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广义函数
基本定义
$C_c^\infty\subset \mathscr{S}\subset C^\infty,$
记为$\mathscr{D}\subset \mathscr{S}\subset \mathscr{E},$
那么$\mathscr{D}'\supset \mathscr{S}'\supset \mathscr{E}'.$
注意这里用到基本空间前者在后者稠密.
广义函数的连续性等价于有界性. 注意这里有界性较为特殊,
因为基本空间不是赋范的. 但是可以取到一列范数.
由该性质可以得到一个重要的对偶性质: $(\mathscr{D}')_c=\mathscr{E}'.$
磨光
$\alpha_\varepsilon(x)=\frac{1}{\varepsilon^n}\alpha(\frac{x}{\varepsilon}).$
称卷积$J_\varepsilon f:=f_\varepsilon=\alpha_\varepsilon\ast f$为$f$的磨光.
由于$\alpha_\varepsilon$为径向函数, 我们有:
$$
\left<{}\alpha_\varepsilon\ast T,\varphi\right>=\left<{}T_x,\left<{}\alpha_\varepsilon(y),\varphi(x+y)\right>\right>=\left<{}T,\alpha_\varepsilon\ast \varphi\right>.
$$
当$T\in \mathscr{D}'$时, $\alpha_\varepsilon\ast T\in C^\infty.$
由Minkowski不等式,
$\Vert\alpha_\varepsilon\ast f\Vert_{L^p}\le \Vert f\Vert_{L^p}.$
Minkowski不等式可由离散情况推得, 或由泛函分析方法证明.
当$\varepsilon\rightarrow 0$时,
我们有$f_\varepsilon\xrightarrow{L^p} f.$
求导
广义函数”差不多”总能写成常义函数的导数. 这里求导为广义导数, 如:
$$
\left<{}T,\varphi\right>=\left<{}D^\alpha f,\varphi\right>=(-1)^{|\alpha|}\left<{}f,D^\alpha \varphi\right>.
$$
求导的定义来源于分部积分, 由Gauss-Green公式推导.
$$
\int_\Omega \partial f\cdot g+f\cdot \partial g=\int_{\Omega} \partial(fg)=\int_{\partial \Omega} fg\cdot \gamma,
$$
当$f,g$有一者在边界上取零时, 右端得零, 建立了关于左端的等式.
Fourier变换
对于速降函数, 有Fourier变换:
$$
F[f] (\xi)=\widehat{f}(\xi):=\int_{\mathbb{R}^n} f(x)e^{-ix\cdot \xi}dx,
$$
有Fourier逆变换:
$$
f(x)=F^{-1}[\widehat{f}] (x):=(2\pi)^{-n}\int_{\mathbb{R}^n}\widehat{f}(\xi)e^{ix\cdot \xi}d\xi,
$$
满足:
$$
\widehat{D^\alpha f}(\xi)=\xi^\alpha \widehat{f}(\xi),\quad D^\alpha=\frac{1}{i^{|\alpha|} }\partial^\alpha,
$$
$$
\widehat{fg}=(2\pi)^{-n}\widehat{f}\ast \widehat{g}.
$$
我们有经典的命题: $F:\mathscr{S}\cong \mathscr{S}$为连续同构.
这直接诱导了$F:\mathscr{S}'\cong \mathscr{S}',$
$\left<{}F[T],\varphi\right>:=\left<{}T,F[\varphi]\right>.$
这是由Fubini定理保证的. 我们还有Plancherel公式:
$$
\int_{\mathbb{R}^n}\widehat{f}\overline{\widehat{g} }\bar{d}\xi=\int_{\mathbb{R}^n} f \overline{g}dx,\quad \bar{d}\xi=(2\pi)^{-n}d\xi.
$$
取$g=f,$ 即得到了连续情形的Parseval恒等式.
Sobolev空间
Sobolev不等式
$$
\Vert u\Vert_{L^q}\le C|u|_{\dot{H}^{m,p} },\quad \frac{n}{q}=\frac{n}{p}-m;
$$
$$
|u|_{C^\alpha}\le C|u|_{\dot{H}^{m,p} },\quad -\alpha=\frac{n}{p}-m,\,\alpha\in (0,1);
$$
$$
\Vert u\Vert_{L^\infty}\le C\Vert u\Vert_{H^{m,p} },\quad 0>\frac{n}{p}-m;
$$
$$
\Vert u\Vert_{L^\infty}\le C\Vert u\Vert_{H^{s} },\quad 0>\frac{n}{2}-s;
$$
$$
\Vert u\Vert_{L^q}\le C\Vert u\Vert_{H^{m,p} },\quad \frac{n}{p}-m=0,\, q\ge p.
$$
关于最后一个不等式, 左侧量纲为$\frac{n}{q},$
右侧量纲为$[\frac{n}{p}-m,\frac{n}{p}]=[0,m].$ 为了使不等式成立,
左侧量纲一定要落在该区间中.
从而$\frac{n}{q}\le \frac{n}{p}=m\,\Rightarrow\,q\ge p$的条件是必要的.
我们还有Poincaré不等式:
$$
\Vert u\Vert_{L^p}\le d\Vert\nabla u\Vert_{L^p}=d|u|_{\dot{H}^{1,p} },\quad \,\forall\,u\in H_0^{1,p}(\Omega).
$$
定理
等价模定理: 由Lions延拓,
$H^{m,p}(\Omega)$中的函数总可以延拓为$H^{m,p}(\mathbb{R}^n)$中的函数.
定义
$$
\Vert f\Vert_{\widetilde{H}^{m,p}(\Omega)}=\inf_{\widetilde{f}|_\Omega =f}\Vert\widetilde{f}\Vert_{H^{m,p}(\mathbb{R}^n)},
$$
则凭借由Lions延拓得到的$\widetilde{f},$ 我们有:
$$
\Vert f\Vert_{H^{m,p}(\Omega)}\le \Vert f\Vert_{\widetilde{H}^{m,p}(\Omega)}\le \Vert\widetilde{f}\Vert_{H^{m,p}(\mathbb{R}^n)}\le C\Vert f\Vert_{H^{m,p}(\Omega)},
$$
紧嵌入定理: 当Sobolev不等式中条件严格时,
即$\frac{n}{q}>\frac{n}{p}-m,$
此时$H^{m,p}(\Omega)\hookrightarrow L^{q}(\Omega)$的嵌入是紧的,
即将有界集映成列紧集.
迹定理:
$$
\Vert u\Vert_{H^{s-\frac{1}{2} }(\partial\Omega)}\le C\Vert u\Vert_{H^s(\Omega)},\quad s>\frac{1}{2}
$$
一个最常应用的简单推论是:
$$
\Vert u\Vert_{L^2(\partial \Omega)}\le C\Vert u\Vert_{H^1(\Omega)}.
$$
关于对偶空间: 需要注意的是, $H^{-1}(\Omega):=(H_0^1(\Omega))^\ast ,$
并不是$H^1(\Omega)$的对偶空间. 仅当$\Omega=\mathbb{R}^n$时,
$H_0^1(\mathbb{R}^n)=H^1(\mathbb{R}^n).$ 同时, 由Riesz表示定理,
$H^{-1}(\Omega)\cong H_0^1(\Omega),$ 当并不相等.
因为前者视为$\mathscr{D}'(\Omega)$的子空间, 以$L^2(\Omega)$为子空间,
泛函取为做$L^2$内积; 而后者以$H^1$内积为泛函, 这导致对偶空间出现了不同,
但总归是同构对应的.
PDE
基本定义
二阶椭圆方程:
$$
Lu=\sum \partial_i(a_{ij}u_j)+\sum b_i u_i+cu.
$$
二阶双曲方程:
$$
Mu=u_{tt}-\sum \partial_i (a_{ij}u_j)-\sum b_{0j}u_{ti}+\sum b_iu_i+b_0 u_t+cu.
$$
可记为$Mu=u_{tt}-M_1u_t+M_2u.$
二阶抛物方程:
$$
Nu=u_{t}-\sum \partial_i (a_{ij}u_j)+\sum b_i u_i+cu.
$$
可记为$Nu=u_t-N_1u.$
系数$a_{ij}$对称且满足椭圆性条件:
$\sum a_{ij}\xi_i\xi_j\ge \alpha|\xi|^2.$ 当涉及时间项时, 条件是一致的.
能量方法
对椭圆型方程, $\left<{}u,Lu\right>$给出Garding不等式;
双曲型方程的$\left<{}u_t,Mu\right>,$
抛物型方程的$\left<{}u,Nu\right>$给出能量估计.
Galerkin方法
取基本空间的一组基$\{\varphi_i\},$ 考虑$u_m=\sum a_i\varphi_i,$ 解
$$
\left<{}Lu_m-f,\varphi_i\right>=0,\quad i=1,\cdots,m.
$$
这化为了线性方程组$Ax=b$的形式, 求解系数即可.
这里可解性由Garding不等式保证. $\frac{A+A^T}{2}$正定时, $A$可逆.
而式中二次型可写为$x^T\frac{A+A^T}{2}x=\left<{}-Lu,u\right>$的形式,
当$C_2\le 0$时, 即推知正定性.
对涉及时间项的Galerkin方法, 需考虑$u_m=\sum a_i(t)\varphi_i.$ 此时
$$
\left<{}Mu_m,\varphi_i\right>=\left<{}f,\varphi_i\right>,\quad i=1,\cdots,m
$$
为$m$个ODE, 总是可解的. 具体地,
$$
\sum \left<{}\varphi_j,\varphi_i\right>a_j''-\left<{}M_1\varphi_j,\varphi_i\right>a_j'+\left<{}M_2\varphi_j,\varphi_i\right>a_j=\left<{}f,\varphi_i\right>.
$$
类似地, 对抛物型方程, 问题为:
$$
\left<{}Nu_m,\varphi_i\right>=\sum \left<{}\varphi_j,\varphi_i\right>a_j'-\left<{}N_1\varphi_j,\varphi_i\right>a_j=\left<{}f,\varphi_i\right>.
$$
算子半群
考虑问题$u_t=N_1u,$ 形式上, $e^{N_1t}u_0$给出了问题解. 具体的,
考虑$N_\lambda:=\lambda N_1(\lambda-N_1)^{-1},$
定义$e^{N_1t}:=\lim\limits_{\lambda\rightarrow+\infty}e^{N_\lambda t}$来将表示严格化.
文章最后更新于 2021-12-26 16:23:44
generalized function
basic definition
$C_c^\infty\subset \mathscr{S}\subset C^\infty,$
Recorded as $\mathscr{D}\subset \mathscr{S}\subset \mathscr{E},$
Then $\mathscr{D}'\supset \mathscr{S}'\supset \mathscr{E}'.$
Note that the basic space used here is that the former is dense in the latter.
The continuity of a generalized function is equivalent to boundedness. Note that boundedness is more special here,
Because the basic space is not normed. But a sequence of norms can be obtained.
An important dual property can be obtained from this property: $(\mathscr{D}')_c=\mathscr{E}'.$
polish
$\alpha_\varepsilon(x)=\frac{1}{\varepsilon^n}\alpha(\frac{x}{\varepsilon}).$
The convolution $J_\varepsilon f:=f_\varepsilon=\alpha_\varepsilon\ast f$ is called the polishing of $f$.
Since $\alpha_\varepsilon$ is a radial function, we have:
$$
\left<{}\alpha_\varepsilon\ast T,\varphi\right>=\left<{}T_x,\left<{}\alpha_\varepsilon(y),\varphi(x+y)\right>\right>=\left<{}T,\alpha_\varepsilon\ast \varphi\right>.
$$
When $T\in \mathscr{D}'$, $\alpha_\varepsilon\ast T\in C^\infty.$
From Minkowski inequality,
$\Vert\alpha_\varepsilon\ast f\Vert_{L^p}\le \Vert f\Vert_{L^p}.$
Minkowski's inequality can be derived from discrete cases or proven by functional analysis methods.
When $\varepsilon\rightarrow 0$,
We have $f_\varepsilon\xrightarrow{L^p} f.$
Derivative
A generalized function can almost always be written as a derivative of a normal function. The derivative here is a generalized derivative, such as:
$$
\left<{}T,\varphi\right>=\left<{}D^\alpha f,\varphi\right>=(-1)^{|\alpha|}\left<{}f,D^\alpha \varphi\right>.
$$
The definition of derivation comes from integration by parts, derived from the Gauss-Green formula.
$$
\int_\Omega \partial f\cdot g+f\cdot \partial g=\int_{\Omega} \partial(fg)=\int_{\partial \Omega} fg\cdot \gamma,
$$
When one of $f,g$ takes zero on the boundary, the right end gets zero, establishing an equation about the left end.
Fourier transform
For the rapid descending function, there is Fourier transformation:
$$
F[f] (\xi)=\widehat{f}(\xi):=\int_{\mathbb{R}^n} f(x)e^{-ix\cdot \xi}dx,
$$
There is an inverse Fourier transformation:
$$
f(x)=F^{-1}[\widehat{f}] (x):=(2\pi)^{-n}\int_{\mathbb{R}^n}\widehat{f}(\xi)e^{ix\cdot \xi}d\xi,
$$
Meet:
$$
\widehat{D^\alpha f}(\xi)=\xi^\alpha \widehat{f}(\xi),\quad D^\alpha=\frac{1}{i^{|\alpha|} }\partial^\alpha,
$$
$$
\widehat{fg}=(2\pi)^{-n}\widehat{f}\ast \widehat{g}.
$$
We have the classic proposition: $F:\mathscr{S}\cong \mathscr{S}$ is a continuous isomorphism.
This directly induces $F:\mathscr{S}'\cong \mathscr{S}',$
$\left<{}F[T],\varphi\right>:=\left<{}T,F[\varphi]\right>.$
This is guaranteed by Fubini's theorem. We also have the Plancherel formula:
$$
\int_{\mathbb{R}^n}\widehat{f}\overline{\widehat{g} }\bar{d}\xi=\int_{\mathbb{R}^n} f \overline{g}dx,\quad \bar{d}\xi=(2\pi)^{-n}d\xi.
$$
Taking $g=f,$ obtains the Parseval identity for the continuous case.
Sobolev space
Sobolev's inequality
$$
\Vert u\Vert_{L^q}\le C|u|_{\dot{H}^{m,p} },\quad \frac{n}{q}=\frac{n}{p}-m;
$$
$$
|u|_{C^\alpha}\le C|u|_{\dot{H}^{m,p} },\quad -\alpha=\frac{n}{p}-m,\,\alpha\in (0,1);
$$
$$
\Vert u\Vert_{L^\infty}\le C\Vert u\Vert_{H^{m,p} },\quad 0>\frac{n}{p}-m;
$$
$$
\Vert u\Vert_{L^\infty}\le C\Vert u\Vert_{H^{s} },\quad 0>\frac{n}{2}-s;
$$
$$
\Vert u\Vert_{L^q}\le C\Vert u\Vert_{H^{m,p} },\quad \frac{n}{p}-m=0,\, q\ge p.
$$
Regarding the last inequality, the left-hand dimension is $\frac{n}{q},$
The right-hand side dimension is $[\frac{n}{p}-m,\frac{n}{p}]=[0,m].$ In order to make the inequality true,
The left-hand dimension must fall within this interval.
Therefore the condition $\frac{n}{q}\le \frac{n}{p}=m\,\Rightarrow\,q\ge p$ is necessary.
We also have Poincaré’s inequality:
$$
\Vert u\Vert_{L^p}\le d\Vert\nabla u\Vert_{L^p}=d|u|_{\dot{H}^{1,p} },\quad \,\forall\,u\in H_0^{1,p}(\Omega).
$$
Theorem
Equivalent modular theorem: by Lions extension,
Functions in $H^{m,p}(\Omega)$ can always be extended to functions in $H^{m,p}(\mathbb{R}^n)$.
definition
$$
\Vert f\Vert_{\widetilde{H}^{m,p}(\Omega)}=\inf_{\widetilde{f}|_\Omega =f}\Vert\widetilde{f}\Vert_{H^{m,p}(\mathbb{R}^n)},
$$
Then with $\widetilde{f},$ obtained by extension of Lions, we have:
$$
\Vert f\Vert_{H^{m,p}(\Omega)}\le \Vert f\Vert_{\widetilde{H}^{m,p}(\Omega)}\le \Vert\widetilde{f}\Vert_{H^{m,p}(\mathbb{R}^n)}\le C\Vert f\Vert_{H^{m,p}(\Omega)},
$$
compact embedding theorem: When the conditions in Sobolev’s inequality are strict,
That is $\frac{n}{q}>\frac{n}{p}-m,$
At this time, the embedding of $H^{m,p}(\Omega)\hookrightarrow L^{q}(\Omega)$ is tight,
That is to map bounded sets into compact sets.
trace theorem:
$$
\Vert u\Vert_{H^{s-\frac{1}{2} }(\partial\Omega)}\le C\Vert u\Vert_{H^s(\Omega)},\quad s>\frac{1}{2}
$$
A simple corollary that is most commonly applied is:
$$
\Vert u\Vert_{L^2(\partial \Omega)}\le C\Vert u\Vert_{H^1(\Omega)}.
$$
Regarding the dual space: It should be noted that $H^{-1}(\Omega):=(H_0^1(\Omega))^\ast ,$
It is not the dual space of $H^1(\Omega)$. Only when $\Omega=\mathbb{R}^n$,
$H_0^1(\mathbb{R}^n)=H^1(\mathbb{R}^n).$ At the same time, expressed by Riesz theorem,
$H^{-1}(\Omega)\cong H_0^1(\Omega),$ When not equal.
Because the former is regarded as the subspace of $\mathscr{D}'(\Omega)$, and $L^2(\Omega)$ is the subspace,
The functional is taken as the inner product of $L^2$; while the latter uses the inner product of $H^1$ as the functional, which leads to a difference in the dual space,
But after all, they are isomorphic correspondences.
PDE
basic definition
Second-order elliptic equation:
$$
Lu=\sum \partial_i(a_{ij}u_j)+\sum b_i u_i+cu.
$$
Second-order hyperbolic equation:
$$
Mu=u_{tt}-\sum \partial_i (a_{ij}u_j)-\sum b_{0j}u_{ti}+\sum b_iu_i+b_0 u_t+cu.
$$
Can be recorded as $Mu=u_{tt}-M_1u_t+M_2u.$
Second order parabolic equation:
$$
Nu=u_{t}-\sum \partial_i (a_{ij}u_j)+\sum b_i u_i+cu.
$$
Can be recorded as $Nu=u_t-N_1u.$
The coefficient $a_{ij}$ is symmetric and satisfies the ellipticity condition:
$\sum a_{ij}\xi_i\xi_j\ge \alpha|\xi|^2.$ The conditions are consistent when time terms are involved.
energy approach
For elliptic equations, $\left<{}u,Lu\right>$ gives Garding's inequality;
$\left<{}u_t,Mu\right>,$ of hyperbolic equation
$\left<{}u,Nu\right>$ of the parabolic equation gives an estimate of the energy.
Galerkin method
Take a set of basis $\{\varphi_i\},$ of the basic space and consider $u_m=\sum a_i\varphi_i,$ solution
$$
\left<{}Lu_m-f,\varphi_i\right>=0,\quad i=1,\cdots,m.
$$
This is converted into the form of a linear system of equations $Ax=b$, and just solve for the coefficients.
The solvability here is guaranteed by Garding's inequality. $\frac{A+A^T}{2}$ is positive definite, $A$ is reversible.
The quadratic form in the formula can be written in the form of $x^T\frac{A+A^T}{2}x=\left<{}-Lu,u\right>$,
When $C_2\le 0$, the positive definiteness is inferred.
For the Galerkin method involving time terms, $u_m=\sum a_i(t)\varphi_i.$ needs to be considered at this time
$$
\left<{}Mu_m,\varphi_i\right>=\left<{}f,\varphi_i\right>,\quad i=1,\cdots,m
$$
is $m$ ODEs, which are always solvable. Specifically,
$$
\sum \left<{}\varphi_j,\varphi_i\right>a_j''-\left<{}M_1\varphi_j,\varphi_i\right>a_j'+\left<{}M_2\varphi_j,\varphi_i\right>a_j=\left<{}f,\varphi_i\right>.
$$
Similarly, for parabolic equations, the question is:
$$
\left<{}Nu_m,\varphi_i\right>=\sum \left<{}\varphi_j,\varphi_i\right>a_j'-\left<{}N_1\varphi_j,\varphi_i\right>a_j=\left<{}f,\varphi_i\right>.
$$
operator semigroup
Consider problem $u_t=N_1u,$. Formally, $e^{N_1t}u_0$ gives the solution to the problem. Specifically,
Consider$N_\lambda:=\lambda N_1(\lambda-N_1)^{-1},$
Define $e^{N_1t}:=\lim\limits_{\lambda\rightarrow+\infty}e^{N_\lambda t}$ to make the representation strict.
The article was last updated on 2021-12-26 16:23:44
