightarrow$唯一性: $\,\forall\,y$使得$A^\ast y=0,$
存在$x$使得$Ax=y.$ 因此,
$$0=(x,A^\ast y)=(Ax,y)=(y,y) \,\Rightarrow\, y=0.$$
唯一性$\Rightarrow$存在性: 对$b\in H,$ 只需找$x$满足:
$$(Ax,y)=(b,y),\quad \forall y\in H.$$
取$A^\ast :H\rightarrow B=\operatorname{Im} A^\ast ,$
为连续线性算子.它是单射因为$A^\ast y=0$只有零解. 由逆算子定理,
$(A^\ast )^{-1}:B\rightarrow H$也是连续线性算子.
定义连续线性泛函$l_b:B\rightarrow \mathbb{R}$:
$$l_b(v)=(b,(A^\ast )^{-1}v)=(b,u),$$
这里$A^\ast u=v.$ 由Riesz表示定理,
存在$x\in B$使得$l_b(v)=(x,v).$ 因此对任意$y\in H:$
$$l_b(A^\ast y)=(b,y)=(x,A^\ast y)=(Ax,y).$$
这说明我们找到的$x\in B$正是$Ax=b$的解.
能量方法
乘子法
由对偶算子, 在大多数情形, 我们只需处理唯一性的证明, 而这是容易的.
通常采用能量方法, 通过求导由能量守恒说明唯一性.
能量方法往往来源于乘子法. 对椭圆方程, 考虑能量形式$\int uLu$(势能);
对波动方程, 考虑能量形式$\int u_t Mu$(动能); 对热传导方程,
考虑能量形式$\int uMu$(势能).
以椭圆方程为例. 对$u\in H_0^1(\Omega),$ 我们有:
$$\begin{aligned}
-\int_\Omega uLudx=&\int_\Omega \sum_{i,j=1}^n a_{ij}(x)u_iu_jdx+\int_\Omega \sum_{i=1}^n\left(\sum_{j=1}^n\frac{\partial {}a_{ij} }{\partial {}x_j}-b_i(x)\right)u_iudx-\int_\Omega c(x)u^2dx\\
\ge &\int_\Omega \alpha \sum_{i=1}^n u_i^2dx-\int_\Omega C'\sum_{i=1}^n \left(\frac{\alpha}{2C'} u_i^2+\frac{C'}{2\alpha}u^2\right)dx-\int_\Omega C''u^2dx\\
\ge &\frac{\alpha}{2}\Vert\nabla u\Vert-C\Vert u\Vert.
\end{aligned}$$
过程中用到了不等式$2ab\le \varepsilon a^2+\frac{1}{\varepsilon} b^2.$
该不等式称为Garding不等式:
定理 2 (Garding不等式). 设$\Omega$为$\mathbb{R}^n$中有界区域, 边界光滑. 若$L$为二阶椭圆型算子, 则$\,\forall\,u\in H_0^1(\Omega),$ 有如下不等式:
$$\operatorname{Re}(-Lu,u)_{L^2(\Omega)}\ge C_1\Vert u\Vert_{1}^2-C_2\Vert u\Vert^2.$$
其中系数可以具体计算出来. 当$C_2\le 0$时, 对问题$Lu=0$考虑唯一性:
$$C_1\Vert u\Vert_{1}^2-C_2\Vert u\Vert^2\le 0\,\Rightarrow\, u\equiv 0.$$
对线性方程, $Lu=0$的唯一性与$Lu=f$的唯一性当然是等价的.
特征问题
一般地, $C_2$不见得非正. 但是对充分大的$\Lambda\in \mathbb{R},$
考虑$(-L+\Lambda)u=0$就可以强行将Garding不等式中右侧项调至非负,
此时问题解具有唯一性. 因此可以在像空间定义逆算子$(-L+\Lambda)^{-1}.$
我们进行如下推导:
$$\begin{aligned}
C_1\Vert u\Vert^2_{1}&\le \left<{}(-L+\Lambda)u,u\right>\\
&\le \Vert(-L+\Lambda)u\Vert\Vert u\Vert\le \Vert(-L+\Lambda)u\Vert\Vert u\Vert_{H^1}\\
\Vert u\Vert_{1}&\le C_3\Vert(-L+\Lambda)u\Vert\\
\Vert(-L+\Lambda)^{-1}v\Vert_{1}&\le C_3 \Vert v\Vert
\end{aligned}$$
因此$(-L+\Lambda)^{-1}$为有界算子.
由于$H_0^1\stackrel{c}\hookrightarrow L^2$为紧嵌入, 它还是紧算子.
其特征值可列, $\mu_i\rightarrow 0.$
考虑特征问题:
$$\begin{aligned}
-Lu&=\lambda u\\
(-L+\Lambda)u&=(\lambda+\Lambda)u\\
\frac{1}{\lambda+\Lambda}u&=(-L+\Lambda)^{-1}u=\mu u\\
\end{aligned}$$
因此$-L$特征值也是可列的,
且$\mu_i=\frac{1}{\lambda_i+\Lambda}\rightarrow 0$ $\Rightarrow$
$\lambda_i\rightarrow \infty,$ 特征值趋于无穷.
文章最后更新于 2021-12-05 20:30:25
classical equations
Recalling mathematical equations, the ones we most often come into contact with are the following equations:
$-\Delta u=f,$ Poisson equation. Discuss boundary value problems,
The boundary conditions are of Dirichlet or Neumann type.
$u_{tt}-\Delta u=f,$ wave equation. Discuss the Cauchy problem on unbounded regions,
The initial boundary value problem is discussed in the bounded area. The initial conditions need to be given for $u$ and $u_t$,
The boundary conditions are of Dirichlet or Neumann type.
$u_t-\Delta u=f,$ Heat conduction equation.Discuss the basic same-wave equation,
The initial conditions only need to be given for $u$.
Considering the generalized form of the above equation, there is the following linear equation,
The problems discussed correspond to the classic problems above and will not be repeated. The coefficients are all set to smooth.
and satisfied ellipticity condition:
$\sum_{ij=1}^n a_{ij(x)}\xi_i\xi_j\ge \alpha \Vert\xi\Vert^2,$
$\,\forall\,\xi\neq 0.$ Let’s assume that the coefficients are symmetric.
$Lu=\sum_{i,j=1}^{n}a_{ij}(x)u_{ij}+\sum_{i=1}^n b_i(x)u_i+c(x)u=f,$
Second order elliptic equation.
$u_{tt}-\sum_{i,j=1}^n\frac{\partial {} }{\partial {}x_i}(a_{ij}(t,x)\frac{\partial {}u}{\partial {}x_j})+\sum_{i=1}^n b_i(t,x)\frac{\partial {}u}{\partial {}x_i}+c(t,x)u=f,$
Second-order hyperbolic equation.
$u_{t}-\sum_{i,j=1}^n\frac{\partial {} }{\partial {}x_i}(a_{ij}(t,x)\frac{\partial {}u}{\partial {}x_j})+\sum_{i=1}^n b_i(t,x)\frac{\partial {}u}{\partial {}x_i}+c(t,x)u=f,$
second order parabolic equation.
We start by considering Classic solution ($C^2$ solution) is transferred to consideration Weak solution (Generalized function solution) (Weak
Formulation). By Stokes formula:
$$\int_\Omega d\omega=\int_{\partial\Omega} \omega,$$
we have Gauss-Green formula:
$$\int_\Omega \partial_i fdx=\int_\Omega \nabla \cdot e_i fdx=\int_{\partial\Omega} n\cdot e_i \gamma fdS=\int_{\partial\Omega} n_i\gamma fdS.$$
where $\gamma$ is the trace, which makes the formula true for more functions. Then we have Integral by parts formula:
$$\int_{\Omega} \partial_i fgdx+\int_\Omega f\partial_i gdx=\int_\Omega{\partial_i (fg)}dx=\int_{\partial\Omega} n_i\gamma(fg)dS.$$
To solve partial differential equations, a commonly used method is operator semigroup method,
This turns the problem into solving an infinite-dimensional ODE.
dual operator
Consider the Dirichlet homogeneous boundary problem, $u\in H_0^1(\Omega).$
Apply the test function $v\in H_0^1(\Omega)$ to the equation:
$$\begin{aligned}
\int_\Omega vLudx=&{\color{blue}\sum_{i,j=1}^n\int_{\Omega}\frac{\partial {} }{\partial {}x_i}(a_{ij}(x)u_jv)dx}-\sum_{i,j=1}^n\int_{\Omega}u_j\frac{\partial {} }{\partial {}x_i}(a_{ij}(x)v)dx\\
&+{\color{blue}\sum_{i=1}^n \int_\Omega \frac{\partial {} }{\partial {}x_i}(b_i(x)uv)dx}-\sum_{i=1}^n \int_\Omega u\frac{\partial {} }{\partial {}x_i}(b_i(x)v)dx+\int_\Omega c(x)uvdx\\
=&-\sum_{i,j=1}^n\int_{\Omega}u_j\frac{\partial {} }{\partial {}x_i}(a_{ij}(x)v)dx-\sum_{i=1}^n \int_\Omega u\frac{\partial {} }{\partial {}x_i}(b_i(x)v)dx+\int_\Omega c(x)uvdx\\
=&{\color{blue}\sum_{i,j=1}^n\int_{\Omega}\frac{\partial {} }{\partial {}x_j}\left(u\frac{\partial {} }{\partial {}x_i}(a_{ij}(x)v)\right)dx}-\sum_{i,j=1}^n\int_{\Omega}u_j\frac{\partial {} }{\partial {}x_i}(a_{ij}(x)v)dx\\\
&-\sum_{i=1}^n \int_\Omega u\frac{\partial {} }{\partial {}x_i}(b_i(x)v)dx+\int_\Omega c(x)uvdx\\
=&\sum_{i,j=1}^n\int_{\Omega}u\frac{\partial^2 {} }{\partial {}x_i\partial {}x_j}(a_{ij}(x)v)dx-\sum_{i=1}^n \int_\Omega u\frac{\partial {} }{\partial {}x_i}(b_i(x)v)dx+\int_\Omega c(x)uvdx\\
=&\int_\Omega uL^*vdx.
\end{aligned}$$
The Gauss-Green formula is used many times in the derivation.
Since the function has homogeneous boundary conditions, all boundary terms (blue terms) disappear.
From this we get the dual operator $L^\ast .$
Similarly, for hyperbolic and parabolic equations, we also have dual operators:
$$M^\ast v=\frac{\partial {}^2v}{\partial {}t^2}-\sum_{i,j=1}^n\frac{\partial {} }{\partial {}x_i}(a_{ij}(x)v_j)-\sum_{i=1}^n\frac{\partial {} }{\partial {}x_i}(b_i(x)v)+c(x)v,$$
$$M^\ast v=-\frac{\partial {}v}{\partial {}t}-\sum_{i,j=1}^n\frac{\partial {} }{\partial {}x_i}(a_{ij}(x)v_j)-\sum_{i=1}^n\frac{\partial {} }{\partial {}x_i}(b_i(x)v)+c(x)v.$$
The importance of the dual operator lies in the following important propositions from functional analysis:
Proposition 1. For Hilbert space $H,$ bounded linear operator $A:H\rightarrow H,$, the existence of problem $Ax=b,$ $\,\forall\,b\in H$ is equivalent to the uniqueness of problem $A^\ast y=0$ (that is, it has only zero solution).
Certificate: Existence $\Rightarrow$ Uniqueness: $\,\forall\,y$ makes $A^\ast y=0,$
There exists $x$ such that $Ax=y.$ Therefore,
$$0=(x,A^\ast y)=(Ax,y)=(y,y) \,\Rightarrow\, y=0.$$
Uniqueness $\Rightarrow$ Existence: For $b\in H,$, just find $x$ that satisfies:
$$(Ax,y)=(b,y),\quad \forall y\in H.$$
Take $A^\ast :H\rightarrow B=\operatorname{Im} A^\ast ,$
is a continuous linear operator. It is injective because $A^\ast y=0$ has only zero solution. According to the inverse operator theorem,
$(A^\ast )^{-1}:B\rightarrow H$ is also a continuous linear operator.
Define the continuous linear functional $l_b:B\rightarrow \mathbb{R}$:
$$l_b(v)=(b,(A^\ast )^{-1}v)=(b,u),$$
Here $A^\ast u=v.$ is expressed by Riesz theorem,
There exists $x\in B$ such that $l_b(v)=(x,v).$ therefore for any $y\in H:$
$$l_b(A^\ast y)=(b,y)=(x,A^\ast y)=(Ax,y).$$
This shows that $x\in B$ we found is exactly the solution of $Ax=b$.
energy approach
multiplier method
With dual operators, in most cases we only have to deal with the proof of uniqueness, which is easy.
Usually the energy method is used, and the uniqueness is explained by energy conservation through derivation.
Energy methods are often derived from the multiplier method. For elliptic equations, consider the energy form $\int uLu$ (potential energy);
For the wave equation, consider the energy form $\int u_t Mu$ (kinetic energy); for the heat conduction equation,
Consider the energy form $\int uMu$ (potential energy).
Take the elliptic equation as an example. For $u\in H_0^1(\Omega),$ we have:
$$\begin{aligned}
-\int_\Omega uLudx=&\int_\Omega \sum_{i,j=1}^n a_{ij}(x)u_iu_jdx+\int_\Omega \sum_{i=1}^n\left(\sum_{j=1}^n\frac{\partial {}a_{ij} }{\partial {}x_j}-b_i(x)\right)u_iudx-\int_\Omega c(x)u^2dx\\
\ge &\int_\Omega \alpha \sum_{i=1}^n u_i^2dx-\int_\Omega C'\sum_{i=1}^n \left(\frac{\alpha}{2C'} u_i^2+\frac{C'}{2\alpha}u^2\right)dx-\int_\Omega C''u^2dx\\
\ge &\frac{\alpha}{2}\Vert\nabla u\Vert-C\Vert u\Vert.
\end{aligned}$$
Inequality $2ab\le \varepsilon a^2+\frac{1}{\varepsilon} b^2.$ is used in the process
This inequality is called Garding's inequality:
Theorem 2 (Garding's inequality). Let $\Omega$ be a bounded area in $\mathbb{R}^n$ with a smooth boundary. If $L$ is a second-order elliptical operator, then $\,\forall\,u\in H_0^1(\Omega),$ has the following inequality:
$$\operatorname{Re}(-Lu,u)_{L^2(\Omega)}\ge C_1\Vert u\Vert_{1}^2-C_2\Vert u\Vert^2.$$
The coefficients can be calculated specifically. When $C_2\le 0$, consider uniqueness for problem $Lu=0$:
$$C_1\Vert u\Vert_{1}^2-C_2\Vert u\Vert^2\le 0\,\Rightarrow\, u\equiv 0.$$
For linear equations, the uniqueness of $Lu=0$ and the uniqueness of $Lu=f$ are of course equivalent.
Characteristic problem
In general, $C_2$ need not be non-positive. However, for sufficiently large $\Lambda\in \mathbb{R},$,
considering $(-L+\Lambda)u=0$ makes the right-hand side of Garding's inequality non-negative,
and the solution is then unique. We can therefore define the inverse operator $(-L+\Lambda)^{-1}.$ on the range.
We obtain the following:
$$\begin{aligned}
C_1\Vert u\Vert^2_{1}&\le \left<{}(-L+\Lambda)u,u\right>\\
&\le \Vert(-L+\Lambda)u\Vert\Vert u\Vert\le \Vert(-L+\Lambda)u\Vert\Vert u\Vert_{H^1}\\
\Vert u\Vert_{1}&\le C_3\Vert(-L+\Lambda)u\Vert\\
\Vert(-L+\Lambda)^{-1}v\Vert_{1}&\le C_3 \Vert v\Vert
\end{aligned}$$
Thus $(-L+\Lambda)^{-1}$ is a bounded operator.
Since $H_0^1\stackrel{c}\hookrightarrow L^2$ is a compact embedding, this operator is also compact.
Its eigenvalues form a countable set, $\mu_i\rightarrow 0.$
Consider feature issues:
$$\begin{aligned}
-Lu&=\lambda u\\
(-L+\Lambda)u&=(\lambda+\Lambda)u\\
\frac{1}{\lambda+\Lambda}u&=(-L+\Lambda)^{-1}u=\mu u\\
\end{aligned}$$
Therefore, $-L$ eigenvalues are also listable,
And $\mu_i=\frac{1}{\lambda_i+\Lambda}\rightarrow 0$ $\Rightarrow$
$\lambda_i\rightarrow \infty,$ Eigenvalues tend to infinity.
The article was last updated on 2021-12-05 20:30:25
