用户: Yao/现代偏微分方程课程重点

1函数的正则化

Take a fixed $\alpha \in C_c^{\infty }({\mathbb{R}}^n)$ supported in the unit ball $B_1(0)$ with ${\int }_{{\mathbb{R}}^n}\alpha =1$. Write$${\alpha }_{ϵ}=\frac{1}{{ϵ}^n}\alpha \left(\frac{x}{ϵ}\right),$$then ${\alpha }_{ϵ}$ is supported in $B_{ϵ}(0)$, with ${\int }_{{\mathbb{R}}^n}{\alpha }_{ϵ}=1$.

Proposition 1.1. For $f\in L^p(\Omega )$ where $1\le p<\infty$, we have

(1)	$f\ast {\alpha }_{ϵ}\in C^{\infty }({\Omega }_{ϵ})$, where ${\Omega }_{ϵ}=\{x\in \Omega \mid d(x,\partial \Omega )>ϵ\}$,
(2)	$\Vert f\ast {\alpha }_{ϵ}{\Vert }_{L^p}\le \Vert f{\Vert }_{L^p}$,
(3)	$f\ast {\alpha }_{ϵ}\stackrel{L^p}{⟶}f$ as $ϵ\to 0$.

Proof.

(1)	The smoothness of $f\ast {\alpha }_{ϵ}$ is derived from interchanging integration and derivatives.
(2)	By Minkowski, $\Vert {\int }_{{\mathbb{R}}^n}f(x-y){\alpha }_{ϵ}(y)dy{\Vert }_{L^p({\mathbb{R}}^n)}\le {\int }_{{\mathbb{R}}^n}\Vert f(x-y){\Vert }_{L^p({\mathbb{R}}^n)}{\alpha }_{ϵ}(y)dy$.
(3)	First show that $f\ast {\alpha }_{ϵ}\stackrel{L^p}{⟶}f$ holds for $f\in C_c^{\infty }$, then use $\Vert f\ast {\alpha }_{ϵ}{\Vert }_{L^p({\mathbb{R}}^n)}\le \Vert f{\Vert }_{L^p({\mathbb{R}}^n)}$ to extend the result to the $L^p$ case. $\square$

Exercise. Show that $f\ast {\alpha }_{ϵ}\to f$ a.e. as $ϵ\to 0$.

Proof. For any Lebesgue point $x$ of $f$, $$\begin{array}{rl}|(f\ast {\alpha }_{ϵ}-f)(x)|& =|{\int }_{B_{ϵ}(x)}{\alpha }_{ϵ}(x-y)(f(y)-f(x)\left)dy\right|\\ & \le {\int }_{B_{ϵ}(x)}\frac{1}{{ϵ}^n}\alpha \left(\frac{x-y}{ϵ}\right)|f(y)-f(x)|dy\\ & \le C{-\int }_{B_{ϵ}(x)}|f(y)-f(x)|dy\to 0.\end{array}$$

$\square$

2广义函数

Definition 2.1. A distribution is a continuous linear functional on the space of test functions like $C_c^{\infty }(\Omega ),\mathcal{S}({\mathbb{R}}^n),C^{\infty }(\Omega )$. The convergence of distributions is defined to be the weak convergence of functionals.

The construction of $C^{\infty }$ functions $T_{ϵ}(x)=⟨T,{\alpha }_{ϵ}(x-y)⟩\to T$ is called the regularization of distributions.

Proposition 2.2.

(1)	A linear functional $T$ on $C_c^{\infty }(\Omega )$ is continuous iff for any compact $K$ there exists $C_K,m_K$ such that $|⟨T,\phi ⟩|\le C_K\underset{|\alpha |\le m_K}{\sup }|{\partial }^{\alpha }\phi |$ for all smooth functions $\phi$ supported in $K$.
(2)	A linear functional $T$ on $C^{\infty }(\Omega )$ is continuous iff there exists compact $K$ and $C,m$ such that $|⟨T,\phi ⟩|\le C\underset{\begin{array}{c}|\alpha |\le m\\ x\in K\end{array}}{\sup }|{\partial }^{\alpha }\phi |$ for all smooth functions $\phi$.

Definition 2.3 (Fourier transform). $$\begin{array}{rl}& F[f]={\int }_{{\mathbb{R}}^n}f(x)e^{-ix\cdot \xi }dx,\\ & F^{-1}[f]=(2\pi {)}^{-n}{\int }_{{\mathbb{R}}^n}f(\xi )e^{ix\cdot \xi }d\xi .\end{array}$$

Proposition 2.4 (Parseval’s relation). $${\int }_{{\mathbb{R}}^n}f\bar{g}dx=(2\pi {)}^{-n}{\int }_{{\mathbb{R}}^n}F[f]\overline{F[g]}d\xi .$$

Definition 2.5 (Calculation on distributions).

(1)	$⟨D^{\alpha }T,\phi ⟩=(-1{)}^{|\alpha |}⟨T,D^{\alpha }\phi ⟩$.
(2)	$⟨\alpha T,\phi ⟩=⟨T,\alpha \phi ⟩$.
(3)	$⟨S\ast T,\phi ⟩=⟨S_x,⟨T_y,\phi (x+y)⟩⟩$.
(4)	$⟨F[T],\phi ⟩=⟨T,F[\phi ]⟩$.
(5)	$⟨F^{-1}[T],\phi ⟩=⟨T,F^{-1}[\phi ]⟩$.

Proposition 2.6.

(1)	$F[D^{\alpha }T]=(i\xi {)}^{\alpha }F[T]$,
(2)	$F[x^{\alpha }T]=(iD{)}^{\alpha }F[T]$,
(3)	$F[T]=⟨T,e^{-ix\cdot \xi }⟩$.

3Sobolev 空间 (上)

Definition 3.1. $H^{m,p}(\Omega )$ is the Banach space of functions with the norm $$\Vert u{\Vert }_{H^{m,p}(\Omega )}=(\sum _{|\alpha |\le m}\Vert D^{\alpha }u{\Vert }_{L^p(\Omega )}^p{)}^{\frac{1}{p}}.$$

Definition 3.2. $$\begin{array}{rl}{H}^{m,2}(\Omega )& =H^m(\Omega ).\\ H_0^{m,p}(\Omega )& =\overline{C_c^{\infty }(\Omega )}.\\ (H_0^{m,p}(\Omega ){)}^{\ast }& =H^{-m,p^{\prime }}(\Omega ).\end{array}$$

Remark 3.3. For a bounded domain $\Omega$, $1\in H^{m,p}(\Omega )$ while $1\notin H_0^{m,p}(\Omega )$, since $H^{m,p}$ convergence implies uniform convergence when $\Omega$ is bounded, and consequently a sequence of functions with compact support converges to a function that also tends to $0$ at $\partial \Omega$.

Definition 3.4. For $s\in \mathbb{R}$, $H^s({\mathbb{R}}^n)$ is the Hilbert space of functions with the norm $$\Vert u{\Vert }_{H^s({\mathbb{R}}^n)}=({\int }_{{\mathbb{R}}^n}(1+|\xi {|}^2{)}^s|\hat{u}{|}^{2}d\xi {)}^{\frac{1}{2}}.$$

Remark 3.5. We may write $⟨\xi ⟩=(1+|\xi {|}^2{)}^{\frac{1}{2}}$, and the expression becomes $\Vert u{\Vert }_{H^s({\mathbb{R}}^n)}=\Vert ⟨\xi {⟩}^s\hat{u}{\Vert }_{L^2}$.

Proposition 3.6 (Trace Theorem). For $f\in H^s({\mathbb{R}}^n)$, $\gamma f=f(x_1,\cdots ,x_{n-1},0)\in H^{s-\frac{1}{2}}({\mathbb{R}}^{n-1})$, and $\gamma$ is a bounded operator.

Proof. Write $\xi =({\xi }^{\prime },{\xi }_n)$. Take the Fourier inversion in ${\xi }_n$ and get $$\widehat{\gamma f}({\xi }^{\prime })=\frac{1}{2\pi }{\int }_{\mathbb{R}}\hat{f}(\xi )e^{i\pi 0}d{\xi }_n=\frac{1}{2\pi }{\int }_{\mathbb{R}}\hat{f}(\xi )d{\xi }_n,$$so$$\begin{array}{rl}|\widehat{\gamma f}({\xi }^{\prime }){|}^2& \le \frac{1}{(2\pi {)}^2}|{\int }_{\mathbb{R}}\hat{f}(\xi )d{\xi }_n{|}^2\\ & \le \frac{1}{(2\pi {)}^2}{\int }_{\mathbb{R}}(1+|\xi {|}^2{)}^s|\hat{f}(\xi ){|}^{2}d{\xi }_n{\int }_{\mathbb{R}}\frac{d{\xi }_n}{(1+|{\xi }^{\prime }{|}^2+|{\xi }_n{|}^2{)}^s}\\ & =C{\int }_{\mathbb{R}}(1+|\xi {|}^2{)}^s|\hat{f}(\xi ){|}^{2}d{\xi }_n\frac{1}{(1+|{\xi }^{\prime }{|}^2{)}^{s-\frac{1}{2}}}.\end{array}$$Then we have $$\begin{array}{rl}\Vert \gamma f{\Vert }_{H^{s-\frac{1}{2}}({\mathbb{R}}^{n-1})}^2=& {\int }_{{\mathbb{R}}^{n-1}}(1+|{\xi }^{\prime }{|}^2{)}^{s-\frac{1}{2}}|\widehat{\gamma f}({\xi }^{\prime }){|}^{2}d{\xi }^{\prime }\\ \le & C{\int }_{{\mathbb{R}}^{n-1}}{\int }_{\mathbb{R}}(1+|\xi {|}^2{)}^s|\hat{f}(\xi ){|}^{2}d{\xi }_{n}d{\xi }^{\prime }\\ =& C\Vert f{\Vert }_{H^s({\mathbb{R}}^n)}^2.\end{array}$$

$\square$

Proposition 3.7 (Lions Extension). For a bounded region $\Omega$ with $C^1$ boundary, $u\in H^{m,p}(\Omega )$ can be extended to $\tilde{u}\in H^{m,p}({\mathbb{R}}^n)$ (which means $\tilde{u}$ equals $u$ in $\Omega$) in such a way that $\Vert \tilde{u}{\Vert }_{H^{m,p}({\mathbb{R}}^n)}\le C\Vert u{\Vert }_{H^{m,p}(\Omega )}$.

Proof. A partition of unity reduces the question to the case that $u$ is compactly supported, and flattening out the the boundary by a coordinate change to the case that $u\in H^{m,p}(B^{+})$ where $B^{+}=\{x\in {\mathbb{R}}^n||x|<1,x_n\ge 0\}$.

Assume that $u\in C^{\infty }(B^{+})$ first. For arbitraty distinct numbers ${\lambda }_1,\cdots ,{\lambda }_m>0$, let$$\tilde{u}=\{\begin{array}{ll}u& x\in B^{+}\\ {\sum }_{i=1}^m{c}_{i}u(x^{\prime },-{\lambda }_i{x}_n)& x=(x^{\prime },x_n)\in B^{-}\end{array}.$$

Let $\{\begin{array}{l}{\sum }_{i=1}^m{c}_i=1\\ {\sum }_{i=1}^m(-{\lambda }_i)c_i=1\\ \cdots \\ {\sum }_{i=1}^m(-{\lambda }_i{)}^{m-1}{c}_i=1\end{array}$ where the coefficient matrix is a nonsingular Vandermonde matrix, so the solution $(c_1,\cdots ,c_m)$ exists. Then $\tilde{u}\in C^{m-1}(B)$ and $D^{\alpha }\tilde{u}$ is continuous for $|\alpha |=m$ on $B^{+}$ and $B^{-}$, so $\tilde{u}\in H^{m,p}(B)$.

Now extend the extension from $C^{\infty }(B^{+})$ to $H^{m,p}(B^{+})$.

$\square$

4Sobolev 空间 (下)

Proposition 4.1. $$\Vert f{\Vert }_{L^{\infty }({\mathbb{R}}^n)}\le C\Vert \hat{f}{\Vert }_{L^1({\mathbb{R}}^n)}\le C\Vert f{\Vert }_{H^s({\mathbb{R}}^n)},s>\frac{n}{2}.$$$$\Vert f{\Vert }_{L^{p^{\ast }}({\mathbb{R}}^n)}\le C\Vert \nabla f{\Vert }_{L^p({\mathbb{R}}^n)},\frac{n}{p^{\ast }}=\frac{n}{p}-1,1\le p<n.$$$$\Vert f{\Vert }_{C^{0,\gamma }({\mathbb{R}}^n)}\le C\Vert \nabla f{\Vert }_{L^p({\mathbb{R}}^n)},\gamma =1-\frac{n}{p},n<p\le \infty .$$$$\Vert f{\Vert }_{L^p({\mathbb{R}}^n)}\le C\Vert \nabla f{\Vert }_{L^n({\mathbb{R}}^n)}^{1-\frac{n}{p}}\Vert f{\Vert }_{L^n({\mathbb{R}}^n)}^{\frac{n}{p}},n<p<\infty .$$$$\Vert f{\Vert }_{L^{\infty }({\mathbb{R}}^n)}\le C\Vert \nabla f{\Vert }_{L^p({\mathbb{R}}^n)}^{\frac{n}{p}}\Vert f{\Vert }_{L^p({\mathbb{R}}^n)}^{1-\frac{n}{p}},n<p\le \infty .$$

Proof. The first is becuase$$\Vert {\hat{f}}^{\vee }{\Vert }_{L^{\infty }({\mathbb{R}}^n)}\le \frac{1}{(2\pi {)}^n}\Vert \hat{f}{\Vert }_{L^1({\mathbb{R}}^n)}\le \frac{1}{(2\pi {)}^n}\Vert (1+|\xi {|}^2{)}^{-\frac{s}{2}}{\Vert }_{L^2({\mathbb{R}}^n)}\Vert (1+|\xi {|}^2{)}^{\frac{s}{2}}\hat{f}{\Vert }_{L^2({\mathbb{R}}^n)}.$$

For the second and the third, see Sobolev inequalities.

The fourth is becuase$$\Vert f{\Vert }_{L^p({\mathbb{R}}^n)}^{\frac{p}{p-n}}=\Vert f^{\frac{p}{p-n}}{\Vert }_{L^{p-n}({\mathbb{R}}^n)}\le C\Vert \nabla f\cdot f^{\frac{n}{p-n}}{\Vert }_{L^{\frac{1}{\frac{1}{n}+\frac{1}{p-n}}}({\mathbb{R}}^n)}\le C\Vert \nabla f{\Vert }_{L^n({\mathbb{R}}^n)}\Vert f^{\frac{n}{p-n}}{\Vert }_{L^{p-n}({\mathbb{R}}^n)}=C\Vert \nabla f{\Vert }_{L^n({\mathbb{R}}^n)}\Vert f{\Vert }_{L^n({\mathbb{R}}^n)}^{\frac{n}{p-n}}.$$

For the fifth, by the second step of Theorem 1.4 of Sobolev inequalities, for appropriate $\lambda$,$$\Vert f(\lambda x){\Vert }_{L^{\infty }({\mathbb{R}}^n)}\le C({\lambda }^{1-\frac{n}{p}}\Vert \nabla f{\Vert }_{L^p({\mathbb{R}}^n)}+{\lambda }^{\frac{n}{p}}\Vert f{\Vert }_{L^p({\mathbb{R}}^n)})=C\Vert \nabla f{\Vert }_{L^p({\mathbb{R}}^n)}^{\frac{n}{p}}\Vert f{\Vert }_{L^p({\mathbb{R}}^n)}^{1-\frac{n}{p}}.$$$\square$

Proposition 4.2 (Rellich-Kondrachov). For a bounded region $\Omega$ with smooth boundaries, the inclusion $H^{m,p}(\Omega )\hookrightarrow L^q(\Omega )$ is compact for $1\le q<\frac{1}{\frac{1}{p}-\frac{m}{n}}$.

Proof. Enough to deal with $m=1$, since the composite of compact operators $H^{m,p}(\Omega )\hookrightarrow H^{m-1,p^{\ast }}(\Omega )\hookrightarrow \cdots \hookrightarrow L^q(\Omega )$ is compact.

For each bounded sequence $\{u_m{\}}_{m\ge 1}$ in $H^{1,p}(\Omega )$, which is also bounded in $L^{p^{\ast }}(\Omega )$, write $u_m^{ϵ}=u_m\ast {\alpha }_{ϵ}$. First, we have $\Vert u_m^{ϵ}-u_m{\Vert }_{L^q}\to 0$ uniformly as $ϵ\to 0$ because $$\Vert u_m^{ϵ}-u_m{\Vert }_{L^q}\le \Vert u_m^{ϵ}-u_m{\Vert }_{L^1}^{\theta }\Vert u_m^{ϵ}-u_m{\Vert }_{L^{p^{\ast }}}^{1-\theta }\le M\Vert u_m^{ϵ}-u_m{\Vert }_{L^1}^{\theta }$$and for $u\in C^{\infty }(\Omega )$, and hence for $u\in H^{1,p}(\Omega )$,$$\Vert u_m^{ϵ}-u_m{\Vert }_{L^1}=|{\int }_{\Omega }\alpha (y)(u_m(x-ϵy)-u_m(x))dy|\le ϵ\Vert D{u}_m{\Vert }_{L^1(\Omega )}\le Cϵ\Vert D{u}_m{\Vert }_{L^p(\Omega )}$$where $C$ is a constant dependent on $\Omega$.

Next, we observe that for fixed $ϵ$, $\{u_m^{ϵ}{\}}_{m\ge 1}$ is compact for it is uniformly bounded and equicontinuous as $\Vert u_m^{ϵ}{\Vert }_{L^{\infty }}\le \Vert {\alpha }_{ϵ}{\Vert }_{L^{\infty }}\Vert u_m{\Vert }_{L^1}$ and $\Vert D{u}_m^{ϵ}{\Vert }_{L^{\infty }}\le \Vert D{\alpha }_{ϵ}{\Vert }_{L^{\infty }}\Vert u_m{\Vert }_{L^1}$.

Now a diagonal argument completes the proof.

$\square$

Proposition 4.3 (Poincaré). $\Omega$ is a domain lying between two parallel hyperplanes of distance $d$. For $u\in H_0^p(\Omega )$, $\Vert u{\Vert }_{L^p}\le d\Vert \nabla u{\Vert }_{L^p}$.

Proof. Let $e$ be a unit vector orthogonal to the hyperplanes. For $u\in C_c^{\infty }(\Omega )$, we have $$\begin{array}{rl}{\int }_{\Omega }|u{|}^{p}dx\le & {\int }_{\Omega }({\int }_{-d}^0|\nabla u(x+te)|dt{)}^{p}dx\\ \le & {\int }_{\Omega }{\int }_{-d}^0|\nabla u(x+te){|}^p{d}^{\frac{p}{p^{\prime }}}dtdx\\ =& d^{\frac{p}{p^{\prime }}}{\int }_{-d}^0{\int }_{\Omega }|\nabla u(x+te){|}^{p}dxdt\\ =& d^p\Vert \nabla u{\Vert }_{L^p}^p.\end{array}$$Then use the density of $C_c^{\infty }(\Omega )$ in $H_0^p(\Omega )$.

$\square$

5椭圆方程

Let $L$ be a second order elliptic partial differential operator $$Lu=-\sum _{i,j=1}^n\frac{\partial }{\partial x_i}(a_{ij}(x)\frac{\partial u}{\partial x_j})+\sum _{i=1}^n{b}_i(x)\frac{\partial u}{\partial x_i}+c(x)u,$$where the matrix $(a_{ij}(x))\ge \lambda I$ for all $x$.

For $u\in H_0^1$, $L$ acts on $u$ in the sense of distributions, and hence $Lu\in H^{-1}$.

Proposition 5.1 (Gårding). For $u\in H_0^1$, $⟨Lu,u⟩\ge C_1\Vert u{\Vert }_{H_0^1}^2-C_2\Vert u{\Vert }_{L^2}^2$.

Proof. $$\begin{array}{rl}⟨Lu,u⟩& =\sum _{i,j}⟨a_{ij}\frac{\partial u}{\partial x_j},\frac{\partial u}{\partial x_i}⟩+\sum _i⟨b_i\frac{\partial u}{\partial x_i},u⟩+⟨cu,u⟩\\ & \ge \lambda \Vert \nabla u{\Vert }_{L^2}^2-C\Vert \nabla u{\Vert }_{L^2}^2\Vert u{\Vert }_{L^2}^2-C\Vert u{\Vert }_{L^2}^2\\ & \ge \frac{\lambda }{2}\Vert \nabla u{\Vert }_{L^2}^2-C\Vert u{\Vert }_{L^2}^2=\frac{\lambda }{2}\Vert u{\Vert }_{H_0^1}^2-C\Vert u{\Vert }_{L^2}^2.\end{array}$$

$\square$

Proposition 5.2. There is a number $\gamma \ge 0$ such that for each $\mu \ge \gamma$, there exists a unique weak solution $u\in H_0^1$ of $(L+\mu )u=f$ for each $f\in H^{-1}$.

Proof. Introduce the formal adjoint operator $L^{\ast }$ defined as (one easily checks $⟨Lu,v⟩=⟨u,L^{\ast }v⟩$) $$L^{\ast }v=-\sum _{i,j=1}^n\frac{\partial }{\partial x_j}(a_{ij}(x)\frac{\partial v}{\partial x_i})-\sum _{i=1}^n{b}_i(x)\frac{\partial v}{\partial x_i}+(-\sum _{i=1}^n\frac{\partial b_i(x)}{\partial x_i}+c(x))v.$$Take $\gamma$ to be $C_2$ of the preceding proposition. Then for $\mu \ge \gamma$ we have $$\begin{array}{r}\Vert u{\Vert }_{H_0^1}^2\le C⟨(L+\mu )u,u⟩\le C\Vert (L+\mu )u{\Vert }_{H^{-1}}\Vert u{\Vert }_{H_0^1},\end{array}$$which is$$\Vert u{\Vert }_{H_0^1}\le C\Vert (L+\mu )u{\Vert }_{H^{-1}}.$$

Apply Lax–Milgram theorem to the form $(u,v)\in H_0^1\times H_0^1\mapsto ⟨(L+\mu )u,v⟩$, we get that the operator $L+\mu :H_0^1\to H^{-1}$ has a bounded inverse. Moreover $\Vert v{\Vert }_{H^{-1}}\le \Vert v{\Vert }_{L^2}$ and $H^1$ is compactly embedded in $L^2$, and we conclude that $(L+\mu {)}^{-1}:L^2\to L^2$ is a compact operator.

$\square$

6Galerkin 方法

Solve the equation$$Lu=-\sum _{i,j=1}^n\frac{\partial }{\partial x_i}(a_{ij}(x)\frac{\partial u}{\partial x_j})+\sum _{i=1}^n{b}_i(x)\frac{\partial u}{\partial x_i}+c(x)u=f(x).$$

Let $\{{\varphi }_I{\}}_{I\ge 1}$ be the eigenfunctions of $-\Delta$, which serves as an orthonormal basis of both $L^2(\Omega )$ and $H_0^1(\Omega )$. (See the last problem of this.)

We look for $u_N(x)=\sum _{I=1}^N{w}_I{\varphi }_I(x)$ such that $L{u}_N-f$ vanishes on $\mathrm{span}\{u_J{\}}_{1\le J\le N}$ (that is, its projection on this subspace is zero), and $u_N$ approximates $u$ in certain sense.

First, we carry out the calculation of $u_N$.

For $J=1,\cdots ,N$, $u_N$ satisfies$$⟨{\varphi }_J,-\sum _{i,j}\frac{\partial }{\partial x_i}\left(a_{ij}\frac{\partial u_N}{\partial x_j}\right)+\sum _i{b}_i\frac{\partial u_N}{\partial x_i}+c{u}_N⟩=⟨{\varphi }_J,f⟩,$$then integration by parts yields$$\sum _{I=1}^N(\sum _{i,j}⟨\frac{\partial {\varphi }_J}{\partial x_i},a_{ij}\frac{\partial {\varphi }_I}{\partial x_j}⟩+\sum _i⟨{\varphi }_J,b_i\frac{\partial {\varphi }_I}{\partial x_i}⟩+c⟨{\varphi }_J,{\varphi }_I⟩)w_I=⟨{\varphi }_J,f⟩,$$which are $N$ linear equations of $w_1,\cdots ,w_N$.

7能量估计

1. For the energy estimate for hyperbolic equations, see the last problem of this.

2. We now establish the energy estimate for the parabolic equation$$u_t-(\sum _{i,j=1}^n(a_{ij}(t,x)u_{x_j}{)}_{x_i}+\sum _{i=1}^n{b}_i(t,x)u_{x_i}+c(t,x)u)=f(x)$$where $(a_{ij})\ge \lambda I$.

We have$${\int }_{\Omega }2u(u_t-\sum _i(a_{ij}{u}_{x_j}{)}_{x_i}-\sum _i{b}_i{u}_{x_i}-cu)={\int }_{\Omega }2uf(x).$$

Let $E(t)={\int }_{\Omega }{u}^2$, then $E^{\prime }(t)={\int }_{\Omega }2u{u}_t$. Since$$-2{\int }_{\Omega }u\sum _i(a_{ij}{u}_{x_j}{)}_{x_i}=2{\int }_{\Omega }\sum _{i=1}^n{a}_{ij}{u}_{x_j}{u}_{x_i}\ge 2\lambda \Vert \nabla u{\Vert }_{L^2}^2,$$$$\left|{\int }_{\Omega }2u\right(-\sum _i{b}_i{u}_{x_i}-cu\left)\right|\le \lambda \Vert \nabla u{\Vert }_{L^2}^2+C\Vert u{\Vert }_{L^2}^2,$$$$\left|{\int }_{\Omega }2uf\right|\le \Vert u{\Vert }_{L^2}^2+\Vert f{\Vert }_{L^2}^2,$$we get that $E^{\prime }(t)\le CE(t)+\Vert f{\Vert }_{L^2}^2$, and by Gronwall we conclude that$$E(t)\le C(E(0)+{\int }_0^t\Vert f{\Vert }_{L^2}^{2}dt).$$

3. Consider the symmetric hyperbolic system$$A_0{u}_t+\sum _{i=1}^n{A}_i{u}_{x_i}+Bu=f$$where $u=(u_1,\cdots ,u_m{)}^T$, $f=(f_1,\cdots ,f_m{)}^T$, $A_0(t,x),A_i(t,x)$ are symmetric matrices, $A_0\ge \lambda I$, and $B$ is a matrix $B(t,x)$.

WLOG, assume that $B-\sum _{i=1}^n\frac{\partial A_i}{\partial x_i}>0$, for we can replace $u$ by $e^{\mu t}u$ for $\mu$ big enough if otherwise. And assume that $\frac{\partial A_0}{\partial t}$ is bounded.

Multiply $2u^T$ on both sides, we get

$$\frac{\partial }{\partial t}(u^T{A}_{0}u)+\frac{\partial }{\partial x_i}(u^T{A}_{0}u)-u^T(\frac{\partial A_0}{\partial t}+\sum _{i=1}^n\frac{\partial A_i}{\partial x_i}-B)u=2u^{T}f,$$so$$\begin{array}{rl}E(t)={\int }_{\Omega }{u}^T{A}_{0}u=& {\int }_0^t(-{\int }_{\Omega }\frac{\partial }{\partial x_i}(u^T{A}_{0}u)+{\int }_{\Omega }{u}^T(\frac{\partial A_0}{\partial t}+\sum _{i=1}^n\frac{\partial A_i}{\partial x_i}-B)u+{\int }_{\Omega }2u^{T}f)\\ \le & C{\int }_0^t{\int }_{\Omega }{u}^{T}u+f^2\\ \le & {\lambda }^{-1}C{\int }_0^{t}E(t)+{\int }_0^t\Vert f{\Vert }_{L^2}^2,\end{array}$$where ${\int }_0^t{\int }_{\Omega }\frac{\partial }{\partial x_i}(u^T{A}_{0}u)\ge 0$ due to Green’s formula and some boundary conditions. Then by Gronwall,$$E(t)\le e^{{\lambda }^{-1}Ct}(E(0)+{\int }_0^t\Vert f{\Vert }_{L^2}^2).$$

8算子半群

$X$ is a Banach space.

Definition 8.1. $\{S(t){\}}_{t\ge 0}\subset B(X)$ is called a semigroup if $S(0)=I$, $S(t+s)=S(t)S(s)$, and for all $u\in X$ the map $t\mapsto S(t)u$ is continuous, and is called a contraction semigroup if in addition $\Vert S(t)\Vert \le 1$ for all $t$.

Definition 8.2. $D(A)=\{u\in X|\underset{t\to 0^{+}}{\lim }\frac{S(t)u-u}{t}\text{ exsits in }X\}$ is called the domain of $\mathbf{A}$.

$A:D(A)\to X$ is called the (infinitesimal) generator of the semigroup.

Theorem 8.3 (Hille-Yosida). For a densely defined closed linear operator $A$ in $X$, the following assertations are equivalent:

(1) $A$ generates a $C_0$-semigroup of contractions on $X$;

(2) $\forall \lambda \in \mathbb{C}$ s.t. $\mathrm{Re}\lambda >0$, $\lambda -A$ is invertible and $\Vert (\lambda -A{)}^{-1}\Vert \le \frac{1}{\mathrm{Re}\lambda }$;

(3) $\forall \lambda \in \mathbb{R}$ s.t. $\lambda >0$, $\lambda -A$ is invertible and $\Vert (\lambda -A{)}^{-1}\Vert \le \frac{1}{\lambda }$.

The textbook of S.X.Chen establishes the equivalence of the first two assertations, while what Y.Zhou teaches is the equivalence of the first and the third.

Define $A_{\lambda }=\frac{\lambda A}{\lambda -A}=-\lambda +\frac{{\lambda }^2}{\lambda -A}$ for $\lambda >0$. Then for $u\in D(A)$, $\Vert A_{\lambda }u-Au\Vert =\Vert \frac{A^{2}u}{\lambda -A}\Vert \le \frac{\Vert A^{2}u\Vert }{\lambda }\to 0$ as $\lambda \to \infty$.

Let $S_{\lambda }(t)=e^{t{A}_{\lambda }}=e^{-\lambda t}{e}^{\frac{{\lambda }^2}{\lambda -A}t}$. Since $\Vert \frac{{\lambda }^2}{\lambda -A}\Vert \le \lambda$, we have $\Vert S_{\lambda }(t)\Vert \le 1$. Since$$\begin{array}{rl}\Vert S_{\lambda }(t)u-S_{\mu }(t)u\Vert =& \Vert {\int }_0^t\frac{d}{ds}(S_{\mu }(t-s)S_{\lambda }(s)u)ds\Vert \\ =& \Vert {\int }_0^t{S}_{\mu }(t-s)S_{\lambda }(s)(A_{\lambda }u-A_{\mu }u)ds\Vert \\ \le & t\Vert A_{\lambda }u-A_{\mu }u\Vert \to 0\end{array}$$as $\lambda ,\mu \to \infty$ for $u\in D(A)$ and hence also for $u\in X$, we can define $S(t)u=\underset{\lambda \to \infty }{\lim }{S}_{\lambda }(t)u$. Since$$S(t)u-u\leftarrow S_{\lambda }(t)u-u={\int }_0^t{S}_{\lambda }(s)A_{\lambda }uds\to {\int }_0^{t}S(s)Auds,$$$A$ is indeed the generator of $\{S(t){\}}_{t\ge 0}$.